{"text": ["ormation. The second way we try to keep numbers simple is to use a consistent set of units\u2014the metric International System of Units, or SI (from the French Syst\u00e8me International d\u2019Unit\u00e9s). The metric system is summarized in Appendix D (see Example 1.2). 16 Chapter 1 Science and the Universe: A Brief Tour Watch this brief PBS animation (https://openstax.org/l/30scinotation) that explains how scientific notation works and why it\u2019s useful. A common unit astronomers use to describe distances in the universe is a light-year, which is the distance light travels during one year. Because light always travels at the same speed, and because its speed turns out to be the fastest possible speed in the universe, it makes a good standard for keeping track of distances. You might be confused because a \u201clight-year\u201d seems to imply that we are measuring time, but this mix-up of time and distance is common in everyday life as well. For example, when your friend asks where the movie theater is located, you might say \u201cabout 20 minutes from downtown.\u201d So, how many kilometers are there in a light-year? Light travels at the amazing pace of 3 \u00d7 105 kilometers per second (km/s), which makes a light-year 9.46 \u00d7 1012 kilometers. You might think that such a large unit would reach the nearest star easily, but the stars are far more remote than our imaginations might lead us to believe. Even the nearest star is 4.3 light-years away\u2014more than 40 trillion kilometers. Other stars visible to the unaided eye are hundreds to thousands of light-years away (Figure 1.4). Figure 1.4. Orion Nebula. This beautiful cloud of cosmic raw material (gas and dust from which new stars and planets are being made) called the Orion Nebula is about 1400 light-years away. That\u2019s a distance of roughly 1.34 \u00d7 1016 kilometers\u2014a pretty big number. The gas and dust in this region are illuminated by the intense light from a few extremely energetic adolescent stars. (credit: NASA, ESA, M. Robberto (Space Telescope Science Institute/ESA) and the Hubble Space Telescope Orion Treasury Project Team Scientific Notation This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 1 Science and the Universe: A Brief Tour 17 In 2015, the richest human being on our planet had a net worth of $79.2 billion. Some might say this is an astronomical sum of money. Express this amount in scientific notation. Solution $79.2 billion can be written $79,200,000,000. Expressed in scientific notation it becomes $7.92 \u00d7 1010 Getting Familiar with a Light-Year How many kilometers are there in a light-year? Solution Light travels 3 \u00d7 105 km in 1 s. So, let\u2019s calculate how far it goes in a year: \u2022 There are 60 (6 \u00d7 101) s in 1 min, and 6 \u00d7 101 min in 1 h. \u2022 Multiply these together and you find that there are 3.6 \u00d7 103 s/h. \u2022 Thus, light covers 3 \u00d7 105 km/s \u00d7 3.6 \u00d7 103 s/h = 1.08 \u00d7 109 km/h. \u2022 There are 24 or 2.4 \u00d7 101 h in a day, and 365.24 (3.65 \u00d7 102) days in 1 y. \u2022 The product of these two numbers is 8.77 \u00d7 103 h/y. \u2022 Multiplying this by 1.08 \u00d7 109 km/h gives 9.46 \u00d7 1012 km/light-year. That\u2019s almost 10,000,000,000,000 km that light covers in a year. To help you imagine how long this distance is, we\u2019ll mention that a string 1 light-year long could fit around the circumference of Earth 236 million times. 1.5 CONSEQUENCES OF LIGHT TRAVEL TIME There is another reason the speed of light is such a natural unit of distance for astronomers. Information about the universe comes to us almost exclusively through various forms of light, and all such light travels at the speed of light\u2014that is, 1 light-year every year. This sets a limit on how quickly we can learn about events in the universe. If a star is 100 light-years away, the light we see from it tonight left that star 100 years ago and is just now arriving in our neighborhood. The soonest we can learn about any changes in that star is 100 years after the fact. For a star 500 light-years away, the light we detect tonight left 500 years ago and is carrying 500-year- old news. Because many of us are accustomed to instant news from the Internet, some might find this frustrating. \u201cYou mean, when I see that star up there,\ufffd", "\ufffd you ask, \u201cI won\u2019t know what\u2019s actually happening there for another 500 years?\u201d But this isn\u2019t the most helpful way to think about the situation. For astronomers, now is when the light reaches us here on Earth. There is no way for us to know anything about that star (or other object) until its light reaches us. 18 Chapter 1 Science and the Universe: A Brief Tour But what at first may seem a great frustration is actually a tremendous benefit in disguise. If astronomers really want to piece together what has happened in the universe since its beginning, they must find evidence about each epoch (or period of time) of the past. Where can we find evidence today about cosmic events that occurred billions of years ago? The delay in the arrival of light provides an answer to this question. The farther out in space we look, the longer the light has taken to get here, and the longer ago it left its place of origin. By looking billions of light-years out into space, astronomers are actually seeing billions of years into the past. In this way, we can reconstruct the history of the cosmos and get a sense of how it has evolved over time. This is one reason why astronomers strive to build telescopes that can collect more and more of the faint light in the universe. The more light we collect, the fainter the objects we can observe. On average, fainter objects are farther away and can, therefore, tell us about periods of time even deeper in the past. Instruments such as the Hubble Space Telescope (Figure 1.5) and the Very Large Telescope in Chile (which you will learn about in the chapter on Astronomical Instruments), are giving astronomers views of deep space and deep time better than any we have had before. Figure 1.5. Telescope in Orbit. The Hubble Space Telescope, shown here in orbit around Earth, is one of many astronomical instruments in space. (credit: modification of work by European Space Agency) 1.6 A TOUR OF THE UNIVERSE We can now take a brief introductory tour of the universe as astronomers understand it today to get acquainted with the types of objects and distances you will encounter throughout the text. We begin at home with Earth, a nearly spherical planet about 13,000 kilometers in diameter (Figure 1.6). A space traveler entering our planetary system would easily distinguish Earth from the other planets in our solar system by the large amount of liquid water that covers some two thirds of its crust. If the traveler had equipment to receive radio or television signals, or came close enough to see the lights of our cities at night, she would soon find signs that this watery planet has sentient life. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 1 Science and the Universe: A Brief Tour 19 Figure 1.6. Humanity\u2019s Home Base. This image shows the Western hemisphere as viewed from space 35,400 kilometers (about 22,000 miles) above Earth. Data about the land surface from one satellite was combined with another satellite\u2019s data about the clouds to create the image. (credit: modification of work by R. Stockli, A. Nelson, F. Hasler, NASA/ GSFC/ NOAA/ USGS) Our nearest astronomical neighbor is Earth\u2019s satellite, commonly called the Moon. Figure 1.7 shows Earth and the Moon drawn to scale on the same diagram. Notice how small we have to make these bodies to fit them on the page with the right scale. The Moon\u2019s distance from Earth is about 30 times Earth\u2019s diameter, or approximately 384,000 kilometers, and it takes about a month for the Moon to revolve around Earth. The Moon\u2019s diameter is 3476 kilometers, about one fourth the size of Earth. Figure 1.7. Earth and Moon, Drawn to Scale. This image shows Earth and the Moon shown to scale for both size and distance. (credit: modification of work by NASA) Light (or radio waves) takes 1.3 seconds to travel between Earth and the Moon. If you\u2019ve seen videos of the Apollo flights to the Moon, you may recall that there was a delay of about 3 seconds between the time Mission Control asked a question and the time the astronauts responded. This was not because the astronomers were thinking slowly, but rather because it took the radio waves almost 3 seconds to make the round trip. Earth revolves around our star, the Sun, which is about 150 million kilometers away\u2014approximately 400 times as far away from us as the Moon. We call the average Earth\u2013Sun distance an astronomical unit (AU) because, in the early days of astronomy, it was the most important measuring standard. Light takes slightly more than 8 minutes to travel 1 astronomical unit, which means the latest news we receive from the Sun is always 8 minutes old. The diameter of the Sun is about 1.5 million kilometers; Earth could fit comfortably inside one of", " the minor eruptions that occurs on the surface of our star. If the Sun were reduced to the size of a basketball, Earth would be a small apple seed about 30 meters from the ball. It takes Earth 1 year (3 \u00d7 107 seconds) to go around the Sun at our distance; to make it around, we must travel at approximately 110,000 kilometers per hour. (If you, like many students, still prefer miles to kilometers, you might find the following trick helpful. To convert kilometers to miles, just multiply kilometers by 0.6. Thus, 20 Chapter 1 Science and the Universe: A Brief Tour 110,000 kilometers per hour becomes 66,000 miles per hour.) Because gravity holds us firmly to Earth and there is no resistance to Earth\u2019s motion in the vacuum of space, we participate in this extremely fast-moving trip without being aware of it day to day. Earth is only one of eight planets that revolve around the Sun. These planets, along with their moons and swarms of smaller bodies such as dwarf planets, make up the solar system (Figure 1.8). A planet is defined as a body of significant size that orbits a star and does not produce its own light. (If a large body consistently produces its own light, it is then called a star.) Later in the book this definition will be modified a bit, but it is perfectly fine for now as you begin your voyage. Figure 1.8. Our Solar Family. The Sun, the planets, and some dwarf planets are shown with their sizes drawn to scale. The orbits of the planets are much more widely separated than shown in this drawing. Notice the size of Earth compared to the giant planets. (credit: modification of work by NASA) We are able to see the nearby planets in our skies only because they reflect the light of our local star, the Sun. If the planets were much farther away, the tiny amount of light they reflect would usually not be visible to us. The planets we have so far discovered orbiting other stars were found from the pull their gravity exerts on their parent stars, or from the light they block from their stars when they pass in front of them. We can\u2019t see most of these planets directly, although a few are now being imaged directly. The Sun is our local star, and all the other stars are also enormous balls of glowing gas that generate vast amounts of energy by nuclear reactions deep within. We will discuss the processes that cause stars to shine in more detail later in the book. The other stars look faint only because they are so very far away. If we continue our basketball analogy, Proxima Centauri, the nearest star beyond the Sun, which is 4.3 light-years away, would be almost 7000 kilometers from the basketball. When you look up at a star-filled sky on a clear night, all the stars visible to the unaided eye are part of a single collection of stars we call the Milky Way Galaxy, or simply the Galaxy. (When referring to the Milky Way, we capitalize Galaxy; when talking about other galaxies of stars, we use lowercase galaxy.) The Sun is one of hundreds of billions of stars that make up the Galaxy; its extent, as we will see, staggers the human imagination. Within a sphere 10 light-years in radius centered on the Sun, we find roughly ten stars. Within a sphere 100 light-years in radius, there are roughly 10,000 (104) stars\u2014far too many to count or name\u2014but we have still This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 1 Science and the Universe: A Brief Tour 21 traversed only a tiny part of the Milky Way Galaxy. Within a 1000-light-year sphere, we find some ten million (107) stars; within a sphere of 100,000 light-years, we finally encompass the entire Milky Way Galaxy. Our Galaxy looks like a giant disk with a small ball in the middle. If we could move outside our Galaxy and look down on the disk of the Milky Way from above, it would probably resemble the galaxy in Figure 1.9, with its spiral structure outlined by the blue light of hot adolescent stars. Figure 1.9. Spiral Galaxy. This galaxy of billions of stars, called by its catalog number NGC 1073, is thought to be similar to our own Milky Way Galaxy. Here we see the giant wheel-shaped system with a bar of stars across its middle. (credit: NASA, ESA) The Sun is somewhat less than 30,000 light-years from the center of the Galaxy, in a location with nothing much to distinguish it. From our position inside the Milky Way Galaxy, we cannot see through to its far rim (at least not with ordinary light) because the space between the stars is not completely empty. It contains a sparse distribution of gas (mostly the simplest element, hydrogen) intermixed with tiny solid particles that we call interstellar dust", ". This gas and dust collect into enormous clouds in many places in the Galaxy, becoming the raw material for future generations of stars. Figure 1.10 shows an image of the disk of the Galaxy as seen from our vantage point. 22 Chapter 1 Science and the Universe: A Brief Tour Figure 1.10. Milky Way Galaxy. Because we are inside the Milky Way Galaxy, we see its disk in cross-section flung across the sky like a great milky white avenue of stars with dark \u201crifts\u201d of dust. In this dramatic image, part of it is seen above Trona Pinnacles in the California desert. (credit: Ian Norman) Typically, the interstellar material is so extremely sparse that the space between stars is a much better vacuum than anything we can produce in terrestrial laboratories. Yet, the dust in space, building up over thousands of light-years, can block the light of more distant stars. Like the distant buildings that disappear from our view on a smoggy day in Los Angeles, the more distant regions of the Milky Way cannot be seen behind the layers of interstellar smog. Luckily, astronomers have found that stars and raw material shine with various forms of light, some of which do penetrate the smog, and so we have been able to develop a pretty good map of the Galaxy. Recent observations, however, have also revealed a rather surprising and disturbing fact. There appears to be more\u2014much more\u2014to the Galaxy than meets the eye (or the telescope). From various investigations, we have evidence that much of our Galaxy is made of material we cannot currently observe directly with our instruments. We therefore call this component of the Galaxy dark matter. We know the dark matter is there by the pull its gravity exerts on the stars and raw material we can observe, but what this dark matter is made of and how much of it exists remain a mystery. Furthermore, this dark matter is not confined to our Galaxy; it appears to be an important part of other star groupings as well. By the way, not all stars live by themselves, as the Sun does. Many are born in double or triple systems with two, three, or more stars revolving about each other. Because the stars influence each other in such close systems, multiple stars allow us to measure characteristics that we cannot discern from observing single stars. In a number of places, enough stars have formed together that we recognized them as star clusters (Figure This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 1 Science and the Universe: A Brief Tour 23 1.11). Some of the largest of the star clusters that astronomers have cataloged contain hundreds of thousands of stars and take up volumes of space hundreds of light-years across. Figure 1.11. Star Cluster. This large star cluster is known by its catalog number, M9. It contains some 250,000 stars and is seen more clearly from space using the Hubble Space Telescope. It is located roughly 25,000 light-years away. (credit: NASA, ESA) You may hear stars referred to as \u201ceternal,\u201d but in fact no star can last forever. Since the \u201cbusiness\u201d of stars is making energy, and energy production requires some sort of fuel to be used up, eventually all stars run out of fuel. This news should not cause you to panic, though, because our Sun still has at least 5 or 6 billion years to go. Ultimately, the Sun and all stars will die, and it is in their death throes that some of the most intriguing and important processes of the universe are revealed. For example, we now know that many of the atoms in our bodies were once inside stars. These stars exploded at the ends of their lives, recycling their material back into the reservoir of the Galaxy. In this sense, all of us are literally made of recycled \u201cstar dust.\u201d 1.7 THE UNIVERSE ON THE LARGE SCALE In a very rough sense, you could think of the solar system as your house or apartment and the Galaxy as your town, made up of many houses and buildings. In the twentieth century, astronomers were able to show that, just as our world is made up of many, many towns, so the universe is made up of enormous numbers of galaxies. (We define the universe to be everything that exists that is accessible to our observations.) Galaxies stretch as far into space as our telescopes can see, many billions of them within the reach of modern instruments. When they were first discovered, some astronomers called galaxies island universes, and the term is aptly descriptive; galaxies do look like islands of stars in the vast, dark seas of intergalactic space. The nearest galaxy, discovered in 1993, is a small one that lies 75,000 light-years from the Sun in the direction of the constellation Sagittarius, where the smog in our own Galaxy makes it especially difficult to discern. (A constellation, we should note, is one of the", " 88 sections into which astronomers divide the sky, each named after a prominent star pattern within it.) Beyond this Sagittarius dwarf galaxy lie two other small galaxies, about 160,000 light-years away. First recorded by Magellan\u2019s crew as he sailed around the world, these are called the Magellanic Clouds (Figure 1.12). All three of these small galaxies are satellites of the Milky Way Galaxy, interacting with it through the force of gravity. Ultimately, all three may even be swallowed by our much larger Galaxy, as other small galaxies have been over the course of cosmic time. 24 Chapter 1 Science and the Universe: A Brief Tour Figure 1.12. Neighbor Galaxies. This image shows both the Large Magellanic Cloud and the Small Magellanic Cloud above the telescopes of the Atacama Large Millimeter/Submillimeter Array (ALMA) in the Atacama Desert of northern Chile. (credit: ESO, C. Malin) The nearest large galaxy is a spiral quite similar to our own, located in the constellation of Andromeda, and is thus called the Andromeda galaxy; it is also known by one of its catalog numbers, M31 (Figure 1.13). M31 is a little more than 2 million light-years away and, along with the Milky Way, is part of a small cluster of more than 50 galaxies referred to as the Local Group. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 1 Science and the Universe: A Brief Tour 25 Figure 1.13. Closest Spiral Galaxy. The Andromeda galaxy (M31) is a spiral-shaped collection of stars similar to our own Milky Way. (credit: Adam Evans) At distances of 10 to 15 million light-years, we find other small galaxy groups, and then at about 50 million light- years there are more impressive systems with thousands of member galaxies. We have discovered that galaxies occur mostly in clusters, both large and small (Figure 1.14). 26 Chapter 1 Science and the Universe: A Brief Tour Figure 1.14. Fornax Cluster of Galaxies. In this image, you can see part of a cluster of galaxies located about 60 million light-years away in the constellation of Fornax. All the objects that are not pinpoints of light in the picture are galaxies of billions of stars. (credit: ESO, J. Emerson, VISTA. Acknowledgment: Cambridge Astronomical Survey Unit) Some of the clusters themselves form into larger groups called superclusters. The Local Group is part of a supercluster of galaxies, called the Virgo Supercluster, which stretches over a diameter of 110 million light- years. We are just beginning to explore the structure of the universe at these enormous scales and are already encountering some unexpected findings. At even greater distances, where many ordinary galaxies are too dim to see, we find quasars. These are brilliant centers of galaxies, glowing with the light of an extraordinarily energetic process. The enormous energy of the quasars is produced by gas that is heated to a temperature of millions of degrees as it falls toward a massive black hole and swirls around it. The brilliance of quasars makes them the most distant beacons we can see in the dark oceans of space. They allow us to probe the universe 10 billion light-years away or more, and thus 10 billion years or more in the past. With quasars we can see way back close to the Big Bang explosion that marks the beginning of time. Beyond the quasars and the most distant visible galaxies, we have detected the feeble glow of the explosion itself, filling the universe and thus coming to us from all directions in space. The discovery of this \u201cafterglow of creation\u201d is considered to be one of the most significant events in twentieth-century science, and we are still exploring the many things it has to tell us about the earliest times of the universe. Measurements of the properties of galaxies and quasars in remote locations require large telescopes, sophisticated light-amplifying devices, and painstaking labor. Every clear night, at observatories around the world, astronomers and students are at work on such mysteries as the birth of new stars and the large-scale structure of the universe, fitting their results into the tapestry of our understanding. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 1 Science and the Universe: A Brief Tour 27 1.8 THE UNIVERSE OF THE VERY SMALL The foregoing discussion has likely impressed on you that the universe is extraordinarily large and extraordinarily empty. On average, it is 10,000 times more empty than our Galaxy. Yet, as we have seen, even the Galaxy is mostly empty space. The air we breathe has about 1019 atoms in each cubic centimeter\u2014and we usually think of air as empty space. In the interstellar gas of", " the Galaxy, there is about one atom in every cubic centimeter. Intergalactic space is filled so sparsely that to find one atom, on average, we must search through a cubic meter of space. Most of the universe is fantastically empty; places that are dense, such as the human body, are tremendously rare. Even our most familiar solids are mostly space. If we could take apart such a solid, piece by piece, we would eventually reach the tiny molecules from which it is formed. Molecules are the smallest particles into which any matter can be divided while still retaining its chemical properties. A molecule of water (H2O), for example, consists of two hydrogen atoms and one oxygen atom bonded together. Molecules, in turn, are built of atoms, which are the smallest particles of an element that can still be identified as that element. For example, an atom of gold is the smallest possible piece of gold. Nearly 100 different kinds of atoms (elements) exist in nature. Most of them are rare, and only a handful account for more than 99% of everything with which we come in contact. The most abundant elements in the cosmos today are listed in Table 1.1; think of this table as the \u201cgreatest hits\u201d of the universe when it comes to elements. The Cosmically Abundant Elements Element[1] Symbol Number of Atoms per Million Hydrogen Atoms Hydrogen Helium Carbon Nitrogen Oxygen Neon Magnesium Silicon Sulfur Iron Table 1.1 H He C N O Ne Mg Si S Fe 1,000,000 80,000 450 92 740 130 40 37 19 32 1 This list of elements is arranged in order of the atomic number, which is the number of protons in each nucleus. 28 Chapter 1 Science and the Universe: A Brief Tour All atoms consist of a central, positively charged nucleus surrounded by negatively charged electrons. The bulk of the matter in each atom is found in the nucleus, which consists of positive protons and electrically neutral neutrons all bound tightly together in a very small space. Each element is defined by the number of protons in its atoms. Thus, any atom with 6 protons in its nucleus is called carbon, any with 50 protons is called tin, and any with 70 protons is called ytterbium. (For a list of the elements, see Appendix K.) The distance from an atomic nucleus to its electrons is typically 100,000 times the size of the nucleus itself. This is why we say that even solid matter is mostly space. The typical atom is far emptier than the solar system out to Neptune. (The distance from Earth to the Sun, for example, is only 100 times the size of the Sun.) This is one reason atoms are not like miniature solar systems. Remarkably, physicists have discovered that everything that happens in the universe, from the smallest atomic nucleus to the largest superclusters of galaxies, can be explained through the action of only four forces: gravity, electromagnetism (which combines the actions of electricity and magnetism), and two forces that act at the nuclear level. The fact that there are four forces (and not a million, or just one) has puzzled physicists and astronomers for many years and has led to a quest for a unified picture of nature To construct an atom, particle by particle, check out this guided animation (https://openstax.org/l/ 30buildanatom) for building an atom. 1.9 A CONCLUSION AND A BEGINNING If you are new to astronomy, you have probably reached the end of our brief tour in this chapter with mixed emotions. On the one hand, you may be fascinated by some of the new ideas you\u2019ve read about and you may be eager to learn more. On the other hand, you may be feeling a bit overwhelmed by the number of topics we have covered, and the number of new words and ideas we have introduced. Learning astronomy is a little like learning a new language: at first it seems there are so many new expressions that you\u2019ll never master them all, but with practice, you soon develop facility with them. At this point you may also feel a bit small and insignificant, dwarfed by the cosmic scales of distance and time. But, there is another way to look at what you have learned from our first glimpses of the cosmos. Let us consider the history of the universe from the Big Bang to today and compress it, for easy reference, into a single year. (We have borrowed this idea from Carl Sagan\u2019s 1997 Pulitzer Prize-winning book, The Dragons of Eden.) On this scale, the Big Bang happened at the first moment of January 1, and this moment, when you are reading this chapter would be the end of the very last second of December 31. When did other events in the development of the universe happen in this \u201ccosmic year?\u201d Our solar system formed around September 10, and the oldest rocks", " we can date on Earth go back to the third week in September (Figure 1.15). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 1 Science and the Universe: A Brief Tour 29 Figure 1.15. Charting Cosmic Time. On a cosmic calendar, where the time since the Big Bang is compressed into 1 year, creatures we would call human do not emerge on the scene until the evening of December 31. (credit: February: modification of work by NASA, JPL-Caltech, W. Reach (SSC/Caltech); March: modification of work by ESA, Hubble and NASA, Acknowledgement: Giles Chapdelaine; April: modification of work by NASA, ESA, CFHT, CXO, M.J. Jee (University of California, Davis), A. Mahdavi (San Francisco State University); May: modification of work by NASA, JPL-Caltech; June: modification of work by NASA/ESA; July: modification of work by NASA, JPL-Caltech, Harvard-Smithsonian; August: modification of work by NASA, JPL-Caltech, R. Hurt (SSC-Caltech); September: modification of work by NASA; October: modification of work by NASA; November: modification of work by D\u00e9nes Em\u0151ke) Where does the origin of human beings fall during the course of this cosmic year? The answer turns out to be the evening of December 31. The invention of the alphabet doesn\u2019t occur until the fiftieth second of 11:59 p.m. on December 31. And the beginnings of modern astronomy are a mere fraction of a second before the New Year. Seen in a cosmic context, the amount of time we have had to study the stars is minute, and our success in piecing together as much of the story as we have is remarkable. Certainly our attempts to understand the universe are not complete. As new technologies and new ideas allow us to gather more and better data about the cosmos, our present picture of astronomy will very likely undergo many changes. Still, as you read our current progress report on the exploration of the universe, take a few minutes every once in a while just to savor how much you have already learned. FOR FURTHER EXPLORATION Books Miller, Ron, and William Hartmann. The Grand Tour: A Traveler\u2019s Guide to the Solar System. 3rd ed. Workman, 2005. This volume for beginners is a colorfully illustrated voyage among the planets. Sagan, Carl. Cosmos. Ballantine, 2013 [1980]. This tome presents a classic overview of astronomy by an astronomer who had a true gift for explaining things clearly. (You can also check out Sagan\u2019s television series Cosmos: A Personal Voyage and Neil DeGrasse Tyson\u2019s current series Cosmos: A Spacetime Odyssey.) 30 Chapter 1 Science and the Universe: A Brief Tour Tyson, Neil DeGrasse, and Don Goldsmith. Origins: Fourteen Billion Years of Cosmic Evolution. Norton, 2004. This book provides a guided tour through the beginnings of the universe, galaxies, stars, planets, and life. Websites If you enjoyed the beautiful images in this chapter (and there are many more fabulous photos to come in other chapters), you may want to know where you can obtain and download such pictures for your own enjoyment. (Many astronomy images are from government-supported instruments or projects, paid for by tax dollars, and therefore are free of copyright laws.) Here are three resources we especially like: \u2022 Astronomy Picture of the Day: apod.nasa.gov/apod/astropix.html. Two space scientists scour the Internet and select one beautiful astronomy image to feature each day. Their archives range widely, from images of planets and nebulae to rockets and space instruments; they also have many photos of the night sky. The search function (see the menu on the bottom of the page) works quite well for finding something specific among the many years\u2019 worth of daily images. \u2022 Hubble Space Telescope Images: www.hubblesite.org/newscenter/archive/browse/images. Starting at this page, you can select from among hundreds of Hubble pictures by subject or by date. Note that many of the images have supporting pictures with them, such as diagrams, animations, or comparisons. Excellent captions and background information are provided. Other ways to approach these images are through the more public-oriented Hubble Gallery (www.hubblesite.org/gallery) and the European homepage (www.spacetelescope.org/images). \u2022 National Aeronautics and Space Administration\u2019s (NASA\u2019s) Planetary Photojournal: photojournal.jpl.nasa.gov. This site features thousands of images from planetary exploration, with captions of varied length. You can select images by world, feature", " name, date, or catalog number, and download images in a number of popular formats. However, only NASA mission images are included. Note the Photojournal Search option on the menu at the top of the homepage to access ways to search their archives. Videos Cosmic Voyage: www.youtube.com/watch?v=qxXf7AJZ73A. This video presents a portion of Cosmic Voyage, narrated by Morgan Freeman (8:34). Powers of Ten: www.youtube.com/watch?v=0fKBhvDjuy0. This classic short video is a much earlier version of Powers of Ten, narrated by Philip Morrison (9:00). The Known Universe: www.youtube.com/watch?v=17jymDn0W6U. This video tour from the American Museum of Natural History has realistic animation, music, and captions (6:30). Wanderers: apod.nasa.gov/apod/ap141208.html. This video provides a tour of the solar system, with narrative by Carl Sagan, imagining other worlds with dramatically realistic paintings (3:50). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 31 2 OBSERVING THE SKY: THE BIRTH OF ASTRONOMY Figure 2.1. Night Sky. In this panoramic photograph of the night sky from the Atacama Desert in Chile, we can see the central portion of the Milky Way Galaxy arcing upward in the center of the frame. On the left, the Large Magellanic Cloud and the Small Magellanic Cloud (smaller galaxies that orbit the Milky Way Galaxy) are easily visible from the Southern Hemisphere. (credit: modification of work by ESO/Y. Beletsky) Chapter Outline 2.1 The Sky Above 2.2 Ancient Astronomy 2.3 Astrology and Astronomy 2.4 The Birth of Modern Astronomy Thinking Ahead Much to your surprise, a member of the Flat Earth Society moves in next door. He believes that Earth is flat and all the NASA images of a spherical Earth are either faked or simply show the round (but flat) disk of Earth from above. How could you prove to your new neighbor that Earth really is a sphere? (When you\u2019ve thought about this on your own, you can check later in the chapter for some suggested answers.) Today, few people really spend much time looking at the night sky. In ancient days, before electric lights robbed so many people of the beauty of the sky, the stars and planets were an important aspect of everyone\u2019s daily life. All the records that we have\u2014on paper and in stone\u2014show that ancient civilizations around the world noticed, worshipped, and tried to understand the lights in the sky and fit them into their own view of the world. These ancient observers found both majestic regularity and never-ending surprise in the motions of the heavens. Through their careful study of the planets, the Greeks and later the Romans laid the foundation of the science of astronomy. 32 Chapter 2 Observing the Sky: The Birth of Astronomy 2.1 THE SKY ABOVE Learning Objectives By the end of this section, you will be able to: Define the main features of the celestial sphere Explain the system astronomers use to describe the sky Describe how motions of the stars appear to us on Earth Describe how motions of the Sun, Moon, and planets appear to us on Earth Understand the modern meaning of the term constellation Our senses suggest to us that Earth is the center of the universe\u2014the hub around which the heavens turn. This geocentric (Earth-centered) view was what almost everyone believed until the European Renaissance. After all, it is simple, logical, and seemingly self-evident. Furthermore, the geocentric perspective reinforced those philosophical and religious systems that taught the unique role of human beings as the central focus of the cosmos. However, the geocentric view happens to be wrong. One of the great themes of our intellectual history is the overthrow of the geocentric perspective. Let us, therefore, take a look at the steps by which we reevaluated the place of our world in the cosmic order. The Celestial Sphere If you go on a camping trip or live far from city lights, your view of the sky on a clear night is pretty much identical to that seen by people all over the world before the invention of the telescope. Gazing up, you get the impression that the sky is a great hollow dome with you at the center (Figure 2.2), and all the stars are an equal distance from you on the surface of the dome. The top of that dome, the point directly above your head, is called the zenith, and where the dome meets Earth is called the horizon. From the sea or a flat pra", "irie, it is easy to see the horizon as a circle around you, but from most places where people live today, the horizon is at least partially hidden by mountains, trees, buildings, or smog. Figure 2.2. The Sky around Us. The horizon is where the sky meets the ground; an observer\u2019s zenith is the point directly overhead. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 33 If you lie back in an open field and observe the night sky for hours, as ancient shepherds and travelers regularly did, you will see stars rising on the eastern horizon (just as the Sun and Moon do), moving across the dome of the sky in the course of the night, and setting on the western horizon. Watching the sky turn like this night after night, you might eventually get the idea that the dome of the sky is really part of a great sphere that is turning around you, bringing different stars into view as it turns. The early Greeks regarded the sky as just such a celestial sphere (Figure 2.3). Some thought of it as an actual sphere of transparent crystalline material, with the stars embedded in it like tiny jewels. Figure 2.3. Circles on the Celestial Sphere. Here we show the (imaginary) celestial sphere around Earth, on which objects are fixed, and which rotates around Earth on an axis. In reality, it is Earth that turns around this axis, creating the illusion that the sky revolves around us. Note that Earth in this picture has been tilted so that your location is at the top and the North Pole is where the N is. The apparent motion of celestial objects in the sky around the pole is shown by the circular arrow. Today, we know that it is not the celestial sphere that turns as night and day proceed, but rather the planet on which we live. We can put an imaginary stick through Earth\u2019s North and South Poles, representing our planet\u2019s axis. It is because Earth turns on this axis every 24 hours that we see the Sun, Moon, and stars rise and set with clockwork regularity. Today, we know that these celestial objects are not really on a dome, but at greatly varying distances from us in space. Nevertheless, it is sometimes still convenient to talk about the celestial dome or sphere to help us keep track of objects in the sky. There is even a special theater, called a planetarium, in which we project a simulation of the stars and planets onto a white dome. As the celestial sphere rotates, the objects on it maintain their positions with respect to one another. A grouping of stars such as the Big Dipper has the same shape during the course of the night, although it turns with the 34 Chapter 2 Observing the Sky: The Birth of Astronomy sky. During a single night, even objects we know to have significant motions of their own, such as the nearby planets, seem fixed relative to the stars. Only meteors\u2014brief \u201cshooting stars\u201d that flash into view for just a few seconds\u2014move appreciably with respect to other objects on the celestial sphere. (This is because they are not stars at all. Rather, they are small pieces of cosmic dust, burning up as they hit Earth\u2019s atmosphere.) We can use the fact that the entire celestial sphere seems to turn together to help us set up systems for keeping track of what things are visible in the sky and where they happen to be at a given time. Celestial Poles and Celestial Equator To help orient us in the turning sky, astronomers use a system that extends Earth\u2019s axis points into the sky. Imagine a line going through Earth, connecting the North and South Poles. This is Earth\u2019s axis, and Earth rotates about this line. If we extend this imaginary line outward from Earth, the points where this line intersects the celestial sphere are called the north celestial pole and the south celestial pole. As Earth rotates about its axis, the sky appears to turn in the opposite direction around those celestial poles (Figure 2.4). We also (in our imagination) throw Earth\u2019s equator onto the sky and call this the celestial equator. It lies halfway between the celestial poles, just as Earth\u2019s equator lies halfway between our planet\u2019s poles. Figure 2.4. Circling the South Celestial Pole. This long-exposure photo shows trails left by stars as a result of the apparent rotation of the celestial sphere around the south celestial pole. (In reality, it is Earth that rotates.) (Credit: ESO/Iztok Bon\u010dina) Now let\u2019s imagine how riding on different parts of our spinning Earth affects our view of the sky. The apparent motion of the celestial sphere depends on your latitude (position north or south of the equator). First of all, notice", " that Earth\u2019s axis is pointing at the celestial poles, so these two points in the sky do not appear to turn. If you stood at the North Pole of Earth, for example, you would see the north celestial pole overhead, at your zenith. The celestial equator, 90\u00b0 from the celestial poles, would lie along your horizon. As you watched the stars during the course of the night, they would all circle around the celestial pole, with none rising or setting. Only that half of the sky north of the celestial equator is ever visible to an observer at the North Pole. Similarly, an observer at the South Pole would see only the southern half of the sky. If you were at Earth\u2019s equator, on the other hand, you see the celestial equator (which, after all, is just an \u201cextension\u201d of Earth\u2019s equator) pass overhead through your zenith. The celestial poles, being 90\u00b0 from the celestial equator, must then be at the north and south points on your horizon. As the sky turns, all stars rise This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 35 and set; they move straight up from the east side of the horizon and set straight down on the west side. During a 24-hour period, all stars are above the horizon exactly half the time. (Of course, during some of those hours, the Sun is too bright for us to see them.) What would an observer in the latitudes of the United States or Europe see? Remember, we are neither at Earth\u2019s pole nor at the equator, but in between them. For those in the continental United States and Europe, the north celestial pole is neither overhead nor on the horizon, but in between. It appears above the northern horizon at an angular height, or altitude, equal to the observer\u2019s latitude. In San Francisco, for example, where the latitude is 38\u00b0 N, the north celestial pole is 38\u00b0 above the northern horizon. For an observer at 38\u00b0 N latitude, the south celestial pole is 38\u00b0 below the southern horizon and, thus, never visible. As Earth turns, the whole sky seems to pivot about the north celestial pole. For this observer, stars within 38\u00b0 of the North Pole can never set. They are always above the horizon, day and night. This part of the sky is called the north circumpolar zone. For observers in the continental United States, the Big Dipper, Little Dipper, and Cassiopeia are examples of star groups in the north circumpolar zone. On the other hand, stars within 38\u00b0 of the south celestial pole never rise. That part of the sky is the south circumpolar zone. To most U.S. observers, the Southern Cross is in that zone. (Don\u2019t worry if you are not familiar with the star groups just mentioned; we will introduce them more formally later on.) The Rotating Sky Lab (https://openstaxcollege.org/l/30rotatingsky) created by the University of Nebraska\u2013Lincoln provides an interactive demonstration that introduces the horizon coordinate system, the apparent rotation of the sky, and allows for exploration of the relationship between the horizon and celestial equatorial coordinate systems. At this particular time in Earth\u2019s history, there happens to be a star very close to the north celestial pole. It is called Polaris, the pole star, and has the distinction of being the star that moves the least amount as the northern sky turns each day. Because it moved so little while the other stars moved much more, it played a special role in the mythology of several Native American tribes, for example (some called it the \u201cfastener of the sky\u201d). What\u2019s Your Angle? Astronomers measure how far apart objects appear in the sky by using angles. By definition, there are 360\u00b0 in a circle, so a circle stretching completely around the celestial sphere contains 360\u00b0. The half- sphere or dome of the sky then contains 180\u00b0 from horizon to opposite horizon. Thus, if two stars are 18\u00b0 apart, their separation spans about 1/10 of the dome of the sky. To give you a sense of how big a degree is, the full Moon is about half a degree across. This is about the width of your smallest finger (pinkie) seen at arm\u2019s length. 36 Chapter 2 Observing the Sky: The Birth of Astronomy Rising and Setting of the Sun We described the movement of stars in the night sky, but what about during the daytime? The stars continue to circle during the day, but the brilliance of the Sun makes them difficult to see. (The Moon can often be seen in the daylight, however.) On any given day, we can think of the Sun as being located at some position on the", " hypothetical celestial sphere. When the Sun rises\u2014that is, when the rotation of Earth carries the Sun above the horizon\u2014sunlight is scattered by the molecules of our atmosphere, filling our sky with light and hiding the stars above the horizon. For thousands of years, astronomers have been aware that the Sun does more than just rise and set. It changes position gradually on the celestial sphere, moving each day about 1\u00b0 to the east relative to the stars. Very reasonably, the ancients thought this meant the Sun was slowly moving around Earth, taking a period of time we call 1 year to make a full circle. Today, of course, we know it is Earth that is going around the Sun, but the effect is the same: the Sun\u2019s position in our sky changes day to day. We have a similar experience when we walk around a campfire at night; we see the flames appear in front of each person seated about the fire in turn. The path the Sun appears to take around the celestial sphere each year is called the ecliptic (Figure 2.5). Because of its motion on the ecliptic, the Sun rises about 4 minutes later each day with respect to the stars. Earth must make just a bit more than one complete rotation (with respect to the stars) to bring the Sun up again. Figure 2.5. Star Circles at Different Latitudes. The turning of the sky looks different depending on your latitude on Earth. (a) At the North Pole, the stars circle the zenith and do not rise and set. (b) At the equator, the celestial poles are on the horizon, and the stars rise straight up and set straight down. (c) At intermediate latitudes, the north celestial pole is at some position between overhead and the horizon. Its angle above the horizon turns out to be equal to the observer\u2019s latitude. Stars rise and set at an angle to the horizon. As the months go by and we look at the Sun from different places in our orbit, we see it projected against different places in our orbit, and thus against different stars in the background (Figure 2.6 and Table 2.1)\u2014or we would, at least, if we could see the stars in the daytime. In practice, we must deduce which stars lie behind and beyond the Sun by observing the stars visible in the opposite direction at night. After a year, when Earth has completed one trip around the Sun, the Sun will appear to have completed one circuit of the sky along the ecliptic. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 37 Figure 2.6. Constellations on the Ecliptic. As Earth revolves around the Sun, we sit on \u201cplatform Earth\u201d and see the Sun moving around the sky. The circle in the sky that the Sun appears to make around us in the course of a year is called the ecliptic. This circle (like all circles in the sky) goes through a set of constellations. The ancients thought these constellations, which the Sun (and the Moon and planets) visited, must be special and incorporated them into their system of astrology. Note that at any given time of the year, some of the constellations crossed by the ecliptic are visible in the night sky; others are in the day sky and are thus hidden by the brilliance of the Sun. Constellations on the Ecliptic Constellation on the Ecliptic Dates When the Sun Crosses It Capricornus Aquarius Pisces Aries Taurus Gemini Cancer Leo Virgo Table 2.1 January 21\u2013February 16 February 16\u2013March 11 March 11\u2013April 18 April 18\u2013May 13 May 13\u2013June 22 June 22\u2013July 21 July 21\u2013August 10 August 10\u2013September 16 September 16\u2013October 31 38 Chapter 2 Observing the Sky: The Birth of Astronomy Constellations on the Ecliptic Constellation on the Ecliptic Dates When the Sun Crosses It Libra Scorpius Ophiuchus Sagittarius Table 2.1 October 31\u2013November 23 November 23\u2013November 29 November 29\u2013December 18 December 18\u2013January 21 The ecliptic does not lie along the celestial equator but is inclined to it at an angle of about 23.5\u00b0. In other words, the Sun\u2019s annual path in the sky is not linked with Earth\u2019s equator. This is because our planet\u2019s axis of rotation is tilted by about 23.5\u00b0 from a vertical line sticking out of the plane of the ecliptic (Figure 2.7). Being tilted from \u201cstraight up\u201d is not at all unusual among celestial bodies; Uranus and Pluto are actually tilted so much that they orbit the Sun \u201con their side.", "\u201d Figure 2.7. The Celestial Tilt. The celestial equator is tilted by 23.5\u00b0 to the ecliptic. As a result, North Americans and Europeans see the Sun north of the celestial equator and high in our sky in June, and south of the celestial equator and low in the sky in December. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 39 The inclination of the ecliptic is the reason the Sun moves north and south in the sky as the seasons change. In Earth, Moon, and Sky, we discuss the progression of the seasons in more detail. Fixed and Wandering Stars The Sun is not the only object that moves among the fixed stars. The Moon and each of the planets that are visible to the unaided eye\u2014Mercury, Venus, Mars, Jupiter, Saturn, and Uranus (although just barely)\u2014also change their positions slowly from day to day. During a single day, the Moon and planets all rise and set as Earth turns, just as the Sun and stars do. But like the Sun, they have independent motions among the stars, superimposed on the daily rotation of the celestial sphere. Noticing these motions, the Greeks of 2000 years ago distinguished between what they called the fixed stars\u2014those that maintain fixed patterns among themselves through many generations\u2014and the wandering stars, or planets. The word \u201cplanet,\u201d in fact, means \u201cwanderer\u201d in ancient Greek. Today, we do not regard the Sun and Moon as planets, but the ancients applied the term to all seven of the moving objects in the sky. Much of ancient astronomy was devoted to observing and predicting the motions of these celestial wanderers. They even dedicated a unit of time, the week, to the seven objects that move on their own; that\u2019s why there are 7 days in a week. The Moon, being Earth\u2019s nearest celestial neighbor, has the fastest apparent motion; it completes a trip around the sky in about 1 month (or moonth). To do this, the Moon moves about 12\u00b0, or 24 times its own apparent width on the sky, each day Angles in the Sky A circle consists of 360 degrees (\u00b0). When we measure the angle in the sky that something moves, we can use this formula: speed = distance time This is true whether the motion is measured in kilometers per hour or degrees per hour; we just need to use consistent units. As an example, let\u2019s say you notice the bright star Sirius due south from your observing location in the Northern Hemisphere. You note the time, and then later, you note the time that Sirius sets below the horizon. You find that Sirius has traveled an angular distance of about 75\u00b0 in 5 h. About how many hours will it take for Sirius to return to its original location? Solution The speed of Sirius is 75\u00b0 5 h = 15\u00b0 1 h. If we want to know the time required for Sirius to return to its original location, we need to wait until it goes around a full circle, or 360\u00b0. Rearranging the formula for speed we were originally given, we find: The actual time is a few minutes shorter than this, and we will explore why in a later chapter. time = distance speed = 360\u00b0 15\u00b0/h = 24 h 40 Chapter 2 Observing the Sky: The Birth of Astronomy Check Your Learning The Moon moves in the sky relative to the background stars (in addition to moving with the stars as a result of Earth\u2019s rotation.) Go outside at night and note the position of the Moon relative to nearby stars. Repeat the observation a few hours later. How far has the Moon moved? (For reference, the diameter of the Moon is about 0.5\u00b0.) Based on your estimate of its motion, how long will it take for the Moon to return to the position relative to the stars in which you first observed it? Answer: The speed of the moon is 0.5\u00b0/1 h. To move a full 360\u00b0, the moon needs 720 h: 0.5\u00b0 1 h = 360\u00b0 720 h. Dividing 720 h by the conversion factor of 24 h/day reveals the lunar cycle is about 30 days. The individual paths of the Moon and planets in the sky all lie close to the ecliptic, although not exactly on it. This is because the paths of the planets about the Sun, and of the Moon about Earth, are all in nearly the same plane, as if they were circles on a huge sheet of paper. The planets, the Sun, and the Moon are thus always found in the sky within a narrow 18-degree-wide belt, centered on the ecliptic, called the zodiac (Figure 2.6). (The root of the term \u201czodiac\u201d is the same as that", " of the word \u201czoo\u201d and means a collection of animals; many of the patterns of stars within the zodiac belt reminded the ancients of animals, such as a fish or a goat.) How the planets appear to move in the sky as the months pass is a combination of their actual motions plus the motion of Earth about the Sun; consequently, their paths are somewhat complex. As we will see, this complexity has fascinated and challenged astronomers for centuries. Constellations The backdrop for the motions of the \u201cwanderers\u201d in the sky is the canopy of stars. If there were no clouds in the sky and we were on a flat plain with nothing to obstruct our view, we could see about 3000 stars with the unaided eye. To find their way around such a multitude, the ancients found groupings of stars that made some familiar geometric pattern or (more rarely) resembled something they knew. Each civilization found its own patterns in the stars, much like a modern Rorschach test in which you are asked to discern patterns or pictures in a set of inkblots. The ancient Chinese, Egyptians, and Greeks, among others, found their own groupings\u2014or constellations\u2014of stars. These were helpful in navigating among the stars and in passing their star lore on to their children. You may be familiar with some of the old star patterns we still use today, such as the Big Dipper, Little Dipper, and Orion the hunter, with his distinctive belt of three stars (Figure 2.8). However, many of the stars we see are not part of a distinctive star pattern at all, and a telescope reveals millions of stars too faint for the eye to see. Therefore, during the early decades of the 20th century, astronomers from many countries decided to establish a more formal system for organizing the sky. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 41 Figure 2.8. Orion. (a) The winter constellation of Orion, the hunter, is surrounded by neighboring constellations, as illustrated in the seventeenth-century atlas by Hevelius. (b) A photograph shows the Orion region in the sky. Note the three blue stars that make up the belt of the hunter. The bright red star above the belt denotes his armpit and is called Betelgeuse (pronounced \u201cBeetel-juice\u201d). The bright blue star below the belt is his foot and is called Rigel. (credit a: modification of work by Johannes Hevelius; b: modification of work by Matthew Spinelli) Today, we use the term constellation to mean one of 88 sectors into which we divide the sky, much as the United States is divided into 50 states. The modern boundaries between the constellations are imaginary lines in the sky running north\u2013south and east\u2013west, so that each point in the sky falls in a specific constellation, although, like the states, not all constellations are the same size. All the constellations are listed in Appendix L. Whenever possible, we have named each modern constellation after the Latin translations of one of the ancient Greek star patterns that lies within it. Thus, the modern constellation of Orion is a kind of box on the sky, which includes, among many other objects, the stars that made up the ancient picture of the hunter. Some people use the term asterism to denote an especially noticeable star pattern within a constellation (or sometimes spanning parts of several constellations). For example, the Big Dipper is an asterism within the constellation of Ursa Major, the Big Bear. Students are sometimes puzzled because the constellations seldom resemble the people or animals for which they were named. In all likelihood, the Greeks themselves did not name groupings of stars because they looked like actual people or subjects (any more than the outline of Washington state resembles George Washington). Rather, they named sections of the sky in honor of the characters in their mythology and then fit the star configurations to the animals and people as best they could This website about objects in the sky (https://openstaxcollege.org/l/30heavensabove) allows users to construct a detailed sky map showing the location and information about the Sun, Moon, planets, stars, constellations, and even satellites orbiting Earth. Begin by setting your observing location using the option in the menu in the upper right corner of the screen. 42 Chapter 2 Observing the Sky: The Birth of Astronomy 2.2 ANCIENT ASTRONOMY Learning Objectives By the end of this section, you will be able to: Describe early examples of astronomy around the world Explain how Greek astronomers were able to deduce that Earth is spherical Explain how Greek astronomers were able to calculate Earth\u2019s size Describe the motion of Earth called precession Describe Ptolemy\u2019s", " geocentric system of planetary motion Let us now look briefly back into history. Much of modern Western civilization is derived in one way or another from the ideas of the ancient Greeks and Romans, and this is true in astronomy as well. However, many other ancient cultures also developed sophisticated systems for observing and interpreting the sky. Astronomy around the World Ancient Babylonian, Assyrian, and Egyptian astronomers knew the approximate length of the year. The Egyptians of 3000 years ago, for example, adopted a calendar based on a 365-day year. They kept careful track of the rising time of the bright star Sirius in the predawn sky, which has a yearly cycle that corresponded with the flooding of the Nile River. The Chinese also had a working calendar; they determined the length of the year at about the same time as the Egyptians. The Chinese also recorded comets, bright meteors, and dark spots on the Sun. (Many types of astronomical objects were introduced in Science and the Universe: A Brief Tour. If you are not familiar with terms like comets and meteors, you may want to review that chapter.) Later, Chinese astronomers kept careful records of \u201cguest stars\u201d\u2014those that are normally too faint to see but suddenly flare up to become visible to the unaided eye for a few weeks or months. We still use some of these records in studying stars that exploded a long time ago. The Mayan culture in Mexico and Central America developed a sophisticated calendar based on the planet Venus, and they made astronomical observations from sites dedicated to this purpose a thousand years ago. The Polynesians learned to navigate by the stars over hundreds of kilometers of open ocean\u2014a skill that enabled them to colonize new islands far away from where they began. In Britain, before the widespread use of writing, ancient people used stones to keep track of the motions of the Sun and Moon. We still find some of the great stone circles they built for this purpose, dating from as far back as 2800 BCE. The best known of these is Stonehenge, which is discussed in Earth, Moon, and Sky. Early Greek and Roman Cosmology Our concept of the cosmos\u2014its basic structure and origin\u2014is called cosmology, a word with Greek roots. Before the invention of telescopes, humans had to depend on the simple evidence of their senses for a picture of the universe. The ancients developed cosmologies that combined their direct view of the heavens with a rich variety of philosophical and religious symbolism. At least 2000 years before Columbus, educated people in the eastern Mediterranean region knew Earth was round. Belief in a spherical Earth may have stemmed from the time of Pythagoras, a philosopher and mathematician who lived 2500 years ago. He believed circles and spheres to be \u201cperfect forms\u201d and suggested that Earth should therefore be a sphere. As evidence that the gods liked spheres, the Greeks cited the fact that the Moon is a sphere, using evidence we describe later. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 43 The writings of Aristotle (384\u2013322 BCE), the tutor of Alexander the Great, summarize many of the ideas of his day. They describe how the progression of the Moon\u2019s phases\u2014its apparent changing shape\u2014results from our seeing different portions of the Moon\u2019s sunlit hemisphere as the month goes by (see Earth, Moon, and Sky). Aristotle also knew that the Sun has to be farther away from Earth than is the Moon because occasionally the Moon passed exactly between Earth and the Sun and hid the Sun temporarily from view. We call this a solar eclipse. Aristotle cited convincing arguments that Earth must be round. First is the fact that as the Moon enters or emerges from Earth\u2019s shadow during an eclipse of the Moon, the shape of the shadow seen on the Moon is always round (Figure 2.9). Only a spherical object always produces a round shadow. If Earth were a disk, for example, there would be some occasions when the sunlight would strike it edge-on and its shadow on the Moon would be a line. Figure 2.9. Earth\u2019s Round Shadow. A lunar eclipse occurs when the Moon moves into and out of Earth\u2019s shadow. Note the curved shape of the shadow\u2014evidence for a spherical Earth that has been recognized since antiquity. (credit: modification of work by Brian Paczkowski) As a second argument, Aristotle explained that travelers who go south a significant distance are able to observe stars that are not visible farther north. And the height of the North Star\u2014the star nearest the north celestial pole\u2014decreases as a traveler moves south. On a flat Earth, everyone would see the same stars overhead. The only possible explanation is that the traveler must have moved over a curved surface on Earth, showing stars from a different angle. (See the How Do We Know Earth Is Round? feature for more ideas on proving Earth is", " round.) One Greek thinker, Aristarchus of Samos (310\u2013230 BCE), even suggested that Earth was moving around the Sun, but Aristotle and most of the ancient Greek scholars rejected this idea. One of the reasons for their conclusion was the thought that if Earth moved about the Sun, they would be observing the stars from different places along Earth\u2019s orbit. As Earth moved along, nearby stars should shift their positions in the sky relative to more distant stars. In a similar way, we see foreground objects appear to move against a more distant background whenever we are in motion. When we ride on a train, the trees in the foreground appear to shift their position relative to distant hills as the train rolls by. Unconsciously, we use this phenomenon all of the time to estimate distances around us. The apparent shift in the direction of an object as a result of the motion of the observer is called parallax. We call the shift in the apparent direction of a star due to Earth\u2019s orbital motion stellar parallax. The Greeks made dedicated efforts to observe stellar parallax, even enlisting the aid of Greek soldiers with the clearest vision, but to no avail. The brighter (and presumably nearer) stars just did not seem to shift as the Greeks observed them in the spring and then again in the fall (when Earth is on the opposite side of the Sun). This meant either that Earth was not moving or that the stars had to be so tremendously far away that the parallax shift was immeasurably small. A cosmos of such enormous extent required a leap of imagination that most ancient philosophers were not prepared to make, so they retreated to the safety of the Earth-centered view, which would dominate Western thinking for nearly two millennia. 44 Chapter 2 Observing the Sky: The Birth of Astronomy How Do We Know Earth Is Round? In addition to the two ways (from Aristotle\u2019s writings) discussed in this chapter, you might also reason as follows: 1. Let\u2019s watch a ship leave its port and sail into the distance on a clear day. On a flat Earth, we would just see the ship get smaller and smaller as it sails away. But this isn\u2019t what we actually observe. Instead, ships sink below the horizon, with the hull disappearing first and the mast remaining visible for a while longer. Eventually, only the top of the mast can be seen as the ship sails around the curvature of Earth. Finally, the ship disappears under the horizon. 2. The International Space Station circles Earth once every 90 minutes or so. Photographs taken from the shuttle and other satellites show that Earth is round from every perspective. 3. Suppose you made a friend in each time zone of Earth. You call all of them at the same hour and ask, \u201cWhere is the Sun?\u201d On a flat Earth, each caller would give you roughly the same answer. But on a round Earth you would find that, for some friends, the Sun would be high in the sky whereas for others it would be rising, setting, or completely out of sight (and this last group of friends would be upset with you for waking them up). Measurement of Earth by Eratosthenes The Greeks not only knew Earth was round, but also they were able to measure its size. The first fairly accurate determination of Earth\u2019s diameter was made in about 200 BCE by Eratosthenes (276\u2013194 BCE), a Greek living in Alexandria, Egypt. His method was a geometric one, based on observations of the Sun. The Sun is so distant from us that all the light rays that strike our planet approach us along essentially parallel lines. To see why, look at Figure 2.10. Take a source of light near Earth\u2014say, at position A. Its rays strike different parts of Earth along diverging paths. From a light source at B, or at C (which is still farther away), the angle between rays that strike opposite parts of Earth is smaller. The more distant the source, the smaller the angle between the rays. For a source infinitely distant, the rays travel along parallel lines. Figure 2.10. Light Rays from Space. The more distant an object, the more nearly parallel the rays of light coming from it. Of course, the Sun is not infinitely far away, but given its distance of 150 million kilometers, light rays striking Earth from a point on the Sun diverge from one another by an angle far too small to be observed with the unaided eye. As a consequence, if people all over Earth who could see the Sun were to point at it, their fingers This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 45 would, essentially, all be parallel to one another. (The same is also true for the planets and stars\u2014an idea we will use in our discussion of how telescopes work.)", " Eratosthenes was told that on the first day of summer at Syene, Egypt (near modern Aswan), sunlight struck the bottom of a vertical well at noon. This indicated that the Sun was directly over the well\u2014meaning that Syene was on a direct line from the center of Earth to the Sun. At the corresponding time and date in Alexandria, Eratosthenes observed the shadow a column made and saw that the Sun was not directly overhead, but was slightly south of the zenith, so that its rays made an angle with the vertical equal to about 1/50 of a circle (7\u00b0). Because the Sun\u2019s rays striking the two cities are parallel to one another, why would the two rays not make the same angle with Earth\u2019s surface? Eratosthenes reasoned that the curvature of the round Earth meant that \u201cstraight up\u201d was not the same in the two cities. And the measurement of the angle in Alexandria, he realized, allowed him to figure out the size of Earth. Alexandria, he saw, must be 1/50 of Earth\u2019s circumference north of Syene (Figure 2.11). Alexandria had been measured to be 5000 stadia north of Syene. (The stadium was a Greek unit of length, derived from the length of the racetrack in a stadium.) Eratosthenes thus found that Earth\u2019s circumference must be 50 \u00d7 5000, or 250,000 stadia. Figure 2.11. How Eratosthenes Measured the Size of Earth. Eratosthenes measured the size of Earth by observing the angle at which the Sun\u2019s rays hit our planet\u2019s surface. The Sun\u2019s rays come in parallel, but because Earth\u2019s surface curves, a ray at Syene comes straight down whereas a ray at Alexandria makes an angle of 7\u00b0 with the vertical. That means, in effect, that at Alexandria, Earth\u2019s surface has curved away from Syene by 7\u00b0 of 360\u00b0, or 1/50 of a full circle. Thus, the distance between the two cities must be 1/50 the circumference of Earth. (credit: modification of work by NOAA Ocean Service Education) It is not possible to evaluate precisely the accuracy of Eratosthenes solution because there is doubt about which of the various kinds of Greek stadia he used as his unit of distance. If it was the common Olympic stadium, his result is about 20% too large. According to another interpretation, he used a stadium equal to 46 Chapter 2 Observing the Sky: The Birth of Astronomy about 1/6 kilometer, in which case his figure was within 1% of the correct value of 40,000 kilometers. Even if his measurement was not exact, his success at measuring the size of our planet by using only shadows, sunlight, and the power of human thought was one of the greatest intellectual achievements in history. Hipparchus and Precession Perhaps the greatest astronomer of antiquity was Hipparchus, born in Nicaea in what is present-day Turkey. He erected an observatory on the island of Rhodes around 150 BCE, when the Roman Republic was expanding its influence throughout the Mediterranean region. There he measured, as accurately as possible, the positions of objects in the sky, compiling a pioneering star catalog with about 850 entries. He designated celestial coordinates for each star, specifying its position in the sky, just as we specify the position of a point on Earth by giving its latitude and longitude. He also divided the stars into apparent magnitudes according to their apparent brightness. He called the brightest ones \u201cstars of the first magnitude\u201d; the next brightest group, \u201cstars of the second magnitude\u201d; and so forth. This rather arbitrary system, in modified form, still remains in use today (although it is less and less useful for professional astronomers). By observing the stars and comparing his data with older observations, Hipparchus made one of his most remarkable discoveries: the position in the sky of the north celestial pole had altered over the previous century and a half. Hipparchus deduced correctly that this had happened not only during the period covered by his observations, but was in fact happening all the time: the direction around which the sky appears to rotate changes slowly but continuously. Recall from the section on celestial poles and the celestial equator that the north celestial pole is just the projection of Earth\u2019s North Pole into the sky. If the north celestial pole is wobbling around, then Earth itself must be doing the wobbling. Today, we understand that the direction in which Earth\u2019s axis points does indeed change slowly but regularly\u2014a motion we call precession. If you have ever watched a spinning top wobble, you observed a similar kind of motion. The top\u2019s axis describes a path in the shape of a cone, as Earth\u2019s gravity tries to topple it (Figure 2.12). This OpenStax book", " is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 47 Figure 2.12. Precession. Just as the axis of a rapidly spinning top wobbles slowly in a circle, so the axis of Earth wobbles in a 26,000-year cycle. Today the north celestial pole is near the star Polaris, but about 5000 years ago it was close to a star called Thuban, and in 14,000 years it will be closest to the star Vega. Because our planet is not an exact sphere, but bulges a bit at the equator, the pulls of the Sun and Moon cause it to wobble like a top. It takes about 26,000 years for Earth\u2019s axis to complete one circle of precession. As a result of this motion, the point where our axis points in the sky changes as time goes on. While Polaris is the star closest to the north celestial pole today (it will reach its closest point around the year 2100), the star Vega in the constellation of Lyra will be the North Star in 14,000 years. Ptolemy\u2019s Model of the Solar System The last great astronomer of the Roman era was Claudius Ptolemy (or Ptolemaeus), who flourished in Alexandria in about the year 140. He wrote a mammoth compilation of astronomical knowledge, which today is called by its Arabic name, Almagest (meaning \u201cThe Greatest\u201d). Almagest does not deal exclusively with Ptolemy\u2019s own work; it includes a discussion of the astronomical achievements of the past, principally those of Hipparchus. Today, it is our main source of information about the work of Hipparchus and other Greek astronomers. Ptolemy\u2019s most important contribution was a geometric representation of the solar system that predicted the positions of the planets for any desired date and time. Hipparchus, not having enough data on hand to solve the problem himself, had instead amassed observational material for posterity to use. Ptolemy supplemented this material with new observations of his own and produced a cosmological model that endured more than a thousand years, until the time of Copernicus. The complicating factor in explaining the motions of the planets is that their apparent wandering in the sky results from the combination of their own motions with Earth\u2019s orbital revolution. As we watch the planets from our vantage point on the moving Earth, it is a little like watching a car race while you are competing in it. Sometimes opponents\u2019 cars pass you, but at other times you pass them, making them appear to move backward for a while with respect to you. 48 Chapter 2 Observing the Sky: The Birth of Astronomy Figure 2.13 shows the motion of Earth and a planet farther from the Sun\u2014in this case, Mars. Earth travels around the Sun in the same direction as the other planet and in nearly the same plane, but its orbital speed is faster. As a result, it overtakes the planet periodically, like a faster race car on the inside track. The figure shows where we see the planet in the sky at different times. The path of the planet among the stars is illustrated in the star field on the right side of the figure. Figure 2.13. Retrograde Motion of a Planet beyond Earth\u2019s Orbit. The letters on the diagram show where Earth and Mars are at different times. By following the lines from each Earth position through each corresponding Mars position, you can see how the retrograde path of Mars looks against the background stars This retrograde simulation of Mars (https://openstaxcollege.org/l/30marsretrograd) illustrates the motion of Mars as seen from Earth as well as Earth\u2019s retrograde motion as seen from Mars. There is also an animation of the movement of the two planets relative to each other that creates the appearance of this motion. Normally, planets move eastward in the sky over the weeks and months as they orbit the Sun, but from positions B to D in Figure 2.13, as Earth passes the planets in our example, it appears to drift backward, moving west in the sky. Even though it is actually moving to the east, the faster-moving Earth has overtaken it and seems, from our perspective, to be leaving it behind. As Earth rounds its orbit toward position E, the planet again takes up its apparent eastward motion in the sky. The temporary apparent westward motion of a planet as Earth swings between it and the Sun is called retrograde motion. Such backward motion is much easier for us to understand today, now that we know Earth is one of the moving planets and not the unmoving center of all creation. But Ptolemy was faced with the far more complex problem of explaining such motion while assuming a stationary Earth. Furthermore, because the Greeks believed that celestial motions had to be circles, Ptolemy had", " to construct his model using circles alone. To do it, he needed dozens of circles, some moving around other circles, in a complex This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 49 structure that makes a modern viewer dizzy. But we must not let our modern judgment cloud our admiration for Ptolemy\u2019s achievement. In his day, a complex universe centered on Earth was perfectly reasonable and, in its own way, quite beautiful. However, as Alfonso X, the King of Castile, was reported to have said after having the Ptolemaic system of planet motions explained to him, \u201cIf the Lord Almighty had consulted me before embarking upon Creation, I should have recommended something simpler.\u201d Ptolemy solved the problem of explaining the observed motions of planets by having each planet revolve in a small orbit called an epicycle. The center of the epicycle then revolved about Earth on a circle called a deferent (Figure 2.14). When the planet is at position x in Figure 2.14 on the epicycle orbit, it is moving in the same direction as the center of the epicycle; from Earth, the planet appears to be moving eastward. When the planet is at y, however, its motion is in the direction opposite to the motion of the epicycle\u2019s center around Earth. By choosing the right combination of speeds and distances, Ptolemy succeeded in having the planet moving westward at the correct speed and for the correct interval of time, thus replicating retrograde motion with his model. Figure 2.14. Ptolemy\u2019s Complicated Cosmological System. Each planet orbits around a small circle called an epicycle. Each epicycle orbits on a larger circle called the deferent. This system is not centered exactly on Earth but on an offset point called the equant. The Greeks needed all this complexity to explain the actual motions in the sky because they believed that Earth was stationary and that all sky motions had to be circular. However, we shall see in Orbits and Gravity that the planets, like Earth, travel about the Sun in orbits that are ellipses, not circles. Their actual behavior cannot be represented accurately by a scheme of uniform circular motions. In order to match the observed motions of the planets, Ptolemy had to center the deferent circles, not on Earth, but at points some distance from Earth. In addition, he introduced uniform circular motion around yet another axis, called the equant point. All of these considerably complicated his scheme. It is a tribute to the genius of Ptolemy as a mathematician that he was able to develop such a complex system to account successfully for the observations of planets. It may be that Ptolemy did not intend for his cosmological model to describe reality, but merely to serve as a mathematical representation that allowed him to predict the positions of the planets at any time. Whatever his thinking, his model, with some modifications, was eventually accepted as authoritative in the Muslim world and (later) in Christian Europe. 2.3 ASTROLOGY AND ASTRONOMY Learning Objectives By the end of this section, you will be able to: 50 Chapter 2 Observing the Sky: The Birth of Astronomy Explain the origins of astrology Explain what a horoscope is Summarize the arguments that invalidate astrology as a scientific practice Many ancient cultures regarded the planets and stars as representatives or symbols of the gods or other supernatural forces that controlled their lives. For them, the study of the heavens was not an abstract subject; it was connected directly to the life-and-death necessity of understanding the actions of the gods and currying favor with them. Before the time of our scientific perspectives, everything that happened in nature\u2014from the weather, to diseases and accidents, to celestial surprises such as eclipses or new comets\u2014was thought to be an expression of the whims or displeasure of the gods. Any signs that helped people understand what these gods had in mind were considered extremely important. The movements of the seven objects that had the power to \u201cwander\u201d through the realm of the sky\u2014the Sun, the Moon, and five planets visible to the unaided eye\u2014clearly must have special significance in such a system of thinking. Most ancient cultures associated these seven objects with various supernatural rulers in their pantheon and kept track of them for religious reasons. Even in the comparatively sophisticated Greece of antiquity, the planets had the names of gods and were credited with having the same powers and influences as the gods whose names they bore. From such ideas was born the ancient system called astrology, still practiced by some people today, in which the positions of these bodies among the stars of the zodiac are thought to hold the key to understanding what we can expect from life. The Beginnings of Astrology Ast", "rology began in Babylonia about two and half millennia ago. The Babylonians, believing the planets and their motions influenced the fortunes of kings and nations, used their knowledge of astronomy to guide their rulers. When the Babylonian culture was absorbed by the Greeks, astrology gradually came to influence the entire Western world and eventually spread to Asia as well. By the 2nd century BCE the Greeks democratized astrology by developing the idea that the planets influence every individual. In particular, they believed that the configuration of the Sun, Moon, and planets at the moment of birth affected a person\u2019s personality and fortune\u2014a doctrine called natal astrology. Natal astrology reached its peak with Ptolemy 400 years later. As famous for his astrology as for his astronomy, Ptolemy compiled the Tetrabiblos, a treatise on astrology that remains the \u201cbible\u201d of the subject. It is essentially this ancient religion, older than Christianity or Islam, that is still practiced by today\u2019s astrologers. The Horoscope The key to natal astrology is the horoscope, a chart showing the positions of the planets in the sky at the moment of an individual\u2019s birth. The word \u201choroscope\u201d comes from the Greek words hora (meaning \u201ctime\u201d) and skopos (meaning a \u201cwatcher\u201d or \u201cmarker\u201d), so \u201choroscope\u201d can literally be translated as \u201cmarker of the hour.\u201d When a horoscope is charted, the planets (including the Sun and Moon, classed as wanderers by the ancients) must first be located in the zodiac. At the time astrology was set up, the zodiac was divided into 12 sectors called signs (Figure 2.15), each 30\u00b0 long. Each sign was named after a constellation in the sky through which the Sun, Moon, and planets were seen to pass\u2014the sign of Virgo after the constellation of Virgo, for example. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 51 Figure 2.15. Zodiac Signs. The signs of the zodiac are shown in a medieval woodcut. When someone today casually asks you your \u201csign,\u201d they are asking for your \u201csun sign\u201d\u2014which zodiac sign the Sun was in at the moment you were born. However, more than 2000 years have passed since the signs received their names from the constellations. Because of precession, the constellations of the zodiac slide westward along the ecliptic, going once around the sky in about 26,000 years. Thus, today the real stars have slipped around by about 1/12 of the zodiac\u2014about the width of one sign. In most forms of astrology, however, the signs have remained assigned to the dates of the year they had when astrology was first set up. This means that the astrological signs and the real constellations are out of step; the sign of Aries, for example, now occupies the constellation of Pisces. When you look up your sun sign in a newspaper astrology column, the name of the sign associated with your birthday is no longer the name of the constellation in which the Sun was actually located when you were born. To know that constellation, you must look for the sign before the one that includes your birthday. A complete horoscope shows the location of not only the Sun, but also the Moon and each planet in the sky by indicating its position in the appropriate sign of the zodiac. However, as the celestial sphere turns (owing to the rotation of Earth), the entire zodiac moves across the sky to the west, completing a circuit of the heavens each day. Thus, the position in the sky (or \u201chouse\u201d in astrology) must also be calculated. There are more or less standardized rules for the interpretation of the horoscope, most of which (at least in Western schools of astrology) are derived from the Tetrabiblos of Ptolemy. Each sign, each house, and each planet\u2014the last acting as a center of force\u2014is supposed to be associated with particular matters in a person\u2019s life. The detailed interpretation of a horoscope is a very complicated business, and there are many schools of astrological thought on how it should be done. Although some of the rules may be standardized, how each rule is to be weighed and applied is a matter of judgment\u2014and \u201cart.\u201d It also means that it is very difficult to tie down astrology to specific predictions or to get the same predictions from different astrologers. 52 Chapter 2 Observing the Sky: The Birth of Astronomy Astrology Today Astrologers today use the same basic principles laid", " down by Ptolemy nearly 2000 years ago. They cast horoscopes (a process much simplified by the development of appropriate computer programs) and suggest interpretations. Sun sign astrology (which you read in the newspapers and many magazines) is a recent, simplified variant of natal astrology. Although even professional astrologers do not place much trust in such a limited scheme, which tries to fit everyone into just 12 groups, sun sign astrology is taken seriously by many people (perhaps because it is discussed so commonly in the media). Today, we know much more about the nature of the planets as physical bodies, as well as about human genetics, than the ancients could. It is hard to imagine how the positions of the Sun, Moon, or planets in the sky at the moment of our birth could have anything to do with our personality or future. There are no known forces, not gravity or anything else, that could cause such effects. (For example, a straightforward calculation shows that the gravitational pull of the obstetrician delivering a newborn baby is greater than that of Mars.) Astrologers thus have to argue there must be unknown forces exerted by the planets that depend on their configurations with respect to one another and that do not vary according to the distance of the planet\u2014forces for which there is no shred of evidence. Another curious aspect of astrology is its emphasis on planet configurations at birth. What about the forces that might influence us at conception? Isn\u2019t our genetic makeup more important for determining our personality than the circumstances of our birth? Would we really be a different person if we had been born a few hours earlier or later, as astrology claims? (Back when astrology was first conceived, birth was thought of as a moment of magic significance, but today we understand a lot more about the long process that precedes it.) Actually, very few well-educated people today buy the claim that our entire lives are predetermined by astrological influences at birth, but many people apparently believe that astrology has validity as an indicator of affinities and personality. A surprising number of Americans make judgments about people\u2014whom they will hire, associate with, and even marry\u2014on the basis of astrological information. To be sure, these are difficult decisions, and you might argue that we should use any relevant information that might help us to make the right choices. But does astrology actually provide any useful information on human personality? This is the kind of question that can be tested using the scientific method (see Testing Astrology). The results of hundreds of tests are all the same: there is no evidence that natal astrology has any predictive power, even in a statistical sense. Why, then, do people often seem to have anecdotes about how well their own astrologer advised them? Effective astrologers today use the language of the zodiac and the horoscope only as the outward trappings of their craft. Mostly they work as amateur therapists, offering simple truths that clients like or need to hear. (Recent studies have shown that just about any sort of short-term therapy makes people feel a little better because the very act of talking about our problems with someone who listens attentively is, in itself, beneficial.) The scheme of astrology has no basis in scientific fact, however; at best, it can be described as a pseudoscience. It is an interesting historical system, left over from prescientific days and best remembered for the impetus it gave people to learn the cycles and patterns of the sky. From it grew the science of astronomy, which is our main subject for discussion. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 53 Testing Astrology In response to modern public interest in astrology, scientists have carried out a wide range of statistical tests to assess its predictive power. The simplest of these examine sun sign astrology to determine whether\u2014as astrologers assert\u2014some signs are more likely than others to be associated with some objective measure of success, such as winning Olympic medals, earning high corporate salaries, or achieving elective office or high military rank. (You can devise such a test yourself by looking up the birth dates of all members of Congress, for example, or all members of the U.S. Olympic team.) Are our political leaders somehow selected at birth by their horoscopes and thus more likely to be Leos, say, than Scorpios? You do not even need to be specific about your prediction in such tests. After all, many schools of astrology disagree about which signs go with which personality characteristics. To demonstrate the validity of the astrological hypothesis, it would be sufficient if the birthdays of all our leaders clustered in any one or two signs in some statistically significant way. Dozens of such tests have been performed, and all have come up completely negative: the birth dates of leaders in all fields tested have been found to be distributed randomly among all", " the signs. Sun sign astrology does not predict anything about a person\u2019s future occupation or strong personality traits. In a fine example of such a test, two statisticians examined the reenlistment records of the United States Marine Corps. We suspect you will agree that it takes a certain kind of personality not only to enlist, but also to reenlist in the Marines. If sun signs can predict strong personality traits\u2014as astrologers claim\u2014then those who reenlisted (with similar personalities) should have been distributed preferentially in those one or few signs that matched the personality of someone who loves being a Marine. However, the reenlisted were distributed randomly among all the signs. More sophisticated studies have also been done, involving full horoscopes calculated for thousands of individuals. The results of all these studies are also negative: none of the systems of astrology has been shown to be at all effective in connecting astrological aspects to personality, success, or finding the right person to love. Other tests show that it hardly seems to matter what a horoscope interpretation says, as long as it is vague enough, and as long as each subject feels it was prepared personally just for him or her. The French statistician Michel Gauquelin, for example, sent the horoscope interpretation for one of the worst mass murderers in history to 150 people, but told each recipient that it was a \u201creading\u201d prepared exclusively for him or her. Ninety-four percent of the readers said they recognized themselves in the interpretation of the mass murderer\u2019s horoscope. Geoffrey Dean, an Australian researcher, reversed the astrological readings of 22 subjects, substituting phrases that were the opposite of what the horoscope actually said. Yet, his subjects said that the resulting readings applied to them just as often (95%) as the people to whom the original phrases were given. 54 Chapter 2 Observing the Sky: The Birth of Astronomy For more on astrology and science from an astronomer\u2019s point of view, read this article (https://openstaxcollege.org/l/30astrosociety) that shines light on the topic through an accessible Q&A. 2.4 THE BIRTH OF MODERN ASTRONOMY Learning Objectives By the end of this section, you will be able to: Explain how Copernicus developed the heliocentric model of the solar system Explain the Copernican model of planetary motion and describe evidence or arguments in favor of it Describe Galileo\u2019s discoveries concerning the study of motion and forces Explain how Galileo\u2019s discoveries tilted the balance of evidence in favor of the Copernican model Astronomy made no major advances in strife-torn medieval Europe. The birth and expansion of Islam after the seventh century led to a flowering of Arabic and Jewish cultures that preserved, translated, and added to many of the astronomical ideas of the Greeks. Many of the names of the brightest stars, for example, are today taken from the Arabic, as are such astronomical terms as \u201czenith.\u201d As European culture began to emerge from its long, dark age, trading with Arab countries led to a rediscovery of ancient texts such as Almagest and to a reawakening of interest in astronomical questions. This time of rebirth (in French, \u201crenaissance\u201d) in astronomy was embodied in the work of Copernicus (Figure 2.16). Figure 2.16. Nicolaus Copernicus (1473\u20131543). Copernicus was a cleric and scientist who played a leading role in the emergence of modern science. Although he could not prove that Earth revolves about the Sun, he presented such compelling arguments for this idea that he turned the tide of cosmological thought and laid the foundations upon which Galileo and Kepler so effectively built in the following century. Copernicus One of the most important events of the Renaissance was the displacement of Earth from the center of the universe, an intellectual revolution initiated by a Polish cleric in the sixteenth century. Nicolaus Copernicus was This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 55 born in Torun, a mercantile town along the Vistula River. His training was in law and medicine, but his main interests were astronomy and mathematics. His great contribution to science was a critical reappraisal of the existing theories of planetary motion and the development of a new Sun-centered, or heliocentric, model of the solar system. Copernicus concluded that Earth is a planet and that all the planets circle the Sun. Only the Moon orbits Earth (Figure 2.17). Figure 2.17. Copernicus\u2019 System. Copernicus developed a heliocentric plan of the solar system. This system was published in the first edition of De Revolutionibus Orbium Coelestium. Notice the word Sol for", " \u201cSun\u201d in the middle. (credit: Nicolai Copernici) Copernicus described his ideas in detail in his book De Revolutionibus Orbium Coelestium (On the Revolution of Celestial Orbs), published in 1543, the year of his death. By this time, the old Ptolemaic system needed significant adjustments to predict the positions of the planets correctly. Copernicus wanted to develop an improved theory from which to calculate planetary positions, but in doing so, he was himself not free of all traditional prejudices. He began with several assumptions that were common in his time, such as the idea that the motions of the heavenly bodies must be made up of combinations of uniform circular motions. But he did not assume (as most people did) that Earth had to be in the center of the universe, and he presented a defense of the heliocentric system that was elegant and persuasive. His ideas, although not widely accepted until more than a century after his death, were much discussed among scholars and, ultimately, had a profound influence on the course of world history. One of the objections raised to the heliocentric theory was that if Earth were moving, we would all sense or feel this motion. Solid objects would be ripped from the surface, a ball dropped from a great height would not strike the ground directly below it, and so forth. But a moving person is not necessarily aware of that motion. We have all experienced seeing an adjacent train, bus, or ship appear to move, only to discover that it is we who are moving. Copernicus argued that the apparent motion of the Sun about Earth during the course of a year could be represented equally well by a motion of Earth about the Sun. He also reasoned that the apparent rotation of the celestial sphere could be explained by assuming that Earth rotates while the celestial sphere is stationary. To the objection that if Earth rotated about an axis it would fly into pieces, Copernicus answered that if such motion 56 Chapter 2 Observing the Sky: The Birth of Astronomy would tear Earth apart, the still faster motion of the much larger celestial sphere required by the geocentric hypothesis would be even more devastating. The Heliocentric Model The most important idea in Copernicus\u2019 De Revolutionibus is that Earth is one of six (then-known) planets that revolve about the Sun. Using this concept, he was able to work out the correct general picture of the solar system. He placed the planets, starting nearest the Sun, in the correct order: Mercury, Venus, Earth, Mars, Jupiter, and Saturn. Further, he deduced that the nearer a planet is to the Sun, the greater its orbital speed. With his theory, he was able to explain the complex retrograde motions of the planets without epicycles and to work out a roughly correct scale for the solar system. Copernicus could not prove that Earth revolves about the Sun. In fact, with some adjustments, the old Ptolemaic system could have accounted, as well, for the motions of the planets in the sky. But Copernicus pointed out that the Ptolemaic cosmology was clumsy and lacking the beauty and symmetry of its successor. In Copernicus\u2019 time, in fact, few people thought there were ways to prove whether the heliocentric or the older geocentric system was correct. A long philosophical tradition, going back to the Greeks and defended by the Catholic Church, held that pure human thought combined with divine revelation represented the path to truth. Nature, as revealed by our senses, was suspect. For example, Aristotle had reasoned that heavier objects (having more of the quality that made them heavy) must fall to Earth faster than lighter ones. This is absolutely incorrect, as any simple experiment dropping two balls of different weights shows. However, in Copernicus\u2019 day, experiments did not carry much weight (if you will pardon the expression); Aristotle\u2019s reasoning was more convincing. In this environment, there was little motivation to carry out observations or experiments to distinguish between competing cosmological theories (or anything else). It should not surprise us, therefore, that the heliocentric idea was debated for more than half a century without any tests being applied to determine its validity. (In fact, in the North American colonies, the older geocentric system was still taught at Harvard University in the first years after it was founded in 1636.) Contrast this with the situation today, when scientists rush to test each new hypothesis and do not accept any ideas until the results are in. For example, when two researchers at the University of Utah announced in 1989 that they had discovered a way to achieve nuclear fusion (the process that powers the stars) at room temperature, other scientists at more than 25 laboratories around the United States attempted to duplicate \u201ccold fusion\u201d within a few weeks\u2014without success, as it turned out. The cold fusion theory soon went down in flames. How would we look at Cop", "ernicus\u2019 model today? When a new hypothesis or theory is proposed in science, it must first be checked for consistency with what is already known. Copernicus\u2019 heliocentric idea passes this test, for it allows planetary positions to be calculated at least as well as does the geocentric theory. The next step is to determine which predictions the new hypothesis makes that differ from those of competing ideas. In the case of Copernicus, one example is the prediction that, if Venus circles the Sun, the planet should go through the full range of phases just as the Moon does, whereas if it circles Earth, it should not (Figure 2.18). Also, we should not be able to see the full phase of Venus from Earth because the Sun would then be between Venus and Earth. But in those days, before the telescope, no one imagined testing these predictions. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 57 Figure 2.18. Phases of Venus. As Venus moves around the Sun, we see changing illumination of its surface, just as we see the face of the Moon illuminated differently in the course of a month This animation (https://openstaxcollege.org/l/30venusphases) shows the phases of Venus. You can also see its distance from Earth as it orbits the Sun. Galileo and the Beginning of Modern Science Many of the modern scientific concepts of observation, experimentation, and the testing of hypotheses through careful quantitative measurements were pioneered by a man who lived nearly a century after Copernicus. Galileo Galilei (Figure 2.19), a contemporary of Shakespeare, was born in Pisa. Like Copernicus, he began training for a medical career, but he had little interest in the subject and later switched to mathematics. He held faculty positions at the University of Pisa and the University of Padua, and eventually became mathematician to the Grand Duke of Tuscany in Florence. 58 Chapter 2 Observing the Sky: The Birth of Astronomy Figure 2.19. Galileo Galilei (1564\u20131642). Galileo advocated that we perform experiments or make observations to ask nature its ways. When Galileo turned the telescope to the sky, he found things were not the way philosophers had supposed. Galileo\u2019s greatest contributions were in the field of mechanics, the study of motion and the actions of forces on bodies. It was familiar to all persons then, as it is to us now, that if something is at rest, it tends to remain at rest and requires some outside influence to start it in motion. Rest was thus generally regarded as the natural state of matter. Galileo showed, however, that rest is no more natural than motion. If an object is slid along a rough horizontal floor, it soon comes to rest because friction between it and the floor acts as a retarding force. However, if the floor and the object are both highly polished, the object, given the same initial speed, will slide farther before stopping. On a smooth layer of ice, it will slide farther still. Galileo reasoned that if all resisting effects could be removed, the object would continue in a steady state of motion indefinitely. He argued that a force is required not only to start an object moving from rest but also to slow down, stop, speed up, or change the direction of a moving object. You will appreciate this if you have ever tried to stop a rolling car by leaning against it, or a moving boat by tugging on a line. Galileo also studied the way objects accelerate\u2014change their speed or direction of motion. Galileo watched objects as they fell freely or rolled down a ramp. He found that such objects accelerate uniformly; that is, in equal intervals of time they gain equal increments in speed. Galileo formulated these newly found laws in precise mathematical terms that enabled future experimenters to predict how far and how fast objects would move in various lengths of time In theory, if Galileo is right, a feather and a hammer, dropped at the same time from a height, should land at the same moment. On Earth, this experiment is not possible because air resistance and air movements make the feather flutter, instead of falling straight down, accelerated only by the force of gravity. For generations, physics teachers had said that the place to try this experiment is somewhere where there is no air, such as the Moon. In 1971, Apollo 15 astronaut David Scott took a hammer and feather to the Moon and tried it, to the delight of physics nerds everywhere. NASA provides the video of This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 59 the hammer and feather (https://openstaxcollege.org/l/30HamVsFeath) as well as a brief explanation. Sometime in the 1590s, Galileo adopted the Copernican", " hypothesis of a heliocentric solar system. In Roman Catholic Italy, this was not a popular philosophy, for Church authorities still upheld the ideas of Aristotle and Ptolemy, and they had powerful political and economic reasons for insisting that Earth was the center of creation. Galileo not only challenged this thinking but also had the audacity to write in Italian rather than scholarly Latin, and to lecture publicly on those topics. For him, there was no contradiction between the authority of the Church in matters of religion and morality, and the authority of nature (revealed by experiments) in matters of science. It was primarily because of Galileo and his \u201cdangerous\u201d opinions that, in 1616, the Church issued a prohibition decree stating that the Copernican doctrine was \u201cfalse and absurd\u201d and not to be held or defended. Galileo\u2019s Astronomical Observations It is not certain who first conceived of the idea of combining two or more pieces of glass to produce an instrument that enlarged images of distant objects, making them appear nearer. The first such \u201cspyglasses\u201d (now called telescopes) that attracted much notice were made in 1608 by the Dutch spectacle maker Hans Lippershey (1570\u20131619). Galileo heard of the discovery and, without ever having seen an assembled telescope, constructed one of his own with a three-power magnification (3\u00d7), which made distant objects appear three times nearer and larger (Figure 2.20). Figure 2.20. Telescope Used by Galileo. The telescope has a wooden tube covered with paper and a lens 26 millimeters across. On August 25, 1609, Galileo demonstrated a telescope with a magnification of 9\u00d7 to government officials of the city-state of Venice. By a magnification of 9\u00d7, we mean the linear dimensions of the objects being viewed appeared nine times larger or, alternatively, the objects appeared nine times closer than they really were. There were obvious military advantages associated with a device for seeing distant objects. For his invention, Galileo\u2019s salary was nearly doubled, and he was granted lifetime tenure as a professor. (His university colleagues were outraged, particularly because the invention was not even original.) Others had used the telescope before Galileo to observe things on Earth. But in a flash of insight that changed the history of astronomy, Galileo realized that he could turn the power of the telescope toward the heavens. Before using his telescope for astronomical observations, Galileo had to devise a stable mount and improve the optics. He increased the magnification to 30\u00d7. Galileo also needed to acquire confidence in the telescope. At that time, human eyes were believed to be the final arbiter of truth about size, shape, and color. Lenses, mirrors, and prisms were known to distort distant images by enlarging, reducing, or inverting them, or spreading the light into a spectrum (rainbow of colors). Galileo undertook repeated experiments to convince himself that what he saw through the telescope was identical to what he saw up close. Only then could he begin to believe that the miraculous phenomena the telescope revealed in the heavens were real. Beginning his astronomical work late in 1609, Galileo found that many stars too faint to be seen with the unaided eye became visible with his telescope. In particular, he found that some nebulous blurs resolved into 60 Chapter 2 Observing the Sky: The Birth of Astronomy many stars, and that the Milky Way\u2014the strip of whiteness across the night sky\u2014was also made up of a multitude of individual stars. Examining the planets, Galileo found four moons revolving about Jupiter in times ranging from just under 2 days to about 17 days. This discovery was particularly important because it showed that not everything has to revolve around Earth. Furthermore, it demonstrated that there could be centers of motion that are themselves in motion. Defenders of the geocentric view had argued that if Earth was in motion, then the Moon would be left behind because it could hardly keep up with a rapidly moving planet. Yet, here were Jupiter\u2019s moons doing exactly that. (To recognize this discovery and honor his work, NASA named a spacecraft that explored the Jupiter system Galileo.) With his telescope, Galileo was able to carry out the test of the Copernican theory mentioned earlier, based on the phases of Venus. Within a few months, he had found that Venus goes through phases like the Moon, showing that it must revolve about the Sun, so that we see different parts of its daylight side at different times (see Figure 2.18.) These observations could not be reconciled with Ptolemy\u2019s model, in which Venus circled about Earth. In Ptolemy\u2019s model, Venus could also show phases, but they were the wrong phases in the wrong order from what Galileo observed. Galileo also observed the Moon and saw craters, mountain ranges, valleys, and flat, dark areas that he thought might be water. These discoveries showed that the Moon might be not so dissimilar to Earth\u2014suggesting that Earth, too,", " could belong to the realm of celestial bodies For more information about the life and work of Galileo, see the Galileo Project (https://openstaxcollege.org/l/30GalProj) at Rice University. After Galileo\u2019s work, it became increasingly difficult to deny the Copernican view, and Earth was slowly dethroned from its central position in the universe and given its rightful place as one of the planets attending the Sun. Initially, however, Galileo met with a great deal of opposition. The Roman Catholic Church, still reeling from the Protestant Reformation, was looking to assert its authority and chose to make an example of Galileo. He had to appear before the Inquisition to answer charges that his work was heretical, and he was ultimately condemned to house arrest. His books were on the Church\u2019s forbidden list until 1836, although in countries where the Roman Catholic Church held less sway, they were widely read and discussed. Not until 1992 did the Catholic Church admit publicly that it had erred in the matter of censoring Galileo\u2019s ideas. The new ideas of Copernicus and Galileo began a revolution in our conception of the cosmos. It eventually became evident that the universe is a vast place and that Earth\u2019s role in it is relatively unimportant. The idea that Earth moves around the Sun like the other planets raised the possibility that they might be worlds themselves, perhaps even supporting life. As Earth was demoted from its position at the center of the universe, so, too, was humanity. The universe, despite what we may wish, does not revolve around us. Most of us take these things for granted today, but four centuries ago such concepts were frightening and heretical for some, immensely stimulating for others. The pioneers of the Renaissance started the European world along the path toward science and technology that we still tread today. For them, nature was rational and ultimately knowable, and experiments and observations provided the means to reveal its secrets. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 61 Observing the Planets At most any time of the night, and at any season, you can spot one or more bright planets in the sky. All five of the planets known to the ancients\u2014Mercury, Venus, Mars, Jupiter, and Saturn\u2014are more prominent than any but the brightest stars, and they can be seen even from urban locations if you know where and when to look. One way to tell planets from bright stars is that planets twinkle less. Venus, which stays close to the Sun from our perspective, appears either as an \u201cevening star\u201d in the west after sunset or as a \u201cmorning star\u201d in the east before sunrise. It is the brightest object in the sky after the Sun and Moon. It far outshines any real star, and under the most favorable circumstances, it can even cast a visible shadow. Some young military recruits have tried to shoot Venus down as an approaching enemy craft or UFO. Mars, with its distinctive red color, can be nearly as bright as Venus is when close to Earth, but normally it remains much less conspicuous. Jupiter is most often the second-brightest planet, approximately equaling in brilliance the brightest stars. Saturn is dimmer, and it varies considerably in brightness, depending on whether its large rings are seen nearly edge-on (faint) or more widely opened (bright). Mercury is quite bright, but few people ever notice it because it never moves very far from the Sun (it\u2019s never more than 28\u00b0 away in the sky) and is always seen against bright twilight skies. True to their name, the planets \u201cwander\u201d against the background of the \u201cfixed\u201d stars. Although their apparent motions are complex, they reflect an underlying order upon which the heliocentric model of the solar system, as described in this chapter, was based. The positions of the planets are often listed in newspapers (sometimes on the weather page), and clear maps and guides to their locations can be found each month in such magazines as Sky & Telescope and Astronomy (available at most libraries and online). There are also a number of computer programs and phone and tablet apps that allow you to display where the planets are on any night. 62 Chapter 2 Observing the Sky: The Birth of Astronomy CHAPTER 2 REVIEW KEY TERMS accelerate to change velocity; to speed up, slow down, or change direction. apparent magnitude a measure of how bright a star looks in the sky; the larger the number, the dimmer the star appears to us astrology the pseudoscience that deals with the supposed influences on human destiny of the configurations and locations in the sky of the Sun, Moon, and planets celestial equator a great circle on the celestial sphere 90\u00b0 from the celestial poles; where the celestial sphere intersects the plane of Earth\u2019s equ", "ator celestial poles points about which the celestial sphere appears to rotate; intersections of the celestial sphere with Earth\u2019s polar axis celestial sphere the apparent sphere of the sky; a sphere of large radius centered on the observer; directions of objects in the sky can be denoted by their position on the celestial sphere circumpolar zone those portions of the celestial sphere near the celestial poles that are either always above or always below the horizon cosmology the study of the organization and evolution of the universe ecliptic the apparent annual path of the Sun on the celestial sphere epicycle the circular orbit of a body in the Ptolemaic system, the center of which revolves about another circle (the deferent) geocentric centered on Earth heliocentric centered on the Sun horizon (astronomical) a great circle on the celestial sphere 90\u00b0 from the zenith; more popularly, the circle around us where the dome of the sky meets Earth horoscope a chart used by astrologers that shows the positions along the zodiac and in the sky of the Sun, Moon, and planets at some given instant and as seen from a particular place on Earth\u2014usually corresponding to the time and place of a person\u2019s birth parallax the apparent displacement of a nearby star that results from the motion of Earth around the Sun planet today, any of the larger objects revolving about the Sun or any similar objects that orbit other stars; in ancient times, any object that moved regularly among the fixed stars precession (of Earth) the slow, conical motion of Earth\u2019s axis of rotation caused principally by the gravitational pull of the Moon and Sun on Earth\u2019s equatorial bulge retrograde motion the apparent westward motion of a planet on the celestial sphere or with respect to the stars year the period of revolution of Earth around the Sun This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 63 zenith the point on the celestial sphere opposite the direction of gravity; point directly above the observer zodiac a belt around the sky about 18\u00b0 wide centered on the ecliptic SUMMARY 2.1 The Sky Above The direct evidence of our senses supports a geocentric perspective, with the celestial sphere pivoting on the celestial poles and rotating about a stationary Earth. We see only half of this sphere at one time, limited by the horizon; the point directly overhead is our zenith. The Sun\u2019s annual path on the celestial sphere is the ecliptic\u2014a line that runs through the center of the zodiac, which is the 18-degree-wide strip of the sky within which we always find the Moon and planets. The celestial sphere is organized into 88 constellations, or sectors. 2.2 Ancient Astronomy Ancient Greeks such as Aristotle recognized that Earth and the Moon are spheres, and understood the phases of the Moon, but because of their inability to detect stellar parallax, they rejected the idea that Earth moves. Eratosthenes measured the size of Earth with surprising precision. Hipparchus carried out many astronomical observations, making a star catalog, defining the system of stellar magnitudes, and discovering precession from the apparent shift in the position of the north celestial pole. Ptolemy of Alexandria summarized classic astronomy in his Almagest; he explained planetary motions, including retrograde motion, with remarkably good accuracy using a model centered on Earth. This geocentric model, based on combinations of uniform circular motion using epicycles, was accepted as authority for more than a thousand years. 2.3 Astrology and Astronomy The ancient religion of astrology, with its main contribution to civilization a heightened interest in the heavens, began in Babylonia. It reached its peak in the Greco-Roman world, especially as recorded in the Tetrabiblos of Ptolemy. Natal astrology is based on the assumption that the positions of the planets at the time of our birth, as described by a horoscope, determine our future. However, modern tests clearly show that there is no evidence for this, even in a broad statistical sense, and there is no verifiable theory to explain what might cause such an astrological influence. 2.4 The Birth of Modern Astronomy Nicolaus Copernicus introduced the heliocentric cosmology to Renaissance Europe in his book De Revolutionibus. Although he retained the Aristotelian idea of uniform circular motion, Copernicus suggested that Earth is a planet and that the planets all circle about the Sun, dethroning Earth from its position at the center of the universe. Galileo was the father of both modern experimental physics and telescopic astronomy. He studied the acceleration of moving objects and, in 1610, began telescopic observations, discovering the nature of the Milky Way, the large-scale features of the Moon, the phases of Venus, and four moons", " of Jupiter. Although he was accused of heresy for his support of heliocentric cosmology, Galileo is credited with observations and brilliant writings that convinced most of his scientific contemporaries of the reality of the Copernican theory. FOR FURTHER EXPLORATION Articles Ancient Astronomy Gingerich, O. \u201cFrom Aristarchus to Copernicus.\u201d Sky & Telescope (November 1983): 410. 64 Chapter 2 Observing the Sky: The Birth of Astronomy Gingerich, O. \u201cIslamic Astronomy.\u201d Scientific American (April 1986): 74. Astronomy and Astrology Fraknoi, A. \u201cYour Astrology Defense Kit.\u201d Sky & Telescope (August 1989): 146. Copernicus and Galileo Gingerich, O. \u201cGalileo and the Phases of Venus.\u201d Sky & Telescope (December 1984): 520. Gingerich, O. \u201cHow Galileo Changed the Rules of Science.\u201d Sky & Telescope (March 1993): 32. Maran, S., and Marschall, L. \u201cThe Moon, the Telescope, and the Birth of the Modern World.\u201d Sky & Telescope (February 2009): 28. Sobel, D. \u201cThe Heretic\u2019s Daughter: A Startling Correspondence Reveals a New Portrait of Galileo.\u201d The New Yorker (September 13, 1999): 52. Websites Ancient Astronomy Aristarchos of Samos: http://adsabs.harvard.edu//full/seri/JRASC/0075//0000029.000.html. By Dr. Alan Batten. Claudius Ptolemy: http://www-history.mcs.st-and.ac.uk/Biographies/Ptolemy.html. An interesting biography. Hipparchus of Rhodes: http://www-history.mcs.st-andrews.ac.uk/Biographies/Hipparchus.html. An interesting biography. Astronomy and Astrology Astrology and Science: http://www.astrology-and-science.com/hpage.htm. The best site for a serious examination of the issues with astrology and the research on whether it works. Real Romance in the Stars: http://www.independent.co.uk/voices/the-real-romance-in-the-stars-1527970.html. 1995 newspaper commentary attacking astrology. Copernicus and Galileo Galileo Galilei: http://www-history.mcs.st-andrews.ac.uk/Biographies/Galileo.html. A good biography with additional links. Galileo Project: http://galileo.rice.edu/. Rice University\u2019s repository of information on Galileo. Nicolaus Copernicus: http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Copernicus.html. A biography including links to photos about his life. Videos Astronomy and Astrology Astrology Debunked: https://www.youtube.com/watch?v=y84HX2pMo5U. A compilation of scientists and magicians commenting skeptically on astrology (9:09). Copernicus and Galileo Galileo: http://www.biography.com/people/galileo-9305220. A brief biography (2:51). Galileo\u2019s Battle for the Heavens: https://www.youtube.com/watch?v=VnEH9rbrIkk. A NOVA episode on PBS (1:48:55) Nicolaus Copernicus: http://www.biography.com/people/nicolaus-copernicus-9256984. An overview of his life and work (2:41). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 65 COLLABORATIVE GROUP ACTIVITIES A. With your group, consider the question with which we began this chapter. How many ways can you think of to prove to a member of the \u201cFlat Earth Society\u201d that our planet is, indeed, round? B. Make a list of ways in which a belief in astrology (the notion that your life path or personality is controlled by the position of the Sun, Moon, and planets at the time of your birth) might be harmful to an individual or to society at large. C. Have members of the group compare their experiences with the night sky. Did you see the Milky Way? Can you identify any constellations? Make a list of reasons why you think so many fewer people know the night sky today than at the time of the ancient Greeks. Discuss reasons for why a person, today, may want to be acquainted with the night sky. D. Constellations commemorate great heroes", ", dangers, or events in the legends of the people who name them. Suppose we had to start from scratch today, naming the patterns of stars in the sky. Whom or what would you choose to commemorate by naming a constellation after it, him, or her and why (begin with people from history; then if you have time, include living people as well)? Can the members of your group agree on any choices? E. Although astronomical mythology no longer holds a powerful sway over the modern imagination, we still find proof of the power of astronomical images in the number of products in the marketplace that have astronomical names. How many can your group come up with? (Think of things like Milky Way candy bars, Eclipse and Orbit gum, or Comet cleanser.) EXERCISES Review Questions 1. From where on Earth could you observe all of the stars during the course of a year? What fraction of the sky can be seen from the North Pole? 2. Give four ways to demonstrate that Earth is spherical. 3. Explain, according to both geocentric and heliocentric cosmologies, why we see retrograde motion of the planets. 4. In what ways did the work of Copernicus and Galileo differ from the views of the ancient Greeks and of their contemporaries? 5. What were four of Galileo\u2019s discoveries that were important to astronomy? 6. Explain the origin of the magnitude designation for determining the brightness of stars. Why does it seem to go backward, with smaller numbers indicating brighter stars? 7. Ursa Minor contains the pole star, Polaris, and the asterism known as the Little Dipper. From most locations in the Northern Hemisphere, all of the stars in Ursa Minor are circumpolar. Does that mean these stars are also above the horizon during the day? Explain. 66 Chapter 2 Observing the Sky: The Birth of Astronomy 8. How many degrees does the Sun move per day relative to the fixed stars? How many days does it take for the Sun to return to its original location relative to the fixed stars? 9. How many degrees does the Moon move per day relative to the fixed stars? How many days does it take for the Moon to return to its original location relative to the fixed stars? 10. Explain how the zodiacal constellations are different from the other constellations. 11. The Sun was once thought to be a planet. Explain why. 12. Is the ecliptic the same thing as the celestial equator? Explain. 13. What is an asterism? Can you name an example? 14. Why did Pythagoras believe that Earth should be spherical? 15. How did Aristotle deduce that the Sun is farther away from Earth than the Moon? 16. What are two ways in which Aristotle deduced that Earth is spherical? 17. How did Hipparchus discover the wobble of Earth\u2019s axis, known as precession? 18. Why did Ptolemy have to introduce multiple circles of motion for the planets instead of a single, simple circle to represent the planet\u2019s motion around the Sun? 19. Why did Copernicus want to develop a completely new system for predicting planetary positions? Provide two reasons. 20. What two factors made it difficult, at first, for astronomers to choose between the Copernican heliocentric model and the Ptolemaic geocentric model? 21. What phases would Venus show if the geocentric model were correct? Thought Questions 22. Describe a practical way to determine in which constellation the Sun is found at any time of the year. 23. What is a constellation as astronomers define it today? What does it mean when an astronomer says, \u201cI saw a comet in Orion last night\u201d? 24. Draw a picture that explains why Venus goes through phases the way the Moon does, according to the heliocentric cosmology. Does Jupiter also go through phases as seen from Earth? Why? 25. Show with a simple diagram how the lower parts of a ship disappear first as it sails away from you on a spherical Earth. Use the same diagram to show why lookouts on old sailing ships could see farther from the masthead than from the deck. Would there be any advantage to posting lookouts on the mast if Earth were flat? (Note that these nautical arguments for a spherical Earth were quite familiar to Columbus and other mariners of his time.) 26. Parallaxes of stars were not observed by ancient astronomers. How can this fact be reconciled with the heliocentric hypothesis? 27. Why do you think so many people still believe in astrology and spend money on it? What psychological needs does such a belief system satisfy? 28. Consider three cosmological perspectives\u2014the geocentric perspective, the heliocentric perspective, and the modern perspective\u2014in which the Sun is a minor star on the outskirts of one galaxy among billions. Discuss some of the cultural and philosophical implications of each point of view", ". This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 2 Observing the Sky: The Birth of Astronomy 67 29. The north celestial pole appears at an altitude above the horizon that is equal to the observer\u2019s latitude. Identify Polaris, the North Star, which lies very close to the north celestial pole. Measure its altitude. (This can be done with a protractor. Alternatively, your fist, extended at arm\u2019s length, spans a distance approximately equal to 10\u00b0.) Compare this estimate with your latitude. (Note that this experiment cannot be performed easily in the Southern Hemisphere because Polaris itself is not visible in the south and no bright star is located near the south celestial pole.) 30. What were two arguments or lines of evidence in support of the geocentric model? 31. Although the Copernican system was largely correct to place the Sun at the center of all planetary motion, the model still gave inaccurate predictions for planetary positions. Explain the flaw in the Copernican model that hindered its accuracy. 32. During a retrograde loop of Mars, would you expect Mars to be brighter than usual in the sky, about average in brightness, or fainter than usual in the sky? Explain. 33. The Great Pyramid of Giza was constructed nearly 5000 years ago. Within the pyramid, archaeologists discovered a shaft leading from the central chamber out of the pyramid, oriented for favorable viewing of the bright star Thuban at that time. Thinking about Earth\u2019s precession, explain why Thuban might have been an important star to the ancient Egyptians. 34. Explain why more stars are circumpolar for observers at higher latitudes. 35. What is the altitude of the north celestial pole in the sky from your latitude? If you do not know your latitude, look it up. If you are in the Southern Hemisphere, answer this question for the south celestial pole, since the north celestial pole is not visible from your location. 36. If you were to drive to some city south of your current location, how would the altitude of the celestial pole in the sky change? 37. Hipparchus could have warned us that the dates associated with each of the natal astrology sun signs would eventually be wrong. Explain why. 38. Explain three lines of evidence that argue against the validity of astrology. 39. What did Galileo discover about the planet Jupiter that cast doubt on exclusive geocentrism? 40. What did Galileo discover about Venus that cast doubt on geocentrism? Figuring For Yourself 41. Suppose Eratosthenes had found that, in Alexandria, at noon on the first day of summer, the line to the Sun makes an angle 30\u00b0 with the vertical. What, then, would he have found for Earth\u2019s circumference? 42. Suppose Eratosthenes\u2019 results for Earth\u2019s circumference were quite accurate. If the diameter of Earth is 12,740 km, what is the length of his stadium in kilometers? 43. Suppose you are on a strange planet and observe, at night, that the stars do not rise and set, but circle parallel to the horizon. Next, you walk in a constant direction for 8000 miles, and at your new location on the planet, you find that all stars rise straight up in the east and set straight down in the west, perpendicular to the horizon. How could you determine the circumference of the planet without any further observations? What is the circumference, in miles, of the planet? 68 Chapter 2 Observing the Sky: The Birth of Astronomy This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 69 3 ORBITS AND GRAVITY Figure 3.1. International Space Station. This space habitat and laboratory orbits Earth once every 90 minutes. (credit: modification of work by NASA) Chapter Outline 3.1 The Laws of Planetary Motion 3.2 Newton\u2019s Great Synthesis 3.3 Newton\u2019s Universal Law of Gravitation 3.4 Orbits in the Solar System 3.5 Motions of Satellites and Spacecraft 3.6 Gravity with More Than Two Bodies Thinking Ahead How would you find a new planet at the outskirts of our solar system that is too dim to be seen with the unaided eye and is so far away that it moves very slowly among the stars? This was the problem confronting astronomers during the nineteenth century as they tried to pin down a full inventory of our solar system. If we could look down on the solar system from somewhere out in space, interpreting planetary motions would be much simpler. But the fact is, we must observe the positions of all the other planets from our own moving planet. Scientists of the Renaissance did not know the details of Earth\u2019s motions any better than the motions of the other planets. Their problem, as we saw in Observing", " the Sky: The Birth of Astronomy, was that they had to deduce the nature of all planetary motion using only their earthbound observations of the other planets\u2019 positions in the sky. To solve this complex problem more fully, better observations and better models of the planetary system were needed. 70 Chapter 3 Orbits and Gravity 3.1 THE LAWS OF PLANETARY MOTION Learning Objectives By the end of this section, you will be able to: Describe how Tycho Brahe and Johannes Kepler contributed to our understanding of how planets move around the Sun Explain Kepler\u2019s three laws of planetary motion At about the time that Galileo was beginning his experiments with falling bodies, the efforts of two other scientists dramatically advanced our understanding of the motions of the planets. These two astronomers were the observer Tycho Brahe and the mathematician Johannes Kepler. Together, they placed the speculations of Copernicus on a sound mathematical basis and paved the way for the work of Isaac Newton in the next century. Tycho Brahe\u2019s Observatory Three years after the publication of Copernicus\u2019 De Revolutionibus, Tycho Brahe was born to a family of Danish nobility. He developed an early interest in astronomy and, as a young man, made significant astronomical observations. Among these was a careful study of what we now know was an exploding star that flared up to great brilliance in the night sky. His growing reputation gained him the patronage of the Danish King Frederick II, and at the age of 30, Brahe was able to establish a fine astronomical observatory on the North Sea island of Hven (Figure 3.2). Brahe was the last and greatest of the pre-telescopic observers in Europe. Figure 3.2. Tycho Brahe (1546\u20131601) and Johannes Kepler (1571\u20131630). measure the altitude of celestial objects above the horizon. The large curved instrument in the foreground allowed him to measure precise angles in the sky. Note that the scene includes hints of the grandeur of Brahe\u2019s observatory at Hven. (b) Kepler was a German mathematician and astronomer. His discovery of the basic laws that describe planetary motion placed the heliocentric cosmology of Copernicus on a firm mathematical basis. (a) A stylized engraving shows Tycho Brahe using his instruments to This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 71 At Hven, Brahe made a continuous record of the positions of the Sun, Moon, and planets for almost 20 years. His extensive and precise observations enabled him to note that the positions of the planets varied from those given in published tables, which were based on the work of Ptolemy. These data were extremely valuable, but Brahe didn\u2019t have the ability to analyze them and develop a better model than what Ptolemy had published. He was further inhibited because he was an extravagant and cantankerous fellow, and he accumulated enemies among government officials. When his patron, Frederick II, died in 1597, Brahe lost his political base and decided to leave Denmark. He took up residence in Prague, where he became court astronomer to Emperor Rudolf of Bohemia. There, in the year before his death, Brahe found a most able young mathematician, Johannes Kepler, to assist him in analyzing his extensive planetary data. Johannes Kepler Johannes Kepler was born into a poor family in the German province of W\u00fcrttemberg and lived much of his life amid the turmoil of the Thirty Years\u2019 War (see Figure 3.2). He attended university at Tubingen and studied for a theological career. There, he learned the principles of the Copernican system and became converted to the heliocentric hypothesis. Eventually, Kepler went to Prague to serve as an assistant to Brahe, who set him to work trying to find a satisfactory theory of planetary motion\u2014one that was compatible with the long series of observations made at Hven. Brahe was reluctant to provide Kepler with much material at any one time for fear that Kepler would discover the secrets of the universal motion by himself, thereby robbing Brahe of some of the glory. Only after Brahe\u2019s death in 1601 did Kepler get full possession of the priceless records. Their study occupied most of Kepler\u2019s time for more than 20 years. Through his analysis of the motions of the planets, Kepler developed a series of principles, now known as Kepler\u2019s three laws, which described the behavior of planets based on their paths through space. The first two laws of planetary motion were published in 1609 in The New Astronomy. Their discovery was a profound step in the development of modern science. The First Two Laws of Planetary Motion The path of an object through space is called its orbit. Kepler initially assumed that the orbits of planets were circles, but doing so did not allow him to find orbits that were consistent with Brahe\ufffd", "\ufffds observations. Working with the data for Mars, he eventually discovered that the orbit of that planet had the shape of a somewhat flattened circle, or ellipse. Next to the circle, the ellipse is the simplest kind of closed curve, belonging to a family of curves known as conic sections (Figure 3.3). Figure 3.3. Conic Sections. The circle, ellipse, parabola, and hyperbola are all formed by the intersection of a plane with a cone. This is why such curves are called conic sections. 72 Chapter 3 Orbits and Gravity You might recall from math classes that in a circle, the center is a special point. The distance from the center to anywhere on the circle is exactly the same. In an ellipse, the sum of the distance from two special points inside the ellipse to any point on the ellipse is always the same. These two points inside the ellipse are called its foci (singular: focus), a word invented for this purpose by Kepler. This property suggests a simple way to draw an ellipse (Figure 3.4). We wrap the ends of a loop of string around two tacks pushed through a sheet of paper into a drawing board, so that the string is slack. If we push a pencil against the string, making the string taut, and then slide the pencil against the string all around the tacks, the curve that results is an ellipse. At any point where the pencil may be, the sum of the distances from the pencil to the two tacks is a constant length\u2014the length of the string. The tacks are at the two foci of the ellipse. The widest diameter of the ellipse is called its major axis. Half this distance\u2014that is, the distance from the center of the ellipse to one end\u2014is the semimajor axis, which is usually used to specify the size of the ellipse. For example, the semimajor axis of the orbit of Mars, which is also the planet\u2019s average distance from the Sun, is 228 million kilometers. Figure 3.4. Drawing an Ellipse. (a) We can construct an ellipse by pushing two tacks (the white objects) into a piece of paper on a drawing board, and then looping a string around the tacks. Each tack represents a focus of the ellipse, with one of the tacks being the Sun. Stretch the string tight using a pencil, and then move the pencil around the tacks. The length of the string remains the same, so that the sum of the distances from any point on the ellipse to the foci is always constant. (b) In this illustration, each semimajor axis is denoted by a. The distance 2a is called the major axis of the ellipse. The shape (roundness) of an ellipse depends on how close together the two foci are, compared with the major axis. The ratio of the distance between the foci to the length of the major axis is called the eccentricity of the ellipse. If the foci (or tacks) are moved to the same location, then the distance between the foci would be zero. This means that the eccentricity is zero and the ellipse is just a circle; thus, a circle can be called an ellipse of zero eccentricity. In a circle, the semimajor axis would be the radius. Next, we can make ellipses of various elongations (or extended lengths) by varying the spacing of the tacks (as long as they are not farther apart than the length of the string). The greater the eccentricity, the more elongated is the ellipse, up to a maximum eccentricity of 1.0, when the ellipse becomes \u201cflat,\u201d the other extreme from a circle. The size and shape of an ellipse are completely specified by its semimajor axis and its eccentricity. Using Brahe\u2019s data, Kepler found that Mars has an elliptical orbit, with the Sun at one focus (the other focus is empty). The eccentricity of the orbit of Mars is only about 0.1; its orbit, drawn to scale, would be practically indistinguishable from a circle, but the difference turned out to be critical for understanding planetary motions. Kepler generalized this result in his first law and said that the orbits of all the planets are ellipses. Here was a decisive moment in the history of human thought: it was not necessary to have only circles in order to have an acceptable cosmos. The universe could be a bit more complex than the Greek philosophers had wanted it to be. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 73 Kepler\u2019s second law deals with the speed", " with which each planet moves along its ellipse, also known as its orbital speed. Working with Brahe\u2019s observations of Mars, Kepler discovered that the planet speeds up as it comes closer to the Sun and slows down as it pulls away from the Sun. He expressed the precise form of this relationship by imagining that the Sun and Mars are connected by a straight, elastic line. When Mars is closer to the Sun (positions 1 and 2 in Figure 3.5), the elastic line is not stretched as much, and the planet moves rapidly. Farther from the Sun, as in positions 3 and 4, the line is stretched a lot, and the planet does not move so fast. As Mars travels in its elliptical orbit around the Sun, the elastic line sweeps out areas of the ellipse as it moves (the colored regions in our figure). Kepler found that in equal intervals of time (t), the areas swept out in space by this imaginary line are always equal; that is, the area of the region B from 1 to 2 is the same as that of region A from 3 to 4. If a planet moves in a circular orbit, the elastic line is always stretched the same amount and the planet moves at a constant speed around its orbit. But, as Kepler discovered, in most orbits that speed of a planet orbiting its star (or moon orbiting its planet) tends to vary because the orbit is elliptical. Figure 3.5. Kepler\u2019s Second Law: The Law of Equal Areas. The orbital speed of a planet traveling around the Sun (the circular object inside the ellipse) varies in such a way that in equal intervals of time (t), a line between the Sun and a planet sweeps out equal areas (A and B). Note that the eccentricities of the planets\u2019 orbits in our solar system are substantially less than shown here. Kepler\u2019s Third Law Kepler\u2019s first two laws of planetary motion describe the shape of a planet\u2019s orbit and allow us to calculate the speed of its motion at any point in the orbit. Kepler was pleased to have discovered such fundamental rules, but they did not satisfy his quest to fully understand planetary motions. He wanted to know why the orbits of the planets were spaced as they are and to find a mathematical pattern in their movements\u2014a \u201charmony of the spheres\u201d as he called it. For many years he worked to discover mathematical relationships governing planetary spacing and the time each planet took to go around the Sun. In 1619, Kepler discovered a basic relationship to relate the planets\u2019 orbits to their relative distances from the Sun. We define a planet\u2019s orbital period, (P), as the time it takes a planet to travel once around the Sun. Also, recall that a planet\u2019s semimajor axis, a, is equal to its average distance from the Sun. The relationship, now known as Kepler\u2019s third law, says that a planet\u2019s orbital period squared is proportional to the semimajor axis of its orbit cubed, or P2 \u221d a3 74 Chapter 3 Orbits and Gravity When P (the orbital period) is measured in years, and a is expressed in a quantity known as an astronomical unit (AU), the two sides of the formula are not only proportional but equal. One AU is the average distance between Earth and the Sun and is approximately equal to 1.5 \u00d7 108 kilometers. In these units, P2 = a3 Kepler\u2019s third law applies to all objects orbiting the Sun, including Earth, and provides a means for calculating their relative distances from the Sun from the time they take to orbit. Let\u2019s look at a specific example to illustrate how useful Kepler\u2019s third law is. For instance, suppose you time how long Mars takes to go around the Sun (in Earth years). Kepler\u2019s third law can then be used to calculate Mars\u2019 average distance from the Sun. Mars\u2019 orbital period (1.88 Earth years) squared, or P2, is 1.882 = 3.53, and according to the equation for Kepler\u2019s third law, this equals the cube of its semimajor axis, or a3. So what number must be cubed to give 3.53? The answer is 1.52 (since 1.52 \u00d7 1.52 \u00d7 1.52 = 3.53). Thus, Mars\u2019 semimajor axis in astronomical units must be 1.52 AU. In other words, to go around the Sun in a little less than two years, Mars must be about 50% (half again) as far from the Sun as Earth is Calculating Periods Imagine an object is traveling around the Sun. What would be the orbital period of the object if its orbit has a semimajor axis of 50 AU? Solution From Kepler\u2019s third law, we know that (when we use units of years and AU) P2 = a3 If the object\ufffd", "\ufffds orbit has a semimajor axis of 4 AU (a = 50), we can cube 50 and then take the square root of the result to get P: P = a3 P = 50 \u00d7 50 \u00d7 50 = 125,000 = 353.6 years Therefore, the orbital period of the object is about 350 years. This would place our hypothetical object beyond the orbit of Pluto. Check Your Learning What would be the orbital period of an asteroid (a rocky chunk between Mars and Jupiter) with a semimajor axis of 3 AU? Answer: P = 3 \u00d7 3 \u00d7 3 = 27 = 5.2 years Kepler\u2019s three laws of planetary motion can be summarized as follows: \u2022 Kepler\u2019s first law: Each planet moves around the Sun in an orbit that is an ellipse, with the Sun at one focus of the ellipse. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 75 \u2022 Kepler\u2019s second law: The straight line joining a planet and the Sun sweeps out equal areas in space in equal intervals of time. \u2022 Kepler\u2019s third law: The square of a planet\u2019s orbital period is directly proportional to the cube of the semimajor axis of its orbit. Kepler\u2019s three laws provide a precise geometric description of planetary motion within the framework of the Copernican system. With these tools, it was possible to calculate planetary positions with greatly improved precision. Still, Kepler\u2019s laws are purely descriptive: they do not help us understand what forces of nature constrain the planets to follow this particular set of rules. That step was left to Isaac Newton Applying Kepler\u2019s Third Law Using the orbital periods and semimajor axes for Venus and Earth that are provided here, calculate P2 and a3, and verify that they obey Kepler\u2019s third law. Venus\u2019 orbital period is 0.62 year, and its semimajor axis is 0.72 AU. Earth\u2019s orbital period is 1.00 year, and its semimajor axis is 1.00 AU. Solution We can use the equation for Kepler\u2019s third law, P2 \u221d a3. For Venus, P2 = 0.62 \u00d7 0.62 = 0.38 year and a3 = 0.72 \u00d7 0.72 \u00d7 0.72 = 0.37 AU (rounding numbers sometimes causes minor discrepancies like this). The orbital period (0.38 year) approximates the semimajor axis (0.37 AU). Therefore, Venus obeys Kepler\u2019s third law. For Earth, P2 = 1.00 \u00d7 1.00 = 1.00 year and a3 = 1.00 \u00d7 1.00 \u00d7 1.00 = 1.00 AU. The orbital period (1.00 year) approximates (in this case, equals) the semimajor axis (1.00 AU). Therefore, Earth obeys Kepler\u2019s third law. Check Your Learning Using the orbital periods and semimajor axes for Saturn and Jupiter that are provided here, calculate P2 and a3, and verify that they obey Kepler\u2019s third law. Saturn\u2019s orbital period is 29.46 years, and its semimajor axis is 9.54 AU. Jupiter\u2019s orbital period is 11.86 years, and its semimajor axis is 5.20 AU. Answer: For Saturn, P2 = 29.46 \u00d7 29.46 = 867.9 years and a3 = 9.54 \u00d7 9.54 \u00d7 9.54 = 868.3 AU. The orbital period (867.9 years) approximates the semimajor axis (868.3 AU). Therefore, Saturn obeys Kepler\u2019s third law In honor of the scientist who first devised the laws that govern the motions of planets, the team that built the first spacecraft to search for planets orbiting other stars decided to name the probe \u201cKepler.\u201d To learn more about Johannes Kepler\u2019s life and his laws of planetary motion, as well as lots of information on the Kepler Mission, visit NASA\u2019s Kepler website (https://openstaxcollege.org/l/30nasakepmiss) and follow the links that interest you. 76 Chapter 3 Orbits and Gravity 3.2 NEWTON\u2019S GREAT SYNTHESIS Learning Objectives By the end of this section, you will be able to: Describe Newton\u2019s three laws of motion Explain how Newton\u2019s three laws of motion relate to momentum Define mass, volume, and density and how they differ Define angular momentum It was the genius of Isaac Newton that found a conceptual framework that completely explained the observations and rules assembled by Galileo, Brahe, Kepler, and others. Newton was born in Lincolnshire, England, in the year after Galileo\u2019s", " death (Figure 3.6). Against the advice of his mother, who wanted him to stay home and help with the family farm, he entered Trinity College at Cambridge in 1661 and eight years later was appointed professor of mathematics. Among Newton\u2019s contemporaries in England were architect Christopher Wren, authors Aphra Behn and Daniel Defoe, and composer G. F. Handel. Figure 3.6. Isaac Newton (1643\u20131727), 1689 Portrait by Sir Godfrey Kneller. Isaac Newton\u2019s work on the laws of motion, gravity, optics, and mathematics laid the foundations for much of physical science. Newton\u2019s Laws of Motion As a young man in college, Newton became interested in natural philosophy, as science was then called. He worked out some of his first ideas on machines and optics during the plague years of 1665 and 1666, when students were sent home from college. Newton, a moody and often difficult man, continued to work on his ideas in private, even inventing new mathematical tools to help him deal with the complexities involved. Eventually, his friend Edmund Halley (profiled in Comets and Asteroids: Debris of the Solar System) prevailed on him to collect and publish the results of his remarkable investigations on motion and gravity. The result was a volume that set out the underlying system of the physical world, Philosophiae Naturalis Principia Mathematica. The Principia, as the book is generally known, was published at Halley\u2019s expense in 1687. At the very beginning of the Principia, Newton proposes three laws that would govern the motions of all objects: This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 77 \u2022 Newton\u2019s first law: Every object will continue to be in a state of rest or move at a constant speed in a straight line unless it is compelled to change by an outside force. \u2022 Newton\u2019s second law: The change of motion of a body is proportional to and in the direction of the force acting on it. \u2022 Newton\u2019s third law: For every action there is an equal and opposite reaction (or: the mutual actions of two bodies upon each other are always equal and act in opposite directions). In the original Latin, the three laws contain only 59 words, but those few words set the stage for modern science. Let us examine them more carefully. Interpretation of Newton\u2019s Laws Newton\u2019s first law is a restatement of one of Galileo\u2019s discoveries, called the conservation of momentum. The law states that in the absence of any outside influence, there is a measure of a body\u2019s motion, called its momentum, that remains unchanged. You may have heard the term momentum used in everyday expressions, such as \u201cThis bill in Congress has a lot of momentum; it\u2019s going to be hard to stop.\u201d Newton\u2019s first law is sometimes called the law of inertia, where inertia is the tendency of objects (and legislatures) to keep doing what they are already doing. In other words, a stationary object stays put, and a moving object keeps moving unless some force intervenes. Let\u2019s define the precise meaning of momentum\u2014it depends on three factors: (1) speed\u2014how fast a body moves (zero if it is stationary), (2) the direction of its motion, and (3) its mass\u2014a measure of the amount of matter in a body, which we will discuss later. Scientists use the term velocity to describe the speed and direction of motion. For example, 20 kilometers per hour due south is velocity, whereas 20 kilometers per hour just by itself is speed. Momentum then can be defined as an object\u2019s mass times its velocity. It\u2019s not so easy to see this rule in action in the everyday world because of the many forces acting on a body at any one time. One important force is friction, which generally slows things down. If you roll a ball along the sidewalk, it eventually comes to a stop because the sidewalk exerts a rubbing force on the ball. But in the space between the stars, where there is so little matter that friction is insignificant, objects can in fact continue to move (to coast) indefinitely. The momentum of a body can change only under the action of an outside influence. Newton\u2019s second law expresses force in terms of its ability to change momentum with time. A force (a push or a pull) has both size and direction. When a force is applied to a body, the momentum changes in the direction of the applied force. This means that a force is required to change either the speed or the direction of a body, or both\u2014that is, to start it moving, to speed it up, to slow it down, to stop it, or to change its direction. As you learned in Observing", " the Sky: The Birth of Astronomy, the rate of change in an object\u2019s velocity is called acceleration. Newton showed that the acceleration of a body was proportional to the force being applied to it. Suppose that after a long period of reading, you push an astronomy book away from you on a long, smooth table. (We use a smooth table so we can ignore friction.) If you push the book steadily, it will continue to speed up as long as you are pushing it. The harder you push the book, the larger its acceleration will be. How much a force will accelerate an object is also determined by the object\u2019s mass. If you kept pushing a pen with the same force with which you pushed the textbook, the pen\u2014having less mass\u2014would be accelerated to a greater speed. Newton\u2019s third law is perhaps the most profound of the rules he discovered. Basically, it is a generalization of the first law, but it also gives us a way to define mass. If we consider a system of two or more objects isolated from outside influences, Newton\u2019s first law says that the total momentum of the objects should remain 78 Chapter 3 Orbits and Gravity constant. Therefore, any change of momentum within the system must be balanced by another change that is equal and opposite so that the momentum of the entire system is not changed. This means that forces in nature do not occur alone: we find that in each situation there is always a pair of forces that are equal to and opposite each other. If a force is exerted on an object, it must be exerted by something else, and the object will exert an equal and opposite force back on that something. We can look at a simple example to demonstrate this. Suppose that a daredevil astronomy student\u2014and avid skateboarder\u2014wants to jump from his second-story dorm window onto his board below (we don\u2019t recommend trying this!). The force pulling him down after jumping (as we will see in the next section) is the force of gravity between him and Earth. Both he and Earth must experience the same total change of momentum because of the influence of these mutual forces. So, both the student and Earth are accelerated by each other\u2019s pull. However, the student does much more of the moving. Because Earth has enormously greater mass, it can experience the same change of momentum by accelerating only a very small amount. Things fall toward Earth all the time, but the acceleration of our planet as a result is far too small to be measured. A more obvious example of the mutual nature of forces between objects is familiar to all who have batted a baseball. The recoil you feel as you swing your bat shows that the ball exerts a force on it during the impact, just as the bat does on the ball. Similarly, when a rifle you are bracing on your shoulder is discharged, the force pushing the bullet out of the muzzle is equal to the force pushing backward upon the gun and your shoulder. This is the principle behind jet engines and rockets: the force that discharges the exhaust gases from the rear of the rocket is accompanied by the force that pushes the rocket forward. The exhaust gases need not push against air or Earth; a rocket actually operates best in a vacuum (Figure 3.7). Figure 3.7. Demonstrating Newton\u2019s Third Law. The U.S. Space Shuttle (here launching Discovery), powered by three fuel engines burning liquid oxygen and liquid hydrogen, with two solid fuel boosters, demonstrates Newton\u2019s third law. (credit: modification of work by NASA For more about Isaac Newton\u2019s life and work, check out this timeline page (https://openstaxcollege.org/l/30IsaacNewTime) with snapshots from his career, produced by the British Broadcasting Corporation (BBC). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 79 Mass, Volume, and Density Before we go on to discuss Newton\u2019s other work, we want to take a brief look at some terms that will be important to sort out clearly. We begin with mass, which is a measure of the amount of material within an object. The volume of an object is the measure of the physical space it occupies. Volume is measured in cubic units, such as cubic centimeters or liters. The volume is the \u201csize\u201d of an object. A penny and an inflated balloon may both have the same mass, but they have very different volumes. The reason is that they also have very different densities, which is a measure of how much mass there is per unit volume. Specifically, density is the mass divided by the volume. Note that in everyday language we often use \u201cheavy\u201d and \u201clight\u201d as indications of density (rather than weight) as, for instance, when we say that iron is heavy or that whipped cream is light. The units of density that will", " be used in this book are grams per cubic centimeter (g/cm3).[1] If a block of some material has a mass of 300 grams and a volume of 100 cm3, its density is 3 g/cm3. Familiar materials span a considerable range in density, from artificial materials such as plastic insulating foam (less than 0.1 g/cm3) to gold (19.3 g/cm3). Table 3.1 gives the densities of some familiar materials. In the astronomical universe, much more remarkable densities can be found, all the way from a comet\u2019s tail (10\u201316 g/cm3) to a collapsed \u201cstar corpse\u201d called a neutron star (1015 g/cm3). Densities of Common Materials Material Density (g/cm3) Gold Lead Iron Earth (bulk) Rock (typical) Water Wood (typical) Insulating foam Silica gel Table 3.1 19.3 11.3 7.9 5.5 2.5 1 0.8 0.1 0.02 To sum up, mass is how much, volume is how big, and density is how tightly packed. 1 Generally we use standard metric (or SI) units in this book. The proper metric unit of density in that system is kg/m3. But to most people, g/cm3 provides a more meaningful unit because the density of water is exactly 1 g/cm3, and this is useful information for comparison. Density expressed in g/cm3 is sometimes called specific density or specific weight. 80 Chapter 3 Orbits and Gravity You can play with a simple animation (https://openstaxcollege.org/l/30phetsimdenmas) demonstrating the relationship between the concepts of density, mass, and volume, and find out why objects like wood float in water. Angular Momentum A concept that is a bit more complex, but important for understanding many astronomical objects, is angular momentum, which is a measure of the rotation of a body as it revolves around some fixed point (an example is a planet orbiting the Sun). The angular momentum of an object is defined as the product of its mass, its velocity, and its distance from the fixed point around which it revolves. If these three quantities remain constant\u2014that is, if the motion of a particular object takes place at a constant velocity at a fixed distance from the spin center\u2014then the angular momentum is also a constant. Kepler\u2019s second law is a consequence of the conservation of angular momentum. As a planet approaches the Sun on its elliptical orbit and the distance to the spin center decreases, the planet speeds up to conserve the angular momentum. Similarly, when the planet is farther from the Sun, it moves more slowly. The conservation of angular momentum is illustrated by figure skaters, who bring their arms and legs in to spin more rapidly, and extend their arms and legs to slow down (Figure 3.8). You can duplicate this yourself on a well-oiled swivel stool by starting yourself spinning slowly with your arms extended and then pulling your arms in. Another example of the conservation of angular momentum is a shrinking cloud of dust or a star collapsing on itself (both are situations that you will learn about as you read on). As material moves to a lesser distance from the spin center, the speed of the material increases to conserve angular momentum. Figure 3.8. Conservation of Angular Momentum. When a spinning figure skater brings in her arms, their distance from her spin center is smaller, so her speed increases. When her arms are out, their distance from the spin center is greater, so she slows down. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 81 3.3 NEWTON\u2019S UNIVERSAL LAW OF GRAVITATION Learning Objectives By the end of this section, you will be able to: Explain what determines the strength of gravity Describe how Newton\u2019s universal law of gravitation extends our understanding of Kepler\u2019s laws Newton\u2019s laws of motion show that objects at rest will stay at rest and those in motion will continue moving uniformly in a straight line unless acted upon by a force. Thus, it is the straight line that defines the most natural state of motion. But the planets move in ellipses, not straight lines; therefore, some force must be bending their paths. That force, Newton proposed, was gravity. In Newton\u2019s time, gravity was something associated with Earth alone. Everyday experience shows us that Earth exerts a gravitational force upon objects at its surface. If you drop something, it accelerates toward Earth as it falls. Newton\u2019s insight was that Earth\u2019s gravity might extend as far as the Moon and produce the force required to curve the Moon\u2019s path from a straight line and keep it in its orbit. He further hypothesized that", " gravity is not limited to Earth, but that there is a general force of attraction between all material bodies. If so, the attractive force between the Sun and each of the planets could keep them in their orbits. (This may seem part of our everyday thinking today, but it was a remarkable insight in Newton\u2019s time.) Once Newton boldly hypothesized that there was a universal attraction among all bodies everywhere in space, he had to determine the exact nature of the attraction. The precise mathematical description of that gravitational force had to dictate that the planets move exactly as Kepler had described them to (as expressed in Kepler\u2019s three laws). Also, that gravitational force had to predict the correct behavior of falling bodies on Earth, as observed by Galileo. How must the force of gravity depend on distance in order for these conditions to be met? The answer to this question required mathematical tools that had not yet been developed, but this did not deter Isaac Newton, who invented what we today call calculus to deal with this problem. Eventually he was able to conclude that the magnitude of the force of gravity must decrease with increasing distance between the Sun and a planet (or between any two objects) in proportion to the inverse square of their separation. In other words, if a planet were twice as far from the Sun, the force would be (1/2)2, or 1/4 as large. Put the planet three times farther away, and the force is (1/3)2, or 1/9 as large. Newton also concluded that the gravitational attraction between two bodies must be proportional to their masses. The more mass an object has, the stronger the pull of its gravitational force. The gravitational attraction between any two objects is therefore given by one of the most famous equations in all of science: Fgravity = G M1 M2 R 2 where Fgravity is the gravitational force between two objects, M1 and M2 are the masses of the two objects, and R is their separation. G is a constant number known as the universal gravitational constant, and the equation itself symbolically summarizes Newton\u2019s universal law of gravitation. With such a force and the laws of motion, Newton was able to show mathematically that the only orbits permitted were exactly those described by Kepler\u2019s laws. Newton\u2019s universal law of gravitation works for the planets, but is it really universal? The gravitational theory should also predict the observed acceleration of the Moon toward Earth as it orbits Earth, as well as of any 82 Chapter 3 Orbits and Gravity object (say, an apple) dropped near Earth\u2019s surface. The falling of an apple is something we can measure quite easily, but can we use it to predict the motions of the Moon? Recall that according to Newton\u2019s second law, forces cause acceleration. Newton\u2019s universal law of gravitation says that the force acting upon (and therefore the acceleration of) an object toward Earth should be inversely proportional to the square of its distance from the center of Earth. Objects like apples at the surface of Earth, at a distance of one Earth-radius from the center of Earth, are observed to accelerate downward at 9.8 meters per second per second (9.8 m/s2). It is this force of gravity on the surface of Earth that gives us our sense of weight. Unlike your mass, which would remain the same on any planet or moon, your weight depends on the local force of gravity. So you would weigh less on Mars and the Moon than on Earth, even though there is no change in your mass. (Which means you would still have to go easy on the desserts in the college cafeteria when you got back!) The Moon is 60 Earth radii away from the center of Earth. If gravity (and the acceleration it causes) gets weaker with distance squared, the acceleration the Moon experiences should be a lot less than for the apple. The acceleration should be (1/60)2 = 1/3600 (or 3600 times less\u2014about 0.00272 m/s2. This is precisely the observed acceleration of the Moon in its orbit. (As we shall see, the Moon does not fall to Earth with this acceleration, but falls around Earth.) Imagine the thrill Newton must have felt to realize he had discovered, and verified, a law that holds for Earth, apples, the Moon, and, as far as he knew, everything in the universe Calculating Weight By what factor would a person\u2019s weight at the surface of Earth change if Earth had its present mass but eight times its present volume? Solution With eight times the volume, Earth\u2019s radius would double. This means the gravitational force at the surface would reduce by a factor of (1/2)2 = 1/4, so a person would weigh only one-fourth as much. Check Your Learning By what factor would a person\u2019s weight at the surface of Earth change if Earth had its present size but only one-third its present mass? Answer: With one-", "third its present mass, the gravitational force at the surface would reduce by a factor of 1/3, so a person would weight only one-third as much. Gravity is a \u201cbuilt-in\u201d property of mass. Whenever there are masses in the universe, they will interact via the force of gravitational attraction. The more mass there is, the greater the force of attraction. Here on Earth, the largest concentration of mass is, of course, the planet we stand on, and its pull dominates the gravitational interactions we experience. But everything with mass attracts everything else with mass anywhere in the universe. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 83 Newton\u2019s law also implies that gravity never becomes zero. It quickly gets weaker with distance, but it continues to act to some degree no matter how far away you get. The pull of the Sun is stronger at Mercury than at Pluto, but it can be felt far beyond Pluto, where astronomers have good evidence that it continuously makes enormous numbers of smaller icy bodies move around huge orbits. And the Sun\u2019s gravitational pull joins with the pull of billions of others stars to create the gravitational pull of our Milky Way Galaxy. That force, in turn, can make other smaller galaxies orbit around the Milky Way, and so on. Why is it then, you may ask, that the astronauts aboard the Space Shuttle appear to have no gravitational forces acting on them when we see images on television of the astronauts and objects floating in the spacecraft? After all, the astronauts in the shuttle are only a few hundred kilometers above the surface of Earth, which is not a significant distance compared to the size of Earth, so gravity is certainly not a great deal weaker that much farther away. The astronauts feel \u201cweightless\u201d (meaning that they don\u2019t feel the gravitational force acting on them) for the same reason that passengers in an elevator whose cable has broken or in an airplane whose engines no longer work feel weightless: they are falling (Figure 3.9).[2] Figure 3.9. Astronauts in Free Fall. While in space, astronauts are falling freely, so they experience \u201cweightlessness.\u201d Clockwise from top left: Tracy Caldwell Dyson (NASA), Naoko Yamzaki (JAXA), Dorothy Metcalf-Lindenburger (NASA), and Stephanie Wilson (NASA). (credit: NASA) When falling, they are in free fall and accelerate at the same rate as everything around them, including their spacecraft or a camera with which they are taking photographs of Earth. When doing so, astronauts experience no additional forces and therefore feel \u201cweightless.\u201d Unlike the falling elevator passengers, however, the astronauts are falling around Earth, not to Earth; as a result they will continue to fall and are said to be \u201cin orbit\u201d around Earth (see the next section for more about orbits). 2 In the film Apollo 13, the scenes in which the astronauts were \u201cweightless\u201d were actually filmed in a falling airplane. As you might imagine, the plane fell for only short periods before the engines engaged again. 84 Chapter 3 Orbits and Gravity Orbital Motion and Mass Kepler\u2019s laws describe the orbits of the objects whose motions are described by Newton\u2019s laws of motion and the law of gravity. Knowing that gravity is the force that attracts planets toward the Sun, however, allowed Newton to rethink Kepler\u2019s third law. Recall that Kepler had found a relationship between the orbital period of a planet\u2019s revolution and its distance from the Sun. But Newton\u2019s formulation introduces the additional factor of the masses of the Sun (M1) and the planet (M2), both expressed in units of the Sun\u2019s mass. Newton\u2019s universal law of gravitation can be used to show mathematically that this relationship is actually where a is the semimajor axis and P is the orbital period. a3 = \u239b \u239dM1 + M2 \u239e \u23a0 \u00d7 P2 How did Kepler miss this factor? In units of the Sun\u2019s mass, the mass of the Sun is 1, and in units of the Sun\u2019s mass, the mass of a typical planet is a negligibly small factor. This means that the sum of the Sun\u2019s mass and a planet\u2019s mass, (M1 + M2), is very, very close to 1. This makes Newton\u2019s formula appear almost the same as Kepler\u2019s; the tiny mass of the planets compared to the Sun is the reason that Kepler did not realize that both masses had to be included in the calculation. There are many situations in astronomy, however, in which we do need to include the two mass terms\u2014for example, when two stars or two galaxies orbit each other. Including the mass term allows us to use this formula in a new", " way. If we can measure the motions (distances and orbital periods) of objects acting under their mutual gravity, then the formula will permit us to deduce their masses. For example, we can calculate the mass of the Sun by using the distances and orbital periods of the planets, or the mass of Jupiter by noting the motions of its moons. Indeed, Newton\u2019s reformulation of Kepler\u2019s third law is one of the most powerful concepts in astronomy. Our ability to deduce the masses of objects from their motions is key to understanding the nature and evolution of many astronomical bodies. We will use this law repeatedly throughout this text in calculations that range from the orbits of comets to the interactions of galaxies Calculating the Effects of Gravity A planet like Earth is found orbiting its star at a distance of 1 AU in 0.71 Earth-year. Can you use Newton\u2019s version of Kepler\u2019s third law to find the mass of the star? (Remember that compared to the mass of a star, the mass of an earthlike planet can be considered negligible.) Solution In the formula a3 = (M1 + M2) \u00d7 P2, the factor M1 + M2 would now be approximately equal to M1 (the mass of the star), since the planet\u2019s mass is so small by comparison. Then the formula becomes a3 = M1 \u00d7 P2, and we can solve for M1: Since a = 1, a3 = 1, so M1 = a3 P2 M1 = 1 P2 = 1 0.712 = 1 0.5 = 2 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 85 So the mass of the star is twice the mass of our Sun. (Remember that this way of expressing the law has units in terms of Earth and the Sun, so masses are expressed in units of the mass of our Sun.) Check Your Learning Suppose a star with twice the mass of our Sun had an earthlike planet that took 4 years to orbit the star. At what distance (semimajor axis) would this planet orbit its star? Answer: Again, we can neglect the mass of the planet. So M1 = 2 and P = 4 years. The formula is a3 = M1 \u00d7 P2, so a3 = 2 \u00d7 42 = 2 \u00d7 16 = 32. So a is the cube root of 32. To find this, you can just ask Google, \u201cWhat is the cube root of 32?\u201d and get the answer 3.2 AU You might like to try a simulation (https://openstaxcollege.org/l/30phetsimsunear) that lets you move the Sun, Earth, Moon, and space station to see the effects of changing their distances on their gravitational forces and orbital paths. You can even turn off gravity and see what happens. 3.4 ORBITS IN THE SOLAR SYSTEM Learning Objectives By the end of this section, you will be able to: Compare the orbital characteristics of the planets in the solar system Compare the orbital characteristics of asteroids and comets in the solar system Recall that the path of an object under the influence of gravity through space is called its orbit, whether that object is a spacecraft, planet, star, or galaxy. An orbit, once determined, allows the future positions of the object to be calculated. Two points in any orbit in our solar system have been given special names. The place where the planet is closest to the Sun (helios in Greek) and moves the fastest is called the perihelion of its orbit, and the place where it is farthest away and moves the most slowly is the aphelion. For the Moon or a satellite orbiting Earth (gee in Greek), the corresponding terms are perigee and apogee. (In this book, we use the word moon for a natural object that goes around a planet and the word satellite to mean a human-made object that revolves around a planet.) Orbits of the Planets Today, Newton\u2019s work enables us to calculate and predict the orbits of the planets with marvelous precision. We know eight planets, beginning with Mercury closest to the Sun and extending outward to Neptune. The average orbital data for the planets are summarized in Table 3.2. (Ceres is the largest of the asteroids, now considered a dwarf planet.) 86 Chapter 3 Orbits and Gravity According to Kepler\u2019s laws, Mercury must have the shortest orbital period (88 Earth-days); thus, it has the highest orbital speed, averaging 48 kilometers per second. At the opposite extreme, Neptune has a period of 165 years and an average orbital speed of just 5 kilometers per second. All the planets have orbits of rather low eccentricity. The most eccentric orbit is that of Mercury (0.21); the rest have eccentricities smaller than 0.1. It is fortunate that among the rest,", " Mars has an eccentricity greater than that of many of the other planets. Otherwise the pre-telescopic observations of Brahe would not have been sufficient for Kepler to deduce that its orbit had the shape of an ellipse rather than a circle. The planetary orbits are also confined close to a common plane, which is near the plane of Earth\u2019s orbit (called the ecliptic). The strange orbit of the dwarf planet Pluto is inclined about 17\u00b0 to the ecliptic, and that of the dwarf planet Eris (orbiting even farther away from the Sun than Pluto) by 44\u00b0, but all the major planets lie within 10\u00b0 of the common plane of the solar system You can use an orbital simulator (https://openstaxcollege.org/l/30phetorbsim) to design your own mini solar system with up to four bodies. Adjust masses, velocities, and positions of the planets, and see what happens to their orbits as a result. Orbits of Asteroids and Comets In addition to the eight planets, there are many smaller objects in the solar system. Some of these are moons (natural satellites) that orbit all the planets except Mercury and Venus. In addition, there are two classes of smaller objects in heliocentric orbits: asteroids and comets. Both asteroids and comets are believed to be small chunks of material left over from the formation process of the solar system. In general, asteroids have orbits with smaller semimajor axes than do comets (Figure 3.10). The majority of them lie between 2.2 and 3.3 AU, in the region known as the asteroid belt (see Comets and Asteroids: Debris of the Solar System). As you can see in Table 3.2, the asteroid belt (represented by its largest member, Ceres) is in the middle of a gap between the orbits of Mars and Jupiter. It is because these two planets are so far apart that stable orbits of small bodies can exist in the region between them. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 87 Figure 3.10. Solar System Orbits. We see the orbits of typical comets and asteroids compared with those of the planets Mercury, Venus, Earth, Mars, and Jupiter (black circles). Shown in red are three comets: Halley, Kopff, and Encke. In blue are the four largest asteroids: Ceres, Pallas, Vesta, and Hygeia. Orbital Data for the Planets Planet Semimajor Axis (AU) Period (y) Eccentricity Mercury Venus Earth Mars ( Ceres) Jupiter Saturn 0.39 0.72 1 1.52 2.77 5.20 9.54 Uranus 19.19 Table 3.2 0.24 0.6 1.00 1.88 4.6 11.86 29.46 84.01 0.21 0.01 0.02 0.09 0.08 0.05 0.06 0.05 88 Chapter 3 Orbits and Gravity Orbital Data for the Planets Planet Semimajor Axis (AU) Period (y) Eccentricity Neptune 30.06 164.82 0.01 Table 3.2 Comets generally have orbits of larger size and greater eccentricity than those of the asteroids. Typically, the eccentricity of their orbits is 0.8 or higher. According to Kepler\u2019s second law, therefore, they spend most of their time far from the Sun, moving very slowly. As they approach perihelion, the comets speed up and whip through the inner parts of their orbits more rapidly. 3.5 MOTIONS OF SATELLITES AND SPACECRAFT Learning Objectives By the end of this section, you will be able to: Explain how an object (such as a satellite) can be put into orbit around Earth Explain how an object (such as a planetary probe) can escape from orbit Newton\u2019s universal law of gravitation and Kepler\u2019s laws describe the motions of Earth satellites and interplanetary spacecraft as well as the planets. Sputnik, the first artificial Earth satellite, was launched by what was then called the Soviet Union on October 4, 1957. Since that time, thousands of satellites have been placed into orbit around Earth, and spacecraft have also orbited the Moon, Venus, Mars, Jupiter, Saturn, and a number of asteroids and comets. Once an artificial satellite is in orbit, its behavior is no different from that of a natural satellite, such as our Moon. If the satellite is high enough to be free of atmospheric friction, it will remain in orbit forever. However, although there is no difficulty in maintaining a satellite once it is in orbit, a great deal of energy is required to lift the spacecraft off Earth and accelerate it to orbital speed. To illustrate how a satellite is launched, imagine a gun firing a bullet horizontally", " from the top of a high mountain, as in Figure 3.11, which has been adapted from a similar diagram by Newton. Imagine, further, that the friction of the air could be removed and that nothing gets in the bullet\u2019s way. Then the only force that acts on the bullet after it leaves the muzzle is the gravitational force between the bullet and Earth. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 89 Figure 3.11. Firing a Bullet into Orbit. (a) For paths a and b, the velocity is not enough to prevent gravity from pulling the bullet back to Earth; in case c, the velocity allows the bullet to fall completely around Earth. (b) This diagram by Newton in his De Mundi Systemate, 1731 edition, illustrates the same concept shown in (a). If the bullet is fired with a velocity we can call va, the gravitational force acting upon it pulls it downward toward Earth, where it strikes the ground at point a. However, if it is given a higher muzzle velocity, vb, its higher speed carries it farther before it hits the ground at point b. If our bullet is given a high enough muzzle velocity, vc, the curved surface of Earth causes the ground to remain the same distance from the bullet so that the bullet falls around Earth in a complete circle. The speed needed to do this\u2014called the circular satellite velocity\u2014is about 8 kilometers per second, or about 17,500 miles per hour in more familiar units. Each year, more than 50 new satellites are launched into orbit by such nations as Russia, the United States, China, Japan, India, and Israel, as well as by the European Space Agency (ESA), a consortium of European nations (Figure 3.12). Today, these satellites are used for weather tracking, ecology, global positioning systems, communications, and military purposes, to name a few uses. Most satellites are launched into low Earth orbit, since this requires the minimum launch energy. At the orbital speed of 8 kilometers per second, they circle the planet in about 90 minutes. Some of the very low Earth orbits are not indefinitely stable because, as Earth\u2019s atmosphere swells from time to time, a frictional drag is generated by the atmosphere on these satellites, eventually leading to a loss of energy and \u201cdecay\u201d of the orbit. 90 Chapter 3 Orbits and Gravity Figure 3.12. Satellites in Earth Orbit. This figure shows the larger pieces of orbital debris that are being tracked by NASA in Earth\u2019s orbit. (credit: NASA/JSC) Interplanetary Spacecraft The exploration of the solar system has been carried out largely by robot spacecraft sent to the other planets. To escape Earth, these craft must achieve escape speed, the speed needed to move away from Earth forever, which is about 11 kilometers per second (about 25,000 miles per hour). After escaping Earth, these craft coast to their targets, subject only to minor trajectory adjustments provided by small thruster rockets on board. In interplanetary flight, these spacecraft follow orbits around the Sun that are modified only when they pass near one of the planets. As it comes close to its target, a spacecraft is deflected by the planet\u2019s gravitational force into a modified orbit, either gaining or losing energy in the process. Spacecraft controllers have actually been able to use a planet\u2019s gravity to redirect a flyby spacecraft to a second target. For example, Voyager 2 used a series of gravity-assisted encounters to yield successive flybys of Jupiter (1979), Saturn (1980), Uranus (1986), and Neptune (1989). The Galileo spacecraft, launched in 1989, flew past Venus once and Earth twice to gain the energy required to reach its ultimate goal of orbiting Jupiter. If we wish to orbit a planet, we must slow the spacecraft with a rocket when the spacecraft is near its destination, allowing it to be captured into an elliptical orbit. Additional rocket thrust is required to bring a vehicle down from orbit for a landing on the surface. Finally, if a return trip to Earth is planned, the landed payload must include enough propulsive power to repeat the entire process in reverse. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 91 3.6 GRAVITY WITH MORE THAN TWO BODIES Learning Objectives By the end of this section, you will be able to: Explain how the gravitational interactions of many bodies can causes perturbations in their motions Explain how the planet Neptune was discovered Until now, we have considered the Sun and a planet (or a planet and one of its moons) as nothing more than a pair of bodies revolving around each other. In fact, all the planets exert gravitational forces upon one another as well. These interplanetary attractions cause slight variations from the orbits than would be expected", " if the gravitational forces between planets were neglected. The motion of a body that is under the gravitational influence of two or more other bodies is very complicated and can be calculated properly only with large computers. Fortunately, astronomers have such computers at their disposal in universities and government research institutes. The Interactions of Many Bodies As an example, suppose you have a cluster of a thousand stars all orbiting a common center (such clusters are quite common, as we shall see in Star Clusters). If we know the exact position of each star at any given instant, we can calculate the combined gravitational force of the entire group on any one member of the cluster. Knowing the force on the star in question, we can therefore find how it will accelerate. If we know how it was moving to begin with, we can then calculate how it will move in the next instant of time, thus tracking its motion. However, the problem is complicated by the fact that the other stars are also moving and thus changing the effect they will have on our star. Therefore, we must simultaneously calculate the acceleration of each star produced by the combination of the gravitational attractions of all the others in order to track the motions of all of them, and hence of any one. Such complex calculations have been carried out with modern computers to track the evolution of hypothetical clusters of stars with up to a million members (Figure 3.13). Figure 3.13. Modern Computing Power. These supercomputers at NASA\u2019s Ames Research Center are capable of tracking the motions of more than a million objects under their mutual gravitation. (credit: NASA Ames Research Center/Tom Trower) Within the solar system, the problem of computing the orbits of planets and spacecraft is somewhat simpler. We have seen that Kepler\u2019s laws, which do not take into account the gravitational effects of the other planets on an orbit, really work quite well. This is because these additional influences are very small in comparison with 92 Chapter 3 Orbits and Gravity the dominant gravitational attraction of the Sun. Under such circumstances, it is possible to treat the effects of other bodies as small perturbations (or disturbances). During the eighteenth and nineteenth centuries, mathematicians developed many elegant techniques for calculating perturbations, permitting them to predict very precisely the positions of the planets. Such calculations eventually led to the prediction and discovery of a new planet in 1846. The Discovery of Neptune The discovery of the eighth planet, Neptune, was one of the high points in the development of gravitational theory. In 1781, William Herschel, a musician and amateur astronomer, accidentally discovered the seventh planet, Uranus. It happens that Uranus had been observed a century before, but in none of those earlier sightings was it recognized as a planet; rather, it was simply recorded as a star. Herschel\u2019s discovery showed that there could be planets in the solar system too dim to be visible to the unaided eye, but ready to be discovered with a telescope if we just knew where to look. By 1790, an orbit had been calculated for Uranus using observations of its motion in the decade following its discovery. Even after allowance was made for the perturbing effects of Jupiter and Saturn, however, it was found that Uranus did not move on an orbit that exactly fit the earlier observations of it made since 1690. By 1840, the discrepancy between the positions observed for Uranus and those predicted from its computed orbit amounted to about 0.03\u00b0\u2014an angle barely discernable to the unaided eye but still larger than the probable errors in the orbital calculations. In other words, Uranus just did not seem to move on the orbit predicted from Newtonian theory. In 1843, John Couch Adams, a young Englishman who had just completed his studies at Cambridge, began a detailed mathematical analysis of the irregularities in the motion of Uranus to see whether they might be produced by the pull of an unknown planet. He hypothesized a planet more distant from the Sun than Uranus, and then determined the mass and orbit it had to have to account for the departures in Uranus\u2019 orbit. In October 1845, Adams delivered his results to George Airy, the British Astronomer Royal, informing him where in the sky to find the new planet. We now know that Adams\u2019 predicted position for the new body was correct to within 2\u00b0, but for a variety of reasons, Airy did not follow up right away. Meanwhile, French mathematician Urbain Jean Joseph Le Verrier, unaware of Adams or his work, attacked the same problem and published its solution in June 1846. Airy, noting that Le Verrier\u2019s predicted position for the unknown planet agreed to within 1\u00b0 with that of Adams, suggested to James Challis, Director of the Cambridge Observatory, that he begin a search for the new object. The Cambridge astronomer, having no up-to-date star charts of the Aquarius region of the sky where the planet was predicted to be, proceeded by recording the positions of all the faint stars he could observe", " with his telescope in that location. It was Challis\u2019 plan to repeat such plots at intervals of several days, in the hope that the planet would distinguish itself from a star by its motion. Unfortunately, he was negligent in examining his observations; although he had actually seen the planet, he did not recognize it. About a month later, Le Verrier suggested to Johann Galle, an astronomer at the Berlin Observatory, that he look for the planet. Galle received Le Verrier\u2019s letter on September 23, 1846, and, possessing new charts of the Aquarius region, found and identified the planet that very night. It was less than a degree from the position Le Verrier predicted. The discovery of the eighth planet, now known as Neptune (the Latin name for the god of the sea), was a major triumph for gravitational theory for it dramatically confirmed the generality of Newton\u2019s laws. The honor for the discovery is properly shared by the two mathematicians, Adams and Le Verrier (Figure 3.14). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 93 Figure 3.14. Mathematicians Who Discovered a Planet. (a) John Couch Adams (1819\u20131892) and (b) Urbain J. J. Le Verrier (1811\u20131877) share the credit for discovering the planet Neptune. We should note that the discovery of Neptune was not a complete surprise to astronomers, who had long suspected the existence of the planet based on the \u201cdisobedient\u201d motion of Uranus. On September 10, 1846, two weeks before Neptune was actually found, John Herschel, son of the discoverer of Uranus, remarked in a speech before the British Association, \u201cWe see [the new planet] as Columbus saw America from the shores of Spain. Its movements have been felt trembling along the far-reaching line of our analysis with a certainty hardly inferior to ocular demonstration.\u201d This discovery was a major step forward in combining Newtonian theory with painstaking observations. Such work continues in our own times with the discovery of planets around other stars For the fuller story of how Neptune was predicted and found (and the effect of the discovery on the search for Pluto), you can read this page (https://openstaxcollege.org/l/30nepplumatdis) on the mathematical discovery of planets Astronomy and the Poets When Copernicus, Kepler, Galileo, and Newton formulated the fundamental rules that underlie everything in the physical world, they changed much more than the face of science. For some, they gave humanity the courage to let go of old superstitions and see the world as rational and manageable; for 94 Chapter 3 Orbits and Gravity others, they upset comforting, ordered ways that had served humanity for centuries, leaving only a dry, \u201cmechanical clockwork\u201d universe in their wake. Poets of the time reacted to such changes in their work and debated whether the new world picture was an appealing or frightening one. John Donne (1573\u20131631), in a poem called \u201cAnatomy of the World,\u201d laments the passing of the old certainties: The new Philosophy [science] calls all in doubt, The element of fire is quite put out; The Sun is lost, and th\u2019 earth, and no man\u2019s wit Can well direct him where to look for it. (Here the \u201celement of fire\u201d refers also to the sphere of fire, which medieval thought placed between Earth and the Moon.) By the next century, however, poets like Alexander Pope were celebrating Newton and the Newtonian world view. Pope\u2019s famous couplet, written upon Newton\u2019s death, goes Nature, and nature\u2019s laws lay hid in night. God said, Let Newton be! And all was light. In his 1733 poem, An Essay on Man, Pope delights in the complexity of the new views of the world, incomplete though they are: Of man, what see we, but his station here, From which to reason, to which refer?... He, who thro\u2019 vast immensity can pierce, See worlds on worlds compose one universe, Observe how system into system runs, What other planets circle other suns, What vary\u2019d being peoples every star, May tell why Heav\u2019n has made us as we are... All nature is but art, unknown to thee; All chance, direction, which thou canst not see; All discord, harmony not understood; All partial evil, universal good: And, in spite of pride, in erring reason\u2019s spite, One truth is clear, whatever is, is right. Poets and philosophers continued to debate whether humanity was exalted or debased by the new views of science. The nineteenth-century", " poet Arthur Hugh Clough (1819\u20131861) cries out in his poem \u201cThe New Sinai\u201d: And as old from Sinai\u2019s top God said that God is one, By science strict so speaks He now to tell us, there is None! Earth goes by chemic forces; Heaven\u2019s a M\u00e9canique Celeste! And heart and mind of humankind a watchwork as the rest! This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 95 (A \u201cm\u00e9canique celeste\u201d is a clockwork model to demonstrate celestial motions.) The twentieth-century poet Robinson Jeffers (whose brother was an astronomer) saw it differently in a poem called \u201cStar Swirls\u201d: There is nothing like astronomy to pull the stuff out of man. His stupid dreams and red-rooster importance: Let him count the star-swirls. 96 Chapter 3 Orbits and Gravity CHAPTER 3 REVIEW KEY TERMS angular momentum the measure of the motion of a rotating object in terms of its speed and how widely the object\u2019s mass is distributed around its axis aphelion the point in its orbit where a planet (or other orbiting object) is farthest from the Sun apogee the point in its orbit where an Earth satellite is farthest from Earth asteroid belt the region of the solar system between the orbits of Mars and Jupiter in which most asteroids are located; the main belt, where the orbits are generally the most stable, extends from 2.2 to 3.3 AU from the Sun astronomical unit (AU) distance is about 1.5 \u00d7 108 kilometers the unit of length defined as the average distance between Earth and the Sun; this density the ratio of the mass of an object to its volume eccentricity in an ellipse, the ratio of the distance between the foci to the major axis ellipse a closed curve for which the sum of the distances from any point on the ellipse to two points inside (called the foci) is always the same escape speed the speed a body must achieve to break away from the gravity of another body focus (plural: foci) one of two fixed points inside an ellipse from which the sum of the distances to any point on the ellipse is constant gravity the mutual attraction of material bodies or particles Kepler\u2019s first law each planet moves around the Sun in an orbit that is an ellipse, with the Sun at one focus of the ellipse Kepler\u2019s second law the straight line joining a planet and the Sun sweeps out equal areas in space in equal intervals of time Kepler\u2019s third law the square of a planet\u2019s orbital period is directly proportional to the cube of the semimajor axis of its orbit major axis the maximum diameter of an ellipse momentum the measure of the amount of motion of a body; the momentum of a body is the product of its mass and velocity; in the absence of an unbalanced force, momentum is conserved Newton\u2019s first law every object will continue to be in a state of rest or move at a constant speed in a straight line unless it is compelled to change by an outside force Newton\u2019s second law on it the change of motion of a body is proportional to and in the direction of the force acting Newton\u2019s third law for every action there is an equal and opposite reaction (or: the mutual actions of two bodies upon each other are always equal and act in opposite directions) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 97 orbit the path of an object that is in revolution about another object or point orbital period (P) the time it takes an object to travel once around the Sun orbital speed the speed at which an object (usually a planet) orbits around the mass of another object; in the case of a planet, the speed at which each planet moves along its ellipse perigee the point in its orbit where an Earth satellite is closest to Earth perihelion the point in its orbit where a planet (or other orbiting object) is nearest to the Sun perturbation a small disturbing effect on the motion or orbit of a body produced by a third body satellite an object that revolves around a planet semimajor axis half of the major axis of a conic section, such as an ellipse velocity the speed and direction a body is moving\u2014for example, 44 kilometers per second toward the north galactic pole SUMMARY 3.1 The Laws of Planetary Motion Tycho Brahe\u2019s accurate observations of planetary positions provided the data used by Johannes Kepler to derive his three fundamental laws of planetary motion. Kepler\u2019s laws describe the behavior of planets in their orbits as follows: (1) planetary orbits are ellipses with the Sun at one focus", "; (2) in equal intervals, a planet\u2019s orbit sweeps out equal areas; and (3) the relationship between the orbital period (P) and the semimajor axis (a) of an orbit is given by P2 = a3 (when a is in units of AU and P is in units of Earth years). 3.2 Newton\u2019s Great Synthesis In his Principia, Isaac Newton established the three laws that govern the motion of objects: (1) objects continue to be at rest or move with a constant velocity unless acted upon by an outside force; (2) an outside force causes an acceleration (and changes the momentum) for an object; and (3) for every action there is an equal and opposite reaction. Momentum is a measure of the motion of an object and depends on both its mass and its velocity. Angular momentum is a measure of the motion of a spinning or revolving object and depends on its mass, velocity, and distance from the point around which it revolves. The density of an object is its mass divided by its volume. 3.3 Newton\u2019s Universal Law of Gravitation Gravity, the attractive force between all masses, is what keeps the planets in orbit. Newton\u2019s universal law of gravitation relates the gravitational force to mass and distance: Fgravity = G M1 M2 R 2 The force of gravity is what gives us our sense of weight. Unlike mass, which is constant, weight can vary depending on the force of gravity (or acceleration) you feel. When Kepler\u2019s laws are reexamined in the light of Newton\u2019s gravitational law, it becomes clear that the masses of both objects are important for the third law, which becomes a3 = (M1 + M2) \u00d7 P2. Mutual gravitational effects permit us to calculate the masses of astronomical objects, from comets to galaxies. 98 Chapter 3 Orbits and Gravity 3.4 Orbits in the Solar System The closest point in a satellite orbit around Earth is its perigee, and the farthest point is its apogee (corresponding to perihelion and aphelion for an orbit around the Sun). The planets follow orbits around the Sun that are nearly circular and in the same plane. Most asteroids are found between Mars and Jupiter in the asteroid belt, whereas comets generally follow orbits of high eccentricity. 3.5 Motions of Satellites and Spacecraft The orbit of an artificial satellite depends on the circumstances of its launch. The circular satellite velocity needed to orbit Earth\u2019s surface is 8 kilometers per second, and the escape speed from our planet is 11 kilometers per second. There are many possible interplanetary trajectories, including those that use gravity- assisted flybys of one object to redirect the spacecraft toward its next target. 3.6 Gravity with More Than Two Bodies Calculating the gravitational interaction of more than two objects is complicated and requires large computers. If one object (like the Sun in our solar system) dominates gravitationally, it is possible to calculate the effects of a second object in terms of small perturbations. This approach was used by John Couch Adams and Urbain Le Verrier to predict the position of Neptune from its perturbations of the orbit of Uranus and thus discover a new planet mathematically. FOR FURTHER EXPLORATION Articles Brahe and Kepler Christianson, G. \u201cThe Celestial Palace of Tycho Brahe.\u201d Scientific American (February 1961): 118. Gingerich, O. \u201cJohannes Kepler and the Rudolphine Tables.\u201d Sky & Telescope (December 1971): 328. Brief article on Kepler\u2019s work. Wilson, C. \u201cHow Did Kepler Discover His First Two Laws?\u201d Scientific American (March 1972): 92. Newton Christianson, G. \u201cNewton\u2019s Principia: A Retrospective.\u201d Sky & Telescope (July 1987): 18. Cohen, I. \u201cNewton\u2019s Discovery of Gravity.\u201d Scientific American (March 1981): 166. Gingerich, O. \u201cNewton, Halley, and the Comet.\u201d Sky & Telescope (March 1986): 230. Sullivant, R. \u201cWhen the Apple Falls.\u201d Astronomy (April 1998): 55. Brief overview. The Discovery of Neptune Sheehan, W., et al. \u201cThe Case of the Pilfered Planet: Did the British Steal Neptune?\u201d Scientific American (December 2004): 92. Websites Brahe and Kepler Johannes Kepler: His Life, His Laws, and Time: http://kepler.nasa.gov/Mission/JohannesKepler/. From NASA\u2019s Kepler mission. Johannes Kepler: http://www.britannica.com/biography/Johannes-Kepler. Encyclopedia Britannica article. This OpenStax book is available for free at http://cn", "x.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 99 Johannes Kepler: http://www-history.mcs.st-andrews.ac.uk/Biographies/Kepler.html. MacTutor article with additional links. Noble Dane: Images of Tycho Brahe: http://www.mhs.ox.ac.uk/tycho/index.htm. A virtual museum exhibit from Oxford. Newton Sir Isaac Newton: http://www-groups.dcs.st-and.ac.uk/~history//Biographies/Newton.html. MacTutor article with additional links. Sir Isaac Newton: http://www.luminarium.org/sevenlit/newton/newtonbio.htm. Newton Biography at the Luminarium. The Discovery of Neptune Adams, Airy, and the Discovery of Neptune: http://www.mikeoates.org/lassell/adams-airy.htm. A defense of Airy\u2019s role by historian Alan Chapman. Mathematical Discovery of Planets: http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/ Neptune_and_Pluto.html. MacTutor article. Videos Brahe and Kepler \u201cHarmony of the Worlds.\u201d This third episode of Carl Sagan\u2019s TV series Cosmos focuses on Kepler and his life and work. Tycho Brahe, Johannes Kepler, and Planetary Motion: https://www.youtube.com/watch?v=x3ALuycrCwI. German-produced video, in English (14:27). Newton Beyond the Big Bang: Sir Isaac Newton\u2019s Law of Gravity: http://www.history.com/topics/enlightenment/videos/ beyond-the-big-bang-sir-isaac-newtons-law-of-gravity. From the History Channel (4:35). Sir Isaac Newton versus Bill Nye: Epic Rap Battles of History: https://www.youtube.com/watch?v=8yis7GzlXNM. (2:47). The Discovery of Neptune Richard Feynman: On the Discovery of Neptune: https://www.youtube.com/watch?v=FgXQffVgZRs. A brief black-and-white Caltech lecture (4:33). COLLABORATIVE GROUP ACTIVITIES A. An eccentric, but very rich, alumnus of your college makes a bet with the dean that if you drop a baseball and a bowling ball from the tallest building on campus, the bowling ball would hit the ground first. Have your group discuss whether you would make a side bet that the alumnus is right. How would you decide who is right? B. Suppose someone in your astronomy class was unhappy about his or her weight. Where could a person go to weigh one-fourth as much as he or she does now? Would changing the unhappy person\u2019s weight have any effect on his or her mass? 100 Chapter 3 Orbits and Gravity C. When the Apollo astronauts landed on the Moon, some commentators commented that it ruined the mystery and \u201cpoetry\u201d of the Moon forever (and that lovers could never gaze at the full moon in the same way again). Others felt that knowing more about the Moon could only enhance its interest to us as we see it from Earth. How do the various members of your group feel? Why? D. Figure 3.12 shows a swarm of satellites in orbit around Earth. What do you think all these satellites do? How many categories of functions for Earth satellites can your group come up with? E. The Making Connections feature box Astronomy and the Poets discusses how poets included the most recent astronomical knowledge in their poetry. Is this still happening today? Can your group members come up with any poems or songs that you know that deal with astronomy or outer space? If not, perhaps you could find some online, or by asking friends or roommates who are into poetry or music. EXERCISES Review Questions 1. State Kepler\u2019s three laws in your own words. 2. Why did Kepler need Tycho Brahe\u2019s data to formulate his laws? 3. Which has more mass: an armful of feathers or an armful of lead? Which has more volume: a kilogram of feathers or a kilogram of lead? Which has higher density: a kilogram of feathers or a kilogram of lead? 4. Explain how Kepler was able to find a relationship (his third law) between the orbital periods and distances of the planets that did not depend on the masses of the planets or the Sun. 5. Write out Newton\u2019s three laws of motion in terms of what happens with the momentum of objects. 6. Which major planet has the largest... A. semimajor axis? B. average", " orbital speed around the Sun? C. orbital period around the Sun? D. eccentricity? 7. Why do we say that Neptune was the first planet to be discovered through the use of mathematics? 8. Why was Brahe reluctant to provide Kepler with all his data at one time? 9. According to Kepler\u2019s second law, where in a planet\u2019s orbit would it be moving fastest? Where would it be moving slowest? 10. The gas pedal, the brakes, and the steering wheel all have the ability to accelerate a car\u2014how? 11. Explain how a rocket can propel itself using Newton\u2019s third law. 12. A certain material has a mass of 565 g while occupying 50 cm3 of space. What is this material? (Hint: Use Table 3.1.) 13. To calculate the momentum of an object, which properties of an object do you need to know? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 3 Orbits and Gravity 101 14. To calculate the angular momentum of an object, which properties of an object do you need to know? 15. What was the great insight Newton had regarding Earth\u2019s gravity that allowed him to develop the universal law of gravitation? 16. Which of these properties of an object best quantifies its inertia: velocity, acceleration, volume, mass, or temperature? 17. Pluto\u2019s orbit is more eccentric than any of the major planets. What does that mean? 18. Why is Tycho Brahe often called \u201cthe greatest naked-eye astronomer\u201d of all time? Thought Questions 19. Is it possible to escape the force of gravity by going into orbit around Earth? How does the force of gravity in the International Space Station (orbiting an average of 400 km above Earth\u2019s surface) compare with that on the ground? 20. What is the momentum of an object whose velocity is zero? How does Newton\u2019s first law of motion include the case of an object at rest? 21. Evil space aliens drop you and your fellow astronomy student 1 km apart out in space, very far from any star or planet. Discuss the effects of gravity on each of you. 22. A body moves in a perfectly circular path at constant speed. Are there forces acting in such a system? How do you know? 23. As friction with our atmosphere causes a satellite to spiral inward, closer to Earth, its orbital speed increases. Why? 24. Use a history book, an encyclopedia, or the internet to find out what else was happening in England during Newton\u2019s lifetime and discuss what trends of the time might have contributed to his accomplishments and the rapid acceptance of his work. 25. Two asteroids begin to gravitationally attract one another. If one asteroid has twice the mass of the other, which one experiences the greater force? Which one experiences the greater acceleration? 26. How does the mass of an astronaut change when she travels from Earth to the Moon? How does her weight change? 27. If there is gravity where the International Space Station (ISS) is located above Earth, why doesn\u2019t the space station get pulled back down to Earth? 28. Compare the density, weight, mass, and volume of a pound of gold to a pound of iron on the surface of Earth. 29. If identical spacecraft were orbiting Mars and Earth at identical radii (distances), which spacecraft would be moving faster? Why? Figuring For Yourself 30. By what factor would a person\u2019s weight be increased if Earth had 10 times its present mass, but the same volume? 31. Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice that of Earth\u2019s). What must be the mass of this planet such that the gravitational force (Fgravity) at the surface would be identical to Earth\u2019s? 102 Chapter 3 Orbits and Gravity 32. What is the semimajor axis of a circle of diameter 24 cm? What is its eccentricity? 33. If 24 g of material fills a cube 2 cm on a side, what is the density of the material? 34. If 128 g of material is in the shape of a brick 2 cm wide, 4 cm high, and 8 cm long, what is the density of the material? 35. If the major axis of an ellipse is 16 cm, what is the semimajor axis? If the eccentricity is 0.8, would this ellipse be best described as mostly circular or very elongated? 36. What is the average distance from the Sun (in astronomical units) of an asteroid with an orbital period of 8 years? 37. What is the average distance from the Sun (in astronomical units) of a planet with an orbital period of 45.66 years? 38. In 1996, astronomers discovered an icy object beyond Pluto that was given the designation 1996 TL 66.", " It has a semimajor axis of 84 AU. What is its orbital period according to Kepler\u2019s third law? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 103 4 EARTH, MOON, AND SKY Figure 4.1. Southern Summer. As captured with a fish-eye lens aboard the Atlantis Space Shuttle on December 9, 1993, Earth hangs above the Hubble Space Telescope as it is repaired. The reddish continent is Australia, its size and shape distorted by the special lens. Because the seasons in the Southern Hemisphere are opposite those in the Northern Hemisphere, it is summer in Australia on this December day. (credit: modification of work by NASA) Chapter Outline 4.1 Earth and Sky 4.2 The Seasons 4.3 Keeping Time 4.4 The Calendar 4.5 Phases and Motions of the Moon 4.6 Ocean Tides and the Moon 4.7 Eclipses of the Sun and Moon Thinking Ahead If Earth\u2019s orbit is nearly a perfect circle (as we saw in earlier chapters), why is it hotter in summer and colder in winter in many places around the globe? And why are the seasons in Australia or Peru the opposite of those in the United States or Europe? The story is told that Galileo, as he left the Hall of the Inquisition following his retraction of the doctrine that Earth rotates and revolves about the Sun, said under his breath, \u201cBut nevertheless it moves.\u201d Historians are not sure whether the story is true, but certainly Galileo knew that Earth was in motion, whatever church authorities said. It is the motions of Earth that produce the seasons and give us our measures of time and date. The Moon\u2019s motions around us provide the concept of the month and the cycle of lunar phases. In this chapter we examine some of the basic phenomena of our everyday world in their astronomical context. 104 Chapter 4 Earth, Moon, and Sky 4.1 EARTH AND SKY Learning Objectives By the end of this section, you will be able to: Describe how latitude and longitude are used to map Earth Explain how right ascension and declination are used to map the sky In order to create an accurate map, a mapmaker needs a way to uniquely and simply identify the location of all the major features on the map, such as cities or natural landmarks. Similarly, astronomical mapmakers need a way to uniquely and simply identify the location of stars, galaxies, and other celestial objects. On Earth maps, we divide the surface of Earth into a grid, and each location on that grid can easily be found using its latitude and longitude coordinate. Astronomers have a similar system for objects on the sky. Learning about these can help us understand the apparent motion of objects in the sky from various places on Earth. Locating Places on Earth Let\u2019s begin by fixing our position on the surface of planet Earth. As we discussed in Observing the Sky: The Birth of Astronomy, Earth\u2019s axis of rotation defines the locations of its North and South Poles and of its equator, halfway between. Two other directions are also defined by Earth\u2019s motions: east is the direction toward which Earth rotates, and west is its opposite. At almost any point on Earth, the four directions\u2014north, south, east, and west\u2014are well defined, despite the fact that our planet is round rather that flat. The only exceptions are exactly at the North and South Poles, where the directions east and west are ambiguous (because points exactly at the poles do not turn). We can use these ideas to define a system of coordinates attached to our planet. Such a system, like the layout of streets and avenues in Manhattan or Salt Lake City, helps us find where we are or want to go. Coordinates on a sphere, however, are a little more complicated than those on a flat surface. We must define circles on the sphere that play the same role as the rectangular grid that you see on city maps. A great circle is any circle on the surface of a sphere whose center is at the center of the sphere. For example, Earth\u2019s equator is a great circle on Earth\u2019s surface, halfway between the North and South Poles. We can also imagine a series of great circles that pass through both the North and South Poles. Each of this circles is called a meridian; they are each perpendicular to the equator, crossing it at right angles. Any point on the surface of Earth will have a meridian passing through it (Figure 4.2). The meridian specifies the east-west location, or longitude, of the place. By international agreement (and it took many meetings for the world\u2019s countries to agree), longitude is defined as the number of degrees of arc along the equator between your meridian and the one passing through Greenwich, England, which has been designated as the Prime", " Meridian. The longitude of the Prime Meridian is defined as 0\u00b0. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 105 Figure 4.2. Latitude and Longitude of Washington, DC. We use latitude and longitude to find cities like Washington, DC, on a globe. Latitude is the number of degrees north or south of the equator, and longitude is the number of degrees east or west of the Prime Meridian. Washington, DC\u2019s coordinates are 38\u00b0 N and 77\u00b0 W. Why Greenwich, you might ask? Every country wanted 0\u00b0 longitude to pass through its own capital. Greenwich, the site of the old Royal Observatory (Figure 4.3), was selected because it was between continental Europe and the United States, and because it was the site for much of the development of the method to measure longitude at sea. Longitudes are measured either to the east or to the west of the Greenwich meridian from 0\u00b0 to 180\u00b0. As an example, the longitude of the clock-house benchmark of the U.S. Naval Observatory in Washington, DC, is 77.066\u00b0 W. Figure 4.3. Royal Observatory in Greenwich, England. At the internationally agreed-upon zero point of longitude at the Royal Observatory Greenwich, tourists can stand and straddle the exact line where longitude \u201cbegins.\u201d(credit left: modification of work by \u201cpdbreen\u201d/Flickr; credit right: modification of work by Ben Sutherland) Your latitude (or north-south location) is the number of degrees of arc you are away from the equator along your meridian. Latitudes are measured either north or south of the equator from 0\u00b0 to 90\u00b0. (The latitude of the equator is 0\u00b0.) As an example, the latitude of the previously mentioned Naval Observatory benchmark is 38.921\u00b0 N. The latitude of the South Pole is 90\u00b0 S, and the latitude of the North Pole is 90\u00b0 N. 106 Chapter 4 Earth, Moon, and Sky Locating Places in the Sky Positions in the sky are measured in a way that is very similar to the way we measure positions on the surface of Earth. Instead of latitude and longitude, however, astronomers use coordinates called declination and right ascension. To denote positions of objects in the sky, it is often convenient to make use of the fictitious celestial sphere. We saw in Observing the Sky: The Birth of Astronomy that the sky appears to rotate about points above the North and South Poles of Earth\u2014points in the sky called the north celestial pole and the south celestial pole. Halfway between the celestial poles, and thus 90\u00b0 from each pole, is the celestial equator, a great circle on the celestial sphere that is in the same plane as Earth\u2019s equator. We can use these markers in the sky to set up a system of celestial coordinates. Declination on the celestial sphere is measured the same way that latitude is measured on the sphere of Earth: from the celestial equator toward the north (positive) or south (negative). So Polaris, the star near the north celestial pole, has a declination of almost +90\u00b0. Right ascension (RA) is like longitude, except that instead of Greenwich, the arbitrarily chosen point where we start counting is the vernal equinox, a point in the sky where the ecliptic (the Sun\u2019s path) crosses the celestial equator. RA can be expressed either in units of angle (degrees) or in units of time. This is because the celestial sphere appears to turn around Earth once a day as our planet turns on its axis. Thus the 360\u00b0 of RA that it takes to go once around the celestial sphere can just as well be set equal to 24 hours. Then each 15\u00b0 of arc is equal to 1 hour of time. For example, the approximate celestial coordinates of the bright star Capella are RA 5h = 75\u00b0 and declination +50\u00b0. One way to visualize these circles in the sky is to imagine Earth as a transparent sphere with the terrestrial coordinates (latitude and longitude) painted on it with dark paint. Imagine the celestial sphere around us as a giant ball, painted white on the inside. Then imagine yourself at the center of Earth, with a bright light bulb in the middle, looking out through its transparent surface to the sky. The terrestrial poles, equator, and meridians will be projected as dark shadows on the celestial sphere, giving us the system of coordinates in the sky You can explore a variety of basic animations about coordinates and motions in the sky at this interactive site (https://openstaxcollege.org/l/30anicoormot) from ClassAction. Click on the \u201cAnimations\u201d tab for a list of options. If you choose the second option", " in the menu, you can play with the celestial sphere and see RA and declination defined visually. The Turning Earth Why do many stars rise and set each night? Why, in other words, does the night sky seem to turn? We have seen that the apparent rotation of the celestial sphere could be accounted for either by a daily rotation of the sky around a stationary Earth or by the rotation of Earth itself. Since the seventeenth century, it has been generally accepted that it is Earth that turns, but not until the nineteenth century did the French physicist Jean Foucault provide an unambiguous demonstration of this rotation. In 1851, he suspended a 60-meter pendulum weighing about 25 kilograms from the dome of the Pantheon in Paris and started the pendulum swinging evenly. If Earth had not been turning, there would have been no alteration of the pendulum\u2019s plane of oscillation, and so it would have continued tracing the same path. Yet after a few minutes Foucault could see that the pendulum\u2019s plane of motion was turning. Foucault explained that it was not the pendulum that was shifting, but rather This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 107 Earth that was turning beneath it (Figure 4.4). You can now find such pendulums in many science centers and planetariums around the world. Figure 4.4. Foucault\u2019s Pendulum. As Earth turns, the plane of oscillation of the Foucault pendulum shifts gradually so that over the course of 12 hours, all the targets in the circle at the edge of the wooden platform are knocked over in sequence. (credit: Manuel M. Vicente) Can you think of other pieces of evidence that indicate that it is Earth and not the sky that is turning? (See Collaborative Group Activity A at the end of this chapter.) 4.2 THE SEASONS Learning Objectives By the end of this section, you will be able to: Describe how the tilt of Earth\u2019s axis causes the seasons Explain how seasonal differences on Earth vary with latitude One of the fundamental facts of life at Earth\u2019s midlatitudes, where most of this book\u2019s readers live, is that there are significant variations in the heat we receive from the Sun during the course of the year. We thus divide the year into seasons, each with its different amount of sunlight. The difference between seasons gets more pronounced the farther north or south from the equator we travel, and the seasons in the Southern Hemisphere are the opposite of what we find on the northern half of Earth. With these observed facts in mind, let us ask what causes the seasons. Many people have believed that the seasons were the result of the changing distance between Earth and the Sun. This sounds reasonable at first: it should be colder when Earth is farther from the Sun. But the facts don\u2019t bear out this hypothesis. Although Earth\u2019s orbit around the Sun is an ellipse, its distance from the Sun varies by only about 3%. That\u2019s not enough to cause significant variations in the Sun\u2019s heating. To make matters worse for people in North America who hold this hypothesis, Earth is actually closest to the Sun in January, when the Northern Hemisphere is in the middle of winter. And if distance were the governing factor, why would the 108 Chapter 4 Earth, Moon, and Sky two hemispheres have opposite seasons? As we shall show, the seasons are actually caused by the 23.5\u00b0 tilt of Earth\u2019s axis. The Seasons and Sunshine Figure 4.5 shows Earth\u2019s annual path around the Sun, with Earth\u2019s axis tilted by 23.5\u00b0. Note that our axis continues to point the same direction in the sky throughout the year. As Earth travels around the Sun, in June the Northern Hemisphere \u201cleans into\u201d the Sun and is more directly illuminated. In December, the situation is reversed: the Southern Hemisphere leans into the Sun, and the Northern Hemisphere leans away. In September and March, Earth leans \u201csideways\u201d\u2014neither into the Sun nor away from it\u2014so the two hemispheres are equally favored with sunshine. Figure 4.5. Seasons. We see Earth at different seasons as it circles the Sun. In June, the Northern Hemisphere \u201cleans into\u201d the Sun, and those in the North experience summer and have longer days. In December, during winter in the Northern Hemisphere, the Southern Hemisphere \u201cleans into\u201d the Sun and is illuminated more directly. In spring and autumn, the two hemispheres receive more equal shares of sunlight.[1] How does the Sun\u2019s favoring one hemisphere translate into making it warmer for us down on the surface of Earth? There are two effects we need to consider. When we lean into the Sun, sunlight hits us at a more direct", " angle and is more effective at heating Earth\u2019s surface (Figure 4.6). You can get a similar effect by shining a flashlight onto a wall. If you shine the flashlight straight on, you get an intense spot of light on the wall. But if you hold the flashlight at an angle (if the wall \u201cleans out\u201d of the beam), then the spot of light is more spread out. Like the straight-on light, the sunlight in June is more direct and intense in the Northern Hemisphere, and hence more effective at heating. 1 Note that the dates indicated for the solstices and equinoxes are approximate; depending on the year, they may occur a day or two earlier or later. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 109 Figure 4.6. The Sun\u2019s Rays in Summer and Winter. (a) In summer, the Sun appears high in the sky and its rays hit Earth more directly, spreading out less. (b) In winter, the Sun is low in the sky and its rays spread out over a much wider area, becoming less effective at heating the ground. The second effect has to do with the length of time the Sun spends above the horizon (Figure 4.7). Even if you\u2019ve never thought about astronomy before, we\u2019re sure you have observed that the hours of daylight increase in summer and decrease in winter. Let\u2019s see why this happens. Figure 4.7. The Sun\u2019s Path in the Sky for Different Seasons. On June 21, the Sun rises north of east and sets north of west. For observers in the Northern Hemisphere of Earth, the Sun spends about 15 hours above the horizon in the United States, meaning more hours of daylight. On December 21, the Sun rises south of east and sets south of west. It spends 9 hours above the horizon in the United States, which means fewer hours of daylight and more hours of night in northern lands (and a strong need for people to hold celebrations to cheer themselves up). On March 21 and September 21, the Sun spends equal amounts of time above and below the horizon in both hemispheres. As we saw in Observing the Sky: The Birth of Astronomy, an equivalent way to look at our path around the Sun each year is to pretend that the Sun moves around Earth (on a circle called the ecliptic). Because Earth\u2019s axis is tilted, the ecliptic is tilted by about 23.5\u00b0 relative to the celestial equator (review Figure 2.7). As a result, where we see the Sun in the sky changes as the year wears on. In June, the Sun is north of the celestial equator and spends more time with those who live in the Northern Hemisphere. It rises high in the sky and is above the horizon in the United States for as long as 15 hours. Thus, the Sun not only heats us with more direct rays, but it also has more time to do it each day. (Notice in Figure 4.7 110 Chapter 4 Earth, Moon, and Sky that the Northern Hemisphere\u2019s gain is the Southern Hemisphere\u2019s loss. There the June Sun is low in the sky, meaning fewer daylight hours. In Chile, for example, June is a colder, darker time of year.) In December, when the Sun is south of the celestial equator, the situation is reversed. Let\u2019s look at what the Sun\u2019s illumination on Earth looks like at some specific dates of the year, when these effects are at their maximum. On or about June 21 (the date we who live in the Northern Hemisphere call the summer solstice or sometimes the first day of summer), the Sun shines down most directly upon the Northern Hemisphere of Earth. It appears about 23\u00b0 north of the equator, and thus, on that date, it passes through the zenith of places on Earth that are at 23\u00b0 N latitude. The situation is shown in detail in Figure 4.8. To a person at 23\u00b0 N (near Hawaii, for example), the Sun is directly overhead at noon. This latitude, where the Sun can appear at the zenith at noon on the first day of summer, is called the Tropic of Cancer. We also see in Figure 4.8 that the Sun\u2019s rays shine down all around the North Pole at the solstice. As Earth turns on its axis, the North Pole is continuously illuminated by the Sun; all places within 23\u00b0 of the pole have sunshine for 24 hours. The Sun is as far north on this date as it can get; thus, 90\u00b0 \u2013 23\u00b0 (or 67\u00b0 N) is the southernmost latitude where the Sun can be seen for a full 24-hour period (sometimes called the \u201cland of the midnight Sun\u201d). That circle of latitude is called the Arctic", " Circle. Figure 4.8. Earth on June 21. This is the date of the summer solstice in the Northern Hemisphere. Note that as Earth turns on its axis (the line connecting the North and South Poles), the North Pole is in constant sunlight while the South Pole is veiled in 24 hours of darkness. The Sun is at the zenith for observers on the Tropic of Cancer. Many early cultures scheduled special events around the summer solstice to celebrate the longest days and thank their gods for making the weather warm. This required people to keep track of the lengths of the days and the northward trek of the Sun in order to know the right day for the \u201cparty.\u201d (You can do the same thing by watching for several weeks, from the same observation point, where the Sun rises or sets relative to a fixed landmark. In spring, the Sun will rise farther and farther north of east, and set farther and farther north of west, reaching the maximum around the summer solstice.) Now look at the South Pole in Figure 4.8. On June 21, all places within 23\u00b0 of the South Pole\u2014that is, south of what we call the Antarctic Circle\u2014do not see the Sun at all for 24 hours. The situation is reversed 6 months later, about December 21 (the date of the winter solstice, or the first day of winter in the Northern Hemisphere), as shown in Figure 4.9. Now it is the Arctic Circle that has the 24-hour night and the Antarctic Circle that has the midnight Sun. At latitude 23\u00b0 S, called the Tropic of Capricorn, the Sun passes through the zenith at noon. Days are longer in the Southern Hemisphere and shorter in the north. In the United States and Southern Europe, there may be only 9 or 10 hours of sunshine during the day. It is winter in the Northern Hemisphere and summer in the Southern Hemisphere. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 111 Figure 4.9. Earth on December 21. This is the date of the winter solstice in the Northern Hemisphere. Now the North Pole is in darkness for 24 hours and the South Pole is illuminated. The Sun is at the zenith for observers on the Tropic of Capricorn and thus is low in the sky for the residents of the Northern Hemisphere Seasonal Variations As you can see in Figure 4.8, the Tropic of Cancer is the latitude for which the Sun is directly overhead on the summer solstice. At this time, the Sun is at a declination of 23\u00b0 N of the celestial equator, and the corresponding latitude on Earth is 23\u00b0 N of the equator. If Earth were tilted a bit less, then the Tropic of Cancer would be at a lower latitude, closer to the equator. The Arctic Circle marks the southernmost latitude for which the day length is 24 hours on the day of the summer solstice. This is located at 90\u00b0 \u2013 23\u00b0 = 67\u00b0 N of Earth\u2019s equator. If Earth were tilted a bit less, then the Arctic Circle would move farther North. In the limit at which Earth is not tilted at all (its axis is perpendicular to the ecliptic), the Tropic of Cancer would be right on Earth\u2019s equator, and the Arctic Circle would simply be the North Pole. Suppose the tilt of Earth\u2019s axis were tilted only 5\u00b0. What would be the effect on the seasons and the locations of the Tropic of Cancer and Arctic Circle? Solution If Earth were tilted less, the seasons would be less extreme. The variation in day length and direct sunlight would be very small over the course of a year, and the Sun\u2019s daily path in the sky would not vary much. If Earth were tilted by 5\u00b0, the Sun\u2019s position on the day of the summer solstice would be 5\u00b0 N of the celestial equator, so the Tropic of Cancer would be at the corresponding latitude on Earth of 5\u00b0 N of the Equator. The Arctic Circle would be located at 90\u00b0 \u2013 5\u00b0 = 85\u00b0 N of the equator. Check Your Learning Suppose the tilt of Earth\u2019s axis were 16\u00b0. What, then, would be the difference in latitude between the Arctic Circle and the Tropic of Cancer? What would be the effect on the seasons compared with that produced by the actual tilt of 23\u00b0? 112 Answer: Chapter 4 Earth, Moon, and Sky The Tropic of Cancer is at a latitude equal to Earth\u2019s tilt, so in this case, it would be at 16\u00b0 N latitude. The Arctic Circle is at a latitude equal to 90\u00b0 minus Earth\u2019s tilt, or 90\u00b0 \u2013 16\u00b0 = 74\u00b0. The difference between these two latitudes is 74\u00b0 \u2013 16\u00b0 = 58\u00b0. Since the tilt of Earth is less,", " there would be less variation in the tilt of Earth and less variation in the Sun\u2019s paths throughout the year, so there would be milder seasonal changes You can see an animation (https://openstaxcollege.org/l/30anisunpath) of the Sun\u2019s path during the seasons alongside a time-lapse view of light and shadow from a camera set up on the University of Nebraska campus. Many cultures that developed some distance north of the equator have a celebration around December 21 to help people deal with the depressing lack of sunlight and the often dangerously cold temperatures. Originally, this was often a time for huddling with family and friends, for sharing the reserves of food and drink, and for rituals asking the gods to return the light and heat and turn the cycle of the seasons around. Many cultures constructed elaborate devices for anticipating when the shortest day of the year was coming. Stonehenge in England, built long before the invention of writing, is probably one such device. In our own time, we continue the winter solstice tradition with various holiday celebrations around that December date. Halfway between the solstices, on about March 21 and September 21, the Sun is on the celestial equator. From Earth, it appears above our planet\u2019s equator and favors neither hemisphere. Every place on Earth then receives roughly 12 hours of sunshine and 12 hours of night. The points where the Sun crosses the celestial equator are called the vernal (spring) and autumnal (fall) equinoxes. The Seasons at Different Latitudes The seasonal effects are different at different latitudes on Earth. Near the equator, for instance, all seasons are much the same. Every day of the year, the Sun is up half the time, so there are approximately 12 hours of sunshine and 12 hours of night. Local residents define the seasons by the amount of rain (wet season and dry season) rather than by the amount of sunlight. As we travel north or south, the seasons become more pronounced, until we reach extreme cases in the Arctic and Antarctic. At the North Pole, all celestial objects that are north of the celestial equator are always above the horizon and, as Earth turns, circle around parallel to it. The Sun is north of the celestial equator from about March 21 to September 21, so at the North Pole, the Sun rises when it reaches the vernal equinox and sets when it reaches the autumnal equinox. Each year there are 6 months of sunshine at each pole, followed by 6 months of darkness. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 113 The Position of the Sun in the Sky The Sun\u2019s coordinates on the celestial sphere range from a declination of 23\u00b0 N of the celestial equator (or +23\u00b0) to a declination 23\u00b0 S of the celestial equator (or \u201323\u00b0). So, the Sun\u2019s altitude at noon, when it crosses the meridian, varies by a total of 46\u00b0. What is the altitude of the Sun at noon on March 21, as seen from a place on Earth\u2019s equator? What is its altitude on June 21, as seen from a place on Earth\u2019s equator? Solution On Earth\u2019s equator, the celestial equator passes through the zenith. On March 21, the Sun is crossing the celestial equator, so it should be found at the zenith (90\u00b0) at noon. On June 21, the Sun is 23\u00b0 N of the celestial equator, so it will be 23\u00b0 away from the zenith at noon. The altitude above the horizon will be 23\u00b0 less than the altitude of the zenith (90\u00b0), so it is 90\u00b0 \u2013 23\u00b0 = 67\u00b0 above the horizon. Check Your Learning What is the altitude of the Sun at noon on December 21, as seen from a place on the Tropic of Cancer? Answer: On the day of the winter solstice, the Sun is located about 23\u00b0 S of the celestial equator. From the Tropic of Cancer, a latitude of 23\u00b0 N, the zenith would be a declination of 23\u00b0 N. The difference in declination between zenith and the position of the Sun is 46\u00b0, so the Sun would be 46\u00b0 away from the zenith. That means it would be at an altitude of 90\u00b0 \u2013 46\u00b0 = 44\u00b0. Clarifications about the Real World In our discussions so far, we have been describing the rising and setting of the Sun and stars as they would appear if Earth had little or no atmosphere. In reality, however, the atmosphere has the curious effect of allowing us to see a little way \u201cover the horizon.\u201d This effect is a result of refraction, the bending of light passing through air or", " water, something we will discuss in Astronomical Instruments. Because of this atmospheric refraction (and the fact that the Sun is not a point of light but a disk), the Sun appears to rise earlier and to set later than it would if no atmosphere were present. In addition, the atmosphere scatters light and provides some twilight illumination even when the Sun is below the horizon. Astronomers define morning twilight as beginning when the Sun is 18\u00b0 below the horizon, and evening twilight extends until the Sun sinks more than 18\u00b0 below the horizon. These atmospheric effects require small corrections in many of our statements about the seasons. At the equinoxes, for example, the Sun appears to be above the horizon for a few minutes longer than 12 hours, and below the horizon for fewer than 12 hours. These effects are most dramatic at Earth\u2019s poles, where the Sun actually can be seen more than a week before it reaches the celestial equator. You probably know that the summer solstice (June 21) is not the warmest day of the year, even if it is the longest. The hottest months in the Northern Hemisphere are July and August. This is because our weather involves the air and water covering Earth\u2019s surface, and these large reservoirs do not heat up instantaneously. You have probably observed this effect for yourself; for example, a pond does not get warm the moment the Sun rises but is warmest late in the afternoon, after it has had time to absorb the Sun\u2019s heat. In the same way, Earth 114 Chapter 4 Earth, Moon, and Sky gets warmer after it has had a chance to absorb the extra sunlight that is the Sun\u2019s summer gift to us. And the coldest times of winter are a month or more after the winter solstice. 4.3 KEEPING TIME Learning Objectives By the end of this section, you will be able to: Explain the difference between the solar day and the sidereal day Explain mean solar time and the reason for time zones The measurement of time is based on the rotation of Earth. Throughout most of human history, time has been reckoned by positions of the Sun and stars in the sky. Only recently have mechanical and electronic clocks taken over this function in regulating our lives. The Length of the Day The most fundamental astronomical unit of time is the day, measured in terms of the rotation of Earth. There is, however, more than one way to define the day. Usually, we think of it as the rotation period of Earth with respect to the Sun, called the solar day. After all, for most people sunrise is more important than the rising time of Arcturus or some other star, so we set our clocks to some version of Sun-time. However, astronomers also use a sidereal day, which is defined in terms of the rotation period of Earth with respect to the stars. A solar day is slightly longer than a sidereal day because (as you can see from Figure 4.10) Earth not only turns but also moves along its path around the Sun in a day. Suppose we start when Earth\u2019s orbital position is at day 1, with both the Sun and some distant star (located in the direction indicated by the long white arrow pointing left), directly in line with the zenith for the observer on Earth. When Earth has completed one rotation with respect to the distant star and is at day 2, the long arrow again points to the same distant star. However, notice that because of the movement of Earth along its orbit from day 1 to 2, the Sun has not yet reached a position above the observer. To complete a solar day, Earth must rotate an additional amount, equal to 1/365 of a full turn. The time required for this extra rotation is 1/365 of a day, or about 4 minutes. So the solar day is about 4 minutes longer than the sidereal day. Figure 4.10. Difference Between a Sidereal Day and a Solar Day. This is a top view, looking down as Earth orbits the Sun. Because Earth moves around the Sun (roughly 1\u00b0 per day), after one complete rotation of Earth relative to the stars, we do not see the Sun in the same position. Because our ordinary clocks are set to solar time, stars rise 4 minutes earlier each day. Astronomers prefer sidereal time for planning their observations because in that system, a star rises at the same time every day. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 115 Sidereal Time and Solar Time The Sun makes a complete circle in the sky approximately every 24 hours, while the stars make a complete circle in the sky in 4 minutes less time, or 23 hours and 56 minutes. This causes the positions of the stars at a given time of day or night to change slightly each day. Since stars rise 4 minutes earlier each day, that works out to about 2 hours per month (4 minutes \u00d7", " 30 = 120 minutes or 2 hours). So, if a particular constellation rises at sunset during the winter, you can be sure that by the summer, it will rise about 12 hours earlier, with the sunrise, and it will not be so easily visible in the night sky. Let\u2019s say that tonight the bright star Sirius rises at 7:00 p.m. from a given location so that by midnight, it is very high in the sky. At what time will Sirius rise in three months? Solution In three months\u2019 time, Sirius will be rising earlier by: 90 days \u00d7 4 minutes day = 360 minutes or 6 hours It will rise at about 1:00 p.m. and be high in the sky at around sunset instead of midnight. Sirius is the brightest star in the constellation of Canis Major (the big dog). So, some other constellation will be prominently visible high in the sky at this later date. Check Your Learning If a star rises at 8:30 p.m. tonight, approximately what time will it rise two months from now? Answer: In two months, the star will rise: 60 days \u00d7 4 minutes day = 240 minutes or 4 hours earlier. This means it will rise at 4:30 p.m. Apparent Solar Time We can define apparent solar time as time reckoned by the actual position of the Sun in the sky (or, during the night, its position below the horizon). This is the kind of time indicated by sundials, and it probably represents the earliest measure of time used by ancient civilizations. Today, we adopt the middle of the night as the starting point of the day and measure time in hours elapsed since midnight. During the first half of the day, the Sun has not yet reached the meridian (the great circle in the sky that passes through our zenith). We designate those hours as before midday (ante meridiem, or a.m.), before the Sun reaches the local meridian. We customarily start numbering the hours after noon over again and designate them by p.m. (post meridiem), after the Sun reaches the local meridian. Although apparent solar time seems simple, it is not really very convenient to use. The exact length of an apparent solar day varies slightly during the year. The eastward progress of the Sun in its annual journey around the sky is not uniform because the speed of Earth varies slightly in its elliptical orbit. Another complication is that Earth\u2019s axis of rotation is not perpendicular to the plane of its revolution. Thus, apparent 116 Chapter 4 Earth, Moon, and Sky solar time does not advance at a uniform rate. After the invention of mechanical clocks that run at a uniform rate, it became necessary to abandon the apparent solar day as the fundamental unit of time. Mean Solar Time and Standard Time Instead, we can consider the mean solar time, which is based on the average value of the solar day over the course of the year. A mean solar day contains exactly 24 hours and is what we use in our everyday timekeeping. Although mean solar time has the advantage of progressing at a uniform rate, it is still inconvenient for practical use because it is determined by the position of the Sun. For example, noon occurs when the Sun is overhead. But because we live on a round Earth, the exact time of noon is different as you change your longitude by moving east or west. If mean solar time were strictly observed, people traveling east or west would have to reset their watches continually as the longitude changed, just to read the local mean time correctly. For instance, a commuter traveling from Oyster Bay on Long Island to New York City would have to adjust the time on the trip through the East River tunnel because Oyster Bay time is actually about 1.6 minutes more advanced than that of Manhattan. (Imagine an airplane trip in which an obnoxious flight attendant gets on the intercom every minute, saying, \u201cPlease reset your watch for local mean time.\u201d) Until near the end of the nineteenth century, every city and town in the United States kept its own local mean time. With the development of railroads and the telegraph, however, the need for some kind of standardization became evident. In 1883, the United States was divided into four standard time zones (now six, including Hawaii and Alaska), each with one system of time within that zone. By 1900, most of the world was using the system of 24 standardized global time zones. Within each zone, all places keep the same standard time, with the local mean solar time of a standard line of longitude running more or less through the middle of each zone. Now travelers reset their watches only when the time change has amounted to a full hour. Pacific standard time is 3 hours earlier than eastern standard time, a fact that becomes painfully obvious in California when someone on the East Coast forgets and calls you at 5:00 a.m. Globally, almost all countries have adopted one or more standard time zones, although one of the largest nations", ", India, has settled on a half-zone, being 5.5 hours from Greenwich standard. Also, several large countries (Russia, China) officially use only one time zone, so all the clocks in that country keep the same time. In Tibet, for example, the Sun rises while the clocks (which keep Beijing time) say it is midmorning already. Daylight saving time is simply the local standard time of the place plus 1 hour. It has been adopted for spring and summer use in most states in the United States, as well as in many countries, to prolong the sunlight into evening hours, on the apparent theory that it is easier to change the time by government action than it would be for individuals or businesses to adjust their own schedules to produce the same effect. It does not, of course, \u201csave\u201d any daylight at all\u2014because the amount of sunlight is not determined by what we do with our clocks\u2014and its observance is a point of legislative debate in some states. The International Date Line The fact that time is always advancing as you move toward the east presents a problem. Suppose you travel eastward around the world. You pass into a new time zone, on the average, about every 15\u00b0 of longitude you travel, and each time you dutifully set your watch ahead an hour. By the time you have completed your trip, you have set your watch ahead a full 24 hours and thus gained a day over those who stayed at home. The solution to this dilemma is the International Date Line, set by international agreement to run approximately along the 180\u00b0 meridian of longitude. The date line runs down the middle of the Pacific Ocean, although it jogs a bit in a few places to avoid cutting through groups of islands and through Alaska (Figure This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 117 4.11). By convention, at the date line, the date of the calendar is changed by one day. Crossing the date line from west to east, thus advancing your time, you compensate by decreasing the date; crossing from east to west, you increase the date by one day. To maintain our planet on a rational system of timekeeping, we simply must accept that the date will differ in different cities at the same time. A good example is the date when the Imperial Japanese Navy bombed Pearl Harbor in Hawaii, known in the United States as Sunday, December 7, 1941, but taught to Japanese students as Monday, December 8. Figure 4.11. Where the Date Changes. The International Date Line is an arbitrarily drawn line on Earth where the date changes. So that neighbors do not have different days, the line is located where Earth\u2019s surface is mostly water. 4.4 THE CALENDAR Learning Objectives By the end of this section, you will be able to: Understand how calendars varied among different cultures Explain the origins of our modern calendar \u201cWhat\u2019s today\u2019s date?\u201d is one of the most common questions you can ask (usually when signing a document or worrying about whether you should have started studying for your next astronomy exam). Long before the era of digital watches, smartphones, and fitness bands that tell the date, people used calendars to help measure the passage of time. The Challenge of the Calendar There are two traditional functions of any calendar. First, it must keep track of time over the course of long spans, allowing people to anticipate the cycle of the seasons and to honor special religious or personal anniversaries. Second, to be useful to a large number of people, a calendar must use natural time intervals that everyone can agree on\u2014those defined by the motions of Earth, the Moon, and sometimes even the planets. The natural units of our calendar are the day, based on the period of rotation of Earth; the month, based on the cycle of the Moon\u2019s phases (see later in this chapter) about Earth; and the year, based on the period of revolution of Earth about the Sun. Difficulties have resulted from the fact that these three periods are not commensurable; that\u2019s a fancy way of saying that one does not divide evenly into any of the others. The rotation period of Earth is, by definition, 1.0000 day (and here the solar day is used, since that is the basis of human experience). The period required by the Moon to complete its cycle of phases, called the lunar month, 118 Chapter 4 Earth, Moon, and Sky is 29.5306 days. The basic period of revolution of Earth, called the tropical year, is 365.2422 days. The ratios of these numbers are not convenient for calculations. This is the historic challenge of the calendar, dealt with in various ways by different cultures. Early Calendars Even the earliest cultures were concerned with the keeping of time and the calendar. Some interesting examples include monuments left by Bronze Age people in northwestern Europe, especially the", " British Isles. The best preserved of the monuments is Stonehenge, about 13 kilometers from Salisbury in southwest England (Figure 4.12). It is a complex array of stones, ditches, and holes arranged in concentric circles. Carbon dating and other studies show that Stonehenge was built during three periods ranging from about 2800 to 1500 BCE. Some of the stones are aligned with the directions of the Sun and Moon during their risings and settings at critical times of the year (such as the summer and winter solstices), and it is generally believed that at least one function of the monument was connected with the keeping of a calendar. Figure 4.12. Stonehenge. The ancient monument known as Stonehenge was used to keep track of the motions of the Sun and Moon. (credit: modification of work by Adriano Aurelio Araujo) The Maya in Central America, who thrived more than a thousand years ago, were also concerned with the keeping of time. Their calendar was as sophisticated as, and perhaps more complex than, contemporary calendars in Europe. The Maya did not attempt to correlate their calendar accurately with the length of the year or lunar month. Rather, their calendar was a system for keeping track of the passage of days and for counting time far into the past or future. Among other purposes, it was useful for predicting astronomical events, such as the position of Venus in the sky (Figure 4.13). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 119 Figure 4.13. El Caracol. This Mayan observatory at Chichen Itza in the Yucatan, Mexico, dates from around the year 1000. (credit: \u201cwiredtourist.com\u201d/Flickr) The ancient Chinese developed an especially complex calendar, largely limited to a few privileged hereditary court astronomer-astrologers. In addition to the motions of Earth and the Moon, they were able to fit in the approximately 12-year cycle of Jupiter, which was central to their system of astrology. The Chinese still preserve some aspects of this system in their cycle of 12 \u201cyears\u201d\u2014the Year of the Dragon, the Year of the Pig, and so on\u2014that are defined by the position of Jupiter in the zodiac. Our Western calendar derives from a long history of timekeeping beginning with the Sumerians, dating back to at least the second millennium BCE, and continuing with the Egyptians and the Greeks around the eighth century BCE. These calendars led, eventually, to the Julian calendar, introduced by Julius Caesar, which approximated the year at 365.25 days, fairly close to the actual value of 365.2422. The Romans achieved this approximation by declaring years to have 365 days each, with the exception of every fourth year. The leap year was to have one extra day, bringing its length to 366 days, and thus making the average length of the year in the Julian calendar 365.25 days. In this calendar, the Romans had dropped the almost impossible task of trying to base their calendar on the Moon as well as the Sun, although a vestige of older lunar systems can be seen in the fact that our months have an average length of about 30 days. However, lunar calendars remained in use in other cultures, and Islamic calendars, for example, are still primarily lunar rather than solar. The Gregorian Calendar Although the Julian calendar (which was adopted by the early Christian Church) represented a great advance, its average year still differed from the true year by about 11 minutes, an amount that accumulates over the centuries to an appreciable error. By 1582, that 11 minutes per year had added up to the point where the first day of spring was occurring on March 11, instead of March 21. If the trend were allowed to continue, eventually the Christian celebration of Easter would be occurring in early winter. Pope Gregory XIII, a contemporary of Galileo, felt it necessary to institute further calendar reform. The Gregorian calendar reform consisted of two steps. First, 10 days had to be dropped out of the calendar to bring the vernal equinox back to March 21; by proclamation, the day following October 4, 1582, became October 15. The second feature of the new Gregorian calendar was a change in the rule for leap year, making the average length of the year more closely approximate the tropical year. Gregory decreed that three of every four century years\u2014all leap years under the Julian calendar\u2014would be common years henceforth. The rule was that only century years divisible by 400 would be leap years. Thus, 1700, 1800, and 1900\u2014all divisible by 4 but not by 400\u2014were not leap years in the Gregorian calendar. On the other hand, the years 1600 and 2000, both 120 Chapter 4 Earth, Moon, and Sky divisible by 400, were leap years. The average length of this Gregorian year, 365", ".2425 mean solar days, is correct to about 1 day in 3300 years. The Catholic countries immediately put the Gregorian reform into effect, but countries of the Eastern Church and most Protestant countries did not adopt it until much later. It was 1752 when England and the American colonies finally made the change. By parliamentary decree, September 2, 1752, was followed by September 14. Although special laws were passed to prevent such abuses as landlords collecting a full month\u2019s rent for September, there were still riots, and people demanded their 12 days back. Russia did not abandon the Julian calendar until the time of the Bolshevik revolution. The Russians then had to omit 13 days to come into step with the rest of the world. The anniversary of the October Revolution (old calendar) of 1917, bringing the communists to power, thus ended up being celebrated in November (new calendar), a difference that is perhaps not so important since the fall of communism. 4.5 PHASES AND MOTIONS OF THE MOON Learning Objectives By the end of this section, you will be able to: Explain the cause of the lunar phases Understand how the Moon rotates and revolves around Earth After the Sun, the Moon is the brightest and most obvious object in the sky. Unlike the Sun, it does not shine under its own power, but merely glows with reflected sunlight. If you were to follow its progress in the sky for a month, you would observe a cycle of phases (different appearances), with the Moon starting dark and getting more and more illuminated by sunlight over the course of about two weeks. After the Moon\u2019s disk becomes fully bright, it begins to fade, returning to dark about two weeks later. These changes fascinated and mystified many early cultures, which came up with marvelous stories and legends to explain the cycle of the Moon. Even in the modern world, many people don\u2019t understand what causes the phases, thinking that they are somehow related to the shadow of Earth. Let us see how the phases can be explained by the motion of the Moon relative to the bright light source in the solar system, the Sun. Lunar Phases Although we know that the Sun moves 1/12 of its path around the sky each month, for purposes of explaining the phases, we can assume that the Sun\u2019s light comes from roughly the same direction during the course of a four-week lunar cycle. The Moon, on the other hand, moves completely around Earth in that time. As we watch the Moon from our vantage point on Earth, how much of its face we see illuminated by sunlight depends on the angle the Sun makes with the Moon. Here is a simple experiment to show you what we mean: stand about 6 feet in front of a bright electric light in a completely dark room (or outdoors at night) and hold in your hand a small round object such as a tennis ball or an orange. Your head can then represent Earth, the light represents the Sun, and the ball the Moon. Move the ball around your head (making sure you don\u2019t cause an eclipse by blocking the light with your head). You will see phases just like those of the Moon on the ball. (Another good way to get acquainted with the phases and motions of the Moon is to follow our satellite in the sky for a month or two, recording its shape, its direction from the Sun, and when it rises and sets.) Let\u2019s examine the Moon\u2019s cycle of phases using Figure 4.14, which depicts the Moon\u2019s behavior for the entire month. The trick to this figure is that you must imagine yourself standing on Earth, facing the Moon in each of This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 121 its phases. So, for the position labeled \u201cNew,\u201d you are on the right side of Earth and it\u2019s the middle of the day; for the position \u201cFull,\u201d you are on the left side of Earth in the middle of the night. Note that in every position on Figure 4.14, the Moon is half illuminated and half dark (as a ball in sunlight should be). The difference at each position has to do with what part of the Moon faces Earth. Figure 4.14. Phases of the Moon. The appearance of the Moon changes over the course of a complete monthly cycle. The pictures of the Moon on the white circle show the perspective from space, with the Sun off to the right in a fixed position. The outer images show how the Moon appears to you in the sky from each point in the orbit. Imagine yourself standing on Earth, facing the Moon at each stage. In the position \u201cNew,\u201d for example, you are facing the Moon from the right side of Earth in the middle of the day. (Note that the distance of the Moon from Earth is not to scale in this diagram: the Moon is roughly 30 Earth", "-diameters away from us.) (credit: modification of work by NASA) The Moon is said to be new when it is in the same general direction in the sky as the Sun (position A). Here, its illuminated (bright) side is turned away from us and its dark side is turned toward us. You might say that the Sun is shining on the \u201cwrong \u201d side of the Moon from our perspective. In this phase the Moon is invisible to us; its dark, rocky surface does not give off any light of its own. Because the new moon is in the same part of the sky as the Sun, it rises at sunrise and sets at sunset. But the Moon does not remain in this phase long because it moves eastward each day in its monthly path around us. Since it takes about 30 days to orbit Earth and there are 360\u00b0 in a circle, the Moon will move about 12\u00b0 in the sky each day (or about 24 times its own diameter). A day or two after the new phase, the thin crescent first appears, as we begin to see a small part of the Moon\u2019s illuminated hemisphere. It has moved into a position where it now reflects a little sunlight toward us along one side. The bright crescent increases in size on successive days as the Moon moves farther and farther around the sky away from the direction of the Sun (position B). Because the Moon is moving eastward away from the Sun, it rises later and later each day (like a student during summer vacation). After about one week, the Moon is one-quarter of the way around its orbit (position C) and so we say it is at the first quarter phase. Half of the Moon\u2019s illuminated side is visible to Earth observers. Because of its eastward motion, the Moon now lags about one-quarter of the day behind the Sun, rising around noon and setting around midnight. During the week after the first quarter phase, we see more and more of the Moon\u2019s illuminated hemisphere (position D), a phase that is called waxing (or growing) gibbous (from the Latin gibbus, meaning hump). 122 Chapter 4 Earth, Moon, and Sky Eventually, the Moon arrives at position E in our figure, where it and the Sun are opposite each other in the sky. The side of the Moon turned toward the Sun is also turned toward Earth, and we have the full phase. When the Moon is full, it is opposite the Sun in the sky. The Moon does the opposite of what the Sun does, rising at sunset and setting at sunrise. Note what that means in practice: the completely illuminated (and thus very noticeable) Moon rises just as it gets dark, remains in the sky all night long, and sets as the Sun\u2019s first rays are seen at dawn. Its illumination throughout the night helps lovers on a romantic stroll and students finding their way back to their dorms after a long night in the library or an off-campus party. And when is the full moon highest in the sky and most noticeable? At midnight, a time made famous in generations of horror novels and films. (Note how the behavior of a vampire like Dracula parallels the behavior of the full Moon: Dracula rises at sunset, does his worst mischief at midnight, and must be back down in his coffin by sunrise. The old legends were a way of personifying the behavior of the Moon, which was a much more dramatic part of people\u2019s lives in the days before electric lights and television.) Folklore has it that more crazy behavior is seen during the time of the full moon (the Moon even gives a name to crazy behavior\u2014\u201clunacy\u201d). But, in fact, statistical tests of this \u201chypothesis\u201d involving thousands of records from hospital emergency rooms and police files do not reveal any correlation of human behavior with the phases of the Moon. For example, homicides occur at the same rate during the new moon or the crescent moon as during the full moon. Most investigators believe that the real story is not that more crazy behavior happens on nights with a full moon, but rather that we are more likely to notice or remember such behavior with the aid of a bright celestial light that is up all night long. During the two weeks following the full moon, the Moon goes through the same phases again in reverse order (points F, G, and H in Figure 4.14), returning to new phase after about 29.5 days. About a week after the full moon, for example, the Moon is at third quarter, meaning that it is three-quarters of the way around (not that it is three-quarters illuminated\u2014in fact, half of the visible side of the Moon is again dark). At this phase, the Moon is now rising around midnight and setting around noon. Note that there is one thing quite misleading about Figure 4.14. If you look at the Moon in position E, although it is full in theory, it appears as if its illumination", " would in fact be blocked by a big fat Earth, and hence we would not see anything on the Moon except Earth\u2019s shadow. In reality, the Moon is nowhere near as close to Earth (nor is its path so identical with the Sun\u2019s in the sky) as this diagram (and the diagrams in most textbooks) might lead you to believe. The Moon is actually 30 Earth-diameters away from us; Science and the Universe: A Brief Tour contains a diagram that shows the two objects to scale. And, since the Moon\u2019s orbit is tilted relative to the path of the Sun in the sky, Earth\u2019s shadow misses the Moon most months. That\u2019s why we regularly get treated to a full moon. The times when Earth\u2019s shadow does fall on the Moon are called lunar eclipses and are discussed in Eclipses of the Sun and Moon Astronomy and the Days of the Week The week seems independent of celestial motions, although its length may have been based on the time between quarter phases of the Moon. In Western culture, the seven days of the week are named after the seven \u201cwanderers\u201d that the ancients saw in the sky: the Sun, the Moon, and the five planets visible to the unaided eye (Mercury, Venus, Mars, Jupiter, and Saturn). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 123 In English, we can easily recognize the names Sun-day (Sunday), Moon-day (Monday), and Saturn-day (Saturday), but the other days are named after the Norse equivalents of the Roman gods that gave their names to the planets. In languages more directly related to Latin, the correspondences are clearer. Wednesday, Mercury\u2019s day, for example, is mercoledi in Italian, mercredi in French, and mi\u00e9rcoles in Spanish. Mars gives its name to Tuesday (martes in Spanish), Jupiter or Jove to Thursday (giovedi in Italian), and Venus to Friday (vendredi in French). There is no reason that the week has to have seven days rather than five or eight. It is interesting to speculate that if we had lived in a planetary system where more planets were visible without a telescope, the Beatles could have been right and we might well have had \u201cEight Days a Week.\u201d View this animation (https://openstaxcollege.org/l/30phamoonearth) to see the phases of the Moon as it orbits Earth and as Earth orbits the Sun. The Moon\u2019s Revolution and Rotation The Moon\u2019s sidereal period\u2014that is, the period of its revolution about Earth measured with respect to the stars\u2014is a little over 27 days: the sidereal month is 27.3217 days to be exact. The time interval in which the phases repeat\u2014say, from full to full\u2014is the solar month, 29.5306 days. The difference results from Earth\u2019s motion around the Sun. The Moon must make more than a complete turn around the moving Earth to get back to the same phase with respect to the Sun. As we saw, the Moon changes its position on the celestial sphere rather rapidly: even during a single evening, the Moon creeps visibly eastward among the stars, traveling its own width in a little less than 1 hour. The delay in moonrise from one day to the next caused by this eastward motion averages about 50 minutes. The Moon rotates on its axis in exactly the same time that it takes to revolve about Earth. As a consequence, the Moon always keeps the same face turned toward Earth (Figure 4.15). You can simulate this yourself by \u201corbiting\u201d your roommate or another volunteer. Start by facing your roommate. If you make one rotation (spin) with your shoulders in the exact same time that you revolve around him or her, you will continue to face your roommate during the whole \u201corbit.\u201d As we will see in coming chapters, our Moon is not the only world that exhibits this behavior, which scientists call synchronous rotation. 124 Chapter 4 Earth, Moon, and Sky Figure 4.15. The Moon without and with Rotation. In this figure, we stuck a white arrow into a fixed point on the Moon to keep track of its sides. (a) If the Moon did not rotate as it orbited Earth, it would present all of its sides to our view; hence the white arrow would point directly toward Earth only in the bottom position on the diagram. (b) Actually, the Moon rotates in the same period that it revolves, so we always see the same side (the white arrow keeps pointing to Earth). The differences in the Moon\u2019s appearance from one night to the next are due to changing illumination by the Sun, not to its own rotation. You sometimes hear", " the back side of the Moon (the side we never see) called the \u201cdark side.\u201d This is a misunderstanding of the real situation: which side is light and which is dark changes as the Moon moves around Earth. The back side is dark no more frequently than the front side. Since the Moon rotates, the Sun rises and sets on all sides of the Moon. With apologies to Pink Floyd, there is simply no regular \u201cDark Side of the Moon.\u201d This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 125 4.6 OCEAN TIDES AND THE MOON Learning Objectives By the end of this section, you will be able to: Describe what causes tides on Earth Explain why the amplitude of tides changes during the course of a month Anyone living near the sea is familiar with the twice-daily rising and falling of the tides. Early in history, it was clear that tides must be related to the Moon because the daily delay in high tide is the same as the daily delay in the Moon\u2019s rising. A satisfactory explanation of the tides, however, awaited the theory of gravity, supplied by Newton. The Pull of the Moon on Earth The gravitational forces exerted by the Moon at several points on Earth are illustrated in Figure 4.16. These forces differ slightly from one another because Earth is not a point, but has a certain size: all parts are not equally distant from the Moon, nor are they all in exactly the same direction from the Moon. Moreover, Earth is not perfectly rigid. As a result, the differences among the forces of the Moon\u2019s attraction on different parts of Earth (called differential forces) cause Earth to distort slightly. The side of Earth nearest the Moon is attracted toward the Moon more strongly than is the center of Earth, which in turn is attracted more strongly than is the side opposite the Moon. Thus, the differential forces tend to stretch Earth slightly into a prolate spheroid (a football shape), with its long diameter pointed toward the Moon. Figure 4.16. Pull of the Moon. The Moon\u2019s differential attraction is shown on different parts of Earth. (Note that the differences have been exaggerated for educational purposes.) If Earth were made of water, it would distort until the Moon\u2019s differential forces over different parts of its surface came into balance with Earth\u2019s own gravitational forces pulling it together. Calculations show that in this case, Earth would distort from a sphere by amounts ranging up to nearly 1 meter. Measurements of the actual deformation of Earth show that the solid Earth does distort, but only about one-third as much as water would, because of the greater rigidity of Earth\u2019s interior. Because the tidal distortion of the solid Earth amounts\u2014at its greatest\u2014to only about 20 centimeters, Earth does not distort enough to balance the Moon\u2019s differential forces with its own gravity. Hence, objects at Earth\u2019s surface experience tiny horizontal tugs, tending to make them slide about. These tide-raising forces are too 126 Chapter 4 Earth, Moon, and Sky insignificant to affect solid objects like astronomy students or rocks in Earth\u2019s crust, but they do affect the waters in the oceans. The Formation of Tides The tide-raising forces, acting over a number of hours, produce motions of the water that result in measurable tidal bulges in the oceans. Water on the side of Earth facing the Moon flows toward it, with the greatest depths roughly at the point below the Moon. On the side of Earth opposite the Moon, water also flows to produce a tidal bulge (Figure 4.17). Figure 4.17. Tidal Bulges in an \u201cIdeal\u201d Ocean. Differences in gravity cause tidal forces that push water in the direction of tidal bulges on Earth You can run this animation (https://openstaxcollege.org/l/30visdemotidal) for a visual demonstration of the tidal bulge. Note that the tidal bulges in the oceans do not result from the Moon\u2019s compressing or expanding the water, nor from the Moon\u2019s lifting the water \u201caway from Earth.\u201d Rather, they result from an actual flow of water over Earth\u2019s surface toward the two regions below and opposite the Moon, causing the water to pile up to greater depths at those places (Figure 4.18). Figure 4.18. High and Low Tides. This is a side-by-side comparison of the Bay of Fundy in Canada at high and low tides. (credit a, b: modification of work by Dylan Kereluk) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 127 In the idealized (and, as we shall see, oversimplified) model just described", ", the height of the tides would be only a few feet. The rotation of Earth would carry an observer at any given place alternately into regions of deeper and shallower water. An observer being carried toward the regions under or opposite the Moon, where the water was deepest, would say, \u201cThe tide is coming in\u201d; when carried away from those regions, the observer would say, \u201cThe tide is going out.\u201d During a day, the observer would be carried through two tidal bulges (one on each side of Earth) and so would experience two high tides and two low tides. The Sun also produces tides on Earth, although it is less than half as effective as the Moon at tide raising. The actual tides we experience are a combination of the larger effect of the Moon and the smaller effect of the Sun. When the Sun and Moon are lined up (at new moon or full moon), the tides produced reinforce each other and so are greater than normal (Figure 4.19). These are called spring tides (the name is connected not to the season but to the idea that higher tides \u201cspring up\u201d). Spring tides are approximately the same, whether the Sun and Moon are on the same or opposite sides of Earth, because tidal bulges occur on both sides. When the Moon is at first quarter or last quarter (at right angles to the Sun\u2019s direction), the tides produced by the Sun partially cancel the tides of the Moon, making them lower than usual. These are called neap tides. Figure 4.19. Tides Caused by Different Alignments of the Sun and Moon. (a) In spring tides, the Sun\u2019s and Moon\u2019s pulls reinforce each other. (b) In neap tides, the Sun and the Moon pull at right angles to each other and the resulting tides are lower than usual. The \u201csimple\u201d theory of tides, described in the preceding paragraphs, would be sufficient if Earth rotated very slowly and were completely surrounded by very deep oceans. However, the presence of land masses stopping the flow of water, the friction in the oceans and between oceans and the ocean floors, the rotation of Earth, the wind, the variable depth of the ocean, and other factors all complicate the picture. This is why, in the real world, some places have very small tides while in other places huge tides become tourist attractions. If you have been in such places, you may know that \u201ctide tables\u201d need to be computed and published for each location; one set of tide predictions doesn\u2019t work for the whole planet. In this introductory chapter, we won\u2019t delve further into these complexities. 128 Chapter 4 Earth, Moon, and Sky George Darwin and the Slowing of Earth The rubbing of water over the face of Earth involves an enormous amount of energy. Over long periods of time, the friction of the tides is slowing down the rotation of Earth. Our day gets longer by about 0.002 second each century. That seems very small, but such tiny changes can add up over millions and billions of years. Although Earth\u2019s spin is slowing down, the angular momentum (see Orbits and Gravity) in a system such as the Earth-Moon system cannot change. Thus, some other spin motion must speed up to take the extra angular momentum. The details of what happens were worked out over a century ago by George Darwin, the son of naturalist Charles Darwin. George Darwin (see Figure 4.20) had a strong interest in science but studied law for six years and was admitted to the bar. However, he never practiced law, returning to science instead and eventually becoming a professor at Cambridge University. He was a prot\u00e9g\u00e9 of Lord Kelvin, one of the great physicists of the nineteenth century, and he became interested in the long-term evolution of the solar system. He specialized in making detailed (and difficult) mathematical calculations of how orbits and motions change over geologic time. Figure 4.20. George Darwin (1845\u20131912). George Darwin is best known for studying Earth\u2019s spin in relation to angular momentum. What Darwin calculated for the Earth-Moon system was that the Moon will slowly spiral outward, away from Earth. As it moves farther away, it will orbit less quickly (just as planets farther from the Sun move more slowly in their orbits). Thus, the month will get longer. Also, because the Moon will be more distant, total eclipses of the Sun will no longer be visible from Earth. Both the day and the month will continue to get longer, although bear in mind that the effects are very gradual. Darwin\u2019s calculations were confirmed by mirrors placed on the Moon by Apollo 11 astronauts. These show that the Moon is moving away by 3.8 centimeters per year, and that ultimately\u2014billions of years in the future\u2014the day and the month will be the same length (about 47 of our present days). At this point the Moon will be stationary in the", " sky over the same spot on Earth, meaning some parts of Earth will see the Moon and its phases and other parts will never see them. This kind of alignment is This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 129 already true for Pluto\u2019s moon Charon (among others). Its rotation and orbital period are the same length as a day on Pluto. 4.7 ECLIPSES OF THE SUN AND MOON Learning Objectives By the end of this section, you will be able to: Describe what causes lunar and solar eclipses Differentiate between a total and partial solar eclipse Explain why lunar eclipses are much more common than solar eclipses One of the coincidences of living on Earth at the present time is that the two most prominent astronomical objects, the Sun and the Moon, have nearly the same apparent size in the sky. Although the Sun is about 400 times larger in diameter than the Moon, it is also about 400 times farther away, so both the Sun and the Moon have the same angular size\u2014about 1/2\u00b0. As a result, the Moon, as seen from Earth, can appear to cover the Sun, producing one of the most impressive events in nature. Any solid object in the solar system casts a shadow by blocking the light of the Sun from a region behind it. This shadow in space becomes apparent whenever another object moves into it. In general, an eclipse occurs whenever any part of either Earth or the Moon enters the shadow of the other. When the Moon\u2019s shadow strikes Earth, people within that shadow see the Sun at least partially covered by the Moon; that is, they witness a solar eclipse. When the Moon passes into the shadow of Earth, people on the night side of Earth see the Moon darken in what is called a lunar eclipse. Let\u2019s look at how these happen in more detail. The shadows of Earth and the Moon consist of two parts: a cone where the shadow is darkest, called the umbra, and a lighter, more diffuse region of darkness called the penumbra. As you can imagine, the most spectacular eclipses occur when an object enters the umbra. Figure 4.21 illustrates the appearance of the Moon\u2019s shadow and what the Sun and Moon would look like from different points within the shadow. 130 Chapter 4 Earth, Moon, and Sky Figure 4.21. Solar Eclipse. (a) The shadow cast by a spherical body (the Moon, for example) is shown. Notice the dark umbra and the lighter penumbra. Four points in the shadow are labeled with numbers. In (b) you see what the Sun and Moon would look like in the sky at the four labeled points. At position 1, you see a total eclipse. At positions 2 and 3, the eclipse is partial. At position 4, the Moon is farther away and thus cannot cover the Sun completely; a ring of light thus shows around the Sun, creating what is called an \u201cannular\u201d eclipse. If the path of the Moon in the sky were identical to the path of the Sun (the ecliptic), we might expect to see an eclipse of the Sun and the Moon each month\u2014whenever the Moon got in front of the Sun or into the shadow of Earth. However, as we mentioned, the Moon\u2019s orbit is tilted relative to the plane of Earth\u2019s orbit about the Sun by about 5\u00b0 (imagine two hula hoops with a common center, but tilted a bit). As a result, during most months, the Moon is sufficiently above or below the ecliptic plane to avoid an eclipse. But when the two paths cross (twice a year), it is then \u201ceclipse season\u201d and eclipses are possible. Eclipses of the Sun The apparent or angular sizes of both the Sun and Moon vary slightly from time to time as their distances from Earth vary. (Figure 4.21 shows the distance of the observer varying at points A\u2013D, but the idea is the same.) Much of the time, the Moon looks slightly smaller than the Sun and cannot cover it completely, even if the two are perfectly aligned. In this type of \u201cannular eclipse,\u201d there is a ring of light around the dark sphere of the Moon. However, if an eclipse of the Sun occurs when the Moon is somewhat nearer than its average distance, the Moon can completely hide the Sun, producing a total solar eclipse. Another way to say it is that a total eclipse of the Sun occurs at those times when the umbra of the Moon\u2019s shadow reaches the surface of Earth. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 131 The geometry of a total solar eclipse is illustrated in Figure 4.22. If the Sun and Moon", " are properly aligned, then the Moon\u2019s darkest shadow intersects the ground at a small point on Earth\u2019s surface. Anyone on Earth within the small area covered by the tip of the Moon\u2019s shadow will, for a few minutes, be unable to see the Sun and will witness a total eclipse. At the same time, observers on a larger area of Earth\u2019s surface who are in the penumbra will see only a part of the Sun eclipsed by the Moon: we call this a partial solar eclipse. Between Earth\u2019s rotation and the motion of the Moon in its orbit, the tip of the Moon\u2019s shadow sweeps eastward at about 1500 kilometers per hour along a thin band across the surface of Earth. The thin zone across Earth within which a total solar eclipse is visible (weather permitting) is called the eclipse path. Within a region about 3000 kilometers on either side of the eclipse path, a partial solar eclipse is visible. It does not take long for the Moon\u2019s shadow to sweep past a given point on Earth. The duration of totality may be only a brief instant; it can never exceed about 7 minutes. Figure 4.22. Geometry of a Total Solar Eclipse. Note that our diagram is not to scale. The Moon blocks the Sun during new moon phase as seen from some parts of Earth and casts a shadow on our planet. Because a total eclipse of the Sun is so spectacular, it is well worth trying to see one if you can. There are some people whose hobby is \u201ceclipse chasing\u201d and who brag about how many they have seen in their lifetimes. Because much of Earth\u2019s surface is water, eclipse chasing can involve lengthy boat trips (and often requires air travel as well). As a result, eclipse chasing is rarely within the budget of a typical college student. Nevertheless, a list of future eclipses is given for your reference in Appendix H, just in case you strike it rich early. (And, as you can see in the Appendix, there will be total eclipses visible in the United States in 2017 and 2024, to which even college students may be able to afford travel.) Appearance of a Total Eclipse What can you see if you are lucky enough to catch a total eclipse? A solar eclipse starts when the Moon just begins to silhouette itself against the edge of the Sun\u2019s disk. A partial phase follows, during which more and more of the Sun is covered by the Moon. About an hour after the eclipse begins, the Sun becomes completely hidden behind the Moon. In the few minutes immediately before this period of totality begins, the sky noticeably darkens, some flowers close up, and chickens may go to roost. As an eerie twilight suddenly descends during the day, other animals (and people) may get disoriented. During totality, the sky is dark enough that planets become visible in the sky, and usually the brighter stars do as well. 132 Chapter 4 Earth, Moon, and Sky As the bright disk of the Sun becomes entirely hidden behind the Moon, the Sun\u2019s remarkable corona flashes into view (Figure 4.23). The corona is the Sun\u2019s outer atmosphere, consisting of sparse gases that extend for millions of miles in all directions from the apparent surface of the Sun. It is ordinarily not visible because the light of the corona is feeble compared with the light from the underlying layers of the Sun. Only when the brilliant glare from the Sun\u2019s visible disk is blotted out by the Moon during a total eclipse is the pearly white corona visible. (We\u2019ll talk more about the corona in the chapter on The Sun: A Garden-Variety Star.) Figure 4.23. The Sun\u2019s Corona. The corona (thin outer atmosphere) of the Sun is visible during a total solar eclipse. (It looks more extensive in photographs than it would to the unaided eye.) (credit: modification of work by Lutfar Rahman Nirjhar) The total phase of the eclipse ends, as abruptly as it began, when the Moon begins to uncover the Sun. Gradually, the partial phases of the eclipse repeat themselves, in reverse order, until the Moon has completely uncovered the Sun. We should make one important safety point here: while the few minutes of the total eclipse are safe to look at, if any part of the Sun is uncovered, you must protect your eyes with safe eclipse glasses[2] or by projecting an image of the Sun (instead of looking at it directly). For more, read the How to Observe Solar Eclipses box in this chapter. Eclipses of the Moon A lunar eclipse occurs when the Moon enters the shadow of Earth. The geometry of a lunar eclipse is shown in Figure 4.24. Earth\u2019s dark shadow is about 1.4 million kilometers long, so at the Moon\u2019s distance (an average of 384,000 kilometers), it could cover about four full moons. Unlike", " a solar eclipse, which is visible only in certain local areas on Earth, a lunar eclipse is visible to everyone who can see the Moon. Because a lunar eclipse can be seen (weather permitting) from the entire night side of Earth, lunar eclipses are observed far more frequently from a given place on Earth than are solar eclipses. 2 Eclipse glasses are available in many planetarium and observatory gift stores, and also from the two main U.S. manufacturers: American Paper Optics and Rainbow Symphony. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 133 Figure 4.24. Geometry of a Lunar Eclipse. The Moon is shown moving through the different parts of Earth\u2019s shadow during a total lunar eclipse. Note that the distance the Moon moves in its orbit during the eclipse has been exaggerated here for clarity. An eclipse of the Moon is total only if the Moon\u2019s path carries it though Earth\u2019s umbra. If the Moon does not enter the umbra completely, we have a partial eclipse of the Moon. But because Earth is larger than the Moon, its umbra is larger, so that lunar eclipses last longer than solar eclipses, as we will discuss below. A lunar eclipse can take place only when the Sun, Earth, and Moon are in a line. The Moon is opposite the Sun, which means the Moon will be in full phase before the eclipse, making the darkening even more dramatic. About 20 minutes before the Moon reaches the dark shadow, it dims somewhat as Earth partly blocks the sunlight. As the Moon begins to dip into the shadow, the curved shape of Earth\u2019s shadow upon it soon becomes apparent. Even when totally eclipsed, the Moon is still faintly visible, usually appearing a dull coppery red. The illumination on the eclipsed Moon is sunlight that has been bent into Earth\u2019s shadow by passing through Earth\u2019s atmosphere. After totality, the Moon moves out of the shadow and the sequence of events is reversed. The total duration of the eclipse depends on how closely the Moon\u2019s path approaches the axis of the shadow. For an eclipse where the Moon goes through the center of Earth\u2019s shadow, each partial phase consumes at least 1 hour, and totality can last as long as 1 hour and 40 minutes. Eclipses of the Moon are much more \u201cdemocratic\u201d than solar eclipses. Since the full moon is visible on the entire night side of Earth, the lunar eclipse is visible for all those who live in that hemisphere. (Recall that a total eclipse of the Sun is visible only in a narrow path where the shadow of the umbra falls.) Total eclipses of the Moon occur, on average, about once every two or three years. A list of future total eclipses of the Moon is in Appendix H. In addition, since the lunar eclipse happens to a full moon, and a full moon is not dangerous to look at, everyone can look at the Moon during all the parts of the eclipse without worrying about safety. Thanks to our understanding of gravity and motion (see Orbits and Gravity), eclipses can now be predicted centuries in advance. We\u2019ve come a long way since humanity stood frightened by the darkening of the Sun or the Moon, fearing the displeasure of the gods. Today, we enjoy the sky show with a healthy appreciation of the majestic forces that keep our solar system running. 134 Chapter 4 Earth, Moon, and Sky How to Observe Solar Eclipses A total eclipse of the Sun is a spectacular sight and should not be missed. However, it is extremely dangerous to look directly at the Sun: even a brief exposure can damage your eyes. Normally, few rational people are tempted to do this because it is painful (and something your mother told you never to do!). But during the partial phases of a solar eclipse, the temptation to take a look is strong. Think before you give in. The fact that the Moon is covering part of the Sun doesn\u2019t make the uncovered part any less dangerous to look at. Still, there are perfectly safe ways to follow the course of a solar eclipse, if you are lucky enough to be in the path of the shadow. The easiest technique is to make a pinhole projector. Take a piece of cardboard with a small (1 millimeter) hole punched in it, and hold it several feet above a light surface, such as a concrete sidewalk or a white sheet of paper, so that the hole is \u201caimed\u201d at the Sun. The hole produces a fuzzy but adequate image of the eclipsed Sun. Alternatively, if it\u2019s the right time of year, you can let the tiny spaces between a tree\u2019s leaves form multiple pinhole images against a wall or sidewalk. Watching hundreds of little crescent Suns dancing in the breeze can be captivating. A kitchen colander also makes an excellent pinhole", " projector. Although there are safe filters for looking at the Sun directly, people have suffered eye damage by looking through improper filters, or no filter at all. For example, neutral density photographic filters are not safe because they transmit infrared radiation that can cause severe damage to the retina. Also unsafe are smoked glass, completely exposed color film, sunglasses, and many other homemade filters. Safe filters include welders\u2019 goggles and specially designed eclipse glasses. You should certainly look at the Sun directly when it is totally eclipsed, even through binoculars or telescopes. Unfortunately, the total phase, as we discussed, is all too brief. But if you know when it is coming and going, be sure you look, for it\u2019s an unforgettably beautiful sight. And, despite the ancient folklore that presents eclipses as dangerous times to be outdoors, the partial phases of eclipses\u2014as long as you are not looking directly at the Sun\u2014are not any more dangerous than being out in sunlight. During past eclipses, unnecessary panic has been created by uninformed public officials acting with the best intentions. There were two marvelous total eclipses in Australia in the twentieth century during which townspeople held newspapers over their heads for protection and schoolchildren cowered indoors with their heads under their desks. What a pity that all those people missed what would have been one of the most memorable experiences of their lifetimes. On August 21, 2017, there will be a total solar eclipse visible across a large swath of the continental United States. The path the Moon\u2019s shadow will cast is shown in Figure 4.25. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 135 Figure 4.25. 2017 Total Solar Eclipse. This map of the United States shows the path of the total solar eclipse of 2017. On August 21, 2017, the shadow will first cross onto the West Coast near Portland, Oregon, traversing the United States and exiting the East Coast in South Carolina approximately 90 minutes later, covering about 3000 miles in the process. (credit: modification of work by NASA) Since the eclipse path is not more than a one-day drive for most people in the United States, this would be a prime opportunity to witness this extraordinary spectacle Check out this useful booklet (https://openstaxcollege.org/l/302017ecliboo) about the 2017 eclipse (with specific times in different locations). 136 Chapter 4 Earth, Moon, and Sky CHAPTER 4 REVIEW KEY TERMS apparent solar time time as measured by the position of the Sun in the sky (the time that would be indicated by a sundial) declination the angular distance north or south of the celestial equator great circle a circle on the surface of a sphere that is the curve of intersection of the sphere with a plane passing through its center International Date Line an arbitrary line on the surface of Earth near longitude 180\u00b0 across which the date changes by one day lunar eclipse an eclipse of the Moon, in which the Moon moves into the shadow of Earth; lunar eclipses can occur only at the time of full moon mean solar time time based on the rotation of Earth; mean solar time passes at a constant rate, unlike apparent solar time meridian a great circle on the terrestrial or celestial sphere that passes through the poles phases of the Moon the different appearance of light and dark on the Moon as seen from Earth during its monthly cycle, from new moon to full moon and back to new moon right ascension the coordinate for measuring the east-west positions of celestial bodies; the angle measured eastward along the celestial equator from the vernal equinox to the hour circle passing through a body sidereal day Earth\u2019s rotation period as defined by the positions of the stars in the sky; the time between successive passages of the same star through the meridian sidereal month the period of the Moon\u2019s revolution about Earth measured with respect to the stars solar day Earth\u2019s rotation period as defined by the position of the Sun in the sky; the time between successive passages of the Sun through the meridian solar eclipse an eclipse of the Sun by the Moon, caused by the passage of the Moon in front of the Sun; solar eclipses can occur only at the time of the new moon solar month the time interval in which the phases repeat\u2014say, from full to full phase synchronous rotation when a body (for example, the Moon) rotates at the same rate that it revolves around another body tides alternate rising and falling of sea level caused by the difference in the strength of the Moon\u2019s gravitational pull on different parts of Earth This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky SUMMARY 4.1 Earth and Sky 137 The terrestrial system of latitude and longitude makes use of the great circles called meridians", ". Longitude is arbitrarily set to 0\u00b0 at the Royal Observatory at Greenwich, England. An analogous celestial coordinate system is called right ascension (RA) and declination, with 0\u00b0 of declination starting at the vernal equinox. These coordinate systems help us locate any object on the celestial sphere. The Foucault pendulum is a way to demonstrate that Earth is turning. 4.2 The Seasons The familiar cycle of the seasons results from the 23.5\u00b0 tilt of Earth\u2019s axis of rotation. At the summer solstice, the Sun is higher in the sky and its rays strike Earth more directly. The Sun is in the sky for more than half of the day and can heat Earth longer. At the winter solstice, the Sun is low in the sky and its rays come in at more of an angle; in addition, it is up for fewer than 12 hours, so those rays have less time to heat. At the vernal and autumnal equinoxes, the Sun is on the celestial equator and we get about 12 hours of day and night. The seasons are different at different latitudes. 4.3 Keeping Time The basic unit of astronomical time is the day\u2014either the solar day (reckoned by the Sun) or the sidereal day (reckoned by the stars). Apparent solar time is based on the position of the Sun in the sky, and mean solar time is based on the average value of a solar day during the year. By international agreement, we define 24 time zones around the world, each with its own standard time. The convention of the International Date Line is necessary to reconcile times on different parts of Earth. 4.4 The Calendar The fundamental problem of the calendar is to reconcile the incommensurable lengths of the day, month, and year. Most modern calendars, beginning with the Roman (Julian) calendar of the first century BCE, neglect the problem of the month and concentrate on achieving the correct number of days in a year by using such conventions as the leap year. Today, most of the world has adopted the Gregorian calendar established in 1582 while finding ways to coexist with the older lunar calendars\u2019 system of months. 4.5 Phases and Motions of the Moon The Moon\u2019s monthly cycle of phases results from the changing angle of its illumination by the Sun. The full moon is visible in the sky only during the night; other phases are visible during the day as well. Because its period of revolution is the same as its period of rotation, the Moon always keeps the same face toward Earth. 4.6 Ocean Tides and the Moon The twice-daily ocean tides are primarily the result of the Moon\u2019s differential force on the material of Earth\u2019s crust and ocean. These tidal forces cause ocean water to flow into two tidal bulges on opposite sides of Earth; each day, Earth rotates through these bulges. Actual ocean tides are complicated by the additional effects of the Sun and by the shape of the coasts and ocean basins. 4.7 Eclipses of the Sun and Moon The Sun and Moon have nearly the same angular size (about 1/2\u00b0). A solar eclipse occurs when the Moon moves between the Sun and Earth, casting its shadow on a part of Earth\u2019s surface. If the eclipse is total, the light from the bright disk of the Sun is completely blocked, and the solar atmosphere (the corona) comes into view. Solar eclipses take place rarely in any one location, but they are among the most spectacular sights in nature. A lunar 138 Chapter 4 Earth, Moon, and Sky eclipse takes place when the Moon moves into Earth\u2019s shadow; it is visible (weather permitting) from the entire night hemisphere of Earth. FOR FURTHER EXPLORATION Articles Bakich, M. \u201cYour Twenty-Year Solar Eclipse Planner.\u201d Astronomy (October 2008): 74. Describes the circumstances of upcoming total eclipses of the Sun. Coco, M. \u201cNot Just Another Pretty Phase.\u201d Astronomy (July 1994): 76. Moon phases explained. Espenak, F., & Anderson, J. \u201cGet Ready for America\u2019s Coast to Coast Experience.\u201d Sky & Telescope (February 2016): 22. Gingerich, O. \u201cNotes on the Gregorian Calendar Reform.\u201d Sky & Telescope (December 1982): 530. Kluepfel, C. \u201cHow Accurate Is the Gregorian Calendar?\u201d Sky & Telescope (November 1982): 417. Krupp, E. \u201cCalendar Worlds.\u201d Sky & Telescope (January 2001): 103. On how the days of the week got their names. Krupp, E. \u201cBehind the Curve.\u201d Sky & Telescope (September 2002): 68. On the reform of the calendar by Pope Gregory XIII. MacRobert, A., & Sinnott, R. \u201cYoung Moon Hunting.\ufffd", "\ufffd Sky & Telescope (February 2005): 75. Hints for finding the Moon as soon after its new phase as possible. Pasachoff, J. \u201cSolar Eclipse Science: Still Going Strong.\u201d Sky & Telescope (February 2001): 40. On what we have learned and are still learning from eclipses. Regas, D. \u201cThe Quest for Totality.\u201d Sky & Telescope (July 2012): 36. On eclipse chasing as a hobby. Schaefer, B. \u201cLunar Eclipses That Changed the World.\u201d Sky & Telescope (December 1992): 639. Schaefer, B. \u201cSolar Eclipses That Changed the World.\u201d Sky & Telescope (May 1994): 36. Websites Ancient Observatories, Timeless Knowledge (Stanford Solar Center): http://solar-center.stanford.edu/AO/. An introduction to ancient sites where the movements of celestial objects were tracked over the years (with a special focus on tracking the Sun). Astronomical Data Services: http://aa.usno.navy.mil/data/index.php. This rich site from the U.S. Naval Observatory has information about Earth, the Moon, and the sky, with tables and online calculators. Calendars through the Ages: http://www.webexhibits.org/calendars/index.html. Like a good museum exhibit on the Web. Calendar Zone: http://www.calendarzone.com/. Everything you wanted to ask or know about calendars and timekeeping, with links from around the world. Eclipse 2017 Information and Safe Viewing Instructions: http://www.nsta.org/publications/press/extras/files/ solarscience/SolarScienceInsert.pdf. Eclipse Maps: http://www.eclipse-maps.com/Eclipse-Maps/Welcome.html. Michael Zeiler specializes in presenting helpful and interactive maps of where solar eclipses will be visible This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 139 Eclipse Predictions: http://astro.unl.edu/classaction/animations/lunarcycles/eclipsetable.html. This visual calendar provides dates for upcoming solar and lunar eclipses through 2029.EclipseWise: http://www.eclipsewise.com/intro.html. An introductory site on future eclipses and eclipse observing by NASA\u2019s Fred Espenak. History of the International Date Line: http://www.staff.science.uu.nl/~gent0113/idl/idl.htm. From R. H. van Gent at Utrecht University in the Netherlands. Lunacy and the Full Moon: http://www.scientificamerican.com/article/lunacy-and-the-full-moon/. This Scientific American article explores whether the Moon\u2019s phase is related to strange behavior. Moon Phase Calculator: https://stardate.org/nightsky/moon. Keep track of the phases of the Moon with this calendar. NASA Eclipse Website: http://eclipse.gsfc.nasa.gov/eclipse.html. This site, by NASA\u2019s eclipse expert Fred Espenak, contains a wealth of information on lunar and solar eclipses, past and future, as well as observing and photography links. Phases of the Moon Gallery and Information: http://astropixels.com/moon/phases/phasesgallery.html. Photographs and descriptions presented by NASA\u2019s Fred Espenak. Time and Date Website: http://www.timeanddate.com/. Comprehensive resource about how we keep time on Earth; has time zone converters and many other historical and mathematical tools. Walk through Time: The Evolution of Time Measurement through the Ages (National Institute of Standards and Technology): http://www.nist.gov/pml/general/time/. Videos Bill Nye, the Science Guy, Explains the Seasons: https://www.youtube.com/watch?v=KUU7IyfR34o. For kids, but college students can enjoy the bad jokes, too (4:45). Geography Lesson Idea: Time Zones: https://www.youtube.com/watch?v=-j-SWKtWEcU. (3:11). How to View a Solar Eclipse: http://www.exploratorium.edu/eclipse/how-to-view-eclipse. (1:35). Shadow of the Moon: https://www.youtube.com/watch?v=XNcfKUJwnjM. This NASA video explains eclipses of the Sun, with discussion and animation, focusing on a 2015 eclipse, and shows what an eclipse looks like from space (1:54).", " Strangest Time Zones in the World: https://www.youtube.com/watch?v=uW6QqcmCfm8. (8:38). Understanding Lunar Eclipses: https://www.youtube.com/watch?v=lNi5UFpales. This NASA video explains why there isn\u2019t an eclipse every month, with good animation (1:58). COLLABORATIVE GROUP ACTIVITIES A. Have your group brainstorm about other ways (besides the Foucault pendulum) you could prove that it is our Earth that is turning once a day, and not the sky turning around us. (Hint: How does the spinning of Earth affect the oceans and the atmosphere?) B. What would the seasons on Earth be like if Earth\u2019s axis were not tilted? Discuss with your group how many things about life on Earth you think would be different. 140 Chapter 4 Earth, Moon, and Sky C. After college and graduate training, members of your U.S. student group are asked to set up a school in New Zealand. Describe some ways your yearly school schedule in the Southern Hemisphere would differ from what students are used to in the Northern Hemisphere. D. During the traditional U.S. Christmas vacation weeks, you are sent to the vicinity of the South Pole on a research expedition (depending on how well you did on your astronomy midterm, either as a research assistant or as a short-order cook!). Have your group discuss how the days and nights will be different there and how these differences might affect you during your stay. E. Discuss with your group all the stories you have heard about the full moon and crazy behavior. Why do members of your group think people associate crazy behavior with the full moon? What other legends besides vampire stories are connected with the phases of the Moon? (Hint: Think Professor Lupin in the Harry Potter stories, for example.) F. Your college town becomes the founding site for a strange new cult that worships the Moon. These true believers gather regularly around sunset and do a dance in which they must extend their arms in the direction of the Moon. Have your group discuss which way their arms will be pointing at sunset when the Moon is new, first quarter, full, and third quarter. G. Changes of the seasons play a large part in our yearly plans and concerns. The seasons have inspired music, stories, poetry, art, and much groaning from students during snowstorms. Search online to come up with some examples of the seasons being celebrated or overcome in fields other than science. H. Use the information in Appendix H and online to figure out when the next eclipse of the Sun or eclipse of the Moon will be visible from where your group is going to college or from where your group members live. What time of day will the eclipse be visible? Will it be a total or partial eclipse? What preparations can you make to have an enjoyable and safe eclipse experience? How do these preparations differ between a solar and lunar eclipse? I. On Mars, a day (often called a sol) is 24 hours and 40 minutes. Since Mars takes longer to go around the Sun, a year is 668.6 sols. Mars has two tiny moons, Phobos and Deimos. Phobos, the inner moon, rises in the west and sets in the east, taking 11 hours from moonrise to the next moonrise. Using your calculators and imaginations, have your group members come up with a calendar for Mars. (After you do your own, and only after, you can search online for the many suggestions that have been made for a martian calendar over the years.) EXERCISES Review Questions 1. Discuss how latitude and longitude on Earth are similar to declination and right ascension in the sky. 2. What is the latitude of the North Pole? The South Pole? Why does longitude have no meaning at the North and South Poles? 3. Make a list of each main phase of the Moon, describing roughly when the Moon rises and sets for each phase. During which phase can you see the Moon in the middle of the morning? In the middle of the afternoon? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 141 4. What are advantages and disadvantages of apparent solar time? How is the situation improved by introducing mean solar time and standard time? 5. What are the two ways that the tilt of Earth\u2019s axis causes the summers in the United States to be warmer than the winters? 6. Why is it difficult to construct a practical calendar based on the Moon\u2019s cycle of phases? 7. Explain why there are two high tides and two low tides each day. Strictly speaking, should the period during which there are two high tides be 24 hours? If not, what should the interval be? 8. What is the phase of the Moon", " during a total solar eclipse? During a total lunar eclipse? 9. On a globe or world map, find the nearest marked latitude line to your location. Is this an example of a great circle? Explain. 10. Explain three lines of evidence that indicate that the seasons in North America are not caused by the changing Earth-Sun distance as a result of Earth\u2019s elliptical orbit around the Sun. 11. What is the origin of the terms \u201ca.m.\u201d and \u201cp.m.\u201d in our timekeeping? 12. Explain the origin of the leap year. Why is it necessary? 13. Explain why the year 1800 was not a leap year, even though years divisible by four are normally considered to be leap years. 14. What fraction of the Moon\u2019s visible face is illuminated during first quarter phase? Why is this phase called first quarter? 15. Why don\u2019t lunar eclipses happen during every full moon? 16. Why does the Moon create tidal bulges on both sides of Earth instead of only on the side of Earth closest to the Moon? 17. Why do the heights of the tides change over the course of a month? 18. Explain how tidal forces are causing Earth to slow down. 19. Explain how tidal forces are causing the Moon to slowly recede from Earth. 20. Explain why the Gregorian calendar modified the nature of the leap year from its original definition in the Julian calendar. 21. The term equinox translates as \u201cequal night.\u201d Explain why this translation makes sense from an astronomical point of view. 22. The term solstice translates as \u201cSun stop.\u201d Explain why this translation makes sense from an astronomical point of view. 23. Why is the warmest day of the year in the United States (or in the Northern Hemisphere temperate zone) usually in August rather than on the day of the summer solstice, in late June? Thought Questions 24. When Earth\u2019s Northern Hemisphere is tilted toward the Sun during June, some would argue that the cause of our seasons is that the Northern Hemisphere is physically closer to the Sun than the Southern Hemisphere, and this is the primary reason the Northern Hemisphere is warmer. What argument or line of evidence could contradict this idea? 142 Chapter 4 Earth, Moon, and Sky 25. Where are you on Earth if you experience each of the following? (Refer to the discussion in Observing the Sky: The Birth of Astronomy as well as this chapter.) A. The stars rise and set perpendicular to the horizon. B. The stars circle the sky parallel to the horizon. C. The celestial equator passes through the zenith. D. In the course of a year, all stars are visible. E. The Sun rises on March 21 and does not set until September 21 (ideally). 26. In countries at far northern latitudes, the winter months tend to be so cloudy that astronomical observations are nearly impossible. Why can\u2019t good observations of the stars be made at those places during the summer months? 27. What is the phase of the Moon if it... A. rises at 3:00 p.m.? B. is highest in the sky at sunrise? C. sets at 10:00 a.m.? 28. A car accident occurs around midnight on the night of a full moon. The driver at fault claims he was blinded momentarily by the Moon rising on the eastern horizon. Should the police believe him? 29. The secret recipe to the ever-popular veggie burgers in the college cafeteria is hidden in a drawer in the director\u2019s office. Two students decide to break in to get their hands on it, but they want to do it a few hours before dawn on a night when there is no Moon, so they are less likely to be caught. What phases of the Moon would suit their plans? 30. Your great-great-grandfather, who often exaggerated events in his own life, once told your relatives about a terrific adventure he had on February 29, 1900. Why would this story make you suspicious? 31. One year in the future, when money is no object, you enjoy your birthday so much that you want to have another one right away. You get into your supersonic jet. Where should you and the people celebrating with you travel? From what direction should you approach? Explain. 32. Suppose you lived in the crater Copernicus on the side of the Moon facing Earth. A. How often would the Sun rise? B. How often would Earth set? C. During what fraction of the time would you be able to see the stars? 33. In a lunar eclipse, does the Moon enter the shadow of Earth from the east or west side? Explain. 34. Describe what an observer at the crater Copernicus would see while the Moon is eclipsed on Earth. What would the same observer see during what would be a total solar eclipse as viewed from Earth? 35. The day on Mars", " is 1.026 Earth-days long. The martian year lasts 686.98 Earth-days. The two moons of Mars take 0.32 Earth-day (for Phobos) and 1.26 Earth-days (for Deimos) to circle the planet. You are given the task of coming up with a martian calendar for a new Mars colony. Would a solar or lunar calendar be better for tracking the seasons? 36. What is the right ascension and declination of the vernal equinox? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 4 Earth, Moon, and Sky 143 37. What is the right ascension and declination of the autumnal equinox? 38. What is the right ascension and declination of the Sun at noon on the summer solstice in the Northern Hemisphere? 39. During summer in the Northern Hemisphere, the North Pole is illuminated by the Sun 24 hours per day. During this time, the temperature often does not rise above the freezing point of water. Explain why. 40. On the day of the vernal equinox, the day length for all places on Earth is actually slightly longer than 12 hours. Explain why. 41. Regions north of the Arctic Circle are known as the \u201cland of the midnight Sun.\u201d Explain what this means from an astronomical perspective. 42. In a part of Earth\u2019s orbit where Earth is moving faster than usual around the Sun, would the length of the sidereal day change? If so, how? Explain. 43. In a part of Earth\u2019s orbit where Earth is moving faster than usual around the Sun, would the length of the solar day change? If so, how? Explain. 44. If Sirius rises at 8:00 p.m. tonight, at what time will it rise tomorrow night, to the nearest minute? Explain. 45. What are three lines of evidence you could use to indicate that the phases of the Moon are not caused by the shadow of Earth falling on the Moon? 46. If the Moon rises at a given location at 6:00 p.m. today, about what time will it rise tomorrow night? 47. Explain why some solar eclipses are total and some are annular. 48. Why do lunar eclipses typically last much longer than solar eclipses? Figuring For Yourself 49. Suppose Earth took exactly 300.0 days to go around the Sun, and everything else (the day, the month) was the same. What kind of calendar would we have? How would this affect the seasons? 50. Consider a calendar based entirely on the day and the month (the Moon\u2019s period from full phase to full phase). How many days are there in a month? Can you figure out a scheme analogous to leap year to make this calendar work? 51. If a star rises at 8:30 p.m. tonight, approximately what time will it rise two months from now? 52. What is the altitude of the Sun at noon on December 22, as seen from a place on the Tropic of Cancer? 53. Show that the Gregorian calendar will be in error by 1 day in about 3300 years. 144 Chapter 4 Earth, Moon, and Sky This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 145 5 RADIATION AND SPECTRA Figure 5.1. Our Sun in Ultraviolet Light. This photograph of the Sun was taken at several different wavelengths of ultraviolet, which our eyes cannot see, and then color coded so it reveals activity in our Sun\u2019s atmosphere that cannot be observed in visible light. This is why it is important to observe the Sun and other astronomical objects in wavelengths other than the visible band of the spectrum. This image was taken by a satellite from above Earth\u2019s atmosphere, which is necessary since Earth\u2019s atmosphere absorbs much of the ultraviolet light coming from space. (credit: modification of work by NASA) Chapter Outline 5.1 The Behavior of Light 5.2 The Electromagnetic Spectrum 5.3 Spectroscopy in Astronomy 5.4 The Structure of the Atom 5.5 Formation of Spectral Lines 5.6 The Doppler Effect Thinking Ahead The nearest star is so far away that the fastest spacecraft humans have built would take almost 100,000 years to get there. Yet we very much want to know what material this neighbor star is composed of and how it differs from our own Sun. How can we learn about the chemical makeup of stars that we cannot hope to visit or sample? In astronomy, most of the objects that we study are completely beyond our reach. The temperature of the Sun is so high that a spacecraft would be fried long before it reached it, and the stars are much too far away to visit in our lifetimes with the technology now available. Even", " light, which travels at a speed of 300,000 kilometers per second (km/s), takes more than 4 years to reach us from the nearest star. If we want to learn about the Sun and stars, we must rely on techniques that allow us to analyze them from a distance. 146 Chapter 5 Radiation and Spectra 5.1 THE BEHAVIOR OF LIGHT Learning Objectives By the end of this section, you will be able to: Explain the evidence for Maxwell\u2019s electromagnetic model of light Describe the relationship between wavelength, frequency, and speed of light Discuss the particle model of light and the definition of photon Explain how and why the amount of light we see from an object depends upon its distance Coded into the light and other kinds of radiation that reach us from objects in the universe is a wide range of information about what those objects are like and how they work. If we can decipher this code and read the messages it contains, we can learn an enormous amount about the cosmos without ever having to leave Earth or its immediate environment. The visible light and other radiation we receive from the stars and planets is generated by processes at the atomic level\u2014by changes in the way the parts of an atom interact and move. Thus, to appreciate how light is generated, we must explore how atoms work. There is a bit of irony in the fact that in order to understand some of the largest structures in the universe, we must become acquainted with some of the smallest. Notice that we have twice used the phrase \u201clight and other radiation.\u201d One of the key ideas explored in this chapter is that visible light is not unique; it is merely the most familiar example of a much larger family of radiation that can carry information to us. The word \u201c radiation\u201d will be used frequently in this book, so it is important to understand what it means. In everyday language, \u201cradiation\u201d is often used to describe certain kinds of energetic subatomic particles released by radioactive materials in our environment. (An example is the kind of radiation used to treat some cancers.) But this is not what we mean when we use the word \u201cradiation\u201d in an astronomy text. Radiation, as used in this book, is a general term for waves (including light waves) that radiate outward from a source. As we saw in Orbits and Gravity, Newton\u2019s theory of gravity accounts for the motions of planets as well as objects on Earth. Application of this theory to a variety of problems dominated the work of scientists for nearly two centuries. In the nineteenth century, many physicists turned to the study of electricity and magnetism, which are intimately connected with the production of light. The scientist who played a role in this field comparable to Newton\u2019s role in the study of gravity was physicist James Clerk Maxwell, born and educated in Scotland (Figure 5.2). Inspired by a number of ingenious experiments that showed an intimate relationship between electricity and magnetism, Maxwell developed a theory that describes both electricity and magnetism with only a small number of elegant equations. It is this theory that gives us important insights into the nature and behavior of light. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 147 Figure 5.2. James Clerk Maxwell (1831\u20131879). Maxwell unified the rules governing electricity and magnetism into a coherent theory. Maxwell\u2019s Theory of Electromagnetism We will look at the structure of the atom in more detail later, but we begin by noting that the typical atom consists of several types of particles, a number of which have not only mass but an additional property called electric charge. In the nucleus (central part) of every atom are protons, which are positively charged; outside the nucleus are electrons, which have a negative charge. Maxwell\u2019s theory deals with these electric charges and their effects, especially when they are moving. In the vicinity of an electron charge, another charge feels a force of attraction or repulsion: opposite charges attract; like charges repel. When charges are not in motion, we observe only this electric attraction or repulsion. If charges are in motion, however (as they are inside every atom and in a wire carrying a current), then we measure another force called magnetism. Magnetism was well known for much of recorded human history, but its cause was not understood until the nineteenth century. Experiments with electric charges demonstrated that magnetism was the result of moving charged particles. Sometimes, the motion is clear, as in the coils of heavy wire that make an industrial electromagnet. Other times, it is more subtle, as in the kind of magnet you buy in a hardware store, in which many of the electrons inside the atoms are spinning in roughly the same direction; it is the alignment of their motion that causes the material to become magnetic. Physicists use the word field to describe the action of forces that one object exerts on other distant objects", ". For example, we say the Sun produces a gravitational field that controls Earth\u2019s orbit, even though the Sun and Earth do not come directly into contact. Using this terminology, we can say that stationary electric charges produce electric fields, and moving electric charges also produce magnetic fields. Actually, the relationship between electric and magnetic phenomena is even more profound. Experiments showed that changing magnetic fields could produce electric currents (and thus changing electric fields), and changing electric currents could in turn produce changing magnetic fields. So once begun, electric and magnetic field changes could continue to trigger each other. Maxwell analyzed what would happen if electric charges were oscillating (moving constantly back and forth) and found that the resulting pattern of electric and magnetic fields would spread out and travel rapidly through space. Something similar happens when a raindrop strikes the surface of water or a frog jumps into a pond. The disturbance moves outward and creates a pattern we call a wave in the water (Figure 5.3). You might, at first, think that there must be very few situations in nature where electric charges oscillate, but this is not at all the case. As we shall see, atoms and molecules (which consist of charged particles) oscillate back and forth all the time. The resulting electromagnetic disturbances are among the most common phenomena in the universe. 148 Chapter 5 Radiation and Spectra Figure 5.3. Making Waves. An oscillation in a pool of water creates an expanding disturbance called a wave. (credit: modification of work by \"vastateparksstaff\"/Flickr) Maxwell was able to calculate the speed at which an electromagnetic disturbance moves through space; he found that it is equal to the speed of light, which had been measured experimentally. On that basis, he speculated that light was one form of a family of possible electromagnetic disturbances called electromagnetic radiation, a conclusion that was again confirmed in laboratory experiments. When light (reflected from the pages of an astronomy textbook, for example) enters a human eye, its changing electric and magnetic fields stimulate nerve endings, which then transmit the information contained in these changing fields to the brain. The science of astronomy is primarily about analyzing radiation from distant objects to understand what they are and how they work. The Wave-Like Characteristics of Light The changing electric and magnetic fields in light are similar to the waves that can be set up in a quiet pool of water. In both cases, the disturbance travels rapidly outward from the point of origin and can use its energy to disturb other things farther away. (For example, in water, the expanding ripples moving away from our frog could disturb the peace of a dragonfly resting on a leaf in the same pool.) In the case of electromagnetic waves, the radiation generated by a transmitting antenna full of charged particles and moving electrons at your local radio station can, sometime later, disturb a group of electrons in your car radio antenna and bring you the news and weather while you are driving to class or work in the morning. The waves generated by charged particles differ from water waves in some profound ways, however. Water waves require water to travel in. The sound waves we hear, to give another example, are pressure disturbances that require air to travel though. But electromagnetic waves do not require water or air: the fields generate each other and so can move through a vacuum (such as outer space). This was such a disturbing idea to nineteenth- century scientists that they actually made up a substance to fill all of space\u2014one for which there was not a single shred of evidence\u2014just so light waves could have something to travel through: they called it the aether. Today, we know that there is no aether and that electromagnetic waves have no trouble at all moving through empty space (as all the starlight visible on a clear night must surely be doing). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 149 The other difference is that all electromagnetic waves move at the same speed in empty space (the speed of light\u2014approximately 300,000 kilometers per second, or 300,000,000 meters per second, which can also be written as 3 \u00d7 108 m/s), which turns out to be the fastest possible speed in the universe. No matter where electromagnetic waves are generated from and no matter what other properties they have, when they are moving (and not interacting with matter), they move at the speed of light. Yet you know from everyday experience that there are different kinds of light. For example, we perceive that light waves differ from one another in a property we call color. Let\u2019s see how we can denote the differences among the whole broad family of electromagnetic waves. The nice thing about a wave is that it is a repeating phenomenon. Whether it is the up-and-down motion of a water wave or the changing electric and magnetic fields in a wave of light, the pattern of disturbance repeats in a cyclical way. Thus, any wave motion can be characterized by a series of crests and", " troughs (Figure 5.4). Moving from one crest through a trough to the next crest completes one cycle. The horizontal length covered by one cycle is called the wavelength. Ocean waves provide an analogy: the wavelength is the distance that separates successive wave crests. Figure 5.4. Characterizing Waves. Electromagnetic radiation has wave-like characteristics. The wavelength (\u03bb) is the distance between crests, the frequency (f) is the number of cycles per second, and the speed (c) is the distance the wave covers during a specified period of time (e.g., kilometers per second). For visible light, our eyes perceive different wavelengths as different colors: red, for example, is the longest visible wavelength, and violet is the shortest. The main colors of visible light from longest to shortest wavelength can be remembered using the mnemonic ROY G BIV\u2014for Red, Orange, Yellow, Green, Blue, Indigo, and Violet. Other invisible forms of electromagnetic radiation have different wavelengths, as we will see in the next section. We can also characterize different waves by their frequency, the number of wave cycles that pass by per second. If you count 10 crests moving by each second, for example, then the frequency is 10 cycles per second (cps). In honor of Heinrich Hertz, the physicist who\u2014inspired by Maxwell\u2019s work\u2014discovered radio waves, a cps is also called a hertz (Hz). Take a look at your radio, for example, and you will see the channel assigned to each radio station is characterized by its frequency, usually in units of KHz (kilohertz, or thousands of hertz) or MHz (megahertz, or millions of hertz). Wavelength (\u03bb) and frequency (f) are related because all electromagnetic waves travel at the same speed. To see how this works, imagine a parade in which everyone is forced by prevailing traffic conditions to move at exactly the same speed. You stand on a corner and watch the waves of marchers come by. First you see row after row of miniature ponies. Because they are not very large and, therefore, have a shorter wavelength, a good number of the ponies can move past you each minute; we can say they have a high frequency. Next, however, come several rows of circus elephants. The elephants are large and marching at the same speed as the ponies, so far fewer of them can march past you per minute: Because they have a wider spacing (longer wavelength), they represent a lower frequency. 150 Chapter 5 Radiation and Spectra The formula for this relationship can be expressed as follows: for any wave motion, the speed at which a wave moves equals the frequency times the wavelength. Waves with longer wavelengths have lower frequencies. Mathematically, we can express this as c = \u03bb f where the Greek letter for \u201cl\u201d\u2014lambda, \u03bb\u2014is used to denote wavelength and c is the scientific symbol for the speed of light. Solving for the wavelength, this is expressed as Deriving and Using the Wave Equation The equation for the relationship between the speed and other characteristics of a wave can be derived from our basic understanding of motion. The average speed of anything that is moving is: average speed = distance time (So, for example, a car on the highway traveling at a speed of 100 km/h covers 100 km during the time of 1 h.) For an electromagnetic wave to travel the distance of one of its wavelengths, \u03bb, at the speed of light, c, we have c = \u03bb/t. The frequency of a wave is the number of cycles per second. If a wave has a frequency of a million cycles per second, then the time for each cycle to go by is a millionth of a second. So, in general, t = 1/f. Substituting into our wave equation, we get c = \u03bb \u00d7 f. Now let\u2019s use this to calculate an example. What is the wavelength of visible light that has a frequency of 5.66 \u00d7 1014 Hz? Solution Solving the wave equation for wavelength, we find: \u03bb = c f Substituting our values gives: \u03bb = 3.00 \u00d7 108 m/s 5.66 \u00d7 1014 Hz = 5.30 \u00d7 10\u20137 m This answer can also be written as 530 nm, which is in the yellow-green part of the visible spectrum (nm stands for nanometers, where the term \u201cnano\u201d means \u201cbillionths\u201d). Check Your Learning \u201cTidal waves,\u201d or tsunamis, are waves caused by earthquakes that travel rapidly through the ocean. If a tsunami travels at the speed of 600 km/h and approaches a shore at a rate of one wave crest every 15 min (4 waves/h), what would be the distance between those wave crests at sea? Answer: \u03bb = 600 km/h 4 waves/h = 150 km This Open", "Stax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 151 Light as a Photon The electromagnetic wave model of light (as formulated by Maxwell) was one of the great triumphs of nineteenth-century science. In 1887, when Heinrich Hertz actually made invisible electromagnetic waves (what today are called radio waves) on one side of a room and detected them on the other side, it ushered in a new era that led to the modern age of telecommunications. His experiment ultimately led to the technologies of television, cell phones, and today\u2019s wireless networks around the globe. However, by the beginning of the twentieth century, more sophisticated experiments had revealed that light behaves in certain ways that cannot be explained by the wave model. Reluctantly, physicists had to accept that sometimes light behaves more like a \u201cparticle\u201d\u2014or at least a self-contained packet of energy\u2014than a wave. We call such a packet of electromagnetic energy a photon. The fact that light behaves like a wave in certain experiments and like a particle in others was a very surprising and unlikely idea. After all, our common sense says that waves and particles are opposite concepts. On one hand, a wave is a repeating disturbance that, by its very nature, is not in only one place, but spreads out. A particle, on the other hand, is something that can be in only one place at any given time. Strange as it sounds, though, countless experiments now confirm that electromagnetic radiation can sometimes behave like a wave and at other times like a particle. Then, again, perhaps we shouldn\u2019t be surprised that something that always travels at the \u201cspeed limit\u201d of the universe and doesn\u2019t need a medium to travel through might not obey our everyday common sense ideas. The confusion that this wave-particle duality of light caused in physics was eventually resolved by the introduction of a more complicated theory of waves and particles, now called quantum mechanics. (This is one of the most interesting fields of modern science, but it is mostly beyond the scope of our book. If you are interested in it, see some of the suggested resources at the end of this chapter.) In any case, you should now be prepared when scientists (or the authors of this book) sometimes discuss electromagnetic radiation as if it consisted of waves and at other times refer to it as a stream of photons. A photon (being a packet of energy) carries a specific amount of energy. We can use the idea of energy to connect the photon and wave models. How much energy a photon has depends on its frequency when you think about it as a wave. A low-energy radio wave has a low frequency as a wave, while a high-energy X-ray at your dentist\u2019s office is a high-frequency wave. Among the colors of visible light, violet-light photons have the highest energy and red-light photons have the lowest. Test whether the connection between photons and waves is clear to you. In the above example, which photon would have the longer wavelength as a wave: the radio wave or the X-ray? If you answered the radio wave, you are correct. Radio waves have a lower frequency, so the wave cycles are longer (they are elephants, not miniature ponies). Propagation of Light Let\u2019s think for a moment about how light from a lightbulb moves through space. As waves expand, they travel away from the bulb, not just toward your eyes but in all directions. They must therefore cover an ever-widening space. Yet the total amount of light available can\u2019t change once the light has left the bulb. This means that, as the same expanding shell of light covers a larger and larger area, there must be less and less of it in any given place. Light (and all other electromagnetic radiation) gets weaker and weaker as it gets farther from its source. The increase in the area that the light must cover is proportional to the square of the distance that the light has traveled (Figure 5.5). If we stand twice as far from the source, our eyes will intercept two-squared (2 \u00d7 2), or four times less light. If we stand 10 times farther from the source, we get 10-squared, or 100 times less light. You can see how this weakening means trouble for sources of light at astronomical distances. One of the nearest stars, 152 Chapter 5 Radiation and Spectra Alpha Centauri A, emits about the same total energy as the Sun. But it is about 270,000 times farther away, and so it appears about 73 billion times fainter. No wonder the stars, which close-up would look more or less like the Sun, look like faint pinpoints of light from far away. Figure 5.5. Inverse Square Law for Light. As light radiates away from its source, it spreads out in such a way that the energy per unit area", " (the amount of energy passing through one of the small squares) decreases as the square of the distance from its source. This idea\u2014that the apparent brightness of a source (how bright it looks to us) gets weaker with distance in the way we have described\u2014is known as the inverse square law for light propagation. In this respect, the propagation of light is similar to the effects of gravity. Remember that the force of gravity between two attracting masses is also inversely proportional to the square of their separation The Inverse Square Law for Light The intensity of a 120-W lightbulb observed from a distance 2 m away is 2.4 W/m2. What would be the intensity if this distance was doubled? Solution If we move twice as far away, then the answer will change according to the inverse square of the distance, so the new intensity will be (1/2)2 = 1/4 of the original intensity, or 0.6 W/m2. Check Your Learning How many times brighter or fainter would a star appear if it were moved to: a. twice its present distance? b. ten times its present distance? c. half its present distance? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 153 Answer: 2 a. 2 \u239b \u239d 1 10 \u239e \u23a0 = 1 100 ; c. 2 \u239b \u239d 1 1/2 \u239e \u23a0 = 4 5.2 THE ELECTROMAGNETIC SPECTRUM Learning Objectives By the end of this section, you will be able to: Understand the bands of the electromagnetic spectrum and how they differ from one another Understand how each part of the spectrum interacts with Earth\u2019s atmosphere Explain how and why the light emitted by an object depends on its temperature Objects in the universe send out an enormous range of electromagnetic radiation. Scientists call this range the electromagnetic spectrum, which they have divided into a number of categories. The spectrum is shown in Figure 5.6, with some information about the waves in each part or band. 154 Chapter 5 Radiation and Spectra Figure 5.6. Radiation and Earth\u2019s Atmosphere. This figure shows the bands of the electromagnetic spectrum and how well Earth\u2019s atmosphere transmits them. Note that high-frequency waves from space do not make it to the surface and must therefore be observed from space. Some infrared and microwaves are absorbed by water and thus are best observed from high altitudes. Low-frequency radio waves are blocked by Earth\u2019s ionosphere. (credit: modification of work by STScI/JHU/NASA) Types of Electromagnetic Radiation Electromagnetic radiation with the shortest wavelengths, no longer than 0.01 nanometer, is categorized as gamma rays (1 nanometer = 10\u20139 meters; see Appendix D). The name gamma comes from the third letter of the Greek alphabet: gamma rays were the third kind of radiation discovered coming from radioactive atoms when physicists first investigated their behavior. Because gamma rays carry a lot of energy, they can be dangerous for living tissues. Gamma radiation is generated deep in the interior of stars, as well as by some of the most violent phenomena in the universe, such as the deaths of stars and the merging of stellar corpses. Gamma rays coming to Earth are absorbed by our atmosphere before they reach the ground (which is a good thing for our health); thus, they can only be studied using instruments in space. Electromagnetic radiation with wavelengths between 0.01 nanometer and 20 nanometers is referred to as X- rays. Being more energetic than visible light, X-rays are able to penetrate soft tissues but not bones, and so allow us to make images of the shadows of the bones inside us. While X-rays can penetrate a short length of human flesh, they are stopped by the large numbers of atoms in Earth\u2019s atmosphere with which they interact. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 155 Thus, X-ray astronomy (like gamma-ray astronomy) could not develop until we invented ways of sending instruments above our atmosphere (Figure 5.7). Figure 5.7. X-Ray Sky. This is a map of the sky tuned to certain types of X-rays (seen from above Earth\u2019s atmosphere). The map tilts the sky so that the disk of our Milky Way Galaxy runs across its center. It was constructed and artificially colored from data gathered by the European ROSAT satellite. Each color (red, yellow, and blue) shows X-rays of different frequencies or energies. For example, red outlines the glow from a hot local bubble of gas all around us, blown by one or more exploding stars in our cosmic vicinity. Yellow and blue show more distant sources of X-rays, such as remnants of other exploded", " stars or the active center of our Galaxy (in the middle of the picture). (credit: modification of work by NASA) Radiation intermediate between X-rays and visible light is ultraviolet (meaning higher energy than violet). Outside the world of science, ultraviolet light is sometimes called \u201cblack light\u201d because our eyes cannot see it. Ultraviolet radiation is mostly blocked by the ozone layer of Earth\u2019s atmosphere, but a small fraction of ultraviolet rays from our Sun do penetrate to cause sunburn or, in extreme cases of overexposure, skin cancer in human beings. Ultraviolet astronomy is also best done from space. Electromagnetic radiation with wavelengths between roughly 400 and 700 nm is called visible light because these are the waves that human vision can perceive. This is also the band of the electromagnetic spectrum that most readily reaches Earth\u2019s surface. These two observations are not coincidental: human eyes evolved to see the kinds of waves that arrive from the Sun most effectively. Visible light penetrates Earth\u2019s atmosphere effectively, except when it is temporarily blocked by clouds. Between visible light and radio waves are the wavelengths of infrared or heat radiation. Astronomer William Herschel first discovered infrared in 1800 while trying to measure the temperatures of different colors of sunlight spread out into a spectrum. He noticed that when he accidently positioned his thermometer beyond the reddest color, it still registered heating due to some invisible energy coming from the Sun. This was the first hint about the existence of the other (invisible) bands of the electromagnetic spectrum, although it would take many decades for our full understanding to develop. A heat lamp radiates mostly infrared radiation, and the nerve endings in our skin are sensitive to this band of the electromagnetic spectrum. Infrared waves are absorbed by water and carbon dioxide molecules, which are more concentrated low in Earth\u2019s atmosphere. For this reason, infrared astronomy is best done from high mountaintops, high-flying airplanes, and spacecraft. After infrared comes the familiar microwave, used in short-wave communication and microwave ovens. (Wavelengths vary from 1 millimeter to 1 meter and are absorbed by water vapor, which makes them effective in heating foods.) The \u201cmicro-\u201d prefix refers to the fact that microwaves are small in comparison to radio waves, the next on the spectrum. You may remember that tea\u2014which is full of water\u2014heats up quickly in 156 Chapter 5 Radiation and Spectra your microwave oven, while a ceramic cup\u2014from which water has been removed by baking\u2014stays cool in comparison. All electromagnetic waves longer than microwaves are called radio waves, but this is so broad a category that we generally divide it into several subsections. Among the most familiar of these are radar waves, which are used in radar guns by traffic officers to determine vehicle speeds, and AM radio waves, which were the first to be developed for broadcasting. The wavelengths of these different categories range from over a meter to hundreds of meters, and other radio radiation can have wavelengths as long as several kilometers. With such a wide range of wavelengths, not all radio waves interact with Earth\u2019s atmosphere in the same way. FM and TV waves are not absorbed and can travel easily through our atmosphere. AM radio waves are absorbed or reflected by a layer in Earth\u2019s atmosphere called the ionosphere (the ionosphere is a layer of charged particles at the top of our atmosphere, produced by interactions with sunlight and charged particles that are ejected from the Sun). We hope this brief survey has left you with one strong impression: although visible light is what most people associate with astronomy, the light that our eyes can see is only a tiny fraction of the broad range of waves generated in the universe. Today, we understand that judging some astronomical phenomenon by using only the light we can see is like hiding under the table at a big dinner party and judging all the guests by nothing but their shoes. There\u2019s a lot more to each person than meets our eye under the table. It is very important for those who study astronomy today to avoid being \u201cvisible light chauvinists\u201d\u2014to respect only the information seen by their eyes while ignoring the information gathered by instruments sensitive to other bands of the electromagnetic spectrum. Table 5.1 summarizes the bands of the electromagnetic spectrum and indicates the temperatures and typical astronomical objects that emit each kind of electromagnetic radiation. While at first, some of the types of radiation listed in the table may seem unfamiliar, you will get to know them better as your astronomy course continues. You can return to this table as you learn more about the types of objects astronomers study. Types of Electromagnetic Radiation Type of Radiation Wavelength Range (nm) Radiated by Objects at This Temperature Typical Sources Gamma rays Less than 0.01 More than 108 K Produced in nuclear reactions; require very high-energy processes X-rays 0.01\u201320 106\u2013108 K Ultraviolet 20\u2013400 Visible 400\u2013700 104\u2013106 K 103\u2013104 K Infrared 103\u2013106 10\u2013103 K Mic", "rowave 106\u2013109 Less than 10 K Table 5.1 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Gas in clusters of galaxies, supernova remnants, solar corona Supernova remnants, very hot stars Stars Cool clouds of dust and gas, planets, moons Active galaxies, pulsars, cosmic background radiation Chapter 5 Radiation and Spectra 157 Types of Electromagnetic Radiation Type of Radiation Wavelength Range (nm) Radiated by Objects at This Temperature Typical Sources Radio More than 109 Less than 10 K Supernova remnants, pulsars, cold gas Table 5.1 Radiation and Temperature Some astronomical objects emit mostly infrared radiation, others mostly visible light, and still others mostly ultraviolet radiation. What determines the type of electromagnetic radiation emitted by the Sun, stars, and other dense astronomical objects? The answer often turns out to be their temperature. At the microscopic level, everything in nature is in motion. A solid is composed of molecules and atoms in continuous vibration: they move back and forth in place, but their motion is much too small for our eyes to make out. A gas consists of atoms and/or molecules that are flying about freely at high speed, continually bumping into one another and bombarding the surrounding matter. The hotter the solid or gas, the more rapid the motion of its molecules or atoms. The temperature of something is thus a measure of the average motion energy of the particles that make it up. This motion at the microscopic level is responsible for much of the electromagnetic radiation on Earth and in the universe. As atoms and molecules move about and collide, or vibrate in place, their electrons give off electromagnetic radiation. The characteristics of this radiation are determined by the temperature of those atoms and molecules. In a hot material, for example, the individual particles vibrate in place or move rapidly from collisions, so the emitted waves are, on average, more energetic. And recall that higher energy waves have a higher frequency. In very cool material, the particles have low-energy atomic and molecular motions and thus generate lower-energy waves Check out the NASA briefing (https://openstax.org/l/30elmagsp1) or NASA\u2019s 5-minute introductory video (https://openstax.org/l/30elmagsp2) to learn more about the electromagnetic spectrum. Radiation Laws To understand, in more quantitative detail, the relationship between temperature and electromagnetic radiation, we imagine an idealized object called a blackbody. Such an object (unlike your sweater or your astronomy instructor\u2019s head) does not reflect or scatter any radiation, but absorbs all the electromagnetic energy that falls onto it. The energy that is absorbed causes the atoms and molecules in it to vibrate or move around at increasing speeds. As it gets hotter, this object will radiate electromagnetic waves until absorption and radiation are in balance. We want to discuss such an idealized object because, as you will see, stars behave in very nearly the same way. The radiation from a blackbody has several characteristics, as illustrated in Figure 5.8. The graph shows the power emitted at each wavelength by objects of different temperatures. In science, the word power means the 158 Chapter 5 Radiation and Spectra energy coming off per second (and it is typically measured in watts, which you are probably familiar with from buying lightbulbs). Figure 5.8. Radiation Laws Illustrated. This graph shows in arbitrary units how many photons are given off at each wavelength for objects at four different temperatures. The wavelengths corresponding to visible light are shown by the colored bands. Note that at hotter temperatures, more energy (in the form of photons) is emitted at all wavelengths. The higher the temperature, the shorter the wavelength at which the peak amount of energy is radiated (this is known as Wien\u2019s law). First of all, notice that the curves show that, at each temperature, our blackbody object emits radiation (photons) at all wavelengths (all colors). This is because in any solid or denser gas, some molecules or atoms vibrate or move between collisions slower than average and some move faster than average. So when we look at the electromagnetic waves emitted, we find a broad range, or spectrum, of energies and wavelengths. More energy is emitted at the average vibration or motion rate (the highest part of each curve), but if we have a large number of atoms or molecules, some energy will be detected at each wavelength. Second, note that an object at a higher temperature emits more power at all wavelengths than does a cooler one. In a hot gas (the taller curves in Figure 5.8), for example, the atoms have more collisions and give off more energy. In the real world of stars, this means that hotter stars give off more energy at every wavelength than do cooler stars. Third, the graph shows us that the higher the temperature, the shorter the wavelength at which the maximum power is emitted. Remember that a shorter wavelength means a higher frequency and energy. It makes", " sense, then, that hot objects give off a larger fraction of their energy at shorter wavelengths (higher energies) than do cool objects. You may have observed examples of this rule in everyday life. When a burner on an electric stove is turned on low, it emits only heat, which is infrared radiation, but does not glow with visible light. If the burner is set to a higher temperature, it starts to glow a dull red. At a still-higher setting, it glows a brighter orange-red (shorter wavelength). At even higher temperatures, which cannot be reached with ordinary stoves, metal can appear brilliant yellow or even blue-white. We can use these ideas to come up with a rough sort of \u201cthermometer\u201d for measuring the temperatures of stars. Because many stars give off most of their energy in visible light, the color of light that dominates a star\u2019s appearance is a rough indicator of its temperature. If one star looks red and another looks blue, which one This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 159 has the higher temperature? Because blue is the shorter-wavelength color, it is the sign of a hotter star. (Note that the temperatures we associate with different colors in science are not the same as the ones artists use. In art, red is often called a \u201chot\u201d color and blue a \u201ccool\u201d color. Likewise, we commonly see red on faucet or air conditioning controls to indicate hot temperatures and blue to indicate cold temperatures. Although these are common uses to us in daily life, in nature, it\u2019s the other way around.) We can develop a more precise star thermometer by measuring how much energy a star gives off at each wavelength and by constructing diagrams like Figure 5.8. The location of the peak (or maximum) in the power curve of each star can tell us its temperature. The average temperature at the surface of the Sun, which is where the radiation that we see is emitted, turns out to be 5800 K. (Throughout this text, we use the kelvin or absolute temperature scale. On this scale, water freezes at 273 K and boils at 373 K. All molecular motion ceases at 0 K. The various temperature scales are described in Appendix D.) There are stars cooler than the Sun and stars hotter than the Sun. The wavelength at which maximum power is emitted can be calculated according to the equation \u03bbmax = 3 \u00d7 106 T where the wavelength is in nanometers (one billionth of a meter) and the temperature is in K. This relationship is called Wien\u2019s law. For the Sun, the wavelength at which the maximum energy is emitted is 520 nanometers, which is near the middle of that portion of the electromagnetic spectrum called visible light. Characteristic temperatures of other astronomical objects, and the wavelengths at which they emit most of their power, are listed in Table 5.1 Calculating the Temperature of a Blackbody We can use Wien\u2019s law to calculate the temperature of a star provided we know the wavelength of peak intensity for its spectrum. If the emitted radiation from a red dwarf star has a wavelength of maximum power at 1200 nm, what is the temperature of this star, assuming it is a blackbody? Solution Solving Wien\u2019s law for temperature gives: T = 3 \u00d7 106 nm K \u03bbmax = 3 \u00d7 106 nm K 1200 nm = 2500 K Check Your Learning What is the temperature of a star whose maximum light is emitted at a much shorter wavelength of 290 nm? Answer: T = 3 \u00d7 106 nm K \u03bbmax = 3 \u00d7 106 nm K 290 nm = 10,300 K 160 Chapter 5 Radiation and Spectra Since this star has a peak wavelength that is at a shorter wavelength (in the ultraviolet part of the spectrum) than that of our Sun (in the visible part of the spectrum), it should come as no surprise that its surface temperature is much hotter than our Sun\u2019s. We can also describe our observation that hotter objects radiate more power at all wavelengths in a mathematical form. If we sum up the contributions from all parts of the electromagnetic spectrum, we obtain the total energy emitted by a blackbody. What we usually measure from a large object like a star is the energy flux, the power emitted per square meter. The word flux means \u201cflow\u201d here: we are interested in the flow of power into an area (like the area of a telescope mirror). It turns out that the energy flux from a blackbody at temperature T is proportional to the fourth power of its absolute temperature. This relationship is known as the Stefan-Boltzmann law and can be written in the form of an equation as F = \u03c3T 4 where F stands for the energy flux and \u03c3 (Greek letter sigma) is a constant number (5.67 \u00d7 108). Notice how impressive this result is.", " Increasing the temperature of a star would have a tremendous effect on the power it radiates. If the Sun, for example, were twice as hot\u2014that is, if it had a temperature of 11,600 K\u2014it would radiate 24, or 16 times more power than it does now. Tripling the temperature would raise the power output 81 times. Hot stars really shine away a tremendous amount of energy Calculating the Power of a Star While energy flux tells us how much power a star emits per square meter, we would often like to know how much total power is emitted by the star. We can determine that by multiplying the energy flux by the number of square meters on the surface of the star. Stars are mostly spherical, so we can use the formula 4\u03c0R2 for the surface area, where R is the radius of the star. The total power emitted by the star (which we call the star\u2019s \u201cabsolute luminosity\u201d) can be found by multiplying the formula for energy flux and the formula for the surface area: Two stars have the same size and are the same distance from us. Star A has a surface temperature of 6000 K, and star B has a surface temperature twice as high, 12,000 K. How much more luminous is star B L = 4\u03c0R2 \u03c3T 4 compared to star A? Solution Take the ratio of the luminosity of Star A to Star B: LA = 4\u03c0RA 2 \u03c3T A 4 and LB = 4\u03c0RB 2 \u03c3T B 4 LB LA = 4\u03c0RB 4\u03c0RA 2 \u03c3T B 2 \u03c3T A 4 4 = RB RA 4 4 2 T B 2 T A Because the two stars are the same size, RA = RB, leaving 4 4 = T B T A (12,000 K)4 (8,000 K)4 = 24 = 16 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 161 Check Your Learning Two stars with identical diameters are the same distance away. One has a temperature of 8700 K and the other has a temperature of 2900 K. Which is brighter? How much brighter is it? Answer: The 5800 K star has triple the temperature, so it is 34 = 81 times brighter. 5.3 SPECTROSCOPY IN ASTRONOMY Learning Objectives By the end of this section, you will be able to: Describe the properties of light Explain how astronomers learn the composition of a gas by examining its spectral lines Discuss the various types of spectra Electromagnetic radiation carries a lot of information about the nature of stars and other astronomical objects. To extract this information, however, astronomers must be able to study the amounts of energy we receive at different wavelengths of light in fine detail. Let\u2019s examine how we can do this and what we can learn. Properties of Light Light exhibits certain behaviors that are important to the design of telescopes and other instruments. For example, light can be reflected from a surface. If the surface is smooth and shiny, as with a mirror, the direction of the reflected light beam can be calculated accurately from knowledge of the shape of the reflecting surface. Light is also bent, or refracted, when it passes from one kind of transparent material into another\u2014say, from the air into a glass lens. Reflection and refraction of light are the basic properties that make possible all optical instruments (devices that help us to see things better)\u2014from eyeglasses to giant astronomical telescopes. Such instruments are generally combinations of glass lenses, which bend light according to the principles of refraction, and curved mirrors, which depend on the properties of reflection. Small optical devices, such as eyeglasses or binoculars, generally use lenses, whereas large telescopes depend almost entirely on mirrors for their main optical elements. We will discuss astronomical instruments and their uses more fully in Astronomical Instruments. For now, we turn to another behavior of light, one that is essential for the decoding of light. In 1672, in the first paper that he submitted to the Royal Society, Sir Isaac Newton described an experiment in which he permitted sunlight to pass through a small hole and then through a prism. Newton found that sunlight, which looks white to us, is actually made up of a mixture of all the colors of the rainbow (Figure 5.9). 162 Chapter 5 Radiation and Spectra Figure 5.9. Action of a Prism. When we pass a beam of white sunlight through a prism, we see a rainbow-colored band of light that we call a continuous spectrum. Figure 5.9 shows how light is separated into different colors with a prism\u2014a piece of glass in the shape of a triangle with refracting surfaces. Upon entering one face of the prism, the path of the light is refracted (bent), but not all of the colors are bent by the same amount. The bending of the beam depends on the wavelength", " of the light as well as the properties of the material, and as a result, different wavelengths (or colors of light) are bent by different amounts and therefore follow slightly different paths through the prism. The violet light is bent more than the red. This phenomenon is called dispersion and explains Newton\u2019s rainbow experiment. Upon leaving the opposite face of the prism, the light is bent again and further dispersed. If the light leaving the prism is focused on a screen, the different wavelengths or colors that make up white light are lined up side by side just like a rainbow (Figure 5.10). (In fact, a rainbow is formed by the dispersion of light though raindrops; see The Rainbow feature box.) Because this array of colors is a spectrum of light, the instrument used to disperse the light and form the spectrum is called a spectrometer. Figure 5.10. Continuous Spectrum. When white light passes through a prism, it is dispersed and forms a continuous spectrum of all the colors. Although it is hard to see in this printed version, in a well-dispersed spectrum, many subtle gradations in color are visible as your eye scans from one end (violet) to the other (red). The Value of Stellar Spectra When Newton described the laws of refraction and dispersion in optics, and observed the solar spectrum, all he could see was a continuous band of colors. If the spectrum of the white light from the Sun and stars were simply a continuous rainbow of colors, astronomers would have little interest in the detailed study of a star\u2019s spectrum once they had learned its average surface temperature. In 1802, however, William Wollaston built an improved spectrometer that included a lens to focus the Sun\u2019s spectrum on a screen. With this device, Wollaston saw that the colors were not spread out uniformly, but instead, some ranges of color were missing, appearing as dark bands in the solar spectrum. He mistakenly attributed these lines to natural boundaries between the colors. In 1815, German physicist Joseph Fraunhofer, upon a more careful examination of the solar spectrum, found about 600 such dark lines (missing colors), which led scientists to rule out the boundary hypothesis (Figure 5.11). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 163 Figure 5.11. Visible Spectrum of the Sun. Our star\u2019s spectrum is crossed by dark lines produced by atoms in the solar atmosphere that absorb light at certain wavelengths. (credit: modification of work by Nigel Sharp, NOAO/National Solar Observatory at Kitt Peak/AURA, and the National Science Foundation) Later, researchers found that similar dark lines could be produced in the spectra (\u201cspectra\u201d is the plural of \u201cspectrum\u201d) of artificial light sources. They did this by passing their light through various apparently transparent substances\u2014usually containers with just a bit of thin gas in them. These gases turned out not to be transparent at all colors: they were quite opaque at a few sharply defined wavelengths. Something in each gas had to be absorbing just a few colors of light and no others. All gases did this, but each different element absorbed a different set of colors and thus showed different dark lines. If the gas in a container consisted of two elements, then light passing through it was missing the colors (showing dark lines) for both of the elements. So it became clear that certain lines in the spectrum \u201cgo with\u201d certain elements. This discovery was one of the most important steps forward in the history of astronomy. What would happen if there were no continuous spectrum for our gases to remove light from? What if, instead, we heated the same thin gases until they were hot enough to glow with their own light? When the gases were heated, a spectrometer revealed no continuous spectrum, but several separate bright lines. That is, these hot gases emitted light only at certain specific wavelengths or colors. When the gas was pure hydrogen, it would emit one pattern of colors; when it was pure sodium, it would emit a different pattern. A mixture of hydrogen and sodium emitted both sets of spectral lines. The colors the gases emitted when they were heated were the very same colors as those they had absorbed when a continuous source of light was behind them. From such experiments, scientists began to see that different substances showed distinctive spectral signatures by which their presence could be detected (Figure 5.12). Just as your signature allows the bank to identify you, the unique pattern of colors for each type of atom (its spectrum) can help us identify which element or elements are in a gas. 164 Chapter 5 Radiation and Spectra Figure 5.12. Continuous Spectrum and Line Spectra from Different Elements. Each type of glowing gas (each element) produces its own unique pattern of lines, so the composition of a gas can be identified by its spectrum. The spectra of sodium, hydrogen", ", calcium, and mercury gases are shown here. Types of Spectra In these experiments, then, there were three different types of spectra. A continuous spectrum (formed when a solid or very dense gas gives off radiation) is an array of all wavelengths or colors of the rainbow. A continuous spectrum can serve as a backdrop from which the atoms of much less dense gas can absorb light. A dark line, or absorption spectrum, consists of a series or pattern of dark lines\u2014missing colors\u2014superimposed upon the continuous spectrum of a source. A bright line, or emission spectrum, appears as a pattern or series of bright lines; it consists of light in which only certain discrete wavelengths are present. (Figure 5.11 shows an absorption spectrum, whereas Figure 5.12 shows the emission spectrum of a number of common elements along with an example of a continuous spectrum.) When we have a hot, thin gas, each particular chemical element or compound produces its own characteristic pattern of spectral lines\u2014its spectral signature. No two types of atoms or molecules give the same patterns. In other words, each particular gas can absorb or emit only certain wavelengths of the light peculiar to that gas. In contrast, absorption spectra occur when passing white light through a cool, thin gas. The temperature and other conditions determine whether the lines are bright or dark (whether light is absorbed or emitted), but the wavelengths of the lines for any element are the same in either case. It is the precise pattern of wavelengths that makes the signature of each element unique. Liquids and solids can also generate spectral lines or bands, but they are broader and less well defined\u2014and hence, more difficult to interpret. Spectral analysis, however, can be quite useful. It can, for example, be applied to light reflected off the surface of a nearby asteroid as well as to light from a distant galaxy. The dark lines in the solar spectrum thus give evidence of certain chemical elements between us and the Sun absorbing those wavelengths of sunlight. Because the space between us and the Sun is pretty empty, astronomers realized that the atoms doing the absorbing must be in a thin atmosphere of cooler gas around the Sun. This outer atmosphere is not all that different from the rest of the Sun, just thinner and cooler. Thus, we can use what we learn about its composition as an indicator of what the whole Sun is made of. Similarly, we This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 165 can use the presence of absorption and emission lines to analyze the composition of other stars and clouds of gas in space. Such analysis of spectra is the key to modern astronomy. Only in this way can we \u201csample\u201d the stars, which are too far away for us to visit. Encoded in the electromagnetic radiation from celestial objects is clear information about the chemical makeup of these objects. Only by understanding what the stars were made of could astronomers begin to form theories about what made them shine and how they evolved. In 1860, German physicist Gustav Kirchhoff became the first person to use spectroscopy to identify an element in the Sun when he found the spectral signature of sodium gas. In the years that followed, astronomers found many other chemical elements in the Sun and stars. In fact, the element helium was found first in the Sun from its spectrum and only later identified on Earth. (The word \u201chelium\u201d comes from helios, the Greek name for the Sun.) Why are there specific lines for each element? The answer to that question was not found until the twentieth century; it required the development of a model for the atom. We therefore turn next to a closer examination of the atoms that make up all matter The Rainbow Rainbows are an excellent illustration of the dispersion of sunlight. You have a good chance of seeing a rainbow any time you are between the Sun and a rain shower, as illustrated in Figure 5.13. The raindrops act like little prisms and break white light into the spectrum of colors. Suppose a ray of sunlight encounters a raindrop and passes into it. The light changes direction\u2014is refracted\u2014when it passes from air to water; the blue and violet light are refracted more than the red. Some of the light is then reflected at the backside of the drop and reemerges from the front, where it is again refracted. As a result, the white light is spread out into a rainbow of colors. Figure 5.13. Rainbow Refraction. (a) This diagram shows how light from the Sun, which is located behind the observer, can be refracted by raindrops to produce (b) a rainbow. (c) Refraction separates white light into its component colors. Note that violet light lies above the red light after it emerges from the raindrop. When you look at a rainbow, however, the red light is higher in the sky. Why? Look again at Figure 5.13. If the observer looks at a raindrop that", " is high in the sky, the violet light passes over her head and the red light enters her eye. Similarly, if the observer looks at a raindrop that is low in the sky, the violet light reaches her eye and the drop appears violet, whereas the red light from that same drop strikes the ground and is not seen. Colors of intermediate wavelengths are refracted to the eye by drops that are intermediate in altitude between 166 Chapter 5 Radiation and Spectra the drops that appear violet and the ones that appear red. Thus, a single rainbow always has red on the outside and violet on the inside. 5.4 THE STRUCTURE OF THE ATOM Learning Objectives By the end of this section, you will be able to: Describe the structure of atoms and the components of nuclei Explain the behavior of electrons within atoms and how electrons interact with light to move among energy levels The idea that matter is composed of tiny particles called atoms is at least 25 centuries old. It took until the twentieth century, however, for scientists to invent instruments that permitted them to probe inside an atom and find that it is not, as had been thought, hard and indivisible. Instead, the atom is a complex structure composed of still smaller particles. Probing the Atom The first of these smaller particles was discovered by British physicist James (J. J.) Thomson in 1897. Named the electron, this particle is negatively charged. (It is the flow of these particles that produces currents of electricity, whether in lightning bolts or in the wires leading to your lamp.) Because an atom in its normal state is electrically neutral, each electron in an atom must be balanced by the same amount of positive charge. The next step was to determine where in the atom the positive and negative charges are located. In 1911, British physicist Ernest Rutherford devised an experiment that provided part of the answer to this question. He bombarded an extremely thin piece of gold foil, only about 400 atoms thick, with a beam of alpha particles (Figure 5.14). Alpha particles (\u03b1 particles) are helium atoms that have lost their electrons and thus are positively charged. Most of these particles passed though the gold foil just as if it and the atoms in it were nearly empty space. About 1 in 8000 of the alpha particles, however, completely reversed direction and bounced backward from the foil. Rutherford wrote, \u201cIt was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.\u201d This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 167 Figure 5.14. Rutherford\u2019s Experiment. (a) When Rutherford allowed \u03b1 particles from a radioactive source to strike a target of gold foil, he found that, although most of them went straight through, some rebounded back in the direction from which they came. (b) From this experiment, he concluded that the atom must be constructed like a miniature solar system, with the positive charge concentrated in the nucleus and the negative charge orbiting in the large volume around the nucleus. Note that this drawing is not to scale; the electron orbits are much larger relative to the size of the nucleus. The only way to account for the particles that reversed direction when they hit the gold foil was to assume that nearly all of the mass, as well as all of the positive charge in each individual gold atom, is concentrated in a tiny center or nucleus. When a positively charged alpha particle strikes a nucleus, it reverses direction, much as a cue ball reverses direction when it strikes another billiard ball. Rutherford\u2019s model placed the other type of charge\u2014the negative electrons\u2014in orbit around this nucleus. Rutherford\u2019s model required that the electrons be in motion. Positive and negative charges attract each other, so stationary electrons would fall into the positive nucleus. Also, because both the electrons and the nucleus are extremely small, most of the atom is empty, which is why nearly all of Rutherford\u2019s particles were able to pass right through the gold foil without colliding with anything. Rutherford\u2019s model was a very successful explanation of the experiments he conducted, although eventually scientists would discover that even the nucleus itself has structure. The Atomic Nucleus The simplest possible atom (and the most common one in the Sun and stars) is hydrogen. The nucleus of ordinary hydrogen contains a single proton. Moving around this proton is a single electron. The mass of an electron is nearly 2000 times smaller than the mass of a proton; the electron carries an amount of charge exactly equal to that of the proton but opposite in sign (Figure 5.15). Opposite charges attract each other, so it is an electromagnetic force that holds the proton and electron together, just as gravity is the force that keeps planets in orbit around the Sun. 168 Chapter 5 Radiation and Spectra Figure 5.15. Hydrogen", " Atom. This is a schematic diagram of a hydrogen atom in its lowest energy state, also called the ground state. The proton and electron have equal but opposite charges, which exert an electromagnetic force that binds the hydrogen atom together. In the illustration, the size of the particles is exaggerated so that you can see them; they are not to scale. They are also shown much closer than they would actually be as it would take more than an entire page to show their actual distance to scale. There are many other types of atoms in nature. Helium, for example, is the second-most abundant element in the Sun. Helium has two protons in its nucleus instead of the single proton that characterizes hydrogen. In addition, the helium nucleus contains two neutrons, particles with a mass comparable to that of the proton but with no electric charge. Moving around this nucleus are two electrons, so the total net charge of the helium atom is also zero (Figure 5.16). Figure 5.16. Helium Atom. Here we see a schematic diagram of a helium atom in its lowest energy state. Two protons are present in the nucleus of all helium atoms. In the most common variety of helium, the nucleus also contains two neutrons, which have nearly the same mass as the proton but carry no charge. Two electrons orbit the nucleus. From this description of hydrogen and helium, perhaps you have guessed the pattern for building up all the elements (different types of atoms) that we find in the universe. The type of element is determined by the number of protons in the nucleus of the atom. For example, any atom with six protons is the element carbon, with eight protons is oxygen, with 26 is iron, and with 92 is uranium. On Earth, a typical atom has the same number of electrons as protons, and these electrons follow complex orbital patterns around the nucleus. Deep inside stars, however, it is so hot that the electrons get loose from the nucleus and (as we shall see) lead separate yet productive lives. The ratio of neutrons to protons increases as the number of protons increases, but each element is unique. The number of neutrons is not necessarily the same for all atoms of a given element. For example, most hydrogen atoms contain no neutrons at all. There are, however, hydrogen atoms that contain one proton and one neutron, and others that contain one proton and two neutrons. The various types of hydrogen nuclei with different numbers of neutrons are called isotopes of hydrogen (Figure 5.17), and all other elements have isotopes as well. You can think of isotopes as siblings in the same element \u201cfamily\u201d\u2014closely related but with different characteristics and behaviors. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 169 Figure 5.17. Isotopes of Hydrogen. A single proton in the nucleus defines the atom to be hydrogen, but there may be zero, one, or two neutrons. The most common isotope of hydrogen is the one with only a single proton and no neutrons To explore the structure of atoms, go to the PhET Build and Atom website (https://openstax.org/l/ 30atombld) where you can add protons, neutrons, or electrons to a model and the name of the element you have created will appear. You can also see the net charge, the mass number, whether it is stable or unstable, and whether it is an ion or a neutral atom. The Bohr Atom Rutherford\u2019s model for atoms has one serious problem. Maxwell\u2019s theory of electromagnetic radiation says that when electrons change either speed or the direction of motion, they must emit energy. Orbiting electrons constantly change their direction of motion, so they should emit a constant stream of energy. Applying Maxwell\u2019s theory to Rutherford\u2019s model, all electrons should spiral into the nucleus of the atom as they lose energy, and this collapse should happen very quickly\u2014in about 10\u201316 seconds. It was Danish physicist Niels Bohr (1885\u20131962) who solved the mystery of how electrons remain in orbit. He was trying to develop a model of the atom that would also explain certain regularities observed in the spectrum of hydrogen. He suggested that the spectrum of hydrogen can be understood if we assume that orbits of only certain sizes are possible for the electron. Bohr further assumed that as long as the electron moves in only one of these allowed orbits, it radiates no energy: its energy would change only if it moved from one orbit to another. This suggestion, in the words of science historian Abraham Pais, was \u201cone of the most audacious hypotheses ever introduced in physics.\u201d If something equivalent were at work in the everyday world, you might find that, as you went for a walk after astronomy class, nature permitted you to walk", " two steps per minute, five steps per minute, and 12 steps per minute, but no speeds in between. No matter how you tried to move your legs, only certain walking speeds would be permitted. To make things more bizarre, it would take no effort to walk at any one of the allowed speeds, but it would be difficult to change from one speed to another. Luckily, no such rules apply at the level of human behavior. But at the microscopic level of the atom, experiment after experiment has confirmed the validity of Bohr\u2019s strange idea. Bohr\u2019s suggestions became one of the foundations of the new (and much more sophisticated) model of the subatomic world called quantum mechanics. In Bohr\u2019s model, if the electron moves from one orbit to another closer to the atomic nucleus, it must give up some energy in the form of electromagnetic radiation. If the electron goes from an inner orbit to one farther from the nucleus, however, it requires some additional energy. One way to obtain the necessary energy is to absorb electromagnetic radiation that may be streaming past the atom from an outside source. 170 Chapter 5 Radiation and Spectra A key feature of Bohr\u2019s model is that each of the permitted electron orbits around a given atom has a certain energy value; we therefore can think of each orbit as an energy level. To move from one orbit to another (which will have its own specific energy value) requires a change in the electron\u2019s energy\u2014a change determined by the difference between the two energy values. If the electron goes to a lower level, the energy difference will be given off; if the electron goes to a higher level, the energy difference must be obtained from somewhere else. Each jump (or transition) to a different level has a fixed and definite energy change associated with it. A crude analogy for this situation might be life in a tower of luxury apartments where the rent is determined by the quality of the view. Such a building has certain, definite numbered levels or floors on which apartments are located. No one can live on floor 5.37 or 22.5. In addition, the rent gets higher as you go up to higher floors. If you want to exchange an apartment on the twentieth floor for one on the second floor, you will not owe as much rent. However, if you want to move from the third floor to the twenty-fifth floor, your rent will increase. In an atom, too, the \u201ccheapest\u201d place for an electron to live is the lowest possible level, and energy is required to move to a higher level. Here we have one of the situations where it is easier to think of electromagnetic radiation as particles (photons) rather than as waves. As electrons move from one level to another, they give off or absorb little packets of energy. When an electron moves to a higher level, it absorbs a photon of just the right energy (provided one is available). When it moves to a lower level, it emits a photon with the exact amount of energy it no longer needs in its \u201clower-cost living situation.\u201d The photon and wave perspectives must be equivalent: light is light, no matter how we look at it. Thus, each photon carries a certain amount of energy that is proportional to the frequency (f) of the wave it represents. The value of its energy (E) is given by the formula E = h f where the constant of proportionality, h, is called Planck\u2019s constant. The constant is named for Max Planck, the German physicist who was one of the originators of the quantum theory (Figure 5.18). If metric units are used (that is, if energy is measured in joules and frequency in hertz), then Planck\u2019s constant has the value h = 6.626 \u00d7 10\u201334 joule-seconds (J-s). Higher-energy photons correspond to higher-frequency waves (which have a shorter wavelength); lower-energy photons are waves of lower frequency. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 171 Figure 5.18. Niels Bohr (1885\u20131962) and Max Planck (1858\u20131947). (a) Bohr, shown at his desk in this 1935 photograph, and (b) Planck helped us understand the energy behavior of photons. To take a specific example, consider a calcium atom inside the Sun\u2019s atmosphere in which an electron jumps from a lower level to a higher level. To do this, it needs about 5 \u00d7 10\u201319 joules of energy, which it can conveniently obtain by absorbing a passing photon of that energy coming from deeper inside the Sun. This photon is equivalent to a wave of light whose frequency is about 7.5 \u00d7 1014 hertz and whose wavelength is about 3.9 \u00d7 10\u20137 meters (393 nanometers), in the deep violet", " part of the visible light spectrum. Although it may seem strange at first to switch from picturing light as a photon (or energy packet) to picturing it as a wave, such switching has become second nature to astronomers and can be a handy tool for doing calculations about spectra The Energy of a Photon Now that we know how to calculate the wavelength and frequency of a photon, we can use this information, along with Planck\u2019s constant, to determine how much energy each photon carries. How much energy does a red photon of wavelength 630 nm have? Solution First, as we learned earlier, we can find the frequency of the photon: f = c \u03bb = 3 \u00d7 108 m/s 630 \u00d7 10\u20139 m = 4.8 \u00d7 1014 Hz Next, we can use Planck\u2019s constant to determine the energy (remember that a Hz is the same as 1/s):.2 \u00d7 10\u201319 J \u239d6.626 \u00d7 10\u201334 J-s \u239d4.8 \u00d7 1014 (1/s) \u23a0 Check Your Learning What is the energy of a yellow photon with a frequency of 5.5 \u00d7 1014 Hz? Chapter 5 Radiation and Spectra 172 Answer: E = h f = \u239d6.626 \u00d7 10\u201334\u239e \u239b \u239d5.5 \u00d7 1014\u239e \u239b \u23a0 \u23a0 = 3.6 \u00d7 10\u201319 J 5.5 FORMATION OF SPECTRAL LINES Learning Objectives By the end of this section, you will be able to: Explain how emission line spectra and absorption line spectra are formed Describe what ions are and how they are formed Explain how spectral lines and ionization levels in a gas can help us determine its temperature We can use Bohr\u2019s model of the atom to understand how spectral lines are formed. The concept of energy levels for the electron orbits in an atom leads naturally to an explanation of why atoms absorb or emit only specific energies or wavelengths of light. The Hydrogen Spectrum Let\u2019s look at the hydrogen atom from the perspective of the Bohr model. Suppose a beam of white light (which consists of photons of all visible wavelengths) shines through a gas of atomic hydrogen. A photon of wavelength 656 nanometers has just the right energy to raise an electron in a hydrogen atom from the second to the third orbit. Thus, as all the photons of different energies (or wavelengths or colors) stream by the hydrogen atoms, photons with this particular wavelength can be absorbed by those atoms whose electrons are orbiting on the second level. When they are absorbed, the electrons on the second level will move to the third level, and a number of the photons of this wavelength and energy will be missing from the general stream of white light. Other photons will have the right energies to raise electrons from the second to the fourth orbit, or from the first to the fifth orbit, and so on. Only photons with these exact energies can be absorbed. All of the other photons will stream past the atoms untouched. Thus, hydrogen atoms absorb light at only certain wavelengths and produce dark lines at those wavelengths in the spectrum we see. Suppose we have a container of hydrogen gas through which a whole series of photons is passing, allowing many electrons to move up to higher levels. When we turn off the light source, these electrons \u201cfall\u201d back down from larger to smaller orbits and emit photons of light\u2014but, again, only light of those energies or wavelengths that correspond to the energy difference between permissible orbits. The orbital changes of hydrogen electrons that give rise to some spectral lines are shown in Figure 5.19. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 173 Figure 5.19. Bohr Model for Hydrogen. In this simplified model of a hydrogen atom, the concentric circles shown represent permitted orbits or energy levels. An electron in a hydrogen atom can only exist in one of these energy levels (or states). The closer the electron is to the nucleus, the more tightly bound the electron is to the nucleus. By absorbing energy, the electron can move to energy levels farther from the nucleus (and even escape if enough energy is absorbed). Similar pictures can be drawn for atoms other than hydrogen. However, because these other atoms ordinarily have more than one electron each, the orbits of their electrons are much more complicated, and the spectra are more complex as well. For our purposes, the key conclusion is this: each type of atom has its own unique pattern of electron orbits, and no two sets of orbits are exactly alike. This means that each type of atom shows its own unique set of spectral lines, produced by electrons moving between its unique set of orbits. Astronomers and physicists have worked hard to learn the lines that go with each element by studying the way atoms absorb and emit light in laboratories here on Earth. Then they", " can use this knowledge to identify the elements in celestial bodies. In this way, we now know the chemical makeup of not just any star, but even galaxies of stars so distant that their light started on its way to us long before Earth had even formed. Energy Levels and Excitation Bohr\u2019s model of the hydrogen atom was a great step forward in our understanding of the atom. However, we know today that atoms cannot be represented by quite so simple a picture. For example, the concept of sharply defined electron orbits is not really correct; however, at the level of this introductory course, the notion that only certain discrete energies are allowable for an atom is very useful. The energy levels we have been discussing can be thought of as representing certain average distances of the electron\u2019s possible orbits from the atomic nucleus. Ordinarily, an atom is in the state of lowest possible energy, its ground state. In the Bohr model of the hydrogen atom, the ground state corresponds to the electron being in the innermost orbit. An atom can absorb energy, which raises it to a higher energy level (corresponding, in the simple Bohr picture, to an electron\u2019s movement to a larger orbit)\u2014this is referred to as excitation. The atom is then said to be in an excited state. Generally, an atom remains excited for only a very brief time. After a short interval, typically a hundred-millionth of a second or so, it drops back spontaneously to its ground state, with the simultaneous emission of light. The atom may return to its lowest state in one jump, or it may make the transition in steps of two or more jumps, stopping at intermediate levels on the way down. With each jump, it emits a photon of the wavelength that corresponds to the energy difference between the levels at the beginning and end of that jump. An energy-level diagram for a hydrogen atom and several possible atomic transitions are shown in Figure 5.20. When we measure the energies involved as the atom jumps between levels, we find that the transitions to or from the ground state, called the Lyman series of lines, result in the emission or absorption of ultraviolet 174 Chapter 5 Radiation and Spectra photons. But the transitions to or from the first excited state (labeled n = 2 in part (a) of Figure 5.20), called the Balmer series, produce emission or absorption in visible light. In fact, it was to explain this Balmer series that Bohr first suggested his model of the atom. Figure 5.20. Energy-Level Diagrams for Hydrogen. (a) Here we follow the emission or absorption of photons by a hydrogen atom according to the Bohr model. Several different series of spectral lines are shown, corresponding to transitions of electrons from or to certain allowed orbits. Each series of lines that terminates on a specific inner orbit is named for the physicist who studied it. At the top, for example, you see the Balmer series, and arrows show electrons jumping from the second orbit (n = 2) to the third, fourth, fifth, and sixth orbits. Each time a \u201cpoor\u201d electron from a lower level wants to rise to a higher position in life, it must absorb energy to do so. It can absorb the energy it needs from passing waves (or photons) of light. The next set of arrows (Lyman series) show electrons falling down to the first orbit from different (higher) levels. Each time a \u201crich\u201d electron goes downward toward the nucleus, it can afford to give off (emit) some energy it no longer needs. (b) At higher and higher energy levels, the levels become more and more crowded together, approaching a limit. The region above the top line represents energies at which the atom is ionized (the electron is no longer attached to the atom). Each series of arrows represents electrons falling from higher levels to lower ones, releasing photons or waves of energy in the process. Atoms that have absorbed specific photons from a passing beam of white light and have thus become excited generally de-excite themselves and emit that light again in a very short time. You might wonder, then, why dark spectral lines are ever produced. In other words, why doesn\u2019t this reemitted light quickly \u201cfill in\u201d the darker absorption lines? Imagine a beam of white light coming toward you through some cooler gas. Some of the reemitted light is actually returned to the beam of white light you see, but this fills in the absorption lines only to a slight extent. The reason is that the atoms in the gas reemit light in all directions, and only a small fraction of the reemitted light is in the direction of the original beam (toward you). In a star, much of the reemitted light actually goes in directions leading back into the star, which does observers outside the star no good whatsoever. Figure 5.21 summarizes the different kinds of spectra we have discussed. An incandescent", " lightbulb produces a continuous spectrum. When that continuous spectrum is viewed through a thinner cloud of gas, an absorption line spectrum can be seen superimposed on the continuous spectrum. If we look only at a cloud of excited gas atoms (with no continuous source seen behind it), we see that the excited atoms give off an emission line spectrum. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 175 Figure 5.21. Three Kinds of Spectra. When we see a lightbulb or other source of continuous radiation, all the colors are present. When the continuous spectrum is seen through a thinner gas cloud, the cloud\u2019s atoms produce absorption lines in the continuous spectrum. When the excited cloud is seen without the continuous source behind it, its atoms produce emission lines. We can learn which types of atoms are in the gas cloud from the pattern of absorption or emission lines. Atoms in a hot gas are moving at high speeds and continually colliding with one another and with any loose electrons. They can be excited (electrons moving to a higher level) and de-excited (electrons moving to a lower level) by these collisions as well as by absorbing and emitting light. The speed of atoms in a gas depends on the temperature. When the temperature is higher, so are the speed and energy of the collisions. The hotter the gas, therefore, the more likely that electrons will occupy the outermost orbits, which correspond to the highest energy levels. This means that the level where electrons start their upward jumps in a gas can serve as an indicator of how hot that gas is. In this way, the absorption lines in a spectrum give astronomers information about the temperature of the regions where the lines originate Use this simulation (https://openstax.org/l/30Hatom) to play with a hydrogen atom and see what happens when electrons move to higher levels and then give off photons as they go to a lower level. Ionization We have described how certain discrete amounts of energy can be absorbed by an atom, raising it to an excited state and moving one of its electrons farther from its nucleus. If enough energy is absorbed, the electron can be completely removed from the atom\u2014this is called ionization. The atom is then said to be ionized. The minimum amount of energy required to remove one electron from an atom in its ground state is called its ionization energy. Still-greater amounts of energy must be absorbed by the now-ionized atom (called an ion) to remove an additional electron deeper in the structure of the atom. Successively greater energies are needed to remove the third, fourth, fifth\u2014and so on\u2014electrons from the atom. If enough energy is available, an atom can become completely ionized, losing all of its electrons. A hydrogen atom, having only one electron to lose, can be ionized 176 Chapter 5 Radiation and Spectra only once; a helium atom can be ionized twice; and an oxygen atom up to eight times. When we examine regions of the cosmos where there is a great deal of energetic radiation, such as the neighborhoods where hot young stars have recently formed, we see a lot of ionization going on. An atom that has become positively ionized has lost a negative charge\u2014the missing electron\u2014and thus is left with a net positive charge. It therefore exerts a strong attraction on any free electron. Eventually, one or more electrons will be captured and the atom will become neutral (or ionized to one less degree) again. During the electron-capture process, the atom emits one or more photons. Which photons are emitted depends on whether the electron is captured at once to the lowest energy level of the atom or stops at one or more intermediate levels on its way to the lowest available level. Just as the excitation of an atom can result from a collision with another atom, ion, or electron (collisions with electrons are usually most important), so can ionization. The rate at which such collisional ionizations occur depends on the speeds of the atoms and hence on the temperature of the gas\u2014the hotter the gas, the more of its atoms will be ionized. The rate at which ions and electrons recombine also depends on their relative speeds\u2014that is, on the temperature. In addition, it depends on the density of the gas: the higher the density, the greater the chance for recapture, because the different kinds of particles are crowded more closely together. From a knowledge of the temperature and density of a gas, it is possible to calculate the fraction of atoms that have been ionized once, twice, and so on. In the Sun, for example, we find that most of the hydrogen and helium atoms in its atmosphere are neutral, whereas most of the calcium atoms, as well as many other heavier atoms, are ionized once. The energy levels of an ionized atom are entirely different from those of the same atom when it is neutral. Each", " time an electron is removed from the atom, the energy levels of the ion, and thus the wavelengths of the spectral lines it can produce, change. This helps astronomers differentiate the ions of a given element. Ionized hydrogen, having no electron, can produce no absorption lines. 5.6 THE DOPPLER EFFECT Learning Objectives By the end of this section, you will be able to: Explain why the spectral lines of photons we observe from an object will change as a result of the object\u2019s motion toward or away from us Describe how we can use the Doppler effect to deduce how astronomical objects are moving through space The last two sections introduced you to many new concepts, and we hope that through those, you have seen one major idea emerge. Astronomers can learn about the elements in stars and galaxies by decoding the information in their spectral lines. There is a complicating factor in learning how to decode the message of starlight, however. If a star is moving toward or away from us, its lines will be in a slightly different place in the spectrum from where they would be in a star at rest. And most objects in the universe do have some motion relative to the Sun. Motion Affects Waves In 1842, Christian Doppler first measured the effect of motion on waves by hiring a group of musicians to play on an open railroad car as it was moving along the track. He then applied what he learned to all waves, including This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 177 light, and pointed out that if a light source is approaching or receding from the observer, the light waves will be, respectively, crowded more closely together or spread out. The general principle, now known as the Doppler effect, is illustrated in Figure 5.22. Figure 5.22. Doppler Effect. (a) A source, S, makes waves whose numbered crests (1, 2, 3, and 4) wash over a stationary observer. (b) The source S now moves toward observer A and away from observer C. Wave crest 1 was emitted when the source was at position S4, crest 2 at position S2, and so forth. Observer A sees waves compressed by this motion and sees a blueshift (if the waves are light). Observer C sees the waves stretched out by the motion and sees a redshift. Observer B, whose line of sight is perpendicular to the source\u2019s motion, sees no change in the waves (and feels left out). In part (a) of the figure, the light source (S) is at rest with respect to the observer. The source gives off a series of waves, whose crests we have labeled 1, 2, 3, and 4. The light waves spread out evenly in all directions, like the ripples from a splash in a pond. The crests are separated by a distance, \u03bb, where \u03bb is the wavelength. The observer, who happens to be located in the direction of the bottom of the image, sees the light waves coming nice and evenly, one wavelength apart. Observers located anywhere else would see the same thing. On the other hand, if the source of light is moving with respect to the observer, as seen in part (b), the situation is more complicated. Between the time one crest is emitted and the next one is ready to come out, the source has moved a bit, toward the bottom of the page. From the point of view of observer A, this motion of the source has decreased the distance between crests\u2014it\u2019s squeezing the crests together, this observer might say. In part (b), we show the situation from the perspective of three observers. The source is seen in four positions, S1, S2, S3, and S4, each corresponding to the emission of one wave crest. To observer A, the waves seem to follow one another more closely, at a decreased wavelength and thus increased frequency. (Remember, all light waves travel at the speed of light through empty space, no matter what. This means that motion cannot affect the speed, but only the wavelength and the frequency. As the wavelength decreases, the frequency must increase. If the waves are shorter, more will be able to move by during each second.) The situation is not the same for other observers. Let\u2019s look at the situation from the point of view of observer C, located opposite observer A in the figure. For her, the source is moving away from her location. As a result, the waves are not squeezed together but instead are spread out by the motion of the source. The crests arrive with an increased wavelength and decreased frequency. To observer B, in a direction at right angles to the motion 178 Chapter 5 Radiation and Spectra of the source, no effect is observed. The wavelength and frequency remain the same as they were in part (a) of", " the figure. We can see from this illustration that the Doppler effect is produced only by a motion toward or away from the observer, a motion called radial velocity. Sideways motion does not produce such an effect. Observers between A and B would observe some shortening of the light waves for that part of the motion of the source that is along their line of sight. Observers between B and C would observe lengthening of the light waves that are along their line of sight. You may have heard the Doppler effect with sound waves. When a train whistle or police siren approaches you and then moves away, you will notice a decrease in the pitch (which is how human senses interpret sound wave frequency) of the sound waves. Compared to the waves at rest, they have changed from slightly more frequent when coming toward you, to slightly less frequent when moving away from you nice example of this change in the sound of a train whistle can be heard at the end of the classic Beach Boys song \u201cCaroline, No\u201d on their album Pet Sounds. To hear this sound, go to this YouTube (https://openstax.org/l/30BBtrain) version of the song. The sound of the train begins at approximately 2:20. Color Shifts When the source of waves moves toward you, the wavelength decreases a bit. If the waves involved are visible light, then the colors of the light change slightly. As wavelength decreases, they shift toward the blue end of the spectrum: astronomers call this a blueshift (since the end of the spectrum is really violet, the term should probably be violetshift, but blue is a more common color). When the source moves away from you and the wavelength gets longer, we call the change in colors a redshift. Because the Doppler effect was first used with visible light in astronomy, the terms \u201c blueshift\u201d and \u201c redshift\u201d became well established. Today, astronomers use these words to describe changes in the wavelengths of radio waves or X-rays as comfortably as they use them to describe changes in visible light. The greater the motion toward or away from us, the greater the Doppler shift. If the relative motion is entirely along the line of sight, the formula for the Doppler shift of light is = v c \u0394\u03bb \u03bb where \u03bb is the wavelength emitted by the source, \u0394\u03bb is the difference between \u03bb and the wavelength measured by the observer, c is the speed of light, and v is the relative speed of the observer and the source in the line of sight. The variable v is counted as positive if the velocity is one of recession, and negative if it is one of approach. Solving this equation for the velocity, we find v = c \u00d7 \u0394\u03bb/\u03bb. If a star approaches or recedes from us, the wavelengths of light in its continuous spectrum appear shortened or lengthened, respectively, as do those of the dark lines. However, unless its speed is tens of thousands of kilometers per second, the star does not appear noticeably bluer or redder than normal. The Doppler shift is thus not easily detected in a continuous spectrum and cannot be measured accurately in such a spectrum. The wavelengths of the absorption lines can be measured accurately, however, and their Doppler shift is relatively simple to detect. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 179 The Doppler Effect We can use the Doppler effect equation to calculate the radial velocity of an object if we know three things: the speed of light, the original (unshifted) wavelength of the light emitted, and the difference between the wavelength of the emitted light and the wavelength we observe. For particular absorption or emission lines, we usually know exactly what wavelength the line has in our laboratories on Earth, where the source of light is not moving. We can measure the new wavelength with our instruments at the telescope, and so we know the difference in wavelength due to Doppler shifting. Since the speed of light is a universal constant, we can then calculate the radial velocity of the star. A particular emission line of hydrogen is originally emitted with a wavelength of 656.3 nm from a gas cloud. At our telescope, we observe the wavelength of the emission line to be 656.6 nm. How fast is this gas cloud moving toward or away from Earth? Solution Because the light is shifted to a longer wavelength (redshifted), we know this gas cloud is moving away from us. The speed can be calculated using the Doppler shift formula: \u03bd = c \u00d7 \u0394\u03bb \u03bb = \u239b \u239e \u239b \u239d3.0 \u00d7 108 m/s \u23a0 \u239d 0.3 nm 656.3 nm \u239e \u23a0 = \u239b 0.3 \u00d7 10\u20139 m \u239e", " \u239b \u239d3.0 \u00d7 108 m/s \u23a0 \u239d 656.3 \u00d7 10\u20139 m \u239e \u23a0 = 140,000 m/s = 140 km/s Check Your Learning Suppose a spectral line of hydrogen, normally at 500 nm, is observed in the spectrum of a star to be at 500.1 nm. How fast is the star moving toward or away from Earth? Answer: Because the light is shifted to a longer wavelength, the star is moving away from us: \u03bd = c \u00d7 \u0394\u03bb \u03bb = \u239b \u239e \u239b \u239d3.0 \u00d7 108 m/s \u23a0 \u239d 0.1 nm 500 nm \u239e \u23a0 = \u239b 0.1 \u00d7 10\u20139 m \u239e \u239b \u239d3.0 \u00d7 108 m/s \u23a0 \u239d 500 \u00d7 10\u20139 m \u239e \u23a0 = 60,000 m/s. Its speed is 60,000 m/s. You may now be asking: if all the stars are moving and motion changes the wavelength of each spectral line, won\u2019t this be a disaster for astronomers trying to figure out what elements are present in the stars? After all, it is the precise wavelength (or color) that tells astronomers which lines belong to which element. And we first measure these wavelengths in containers of gas in our laboratories, which are not moving. If every line in a star\u2019s spectrum is now shifted by its motion to a different wavelength (color), how can we be sure which lines and which elements we are looking at in a star whose speed we do not know? Take heart. This situation sounds worse than it really is. Astronomers rarely judge the presence of an element in an astronomical object by a single line. It is the pattern of lines unique to hydrogen or calcium that enables us to determine that those elements are part of the star or galaxy we are observing. The Doppler effect does not change the pattern of lines from a given element\u2014it only shifts the whole pattern slightly toward redder 180 Chapter 5 Radiation and Spectra or bluer wavelengths. The shifted pattern is still quite easy to recognize. Best of all, when we do recognize a familiar element\u2019s pattern, we get a bonus: the amount the pattern is shifted can enable us to determine the speed of the objects in our line of sight. The training of astronomers includes much work on learning to decode light (and other electromagnetic radiation). A skillful \u201cdecoder\u201d can learn the temperature of a star, what elements are in it, and even its speed in a direction toward us or away from us. That\u2019s really an impressive amount of information for stars that are light-years away. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 181 CHAPTER 5 REVIEW KEY TERMS absorption spectrum a series or pattern of dark lines superimposed on a continuous spectrum blackbody an idealized object that absorbs all electromagnetic energy that falls onto it continuous spectrum a spectrum of light composed of radiation of a continuous range of wavelengths or colors, rather than only certain discrete wavelengths dispersion separation of different wavelengths of white light through refraction of different amounts Doppler effect the apparent change in wavelength or frequency of the radiation from a source due to its relative motion away from or toward the observer electromagnetic radiation radiation consisting of waves propagated through regularly varying electric and magnetic fields and traveling at the speed of light electromagnetic spectrum the whole array or family of electromagnetic waves, from radio to gamma rays emission spectrum a series or pattern of bright lines superimposed on a continuous spectrum energy flux the amount of energy passing through a unit area (for example, 1 square meter) per second; the units of flux are watts per square meter energy level a particular level, or amount, of energy possessed by an atom or ion above the energy it possesses in its least energetic state; also used to refer to the states of energy an electron can have in an atom excitation the process of giving an atom or an ion an amount of energy greater than it has in its lowest energy (ground) state frequency the number of waves that cross a given point per unit time (in radiation) gamma rays photons (of electromagnetic radiation) of energy with wavelengths no longer than 0.01 nanometer; the most energetic form of electromagnetic radiation ground state the lowest energy state of an atom infrared electromagnetic radiation of wavelength 103\u2013106 nanometers; longer than the longest (red) wavelengths that can be perceived by the eye, but shorter than radio wavelengths inverse square law (for light) the amount of energy (light) flowing through a given area in a given time decreases in proportion to the square of the distance from the source of energy or light ion an atom that has become electrically charged by the addition or loss of one or more electrons ionization the process by", " which an atom gains or loses electrons isotope any of two or more forms of the same element whose atoms have the same number of protons but different numbers of neutrons microwave electromagnetic radiation of wavelengths from 1 millimeter to 1 meter; longer than infrared but shorter than radio waves 182 Chapter 5 Radiation and Spectra nucleus (of an atom) the massive part of an atom, composed mostly of protons and neutrons, and about which the electrons revolve photon a discrete unit (or \u201cpacket\u201d) of electromagnetic energy radial velocity motion toward or away from the observer; the component of relative velocity that lies in the line of sight radio waves all electromagnetic waves longer than microwaves, including radar waves and AM radio waves spectrometer an instrument for obtaining a spectrum; in astronomy, usually attached to a telescope to record the spectrum of a star, galaxy, or other astronomical object Stefan-Boltzmann law a formula from which the rate at which a blackbody radiates energy can be computed; the total rate of energy emission from a unit area of a blackbody is proportional to the fourth power of its absolute temperature: F = \u03c3T4 ultraviolet electromagnetic radiation of wavelengths 10 to 400 nanometers; shorter than the shortest visible wavelengths visible light electromagnetic radiation with wavelengths of roughly 400\u2013700 nanometers; visible to the human eye wavelength the distance from crest to crest or trough to trough in a wave Wien\u2019s law formula that relates the temperature of a blackbody to the wavelength at which it emits the greatest intensity of radiation X-rays electromagnetic radiation with wavelengths between 0.01 nanometer and 20 nanometers; intermediate between those of ultraviolet radiation and gamma rays SUMMARY 5.1 The Behavior of Light James Clerk Maxwell showed that whenever charged particles change their motion, as they do in every atom and molecule, they give off waves of energy. Light is one form of this electromagnetic radiation. The wavelength of light determines the color of visible radiation. Wavelength (\u03bb) is related to frequency (f) and the speed of light (c) by the equation c = \u03bbf. Electromagnetic radiation sometimes behaves like waves, but at other times, it behaves as if it were a particle\u2014a little packet of energy, called a photon. The apparent brightness of a source of electromagnetic energy decreases with increasing distance from that source in proportion to the square of the distance\u2014a relationship known as the inverse square law. 5.2 The Electromagnetic Spectrum The electromagnetic spectrum consists of gamma rays, X-rays, ultraviolet radiation, visible light, infrared, and radio radiation. Many of these wavelengths cannot penetrate the layers of Earth\u2019s atmosphere and must be observed from space, whereas others\u2014such as visible light, FM radio and TV\u2014can penetrate to Earth\u2019s surface. The emission of electromagnetic radiation is intimately connected to the temperature of the source. The higher the temperature of an idealized emitter of electromagnetic radiation, the shorter is the wavelength at which the maximum amount of radiation is emitted. The mathematical equation describing this relationship is known as Wien\u2019s law: \u03bbmax = (3 \u00d7 106)/T. The total power emitted per square meter increases with increasing temperature. The relationship between emitted energy flux and temperature is known as the Stefan-Boltzmann law: F = \u03c3T4. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 183 5.3 Spectroscopy in Astronomy A spectrometer is a device that forms a spectrum, often utilizing the phenomenon of dispersion. The light from an astronomical source can consist of a continuous spectrum, an emission (bright line) spectrum, or an absorption (dark line) spectrum. Because each element leaves its spectral signature in the pattern of lines we observe, spectral analyses reveal the composition of the Sun and stars. 5.4 The Structure of the Atom Atoms consist of a nucleus containing one or more positively charged protons. All atoms except hydrogen can also contain one or more neutrons in the nucleus. Negatively charged electrons orbit the nucleus. The number of protons defines an element (hydrogen has one proton, helium has two, and so on) of the atom. Nuclei with the same number of protons but different numbers of neutrons are different isotopes of the same element. In the Bohr model of the atom, electrons on permitted orbits (or energy levels) don\u2019t give off any electromagnetic radiation. But when electrons go from lower levels to higher ones, they must absorb a photon of just the right energy, and when they go from higher levels to lower ones, they give off a photon of just the right energy. The energy of a photon is connected to the frequency of the electromagnetic wave it represents by Planck\u2019s formula, E = hf. 5.5 Formation of Spectral Lines When electrons move from a higher energy level to a lower one, photons are emitted,", " and an emission line can be seen in the spectrum. Absorption lines are seen when electrons absorb photons and move to higher energy levels. Since each atom has its own characteristic set of energy levels, each is associated with a unique pattern of spectral lines. This allows astronomers to determine what elements are present in the stars and in the clouds of gas and dust among the stars. An atom in its lowest energy level is in the ground state. If an electron is in an orbit other than the least energetic one possible, the atom is said to be excited. If an atom has lost one or more electrons, it is called an ion and is said to be ionized. The spectra of different ions look different and can tell astronomers about the temperatures of the sources they are observing. 5.6 The Doppler Effect If an atom is moving toward us when an electron changes orbits and produces a spectral line, we see that line shifted slightly toward the blue of its normal wavelength in a spectrum. If the atom is moving away, we see the line shifted toward the red. This shift is known as the Doppler effect and can be used to measure the radial velocities of distant objects. FOR FURTHER EXPLORATION Articles Augensen, H. & Woodbury, J. \u201cThe Electromagnetic Spectrum.\u201d Astronomy (June 1982): 6. Darling, D. \u201cSpectral Visions: The Long Wavelengths.\u201d Astronomy (August 1984): 16; \u201cThe Short Wavelengths.\u201d Astronomy (September 1984): 14. Gingerich, O. \u201cUnlocking the Chemical Secrets of the Cosmos.\u201d Sky & Telescope (July 1981): 13. Stencil, R. et al. \u201cAstronomical Spectroscopy.\u201d Astronomy (June 1978): 6. 184 Websites Chapter 5 Radiation and Spectra Doppler Effect: http://www.physicsclassroom.com/class/waves/Lesson-3/The-Doppler-Effect. A shaking bug and the Doppler Effect explained. Electromagnetic Spectrum: http://imagine.gsfc.nasa.gov/science/toolbox/emspectrum1.html. An introduction to the electromagnetic spectrum from NASA\u2019s Imagine the Universe; note that you can click the \u201cAdvanced\u201d button near the top and get a more detailed discussion. Rainbows: How They Form and How to See Them: http://www.livescience.com/30235-rainbows-formation- explainer.html. By meteorologist and amateur astronomer Joe Rao. Videos Doppler Effect: http://www.esa.int/spaceinvideos/Videos/2014/07/Doppler_effect_- _classroom_demonstration_video_VP05. ESA video with Doppler ball demonstration and Doppler effect and satellites (4:48). How a Prism Works to Make Rainbow Colors: https://www.youtube.com/watch?v=JGqsi_LDUn0. Short video on how a prism bends light to make a rainbow of colors (2:44). Tour of the Electromagnetic Spectrum: https://www.youtube.com/watch?v=HPcAWNlVl-8. NASA Mission Science video tour of the bands of the electromagnetic spectrum (eight short videos). Introductions To Quantum Mechanics Ford, Kenneth. The Quantum World. 2004. A well-written recent introduction by a physicist/educator. Gribbin, John. In Search of Schroedinger\u2019s Cat. 1984. Clear, very basic introduction to the fundamental ideas of quantum mechanics, by a British physicist and science writer. Rae, Alastair. Quantum Physics: A Beginner\u2019s Guide. 2005. Widely praised introduction by a British physicist. COLLABORATIVE GROUP ACTIVITIES A. Have your group make a list of all the electromagnetic wave technology you use during a typical day. B. How many applications of the Doppler effect can your group think of in everyday life? For example, why would the highway patrol find it useful? C. Have members of your group go home and \u201cread\u201d the face of your radio set and then compare notes. If you do not have a radio, research \u201cbroadcast radio frequencies\u201d to find answers to the following questions. What do all the words and symbols mean? What frequencies can your radio tune to? What is the frequency of your favorite radio station? What is its wavelength? D. If your instructor were to give you a spectrometer, what kind of spectra does your group think you would see from each of the following: (1) a household lightbulb, (2) the Sun, (3) the \u201cneon lights of Broadway,\u201d (4) an ordinary household flashlight, and (5) a streetlight on a busy shopping street", "? E. Suppose astronomers want to send a message to an alien civilization that is living on a planet with an atmosphere very similar to that of Earth\u2019s. This message must travel through space, make it through the other planet\u2019s atmosphere, and be noticeable to the residents of that planet. Have your group discuss This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 185 what band of the electromagnetic spectrum might be best for this message and why. (Some people, including noted physicist Stephen Hawking, have warned scientists not to send such messages and reveal the presence of our civilization to a possible hostile cosmos. Do you agree with this concern?) EXERCISES Review Questions 1. What distinguishes one type of electromagnetic radiation from another? What are the main categories (or bands) of the electromagnetic spectrum? 2. What is a wave? Use the terms wavelength and frequency in your definition. 3. Is your textbook the kind of idealized object (described in section on radiation laws) that absorbs all the radiation falling on it? Explain. How about the black sweater worn by one of your classmates? 4. Where in an atom would you expect to find electrons? Protons? Neutrons? 5. Explain how emission lines and absorption lines are formed. In what sorts of cosmic objects would you expect to see each? 6. Explain how the Doppler effect works for sound waves and give some familiar examples. 7. What kind of motion for a star does not produce a Doppler effect? Explain. 8. Describe how Bohr\u2019s model used the work of Maxwell. 9. Explain why light is referred to as electromagnetic radiation. 10. Explain the difference between radiation as it is used in most everyday language and radiation as it is used in an astronomical context. 11. What are the differences between light waves and sound waves? 12. Which type of wave has a longer wavelength: AM radio waves (with frequencies in the kilohertz range) or FM radio waves (with frequencies in the megahertz range)? Explain. 13. Explain why astronomers long ago believed that space must be filled with some kind of substance (the \u201caether\u201d) instead of the vacuum we know it is today. 14. Explain what the ionosphere is and how it interacts with some radio waves. 15. Which is more dangerous to living things, gamma rays or X-rays? Explain. 16. Explain why we have to observe stars and other astronomical objects from above Earth\u2019s atmosphere in order to fully learn about their properties. 17. Explain why hotter objects tend to radiate more energetic photons compared to cooler objects. 18. Explain how we can deduce the temperature of a star by determining its color. 19. Explain what dispersion is and how astronomers use this phenomenon to study a star\u2019s light. 20. Explain why glass prisms disperse light. 21. Explain what Joseph Fraunhofer discovered about stellar spectra. 186 Chapter 5 Radiation and Spectra 22. Explain how we use spectral absorption and emission lines to determine the composition of a gas. 23. Explain the results of Rutherford\u2019s gold foil experiment and how they changed our model of the atom. 24. Is it possible for two different atoms of carbon to have different numbers of neutrons in their nuclei? Explain. 25. What are the three isotopes of hydrogen, and how do they differ? 26. Explain how electrons use light energy to move among energy levels within an atom. 27. Explain why astronomers use the term \u201cblueshifted\u201d for objects moving toward us and \u201credshifted\u201d for objects moving away from us. 28. If spectral line wavelengths are changing for objects based on the radial velocities of those objects, how can we deduce which type of atom is responsible for a particular absorption or emission line? Thought Questions 29. Make a list of some of the many practical consequences of Maxwell\u2019s theory of electromagnetic waves (television is one example). 30. With what type of electromagnetic radiation would you observe: A. A star with a temperature of 5800 K? B. A gas heated to a temperature of one million K? C. A person on a dark night? 31. Why is it dangerous to be exposed to X-rays but not (or at least much less) dangerous to be exposed to radio waves? 32. Go outside on a clear night, wait 15 minutes for your eyes to adjust to the dark, and look carefully at the brightest stars. Some should look slightly red and others slightly blue. The primary factor that determines the color of a star is its temperature. Which is hotter: a blue star or a red one? Explain 33. Water faucets are often labeled with a red dot for hot water and a blue dot for cold. Given Wien\u2019s law, does this labeling make sense? 34. Suppose you are standing at the exact", " center of a park surrounded by a circular road. An ambulance drives completely around this road, with siren blaring. How does the pitch of the siren change as it circles around you? 35. How could you measure Earth\u2019s orbital speed by photographing the spectrum of a star at various times throughout the year? (Hint: Suppose the star lies in the plane of Earth\u2019s orbit.) 36. Astronomers want to make maps of the sky showing sources of X-rays or gamma rays. Explain why those X-rays and gamma rays must be observed from above Earth\u2019s atmosphere. 37. The greenhouse effect can be explained easily if you understand the laws of blackbody radiation. A greenhouse gas blocks the transmission of infrared light. Given that the incoming light to Earth is sunlight with a characteristic temperature of 5800 K (which peaks in the visible part of the spectrum) and the outgoing light from Earth has a characteristic temperature of about 300 K (which peaks in the infrared part of the spectrum), explain how greenhouse gases cause Earth to warm up. As part of your answer, discuss that greenhouse gases block both incoming and outgoing infrared light. Explain why these two effects don\u2019t simply cancel each other, leading to no net temperature change. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 5 Radiation and Spectra 187 38. An idealized radiating object does not reflect or scatter any radiation but instead absorbs all of the electromagnetic energy that falls on it. Can you explain why astronomers call such an object a blackbody? Keep in mind that even stars, which shine brightly in a variety of colors, are considered blackbodies. Explain why. 39. Why are ionized gases typically only found in very high-temperature environments? 40. Explain why each element has a unique spectrum of absorption or emission lines. Figuring For Yourself 41. What is the wavelength of the carrier wave of a campus radio station, broadcasting at a frequency of 97.2 MHz (million cycles per second or million hertz)? 42. What is the frequency of a red laser beam, with a wavelength of 670 nm, which your astronomy instructor might use to point to slides during a lecture on galaxies? 43. You go to a dance club to forget how hard your astronomy midterm was. What is the frequency of a wave of ultraviolet light coming from a blacklight in the club, if its wavelength is 150 nm? 44. What is the energy of the photon with the frequency you calculated in Exercise 5.43? 45. If the emitted infrared radiation from Pluto, has a wavelength of maximum intensity at 75,000 nm, what is the temperature of Pluto assuming it follows Wien\u2019s law? 46. What is the temperature of a star whose maximum light is emitted at a wavelength of 290 nm? 188 Chapter 5 Radiation and Spectra This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 189 6 ASTRONOMICAL INSTRUMENTS Figure 6.1. Hubble Space Telescope (HST). This artist\u2019s impression shows the Hubble above Earth, with the rectangular solar panels that provide it with power seen to the left and right. Chapter Outline 6.1 Telescopes 6.2 Telescopes Today 6.3 Visible-Light Detectors and Instruments 6.4 Radio Telescopes 6.5 Observations outside Earth\u2019s Atmosphere 6.6 The Future of Large Telescopes Thinking Ahead If you look at the sky when you are far away from city lights, there seem to be an overwhelming number of stars up there. In reality, only about 9000 stars are visible to the unaided eye (from both hemispheres of our planet). The light from most stars is so weak that by the time it reaches Earth, it cannot be detected by the human eye. How can we learn about the vast majority of objects in the universe that our unaided eyes simply cannot see? In this chapter, we describe the tools astronomers use to extend their vision into space. We have learned almost everything we know about the universe from studying electromagnetic radiation, as discussed in the chapter on Radiation and Spectra. In the twentieth century, our exploration of space made it possible to detect electromagnetic radiation at all wavelengths, from gamma rays to radio waves. The different wavelengths carry different kinds of information, and the appearance of any given object often depends on the wavelength at which the observations are made. 190 Chapter 6 Astronomical Instruments 6.1 TELESCOPES Learning Objectives By the end of this section, you will be able to: Describe the three basic components of a modern system for measuring astronomical sources Describe the main functions of a telescope Describe the two basic types of visible-light telescopes and how they form images Systems for Measuring Radiation There are three basic components of a modern system for measuring radiation from astronomical sources. First, there is a telescope", ", which serves as a \u201cbucket\u201d for collecting visible light (or radiation at other wavelengths, as shown in (Figure 6.2). Just as you can catch more rain with a garbage can than with a coffee cup, large telescopes gather much more light than your eye can. Second, there is an instrument attached to the telescope that sorts the incoming radiation by wavelength. Sometimes the sorting is fairly crude. For example, we might simply want to separate blue light from red light so that we can determine the temperature of a star. But at other times, we want to see individual spectral lines to determine what an object is made of, or to measure its speed (as explained in the Radiation and Spectra chapter). Third, we need some type of detector, a device that senses the radiation in the wavelength regions we have chosen and permanently records the observations. Figure 6.2. Orion Region at Different Wavelengths. The same part of the sky looks different when observed with instruments that are sensitive to different bands of the spectrum. (a) Visible light: this shows part of the Orion region as the human eye sees it, with dotted lines added to show the figure of the mythical hunter, Orion. (b) X-rays: here, the view emphasizes the point-like X-ray sources nearby. The colors are artificial, changing from yellow to white to blue with increasing energy of the X-rays. The bright, hot stars in Orion are still seen in this image, but so are many other objects located at very different distances, including other stars, star corpses, and galaxies at the edge of the observable universe. (c) Infrared radiation: here, we mainly see the glowing dust in this region. (credit a: modification of work by Howard McCallon/NASA/ IRAS; credit b: modification of work by Howard McCallon/NASA/IRAS; credit c: modification of work by Michael F. Corcoran) The history of the development of astronomical telescopes is about how new technologies have been applied to improve the efficiency of these three basic components: the telescopes, the wavelength-sorting device, and the detectors. Let\u2019s first look at the development of the telescope. Many ancient cultures built special sites for observing the sky (Figure 6.3). At these ancient observatories, they could measure the positions of celestial objects, mostly to keep track of time and date. Many of these ancient This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 191 observatories had religious and ritual functions as well. The eye was the only device available to gather light, all of the colors in the light were observed at once, and the only permanent record of the observations was made by human beings writing down or sketching what they saw. Figure 6.3. Two Pre-Telescopic Observatories. (a) Machu Picchu is a fifteenth century Incan site located in Peru. (b) Stonehenge, a prehistoric site (3000\u20132000 BCE), is located in England. (credit a: modification of work by Allard Schmidt) While Hans Lippershey, Zaccharias Janssen, and Jacob Metius are all credited with the invention of the telescope around 1608\u2014applying for patents within weeks of each other\u2014it was Galileo who, in 1610, used this simple tube with lenses (which he called a spyglass) to observe the sky and gather more light than his eyes alone could. Even his small telescope\u2014used over many nights\u2014revolutionized ideas about the nature of the planets and the position of Earth. How Telescopes Work Telescopes have come a long way since Galileo\u2019s time. Now they tend to be huge devices; the most expensive cost hundreds of millions to billions of dollars. (To provide some reference point, however, keep in mind that just renovating college football stadiums typically costs hundreds of millions of dollars\u2014with the most expensive recent renovation, at Texas A&M University\u2019s Kyle Field, costing $450 million.) The reason astronomers keep building bigger and bigger telescopes is that celestial objects\u2014such as planets, stars, and galaxies\u2014send much more light to Earth than any human eye (with its tiny opening) can catch, and bigger telescopes can detect fainter objects. If you have ever watched the stars with a group of friends, you know that there\u2019s plenty of starlight to go around; each of you can see each of the stars. If a thousand more people were watching, each of them would also catch a bit of each star\u2019s light. Yet, as far as you are concerned, the light not shining into your eye is wasted. It would be great if some of this \u201cwasted\u201d light could also be captured and brought to your eye. This is precisely what a telescope does. The most important functions of a telescope are (1) to collect the faint light", " from an astronomical source and (2) to focus all the light into a point or an image. Most objects of interest to astronomers are extremely faint: the more light we can collect, the better we can study such objects. (And remember, even though we are focusing on visible light first, there are many telescopes that collect other kinds of electromagnetic radiation.) Telescopes that collect visible radiation use a lens or mirror to gather the light. Other types of telescopes may use collecting devices that look very different from the lenses and mirrors with which we are familiar, but they serve the same function. In all types of telescopes, the light-gathering ability is determined by the area of the device acting as the light-gathering \u201cbucket.\u201d Since most telescopes have mirrors or lenses, we can compare 192 Chapter 6 Astronomical Instruments their light-gathering power by comparing the apertures, or diameters, of the opening through which light travels or reflects. The amount of light a telescope can collect increases with the size of the aperture. A telescope with a mirror that is 4 meters in diameter can collect 16 times as much light as a telescope that is 1 meter in diameter. (The diameter is squared because the area of a circle equals \u03c0d2/4, where d is the diameter of the circle.) Calculating the Light-Collecting Area What is the area of a 1-m diameter telescope? A 4-m diameter one? Solution Using the equation for the area of a circle, the area of a 1-m telescope is and the area of a 4-m telescope is A = \u03c0d 2 4 \u03c0d 2 4 = \u03c0(1 m)2 4 = 0.79 m2 \u03c0d 2 4 = \u03c0(4 m)2 4 = 12.6 m2 Check Your Learning Show that the ratio of the two areas is 16:1. Answer: 12.6 m2 0.79 m2 = 16. Therefore, with 16 times the area, a 4-m telescope collects 16 times the light of a 1-m telescope. After the telescope forms an image, we need some way to detect and record it so that we can measure, reproduce, and analyze the image in various ways. Before the nineteenth century, astronomers simply viewed images with their eyes and wrote descriptions of what they saw. This was very inefficient and did not lead to a very reliable long-term record; you know from crime shows on television that eyewitness accounts are often inaccurate. In the nineteenth century, the use of photography became widespread. In those days, photographs were a chemical record of an image on a specially treated glass plate. Today, the image is generally detected with sensors similar to those in digital cameras, recorded electronically, and stored in computers. This permanent record can then be used for detailed and quantitative studies. Professional astronomers rarely look through the large telescopes that they use for their research. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 193 Formation of an Image by a Lens or a Mirror Whether or not you wear glasses, you see the world through lenses; they are key elements of your eyes. A lens is a transparent piece of material that bends the rays of light passing through it. If the light rays are parallel as they enter, the lens brings them together in one place to form an image (Figure 6.4). If the curvatures of the lens surfaces are just right, all parallel rays of light (say, from a star) are bent, or refracted, in such a way that they converge toward a point, called the focus of the lens. At the focus, an image of the light source appears. In the case of parallel light rays, the distance from the lens to the location where the light rays focus, or image, behind the lens is called the focal length of the lens. Figure 6.4. Formation of an Image by a Simple Lens. Parallel rays from a distant source are bent by the convex lens so that they all come together in a single place (the focus) to form an image. As you look at Figure 6.4, you may ask why two rays of light from the same star would be parallel to each other. After all, if you draw a picture of star shining in all directions, the rays of light coming from the star don\u2019t look parallel at all. But remember that the stars (and other astronomical objects) are all extremely far away. By the time the few rays of light pointed toward us actually arrive at Earth, they are, for all practical purposes, parallel to each other. Put another way, any rays that were not parallel to the ones pointed at Earth are now heading in some very different direction in the universe. To view the image formed by the lens in a telescope, we use an additional lens called an eyepiece. The eyepiece focuses the image at a distance that is either directly viewable by a human", " or at a convenient place for a detector. Using different eyepieces, we can change the magnification (or size) of the image and also redirect the light to a more accessible location. Stars look like points of light, and magnifying them makes little difference, but the image of a planet or a galaxy, which has structure, can often benefit from being magnified. Many people, when thinking of a telescope, picture a long tube with a large glass lens at one end. This design, which uses a lens as its main optical element to form an image, as we have been discussing, is known as a refractor (Figure 6.5), and a telescope based on this design is called a refracting telescope. Galileo\u2019s telescopes were refractors, as are today\u2019s binoculars and field glasses. However, there is a limit to the size of a refracting telescope. The largest one ever built was a 49-inch refractor built for the Paris 1900 Exposition, and it was dismantled after the Exposition. Currently, the largest refracting telescope is the 40-inch refractor at Yerkes Observatory in Wisconsin. 194 Chapter 6 Astronomical Instruments Figure 6.5. Refracting and Reflecting Telescopes. Light enters a refracting telescope through a lens at the upper end, which focuses the light near the bottom of the telescope. An eyepiece then magnifies the image so that it can be viewed by the eye, or a detector like a photographic plate can be placed at the focus. The upper end of a reflecting telescope is open, and the light passes through to the mirror located at the bottom of the telescope. The mirror then focuses the light at the top end, where it can be detected. Alternatively, as in this sketch, a second mirror may reflect the light to a position outside the telescope structure, where an observer can have easier access to it. Professional astronomers\u2019 telescopes are more complicated than this, but they follow the same principles of reflection and refraction. One problem with a refracting telescope is that the light must pass through the lens of a refractor. That means the glass must be perfect all the way through, and it has proven very difficult to make large pieces of glass without flaws and bubbles in them. Also, optical properties of transparent materials change a little bit with the wavelengths (or colors) of light, so there is some additional distortion, known as chromatic aberration. Each wavelength focuses at a slightly different spot, causing the image to appear blurry. In addition, since the light must pass through the lens, the lens can only be supported around its edges (just like the frames of our eyeglasses). The force of gravity will cause a large lens to sag and distort the path of the light rays as they pass through it. Finally, because the light passes through it, both sides of the lens must be manufactured to precisely the right shape in order to produce a sharp image. A different type of telescope uses a concave primary mirror as its main optical element. The mirror is curved like the inner surface of a sphere, and it reflects light in order to form an image (Figure 6.5). Telescope mirrors are coated with a shiny metal, usually silver, aluminum, or, occasionally, gold, to make them highly reflective. If the mirror has the correct shape, all parallel rays are reflected back to the same point, the focus of the mirror. Thus, images are produced by a mirror exactly as they are by a lens. Telescopes designed with mirrors avoid the problems of refracting telescopes. Because the light is reflected from the front surface only, flaws and bubbles within the glass do not affect the path of the light. In a telescope designed with mirrors, only the front surface has to be manufactured to a precise shape, and the mirror can be supported from the back. For these reasons, most astronomical telescopes today (both amateur and professional) use a mirror rather than a lens to form an image; this type of telescope is called a reflecting telescope. The first successful reflecting telescope was built by Isaac Newton in 1668. In a reflecting telescope, the concave mirror is placed at the bottom of a tube or open framework. The mirror reflects the light back up the tube to form an image near the front end at a location called the prime focus. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 195 The image can be observed at the prime focus, or additional mirrors can intercept the light and redirect it to a position where the observer can view it more easily (Figure 6.6). Since an astronomer at the prime focus can block much of the light coming to the main mirror, the use of a small secondary mirror allows more light to get through the system. Figure 6.6. Focus Arrangements for Reflecting Telescopes. Reflecting telescopes have different options for where the light is brought to a focus. With prime", " focus, light is detected where it comes to a focus after reflecting from the primary mirror. With Newtonian focus, light is reflected by a small secondary mirror off to one side, where it can be detected (see also Figure 6.5). Most large professional telescopes have a Cassegrain focus in which light is reflected by the secondary mirror down through a hole in the primary mirror to an observing station below the telescope Choosing Your Own Telescope If the astronomy course you are taking whets your appetite for exploring the sky further, you may be thinking about buying your own telescope. Many excellent amateur telescopes are available, and some research is required to find the best model for your needs. Some good sources of information about personal telescopes are the two popular US magazines aimed at amateur astronomers: Sky & Telescope and Astronomy. Both carry regular articles with advice, reviews, and advertisements from reputable telescope dealers. Some of the factors that determine which telescope is right for you depend upon your preferences: \u2022 Will you be setting up the telescope in one place and leaving it there, or do you want an instrument that is portable and can come with you on outdoor excursions? How portable should it be, in terms of size and weight? \u2022 Do you want to observe the sky with your eyes only, or do you want to take photographs? (Long- exposure photography, for example, requires a good clock drive to turn your telescope to compensate for Earth\u2019s rotation.) 196 Chapter 6 Astronomical Instruments \u2022 What types of objects will you be observing? Are you interested primarily in comets, planets, star clusters, or galaxies, or do you want to observe all kinds of celestial sights? You may not know the answers to some of these questions yet. For this reason, you may want to \u201ctest- drive\u201d some telescopes first. Most communities have amateur astronomy clubs that sponsor star parties open to the public. The members of those clubs often know a lot about telescopes and can share their ideas with you. Your instructor may know where the nearest amateur astronomy club meets; or, to find a club near you, use the websites suggested in Appendix B. Furthermore, you may already have an instrument like a telescope at home (or have access to one through a relative or friend). Many amateur astronomers recommend starting your survey of the sky with a good pair of binoculars. These are easily carried around and can show you many objects not visible (or clear) to the unaided eye. When you are ready to purchase a telescope, you might find the following ideas useful: \u2022 The key characteristic of a telescope is the aperture of the main mirror or lens; when someone says they have a 6-inch or 8-inch telescope, they mean the diameter of the collecting surface. The larger the aperture, the more light you can gather, and the fainter the objects you can see or photograph. \u2022 Telescopes of a given aperture that use lenses (refractors) are typically more expensive than those using mirrors (reflectors) because both sides of a lens must be polished to great accuracy. And, because the light passes through it, the lens must be made of high-quality glass throughout. In contrast, only the front surface of a mirror must be accurately polished. \u2022 Magnification is not one of the criteria on which to base your choice of a telescope. As we discussed, the magnification of the image is done by a smaller eyepiece, so the magnification can be adjusted by changing eyepieces. However, a telescope will magnify not only the astronomical object you are viewing but also the turbulence of Earth\u2019s atmosphere. If the magnification is too high, your image will shimmer and shake and be difficult to view. A good telescope will come with a variety of eyepieces that stay within the range of useful magnification. \u2022 The mount of a telescope (the structure on which it rests) is one of its most critical elements. Because a telescope shows a tiny field of view, which is magnified significantly, even the smallest vibration or jarring of the telescope can move the object you are viewing around or out of your field of view. A sturdy and stable mount is essential for serious viewing or photography (although it clearly affects how portable your telescope can be). \u2022 A telescope requires some practice to set up and use effectively. Don\u2019t expect everything to go perfectly on your first try. Take some time to read the instructions. If a local amateur astronomy club is nearby, use it as a resource. 6.2 TELESCOPES TODAY Learning Objectives By the end of this section, you will be able to: Recognize the largest visible-light and infrared telescopes in operation today This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 197 Discuss the factors relevant to choosing an appropriate telescope site Define the technique of adaptive optics and describe the effects of the atmosphere on astronomical observations Since Newton\u2019s time, when the sizes of the mirrors in", " telescopes were measured in inches, reflecting telescopes have grown ever larger. In 1948, US astronomers built a telescope with a 5-meter (200-inch) diameter mirror on Palomar Mountain in Southern California. It remained the largest visible-light telescope in the world for several decades. The giants of today, however, have primary mirrors (the largest mirrors in the telescope) that are 8- to 10-meters in diameter, and larger ones are being built (Figure 6.7). Figure 6.7. Large Telescope Mirror. This image shows one of the primary mirrors of the European Southern Observatory\u2019s Very Large Telescope, named Yepun, just after it was recoated with aluminum. The mirror is a little over 8 meters in diameter. (credit: ESO/G. Huedepohl) Modern Visible-Light and Infrared Telescopes The decades starting in 1990 saw telescope building around the globe grow at an unprecedented rate. (See Table 6.1, which also includes websites for each telescope in case you want to visit or learn more about them.) Technological advancements had finally made it possible to build telescopes significantly larger than the 5-meter telescope at Palomar at a reasonable cost. New technologies have also been designed to work well in the infrared, and not just visible, wavelengths. Large Single-Dish Visible-Light and Infrared Telescopes Telescope Name Location Status Website European Extremely Large Telescope (EELT) Cerro Armazonas, Chile First light 2025 (estimated) www.eso.org/sci/ facilities/eelt Thirty-Meter Telescope (TMT) Mauna Kea, HI First light 2025 (estimated) www.tmt.org Giant Magellan Telescope (GMT) Las Campanas Observatory, Chile First light 2025 (estimated) www.gmto.org Aperture (m) 39 30 24.5 Table 6.1 198 Chapter 6 Astronomical Instruments Large Single-Dish Visible-Light and Infrared Telescopes Telescope Name Location Status Website Southern African Large Telescope (SALT) Sutherland, South Africa 2005 www.salt.ac.za Gran Telescopio Canarias (GTC) La Palma, Canary Islands First light 2007 http://www.gtc.iac.es Keck I and II (two telescopes) Mauna Kea, HI Completed 1993\u201396 www.keckobservatory.org Hobby\u2013Eberly Telescope (HET) Mount Locke, TX Completed 1997 www.as.utexas.edu/ mcdonald/het Large Binocular Telescope (LBT) (two telescopes) Large Synoptic Survey Telescope (LSST) Mount Graham, AZ First light 2004 www.lbto.org The Cerro Pach\u00f3n, Chile First light 2021 www.lsst.org Subaru Telescope Mauna Kea, HI First light 1998 www.naoj.org Very Large Telescope (VLT) Cerro Paranal, Chile All four telescopes completed 2000 www.eso.org/public/ teles-instr/paranal Gemini North and Gemini South Mauna Kea, HI (North) and Cerro Pach\u00f3n, Chile (South) First light 1999 (North), First light 2000 (South) www.gemini.edu Magellan Telescopes (two telescopes: Baade and Landon Clay) Las Campanas, Chile First light 2000 and 2002 obs.carnegiescience.edu/ Magellan Multi-Mirror Telescope (MMT) Mount Hopkins, AZ Completed 1979 www.mmto.org Big Telescope Altazimuth (BTA-6) Mount Pastukhov, Russia Completed 1976 w0.sao.ru/Doc-en/ Telescopes/bta/ descrip.html Aperture (m) 11.1 \u00d7 9.9 10.4 10.0 9.1 8.4 8.4 8.3 8.2 8.1 6.5 6.5 6.0 Table 6.1 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 199 Large Single-Dish Visible-Light and Infrared Telescopes Aperture (m) Telescope Name Location Status Website 5.1 Hale Telescope Mount Palomar, CA Completed 1948 www.astro.caltech.edu/ palomar/about/ telescopes/hale.html Table 6.1 The differences between the Palomar telescope and the modern Gemini North telescope (to take an example) are easily seen in Figure 6.8. The Palomar telescope is a massive steel structure designed to hold the 14.5-ton primary mirror with a 5-meter diameter. Glass tends to sag under its own weight; hence, a huge steel structure is needed to hold the mirror. A mirror 8 meters in diameter, the size of the Gemini North telescope, if it were built using the same technology as the Palomar", " telescope, would have to weigh at least eight times as much and would require an enormous steel structure to support it. Figure 6.8. Modern Reflecting Telescopes. (a) The Palomar 5-meter reflector: The Hale telescope on Palomar Mountain has a complex mounting structure that enables the telescope (in the open \u201ctube\u201d pointing upward in this photo) to swing easily into any position. (b) The Gemini North 8-meter telescope: The Gemini North mirror has a larger area than the Palomar mirror, but note how much less massive the whole instrument seems. (credit a: modification of work by Caltech/Palomar Observatory; credit b: modification of work by Gemini Observatory/AURA) The 8-meter Gemini North telescope looks like a featherweight by contrast, and indeed it is. The mirror is only about 8 inches thick and weighs 24.5 tons, less than twice as much as the Palomar mirror. The Gemini North telescope was completed about 50 years after the Palomar telescope. Engineers took advantage of new technologies to build a telescope that is much lighter in weight relative to the size of the primary mirror. The 200 Chapter 6 Astronomical Instruments Gemini mirror does sag, but with modern computers, it is possible to measure that sag many times each second and apply forces at 120 different locations to the back of the mirror to correct the sag, a process called active control. Seventeen telescopes with mirrors 6.5 meters in diameter and larger have been constructed since 1990. The twin 10-meter Keck telescopes on Mauna Kea, which were the first of these new-technology instruments, use precision control in an entirely novel way. Instead of a single primary mirror 10 meters in diameter, each Keck telescope achieves its larger aperture by combining the light from 36 separate hexagonal mirrors, each 1.8 meters wide (Figure 6.9). Computer-controlled actuators (motors) constantly adjust these 36 mirrors so that the overall reflecting surface acts like a single mirror with just the right shape to collect and focus the light into a sharp image. Figure 6.9. Thirty-Six Eyes Are Better Than One. The mirror of the 10-meter Keck telescope is composed of 36 hexagonal sections. (credit: NASA Learn more about the Keck Observatory on Mauna Kea (https://openstaxcollege.org/l/ 30KeckObserv) through this History Channel clip on the telescopes and the work that they do. In addition to holding the mirror, the steel structure of a telescope is designed so that the entire telescope can be pointed quickly toward any object in the sky. Since Earth is rotating, the telescope must have a motorized drive system that moves it very smoothly from east to west at exactly the same rate that Earth is rotating from west to east, so it can continue to point at the object being observed. All this machinery must be housed in a dome to protect the telescope from the elements. The dome has an opening in it that can be positioned in front of the telescope and moved along with it, so that the light from the objects being observed is not blocked. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 201 George Ellery Hale: Master Telescope Builder George Ellery Hale (Figure 6.10) was a giant among early telescope builders. Not once, but four times, he initiated projects that led to the construction of what was the world\u2019s largest telescope at the time. And he was a master at winning over wealthy benefactors to underwrite the construction of these new instruments. Figure 6.10. George Ellery Hale (1868\u20131938). Hale\u2019s work led to the construction of several major telescopes, including the 40-inch refracting telescope at Yerkes Observatory, and three reflecting telescopes: the 60-inch Hale and 100-inch Hooker telescopes at Mount Wilson Observatory, and the 200-inch Hale Telescope at Palomar Observatory. Hale\u2019s training and early research were in solar physics. In 1892, at age 24, he was named associate professor of astral physics and director of the astronomical observatory at the University of Chicago. At the time, the largest telescope in the world was the 36-inch refractor at the Lick Observatory near San Jose, California. Taking advantage of an existing glass blank for a 40-inch telescope, Hale set out to raise money for a larger telescope than the one at Lick. One prospective donor was Charles T. Yerkes, who, among other things, ran the trolley system in Chicago. Hale wrote to Yerkes, encouraging him to support the construction of the giant telescope by saying that \u201cthe donor could have no more enduring monument. It is certain that Mr. Lick\u2019s name would not have been nearly so widely known today were it not for the famous observatory established as a result of his", " munificence.\u201d Yerkes agreed, and the new telescope was completed in May 1897; it remains the largest refractor in the world (Figure 6.11). 202 Chapter 6 Astronomical Instruments Figure 6.11. World\u2019s Largest Refractor. The Yerkes 40-inch (1-meter) telescope. Even before the completion of the Yerkes refractor, Hale was not only dreaming of building a still larger telescope but was also taking concrete steps to achieve that goal. In the 1890s, there was a major controversy about the relative quality of refracting and reflecting telescopes. Hale realized that 40 inches was close to the maximum feasible aperture for refracting telescopes. If telescopes with significantly larger apertures were to be built, they would have to be reflecting telescopes. Using funds borrowed from his own family, Hale set out to construct a 60-inch reflector. For a site, he left the Midwest for the much better conditions on Mount Wilson\u2014at the time, a wilderness peak above the small city of Los Angeles. In 1904, at the age of 36, Hale received funds from the Carnegie Foundation to establish the Mount Wilson Observatory. The 60-inch mirror was placed in its mount in December 1908. Two years earlier, in 1906, Hale had already approached John D. Hooker, who had made his fortune in hardware and steel pipe, with a proposal to build a 100-inch telescope. The technological risks were substantial. The 60-inch telescope was not yet complete, and the usefulness of large reflectors for astronomy had yet to be demonstrated. George Ellery Hale\u2019s brother called him \u201cthe greatest gambler in the world.\u201d Once again, Hale successfully obtained funds, and the 100-inch telescope was completed in November 1917. (It was with this telescope that Edwin Hubble was able to establish that the spiral nebulae were separate islands of stars\u2014or galaxies\u2014quite removed from our own Milky Way.) Hale was not through dreaming. In 1926, he wrote an article in Harper\u2019s Magazine about the scientific value of a still larger telescope. This article came to the attention of the Rockefeller Foundation, which granted $6 million for the construction of a 200-inch telescope. Hale died in 1938, but the 200-inch (5-meter) telescope on Palomar Mountain was dedicated 10 years later and is now named in Hale\u2019s honor. Picking the Best Observing Sites A telescope like the Gemini or Keck telescope costs about $100 million to build. That kind of investment demands that the telescope be placed in the best possible site. Since the end of the nineteenth century, astronomers have realized that the best observatory sites are on mountains, far from the lights and pollution of cities. Although a number of urban observatories remain, especially in the large cities of Europe, they have This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 203 become administrative centers or museums. The real action takes place far away, often on desert mountains or isolated peaks in the Atlantic and Pacific Oceans, where we find the staff\u2019s living quarters, computers, electronic and machine shops, and of course the telescopes themselves. A large observatory today requires a supporting staff of 20 to 100 people in addition to the astronomers. The performance of a telescope is determined not only by the size of its mirror but also by its location. Earth\u2019s atmosphere, so vital to life, presents challenges for the observational astronomer. In at least four ways, our air imposes limitations on the usefulness of telescopes: 1. The most obvious limitation is weather conditions such as clouds, wind, and rain. At the best sites, the weather is clear as much as 75% of the time. 2. Even on a clear night, the atmosphere filters out a certain amount of starlight, especially in the infrared, where the absorption is due primarily to water vapor. Astronomers therefore prefer dry sites, generally found at high altitudes. 3. The sky above the telescope should be dark. Near cities, the air scatters the glare from lights, producing an illumination that hides the faintest stars and limits the distances that can be probed by telescopes. (Astronomers call this effect light pollution.) Observatories are best located at least 100 miles from the nearest large city. 4. Finally, the air is often unsteady; light passing through this turbulent air is disturbed, resulting in blurred star images. Astronomers call these effects \u201cbad seeing.\u201d When seeing is bad, images of celestial objects are distorted by the constant twisting and bending of light rays by turbulent air. The best observatory sites are therefore high, dark, and dry. The world\u2019s largest telescopes are found in such remote mountain locations as the Andes Mountains of Chile (Figure 6.12), the desert peaks of Arizona, the Canary Islands in the Atlantic Ocean, and Mauna Kea in", " Hawaii, a dormant volcano with an altitude of 13,700 feet (4200 meters). Light pollution is a problem not just for professional astronomers but for everyone who wants to enjoy the beauty of the night sky. In addition research is now showing that it can disrupt the life cycle of animals with whom we share the urban and suburban landscape. And the light wasted shining into the sky leads to unnecessary municipal expenses and use of fossil fuels. Concerned people have formed an organization, the International Dark-Sky Association, whose website (https://openstaxcollege.org/l/ 30IntDSA) is full of good information. A citizen science project called Globe at Night (https://openstaxcollege.org/l/30GlbatNght) allows you to measure the light levels in your community by counting stars and to compare it to others around the world. And, if you get interested in this topic and want to do a paper for your astronomy course or another course while you are in college, the Dark Night Skies guide (https://openstaxcollege.org/l/30DNSGuide) can point you to a variety of resources on the topic. 204 Chapter 6 Astronomical Instruments Figure 6.12. High and Dry Site. Cerro Paranal, a mountain summit 2.7 kilometers above sea level in Chile\u2019s Atacama Desert, is the site of the European Southern Observatory\u2019s Very Large Telescope. This photograph shows the four 8-meter telescope buildings on the site and vividly illustrates that astronomers prefer high, dry sites for their instruments. The 4.1-meter Visible and Infrared Survey Telescope for Astronomy (VISTA) can be seen in the distance on the next mountain peak. (credit: ESO) The Resolution of a Telescope In addition to gathering as much light as they can, astronomers also want to have the sharpest images possible. Resolution refers to the precision of detail present in an image: that is, the smallest features that can be distinguished. Astronomers are always eager to make out more detail in the images they study, whether they are following the weather on Jupiter or trying to peer into the violent heart of a \u201ccannibal galaxy\u201d that recently ate its neighbor for lunch. One factor that determines how good the resolution will be is the size of the telescope. Larger apertures produce sharper images. Until very recently, however, visible-light and infrared telescopes on Earth\u2019s surface could not produce images as sharp as the theory of light said they should. The problem\u2014as we saw earlier in this chapter\u2014is our planet\u2019s atmosphere, which is turbulent. It contains many small-scale blobs or cells of gas that range in size from inches to several feet. Each cell has a slightly different temperature from its neighbor, and each cell acts like a lens, bending (refracting) the path of the light by a small amount. This bending slightly changes the position where each light ray finally reaches the detector in a telescope. The cells of air are in motion, constantly being blown through the light path of the telescope by winds, often in different directions at different altitudes. As a result, the path followed by the light is constantly changing. For an analogy, think about watching a parade from a window high up in a skyscraper. You decide to throw some confetti down toward the marchers. Even if you drop a handful all at the same time and in the same direction, air currents will toss the pieces around, and they will reach the ground at different places. As we described earlier, we can think of the light from the stars as a series of parallel beams, each making its way through the atmosphere. Each path will be slightly different, and each will reach the detector of the telescope at a slightly different place. The result is a blurred image, and because the cells are being blown by the wind, the nature of the blur will change many times each second. You have probably noticed this effect as the \u201ctwinkling\u201d of stars seen from Earth. The light beams are bent enough that part of the time they reach your eye, and part of the time some of them miss, thereby making the star seem to vary in brightness. In space, however, the light of the stars is steady. Astronomers search the world for locations where the amount of atmospheric blurring, or turbulence, is as small as possible. It turns out that the best sites are in coastal mountain ranges and on isolated volcanic peaks in the middle of an ocean. Air that has flowed long distances over water before it encounters land is especially stable. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 205 The resolution of an image is measured in units of angle on the sky, typically in units of arcseconds. One arcsecond is 1/3600 degree, and there are 360 degrees in a full circle. So we are talking about tiny angles", " on the sky. To give you a sense of just how tiny, we might note that 1 arcsecond is how big a quarter would look when seen from a distance of 5 kilometers. The best images obtained from the ground with traditional techniques reveal details as small as several tenths of an arcsecond across. This image size is remarkably good. One of the main reasons for launching the Hubble Space Telescope was to escape Earth\u2019s atmosphere and obtain even sharper images. But since we can\u2019t put every telescope into space, astronomers have devised a technique called adaptive optics that can beat Earth\u2019s atmosphere at its own game of blurring. This technique (which is most effective in the infrared region of the spectrum with our current technology) makes use of a small flexible mirror placed in the beam of a telescope. A sensor measures how much the atmosphere has distorted the image, and as often as 500 times per second, it sends instructions to the flexible mirror on how to change shape in order to compensate for distortions produced by the atmosphere. The light is thus brought back to an almost perfectly sharp focus at the detector. Figure 6.13 shows just how effective this technique is. With adaptive optics, ground-based telescopes can achieve resolutions of 0.1 arcsecond or a little better in the infrared region of the spectrum. This impressive figure is the equivalent of the resolution that the Hubble Space Telescope achieves in the visible-light region of the spectrum. Figure 6.13. Power of Adaptive Optics. One of the clearest pictures of Jupiter ever taken from the ground, this image was produced with adaptive optics using an 8-meter-diameter telescope at the Very Large Telescope in Chile. Adaptive optics uses infrared wavelengths to remove atmospheric blurring, resulting in a much clearer image. (credit: modification of work by ESO, F.Marchis, M.Wong (UC Berkeley); E.Marchetti, P.Amico, S.Tordo (ESO)) How Astronomers Really Use Telescopes In the popular view (and some bad movies), an astronomer spends most nights in a cold observatory peering through a telescope, but this is not very accurate today. Most astronomers do not live at 206 Chapter 6 Astronomical Instruments observatories, but near the universities or laboratories where they work. An astronomer might spend only a week or so each year observing at the telescope and the rest of the time measuring or analyzing the data acquired from large project collaborations and dedicated surveys. Many astronomers use radio telescopes for space experiments, which work just as well during the daylight hours. Still others work at purely theoretical problems using supercomputers and never observe at a telescope of any kind. Even when astronomers are observing with large telescopes, they seldom peer through them. Electronic detectors permanently record the data for detailed analysis later. At some observatories, observations may be made remotely, with the astronomer sitting at a computer thousands of miles away from the telescope. Time on major telescopes is at a premium, and an observatory director will typically receive many more requests for telescope time than can be accommodated during the year. Astronomers must therefore write a convincing proposal explaining how they would like to use the telescope and why their observations will be important to the progress of astronomy. A committee of astronomers is then asked to judge and rank the proposals, and time is assigned only to those with the greatest merit. Even if your proposal is among the high-rated ones, you may have to wait many months for your turn. If the skies are cloudy on the nights you have been assigned, it may be more than a year before you get another chance. Some older astronomers still remember long, cold nights spent alone in an observatory dome, with only music from a tape recorder or an all-night radio station for company. The sight of the stars shining brilliantly hour after hour through the open slit in the observatory dome was unforgettable. So, too, was the relief as the first pale light of dawn announced the end of a 12-hour observation session. Astronomy is much easier today, with teams of observers working together, often at their computers, in a warm room. Those who are more nostalgic, however, might argue that some of the romance has gone from the field, too. 6.3 VISIBLE-LIGHT DETECTORS AND INSTRUMENTS Learning Objectives By the end of this section, you will be able to: Describe the difference between photographic plates and charge-coupled devices Describe the unique difficulties associated with infrared observations and their solutions Describe how a spectrometer works After a telescope collects radiation from an astronomical source, the radiation must be detected and measured. The first detector used for astronomical observations was the human eye, but it suffers from being connected to an imperfect recording and retrieving device\u2014the human brain. Photography and modern electronic detectors have eliminated the quirks of human memory by making a permanent record of the information from the cosmos. The eye also suffers from having a very short integration time; it takes only a fraction of a second to add light energy together", " before sending the image to the brain. One important advantage of modern detectors is that the light from astronomical objects can be collected by the detector over longer periods of time; this technique is called \u201ctaking a long exposure.\u201d Exposures of several hours are required to detect very faint objects in the cosmos. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 207 Before the light reaches the detector, astronomers today normally use some type of instrument to sort the light according to wavelength. The instrument may be as simple as colored filters, which transmit light within a specified range of wavelengths. A red transparent plastic is an everyday example of a filter that transmits only the red light and blocks the other colors. After the light passes through a filter, it forms an image that astronomers can then use to measure the apparent brightness and color of objects. We will show you many examples of such images in the later chapters of this book, and we will describe what we can learn from them. Alternatively, the instrument between telescope and detector may be one of several devices that spread the light out into its full rainbow of colors so that astronomers can measure individual lines in the spectrum. Such an instrument (which you learned about in the chapter on Radiation and Spectra) is called a spectrometer because it allows astronomers to measure (to meter) the spectrum of a source of radiation. Whether a filter or a spectrometer, both types of wavelength-sorting instruments still have to use detectors to record and measure the properties of light. Photographic and Electronic Detectors Throughout most of the twentieth century, photographic film or glass plates served as the prime astronomical detectors, whether for photographing spectra or direct images of celestial objects. In a photographic plate, a light-sensitive chemical coating is applied to a piece of glass that, when developed, provides a lasting record of the image. At observatories around the world, vast collections of photographs preserve what the sky has looked like during the past 100 years. Photography represents a huge improvement over the human eye, but it still has limitations. Photographic films are inefficient: only about 1% of the light that actually falls on the film contributes to the chemical change that makes the image; the rest is wasted. Astronomers today have much more efficient electronic detectors to record astronomical images. Most often, these are charge-coupled devices (CCDs), which are similar to the detectors used in video camcorders or in digital cameras (like the one more and more students have on their cell phones) (see Figure 6.14). In a CCD, photons of radiation hitting any part of the detector generate a stream of charged particles (electrons) that are stored and counted at the end of the exposure. Each place where the radiation is counted is called a pixel (picture element), and modern detectors can count the photons in millions of pixels (megapixels, or MPs). Figure 6.14. Charge-Coupled Devices (CCDs). (a) This CCD is a mere 300-micrometers thick (thinner than a human hair) yet holds more than 21 million pixels. (b) This matrix of 42 CCDs serves the Kepler telescope. (credit a: modification of work by US Department of Energy; credit b: modification of work by NASA and Ball Aerospace) 208 Chapter 6 Astronomical Instruments Because CCDs typically record as much as 60\u201370% of all the photons that strike them, and the best silicon and infrared CCDs exceed 90% sensitivity, we can detect much fainter objects. Among these are many small moons around the outer planets, icy dwarf planets beyond Pluto, and dwarf galaxies of stars. CCDs also provide more accurate measurements of the brightness of astronomical objects than photography, and their output is digital\u2014in the form of numbers that can go directly into a computer for analysis. Infrared Observations Observing the universe in the infrared band of the spectrum presents some additional challenges. The infrared region extends from wavelengths near 1 micrometer (\u00b5m), which is about the long wavelength sensitivity limit of both CCDs and photography, to 100 micrometers or longer. Recall from the discussion on radiation and spectra that infrared is \u201cheat radiation\u201d (given off at temperatures that we humans are comfortable with). The main challenge to astronomers using infrared is to distinguish between the tiny amount of heat radiation that reaches Earth from stars and galaxies, and the much greater heat radiated by the telescope itself and our planet\u2019s atmosphere. Typical temperatures on Earth\u2019s surface are near 300 K, and the atmosphere through which observations are made is only a little cooler. According to Wien\u2019s law (from the chapter on Radiation and Spectra), the telescope, the observatory, and even the sky are radiating infrared energy with a peak wavelength of about 10 micrometers. To infrared eyes, everything on Earth is brightly aglow\u2014including the telescope and", " camera (Figure 6.15). The challenge is to detect faint cosmic sources against this sea of infrared light. Another way to look at this is that an astronomer using infrared must always contend with the situation that a visible-light observer would face if working in broad daylight with a telescope and optics lined with bright fluorescent lights. Figure 6.15. Infrared Eyes. Infrared waves can penetrate places in the universe from which light is blocked, as shown in this infrared image where the plastic bag blocks visible light but not infrared. (credit: NASA/JPL-Caltech/R. Hurt (SSC)) To solve this problem, astronomers must protect the infrared detector from nearby radiation, just as you would shield photographic film from bright daylight. Since anything warm radiates infrared energy, the detector must be isolated in very cold surroundings; often, it is held near absolute zero (1 to 3 K) by immersing it in liquid helium. The second step is to reduce the radiation emitted by the telescope structure and optics, and to block this heat from reaching the infrared detector. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 209 Check out The Infrared Zoo (https://openstaxcollege.org/l/30IFZoo) to get a sense of what familiar objects look like with infrared radiation. Slide the slider to change the wavelength of radiation for the picture, and click the arrow to see other animals. Spectroscopy Spectroscopy is one of the astronomer\u2019s most powerful tools, providing information about the composition, temperature, motion, and other characteristics of celestial objects. More than half of the time spent on most large telescopes is used for spectroscopy. The many different wavelengths present in light can be separated by passing them through a spectrometer to form a spectrum. The design of a simple spectrometer is illustrated in Figure 6.16. Light from the source (actually, the image of a source produced by the telescope) enters the instrument through a small hole or narrow slit, and is collimated (made into a beam of parallel rays) by a lens. The light then passes through a prism, producing a spectrum: different wavelengths leave the prism in different directions because each wavelength is bent by a different amount when it enters and leaves the prism. A second lens placed behind the prism focuses the many different images of the slit or entrance hole onto a CCD or other detecting device. This collection of images (spread out by color) is the spectrum that astronomers can then analyze at a later point. As spectroscopy spreads the light out into more and more collecting bins, fewer photons go into each bin, so either a larger telescope is needed or the integration time must be greatly increased\u2014usually both. Figure 6.16. Prism Spectrometer. The light from the telescope is focused on a slit. A prism (or grating) disperses the light into a spectrum, which is then photographed or recorded electronically. In practice, astronomers today are more likely to use a different device, called a grating, to disperse the spectrum. A grating is a piece of material with thousands of grooves on its surface. While it functions completely differently, a grating, like a prism, also spreads light out into a spectrum. 210 Chapter 6 Astronomical Instruments 6.4 RADIO TELESCOPES Learning Objectives By the end of this section, you will be able to: Describe how radio waves from space are detected Identify the world\u2019s largest radio telescopes Define the technique of interferometry and discuss the benefits of interferometers over single-dish telescopes In addition to visible and infrared radiation, radio waves from astronomical objects can also be detected from the surface of Earth. In the early 1930s, Karl G. Jansky, an engineer at Bell Telephone Laboratories, was experimenting with antennas for long-range radio communication when he encountered some mysterious static\u2014radio radiation coming from an unknown source (Figure 6.17). He discovered that this radiation came in strongest about four minutes earlier on each successive day and correctly concluded that since Earth\u2019s sidereal rotation period (how long it takes us to rotate relative to the stars) is four minutes shorter than a solar day, the radiation must be originating from some region fixed on the celestial sphere. Subsequent investigation showed that the source of this radiation was part of the Milky Way Galaxy; Jansky had discovered the first source of cosmic radio waves. Figure 6.17. First Radio Telescope. This rotating radio antenna was used by Jansky in his serendipitous discovery of radio radiation from the Milky Way. In 1936, Grote Reber, who was an amateur astronomer interested in radio communications, used galvanized iron and wood to build the first antenna specifically designed to receive cosmic radio waves. Over the years, Reber built several such antennas and used them to carry out pioneering surveys of the sky for celestial radio sources; he remained active", " in radio astronomy for more than 30 years. During the first decade, he worked practically alone because professional astronomers had not yet recognized the vast potential of radio astronomy. Detection of Radio Energy from Space It is important to understand that radio waves cannot be \u201cheard\u201d: they are not the sound waves you hear coming out of the radio receiver in your home or car. Like light, radio waves are a form of electromagnetic radiation, but unlike light, we cannot detect them with our senses\u2014we must rely on electronic equipment to This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 211 pick them up. In commercial radio broadcasting, we encode sound information (music or a newscaster\u2019s voice) into radio waves. These must be decoded at the other end and then turned back into sound by speakers or headphones. The radio waves we receive from space do not, of course, have music or other program information encoded in them. If cosmic radio signals were translated into sound, they would sound like the static you hear when scanning between stations. Nevertheless, there is information in the radio waves we receive\u2014information that can tell us about the chemistry and physical conditions of the sources of the waves. Just as vibrating charged particles can produce electromagnetic waves (see the Radiation and Spectra chapter), electromagnetic waves can make charged particles move back and forth. Radio waves can produce a current in conductors of electricity such as metals. An antenna is such a conductor: it intercepts radio waves, which create a feeble current in it. The current is then amplified in a radio receiver until it is strong enough to measure or record. Like your television or radio, receivers can be tuned to select a single frequency (channel). In astronomy, however, it is more common to use sophisticated data-processing techniques that allow thousands of separate frequency bands to be detected simultaneously. Thus, the astronomical radio receiver operates much like a spectrometer on a visible-light or infrared telescope, providing information about how much radiation we receive at each wavelength or frequency. After computer processing, the radio signals are recorded on magnetic disks for further analysis. Radio waves are reflected by conducting surfaces, just as light is reflected from a shiny metallic surface, and according to the same laws of optics. A radio-reflecting telescope consists of a concave metal reflector (called a dish), analogous to a telescope mirror. The radio waves collected by the dish are reflected to a focus, where they can then be directed to a receiver and analyzed. Because humans are such visual creatures, radio astronomers often construct a pictorial representation of the radio sources they observe. Figure 6.18 shows such a radio image of a distant galaxy, where radio telescopes reveal vast jets and complicated regions of radio emissions that are completely invisible in photographs taken with light. Figure 6.18. Radio Image. This image has been constructed of radio observations at the Very Large Array of a galaxy called Cygnus A. Colors have been added to help the eye sort out regions of different radio intensities. Red regions are the most intense, blue the least. The visible galaxy would be a small dot in the center of the image. The radio image reveals jets of expelled material (more than 160,000 light-years long) on either side of the galaxy. (credit: NRAO/AUI) Radio astronomy is a young field compared with visible-light astronomy, but it has experienced tremendous growth in recent decades. The world\u2019s largest radio reflectors that can be pointed to any direction in the sky have apertures of 100 meters. One of these has been built at the US National Radio Astronomy Observatory in West Virginia (Figure 6.19). Table 6.2 lists some of the major radio telescopes of the world. 212 Chapter 6 Astronomical Instruments Figure 6.19. Robert C. Byrd Green Bank Telescope. This fully steerable radio telescope in West Virginia went into operation in August 2000. Its dish is about 100 meters across. (credit: modification of work by \u201cb3nscott\u201d/Flickr) Major Radio Observatories of the World Observatory Location Description Website Individual Radio Dishes Arecibo Observatory Arecibo, Puerto Rico 305-m fixed dish www.naic.edu Green Bank Telescope (GBT) Green Bank, WV 110 \u00d7 100-m steerable dish www.science.nrao.edu/facilities/ gbt Effelsberg 100-m Telescope Lovell Telescope Canberra Deep Space Communication Complex (CDSCC) Goldstone Deep Space Communications Complex (GDSCC) Bonn, Germany 100-m steerable dish www.mpifr-bonn.mpg.de/en/ effelsberg Manchester, England Tidbinbilla, Australia 76-m steerable dish www.jb.man.ac.uk/aboutus/ lovell 70-m steerable dish www.cdscc.nasa.gov Barstow, CA 70-m", " steerable dish www.gdscc.nasa.gov Parkes Observatory Parkes, Australia 64-m steerable dish www.parkes.atnf.csiro.au Arrays of Radio Dishes Table 6.2 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 213 Major Radio Observatories of the World Observatory Location Description Website Square Kilometre Array (SKA) South Africa and Western Australia Thousands of dishes, km2 collecting area, partial array in 2020 www.skatelescope.org Atacama Large Millimeter/ submillimeter Array (ALMA) Very Large Array (VLA) Atacama desert, Northern Chile 66 7-m and 12-m dishes www.almaobservatory.org Socorro, New Mexico 27-element array of 25-m dishes (36-km baseline) www.science.nrao.edu/facilities/ vla Westerbork Synthesis Radio Telescope (WSRT) Westerbork, the Netherlands 12-element array of 25-m dishes (1.6-km baseline) www.astron.nl/radioobservatory/public/public-0 Very Long Baseline Array (VLBA) Ten US sites, HI to the Virgin Islands 10-element array of 25-m dishes (9000 km baseline) www.science.nrao.edu/facilities/ vlba Australia Telescope Compact Array (ATCA) Several sites in Australia Multi-Element Radio Linked Interferometer Network (MERLIN) Cambridge, England, and other British sites Millimeter-wave Telescopes 8-element array (seven 22-m dishes plus Parkes 64 m) Network of seven dishes (the largest is 32 m) www.narrabri.atnf.csiro.au www.e-merlin.ac.uk IRAM Granada, Spain 30-m steerable mmwave dish www.iram-institute.org James Clerk Maxwell Telescope (JCMT) Mauna Kea, HI 15-m steerable mmwave dish www.eaobservatory.org/jcmt Nobeyama Radio Observatory (NRO) Minamimaki, Japan 6-element array of 10-m wave dishes www.nro.nao.ac.jp/en Hat Creek Radio Observatory (HCRO) Cassel, CA 6-element array of 5-m wave dishes www.sri.com/researchdevelopment/specializedfacilities/hat-creek-radioobservatory Table 6.2 214 Chapter 6 Astronomical Instruments Radio Interferometry As we discussed earlier, a telescope\u2019s ability to show us fine detail (its resolution) depends upon its aperture, but it also depends upon the wavelength of the radiation that the telescope is gathering. The longer the waves, the harder it is to resolve fine detail in the images or maps we make. Because radio waves have such long wavelengths, they present tremendous challenges for astronomers who need good resolution. In fact, even the largest radio dishes on Earth, operating alone, cannot make out as much detail as the typical small visible- light telescope used in a college astronomy lab. To overcome this difficulty, radio astronomers have learned to sharpen their images by linking two or more radio telescopes together electronically. Two or more telescopes linked together in this way are called an interferometer. \u201cInterferometer\u201d may seem like a strange term because the telescopes in an interferometer work cooperatively; they don\u2019t \u201cinterfere\u201d with each other. Interference, however, is a technical term for the way that multiple waves interact with each other when they arrive in our instruments, and this interaction allows us to coax more detail out of our observations. The resolution of an interferometer depends upon the separation of the telescopes, not upon their individual apertures. Two telescopes separated by 1 kilometer provide the same resolution as would a single dish 1 kilometer across (although they are not, of course, able to collect as much radiation as a radio-wave bucket that is 1 kilometer across). To get even better resolution, astronomers combine a large number of radio dishes into an interferometer array. In effect, such an array works like a large number of two-dish interferometers, all observing the same part of the sky together. Computer processing of the results permits the reconstruction of a high-resolution radio image. The most extensive such instrument in the United States is the National Radio Astronomy Observatory\u2019s Very Large Array (VLA) near Socorro, New Mexico. It consists of 27 movable radio telescopes (on railroad tracks), each having an aperture of 25 meters, spread over a total span of about 36 kilometers. By electronically combining the signals from all of its individual telescopes, this array permits the radio astronomer to make pictures of the sky", " at radio wavelengths comparable to those obtained with a visible-light telescope, with a resolution of about 1 arcsecond. The Atacama Large Millimeter/submillimeter array (ALMA) in the Atacama Desert of Northern Chile (Figure 6.20), at an altitude of 16,400 feet, consists of 12 7-meter and 54 12-meter telescopes, and can achieve baselines up to 16 kilometers. Since it became operational in 2013, it has made observations at resolutions down to 6 milliarcseconds (0.006 arcseconds), a remarkable achievement for radio astronomy. Figure 6.20. Atacama Large Millimeter/Submillimeter Array (ALMA). Located in the Atacama Desert of Northern Chile, ALMA currently provides the highest resolution for radio observations. (credit: ESO/S. Guisard) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 215 Watch this documentary (https://openstax.org/l/30ALMAdoc) that explains the work that went into designing and building ALMA, discusses some of its first images, and explores its future. Initially, the size of interferometer arrays was limited by the requirement that all of the dishes be physically wired together. The maximum dimensions of the array were thus only a few tens of kilometers. However, larger interferometer separations can be achieved if the telescopes do not require a physical connection. Astronomers, with the use of current technology and computing power, have learned to time the arrival of electromagnetic waves coming from space very precisely at each telescope and combine the data later. If the telescopes are as far apart as California and Australia, or as West Virginia and Crimea in Ukraine, the resulting resolution far surpasses that of visible-light telescopes. The United States operates the Very Long Baseline Array (VLBA), made up of 10 individual telescopes stretching from the Virgin Islands to Hawaii (Figure 6.21). The VLBA, completed in 1993, can form astronomical images with a resolution of 0.0001 arcseconds, permitting features as small as 10 astronomical units (AU) to be distinguished at the center of our Galaxy. Figure 6.21. Very Long Baseline Array. This map shows the distribution of 10 antennas that constitute an array of radio telescopes stretching across the United States and its territories. Recent advances in technology have also made it possible to do interferometry at visible-light and infrared wavelengths. At the beginning of the twenty-first century, three observatories with multiple telescopes each began using their dishes as interferometers, combining their light to obtain a much greater resolution. In addition, a dedicated interferometric array was built on Mt. Wilson in California. Just as in radio arrays, these observations allow astronomers to make out more detail than a single telescope could provide. Visible-Light Interferometers Telescope Name Location Mirrors Status CHARA Array (Center for High Angular Resolution Astronomy) Mount Wilson, CA Six 1-m telescopes Operational since 2004 Longest Baseline (m) 400 Table 6.3 216 Chapter 6 Astronomical Instruments Visible-Light Interferometers Longest Baseline (m) Telescope Name Location Mirrors Status 200 Very Large Telescope 85 Keck I and II telescopes 22.8 Large Binocular Telescope Cerro Paranal, Chile Mauna Kea, HI Mount Graham, AZ Four 8.2-m telescopes Completed 2000 Two 10-m telescopes Operated from 2001 to 2012 Two 8.4-m telescopes First light 2004 Table 6.3 Radar Astronomy Radar is the technique of transmitting radio waves to an object in our solar system and then detecting the radio radiation that the object reflects back. The time required for the round trip can be measured electronically with great precision. Because we know the speed at which radio waves travel (the speed of light), we can determine the distance to the object or a particular feature on its surface (such as a mountain). Radar observations have been used to determine the distances to planets and how fast things are moving in the solar system (using the Doppler effect, discussed in the Radiation and Spectra chapter). Radar waves have played important roles in navigating spacecraft throughout the solar system. In addition, as will be discussed in later chapters, radar observations have determined the rotation periods of Venus and Mercury, probed tiny Earth-approaching asteroids, and allowed us to investigate the mountains and valleys on the surfaces of Mercury, Venus, Mars, and the large moons of Jupiter. Any radio dish can be used as a radar telescope if it is equipped with a powerful transmitter as well as a receiver. The most spectacular facility in the world for radar astronomy is the 1000-foot (305-meter) telescope at Arecibo in Puerto Rico (Figure 6.22). The Arecibo telescope is too large to be pointed directly at different parts of the sky. Instead, it is constructed in a huge natural \u201cbowl\u201d (more than a mere", " dish) formed by several hills, and it is lined with reflecting metal panels. A limited ability to track astronomical sources is achieved by moving the receiver system, which is suspended on cables 100 meters above the surface of the bowl. An even larger (500-meter) radar telescope is currently under construction. It is the Five-hundred-meter Aperture Spherical Telescope (FAST) in China and is expected to be completed in 2016. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 217 Figure 6.22. Largest Radio and Radar Dish. The Arecibo Observatory, with its 1000-foot radio dish-filling valley in Puerto Rico, is part of the National Astronomy and Ionosphere Center, operated by SRI International, USRA, and UMET under a cooperative agreement with the National Science Foundation. (credit: National Astronomy and Ionosphere Center, Cornell U., NSF) 6.5 OBSERVATIONS OUTSIDE EARTH\u2019S ATMOSPHERE Learning Objectives By the end of this section, you will be able to: List the advantages of making astronomical observations from space Explain the importance of the Hubble Space Telescope Describe some of the major space-based observatories astronomers use Earth\u2019s atmosphere blocks most radiation at wavelengths shorter than visible light, so we can only make direct ultraviolet, X-ray, and gamma ray observations from space (though indirect gamma ray observations can be made from Earth). Getting above the distorting effects of the atmosphere is also an advantage at visible and infrared wavelengths. The stars don\u2019t \u201ctwinkle\u201d in space, so the amount of detail you can observe is limited only by the size of your instrument. On the other hand, it is expensive to place telescopes into space, and repairs can present a major challenge. This is why astronomers continue to build telescopes for use on the ground as well as for launching into space. Airborne and Space Infrared Telescopes Water vapor, the main source of atmospheric interference for making infrared observations, is concentrated in the lower part of Earth\u2019s atmosphere. For this reason, a gain of even a few hundred meters in elevation can make an important difference in the quality of an infared observatory site. Given the limitations of high mountains, most of which attract clouds and violent storms, and the fact that the ability of humans to perform complex tasks degrades at high altitudes, it was natural for astronomers to investigate the possibility of observing infrared waves from airplanes and ultimately from space. Infrared observations from airplanes have been made since the 1960s, starting with a 15-centimeter telescope on board a Learjet. From 1974 through 1995, NASA operated a 0.9-meter airborne telescope flying regularly out of the Ames Research Center south of San Francisco. Observing from an altitude of 12 kilometers, the telescope was above 99% of the atmospheric water vapor. More recently, NASA (in partnership with the German Aerospace Center) has constructed a much larger 2.5-meter telescope, called the Stratospheric Observatory for Infrared Astronomy (SOFIA), which flies in a modified Boeing 747SP (Figure 6.23). 218 Chapter 6 Astronomical Instruments Figure 6.23. Stratospheric Observatory for Infrared Astronomy (SOFIA). SOFIA allows observations to be made above most of Earth\u2019s atmospheric water vapor. (credit: NASA To find out more about SOFIA, watch this video (https://openstaxcollege.org/l/30SOFIAvid) provided by NASA\u2019s Armstrong Flight Research Center. Getting even higher and making observations from space itself have important advantages for infrared astronomy. First is the elimination of all interference from the atmosphere. Equally important is the opportunity to cool the entire optical system of the instrument in order to nearly eliminate infrared radiation from the telescope itself. If we tried to cool a telescope within the atmosphere, it would quickly become coated with condensing water vapor and other gases, making it useless. Only in the vacuum of space can optical elements be cooled to hundreds of degrees below freezing and still remain operational. The first orbiting infrared observatory, launched in 1983, was the Infrared Astronomical Satellite (IRAS), built as a joint project by the United States, the Netherlands, and Britain. IRAS was equipped with a 0.6-meter telescope cooled to a temperature of less than 10 K. For the first time, the infrared sky could be seen as if it were night, rather than through a bright foreground of atmospheric and telescope emissions. IRAS carried out a rapid but comprehensive survey of the entire infrared sky over a 10-month period, cataloging about 350,000 sources of infrared radiation. Since then, several other infrared telescopes have operated in space with much better sensitivity and resolution due to improvements in infrared detectors. The most powerful of these infrared telescopes is the 0.85-meter Spitzer Space Telescope,", " which launched in 2003. A few of its observations are shown in Figure 6.24. With infrared observations, astronomers can detect cooler parts of cosmic objects, such as the dust clouds around star nurseries and the remnants of dying stars, that visible-light images don\u2019t reveal. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 219 Figure 6.24. Observations from the Spitzer Space Telescope (SST). These infrared images\u2014a region of star formation, the remnant of an exploded star, and a region where an old star is losing its outer shell\u2014show just a few of the observations made and transmitted back to Earth from the SST. Since our eyes are not sensitive to infrared rays, we don\u2019t perceive colors from them. The colors in these images have been selected by astronomers to highlight details like the composition or temperature in these regions. (credit \u201cFlame nebula\u201d: modification of work by NASA (X-ray: NASA/CXC/PSU/K.Getman, E.Feigelson, M.Kuhn & the MYStIX team; Infrared:NASA/JPL-Caltech); credit \u201cCassiopeia A\u201d: modification of work by NASA/JPL-Caltech; credit \u201cHelix nebula\u201d: modification of work by NASA/JPL-Caltech) Hubble Space Telescope In April 1990, a great leap forward in astronomy was made with the launch of the Hubble Space Telescope (HST). With an aperture of 2.4 meters, this is the largest telescope put into space so far. (Its aperture was limited by the size of the payload bay in the Space Shuttle that served as its launch vehicle.) It was named for Edwin Hubble, the astronomer who discovered the expansion of the universe in the 1920s (whose work we will discuss in the chapters on Galaxies). HST is operated jointly by NASA\u2019s Goddard Space Flight Center and the Space Telescope Science Institute in Baltimore. It was the first orbiting observatory designed to be serviced by Shuttle astronauts and, over the years since it was launched, they made several visits to improve or replace its initial instruments and to repair some of the systems that operate the spacecraft (Figure 6.1)\u2014though this repair program has now been discontinued, and no more visits or improvements will be made. With the Hubble, astronomers have obtained some of the most detailed images of astronomical objects from the solar system outward to the most distant galaxies. Among its many great achievements is the Hubble Ultra- Deep Field, an image of a small region of the sky observed for almost 100 hours. It contains views of about 10,000 galaxies, some of which formed when the universe was just a few percent of its current age (Figure 6.25). 220 Chapter 6 Astronomical Instruments Figure 6.25. Hubble Ultra-Deep Field (HUDF). The Hubble Space Telescope has provided an image of a specific region of space built from data collected between September 24, 2003, and January 16, 2004. These data allow us to search for galaxies that existed approximately 13 billion years ago. (credit: modification of work by NASA) The HST\u2019s mirror was ground and polished to a remarkable degree of accuracy. If we were to scale up its 2.4-meter mirror to the size of the entire continental United States, there would be no hill or valley larger than about 6 centimeters in its smooth surface. Unfortunately, after it was launched, scientists discovered that the primary mirror had a slight error in its shape, equal to roughly 1/50 the width of a human hair. Small as that sounds, it was enough to ensure that much of the light entering the telescope did not come to a clear focus and that all the images were blurry. (In a misplaced effort to save money, a complete test of the optical system had not been carried out before launch, so the error was not discovered until HST was in orbit.) The solution was to do something very similar to what we do for astronomy students with blurry vision: put corrective optics in front of their eyes. In December 1993, in one of the most exciting and difficult space missions ever flown, astronauts captured the orbiting telescope and brought it back into the shuttle payload bay. There they installed a package containing compensating optics as well as a new, improved camera before releasing HST back into orbit. The telescope now works as it was intended to, and further missions to it were able to install even more advanced instruments to take advantage of its capabilities. High-Energy Observatories Ultraviolet, X-ray, and direct gamma-ray (high-energy electromagnetic wave) observations can be made only from space. Such observations first became possible in 1946, with V2 rockets captured from Germany after World War II. The US Naval Research Laboratory put instruments on these rockets for a series of pioneering flights, used initially to detect ultraviolet radiation from the Sun.", " Since then, many other rockets have been launched to make X-ray and ultraviolet observations of the Sun, and later of other celestial objects. Beginning in the 1960s, a steady stream of high-energy observatories has been launched into orbit to reveal and explore the universe at short wavelengths. Among recent X-ray telescopes is the Chandra X-ray Observatory, which was launched in 1999 (Figure 6.26). It is producing X-ray images with unprecedented resolution and sensitivity. Designing instruments that can collect and focus energetic radiation like X-rays and gamma rays This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 221 is an enormous technological challenge. The 2002 Nobel Prize in physics was awarded to Riccardo Giacconi, a pioneer in the field of building and launching sophisticated X-ray instruments. In 2008, NASA launched the Fermi Gamma-ray Space Telescope, designed to measure cosmic gamma rays at energies greater than any previous telescope, and thus able to collect radiation from some of the most energetic events in the universe. Figure 6.26. Chandra X-Ray Satellite. Chandra, the world\u2019s most powerful X-ray telescope, was developed by NASA and launched in July 1999. (credit: modification of work by NASA) One major challenge is to design \u201cmirrors\u201d to reflect such penetrating radiation as X-rays and gamma rays, which normally pass straight through matter. However, although the technical details of design are more complicated, the three basic components of an observing system, as we explained earlier in this chapter, are the same at all wavelengths: a telescope to gather up the radiation, filters or instruments to sort the radiation according to wavelength, and some method of detecting and making a permanent record of the observations. Table 6.4 lists some of the most important active space observatories that humanity has launched. Gamma-ray detections can also be made from Earth\u2019s surface by using the atmosphere as the primary detector. When a gamma ray hits our atmosphere, it accelerates charged particles (mostly electrons) in the atmosphere. Those energetic particles hit other particles in the atmosphere and give off their own radiation. The effect is a cascade of light and energy that can be detected on the ground. The VERITAS array in Arizona and the H.E.S.S. array in Namibia are two such ground-based gamma-ray observatories. Recent Observatories in Space Observatory Date Operation Began Bands of the Spectrum Notes Website Hubble Space Telescope (HST) 1990 visible, UV, IR 2.4-m mirror; images and spectra www.hubblesite.org Table 6.4 222 Chapter 6 Astronomical Instruments Recent Observatories in Space Observatory Date Operation Began Bands of the Spectrum Notes Website Chandra X-Ray Observatory 1999 X-rays X-ray images and spectra www.chandra.si.edu XMM-Newton 1999 X-rays X-ray spectroscopy http://www.cosmos.esa.int/ web/xmm-newton International Gamma-Ray Astrophysics Laboratory (INTEGRAL) 2002 X- and gammarays Spitzer Space Telescope 2003 IR higher resolution gamma-ray images 0.85-m telescope http://sci.esa.int/integral/ www.spitzer.caltech.edu Fermi Gamma-ray Space Telescope 2008 gammarays first high-energy gamma-ray observations fermi.gsfc.nasa.gov Kepler 2009 visiblelight planet finder http://kepler.nasa.gov Wide-field Infrared Survey Explorer (WISE) 2009 IR whole-sky map, asteroid searches www.nasa.gov/ mission_pages/WISE/main Gaia 2013 visiblelight Precise map of the Milky Way http://sci.esa.int/gaia/ Table 6.4 6.6 THE FUTURE OF LARGE TELESCOPES Learning Objectives By the end of this section, you will be able to: Describe the next generation of ground- and space-based observatories Explain some of the challenges involved in building these observatories If you\u2019ve ever gone on a hike, you have probably been eager to see what lies just around the next bend in the path. Researchers are no different, and astronomers and engineers are working on the technologies that will allow us to explore even more distant parts of the universe and to see them more clearly. The premier space facility planned for the next decade is the James Webb Space Telescope (Figure 6.27), which (in a departure from tradition) is named after one of the early administrators of NASA instead of a scientist. This telescope will have a mirror 6 meters in diameter, made up, like the Keck telescopes, of 36 small hexagons. These will have to unfold into place once the telescope reaches its stable orbit point, some 1.5 million kilometers This OpenStax book is available for free at", " http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 223 from Earth (where no astronauts can currently travel if it needs repair.) The telescope is scheduled for launch in 2018 and should have the sensitivity needed to detect the very first generation of stars, formed when the universe was only a few hundred million years old. With the ability to measure both visible and infrared wavelengths, it will serve as the successor to both HST and the Spitzer Space Telescope. Figure 6.27. James Webb Space Telescope (JWST). This image shows some of the mirrors of the JWST as they underwent cryogenic testing. The mirrors were exposed to extreme temperatures in order to gather accurate measurements on changes in their shape as they heated and cooled. (credit: NASA/MSFC/David Higginbotham/Emmett Given Watch this video (https://openstaxcollege.org/l/30JWSTvid) to learn more about the James Webb Space Telescope and how it will build upon the work that Hubble has allowed us to begin in exploring the universe. On the ground, astronomers have started building the Large Synoptic Survey Telescope (LSST), an 8.4-meter telescope with a significantly larger field of view than any existing telescopes. It will rapidly scan the sky to find transients, phenomena that change quickly, such as exploding stars and chunks of rock that orbit near Earth. The LSST is expected to see first light in 2021. The international gamma-ray community is planning the Cherenkov Telescope Array (CTA), two arrays of telescopes, one in each hemisphere, which will indirectly measure gamma rays from the ground. The CTA will measure gamma-ray energies a thousand times as great as the Fermi telescope can detect. Several groups of astronomers around the globe interested in studying visible light and infrared are exploring the feasibility of building ground-based telescopes with mirrors larger than 30 meters across. Stop and think what this means: 30 meters is one-third the length of a football field. It is technically impossible to build and transport a single astronomical mirror that is 30 meters or larger in diameter. The primary mirror of these giant telescopes will consist of smaller mirrors, all aligned so that they act as a very large mirror in combination. These include the Thirty-Meter Telescope for which construction has begun at the top of Mauna Kea in Hawaii. The most ambitious of these projects is the European Extremely Large Telescope (E-ELT) (Figure 6.28). (Astronomers try to outdo each other not only with the size of these telescopes, but also their names!) The design of the E-ELT calls for a 39.3-meter primary mirror, which will follow the Keck design and be made up of 798 hexagonal mirrors, each 1.4 meters in diameter and all held precisely in position so that they form a continuous surface. 224 Chapter 6 Astronomical Instruments Construction on the site in the Atacama Desert in Northern Chile started in 2014. The E-ELT, along with the Thirty Meter Telescope and the Giant Magellan Telescope, which are being built by international consortia led by US astronomers, will combine light-gathering power with high-resolution imaging. These powerful new instruments will enable astronomers to tackle many important astronomical problems. For example, they should be able to tell us when, where, and how often planets form around other stars. They should even be able to provide us images and spectra of such planets and thus, perhaps, give us the first real evidence (from the chemistry of these planets\u2019 atmospheres) that life exists elsewhere. Figure 6.28. Artist\u2019s Conception of the European Extremely Large Telescope. The primary mirror in this telescope is 39.3 meters across. The telescope is under construction in the Atacama Desert in Northern Chile. (credit: ESO/L. Cal\u00e7ada Check out this fun diagram (https://openstaxcollege.org/l/30JWSTdiag) comparing the sizes of the largest planned and existing telescopes to a regulation basketball and tennis court. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 225 CHAPTER 6 REVIEW KEY TERMS adaptive optics systems used with telescopes that can compensate for distortions in an image introduced by the atmosphere, thus resulting in sharper images aperture diameter of the primary lens or mirror of a telescope charge-coupled device (CCD) array of high-sensitivity electronic detectors of electromagnetic radiation, used at the focus of a telescope (or camera lens) to record an image or spectrum chromatic aberration distortion that causes an image to appear fuzzy when each wavelength coming into a transparent material focuses at a different spot detector device sensitive to electromagnetic radiation that makes a record of astronomical observations eyepiece magnifying lens used to view the image produced by the objective lens or primary mirror of a telescope focus", " (of telescope) point where the rays of light converged by a mirror or lens meet interference process in which waves mix together such that their crests and troughs can alternately reinforce and cancel one another interferometer instrument that combines electromagnetic radiation from one or more telescopes to obtain a resolution equivalent to what would be obtained with a single telescope with a diameter equal to the baseline separating the individual separate telescopes interferometer array combination of multiple radio dishes to, in effect, work like a large number of two-dish interferometers prime focus point in a telescope where the objective lens or primary mirror focuses the light radar technique of transmitting radio waves to an object and then detecting the radiation that the object reflects back to the transmitter; used to measure the distance to, and motion of, a target object or to form images of it reflecting telescope telescope in which the principal light collector is a concave mirror refracting telescope telescope in which the principal light collector is a lens or system of lenses resolution detail in an image; specifically, the smallest angular (or linear) features that can be distinguished seeing unsteadiness of Earth\u2019s atmosphere, which blurs telescopic images; good seeing means the atmosphere is steady telescope instrument for collecting visible-light or other electromagnetic radiation 226 Chapter 6 Astronomical Instruments SUMMARY 6.1 Telescopes A telescope collects the faint light from astronomical sources and brings it to a focus, where an instrument can sort the light according to wavelength. Light is then directed to a detector, where a permanent record is made. The light-gathering power of a telescope is determined by the diameter of its aperture, or opening\u2014that is, by the area of its largest or primary lens or mirror. The primary optical element in a telescope is either a convex lens (in a refracting telescope) or a concave mirror (in a reflector) that brings the light to a focus. Most large telescopes are reflectors; it is easier to manufacture and support large mirrors because the light does not have to pass through glass. 6.2 Telescopes Today New technologies for creating and supporting lightweight mirrors have led to the construction of a number of large telescopes since 1990. The site for an astronomical observatory must be carefully chosen for clear weather, dark skies, low water vapor, and excellent atmospheric seeing (low atmospheric turbulence). The resolution of a visible-light or infrared telescope is degraded by turbulence in Earth\u2019s atmosphere. The technique of adaptive optics, however, can make corrections for this turbulence in real time and produce exquisitely detailed images. 6.3 Visible-Light Detectors and Instruments Visible-light detectors include the human eye, photographic film, and charge-coupled devices (CCDs). Detectors that are sensitive to infrared radiation must be cooled to very low temperatures since everything in and near the telescope gives off infrared waves. A spectrometer disperses the light into a spectrum to be recorded for detailed analysis. 6.4 Radio Telescopes In the 1930s, radio astronomy was pioneered by Karl G. Jansky and Grote Reber. A radio telescope is basically a radio antenna (often a large, curved dish) connected to a receiver. Significantly enhanced resolution can be obtained with interferometers, including interferometer arrays like the 27-element VLA and the 66-element ALMA. Expanding to very long baseline interferometers, radio astronomers can achieve resolutions as precise as 0.0001 arcsecond. Radar astronomy involves transmitting as well as receiving. The largest radar telescope currently in operation is a 305-meter bowl at Arecibo. 6.5 Observations outside Earth\u2019s Atmosphere Infrared observations are made with telescopes aboard aircraft and in space, as well as from ground-based facilities on dry mountain peaks. Ultraviolet, X-ray, and gamma-ray observations must be made from above the atmosphere. Many orbiting observatories have been flown to observe in these bands of the spectrum in the last few decades. The largest-aperture telescope in space is the Hubble Space telescope (HST), the most significant infrared telescope is Spitzer, and Chandra and Fermi are the premier X-ray and gamma-ray observatories, respectively. 6.6 The Future of Large Telescopes New and even larger telescopes are on the drawing boards. The James Webb Space Telescope, a 6-meter successor to Hubble, is currently scheduled for launch in 2018. Gamma-ray astronomers are planning to build the CTA to measure very energetic gamma rays. Astronomers are building the LSST to observe with an unprecedented field of view and a new generation of visible-light/infrared telescopes with apertures of 24.5 to 39 meters in diameter. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 227 FOR FURTHER EXPLORATION Articles Blades, J. C. \u201cFixing the Hubble One Last Time.\u201d Sky & Telescope (October 2008", "): 26. On the last Shuttle service mission and what the Hubble was then capable of doing. Brown, A. \u201cHow Gaia will Map a Billion Stars.\u201d Astronomy (December 2014): 32. Nice review of the mission to do photometry and spectroscopy of all stars above a certain brightness. Irion, R. \u201cPrime Time.\u201d Astronomy (February 2001): 46. On how time is allotted on the major research telescopes. Jedicke, Peter & Robert. \u201cThe Coming Giant Sky Patrols.\u201d Sky & Telescope (September 2008): 30. About giant telescopes to survey the sky continuously. Lazio, Joseph, et al. \u201cTuning in to the Universe: 21st Century Radio Astronomy.\u201d Sky & Telescope (July 2008): 21. About ALMA and the Square Kilometer Array. Lowe, Jonathan. \u201cMirror, Mirror.\u201d Sky & Telescope (December 2007): 22. On the Large Binocular Telescope in Arizona. Lowe, Jonathan. \u201cNext Light: Tomorrow\u2019s Monster Telescopes.\u201d Sky & Telescope (April 2008): 20. About plans for extremely large telescopes on the ground. Mason, Todd & Robin. \u201cPalomar\u2019s Big Eye.\u201d Sky & Telescope (December 2008): 36. On the Hale 200-inch telescope. Subinsky, Raymond. \u201cWho Really Invented the Telescope.\u201d Astronomy (August 2008): 84. Brief historical introduction, focusing on Hans Lippershey. Websites Websites for major telescopes are given in Table 6.1, Table 6.2, Table 6.3, and Table 6.4. Videos Astronomy from the Stratosphere: SOFIA: https://www.youtube.com/watch?v=NV98BcBBA9c. A talk by Dr. Dana Backman (1:15:32) Galaxies Viewed in Full Spectrum of Light: https://www.youtube.com/watch?v=368K0iQv8nE. Scientists with the Spitzer Observatory show how a galaxy looks different at different wavelengths (6:22) Lifting the Cosmic Veil: Highlights from a Decade of the Spitzer Space Telescope: https://www.youtube.com/ watch?v=nkrNQcwkY78. A talk by Dr. Michael Bicay (1:42:44) COLLABORATIVE GROUP ACTIVITIES A. Most large telescopes get many more proposals for observing projects than there is night observing time available in a year. Suppose your group is the telescope time allocation committee reporting to an observatory director. What criteria would you use in deciding how to give out time on the telescope? What steps could you take to make sure all your colleagues thought the process was fair and people would still talk to you at future astronomy meetings? 228 Chapter 6 Astronomical Instruments B. Your group is a committee of nervous astronomers about to make a proposal to the government ministers of your small European country to chip in with other countries to build the world\u2019s largest telescope in the high, dry desert of the Chilean Andes Mountains. You expect the government ministers to be very skeptical about supporting this project. What arguments would you make to convince them to participate? C. The same government ministers we met in the previous activity ask you to draw up a list of the pros and cons of having the world\u2019s largest telescope in the mountains of Chile (instead of a mountain in Europe). What would your group list in each column? D. Your group should discuss and make a list of all the ways in which an observing session at a large visible- light telescope and a large radio telescope might differ. (Hint: Bear in mind that because the Sun is not especially bright at many radio wavelengths, observations with radio telescopes can often be done during the day.) E. Another \u201cenvironmental threat\u201d to astronomy (besides light pollution) comes from the spilling of terrestrial communications into the \u201cchannels\u201d\u2014wavelengths and frequencies\u2014previously reserved for radio astronomy. For example, the demand for cellular phones means that more and more radio channels will be used for this purpose. The faint signals from cosmic radio sources could be drowned in a sea of earthly conversation (translated and sent as radio waves). Assume your group is a congressional committee being lobbied by both radio astronomers, who want to save some clear channels for doing astronomy, and the companies that stand to make a lot of money from expanding cellular phone use. What arguments would sway you to each side? F. When the site for the new Thirty-Meter Telescope on Hawaii\u2019s Mauna Kea was dedicated, a group of native Hawaiians announced opposition to the project because astronomers were building too many telescopes on a mountain that native Hawaiians consider a sacred site. You can read more about this controversy at http://www.nytimes.com/2015/12/04", "/science/space/hawaii-court-rescinds-permit-to-build-thirty-meter- telescope.html?_r=0 and at http://www.nature.com/news/the-mountain-top-battle-over-the-thirty-meter- telescope-1.18446. Once your group has the facts, discuss the claims of each side in the controversy. How do you think it should be resolved? G. If you could propose to use a large modern telescope, what would you want to find out? What telescope would you use and why? H. Light pollution (spilled light in the night sky making it difficult to see the planets and stars) used to be an issue that concerned mostly astronomers. Now spilled light at night is also of concern to environmentalists and those worrying about global warming. Can your group come up with some non-astronomical reasons to be opposed to light pollution? EXERCISES Review Questions 1. What are the three basic components of a modern astronomical instrument? Describe each in one to two sentences. 2. Name the two spectral windows through which electromagnetic radiation easily reaches the surface of Earth and describe the largest-aperture telescope currently in use for each window. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 229 3. List the largest-aperture single telescope currently in use in each of the following bands of the electromagnetic spectrum: radio, X-ray, gamma ray. 4. When astronomers discuss the apertures of their telescopes, they say bigger is better. Explain why. 5. The Hooker telescope at Palomar Observatory has a diameter of 5 m, and the Keck I telescope has a diameter of 10 m. How much more light can the Keck telescope collect than the Hooker telescope in the same amount of time? 6. What is meant by \u201creflecting\u201d and \u201crefracting\u201d telescopes? 7. Why are the largest visible-light telescopes in the world made with mirrors rather than lenses? 8. Compare the eye, photographic film, and CCDs as detectors for light. What are the advantages and disadvantages of each? 9. What is a charge-coupled device (CCD), and how is it used in astronomy? 10. Why is it difficult to observe at infrared wavelengths? What do astronomers do to address this difficulty? 11. Radio and radar observations are often made with the same antenna, but otherwise they are very different techniques. Compare and contrast radio and radar astronomy in terms of the equipment needed, the methods used, and the kind of results obtained. 12. Look back at Figure 6.18 of Cygnus A and read its caption again. The material in the giant lobes at the edges of the image had to have been ejected from the center at least how many years ago? 13. Why do astronomers place telescopes in Earth\u2019s orbit? What are the advantages for the different regions of the spectrum? 14. What was the problem with the Hubble Space Telescope and how was it solved? 15. Describe the techniques radio astronomers use to obtain a resolution comparable to what astronomers working with visible light can achieve. 16. What kind of visible-light and infrared telescopes on the ground are astronomers planning for the future? Why are they building them on the ground and not in space? 17. Describe one visible-light or infrared telescope that astronomers are planning to launch into space in the future. Thought Questions 18. What happens to the image produced by a lens if the lens is \u201cstopped down\u201d (the aperture reduced, thereby reducing the amount of light passing through the lens) with an iris diaphragm\u2014a device that covers its periphery? 19. What would be the properties of an ideal astronomical detector? How closely do the actual properties of a CCD approach this ideal? 20. Many decades ago, the astronomers on the staff of Mount Wilson and Palomar Observatories each received about 60 nights per year for their observing programs. Today, an astronomer feels fortunate to get 10 nights per year on a large telescope. Can you suggest some reasons for this change? 21. The largest observatory complex in the world is on Mauna Kea, the tallest mountain on Earth. What are some factors astronomers consider when selecting an observatory site? Don\u2019t forget practical ones. Should astronomers, for example, consider building an observatory on Denali (Mount McKinley) or Mount Everest? 230 Chapter 6 Astronomical Instruments 22. Suppose you are looking for sites for a visible-light observatory, an infrared observatory, and a radio observatory. What are the main criteria of excellence for each? What sites are actually considered the best today? 23. Radio astronomy involves wavelengths much longer than those of visible light, and many orbiting observatories have probed the universe for radiation of very short wavelengths.", " What sorts of objects and physical conditions would you expect to be associated with emission of radiation at very long and very short wavelengths? 24. The dean of a university located near the ocean (who was not a science major in college) proposes building an infrared telescope right on campus and operating it in a nice heated dome so that astronomers will be comfortable on cold winter nights. Criticize this proposal, giving your reasoning. Figuring For Yourself 25. What is the area, in square meters, of a 10-m telescope? 26. Approximately 9000 stars are visible to the naked eye in the whole sky (imagine that you could see around the entire globe and both the northern and southern hemispheres), and there are about 41,200 square degrees on the sky. How many stars are visible per square degree? Per square arcsecond? 27. Theoretically (that is, if seeing were not an issue), the resolution of a telescope is inversely proportional to its diameter. How much better is the resolution of the ALMA when operating at its longest baseline than the resolution of the Arecibo telescope? 28. In broad daylight, the size of your pupil is typically 3 mm. In dark situations, it expands to about 7 mm. How much more light can it gather? 29. How much more light can be gathered by a telescope that is 8 m in diameter than by your fully dark- adapted eye at 7 mm? 30. How much more light can the Keck telescope (with its 10-m diameter mirror) gather than an amateur telescope whose mirror is 25 cm (0.25 m) across? 31. People are often bothered when they discover that reflecting telescopes have a second mirror in the middle to bring the light out to an accessible focus where big instruments can be mounted. \u201cDon\u2019t you lose light?\u201d people ask. Well, yes, you do, but there is no better alternative. You can estimate how much light is lost by such an arrangement. The primary mirror (the one at the bottom in Figure 6.6) of the Gemini North telescope is 8 m in diameter. The secondary mirror at the top is about 1 m in diameter. Use the formula for the area of a circle to estimate what fraction of the light is blocked by the secondary mirror. 32. Telescopes can now be operated remotely from a warm room, but until about 25 years ago, astronomers worked at the telescope to guide it so that it remained pointed in exactly the right place. In a large telescope, like the Palomar 200-inch telescope, astronomers sat in a cage at the top of the telescope, where the secondary mirror is located, as shown in Figure 6.6. Assume for the purpose of your calculation that the diameter of this cage was 40 inches. What fraction of the light is blocked? 33. The HST cost about $1.7 billion for construction and $300 million for its shuttle launch, and it costs $250 million per year to operate. If the telescope lasts for 20 years, what is the total cost per year? Per day? If the telescope can be used just 30% of the time for actual observations, what is the cost per hour and per minute for the astronomer\u2019s observing time on this instrument? What is the cost per person in the United States? Was your investment in the Hubble Space telescope worth it? 34. How much more light can the James Webb Space Telescope (with its 6-m diameter mirror) gather than the Hubble Space Telescope (with a diameter of 2.4 m)? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 6 Astronomical Instruments 231 35. The Palomar telescope\u2019s 5-m mirror weighs 14.5 tons. If a 10-m mirror were constructed of the same thickness as Palomar\u2019s (only bigger), how much would it weigh? 232 Chapter 6 Astronomical Instruments This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 233 7 OTHER WORLDS: AN INTRODUCTION TO THE SOLAR SYSTEM Figure 7.1. \u201cSelf-Portrait\u201d of Mars. This picture was taken by the Curiosity Rover on Mars in 2012. The image is reconstructed digitally from 55 different images taken by a camera on the rover\u2019s extended mast, so that the many positions of the mast (which acted like a selfie stick) are edited out. (credit: modification of work by NASA/JPL-Caltech/MSSS) Chapter Outline 7.1 Overview of Our Planetary System 7.2 Composition and Structure of Planets 7.3 Dating Planetary Surfaces 7.4 Origin of the Solar System Thinking Ahead Surrounding the Sun is a complex system of worlds with a wide range of conditions: eight major planets, many dwarf planets, hundreds of moons,", " and countless smaller objects. Thanks largely to visits by spacecraft, we can now envision the members of the solar system as other worlds like our own, each with its own chemical and geological history, and unique sights that interplanetary tourists may someday visit. Some have called these past few decades the \u201cgolden age of planetary exploration,\u201d comparable to the golden age of exploration in the fifteenth century, when great sailing ships plied Earth\u2019s oceans and humanity became familiar with our own planet\u2019s surface. In this chapter, we discuss our planetary system and introduce the idea of comparative planetology\u2014studying how the planets work by comparing them with one another. We want to get to know the planets not only for what we can learn about them, but also to see what they can tell us about the origin and evolution of the entire solar system. In the upcoming chapters, we describe the better-known members of the solar system and begin to compare them to the thousands of planets that have been discovered recently, orbiting other stars. 234 Chapter 7 Other Worlds: An Introduction to the Solar System 7.1 OVERVIEW OF OUR PLANETARY SYSTEM Learning Objectives By the end of this section, you will be able to: Describe how the objects in our solar system are identified, explored, and characterized Describe the types of small bodies in our solar system, their locations, and how they formed Model the solar system with distances from everyday life to better comprehend distances in space The solar system[1] consists of the Sun and many smaller objects: the planets, their moons and rings, and such \u201cdebris\u201d as asteroids, comets, and dust. Decades of observation and spacecraft exploration have revealed that most of these objects formed together with the Sun about 4.5 billion years ago. They represent clumps of material that condensed from an enormous cloud of gas and dust. The central part of this cloud became the Sun, and a small fraction of the material in the outer parts eventually formed the other objects. During the past 50 years, we have learned more about the solar system than anyone imagined before the space age. In addition to gathering information with powerful new telescopes, we have sent spacecraft directly to many members of the planetary system. (Planetary astronomy is the only branch of our science in which we can, at least vicariously, travel to the objects we want to study.) With evocative names such as Voyager, Pioneer, Curiosity, and Pathfinder, our robot explorers have flown past, orbited, or landed on every planet, returning images and data that have dazzled both astronomers and the public. In the process, we have also investigated two dwarf planets, hundreds of fascinating moons, four ring systems, a dozen asteroids, and several comets (smaller members of our solar system that we will discuss later). Our probes have penetrated the atmosphere of Jupiter and landed on the surfaces of Venus, Mars, our Moon, Saturn\u2019s moon Titan, the asteroids Eros and Itokawa, and the Comet Churyumov-Gerasimenko (usually referred to as 67P). Humans have set foot on the Moon and returned samples of its surface soil for laboratory analysis (Figure 7.2). We have even discovered other places in our solar system that might be able to support some kind of life. 1 The generic term for a group of planets and other bodies circling a star is planetary system. Ours is called the solar system because our Sun is sometimes called Sol. Strictly speaking, then, there is only one solar system; planets orbiting other stars are in planetary systems. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 235 Figure 7.2. Astronauts on the Moon. The lunar lander and surface rover from the Apollo 15 mission are seen in this view of the one place beyond Earth that has been explored directly by humans. (credit: modification of work by David R. Scott, NASA View this gallery of NASA images (https://openstaxcollege.org/l/30projapolloarc) that trace the history of the Apollo mission. An Inventory The Sun, a star that is brighter than about 80% of the stars in the Galaxy, is by far the most massive member of the solar system, as shown in Table 7.1. It is an enormous ball about 1.4 million kilometers in diameter, with surface layers of incandescent gas and an interior temperature of millions of degrees. The Sun will be discussed in later chapters as our first, and best-studied, example of a star. Mass of Members of the Solar System Object Percentage of Total Mass of Solar System Sun Jupiter Comets 99.80 0.10 0.0005\u20130.03 (estimate) All other planets and dwarf planets 0.04 Moons and rings 0.00005 Table 7.1 236 Chapter 7 Other Worlds: An Introduction to the Solar", " System Mass of Members of the Solar System Object Percentage of Total Mass of Solar System Asteroids Cosmic dust Table 7.1 0.000002 (estimate) 0.0000001 (estimate) Table 7.1 also shows that most of the material of the planets is actually concentrated in the largest one, Jupiter, which is more massive than all the rest of the planets combined. Astronomers were able to determine the masses of the planets centuries ago using Kepler\u2019s laws of planetary motion and Newton\u2019s law of gravity to measure the planets\u2019 gravitational effects on one another or on moons that orbit them (see Orbits and Gravity). Today, we make even more precise measurements of their masses by tracking their gravitational effects on the motion of spacecraft that pass near them. Beside Earth, five other planets were known to the ancients\u2014Mercury, Venus, Mars, Jupiter, and Saturn\u2014and two were discovered after the invention of the telescope: Uranus and Neptune. The eight planets all revolve in the same direction around the Sun. They orbit in approximately the same plane, like cars traveling on concentric tracks on a giant, flat racecourse. Each planet stays in its own \u201ctraffic lane,\u201d following a nearly circular orbit about the Sun and obeying the \u201ctraffic\u201d laws discovered by Galileo, Kepler, and Newton. Besides these planets, we have also been discovering smaller worlds beyond Neptune that are called trans-Neptunian objects or TNOs (see Figure 7.3). The first to be found, in 1930, was Pluto, but others have been discovered during the twenty- first century. One of them, Eris, is about the same size as Pluto and has at least one moon (Pluto has five known moons.) The largest TNOs are also classed as dwarf planets, as is the largest asteroid, Ceres. (Dwarf planets will be discussed further in the chapter on Rings, Moons, and Pluto). To date, more than 1750 of these TNOs have been discovered. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 237 Figure 7.3. Orbits of the Planets. All eight major planets orbit the Sun in roughly the same plane. The five currently known dwarf planets are also shown: Eris, Haumea, Pluto, Ceres, and Makemake. Note that Pluto\u2019s orbit is not in the plane of the planets. Each of the planets and dwarf planets also rotates (spins) about an axis running through it, and in most cases the direction of rotation is the same as the direction of revolution about the Sun. The exceptions are Venus, which rotates backward very slowly (that is, in a retrograde direction), and Uranus and Pluto, which also have strange rotations, each spinning about an axis tipped nearly on its side. We do not yet know the spin orientations of Eris, Haumea, and Makemake. The four planets closest to the Sun (Mercury through Mars) are called the inner or terrestrial planets. Often, the Moon is also discussed as a part of this group, bringing the total of terrestrial objects to five. (We generally call Earth\u2019s satellite \u201cthe Moon,\u201d with a capital M, and the other satellites \u201cmoons,\u201d with lowercase m\u2019s.) The terrestrial planets are relatively small worlds, composed primarily of rock and metal. All of them have solid surfaces that bear the records of their geological history in the forms of craters, mountains, and volcanoes (Figure 7.4). 238 Chapter 7 Other Worlds: An Introduction to the Solar System Figure 7.4. Surface of Mercury. The pockmarked face of the terrestrial world of Mercury is more typical of the inner planets than the watery surface of Earth. This black-and-white image, taken with the Mariner 10 spacecraft, shows a region more than 400 kilometers wide. (credit: modification of work by NASA/John Hopkins University Applied Physics Laboratory/Carnegie Institution of Washington) The next four planets (Jupiter through Neptune) are much larger and are composed primarily of lighter ices, liquids, and gases. We call these four the jovian planets (after \u201cJove,\u201d another name for Jupiter in mythology) or giant planets\u2014a name they richly deserve (Figure 7.5). More than 1400 Earths could fit inside Jupiter, for example. These planets do not have solid surfaces on which future explorers might land. They are more like vast, spherical oceans with much smaller, dense cores. Figure 7.5. The Four Giant Planets. This montage shows the four giant planets: Jupiter, Saturn, Uranus, and Neptune. Below them, Earth is shown to scale. (credit: modification of work by NASA, Solar System Exploration) Near the outer", " edge of the system lies Pluto, which was the first of the distant icy worlds to be discovered beyond Neptune (Pluto was visited by a spacecraft, the NASA New Horizons mission, in 2015 [see Figure 7.6]). Table 7.2 summarizes some of the main facts about the planets. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 239 Figure 7.6. Pluto Close-up. This intriguing image from the New Horizons spacecraft, taken when it flew by the dwarf planet in July 2015, shows some of its complex surface features. The rounded white area is temporarily being called the Sputnik Plain, after humanity\u2019s first spacecraft. (credit: modification of work by NASA/Johns Hopkins University Applied Physics Laboratory/Southwest Research Institute) The Planets Name Distance from Sun (AU)[2] Revolution Period (y) Diameter (km) Mass (1023 kg) Density (g/cm3)[3] Mercury 0.39 Venus Earth Mars 0.72 1.00 1.52 Jupiter 5.20 Saturn 9.54 Uranus 19.18 0.24 0.62 1.00 1.88 11.86 29.46 84.07 4,878 12,120 12,756 6,787 3.3 48.7 59.8 6.4 142,984 18,991 120,536 5686 51,118 866 Neptune 30.06 164.82 49,660 1030 5.4 5.2 5.5 3.9 1.3 0.7 1.3 1.6 Table 7. Comparing Densities Let\u2019s compare the densities of several members of the solar system. The density of an object equals its mass divided by its volume. The volume (V) of a sphere (like a planet) is calculated using the equation An AU (or astronomical unit) is the distance from Earth to the Sun. 2 3 We give densities in units where the density of water is 1 g/cm3. To get densities in units of kg/m3, multiply the given value by 1000. 240 Chapter 7 Other Worlds: An Introduction to the Solar System V = 4 3 \u03c0R3 where \u03c0 (the Greek letter pi) has a value of approximately 3.14. Although planets are not perfect spheres, this equation works well enough. The masses and diameters of the planets are given in Table 7.2. For data on selected moons, see Appendix G. Let\u2019s use Saturn\u2019s moon Mimas as our example, with a mass of 4 \u00d7 1019 kg and a diameter of approximately 400 km (radius, 200 km = 2 \u00d7 105 m). Solution The volume of Mimas is \u00d7 3.14 \u00d7 \u239e \u239b \u239d2 \u00d7 105 m \u23a0 3 = 3.3 \u00d7 1016 m3. 4 3 Density is mass divided by volume: 4 \u00d7 1019 kg 3.3 \u00d7 1016 m3 = 1.2 \u00d7 103 kg/m3. Note that the density of water in these units is 1000 kg/m3, so Mimas must be made mainly of ice, not rock. (Note that the density of Mimas given in Appendix G is 1.2, but the units used there are different. In that table, we give density in units of g/cm3, for which the density of water equals 1. Can you show, by converting units, that 1 g/cm3 is the same as 1000 kg/m3?) Check Your Learning Calculate the average density of our own planet, Earth. Show your work. How does it compare to the density of an ice moon like Mimas? See Table 7.2 for data. Answer: For a sphere, density = mass 3\u03c0R3\u239e For Earth, then, \u239b \u239d 4 \u23a0 kg/m3. density = 6 \u00d7 1024 kg 4.2 \u00d7 2.6 \u00d7 1020 m3 = 5.5 \u00d7 103 kg/m3. This density is four to five times greater than Mimas\u2019. In fact, Earth is the densest of the planets Learn more about NASA\u2019s mission to Pluto (https://openstaxcollege.org/l/30NASAmisspluto) and see high-resolution images of Pluto\u2019s moon Charon. Smaller Members of the Solar System Most of the planets are accompanied by one or more moons; only Mercury and Venus move through space alone. There are more than 180 known moons orbiting planets and dwarf planets (see Appendix G for a listing This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 241 of the larger ones), and undoubtedly many other small ones", " remain undiscovered. The largest of the moons are as big as small planets and just as interesting. In addition to our Moon, they include the four largest moons of Jupiter (called the Galilean moons, after their discoverer) and the largest moons of Saturn and Neptune (confusingly named Titan and Triton). Each of the giant planets also has rings made up of countless small bodies ranging in size from mountains to mere grains of dust, all in orbit about the equator of the planet. The bright rings of Saturn are, by far, the easiest to see. They are among the most beautiful sights in the solar system (Figure 7.7). But, all four ring systems are interesting to scientists because of their complicated forms, influenced by the pull of the moons that also orbit these giant planets. Figure 7.7. Saturn and Its Rings. This 2007 Cassini image shows Saturn and its complex system of rings, taken from a distance of about 1.2 million kilometers. This natural-color image is a composite of 36 images taken over the course of 2.5 hours. (credit: modification of work by NASA/JPL/Space Science Institute) The solar system has many other less-conspicuous members. Another group is the asteroids, rocky bodies that orbit the Sun like miniature planets, mostly in the space between Mars and Jupiter (although some do cross the orbits of planets like Earth\u2014see Figure 7.8). Most asteroids are remnants of the initial population of the solar system that existed before the planets themselves formed. Some of the smallest moons of the planets, such as the moons of Mars, are very likely captured asteroids. Figure 7.8. Asteroid Eros. This small Earth-crossing asteroid image was taken by the NEAR-Shoemaker spacecraft from an altitude of about 100 kilometers. This view of the heavily cratered surface is about 10 kilometers wide. The spacecraft orbited Eros for a year before landing gently on its surface. (credit: modification of work by NASA/JHUAPL) Another class of small bodies is composed mostly of ice, made of frozen gases such as water, carbon dioxide, and carbon monoxide; these objects are called comets (see Figure 7.9). Comets also are remnants from the formation of the solar system, but they were formed and continue (with rare exceptions) to orbit the Sun in 242 Chapter 7 Other Worlds: An Introduction to the Solar System distant, cooler regions\u2014stored in a sort of cosmic deep freeze. This is also the realm of the larger icy worlds, called dwarf planets. Figure 7.9. Comet Churyumov-Gerasimenko (67P). This image shows Comet Churyumov-Gerasimenko, also known as 67P, near its closest approach to the Sun in 2015, as seen from the Rosetta spacecraft. Note the jets of gas escaping from the solid surface. (credit: modification of work by ESA/Rosetta/NAVACAM, CC BY-SA IGO 3.0 (http://creativecommons.org/licenses/by-sa/3.0/igo/) ) Finally, there are countless grains of broken rock, which we call cosmic dust, scattered throughout the solar system. When these particles enter Earth\u2019s atmosphere (as millions do each day) they burn up, producing a brief flash of light in the night sky known as a meteor (meteors are often referred to as shooting stars). Occasionally, some larger chunk of rocky or metallic material survives its passage through the atmosphere and lands on Earth. Any piece that strikes the ground is known as a meteorite. (You can see meteorites on display in many natural history museums and can sometimes even purchase pieces of them from gem and mineral dealers.) Carl Sagan: Solar System Advocate The best-known astronomer in the world during the 1970s and 1980s, Carl Sagan devoted most of his professional career to studying the planets and considerable energy to raising public awareness of what we can learn from exploring the solar system (see Figure 7.10). Born in Brooklyn, New York, in 1934, Sagan became interested in astronomy as a youngster; he also credits science fiction stories for sustaining his fascination with what\u2019s out in the universe. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 243 Figure 7.10. Carl Sagan (1934\u20131996) and Neil deGrasse Tyson. Sagan was Tyson\u2019s inspiration to become a scientist. (credit \u201cSagan\u201d: modification of work by NASA, JPL; credit \u201cTyson\u201d: modification of work by Bruce F. Press) In the early 1960s, when many scientists still thought Venus might turn out to be a hospitable place, Sagan calculated that the thick atmosphere of Venus could act like a giant greenhouse, keeping the heat in and raising the temperature enormously.", " He showed that the seasonal changes astronomers had seen on Mars were caused, not by vegetation, but by wind-blown dust. He was a member of the scientific teams for many of the robotic missions that explored the solar system and was instrumental in getting NASA to put a message-bearing plaque aboard the Pioneer spacecraft, as well as audio-video records on the Voyager spacecraft\u2014all of them destined to leave our solar system entirely and send these little bits of Earth technology out among the stars. To encourage public interest and public support of planetary exploration, Sagan helped found The Planetary Society, now the largest space-interest organization in the world. He was a tireless and eloquent advocate of the need to study the solar system close-up and the value of learning about other worlds in order to take better care of our own. Sagan simulated conditions on early Earth to demonstrate how some of life\u2019s fundamental building blocks might have formed from the \u201cprimordial soup\u201d of natural compounds on our planet. In addition, he and his colleagues developed computer models showing the consequences of nuclear war for Earth would be even more devastating than anyone had thought (this is now called the nuclear winter hypothesis) and demonstrating some of the serious consequences of continued pollution of our atmosphere. Sagan was perhaps best known, however, as a brilliant popularizer of astronomy and the author of many books on science, including the best-selling Cosmos, and several evocative tributes to solar system exploration such as The Cosmic Connection and Pale Blue Dot. His book The Demon Haunted World, completed just before his death in 1996, is perhaps the best antidote to fuzzy thinking about pseudo- science and irrationality in print today. An intriguing science fiction novel he wrote, titled Contact, which became a successful film as well, is still recommended by many science instructors as a scenario for making contact with life elsewhere that is much more reasonable than most science fiction. Sagan was a master, too, of the television medium. His 13-part public television series, Cosmos, was seen by an estimated 500 million people in 60 countries and has become one of the most-watched series in the 244 Chapter 7 Other Worlds: An Introduction to the Solar System history of public broadcasting. A few astronomers scoffed at a scientist who spent so much time in the public eye, but it is probably fair to say that Sagan\u2019s enthusiasm and skill as an explainer won more friends for the science of astronomy than anyone or anything else in the second half of the twentieth century. In the two decades since Sagan\u2019s death, no other scientist has achieved the same level of public recognition. Perhaps closest is the director of the Hayden Planetarium, Neil deGrasse Tyson, who followed in Sagan\u2019s footsteps by making an updated version of the Cosmos program in 2014. Tyson is quick to point out that Sagan was his inspiration to become a scientist, telling how Sagan invited him to visit for a day at Cornell when he was a high school student looking for a career. However, the media environment has fragmented a great deal since Sagan\u2019s time. It is interesting to speculate whether Sagan could have adapted his communication style to the world of cable television, Twitter, Facebook, and podcasts Two imaginative videos provide a tour of the solar system objects we have been discussing. Shane Gellert\u2019s I Need Some Space (https://openstaxcollege.org/l/30needsomespace) uses NASA photography and models to show the various worlds with which we share our system. In the more science fiction-oriented Wanderers (https://openstaxcollege.org/l/30wanderers) video, we see some of the planets and moons as tourist destinations for future explorers, with commentary taken from recordings by Carl Sagan. A Scale Model of the Solar System Astronomy often deals with dimensions and distances that far exceed our ordinary experience. What does 1.4 billion kilometers\u2014the distance from the Sun to Saturn\u2014really mean to anyone? It can be helpful to visualize such large systems in terms of a scale model. In our imaginations, let us build a scale model of the solar system, adopting a scale factor of 1 billion (109)\u2014that is, reducing the actual solar system by dividing every dimension by a factor of 109. Earth, then, has a diameter of 1.3 centimeters, about the size of a grape. The Moon is a pea orbiting this at a distance of 40 centimeters, or a little more than a foot away. The Earth-Moon system fits into a standard backpack. In this model, the Sun is nearly 1.5 meters in diameter, about the average height of an adult, and our Earth is at a distance of 150 meters\u2014about one city block\u2014from the Sun. Jupiter is five blocks away from the Sun, and its diameter is 15 centimeters, about the size of a very large grapefruit. Saturn is 10 blocks from the Sun; Uranus, 20 blocks; and Neptune, 30 blocks. Pluto, with a distance that varies quite a bit during its 249-year orbit", ", is currently just beyond 30 blocks and getting farther with time. Most of the moons of the outer solar system are the sizes of various kinds of seeds orbiting the grapefruit, oranges, and lemons that represent the outer planets. In our scale model, a human is reduced to the dimensions of a single atom, and cars and spacecraft to the size of molecules. Sending the Voyager spacecraft to Neptune involves navigating a single molecule from the Earth\u2013grape toward a lemon 5 kilometers away with an accuracy equivalent to the width of a thread in a spider\u2019s web. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 245 If that model represents the solar system, where would the nearest stars be? If we keep the same scale, the closest stars would be tens of thousands of kilometers away. If you built this scale model in the city where you live, you would have to place the representations of these stars on the other side of Earth or beyond. By the way, model solar systems like the one we just presented have been built in cities throughout the world. In Sweden, for example, Stockholm\u2019s huge Globe Arena has become a model for the Sun, and Pluto is represented by a 12-centimeter sculpture in the small town of Delsbo, 300 kilometers away. Another model solar system is in Washington on the Mall between the White House and Congress (perhaps proving they are worlds apart?). Names in the Solar System We humans just don\u2019t feel comfortable until something has a name. Types of butterflies, new elements, and the mountains of Venus all need names for us to feel we are acquainted with them. How do we give names to objects and features in the solar system? Planets and moons are named after gods and heroes in Greek and Roman mythology (with a few exceptions among the moons of Uranus, which have names drawn from English literature). When William Herschel, a German immigrant to England, first discovered the planet we now call Uranus, he wanted to name it Georgium Sidus (George\u2019s star) after King George III of his adopted country. This caused such an outcry among astronomers in other nations, however, that the classic tradition was upheld\u2014and has been maintained ever since. Luckily, there were a lot of minor gods in the ancient pantheon, so plenty of names are left for the many small moons we are discovering around the giant planets. (Appendix G lists the larger moons). Comets are often named after their discoverers (offering an extra incentive to comet hunters). Asteroids are named by their discoverers after just about anyone or anything they want. Recently, asteroid names have been used to recognize people who have made significant contributions to astronomy, including the three original authors of this book. That was pretty much all the naming that was needed while our study of the solar system was confined to Earth. But now, our spacecraft have surveyed and photographed many worlds in great detail, and each world has a host of features that also need names. To make sure that naming things in space remains multinational, rational, and somewhat dignified, astronomers have given the responsibility of approving names to a special committee of the International Astronomical Union (IAU), the body that includes scientists from every country that does astronomy. This IAU committee has developed a set of rules for naming features on other worlds. For example, craters on Venus are named for women who have made significant contributions to human knowledge and welfare. Volcanic features on Jupiter\u2019s moon Io, which is in a constant state of volcanic activity, are named after gods of fire and thunder from the mythologies of many cultures. Craters on Mercury commemorate famous novelists, playwrights, artists, and composers. On Saturn\u2019s moon Tethys, all the features are named after characters and places in Homer\u2019s great epic poem, The Odyssey. As we explore further, it may well turn out that more places in the solar system need names than Earth history can provide. Perhaps by then, explorers and settlers on these worlds will be ready to develop their own names for the places they may (if but for a while) call home. 246 Chapter 7 Other Worlds: An Introduction to the Solar System You may be surprised to know that the meaning of the word planet has recently become controversial because we have discovered many other planetary systems that don\u2019t look very much like our own. Even within our solar system, the planets differ greatly in size and chemical properties. The biggest dispute concerns Pluto, which is much smaller than the other eight major planets. The category of dwarf planet was invented to include Pluto and similar icy objects beyond Neptune. But is a dwarf planet also a planet? Logically, it should be, but even this simple issue of grammar has been the subject of heated debate among both astronomers and the general public. 7.2 COMPOSITION AND STRUCTURE OF PLANETS Learning", " Objectives By the end of this section, you will be able to: Describe the characteristics of the giant planets, terrestrial planets, and small bodies in the solar system Explain what influences the temperature of a planet\u2019s surface Explain why there is geological activity on some planets and not on others The fact that there are two distinct kinds of planets\u2014the rocky terrestrial planets and the gas-rich jovian planets\u2014leads us to believe that they formed under different conditions. Certainly their compositions are dominated by different elements. Let us look at each type in more detail. The Giant Planets The two largest planets, Jupiter and Saturn, have nearly the same chemical makeup as the Sun; they are composed primarily of the two elements hydrogen and helium, with 75% of their mass being hydrogen and 25% helium. On Earth, both hydrogen and helium are gases, so Jupiter and Saturn are sometimes called gas planets. But, this name is misleading. Jupiter and Saturn are so large that the gas is compressed in their interior until the hydrogen becomes a liquid. Because the bulk of both planets consists of compressed, liquefied hydrogen, we should really call them liquid planets. Under the force of gravity, the heavier elements sink toward the inner parts of a liquid or gaseous planet. Both Jupiter and Saturn, therefore, have cores composed of heavier rock, metal, and ice, but we cannot see these regions directly. In fact, when we look down from above, all we see is the atmosphere with its swirling clouds (Figure 7.11). We must infer the existence of the denser core inside these planets from studies of each planet\u2019s gravity. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 247 Figure 7.11. Jupiter. This true-color image of Jupiter was taken from the Cassini spacecraft in 2000. (credit: modification of work by NASA/JPL/ University of Arizona) Uranus and Neptune are much smaller than Jupiter and Saturn, but each also has a core of rock, metal, and ice. Uranus and Neptune were less efficient at attracting hydrogen and helium gas, so they have much smaller atmospheres in proportion to their cores. Chemically, each giant planet is dominated by hydrogen and its many compounds. Nearly all the oxygen present is combined chemically with hydrogen to form water (H2O). Chemists call such a hydrogen-dominated composition reduced. Throughout the outer solar system, we find abundant water (mostly in the form of ice) and reducing chemistry. The Terrestrial Planets The terrestrial planets are quite different from the giants. In addition to being much smaller, they are composed primarily of rocks and metals. These, in turn, are made of elements that are less common in the universe as a whole. The most abundant rocks, called silicates, are made of silicon and oxygen, and the most common metal is iron. We can tell from their densities (see Table 7.2) that Mercury has the greatest proportion of metals (which are denser) and the Moon has the lowest. Earth, Venus, and Mars all have roughly similar bulk compositions: about one third of their mass consists of iron-nickel or iron-sulfur combinations; two thirds is made of silicates. Because these planets are largely composed of oxygen compounds (such as the silicate minerals of their crusts), their chemistry is said to be oxidized. When we look at the internal structure of each of the terrestrial planets, we find that the densest metals are in a central core, with the lighter silicates near the surface. If these planets were liquid, like the giant planets, we could understand this effect as the result the sinking of heavier elements due to the pull of gravity. This leads us to conclude that, although the terrestrial planets are solid today, at one time they must have been hot enough to melt. Differentiation is the process by which gravity helps separate a planet\u2019s interior into layers of different compositions and densities. The heavier metals sink to form a core, while the lightest minerals float to the surface to form a crust. Later, when the planet cools, this layered structure is preserved. In order for a rocky planet to differentiate, it must be heated to the melting point of rocks, which is typically more than 1300 K. 248 Chapter 7 Other Worlds: An Introduction to the Solar System Moons, Asteroids, and Comets Chemically and structurally, Earth\u2019s Moon is like the terrestrial planets, but most moons are in the outer solar system, and they have compositions similar to the cores of the giant planets around which they orbit. The three largest moons\u2014Ganymede and Callisto in the jovian system, and Titan in the saturnian system\u2014are composed half of frozen water, and half of rocks and metals. Most of these moons differentiated during formation, and today they have cores of rock and metal, with upper layers", " and crusts of very cold and\u2014thus very hard\u2014ice (Figure 7.12). Figure 7.12. Ganymede. This view of Jupiter\u2019s moon Ganymede was taken in June 1996 by the Galileo spacecraft. The brownish gray color of the surface indicates a dusty mixture of rocky material and ice. The bright spots are places where recent impacts have uncovered fresh ice from underneath. (credit: modification of work by NASA/JPL) Most of the asteroids and comets, as well as the smallest moons, were probably never heated to the melting point. However, some of the largest asteroids, such as Vesta, appear to be differentiated; others are fragments from differentiated bodies. Because most asteroids and comets retain their original composition, they represent relatively unmodified material dating back to the time of the formation of the solar system. In a sense, they act as chemical fossils, helping us to learn about a time long ago whose traces have been erased on larger worlds. Temperatures: Going to Extremes Generally speaking, the farther a planet or moon is from the Sun, the cooler its surface. The planets are heated by the radiant energy of the Sun, which gets weaker with the square of the distance. You know how rapidly the heating effect of a fireplace or an outdoor radiant heater diminishes as you walk away from it; the same effect applies to the Sun. Mercury, the closest planet to the Sun, has a blistering surface temperature that ranges from 280\u2013430 \u00b0C on its sunlit side, whereas the surface temperature on Pluto is only about \u2013220 \u00b0C, colder than liquid air. Mathematically, the temperatures decrease approximately in proportion to the square root of the distance from the Sun. Pluto is about 30 AU at its closest to the Sun (or 100 times the distance of Mercury) and about 49 AU at its farthest from the Sun. Thus, Pluto\u2019s temperature is less than that of Mercury by the square root of 100, or a factor of 10: from 500 K to 50 K. In addition to its distance from the Sun, the surface temperature of a planet can be influenced strongly by its atmosphere. Without our atmospheric insulation (the greenhouse effect, which keeps the heat in), the oceans of Earth would be permanently frozen. Conversely, if Mars once had a larger atmosphere in the past, This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 249 it could have supported a more temperate climate than it has today. Venus is an even more extreme example, where its thick atmosphere of carbon dioxide acts as insulation, reducing the escape of heat built up at the surface, resulting in temperatures greater than those on Mercury. Today, Earth is the only planet where surface temperatures generally lie between the freezing and boiling points of water. As far as we know, Earth is the only planet to support life There\u2019s No Place Like Home In the classic film The Wizard of Oz, Dorothy, the heroine, concludes after her many adventures in \u201calien\u201d environments that \u201cthere\u2019s no place like home.\u201d The same can be said of the other worlds in our solar system. There are many fascinating places, large and small, that we might like to visit, but humans could not survive on any without a great deal of artificial assistance. A thick carbon dioxide atmosphere keeps the surface temperature on our neighbor Venus at a sizzling 700 K (near 900 \u00b0F). Mars, on the other hand, has temperatures generally below freezing, with air (also mostly carbon dioxide) so thin that it resembles that found at an altitude of 30 kilometers (100,000 feet) in Earth\u2019s atmosphere. And the red planet is so dry that it has not had any rain for billions of years. The outer layers of the jovian planets are neither warm enough nor solid enough for human habitation. Any bases we build in the systems of the giant planets may well have to be in space or one of their moons\u2014none of which is particularly hospitable to a luxury hotel with a swimming pool and palm trees. Perhaps we will find warmer havens deep inside the clouds of Jupiter or in the ocean under the frozen ice of its moon Europa. All of this suggests that we had better take good care of Earth because it is the only site where life as we know it could survive. Recent human activity may be reducing the habitability of our planet by adding pollutants to the atmosphere, especially the potent greenhouse gas carbon dioxide. Human civilization is changing our planet dramatically, and these changes are not necessarily for the better. In a solar system that seems unready to receive us, making Earth less hospitable to life may be a grave mistake. Geological Activity The crusts of all of the terrestrial planets, as well as of the larger moons, have been modified over their histories by both internal and external forces. Externally, each has been battered by a slow rain of projectiles", " from space, leaving their surfaces pockmarked by impact craters of all sizes (see Figure 7.4). We have good evidence that this bombardment was far greater in the early history of the solar system, but it certainly continues to this day, even if at a lower rate. The collision of more than 20 large pieces of Comet Shoemaker\u2013Levy 9 with Jupiter in the summer of 1994 (see Figure 7.13) is one dramatic example of this process. 250 Chapter 7 Other Worlds: An Introduction to the Solar System Figure 7.13. Comet Shoemaker\u2013Levy 9. In this image of Comet Shoemaker\u2013Levy 9 taken on May 17, 1994, by NASA\u2019s Hubble Space Telescope, you can see about 20 icy fragments into which the comet broke. The comet was approximately 660 million kilometers from Earth, heading on a collision course with Jupiter. (credit: modification of work by NASA, ESA, H. Weaver (STScl), E. Smith (STScl)) Figure 7.14 shows the aftermath of these collisions, when debris clouds larger than Earth could be seen in Jupiter\u2019s atmosphere. Figure 7.14. Jupiter with Huge Dust Clouds. The Hubble Space Telescope took this sequence of images of Jupiter in summer 1994, when fragments of Comet Shoemaker\u2013Levy 9 collided with the giant planet. Here we see the site hit by fragment G, from five minutes to five days after impact. Several of the dust clouds generated by the collisions became larger than Earth. (credit: modification of work by H. Hammel, NASA) During the time all the planets have been subject to such impacts, internal forces on the terrestrial planets have buckled and twisted their crusts, built up mountain ranges, erupted as volcanoes, and generally reshaped the surfaces in what we call geological activity. (The prefix geo means \u201cEarth,\u201d so this is a bit of an \u201cEarth-chauvinist\u201d term, but it is so widely used that we bow to tradition.) Among the terrestrial planets, Earth and Venus have experienced the most geological activity over their histories, although some of the moons in the outer solar system are also surprisingly active. In contrast, our own Moon is a dead world where geological activity ceased billions of years ago. Geological activity on a planet is the result of a hot interior. The forces of volcanism and mountain building are driven by heat escaping from the interiors of planets. As we will see, each of the planets was heated at the time of its birth, and this primordial heat initially powered extensive volcanic activity, even on our Moon. But, small objects such as the Moon soon cooled off. The larger the planet or moon, the longer it retains its internal heat, and therefore the more we expect to see surface evidence of continuing geological activity. The effect is similar to our own experience with a hot baked potato: the larger the potato, the more slowly it cools. If we want a potato to cool quickly, we cut it into small pieces. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 251 For the most part, the history of volcanic activity on the terrestrial planets conforms to the predictions of this simple theory. The Moon, the smallest of these objects, is a geologically dead world. Although we know less about Mercury, it seems likely that this planet, too, ceased most volcanic activity about the same time the Moon did. Mars represents an intermediate case. It has been much more active than the Moon, but less so than Earth. Earth and Venus, the largest terrestrial planets, still have molten interiors even today, some 4.5 billion years after their birth. 7.3 DATING PLANETARY SURFACES Learning Objectives By the end of this section, you will be able to: Explain how astronomers can tell whether a planetary surface is geologically young or old Describe different methods for dating planets How do we know the age of the surfaces we see on planets and moons? If a world has a surface (as opposed to being mostly gas and liquid), astronomers have developed some techniques for estimating how long ago that surface solidified. Note that the age of these surfaces is not necessarily the age of the planet as a whole. On geologically active objects (including Earth), vast outpourings of molten rock or the erosive effects of water and ice, which we call planet weathering, have erased evidence of earlier epochs and present us with only a relatively young surface for investigation. Counting the Craters One way to estimate the age of a surface is by counting the number of impact craters. This technique works because the rate at which impacts have occurred in the solar system has been roughly constant for several billion years. Thus, in the absence of forces to eliminate craters, the number of craters is simply proportional to the length of time the surface has been exposed. This technique has", " been applied successfully to many solid planets and moons (Figure 7.15). Figure 7.15. Our Cratered Moon. This composite image of the Moon\u2019s surface was made from many smaller images taken between November 2009 and February 2011 by the Lunar Reconnaissance Orbiter (LRO) and shows craters of many different sizes. (credit: modification of work by NASA/GSFC/Arizona State University) 252 Chapter 7 Other Worlds: An Introduction to the Solar System Bear in mind that crater counts can tell us only the time since the surface experienced a major change that could modify or erase preexisting craters. Estimating ages from crater counts is a little like walking along a sidewalk in a snowstorm after the snow has been falling steadily for a day or more. You may notice that in front of one house the snow is deep, while next door the sidewalk may be almost clear. Do you conclude that less snow has fallen in front of Ms. Jones\u2019 house than Mr. Smith\u2019s? More likely, you conclude that Jones has recently swept the walk clean and Smith has not. Similarly, the numbers of craters indicate how long it has been since a planetary surface was last \u201cswept clean\u201d by ongoing lava flows or by molten materials ejected when a large impact happened nearby. Still, astronomers can use the numbers of craters on different parts of the same world to provide important clues about how regions on that world evolved. On a given planet or moon, the more heavily cratered terrain will generally be older (that is, more time will have elapsed there since something swept the region clean). Radioactive Rocks Another way to trace the history of a solid world is to measure the age of individual rocks. After samples were brought back from the Moon by Apollo astronauts, the techniques that had been developed to date rocks on Earth were applied to rock samples from the Moon to establish a geological chronology for the Moon. Furthermore, a few samples of material from the Moon, Mars, and the large asteroid Vesta have fallen to Earth as meteorites and can be examined directly (see the chapter on Cosmic Samples and the Origin of the Solar System). Scientists measure the age of rocks using the properties of natural radioactivity. Around the beginning of the twentieth century, physicists began to understand that some atomic nuclei are not stable but can split apart (decay) spontaneously into smaller nuclei. The process of radioactive decay involves the emission of particles such as electrons, or of radiation in the form of gamma rays (see the chapter on Radiation and Spectra). For any one radioactive nucleus, it is not possible to predict when the decay process will happen. Such decay is random in nature, like the throw of dice: as gamblers have found all too often, it is impossible to say just when the dice will come up 7 or 11. But, for a very large number of dice tosses, we can calculate the odds that 7 or 11 will come up. Similarly, if we have a very large number of radioactive atoms of one type (say, uranium), there is a specific time period, called its half-life, during which the chances are fifty-fifty that decay will occur for any of the nuclei. A particular nucleus may last a shorter or longer time than its half-life, but in a large sample, almost exactly half of the nuclei will have decayed after a time equal to one half-life. Half of the remaining nuclei will have decayed after two half-lives pass, leaving only one half of a half\u2014or one quarter\u2014of the original sample (Figure 7.16). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 253 Figure 7.16. Radioactive Decay. This graph shows (in pink) the amount of a radioactive sample that remains after several half-lives have passed. After one half-life, half the sample is left; after two half-lives, one half of the remainder (or one quarter) is left; and after three half-lives, one half of that (or one eighth) is left. Note that, in reality, the decay of radioactive elements in a rock sample would not cause any visible change in the appearance of the rock; the splashes of color are shown here for conceptual purposes only. If you had 1 gram of pure radioactive nuclei with a half-life of 100 years, then after 100 years you would have 1/2 gram; after 200 years, 1/4 gram; after 300 years, only 1/8 gram; and so forth. However, the material does not disappear. Instead, the radioactive atoms are replaced with their decay products. Sometimes the radioactive atoms are called parents and the decay products are called daughter elements. In this way, radioactive elements with half-lives we have determined can provide accurate nuclear clocks. By comparing how much of a radioactive parent element is", " left in a rock to how much of its daughter products have accumulated, we can learn how long the decay process has been going on and hence how long ago the rock formed. Table 7.3 summarizes the decay reactions used most often to date lunar and terrestrial rocks. Radioactive Decay Reaction Used to Date Rocks[4] Parent Daughter Half-Life (billions of years) Samarium-147 Neodymium-143 Rubidium-87 Strontium-87 Thorium-232 Lead-208 Uranium-238 Lead-206 Potassium-40 Argon-40 Table 7.3 106 48.8 14.0 4.47 1.31 4 The number after each element is its atomic weight, equal to the number of protons plus neutrons in its nucleus. This specifies the isotope of the element; different isotopes of the same element differ in the number of neutrons. 254 Chapter 7 Other Worlds: An Introduction to the Solar System PBS provides an evolution series excerpt (https://openstaxcollege.org/l/30pbsradiomat) that explains how we use radioactive elements to date Earth. This Science Channel video (https://openstaxcollege.org/l/30billnyevideo) features Bill Nye the Science Guy showing how scientists have used radioactive dating to determine the age of Earth. When astronauts first flew to the Moon, one of their most important tasks was to bring back lunar rocks for radioactive age-dating. Until then, astronomers and geologists had no reliable way to measure the age of the lunar surface. Counting craters had let us calculate relative ages (for example, the heavily cratered lunar highlands were older than the dark lava plains), but scientists could not measure the actual age in years. Some thought that the ages were as young as those of Earth\u2019s surface, which has been resurfaced by many geological events. For the Moon\u2019s surface to be so young would imply active geology on our satellite. Only in 1969, when the first Apollo samples were dated, did we learn that the Moon is an ancient, geologically dead world. Using such dating techniques, we have been able to determine the ages of both Earth and the Moon: each was formed about 4.5 billion years ago (although, as we shall see, Earth probably formed earlier). We should also note that the decay of radioactive nuclei generally releases energy in the form of heat. Although the energy from a single nucleus is not very large (in human terms), the enormous numbers of radioactive nuclei in a planet or moon (especially early in its existence) can be a significant source of internal energy for that world. Geologists estimate that about half of Earth\u2019s current internal heat budget comes from the decay of radioactive isotopes in its interior. 7.4 ORIGIN OF THE SOLAR SYSTEM Learning Objectives By the end of this section, you will be able to: Describe the characteristics of planets that are used to create formation models of the solar system Describe how the characteristics of extrasolar systems help us to model our own solar system Explain the importance of collisions in the formation of the solar system Much of astronomy is motivated by a desire to understand the origin of things: to find at least partial answers to age-old questions of where the universe, the Sun, Earth, and we ourselves came from. Each planet and moon is a fascinating place that may stimulate our imagination as we try to picture what it would be like to visit. Taken together, the members of the solar system preserve patterns that can tell us about the formation of the entire system. As we begin our exploration of the planets, we want to introduce our modern picture of how the solar system formed. The recent discovery of hundreds of planets in orbit around other stars has shown astronomers that many exoplanetary systems can be quite different from our own solar system. For example, it is common for these systems to include planets intermediate in size between our terrestrial and giant planets. These are often called superearths. Some exoplanet systems even have giant planets close to the star, reversing the order we see in our system. In The Birth of Stars and the Discovery of Planets outside the Solar System, we will look at This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 255 these exoplanet systems. But for now, let us focus on theories of how our own particular system has formed and evolved. Looking for Patterns One way to approach our question of origin is to look for regularities among the planets. We found, for example, that all the planets lie in nearly the same plane and revolve in the same direction around the Sun. The Sun also spins in the same direction about its own axis. Astronomers interpret this pattern as evidence that the Sun and planets formed together from a spinning cloud of gas and dust that we call the solar nebula (Figure 7.17). Figure 7", ".17. Solar Nebula. This artist\u2019s conception of the solar nebula shows the flattened cloud of gas and dust from which our planetary system formed. Icy and rocky planetesimals (precursors of the planets) can be seen in the foreground. The bright center is where the Sun is forming. (credit: William K. Hartmann, Planetary Science Institute) The composition of the planets gives another clue about origins. Spectroscopic analysis allows us to determine which elements are present in the Sun and the planets. The Sun has the same hydrogen-dominated composition as Jupiter and Saturn, and therefore appears to have been formed from the same reservoir of material. In comparison, the terrestrial planets and our Moon are relatively deficient in the light gases and the various ices that form from the common elements oxygen, carbon, and nitrogen. Instead, on Earth and its neighbors, we see mostly the rarer heavy elements such as iron and silicon. This pattern suggests that the processes that led to planet formation in the inner solar system must somehow have excluded much of the lighter materials that are common elsewhere. These lighter materials must have escaped, leaving a residue of heavy stuff. The reason for this is not hard to guess, bearing in mind the heat of the Sun. The inner planets and most of the asteroids are made of rock and metal, which can survive heat, but they contain very little ice or gas, which evaporate when temperatures are high. (To see what we mean, just compare how long a rock and an ice cube survive when they are placed in the sunlight.) In the outer solar system, where it has always been cooler, the planets and their moons, as well as icy dwarf planets and comets, are composed mostly of ice and gas. 256 Chapter 7 Other Worlds: An Introduction to the Solar System The Evidence from Far Away A second approach to understanding the origins of the solar system is to look outward for evidence that other systems of planets are forming elsewhere. We cannot look back in time to the formation of our own system, but many stars in space are much younger than the Sun. In these systems, the processes of planet formation might still be accessible to direct observation. We observe that there are many other \u201csolar nebulas\u201d or circumstellar disks\u2014flattened, spinning clouds of gas and dust surrounding young stars. These disks resemble our own solar system\u2019s initial stages of formation billions of years ago (Figure 7.18). Figure 7.18. Atlas of Planetary Nurseries. These Hubble Space Telescope photos show sections of the Orion Nebula, a relatively close-by region where stars are currently forming. Each image shows an embedded circumstellar disk orbiting a very young star. Seen from different angles, some are energized to glow by the light of a nearby star while others are dark and seen in silhouette against the bright glowing gas of the Orion Nebula. Each is a contemporary analog of our own solar nebula\u2014a location where planets are probably being formed today. (credit: modification of work by NASA/ESA, L. Ricci (ESO)) Building Planets Circumstellar disks are a common occurrence around very young stars, suggesting that disks and stars form together. Astronomers can use theoretical calculations to see how solid bodies might form from the gas and dust in these disks as they cool. These models show that material begins to coalesce first by forming smaller objects, precursors of the planets, which we call planetesimals. Today\u2019s fast computers can simulate the way millions of planetesimals, probably no larger than 100 kilometers in diameter, might gather together under their mutual gravity to form the planets we see today. We are beginning to understand that this process was a violent one, with planetesimals crashing into each other and sometimes even disrupting the growing planets themselves. As a consequence of those violent impacts (and the heat from radioactive elements in them), all the planets were heated until they were liquid and gas, and therefore differentiated, which helps explain their present internal structures. The process of impacts and collisions in the early solar system was complex and, apparently, often random. The solar nebula model can explain many of the regularities we find in the solar system, but the random collisions of massive collections of planetesimals could be the reason for some exceptions to the \u201crules\u201d of solar system behavior. For example, why do the planets Uranus and Pluto spin on their sides? Why does Venus spin slowly and in the opposite direction from the other planets? Why does the composition of the Moon resemble Earth in many ways and yet exhibit substantial differences? The answers to such questions probably lie in enormous collisions that took place in the solar system long before life on Earth began. Today, some 4.5 billion years after its origin, the solar system is\u2014thank goodness\u2014a much less violent place. As we will see, however, some planetesimals have continued to interact and collide, and their fragments move about the solar system as roving \u201ctransients\ufffd", "\ufffd that can make trouble for the established members of the Sun\u2019s family, such as our own Earth. (We discuss this \u201ctroublemaking\u201d in Comets and Asteroids: Debris of the Solar System.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 257 great variety of infographics (https://openstaxcollege.org/l/30worldsinsolar) at space.com let you explore what it would be like to live on various worlds in the solar system. 258 Chapter 7 Other Worlds: An Introduction to the Solar System CHAPTER 7 REVIEW KEY TERMS asteroid a stony or metallic object orbiting the Sun that is smaller than a major planet but that shows no evidence of an atmosphere or of other types of activity associated with comets comet a small body of icy and dusty matter that revolves about the Sun; when a comet comes near the Sun, some of its material vaporizes, forming a large head of tenuous gas and often a tail differentiation gravitational separation of materials of different density into layers in the interior of a planet or moon giant planet any of the planets Jupiter, Saturn, Uranus, and Neptune in our solar system, or planets of roughly that mass and composition in other planetary systems half-life time required for half of the radioactive atoms in a sample to disintegrate meteor a small piece of solid matter that enters Earth\u2019s atmosphere and burns up, popularly called a shooting star because it is seen as a small flash of light meteorite a portion of a meteor that survives passage through an atmosphere and strikes the ground planetesimals objects, from tens to hundreds of kilometers in diameter, that formed in the solar nebula as an intermediate step between tiny grains and the larger planetary objects we see today; the comets and some asteroids may be leftover planetesimals radioactivity process by which certain kinds of atomic nuclei decay naturally, with the spontaneous emission of subatomic particles and gamma rays solar nebula the cloud of gas and dust from which the solar system formed terrestrial planet any of the planets Mercury, Venus, Earth, or Mars; sometimes the Moon is included in the list SUMMARY 7.1 Overview of Our Planetary System Our solar system currently consists of the Sun, eight planets, five dwarf planets, nearly 200 known moons, and a host of smaller objects. The planets can be divided into two groups: the inner terrestrial planets and the outer giant planets. Pluto, Eris, Haumea, and Makemake do not fit into either category; as icy dwarf planets, they exist in an ice realm on the fringes of the main planetary system. The giant planets are composed mostly of liquids and gases. Smaller members of the solar system include asteroids (including the dwarf planet Ceres), which are rocky and metallic objects found mostly between Mars and Jupiter; comets, which are made mostly of frozen gases and generally orbit far from the Sun; and countless smaller grains of cosmic dust. When a meteor survives its passage through our atmosphere and falls to Earth, we call it a meteorite. 7.2 Composition and Structure of Planets The giant planets have dense cores roughly 10 times the mass of Earth, surrounded by layers of hydrogen and helium. The terrestrial planets consist mostly of rocks and metals. They were once molten, which allowed their structures to differentiate (that is, their denser materials sank to the center). The Moon resembles the terrestrial This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 259 planets in composition, but most of the other moons\u2014which orbit the giant planets\u2014have larger quantities of frozen ice within them. In general, worlds closer to the Sun have higher surface temperatures. The surfaces of terrestrial planets have been modified by impacts from space and by varying degrees of geological activity. 7.3 Dating Planetary Surfaces The ages of the surfaces of objects in the solar system can be estimated by counting craters: on a given world, a more heavily cratered region will generally be older than one that is less cratered. We can also use samples of rocks with radioactive elements in them to obtain the time since the layer in which the rock formed last solidified. The half-life of a radioactive element is the time it takes for half the sample to decay; we determine how many half-lives have passed by how much of a sample remains the radioactive element and how much has become the decay product. In this way, we have estimated the age of the Moon and Earth to be roughly 4.5 billion years. 7.4 Origin of the Solar System Regularities among the planets have led astronomers to hypothesize that the Sun and the planets formed together in a giant, spinning cloud of gas and dust called the solar nebula. Astronomical observations show tantalizingly similar circumstellar disks around other stars. Within the", " solar nebula, material first coalesced into planetesimals; many of these gathered together to make the planets and moons. The remainder can still be seen as comets and asteroids. Probably all planetary systems have formed in similar ways, but many exoplanet systems have evolved along quite different paths, as we will see in Cosmic Samples and the Origin of the Solar System. FOR FURTHER EXPLORATION Articles Davidson, K. \u201cCarl Sagan\u2019s Coming of Age.\u201d Astronomy. (November 1999): 40. About the noted popularizer of science and how he developed his interest in astronomy. Garget, J. \u201cMysterious Microworlds.\u201d Astronomy. (July 2005): 32. A quick tour of a number of the moons in the solar system. Hartmann, W. \u201cThe Great Solar System Revision.\u201d Astronomy. (August 1998): 40. How our views have changed over the past 25 years. Kross, J. \u201cWhat\u2019s in a Name?\u201d Sky & Telescope. (May 1995): 28. How worlds are named. Rubin, A. \u201cSecrets of Primitive Meteorites.\u201d Scientific American. (February 2013): 36. What meteorites can teach us about the environment in which the solar system formed. Soter, S. \u201cWhat Is a Planet?\u201d Scientific American. (January 2007): 34. The IAU\u2019s new definition of a planet in our solar system, and what happened to Pluto as a result. Talcott, R. \u201cHow the Solar System Came to Be.\u201d Astronomy. (November 2012): 24. On the formation period of the Sun and the planets. Wood, J. \u201cForging the Planets: The Origin of our Solar System.\u201d Sky & Telescope. (January 1999): 36. Good overview. 260 Websites Chapter 7 Other Worlds: An Introduction to the Solar System Gazetteer of Planetary Nomenclature: http://planetarynames.wr.usgs.gov/. Outlines the rules for naming bodies and features in the solar system. Planetary Photojournal: http://photojournal.jpl.nasa.gov/index.html. This NASA site features thousands of the best images from planetary exploration, with detailed captions and excellent indexing. You can find images by world, feature name, or mission, and download them in a number of formats. And the images are copyright- free because your tax dollars paid for them. The following sites present introductory information and pictures about each of the worlds of our solar system: \u2022 NASA/JPL Solar System Exploration pages: http://solarsystem.nasa.gov/index.cfm. \u2022 National Space Science Data Center Lunar and Planetary Science pages: http://nssdc.gsfc.nasa.gov/ planetary/. \u2022 Nine [now 8] Planets Solar System Tour: http://www.nineplanets.org/. \u2022 Planetary Society solar system pages: http://www.planetary.org/explore/space-topics/compare/. \u2022 Views of the Solar System by Calvin J. Hamilton: http://www.solarviews.com/eng/homepage.htm. Videos Brown Dwarfs and Free Floating Planets: When You Are Just Too Small to Be a Star: https://www.youtube.com/ watch?v=zXCDsb4n4KU. A nontechnical talk by Gibor Basri of the University of California at Berkeley, discussing some of the controversies about the meaning of the word \u201cplanet\u201d (1:32:52). In the Land of Enchantment: The Epic Story of the Cassini Mission to Saturn: https://www.youtube.com/ watch?v=Vx135n8VFxY. A public lecture by Dr. Carolyn Porco that focuses mainly on the exploration of Saturn and its moons, but also presents an eloquent explanation of why we explore the solar system (1:37:52). Origins of the Solar System: http://www.pbs.org/wgbh/nova/space/origins-solar-system.html. A video from PBS that focuses on the evidence from meteorites, narrated by Neil deGrasse Tyson (13:02). To Scale: The Solar System: https://www.youtube.com/watch?t=84&v=zR3Igc3Rhfg. Constructing a scale model of the solar system in the Nevada desert (7:06). COLLABORATIVE GROUP ACTIVITIES A. Discuss and make a list of the reasons why we humans might want to explore the other worlds in the solar system. Does your group think such missions of exploration are worth the investment? Why? B. Your instructor will assign each group a world. Your task is to think about what it would be like to be there.", " (Feel free to look ahead in the book to the relevant chapters.) Discuss where on or around your world we would establish a foothold and what we would need to survive there. C. In the There\u2019s No Place Like Home feature, we discuss briefly how human activity is transforming our planet\u2019s overall environment. Can you think of other ways that this is happening? D. Some scientists criticized Carl Sagan for \u201cwasting his research time\u201d popularizing astronomy. To what extent do you think scientists should spend their time interpreting their field of research for the public? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 261 Why or why not? Are there ways that scientists who are not as eloquent or charismatic as Carl Sagan or Neil deGrasse Tyson can still contribute to the public understanding of science? E. Your group has been named to a special committee by the International Astronomical Union to suggest names of features (such as craters, trenches, and so on) on a newly explored asteroid. Given the restriction that any people after whom features are named must no longer be alive, what names or types of names would you suggest? (Keep in mind that you are not restricted to names of people, by the way.) F. A member of your group has been kidnapped by a little-known religious cult that worships the planets. They will release him only if your group can tell which of the planets are currently visible in the sky during the evening and morning. You are forbidden from getting your instructor involved. How and where else could you find out the information you need? (Be as specific as you can. If your instructor says it\u2019s okay, feel free to answer this question using online or library resources.) G. In the Carl Sagan: Solar System Advocate feature, you learned that science fiction helped spark and sustain his interest in astronomy. Did any of the members of your group get interested in astronomy as a result of a science fiction story, movie, or TV show? Did any of the stories or films you or your group members saw take place on the planets of our solar system? Can you remember any specific ones that inspired you? If no one in the group is into science fiction, perhaps you can interview some friends or classmates who are and report back to the group. H. A list of NASA solar system spacecraft missions can be found at http://www.nasa.gov/content/solar- missions-list. Your instructor will assign each group a mission. Look up when the mission was launched and executed, and describe the mission goals, the basic characteristics of the spacecraft (type of instruments, propellant, size, and so on), and what was learned from the mission. If time allows, each group should present its findings to the rest of the class. I. What would be some of the costs or risks of developing a human colony or base on another planetary body? What technologies would need to be developed? What would people need to give up to live on a different world in our solar system? EXERCISES Review Questions 1. Venus rotates backward and Uranus and Pluto spin about an axis tipped nearly on its side. Based on what you learned about the motion of small bodies in the solar system and the surfaces of the planets, what might be the cause of these strange rotations? 2. What is the difference between a differentiated body and an undifferentiated body, and how might that influence a body\u2019s ability to retain heat for the age of the solar system? 3. What does a planet need in order to retain an atmosphere? How does an atmosphere affect the surface of a planet and the ability of life to exist? 4. Which type of planets have the most moons? Where did these moons likely originate? 5. What is the difference between a meteor and a meteorite? 6. Explain our ideas about why the terrestrial planets are rocky and have less gas than the giant planets. 262 Chapter 7 Other Worlds: An Introduction to the Solar System 7. Do all planetary systems look the same as our own? 8. What is comparative planetology and why is it useful to astronomers? 9. What changed in our understanding of the Moon and Moon-Earth system as a result of humans landing on the Moon\u2019s surface? 10. If Earth was to be hit by an extraterrestrial object, where in the solar system could it come from and how would we know its source region? 11. List some reasons that the study of the planets has progressed more in the past few decades than any other branch of astronomy. 12. Imagine you are a travel agent in the next century. An eccentric billionaire asks you to arrange a \u201cGuinness Book of Solar System Records\u201d kind of tour. Where would you direct him to find the following (use this chapter and Appendix F and Appendix G): A. the least-dense planet B.", " the densest planet C. the largest moon in the solar system D. excluding the jovian planets, the planet where you would weigh the most on its surface (Hint: Weight is directly proportional to surface gravity.) E. the smallest planet F. the planet that takes the longest time to rotate G. the planet that takes the shortest time to rotate H. the planet with a diameter closest to Earth\u2019s I. the moon with the thickest atmosphere J. the densest moon K. the most massive moon 13. What characteristics do the worlds in our solar system have in common that lead astronomers to believe that they all formed from the same \u201cmother cloud\u201d (solar nebula)? 14. How do terrestrial and giant planets differ? List as many ways as you can think of. 15. Why are there so many craters on the Moon and so few on Earth? 16. How do asteroids and comets differ? 17. How and why is Earth\u2019s Moon different from the larger moons of the giant planets? 18. Where would you look for some \u201coriginal\u201d planetesimals left over from the formation of our solar system? 19. Describe how we use radioactive elements and their decay products to find the age of a rock sample. Is this necessarily the age of the entire world from which the sample comes? Explain. 20. What was the solar nebula like? Why did the Sun form at its center? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 7 Other Worlds: An Introduction to the Solar System 263 Thought Questions 21. What can we learn about the formation of our solar system by studying other stars? Explain. 22. Earlier in this chapter, we modeled the solar system with Earth at a distance of about one city block from the Sun. If you were to make a model of the distances in the solar system to match your height, with the Sun at the top of your head and Pluto at your feet, which planet would be near your waist? How far down would the zone of the terrestrial planets reach? 23. Seasons are a result of the inclination of a planet\u2019s axial tilt being inclined from the normal of the planet\u2019s orbital plane. For example, Earth has an axis tilt of 23.4\u00b0 (Appendix F). Using information about just the inclination alone, which planets might you expect to have seasonal cycles similar to Earth, although different in duration because orbital periods around the Sun are different? 24. Again using Appendix F, which planet(s) might you expect not to have significant seasonal activity? Why? 25. Again using Appendix F, which planets might you expect to have extreme seasons? Why? 26. Using some of the astronomical resources in your college library or the Internet, find five names of features on each of three other worlds that are named after real people. In a sentence or two, describe each of these people and what contributions they made to the progress of science or human thought. 27. Explain why the planet Venus is differentiated, but asteroid Fraknoi, a very boring and small member of the asteroid belt, is not. 28. Would you expect as many impact craters per unit area on the surface of Venus as on the surface of Mars? Why or why not? 29. Interview a sample of 20 people who are not taking an astronomy class and ask them if they can name a living astronomer. What percentage of those interviewed were able to name one? Typically, the two living astronomers the public knows these days are Stephen Hawking and Neil deGrasse Tyson. Why are they better known than most astronomers? How would your result have differed if you had asked the same people to name a movie star or a professional basketball player? 30. Using Appendix G, complete the following table that describes the characteristics of the Galilean moons of Jupiter, starting from Jupiter and moving outward in distance. Semimajor Axis (km3) Diameter Density (g/cm3) Moon Io Europa Ganymede Callisto Table A This system has often been described as a mini solar system. Why might this be so? If Jupiter were to represent the Sun and the Galilean moons represented planets, which moons could be considered more terrestrial in nature and which ones more like gas/ice giants? Why? (Hint: Use the values in your table to help explain your categorization.) Figuring For Yourself 31. Calculate the density of Jupiter. Show your work. Is it more or less dense than Earth? Why? 264 Chapter 7 Other Worlds: An Introduction to the Solar System 32. Calculate the density of Saturn. Show your work. How does it compare with the density of water? Explain how this can be. 33. What is the density of Jupiter\u2019s moon Europa (see Appendix G for data on moons)? Show your work. 34. Look at Appendix F and Appendix G and indicate the moon with a diameter that is the largest", " fraction of the diameter of the planet or dwarf planet it orbits. 35. Barnard\u2019s Star, the second closest star to us, is about 56 trillion (5.6 \u00d7 1012) km away. Calculate how far it would be using the scale model of the solar system given in Overview of Our Planetary System. 36. A radioactive nucleus has a half-life of 5 \u00d7 108 years. Assuming that a sample of rock (say, in an asteroid) solidified right after the solar system formed, approximately what fraction of the radioactive element should be left in the rock today? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 265 8 EARTH AS A PLANET Figure 8.1. Active Geology. This image, taken from the International Space Station in 2006, shows a plume of ash coming from the Cleveland Volcano in the Aleutian Islands. Although the plume was only visible for around two hours, such events are a testament to the dynamic nature of Earth\u2019s crust. (credit: modification of work by NASA) Chapter Outline 8.1 The Global Perspective 8.2 Earth\u2019s Crust 8.3 Earth\u2019s Atmosphere 8.4 Life, Chemical Evolution, and Climate Change 8.5 Cosmic Influences on the Evolution of Earth Thinking Ahead Airless worlds in our solar system seem peppered with craters large and small. Earth, on the other hand, has few craters, but a thick atmosphere and much surface activity. Although impacts occurred on Earth at the same rate, craters have since been erased by forces in the planet\u2019s crust and atmosphere. What can the comparison between the obvious persistent cratering on so many other worlds, and the different appearance of Earth, tell us about the history of our planet? As our first step in exploring the solar system in more detail, we turn to the most familiar planet, our own Earth. The first humans to see Earth as a blue sphere floating in the blackness of space were the astronauts who made the first voyage around the Moon in 1968. For many people, the historic images showing our world as a small, distant globe represent a pivotal moment in human history, when it became difficult for educated human beings to view our world without a global perspective. In this chapter, we examine the composition and structure of our planet with its envelope of ocean and atmosphere. We ask how our terrestrial environment came to be the way it is today, and how it compares with other planets. 266 Chapter 8 Earth as a Planet 8.1 THE GLOBAL PERSPECTIVE Learning Objectives By the end of this section, you will be able to: Describe the components of Earth\u2019s interior and explain how scientists determined its structure Specify the origin, size, and extent of Earth\u2019s magnetic field Earth is a medium-size planet with a diameter of approximately 12,760 kilometers (Figure 8.2). As one of the inner or terrestrial planets, it is composed primarily of heavy elements such as iron, silicon, and oxygen\u2014very different from the composition of the Sun and stars, which are dominated by the light elements hydrogen and helium. Earth\u2019s orbit is nearly circular, and Earth is warm enough to support liquid water on its surface. It is the only planet in our solar system that is neither too hot nor too cold, but \u201cjust right\u201d for the development of life as we know it. Some of the basic properties of Earth are summarized in Table 8.1. Figure 8.2. Blue Marble. This image of Earth from space, taken by the Apollo 17 astronauts, is known as the \u201cBlue Marble.\u201d This is one of the rare images of a full Earth taken during the Apollo program; most images show only part of Earth\u2019s disk in sunlight. (credit: modification of work by NASA) Some Properties of Earth Semimajor axis Period Mass Diameter Radius Table 8.1 Property Measurement 1.00 AU 1.00 year 5.98 \u00d7 1024 kg 12,756 km 6378 km This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 267 Property Measurement 11.2 km/s 23 h 56 m 4 s 5.1 \u00d7 108 km2 5.514 g/cm3 1.00 bar Some Properties of Earth Escape velocity Rotational period Surface area Density Atmospheric pressure Table 8.1 Earth\u2019s Interior The interior of a planet\u2014even our own Earth\u2014is difficult to study, and its composition and structure must be determined indirectly. Our only direct experience is with the outermost skin of Earth\u2019s crust, a layer no more than a few kilometers deep. It is important to remember that, in many ways, we know less about our own planet 5 kilometers beneath our feet than we do about the surfaces of Venus and", " Mars. Earth is composed largely of metal and silicate rock (see the Composition and Structure of Planets section). Most of this material is in a solid state, but some of it is hot enough to be molten. The structure of material in Earth\u2019s interior has been probed in considerable detail by measuring the transmission of seismic waves through Earth. These are waves that spread through the interior of Earth from earthquakes or explosion sites. Seismic waves travel through a planet rather like sound waves through a struck bell. Just as the sound frequencies vary depending on the material the bell is made of and how it is constructed, so a planet\u2019s response depends on its composition and structure. By monitoring the seismic waves in different locations, scientists can learn about the layers through which the waves have traveled. Some of these vibrations travel along the surface; others pass directly through the interior. Seismic studies have shown that Earth\u2019s interior consists of several distinct layers with different compositions, illustrated in Figure 8.3. As waves travel through different materials in Earth\u2019s interior, the waves\u2014just like light waves in telescope lenses\u2014bend (or refract) so that some seismic stations on Earth receive the waves and others are in \u201cshadows.\u201d Detecting the waves in a network of seismographs helps scientists construct a model of Earth\u2019s interior, showing liquid and solid layers. This type of seismic imaging is not unlike that used in ultrasound, a type of imaging used to see inside the body. 268 Chapter 8 Earth as a Planet Figure 8.3. Interior Structure of Earth. The crust, mantle, and inner and outer cores (liquid and solid, respectively) as shown as revealed by seismic studies. The top layer is the crust, the part of Earth we know best (Figure 8.4). Oceanic crust covers 55% of Earth\u2019s surface and lies mostly submerged under the oceans. It is typically about 6 kilometers thick and is composed of volcanic rocks called basalt. Produced by the cooling of volcanic lava, basalts are made primarily of the elements silicon, oxygen, iron, aluminum, and magnesium. The continental crust covers 45% of the surface, some of which is also beneath the oceans. The continental crust is 20 to 70 kilometers thick and is composed predominantly of a different volcanic class of silicates (rocks made of silicon and oxygen) called granite. These crustal rocks, both oceanic and continental, typically have densities of about 3 g/cm3. (For comparison, the density of water is 1 g/cm3.) The crust is the easiest layer for geologists to study, but it makes up only about 0.3% of the total mass of Earth. Figure 8.4. Earth\u2019s Crust. This computer-generated image shows the surface of Earth\u2019s crust as determined from satellite images and ocean floor radar mapping. Oceans and lakes are shown in blue, with darker areas representing depth. Dry land is shown in shades of green and brown, and the Greenland and Antarctic ice sheets are depicted in shades of white. (credit: modification of work by C. Amante, B. W. Eakins, National Geophysical Data Center, NOAA) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 269 The largest part of the solid Earth, called the mantle, stretches from the base of the crust downward to a depth of 2900 kilometers. The mantle is more or less solid, but at the temperatures and pressures found there, mantle rock can deform and flow slowly. The density in the mantle increases downward from about 3.5 g/cm3 to more than 5 g/cm3 as a result of the compression produced by the weight of overlying material. Samples of upper mantle material are occasionally ejected from volcanoes, permitting a detailed analysis of its chemistry. Beginning at a depth of 2900 kilometers, we encounter the dense metallic core of Earth. With a diameter of 7000 kilometers, our core is substantially larger than the entire planet Mercury. The outer core is liquid, but the innermost part of the core (about 2400 kilometers in diameter) is probably solid. In addition to iron, the core probably also contains substantial quantities of nickel and sulfur, all compressed to a very high density. The separation of Earth into layers of different densities is an example of differentiation, the process of sorting the major components of a planet by density. The fact that Earth is differentiated suggests that it was once warm enough for its interior to melt, permitting the heavier metals to sink to the center and form the dense core. Evidence for differentiation comes from comparing the planet\u2019s bulk density (5.5 g/cm3) with the surface materials (3 g/cm3) to suggest that denser material must be buried in the core. Magnetic Field and Magnetosphere We can find additional clues about Earth\u2019s interior from its magnetic field.", " Our planet behaves in some ways as if a giant bar magnet were inside it, aligned approximately with the rotational poles of Earth. This magnetic field is generated by moving material in Earth\u2019s liquid metallic core. As the liquid metal inside Earth circulates, it sets up a circulating electric current. When many charged particles are moving together like that\u2014in the laboratory or on the scale of an entire planet\u2014they produce a magnetic field. Earth\u2019s magnetic field extends into surrounding space. When a charged particle encounters a magnetic field in space, it becomes trapped in the magnetic zone. Above Earth\u2019s atmosphere, our field is able to trap small quantities of electrons and other atomic particles. This region, called the magnetosphere, is defined as the zone within which Earth\u2019s magnetic field dominates over the weak interplanetary magnetic field that extends outward from the Sun (Figure 8.5). Figure 8.5. Earth\u2019s Magnetosphere. A cross-sectional view of our magnetosphere (or zone of magnetic influence), as revealed by numerous spacecraft missions. Note how the wind of charged particles from the Sun \u201cblows\u201d the magnetic field outward like a wind sock. 270 Chapter 8 Earth as a Planet Where do the charged particles trapped in our magnetosphere come from? They flow outward from the hot surface of the Sun; this is called the solar wind. It not only provides particles for Earth\u2019s magnetic field to trap, it also stretches our field in the direction pointing away from the Sun. Typically, Earth\u2019s magnetosphere extends about 60,000 kilometers, or 10 Earth radii, in the direction of the Sun. But, in the direction away from the Sun, the magnetic field can reach as far as the orbit of the Moon, and sometimes farther. The magnetosphere was discovered in 1958 by instruments on the first US Earth satellite, Explorer 1, which recorded the ions (charged particles) trapped in its inner part. The regions of high-energy ions in the magnetosphere are often called the Van Allen belts in recognition of the University of Iowa professor who built the scientific instrumentation for Explorer 1. Since 1958, hundreds of spacecraft have explored various regions of the magnetosphere. You can read more about its interaction with the Sun in a later chapter. 8.2 EARTH\u2019S CRUST Learning Objectives By the end of this section, you will be able to: Denote the primary types of rock that constitute Earth\u2019s crust Explain the theory of plate tectonics Describe the difference between rift and subduction zones Describe the relationship between fault zones and mountain building Explain the various types of volcanic activity occurring on Earth Let us now examine our planet\u2019s outer layers in more detail. Earth\u2019s crust is a dynamic place. Volcanic eruptions, erosion, and large-scale movements of the continents rework the surface of our planet constantly. Geologically, ours is the most active planet. Many of the geological processes described in this section have taken place on other planets as well, but usually in their distant pasts. Some of the moons of the giant planets also have impressive activity levels. For example, Jupiter\u2019s moon Io has a remarkable number of active volcanoes. Composition of the Crust Earth\u2019s crust is largely made up of oceanic basalt and continental granite. These are both igneous rock, the term used for any rock that has cooled from a molten state. All volcanically produced rock is igneous (Figure 8.6). Figure 8.6. Formation of Igneous Rock as Liquid Lava Cools and Freezes. This is a lava flow from a basaltic eruption. Basaltic lava flows quickly and can move easily over distances of more than 20 kilometers. (credit: USGS) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 271 Two other kinds of rock are familiar to us on Earth, although it turns out that neither is common on other planets. Sedimentary rocks are made of fragments of igneous rock or the shells of living organisms deposited by wind or water and cemented together without melting. On Earth, these rocks include the common sandstones, shales, and limestones. Metamorphic rocks are produced when high temperature or pressure alters igneous or sedimentary rock physically or chemically (the word metamorphic means \u201cchanged in form\u201d). Metamorphic rocks are produced on Earth because geological activity carries surface rocks down to considerable depths and then brings them back up to the surface. Without such activity, these changed rocks would not exist at the surface. There is a fourth very important category of rock that can tell us much about the early history of the planetary system: primitive rock, which has largely escaped chemical modification by heating. Primitive rock represents the original material out of which the planetary system was made. No primitive material is left on Earth because the entire planet was heated", " early in its history. To find primitive rock, we must look to smaller objects such as comets, asteroids, and small planetary moons. We can sometimes see primitive rock in samples that fall to Earth from these smaller objects. A block of quartzite on Earth is composed of materials that have gone through all four of these states. Beginning as primitive material before Earth was born, it was heated in the early Earth to form igneous rock, transformed chemically and redeposited (perhaps many times) to form sedimentary rock, and finally changed several kilometers below Earth\u2019s surface into the hard, white metamorphic stone we see today. Plate Tectonics Geology is the study of Earth\u2019s crust and the processes that have shaped its surface throughout history. (Although geo- means \u201crelated to Earth,\u201d astronomers and planetary scientists also talk about the geology of other planets.) Heat escaping from the interior provides energy for the formation of our planet\u2019s mountains, valleys, volcanoes, and even the continents and ocean basins themselves. But not until the middle of the twentieth century did geologists succeed in understanding just how these landforms are created. Plate tectonics is a theory that explains how slow motions within the mantle of Earth move large segments of the crust, resulting in a gradual \u201cdrifting\u201d of the continents as well as the formation of mountains and other large-scale geological features. Plate tectonics is a concept as basic to geology as evolution by natural selection is to biology or gravity is to understanding the orbits of planets. Looking at it from a different perspective, plate tectonics is a mechanism for Earth to transport heat efficiently from the interior, where it has accumulated, out to space. It is a cooling system for the planet. All planets develop a heat transfer process as they evolve; mechanisms may differ from that on Earth as a result of chemical makeup and other constraints. Earth\u2019s crust and upper mantle (to a depth of about 60 kilometers) are divided into about a dozen tectonic plates that fit together like the pieces of a jigsaw puzzle (Figure 8.7). In some places, such as the Atlantic Ocean, the plates are moving apart; in others, such as off the western coast of South America, they are being forced together. The power to move the plates is provided by slow convection of the mantle, a process by which heat escapes from the interior through the upward flow of warmer material and the slow sinking of cooler material. (Convection, in which energy is transported from a warm region, such as the interior of Earth, to a cooler region, such as the upper mantle, is a process we encounter often in astronomy\u2014in stars as well as planets. It is also important in boiling water for coffee while studying for astronomy exams.) 272 Chapter 8 Earth as a Planet Figure 8.7. Earth\u2019s Continental Plates. This map shows the major plates into which the crust of Earth is divided. Arrows indicate the motion of the plates at average speeds of 4 to 5 centimeters per year, similar to the rate at which your hair grows The US Geological Survey provides a map of recent earthquakes (https://openstax.org/l/ 30geosurmapeart) and shows the boundaries of the tectonic plates and where earthquakes occur in relation to these boundaries. You can look close-up at the United States or zoom out for a global view. As the plates slowly move, they bump into each other and cause dramatic changes in Earth\u2019s crust over time. Four basic kinds of interactions between crustal plates are possible at their boundaries: (1) they can pull apart, (2) one plate can burrow under another, (3) they can slide alongside each other, or (4) they can jam together. Each of these activities is important in determining the geology of Earth Alfred Wegener: Catching the Drift of Plate Tectonics When studying maps or globes of Earth, many students notice that the coast of North and South America, with only minor adjustments, could fit pretty well against the coast of Europe and Africa. It seems as if these great landmasses could once have been together and then were somehow torn apart. The same idea had occurred to others (including Francis Bacon as early as 1620), but not until the twentieth century could such a proposal be more than speculation. The scientist who made the case for continental drift in 1920 was a German meteorologist and astronomer named Alfred Wegener (Figure 8.8). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 273 Figure 8.8. Alfred Wegener (1880\u20131930). Wegener proposed a scientific theory for the slow shifting of the continents. Born in Berlin in 1880, Wegener was, from an early age, fascinated by Greenland, the world\u2019s largest island, which he dreamed of exploring. He studied", " at the universities in Heidelberg, Innsbruck, and Berlin, receiving a doctorate in astronomy by reexamining thirteenth-century astronomical tables. But, his interests turned more and more toward Earth, particularly its weather. He carried out experiments using kites and balloons, becoming so accomplished that he and his brother set a world record in 1906 by flying for 52 hours in a balloon. Wegener first conceived of continental drift in 1910 while examining a world map in an atlas, but it took 2 years for him to assemble sufficient data to propose the idea in public. He published the results in book form in 1915. Wegener\u2019s evidence went far beyond the congruence in the shapes of the continents. He proposed that the similarities between fossils found only in South America and Africa indicated that these two continents were joined at one time. He also showed that resemblances among living animal species on different continents could best be explained by assuming that the continents were once connected in a supercontinent he called Pangaea (from Greek elements meaning \u201call land\u201d). Wegener\u2019s suggestion was met with a hostile reaction from most scientists. Although he had marshaled an impressive list of arguments for his hypothesis, he was missing a mechanism. No one could explain how solid continents could drift over thousands of miles. A few scientists were sufficiently impressed by Wegener\u2019s work to continue searching for additional evidence, but many found the notion of moving continents too revolutionary to take seriously. Developing an understanding of the mechanism (plate tectonics) would take decades of further progress in geology, oceanography, and geophysics. Wegener was disappointed in the reception of his suggestion, but he continued his research and, in 1924, he was appointed to a special meteorology and geophysics professorship created especially for him at the University of Graz (where he was, however, ostracized by most of the geology faculty). Four years later, on his fourth expedition to his beloved Greenland, he celebrated his fiftieth birthday with colleagues and then set off on foot toward a different camp on the island. He never made it; he was found a few days later, dead of an apparent heart attack. 274 Chapter 8 Earth as a Planet Critics of science often point to the resistance to the continental drift hypothesis as an example of the flawed way that scientists regard new ideas. (Many people who have advanced crackpot theories have claimed that they are being ridiculed unjustly, just as Wegener was.) But we think there is a more positive light in which to view the story of Wegener\u2019s suggestion. Scientists in his day maintained a skeptical attitude because they needed more evidence and a clear mechanism that would fit what they understood about nature. Once the evidence and the mechanism were clear, Wegener\u2019s hypothesis quickly became the centerpiece of our view of a dynamic Earth See how the drift of the continents (https://openstax.org/l/30contintdrift) has changed the appearance of our planet\u2019s crust. Rift and Subduction Zones Plates pull apart from each other along rift zones, such as the Mid-Atlantic ridge, driven by upwelling currents in the mantle (Figure 8.9). A few rift zones are found on land. The best known is the central African rift\u2014an area where the African continent is slowly breaking apart. Most rift zones, however, are in the oceans. Molten rock rises from below to fill the space between the receding plates; this rock is basaltic lava, the kind of igneous rock that forms most of the ocean basins. Figure 8.9. Rift Zone and Subduction Zone. Rift and subduction zones are the regions (mostly beneath the oceans) where new crust is formed and old crust is destroyed as part of the cycle of plate tectonics. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 275 From a knowledge of how the seafloor is spreading, we can calculate the average age of the oceanic crust. About 60,000 kilometers of active rifts have been identified, with average separation rates of about 4 centimeters per year. The new area added to Earth each year is about 2 square kilometers, enough to renew the entire oceanic crust in a little more than 100 million years. This is a very short interval in geological time\u2014less than 3% of the age of Earth. The present ocean basins thus turn out to be among the youngest features on our planet. As new crust is added to Earth, the old crust must go somewhere. When two plates come together, one plate is often forced beneath another in what is called a subduction zone (Figure 8.9). In general, the thick continental masses cannot be subducted, but the thinner oceanic plates can be rather readily thrust down into the upper mantle. Often a subduction zone is marked by an", " ocean trench; a fine example of this type of feature is the deep Japan trench along the coast of Asia. The subducted plate is forced down into regions of high pressure and temperature, eventually melting several hundred kilometers below the surface. Its material is recycled into a downward-flowing convection current, ultimately balancing the flow of material that rises along rift zones. The amount of crust destroyed at subduction zones is approximately equal to the amount formed at rift zones. All along the subduction zone, earthquakes and volcanoes mark the death throes of the plate. Some of the most destructive earthquakes in history have taken place along subduction zones, including the 1923 Yokohama earthquake and fire that killed 100,000 people, the 2004 Sumatra earthquake and tsunami that killed more than 200,000 people, and the 2011 Tohoku earthquake that resulted in the meltdown of three nuclear power reactors in Japan. Fault Zones and Mountain Building Along much of their length, the crustal plates slide parallel to each other. These plate boundaries are marked by cracks or faults. Along active fault zones, the motion of one plate with respect to the other is several centimeters per year, about the same as the spreading rates along rifts. One of the most famous faults is the San Andreas Fault in California, which lies at the boundary between the Pacific plate and the North American plate (Figure 8.10). This fault runs from the Gulf of California to the Pacific Ocean northwest of San Francisco. The Pacific plate, to the west, is moving northward, carrying Los Angeles, San Diego, and parts of the southern California coast with it. In several million years, Los Angeles may be an island off the coast of San Francisco. 276 Chapter 8 Earth as a Planet Figure 8.10. San Andreas Fault. We see part of a very active region in California where one crustal plate is sliding sideways with respect to the other. The fault is marked by the valley running up the right side of the photo. Major slippages along this fault can produce extremely destructive earthquakes. (credit: John Wiley) Unfortunately for us, the motion along fault zones does not take place smoothly. The creeping motion of the plates against each other builds up stresses in the crust that are released in sudden, violent slippages that generate earthquakes. Because the average motion of the plates is constant, the longer the interval between earthquakes, the greater the stress and the more energy released when the surface finally moves. For example, the part of the San Andreas Fault near the central California town of Parkfield has slipped every 25 years or so during the past century, moving an average of about 1 meter each time. In contrast, the average interval between major earthquakes in the Los Angeles region is about 150 years, and the average motion is about 7 meters. The last time the San Andreas fault slipped in this area was in 1857; tension has been building ever since, and sometime soon it is bound to be released. Sensitive instruments placed within the Los Angeles basin show that the basin is distorting and contracting in size as these tremendous pressures build up beneath the surface Fault Zones and Plate Motion After scientists mapped the boundaries between tectonic plates in Earth\u2019s crust and measured the annual rate at which the plates move (which is about 5 cm/year), we could estimate quite a lot about the rate at which the geology of Earth is changing. As an example, let\u2019s suppose that the next slippage along the San Andreas Fault in southern California takes place in the year 2017 and that it completely relieves the accumulated strain in this region. How much slippage is required for this to occur? Solution The speed of motion of the Pacific plate relative to the North American plate is 5 cm/y. That\u2019s 500 cm (or 5 m) per century. The last southern California earthquake was in 1857. The time from 1857 to 2017 is This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 277 160 y, or 1.6 centuries, so the slippage to relieve the strain completely would be 5 m/century \u00d7 1.6 centuries = 8.0 m. Check Your Learning If the next major southern California earthquake occurs in 2047 and only relieves one-half of the accumulated strain, how much slippage will occur? Answer: The difference in time from 1857 to 2047 is 190 y, or 1.9 centuries. Because only half the strain is released, this is equivalent to half the annual rate of motion. The total slippage comes to 0.5 \u00d7 5 m/century \u00d7 1.9 centuries = 4.75 m. When two continental masses are moving on a collision course, they push against each other under great pressure. Earth buckles and folds, dragging some rock deep below the surface and raising other folds to heights of many kilometers. This is the way many, but not all, of the mountain ranges on Earth were formed", ". The Alps, for example, are a result of the African plate bumping into the Eurasian plate. As we will see, however, quite different processes produced the mountains on other planets. Once a mountain range is formed by upthrusting of the crust, its rocks are subject to erosion by water and ice. The sharp peaks and serrated edges have little to do with the forces that make the mountains initially. Instead, they result from the processes that tear down mountains. Ice is an especially effective sculptor of rock (Figure 8.11). In a world without moving ice or running water (such as the Moon or Mercury), mountains remain smooth and dull. Figure 8.11. Mountains on Earth. The Torres del Paine are a young region of Earth\u2019s crust where sharp mountain peaks are being sculpted by glaciers. We owe the beauty of our young, steep mountains to the erosion by ice and water. (credit: David Morrison) Volcanoes Volcanoes mark locations where lava rises to the surface. One example is mid ocean ridges, which are long undersea mountain ranges formed by lava rising from Earth\u2019s mantle at plate boundaries. A second major kind of volcanic activity is associated with subduction zones, and volcanoes sometimes also appear in regions where continental plates are colliding. In each case, the volcanic activity gives us a way to sample some of the material from deeper within our planet. 278 Chapter 8 Earth as a Planet Other volcanic activity occurs above mantle \u201chot spots\u201d\u2014areas far from plate boundaries where heat is nevertheless rising from the interior of Earth. One of the best-known hot spot is under the island of Hawaii, where it currently supplies the heat to maintain three active volcanoes, two on land and one under the ocean. The Hawaii hot spot has been active for at least 100 million years. As Earth\u2019s plates have moved during that time, the hot spot has generated a 3500-kilometer-long chain of volcanic islands. The tallest Hawaiian volcanoes are among the largest individual mountains on Earth, more than 100 kilometers in diameter and rising 9 kilometers above the ocean floor. One of the Hawaiian volcanic mountains, the now-dormant Mauna Kea, has become one of the world\u2019s great sites for doing astronomy The US Geological Service provides an interactive map (https://openstax.org/l/30mapringoffire) of the famous \u201cring of fire,\u201d which is the chain of volcanoes surrounding the Pacific Ocean, and shows the Hawaiian \u201chot spot\u201d enclosed within. Not all volcanic eruptions produce mountains. If lava flows rapidly from long cracks, it can spread out to form lava plains. The largest known terrestrial eruptions, such as those that produced the Snake River basalts in the northwestern United States or the Deccan plains in India, are of this type. Similar lava plains are found on the Moon and the other terrestrial planets. 8.3 EARTH\u2019S ATMOSPHERE Learning Objectives By the end of this section, you will be able to: Differentiate between Earth\u2019s various atmospheric layers Describe the chemical composition and possible origins of our atmosphere Explain the difference between weather and climate We live at the bottom of the ocean of air that envelops our planet. The atmosphere, weighing down upon Earth\u2019s surface under the force of gravity, exerts a pressure at sea level that scientists define as 1 bar (a term that comes from the same root as barometer, an instrument used to measure atmospheric pressure). A bar of pressure means that each square centimeter of Earth\u2019s surface has a weight equivalent to 1.03 kilograms pressing down on it. Humans have evolved to live at this pressure; make the pressure a lot lower or higher and we do not function well. The total mass of Earth\u2019s atmosphere is about 5 \u00d7 1018 kilograms. This sounds like a large number, but it is only about a millionth of the total mass of Earth. The atmosphere represents a smaller fraction of Earth than the fraction of your mass represented by the hair on your head. Structure of the Atmosphere The structure of the atmosphere is illustrated in Figure 8.12. Most of the atmosphere is concentrated near the surface of Earth, within about the bottom 10 kilometers where clouds form and airplanes fly. Within this region\u2014called the troposphere\u2014warm air, heated by the surface, rises and is replaced by descending currents of cooler air; this is an example of convection. This circulation generates clouds and wind. Within This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 279 the troposphere, temperature decreases rapidly with increasing elevation to values near 50 \u00b0C below freezing at its upper boundary, where the stratosphere begins. Most of the stratosphere, which extends to about 50 kilometers above the surface, is cold and free of clouds. Figure 8.12. Structure of", " Earth\u2019s Atmosphere. Height increases up the left side of the diagram, and the names of the different atmospheric layers are shown at the right. In the upper ionosphere, ultraviolet radiation from the Sun can strip electrons from their atoms, leaving the atmosphere ionized. The curving red line shows the temperature (see the scale on the x-axis). Near the top of the stratosphere is a layer of ozone (O3), a heavy form of oxygen with three atoms per molecule instead of the usual two. Because ozone is a good absorber of ultraviolet light, it protects the surface from some of the Sun\u2019s dangerous ultraviolet radiation, making it possible for life to exist on Earth. The breakup of ozone adds heat to the stratosphere, reversing the decreasing temperature trend in the troposphere. Because ozone is essential to our survival, we reacted with justifiable concern to evidence that became clear in the 1980s that atmospheric ozone was being destroyed by human activities. By international agreement, the production of industrial chemicals that cause ozone depletion, called chlorofluorocarbons, or CFCs, has been phased out. As a result, ozone loss has stopped and the \u201cozone hole\u201d over the Antarctic is shrinking gradually. This is an example of how concerted international action can help maintain the habitability of Earth Visit NASA\u2019s scientific visualization studio for a short video (https://openstax.org/l/302065regcfc) of what would have happened to Earth\u2019s ozone layer by 2065 if CFCs had not been regulated. At heights above 100 kilometers, the atmosphere is so thin that orbiting satellites can pass through it with very little friction. Many of the atoms are ionized by the loss of an electron, and this region is often called the ionosphere. At these elevations, individual atoms can occasionally escape completely from the gravitational field of Earth. There is a continuous, slow leaking of atmosphere\u2014especially of lightweight atoms, which move faster than heavy ones. Earth\u2019s atmosphere cannot, for example, hold on for long to hydrogen or helium, which 280 Chapter 8 Earth as a Planet escape into space. Earth is not the only planet to experience atmosphere leakage. Atmospheric leakage also created Mars\u2019 thin atmosphere. Venus\u2019 dry atmosphere evolved because its proximity to the Sun vaporized and dissociated any water, with the component gases lost to space. Atmospheric Composition and Origin At Earth\u2019s surface, the atmosphere consists of 78% nitrogen (N2), 21% oxygen (O2), and 1% argon (Ar), with traces of water vapor (H2O), carbon dioxide (CO2), and other gases. Variable amounts of dust particles and water droplets are also found suspended in the air. A complete census of Earth\u2019s volatile materials, however, should look at more than the gas that is now present. Volatile materials are those that evaporate at a relatively low temperature. If Earth were just a little bit warmer, some materials that are now liquid or solid might become part of the atmosphere. Suppose, for example, that our planet were heated to above the boiling point of water (100 \u00b0C, or 373 K); that\u2019s a large change for humans, but a small change compared to the range of possible temperatures in the universe. At 100 \u00b0C, the oceans would boil and the resulting water vapor would become a part of the atmosphere. To estimate how much water vapor would be released, note that there is enough water to cover the entire Earth to a depth of about 300 meters. Because the pressure exerted by 10 meters of water is equal to about 1 bar, the average pressure at the ocean floor is about 300 bars. Water weighs the same whether in liquid or vapor form, so if the oceans boiled away, the atmospheric pressure of the water would still be 300 bars. Water would therefore greatly dominate Earth\u2019s atmosphere, with nitrogen and oxygen reduced to the status of trace constituents. On a warmer Earth, another source of additional atmosphere would be found in the sedimentary carbonate rocks of the crust. These minerals contain abundant carbon dioxide. If all these rocks were heated, they would release about 70 bars of CO2, far more than the current CO2 pressure of only 0.0005 bar. Thus, the atmosphere of a warm Earth would be dominated by water vapor and carbon dioxide, with a surface pressure nearing 400 bars. Several lines of evidence show that the composition of Earth\u2019s atmosphere has changed over our planet\u2019s history. Scientists can infer the amount of atmospheric oxygen, for example, by studying the chemistry of minerals that formed at various times. We examine this issue in more detail later in this chapter. Today we see that CO2, H2O, sulfur dioxide (SO2), and other gases are released from deeper within Earth through the action of volcanoes. (For CO2, the primary source today is the burning of fossil fuels, which releases far more CO2 than that from volcanic eru", "ptions.) Much of this apparently new gas, however, is recycled material that has been subducted through plate tectonics. But where did our planet\u2019s original atmosphere come from? Three possibilities exist for the original source of Earth\u2019s atmosphere and oceans: (1) the atmosphere could have been formed with the rest of Earth as it accumulated from debris left over from the formation of the Sun; (2) it could have been released from the interior through volcanic activity, subsequent to the formation of Earth; or (3) it may have been derived from impacts by comets and asteroids from the outer parts of the solar system. Current evidence favors a combination of the interior and impact sources. Weather and Climate All planets with atmospheres have weather, which is the name we give to the circulation of the atmosphere. The energy that powers the weather is derived primarily from the sunlight that heats the surface. Both the rotation of the planet and slower seasonal changes cause variations in the amount of sunlight striking different parts of Earth. The atmosphere and oceans redistribute the heat from warmer to cooler areas. Weather on any planet This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 281 represents the response of its atmosphere to changing inputs of energy from the Sun (see Figure 8.13 for a dramatic example). Figure 8.13. Storm from Space. This satellite image shows Hurricane Irene in 2011, shortly before the storm hit land in New York City. The combination of Earth\u2019s tilted axis of rotation, moderately rapid rotation, and oceans of liquid water can lead to violent weather on our planet. (credit: NASA/NOAA GOES Project) Climate is a term used to refer to the effects of the atmosphere that last through decades and centuries. Changes in climate (as opposed to the random variations in weather from one year to the next) are often difficult to detect over short time periods, but as they accumulate, their effect can be devastating. One saying is that \u201cClimate is what you expect, and weather is what you get.\u201d Modern farming is especially sensitive to temperature and rainfall; for example, calculations indicate that a drop of only 2 \u00b0C throughout the growing season would cut the wheat production by half in Canada and the United States. At the other extreme, an increase of 2 \u00b0C in the average temperature of Earth would be enough to melt many glaciers, including much of the ice cover of Greenland, raising sea level by as much as 10 meters, flooding many coastal cities and ports, and putting small islands completely under water. The best documented changes in Earth\u2019s climate are the great ice ages, which have lowered the temperature of the Northern Hemisphere periodically over the past half million years or so (Figure 8.14). The last ice age, which ended about 14,000 years ago, lasted some 20,000 years. At its height, the ice was almost 2 kilometers thick over Boston and stretched as far south as New York City. 282 Chapter 8 Earth as a Planet Figure 8.14. Ice Age. This computer-generated image shows the frozen areas of the Northern Hemisphere during past ice ages from the vantage point of looking down on the North Pole. The area in black indicates the most recent glaciation (coverage by glaciers), and the area in gray shows the maximum level of glaciation ever reached. (credit: modification of work by Hannes Grobe/AWI) These ice ages were primarily the result of changes in the tilt of Earth\u2019s rotational axis, produced by the gravitational effects of the other planets. We are less certain about evidence that at least once (and perhaps twice) about a billion years ago, the entire ocean froze over, a situation called snowball Earth. The development and evolution of life on Earth has also produced changes in the composition and temperature of our planet\u2019s atmosphere, as we shall see in the next section Watch this short excerpt (https://openstax.org/l/30natgeoearth) from the National Geographic documentary Earth: The Biography. In this segment, Dr. Iain Stewart explains the fluid nature of our atmosphere. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 283 8.4 LIFE, CHEMICAL EVOLUTION, AND CLIMATE CHANGE Learning Objectives By the end of this section, you will be able to: Outline the origins and subsequent diversity of life on Earth Explain the ways that life and geological activity have influenced the evolution of the atmosphere Describe the causes and effects of the atmospheric greenhouse effect and global warming Describe the impact of human activity on our planet\u2019s atmosphere and ecology As far as we know, Earth seems to be the only planet in the solar system with life. The origin and development of life are an important part of our planet\u2019s story. Life arose early in Earth\u2019", "s history, and living organisms have been interacting with their environment for billions of years. We recognize that life-forms have evolved to adapt to the environment on Earth, and we are now beginning to realize that Earth itself has been changed in important ways by the presence of living matter. The study of the coevolution of life and our planet is one of the subjects of the modern science of astrobiology. The Origin of Life The record of the birth of life on Earth has been lost in the restless motions of the crust. According to chemical evidence, by the time the oldest surviving rocks were formed about 3.9 billion years ago, life already existed. At 3.5 billion years ago, life had achieved the sophistication to build large colonies called stromatolites, a form so successful that stromatolites still grow on Earth today (Figure 8.15). But, few rocks survive from these ancient times, and abundant fossils have been preserved only during the past 600 million years\u2014less than 15% of our planet\u2019s history. Figure 8.15. Cross-Sections of Fossil Stromatolites. This polished cross-section of a fossilized colony of stromatolites dates to the Precambrian Era. The layered, domelike structures are mats of sediment trapped in shallow waters by large numbers of blue-green bacteria that can photosynthesize. Such colonies of microorganisms date back more than 3 billion years. (credit: James St. John) There is little direct evidence about the actual origin of life. We know that the atmosphere of early Earth, unlike today\u2019s, contained abundant carbon dioxide and some methane, but no oxygen gas. In the absence of oxygen, many complex chemical reactions are possible that lead to the production of amino acids, proteins, and other chemical building blocks of life. Therefore, it seems likely that these chemical building blocks were available very early in Earth\u2019s history and they would have combined to make living organisms. 284 Chapter 8 Earth as a Planet For tens of millions of years after Earth\u2019s formation, life (perhaps little more than large molecules, like the viruses of today) probably existed in warm, nutrient-rich seas, living off accumulated organic chemicals. When this easily accessible food became depleted, life began the long evolutionary road that led to the vast numbers of different organisms on Earth today. As it did so, life began to influence the chemical composition of the atmosphere. In addition to the study of life\u2019s history as revealed by chemical and fossil evidence in ancient rocks, scientists use tools from the rapidly advancing fields of genetics and genomics\u2014the study of the genetic code that is shared by all life on Earth. While each individual has a unique set of genes (which is why genetic \u201cfingerprinting\u201d is so useful for the study of crime), we also have many genetic traits in common. Your genome, the complete map of the DNA in your body, is identical at the 99.9% level to that of Julius Caesar or Marie Curie. At the 99% level, human and chimpanzee genomes are the same. By looking at the gene sequences of many organisms, we can determine that all life on Earth is descended from a common ancestor, and we can use the genetic variations among species as a measure of how closely different species are related. These genetic analysis tools have allowed scientists to construct what is called the \u201c tree of life\u201d (Figure 8.16). This diagram illustrates the way organisms are related by examining one sequence of the nucleic acid RNA that all species have in common. This figure shows that life on Earth is dominated by microscopic creatures that you have probably never heard of. Note that the plant and animal kingdoms are just two little branches at the far right. Most of the diversity of life, and most of our evolution, has taken place at the microbial level. Indeed, it may surprise you to know that there are more microbes in a bucket of soil than there are stars in the Galaxy. You may want to keep this in mind when, later in this book, we turn to the search for life on other worlds. The \u201caliens\u201d that are most likely to be out there are microbes. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 285 Figure 8.16. Tree of Life. This chart shows the main subdivisions of life on Earth and how they are related. Note that the animal and plant kingdoms are just short branches on the far right, along with the fungi. The most fundamental division of Earth\u2019s living things is onto three large domains called bacteria, archaea, and eukarya. Most of the species listed are microscopic. (credit: modification of work by Eric Gaba) Such genetic studies lead to other interesting conclusions as well. For example, it appears that the earliest surviving terrestrial life-forms were all adapted to live at high temperatures. Some", " biologists think that life might actually have begun in locations on our planet that were extremely hot. Yet another intriguing possibility is that life began on Mars (which cooled sooner) rather than Earth and was \u201cseeded\u201d onto our planet by meteorites traveling from Mars to Earth. Mars rocks are still making their way to Earth, but so far none has shown evidence of serving as a \u201cspaceship\u201d to carry microorganisms from Mars to Earth. The Evolution of the Atmosphere One of the key steps in the evolution of life on Earth was the development of blue-green algae, a very successful life-form that takes in carbon dioxide from the environment and releases oxygen as a waste product. These successful microorganisms proliferated, giving rise to all the lifeforms we call plants. Since the energy for making new plant material from chemical building blocks comes from sunlight, we call the process photosynthesis. Studies of the chemistry of ancient rocks show that Earth\u2019s atmosphere lacked abundant free oxygen until about 2 billion years ago, despite the presence of plants releasing oxygen by photosynthesis. Apparently, chemical reactions with Earth\u2019s crust removed the oxygen gas as quickly as it formed. Slowly, however, the increasing evolutionary sophistication of life led to a growth in the plant population and thus increased oxygen production. At the same time, it appears that increased geological activity led to heavy erosion on our planet\u2019s surface. Tthis buried much of the plant carbon before it could recombine with oxygen to form CO2. 286 Chapter 8 Earth as a Planet Free oxygen began accumulating in the atmosphere about 2 billion years ago, and the increased amount of this gas led to the formation of Earth\u2019s ozone layer (recall that ozone is a triple molecule of oxygen, O3), which protects the surface from deadly solar ultraviolet light. Before that, it was unthinkable for life to venture outside the protective oceans, so the landmasses of Earth were barren. The presence of oxygen, and hence ozone, thus allowed colonization of the land. It also made possible a tremendous proliferation of animals, which lived by taking in and using the organic materials produced by plants as their own energy source. As animals evolved in an environment increasingly rich in oxygen, they were able to develop techniques for breathing oxygen directly from the atmosphere. We humans take it for granted that plenty of free oxygen is available in Earth\u2019s atmosphere, and we use it to release energy from the food we take in. Although it may seem funny to think of it this way, we are lifeforms that have evolved to breathe in the waste product of plants. It is plants and related microbes that are the primary producers, using sunlight to create energy-rich \u201cfood\u201d for the rest of us. On a planetary scale, one of the consequences of life has been a decrease in atmospheric carbon dioxide. In the absence of life, Earth would probably have an atmosphere dominated by CO2, like Mars or Venus. But living things, in combination with high levels of geological activity, have effectively stripped our atmosphere of most of this gas. The Greenhouse Effect and Global Warming We have a special interest in the carbon dioxide content of the atmosphere because of the key role this gas plays in retaining heat from the Sun through a process called the greenhouse effect. To understand how the greenhouse effect works, consider the fate of sunlight that strikes the surface of Earth. The light penetrates our atmosphere, is absorbed by the ground, and heats the surface layers. At the temperature of Earth\u2019s surface, that energy is then reemitted as infrared or heat radiation (Figure 8.17). However, the molecules of our atmosphere, which allow visible light through, are good at absorbing infrared energy. As a result, CO2 (along with methane and water vapor) acts like a blanket, trapping heat in the atmosphere and impeding its flow back to space. To maintain an energy balance, the temperature of the surface and lower atmosphere must increase until the total energy radiated by Earth to space equals the energy received from the Sun. The more CO2 there is in our atmosphere, the higher the temperature at which Earth\u2019s surface reaches a new balance. Figure 8.17. How the Greenhouse Effect Works. Sunlight that penetrates to Earth\u2019s lower atmosphere and surface is reradiated as infrared or heat radiation, which is trapped by greenhouse gases such as water vapor, methane, and CO2 in the atmosphere. The result is a higher surface temperature for our planet. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 287 The greenhouse effect in a planetary atmosphere is similar to the heating of a gardener\u2019s greenhouse or the inside of a car left out in the Sun with the windows rolled up. In these examples, the window glass plays the role of greenhouse gases, letting sunlight in but reducing the outward flow of heat radiation. As a result, a greenhouse or car interior winds up much hotter", " than would be expected from the heating of sunlight alone. On Earth, the current greenhouse effect elevates the surface temperature by about 23 \u00b0C. Without this greenhouse effect, the average surface temperature would be well below freezing and Earth would be locked in a global ice age. That\u2019s the good news; the bad news is that the heating due to the greenhouse effect is increasing. Modern industrial society depends on energy extracted from burning fossil fuels. In effect, we are exploiting the energy- rich material created by photosynthesis tens of millions of years ago. As these ancient coal and oil deposits are oxidized (burned using oxygen), large quantities of carbon dioxide are released into the atmosphere. The problem is exacerbated by the widespread destruction of tropical forests, which we depend on to extract CO2 from the atmosphere and replenish our supply of oxygen. In the past century of increased industrial and agricultural development, the amount of CO2 in the atmosphere increased by about 30% and continues to rise at more than 0.5% per year. Before the end of the present century, Earth\u2019s CO2 level is predicted to reach twice the value it had before the industrial revolution (Figure 8.18). The consequences of such an increase for Earth\u2019s surface and atmosphere (and the creatures who live there) are likely to be complex changes in climate, and may be catastrophic for many species. Many groups of scientists are now studying the effects of such global warming with elaborate computer models, and climate change has emerged as the greatest known threat (barring nuclear war) to both industrial civilization and the ecology of our planet. Figure 8.18. Increase of Atmospheric Carbon Dioxide over Time. Scientists expect that the amount of CO2 will double its preindustrial level before the end of the twenty-first century. Measurements of the isotopic signatures of this added CO2 demonstrate that it is mostly coming from burning fossil fuels. (credit: modification of work by NOAA This short PBS video (https://openstax.org/l/30pbsgreengas) explains the physics of the greenhouse effect. 288 Chapter 8 Earth as a Planet Already climate change is widely apparent. Around the world, temperature records are constantly set and broken; all but one of the hottest recorded years have taken place since 2000. Glaciers are retreating, and the Arctic Sea ice is now much thinner than when it was first explored with nuclear submarines in the 1950s. Rising sea levels (from both melting glaciers and expansion of the water as its temperature rises) pose one of the most immediate threats, and many coastal cities have plans to build dikes or seawalls to hold back the expected flooding. The rate of temperature increase is without historical precedent, and we are rapidly entering \u201cunknown territory\u201d where human activities are leading to the highest temperatures on Earth in more than 50 million years. Human Impacts on Our Planet Earth is so large and has been here for so long that some people have trouble accepting that humans are really changing the planet, its atmosphere, and its climate. They are surprised to learn, for example, that the carbon dioxide released from burning fossil fuels is 100 times greater than that emitted by volcanoes. But, the data clearly tell the story that our climate is changing rapidly, and that almost all of the change is a result of human activity. This is not the first time that humans have altered our environment dramatically. Some of the greatest changes were caused by our ancestors, before the development of modern industrial society. If aliens had visited Earth 50,000 years ago, they would have seen much of the planet supporting large animals of the sort that now survive only in Africa. The plains of Australia were occupied by giant marsupials such as diprododon and zygomaturus (the size of our elephants today), and a species of kangaroo that stood 10 feet high. North America and North Asia hosted mammoths, saber tooth cats, mastodons, giant sloths, and even camels. The Islands of the Pacific teemed with large birds, and vast forests covered what are now the farms of Europe and China. Early human hunters killed many large mammals and marsupials, early farmers cut down most of the forests, and the Polynesian expansion across the Pacific doomed the population of large birds. An even greater mass extinction is underway as a result of rapid climate change. In recognition of our impact on the environment, scientists have proposed giving a new name to the current epoch, the anthropocine, when human activity started to have a significant global impact. Although not an officially approved name, the concept of \u201canthropocine\u201d is useful for recognizing that we humans now represent the dominant influence on our planet\u2019s atmosphere and ecology, for better or for worse. 8.5 COSMIC INFLUENCES ON THE EVOLUTION OF EARTH Learning Objectives By the end of this section, you will be able to: Explain the scarcity of impact craters on Earth compared with other planets and moons Describe the evidence for recent impacts", " on Earth Detail how a massive impact changed the conditions for life on Earth, leading to the extinction of the dinosaurs Describe how impacts have influenced the evolution of life on Earth Discuss the search for objects that could potentially collide with our planet In discussing Earth\u2019s geology earlier in this chapter, we dealt only with the effects of internal forces, expressed through the processes of plate tectonics and volcanism. On the Moon, in contrast, we see primarily craters, produced by the impacts of interplanetary debris such as asteroids and comets. Why don\u2019t we see more evidence on Earth of the kinds of impact craters that are so prominent on the Moon and other worlds? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 289 Where Are the Craters on Earth? It is not possible that Earth escaped being struck by the interplanetary debris that has pockmarked the Moon. From a cosmic perspective, the Moon is almost next door. Our atmosphere does make small pieces of cosmic debris burn up (which we see as meteors\u2014commonly called shooting stars). But, the layers of our air provide no shield against the large impacts that form craters several kilometers in diameter and are common on the Moon. In the course of its history, Earth must therefore have been impacted as heavily as the Moon. The difference is that, on Earth, these craters are destroyed by our active geology before they can accumulate. As plate tectonics constantly renews our crust, evidence of past cratering events is slowly erased. Only in the past few decades have geologists succeeded in identifying the eroded remnants of many impact craters (Figure 8.19). Even more recent is our realization that, over the history of Earth, these impacts have had an important influence on the evolution of life. Figure 8.19. Ouarkziz Impact Crater. Located in Algeria, this crater (the round feature in the center) is the result of a meteor impact during the Cretaceous period. Although the crater has experienced heavy erosion, this image from the International Space Station shows the circular pattern resulting from impact. (credit: modification of work by NASA) Recent Impacts The collision of interplanetary debris with Earth is not a hypothetical idea. Evidence of relatively recent impacts can be found on our planet\u2019s surface. One well-studied historic collision took place on June 30, 1908, near the Tunguska River in Siberia. In this desolate region, there was a remarkable explosion in the atmosphere about 8 kilometers above the surface. The shock wave flattened more than a thousand square kilometers of forest (Figure 8.20). Herds of reindeer and other animals were killed, and a man at a trading post 80 kilometers from the blast was thrown from his chair and knocked unconscious. The blast wave spread around the world, as recorded by instruments designed to measure changes in atmospheric pressure. 290 Chapter 8 Earth as a Planet Figure 8.20. Aftermath of the Tunguska Explosion. This photograph, taken 21 years after the blast, shows a part of the forest that was destroyed by the 5-megaton explosion, resulting when a stony projectile about the size of a small office building (40 meters in diameter) collided with our planet. (credit: modification of work by Leonid Kulik) Despite this violence, no craters were formed by the Tunguska explosion. Shattered by atmospheric pressure, the stony projectile with a mass of approximately 10,000 tons disintegrated above our planet\u2019s surface to create a blast equivalent to a 5-megaton nuclear bomb. Had it been smaller or more fragile, the impacting body would have dissipated its energy at high altitude and probably attracted no attention. Today, such high-altitude atmospheric explosions are monitored regularly by military surveillance systems. If it had been larger or made of stronger material (such as metal), the Tunguska projectile would have penetrated all the way to the surface of Earth and exploded to form a crater. Instead, only the heat and shock of the atmospheric explosion reached the surface, but the devastation it left behind in Siberia bore witness to the power of such impacts. Imagine if the same rocky impactor had exploded over New York City in 1908; history books might today record it as one of the most deadly events in human history. Tens of thousands of people witnessed directly the explosion of a smaller (20-meter) projectile over the Russian city of Chelyabinsk on an early winter morning in 2013. It exploded at a height of 21 kilometers in a burst of light brighter than the Sun, and the shockwave of the 0.5-megaton explosion broke tens of thousands of windows and sent hundreds of people to the hospital. Rock fragments (meteorites) were easily collected by people in the area after the blast because they landed on fresh snow Dr. David Morrison, one of the original authors of this textbook, provides a", " nontechnical talk (https://openstax.org/l/30chelyabinskex) about the Chelyabinsk explosion, and impacts in general. The best-known recent crater on Earth was formed about 50,000 years ago in Arizona. The projectile in this case was a lump of iron about 40 meters in diameter. Now called Meteor Crater and a major tourist attraction on the way to the Grand Canyon, the crater is about a mile across and has all the features associated with similar- This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 291 size lunar impact craters (Figure 8.21). Meteor Crater is one of the few impact features on Earth that remains relatively intact; some older craters are so eroded that only a trained eye can distinguish them. Nevertheless, more than 150 have been identified. (See the list of suggested online sites at the end of this chapter if you want to find out more about these other impact scars.) Figure 8.21. Meteor Crater in Arizona. Here we see a 50,000-year-old impact crater made by the collision of a 40-meter lump of iron with our planet. Although impact craters are common on less active bodies such as the Moon, this is one of the very few well-preserved craters on Earth. (modification of work by D. Roddy/USGS) Mass Extinction The impact that produced Meteor Crater would have been dramatic indeed to any humans who witnessed it (from a safe distance) since the energy release was equivalent to a 10-megaton nuclear bomb. But such explosions are devastating only in their local areas; they have no global consequences. Much larger (and rarer) impacts, however, can disturb the ecological balance of the entire planet and thus influence the course of evolution. The best-documented large impact took place 65 million years ago, at the end of what is now called the Cretaceous period of geological history. This time in the history of life on Earth was marked by a mass extinction, in which more than half of the species on our planet died out. There are a dozen or more mass extinctions in the geological record, but this particular event (nicknamed the \u201cgreat dying\u201d) has always intrigued paleontologists because it marks the end of the dinosaur age. For tens of millions of years these great creatures had flourished and dominated. Then, they suddenly disappeared (along with many other species), and thereafter mammals began the development and diversification that ultimately led to all of us. The object that collided with Earth at the end of the Cretaceous period struck a shallow sea in what is now the Yucat\u00e1n peninsula of Mexico. Its mass must have been more than a trillion tons, determined from study of a worldwide layer of sediment deposited from the dust cloud that enveloped the planet after its impact. First identified in 1979, this sediment layer is rich in the rare metal iridium and other elements that are relatively abundant in asteroids and comets, but exceedingly rare in Earth\u2019s crust. Even though it was diluted by the material that the explosion excavated from the surface of Earth, this cosmic component can still be identified. In addition, this layer of sediment contains many minerals characteristic of the temperatures and pressures of a gigantic explosion. 292 Chapter 8 Earth as a Planet The impact that led to the extinction of dinosaurs released energy equivalent to 5 billion Hiroshima-size nuclear bombs and excavated a crater 200 kilometers across and deep enough to penetrate through Earth\u2019s crust. This large crater, named Chicxulub for a small town near its center, has subsequently been buried in sediment, but its outlines can still be identified (Figure 8.22). The explosion that created the Chicxulub crater lifted about 100 trillion tons of dust into the atmosphere. We can determine this amount by measuring the thickness of the sediment layer that formed when this dust settled to the surface. Figure 8.22. Site of the Chicxulub Crater. This map shows the location of the impact crater created 65 million years ago on Mexico\u2019s Yucat\u00e1n peninsula. The crater is now buried under more than 500 meters of sediment. (credit: modification of work by \u201cCarport\u201d/Wikimedia) Such a quantity of airborne material would have blocked sunlight completely, plunging Earth into a period of cold and darkness that lasted several months. Many plants dependent on sunlight would have died, leaving plant-eating animals without a food supply. Other worldwide effects included large-scale fires (started by the hot, flying debris from the explosion) that destroyed much of the planet\u2019s forests and grasslands, and a long period in which rainwater around the globe was acidic. It was these environmental effects, rather than the explosion itself, that were responsible for the mass extinction, including the demise of the dinosaurs. Impacts and the Evolution of Life It is becoming clear that many\u2014", "perhaps most\u2014mass extinctions in Earth\u2019s long history resulted from a variety of other causes, but in the case of the dinosaur killer, the cosmic impact certainly played a critical role and may have been the \u201cfinal straw\u201d in a series of climactic disturbances that resulted in the \u201cgreat dying.\u201d A catastrophe for one group of living things, however, may create opportunities for another group. Following each mass extinction, there is a sudden evolutionary burst as new species develop to fill the ecological niches opened by the event. Sixty-five million years ago, our ancestors, the mammals, began to thrive when so many other species died out. We are the lucky beneficiaries of this process. Impacts by comets and asteroids represent the only mechanisms we know of that could cause truly global catastrophes and seriously influence the evolution of life all over the planet. As paleontologist Stephen Jay Gould of Harvard noted, such a perspective changes fundamentally our view of biological evolution. The central This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 293 issues for the survival of a species must now include more than just its success in competing with other species and adapting to slowly changing environments, as envisioned by Darwin\u2019s idea of natural selection. Also required is an ability to survive random global catastrophes due to impacts. Still earlier in its history, Earth was subject to even larger impacts from the leftover debris of planet formation. We know that the Moon was struck repeatedly by objects larger than 100 kilometers in diameter\u20141000 times more massive than the object that wiped out most terrestrial life 65 million years ago. Earth must have experienced similar large impacts during its first 700 million years of existence. Some of them were probably violent enough to strip the planet of most its atmosphere and to boil away its oceans. Such events would sterilize the planet, destroying any life that had begun. Life may have formed and been wiped out several times before our own microbial ancestors took hold sometime about 4 billion years ago. The fact that the oldest surviving microbes on Earth are thermophiles (adapted to very high temperatures) can also be explained by such large impacts. An impact that was just a bit too small to sterilize the planet would still have destroyed anything that lived in what we consider \u201cnormal\u201d environments, and only the creatures adapted to high temperatures would survive. Thus, the oldest surviving terrestrial lifeforms are probably the remnants of a sort of evolutionary bottleneck caused by repeated large impacts early in the planet\u2019s history. Impacts in Our Future? The impacts by asteroids and comets that have had such a major influence on life are not necessarily a thing of the past. In the full scope of planetary history, 65 million years ago was just yesterday. Earth actually orbits the Sun within a sort of cosmic shooting gallery, and although major impacts are rare, they are by no means over. Humanity could suffer the same fate as the dinosaurs, or lose a city to the much more frequent impacts like the one over Tunguska, unless we figure out a way to predict the next big impact and to protect our planet. The fact that our solar system is home to some very large planets in outer orbits may be beneficial to us; the gravitational fields of those planets can be very effective at pulling in cosmic debris and shielding us from larger, more frequent impacts. Beginning in the 1990s, a few astronomers began to analyze the cosmic impact hazard and to persuade the government to support a search for potentially hazardous asteroids. Several small but sophisticated wide-field telescopes are now used for this search, which is called the NASA Spaceguard Survey. Already we know that there are currently no asteroids on a collision course with Earth that are as big (10\u201315 kilometers) as the one that killed the dinosaurs. The Spaceguard Survey now concentrates on finding smaller potential impactors. By 2015, the search had netted more than 15,000 near-Earth-asteroids, including most of those larger than 1 kilometer. None of those discovered so far poses any danger to us. Of course, we cannot make a similar statement about the asteroids that have not yet been discovered, but these will be found and evaluated one by one for their potential hazard. These asteroid surveys are one of the few really life-and-death projects carried out by astronomers, with a potential to help to save our planet from future major impacts The Torino Impact Hazard Scale (https://openstax.org/l/30torhazscale) is a method for categorizing the impact hazard associated with near-Earth objects such as asteroids and comets. It is a communication tool for astronomers and the public to assess the seriousness of collision predictions by combining probability statistics and known kinetic damage potentials into a single threat value. 294 Chapter 8 Earth as a Planet Purdue University\u2019s \u201cImpact: Earth\u201d calculator (https://openstax.org/l/30purimpearcal)", " lets you input the characteristics of an approaching asteroid to determine the effect of its impact on our planet. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 295 CHAPTER 8 REVIEW KEY TERMS bar a force of 100,000 Newtons acting on a surface area of 1 square meter; the average pressure of Earth\u2019s atmosphere at sea level is 1.013 bars basalt igneous rock produced by the cooling of lava; makes up most of Earth\u2019s oceanic crust and is found on other planets that have experienced extensive volcanic activity convection movement caused within a gas or liquid by the tendency of hotter, and therefore less dense material, to rise and colder, denser material to sink under the influence of gravity, which consequently results in transfer of heat core the central part of the planet; consists of higher density material crust the outer layer of a terrestrial planet fault in geology, a crack or break in the crust of a planet along which slippage or movement can take place, accompanied by seismic activity granite a type of igneous silicate rock that makes up most of Earth\u2019s continental crust greenhouse effect by CO2 in its atmosphere the blanketing (absorption) of infrared radiation near the surface of a planet\u2014for example, greenhouse gas a gas in an atmosphere that absorbs and emits radiation within the thermal infrared range; on Earth, these atmospheric gases primarily include carbon dioxide, methane, and water vapor igneous rock rock produced by cooling from a molten state magnetosphere the region around a planet in which its intrinsic magnetic field dominates the interplanetary field carried by the solar wind; hence, the region within which charged particles can be trapped by the planetary magnetic field mantle the largest part of Earth\u2019s interior; lies between the crust and the core mass extinction the sudden disappearance in the fossil record of a large number of species of life, to be replaced by fossils of new species in subsequent layers; mass extinctions are indicators of catastrophic changes in the environment, such as might be produced by a large impact on Earth metamorphic rock rock produced by physical and chemical alteration (without melting) under high temperature and pressure ozone (O3) a heavy molecule of oxygen that contains three atoms rather than the more normal two photosynthesis a complex sequence of chemical reactions through which some living things can use sunlight to manufacture products that store energy (such as carbohydrates), releasing oxygen as one by-product plate tectonics the motion of segments or plates of the outer layer of a planet over the underlying mantle primitive rock rock that has not experienced great heat or pressure and therefore remains representative of the original condensed materials from the solar nebula 296 Chapter 8 Earth as a Planet rift zone in geology, a place where the crust is being torn apart by internal forces generally associated with the injection of new material from the mantle and with the slow separation of tectonic plates sedimentary rock rock formed by the deposition and cementing of fine grains of material, such as pieces of igneous rock or the shells of living things seismic wave a vibration that travels through the interior of Earth or any other object; on Earth, these are generally caused by earthquakes stratosphere the layer of Earth\u2019s atmosphere above the troposphere and below the ionosphere subduction the sideways and downward movement of the edge of a plate of Earth\u2019s crust into the mantle beneath another plate troposphere the lowest level of Earth\u2019s atmosphere, where most weather takes place volcano a place where material from a planet\u2019s mantle erupts on its surface SUMMARY 8.1 The Global Perspective Earth is the prototype terrestrial planet. Its interior composition and structure are probed using seismic waves. Such studies reveal that Earth has a metal core and a silicate mantle. The outer layer, or crust, consists primarily of oceanic basalt and continental granite. A global magnetic field, generated in the core, produces Earth\u2019s magnetosphere, which can trap charged atomic particles. 8.2 Earth\u2019s Crust Terrestrial rocks can be classified as igneous, sedimentary, or metamorphic. A fourth type, primitive rock, is not found on Earth. Our planet\u2019s geology is dominated by plate tectonics, in which crustal plates move slowly in response to mantle convection. The surface expression of plate tectonics includes continental drift, recycling of the ocean floor, mountain building, rift zones, subduction zones, faults, earthquakes, and volcanic eruptions of lava from the interior. 8.3 Earth\u2019s Atmosphere The atmosphere has a surface pressure of 1 bar and is composed primarily of N2 and O2, plus such important trace gases as H2O, CO2, and O3. Its structure consists of the troposphere, stratosphere, mesosphere, and ionosphere. Changing the composition of the atmosphere also influences the temperature. Atmospheric circulation (weather) is driven by seasonally changing deposition of sunlight. Many longer term climate variations", ", such as the ice ages, are related to changes in the planet\u2019s orbit and axial tilt. 8.4 Life, Chemical Evolution, and Climate Change Life originated on Earth at a time when the atmosphere lacked O2 and consisted mostly of CO2. Later, photosynthesis gave rise to free oxygen and ozone. Modern genomic analysis lets us see how the wide diversity of species on the planet are related to each other. CO2 and methane in the atmosphere heat the surface through the greenhouse effect; today, increasing amounts of atmospheric CO2 are leading to the global warming of our planet. 8.5 Cosmic Influences on the Evolution of Earth Earth, like the Moon and other planets, has been influenced by the impacts of cosmic debris, including such recent examples as Meteor Crater and the Tunguska explosion. Larger past impacts are implicated in some mass extinctions, including the large impact 65 million years ago at the end of the Cretaceous period that This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 297 wiped out the dinosaurs and many other species. Today, astronomers are working to predict the next impact in advance, while other scientists are coming to grips with the effect of impacts on the evolution and diversity of life on Earth. FOR FURTHER EXPLORATION Articles Earth Collins, W., et al. \u201cThe Physical Science behind Climate Change.\u201d Scientific American (August 2007): 64. Why scientists are now confident that human activities are changing our planet\u2019s climate. Glatzmaier, G., & Olson, P. \u201cProbing the Geodynamo.\u201d Scientific American (April 2005): 50. Experiments and modeling that tell us about the source and reversals of Earth\u2019s magnetic field. Gurnis, M. \u201cSculpting the Earth from Inside Out.\u201d Scientific American (March 2001): 40. On motions that lift and lower the continents. Hartmann, W. \u201cPiecing Together Earth\u2019s Early History.\u201d Astronomy (June 1989): 24. Jewitt, D., & Young, E. \u201cOceans from the Skies.\u201d Scientific American (March 2015): 36. How did Earth get its water after its initial hot period? Impacts Boslaugh, M. \u201cIn Search of Death-Plunge Asteroids.\u201d Astronomy (July 2015): 28. On existing and proposed programs to search for earth-crossing asteroids. Brusatte, S. \u201cWhat Killed the Dinosaurs?\u201d Scientific American (December 2015): 54. The asteroid hit Earth at an already vulnerable time. Chyba, C. \u201cDeath from the Sky: Tunguska.\u201d Astronomy (December 1993): 38. Excellent review article. Durda, D. \u201cThe Chelyabinsk Super-Meteor.\u201d Sky & Telescope (June 2013): 24. A nice summary with photos and eyewitness reporting. Gasperini, L., et al. \u201cThe Tunguska Mystery.\u201d Scientific American (June 2008): 80. A more detailed exploration of the site of the 1908 impact over Siberia. Kring, D. \u201cBlast from the Past.\u201d Astronomy (August 2006): 46. Six-page introduction to Arizona\u2019s meteor crater. Websites Earth Astronaut Photography of Earth from Space: http://earth.jsc.nasa.gov/. A site with many images and good information. Exploration of the Earth\u2019s Magnetosphere: http://phy6.org/Education/Intro.html. An educational website by Dr. Daniel Stern. NASA Goddard: Earth from Space: Fifteen Amazing Things in 15 Years: https://www.nasa.gov/content/goddard/ earth-from-space-15-amazing-things-in-15-years. Images and videos that reveal things about our planet and its atmosphere. U.S. Geological Survey: Earthquake Information Center: http://earthquake.usgs.gov/learn/ Views of the Solar System: http://www.solarviews.com/eng/earth.htm. Overview of Earth. 298 Impacts Chapter 8 Earth as a Planet B612 Foundation : https://b612foundation.org/. Set up by several astronauts for research and education about the asteroid threat to Earth and to build a telescope in space to search for dangerous asteroids. Lunar and Planetary Institute: Introduction to Terrestrial Impact Craters: http://www.lpi.usra.edu/publications/ slidesets/craters/. Includes images. Meteor Crater Tourist Site: http://meteorcrater.com/. NASA/Jet Propulsion Lab Near Earth Object Program: http://neo.jpl.nasa.gov/neo/. What Are Near-Earth-Objects", ": http://spaceguardcentre.com/what-are-neos/. From the British Spaceguard Centre. Videos Earth All Alone in the Night: http://apod.nasa.gov/apod/ap120305.html. Flying over Earth at night (2:30). Earth Globes Movies (including Earth at night): http://astro.uchicago.edu/cosmus/projects/earth/. Earth: The Operator\u2019s Manual: http://earththeoperatorsmanual.com/feature-video/earth-the-operators- manual. A National Science Foundation\u2013sponsored miniseries on climate change and energy, with geologist Richard Alley (53:43). PBS NOVA Videos about Earth: http://www.pbs.org/wgbh/nova/earth/. Programs and information about planet Earth. Click full episodes on the menu at left to be taken to a nice array of videos. U. S. National Weather Service: http://earth.nullschool.net. Real Time Globe of Earth showing wind patterns which can be zoomed and moved to your preferred view. Impacts Chelyabinsk Meteor: Can We Survive a Bigger Impact?: https://www.youtube.com/watch?v=Y-e6xyUZLLs. Talk by Dr. David Morrison (1:34:48). Large Asteroid Impact Simulation: https://www.youtube.com/watch?v=bU1QPtOZQZU. Large asteroid impact simulation from the Discovery Channel (4:45). Meteor Hits Russia February 15, 2013: https://www.youtube.com/watch?v=dpmXyJrs7iU. Archive of eyewitness footage (10:11). Sentinel Mission: Finding an Asteroid Headed for Earth: https://www.youtube.com/watch?v=efz8c3ijD_A. Public lecture by astronaut Ed Lu (1:08:57). COLLABORATIVE GROUP ACTIVITIES A. If we can predict that lots of ground movement takes place along subduction zones and faults, then why do so many people live there? Should we try to do anything to discourage people from living in these areas? What inducement would your group offer people to move? Who would pay for the relocation? (Note that two of the original authors of this book live quite close to the San Andreas and Hayward faults. If they wrote this chapter and haven\u2019t moved, what are the chances others living in these kinds of areas will move?) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 299 B. After your group reads the feature box on Alfred Wegener: Catching the Drift of Plate Tectonics, discuss some reasons his idea did not catch on right away among scientists. From your studies in this course and in other science courses (in college and before), can you cite other scientific ideas that we now accept but that had controversial beginnings? Can you think of any scientific theories that are still controversial today? If your group comes up with some, discuss ways scientists could decide whether each theory on your list is right. C. Suppose we knew that a large chunk of rock or ice (about the same size as the one that hit 65 million years ago) will impact Earth in about 5 years. What could or should we do about it? (The film Deep Impact dealt with this theme.) Does your group think that the world as a whole should spend more money to find and predict the orbits of cosmic debris near Earth? D. Carl Sagan pointed out that any defensive weapon that we might come up with to deflect an asteroid away from Earth could be used as an offensive weapon by an unstable dictator in the future to cause an asteroid not heading our way to come toward Earth. The history of human behavior, he noted, has shown that most weapons that are built (even with the best of motives) seem to wind up being used. Bearing this in mind, does your group think we should be building weapons to protect Earth from asteroid or comet impact? Can we afford not to build them? How can we safeguard against these collisions? E. Is there evidence of climate change in your area over the past century? How would you distinguish a true climate change from the random variations in weather that take place from one year to the next? EXERCISES Review Questions 1. What is the thickest interior layer of Earth? The thinnest? 2. What are Earth\u2019s core and mantle made of? Explain how we know. 3. Describe the differences among primitive, igneous, sedimentary, and metamorphic rock, and relate these differences to their origins. 4. Explain briefly how the following phenomena happen on Earth, relating your answers to the theory of plate tectonics A. earthquakes B. continental drift C. mountain building D. volcanic eruptions E", ". creation of the Hawaiian island chain 5. What is the source of Earth\u2019s magnetic field? 6. Why is the shape of the magnetosphere not spherical like the shape of Earth? 7. Although he did not present a mechanism, what were the key points of Alfred Wegener\u2019s proposal for the concept of continental drift? 300 Chapter 8 Earth as a Planet 8. List the possible interactions between Earth\u2019s crustal plates that can occur at their boundaries. 9. List, in order of decreasing altitude, the principle layers of Earth\u2019s atmosphere. 10. In which atmospheric layer are almost all water-based clouds formed? 11. What is, by far, the most abundant component of Earth\u2019s atmosphere? 12. In which domain of living things do you find humankind? 13. Describe three ways in which the presence of life has affected the composition of Earth\u2019s atmosphere. 14. Briefly describe the greenhouse effect. 15. How do impacts by comets and asteroids influence Earth\u2019s geology, its atmosphere, and the evolution of life? 16. Why are there so many impact craters on our neighbor world, the Moon, and so few on Earth? 17. Detail some of the anthropogenic changes to Earth\u2019s climate and their potential impact on life. Thought Questions 18. If you wanted to live where the chances of a destructive earthquake were small, would you pick a location near a fault zone, near a mid ocean ridge, near a subduction zone, or on a volcanic island such as Hawaii? What are the relative risks of earthquakes at each of these locations? 19. Which type of object would likely cause more damage if it struck near an urban area: a small metallic object or a large stony/icy one? 20. If all life were destroyed on Earth by a large impact, would new life eventually form to take its place? Explain how conditions would have to change for life to start again on our planet. 21. Why is a decrease in Earth\u2019s ozone harmful to life? 22. Why are we concerned about the increases in CO2 and other gases that cause the greenhouse effect in Earth\u2019s atmosphere? What steps can we take in the future to reduce the levels of CO2 in our atmosphere? What factors stand in the way of taking the steps you suggest? (You may include technological, economic, and political factors in your answer.) 23. Do you think scientists should make plans to defend Earth from future asteroid impacts? Is it right to intervene in the same evolutionary process that made the development of mammals (including us) possible after the big impact 65 million years ago? Figuring For Yourself 24. Europe and North America are moving apart by about 5 m per century. As the continents separate, new ocean floor is created along the mid-Atlantic Rift. If the rift is 5000 km long, what is the total area of new ocean floor created in the Atlantic each century? (Remember that 1 km = 1000 m.) 25. Over the entire Earth, there are 60,000 km of active rift zones, with average separation rates of 5 m/ century. How much area of new ocean crust is created each year over the entire planet? (This area is approximately equal to the amount of ocean crust that is subducted since the total area of the oceans remains about the same.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 8 Earth as a Planet 301 26. With the information from Exercise 8.25, you can calculate the average age of the ocean floor. First, find the total area of the ocean floor (equal to about 60% of the surface area of Earth). Then compare this with the area created (or destroyed) each year. The average lifetime is the ratio of these numbers: the total area of ocean crust compared to the amount created (or destroyed) each year. 27. What is the volume of new oceanic basalt added to Earth\u2019s crust each year? Assume that the thickness of the new crust is 5 km, that there are 60,000 km of rifts, and that the average speed of plate motion is 4 cm/y. What fraction of Earth\u2019s entire volume does this annual addition of new material represent? 28. Suppose a major impact that produces a mass extinction takes place on Earth once every 5 million years. Suppose further that if such an event occurred today, you and most other humans would be killed (this would be true even if the human species as a whole survived). Such impact events are random, and one could take place at any time. Calculate the probability that such an impact will occur within the next 50 years (within your lifetime). 29. How do the risks of dying from the impact of an asteroid or comet compare with other risks we are concerned about, such as dying in a car accident or from heart disease or some other natural cause? (Hint: To find the annual risk", ", go to the library or internet and look up the annual number of deaths from a particular cause in a particular country, and then divide by the population of that country.) 30. What fraction of Earth\u2019s volume is taken up by the core? 31. Approximately what percentage of Earth\u2019s radius is represented by the crust? 32. What is the drift rate of the Pacific plate over the Hawaiian hot spot? 33. What is the percent increase of atmospheric CO2 in the past 20 years? 34. Estimate the mass of the object that formed Meteor Crater in Arizona. 302 Chapter 8 Earth as a Planet This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 303 9 CRATERED WORLDS Figure 9.1. Apollo 11 Astronaut Edwin \u201cBuzz\u201d Aldrin on the Surface of the Moon. Because there is no atmosphere, ocean, or geological activity on the Moon today, the footprints you see in the image will likely be preserved in the lunar soil for millions of years (credit: modification of work by NASA/ Neil A. Armstrong). Chapter Outline 9.1 General Properties of the Moon 9.2 The Lunar Surface 9.3 Impact Craters 9.4 The Origin of the Moon 9.5 Mercury Thinking Ahead The Moon is the only other world human beings have ever visited. What is it like to stand on the surface of our natural satellite? And what can we learn from going there and bringing home pieces of a different world? We begin our discussion of the planets as cratered worlds with two relatively simple objects: the Moon and Mercury. Unlike Earth, the Moon is geologically dead, a place that has exhausted its internal energy sources. Because its airless surface preserves events that happened long ago, the Moon provides a window on earlier epochs of solar system history. The planet Mercury is in many ways similar to the Moon, which is why the two are discussed together: both are relatively small, lacking in atmospheres, deficient in geological activity, and dominated by the effects of impact cratering. Still, the processes that have molded their surfaces are not unique to these two worlds. We shall see that they have acted on many other members of the planetary system as well. 9.1 GENERAL PROPERTIES OF THE MOON Learning Objectives By the end of this section, you will be able to: 304 Chapter 9 Cratered Worlds Discuss what has been learned from both manned and robotic lunar exploration Describe the composition and structure of the Moon The Moon has only one-eightieth the mass of Earth and about one-sixth Earth\u2019s surface gravity\u2014too low to retain an atmosphere (Figure 9.2). Moving molecules of a gas can escape from a planet just the way a rocket does, and the lower the gravity, the easier it is for the gas to leak away into space. While the Moon can acquire a temporary atmosphere from impacting comets, this atmosphere is quickly lost by freezing onto the surface or by escape to surrounding space. The Moon today is dramatically deficient in a wide range of volatiles, those elements and compounds that evaporate at relatively low temperatures. Some of the Moon\u2019s properties are summarized in Table 9.1, along with comparative values for Mercury. Figure 9.2. Two Sides of the Moon. The left image shows part of the hemisphere that faces Earth; several dark maria are visible. The right image shows part of the hemisphere that faces away from Earth; it is dominated by highlands. The resolution of this image is several kilometers, similar to that of high-powered binoculars or a small telescope. (credit: modification of work by NASA/GSFC/Arizona State University) Properties of the Moon and Mercury Property Moon Mercury Mass (Earth = 1) Diameter (km) Density (g/cm3) Surface gravity (Earth = 1) Escape velocity (km/s) Rotation period (days) Table 9.1 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 0.0123 3476 3.3 0.17 2.4 27.3 0.055 4878 5.4 0.38 4.3 58.65 Chapter 9 Cratered Worlds 305 Properties of the Moon and Mercury Property Moon Mercury Surface area (Earth = 1) 0.27 0.38 Table 9.1 Exploration of the Moon Most of what we know about the Moon today derives from the US Apollo program, which sent nine piloted spacecraft to our satellite between 1968 and 1972, landing 12 astronauts on its surface (Figure 9.1). Before the era of spacecraft studies, astronomers had mapped the side of the Moon that faces Earth with telescopic resolution of about 1 kilometer, but lunar geology hardly existed as a scientific subject. All that changed beginning in the early 1960s. Initially, Russia took the lead in lunar exploration with Luna 3, which returned", " the first photos of the lunar far side in 1959, and then with Luna 9, which landed on the surface in 1966 and transmitted pictures and other data to Earth. However, these efforts were overshadowed on July 20, 1969, when the first American astronaut set foot on the Moon. Table 9.2 summarizes the nine Apollo flights: six that landed and three others that circled the Moon but did not land. The initial landings were on flat plains selected for safety reasons. But with increasing experience and confidence, NASA targeted the last three missions to more geologically interesting locales. The level of scientific exploration also increased with each mission, as the astronauts spent longer times on the Moon and carried more elaborate equipment. Finally, on the last Apollo landing, NASA included one scientist, geologist Jack Schmitt, among the astronauts (Figure 9.3). Apollo Flights to the Moon Flight Date Landing Site Main Accomplishment Apollo 8 Dec. 1968 \u2014 First humans to fly around the Moon Apollo 10 May 1969 \u2014 First spacecraft rendezvous in lunar orbit Apollo 11 July 1969 Mare Tranquillitatis First human landing on the Moon; 22 kilograms of samples returned Apollo 12 Nov. 1969 Oceanus Procellarum First Apollo Lunar Surface Experiment Package (ALSEP); visit to Surveyor 3 lander Apollo 13 Apr. 1970 \u2014 Landing aborted due to explosion in command module Apollo 14 Jan. 1971 Mare Nubium First \u201crickshaw\u201d on the Moon Apollo 15 July 1971 Mare Imbrium/ Hadley First \u201crover;\u201d visit to Hadley Rille; astronauts traveled 24 kilometers Table 9.2 306 Chapter 9 Cratered Worlds Apollo Flights to the Moon Flight Date Landing Site Main Accomplishment Apollo 16 Apr. 1972 Descartes First landing in highlands; 95 kilograms of samples returned Apollo 17 Dec. 1972 Taurus-Littrow highlands Geologist among the crew; 111 kilograms of samples returned Table 9.2 Figure 9.3. Scientist on the Moon. Geologist (and later US senator) Harrison \u201cJack\u201d Schmitt in front of a large boulder in the Littrow Valley at the edge of the lunar highlands. Note how black the sky is on the airless Moon. No stars are visible because the surface is brightly lit by the Sun, and the exposure therefore is not long enough to reveal stars. In addition to landing on the lunar surface and studying it at close range, the Apollo missions accomplished three objectives of major importance for lunar science. First, the astronauts collected nearly 400 kilograms of samples for detailed laboratory analysis on Earth (Figure 9.4). These samples have revealed as much about the Moon and its history as all other lunar studies combined. Second, each Apollo landing after the first one deployed an Apollo Lunar Surface Experiment Package (ALSEP), which continued to operate for years after the astronauts departed. Third, the orbiting Apollo command modules carried a wide range of instruments to photograph and analyze the lunar surface from above. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 307 Figure 9.4. Handling Moon Rocks. Lunar samples collected in the Apollo Project are analyzed and stored in NASA facilities at the Johnson Space Center in Houston, Texas. Here, a technician examines a rock sample using gloves in a sealed environment to avoid contaminating the sample. (credit: NASA JSC) The last human left the Moon in December 1972, just a little more than three years after Neil Armstrong took his \u201cgiant leap for mankind.\u201d The program of lunar exploration was cut off midstride due to political and economic pressures. It had cost just about $100 per American, spread over 10 years\u2014the equivalent of one large pizza per person per year. Yet for many people, the Moon landings were one of the central events in twentieth-century history. The giant Apollo rockets built to travel to the Moon were left to rust on the lawns of NASA centers in Florida, Texas, and Alabama, although recently, some have at least been moved indoors to museums (Figure 9.5). Today, neither NASA nor Russia have plans to send astronauts to the Moon, and China appears to be the nation most likely to attempt this feat. (In a bizarre piece of irony, a few people even question whether we went to the Moon at all, proposing instead that the Apollo program was a fake, filmed on a Hollywood sound stage. See the Link to Learning box below for some scientists\u2019 replies to such claims.) However, scientific interest in the Moon is stronger than ever, and more than half a dozen scientific spacecraft\u2014sent from NASA, ESA, Japan, India, and China\u2014have orbited or landed on our nearest neighbor during the past decade Read The Great Moon Hoax (https://openstax.org/l/30greatmoonhoax) about the claim that NASA never succeeded in putting people on the Moon. 308 Chapter 9 Cratered Worlds Figure 9.5. Moon Rocket on Display. One", " of the unused Saturn 5 rockets built to go to the Moon is now a tourist attraction at NASA\u2019s Johnson Space Center in Houston, although it has been moved indoors since this photo was taken. (credit: modification of work by David Morrison) Lunar exploration has become an international enterprise with many robotic spacecraft focusing on lunar science. The USSR sent a number in the 1960s, including robot sample returns. Table 9.3 lists some of the most recent lunar missions. Some International Missions to the Moon Launch Year Spacecraft Type of Mission Agency 1994 1998 2003 2007 2007 2008 2009 2009 Table 9.3 Clementine Orbiter US (USAF/NASA) Lunar Prospector Orbiter US (NASA) SMART-1 SELENE 1 Chang\u2019e 1 Chandrayaan-1 LRO LCROSS Orbiter Orbiter Orbiter Orbiter Orbiter Impactor Europe (ESA) Japan (JAXA) China (CNSA) India (ISRO) US (NASA) US (NASA) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 309 Some International Missions to the Moon Launch Year Spacecraft Type of Mission Agency 2010 2011 2013 2013 Table 9.3 Chang\u2019e 2 Orbiter China (CNSA) GRAIL LADEE Twin orbiters US (NASA) Orbiter US (NASA) Chang\u2019e 3 Lander/Rover China (CNSA) Composition and Structure of the Moon The composition of the Moon is not the same as that of Earth. With an average density of only 3.3 g/cm3, the Moon must be made almost entirely of silicate rock. Compared to Earth, it is depleted in iron and other metals. It is as if the Moon were composed of the same silicates as Earth\u2019s mantle and crust, with the metals and the volatiles selectively removed. These differences in composition between Earth and Moon provide important clues about the origin of the Moon, a topic we will cover in detail later in this chapter. Studies of the Moon\u2019s interior carried out with seismometers taken to the Moon as part of the Apollo program confirm the absence of a large metal core. The twin GRAIL spacecraft launched into lunar orbit in 2011 provided even more precise tracking of the interior structure. We also know from the study of lunar samples that water and other volatiles have been depleted from the lunar crust. The tiny amounts of water detected in these samples were originally attributed to small leaks in the container seal that admitted water vapor from Earth\u2019s atmosphere. However, scientists have now concluded that some chemically bound water is present in the lunar rocks. Most dramatically, water ice has been detected in permanently shadowed craters near the lunar poles. In 2009, NASA crashed a small spacecraft called the Lunar Crater Observation and Sensing Satellite (LCROSS) into the crater Cabeus near the Moon\u2019s south pole. The impact at 9,000 kilometers per hour released energy equivalent to 2 tons of dynamite, blasting a plume of water vapor and other chemicals high above the surface. This plume was visible to telescopes in orbit around the Moon, and the LCROSS spacecraft itself made measurements as it flew through the plume. A NASA spacecraft called the Lunar Reconnaissance Orbiter (LRO) also measured the very low temperatures inside several lunar craters, and its sensitive cameras were even able to image crater interiors by starlight. The total quantity of water ice in the Moon\u2019s polar craters is estimated to be hundreds of billions of tons. As liquid, this would only be enough water to fill a lake 100 miles across, but compared with the rest of the dry lunar crust, so much water is remarkable. Presumably, this polar water was carried to the Moon by comets and asteroids that hit its surface. Some small fraction of the water froze in a few extremely cold regions (cold traps) where the Sun never shines, such as the bottom of deep craters at the Moon\u2019s poles. One reason this discovery could be important is that it raises the possibility of future human habitation near the lunar poles, or even of a lunar base as a way-station on routes to Mars and the rest of the solar system. If the ice could be mined, it would yield both water and oxygen for human support, and it could be broken down into hydrogen and oxygen, a potent rocket fuel. 310 Chapter 9 Cratered Worlds 9.2 THE LUNAR SURFACE Learning Objectives By the end of this section, you will be able to: Differentiate between the major surface features of the Moon Describe the history of the lunar surface Describe the properties of the lunar \u201csoil\u201d General Appearance If you look at the Moon through a telescope, you can see that it is covered by impact craters of all sizes. The most conspicuous of the Moon\u2019s surface features\u2014those that can be seen with the un", "aided eye and that make up the feature often called \u201cthe man in the Moon\u201d\u2014are vast splotches of darker lava flows. Centuries ago, early lunar observers thought that the Moon had continents and oceans and that it was a possible abode of life. They called the dark areas \u201cseas\u201d (maria in Latin, or mare in the singular, pronounced \u201cmah ray\u201d). Their names, Mare Nubium (Sea of Clouds), Mare Tranquillitatis (Sea of Tranquility), and so on, are still in use today. In contrast, the \u201cland\u201d areas between the seas are not named. Thousands of individual craters have been named, however, mostly for great scientists and philosophers (Figure 9.6). Among the most prominent craters are those named for Plato, Copernicus, Tycho, and Kepler. Galileo only has a small crater, however, reflecting his low standing among the Vatican scientists who made some of the first lunar maps. We know today that the resemblance of lunar features to terrestrial ones is superficial. Even when they look somewhat similar, the origins of lunar features such as craters and mountains are very different from their terrestrial counterparts. The Moon\u2019s relative lack of internal activity, together with the absence of air and water, make most of its geological history unlike anything we know on Earth. Figure 9.6. Sunrise on the Central Mountain Peaks of Tycho Crater, as Imaged by the NASA Lunar Reconnaissance Orbiter. Tycho, about 82 kilometers in diameter, is one of the youngest of the very large lunar craters. The central mountain rises 12 kilometers above the crater floor. (credit: modification of work by NASA/Goddard/Arizona State University) Lunar History To trace the detailed history of the Moon or of any planet, we must be able to estimate the ages of individual rocks. Once lunar samples were brought back by the Apollo astronauts, the radioactive dating techniques that had been developed for Earth were applied to them. The solidification ages of the samples ranged from about This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 311 3.3 to 4.4 billion years old, substantially older than most of the rocks on Earth. For comparison, as we saw in the chapter on Earth, Moon, and Sky, both Earth and the Moon were formed between 4.5 and 4.6 billion years ago. Most of the crust of the Moon (83%) consists of silicate rocks called anorthosites; these regions are known as the lunar highlands. They are made of relatively low-density rock that solidified on the cooling Moon like slag floating on the top of a smelter. Because they formed so early in lunar history (between 4.1 and 4.4 billion years ago), the highlands are also extremely heavily cratered, bearing the scars of all those billions of years of impacts by interplanetary debris (Figure 9.7). Figure 9.7. Lunar Highlands. The old, heavily cratered lunar highlands make up 83% of the Moon\u2019s surface. (credit: Apollo 11 Crew, NASA) Unlike the mountains on Earth, the Moon\u2019s highlands do not have any sharp folds in their ranges. The highlands have low, rounded profiles that resemble the oldest, most eroded mountains on Earth (Figure 9.8). Because there is no atmosphere or water on the Moon, there has been no wind, water, or ice to carve them into cliffs and sharp peaks, the way we have seen them shaped on Earth. Their smooth features are attributed to gradual erosion, mostly due to impact cratering from meteorites. 312 Chapter 9 Cratered Worlds Figure 9.8. Lunar Mountain. This photo of Mt. Hadley on the edge of Mare Imbrium was taken by Dave Scott, one of the Apollo 15 astronauts. Note the smooth contours of the lunar mountains, which have not been sculpted by water or ice. (credit: NASA/Apollo Lunar Surface Journal) The maria are much less cratered than the highlands, and cover just 17% of the lunar surface, mostly on the side of the Moon that faces Earth (Figure 9.9). Figure 9.9. Lunar Maria. About 17% of the Moon\u2019s surface consists of the maria\u2014flat plains of basaltic lava. This view of Mare Imbrium also shows numerous secondary craters and evidence of material ejected from the large crater Copernicus on the upper horizon. Copernicus is an impact crater almost 100 kilometers in diameter that was formed long after the lava in Imbrium had already been deposited. (credit: NASA, Apollo 17) Today, we know that the maria consist mostly of dark-colored basalt (volcanic lava) laid down in volcanic eruptions billions of years ago. Eventually, these lava", " flows partly filled the huge depressions called impact basins, which had been produced by collisions of large chunks of material with the Moon relatively early in its history. The basalt on the Moon (Figure 9.10) is very similar in composition to the crust under the oceans of This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 313 Earth or to the lavas erupted by many terrestrial volcanoes. The youngest of the lunar impact basins is Mare Orientale, shown in Figure 9.11. Figure 9.10. Rock from a Lunar Mare. In this sample of basalt from the mare surface, you can see the holes left by gas bubbles, which are characteristic of rock formed from lava. All lunar rocks are chemically distinct from terrestrial rocks, a fact that has allowed scientists to identify a few lunar samples among the thousands of meteorites that reach Earth. (credit: modification of work by NASA) Figure 9.11. Mare Orientale. The youngest of the large lunar impact basins is Orientale, formed 3.8 billion years ago. Its outer ring is about 1000 kilometers in diameter, roughly the distance between New York City and Detroit, Michigan. Unlike most of the other basins, Orientale has not been completely filled in with lava flows, so it retains its striking \u201cbull\u2019s-eye\u201d appearance. It is located on the edge of the Moon as seen from Earth. (credit: NASA) Volcanic activity may have begun very early in the Moon\u2019s history, although most evidence of the first half billion years is lost. What we do know is that the major mare volcanism, which involved the release of lava from hundreds of kilometers below the surface, ended about 3.3 billion years ago. After that, the Moon\u2019s interior cooled, and volcanic activity was limited to a very few small areas. The primary forces altering the surface come from the outside, not the interior. 314 Chapter 9 Cratered Worlds On the Lunar Surface \u201cThe surface is fine and powdery. I can pick it up loosely with my toe. But I can see the footprints of my boots and the treads in the fine sandy particles.\u201d \u2014Neil Armstrong, Apollo 11 astronaut, immediately after stepping onto the Moon for the first time. The surface of the Moon is buried under a fine-grained soil of tiny, shattered rock fragments. The dark basaltic dust of the lunar maria was kicked up by every astronaut footstep, and thus eventually worked its way into all of the astronauts\u2019 equipment. The upper layers of the surface are porous, consisting of loosely packed dust into which their boots sank several centimeters (Figure 9.12). This lunar dust, like so much else on the Moon, is the product of impacts. Each cratering event, large or small, breaks up the rock of the lunar surface and scatters the fragments. Ultimately, billions of years of impacts have reduced much of the surface layer to particles about the size of dust or sand. Figure 9.12. Footprint on Moon Dust. Apollo photo of an astronaut\u2019s boot print in the lunar soil. (credit: NASA) In the absence of any air, the lunar surface experiences much greater temperature extremes than the surface of Earth, even though Earth is virtually the same distance from the Sun. Near local noon, when the Sun is highest in the sky, the temperature of the dark lunar soil rises above the boiling point of water. During the long lunar night (which, like the lunar day, lasts two Earth weeks[1]), the temperature drops to about 100 K (\u2013173 \u00b0C). The extreme cooling is a result not only of the absence of air but also of the porous nature of the Moon\u2019s dusty soil, which cools more rapidly than solid rock would Learn how the moon\u2019s craters and maria were formed by watching a video produced by NASA\u2019s Lunar Reconnaissance Orbiter (LRO) team (https://openstax.org/l/30mooncratersfo) about the evolution of the Moon, tracing it from its origin about 4.5 billion years ago to the Moon we see today. See a simulation of how the Moon\u2019s craters and maria were formed through periods of impact, volcanic activity, and heavy bombardment. 1 You can see the cycle of day and night on the side of the Moon facing us in the form of the Moon\u2019s phases. It takes about 14 days for the side of the Moon facing us to go from full moon (all lit up) to new moon (all dark). There is more on this in Chapter 4: Earth, Moon, and Sky. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 315 9.3 IMPACT CRATERS Learning Objectives By the end", " of this section, you will be able to: Compare and contrast ideas about how lunar craters form Explain the process of impact crater formation Discuss the use of crater counts to determine relative ages of lunar landforms The Moon provides an important benchmark for understanding the history of our planetary system. Most solid worlds show the effects of impacts, often extending back to the era when a great deal of debris from our system\u2019s formation process was still present. On Earth, this long history has been erased by our active geology. On the Moon, in contrast, most of the impact history is preserved. If we can understand what has happened on the Moon, we may be able to apply this knowledge to other worlds. The Moon is especially interesting because it is not just any moon, but our Moon\u2014a nearby world that has shared the history of Earth for more than 4 billion years and preserved a record that, for Earth, has been destroyed by our active geology. Volcanic Versus Impact Origin of Craters Until the middle of the twentieth century, scientists did not generally recognize that lunar craters were the result of impacts. Since impact craters are extremely rare on Earth, geologists did not expect them to be the major feature of lunar geology. They reasoned (perhaps unconsciously) that since the craters we have on Earth are volcanic, the lunar craters must have a similar origin. One of the first geologists to propose that lunar craters were the result of impacts was Grove K. Gilbert, a scientist with the US Geological Survey in the 1890s. He pointed out that the large lunar craters\u2014mountain- rimmed, circular features with floors generally below the level of the surrounding plains\u2014are larger and have different shapes from known volcanic craters on Earth. Terrestrial volcanic craters are smaller and deeper and almost always occur at the tops of volcanic mountains (Figure 9.13). The only alternative to explain the Moon\u2019s craters was an impact origin. His careful reasoning, although not accepted at the time, laid the foundations for the modern science of lunar geology. Figure 9.13. Volcanic and Impact Craters. Profiles of a typical terrestrial volcanic crater and a typical lunar impact crater are quite different. Gilbert concluded that the lunar craters were produced by impacts, but he didn\u2019t understand why all of them were circular and not oval. The reason lies in the escape velocity, the minimum speed that a body must reach to permanently break away from the gravity of another body; it is also the minimum speed that a projectile approaching Earth or the Moon will hit with. Attracted by the gravity of the larger body, the incoming chunk strikes with at least escape velocity, which is 11 kilometers per second for Earth and 2.4 kilometers per second (5400 miles per hour) for the Moon. To this escape velocity is added whatever speed the projectile already had with respect to Earth or Moon, typically 10 kilometers per second or more. At these speeds, the energy of impact produces a violent explosion that excavates a large volume of material in a symmetrical way. Photographs of bomb and shell craters on Earth confirm that explosion craters are always 316 Chapter 9 Cratered Worlds essentially circular. Only following World War I did scientists recognize the similarity between impact craters and explosion craters, but, sadly, Gilbert did not live to see his impact hypothesis widely accepted. The Cratering Process Let\u2019s consider how an impact at these high speeds produces a crater. When such a fast projectile strikes a planet, it penetrates two or three times its own diameter before stopping. During these few seconds, its energy of motion is transferred into a shock wave (which spreads through the target body) and into heat (which vaporizes most of the projectile and some of the surrounding target). The shock wave fractures the rock of the target, while the expanding silicate vapor generates an explosion similar to that of a nuclear bomb detonated at ground level (Figure 9.14). The size of the excavated crater depends primarily on the speed of impact, but generally it is 10 to 15 times the diameter of the projectile. Figure 9.14. Stages in the Formation of an Impact Crater. (a) The impact occurs. (b) The projectile vaporizes and a shock wave spreads through the lunar rock. (c) Ejecta are thrown out of the crater. (d) Most of the ejected material falls back to fill the crater, forming an ejecta blanket. An impact explosion of the sort described above leads to a characteristic kind of crater, as shown in Figure 9.15. The central cavity is initially bowl-shaped (the word \u201ccrater\u201d comes from the Greek word for \u201cbowl\u201d), but the rebound of the crust partially fills it in, producing a flat floor and sometimes creating a central peak. Around the rim, landslides create a series of terraces. Figure 9.15. Typical Impact Crater. King Crater on the far side of the Moon, a fairly recent lunar", " crater 75 kilometers in diameter, shows most of the features associated with large impact structures. (credit: NASA/JSC/Arizona State University) The rim of the crater is turned up by the force of the explosion, so it rises above both the floor and the adjacent terrain. Surrounding the rim is an ejecta blanket consisting of material thrown out by the explosion. This debris falls back to create a rough, hilly region, typically about as wide as the crater diameter. Additional, higher- This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 317 speed ejecta fall at greater distances from the crater, often digging small secondary craters where they strike the surface (Figure 9.9). Some of these streams of ejecta can extend for hundreds or even thousands of kilometers from the crater, creating the bright crater rays that are prominent in lunar photos taken near full phase. The brightest lunar crater rays are associated with large young craters such as Kepler and Tycho Observing the Moon The Moon is one of the most beautiful sights in the sky, and it is the only object close enough to reveal its topography (surface features such as mountains and valleys) without a visit from a spacecraft. A fairly small amateur telescope easily shows craters and mountains on the Moon as small as a few kilometers across. Even as seen through a good pair of binoculars, we can observe that the appearance of the Moon\u2019s surface changes dramatically with its phase. At full phase, it shows almost no topographic detail, and you must look closely to see more than a few craters. This is because sunlight illuminates the surface straight on, and in this flat lighting, no shadows are cast. Much more revealing is the view near first or third quarter, when sunlight streams in from the side, causing topographic features to cast sharp shadows. It is almost always more rewarding to study a planetary surface under such oblique lighting, when the maximum information about surface relief can be obtained. The flat lighting at full phase does, however, accentuate brightness contrasts on the Moon, such as those between the maria and highlands. Notice in Figure 9.16 that several of the large mare craters seem to be surrounded by white material and that the light streaks or rays that can stretch for hundreds of kilometers across the surface are clearly visible. These lighter features are ejecta, splashed out from the crater-forming impact. 318 Chapter 9 Cratered Worlds Figure 9.16. Appearance of the Moon at Different Phases. (a) Illumination from the side brings craters and other topographic features into sharp relief, as seen on the far left side. (b) At full phase, there are no shadows, and it is more difficult to see such features. However, the flat lighting at full phase brings out some surface features, such as the bright rays of ejecta that stretch out from a few large young craters. (credit: modification of work by Luc Viatour) By the way, there is no danger in looking at the Moon with binoculars or telescopes. The reflected sunlight is never bright enough to harm your eyes. In fact, the sunlit surface of the Moon has about the same brightness as a sunlit landscape of dark rock on Earth. Although the Moon looks bright in the night sky, its surface is, on average, much less reflective than Earth\u2019s, with its atmosphere and white clouds. This difference is nicely illustrated by the photo of the Moon passing in front of Earth taken from the Deep Space Climate Observatory spacecraft (Figure 9.17). Since the spacecraft took the image from a position inside the orbit of Earth, we see both objects fully illuminated (full Moon and full Earth). By the way, you cannot see much detail on the Moon because the exposure has been set to give a bright image of Earth, not the Moon. Figure 9.17. The Moon Crossing the Face of Earth. In this 2015 image from the Deep Space Climate Observatory spacecraft, both objects are fully illuminated, but the Moon looks darker because it has a much lower average reflectivity than Earth. (credit: modification of work by NASA, DSCOVR EPIC team) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 319 One interesting thing about the Moon that you can see without binoculars or telescopes is popularly called \u201cthe new Moon in the old Moon\u2019s arms.\u201d Look at the Moon when it is a thin crescent, and you can often make out the faint circle of the entire lunar disk, even though the sunlight shines on only the crescent. The rest of the disk is illuminated not by sunlight but by earthlight\u2014sunlight reflected from Earth. The light of the full Earth on the Moon is about 50 times brighter than that of the full Moon shining on Earth. Using Crater", " Counts If a world has had little erosion or internal activity, like the Moon during the past 3 billion years, it is possible to use the number of impact craters on its surface to estimate the age of that surface. By \u201cage\u201d here we mean the time since a major disturbance occurred on that surface (such as the volcanic eruptions that produced the lunar maria). We cannot directly measure the rate at which craters are being formed on Earth and the Moon, since the average interval between large crater-forming impacts is longer than the entire span of human history. Our best-known example of such a large crater, Meteor Crater in Arizona (Figure 9.18), is about 50,000 years old. However, the cratering rate can be estimated from the number of craters on the lunar maria or calculated from the number of potential \u201cprojectiles\u201d (asteroids and comets) present in the solar system today. Both lines of reasoning lead to about the same estimations. Figure 9.18. Meteor Crater. This aerial photo of Meteor Crater in Arizona shows the simple form of a meteorite impact crater. The crater\u2019s rim diameter is about 1.2 kilometers. (credit: Shane Torgerson) For the Moon, these calculations indicate that a crater 1 kilometer in diameter should be produced about every 200,000 years, a 10-kilometer crater every few million years, and one or two 100-kilometer craters every billion years. If the cratering rate has stayed the same, we can figure out how long it must have taken to make all the craters we see in the lunar maria. Our calculations show that it would have taken several billion years. This result is similar to the age determined for the maria from radioactive dating of returned samples\u20143.3 to 3.8 billion years old. The fact that these two calculations agree suggests that astronomers\u2019 original assumption was right: comets and asteroids in approximately their current numbers have been impacting planetary surfaces for billions of years. Calculations carried out for other planets (and their moons) indicate that they also have been subject to about the same number of interplanetary impacts during this time. 320 Chapter 9 Cratered Worlds We have good reason to believe, however, that earlier than 3.8 billion years ago, the impact rates must have been a great deal higher. This becomes immediately evident when comparing the numbers of craters on the lunar highlands with those on the maria. Typically, there are 10 times more craters on the highlands than on a similar area of maria. Yet the radioactive dating of highland samples showed that they are only a little older than the maria, typically 4.2 billion years rather than 3.8 billion years. If the rate of impacts had been constant throughout the Moon\u2019s history, the highlands would have had to be at least 10 times older. They would thus have had to form 38 billion years ago\u2014long before the universe itself began. In science, when an assumption leads to an implausible conclusion, we must go back and re-examine that assumption\u2014in this case, the constant impact rate. The contradiction is resolved if the impact rate varied over time, with a much heavier bombardment earlier than 3.8 billion years ago (Figure 9.19). This \u201cheavy bombardment\u201d produced most of the craters we see today in the highlands. Figure 9.19. Cratering Rates over Time. The number of craters being made on the Moon\u2019s surface has varied with time over the past 4.3 billion years. This idea we have been exploring\u2014that large impacts (especially during the early history of the solar system) played a major role in shaping the worlds we see\u2014is not unique to our study of the Moon. As you read through the other chapters about the planets, you will see further indications that a number of the present-day characteristics of our system may be due to its violent past. 9.4 THE ORIGIN OF THE MOON Learning Objectives By the end of this section, you will be able to: Describe the top three early hypotheses of the formation of the Moon Summarize the current \u201cgiant impact\u201d concept of how the Moon formed It is characteristic of modern science to ask how things originated. Understanding the origin of the Moon has proven to be challenging for planetary scientists, however. Part of the difficulty is simply that we know so much This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 321 about the Moon (quite the opposite of our usual problem in astronomy). As we will see, one key problem is that the Moon is both tantalizingly similar to Earth and frustratingly different. Ideas for the Origin of the Moon Most of the earlier hypotheses for the Moon\u2019s origin followed one of three general ideas: 1. The fission theory\u2014the Moon", " was once part of Earth, but somehow separated from it early in their history. 2. The sister theory\u2014the Moon formed together with (but independent of) Earth, as we believe many moons of the outer planets formed. 3. The capture theory\u2014the Moon formed elsewhere in the solar system and was captured by Earth. Unfortunately, there seem to be fundamental problems with each of these ideas. Perhaps the easiest hypothesis to reject is the capture theory. Its primary drawback is that no one knows of any way that early Earth could have captured such a large moon from elsewhere. One body approaching another cannot go into orbit around it without a substantial loss of energy; this is the reason that spacecraft destined to orbit other planets are equipped with retro-rockets. Furthermore, if such a capture did take place, the captured object would go into a very eccentric orbit rather than the nearly circular orbit our Moon occupies today. Finally, there are too many compositional similarities between Earth and the Moon, particularly an identical fraction of the major isotopes[2] of oxygen, to justify seeking a completely independent origin. The fission hypothesis, which states that the Moon separated from Earth, was suggested in the late nineteenth century. Modern calculations have shown that this sort of spontaneous fission or splitting is impossible. Furthermore, it is difficult to understand how a Moon made out of terrestrial material in this way could have developed the many distinctive chemical differences now known to characterize our neighbor. Scientists were therefore left with the sister hypothesis\u2014that the Moon formed alongside Earth\u2014or with some modification of the fission hypothesis that can find a more acceptable way for the lunar material to have separated from Earth. But the more we learned about our Moon, the less these old ideas seem to fit the bill. The Giant Impact Hypothesis In an effort to resolve these apparent contradictions, scientists developed a fourth hypothesis for the origin of the Moon, one that involves a giant impact early in Earth\u2019s history. There is increasing evidence that large chunks of material\u2014objects of essentially planetary mass\u2014were orbiting in the inner solar system at the time that the terrestrial planets formed. The giant impact hypothesis envisions Earth being struck obliquely by an object approximately one-tenth Earth\u2019s mass\u2014a \u201cbullet\u201d about the size of Mars. This is very nearly the largest impact Earth could experience without being shattered. Such an impact would disrupt much of Earth and eject a vast amount of material into space, releasing almost enough energy to break the planet apart. Computer simulations indicate that material totaling several percent of Earth\u2019s mass could be ejected in such an impact. Most of this material would be from the stony mantles of Earth and the impacting body, not from their metal cores. This ejected rock vapor then cooled and formed a ring of material orbiting Earth. It was this ring that ultimately condensed into the Moon. While we do not have any current way of showing that the giant impact hypothesis is the correct model of the Moon\u2019s origin, it does offer potential solutions to most of the major problems raised by the chemistry of the Moon. First, since the Moon\u2019s raw material is derived from the mantles of Earth and the projectile, the absence of metals is easily understood. Second, most of the volatile elements would have been lost during the high- 2 Remember from the Radiation and Spectra chapter that the term isotope means a different \u201cversion\u201d of an element. Specifically, different isotopes of the same element have equal numbers of protons but different numbers of neutrons (as in carbon-12 versus carbon-14.) 322 Chapter 9 Cratered Worlds temperature phase following the impact, explaining the lack of these materials on the Moon. Yet, by making the Moon primarily of terrestrial mantle material, it is also possible to understand similarities such as identical abundances of various oxygen isotopes. 9.5 MERCURY Learning Objectives By the end of this section, you will be able to: Characterize the orbit of Mercury around the Sun Describe Mercury\u2019s structure and composition Explain the relationship between Mercury\u2019s orbit and rotation Describe the topography and features of Mercury\u2019s surface Summarize our ideas about the origin and evolution of Mercury The planet Mercury is similar to the Moon in many ways. Like the Moon, it has no atmosphere, and its surface is heavily cratered. As described later in this chapter, it also shares with the Moon the likelihood of a violent birth. Mercury\u2019s Orbit Mercury is the nearest planet to the Sun, and, in accordance with Kepler\u2019s third law, it has the shortest period of revolution about the Sun (88 of our days) and the highest average orbital speed (48 kilometers per second). It is appropriately named for the fleet-footed messenger god of the Romans. Because Mercury remains close to the Sun, it can be difficult to pick out in the sky. As you might expect, it\u2019s best seen when its eccentric orbit takes it as far from the Sun as possible. The semim", "ajor axis of Mercury\u2019s orbit\u2014that is, the planet\u2019s average distance from the Sun\u2014is 58 million kilometers, or 0.39 AU. However, because its orbit has the high eccentricity of 0.206, Mercury\u2019s actual distance from the Sun varies from 46 million kilometers at perihelion to 70 million kilometers at aphelion (the ideas and terms that describe orbits were introduced in Orbits and Gravity). Composition and Structure Mercury\u2019s mass is one-eighth that of Earth, making it the smallest terrestrial planet. Mercury is the smallest planet (except for the dwarf planets), having a diameter of 4878 kilometers, less than half that of Earth. Mercury\u2019s density is 5.4 g/cm3, much greater than the density of the Moon, indicating that the composition of those two objects differs substantially. Mercury\u2019s composition is one of the most interesting things about it and makes it unique among the planets. Mercury\u2019s high density tells us that it must be composed largely of heavier materials such as metals. The most likely models for Mercury\u2019s interior suggest a metallic iron-nickel core amounting to 60% of the total mass, with the rest of the planet made up primarily of silicates. The core has a diameter of 3500 kilometers and extends out to within 700 kilometers of the surface. We could think of Mercury as a metal ball the size of the Moon surrounded by a rocky crust 700 kilometers thick (Figure 9.20). Unlike the Moon, Mercury does have a weak magnetic field. The existence of this field is consistent with the presence of a large metal core, and it suggests that at least part of the core must be liquid in order to generate the observed magnetic field.[3] 3 Recall from the Radiation and Spectra chapter that magnetism is an effect of moving electric charges. In atoms of metals, the outer electrons are easier to dislodge and they can form a current when the metal is in liquid form and can flow. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 323 Figure 9.20. Mercury\u2019s Internal Structure. The interior of Mercury is dominated by a metallic core about the same size as our Moon Densities of Worlds The average density of a body equals its mass divided by its volume. For a sphere, density is: density = mass 3\u03c0R3 4 Astronomers can measure both mass and radius accurately when a spacecraft flies by a body. Using the information in this chapter, we can calculate the approximate average density of the Moon. Solution For a sphere, density = mass 3\u03c0R3 = 4 7.35 \u00d7 1022 kg 4.2 \u00d7 5.2 \u00d7 1018 m3 = 3.4 \u00d7 103 kg/m3 Table 9.1 gives a value of 3.3 g/cm3, which is 3.3 \u00d7 103 kg/m3. Check Your Learning Using the information in this chapter, calculate the average density of Mercury. Show your work. Does your calculation agree with the figure we give in this chapter? Answer: density = mass 3\u03c0R3 = 4 3.3 \u00d7 1023 kg 4.2 \u00d7 1.45 \u00d7 1019 m3 = 5.4 \u00d7 103 kg/m3 That matches the value given in Table 9.1 when g/cm3 is converted into kg/m3. 324 Chapter 9 Cratered Worlds Mercury\u2019s Strange Rotation Visual studies of Mercury\u2019s indistinct surface markings were once thought to indicate that the planet kept one face to the Sun (as the Moon does to Earth). Thus, for many years, it was widely believed that Mercury\u2019s rotation period was equal to its revolution period of 88 days, making one side perpetually hot while the other was always cold. Radar observations of Mercury in the mid-1960s, however, showed conclusively that Mercury does not keep one side fixed toward the Sun. If a planet is turning, one side seems to be approaching Earth while the other is moving away from it. The resulting Doppler shift spreads or broadens the precise transmitted radar-wave frequency into a range of frequencies in the reflected signal (Figure 9.21). The degree of broadening provides an exact measurement of the rotation rate of the planet. Figure 9.21. Doppler Radar Measures Rotation. When a radar beam is reflected from a rotating planet, the motion of one side of the planet\u2019s disk toward us and the other side away from us causes Doppler shifts in the reflected signal. The effect is to cause both a redshift and a blueshift, widening the spread of frequencies in the radio beam. Mercury\u2019s period of rotation (how long it takes to turn with respect to the distant stars) is 59 days, which is just two-thirds of the planet\u2019s period of revolution. Subsequently, astronomers found that a situation where", " the spin and the orbit of a planet (its year) are in a 2:3 ratio turns out to be stable. (See What a Difference a Day Makes for more on the effects of having such a long day on Mercury.) Mercury, being close to the Sun, is very hot on its daylight side; but because it has no appreciable atmosphere, it gets surprisingly cold during the long nights. The temperature on the surface climbs to 700 K (430 \u00b0C) at noontime. After sunset, however, the temperature drops, reaching 100 K (\u2013170 \u00b0C) just before dawn. (It is even colder in craters near the poles that receive no sunlight at all.) The range in temperature on Mercury is thus 600 K (or 600 \u00b0C), a greater difference than on any other planet. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 325 What a Difference a Day Makes Mercury rotates three times for each two orbits around the Sun. It is the only planet that exhibits this relationship between its spin and its orbit, and there are some interesting consequences for any observers who might someday be stationed on the surface of Mercury. Here on Earth, we take for granted that days are much shorter than years. Therefore, the two astronomical ways of defining the local \u201cday\u201d\u2014how long the planet takes to rotate and how long the Sun takes to return to the same position in the sky\u2014are the same on Earth for most practical purposes. But this is not the case on Mercury. While Mercury rotates (spins once) in 59 Earth days, the time for the Sun to return to the same place in Mercury\u2019s sky turns out to be two Mercury years, or 176 Earth days. (Note that this result is not intuitively obvious, so don\u2019t be upset if you didn\u2019t come up with it.) Thus, if one day at noon a Mercury explorer suggests to her companion that they should meet at noon the next day, this could mean a very long time apart! To make things even more interesting, recall that Mercury has an eccentric orbit, meaning that its distance from the Sun varies significantly during each mercurian year. By Kepler\u2019s law, the planet moves fastest in its orbit when closest to the Sun. Let\u2019s examine how this affects the way we would see the Sun in the sky during one 176-Earth-day cycle. We\u2019ll look at the situation as if we were standing on the surface of Mercury in the center of a giant basin that astronomers call Caloris (Figure 9.23). At the location of Caloris, Mercury is most distant from the Sun at sunrise; this means the rising Sun looks smaller in the sky (although still more than twice the size it appears from Earth). As the Sun rises higher and higher, it looks bigger and bigger; Mercury is now getting closer to the Sun in its eccentric orbit. At the same time, the apparent motion of the Sun slows down as Mercury\u2019s faster motion in orbit begins to catch up with its rotation. At noon, the Sun is now three times larger than it looks from Earth and hangs almost motionless in the sky. As the afternoon wears on, the Sun appears smaller and smaller, and moves faster and faster in the sky. At sunset, a full Mercury year (or 88 Earth days after sunrise), the Sun is back to its smallest apparent size as it dips out of sight. Then it takes another Mercury year before the Sun rises again. (By the way, sunrises and sunsets are much more sudden on Mercury, since there is no atmosphere to bend or scatter the rays of sunlight.) Astronomers call locations like the Caloris Basin the \u201chot longitudes\u201d on Mercury because the Sun is closest to the planet at noon, just when it is lingering overhead for many Earth days. This makes these areas the hottest places on Mercury. We bring all this up not because the exact details of this scenario are so important but to illustrate how many of the things we take for granted on Earth are not the same on other worlds. As we\u2019ve mentioned before, one of the best things about taking an astronomy class should be ridding you forever of any \u201cEarth chauvinism\u201d you might have. The way things are on our planet is just one of the many ways nature can arrange reality. 326 Chapter 9 Cratered Worlds The Surface of Mercury The first close-up look at Mercury came in 1974, when the US spacecraft Mariner 10 passed 9500 kilometers from the surface of the planet and transmitted more than 2000 photographs to Earth, revealing details with a resolution down to 150 meters. Subsequently, the planet was mapped in great detail by the MESSENGER spacecraft, which was launched in 2004 and made multiple flybys of Earth, Venus, and Mercury before settling into orbit around Mercury in 2011. It ended its life in 2015, when it was", " commanded to crash into the surface of the planet. Mercury\u2019s surface strongly resembles the Moon in appearance (Figure 9.22 and Figure 9.23). It is covered with thousands of craters and larger basins up to 1300 kilometers in diameter. Some of the brighter craters are rayed, like Tycho and Copernicus on the Moon, and many have central peaks. There are also scarps (cliffs) more than a kilometer high and hundreds of kilometers long, as well as ridges and plains. MESSENGER instruments measured the surface composition and mapped past volcanic activity. One of its most important discoveries was the verification of water ice (first detected by radar) in craters near the poles, similar to the situation on the Moon, and the unexpected discovery of organic (carbon-rich) compounds mixed with the water ice Scientists working with data from the MESSENGER mission (https://openstax.org/l/30MESSmercuryrt) put together a rotating globe of Mercury, in false color, showing some of the variations in the composition of the planet\u2019s surface. You can watch it spin. Figure 9.22. Mercury\u2019s Topography. The topography of Mercury\u2019s northern hemisphere is mapped in great detail from MESSENGER data. The lowest regions are shown in purple and blue, and the highest regions are shown in red. The difference in elevation between the lowest and highest regions shown here is roughly 10 kilometers. The permanently shadowed low-lying craters near the north pole contain radar-bright water ice. (credit: modification of work by NASA/Johns Hopkins University Applied Physics Laboratory/Carnegie Institution of Washington) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 327 Figure 9.23. Caloris Basin. This partially flooded impact basin is the largest known structural feature on Mercury. The smooth plains in the interior of the basin have an area of almost two million square kilometers. Compare this photo with Figure 9.11, the Orientale Basin on the Moon. (credit: NASA/Johns Hopkins University Applied Physics Laboratory/Carnegie Institution of Washington) Most of the mercurian features have been named in honor of artists, writers, composers, and other contributors to the arts and humanities, in contrast with the scientists commemorated on the Moon. Among the named craters are Bach, Shakespeare, Tolstoy, Van Gogh, and Scott Joplin. There is no evidence of plate tectonics on Mercury. However, the planet\u2019s distinctive long scarps can sometimes be seen cutting across craters; this means the scarps must have formed later than the craters (Figure 9.24). These long, curved cliffs appear to have their origin in the slight compression of Mercury\u2019s crust. Apparently, at some point in its history, the planet shrank, wrinkling the crust, and it must have done so after most of the craters on its surface had already formed. If the standard cratering chronology applies to Mercury, this shrinkage must have taken place during the last 4 billion years and not during the solar system\u2019s early period of heavy bombardment. Figure 9.24. Discovery Scarp on Mercury. This long cliff, nearly 1 kilometer high and more than 100 kilometers long, cuts across several craters. Astronomers conclude that the compression that made \u201cwrinkles\u201d like this in the plank\u2019s surface must have taken place after the craters were formed. (credit: modification of work by NASA/JPL/Northwestern University) 328 Chapter 9 Cratered Worlds The Origin of Mercury The problem with understanding how Mercury formed is the reverse of the problem posed by the composition of the Moon. We have seen that, unlike the Moon, Mercury is composed mostly of metal. However, astronomers think that Mercury should have formed with roughly the same ratio of metal to silicate as that found on Earth or Venus. How did it lose so much of its rocky material? The most probable explanation for Mercury\u2019s silicate loss may be similar to the explanation for the Moon\u2019s lack of a metal core. Mercury is likely to have experienced several giant impacts very early in its youth, and one or more of these may have torn away a fraction of its mantle and crust, leaving a body dominated by its iron core You can follow some of NASA\u2019s latest research on Mercury (https://openstax.org/l/ 30NASAresmercu) and see some helpful animations on the MESSENGER web page. Today, astronomers recognize that the early solar system was a chaotic place, with the final stages of planet formation characterized by impacts of great violence. Some objects of planetary mass have been destroyed, whereas others could have fragmented and then re-formed, perhaps more than once. Both the Moon and Mercury, with their strange compositions, bear testimony to", " the catastrophes that must have characterized the solar system during its youth. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 329 CHAPTER 9 REVIEW KEY TERMS highlands the lighter, heavily cratered regions of the Moon, which are generally several kilometers higher than the maria mare (plural: maria) Latin for \u201csea;\u201d the name applied to the dark, relatively smooth features that cover 17% of the Moon\u2019s surface SUMMARY 9.1 General Properties of the Moon Most of what we know about the Moon derives from the Apollo program, including 400 kilograms of lunar samples still being intensively studied. The Moon has one-eightieth the mass of Earth and is severely depleted in both metals and volatile materials. It is made almost entirely of silicates like those in Earth\u2019s mantle and crust. However, more recent spacecraft have found evidence of a small amount of water near the lunar poles, most likely deposited by comet and asteroid impacts. 9.2 The Lunar Surface The Moon, like Earth, was formed about 4.5 billion year ago. The Moon\u2019s heavily cratered highlands are made of rocks more than 4 billion years old. The darker volcanic plains of the maria were erupted primarily between 3.3 and 3.8 billion years ago. Generally, the surface is dominated by impacts, including continuing small impacts that produce its fine-grained soil. 9.3 Impact Craters A century ago, Grove Gilbert suggested that the lunar craters were caused by impacts, but the cratering process was not well understood until more recently. High-speed impacts produce explosions and excavate craters 10 to 15 times the size of the impactor with raised rims, ejecta blankets, and often central peaks. Cratering rates have been roughly constant for the past 3 billion years but earlier were much greater. Crater counts can be used to derive approximate ages for geological features on the Moon and other worlds with solid surfaces. 9.4 The Origin of the Moon The three standard hypotheses for the origin of the Moon were the fission hypothesis, the sister hypothesis, and the capture hypothesis. All have problems, and they have been supplanted by the giant impact hypothesis, which ascribes the origin of the Moon to the impact of a Mars-sized projectile with Earth 4.5 billion years ago. The debris from the impact made a ring around Earth which condensed and formed the Moon. 9.5 Mercury Mercury is the nearest planet to the Sun and the fastest moving. Mercury is similar to the Moon in having a heavily cratered surface and no atmosphere, but it differs in having a very large metal core. Early in its evolution, it apparently lost part of its silicate mantle, probably due to one or more giant impacts. Long scarps on its surface testify to a global compression of Mercury\u2019s crust during the past 4 billion years. 330 Chapter 9 Cratered Worlds FOR FURTHER EXPLORATION Articles The Moon Bakich, Michael. \u201cAsia\u2019s New Assault on the Moon.\u201d Astronomy (August 2009): 50. The Japanese Selene and Chinese Chang\u2019e 1 missions. Beatty, J. \u201cNASA Slams the Moon.\u201d Sky & Telescope (February 2010): 28. The impact of the LCROSS mission on the Moon and what we learned from it. Bell, T. \u201cWarning: Dust Ahead.\u201d Astronomy (March 2006): 46. What we know about lunar dust and the problems it can cause. Dorminey, B. \u201cSecrets beneath the Moon\u2019s Surface.\u201d Astronomy (March 2011): 24. A nice timeline of the Moon\u2019s evolution and the story of how we are finding out more about its internal structure. Jayawardhana, R. \u201cDeconstructing the Moon.\u201d Astronomy (September 1998): 40. An update on the giant impact hypothesis for forming the Moon. Register, B. \u201cThe Fate of the Moon Rocks.\u201d Astronomy (December 1985): 15. What was done with the rocks the astronauts brought back from the Moon. Schmitt, H. \u201cExploring Taurus\u2013Littrow: Apollo 17.\u201d National Geographic (September 1973). First-person account given by the only scientist to walk on the Moon. Schmitt, H. \u201cFrom the Moon to Mars.\u201d Scientific American (July 2009): 36. The only scientist to walk on the Moon reflects on the science from Apollo and future missions to Mars. Schultz, P. \u201cNew Clues to the Moon\u2019s Distant Past.\u201d Astronomy (December 2011): 34. Summary of results and ideas from the LCROSS and LRO missions. Shirao, M. \u201cKayuga\u2019s High Def Highlights.\u201d Sky & Telescope", " (February 2010): 20. Results from the Japanese mission to the Moon, with high definition TV cameras. Wadhwa, M. \u201cWhat Are We Learning from the Moon Rocks?\u201d Astronomy (June 2013): 54. Very nice discussion of how the rocks tell us about Moon\u2019s composition, age, and origin. Wood, Charles. \u201cThe Moon\u2019s Far Side: Nearly a New World.\u201d Sky & Telescope (January 2007): 48. This article compares what we know about the two sides and why they are different. Zimmerman, R. \u201cHow Much Water is on the Moon?\u201d Astronomy (January 2014): 50. Results from the LRO\u2019s instruments and good overview of issue. Mercury Beatty, J. \u201cMercury Gets a Second Look.\u201d Sky & Telescope (March 2009): 26. The October 2008 MESSENGER mission flyby. Beatty, J. \u201cReunion with Mercury.\u201d Sky & Telescope (May 2008): 24. The January 2008 MESSENGER encounter with Mercury. \u201cMercury: Meet the Planet Nearest the Sun.\u201d Sky & Telescope (March 2014): 39. Four-page pictorial introduction, including the new MESSENGER probe full map of the planet provided. Oberg, J. \u201cTorrid Mercury\u2019s Icy Poles.\u201d Astronomy (December 2013): 30. A nice overview of results from MESSENGER mission, including the ice in polar craters. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 331 Sheehan, W., and Dobbins, T. \u201cMesmerized by Mercury.\u201d Sky & Telescope (June 2000): 109. History of Mercury observations and how amateur astronomers can contribute. Talcott, R. \u201cSurprises from MESSENGER\u2019s Historic Mercury Fly-by.\u201d Astronomy (March 2009): 28. Talcott, R. \u201cMercury Reveals its Hidden Side.\u201d Astronomy (May 2008): 26. Results and image from the MESSENGER mission flyby of January 2008. Websites The Moon Apollo Lunar Surface Journal: http://www.hq.nasa.gov/office/pao/History/alsj/. Information, interviews, maps, photos, video and audio clips, and much more on each of the Apollo landing missions. Lunar & Planetary Institute: http://www.lpi.usra.edu/lunar/missions/. Lunar Science and Exploration web pages. Lunar Reconnaissance Orbiter Mission Page: http://lro.gsfc.nasa.gov/. NASA\u2019s Guide to Moon Missions and Information: http://nssdc.gsfc.nasa.gov/planetary/planets/ moonpage.html. Origin of the Moon: http://www.psi.edu/projects/moon/moon.html. By William Hartmann, who, with a colleague, first suggested the giant impact hypothesis for how the Moon formed, in 1975. Sky & Telescope magazine\u2019s observing guides and articles about the Moon: http://www.skyandtelescope.com/ observing/celestial-objects-to-watch/moon/. To the Moon: http://www.pbs.org/wgbh/nova/tothemoon/. PBS program on the Apollo landings. We Choose the Moon: http://wechoosethemoon.org/. A recreation of the Apollo 11 mission. Mercury Mercury Unveiled by G. Jeffrey Taylor (summarizing the Mariner 10 Mission): http://www.psrd.hawaii.edu/ Jan97/MercuryUnveiled.html. MESSENGER Mission Website: http://messenger.jhuapl.edu/. NASA Planetary Data Center Mercury Page: http://nssdc.gsfc.nasa.gov/planetary/planets/mercurypage.html. Views of the Solar System Mercury Page: http://solarviews.com/eng/mercury.htm. COLLABORATIVE GROUP ACTIVITIES A. We mentioned that no nation on Earth now has the capability to send a human being to the Moon, even though the United States once sent 12 astronauts to land there. What does your group think about this? Should we continue the exploration of space with human beings? Should we put habitats on the Moon? Should we go to Mars? Does humanity have a \u201cdestiny in space?\u201d Whatever your answer to these questions, make a list of the arguments and facts that support your position. B. When they hear about the giant impact hypothesis for the origin of the Moon, many students are intrigued and wonder why we can\u2019t cite more evidence for it. In your group, make a", " list of reasons we cannot find any traces on Earth of the great impact that formed the Moon? 332 Chapter 9 Cratered Worlds C. We discussed that the ice (mixed into the soil) that is found on the Moon was most likely delivered by comets. Have your group make a list of all the reasons the Moon would not have any ice of its own left over from its early days. D. Can your group make a list of all the things that would be different if Earth had no Moon? Don\u2019t restrict your answer to astronomy and geology. Think about our calendars and moonlit romantic strolls, for example. (You may want to review Earth, Moon, and Sky.) E. If, one day, humanity decides to establish a colony on the Moon, where should we put it? Make a list of the advantages and disadvantages of locating such a human habitat on the near side, the far side, or at the poles. What site would be best for doing visible-light and radio astronomy from observatories on the Moon? F. A member of the class (but luckily, not a member of your group) suggests that he has always dreamed of building a vacation home on the planet Mercury. Can your group make a list of all reasons such a house would be hard to build and keep in good repair? G. As you\u2019ve read in this chapter, craters on the Moon are (mostly) named after scientists. (See the official list at: http://planetarynames.wr.usgs.gov/SearchResults?target=MOON&featureType=Crater,%20craters). The craters on Mercury, on the other hand, are named for writers, artists, composers, and others in the humanities. See the official list at: http://planetarynames.wr.usgs.gov/SearchResults? target=MERCURY&featureType=Crater,%20craters). Living persons are not eligible. Can each person in your group think of a scientist or someone in the arts whom they especially respect? Now check to see if they are listed. Are there scientists or people in the arts who should have their names on the Moon or Mercury and do not? H. Imagine that a distant relative, hearing you are taking an astronomy course, calls you up and tells you that NASA faked the Moon landings. His most significant argument is that all the photos of the Moon show black skies, but none of them have any stars showing. This proves that the photos were taken against a black backdrop in a studio and not on the Moon. Based on your reading in this chapter, what arguments can your group come up with to rebut this idea? EXERCISES Review Questions 1. What is the composition of the Moon, and how does it compare to the composition of Earth? Of Mercury? 2. Why does the Moon not have an atmosphere? 3. What are the principal features of the Moon observable with the unaided eye? 4. Frozen water exists on the lunar surface primarily in which location? Why? 5. Outline the main events in the Moon\u2019s geological history. 6. What are the maria composed of? Is this material found elsewhere in the solar system? 7. The mountains on the Moon were formed by what process? 8. With no wind or water erosion of rocks, what is the mechanism for the creation of the lunar \u201csoil?\u201d This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 9 Cratered Worlds 333 9. What differences did Grove K. Gilbert note between volcanic craters on Earth and lunar craters? 10. Explain how high-speed impacts form circular craters. How can this explanation account for the various characteristic features of impact craters? 11. Explain the evidence for a period of heavy bombardment on the Moon about 4 billion years ago. 12. How did our exploration of the Moon differ from that of Mercury (and the other planets)? 13. Summarize the four main hypotheses for the origin of the Moon. 14. What are the difficulties with the capture hypothesis of the Moon\u2019s origin? 15. What is the main consequence of Mercury\u2019s orbit being so highly eccentric? 16. Describe the basic internal structure of Mercury. 17. How was the rotation rate of Mercury determined? 18. What is the relationship between Mercury\u2019s rotational period and orbital period? 19. The features of Mercury are named in honor of famous people in which fields of endeavor? 20. What do our current ideas about the origins of the Moon and Mercury have in common? How do they differ? Thought Questions 21. One of the primary scientific objectives of the Apollo program was the return of lunar material. Why was this so important? What can be learned from samples? Are they still of value now? 22. Apollo astronaut David Scott dropped a hammer and a feather together on the Moon, and both", " reached the ground at the same time. What are the two distinct advantages that this experiment on the Moon had over the same kind of experiment as performed by Galileo on Earth? 23. Galileo thought the lunar maria might be seas of water. If you had no better telescope than the one he had, could you demonstrate that they are not composed of water? 24. Why did it take so long for geologists to recognize that the lunar craters had an impact origin rather than a volcanic one? 25. How might a crater made by the impact of a comet with the Moon differ from a crater made by the impact of an asteroid? 26. Why are the lunar mountains smoothly rounded rather than having sharp, pointed peaks (as they were almost always depicted in science-fiction illustrations and films before the first lunar landings)? 27. The lunar highlands have about ten times more craters in a given area than do the maria. Does this mean that the highlands are 10 times older? Explain your reasoning. 28. At the end of the section on the lunar surface, your authors say that lunar night and day each last about two Earth weeks. After looking over the information in Earth, Moon, and Sky and this chapter about the motions of the Moon, can you explain why? (It helps to draw a diagram for yourself.) 29. Give several reasons Mercury would be a particularly unpleasant place to build an astronomical observatory. 334 Chapter 9 Cratered Worlds 30. If, in the remote future, we establish a base on Mercury, keeping track of time will be a challenge. Discuss how to define a year on Mercury, and the two ways to define a day. Can you come up with ways that humans raised on Earth might deal with time cycles on Mercury? 31. The Moon has too little iron, Mercury too much. How can both of these anomalies be the result of giant impacts? Explain how the same process can yield such apparently contradictory results. Figuring For Yourself 32. In the future, astronomers discover a solid moon around a planet orbiting one of the nearest stars. This moon has a diameter of 1948 km and a mass of 1.6 \u00d7 1022 kg. What is its density? 33. The Moon was once closer to Earth than it is now. When it was at half its present distance, how long was its period of revolution? (See Orbits and Gravity for the formula to use.) 34. Astronomers believe that the deposit of lava in the giant mare basins did not happen in one flow but in many different eruptions spanning some time. Indeed, in any one mare, we find a variety of rock ages, typically spanning about 100 million years. The individual lava flows as seen in Hadley Rille by the Apollo 15 astronauts were about 4 m thick. Estimate the average time interval between the beginnings of successive lava flows if the total depth of the lava in the mare is 2 km. 35. The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler\u2019s third law, as modified by Newton, to calculate the mass of Earth relative to the Sun. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 335 10 EARTHLIKE PLANETS: VENUS AND MARS Figure 10.1. Spirit Rover on Mars. This May 2004 image shows the tracks made by the Mars Exploration Spirit rover on the surface of the red planet. Spirit was active on Mars between 2004 and 2010, twenty times longer than its planners had expected. It \u201cdrove\u201d over 7.73 kilometers in the process of examining the martian landscape. (credit: modification of work by NASA/JPL/Cornell) Chapter Outline 10.1 The Nearest Planets: An Overview 10.2 The Geology of Venus 10.3 The Massive Atmosphere of Venus 10.4 The Geology of Mars 10.5 Water and Life on Mars 10.6 Divergent Planetary Evolution Thinking Ahead The Moon and Mercury are geologically dead. In contrast, the larger terrestrial planets\u2014Earth, Venus, and Mars\u2014are more active and interesting worlds. We have already discussed Earth, and we now turn to Venus and Mars. These are the nearest planets and the most accessible to spacecraft. Not surprisingly, the greatest effort in planetary exploration has been devoted to these fascinating worlds. In the chapter, we discuss some of the results of more than four decades of scientific exploration of Mars and Venus. Mars is exceptionally interesting, with evidence that points to habitable conditions in the past. Even today, we are discovering things about Mars that make it the most likely place where humans might set up a habitat in the future. However, our robot explorers have clearly shown that neither Venus nor Mars has conditions similar to Earth. How did it happen that these three neighboring terrestrial planets have diverged so dramatically in", " their evolution? 10.1 THE NEAREST PLANETS: AN OVERVIEW Learning Objectives By the end of this section, you will be able to: 336 Chapter 10 Earthlike Planets: Venus and Mars Explain why it\u2019s difficult to learn about Venus from Earth-based observation alone Describe the history of our interest in Mars before the Space Age Compare the basic physical properties of Earth, Mars, and Venus, including their orbits As you might expect from close neighbors, Mars and Venus are among the brightest objects in the night sky. The average distance of Mars from the Sun is 227 million kilometers (1.52 AU), or about half again as far from the Sun as Earth. Venus\u2019 orbit is very nearly circular, at a distance of 108 million kilometers (0.72 AU) from the Sun. Like Mercury, Venus sometimes appears as an \u201cevening star\u201d and sometimes as a \u201cmorning star.\u201d Venus approaches Earth more closely than does any other planet: at its nearest, it is only 40 million kilometers from us. The closest Mars ever gets to Earth is about 56 million kilometers. Appearance Venus appears very bright, and even a small telescope reveals that it goes through phases like the Moon. Galileo discovered that Venus displays a full range of phases, and he used this as an argument to show that Venus must circle the Sun and not Earth. The planet\u2019s actual surface is not visible because it is shrouded by dense clouds that reflect about 70% of the sunlight that falls on them, frustrating efforts to study the underlying surface, even with cameras in orbit around the planet (Figure 10.2). Figure 10.2. Venus as Photographed by the Pioneer Venus Orbiter. This ultraviolet image shows an upper-atmosphere cloud structure that would be invisible at visible wavelengths. Note that there is not even a glimpse of the planet\u2019s surface. (credit: modification of work by NASA) In contrast, Mars is more tantalizing as seen through a telescope (Figure 10.3). The planet is distinctly red, due (as we now know) to the presence of iron oxides in its soil. This color may account for its association with war (and blood) in the legends of early cultures. The best resolution obtainable from telescopes on the ground is about 100 kilometers, or about the same as what we can see on the Moon with the unaided eye. At this resolution, no hint of topographic structure can be detected: no mountains, no valleys, not even impact craters. On the other hand, bright polar ice caps can be seen easily, together with dusky surface markings that sometimes change in outline and intensity from season to season. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 337 Figure 10.3. Mars as Seen from Earth\u2019s Surface. These are among the best Earth-based photos of Mars, taken in 1988 when the planet was exceptionally close to Earth. The polar caps and dark surface markings are evident, but not topographic features. (credit: modification of work by Steve Larson, Lunar and Planetary Laboratory, University of Arizona) For a few decades around the turn of the twentieth century, some astronomers believed that they saw evidence of an intelligent civilization on Mars. The controversy began in 1877, when Italian astronomer Giovanni Schiaparelli (1835\u20131910) announced that he could see long, faint, straight lines on Mars that he called canale, or channels. In English-speaking countries, the term was mistakenly translated as \u201ccanals,\u201d implying an artificial origin. Even before Schiaparelli\u2019s observations, astronomers had watched the bright polar caps change size with the seasons and had seen variations in the dark surface features. With a little imagination, it was not difficult to picture the canals as long fields of crops bordering irrigation ditches that brought water from the melting polar ice to the parched deserts of the red planet. (They assumed the polar caps were composed of water ice, which isn\u2019t exactly true, as we will see shortly.) Until has death in 1916, the most effective proponent of intelligent life on Mars was Percival Lowell, a self-made American astronomer and member of the wealthy Lowell family of Boston (see the feature box on Percival Lowell: Dreaming of an Inhabited Mars). A skilled author and speaker, Lowell made what seemed to the public to be a convincing case for intelligent Martians, who had constructed the huge canals to preserve their civilization in the face of a deteriorating climate (Figure 10.4). 338 Chapter 10 Earthlike Planets: Venus and Mars Figure 10.4. Lowell\u2019s Mars Globe. One of the remarkable globes of Mars prepared by Percival Lowell, showing a network of dozens of canals, oases, and triangular water reservoirs that he claimed were visible on the red", " planet. The argument for a race of intelligent Martians, however, hinged on the reality of the canals, a matter that remained in serious dispute among astronomers. The canal markings were always difficult to study, glimpsed only occasionally because atmospheric conditions caused the tiny image of Mars to shimmer in the telescope. Lowell saw canals everywhere (even a few on Venus), but many other observers could not see them at all and remained unconvinced of their existence. When telescopes larger than Lowell\u2019s failed to confirm the presence of canals, the skeptics felt vindicated. Now it is generally accepted that the straight lines were an optical illusion, the result of the human mind\u2019s tendency to see order in random features that are glimpsed dimly at the limits of the eye\u2019s resolution. When we see small, dim dots of surface markings, our minds tend to connect those dots into straight lines Percival Lowell: Dreaming of an Inhabited Mars Percival Lowell was born into the well-to-do Massachusetts family about whom John Bossidy made the famous toast: And this is good old Boston, The home of the bean and the cod, Where the Lowells talk to the Cabots And the Cabots talk only to God. Percival\u2019s brother Lawrence became president of Harvard University, and his sister, Amy, became a distinguished poet. Percival was already interested in astronomy as a boy: he made observations of Mars at age 13. His undergraduate thesis at Harvard dealt with the origin of the solar system, but he did not pursue this interest immediately. Instead, he entered the family business and traveled extensively in Asia. In 1892, however, he decided to dedicate himself to carrying on Schiaparelli\u2019s work and solving the mysteries of the martian canals. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 339 In 1894, with the help of astronomers at Harvard but using his own funds, Lowell built an observatory on a high plateau in Flagstaff, Arizona, where he hoped the seeing would be clear enough to show him Mars in unprecedented detail. He and his assistants quickly accumulated a tremendous number of drawings and maps, purporting to show a vast network of martian canals (see Figure 10.4). He elaborated his ideas about the inhabitants of the red planet in several books, including Mars (1895) and Mars and Its Canals (1906), and in hundreds of articles and speeches. As Lowell put it, A mind of no mean order would seem to have presided over the system we see\u2014a mind certainly of considerably more comprehensiveness than that which presides over the various departments of our own public works. Party politics, at all events, have had no part in them; for the system is planet-wide.... Certainly what we see hints at the existence of beings who are in advance of, not behind us, in the journey of life. Lowell\u2019s views captured the public imagination and inspired many novels and stories, the most famous of which was H. G. Wells\u2019 War of the Worlds (1897). In this famous \u201cinvasion\u201d novel, the thirsty inhabitants of a dying planet Mars (based entirely on Lowell\u2019s ideas) come to conquer Earth with advanced technology. Although the Lowell Observatory first became famous for its work on the martian canals, both Lowell and the observatory eventually turned to other projects as well. He became interested in the search for a ninth (and then undiscovered) planet in the solar system. In 1930, Pluto was found at the Lowell Observatory, and it is not a coincidence that the name selected for the new planet starts with Lowell\u2019s initials. It was also at the Lowell Observatory that the first measurements were made of the great speed at which galaxies are moving away from us, observations that would ultimately lead to our modern view of an expanding universe. Lowell (Figure 10.5) continued to live at his observatory, marrying at age 53 and publishing extensively. He relished the debate his claims about Mars caused far more than the astronomers on the other side, who often complained that Lowell\u2019s work was making planetary astronomy a less respectable field. At the same time, the public fascination with the planets fueled by Lowell\u2019s work (and its interpreters) may, several generations later, have helped fan support for the space program and the many missions whose results grace the pages of our text. 340 Chapter 10 Earthlike Planets: Venus and Mars Figure 10.5. Percival Lowell (1855\u20131916). This 1914 photograph shows Percival Lowell observing Venus with his 24-inch telescope at Flagstaff, Arizona In October 1938, the Mercury Theater of the Air on radio dramatized The War of the Worlds as a series of radio news reports. This broadcast (https://openstax.org", "/l/30WarofWorlds) scared many people into thinking that Lowell\u2019s Martians were really invading New Jersey, and caused something of a panic. You can listen to the original radio broadcast if you scroll down to \u201cWar of the Worlds.\u201d Rotation of the Planets Astronomers have determined the rotation period of Mars with great accuracy by watching the motion of permanent surface markings; its sidereal day is 24 hours 37 minutes 23 seconds, just a little longer than the rotation period of Earth. This high precision is not obtained by watching Mars for a single rotation, but by noting how many turns it makes over a long period of time. Good observations of Mars date back more than 200 years, a period during which tens of thousands of martian days have passed. As a result, the rotation period can be calculated to within a few hundredths of a second. The rotational axis of Mars has a tilt of about 25\u00b0, similar to the tilt of Earth\u2019s axis. Thus, Mars experiences seasons very much like those on Earth. Because of the longer martian year (almost two Earth years), however, each season there lasts about six of our months. The situation with Venus is different. Since no surface detail can be seen through Venus\u2019 clouds, its rotation period can be found only by bouncing radar signals off the planet (as explained for Mercury in the Cratered Worlds chapter). The first radar observations of Venus\u2019 rotation were made in the early 1960s. Later, topographical surface features were identified on the planet that showed up in the reflected radar signals. The rotation period of Venus, precisely determined from the motion of such \u201cradar features\u201d across its disk, is 243 days. Even more surprising than how long Venus takes to rotate is the fact that it spins in a backward or retrograde direction (east to west). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 341 Stop for a moment and think about how odd this slow rotation makes the calendar on Venus. The planet takes 225 Earth days to orbit the Sun and 243 Earth days to spin on its axis. So the day on Venus (as defined by its spinning once) is longer than the year! As a result, the time the Sun takes to return to the same place in Venus\u2019 sky\u2014another way we might define the meaning of a day\u2014turns out to be 117 Earth days. (If you say \u201cSee you tomorrow\u201d on Venus, you\u2019ll have a long time to wait.) Although we do not know the reason for Venus\u2019 slow backward rotation, we can guess that it may have suffered one or more extremely powerful collisions during the formation process of the solar system. Basic Properties of Venus and Mars Before discussing each planet individually, let us compare some of their basic properties with each other and with Earth (Table 10.1). Venus is in many ways Earth\u2019s twin, with a mass 0.82 times the mass of Earth and an almost identical density. The average amount of geological activity has been also relatively high, almost as high as on Earth. On the other hand, with a surface pressure nearly 100 times greater than ours, Venus\u2019 atmosphere is not at all like that of Earth. The surface of Venus is also remarkably hot, with a temperature of 730 K (over 850 \u00b0F), hotter than the self-cleaning cycle of your oven. One of the major challenges presented by Venus is to understand why the atmosphere and surface environment of this twin have diverged so sharply from those of our own planet. Properties of Earth, Venus, and Mars Property Earth Venus Mars Semimajor axis (AU) Period (year) Mass (Earth = 1) Diameter (km) Density (g/cm3) Surface gravity (Earth = 1) Escape velocity (km/s) 1.00 1.00 1.00 0.72 0.61 0.82 1.52 1.88 0.11 12,756 12,102 6,790 5.5 1.00 11.2 5.3 0.91 10.4 3.9 0.38 5.0 Rotation period (hours or days) 23.9 h 243 d 24.6 h Surface area (Earth = 1) Atmospheric pressure (bar) Table 10.1 1.00 1.00 0.90 90 0.28 0.007 Mars, by contrast, is rather small, with a mass only 0.11 times the mass of Earth. It is larger than either the Moon or Mercury, however, and, unlike them, it retains a thin atmosphere. Mars is also large enough to have supported considerable geological activity in the distant past. But the most fascinating thing about Mars is that long ago it probably had a thick atmosphere and seas of liquid water\u2014the conditions we associate with 342 Chapter 10 Earthlike Planets: Venus and Mars development", " of life. There is even a chance that some form of life persists today in protected environments below the martian surface. 10.2 THE GEOLOGY OF VENUS Learning Objectives By the end of this section, you will be able to: Describe the general features of the surface of Venus Explain what the study of craters on Venus tells us about the age of its surface Compare tectonic activity and volcanoes on Venus with those of Earth Explain why the surface of Venus is inhospitable to human life Since Venus has about the same size and composition as Earth, we might expect its geology to be similar. This is partly true, but Venus does not exhibit the same kind of plate tectonics as Earth, and we will see that its lack of erosion results in a very different surface appearance. Spacecraft Exploration of Venus Nearly 50 spacecraft have been launched to Venus, but only about half were successful. Although the 1962 US Mariner 2 flyby was the first, the Soviet Union launched most of the subsequent missions to Venus. In 1970, Venera 7 became the first probe to land and broadcast data from the surface of Venus. It operated for 23 minutes before succumbing to the high surface temperature. Additional Venera probes and landers followed, photographing the surface and analyzing the atmosphere and soil. To understand the geology of Venus, however, we needed to make a global study of its surface, a task made very difficult by the perpetual cloud layers surrounding the planet. The problem resembles the challenge facing air traffic controllers at an airport, when the weather is so cloudy or smoggy that they can\u2019t locate the incoming planes visually. The solution is similar in both cases: use a radar instrument to probe through the obscuring layer. The first global radar map was made by the US Pioneer Venus orbiter in the late 1970s, followed by better maps from the twin Soviet Venera 15 and 16 radar orbiters in the early 1980s. However, most of our information on the geology of Venus is derived from the US Magellan spacecraft, which mapped Venus with a powerful imaging radar. Magellan produced images with a resolution of 100 meters, much better than that of previous missions, yielding our first detailed look at the surface of our sister planet (Figure 10.6). (The Magellan spacecraft returned more data to Earth than all previous planetary missions combined; each 100 minutes of data transmission from the spacecraft provided enough information, if translated into characters, to fill two 30-volume encyclopedias.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 343 Figure 10.6. Radar Map of Venus. This composite image has a resolution of about 3 kilometers. Colors have been added to indicate elevation, with blue meaning low and brown and white high. The large continent Aphrodite stretches around the equator, where the bright (therefore rough) surface has been deformed by tectonic forces in the crust of Venus. (credit: modification of work by NASA/JPL/USGS) Consider for a moment how good Magellan\u2019s resolution of 100 meters really is. It means the radar images from Venus can show anything on the surface larger than a football field. Suddenly, a whole host of topographic features on Venus became accessible to our view. As you look at the radar images throughout this chapter, bear in mind that these are constructed from radar reflections, not from visible-light photographs. For example, bright features on these radar images are an indication of rough terrain, whereas darker regions are smoother. Probing Through the Clouds of Venus The radar maps of Venus reveal a planet that looks much the way Earth might look if our planet\u2019s surface were not constantly being changed by erosion and deposition of sediment. Because there is no water or ice on Venus and the surface wind speeds are low, almost nothing obscures or erases the complex geological features produced by the movements of Venus\u2019 crust, by volcanic eruptions, and by impact craters. Having finally penetrated below the clouds of Venus, we find its surface to be naked, revealing the history of hundreds of millions of years of geological activity. About 75% of the surface of Venus consists of lowland lava plains. Superficially, these plains resemble the basaltic ocean basins of Earth, but they were not produced in quite the same way. There is no evidence of subduction zones on Venus, indicating that, unlike Earth, this planet never experienced plate tectonics. Although convection (the rising of hot materials) in its mantle generated great stresses in the crust of Venus, they did not start large continental plates moving. The formation of the lava plains of Venus more nearly resembles that of the lunar maria. Both were the result of widespread lava eruptions without the crustal spreading associated with plate tectonics. Rising above the lowland lava plains are two full-scale continents of mountainous terrain", ". The largest continent on Venus, called Aphrodite, is about the size of Africa (you can see it stand out in Figure 10.6). Aphrodite stretches along the equator for about one-third of the way around the planet. Next in size is the northern highland region Ishtar, which is about the size of Australia. Ishtar contains the highest region on the planet, the Maxwell Mountains, which rise 11 kilometers above the surrounding lowlands. (The Maxwell Mountains are the only feature on Venus named after a man. They commemorate James Clerk Maxwell, whose theory of 344 Chapter 10 Earthlike Planets: Venus and Mars electromagnetism led to the invention of radar. All other features are named for women, either from history or mythology.) Craters and the Age of the Venus Surface One of the first questions astronomers addressed with the high-resolution Magellan images was the age of the surface of Venus. Remember that the age of a planetary surface is rarely the age of the world it is on. A young age merely implies an active geology in that location. Such ages can be derived from counting impact craters. Figure 10.7 is an example of what these craters look like on the Venus radar images. The more densely cratered the surface, the greater its age. The largest crater on Venus (called Mead) is 275 kilometers in diameter, slightly larger than the largest known terrestrial crater (Chicxulub), but much smaller than the lunar impact basins. Figure 10.7. Impact Craters on Venus. (a) These large impact craters are in the Lavinia region of Venus. Because they are rough, the crater rims and ejecta appear brighter in these radar images than do the smoother surrounding lava plains. The largest of these craters has a diameter of 50 kilometers. (b) This small, complex crater is named after writer Gertrude Stein. The triple impact was caused by the breaking apart of the incoming asteroid during its passage through the thick atmosphere of Venus. The projectile had an initial diameter of between 1 and 2 kilometers. (credit a: modification of work by NASA/JPL; credit b: modification of work by NASA/JPL) You might think that the thick atmosphere of Venus would protect the surface from impacts, burning up the projectiles long before they could reach the surface. But this is the case for only smaller projectiles. Crater statistics show very few craters less than 10 kilometers in diameter, indicating that projectiles smaller than about 1 kilometer (the size that typically produces a 10-kilometer crater) were stopped by the atmosphere. Those craters with diameters from 10 to 30 kilometers are frequently distorted or multiple, apparently because the incoming projectile broke apart in the atmosphere before it could strike the ground as shown in the Stein crater in Figure 10.7. If we limit ourselves to impacts that produce craters with diameters of 30 kilometers or larger, however, then crater counts are as useful on Venus for measuring surface age as they are on airless bodies such as the Moon. The large craters in the venusian plains indicate an average surface age that is only between 300 and 600 million years. These results indicate that Venus is indeed a planet with persistent geological activity, intermediate between that of Earth\u2019s ocean basins (which are younger and more active) and that of its continents (which are older and less active). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 345 Almost all of the large craters on Venus look fresh, with little degradation or filling in by either lava or windblown dust. This is one way we know that the rates of erosion or sediment deposition are very low. We have the impression that relatively little has happened since the venusian plains were last resurfaced by large-scale volcanic activity. Apparently Venus experienced some sort of planet-wide volcanic convulsion between 300 and 600 million years ago, a mysterious event that is unlike anything in terrestrial history. Volcanoes on Venus Like Earth, Venus is a planet that has experienced widespread volcanism. In the lowland plains, volcanic eruptions are the principal way the surface is renewed, with large flows of highly fluid lava destroying old craters and generating a fresh surface. In addition, numerous younger volcanic mountains and other structures are associated with surface hot spots\u2014places where convection in the planet\u2019s mantle transports the interior heat to the surface. The largest individual volcano on Venus, called Sif Mons, is about 500 kilometers across and 3 kilometers high\u2014broader but lower than the Hawaiian volcano Mauna Loa. At its top is a volcanic crater, or caldera, about 40 kilometers across, and its slopes show individual lava flows up to 500 kilometers long. Thousands of smaller volcanoes dot the surface, down to the limit of visibility of the Magellan images, which correspond to cones or domes about the size of a shopping mall parking", " lot. Most of these seem similar to terrestrial volcanoes. Other volcanoes have unusual shapes, such as the \u201cpancake domes\u201d illustrated in Figure 10.8. Figure 10.8. Pancake-Shaped Volcanoes on Venus. These remarkable circular domes, each about 25 kilometers across and about 2 kilometers tall, are the result of eruptions of highly viscous (sludgy) lava that spreads out evenly in all directions. (credit: modification of work by NASA/JPL) All of the volcanism is the result of eruption of lava onto the surface of the planet. But the hot lava rising from the interior of a planet does not always make it to the surface. On both Earth and Venus, this upwelling lava can collect to produce bulges in the crust. Many of the granite mountain ranges on Earth, such as the Sierra Nevada in California, involve such subsurface volcanism. These bulges are common on Venus, where they produce large circular or oval features called coronae (singular: corona) (Figure 10.9). 346 Chapter 10 Earthlike Planets: Venus and Mars Figure 10.9. The \u201cMiss Piggy\u201d Corona. Fotla Corona is located in the plains to the south of Aphrodite Terra. Curved fracture patterns show where the material beneath has put stress on the surface. A number of pancake and dome volcanoes are also visible. Fotla was a Celtic fertility goddess. Some students see a resemblance between this corona and Miss Piggy of the Muppets (her left ear, at the top of the picture, is the pancake volcano in the upper center of the image). (credit: NASA/JPL) Tectonic Activity Convection currents of molten material in the mantle of Venus push and stretch the crust. Such forces are called tectonic, and the geological features that result from these forces are called tectonic features. On Venus\u2019 lowland plains, tectonic forces have broken the lava surface to create remarkable patterns of ridges and cracks (Figure 10.10). In a few places, the crust has even torn apart to generate rift valleys. The circular features associated with coronae are tectonic ridges and cracks, and most of the mountains of Venus also owe their existence to tectonic forces. Figure 10.10. Ridges and Cracks. This region of the Lakshmi Plains on Venus has been fractured by tectonic forces to produce a cross-hatched grid of cracks and ridges. Be sure to notice the fainter linear features that run perpendicular to the brighter ones. As this is a radar image, the brightness of the ridges indicates their relative height. This image shows a region about 80 kilometers wide and 37 kilometers high. Lakshmi is a Hindu goddess of prosperity. (credit: modification of work by Magellan Team, JPL, NASA) The Ishtar continent, which has the highest elevations on Venus, is the most dramatic product of these tectonic forces. Ishtar and its tall Maxwell Mountains resemble the Tibetan Plateau and Himalayan Mountains on Earth. Both are the product of compression of the crust, and both are maintained by the continuing forces of mantle convection. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 347 On Venus\u2019 Surface The successful Venera landers of the 1970s found themselves on an extraordinarily inhospitable planet, with a surface pressure of 90 bars and a temperature hot enough to melt lead and zinc. Despite these unpleasant conditions, the spacecraft were able to photograph their surroundings and collect surface samples for chemical analysis before their instruments gave out. The diffuse sunlight striking the surface was tinted red by the clouds, and the illumination level was equivalent to a heavy overcast on Earth. The probes found that the rock in the landing areas is igneous, primarily basalts. Examples of the Venera photographs are shown in Figure 10.11. Each picture shows a flat, desolate landscape with a variety of rocks, some of which may be ejecta from impacts. Other areas show flat, layered lava flows. There have been no further landings on Venus since the 1970s. Figure 10.11. Surface of Venus. These views of the surface of Venus are from the Venera 13 spacecraft. Everything is orange because the thick atmosphere of Venus absorbs the bluer colors of light. The horizon is visible in the upper corner of each image. (credit: NASA) 10.3 THE MASSIVE ATMOSPHERE OF VENUS Learning Objectives By the end of this section, you will be able to: Describe the general composition and structure of the atmosphere on Venus Explain how the greenhouse effect has led to high temperatures on Venus The thick atmosphere of Venus produces the high surface temperature and shrouds the surface in a perpetual red twilight. Sunlight does", " not penetrate directly through the heavy clouds, but the surface is fairly well lit by diffuse light (about the same as the light on Earth under a heavy overcast). The weather at the bottom of this deep atmosphere remains perpetually hot and dry, with calm winds. Because of the heavy blanket of clouds and atmosphere, one spot on the surface of Venus is similar to any other as far as weather is concerned. Composition and Structure of the Atmosphere The most abundant gas on Venus is carbon dioxide (CO2), which accounts for 96% of the atmosphere. The second most common gas is nitrogen. The predominance of carbon dioxide over nitrogen is not surprising when you recall that Earth\u2019s atmosphere would also be mostly carbon dioxide if this gas were not locked up in marine sediments (see the discussion of Earth\u2019s atmosphere in Earth as a Planet). Table 10.2 compares the compositions of the atmospheres of Venus, Mars, and Earth. Expressed in this way, as percentages, the proportions of the major gases are very similar for Venus and Mars, but in total quantity, their atmospheres are dramatically different. With its surface pressure of 90 bars, the venusian atmosphere is more than 10,000 times more massive than its martian counterpart. Overall, the atmosphere of Venus is very dry; the absence of water is one of the important ways that Venus differs from Earth. 348 Chapter 10 Earthlike Planets: Venus and Mars Atmospheric Composition of Earth, Venus, and Mars Gas Earth Venus Mars Carbon dioxide (CO2) Nitrogen (N2) Argon (Ar) Oxygen (O2) Neon (Ne) Table 10.2 0.03% 78.1% 0.93% 21.0% 0.002% 96% 3.5% 0.006% 0.003% 0.001% 95.3% 2.7% 1.6% 0.15% 0.0003% The atmosphere of Venus has a huge troposphere (region of convection) that extends up to at least 50 kilometers above the surface (Figure 10.12). Within the troposphere, the gas is heated from below and circulates slowly, rising near the equator and descending over the poles. Being at the base of the atmosphere of Venus is something like being a kilometer or more below the ocean surface on Earth. There, the mass of water evens out temperature variations and results in a uniform environment\u2014the same effect the thick atmosphere has on Venus. Figure 10.12. Venus\u2019 Atmosphere. The layers of the massive atmosphere of Venus shown here are based on data from the Pioneer and Venera entry probes. Height is measured along the left axis, the bottom scale shows temperature, and the red line allows you to read off the temperature at each height. Notice how steeply the temperature rises below the clouds, thanks to the planet\u2019s huge greenhouse effect. In the upper troposphere, between 30 and 60 kilometers above the surface, a thick cloud layer is composed primarily of sulfuric acid droplets. Sulfuric acid (H2SO4) is formed from the chemical combination of sulfur dioxide (SO2) and water (H2O). In the atmosphere of Earth, sulfur dioxide is one of the primary gases emitted by volcanoes, but it is quickly diluted and washed out by rainfall. In the dry atmosphere of Venus, this unpleasant substance is apparently stable. Below 30 kilometers, the Venus atmosphere is clear of clouds. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 349 Surface Temperature on Venus The high surface temperature of Venus was discovered by radio astronomers in the late 1950s and confirmed by the Mariner and Venera probes. How can our neighbor planet be so hot? Although Venus is somewhat closer to the Sun than is Earth, its surface is hundreds of degrees hotter than you would expect from the extra sunlight it receives. Scientists wondered what could be heating the surface of Venus to a temperature above 700 K. The answer turned out to be the greenhouse effect. The greenhouse effect works on Venus just as it does on Earth, but since Venus has so much more CO2\u2014almost a million times more\u2014the effect is much stronger. The thick CO2 acts as a blanket, making it very difficult for the infrared (heat) radiation from the ground to get back into space. As a result, the surface heats up. The energy balance is only restored when the planet is radiating as much energy as it receives from the Sun, but this can happen only when the temperature of the lower atmosphere is very high. One way of thinking of greenhouse heating is that it must raise the surface temperature of Venus until this energy balance is achieved. Has Venus always had such a massive atmosphere and high surface temperature, or might it have evolved to such conditions from a climate that was once more nearly earthlike? The answer to this question is of particular interest to us as we", " look at the increasing levels of CO2 in Earth\u2019s atmosphere. As the greenhouse effect becomes stronger on Earth, are we in any danger of transforming our own planet into a hellish place like Venus? Let us try to reconstruct the possible evolution of Venus from an earthlike beginning to its present state. Venus may once have had a climate similar to that of Earth, with moderate temperatures, water oceans, and much of its CO2 dissolved in the ocean or chemically combined with the surface rocks. Then we allow for modest additional heating\u2014by gradual increase in the energy output of the Sun, for example. When we calculate how Venus\u2019 atmosphere would respond to such effects, it turns out that even a small amount of extra heat can lead to increased evaporation of water from the oceans and the release of gas from surface rocks. This in turn means a further increase in the atmospheric CO2 and H2O, gases that would amplify the greenhouse effect in Venus\u2019 atmosphere. That would lead to still more heat near Venus\u2019 surface and the release of further CO2 and H2O. Unless some other processes intervene, the temperature thus continues to rise. Such a situation is called the runaway greenhouse effect. We want to emphasize that the runaway greenhouse effect is not just a large greenhouse effect; it is an evolutionary process. The atmosphere evolves from having a small greenhouse effect, such as on Earth, to a situation where greenhouse warming is a major factor, as we see today on Venus. Once the large greenhouse conditions develop, the planet establishes a new, much hotter equilibrium near its surface. Reversing the situation is difficult because of the role water plays. On Earth, most of the CO2 is either chemically bound in the rocks of our crust or dissolved by the water in our oceans. As Venus got hotter and hotter, its oceans evaporated, eliminating that safety valve. But the water vapor in the planet\u2019s atmosphere will not last forever in the presence of ultraviolet light from the Sun. The light element hydrogen can escape from the atmosphere, leaving the oxygen behind to combine chemically with surface rock. The loss of water is therefore an irreversible process: once the water is gone, it cannot be restored. There is evidence that this is just what happened to the water once present on Venus. We don\u2019t know if the same runaway greenhouse effect could one day happen on Earth. Although we are uncertain about the point at which a stable greenhouse effect breaks down and turns into a runaway greenhouse effect, Venus stands as clear testament to the fact that a planet cannot continue heating indefinitely without a major change in its oceans and atmosphere. It is a conclusion that we and our descendants will surely want to pay close attention to. 350 Chapter 10 Earthlike Planets: Venus and Mars 10.4 THE GEOLOGY OF MARS Learning Objectives By the end of this section, you will be able to: Discuss the main missions that have explored Mars Explain what we have learned from examination of meteorites from Mars Describe the various features found on the surface of Mars Compare the volcanoes and canyons on Mars with those of Earth Describe the general conditions on the surface of Mars Mars is more interesting to most people than Venus because it is more hospitable. Even from the distance of Earth, we can see surface features on Mars and follow the seasonal changes in its polar caps (Figure 10.13). Although the surface today is dry and cold, evidence collected by spacecraft suggests that Mars once had blue skies and lakes of liquid water. Even today, it is the sort of place we can imagine astronauts visiting and perhaps even setting up permanent bases. Figure 10.13. Mars Photographed by the Hubble Space Telescope. This is one of the best photos of Mars taken from our planet, obtained in June 2001 when Mars was only 68 million kilometers away. The resolution is about 20 kilometers\u2014much better than can be obtained with ground-based telescopes but still insufficient to reveal the underlying geology of Mars. (credit: modification of work by NASA and the Hubble Heritage Team (STScI/AURA)) Spacecraft Exploration of Mars Mars has been intensively investigated by spacecraft. More than 50 spacecraft have been launched toward Mars, but only about half were fully successful. The first visitor was the US Mariner 4, which flew past Mars in 1965 and transmitted 22 photos to Earth. These pictures showed an apparently bleak planet with abundant impact craters. In those days, craters were unexpected; some people who were romantically inclined still hoped to see canals or something like them. In any case, newspaper headlines sadly announced that Mars was a \u201cdead planet.\u201d This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 351 In 1971, NASA\u2019s Mariner 9 became the first spacecraft to orbit another planet, mapping the entire surface of Mars at a resolution of about 1 kilometer and discovering a great variety of geological features, including volcanoes,", " huge canyons, intricate layers on the polar caps, and channels that appeared to have been cut by running water. Geologically, Mars didn\u2019t look so dead after all. The twin Viking spacecraft of the 1970s were among the most ambitious and successful of all planetary missions. Two orbiters surveyed the planet and served to relay communications for two landers on the surface. After an exciting and sometimes frustrating search for a safe landing spot, the Viking 1 lander touched down on the surface of Chryse Planitia (the Plains of Gold) on July 20, 1976, exactly 7 years after Neil Armstrong\u2019s historic first step on the Moon. Two months later, Viking 2 landed with equal success in another plain farther north, called Utopia. The landers photographed the surface with high resolution and carried out complex experiments searching for evidence of life, while the orbiters provided a global perspective on Mars geology. Mars languished unvisited for two decades after Viking. Two more spacecraft were launched toward Mars, by NASA and the Russian Space Agency, but both failed before reaching the planet. The situation changed in the 1990s as NASA began a new exploration program using spacecraft that were smaller and less expensive than Viking. The first of the new missions, appropriately called Pathfinder, landed the first wheeled, solar-powered rover on the martian surface on July 4, 1997 (Figure 10.14). An orbiter called Mars Global Surveyor (MGS) arrived a few months later and began high-resolution photography of the entire surface over more than one martian year. The most dramatic discovery by this spacecraft, which is still operating, was evidence of gullies apparently cut by surface water, as we will discuss later. These missions were followed in 2003 by the NASA Mars Odyssey orbiter, and the ESA Mars Express orbiter, both carrying high-resolution cameras. A gamma-ray spectrometer on Odyssey discovered a large amount of subsurface hydrogen (probably in the form of frozen water). Subsequent orbiters included the NASA Mars Reconnaissance Orbiter to evaluate future landing sites, MAVEN to study the upper atmosphere, and India\u2019s Mangalayaan, also focused on study of Mars\u2019 thin layers of air. Several of these orbiters are also equipped to communicate with landers and rovers on the surface and serve as data relays to Earth. Figure 10.14. Surface View from Mars Pathfinder. The scene from the Pathfinder lander shows a windswept plain, sculpted long ago when water flowed out of the martian highlands and into the depression where the spacecraft landed. The Sojourner rover, the first wheeled vehicle on Mars, is about the size of a microwave oven. Its flat top contains solar cells that provided electricity to run the vehicle. You can see the ramp from the lander and the path the rover took to the larger rock that the mission team nicknamed \u201cYogi.\u201d (credit: NASA/JPL) In 2003, NASA began a series of highly successful Mars landers. Twin Mars Exploration Rovers (MER), named Spirit and Opportunity, have been successful far beyond their planned lifetimes. The design goal for the rovers was 600 meters of travel; in fact, they have traveled jointly more than 50 kilometers. After scouting around its 352 Chapter 10 Earthlike Planets: Venus and Mars rim, Opportunity drove down the steep walls into an impact crater called Victoria, then succeeded with some difficulty in climbing back out to resume its route (Figure 10.15). Dust covering the rovers\u2019 solar cells caused a drop in power, but when a seasonal dust storm blew away the dust, the rovers resumed full operation. In order to survive winter, the rovers were positioned on slopes to maximize solar heating and power generation. In 2006, Spirit lost power on one of its wheels, and subsequently became stuck in the sand, where it continued operation as a fixed ground station. Meanwhile, in 2008, Phoenix (a spacecraft \u201creborn\u201d of spare parts from a previous Mars mission that had failed) landed near the edge of the north polar cap, at latitude 68\u00b0, and directly measured water ice in the soil. Figure 10.15. Victoria Crater. (a) This crater in Meridiani Planum is 800 meters wide, making it slightly smaller than Meteor crater on Earth. Note the dune field in the interior. (b) This image shows the view from the Opportunity rover as it scouted the rim of Victoria crater looking for a safe route down into the interior. (credit a: modification of work by NASA/JPL-Caltech/University of Arizona/Cornell/Phio State University; credit b: modification of work by NASA/JPL/Cornell) In 2011, NASA launched its largest (and most expensive) Mars mission since Viking (see Figure 10.1). The 1-ton rover Curiosity, the size of a subcompact car, has plutonium-powered electrical generators, so that it is not dependent", " on sunlight for power. Curiosity made a pinpoint landing on the floor of Gale crater, a site selected for its complex geology and evidence that it had been submerged by water in the past. Previously, Mars landers had been sent to flat terrains with few hazards, as required by their lower targeting accuracy. The scientific goals of Curiosity include investigations of climate and geology, and assessment of the habitability of past and present Mars environments. It does not carry a specific life detection instrument, however. So far, scientists have not been able to devise a simple instrument that could distinguish living from nonliving materials on Mars The Curiosity rover required a remarkably complex landing sequence and NASA made a video (https://openstax.org/l/30Curiosityrove) about it called \u201c7 Minutes of Terror\u201d that went viral on the Internet. A dramatic video summary (https://openstax.org/l/30MarsSurface) of the first two years of Curiosity\u2019s exploration of the martian surface can be viewed as well. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 353 Martian Samples Much of what we know of the Moon, including the circumstances of its origin, comes from studies of lunar samples, but spacecraft have not yet returned martian samples to Earth for laboratory analysis. It is with great interest, therefore, that scientists have discovered that samples of martian material are nevertheless already here on Earth, available for study. These are all members of a rare class of meteorites (Figure 10.16)\u2014rocks that have fallen from space. Figure 10.16. Martian Meteorite. This fragment of basalt, ejected from Mars in a crater-forming impact, eventually arrived on Earth\u2019s surface. (credit: NASA) How would rocks have escaped from Mars? Many impacts have occurred on the red planet, as shown by its heavily cratered surface. Fragments blasted from large impacts can escape from Mars, whose surface gravity is only 38% of Earth\u2019s. A long time later (typically a few million years), a very small fraction of these fragments collide with Earth and survive their passage through our atmosphere, just like other meteorites. (We\u2019ll discuss meteorites in more detail in the chapter on Cosmic Samples and the Origin of the Solar System.) By the way, rocks from the Moon have also reached our planet as meteorites, although we were able to demonstrate their lunar origin only by comparison with samples returned by the Apollo missions Most of the martian meteorites are volcanic basalts; most of them are also relatively young\u2014about 1.3 billion years old. We know from details of their composition that they are not from Earth or the Moon. Besides, there was no volcanic activity on the Moon to form them as recently as 1.3 billon years ago. It would be very difficult for ejecta from impacts on Venus to escape through its thick atmosphere. By the process of elimination, the only reasonable origin seems to be Mars, where the Tharsis volcanoes were active at that time. The martian origin of these meteorites was confirmed by the analysis of tiny gas bubbles trapped inside several of them. These bubbles match the atmospheric properties of Mars as first measured directly by Viking. It appears that some atmospheric gas was trapped in the rock by the shock of the impact that ejected it from Mars and started it on its way toward Earth. One of the most exciting results from analysis of these martian samples has been the discovery of both water and organic (carbon-based) compounds in them, which suggests that Mars may once have had oceans and perhaps even life on its surface. As we have already hinted, there is other evidence for the presence of flowing water on Mars in the remote past, and even extending to the present. In this and the following sections, we will summarize the picture of Mars as revealed by all these exploratory missions and by about 40 samples from Mars. Global Properties of Mars Mars has a diameter of 6790 kilometers, just over half the diameter of Earth, giving it a total surface area very nearly equal to the continental (land) area of our planet. Its overall density of 3.9 g/cm3 suggests a composition consisting primarily of silicates but with a small metal core. The planet has no global magnetic field, although 354 Chapter 10 Earthlike Planets: Venus and Mars there are areas of strong surface magnetization that indicate that there was a global field billions of years ago. Apparently, the red planet has no liquid material in its core today that would conduct electricity. Thanks to the Mars Global Surveyor, we have mapped the entire planet, as shown in Figure 10.17. A laser altimeter on board made millions of separate measurements of the surface topography to a precision of a few meters\u2014good enough to show even the annual deposition and evaporation of the polar caps. Like Earth, the Moon, and Venus", ", the surface of Mars has continental or highland areas as well as widespread volcanic plains. The total range in elevation from the top of the highest mountain ( Olympus Mons) to the bottom of the deepest basin (Hellas) is 31 kilometers. Figure 10.17. Mars Map from Laser Ranging. These globes are highly precise topographic maps, reconstructed from millions of individual elevation measurements made with the Mars Global Surveyor. Color is used to indicate elevation. The hemisphere on the left includes the Tharsis bulge and Olympus Mons, the highest mountain on Mars; the hemisphere on the right includes the Hellas basin, which has the lowest elevation on Mars. (credit: modification of work by NASA/JPL) Approximately half the planet consists of heavily cratered highland terrain, found primarily in the southern hemisphere. The other half, which is mostly in the north, contains younger, lightly cratered volcanic plains at an average elevation about 5 kilometers lower than the highlands. Remember that we saw a similar pattern on Earth, the Moon, and Venus. A geological division into older highlands and younger lowland plains seems to be characteristic of all the terrestrial planets except Mercury. Lying across the north-south division of Mars is an uplifted continent the size of North America. This is the 10-kilometer-high Tharsis bulge, a volcanic region crowned by four great volcanoes that rise still higher into the martian sky. Volcanoes on Mars The lowland plains of Mars look very much like the lunar maria, and they have about the same density of impact craters. Like the lunar maria, they probably formed between 3 and 4 billion years ago. Apparently, Mars experienced extensive volcanic activity at about the same time the Moon did, producing similar basaltic lavas. The largest volcanic mountains of Mars are found in the Tharsis area (you can see them in Figure 10.17), although smaller volcanoes dot much of the surface. The most dramatic volcano on Mars is Olympus Mons (Mount Olympus), with a diameter larger than 500 kilometers and a summit that towers more than 20 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 355 kilometers above the surrounding plains\u2014three times higher than the tallest mountain on Earth (Figure 10.18). The volume of this immense volcano is nearly 100 times greater than that of Mauna Loa in Hawaii. Placed on Earth\u2019s surface, Olympus would more than cover the entire state of Missouri. Figure 10.18. Olympus Mons. The largest volcano on Mars, and probably the largest in the solar system, is Olympus Mons, illustrated in this computer-generated rendering based on data from the Mars Global Surveyor\u2019s laser altimeter. Placed on Earth, the base of Olympus Mons would completely cover the state of Missouri; the caldera, the circular opening at the top, is 65 kilometers across, about the size of Los Angeles. (credit: NASA/Corbis) Images taken from orbit allow scientists to search for impact craters on the slopes of these volcanoes in order to estimate their age. Many of the volcanoes show a fair number of such craters, suggesting that they ceased activity a billion years or more ago. However, Olympus Mons has very, very few impact craters. Its present surface cannot be more than about 100 million years old; it may even be much younger. Some of the fresh- looking lava flows might have been formed a hundred years ago, or a thousand, or a million, but geologically speaking, they are quite young. This leads geologists to the conclusion that Olympus Mons possibly remains intermittently active today\u2014something future Mars land developers may want to keep in mind. Martian Cracks and Canyons The Tharsis bulge has many interesting geological features in addition to its huge volcanoes. In this part of the planet, the surface itself has bulged upward, forced by great pressures from below, resulting in extensive tectonic cracking of the crust. Among the most spectacular tectonic features on Mars are the canyons called the Valles Marineris (or Mariner Valleys, named after Mariner 9, which first revealed them to us), which are shown in Figure 10.19. They extend for about 5000 kilometers (nearly a quarter of the way around Mars) along the slopes of the Tharsis bulge. If it were on Earth, this canyon system would stretch all the way from Los Angeles to Washington, DC. The main canyon is about 7 kilometers deep and up to 100 kilometers wide, large enough for the Grand Canyon of the Colorado River to fit comfortably into one of its side canyons. 356 Chapter 10 Earthlike Planets: Venus and Mars Figure 10.19. Heavily Eroded Canyonlands on Mars. This image shows the Valles Marineris canyon complex, which is 3000 kilometers wide and 8 kilometers deep.", " (credit: NASA/JPL/USGS An excellent 4-minute video tour (https://openstax.org/l/30VallesMariner) of Valles Marineris, narrated by planetary scientist Phil Christensen, is available for viewing. The term \u201ccanyon\u201d is somewhat misleading here because the Valles Marineris canyons have no outlets and were not cut by running water. They are basically tectonic cracks, produced by the same crustal tensions that caused the Tharsis uplift. However, water has played a later role in shaping the canyons, primarily by seeping from deep springs and undercutting the cliffs. This undercutting led to landslides that gradually widened the original cracks into the great valleys we see today (Figure 10.20). Today, the primary form of erosion in the canyons is probably wind. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 357 Figure 10.20. Martian Landslides. This Viking orbiter image shows Ophir Chasma, one of the connected valleys of the Valles Marineris canyon system. Look carefully and you can see enormous landslides whose debris is piled up underneath the cliff wall, which tower up to 10 kilometers above the canyon floor. (credit: modification of work by NASA/JPL/USGS) While the Tharsis bulge and Valles Marineris are impressive, in general, we see fewer tectonic structures on Mars than on Venus. In part, this may reflect a lower general level of geological activity, as would be expected for a smaller planet. But it is also possible that evidence of widespread faulting has been buried by wind- deposited sediment over much of Mars. Like Earth, Mars may have hidden part of its geological history under a cloak of soil. The View on the Martian Surface The first spacecraft to land successfully on Mars were Vikings 1 and 2 and Mars Pathfinder. All sent back photos that showed a desolate but strangely beautiful landscape, including numerous angular rocks interspersed with dune like deposits of fine-grained, reddish soil (Figure 10.21). Figure 10.21. Three Martian Landing Sites. The Mars landers Viking 1 in Chryse, Pathfinder in Ares Valley, and Viking 2 in Utopia, all photographed their immediate surroundings. It is apparent from the similarity of these three photos that each spacecraft touched down on a flat, windswept plain littered with rocks ranging from tiny pebbles up to meter-size boulders. It is probable that most of Mars looks like this on the surface. (credit \u201cViking 1\u201d: modification of work by Van der Hoorn/NASA; credit \u201cPathfinder\u201d: modification of work by NASA; credit \u201cViking 2\u201d: modification of work by NASA; credit Mars: modification of work by NASA/Goddard Space Flight Center) All three of these landers were targeted to relatively flat, lowland terrain. Instruments on the landers found that the soil consisted of clays and iron oxides, as had long been expected from the red color of the planet. All 358 Chapter 10 Earthlike Planets: Venus and Mars the rocks measured appeared to be of volcanic origin and roughly the same composition. Later landers were targeted to touch down in areas that apparently were flooded sometime in the past, where sedimentary rock layers, formed in the presence of water, are common. (Although we should note that nearly all the planet is blanketed in at least a thin layer of wind-blown dust). The Viking landers included weather stations that operated for several years, providing a perspective on martian weather. The temperatures they measured varied greatly with the seasons, due to the absence of moderating oceans and clouds. Typically, the summer maximum at Viking 1 was 240 K (\u201333 \u00b0C), dropping to 190 K (\u201383 \u00b0C) at the same location just before dawn. The lowest air temperatures, measured farther north by Viking 2, were about 173 K (\u2013100 \u00b0C). During the winter, Viking 2 also photographed water frost deposits on the ground (Figure 10.22). We make a point of saying \u201cwater frost\u201d here because at some locations on Mars, it gets cold enough for carbon dioxide (dry ice) to freeze out of the atmosphere as well. Figure 10.22. Water Frost in Utopia. This image of surface frost was photographed at the Viking 2 landing site during late winter. (credit: NASA/JPL) Most of the winds measured on Mars are only a few kilometers per hour. However, Mars is capable of great windstorms that can shroud the entire planet with windblown dust. Such high winds can strip the surface of some of its loose, fine dust, leaving the rock exposed. The later rovers found that each sunny afternoon the atmosphere became turbulent as heat rose off the", " surface. This turbulence generated dust devils, which play an important role in lifting the fine dust into the atmosphere. As the dust devils strip off the top layer of light dust and expose darker material underneath, they can produce fantastic patterns on the ground (Figure 10.23). Wind on Mars plays an important role in redistributing surface material. Figure 10.23 shows a beautiful area of dark sand dunes on top of lighter material. Much of the material stripped out of the martian canyons has been dumped in extensive dune fields like this, mostly at high latitudes. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 359 Figure 10.23. Dust Devil Tracks and Sand Dunes. (a) This high-resolution photo from the Mars Global Surveyor shows the dark tracks of several dust devils that have stripped away a thin coating of light-colored dust. This view is of an area about 3 kilometers across. Dust devils are one of the most important ways that dust gets redistributed by the martian winds. They may also help keep the solar panels of our rovers free of dust. (b) These windblown sand dunes on Mars overlay a lighter sandy surface. Each dune in this high-resolution view is about 1 kilometer across. (credit a: modification of work by NASA/JPL/University of Arizona; credit b: modification of work by NASA/JPL-Caltech/University of Arizona) 10.5 WATER AND LIFE ON MARS Learning Objectives By the end of this section, you will be able to: Describe the general composition of the atmosphere on Mars Explain what we know about the polar ice caps on Mars and how we know it Describe the evidence for the presence of water in the past history of Mars Summarize the evidence for and against the possibility of life on Mars Of all the planets and moons in the solar system, Mars seems to be the most promising place to look for life, both fossil microbes and (we hope) some forms of life deeper underground that still survive today. But where (and how) should we look for life? We know that the one requirement shared by all life on Earth is liquid water. Therefore, the guiding principle in assessing habitability on Mars and elsewhere has been to \u201cfollow the water.\u201d That is the perspective we take in this section, to follow the water on the red planet and hope it will lead us to life. Atmosphere and Clouds on Mars The atmosphere of Mars today has an average surface pressure of only 0.007 bar, less than 1% that of Earth. (This is how thin the air is about 30 kilometers above Earth\u2019s surface.) Martian air is composed primarily of carbon dioxide (95%), with about 3% nitrogen and 2% argon. The proportions of different gases are similar to those in the atmosphere of Venus (see Table 10.2), but a lot less of each gas is found in the thin air on Mars. While winds on Mars can reach high speeds, they exert much less force than wind of the same velocity would on Earth because the atmosphere is so thin. The wind is able, however, to loft very fine dust particles, which can sometimes develop planet-wide dust storms. It is this fine dust that coats almost all the surface, giving Mars its distinctive red color. In the absence of surface water, wind erosion plays a major role in sculpting the martian surface (Figure 10.24). 360 Chapter 10 Earthlike Planets: Venus and Mars Figure 10.24. Wind Erosion on Mars. These long straight ridges, called yardangs, are aligned with the dominant wind direction. This is a highresolution image from the Mars Reconnaissance Orbiter and is about 1 kilometer wide. (credit: NASA/JPL-Caltech/University of Arizona The issue of how strong the winds on Mars can be plays a big role in the 2015 hit movie The Martian (https://openstax.org/l/30TheMartian) in which the main character is stranded on Mars after being buried in the sand in a windstorm so great that his fellow astronauts have to leave the planet so their ship is not damaged. Astronomers have noted that the martian winds could not possibly be as forceful as depicted in the film. In most ways, however, the depiction of Mars in this movie is remarkably accurate. Although the atmosphere contains small amounts of water vapor and occasional clouds of water ice, liquid water is not stable under present conditions on Mars. Part of the problem is the low temperatures on the planet. But even if the temperature on a sunny summer day rises above the freezing point, the low pressure means that liquid water still cannot exist on the surface, except at the lowest elevations. At a pressure of less than 0.006 bar, the boiling point is as low or lower than the freezing point, and water changes directly from solid to", " vapor without an intermediate liquid state (as does \u201cdry ice,\u201d carbon dioxide, on Earth). However, salts dissolved in water lower its freezing point, as we know from the way salt is used to thaw roads after snow and ice forms during winter on Earth. Salty water is therefore sometimes able to exist in liquid form on the martian surface, under the right conditions. Several types of clouds can form in the martian atmosphere. First there are dust clouds, discussed above. Second are water-ice clouds similar to those on Earth. These often form around mountains, just as happens on our planet. Finally, the CO2 of the atmosphere can itself condense at high altitudes to form hazes of dry ice crystals. The CO2 clouds have no counterpart on Earth, since on our planet temperatures never drop low enough (down to about 150 K or about 125 \u00b0C) for this gas to condense. The Polar Caps Through a telescope, the most prominent surface features on Mars are the bright polar caps, which change with the seasons, similar to the seasonal snow cover on Earth. We do not usually think of the winter snow in northern latitudes as a part of our polar caps, but seen from space, the thin winter snow merges with Earth\u2019s thick, permanent ice caps to create an impression much like that seen on Mars (Figure 10.25). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 361 Figure 10.25. Martian North Polar Cap. (a) This is a composite image of the north pole in summer, obtained in October 2006 by the Mars Reconnaissance Orbiter. It shows the mostly water-ice residual cap sitting atop light, tan-colored, layered sediments. Note that although the border of this photo is circular, it shows only a small part of the planet. (b) Here we see a small section of the layered terrain near the martian north pole. There is a mound about 40 meters high that is sticking out of a trough in the center of the picture. (credit a: modification of work by NASA/JPL/MSSS; credit b: modification of work by NASA/JPL-Caltech/University of Arizona) The seasonal caps on Mars are composed not of ordinary snow but of frozen CO2 (dry ice). These deposits condense directly from the atmosphere when the surface temperature drops below about 150 K. The caps develop during the cold martian winters and extend down to about 50\u00b0 latitude by the start of spring. Quite distinct from these thin seasonal caps of CO2 are the permanent or residual caps that are always present near the poles. The southern permanent cap has a diameter of 350 kilometers and is composed of frozen CO2 deposits together with a great deal of water ice. Throughout the southern summer, it remains at the freezing point of CO2, 150 K, and this cold reservoir is thick enough to survive the summer heat intact. The northern permanent cap is different. It is much larger, never shrinking to a diameter less than 1000 kilometers, and is composed of water ice. Summer temperatures in the north are too high for the frozen CO2 to be retained. Measurements from the Mars Global Surveyor have established the exact elevations in the north polar region of Mars, showing that it is a large basin about the size of our own Arctic Ocean basin. The ice cap itself is about 3 kilometers thick, with a total volume of about 10 million km3 (similar to that of Earth\u2019s Mediterranean Sea). If Mars ever had extensive liquid water, this north polar basin would have contained a shallow sea. There is some indication of ancient shorelines visible, but better images will be required to verify this suggestion. Images taken from orbit also show a distinctive type of terrain surrounding the permanent polar caps, as shown in Figure 10.25. At latitudes above 80\u00b0 in both hemispheres, the surface consists of recent layered deposits that cover the older cratered ground below. Individual layers are typically ten to a few tens of meters thick, marked by alternating light and dark bands of sediment. Probably the material in the polar deposits includes dust carried by wind from the equatorial regions of Mars. What do these terraced layers tell us about Mars? Some cyclic process is depositing dust and ice over periods of time. The time scales represented by the polar layers are tens of thousands of years. Apparently the martian climate experiences periodic changes at intervals similar to those between ice ages on Earth. Calculations indicate that the causes are probably also similar: the gravitational pull of the other planets produces variations in Mars\u2019 orbit and tilt as the great clockwork of the solar system goes through its paces. 362 Chapter 10 Earthlike Planets: Venus and Mars The Phoenix spacecraft landed near the north polar cap in summer (Figure 10.26). Controllers knew that is would not be able to survive a polar winter, but directly measuring the characteristics", " of the polar region was deemed important enough to send a dedicated mission. The most exciting discovery came when the spacecraft tried to dig a shallow trench under the spacecraft. When the overlying dust was stripped off, they saw bright white material, apparently some kind of ice. From the way this ice sublimated over the next few days, it was clear that it was frozen water. Figure 10.26. Evaporating Ice on Mars. We see a trench dug by the Phoenix lander in the north polar region four martian days apart in June 2008. If you look at the shadowed region in the bottom left of the trench, you can see three spots of ice in the left image which have sublimated away in the right image. (credit: modification of work by NASA/JPL-Caltech/University of Arizona/Texas A&M University Comparing the Amount of Water on Mars and Earth It is interesting to estimate the amount of water (in the form of ice) on Mars and to compare this with the amount of water on Earth. In each case, we can find the total volume of a layer on a sphere by multiplying the area of the sphere (4\u03c0R2) by the thickness of the layer. For Earth, the ocean water is equivalent to a layer 3 km thick spread over the entire planet, and the radius of Earth is 6.378 \u00d7 106 m (see Appendix F). For Mars, most of the water we are sure of is in the form of ice near the poles. We can calculate the amount of ice in one of the residual polar caps if it is (for example) 2 km thick and has a radius of 400 km (the area of a circle is \u03c0R2). Solution The volume of Earth\u2019s water is therefore the area 4\u03c0R2 \u239e \u239b \u239d6.378 \u00d7 106 m 4\u03c0 \u23a0 2 = 5.1 \u00d7 1014 m2 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 363 multiplied by the thickness of 3000 m: 5.1 \u00d7 1014 m2 \u00d7 3000 m = 1.5 \u00d7 1018 m3 This gives 1.5 \u00d7 1018 m3 of water. Since water has a density of 1 ton per cubic meter (1000 kg/m3), we can calculate the mass: For Mars, the ice doesn\u2019t cover the whole planet, only the caps; the polar cap area is 1.5 \u00d7 1018 m3 \u00d7 1 ton/m3 = 1.5 \u00d7 1018 tons \u03c0R2 = \u03c0(4 \u00d7 105 m)2 = 5 \u00d7 1011 m2 (Note that we converted kilometers to meters.) The volume = area \u00d7 height, so we have: (2 \u00d7 103 m)(5 \u00d7 1011 m2) = 1 \u00d7 1015 m3 = 1015 m3 Therefore, the mass is: 1015 m3 \u00d7 1 ton/m3 = 1015 tons This is about 0.1% that of Earth\u2019s oceans. Check Your Learning A better comparison might be to compare the amount of ice in the Mars polar ice caps to the amount of ice in the Greenland ice sheet on Earth, which has been estimated as 2.85 \u00d7 1015 m3. How does this compare with the ice on Mars? Answer: The Greenland ice sheet has about 2.85 times as much ice as in the polar ice caps on Mars. They are about the same to the nearest power of 10. Channels and Gullies on Mars Although no bodies of liquid water exist on Mars today, evidence has accumulated that rivers flowed on the red planet long ago. Two kinds of geological features appear to be remnants of ancient watercourses, while a third class\u2014smaller gullies\u2014suggests intermittent outbreaks of liquid water even today. We will examine each of these features in turn. In the highland equatorial plains, there are multitudes of small, sinuous (twisting) channels\u2014typically a few meters deep, some tens of meters wide, and perhaps 10 or 20 kilometers long (Figure 10.27). They are called runoff channels because they look like what geologists would expect from the surface runoff of ancient rain storms. These runoff channels seem to be telling us that the planet had a very different climate long ago. To estimate the age of these channels, we look at the cratering record. Crater counts show that this part of the planet is more cratered than the lunar maria but less cratered than the lunar highlands. Thus, the runoff channels are probably older than the lunar maria, presumably about 4 billion years old. The second set of water-related features we see are outflow channels (Figure 10.27) are much larger than the runoff channels. The largest of these, which drain into the Chryse basin where Pathfinder landed, are 10 kilometers or more wide", " and hundreds of kilometers long. Many features of these outflow channels have 364 Chapter 10 Earthlike Planets: Venus and Mars convinced geologists that they were carved by huge volumes of running water, far too great to be produced by ordinary rainfall. Where could such floodwater have come from on Mars? Figure 10.27. Runoff and Outflow Channels. (a) These runoff channels in the old martian highlands are interpreted as the valleys of ancient rivers fed by either rain or underground springs. The width of this image is about 200 kilometers. (b) This intriguing channel, called Nanedi Valles, resembles Earth riverbeds in some (but not all) ways. The tight curves and terraces seen in the channel certainly suggest the sustained flow of a fluid like water. The channel is about 2.5 kilometers across. (credit a: modification of work by Jim Secosky/NASA; credit b: modification of work by Jim Secosky/NASA) As far we can tell, the regions where the outflow channels originate contained abundant water frozen in the soil as permafrost. Some local source of heating must have released this water, leading to a period of rapid and catastrophic flooding. Perhaps this heating was associated with the formation of the volcanic plains on Mars, which date back to roughly the same time as the outflow channels. Note that neither the runoff channels nor the outflow channels are wide enough to be visible from Earth, nor do they follow straight lines. They could not have been the \u201ccanals\u201d Percival Lowell imagined seeing on the red planet. The third type of water feature, the smaller gullies, was discovered by the Mars Global Surveyor (Figure 10.28). The Mars Global Surveyor\u2019s camera images achieved a resolution of a few meters, good enough to see something as small as a truck or bus on the surface. On the steep walls of valleys and craters at high latitudes, there are many erosional features that look like gullies carved by flowing water. These gullies are very young: not only are there no superimposed impact craters, but in some instances, the gullies seem to cut across recent wind- deposited dunes. Perhaps there is liquid water underground that can occasionally break out to produce short- lived surface flows before the water can freeze or evaporate. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 365 Figure 10.28. Gullies on the Wall of Garni Crater. This high-resolution image is from the Mars Reconnaissance Orbiter. The dark streaks, which are each several hundred meters long, change in a seasonal pattern that suggests they are caused by the temporary flow of surface water. (credit: NASA/JPL-Caltech/University of Arizona) The gullies also have the remarkable property of changing regularly with the martian seasons. Many of the dark streaks (visible in Figure 10.28) elongate within a period of a few days, indicating that something is flowing downhill\u2014either water or dark sediment. If it is water, it requires a continuing source, either from the atmosphere or from springs that tap underground water layers (aquifers.) Underground water would be the most exciting possibility, but this explanation seems inconsistent with the fact that many of the dark streaks start at high elevations on the walls of craters. Additional evidence that the dark streaks (called by the scientists recurring slope lineae) are caused by water was found in 2015 when spectra were obtained of the dark streaks (Figure 10.29). These showed the presence of hydrated salts produced by the evaporation of salty water. If the water is salty, it could remain liquid long enough to flow downstream for distances of a hundred meters or more, before it either evaporates or soaks into the ground. However, this discovery still does not identify the ultimate source of the water. 366 Chapter 10 Earthlike Planets: Venus and Mars Figure 10.29. Evidence for Liquid Water on Mars. The dark streaks in Horowitz crater, which move downslope, have been called recurring slope lineae. The streaks in the center of the image go down the wall of the crater for about a distance of 100 meters. Spectra taken of this region indicate that these are locations where salty liquid water flows on or just below the surface of Mars. (The vertical dimension is exaggerated by a factor of 1.5 compared to horizontal dimensions.) (credit: NASA/JPL-Caltech/University of Arizona) Ancient Lakes The rovers (Spirit, Opportunity, and Curiosity) that have operated on the surface of Mars have been used to hunt for additional evidence of water. They could not reach the most interesting sites, such as the gullies, which are located on steep slopes. Instead, they explored sites that might be dried-out lake beds, dating back to a time when the climate on Mars was warmer and", " the atmosphere thicker\u2014allowing water to be liquid on the surface. Spirit was specifically targeted to explore what looked like an ancient lake-bed in Gusev crater, with an outflow channel emptying into it. However, when the spacecraft landed, it found that the former lakebed had been covered by thin lava flows, blocking the rover from access to the sedimentary rocks it had hoped to find. However, Opportunity had better luck. Peering at the walls of a small crater, it detected layered sedimentary rock. These rocks contained chemical evidence of evaporation, suggesting there had been a shallow salty lake in that location. In these sedimentary rocks were also small spheres that were rich in the mineral hematite, which forms only in watery environments. Apparently this very large basin had once been underwater The small spherical rocks were nicknamed \u201cblueberries\u201d by the science team and the discovery of a whole \u201cberry-bowl\u201d of them was announced in this interesting news release (https://openstax.org/l/ 30berrybowl) from NASA. The Curiosity rover landed inside Gale crater, where photos taken from orbit also suggested past water erosion. It discovered numerous sedimentary rocks, some in the form of mudstones from an ancient lakebed; it also This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 367 found indications of rocks formed by the action of shallow water at the time the sediment formed (Figure 10.30). Figure 10.30. Gale Crater. (a) This scene, photographed by the Curiosity rover, shows an ancient lakebed of cracked mudstones. (b) Geologists working with the Curiosity rover interpret this image of cross-bedded sandstone in Gale crater as evidence of liquid water passing over a loose bed of sediment at the time this rock formed. (credit a: modification of work by NASA/JPL-Caltech/MSSS; credit b: modification of work by NASA/ JPL-Caltech/MSSS Astronomy and Pseudoscience: The \u201cFace on Mars\u201d People like human faces. We humans have developed great skill in recognizing people and interpreting facial expressions. We also have a tendency to see faces in many natural formations, from clouds to the man in the Moon. One of the curiosities that emerged from the Viking orbiters\u2019 global mapping of Mars was the discovery of a strangely shaped mesa in the Cydonia region that resembled a human face. Despite later rumors of a cover-up, the \u201cFace on Mars\u201d was, in fact, recognized by Viking scientists and included in one of the early mission press releases. At the low resolution and oblique lighting under which the Viking image was obtained, the mile-wide mesa had something of a Sphinx-like appearance. Unfortunately, a small band of individuals decided that this formation was an artificial, carved sculpture of a human face placed on Mars by an ancient civilization that thrived there hundreds of thousands of years ago. A band of \u201ctrue believers\u201d grew around the face and tried to deduce the nature of the \u201csculptors\u201d who made it. This group also linked the face to a variety of other pseudoscientific phenomena such as crop circles (patterns in fields of grain, mostly in Britain, now known to be the work of pranksters). Members of this group accused NASA of covering up evidence of intelligent life on Mars, and they received a great deal of help in publicizing their perspective from tabloid media. Some of the believers picketed the Jet Propulsion Laboratory at the time of the failure of the Mars Observer spacecraft, circulating stories that the \u201cfailure\u201d of the Mars Observer was itself a fake, and that its true (secret) mission was to photograph the face. The high-resolution Mars Observer camera (MOC) was reflown on the Mars Global Surveyor mission, which arrived at Mars in 1997. On April 5, 1998, in Orbit 220, the MOC obtained an oblique image of the face at a resolution of 4 meters per pixel, a factor-of-10 improvement in resolution over the Viking image. Another image in 2001 had even higher resolution. Immediately released by NASA, the new images showed a low mesa-like hill cut crossways by several roughly linear ridges and depressions, which were misidentified in 368 Chapter 10 Earthlike Planets: Venus and Mars the 1976 photo as the eyes and mouth of a face. Only with an enormous dose of imagination can any resemblance to a face be seen in the new images, demonstrating how dramatically our interpretation of geology can change with large improvements in resolution. The original and the higher resolution images can be seen in Figure 10.31. Figure 10.31. Face on Mars. The so-called \u201cFace on Mars\u201d is seen (a) in low resolution", " from Viking (the \u201cface\u201d is in the upper part of the picture) and (b) with 20 times better resolution from the Mars Global Surveyor. (credit a: modification of work NASA/JPL; credit b: modification of work by NASA/JPL/MSSS) After 20 years of promoting pseudoscientific interpretations and various conspiracy theories, can the \u201cFace on Mars\u201d believers now accept reality? Unfortunately, it does not seem so. They have accused NASA of faking the new picture. They also suggest that the secret mission of the Mars Observer included a nuclear bomb used to destroy the face before it could be photographed in greater detail by the Mars Global Surveyor. Space scientists find these suggestions incredible. NASA is spending increasing sums for research on life in the universe, and a major objective of current and upcoming Mars missions is to search for evidence of past microbial life on Mars. Conclusive evidence of extraterrestrial life would be one of the great discoveries of science and incidentally might well lead to increased funding for NASA. The idea that NASA or other government agencies would (or could) mount a conspiracy to suppress such welcome evidence is truly bizarre. Alas, the \u201cFace on Mars\u201d story is only one example of a whole series of conspiracy theories that are kept before the public by dedicated believers, by people out to make a fast buck, and by irresponsible media attention. Others include the \u201curban legend\u201d that the Air Force has the bodies of extraterrestrials at a secret base, the widely circulated report that UFOs crashed near Roswell, New Mexico (actually it was a balloon carrying scientific instruments to find evidence of Soviet nuclear tests), or the notion that alien astronauts helped build the Egyptian pyramids and many other ancient monuments because our ancestors were too stupid to do it alone. In response to the increase in publicity given to these \u201cfiction science\u201d ideas, a group of scientists, educators, scholars, and magicians (who know a good hoax when they see one) have formed the Committee for Skeptical Inquiry. Two of the original authors of your book are active on the committee. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 369 For more information about its work delving into the rational explanations for paranormal claims, see their excellent magazine, The Skeptical Inquirer, or check out their website at www.csicop.org/. Climate Change on Mars The evidence about ancient rivers and lakes of water on Mars discussed so far suggests that, billions of years ago, martian temperatures must have been warmer and the atmosphere must have been more substantial than it is today. But what could have changed the climate on Mars so dramatically? We presume that, like Earth and Venus, Mars probably formed with a higher surface temperature thanks to the greenhouse effect. But Mars is a smaller planet, and its lower gravity means that atmospheric gases could escape more easily than from Earth and Venus. As more and more of the atmosphere escaped into space, the temperature on the surface gradually fell. Eventually Mars became so cold that most of the water froze out of the atmosphere, further reducing its ability to retain heat. The planet experienced a sort of runaway refrigerator effect, just the opposite of the runaway greenhouse effect that occurred on Venus. Probably, this loss of atmosphere took place within less than a billion years after Mars formed. The result is the cold, dry Mars we see today. Conditions a few meters below the martian surface, however, may be much different. There, liquid water (especially salty water) might persist, kept warm by the internal heat of Mars or the insulating layers solid and rock. Even on the surface, there may be ways to change the martian atmosphere temporarily. Mars is likely to experience long-term climate cycles, which may be caused by the changing orbit and tilt of the planet. At times, one or both of the polar caps might melt, releasing a great deal of water vapor into the atmosphere. Perhaps an occasional impact by a comet might produce a temporary atmosphere that is thick enough to permit liquid water on the surface for a few weeks or months. Some have even suggested that future technology might allow us to terraform Mars\u2014that is, to engineer its atmosphere and climate in ways that might make the planet more hospitable for long-term human habitation. The Search for Life on Mars If there was running water on Mars in the past, perhaps there was life as well. Could life, in some form, remain in the martian soil today? Testing this possibility, however unlikely, was one of the primary objectives of the Viking landers in 1976. These landers carried miniature biological laboratories to test for microorganisms in the martian soil. Martian soil was scooped up by the spacecraft\u2019s long arm and placed into the experimental chambers, where it was isolated and incubated in contact with a variety of gases, radioactive isotopes, and nutrients to", " see what would happen. The experiments looked for evidence of respiration by living animals, absorption of nutrients offered to organisms that might be present, and an exchange of gases between the soil and its surroundings for any reason whatsoever. A fourth instrument pulverized the soil and analyzed it carefully to determine what organic (carbon-bearing) material it contained. The Viking experiments were so sensitive that, had one of the spacecraft landed anywhere on Earth (with the possible exception of Antarctica), it would easily have detected life. But, to the disappointment of many scientists and members of the public, no life was detected on Mars. The soil tests for absorption of nutrients and gas exchange did show some activity, but this was most likely caused by chemical reactions that began as water was added to the soil and had nothing to do with life. In fact, these experiments showed that martian soil seems much more chemically active than terrestrial soils because of its exposure to solar ultraviolet radiation (since Mars has no ozone layer). 370 Chapter 10 Earthlike Planets: Venus and Mars The organic chemistry experiment showed no trace of organic material, which is apparently destroyed on the martian surface by the sterilizing effect of this ultraviolet light. While the possibility of life on the surface has not been eliminated, most experts consider it negligible. Although Mars has the most earthlike environment of any planet in the solar system, the sad fact is that nobody seems to be home today, at least on the surface. However, there is no reason to think that life could not have begun on Mars about 4 billion years ago, at the same time it started on Earth. The two planets had very similar surface conditions then. Thus, the attention of scientists has shifted to the search for fossil life on Mars. One of the primary questions to be addressed by future spacecraft is whether Mars once supported its own life forms and, if so, how this martian life compared with that on our own planet. Future missions will include the return of martian samples selected from sedimentary rocks at sites that once held water and thus perhaps ancient life. The most powerful searches for martian life (past or present) will thus be carried out in our laboratories here on Earth Planetary Protection When scientists begin to search for life on another planet, they must make sure that we do not contaminate the other world with life carried from Earth. At the very beginning of spacecraft exploration on Mars, an international agreement specified that all landers were to be carefully sterilized to avoid accidentally transplanting terrestrial microbes to Mars. In the case of Viking, we know the sterilization was successful. Viking\u2019s failure to detect martian organisms also implies that these experiments did not detect hitchhiking terrestrial microbes. As we have learned more about the harsh conditions on the martian surface, the sterilization requirements have been somewhat relaxed. It is evident that no terrestrial microbes could grow on the martian surface, with its low temperature, absence of water, and intense ultraviolet radiation. Microbes from Earth might survive in a dormant, dried state, but they cannot grow and proliferate on Mars. The problem of contaminating Mars will become more serious, however, as we begin to search for life below the surface, where temperatures are higher and no ultraviolet light penetrates. The situation will be even more daunting if we consider human flights to Mars. Any humans will carry with them a multitude of terrestrial microbes of all kinds, and it is hard to imagine how we can effectively keep the two biospheres isolated from each other if Mars has indigenous life. Perhaps the best situation could be one in which the two life-forms are so different that each is effectively invisible to the other\u2014not recognized on a chemical level as living or as potential food. The most immediate issue of public concern is not with the contamination of Mars but with any dangers associated with returning Mars samples to Earth. NASA is committed to the complete biological isolation of returned samples until they are demonstrated to be safe. Even though the chances of contamination are extremely low, it is better to be safe than sorry. Most likely there is no danger, even if there is life on Mars and alien microbes hitch a ride to Earth inside some of the returned samples. In fact, Mars is sending samples to Earth all the time in the form of the Mars meteorites. Since some of these microbes (if they exist) could probably survive the trip to Earth inside their rocky home, we may have been exposed many times over to martian microbes. Either they do not interact with our terrestrial life, or in effect our planet has already been inoculated against such alien bugs. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 371 More than any other planet, Mars has inspired science fiction writers over the years. You can find scientifically reasonable stories about Mars in a subject index of such stories online. If you click on Mars (https://openstax.org/l/30MarsStories) as a topic, you will find stories", " by a number of space scientists, including William Hartmann, Geoffrey Landis, and Ludek Pesek. 10.6 DIVERGENT PLANETARY EVOLUTION Learning Objectives By the end of this section, you will be able to: Compare the planetary evolution of Venus, Earth, and Mars Venus, Mars, and our own planet Earth form a remarkably diverse triad of worlds. Although all three orbit in roughly the same inner zone around the Sun and all apparently started with about the same chemical mix of silicates and metals, their evolutionary paths have diverged. As a result, Venus became hot and dry, Mars became cold and dry, and only Earth ended up with what we consider a hospitable climate. We have discussed the runaway greenhouse effect on Venus and the runaway refrigerator effect on Mars, but we do not understand exactly what started these two planets down these separate evolutionary paths. Was Earth ever in danger of a similar fate? Or might it still be diverted onto one of these paths, perhaps due to stress on the atmosphere generated by human pollutants? One of the reasons for studying Venus and Mars is to seek insight into these questions. Some people have even suggested that if we understood the evolution of Mars and Venus better, we could possibly reverse their evolution and restore more earthlike environments. While it seems unlikely that humans could ever make either Mars or Venus into a replica of Earth, considering such possibilities is a useful part of our more general quest to understand the delicate environmental balance that distinguishes our planet from its two neighbors. In Cosmic Samples and the Origin of the Solar System, we return to the comparative study of the terrestrial planets and their divergent evolutionary histories. 372 Chapter 10 Earthlike Planets: Venus and Mars CHAPTER 10 REVIEW KEY TERMS runaway greenhouse effect the process by which the greenhouse effect, rather than remaining stable or being lessened through intervention, continues to grow at an increasing rate tectonic geological features that result from stresses and pressures in the crust of a planet; tectonic forces can lead to earthquakes and motion of the crust SUMMARY 10.1 The Nearest Planets: An Overview Venus, the nearest planet, is a great disappointment through the telescope because of its impenetrable cloud cover. Mars is more tantalizing, with dark markings and polar caps. Early in the twentieth century, it was widely believed that the \u201ccanals\u201d of Mars indicated intelligent life there. Mars has only 11% the mass of Earth, but Venus is nearly our twin in size and mass. Mars rotates in 24 hours and has seasons like Earth; Venus has a retrograde rotation period of 243 days. Both planets have been extensively explored by spacecraft. 10.2 The Geology of Venus Venus has been mapped by radar, especially with the Magellan spacecraft. Its crust consists of 75% lowland lava plains, numerous volcanic features, and many large coronae, which are the expression of subsurface volcanism. The planet has been modified by widespread tectonics driven by mantle convection, forming complex patterns of ridges and cracks and building high continental regions such as Ishtar. The surface is extraordinarily inhospitable, with pressure of 90 bars and temperature of 730 K, but several Russian Venera landers investigated it successfully. 10.3 The Massive Atmosphere of Venus The atmosphere of Venus is 96% CO2. Thick clouds at altitudes of 30 to 60 kilometers are made of sulfuric acid, and a CO2 greenhouse effect maintains the high surface temperature. Venus presumably reached its current state from more earthlike initial conditions as a result of a runaway greenhouse effect, which included the loss of large quantities of water. 10.4 The Geology of Mars Most of what we know about Mars is derived from spacecraft: highly successful orbiters, landers, and rovers. We have also been able to study a few martian rocks that reached Earth as meteorites. Mars has heavily cratered highlands in its southern hemisphere, but younger, lower volcanic plains over much of its northern half. The Tharsis bulge, as big as North America, includes several huge volcanoes; Olympus Mons is more than 20 kilometers high and 500 kilometers in diameter. The Valles Marineris canyons are tectonic features widened by erosion. Early landers revealed only barren, windswept plains, but later missions have visited places with more geological (and scenic) variety. Landing sites have been selected in part to search for evidence of past water. 10.5 Water and Life on Mars The martian atmosphere has a surface pressure of less than 0.01 bar and is 95% CO2. It has dust clouds, water clouds, and carbon dioxide (dry ice) clouds. Liquid water on the surface is not possible today, but there is subsurface permafrost at high latitudes. Seasonal polar caps are made of dry ice; the northern residual cap This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter", " 10 Earthlike Planets: Venus and Mars 373 is water ice, whereas the southern permanent ice cap is made predominantly of water ice with a covering of carbon dioxide ice. Evidence of a very different climate in the past is found in water erosion features: both runoff channels and outflow channels, the latter carved by catastrophic floods. Our rovers, exploring ancient lakebeds and places where sedimentary rock has formed, have found evidence for extensive surface water in the past. Even more exciting are the gullies that seem to show the presence of flowing salty water on the surface today, hinting at near-surface aquifers. The Viking landers searched for martian life in 1976, with negative results, but life might have flourished long ago. We have found evidence of water on Mars, but following the water has not yet led us to life on that planet. 10.6 Divergent Planetary Evolution Earth, Venus, and Mars have diverged in their evolution from what may have been similar beginnings. We need to understand why if we are to protect the environment of Earth. FOR FURTHER EXPLORATION Articles Venus Dorminey, B. \u201cCool Science on a Hot World.\u201d Astronomy (February 2006): 46. Five-page overview of Venus and the Venus Express mission plans. Kargel, J. \u201cRivers of Venus.\u201d Sky & Telescope (August 1997): 32. On lava channels. Robertson, D. \u201cParched Planet.\u201d Sky & Telescope (April 2008): 26. Overview of our understanding of the planet. Robinson, C. \u201cMagellan Reveals Venus.\u201d Astronomy (February 1995): 32. Stofan, E. \u201cThe New Face of Venus.\u201d Sky & Telescope (August 1993): 22. Zimmerman, R. \u201cTaking Venus by Storm.\u201d Astronomy (October 2008): 66. On results from the Venus Express mission. Mars Albee, A. \u201cThe Unearthly Landscapes of Mars.\u201d Scientific American (June 2003): 44. Results from the Mars Global Surveyor and Mars Odyssey missions and an overview. Bell, J. \u201cA Fresh Look at Mars.\u201d Astronomy (August 2015): 28. Nice summary of recent spacecraft results and how they are revising our understanding of Mars. Bell, J. \u201cUncovering Mars\u2019 Secret Past.\u201d Sky & Telescope (July 2009): 22. How rovers and orbiters are helping us to understand Mars history and the role of water. Bell, J. \u201cThe Red Planet\u2019s Watery Past.\u201d Scientific American (December 2006): 62. Rovers are furnishing proof that ancient Mars was wet. Burnham, R. \u201cRed Planet Rendezvous.\u201d Astronomy (May 2006): 68. About Mariner Valley and a flyover film constructed from many still images. Christensen, P. \u201cThe Many Faces of Mars.\u201d Scientific American (July 2005): 32. Results from the Rover mission; evidence that Mars was once wet in places. Lakdawalla, E. \u201cThe History of Water on Mars.\u201d Sky & Telescope (September 2013): 16. Clear review of our current understanding of the role of water on Mars in different epochs. 374 Chapter 10 Earthlike Planets: Venus and Mars Malin, M. \u201cVisions of Mars.\u201d Sky & Telescope (April 1999): 42. A geological tour of the red planet, with new Mars Global Surveyor images. McEwen, A. \u201cMars in Motion.\u201d Scientific American (May 2013): 58. On gullies and other surface changes. McKay, C. & Garcia, V. \u201cHow to Search for Life on Mars.\u201d Scientific American (June 2014): 44. Experiments future probes could perform. Naeye, R. \u201cEurope\u2019s Eye on Mars.\u201d Sky & Telescope (December 2005): 30. On the Mars Express mission and the remarkable close-up images it is sending. Talcott, R. \u201cSeeking Ground Truth on Mars.\u201d Astronomy (October 2009): 34. How rovers and orbiters are helping scientists understand the red planet\u2019s surface. Websites European Space Agency Mars Express Page: http://www.esa.int/Our_Activities/Space_Science/Mars_Express. European Space Agency Venus Express Page: http://www.esa.int/Our_Activities/Space_Science/Venus_Express. High Resolution Imaging Science Experiment: http://hirise.lpl.arizona.edu/. Jet Propulsion Lab Mars Exploration Page: http://mars.jpl.nasa.gov/. Mars Globe HD app: https://itunes.apple.com/us/app/mars-globe-hd/id376020224?mt=8. Mars", " Rover 360\u00b0 Panorama: http://www.360cities.net/image/curiosity-rover-martian-solar- day-2#171.10,26.50,70.0. Interactive. NASA Center for Mars Exploration: http://www.nasa.gov/mission_pages/mars/main/index.html. NASA Solar System Exploration Mars Page: http://solarsystem.nasa.gov/planets/mars. NASA Solar System Exploration Venus Page: http://solarsystem.nasa.gov/planets/venus. NASA\u2019s apps about Mars for phones and tablets can be found at: http://mars.nasa.gov/mobile/info/. NASA\u2019s Magellan Mission to Venus: http://www2.jpl.nasa.gov/magellan/. Russian (Soviet) Venus Missions and Images: http://mentallandscape.com/C_CatalogVenus.htm. Venus Atlas app: https://itunes.apple.com/us/app/venus-atlas/id317310503?mt=8. Venus Express Results Article: http://www.mpg.de/798302/F002_Focus_026-033.pdf. Videos 50 Years of Mars Exploration: http://www.jpl.nasa.gov/video/details.php?id=1395. NASA\u2019s summary of all missions through MAVEN; good quick overview (4:08). Being a Mars Rover: What It\u2019s Like to be an Interplanetary Explorer: https://www.youtube.com/ watch?v=nRpCOEsPD54. 2013 talk by Dr. Lori Fenton about what it\u2019s like on the surface of Mars (1:07:24). Magellan Maps Venus: http://www.bbc.co.uk/science/space/solarsystem/space_missions/ magellan_probe#p005y07s. BBC clip with Dr. Ellen Stofan on the radar images of Venus and what they tell us (3:06). Our Curiosity: https://www.youtube.com/watch?v=XczKXWvokm4. Mars Curiosity rover 2-year anniversary video narrated by Neil deGrasse Tyson and Felicia Day (6:01). Planet Venus: The Deadliest Planet, Venus Surface and Atmosphere: https://www.youtube.com/ watch?v=HqFVxWfVtoo. Quick tour of Venus\u2019 atmosphere and surface (2:04). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 375 Planetary Protection and Hitchhikers in the Solar System: The Danger of Mingling Microbes: https://www.youtube.com/watch?v=6iGC3uO7jBI. 2009 talk by Dr. Margaret Race on preventing contamination between worlds (1:28:50). COLLABORATIVE GROUP ACTIVITIES A. Your group has been asked by high NASA officials to start planning the first human colony on Mars. Begin by making a list of what sorts of things humans would need to bring along to be able to survive for years on the surface of the red planet. B. As a publicity stunt, the mayor of Venus, Texas (there really is such a town), proposes that NASA fund a mission to Venus with humans on board. Clearly, the good mayor neglected to take an astronomy course in college. Have your group assemble a list of as many reasons as possible why it is unlikely that humans will soon land on the surface of Venus. C. Even if humans would have trouble surviving on the surface of Venus, this does not mean we could not learn a lot more about our veiled sister planet. Have your group brainstorm a series of missions (pretend cost is no object) that would provide us with more detailed information about Venus\u2019 atmosphere, surface, and interior. D. Sometime late in the twenty-first century, when travel to Mars has become somewhat routine, a very wealthy couple asks you to plan a honeymoon tour of Mars that includes the most spectacular sights on the red planet. Constitute your group as the Percival Lowell Memorial Tourist Agency, and come up with a list of not-to-be missed tourist stops on Mars. E. In the popular book and film, called The Martian, the drama really begins when our hero is knocked over and loses consciousness as he is half buried by an intense wind storm on Mars. Given what you have learned about Mars\u2019 atmosphere in this chapter, have your group discuss how realistic that scenario is. (By the way, the author of the book has himself genially acknowledged in interviews and talks that this is a reasonable question to ask.) F. Astronomers have been", " puzzled and annoyed about the extensive media publicity that was given the small group of \u201ctrue believers\u201d who claimed the \u201cFace on Mars\u201d was not a natural formation (see the Astronomy and Pseudoscience: The \u201cFace on Mars\u201d feature box). Have your group make a list of the reasons many of the media were so enchanted by this story. What do you think astronomers could or should do to get the skeptical, scientific perspective about such issues before the public? G. Your group is a special committee of scientists set up by the United Nations to specify how any Mars samples should be returned to Earth so that possible martian microbes do not harm Earth life. What precautions would you recommend, starting at Mars and going all the way to the labs that analyze the martian samples back on Earth? H. Have your group brainstorm about Mars in popular culture. How many movies, songs or other music, and products can you think of connected with Mars? What are some reasons that Mars would be a popular theme for filmmakers, songwriters, and product designers? 376 Chapter 10 Earthlike Planets: Venus and Mars EXERCISES Review Questions 1. List several ways that Venus, Earth, and Mars are similar, and several ways they are different. 2. Compare the current atmospheres of Earth, Venus, and Mars in terms of composition, thickness (and pressure at the surface), and the greenhouse effect. 3. How might Venus\u2019 atmosphere have evolved to its present state through a runaway greenhouse effect? 4. Describe the current atmosphere on Mars. What evidence suggests that it must have been different in the past? 5. Explain the runaway refrigerator effect and the role it may have played in the evolution of Mars. 6. What evidence do we have that there was running (liquid) water on Mars in the past? What evidence is there for water coming out of the ground even today? 7. What evidence is there that Venus was volcanically active about 300\u2013600 million years ago? 8. Why is Mars red? 9. What is the composition of clouds on Mars? 10. What is the composition of the polar caps on Mars? 11. Describe two anomalous features of the rotation of Venus and what might account for them. 12. How was the Mars Odyssey spacecraft able to detect water on Mars without landing on it? Thought Questions 13. What are the advantages of using radar imaging rather than ordinary cameras to study the topography of Venus? What are the relative advantages of these two approaches to mapping Earth or Mars? 14. Venus and Earth are nearly the same size and distance from the Sun. What are the main differences in the geology of the two planets? What might be some of the reasons for these differences? 15. Why is there so much more carbon dioxide in the atmosphere of Venus than in that of Earth? Why so much more carbon dioxide than on Mars? 16. If the Viking missions were such a rich source of information about Mars, why have we sent the Pathfinder, Global Surveyor, and other more recent spacecraft to Mars? Make a list of questions about Mars that still puzzle astronomers. 17. Compare Mars with Mercury and the Moon in terms of overall properties. What are the main similarities and differences? 18. Contrast the mountains on Mars and Venus with those on Earth and the Moon. 19. We believe that all of the terrestrial planets had similar histories when it comes to impacts from space. Explain how this idea can be used to date the formation of the martian highlands, the martian basins, and the Tharsis volcanoes. How certain are the ages derived for these features (in other words, how do we check the ages we derive from this method)? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 10 Earthlike Planets: Venus and Mars 377 20. Is it likely that life ever existed on either Venus or Mars? Justify your answer in each case. 21. Suppose that, decades from now, NASA is considering sending astronauts to Mars and Venus. In each case, describe what kind of protective gear they would have to carry, and what their chances for survival would be if their spacesuits ruptured. 22. We believe that Venus, Earth, and Mars all started with a significant supply of water. Explain where that water is now for each planet. 23. One source of information about Mars has been the analysis of meteorites from Mars. Since no samples from Mars have ever been returned to Earth from any of the missions we sent there, how do we know these meteorites are from Mars? What information have they revealed about Mars? 24. The runaway greenhouse effect and its inverse, the runaway refrigerator effect, have led to harsh, uninhabitable conditions on Venus and Mars. Does the greenhouse effect always cause climate changes leading to loss of water and life? Give a reason for your answer. 25. In what way is the high surface temperature of Venus relevant to concerns about global warming", " on Earth today? 26. What is a dust devil? Would you expect to feel more of a breeze from a dust devil on Mars or on Earth? Explain. 27. Near the martian equator, temperatures at the same spot can vary from an average of \u2013135 \u00b0C at night to an average of 30 \u00b0C during the day. How can you explain such a wide difference in temperature compared to that on Earth? Figuring For Yourself 28. Estimate the amount of water there could be in a global (planet-wide) region of subsurface permafrost on Mars (do the calculations for two permafrost thicknesses, 1 and 10 km, and a concentration of ice in the permafrost of 10% by volume). Compare the two results you get with the amount of water in Earth\u2019s oceans calculated in Example 10.1. 29. At its nearest, Venus comes within about 41 million km of Earth. How distant is it at its farthest? 30. If you weigh 150 lbs. on the surface of Earth, how much would you weigh on Venus? On Mars? 31. Calculate the relative land area\u2014that is, the amount of the surface not covered by liquids\u2014of Earth, the Moon, Venus, and Mars. (Assume that 70% of Earth is covered with water.) 32. The closest approach distance between Mars and Earth is about 56 million km. Assume you can travel in a spaceship at 58,000 km/h, which is the speed achieved by the New Horizons space probe that went to Pluto and is the fastest speed so far of any space vehicle launched from Earth. How long would it take to get to Mars at the time of closest approach? 378 Chapter 10 Earthlike Planets: Venus and Mars This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 379 11 THE GIANT PLANETS Figure 11.1. Giant Planets. The four giant planets in our solar system all have hydrogen atmospheres, but the warm gas giants, Jupiter and Saturn, have tan, beige, red, and white clouds that are thought to be composed of ammonia ice particles with various colorants called \u201cchromophores.\u201d The blue-tinted ice giants, Uranus and Neptune, are much colder and covered in methane ice clouds. (credit: modification of work by Lunar and Planetary Institute, NASA) Chapter Outline 11.1 Exploring the Outer Planets 11.2 The Giant Planets 11.3 Atmospheres of the Giant Planets Thinking Ahead \u201cWhat do we learn about the Earth by studying the planets? Humility.\u201d\u2014Andrew Ingersoll discussing the results of the Voyager mission in 1986. Beyond Mars and the asteroid belt, we encounter a new region of the solar system: the realm of the giants. Temperatures here are lower, permitting water and other volatiles to condense as ice. The planets are much larger, distances between them are much greater, and each giant world is accompanied by an extensive system of moons and rings. From many perspectives, the outer solar system is where the action is, and the giant planets are the most important members of the Sun\u2019s family. When compared to these outer giants, the little cinders of rock and metal that orbit closer to the Sun can seem insignificant. These four giant worlds\u2014Jupiter, Saturn, Uranus, Neptune\u2014are the subjects of this chapter. Their rings, moons, and the dwarf planet Pluto are discussed in a later chapter. 11.1 EXPLORING THE OUTER PLANETS Learning Objectives By the end of this section, you will be able to: 380 Chapter 11 The Giant Planets Provide an overview of the composition of the giant planets Chronicle the robotic exploration of the outer solar system Summarize the missions sent to orbit the gas giants The giant planets hold most of the mass in our planetary system. Jupiter alone exceeds the mass of all the other planets combined (Figure 11.2). The material available to build these planets can be divided into three classes by what they are made of: \u201cgases,\u201d \u201cices,\u201d and \u201crocks\u201d (see Table 11.1). The \u201cgases\u201d are primarily hydrogen and helium, the most abundant elements in the universe. The way it is used here, the term \u201cices\u201d refers to composition only and not whether a substance is actually in a solid state. \u201cIces\u201d means compounds that form from the next most abundant elements: oxygen, carbon, and nitrogen. Common ices are water, methane, and ammonia, but ices may also include carbon monoxide, carbon dioxide, and others. \u201cRocks\u201d are even less abundant than ices, and include everything else: magnesium, silicon, iron, and so on. Figure 11.2. Jupiter. The Cassini spacecraft imaged Jupiter on", " its way to Saturn in 2012. The giant storm system called the Great Red Spot is visible to the lower right. The dark spot to the lower left is the shadow of Jupiter\u2019s moon Europa. (credit: modification of work by NASA/JPL) Abundances in the Outer Solar System Type of Material Name Approximate % (by Mass) Gas Gas Ice Ice Ice Hydrogen (H2) Helium (He) Water (H2O) Methane (CH4) Ammonia (NH3) Rock Magnesium (Mg), iron (Fe), silicon (Si) Table 11.1 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 75 24 0.6 0.4 0.1 0.3 Chapter 11 The Giant Planets 381 In the outer solar system, gases dominate the two largest planets, Jupiter and Saturn, hence their nickname \u201cgas giants.\u201d Uranus and Neptune are called \u201cice giants\u201d because their interiors contain far more of the \u201cice\u201d component than their larger cousins. The chemistry for all four giant planet atmospheres is dominated by hydrogen. This hydrogen caused the chemistry of the outer solar system to become reducing, meaning that other elements tend to combine with hydrogen first. In the early solar system, most of the oxygen combined with hydrogen to make H2O and was thus unavailable to form the kinds of oxidized compounds with other elements that are more familiar to us in the inner solar system (such as CO2). As a result, the compounds detected in the atmosphere of the giant planets are mostly hydrogen-based gases such as methane (CH4) and ammonia (NH3), or more complex hydrocarbons (combinations of hydrogen and carbon) such as ethane (C2H6) and acetylene (C2H2). Exploration of the Outer Solar System So Far Eight spacecraft, seven from the United States and one from Europe, have penetrated beyond the asteroid belt into the realm of the giants. Table 11.2 summarizes the spacecraft missions to the outer solar system. Missions to the Giant Planets Planet Spacecraft[1] Encounter Date Type Jupiter Pioneer 10 December 1973 Pioneer 11 December 1974 Voyager 1 March 1979 Voyager 2 July 1979 Flyby Flyby Flyby Flyby Ulysses February 1992 Flyby during gravity assist Galileo December 1995 Orbiter and probe Cassini December 2002 Flyby New Horizons February 2007 Flyby during gravity assist Juno July 2016 Saturn Pioneer 11 September 1979 Voyager 1 November 1980 Voyager 2 August 1981 Orbiter Flyby Flyby Flyby Cassini July 2004 (Saturn orbit injection 2000) Orbiter Table 11.2 1 Both the Ulysses and the New Horizons spacecraft (designed to study the Sun and Pluto, respectively) flew past Jupiter for a gravity boost (gaining energy by \u201cstealing\u201d a little bit from the giant planet\u2019s rotation). 382 Chapter 11 The Giant Planets Missions to the Giant Planets Planet Spacecraft Encounter Date Type Uranus Voyager 2 January 1986 Neptune Voyager 2 August 1989 Flyby Flyby Table 11.2 The challenges of exploring so far away from Earth are considerable. Flight times to the giant planets are measured in years to decades, rather than the months required to reach Venus or Mars. Even at the speed of light, messages take hours to pass between Earth and the spacecraft. If a problem develops near Saturn, for example, a wait of hours for the alarm to reach Earth and for instructions to be routed back to the spacecraft could spell disaster. Spacecraft to the outer solar system must therefore be highly reliable and capable of a greater degree of independence and autonomy. Outer solar system missions also must carry their own power sources since the Sun is too far away to provide enough energy. Heaters are required to keep instruments at proper operating temperatures, and spacecraft must have radio transmitters powerful enough to send their data to receivers on distant Earth. The first spacecraft to investigate the regions past Mars were the NASA Pioneers 10 and 11, launched in 1972 and 1973 as pathfinders to Jupiter. One of their main objectives was simply to determine whether a spacecraft could actually navigate through the belt of asteroids that lies beyond Mars without getting destroyed by collisions with asteroidal dust. Another objective was to measure the radiation hazards in the magnetosphere (or zone of magnetic influence) of Jupiter. Both spacecraft passed through the asteroid belt without incident, but the energetic particles in Jupiter\u2019s magnetic field nearly wiped out their electronics, providing information necessary for the safe design of subsequent missions. Pioneer 10 flew past Jupiter in 1973, after which it sped outward toward the limits of the solar system. Pioneer 11 undertook a more ambitious program, using the gravity of Jupiter to aim for Saturn, which it reached in 1979. The twin Voyager spacecraft launched the next wave of outer planet exploration in 1977. Voyagers 1 and 2 each carried 11 scientific instruments, including cameras and spectrometers, as well as devices to measure the characteristics of planetary magnetospheres. Since", " they kept going outward after their planetary encounters, these are now the most distant spacecraft ever launched by humanity. Voyager 1 reached Jupiter in 1979 and used a gravity assist from that planet to take it on to Saturn in 1980. Voyager 2 arrived at Jupiter four months later, but then followed a different path to visit all the outer planets, reaching Saturn in 1981, Uranus in 1986, and Neptune in 1989. This trajectory was made possible by the approximate alignment of the four giant planets on the same side of the Sun. About once every 175 years, these planets are in such a position, and it allows a single spacecraft to visit them all by using gravity-assisted flybys to adjust course for each subsequent encounter; such a maneuver has been nicknamed a \u201cGrand Tour\u201d by astronomers. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 383 The Jet Propulsion Laboratory has a nice video called Voyager: The Grand Tour (https://openstax.org/ l/30JPLGrandT) that describes the Voyager mission and what it found Engineering and Space Science: Teaching an Old Spacecraft New Tricks By the time Voyager 2 arrived at Neptune in 1989, 12 years after its launch, the spacecraft was beginning to show signs of old age. The arm on which the camera and other instruments were located was \u201carthritic\u201d: it could no longer move easily in all directions. The communications system was \u201chard of hearing\u201d: part of its radio receiver had stopped working. The \u201cbrains\u201d had significant \u201cmemory loss\u201d: some of the onboard computer memory had failed. And the whole spacecraft was beginning to run out of energy: its generators had begun showing serious signs of wear. To make things even more of a challenge, Voyager\u2019s mission at Neptune was in many ways the most difficult of all four flybys. For example, since sunlight at Neptune is 900 times weaker than at Earth, the onboard camera had to take much longer exposures in this light-starved environment. This was a nontrivial requirement, given that the spacecraft was hurtling by Neptune at ten times the speed of a rifle bullet. The solution was to swivel the camera backward at exactly the rate that would compensate for the forward motion of the spacecraft. Engineers had to preprogram the ship\u2019s computer to execute an incredibly complex series of maneuvers for each image. The beautiful Voyager images of Neptune are a testament to the ingenuity of spacecraft engineers. The sheer distance of the craft from its controllers on Earth was yet another challenge. Voyager 2 received instructions and sent back its data via on-board radio transmitter. The distance from Earth to Neptune is about 4.8 billion kilometers. Over this vast distance, the power that reached us from Voyager 2 at Neptune was approximately10\u201316 watts, or 20 billion times less power than it takes to operate a digital watch. Thirty-eight different antennas on four continents were used by NASA to collect the faint signals from the spacecraft and decode the precious information about Neptune that they contained. Enter the Orbiters: Galileo and Cassini The Pioneer and Voyager missions were flybys of the giant planets: they each produced only quick looks before the spacecraft sped onward. For more detailed studies of these worlds, we require spacecraft that can go into orbit around a planet. For Jupiter and Saturn, these orbiters were the Galileo and Cassini spacecraft, respectively. To date, no orbiter missions have been started for Uranus and Neptune, although planetary scientists have expressed keen interest. The Galileo spacecraft was launched toward Jupiter in 1989 and arrived in 1995. Galileo began its investigations by deploying an entry probe into Jupiter, for the first direct studies of the planet\u2019s outer atmospheric layers. 384 Chapter 11 The Giant Planets The probe plunged at a shallow angle into Jupiter\u2019s atmosphere, traveling at a speed of 50 kilometers per second\u2014that\u2019s fast enough to fly from New York to San Francisco in 100 seconds! This was the highest speed at which any probe has so far entered the atmosphere of a planet, and it put great demands on the heat shield protecting it. The high entry speed was a result of acceleration by the strong gravitational attraction of Jupiter. Atmospheric friction slowed the probe within 2 minutes, producing temperatures at the front of its heat shield as high as 15,000 \u00b0C. As the probe\u2019s speed dropped to 2500 kilometers per hour, the remains of the glowing heat shield were jettisoned, and a parachute was deployed to lower the instrumented probe spacecraft more gently into the atmosphere (Figure 11.3). The data from the probe instruments were relayed to Earth via the main Galileo spacecraft. Figure 11.3. Galileo Probe Falling into Jupiter. This artist\u2019s depiction shows the Galileo probe descending into the clouds via parachute just after the protective heat shield separated. The probe made its measurements of Jupiter\u2019s atmosphere on December 7, 1995. (credit: modification of work by NASA", "/Ames Research Center) The probe continued to operate for an hour, descending 200 kilometers into the atmosphere. A few minutes later the polyester parachute melted, and within a few hours the main aluminum and titanium structure of the probe vaporized to become a part of Jupiter itself. About 2 hours after receipt of the final probe data, the main spacecraft fired its retro-rockets so it could be captured into orbit around the planet, where its primary objectives were to study Jupiter\u2019s large and often puzzling moons. The Cassini mission to Saturn (Figure 11.4), a cooperative venture between NASA and the European Space Agency, was similar to Galileo in its two-fold approach. Launched in 1997, Cassini arrived in 2004 and went into orbit around Saturn, beginning extensive studies of its rings and moons, as well as the planet itself. In January 2005, Cassini deployed an entry probe into the atmosphere of Saturn\u2019s large moon, Titan, where it successfully landed on the surface. (We\u2019ll discuss the probe and what it found in the chapter on Rings, Moons, and Pluto.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 385 Figure 11.4. Earth as Seen from Saturn. This popular Cassini image shows Earth as a tiny dot (marked with an arrow) seen below Saturn\u2019s rings. It was taken in July 2013, when Saturn was 1.4 billion kilometers from Earth. (credit: modification of work by NASA/JPL-Caltech/Space Science Institute) 11.2 THE GIANT PLANETS Learning Objectives By the end of this section, you will be able to: Describe the basic physical characteristics, general appearance, and rotation of the giant planets Describe the composition and structure of Jupiter, Saturn, Uranus, and Neptune Compare and contrast the internal heat sources of the giant planets Describe the discovery and characteristics of the giant planets\u2019 magnetic fields Let us now examine the four giant (or jovian) planets in more detail. Our approach is not just to catalog their characteristics, but to compare them with each other, noting their similarities and differences and attempting to relate their properties to their differing masses and distances from the Sun. Basic Characteristics The giant planets are very far from the Sun. Jupiter is more than five times farther from the Sun than Earth\u2019s distance (5 AU), and takes just under 12 years to circle the Sun. Saturn is about twice as far away as Jupiter (almost 10 AU) and takes nearly 30 years to complete one orbit. Uranus orbits at 19 AU with a period of 84 years, while Neptune, at 30 AU, requires 165 years for each circuit of the Sun. These long timescales make it difficult for us short-lived humans to study seasonal change on the outer planets. Jupiter and Saturn have many similarities in composition and internal structure, although Jupiter is nearly four times more massive. Uranus and Neptune are smaller and differ in composition and internal structure from their large siblings. Some of the main properties of these four planets are summarized in Table 11.3. 386 Chapter 11 The Giant Planets Basic Properties of the Jovian Planets Planet Distance (AU) Period (years) Diameter (km) Mass (Earth = 1) Density (g/cm3) Rotation (hours) Jupiter Saturn 5.2 9.5 Uranus 19.2 11.9 29.5 84.1 142,800 318 120,540 51,200 95 14 17 1.3 0.7 1.3 1.6 9.9 10.7 17.2 16.1 Neptune 30.0 164.8 49,500 Table 11.3 Jupiter, the giant among giants, has enough mass to make 318 Earths. Its diameter is about 11 times that of Earth (and about one tenth that of the Sun). Jupiter\u2019s average density is 1.3 g/cm3, much lower than that of any of the terrestrial planets. (Recall that water has a density of 1 g/cm3.) Jupiter\u2019s material is spread out over a volume so large that more than 1400 Earths could fit within it. Saturn\u2019s mass is 95 times that of Earth, and its average density is only 0.7 g/cm3\u2014the lowest of any planet. Since this is less than the density of water, Saturn would be light enough to float. Uranus and Neptune each have a mass about 15 times that of Earth and, hence, are only 5% as massive as Jupiter. Their densities of 1.3 g/cm3 and 1.6 g/cm3, respectively, are much higher than that of Saturn. This is one piece of evidence that tells us that their composition must differ fundamentally from the gas giants. When astronomers began to discover other planetary systems (exoplanets), we found that planets the size of Uranus and Neptune are common,", " and that there are even more exoplanets intermediate in size between Earth and these ice giants, a type of planet not found in our solar system. Appearance and Rotation When we look at the planets, we see only their atmospheres, composed primarily of hydrogen and helium gas (see Figure 11.1). The uppermost clouds of Jupiter and Saturn, the part we see when looking down at these planets from above, are composed of ammonia crystals. On Neptune, the upper clouds are made of methane. On Uranus, we see no obvious cloud layer at all, but only a deep and featureless haze. Seen through a telescope, Jupiter is a colorful and dynamic planet. Distinct details in its cloud patterns allow us to determine the rotation rate of its atmosphere at the cloud level, although such atmosphere rotation may have little to do with the spin of the underlying planet. Much more fundamental is the rotation of the mantle and core; these can be determined by periodic variations in radio waves coming from Jupiter, which are controlled by its magnetic field. Since the magnetic field (which we will discuss below) originates deep inside the planet, it shares the rotation of the interior. The rotation period we measure in this way is 9 hours 56 minutes, which gives Jupiter the shortest \u201cday\u201d of any planet. In the same way, we can measure that the underlying rotation period of Saturn is 10 hours 40 minutes. Uranus and Neptune have slightly longer rotation periods of about 17 hours, also determined from the rotation of their magnetic fields. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 387 brief video made from Hubble Space Telescope photos shows the rotation of Jupiter (https://openstax.org/l/30HSTJupRot) with its many atmospheric features. Remember that Earth and Mars have seasons because their spin axes, instead of \u201cstanding up straight,\u201d are tilted relative to the orbital plane of the solar system. This means that as Earth revolves around the Sun, sometimes one hemisphere and sometimes the other \u201cleans into\u201d the Sun. What are the seasons like for the giant planets? The spin axis of Jupiter is tilted by only 3\u00b0, so there are no seasons to speak of. Saturn, however, does have seasons, since its spin axis is inclined at 27\u00b0 to the perpendicular to its orbit. Neptune has about the same tilt as Saturn (29\u00b0); therefore, it experiences similar seasons (only more slowly). The strangest seasons of all are on Uranus, which has a spin axis tilted by 98\u00b0 with respect to the north direction. Practically speaking, we can say that Uranus orbits on its side, and its ring and moon system follow along, orbiting about Uranus\u2019 equator (Figure 11.5). Figure 11.5. Infrared Image of Uranus. The infrared camera on the Hubble Space Telescope took these false-color images of the planet Uranus, its ring system, and moons in 1997. The south pole of the planet (marked with a \u201c+\u201d on the right image) faces the Sun; its green color shows a strong local haze. The two images were taken 90 minutes apart, and during that time the five reddish clouds can be seen to rotate around the parallel to the equator. The rings (which are very faint in the visible light, but prominent in infrared) and eight moons can be seen around the equator. This was the \u201cbull\u2019s eye\u201d arrangement that Voyager saw as it approached Uranus in 1986. (credit: modification of work by Erich Karkoschka (University of Arizona), and NASA/ESA) We don\u2019t know what caused Uranus to be tipped over like this, but one possibility is a collision with a large planetary body when our system was first forming. Whatever the cause, this unusual tilt creates dramatic seasons. When Voyager 2 arrived at Uranus, its south pole was facing directly into the Sun. The southern hemisphere was experiencing a 21-year sunlit summer, while during that same period the northern hemisphere was plunged into darkness. For the next 21-year season, the Sun shines on Uranus\u2019 equator, and both hemispheres go through cycles of light and dark as the planet rotates (Figure 11.6). Then there are 21 years of an illuminated northern hemisphere and a dark southern hemisphere. After that the pattern of alternating day and night repeats. 388 Chapter 11 The Giant Planets Just as on Earth, the seasons are even more extreme at the poles. If you were to install a floating platform at the south pole of Uranus, for example, it would experience 42 years of light and 42 years of darkness. Any future astronauts crazy enough to set up camp there could spend most of their lives without ever seeing the Sun. Figure 11.6. Strange Seasons on Uranus. (a) This diagram shows the orbit of Uranus as seen from above", ". At the time Voyager 2 arrived (position 1), the South Pole was facing the Sun. As we move counterclockwise in the diagram, we see the planet 21 years later at each step. (b) This graph compares the amount of sunlight seen at the poles and the equator of Uranus over the course of its 84-year revolution around the Sun. Composition and Structure Although we cannot see into these planets, astronomers are confident that the interiors of Jupiter and Saturn are composed primarily of hydrogen and helium. Of course, these gases have been measured only in their atmosphere, but calculations first carried out more than 50 years ago showed that these two light gases are the only possible materials out of which a planet with the observed masses and densities of Jupiter and Saturn could be constructed. The deep internal structures of these two planets are difficult to predict. This is mainly because these planets are so big that the hydrogen and helium in their centers become tremendously compressed and behave in ways that these gases can never behave on Earth. The best theoretical models we have of Jupiter\u2019s structure predict a central pressure greater than 100 million bars and a central density of about 31 g/cm3. (By contrast, Earth\u2019s core has a central pressure of 4 million bars and a central density of 17 g/cm3.) At the pressures inside the giant planets, familiar materials can take on strange forms. A few thousand kilometers below the visible clouds of Jupiter and Saturn, pressures become so great that hydrogen changes from a gaseous to a liquid state. Still deeper, this liquid hydrogen is further compressed and begins to act like a metal, something it never does on Earth. (In a metal, electrons are not firmly attached to their parent nuclei but can wander around. This is why metals are such good conductors of electricity.) On Jupiter, the greater part of the interior is liquid metallic hydrogen. Because Saturn is less massive, it has only a small volume of metallic hydrogen, but most of its interior is liquid. Uranus and Neptune are too small to reach internal pressures sufficient to liquefy hydrogen. We will return to the discussion of the metallic hydrogen layers when we examine the magnetic fields of the giant planets. Each of these planets has a core composed of heavier materials, as demonstrated by detailed analyses of their gravitational fields. Presumably these cores are the original rock-and-ice bodies that formed before the capture of gas from the surrounding nebula. The cores exist at pressures of tens of millions of bars. While scientists speak of the giant planet cores being composed of rock and ice, we can be sure that neither rock nor ice This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 389 assumes any familiar forms at such pressures and temperatures. Remember that what is really meant by \u201crock\u201d is any material made up primarily of iron, silicon, and oxygen, while the term \u201cice\u201d in this chapter denotes materials composed primarily of the elements carbon, nitrogen, and oxygen in combination with hydrogen. Figure 11.7 illustrates the likely interior structures of the four jovian planets. It appears that all four have similar cores of rock and ice. On Jupiter and Saturn, the cores constitute only a few percent of the total mass, consistent with the initial composition of raw materials shown in Table 11.1. However, most of the mass of Uranus and Neptune resides in these cores, demonstrating that the two outer planets were unable to attract massive quantities of hydrogen and helium when they were first forming. Figure 11.7. Internal Structures of the Jovian Planets. Jupiter and Saturn are composed primarily of hydrogen and helium (but hydrogen dominates), but Uranus and Neptune consist in large part of compounds of carbon, nitrogen, and oxygen. (The diagrams are drawn to scale; numbers show radii in thousands of kilometers.) Internal Heat Sources Because of their large sizes, all the giant planets were strongly heated during their formation by the collapse of surrounding material onto their cores. Jupiter, being the largest, was the hottest. Some of this primordial heat can still remain inside such large planets. In addition, it is possible for giant, largely gaseous planets to generate heat after formation by slowly contracting. (With so large a mass, even a minuscule amount of shrinking can generate significant heat.) The effect of these internal energy sources is to raise the temperatures in the interiors and atmospheres of the planets higher than we would expect from the heating effect of the Sun alone. Jupiter has the largest internal energy source, amounting to 4 \u00d7 1017 watts; that is, it is heated from inside with energy equivalent to 4 million billion 100-watt lightbulbs. This energy is about the same as the total solar energy absorbed by Jupiter. The atmosphere of Jupiter is therefore something of a cross between a normal planetary atmosphere (like Earth\u2019s), which obtains most of its energy from the Sun, and the atmosphere of a star,", " which is entirely heated by an internal energy source. Most of the internal energy of Jupiter is primordial heat, left over from the formation of the planet 4.5 billon years ago. Saturn has an internal energy source about half as large as that of Jupiter, which means (since its mass is only about one quarter as great) that it is producing twice as much energy per kilogram of material as does Jupiter. Since Saturn is expected to have much less primordial heat, there must be another source at work generating most of this 2 \u00d7 1017 watts of power. This source is the separation of helium from hydrogen in 390 Chapter 11 The Giant Planets Saturn\u2019s interior. In the liquid hydrogen mantle, the heavier helium forms droplets that sink toward the core, releasing gravitational energy. In effect, Saturn is still differentiating\u2014letting lighter material rise and heavier material fall. Uranus and Neptune are different. Neptune has a small internal energy source, while Uranus does not emit a measurable amount of internal heat. As a result, these two planets have almost the same atmospheric temperature, in spite of Neptune\u2019s greater distance from the Sun. No one knows why these two planets differ in their internal heat, but all this shows how nature can contrive to make each world a little bit different from its neighbors. Magnetic Fields Each of the giant planets has a strong magnetic field, generated by electric currents in its rapidly spinning interior. Associated with the magnetic fields are the planets\u2019 magnetospheres, which are regions around the planet within which the planet\u2019s own magnetic field dominates over the general interplanetary magnetic field. The magnetospheres of these planets are their largest features, extending millions of kilometers into space. In the late 1950s, astronomers discovered that Jupiter was a source of radio waves that got more intense at longer rather than at shorter wavelengths\u2014just the reverse of what is expected from thermal radiation (radiation caused by the normal vibrations of particles within all matter). Such behavior is typical, however, of the radiation emitted when high-speed electrons are accelerated by a magnetic field. We call this synchrotron radiation because it was first observed on Earth in particle accelerators, called synchrotrons. This was our first hint that Jupiter must have a strong magnetic field. Later observations showed that the radio waves are coming from a region surrounding Jupiter with a diameter several times that of the planet itself (Figure 11.8). The evidence suggested that a vast number of charged atomic particles must be circulating around Jupiter, spiraling around the lines of force of a magnetic field associated with the planet. This is just what we observe happening, but on a smaller scale, in the Van Allen belt around Earth. The magnetic fields of Saturn, Uranus, and Neptune, discovered by the spacecraft that first passed close to these planets, work in a similar way, but are not as strong. Figure 11.8. Jupiter in Radio Waves. This false-color image of Jupiter was made with the Very Large Array (of radio telescopes) in New Mexico. We see part of the magnetosphere, brightest in the middle because the largest number of charged particles are in the equatorial zone of Jupiter. The planet itself is slightly smaller than the green oval in the center. Different colors are used to indicate different intensities of synchrotron radiation. (credit: modification of work by I. de Pater (UC Berkeley) NRAO, AUI, NSF) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 391 Learn more about the magnetosphere of Jupiter (https://openstax.org/l/30NASAJupMag) and why we continue to be interested in it from this brief NASA video. Inside each magnetosphere, charged particles spiral around in the magnetic field; as a result, they can be accelerated to high energies. These charged particles can come from the Sun or from the neighborhood of the planet itself. In Jupiter\u2019s case, Io, one of its moons, turns out to have volcanic eruptions that blast charged particles into space and right into the jovian magnetosphere. The axis of Jupiter\u2019s magnetic field (the line that connects the magnetic north pole with the magnetic south pole) is not aligned exactly with the axis of rotation of the planet; rather, it is tipped by about 10\u00b0. Uranus and Neptune have even greater magnetic tilts, of 60\u00b0 and 55\u00b0, respectively. Saturn\u2019s field, on the other hand, is perfectly aligned with its rotation axis. Why different planets have such different magnetic tilts is not well understood. The physical processes around the jovian planets turn out to be milder versions of what astronomers find in many distant objects, from the remnants of dead stars to the puzzling distant powerhouses we call quasars. One reason to study the magnetospheres of the giant planets and Earth is that they provide nearby accessible analogues of", " more energetic and challenging cosmic processes. 11.3 ATMOSPHERES OF THE GIANT PLANETS Learning Objectives By the end of this section, you will be able to: Discuss the atmospheric composition of the giant planets Describe the cloud formation and atmospheric structure of the gas giants Characterize the giant planets\u2019 wind and weather patterns Understand the scale and longevity of storms on the giant planets The atmospheres of the jovian planets are the parts we can observe or measure directly. Since these planets have no solid surfaces, their atmospheres are more representative of their general compositions than is the case with the terrestrial planets. These atmospheres also present us with some of the most dramatic examples of weather patterns in the solar system. As we will see, storms on these planets can grow bigger than the entire planet Earth. Atmospheric Composition When sunlight reflects from the atmospheres of the giant planets, the atmospheric gases leave their \u201cfingerprints\u201d in the spectrum of light. Spectroscopic observations of the jovian planets began in the nineteenth century, but for a long time, astronomers were not able to interpret the spectra they observed. As late as the 1930s, the most prominent features photographed in these spectra remained unidentified. Then better spectra revealed the presence of molecules of methane (CH4) and ammonia (NH3) in the atmospheres of Jupiter and Saturn. 392 Chapter 11 The Giant Planets At first astronomers thought that methane and ammonia might be the main constituents of these atmospheres, but now we know that hydrogen and helium are actually the dominant gases. The confusion arose because neither hydrogen nor helium possesses easily detected spectral features in the visible spectrum. It was not until the Voyager spacecraft measured the far-infrared spectra of Jupiter and Saturn that a reliable abundance for the elusive helium could be found. The compositions of the two atmospheres are generally similar, except that on Saturn there is less helium as the result of the precipitation of helium that contributes to Saturn\u2019s internal energy source. The most precise measurements of composition were made on Jupiter by the Galileo entry probe in 1995; as a result, we know the abundances of some elements in the jovian atmosphere even better than we know those in the Sun James Van Allen: Several Planets under His Belt The career of physicist James Van Allen spanned the birth and growth of the space age, and he played a major role in its development. Born in Iowa in 1914, Van Allen received his PhD from the University of Iowa. He then worked for several research institutions and served in the Navy during World War II. After the war, Van Allen (Figure 11.9) was appointed Professor of Physics at the University of Iowa. He and his collaborators began using rockets to explore cosmic radiation in Earth\u2019s outer atmosphere. To reach extremely high altitudes, Van Allen designed a technique in which a balloon lifts and then launches a small rocket (the rocket is nicknamed \u201cthe rockoon\u201d). Figure 11.9. James Van Allen (1914\u20132006). In this 1950s photograph, Van Allen holds a \u201crockoon.\u201d (credit: modification of work by Frederick W. Kent Collection, University of Iowa Archives) Over dinner one night in 1950, Van Allen and several colleagues came up with the idea of the International Geophysical Year (IGY), an opportunity for scientists around the world to coordinate their investigations of the physics of Earth, especially research done at high altitudes. In 1955, the United States and the Soviet Union each committed themselves to launching an Earth-orbiting satellite during IGY, a competition that began what came to be known as the space race. The IGY (stretched to 18 months) took place between July 1957 and December 1958. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 393 The Soviet Union won the first lap of the race by launching Sputnik 1 in October 1957. The US government spurred its scientists and engineers to even greater efforts to get something into space to maintain the country\u2019s prestige. However, the primary US satellite program, Vanguard, ran into difficulties: each of its early launches crashed or exploded. Simultaneously, a second team of rocket engineers and scientists had quietly been working on a military launch vehicle called Jupiter-C. Van Allen spearheaded the design of the instruments aboard a small satellite that this vehicle would carry. On January 31, 1958, Van Allen\u2019s Explorer 1 became the first US satellite in space. Unlike Sputnik, Explorer 1 was equipped to make scientific measurements of high-energy charged particles above the atmosphere. Van Allen and his team discovered a belt of highly charged particles surrounding Earth, and these belts now bear his name. This first scientific discovery of the space program made Van Allen\u2019s name known around the world. Van Allen and his colleagues continued to measure the magnetic and particle environment around planets with increasingly sophisticated spacecraft, including Pioneers 10 and 11, which made exploratory", " surveys of the environments of Jupiter and Saturn. Some scientists refer to the charged-particle zones around those planets as Van Allen belts as well. (Once, when Van Allen was giving a lecture at the University of Arizona, the graduate students in planetary science asked him if he would leave his belt at the school. It is now proudly displayed as the university\u2019s \u201cVan Allen belt.\u201d) Van Allen was a strong supporter of space science and an eloquent senior spokesperson for the American scientific community, warning NASA not to put all its efforts into human spaceflight, but to also use robotic spacecraft as productive tools for space exploration. Clouds and Atmospheric Structure The clouds of Jupiter (Figure 11.10) are among the most spectacular sights in the solar system, much beloved by makers of science-fiction films. They range in color from white to orange to red to brown, swirling and twisting in a constantly changing kaleidoscope of patterns. Saturn shows similar but much more subdued cloud activity; instead of vivid colors, its clouds have a nearly uniform butterscotch hue (Figure 11.11). 394 Chapter 11 The Giant Planets Figure 11.10. Jupiter\u2019s Colorful Clouds. The vibrant colors of the clouds on Jupiter present a puzzle to astronomers: given the cool temperatures and the composition of nearly 90% hydrogen, the atmosphere should be colorless. One hypothesis suggests that perhaps colorful hydrogen compounds rise from warm areas. The actual colors are a bit more muted, as shown in Figure 11.2. (credit: modification of work by Voyager Project, JPL, and NASA) Different gases freeze at different temperatures. At the temperatures and pressures of the upper atmospheres of Jupiter and Saturn, methane remains a gas, but ammonia can condense and freeze. (Similarly, water vapor condenses high in Earth\u2019s atmosphere to produce clouds of ice crystals.) The primary clouds that we see around these planets, whether from a spacecraft or through a telescope, are composed of frozen ammonia crystals. The ammonia clouds mark the upper edge of the planets\u2019 tropospheres; above that is the stratosphere, the coldest part of the atmosphere. (These layers were initially defined in Earth as a Planet.) Figure 11.11. Saturn over Five Years. These beautiful images of Saturn were recorded by the Hubble Space Telescope between 1996 and 2000. Since Saturn is tilted by 27\u00b0, we see the orientation of Saturn\u2019s rings around its equator change as the planet moves along its orbit. Note the horizontal bands in the atmosphere. (credit: modification of work by NASA and The Hubble Heritage Team (STScI/AURA)) The diagrams in Figure 11.12 show the structure and clouds in the atmospheres of all four jovian planets. On both Jupiter and Saturn, the temperature near the cloud tops is about 140 K (only a little cooler than the polar caps of Mars). On Jupiter, this cloud level is at a pressure of about 0.1 bar (one tenth the atmospheric pressure at the surface of Earth), but on Saturn it occurs lower in the atmosphere, at about 1 bar. Because the ammonia This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 395 clouds lie so much deeper on Saturn, they are more difficult to see, and the overall appearance of the planet is much blander than is Jupiter\u2019s appearance. Figure 11.12. Atmospheric Structure of the Jovian Planets. In each diagram, the yellow line shows how the temperature (see the scale on the bottom) changes with altitude (see the scale at the left). The location of the main layers on each planet is also shown. Within the tropospheres of these planets, the temperature and pressure both increase with depth. Through breaks in the ammonia clouds, we can see tantalizing glimpses of other cloud layers that can form in these deeper regions of the atmosphere\u2014regions that were sampled directly for Jupiter by the Galileo probe that fell into the planet. As it descended to a pressure of 5 bars, the probe should have passed into a region of frozen water clouds, then below that into clouds of liquid water droplets, perhaps similar to the common clouds of the terrestrial troposphere. At least this is what scientists expected. But the probe saw no water clouds, and it measured a surprisingly low abundance of water vapor in the atmosphere. It soon became clear to the Galileo scientists that the probe happened to descend through an unusually dry, cloud-free region of the atmosphere\u2014a giant downdraft of cool, dry gas. Andrew Ingersoll of Caltech, a member of the Galileo team, called this entry site the \u201cdesert\u201d of Jupiter. It\u2019s a pity that the probe did not enter a more representative region, but that\u2019s the luck of the cosmic draw. The probe continued to make measurements to a pressure of 22 bars but found no other cloud layers before its instruments stopped working. It also detected", " lightning storms, but only at great distances, further suggesting that the probe itself was in a region of clear weather. Above the visible ammonia clouds in Jupiter\u2019s atmosphere, we find the clear stratosphere, which reaches a minimum temperature near 120 K. At still higher altitudes, temperatures rise again, just as they do in the upper atmosphere of Earth, because here the molecules absorb ultraviolet light from the Sun. The cloud colors are due to impurities, the product of chemical reactions among the atmospheric gases in a process we call photochemistry. In Jupiter\u2019s upper atmosphere, photochemical reactions create a variety of fairly complex compounds of hydrogen and carbon that form a thin layer of smog far above the visible clouds. We show this smog as a fuzzy orange region in Figure 11.12; however, this thin layer does not block our view of the clouds beneath it. The visible atmosphere of Saturn is composed of approximately 75% hydrogen and 25% helium, with trace amounts of methane, ethane, propane, and other hydrocarbons. The overall structure is similar to that of Jupiter. Temperatures are somewhat colder, however, and the atmosphere is more extended due to Saturn\u2019s 396 Chapter 11 The Giant Planets lower surface gravity. Thus, the layers are stretched out over a longer distance, as you can see in Figure 11.12. Overall, though, the same atmospheric regions, condensation cloud, and photochemical reactions that we see on Jupiter should be present on Saturn (Figure 11.13). Figure 11.13. Cloud Structure on Saturn. In this Cassini image, colors have been intensified, so we can see the bands and zones and storms in the atmosphere. The dark band is the shadow of the rings on the planet. (credit: NASA/JPL-Caltech/Space Science Institute) Saturn has one anomalous cloud structure that has mystified scientists: a hexagonal wave pattern around the north pole, shown in Figure 11.14. The six sides of the hexagon are each longer than the diameter of Earth. Winds are also extremely high on Saturn, with speeds of up to 1800 kilometers per hour measured near the equator. Figure 11.14. Hexagon Pattern on Saturn\u2019s North Pole. In this infrared nighttime image from the Cassini mission, the path of Saturn\u2019s hexagonal jet stream is visible as the planet\u2019s north pole emerges from the darkness of winter. (credit: NASA/JPL/University of Arizona) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 397 See images of Saturn\u2019s hexagon (https://openstax.org/l/30Hexagon) with exaggerated color in this brief NASA video. Unlike Jupiter and Saturn, Uranus is almost entirely featureless as seen at wavelengths that range from the ultraviolet to the infrared (see its rather boring image in Figure 11.1). Calculations indicate that the basic atmospheric structure of Uranus should resemble that of Jupiter and Saturn, although its upper clouds (at the 1-bar pressure level) are composed of methane rather than ammonia. However, the absence of an internal heat source suppresses up-and-down movement and leads to a very stable atmosphere with little visible structure. Neptune differs from Uranus in its appearance, although their basic atmospheric temperatures are similar. The upper clouds are composed of methane, which forms a thin cloud layer near the top of the troposphere at a temperature of 70 K and a pressure of 1.5 bars. Most the atmosphere above this level is clear and transparent, with less haze than is found on Uranus. The scattering of sunlight by gas molecules lends Neptune a pale blue color similar to that of Earth\u2019s atmosphere (Figure 11.15). Another cloud layer, perhaps composed of hydrogen sulfide ice particles, exists below the methane clouds at a pressure of 3 bars. Figure 11.15. Neptune. The planet Neptune is seen here as photographed by Voyager in 1989. The blue color, exaggerated with computer processing, is caused by the scattering of sunlight in the planet\u2019s upper atmosphere. (credit: modification of work by NASA) Unlike Uranus, Neptune has an atmosphere in which convection currents\u2014vertical drafts of gas\u2014emanate from the interior, powered by the planet\u2019s internal heat source. These currents carry warm gas above the 1.5-bar cloud level, forming additional clouds at elevations about 75 kilometers higher. These high-altitude clouds form bright white patterns against the blue planet beneath. Voyager photographed distinct shadows on the methane cloud tops, permitting the altitudes of the high clouds to be calculated. Figure 11.16 is a remarkable close-up of Neptune\u2019s outer layers that could never have been obtained from Earth. 398 Chapter 11 The Giant Planets Figure 11.16. High Clouds in the Atmosphere of Neptune. These bright, narrow cirrus clouds are made of methane ice crystals", ". From the shadows they cast on the thicker cloud layer below, we can measure that they are about 75 kilometers higher than the main clouds. (credit: modification of work by NASA/JPL) Winds and Weather The atmospheres of the jovian planets have many regions of high pressure (where there is more air) and low pressure (where there is less). Just as it does on Earth, air flows between these regions, setting up wind patterns that are then distorted by the rotation of the planet. By observing the changing cloud patterns on the jovian planets, we can measure wind speeds and track the circulation of their atmospheres. The atmospheric motions we see on these planets are fundamentally different from those on the terrestrial planets. The giants spin faster, and their rapid rotation tends to smear out of the circulation into horizontal (east-west) patterns parallel to the equator. In addition, there is no solid surface below the atmosphere against which the circulation patterns can rub and lose energy (which is how tropical storms on Earth ultimately die out when they come over land). As we have seen, on all the giants except Uranus, heat from the inside contributes about as much energy to the atmosphere as sunlight from the outside. This means that deep convection currents of rising hot air and falling cooler air circulate throughout the atmospheres of the planets in the vertical direction. The main features of Jupiter\u2019s visible clouds (see Figure 11.2 and Figure 11.10, for example) are alternating dark and light bands that stretch around the planet parallel to the equator. These bands are semi-permanent features, although they shift in intensity and position from year to year. Consistent with the small tilt of Jupiter\u2019s axis, the pattern does not change with the seasons. More fundamental than these bands are underlying east-west wind patterns in the atmosphere, which do not appear to change at all, even over many decades. These are illustrated in Figure 11.17, which indicates how strong the winds are at each latitude for the giant planets. At Jupiter\u2019s equator, a jet stream flows eastward with a speed of about 90 meters per second (300 kilometers per hour), similar to the speed of jet streams in Earth\u2019s upper atmosphere. At higher latitudes there are alternating east- and west-moving streams, with each hemisphere an almost perfect mirror image of the other. Saturn shows a similar pattern, but with a much stronger equatorial jet stream, as we noted earlier. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 399 Figure 11.17. Winds on the Giant Planets. This image compares the winds of the giant planets, illustrating that wind speed (shown on the horizontal axis) and wind direction vary with latitude (shown on the vertical axis). Winds are measured relative to a planet\u2019s internal rotation speed. A positive velocity means that the winds are blowing in the same direction as, but faster than, the planet\u2019s internal rotation. A negative velocity means that the winds are blowing more slowly than the planet\u2019s internal rotation. Note that Saturn\u2019s winds move faster than those of the other planets. The light zones on Jupiter are regions of upwelling air capped by white ammonia cirrus clouds. They apparently represent the tops of upward-moving convection currents.[2] The darker belts are regions where the cooler atmosphere moves downward, completing the convection cycle; they are darker because fewer ammonia clouds mean we can see deeper into the atmosphere, perhaps down to a region of ammonium hydrosulfide (NH4SH) clouds. The Galileo probe sampled one of the clearest of these dry downdrafts. In spite of the strange seasons induced by the 98\u00b0 tilt of its axis, Uranus\u2019 basic circulation is parallel with its equator, as is the case on Jupiter and Saturn. The mass of the atmosphere and its capacity to store heat are so great that the alternating 42-year periods of sunlight and darkness have little effect. In fact, Voyager measurements show that the atmospheric temperature is even a few degrees higher on the dark winter side than on the hemisphere facing the Sun. This is another indication that the behavior of such giant planet atmospheres is a complex problem that we do not fully understand. Neptune\u2019s weather is characterized by strong east-west winds generally similar to those observed on Jupiter and Saturn. The highest wind speeds near its equator reach 2100 kilometers per hour, even higher than the peak winds on Saturn. The Neptune equatorial jet stream actually approaches supersonic speeds (faster than the speed of sound in Neptune\u2019s air). Giant Storms on Giant Planets Superimposed on the regular atmospheric circulation patterns we have just described are many local disturbances\u2014weather systems or storms, to borrow the term we use on Earth. The most prominent of these are large, oval-shaped, high-pressure regions on both Jupiter (Figure 11.18) and", " Neptune. 2 Recall from earlier chapters that convection is a process in which liquids, heated from underneath, have regions where hot material rises and cooler material descends. You can see convection at work if you heat oatmeal on a stovetop or watch miso soup boil. 400 Chapter 11 The Giant Planets Figure 11.18. Storms on Jupiter. Two examples of storms on Jupiter illustrate the use of enhanced color and contrast to bring out faint features. (a) The three oval-shaped white storms below and to the left of Jupiter\u2019s Great Red Spot are highly active, and moved closer together over the course of seven months between 1994 and 1995. (b) The clouds of Jupiter are turbulent and ever-changing, as shown in this Hubble Space Telescope image from 2007. (credit a: modification of work by Reta Beebe, Amy Simon (New Mexico State Univ.), and NASA; credit b: modification of work by NASA, ESA, and A. Simon-Miller (NASA Goddard Space Flight Center)) The largest and most famous of Jupiter\u2019s storms is the Great Red Spot, a reddish oval in the southern hemisphere that changes slowly; it was 25,000 kilometers long when Voyager arrived in 1979, but it had shrunk to 20,000 kilometers by the end of the Galileo mission in 2000 (Figure 11.19). The giant storm has persisted in Jupiter\u2019s atmosphere ever since astronomers were first able to observe it after the invention of the telescope, more than 300 years ago. However, it has continued to shrink, raising speculation that we may see its end within a few decades. Figure 11.19. Jupiter\u2019s Great Red Spot. This is the largest storm system on Jupiter, as seen during the Voyager spacecraft flyby. Below and to the right of the Red Spot is one of the white ovals, which are similar but smaller high-pressure features. The white oval is roughly the size of planet Earth, to give you a sense of the huge scale of the weather patterns we are seeing. The colors on the Jupiter image have been somewhat exaggerated here so astronomers (and astronomy students) can study their differences more effectively. See Figure 11.2 to get a better sense of the colors your eye would actually see near Jupiter. (credit: NASA/JPL) In addition to its longevity, the Red Spot differs from terrestrial storms in being a high-pressure region; on our planet, such storms are regions of lower pressure. The Red Spot\u2019s counterclockwise rotation has a period of six days. Three similar but smaller disturbances (about as big as Earth) formed on Jupiter in the 1930s. They look like white ovals, and one can be seen clearly below and to the right of the Great Red Spot in Figure 11.19. In 1998, the Galileo spacecraft watched as two of these ovals collided and merged into one. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 401 We don\u2019t know what causes the Great Red Spot or the white ovals, but we do have an idea how they can last so long once they form. On Earth, the lifetime of a large oceanic hurricane or typhoon is typically a few weeks, or even less when it moves over the continents and encounters friction with the land. Jupiter has no solid surface to slow down an atmospheric disturbance; furthermore, the sheer size of the disturbances lends them stability. We can calculate that on a planet with no solid surface, the lifetime of anything as large as the Red Spot should be measured in centuries, while lifetimes for the white ovals should be measured in decades, which is pretty much what we have observed. Despite Neptune\u2019s smaller size and different cloud composition, Voyager showed that it had an atmospheric feature surprisingly similar to Jupiter\u2019s Great Red Spot. Neptune\u2019s Great Dark Spot was nearly 10,000 kilometers long (Figure 11.15). On both planets, the giant storms formed at latitude 20\u00b0 S, had the same shape, and took up about the same fraction of the planet\u2019s diameter. The Great Dark Spot rotated with a period of 17 days, versus about 6 days for the Great Red Spot. When the Hubble Space Telescope examined Neptune in the mid-1990s, however, astronomers could find no trace of the Great Dark Spot on their images. Although many of the details of the weather on the jovian planets are not yet understood, it is clear that if you are a fan of dramatic weather, these worlds are the place to look. We study the features in these atmospheres not only for what they have to teach us about conditions in the jovian planets, but also because we hope they can help us understand the weather on Earth just a bit better Storms and Winds The wind speeds in circular storm systems can be formidable on both Earth and the giant planets. Think about our big terrestrial hurricanes. If you watch their behavior in satellite images", " shown on weather outlets, you will see that they require about one day to rotate. If a storm has a diameter of 400 km and rotates once in 24 h, what is the wind speed? Solution Speed equals distance divided by time. The distance in this case is the circumference (2\u03c0R or \u03c0d), or approximately 1250 km, and the time is 24 h, so the speed at the edge of the storm would be about 52 km/h. Toward the center of the storm, the wind speeds can be much higher. Check Your Learning Jupiter\u2019s Great Red Spot rotates in 6 d and has a circumference equivalent to a circle with radius 10,000 km. Calculate the wind speed at the outer edge of the spot. Answer: For the Great Red Spot of Jupiter, the circumference (2\u03c0R) is about 63,000 km. Six d equals 144 h, suggesting a speed of about 436 km/h. This is much faster than wind speeds on Earth. 402 Chapter 11 The Giant Planets CHAPTER 11 REVIEW KEY TERMS photochemistry chemical changes caused by electromagnetic radiation synchrotron radiation the radiation emitted by charged particles being accelerated in magnetic fields and moving at speeds near that of light SUMMARY 11.1 Exploring the Outer Planets The outer solar system contains the four giant planets: Jupiter, Saturn, Uranus, and Neptune. The gas giants Jupiter and Saturn have overall compositions similar to that of the Sun. These planets have been explored by the Pioneer, Voyager, Galileo, and Cassini spacecraft. Voyager 2, perhaps the most successful of all space-science missions, explored Jupiter (1979), Saturn (1981), Uranus (1986), and Neptune (1989)\u2014a grand tour of the giant planets\u2014and these flybys have been the only explorations to date of the ice giants Uranus and Neptune. The Galileo and Cassini missions were long-lived orbiters, and each also deployed an entry probe, one into Jupiter and one into Saturn\u2019s moon Titan. 11.2 The Giant Planets Jupiter is 318 times more massive than Earth. Saturn is about 25% as massive as Jupiter, and Uranus and Neptune are only 5% as massive. All four have deep atmospheres and opaque clouds, and all rotate quickly with periods from 10 to 17 hours. Jupiter and Saturn have extensive mantles of liquid hydrogen. Uranus and Neptune are depleted in hydrogen and helium relative to Jupiter and Saturn (and the Sun). Each giant planet has a core of \u201cice\u201d and \u201crock\u201d of about 10 Earth masses. Jupiter, Saturn, and Neptune have major internal heat sources, obtaining as much (or more) energy from their interiors as by radiation from the Sun. Uranus has no measurable internal heat. Jupiter has the strongest magnetic field and largest magnetosphere of any planet, first discovered by radio astronomers from observations of synchrotron radiation. 11.3 Atmospheres of the Giant Planets The four giant planets have generally similar atmospheres, composed mostly of hydrogen and helium. Their atmospheres contain small quantities of methane and ammonia gas, both of which also condense to form clouds. Deeper (invisible) cloud layers consist of water and possibly ammonium hydrosulfide (Jupiter and Saturn) and hydrogen sulfide (Neptune). In the upper atmospheres, hydrocarbons and other trace compounds are produced by photochemistry. We do not know exactly what causes the colors in the clouds of Jupiter. Atmospheric motions on the giant planets are dominated by east-west circulation. Jupiter displays the most active cloud patterns, with Neptune second. Saturn is generally bland, in spite of its extremely high wind speeds, and Uranus is featureless (perhaps due to its lack of an internal heat source). Large storms (oval-shaped high- pressure systems such as the Great Red Spot on Jupiter and the Great Dark Spot on Neptune) can be found in some of the planet atmospheres. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 403 FOR FURTHER EXPLORATION Articles Jupiter Aguirre, Edwin. \u201cHubble Zooms in on Jupiter\u2019s New Red Spot.\u201d Sky & Telescope (August 2006): 26. Beatty, J. \u201cInto the Giant.\u201d Sky & Telescope (April 1996): 20. On the Galileo probe. Beebe, R. \u201cQueen of the Giant Storms.\u201d Sky & Telescope (October 1990): 359. Excellent review of the Red Spot. Johnson, T. \u201cThe Galileo Mission to Jupiter and Its Moons.\u201d Scientific American (February 2000): 40. Results about Jupiter, Io, Ganymede, and Callisto. Simon, A. \u201cThe Not-So-Great Red Spot.\u201d Sky & Telescope (March 2016): 18. On how the huge storm on Jupiter is evolving with time. Smith", ", B. \u201cVoyage of the Century.\u201d National Geographic (August 1990): 48. Beautiful summary of the Voyager mission to all four outer planets. Stern, S. \u201cJupiter Up Close and Personal.\u201d Astronomy (August 2007): 28. On the New Horizons mission flyby in February 2007. Saturn Gore, R. \u201cThe Riddle of the Rings.\u201d National Geographic (July 1981): 3. Colorful report on the Voyager mission. McEwen, A. \u201cCassini Unveils Saturn.\u201d Astronomy (July 2006): 30. A report on the first two years of discoveries in the Saturn system. Spilker, L. \u201cSaturn Revolution.\u201d Astronomy (October 2008): 34. On results from the Cassini mission. Talcott, R. \u201cSaturn\u2019s Sweet Surprises.\u201d Astronomy (June 2007): 52. On Cassini mission results. Uranus and Neptune Cowling, T. \u201cBig Blue: The Twin Worlds of Uranus and Neptune.\u201d Astronomy (October 1990): 42. Nice, long review of the two planets. Gore, R. \u201cNeptune: Voyager\u2019s Last Picture Show.\u201d National Geographic (August 1990): 35. Lunine, J. \u201cNeptune at 150.\u201d Sky & Telescope (September 1996): 38. Nice review. Websites Jupiter NASA Solar System Exploration: http://Solarsystem.nasa.gov/planets/jupiter Nine Planets Site: http://nineplanets.org/jupiter.html Planetary Sciences Site: http://nssdc.gsfc.nasa.gov/planetary/planets/jupiterpage.html Saturn NASA Solar System Exploration: http://Solarsystem.nasa.gov/planets/saturn Nine Planets Site: http://nineplanets.org/saturn.html Planetary Sciences Site: http://nssdc.gsfc.nasa.gov/planetary/planets/saturnpage.html 404 Uranus Chapter 11 The Giant Planets NASA Solar System Exploration: http://Solarsystem.nasa.gov/planets/uranus Nine Planets Site: http://nineplanets.org/uranus.html Planetary Sciences Site: http://nssdc.gsfc.nasa.gov/planetary/planets/uranuspage.html Neptune NASA Solar System Exploration: http://Solarsystem.nasa.gov/planets/neptune Nine Planets Site: http://nineplanets.org/neptune.html Planetary Sciences Site: http://nssdc.gsfc.nasa.gov/planetary/planets/neptunepage.html Missions Cassini Mission Site at the Jet Propulsion Lab: http://saturn.jpl.nasa.gov/index.cfm Cassini-Huygens Mission Site at European Space Agency: http://sci.esa.int/cassini-huygens/ NASA Galileo Mission Site: http://Solarsystem.nasa.gov/galileo/ NASA\u2019s Juno Mission to Jupiter: http://www.nasa.gov/mission_pages/juno/main/index.html Voyager Mission Site at the Jet Propulsion Lab: http://voyager.jpl.nasa.gov/ Videos Cassini: 15 Years of Exploration: https://www.youtube.com/watch?v=2z8fzz_MBAw. Quick visual summary of mission highlights (2:29). In the Land of Enchantment: The Epic Story of the Cassini Mission to Saturn: https://www.youtube.com/ watch?v=Vx135n8VFxY. An inspiring illustrated lecture by Cassini Mission Imagining Lead Scientist Carolyn Porco (1:37:52). Jupiter: The Largest Planet: http://www.youtube.com/watch?v=s56pxa9lpvo. Produced by NASA\u2019s Goddard Space Flight Center and Science on a Sphere (7:29). COLLABORATIVE GROUP ACTIVITIES A. A new member of Congress has asked your group to investigate why the Galileo probe launched into the Jupiter atmosphere in 1995 survived only 57 minutes and whether this was an example of a terrible scandal. Make a list of all the reasons the probe did not last longer, and why it was not made more durable. (Remember that the probe had to hitch a ride to Jupiter!) B. Select one of the jovian planets and organize your group to write a script for an evening news weather report for the planet you chose. Be sure you specify roughly how high in the atmosphere the region lies for which you are giving the report. C. What does your group think should be the next step to learn more about the giant", " planets? Put cost considerations aside for a moment: What kind of mission would you recommend to NASA to learn more about these giant worlds? Which world or worlds should get the highest priority and why? D. Suppose that an extremely dedicated (and slightly crazy) astronomer volunteers to become a human probe into Jupiter (and somehow manages to survive the trip through Jupiter\u2019s magnetosphere alive). As she This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 11 The Giant Planets 405 enters the upper atmosphere of Jupiter, would she fall faster or slower than she would fall doing the same suicidal jump into the atmosphere of solid Earth? Groups that have some algebra background could even calculate the force she would feel compared to the force on Earth. (Bonus question: If she were in a capsule, falling into Jupiter feet first, and the floor of the capsule had a scale, what would the scale show as her weight compared to her weight on Earth?) E. Would you or anyone in your group volunteer for a one-way, life-long mission to a space station orbiting any of the gas giants without ever being able to return to Earth? What are the challenges of such a mission? Should we leave all exploration of the outer solar system to unmanned space probes? EXERCISES Review Questions 1. What are the main challenges involved in sending probes to the giant planets? 2. Why is it difficult to drop a probe like Galileo? How did engineers solve this problem? 3. Explain why visual observation of the gas giants is not sufficient to determine their rotation periods, and what evidence was used to deduce the correct periods. 4. What are the seasons like on Jupiter? 5. What is the consequence of Uranus\u2019 spin axis being 98\u00b0 away from perpendicular to its orbital plane? 6. Describe the seasons on the planet Uranus. 7. At the pressures in Jupiter\u2019s interior, describe the physical state of the hydrogen found there. 8. Which of the gas giants has the largest icy/rocky core compared to its overall size? 9. In the context of the giant planets and the conditions in their interiors, what is meant by \u201crock\u201d and \u201cice\u201d? 10. What is the primary source of Jupiter\u2019s internal heat? 11. Describe the interior heat source of Saturn. 12. Which planet has the strongest magnetic field, and hence the largest magnetosphere? What is its source? 13. What are the visible clouds on the four giant planets composed of, and why are they different from each other? 14. Compare the atmospheric circulation (weather) of the four giant planets. 15. What are the main atmospheric heat sources of each of the giant planets? 16. Why do the upper levels of Neptune\u2019s atmosphere appear blue? 17. How do storms on Jupiter differ from storm systems on Earth? Thought Questions 18. Describe the differences in the chemical makeup of the inner and outer parts of the solar system. What is the relationship between what the planets are made of and the temperature where they formed? 19. How did the giant planets grow to be so large? 406 Chapter 11 The Giant Planets 20. Jupiter is denser than water, yet composed for the most part of two light gases, hydrogen and helium. What makes Jupiter as dense as it is? 21. Would you expect to find free oxygen gas in the atmospheres of the giant planets? Why or why not? 22. Why would a tourist brochure (of the future) describing the most dramatic natural sights of the giant planets have to be revised more often than one for the terrestrial planets? 23. The water clouds believed to be present on Jupiter and Saturn exist at temperatures and pressures similar to those in the clouds of the terrestrial atmosphere. What would it be like to visit such a location on Jupiter or Saturn? In what ways would the environment differ from that in the clouds of Earth? 24. Describe the different processes that lead to substantial internal heat sources for Jupiter and Saturn. Since these two objects generate much of their energy internally, should they be called stars instead of planets? Justify your answer. 25. Research the Galileo mission. What technical problems occurred between the mission launch and the arrival of the craft in Jupiter\u2019s system, and how did the mission engineers deal with them? (Good sources of information include Astronomy and Sky & Telescope articles, plus the mission website.) Figuring For Yourself 26. How many times more pressure exists in the interior of Jupiter compared to that of Earth? 27. Calculate the wind speed at the edge of Neptune\u2019s Great Dark Spot, which was 10,000 km in diameter and rotated in 17 d. 28. Calculate how many Earths would fit into the volumes of Saturn, Uranus, and Neptune. 29. As the Voyager spacecraft penetrated into the outer solar system, the illumination from the Sun declined. Relative to the situation at Earth, how bright is the sunlight at each of the", " jovian planets? 30. The ions in the inner parts of Jupiter\u2019s magnetosphere rotate with the same period as Jupiter. Calculate how fast they are moving at the orbit of Jupiter\u2019s moon Io (see Appendix G). Will these ions strike Io from behind or in front as it moves about Jupiter? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 407 12 RINGS, MOONS, AND PLUTO Figure 12.1. Jupiter Family. This montage, assembled from individual Galileo and Voyager images, shows a \u201cfamily portrait\u201d of Jupiter (with its giant red spot) and its four large moons. From top to bottom, we see Io, Europa, Ganymede, and Callisto. The colors are exaggerated by image processing to emphasize contrasts. (credit: modification of work by NASA) Chapter Outline 12.1 Ring and Moon Systems Introduced 12.2 The Galilean Moons of Jupiter 12.3 Titan and Triton 12.4 Pluto and Charon 12.5 Planetary Rings Thinking Ahead \u201cOur imaginations always fall short of anticipating the beauty we find in nature.\u201d\u2014Geologist Laurence Soderblom, discussing the 1989 Voyager encounter with Neptune\u2019s moons All four giant planets are accompanied by moons that orbit about them like planets in a miniature solar system. Nearly 200 moons are known in the outer solar system\u2014too many to name individually or discuss in any detail. Astronomers anticipate that additional small moons await future discovery. We have also discovered a fascinating variety of rings around each of the jovian planets. Before the Voyager missions, even the largest of the outer-planet moons were mere points of light in our telescopes. Then, in less than a decade, we had close-up images, and these moons became individual worlds for us, each with unique features and peculiarities. The Galileo mission added greatly to our knowledge of the moons of Jupiter, and the Cassini mission has done the same for the Saturn system. In 2015, the NASA New Horizons spacecraft completed the initial exploration of the \u201cclassical\u201d planets in the solar system with its flyby of Pluto and its moons. We include Pluto here because in some ways it resembles some of the larger moons in the outer solar system. Each new spacecraft mission has revealed many surprises, as the objects in the outer solar system are much more varied and geologically active than scientists had anticipated. 408 Chapter 12 Rings, Moons, and Pluto 12.1 RING AND MOON SYSTEMS INTRODUCED Learning Objectives By the end of this section, you will be able to: Name the major moons of each of the jovian planets Describe the basic composition of each jovian planet\u2019s ring system The rings and moons (see the moons in Figure 12.2) of the outer solar system are not composed of the same materials as the mostly rocky objects in the inner solar system. We should expect this, since they formed in regions of lower temperature, cool enough so that large quantities of water ice were available as building materials. Most of these objects also contain dark, organic compounds mixed with their ice and rock. Don\u2019t be surprised, therefore, to find that many objects in the ring and moon systems are both icy and dark. Roughly a third of the moons in the outer solar system are in direct or regular orbits; that is, they revolve about their parent planet in a west-to-east direction and in the plane of the planet\u2019s equator. The majority are irregular moons that orbit in a retrograde (east-to-west) direction or else have orbits of high eccentricity (more elliptical than circular) or high inclination (moving in and out of the planet\u2019s equatorial plane). These irregular moons are mostly located relatively far from their planet; they were probably formed elsewhere and subsequently captured by the planet they now orbit. (Perhaps the fact that they were not born locally will excuse their ill- mannered behavior.) Figure 12.2. Moons of the Solar System. This image shows some selected moons of our solar system and their comparison to the size of Earth\u2019s Moon and Earth itself. (credit: modification of work by NASA) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 409 The Jupiter System Jupiter has 67 known moons (that\u2019s the number as we write) and a faint ring. These include four large moons\u2014 Callisto, Ganymede, Europa, and Io (see Figure 12.1)\u2014discovered in 1610 by Galileo and therefore often called the Galilean moons. The smaller of these, Europa and Io, are about the size of our Moon, while the larger, Ganymede and Callisto, are about the", " same size as the planet Mercury. Most of Jupiter\u2019s moons are much smaller. The majority are in retrograde orbits more than 20 million kilometers from Jupiter; these are very likely small captured asteroids. The Saturn System Saturn has at least 62 known moons in addition to a magnificent set of rings. The largest of the moons, Titan, is almost as big as Ganymede in Jupiter\u2019s system, and it is the only moon with a substantial atmosphere and lakes or seas of liquid hydrocarbons (such as methane and ethane) on the surface. Saturn has six other large regular moons with diameters between 400 and 1600 kilometers, a collection of small moons orbiting in or near the rings, and many captured strays similar to those of Jupiter. Mysteriously, one of Saturn\u2019s smaller moons, Enceladus, has active geysers of water being expelled into space. The rings of Saturn, one of the most impressive sights in the solar system, are broad and flat, with a few major and many minor gaps. They are not solid, but rather a huge collection of icy fragments, all orbiting the equator of Saturn in a traffic pattern that makes rush hour in a big city look simple by comparison. Individual ring particles are composed primarily of water ice and are typically the size of ping-pong balls, tennis balls, and basketballs. The Uranus System The ring and moon system of Uranus is tilted at 98\u00b0, just like the planet itself. It consists of 11 rings and 27 currently known moons. The five largest moons are similar in size to the six regular moons of Saturn, with diameters of 500 to 1600 kilometers. Discovered in 1977, the rings of Uranus are narrow ribbons of dark material with broad gaps in between. Astronomers suppose that the ring particles are confined to these narrow paths by the gravitational effects of numerous small moons, many of which we have not yet glimpsed. The Neptune System Neptune has 14 known moons. The most interesting of these is Triton, a relatively large moon in a retrograde orbit\u2014which is unusual. Triton has a very thin atmosphere, and active eruptions were discovered there by Voyager in its 1989 flyby. To explain its unusual characteristics, astronomers have suggested that Triton may have originated beyond the Neptune system, as a dwarf planet like Pluto. The rings of Neptune are narrow and faint. Like those of Uranus, they are composed of dark materials and are thus not easy to see. 12.2 THE GALILEAN MOONS OF JUPITER Learning Objectives By the end of this section, you will be able to: Describe the major features we can observe about Callisto and what we can deduce from them Explain the evidence for tectonic and volcanic activity on Ganymede Explain what may be responsible for the unusual features on the icy surface of Europa Describe the major distinguishing characteristic of Io Explain how tidal forces generate the geological activity we see on Europa and Io 410 Chapter 12 Rings, Moons, and Pluto From 1996 to 1999, the Galileo spacecraft careered through the jovian system on a complex but carefully planned trajectory that provided repeated close encounters with the large Galilean moons. (Beginning in 2004, we received an even greater bonanza of information about Titan, obtained from the Cassini spacecraft and its Huygens probe, which landed on its surface. We include Titan, Saturn\u2019s one big moon, here for comparison.) Table 12.1 summarizes some basic facts about these large moons (plus our own Moon for comparison). The Largest Moons Name Diameter (km) Mass (Earth\u2019s Moon = 1) Density (g/cm3) Reflectivity (%) Moon Callisto Ganymede Europa Io Titan Table 12.1 3476 4820 5270 3130 3640 5150 1.0 1.5 2.0 0.7 1.2 1.9 3.3 1.8 1.9 3.0 3.5 1.9 12 20 40 70 60 20 Callisto: An Ancient, Primitive World We begin our discussion of the Galilean moons with the outermost one, Callisto, not because it is remarkable but because it is not. This makes it a convenient object with which other, more active, worlds can be compared. Its distance from Jupiter is about 2 million kilometers, and it orbits the planet in 17 days. Like our own Moon, Callisto rotates in the same period as it revolves, so it always keeps the same face toward Jupiter. Callisto\u2019s day thus equals its month: 17 days. Its noontime surface temperature is only 130 K (about 140 \u00b0C below freezing), so that water ice is stable (it never evaporates) on its surface year round. Callisto has a diameter of 4820 kilometers, almost the same as the planet Mercury (Figure 12.3). Yet its mass is only one-third as great, which", " means its density (the mass divided by the volume) must be only one-third as great as well. This tells us that Callisto has far less of the rocky and metallic materials found in the inner planets and must instead be an icy body through much of its interior. Callisto can show us how the geology of an icy object compares with those made primarily of rock. Unlike the worlds we have studied so far, Callisto has not fully differentiated (separated into layers of different density materials). We can tell that it lacks a dense core from the details of its gravitational pull on the Galileo spacecraft. This surprised scientists, who expected that all the big icy moons would be differentiated. It should be easier for an icy body to differentiate than for a rocky one because the melting temperature of ice is so low. Only a little heating will soften the ice and get the process started, allowing the rock and metal to sink to the center while the slushy ice floats to the surface. Yet Callisto seems to have frozen solid before the process of differentiation was complete. The surface of Callisto is covered with impact craters, like the lunar highlands. The survival of these craters tells us that an icy object can retain impact craters on its surface. Callisto is unique among the planet-sized objects This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 411 of the solar system in the apparent absence of interior forces to drive geological change. You might say that this moon was stillborn, and it has remained geologically dead for more than 4 billion years (Figure 12.3). Figure 12.3. Callisto. (a) Jupiter\u2019s outermost large moon shows a heavily cratered surface. Astronomers believe that the bright areas are mostly ice, while the darker areas are more eroded, ice-poor material. (b) These high-resolution images, taken by NASA\u2019s Galileo spacecraft in May 2001, show the icy spires (top) on Callisto\u2019s surface, with darker dust that has slid down as the ice erodes, collecting in the low-lying areas. The spires are about 80 to 100 meters tall. As the surface erodes even further, the icy spires eventually disappear, leaving impact craters exposed, as shown in the lower image. (credit a: modification of work by NASA/JPL/DLR; credit b: modification of work by NASA/JPL/Arizona State University, Academic Research Lab) In thinking about ice so far from the Sun, we must take care not to judge its behavior from the much warmer ice we know and love on Earth. At the temperatures of the outer solar system, ice on the surface is nearly as hard as rock, and it behaves similarly. Ice on Callisto does not deform or flow like ice in glaciers on Earth. Ganymede, the Largest Moon Ganymede, the largest moon in the solar system, also shows a great deal of cratering (Figure 12.4). Recall from Other Worlds: An Introduction to the Solar System) that we can use crater counts on solid worlds to estimate the age of the surface. The more craters, the longer the surface has been exposed to battering from space, and the older it must therefore be. About one-quarter of Ganymede\u2019s surface seems to be as old and heavily cratered as that of Callisto; the rest formed more recently, as we can tell by the sparse covering of impact craters as well as the relative freshness of those craters. If we judge from crater counts, this fresher terrain on Ganymede is somewhat younger than the lunar maria or the martian volcanic plains, perhaps 2 to 3 billion years old. The differences between Ganymede and Callisto are more than skin deep. Ganymede is a differentiated world, like the terrestrial planets. Measurements of its gravity field tell us that the rock sank to form a core about the size of our Moon, with a mantle and crust of ice \u201cfloating\u201d above it. In addition, the Galileo spacecraft discovered that Ganymede has a magnetic field, the sure signature of a partially molten interior. There is very likely liquid water trapped within the interior. Thus, Ganymede is not a dead world but rather a place of intermittent geological activity powered by an internal heat source. Some surface features could be as young as the surface of Venus (a few hundred million years). The younger terrain was formed by tectonic and volcanic forces (Figure 12.4). In some places, the crust apparently cracked, flooding many of the craters with water from the interior. Extensive mountain ranges were formed from compression of the crust, forming long ridges with parallel valleys spaced a few kilometers apart. 412 Chapter 12 Rings, Moons, and Pluto In", " some areas, older impact craters were split and pulled apart. There are even indications of large-scale crustal movements that are similar to the plate tectonics of Earth. Figure 12.4. Ganymede. (a) This global view of Ganymede, the largest moon in the solar system, was taken by Voyager 2. The colors are enhanced to make spotting differences easier. Darker places are older, more heavily cratered regions; the lighter areas are younger (the reverse of our Moon). The brightest spots are sites of geologically recent impacts. (b) This close-up of Nicholson Regio on Ganymede shows an old impact crater (on the lower left-hand side) that has been split and pulled apart by tectonic forces. Against Ganymede\u2019s dark terrain, a line of grooves and ridges appears to cut through the crater, deforming its circular shape. (credit a: modification of work by NASA/JPL/DLR; credit b: modification of work by NASA/JPL/Brown University) Why is Ganymede so different from Callisto? Possibly the small difference in size and internal heating between the two led to this divergence in their evolution. But more likely the gravity of Jupiter is to blame for Ganymede\u2019s continuing geological activity. Ganymede is close enough to Jupiter that tidal forces from the giant planet may have episodically heated its interior and triggered major convulsions on its crust. A tidal force results from the unequal gravitational pull on two sides of a body. In a complex kind of modern dance, the large moons of Jupiter are caught in the varying gravity grip of both the giant planet and each other. This leads to gravitational flexing or kneading in their centers, which can heat them\u2014an effect called tidal heating. (A fuller explanation is given in the section on Io.) We will see as we move inward to Europa and Io that the role of jovian tides becomes more important for moons close to the planet. Europa, a Moon with an Ocean Europa and Io, the inner two Galilean moons, are not icy worlds like most of the moons of the outer planets. With densities and sizes similar to our Moon, they appear to be predominantly rocky objects. How did they fail to acquire a majority share of the ice that must have been plentiful in the outer solar system at the time of their formation? The most probable cause is Jupiter itself, which was hot enough to radiate a great deal of infrared energy during the first few million years after its formation. This infrared radiation would have heated the disk of material near the planet that would eventually coalesce into the closer moons. Thus, any ice near Jupiter was vaporized, leaving Europa and Io with compositions similar to planets in the inner solar system. Despite its mainly rocky composition, Europa has an ice-covered surface, as astronomers have long known from examining spectra of sunlight reflected from it. In this it resembles Earth, which has a layer of water on its surface, but in Europa\u2019s case the water is capped by a thick crust of ice. There are very few impact craters in this This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 413 ice, indicating that the surface of Europa is in a continual state of geological self-renewal. Judging from crater counts, the surface must be no more than a few million years old, and perhaps substantially less. In terms of its ability to erase impact craters, Europa is more geologically active than Earth. When we look at close-up photos of Europa, we see a strange, complicated surface (Figure 12.5). For the most part, the icy crust is extremely smooth, but it is crisscrossed with cracks and low ridges that often stretch for thousands of kilometers. Some of these long lines are single, but most are double or multiple, looking rather like the remnants of a colossal freeway system. Figure 12.5. Evidence for an Ocean on Europa. (a) A close-up of an area called Conamara Chaos is shown here with enhanced color. This view is 70 kilometers wide in its long dimension. It appears that Conamara is a region where Europa\u2019s icy crust is (or recently was) relatively thin and there is easier access to the possible liquid or slushy ocean beneath. Not anchored to solid crust underneath, many of the ice blocks here seem to have slid or rotated from their original positions. In fact, the formations seen here look similar to views of floating sea-ice and icebergs in Earth\u2019s Arctic Ocean. (b) In this high-resolution view, the ice is wrinkled and crisscrossed by long ridges. Where these ridges intersect, we can see which ones are older and which younger; the younger ones cross over the older ones. While", " superficially this system of ridges resembles a giant freeway system on Europa, the ridges are much wider than our freeways and are a natural result of the flexing of the moon. (credit a: modification of work by NASA/JPL/University of Arizona; credit b: modification of work by NASA/JPL) It is very difficult to make straight lines on a planetary surface. In discussing Mars, we explained that when Percival Lowell saw what appeared to him to be straight lines (the so-called martian \u201ccanals\u201d), he attributed them to the engineering efforts of intelligent beings. We now know the lines on Mars were optical illusions, but the lines on Europa are real. These long cracks can form in the icy crust if it is floating without much friction on an ocean of liquid water (Figure 12.6). Figure 12.6. Very High-Resolution Galileo Image of One Young Double Ridge on Europa. The area in this picture is only 15 kilometers across. It appears to have formed when viscous icy material was forced up through a long, straight crack in the crust. Note how the young ridge going from top left toward bottom right lies on top of older features, which are themselves on top of even older ones. (credit: modification of work by NASA/JPL) 414 Chapter 12 Rings, Moons, and Pluto The close-up Galileo images appear to confirm the existence of a global ocean. In many places, the surface of Europa looks just as we would expect for a thick layer of ice that was broken up into giant icebergs and ice floes and then refrozen in place. When the ice breaks, water or slush from below may be able to seep up through the cracks and make the ridges and multiple-line features we observe. Many episodes of ice cracking, shifting, rotating, and refreezing are required to explain the complexity we see. The icy crust might vary in thickness from a kilometer or so up to 20 kilometers. Further confirmation that a liquid ocean exists below the ice comes from measurements of the small magnetic field induced by Europa\u2019s interactions with the magnetosphere of Jupiter. The \u201cmagnetic signature\u201d of Europa is that of a liquid water ocean, not one of ice or rock. If Europa really has a large ocean of liquid water under its ice, then it may be the only place in the solar system, other than Earth, with really large amounts of liquid water.[1] To remain liquid, this ocean must be warmed by heat escaping from the interior of Europa. Hot (or at least warm) springs might be active there, analogous to those we have discovered in the deep oceans of Earth. The necessary internal heat is generated by tidal heating (see the discussion later in this chapter). short film (https://openstax.org/l/30Europa) with planetary scientist Kevin Hand explains why Europa is so interesting for future exploration. Or listen to this more in-depth talk (https://openstax.org/l/30Europa2) on Europa. What makes the idea of an ocean with warm springs exciting is the discovery in Earth\u2019s oceans of large ecosystems clustered around deep ocean hot springs. Such life derives all its energy from the mineral-laden water and thrives independent of the sunlight shining on Earth\u2019s surface. Is it possible that similar ecosystems could exist today under the ice of Europa? Many scientists now think that Europa is the most likely place beyond Earth to find life in the solar system. In response, NASA is designing a Europa mission to characterize its liquid ocean and its ice crust, and to identify locations where material from inside has risen to the surface. Such interior material might reveal direct evidence for microbial life. In planning a future mission, it may be possible to include a small lander craft as well. lo, a Volcanic Moon Io, the innermost of Jupiter\u2019s Galilean moons, is in many ways a close twin of our Moon, with nearly the same size and density. We might therefore expect it to have experienced a similar history. Its appearance, as photographed from space, tells us another story, however (Figure 12.7). Instead of being a dead cratered world, Io turns out to have the highest level of volcanism in the solar system, greatly exceeding that of Earth. 1 Ganymede and Saturn\u2019s moon Enceladus may have smaller amounts of liquid water under their surfaces. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 415 Figure 12.7. Two Sides of Io. This composite image shows both sides of the volcanically active moon Io. The orange deposits are sulfur snow; the white is sulfur dioxide. (Carl Sagan once quipped that Io looks as if it desperately needs a shot of penicillin.) (credit: modification of work by NASA/JPL/USGS) Io", "\u2019s active volcanism was discovered by the Voyager spacecraft. Eight volcanoes were seen erupting when Voyager 1 passed in March 1979, and six of these were still active four months later when Voyager 2 passed. With the improved instruments carried by the Galileo spacecraft, more than 50 eruptions were found during 1997 alone. Many of the eruptions produce graceful plumes that extend hundreds of kilometers out into space (Figure 12.8). Figure 12.8. Volcanic Eruptions on Io. This composite image from NASA\u2019s Galileo spacecraft shows close-ups (the two inset photos) of two separate volcanic eruptions on Jupiter\u2019s volcanic moon, Io. In the upper inset image, you can see a close up of a bluish plume rising about 140 kilometers above the surface of the volcano. In the lower inset image is the Prometheus plume, rising about 75 kilometers from Io\u2019s surface. The Prometheus plume is named for the Greek god of fire. (credit: modification of work by NASA/JPL) 416 Chapter 12 Rings, Moons, and Pluto Watch a brief movie (https://openstax.org/l/30IoSurf) made from Voyager and Galileo data, showing a rotating Io with its dramatic surface features. The Galileo data show that most of the volcanism on Io consists of hot silicate lava, like the volcanoes on Earth. Sometimes the hot lava encounters frozen deposits of sulfur and sulfur dioxide. When these icy deposits are suddenly heated, the result is great eruptive plumes far larger than any ejected from terrestrial volcanoes. As the rising plumes cool, the sulfur and sulfur dioxide recondense as solid particles that fall back to the surface in colorful \u201csnowfalls\u201d that extend as much as a thousand kilometers from the vent. Major new surface features were even seen to appear between Galileo orbits, as shown in Figure 12.9. Figure 12.9. Volcanic Changes on Io. These three images were taken of the same 1700-kilometer-square region of Io in April 1997, September 1997, and July 1999. The dark volcanic center called Pillan Patera experienced a huge eruption, producing a dark deposit some 400 kilometers across (seen as the grey area in the upper center of the middle image). In the right image, however, some of the new dark deposit is already being covered by reddish material from the volcano Pele. Also, a small unnamed volcano to the right of Pillan has erupted since 1997, and some of its dark deposit and a yellow ring around it are visible on the right image (to the right of the grey spot). The color range is exaggerated in these images. (credit: modification of work by NASA/JPL/University of Arizona) As the Galileo mission drew to a close, controllers were willing to take risks in getting close to Io. Approaching this moon is a dangerous maneuver because the belts of atomic particles trapped in Jupiter\u2019s magnetic environment are at their most intense near Io\u2019s orbit. Indeed, in its very first pass by Io, the spacecraft absorbed damaging radiation beyond its design levels. To keep the system working at all, controllers had to modify or disable various fault-protection software routines in the onboard computers. In spite of these difficulties, the spacecraft achieved four successful Io flybys, obtaining photos and spectra of the surface with unprecedented resolution. Maps of Io reveal more than 100 recently active volcanoes. Huge flows spread out from many of these vents, covering about 25% of the moon\u2019s total surface with still-warm lava. From these measurements, it seems clear that the bright surface colors that first attracted attention to Io are the result of a thin veneer of sulfur compounds. The underlying volcanism is driven by eruptions of molten silicates, just like on Earth (Figure 12.10). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 417 Figure 12.10. Lava Fountains on Io. Galileo captured a number of eruptions along the chain of huge volcanic calderas (or pits) on Io called Tvashtar Catena in this false-color image combining infrared and visible light. The bright orange-yellow areas at left are places where fresh, hot lava is erupting from below ground. (credit: modification of work by NASA/JPL) Tidal Heating How can Io remain volcanically active in spite of its small size? The answer, as we hinted earlier, lies in the effect of gravity, through tidal heating. Io is about the same distance from Jupiter as our Moon is from Earth. Yet Jupiter is more than 300 times more massive than Earth, causing forces that pull Io into an elongated shape, with a several-kilometer-high bulge extending toward Jupiter. If Io always kept exactly the same face turned toward Jupiter, this bulge would not generate heat", ". However, Io\u2019s orbit is not exactly circular due to gravitational perturbations (tugs) from Europa and Ganymede. In its slightly eccentric orbit, Io twists back and forth with respect to Jupiter, at the same time moving nearer and farther from the planet on each revolution. The twisting and flexing heat Io, much as repeated flexing of a wire coat hanger heats the wire. After billions of years, this constant flexing and heating have taken their toll on Io, driving away water and carbon dioxide and other gases, so that now sulfur and sulfur compounds are the most volatile materials remaining. Its interior is entirely melted, and the crust itself is constantly recycled by volcanic activity. In moving inward toward Jupiter from Callisto to Io, we have encountered more and more evidence of geological activity and internal heating, culminating in the violent volcanism on Io. Three of these surfaces are compared in Figure 12.11. Just as the character of the planets in our solar system depends in large measure on their distance from the Sun (and on the amount of heat they receive), so it appears that distance from a giant planet like Jupiter can play a large role in the composition and evolution of its moons (at least partly due to differences in internal heating of each moon by Jupiter\u2019s unrelenting tidal forces). 418 Chapter 12 Rings, Moons, and Pluto Figure 12.11. Three Icy Moons. These Galileo images compare the surfaces of Europa, Ganymede, and Callisto at the same resolution. Note that the number of craters (and thus the age of the surface we see) increases as we go from Europa to Ganymede to Callisto. The Europa image is one of those where the system of cracks and ridges resembles a freeway system. (credit: modification of work by NASA/JPL/DLR) 12.3 TITAN AND TRITON Learning Objectives By the end of this section, you will be able to: Explain how the thick atmosphere of Titan makes bodies of liquid on its surface possible Describe what we learned from the landing on Titan with the Huygens probe Discuss the features we observed on the surface of Triton when Voyager 2 flew by We shift our attention now to small worlds in the more distant parts of the solar system. Saturn\u2019s large moon Titan turns out to be a weird cousin of Earth, with many similarities in spite of frigid temperatures. The Cassini observations of Titan have provided some of the most exciting recent discoveries in planetary science. Neptune\u2019s moon Triton also has unusual characteristics and resembles Pluto, which we will discuss in the following section. Titan, a Moon with Atmosphere and Hydrocarbon Lakes Titan, first seen in 1655 by the Dutch astronomer Christiaan Huygens, was the first moon discovered after Galileo saw the four large moons of Jupiter. Titan has roughly the same diameter, mass, and density as Callisto or Ganymede. Presumably it also has a similar composition\u2014about half ice and half rock. However, Titan is unique among moons, with a thick atmosphere and lakes and rivers and falling rain (although these are not composed of water but of hydrocarbons such as ethane and methane, which can stay liquid at the frigid temperatures on Titan). The 1980 Voyager flyby of Titan determined that the surface density of its atmosphere is four times greater than that on Earth. The atmospheric pressure on this moon is 1.6 bars, higher than that on any other moon and, remarkably, even higher than that of the terrestrial planets Mars and Earth. The atmospheric composition is primarily nitrogen, an important way in which Titan\u2019s atmosphere resembles Earth\u2019s. Also detected in Titan\u2019s atmosphere were carbon monoxide (CO), hydrocarbons (compounds of hydrogen and carbon) such as methane (CH4), ethane (C2H6), and propane (C3H8), and nitrogen compounds such as hydrogen cyanide (HCN), cyanogen (C2N2), and cyanoacetylene (HC3N). Their presence indicates an active chemistry in This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 419 which sunlight interacts with atmospheric nitrogen and methane to create a rich mix of organic molecules. There are also multiple layers of hydrocarbon haze and clouds in the atmosphere, as illustrated in Figure 12.12. Figure 12.12. Structure of Titan\u2019s Atmosphere. Some characteristics of Titan\u2019s atmosphere resemble those of Earth\u2019s atmosphere, although it is much colder than our planet. The red line indicates the temperature of Titan\u2019s atmosphere at different altitudes. These Voyager discoveries motivated a much more ambitious exploration program using the NASA Cassini Saturn orbiter and a probe to land on Titan called Huygens, built by the European Space Agency. The orbiter, which included several cameras, spect", "rometers, and a radar imaging system, made dozens of close flybys of Titan between 2004 and 2015, each yielding radar and infrared images of portions of the surface (see Exploring the Outer Planets). The Huygens probe successfully descended by parachute through the atmosphere, photographing the surface from below the clouds, and landing on January 14, 2005. This was the first (and so far the only) spacecraft landing on a moon in the outer solar system. At the end of its parachute descent, the 319-kilogram Huygens probe safely touched down, slid a short distance, and began sending data back to Earth, including photos and analyses of the atmosphere. It appeared to have landed on a flat, boulder-strewn plain, but both the surface and the boulders were composed of water ice, which is as hard as rock at the temperature of Titan (see Figure 12.13). The photos taken during descent showed a variety of features, including drainage channels, suggesting that Huygens had landed on the shore of an ancient hydrocarbon lake. The sky was deep orange, and the brightness of the Sun was a thousand times less than sunlight on Earth (but still more than a hundred times brighter than under the full moon on Earth). Titan\u2019s surface temperature was 94 K (\u2212179 \u00b0C). The warmer spacecraft heated enough of the ice where it landed for its instruments to measure released hydrocarbon gas. Measurements on the surface continued for more than an hour before the probe succumbed to the frigid temperature. 420 Chapter 12 Rings, Moons, and Pluto Figure 12.13. Views of the Surface of Titan. The left image shows the views of Titan from the descent camera, in a flattened projection, at different altitudes. The right image, taken after landing, shows a boulder-strewn surface illuminated by faint reddish sunlight. The boulders are composed of water ice. (credit left: modification of work by ESA/NASA/JPL/University of Arizona; credit right: modification of work by ESA/NASA/ JPL/University of Arizona; processed by Andrey Pivovarov) Radar and infrared imaging of Titan from the Cassini orbiter gradually built up a picture of a remarkably active surface on this moon, complex and geologically young (Figure 12.14). There are large methane lakes near the polar regions that interact with the methane in the atmosphere, much as Earth\u2019s water oceans interact with the water vapor in our atmosphere. The presence of many erosional features indicates that atmospheric methane can condense and fall as rain, then flow down valleys to the big lakes. Thus, Titan has a low-temperature equivalent of the water cycle on Earth, with liquid on the surface that evaporates, forms clouds, and then condenses to fall as rain\u2014but on Titan the liquid is a combination of methane, ethane, and a trace of other hydrocarbons. It is a weirdly familiar and yet utterly alien landscape. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 421 Figure 12.14. Titan\u2019s Lakes. (a) This Cassini image from a September 2006 flyby shows the liquid lakes on Titan. Their composition is most likely a combination of methane and ethane. (Since this is a radar image, the colors are artificially added. The dark blue areas are the smooth surfaces of the liquid lakes, and yellow is the rougher solid terrain around them.) (b) This mosaic of Titan\u2019s surface from the Cassini-Huygens mission shows in detail a high ridge area and many narrow, sinuous erosion channels that appear to be part of a widespread network of \u201crivers\u201d carved by flowing hydrocarbons. (credit a: modification of work by NASA/JPL-Caltech/USGS; credit b; modification of work by NASA/JPL/ ESA/University of Arizona) These discoveries raise the question of whether there could be life on Titan. Hydrocarbons are fundamental for the formation of the large carbon molecules that are essential to life on our planet. However, the temperature on Titan is far too low for liquid water or for many of the chemical processes that are essential to life as we know it. There remains, though, an intriguing possibility that Titan might have developed a different form of low-temperature carbon-based life that could operate with liquid hydrocarbons playing the role of water. The discovery of such \u201clife as we don\u2019t know it\u201d could be even more exciting than finding life like ours on Mars. If such a truly alien life is present on Titan, its existence would greatly expand our understanding of the nature of life and of habitable environments The Cassini mission scientists and the visual presentation specialists at NASA\u2019s Jet Propulsion Laboratory have put together some nice films from the images taken by Cassini and Huygens. See", ", for example, the Titan approach (https://openstax.org/l/30Titan) and the flyover (https://openstax.org/l/30Titan2) of the Northern lakes district. Triton and Its Volcanoes Neptune\u2019s largest moon Triton (don\u2019t get its name confused with Titan) has a diameter of 2720 kilometers and a density of 2.1 g/cm3, indicating that it\u2019s probably composed of about 75% rock mixed with 25% water ice. Measurements indicate that Triton\u2019s surface has the coldest temperature of any of the worlds our robot 422 Chapter 12 Rings, Moons, and Pluto representatives have visited. Because its reflectivity is so high (about 80%), Triton reflects most of the solar energy that falls on it, resulting in a surface temperature between 35 and 40 K. The surface material of Triton is made of frozen water, nitrogen, methane, and carbon monoxide. Methane and nitrogen exist as gas in most of the solar system, but they are frozen at Triton\u2019s temperatures. Only a small quantity of nitrogen vapor persists to form an atmosphere. Although the surface pressure of this atmosphere is only 16 millionths of a bar, this is sufficient to support thin haze or cloud layers. Triton\u2019s surface, like that of many other moons in the outer solar system, reveals a long history of geological evolution (Figure 12.15). Although some impact craters are found, many regions have been flooded fairly recently by the local version of \u201clava\u201d (perhaps water or water-ammonia mixtures). There are also mysterious regions of jumbled or mountainous terrain. Figure 12.15. Neptune\u2019s Moon Triton. This mosaic of Voyager 2 images of Triton shows a wide range of surface features. The pinkish area at the bottom is Triton\u2019s large southern polar cap. The south pole of Triton faces the Sun here, and the slight heating effect is driving some of the material northward, where it is colder. (credit: modification of work by NASA/JPL/USGS) The Voyager flyby of Triton took place at a time when the moon\u2019s southern pole was tipped toward the Sun, allowing this part of the surface to enjoy a period of relative warmth. (Remember that \u201cwarm\u201d on Triton is still outrageously colder than anything we experience on Earth.) A polar cap covers much of Triton\u2019s southern hemisphere, apparently evaporating along the northern edge. This polar cap may consist of frozen nitrogen that was deposited during the previous winter. Remarkably, the Voyager images showed that the evaporation of Triton\u2019s polar cap generates geysers or volcanic plumes of nitrogen gas (see Figure 12.16). (Fountains of such gas rose about 10 kilometers high, visible in the thin atmosphere because dust from the surface rose with them and colored them dark.) These plumes differ from the volcanic plumes of Io in their composition and also in that they derive their energy from sunlight warming the surface rather than from internal heat. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 423 Figure 12.16. Triton\u2019s Geysers. This close-up view shows some of the geysers on Neptune\u2019s moon Triton, with the long trains of dust pointing to the lower right in this picture. (credit: modification of work by NASA/JPL) 12.4 PLUTO AND CHARON Learning Objectives By the end of this section, you will be able to: Compare the orbital characteristics of Pluto with those of the planets Describe information about Pluto\u2019s surface deduced from the New Horizons images Note some distinguishing characteristics of Pluto\u2019s large moon Charon Pluto is not a moon, but we discuss it here because its size and composition are similar to many moons in the outer solar system. Our understanding of Pluto (and its large moon Charon) have changed dramatically as a result of the New Horizons flyby in 2015. Is Pluto a Planet? Pluto was discovered through a careful, systematic search, unlike Neptune, whose position was calculated from gravitational theory. Nevertheless, the history of the search for Pluto began with indications that Uranus had slight departures from its predicted orbit, departures that could be due to the gravitation of an undiscovered \u201cPlanet X.\u201d Early in the twentieth century, several astronomers, most notably Percival Lowell, then at the peak of his fame as an advocate of intelligent life on Mars, became interested in searching for this ninth planet. Lowell and his contemporaries based their calculations primarily on tiny unexplained irregularities in the motion of Uranus. Lowell\u2019s computations indicated two possible locations for", " a perturbing Planet X; the more likely of the two was in the constellation Gemini. He predicted a mass for the planet intermediate between the masses of Earth and Neptune (his calculations gave about 6 Earth masses). Other astronomers, however, obtained other solutions from the tiny orbital irregularities, even including one model that indicated two planets beyond Neptune. 424 Chapter 12 Rings, Moons, and Pluto At his Arizona observatory, Lowell searched without success for the unknown planet from 1906 until his death in 1916, and the search was not renewed until 1929. In February 1930, a young observing assistant named Clyde Tombaugh (see the Clyde Tombaugh: From the Farm to Fame feature box), comparing photographs he made on January 23 and 29 of that year, found a faint object whose motion appeared to be about right for a planet far beyond the orbit of Neptune (Figure 12.17). The new planet was named for Pluto, the Roman god of the underworld, who dwelt in remote darkness, just like the new planet. The choice of this name, among hundreds suggested, was helped by the fact that the first two letters were Percival Lowell\u2019s initials. Figure 12.17. Pluto\u2019s Motion. Portions of the two photographs by which Clyde Tombaugh discovered Pluto in 1930. The left one was taken on January 23 and the right on January 29. Note that Pluto, indicated by an arrow, has moved among the stars during those six nights. If we hadn\u2019t put an arrow next to it, though, you probably would never have spotted the dot that moved. (credit: modification of work by the Lowell Observatory Archives) Although the discovery of Pluto appeared initially to be a vindication of gravitational theory similar to the earlier triumph of Adams and Le Verrier in predicting the position of Neptune, we now know that Lowell\u2019s calculations were wrong. When its mass and size were finally measured, it was found that Pluto could not possibly have exerted any measurable pull on either Uranus or Neptune. Astronomers are now convinced that the reported small anomalies in the motions of Uranus are not, and never were, real. From the time of its discovery, it was clear that Pluto was not a giant like the other four outer solar system planets. For a long time, it was thought that the mass of Pluto was similar to that of Earth, so that it was classed as a fifth terrestrial planet, somehow misplaced in the far outer reaches of the solar system. There were other anomalies, however, as Pluto\u2019s orbit was more eccentric and inclined to the plane of our solar system than that of any other planet. Only after the discovery of its moon Charon in 1978 could the mass of Pluto be measured, and it turned out to be far less than the mass of Earth. In addition to Charon, Pluto has four small moons. Subsequent observations of Charon showed that this moon is in a retrograde orbit and has a diameter of about 1200 kilometers, more than half the size of Pluto itself (Figure 12.18). This makes Charon the moon whose size is the largest fraction of its parent planet. We could even think of Pluto and Charon as a double world. Seen from Pluto, Charon would be as large as eight full moons on Earth. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 425 Figure 12.18. Comparison of the Sizes of Pluto and Its Moon Charon with Earth. This graphic vividly shows how tiny Pluto is relative to a terrestrial planet like Earth. That is the primary justification for putting Pluto in the class of dwarf planets rather than terrestrial planets. (credit: modification of work by NASA) To many astronomers, Pluto seemed like the odd cousin that everyone hopes will not show up at the next family reunion. Neither its path around the Sun nor its size resembles either the giant planets or the terrestrial planets. In the 1990s, astronomers began to discover additional small objects in the far outer solar system, showing that Pluto was not unique. We will discuss these trans-neptunian objects later with other small bodies, in the chapter on Comets and Asteroids: Debris of the Solar System. One of them (called Eris) is nearly the same size as Pluto, and another ( Makemake) is substantially smaller. It became clear to astronomers that Pluto was so different from the other planets that it needed a new classification. Therefore, it was called a dwarf planet, meaning a planet much smaller than the terrestrial planets. We now know of many small objects in the vicinity of Pluto and we have classified several as dwarf planets. A similar history was associated with the discovery of the asteroids. When the first asteroid ( Ceres) was discovered at the beginning of the nineteenth century, it was hailed as a new planet. In the following years, however, other objects were found with similar orbits to Ceres. Astronomers decided that these should not all be considered planets, so they invented a new", " class of objects, called minor planets or asteroids. Today, Ceres is also called a dwarf planet. Both minor planets and dwarf planets are part of a whole belt or zones of similar objects (as we will discuss in Comets and Asteroids: Debris of the Solar System). So, is Pluto a planet? Our answer is yes, but it is a dwarf planet, clearly not in the same league with the eight major planets (four giants and four terrestrials). While some people were upset when Pluto was reclassified, we might point out that a dwarf tree is still a type of tree and (as we shall see) a dwarf galaxy is still a type of galaxy. 426 Chapter 12 Rings, Moons, and Pluto Clyde Tombaugh: From the Farm to Fame Clyde Tombaugh discovered Pluto when he was 24 years old, and his position as staff assistant at the Lowell Observatory was his first paying job. Tombaugh had been born on a farm in Illinois, but when he was 16, his family moved to Kansas. There, with his uncle\u2019s encouragement, he observed the sky through a telescope the family had ordered from the Sears catalog. Tombaugh later constructed a larger telescope on his own and devoted his nights (when he wasn\u2019t too tired from farm work) to making detailed sketches of the planets (Figure 12.19). Figure 12.19. Clyde Tombaugh (1906\u20131997). (a) Tombaugh is pictured on his family farm in 1928 with a 9-inch telescope he built. (b) Here Tombaugh is looking through an eyepiece at the Lowell Observatory. (credit b: modification of work by NASA) In 1928, after a hailstorm ruined the crop, Tombaugh decided he needed a job to help support his family. Although he had only a high school education, he thought of becoming a telescope builder. He sent his planet sketches to the Lowell Observatory, seeking advice about whether such a career choice was realistic. By a wonderful twist of fate, his query arrived just when the Lowell astronomers realized that a renewed search for a ninth planet would require a very patient and dedicated observer. The large photographic plates (pieces of glass with photographic emulsion on them) that Tombaugh was hired to take at night and search during the day contained an average of about 160,000 star images each. How to find Pluto among them? The technique involved taking two photographs about a week apart. During that week, a planet would move a tiny bit, while the stars remained in the same place relative to each other. A new instrument called a \u201cblink comparator\u201d could quickly alternate the two images in an eyepiece. The stars, being in the same position on the two plates, would not appear to change as the two images were \u201cblinked.\u201d But a moving object would appear to wiggle back and forth as the plates were alternated. After examining more than 2 million stars (and many false alarms), Tombaugh found his planet on February 18, 1930. The astronomers at the observatory checked his results carefully, and the find was announced on March 13, the 149th anniversary of the discovery of Uranus. Congratulations and requests for interviews poured in from around the world. Visitors descended on the observatory in scores, wanting to see the place where the first new planet in almost a century had been discovered, as well as the person who had discovered it. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 427 In 1932, Tombaugh took leave from Lowell, where he had continued to search and blink, to get a college degree. Eventually, he received a master\u2019s degree in astronomy and taught navigation for the Navy during World War II. In 1955, after working to develop a rocket-tracking telescope, he became a professor at New Mexico State University, where he helped found the astronomy department. He died in 1997; some of his ashes were placed inside the New Horizons spacecraft to Pluto Here is a touching video (https://openstax.org/l/30Tbaugh) about Tombaugh\u2019s life as described by his children. The Nature of Pluto Using data from the New Horizons probe, astronomers have measured the diameter of Pluto as 2370 kilometers, only 60 perent as large as our Moon. From the diameter and mass, we find a density of 1.9 g/cm3, suggesting that Pluto is a mixture of rocky materials and water ice in about the same proportions as many outer-planet moons. Parts of Pluto\u2019s surface are highly reflective, and its spectrum demonstrates the presence on its surface of frozen methane, carbon monoxide, and nitrogen. The maximum surface temperature ranges from about 50 K when Pluto is farthest from the Sun to 60 K when it is closest. Even this small difference is enough to cause a partial sublimation (going from solid to gas) of the methane and nitrogen ice. This generates", " an atmosphere when Pluto is close to the Sun, and it freezes out when Pluto is farther away. Observations of distant stars seen through this thin atmosphere indicate that the surface pressure is about a ten-thousandth of Earth\u2019s. Because Pluto is a few degrees warmer than Triton, its atmospheric pressure is about ten times greater. This atmosphere contains several distinct haze layers, presumably caused by photochemical reactions, like those in Titan\u2019s atmosphere (Figure 12.20). Figure 12.20. Haze Layers in the Atmosphere of Pluto. This is one of the highest-resolution photos of Pluto, taken by the New Horizons spacecraft 15 minutes after its closest approach. It shows 12 layers of haze. Note also the range of mountains with heights up to 3500 meters. (credit: modification of work by NASA/Johns Hopkins University Applied Physics Laboratory/Southwest Research Institute) Reaching Pluto with a spacecraft was a major challenge, especially in an era when reduced NASA budgets could not support large, expensive missions like Galileo and Cassini. Yet like Galileo and Cassini, a Pluto mission 428 Chapter 12 Rings, Moons, and Pluto would require a nuclear electric system that used the heat from plutonium to generate the energy to power the instruments and keep them operating far from the warmth of the Sun. NASA made available one of the last of its nuclear generators for such a mission. Assuming an affordable but highly capable spacecraft could be built, there was still the problem of getting to Pluto, nearly 5 billion kilometers from Earth, without waiting decades. The answer was to use Jupiter\u2019s gravity to slingshot the spacecraft toward Pluto. The 2006 launch of New Horizons started the mission with a high speed, and the Jupiter flyby just a year later gave it the required additional boost. The New Horizons spacecraft arrived at Pluto in July 2015, traveling at a relative speed of 14 kilometers per second (or about 50,000 kilometers per hour). With this high speed, the entire flyby sequence was compressed into just one day. Most of the data recorded near closest approach could not be transmitted to Earth until many months later, but when it finally arrived, astronomers were rewarded with a treasure trove of images and data. First Close-up Views of Pluto Pluto is not the geologically dead world that many anticipated for such a small object\u2014far from it. The division of the surface into areas with different composition and surface texture is apparent in the global color photo shown in Figure 12.21. The reddish color is enhanced in this image to bring out differences in color more clearly. The darker parts of the surface appear to be cratered, but adjacent to them is a nearly featureless light area in the lower right quadrant of this image. The dark areas show the colors of photochemical haze or smog similar to that in the atmosphere of Titan. The dark material that is staining these old surfaces could come from Pluto\u2019s atmospheric haze or from chemical reactions taking place at the surface due to the action of sunlight. The light areas in the photo are lowland basins. These are apparently seas of frozen nitrogen, perhaps many kilometers deep. Both nitrogen and methane gas are able to escape from Pluto when it is in the part of its orbit close to the Sun, but only very slowly, so there is no reason that a vast bowl of frozen nitrogen could not persist for a long time. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 429 Figure 12.21. Global Color Image of Pluto. This New Horizons image clearly shows the variety of terrains on Pluto. The dark area in the lower left is covered with impact craters, while the large light area in the center and lower right is a flat basin devoid of craters. The colors you see are somewhat enhanced to bring out subtle differences. (credit: modification of work by NASA/Johns Hopkins University Applied Physics Laboratory/Southwest Research Institute) Figure 12.22 shows some of the remarkable variety of surface features New Horizons revealed. At the right of this image we see the \u201cshoreline\u201d of the vast bowl of nitrogen ice we saw as the smooth region in Figure 12.21. Temporarily nicknamed the \u201cSputnik Plains,\u201d after the first human object to get into space, this round region is roughly a thousand kilometers wide and shows intriguing cells or polygons that have an average width of more than 30 kilometers. The mountains in the middle are great blocks of frozen water ice, some reaching heights of 2 to 3 kilometers. Figure 12.22. Diversity of Terrain on Pluto. This enhanced color view of a strip of Pluto\u2019s surface about 80 kilometers long shows a variety of different surface features. From left to right, we first cross a region of \u201cbadlands\u201d with some craters showing, and then move across a wide range of mountains made of water ice and coated with the redder material we saw in the", " previous image. Then, at right, we arrive at the \u201cshoreline\u201d of the great sea of frozen nitrogen that the mission scientists have nicknamed the \u201cSputnik Plains.\u201d This nitrogen sea is divided into mysterious cells or segments that are many kilometers across. (credit: modification of work by NASA/Johns Hopkins University Applied Physics Laboratory/Southwest Research Institute) 430 Chapter 12 Rings, Moons, and Pluto Figure 12.23 shows another view of the boundary between different types of geology. The width of this image is 250 kilometers, and it shows dark, ancient, heavily cratered terrain; dark, uncratered terrain with a hilly surface; smooth, geologically young terrain; and a small cluster of mountains more than 3000 meters high. In the best images, the light areas of nitrogen ice seem to have flowed much like glaciers on Earth, covering some of the older terrain underneath them. The isolated mountains in the midst of the smooth nitrogen plains are probably also made of water ice, which is very hard at the temperatures on Pluto and can float on frozen nitrogen. Additional mountains, and some hilly terrain that reminded the mission scientists of snakeskin, are visible in part (b) of Figure 12.23. These are preliminary interpretations from just the first data coming back from New Horizons in 2015 and early 2016. As time goes on, scientists will have a better understanding of the unique geology of Pluto. Figure 12.23. Diversity of Terrains on Pluto. (a) In this photo, about 250 kilometers across, we can see many different kinds of terrain. At the bottom are older, cratered highlands; a V-shaped region of hills without cratering points toward the bottom of the image. Surrounding the Vshaped dark region is the smooth, brighter frozen nitrogen plain, acting as glaciers on Earth do. Some isolated mountains, made of frozen water ice, are floating in the nitrogen near the top of the picture. (b) This scene is about 390 kilometers across. The rounded mountains, quite different from those we know on Earth, are named Tartarus Dorsa. The patterns, made of repeating ridges with the more reddish terrain between them, are not yet understood. (credit a, b: modification of work by NASA/Johns Hopkins University Applied Physics Laboratory/ Southwest Research Institute) A Quick Look at Charon To add to the mysteries of Pluto, we show in Figure 12.24 one of the best New Horizons images of Pluto\u2019s large moon Charon. Recall from earlier that Charon is roughly half Pluto\u2019s size (its diameter is about the size of Texas). Charon keeps the same side toward Pluto, just as our Moon keeps the same side toward Earth. What is unique about the Pluto-Charon system, however, is that Pluto also keeps its same face toward Charon. Like two dancers embracing, these two constantly face each other as they spin across the celestial dance floor. Astronomers call this a double tidal lock. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 431 Figure 12.24. Pluto\u2019s Large Moon Charon. (a) In this New Horizons image, the color has been enhanced to bring out the color of the moon\u2019s strange red polar cap. Charon has a diameter of 1214 kilometers, and the resolution of this image is 3 kilometers. (b) Here we see the moon from a slightly different angle, in true color. The inset shows an area about 390 kilometers from top to bottom. Near the top left is an intriguing feature\u2014what appears to be a mountain in the middle of a depression or moat. (credit a, b: modification of work by NASA/JHUAPL/SwRI) What New Horizons showed was another complex world. There are scattered craters in the lower part of the image, but much of the rest of the surface appears smooth. Crossing the center of the image is a belt of rough terrain, including what appear to be tectonic valleys, as if some forces had tried to split Charon apart. Topping off this strange image is a distinctly red polar cap, of unknown composition. Many features on Charon are not yet understood, including what appears to be a mountain in the midst of a low-elevation region. 12.5 PLANETARY RINGS Learning Objectives By the end of this section, you will be able to: Describe the two theories of planetary ring formation Compare the major rings of Saturn and explain the role of the moon Enceladus in the formation of the E ring Explain how the rings of Uranus and Neptune differ in composition and appearance from the rings of Saturn Describe how ring structure is affected by the presence of moons In addition to their moons, all four of the giant planets have rings, with each ring system consisting of billions of small particles or \u201c", "moonlets\u201d orbiting close to their planet. Each of these rings displays a complicated structure that is related to interactions between the ring particles and the larger moons. However, the four ring systems are very different from each other in mass, structure, and composition, as outlined in Table 12.2. 432 Chapter 12 Rings, Moons, and Pluto Properties of the Ring Systems Planet Outer Radius (km) Outer Radius (Rplanet) Jupiter 128,000 Saturn 140,000 Uranus 51,000 Neptune 63,000 Table 12.2 1.8 2.3 2.2 2.5 Reflectivity (%) Mass (kg) 1010(?) 1019 1014 1012? 60 5 5 Saturn\u2019s large ring system is made up of icy particles spread out into several vast, flat rings containing a great deal of fine structure. The Uranus and Neptune ring systems, on the other hand, are nearly the reverse of Saturn\u2019s: they consist of dark particles confined to a few narrow rings with broad empty gaps in between. Jupiter\u2019s ring and at least one of Saturn\u2019s are merely transient dust bands, constantly renewed by dust grains eroded from small moons. In this section, we focus on the two most massive ring systems, those of Saturn and Uranus. What Causes Rings? A ring is a collection of vast numbers of particles, each like a tiny moon obeying Kepler\u2019s laws as it follows its own orbit around the planet. Thus, the inner particles revolve faster than those farther out, and the ring as a whole does not rotate as a solid body. In fact, it is better not to think of a ring rotating at all, but rather to consider the revolution (or motion in orbit) of its individual moonlets. If the ring particles were widely spaced, they would move independently, like separate moonlets. However, in the main rings of Saturn and Uranus the particles are close enough to exert mutual gravitational influence, and occasionally even to rub together or bounce off each other in low-speed collisions. Because of these interactions, we see phenomena such as waves that move across the rings\u2014just the way water waves move over the surface of the ocean. There are two basic ideas of how such rings come to be. First is the breakup hypothesis, which suggests that the rings are the remains of a shattered moon. A passing comet or asteroid might have collided with the moon, breaking it into pieces. Tidal forces then pulled the fragments apart, and they dispersed into a disk. The second hypothesis, which takes the reverse perspective, suggests that the rings are made of particles that were unable to come together to form a moon in the first place. In either theory, the gravity of the planet plays an important role. Close to the planet (see Figure 12.25), tidal forces can tear bodies apart or inhibit loose particles from coming together. We do not know which explanation holds for any given ring, although many scientists have concluded that at least a few of the rings are relatively young and must therefore be the result of breakup. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 433 Figure 12.25. Four Ring Systems. This diagram shows the locations of the ring systems of the four giant planets. The left axis represents the planet\u2019s surface. The dotted vertical line is the limit inside which gravitational forces can break up moons (each planet\u2019s system is drawn to a different scale, so that this stability limit lines up for all four of them). The black dots are the inner moons of each planet on the same scale as its rings. Notice that only really small moons survive inside the stability limit. Rings of Saturn Saturn\u2019s rings are one of the most beautiful sights in the solar system (Figure 12.26). From outer to inner, the three brightest rings are labeled with the extremely unromantic names of A, B, and C Rings. Table 12.3 gives the dimensions of the rings in both kilometers and units of the radius of Saturn, RSaturn. The B Ring is the brightest and has the most closely packed particles, whereas the A and C Rings are translucent. The total mass of the B Ring, which is probably close to the mass of the entire ring system, is about equal to that of an icy moon 250 kilometers in diameter (suggesting that the ring could have originated in the breakup of such a moon). Between the A and B Rings is a wide gap named the Cassini Division after Gian Domenico Cassini, who first glimpsed it through a telescope in 1675 and whose name planetary scientists have also given to the Cassini spacecraft exploring the Saturn system. 434 Chapter 12 Rings, Moons, and Pluto Figure 12.26. Saturn\u2019s Rings as Seen from Above and Below. (a) The view from above is illuminated by direct sunlight. (b) The illumination seen from below is sunlight that has diffused through gaps in the", " rings. (credit a, b: modification of work by NASA/JPL-Caltech/Space Science Institute) Selected Features in the Rings of Saturn Ring Name[2] Outer Edge (RSaturn) Outer Edge (km) Width (km) F A Cassini Division B C Table 12.3 2.324 2.267 2.025 1.949 1.525 140,180 136,780 122,170 117,580 92,000 90 14,600 4590 25,580 17,490 Saturn\u2019s rings are very broad and very thin. The width of the main rings is 70,000 kilometers, yet their average thickness is only 20 meters. If we made a scale model of the rings out of paper, we would have to make them 1 kilometer across. On this scale, Saturn itself would loom as high as an 80-story building. The ring particles are composed primarily of water ice, and they range from grains the size of sand up to house-sized boulders. An insider\u2019s view of the rings would probably resemble a bright cloud of floating snowflakes and hailstones, with a few snowballs and larger objects, many of them loose aggregates of smaller particles (Figure 12.27). 2 The ring letters are assigned in the order of their discovery. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 435 Figure 12.27. Artist\u2019s Idealized Impression of the Rings of Saturn as Seen from the Inside. Note that the rings are mostly made of pieces of water ice of different sizes. At the end of its mission, the Cassini spacecraft is planning to cut through one of the gaps in Saturn\u2019s rings, but it won\u2019t get this close. (credit: modification of work by NASA/JPL/University of Colorado) In addition to the broad A, B, and C Rings, Saturn has a handful of very narrow rings no more than 100 kilometers wide. The most substantial of these, which lies just outside the A Ring, is called the F Ring; its surprising appearance is discussed below. In general, Saturn\u2019s narrow rings resemble the rings of Uranus and Neptune. There is also a very faint, tenuous ring, called the E Ring, associated with Saturn\u2019s small icy moon Enceladus. The particles in the E Ring are very small and composed of water ice. Since such a tenuous cloud of ice crystals will tend to dissipate, the ongoing existence of the E Ring strongly suggests that it is being continually replenished by a source at Enceladus. This icy moon is very small\u2014only 500 kilometers in diameter\u2014but the Voyager images showed that the craters on about half of its surface have been erased, indicating geological activity sometime in the past few million years. It was with great anticipation that the Cassini scientists maneuvered the spacecraft orbit to allow multiple close flybys of Enceladus starting in 2005. Those awaiting the Cassini flyby results were not disappointed. High-resolution images showed long, dark stripes of smooth ground near its south pole, which were soon nicknamed \u201ctiger stripes\u201d (Figure 12.28). Infrared measurements revealed that these tiger stripes are warmer than their surroundings. Best of all, dozens of cryovolcanic vents on the tiger stripes were seen to be erupting geysers of salty water and ice (Figure 12.29). Estimates suggested that 200 kilograms of material were shooting into space each second\u2014not a lot, but enough for the spacecraft to sample. 436 Chapter 12 Rings, Moons, and Pluto Figure 12.28. Enceladus. (a) This image shows both smooth and cratered terrain on Saturn\u2019s moon, and also \u201ctiger stripes\u201d in the south polar region (lower part of image). These dark stripes (shown here in exaggerated color) have elevated temperatures and are the source of the many geysers discovered on Enceladus. They are about 130 kilometers long and 40 kilometers apart. (b) Here Enceladus is shown to scale with Great Britain and the coast of Western Europe, to emphasize that it is a small moon, only about 500 kilometers in diameter. (credit a, b: modification of work by NASA/JPL/Space Science Institute) When Cassini was directed to fly into the plumes, it measured their composition and found them to be similar to material we see liberated from comets (see Comets and Asteroids: Debris of the Solar System). The vapor and ice plumes consisted mostly of water, but with trace amounts of nitrogen, ammonia, methane, and other hydrocarbons. Minerals found in the geysers in trace amounts included ordinary salt, meaning that the geyser plumes were high-pressure sprays of salt water. Based on the continuing study of Enceladus", "\u2019 bulk properties and the ongoing geysers, in 2015 the Cassini mission scientists tentatively identified a subsurface ocean of water feeding the geysers. These discoveries suggested that in spite of its small size, Enceladus should be added to the list of worlds that we would like to explore for possible life. Since its subsurface ocean is conveniently escaping into space, it might be much easier to sample than the ocean of Europa, which is deeply buried below its thick crust of ice. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 437 Figure 12.29. Geysers on Enceladus. This Cassini image shows a number of water geysers on Saturn\u2019s small moon Enceladus, apparently salty water from a subsurface source escaping through cracks in the surface. You can see curved lines of geysers along the four \u201ctiger stripes\u201d on the surface. (credit: modification of work by NASA/JPL/Space Science Institute) Rings of Uranus and Neptune Uranus\u2019 rings are narrow and black, making them almost invisible from Earth. The nine main rings were discovered in 1977 from observations made of a star as Uranus passed in front of it. We call such a passage of one astronomical object in front of another an occultation. During the 1977 occultation, astronomers expected the star\u2019s light to disappear as the planet moved across it. But in addition, the star dimmed briefly several times before Uranus reached it, as each narrow ring passed between the star and the telescope. Thus, the rings were mapped out in detail even though they could not be seen or photographed directly, like counting the number of cars in a train at night by watching the blinking of a light as the cars successively pass in front of it. When Voyager approached Uranus in 1986, it was able to study the rings at close range; the spacecraft also photographed two new rings (Figure 12.30). 438 Chapter 12 Rings, Moons, and Pluto Figure 12.30. Rings of Uranus. The Voyager team had to expose this image for a long time to get a glimpse of Uranus\u2019 narrow dark rings. You can see the grainy structure of \u201cnoise\u201d in the electronics of the camera in the picture background. (credit: modification of work by NASA/JPL) The outermost and most massive of Uranus\u2019 rings is called the Epsilon Ring. It is only about 100 kilometers wide and probably no more than 100 meters thick (similar to the F Ring of Saturn). The Epsilon Ring encircles Uranus at a distance of 51,000 kilometers, about twice the radius of Uranus. This ring probably contains as much mass as all of Uranus\u2019 other ten rings combined; most of them are narrow ribbons less than 10 kilometers wide, just the reverse of the broad rings of Saturn. The individual particles in the uranian rings are nearly as black as lumps of coal. While astronomers do not understand the composition of this material in detail, it seems to consist in large part of carbon and hydrocarbon compounds. Organic material of this sort is rather common in the outer solar system. Many of the asteroids and comets are also composed of dark, tarlike materials. In the case of Uranus, its ten small inner moons have a similar composition, suggesting that one or more moons might have broken up to make the rings. Neptune\u2019s rings are generally similar to those of Uranus but even more tenuous (Figure 12.31). There are only four of them, and the particles are not uniformly distributed along their lengths. Because these rings are so difficult to investigate from Earth, it will probably be a long time before we understand them very well. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 439 Figure 12.31. Rings of Neptune. This long exposure of Neptune\u2019s rings was photographed by Voyager 2. Note the two denser regions of the outer ring. (credit: modification of work by NASA/JPL Mark Showalter (of the SETI Institute) and his colleagues maintain the NASA\u2019s Planetary Ring Node (https://openstax.org/l/30NASArings) website. It is full of information about the rings and their interactions with moons; check out their press-release images of the Saturn ring system, for example. And Showalter gives an entertaining illustrated talk (https://openstax.org/l/30StrnRngs) about Saturn\u2019s ring and moon system Resolution of Planetary Rings Using the occultations of stars by the rings of Saturn, astronomers have been able to measure details in the ring structure to a resolution of 10 km. This is a much higher resolution", " than can be obtained in a conventional photo of the rings. Let\u2019s figure out what angular resolution (in arcsec) a space telescope in Earth orbit would have to achieve to obtain equal resolution. Solution To solve this problem, we use the \u201csmall-angle formula\u201d to relate angular and linear diameters in the sky. For angles in the sky that are small, the formula is usually written as angular diameter 206,265 arcsec = linear diameter distance where angular diameter is expressed in arcsec. The distance of Saturn near opposition is about 9 AU = 1.4 \u00d7 109 km. Substituting in the above formula and solving for the angular resolution, we get 440 Chapter 12 Rings, Moons, and Pluto angular resolution = 206,265 arcsec \u00d7 10 1.4 \u00d7 109 km which is about 10\u22123 arcsec, or a milliarcsec. This is not possible for our telescopes to achieve. For comparison, the best resolution from either the Hubble Space Telescope or ground-based telescopes is about 0.1 arcsec, or 100 times worse than what we would need. This is why such occultation measurements are so useful for astronomers. Check Your Learning How close to Saturn would a spacecraft have to be to make out detail in its rings as small as 20 km, if its camera has an angular resolution of 5 arcsec? Answer: Using our formula, angular diameter 206,265 arcsec = linear diameter distance we get 5 arcsec 206,265 arcsec = 20 km distance. So, the distance is about 825,000 km. Interactions between Rings and Moons Much of our fascination with planetary rings is a result of their intricate structures, most of which owe their existence to the gravitational effect of moons, without which the rings would be flat and featureless. Indeed, it is becoming clear that without moons there would probably be no rings at all because, left to themselves, thin disks of small particles gradually spread and dissipate. Most of the gaps in Saturn\u2019s rings, and also the location of the outer edge of the A Ring, result from gravitational resonances with small inner moons. A resonance takes place when two objects have orbital periods that are exact ratios of each other, such as 1:2 or 2:3. For example, any particle in the gap at the inner side of the Cassini Division of Saturn\u2019s rings would have a period equal to one-half that of Saturn\u2019s moon Mimas. Such a particle would be nearest Mimas in the same part of its orbit every second revolution. The repeated gravitational tugs of Mimas, acting always in the same direction, would perturb it, forcing it into a new orbit outside the gap. In this way, the Cassini Division became depleted of ring material over long periods of time. The Cassini mission revealed a great deal of fine structure in Saturn\u2019s rings. Unlike the earlier Voyager flybys, Cassini was able to observe the rings for more than a decade, revealing a remarkable range of changes, on time scales from a few minutes to several years. Many of the features newly seen in Cassini data indicated the presence of condensations or small moons only a few tens of meters across imbedded in the rings. As each small moon moves, it produces waves in the surrounding ring material like the wake left by a moving ship. Even when the moon is too small to be resolved, its characteristic waves could be photographed by Cassini. One of the most interesting rings of Saturn is the narrow F Ring, which contains several apparent ringlets within its 90-kilometer width. In places, the F Ring breaks up into two or three parallel strands that sometimes show bends or kinks. Most of the rings of Uranus and Neptune are also narrow ribbons like the F Ring of Saturn. Clearly, the gravity of some objects must be keeping the particles in these thin rings from spreading out. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 441 As we have seen, the largest features in the rings of Saturn are produced by gravitational resonances with the inner moons, while much of the fine structure is caused by smaller embedded moons. In the case of Saturn\u2019s F Ring, close-up images revealed that it is bounded by the orbits of two moons, called Pandora and Prometheus (Figure 12.32). These two small moons (each about 100 kilometers in diameter) are referred to as shepherd moons, since their gravitation serves to \u201cshepherd\u201d the ring particles and keep them confined to a narrow ribbon. A similar situation applies to the Epsilon Ring of Uranus, which is shepherded by the moons Cordelia and Ophelia. These two shepherds, each about 50 kilometers in diameter, orbit about 2000 kilometers inside and outside the ring. Figure 12.32. Saturn\u2019s F Ring and Its Shepherd Moons. (a) This", " Cassini image shows the narrow, complex F Ring of Saturn, with its two small shepherd moons Pandora (left) and Prometheus (right). (b) In this closer view, the shepherd moon Pandora (84 kilometers across) is seen next to the F ring, in which the moon is perturbing the main (brightest) strand of ring particles as it passes. You can see the dark side of Pandora on this image because it is being illuminated by the light reflected from Saturn. (credit a, b: modification of work by NASA/JPL/Space Science Institute You can download a movie (https://openstax.org/l/30ShprdMns) showing the two shepherd moons on either side of Saturn\u2019s F ring. Theoretical calculations suggest that the other narrow rings in the uranian and neptunian systems should also be controlled by shepherd moons, but none has been located. The calculated diameter for such shepherds (about 10 kilometers) was just at the limit of detectability for the Voyager cameras, so it is impossible to say whether they are present or not. (Given all the narrow rings we see, some scientists still hope to find another more satisfactory mechanism for keeping them confined.) One of the outstanding problems with understanding the rings is determining their ages. Have the giant planets always had the ring systems we see today, or might these be a recent or transient addition to the solar system? In the case of the main rings of Saturn, their mass is about the same as that of the inner moon 442 Chapter 12 Rings, Moons, and Pluto Mimas. Thus, they could have been formed by the break-up of a Mimas-sized moon, perhaps very early in solar system history, when there were many interplanetary projectiles left over from planet formation. It is harder to understand how such a catastrophic event could have taken place recently, when the solar system had become a more stable place. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 443 CHAPTER 12 REVIEW KEY TERMS resonance an orbital condition in which one object is subject to periodic gravitational perturbations by another, most commonly arising when two objects orbiting a third have periods of revolution that are simple multiples or fractions of each other tidal heating the heating of a planet or moon\u2019s interior by variable tidal forces caused by changing gravitational pull from a nearby planet or moon SUMMARY 12.1 Ring and Moon Systems Introduced The four jovian planets are accompanied by impressive systems of moons and rings. Nearly 200 moons have been discovered in the outer solar system. Of the four ring systems, Saturn\u2019s is the largest and is composed primarily of water ice; in contrast, Uranus and Neptune have narrow rings of dark material, and Jupiter has a tenuous ring of dust. 12.2 The Galilean Moons of Jupiter Jupiter\u2019s largest moons are Ganymede and Callisto, both low-density objects that are composed of more than half water ice. Callisto has an ancient cratered surface, while Ganymede shows evidence of extensive tectonic and volcanic activity, persisting until perhaps a billion years ago. Io and Europa are denser and smaller, each about the size of our Moon. Io is the most volcanically active object in the solar system. Various lines of evidence indicate that Europa has a global ocean of liquid water under a thick ice crust. Many scientists think that Europa may offer the most favorable environment in the solar system to search for life. 12.3 Titan and Triton Saturn\u2019s moon Titan has an atmosphere that is thicker than that of Earth. There are lakes and rivers of liquid hydrocarbons, and evidence of a cycle of evaporation, condensation, and return to the surface that is similar to the water cycle on Earth (but with liquid methane and ethane). The Cassini-Huygens lander set down on Titan and showed a scene with boulders, made of water ice, frozen harder than rock. Neptune\u2019s cold moon Triton has a very thin atmosphere and nitrogen gas geysers. 12.4 Pluto and Charon Pluto and Charon have been revealed by the New Horizons spacecraft to be two of the most fascinating objects in the outer solar system. Pluto is small (a dwarf planet) but also surprisingly active, with contrasting areas of dark cratered terrain, light-colored basins of nitrogen ice, and mountains of frozen water that may be floating in the nitrogen ice. Even Pluto\u2019s largest moon Charon shows evidence of geological activity. Both Pluto and Charon turn out to be far more dynamic and interesting than could have been imagined before the New Horizons mission. 12.5 Planetary Rings Rings are composed of vast numbers of individual particles orbiting so close to a planet that its gravitational forces could have broken larger pieces apart or kept small pieces from gathering together. Saturn\u2019", "s rings are broad, flat, and nearly continuous, except for a handful of gaps. The particles are mostly water ice, with typical dimensions of a few centimeters. One Saturn moon, Enceladus, is today erupting geysers of water to maintain 444 Chapter 12 Rings, Moons, and Pluto the tenuous E Ring, which is composed of very small ice crystals. The rings of Uranus are narrow ribbons separated by wide gaps and contain much less mass. Neptune\u2019s rings are similar but contain even less material. Much of the complex structure of the rings is due to waves and resonances induced by moons within the rings or orbiting outside them. The origin and age of each of these ring systems is still a mystery. FOR FURTHER EXPLORATION Articles Moons Carroll, M. \u201cTitan: What We\u2019ve Learned about a Strange New World.\u201d Astronomy (March 2010): 30. Nice review of Cassini mission results. Elliot, J. \u201cThe Warming Wisps of Triton.\u201d Sky & Telescope (February 1999): 42. About Neptune\u2019s intriguing moon. Hayes, A., \u201cSecrets from Titan\u2019s Seas.\u201d Astronomy (October 2015): 24. Good review of what we now know and what puzzles us about the hydrocarbon lakes of Titan. Jewitt, D., et al. \u201cThe Strangest Satellites in the Solar System.\u201d Scientific American (August 2006): 40. Small irregular moons in the outer solar system. Lakdawalla, E. \u201cIce Worlds of the Ringed Planet.\u201d Sky & Telescope (June 2009): 27. On the Cassini mission exploration of Enceladus, Iapetus, and other moons. Mackenzie, D. \u201cIs There Life under the Ice?\u201d Astronomy (August 2001): 32. On future exploration of Europa. Robertson, D. \u201cWhere Goes the Rain?\u201d Sky & Telescope (March 2013): 26. About the methane weather cycle on Titan and what Cassini experiments are telling us. Scharf, C. \u201cA Universe of Dark Oceans.\u201d Sky & Telescope (December 2014): 20. Subsurface oceans on Europa, Ganymede, Enceladus, and Titan. Showalter, M. \u201cHow to Catch a Moon (or Two) of Pluto.\u201d Astronomy Beat (December 2012): http://www.astrosociety.org/wp-content/uploads/2013/02/ab2012-106.pdf. On the discovery of small moons around Pluto, written by the person who discovered two of them. Spencer, J. \u201cGalileo\u2019s Closest Look at Io.\u201d Sky & Telescope (May 2001): 40. Talcott, R. \u201cCassini Flies through Enceladus\u2019 Geysers.\u201d Astronomy (March 2009): 32. Zimmerman, R. \u201cDoes Methane Flow on Titan?\u201d Astronomy (February 2014): 22. Ideas about lakes, channels, and rain. Pluto Stern, A. \u201cPluto: Up Close and Personal.\u201d Astronomy (July 2015): 22. Good summary of the history of understanding Pluto and our current knowledge on the eve of the New Horizons encounter. Stern, A. \u201cThe Pluto System Explored.\u201d Astronomy (November 2015): 24. Fine review of what the team learned from the first few data downloads from New Horizons. Tombaugh, C. \u201cHow I Found Pluto\u201d Astronomy Beat (May 2009): http://astrosociety.org/wp-content/uploads/ 2013/02/ab2009-23.pdf. Rings Beatty, J. \u201cSaturn\u2019s Amazing Rings.\u201d Sky & Telescope (May 2013): 18. Good 7-page summary of what we know. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 445 Burns, J., et al. \u201cBejeweled Worlds.\u201d Scientific American (February 2002): 64. On rings throughout the solar system. Elliot, J., et al. \u201cDiscovering the Rings of Uranus.\u201d Sky & Telescope (June 1977): 412. Esposito, L. \u201cThe Changing Shape of Planetary Rings.\u201d Astronomy (September 1987): 6. Sobel, D. \u201cSecrets of the Rings.\u201d Discover (April 1994): 86. Discusses the outer planet ring systems. Tiscareno, M. \u201cRingworld Revelations.\u201d Sky & Telescope (February 2007): 32. Cassini results about the rings of Saturn. Websites Note: Many of the sites about planets and planetary missions listed for Other Worlds: An Introduction to the Solar System", " and The Giant Planets also include good information about the moons of the planets. Cassini Mission to Saturn: http://saturn.jpl.nasa.gov/ and http://www.esa.int/SPECIALS/Cassini-Huygens/ index.html and http://ciclops.org Jupiter\u2019s Moons, at JPL: http://solarsystem.nasa.gov/planets/jupiter/moons Neptune\u2019s Moons, at JPL: http://solarsystem.nasa.gov/planets/neptune/moons New Horizons Mission: http://pluto.jhuapl.edu. Gives the latest news bulletins and images from the Pluto encounter, plus lots of background information. Pluto, at JPL: http://solarsystem.nasa.gov/planets/pluto Saturn\u2019s Moons, at JPL: http://solarsystem.nasa.gov/planets/saturn/moons Uranus\u2019 Moons, at JPL: http://solarsystem.nasa.gov/planets/uranus/moons Apps Two apps you can buy for iPhones or iPads can show you the positions and features of the moons of Jupiter and Saturn for any selected date: \u2022 Jupiter Atlas: https://itunes.apple.com/us/app/ju[iter-atlas/id352033947?mt=8 \u2022 Saturn Atlas: https://itunes.apple.com/us/app/saturn-atlas/id352038051?mt=8 Videos Amazing Moons: https://www.youtube.com/watch?v=CQjZf2bW9XQ. 2016 NASA video on intriguing moons in our solar system (4:16). Briny Breath of Enceladus: http://www.jpl.nasa.gov/video/details.php?id=846. Brief 2009 JPL film on the geysers of Enceladus (2:36). Dr. Carolyn Porco\u2019s TED Talk on Enceladus: https://www.youtube.com/watch?v=TRQdHrGuVgI (3:26). Titan: http://www.youtube.com/watch?v=iTrOFefYxFg. Video from Open University, with interviews, animations, and images (8:11). Europa Mission: http://www.jpl.nasa.gov/events/lectures_archive.php?year=2016&month=2. 2016 talk by two JPL scientists on NASA\u2019s plans for a mission to Jupiter\u2019s moon, which may have an underground liquid ocean (1:26:22). 446 Chapter 12 Rings, Moons, and Pluto Great Planet Debate: http://gpd.jhuapl.edu/debate/debateStream.php OR https://www.youtube.com/ watch?v=RJ8EErV6-6Q. Neil deGrasse Tyson debates Mark Sykes about how to characterize Pluto, in 2008 (1:14:11). How I Killed Pluto and Why It Had It Coming: http://www.youtube.com/watch?v=7pbj_llmiMg. 2011 Silicon Valley Astronomy Lecture by Michael Brown on the \u201cdemotion\u201d of Pluto to a dwarf planet (1:27:13). Seeking Pluto\u2019s Frigid Heart: https://www.youtube.com/watch?v=jIxQXGTl_mo. Dramatic 2016 New York Times production, narrated by Dennis Overbye (7:43). Saturn\u2019s Restless Rings: https://www.youtube.com/watch?v=X5zcrEze8L4. 2013 talk by Mark Showalter in the Silicon Valley Astronomy Lecture Series (1:30:59). COLLABORATIVE GROUP ACTIVITIES A. Imagine it\u2019s the distant future and humans can now travel easily among the planets. Your group is a travel agency, with the task of designing a really challenging tour of the Galilean moons for a group of sports enthusiasts. What kinds of activities are possible on each world? How would rock climbing on Ganymede, for example, differ from rock climbing on Earth? (If you design an activity for Io, you had better bring along very strong radiation shielding. Why?) B. In the same spirit as Activity A, have your agency design a tour that includes the seven most spectacular sights of any kind on all the moons or rings covered in this chapter. What are the not-to-be-missed destinations that future tourists will want to visit and why? Which of the sights you pick are going to be spectacular if you are on the moon\u2019s surface or inside the ring, and which", " would look interesting only from far away in space? C. In this chapter we could cover only a few of the dozens of moons in the outer solar system. Using the Internet or your college library, organize your group into a research team and find out more about one of the moons we did not cover in detail. Our favorites include Uranus\u2019 Miranda, with its jigsaw puzzle surface; Saturn\u2019s Mimas, with a \u201cknockout\u201d crater called Herschel; and Saturn\u2019s Iapetus, whose two hemispheres differ significantly. Prepare a report to attract tourists to the world you selected. D. In a novel entitled 2010, science fiction writer Arthur C. Clarke, inspired by the information coming back from the Voyager spacecraft, had fun proposing a life form under the ice of Europa that was evolving toward intelligence. Suppose future missions do indeed find some sort of life (not necessarily intelligent but definitely alive) under the ice of Europa\u2014life that evolved completely independently from life on Earth. Have your group discuss what effect such a discovery would have on humanity\u2019s view of itself. What should be our attitude toward such a life form? Do we have an obligation to guard it against contamination by our microbes and viruses? Or, to take an extreme position, should we wipe it out before it becomes competitive with Earth life or contaminates our explorers with microorganisms we are not prepared to deal with? Who should be in charge of making such decisions? E. In the same spirit as Activity D, your group may want to watch the 2013 science fiction film Europa Report. The producers tried to include good science in depicting what it would be like for astronauts to visit that jovian moon. How well does your group think they did? F. A number of modern science fiction writers (especially those with training in science) have written short stories that take place on the moons of Jupiter and Saturn. There is a topical listing of science fiction stories This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 12 Rings, Moons, and Pluto 447 with good astronomy at http://www.astrosociety.org/scifi. Members of your group can look under \u201cJupiter\u201d or \u201cSaturn\u201d and find a story that interests you and then report on it to the whole class. G. Work together to make a list of all the reasons it is hard to send a mission to Pluto. What compromises had to be made so that the New Horizons mission was affordable? How would you design a second mission to learn more about the Pluto system? H. Your group has been asked by NASA to come up with one or more missions to learn about Europa. Review what we know about this moon so far and then design a robotic mission that would answer some of the questions we have. You can assume that budget is not a factor, but your instruments have to be realistic. (Bear in mind that Europa is cold and far from the Sun.) I. Imagine your group is the first landing party on Pluto (let\u2019s hope you remembered to bring long underwear!). You land in a place where Charon is visible in the sky and you observe Charon for one Earth week. Describe what Charon will look like during that week. Now you move your camp to the opposite hemisphere of Pluto. What will Charon look like there during the course of a week? J. When, in 2006, the International Astronomical Union (IAU) decided that Pluto should be called a dwarf planet and not a planet, they set up three criteria that a world must meet to be called a planet. Your group should use the Internet to find these criteria. Which of them did Pluto not meet? Read a little bit about the reaction to the IAU\u2019s decision among astronomers and the public. How do members of your group feel about Pluto\u2019s new classification? (After you have discussed it within the group, you may want to watch The Great Planet Debate video recommended in \u201cFor Further Exploration.\u201d) EXERCISES Review Questions 1. What are the moons of the outer planets made of, and how is their composition different from that of our Moon? 2. Compare the geology of Callisto, Ganymede, and Titan. 3. What is the evidence for a liquid water ocean on Europa, and why is this interesting to scientists searching for extraterrestrial life? 4. Explain the energy source that powers the volcanoes of Io. 5. Compare the properties of Titan\u2019s atmosphere with those of Earth\u2019s atmosphere. 6. How was Pluto discovered? Why did it take so long to find it? 7. How are Triton and Pluto similar? 8. Describe and compare the rings of Saturn and Uranus, including their possible origins. 9. Why were the rings of Uranus not observed directly from telescopes on the ground on Earth? How were they discovered? 10.", " List at least three major differences between Pluto and the terrestrial planets. 448 Chapter 12 Rings, Moons, and Pluto 11. The Hubble Space Telescope images of Pluto in 2002 showed a bright spot and some darker areas around it. Now that we have the close-up New Horizons images, what did the large bright region on Pluto turn out to be? 12. Saturn\u2019s E ring is broad and thin, and far from Saturn. It requires fresh particles to sustain itself. What is the source of new E-ring particles? Thought Questions 13. Why do you think the outer planets have such extensive systems of rings and moons, while the inner planets do not? 14. Ganymede and Callisto were the first icy objects to be studied from a geological point of view. Summarize the main differences between their geology and that of the rocky terrestrial planets. 15. Compare the properties of the volcanoes on Io with those of terrestrial volcanoes. Give at least two similarities and two differences. 16. Would you expect to find more impact craters on Io or Callisto? Why? 17. Why is it unlikely that humans will be traveling to Io? (Hint: Review the information about Jupiter\u2019s magnetosphere in The Giant Planets.) 18. Why do you suppose the rings of Saturn are made of bright particles, whereas the particles in the rings of Uranus and Neptune are black? 19. Suppose you miraculously removed all of Saturn\u2019s moons. What would happen to its rings? 20. We have a lot of good images of the large moons of Jupiter and Saturn from the Galileo and Cassini spacecraft missions (check out NASA\u2019s Planetary Photojournal site, at http://photojournal.jpl.nasa.gov, to see the variety). Now that the New Horizons mission has gone to Pluto, why don\u2019t we have as many good images of all sides of Pluto and Charon? 21. In the Star Wars movie Star Wars Episode VI: Return of the Jedi, a key battle takes place on the inhabited \u201cforest moon\u201d Endor, which supposedly orbits around a gas giant planet. From what you have learned about planets and moons of the solar system, why would this be an unusual situation? Figuring For Yourself 22. Which would have the longer orbital period: a moon 1 million km from the center of Jupiter, or a moon 1 million km from the center of Earth? Why? 23. How close to Uranus would a spacecraft have to get to obtain the same resolution as in Example 12.1 with a camera that has an angular resolution of 2 arcsec? 24. Saturn\u2019s A, B, and C Rings extend 75,000 to 137,000 km from the center of the planet. Use Kepler\u2019s third law to calculate the difference between how long a particle at the inner edge and a particle at the outer edge of the three-ring system would take to revolve about the planet. 25. Use the information in Appendix G to calculate what you would weigh on Titan, Io, and Uranus\u2019 moon Miranda. 26. The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much longer does it take Titan to orbit Saturn compared to Enceladus? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 449 13 COMETS AND ASTEROIDS: DEBRIS OF THE SOLAR SYSTEM Figure 13.1. Hale-Bopp. Comet Hale-Bopp was one of the most attractive and easily visible comets of the twentieth century. It is shown here as it appeared in the sky in March 1997. You can see the comet\u2019s long blue ion tail and the shorter white dust tail. You will learn about these two types of comet tails, and how they form, in this chapter. (credit: modification of work by ESO/E. Slawik) Chapter Outline 13.1 Asteroids 13.2 Asteroids and Planetary Defense 13.3 The \u201cLong-Haired\u201d Comets 13.4 The Origin and Fate of Comets and Related Objects Thinking Ahead Hundreds of smaller members of the solar system\u2014asteroids and comets\u2014are known to have crossed Earth\u2019s orbit in the past, and many others will do so in centuries ahead. What could we do if we knew a few years in advance that one of these bodies would hit Earth? To understand the early history of life on Earth, scientists study ancient fossils. To reconstruct the early history of the solar system, we need cosmic fossils\u2014materials that formed when our system was very young. However, reconstructing the early history of the solar system by looking just at the planets is almost as difficult as determining the circumstances of human birth by merely looking at an adult", ". Instead, we turn to the surviving remnants of the creation process\u2014ancient but smaller objects in our cosmic neighborhood. Asteroids are rocky or metallic and contain little volatile (easily evaporated) material. Comets are small icy objects that contain frozen water and other volatile materials but with solid grains mixed in. In the deep freeze beyond Neptune, we also have a large reservoir of material unchanged since the formation of the solar system, as well as a number of dwarf planets. 450 Chapter 13 Comets and Asteroids: Debris of the Solar System 13.1 ASTEROIDS Learning Objectives By the end of this section, you will be able to: Outline the story of the discovery of asteroids and describe their typical orbits Describe the composition and classification of the various types of asteroids Discuss what was learned from spacecraft missions to several asteroids The asteroids are mostly found in the broad space between Mars and Jupiter, a region of the solar system called the asteroid belt. Asteroids are too small to be seen without a telescope; the first of them was not discovered until the beginning of the nineteenth century. Discovery and Orbits of the Asteroids In the late 1700s, many astronomers were hunting for an additional planet they thought should exist in the gap between the orbits of Mars and Jupiter. The Sicilian astronomer Giovanni Piazzi thought he had found this missing planet in 1801, when he discovered the first asteroid (or as it was later called, \u201cminor planet\u201d) orbiting at 2.8 AU from the Sun. His discovery, which he named Ceres, was quickly followed by the detection of three other little planets in similar orbits. Clearly, there was not a single missing planet between Mars and Jupiter but rather a whole group of objects, each much smaller than our Moon. (An analogous discovery history has played out in slow motion in the outer solar system. Pluto was discovered beyond Neptune in 1930 and was initially called a planet, but early in the twenty-first century, several other similar objects were found. We now call all of them dwarf planets.) By 1890, more than 300 of these minor planets or asteroids had been discovered by sharp-eyed observers. In that year, Max Wolf at Heidelberg introduced astronomical photography to the search for asteroids, greatly accelerating the discovery of these dim objects. In the twenty-first century, searchers use computer-driven electronic cameras, another leap in technology. More than half a million asteroids now have well-determined orbits. Asteroids are given a number (corresponding to the order of discovery) and sometimes also a name. Originally, the names of asteroids were chosen from goddesses in Greek and Roman mythology. After exhausting these and other female names (including, later, those of spouses, friends, flowers, cities, and others), astronomers turned to the names of colleagues (and other people of distinction) whom they wished to honor. For example, asteroids 2410, 4859, and 68448 are named Morrison, Fraknoi, and Sidneywolff, for the three original authors of this textbook. The largest asteroid is Ceres (numbered 1), with a diameter just less than 1000 kilometers. As we saw, Ceres was considered a planet when it was discovered but later was called an asteroid (the first of many.) Now, it has again been reclassified and is considered one of the dwarf planets, like Pluto (see the chapter on Moons, Rings and Pluto). We still find it convenient, however, to discuss Ceres as the largest of the asteroids. Two other asteroids, Pallas and Vesta, have diameters of about 500 kilometers, and about 15 more are larger than 250 kilometers (see Table 13.1). The number of asteroids increases rapidly with decreasing size; there are about 100 times more objects 10 kilometers across than there are 100 kilometers across. By 2016, nearly a million asteroids have been discovered by astronomers. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 451 The Minor Planet Center (https://openstaxcollege.org/l/30minplancen) is a worldwide repository of data on asteroids. Visit it online to find out about the latest discoveries related to the small bodies in our solar system. (Note that some of the material on this site is technical; it\u2019s best to click on the menu tab for the \u201cpublic\u201d for information more at the level of this textbook.) The Largest Asteroids # Name Year of Discovery Orbit\u2019s Semimajor Axis (AU) Diameter (km) Compositional Class 1 2 3 4 10 16 31 52 65 87 Ceres Pallas Juno Vesta 1801 1802 1804 1807 Hygiea 1849 Psyche 1852 Euphrosyne 1854 Europa 1858 Cybele Sylvia 1861 1866 451 Patientia 1899 511 Davida 1903 704 Interamnia 1910 Table 13.1 2.77 2.77", " 2.67 2.36 3.14 2.92 3.15 3.10 3.43 3.48 3.06 3.16 3.06 940 540 265 510 410 265 250 280 280 275 260 310 310 C (carbonaceous) C (carbonaceous) S (stony) basaltic C (carbonaceous) M (metallic) C (carbonaceous) C (carbonaceous) C (carbonaceous) C (carbonaceous) C (carbonaceous) C (carbonaceous) C (carbonaceous) The asteroids all revolve about the Sun in the same direction as the planets, and most of their orbits lie near the plane in which Earth and other planets circle. The majority of asteroids are in the asteroid belt, the region between Mars and Jupiter that contains all asteroids with orbital periods between 3.3 to 6 years (Figure 13.2). Although more than 75% of the known asteroids are in the belt, they are not closely spaced (as they are sometimes depicted in science fiction movies). The volume of the belt is actually very large, and the typical spacing between objects (down to 1 kilometer in size) is several million kilometers. (This was fortunate for 452 Chapter 13 Comets and Asteroids: Debris of the Solar System spacecraft like Galileo, Cassini, Rosetta, and New Horizons, which needed to travel through the asteroid belt without a collision.) Figure 13.2. Asteroids in the Solar System. This computer-generated diagram shows the positions of the asteroids known in 2006. If the asteroid sizes were drawn to scale, none of the dots representing an asteroid would be visible. Here, the asteroid dots are too big and give a false impression of how crowded the asteroid belt would look if you were in it. Note that in addition to those in the asteroid belt, there are also asteroids in the inner solar system and some along Jupiter\u2019s orbit (such as the Trojans and Greeks groups), controlled by the giant planet\u2019s gravity. Still, over the long history of our solar system, there have been a good number of collisions among the asteroids themselves. In 1918, the Japanese astronomer Kiyotsugu Hirayama found that some asteroids fall into families, groups with similar orbital characteristics. He hypothesized that each family may have resulted from the breakup of a larger body or, more likely, from the collision of two asteroids. Slight differences in the speeds with which the various fragments left the collision scene account for the small spread in orbits now observed for the different asteroids in a given family. Several dozen such families exist, and observations have shown that individual members of most families have similar compositions, as we would expect if they were fragments of a common parent. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 453 You can see a dramatic animated video (https://openstaxcollege.org/l/30anividastorb) showing the orbits of 100,000 asteroids found by one sky survey. As the 3-minute video goes on, you get to see the orbits of the planets and how the asteroids are distributed in the solar system. But note that all such videos are misleading in one sense. The asteroids themselves are really small compared to the distances covered, so they have to be depicted as larger points to be visible. If you were in the asteroid belt, there would be far more empty space than asteroids. Composition and Classification Asteroids are as different as black and white. The majority are very dark, with reflectivity of only 3 to 4%, like a lump of coal. However, another large group has a typical reflectivity of 15%. To understand more about these differences and how they are related to chemical composition, astronomers study the spectrum of the light reflected from asteroids for clues about their composition. The dark asteroids are revealed from spectral studies to be primitive bodies (those that have changed little chemically since the beginning of the solar system) composed of silicates mixed with dark, organic carbon compounds. These are known as C-type asteroids (\u201cC\u201d for carbonaceous). Two of the largest asteroids, Ceres and Pallas, are primitive, as are almost all of the asteroids in the outer part of the belt. The second most populous group is the S-type asteroids, where \u201cS\u201d stands for a stony or silicate composition. Here, the dark carbon compounds are missing, resulting in higher reflectivity and clearer spectral signatures of silicate minerals. The S-type asteroids are also chemically primitive, but their different composition indicates that they were probably formed in a different location in the solar system from the C-type asteroids. Asteroids of a third class, much less numerous than those of the first two, are composed primarily of metal and are called M-type asteroids (\u201cM\u201d for metallic). Spectroscopically, the identification of metal is difficult, but for at least", " the largest M-type asteroid, Psyche, this identification has been confirmed by radar. Since a metal asteroid, like an airplane or ship, is a much better reflector of radar than is a stony object, Psyche appears bright when we aim a radar beam at it. How did such metal asteroids come to be? We suspect that each came from a parent body large enough for its molten interior to settle out or differentiate, and the heavier metals sank to the center. When this parent body shattered in a later collision, the fragments from the core were rich in metals. There is enough metal in even a 1-kilometer M-type asteroid to supply the world with iron and many other industrial metals for the foreseeable future, if we could bring one safely to Earth. In addition to the M-type asteroids, a few other asteroids show signs of early heating and differentiation. These have basaltic surfaces like the volcanic plains of the Moon and Mars; the large asteroid Vesta (discussed in a moment) is in this last category. The different classes of asteroids are found at different distances from the Sun (Figure 13.3). By tracing how asteroid compositions vary with distance from the Sun, we can reconstruct some of the properties of the solar nebula from which they originally formed. 454 Chapter 13 Comets and Asteroids: Debris of the Solar System Figure 13.3. Where Different Types of Asteroids Are Found. Asteroids of different composition are distributed at different distances from the Sun. The S-type and C-type are both primitive; the M-type consists of cores of differentiated parent bodies. Vesta: A Differentiated Asteroid Vesta is one of the most interesting of the asteroids. It orbits the Sun with a semi-major axis of 2.4 AU in the inner part of the asteroid belt. Its relatively high reflectivity of almost 30% makes it the brightest asteroid, so bright that it is actually visible to the unaided eye if you know just where to look. But its real claim to fame is that its surface is covered with basalt, indicating that Vesta is a differentiated object that must once have been volcanically active, in spite of its small size (about 500 kilometers in diameter). Meteorites from Vesta\u2019s surface (Figure 13.4), identified by comparing their spectra with that of Vesta itself, have landed on Earth and are available for direct study in the laboratory. We thus know a great deal about this asteroid. The age of the lava flows from which these meteorites derived has been measured at 4.4 to 4.5 billion years, very soon after the formation of the solar system. This age is consistent with what we might expect for volcanoes on Vesta; whatever process heated such a small object was probably intense and short-lived. In 2016, a meteorite fell in Turkey that could be identified with a particular lava flow as revealed by the orbiting Dawn spacecraft. Figure 13.4. Piece of Vesta. This meteorite (rock that fell from space) has been identified as a volcanic fragment from the crust of asteroid Vesta. (credit: modification of work by R. Kempton (New England Meteoritical Services)) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 455 Asteroids Up Close On the way to its 1995 encounter with Jupiter, the Galileo spacecraft was targeted to fly close to two main-belt S- type asteroids called Gaspra and Ida. The Galileo camera revealed both as long and highly irregular (resembling a battered potato), as befits fragments from a catastrophic collision (Figure 13.5). Figure 13.5. Mathilde, Gaspra, and Ida. The first three asteroids photographed from spacecraft flybys, printed to the same scale. Gaspra and Ida are S-type and were investigated by the Galileo spacecraft; Mathilde is C-type and was a flyby target for the NEAR-Shoemaker spacecraft. (credit: modification of work by NEAR Project, Galileo Project, NASA) The detailed images allowed us to count the craters on Gaspra and Ida, and to estimate the length of time their surfaces have been exposed to collisions. The Galileo scientists concluded that these asteroids are only about 200 million years old (that is, the collisions that formed them took place about 200 million years ago). Calculations suggest that an asteroid the size of Gaspra or Ida can expect another catastrophic collision sometime in the next billion years, at which time it will be disrupted to form another generation of still-smaller fragments. The greatest surprise of the Galileo flyby of Ida was the discovery of a moon (which was then named Dactyl), in orbit about the asteroid (Figure 13.6). Although only 1.5 kilometers in diameter, smaller than many college campuses, Dactyl provides scientists with something otherwise", " beyond their reach\u2014a measurement of the mass and density of Ida using Kepler\u2019s laws. The moon\u2019s distance of about 100 kilometers and its orbital period of about 24 hours indicate that Ida has a density of approximately 2.5 g/cm3, which matches the density of primitive rocks. Subsequently, both large visible-light telescopes and high-powered planetary radar have discovered many other asteroid moons, so that we are now able to accumulate valuable data on asteroid masses and densities. 456 Chapter 13 Comets and Asteroids: Debris of the Solar System Figure 13.6. Ida and Dactyl. The asteroid Ida and its tiny moon Dactyl (the small body off to its right), were photographed by the Galileo spacecraft in 1993. Irregularly shaped Ida is 56 kilometers in its longest dimension, while Dactyl is about 1.5 kilometers across. The colors have been intensified in this image; to the eye, all asteroids look basically gray. (credit: modification of work by NASA/JPL) By the way, Phobos and Deimos, the two small moons of Mars, are probably captured asteroids (Figure 13.7). They were first studied at close range by the Viking orbiters in 1977 and later by Mars Global Surveyor. Both are irregular, somewhat elongated, and heavily created, resembling other smaller asteroids. Their largest dimensions are about 26 kilometers and 16 kilometers, respectively. The small outer moons of Jupiter and Saturn were probably also captured from passing asteroids, perhaps early in the history of the solar system. Figure 13.7. Moons of Mars. The two small moons of Mars, (a) Phobos and (b) Deimos, were discovered in 1877 by American astronomer Asaph Hall. Their surface materials are similar to many of the asteroids in the outer asteroid belt, leading astronomers to believe that the two moons may be captured asteroids. (credit a: modification of work by NASA; credit b: modification of work by NASA/JPL-Caltech/University of Arizona) Beginning in the 1990s, spacecraft have provided close looks at several more asteroids. The Near Earth Asteroid Rendezvous (NEAR) spacecraft went into orbit around the S-type asteroid Eros, becoming a temporary moon of this asteroid. On its way to Eros, the NEAR spacecraft was renamed after planetary geologist Eugene Shoemaker, a pioneer in our understanding of craters and impacts. For a year, the NEAR-Shoemaker spacecraft orbited the little asteroid at various altitudes, measuring its surface and interior composition as well as mapping Eros from all sides (Figure 13.8). The data showed that Eros is made of some of the most chemically primitive materials in the solar system. Several other asteroids have been revealed as made of loosely bound rubble throughout, but not Eros. Its uniform density (about the same as that of Earth\u2019s crust) and extensive global-scale grooves and ridges show that it is a cracked but solid rock. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 457 Figure 13.8. Looking Down on the North Pole of Eros. This view was constructed from six images of the asteroid taken from an altitude of 200 kilometers. The large crater at the top has been named Psyche (after the maiden who was Eros\u2019 lover in classical mythology) and is about 5.3 kilometers wide. A saddle-shaped region can be seen directly below it. Craters of many different sizes are visible. (credit: modification of work by NASA/JHUPL) Eros has a good deal of loose surface material that appears to have slid down toward lower elevations. In some places, the surface rubble layer is 100 meters deep. The top of loose soil is dotted with scattered, half-buried boulders. There are so many of these boulders that they are more numerous than the craters. Of course, with the gravity so low on this small world, a visiting astronaut would find loose boulders rolling toward her pretty slowly and could easily leap high enough to avoid being hit by one. Although the NEAR-Shoemaker spacecraft was not constructed as a lander, at the end of its orbital mission in 2000, it was allowed to fall gently to the surface, where it continued its chemical analysis for another week. In 2003, Japan\u2019s Hayabusa 1 mission not only visited a small asteroid but also brought back samples to study in laboratories on Earth. The target S-type asteroid, Itokawa (shown in Figure 13.9), is much smaller than Eros, only about 500 meters long. This asteroid is elongated and appears to be the result of the collision of two separate asteroids long ago. There are almost no impact craters, but an abundance of boulders (like a pile of rubble", ") on the surface. 458 Chapter 13 Comets and Asteroids: Debris of the Solar System Figure 13.9. Asteroid Itokawa. The surface of asteroid Itokawa appears to have no craters. Astronomers have hypothesized that its surface consists of rocks and ice chunks held together by a small amount of gravity, and its interior is probably also a similar rubble pile. (credit: JAXA) The Hayabusa spacecraft was designed not to land, but to touch the surface just long enough to collect a small sample. This tricky maneuver failed on its first try, with the spacecraft briefly toppling over on its side. Eventually, the controllers were successful in picking up a few grains of surface material and transferring them into the return capsule. The 2010 reentry into Earth\u2019s atmosphere over Australia was spectacular (Figure 13.10), with a fiery breakup of the spacecraft, while a small return capsule successfully parachuted to the surface. Months of careful extraction and study of more than a thousand tiny dust particles confirmed that the surface of Itokawa had a composition similar to a well-known class of primitive meteorites. We estimate that the dust grains Hayabusa picked up had been exposed on the surface of the asteroid for about 8 million years. Figure 13.10. Hayabusa Return. This dramatic image shows the Hayabusa probe breaking up upon reentry. The return capsule, which separated from the main spacecraft and parachuted to the surface, glows at the bottom right. (credit: modification of work by NASA Ames/Jesse Carpenter/Greg Merkes) The most ambitious asteroid space mission (called Dawn) has visited the two largest main belt asteroids, Ceres and Vesta, orbiting each for about a year (Figure 13.11). Their large sizes (diameters of about 1000 and 500 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 459 kilometers, respectively) make them appropriate for comparison with the planets and large moons. Both turned out to be heavily cratered, implying their surfaces are old. On Vesta, we have now actually located the large impact craters that ejected the basaltic meteorites previously identified as coming from this asteroid. These craters are so large that they sample several layers of Vesta\u2019s crustal material. Figure 13.11. Vesta and Ceres. The NASA Dawn spacecraft took these images of the large asteroids (a) Vesta and (b) Ceres. (a) Note that Vesta is not round, as Ceres (which is considered a dwarf planet) is. A mountain twice the height of Mt. Everest on Earth is visible at the very bottom of the Vesta image. (b) The image of Ceres has its colors exaggerated to bring out differences in composition. You can see a white feature in Occator crater near the center of the image. (credit a, b: modification of work by NASA/JPL-Caltech/UCLA/MPS/DLR/IDA) Ceres has not had a comparable history of giant impacts, so its surface is covered with craters that look more like those from the lunar highlands. The big surprise at Ceres is the presence of very bright white spots, associated primarily with the central peaks of large craters (Figure 13.12). The light-colored mineral is some kind of salt, either produced when these craters were formed or subsequently released from the interior. Figure 13.12. White Spots in a Larger Crater on Ceres. These bright features appear to be salt deposits in a Ceres crater called Occator, which is 92 kilometers across. (credit: modification of work by NASA/JPL-Caltech/UCLA/MPS/DLR/IDA) 460 Chapter 13 Comets and Asteroids: Debris of the Solar System The space agencies involved with the Dawn mission have produced nice animated \u201cflyover\u201d videos of Vesta (https://openstaxcollege.org/l/30vestaflyover) and Ceres (https://openstaxcollege.org/l/ 30ceresflyover) available online. 13.2 ASTEROIDS AND PLANETARY DEFENSE Learning Objectives By the end of this section, you will be able to: Recognize the threat that near-Earth objects represent for Earth Discuss possible defensive strategies to protect our planet Not all asteroids are in the main asteroid belt. In this section, we consider some special groups of asteroids with orbits that approach or cross the orbit of Earth. These pose the risk of a catastrophic collision with our planet, such as the collision 65 million years ago that killed the dinosaurs. Earth-Approaching Asteroids Asteroids that stray far outside the main belt are of interest mostly to astronomers. But asteroids that come inward, especially those with orbits that come close to or cross the orbit of Earth, are of interest to political leaders, military planners", "\u2014indeed, everyone alive on Earth. Some of these asteroids briefly become the closest celestial object to us. In 1994, a 1-kilometer object was picked up passing closer than the Moon, causing a stir of interest in the news media. Today, it is routine to read of small asteroids coming this close to Earth. (They were always there, but only in recent years have astronomers been able to detect such faint objects.) In 2013, a small asteroid hit our planet, streaking across the sky over the Russian city of Chelyabinsk and exploding with the energy of a nuclear bomb (Figure 13.13). The impactor was a stony object about 20 meters in diameter, exploding about 30 kilometers high with an energy of 500 kilotons (about 30 times larger than the nuclear bombs dropped on Japan in World War II). No one was hurt by the blast itself, although it briefly became as bright as the Sun, drawing many spectators to the windows in their offices and homes. When the blast wave from the explosion then reached the town, it blew out the windows. About 1500 people had to seek medical attention from injuries from the shattered glass. A much larger atmospheric explosion took place in Russia in 1908, caused by an asteroid about 40 meters in diameter, releasing an energy of 5 megatons, as large the most powerful nuclear weapons of today. Fortunately, the area directly affected, on the Tunguska River in Siberia, was unpopulated, and no one was killed. However, the area of forest destroyed by the blast was large equal to the size of a major city (Figure 13.13). Together with any comets that come close to our planet, such asteroids are known collectively as near-Earth objects (NEOs). As we will see (and as the dinosaurs found out 65 million years ago,) the collision of a significant-sized NEO could be a catastrophe for life on our planet. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 461 Figure 13.13. Impacts with Earth. (a) As the Chelyabinsk meteor passed through the atmosphere, it left a trail of smoke and briefly became as bright as the Sun. (b) Hundreds of kilometers of forest trees were knocked down and burned at the Tunguska impact site. (credit a: modification of work by Alex Alishevskikh Visit the video compilation (https://openstaxcollege.org/l/30vidcomchelmet) of the Chelyabinsk meteor streaking through the sky over the city on February 15, 2013, as taken by people who were in the area when it occurred View this video of a non-technical talk by David Morrison (https://openstaxcollege.org/l/ 30davmorrison) to watch \u201cThe Chelyabinsk Meteor: Can We Survive a Bigger Impact?\u201d Dr. Morrison (SETI Institute and NASA Ames Research Center) discusses the Chelyabinsk impact and how we learn about NEOs and protect ourselves; the talk is from the Silicon Valley Astronomy Lectures series. Astronomers have urged that the first step in protecting Earth from future impacts by NEOs must be to learn what potential impactors are out there. In 1998, NASA began the Spaceguard Survey, with the goal to discover and track 90% of Earth-approaching asteroids greater than 1 kilometer in diameter. The size of 1 kilometer was selected to include all asteroids capable of causing global damage, not merely local or regional effects. At 1 kilometer or larger, the impact could blast so much dust into the atmosphere that the sunlight would be dimmed for months, causing global crop failures\u2014an event that could threaten the survival of our civilization. The Spaceguard goal of 90% was reached in 2012 when nearly a thousand of these 1-kilometer near-Earth asteroids (NEAs) had been found, along with more than 10,000 smaller asteroids. Figure 13.14 shows how the pace of NEA discoveries has been increasing over recent years. 462 Chapter 13 Comets and Asteroids: Debris of the Solar System Figure 13.14. Discovery of Near-Earth Asteroids. The accelerating rate of discovery of NEAs is illustrated in this graph, which shows the total number of known NEAs, the number over 140 kilometers in diameter, and the number over 1 kilometer in diameter, the size that poses the dominant impact risk on Earth. How did astronomers know when they had discovered 90% of these asteroids? There are several ways to estimate the total number, even before they were individually located. One way is to look at the numbers of large craters on the dark lunar maria. Remember that these craters were made by impacts just like the ones we are considering. They are preserved on the Moon\u2019s airless surface, whereas Earth soon erases the imprint", "s of past impacts. Thus, the number of large craters on the Moon allows us to estimate how often impacts have occurred on both the Moon and Earth over the past several billion years. The number of impacts is directly related to the number of asteroids and comets on Earth-crossing orbits. Another approach is to see how often the surveys (which are automated searches for faint points of light that move among the stars) rediscover a previously known asteroid. At the beginning of a survey, all the NEAs it finds will be new. But as the survey becomes more complete, more and more of the moving points the survey cameras record will be rediscoveries. The more rediscoveries each survey experiences, the more complete our inventory of these asteroids must be. We have been relieved to find that none of the NEAs discovered so far is on a trajectory that will impact Earth within the foreseeable future. However, we can\u2019t speak for the handful of asteroids larger than 1 kilometer that have not yet been found, or for the much more numerous smaller ones. It is estimated that there are a million NEAs capable of hitting Earth that are smaller than 1 kilometer but still large enough to destroy a city, and our surveys have found fewer than 10% of them. Researchers who work with asteroid orbits estimate that for smaller (and therefore fainter) asteroids we are not yet tracking, we will have about a 5-second warning that one is going to hit Earth\u2014in other words, we won\u2019t see it until it enters the atmosphere. Clearly, this estimate gives us a lot of motivation to continue these surveys to track as many asteroids as possible. Though entirely predictable over times of a few centuries, the orbits of Earth-approaching asteroids are unstable over long time spans as they are tugged by the gravitational attractions of the planets. These objects will eventually meet one of two fates: either they will impact one of the terrestrial planets or the Sun, or they will be ejected gravitationally from the inner solar system due to a near-encounter with a planet. The probabilities of these two outcomes are about the same. The timescale for impact or ejection is only about a hundred million years, very short compared with the 4-billion-year age of the solar system. Calculations show that only approximately one quarter of the current Earth-approaching asteroids will eventually end up colliding with Earth itself. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 463 If most of the current population of Earth-approaching asteroids will be removed by impact or ejection in a hundred million years, there must be a continuing source of new objects to replenish our supply of NEAs. Most of them come from the asteroid belt between Mars and Jupiter, where collisions between asteroids can eject fragments into Earth-crossing orbits (see Figure 13.15). Others may be \u201cdead\u201d comets that have exhausted their volatile materials (which we\u2019ll discuss in the next section). Figure 13.15. Near-Earth Asteroid. Toutatis is a 5-kilometer long NEA that approached within 3 million kilometers of Earth in 1992. This series of images is a reconstruction its size and shape obtained from bouncing radar waves off the asteroid during its close flyby. Toutatis appears to consist of two irregular, lumpy bodies rotating in contact with each other. (Note that the color has been artificially added.) (credit: modification of work by NASA) One reason scientists are interested in the composition and interior structure of NEAs is that humans will probably need to defend themselves against an asteroid impact someday. If we ever found one of these asteroids on a collision course with us, we would need to deflect it so it would miss Earth. The most straightforward way to deflect it would be to crash a spacecraft into it, either slowing it or speeding it up, slightly changing its orbital period. If this were done several years before the predicted collision, the asteroid would miss the planet entirely\u2014making an asteroid impact the only natural hazard that we could eliminate completely by the application of technology. Alternatively, such deflection could be done by exploding a nuclear bomb near the asteroid to nudge it off course. To achieve a successful deflection by either technique, we need to know more about the density and interior structure of the asteroid. A spacecraft impact or a nearby explosion would have a greater effect on a solid rocky asteroid such as Eros than on a loose rubble pile. Think of climbing a sand dune compared to climbing a rocky hill with the same slope. On the dune, much of our energy is absorbed in the slipping sand, so the climb is much more difficult and takes more energy. There is increasing international interest in the problem of asteroid impacts. The United Nations has formed two technical committees on planetary defense, recognizing that the entire planet is at risk from asteroid impacts. However, the fundamental", " problem remains one of finding NEAs in time for defensive measures to be taken. We must be able to find the next impactor before it finds us. And that\u2019s a job for the astronomers. 13.3 THE \u201cLONG-HAIRED\u201d COMETS Learning Objectives By the end of this section, you will be able to: Characterize the general physical appearance of comets Explain the range of cometary orbits Describe the size and composition of a typical comet\u2019s nucleus Discuss the atmospheres of comets Summarize the discoveries of the Rosetta mission 464 Chapter 13 Comets and Asteroids: Debris of the Solar System Comets differ from asteroids primarily in their icy composition, a difference that causes them to brighten dramatically as they approach the Sun, forming a temporary atmosphere. In some early cultures, these so- called \u201chairy stars\u201d were considered omens of disaster. Today, we no longer fear comets, but eagerly anticipate those that come close enough to us to put on a good sky show. Appearance of Comets A comet is a are relatively small chunk of icy material (typically a few kilometers across) that develops an atmosphere as it approaches the Sun. Later, there may be a very faint, nebulous tail, extending several million kilometers away from the main body of the comet. Comets have been observed from the earliest times: accounts of comets are found in the histories of virtually all ancient civilizations. The typical comet, however, is not spectacular in our skies, instead having the appearance of a rather faint, diffuse spot of light somewhat smaller than the Moon and many times less brilliant. (Comets seemed more spectacular to people before the invention of artificial lighting, which compromises our view of the night sky.) Like the Moon and planets, comets appear to wander among the stars, slowly shifting their positions in the sky from night to night. Unlike the planets, however, most comets appear at unpredictable times, which perhaps explain why they frequently inspired fear and superstition in earlier times. Comets typically remain visible for periods that vary from a couple of weeks to several months. We\u2019ll say more about what they are made of and how they become visible after we discuss their motions. Note that still images of comets give the impression that they are moving rapidly across the sky, like a bright meteor or shooting star. Looking only at such images, it is easy to confuse comets and meteors. But seen in the real sky, they are very different: the meteor burns up in our atmosphere and is gone in a few seconds, whereas the comet may be visible for weeks in nearly the same part of the sky. Comet Orbits The study of comets as members of the solar system dates from the time of Isaac Newton, who first suggested that they orbited the Sun on extremely elongated ellipses. Newton\u2019s colleague Edmund Halley (see the Edmund Halley: Astronomy\u2019s Renaissance Man feature box) developed these ideas, and in 1705, he published calculations of 24 comet orbits. In particular, he noted that the orbits of the bright comets that had appeared in the years 1531, 1607, and 1682 were so similar that the three could well be the same comet, returning to perihelion (closest approach to the Sun) at average intervals of 76 years. If so, he predicted that the object should next return about 1758. Although Halley had died by the time the comet appeared as he predicted, it was given the name Comet Halley (rhymes with \u201cvalley\u201d) in honor of the astronomer who first recognized it as a permanent member of our solar system, orbiting around the Sun. Its aphelion (furthest point from the Sun) is beyond the orbit of Neptune. We now know from historical records that Comet Halley has actually been observed and recorded on every passage near the Sun since 239 BCE at intervals ranging from 74 to 79 years. The period of its return varies somewhat because of orbital changes produced by the pull of the giant planets. In 1910, Earth was brushed by the comet\u2019s tail, causing much needless public concern. Comet Halley last appeared in our skies in 1986 (Figure 13.16), when it was met by several spacecraft that gave us a wealth of information about its makeup; it will return in 2061. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 465 Figure 13.16. Comet Halley. This composite of three images (one in red, one in green, one in blue) shows Comet Halley as seen with a large telescope in Chile in 1986. During the time the three images were taken in sequence, the comet moved among the stars. The telescope was moved to keep the image of the comet steady, causing the stars to appear in triplicate (once", " in each color) in the background. (credit: modification of work by ESO Edmund Halley: Astronomy\u2019s Renaissance Man Edmund Halley (Figure 13.17), a brilliant astronomer who made contributions in many fields of science and statistics, was by all accounts a generous, warm, and outgoing person. In this, he was quite the opposite of his good friend Isaac Newton, whose great work, the Principia (see Orbits and Gravity), Halley encouraged, edited, and helped pay to publish. Halley himself published his first scientific paper at age 20, while still in college. As a result, he was given a royal commission to go to Saint Helena (a remote island off the coast of Africa where Napoleon would later be exiled) to make the first telescopic survey of the southern sky. After returning, he received the equivalent of a master\u2019s degree and was elected to the prestigious Royal Society in England, all at the age of 22. In addition to his work on comets, Halley was the first astronomer to recognize that the so-called \u201cfixed\u201d stars move relative to each other, by noting that several bright stars had changed their positions since Ptolemy\u2019s publication of the ancient Greek catalogs. He wrote a paper on the possibility of an infinite universe, proposed that some stars may be variable, and discussed the nature and size of nebulae (glowing cloudlike structures visible in telescopes). While in Saint Helena, Halley observed the planet 466 Chapter 13 Comets and Asteroids: Debris of the Solar System Mercury going across the face of the Sun and developed the mathematics of how such transits could be used to establish the size of the solar system. In other fields, Halley published the first table of human life expectancies (the precursor of life-insurance statistics); wrote papers on monsoons, trade winds, and tides (charting the tides in the English Channel for the first time); laid the foundations for the systematic study of Earth\u2019s magnetic field; studied evaporation and how inland waters become salty; and even designed an underwater diving bell. He served as a British diplomat, advising the emperor of Austria and squiring the future czar of Russia around England (avidly discussing, we are told, both the importance of science and the quality of local brandy). In 1703, Halley became a professor of geometry at Oxford, and in 1720, he was appointed Astronomer Royal of England. He continued observing Earth and the sky and publishing his ideas for another 20 years, until death claimed him at age 85. Figure 13.17. Edmund Halley (1656\u20131742). Halley was a prolific contributor to the sciences. His study of comets at the turn of the eighteenth century helped predict the orbit of the comet that now bears his name. Only a few comets return in a time measureable in human terms (shorter than a century), like Comet Halley does; these are called short-period comets. Many short-period comets have had their orbits changed by coming too close to one of the giant planets\u2014most often Jupiter (and they are thus sometimes called Jupiter-family comets). Most comets have long periods and will take thousands of years to return, if they return at all. As we will see later in this chapter, most Jupiter-family comets come from a different source than the long-period comets (those with orbital periods longer than about a century). Observational records exist for thousands of comets. We were visited by two bright comets in recent decades. First, in March 1996, came Comet Hyakutake, with a very long tail. A year later, Comet Hale-Bopp appeared; it was as bright as the brightest stars and remained visible for several weeks, even in urban areas (see the image that opens this chapter, Figure 13.1). Table 13.2 lists some well-known comets whose history or appearance is of special interest. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 467 Some Interesting Comets Name Period Significance Great Comet of 1577 Long Tycho Brahe showed it was beyond the Moon (a big step in our understanding) Great Comet of 1843 Long Brightest recorded comet; visible in daytime Daylight Comet of 1910 Long Brightest comet of the twentieth century West Long Nucleus broke into pieces (1976) Hyakutake Long Passed within 15 million km of Earth (1996) Hale\u2013Bopp Long Brightest recent comet (1997) Swift-Tuttle 133 years Parent comet of Perseid meteor shower Halley Borrelly Biela ChuryumovGerasimenko 76 years First comet found to be periodic; explored by spacecraft in 1986 6.8 years Flyby by Deep Space 1 spacecraft (2000) 6.7", " years Broke up in 1846 and not seen again 6.5 years Target of Rosetta mission (2014\u201316) Wild 2 6.4 years Target of Stardust sample return mission (2004) Tempel 1 5.7 years Target of Deep Impact mission (2005) Encke 3.3 years Shortest known period Table 13.2 The Comet\u2019s Nucleus When we look at an active comet, all we normally see is its temporary atmosphere of gas and dust illuminated by sunlight. This atmosphere is called the comet\u2019s head or coma. Since the gravity of such small bodies is very weak, the atmosphere is rapidly escaping all the time; it must be replenished by new material, which has to come from somewhere. The source is the small, solid nucleus inside, just a few kilometers across, usually hidden by the glow from the much-larger atmosphere surrounding it. The nucleus is the real comet, the fragment of ancient icy material responsible for the atmosphere and the tail (Figure 13.18). 468 Chapter 13 Comets and Asteroids: Debris of the Solar System Figure 13.18. Parts of a Comet. This schematic illustration shows the main parts of a comet. Note that the different structures are not to scale. The modern theory of the physical and chemical nature of comets was first proposed by Harvard astronomer Fred Whipple in 1950. Before Whipple\u2019s work, many astronomers thought that a comet\u2019s nucleus might be a loose aggregation of solids, sort of an orbiting \u201cgravel bank,\u201d Whipple proposed instead that the nucleus is a solid object a few kilometers across, composed in substantial part of water ice (but with other ices as well) mixed with silicate grains and dust. This proposal became known as the \u201cdirty snowball\u201d model. The water vapor and other volatiles that escape from the nucleus when it is heated can be detected in the comet\u2019s head and tail, and therefore, we can use spectra to analyze what atoms and molecules the nucleus ice consists of. However, we are somewhat less certain of the non-icy component. We have never identified a fragment of solid matter from a comet that has survived passage through Earth\u2019s atmosphere. However, spacecraft that have approached comets have carried dust detectors, and some comet dust has even been returned to Earth (see Figure 13.19). It seems that much of the \u201cdirt\u201d in the dirty snowball is dark, primitive hydrocarbons and silicates, rather like the material thought to be present on the dark, primitive asteroids. Figure 13.19. Captured Comet Dust. This particle (seen through a microscope) is believed to be a tiny fragment of cometary dust, collected in the upper atmosphere of Earth. It measures about 10 microns, or 1/100 of a millimeter, across. (credit: NASA/JPL) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 469 Since the nuclei of comets are small and dark, they are difficult to study from Earth. Spacecraft did obtain direct measurements of a comet nucleus, however, in 1986, when three spacecraft swept past Comet Halley at close range (see Figure 13.20). Subsequently, other spacecraft have flown close to other comets. In 2005, the NASA Deep Impact spacecraft even carried a probe for a high-speed impact with the nucleus of Comet Tempel 1. But by far, the most productive study of a comet has been by the 2015 Rosetta mission, which we will discuss shortly. Figure 13.20. Close-up of Comet Halley. This historic photograph of the black, irregularly shaped nucleus of Comet Halley was obtained by the ESA Giotto spacecraft from a distance of about 1000 kilometers. The bright areas are jets of material escaping from the surface. The length of the nucleus is 10 kilometers, and details as small as 1 kilometer can be made out. (credit: modification of work by ESA) The Comet\u2019s Atmosphere The spectacular activity that allows us to see comets is caused by the evaporation of cometary ices heated by sunlight. Beyond the asteroid belt, where comets spend most of their time, these ices are solidly frozen. But as a comet approaches the Sun, it begins to warm up. If water (H2O) is the dominant ice, significant quantities vaporize as sunlight heats the surface above 200 K. This happens for the typical comet somewhat beyond the orbit of Mars. The evaporating H2O in turn releases the dust that was mixed with the ice. Since the comet\u2019s nucleus is so small, its gravity cannot hold back either the gas or the dust, both of which flow away into space at speeds of about 1 kilometer per second. The comet continues to absorb energy as it approaches the Sun. A great deal", " of this energy goes into the evaporation of its ice, as well as into heating the surface. However, recent observations of many comets indicate that the evaporation is not uniform and that most of the gas is released in sudden spurts, perhaps confined to a few areas of the surface. Expanding into space at a speed of about 1 kilometer per second, the comet\u2019s atmosphere can reach an enormous size. The diameter of a comet\u2019s head is often as large as Jupiter, and it can sometimes approach a diameter of a million kilometers (Figure 13.21). Figure 13.21. Head of Comet Halley. Here we see the cloud of gas and dust that make up the head, or coma, of Comet Halley in 1986. On this scale, the nucleus (hidden inside the cloud) would be a dot too small to see. (credit: modification of work by NASA/W. Liller) 470 Chapter 13 Comets and Asteroids: Debris of the Solar System Most comets also develop tails as they approach the Sun. A comet\u2019s tail is an extension of its atmosphere, consisting of the same gas and dust that make up its head. As early as the sixteenth century, observers realized that comet tails always point away from the Sun (Figure 13.22), not back along the comet\u2019s orbit. Newton proposed that comet tails are formed by a repulsive force of sunlight driving particles away from the head\u2014an idea close to our modern view. Figure 13.22. Comet Orbit and Tail. The orientation of a typical comet tail changes as the comet passes perihelion. Approaching the Sun, the tail is behind the incoming comet head, but on the way out, the tail precedes the head. The two different components that make up the tail (the dust and gas) act somewhat differently. The brightest part of the tail is called the dust tail, to differentiate it from a fainter, straight tail made of ionized gas, called the ion tail. The ion tail is carried outward by streams of ions (charged particles) emitted by the Sun. As you can see in Figure 13.23, the smoother dust tail curves a bit, as individual dust particles spread out along the comet\u2019s orbit, whereas the straight ion is tail pushed more directly outward from the Sun by our star\u2019s wind of charged particles This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 471 Figure 13.23. Comet Tails. (a) As a comet nears the Sun, its features become more visible. In this illustration from NASA showing Comet HaleBopp, you can see a comet\u2019s two tails: the more easily visible dust tail, which can be up to 10 million kilometers long, and the fainter gas tail (or ion tail), which is up to hundreds of millions of kilometers long. The grains that make up the dust tail are the size of smoke particles. (b) Comet Mrkos was photographed in 1957 with a wide-field telescope at Palomar Observatory and also shows a clear distinction between the straight gas tail and the curving dust tail. (credit a: modification of work by ESO/E. Slawik; credit b: modification of work by Charles Kearns, George O. Abell, and Byron Hill These days, comets close to the Sun can be found with spacecraft designed to observe our star. For example, in early July, 2011, astronomers at the ESA/NASA\u2019s Solar and Heliospheric Observatory (SOHO) witnessed a comet (https://openstaxcollege.org/l/30ESANASAcomet) streaking toward the Sun, one of almost 3000 such sightings. You can also watch a brief video by NASA entitled \u201cWhy Are We Seeing So Many Sungrazing Comets?\u201d The Rosetta Comet Mission In the 1990s, European scientists decided to design a much more ambitious mission that would match orbits with an incoming comet and follow it as it approached the Sun. They also proposed that a smaller spacecraft would actually try to land on the comet. The 2-ton main spacecraft was named Rosetta, carrying a dozen scientific instruments, and its 100-kilogram lander with nine more instruments was named Philae. The Rosetta mission was launched in 2004. Delays with the launch rocket caused it to miss its original target comet, so an alternate destination was picked, Comet Churyumov-Gerasimenko (named after the two discoverers, but generally denoted 67P). This comet\u2019s period of revolution is 6.45 years, making it a Jupiter- family comet. Since the European Space Agency did not have access to the plutonium-fueled nuclear power sources used by NASA for deep space missions, Rosetta had to be solar powered,", " requiring especially large solar panels. Even these were not enough to keep the craft operating as it matched orbits with 67P near the comet\u2019s aphelion. The only solution was to turn off all the spacecraft systems and let it coast for several years toward the Sun, out of 472 Chapter 13 Comets and Asteroids: Debris of the Solar System contact with controllers on Earth until solar energy was stronger. The success of the mission depended on an automatic timer to turn the power back on as it neared the Sun. Fortunately, this strategy worked. In August 2014, Rosetta began a gradual approach to the comet nucleus, which is a strangely misshapen object about 5 kilometers across, quite different from the smooth appearance of Halley\u2019s nucleus (but equally dark). Its rotation period is 12 hours. On November 12, 2014, the Philae lander was dropped, descending slowly for 7 hours before gently hitting the surface. It bounced and rolled, coming to rest under an overhang where there was not enough sunlight to keep its batteries charged. After operating for a few hours and sending data back to the orbiter, Philae went silent. The main Rosetta spacecraft continued operations, however, as the level of comet activity increased, with steamers of gas jetting from the surface. As the comet approached perihelion in September 2015, the spacecraft backed off to ensure its safety. The extent of the Rosetta images (and data from other instruments) far exceeds anything astronomers had seen before from a comet. The best imaging resolution was nearly a factor of 100 greater than in the best Halley images. At this scale, the comet appears surprisingly rough, with sharp angles, deep pits, and overhangs (Figure 13.24). Figure 13.24. Comet 67P\u2019s Strange Shape and Surface Features. (a) This image from the Rosetta camera was taken from a distance of 285 kilometers. The resolution is 5 meters. You can see that the comet consists of two sections with a connecting \u201cneck\u201d between them. (b) This close-up view of Comet Churyumov-Gerasimenko is from the Philae lander. One of the lander\u2019s three feet is visible in the foreground. The lander itself is mostly in shadow. (credit a: modification of work by ESA/Rosetta/MPS for OSIRIS Team MPS/UPD/LAM/IAA/SSO/INTA/UPM/DASP/IDA; credit b: modification of work by ESA/Rosetta/Philae/CIVA) The double-lobed shape of 67P\u2019s nucleus has been tentatively attributed to the collision and merger of two independent comet nuclei long ago. The spacecraft verified that the comet\u2019s dark surface was covered with organic carbon-rich compounds, mixed with sulfides and iron-nickel grains. 67P has an average density of only 0.5 g/cm3 (recall water in these units has a density of 1 g/cm3.) This low density indicates that the comet is quite porous, that is, there is a large amount of empty space among its materials. We already knew that the evaporation of comet ices was sporadic and limited to small jets, but in comet 67P, this was carried to an extreme. At any one time, more than 99% of the surface is inactive. The active vents are only a few meters across, with the material confined to narrow jets that persist for just a few minutes (Figure 13.25). The level of activity is strongly dependent on solar heating, and between July and August 2015, it increased by a factor of 10. Isotopic analysis of deuterium in the water ejected by the comet shows that it is different from the water found on Earth. Thus, apparently comets like 67P did not contribute to the origin of our oceans or the water in our bodies, as some scientists had thought. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 473 Figure 13.25. Gas Jets on Comet 67P. (a) This activity was photographed by the Rosetta spacecraft near perihelion. You can see a jet suddenly appearing; it was active for only a few minutes. (b) This spectacular photo, taken near perihelion, shows the active comet surrounded by multiple jets of gas and dust. (credit a, b: modification of work by ESA/Rosetta/MPS; credit c: modification of work by ESA/Rosetta/NAVCAM The European Space Agency is continuing to make interesting short videos (https://openstaxcollege.org/l/30ESAvideoros) illustrating the challenges and results of the Rosetta and Philae missions. For example, watch \u201cRosetta\u2019s", " Moment in the Sun\u201d to see some of the images of the comet generating plumes of gas and dust and hear about some of the dangers an active comet poses for the spacecraft. 13.4 THE ORIGIN AND FATE OF COMETS AND RELATED OBJECTS Learning Objectives By the end of this section, you will be able to: Describe the traits of the centaur objects Chronicle the discovery and describe the composition of the Oort cloud Describe trans-Neptunian and Kuiper-belt objects Explain the proposed fate of comets that enter the inner solar system The comets we notice when they come near Earth (especially the ones coming for the first time) are probably the most primitive objects we can study, preserved unchanged for billions of years in the deep freeze of the outer solar system. However, astronomers have discovered many other objects that orbit the Sun beyond the planets. Centaurs In the outer solar system, where most objects contain large amounts of water ice, the distinction between asteroids and comets breaks down. Astronomers initially still used the name \u201casteroids\u201d for new objects discovered going around the Sun with orbits that carry them far beyond Jupiter. The first of these objects is Chiron, found in 1977 on a path that carries it from just inside the orbit of Saturn at its closest approach to the 474 Chapter 13 Comets and Asteroids: Debris of the Solar System Sun out to almost the distance of Uranus (Figure 13.26). The diameter of Chiron is estimated to be about 200 kilometers, much larger than any known comet. Figure 13.26. Chiron\u2019s Orbit. Chiron orbits the Sun every 50 years, with its closest approach being inside the orbit of Saturn and its farthest approach out to the orbit of Uranus. In 1992, a still-more-distant object named Pholus was discovered with an orbit that takes it 33 AU from the Sun, beyond the orbit of Neptune. Pholus has the reddest surface of any object in the solar system, indicating a strange (and still unknown) surface composition. As more objects are discovered in these distant reaches, astronomers decided that they will be given the names of centaurs from classical mythology; this is because the centaurs were half human, half horse, and these new objects display some of the properties of both asteroids and comets. Beyond the orbit of Neptune lies a cold, dark realm populated by objects called simply trans-Neptunian objects (TNOs). The first discovered, and best known, of these TNOs is the dwarf planet Pluto. We discussed Pluto and the New Horizons spacecraft encounter with it in Rings, Moons, and Pluto. The second TNO was discovered in 1992, and now more than a thousand are known, most of them smaller than Pluto. The largest ones after Pluto\u2014named Eris, Makemake, and Haumea\u2014are also classed as dwarf planets. Except for their small size, dwarf planets have many properties in common with the larger planets. Pluto has five moons, and two moons have been discovered orbiting Haumea and one each circling Eris and Makemake. The Kuiper Belt and the Oort Cloud TNOs are a part of what is called the Kuiper belt, a large area of space beyond Neptune that is also the source of many comets. Astronomers study the Kuiper belt in two ways. New, more powerful telescopes allow us to discover many of the larger members of the Kuiper belt directly. We can also measure the composition of comets that come from the Kuiper belt. More than a thousand Kuiper belt objects have been discovered, and astronomers estimate that there are more than 100,000 with diameters large than 100 kilometers, in a disk extending out to about 50 AU from the Sun. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 475 The short-period comets (such as Halley) are thought to originate in the Kuiper belt, where small gravitational perturbations from Neptune can gradually shift their orbits until they can penetrate the inner solar system. The long-period comets, however, come from a much more distant reservoir of icy objects, called the Oort cloud. Careful studies of the orbits of long-period comets revealed that they come initially from very great distances. By following their orbits backward, we can calculate that the aphelia (points farthest from the Sun) of newly discovered comets typically have values near 50,000 AU (more than a thousand times farther than Pluto). This clustering of aphelion distances was first noted by Dutch astronomer Jan Oort, who, in 1950, proposed an idea for the origin of those comets that is still accepted today (Figure 13.27). Figure 13.27. Jan Oort (1900\u20131992).", " Jan Oort first suggested that there might be a reservoir of frozen chunks, potential comet nuclei, at the edge of the region of the Sun\u2019s gravitational influence. (credit: The Leiden Observatory) It is possible to calculate that a star\u2019s gravitational sphere of influence\u2014the distance within which it can exert sufficient gravitation to hold onto orbiting objects\u2014is about one third of its distance to the nearest other stars. Stars in the vicinity of the Sun are spaced in such a way that the Sun\u2019s sphere of influence extends a little beyond 50,000 AU, or about 1 light-year. At such great distances, however, objects in orbit about the Sun can be perturbed by the gravity of passing stars. Some of the perturbed objects can then take on orbits that bring them much closer to the Sun (while others might be lost to the solar system forever). Oort suggested, therefore, that the new comets we were seeing were examples of objects orbiting the Sun near the edge of its sphere of influence, whose orbits had been disturbed by nearby stars, eventually bringing them close to the Sun where we can see them. The reservoir of ancient icy objects from which such comets are derived is now called the Oort cloud. Astronomers estimate that there are about a trillion (1012) comets in the Oort cloud. In addition, we estimate that about 10 times this number of icy objects could be orbiting the Sun in the volume of space between the Kuiper belt (which is gravitationally linked to Neptune) and the Oort cloud. These objects remain undiscovered because they are too faint to be seen directly and their orbits are too stable to permit any of them to be deflected inward close to the Sun. The total number of icy or cometary objects in the outer reaches of our solar system could thus be on the order of 10 trillion (1013), a very large number indeed. What is the mass represented by 1013 comets? We can make an estimate if we assume something about comet sizes and masses. Let us suppose that the nucleus of Comet Halley is typical. Its observed volume is about 600 476 Chapter 13 Comets and Asteroids: Debris of the Solar System km3. If the primary constituent is water ice with a density of about 1 g/cm3, then the total mass of Halley\u2019s nucleus must be about 6 \u00d7 1014 kilograms. This is about one ten billionth (10\u201310) of the mass of Earth. If our estimate is reasonable and there are 1013 comets with this mass out there, their total mass would be equal to about 1000 Earths\u2014comparable to the mass of all the planets put together. Therefore, icy, cometary material could be the most important constituent of the solar system after the Sun itself Mass of the Oort Cloud Comets Suppose the Oort cloud contains 1012 comets with an average diameter of 10 km each. Let\u2019s estimate the mass of the total Oort cloud. Solution We can start by assuming that typical comets are about the size of Comets Halley and Borrelly, with a diameter of 10 km and a density appropriate to water ice, which is about 1 g/cm3 or 1000 kg/m3. We know that density = mass/volume, the volume of a sphere, V = 4 3 \u03c0R3, and the radius, R = 1 2 D. Therefore, for each comet, mass = density \u00d7 volume = density \u00d7 4 3 \u239b \u03c0 \u239d 1 2 D\u239e \u23a0 3 Given that 10 km = 104 m, each comet\u2019s mass is mass = 1000 kg/m3 \u00d7 4 3 \u00d7 3.14 \u00d7 1 8 \u239b \u239d104\u239e \u23a0 \u00d7 3 m3 \u2248 1015 kg = 1012 tons To calculate the total mass of the cloud, we multiply this typical mass for one comet by the number of comets: Check Your Learning total mass = 1015 kg/comet \u00d7 1012 comets = 1027 kg How does the total mass we calculated above compare to the mass of Jupiter? To the mass of the Sun? (Give a numerical answer.) Answer: The mass of Jupiter is about 1.9 \u00d7 1027 kg. The mass of the Oort cloud calculated above is 1027 kg. So the cloud would contain about half a Jupiter of mass. The mass of the Sun is 2 \u00d7 1030 kg. This means the Oort cloud would be 1027 kg \u239b \u239d2 \u00d7 1030 kg \u239e \u23a0 = 0.0005 \u00d7 the mass of the Sun This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 477 Early Evolution of the Planetary System Comets from the Oort cloud help us sample material that formed very", " far from the Sun, whereas the short- period comets from the Kuiper belt sample materials that were planetesimals in the solar nebula disk but did not form planets. Studies of the Kuiper belt also are influencing our understanding of the early evolution of our planetary system. The objects in the Oort cloud and the Kuiper belt have different histories, and they may therefore have different compositions. Astronomers are therefore very interested in comparing detailed measurements of the comets derived from these two source regions. Most of the bright comets that have been studied in the past (Halley, Hyakutake, Hale-Bopp) are Oort cloud comets, but P67 and several other comets targeted for spacecraft measurements in the next decade are Jupiter-family comets from the Kuiper belt (see Table 13.2). The Kuiper belt is made up of ice-and rock planetesimals, a remnant of the building blocks of the planets. Since it is gravitationally linked to Neptune, it can help us understand the formation and history of the solar system. As the giant planets formed, their gravity profoundly influenced the orbits of Kuiper belt objects. Computer simulations of the early evolution of the planetary system suggest that the gravitational interactions between the giant planets and the remaining planetesimals caused the orbit of Jupiter to drift inward, whereas the orbits of Saturn, Uranus, and Neptune all expanded, carrying the Kuiper belt with them. Another hypotheses involves a fifth giant planet that was expelled from the solar system entirely as the planetary orbits shifted. Neptune\u2019s retrograde (backward-orbiting) moon Triton (which is nearly as large as Pluto) may have been a Kuiper belt object captured by Neptune during the period of shifting orbits. It clearly seems that the Kuiper belt may carry important clues to the way our solar system reached its present planetary configuration Comet Hunting as a Hobby When amateur astronomer David Levy (Figure 13.28), the co-discoverer of Comet Shoemaker-Levy 9, found his first comet, he had already spent 928 fruitless hours searching through the dark night sky. But the discovery of the first comet only whetted his appetite. Since then, he has found 8 others on his own and 13 more working with others. Despite this impressive record, he ranks only third in the record books for number of comet discoveries. But David hopes to break the record someday. All around the world, dedicated amateur observers spend countless nights scanning the sky for new comets. Astronomy is one of the very few fields of science where amateurs can still make a meaningful contribution, and the discovery of a comet is one of the most exciting ways they can establish their place in astronomical history. Don Machholz, a California amateur (and comet hunter) who has been making a study of comet discoveries, reported that between 1975 and 1995, 38% of all comets discovered were found by amateurs. Those 20 years yielded 67 comets for amateurs, or almost 4 per year. That might sound pretty encouraging to new comet hunters, until they learn that the average number of hours the typical amateur spent searching for a comet before finding one was about 420. Clearly, this is not an activity for impatient personalities. What do comet hunters do if they think they have found a new comet? First, they must check the object\u2019s location in an atlas of the sky to make sure it really is a comet. Since the first sighting of a comet usually occurs when it is still far from the Sun and before it sports a significant tail, it will look like only a small, 478 Chapter 13 Comets and Asteroids: Debris of the Solar System fuzzy patch. And through most amateur telescopes, so will nebulae (clouds of cosmic gas and dust) and galaxies (distant groupings of stars). Next, they must check that they have not come across a comet that is already known, in which case, they will only get a pat on the back instead of fame and glory. Then they must re-observe or re-image it sometime later to see whether its motion in the sky is appropriate for comets. Often, comet hunters who think they have made a discovery get another comet hunter elsewhere in the country to confirm it. If everything checks out, the place they contact is the Central Bureau for Astronomical Telegrams at the Harvard-Smithsonian Center for Astrophysics in Cambridge, Massachusetts (http://www.cbat.eps.harvard.edu/). If the discovery is confirmed, the bureau will send the news out to astronomers and observatories around the world. One of the unique rewards of comet hunting is that the discoverer\u2019s name becomes associated with the new comet\u2014a bit of cosmic fame that few hobbies can match. Figure 13.28. David Levy. Amateur astronomer David Levy ranks third in the world for comet discoveries. (credit: Andrew Fraknoi", ") The Fate of Comets Any comet we see today will have spent nearly its entire existence in the Oort cloud or the Kuiper belt at a temperature near absolute zero. But once a comet enters the inner solar system, its previously uneventful life history begins to accelerate. It may, of course, survive its initial passage near the Sun and return to the cold reaches of space where it spent the previous 4.5 billion years. At the other extreme, it may collide with the Sun or come so close that it is destroyed on its first perihelion passage (several such collisions have been observed with space telescopes that monitor the Sun). Sometimes, however, the new comet does not come that close to the Sun but instead interacts with one or more of the planets SOHO (the Solar and Heliospheric Observatory) has an excellent collection of videos of comets (https://openstaxcollege.org/l/30SOHOcomvid) that come near the Sun. At this site, comet ISON approaches the Sun and is believed to be destroyed in its passage. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 479 A comet that comes within the gravitational influence of a planet has three possible fates. It can (1) impact the planet, ending the story at once; (2) speed up and be ejected, leaving the solar system forever; or (3) be perturbed into an orbit with a shorter period. In the last case, its fate is sealed. Each time it approaches the Sun, it loses part of its material and also has a significant chance of collision with a planet. Once the comet is in this kind of short-period orbit, its lifetime starts being measured in thousands, not billions, of years. A few comets end their lives catastrophically by breaking apart (sometimes for no apparent reason) (Figure 13.29). Especially spectacular was the fate of the faint Comet Shoemaker-Levy 9, which broke into about 20 pieces when it passed close to Jupiter in July 1992. The fragments of Shoemaker-Levy were actually captured into a very elongated, two-year orbit around Jupiter, more than doubling the number of known jovian moons. This was only a temporary enrichment of Jupiter\u2019s family, however, because in July 1994, all the comet fragments crashed unto Jupiter, releasing energy equivalent to millions of megatons of TNT. Figure 13.29. Breakup of Comet LINEAR. (a) A ground-based view with much less detail and (b) a much more detailed photo with the Hubble Space Telescope, showing the multiple fragments of the nucleus of Comet LINEAR. The comet disintegrated in July 2000 for no apparent reason. (Note in the left view, the fragments all blend their light together, and can\u2019t be distinguished. The short diagonal white lines are stars that move in the image, which is keeping track of the moving comet.) (credit a: modification of work by the University of Hawaii; credit b: modification of work by NASA, Harold Weaver (the Johns Hopkins University), and the HST Comet LINEAR Investigation Team) As each cometary fragment streaked into the jovian atmosphere at a speed of 60 kilometers per second, it disintegrated and exploded, producing a hot fireball that carried the comet dust as well as atmospheric gases to high altitudes. These fireballs were clearly visible in profile, with the actual point of impact just beyond the jovian horizon as viewed from Earth (Figure 13.30). As each explosive plume fell back into Jupiter, a region of the upper atmosphere larger than Earth was heated to incandescence and glowed brilliantly for about 15 minutes, a glow we could detect with infrared-sensitive telescopes. 480 Chapter 13 Comets and Asteroids: Debris of the Solar System Figure 13.30. Comet Impact on Jupiter. (a) The \u201cstring\u201d of white objects are fragments of Comet Shoemaker-Levy 9 approaching Jupiter. (b) The first fragment of the comet impacts Jupiter, with the point of contact on the bottom left side in this image. On the right is Jupiter\u2019s moon, Io. The equally bright spot in the top image is the comet fragment flaring to maximum brightness. The bottom image, taken about 20 minutes later, shows the lingering flare from the impact. The Great Red Spot is visible near the center of Jupiter. These infrared images were taken with a German-Spanish telescope on Calar Alto in southern Spain. (credit a: modification of work by ESA; credit b: modification of work by Tom Herbst, Max-Planck-Institut fuer Astronomie, Heidelberg, Doug Hamilton, Max-Planck-Institut fuer Kernphysik, Heidelberg, Hermann Boehnhardt, Universitaets-S", "ternewarte, Muenchen, and Jose Luis Ortiz Moreno, Instituto de Astrofisica de Andalucia, Granada) After this event, dark clouds of debris settled into the stratosphere of Jupiter, producing long-lived \u201cbruises\u201d (each still larger than Earth) that could be easily seen through even small telescopes (Figure 13.31). Millions of people all over the world peered at Jupiter through telescopes or followed the event via television or online. Another impact feature was seen on Jupiter in summer 2009, indicating that the 1994 events were by no means unique. Seeing these large, impact explosions on Jupiter helps us to appreciate the disaster that would happen to our planet if we were hit by a comet or asteroid. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 481 Figure 13.31. Impact Dust Cloud on Jupiter. These features result from the impact of Comet Shoemaker-Levy 9 with Jupiter, seen with the Hubble Space Telescope 105 minutes after the impact that produced the dark rings (the compact back dot came from another fragment). The inner edge of the diffuse, outer ring is about the same size as Earth. Later, the winds on Jupiter blended these features into a broad spot that remained visible for more than a month. (credit: modification of work by H. Hammel, MIT, and NASA/ESA) For comets that do not meet so dramatic an end, measurements of the amount of gas and dust in their atmospheres permit us to estimate the total losses during one orbit. Typical loss rates are up to a million tons per day from an active comet near the Sun, adding up to some tens of millions of tons per orbit. At that rate, a typical comet will be gone after a few thousand orbits. This will probably be the fate of Comet Halley in the long run This History Channel video (https://openstaxcollege.org/l/30hischanUNIV) shows a short discussion and animation from the TV documentary series Universe, showing the collision of Comet Shoemaker-Levy 9 with Jupiter. 482 Chapter 13 Comets and Asteroids: Debris of the Solar System CHAPTER 13 REVIEW KEY TERMS asteroid a stony or metallic object orbiting the Sun that is smaller than a major planet but that shows no evidence of an atmosphere or of other types of activity associated with comets asteroid belt the region of the solar system between the orbits of Mars and Jupiter in which most asteroids are located; the main belt, where the orbits are generally the most stable, extends from 2.2 to 3.3 AU from the Sun comet a small body of icy and dusty matter that revolves about the Sun; when a comet comes near the Sun, some of its material vaporizes, forming a large head of tenuous gas and often a tail Kuiper belt a region of space beyond Neptune that is dynamically stable (like the asteroid belt); the source region for most short-period comets near-Earth asteroid (NEA) an Earth-approaching asteroid, one whose orbit could bring it on a collision course with our planet near-Earth object (NEO) a comet or asteroid whose path intersects the orbit of Earth nucleus (of a comet) the solid chunk of ice and dust in the head of a comet Oort cloud the large spherical region around the Sun from which most \u201cnew\u201d comets come; a reservoir of objects with aphelia at about 50,000 AU tail (of a comet) a tail consisting of two parts: the dust tail is made of dust loosened by the sublimation of ice in a comet that is then pushed by photons from the Sun into a curved stream; the ion tail is a stream of ionized particles evaporated from a comet and then swept away from the Sun by the solar wind SUMMARY 13.1 Asteroids The solar system includes many objects that are much smaller than the planets and their larger moons. The rocky ones are generally called asteroids. Ceres is the largest asteroid; about 15 are larger than 250 kilometers and about 100,000 are larger than 1 kilometer. Most are in the asteroid belt between Mars and Jupiter. The presence of asteroid families in the belt indicates that many asteroids are the remnants of ancient collisions and fragmentation. The asteroids include both primitive and differentiated objects. Most asteroids are classed as C-type, meaning they are composed of carbonaceous materials. Dominating the inner belt are S-type (stony) asteroids, with a few M-type (metallic) ones. We have spacecraft images of several asteroids and returned samples from asteroid Itokawa. Recent observations have detected a number of asteroid moons, making it possible to measure the masses and densities of the asteroids they orbit. The two largest asteroids, Ceres and Vesta, have been extensively studied from orbit by the Dawn spacecraft. 13.2 Asteroids and Planetary", " Defense Near-Earth asteroids (NEAs), and near-Earth objects (NEOs) in general, are of interest in part because of their potential to hit Earth. They are on unstable orbits, and on timescales of 100 million years, they will either impact one of the terrestrial planets or the Sun, or be ejected. Most of them probably come from the asteroid belt, but some may be dead comets. NASA\u2019s Spaceguard Survey has found 90% of the NEAs larger than 1 kilometer, and This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 483 none of the ones found so far are on a collision course with Earth. Scientists are actively working on possible technologies for planetary defense in case any NEOs are found on a collision course with Earth years in advance. For now, the most important task is to continue our surveys, so we can find the next Earth impactor before it finds us. 13.3 The \u201cLong-Haired\u201d Comets Halley first showed that some comets are on closed orbits and return periodically to swing around the Sun. The heart of a comet is its nucleus, a few kilometers in diameter and composed of volatiles (primarily frozen H2O) and solids (including both silicates and carbonaceous materials). Whipple first suggested this \u201cdirty snowball\u201d model in 1950; it has been confirmed by spacecraft studies of several comets. As the nucleus approaches the Sun, its volatiles evaporate (perhaps in localized jets or explosions) to form the comet\u2019s head or atmosphere, which escapes at about 1 kilometer per second. The atmosphere streams away from the Sun to form a long tail. The ESA Rosetta mission to Comet P67 (Churyumov-Gerasimenko) has greatly increased our knowledge of the nature of the nucleus and of the process by which comets release water and other volatiles when heated by sunlight. 13.4 The Origin and Fate of Comets and Related Objects Oort proposed in 1950 that long-period comets are derived from what we now call the Oort cloud, which surrounds the Sun out to about 50,000 AU (near the limit of the Sun\u2019s gravitational sphere of influence) and contains between 1012 and 1013 comets. Comets also come from the Kuiper belt, a disk-shaped region beyond the orbit of Neptune, extending to 50 AU from the Sun. Comets are primitive bodies left over from the formation of the outer solar system. Once a comet is diverted into the inner solar system, it typically survives no more than a few thousand perihelion passages before losing all its volatiles. Some comets die spectacular deaths: Shoemaker-Levy 9, for example, broke into 20 pieces before colliding with Jupiter in 1994. FOR FURTHER EXPLORATION Articles Asteroids Asphang, E. \u201cThe Small Planets.\u201d Scientific American (May 2000): 46. On asteroids, including results from the NEAR mission. Beatty, J. \u201cThe Falcon\u2019s Wild Flight.\u201d Sky & Telescope (September 2006): 34. On the Japanese mission to asteroid Itakawa. Beatty, J. \u201cNEAR Falls for Eros.\u201d Sky & Telescope (May 2001): 35. On the first landing on an asteroid. Betz, E. \u201cDawn Mission Reveals Dwarf Planet Ceres.\u201d Astronomy (January 2016): 44. First images and discoveries. Binzel, R. \u201cA New Century for Asteroids.\u201d Sky & Telescope (July 2001): 44. Nice overview. Boslaugh, M. \u201cIn Search of Death-Plunge Asteroids.\u201d Astronomy (July 2015): 28. On existing and proposed programs to search for Earth-crossing asteroids. Cooke, B. \u201cFatal Attraction.\u201d Astronomy (May 2006): 46. On near-Earth asteroid Apophis, its orbit, and what we can learn from it. Durda, D. \u201cOdd Couples.\u201d Astronomy (December 2005): 54. On binary asteroids. Durda, D. \u201cAll in the Family.\u201d Astronomy (February 1993): 36. Discusses asteroid families. Oberg, J. \u201c2013\u2019s Historic Russian Meteorite Fall\u201d Astronomy (June 2012): 18. On the Chelyabinsk event. 484 Chapter 13 Comets and Asteroids: Debris of the Solar System Sheppard, S. \u201cDancing with the Planets.\u201d Sky & Telescope (June 2016): 16. On Trojan asteroids that \u201cfollow\u201d planets like Jupiter. Talcott, R. \u201cGalileo Views Gaspra.", "\u201d Astronomy (February 1992): 52. Yeomans, D. \u201cJapan Visits an Asteroid.\u201d Astronomy (March 2006): 32. On the Hayabusa probe exploration of asteroid Itakawa. Zimmerman, R. \u201cIce Cream Sundaes and Mashed Potatoes.\u201d Astronomy (February 1999): 54. On the NEAR mission. Comets Aguirre, E. \u201cThe Great Comet of 1997.\u201d Sky & Telescope (July 1997): 50. On Comet Hale-Bopp. Bakich, M. \u201cHow to Observe Comets.\u201d Astronomy (December 2009): 50. A guide for amateur astronomers. Gore, R. \u201cHalley\u2019s Comet \u201986: Much More Than Met the Eye.\u201d National Geographic (December 1986): 758. (Also, the March 1987 issue of Sky & Telescope was devoted to what we learned from Halley\u2019s Comet in 1986.) Hale, A. \u201cHale-Bopp Plus Ten.\u201d Astronomy (July 2005): 76. The co-discoverer of a naked-eye comet tells the story of the discovery and what followed. Jewett, D. \u201cMysterious Travelers: Comet Science.\u201d Sky & Telescope (December. 2013): 18. Nice summary of what we know about comets and questions we have. Rao, J. \u201cHow Often do Bright Comets Appear?\u201d Sky & Telescope (November 2013): 30. Nice summary of bright comets in the last century and what factors make a comet spectacular in our skies. Sekanina, Z. \u201cSungrazing Comets.\u201d Astronomy (March 2006): 36. Sheppard, S. \u201cBeyond the Kuiper Belt.\u201d Sky & Telescope (March 2015): 26. On Sedna and the Oort cloud. Stern, S. \u201cEvolution at the Edge.\u201d Astronomy (September 2005): 46. How comet nuclei evolve with time. Talcott, R. \u201cRendezvous with an Evolving Comet [Rosetta at Comet 67P/C-G].\u201d Astronomy (September 2015): 44. Tytell, D. \u201cDeep Impact\u2019s Hammer Throw.\u201d Sky & Telescope (October 2006): 34. On the mission that threw a probe at the nucleus of a comet. See also (June 2005): 40. Weissman, P. \u201cA Comet Tale.\u201d Sky & Telescope (February 2006): 36. A nice review of what we know and don\u2019t know about the physical nature of comets. Websites Asteroids Dawn Mission: http://dawn.jpl.nasa.gov. Discover more about this mission to the largest asteroids. NEAR-Shoemaker Mission: http://near.jhuapl.edu/. Review background information and see great images from the mission that went by Mathilde and Eros. Comets Deep Impact Mission: http://www.nasa.gov/mission_pages/deepimpact/main/. Kuiper Belt: http://www2.ess.ucla.edu/~jewitt/kb.html. David Jewitt of the University of Hawaii keeps track of the objects that have been discovered. Missions to Comets: http://solarsystem.nasa.gov/missions/target/comets. Read about NASA\u2019s current and past missions to comets. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 485 Stardust Mission: http://stardust.jpl.nasa.gov/home/index.html. Learn about this mission to collect a sample of a comet and bring it back to Earth. Videos Asteroids Sweating the Small Stuff: The Fear and Fun of Near-Earth Asteroids: https://www.youtube.com/ watch?v=5gyAvc5OhII. Harvard Observatory Night Lecture by Jose-Luis Galache (1:18:07). Unveiling Dwarf Planet Ceres: https://www.youtube.com/watch?v=_G9LudkLWOY. A vonKarman Lecture by Dr. Carol Raymond, Oct. 2015, also includes Vesta results (1:18:38). Comets Great Comets, Comets in General, and Comet ISON: https://www.youtube.com/watch?v=DiBkYAnQ_C. Talk by Frank Summers, Space Telescope Science Institute (1:01:10). Press Conference on the Impact of Comet Shoemaker-Levy 9 with Jupiter: https://www.youtube.com/ watch?v=B-tUP8afEIo. Day 2 after", " impact; July 17, 1994; with the discoverers and Heidi Hammel (1:22:29). Rosetta: The Story So Far: https://www.ras.org.uk/events-and-meetings/public-lectures/public-lecture-videos/ 2726-rosetta-the-story-so-far. Royal Astronomical Society Lecture by Dr. Ian Wright (1:00:29). COLLABORATIVE GROUP ACTIVITIES A. Your group is a congressional committee charged with evaluating the funding for an effort to find all the NEAs (near-Earth asteroids) that are larger than 0.5 kilometers across. Make a list of reasons it would be useful to humanity to find such objects. What should we (could we) do if we found one that will hit Earth in a few years? B. Many cultures considered comets bad omens. Legends associate comets with the deaths of kings, losses in war, or ends of dynasties. Did any members of your group ever hear about such folktales? Discuss reasons why comets in earlier times may have gotten this bad reputation. C. Because asteroids have a variety of compositions and a low gravity that makes the removal of materials quite easy, some people have suggested that mining asteroids may be a way to get needed resources in the future. Make a list of materials in asteroids (and comets that come to the inner solar system) that may be valuable to a space-faring civilization. What are the pros and cons of undertaking mining operations on these small worlds? D. As discussed in the feature box on Comet Hunting as a Hobby, amateur comet hunters typically spend more than 400 hours scanning the skies with their telescopes to find a comet. That\u2019s a lot of time to spend (usually alone, usually far from city lights, usually in the cold, and always in the dark). Discuss with members of your group whether you can see yourself being this dedicated. Why do people undertake such quests? Do you envy their dedication? E. The largest Kuiper belt objects known are also called dwarf planets. All the planets (terrestrial, jovian, and dwarf) in our solar system have so far been named after mythological gods. (The dwarf planet names have moved away from Roman mythology to include the gods of other cultures.) Have your group discuss whether we should continue this naming tradition with newly discovered dwarf planets. Why or why not? 486 Chapter 13 Comets and Asteroids: Debris of the Solar System F. The total cost of the Rosetta mission to match courses with a comet was about 1.4 billion Euros (about $1.6 billion US). Have your group discuss whether this investment was worth it, giving reasons for whichever side you choose. (On the European Space Agency website, they put this cost in context by saying, \u201cThe figure is barely half the price of a modern submarine, or three Airbus 380 jumbo jets, and covers a period of almost 20 years, from the start of the project in 1996 through the end of the mission in 2015.\u201d) G. If an Earth-approaching asteroid were discovered early enough, humanity could take measures to prevent a collision. Discuss possible methods for deflecting or even destroying an asteroid or comet. Go beyond the few methods mentioned in the text and use your creativity. Give pros and cons for each method. EXERCISES Review Questions 1. Why are asteroids and comets important to our understanding of solar system history? 2. Give a brief description of the asteroid belt. 3. Describe the main differences between C-type and S-type asteroids. 4. In addition to the ones mentioned in Exercise 13.3, what is the third, rarer class of asteroids? 5. Vesta is unusual as it contains what mineral on its surface? What does the presence of this material indicate? 6. Compare asteroids of the asteroid belt with Earth-approaching asteroids. What is the main difference between the two groups? 7. Briefly describe NASA\u2019s Spaceguard Survey. How many objects have been found in this survey? 8. Who first calculated the orbits of comets based on historical records dating back to antiquity? 9. Describe the nucleus of a typical comet and compare it with an asteroid of similar size. 10. Describe the two types of comet tails and how each are formed. 11. What classification is given to objects such as Pluto and Eris, which are large enough to be round, and whose orbits lie beyond that of Neptune? 12. Describe the origin and eventual fate of the comets we see from Earth. 13. What evidence do we have for the existence of the Kuiper belt? What kind of objects are found there? 14. Give brief descriptions of both the Kuiper belt and the Oort cloud. Thought Questions 15. Give at least two reasons today\u2019s astronomers are so interested in the discovery of additional Earth- approaching asteroids. 16.", " Suppose you were designing a spacecraft that would match course with an asteroid and follow along its orbit. What sorts of instruments would you put on board to gather data, and what would you like to learn? 17. Suppose you were designing a spacecraft that would match course with a comet and move with it for a while. What sorts of instruments would you put on board to gather data, and what would you like to learn? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 13 Comets and Asteroids: Debris of the Solar System 487 18. Suppose a comet were discovered approaching the Sun, one whose orbit would cause it to collide with Earth 20 months later, after perihelion passage. (This is approximately the situation described in the science-fiction novel Lucifer\u2019s Hammer by Larry Niven and Jerry Pournelle.) What could we do? Would there be any way to protect ourselves from a catastrophe? 19. We believe that chains of comet fragments like Comet Shoemaker-Levy 9\u2019s have collided not only with the jovian planets, but occasionally with their moons. What sort of features would you look for on the outer planet moons to find evidence of such collisions? (As an extra bonus, can you find any images of such features on a moon like Callisto? You can use an online site of planetary images, such as the Planetary Photojournal, at photojournal.jpl.nasa.gov.) 20. Why have we found so many objects in the Kuiper belt in the last two decades and not before then? 21. Why is it hard to give exact diameters for even the larger objects in the Kuiper belt? Figuring For Yourself 22. Refer to Example 13.1. How would the calculation change if a typical comet in the Oort cloud is only 1 km in diameter? 23. Refer to Example 13.1. How would the calculation change if a typical comet in the Oort cloud is larger\u2014say, 50 km in diameter? 24. The calculation in Example 13.1 refers to the known Oort cloud, the source for most of the comets we see. If, as some astronomers suspect, there are 10 times this many cometary objects in the solar system, how does the total mass of cometary matter compare with the mass of Jupiter? 25. If the Oort cloud contains 1012 comets, and ten new comets are discovered coming close to the Sun each year, what percentage of the comets have been \u201cused up\u201d since the beginning of the solar system? 26. The mass of the asteroids is found mostly in the larger asteroids, so to estimate the total mass we need to consider only the larger objects. Suppose the three largest asteroids\u2014Ceres (1000 km in diameter), Pallas (500 km in diameter), and Vesta (500 km in diameter)\u2014account for half the total mass. Assume that each of these three asteroids has a density of 3 g/cm3 and calculate their total mass. Multiply your result by 2 to obtain an estimate for the mass of the total asteroid belt. How does this compare with the mass of the Oort cloud? 27. Make a similar estimate for the mass of the Kuiper belt. The three largest objects are Pluto, Eris, and Makemake (each roughly 2000 km). In addition, assume there are eight objects (including Haumea, Orcus, Quaoar, Ixion, Varuna, and Charon, and objects that have not been named yet) with diameters of about 1000 km. Assume that all objects have Pluto\u2019s density of 2 g/cm3. Calculate twice the mass of the largest 13 objects and compare it to the mass of the main asteroid belt. 28. What is the period of revolution about the Sun for an asteroid with a semi-major axis of 3 AU in the middle of the asteroid belt? 29. What is the period of revolution for a comet with aphelion at 5 AU and perihelion at the orbit of Earth? 488 Chapter 13 Comets and Asteroids: Debris of the Solar System This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 489 14 COSMIC SAMPLES AND THE ORIGIN OF THE SOLAR SYSTEM Figure 14.1. Planetesimals. This illustration depicts a disk of dust and gas around a new star. Material in this disk comes together to form planetesimals. (credit: modification of work by University of Copenhagen/Lars Buchhave, NASA) Chapter Outline 14.1 Meteors 14.2 Meteorites: Stones from Heaven 14.3 Formation of the Solar System 14.4 Comparison with Other Planetary Systems 14.5 Planetary Evolution Thinking Ahead Imagine you are a scientist", " examining a sample of rock that had fallen from space a few days earlier and you find within it some of the chemical building blocks of life. How could you determine whether those \u201corganic\u201d materials came from space or were merely the result of earthly contamination? We conclude our survey of the solar system with a discussion of its origin and evolution. Some of these ideas were introduced in Other Worlds: An Introduction to the Solar System; we now return to them, using the information we have learned about individual planets and smaller members of the solar system. In addition, astronomers have recently discovered several thousand planets around other stars, including numerous multiplanet systems. This is an important new source of data, providing us a perspective that extends beyond our own particular (and perhaps atypical) solar system. But first, we want to look at another crucial way that astronomers learn about the ancient history of the solar system: by examining samples of primitive matter, the debris of the processes that formed the solar system some 4.5 billion years ago. Unlike the Apollo Moon rocks, these samples of cosmic material come to us free of charge\u2014they literally fall from the sky. We call this material cosmic dust and meteorites. 490 Chapter 14 Cosmic Samples and the Origin of the Solar System 14.1 METEORS Learning Objectives By the end of this section, you will be able to: Explain what a meteor is and why it is visible in the night sky Describe the origins of meteor showers As we saw in Comets and Asteroids: Debris of the Solar System, the ices in comets evaporate when they get close to the Sun, together spraying millions of tons of rock and dust into the inner solar system. There is also dust from asteroids that have collided and broken up. Earth is surrounded by this material. As each of the larger dust or rock particles enters Earth\u2019s atmosphere, it creates a brief fiery trail; this is often called a shooting star, but it is properly known as a meteor. Observing Meteors Meteors are tiny solid particles that enter Earth\u2019s atmosphere from interplanetary space. Since the particles move at speeds of many kilometers per second, friction with the air vaporizes them at altitudes between 80 and 130 kilometers. The resulting flashes of light fade out within a few seconds. These \u201cshooting stars\u201d got their name because at night their luminous vapors look like stars moving rapidly across the sky. To be visible, a meteor must be within about 200 kilometers of the observer. On a typical dark, moonless night, an alert observer can see half a dozen meteors per hour. These sporadic meteors\u2014those not associated with a meteor shower (explained in the next section)\u2014are random occurrences. Over the entire Earth, the total number of meteors bright enough to be visible totals about 25 million per day. The typical meteor is produced by a particle with a mass of less than 1 gram\u2014no larger than a pea. How can we see such a small particle? The light you see comes from the much larger region of heated, glowing gas surrounding this little grain of interplanetary material. Because of its high speed, the energy in a pea- sized meteor is as great as that of an artillery shell fired on Earth, but this energy is dispersed high in Earth\u2019s atmosphere. (When these tiny projectiles hit an airless body like the Moon, they do make small craters and generally pulverize the surface.) If a particle the size of a golf ball strikes our atmosphere, it produces a much brighter trail called a fireball (Figure 14.2). A piece as large as a bowling ball has a fair chance of surviving its fiery entry if its approach speed is not too high. The total mass of meteoric material entering Earth\u2019s atmosphere is estimated to be about 100 tons per day (which seems like a lot if you imagine it all falling in one place, but remember it is spread out all over our planet\u2019s surface). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 491 Figure 14.2. Fireball. When a larger piece of cosmic material strikes Earth\u2019s atmosphere, it can make a bright fireball. This time-lapse meteor image was captured in April 2014 at the Atacama Large Millimeter/Submillimeter Array (ALMA). The visible trail results from the burning gas around the particle. (credit: modification of work by ESO/C Malin While it is difficult to capture images of fireballs and other meteors with still photography, it\u2019s easy to capture the movement of these objects on video. The American Meteor Society maintains a website (https://openstaxcollege.org/l/30ammetsocweb) on which their members can share such videos. Meteor Showers Many\u2014perhaps most\u2014of the meteors that strike Earth", " are associated with specific comets. Some of these periodic comets still return to our view; others have long ago fallen apart, leaving only a trail of dust behind them. The dust particles from a given comet retain approximately the orbit of their parent, continuing to move together through space but spreading out over the orbit with time. When Earth, in its travels around the Sun, crosses such a dust stream, we see a sudden burst of meteor activity that usually lasts several hours; such an event is called a meteor shower. The dust particles and pebbles that produce meteor showers are moving together in space before they encounter Earth. Thus, as we look up at the atmosphere, their parallel paths seem to come toward us from a place in the sky called the radiant. This is the direction in space from which the meteor stream seems to be diverging, just as long railroad tracks seem to diverge from a single spot on the horizon (Figure 14.3). Meteor showers are often designated by the constellation in which this radiant is located: for example, the Perseid meteor shower has its radiant in the constellation of Perseus. But you are likely to see shower meteors anywhere in the sky, not just in the constellation of the radiant. The characteristics of some of the more famous meteor showers are summarized in Table 14.1. 492 Chapter 14 Cosmic Samples and the Origin of the Solar System Figure 14.3. Radiant of a Meteor Shower. The tracks of the meteors diverge from a point in the distance, just as long, parallel railroad tracks appear to do. (credit \u201ctracks\u201d: Nathan Vaughn) Major Annual Meteor Showers Shower Name Date of Maximum Associated Parent Object Comet\u2019s Period (years) Quadrantid January 3\u20134 2003EH (asteroid) Lyrid April 22 Comet Thatcher Eta Aquarid May 4\u20135 Comet Halley Delta Aquarid July 29\u201330 Comet Machholz Perseid August 11\u201312 Comet Swift-Tuttle Orionid October 20\u201321 Comet Halley Southern Taurid October 31 Comet Encke Leonid November 16\u201317 Comet Tempel-Tuttle Geminid December 13 Phaethon (asteroid) Table 14.1 \u2014 415 76 \u2014 133 76 3 33 1.4 The meteoric dust is not always evenly distributed along the orbit of the comet, so during some years more meteors are seen when Earth intersects the dust stream, and in other years fewer. For example, a very clumpy distribution is associated with the Leonid meteors, which in 1833 and again in 1866 (after an interval of 33 years\u2014the period of the comet) yielded the most spectacular showers (sometimes called meteor storms) ever recorded (Figure 14.4). During the Leonid storm on November 17, 1866, up to a hundred meteors were observed per second in some locations. The Leonid shower of 2001 was not this intense, but it peaked at nearly a thousand meteors per hour\u2014one every few seconds\u2014observable from any dark viewing site. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 493 Figure 14.4. Leonid Meteor Storm. A painting depicts the great meteor shower or storm of 1833, shown with a bit of artistic license. The most dependable annual meteor display is the Perseid shower, which appears each year for about three nights near August 11. In the absence of bright moonlight, you can see one meteor every few minutes during a typical Perseid shower. Astronomers estimate that the total combined mass of the particles in the Perseid swarm is nearly a billion tons; the comet that gave rise to the particles in that swarm, called Swift-Tuttle, must originally have had at least that much mass. However, if its initial mass were comparable to the mass measured for Comet Halley, then Swift-Tuttle would have contained several hundred billion tons, suggesting that only a very small fraction of the original cometary material survives in the meteor stream The California Academy of Sciences has a short animated guide (https://openstaxcollege.org/l/ 30howobsmetsho) on \u201cHow to Observe a Meteor Shower.\u201d No shower meteor has ever survived its flight through the atmosphere and been recovered for laboratory analysis. However, there are other ways to investigate the nature of these particles and thereby gain additional insight into the comets from which they are derived. Analysis of the flight paths of meteors shows that most of them are very light or porous, with densities typically less than 1.0 g/cm3. If you placed a fist-sized lump of meteor material on a table in Earth\u2019s gravity, it might well fall apart under its own weight. Such light particles break up very easily in the atmosphere, accounting for the failure of even relatively large shower meteors to reach", " the ground. Comet dust is apparently fluffy, rather inconsequential stuff. NASA\u2019s Stardust mission used a special substance, called aerogel, to collect these particles. We can also infer this from the tiny comet particles recovered in Earth\u2019s atmosphere with high-flying aircraft (see Figure 13.19). This fluff, by its very nature, cannot reach Earth\u2019s surface intact. However, more substantial fragments from asteroids do make it into our laboratories, as we will see in the next section. 494 Chapter 14 Cosmic Samples and the Origin of the Solar System Showering with the Stars Observing a meteor shower is one of the easiest and most enjoyable astronomy activities for beginners (Figure 14.5). The best thing about it is that you don\u2019t need a telescope or binoculars\u2014in fact, they would positively get in your way. What you do need is a site far from city lights, with an unobstructed view of as much sky as possible. While the short bright lines in the sky made by individual meteors could, in theory, be traced back to a radiant point (as shown in Figure 14.3), the quick blips of light that represent the end of the meteor could happen anywhere above you. Figure 14.5. Perseid Meteor Shower. This twenty-second exposure shows a meteor during the 2015 Perseid meteor shower. (credit: NASA/Bill Ingalls) The key to observing meteor showers is not to restrict your field of view, but to lie back and scan the sky alertly. Try to select a good shower (see the list in Table 14.1) and a night when the Moon will not be bright at the time you are observing. The Moon, street lights, vehicle headlights, bright flashlights, and cell phone and tablet screens will all get in the way of your seeing the faint meteor streaks. You will see more meteors after midnight, when you are on the hemisphere of Earth that faces forward\u2014in the direction of Earth\u2019s revolution around the Sun. Before midnight, you are observing from the \u201cback side\u201d of Earth, and the only meteors you see will be those that traveled fast enough to catch up with Earth\u2019s orbital motion. When you\u2019ve gotten away from all the lights, give your eyes about 15 minutes to get \u201cdark adapted\u201d\u2014that is, for the pupils of your eyes to open up as much as possible. (This adaptation is the same thing that happens in a dark movie theater. When you first enter, you can\u2019t see a thing, but eventually, as your pupils open wider, you can see pretty clearly by the faint light of the screen\u2014and notice all that spilled popcorn on the floor.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 495 Seasoned meteor observers find a hill or open field and make sure to bring warm clothing, a blanket, and a thermos of hot coffee or chocolate with them. (It\u2019s also nice to take along someone with whom you enjoy sitting in the dark.) Don\u2019t expect to see fireworks or a laser show: meteor showers are subtle phenomena, best approached with a patience that reflects the fact that some of the dust you are watching burn up may first have been gathered into its parent comet more than 4.5 billion years ago, as the solar system was just forming. 14.2 METEORITES: STONES FROM HEAVEN Learning Objectives By the end of this section, you will be able to: Explain the origin of meteorites and the difference between a meteor and a meteorite Describe how most meteorites have been found Explain how primitive stone meteorites are significantly different from other types Explain how the study of meteorites informs our understanding of the age of the solar system. Any fragment of interplanetary debris that survives its fiery plunge through Earth\u2019s atmosphere is called a meteorite. Meteorites fall only very rarely in any one locality, but over the entire Earth thousands fall each year. Some meteorites are loners, but many are fragments from the breakup in the atmosphere of a single larger object. These rocks from the sky carry a remarkable record of the formation and early history of the solar system. Extraterrestrial Origin of Meteorites Occasional meteorites have been found throughout history, but their extraterrestrial origin was not accepted by scientists until the beginning of the nineteenth century. Before that, these strange stones were either ignored or considered to have a supernatural origin. The falls of the earliest recovered meteorites are lost in the fog of mythology. A number of religious texts speak of stones from heaven, which sometimes arrived at opportune moments to smite the enemies of the authors of those texts. At least one sacred meteorite has apparently survived in the form of the Ka\u2019aba, the holy black stone", " in Mecca that is revered by Islam as a relic from the time of the Patriarchs\u2014although understandably, no chip from this sacred stone has been subject to detailed chemical analysis. The modern scientific history of the meteorites begins in the late eighteenth century, when a few scientists suggested that some strange-looking stones had such peculiar composition and structure that they were probably not of terrestrial origin. The idea that indeed \u201cstones fall from the sky\u201d was generally accepted only after a scientific team led by French physicist Jean-Baptiste Biot investigated a well-observed fall in 1803. Meteorites sometimes fall in groups or showers. Such a fall occurs when a single larger object breaks up during its violent passage through the atmosphere. It is important to remember that such a shower of meteorites has nothing to do with a meteor shower. No meteorites have ever been recovered in association with meteor showers. Whatever the ultimate source of the meteorites, they do not appear to come from the comets or their associated particle streams. 496 Chapter 14 Cosmic Samples and the Origin of the Solar System Meteorite Falls and Finds Meteorites are found in two ways. First, sometimes bright meteors (fireballs) are observed to penetrate the atmosphere to low altitudes. If we search the area beneath the point where the fireball burned out, we may find one or more remnants that reached the ground. Observed meteorite falls, in other words, may lead to the recovery of fallen meteorites. (A few meteorites have even hit buildings or, very rarely, people; see Making Connections: Some Striking Meteorites). The 2013 Chelyabinsk fireball, which we discussed in the chapter on Comets and Asteroids: Debris of the Solar System, produced tens of thousands of small meteorites, many of them easy to find because these dark stones fell on snow. There are, however, many false alarms about meteorite falls. Most observers of a bright fireball conclude that part of it hit the ground, but that is rarely the case. Every few months news outlets report that a meteorite has been implicated in the start of a fire. Such stories have always proved to be wrong. The meteorite is ice-cold in space, and most of its interior remains cold even after its brief fiery plunge through the atmosphere. A freshly fallen meteorite is more likely to acquire a coating of frost than to start a fire. People sometimes discover unusual-looking rocks that turn out to be meteoritic; these rocks are termed meteorite finds. Now that the public has become meteorite-conscious, many unusual fragments, not all of which turn out to be from space, are sent to experts each year. Some scientists divide these objects into two categories: \u201cmeteorites\u201d and \u201cmeteorwrongs.\u201d Outside Antarctica (see the next paragraph), genuine meteorites turn up at an average rate of 25 or so per year. Most of these end up in natural history museums or specialized meteoritical laboratories throughout the world (Figure 14.6). Figure 14.6. Meteorite Find. (a) This early twentieth century photo shows a 15-ton iron meteorite found in the Willamette Valley in Oregon. Although known to Native Americans in the area, it was \u201cdiscovered\u201d by an enterprising local farmer in 1902, who proceeded to steal it and put it on display. (b) It was eventually purchased for the American Museum of Natural History and is now on display in the museum\u2019s Rose Center in New York City as the largest iron meteorite in the United States. In this 1911 photo, two young boys are perched in the meteor\u2019s crevices. Since the 1980s, sources in the Antarctic have dramatically increased our knowledge of meteorites. More than ten thousand meteorites have been recovered from the Antarctic as a result of the motion of the ice in some parts of that continent (Figure 14.7). Meteorites that fall in regions where ice accumulates are buried and then carried slowly to other areas where the ice is gradually worn away. After thousands of years, the rock again finds itself on the surface, along with other meteorites carried to these same locations. The ice thus concentrates the meteorites that have fallen both over a large area and over a long period of time. Once on the surface, the rocks stand out in contrast to the ice and are thus easier to spot than in other places on our rocky planet. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 497 Figure 14.7. Antarctic Meteorite. (a) The US Antarctic Search for Meteorites (ANSMET) team recovers a meteorite from the Antarctic ice during a 2001\u20132002 mission. (b) The team is shown with some of the equipment used in the search. (credit a, b: modification of work by NASA Some Striking Meteorites", " Although meteorites fall regularly onto Earth\u2019s surface, few of them have much of an impact on human civilization. There is so much water and uninhabited land on our planet that rocks from space typically fall where no one even sees them come down. But given the number of meteorites that land each year, you may not be surprised that a few have struck buildings, cars, and even people. In September 1938, for example, a meteorite plunged through the roof of Edward McCain\u2019s garage, where it became embedded in the seat of his Pontiac Coupe (Figure 14.8). In November 1982, Robert and Wanda Donahue of Wethersfield, Connecticut, were watching M*A*S*H* on television when a 6-pound meteorite came thundering through their roof, making a hole in the living room ceiling. After bouncing, it finally came to rest under their dining room table. Eighteen-year-old Michelle Knapp of Peekskill, New York, got quite a surprise one morning in October 1992. She had just purchased her very first car, her grandmother\u2019s 1980 Chevy Malibu. But she awoke to find its rear end mangled and a crater in the family driveway\u2014thanks to a 3-pound meteorite. Michelle was not sure whether to be devastated by the loss of her car or thrilled by all the media attention. In June 1994, Jose Martin and his wife were driving from Madrid, Spain, to a golfing vacation when a fist- sized meteorite crashed through the windshield of their car, bounced off the dashboard, broke Jose\u2019s little finger, and then landed in the back seat. Before Martin, the most recent person known to have been struck by a meteorite was Annie Hodges of Sylacauga, Alabama. In November 1954, she was napping on a couch when a meteorite came through the roof, bounced off a large radio set, and hit her first on the arm and then on the leg. The fireball that exploded at an altitude of about 20 kilometers near the Russian city of Chelyabinsk on February 15, 2013, produced a very large meteorite shower, and quite a few of the small rocks hit 498 Chapter 14 Cosmic Samples and the Origin of the Solar System buildings. None is known to have hit people, however, and the individual meteorites were so small that they did not do much damage\u2014much less than the shockwave from the exploding fireball, which broke the glass in thousands of windows. Figure 14.8. Benld Meteorite. A meteorite (inset) left a hole in the seat cushion of Edward McCain\u2019s car. (credit: \u201cShsilver\u201d/Wikimedia Commons) Meteorite Classification The meteorites in our collections have a wide range of compositions and histories, but traditionally they have been placed into three broad classes. First are the irons, composed of nearly pure metallic nickel-iron. Second are the stones, the term used for any silicate or rocky meteorite. Third are the rarer stony-irons, made (as the name implies) of mixtures of stone and metallic iron (Figure 14.9). Figure 14.9. Meteorite Types. (a) This piece of the Allende carbonaceous meteorite has white inclusions that may date back to before the formation of the solar nebula. (b) This fragment is from the iron meteorite responsible for the formation of Meteor Crater in Arizona. (c) This piece of the Imilac stony-iron meteorite is a beautiful mixture of green olivine crystals and metallic iron. (credit a: modification of work by James St. John; credit b: modification of work by \u201cTaty2007\u201d/Wikimedia Commons; credit c: modification of work by Juan Manuel Flux\u00e0) Of these three types, the irons and stony-irons are the most obviously extraterrestrial because of their metallic content. Pure iron almost never occurs naturally on Earth; it is generally found here as an oxide (chemically combined with oxygen) or other mineral ore. Therefore, if you ever come across a chunk of metallic iron, it is sure to be either man-made or a meteorite. The stones are much more common than the irons but more difficult to recognize. Often laboratory analysis is required to demonstrate that a particular sample is really of extraterrestrial origin, especially if it has lain on This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 499 the ground for some time and been subject to weathering. The most scientifically valuable stones are those collected immediately after they fall, or the Antarctic samples preserved in a nearly pristine state by ice. Table 14.2 summarizes the frequencies of occurrence of the different classes of meteorites among the fall, find, and Antarctic categories. Frequency of", " Occurrence of Meteorite Classes Class Falls (%) Finds (%) Antarctic (%) Primitive stones Differentiated stones Irons Stony-irons Table 14.2 88 8 3 1 51 2 42 5 85 12 2 1 Ages and Compositions of Meteorites It was not until the ages of meteorites were measured and their compositions analyzed in detail that scientists appreciated their true significance. The meteorites include the oldest and most primitive materials available for direct study in the laboratory. The ages of stony meteorites can be determined from the careful measurement of radioactive isotopes and their decay products. Almost all meteorites have radioactive ages between 4.50 and 4.56 billion years, as old as any ages we have measured in the solar system. The few younger exceptions are igneous rocks that have been ejected from cratering events on the Moon or Mars (and have made their way to Earth). The average age for the most primitive meteorites, calculated using the most accurate values now available for radioactive half-lives, is 4.56 billion years, with an uncertainly of less than 0.01 billion years. This value (which we round off to 4.5 billion years in this book) is taken to represent the age of the solar system\u2014the time since the first solids condensed and began to form into larger bodies. The traditional classification of meteorites into irons, stones, and stony-irons is easy to use because it is obvious from inspection which category a meteorite falls into (although it may be much more difficult to distinguish a meteoritic stone from a terrestrial rock). More scientifically significant, however, is the distinction between primitive and differentiated meteorites. The differentiated meteorites are fragments of larger parent bodies that were molten before they broke up, allowing the denser materials (such as metals) to sink to their centers. Like many rocks on Earth, they have been subject to a degree of chemical reshuffling, with the different materials sorted according to density. Differentiated meteorites include the irons, which come from the metal cores of their parent bodies; stony-irons, which probably originate in regions between a metal core and a stony mantle; and some stones that are composed of mantle or crust material from the their differentiated parent bodies. The Most Primitive Meteorites For information on the earliest history of the solar system, we turn to the primitive meteorites\u2014those made of materials that have not been subject to great heat or pressure since their formation. We can look at the spectrum of sunlight reflected from asteroids and compare their compositions with those of primitive meteorites. Such analysis indicates that their parent bodies are almost certainly asteroids. Since asteroids are 500 Chapter 14 Cosmic Samples and the Origin of the Solar System believed to be fragments left over from the formation process of the solar system, it makes sense that they should be the parent bodies of the primitive meteorites. The great majority of the meteorites that reach Earth are primitive stones. Many of them are composed of light-colored gray silicates with some metallic grains mixed in, but there is also an important group of darker stones called carbonaceous meteorites. As their name suggests, these meteorites contain carbon, but we also find various complex organic molecules in them\u2014chemicals based on carbon, which on Earth are the chemical building blocks of life. In addition, some of them contain chemically bound water, and many are depleted in metallic iron. The carbonaceous (or C-type) asteroids are concentrated in the outer part of the asteroid belt. Among the most useful of these meteorites have been the Allende meteorite that fell in Mexico (see Figure 14.9), the Murchison meteorite that fell in Australia (both in 1969), and the Tagish Lake meteorite that landed in a winter snowdrift on Tagish Lake, Canada, in 2000. (The fragile bits of dark material from the Tagish Lake meteorite were readily visible against the white snow, although at first they were mistaken for wolf droppings.) The Murchison meteorite (Figure 14.10) is known for the variety of organic chemicals it has yielded. Most of the carbon compounds in carbonaceous meteorites are complex, tarlike substances that defy exact analysis. Murchison also contains 16 amino acids (the building blocks of proteins), 11 of which are rare on Earth. The most remarkable thing about these organic molecules is that they include equal numbers with right-handed and left-handed molecular symmetry. Amino acids can have either kind of symmetry, but all life on Earth has evolved using only the left-handed versions to make proteins. The presence of both kinds of amino acids clearly demonstrates that the ones in the meteorites had an extraterrestrial origin. Figure 14.10. Murchison Meteorite. A fragment of the meteorite that fell near the small town of Murchison, Australia, is shown next to a small sample of its material in a test tube, used for analysis of its chemical makeup. These naturally occurring amino acids and other complex organic molecules in Murchison\u2014formed without the benefit", " of the sheltering environment of planet Earth\u2014show that a great deal of interesting chemistry must have taken place when the solar system was forming. If so, then perhaps some of the molecular building blocks of life on Earth were first delivered by primitive meteorites and comets. This is an interesting idea because our planet was probably much too hot for any organic materials to survive its earliest history. But after Earth\u2019s surface cooled, the asteroid and comet fragments that pelted it could have refreshed its supply of organic materials. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 501 14.3 FORMATION OF THE SOLAR SYSTEM Learning Objectives By the end of this section, you will be able to: Describe the motion, chemical, and age constraints that must be met by any theory of solar system formation Summarize the physical and chemical changes during the solar nebula stage of solar system formation Explain the formation process of the terrestrial and giant planets Describe the main events of the further evolution of the solar system As we have seen, the comets, asteroids, and meteorites are surviving remnants from the processes that formed the solar system. The planets, moons, and the Sun, of course, also are the products of the formation process, although the material in them has undergone a wide range of changes. We are now ready to put together the information from all these objects to discuss what is known about the origin of the solar system. Observational Constraints There are certain basic properties of the planetary system that any theory of its formation must explain. These may be summarized under three categories: motion constraints, chemical constraints, and age constraints. We call them constraints because they place restrictions on our theories; unless a theory can explain the observed facts, it will not survive in the competitive marketplace of ideas that characterizes the endeavor of science. Let\u2019s take a look at these constraints one by one. There are many regularities to the motions in the solar system. We saw that the planets all revolve around the Sun in the same direction and approximately in the plane of the Sun\u2019s own rotation. In addition, most of the planets rotate in the same direction as they revolve, and most of the moons also move in counterclockwise orbits (when seen from the north). With the exception of the comets and other trans-neptunian objects, the motions of the system members define a disk or Frisbee shape. Nevertheless, a full theory must also be prepared to deal with the exceptions to these trends, such as the retrograde rotation (not revolution) of Venus. In the realm of chemistry, we saw that Jupiter and Saturn have approximately the same composition\u2014dominated by hydrogen and helium. These are the two largest planets, with sufficient gravity to hold on to any gas present when and where they formed; thus, we might expect them to be representative of the original material out of which the solar system formed. Each of the other members of the planetary system is, to some degree, lacking in the light elements. A careful examination of the composition of solid solar-system objects shows a striking progression from the metal-rich inner planets, through those made predominantly of rocky materials, out to objects with ice-dominated compositions in the outer solar system. The comets in the Oort cloud and the trans-neptunian objects in the Kuiper belt are also icy objects, whereas the asteroids represent a transitional rocky composition with abundant dark, carbon-rich material. As we saw in Other Worlds: An Introduction to the Solar System, this general chemical pattern can be interpreted as a temperature sequence: hot near the Sun and cooler as we move outward. The inner parts of the system are generally missing those materials that could not condense (form a solid) at the high temperatures found near the Sun. However, there are (again) important exceptions to the general pattern. For example, it is difficult to explain the presence of water on Earth and Mars if these planets formed in a region where the temperature was too hot for ice to condense, unless the ice or water was brought in later from cooler regions. The extreme example is the observation that there are polar deposits of ice on both Mercury and the Moon; these are almost certainly formed and maintained by occasional comet impacts. 502 Chapter 14 Cosmic Samples and the Origin of the Solar System As far as age is concerned, we discussed that radioactive dating demonstrates that some rocks on the surface of Earth have been present for at least 3.8 billion years, and that certain lunar samples are 4.4 billion years old. The primitive meteorites all have radioactive ages near 4.5 billion years. The age of these unaltered building blocks is considered the age of the planetary system. The similarity of the measured ages tells us that planets formed and their crusts cooled within a few tens of millions of years (at most) of the beginning of the solar system. Further", ", detailed examination of primitive meteorites indicates that they are made primarily from material that condensed or coagulated out of a hot gas; few identifiable fragments appear to have survived from before this hot-vapor stage 4.5 billion years ago. The Solar Nebula All the foregoing constraints are consistent with the general idea, introduced in Other Worlds: An Introduction to the Solar System, that the solar system formed 4.5 billion years ago out of a rotating cloud of vapor and dust\u2014which we call the solar nebula\u2014with an initial composition similar to that of the Sun today. As the solar nebula collapsed under its own gravity, material fell toward the center, where things became more and more concentrated and hot. Increasing temperatures in the shrinking nebula vaporized most of the solid material that was originally present. At the same time, the collapsing nebula began to rotate faster through the conservation of angular momentum (see the Orbits and Gravity and Earth, Moon, and Sky chapters). Like a figure skater pulling her arms in to spin faster, the shrinking cloud spun more quickly as time went on. Now, think about how a round object spins. Close to the poles, the spin rate is slow, and it gets faster as you get closer to the equator. In the same way, near the poles of the nebula, where orbits were slow, the nebular material fell directly into the center. Faster moving material, on the other hand, collapsed into a flat disk revolving around the central object (Figure 14.11). The existence of this disk-shaped rotating nebula explains the primary motions in the solar system that we discussed in the previous section. And since they formed from a rotating disk, the planets all orbit the same way. Figure 14.11. Steps in Forming the Solar System. This illustration shows the steps in the formation of the solar system from the solar nebula. As the nebula shrinks, its rotation causes it to flatten into a disk. Much of the material is concentrated in the hot center, which will ultimately become a star. Away from the center, solid particles can condense as the nebula cools, giving rise to planetesimals, the building blocks of the planets and moons. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 503 Picture the solar nebula at the end of the collapse phase, when it was at its hottest. With no more gravitational energy (from material falling in) to heat it, most of the nebula began to cool. The material in the center, however, where it was hottest and most crowded, formed a star that maintained high temperatures in its immediate neighborhood by producing its own energy. Turbulent motions and magnetic fields within the disk can drain away angular momentum, robbing the disk material of some of its spin. This allowed some material to continue to fall into the growing star, while the rest of the disk gradually stabilized. The temperature within the disk decreased with increasing distance from the Sun, much as the planets\u2019 temperatures vary with position today. As the disk cooled, the gases interacted chemically to produce compounds; eventually these compounds condensed into liquid droplets or solid grains. This is similar to the process by which raindrops on Earth condense from moist air as it rises over a mountain. Let\u2019s look in more detail at how material condensed at different places in the maturing disk (Figure 14.12). The first materials to form solid grains were the metals and various rock-forming silicates. As the temperature dropped, these were joined throughout much of the solar nebula by sulfur compounds and by carbon- and water-rich silicates, such as those now found abundantly among the asteroids. However, in the inner parts of the disk, the temperature never dropped low enough for such materials as ice or carbonaceous organic compounds to condense, so they were lacking on the innermost planets. Figure 14.12. Chemical Condensation Sequence in the Solar Nebula. The scale along the bottom shows temperature; above are the materials that would condense out at each temperature under the conditions expected to prevail in the nebula. Far from the Sun, cooler temperatures allowed the oxygen to combine with hydrogen and condense in the form of water (H2O) ice. Beyond the orbit of Saturn, carbon and nitrogen combined with hydrogen to make ices such as methane (CH4) and ammonia (NH3). This sequence of events explains the basic chemical composition differences among various regions of the solar system. 504 Chapter 14 Cosmic Samples and the Origin of the Solar System Rotation of the Solar Nebula We can use the concept of angular momentum to trace the evolution of the collapsing solar nebula. The angular momentum of an object is proportional to the square of its size (diameter) times its period of rotation (D2/P). If angular momentum is conserved, then any change in the size of a nebula must be compensated for by", " a proportional change in period, in order to keep D2/P constant. Suppose the solar nebula began with a diameter of 10,000 AU and a rotation period of 1 million years. What is its rotation period when it has shrunk to the size of Pluto\u2019s orbit, which Appendix F tells us has a radius of about 40 AU? Solution We are given that the final diameter of the solar nebula is about 80 AU. Noting the initial state before the collapse and the final state at Pluto\u2019s orbit, then Pfina Pinitial = \u239b \u239d Dfina Dinitial \u239e \u23a0 2 = \u239b \u239d 80 10,000 2 \u239e \u23a0 = (0.008)2 = 0.000064 With Pinitial equal to 1,000,000 years, Pfinal, the new rotation period, is 64 years. This is a lot shorter than the actual time Pluto takes to go around the Sun, but it gives you a sense of the kind of speeding up the conservation of angular momentum can produce. As we noted earlier, other mechanisms helped the material in the disk lose angular momentum before the planets fully formed. Check Your Learning What would the rotation period of the nebula in our example be when it had shrunk to the size of Jupiter\u2019s orbit? Answer: The period of the rotating nebula is inversely proportional to D2. As we have just seen, 2 = \u239e \u23a0 \u239b \u239d Dfina Dinitial Pfina Pinitial years) is given by Pfina = 0.01Dfina The period is then 1.08 years. 2. Initially, we have Pinitial = 106 yr and Dinitial = 104 AU. Then, if Dfinal is in AU, Pfinal (in. If Jupiter\u2019s orbit has a radius of 5.2 AU, then the diameter is 10.4 AU. Formation of the Terrestrial Planets The grains that condensed in the solar nebula rather quickly joined into larger and larger chunks, until most of the solid material was in the form of planetesimals, chunks a few kilometers to a few tens of kilometers in diameter. Some planetesimals still survive today as comets and asteroids. Others have left their imprint on the cratered surfaces of many of the worlds we studied in earlier chapters. A substantial step up in size is required, however, to go from planetesimal to planet. Some planetesimals were large enough to attract their neighbors gravitationally and thus to grow by the process called accretion. While the intermediate steps are not well understood, ultimately several dozen centers of accretion seem to have grown in the inner solar system. Each of these attracted surrounding This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 505 planetesimals until it had acquired a mass similar to that of Mercury or Mars. At this stage, we may think of these objects as protoplanets\u2014\u201cnot quite ready for prime time\u201d planets. Each of these protoplanets continued to grow by the accretion of planetesimals. Every incoming planetesimal was accelerated by the gravity of the protoplanet, striking with enough energy to melt both the projectile and a part of the impact area. Soon the entire protoplanet was heated to above the melting temperature of rocks. The result was planetary differentiation, with heavier metals sinking toward the core and lighter silicates rising toward the surface. As they were heated, the inner protoplanets lost some of their more volatile constituents (the lighter gases), leaving more of the heavier elements and compounds behind. Formation of the Giant Planets In the outer solar system, where the available raw materials included ices as well as rocks, the protoplanets grew to be much larger, with masses ten times greater than Earth. These protoplanets of the outer solar system were so large that they were able to attract and hold the surrounding gas. As the hydrogen and helium rapidly collapsed onto their cores, the giant planets were heated by the energy of contraction. But although these giant planets got hotter than their terrestrial siblings, they were far too small to raise their central temperatures and pressures to the point where nuclear reactions could begin (and it is such reactions that give us our definition of a star). After glowing dull red for a few thousand years, the giant planets gradually cooled to their present state (Figure 14.13). Figure 14.13. Saturn Seen in Infrared. This image from the Cassini spacecraft is stitched together from 65 individual observations. Sunlight reflected at a wavelength of 2 micrometers is shown as blue, sunlight reflected at 3 micrometers is shown as green, and heat radiated from Saturn\u2019s interior at 5 micrometers is red. For example, Saturn\u2019s rings reflect sunlight at 2 micrometers", ", but not at 3 and 5 micrometers, so they appear blue. Saturn\u2019s south polar regions are seen glowing with internal heat. (credit: modification of work by NASA/JPL/University of Arizona) The collapse of gas from the nebula onto the cores of the giant planets explains how these objects acquired nearly the same hydrogen-rich composition as the Sun. The process was most efficient for Jupiter and Saturn; hence, their compositions are most nearly \u201ccosmic.\u201d Much less gas was captured by Uranus and Neptune, which is why these two planets have compositions dominated by the icy and rocky building blocks that made up their large cores rather than by hydrogen and helium. The initial formation period ended when much of the available raw material was used up and the solar wind (the flow of atomic particles) from the young Sun blew away the remaining supply of lighter gases. Further Evolution of the System All the processes we have just described, from the collapse of the solar nebula to the formation of protoplanets, took place within a few million years. However, the story of the formation of the solar system was not complete 506 Chapter 14 Cosmic Samples and the Origin of the Solar System at this stage; there were many planetesimals and other debris that did not initially accumulate to form the planets. What was their fate? The comets visible to us today are merely the tip of the cosmic iceberg (if you\u2019ll pardon the pun). Most comets are believed to be in the Oort cloud, far from the region of the planets. Additional comets and icy dwarf planets are in the Kuiper belt, which stretches beyond the orbit of Neptune. These icy pieces probably formed near the present orbits of Uranus and Neptune but were ejected from their initial orbits by the gravitational influence of the giant planets. In the inner parts of the system, remnant planetesimals and perhaps several dozen protoplanets continued to whiz about. Over the vast span of time we are discussing, collisions among these objects were inevitable. Giant impacts at this stage probably stripped Mercury of part of its mantle and crust, reversed the rotation of Venus, and broke off part of Earth to create the Moon (all events we discussed in other chapters). Smaller-scale impacts also added mass to the inner protoplanets. Because the gravity of the giant planets could \u201cstir up\u201d the orbits of the planetesimals, the material impacting on the inner protoplanets could have come from almost anywhere within the solar system. In contrast to the previous stage of accretion, therefore, this new material did not represent just a narrow range of compositions. As a result, much of the debris striking the inner planets was ice-rich material that had condensed in the outer part of the solar nebula. As this comet-like bombardment progressed, Earth accumulated the water and various organic compounds that would later be critical to the formation of life. Mars and Venus probably also acquired abundant water and organic materials from the same source, as Mercury and the Moon are still doing to form their icy polar caps. Gradually, as the planets swept up or ejected the remaining debris, most of the planetesimals disappeared. In two regions, however, stable orbits are possible where leftover planetesimals could avoid impacting the planets or being ejected from the system. These regions are the asteroid belt between Mars and Jupiter and the Kuiper belt beyond Neptune. The planetesimals (and their fragments) that survive in these special locations are what we now call asteroids, comets, and trans-neptunian objects. Astronomers used to think that the solar system that emerged from this early evolution was similar to what we see today. Detailed recent studies of the orbits of the planets and asteroids, however, suggest that there were more violent events soon afterward, perhaps involving substantial changes in the orbits of Jupiter and Saturn. These two giant planets control, through their gravity, the distribution of asteroids. Working backward from our present solar system, it appears that orbital changes took place during the first few hundred million years. One consequence may have been scattering of asteroids into the inner solar system, causing the period of \u201cheavy bombardment\u201d recorded in the oldest lunar craters. 14.4 COMPARISON WITH OTHER PLANETARY SYSTEMS Learning Objectives By the end of this section, you will be able to: Describe how the observations of protoplanetary disks provides evidence for the existence of other planetary systems Explain the two primary methods for detection of exoplanets Compare the main characteristics of other planetary systems with the features of the solar system Until the middle 1990s, the practical study of the origin of planets focused on our single known example\u2014the solar system. Although there had been a great deal of speculation about planets circling other stars, none had This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 507 actually been detected", ". Logically enough, in the absence of data, most scientists assumed that our own system was likely to be typical. They were in for a big surprise. Discovery of Other Planetary Systems In The Birth of Stars and the Discovery of Planets outside the Solar System, we discuss the formation of stars and planets in some detail. Stars like our Sun are formed when dense regions in a molecular cloud (made of gas and dust) feel an extra gravitational force and begin to collapse. This is a runaway process: as the cloud collapses, the gravitational force gets stronger, concentrating material into a protostar. Roughly half of the time, the protostar will fragment or be gravitationally bound to other protostars, forming a binary or multiple star system\u2014stars that are gravitationally bound and orbit each other. The rest of the time, the protostar collapses in isolation, as was the case for our Sun. In all cases, as we saw, conservation of angular momentum results in a spin-up of the collapsing protostar, with surrounding material flattened into a disk. Today, this kind of structure can actually be observed. The Hubble Space Telescope, as well as powerful new ground-based telescopes, enable astronomers to study directly the nearest of these circumstellar disks in regions of space where stars are being born today, such as the Orion Nebula (Figure 14.14) or the Taurus star-forming region. Figure 14.14. Protoplanetary Disk in the Orion Nebula. The Hubble Space Telescope imaged this protoplanetary disk in the Orion Nebula, a region of active star formation, using two different filters. The disk, about 17 times the size of our solar system, is in an edge-on orientation to us, and the newly formed star is shining at the center of the flattened dust cloud. The dark areas indicate absorption, not an absence of material. In the left image we see the light of the nebula and the dark cloud; in the right image, a special filter was used to block the light of the background nebula. You can see gas above and below the disk set to glow by the light of the newborn star hidden by the disk. (credit: modification of work by Mark McCaughrean (Max-Planck-Institute for Astronomy), C. Robert O\u2019Dell (Rice University), and NASA) Many of the circumstellar disks we have discovered show internal structure. The disks appear to be donut- shaped, with gaps close to the star. Such gaps indicate that the gas and dust in the disk have already collapsed to form large planets (Figure 14.15). The newly born protoplanets are too small and faint to be seen directly, but the depletion of raw materials in the gaps hints at the presence of something invisible in the inner part of the circumstellar disk\u2014and that something is almost certainly one or more planets. Theoretical models of planet formation, like the one seen at right in Figure 14.15, have long supported the idea that planets would clear gaps as they form in disks. 508 Chapter 14 Cosmic Samples and the Origin of the Solar System Figure 14.15. Protoplanetary Disk around HL Tau. (a) This image of a protoplanetary disk around HL Tau was taken with the Atacama Large Millimeter/submillimeter Array (ALMA), which allows astronomers to construct radio images that rival those taken with visible light. (b) Newly formed planets that orbit the central star clear out dust lanes in their paths, just as our theoretical models predict. This computer simulation shows the empty lane and spiral density waves that result as a giant planet is forming within the disk. The planet is not shown to scale. (credit a: modification of work by ALMA (ESO/NAOJ/NRAO); credit b: modification of work by NASA/ESA and A. Feild (STScI)) Our figure shows HL Tau, a one-million-year-old \u201cnewborn\u201d star in the Taurus star-forming region. The star is embedded in a shroud of dust and gas that obscures our visible-light view of a circumstellar disk around the star. In 2014 astronomers obtained a dramatic view of the HL Tau circumstellar disk using millimeter waves, which pierce the cocoon of dust around the star, showing dust lanes being carved out by several newly formed protoplanets. As the mass of the protoplanets increases, they travel in their orbits at speeds that are faster than the dust and gas in the circumstellar disk. As the protoplanets plow through the disk, their gravitational reach begins to exceed their cross-sectional area, and they become very efficient at sweeping up material and growing until they clear a gap in the disk. The image of Figure 14.15 shows us that a number of protoplanets are forming in the disk and that they were able to form faster than our earlier ideas had suggested\u2014all in the first million years of star", " formation For an explanation of ALMA\u2019s ground-breaking observations of HL Tau and what they reveal about plant formation, watch this videocast (https://openstaxcollege.org/l/30eusoobhltavid) from the European Southern Observatory. Discovering Exoplanets You might think that with the advanced telescopes and detectors astronomers have today, they could directly image planets around nearby stars (which we call exoplanets). This has proved extremely difficult, however, not only because the exoplanets are faint, but also because they are generally lost in the brilliant glare of the star they orbit. As we discuss in more detail in The Birth of Stars and the Discovery of Planets outside the Solar System, the detection techniques that work best are indirect: they observe the effects of the planet on the star it orbits, rather than seeing the planet itself. The first technique that yielded many planet detections is very high-resolution stellar spectroscopy. The Doppler effect lets astronomers measure the star\u2019s radial velocity: that is, the speed of the star, toward us or away from us, relative to the observer. If there is a massive planet in orbit around the star, the gravity of the planet causes This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 509 the star to wobble, changing its radial velocity by a small but detectable amount. The distance of the star does not matter, as long as it is bright enough for us to take very high quality spectra. Measurements of the variation in the star\u2019s radial velocity as the planet goes around the star can tell us the mass and orbital period of the planet. If there are several planets present, their effects on the radial velocity can be disentangled, so the entire planetary system can be deciphered\u2014as long as the planets are massive enough to produce a measureable Doppler effect. This detection technique is most sensitive to large planets orbiting close to the star, since these produce the greatest wobble in their stars. It has been used on large ground-based telescopes to detect hundreds of planets, including one around Proxima Centauri, the nearest star to the Sun. The second indirect technique is based on the slight dimming of a star when one of its planets transits, or crosses over the face of the star, as seen from Earth. Astronomers do not see the planet, but only detect its presence from careful measurements of a change in the brightness of the star over long periods of time. If the slight dips in brightness repeat at regular intervals, we can determine the orbital period of the planet. From the amount of starlight obscured, we can measure the planet\u2019s size. While some transits have been measured from Earth, large-scale application of this transit technique requires a telescope in space, above the atmosphere and its distortions of the star images. It has been most successfully applied from the NASA Kepler space observatory, which was built for the sole purpose of \u201cstaring\u201d for 5 years at a single part of the sky, continuously monitoring the light from more than 150,000 stars. The primary goal of Kepler was to determine the frequency of occurrence of exoplanets of different sizes around different classes of stars. Like the Doppler technique, the transit observations favor discovery of large planets and short-period orbits. Recent detection of exoplanets using both the Doppler and transit techniques has been incredibly successful. Within two decades, we went from no knowledge of other planetary systems to a catalog of thousands of exoplanets. Most of the exoplanets found so far are more massive than or larger in size than Earth. It is not that Earth analogs do not exist. Rather, the shortage of small rocky planets is an observational bias: smaller planets are more difficult to detect. Analyses of the data to correct for such biases or selection effects indicate that small planets (like the terrestrial planets in our system) are actually much more common than giant planets. Also relatively common are \u201csuper Earths,\u201d planets with two to ten times the mass of our planet (Figure 14.16). We don\u2019t have any of these in our solar system, but nature seems to have no trouble making them elsewhere. Overall, the Kepler data suggest that approximately one quarter of stars have exoplanet systems, implying the existence of at least 50 billion planets in our Galaxy alone. 510 Chapter 14 Cosmic Samples and the Origin of the Solar System Figure 14.16. Transiting Planets by Size. This bar graph shows the planets found so far using the transit method (the vast majority found by the Kepler mission). The orange parts of each bar indicate the planets announced by the Kepler team in May 2016. Note that the largest number of planets found so far are in two categories that we don\u2019t have in our own solar system\u2014planets whose size is", " between Earth\u2019s and Neptune\u2019s. (credit: modification of work by NASA) The Configurations of Other Planetary Systems Let\u2019s look more closely at the progress in the detection of exoplanets. Figure 14.17 shows the planets that were discovered each year by the two techniques we discussed. In the early years of exoplanet discovery, most of the planets were similar in mass to Jupiter. This is because, as mentioned above, the most massive planets were easiest to detect. In more recent years, planets smaller than Neptune and even close to the size of Earth have been detected. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 511 Figure 14.17. Masses of Exoplanets Discovered by Year. Horizontal lines are drawn to reference the masses of Jupiter, Saturn, Neptune, and Earth. The gray dots indicate planets discovered by measuring the radial velocity of the star, and the red dots are for planets that transit their stars. In the early years, the only planets that could be detected were similar in mass to Jupiter. Improvements in technology and observing strategies enabled the detection of lower mass planets as time went on, and now even smaller worlds are being found. (Note that this tally ends in 2014.) We also know that many exoplanets are in multiplanet systems. This is one characteristic that our solar system shares with exosystems. Looking back at Figure 14.15 and seeing how such large disks can give rise to more than one center of condensation, it is not too surprising that multiplanet systems are a typical outcome of planet formation. Astronomers have tried to measure whether multiple planet systems all lie in the same plane using astrometry. This is a difficult measurement to make with current technology, but it is an important measurement that could help us understand the origin and evolution of planetary systems. Comparison between Theory and Data Many of the planetary systems discovered so far do not resemble our own solar system. Consequently, we have had to reassess some aspects of the \u201cstandard models\u201d for the formation of planetary systems. Science sometimes works in this way, with new data contradicting our expectations. The press often talks about a scientist making experiments to \u201cconfirm\u201d a theory. Indeed, it is comforting when new data support a hypothesis or theory and increase our confidence in an earlier result. But the most exciting and productive moments in science often come when new data don\u2019t support existing theories, forcing scientists to rethink their position and develop new and deeper insights into the way nature works. Nothing about the new planetary systems contradicts the basic idea that planets form from the aggregation (clumping) of material within circumstellar disks. However, the existence of \u201c hot Jupiters\u201d\u2014planets of jovian mass that are closer to their stars than the orbit of Mercury\u2014poses the biggest problem. As far as we know, a giant planet cannot be formed without the condensation of water ice, and water ice is not stable so close to the heat of a star. It seems likely that all the giant planets, \u201chot\u201d or \u201cnormal,\u201d formed at a distance of several astronomical units from the star, but we now see that they did not necessarily stay there. This discovery has led to a revision in our understanding of planet formation that now includes \u201cplanet migrations\u201d within the protoplanetary disk, or later gravitational encounters between sibling planets that scatter one of the planets inward. Many exoplanets have large orbital eccentricity (recall this means the orbits are not circular). High eccentricities were not expected for planets that form in a disk. This discovery provides further support for the scattering 512 Chapter 14 Cosmic Samples and the Origin of the Solar System of planets when they interact gravitationally. When planets change each other\u2019s motions, their orbits could become much more eccentric than the ones with which they began. There are several suggestions for ways migration might have occurred. Most involve interactions between the giant planets and the remnant material in the circumstellar disk from which they formed. These interactions would have taken place when the system was very young, while material still remained in the disk. In such cases, the planet travels at a faster velocity than the gas and dust and feels a kind of \u201cheadwind\u201d (or friction) that causes it to lose energy and spiral inward. It is still unclear how the spiraling planet stops before it plunges into the star. Our best guess is that this plunge into the star is the fate for many protoplanets; however, clearly some migrating planets can stop their inward motions and escape this destruction, since we find hot Jupiters in many mature planetary systems. 14.5 PLANETARY EVOLUTION Learning Objectives By the end of this section, you will be able to: Describe the geological activity during the evolution of the planets,", " particularly on the terrestrial planets Describe the factors that affect differences in elevation on the terrestrial planets Explain how the differences in atmosphere on Venus, Earth, and Mars evolved from similar starting points in the early history of the solar system While we await more discoveries and better understanding of other planetary systems, let us look again at the early history of our own solar system, after the dissipation of our dust disk. The era of giant impacts was probably confined to the first 100 million years of solar system history, ending by about 4.4 billion years ago. Shortly thereafter, the planets cooled and began to assume their present aspects. Up until about 4 billion years ago, they continued to acquire volatile materials, and their surfaces were heavily cratered from the remaining debris that hit them. However, as external influences declined, all the terrestrial planets as well as the moons of the outer planets began to follow their own evolutionary courses. The nature of this evolution depended on each object\u2019s composition, mass, and distance from the Sun. Geological Activity We have seen a wide range in the level of geological activity on the terrestrial planets and icy moons. Internal sources of such activity (as opposed to pummeling from above) require energy, either in the form of primordial heat left over from the formation of a planet or from the decay of radioactive elements in the interior. The larger the planet or moon, the more likely it is to retain its internal heat and the more slowly it cools\u2014this is the \u201cbaked potato effect\u201d mentioned in Other Worlds: An Introduction to the Solar System. Therefore, we are more likely to see evidence of continuing geological activity on the surface of larger (solid) worlds (Figure 14.18). Jupiter\u2019s moon Io is an interesting exception to this rule; we saw that it has an unusual source of heat from the gravitational flexing of its interior by the tidal pull of Jupiter. Europa is probably also heated by jovian tides. Saturn may be having a similar effect on its moon Enceladus. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 513 Figure 14.18. Stages in the Geological History of a Terrestrial Planet. In this image, time increases downward along the left side, where the stages are described. Each planet is shown roughly in its present stage. The smaller the planet, the more quickly it passes through these stages. The Moon, the smallest of the terrestrial worlds, was internally active until about 3.3 billion years ago, when its major volcanism ceased. Since that time, its mantle has cooled and become solid, and today even internal seismic activity has declined to almost zero. The Moon is a geologically dead world. Although we know much less about Mercury, it seems likely that this planet, too, ceased most volcanic activity about the same time the Moon did. Mars represents an intermediate case, and it has been much more active than the Moon. The southern hemisphere crust had formed by 4 billion years ago, and the northern hemisphere volcanic plains seem to be contemporary with the lunar maria. However, the Tharsis bulge formed somewhat later, and activity in the large Tharsis volcanoes has apparently continued on and off to the present era. Earth and Venus are the largest and most active terrestrial planets. Our planet experiences global plate tectonics driven by convection in its mantle. As a result, our surface is continually reworked, and most of Earth\u2019s surface material is less than 200 million years old. Venus has generally similar levels of volcanic activity, but unlike Earth, it has not experienced plate tectonics. Most of its surface appears to be no more than 500 million years old. We did see that the surface of our sister planet is being modified by a kind of \u201cblob tectonics\u201d\u2014where hot material from below puckers and bursts through the surface, leading to coronae, pancake volcanoes, and other such features. A better understanding of the geological differences between Venus and Earth is a high priority for planetary geologists. The geological evolution of the icy moons and Pluto has been somewhat different from that of the terrestrial planets. Tidal energy sources have been active, and the materials nature has to work with are not the same. On these outer worlds, we see evidence of low-temperature volcanism, with the silicate lava of the inner planets being supplemented by sulfur compounds on Io, and replaced by water and other ices on Pluto and other outer- planet moons. Elevation Differences Let\u2019s look at some specific examples of how planets differ. The mountains on the terrestrial planets owe their origins to different processes. On the Moon and Mercury, the major mountains are ejecta thrown up by the large basin-forming impacts that took place billions of years ago. Most large mountains on Mars are volcanoes, produced by repeated eruptions of lava from the same vents. There are similar (but", " smaller) volcanoes on Earth and Venus. However, the highest mountains on Earth and Venus are the result of compression and uplift of the surface. On Earth, this crustal compression results from collisions of one continental plate with another. 514 Chapter 14 Cosmic Samples and the Origin of the Solar System It is interesting to compare the maximum heights of the volcanoes on Earth, Venus, and Mars (Figure 14.19). On Venus and Earth, the maximum elevation differences between these mountains and their surroundings are about 10 kilometers. Olympus Mons, in contrast, towers more than 20 kilometers above its surroundings and nearly 30 kilometers above the lowest elevation areas on Mars. Figure 14.19. Highest Mountains on Mars, Venus, and Earth. Mountains can rise taller on Mars because Mars has less surface gravity and no moving plates. The vertical scale is exaggerated by a factor of three to make comparison easier. The label \u201csea level\u201d refers only to Earth, of course, since the other two planets don\u2019t have oceans. Mauna Loa and Mt. Everest are on Earth, Olympus Mons is on Mars, and the Maxwell Mountains are on Venus. One reason Olympus Mons (Figure 14.20) is so much higher than its terrestrial counterparts is that the crustal plates on Earth never stop moving long enough to let a really large volcano grow. Instead, the moving plate creates a long row of volcanoes like the Hawaiian Islands. On Mars (and perhaps Venus) the crust remains stationary with respect to the underlying hot spot, and so a single volcano can continue to grow for hundreds of millions of years. Figure 14.20. Olympus Mons. The largest martian volcano is seen from above in this spectacular composite image created from many Viking orbiter photographs. The volcano is nearly 500 kilometers wide at its base and more than 20 kilometers high. (Its height is almost three times the height of the tallest mountain on Earth.) (credit: modification of work by NASA/USGS) A second difference relates to the strength of gravity on the three planets. The surface gravity on Venus is nearly the same as that on Earth, but on Mars it is only about one third as great. In order for a mountain to survive, its internal strength must be great enough to support its weight against the force of gravity. Volcanic rocks have known strengths, and we can calculate that on Earth, 10 kilometers is about the limit. For instance, when new lava is added to the top of Mauna Loa in Hawaii, the mountain slumps downward under its own weight. The same height limit applies on Venus, where the force of gravity is the same as Earth\u2019s. On Mars, however, with its lesser surface gravity, much greater elevation differences can be supported, which helps explain why Olympus Mons is more than twice as high as the tallest mountains of Venus or Earth. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 515 By the way, the same kind of calculation that determines the limiting height of a mountain can be used to ascertain the largest body that can have an irregular shape. Gravity, if it can, pulls all objects into the most \u201cefficient\u201d shape (where all the outside points are equally distant from the center). All the planets and larger moons are nearly spherical, due to the force of their own gravity pulling them into a sphere. But the smaller the object, the greater the departure from spherical shape that the strength of its rocks can support. For silicate bodies, the limiting diameter is about 400 kilometers; larger objects will always be approximately spherical, while smaller ones can have almost any shape (as we see in photographs of asteroids, such as Figure 14.21). Figure 14.21. Irregular Asteroid. Small objects such as asteroid Ida (shown here in multiple views taken by the Galileo spacecraft camera as it flew past) are generally irregular or elongated; they do not have strong enough gravity to pull them into a spherical shape. Ida is about 60 kilometers long in its longest dimension. (credit: modification of work by NASA/JPL) Atmospheres The atmospheres of the planets were formed by a combination of gas escaping from their interiors and the impacts of volatile-rich debris from the outer solar system. Each of the terrestrial planets must have originally had similar atmospheres, but Mercury was too small and too hot to retain its gas. The Moon probably never had an atmosphere since the material composing it was depleted in volatile materials. The predominant volatile gas on the terrestrial planets is now carbon dioxide (CO2), but initially there were probably also hydrogen-containing gases. In this more chemically reduced (hydrogen-dominated) environment, there should have been large amounts of carbon monoxide (CO) and traces of ammonia (NH3) and methane (CH4). Ultraviolet light from the Sun split apart the molecules of reducing gases in the inner solar system, leaving behind the oxidized (", "oxygen-dominated) however. Most of the light hydrogen atoms escaped, atmospheres we see today on Earth, Venus, and Mars. The fate of water was different on each of these three planets, depending on its size and distance from the Sun. Early in its history, Mars apparently had a thick atmosphere with abundant liquid water, but it could not retain those conditions. The CO2 necessary for a substantial greenhouse effect was lost, the temperature dropped, and eventually the remaining water froze. On Venus the reverse process took place, with a runaway greenhouse effect leading to the permanent loss of water. Only Earth managed to maintain the delicate balance that permits liquid water to persist on its surface. With the water gone, Venus and Mars each ended up with an atmosphere of about 96 percent carbon dioxide and a few percent nitrogen. On Earth, the presence first of water and then of life led to a very different kind of atmosphere. The CO2 was removed and deposited in marine sediment. The proliferation of life forms that could photosynthesize eventually led to the release of more oxygen than natural chemical reactions can remove from 516 Chapter 14 Cosmic Samples and the Origin of the Solar System the atmosphere. As a result, thanks to the life on its surface, Earth finds itself with a great deficiency of CO2, with nitrogen as the most abundant gas, and the only planetary atmosphere that contains free oxygen. In the outer solar system, Titan is the only moon with a substantial atmosphere. This object must have contained sufficient volatiles\u2014such as ammonia, methane, and nitrogen\u2014to form an atmosphere. Thus, today Titan\u2019s atmosphere consists primarily of nitrogen. Compared with those on the inner planets, temperatures on Titan are too low for either carbon dioxide or water to be in vapor form. With these two common volatiles frozen solid, it is perhaps not too surprising that nitrogen has ended up as the primary atmospheric constituent. We see that nature, starting with one set of chemical constituents, can fashion a wide range of final atmospheres appropriate to the conditions and history of each world. The atmosphere we have on Earth is the result of many eons of evolution and adaptation. And, as we saw, it can be changed by the actions of the life forms that inhabit the planet. One of the motivations for exploration of our planetary system is the search for life, beginning with a survey for potentially habitable environments. Mercury, Venus, and the Moon are not suitable; neither are most of the moons in the outer solar system. The giant planets, which do not have solid surfaces, also fail the test for habitability. So far, the search for habitable environments has focused on the presence of liquid water. Earth and Europa both have large oceans, although Europa\u2019s ocean is covered with a thick crust of ice. Mars has a long history of liquid water on its surface, although the surface today is mostly dry and cold. However, there is strong evidence for subsurface water on Mars, and even today water flows briefly on the surface under the right conditions. Enceladus may have the most accessible liquid water, which is squirting into space by means of the geysers observed with our Cassini spacecraft. Titan is in many ways the most interesting world we have explored. It is far too cold for liquid water, but with its thick atmosphere and hydrocarbon lakes, it may be the best place to search for \u201clife as we don\u2019t know it.\u201d We now come to the end of our study of the planetary system. Although we have learned a great deal about the other planets during the past few decades of spacecraft exploration, much remains unknown. Discoveries in recent years of geological activity on Titan and Enceladus were unexpected, as was the complex surface of Pluto revealed by New Horizons. The study of exoplanetary systems provides a new perspective, teaching us that there is much more variety among planetary systems than scientists had imagined a few decades ago. The exploration of the solar system is one of the greatest human adventures, and, in many ways, it has just begun. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 517 CHAPTER 14 REVIEW KEY TERMS accretion the gradual accumulation of mass, as by a planet forming from colliding particles in the solar nebula exoplanet a planet orbiting a star other than our Sun iron meteorite a meteorite composed primarily of iron and nickel meteor a small piece of solid matter that enters Earth\u2019s atmosphere and burns up, popularly called a shooting star because it is seen as a small flash of light meteor shower many meteors appearing to radiate from one point in the sky; produced when Earth passes through a cometary dust stream meteorite a portion of a meteor that survives passage through the atmosphere and strikes the ground stony meteorite a meteorite composed mostly of stony material, either primitive or differentiated stony-iron meteorite a type", " of differentiated meteorite that is a blend of nickel-iron and silicate materials SUMMARY 14.1 Meteors When a fragment of interplanetary dust strikes Earth\u2019s atmosphere, it burns up to create a meteor. Streams of dust particles traveling through space together produce meteor showers, in which we see meteors diverging from a spot in the sky called the radiant of the shower. Many meteor showers recur each year and are associated with particular comets that have left dust behind as they come close to the Sun and their ices evaporate (or have broken up into smaller pieces). 14.2 Meteorites: Stones from Heaven Meteorites are the debris from space (mostly asteroid fragments) that survive to reach the surface of Earth. Meteorites are called finds or falls according to how they are discovered; the most productive source today is the Antarctic ice cap. Meteorites are classified as irons, stony-irons, or stones accordingly to their composition. Most stones are primitive objects, dated to the origin of the solar system 4.5 billion years ago. The most primitive are the carbonaceous meteorites, such as Murchison and Allende. These can contain a number of organic (carbon- rich) molecules. 14.3 Formation of the Solar System A viable theory of solar system formation must take into account motion constraints, chemical constraints, and age constraints. Meteorites, comets, and asteroids are survivors of the solar nebula out of which the solar system formed. This nebula was the result of the collapse of an interstellar cloud of gas and dust, which contracted (conserving its angular momentum) to form our star, the Sun, surrounded by a thin, spinning disk of dust and vapor. Condensation in the disk led to the formation of planetesimals, which became the building blocks of the planets. Accretion of infalling materials heated the planets, leading to their differentiation. The giant planets were also able to attract and hold gas from the solar nebula. After a few million years of violent impacts, most of the debris was swept up or ejected, leaving only the asteroids and cometary remnants surviving to the present. 518 Chapter 14 Cosmic Samples and the Origin of the Solar System 14.4 Comparison with Other Planetary Systems The first planet circling a distant solar-type star was announced in 1995. Twenty years later, thousands of exoplanets have been identified, including planets with sizes and masses between Earth\u2019s and Neptune\u2019s, which we don\u2019t have in our own solar system. A few percent of exoplanet systems have \u201chot Jupiters,\u201d massive planets that orbit close to their stars, and many exoplanets are also in eccentric orbits. These two characteristics are fundamentally different from the attributes of gas giant planets in our own solar system and suggest that giant planets can migrate inward from their place of formation where it is cold enough for ice to form. Current data indicate that small (terrestrial type) rocky planets are common in our Galaxy; indeed, there must be tens of billions of such earthlike planets. 14.5 Planetary Evolution After their common beginning, each of the planets evolved on its own path. Different possible outcomes are illustrated by comparison of the terrestrial planets (Earth, Venus, Mars, Mercury, and the Moon). All are rocky, differentiated objects. The level of geological activity is proportional to mass: greatest for Earth and Venus, less for Mars, and absent for the Moon and Mercury. However, tides from another nearby world can also generate heat to drive geological activity, as shown by Io, Europa, and Enceladus. Pluto is also active, to the surprise of planetary scientists. On the surfaces of solid worlds, mountains can result from impacts, volcanism, or uplift. Whatever their origin, higher mountains can be supported on smaller planets that have less surface gravity. The atmospheres of the terrestrial planets may have acquired volatile materials from comet impacts. The Moon and Mercury lost their atmospheres; most volatiles on Mars are frozen due to its greater distance from the Sun and its thinner atmosphere; and Venus retained CO2 but lost H2O when it developed a massive greenhouse effect. Only Earth still has liquid water on its surface and hence can support life. FOR FURTHER EXPLORATION Note: Resources about exoplanets are provided in The Birth of Stars and the Discovery of Planets outside the Solar System. Articles Meteors and Meteorites Alper, J. \u201cIt Came from Outer Space.\u201d Astronomy (November 2002): 36. On the analysis of organic materials in meteorites. Beatty, J. \u201cCatch a Fallen Star.\u201d Sky & Telescope (August 2009): 22. On the recovery of meteorites from an impact that was seen in the sky. Durda, D. \u201cThe Chelyabinsk Super-Meteor.\u201d Sky & Telescope (June 2013): 24. A nice summary, with photos and eyewitness reporting. Garcia", ", R., & Notkin, G. \u201cTouching the Stars without Leaving Home.\u201d Sky & Telescope (October 2008): 32. Hunting and collecting meteorites. Kring, D. \u201cUnlocking the Solar System\u2019s Past.\u201d Astronomy (August 2006): 32. Part of a special issue devoted to meteorites. Rubin, A. \u201cSecrets of Primitive Meteorites.\u201d Scientific American (February 2013): 36. What they can teach us about the environment in which the solar system formed. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 519 Evolution of the Solar System and Protoplanetary Disks Jewitt, D., & Young, E. \u201cOceans from the Skies.\u201d Scientific American (March 2015): 36\u201343. How did Earth and the other inner planets get their water after the initial hot period? Talcott, R. \u201cHow the Solar System Came to Be.\u201d Astronomy (November 2012): 24. On the formation period of the Sun and the planets. Young, E. \u201cCloudy with a Chance of Stars.\u201d Scientific American (February 2010): 34. On how clouds of interstellar matter turn into star systems. Websites Meteors and Meteorites American Meteor Society: http://www.amsmeteors.org/. For serious observers. British and Irish Meteorite Society: http://www.bimsociety.org/meteorites1.shtml. Meteor Showers Online: http://meteorshowersonline.com/. By Gary Kronk. Meteorite Information: http://www.meteorite-information.com/. A great collection of links for understanding and even collecting meteorites. Meteorites from Mars: http://www2.jpl.nasa.gov/snc/. A listing and links from the Jet Propulsion Lab. Meteors and Meteor Showers: http://www.astronomy.com/observing/observe-the-solar-system/2010/04/ meteors-and-meteor-showers. From Astronomy magazine. Meteors: http://www.skyandtelescope.com/observing/celestial-objects-to-watch/meteors/. A collection of articles on meteor observing from Sky & Telescope magazine. Nine Planets Meteorites and Meteors Page: http://nineplanets.org/meteorites.html. Some Interesting Meteorite Falls of the Last Two Centuries: http://www.icq.eps.harvard.edu/meteorites-1.html. Evolution of the Solar System and Protoplanetary Disks Circumstellar Disk Learning Site: http://www.disksite.com/. By Dr. Paul Kalas. Disk Detective Project: http://www.diskdetective.org/. The WISE mission is asking the public to help them find protoplanetary disks in their infrared data. Videos Meteors and Meteorites Meteorites and Meteor-wrongs: https://www.youtube.com/watch?v=VQO335Y3zXo. Video with Dr. Randy Korotev of Washington U. in St. Louis (7:05). Rare Meteorites from London\u2019s Natural History Museum: https://www.youtube.com/watch?v=w-Rsk-ywN44. A tour of the meteorite collection with curator Caroline Smith (18:22). Also see a short news piece about a martian meteorite: https://www.youtube.com/watch?v=1EMR2r53f2s (2:54). What Is a Meteor Shower (and How to Watch Them): https://www.youtube.com/watch?v=xNmgvlwInCA. Top tips for watching meteor showers from the At-Bristol Science Center (3:18). Evolution of the Solar System and Protoplanetary Disks Origins of the Solar System: http://www.pbs.org/wgbh/nova/space/origins-solar-system.html. Video from Nova ScienceNow narrated by Neil deGrasse Tyson (13:02). 520 Chapter 14 Cosmic Samples and the Origin of the Solar System Where Do Planets Come From?: https://www.youtube.com/watch?v=zdIJUdZWlXo. Public talk by Anjali Tripathi in March 2016 in the Center for Astrophysics Observatory Nights Series (56:14). COLLABORATIVE GROUP ACTIVITIES A. Ever since the true (cosmic) origin of meteorites was understood, people have tried to make money selling them to museums and planetariums. More recently, a growing number of", " private collectors have been interested in purchasing meteorite fragments, and a network of dealers (some more reputable than others) has sprung up to meet this need. What does your group think of all this? Who should own a meteorite? The person on whose land it falls, the person who finds it, or the local, state, or federal government where it falls? What if it falls on public land? Should there be any limit to what people charge for meteorites? Or should all meteorites be the common property of humanity? (If you can, try to research what the law is now in your area. See, for example, http://www.space.com/18009-meteorite-collectors-public-lands-rules.html.) B. Your group has been formed to advise a very rich person who wants to buy some meteorites but is afraid of being cheated and sold some Earth rocks. How would you advise your client to make sure that the meteorites she buys are authentic? C. Your group is a committee set up to give advice to NASA about how to design satellites and telescopes in space to minimize the danger of meteor impacts. Remember that the heavier a satellite is, the harder (more expensive) it is to launch. What would you include in your recommendations? D. Discuss what you would do if you suddenly found that a small meteorite had crashed in or near your home. Whom would you call first, second, third? What would you do with the sample? (And would any damage to your home be covered by your insurance?) E. A friend of your group really wants to see a meteor shower. The group becomes a committee to assist her in fulfilling this desire. What time of year would be best? What equipment would you recommend she gets? What advice would you give her? F. Work with your group to find a table of the phases of the Moon for the next calendar year. Then look at the table of well-known meteor showers in this chapter and report on what phase the Moon will be in during each shower. (The brighter the Moon is in the night sky, the harder it is to see the faint flashes of meteors.) G. Thinking that all giant planets had to be far from their stars (because the ones in our solar system are) is an example of making theories without having enough data (or examples). Can your group make a list of other instances in science (and human relations) where we have made incorrect judgments without having explored enough examples? H. Have your group list and then discuss several ways in which the discovery of a diverse group of exoplanets (planets orbiting other stars) has challenged our conventional view of the formation of planetary systems like our solar system. EXERCISES This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 14 Cosmic Samples and the Origin of the Solar System 521 Review Questions 1. A friend of yours who has not taken astronomy sees a meteor shower (she calls it a bunch of shooting stars). The next day she confides in you that she was concerned that the stars in the Big Dipper (her favorite star pattern) might be the next ones to go. How would you put her mind at ease? 2. In what ways are meteorites different from meteors? What is the probable origin of each? 3. How are comets related to meteor showers? 4. What do we mean by primitive material? How can we tell if a meteorite is primitive? 5. Describe the solar nebula, and outline the sequence of events within the nebula that gave rise to the planetesimals. 6. Why do the giant planets and their moons have compositions different from those of the terrestrial planets? 7. How do the planets discovered so far around other stars differ from those in our own solar system? List at least two ways. 8. Explain the role of impacts in planetary evolution, including both giant impacts and more modest ones. 9. Why are some planets and moons more geologically active than others? 10. Summarize the origin and evolution of the atmospheres of Venus, Earth, and Mars. 11. Why do meteors in a meteor shower appear to come from just one point in the sky? Thought Questions 12. What methods do scientists use to distinguish a meteorite from terrestrial material? 13. Why do iron meteorites represent a much higher percentage of finds than of falls? 14. Why is it more useful to classify meteorites according to whether they are primitive or differentiated rather than whether they are stones, irons, or stony-irons? 15. Which meteorites are the most useful for defining the age of the solar system? Why? 16. Suppose a new primitive meteorite is discovered (sometime after it falls in a field of soybeans) and analysis reveals that it contains a trace of amino acids, all of which show the same rotational symmetry (unlike the Murchison meteorite).", " What might you conclude from this finding? 17. How do we know when the solar system formed? Usually we say that the solar system is 4.5 billion years old. To what does this age correspond? 18. We have seen how Mars can support greater elevation differences than Earth or Venus. According to the same arguments, the Moon should have higher mountains than any of the other terrestrial planets, yet we know it does not. What is wrong with applying the same line of reasoning to the mountains on the Moon? 19. Present theory suggests that giant planets cannot form without condensation of water ice, which becomes vapor at the high temperatures close to a star. So how can we explain the presence of jovian-sized exoplanets closer to their star than Mercury is to our Sun? 20. Why are meteorites of primitive material considered more important than other meteorites? Why have most of them been found in Antarctica? 522 Chapter 14 Cosmic Samples and the Origin of the Solar System Figuring For Yourself 21. How long would material take to go around if the solar nebula in Example 14.1 became the size of Earth\u2019s orbit? 22. Consider the differentiated meteorites. We think the irons are from the cores, the stony-irons are from the interfaces between mantles and cores, and the stones are from the mantles of their differentiated parent bodies. If these parent bodies were like Earth, what fraction of the meteorites would you expect to consist of irons, stony-irons, and stones? Is this consistent with the observed numbers of each? (Hint: You will need to look up what percent of the volume of Earth is taken up by its core, mantle, and crust.) 23. Estimate the maximum height of the mountains on a hypothetical planet similar to Earth but with twice the surface gravity of our planet. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 523 15 THE SUN: A GARDEN-VARIETY STAR Figure 15.1. Our Star. The Sun\u2014our local star\u2014is quite average in many ways. However, that does not stop it from being a fascinating object to study. From solar flares and coronal mass ejections, like the one seen coming from the Sun in the top right of this image, the Sun is a highly dynamic body at the center of our solar system. This image combines two separate satellite pictures of the Sun\u2014the inner one from the Solar Dynamics Observatory and the outer one from the Solar and Heliospheric Observatory. (credit: modification of work by ESA/NASA) Chapter Outline 15.1 The Structure and Composition of the Sun 15.2 The Solar Cycle 15.3 Solar Activity above the Photosphere 15.4 Space Weather Thinking Ahead \u201cSpace weather\u201d may sound like a contradiction. How can there be weather in the vacuum of space? Yet space weather, which refers to changing conditions in space, is an active field of research and can have profound effects on Earth. We are all familiar with the ups and downs of weather on Earth, and how powerful storms can be devastating for people and vegetation. Although we are separated from the Sun by a large distance as well as by the vacuum of space, we now understand that great outbursts on the Sun (solar storms, in effect) can cause changes in the atmosphere and magnetic field of Earth, sometimes even causing serious problems on the ground. In this chapter, we will explore the nature of the Sun\u2019s outer layers, the changing conditions and activity there, and the ways that the Sun affects Earth. By studying the Sun, we also learn much that helps us understand stars in general. The Sun is, in astronomical terms, a rather ordinary star\u2014not unusually hot or cold, old or young, large or small. Indeed, we are lucky that the Sun is typical. Just as studies of Earth help us understand observations of the more distant planets, so too does the Sun serve as a guide to astronomers in interpreting the messages contained in the light we receive from distant stars. As you will learn, the Sun is dynamic, continuously undergoing change, balancing the forces of nature to keep itself in equilibrium. In this chapter, we describe the components of the Sun, how it changes with time, and how those changes affect Earth. 524 Chapter 15 The Sun: A Garden-Variety Star 15.1 THE STRUCTURE AND COMPOSITION OF THE SUN Learning Objectives By the end of this section, you will be able to: Explain how the composition of the Sun differs from that of Earth Describe the various layers of the Sun and their functions Explain what happens in the different parts of the Sun\u2019s atmosphere The Sun, like all stars, is an enormous ball of extremely hot, largely ionized gas, shining under its own power. And we do mean enormous. The Sun could fit 109", " Earths side-by-side across its diameter, and it has enough volume (takes up enough space) to hold about 1.3 million Earths. The Sun does not have a solid surface or continents like Earth, nor does it have a solid core (Figure 15.2). However, it does have a lot of structure and can be discussed as a series of layers, not unlike an onion. In this section, we describe the huge changes that occur in the Sun\u2019s extensive interior and atmosphere, and the dynamic and violent eruptions that occur daily in its outer layers. Figure 15.2. Earth and the Sun. Here, Earth is shown to scale with part of the Sun and a giant loop of hot gas erupting from its surface. The inset shows the entire Sun, smaller. (credit: modification of work by SOHO/EIT/ESA) Some of the basic characteristics of the Sun are listed in Table 15.1. Although some of the terms in that table may be unfamiliar to you right now, you will get to know them as you read further. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 525 Characteristics of the Sun Characteristic How Found Value Mean distance Radar reflection from planets 1 AU (149,597,892 km) Maximum distance from Earth Minimum distance from Earth Mass Orbit of Earth 1.521 \u00d7 108 km 1.471 \u00d7 108 km 333,400 Earth masses (1.99 \u00d7 1030 kg) Mean angular diameter Direct measure 31\u00b459\u00b4\u00b4.3 Diameter of photosphere Angular size and distance Mean density Mass/volume Gravitational acceleration at photosphere (surface gravity) GM/R2 109.3 \u00d7 Earth diameter (1.39 \u00d7 106 km) 1.41 g/cm3 (1400 kg/m3) 27.9 \u00d7 Earth surface gravity = 273 m/s2 Solar constant Luminosity Instrument sensitive to radiation at all wavelengths 1370 W/m2 Solar constant \u00d7 area of spherical surface 1 AU in radius 3.8 \u00d7 1026 W Spectral class Spectrum Effective temperature Derived from luminosity and radius of the Sun G2V 5800 K Rotation period at equator Sunspots and Doppler shift in spectra taken at the edge of the Sun 24 days 16 hours Inclination of equator to ecliptic Motions of sunspots 7\u00b010\u00b4.5 Table 15.1 Composition of the Sun\u2019s Atmosphere Let\u2019s begin by asking what the solar atmosphere is made of. As explained in Radiation and Spectra, we can use a star\u2019s absorption line spectrum to determine what elements are present. It turns out that the Sun contains the same elements as Earth but not in the same proportions. About 73% of the Sun\u2019s mass is hydrogen, and another 25% is helium. All the other chemical elements (including those we know and love in our own bodies, such as carbon, oxygen, and nitrogen) make up only 2% of our star. The 10 most abundant gases in the Sun\u2019s visible surface layer are listed in Table 15.2. Examine that table and notice that the composition of the Sun\u2019s outer layer is very different from Earth\u2019s crust, where we live. (In our planet\u2019s crust, the three most abundant 526 Chapter 15 The Sun: A Garden-Variety Star elements are oxygen, silicon, and aluminum.) Although not like our planet\u2019s, the makeup of the Sun is quite typical of stars in general. The Abundance of Elements in the Sun Element Percentage by Number of Atoms Percentage By Mass Hydrogen Helium Carbon 92.0 7.8 0.02 Nitrogen 0.008 Oxygen Neon Magnesium Silicon Sulfur Iron Table 15.2 0.06 0.01 0.003 0.004 0.002 0.003 73.4 25.0 0.20 0.09 0.80 0.16 0.06 0.09 0.05 0.14 The fact that our Sun and the stars all have similar compositions and are made up of mostly hydrogen and helium was first shown in a brilliant thesis in 1925 by Cecilia Payne-Gaposchkin, the first woman to get a PhD in astronomy in the United States (Figure 15.3). However, the idea that the simplest light gases\u2014hydrogen and helium\u2014were the most abundant elements in stars was so unexpected and so shocking that she assumed her analysis of the data must be wrong. At the time, she wrote, \u201cThe enormous abundance derived for these elements in the stellar atmosphere is almost certainly not real.\u201d Even scientists sometimes find it hard to accept new ideas that do not agree with what everyone \u201cknows\u201d to be right. Figure 15.3. Cecilia Payne-Gap", "oschkin (1900\u20131979). Her 1925 doctoral thesis laid the foundations for understanding the composition of the Sun and the stars. Yet, being a woman, she was not given a formal appointment at Harvard, where she worked, until 1938 and was not appointed a professor until 1956. (credit: Smithsonian Institution) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 527 Before Payne-Gaposchkin\u2019s work, everyone assumed that the composition of the Sun and stars would be much like that of Earth. It was 3 years after her thesis that other studies proved beyond a doubt that the enormous abundance of hydrogen and helium in the Sun is indeed real. (And, as we will see, the composition of the Sun and the stars is much more typical of the makeup of the universe than the odd concentration of heavier elements that characterizes our planet.) Most of the elements found in the Sun are in the form of atoms, with a small number of molecules, all in the form of gases: the Sun is so hot that no matter can survive as a liquid or a solid. In fact, the Sun is so hot that many of the atoms in it are ionized, that is, stripped of one or more of their electrons. This removal of electrons from their atoms means that there is a large quantity of free electrons and positively charged ions in the Sun, making it an electrically charged environment\u2014quite different from the neutral one in which you are reading this text. (Scientists call such a hot ionized gas a plasma.) In the nineteenth century, scientists observed a spectral line at 530.3 nanometers in the Sun\u2019s outer atmosphere, called the corona (a layer we will discuss in a minute.) This line had never been seen before, and so it was assumed that this line was the result of a new element found in the corona, quickly named coronium. It was not until 60 years later that astronomers discovered that this emission was in fact due to highly ionized iron\u2014iron with 13 of its electrons stripped off. This is how we first discovered that the Sun\u2019s atmosphere had a temperature of more than a million degrees. The Layers of the Sun beneath the Visible Surface Figure 15.4 shows what the Sun would look like if we could see all parts of it from the center to its outer atmosphere; the terms in the figure will become familiar to you as you read on. Figure 15.4. Parts of the Sun. This illustration shows the different parts of the Sun, from the hot core where the energy is generated through regions where energy is transported outward, first by radiation, then by convection, and then out through the solar atmosphere. The parts of the atmosphere are also labeled the photosphere, chromosphere, and corona. Some typical features in the atmosphere are shown, such as coronal holes and prominences. (credit: modification of work by NASA/Goddard) 528 Chapter 15 The Sun: A Garden-Variety Star The Sun\u2019s layers are different from each other, and each plays a part in producing the energy that the Sun ultimately emits. We will begin with the core and work our way out through the layers. The Sun\u2019s core is extremely dense and is the source of all of its energy. Inside the core, nuclear energy is being released (in ways we will discuss in The Sun: A Nuclear Powerhouse). The core is approximately 20% of the size of the solar interior and is thought to have a temperature of approximately 15 million K, making it the hottest part of the Sun. Above the core is a region known as the radiative zone\u2014named for the primary mode of transporting energy across it. This region starts at about 25% of the distance to the solar surface and extends up to about 70% of the way to the surface. The light generated in the core is transported through the radiative zone very slowly, since the high density of matter in this region means a photon cannot travel too far without encountering a particle, causing it to change direction and lose some energy. The convective zone is the outermost layer of the solar interior. It is a thick layer approximately 200,000 kilometers deep that transports energy from the edge of the radiative zone to the surface through giant convection cells, similar to a pot of boiling oatmeal. The plasma at the bottom of the convective zone is extremely hot, and it bubbles to the surface where it loses its heat to space. Once the plasma cools, it sinks back to the bottom of the convective zone. Now that we have given a quick overview of the structure of the whole Sun, in this section, we will embark on a journey through the visible layers of the Sun, beginning with the photosphere\u2014the visible surface. The Solar Photosphere Earth\u2019s air is generally transparent. But on a smoggy day in many cities", ", it can become opaque, which prevents us from seeing through it past a certain point. Something similar happens in the Sun. Its outer atmosphere is transparent, allowing us to look a short distance through it. But when we try to look through the atmosphere deeper into the Sun, our view is blocked. The photosphere is the layer where the Sun becomes opaque and marks the boundary past which we cannot see (Figure 15.5). Figure 15.5. Solar Photosphere plus Sunspots. This photograph shows the photosphere\u2014the visible surface of the Sun. Also shown is an enlarged image of a group of sunspots; the size of Earth is shown for comparison. Sunspots appear darker because they are cooler than their surroundings. The typical temperature at the center of a large sunspot is about 3800 K, whereas the photosphere has a temperature of about 5800 K. (credit: modification of work by NASA/SDO) As we saw, the energy that emerges from the photosphere was originally generated deep inside the Sun (more on this in The Sun: A Nuclear Powerhouse). This energy is in the form of photons, which make their way slowly toward the solar surface. Outside the Sun, we can observe only those photons that are emitted into the solar photosphere, where the density of atoms is sufficiently low and the photons can finally escape from the Sun without colliding with another atom or ion. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 529 As an analogy, imagine that you are attending a big campus rally and have found a prime spot near the center of the action. Your friend arrives late and calls you on your cell phone to ask you to join her at the edge of the crowd. You decide that friendship is worth more than a prime spot, and so you work your way out through the dense crowd to meet her. You can move only a short distance before bumping into someone, changing direction, and trying again, making your way slowly to the outside edge of the crowd. All this while, your efforts are not visible to your waiting friend at the edge. Your friend can\u2019t see you until you get very close to the edge because of all the bodies in the way. So too photons making their way through the Sun are constantly bumping into atoms, changing direction, working their way slowly outward, and becoming visible only when they reach the atmosphere of the Sun where the density of atoms is too low to block their outward progress. Astronomers have found that the solar atmosphere changes from almost perfectly transparent to almost completely opaque in a distance of just over 400 kilometers; it is this thin region that we call the photosphere, a word that comes from the Greek for \u201clight sphere.\u201d When astronomers speak of the \u201cdiameter\u201d of the Sun, they mean the size of the region surrounded by the photosphere. The photosphere looks sharp only from a distance. If you were falling into the Sun, you would not feel any surface but would just sense a gradual increase in the density of the gas surrounding you. It is much the same as falling through a cloud while skydiving. From far away, the cloud looks as if it has a sharp surface, but you do not feel a surface as you fall into it. (One big difference between these two scenarios, however, is temperature. The Sun is so hot that you would be vaporized long before you reached the photosphere. Skydiving in Earth\u2019s atmosphere is much safer.) We might note that the atmosphere of the Sun is not a very dense layer compared to the air in the room where you are reading this text. At a typical point in the photosphere, the pressure is less than 10% of Earth\u2019s pressure at sea level, and the density is about one ten-thousandth of Earth\u2019s atmospheric density at sea level. Observations with telescopes show that the photosphere has a mottled appearance, resembling grains of rice spilled on a dark tablecloth or a pot of boiling oatmeal. This structure of the photosphere is called granulation (see Figure 15.6). Granules, which are typically 700 to 1000 kilometers in diameter (about the width of Texas), appear as bright areas surrounded by narrow, darker (cooler) regions. The lifetime of an individual granule is only 5 to 10 minutes. Even larger are supergranules, which are about 35,000 kilometers across (about the size of two Earths) and last about 24 hours. Figure 15.6. Granulation Pattern. The surface markings of the convection cells create a granulation pattern on this dramatic image (left) taken from the Japanese Hinode spacecraft. You can see the same pattern when you heat up miso soup. The right image shows an irregular-shaped sunspot and granules on the Sun\u2019s surface, seen with the Swedish Solar Telescope on August", " 22, 2003. (credit left: modification of work by Hinode JAXA/NASA/PPARC; credit right: ISP/SST/Oddbjorn Engvold, Jun Elin Wiik, Luc Rouppe van der Voort) 530 Chapter 15 The Sun: A Garden-Variety Star The motions of the granules can be studied by examining the Doppler shifts in the spectra of gases just above them (see The Doppler Effect). The bright granules are columns of hotter gases rising at speeds of 2 to 3 kilometers per second from below the photosphere. As this rising gas reaches the photosphere, it spreads out, cools, and sinks down again into the darker regions between the granules. Measurements show that the centers of the granules are hotter than the intergranular regions by 50 to 100 K See the \u201cboiling\u201d action of granulation in this 30-second time-lapse video (https://openstax.org/l/ 30SolarGran) from the Swedish Institute for Solar Physics. The Chromosphere The Sun\u2019s outer gases extend far beyond the photosphere (Figure 15.7). Because they are transparent to most visible radiation and emit only a small amount of light, these outer layers are difficult to observe. The region of the Sun\u2019s atmosphere that lies immediately above the photosphere is called the chromosphere. Until this century, the chromosphere was visible only when the photosphere was concealed by the Moon during a total solar eclipse (see the chapter on Earth, Moon, and Sky). In the seventeenth century, several observers described what appeared to them as a narrow red \u201cstreak\u201d or \u201cfringe\u201d around the edge of the Moon during a brief instant after the Sun\u2019s photosphere had been covered. The name chromosphere, from the Greek for \u201ccolored sphere,\u201d was given to this red streak. Figure 15.7. The Sun\u2019s Atmosphere. Composite image showing the three components of the solar atmosphere: the photosphere or surface of the Sun taken in ordinary light; the chromosphere, imaged in the light of the strong red spectral line of hydrogen (H-alpha); and the corona as seen with X-rays. (credit: modification of work by NASA) Observations made during eclipses show that the chromosphere is about 2000 to 3000 kilometers thick, and its spectrum consists of bright emission lines, indicating that this layer is composed of hot gases emitting light at discrete wavelengths. The reddish color of the chromosphere arises from one of the strongest emission lines in the visible part of its spectrum\u2014the bright red line caused by hydrogen, the element that, as we have already seen, dominates the composition of the Sun. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 531 In 1868, observations of the chromospheric spectrum revealed a yellow emission line that did not correspond to any previously known element on Earth. Scientists quickly realized they had found a new element and named it helium (after helios, the Greek word for \u201cSun\u201d). It took until 1895 for helium to be discovered on our planet. Today, students are probably most familiar with it as the light gas used to inflate balloons, although it turns out to be the second-most abundant element in the universe. The temperature of the chromosphere is about 10,000 K. This means that the chromosphere is hotter than the photosphere, which should seem surprising. In all the situations we are familiar with, temperatures fall as one moves away from the source of heat, and the chromosphere is farther from the center of the Sun than the photosphere is. The Transition Region The increase in temperature does not stop with the chromosphere. Above it is a region in the solar atmosphere where the temperature changes from 10,000 K (typical of the chromosphere) to nearly a million degrees. The hottest part of the solar atmosphere, which has a temperature of a million degrees or more, is called the corona. Appropriately, the part of the Sun where the rapid temperature rise occurs is called the transition region. It is probably only a few tens of kilometers thick. Figure 15.8 summarizes how the temperature of the solar atmosphere changes from the photosphere outward. Figure 15.8. Temperatures in the Solar Atmosphere. On this graph, temperature is shown increasing upward, and height above the photosphere is shown increasing to the right. Note the very rapid increase in temperature over a very short distance in the transition region between the chromosphere and the corona. In 2013, NASA launched the Interface Region Imaging Spectrograph (IRIS) to study the transition region to understand better how and why this sharp temperature increase occurs. IRIS is the first space mission that is able to obtain high spatial resolution images of the different features produced", " over this wide temperature range and to see how they change with time and location (Figure 15.9). 532 Chapter 15 The Sun: A Garden-Variety Star Figure 15.9. Portion of the Transition Region. This image shows a giant ribbon of relatively cool gas threading through the lower portion of the hot corona. This ribbon (the technical term is filament) is made up of many individual threads. Time-lapse movies of this filament showed that it gradually heated as it moved through the corona. Scientists study events like this in order to try to understand what heats the chromosphere and corona to high temperatures. The \u201cwhiskers\u201d at the edge of the Sun are spicules, jets of gas that shoot material up from the Sun\u2019s surface and disappear after only a few minutes. This single image gives a hint of just how complicated it is to construct a model of the all the different structures and heating mechanisms in the solar atmosphere. (credit: JAXA/NASA/Hinode) Figure 15.4 and the red graph in Figure 15.8 make the Sun seem rather like an onion, with smooth spherical shells, each one with a different temperature. For a long time, astronomers did indeed think of the Sun this way. However, we now know that while this idea of layers\u2014photosphere, chromosphere, transition region, corona\u2014describes the big picture fairly well, the Sun\u2019s atmosphere is really more complicated, with hot and cool regions intermixed. For example, clouds of carbon monoxide gas with temperatures colder than 4000 K have now been found at the same height above the photosphere as the much hotter gas of the chromosphere. The Corona The outermost part of the Sun\u2019s atmosphere is called the corona. Like the chromosphere, the corona was first observed during total eclipses (Figure 15.10). Unlike the chromosphere, the corona has been known for many centuries: it was referred to by the Roman historian Plutarch and was discussed in some detail by Kepler. The corona extends millions of kilometers above the photosphere and emits about half as much light as the full moon. The reason we don\u2019t see this light until an eclipse occurs is the overpowering brilliance of the photosphere. Just as bright city lights make it difficult to see faint starlight, so too does the intense light from the photosphere hide the faint light from the corona. While the best time to see the corona from Earth is during a total solar eclipse, it can be observed easily from orbiting spacecraft. Its brighter parts can now be photographed with a special instrument\u2014a coronagraph\u2014that removes the Sun\u2019s glare from the image with an occulting disk (a circular piece of material held so it is just in front of the Sun). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 533 Figure 15.10. Coronagraph. This image of the Sun was taken March 2, 2016. The larger dark circle in the center is the disk the blocks the Sun\u2019s glare, allowing us to see the corona. The smaller inner circle is where the Sun would be if it were visible in this image. (credit: modification of work by NASA/SOHO) Studies of its spectrum show the corona to be very low in density. At the bottom of the corona, there are only about 109 atoms per cubic centimeter, compared with about 1016 atoms per cubic centimeter in the upper photosphere and 1019 molecules per cubic centimeter at sea level in Earth\u2019s atmosphere. The corona thins out very rapidly at greater heights, where it corresponds to a high vacuum by Earth laboratory standards. The corona extends so far into space\u2014far past Earth\u2014that here on our planet, we are technically living in the Sun\u2019s atmosphere. The Solar Wind One of the most remarkable discoveries about the Sun\u2019s atmosphere is that it produces a stream of charged particles (mainly protons and electrons) that we call the solar wind. These particles flow outward from the Sun into the solar system at a speed of about 400 kilometers per second (almost 1 million miles per hour)! The solar wind exists because the gases in the corona are so hot and moving so rapidly that they cannot be held back by solar gravity. (This wind was actually discovered by its effects on the charged tails of comets; in a sense, we can see the comet tails blow in the solar breeze the way wind socks at an airport or curtains in an open window flutter on Earth.) Although the solar wind material is very, very rarified (i.e., extremely low density), the Sun has an enormous surface area. Astronomers estimate that the Sun is losing about 10 million tons of material each year through this wind. While this amount of lost mass", " seems large by Earth standards, it is completely insignificant for the Sun. From where in the Sun does the solar wind emerge? In visible photographs, the solar corona appears fairly uniform and smooth. X-ray and extreme ultraviolet pictures, however, show that the corona has loops, plumes, and both bright and dark regions. Large dark regions of the corona that are relatively cool and quiet are called coronal holes (Figure 15.11). In these regions, magnetic field lines stretch far out into space away from the Sun, rather than looping back to the surface. The solar wind comes predominantly from coronal holes, where gas can stream away from the Sun into space unhindered by magnetic fields. Hot coronal gas, on the other hand, is present mainly where magnetic fields have trapped and concentrated it. 534 Chapter 15 The Sun: A Garden-Variety Star Figure 15.11. Coronal Hole. The dark area visible near the Sun\u2019s south pole on this Solar Dynamics Observer spacecraft image is a coronal hole. (credit: modification of work by NASA/SDO) At the surface of Earth, we are protected to some degree from the solar wind by our atmosphere and Earth\u2019s magnetic field (see Earth as a Planet). However, the magnetic field lines come into Earth at the north and south magnetic poles. Here, charged particles accelerated by the solar wind can follow the field down into our atmosphere. As the particles strike molecules of air, they cause them to glow, producing beautiful curtains of light called the auroras, or the northern and southern lights (Figure 15.12). Figure 15.12. Aurora. The colorful glow in the sky results from charged particles in a solar wind interacting with Earth\u2019s magnetic fields. The stunning display captured here occurred over Jokulsarlon Lake in Iceland in 2013. (credit: Moyan Brenn) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 535 This NASA video (https://openstax.org/l/30Aurora) explains and demonstrates the nature of the auroras and their relationship to Earth\u2019s magnetic field. 15.2 THE SOLAR CYCLE Learning Objectives By the end of this section, you will be able to: Describe the sunspot cycle and, more generally, the solar cycle Explain how magnetism is the source of solar activity Before the invention of the telescope, the Sun was thought to be an unchanging and perfect sphere. We now know that the Sun is in a perpetual state of change: its surface is a seething, bubbling cauldron of hot gas. Areas that are darker and cooler than the rest of the surface come and go. Vast plumes of gas erupt into the chromosphere and corona. Occasionally, there are even giant explosions on the Sun that send enormous streamers of charged particles and energy hurtling toward Earth. When they arrive, these can cause power outages and other serious effects on our planet. Sunspots The first evidence that the Sun changes came from studies of sunspots, which are large, dark features seen on the surface of the Sun caused by increased magnetic activity. They look darker because the spots are typically at a temperature of about 3800 K, whereas the bright regions that surround them are at about 5800 K (Figure 15.13). Occasionally, these spots are large enough to be visible to the unaided eye, and we have records going back over a thousand years from observers who noticed them when haze or mist reduced the Sun\u2019s intensity. (We emphasize what your parents have surely told you: looking at the Sun for even a brief time can cause permanent eye damage. This is the one area of astronomy where we don\u2019t encourage you to do your own observing without getting careful instructions or filters from your instructor.) Figure 15.13. Sunspots. This image of sunspots, cooler and thus darker regions on the Sun, was taken in July 2012. You can see the dark, central region of each sunspot (called the umbra) surrounded by a less dark region (the penumbra). The largest spot shown here is about 11 Earths wide. Although sunspots appear dark when seen next to the hotter gases of the photosphere, an average sunspot, cut out of the solar surface and left standing in the night sky, would be about as bright as the full moon. The mottled appearance of the Sun\u2019s surface is granulation. (credit: NASA Goddard Space Flight Center, Alan Friedman) 536 Chapter 15 The Sun: A Garden-Variety Star While we understand that sunspots look darker because they are cooler, they are nevertheless hotter than the surfaces of many stars. If they could be removed from the Sun, they would shine brightly. They appear dark only in contrast with the hotter, brighter photosphere around them", ". Individual sunspots come and go, with lifetimes that range from a few hours to a few months. If a spot lasts and develops, it usually consists of two parts: an inner darker core, the umbra, and a surrounding less dark region, the penumbra. Many spots become much larger than Earth, and a few, like the largest one shown in Figure 15.13, have reached diameters over 140,000 kilometers. Frequently, spots occur in groups of 2 to 20 or more. The largest groups are very complex and may have over 100 spots. Like storms on Earth, sunspots are not fixed in position, but they drift slowly compared with the Sun\u2019s rotation. By recording the apparent motions of the sunspots as the turning Sun carried them across its disk (Figure 15.14), Galileo, in 1612, demonstrated that the Sun rotates on its axis with a rotation period of approximately 1 month. Our star turns in a west-to-east direction, like the orbital motions of the planets. The Sun, however, is a gas and does not have to rotate rigidly, the way a solid body like Earth does. Modern observations show that the speed of rotation of the Sun varies according to latitude, that is, it\u2019s different as you go north or south of the Sun\u2019s equator. The rotation period is about 25 days at the equator, 28 days at latitude 40\u00b0, and 36 days at latitude 80\u00b0. We call this behavior differential rotation. Figure 15.14. Sunspots Rotate Across Sun\u2019s Surface. This sequence of photographs of the Sun\u2019s surface tracks the movement of sunspots across the visible hemisphere of the Sun. On March 30, 2001, this group of sunspots extended across an area about 13 times the diameter of Earth. This region produced many flares and coronal mass ejections. (credit: modification of work by SOHO/NASA/ESA) The Sunspot Cycle Between 1826 and 1850, Heinrich Schwabe, a German pharmacist and amateur astronomer, kept daily records of the number of sunspots. What he was really looking for was a planet inside the orbit of Mercury, which he hoped to find by observing its dark silhouette as it passed between the Sun and Earth. He failed to find the hoped-for planet, but his diligence paid off with an even-more important discovery: the sunspot cycle. He found that the number of sunspots varied systematically, in cycles about a decade long. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 537 What Schwabe observed was that, although individual spots are short lived, the total number visible on the Sun at any one time was likely to be very much greater at certain times\u2014the periods of sunspot maximum\u2014than at other times\u2014the periods of sunspot minimum. We now know that sunspot maxima occur at an average interval of 11 years, but the intervals between successive maxima have ranged from as short as 9 years to as long as 14 years. During sunspot maxima, more than 100 spots can often be seen at once. Even then, less than one-half of one percent of the Sun\u2019s surface is covered by spots (Figure 15.22). During sunspot minima, sometimes no spots are visible. The Sun\u2019s activity reached its most recent maximum in 2014 Watch this brief video (https://openstax.org/l/30SolarCyc) from NASA\u2019s Goddard Space Flight Center that explains the sunspot cycle. Magnetism and the Solar Cycle Now that we have discussed the Sun\u2019s activity cycle, you might be asking, \u201cWhy does the Sun change in such a regular way?\u201d Astronomers now understand that it is the Sun\u2019s changing magnetic field that drives solar activity. The solar magnetic field is measured using a property of atoms called the Zeeman effect. Recall from Radiation and Spectra that an atom has many energy levels and that spectral lines are formed when electrons shift from one level to another. If each energy level is precisely defined, then the difference between them is also quite precise. As an electron changes levels, the result is a sharp, narrow spectral line (either an absorption or emission line, depending on whether the electron\u2019s energy increases or decreases in the transition). In the presence of a strong magnetic field, however, each energy level is separated into several levels very close to one another. The separation of the levels is proportional to the strength of the field. As a result, spectral lines formed in the presence of a magnetic field are not single lines but a series of very closely spaced lines corresponding to the subdivisions of the atomic energy levels. This splitting of lines in the presence of a magnetic field is what we call the Zeeman effect (after the Dutch scientist who", " first discovered it in 1896). Measurements of the Zeeman effect in the spectra of the light from sunspot regions show them to have strong magnetic fields (Figure 15.15). Bear in mind that magnets always have a north pole and a south pole. Whenever sunspots are observed in pairs, or in groups containing two principal spots, one of the spots usually has the magnetic polarity of a north-seeking magnetic pole and the other has the opposite polarity. Moreover, during a given cycle, the leading spots of pairs (or leading principle spots of groups) in the Northern Hemisphere all tend to have the same polarity, whereas those in the Southern Hemisphere all tend to have the opposite polarity. 538 Chapter 15 The Sun: A Garden-Variety Star Figure 15.15. Zeeman Effect. These photographs show how magnetic fields in sunspots are measured by means of the Zeeman effect. (left) The vertical black line indicates the position of the spectrograph slit through which light is passed to obtain the spectrum in (right). (credit: modification of work by NSO/AURA/NSF) During the next sunspot cycle, however, the polarity of the leading spots is reversed in each hemisphere. For example, if during one cycle, the leading spots in the Northern Hemisphere all had the polarity of a north- seeking pole, then the leading spots in the Southern Hemisphere would have the polarity of a south-seeking pole. During the next cycle, the leading spots in the Northern Hemisphere would have south-seeking polarity, whereas those in the Southern Hemisphere would have north-seeking polarity. Therefore, strictly speaking, the sunspot cycle does not repeat itself in regard to magnetic polarity until two 11-year cycles have passed. A visual representation of the Sun\u2019s magnetic fields, called a magnetogram, can be used to see the relationship between sunspots and the Sun\u2019s magnetic field (Figure 15.16). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 539 Figure 15.16. Magnetogram and Solar Cycle. In the image on the left, called a magnetogram, we see the magnetic polarity of sunspots. The black areas are where the magnetism is pointing toward the Sun\u2019s core, whereas the white regions are where it is pointing away from the core, toward us. This dramatic sequence on the right shows the activity cycle of the Sun. The 10 maps of the magnetic field on the surface of the Sun span a period of 7.5 years. The two magnetic polarities (N and S) of the magnetic field are shown against a blue disk as dark blue to black (N) and as light blue to white (S). The earliest image, taken on January 8, 1992, is at the lower left and was taken just after solar maximum. Each image, from left to right around the arc, was taken one-half to one year after the preceding one. The last image was taken on July 25, 1999, as the Sun was approaching the next solar maximum. Note a few striking patterns in the magnetic maps: the direction from white to black polarity in the Southern Hemisphere is opposite from that in the Northern Hemisphere. (credit left: modification of work by NASA/SDO; credit right: modification of work by NASA/SOHO) Why is the Sun such a strong and complicated magnet? Astronomers have found that it is the Sun\u2019s dynamo that generates the magnetic field. A dynamo is a machine that converts kinetic energy (i.e., the energy of motion) into electricity. On Earth, dynamos are found in power plants where, for example, the energy from wind or flowing water is used to cause turbines to rotate. In the Sun, the source of kinetic energy is the churning of turbulent layers of ionized gas within the Sun\u2019s interior that we mentioned earlier. These generate electric currents\u2014moving electrons\u2014which in turn generate magnetic fields. Most solar researchers agree that the solar dynamo is located in the convection zone or in the interface layer between the convection zone and the radiative zone below it. As the magnetic fields from the Sun\u2019s dynamo interact, they break, reconnect, and rise through the Sun\u2019s surface. We should say that, although we have good observations that show us how the Sun changes during each solar cycle, it is still very difficult to build physical models of something as complicated as the Sun that can account satisfactorily for why it changes. Researchers have not yet developed a generally accepted model that describes in detail the physical processes that control the solar cycle. Calculations do show that differential rotation (the idea that the Sun rotates at different rates at different latitudes) and convection just below the solar surface can twist and distort the magnetic fields. This causes them to grow and then", " decay, regenerating with opposite polarity approximately every 11 years. The calculations also show that as the fields grow stronger near solar maximum, they flow from the interior of the Sun toward its surface in the form of loops. When a large loop emerges from the solar surface, it creates regions of sunspot activity (Figure 15.17). 540 Chapter 15 The Sun: A Garden-Variety Star Figure 15.17. Magnetic Field Lines Wind Up. Because the Sun spins faster at the equator than near the poles, the magnetic fields in the Sun tend to wind up as shown, and after a while make loops. This is an idealized diagram; the real situation is much more complex. This idea of magnetic loops offers a natural explanation of why the leading and trailing sunspots in an active region have opposite polarity. The leading sunspot coincides with one end of the loop and the trailing spot with the other end. Magnetic fields also hold the key to explaining why sunspots are cooler and darker than the regions without strong magnetic fields. The forces produced by the magnetic field resist the motions of the bubbling columns of rising hot gases. Since these columns carry most of the heat from inside the Sun to the surface by means of convection, and strong magnetic fields inhibit this convection, the surface of the Sun is allowed to cool. As a result, these regions are seen as darker, cooler sunspots. Beyond this general picture, researchers are still trying to determine why the magnetic fields are as large as they are, why the polarity of the field in each hemisphere flips from one cycle to the next, why the length of the solar cycle can vary from one cycle to the next, and why events like the Maunder Minimum occur In this video (https://openstax.org/l/30MagField) solar scientist Holly Gilbert discusses the Sun\u2019s magnetic field. 15.3 SOLAR ACTIVITY ABOVE THE PHOTOSPHERE Learning Objectives By the end of this section, you will be able to: Describe the various ways in which the solar activity cycle manifests itself, including flares, coronal mass ejections, prominences, and plages Sunspots are not the only features that vary during a solar cycle. There are dramatic changes in the chromosphere and corona as well. To see what happens in the chromosphere, we must observe the emission This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 541 lines from elements such as hydrogen and calcium, which emit useful spectral lines at the temperatures in that layer. The hot corona, on the other hand, can be studied by observations of X-rays and of extreme ultraviolet and other wavelengths at high energies. Plages and Prominences As we saw, emission lines of hydrogen and calcium are produced in the hot gases of the chromosphere. Astronomers routinely photograph the Sun through filters that transmit light only at the wavelengths that correspond to these emission lines. Pictures taken through these special filters show bright \u201cclouds\u201d in the chromosphere around sunspots; these bright regions are known as plages (Figure 15.18). These are regions within the chromosphere that have higher temperature and density than their surroundings. The plages actually contain all of the elements in the Sun, not just hydrogen and calcium. It just happens that the spectral lines of hydrogen and calcium produced by these clouds are bright and easy to observe. Figure 15.18. Plages on the Sun. This image of the Sun was taken with a filter that transmits only the light of the spectral line produced by singly ionized calcium. The bright cloud-like regions are the plages. (credit: modification of work by NASA) Moving higher into the Sun\u2019s atmosphere, we come to the spectacular phenomena called prominences (Figure 15.19), which usually originate near sunspots. Eclipse observers often see prominences as red features rising above the eclipsed Sun and reaching high into the corona. Some, the quiescent prominences, are graceful loops of plasma (ionized gas) that can remain nearly stable for many hours or even days. The relatively rare eruptive prominences appear to send matter upward into the corona at high speeds, and the most active surge prominences may move as fast as 1300 kilometers per second (almost 3 million miles per hour). Some eruptive prominences have reached heights of more than 1 million kilometers above the photosphere; Earth would be completely lost inside one of those awesome displays (Figure 15.19). 542 Chapter 15 The Sun: A Garden-Variety Star Figure 15.19. Prominences. (a) This image of an eruptive prominence was taken in the light of singly ionized helium in the extreme ultraviolet part of the spectrum. The prominence is a particularly large one. An image of Earth is shown at the same scale", " for comparison. (b) A prominence is a huge cloud of relatively cool (about 60,000 K in this case), fairly dense gas suspended in the much hotter corona. These pictures, taken in ultraviolet, are color coded so that white corresponds to the hottest temperatures and dark red to cooler ones. The four images were taken, moving clockwise from the upper left, on May 15, 2001; March 28, 2000; January 18, 2000; and February 2, 2001. (credit a: modification of work by NASA/SOHO; credit b: modification of work by NASA/SDO) Flares and Coronal Mass Ejections The most violent event on the surface of the Sun is a rapid eruption called a solar flare (Figure 15.20). A typical flare lasts for 5 to 10 minutes and releases a total amount of energy equivalent to that of perhaps a million hydrogen bombs. The largest flares last for several hours and emit enough energy to power the entire United States at its current rate of electrical consumption for 100,000 years. Near sunspot maximum, small flares occur several times per day, and major ones may occur every few weeks. Figure 15.20. Solar Flare. The bright white area seen on the right side of the Sun in this image from the Solar Dynamics Observer spacecraft is a solar flare that was observed on June 25, 2015. (credit: NASA/SDO) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 543 Flares, like the one shown in Figure 15.21, are often observed in the red light of hydrogen, but the visible emission is only a tiny fraction of the energy released when a solar flare explodes. At the moment of the explosion, the matter associated with the flare is heated to temperatures as high as 10 million K. At such high temperatures, a flood of X-ray and ultraviolet radiation is emitted. Flares seem to occur when magnetic fields pointing in opposite directions release energy by interacting with and destroying each other\u2014much as a stretched rubber band releases energy when it breaks. What is different about flares is that their magnetic interactions cover a large volume in the solar corona and release a tremendous amount of electromagnetic radiation. In some cases, immense quantities of coronal material\u2014mainly protons and electrons\u2014may also be ejected at high speeds (500\u20131000 kilometers per second) into interplanetary space. Such a coronal mass ejection (CME) can affect Earth in a number of ways (which we will discuss in the section on space weather). Figure 15.21. Flare and Coronal Mass Ejection. This sequence of four images shows the evolution over time of a giant eruption on the Sun. (a) The event began at the location of a sunspot group, and (b) a flare is seen in far-ultraviolet light. (c) Fourteen hours later, a CME is seen blasting out into space. (d) Three hours later, this CME has expanded to form a giant cloud of particles escaping from the Sun and is beginning the journey out into the solar system. The white circle in (c) and (d) shows the diameter of the solar photosphere. The larger dark area shows where light from the Sun has been blocked out by a specially designed instrument to make it possible to see the faint emission from the corona. (credit a, b, c, d: modification of work by SOHO/EIT, SOHO/LASCO, SOHO/MDI (ESA & NASA)) See a coronal mass ejection (https://openstax.org/l/30CorMaEj) recorded by the Solar Dynamics Observatory. Active Regions To bring the discussion of the last two sections together, astronomers now realize that sunspots, flares, and bright regions in the chromosphere and corona tend to occur together on the Sun in time and space. That is, they all tend to have similar longitudes and latitudes, but they are located at different heights in the atmosphere. Because they all occur together, they vary with the sunspot cycle. 544 Chapter 15 The Sun: A Garden-Variety Star Figure 15.22. Solar Cycle. This dramatic sequence of images taken from the SOHO satellite over a period of 11 years shows how active regions change during the solar cycle. The images were taken in the ultraviolet region of the spectrum and show that active regions on the Sun increase and decrease during the cycle. Sunspots are located in the cooler photosphere, beneath the hot gases shown in this image, and vary in phase with the emission from these hot gases\u2014more sunspots and more emission from hot gases occur together. (credit: modification of work by ESA/NASA/SOHO) For example, flares are more likely to occur near sunspot maximum, and the corona is much more conspicuous at that time", " (see Figure 15.22). A place on the Sun where a number of these phenomena are seen is called an active region (Figure 15.23). As you might deduce from our earlier discussion, active regions are always associated with strong magnetic fields. Figure 15.23. Solar Active Region Observed at Different Heights in the Sun\u2019s Atmosphere. These four images of a solar flare on October 22, 2012, show from the left: light from the Sun at a wavelength of 171 angstroms, which shows the structure of loops of solar material in the corona; ultraviolet at 304 angstroms, which shows light from the region of the Sun\u2019s atmosphere where flares originate; light at 335 angstroms, which highlights radiation from active regions in the corona; a magnetogram, which shows magnetically active regions on the Sun. Note how these different types of activity all occur above a sunspot region with a strong magnetic field. (credit: modification of work by NASA/SDO/ Goddard) 15.4 SPACE WEATHER Learning Objectives By the end of this section, you will be able to: This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 545 Explain what space weather is and how it affects Earth In the previous sections, we have seen that some of the particles coming off the Sun\u2014either steadily as in the solar wind or in great bursts like CMEs\u2014will reach Earth and its magnetosphere (the zone of magnetic influence that surrounds our planet). As if scientists did not have enough trouble trying to predict weather on Earth, this means that they are now facing the challenge of predicting the effects of solar storms on Earth. This field of research is called space weather; when that weather turns stormy, our technology turns out to be at risk. With thousands of satellites in orbit, astronauts taking up long-term residence in the International Space Station, millions of people using cell phones, GPS, and wireless communication, and nearly everyone relying on the availability of dependable electrical power, governments are now making major investments in trying to learn how to predict when solar storms will occur and how strongly they will affect Earth. Some History What we now study as space weather was first recognized (though not yet understood) in 1859, in what is now known as the Carrington Event. In early September of that year, two amateur astronomers, including Richard Carrington in England, independently observed a solar flare. This was followed a day or two later by a significant solar storm reaching the region of Earth\u2019s magnetic field, which was soon overloaded with charged particles (see Earth as a Planet). As a result, aurora activity was intense and the northern lights were visible well beyond their normal locations near the poles\u2014as far south as Hawaii and the Caribbean. The glowing lights in the sky were so intense that some people reported getting up in the middle of the night, thinking it must be daylight. The 1859 solar storm happened at a time when a new technology was beginning to tie people in the United States and some other countries together: the telegraph system. This was a machine and network for sending messages in code through overhead electrical wires (a bit like a very early version of the internet). The charged particles that overwhelmed Earth\u2019s magnetic field descended toward our planet\u2019s surface and affected the wires of the telegraph system. Sparks were seen coming out of exposed wires and out of the telegraph machines in the system\u2019s offices. The observation of the bright flare that preceded these effects on Earth led to scientific speculation that a connection existed between solar activity and impacts on Earth\u2014this was the beginning of our understanding of what today we call space weather Watch NASA scientists answer some questions (https://openstax.org/l/30SpcWeath) about space weather, and discuss (https://openstax.org/l/30SpcWeath2) some effects it can have in space and on Earth. Sources of Space Weather Three solar phenomena\u2014 coronal holes, solar flares, and CMEs\u2014account for most of the space weather we experience. Coronal holes allow the solar wind to flow freely away from the Sun, unhindered by solar magnetic fields. When the solar wind reaches Earth, as we saw, it causes Earth\u2019s magnetosphere to contract and then expand after the solar wind passes by. These changes can cause (usually mild) electromagnetic disturbances on Earth. 546 Chapter 15 The Sun: A Garden-Variety Star More serious are solar flares, which shower the upper atmosphere of Earth with X-rays, energetic particles, and intense ultraviolet radiation. The X-rays and ultraviolet radiation can ionize atoms in Earth\u2019s upper atmosphere, and the freed electrons can build up a charge on the surface of a spacecraft. When this static charge discharges, it can damage the electronics in the spacecraft\u2014just as you can receive a", " shock when you walk across a carpet in your stocking feet in a dry climate and then touch a light switch or some other metal object. Most disruptive are coronal mass ejections. A CME is an erupting bubble of tens of millions of tons of gas blown away from the Sun into space. When this bubble reaches Earth a few days after leaving the Sun, it heats the ionosphere, which expands and reaches farther into space. As a consequence, friction between the atmosphere and spacecraft increases, dragging satellites to lower altitudes. At the time of a particularly strong flare and CME in March 1989, the system responsible for tracking some 19,000 objects orbiting Earth temporarily lost track of 11,000 of them because their orbits were changed by the expansion of Earth\u2019s atmosphere. During solar maximum, a number of satellites are brought to such a low altitude that they are destroyed by friction with the atmosphere. Both the Hubble Space Telescope and the International Space Station (Figure 15.24) require reboosts to higher altitude so that they can remain in orbit. Figure 15.24. International Space Station. The International Space Station is see above Earth, as photographed in 2010 by the departing crew of the Space Shuttle Atlantis. (credit: NASA) Solar Storm Damage on Earth When a CME reaches Earth, it distorts Earth\u2019s magnetic field. Since a changing magnetic field induces electrical current, the CME accelerates electrons, sometimes to very high speeds. These \u201ckiller electrons\u201d can penetrate deep into satellites, sometimes destroying their electronics and permanently disabling operation. This has happened with some communications satellites. Disturbances in Earth\u2019s magnetic field can cause disruptions in communications, especially cell phone and wireless systems. In fact, disruptions can be expected to occur several times a year during solar maximum. Changes in Earth\u2019s magnetic field due to CMEs can also cause surges in power lines large enough to burn out transformers and cause major power outages. For example, in 1989, parts of Montreal and Quebec Province in Canada were without power for up to 9 hours as a result of a major solar storm. Electrical outages due to CMEs are more likely to occur in North America than in Europe because North America is closer to Earth\u2019s magnetic pole, where the currents induced by CMEs are strongest. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 547 Besides changing the orbits of satellites, CMEs can also distort the signals sent by them. These effects can be large enough to reduce the accuracy of GPS-derived positions so that they cannot meet the limits required for airplane systems, which must know their positions to within 160 feet. Such disruptions caused by CMEs have occasionally forced the Federal Aviation Administration to restrict flights for minutes or, in a few cases, even days. Solar storms also expose astronauts, passengers in high-flying airplanes, and even people on the surface of Earth to increased amounts of radiation. Astronauts, for example, are limited in the total amount of radiation to which they can be exposed during their careers. A single ill-timed solar outburst could end an astronaut\u2019s career. This problem becomes increasingly serious as astronauts spend more time in space. For example, the typical daily dose of radiation aboard the Russian Mir space station was equivalent to about eight chest X-rays. One of the major challenges in planning the human exploration of Mars is devising a way to protect astronauts from high-energy solar radiation. Advance warning of solar storms would help us minimize their disruptive effects. Power networks could be run at less than their full capacity so that they could absorb the effects of power surges. Communications networks could be prepared for malfunctions and have backup plans in place. Spacewalks could be timed to avoid major solar outbursts. Scientists are now trying to find ways to predict where and when flares and CMEs will occur, and whether they will be big, fast events or small, slow ones with little consequence for Earth. The strategy is to relate changes in the appearance of small, active regions and changes in local magnetic<|endoftext|> fields on the Sun to subsequent eruptions. However, right now, our predictive capability is still poor, and so the only real warning we have is from actually seeing CMEs and flares occur. Since a CME travels outward at about 500 kilometers per second, an observation of an eruption provides several days warning at the distance of Earth. However, the severity of the impact on Earth depends on how the magnetic field associated with the CME is oriented relative to Earth\u2019s magnetic field. The orientation can be measured only when the CME flows past a satellite we have put up for this purpose. However, it is located only about an hour upstream from Earth. Space weather predictions are now available online to scientists and the public. Outlooks are given a week ahead, bulletins are issued when there is an", " event that is likely to be of interest to the public, and warnings and alerts are posted when an event is imminent or already under way (Figure 15.25). Figure 15.25. NOAA Space Weather Prediction Operations Center. Bill Murtagh, a space weather forecaster, leads a workshop on preparedness for events like geomagnetic storms. (credit: modification of work by FEMA/Jerry DeFelice) 548 Chapter 15 The Sun: A Garden-Variety Star To find public information and alerts about space weather, you can turn to the National Space Weather Prediction Center (https://openstax.org/l/30NSWPC) or SpaceWeather (https://openstax.org/l/ 30SpcWeath3) for consolidated information from many sources. Fortunately, we can expect calmer space weather for the next few years, since the most recent solar maximum, which was relatively weak, occurred in 2014, and scientists believe the current solar cycle to be one of the least active in recent history. We expect more satellites to be launched that will allow us to determine whether CMEs are headed toward Earth and how big they are. Models are being developed that will then allow scientists to use early information about the CME to predict its likely impact on Earth. The hope is that by the time of the next maximum, solar weather forecasting will have some of the predictive capability that meteorologists have achieved for terrestrial weather at Earth\u2019s surface. However, the most difficult events to predict are the largest and most damaging storms\u2014hurricanes on Earth and extreme, rare storm events on the Sun. Thus, it is inevitable that the Sun will continue to surprise us The Timing of Solar Events A basic equation is useful in figuring out when events on the Sun will impact Earth: distance = velocity \u00d7 time, or D = v \u00d7 t Dividing both sides by v, we get T = D/v Suppose you observe a major solar flare while astronauts are orbiting Earth. If the average speed of solar wind is 400 km/s and the distance to the Sun as 1.496 \u00d7 108 km, how long it will before the charged particles ejected from the Sun during the flare reach the space station? Solution The time required for solar wind particles to reach Earth is T = D/v. 1.496 \u00d7 108 km 400 km/s = 3.74 \u00d7 105 s, or 3.74 \u00d7105 s 60 s/min \u00d7 60 min/h \u00d7 24 h/d = 4.3 d Check Your Learning How many days would it take for the particles to reach Earth if the solar wind speed increased to 500 km/s? Answer: 1.496 \u00d7 108 km 500 km/s = 2.99 \u00d7 105 s, or 2.99 \u00d7105 s 60 s/min \u00d7 60 min/h \u00d7 24 h/d = 3.46 d This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 549 Earth\u2019s Climate and the Sunspot Cycle: Is There a Connection? While the Sun rises faithfully every day at a time that can be calculated precisely, scientists have determined that the Sun\u2019s energy output is not truly constant but varies over the centuries by a small amount\u2014probably less than 1%. We\u2019ve seen that the number of sunspots varies, with the time between sunspot maxima of about 11 years, and that the number of sunspots at maximum is not always the same. Considerable evidence shows that between the years 1645 and 1715, the number of sunspots, even at sunspot maximum, was much lower than it is now. This interval of significantly low sunspot numbers was first noted by Gustav Sp\u04e7rer in 1887 and then by E. W. Maunder in 1890; it is now called the Maunder Minimum. The variation in the number of sunspots over the past three centuries is shown in Figure 15.26. Besides the Maunder Minimum in the seventeenth century, sunspot numbers were somewhat lower during the first part of the nineteenth century than they are now; this period is called the Little Maunder Minimum. Figure 15.26. Numbers of Sunspots over Time. This diagram shows how the number of sunspots has changed with time since counts of the numbers of spots began to be recorded on a consistent scale. Note the low number of spots during the early years of the nineteenth century, the Little Maunder Minimum. (credit: modification of work by NASA/ARC) When the number of sunspots is high, the Sun is active in various other ways as well, and, as we will see in several sections below, some of this activity affects Earth directly. For example, there are more auroral displays when the sunspot number is high. Auroras are caused when energetically charged particles from the Sun interact with Earth\u2019s magnetosphere, and", " the Sun is more likely to spew out particles when it is active and the sunspot number is high. Historical accounts also indicate that auroral activity was abnormally low throughout the several decades of the Maunder Minimum. The Maunder Minimum was a time of exceptionally low temperatures in Europe\u2014so low that this period is described as the Little Ice Age. This coincidence in time caused scientists to try to understand whether small changes in the Sun could affect the climate on Earth. There is clear evidence that it was unusually cold in Europe during part of the seventeenth century. The River Thames in London froze at least 11 times, ice appeared in the oceans off the coasts of southeast England, and low summer temperatures led to short growing seasons and poor harvests. However, whether and how changes on the Sun on this timescale influence Earth\u2019s climate is still a matter of debate among scientists. Other small changes in climate like the Little Ice Age have occurred and have had their impacts on human history. For example, explorers from Norway first colonized Iceland and then reached Greenland by 986. From there, they were able to make repeated visits to the northeastern coasts of North America, including Newfoundland, between about 1000 and 1350. (The ships of the time did not allow the Norse explorers to travel all the way to North America directly, but only from Greenland, which served as a station for further exploration.) Most of Greenland is covered by ice, and the Greenland station was never self-sufficient; rather, it depended on imports of food and other goods from Norway for its survival. When a little ice age began in the thirteenth century, voyaging became very difficult, and support of the Greenland colony was no longer possible. The last- 550 Chapter 15 The Sun: A Garden-Variety Star known contact with it was made by a ship from Iceland blown off course in 1410. When European ships again began to visit Greenland in 1577, the entire colony there had disappeared. The estimated dates for these patterns of migration follow what we know about solar activity. Solar activity was unusually high between 1100 and 1250, which includes the time when the first European contacts were made with North America. Activity was low from 1280 to 1340 and there was a little ice age, which was about the time regular contact with North America and between Greenland and Europe stopped. One must be cautious, however, about assuming that low sunspot numbers or variations in the Sun\u2019s output of energy caused the Little Ice Age. There is no satisfactory model that can explain how a reduction in solar activity might cause cooler temperatures on Earth. An alternative possibility is that the cold weather during the Little Ice Age was related to volcanic activity. Volcanoes can eject aerosols (tiny droplets or particles) into the atmosphere that efficiently reflect sunlight. Observations show, for example, that the Pinatubo eruption in 1991 ejected SO2 aerosols into the atmosphere, which reduced the amount of sunlight reaching Earth\u2019s surface enough to lower global temperatures by 0.4 \u00b0C. Satellite data show that the energy output from the Sun during a solar cycle varies by only about 0.1%. We know of no physical process that would explain how such a small variation could cause global temperature changes. The level of solar activity may, however, have other effects. For example, although the Sun\u2019s total energy output varies by only 0.1% during a solar cycle, its extreme ultraviolet radiation is 10 times higher at times of solar maximum than at solar minimum. This large variation can affect the chemistry and temperature structure of the upper atmosphere. One effect might be a reduction in the ozone layer and a cooling of the stratosphere near Earth\u2019s poles. This, in turn, could change the circulation patterns of winds aloft and, hence, the tracks of storms. There is some recent evidence that variations in regional rainfall correlate better with solar activity than does the global temperature of Earth. But, as you can see, the relationship between what happens on the Sun and what happens to Earth\u2019s climate over the short term is still an area that scientists are investigating and debating. Whatever the effects of solar activity may be on local rainfall or temperature patterns, we want to emphasize one important idea: Our climate change data and the models developed to account for the data consistently show that solar variability is not the cause of the global warming that has occurred during the past 50 years. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 551 CHAPTER 15 REVIEW KEY TERMS active region an area on the Sun where magnetic fields are concentrated; sunspots, prominences, flares, and CMEs all tend to occur in active regions aurora light radiated by atoms and ions in the ionosphere excited by charged particles from the Sun, mostly seen in the magnetic polar regions chromosphere the part of the solar atmosphere that lies immediately above the photospheric layers corona (", "of the Sun) the outer (hot) atmosphere of the Sun coronal hole a region in the Sun\u2019s outer atmosphere that appears darker because there is less hot gas there coronal mass ejection (CME) a solar flare in which immense quantities of coronal material\u2014mainly protons and electrons\u2014is ejected at high speeds (500\u20131000 kilometers per second) into interplanetary space differential rotation the phenomenon that occurs when different parts of a rotating object rotate at different rates at different latitudes granulation the rice-grain-like structure of the solar photosphere; granulation is produced by upwelling currents of gas that are slightly hotter, and therefore brighter, than the surrounding regions, which are flowing downward into the Sun Maunder Minimum a period during the eighteenth century when the number of sunspots seen throughout the solar cycle was unusually low photosphere the region of the solar (or stellar) atmosphere from which continuous radiation escapes into space plage a bright region of the solar surface observed in the light of some spectral line plasma a hot ionized gas prominence a large, bright, gaseous feature that appears above the surface of the Sun and extends into the corona solar flare a sudden and temporary outburst of electromagnetic radiation from an extended region of the Sun\u2019s surface solar wind a flow of hot, charged particles leaving the Sun sunspot large, dark features seen on the surface of the Sun caused by increased magnetic activity sunspot cycle the semiregular 11-year period with which the frequency of sunspots fluctuates transition region relatively low temperatures that characterize the chromosphere to the high temperatures of the corona the region in the Sun\u2019s atmosphere where the temperature rises very rapidly from the 552 Chapter 15 The Sun: A Garden-Variety Star SUMMARY 15.1 The Structure and Composition of the Sun The Sun, our star, has several layers beneath the visible surface: the core, radiative zone, and convective zone. These, in turn, are surrounded by a number of layers that make up the solar atmosphere. In order of increasing distance from the center of the Sun, they are the photosphere, with a temperature that ranges from 4500 K to about 6800 K; the chromosphere, with a typical temperature of 104 K; the transition region, a zone that may be only a few kilometers thick, where the temperature increases rapidly from 104 K to 106 K; and the corona, with temperatures of a few million K. The Sun\u2019s surface is mottled with upwelling convection currents seen as hot, bright granules. Solar wind particles stream out into the solar system through coronal holes. When such particles reach the vicinity of Earth, they produce auroras, which are strongest near Earth\u2019s magnetic poles. Hydrogen and helium together make up 98% of the mass of the Sun, whose composition is much more characteristic of the universe at large than is the composition of Earth. 15.2 The Solar Cycle Sunspots are dark regions where the temperature is up to 2000 K cooler than the surrounding photosphere. Their motion across the Sun\u2019s disk allows us to calculate how fast the Sun turns on its axis. The Sun rotates more rapidly at its equator, where the rotation period is about 25 days, than near the poles, where the period is slightly longer than 36 days. The number of visible sunspots varies according to a sunspot cycle that averages 11 years in length. Spots frequently occur in pairs. During a given 11-year cycle, all leading spots in the Northern Hemisphere have the same magnetic polarity, whereas all leading sports in the Southern Hemisphere have the opposite polarity. In the subsequent 11-year cycle, the polarity reverses. For this reason, the magnetic activity cycle of the Sun is understood to last for 22 years. This activity cycle is connected with the behavior of the Sun\u2019s magnetic field, but the exact mechanism is not yet understood. 15.3 Solar Activity above the Photosphere Signs of more intense solar activity, an increase in the number of sunspots, as well as prominences, plages, solar flares, and coronal mass ejections, all tend to occur in active regions\u2014that is, in places on the Sun with the same latitude and longitude but at different heights in the atmosphere. Active regions vary with the solar cycle, just like sunspots do. 15.4 Space Weather Space weather is the effect of solar activity on our own planet, both in our magnetosphere and on Earth\u2019s surface. Coronal holes allow more of the Sun\u2019s material to flow out into space. Solar flares and coronal mass ejections can cause auroras, disrupt communications, damage satellites, and cause power outages on Earth. FOR FURTHER EXPLORATION Articles Berman, B. \u201cHow Solar Storms Could Shut Down Earth.\u201d Astronomy (September 2013): 22. Up-to-date review of how events on the", " Sun can hurt our civilization. Frank, A. \u201cBlowin\u2019 in the Solar Wind.\u201d Astronomy (October 1998): 60. On results from the SOHO spacecraft. Holman, G. \u201cThe Mysterious Origins of Solar Flares.\u201d Scientific American (April 2006): 38. New ideas involving magnetic reconnection and new observations of flares. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 553 James, C. \u201cSolar Forecast: Storm Ahead.\u201d Sky & Telescope (July 2007): 24. Nice review on the effects of the Sun\u2019s outbursts and on Earth and how we monitor \u201cspace weather.\u201d Schaefer, B. \u201cSunspots That Changed the World.\u201d Sky & Telescope (April 1997): 34. Historical events connected with sunspots and solar activity. Schrijver, C. and Title, A. \u201cToday\u2019s Science of the Sun.\u201d Sky & Telescope (February 2001): 34; (March 2001): 34. Excellent reviews of recent results about the solar atmosphere. Wadhwa, M. \u201cOrder from Chaos: Genesis Samples the Solar Wind.\u201d Astronomy (October 2013): 54. On a satellite that returned samples of the Sun\u2019s wind. Websites Dr. Sten Odenwald\u2019s \u201cSolar Storms\u201d site: http://www.solarstorms.org/. ESA/NASA\u2019s Solar & Heliospheric Observatory: http://sohowww.nascom.nasa.gov. A satellite mission with a rich website to explore. High Altitude Observatory Introduction to the Sun: http://www.hao.ucar.edu/education/basic.php. For beginners. NASA\u2019s Solar Missions: https://www.nasa.gov/mission_pages/sunearth/missions/index.html. Good summary of the many satellites and missions NASA has. NOAA Profile of Space Weather: http://www.swpc.noaa.gov/sites/default/files/images/u33/ primer_2010_new.pdf. A primer. NOAA Space Weather Prediction Center Information Pages: http://www.swpc.noaa.gov/content/education- and-outreach. Includes primers, videos, a curriculum and training modules. Nova Sun Lab: http://www.pbs.org/wgbh/nova/labs/lab/sun/. Videos, scientist profiles, a research challenge related to the active Sun from the PBS science program. Space Weather: Storms on the Sun: http://www.swpc.noaa.gov/sites/default/files/images/u33/ swx_booklet.pdf. An illustrated booklet from NOAA. Stanford Solar Center: http://solar-center.stanford.edu/. An excellent site with information for students and teachers. Apps These can tell you and your students more about what\u2019s happening on the Sun in real time. NASA\u2019s 3-D Sun: http://3dsun.org/. NASA Space Weather: https://itunes.apple.com/us/app/nasa-space-weather/id422621403?mt=8. Solaris Alpha: https://play.google.com/store/apps/details?id=com.tomoreilly.solarisalpha. Solar Monitor Pro: http://www.solarmonitor.eu/. Videos Journey into the Sun: https://www.youtube.com/watch?v=fqKFQ7z0Nuk. 2010 KQED Quest TV Program mostly about the Solar Dynamics Observatory spacecraft, its launch and capabilities, but with good general information on how the Sun works (12:24). 554 Chapter 15 The Sun: A Garden-Variety Star NASA | SDO: Three Years in Three Minutes--With Expert Commentary: https://www.youtube.com/ watch?v=QaCG0wAjJSY&src. Video of 3 years of observations of the Sun by the Solar Dynamics Observatory made into a speeded up movie, with commentary by solar physicist Alex Young (5:03). Our Explosive Sun: http://www.youtube.com/watch?v=kI6YGSIJqrE. Video of a 2011 public lecture in the Silicon Valley Astronomy Lecture Series by Dr. Thomas Berger about solar activity and recent satellite missions to observe and understand it (1:20:22). Out There Raining Fire: http://www.nytimes.com/video/science/100000003489464/out-there-raining- fire.html?emc=eta1. Nice overview and introduction to the Sun by science reporter Dennis Over", "bye of the NY Times (2:28) Space Weather Impacts: http://www.swpc.noaa.gov/content/education-and-outreach. Video from NOAA (2:47); https://www.youtube.com/playlist?list=PLBdd8cMH5jFmvVR2sZubIUzBO6JI0Pvx0. Videos from the National Weather Service (four short videos) (14:41). Space Weather: Storms on the Sun: http://www.youtube.com/watch?v=vWsmp4o-qVg. Science bulletin from the American Museum of Natural History, giving the background to what happens on the Sun to cause space weather (6:10). Sun Storms: http://www.livescience.com/11754-sun-storms-havoc-electronic-world.html. From the Starry Night company about storms from the Sun now and in the past (4:49). Sunspot Group AR 2339 Crosses the Sun: http://apod.nasa.gov/apod/ap150629.html. Short video (with music) animates Solar Dynamics Observatory images of an especially large sunspot group going across the Sun\u2019s face (1:15). What Happens on the Sun Doesn\u2019t Stay on the Sun: https://www.youtube.com/watch?v=bg_gD2-ujCk. From the National Oceanic and Atmospheric Administration: introduction to the Sun, space weather, its effects, and how we monitor it (4:56). COLLABORATIVE GROUP ACTIVITIES A. Have your group make a list of all the ways the Sun personally affects your life on Earth. (Consider the everyday effects as well as the unusual effects due to high solar activity.) B. Long before the nature of the Sun was fully understood, astronomer (and planet discoverer) William Herschel (1738\u20131822) proposed that the hot Sun may have a cool interior and may be inhabited. Have your group discuss this proposal and come up with modern arguments against it. C. We discussed how the migration of Europeans to North America was apparently affected by short-term climate change. If Earth were to become significantly hotter, either because of changes in the Sun or because of greenhouse warming, one effect would be an increase in the rate of melting of the polar ice caps. How would this affect modern civilization? D. Suppose we experience another Maunder Minimum on Earth, and it is accompanied by a drop in the average temperature like the Little Ice Age in Europe. Have your group discuss how this would affect civilization and international politics. Make a list of the most serious effects that you can think of. E. Watching sunspots move across the disk of the Sun is one way to show that our star rotates on its axis. Can your group come up with other ways to show the Sun\u2019s rotation? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 555 F. Suppose in the future, we are able to forecast space weather as well as we forecast weather on Earth. And suppose we have a few days of warning that a big solar storm is coming that will overload Earth\u2019s magnetosphere with charged particles and send more ultraviolet and X-rays toward our planet. Have your group discuss what steps we might take to protect our civilization? G. Have your group members research online to find out what satellites are in space to help astronomers study the Sun. In addition to searching for NASA satellites, you might also check for satellites launched by the European Space Agency and the Japanese Space Agency. H. Some scientists and engineers are thinking about building a \u201csolar sail\u201d\u2014something that can use the Sun\u2019s wind or energy to propel a spacecraft away from the Sun. The Planetary Society is a nonprofit organization that is trying to get solar sails launched, for example. Have your group do a report on the current state of solar-sail projects and what people are dreaming about for the future. EXERCISES Review Questions 1. Describe the main differences between the composition of Earth and that of the Sun. 2. Describe how energy makes its way from the nuclear core of the Sun to the atmosphere. Include the name of each layer and how energy moves through the layer. 3. Make a sketch of the Sun\u2019s atmosphere showing the locations of the photosphere, chromosphere, and corona. What is the approximate temperature of each of these regions? 4. Why do sunspots look dark? 5. Which aspects of the Sun\u2019s activity cycle have a period of about 11 years? Which vary during intervals of about 22 years? 6. Summarize the evidence indicating that over several hundreds of years or more there have been variations", " in the level of the solar activity. 7. What it the Zeeman effect and what does it tell us about the Sun? 8. Explain how the theory of the Sun\u2019s dynamo results in an average 22-year solar activity cycle. Include the location and mechanism for the dynamo. 9. Compare and contrast the four different types of solar activity above the photosphere. 10. What are the two sources of particles coming from the Sun that cause space weather? How are they different? 11. How does activity on the Sun affect human technology on Earth and in the rest of the solar system? 12. How does activity on the Sun affect natural phenomena on Earth? Thought Questions 13. Table 15.1 indicates that the density of the Sun is 1.41 g/cm3. Since other materials, such as ice, have similar densities, how do you know that the Sun is not made of ice? 556 Chapter 15 The Sun: A Garden-Variety Star 14. Starting from the core of the Sun and going outward, the temperature decreases. Yet, above the photosphere, the temperature increases. How can this be? 15. Since the rotation period of the Sun can be determined by observing the apparent motions of sunspots, a correction must be made for the orbital motion of Earth. Explain what the correction is and how it arises. Making some sketches may help answer this question. 16. Suppose an (extremely hypothetical) elongated sunspot forms that extends from a latitude of 30\u00b0 to a latitude of 40\u00b0 along a fixed of longitude on the Sun. How will the appearance of that sunspot change as the Sun rotates? (Figure 15.17 should help you figure this out.) 17. The text explains that plages are found near sunspots, but Figure 15.18 shows that they appear even in areas without sunspots. What might be the explanation for this? 18. Why would a flare be observed in visible light, when they are so much brighter in X-ray and ultraviolet light? 19. How can the prominences, which are so big and \u2018float\u2019 in the corona, stay gravitationally attached to the Sun while flares can escape? 20. If you were concerned about space weather and wanted to avoid it, where would be the safest place on Earth for you to live? 21. Suppose you live in northern Canada and an extremely strong flare is reported on the Sun. What precautions might you take? What might be a positive result? Figuring For Yourself 22. The edge of the Sun doesn\u2019t have to be absolutely sharp in order to look that way to us. It just has to go from being transparent to being completely opaque in a distance that is smaller than your eye can resolve. Remember from Astronomical Instruments that the ability to resolve detail depends on the size of the telescope\u2019s aperture. The pupil of your eye is very small relative to the size of a telescope and therefore is very limited in the amount of detail you can see. In fact, your eye cannot see details that are smaller than 1/30 of the diameter of the Sun (about 1 arcminute). Nearly all the light from the Sun emerges from a layer that is only about 400 km thick. What fraction is this of the diameter of the Sun? How does this compare with the ability of the human eye to resolve detail? Suppose we could see light emerging directly from a layer that was 300,000 km thick. Would the Sun appear to have a sharp edge? 23. Show that the statement that 92% of the Sun\u2019s atoms are hydrogen is consistent with the statement that 73% of the Sun\u2019s mass is made up of hydrogen, as found in Table 15.2. (Hint: Make the simplifying assumption, which is nearly correct, that the Sun is made up entirely of hydrogen and helium.) 24. From Doppler shifts of the spectral lines in the light coming from the east and west edges of the Sun, astronomers find that the radial velocities of the two edges differ by about 4 km/s, meaning that the Sun\u2019s rotation rate is 2 km/s. Find the approximate period of rotation of the Sun in days. The circumference of a sphere is given by 2\u03c0R, where R is the radius of the sphere. 25. Assuming an average sunspot cycle of 11 years, how many revolutions does the equator of the Sun make during that one cycle? Do higher latitudes make more or fewer revolutions compared to the equator? 26. This chapter gives the average sunspot cycle as 11 years. Verify this using Figure 15.26. 27. The escape velocity from any astronomical object can be calculated as vescape = 2GM/R. Using the data in Appendix E, calculate the escape velocity from the photosphere of the Sun. Since coronal mass ejections escape from the corona, would the escape velocity from there be more or less than from the photosphere? This OpenStax book", " is available for free at http://cnx.org/content/col11992/1.8 Chapter 15 The Sun: A Garden-Variety Star 557 28. Suppose you observe a major solar flare while astronauts are orbiting Earth. Use the data in the text to calculate how long it will before the charged particles ejected from the Sun during the flare reach them. 29. Suppose an eruptive prominence rises at a speed of 150 km/s. If it does not change speed, how far from the photosphere will it extend after 3 hours? How does this distance compare with the diameter of Earth? 30. From the information in Figure 15.21, estimate the speed with which the particles in the CME in parts (c) and (d) are moving away from the Sun. 558 Chapter 15 The Sun: A Garden-Variety Star This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 559 16 THE SUN: A NUCLEAR POWERHOUSE Figure 16.1. The Sun. It takes an incredible amount of energy for the Sun to shine, as it has and will continue to do for billions of years. (credit: modification of work by Ed Dunens) Chapter Outline 16.1 Sources of Sunshine: Thermal and Gravitational Energy 16.2 Mass, Energy, and the Theory of Relativity 16.3 The Solar Interior: Theory 16.4 The Solar Interior: Observations Thinking Ahead The Sun puts out an incomprehensible amount of energy\u2014so much that its ultraviolet radiation can cause sunburns from 93 million miles away. It is also very old. As you learned earlier, evidence shows that the Sun formed about 4.5 billion years ago and has been shining ever since. How can the Sun produce so much energy for so long? The Sun\u2019s energy output is about 4 \u00d7 1026 watts. This is unimaginably bright: brighter than a trillion cities together each with a trillion 100-watt light bulbs. Most known methods of generating energy fall far short of the capacity of the Sun. The total amount of energy produced over the entire life of the Sun is staggering, since the Sun has been shining for billions of years. Scientists were unable to explain the seemingly unlimited energy of stars like the Sun prior to the twentieth century. 16.1 SOURCES OF SUNSHINE: THERMAL AND GRAVITATIONAL ENERGY Learning Objectives By the end of this section, you will be able to: Identify different forms of energy Understand the law of conservation of energy 560 Chapter 16 The Sun: A Nuclear Powerhouse Explain ways that energy can be transformed Energy is a challenging concept to grasp because it exists in so many different forms that it defies any single simple explanation. In many ways, comprehending energy is like comprehending wealth: There are very different forms of wealth and they follow different rules, depending on if they are the stock market, real estate, a collection of old comic books, great piles of cash, or one of the many other ways to make and lose money. It is easier to discuss one or two forms of wealth\u2014or energy\u2014than to discuss that concept in general. When striving to understand how the Sun can put out so much energy for so long, scientists considered many different types of energy. Nineteenth-century scientists knew of two possible sources for the Sun\u2019s energy: chemical and gravitational energy. The source of chemical energy most familiar to them was the burning (the chemical term is oxidation) of wood, coal, gasoline, or other fuel. We know exactly how much energy the burning of these materials can produce. We can thus calculate that even if the immense mass of the Sun consisted of a burnable material like coal or wood, our star could not produce energy at its present rate for more than few thousand years. However, we know from geologic evidence that water was present on Earth\u2019s surface nearly 4 billion years ago, so the Sun must have been shining brightly (and making Earth warm) at least as long as that. Today, we also know that at the temperatures found in the Sun, nothing like solid wood or coal could survive What\u2019s Watt? Just a word about the units we are using. A watt (W) is a unit of power, which is energy used or given off per unit time. It is measured in joules per second (J/s). You know from your everyday experience that it is not just how much energy you expend, but how long you take to do it. (Burning 10 Calories in 10 minutes requires a very different kind of exercise than burning those 10 Calories in an hour.) Watts tell you the rate at which energy is being used; for example, a 100-watt bulb uses 100 joules (J) of energy every second. And how big is a joule? A 73-kilogram (160-pound) astronomy instructor", " running at about 4.4 meters per second (10 miles per hour) because he is late for class has a motion energy of about 700 joules. Conservation of Energy Other nineteenth-century attempts to determine what makes the Sun shine used the law of conservation of energy. Simply stated, this law says that energy cannot be created or destroyed, but can be transformed from one type to another, such as from heat to mechanical energy. The steam engine, which was key to the Industrial Revolution, provides a good example. In this type of engine, the hot steam from a boiler drives the movement of a piston, converting heat energy into motion energy. Conversely, motion can be transformed into heat. If you clap your hands vigorously at the end of an especially good astronomy lecture, your palms become hotter. If you rub ice on the surface of a table, the heat produced by friction melts the ice. The brakes on cars use friction to reduce speed, and in the process, transform motion energy into heat energy. That is why after bringing a car to a stop, the brakes can be very hot; this also explains why brakes can overheat when used carelessly while descending long mountain roads. In the nineteenth century, scientists thought that the source of the Sun\u2019s heat might be the mechanical motion of meteorites falling into it. Their calculations showed, however, that in order to produce the total amount of This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 561 energy emitted by the Sun, the mass in meteorites that would have to fall into the Sun every 100 years would equal the mass of Earth. The resulting increase in the Sun\u2019s mass would, according to Kepler\u2019s third law, change the period of Earth\u2019s orbit by 2 seconds per year. Such a change would be easily measurable and was not, in fact, occurring. Scientists could then disprove this as the source of the Sun\u2019s energy. Gravitational Contraction as a Source of Energy Proposing an alternative explanation, British physicist Lord Kelvin and German scientist Hermann von Helmholtz (Figure 16.2), in about the middle of the nineteenth century, proposed that the Sun might produce energy by the conversion of gravitational energy into heat. They suggested that the outer layers of the Sun might be \u201cfalling\u201d inward because of the force of gravity. In other words, they proposed that the Sun could be shrinking in size, staying hot and bright as a result. Figure 16.2. Kelvin (1824\u20131907) and Helmholtz (1821\u20131894). (a) British physicist William Thomson (Lord Kelvin) and (b) German scientist Hermann von Helmholtz proposed that the contraction of the Sun under its own gravity might account for its energy. (credit a: modification of work by Wellcome Library, London; credit b: modification of work by Wellcome Library, London) To imagine what would happen if this hypothesis were true, picture the outer layer of the Sun starting to fall inward. This outer layer is a gas made up of individual atoms, all moving about in random directions. If a layer falls inward, the atoms acquire an additional speed because of falling motion. As the outer layer falls inward, it also contracts, moving the atoms closer together. Collisions become more likely, and some of them transfer the extra speed associated with the falling motion to other atoms. This, in turn, increases the speeds of those atoms. The temperature of a gas is a measure of the kinetic energy (motion) of the atoms within it; hence, the temperature of this layer of the Sun increases. Collisions also excite electrons within the atoms to higher-energy orbits. When these electrons return to their normal orbits, they emit photons, which can then escape from the Sun (see Radiation and Spectra). Kelvin and Helmholtz calculated that a contraction of the Sun at a rate of only about 40 meters per year would be enough to produce the amount of energy that it is now radiating. Over the span of human history, the decrease in the Sun\u2019s size from such a slow contraction would be undetectable. If we assume that the Sun began its life as a large, diffuse cloud of gas, then we can calculate how much energy has been radiated by the Sun during its entire lifetime as it has contracted from a very large diameter to its present size. The amount of energy is on the order of 1042 joules. Since the solar luminosity is 4 \u00d7 1026 watts 562 Chapter 16 The Sun: A Nuclear Powerhouse (joules/second) or about 1034 joules per year, contraction could keep the Sun shining at its present rate for roughly 100 million years. In the nineteenth century, 100 million years at first seemed plenty long enough, since Earth was then widely thought to be much younger than this. But toward the end of that century and", " into the twentieth, geologists and physicists showed that Earth (and, hence, the Sun) is actually much older. Contraction therefore cannot be the primary source of solar energy (although, as we shall see in The Birth of Stars and the Discovery of Planets Outside the Solar System, contraction is an important source of energy for a while in stars that are just being born). Scientists were thus confronted with a puzzle of enormous proportions. Either an unknown type of energy was responsible for the most important energy source known to humanity, or estimates of the age of the solar system (and life on Earth) had to be seriously modified. Charles Darwin, whose theory of evolution required a longer time span than the theories of the Sun seemed to permit, was discouraged by these results and continued to worry about them until his death in 1882. It was only in the twentieth century that the true source of the Sun\u2019s energy was identified. The two key pieces of information required to solve the puzzle were the structure of the nucleus of the atom and the fact that mass can be converted into energy. 16.2 MASS, ENERGY, AND THE THEORY OF RELATIVITY Learning Objectives By the end of this section, you will be able to: Explain how matter can be converted into energy Describe the particles that make up atoms Describe the nucleus of an atom Understand the nuclear forces that hold atoms together Trace the nuclear reactions in the solar interior As we have seen, energy cannot be created or destroyed, but only converted from one form to another. One of the remarkable conclusions derived by Albert Einstein (see Albert Einstein) when he developed his theory of relativity is that matter can be considered a form of energy too and can be converted into energy. Furthermore, energy can also be converted into matter. This seemed to contradict what humans had learned over thousands of years by studying nature. Matter is something we can see and touch, whereas energy is something objects have when they do things like move or heat up. The idea that matter or energy can be converted into each other seemed as outrageous as saying you could accelerate a car by turning the bumper into more speed, or that you could create a bigger front seat by slowing down your car. That would be pretty difficult to believe; yet, the universe actually works somewhat like that. Converting Matter into Energy The remarkable equivalence between matter and energy is given in one of the most famous equations: E = mc2 In this equation, E stands for energy, m stands for mass, and c, the constant that relates the two, is the speed of light (3 \u00d7 108 meters per second). Note that mass is a measure of the quantity of matter, so the significance of this equation is that matter can be converted into energy and energy can be converted into matter. Let\u2019s compare this equation of converting matter and energy to some common conversion equations that have the same form: This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 563 inches = feet \u00d7 12, or cents = dollars \u00d7 100 Just as each conversion formula allows you to calculate the conversion of one thing into another, when we convert matter into energy, we consider how much mass the matter has. The conversion factor in this case turns out not to be either 12 or 100, as in our examples, but another constant quantity: the speed of light squared. Note that matter does not have to travel at the speed of light (or the speed of light squared) for this conversion to occur. The factor of c2 is just the number that Einstein showed must be used to relate mass and energy. Notice that this formula does not tell us how to convert mass into energy, just as the formula for cents does not tell us where to exchange coins for a dollar bill. The formulas merely tell us what the equivalent values are if we succeed in making the conversion. When Einstein first derived his formula in 1905, no one had the faintest idea how to convert mass into energy in any practical way. Einstein himself tried to discourage speculation that the large-scale conversion of atomic mass into energy would be feasible in the near future. Today, as a result of developments in nuclear physics, we regularly convert mass into energy in power plants, nuclear weapons, and high-energy physics experiments in particle accelerators. Because the speed of light squared (c2) is a very large quantity, the conversion of even a small amount of mass results in a very large amount of energy. For example, the complete conversion of 1 gram of matter (about 1/ 28 ounce, or approximately 1 paperclip) would produce as much energy as the burning of 15,000 barrels of oil. Scientists soon realized that the conversion of mass into energy is the source of the Sun\u2019s heat and light. With Einstein\u2019s equation of E = mc2, we can calculate that the amount of energy radiated by the Sun could be produced by the complete conversion of about 4 million tons", " of matter into energy inside the Sun each second. Destroying 4 million tons per second sounds like a lot when compared to earthly things, but bear in mind that the Sun is a very big reservoir of matter. In fact, we will see that the Sun contains more than enough mass to destroy such huge amounts of matter and still continue shining at its present rate for billions of years. But knowing all that still does not tell us how mass can be converted into energy. To understand the process that actually occurs in the Sun, we need to explore the structure of the atom a bit further Albert Einstein For a large part of his life, Albert Einstein (Figure 16.3) was one of the most recognized celebrities of his day. Strangers stopped him on the street, and people all over the world asked him for endorsements, advice, and assistance. In fact, when Einstein and the great film star Charlie Chaplin met in California, they found they shared similar feelings about the loss of privacy that came with fame. Einstein\u2019s name was a household word despite the fact that most people did not understand the ideas that had made him famous. Einstein was born in 1879 in Ulm, Germany. Legend has it that he did not do well in school (even in arithmetic), and thousands of students have since attempted to justify a bad grade by referring to this story. Alas, like many legends, this one is not true. Records indicate that although he tended to rebel against the authoritarian teaching style in vogue in Germany at that time, Einstein was a good student. After graduating from the Federal Polytechnic Institute in Zurich, Switzerland, Einstein at first had trouble getting a job (even as a high school teacher), but he eventually became an examiner in the Swiss Patent Office. Working in his spare time, without the benefit of a university environment but using his superb 564 Chapter 16 The Sun: A Nuclear Powerhouse physical intuition, he wrote four papers in 1905 that would ultimately transform the way physicists looked at the world. One of these, which earned Einstein the Nobel Prize in 1921, set part of the foundation of quantum mechanics\u2014the rich, puzzling, and remarkable theory of the subatomic realm. But his most important paper presented the special theory of relativity, a reexamination of space, time, and motion that added a whole new level of sophistication to our understanding of those concepts. The famed equation E = mc2 was actually a relatively minor part of this theory, added in a later paper. In 1916, Einstein published his general theory of relativity, which was, among other things, a fundamentally new description of gravity (see Black Holes and Curved Spacetime). When this theory was confirmed by measurements of the \u201cbending of starlight\u201d during a 1919 eclipse (The New York Times headline read, \u201cLights All Askew in the Heavens\u201d), Einstein became world famous. In 1933, to escape Nazi persecution, Einstein left his professorship in Berlin and settled in the United States at the newly created Institute for Advanced Studies at Princeton. He remained there until his death in 1955, writing, lecturing, and espousing a variety of intellectual and political causes. For example, he agreed to sign a letter written by Leo Szilard and other scientists in 1939, alerting President Roosevelt to the dangers of allowing Nazi Germany to develop the atomic bomb first. And in 1952, Einstein was offered the second presidency of Israel. In declining the position, he said, \u201cI know a little about nature and hardly anything about men.\u201d Figure 16.3. Albert Einstein (1879\u20131955). This portrait of Einstein was taken in 1912. (credit: modification of work by J. F. Langhans) Elementary Particles The fundamental components of atoms are the proton, neutron, and electron (see The Structure of the Atom). Protons, neutrons, and electrons are by no means all the particles that exist. First, for each kind of particle, there is a corresponding but opposite antiparticle. If the particle carries a charge, its antiparticle has the opposite charge. The antielectron is the positron, which has the same mass as the electron but is positively charged. Similarly, the antiproton has a negative charge. The remarkable thing about such antimatter is that when a particle comes into contact with its antiparticle, the original particles are annihilated, and substantial amounts of energy in the form of photons are produced. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 565 Since our world is made exclusively of ordinary particles of matter, antimatter cannot survive for very long. But individual antiparticles are found in cosmic rays (particles that arrive at the top of Earth\u2019s atmosphere from space) and can be created in particle accelerators. And, as we will see in a moment, antimatter is created in the core of the Sun and other stars. Science fiction", " fans may be familiar with antimatter from the Star Trek television series and films. The Starship Enterprise is propelled by the careful combining of matter and antimatter in the ship\u2019s engine room. According to E = mc2, the annihilation of matter and antimatter can produce a huge amount of energy, but keeping the antimatter fuel from touching the ship before it is needed must be a big problem. No wonder Scotty, the chief engineer in the original TV show, always looked worried! In 1933, physicist Wolfgang Pauli (Figure 16.4) suggested that there might be another type of elementary particle. Energy seemed to disappear when certain types of nuclear reactions took place, violating the law of conservation of energy. Pauli was reluctant to accept the idea that one of the basic laws of physics was wrong, and he suggested a \u201cdesperate remedy.\u201d Perhaps a so-far-undetected particle, which was given the name neutrino (\u201clittle neutral one\u201d), carried away the \u201cmissing\u201d energy. He suggested that neutrinos were particles with zero mass, and that like photons, they moved with the speed of light. Figure 16.4. Wolfgang Pauli in 1945. Pauli is considered the \u201cfather\u201d of the neutrino, having conceived of it in 1933. The elusive neutrino was not detected until 1956. The reason it was so hard to find is that neutrinos interact very weakly with other matter and therefore are very difficult to detect. Earth is more transparent to a neutrino than the thinnest and cleanest pane of glass is to a photon of light. In fact, most neutrinos can pass completely through a star or planet without being absorbed. As we shall see, this behavior of neutrinos makes them a very important tool for studying the Sun. Since Pauli\u2019s prediction, scientists have learned a lot more about the neutrino. We now know that there are three different types of neutrinos, and in 1998, neutrinos were discovered to have a tiny amount of mass. Indeed, it is so tiny that electrons are at least 500,000 times more massive. Ongoing research is focused on determining the mass of neutrinos more precisely, and it may still turn out that one of the three types is massless. We will return to the subject of neutrinos later in this chapter. Some of the properties of the proton, electron, neutron, and neutrino are summarized in Table 16.1. (Other subatomic particles have been produced by experiments with particle accelerators, but they do not play a role in the generation of solar energy.) 566 Chapter 16 The Sun: A Nuclear Powerhouse Properties of Some Common Particles Particle Mass (kg) Charge Proton Neutron Electron Neutrino Table 16.1 1.67265 \u00d7 10\u221227 1.67495 \u00d7 10\u221227 9.11 \u00d7 10\u221231 <2 \u00d7 10\u221236 (uncertain) +1 0 \u22121 0 The Atomic Nucleus The nucleus of an atom is not just a loose collection of elementary particles. Inside the nucleus, particles are held together by a very powerful force called the strong nuclear force. This is short-range force, only capable of acting over distances about the size of the atomic nucleus. A quick thought experiment shows how important this force is. Take a look at your finger and consider the atoms composing it. Among them is carbon, one of the basic elements of life. Focus your imagination on the nucleus of one of your carbon atoms. It contains six protons, which have a positive charge, and six neutrons, which are neutral. Thus, the nucleus has a net charge of six positives. If only the electrical force were acting, the protons in this and every carbon atom would find each other very repulsive and fly apart. The strong nuclear force is an attractive force, stronger than the electrical force, and it keeps the particles of the nucleus tightly bound together. We saw earlier that if under the force of gravity a star \u201cshrinks\u201d\u2014bringing its atoms closer together\u2014gravitational energy is released. In the same way, if particles come together under the strong nuclear force and unite to form an atomic nucleus, some of the nuclear energy is released. The energy given up in such a process is called the binding energy of the nucleus. When such binding energy is released, the resulting nucleus has slightly less mass than the sum of the masses of the particles that came together to form it. In other words, the energy comes from the loss of mass. This slight deficit in mass is only a small fraction of the mass of one proton. But because each bit of lost mass can provide a lot of energy (remember, E = mc2), this nuclear energy release can be quite substantial. Measurements show that the binding energy is greatest for atoms with a mass near that of the iron nucleus (with a combined number of prot", "ons and neutrons equal to 56) and less for both the lighter and the heavier nuclei. Iron, therefore, is the most stable element: since it gives up the most energy when it forms, it would require the most energy to break it back down into its component particles. What this means is that, in general, when light atomic nuclei come together to form a heavier one (up to iron), mass is lost and energy is released. This joining together of atomic nuclei is called nuclear fusion. Energy can also be produced by breaking up heavy atomic nuclei into lighter ones (down to iron); this process is called nuclear fission. Nuclear fission was the process we learned to use first\u2014in atomic bombs and in nuclear reactors used to generate electrical power\u2014and it may therefore be more familiar to you. Fission also sometimes occurs spontaneously in some unstable nuclei through the process of natural radioactivity. But fission requires big, complex nuclei, whereas we know that the stars are made up predominantly of small, simple nuclei. So we must look to fusion first to explain the energy of the Sun and the stars (Figure 16.5). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 567 Figure 16.5. Fusion and Fission. (a) In fusion, light atomic nuclei join together to form a heavier nuclei, releasing energy in the process. (b) In fission, energy is produced by the breaking up of heavy, complex nuclei into lighter ones. Nuclear Attraction versus Electrical Repulsion So far, we seem to have a very attractive prescription for producing the energy emitted by the Sun: \u201croll\u201d some nuclei together and join them via nuclear fusion. This will cause them to lose some of their mass, which then turns into energy. However, every nucleus, even simple hydrogen, has protons\u2014and protons all have positive charges. Since like charges repel via the electrical force, the closer we get two nuclei to each other, the more they repel. It\u2019s true that if we can get them within \u201cstriking distance\u201d of the nuclear force, they will then come together with a much stronger attraction. But that striking distance is very tiny, about the size of a nucleus. How can we get nuclei close enough to participate in fusion? The answer turns out to be heat\u2014tremendous heat\u2014which speeds the protons up enough to overcome the electrical forces that try to keep protons apart. Inside the Sun, as we saw, the most common element is hydrogen, whose nucleus contains only a single proton. Two protons can fuse only in regions where the temperature is greater than about 12 million K, and the speed of the protons average around 1000 kilometers per second or more. (In old-fashioned units, that\u2019s over 2 million miles per hour!) In our Sun, such extreme temperatures are reached only in the regions near its center, which has a temperature of 15 million K. Calculations show that nearly all of the Sun\u2019s energy is generated within about 150,000 kilometers of its core, or within less than 10% of its total volume. Even at these high temperatures, it is exceedingly difficult to force two protons to combine. On average, a proton will rebound from other protons in the Sun\u2019s crowded core for about 14 billion years, at the rate of 100 million collisions per second, before it fuses with a second proton. This is, however, only the average waiting time. Some of the enormous numbers of protons in the Sun\u2019s inner region are \u201clucky\u201d and take only a few collisions to achieve a fusion reaction: they are the protons responsible for producing the energy radiated by the Sun. Since the Sun is about 4.5 billion years old, most of its protons have not yet been involved in fusion reactions. Nuclear Reactions in the Sun\u2019s Interior The Sun, then, taps the energy contained in the nuclei of atoms through nuclear fusion. Let\u2019s look at what happens in more detail. Deep inside the Sun, a three-step process takes four hydrogen nuclei and fuses them together to form a single helium nucleus. The helium nucleus is slightly less massive than the four hydrogen nuclei that combine to form it, and that mass is converted into energy. The initial step required to form one helium nucleus from four hydrogen nuclei is shown in Figure 16.6. At the high temperatures inside the Sun\u2019s core, two protons combine to make a deuterium nucleus, which is an isotope (or version) of hydrogen that contains one proton and one neutron. In effect, one of the original protons has been converted into a neutron in the fusion reaction. Electric charge has to be conserved in nuclear reactions, 568 Chapter 16", " The Sun: A Nuclear Powerhouse and it is conserved in this one. A positron (antimatter electron) emerges from the reaction and carries away the positive charge originally associated with one of the protons. Figure 16.6. Proton-Proton Chain, Step 1. This is the first step in the process of fusing hydrogen into helium in the Sun. High temperatures are required because this reaction starts with two hydrogen nuclei, which are protons (shown in blue at left) that must overcome electrical repulsion to combine, forming a hydrogen nucleus with a proton and a neutron (shown in red). Note that hydrogen containing one proton and one neutron is given its own name: deuterium. Also produced in this reaction are a positron, which is an antielectron, and an elusive particle named the neutrino. Since it is antimatter, this positron will instantly collide with a nearby electron, and both will be annihilated, producing electromagnetic energy in the form of gamma-ray photons. This gamma ray, which has been created in the center of the Sun, finds itself in a world crammed full of fast-moving nuclei and electrons. The gamma ray collides with particles of matter and transfers its energy to one of them. The particle later emits another gamma-ray photon, but often the emitted photon has a bit less energy than the one that was absorbed. Such interactions happen to gamma rays again and again and again as they make their way slowly toward the outer layers of the Sun, until their energy becomes so reduced that they are no longer gamma rays but X- rays (recall what you learned in The Electromagnetic Spectrum). Later, as the photons lose still more energy through collisions in the crowded center of the Sun, they become ultraviolet photons. By the time they reach the Sun\u2019s surface, most of the photons have given up enough energy to be ordinary light\u2014and they are the sunlight we see coming from our star. (To be precise, each gamma-ray photon is ultimately converted into many separate lower-energy photons of sunlight.) So, the sunlight given off by the Sun today had its origin as a gamma ray produced by nuclear reactions deep in the Sun\u2019s core. The length of time that photons require to reach the surface depends on how far a photon on average travels between collisions, and the travel time depends on what model of the complicated solar interior we accept. Estimates are somewhat uncertain but indicate that the emission of energy from the surface of the Sun can lag its production in the interior by 100,000 years to as much as 1,000,000 years. In addition to the positron, the fusion of two hydrogen atoms to form deuterium results in the emission of a neutrino. Because neutrinos interact so little with ordinary matter, those produced by fusion reactions near the center of the Sun travel directly to the Sun\u2019s surface and then out into space, in all directions. Neutrinos move at nearly the speed of light, and they escape the Sun about two seconds after they are created. Figure 16.7. Proton-Proton Chain, Step 2. This is the second step of the proton-proton chain, the fusion reaction that converts hydrogen into helium in the Sun. This step combines one hydrogen nucleus, which is a proton (shown in blue), with the deuterium nucleus from the previous step (shown as a red and blue particle). The product of this is an isotope of helium with two protons (blue) and one neutron (red) and energy in the form of gamma-ray radiation. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 569 The second step in forming helium from hydrogen is to add another proton to the deuterium nucleus to create a helium nucleus that contains two protons and one neutron (Figure 16.7). In the process, some mass is again lost and more gamma radiation is emitted. Such a nucleus is helium because an element is defined by its number of protons; any nucleus with two protons is called helium. But this form of helium, which we call helium-3 (and write in shorthand as 3He) is not the isotope we see in the Sun\u2019s atmosphere or on Earth. That helium has two neutrons and two protons and hence is called helium-4 (4He). To get to helium-4 in the Sun, helium-3 must combine with another helium-3 in the third step of fusion (illustrated in Figure 16.8). Note that two energetic protons are left over from this step; each of them comes out of the reaction ready to collide with other protons and to start step 1 in the chain of reactions all over again. Figure 16.8. Proton-Proton Chain, Step 3. This is the third step in the fusion of", " hydrogen into helium in the Sun. Note that the two helium-3 nuclei from the second step (see Figure 16.7) must combine before the third step becomes possible. The two protons that come out of this step have the energy to collide with other protons in the Sun and start step one again These animations of proton-proton reactions (https://openstaxcollege.org/l/30proproreactio) show the steps required for fusion of hydrogen into helium in the Sun Visit the Tokamak Fusion Reactor (https://openstaxcollege.org/l/30tokamakfusrea) at the General Atomics Lab in San Diego, CA, for an 8-minute tour. The Proton-Proton Chain The nuclear reactions in the Sun that we have been discussing can be described succinctly through the following nuclear formulas: 1 H + 1 H\u27f6 2 H + e+ + v 2 H + 1 H\u27f6 3 He + \u03b3 3 He + 3 He\u27f6 4 He + 1 H + 1 H Here, the superscripts indicate the total number of neutrons plus protons in the nucleus, e+ is the positron, v is the neutrino, and \u03b3 indicates that gamma rays are emitted. Note that the third step requires two helium-3 nuclei to start; the first two steps must happen twice before the third step can occur. 570 Chapter 16 The Sun: A Nuclear Powerhouse Although, as we discussed, the first step in this chain of reactions is very difficult and generally takes a long time, the other steps happen more quickly. After the deuterium nucleus is formed, it survives an average of only about 6 seconds before being converted into 3He. About a million years after that (on average), the 3He nucleus will combine with another to form 4He. We can compute the amount of energy these reactions generate by calculating the difference in the initial and final masses. The masses of hydrogen and helium atoms in the units normally used by scientists are 1.007825 u and 4.00268 u, respectively. (The unit of mass, u, is defined to be 1/12 the mass of an atom of carbon, or approximately the mass of a proton.) Here, we include the mass of the entire atom, not just the nucleus, because electrons are involved as well. When hydrogen is converted into helium, two positrons are created (remember, the first step happens twice), and these are annihilated with two free electrons, adding to the energy produced. 4 \u00d7 1.007825 = 4.03130 u (mass of initial hydrogen atoms) \u2212 4.00268 u (mass of final helium a oms) = 0.02862 u (mass lost in the transformation) The mass lost, 0.02862 u, is 0.71% of the mass of the initial hydrogen. Thus, if 1 kilogram of hydrogen is converted into helium, then the mass of the helium is only 0.9929 kilogram, and 0.0071 kilogram of material is converted into energy. The speed of light (c) is 3 \u00d7 108 meters per second, so the energy released by the conversion of just 1 kilogram of hydrogen into helium is: E = mc2 E = 0.0071 kg \u00d7 \u239e \u239b \u239d3 \u00d7 108 m/s \u23a0 2 = 6.4 \u00d7 1014 J This amount, the energy released when a single kilogram (2.2 pounds) of hydrogen undergoes fusion, would supply all of the electricity used in the United States for about 2 weeks. To produce the Sun\u2019s luminosity of 4 \u00d7 1026 watts, some 600 million tons of hydrogen must be converted into helium each second, of which about 4 million tons are converted from matter into energy. As large as these numbers are, the store of hydrogen (and thus of nuclear energy) in the Sun is still more enormous, and can last a long time\u2014billions of years, in fact. At the temperatures inside the stars with masses smaller than about 1.2 times the mass of our Sun (a category that includes the Sun itself), most of the energy is produced by the reactions we have just described, and this set of reactions is called the proton-proton chain (or sometimes, the p-p chain). In the proton-proton chain, protons collide directly with other protons to form helium nuclei. In hotter stars, another set of reactions, called the carbon-nitrogen-oxygen (CNO) cycle, accomplishes the same net result. In the CNO cycle, carbon and hydrogen nuclei collide to initiate a series of reactions that form nitrogen, oxygen, and ultimately, helium. The nitrogen and oxygen nuclei do not survive but interact to form carbon again. Therefore, the outcome is the same as in the proton-proton chain: four hydrogen atoms", " disappear, and in their place, a single helium atom is created. The CNO cycle plays only a minor role in the Sun but is the main source of energy for stars with masses greater than about the mass of the Sun. So you can see that we have solved the puzzle that so worried scientists at the end of the nineteenth century. The Sun can maintain its high temperature and energy output for billions of years through the fusion of the simplest element in the universe, hydrogen. Because most of the Sun (and the other stars) is made of hydrogen, it is an ideal \u201cfuel\u201d for powering a star. As will be discussed in the following chapters, we can define a star as a ball of gas capable of getting its core hot enough to initiate the fusion of hydrogen. There are balls of gas that This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 571 lack the mass required to do this (Jupiter is a local example); like so many hopefuls in Hollywood, they will never be stars Fusion on Earth Wouldn\u2019t it be wonderful if we could duplicate the Sun\u2019s energy mechanism in a controlled way on Earth? (We have already duplicated it in an uncontrolled way in hydrogen bombs, but we hope our storehouses of these will never be used.) Fusion energy would have many advantages: it would use hydrogen (or deuterium, which is heavy hydrogen) as fuel, and there is abundant hydrogen in Earth\u2019s lakes and oceans. Water is much more evenly distributed around the world than oil or uranium, meaning that a few countries would no longer hold an energy advantage over the others. And unlike fission, which leaves dangerous byproducts, the nuclei that result from fusion are perfectly safe. The problem is that, as we saw, it takes extremely high temperatures for nuclei to overcome their electrical repulsion and undergo fusion. When the first hydrogen bombs were exploded in tests in the 1950s, the \u201cfuses\u201d to get them hot enough were fission bombs. Interactions at such temperatures are difficult to sustain and control. To make fusion power on Earth, after all, we have to do what the Sun does: produce temperatures and pressures high enough to get hydrogen nuclei on intimate terms with one another. The European Union, the United States, South Korea, Japan, China, Russia, Switzerland, and India are collaborating on the International Thermonuclear Experimental Reactor (ITER), a project to demonstrate the feasibility of controlled fusion (Figure 16.9). The facility is being built in France. Construction will require over 10,000,000 components and 2000 workers for assembly. The date for the start of operations is yet to be determined. ITER is based on the Tokamak design, in which a large doughnut-shaped container is surrounded by superconducting magnets to confine and control the hydrogen nuclei in a strong magnetic field. Previous fusion experiments have produced about 15 million watts of energy, but only for a second or two, and they have required 100 million watts to produce the conditions necessary to achieve fusion. The goal of ITER is to build the first fusion device capable of producing 500 million watts of fusion energy for up to 1000 seconds. The challenge is keeping the deuterium and tritium\u2014which will participate in fusion reactions\u2014hot enough and dense enough, for a long enough time to produce energy. 572 Chapter 16 The Sun: A Nuclear Powerhouse Figure 16.9. ITER Design. The bright yellow areas in this model show where the superconducting magnets will circle the chamber within which fusion will take place. A huge magnet will keep the charged nuclei of heavy hydrogen confined. The goal is to produce 500 megawatts of energy. (credit: modification of work by Stephan Mosel) 16.3 THE SOLAR INTERIOR: THEORY Learning Objectives By the end of this section, you will be able to: Describe the state of equilibrium of the Sun Understand the energy balance of the Sun Explain how energy moves outward through the Sun Describe the structure of the solar interior Fusion of protons can occur in the center of the Sun only if the temperature exceeds 12 million K. How do we know that the Sun is actually this hot? To determine what the interior of the Sun might be like, it is necessary to resort to complex calculations. Since we can\u2019t see the interior of the Sun, we have to use our understanding of physics, combined with what we see at the surface, to construct a mathematical model of what must be happening in the interior. Astronomers use observations to build a computer program containing everything they think they know about the physical processes going on in the Sun\u2019s interior. The computer then calculates the temperature and pressure at every point inside the Sun and determines what nuclear reactions, if any, are taking place. For some calculations, we can use observations to determine whether the computer program is", " producing results that match what we see. In this way, the program evolves with ever-improving observations. The computer program can also calculate how the Sun will change with time. After all, the Sun must change. In its center, the Sun is slowly depleting its supply of hydrogen and creating helium instead. Will the Sun get hotter? Cooler? Larger? Smaller? Brighter? Fainter? Ultimately, the changes in the center could be catastrophic, This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 573 since eventually all the hydrogen fuel hot enough for fusion will be exhausted. Either a new source of energy must be found, or the Sun will cease to shine. We will describe the ultimate fate of the Sun in later chapters. For now, let\u2019s look at some of the things we must teach the computer about the Sun in order to carry out such calculations. The Sun Is a Plasma The Sun is so hot that all of the material in it is in the form of an ionized gas, called a plasma. Plasma acts much like a hot gas, which is easier to describe mathematically than either liquids or solids. The particles that constitute a gas are in rapid motion, frequently colliding with one another. This constant bombardment is the pressure of the gas (Figure 16.10). Figure 16.10. Gas Pressure. The particles in a gas are in rapid motion and produce pressure through collisions with the surrounding material. Here, particles are shown bombarding the sides of an imaginary container. More particles within a given volume of gas produce more pressure because the combined impact of the moving particles increases with their number. The pressure is also greater when the molecules or atoms are moving faster. Since the molecules move faster when the temperature is hotter, higher temperatures produce higher pressure. The Sun Is Stable The Sun, like the majority of other stars, is stable; it is neither expanding nor contracting. Such a star is said to be in a condition of equilibrium. All the forces within it are balanced, so that at each point within the star, the temperature, pressure, density, and so on are maintained at constant values. We will see in later chapters that even these stable stars, including the Sun, are changing as they evolve, but such evolutionary changes are so gradual that, for all intents and purposes, the stars are still in a state of equilibrium at any given time. The mutual gravitational attraction between the masses of various regions within the Sun produces tremendous forces that tend to collapse the Sun toward its center. Yet we know from the history of Earth that the Sun has been emitting roughly the same amount of energy for billions of years, so clearly it has managed to resist collapse for a very long time. The gravitational forces must therefore be counterbalanced by some other force. That force is due to the pressure of gases within the Sun (Figure 16.11). Calculations show that, in order to exert enough pressure to prevent the Sun from collapsing due to the force of gravity, the gases at its center must be maintained at a temperature of 15 million K. Think about what this tells us. Just from the fact that the Sun is not contracting, we can conclude that its temperature must indeed be high enough at the center for protons to undergo fusion. 574 Chapter 16 The Sun: A Nuclear Powerhouse Figure 16.11. Hydrostatic Equilibrium. In the interior of a star, the inward force of gravity is exactly balanced at each point by the outward force of gas pressure. The Sun maintains its stability in the following way. If the internal pressure in such a star were not great enough to balance the weight of its outer parts, the star would collapse somewhat, contracting and building up the pressure inside. On the other hand, if the pressure were greater than the weight of the overlying layers, the star would expand, thus decreasing the internal pressure. Expansion would stop, and equilibrium would again be reached when the pressure at every internal point equaled the weight of the stellar layers above that point. An analogy is an inflated balloon, which will expand or contract until an equilibrium is reached between the pressure of the air inside and outside. The technical term for this condition is hydrostatic equilibrium. Stable stars are all in hydrostatic equilibrium; so are the oceans of Earth as well as Earth\u2019s atmosphere. The air\u2019s own pressure keeps it from falling to the ground. The Sun Is Not Cooling Down As everyone who has ever left a window open on a cold winter night knows, heat always flows from hotter to cooler regions. As energy filters outward toward the surface of a star, it must be flowing from inner, hotter regions. The temperature cannot ordinarily get cooler as we go inward in a star, or energy would flow in and heat up those regions until they were at least as hot as the outer ones. Scientists conclude that the temperature is highest at the center of a star, dropping to lower", " and lower values toward the stellar surface. (The high temperature of the Sun\u2019s chromosphere and corona may therefore appear to be a paradox. But remember from The Sun: A Garden-Variety Star that these high temperatures are maintained by magnetic effects, which occur in the Sun\u2019s atmosphere.) The outward flow of energy through a star robs it of its internal heat, and the star would cool down if that energy were not replaced. Similarly, a hot iron begins to cool as soon as it is unplugged from its source of electric energy. Therefore, a source of fresh energy must exist within each star. In the Sun\u2019s case, we have seen that this energy source is the ongoing fusion of hydrogen to form helium. Heat Transfer in a Star Since the nuclear reactions that generate the Sun\u2019s energy occur deep within it, the energy must be transported from the center of the Sun to its surface\u2014where we see it in the form of both heat and light. There are three ways in which energy can be transferred from one place to another. In conduction, atoms or molecules pass on their energy by colliding with others nearby. This happens, for example, when the handle of a metal spoon This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 575 heats up as you stir a cup of hot coffee. In convection, currents of warm material rise, carrying their energy with them to cooler layers. A good example is hot air rising from a fireplace. In radiation, energetic photons move away from hot material and are absorbed by some material to which they convey some or all of their energy. You can feel this when you put your hand close to the coils of an electric heater, allowing infrared photons to heat up your hand. Conduction and convection are both important in the interiors of planets. In stars, which are much more transparent, radiation and convection are important, whereas conduction can usually be ignored. Stellar convection occurs as currents of hot gas flow up and down through the star (Figure 16.12). Such currents travel at moderate speeds and do not upset the overall stability of the star. They don\u2019t even result in a net transfer of mass either inward or outward because, as hot material rises, cool material falls and replaces it. This results in a convective circulation of rising and falling cells as seen in Figure 16.12. In much the same way, heat from a fireplace can stir up air currents in a room, some rising and some falling, without driving any air into or out the room. Convection currents carry heat very efficiently outward through a star. In the Sun, convection turns out to be important in the central regions and near the surface. Figure 16.12. Convection. Rising convection currents carry heat from the Sun\u2019s interior to its surface, whereas cooler material sinks downward. Of course, nothing in a real star is as simple as diagrams in textbooks suggest. Unless convection occurs, the only significant mode of energy transport through a star is by electromagnetic radiation. Radiation is not an efficient means of energy transport in stars because gases in stellar interiors are very opaque, that is, a photon does not go far (in the Sun, typically about 0.01 meter) before it is absorbed. (The processes by which atoms and ions can interrupt the outward flow of photons\u2014such as becoming ionized\u2014were discussed in the section on the Formation of Spectral Lines.) The absorbed energy is always reemitted, but it can be reemitted in any direction. A photon absorbed when traveling outward in a star has almost as good a chance of being radiated back toward the center of the star as toward its surface. A particular quantity of energy, therefore, zigzags around in an almost random manner and takes a long time to work its way from the center of a star to its surface (Figure 16.13). Estimates are somewhat uncertain, but in the Sun, as we saw, the time required is probably between 100,000 and 1,000,000 years. If the photons were not absorbed and reemitted along the way, they would travel at the speed of light and could reach the surface in a little over 2 seconds, just as neutrinos do (Figure 16.14). 576 Chapter 16 The Sun: A Nuclear Powerhouse Figure 16.13. Photons Deep in the Sun. A photon moving through the dense gases in the solar interior travels only a short distance before it interacts with one of the surrounding atoms. The photon usually has a lower energy after each interaction and may then travel in any random direction. Figure 16.14. Photon and Neutrino Paths in the Sun. (a) Because photons generated by fusion reactions in the solar interior travel only a short distance before being absorbed or scattered by atoms and sent off in random directions, estimates are that it takes", " between 100,000 and 1,000,000 years for energy to make its way from the center of the Sun to its surface. (b) In contrast, neutrinos do not interact with matter but traverse straight through the Sun at the speed of light, reaching the surface in only a little more than 2 seconds Heat Transfer and Cooking The three ways that heat energy moves from higher-temperature regions to cooler regions are all used in cooking, and this is important to all of us who enjoy making or eating food. Conduction is heat transfer by physical contact during which the energetic motion of particles in one region spread to other regions and even to adjacent objects in close contact. A tasty example of this is cooking a steak on a hot iron skillet. When a flame makes the bottom of a skillet hot, the particles in it vibrate actively and collide with neighboring particles, spreading the heat energy throughout the skillet (the ability to spread heat uniformly is a key criterion for selecting materials for cookware). A steak sitting on the surface of the skillet picks up heat energy by the particles in the surface of the skillet colliding with particles on the surface of the steak. Many cooks will put a little oil on the pan, and this layer of oil, This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 577 besides preventing sticking, increases heat transfer by filling in gaps and increasing the contact surface area. Convection is heat transfer by the motion of matter that rises because it is hot and less dense. Heating a fluid makes it expand, which makes it less dense, so it rises. An oven is a great example of this: the fire is at the bottom of the oven and heats the air down there, causing it to expand (becoming less dense), so it rises up to where the food is. The rising hot air carries the heat from the fire to the food by convection. This is how conventional ovens work. You may also be familiar with convection ovens that use a fan to circulate hot air for more even cooking. A scientist would object to that name because normal non-fan ovens that rely on hot air rising to circulate the heat are convection ovens; technically, the ovens that use fans to help move heat are \u201cadvection\u201d ovens. (You may not have heard about this because the scientists who complain loudly about misusing the terms convection and advection don\u2019t get out much.) Radiation is the transfer of heat energy by electromagnetic radiation. Although microwave ovens are an obvious example of using radiation to heat food, a simpler example is a toy oven. Toy ovens are powered by a very bright light bulb. The child-chefs prepare a mix for brownies or cookies, put it into a tray, and place it in the toy oven under the bright light bulb. The light and heat from the bulb hit the brownie mix and cook it. If you have ever put your hand near a bright light, you have undoubtedly noticed your hand getting warmed by the light. Model Stars Scientists use the principles we have just described to calculate what the Sun\u2019s interior is like. These physical ideas are expressed as mathematical equations that are solved to determine the values of temperature, pressure, density, the efficiency with which photons are absorbed, and other physical quantities throughout the Sun. The solutions obtained, based on a specific set of physical assumptions, provide a theoretical model for the interior of the Sun. Figure 16.15 schematically illustrates the predictions of a theoretical model for the Sun\u2019s interior. Energy is generated through fusion in the core of the Sun, which extends only about one-quarter of the way to the surface but contains about one-third of the total mass of the Sun. At the center, the temperature reaches a maximum of approximately 15 million K, and the density is nearly 150 times that of water. The energy generated in the core is transported toward the surface by radiation until it reaches a point about 70% of the distance from the center to the surface. At this point, convection begins, and energy is transported the rest of the way, primarily by rising columns of hot gas. 578 Chapter 16 The Sun: A Nuclear Powerhouse Figure 16.15. Interior Structure of the Sun. Energy is generated in the core by the fusion of hydrogen to form helium. This energy is transmitted outward by radiation\u2014that is, by the absorption and reemission of photons. In the outermost layers, energy is transported mainly by convection. (credit: modification of work by NASA/Goddard) Figure 16.16 shows how the temperature, density, rate of energy generation, and composition vary from the center of the Sun to its surface. Figure 16.16. Interior of the Sun. Diagrams showing how temperature, density, rate of energy generation, and the percentage (by mass) abundance of hydrogen vary inside the Sun", ". The horizontal scale shows the fraction of the Sun\u2019s radius: the left edge is the very center, and the right edge is the visible surface of the Sun, which is called the photosphere. 16.4 THE SOLAR INTERIOR: OBSERVATIONS Learning Objectives By the end of this section, you will be able to: This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 579 Explain how the Sun pulsates Explain what helioseismology is and what it can tell us about the solar interior Discuss how studying neutrinos from the Sun has helped understand neutrinos Recall that when we observe the Sun\u2019s photosphere (the surface layer we see from the outside), we are not seeing very deeply into our star, certainly not into the regions where energy is generated. That\u2019s why the title of this section\u2014observations of the solar interior\u2014should seem very surprising. However, astronomers have indeed devised two types of measurements that can be used to obtain information about the inner parts of the Sun. One technique involves the analysis of tiny changes in the motion of small regions at the Sun\u2019s surface. The other relies on the measurement of the neutrinos emitted by the Sun. Solar Pulsations Astronomers discovered that the Sun pulsates\u2014that is, it alternately expands and contracts\u2014just as your chest expands and contracts as you breathe. This pulsation is very slight, but it can be detected by measuring the radial velocity of the solar surface\u2014the speed with which it moves toward or away from us. The velocities of small regions on the Sun are observed to change in a regular way, first toward Earth, then away, then toward, and so on. It is as if the Sun were \u201cbreathing\u201d through thousands of individual lungs, each having a size in the range of 4000 to 15,000 kilometers, each fluctuating back and forth (Figure 16.17). Figure 16.17. Oscillations in the Sun. New observational techniques permit astronomers to measure small differences in velocity at the Sun\u2019s surface to infer what the deep solar interior is like. In this computer simulation, red shows surface regions that are moving away from the observer (inward motion); blue marks regions moving toward the observer (outward motion). Note that the velocity changes penetrate deep into the Sun\u2019s interior. (credit: modification of work by GONG, NOAO) The typical velocity of one of the oscillating regions on the Sun is only a few hundred meters per second, and it takes about 5 minutes to complete a full cycle from maximum to minimum velocity and back again. The change in the size of the Sun measured at any given point is no more than a few kilometers. The remarkable thing is that these small velocity variations can be used to determine what the interior of the Sun is like. The motion of the Sun\u2019s surface is caused by waves that reach it from deep in the interior. Study of the amplitude and cycle length of velocity changes provides information about the temperature, density, and composition of the layers through which the waves passed before they reached the surface. The situation is somewhat analogous to the use of seismic waves generated by earthquakes to infer the properties 580 Chapter 16 The Sun: A Nuclear Powerhouse of Earth\u2019s interior. For this reason, studies of solar oscillations (back-and-forth motions) are referred to as helioseismology. It takes a little over an hour for waves to traverse the Sun from center to surface, so the waves, like neutrinos, provide information about what the solar interior is like at the present time. In contrast, remember that the sunlight we see today emerging from the Sun was actually generated in the core several hundred thousand years ago. Helioseismology has shown that convection extends inward from the surface 30% of the way toward the center; we have used this information in drawing Figure 16.15. Pulsation measurements also show that the differential rotation that we see at the Sun\u2019s surface, with the fastest rotation occurring at the equator, persists down through the convection zone. Below the convection zone, however, the Sun, even though it is gaseous throughout, rotates as if it were a solid body like a bowling ball. Another finding from helioseismology is that the abundance of helium inside the Sun, except in the center where nuclear reactions have converted hydrogen into helium, is about the same as at its surface. That result is important to astronomers because it means we are correct when we use the abundance of the elements measured in the solar atmosphere to construct models of the solar interior. Helioseismology also allows scientists to look beneath a sunspot and see how it works. In The Sun: A Garden- Variety Star, we said that sunspots are cool because strong magnetic fields block", " the outward flow of energy. Figure 16.18 shows how gas moves around underneath a sunspot. Cool material from the sunspot flows downward, and material surrounding the sunspot is pulled inward, carrying magnetic field with it and thus maintaining the strong field that is necessary to form a sunspot. As the new material enters the sunspot region, it too cools, becomes denser, and sinks, thus setting up a self-perpetuating cycle that can last for weeks. Figure 16.18. Sunspot Structure. This drawing shows our new understanding, from helioseismology, of what lies beneath a sunspot. The black arrows show the direction of the flow of material. The intense magnetic field associated with the sunspot stops the upward flow of hot material and creates a kind of plug that blocks the hot gas. As the material above the plug cools (shown in blue), it becomes denser and plunges inward, drawing more gas and more magnetic field behind it into the spot. The concentrated magnetic field causes more cooling, thereby setting up a self-perpetuating cycle that allows a spot to survive for several weeks. Since the plug keeps hot material from flowing up into the sunspot, the region below the plug, represented by red in this picture, becomes hotter. This material flows sideways and then upward, eventually reaching the solar surface in the area surrounding the sunspot. (credit: modification of work by NASA, SDO) The downward-flowing cool material acts as a kind of plug that block the upward flow of hot material, which is then diverted sideways and eventually reaches the solar surface in the region around the sunspot. This outward This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 581 flow of hot material accounts for the paradox that we described in The Sun: A Garden-Variety Star\u2014namely, that the Sun emits slightly more energy when more of its surface is covered by cool sunspots. Helioseismology has become an important tool for predicting solar storms that might impact Earth. Active regions can appear and grow large in only a few days. The solar rotation period is about 28 days. Therefore, regions capable of producing solar flares and coronal mass ejections can develop on the far side of the Sun, where, for a long time, we couldn\u2019t see them directly. Fortunately, we now have space telescopes monitoring the Sun from all angles, so we know if there are sunspots forming on the opposite side of the Sun. Moreover, sound waves travel slightly faster in regions of high magnetic field, and waves generated in active regions traverse the Sun about 6 seconds faster than waves generated in quiet regions. By detecting this subtle difference, scientists can provide warnings of a week or more to operators of electric utilities and satellites about when a potentially dangerous active region might rotate into view. With this warning, it is possible to plan for disruptions, put key instruments into safe mode, or reschedule spacewalks in order to protect astronauts. Solar Neutrinos The second technique for obtaining information about the Sun\u2019s interior involves the detection of a few of those elusive neutrinos created during nuclear fusion. Recall from our earlier discussion that neutrinos created in the center of the Sun make their way directly out of the Sun and travel to Earth at nearly the speed of light. As far as neutrinos are concerned, the Sun is transparent. About 3% of the total energy generated by nuclear fusion in the Sun is carried away by neutrinos. So many protons react and form neutrinos inside the Sun\u2019s core that, scientists calculate, 35 million billion (3.5 \u00d7 1016) solar neutrinos pass through each square meter of Earth\u2019s surface every second. If we can devise a way to detect even a few of these solar neutrinos, then we can obtain information directly about what is going on in the center of the Sun. Unfortunately for those trying to \u201ccatch\u201d some neutrinos, Earth and everything on it are also nearly transparent to passing neutrinos, just like the Sun. On very, very rare occasions, however, one of the billions and billions of solar neutrinos will interact with another atom. The first successful detection of solar neutrinos made use of cleaning fluid (C2Cl4), which is the least expensive way to get a lot of chlorine atoms together. The nucleus of a chlorine (Cl) atom in the cleaning fluid can be turned into a radioactive argon nucleus by an interaction with a neutrino. Because the argon is radioactive, its presence can be detected. However, since the interaction of a neutrino with chlorine happens so rarely, a huge amount of chlorine is needed. Raymond Davis, Jr. (Figure 16.19) and his colleagues at Brookhaven National Laboratory, placed a tank containing nearly 400,000 liters of cleaning fluid 1", ".5 kilometers beneath Earth\u2019s surface in a gold mine at Lead, South Dakota. A mine was chosen so that the surrounding material of Earth would keep cosmic rays (high- energy particles from space) from reaching the cleaning fluid and creating false signals. (Cosmic-ray particles are stopped by thick layers of Earth, but neutrinos find them of no significance.) Calculations show that solar neutrinos should produce about one atom of radioactive argon in the tank each day. 582 Chapter 16 The Sun: A Nuclear Powerhouse Figure 16.19. Davis Experiment. (a) Raymond Davis received the Nobel Prize in physics in 2002. (b) Davis\u2019 experiment at the bottom of an abandoned gold mine first revealed problems with our understanding of neutrinos. (credit a: modification of work by Brookhaven National Laboratory; credit b: modification of work by the United States Department of Energy) This was an amazing project: they counted argon atoms about once per month\u2014and remember, they were looking for a tiny handful of argon atoms in a massive tank of chlorine atoms. When all was said and done, Davis\u2019 experiment, begun in 1970, detected only about one-third as many neutrinos as predicted by solar models! This was a shocking result because astronomers thought they had a pretty good understanding of both neutrinos and the Sun\u2019s interior. For many years, astronomers and physicists wrestled with Davis\u2019 results, trying to find a way out of the dilemma of the \u201cmissing\u201d neutrinos. Eventually Davis\u2019 result was explained by the surprising discovery that there are actually three types of neutrinos. Solar fusion produces only one type of neutrino, the so-called electron neutrino, and the initial experiments to detect solar neutrinos were designed to detect this one type. Subsequent experiments showed that these neutrinos change to a different type during their journey from the center of the Sun through space to Earth in a process called neutrino oscillation. An experiment, conducted at the Sudbury Neutrino Observatory in Canada, was the first one designed to capture all three types of neutrinos (Figure 16.20). The experiment was located in a mine 2 kilometers underground. The neutrino detector consisted of a 12-meter-diameter transparent acrylic plastic sphere, which contained 1000 metric tons of heavy water. Remember that an ordinary water nucleus contains two hydrogen atoms and one oxygen atom. Heavy water instead contains two deuterium atoms and one oxygen atom, and incoming neutrinos can occasionally break up the loosely bound proton and neutron that make up the deuterium nucleus. The sphere of heavy water was surrounded by a shield of 1700 metric tons of very pure water, which in turn was surrounded by 9600 photomultipliers, devices that detect flashes of light produced after neutrons interact with the heavy water. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 583 Figure 16.20. Sudbury Neutrino Detector. The 12-meter sphere of the Sudbury Neutrino Detector lies more than 2 kilometers underground and holds 1000 metric tons of heavy water. (credit: A.B. McDonald (Queen\u2019s University) et al., The Sudbury Neutrino Observatory Institute) To the enormous relief of astronomers who make models of the Sun, the Sudbury experiment detected about 1 neutrino per hour and has shown that the total number of neutrinos reaching the heavy water is just what solar models predict. Only one-third of these, however, are electron neutrinos. It appears that two-thirds of the electron neutrinos produced by the Sun transform themselves into one of the other types of neutrinos as they make their way from the core of the Sun to Earth. This is why the earlier experiments saw only one-third the number of neutrinos expected. Although it is not intuitively obvious, such neutrino oscillations can happen only if the mass of the electron neutrino is not zero. Other experiments indicate that its mass is tiny (even compared to the electron). The 2015 Nobel Prize in physics was awarded to researchers Takaaki Kajita and Arthur B. McDonald for their work establishing the changeable nature of neutrinos. (Raymond Davis shared the 2002 Nobel Prize with Japan\u2019s Masatoshi Koshiba for the experiments that led to our understanding of the neutrino problem in the first place.) But the fact that the neutrino has mass at all has deep implications for both physics and astronomy. For example, we will look at the role that neutrinos play in the inventory of the mass of the universe in The Big Bang. The Borexino experiment, an international experiment conducted in Italy, detected neutrinos coming from the Sun that were identified as coming from different reactions. Whereas the", " p-p chain is the reaction producing most of the Sun\u2019s energy, it is not the only nuclear reaction occurring in the Sun\u2019s core. There are side reactions involving nuclei of such elements as beryllium and boron. By probing the number of neutrinos that come from each reaction, the Borexino experiment has helped us confirm in detail our understanding of nuclear fusion in the Sun. In 2014, the Borexino experiment also identified neutrinos that were produced by the first step in the p-p chain, confirming the models of solar astronomers. It\u2019s amazing that a series of experiments that began with enough cleaning fluid to fill a swimming pool brought down the shafts of an old gold mine is now teaching us about the energy source of the Sun and the properties of matter! This is a good example of how experiments in astronomy and physics, coupled with the best theoretical models we can devise, continue to lead to fundamental changes in our understanding of nature. 584 Chapter 16 The Sun: A Nuclear Powerhouse CHAPTER 16 REVIEW KEY TERMS conduction process by which heat is directly transmitted through a substance when there is a difference of temperature between adjoining regions caused by atomic or molecular collisions convection movement caused within a gas or liquid by the tendency of hotter, and therefore less dense material, to rise and colder, denser material to sink under the influence of gravity, which consequently results in transfer of heat fission breaking up of heavier atomic nuclei into lighter ones fusion building up of heavier atomic nuclei from lighter ones helioseismology study of pulsations or oscillations of the Sun in order to determine the characteristics of the solar interior hydrostatic equilibrium balance between the weights of various layers, as in a star or Earth\u2019s atmosphere, and the pressures that support them neutrino fundamental particle that has no charge and a mass that is tiny relative to an electron; it rarely interacts with ordinary matter and comes in three different types positron particle with the same mass as an electron, but positively charged proton-proton chain series of thermonuclear reactions by which nuclei of hydrogen are built up into nuclei of helium radiation emission of energy as electromagnetic waves or photons also the transmitted energy itself SUMMARY 16.1 Sources of Sunshine: Thermal and Gravitational Energy The Sun produces an enormous amount of energy every second. Since Earth and the solar system are roughly 4.5 billion years old, this means that the Sun has been producing vast amounts for energy for a very, very long time. Neither chemical burning nor gravitational contraction can account for the total amount of energy radiated by the Sun during all this time. 16.2 Mass, Energy, and the Theory of Relativity Solar energy is produced by interactions of particles\u2014that is, protons, neutrons, electrons, positrons, and neutrinos. Specifically, the source of the Sun\u2019s energy is the fusion of hydrogen to form helium. The series of reactions required to convert hydrogen to helium is called the proton-proton chain. A helium atom is about 0.71% less massive than the four hydrogen atoms that combine to form it, and that lost mass is converted to energy (with the amount of energy given by the formula E = mc2). 16.3 The Solar Interior: Theory Even though we cannot see inside the Sun, it is possible to calculate what its interior must be like. As input for these calculations, we use what we know about the Sun. It is made entirely of hot gas. Apart from some very tiny changes, the Sun is neither expanding nor contracting (it is in hydrostatic equilibrium) and puts out energy This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 585 at a constant rate. Fusion of hydrogen occurs in the center of the Sun, and the energy generated is carried to the surface by radiation and then convection. A solar model describes the structure of the Sun\u2019s interior. Specifically, it describes how pressure, temperature, mass, and luminosity depend on the distance from the center of the Sun. 16.4 The Solar Interior: Observations Studies of solar oscillations (helioseismology) and neutrinos can provide observational data about the Sun\u2019s interior. The technique of helioseismology has so far shown that the composition of the interior is much like that of the surface (except in the core, where some of the original hydrogen has been converted into helium), and that the convection zone extends about 30% of the way from the Sun\u2019s surface to its center. Helioseismology can also detect active regions on the far side of the Sun and provide better predictions of solar storms that may affect Earth. Neutrinos from the Sun call tell us about what is happening in the solar interior. A recent experiment has shown that solar models do predict accurately the number of electron neut", "rinos produced by nuclear reactions in the core of the Sun. However, two-thirds of these neutrinos are converted into different types of neutrinos during their long journey from the Sun to Earth, a result that also indicates that neutrinos are not massless particles. FOR FURTHER EXPLORATION Articles Harvey, J. et al. \u201cGONG: To See Inside Our Sun.\u201d Sky & Telescope (November 1987): 470. Hathaway, D. \u201cJourney to the Heart of the Sun.\u201d Astronomy (January 1995): 38. Kennedy, J. \u201cGONG: Probing the Sun\u2019s Hidden Heart.\u201d Sky & Telescope (October 1996): 20. A discussion on hydroseismology. LoPresto, J. \u201cLooking Inside the Sun.\u201d Astronomy (March 1989): 20. A discussion on hydroseismology. McDonald, A. et al. \u201cSolving the Solar Neutrino Problem.\u201d Scientific American (April 2003): 40. A discussion on how underground experiments with neutrino detectors helped explain the seeming absence of neutrinos from the Sun. Trefil, J. \u201cHow Stars Shine.\u201d Astronomy (January 1998): 56. Websites Albert Einstein Online: http://www.westegg.com/einstein/. Ghost Particle: http://www.pbs.org/wgbh/nova/neutrino/. GONG Project Site: http://gong.nso.edu/. Helioseismology: http://solar-center.stanford.edu/about/helioseismology.html. Princeton Plasma Physics Lab: http://www.pppl.gov/. Solving the Mystery of the Solar Neutrinos: http://www.nobelprize.org/nobel_prizes/themes/physics/bahcall/. Super Kamiokande Neutrino Mass Page: http://www.ps.uci.edu/~superk/. 586 Videos Chapter 16 The Sun: A Nuclear Powerhouse Deep Secrets of the Neutrino: Physics Underground: https://www.youtube.com/watch?v=Ar9ydagYkYg. 2010 Public Lecture by Peter Rowson at the Stanford Linear Accelerator Center (1:22:00). The Elusive Neutrino and the Nature of Physics: https://www.youtube.com/watch?v=CBfUHzkcaHQ. Panel at the 2014 World Science Festival (1:30:00). The Ghost Particle: http://www.dailymotion.com/video/x20rn7s_nova-the-ghost-particle-discovery-science- universe-documentary_tv. 2006 NOVA episode (52:49). COLLABORATIVE GROUP ACTIVITIES A. In this chapter, we learned that meteorites falling into the Sun could not be the source of the Sun\u2019s energy because the necessary increase in the mass of the Sun would lengthen Earth\u2019s orbital period by 2 seconds per year. Have your group discuss what effects this would cause for our planet and for us as the centuries went on. B. Solar astronomers can learn more about the Sun\u2019s interior if they can observe the Sun\u2019s oscillations 24 hours each day. This means that they cannot have their observations interrupted by the day/night cycle. Such an experiment, called the GONG (Global Oscillation Network Group) project, was first set up in the 1990s. To save money, this experiment was designed to make use of the minimum possible number of telescopes. It turns out that if the sites are selected carefully, the Sun can be observed all but about 10% of the time with only six observing stations. What factors do you think have to be taken into consideration in selecting the observing sites? Can your group suggest six general geographic locations that would optimize the amount of time that the Sun can be observed? Check your answer by looking at the GONG website. C. What would it be like if we actually manage to get controlled fusion on Earth to be economically feasible? If the hydrogen in water becomes the fuel for releasing enormous amounts of energy (instead of fossil fuels), have your group discuss how this affects the world economy and international politics. (Think of the role that oil and natural gas deposits now play on the world scene and in international politics.) D. Your group is a delegation sent to the city council of a small mining town to explain why the government is putting a swimming-pool-sized vat of commercial cleaning fluid down one of the shafts of an old gold mine. How would you approach this meeting? Assuming that the members of the city council do not have much science background, how would you explain the importance of the project to them? Suggest some visual aids you could use. E. When Raymond Davis first suggested", " his experiment in the underground gold mine, which had significant costs associated with it, some people said it wasn\u2019t worth the expense since we already understood the conditions and reactions in the core of the Sun. Yet his experiment led to a major change in our understanding of neutrinos and the physics of subatomic particles. Can your group think of other \u201cexpensive\u201d experiments in astronomy that led to fundamental improvements in our understanding of nature? EXERCISES This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 587 Review Questions 1. How do we know the age of the Sun? 2. Explain how we know that the Sun\u2019s energy is not supplied either by chemical burning, as in fires here on Earth, or by gravitational contraction (shrinking). 3. What is the ultimate source of energy that makes the Sun shine? 4. What are the formulas for the three steps in the proton-proton chain? 5. How is a neutrino different from a neutron? List all the ways you can think of. 6. Describe in your own words what is meant by the statement that the Sun is in hydrostatic equilibrium. 7. Two astronomy students travel to South Dakota. One stands on Earth\u2019s surface and enjoys some sunshine. At the same time, the other descends into a gold mine where neutrinos are detected, arriving in time to detect the creation of a new radioactive argon nucleus. Although the photon at the surface and the neutrinos in the mine arrive at the same time, they have had very different histories. Describe the differences. 8. What do measurements of the number of neutrinos emitted by the Sun tell us about conditions deep in the solar interior? 9. Do neutrinos have mass? Describe how the answer to this question has changed over time and why. 10. Neutrinos produced in the core of the Sun carry energy to its exterior. Is the mechanism for this energy transport conduction, convection, or radiation? 11. What conditions are required before proton-proton chain fusion can start in the Sun? 12. Describe the two main ways that energy travels through the Sun. Thought Questions 13. Someone suggests that astronomers build a special gamma-ray detector to detect gamma rays produced during the proton-proton chain in the core of the Sun, just like they built a neutrino detector. Explain why this would be a fruitless effort. 14. Earth contains radioactive elements whose decay produces neutrinos. How might we use neutrinos to determine how these elements are distributed in Earth\u2019s interior? 15. The Sun is much larger and more massive than Earth. Do you think the average density of the Sun is larger or smaller than that of Earth? Write down your answer before you look up the densities. Now find the values of the densities elsewhere in this text. Were you right? Explain clearly the meanings of density and mass. 16. A friend who has not had the benefit of an astronomy course suggests that the Sun must be full of burning coal to shine as brightly as it does. List as many arguments as you can against this hypothesis. 17. Which of the following transformations is (are) fusion and which is (are) fission: helium to carbon, carbon to iron, uranium to lead, boron to carbon, oxygen to neon? (See Appendix K for a list of the elements.) 18. Why is a higher temperature required to fuse hydrogen to helium by means of the CNO cycle than is required by the process that occurs in the Sun, which involves only isotopes of hydrogen and helium? 588 Chapter 16 The Sun: A Nuclear Powerhouse 19. Earth\u2019s atmosphere is in hydrostatic equilibrium. What this means is that the pressure at any point in the atmosphere must be high enough to support the weight of air above it. How would you expect the pressure on Mt. Everest to differ from the pressure in your classroom? Explain why. 20. Explain what it means when we say that Earth\u2019s oceans are in hydrostatic equilibrium. Now suppose you are a scuba diver. Would you expect the pressure to increase or decrease as you dive below the surface to a depth of 200 feet? Why? 21. What mechanism transfers heat away from the surface of the Moon? If the Moon is losing energy in this way, why does it not simply become colder and colder? 22. Suppose you are standing a few feet away from a bonfire on a cold fall evening. Your face begins to feel hot. What is the mechanism that transfers heat from the fire to your face? (Hint: Is the air between you and the fire hotter or cooler than your face?) 23. Give some everyday examples of the transport of heat by convection and by radiation. 24. Suppose the proton-proton cycle in the Sun were to slow", " down suddenly and generate energy at only 95% of its current rate. Would an observer on Earth see an immediate decrease in the Sun\u2019s brightness? Would she immediately see a decrease in the number of neutrinos emitted by the Sun? 25. Do you think that nuclear fusion takes place in the atmospheres of stars? Why or why not? 26. Why is fission not an important energy source in the Sun? 27. Why do you suppose so great a fraction of the Sun\u2019s energy comes from its central regions? Within what fraction of the Sun\u2019s radius does practically all of the Sun\u2019s luminosity originate (see Figure 16.16)? Within what radius of the Sun has its original hydrogen been partially used up? Discuss what relationship the answers to these questions bear to one another. 28. Explain how mathematical computer models allow us to understand what is going on inside of the Sun. Figuring For Yourself 29. Estimate the amount of mass that is converted to energy when a proton combines with a deuterium nucleus to form tritium, 3He. 30. How much energy is released when a proton combines with a deuterium nucleus to produce tritium, 3He? 31. The Sun converts 4 \u00d7 109 kg of mass to energy every second. How many years would it take the Sun to convert a mass equal to the mass of Earth to energy? 32. Assume that the mass of the Sun is 75% hydrogen and that all of this mass could be converted to energy according to Einstein\u2019s equation E = mc2. How much total energy could the Sun generate? If m is in kg and c is in m/s, then E will be expressed in J. (The mass of the Sun is given in Appendix E.) 33. In fact, the conversion of mass to energy in the Sun is not 100% efficient. As we have seen in the text, the conversion of four hydrogen atoms to one helium atom results in the conversion of about 0.02862 times the mass of a proton to energy. How much energy in joules does one such reaction produce? (See Appendix E for the mass of the hydrogen atom, which, for all practical purposes, is the mass of a proton.) 34. Now suppose that all of the hydrogen atoms in the Sun were converted into helium. How much total energy would be produced? (To calculate the answer, you will have to estimate how many hydrogen atoms are in the Sun. This will give you good practice with scientific notation, since the numbers involved are very large! See Appendix C for a review of scientific notation.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 16 The Sun: A Nuclear Powerhouse 589 35. Models of the Sun indicate that only about 10% of the total hydrogen in the Sun will participate in nuclear reactions, since it is only the hydrogen in the central regions that is at a high enough temperature. Use the total energy radiated per second by the Sun, 3.8 \u00d7 1026 watts, alongside the exercises and information given here to estimate the lifetime of the Sun. (Hint: Make sure you keep track of the units: if the luminosity is the energy radiated per second, your answer will also be in seconds. You should convert the answer to something more meaningful, such as years.) 36. Show that the statement in the text is correct: namely, that roughly 600 million tons of hydrogen must be converted to helium in the Sun each second to explain its energy output. (Hint: Recall Einstein\u2019s most famous formula, and remember that for each kg of hydrogen, 0.0071 kg of mass is converted into energy.) How long will it be before 10% of the hydrogen is converted into helium? Does this answer agree with the lifetime you calculated in Exercise 16.35? 37. Every second, the Sun converts 4 million tons of matter to energy. How long will it take the Sun to reduce its mass by 1% (the mass of the Sun is 2 \u00d7 1030)? Compare your answer with the lifetime of the Sun so far. 38. Raymond Davis Jr.\u2019s neutrino detector contained approximately 1030 chlorine atoms. During his experiment, he found that one neutrino reacted with a chlorine atom to produce one argon atom each day. A. How many days would he have to run the experiment for 1% of his tank to be filled with argon atoms? B. Convert your answer from A. into years. C. Compare this answer to the age of the universe, which is approximately 14 billion years (1.4 \u00d7 1010 y). D. What does this tell you about how frequently neutrinos interact with matter? 590 Chapter 16 The Sun: A Nuclear Powerhouse This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing", " Starlight 591 17 ANALYZING STARLIGHT Figure 17.1. Star Colors. This long time exposure shows the colors of the stars. The circular motion of the stars across the image is provided by Earth\u2019s rotation. The various colors of the stars are caused by their different temperatures. (credit: modification of work by ESO/A.Santerne) Chapter Outline 17.1 The Brightness of Stars 17.2 Colors of Stars 17.3 The Spectra of Stars (and Brown Dwarfs) 17.4 Using Spectra to Measure Stellar Radius, Composition, and Motion Thinking Ahead Everything we know about stars\u2014how they are born, what they are made of, how far away they are, how long they live, and how they will die\u2014we learn by decoding the messages contained in the light and radiation that reaches Earth. What questions should we ask, and how do we find the answers? We can begin our voyage to the stars by looking at the night sky. It is obvious that stars do not all appear equally bright, nor are they all the same color. To understand the stars, we must first determine their basic properties, such as what their temperatures are, how much material they contain (their masses), and how much energy they produce. Since our Sun is a star, of course the same techniques, including spectroscopy, used to study the Sun can be used to find out what stars are like. As we learn more about the stars, we will use these characteristics to begin assembling clues to the main problems we are interested in solving: How do stars form? How long do they survive? What is their ultimate fate? 17.1 THE BRIGHTNESS OF STARS Learning Objectives By the end of this section, you will be able to: Explain the difference between luminosity and apparent brightness 592 Chapter 17 Analyzing Starlight Understand how astronomers specify brightness with magnitudes Luminosity Perhaps the most important characteristic of a star is its luminosity\u2014the total amount of energy at all wavelengths that it emits per second. Earlier, we saw that the Sun puts out a tremendous amount of energy every second. (And there are stars far more luminous than the Sun out there.) To make the comparison among stars easy, astronomers express the luminosity of other stars in terms of the Sun\u2019s luminosity. For example, the luminosity of Sirius is about 25 times that of the Sun. We use the symbol LSun to denote the Sun\u2019s luminosity; hence, that of Sirius can be written as 25 LSun. In a later chapter, we will see that if we can measure how much energy a star emits and we also know its mass, then we can calculate how long it can continue to shine before it exhausts its nuclear energy and begins to die. Apparent Brightness Astronomers are careful to distinguish between the luminosity of the star (the total energy output) and the amount of energy that happens to reach our eyes or a telescope on Earth. Stars are democratic in how they produce radiation; they emit the same amount of energy in every direction in space. Consequently, only a minuscule fraction of the energy given off by a star actually reaches an observer on Earth. We call the amount of a star\u2019s energy that reaches a given area (say, one square meter) each second here on Earth its apparent brightness. If you look at the night sky, you see a wide range of apparent brightnesses among the stars. Most stars, in fact, are so dim that you need a telescope to detect them. If all stars were the same luminosity\u2014if they were like standard bulbs with the same light output\u2014we could use the difference in their apparent brightnesses to tell us something we very much want to know: how far away they are. Imagine you are in a big concert hall or ballroom that is dark except for a few dozen 25-watt bulbs placed in fixtures around the walls. Since they are all 25-watt bulbs, their luminosity (energy output) is the same. But from where you are standing in one corner, they do not have the same apparent brightness. Those close to you appear brighter (more of their light reaches your eye), whereas those far away appear dimmer (their light has spread out more before reaching you). In this way, you can tell which bulbs are closest to you. In the same way, if all the stars had the same luminosity, we could immediately infer that the brightest-appearing stars were close by and the dimmest-appearing ones were far away. To pin down this idea more precisely, recall from the Radiation and Spectra chapter that we know exactly how light fades with increasing distance. The energy we receive is inversely proportional to the square of the distance. If, for example, we have two stars of the same luminosity and one is twice as far away as the other, it will look four times dimmer than the closer one. If it is", " three times farther away, it will look nine (three squared) times dimmer, and so forth. Alas, the stars do not all have the same luminosity. (Actually, we are pretty glad about that because having many different types of stars makes the universe a much more interesting place.) But this means that if a star looks dim in the sky, we cannot tell whether it appears dim because it has a low luminosity but is relatively nearby, or because it has a high luminosity but is very far away. To measure the luminosities of stars, we must first compensate for the dimming effects of distance on light, and to do that, we must know how far away they are. Distance is among the most difficult of all astronomical measurements. We will return to how it is determined after we have learned more about the stars. For now, we will describe how astronomers specify the apparent brightness of stars. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 593 The Magnitude Scale The process of measuring the apparent brightness of stars is called photometry (from the Greek photo meaning \u201clight\u201d and \u2013metry meaning \u201cto measure\u201d). As we saw Observing the Sky: The Birth of Astronomy, astronomical photometry began with Hipparchus. Around 150 B.C.E., he erected an observatory on the island of Rhodes in the Mediterranean. There he prepared a catalog of nearly 1000 stars that included not only their positions but also estimates of their apparent brightnesses. Hipparchus did not have a telescope or any instrument that could measure apparent brightness accurately, so he simply made estimates with his eyes. He sorted the stars into six brightness categories, each of which he called a magnitude. He referred to the brightest stars in his catalog as first-magnitudes stars, whereas those so faint he could barely see them were sixth-magnitude stars. During the nineteenth century, astronomers attempted to make the scale more precise by establishing exactly how much the apparent brightness of a sixth- magnitude star differs from that of a first-magnitude star. Measurements showed that we receive about 100 times more light from a first-magnitude star than from a sixth-magnitude star. Based on this measurement, astronomers then defined an accurate magnitude system in which a difference of five magnitudes corresponds exactly to a brightness ratio of 100:1. In addition, the magnitudes of stars are decimalized; for example, a star isn\u2019t just a \u201csecond-magnitude star,\u201d it has a magnitude of 2.0 (or 2.1, 2.3, and so forth). So what number is it that, when multiplied together five times, gives you this factor of 100? Play on your calculator and see if you can get it. The answer turns out to be about 2.5, which is the fifth root of 100. This means that a magnitude 1.0 star and a magnitude 2.0 star differ in brightness by a factor of about 2.5. Likewise, we receive about 2.5 times as much light from a magnitude 2.0 star as from a magnitude 3.0 star. What about the difference between a magnitude 1.0 star and a magnitude 3.0 star? Since the difference is 2.5 times for each \u201cstep\u201d of magnitude, the total difference in brightness is 2.5 \u00d7 2.5 = 6.25 times. Here are a few rules of thumb that might help those new to this system. If two stars differ by 0.75 magnitudes, they differ by a factor of about 2 in brightness. If they are 2.5 magnitudes apart, they differ in brightness by a factor of 10, and a 4-magnitude difference corresponds to a difference in brightness of a factor of 40.You might be saying to yourself at this point, \u201cWhy do astronomers continue to use this complicated system from more than 2000 years ago?\u201d That\u2019s an excellent question and, as we shall discuss, astronomers today can use other ways of expressing how bright a star looks. But because this system is still used in many books, star charts, and computer apps, we felt we had to introduce students to it (even though we were very tempted to leave it out.) The brightest stars, those that were traditionally referred to as first-magnitude stars, actually turned out (when measured accurately) not to be identical in brightness. For example, the brightest star in the sky, Sirius, sends us about 10 times as much light as the average first-magnitude star. On the modern magnitude scale, Sirius, the star with the brightest apparent magnitude, has been assigned a magnitude of \u22121.5. Other objects in the sky can appear even brighter. Venus at its brightest is of magnitude \u22124.4, while the Sun has a magnitude of \u221226.8. Figure", " 17.2 shows the range of observed magnitudes from the brightest to the faintest, along with the actual magnitudes of several well-known objects. The important fact to remember when using magnitude is that the system goes backward: the larger the magnitude, the fainter the object you are observing. 594 Chapter 17 Analyzing Starlight Figure 17.2. Apparent Magnitudes of Well-Known Objects. The faintest magnitudes that can be detected by the unaided eye, binoculars, and large telescopes are also shown The Magnitude Equation Even scientists can\u2019t calculate fifth roots in their heads, so astronomers have summarized the above discussion in an equation to help calculate the difference in brightness for stars with different magnitudes. If m1 and m2 are the magnitudes of two stars, then we can calculate the ratio of their brightness \u239b b2 \u239d b1 \u239e \u23a0 using this equation: \u239b b2 m1 \u2212 m2 = 2.5 log \u239d b1 \u239e \u23a0 or m1 \u2212 m2 = 2.5 b2 b1 Here is another way to write this equation: m1 \u2212 m2 b2 b1 \u239d1000.2\u239e \u239b \u23a0 = Let\u2019s do a real example, just to show how this works. Imagine that an astronomer has discovered something special about a dim star (magnitude 8.5), and she wants to tell her students how much dimmer the star is than Sirius. Star 1 in the equation will be our dim star and star 2 will be Sirius. Solution Remember, Sirius has a magnitude of \u22121.5. In that case: \u239d1000.2\u239e \u239b \u23a0 = 8.5 \u2212 (\u22121.5) 10 \u239d1000.2\u239e \u239b \u23a0 = b2 b1 = (100)2 = 100 \u00d7 100 = 10,000 Check Your Learning This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 595 It is a common misconception that Polaris (magnitude 2.0) is the brightest star in the sky, but, as we saw, that distinction actually belongs to Sirius (magnitude \u22121.5). How does Sirius\u2019 apparent brightness compare to that of Polaris? Answer: bSirius bPolaris \u239d1000.2\u239e \u239b \u23a0 = 2.0 \u2212 (\u22121.5) 3.5 \u239d1000.2\u239e \u239b \u23a0 = = 1000.7 = 25 (Hint: If you only have a basic calculator, you may wonder how to take 100 to the 0.7th power. But this is something you can ask Google to do. Google now accepts mathematical questions and will answer them. So try it for yourself. Ask Google, \u201cWhat is 100 to the 0.7th power?\u201d) Our calculation shows that Sirius\u2019 apparent brightness is 25 times greater than Polaris\u2019 apparent brightness. Other Units of Brightness Although the magnitude scale is still used for visual astronomy, it is not used at all in newer branches of the field. In radio astronomy, for example, no equivalent of the magnitude system has been defined. Rather, radio astronomers measure the amount of energy being collected each second by each square meter of a radio telescope and express the brightness of each source in terms of, for example, watts per square meter. Similarly, most researchers in the fields of infrared, X-ray, and gamma-ray astronomy use energy per area per second rather than magnitudes to express the results of their measurements. Nevertheless, astronomers in all fields are careful to distinguish between the luminosity of the source (even when that luminosity is all in X- rays) and the amount of energy that happens to reach us on Earth. After all, the luminosity is a really important characteristic that tells us a lot about the object in question, whereas the energy that reaches Earth is an accident of cosmic geography. To make the comparison among stars easy, in this text, we avoid the use of magnitudes as much as possible and will express the luminosity of other stars in terms of the Sun\u2019s luminosity. For example, the luminosity of Sirius is 25 times that of the Sun. We use the symbol LSun to denote the Sun\u2019s luminosity; hence, that of Sirius can be written as 25 LSun. 17.2 COLORS OF STARS Learning Objectives By the end of this section, you will be able to: Compare the relative temperatures of stars based on their colors Understand how astronomers use color indexes to measure the temperatures of stars Look at the beautiful picture of the stars in the Sagittarius Star Cloud shown in Figure 17.3. The stars show a", " multitude of colors, including red, orange, yellow, white, and blue. As we have seen, stars are not all the same color because they do not all have identical temperatures. To define color precisely, astronomers have devised quantitative methods for characterizing the color of a star and then using those colors to determine stellar 596 Chapter 17 Analyzing Starlight temperatures. In the chapters that follow, we will provide the temperature of the stars we are describing, and this section tells you how those temperatures are determined from the colors of light the stars give off. Figure 17.3. Sagittarius Star Cloud. This image, which was taken by the Hubble Space Telescope, shows stars in the direction toward the center of the Milky Way Galaxy. The bright stars glitter like colored jewels on a black velvet background. The color of a star indicates its temperature. Blue-white stars are much hotter than the Sun, whereas red stars are cooler. On average, the stars in this field are at a distance of about 25,000 light-years (which means it takes light 25,000 years to traverse the distance from them to us) and the width of the field is about 13.3 light-years. (credit: Hubble Heritage Team (AURA/STScI/NASA)) Color and Temperature As we learned in The Electromagnetic Spectrum section, Wien\u2019s law relates stellar color to stellar temperature. Blue colors dominate the visible light output of very hot stars (with much additional radiation in the ultraviolet). On the other hand, cool stars emit most of their visible light energy at red wavelengths (with more radiation coming off in the infrared) (Table 17.1). The color of a star therefore provides a measure of its intrinsic or true surface temperature (apart from the effects of reddening by interstellar dust, which will be discussed in Between the Stars: Gas and Dust in Space). Color does not depend on the distance to the object. This should be familiar to you from everyday experience. The color of a traffic signal, for example, appears the same no matter how far away it is. If we could somehow take a star, observe it, and then move it much farther away, its apparent brightness (magnitude) would change. But this change in brightness is the same for all wavelengths, and so its color would remain the same. Example Star Colors and Corresponding Approximate Temperatures Star Color Approximate Temperature Example Blue White Yellow Table 17.1 25,000 K 10,000 K 6000 K This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Spica Vega Sun Chapter 17 Analyzing Starlight 597 Example Star Colors and Corresponding Approximate Temperatures Star Color Approximate Temperature Example Orange Red Table 17.1 4000 K 3000 Aldebaran Betelgeuse Go to this interactive simulation from the University of Colorado (https://openstax.org/l/ 30UofCsimstar) to see the color of a star changing as the temperature is changed. The hottest stars have temperatures of over 40,000 K, and the coolest stars have temperatures of about 2000 K. Our Sun\u2019s surface temperature is about 6000 K; its peak wavelength color is a slightly greenish-yellow. In space, the Sun would look white, shining with about equal amounts of reddish and bluish wavelengths of light. It looks somewhat yellow as seen from Earth\u2019s surface because our planet\u2019s nitrogen molecules scatter some of the shorter (i.e., blue) wavelengths out of the beams of sunlight that reach us, leaving more long wavelength light behind. This also explains why the sky is blue: the blue sky is sunlight scattered by Earth\u2019s atmosphere. Color Indices In order to specify the exact color of a star, astronomers normally measure a star\u2019s apparent brightness through filters, each of which transmits only the light from a particular narrow band of wavelengths (colors). A crude example of a filter in everyday life is a green-colored, plastic, soft drink bottle, which, when held in front of your eyes, lets only the green colors of light through. One commonly used set of filters in astronomy measures stellar brightness at three wavelengths corresponding to ultraviolet, blue, and yellow light. The filters are named: U (ultraviolet), B (blue), and V (visual, for yellow). These filters transmit light near the wavelengths of 360 nanometers (nm), 420 nm, and 540 nm, respectively. The brightness measured through each filter is usually expressed in magnitudes. The difference between any two of these magnitudes\u2014say, between the blue and the visual magnitudes (B\u2013V)\u2014is called a color index Go to this light and filters simulator (https://openstax.org/l/30lightfiltsim) for a demonstration of how different light sources and filters can combine to determine the observed spectrum. You can also see how the perceived colors are associated with the spectrum. By agreement among", " astronomers, the ultraviolet, blue, and visual magnitudes of the UBV system are adjusted to give a color index of 0 to a star with a surface temperature of about 10,000 K, such as Vega. The B\u2013V color 598 Chapter 17 Analyzing Starlight indexes of stars range from \u22120.4 for the bluest stars, with temperatures of about 40,000 K, to +2.0 for the reddest stars, with temperatures of about 2000 K. The B\u2013V index for the Sun is about +0.65. Note that, by convention, the B\u2013V index is always the \u201cbluer\u201d minus the \u201credder\u201d color. Why use a color index if it ultimately implies temperature? Because the brightness of a star through a filter is what astronomers actually measure, and we are always more comfortable when our statements have to do with measurable quantities. 17.3 THE SPECTRA OF STARS (AND BROWN DWARFS) Learning Objectives By the end of this section, you will be able to: Describe how astronomers use spectral classes to characterize stars Explain the difference between a star and a brown dwarf Measuring colors is only one way of analyzing starlight. Another way is to use a spectrograph to spread out the light into a spectrum (see the Radiation and Spectra and the Astronomical Instruments chapters). In 1814, the German physicist Joseph Fraunhofer observed that the spectrum of the Sun shows dark lines crossing a continuous band of colors. In the 1860s, English astronomers Sir William Huggins and Lady Margaret Huggins (Figure 17.4) succeeded in identifying some of the lines in stellar spectra as those of known elements on Earth, showing that the same chemical elements found in the Sun and planets exist in the stars. Since then, astronomers have worked hard to perfect experimental techniques for obtaining and measuring spectra, and they have developed a theoretical understanding of what can be learned from spectra. Today, spectroscopic analysis is one of the cornerstones of astronomical research. Figure 17.4. William Huggins (1824\u20131910) and Margaret Huggins (1848\u20131915). William and Margaret Huggins were the first to identify the lines in the spectrum of a star other than the Sun; they also took the first spectrogram, or photograph of a stellar spectrum. Formation of Stellar Spectra When the spectra of different stars were first observed, astronomers found that they were not all identical. Since the dark lines are produced by the chemical elements present in the stars, astronomers first thought that the spectra differ from one another because stars are not all made of the same chemical elements. This hypothesis turned out to be wrong. The primary reason that stellar spectra look different is because the stars have different temperatures. Most stars have nearly the same composition as the Sun, with only a few exceptions. Hydrogen, for example, is by far the most abundant element in most stars. However, lines of hydrogen are not seen in the spectra of the hottest and the coolest stars. In the atmospheres of the hottest stars, hydrogen atoms This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 599 are completely ionized. Because the electron and the proton are separated, ionized hydrogen cannot produce absorption lines. (Recall from the Formation of Spectral Lines section, the lines are the result of electrons in orbit around a nucleus changing energy levels.) In the atmospheres of the coolest stars, hydrogen atoms have their electrons attached and can switch energy levels to produce lines. However, practically all of the hydrogen atoms are in the lowest energy state (unexcited) in these stars and thus can absorb only those photons able to lift an electron from that first energy level to a higher level. Photons with enough energy to do this lie in the ultraviolet part of the electromagnetic spectrum, and there are very few ultraviolet photons in the radiation from a cool star. What this means is that if you observe the spectrum of a very hot or very cool star with a typical telescope on the surface of Earth, the most common element in that star, hydrogen, will show very weak spectral lines or none at all. The hydrogen lines in the visible part of the spectrum (called Balmer lines) are strongest in stars with intermediate temperatures\u2014not too hot and not too cold. Calculations show that the optimum temperature for producing visible hydrogen lines is about 10,000 K. At this temperature, an appreciable number of hydrogen atoms are excited to the second energy level. They can then absorb additional photons, rise to still-higher levels of excitation, and produce a dark absorption line. Similarly, every other chemical element, in each of its possible stages of ionization, has a characteristic temperature at which it is most effective in producing absorption lines in any particular part of the spectrum. Classification of Stellar Spectra Astronomers use the patterns of lines", " observed in stellar spectra to sort stars into a spectral class. Because a star\u2019s temperature determines which absorption lines are present in its spectrum, these spectral classes are a measure of its surface temperature. There are seven standard spectral classes. From hottest to coldest, these seven spectral classes are designated O, B, A, F, G, K, and M. Recently, astronomers have added three additional classes for even cooler objects\u2014L, T, and Y. At this point, you may be looking at these letters with wonder and asking yourself why astronomers didn\u2019t call the spectral types A, B, C, and so on. You will see, as we tell you the history, that it\u2019s an instance where tradition won out over common sense. In the 1880s, Williamina Fleming devised a system to classify stars based on the strength of hydrogen absorption lines. Spectra with the strongest lines were classified as \u201cA\u201d stars, the next strongest \u201cB,\u201d and so on down the alphabet to \u201cO\u201d stars, in which the hydrogen lines were very weak. But we saw above that hydrogen lines alone are not a good indicator for classifying stars, since their lines disappear from the visible light spectrum when the stars get too hot or too cold. In the 1890s, Annie Jump Cannon revised this classification system, focusing on just a few letters from the original system: A, B, F, G, K, M, and O. Instead of starting over, Cannon also rearranged the existing classes\u2014in order of decreasing temperature\u2014into the sequence we have learned: O, B, A, F, G, K, M. As you can read in the feature on Annie Cannon: Classifier of the Stars in this chapter, she classified around 500,000 stars over her lifetime, classifying up to three stars per minute by looking at the stellar spectra For a deep dive into spectral types, explore the interactive project at the Sloan Digital Sky Survey (https://openstax.org/l/30sloandigsky) in which you can practice classifying stars yourself. 600 Chapter 17 Analyzing Starlight To help astronomers remember this crazy order of letters, Cannon created a mnemonic, \u201cOh Be A Fine Girl, Kiss Me.\u201d (If you prefer, you can easily substitute \u201cGuy\u201d for \u201cGirl.\u201d) Other mnemonics, which we hope will not be relevant for you, include \u201cOh Brother, Astronomers Frequently Give Killer Midterms\u201d and \u201cOh Boy, An F Grade Kills Me!\u201d With the new L, T, and Y spectral classes, the mnemonic might be expanded to \u201cOh Be A Fine Girl (Guy), Kiss Me Like That, Yo!\u201d Each of these spectral classes, except possibly for the Y class which is still being defined, is further subdivided into 10 subclasses designated by the numbers 0 through 9. A B0 star is the hottest type of B star; a B9 star is the coolest type of B star and is only slightly hotter than an A0 star. And just one more item of vocabulary: for historical reasons, astronomers call all the elements heavier than helium metals, even though most of them do not show metallic properties. (If you are getting annoyed at the peculiar jargon that astronomers use, just bear in mind that every field of human activity tends to develop its own specialized vocabulary. Just try reading a credit card or social media agreement form these days without training in law!) Let\u2019s take a look at some of the details of how the spectra of the stars change with temperature. (It is these details that allowed Annie Cannon to identify the spectral types of stars as quickly as three per minute!) As Figure 17.5 shows, in the hottest O stars (those with temperatures over 28,000 K), only lines of ionized helium and highly ionized atoms of other elements are conspicuous. Hydrogen lines are strongest in A stars with atmospheric temperatures of about 10,000 K. Ionized metals provide the most conspicuous lines in stars with temperatures from 6000 to 7500 K (spectral type F). In the coolest M stars (below 3500 K), absorption bands of titanium oxide and other molecules are very strong. By the way, the spectral class assigned to the Sun is G2. The sequence of spectral classes is summarized in Table 17.2. Figure 17.5. Absorption Lines in Stars of Different Temperatures. This graph shows the strengths of absorption lines of different chemical species (atoms, ions, molecules) as we move from hot (left) to cool (right) stars. The sequence of spectral types is also shown. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 601 Spectral Classes for Stars Spectral Class Color Approximate Temperature (K) Principal Features Examples", " O B A F G K Blue > 30,000 Neutral and ionized helium lines, weak hydrogen lines 10 Lacertae Bluewhite 10,000\u201330,000 Neutral helium lines, strong hydrogen lines Rigel, Spica White 7500\u201310,000 Strongest hydrogen lines, weak ionized calcium lines, weak ionized metal (e.g., iron, magnesium) lines Sirius, Vega Yellowwhite 6000\u20137500 Strong hydrogen lines, strong ionized calcium lines, weak sodium lines, many ionized metal lines Canopus, Procyon Yellow 5200\u20136000 Weaker hydrogen lines, strong ionized calcium lines, strong sodium lines, many lines of ionized and neutral metals Sun, Capella Orange 3700\u20135200 Very weak hydrogen lines, strong ionized calcium lines, strong sodium lines, many lines of neutral metals Arcturus, Aldebaran M Red 2400\u20133700 Strong lines of neutral metals and molecular bands of titanium oxide dominate Betelgeuse, Antares Red 1300\u20132400 Metal hydride lines, alkali metal lines (e.g., sodium, potassium, rubidium) Teide 1 Magenta 700\u20131300 Methane lines Infrared[1] < 700 Ammonia lines Gliese 229B WISE 1828+2650 L T Y Table 17.2 To see how spectral classification works, let\u2019s use Figure 17.5. Suppose you have a spectrum in which the hydrogen lines are about half as strong as those seen in an A star. Looking at the lines in our figure, you see that the star could be either a B star or a G star. But if the spectrum also contains helium lines, then it is a B star, whereas if it contains lines of ionized iron and other metals, it must be a G star. If you look at Figure 17.6, you can see that you, too, could assign a spectral class to a star whose type was not already known. All you have to do is match the pattern of spectral lines to a standard star (like the ones shown in the figure) whose type has already been determined. 1 Absorption by sodium and potassium atoms makes Y dwarfs appear a bit less red than L dwarfs. 602 Chapter 17 Analyzing Starlight Figure 17.6. Spectra of Stars with Different Spectral Classes. This image compares the spectra of the different spectral classes. The spectral class assigned to each of these stellar spectra is listed at the left of the picture. The strongest four lines seen at spectral type A1 (one in the red, one in the blue-green, and two in the blue) are Balmer lines of hydrogen. Note how these lines weaken at both higher and lower temperatures, as Figure 17.5 also indicates. The strong pair of closely spaced lines in the yellow in the cool stars is due to neutral sodium (one of the neutral metals in Figure 17.5). (Credit: modification of work by NOAO/AURA/NSF) Both colors and spectral classes can be used to estimate the temperature of a star. Spectra are harder to measure because the light has to be bright enough to be spread out into all colors of the rainbow, and detectors must be sensitive enough to respond to individual wavelengths. In order to measure colors, the detectors need only respond to the many wavelengths that pass simultaneously through the colored filters that have been chosen\u2014that is, to all the blue light or all the yellow-green light Annie Cannon: Classifier of the Stars Annie Jump Cannon was born in Delaware in 1863 (Figure 17.7). In 1880, she went to Wellesley College, one of the new breed of US colleges opening up to educate young women. Wellesley, only 5 years old at the time, had the second student physics lab in the country and provided excellent training in basic science. After college, Cannon spent a decade with her parents but was very dissatisfied, longing to do scientific work. After her mother\u2019s death in 1893, she returned to Wellesley as a teaching assistant and also to take courses at Radcliffe, the women\u2019s college associated with Harvard. Figure 17.7. Annie Jump Cannon (1863\u20131941). Cannon is well-known for her classifications of stellar spectra. (credit: modification of work by Smithsonian Institution) In the late 1800s, the director of the Harvard Observatory, Edward C. Pickering, needed lots of help with his ambitious program of classifying stellar spectra. The basis for these studies was a monumental collection of nearly a million photographic spectra of stars, obtained from many years of observations This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 603 made at Harvard College Observatory in Massachusetts as well as at its remote observing stations in South America and South Africa. Pickering quickly discovered that educated young women could be hired as assistants for one-third or one-", "fourth the salary paid to men, and they would often put up with working conditions and repetitive tasks that men with the same education would not tolerate. These women became known as the Harvard Computers. (We should emphasize that astronomers were not alone in reaching such conclusions about the relatively new idea of upper-class, educated women working outside the home: women were exploited and undervalued in many fields. This is a legacy from which our society is just beginning to emerge.) Cannon was hired by Pickering as one of the \u201ccomputers\u201d to help with the classification of spectra. She became so good at it that she could visually examine and determine the spectral types of several hundred stars per hour (dictating her conclusions to an assistant). She made many discoveries while investigating the Harvard photographic plates, including 300 variable stars (stars whose luminosity changes periodically). But her main legacy is a marvelous catalog of spectral types for hundreds of thousands of stars, which served as a foundation for much of twentieth-century astronomy. In 1911, a visiting committee of astronomers reported that \u201cshe is the one person in the world who can do this work quickly and accurately\u201d and urged Harvard to give Cannon an official appointment in keeping with her skill and renown. Not until 1938, however, did Harvard appoint her an astronomer at the university; she was then 75 years old. Cannon received the first honorary degree Oxford awarded to a woman, and she became the first woman to be elected an officer of the American Astronomical Society, the main professional organization of astronomers in the US. She generously donated the money from one of the major prizes she had won to found a special award for women in astronomy, now known as the Annie Jump Cannon Prize. True to form, she continued classifying stellar spectra almost to the very end of her life in 1941. Spectral Classes L, T, and Y The scheme devised by Cannon worked well until 1988, when astronomers began to discover objects even cooler than M9-type stars. We use the word object because many of the new discoveries are not true stars. A star is defined as an object that during some part of its lifetime derives 100% of its energy from the same process that makes the Sun shine\u2014the fusion of hydrogen nuclei (protons) into helium. Objects with masses less than about 7.5% of the mass of our Sun (about 0.075 MSun) do not become hot enough for hydrogen fusion to take place. Even before the first such \u201cfailed star\u201d was found, this class of objects, with masses intermediate between stars and planets, was given the name brown dwarfs. Brown dwarfs are very difficult to observe because they are extremely faint and cool, and they put out most of their light in the infrared part of the spectrum. It was only after the construction of very large telescopes, like the Keck telescopes in Hawaii, and the development of very sensitive infrared detectors, that the search for brown dwarfs succeeded. The first brown dwarf was discovered in 1988, and, as of the summer of 2015, there are more than 2200 known brown dwarfs. Initially, brown dwarfs were given spectral classes like M10+ or \u201cmuch cooler than M9,\u201d but so many are now known that it is possible to begin assigning spectral types. The hottest brown dwarfs are given types L0\u2013L9 (temperatures in the range 2400\u20131300 K), whereas still cooler (1300\u2013700 K) objects are given types T0\u2013T9 (see Figure 17.8). In class L brown dwarfs, the lines of titanium oxide, which are strong in M stars, have disappeared. This is because the L dwarfs are so cool that atoms and molecules can gather together into dust particles in their atmospheres; the titanium is locked up in the dust grains rather than being available to form molecules of 604 Chapter 17 Analyzing Starlight titanium oxide. Lines of steam (hot water vapor) are present, along with lines of carbon monoxide and neutral sodium, potassium, cesium, and rubidium. Methane (CH4) lines are strong in class-T brown dwarfs, as methane exists in the atmosphere of the giant planets in our own solar system. In 2009, astronomers discovered ultra-cool brown dwarfs with temperatures of 500\u2013600 K. These objects exhibited absorption lines due to ammonia (NH3), which are not seen in T dwarfs. A new spectral class, Y, was created for these objects. As of 2015, over two dozen brown dwarfs belonging to spectral class Y have been discovered, some with temperatures comparable to that of the human body (about 300 K). Figure 17.8. Brown Dwarfs. This illustration shows the sizes and surface temperatures of brown dwarfs Teide 1, Gliese 229B, and WISE1828 in relation to the Sun, a red dwarf star (Gliese 229A), and Jupiter. (credit: modification of work by MPIA/V. Joergens", ") Most brown dwarfs start out with atmospheric temperatures and spectra like those of true stars with spectral classes of M6.5 and later, even though the brown dwarfs are not hot and dense enough in their interiors to fuse hydrogen. In fact, the spectra of brown dwarfs and true stars are so similar from spectral types late M through L that it is not possible to distinguish the two types of objects based on spectra alone. An independent measure of mass is required to determine whether a specific object is a brown dwarf or a very low mass star. Since brown dwarfs cool steadily throughout their lifetimes, the spectral type of a given brown dwarf changes with time over a billion years or more from late M through L, T, and Y spectral types. Low-Mass Brown Dwarfs vs. High-Mass Planets An interesting property of brown dwarfs is that they are all about the same radius as Jupiter, regardless of their masses. Amazingly, this covers a range of masses from about 13 to 80 times the mass of Jupiter (MJ). This can make distinguishing a low-mass brown dwarf from a high-mass planet very difficult. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 605 So, what is the difference between a low-mass brown dwarf and a high-mass planet? The International Astronomical Union considers the distinctive feature to be deuterium fusion. Although brown dwarfs do not sustain regular (proton-proton) hydrogen fusion, they are capable of fusing deuterium (a rare form of hydrogen with one proton and one neutron in its nucleus). The fusion of deuterium can happen at a lower temperature than the fusion of hydrogen. If an object has enough mass to fuse deuterium (about 13 MJ or 0.012 MSun), it is a brown dwarf. Objects with less than 13 MJ do not fuse deuterium and are usually considered planets. 17.4 USING SPECTRA TO MEASURE STELLAR RADIUS, COMPOSITION, AND MOTION Learning Objectives By the end of this section, you will be able to: Understand how astronomers can learn about a star\u2019s radius and composition by studying its spectrum Explain how astronomers can measure the motion and rotation of a star using the Doppler effect Describe the proper motion of a star and how it relates to a star\u2019s space velocity Analyzing the spectrum of a star can teach us all kinds of things in addition to its temperature. We can measure its detailed chemical composition as well as the pressure in its atmosphere. From the pressure, we get clues about its size. We can also measure its motion toward or away from us and estimate its rotation. Clues to the Size of a Star As we shall see in The Stars: A Celestial Census, stars come in a wide variety of sizes. At some periods in their lives, stars can expand to enormous dimensions. Stars of such exaggerated size are called giants. Luckily for the astronomer, stellar spectra can be used to distinguish giants from run-of-the-mill stars (such as our Sun). Suppose you want to determine whether a star is a giant. A giant star has a large, extended photosphere. Because it is so large, a giant star\u2019s atoms are spread over a great volume, which means that the density of particles in the star\u2019s photosphere is low. As a result, the pressure in a giant star\u2019s photosphere is also low. This low pressure affects the spectrum in two ways. First, a star with a lower-pressure photosphere shows narrower spectral lines than a star of the same temperature with a higher-pressure photosphere (Figure 17.9). The difference is large enough that careful study of spectra can tell which of two stars at the same temperature has a higher pressure (and is thus more compressed) and which has a lower pressure (and thus must be extended). This effect is due to collisions between particles in the star\u2019s photosphere\u2014more collisions lead to broader spectral lines. Collisions will, of course, be more frequent in a higher-density environment. Think about it like traffic\u2014collisions are much more likely during rush hour, when the density of cars is high. Second, more atoms are ionized in a giant star than in a star like the Sun with the same temperature. The ionization of atoms in a star\u2019s outer layers is caused mainly by photons, and the amount of energy carried by photons is determined by temperature. But how long atoms stay ionized depends in part on pressure. Compared with what happens in the Sun (with its relatively dense photosphere), ionized atoms in a giant star\u2019s photosphere are less likely to pass close enough to electrons to interact and combine with one or more of them, thereby becoming neutral again. Ionized atoms, as we discussed earlier, have different spectra from", " atoms that are neutral. 606 Chapter 17 Analyzing Starlight Figure 17.9. Spectral Lines. This figure illustrates one difference in the spectral lines from stars of the same temperature but different pressures. A giant star with a very-low-pressure photosphere shows very narrow spectral lines (bottom), whereas a smaller star with a higherpressure photosphere shows much broader spectral lines (top). (credit: modification of work by NASA, ESA, A. Field, and J. Kalirai (STScI)) Abundances of the Elements Absorption lines of a majority of the known chemical elements have now been identified in the spectra of the Sun and stars. If we see lines of iron in a star\u2019s spectrum, for example, then we know immediately that the star must contain iron. Note that the absence of an element\u2019s spectral lines does not necessarily mean that the element itself is absent. As we saw, the temperature and pressure in a star\u2019s atmosphere will determine what types of atoms are able to produce absorption lines. Only if the physical conditions in a star\u2019s photosphere are such that lines of an element should (according to calculations) be there can we conclude that the absence of observable spectral lines implies low abundance of the element. Suppose two stars have identical temperatures and pressures, but the lines of, say, sodium are stronger in one than in the other. Stronger lines mean that there are more atoms in the stellar photosphere absorbing light. Therefore, we know immediately that the star with stronger sodium lines contains more sodium. Complex calculations are required to determine exactly how much more, but those calculations can be done for any element observed in any star with any temperature and pressure. Of course, astronomy textbooks such as ours always make these things sound a bit easier than they really are. If you look at the stellar spectra such as those in Figure 17.6, you may get some feeling for how hard it is to decode all of the information contained in the thousands of absorption lines. First of all, it has taken many years of careful laboratory work on Earth to determine the precise wavelengths at which hot gases of each element have their spectral lines. Long books and computer databases have been compiled to show the lines of each element that can be seen at each temperature. Second, stellar spectra usually have many lines from a number of elements, and we must be careful to sort them out correctly. Sometimes nature is unhelpful, and lines of different elements have identical wavelengths, thereby adding to the confusion. And third, as we saw in the chapter on Radiation and Spectra, the motion of the star can change the observed wavelength of each of the lines. So, the observed wavelengths may not match laboratory measurements exactly. In practice, analyzing stellar spectra is a demanding, sometimes frustrating task that requires both training and skill. Studies of stellar spectra have shown that hydrogen makes up about three-quarters of the mass of most stars. Helium is the second-most abundant element, making up almost a quarter of a star\u2019s mass. Together, hydrogen and helium make up from 96 to 99% of the mass; in some stars, they amount to more than 99.9%. Among the 4% or less of \u201cheavy elements,\u201d oxygen, carbon, neon, iron, nitrogen, silicon, magnesium, and sulfur are among This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 607 the most abundant. Generally, but not invariably, the elements of lower atomic weight are more abundant than those of higher atomic weight. Take a careful look at the list of elements in the preceding paragraph. Two of the most abundant are hydrogen and oxygen (which make up water); add carbon and nitrogen and you are starting to write the prescription for the chemistry of an astronomy student. We are made of elements that are common in the universe\u2014just mixed together in a far more sophisticated form (and a much cooler environment) than in a star. As we mentioned in The Spectra of Stars (and Brown Dwarfs) section, astronomers use the term \u201cmetals\u201d to refer to all elements heavier than hydrogen and helium. The fraction of a star\u2019s mass that is composed of these elements is referred to as the star\u2019s metallicity. The metallicity of the Sun, for example, is 0.02, since 2% of the Sun\u2019s mass is made of elements heavier than helium. Appendix K lists how common each element is in the universe (compared to hydrogen); these estimates are based primarily on investigation of the Sun, which is a typical star. Some very rare elements, however, have not been detected in the Sun. Estimates of the amounts of these elements in the universe are based on laboratory measurements of their abundance in primitive meteorites, which are considered representative of unaltered material condensed from the solar nebula (see the Cosmic Samples and the", " Origin of the Solar System chapter). Radial Velocity When we measure the spectrum of a star, we determine the wavelength of each of its lines. If the star is not moving with respect to the Sun, then the wavelength corresponding to each element will be the same as those we measure in a laboratory here on Earth. But if stars are moving toward or away from us, we must consider the Doppler effect (see The Doppler Effect section). We should see all the spectral lines of moving stars shifted toward the red end of the spectrum if the star is moving away from us, or toward the blue (violet) end if it is moving toward us (Figure 17.10). The greater the shift, the faster the star is moving. Such motion, along the line of sight between the star and the observer, is called radial velocity and is usually measured in kilometers per second. Figure 17.10. Doppler-Shifted Stars. When the spectral lines of a moving star shift toward the red end of the spectrum, we know that the star is moving away from us. If they shift toward the blue end, the star is moving toward us. 608 Chapter 17 Analyzing Starlight William Huggins, pioneering yet again, in 1868 made the first radial velocity determination of a star. He observed the Doppler shift in one of the hydrogen lines in the spectrum of Sirius and found that this star is moving toward the solar system. Today, radial velocity can be measured for any star bright enough for its spectrum to be observed. As we will see in The Stars: A Celestial Census, radial velocity measurements of double stars are crucial in deriving stellar masses. Proper Motion There is another type of motion stars can have that cannot be detected with stellar spectra. Unlike radial motion, which is along our line of sight (i.e., toward or away from Earth), this motion, called proper motion, is transverse: that is, across our line of sight. We see it as a change in the relative positions of the stars on the celestial sphere (Figure 17.11). These changes are very slow. Even the star with the largest proper motion takes 200 years to change its position in the sky by an amount equal to the width of the full Moon, and the motions of other stars are smaller yet. Figure 17.11. Large Proper Motion. Three photographs of Barnard\u2019s star, the star with the largest known proper motion, show how this faint star has moved over a period of 20 years. (modification of work by Steve Quirk) For this reason, with our naked eyes, we do not notice any change in the positions of the bright stars during the course of a human lifetime. If we could live long enough, however, the changes would become obvious. For example, some 50,000 years from now, terrestrial observers will find the handle of the Big Dipper unmistakably more bent than it is now (Figure 17.12). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 609 Figure 17.12. Changes in the Big Dipper. This figure shows changes in the appearance of the Big Dipper due to proper motion of the stars over 100,000 years. We measure the proper motion of a star in arcseconds (1/3600 of a degree) per year. That is, the measurement of proper motion tells us only by how much of an angle a star has changed its position on the celestial sphere. If two stars at different distances are moving at the same velocity perpendicular to our line of sight, the closer one will show a larger shift in its position on the celestial sphere in a year\u2019s time. As an analogy, imagine you are standing at the side of a freeway. Cars will appear to whiz past you. If you then watch the traffic from a vantage point half a mile away, the cars will move much more slowly across your field of vision. In order to convert this angular motion to a velocity, we need to know how far away the star is. To know the true space velocity of a star\u2014that is, its total speed and the direction in which it is moving through space relative to the Sun\u2014we must know its radial velocity, proper motion, and distance (Figure 17.13). A star\u2019s space velocity can also, over time, cause its distance from the Sun to change significantly. Over several hundred thousand years, these changes can be large enough to affect the apparent brightnesses of nearby stars. Today, Sirius, in the constellation Canis Major (the Big Dog) is the brightest star in the sky, but 100,000 years ago, the star Canopus in the constellation Carina (the Keel) was the brightest one. A little over 200,000 years from now, Sirius will have moved away and faded somewhat, and Vega, the bright blue star in Lyra, will take over its place of honor as", " the brightest star in Earth\u2019s skies. 610 Chapter 17 Analyzing Starlight Figure 17.13. Space Velocity and Proper Motion. This figure shows the true space velocity of a star. The radial velocity is the component of the space velocity projected along the line of sight from the Sun to a star. The transverse velocity is a component of the space velocity projected on the sky. What astronomers measure is proper motion (\u03bc), which is the change in the apparent direction on the sky measured in fractions of a degree. To convert this change in direction to a speed in, say, kilometers per second, it is necessary to also know the distance (d) from the Sun to the star. Rotation We can also use the Doppler effect to measure how fast a star rotates. If an object is rotating, then one of its sides is approaching us while the other is receding (unless its axis of rotation happens to be pointed exactly toward us). This is clearly the case for the Sun or a planet; we can observe the light from either the approaching or receding edge of these nearby objects and directly measure the Doppler shifts that arise from the rotation. Stars, however, are so far away that they all appear as unresolved points. The best we can do is to analyze the light from the entire star at once. Due to the Doppler effect, the lines in the light that come from the side of the star rotating toward us are shifted to shorter wavelengths and the lines in the light from the opposite edge of the star are shifted to longer wavelengths. You can think of each spectral line that we observe as the sum or composite of spectral lines originating from different speeds with respect to us. Each point on the star has its own Doppler shift, so the absorption line we see from the whole star is actually much wider than it would be if the star were not rotating. If a star is rotating rapidly, there will be a greater spread of Doppler shifts and all its spectral lines should be quite broad. In fact, astronomers call this effect line broadening, and the amount of broadening can tell us the speed at which the star rotates (Figure 17.14). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 611 Figure 17.14. Using a Spectrum to Determine Stellar Rotation. A rotating star will show broader spectral lines than a nonrotating star. Measurements of the widths of spectral lines show that many stars rotate faster than the Sun, some with periods of less than a day! These rapid rotators spin so fast that their shapes are \u201cflattened\u201d into what we call oblate spheroids. An example of this is the star Vega, which rotates once every 12.5 hours. Vega\u2019s rotation flattens its shape so much that its diameter at the equator is 23% wider than its diameter at the poles (Figure 17.15). The Sun, with its rotation period of about a month, rotates rather slowly. Studies have shown that stars decrease their rotational speed as they age. Young stars rotate very quickly, with rotational periods of days or less. Very old stars can have rotation periods of several months. 612 Chapter 17 Analyzing Starlight Figure 17.15. Comparison of Rotating Stars. This illustration compares the more rapidly rotating star Altair to the slower rotating Sun. As you can see, spectroscopy is an extremely powerful technique that helps us learn all kinds of information about stars that we simply could not gather any other way. We will see in later chapters that these same techniques can also teach us about galaxies, which are the most distant objects that can we observe. Without spectroscopy, we would know next to nothing about the universe beyond the solar system Astronomy and Philanthropy Throughout the history of astronomy, contributions from wealthy patrons of the science have made an enormous difference in building new instruments and carrying out long-term research projects. Edward Pickering\u2019s stellar classification project, which was to stretch over several decades, was made possible by major donations from Anna Draper. She was the widow of Henry Draper, a physician who was one of the most accomplished amateur astronomers of the nineteenth century and the first person to successfully photograph the spectrum of a star. Anna Draper gave several hundred thousand dollars to Harvard Observatory. As a result, the great spectroscopic survey is still known as the Henry Draper Memorial, and many stars are still referred to by their \u201cHD\u201d numbers in that catalog (such as HD 209458). In the 1870s, the eccentric piano builder and real estate magnate James Lick (Figure 17.16) decided to leave some of his fortune to build the world\u2019s largest telescope. When, in 1887, the pier to house the telescope was finished, Lick\u2019s body was entombed in it. Atop the foundation rose a 36-inch", " refractor, which for many years was the main instrument at the Lick Observatory near San Jose. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 613 Figure 17.16. Henry Draper (1837\u20131882) and James Lick (1796\u20131876). (a) Draper stands next to a telescope used for photography. After his death, his widow funded further astronomy work in his name. (b) Lick was a philanthropist who provided funds to build a 36-inch refractor not only as a memorial to himself but also to aid in further astronomical research. The Lick telescope remained the largest in the world until 1897, when George Ellery Hale persuaded railroad millionaire Charles Yerkes to finance the construction of a 40-inch telescope near Chicago. More recently, Howard Keck, whose family made its fortune in the oil industry, gave $70 million from his family foundation to the California Institute of Technology to help build the world\u2019s largest telescope atop the 14,000-foot peak of Mauna Kea in Hawaii (see the chapter on Astronomical Instruments to learn more about these telescopes). The Keck Foundation was so pleased with what is now called the Keck telescope that they gave $74 million more to build Keck II, another 10-meter reflector on the same volcanic peak. Now, if any of you become millionaires or billionaires, and astronomy has sparked your interest, do keep an astronomical instrument or project in mind as you plan your estate. But frankly, private philanthropy could not possibly support the full enterprise of scientific research in astronomy. Much of our exploration of the universe is financed by federal agencies such as the National Science Foundation and NASA in the United States, and by similar government agencies in the other countries. In this way, all of us, through a very small share of our tax dollars, are philanthropists for astronomy. 614 Chapter 17 Analyzing Starlight CHAPTER 17 REVIEW KEY TERMS apparent brightness a measure of the amount of light received by Earth from a star or other object\u2014that is, how bright an object appears in the sky, as contrasted with its luminosity brown dwarf an object intermediate in size between a planet and a star; the approximate mass range is from about 1/100 of the mass of the Sun up to the lower mass limit for self-sustaining nuclear reactions, which is about 0.075 the mass of the Sun; brown dwarfs are capable of deuterium fusion, but not hydrogen fusion color index difference between the magnitudes of a star or other object measured in light of two different spectral regions\u2014for example, blue minus visual (B\u2013V) magnitudes giant a star of exaggerated size with a large, extended photosphere luminosity the rate at which a star or other object emits electromagnetic energy into space; the total power output of an object magnitude an older system of measuring the amount of light we receive from a star or other luminous object; the larger the magnitude, the less radiation we receive from the object proper motion the angular change per year in the direction of a star as seen from the Sun radial velocity motion toward or away from the observer; the component of relative velocity that lies in the line of sight space velocity the total (three-dimensional) speed and direction with which an object is moving through space relative to the Sun spectral class (or spectral type) the classification of stars according to their temperatures using the characteristics of their spectra; the types are O, B, A, F, G, K, and M with L, T, and Y added recently for cooler star- like objects that recent survey have revealed SUMMARY 17.1 The Brightness of Stars The total energy emitted per second by a star is called its luminosity. How bright a star looks from the perspective of Earth is its apparent brightness. The apparent brightness of a star depends on both its luminosity and its distance from Earth. Thus, the determination of apparent brightness and measurement of the distance to a star provide enough information to calculate its luminosity. The apparent brightnesses of stars are often expressed in terms of magnitudes, which is an old system based on how human vision interprets relative light intensity. 17.2 Colors of Stars Stars have different colors, which are indicators of temperature. The hottest stars tend to appear blue or blue- white, whereas the coolest stars are red. A color index of a star is the difference in the magnitudes measured at any two wavelengths and is one way that astronomers measure and express the temperature of stars. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 615 17.3 The Spectra of Stars (and Brown Dwarfs) The differences in the spectra of stars are principally due to differences in temperature, not composition. The spectra of stars are described in terms of spectral classes. In order of", " decreasing temperature, these spectral classes are O, B, A, F, G, K, M, L, T, and Y. These are further divided into subclasses numbered from 0 to 9. The classes L, T, and Y have been added recently to describe newly discovered star-like objects\u2014mainly brown dwarfs\u2014that are cooler than M9. Our Sun has spectral type G2. 17.4 Using Spectra to Measure Stellar Radius, Composition, and Motion Spectra of stars of the same temperature but different atmospheric pressures have subtle differences, so spectra can be used to determine whether a star has a large radius and low atmospheric pressure (a giant star) or a small radius and high atmospheric pressure. Stellar spectra can also be used to determine the chemical composition of stars; hydrogen and helium make up most of the mass of all stars. Measurements of line shifts produced by the Doppler effect indicate the radial velocity of a star. Broadening of spectral lines by the Doppler effect is a measure of rotational velocity. A star can also show proper motion, due to the component of a star\u2019s space velocity across the line of sight. FOR FURTHER EXPLORATION Articles Berman, B. \u201cMagnitude Cum Laude.\u201d Astronomy (December 1998): 92. How we measure the apparent brightnesses of stars is discussed. Dvorak, J. \u201cThe Women Who Created Modern Astronomy [including Annie Cannon].\u201d Sky & Telescope (August 2013): 28. Hearnshaw, J. \u201cOrigins of the Stellar Magnitude Scale.\u201d Sky & Telescope (November 1992): 494. A good history of how we have come to have this cumbersome system is discussed. Hirshfeld, A. \u201cThe Absolute Magnitude of Stars.\u201d Sky & Telescope (September 1994): 35. Kaler, J. \u201cStars in the Cellar: Classes Lost and Found.\u201d Sky & Telescope (September 2000): 39. An introduction is provided for spectral types and the new classes L and T. Kaler, J. \u201cOrigins of the Spectral Sequence.\u201d Sky & Telescope (February 1986): 129. Skrutskie, M. \u201c2MASS: Unveiling the Infrared Universe.\u201d Sky & Telescope (July 2001): 34. This article focuses on an all-sky survey at 2 microns. Sneden, C. \u201cReading the Colors of the Stars.\u201d Astronomy (April 1989): 36. This article includes a discussion of what we learn from spectroscopy. Steffey, P. \u201cThe Truth about Star Colors.\u201d Sky & Telescope (September 1992): 266. The color index and how the eye and film \u201csee\u201d colors are discussed. Tomkins, J. \u201cOnce and Future Celestial Kings.\u201d Sky & Telescope (April 1989): 59. Calculating the motion of stars and determining which stars were, are, and will be brightest in the sky are discussed. Websites Discovery of Brown Dwarfs: http://w.astro.berkeley.edu/~basri/bdwarfs/SciAm-book.pdf. Listing of Nearby Brown Dwarfs: http://www.solstation.com/stars/pc10bd.htm. 616 Chapter 17 Analyzing Starlight Spectral Types of Stars: http://www.skyandtelescope.com/astronomy-equipment/the-spectral-types-of-stars/. Stellar Velocities https://www.e-education.psu.edu/astro801/content/l4_p7.html. Unheard Voices! The Contributions of Women to Astronomy: A Resource Guide: http://multiverse.ssl.berkeley.edu/women and http://www.astrosociety.org/education/astronomy-resource- guides/women-in-astronomy-an-introductory-resource-guide/. Videos When You Are Just Too Small to be a Star: https://www.youtube.com/watch?v=zXCDsb4n4KU. 2013 Public Talk on Brown Dwarfs and Planets by Dr. Gibor Basri of the University of California\u2013Berkeley (1:32:52). COLLABORATIVE GROUP ACTIVITIES A. The Voyagers in Astronomy feature on Annie Cannon: Classifier of the Stars discusses some of the difficulties women who wanted to do astronomy faced in the first half of the twentieth century. What does your group think about the situation for women today? Do men and women have an equal chance to become scientists? Discuss with your group whether, in your experience, boys and girls were equally encouraged to do science and math where you went to school. B. In the section on magnitudes in The Brightness of Stars, we discussed how this", " old system of classifying how bright different stars appear to the eye first developed. Your authors complained about the fact that this old system still has to be taught to every generation of new students. Can your group think of any other traditional systems of doing things in science and measurement where tradition rules even though common sense says a better system could certainly be found. Explain. (Hint: Try Daylight Savings Time, or metric versus English units.) C. Suppose you could observe a star that has only one spectral line. Could you tell what element that spectral line comes from? Make a list of reasons with your group about why you answered yes or no. D. A wealthy alumnus of your college decides to give $50 million to the astronomy department to build a world-class observatory for learning more about the characteristics of stars. Have your group discuss what kind of equipment they would put in the observatory. Where should this observatory be located? Justify your answers. (You may want to refer back to the Astronomical Instruments chapter and to revisit this question as you learn more about the stars and equipment for observing them in future chapters.) E. For some astronomers, introducing a new spectral type for the stars (like the types L, T, and Y discussed in the text) is similar to introducing a new area code for telephone calls. No one likes to disrupt the old system, but sometimes it is simply necessary. Have your group make a list of steps an astronomer would have to go through to persuade colleagues that a new spectral class is needed. EXERCISES Review Questions 1. What two factors determine how bright a star appears to be in the sky? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 617 2. Explain why color is a measure of a star\u2019s temperature. 3. What is the main reason that the spectra of all stars are not identical? Explain. 4. What elements are stars mostly made of? How do we know this? 5. What did Annie Cannon contribute to the understanding of stellar spectra? 6. Name five characteristics of a star that can be determined by measuring its spectrum. Explain how you would use a spectrum to determine these characteristics. 7. How do objects of spectral types L, T, and Y differ from those of the other spectral types? 8. Do stars that look brighter in the sky have larger or smaller magnitudes than fainter stars? 9. The star Antares has an apparent magnitude of 1.0, whereas the star Procyon has an apparent magnitude of 0.4. Which star appears brighter in the sky? 10. Based on their colors, which of the following stars is hottest? Which is coolest? Archenar (blue), Betelgeuse (red), Capella (yellow). 11. Order the seven basic spectral types from hottest to coldest. 12. What is the defining difference between a brown dwarf and a true star? Thought Questions 13. If the star Sirius emits 23 times more energy than the Sun, why does the Sun appear brighter in the sky? 14. How would two stars of equal luminosity\u2014one blue and the other red\u2014appear in an image taken through a filter that passes mainly blue light? How would their appearance change in an image taken through a filter that transmits mainly red light? 15. Table 17.2 lists the temperature ranges that correspond to the different spectral types. What part of the star do these temperatures refer to? Why? 16. Suppose you are given the task of measuring the colors of the brightest stars, listed in Appendix J, through three filters: the first transmits blue light, the second transmits yellow light, and the third transmits red light. If you observe the star Vega, it will appear equally bright through each of the three filters. Which stars will appear brighter through the blue filter than through the red filter? Which stars will appear brighter through the red filter? Which star is likely to have colors most nearly like those of Vega? 17. Star X has lines of ionized helium in its spectrum, and star Y has bands of titanium oxide. Which is hotter? Why? The spectrum of star Z shows lines of ionized helium and also molecular bands of titanium oxide. What is strange about this spectrum? Can you suggest an explanation? 18. The spectrum of the Sun has hundreds of strong lines of nonionized iron but only a few, very weak lines of helium. A star of spectral type B has very strong lines of helium but very weak iron lines. Do these differences mean that the Sun contains more iron and less helium than the B star? Explain. 618 Chapter 17 Analyzing Starlight 19. What are the approximate spectral classes of stars with the following characteristics? A. Balmer lines of hydrogen are very strong; some lines of ionized metals are present. B. The strongest lines are those of ionized helium. C. Lines of ionized calcium are the strongest", " in the spectrum; hydrogen lines show only moderate strength; lines of neutral and metals are present. D. The strongest lines are those of neutral metals and bands of titanium oxide. 20. Look at the chemical elements in Appendix K. Can you identify any relationship between the abundance of an element and its atomic weight? Are there any obvious exceptions to this relationship? 21. Appendix I lists some of the nearest stars. Are most of these stars hotter or cooler than the Sun? Do any of them emit more energy than the Sun? If so, which ones? 22. Appendix J lists the stars that appear brightest in our sky. Are most of these hotter or cooler than the Sun? Can you suggest a reason for the difference between this answer and the answer to the previous question? (Hint: Look at the luminosities.) Is there any tendency for a correlation between temperature and luminosity? Are there exceptions to the correlation? 23. What star appears the brightest in the sky (other than the Sun)? The second brightest? What color is Betelgeuse? Use Appendix J to find the answers. 24. Suppose hominids one million years ago had left behind maps of the night sky. Would these maps represent accurately the sky that we see today? Why or why not? 25. Why can only a lower limit to the rate of stellar rotation be determined from line broadening rather than the actual rotation rate? (Refer to Figure 17.14.) 26. Why do you think astronomers have suggested three different spectral types (L, T, and Y) for the brown dwarfs instead of M? Why was one not enough? 27. Sam, a college student, just bought a new car. Sam\u2019s friend Adam, a graduate student in astronomy, asks Sam for a ride. In the car, Adam remarks that the colors on the temperature control are wrong. Why did he say that? Figure 17.17. (credit: modification of work by Michael Sheehan) 28. Would a red star have a smaller or larger magnitude in a red filter than in a blue filter? 29. Two stars have proper motions of one arcsecond per year. Star A is 20 light-years from Earth, and Star B is 10 light-years away from Earth. Which one has the faster velocity in space? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 17 Analyzing Starlight 619 30. Suppose there are three stars in space, each moving at 100 km/s. Star A is moving across (i.e., perpendicular to) our line of sight, Star B is moving directly away from Earth, and Star C is moving away from Earth, but at a 30\u00b0 angle to the line of sight. From which star will you observe the greatest Doppler shift? From which star will you observe the smallest Doppler shift? 31. What would you say to a friend who made this statement, \u201cThe visible-light spectrum of the Sun shows weak hydrogen lines and strong calcium lines. The Sun must therefore contain more calcium than hydrogen.\u201d? Figuring For Yourself 32. In Appendix J, how much more luminous is the most luminous of the stars than the least luminous? For Exercise 17.33 through Exercise 17.38, use the equations relating magnitude and apparent brightness given in the section on the magnitude scale in The Brightness of Stars and Example 17.1. 33. Verify that if two stars have a difference of five magnitudes, this corresponds to a factor of 100 in the ratio \u239b b2 \u239d b1 \u239e \u23a0; that 2.5 magnitudes corresponds to a factor of 10; and that 0.75 magnitudes corresponds to a factor of 2. 34. As seen from Earth, the Sun has an apparent magnitude of about \u221226.7. What is the apparent magnitude of the Sun as seen from Saturn, about 10 AU away? (Remember that one AU is the distance from Earth to the Sun and that the brightness decreases as the inverse square of the distance.) Would the Sun still be the brightest star in the sky? 35. An astronomer is investigating a faint star that has recently been discovered in very sensitive surveys of the sky. The star has a magnitude of 16. How much less bright is it than Antares, a star with magnitude roughly equal to 1? 36. The center of a faint but active galaxy has magnitude 26. How much less bright does it look than the very faintest star that our eyes can see, roughly magnitude 6? 37. You have enough information from this chapter to estimate the distance to Alpha Centauri, the second nearest star, which has an apparent magnitude of 0. Since it is a G2 star, like the Sun, assume it has the same luminosity as the Sun and the difference in magnitudes is a result only of the difference in distance. Estimate how far away Alpha Centauri is. Describe the necessary steps in", " words and then do the calculation. (As we will learn in the Celestial Distances chapter, this method\u2014namely, assuming that stars with identical spectral types emit the same amount of energy\u2014is actually used to estimate distances to stars.) If you assume the distance to the Sun is in AU, your answer will come out in AU. 38. Do the previous problem again, this time using the information that the Sun is 150,000,000 km away. You will get a very large number of km as your answer. To get a better feeling for how the distances compare, try calculating the time it takes light at a speed of 299,338 km/s to travel from the Sun to Earth and from Alpha Centauri to Earth. For Alpha Centauri, figure out how long the trip will take in years as well as in seconds. 39. Star A and Star B have different apparent brightnesses but identical luminosities. If Star A is 20 light-years away from Earth and Star B is 40 light-years away from Earth, which star appears brighter and by what factor? 40. Star A and Star B have different apparent brightnesses but identical luminosities. Star A is 10 light-years away from Earth and appears 36 times brighter than Star B. How far away is Star B? 620 Chapter 17 Analyzing Starlight 41. The star Sirius A has an apparent magnitude of \u22121.5. Sirius A has a dim companion, Sirius B, which is 10,000 times less bright than Sirius A. What is the apparent magnitude of Sirius B? Can Sirius B be seen with the naked eye? 42. Our Sun, a type G star, has a surface temperature of 5800 K. We know, therefore, that it is cooler than a type O star and hotter than a type M star. Given what you learned about the temperature ranges of these types of stars, how many times hotter than our Sun is the hottest type O star? How many times cooler than our Sun is the coolest type M star? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 621 18 THE STARS: A CELESTIAL CENSUS Figure 18.1. Variety of Stars. Stars come in a variety of sizes, masses, temperatures, and luminosities. This image shows part of a cluster of stars in the Small Magellanic Cloud (catalog number NGC 290). Located about 200,000 light-years away, NGC 290 is about 65 light-years across. Because the stars in this cluster are all at about the same distance from us, the differences in apparent brightness correspond to differences in luminosity; differences in temperature account for the differences in color. The various colors and luminosities of these stars provide clues about their life stories. (credit: modification of work by E. Olszewski (University of Arizona), European Space Agency, NASA) Chapter Outline 18.1 A Stellar Census 18.2 Measuring Stellar Masses 18.3 Diameters of Stars 18.4 The H\u2013R Diagram Thinking Ahead How do stars form? How long do they live? And how do they die? Stop and think how hard it is to answer these questions. Stars live such a long time that nothing much can be gained from staring at one for a human lifetime. To discover how stars evolve from birth to death, it was necessary to measure the characteristics of many stars (to take a celestial census, in effect) and then determine which characteristics help us understand the stars\u2019 life stories. Astronomers tried a variety of hypotheses about stars until they came up with the right approach to understanding their development. But the key was first making a thorough census of the stars around us. 18.1 A STELLAR CENSUS Learning Objectives By the end of this section, you will be able to: Explain why the stars visible to the unaided eye are not typical Describe the distribution of stellar masses found close to the Sun 622 Chapter 18 The Stars: A Celestial Census Before we can make our own survey, we need to agree on a unit of distance appropriate to the objects we are studying. The stars are all so far away that kilometers (and even astronomical units) would be very cumbersome to use; so\u2014as discussed in Science and the Universe: A Brief Tour\u2014astronomers use a much larger \u201cmeasuring stick\u201d called the light-year. A light-year is the distance that light (the fastest signal we know) travels in 1 year. Since light covers an astounding 300,000 kilometers per second, and since there are a lot of seconds in 1 year, a light-year is a very large quantity: 9.5 trillion (9.5 \u00d7 1012) kilometers to be exact. (Bear in mind that the light-year is a unit of distance even though the term year appears in it.) If you drove at the legal US speed limit", " without stopping for food or rest, you would not arrive at the end of a light-year in space until roughly 12 million years had passed. And the closest star is more than 4 light-years away. Notice that we have not yet said much about how such enormous distances can be measured. That is a complicated question, to which we will return in Celestial Distances. For now, let us assume that distances have been measured for stars in our cosmic vicinity so that we can proceed with our census. Small Is Beautiful\u2014Or at Least More Common When we do a census of people in the United States, we count the inhabitants by neighborhood. We can try the same approach for our stellar census and begin with our own immediate neighborhood. As we shall see, we run into two problems\u2014just as we do with a census of human beings. First, it is hard to be sure we have counted all the inhabitants; second, our local neighborhood may not contain all possible types of people. Table 18.1 shows an estimate of the number of stars of each spectral type[1] in our own local neighborhood\u2014within 21 light-years of the Sun. (The Milky Way Galaxy, in which we live, is about 100,000 light- years in diameter, so this figure really applies to a very local neighborhood, one that contains a tiny fraction of all the billions of stars in the Milky Way.) You can see that there are many more low-luminosity (and hence low mass) stars than high-luminosity ones. Only three of the stars in our local neighborhood (one F type and two A types) are significantly more luminous and more massive than the Sun. This is truly a case where small triumphs over large\u2014at least in terms of numbers. The Sun is more massive than the vast majority of stars in our vicinity. Stars within 21 Light-Years of the Sun Spectral Type Number of Stars A F G K M White dwarfs Brown dwarfs Table 18.1 2 1 7 17 94 8 33 1 The spectral types of stars were defined and discussed in Analyzing Starlight. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 623 This table is based on data published through 2015, and it is likely that more faint objects remain to be discovered (see Figure 18.2). Along with the L and T brown dwarfs already observed in our neighborhood, astronomers expect to find perhaps hundreds of additional T dwarfs. Many of these are likely to be even cooler than the coolest currently known T dwarf. The reason the lowest-mass dwarfs are so hard to find is that they put out very little light\u2014ten thousand to a million times less light than the Sun. Only recently has our technology progressed to the point that we can detect these dim, cool objects. Figure 18.2. Dwarf Simulation. This computer simulation shows the stars in our neighborhood as they would be seen from a distance of 30 light-years away. The Sun is in the center. All the brown dwarfs are circled; those found earlier are circled in blue, the ones found recently with the WISE infrared telescope in space (whose scientists put this diagram together) are circled in red. The common M stars, which are red and faint, are made to look brighter than they really would be so that you can see them in the simulation. Note that luminous hot stars like our Sun are very rare. (credit: modification of work by NASA/ JPL-Caltech) To put all this in perspective, we note that even though the stars counted in the table are our closest neighbors, you can\u2019t just look up at the night sky and see them without a telescope; stars fainter than the Sun cannot be seen with the unaided eye unless they are very nearby. For example, stars with luminosities ranging from 1/100 to 1/10,000 the luminosity of the Sun (LSun) are very common, but a star with a luminosity of 1/100 LSun would have to be within 5 light-years to be visible to the naked eye\u2014and only three stars (all in one system) are this close to us. The nearest of these three stars, Proxima Centauri, still cannot be seen without a telescope because it has such a low luminosity. Astronomers are working hard these days to complete the census of our local neighborhood by finding our faintest neighbors. Recent discoveries of nearby stars have relied heavily upon infrared telescopes that are able to find these many cool, low-mass stars. You should expect the number of known stars within 26 light-years of the Sun to keep increasing as more and better surveys are undertaken. Bright Does Not Necessarily Mean Close If we confine our census to the local neighborhood, we will miss many of the most interesting kinds of stars. After all, the neighborhood in which you live does not contain all the types of people\u2014", "distinguished according to age, education, income, race, and so on\u2014that live in the entire country. For example, a few people do live to be over 100 years old, but there may be no such individual within several miles of where you live. In order to sample the full range of the human population, you would have to extend your census to a much larger area. Similarly, some types of stars simply are not found nearby. 624 Chapter 18 The Stars: A Celestial Census A clue that we are missing something in our stellar census comes from the fact that only six of the 20 stars that appear brightest in our sky\u2014 Sirius, Vega, Altair, Alpha Centauri, Fomalhaut, and Procyon\u2014are found within 26 light-years of the Sun (Figure 18.3). Why are we missing most of the brightest stars when we take our census of the local neighborhood? Figure 18.3. The Closest Stars. (a) This image, taken with a wide-angle telescope at the European Southern Observatory in Chile, shows the system of three stars that is our nearest neighbor. (b) Two bright stars that are close to each other ( Alpha Centauri A and B) blend their light together. (c) Indicated with an arrow (since you\u2019d hardly notice it otherwise) is the much fainter Proxima Centauri star, which is spectral type M. (credit: modification of work by ESO) The answer, interestingly enough, is that the stars that appear brightest are not the ones closest to us. The brightest stars look the way they do because they emit a very large amount of energy\u2014so much, in fact, that they do not have to be nearby to look brilliant. You can confirm this by looking at Appendix J, which gives distances for the 20 stars that appear brightest from Earth. The most distant of these stars is more than 1000 light-years from us. In fact, it turns out that most of the stars visible without a telescope are hundreds of light- years away and many times more luminous than the Sun. Among the 6000 stars visible to the unaided eye, only about 50 are intrinsically fainter than the Sun. Note also that several of the stars in Appendix J are spectral type B, a type that is completely missing from Table 18.1. The most luminous of the bright stars listed in Appendix J emit more than 50,000 times more energy than does the Sun. These highly luminous stars are missing from the solar neighborhood because they are very rare. None of them happens to be in the tiny volume of space immediately surrounding the Sun, and only this small volume was surveyed to get the data shown in Table 18.1. For example, let\u2019s consider the most luminous stars\u2014those 100 or more times as luminous as the Sun. Although such stars are rare, they are visible to the unaided eye, even when hundreds to thousands of light-years away. A star with a luminosity 10,000 times greater than that of the Sun can be seen without a telescope out to a distance of 5000 light-years. The volume of space included within a distance of 5000 light-years, however, is enormous; so even though highly luminous stars are intrinsically rare, many of them are readily visible to our unaided eye. The contrast between these two samples of stars, those that are close to us and those that can be seen with the unaided eye, is an example of a selection effect. When a population of objects (stars in this example) includes a great variety of different types, we must be careful what conclusions we draw from an examination of any particular subgroup. Certainly we would be fooling ourselves if we assumed that the stars visible to the unaided eye are characteristic of the general stellar population; this subgroup is heavily weighted to the most luminous stars. It requires much more effort to assemble a complete data set for the nearest stars, since most are so faint that they can be observed only with a telescope. However, it is only by doing so that astronomers are able to This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 625 know about the properties of the vast majority of the stars, which are actually much smaller and fainter than our own Sun. In the next section, we will look at how we measure some of these properties. 18.2 MEASURING STELLAR MASSES Learning Objectives By the end of this section, you will be able to: Distinguish the different types of binary star systems Understand how we can apply Newton\u2019s version of Kepler\u2019s third law to derive the sum of star masses in a binary star system Apply the relationship between stellar mass and stellar luminosity to determine the physical characteristics of a star The mass of a star\u2014how much material it contains\u2014is one of its most", " important characteristics. If we know a star\u2019s mass, as we shall see, we can estimate how long it will shine and what its ultimate fate will be. Yet the mass of a star is very difficult to measure directly. Somehow, we need to put a star on the cosmic equivalent of a scale. Luckily, not all stars live like the Sun, in isolation from other stars. About half the stars are binary stars\u2014two stars that orbit each other, bound together by gravity. Masses of binary stars can be calculated from measurements of their orbits, just as the mass of the Sun can be derived by measuring the orbits of the planets around it (see Orbits and Gravity). Binary Stars Before we discuss in more detail how mass can be measured, we will take a closer look at stars that come in pairs. The first binary star was discovered in 1650, less than half a century after Galileo began to observe the sky with a telescope. John Baptiste Riccioli (1598\u20131671), an Italian astronomer, noted that the star Mizar, in the middle of the Big Dipper\u2019s handle, appeared through his telescope as two stars. Since that discovery, thousands of binary stars have been cataloged. (Astronomers call any pair of stars that appear to be close to each other in the sky double stars, but not all of these form a true binary, that is, not all of them are physically associated. Some are just chance alignments of stars that are actually at different distances from us.) Although stars most commonly come in pairs, there are also triple and quadruple systems. One well-known binary star is Castor, located in the constellation of Gemini. By 1804, astronomer William Herschel, who also discovered the planet Uranus, had noted that the fainter component of Castor had slightly changed its position relative to the brighter component. (We use the term \u201ccomponent\u201d to mean a member of a star system.) Here was evidence that one star was moving around another. It was actually the first evidence that gravitational influences exist outside the solar system. The orbital motion of a binary star is shown in Figure 18.4. A binary star system in which both of the stars can be seen with a telescope is called a visual binary. 626 Chapter 18 The Stars: A Celestial Census Figure 18.4. Revolution of a Binary Star. This figure shows seven observations of the mutual revolution of two stars, one a brown dwarf and one an ultra-cool L dwarf. Each red dot on the orbit, which is shown by the blue ellipse, corresponds to the position of one of the dwarfs relative to the other. The reason that the pair of stars looks different on the different dates is that some images were taken with the Hubble Space Telescope and others were taken from the ground. The arrows point to the actual observations that correspond to the positions of each red dot. From these observations, an international team of astronomers directly measured the mass of an ultra-cool brown dwarf star for the first time. Barely the size of the planet Jupiter, the dwarf star weighs in at just 8.5% of the mass of our Sun. (credit: modification of work by ESA/NASA and Herve Bouy (Max-Planck-Institut f\u00fcr Extraterrestrische Physik/ESO, Germany)) Edward C. Pickering (1846\u20131919), at Harvard, discovered a second class of binary stars in 1889\u2014a class in which only one of the stars is actually seen directly. He was examining the spectrum of Mizar and found that the dark absorption lines in the brighter star\u2019s spectrum were usually double. Not only were there two lines where astronomers normally saw only one, but the spacing of the lines was constantly changing. At times, the lines even became single. Pickering correctly deduced that the brighter component of Mizar, called Mizar A, is itself really two stars that revolve about each other in a period of 104 days. A star like Mizar A, which appears as a single star when photographed or observed visually through the telescope, but which spectroscopy shows really to be a double star, is called a spectroscopic binary. Mizar, by the way, is a good example of just how complex such star systems can be. Mizar has been known for centuries to have a faint companion called Alcor, which can be seen without a telescope. Mizar and Alcor form an optical double\u2014a pair of stars that appear close together in the sky but do not orbit each other. Through a telescope, as Riccioli discovered in 1650, Mizar can be seen to have another, closer companion that does orbit it; Mizar is thus a visual binary. The two components that make up this visual binary, known as Mizar A and Mizar B, are both spectroscopic binaries. So, Mizar is really a quadruple system of stars. Strictly speaking, it is not correct", " to describe the motion of a binary star system by saying that one star orbits the other. Gravity is a mutual attraction. Each star exerts a gravitational force on the other, with the result that both stars orbit a point between them called the center of mass. Imagine that the two stars are seated at either end of a seesaw. The point at which the fulcrum would have to be located in order for the seesaw to balance is the center of mass, and it is always closer to the more massive star (Figure 18.5). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 627 Figure 18.5. Binary Star System. In a binary star system, both stars orbit their center of mass. The image shows the relative positions of two, different-mass stars from their center of mass, similar to how two masses would have to be located on a seesaw in order to keep it level. The star with the higher mass will be found closer to the center of mass, while the star with the lower mass will be farther from it. Figure 18.6 shows two stars (A and B) moving around their center of mass, along with one line in the spectrum of each star that we observe from the system at different times. When one star is approaching us relative to the center of mass, the other star is receding from us. In the top left illustration, star A is moving toward us, so the line in its spectrum is Doppler-shifted toward the blue end of the spectrum. Star B is moving away from us, so its line shows a redshift. When we observe the composite spectrum of the two stars, the line appears double. When the two stars are both moving across our line of sight (neither away from nor toward us), they both have the same radial velocity (that of the pair\u2019s center of mass); hence, the spectral lines of the two stars come together. This is shown in the two bottom illustrations in Figure 18.6. Figure 18.6. Motions of Two Stars Orbiting Each Other and What the Spectrum Shows. We see changes in velocity because when one star is moving toward Earth, the other is moving away; half a cycle later, the situation is reversed. Doppler shifts cause the spectral lines to move back and forth. In diagrams 1 and 3, lines from both stars can be seen well separated from each other. When the two stars are moving perpendicular to our line of sight (that is, they are not moving either toward or away from us), the two lines are exactly superimposed, and so in diagrams 2 and 4, we see only a single spectral line. Note that in the diagrams, the orbit of the star pair is tipped slightly with respect to the viewer (or if the viewer were looking at it in the sky, the orbit would be tilted with respect to the viewer\u2019s line of sight). If the orbit were exactly in the plane of the page or screen (or the sky), then it would look nearly circular, but we would see no change in radial velocity (no part of the motion would be toward us or away from us.) If the orbit were perpendicular to the plane of the page or screen, then the stars would appear to move back and forth in a straight line, and we would see the largest-possible radial velocity variations. 628 Chapter 18 The Stars: A Celestial Census A plot showing how the velocities of the stars change with time is called a radial velocity curve; the curve for the binary system in Figure 18.6 is shown in Figure 18.7. Figure 18.7. Radial Velocities in a Spectroscopic Binary System. These curves plot the radial velocities of two stars in a spectroscopic binary system, showing how the stars alternately approach and recede from Earth. Note that positive velocity means the star is moving away from us relative to the center of mass of the system, which in this case is 40 kilometers per second. Negative velocity means the star is moving toward us relative to the center of mass. The positions on the curve corresponding to the illustrations in Figure 18.6 are marked with the diagram number (1\u20134). This animation (https://openstax.org/l/30binstaranim) lets you follow the orbits of a binary star system in various combinations of the masses of the two stars. Masses from the Orbits of Binary Stars We can estimate the masses of binary star systems using Newton\u2019s reformulation of Kepler\u2019s third law (discussed in Newton\u2019s Universal Law of Gravitation). Kepler found that the time a planet takes to go around the Sun is related by a specific mathematical formula to its distance from the Sun. In our binary star situation, if two objects are in mutual revolution, then the period (P) with which they go around each other is related to the", " semimajor axis (D) of the orbit of one with respect to the other, according to this equation D3 = (M1 + M2)P2 where D is in astronomical units, P is measured in years, and M1 + M2 is the sum of the masses of the two stars in units of the Sun\u2019s mass. This is a very useful formula for astronomers; it says that if we can observe the size of the orbit and the period of mutual revolution of the stars in a binary system, we can calculate the sum of their masses. Most spectroscopic binaries have periods ranging from a few days to a few months, with separations of usually less than 1 AU between their member stars. Recall that an AU is the distance from Earth to the Sun, so this is a This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 629 small separation and very hard to see at the distances of stars. This is why many of these systems are known to be double only through careful study of their spectra. We can analyze a radial velocity curve (such as the one in Figure 18.7) to determine the masses of the stars in a spectroscopic binary. This is complex in practice but not hard in principle. We measure the speeds of the stars from the Doppler effect. We then determine the period\u2014how long the stars take to go through an orbital cycle\u2014from the velocity curve. Knowing how fast the stars are moving and how long they take to go around tells us the circumference of the orbit and, hence, the separation of the stars in kilometers or astronomical units. From Kepler\u2019s law, the period and the separation allow us to calculate the sum of the stars\u2019 masses. Of course, knowing the sum of the masses is not as useful as knowing the mass of each star separately. But the relative orbital speeds of the two stars can tell us how much of the total mass each star has. As we saw in our seesaw analogy, the more massive star is closer to the center of mass and therefore has a smaller orbit. Therefore, it moves more slowly to get around in the same time compared to the more distant, lower-mass star. If we sort out the speeds relative to each other, we can sort out the masses relative to each other. In practice, we also need to know how the binary system is oriented in the sky to our line of sight, but if we do, and the just-described steps are carried out carefully, the result is a calculation of the masses of each of the two stars in the system. To summarize, a good measurement of the motion of two stars around a common center of mass, combined with the laws of gravity, allows us to determine the masses of stars in such systems. These mass measurements are absolutely crucial to developing a theory of how stars evolve. One of the best things about this method is that it is independent of the location of the binary system. It works as well for stars 100 light-years away from us as for those in our immediate neighborhood. To take a specific example, Sirius is one of the few binary stars in Appendix J for which we have enough information to apply Kepler\u2019s third law: D3 = (M1 + M2)P2 In this case, the two stars, the one we usually call Sirius and its very faint companion, are separated by about 20 AU and have an orbital period of about 50 years. If we place these values in the formula we would have (20)3 = (M1 + M2)(50)2 8000 = (M1 + M2)(2500) This can be solved for the sum of the masses: M1 + M2 = 8000 2500 = 3.2 Therefore, the sum of masses of the two stars in the Sirius binary system is 3.2 times the Sun\u2019s mass. In order to determine the individual mass of each star, we would need the velocities of the two stars and the orientation of the orbit relative to our line of sight. The Range of Stellar Masses How large can the mass of a star be? Stars more massive than the Sun are rare. None of the stars within 30 light-years of the Sun has a mass greater than four times that of the Sun. Searches at large distances from the Sun have led to the discovery of a few stars with masses up to about 100 times that of the Sun, and a handful of stars (a few out of several billion) may have masses as large as 250 solar masses. However, most stars have less mass than the Sun. According to theoretical calculations, the smallest mass that a true star can have is about 1/12 that of the Sun. By a \u201ctrue\u201d star, astronomers mean one that becomes hot enough to fuse protons to form helium (as discussed 630 Chapter 18 The Stars: A Celestial Census in The Sun:", " A Nuclear Powerhouse). Objects with masses between roughly 1/100 and 1/12 that of the Sun may produce energy for a brief time by means of nuclear reactions involving deuterium, but they do not become hot enough to fuse protons. Such objects are intermediate in mass between stars and planets and have been given the name brown dwarfs (Figure 18.8). Brown dwarfs are similar to Jupiter in radius but have masses from approximately 13 to 80 times larger than the mass of Jupiter.[2] Figure 18.8. Brown Dwarfs in Orion. These images, taken with the Hubble Space Telescope, show the region surrounding the Trapezium star cluster inside the star-forming region called the Orion Nebula. (a) No brown dwarfs are seen in the visible light image, both because they put out very little light in the visible and because they are hidden within the clouds of dust in this region. (b) This image was taken in infrared light, which can make its way to us through the dust. The faintest objects in this image are brown dwarfs with masses between 13 and 80 times the mass of Jupiter. (credit a: NASA, C.R. O\u2019Dell and S.K. Wong (Rice University); credit b: NASA; K.L. Luhman (Harvard-Smithsonian Center for Astrophysics) and G. Schneider, E. Young, G. Rieke, A. Cotera, H. Chen, M. Rieke, R. Thompson (Steward Observatory)) Still-smaller objects with masses less than about 1/100 the mass of the Sun (or 10 Jupiter masses) are called planets. They may radiate energy produced by the radioactive elements that they contain, and they may also radiate heat generated by slowly compressing under their own weight (a process called gravitational contraction). However, their interiors will never reach temperatures high enough for any nuclear reactions, to take place. Jupiter, whose mass is about 1/1000 the mass of the Sun, is unquestionably a planet, for example. Until the 1990s, we could only detect planets in our own solar system, but now we have thousands of them elsewhere as well. (We will discuss these exciting observations in The Birth of Stars and the Discovery of Planets outside the Solar System.) The Mass-Luminosity Relation Now that we have measurements of the characteristics of many different types of stars, we can search for relationships among the characteristics. For example, we can ask whether the mass and luminosity of a star are related. It turns out that for most stars, they are: The more massive stars are generally also the more luminous. This relationship, known as the mass-luminosity relation, is shown graphically in Figure 18.9. Each point represents a star whose mass and luminosity are both known. The horizontal position on the graph shows 2 Exactly where to put the dividing line between planets and brown dwarfs is a subject of some debate among astronomers as we write this book (as is, in fact, the exact definition of each of these objects). Even those who accept deuterium fusion (see The Birth of Stars and the Discovery of Planets outside the Solar System) as the crucial issue for brown dwarfs concede that, depending on the composition of the star and other factors, the lowest mass for such a dwarf could be anywhere from 11 to 16 Jupiter masses. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 631 the star\u2019s mass, given in units of the Sun\u2019s mass, and the vertical position shows its luminosity in units of the Sun\u2019s luminosity. Figure 18.9. Mass-Luminosity Relation. The plotted points show the masses and luminosities of stars. The three points lying below the sequence of points are all white dwarf stars. We can also say this in mathematical terms. L ~ M 3.9 It\u2019s a reasonably good approximation to say that luminosity (expressed in units of the Sun\u2019s luminosity) varies as the fourth power of the mass (in units of the Sun\u2019s mass). (The symbol ~ means the two quantities are proportional.) If two stars differ in mass by a factor of 2, then the more massive one will be 24, or about 16 times brighter; if one star is 1/3 the mass of another, it will be approximately 81 times less luminous Calculating the Mass from the Luminosity of a Star The mass-luminosity formula can be rewritten so that a value of mass can be determined if the luminosity is known. Solution First, we must get our units right by expressing both the mass and the luminosity of a star in units of the Sun\u2019s mass and luminosity: L/LSun = \u239b \u239dM/MSun \u239e", " \u23a0 4 Now we can take the 4th root of both sides, which is equivalent to taking both sides to the 1/4 = 0.25 power. The formula in this case would be: Check Your Learning M/MSun = \u239b \u239dL/LSun \u239e \u23a0 0.25 = \u239b \u239dL/LSun 0.25 \u239e \u23a0 632 Chapter 18 The Stars: A Celestial Census In the previous section, we determined the sum of the masses of the two stars in the Sirius binary system (Sirius and its faint companion) using Kepler\u2019s third law to be 3.2 solar masses. Using the mass- luminosity relationship, calculate the mass of each individual star. Answer: In Appendix J, Sirius is listed with a luminosity 23 times that of the Sun. This value can be inserted into the mass-luminosity relationship to get the mass of Sirius: M/MSun = 230.25 = 2.2 The mass of the companion star to Sirius is then 3.2 \u2013 2.2 = 1.0 solar mass. Notice how good this mass-luminosity relationship is. Most stars (see Figure 18.9) fall along a line running from the lower-left (low mass, low luminosity) corner of the diagram to the upper-right (high mass, high luminosity) corner. About 90% of all stars obey the mass-luminosity relation. Later, we will explore why such a relationship exists and what we can learn from the roughly 10% of stars that \u201cdisobey\u201d it. 18.3 DIAMETERS OF STARS Learning Objectives By the end of this section, you will be able to: Describe the methods used to determine star diameters Identify the parts of an eclipsing binary star light curve that correspond to the diameters of the individual components It is easy to measure the diameter of the Sun. Its angular diameter\u2014that is, its apparent size on the sky\u2014is about 1/2\u00b0. If we know the angle the Sun takes up in the sky and how far away it is, we can calculate its true (linear) diameter, which is 1.39 million kilometers, or about 109 times the diameter of Earth. Unfortunately, the Sun is the only star whose angular diameter is easily measured. All the other stars are so far away that they look like pinpoints of light through even the largest ground-based telescopes. (They often seem to be bigger, but that is merely distortion introduced by turbulence in Earth\u2019s atmosphere.) Luckily, there are several techniques that astronomers can use to estimate the sizes of stars. Stars Blocked by the Moon One technique, which gives very precise diameters but can be used for only a few stars, is to observe the dimming of light that occurs when the Moon passes in front of a star. What astronomers measure (with great precision) is the time required for the star\u2019s brightness to drop to zero as the edge of the Moon moves across the star\u2019s disk. Since we know how rapidly the Moon moves in its orbit around Earth, it is possible to calculate the angular diameter of the star. If the distance to the star is also known, we can calculate its diameter in kilometers. This method works only for fairly bright stars that happen to lie along the zodiac, where the Moon (or, much more rarely, a planet) can pass in front of them as seen from Earth. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 633 Eclipsing Binary Stars Accurate sizes for a large number of stars come from measurements of eclipsing binary star systems, and so we must make a brief detour from our main story to examine this type of star system. Some binary stars are lined up in such a way that, when viewed from Earth, each star passes in front of the other during every revolution (Figure 18.10). When one star blocks the light of the other, preventing it from reaching Earth, the luminosity of the system decreases, and astronomers say that an eclipse has occurred. Figure 18.10. Light Curve of an Eclipsing Binary. The light curve of an eclipsing binary star system shows how the combined light from both stars changes due to eclipses over the time span of an orbit. This light curve shows the behavior of a hypothetical eclipsing binary star with total eclipses (one star passes directly in front of and behind the other). The numbers indicate parts of the light curve corresponding to various positions of the smaller star in its orbit. In this diagram, we have assumed that the smaller star is also the hotter one so that it emits more flux (energy per second per square meter) than the larger one. When the smaller, hotter star goes behind the larger one, its light is completely blocked, and so there", " is a strong dip in the light curve. When the smaller star goes in front of the bigger one, a small amount of light from the bigger star is blocked, so there is a smaller dip in the light curve. The discovery of the first eclipsing binary helped solve a long-standing puzzle in astronomy. The star Algol, in the constellation of Perseus, changes its brightness in an odd but regular way. Normally, Algol is a fairly bright star, but at intervals of 2 days, 20 hours, 49 minutes, it fades to one-third of its regular brightness. After a few hours, it brightens to normal again. This effect is easily seen, even without a telescope, if you know what to look for. In 1783, a young English astronomer named John Goodricke (1764\u20131786) made a careful study of Algol (see the feature on John Goodricke for a discussion of his life and work). Even though Goodricke could neither hear nor speak, he made a number of major discoveries in the 21 years of his brief life. He suggested that Algol\u2019s unusual brightness variations might be due to an invisible companion that regularly passes in front of the brighter star and blocks its light. Unfortunately, Goodricke had no way to test this idea, since it was not until about a century later that equipment became good enough to measure Algol\u2019s spectrum. In 1889, the German astronomer Hermann Vogel (1841\u20131907) demonstrated that, like Mizar, Algol is a spectroscopic binary. The spectral lines of Algol were not observed to be double because the fainter star of the pair gives off too-little light compared with the brighter star for its lines to be conspicuous in the composite spectrum. Nevertheless, the periodic shifting back and forth of the brighter star\u2019s lines gave evidence that it was revolving about an unseen companion. (The lines of both components need not be visible for a star to be recognized as a spectroscopic binary.) The discovery that Algol is a spectroscopic binary verified Goodricke\u2019s hypothesis. The plane in which the stars revolve is turned nearly edgewise to our line of sight, and each star passes in front of the other during every revolution. (The eclipse of the fainter star in the Algol system is not very noticeable because the part of it that 634 Chapter 18 The Stars: A Celestial Census is covered contributes little to the total light of the system. This second eclipse can, however, be detected by careful measurements.) Any binary star produces eclipses if viewed from the proper direction, near the plane of its orbit, so that one star passes in front of the other (see Figure 18.10). But from our vantage point on Earth, only a few binary star systems are oriented in this way Astronomy and Mythology: Algol the Demon Star and Perseus the Hero The name Algol comes from the Arabic Ras al Ghul, meaning \u201cthe demon\u2019s head.\u201d[3] The word \u201cghoul\u201d in English has the same derivation. As discussed in Observing the Sky: The Birth of Astronomy, many of the bright stars have Arabic names because during the long dark ages in medieval Europe, it was Arabic astronomers who preserved and expanded the Greek and Roman knowledge of the skies. The reference to the demon is part of the ancient Greek legend of the hero Perseus, who is commemorated by the constellation in which we find Algol and whose adventures involve many of the characters associated with the northern constellations. Perseus was one of the many half-god heroes fathered by Zeus (Jupiter in the Roman version), the king of the gods in Greek mythology. Zeus had, to put it delicately, a roving eye and was always fathering somebody or other with a human maiden who caught his fancy. (Perseus derives from Per Zeus, meaning \u201cfathered by Zeus.\u201d) Set adrift with his mother by an (understandably) upset stepfather, Perseus grew up on an island in the Aegean Sea. The king there, taking an interest in Perseus\u2019 mother, tried to get rid of the young man by assigning him an extremely difficult task. In a moment of overarching pride, a beautiful young woman named Medusa had compared her golden hair to that of the goddess Athena (Minerva for the Romans). The Greek gods did not take kindly to being compared to mere mortals, and Athena turned Medusa into a gorgon: a hideous, evil creature with writhing snakes for hair and a face that turned anyone who looked at it into stone. Perseus was given the task of slaying this demon, which seemed like a pretty sure way to get him out of the way forever. But because Perseus had a god for a father, some of the other gods gave him tools for", " the job, including Athena\u2019s reflective shield and the winged sandals of Hermes (Mercury in the Roman story). By flying over her and looking only at her reflection, Perseus was able to cut off Medusa\u2019s head without ever looking at her directly. Taking her head (which, conveniently, could still turn onlookers to stone even without being attached to her body) with him, Perseus continued on to other adventures. He next came to a rocky seashore, where boasting had gotten another family into serious trouble with the gods. Queen Cassiopeia had dared to compare her own beauty to that of the Nereids, sea nymphs who were daughters of Poseidon (Neptune in Roman mythology), the god of the sea. Poseidon was so offended that he created a sea-monster named Cetus to devastate the kingdom. King Cepheus, Cassiopeia\u2019s beleaguered husband, consulted the oracle, who told him that he must sacrifice his beautiful daughter Andromeda to the monster. When Perseus came along and found Andromeda chained to a rock near the sea, awaiting her fate, he rescued her by turning the monster to stone. (Scholars of mythology actually trace the essence of this story back to far-older legends from ancient Mesopotamia, in which the god-hero Marduk vanquishes a monster named Tiamat. Symbolically, a hero like Perseus or Marduk is usually associated with the Sun, This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 635 the monster with the power of night, and the beautiful maiden with the fragile beauty of dawn, which the Sun releases after its nightly struggle with darkness.) Many of the characters in these Greek legends can be found as constellations in the sky, not necessarily resembling their namesakes but serving as reminders of the story. For example, vain Cassiopeia is sentenced to be very close to the celestial pole, rotating perpetually around the sky and hanging upside down every winter. The ancients imagined Andromeda still chained to her rock (it is much easier to see the chain of stars than to recognize the beautiful maiden in this star grouping). Perseus is next to her with the head of Medusa swinging from his belt. Algol represents this gorgon head and has long been associated with evil and bad fortune in such tales. Some commentators have speculated that the star\u2019s change in brightness (which can be observed with the unaided eye) may have contributed to its unpleasant reputation, with the ancients regarding such a change as a sort of evil \u201cwink.\u201d Diameters of Eclipsing Binary Stars We now turn back to the main thread of our story to discuss how all this can be used to measure the sizes of stars. The technique involves making a light curve of an eclipsing binary, a graph that plots how the brightness changes with time. Let us consider a hypothetical binary system in which the stars are very different in size, like those illustrated in Figure 18.11. To make life easy, we will assume that the orbit is viewed exactly edge-on. Even though we cannot see the two stars separately in such a system, the light curve can tell us what is happening. When the smaller star just starts to pass behind the larger star (a point we call first contact), the brightness begins to drop. The eclipse becomes total (the smaller star is completely hidden) at the point called second contact. At the end of the total eclipse (third contact), the smaller star begins to emerge. When the smaller star has reached last contact, the eclipse is completely over. To see how this allows us to measure diameters, look carefully at Figure 18.11. During the time interval between the first and second contacts, the smaller star has moved a distance equal to its own diameter. During the time interval from the first to third contacts, the smaller star has moved a distance equal to the diameter of the larger star. If the spectral lines of both stars are visible in the spectrum of the binary, then the speed of the smaller star with respect to the larger one can be measured from the Doppler shift. But knowing the speed with which the smaller star is moving and how long it took to cover some distance can tell the span of that distance\u2014in this case, the diameters of the stars. The speed multiplied by the time interval from the first to second contact gives the diameter of the smaller star. We multiply the speed by the time between the first and third contacts to get the diameter of the larger star. 3 Fans of Batman comic books and movies will recognize that this name was given to an archvillain in the series. 636 Chapter 18 The Stars: A Celestial Census Figure 18.11. Light Curve of an Edge-On Eclipsing Binary. Here we see the light curve of a hypothetical eclipsing binary", " star whose orbit we view exactly edge-on, in which the two stars fully eclipse each other. From the time intervals between contacts, it is possible to estimate the diameters of the two stars. In actuality, the situation with eclipsing binaries is often a bit more complicated: orbits are generally not seen exactly edge-on, and the light from each star may be only partially blocked by the other. Furthermore, binary star orbits, just like the orbits of the planets, are ellipses, not circles. However, all these effects can be sorted out from very careful measurements of the light curve. Using the Radiation Law to Get the Diameter Another method for measuring star diameters makes use of the Stefan-Boltzmann law for the relationship between energy radiated and temperature (see Radiation and Spectra). In this method, the energy flux (energy emitted per second per square meter by a blackbody, like the Sun) is given by where \u03c3 is a constant and T is the temperature. The surface area of a sphere (like a star) is given by F = \u03c3T 4 A = 4\u03c0R2 The luminosity (L) of a star is then given by its surface area in square meters times the energy flux: L = (A \u00d7 F) Previously, we determined the masses of the two stars in the Sirius binary system. Sirius gives off 8200 times more energy than its fainter companion star, although both stars have nearly identical temperatures. The extremely large difference in luminosity is due to the difference in radius, since the temperatures and hence the energy fluxes for the two stars are nearly the same. To determine the relative sizes of the two stars, we take the ratio of the corresponding luminosities: This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 637 LSirius Lcompanion = = ASirius Acompanion \u239b \u239e \u23a0 \u239dASirius \u00d7 FSirius \u239e \u239b \u239dAcompanion \u00d7 Fcompanion \u23a0 R2 4\u03c0R2 Sirius = = 4\u03c0R2 companion R2 companion Sirius LSirius Lcompanion = 8200 = R2 Sirius R2 companion Therefore, the relative sizes of the two stars can be found by taking the square root of the relative luminosity. Since 8200 = 91, the radius of Sirius is 91 times larger than the radium of its faint companion. The method for determining the radius shown here requires both stars be visible, which is not always the case. Stellar Diameters The results of many stellar size measurements over the years have shown that most nearby stars are roughly the size of the Sun, with typical diameters of a million kilometers or so. Faint stars, as we might have expected, are generally smaller than more luminous stars. However, there are some dramatic exceptions to this simple generalization. A few of the very luminous stars, those that are also red (indicating relatively low surface temperatures), turn out to be truly enormous. These stars are called, appropriately enough, giant stars or supergiant stars. An example is Betelgeuse, the second brightest star in the constellation of Orion and one of the dozen brightest stars in our sky. Its diameter, remarkably, is greater than 10 AU (1.5 billion kilometers!), large enough to fill the entire inner solar system almost as far out as Jupiter. In Stars from Adolescence to Old Age, we will look in detail at the evolutionary process that leads to the formation of such giant and supergiant stars Watch this star size comparison video (https://openstax.org/l/30starsizecomp) for a striking visual that highlights the size of stars versus planets and the range of sizes among stars. 18.4 THE H\u2013R DIAGRAM Learning Objectives By the end of this section, you will be able to: Identify the physical characteristics of stars that are used to create an H\u2013R diagram, and describe how those characteristics vary among groups of stars Discuss the physical properties of most stars found at different locations on the H\u2013R diagram, such as radius, and for main sequence stars, mass In this chapter and Analyzing Starlight, we described some of the characteristics by which we might classify stars and how those are measured. These ideas are summarized in Table 18.2. We have also given an example of a relationship between two of these characteristics in the mass-luminosity relation. When the characteristics 638 Chapter 18 The Stars: A Celestial Census of large numbers of stars were measured at the beginning of the twentieth century, astronomers were able to begin a deeper search for patterns and relationships in these data. Measuring the Characteristics of Stars Characteristic Technique Surface temperature 1. Determine the color (very rough). 2. Measure the spectrum and get the spectral type. Chemical composition Determine which", " lines are present in the spectrum. Luminosity Measure the apparent brightness and compensate for distance. Radial velocity Measure the Doppler shift in the spectrum. Rotation Measure the width of spectral lines. Mass Diameter Table 18.2 Measure the period and radial velocity curves of spectroscopic binary stars. 1. Measure the way a star\u2019s light is blocked by the Moon. 2. Measure the light curves and Doppler shifts for eclipsing binary stars. To help understand what sorts of relationships might be found, let\u2019s look briefly at a range of data about human beings. If you want to understand humans by comparing and contrasting their characteristics\u2014without assuming any previous knowledge of these strange creatures\u2014you could try to determine which characteristics lead you in a fruitful direction. For example, you might plot the heights of a large sample of humans against their weights (which is a measure of their mass). Such a plot is shown in Figure 18.12 and it has some interesting features. In the way we have chosen to present our data, height increases upward, whereas weight increases to the left. Notice that humans are not randomly distributed in the graph. Most points fall along a sequence that goes from the upper left to the lower right. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 639 Figure 18.12. Height versus Weight. The plot of the heights and weights of a representative group of human beings. Most points lie along a \u201cmain sequence\u201d representing most people, but there are a few exceptions. We can conclude from this graph that human height and weight are related. Generally speaking, taller human beings weigh more, whereas shorter ones weigh less. This makes sense if you are familiar with the structure of human beings. Typically, if we have bigger bones, we have more flesh to fill out our larger frame. It\u2019s not mathematically exact\u2014there is a wide range of variation\u2014but it\u2019s not a bad overall rule. And, of course, there are some dramatic exceptions. You occasionally see a short human who is very overweight and would thus be more to the bottom left of our diagram than the average sequence of people. Or you might have a very tall, skinny fashion model with great height but relatively small weight, who would be found near the upper right. A similar diagram has been found extremely useful for understanding the lives of stars. In 1913, American astronomer Henry Norris Russell plotted the luminosities of stars against their spectral classes (a way of denoting their surface temperatures). This investigation, and a similar independent study in 1911 by Danish astronomer Ejnar Hertzsprung, led to the extremely important discovery that the temperature and luminosity of stars are related (Figure 18.13). Figure 18.13. Hertzsprung (1873\u20131967) and Russell (1877\u20131957). (a) Ejnar Hertzsprung and (b) Henry Norris Russell independently discovered the relationship between the luminosity and surface temperature of stars that is summarized in what is now called the H\u2013R diagram. 640 Chapter 18 The Stars: A Celestial Census Henry Norris Russell When Henry Norris Russell graduated from Princeton University, his work had been so brilliant that the faculty decided to create a new level of honors degree beyond \u201csumma cum laude\u201d for him. His students later remembered him as a man whose thinking was three times faster than just about anybody else\u2019s. His memory was so phenomenal, he could correctly quote an enormous number of poems and limericks, the entire Bible, tables of mathematical functions, and almost anything he had learned about astronomy. He was nervous, active, competitive, critical, and very articulate; he tended to dominate every meeting he attended. In outward appearance, he was an old-fashioned product of the nineteenth century who wore high-top black shoes and high starched collars, and carried an umbrella every day of his life. His 264 papers were enormously influential in many areas of astronomy. Born in 1877, the son of a Presbyterian minister, Russell showed early promise. When he was 12, his family sent him to live with an aunt in Princeton so he could attend a top preparatory school. He lived in the same house in that town until his death in 1957 (interrupted only by a brief stay in Europe for graduate work). He was fond of recounting that both his mother and his maternal grandmother had won prizes in mathematics, and that he probably inherited his talents in that field from their side of the family. Before Russell, American astronomers devoted themselves mainly to surveying the stars and making impressive catalogs of their properties, especially their spectra (as described in Analyzing Starlight. Russell began to see that interpreting the spectra of stars required a much more sophisticated understanding of the physics of the atom, a subject that was being developed by European physicists in the 1910s and 1920s. Russell embarked on a lifelong quest to", " ascertain the physical conditions inside stars from the clues in their spectra; his work inspired, and was continued by, a generation of astronomers, many trained by Russell and his collaborators. Russell also made important contributions in the study of binary stars and the measurement of star masses, the origin of the solar system, the atmospheres of planets, and the measurement of distances in astronomy, among other fields. He was an influential teacher and popularizer of astronomy, writing a column on astronomical topics for Scientific American magazine for more than 40 years. He and two colleagues wrote a textbook for college astronomy classes that helped train astronomers and astronomy enthusiasts over several decades. That book set the scene for the kind of textbook you are now reading, which not only lays out the facts of astronomy but also explains how they fit together. Russell gave lectures around the country, often emphasizing the importance of understanding modern physics in order to grasp what was happening in astronomy. Harlow Shapley, director of the Harvard College Observatory, called Russell \u201cthe dean of American astronomers.\u201d Russell was certainly regarded as the leader of the field for many years and was consulted on many astronomical problems by colleagues from around the world. Today, one of the highest recognitions that an astronomer can receive is an award from the American Astronomical Society called the Russell Prize, set up in his memory. Features of the H\u2013R Diagram Following Hertzsprung and Russell, let us plot the temperature (or spectral class) of a selected group of nearby stars against their luminosity and see what we find (Figure 18.14). Such a plot is frequently called This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 641 the Hertzsprung\u2013Russell diagram, abbreviated H\u2013R diagram. It is one of the most important and widely used diagrams in astronomy, with applications that extend far beyond the purposes for which it was originally developed more than a century ago. Figure 18.14. H\u2013R Diagram for a Selected Sample of Stars. In such diagrams, luminosity is plotted along the vertical axis. Along the horizontal axis, we can plot either temperature or spectral type (also sometimes called spectral class). Several of the brightest stars are identified by name. Most stars fall on the main sequence. It is customary to plot H\u2013R diagrams in such a way that temperature increases toward the left and luminosity toward the top. Notice the similarity to our plot of height and weight for people (Figure 18.12). Stars, like people, are not distributed over the diagram at random, as they would be if they exhibited all combinations of luminosity and temperature. Instead, we see that the stars cluster into certain parts of the H\u2013R diagram. The great majority are aligned along a narrow sequence running from the upper left (hot, highly luminous) to the lower right (cool, less luminous). This band of points is called the main sequence. It represents a relationship between temperature and luminosity that is followed by most stars. We can summarize this relationship by saying that hotter stars are more luminous than cooler ones. A number of stars, however, lie above the main sequence on the H\u2013R diagram, in the upper-right region, where stars have low temperature and high luminosity. How can a star be at once cool, meaning each square meter on the star does not put out all that much energy, and yet very luminous? The only way is for the star to be enormous\u2014to have so many square meters on its surface that the total energy output is still large. These stars must be giants or supergiants, the stars of huge diameter we discussed earlier. There are also some stars in the lower-left corner of the diagram, which have high temperature and low luminosity. If they have high surface temperatures, each square meter on that star puts out a lot of energy. How then can the overall star be dim? It must be that it has a very small total surface area; such stars are known as white dwarfs (white because, at these high temperatures, the colors of the electromagnetic radiation that 642 Chapter 18 The Stars: A Celestial Census they emit blend together to make them look bluish-white). We will say more about these puzzling objects in a moment. Figure 18.15 is a schematic H\u2013R diagram for a large sample of stars, drawn to make the different types more apparent. Figure 18.15. Schematic H\u2013R Diagram for Many Stars. Ninety percent of all stars on such a diagram fall along a narrow band called the main sequence. A minority of stars are found in the upper right; they are both cool (and hence red) and bright, and must be giants. Some stars fall in the lower left of the diagram; they are both hot and dim, and must be white dwarfs. Now, think back to our discussion of star surveys. It is difficult to plot an H\u2013R diagram that", " is truly representative of all stars because most stars are so faint that we cannot see those outside our immediate neighborhood. The stars plotted in Figure 18.14 were selected because their distances are known. This sample omits many intrinsically faint stars that are nearby but have not had their distances measured, so it shows fewer faint main- sequence stars than a \u201cfair\u201d diagram would. To be truly representative of the stellar population, an H\u2013R diagram should be plotted for all stars within a certain distance. Unfortunately, our knowledge is reasonably complete only for stars within 10 to 20 light-years of the Sun, among which there are no giants or supergiants. Still, from many surveys (and more can now be done with new, more powerful telescopes), we estimate that about 90% of the true stars overall (excluding brown dwarfs) in our part of space are main-sequence stars, about 10% are white dwarfs, and fewer than 1% are giants or supergiants. These estimates can be used directly to understand the lives of stars. Permit us another quick analogy with people. Suppose we survey people just like astronomers survey stars, but we want to focus our attention on the location of young people, ages 6 to 18 years. Survey teams fan out and take data about where such youngsters are found at all times during a 24-hour day. Some are found in the local pizza parlor, others are asleep at home, some are at the movies, and many are in school. After surveying a very large number of young people, one of the things that the teams determine is that, averaged over the course of the 24 hours, one-third of all youngsters are found in school. How can they interpret this result? Does it mean that two-thirds of students are truants and the remaining one-third spend all their time in school? No, we must bear in mind that the survey teams counted youngsters throughout the full 24-hour day. Some survey teams worked at night, when most youngsters were at home asleep, and others worked in the late afternoon, when most youngsters were on their way home from school (and more likely to be enjoying a pizza). If the survey was truly representative, we can conclude, however, that if an average of one-third of all youngsters are found in school, then humans ages 6 to 18 years must spend about one-third of their time in school. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 643 We can do something similar for stars. We find that, on average, 90% of all stars are located on the main sequence of the H\u2013R diagram. If we can identify some activity or life stage with the main sequence, then it follows that stars must spend 90% of their lives in that activity or life stage. Understanding the Main Sequence In The Sun: A Nuclear Powerhouse, we discussed the Sun as a representative star. We saw that what stars such as the Sun \u201cdo for a living\u201d is to convert protons into helium deep in their interiors via the process of nuclear fusion, thus producing energy. The fusion of protons to helium is an excellent, long-lasting source of energy for a star because the bulk of every star consists of hydrogen atoms, whose nuclei are protons. Our computer models of how stars evolve over time show us that a typical star will spend about 90% of its life fusing the abundant hydrogen in its core into helium. This then is a good explanation of why 90% of all stars are found on the main sequence in the H\u2013R diagram. But if all the stars on the main sequence are doing the same thing (fusing hydrogen), why are they distributed along a sequence of points? That is, why do they differ in luminosity and surface temperature (which is what we are plotting on the H\u2013R diagram)? To help us understand how main-sequence stars differ, we can use one of the most important results from our studies of model stars. Astrophysicists have been able to show that the structure of stars that are in equilibrium and derive all their energy from nuclear fusion is completely and uniquely determined by just two quantities: the total mass and the composition of the star. This fact provides an interpretation of many features of the H\u2013R diagram. Imagine a cluster of stars forming from a cloud of interstellar \u201craw material\u201d whose chemical composition is similar to the Sun\u2019s. (We\u2019ll describe this process in more detail in The Birth of Stars and Discovery of Planets outside the Solar System, but for now, the details will not concern us.) In such a cloud, all the clumps of gas and dust that become stars begin with the same chemical composition and differ from one another only in mass. Now suppose that we compute a model of each of these stars for the time at which it becomes stable and derives its energy from nuclear reactions, but before", " it has time to alter its composition appreciably as a result of these reactions. The models calculated for these stars allow us to determine their luminosities, temperatures, and sizes. If we plot the results from the models\u2014one point for each model star\u2014on the H\u2013R diagram, we get something that looks just like the main sequence we saw for real stars. And here is what we find when we do this. The model stars with the largest masses are the hottest and most luminous, and they are located at the upper left of the diagram. The least-massive model stars are the coolest and least luminous, and they are placed at the lower right of the plot. The other model stars all lie along a line running diagonally across the diagram. In other words, the main sequence turns out to be a sequence of stellar masses. This makes sense if you think about it. The most massive stars have the most gravity and can thus compress their centers to the greatest degree. This means they are the hottest inside and the best at generating energy from nuclear reactions deep within. As a result, they shine with the greatest luminosity and have the hottest surface temperatures. The stars with lowest mass, in turn, are the coolest inside and least effective in generating energy. Thus, they are the least luminous and wind up being the coolest on the surface. Our Sun lies somewhere in the middle of these extremes (as you can see in Figure 18.14). The characteristics of representative main-sequence stars (excluding brown dwarfs, which are not true stars) are listed in Table 18.3. 644 Chapter 18 The Stars: A Celestial Census Characteristics of Main-Sequence Stars Spectral Type Mass (Sun = 1) Luminosity (Sun = 1) Temperature Radius (Sun = 1) O5 B0 A0 F0 G0 K0 M0 Table 18.3 40 16 3.3 1.7 1.1 0.8 0.4 7 \u00d7 105 2.7 \u00d7 105 55 5 1.4 0.35 0.05 40,000 K 28,000 K 10,000 K 7500 K 6000 K 5000 K 3500 K 18 7 2.5 1.4 1.1 0.8 0.6 Note that we\u2019ve seen this 90% figure come up before. This is exactly what we found earlier when we examined the mass-luminosity relation (Figure 18.9). We observed that 90% of all stars seem to follow the relationship; these are the 90% of all stars that lie on the main sequence in our H\u2013R diagram. Our models and our observations agree. What about the other stars on the H\u2013R diagram\u2014the giants and supergiants, and the white dwarfs? As we will see in the next few chapters, these are what main-sequence stars turn into as they age: They are the later stages in a star\u2019s life. As a star consumes its nuclear fuel, its source of energy changes, as do its chemical composition and interior structure. These changes cause the star to alter its luminosity and surface temperature so that it no longer lies on the main sequence on our diagram. Because stars spend much less time in these later stages of their lives, we see fewer stars in those regions of the H\u2013R diagram. Extremes of Stellar Luminosities, Diameters, and Densities We can use the H\u2013R diagram to explore the extremes in size, luminosity, and density found among the stars. Such extreme stars are not only interesting to fans of the Guinness Book of World Records; they can teach us a lot about how stars work. For example, we saw that the most massive main-sequence stars are the most luminous ones. We know of a few extreme stars that are a million times more luminous than the Sun, with masses that exceed 100 times the Sun\u2019s mass. These superluminous stars, which are at the upper left of the H\u2013R diagram, are exceedingly hot, very blue stars of spectral type O. These are the stars that would be the most conspicuous at vast distances in space. The cool supergiants in the upper corner of the H\u2013R diagram are as much as 10,000 times as luminous as the Sun. In addition, these stars have diameters very much larger than that of the Sun. As discussed above, some supergiants are so large that if the solar system could be centered in one, the star\u2019s surface would lie beyond the orbit of Mars (see Figure 18.16). We will have to ask, in coming chapters, what process can make a star swell up to such an enormous size, and how long these \u201cswollen\u201d stars can last in their distended state. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 645 Figure 18.16. The Sun", " and a Supergiant. Here you see how small the Sun looks in comparison to one of the largest known stars: VY Canis Majoris, a supergiant. In contrast, the very common red, cool, low-luminosity stars at the lower end of the main sequence are much smaller and more compact than the Sun. An example of such a red dwarf is Ross 614B, with a surface temperature of 2700 K and only 1/2000 of the Sun\u2019s luminosity. We call such a star a dwarf because its diameter is only 1/10 that of the Sun. A star with such a low luminosity also has a low mass (about 1/12 that of the Sun). This combination of mass and diameter means that it is so compressed that the star has an average density about 80 times that of the Sun. Its density must be higher, in fact, than that of any known solid found on the surface of Earth. (Despite this, the star is made of gas throughout because its center is so hot.) The faint, red, main-sequence stars are not the stars of the most extreme densities, however. The white dwarfs, at the lower-left corner of the H\u2013R diagram, have densities many times greater still. The White Dwarfs The first white dwarf star was detected in 1862. Called Sirius B, it forms a binary system with Sirius A, the brightest-appearing star in the sky. It eluded discovery and analysis for a long time because its faint light tends to be lost in the glare of nearby Sirius A (Figure 18.17). (Since Sirius is often called the Dog Star\u2014being the brightest star in the constellation of Canis Major, the big dog\u2014Sirius B is sometimes nicknamed the Pup.) 646 Chapter 18 The Stars: A Celestial Census Figure 18.17. Two Views of Sirius and Its White Dwarf Companion. (a) The (visible light) image, taken with the Hubble Space Telescope, shows bright Sirius A, and, below it and off to its left, faint Sirius B. (b) This image of the Sirius star system was taken with the Chandra X-Ray Telescope. Now, the bright object is the white dwarf companion, Sirius B. Sirius A is the faint object above it; what we are seeing from Sirius is probably not actually X-ray radiation but rather ultraviolet light that has leaked into the detector. Note that the ultraviolet intensities of these two objects are completely reversed from the situation in visible light because Sirius B is hotter and emits more higher-frequency radiation. (credit a: modification of work by NASA, H.E. Bond and E. Nelan (Space Telescope Science Institute), M. Barstow and M. Burleigh (University of Leicester) and J.B. Holberg (University of Arizona); credit b: modification of work by NASA/SAO/CXC) We have now found thousands of white dwarfs. Table 18.1 shows that about 7% of the true stars (spectral types O\u2013M) in our local neighborhood are white dwarfs. A good example of a typical white dwarf is the nearby star 40 Eridani B. Its surface temperature is a relatively hot 12,000 K, but its luminosity is only 1/275 LSun. Calculations show that its radius is only 1.4% of the Sun\u2019s, or about the same as that of Earth, and its volume is 2.5 \u00d7 10\u20136 that of the Sun. Its mass, however, is 0.43 times the Sun\u2019s mass, just a little less than half. To fit such a substantial mass into so tiny a volume, the star\u2019s density must be about 170,000 times the density of the Sun, or more than 200,000 g/cm3. A teaspoonful of this material would have a mass of some 50 tons! At such enormous densities, matter cannot exist in its usual state; we will examine the particular behavior of this type of matter in The Death of Stars. For now, we just note that white dwarfs are dying stars, reaching the end of their productive lives and ready for their stories to be over. The British astrophysicist (and science popularizer) Arthur Eddington (1882\u20131944) described the first known white dwarf this way: The message of the companion of Sirius, when decoded, ran: \u201cI am composed of material three thousand times denser than anything you\u2019ve ever come across. A ton of my material would be a little nugget you could put in a matchbox.\u201d What reply could one make to something like that? Well, the reply most of us made in 1914 was, \u201cShut up; don\u2019t talk nonsense.\u201d Today, however, astronomers not only accept that stars as dense as white dwarfs exist but (as we will see) have found even denser and stranger", " objects in their quest to understand the evolution of different types of stars. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 647 CHAPTER 18 REVIEW KEY TERMS binary stars two stars that revolve about each other brown dwarf an object intermediate in size between a planet and a star; the approximate mass range is from about 1/100 of the mass of the Sun up to the lower mass limit for self-sustaining nuclear reactions, which is about 1/12 the mass of the Sun eclipsing binary a binary star in which the plane of revolution of the two stars is nearly edge-on to our line of sight, so that the light of one star is periodically diminished by the other passing in front of it H\u2013R diagram (Hertzsprung\u2013Russell diagram) a plot of luminosity against surface temperature (or spectral type) for a group of stars main sequence a sequence of stars on the Hertzsprung\u2013Russell diagram, containing the majority of stars, that runs diagonally from the upper left to the lower right mass-luminosity relation the observed relation between the masses and luminosities of many (90% of all) stars selection effect the selection of sample data in a nonrandom way, causing the sample data to be unrepresentative of the entire data set spectroscopic binary a binary star in which the components are not resolved but whose binary nature is indicated by periodic variations in radial velocity, indicating orbital motion visual binary a binary star in which the two components are telescopically resolved white dwarf a low-mass star that has exhausted most or all of its nuclear fuel and has collapsed to a very small size; such a star is near its final state of life SUMMARY 18.1 A Stellar Census To understand the properties of stars, we must make wide-ranging surveys. We find the stars that appear brightest to our eyes are bright primarily because they are intrinsically very luminous, not because they are the closest to us. Most of the nearest stars are intrinsically so faint that they can be seen only with the aid of a telescope. Stars with low mass and low luminosity are much more common than stars with high mass and high luminosity. Most of the brown dwarfs in the local neighborhood have not yet been discovered. 18.2 Measuring Stellar Masses The masses of stars can be determined by analysis of the orbit of binary stars\u2014two stars that orbit a common center of mass. In visual binaries, the two stars can be seen separately in a telescope, whereas in a spectroscopic binary, only the spectrum reveals the presence of two stars. Stellar masses range from about 1/12 to more than 100 times the mass of the Sun (in rare cases, going to 250 times the Sun\u2019s mass). Objects with masses between 1/12 and 1/100 that of the Sun are called brown dwarfs. Objects in which no nuclear reactions can take place are planets. The most massive stars are, in most cases, also the most luminous, and this correlation is known as the mass-luminosity relation. 648 Chapter 18 The Stars: A Celestial Census 18.3 Diameters of Stars The diameters of stars can be determined by measuring the time it takes an object (the Moon, a planet, or a companion star) to pass in front of it and block its light. Diameters of members of eclipsing binary systems (where the stars pass in front of each other) can be determined through analysis of their orbital motions. 18.4 The H\u2013R Diagram The Hertzsprung\u2013Russell diagram, or H\u2013R diagram, is a plot of stellar luminosity against surface temperature. Most stars lie on the main sequence, which extends diagonally across the H\u2013R diagram from high temperature and high luminosity to low temperature and low luminosity. The position of a star along the main sequence is determined by its mass. High-mass stars emit more energy and are hotter than low-mass stars on the main sequence. Main-sequence stars derive their energy from the fusion of protons to helium. About 90% of the stars lie on the main sequence. Only about 10% of the stars are white dwarfs, and fewer than 1% are giants or supergiants. FOR FURTHER EXPLORATION Articles Croswell, K. \u201cThe Periodic Table of the Cosmos.\u201d Scientific American (July 2011):45\u201349. A brief introduction to the history and uses of the H\u2013R diagram. Davis, J. \u201cMeasuring the Stars.\u201d Sky & Telescope (October 1991): 361. The article explains direct measurements of stellar diameters. DeVorkin, D. \u201cHenry Norris Russell.\u201d Scientific American (May 1989): 126. Kaler, J. \u201cJourneys on the H\u2013R Diagram.\u201d Sky & Telescope (May 1988", "): 483. McAllister, H. \u201cTwenty Years of Seeing Double.\u201d Sky & Telescope (November 1996): 28. An update on modern studies of binary stars. Parker, B. \u201cThose Amazing White Dwarfs.\u201d Astronomy (July 1984): 15. The article focuses on the history of their discovery. Pasachoff, J. \u201cThe H\u2013R Diagram\u2019s 100th Anniversary.\u201d Sky & Telescope (June 2014): 32. Roth, J., and Sinnott, R. \u201cOur Studies of Celestial Neighbors.\u201d Sky & Telescope (October 1996): 32. A discussion is provided on finding the nearest stars. Websites Eclipsing Binary Stars: http://www.midnightkite.com/index.aspx?URL=Binary. Dan Bruton at Austin State University has created this collection of animations, articles, and links showing how astronomers use eclipsing binary light curves. Henry Norris Russell: http://www.nasonline.org/publications/biographical-memoirs/memoir-pdfs/russell- henry-n.pdf. A biographic memoir by Harlow Shapley. Henry Norris Russell: http://www.phys-astro.sonoma.edu/brucemedalists/russell/RussellBio.pdf. A Bruce Medal profile of Russell. Hertzsprung\u2013Russell Diagram: http://skyserver.sdss.org/dr1/en/proj/advanced/hr/. This site from the Sloan Digital Sky Survey introduces the H\u2013R diagram and gives you information for making your own. You can go step This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 649 by step by using the menu at the left. Note that in the project instructions, the word \u201chere\u201d is a link and takes you to the data you need. Stars of the Week: http://stars.astro.illinois.edu/sow/sowlist.html. Astronomer James Kaler does \u201cbiographical summaries\u201d of famous stars\u2014not the Hollywood type, but ones in the real sky. Videos WISE Mission Surveys Nearby Stars: http://www.jpl.nasa.gov/video/details.php?id=1089. Short video about the WISE telescope survey of brown dwarfs and M dwarfs in our immediate neighborhood (1:21). COLLABORATIVE GROUP ACTIVITIES A. Two stars are seen close together in the sky, and your group is given the task of determining whether they are a visual binary or whether they just happen to be seen in nearly the same direction. You have access to a good observatory. Make a list of the types of measurements you would make to determine whether they orbit each other. B. Your group is given information about five main sequence stars that are among the brightest-appearing stars in the sky and yet are pretty far away. Where would these stars be on the H\u2013R diagram and why? Next, your group is given information about five main-sequence stars that are typical of the stars closest to us. Where would these stars be on the H\u2013R diagram and why? C. A very wealthy (but eccentric) alumnus of your college donates a lot of money for a fund that will help in the search for more brown dwarfs. Your group is the committee in charge of this fund. How would you spend the money? (Be as specific as you can, listing instruments and observing programs.) D. Use the internet to search for information about the stars with the largest known diameter. What star is considered the record holder (this changes as new measurements are made)? Read about some of the largest stars on the web. Can your group list some reasons why it might be hard to know which star is the largest? E. Use the internet to search for information about stars with the largest mass. What star is the current \u201cmass champion\u201d among stars? Try to research how the mass of one or more of the most massive stars was measured, and report to the group or the whole class. EXERCISES Review Questions 1. How does the mass of the Sun compare with that of other stars in our local neighborhood? 2. Name and describe the three types of binary systems. 3. Describe two ways of determining the diameter of a star. 4. What are the largest- and smallest-known values of the mass, luminosity, surface temperature, and diameter of stars (roughly)? 650 Chapter 18 The Stars: A Celestial Census 5. You are able to take spectra of both stars in an eclipsing binary system. List all properties of the stars that can be measured from their spectra and light curves. 6. Sketch an H\u2013R diagram. Label the", " axes. Show where cool supergiants, white dwarfs, the Sun, and main- sequence stars are found. 7. Describe what a typical star in the Galaxy would be like compared to the Sun. 8. How do we distinguish stars from brown dwarfs? How do we distinguish brown dwarfs from planets? 9. Describe how the mass, luminosity, surface temperature, and radius of main-sequence stars change in value going from the \u201cbottom\u201d to the \u201ctop\u201d of the main sequence. 10. One method to measure the diameter of a star is to use an object like the Moon or a planet to block out its light and to measure the time it takes to cover up the object. Why is this method used more often with the Moon rather than the planets, even though there are more planets? 11. We discussed in the chapter that about half of stars come in pairs, or multiple star systems, yet the first eclipsing binary was not discovered until the eighteenth century. Why? Thought Questions 12. Is the Sun an average star? Why or why not? 13. Suppose you want to determine the average educational level of people throughout the nation. Since it would be a great deal of work to survey every citizen, you decide to make your task easier by asking only the people on your campus. Will you get an accurate answer? Will your survey be distorted by a selection effect? Explain. 14. Why do most known visual binaries have relatively long periods and most spectroscopic binaries have relatively short periods? 15. Figure 18.11 shows the light curve of a hypothetical eclipsing binary star in which the light of one star is completely blocked by another. What would the light curve look like for a system in which the light of the smaller star is only partially blocked by the larger one? Assume the smaller star is the hotter one. Sketch the relative positions of the two stars that correspond to various portions of the light curve. 16. There are fewer eclipsing binaries than spectroscopic binaries. Explain why. 17. Within 50 light-years of the Sun, visual binaries outnumber eclipsing binaries. Why? 18. Which is easier to observe at large distances\u2014a spectroscopic binary or a visual binary? 19. The eclipsing binary Algol drops from maximum to minimum brightness in about 4 hours, remains at minimum brightness for 20 minutes, and then takes another 4 hours to return to maximum brightness. Assume that we view this system exactly edge-on, so that one star crosses directly in front of the other. Is one star much larger than the other, or are they fairly similar in size? (Hint: Refer to the diagrams of eclipsing binary light curves.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 651 20. Review this spectral data for five stars. Table A Star Spectrum 1 2 3 4 5 G, main sequence K, giant K, main sequence O, main sequence M, main sequence Which is the hottest? Coolest? Most luminous? Least luminous? In each case, give your reasoning. 21. Which changes by the largest factor along the main sequence from spectral types O to M\u2014mass or luminosity? 22. Suppose you want to search for brown dwarfs using a space telescope. Will you design your telescope to detect light in the ultraviolet or the infrared part of the spectrum? Why? 23. An astronomer discovers a type-M star with a large luminosity. How is this possible? What kind of star is it? 24. Approximately 6000 stars are bright enough to be seen without a telescope. Are any of these white dwarfs? Use the information given in this chapter to explain your reasoning. 25. Use the data in Appendix J to plot an H\u2013R diagram for the brightest stars. Use the data from Table 18.3 to show where the main sequence lies. Do 90% of the brightest stars lie on or near the main sequence? Explain why or why not. 26. Use the diagram you have drawn for Exercise 18.25 to answer the following questions: Which star is more massive\u2014Sirius or Alpha Centauri? Rigel and Regulus have nearly the same spectral type. Which is larger? Rigel and Betelgeuse have nearly the same luminosity. Which is larger? Which is redder? 27. Use the data in Appendix I (http://cnx.org/content/The Nearest Stars/latest/) to plot an H\u2013R diagram for this sample of nearby stars. How does this plot differ from the one for the brightest stars in Exercise 18.25? Why? 28. If a visual binary system were to have two equal-mass stars, how would they be located relative to the center of the mass of the system? What would you observe as you watched these stars as they orbited the center of mass, assuming very circular orbits, and assuming the orbit was face", " on to your view? 29. Two stars are in a visual binary star system that we see face on. One star is very massive whereas the other is much less massive. Assuming circular orbits, describe their relative orbits in terms of orbit size, period, and orbital velocity. 30. Describe the spectra for a spectroscopic binary for a system comprised of an F-type and L-type star. Assume that the system is too far away to be able to easily observe the L-type star. 31. Figure 18.7 shows the velocity of two stars in a spectroscopic binary system. Which star is the most massive? Explain your reasoning. 652 Chapter 18 The Stars: A Celestial Census 32. You go out stargazing one night, and someone asks you how far away the brightest stars we see in the sky without a telescope are. What would be a good, general response? (Use Appendix J for more information.) 33. If you were to compare three stars with the same surface temperature, with one star being a giant, another a supergiant, and the third a main-sequence star, how would their radii compare to one another? 34. Are supergiant stars also extremely massive? Explain the reasoning behind your answer. 35. Consider the following data on four stars: Table B Star Luminosity (in LSun) Type 1 2 3 4 100 1/100 1/100 100 B, main sequence B, white dwarf M, main sequence M, giant Which star would have the largest radius? Which star would have the smallest radius? Which star is the most common in our area of the Galaxy? Which star is the least common? Figuring For Yourself 36. If two stars are in a binary system with a combined mass of 5.5 solar masses and an orbital period of 12 years, what is the average distance between the two stars? 37. It is possible that stars as much as 200 times the Sun\u2019s mass or more exist. What is the luminosity of such a star based upon the mass-luminosity relation? 38. The lowest mass for a true star is 1/12 the mass of the Sun. What is the luminosity of such a star based upon the mass-luminosity relationship? 39. Spectral types are an indicator of temperature. For the first 10 stars in Appendix J, the list of the brightest stars in our skies, estimate their temperatures from their spectral types. Use information in the figures and/or tables in this chapter and describe how you made the estimates. 40. We can estimate the masses of most of the stars in Appendix J from the mass-luminosity relationship in Figure 18.9. However, remember this relationship works only for main sequence stars. Determine which of the first 10 stars in Appendix J are main sequence stars. Use one of the figures in this chapter. Make a table of stars\u2019 masses. 41. In Diameters of Stars, the relative diameters of the two stars in the Sirius system were determined. Let\u2019s use this value to explore other aspects of this system. This will be done through several steps, each in its own exercise. Assume the temperature of the Sun is 5800 K, and the temperature of Sirius A, the larger star of the binary, is 10,000 K. The luminosity of Sirius A can be found in Appendix J, and is given as about 23 times that of the Sun. Using the values provided, calculate the radius of Sirius A relative to that of the Sun. 42. Now calculate the radius of Sirius\u2019 white dwarf companion, Sirius B, to the Sun. 43. How does this radius of Sirius B compare with that of Earth? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 18 The Stars: A Celestial Census 653 44. From the previous calculations and the results from Diameters of Stars, it is possible to calculate the density of Sirius B relative to the Sun. It is worth noting that the radius of the companion is very similar to that of Earth, whereas the mass is very similar to the Sun\u2019s. How does the companion\u2019s density compare to that of the Sun? Recall that density = mass/volume, and the volume of a sphere = (4/3)\u03c0R3. How does this density compare with that of water and other materials discussed in this text? Can you see why astronomers were so surprised and puzzled when they first determined the orbit of the companion to Sirius? 45. How much would you weigh if you were suddenly transported to the white dwarf Sirius B? You may use your own weight (or if don\u2019t want to own up to what it is, assume you weigh 70 kg or 150 lb). In this case, assume that the companion to Sirius has a mass equal to that of the Sun and a radius equal to that of Earth. Remember Newton\u2019s law of gravity", ": F = GM1 M2 /R2 and that your weight is proportional to the force that you feel. What kind of star should you travel to if you want to lose weight (and not gain it)? 46. The star Betelgeuse has a temperature of 3400 K and a luminosity of 13,200 LSun. Calculate the radius of Betelgeuse relative to the Sun. 47. Using the information provided in Table 18.1, what is the average stellar density in our part of the Galaxy? Use only the true stars (types O\u2013M) and assume a spherical distribution with radius of 26 light-years. 48. Confirm that the angular diameter of the Sun of 1/2\u00b0 corresponds to a linear diameter of 1.39 million km. Use the average distance of the Sun and Earth to derive the answer. (Hint: This can be solved using a trigonometric function.) 49. An eclipsing binary star system is observed with the following contact times for the main eclipse: Table C Contact First contact Second contact Third contact Fourth contact Time 12:00 p.m. 4:00 p.m. 9:00 a.m. 1:00 p.m. Date March 12 March 13 March 18 March 19 The orbital velocity of the smaller star relative to the larger is 62,000 km/h. Determine the diameters for each star in the system. 50. If a 100 solar mass star were to have a luminosity of 107 times the Sun\u2019s luminosity, how would such a star\u2019s density compare when it is on the main sequence as an O-type star, and when it is a cool supergiant (M-type)? Use values of temperature from Figure 18.14 or Figure 18.15 and the relationship between luminosity, radius, and temperature as given in Exercise 18.47. 51. If Betelgeuse had a mass that was 25 times that of the Sun, how would its average density compare to that of the Sun? Use the definition of density = mass volume, where the volume is that of a sphere. 654 Chapter 18 The Stars: A Celestial Census This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 655 19 CELESTIAL DISTANCES Figure 19.1. Globular Cluster M80. This beautiful image shows a giant cluster of stars called Messier 80, located about 28,000 light-years from Earth. Such crowded groups, which astronomers call globular clusters, contain hundreds of thousands of stars, including some of the RR Lyrae variables discussed in this chapter. Especially obvious in this picture are the bright red giants, which are stars similar to the Sun in mass that are nearing the ends of their lives. (credit: modification of work by The Hubble Heritage Team (AURA/ STScI/ NASA)) Chapter Outline 19.1 Fundamental Units of Distance 19.2 Surveying the Stars 19.3 Variable Stars: One Key to Cosmic Distances 19.4 The H\u2013R Diagram and Cosmic Distances Thinking Ahead How large is the universe? What is the most distant object we can see? These are among the most fundamental questions astronomers can ask. But just as babies must crawl before they can take their first halting steps, so too must we start with a more modest question: How far away are the stars? And even this question proves to be very hard to answer. After all, stars are mere points of light. Suppose you see a point of light in the darkness when you are driving on a country road late at night. How can you tell whether it is a nearby firefly, an oncoming motorcycle some distance away, or the porchlight of a house much farther down the road? It\u2019s not so easy, is it? Astronomers faced an even more difficult problem when they tried to estimate how far away the stars are. In this chapter, we begin with the fundamental definitions of distances on Earth and then extend our reach outward to the stars. We will also examine the newest satellites that are surveying the night sky and discuss the special types of stars that can be used as trail markers to distant galaxies. 19.1 FUNDAMENTAL UNITS OF DISTANCE Learning Objectives By the end of this section, you will be able to: 656 Chapter 19 Celestial Distances Understand the importance of defining a standard distance unit Explain how the meter was originally defined and how it has changed over time Discuss how radar is used to measure distances to the other members of the solar system The first measures of distances were based on human dimensions\u2014the inch as the distance between knuckles on the finger, or the yard as the span from the extended index finger to the nose of the British king. Later, the requirements of commerce led to some standardization of such units, but each nation tended to set up its own definitions. It was not until the middle of the", " eighteenth century that any real efforts were made to establish a uniform, international set of standards. The Metric System One of the enduring legacies of the era of the French emperor Napoleon is the establishment of the metric system of units, officially adopted in France in 1799 and now used in most countries around the world. The fundamental metric unit of length is the meter, originally defined as one ten-millionth of the distance along Earth\u2019s surface from the equator to the pole. French astronomers of the seventeenth and eighteenth centuries were pioneers in determining the dimensions of Earth, so it was logical to use their information as the foundation of the new system. Practical problems exist with a definition expressed in terms of the size of Earth, since anyone wishing to determine the distance from one place to another can hardly be expected to go out and re-measure the planet. Therefore, an intermediate standard meter consisting of a bar of platinum-iridium metal was set up in Paris. In 1889, by international agreement, this bar was defined to be exactly one meter in length, and precise copies of the original meter bar were made to serve as standards for other nations. Other units of length are derived from the meter. Thus, 1 kilometer (km) equals 1000 meters, 1 centimeter (cm) equals 1/100 meter, and so on. Even the old British and American units, such as the inch and the mile, are now defined in terms of the metric system. Modern Redefinitions of the Meter In 1960, the official definition of the meter was changed again. As a result of improved technology for generating spectral lines of precisely known wavelengths (see the chapter on Radiation and Spectra), the meter was redefined to equal 1,650,763.73 wavelengths of a particular atomic transition in the element krypton-86. The advantage of this redefinition is that anyone with a suitably equipped laboratory can reproduce a standard meter, without reference to any particular metal bar. In 1983, the meter was defined once more, this time in terms of the velocity of light. Light in a vacuum can travel a distance of one meter in 1/299,792,458.6 second. Today, therefore, light travel time provides our basic unit of length. Put another way, a distance of one light-second (the amount of space light covers in one second) is defined to be 299,792,458.6 meters. That\u2019s almost 300 million meters that light covers in just one second; light really is very fast! We could just as well use the light-second as the fundamental unit of length, but for practical reasons (and to respect tradition), we have defined the meter as a small fraction of the light-second. Distance within the Solar System The work of Copernicus and Kepler established the relative distances of the planets\u2014that is, how far from the Sun one planet is compared to another (see Observing the Sky: The Birth of Astronomy and Orbits and Gravity). But their work could not establish the absolute distances (in light-seconds or meters or other standard units of length). This is like knowing the height of all the students in your class only as compared to the height of your astronomy instructor, but not in inches or centimeters. Somebody\u2019s height has to be measured directly. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 657 Similarly, to establish absolute distances, astronomers had to measure one distance in the solar system directly. Generally, the closer to us the object is, the easier such a measurement would be. Estimates of the distance to Venus were made as Venus crossed the face of the Sun in 1761 and 1769, and an international campaign was organized to estimate the distance to the asteroid Eros in the early 1930s, when its orbit brought it close to Earth. More recently, Venus crossed (or transited) the surface of the Sun in 2004 and 2012, and allowed us to make a modern distance estimate, although, as we will see below, by then it wasn\u2019t needed (Figure 19.2). If you would like more information on just how the motion of Venus across the Sun helped us pin down distances in the solar system, you can turn to a nice explanation (https://openstaxcollege.org/l/ 30VenusandSun) by a NASA astronomer. Figure 19.2. Venus Transits the Sun, 2012. This striking \u201cpicture\u201d of Venus crossing the face of the Sun (it\u2019s the black dot at about 2 o\u2019clock) is more than just an impressive image. Taken with the Solar Dynamics Observatory spacecraft and special filters, it shows a modern transit of Venus. Such events allowed astronomers in the 1800s to estimate the distance to Venus. They measured the time it took Venus to cross the face of the Sun from different latitudes on Earth. The differences in times can be used to estimate the", " distance to the planet. Today, radar is used for much more precise distance estimates. (credit: modification of work by NASA/SDO, AIA) The key to our modern determination of solar system dimensions is radar, a type of radio wave that can bounce off solid objects (Figure 19.3). As discussed in several earlier chapters, by timing how long a radar beam (traveling at the speed of light) takes to reach another world and return, we can measure the distance involved very accurately. In 1961, radar signals were bounced off Venus for the first time, providing a direct measurement of the distance from Earth to Venus in terms of light-seconds (from the roundtrip travel time of the radar signal). Subsequently, radar has been used to determine the distances to Mercury, Mars, the satellites of Jupiter, the rings of Saturn, and several asteroids. Note, by the way, that it is not possible to use radar to measure the distance to the Sun directly because the Sun does not reflect radar very efficiently. But we can measure the distance to many other solar system objects and use Kepler\u2019s laws to give us the distance to the Sun. 658 Chapter 19 Celestial Distances Figure 19.3. Radar Telescope. This dish-shaped antenna, part of the NASA Deep Space Network in California\u2019s Mojave Desert, is 70 meters wide. Nicknamed the \u201cMars antenna,\u201d this radar telescope can send and receive radar waves, and thus measure the distances to planets, satellites, and asteroids. (credit: NASA/JPL-Caltech) From the various (related) solar system distances, astronomers selected the average distance from Earth to the Sun as our standard \u201cmeasuring stick\u201d within the solar system. When Earth and the Sun are closest, they are about 147.1 million kilometers apart; when Earth and the Sun are farthest, they are about 152.1 million kilometers apart. The average of these two distances is called the astronomical unit (AU). We then express all the other distances in the solar system in terms of the AU. Years of painstaking analyses of radar measurements have led to a determination of the length of the AU to a precision of about one part in a billion. The length of 1 AU can be expressed in light travel time as 499.004854 light-seconds, or about 8.3 light-minutes. If we use the definition of the meter given previously, this is equivalent to 1 AU = 149,597,870,700 meters. These distances are, of course, given here to a much higher level of precision than is normally needed. In this text, we are usually content to express numbers to a couple of significant places and leave it at that. For our purposes, it will be sufficient to round off these numbers: speed of light: c = 3 \u00d7 108 m/s = 3 \u00d7 105 km/s length of light-second: ls = 3 \u00d7 108 m = 3 \u00d7 105 km astronomical unit: AU = 1.50 \u00d7 1011 m = 1.50 \u00d7 108 km = 500 light-seconds We now know the absolute distance scale within our own solar system with fantastic accuracy. This is the first link in the chain of cosmic distances The distances between the celestial bodies in our solar system are sometimes difficult to grasp or put into perspective. This interactive website (https://openstaxcollege.org/l/30DistanceScale) provides a \u201cmap\u201d that shows the distances by using a scale at the bottom of the screen and allows you to scroll This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 659 (using your arrow keys) through screens of \u201cempty space\u201d to get to the next planet\u2014all while your current distance from the Sun is visible on the scale. 19.2 SURVEYING THE STARS Learning Objectives By the end of this section, you will be able to: Understand the concept of triangulating distances to distant objects, including stars Explain why space-based satellites deliver more precise distances than ground-based methods Discuss astronomers\u2019 efforts to study the stars closest to the Sun It is an enormous step to go from the planets to the stars. For example, our Voyager 1 probe, which was launched in 1977, has now traveled farther from Earth than any other spacecraft. As this is written in 2016, Voyager 1 is 134 AU from the Sun.[1] The nearest star, however, is hundreds of thousands of AU from Earth. Even so, we can, in principle, survey distances to the stars using the same technique that a civil engineer employs to survey the distance to an inaccessible mountain or tree\u2014the method of triangulation. Triangulation in Space A practical example of triangulation is your own depth perception. As you are pleased to discover every morning when you look in the mirror, your two eyes are located some distance apart. You therefore view the world from two different vantage points", ", and it is this dual perspective that allows you to get a general sense of how far away objects are. To see what we mean, take a pen and hold it a few inches in front of your face. Look at it first with one eye (closing the other) and then switch eyes. Note how the pen seems to shift relative to objects across the room. Now hold the pen at arm\u2019s length: the shift is less. If you play with moving the pen for a while, you will notice that the farther away you hold it, the less it seems to shift. Your brain automatically performs such comparisons and gives you a pretty good sense of how far away things in your immediate neighborhood are. If your arms were made of rubber, you could stretch the pen far enough away from your eyes that the shift would become imperceptible. This is because our depth perception fails for objects more than a few tens of meters away. In order to see the shift of an object a city block or more from you, your eyes would need to be spread apart a lot farther. Let\u2019s see how surveyors take advantage of the same idea. Suppose you are trying to measure the distance to a tree across a deep river (Figure 19.4). You set up two observing stations some distance apart. That distance (line AB in Figure 19.4) is called the baseline. Now the direction to the tree (C in the figure) in relation to the baseline is observed from each station. Note that C appears in different directions from the two stations. This apparent change in direction of the remote object due to a change in vantage point of the observer is called parallax. 1 To have some basis for comparison, the dwarf planet Pluto orbits at an average distance of 40 AU from the Sun, and the dwarf planet Eris is currently roughly 96 AU from the Sun. 660 Chapter 19 Celestial Distances Figure 19.4. Triangulation. Triangulation allows us to measure distances to inaccessible objects. By getting the angle to a tree from two different vantage points, we can calculate the properties of the triangle they make and thus the distance to the tree. The parallax is also the angle that lines AC and BC make\u2014in mathematical terms, the angle subtended by the baseline. A knowledge of the angles at A and B and the length of the baseline, AB, allows the triangle ABC to be solved for any of its dimensions\u2014say, the distance AC or BC. The solution could be reached by constructing a scale drawing or by using trigonometry to make a numerical calculation. If the tree were farther away, the whole triangle would be longer and skinnier, and the parallax angle would be smaller. Thus, we have the general rule that the smaller the parallax, the more distant the object we are measuring must be. In practice, the kinds of baselines surveyors use for measuring distances on Earth are completely useless when we try to gauge distances in space. The farther away an astronomical object lies, the longer the baseline has to be to give us a reasonable chance of making a measurement. Unfortunately, nearly all astronomical objects are very far away. To measure their distances requires a very large baseline and highly precise angular measurements. The Moon is the only object near enough that its distance can be found fairly accurately with measurements made without a telescope. Ptolemy determined the distance to the Moon correctly to within a few percent. He used the turning Earth itself as a baseline, measuring the position of the Moon relative to the stars at two different times of night. With the aid of telescopes, later astronomers were able to measure the distances to the nearer planets and asteroids using Earth\u2019s diameter as a baseline. This is how the AU was first established. To reach for the stars, however, requires a much longer baseline for triangulation and extremely sensitive measurements. Such a baseline is provided by Earth\u2019s annual trip around the Sun. Distances to Stars As Earth travels from one side of its orbit to the other, it graciously provides us with a baseline of 2 AU, or about 300 million kilometers. Although this is a much bigger baseline than the diameter of Earth, the stars are so far away that the resulting parallax shift is still not visible to the naked eye\u2014not even for the closest stars. In the chapter on Observing the Sky: The Birth of Astronomy, we discussed how this dilemma perplexed the ancient Greeks, some of whom had actually suggested that the Sun might be the center of the solar system, with Earth in motion around it. Aristotle and others argued, however, that Earth could not be revolving about This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 661 the Sun. If it were, they said, we would surely observe the parallax of the nearer stars against the background of more distant objects as we viewed the sky from different parts of Earth\u2019s orbit (", "Figure 19.6). Tycho Brahe (1546\u20131601) advanced the same faulty argument nearly 2000 years later, when his careful measurements of stellar positions with the unaided eye revealed no such shift. These early observers did not realize how truly distant the stars were and how small the change in their positions therefore was, even with the entire orbit of Earth as a baseline. The problem was that they did not have tools to measure parallax shifts too small to be seen with the human eye. By the eighteenth century, when there was no longer serious doubt about Earth\u2019s revolution, it became clear that the stars must be extremely distant. Astronomers equipped with telescopes began to devise instruments capable of measuring the tiny shifts of nearby stars relative to the background of more distant (and thus unshifting) celestial objects. This was a significant technical challenge, since, even for the nearest stars, parallax angles are usually only a fraction of a second of arc. Recall that one second of arc (arcsec) is an angle of only 1/3600 of a degree. A coin the size of a US quarter would appear to have a diameter of 1 arcsecond if you were viewing it from a distance of about 5 kilometers (3 miles). Think about how small an angle that is. No wonder it took astronomers a long time before they could measure such tiny shifts. The first successful detections of stellar parallax were in the year 1838, when Friedrich Bessel in Germany (Figure 19.5), Thomas Henderson, a Scottish astronomer working at the Cape of Good Hope, and Friedrich Struve in Russia independently measured the parallaxes of the stars 61 Cygni, Alpha Centauri, and Vega, respectively. Even the closest star, Alpha Centauri, showed a total displacement of only about 1.5 arcseconds during the course of a year. Figure 19.5. Friedrich Wilhelm Bessel (1784\u20131846), Thomas J. Henderson (1798\u20131844), and Friedrich Struve (1793\u20131864). (a) Bessel made the first authenticated measurement of the distance to a star (61 Cygni) in 1838, a feat that had eluded many dedicated astronomers for almost a century. But two others, (b) Scottish astronomer Thomas J. Henderson and (c) Friedrich Struve, in Russia, were close on his heels. Figure 19.6 shows how such measurements work. Seen from opposite sides of Earth\u2019s orbit, a nearby star shifts position when compared to a pattern of more distant stars. Astronomers actually define parallax to be one-half the angle that a star shifts when seen from opposite sides of Earth\u2019s orbit (the angle labeled P in Figure 19.6). The reason for this definition is just that they prefer to deal with a baseline of 1 AU instead of 2 AU. 662 Chapter 19 Celestial Distances Figure 19.6. Parallax. As Earth revolves around the Sun, the direction in which we see a nearby star varies with respect to distant stars. We define the parallax of the nearby star to be one half of the total change in direction, and we usually measure it in arcseconds. Units of Stellar Distance With a baseline of one AU, how far away would a star have to be to have a parallax of 1 arcsecond? The answer turns out to be 206,265 AU, or 3.26 light-years. This is equal to 3.1 \u00d7 1013 kilometers (in other words, 31 trillion kilometers). We give this unit a special name, the parsec (pc)\u2014derived from \u201cthe distance at which we have a parallax of one second.\u201d The distance (D) of a star in parsecs is just the reciprocal of its parallax (p) in arcseconds; that is, D = 1 p Thus, a star with a parallax of 0.1 arcsecond would be found at a distance of 10 parsecs, and one with a parallax of 0.05 arcsecond would be 20 parsecs away. Back in the days when most of our distances came from parallax measurements, a parsec was a useful unit of distance, but it is not as intuitive as the light-year. One advantage of the light-year as a unit is that it emphasizes the fact that, as we look out into space, we are also looking back into time. The light that we see from a star 100 light-years away left that star 100 years ago. What we study is not the star as it is now, but rather as it was in the past. The light that reaches our telescopes today from distant galaxies left them before Earth even existed. In this text, we will use light-years as our unit of distance, but many astronomers still use parsecs when they write technical papers or talk with each other at meetings. To convert between the two distance units, just bear in mind: 1 par", "sec = 3.26 light-year, and 1 light-year = 0.31 parsec How Far Is a Light-Year? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 663 A light-year is the distance light travels in 1 year. Given that light travels at a speed of 300,000 km/s, how many kilometers are there in a light-year? Solution We learned earlier that speed = distance/time. We can rearrange this equation so that distance = velocity \u00d7 time. Now, we need to determine the number of seconds in a year. There are approximately 365 days in 1 year. To determine the number of seconds, we must estimate the number of seconds in 1 day. We can change units as follows (notice how the units of time cancel out): 1 day \u00d7 24 hr/day \u00d7 60 min/hr \u00d7 60 s/min = 86,400 s/day Next, to get the number of seconds per year: 365 days/year \u00d7 86,400 s/day = 31,536,000 s/year Now we can multiply the speed of light by the number of seconds per year to get the distance traveled by light in 1 year: distance = velocity \u00d7 time = 300,000 km/s \u00d7 31,536,000 s = 9.46 \u00d7 1012 km That\u2019s almost 10,000,000,000,000 km that light covers in a year. To help you imagine how long this distance is, we\u2019ll mention that a string 1 light-year long could fit around the circumference of Earth 236 million times. Check Your Learning The number above is really large. What happens if we put it in terms that might be a little more understandable, like the diameter of Earth? Earth\u2019s diameter is about 12,700 km. Answer: 1 light-year = 9.46 \u00d7 1012 km = 9.46 \u00d7 1012 km \u00d7 1 Earth diameter 12,700 km = 7.45 \u00d7 108 Earth diameters That means that 1 light-year is about 745 million times the diameter of Earth Naming Stars You may be wondering why stars have such a confusing assortment of names. Just look at the first three stars to have their parallaxes measured: 61 Cygni, Alpha Centauri, and Vega. Each of these names comes from a different tradition of designating stars. 664 Chapter 19 Celestial Distances The brightest stars have names that derive from the ancients. Some are from the Greek, such as Sirius, which means \u201cthe scorched one\u201d\u2014a reference to its brilliance. A few are from Latin, but many of the best-known names are from Arabic because, as discussed in Observing the Sky: The Birth of Astronomy, much of Greek and Roman astronomy was \u201crediscovered\u201d in Europe after the Dark Ages by means of Arabic translations. Vega, for example, means \u201cswooping Eagle,\u201d and Betelgeuse (pronounced \u201cBeetle-juice\u201d) means \u201cright hand of the central one.\u201d In 1603, German astronomer Johann Bayer (1572\u20131625) introduced a more systematic approach to naming stars. For each constellation, he assigned a Greek letter to the brightest stars, roughly in order of brightness. In the constellation of Orion, for example, Betelgeuse is the brightest star, so it got the first letter in the Greek alphabet\u2014alpha\u2014and is known as Alpha Orionis. (\u201cOrionis\u201d is the possessive form of Orion, so Alpha Orionis means \u201cthe first of Orion.\u201d) A star called Rigel, being the second brightest in that constellation, is called Beta Orionis (Figure 19.7). Since there are 24 letters in the Greek alphabet, this system allows the labeling of 24 stars in each constellation, but constellations have many more stars than that. Figure 19.7. Objects in Orion. (a) This image shows the brightest objects in or near the star pattern of Orion, the hunter (of Greek mythology), in the constellation of Orion. (b) Note the Greek letters of Bayer\u2019s system in this diagram of the Orion constellation. The objects denoted M42, M43, and M78 are not stars but nebulae\u2014clouds of gas and dust; these numbers come from a list of \u201cfuzzy objects\u201d made by Charles Messier in 1781. (credit a: modification of work by Matthew Spinelli; credit b: modification of work by ESO, IAU and Sky & Telescope) In 1725, the English Astronomer Royal John Flamsteed introduced yet another system, in which the brighter stars eventually got a number in each constellation in order of their location in the sky or, more precisely, their right ascension. (The system of sky coordinates that", " includes right ascension was discussed in Earth, Moon, and Sky.) In this system, Betelgeuse is called 58 Orionis and 61 Cygni is the 61st star in the constellation of Cygnus, the swan. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 665 It gets worse. As astronomers began to understand more and more about stars, they drew up a series of specialized star catalogs, and fans of those catalogs began calling stars by their catalog numbers. If you look at Appendix I\u2014our list of the nearest stars (many of which are much too faint to get an ancient name, Bayer letter, or Flamsteed number)\u2014you will see references to some of these catalogs. An example is a set of stars labeled with a BD number, for \u201cBonner Durchmusterung.\u201d This was a mammoth catalog of over 324,000 stars in a series of zones in the sky, organized at the Bonn Observatory in the 1850s and 1860s. Keep in mind that this catalog was made before photography or computers came into use, so the position of each star had to be measured (at least twice) by eye, a daunting undertaking. There is also a completely different system for keeping track of stars whose luminosity varies, and another for stars that brighten explosively at unpredictable times. Astronomers have gotten used to the many different star-naming systems, but students often find them bewildering and wish astronomers would settle on one. Don\u2019t hold your breath: in astronomy, as in many fields of human thought, tradition holds a powerful attraction. Still, with high-speed computer databases to aid human memory, names may become less and less necessary. Today\u2019s astronomers often refer to stars by their precise locations in the sky rather than by their names or various catalog numbers. The Nearest Stars No known star (other than the Sun) is within 1 light-year or even 1 parsec of Earth. The stellar neighbors nearest the Sun are three stars in the constellation of Centaurus. To the unaided eye, the brightest of these three stars is Alpha Centauri, which is only 30\u25cb from the south celestial pole and hence not visible from the mainland United States. Alpha Centauri itself is a binary star\u2014two stars in mutual revolution\u2014too close together to be distinguished without a telescope. These two stars are 4.4 light-years from us. Nearby is a third faint star, known as Proxima Centauri. Proxima, with a distance of 4.3 light-years, is slightly closer to us than the other two stars. If Proxima Centauri is part of a triple star system with the binary Alpha Centauri, as seems likely, then its orbital period may be longer than 500,000 years. Proxima Centauri is an example of the most common type of star, and our most common type of stellar neighbor (as we saw in Stars: A Celestial Census.) Low-mass red M dwarfs make up about 70% of all stars and dominate the census of stars within 10 parsecs of the Sun. The latest survey of the solar neighborhood has counted 357 stars and brown dwarfs within 10 parsecs, and 248 of these are red dwarfs. Yet, if you wanted to see an M dwarf with your naked eye, you would be out of luck. These stars only produce a fraction of the Sun\u2019s light, and nearly all of them require a telescope to be detected. The nearest star visible without a telescope from most of the United States is the brightest appearing of all the stars, Sirius, which has a distance of a little more than 8 light-years. It too is a binary system, composed of a faint white dwarf orbiting a bluish-white, main-sequence star. It is an interesting coincidence of numbers that light reaches us from the Sun in about 8 minutes and from the next brightest star in the sky in about 8 years Calculating the Diameter of the Sun 666 Chapter 19 Celestial Distances For nearby stars, we can measure the apparent shift in their positions as Earth orbits the Sun. We wrote earlier that an object must be 206,265 AU distant to have a parallax of one second of arc. This must seem like a very strange number, but you can figure out why this is the right value. We will start by estimating the diameter of the Sun and then apply the same idea to a star with a parallax of 1 arcsecond. Make a sketch that has a round circle to represent the Sun, place Earth some distance away, and put an observer on it. Draw two lines from the point where the observer is standing, one to each side of the Sun. Sketch a circle centered at Earth with its circumference passing through the center of the Sun. Now think about proportions. The Sun spans about half a degree on the sky. A full circle has 360\ufffd", "\ufffd. The circumference of the circle centered on Earth and passing through the Sun is given by: Then, the following two ratios are equal: circumference = 2\u03c0 \u00d7 93,000,000 miles 0.5\u00b0 360\u00b0 = diameter of Sun 2\u03c0 \u00d7 93,000,000 Calculate the diameter of the Sun. How does your answer compare to the actual diameter? Solution To solve for the diameter of the Sun, we can evaluate the expression above. diameter of the sun = 0.5\u00b0 360\u00b0 \u00d7 2\u03c0 \u00d7 93,000,000 miles = 811,577 miles This is very close to the true value of about 848,000 miles. Check Your Learning Now apply this idea to calculating the distance to a star that has a parallax of 1 arcsec. Draw a picture similar to the one we suggested above and calculate the distance in AU. (Hint: Remember that the parallax angle is defined by 1 AU, not 2 AU, and that 3600 arcseconds = 1 degree.) Answer: 206,265 AU Measuring Parallaxes in Space The measurements of stellar parallax were revolutionized by the launch of the spacecraft Hipparcos in 1989, which measured distances for thousands of stars out to about 300 light-years with an accuracy of 10 to 20% (see Figure 19.8 and the feature on Parallax and Space Astronomy). However, even 300 light-years are less than 1% the size of our Galaxy\u2019s main disk. In December 2013, the successor to Hipparcos, named Gaia, was launched by the European Space Agency. Gaia is expected to measure the position and distances to almost one billion stars with an accuracy of a few ten- millionths of an arcsecond. Gaia\u2019s distance limit will extend well beyond Hipparcos, studying stars out to 30,000 light-years (100 times farther than Hipparcos, covering nearly 1/3 of the galactic disk). Gaia will also be able to measure proper motions[2] for thousands of stars in the halo of the Milky Way\u2014something that can only be done for the brightest stars right now. At the end of Gaia\u2019s mission, we will not only have a three-dimensional 2 Proper motion (as discussed in Analyzing Starlight, is the motion of a star across the sky (perpendicular to our line of sight.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 667 map of a large fraction of our own Milky Way Galaxy, but we will also have a strong link in the chain of cosmic distances that we are discussing in this chapter. Yet, to extend this chain beyond Gaia\u2019s reach and explore distances to nearby galaxies, we need some completely new techniques. Figure 19.8. H\u2013R Diagram of Stars Measured by Gaia and Hipparcos. This plot includes 16,631 stars for which the parallaxes have an accuracy of 10% or better. The colors indicate the numbers of stars at each point of the diagram, with red corresponding to the largest number and blue to the lowest. Luminosity is plotted along the vertical axis, with luminosity increasing upward. An infrared color is plotted as a proxy for temperature, with temperature decreasing to the right. Most of the data points are distributed along the diagonal running from the top left corner (high luminosity, high temperature) to the bottom right (low temperature, low luminosity). These are main sequence stars. The large clump of data points above the main sequence on the right side of the diagram is composed of red giant stars. (credit: modification of work by the European Space Agency Parallax and Space Astronomy One of the most difficult things about precisely measuring the tiny angles of parallax shifts from Earth is that you have to observe the stars through our planet\u2019s atmosphere. As we saw in Astronomical Instruments, the effect of the atmosphere is to spread out the points of starlight into fuzzy disks, making exact measurements of their positions more difficult. Astronomers had long dreamed of being 668 Chapter 19 Celestial Distances able to measure parallaxes from space, and two orbiting observatories have now turned this dream into reality. The name of the Hipparcos satellite, launched in 1989 by the European Space Agency, is both an abbreviation for High Precision Parallax Collecting Satellite and a tribute to Hipparchus, the pioneering Greek astronomer whose work we discussed in the Observing the Sky: The Birth of Astronomy. The satellite was designed to make the most accurate parallax measurements in history, from 36,000 kilometers above Earth. However, its onboard rocket motor failed to fire, which meant it did not get the needed boost to reach the desired altitude. Hipparcos ended up spending its 4-year life in an elliptical orbit that varied from 500 to 36,000 kilometers high. In this orbit, the satellite plunged", " into Earth\u2019s radiation belts every 5 hours or so, which finally took its toll on the solar panels that provided energy to power the instruments. Nevertheless, the mission was successful, resulting in two catalogs. One gives positions of 120,000 stars to an accuracy of one-thousandth of an arcsecond\u2014about the diameter of a golf ball in New York as viewed from Europe. The second catalog contains information for more than a million stars, whose positions have been measured to thirty-thousandths of an arcsecond. We now have accurate parallax measurements of stars out to distances of about 300 light-years. (With ground-based telescopes, accurate measurements were feasible out to only about 60 light-years.) In order to build on the success of Hipparcos, in 2013, the European Space Agency launched a new satellite called Gaia. The Gaia mission is scheduled to last for 5 years. Because Gaia carries larger telescopes than Hipparcos, it can observe fainter stars and measure their positions 200 times more accurately. The main goal of the Gaia mission is to make an accurate three-dimensional map of that portion of the Galaxy within about 30,000 light-years by observing 1 billion stars 70 times each, measuring their positions and hence their parallaxes as well as their brightnesses. For a long time, the measurement of parallaxes and accurate stellar positions was a backwater of astronomical research\u2014mainly because the accuracy of measurements did not improve much for about 100 years. However, the ability to make measurements from space has revolutionized this field of astronomy and will continue to provide a critical link in our chain of cosmic distances The European Space Agency (ESA) maintains a Gaia mission website (https://openstaxcollege.org/l/ 30GaiaMission) where you can learn more about the Gaia mission and to get the latest news on Gaia observations. To learn more about Hipparcos, explore this European Space Agency webpage (https://openstaxcollege.org/l/30Hipparcos) with an ESA vodcast Charting the Galaxy\u2014from Hipparcos to Gaia. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 669 19.3 VARIABLE STARS: ONE KEY TO COSMIC DISTANCES Learning Objectives By the end of this section, you will be able to: Describe how some stars vary their light output and why such stars are important Explain the importance of pulsating variable stars, such as cepheids and RR Lyrae-type stars, to our study of the universe Let\u2019s briefly review the key reasons that measuring distances to the stars is such a struggle. As discussed in The Brightness of Stars, our problem is that stars come in a bewildering variety of intrinsic luminosities. (If stars were light bulbs, we\u2019d say they come in a wide range of wattages.) Suppose, instead, that all stars had the same \u201cwattage\u201d or luminosity. In that case, the more distant ones would always look dimmer, and we could tell how far away a star is simply by how dim it appeared. In the real universe, however, when we look at a star in our sky (with eye or telescope) and measure its apparent brightness, we cannot know whether it looks dim because it\u2019s a low-wattage bulb or because it is far away, or perhaps some of each. Astronomers need to discover something else about the star that allows us to \u201cread off\u201d its intrinsic luminosity\u2014in effect, to know what the star\u2019s true wattage is. With this information, we can then attribute how dim it looks from Earth to its distance. Recall that the apparent brightness of an object decreases with the square of the distance to that object. If two objects have the same luminosity but one is three times farther than the other, the more distant one will look nine times fainter. Therefore, if we know the luminosity of a star and its apparent brightness, we can calculate how far away it is. Astronomers have long searched for techniques that would somehow allow us to determine the luminosity of a star\u2014and it is to these techniques that we turn next. Variable Stars The breakthrough in measuring distances to remote parts of our Galaxy, and to other galaxies as well, came from the study of variable stars. Most stars are constant in their luminosity, at least to within a percent or two. Like the Sun, they generate a steady flow of energy from their interiors. However, some stars are seen to vary in brightness and, for this reason, are called variable stars. Many such stars vary on a regular cycle, like the flashing bulbs that decorate stores and homes during the winter holidays. Let\u2019s define some tools to help us keep track of how a star varies. A graph that", " shows how the brightness of a variable star changes with time is called a light curve (Figure 19.9). The maximum is the point of the light curve where the star has its greatest brightness; the minimum is the point where it is faintest. If the light variations repeat themselves periodically, the interval between the two maxima is called the period of the star. (If this kind of graph looks familiar, it is because we introduced it in Diameters of Stars.) Figure 19.9. Cepheid Light Curve. This light curve shows how the brightness changes with time for a typical cepheid variable, with a period of about 6 days. 670 Chapter 19 Celestial Distances Pulsating Variables There are two special types of variable stars for which\u2014as we will see\u2014measurements of the light curve give us accurate distances. These are called cepheid and RR Lyrae variables, both of which are pulsating variable stars. Such a star actually changes its diameter with time\u2014periodically expanding and contracting, as your chest does when you breathe. We now understand that these stars are going through a brief unstable stage late in their lives. The expansion and contraction of pulsating variables can be measured by using the Doppler effect. The lines in the spectrum shift toward the blue as the surface of the star moves toward us and then shift to the red as the surface shrinks back. As the star pulsates, it also changes its overall color, indicating that its temperature is also varying. And, most important for our purposes, the luminosity of the pulsating variable also changes in a regular way as it expands and contracts. Cepheid Variables Cepheids are large, yellow, pulsating stars named for the first-known star of the group, Delta Cephei. This, by the way, is another example of how confusing naming conventions get in astronomy; here, a whole class of stars in named after the constellation in which the first one happened to be found. (We textbook authors can only apologize to our students for the whole mess!) The variability of Delta Cephei was discovered in 1784 by the young English astronomer John Goodricke (see John Goodricke). The star rises rather rapidly to maximum light and then falls more slowly to minimum light, taking a total of 5.4 days for one cycle. The curve in Figure 19.9 represents a simplified version of the light curve of Delta Cephei. Several hundred cepheid variables are known in our Galaxy. Most cepheids have periods in the range of 3 to 50 days and luminosities that are about 1000 to 10,000 times greater than that of the Sun. Their variations in luminosity range from a few percent to a factor of 10. Polaris, the North Star, is a cepheid variable that, for a long time, varied by one tenth of a magnitude, or by about 10% in visual luminosity, in a period of just under 4 days. Recent measurements indicate that the amount by which the brightness of Polaris changes is decreasing and that, sometime in the future, this star will no longer be a pulsating variable. This is just one more piece of evidence that stars really do evolve and change in fundamental ways as they age, and that being a cepheid variable represents a stage in the life of the star. The Period-Luminosity Relation The importance of cepheid variables lies in the fact that their periods and average luminosities turn out to be directly related. The longer the period (the longer the star takes to vary), the greater the luminosity. This period- luminosity relation was a remarkable discovery, one for which astronomers still (pardon the expression) thank their lucky stars. The period of such a star is easy to measure: a good telescope and a good clock are all you need. Once you have the period, the relationship (which can be put into precise mathematical terms) will give you the luminosity of the star. Let\u2019s be clear on what that means. The relation allows you to essentially \u201cread off\u201d how bright the star really is (how much energy it puts out). Astronomers can then compare this intrinsic brightness with the apparent brightness of the star. As we saw, the difference between the two allows them to calculate the distance. The relation between period and luminosity was discovered in 1908 by Henrietta Leavitt (Figure 19.10), a staff member at the Harvard College Observatory (and one of a number of women working for low wages assisting Edward Pickering, the observatory\u2019s director; see Annie Cannon: Classifier of the Stars). Leavitt discovered hundreds of variable stars in the Large Magellanic Cloud and Small Magellanic Cloud, two great star systems This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 671 that are actually neighboring galaxies (although they were not known to be galaxies then", "). A small fraction of these variables were cepheids (Figure 19.11). Figure 19.10. Henrietta Swan Leavitt (1868\u20131921). Leavitt worked as an astronomer at the Harvard College Observatory. While studying photographs of the Magellanic Clouds, she found over 1700 variable stars, including 20 cepheids. Since all the cepheids in these systems were at roughly the same distance, she was able to compare their luminosities and periods of variation. She thus discovered a fundamental relationship between these characteristics that led to a new and much better way of estimating cosmic distances. (credit: modification of work by AIP) These systems presented a wonderful opportunity to study the behavior of variable stars independent of their distance. For all practical purposes, the Magellanic Clouds are so far away that astronomers can assume that all the stars in them are at roughly the same distance from us. (In the same way, all the suburbs of Los Angeles are roughly the same distance from New York City. Of course, if you are in Los Angeles, you will notice annoying distances between the suburbs, but compared to how far away New York City is, the differences seem small.) If all the variable stars in the Magellanic Clouds are at roughly the same distance, then any difference in their apparent brightnesses must be caused by differences in their intrinsic luminosities. 672 Chapter 19 Celestial Distances Figure 19.11. Large Magellanic Cloud. The Large Magellanic Cloud (so named because Magellan\u2019s crew were the first Europeans to record it) is a small, irregularly shaped galaxy near our own Milky Way. It was in this galaxy that Henrietta Leavitt discovered the cepheid periodluminosity relation. (credit: ESO) Leavitt found that the brighter-appearing cepheids always have the longer periods of light variation. Thus, she reasoned, the period must be related to the luminosity of the stars. When Leavitt did this work, the distance to the Magellanic Clouds was not known, so she was only able to show that luminosity was related to period. She could not determine exactly what the relationship is. To define the period-luminosity relation with actual numbers (to calibrate it), astronomers first had to measure the actual distances to a few nearby cepheids in another way. (This was accomplished by finding cepheids associated in clusters with other stars whose distances could be estimated from their spectra, as discussed in the next section of this chapter.) But once the relation was thus defined, it could give us the distance to any cepheid, wherever it might be located (Figure 19.12). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 673 Figure 19.12. How to Use a Cepheid to Measure Distance. (a) Find a cepheid variable star and measure its period. (b) Use the periodluminosity relation to calculate the star\u2019s luminosity. (c) Measure the star\u2019s apparent brightness. (d) Compare the luminosity with the apparent brightness to calculate the distance. Here at last was the technique astronomers had been searching for to break the confines of distance that parallax imposed on them. Cepheids can be observed and monitored, it turns out, in many parts of our own Galaxy and in other nearby galaxies as well. Astronomers, including Ejnar Hertzsprung and Harvard\u2019s Harlow Shapley, immediately saw the potential of the new technique; they and many others set to work exploring more distant reaches of space using cepheids as signposts. In the 1920s, Edwin Hubble made one of the most significant astronomical discoveries of all time using cepheids, when he observed them in nearby galaxies and discovered the expansion of the universe. As we will see, this work still continues, as the Hubble Space Telescope and other modern instruments try to identify and measure individual cepheids in galaxies farther and farther away. The most distant known variable stars are all cepheids, with some about 60 million light-years away John Goodricke The brief life of John Goodricke (Figure 19.13) is a testament to the human spirit under adversity. Born deaf and unable to speak, Goodricke nevertheless made a number of pioneering discoveries in astronomy through patient and careful observations of the heavens. 674 Chapter 19 Celestial Distances Figure 19.13. John Goodricke (1764\u20131786). This portrait of Goodricke by artist J. Scouler hangs in the Royal Astronomical Society in London. There is some controversy about whether this is actually what Goodricke looked like or whether the painting was much retouched to please his family. (credit: James Scouler) Born in Holland, where his father was on", " a diplomatic mission, Goodricke was sent back to England at age eight to study at a special school for the deaf. He did sufficiently well to enter Warrington Academy, a secondary school that offered no special assistance for students with handicaps. His mathematics teacher there inspired an interest in astronomy, and in 1781, at age 17, Goodricke began observing the sky at his family home in York, England. Within a year, he had discovered the brightness variations of the star Algol (discussed in The Stars: A Celestial Census) and suggested that an unseen companion star was causing the changes, a theory that waited over 100 years for proof. His paper on the subject was read before the Royal Society (the main British group of scientists) in 1783 and won him a medal from that distinguished group. In the meantime, Goodricke had discovered two other stars that varied regularly, Beta Lyrae and Delta Cephei, both of which continued to interest astronomers for years to come. Goodricke shared his interest in observing with his older cousin, Edward Pigott, who went on to discover other variable stars during his much longer life. But Goodricke\u2019s time was quickly drawing to a close; at age 21, only 2 weeks after he was elected to the Royal Society, he caught a cold while making astronomical observations and never recovered. Today, the University of York has a building named Goodricke Hall and a plaque that honors his contributions to science. Yet if you go to the churchyard cemetery where he is buried, an overgrown tombstone has only the initials \u201cJ. G.\u201d to show where he lies. Astronomer Zdenek Kopal, who looked carefully into Goodricke\u2019s life, speculated on why the marker is so modest: perhaps the rather staid Goodricke relatives were ashamed of having a \u201cdeaf-mute\u201d in the family and could not sufficiently appreciate how much a man who could not hear could nevertheless see. RR Lyrae Stars A related group of stars, whose nature was understood somewhat later than that of the cepheids, are called RR Lyrae variables, named for the star RR Lyrae, the best-known member of the group. More common than the cepheids, but less luminous, thousands of these pulsating variables are known in our Galaxy. The periods of RR Lyrae stars are always less than 1 day, and their changes in brightness are typically less than about a factor of two. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 675 Astronomers have observed that the RR Lyrae stars occurring in any particular cluster all have about the same apparent brightness. Since stars in a cluster are all at approximately the same distance, it follows that RR Lyrae variables must all have nearly the same intrinsic luminosity, which turns out to be about 50 LSun. In this sense, RR Lyrae stars are a little bit like standard light bulbs and can also be used to obtain distances, particularly within our Galaxy. Figure 19.14 displays the ranges of periods and luminosities for both the cepheids and the RR Lyrae stars. Figure 19.14. Period-Luminosity Relation for Cepheid Variables. In this class of variable stars, the time the star takes to go through a cycle of luminosity changes is related to the average luminosity of the star. Also shown are the period and luminosity for RR Lyrae stars. 19.4 THE H\u2013R DIAGRAM AND COSMIC DISTANCES Learning Objectives By the end of this section, you will be able to: Understand how spectral types are used to estimate stellar luminosities Examine how these techniques are used by astronomers today Variable stars are not the only way that we can estimate the luminosity of stars. Another way involves the H\u2013R diagram, which shows that the intrinsic brightness of a star can be estimated if we know its spectral type. Distances from Spectral Types As satisfying and productive as variable stars have been for distance measurement, these stars are rare and are not found near all the objects to which we wish to measure distances. Suppose, for example, we need the distance to a star that is not varying, or to a group of stars, none of which is a variable. In this case, it turns out the H\u2013R diagram can come to our rescue. If we can observe the spectrum of a star, we can estimate its distance from our understanding of the H\u2013R diagram. As discussed in Analyzing Starlight, a detailed examination of a stellar spectrum allows astronomers to classify the star into one of the spectral types indicating surface temperature. (The types are O, B, A, F, G, K, M, L, T, and Y; each of these can be divided into numbered subgroups.)", " In general, however, the spectral type alone is not enough to allow us to estimate luminosity. Look again at Figure 18.15. A G2 star could be a main- sequence star with a luminosity of 1 LSun, or it could be a giant with a luminosity of 100 LSun, or even a supergiant with a still higher luminosity. 676 Chapter 19 Celestial Distances We can learn more from a star\u2019s spectrum, however, than just its temperature. Remember, for example, that we can detect pressure differences in stars from the details of the spectrum. This knowledge is very useful because giant stars are larger (and have lower pressures) than main-sequence stars, and supergiants are still larger than giants. If we look in detail at the spectrum of a star, we can determine whether it is a main-sequence star, a giant, or a supergiant. Suppose, to start with the simplest example, that the spectrum, color, and other properties of a distant G2 star match those of the Sun exactly. It is then reasonable to conclude that this distant star is likely to be a main- sequence star just like the Sun and to have the same luminosity as the Sun. But if there are subtle differences between the solar spectrum and the spectrum of the distant star, then the distant star may be a giant or even a supergiant. The most widely used system of star classification divides stars of a given spectral class into six categories called luminosity classes. These luminosity classes are denoted by Roman numbers as follows: \u2022 \u2022 \u2022 \u2022 \u2022 Ia: Brightest supergiants Ib: Less luminous supergiants II: Bright giants III: Giants IV: Subgiants (intermediate between giants and main-sequence stars) \u2022 V: Main-sequence stars The full spectral specification of a star includes its luminosity class. For example, a main-sequence star with spectral class F3 is written as F3 V. The specification for an M2 giant is M2 III. Figure 19.15 illustrates the approximate position of stars of various luminosity classes on the H\u2013R diagram. The dashed portions of the lines represent regions with very few or no stars. Figure 19.15. Luminosity Classes. Stars of the same temperature (or spectral class) can fall into different luminosity classes on the Hertzsprung-Russell diagram. By studying details of the spectrum for each star, astronomers can determine which luminosity class they fall in (whether they are main-sequence stars, giant stars, or supergiant stars). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 677 With both its spectral and luminosity classes known, a star\u2019s position on the H\u2013R diagram is uniquely determined. Since the diagram plots luminosity versus temperature, this means we can now read off the star\u2019s luminosity (once its spectrum has helped us place it on the diagram). As before, if we know how luminous the star really is and see how dim it looks, the difference allows us to calculate its distance. (For historical reasons, astronomers sometimes call this method of distance determination spectroscopic parallax, even though the method has nothing to do with parallax.) The H\u2013R diagram method allows astronomers to estimate distances to nearby stars, as well as some of the most distant stars in our Galaxy, but it is anchored by measurements of parallax. The distances measured using parallax are the gold standard for distances: they rely on no assumptions, only geometry. Once astronomers take a spectrum of a nearby star for which we also know the parallax, we know the luminosity that corresponds to that spectral type. Nearby stars thus serve as benchmarks for more distant stars because we can assume that two stars with identical spectra have the same intrinsic luminosity. A Few Words about the Real World Introductory textbooks such as ours work hard to present the material in a straightforward and simplified way. In doing so, we sometimes do our students a disservice by making scientific techniques seem too clean and painless. In the real world, the techniques we have just described turn out to be messy and difficult, and often give astronomers headaches that last long into the day. For example, the relationships we have described such as the period-luminosity relation for certain variable stars aren\u2019t exactly straight lines on a graph. The points representing many stars scatter widely when plotted, and thus, the distances derived from them also have a certain built-in scatter or uncertainty. The distances we measure with the methods we have discussed are therefore only accurate to within a certain percentage of error\u2014sometimes 10%, sometimes 25%, sometimes as much as 50% or more. A 25% error for a star estimated to be 10,000 light-years away means it could be anywhere from 7500 to 12,500 light-years away. This would", " be an unacceptable uncertainty if you were loading fuel into a spaceship for a trip to the star, but it is not a bad first figure to work with if you are an astronomer stuck on planet Earth. Nor is the construction of H\u2013R diagrams as easy as you might think at first. To make a good diagram, one needs to measure the characteristics and distances of many stars, which can be a time-consuming task. Since our own solar neighborhood is already well mapped, the stars astronomers most want to study to advance our knowledge are likely to be far away and faint. It may take hours of observing to obtain a single spectrum. Observers may have to spend many nights at the telescope (and many days back home working with their data) before they get their distance measurement. Fortunately, this is changing because surveys like Gaia will study billions of stars, producing public datasets that all astronomers can use. Despite these difficulties, the tools we have been discussing allow us to measure a remarkable range of distances\u2014parallaxes for the nearest stars, RR Lyrae variable stars; the H\u2013R diagram for clusters of stars in our own and nearby galaxies; and cepheids out to distances of 60 million light-years. Table 19.1 describes the distance limits and overlap of each method. Each technique described in this chapter builds on at least one other method, forming what many call the cosmic distance ladder. Parallaxes are the foundation of all stellar distance estimates, spectroscopic methods use nearby stars to calibrate their H\u2013R diagrams, and RR Lyrae and cepheid distance estimates are grounded in H\u2013R diagram distance estimates (and even in a parallax measurement to a nearby cepheid, Delta Cephei). This chain of methods allows astronomers to push the limits when looking for even more distant stars. Recent work, for example, has used RR Lyrae stars to identify dim companion galaxies to our own Milky Way out at 678 Chapter 19 Celestial Distances distances of 300,000 light-years. The H\u2013R diagram method was recently used to identify the two most distant stars in the Galaxy: red giant stars way out in the halo of the Milky Way with distances of almost 1 million light- years. We can combine the distances we find for stars with measurements of their composition, luminosity, and temperature\u2014made with the techniques described in Analyzing Starlight and The Stars: A Celestial Census. Together, these make up the arsenal of information we need to trace the evolution of stars from birth to death, the subject to which we turn in the chapters that follow. Distance Range of Celestial Measurement Methods Method Distance Range Trigonometric parallax 4\u201330,000 light-years when the Gaia mission is complete RR Lyrae stars Out to 300,000 light-years H\u2013R diagram and spectroscopic distances Out to 1,200,000 light-years Cepheid stars Out to 60,000,000 light-years Table 19.1 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 679 CHAPTER 19 REVIEW KEY TERMS cepheid a star that belongs to a class of yellow supergiant pulsating stars; these stars vary periodically in brightness, and the relationship between their periods and luminosities is useful in deriving distances to them light curve a graph that displays the time variation of the light from a variable or eclipsing binary star or, more generally, from any other object whose radiation output changes with time luminosity class a classification of a star according to its luminosity within a given spectral class; our Sun, a G2V star, has luminosity class V, for example parallax an apparent displacement of a nearby star that results from the motion of Earth around the Sun parsec a unit of distance in astronomy, equal to 3.26 light-years; at a distance of 1 parsec, a star has a parallax of 1 arcsecond period-luminosity relation an empirical relation between the periods and luminosities of certain variable stars pulsating variable star a variable star that pulsates in size and luminosity RR Lyrae one of a class of giant pulsating stars with periods shorter than 1 day, useful for finding distances SUMMARY 19.1 Fundamental Units of Distance Early measurements of length were based on human dimensions, but today, we use worldwide standards that specify lengths in units such as the meter. Distances within the solar system are now determined by timing how long it takes radar signals to travel from Earth to the surface of a planet or other body and then return. 19.2 Surveying the Stars For stars that are relatively nearby, we can \u201ctriangulate\u201d the distances from a baseline created by Earth\u2019s annual motion around the Sun. Half the shift in a nearby star\u2019s position relative to very distant background stars, as viewed from opposite sides of Earth\u2019s orbit,", " is called the parallax of that star and is a measure of its distance. The units used to measure stellar distance are the light-year, the distance light travels in 1 year, and the parsec (pc), the distance of a star with a parallax of 1 arcsecond (1 parsec = 3.26 light-years). The closest star, a red dwarf, is over 1 parsec away. The first successful measurements of stellar parallaxes were reported in 1838. Parallax measurements are a fundamental link in the chain of cosmic distances. The Hipparcos satellite has allowed us to measure accurate parallaxes for stars out to about 300 light-years, and the Gaia mission will result in parallaxes out to 30,000 light-years. 19.3 Variable Stars: One Key to Cosmic Distances Cepheids and RR Lyrae stars are two types of pulsating variable stars. Light curves of these stars show that their luminosities vary with a regularly repeating period. RR Lyrae stars can be used as standard bulbs, and cepheid variables obey a period-luminosity relation, so measuring their periods can tell us their luminosities. Then, we can calculate their distances by comparing their luminosities with their apparent brightnesses, and this can allow us to measure distances to these stars out to over 60 million light-years. 680 Chapter 19 Celestial Distances 19.4 The H\u2013R Diagram and Cosmic Distances Stars with identical temperatures but different pressures (and diameters) have somewhat different spectra. Spectral classification can therefore be used to estimate the luminosity class of a star as well as its temperature. As a result, a spectrum can allow us to pinpoint where the star is located on an H\u2013R diagram and establish its luminosity. This, with the star\u2019s apparent brightness, again yields its distance. The various distance methods can be used to check one against another and thus make a kind of distance ladder which allows us to find even larger distances. FOR FURTHER EXPLORATION Articles Adams, A. \u201cThe Triumph of Hipparcos.\u201d Astronomy (December 1997): 60. Brief introduction. Dambeck, T. \u201cGaia\u2019s Mission to the Milky Way.\u201d Sky & Telescope (March 2008): 36\u201339. An introduction to the mission to measure distances and positions of stars with unprecedented accuracy. Hirshfeld, A. \u201cThe Absolute Magnitude of Stars.\u201d Sky & Telescope (September 1994): 35. Good review of how we measure luminosity, with charts. Hirshfeld, A. \u201cThe Race to Measure the Cosmos.\u201d Sky & Telescope (November 2001): 38. On parallax. Trefil, J. Puzzling Out Parallax.\u201d Astronomy (September 1998): 46. On the concept and history of parallax. Turon, C. \u201cMeasuring the Universe.\u201d Sky & Telescope (July 1997): 28. On the Hipparcos mission and its results. Zimmerman, R. \u201cPolaris: The Code-Blue Star.\u201d Astronomy (March 1995): 45. On the famous cepheid variable and how it is changing. Websites ABCs of Distance: http://www.astro.ucla.edu/~wright/distance.htm. Astronomer Ned Wright (UCLA) gives a concise primer on many different methods of obtaining distances. This site is at a higher level than our textbook, but is an excellent review for those with some background in astronomy. American Association of Variable Star Observers (AAVSO): https://www.aavso.org/. This organization of amateur astronomers helps to keep track of variable stars; its site has some background material, observing instructions, and links. Friedrich Wilhelm Bessel: http://messier.seds.org/xtra/Bios/bessel.html. A brief site about the first person to detect stellar parallax, with references and links. Gaia: http://sci.esa.int/gaia/. News from the Gaia mission, including images and a blog of the latest findings. Hipparchos: http://sci.esa.int/hipparcos/. Background, results, catalogs of data, and educational resources from the Hipparchos mission to observe parallaxes from space. Some sections are technical, but others are accessible to students. John Goodricke: The Deaf Astronomer: http://www.bbc.com/news/magazine-20725639. A biographical article from the BBC. Women in Astronomy: http://www.astrosociety.org/education/astronomy-resource-guides/women-in- astronomy-an-introductory-resource-guide/. More about Henrietta Leavitt\u2019s and other women\u2019s contributions to astronomy", " and the obstacles they faced. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances Videos 681 Gaia\u2019s Mission: Solving the Celestial Puzzle: https://www.youtube.com/watch?v=oGri4YNggoc. Describes the Gaia mission and what scientists hope to learn, from Cambridge University (19:58). Hipparcos: Route Map to the Stars: https://www.youtube.com/watch?v=4d8a75fs7KI. This ESA video describes the mission to measure parallax and its results (14:32) How Big Is the Universe: https://www.youtube.com/watch?v=K_xZuopg4Sk. Astronomer Pete Edwards from the British Institute of Physics discusses the size of the universe and gives a step-by-step introduction to the concepts of distances (6:22) Search for Miss Leavitt: http://perimeterinstitute.ca/videos/search-miss-leavitt., Video of talk by George Johnson on his search for Miss Leavitt (55:09). Women in Astronomy: http://www.youtube.com/watch?v=5vMR7su4fi8. Emily Rice (CUNY) gives a talk on the contributions of women to astronomy, with many historical and contemporary examples, and an analysis of modern trends (52:54). COLLABORATIVE GROUP ACTIVITIES A. In this chapter, we explain the various measurements that have been used to establish the size of a standard meter. Your group should discuss why we have changed the definitions of our standard unit of measurement in science from time to time. What factors in our modern society contribute to the growth of technology? Does technology \u201cdrive\u201d science, or does science \u201cdrive\u201d technology? Or do you think the two are so intertwined that it\u2019s impossible to say which is the driver? B. Cepheids are scattered throughout our own Milky Way Galaxy, but the period-luminosity relation was discovered from observations of the Magellanic Clouds, a satellite galaxy now known to be about 160,000 light-years away. What reasons can you give to explain why the relation was not discovered from observations of cepheids in our own Galaxy? Would your answer change if there were a small cluster in our own Galaxy that contained 20 cepheids? Why or why not? C. You want to write a proposal to use the Hubble Space Telescope to look for the brightest cepheids in galaxy M100 and estimate their luminosities. What observations would you need to make? Make a list of all the reasons such observations are harder than it first might appear. D. Why does your group think so many different ways of naming stars developed through history? (Think back to the days before everyone connected online.) Are there other fields where things are named confusingly and arbitrarily? How do stars differ from other phenomena that science and other professions tend to catalog? E. Although cepheids and RR Lyrae variable stars tend to change their brightness pretty regularly (while they are in that stage of their lives), some variable stars are unpredictable or change their their behavior even during the course of a single human lifetime. Amateur astronomers all over the world follow such variable stars patiently and persistently, sending their nightly observations to huge databases that are being kept on the behavior of many thousands of stars. None of the hobbyists who do this get paid for making such painstaking observations. Have your group discuss why they do it. Would you ever consider a hobby that involves so much work, long into the night, often on work nights? If observing variable stars doesn\u2019t pique 682 Chapter 19 Celestial Distances your interest, is there something you think you could do as a volunteer after college that does excite you? Why? F. In Figure 19.8, the highest concentration of stars occurs in the middle of the main sequence. Can your group give reasons why this might be so? Why are there fewer very hot stars and fewer very cool stars on this diagram? G. In this chapter, we discuss two astronomers who were differently abled than their colleagues. John Goodricke could neither hear nor speak, and Henrietta Leavitt struggled with hearing impairment for all of her adult life. Yet they each made fundamental contributions to our understanding of the universe. Does your group know people who are handling a disability? What obstacles would people with different disabilities face in trying to do astronomy and what could be done to ease their way? For a set of resources in this area, see http://astronomerswithoutborders.org/gam2013/programs/1319-people-with-disabilities- astronomy-resources.html. EXERCISES Review Questions 1. Explain how parallax measurements can be used to determine distances to stars", ". Why can we not make accurate measurements of parallax beyond a certain distance? 2. Suppose you have discovered a new cepheid variable star. What steps would you take to determine its distance? 3. Explain how you would use the spectrum of a star to estimate its distance. 4. Which method would you use to obtain the distance to each of the following? A. An asteroid crossing Earth\u2019s orbit B. A star astronomers believe to be no more than 50 light-years from the Sun C. A tight group of stars in the Milky Way Galaxy that includes a significant number of variable stars D. A star that is not variable but for which you can obtain a clearly defined spectrum 5. What are the luminosity class and spectral type of a star with an effective temperature of 5000 K and a luminosity of 100 LSun? Thought Questions 6. The meter was redefined as a reference to Earth, then to krypton, and finally to the speed of light. Why do you think the reference point for a meter continued to change? 7. While a meter is the fundamental unit of length, most distances traveled by humans are measured in miles or kilometers. Why do you think this is? 8. Most distances in the Galaxy are measured in light-years instead of meters. Why do you think this is the case? 9. The AU is defined as the average distance between Earth and the Sun, not the distance between Earth and the Sun. Why does this need to be the case? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 683 10. What would be the advantage of making parallax measurements from Pluto rather than from Earth? Would there be a disadvantage? 11. Parallaxes are measured in fractions of an arcsecond. One arcsecond equals 1/60 arcmin; an arcminute is, in turn, 1/60th of a degree (\u00b0). To get some idea of how big 1\u00b0 is, go outside at night and find the Big Dipper. The two pointer stars at the ends of the bowl are 5.5\u00b0 apart. The two stars across the top of the bowl are 10\u00b0 apart. (Ten degrees is also about the width of your fist when held at arm\u2019s length and projected against the sky.) Mizar, the second star from the end of the Big Dipper\u2019s handle, appears double. The fainter star, Alcor, is about 12 arcmin from Mizar. For comparison, the diameter of the full moon is about 30 arcmin. The belt of Orion is about 3\u00b0 long. Keeping all this in mind, why did it take until 1838 to make parallax measurements for even the nearest stars? 12. For centuries, astronomers wondered whether comets were true celestial objects, like the planets and stars, or a phenomenon that occurred in the atmosphere of Earth. Describe an experiment to determine which of these two possibilities is correct. 13. The Sun is much closer to Earth than are the nearest stars, yet it is not possible to measure accurately the diurnal parallax of the Sun relative to the stars by measuring its position relative to background objects in the sky directly. Explain why. 14. Parallaxes of stars are sometimes measured relative to the positions of galaxies or distant objects called quasars. Why is this a good technique? 15. Estimating the luminosity class of an M star is much more important than measuring it for an O star if you are determining the distance to that star. Why is that the case? 16. Figure 19.9 is the light curve for the prototype cepheid variable Delta Cephei. How does the luminosity of this star compare with that of the Sun? 17. Which of the following can you determine about a star without knowing its distance, and which can you not determine: radial velocity, temperature, apparent brightness, or luminosity? Explain. 18. A G2 star has a luminosity 100 times that of the Sun. What kind of star is it? How does its radius compare with that of the Sun? 19. A star has a temperature of 10,000 K and a luminosity of 10\u20132 LSun. What kind of star is it? 20. What is the advantage of measuring a parallax distance to a star as compared to our other distance measuring methods? 21. What is the disadvantage of the parallax method, especially for studying distant parts of the Galaxy? 22. Luhman 16 and WISE 0720 are brown dwarfs, also known as failed stars, and are some of the new closest neighbors to Earth, but were only discovered in the last decade. Why do you think they took so long to be discovered? 23. Most stars close to the Sun are red dwarfs. What does this tell us about the average star formation event in our Galaxy? 24. Why would it be easier to measure the", " characteristics of intrinsically less luminous cepheids than more luminous ones? 25. When Henrietta Leavitt discovered the period-luminosity relationship, she used cepheid stars that were all located in the Large Magellanic Cloud. Why did she need to use stars in another galaxy and not cepheids located in the Milky Way? 684 Chapter 19 Celestial Distances Figuring For Yourself 26. A radar astronomer who is new at the job claims she beamed radio waves to Jupiter and received an echo exactly 48 min later. Should you believe her? Why or why not? 27. The New Horizons probe flew past Pluto in July 2015. At the time, Pluto was about 32 AU from Earth. How long did it take for communication from the probe to reach Earth, given that the speed of light in km/hr is 1.08 \u00d7 109? 28. Estimate the maximum and minimum time it takes a radar signal to make the round trip between Earth and Venus, which has a semimajor axis of 0.72 AU. 29. The Apollo program (not the lunar missions with astronauts) being conducted at the Apache Point Observatory uses a 3.5-m telescope to direct lasers at retro-reflectors left on the Moon by the Apollo astronauts. If the Moon is 384,472 km away, approximately how long do the operators need to wait to see the laser light return to Earth? 30. In 1974, the Arecibo Radio telescope in Puerto Rico was used to transmit a signal to M13, a star cluster about 25,000 light-years away. How long will it take the message to reach M13, and how far has the message travelled so far (in light-years)? 31. Demonstrate that 1 pc equals 3.09 \u00d7 1013 km and that it also equals 3.26 light-years. Show your calculations. 32. The best parallaxes obtained with Hipparcos have an accuracy of 0.001 arcsec. If you want to measure the distance to a star with an accuracy of 10%, its parallax must be 10 times larger than the typical error. How far away can you obtain a distance that is accurate to 10% with Hipparcos data? The disk of our Galaxy is 100,000 light-years in diameter. What fraction of the diameter of the Galaxy\u2019s disk is the distance for which we can measure accurate parallaxes? 33. Astronomers are always making comparisons between measurements in astronomy and something that might be more familiar. For example, the Hipparcos web pages tell us that the measurement accuracy of 0.001 arcsec is equivalent to the angle made by a golf ball viewed from across the Atlantic Ocean, or to the angle made by the height of a person on the Moon as viewed from Earth, or to the length of growth of a human hair in 10 sec as seen from 10 meters away. Use the ideas in Example 19.2 to verify one of the first two comparisons. 34. Gaia will have greatly improved precision over the measurements of Hipparcos. The average uncertainty for most Gaia parallaxes will be about 50 microarcsec, or 0.00005 arcsec. How many times better than Hipparcos (see Exercise 19.32) is this precision? 35. Using the same techniques as used in Exercise 19.32, how far away can Gaia be used to measure distances with an uncertainty of 10%? What fraction of the Galactic disk does this correspond to? 36. The human eye is capable of an angular resolution of about one arcminute, and the average distance between eyes is approximately 2 in. If you blinked and saw something move about one arcmin across, how far away from you is it? (Hint: You can use the setup in Example 19.2 as a guide.) 37. How much better is the resolution of the Gaia spacecraft compared to the human eye (which can resolve about 1 arcmin)? 38. The most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years. How many parsecs is this? 39. What would the parallax of Luhman 16 (see Exercise 19.38) be as measured from Earth? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 19 Celestial Distances 685 40. The New Horizons probe that passed by Pluto during July 2015 is one of the fastest spacecraft ever assembled. It was moving at about 14 km/s when it went by Pluto. If it maintained this speed, how long would it take New Horizons to reach the nearest star, Proxima Centauri, which is about 4.3 light-years away? (Note: It isn\u2019t headed in that direction, but you can pretend that it is.) 41. What physical properties are different for an M giant with a luminosity of 1000 LSun and an", " M dwarf with a luminosity of 0.5 LSun? What physical properties are the same? 686 Chapter 19 Celestial Distances This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 687 20 BETWEEN THE STARS: GAS AND DUST IN SPACE Figure 20.1. NGC 3603 and Its Parent Cloud. This image, taken by the Hubble Space Telescope, shows the young star cluster NGC 3603 interacting with the cloud of gas from which it recently formed. The bright blue stars of the cluster have blown a bubble in the gas cloud. The remains of this cloud can be seen in the lower right part of the frame, glowing in response to the starlight illuminating it. In its darker parts, shielded from the harsh light of NGC 3603, new stars continue to form. Although the stars of NGC 3603 formed only recently, the most massive of them are already dying and ejecting their mass, producing the blue ring and streak features visible in the upper left part of the image. Thus, this image shows the full life cycle of stars, from formation out of interstellar gas, through life on the main sequence, to death and the return of stellar matter to interstellar space. (credit: modification of work by NASA, Wolfgang Brandner (JPL/IPAC), Eva K. Grebel (University of Washington), You-Hua Chu (University of Illinois Urbana-Champaign)) Chapter Outline 20.1 The Interstellar Medium 20.2 Interstellar Gas 20.3 Cosmic Dust 20.4 Cosmic Rays 20.5 The Life Cycle of Cosmic Material 20.6 Interstellar Matter around the Sun Thinking Ahead Where do stars come from? We already know from earlier chapters that stars must die because ultimately they exhaust their nuclear fuel. We might hypothesize that new stars come into existence to replace the ones that die. In order to form new stars, however, we need the raw material to make them. It also turns out that stars eject mass throughout their lives (a kind of wind blows from their surface layers) and that material must go somewhere. What does this \u201craw material\u201d of stars look like? How would you detect it, especially if it is not yet in the form of stars and cannot generate its own energy? One of the most exciting discoveries of twentieth-century astronomy was that our Galaxy contains vast quantities of this \u201craw material\u201d\u2014atoms or molecules of gas and tiny solid dust particles found between the stars. Studying this diffuse matter between the stars helps us understand how new stars form and gives us important clues about our own origins billions of years ago. 688 Chapter 20 Between the Stars: Gas and Dust in Space 20.1 THE INTERSTELLAR MEDIUM Learning Objectives By the end of this section, you will be able to: Explain how much interstellar matter there is in the Milky Way, and what its typical density is Describe how the interstellar medium is divided into gaseous and solid components Astronomers refer to all the material between stars as interstellar matter; the entire collection of interstellar matter is called the interstellar medium (ISM). Some interstellar material is concentrated into giant clouds, each of which is known as a nebula (plural \u201cnebulae,\u201d Latin for \u201cclouds\u201d). The best-known nebulae are the ones that we can see glowing or reflecting visible light; there are many pictures of these in this chapter. Interstellar clouds do not last for the lifetime of the universe. Instead, they are like clouds on Earth, constantly shifting, merging with each other, growing, or dispersing. Some become dense and massive enough to collapse under their own gravity, forming new stars. When stars die, they, in turn, eject some of their material into interstellar space. This material can then form new clouds and begin the cycle over again. About 99% of the material between the stars is in the form of a gas\u2014that is, it consists of individual atoms or molecules. The most abundant elements in this gas are hydrogen and helium (which we saw are also the most abundant elements in the stars), but the gas also includes other elements. Some of the gas is in the form of molecules\u2014combinations of atoms. The remaining 1% of the interstellar material is solid\u2014frozen particles consisting of many atoms and molecules that are called interstellar grains or interstellar dust (Figure 20.2). A typical dust grain consists of a core of rocklike material (silicates) or graphite surrounded by a mantle of ices; water, methane, and ammonia are probably the most abundant ices. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 689 Figure 20.2. Various Types of Interstellar Matter. The reddish nebulae in this spectacular photograph glow with", " light emitted by hydrogen atoms. The darkest areas are clouds of dust that block the light from stars behind them. The upper part of the picture is filled with the bluish glow of light reflected from hot stars embedded in the outskirts of a huge, cool cloud of dust and gas. The cool supergiant star Antares can be seen as a big, reddish patch in the lower-left part of the picture. The star is shedding some of its outer atmosphere and is surrounded by a cloud of its own making that reflects the red light of the star. The red nebula in the middle right partially surrounds the star Sigma Scorpii. (To the right of Antares, you can see M4, a much more distant cluster of extremely old stars.) (credit: modification of work by ESO/Digitized Sky Survey 2) If all the interstellar gas within the Galaxy were spread out smoothly, there would be only about one atom of gas per cm3 in interstellar space. (In contrast, the air in the room where you are reading this book has roughly 1019 atoms per cm3.) The dust grains are even scarcer. A km3 of space would contain only a few hundred to a few thousand tiny grains, each typically less than one ten-thousandth of a millimeter in diameter. These numbers are just averages, however, because the gas and dust are distributed in a patchy and irregular way, much as water vapor in Earth\u2019s atmosphere is often concentrated into clouds. In some interstellar clouds, the density of gas and dust may exceed the average by as much as a thousand times or more, but even this density is more nearly a vacuum than any we can make on Earth. To show what we mean, let\u2019s imagine a vertical tube of air reaching from the ground to the top of Earth\u2019s atmosphere with a cross-section of 1 square meter. Now let us extend the same-size tube from the top of the atmosphere all the way to the edge of the observable universe\u2014over 10 billion light-years away. Long though it is, the second tube would still contain fewer atoms than the one in our planet\u2019s atmosphere. While the density of interstellar matter is very low, the volume of space in which such matter is found is huge, and so its total mass is substantial. To see why, we must bear in mind that stars occupy only a tiny fraction of the volume of the Milky Way Galaxy. For example, it takes light only about four seconds to travel a distance equal to the diameter of the Sun, but more than four years to travel from the Sun to the nearest star. Even though the spaces among the stars are sparsely populated, there\u2019s a lot of space out there! Astronomers estimate that the total mass of gas and dust in the Milky Way Galaxy is equal to about 15% of the mass contained in stars. This means that the mass of the interstellar matter in our Galaxy amounts to about 10 690 Chapter 20 Between the Stars: Gas and Dust in Space billion times the mass of the Sun. There is plenty of raw material in the Galaxy to make generations of new stars and planets (and perhaps even astronomy students). Estimating Interstellar Mass You can make a rough estimate of how much interstellar mass our Galaxy contains and also how many new stars could be made from this interstellar matter. All you need to know is how big the Galaxy is and the average density using this formula: total mass = volume \u00d7 density of atoms \u00d7 mass per atom You have to remember to use consistent units\u2014such as meters and kilograms. We will assume that our Galaxy is shaped like a cylinder; the volume of a cylinder equals the area of its base times its height where R is the radius of the cylinder and h is its height. V = \u03c0R2 h Suppose that the average density of hydrogen gas in our Galaxy is one atom per cm3. Each hydrogen atom has a mass of 1.7 \u00d7 10\u221227 kg. If the Galaxy is a cylinder with a diameter of 100,000 light-years and a height of 300 light-years, what is the mass of this gas? How many solar-mass stars (2.0 \u00d7 1030 kg) could be produced from this mass of gas if it were all turned into stars? Solution Recall that 1 light-year = 9.5 \u00d7 1012 km = 9.5 \u00d7 1017 cm, so the volume of the Galaxy is V = \u03c0R2 h = \u03c0(100,000 \u00d7 9.5 \u00d7 1017 cm)2 (300 \u00d7 9.5 \u00d7 1017 cm) = 8.0 \u00d7 1066 cm3 The total mass is therefore M = V \u00d7 density of atoms \u00d7 mass per atom 8.0 \u00d7 1066 cm3 \u00d7 (1 atom/cm3) \u00d7 1.7 \u00d7 10\u201327 kg = 1.4 \u00d7 1040 kg This is sufficient to make N = M (2.0 \u00d7 1030 kg) = 6.9 \u00d7 109 stars equal in mass", " to the Sun. That\u2019s roughly 7 billion stars. Check Your Learning You can use the same method to estimate the mass of interstellar gas around the Sun. The distance from the Sun to the nearest other star, Proxima Centauri, is 4.2 light-years. We will see in Interstellar Matter around the Sun that the gas in the immediate vicinity of the Sun is less dense than average, about 0.1 atoms per cm3. What is the total mass of interstellar hydrogen in a sphere centered on the Sun and extending out to Proxima Centauri? How does this compare to the mass of the Sun? It is helpful to remember that the volume of a sphere is related to its radius: V = (4/3)\u03c0R3 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 691 Answer: The volume of a sphere stretching from the Sun to Proxima Centauri is: \u239e \u239b V = (4/3)\u03c0R3 = (4/3)\u03c0 \u239d4.2 \u00d7 9.5 \u00d7 1017 cm \u23a0 3 = 2.7 \u00d7 1056 cm3 Therefore, the mass of hydrogen in this sphere is: M = V \u00d7 \u239d0.1 atom/cm3\u239e \u239b \u23a0 \u00d7 1.7 \u00d7 10\u201327 kg = 4.5 \u00d7 1028 kg This is only (4.5 \u00d7 1028 kg)/(2.0 \u00d7 1030 kg) = 2.2% the mass of the Sun Naming the Nebulae As you look at the captions for some of the spectacular photographs in this chapter and The Birth of Stars and the Discovery of Planets outside the Solar System, you will notice the variety of names given to the nebulae. A few, which in small telescopes look like something recognizable, are sometimes named after the creatures or objects they resemble. Examples include the Crab, Tarantula, and Keyhole Nebulae. But most have only numbers that are entries in a catalog of astronomical objects. Perhaps the best-known catalog of nebulae (as well as star clusters and galaxies) was compiled by the French astronomer Charles Messier (1730\u20131817). Messier\u2019s passion was discovering comets, and his devotion to this cause earned him the nickname \u201cThe Comet Ferret\u201d from King Louis XV. When comets are first seen coming toward the Sun, they look like little fuzzy patches of light; in small telescopes, they are easy to confuse with nebulae or with groupings of many stars so far away that their light is all blended together. Time and again, Messier\u2019s heart leapt as he thought he had discovered one of his treasured comets, only to find that he had \u201cmerely\u201d observed a nebula or cluster. In frustration, Messier set out to catalog the position and appearance of over 100 objects that could be mistaken for comets. For him, this list was merely a tool in the far more important work of comet hunting. He would be very surprised if he returned today to discover that no one recalls his comets anymore, but his catalog of \u201cfuzzy things that are not comets\u201d is still widely used. When Figure 20.2 refers to M4, it denotes the fourth entry in Messier\u2019s list. A far more extensive listing was compiled under the title of the New General Catalog (NGC) of Nebulae and Star Clusters in 1888 by John Dreyer, working at the observatory in Armagh, Ireland. He based his compilation on the work of William Herschel and his son John, plus many other observers who followed them. With the addition of two further listings (called the Index Catalogs), Dreyer\u2019s compilation eventually included 13,000 objects. Astronomers today still use his NGC numbers when referring to most nebulae and star groups. 692 Chapter 20 Between the Stars: Gas and Dust in Space 20.2 INTERSTELLAR GAS Learning Objectives By the end of this section, you will be able to: Name the major types of interstellar gas Discuss how we can observe each type Describe the temperature and other major properties of each type Interstellar gas, depending on where it is located, can be as cold as a few degrees above absolute zero or as hot as a million degrees or more. We will begin our voyage through the interstellar medium by exploring the different conditions under which we find gas. Ionized Hydrogen (H II) Regions\u2014Gas Near Hot Stars Some of the most spectacular astronomical photographs show interstellar gas located near hot stars (Figure 20.3). The strongest line in the visible region of the hydrogen spectrum is the red line in the Balmer series[1] (as explained in the chapter on Radiation and Spectra); this emission", " line accounts for the characteristic red glow in images like Figure 20.3. Figure 20.3. Orion Nebula. The red glow that pervades the great Orion Nebula is produced by the first line in the Balmer series of hydrogen. Hydrogen emission indicates that there are hot young stars nearby that ionize these clouds of gas. When electrons then recombine with protons and move back down into lower energy orbits, emission lines are produced. The blue color seen at the edges of some of the clouds is produced by small particles of dust that scatter the light from the hot stars. Dust can also be seen silhouetted against the glowing gas. (credit: NASA,ESA, M. Robberto (Space Telescope Science Institute/ESA) and the Hubble Space Telescope Orion Treasury Project Team) Hot stars are able to heat nearby gas to temperatures close to 10,000 K. The ultraviolet radiation from the stars also ionizes the hydrogen (remember that during ionization, the electron is stripped completely away from the proton). Such a detached proton won\u2019t remain alone forever when attractive electrons are around; it will capture a free electron, becoming a neutral hydrogen once more. However, such a neutral atom can then 1 Scientists also call this red Balmer line the H-alpha line, with alpha meaning it is the first spectral line in the Balmer series. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 693 absorb ultraviolet radiation again and start the cycle over. At a typical moment, most of the atoms near a hot star are in the ionized state. Since hydrogen is the main constituent of interstellar gas, we often characterize a region of space according to whether its hydrogen is neutral or ionized. A cloud of ionized hydrogen is called an H II region. (Scientists who work with spectra use the Roman numeral I to indicate that an atom is neutral; successively higher Roman numerals are used for each higher stage of ionization. H II thus refers to hydrogen that has lost its one electron; Fe III is iron with two electrons missing.) The electrons that are captured by the hydrogen nuclei cascade down through the various energy levels of the hydrogen atoms on their way to the lowest level, or ground state. During each transition downward, they give up energy in the form of light. The process of converting ultraviolet radiation into visible light is called fluorescence. Interstellar gas contains other elements besides hydrogen. Many of them are also ionized in the vicinity of hot stars; they then capture electrons and emit light, just as hydrogen does, allowing them to be observed by astronomers. But generally, the red hydrogen line is the strongest, and that is why H II regions look red. A fluorescent light on Earth works using the same principles as a fluorescent H II region. When you turn on the current, electrons collide with atoms of mercury vapor in the tube. The mercury is excited to a high-energy state because of these collisions. When the electrons in the mercury atoms return to lower-energy levels, some of the energy they emit is in the form of ultraviolet photons. These, in turn, strike a phosphor-coated screen on the inner wall of the light tube. The atoms in the screen absorb the ultraviolet photons and emit visible light as they cascade downward among the energy levels. (The difference is that these atoms give off a wider range of light colors, which mix to give the characteristic white glow of fluorescent lights, whereas the hydrogen atoms in an H II region give off a more limited set of colors.) Neutral Hydrogen Clouds The very hot stars required to produce H II regions are rare, and only a small fraction of interstellar matter is close enough to such hot stars to be ionized by them. Most of the volume of the interstellar medium is filled with neutral (nonionized) hydrogen. How do we go about looking for it? Unfortunately, neutral hydrogen atoms at temperatures typical of the gas in interstellar space neither emit nor absorb light in the visible part of the spectrum. Nor, for the most part, do the other trace elements that are mixed with the interstellar hydrogen. However, some of these other elements can absorb visible light even at typical interstellar temperatures. This means that when we observe a bright source such as a hot star or a galaxy, we can sometimes see additional lines in its spectrum produced when interstellar gas absorbs light at particular frequencies (see Figure 20.4). Some of the strongest interstellar absorption lines are produced by calcium and sodium, but many other elements can be detected as well in sufficiently sensitive observations (as discussed in Radiation and Spectra). 694 Chapter 20 Between the Stars: Gas and Dust in Space Figure 20.4. Absorption Lines though an Interstellar Dust Cloud. When there is a significant amount of cool interstellar matter between us and a star, we can see the absorption lines of the gas in the star\u2019s spectrum. We can distinguish the two kinds of lines because, whereas the", " star\u2019s lines are broad, the lines from the gas are narrower. The first evidence for absorption by interstellar clouds came from the analysis of a spectroscopic binary star (see The Stars: A Celestial Census), published in 1904. While most of the lines in the spectrum of this binary shifted alternately from longer to shorter wavelengths and back again, as we would expect from the Doppler effect for stars in orbit around each other, a few lines in the spectrum remained fixed in wavelength. Since both stars are moving in a binary system, lines that showed no motion puzzled astronomers. The lines were also peculiar in that they were much, much narrower than the rest of the lines, indicating that the gas producing them was at a very low pressure. Subsequent work demonstrated that these lines were not formed in the star\u2019s atmosphere at all, but rather in a cold cloud of gas located between Earth and the binary star. While these and similar observations proved there was interstellar gas, they could not yet detect hydrogen, the most common element, due to its lack of spectral features in the visible part of the spectrum. (The Balmer line of hydrogen is in the visible range, but only excited hydrogen atoms produce it. In the cold interstellar medium, the hydrogen atoms are all in the ground state and no electrons are in the higher-energy levels required to produce either emission or absorption lines in the Balmer series.) Direct detection of hydrogen had to await the development of telescopes capable of seeing very-low-energy changes in hydrogen atoms in other parts of the spectrum. The first such observations were made using radio telescopes, and radio emission and absorption by interstellar hydrogen remains one of our main tools for studying the vast amounts of cold hydrogen in the universe to this day. In 1944, while he was still a student, the Dutch astronomer Hendrik van de Hulst predicted that hydrogen would produce a strong line at a wavelength of 21 centimeters. That\u2019s quite a long wavelength, implying that the wave has such a low frequency and low energy that it cannot come from electrons jumping between energy levels (as we discussed in Radiation and Spectra). Instead, energy is emitted when the electron does a flip, something like an acrobat in a circus flipping upright after standing on his head. The flip works like this: a hydrogen atom consists of a proton and an electron bound together. Both the proton and the electron act is if they were spinning like tops, and spin axes of the two tops can either be pointed in the same direction (aligned) or in opposite directions (anti-aligned). If the proton and electron were spinning in opposite directions, the atom as a whole would have a very slightly lower energy than if the two spins were aligned (Figure 20.5). If an atom in the lower-energy state (spins opposed) acquired a small amount of energy, then the spins of the proton and electron could be aligned, leaving the atom in a slightly excited state. If the atom then lost that same amount of energy again, it would return to its ground state. The amount of energy involved corresponds to a wave with a wavelength of 21 centimeters; hence, it is known as the 21-centimeter line. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 695 Figure 20.5. Formation of the 21-Centimeter Line. When the electron in a hydrogen atom is in the orbit closest to the nucleus, the proton and the electron may be spinning either (a) in the same direction or (b) in opposite directions. When the electron flips over, the atom gains or loses a tiny bit of energy by either absorbing or emitting electromagnetic energy with a wavelength of 21 centimeters. Neutral hydrogen atoms can acquire small amounts of energy through collisions with other hydrogen atoms or with free electrons. Such collisions are extremely rare in the sparse gases of interstellar space. An individual atom may wait centuries before such an encounter aligns the spins of its proton and electron. Nevertheless, over many millions of years, a significant fraction of the hydrogen atoms are excited by a collision. (Out there in cold space, that\u2019s about as much excitement as an atom typically experiences.) An excited atom can later lose its excess energy either by colliding with another particle or by giving off a radio wave with a wavelength of 21 centimeters. If there are no collisions, an excited hydrogen atom will wait an average of about 10 million years before emitting a photon and returning to its state of lowest energy. Even though the probability that any single atom will emit a photon is low, there are so many hydrogen atoms in a typical gas cloud that collectively they will produce an observable line at 21 centimeters. Equipment sensitive enough to detect the 21-cm line of neutral hydrogen became available in 1951. Dutch astronomers had built an instrument to detect the 21-cm waves that they had predicted, but a fire destroyed it. As a result, two Harvard physicists, Harold Ewen", " and Edward Purcell, made the first detection (Figure 20.6), soon followed by confirmations from the Dutch and a group in Australia. Since the detection of the 21-cm line, many other radio lines produced by both atoms and molecules have been discovered (as we will discuss in a moment), and these have allowed astronomers to map out the neutral gas throughout our home Galaxy. Astronomers have also detected neutral interstellar gas, including hydrogen, at many other wavelengths from the infrared to the ultraviolet. 696 Chapter 20 Between the Stars: Gas and Dust in Space Figure 20.6. Harold Ewen (1922\u20132015) and Edward Purcell (1912\u20131997). We see Harold Ewen in 1952 working with the horn antenna (atop the physics laboratory at Harvard) that made the first detection of interstellar 21-cm radiation. The inset shows Edward Purcell, the winner of the 1952 Nobel Prize in physics, a few years later. (credit: modification of work by NRAO) Modern radio observations show that most of the neutral hydrogen in our Galaxy is confined to an extremely flat layer, less than 300 light-years thick, that extends throughout the disk of the Milky Way Galaxy. This gas has densities ranging from about 0.1 to about 100 atoms per cm3, and it exists at a wide range of temperatures, from as low as about 100 K (\u2013173 \u00b0C) to as high as about 8000 K. These regions of warm and cold gas are interspersed with each other, and the density and temperature at any particular point in space are constantly changing. Ultra-Hot Interstellar Gas While the temperatures of 10,000 K found in H II regions might seem warm, they are not the hottest phase of the interstellar medium. Some of the interstellar gas is at a temperature of a million degrees, even though there is no visible source of heat nearby. The discovery of this ultra-hot interstellar gas was a big surprise. Before the launch of astronomical observatories into space, which could see radiation in the ultraviolet and X-ray parts of the spectrum, astronomers assumed that most of the region between stars was filled with hydrogen at temperatures no warmer than those found in H II regions. But telescopes launched above Earth\u2019s atmosphere obtained ultraviolet spectra that contained interstellar lines produced by oxygen atoms that have been ionized five times. To strip five electrons from their orbits around an oxygen nucleus requires a lot of energy. Subsequent observations with orbiting X-ray telescopes revealed that the Galaxy is filled with numerous bubbles of X-ray-emitting gas. To emit X-rays, and to contain oxygen atoms that have been ionized five times, gas must be heated to temperatures of a million degrees or more. Theorists have now shown that the source of energy producing these remarkable temperatures is the explosion of massive stars at the ends of their lives (Figure 20.7). Such explosions, called supernovae, will be discussed in detail in the chapter on The Death of Stars. For now, we\u2019ll just say that some stars, nearing the ends of their lives, become unstable and literally explode. These explosions launch gas into interstellar space at velocities of tens of thousands of kilometers per second (up to about 30% the speed of light). When this ejected gas collides with interstellar gas, it produces shocks that heat the gas to millions or tens of millions of degrees. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 697 Figure 20.7. Vela Supernova Remnant. About 11,000 years ago, a dying star in the constellation of Vela exploded, becoming as bright as the full moon in Earth\u2019s skies. You can see the faint rounded filaments from that explosion in the center of this colorful image. The edges of the remnant are colliding with the interstellar medium, heating the gas they plow through to temperatures of millions of K. Telescopes in space also reveal a glowing sphere of X-ray radiation from the remnant. (credit: Digitized Sky Survey, ESA/ESO/NASA FITS Liberator, Davide De Martin) Astronomers estimate that one supernova explodes roughly every 100 years somewhere in the Galaxy. On average, shocks launched by supernovae sweep through any given point in the Galaxy about once every few million years. These shocks keep some interstellar space filled with gas at temperatures of millions of degrees, and they continually disturb the colder gas, keeping it in constant, turbulent motion. Molecular Clouds A few simple molecules out in space, such as CN and CH, were discovered decades ago because they produce absorption lines in the visible-light spectra of stars behind them. When more sophisticated equipment for obtaining spectra in radio and infrared wavelengths became available, astronomers\u2014to their surprise\u2014found much more complex molecules in interstellar clouds as well. Just as atoms leave their \u201cfingerprints\u201d in the spectrum of visible light, so the vibration", " and rotation of atoms within molecules can leave spectral fingerprints in radio and infrared waves. If we spread out the radiation at such longer wavelengths, we can detect emission or absorption lines in the spectra that are characteristic of specific molecules. Over the years, experiments in our laboratories have shown us the exact wavelengths associated with changes in the rotation and vibration of many common molecules, giving us a template of possible lines against which we can now compare our observations of interstellar matter. The discovery of complex molecules in space came as a surprise because most of interstellar space is filled with ultraviolet light from stars, and this light is capable of dissociating molecules (breaking them apart into individual atoms). In retrospect, however, the presence of molecules is not surprising. As we will discuss further in the next section, and have already seen above, interstellar space also contains significant amounts of dust capable of blocking out starlight. When this dust accumulates in a single location, the result is a dark cloud where ultraviolet starlight is blocked and molecules can survive. The largest of these structures are created where gravity pulls interstellar gas together to form giant molecular clouds, structures as massive as a million times the mass of the Sun. Within these, most of the interstellar hydrogen has formed the molecule H2 (molecular hydrogen). Other, more complex molecules are also present in much smaller quantities. 698 Chapter 20 Between the Stars: Gas and Dust in Space Giant molecular clouds have densities of hundreds to thousands of atoms per cm3, much denser than interstellar space is on average. As a result, though they account for a very small fraction of the volume of interstellar space, they contain a significant fraction\u201420\u201330%\u2014of the total mass of the Milky Way\u2019s gas. Because of their high density, molecular clouds block ultraviolet starlight, the main agent for heating most interstellar gas. As a result, they tend to be extremely cold, with typical temperatures near 10 K (\u2212263 \u00b0C). Giant molecular clouds are also the sites where new stars form, as we will discuss below. It is in these dark regions of space, protected from starlight, that molecules can form. Chemical reactions occurring both in the gas and on the surface of dust grains lead to much more complex compounds, hundreds of which have been identified in interstellar space. Among the simplest of these are water (H2O), carbon monoxide (CO), which is produced by fires on Earth, and ammonia (NH3), whose smell you recognize in strong home cleaning products. Carbon monoxide is particularly abundant in interstellar space and is the primary tool that astronomers use to study giant molecular clouds. Unfortunately, the most abundant molecule, H2, is particularly difficult to observe directly because in most giant molecular clouds, it is too cold to emit even at radio wavelengths. CO, which tends to be present wherever H2 is found, is a much better emitter and is often used by astronomers to trace molecular hydrogen. The more complex molecules astronomers have found are mostly combinations of hydrogen, oxygen, carbon, nitrogen, and sulfur atoms. Many of these molecules are organic (those that contain carbon and are associated with the carbon chemistry of life on Earth.) They include formaldehyde (used to preserve living tissues), alcohol (see the feature box on Cocktails in Space), and antifreeze. In 1996, astronomers discovered acetic acid (the prime ingredient of vinegar) in a cloud lying in the direction of the constellation of Sagittarius. To balance the sour with the sweet, a simple sugar (glycolaldehyde) has also been found. The largest compounds yet discovered in interstellar space are fullerenes, molecules in which 60 or 70 carbon atoms are arranged in a cage-like configuration (see Figure 20.8). See Table 20.1 for a list of a few of the more interesting interstellar molecules that have been found so far. Figure 20.8. Fullerene C60. This three-dimensional perspective shows the characteristic cage-like arrangement of the 60 carbon atoms in a molecule of fullerene C60. Fullerene C60 is also known as a \u201cbuckyball,\u201d or as its full name, buckminsterfullerene, because of its similarity to the multisided architectural domes designed by American inventor R. Buckminster Fuller. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 699 Some Interesting Interstellar Molecules Name Chemical Formula Use on Earth Ammonia Formaldehyde Acetylene NH3 H2CO HC2H Household cleansers Embalming fluid Fuel for a welding torch Acetic acid C2H2O4 The essence of vinegar Ethyl alcohol CH3CH2OH End-of-semester parties Ethylene glycol HOCH2CH2OH Antifreeze ingredient Benzene C6H6 Carbon ring, ingredient in varnishes and dyes Table 20.1 The cold interstellar clouds also contain cyanoacet", "ylene (HC3N) and acetaldehyde (CH3CHO), generally regarded as starting points for amino acid formation. These are building blocks of proteins, which are among the fundamental chemicals from which living organisms on Earth are constructed. The presence of these organic molecules does not imply that life exists in space, but it does show that the chemical building blocks of life can form under a wide range of conditions in the universe. As we learn more about how complex molecules are produced in interstellar clouds, we gain an increased understanding of the kinds of processes that preceded the beginnings of life on Earth billions of years ago Interested in learning more about fullerenes, buckyballs, or buckminsterfullerenes (as they\u2019re called)? Watch a brief video from NASA\u2019s Jet Propulsion Laboratory (https://openstax.org/l/30NASAjetprop) that explains what they are and illustrates how they were discovered in space Cocktails in Space Among the molecules astronomers have identified in interstellar clouds is alcohol, which comes in two varieties: methyl (or wood) alcohol and ethyl alcohol (the kind you find in cocktails). Ethyl alcohol is a pretty complex molecule, written by chemists as C2H5OH. It is quite plentiful in space (relatively speaking). In clouds where it has been identified, we detect up to one molecule for every m3. The largest 700 Chapter 20 Between the Stars: Gas and Dust in Space of the clouds (which can be several hundred light-years across) have enough ethyl alcohol to make 1028 fifths of liquor. We need not fear, however, that future interstellar astronauts will become interstellar alcoholics. Even if a spaceship were equipped with a giant funnel 1 kilometer across and could scoop it through such a cloud at the speed of light, it would take about a thousand years to gather up enough alcohol for one standard martini. Furthermore, the very same clouds also contain water (H2O) molecules. Your scoop would gather them up as well, and there are a lot more of them because they are simpler and thus easier to form. For the fun of it, one astronomical paper actually calculated the proof of a typical cloud. Proof is the ratio of alcohol to water in a drink, where 0 proof means all water, 100 proof means half alcohol and half water, and 200 proof means all alcohol. The proof of the interstellar cloud was only 0.2, not enough to qualify as a stiff drink 20.3 COSMIC DUST Learning Objectives By the end of this section, you will be able to: Describe how we can detect interstellar dust Understand the role and importance of infrared observations in studying dust Explain the terms extinction and interstellar reddening Figure 20.9 shows a striking example of what is actually a common sight through large telescopes: a dark region on the sky that appears to be nearly empty of stars. For a long time, astronomers debated whether these dark regions were empty \u201ctunnels\u201d through which we looked beyond the stars of the Milky Way Galaxy into intergalactic space, or clouds of some dark material that blocked the light of the stars beyond. The astronomer William Herschel (discoverer of the planet Uranus) thought it was the former, once remarking after seeing one, \u201cHere truly is a hole in heaven!\u201d However, American astronomer E. E. Barnard is generally credited with showing from his extensive series of nebula photographs that the latter interpretation is the correct one (see the feature box on Edward Emerson Barnard). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 701 Figure 20.9. Barnard 68. This object, first catalogued by E. E. Barnard, is a dark interstellar cloud. Its striking appearance is due to the fact that, since it is relatively close to Earth, there are no bright stars between us and it, and its dust obscures the light from the stars behind it. (It looks a little bit like a sideways heart; one astronomers sent a photo of this object to his sweetheart as a valentine.) (credit: modification of work by ESO) Dusty clouds in space betray their presence in several ways: by blocking the light from distant stars, by emitting energy in the infrared part of the spectrum, by reflecting the light from nearby stars, and by making distant stars look redder than they really are Edward Emerson Barnard Born in 1857 in Nashville, Tennessee, two months after his father died, Edward Barnard (Figure 20.10) grew up in such poor circumstances that he had to drop out of school at age nine to help support his ailing mother. He soon became an assistant to a local photographer, where he learned to love both photography and astronomy, destined to become the dual passions of his life. He worked as a photographer\u2019s aide for 17 years, studying astronomy on his", " own. In 1883, he obtained a job as an assistant at the Vanderbilt University Observatory, which enabled him at last to take some astronomy courses. Married in 1881, Barnard built a house for his family that he could ill afford. But as it happened, a patent medicine manufacturer offered a $200 prize (a lot of money in those days) for the discovery of any new comet. With the determination that became characteristic of him, Barnard spent every clear night searching for comets. He discovered seven of them between 1881 and 1887, earning enough money to make the payments on his home; this \u201cComet House\u201d later became a local attraction. (By the end of his life, Barnard had found 17 comets through diligent observation.) In 1887, Barnard got a position at the newly founded Lick Observatory, where he soon locked horns with the director, Edward Holden, a blustering administrator who made Barnard\u2019s life miserable. (To be fair, Barnard soon tried to do the same for him.) Despite being denied the telescope time that he needed for his photographic work, in 1892, Barnard managed to discover the first new moon found around Jupiter since Galileo\u2019s day, a stunning observational feat that earned him world renown. Now in a position to 702 Chapter 20 Between the Stars: Gas and Dust in Space demand more telescope time, he perfected his photographic techniques and soon began to publish the best images of the Milky Way taken up to that time. It was during the course of this work that he began to examine the dark regions among the crowded star lanes of the Galaxy and to realize that they must be vast clouds of obscuring material (rather than \u201choles\u201d in the distribution of stars). Astronomer-historian Donald Osterbrock has called Barnard an \u201cobservaholic:\u201d his daily mood seemed to depend entirely on how clear the sky promised to be for his night of observing. He was a driven, neurotic man, concerned about his lack of formal training, fearful of being scorned, and afraid that he might somehow slip back into the poverty of his younger days. He had difficulty taking vacations and lived for his work: only serious illness could deter him from making astronomical observations. In 1895, Barnard, having had enough of the political battles at Lick, accepted a job at the Yerkes Observatory near Chicago, where he remained until his death in 1923. He continued his photographic work, publishing compilations of his images that became classic photographic atlases, and investigating the varieties of nebulae revealed in his photographs. He also made measurements of the sizes and features of planets, participated in observations of solar eclipses, and carefully cataloged dark nebulae (see Figure 20.9). In 1916, he discovered the star with the largest proper motion, the second-closest star system to our own (see Analyzing Starlight). It is now called Barnard\u2019s Star in his honor. Figure 20.10. Edward Emerson Barnard (1857\u20131923). Barnard\u2019s observations provided information that furthered many astronomical explorations. (credit: The Lick Observatory) Detecting Dust The dark cloud seen in Figure 20.9 blocks the light of the many stars that lie behind it; note how the regions in other parts of the photograph are crowded with stars. Barnard 68 is an example of a relatively dense cloud or dark nebula containing tiny, solid dust grains. Such opaque clouds are conspicuous on any photograph of the Milky Way, the galaxy in which the Sun is located (see the figures in The Milky Way Galaxy). The \u201cdark rift,\u201d which runs lengthwise down a long part of the Milky Way in our sky and appears to split it in two, is produced by a collection of such obscuring clouds. While dust clouds are too cold to radiate a measurable amount of energy in the visible part of the spectrum, they glow brightly in the infrared (Figure 20.11). The reason is that small dust grains absorb visible light and This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 703 ultraviolet radiation very efficiently. The grains are heated by the absorbed radiation, typically to temperatures from 10 to about 500 K, and re-radiate this heat at infrared wavelengths. Figure 20.11. Visible and Infrared Images of the Horsehead Nebula in Orion. This dark cloud is one of the best-known images in astronomy, probably because it really does resemble a horse\u2019s head. The horse-head shape is an extension of a large cloud of dust that fills the lower part of the picture. (a) Seen in visible light, the dust clouds are especially easy to see against the bright background. (b) This infrared radiation image from the region of the horse head was recorded by NASA\u2019s Wide-", "Field Infrared Survey Explorer. Note how the regions that appear dark in visible light appear bright in the infrared. The dust is heated by nearby stars and re-radiates this heat in the infrared. Only the top of the horse\u2019s head is visible in the infrared image. Bright dots seen in the nebula below and to the left and at the top of the horse head are young, newly formed stars. The insets show the horse head and the bright nebula in more detail. (credit a: modification of work by ESO and Digitized Sky Survey; credit b: modification of work by NASA/JPL-Caltech) Thanks to their small sizes and low temperatures, interstellar grains radiate most of their energy at infrared to microwave frequencies, with wavelengths of tens to hundreds of microns. Earth\u2019s atmosphere is opaque to radiation at these wavelengths, so emission by interstellar dust is best measured from space. Observations from above Earth\u2019s atmosphere show that dust clouds are present throughout the plane of the Milky Way (Figure 20.12). Figure 20.12. Infrared Emission from the Plane of the Milky Way. This infrared image taken by the Spitzer Space Telescope shows a field in the plane of the Milky Way Galaxy. (Our Galaxy is in the shape of a frisbee; the plane of the Milky Way is the flat disk of that frisbee. Since the Sun, Earth, and solar system are located in the plane of the Milky Way and at a large distance from its center, we view the Galaxy edge on, much as we might look at a glass plate from its edge.) This emission is produced by tiny dust grains, which emit at 3.6 microns (blue in this image), 8.0 microns (green), and 24 microns (red). The densest regions of dust are so cold and opaque that they appear as dark clouds even at these infrared wavelengths. The red bubbles visible throughout indicate regions where the dust has been warmed up by young stars. This heating increases the emission at 24 microns, leading to the redder color in this image. (credit: modification of work by NASA/JPL-Caltech/University of Wisconsin) 704 Chapter 20 Between the Stars: Gas and Dust in Space Some dense clouds of dust are close to luminous stars and scatter enough starlight to become visible. Such a cloud of dust, illuminated by starlight, is called a reflection nebula, since the light we see is starlight reflected off the grains of dust. One of the best-known examples is the nebulosity around each of the brightest stars in the Pleiades cluster (see Figure 20.1). The dust grains are small, and such small particles turn out to scatter light with blue wavelengths more efficiently than light at red wavelengths. A reflection nebula, therefore, usually appears bluer than its illuminating star (Figure 20.13). Figure 20.13. Pleiades Star Cluster. The bluish light surrounding the stars in this image is an example of a reflection nebula. Like fog around a street lamp, a reflection nebula shines only because the dust within it scatters light from a nearby bright source. The Pleiades cluster is currently passing through an interstellar cloud that contains dust grains, which scatter the light from the hot blue stars in the cluster. The Pleiades cluster is about 400 light-years from the Sun. (credit: NASA, ESA and AURA/Caltech) Gas and dust are generally intermixed in space, although the proportions are not exactly the same everywhere. The presence of dust is apparent in many photographs of emission nebulae in the constellation of Sagittarius, where we see an H II region surrounded by a blue reflection nebula. Which type of nebula appears brighter depends on the kinds of stars that cause the gas and dust to glow. Stars cooler than about 25,000 K have so little ultraviolet radiation of wavelengths shorter than 91.2 nanometers\u2014which is the wavelength required to ionize hydrogen\u2014that the reflection nebulae around such stars outshine the emission nebulae. Stars hotter than 25,000 K emit enough ultraviolet energy that the emission nebulae produced around them generally outshine the reflection nebulae. Interstellar Reddening The tiny interstellar dust grains absorb some of the starlight they intercept. But at least half of the starlight that interacts with a grain is merely scattered, that is, it is redirected rather than absorbed. Since neither the absorbed nor the scattered starlight reaches us directly, both absorption and scattering make stars look dimmer. The effects of both processes are called interstellar extinction (Figure 20.14). Astronomers first came to understand interstellar extinction around the early 1930s, as the explanation of a puzzling observation. In the early part of the twentieth century, astronomers discovered that some stars look red even though their spectral lines indicate that they must be extremely hot (and thus should look blue). The solution to this seeming contradiction turned out to be that the light from these hot", " stars is not only dimmed but also reddened by interstellar dust, a phenomenon known as interstellar reddening. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 705 Figure 20.14. Barnard 68 in Infrared. In this image, we see Barnard 68, the same object shown in Figure 20.9. The difference is that, in the previous image, the blue, green, and red channels showed light in the visible (or very nearly visible) part of the spectrum. In this image, the red color shows radiation emitted in the infrared at a wavelength of 2.2 microns. Interstellar extinction is much smaller at infrared than at visible wavelengths, so the stars behind the cloud become visible in the infrared channel. (credit: ESO) Dust does not interact with all the colors of light the same way. Much of the violet, blue, and green light from these stars has been scattered or absorbed by dust, so it does not reach Earth. Some of their orange and red light, with longer wavelengths, on the other hand, more easily penetrates the intervening dust and completes its long journey through space to enter Earth-based telescopes (Figure 20.15). Thus, the star looks redder from Earth than it would if you could see it from nearby. (Strictly speaking, reddening is not the most accurate term for this process, since no red color is added; instead, blues and related colors are subtracted, so it should more properly be called \u201cdeblueing.\u201d) In the most extreme cases, stars can be so reddened that they are entirely undetectable at visible wavelengths and can be seen only at infrared or longer wavelengths (Figure 20.14). Figure 20.15. Scattering of Light by Dust. Interstellar dust scatters blue light more efficiently than red light, thereby making distant stars appear redder and giving clouds of dust near stars a bluish hue. Here, a red ray of light from a star comes straight through to the observer, whereas a blue ray is shown scattering. A similar scattering process makes Earth\u2019s sky look blue. 706 Chapter 20 Between the Stars: Gas and Dust in Space We have all seen an example of reddening on Earth. The Sun appears much redder at sunset than it does at noon. The lower the Sun is in the sky, the longer the path its light must travel through the atmosphere. Over this greater distance, there is a greater chance that sunlight will be scattered. Since red light is less likely to be scattered than blue light, the Sun appears more and more red as it approaches the horizon. By the way, scattering of sunlight is also what causes our sky to look blue, even though the gases that make up Earth\u2019s atmosphere are transparent. As sunlight comes in, it scatters from the molecules of air. The small size of the molecules means that the blue colors scatter much more efficiently than the greens, yellows, and reds. Thus, the blue in sunlight is scattered out of the beam and all over the sky. The light from the Sun that comes to your eye, on the other hand, is missing some of its blue, so the Sun looks a bit yellower, even when it is high in the sky, than it would from space. The fact that starlight is reddened by interstellar dust means that long-wavelength radiation is transmitted through the Galaxy more efficiently than short-wavelength radiation. Consequently, if we wish to see farther in a direction with considerable interstellar material, we should look at long wavelengths. This simple fact provides one of the motivations for the development of infrared astronomy. In the infrared region at 2 microns (2000 nanometers), for example, the obscuration is only one-sixth as great as in the visible region (500 nanometers), and we can therefore study stars that are more than twice as distant before their light is blocked by interstellar dust. This ability to see farther by observing in the infrared portion of the spectrum represents a major gain for astronomers trying to understand the structure of our Galaxy or probing its puzzling, but distant, center (see The Milky Way Galaxy). Interstellar Grains Before we get to the details about interstellar dust, we should perhaps get one concern out of the way. Why couldn\u2019t it be the interstellar gas that reddens distant stars and not the dust? We already know from everyday experience that atomic or molecular gas is almost transparent. Consider Earth\u2019s atmosphere. Despite its very high density compared with that of interstellar gas, it is so transparent as to be practically invisible. (Gas does have a few specific spectral lines, but they absorb only a tiny fraction of the light as it passes through.)The quantity of gas required to produce the observed absorption of light in interstellar space would have to be enormous. The gravitational attraction of so great a mass of gas would affect the motions of stars in ways that could", " easily be detected. Such motions are not observed, and thus, the interstellar absorption cannot be the result of gases. Although gas does not absorb much light, we know from everyday experience that tiny solid or liquid particles can be very efficient absorbers. Water vapor in the air is quite invisible. When some of that vapor condenses into tiny water droplets, however, the resulting cloud is opaque. Dust storms, smoke, and smog offer familiar examples of the efficiency with which solid particles absorb light. On the basis of arguments like these, astronomers have concluded that widely scattered solid particles in interstellar space must be responsible for the observed dimming of starlight. What are these particles made of? And how did they form? Observations like the pictures in this chapter show that a great deal of this dust exists; hence, it must be primarily composed of elements that are abundant in the universe (and in interstellar matter). After hydrogen and helium, the most abundant elements are oxygen, carbon, and nitrogen. These three elements, along with magnesium, silicon, iron\u2014and perhaps hydrogen itself\u2014turn out to be the most important components of interstellar dust. Many of the dust particles can be characterized as sootlike (rich in carbon) or sandlike (containing silicon and oxygen). Grains of interstellar dust are found in meteorites and can be identified because the abundances of certain isotopes are different from what we see in other solar system material. Several different interstellar This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 707 dust substances have been identified in this way in the laboratory, including graphite and diamonds. (Don\u2019t get excited; these diamonds are only a billionth of a meter in size and would hardly make an impressive engagement ring!) The most widely accepted model pictures the grains with rocky cores that are either like soot (rich in carbon) or like sand (rich in silicates). In the dark clouds where molecules can form, these cores are covered by icy mantles (Figure 20.16). The most common ices in the grains are water (H2O), methane (CH4), and ammonia (NH3)\u2014all built out of atoms that are especially abundant in the realm of the stars. The ice mantles, in turn, are sites for some of the chemical reactions that produce complex organic molecules. Figure 20.16. Model of an Interstellar Dust Grain. A typical interstellar grain is thought to consist of a core of rocky material (silicates) or graphite, surrounded by a mantle of ices. Typical grain sizes are 10\u20138 to 10\u20137 meters. (This is from 1/100 to 1/10 of a micron; by contrast, human hair is about 10\u2013200 microns wide.) Typical individual grains must be just slightly smaller than the wavelength of visible light. If the grains were a lot smaller, they would not block the light efficiently, as Figure 20.13 and other images in this chapter show that it does. On the other hand, if the dust grains were much larger than the wavelength of light, then starlight would not be reddened. Things that are much larger than the wavelength of light would block both blue and red light with equal efficiency. In this way we can deduce that a characteristic interstellar dust grain contains 106 to 109 atoms and has a diameter of 10\u20138 to 10\u20137 meters (10 to 100 nanometers). This is actually more like the specks of solid matter in cigarette smoke than the larger grains of dust you might find under your desk when you are too busy studying astronomy to clean properly. 20.4 COSMIC RAYS Learning Objectives By the end of this section, you will be able to: Define cosmic rays and describe their composition 708 Chapter 20 Between the Stars: Gas and Dust in Space Explain why it is hard to study the origin of cosmic rays, and the current leading hypotheses about where they might come from In addition to gas and dust, a third class of particles, noteworthy for the high speeds with which they travel, is found in interstellar space. Cosmic rays were discovered in 1911 by an Austrian physicist, Victor Hess, who flew simple instruments aboard balloons and showed that high-speed particles arrive at Earth from space (Figure 20.17). The term \u201ccosmic ray\u201d is misleading, implying it might be like a ray of light, but we are stuck with the name. They are definitely particles and have nearly the same composition as ordinary interstellar gas. Their behavior, however, is radically different from the gas we have discussed so far. Figure 20.17. Victor Hess (1883\u20131964). Cosmic-ray pioneer Victor Hess returns from a 1912 balloon flight that reached an altitude of 5.3 kilometers. It was on such balloon flights that Hess discovered cosmic rays. The Nature of Cosmic Rays Cosmic rays are mostly high-speed atomic nuclei and electrons. Speeds equal to 90%", " of the speed of light are typical. Almost 90% of the cosmic rays are hydrogen nuclei (protons) stripped of their accompanying electron. Helium and heavier nuclei constitute about 9% more. About 1% of cosmic rays have masses equal to the mass of the electron, and 10\u201320% of these carry positive charge rather than the negative charge that characterizes electrons. A positively charged particle with the mass of an electron is called a positron and is a form of antimatter (we discussed antimatter in The Sun: A Nuclear Powerhouse). The abundances of various atomic nuclei in cosmic rays mirror the abundances in stars and interstellar gas, with one important exception. The light elements lithium, beryllium, and boron are far more abundant in cosmic rays than in the Sun and stars. These light elements are formed when high-speed, cosmic-ray nuclei of carbon, nitrogen, and oxygen collide with protons in interstellar space and break apart. (By the way, if you, like most readers, have not memorized all the elements and want to see how any of those we mention fit into the sequence of elements, you will find them all listed in Appendix K in order of the number of protons they contain.) Cosmic rays reach Earth in substantial numbers, and we can determine their properties either by capturing them directly or by observing the reactions that occur when they collide with atoms in our atmosphere. The total energy deposited by cosmic rays in Earth\u2019s atmosphere is only about one-billionth the energy received from the Sun, but it is comparable to the energy received in the form of starlight. Some of the cosmic rays come to Earth from the surface of the Sun, but most come from outside the solar system. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 709 Where Do They Come From? There is a serious problem in identifying the source of cosmic rays. Since light travels in straight lines, we can tell where it comes from simply by looking. Cosmic rays are charged particles, and their direction of motion can be changed by magnetic fields. The paths of cosmic rays are curved both by magnetic fields in interstellar space and by Earth\u2019s own field. Calculations show that low-energy cosmic rays may spiral many times around Earth before entering the atmosphere where we can detect them. If an airplane circles an airport many times before landing, it is difficult for an observer to determine the direction from which it originated. So, too, after a cosmic ray circles Earth several times, it is impossible to know where its journey began. There are a few clues, however, about where cosmic rays might be generated. We know, for example, that magnetic fields in interstellar space are strong enough to keep all but the most energetic cosmic rays from escaping the Galaxy. It therefore seems likely that they are produced somewhere inside the Galaxy. The only likely exceptions are those with the very highest energy. Such cosmic rays move so rapidly that they are not significantly influenced by interstellar magnetic fields, and thus, they could escape our Galaxy. By analogy, they could escape other galaxies as well, so some of the highest-energy cosmic rays that we detect may have been created in some distant galaxy. Still, most cosmic rays must have their source inside the Milky Way Galaxy. We can also estimate how far typical cosmic rays travel before striking Earth. The light elements lithium, beryllium, and boron hold the key. Since these elements are formed when carbon, nitrogen, and oxygen strike interstellar protons, we can calculate how long, on average, cosmic rays must travel through space in order to experience enough collisions to account for the amount of lithium and the other light elements that they contain. It turns out that the required distance is about 30 times around the Galaxy. At speeds near the speed of light, it takes perhaps 3\u201310 million years for the average cosmic ray to travel this distance. This is only a small fraction of the age of the Galaxy or the universe, so cosmic rays must have been created fairly recently on a cosmic timescale. The best candidates for a source of cosmic rays are the supernova explosions, which mark the violent deaths of some stars (and which we will discuss in The Death of Stars). The material ejected by the explosion produces a shock wave, which travels through the interstellar medium. Charged particles can become trapped, bouncing back and forth across the front of the shock wave many times. With each pass through the shock, the magnetic fields inside it accelerate the particles more and more. Eventually, they are traveling at close to the speed of light and can escape from the shock to become cosmic rays. Some collapsed stars (including star remnants left over from supernova explosions) may, under the right circumstances, also serve as accelerators of particles. In any case, we again find that the raw material of the Galaxy is enriched by the life cycle of stars. In the next section, we", " will look at this enrichment process in more detail You can watch a brief video about the Calorimetric Electron Telescope (CALET) (https://openstax.org/l/30CALETvid) mission, a cosmic ray detector at the International Space Station. The link takes you to NASA Johnson\u2019s \u201cSpace Station Live: Cosmic Ray Detector for ISS.\u201d 710 Chapter 20 Between the Stars: Gas and Dust in Space 20.5 THE LIFE CYCLE OF COSMIC MATERIAL Learning Objectives By the end of this section, you will be able to: Explain how interstellar matter flows into and out of our Galaxy and transforms from one phase to another, and understand how star formation and evolution affects the properties of the interstellar medium Explain how the heavy elements and dust grains found in interstellar space got there and describe how dust grains help produce molecules that eventually find their way into planetary systems Flows of Interstellar Gas The most important thing to understand about the interstellar medium is that it is not static. Interstellar gas orbits through the Galaxy, and as it does so, it can become more or less dense, hotter and colder, and change its state of ionization. A particular parcel of gas may be neutral hydrogen at some point, then find itself near a young, hot star and become part of an H II region. The star may then explode as a supernova, heating the nearby gas up to temperatures of millions of degrees. Over millions of years, the gas may cool back down and become neutral again, before it collects into a dense region that gravity gathers into a giant molecular cloud (Figure 20.18) Figure 20.18. Large-Scale Distribution of Interstellar Matter. This image is from a computer simulation of the Milky Way Galaxy\u2019s interstellar medium as a whole. The majority of gas, visible in greenish colors, is neutral hydrogen. In the densest regions in the spiral arms, shown in yellow, the gas is collected into giant molecular clouds. Low-density holes in the spiral arms, shown in blue, are the result of supernova explosions. (credit: modification of work by Mark Krumholz) At any given time in the Milky Way, the majority of the interstellar gas by mass and volume is in the form of atomic hydrogen. The much-denser molecular clouds occupy a tiny fraction of the volume of interstellar space but add roughly 30% to the total mass of gas between the stars. Conversely, the hot gas produced by supernova explosions contributes a negligible mass but occupies a significant fraction of the volume of interstellar space. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 711 H II regions, though they are visually spectacular, constitute only a very small fraction of either the mass or volume of interstellar material. However, the interstellar medium is not a closed system. Gas from intergalactic space constantly falls onto the Milky Way due to its gravity, adding new gas to the interstellar medium. Conversely, in giant molecular clouds where gas collects together due to gravity, the gas can collapse to form new stars, as discussed in The Birth of Stars and the Discovery of Planets outside the Solar System. This process locks interstellar matter into stars. As the stars age, evolve, and eventually die, massive stars lose a large fraction of their mass, and low- mass stars lose very little. On average, roughly one-third of the matter incorporated into stars goes back into interstellar space. Supernova explosions have so much energy that they can drive interstellar mass out of the Galaxy and back into intergalactic space. Thus, the total amount of mass of the interstellar medium is set by a competition between the gain of mass from intergalactic space, the conversion of interstellar mass into stars, and the loss of interstellar mass back into intergalactic space due to supernovae. This entire process is known as the baryon cycle\u2014baryon is from the Latin word for \u201cheavy,\u201d and the cycle has this name because it is the repeating process that the heavier components of the universe\u2014the atoms\u2014undergo. The Cycle of Dust and Heavy Elements While much of the mass of the interstellar medium is material accreted during the last few billion years from intergalactic space, this is not true of the elements heavier than hydrogen and helium, or of the dust. Instead, these components of the interstellar medium were made inside stars in the Milky Way, which returned them to the interstellar medium at the end of their lives. We will talk more about this process in later chapters, but for now just bear in mind what we learned in The Sun: A Nuclear Powerhouse. What stars \u201cdo for a living\u201d is fuse heavier elements from lighter ones, producing energy in the process. As stars mature, they begin to lose some of the newly made elements to the reservoir of interstellar matter. The same is true of dust grains.", " Dust forms when grains can condense in regions where gas is dense and cool. One place where the right conditions are found is the winds from luminous cool stars (the red giants and supergiants we discussed in The Stars: A Celestial Census). Grains can also condense in the matter thrown off by a supernova explosion as the ejected gases begin to cool. The dust grains produced by stars may grow even further when they spend time in the dense parts of the interstellar medium, inside molecular clouds. In these environments, grains can stick together or gather additional atoms from the gas around them, growing larger. They also facilitate the production of other compounds, including some of the more complex molecules we discussed earlier. The surfaces of the dust grains (see Cosmic Dust)\u2014which would seem very large if you were an atom\u2014provide \u201cnooks and crannies\u201d where these atoms can stick long enough to find partners and form molecules. (Think of the dust grains as \u201cinterstellar social clubs\u201d where lonely atoms can meet and form meaningful relationships.) Eventually, the dust grains become coated with ices. The presence of the dust shields the molecules inside the clouds from ultraviolet radiation and cosmic rays that would break them up. When stars finally begin to form within the cloud, they heat the grains and evaporate the ices. The gravitational attraction of the newly forming stars also increases the density of the surrounding cloud material. Many more chemical reactions take place on the surfaces of grains in the gas surrounding the newly forming stars, and these areas are where organic molecules are formed. These molecules can be incorporated into newly formed planetary systems, and the early Earth may have been seeded in just such a way. Indeed, scientists speculate that some of the water on Earth may have come from interstellar grains. Recent observations from space have shown that water is abundant in dense interstellar clouds. Since stars are formed from this material, water must be present when solar systems, including our own, come into existence. The 712 Chapter 20 Between the Stars: Gas and Dust in Space water in our oceans and lakes may have come initially from water locked into the rocky material that accreted to form Earth. Alternatively, the water may have been brought to Earth when asteroids and comets (formed from the same cloud that made the planets) later impacted it. Scientists estimate that one comet impact every thousand years during Earth\u2019s first billion years would have been enough to account for the water we see today. Of course, both sources may have contributed to the water we now enjoy drinking and swimming in. Any interstellar grains that are incorporated into newly forming stars (instead of the colder planets and smaller bodies around them) will be destroyed by their high temperatures. But eventually, each new generation of stars will evolve to become red giants, with stellar winds of their own. Some of these stars will also become supernovae and explode. Thus, the process of recycling cosmic material can start all over again. 20.6 INTERSTELLAR MATTER AROUND THE SUN Learning Objectives By the end of this section, you will be able to: Describe how interstellar matter is arranged around our solar system Explain why scientists think that the Sun is located in a hot bubble We want to conclude our discussion of interstellar matter by asking how this material is organized in our immediate neighborhood. As we discussed above, orbiting X-ray observatories have shown that the Galaxy is full of bubbles of hot, X-ray-emitting gas. They also revealed a diffuse background of X-rays that appears to fill the entire sky from our perspective (Figure 20.19). While some of this emission comes from the interaction of the solar wind with the interstellar medium, a majority of it comes from beyond the solar system. The natural explanation for why there is X-ray-emitting gas all around us is that the Sun is itself inside one of the bubbles. We therefore call our \u201cneighborhood\u201d the Local Hot Bubble, or Local Bubble for short. The Local Bubble is much less dense\u2014an average of approximately 0.01 atoms per cm3\u2014than the average interstellar density of about 1 atom per cm3. This local gas has a temperature of about a million degrees, just like the gas in the other superbubbles that spread throughout our Galaxy, but because there is so little hot material, this high temperature does not affect the stars or planets in the area in any way. What caused the Local Bubble to form? Scientists are not entirely sure, but the leading candidate is winds from stars and supernova explosions. In a nearby region in the direction of the constellations Scorpius and Centaurus, a lot of star formation took place about 15 million years ago. The most massive of these stars evolved very quickly until they produced strong winds, and some ended their lives by exploding. These processes filled the region around the Sun with hot gas, driving away cooler, denser gas. The rim of this expanding superbubble reached the Sun about 7.6 million years ago and now lies more than 200 light-", "years past the Sun in the general direction of the constellations of Orion, Perseus, and Auriga. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 713 Figure 20.19. Sky in X-Rays. This image, made by the ROSAT satellite, shows the whole sky in X-rays as seen from Earth. Different colors indicate different X-ray energies: red is 0.25 kiloelectron volts, green is 0.75 kiloelectron volts, and blue is 1.5 kiloelectron volts. The image is oriented so the plane of the Galaxy runs across the middle of the image. The ubiquitous red color, which does not disappear completely even in the galactic plane, is evidence for a source of X-rays all around the Sun. (credit: modification of work by NASA) A few clouds of interstellar matter do exist within the Local Bubble. The Sun itself seems to have entered a cloud about 10,000 years ago. This cloud is warm (with a temperature of about 7000 K) and has a density of 0.3 hydrogen atom per cm3\u2014higher than most of the Local Bubble but still so tenuous that it is also referred to as Local Fluff (Figure 20.20). (Aren\u2019t these astronomical names fun sometimes?) While this is a pretty thin cloud, we estimate that it contributes 50 to 100 times more particles than the solar wind to the diffuse material between the planets in our solar system. These interstellar particles have been detected and their numbers counted by the spacecraft traveling between the planets. Perhaps someday, scientists will devise a way to collect them without destroying them and to return them to Earth, so that we can touch\u2014or at least study in our laboratories\u2014these messengers from distant stars. 714 Chapter 20 Between the Stars: Gas and Dust in Space Figure 20.20. Local Fluff. The Sun and planets are currently moving through the Local Interstellar Cloud, which is also called the Local Fluff. Fluff is an appropriate description because the density of this cloud is only about 0.3 atom per cm3. In comparison, Earth\u2019s atmosphere at the edge of space has around 1.2 \u00d7 1013 molecules per cm3. This image shows the patches of interstellar matter (mostly hydrogen gas) within about 20 light-years of the Sun. The temperature of the Local Interstellar Cloud is about 7,000 K. The arrows point toward the directions that different parts of the cloud are moving. The names associated with each arrow indicate the constellations located on the sky toward which the parts of the cloud are headed. The solar system is thought to have entered the Local Interstellar Cloud, which is a small cloud located within a much larger superbubble that is expanding outward from the Scorpius-Centaurus region of the sky, at some point between 44,000 and 150,000 years ago and is expected to remain within it for another 10,000 to 20,000 years. (credit: modification of work by NASA/Goddard/Adler/ University Chicago/Wesleyan) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 715 CHAPTER 20 REVIEW KEY TERMS baryon cycle the cycling of mass in and out of the interstellar medium, including accretion of gas from intergalactic space, loss of gas back into intergalactic space, and conversion of interstellar gas into stars cosmic rays atomic nuclei (mostly protons) and electrons that are observed to strike Earth\u2019s atmosphere with exceedingly high energies. H II region the region of ionized hydrogen in interstellar space interstellar dust tiny solid grains in interstellar space thought to consist of a core of rocklike material (silicates) or graphite surrounded by a mantle of ices; water, methane, and ammonia are probably the most abundant ices interstellar extinction the attenuation or absorption of light by dust in the interstellar medium interstellar medium (ISM) (or interstellar matter) the gas and dust between the stars in a galaxy Local Bubble (or Local Hot Bubble) a region of low-density, million degree gas in which the Sun and solar system are currently located Local Fluff a slightly denser cloud inside the Local Bubble, inside which the Sun also lies molecular cloud a large, dense, cold interstellar cloud; because of its size and density, this type of cloud can keep ultraviolet radiation from reaching its interior, where molecules are able to form nebula a cloud of interstellar gas or dust; the term is most often used for clouds that are seen to glow with visible light or infrared reddening (interstellar) the reddening of starlight passing through interstellar dust because dust scatters blue light more effectively than red SUMMARY 20.1 The Interstellar Medium About 15% of the visible matter in the Galaxy is", " in the form of gas and dust, serving as the raw material for new stars. About 99% of this interstellar matter is in the form of gas\u2014individual atoms or molecules. The most abundant elements in the interstellar gas are hydrogen and helium. About 1% of the interstellar matter is in the form of solid interstellar dust grains. 20.2 Interstellar Gas Interstellar gas may be hot or cold. Gas found near hot stars emits light by fluorescence, that is, light is emitted when an electron is captured by an ion and cascades down to lower-energy levels. Glowing clouds (nebulae) of ionized hydrogen are called H II regions and have temperatures of about 10,000 K. Most hydrogen in interstellar space is not ionized and can best be studied by radio measurements of the 21-centimeter line. Some of the gas in interstellar space is at a temperature of a million degrees, even though it is far away in hot stars; this ultra-hot gas is probably heated when rapidly moving gas ejected in supernova explosions sweeps through space. In some places, gravity gathers interstellar gas into giant clouds, within which the gas is protected from 716 Chapter 20 Between the Stars: Gas and Dust in Space starlight and can form molecules; more than 200 different molecules have been found in space, including the basic building blocks of proteins, which are fundamental to life as we know it here on Earth. 20.3 Cosmic Dust Interstellar dust can be detected: (1) when it blocks the light of stars behind it, (2) when it scatters the light from nearby stars, and (3) because it makes distant stars look both redder and fainter. These effects are called reddening and interstellar extinction, respectively. Dust can also be detected in the infrared because it emits heat radiation. Dust is found throughout the plane of the Milky Way. The dust particles are about the same size as the wavelength of light and consist of rocky cores that are either sootlike (carbon-rich) or sandlike (silicates) with mantles made of ices such as water, ammonia, and methane. 20.4 Cosmic Rays Cosmic rays are particles that travel through interstellar space at a typical speed of 90% of the speed of light. The most abundant elements in cosmic rays are the nuclei of hydrogen and helium, but electrons and positrons are also found. It is likely that many cosmic rays are produced in supernova shocks. 20.5 The Life Cycle of Cosmic Material Interstellar matter is constantly flowing through the Galaxy and changing from one phase to another. At the same time, gas is constantly being added to the Galaxy by accretion from extragalactic space, while mass is removed from the interstellar medium by being locked in stars. Some of the mass in stars is, in turn, returned to the interstellar medium when those stars evolve and die. In particular, the heavy elements in interstellar space were all produced inside stars, while the dust grains are made in the outer regions of stars that have swelled to be giants. These elements and grains, in turn, can then be incorporated into new stars and planetary systems that form out of the interstellar medium. 20.6 Interstellar Matter around the Sun The Sun is located at the edge of a low-density cloud called the Local Fluff. The Sun and this cloud are located within the Local Bubble, a region extending to at least 300 light-years from the Sun, within which the density of interstellar material is extremely low. Astronomers think this bubble was blown by some nearby stars that experienced a strong wind and some supernova explosions. FOR FURTHER EXPLORATION Articles Goodman, A. \u201cRecycling the Universe.\u201d Sky & Telescope November (2000): 44. Review of how stellar evolution, the interstellar medium, and supernovae all work together to recycle cosmic material. Greenberg, J. \u201cThe Secrets of Stardust.\u201d Scientific American December (2000): 70. The makeup and evolutionary role of solid particles between the stars. Knapp, G. \u201cThe Stuff between the Stars.\u201d Sky & Telescope May (1995): 20. An introduction to the interstellar medium. Nadis, S. \u201cSearching for the Molecules of Life in Space.\u201d Sky & Telescope January (2002): 32. Recent observations of water in the interstellar medium by satellite telescopes. Olinto, A. \u201cSolving the Mystery of Cosmic Rays.\u201d Astronomy April (2014): 30. What accelerates them to such high energies. Reynolds, R. \u201cThe Gas between the Stars.\u201d Scientific American January (2002): 34. On the interstellar medium. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 717 Websites And Apps Barnard, E. E., Biographical Memoir: http://www.nasonline.org/publications/biographical-memoirs/mem", "oir- pdfs/barnard-edward.pdf. Cosmicopia: http://helios.gsfc.nasa.gov/cosmic.html. NASA\u2019s learning site explains about the history and modern understanding of cosmic rays. DECO: https://wipac.wisc.edu/deco. A smart-phone app for turning your phone into a cosmic-ray detector. Hubble Space Telescope Images of Nebulae: http://hubblesite.org/gallery/album/nebula/. Click on any of the beautiful images in this collection, and you are taken to a page with more information; while looking at these images, you may also want to browse through the slide sequence on the meaning of colors in the Hubble pictures (http://hubblesite.org/gallery/behind_the_pictures/meaning_of_color/). Interstellar Medium Online Tutorial: http://www-ssg.sr.unh.edu/ism/intro.htm. Nontechnical introduction to the interstellar medium (ISM) and how we study it; by the University of New Hampshire astronomy department. Messier Catalog of Nebulae, Clusters, and Galaxies: http://astropixels.com/messier/messiercat.html. Astronomer Fred Espenak provides the full catalog, with information and images. (The Wikipedia list does something similar: https://en.wikipedia.org/wiki/List_of_Messier_objects.) Nebulae: What Are They?: http://www.universetoday.com/61103/what-is-a-nebula/. Concise introduction by Matt Williams. Videos Barnard 68: The Hole in the Sky: https://www.youtube.com/watch?v=8No6I0Uc3No. About this dark cloud and dark clouds in interstellar space in general (02:08). Horsehead Nebula in New Light: http://www.esa.int/spaceinvideos/Videos/2013/04/ The_Horsehead_Nebula_in_new_light. Tour of the dark nebula in different wavelengths; no audio narration, just music, but explanatory material appears on the screen (03:03). Hubblecast 65: A Whole New View of the Horsehead Nebula: http://www.spacetelescope.org/videos/ heic1307a/. Report on nebulae in general and about the Horsehead specifically, with ESO astronomer Joe Liske (06:03). Interstellar Reddening: https://www.youtube.com/watch?v=H2M80RAQB6k. Video demonstrating how reddening works, with Scott Miller of Penn State; a bit nerdy but useful (03:45). COLLABORATIVE GROUP ACTIVITIES A. The Sun is located in a region where the density of interstellar matter is low. Suppose that instead it were located in a dense cloud 20 light-years in diameter that dimmed the visible light from stars lying outside it by a factor of 100. Have your group discuss how this would have affected the development of civilization on Earth. For example, would it have presented a problem for early navigators? B. Your group members should look through the pictures in this chapter. How big are the nebulae you see in the images? Are there any clues either in the images or in the captions? Are the clouds they are part of significantly bigger than the nebulae we can see? Why? Suggest some ways that we can determine the sizes of nebulae. 718 Chapter 20 Between the Stars: Gas and Dust in Space C. How do the members of your group think astronomers are able to estimate the distances of such nebulae in our own Galaxy? (Hint: Look at the images. Can you see anything between us and the nebula in some cases. Review Celestial Distances, if you need to remind yourself about methods of measuring distances.) D. The text suggests that a tube of air extending from the surface of Earth to the top of the atmosphere contains more atoms than a tube of the same diameter extending from the top of the atmosphere to the edge of the observable universe. Scientists often do what they call \u201cback of the envelope calculations,\u201d in which they make very rough approximations just to see whether statements or ideas are true. Try doing such a \u201cquick and dirty\u201d estimate for this statement with your group. What are the steps in comparing the numbers of atoms contained in the two different tubes? What information do you need to make the approximations? Can you find it in this text? And is the statement true? E. If your astronomy course has involved learning about the solar system before you got to this chapter, have your group discuss where else besides interstellar clouds astronomers have been discovering organic molecules (the chemical building blocks of life). How might the discoveries of such molecules in our own solar", " system be related to the molecules in the clouds discussed in this chapter? F. Two stars both have a reddish appearance in telescopes. One star is actually red; the other\u2019s light has been reddened by interstellar dust on its way to us. Have your group make a list of the observations you could perform to determine which star is which. G. You have been asked to give a talk to your little brother\u2019s middle school class on astronomy, and you decide to talk about how nature recycles gas and dust. Have your group discuss what images from this book you would use in your talk. In what order? What is the one big idea you would like the students to remember when the class is over? H. This chapter and the next (on The Birth of Stars) include some of the most beautiful images of nebulae that glow with the light produced when starlight interacts with gas and dust. Have your group select one to four of your favorite such nebulae and prepare a report on them to share with the rest of the class. (Include such things as their location, distance, size, way they are glowing, and what is happening within them.) EXERCISES Review Questions 1. Identify several dark nebulae in photographs in this chapter. Give the figure numbers of the photographs, and specify where the dark nebulae are to be found on them. 2. Why do nebulae near hot stars look red? Why do dust clouds near stars usually look blue? 3. Describe the characteristics of the various kinds of interstellar gas (HII regions, neutral hydrogen clouds, ultra-hot gas clouds, and molecular clouds). 4. Prepare a table listing the different ways in which dust and gas can be detected in interstellar space. 5. Describe how the 21-cm line of hydrogen is formed. Why is this line such an important tool for understanding the interstellar medium? 6. Describe the properties of the dust grains found in the space between stars. 7. Why is it difficult to determine where cosmic rays come from? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 719 8. What causes reddening of starlight? Explain how the reddish color of the Sun\u2019s disk at sunset is caused by the same process. 9. Why do molecules, including H2 and more complex organic molecules, only form inside dark clouds? Why don\u2019t they fill all interstellar space? 10. Why can\u2019t we use visible light telescopes to study molecular clouds where stars and planets form? Why do infrared or radio telescopes work better? 11. The mass of the interstellar medium is determined by a balance between sources (which add mass) and sinks (which remove it). Make a table listing the major sources and sinks, and briefly explain each one. 12. Where does interstellar dust come from? How does it form? Thought Questions 13. Figure 20.2 shows a reddish glow around the star Antares, and yet the caption says that is a dust cloud. What observations would you make to determine whether the red glow is actually produced by dust or whether it is produced by an H II region? 14. If the red glow around Antares is indeed produced by reflection of the light from Antares by dust, what does its red appearance tell you about the likely temperature of Antares? Look up the spectral type of Antares in Appendix J. Was your estimate of the temperature about right? In most of the images in this chapter, a red glow is associated with ionized hydrogen. Would you expect to find an H II region around Antares? Explain your answer. 15. Even though neutral hydrogen is the most abundant element in interstellar matter, it was detected first with a radio telescope, not a visible light telescope. Explain why. (The explanation given in Analyzing Starlight for the fact that hydrogen lines are not strong in stars of all temperatures may be helpful.) 16. The terms H II and H2 are both pronounced \u201cH two.\u201d What is the difference in meaning of those two terms? Can there be such a thing as H III? 17. Suppose someone told you that she had discovered H II around the star Aldebaran. Would you believe her? Why or why not? 18. Describe the spectrum of each of the following: A. starlight reflected by dust, B. a star behind invisible interstellar gas, and C. an emission nebula. 19. According to the text, a star must be hotter than about 25,000 K to produce an H II region. Both the hottest white dwarfs and main-sequence O stars have temperatures hotter than 25,000 K. Which type of star can ionize more hydrogen? Why? 20. From the comments in the text about which kinds of stars produce emission nebulae and which kinds are associated with reflection nebulae, what can you say about the", " temperatures of the stars that produce NGC 1999 (Figure 20.13)? 21. One way to calculate the size and shape of the Galaxy is to estimate the distances to faint stars just from their observed apparent brightnesses and to note the distance at which stars are no longer observable. The first astronomers to try this experiment did not know that starlight is dimmed by interstellar dust. Their estimates of the size of the Galaxy were much too small. Explain why. 720 Chapter 20 Between the Stars: Gas and Dust in Space 22. New stars form in regions where the density of gas and dust is relatively high. Suppose you wanted to search for some recently formed stars. Would you more likely be successful if you observed at visible wavelengths or at infrared wavelengths? Why? 23. Thinking about the topics in this chapter, here is an Earth analogy. In big cities, you can see much farther on days without smog. Why? 24. Stars form in the Milky Way at a rate of about 1 solar mass per year. At this rate, how long would it take for all the interstellar gas in the Milky Way to be turned into stars if there were no fresh gas coming in from outside? How does this compare to the estimated age of the universe, 14 billion years? What do you conclude from this? 25. The 21-cm line can be used not just to find out where hydrogen is located in the sky, but also to determine how fast it is moving toward or away from us. Describe how this might work. 26. Astronomers recently detected light emitted by a supernova that was originally observed in 1572, just reaching Earth now. This light was reflected off a dust cloud; astronomers call such a reflected light a \u201clight echo\u201d (just like reflected sound is called an echo). How would you expect the spectrum of the light echo to compare to that of the original supernova? 27. We can detect 21-cm emission from other galaxies as well as from our own Galaxy. However, 21-cm emission from our own Galaxy fills most of the sky, so we usually see both at once. How can we distinguish the extragalactic 21-cm emission from that arising in our own Galaxy? (Hint: Other galaxies are generally moving relative to the Milky Way.) 28. We have said repeatedly that blue light undergoes more extinction than red light, which is true for visible and shorter wavelengths. Is the same true for X-rays? Look at Figure 20.19. The most dust is in the galactic plane in the middle of the image, and the red color in the image corresponds to the reddest (lowestenergy) light. Based on what you see in the galactic plane, are X-rays experiencing more extinction at redder or bluer colors? You might consider comparing Figure 20.19 to Figure 20.14. 29. Suppose that, instead of being inside the Local Bubble, the Sun were deep inside a giant molecular cloud. What would the night sky look like as seen from Earth at various wavelengths? 30. Suppose that, instead of being inside the Local Bubble, the Sun were inside an H II region. What would the night sky look like at various wavelengths? Figuring For Yourself 31. A molecular cloud is about 1000 times denser than the average of the interstellar medium. Let\u2019s compare this difference in densities to something more familiar. Air has a density of about 1 kg/m3, so something 1000 times denser than air would have a density of about 1000 kg/m3. How does this compare to the typical density of water? Of granite? (You can find figures for these densities on the internet.) Is the density difference between a molecular cloud and the interstellar medium larger or smaller than the density difference between air and water or granite? 32. Would you expect to be able to detect an H II region in X-ray emission? Why or why not? (Hint: You might apply Wien\u2019s law) 33. Suppose that you gathered a ball of interstellar gas that was equal to the size of Earth (a radius of about 6000 km). If this gas has a density of 1 hydrogen atom per cm3, typical of the interstellar medium, how would its mass compare to the mass of a bowling ball (5 or 6 kg)? How about if it had the typical density of the Local Bubble, about 0.01 atoms per cm3? The volume of a sphere is V = (4/3)\u03c0R3. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 20 Between the Stars: Gas and Dust in Space 721 34. At the average density of the interstellar medium, 1 atom per cm3, how big a volume of material must be used to make a star with the mass of the Sun? What is the radius of a sphere this size? Express your answer in light-years. 35. Consider a grain of sand that contains 1 mg of oxygen (", "a typical amount for a medium-sized sand grain, since sand is mostly SiO2). How many oxygen atoms does the grain contain? What is the radius of the sphere you would have to spread them out over if you wanted them to have the same density as the interstellar medium, about 1 atom per cm3? You can look up the mass of an oxygen atom. 36. H II regions can exist only if there is a nearby star hot enough to ionize hydrogen. Hydrogen is ionized only by radiation with wavelengths shorter than 91.2 nm. What is the temperature of a star that emits its maximum energy at 91.2 nm? (Use Wien\u2019s law from Radiation and Spectra.) Based on this result, what are the spectral types of those stars likely to provide enough energy to produce H II regions? 37. In the text, we said that the five-times ionized oxygen (OVI) seen in hot gas must have been produced by supernova shocks that heated the gas to millions of degrees, and not by starlight, the way H II is produced. Producing OVI by light requires wavelengths shorter than 10.9 nm. The hottest observed stars have surface temperatures of about 50,000 K. Could they produce OVI? 38. Dust was originally discovered because the stars in certain clusters seemed to be fainter than expected. Suppose a star is behind a cloud of dust that dims its brightness by a factor of 100. Suppose you do not realize the dust is there. How much in error will your distance estimate be? Can you think of any measurement you might make to detect the dust? 39. How would the density inside a cold cloud (T = 10 K) compare with the density of the ultra-hot interstellar gas (T = 106 K) if they were in pressure equilibrium? (It takes a large cloud to be able to shield its interior from heating so that it can be at such a low temperature.) (Hint: In pressure equilibrium, the two regions must have nT equal, where n is the number of particles per unit volume and T is the temperature.) Which region do you think is more suitable for the creation of new stars? Why? 40. The text says that the Local Fluff, which surrounds the Sun, has a temperature of 7500 K and a density 0.1 atom per cm3. The Local Fluff is embedded in hot gas with a temperature of 106 K and a density of about 0.01 atom per cm3. Are they in equilibrium? (Hint: In pressure equilibrium, the two regions must have nT equal, where n is the number of particles per unit volume and T is the temperature.) What is likely to happen to the Local Fluff? 722 Chapter 20 Between the Stars: Gas and Dust in Space This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 723 21 THE BIRTH OF STARS AND THE DISCOVERY OF PLANETS OUTSIDE THE SOLAR SYSTEM Figure 21.1. Where Stars Are Born. We see a close-up of part of the Carina Nebula taken with the Hubble Space Telescope. This image reveals jets powered by newly forming stars embedded in a great cloud of gas and dust. Parts of the clouds are glowing from the energy of very young stars recently formed within them. (credit: modification of work by NASA, ESA, and M. Livio and the Hubble 20th Anniversary Team (STScI)) Chapter Outline 21.1 Star Formation 21.2 The H\u2013R Diagram and the Study of Stellar Evolution 21.3 Evidence That Planets Form around Other Stars 21.4 Planets beyond the Solar System: Search and Discovery 21.5 Exoplanets Everywhere: What We Are Learning 21.6 New Perspectives on Planet Formation Thinking Ahead \u201cThere are countless suns and countless earths all rotating round their suns in exactly the same way as the planets of our system. We see only the suns because they are the largest bodies and are luminous, but their planets remain invisible to us because they are smaller and non-luminous.... The unnumbered worlds in the universe are all similar in form and rank and subject to the same forces and the same laws.\u201d\u2014Giordano Bruno in On the Infinite Universe and Worlds (1584). Bruno was tried for heresy by the Roman Inquisition and burned at the stake in 1600. We\u2019ve discussed stars as nuclear furnaces that convert light elements into heavier ones. A star\u2019s nuclear evolution begins when hydrogen is fused into helium, but that can only occur when the core temperature exceeds 10 to 12 million K. Since stars form from cold interstellar material, we must understand how they collapse and eventually reach this \u201cignition temperature\u201d to explain the birth of stars. Star formation is a continuous process, from", " the birth of our Galaxy right up to today. We estimate that every year in our Galaxy, on average, three solar masses of interstellar matter are converted into stars. This may sound like a small amount of mass for an object as large as a galaxy, but only three new stars (out of billions in the Galaxy) are formed each year. 724 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System Do planets orbit other stars or is ours the only planetary system? In the past few decades, new technology has enabled us to answer that question by revealing nearly 3500 exoplanets in over 2600 planetary systems. Even before planets were detected, astronomers had predicted that planetary systems were likely to be byproducts of the star-formation process. In this chapter, we look at how interstellar matter is transformed into stars and planets. 21.1 STAR FORMATION Learning Objectives By the end of this section, you will be able to: Identify the sometimes-violent processes by which parts of a molecular cloud collapse to produce stars Recognize some of the structures seen in images of molecular clouds like the one in Orion Explain how the environment of a molecular cloud enables the formation of stars Describe how advancing waves of star formation cause a molecular cloud to evolve As we begin our exploration of how stars are formed, let\u2019s review some basics about stars discussed in earlier chapters: \u2022 Stable (main-sequence) stars such as our Sun maintain equilibrium by producing energy through nuclear fusion in their cores. The ability to generate energy by fusion defines a star. \u2022 Each second in the Sun, approximately 600 million tons of hydrogen undergo fusion into helium, with about 4 million tons turning into energy in the process. This rate of hydrogen use means that eventually the Sun (and all other stars) will run out of central fuel. \u2022 Stars come with many different masses, ranging from 1/12 solar masses (MSun) to roughly 100\u2013200 MSun. There are far more low-mass than high-mass stars. \u2022 The most massive main-sequence stars (spectral type O) are also the most luminous and have the highest surface temperature. The lowest-mass stars on the main sequence (spectral type M or L) are the least luminous and the coolest. \u2022 A galaxy of stars such as the Milky Way contains enormous amounts of gas and dust\u2014enough to make billions of stars like the Sun. If we want to find stars still in the process of formation, we must look in places that have plenty of the raw material from which stars are assembled. Since stars are made of gas, we focus our attention (and our telescopes) on the dense and cold clouds of gas and dust that dot the Milky Way (see Figure 21.1 and Figure 21.2). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 725 Figure 21.2. Pillars of Dust and Dense Globules in M16. (a) This Hubble Space Telescope image of the central regions of M16 (also known as the Eagle Nebula) shows huge columns of cool gas, (including molecular hydrogen, H2) and dust. These columns are of higher density than the surrounding regions and have resisted evaporation by the ultraviolet radiation from a cluster of hot stars just beyond the upper-right corner of this image. The tallest pillar is about 1 light-year long, and the M16 region is about 7000 light-years away from us. (b) This close-up view of one of the pillars shows some very dense globules, many of which harbor embryonic stars. Astronomers coined the term evaporating gas globules (EGGs) for these structures, in part so that they could say we found EGGs inside the Eagle Nebula. It is possible that because these EGGs are exposed to the relentless action of the radiation from nearby hot stars, some may not yet have collected enough material to form a star. (credit a : modification of work by NASA, ESA, and the Hubble Heritage Team (STScI/AURA); credit b: modification of work by NASA, ESA, STScI, J. Hester and P. Scowen (Arizona State University)) Molecular Clouds: Stellar Nurseries As we saw in Between the Stars: Gas and Dust in Space, the most massive reservoirs of interstellar matter\u2014and some of the most massive objects in the Milky Way Galaxy\u2014are the giant molecular clouds. These clouds have cold interiors with characteristic temperatures of only 10\u201320 K; most of their gas atoms are bound into molecules. These clouds turn out to be the birthplaces of most stars in our Galaxy. The masses of molecular clouds range from a thousand times the mass of the Sun to about 3 million solar masses. Molecular clouds have a complex filamentary structure, similar to cirrus clouds in Earth\u2019s atmosphere, but much less dense. The molecular cloud", " filaments can be up to 1000 light-years long. Within the clouds are cold, dense regions with typical masses of 50 to 500 times the mass of the Sun; we give these regions the highly technical name clumps. Within these clumps, there are even denser, smaller regions called cores. The cores are the embryos of stars. The conditions in these cores\u2014low temperature and high density\u2014are just what is required to make stars. Remember that the essence of the life story of any star is the ongoing competition between two forces: gravity and pressure. The force of gravity, pulling inward, tries to make a star collapse. Internal pressure produced by the motions of the gas atoms, pushing outward, tries to force the star to expand. When a star is first forming, low temperature (and hence, low pressure) and high density (hence, greater gravitational attraction) both work to give gravity the advantage. In order to form a star\u2014that is, a dense, hot ball of matter capable of starting nuclear reactions deep within\u2014we need a typical core of interstellar atoms and molecules to shrink in radius and increase in density by a factor of nearly 1020. It is the force of gravity that produces this drastic collapse. 726 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System The Orion Molecular Cloud Let\u2019s discuss what happens in regions of star formation by considering a nearby site where stars are forming right now. One of the best-studied stellar nurseries is in the constellation of Orion, The Hunter, about 1500 light- years away (Figure 21.3). The pattern of the hunter is easy to recognize by the conspicuous \u201cbelt\u201d of three stars that mark his waist. The Orion molecular cloud is much larger than the star pattern and is truly an impressive structure. In its long dimension, it stretches over a distance of about 100 light-years. The total quantity of molecular gas is about 200,000 times the mass of the Sun. Most of the cloud does not glow with visible light but betrays its presence by the radiation that the dusty gas gives off at infrared and radio wavelengths. Figure 21.3. Orion in Visible and Infrared. (a) The Orion star group was named after the legendary hunter in Greek mythology. Three stars close together in a link mark Orion\u2019s belt. The ancients imagined a sword hanging from the belt; the object at the end of the blue line in this sword is the Orion Nebula. (b) This wide-angle, infrared view of the same area was taken with the Infrared Astronomical Satellite. Heated dust clouds dominate in this false-color image, and many of the stars that stood out on part (a) are now invisible. An exception is the cool, red-giant star Betelgeuse, which can be seen as a yellowish point at the left vertex of the blue triangle (at Orion\u2019s left armpit). The large, yellow ring to the right of Betelgeuse is the remnant of an exploded star. The infrared image lets us see how large and full of cooler material the Orion molecular cloud really is. On the visible-light image at left, you see only two colorful regions of interstellar matter\u2014the two, bright yellow splotches at the left end of and below Orion\u2019s belt. The lower one is the Orion Nebula and the higher one is the region of the Horsehead Nebula. (credit: modification of work by NASA, visible light: Akira Fujii; infrared: Infrared Astronomical Satellite) The stars in Orion\u2019s belt are typically about 5 million years old, whereas the stars near the middle of the \u201csword\u201d hanging from Orion\u2019s belt are only 300,000 to 1 million years old. The region about halfway down the sword where star formation is still taking place is called the Orion Nebula. About 2200 young stars are found in this region, which is only slightly larger than a dozen light-years in diameter. The Orion Nebula also contains a tight cluster of stars called the Trapezium (Figure 21.4). The brightest Trapezium stars can be seen easily with a small telescope. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 727 Figure 21.4. Orion Nebula. (a) The Orion Nebula is shown in visible light. (b) With near-infrared radiation, we can see more detail within the dusty nebula since infrared can penetrate dust more easily than can visible light. (credit a: modification of work by Filip Loli\u0107; credit b: modification of work by NASA/JPL-Caltech/T. Megeath (University of Toledo, Ohio)) Compare this with our own solar neighborhood, where the typical spacing between stars is about 3 light-years. Only a small number of stars in the Orion cluster", " can be seen with visible light, but infrared images\u2014which penetrate the dust better\u2014detect the more than 2000 stars that are part of the group (Figure 21.5). Figure 21.5. Central Region of the Orion Nebula. The Orion Nebula harbors some of the youngest stars in the solar neighborhood. At the heart of the nebula is the Trapezium cluster, which includes four very bright stars that provide much of the energy that causes the nebula to glow so brightly. In these images, we see a section of the nebula in (a) visible light and (b) infrared. The four bright stars in the center of the visible-light image are the Trapezium stars. Notice that most of the stars seen in the infrared are completely hidden by dust in the visible-light image. (credit a: modification of work by NASA, C.R. O\u2019Dell and S.K. Wong (Rice University); credit b: modification of work by NASA; K.L. Luhman (Harvard-Smithsonian Center for Astrophysics); and G. Schneider, E. Young, G. Rieke, A. Cotera, H. Chen, M. Rieke, R. Thompson (Steward Observatory, University of Arizona)) Studies of Orion and other star-forming regions show that star formation is not a very efficient process. In the region of the Orion Nebula, about 1% of the material in the cloud has been turned into stars. That is why we still see a substantial amount of gas and dust near the Trapezium stars. The leftover material is eventually heated, 728 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System either by the radiation and winds from the hot stars that form or by explosions of the most massive stars. (We will see in later chapters that the most massive stars go through their lives very quickly and end by exploding.) Take a journey through the Orion Nebula (https://openstaxcollege.org/l/30OriNebula) to view a nice narrated video tour of this region. Whether gently or explosively, the material in the neighborhood of the new stars is blown away into interstellar space. Older groups or clusters of stars can now be easily observed in visible light because they are no longer shrouded in dust and gas (Figure 21.6). Figure 21.6. Westerlund 2. This young cluster of stars known as Westerlund 2 formed within the Carina star-forming region about 2 million years ago. Stellar winds and pressure produced by the radiation from the hot stars within the cluster are blowing and sculpting the surrounding gas and dust. The nebula still contains many globules of dust. Stars are continuing to form within the denser globules and pillars of the nebula. This Hubble Space Telescope image includes near-infrared exposures of the star cluster and visible-light observations of the surrounding nebula. Colors in the nebula are dominated by the red glow of hydrogen gas, and blue-green emissions from glowing oxygen. (credit: NASA, ESA, the Hubble Heritage Team (STScI/AURA), A. Nota (ESA/STScI), and the Westerlund 2 Science Team) Although we do not know what initially caused stars to begin forming in Orion, there is good evidence that the first generation of stars triggered the formation of additional stars, which in turn led to the formation of still more stars (Figure 21.7). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 729 Figure 21.7. Propagating Star Formation. Star formation can move progressively through a molecular cloud. The oldest group of stars lies to the left of the diagram and has expanded because of the motions of individual stars. Eventually, the stars in the group will disperse and no longer be recognizable as a cluster. The youngest group of stars lies to the right, next to the molecular cloud. This group of stars is only 1 to 2 million years old. The pressure of the hot, ionized gas surrounding these stars compresses the material in the nearby edge of the molecular cloud and initiates the gravitational collapse that will lead to the formation of more stars. The basic idea of triggered star formation is this: when a massive star is formed, it emits a large amount of ultraviolet radiation and ejects high-speed gas in the form of a stellar wind. This injection of energy heats the gas around the stars and causes it to expand. When massive stars exhaust their supply of fuel, they explode, and the energy of the explosion also heats the gas. The hot gases pile into the surrounding cold molecular cloud, compressing the material in it and increasing its density. If this increase in density is large enough, gravity will overcome pressure, and stars will begin to form in the compressed gas. Such a chain reaction\u2014where the brightest and hottest", " stars of one area become the cause of star formation \u201cnext door\u201d\u2014seems to have occurred not only in Orion but also in many other molecular clouds. There are many molecular clouds that form only (or mainly) low-mass stars. Because low-mass stars do not have strong winds and do not die by exploding, triggered star formation cannot occur in these clouds. There are also stars that form in relative isolation in small cores. Therefore, not all star formation is originally triggered by the death of massive stars. However, there are likely to be other possible triggers, such as spiral density waves and other processes we do not yet understand. The Birth of a Star Although regions such as Orion give us clues about how star formation begins, the subsequent stages are still shrouded in mystery (and a lot of dust). There is an enormous difference between the density of a molecular cloud core and the density of the youngest stars that can be detected. Direct observations of this collapse to higher density are nearly impossible for two reasons. First, the dust-shrouded interiors of molecular clouds where stellar births take place cannot be observed with visible light. Second, the timescale for the initial collapse\u2014thousands of years\u2014is very short, astronomically speaking. Since each star spends such a tiny fraction of its life in this stage, relatively few stars are going through the collapse process at any given time. Nevertheless, through a combination of theoretical calculations and the limited observations available, astronomers have pieced together a picture of what the earliest stages of stellar evolution are likely to be. The first step in the process of creating stars is the formation of dense cores within a clump of gas and dust (Figure 21.8(a)). It is generally thought that all the material for the star comes from the core, the larger structure surrounding the forming star. Eventually, the gravitational force of the infalling gas becomes strong enough to overwhelm the pressure exerted by the cold material that forms the dense cores. The material then undergoes a rapid collapse, and the density of the core increases greatly as a result. During the time a dense 730 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System core is contracting to become a true star, but before the fusion of protons to produce helium begins, we call the object a protostar. Figure 21.8. Formation of a Star. (a) Dense cores form within a molecular cloud. (b) A protostar with a surrounding disk of material forms at the center of a dense core, accumulating additional material from the molecular cloud through gravitational attraction. (c) A stellar wind breaks out but is confined by the disk to flow out along the two poles of the star. (d) Eventually, this wind sweeps away the cloud material and halts the accumulation of additional material, and a newly formed star, surrounded by a disk, becomes observable. These sketches are not drawn to the same scale. The diameter of a typical envelope that is supplying gas to the newly forming star is about 5000 AU. The typical diameter of the disk is about 100 AU or slightly larger than the diameter of the orbit of Pluto. The natural turbulence inside a clump tends to give any portion of it some initial spinning motion (even if it is very slow). As a result, each collapsing core is expected to spin. According to the law of conservation of angular momentum (discussed in the chapter on Orbits and Gravity), a rotating body spins more rapidly as it decreases in size. In other words, if the object can turn its material around a smaller circle, it can move that material more quickly\u2014like a figure skater spinning more rapidly as she brings her arms in tight to her body. This is exactly what happens when a core contracts to form a protostar: as it shrinks, its rate of spin increases. But all directions on a spinning sphere are not created equal. As the protostar rotates, it is much easier for material to fall right onto the poles (which spin most slowly) than onto the equator (where material moves around most rapidly). Therefore, gas and dust falling in toward the protostar\u2019s equator are \u201cheld back\u201d by the rotation and form a whirling extended disk around the equator (part b in Figure 21.8). You may have observed this same \u201cequator effect\u201d on the amusement park ride in which you stand with your back to a cylinder that is spun faster and faster. As you spin really fast, you are pushed against the wall so strongly that you cannot possibly fall toward the center of the cylinder. Gas can, however, fall onto the protostar easily from directions away from the star\u2019s equator. The protostar and disk at this stage are embedded in an envelope of dust and gas from which material is still falling onto the protostar. This dusty envelope blocks visible light, but infrared radiation can get through. As a result, in this phase of its evolution, the prot", "ostar itself is emitting infrared radiation and so is observable only in the infrared region of the spectrum. Once almost all of the available material has been accreted and the central protostar has reached nearly its final mass, it is given a special name: it is called a T Tauri star, named after one of the best studied and brightest members of this class of stars, which was discovered in the constellation of Taurus. (Astronomers have a tendency to name types of stars after the first example they discover or come to understand. It\u2019s not an elegant system, but it works.) Only stars with masses less than or similar to the mass of the Sun become T Tauri stars. Massive stars do not go through this stage, although they do appear to follow the formation scenario illustrated in Figure 21.8. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 731 Winds and Jets Recent observations suggest that T Tauri stars may actually be stars in a middle stage between protostars and hydrogen-fusing stars such as the Sun. High-resolution infrared images have revealed jets of material as well as stellar winds coming from some T Tauri stars, proof of interaction with their environment. A stellar wind consists mainly of protons (hydrogen nuclei) and electrons streaming away from the star at speeds of a few hundred kilometers per second (several hundred thousand miles per hour). When the wind first starts up, the disk of material around the star\u2019s equator blocks the wind in its direction. Where the wind particles can escape most effectively is in the direction of the star\u2019s poles. Astronomers have actually seen evidence of these beams of particles shooting out in opposite directions from the popular regions of newly formed stars. In many cases, these beams point back to the location of a protostar that is still so completely shrouded in dust that we cannot yet see it (Figure 21.9). Figure 21.9. Gas Jets Flowing away from a Protostar. Here we see the neighborhood of a protostar, known to us as HH 34 because it is a Herbig-Haro object. The star is about 450 light-years away and only about 1 million years old. Light from the star itself is blocked by a disk, which is larger than 60 billion kilometers in diameter and is seen almost edge-on. Jets are seen emerging perpendicular to the disk. The material in these jets is flowing outward at speeds up to 580,000 kilometers per hour. The series of three images shows changes during a period of 5 years. Every few months, a compact clump of gas is ejected, and its motion outward can be followed. The changes in the brightness of the disk may be due to motions of clouds within the disk that alternately block some of the light and then let it through. This image corresponds to the stage in the life of a protostar shown in part (c) of Figure 21.8. (credit: modification of work by Hubble Space Telescope, NASA, ESA) On occasion, the jets of high-speed particles streaming away from the protostar collide with a somewhat-denser lump of gas nearby, excite its atoms, and cause them to emit light. These glowing regions, each of which is known as a Herbig-Haro (HH) object after the two astronomers who first identified them, allow us to trace the progress of the jet to a distance of a light-year or more from the star that produced it. Figure 21.10 shows two spectacular images of HH objects. 732 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System Figure 21.10. Outflows from Protostars. These images were taken with the Hubble Space Telescope and show jets flowing outward from newly formed stars. In the HH47 image, a protostar 1500 light-years away (invisible inside a dust disk at the left edge of the image) produces a very complicated jet. The star may actually be wobbling, perhaps because it has a companion. Light from the star illuminates the white region at the left because light can emerge perpendicular to the disk (just as the jet does). At right, the jet is plowing into existing clumps of interstellar gas, producing a shock wave that resembles an arrowhead. The HH1/2 image shows a double-beam jet emanating from a protostar (hidden in a dust disk in the center) in the constellation of Orion. Tip to tip, these jets are more than 1 light-year long. The bright regions (first identified by Herbig and Haro) are places where the jet is a slamming into a clump of interstellar gas and causing it to glow. (credit \u201cHH 47\u201d: modification of work by NASA, ESA, and P. Hartigan (Rice University); credit \u201cHH", " 1 and HH 2: modification of work by J. Hester, WFPC2 Team, NASA) The wind from a forming star will ultimately sweep away the material that remains in the obscuring envelope of dust and gas, leaving behind the naked disk and protostar, which can then be seen with visible light. We should note that at this point, the protostar itself is still contracting slowly and has not yet reached the main- sequence stage on the H\u2013R diagram (a concept introduced in the chapter The Stars: A Celestial Census). The disk can be detected directly when observed at infrared wavelengths or when it is seen silhouetted against a bright background (Figure 21.11). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 733 Figure 21.11. Disks around Protostars. These Hubble Space Telescope infrared images show disks around young stars in the constellation of Taurus, in a region about 450 light-years away. In some cases, we can see the central star (or stars\u2014some are binaries). In other cases, the dark, horizontal bands indicate regions where the dust disk is so thick that even infrared radiation from the star embedded within it cannot make its way through. The brightly glowing regions are starlight reflected from the upper and lower surfaces of the disk, which are less dense than the central, dark regions. (Credit: modification of work by D. Padgett (IPAC/Caltech), W. Brandner (IPAC), K. Stapelfeldt (JPL) and NASA) This description of a protostar surrounded by a rotating disk of gas and dust sounds very much like what happened in our solar system when the Sun and planets formed. Indeed, one of the most important discoveries from the study of star formation in the last decade of the twentieth century was that disks are an inevitable byproduct of the process of creating stars. The next questions that astronomers set out to answer was: will the disks around protostars also form planets? And if so, how often? We will return to these questions later in this chapter. To keep things simple, we have described the formation of single stars. Many stars, however, are members of binary or triple systems, where several stars are born together. In this case, the stars form in nearly the same way. Widely separated binaries may each have their own disk; close binaries may share a single disk. 21.2 THE H\u2013R DIAGRAM AND THE STUDY OF STELLAR EVOLUTION Learning Objectives By the end of this section, you will be able to: Determine the age of a protostar using an H\u2013R diagram and the protostar\u2019s luminosity and temperature Explain the interplay between gravity and pressure, and how the contracting protostar changes its position in the H\u2013R diagram as a result One of the best ways to summarize all of these details about how a star or protostar changes with time is to use a Hertzsprung-Russell (H\u2013R) diagram. Recall from The Stars: A Celestial Census that, when looking at an H\u2013R diagram, the temperature (the horizontal axis) is plotted increasing toward the left. As a star goes through the stages of its life, its luminosity and temperature change. Thus, its position on the H\u2013R diagram, in which luminosity is plotted against temperature, also changes. As a star ages, we must replot it in different places on the diagram. Therefore, astronomers often speak of a star moving on the H\u2013R diagram, or of its evolution 734 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System tracing out a path on the diagram. In this context, \u201ctracing out a path\u201d has nothing to do with the star\u2019s motion through space; this is just a shorthand way of saying that its temperature and luminosity change as it evolves Watch an animation (https://openstaxcollege.org/l/30aniomelen) of the stars in the Omega Centauri cluster as they rearrange according to luminosity and temperature, forming a Hertzsprung-Russell (H\u2013R) diagram. To estimate just how much the luminosity and temperature of a star change as it ages, we must resort to calculations. Theorists compute a series of models for a star, with each successive model representing a later point in time. Stars may change for a variety of reasons. Protostars, for example, change in size because they are contracting, and their temperature and luminosity change as they do so. After nuclear fusion begins in the star\u2019s core (see Stars from Adolescence to Old Age), main-sequence stars change because they are using up their nuclear fuel. Given a model that represents a star at one stage of", " its evolution, we can calculate what it will be like at a slightly later time. At each step, the model predicts the luminosity and size of the star, and from these values, we can figure out its surface temperature. A series of points on an H\u2013R diagram, calculated in this way, allows us to follow the life changes of a star and hence is called its evolutionary track. Evolutionary Tracks Let\u2019s now use these ideas to follow the evolution of protostars that are on their way to becoming main- sequence stars. The evolutionary tracks of newly forming stars with a range of stellar masses are shown in Figure 21.12. These young stellar objects are not yet producing energy by nuclear reactions, but they derive energy from gravitational contraction\u2014through the sort of process proposed for the Sun by Helmhotz and Kelvin in this last century (see the chapter on The Sun: A Nuclear Powerhouse). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 735 Figure 21.12. Evolutionary Tracks for Contracting Protostars. Tracks are plotted on the H\u2013R diagram to show how stars of different masses change during the early parts of their lives. The number next to each dark point on a track is the rough number of years it takes an embryo star to reach that stage (the numbers are the result of computer models and are therefore not well known). Note that the surface temperature (K) on the horizontal axis increases toward the left. You can see that the more mass a star has, the shorter time it takes to go through each stage. Stars above the dashed line are typically still surrounded by infalling material and are hidden by it. Initially, a protostar remains fairly cool with a very large radius and a very low density. It is transparent to infrared radiation, and the heat generated by gravitational contraction can be radiated away freely into space. Because heat builds up slowly inside the protostar, the gas pressure remains low, and the outer layers fall almost unhindered toward the center. Thus, the protostar undergoes very rapid collapse, a stage that corresponds to the roughly vertical lines at the right of Figure 21.12. As the star shrinks, its surface area gets smaller, and so its total luminosity decreases. The rapid contraction stops only when the protostar becomes dense and opaque enough to trap the heat released by gravitational contraction. When the star begins to retain its heat, the contraction becomes much slower, and changes inside the contracting star keep the luminosity of stars like our Sun roughly constant. The surface temperatures start to build up, and the star \u201cmoves\u201d to the left in the H\u2013R diagram. Stars first become visible only after the stellar wind described earlier clears away the surrounding dust and gas. This can happen during the rapid-contraction phase for low-mass stars, but high-mass stars remain shrouded in dust until they end their early phase of gravitational contraction (see the dashed line in Figure 21.12). To help you keep track of the various stages that stars go through in their lives, it can be useful to compare the development of a star to that of a human being. (Clearly, you will not find an exact correspondence, but thinking through the stages in human terms may help you remember some of the ideas we are trying to emphasize.) Protostars might be compared to human embryos\u2014as yet unable to sustain themselves but drawing resources from their environment as they grow. Just as the birth of a child is the moment it is called upon to produce its own energy (through eating and breathing), so astronomers say that a star is born when it is able to sustain itself through nuclear reactions (by making its own energy.) When the star\u2019s central temperature becomes high enough (about 10 million K) to fuse hydrogen into helium, we say that the star has reached the main sequence (a concept introduced in The Stars: A Celestial Census). It 736 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System is now a full-fledged star, more or less in equilibrium, and its rate of change slows dramatically. Only the gradual depletion of hydrogen as it is transformed into helium in the core slowly changes the star\u2019s properties. The mass of a star determines exactly where it falls on the main sequence. As Figure 21.12 shows, massive stars on the main sequence have high temperatures and high luminosities. Low-mass stars have low temperatures and low luminosities. Objects of extremely low mass never achieve high-enough central temperatures to ignite nuclear reactions. The lower end of the main sequence stops where stars have a mass just barely great enough to sustain nuclear reactions at a sufficient rate to stop gravitational contraction. This critical mass is calculated to be about 0.075 times the mass of the Sun. As we discussed in the chapter on Analyzing Starlight, objects below this", " critical mass are called either brown dwarfs or planets. At the other extreme, the upper end of the main sequence terminates at the point where the energy radiated by the newly forming massive star becomes so great that it halts the accretion of additional matter. The upper limit of stellar mass is between 100 and 200 solar masses. Evolutionary Timescales How long it takes a star to form depends on its mass. The numbers that label the points on each track in Figure 21.12 are the times, in years, required for the embryo stars to reach the stages we have been discussing. Stars of masses much higher than the Sun\u2019s reach the main sequence in a few thousand to a million years. The Sun required millions of years before it was born. Tens of millions of years are required for stars of lower mass to evolve to the lower main sequence. (We will see that this turns out to be a general principle: massive stars go through all stages of evolution faster than low-mass stars do.) We will take up the subsequent stages in the life of a star in Stars from Adolescence to Old Age, examining what happens after stars arrive in the main sequence and begin a \u201cprolonged adolescence\u201d and \u201cadulthood\u201d of fusing hydrogen to form helium. But now we want to examine the connection between the formation of stars and planets. 21.3 EVIDENCE THAT PLANETS FORM AROUND OTHER STARS Learning Objectives By the end of this section, you will be able to: Trace the evolution of dust surrounding a protostar, leading to the development of rocky planets and gas giants Estimate the timescale for growth of planets using observations of the disks surrounding young stars Evaluate evidence for planets around forming stars based on the structures seen in images of the circumstellar dust disks Having developed on a planet and finding it essential to our existence, we have a special interest in how planets fit into the story of star formation. Yet planets outside the solar system are extremely difficult to detect. Recall that we see planets in our own system only because they reflect sunlight and are close by. When we look to the other stars, we find that the amount of light a planet reflects is a depressingly tiny fraction of the light its star gives off. Furthermore, from a distance, planets are lost in the glare of their much-brighter parent stars. Disks around Protostars: Planetary Systems in Formation It is a lot easier to detect the spread-out raw material from which planets might be assembled than to detect planets after they are fully formed. From our study of the solar system, we understand that planets form by the This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 737 gathering together of gas and dust particles in orbit around a newly created star. Each dust particle is heated by the young protostar and radiates in the infrared region of the spectrum. Before any planets form, we can detect such radiation from all of the spread-out individual dust particles that are destined to become parts of planets. We can also detect the silhouette of the disk if it blocks bright light coming from a source behind it (Figure 21.13). Figure 21.13. Disks around Protostars. These Hubble Space Telescope images show four disks around young stars in the Orion Nebula. The dark, dusty disks are seen silhouetted against the bright backdrop of the glowing gas in the nebula. The size of each image is about 30 times the diameter of our planetary system; this means the disks we see here range in size from two to eight times the orbit of Pluto. The red glow at the center of each disk is a young star, no more than a million years old. These images correspond to the stage in the life of a protostar shown in part (d) of Figure 21.8. (credit: modification of work by Mark McCaughrean (Max-Planck-Institute for Astronomy), C. Robert O\u2019Dell (Rice University), and NASA) Once the dust particles gather together and form a few planets (and maybe some moons), the overwhelming majority of the dust is hidden in the interiors of the planets where we cannot see it. All we can now detect is the radiation from the outside surfaces, which cover a drastically smaller area than the huge, dusty disk from which they formed. The amount of infrared radiation is therefore greatest before the dust particles combine into planets. For this reason, our search for planets begins with a search for infrared radiation from the material required to make them. A disk of gas and dust appears to be an essential part of star formation. Observations show that nearly all very young protostars have disks and that the disks range in size from 10 to 1000 AU. (For comparison, the average diameter of the orbit of Pluto, which can be considered the rough size of our own planetary", " system, is 80 AU, whereas the outer diameter of the Kuiper belt of smaller icy bodies is about 100 AU.) The mass contained in these disks is typically 1\u201310% of the mass of our own Sun, which is more than the mass of all the planets in our solar system put together. Such observations already demonstrate that a large fraction of stars begin their lives with enough material in the right place to form a planetary system. The Timing of Planet Formation and Growth We can use observations of how the disks change with time to estimate how long it takes for planets to form. If we measure the temperature and luminosity of a protostar, then, as we saw, we can place it in an H\u2013R diagram like the one shown in Figure 21.12. By comparing the real star with our models of how protostars should evolve with time, we can estimate its age. We can then look at how the disks we observe change with the ages of the stars that they surround. What such observations show is that if a protostar is less than about 1 to 3 million years old, its disk extends all the way from very close to the surface of the star out to tens or hundreds of AU away. In older stars, we find disks with outer parts that still contain large amounts of dust, but the inner regions have lost most of their dust. In these objects, the disk looks like a donut, with the protostar centered in its hole. The inner, dense parts of most disks have disappeared by the time the stars are 10 million years old (Figure 21.14). 738 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System Figure 21.14. Protoplanetary Disks around Two Stars. The left view of each star shows infrared observations by the Hubble Space Telescope of their protoplanetary disks. The central star is much brighter than the surrounding disk, so the instrument includes a coronograph, which has a small shield that blocks the light of the central star but allows the surrounding disk to be imaged. The right image of each star shows models of the disks based on the observations. The star HD 141943 has an age of about 17 million years, while HD 191089 is about 12 million years old. (credit: modification of work by NASA, ESA, R. Soummer and M. Perrin (STScI), L. Pueyo (STScI/Johns Hopkins University), C. Chen and D. Golimowski (STScI), J.B. Hagan (STScI/Purdue University), T. Mittal (University of California, Berkeley/Johns Hopkins University), E. Choquet, M. Moerchen, and M. N\u2019Diaye (STScI), A. Rajan (Arizona State University), S. Wolff (STScI/Purdue University), J. Debes and D. Hines (STScI), and G. Schneider (Steward Observatory/University of Arizona)) Calculations show that the formation of one or more planets could produce such a donut-like distribution of dust. Suppose a planet forms a few AU away from the protostar, presumably due to the gathering together of matter from the disk. As the planet grows in mass, the process clears out a dust-free region in its immediate neighborhood. Calculations also show that any small dust particles and gas that were initially located in the region between the protostar and the planet, and that are not swept up by the planet, will then fall onto the star very quickly in about 50,000 years. Matter lying outside the planet\u2019s orbit, in contrast, is prevented from moving into the hole by the gravitational forces exerted by the planet. (We saw something similar in Saturn\u2019s rings, where the action of the shepherd moons keeps the material near the edge of the rings from spreading out.) If the formation of a planet is indeed what produces and sustains holes in the disks that surround very young stars, then planets must form in 3 to 30 million years. This is a short period compared with the lifetimes of most stars and shows that the formation of planets may be a quick byproduct of the birth of stars. Calculations show that accretion can drive the rapid growth of planets\u2014small, dust-grain-size particles orbiting in the disk collide and stick together, with the larger collections growing more rapidly as they attract and capture smaller ones. Once these clumps grow to about 10 centimeters in size or so, they enter a perilous stage in their development. At that size, unless they can grow to larger than about 100 meters in diameter, they are subject to drag forces produced by friction with the gas in the disk\u2014and their orbits can rapidly decay, plunging them into the host star. Therefore, these bodies must rapidly grow to nearly 1 kilometer in size in diameter to avoid a fiery fate. At this stage, they are considered planetesimals (", "the small chunks of solid matter\u2014ice and dust particles\u2014that you learned about in Other Worlds: An Introduction to the Solar System). Once they survive to those sizes, the largest survivors will continue to grow by accreting smaller planetesimals; ultimately, this process results in a few large planets. If the growing planets reach a mass bigger than about 10 times the mass of Earth, their gravity is strong enough to capture and hold on to hydrogen gas that remains in the disk. At that point, they will grow in mass and radius rapidly, reaching giant planet dimensions. However, to do so requires that the rapidly evolving central star hasn\u2019t yet driven away the gas in the disk with its increasingly vigorous wind (see the earlier section on Star Formation). From observations, we see that the disk can be blown away within 10 million years, so growth of a giant planet must also be a very fast process, astronomically speaking. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 739 Debris Disks and Shepherd Planets The dust around newly formed stars is gradually either incorporated into the growing planets in the newly forming planetary system or ejected through gravitational interactions with the planets into space. The dust will disappear after about 30 million years unless the disk is continually supplied with new material. Local comets and asteroids are the most likely sources of new dust. As the planet-size bodies grow, they stir up the orbits of smaller objects in the area. These small bodies collide at high speeds, shatter, and produce tiny particles of silicate dust and ices that can keep the disk supplied with the debris from these collisions. Over several hundred million years, the comets and asteroids will gradually be reduced in number, the frequency of collisions will go down, and the supply of fresh dust will diminish. Remember that the heavy bombardment in the early solar system ended when the Sun was only about 500 million years old. Observations show that the dusty \u201cdebris disks\u201d around stars also become largely undetectable by the time the stars reach an age of 400 to 500 million years. It is likely, however, that some small amount of cometary material will remain in orbit, much like our Kuiper belt, a flattened disk of comets outside the orbit of Neptune. In a young planetary system, even if we cannot see the planets directly, the planets can concentrate the dust particles into clumps and arcs that are much larger than the planets themselves and more easily imaged. This is similar to how the tiny moons of Saturn shepherd the particles in the rings and produce large arcs and structures in Saturn\u2019s rings. Debris disks\u2014many with just such clumps and arcs\u2014have now been found around many stars, such as HL Tau, located about 450 light-years from Earth in the constellation Taurus (Figure 21.15). In some stars, the brightness of the rings varies with position; around other stars, there are bright arcs and gaps in the rings. The brightness indicates the relative concentration of dust, since what we are seeing is infrared (heat radiation) from the dust particles in the rings. More dust means more radiation. Figure 21.15. Dust Ring around a Young Star. This image was made by ALMA (the Atacama Large Millimeter/Submillimeter Array) at a wavelength of 1.3 millimeters and shows the young star HL Tau and its protoplanetary disk. It reveals multiple rings and gaps that indicate the presence of emerging planets, which are sweeping their orbits clear of dust and gas. (credit: modification of work by ALMA (ESO/NAOJ/NRAO)) 740 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System Watch a short video clip (https://openstaxcollege.org/l/30vidNRAOdir) of the director of NRAO (National Radio Astronomy Observatory) describing the high-resolution observations of the young star HL Tau. While you\u2019re there, watch an artist\u2019s animation of a protoplanetary disk to see newly formed planets traveling around a host (parent) star. 21.4 PLANETS BEYOND THE SOLAR SYSTEM: SEARCH AND DISCOVERY Learning Objectives By the end of this section, you will be able to: Describe the orbital motion of planets in our solar system using Kepler\u2019s laws Compare the indirect and direct observational techniques for exoplanet detection For centuries, astronomers have dreamed of finding planets around other stars, including other planets like Earth. Direct observations of such distant planets are very difficult, however. You might compare a planet orbiting a star to a mosquito flying around one of those giant spotlights at a shopping center opening. From close up, you might spot the mosquito. But imagine viewing the scene from some distance away\u2014say, from an airplane. You could see", " the spotlight just fine, but what are your chances of catching the mosquito in that light? Instead of making direct images, astronomers have relied on indirect observations and have now succeeded in detecting a multitude of planets around other stars. In 1995, after decades of effort, we found the first such exoplanet (a planet outside our solar system) orbiting a main-sequence star, and today we know that most stars form with planets. This is an example of how persistence and new methods of observation advance the knowledge of humanity. By studying exoplanets, astronomers hope to better understand our solar system in context of the rest of the universe. For instance, how does the arrangement of our solar system compare to planetary systems in the rest of the universe? What do exoplanets tell us about the process of planet formation? And how does knowing the frequency of exoplanets influence our estimates of whether there is life elsewhere? Searching for Orbital Motion Most exoplanet detections are made using techniques where we observe the effect that the planet exerts on the host star. For example, the gravitational tug of an unseen planet will cause a small wobble in the host star. Or, if its orbit is properly aligned, a planet will periodically cross in front of the star, causing the brightness of the star to dim. To understand how a planet can move its host star, consider a single Jupiter-like planet. Both the planet and the star actually revolve about their common center of mass. Remember that gravity is a mutual attraction. The star and the planet each exert a force on the other, and we can find a stable point, the center of mass, between them about which both objects move. The smaller the mass of a body in such a system, the larger its orbit. A massive star barely swings around the center of mass, while a low-mass planet makes a much larger \u201ctour.\u201d Suppose the planet is like Jupiter and has a mass about one-thousandth that of its star; in this case, the size of the star\u2019s orbit is one-thousandth the size of the planet\u2019s. To get a sense of how difficult observing such motion This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 741 might be, let\u2019s see how hard Jupiter would be to detect in this way from the distance of a nearby star. Consider an alien astronomer trying to observe our own system from Alpha Centauri, the closest star system to our own (about 4.3 light-years away). There are two ways this astronomer could try to detect the orbital motion of the Sun. One way would be to look for changes in the Sun\u2019s position on the sky. The second would be to use the Doppler effect to look for changes in its velocity. Let\u2019s discuss each of these in turn. The diameter of Jupiter\u2019s apparent orbit viewed from Alpha Centauri is 10 seconds of arc, and that of the Sun\u2019s orbit is 0.010 seconds of arc. (Remember, 1 second of arc is 1/3600 degree.) If they could measure the apparent position of the Sun (which is bright and easy to detect) to sufficient precision, they would describe an orbit of diameter 0.010 seconds of arc with a period equal to that of Jupiter, which is 12 years. In other words, if they watched the Sun for 12 years, they would see it wiggle back and forth in the sky by this minuscule fraction of a degree. From the observed motion and the period of the \u201cwiggle,\u201d they could deduce the mass of Jupiter and its distance using Kepler\u2019s laws. (To refresh your memory about these laws, see the chapter on Orbits and Gravity.) Measuring positions in the sky this accurately is extremely difficult, and so far, astronomers have not made any confirmed detections of planets using this technique. However, we have been successful in using spectrometers to measure the changing velocity of stars with planets around them. As the star and planet orbit each other, part of their motion will be in our line of sight (toward us or away from us). Such motion can be measured using the Doppler effect and the star\u2019s spectrum. As the star moves back and forth in orbit around the system\u2019s center of mass in response to the gravitational tug of an orbiting planet, the lines in its spectrum will shift back and forth. Let\u2019s again consider the example of the Sun. Its radial velocity (motion toward or away from us) changes by about 13 meters per second with a period of 12 years because of the gravitational pull of Jupiter. This corresponds to about 30 miles per hour, roughly the speed at which many of us drive around town. Detecting motion at this level in a star\u2019s spectrum presents an enormous technical challenge,", " but several groups of astronomers around the world, using specialized spectrographs designed for this purpose, have succeeded. Note that the change in speed does not depend on the distance of the star from the observer. Using the Doppler effect to detect planets will work at any distance, as long as the star is bright enough to provide a good spectrum and a large telescope is available to make the observations (Figure 21.16). 742 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System Figure 21.16. Doppler Method of Detecting Planets. The motion of a star around a common center of mass with an orbiting planet can be detected by measuring the changing speed of the star. When the star is moving away from us, the lines in its spectrum show a tiny redshift; when it is moving toward us, they show a tiny blueshift. The change in color (wavelength) has been exaggerated here for illustrative purposes. In reality, the Doppler shifts we measure are extremely small and require sophisticated equipment to be detected. The first successful use of the Doppler effect to find a planet around another star was in 1995. Michel Mayor and Didier Queloz of the Geneva Observatory (Figure 21.17) used this technique to find a planet orbiting a star resembling our Sun called 51 Pegasi, about 40 light-years away. (The star can be found in the sky near the great square of Pegasus, the flying horse of Greek mythology, one of the easiest-to-find star patterns.) To everyone\u2019s surprise, the planet takes a mere 4.2 days to orbit around the star. (Remember that Mercury, the innermost planet in our solar system, takes 88 days to go once around the Sun, so 4.2 days seems fantastically short.) Figure 21.17. Planet Discoverers. In 1995, Didier Queloz and Michel Mayor of the Geneva Observatory were the first to discover a planet around a regular star (51 Pegasi). They are seen here at an observatory in Chile where they are continuing their planet hunting. (credit: Weinstein/Ciel et Espace Photos) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 743 Mayor and Queloz\u2019s findings mean the planet must be very close to 51 Pegasi, circling it about 7 million kilometers away (Figure 21.18). At that distance, the energy of the star should heat the planet\u2019s surface to a temperature of a few thousand degrees Celsius (a bit hot for future tourism). From its motion, astronomers calculate that it has at least half the mass of Jupiter[1], making it clearly a jovian and not a terrestrial-type planet. Figure 21.18. Hot Jupiter. Artist Greg Bacon painted this impression of a hot, Jupiter-type planet orbiting close to a sunlike star. The artist shows bands on the planet like Jupiter, but we only estimate the mass of most hot, Jupiter-type planets from the Doppler method and don\u2019t know what conditions on the planet are like. (credit: ESO) Since that initial planet discovery, the rate of progress has been breathtaking. Hundreds of giant planets have been discovered using the Doppler technique. Many of these giant planets are orbiting close to their stars\u2014astronomers have called these hot Jupiters. The existence of giant planets so close to their stars was a surprise, and these discoveries have forced us to rethink our ideas about how planetary systems form. But for now, bear in mind that the Doppler-shift method\u2014which relies on the pull of a planet making its star \u201cwiggle\u201d back and forth around the center of mass\u2014is most effective at finding planets that are both close to their stars and massive. These planets cause the biggest \u201cwiggles\u201d in the motion of their stars and the biggest Doppler shifts in the spectrum. Plus, they will be found sooner, since astronomers like to monitor the star for at least one full orbit (and perhaps more) and hot Jupiters take the shortest time to complete their orbit. So if such planets exist, we would expect to be finding this type first. Scientists call this a selection effect\u2014where our technique of discovery selects certain kinds of objects as \u201ceasy finds.\u201d As an example of a selection effect in everyday life, imagine you decide you are ready for a new romantic relationship in your life. To begin with, you only attend social events on campus, all of which require a student ID to get in. Your selection of possible partners will then be limited to students at your college. That may not give you as diverse a group to choose from as you want. In the same way, when we first used the Doppler technique, it selected massive planets close 1 The Do", "ppler method only allows us to find the minimum mass of a planet. To determine the exact mass using the Doppler shift and Kepler\u2019s laws, we must also have the angle at which the planet\u2019s orbit is oriented to our view\u2014something we don\u2019t have any independent way of knowing in most cases. Still, if the minimum mass is half of Jupiter\u2019s, the actual mass can only be larger than that, and we are sure that we are dealing with a jovian planet. 744 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System to their stars as the most likely discoveries. As we spend longer times watching target stars and as our ability to measure smaller Doppler shifts improves, this technique can reveal more distant and less massive planets too View a series of animations (https://openstaxcollege.org/l/30keplawsolarani) demonstrating solar system motion and Kepler\u2019s laws, and select animation 1 (Kepler\u2019s laws) from the dropdown playlist. To view an animation demonstrating the radial velocity curve for an exoplanet, select animation 29 (radial velocity curve for an exoplanet) and animation 30 (radial velocity curve for an exoplanet\u2014elliptical orbit) from the dropdown playlist. Transiting Planets The second method for indirect detection of exoplanets is based not on the motion of the star but on its brightness. When the orbital plane of the planet is tilted or inclined so that it is viewed edge-on, we will see the planet cross in front of the star once per orbit, causing the star to dim slightly; this event is known as transit. Figure 21.19 shows a sketch of the transit at three time steps: (1) out of transit, (2) the start of transit, and (3) full transit, along with a sketch of the light curve, which shows the drop in the brightness of the host star. The amount of light blocked\u2014the depth of the transit\u2014depends on the area of the planet (its size) compared to the star. If we can determine the size of the star, the transit method tells us the size of the planet. Figure 21.19. Planet Transits. As the planet transits, it blocks out some of the light from the star, causing a temporary dimming in the brightness of the star. The top figure shows three moments during the transit event and the bottom panel shows the corresponding light curve: (1) out of transit, (2) transit ingress, and (3) the full drop in brightness. The interval between successive transits is the length of the year for that planet, which can be used (again using Kepler\u2019s laws) to find its distance from the star. Larger planets like Jupiter block out more starlight than small earthlike planets, making transits by giant planets easier to detect, even from ground-based observatories. But by going into space, above the distorting effects of Earth\u2019s atmosphere, the transit technique has been extended to exoplanets as small as Mars. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 745 Transit Depth In a transit, the planet\u2019s circular disk blocks the light of the star\u2019s circular disk. The area of a circle is \u03c0R2. The amount of light the planet blocks, called the transit depth, is then given by \u03c0R2 \u03c0R2 planet = star R2 R2 planet star \u239b \u239d Rplanet Rstar 2 \u239e \u23a0 = Now calculate the transit depth for a star the size of the Sun with a gas giant planet the size of Jupiter. Solution The radius of Jupiter is 71,400 km, while the radius of the Sun is 695,700 km. Substituting into the equation, we get \u239b \u239d = 0.01 or 1%, which can easily be detected with the 71,400 km 695,700 km 2 \u239e \u23a0 Rplanet Rstar \u239e \u23a0 \u239b \u239d = 2 instruments on board the Kepler spacecraft. Check Your Learning What is the transit depth for a star half the size of the Sun with a much smaller planet, like the size of Earth? Answer: The radius of Earth is 6371 km. Therefore, \u239b \u239d Rplanet Rstar 2 \u239e \u23a0 = \u239b \u239d 6371 km 695,700/2 km 2 \u239e \u23a0 = \u239b \u239d 6371 km 347,850 km 2 \u239e \u23a0 = 0.0003, or significantly less than 1%. The Doppler method allows us to estimate the mass", " of a planet. If the same object can be studied by both the Doppler and transit techniques, we can measure both the mass and the size of the exoplanet. This is a powerful combination that can be used to derive the average density (mass/volume) of the planet. In 1999, using measurements from ground-based telescopes, the first transiting planet was detected orbiting the star HD 209458. The planet transits its parent star for about 3 hours every 3.5 days as we view it from Earth. Doppler measurements showed that the planet around HD 209458 has about 70% the mass of Jupiter, but its radius is about 35% larger than Jupiter\u2019s. This was the first case where we could determine what an exoplanet was made of\u2014with that mass and radius, HD 209458 must be a gas and liquid world like Jupiter or Saturn. It is even possible to learn something about the planet\u2019s atmosphere. When the planet passes in front of HD 209458, the atoms in the planet\u2019s atmosphere absorb starlight. Observations of this absorption were first made at the wavelengths of yellow sodium lines and showed that the atmosphere of the planet contains sodium; now, other elements can be measured as well. 746 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System Try a transit simulator (https://openstaxcollege.org/l/30transimul) that demonstrates how a planet passing in front of its parent star can lead to the planet\u2019s detection. Follow the instructions to run the animation on your computer. Transiting planets reveal such a wealth of information that the French Space Agency (CNES) and the European Space Agency (ESA) launched the CoRoT space telescope in 2007 to detect transiting exoplanets. CoRoT discovered 32 transiting exoplanets, including the first transiting planet with a size and density similar to Earth. In 2012, the spacecraft suffered an onboard computer failure, ending the mission. Meanwhile, NASA built a much more powerful transit observatory called Kepler. In 2009, NASA launched the Kepler space telescope, dedicated to the discovery of transiting exoplanets. This spacecraft stared continuously at more than 150,000 stars in a small patch of sky near the constellation of Cygnus\u2014just above the plane of our Milky Way Galaxy (Figure 21.20). Kepler\u2019s cameras and ability to measure small changes in brightness very precisely enabled the discovery of thousands of exoplanets, including many multi-planet systems. The spacecraft required three reaction wheels\u2014a type of wheel used to help control slight rotation of the spacecraft\u2014to stabilize the pointing of the telescope and monitor the brightness of the same group of stars over and over again. Kepler was launched with four reaction wheels (one a spare), but by May 2013, two wheels had failed and the telescope could no longer be accurately pointed toward the target area. Kepler had been designed to operate for 4 years, and ironically, the pointing failure occurred exactly 4 years and 1 day after it began observing. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 747 Figure 21.20. Kepler\u2019s Field of View. The boxes show the region where the Kepler spacecraft cameras took images of over 150,000 stars regularly, to find transiting planets. (credit \u201cfield of view\u201d: modification of work by NASA/Kepler mission; credit \u201cspacecraft\u201d: modification of work by NASA/Kepler mission/Wendy Stenzel) What do we mean, exactly, by \u201cdiscovery\u201d of transiting exoplanets? A single transit shows up as a very slight drop in the brightness of the star, lasting several hours. However, astronomers must be on guard against other factors that might produce a false transit, especially when working at the limit of precision of the telescope. We must wait for a second transit of similar depth. But when another transit is observed, we don\u2019t initially know whether it might be due to another planet in a different orbit. The \u201cdiscovery\u201d occurs only when a third transit is found with similar depth and the same spacing in time as the first pair. Computers normally conduct the analysis, which involves searching for tiny, periodic dips in the light from each star, extending over 4 years of observation. But the Kepler mission also has a program in which non- astronomers\u2014citizen scientists\u2014can examine the data. These dedicated volunteers have found several transits that were missed by the computer analyses, showing that the human eye and brain sometimes recognize unusual events that a computer was not programmed to look for. Measuring three or four evenly spaced transits is normally enough to \u201cdiscover\u201d an exoplanet. But in a new field like exoplanet research, we would like to find further independent verification.", " The strongest confirmation happens when ground-based telescopes are also able to detect a Doppler shift with the same period as the transits. However, this is generally not possible for Earth-size planets. One of the most convincing ways to verify that a dip in brightness is due to a planet is to find more planets orbiting the same star\u2014a planetary system. Multi-planet systems also provide alternative ways to estimate the masses of the planets, as we will discuss in the next section. The selection effects (or biases) in the Kepler data are similar to those in Doppler observations. Large planets are easier to find than small ones, and short-period planets are easier than long-period planets. If we require three transits to establish the presence of a planet, we are of course limited to discovering planets with orbital 748 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System periods less than one-third of the observing interval. Thus, it was only in its fourth and final year of operation that Kepler was able to find planets with orbits like Earth\u2019s that require 1 year to go around their star. Direct Detection The best possible evidence for an earthlike planet elsewhere would be an image. After all, \u201cseeing is believing\u201d is a very human prejudice. But imaging a distant planet is a formidable challenge indeed. Suppose, for example, you were a great distance away and wished to detect reflected light from Earth. Earth intercepts and reflects less than one billionth of the Sun\u2019s radiation, so its apparent brightness in visible light is less than one billionth that of the Sun. Compounding the challenge of detecting such a faint speck of light, the planet is swamped by the blaze of radiation from its parent star. Even today, the best telescope mirrors\u2019 optics have slight imperfections that prevent the star\u2019s light from coming into focus in a completely sharp point. Direct imaging works best for young gas giant planets that emit infrared light and reside at large separations from their host stars. Young giant planets emit more infrared light because they have more internal energy, stored from the process of planet formation. Even then, clever techniques must be employed to subtract out the light from the host star. In 2008, three such young planets were discovered orbiting HR 8799, a star in the constellation of Pegasus (Figure 21.21). Two years later, a fourth planet was detected closer to the star. Additional planets may reside even closer to HR 8799, but if they exist, they are currently lost in the glare of the star. Since then, a number of planets around other stars have been found using direct imaging. However, one challenge is to tell whether the objects we are seeing are indeed planets or if they are brown dwarfs (failed stars) in orbit around a star. Figure 21.21. Exoplanets around HR 8799. This image shows Keck telescope observations of four directly imaged planets orbiting HR 8799. A size scale for the system gives the distance in AU (remember that one astronomical unit is the distance between Earth and the Sun.) (credit: modification of work by Ben Zuckerman) Direct imaging is an important technique for characterizing an exoplanet. The brightness of the planet can be measured at different wavelengths. These observations provide an estimate for the temperature of the planet\u2019s atmosphere; in the case of HR 8799 planet 1, the color suggests the presence of thick clouds. Spectra can also be This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 749 obtained from the faint light to analyze the atmospheric constituents. A spectrum of HR 8799 planet 1 indicates a hydrogen-rich atmosphere, while the closer planet 4 shows evidence for methane in the atmosphere. Another way to overcome the blurring effect of Earth\u2019s atmosphere is to observe from space. Infrared may be the optimal wavelength range in which to observe because planets get brighter in the infrared while stars like our Sun get fainter, thereby making it easier to detect a planet against the glare of its star. Special optical techniques can be used to suppress the light from the central star and make it easier to see the planet itself. However, even if we go into space, it will be difficult to obtain images of Earth-size planets. 21.5 EXOPLANETS EVERYWHERE: WHAT WE ARE LEARNING Learning Objectives By the end of this section, you will be able to: Explain what we have learned from our discovery of exoplanets Identify which kind of exoplanets appear to be the most common in the Galaxy Discuss the kinds of planetary systems we are finding around other stars Before the discovery of exoplanets, most astronomers expected that other planetary systems would be much like our own\u2014planets following roughly circular orbits, with the most massive planets several AU from their parent star. Such systems do exist in", " large numbers, but many exoplanets and planetary systems are very different from those in our solar system. Another surprise is the existence of whole classes of exoplanets that we simply don\u2019t have in our solar system: planets with masses between the mass of Earth and Neptune, and planets that are several times more massive than Jupiter. Kepler Results The Kepler telescope has been responsible for the discovery of most exoplanets, especially at smaller sizes, as illustrated in Figure 21.22, where the Kepler discoveries are plotted in yellow. You can see the wide range of sizes, including planets substantially larger than Jupiter and smaller than Earth. The absence of Kepler- discovered exoplanets with orbital periods longer than a few hundred days is a consequence of the 4-year lifetime of the mission. (Remember that three evenly spaced transits must be observed to register a discovery.) At the smaller sizes, the absence of planets much smaller than one earth radius is due to the difficulty of detecting transits by very small planets. In effect, the \u201cdiscovery space\u201d for Kepler was limited to planets with orbital periods less than 400 days and sizes larger than Mars. 750 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System Figure 21.22. Exoplanet Discoveries through 2015. The vertical axis shows the radius of each planet compared to Earth. Horizontal lines show the size of Earth, Neptune, and Jupiter. The horizontal axis shows the time each planet takes to make one orbit (and is given in Earth days). Recall that Mercury takes 88 days and Earth takes a little more than 365 days to orbit the Sun. The yellow and red dots show planets discovered by transits, and the blue dots are the discoveries by the radial velocity (Doppler) technique. (credit: modification of work by NASA/Kepler mission) One of the primary objectives of the Kepler mission was to find out how many stars hosted planets and especially to estimate the frequency of earthlike planets. Although Kepler looked at only a very tiny fraction of the stars in the Galaxy, the sample size was large enough to draw some interesting conclusions. While the observations apply only to the stars observed by Kepler, those stars are reasonably representative, and so astronomers can extrapolate to the entire Galaxy. Figure 21.23 shows that the Kepler discoveries include many rocky, Earth-size planets, far more than Jupiter- size gas planets. This immediately tells us that the initial Doppler discovery of many hot Jupiters was a biased sample, in effect, finding the odd planetary systems because they were the easiest to detect. However, there is one huge difference between this observed size distribution and that of planets in our solar system. The most common planets have radii between 1.4 and 2.8 that of Earth, sizes for which we have no examples in the solar system. These have been nicknamed super-Earths, while the other large group with sizes between 2.8 and 4 that of Earth are often called mini-Neptunes. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 751 Figure 21.23. Kepler Discoveries. This bar graph shows the number of planets of each size range found among the first 2213 Kepler planet discoveries. Sizes range from half the size of Earth to 20 times that of Earth. On the vertical axis, you can see the fraction that each size range makes up of the total. Note that planets that are between 1.4 and 4 times the size of Earth make up the largest fractions, yet this size range is not represented among the planets in our solar system. (credit: modification of work by NASA/Kepler mission) What a remarkable discovery it is that the most common types of planets in the Galaxy are completely absent from our solar system and were unknown until Kepler\u2019s survey. However, recall that really small planets were difficult for the Kepler instruments to find. So, to estimate the frequency of Earth-size exoplanets, we need to correct for this sampling bias. The result is the corrected size distribution shown in Figure 21.24. Notice that in this graph, we have also taken the step of showing not the number of Kepler detections but the average number of planets per star for solar-type stars (spectral types F, G, and K). 752 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System Figure 21.24. Size Distribution of Planets for Stars Similar to the Sun. We show the average number of planets per star in each planet size range. (The average is less than one because some stars will have zero planets of that size range.) This distribution, corrected for biases in the Kepler data, shows that Earth-size planets may actually be the most common type of exoplanets. (credit: modification of work by NASA/Kepler mission) We see that the most common planet sizes", " of are those with radii from 1 to 3 times that of Earth\u2014what we have called \u201cEarths\u201d and \u201csuper-Earths.\u201d Each group occurs in about one-third to one-quarter of stars. In other words, if we group these sizes together, we can conclude there is nearly one such planet per star! And remember, this census includes primarily planets with orbital periods less than 2 years. We do not yet know how many undiscovered planets might exist at larger distances from their star. To estimate the number of Earth-size planets in our Galaxy, we need to remember that there are approximately 100 billion stars of spectral types F, G, and K. Therefore, we estimate that there are about 30 billion Earth-size planets in our Galaxy. If we include the super-Earths too, then there could be one hundred billion in the whole Galaxy. This idea\u2014that planets of roughly Earth\u2019s size are so numerous\u2014is surely one of the most important discoveries of modern astronomy. Planets with Known Densities For several hundred exoplanets, we have been able to measure both the size of the planet from transit data and its mass from Doppler data, yielding an estimate of its density. Comparing the average density of exoplanets to the density of planets in our solar system helps us understand whether they are rocky or gaseous in nature. This has been particularly important for understanding the structure of the new categories of super-Earths and mini-Neptunes with masses between 3\u201310 times the mass of Earth. A key observation so far is that planets This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 753 that are more than 10 times the mass of Earth have substantial gaseous envelopes (like Uranus and Neptune) whereas lower-mass planets are predominately rocky in nature (like the terrestrial planets). Figure 21.25 compares all the exoplanets that have both mass and radius measurements. The dependence of the radius on planet mass is also shown for a few illustrative cases\u2014hypothetical planets made of pure iron, rock, water, or hydrogen. Figure 21.25. Exoplanets with Known Densities. Exoplanets with known masses and radii (red circles) are plotted along with solid lines that show the theoretical size of pure iron, rock, water, and hydrogen planets with increasing mass. Masses are given in multiples of Earth\u2019s mass. (For comparison, Jupiter contains enough mass to make 320 Earths.) The green triangles indicate planets in our solar system. At lower masses, notice that as the mass of these hypothetical planets increases, the radius also increases. That makes sense\u2014if you were building a model of a planet out of clay, your toy planet would increase in size as you added more clay. However, for the highest mass planets (M > 1000 MEarth) in Figure 21.25, notice that the radius stops increasing and the planets with greater mass are actually smaller. This occurs because increasing the mass also increases the gravity of the planet, so that compressible materials (even rock is compressible) will become more tightly packed, shrinking the size of the more massive planet. In reality, planets are not pure compositions like the hypothetical water or iron planet. Earth is composed of a solid iron core, an outer liquid-iron core, a rocky mantle and crust, and a relatively thin atmospheric layer. Exoplanets are similarly likely to be differentiated into compositional layers. The theoretical lines in Figure 21.25 are simply guides that suggest a range of possible compositions. Astronomers who work on the complex modeling of the interiors of rocky planets make the simplifying assumption that the planet consists of two or three layers. This is not perfect, but it is a reasonable 754 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System approximation and another good example of how science works. Often, the first step in understanding something new is to narrow down the range of possibilities. This sets the stage for refining and deepening our knowledge. In Figure 21.25, the two green triangles with roughly 1 MEarth and 1 REarth represent Venus and Earth. Notice that these planets fall between the models for a pure iron and a pure rock planet, consistent with what we would expect for the known mixed-chemical composition of Venus and Earth. In the case of gaseous planets, the situation is more complex. Hydrogen is the lightest element in the periodic table, yet many of the detected exoplanets in Figure 21.25 with masses greater than 100 MEarth have radii that suggest they are lower in density than a pure hydrogen planet. Hydrogen is the lightest element, so what is happening here? Why do some gas giant planets have inflated radii that are larger than the fictitious pure hydrogen planet? Many of these planets", " reside in short-period orbits close to the host star where they intercept a significant amount of radiated energy. If this energy is trapped deep in the planet atmosphere, it can cause the planet to expand. Planets that orbit close to their host stars in slightly eccentric orbits have another source of energy: the star will raise tides in these planets that tend to circularize the orbits. This process also results in tidal dissipation of energy that can inflate the atmosphere. It would be interesting to measure the size of gas giant planets in wider orbits where the planets should be cooler\u2014the expectation is that unless they are very young, these cooler gas giant exoplanets (sometimes called \u201ccold Jupiters\u201d) should not be inflated. But we don\u2019t yet have data on these more distant exoplanets. Exoplanetary Systems As we search for exoplanets, we don\u2019t expect to find only one planet per star. Our solar system has eight major planets, half a dozen dwarf planets, and millions of smaller objects orbiting the Sun. The evidence we have of planetary systems in formation also suggest that they are likely to produce multi-planet systems. The first planetary system was found around the star Upsilon Andromedae in 1999 using the Doppler method, and many others have been found since then (about 2600 as of 2016). If such exoplanetary system are common, let\u2019s consider which systems we expect to find in the Kepler transit data. A planet will transit its star only if Earth lies in the plane of the planet\u2019s orbit. If the planets in other systems do not have orbits in the same plane, we are unlikely to see multiple transiting objects. Also, as we have noted before, Kepler was sensitive only to planets with orbital periods less than about 4 years. What we expect from Kepler data, then, is evidence of coplanar planetary systems confined to what would be the realm of the terrestrial planets in our solar system. In fact, today we have data on about 2600 such exoplanet systems. Many have only two known planets, but a few have as many as five. For the most part, these are very compact systems with most of their planets closer to their star than Mercury is to the Sun. The figure below shows one of the largest exoplanet systems: that of the star called Kepler-62 (Figure 21.26). Our solar system is shown to the same scale, for comparison. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 755 Figure 21.26. Exoplanet System Kepler-62, with the Solar System Shown to the Same Scale. The green areas are the \u201chabitable zones,\u201d the range of distance from the star where surface temperatures are likely to be consistent with liquid water. (credit: modification of work by NASA/ Ames/JPL-Caltech) All but one of the planets in the K-62 system are larger than Earth. These are super-Earths, and one of them (62d) is in the size range of a mini-Neptune, where it is likely to be largely gaseous. The smallest planet in this system is about the size of Mars. The three inner planets orbit very close to their star, and only the outer two have orbits larger than Mercury in our system. The green areas represent each star\u2019s \u201chabitable zone,\u201d which is the distance from the star where we calculate that surface temperatures would be consistent with liquid water. The Kepler-62 habitable zone is much smaller than that of the Sun because the star is intrinsically fainter. With closely spaced systems like this, the planets can interact gravitationally with each other. The result is that the observed transits occur a few minutes earlier or later than would be predicted from simple orbits. These gravitational interactions have allowed the Kepler scientists to calculate masses for the planets, providing another way to learn about exoplanets. Kepler has discovered some interesting and unusual planetary systems. For example, most astronomers expected planets to be limited to single stars. But we have found planets orbiting close double stars, so that the planet would see two suns in its sky, like those of the fictional planet Tatooine in the Star Wars films. At the opposite extreme, planets can orbit one star of a wide, double-star system without major interference from the second star. 21.6 NEW PERSPECTIVES ON PLANET FORMATION Learning Objectives By the end of this section, you will be able to: 756 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System Explain how exoplanet discoveries have revised our understanding of planet formation Discuss how planetary systems quite different from our solar system might have come about Traditionally, astronomers have assumed that the planets in our solar system formed at about their current distances from the", " Sun and have remained there ever since. The first step in the formation of a giant planet is to build up a solid core, which happens when planetesimals collide and stick. Eventually, this core becomes massive enough to begin sweeping up gaseous material in the disk, thereby building the gas giants Jupiter and Saturn. How to Make a Hot Jupiter The traditional model for the formation of planets works only if the giant planets are formed far from the central star (about 5\u201310 AU), where the disk is cold enough to have a fairly high density of solid matter. It cannot explain the hot Jupiters, which are located very close to their stars where any rocky raw material would be completely vaporized. It also cannot explain the elliptical orbits we observe for some exoplanets because the orbit of a protoplanet, whatever its initial shape, will quickly become circular through interactions with the surrounding disk of material and will remain that way as the planet grows by sweeping up additional matter. So we have two options: either we find a new model for forming planets close to the searing heat of the parent star, or we find a way to change the orbits of planets so that cold Jupiters can travel inward after they form. Most research now supports the latter explanation. Calculations show that if a planet forms while a substantial amount of gas remains in the disk, then some of the planet\u2019s orbital angular momentum can be transferred to the disk. As it loses momentum (through a process that reminds us of the effects of friction), the planet will spiral inward. This process can transport giant planets, initially formed in cold regions of the disk, closer to the central star\u2014thereby producing hot Jupiters. Gravitational interactions between planets in the chaotic early solar system can also cause planets to slingshot inward from large distances. But for this to work, the other planet has to carry away the angular momentum and move to a more distant orbit. In some cases, we can use the combination of transit plus Doppler measurements to determine whether the planets orbit in the same plane and in the same direction as the star. For the first few cases, things seemed to work just as we anticipated: like the solar system, the gas giant planets orbited in their star\u2019s equatorial plane and in the same direction as the spinning star. Then, some startling discoveries were made of gas giant planets that orbited at right angles or even in the opposite sense as the spin of the star. How could this happen? Again, there must have been interactions between planets. It\u2019s possible that before the system settled down, two planets came close together, so that one was kicked into an usual orbit. Or perhaps a passing star perturbed the system after the planets were newly formed. Forming Planetary Systems When the Milky Way Galaxy was young, the stars that formed did not contain many heavy elements like iron. Several generations of star formation and star death were required to enrich the interstellar medium for subsequent generations of stars. Since planets seem to form \u201cinside out,\u201d starting with the accretion of the materials that can make the rocky cores with which planets start, astronomers wondered when in the history of the Galaxy, planet formation would turn on. The star Kepler-444 has shed some light on this question. This is a tightly packed system of five planets\u2014the smallest comparable in size to Mercury and the largest similar in size to Venus. All five planets were detected with the Kepler spacecraft as they transited their parent star. All five planets orbit their host star in less than the This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 757 time it takes Mercury to complete one orbit about the Sun. Remarkably, the host star Kepler-444 is more than 11 billion years old and formed when the Milky Way was only 2 billion years old. So the heavier elements needed to make rocky planets must have already been available then. This ancient planetary system sets the clock on the beginning of rocky planet formation to be relatively soon after the formation of our Galaxy. Kepler data demonstrate that while rocky planets inside Mercury\u2019s orbit are missing from our solar system, they are common around other stars, like Kepler-444. When the first systems packed with close-in rocky planets were discovered, we wondered why they were so different from our solar system. When many such systems were discovered, we began to wonder if it was our solar system that was different. This led to speculation that additional rocky planets might once have existed close to the Sun in our solar system. There is some evidence from the motions in the outer solar system that Jupiter may have migrated inward long ago. If correct, then gravitational perturbations from Jupiter could have dislodged the orbits of close-in rocky planets, causing them to fall into the Sun. Consistent with this picture, astronomers now think that Uranus and Neptune", " probably did not form at their present distances from the Sun but rather closer to where Jupiter and Saturn are now. The reason for this idea is that density in the disk of matter surrounding the Sun at the time the planets formed was so low outside the orbit of Saturn that it would take several billion years to build up Uranus and Neptune. Yet we saw earlier in the chapter that the disks around protostars survive only a few million years. Therefore, scientists have developed computer models demonstrating that Uranus and Neptune could have formed near the current locations of Jupiter and Saturn, and then been kicked out to larger distances through gravitational interactions with their neighbors. All these wonderful new observations illustrate how dangerous it can be to draw conclusions about a phenomenon in science (in this case, how planetary systems form and arrange themselves) when you are only working with a single example. Exoplanets have given rise to a new picture of planetary system formation\u2014one that is much more chaotic than we originally thought. If we think of the planets as being like skaters in a rink, our original model (with only our own solar system as a guide) assumed that the planets behaved like polite skaters, all obeying the rules of the rink and all moving in nearly the same direction, following roughly circular paths. The new picture corresponds more to a roller derby, where the skaters crash into one another, change directions, and sometimes are thrown entirely out of the rink. Habitable Exoplanets While thousands of exoplanets have been discovered in the past two decades, every observational technique has fallen short of finding more than a few candidates that resemble Earth (Figure 21.27). Astronomers are not sure exactly what properties would define another Earth. Do we need to find a planet that is exactly the same size and mass as Earth? That may be difficult and may not be important from the perspective of habitability. After all, we have no reason to think that life could not have arisen on Earth if our planet had been a little bit smaller or larger. And, remember that how habitable a planet is depends on both its distance from its star and the nature of its atmosphere. The greenhouse effect can make some planets warmer (as it did for Venus and is doing more and more for Earth). 758 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System Figure 21.27. Many Earthlike Planets. This painting, commissioned by NASA, conveys the idea that there may be many planets resembling Earth out there as our methods for finding them improve. (credit: NASA/JPL-Caltech/R. Hurt (SSC-Caltech)) We can ask other questions to which we don\u2019t yet know the answers. Does this \u201ctwin\u201d of Earth need to orbit a solar-type star, or can we consider as candidates the numerous exoplanets orbiting K- and M-class stars? (In the summer of 2016, astronomers reported the discovery of a planet with at least 1.3 times the mass of Earth around the nearest star, Proxima Centauri, which is spectral type M and located 4.2 light years from us.) We have a special interest in finding planets that could support life like ours, in which case, we need to find exoplanets within their star\u2019s habitable zone, where surface temperatures are consistent with liquid water on the surface. This is probably the most important characteristic defining an Earth-analog exoplanet. The search for potentially habitable worlds is one of the prime drivers for exoplanet research in the next decade. Astronomers are beginning to develop realistic plans for new instruments that can even look for signs of life on distant worlds (examining their atmospheres for gases associated with life, for example). If we require telescopes in space to find such worlds, we need to recognize that years are required to plan, build, and launch such space observatories. The discovery of exoplanets and the knowledge that most stars have planetary systems are transforming our thinking about life beyond Earth. We are closer than ever to knowing whether habitable (and inhabited) planets are common. This work lends a new spirit of optimism to the search for life elsewhere, a subject to which we will return in Life in the Universe Check out the habitability of various stars and planets by trying out the interactive Circumstellar Habitable Zone Simulator (https://openstaxcollege.org/l/30cirhabzonsim) and select a star system to investigate. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 759 CHAPTER 21 REVIEW KEY TERMS exoplanet a planet orbiting a star other than our Sun giant molecular clouds masses of 105 solar masses; found in the spiral arms of galaxies, these clouds are where stars form large, cold interstellar clouds with diameters of dozens of light-years and typical Herbig-Haro (HH)", " object luminous knots of gas in an area of star formation that are set to glow by jets of material from a protostar mini-Neptune a planet that is intermediate between the largest terrestrial planet in our solar system (Earth) and the smallest jovian planet (Neptune); generally, mini-Neptunes have sizes between 2.8 and 4 times Earth\u2019s size protostar a very young star still in the process of formation, before nuclear fusion begins stellar wind the outflow of gas, sometimes at speeds as high as hundreds of kilometers per second, from a star super-Earth a planet larger than Earth, generally between 1.4 and 2.8 times the size of our planet transit when one astronomical object moves in front of another SUMMARY 21.1 Star Formation Most stars form in giant molecular clouds with masses as large as 3 \u00d7 106 solar masses. The most well-studied molecular cloud is Orion, where star formation is currently taking place. Molecular clouds typically contain regions of higher density called clumps, which in turn contain several even-denser cores of gas and dust, each of which may become a star. A star can form inside a core if its density is high enough that gravity can overwhelm the internal pressure and cause the gas and dust to collapse. The accumulation of material halts when a protostar develops a strong stellar wind, leading to jets of material being observed coming from the star. These jets of material can collide with the material around the star and produce regions that emit light that are known as Herbig-Haro objects. 21.2 The H\u2013R Diagram and the Study of Stellar Evolution The evolution of a star can be described in terms of changes in its temperature and luminosity, which can best be followed by plotting them on an H\u2013R diagram. Protostars generate energy (and internal heat) through gravitational contraction that typically continues for millions of years, until the star reaches the main sequence. 21.3 Evidence That Planets Form around Other Stars Observational evidence shows that most protostars are surrounded by disks with large-enough diameters and enough mass (as much as 10% that of the Sun) to form planets. After a few million years, the inner part of the disk is cleared of dust, and the disk is then shaped like a donut with the protostar centered in the hole\u2014something that can be explained by the formation of planets in that inner zone. Around a few older stars, we see disks formed from the debris produced when small bodies (comets and asteroids) collide with each other. The distribution of material in the rings of debris disks is probably determined by shepherd planets, just as Saturn\u2019s shepherd moons affect the orbits of the material in its rings. Protoplanets that grow to be 10 times 760 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System the mass of Earth or bigger while there is still considerable gas in their disk can then capture more of that gas and become giant planets like Jupiter in the solar system. 21.4 Planets beyond the Solar System: Search and Discovery Several observational techniques have successfully detected planets orbiting other stars. These techniques fall into two general categories\u2014direct and indirect detection. The Doppler and transit techniques are our most powerful indirect tools for finding exoplanets. Some planets are also being found by direct imaging. 21.5 Exoplanets Everywhere: What We Are Learning Although the Kepler mission is finding thousands of new exoplanets, these are limited to orbital periods of less than 400 days and sizes larger than Mars. Still, we can use the Kepler discoveries to extrapolate the distribution of planets in our Galaxy. The data so far imply that planets like Earth are the most common type of planet, and that there may be 100 billion Earth-size planets around Sun-like stars in the Galaxy. About 2600 planetary systems have been discovered around other stars. In many of them, planets are arranged differently than in our solar system. 21.6 New Perspectives on Planet Formation The ensemble of exoplanets is incredibly diverse and has led to a revision in our understanding of planet formation that includes the possibility of vigorous, chaotic interactions, with planet migration and scattering. It is possible that the solar system is unusual (and not representative) in how its planets are arranged. Many systems seem to have rocky planets farther inward than we do, for example, and some even have \u201chot Jupiters\u201d very close to their star. Ambitious space experiments should make it possible to image earthlike planets outside the solar system and even to obtain information about their habitability as we search for life elsewhere. FOR FURTHER EXPLORATION Articles Star Formation Blaes, O. \u201cA Universe of Disks.\u201d Scientific American (October 2004): 48. On accretion disks and jets around young stars and black holes. Croswell, K. \u201cThe Dust Belt Next Door [Tau Ceti].\u201d Scientific", " American (January 2015): 24. Short intro to recent observations of planets and a wide dust belt. Frank, A. \u201cStarmaker: The New Story of Stellar Birth.\u201d Astronomy (July 1996): 52. Jayawardhana, R. \u201cSpying on Stellar Nurseries.\u201d Astronomy (November 1998): 62. On protoplanetary disks. O\u2019Dell, C. R. \u201cExploring the Orion Nebula.\u201d Sky & Telescope (December 1994): 20. Good review with Hubble results. Ray, T. \u201cFountains of Youth: Early Days in the Life of a Star.\u201d Scientific American (August 2000): 42. On outflows from young stars. Young, E. \u201cCloudy with a Chance of Stars.\u201d Scientific American (February 2010): 34. On how clouds of interstellar matter turn into star systems. Young, Monica \u201cMaking Massive Stars.\u201d Sky & Telescope (October 2015): 24. Models and observations on how the most massive stars form. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 761 Exoplanets Billings, L. \u201cIn Search of Alien Jupiters.\u201d Scientific American (August 2015): 40\u201347. The race to image jovian planets with current instruments and why a direct image of a terrestrial planet is still in the future. Heller, R. \u201cBetter Than Earth.\u201d Scientific American (January 2015): 32\u201339. What kinds of planets may be habitable; super-Earths and jovian planet moons should also be considered. Laughlin, G. \u201cHow Worlds Get Out of Whack.\u201d Sky & Telescope (May 2013): 26. On how planets can migrate from the places they form in a star system. Marcy, G. \u201cThe New Search for Distant Planets.\u201d Astronomy (October 2006): 30. Fine brief overview. (The same issue has a dramatic fold-out visual atlas of extrasolar planets, from that era.) Redd, N. \u201cWhy Haven\u2019t We Found Another Earth?\u201d Astronomy (February 2016): 25. Looking for terrestrial planets in the habitable zone with evidence of life. Seager, S. \u201cExoplanets Everywhere.\u201d Sky & Telescope (August 2013): 18. An excellent discussion of some of the frequently asked questions about the nature and arrangement of planets out there. Seager, S. \u201cThe Hunt for Super-Earths.\u201d Sky & Telescope (October 2010): 30. The search for planets that are up to 10 times the mass of Earth and what they can teach us. Villard, R. \u201cHunting for Earthlike Planets.\u201d Astronomy (April 2011): 28. How we expect to find and characterize super-Earth (planets somewhat bigger than ours) using new instruments and techniques that could show us what their atmospheres are made of. Websites Exoplanet Exploration: http://planetquest.jpl.nasa.gov/. PlanetQuest (from the Navigator Program at the Jet Propulsion Lab) is probably the best site for students and beginners, with introductory materials and nice illustrations; it focuses mostly on NASA work and missions. Exoplanets: http://www.planetary.org/exoplanets/. Planetary Society\u2019s exoplanets pages with a dynamic catalog of planets found and good explanations. Exoplanets: The Search for Planets beyond Our Solar System: http://www.iop.org/publications/iop/2010/ page_42551.html. From the British Institute of Physics in 2010. Extrasolar Planets Encyclopedia: http://exoplanet.eu/. Maintained by Jean Schneider of the Paris Observatory, has the largest catalog of planet discoveries and useful background material (some of it more technical). Formation of Stars: https://www.spacetelescope.org/science/formation_of_stars/. Star Formation page from the Hubble Space Telescope, with links to images and information. Kepler Mission: http://kepler.nasa.gov/. The public website for the remarkable telescope in space that is searching planets using the transit technique and is our best hope for finding earthlike planets. Proxima Centauri Planet Discovery: http://www.eso.org/public/news/eso1629/. Apps Exoplanet: http://itunes.apple.com/us/app/exoplanet/id327702034?mt=8. Allows you to browse through a regularly updated visual catalog of exoplanets that have been found so far. Journey to the Exoplanets: http://itunes.apple.com/us/app/journey-to-the-exoplanets/", "id463532472?mt=8. Produced by the staff of Scientific American, with input from scientists and space artists; gives background information and visual tours of the nearer star systems with planets. 762 Videos Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System A Star Is Born: http://www.discovery.com/tv-shows/other-shows/videos/how-the-universe-works-a-star-is- born/. Discovery Channel video with astronomer Michelle Thaller (2:25). Are We Alone: An Evening Dialogue with the Kepler Mission Leaders: http://www.youtube.com/ watch?v=O7ItAXfl0Lw. A non-technical panel discussion on Kepler results and ideas about planet formation with Bill Borucki, Natalie Batalha, and Gibor Basri (moderated by Andrew Fraknoi) at the University of California, Berkeley (2:07:01). Finding the Next Earth: The Latest Results from Kepler: https://www.youtube.com/watch?v=ZbijeR_AALo. Natalie Batalha (San Jose State University & NASA Ames) public talk in the Silicon Valley Astronomy Lecture Series (1:28:38). From Hot Jupiters to Habitable Worlds: https://vimeo.com/37696087 (Part 1) and https://vimeo.com/37700700 (Part 2). Debra Fischer (Yale University) public talk in Hawaii sponsored by the Keck Observatory (15:20 Part 1, 21:32 Part 2). Search for Habitable Exoplanets: http://www.youtube.com/watch?v=RLWb_T9yaDU. Sara Seeger (MIT) public talk at the SETI Institute, with Kepler results (1:10:35). Strange Planetary Vistas: http://www.youtube.com/watch?v=_8ww9eLRSCg. Josh Carter (CfA) public talk at Harvard\u2019s Center for Astrophysics with a friendly introduction to exoplanets for non-specialists (46:35). COLLABORATIVE GROUP ACTIVITIES A. Your group is a subcommittee of scientists examining whether any of the \u201chot Jupiters\u201d (giant planets closer to their stars than Mercury is to the Sun) could have life on or near them. Can you come up with places on, in, or near such planets where life could develop or where some forms of life might survive? B. A wealthy couple (who are alumni of your college or university and love babies) leaves the astronomy program several million dollars in their will, to spend in the best way possible to search for \u201cinfant stars in our section of the Galaxy.\u201d Your group has been assigned the task of advising the dean on how best to spend the money. What kind of instruments and search programs would you recommend, and why? C. Some people consider the discovery of any planets (even hot Jupiters) around other stars one of the most important events in the history of astronomical research. Some astronomers have been surprised that the public is not more excited about the planet discoveries. One reason that has been suggested for this lack of public surprise and excitement is that science fiction stories have long prepared us for there being planets around other stars. (The Starship Enterprise on the 1960s Star Trek TV series found some in just about every weekly episode.) What does your group think? Did you know about the discovery of planets around other stars before taking this course? Do you consider it exciting? Were you surprised to hear about it? Are science fiction movies and books good or bad tools for astronomy education in general, do you think? D. What if future space instruments reveal an earthlike exoplanet with significant amounts of oxygen and methane in its atmosphere? Suppose the planet and its star are 50 light-years away. What does your group suggest astronomers do next? How much effort and money would you recommend be put into finding out more about this planet and why? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 763 E. Discuss with your group the following question: which is easier to find orbiting a star with instruments we have today: a jovian planet or a proto-planetary disk? Make a list of arguments for each side of this question. F. (This activity should be done when your group has access to the internet.) Go to the page which indexes all the publicly released Hubble Space Telescope images by subject: http://hubblesite.org/newscenter/ archive/browse/image/. Under \u201cStar,\u201d go to \u201cProtoplanetary Disk\u201d and find a system\u2014not mentioned in this", " chapter\u2014that your group likes, and prepare a short report to the class about why you find it interesting. Then, under \u201cNebula,\u201d go to \u201cEmission\u201d and find a region of star formation not mentioned in this chapter, and prepare a short report to the class about what you find interesting about it. G. There is a \u201ccitizen science\u201d website called Planet Hunters (http://www.planethunters.org/) where you can participate in identifying exoplanets from the data that Kepler provided. Your group should access the site, work together to use it, and classify two light curves. Report back to the class on what you have done. H. Yuri Milner, a Russian-American billionaire, recently pledged $100 million to develop the technology to send many miniaturized probes to a star in the Alpha Centauri triple star system (which includes Proxima Centauri, the nearest star to us, now known to have at least one planet.) Each tiny probe will be propelled by powerful lasers at 20% the speed of light, in the hope that one or more might arrive safely and be able to send back information about what it\u2019s like there. Your group should search online for more information about this project (called \u201cBreakthrough: Starshot\u201d) and discuss your reactions to this project. Give specific reasons for your arguments. EXERCISES Review Questions 1. Give several reasons the Orion molecular cloud is such a useful \u201claboratory\u201d for studying the stages of star formation. 2. Why is star formation more likely to occur in cold molecular clouds than in regions where the temperature of the interstellar medium is several hundred thousand degrees? 3. Why have we learned a lot about star formation since the invention of detectors sensitive to infrared radiation? 4. Describe what happens when a star forms. Begin with a dense core of material in a molecular cloud and trace the evolution up to the time the newly formed star reaches the main sequence. 5. Describe how the T Tauri star stage in the life of a low-mass star can lead to the formation of a Herbig- Haro (H-H) object. 6. Look at the four stages shown in Figure 21.8. In which stage(s) can we see the star in visible light? In infrared radiation? 7. The evolutionary track for a star of 1 solar mass remains nearly vertical in the H\u2013R diagram for a while (see Figure 21.12). How is its luminosity changing during this time? Its temperature? Its radius? 8. Two protostars, one 10 times the mass of the Sun and one half the mass of the Sun are born at the same time in a molecular cloud. Which one will be first to reach the main sequence stage, where it is stable and getting energy from fusion? 764 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 9. Compare the scale (size) of a typical dusty disk around a forming star with the scale of our solar system. 10. Why is it so hard to see planets around other stars and so easy to see them around our own? 11. Why did it take astronomers until 1995 to discover the first exoplanet orbiting another star like the Sun? 12. Which types of planets are most easily detected by Doppler measurements? By transits? 13. List three ways in which the exoplanets we have detected have been found to be different from planets in our solar system. 14. List any similarities between discovered exoplanets and planets in our solar system. 15. What revisions to the theory of planet formation have astronomers had to make as a result of the discovery of exoplanets? 16. Why are young Jupiters easier to see with direct imaging than old Jupiters? Thought Questions 17. A friend of yours who did not do well in her astronomy class tells you that she believes all stars are old and none could possibly be born today. What arguments would you use to persuade her that stars are being born somewhere in the Galaxy during your lifetime? 18. Observations suggest that it takes more than 3 million years for the dust to begin clearing out of the inner regions of the disks surrounding protostars. Suppose this is the minimum time required to form a planet. Would you expect to find a planet around a 10-MSun star? (Refer to Figure 21.12.) 19. Suppose you wanted to observe a planet around another star with direct imaging. Would you try to observe in visible light or in the infrared? Why? Would the planet be easier to see if it were at 1 AU or 5 AU from its star? 20. Why were giant planets close to their stars the first ones to be discovered? Why has the same technique not been used yet to discover giant planets at the distance of Saturn? 21. Exoplanets in eccentric orbits experience large temperature swings during their orbits. Suppose you had to plan for a mission to such a", " planet. Based on Kepler\u2019s second law, does the planet spend more time closer or farther from the star? Explain. Figuring For Yourself 22. When astronomers found the first giant planets with orbits of only a few days, they did not know whether those planets were gaseous and liquid like Jupiter or rocky like Mercury. The observations of HD 209458 settled this question because observations of the transit of the star by this planet made it possible to determine the radius of the planet. Use the data given in the text to estimate the density of this planet, and then use that information to explain why it must be a gas giant. 23. An exoplanetary system has two known planets. Planet X orbits in 290 days and Planet Y orbits in 145 days. Which planet is closest to its host star? If the star has the same mass as the Sun, what is the semi-major axis of the orbits for Planets X and Y? 24. Kepler\u2019s third law says that the orbital period (in years) is proportional to the square root of the cube of the mean distance (in AU) from the Sun (P \u221d a1.5). For mean distances from 0.1 to 32 AU, calculate and plot a curve showing the expected Keplerian period. For each planet in our solar system, look up the mean distance from the Sun in AU and the orbital period in years and overplot these data on the theoretical Keplerian curve. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System 765 25. Calculate the transit depth for an M dwarf star that is 0.3 times the radius of the Sun with a gas giant planet the size of Jupiter. 26. If a transit depth of 0.00001 can be detected with the Kepler spacecraft, what is the smallest planet that could be detected around a 0.3 Rsun M dwarf star? 27. What fraction of gas giant planets seems to have inflated radii? 766 Chapter 21 The Birth of Stars and the Discovery of Planets outside the Solar System This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 767 22 STARS FROM ADOLESCENCE TO OLD AGE Figure 22.1. Ant Nebula. During the later phases of stellar evolution, stars expel some of their mass, which returns to the interstellar medium to form new stars. This Hubble Space Telescope image shows a star losing mass. Known as Menzel 3, or the Ant Nebula, this beautiful region of expelled gas is about 3000 light-years away from the Sun. We see a central star that has ejected mass preferentially in two opposite directions. The object is about 1.6 light-years long. The image is color coded\u2014red corresponds to an emission line of sulfur, green to nitrogen, blue to hydrogen, and blue/violet to oxygen. (credit: modification of work by NASA, ESA and The Hubble Heritage Team (STScI/AURA)) Chapter Outline 22.1 Evolution from the Main Sequence to Red Giants 22.2 Star Clusters 22.3 Checking Out the Theory 22.4 Further Evolution of Stars 22.5 The Evolution of More Massive Stars Thinking Ahead The Sun and other stars cannot last forever. Eventually they will exhaust their nuclear fuel and cease to shine. But how do they change during their long lifetimes? And what do these changes mean for the future of Earth? We now turn from the birth of stars to the rest of their life stories. This is not an easy task since stars live much longer than astronomers. Thus, we cannot hope to see the life story of any single star unfold before our eyes or telescopes. To learn about their lives, we must survey as many of the stellar inhabitants of the Galaxy as possible. With thoroughness and a little luck, we can catch at least a few of them in each stage of their lives. As you\u2019ve learned, stars have many different characteristics, with the differences sometimes resulting from their different masses, temperatures, and luminosities, and at other times derived from changes that occur as they age. Through a combination of observation and theory, we can use these differences to piece together the life story of a star. 768 Chapter 22 Stars from Adolescence to Old Age 22.1 EVOLUTION FROM THE MAIN SEQUENCE TO RED GIANTS Learning Objectives By the end of this section, you will be able to: Explain the zero-age main sequence Describe what happens to main-sequence stars of various masses as they exhaust their hydrogen supply One of the best ways to get a \u201csnapshot\u201d of a group of stars is by plotting their properties on an H\u2013R diagram. We have already used the H\u2013R diagram to follow the evolution of prot", "ostars up to the time they reach the main sequence. Now we\u2019ll see what happens next. Once a star has reached the main-sequence stage of its life, it derives its energy almost entirely from the conversion of hydrogen to helium via the process of nuclear fusion in its core (see The Sun: A Nuclear Powerhouse). Since hydrogen is the most abundant element in stars, this process can maintain the star\u2019s equilibrium for a long time. Thus, all stars remain on the main sequence for most of their lives. Some astronomers like to call the main-sequence phase the star\u2019s \u201cprolonged adolescence\u201d or \u201cadulthood\u201d (continuing our analogy to the stages in a human life). The left-hand edge of the main-sequence band in the H\u2013R diagram is called the zero-age main sequence (see Figure 18.15). We use the term zero-age to mark the time when a star stops contracting, settles onto the main sequence, and begins to fuse hydrogen in its core. The zero-age main sequence is a continuous line in the H\u2013R diagram that shows where stars of different masses but similar chemical composition can be found when they begin to fuse hydrogen. Since only 0.7% of the hydrogen used in fusion reactions is converted into energy, fusion does not change the total mass of the star appreciably during this long period. It does, however, change the chemical composition in its central regions where nuclear reactions occur: hydrogen is gradually depleted, and helium accumulates. This change of composition changes the luminosity, temperature, size, and interior structure of the star. When a star\u2019s luminosity and temperature begin to change, the point that represents the star on the H\u2013R diagram moves away from the zero-age main sequence. Calculations show that the temperature and density in the inner region slowly increase as helium accumulates in the center of a star. As the temperature gets hotter, each proton acquires more energy of motion on average; this means it is more likely to interact with other protons, and as a result, the rate of fusion also increases. For the proton-proton cycle described in The Sun: A Nuclear Powerhouse, the rate of fusion goes up roughly as the temperature to the fourth power. If the rate of fusion goes up, the rate at which energy is being generated also increases, and the luminosity of the star gradually rises. Initially, however, these changes are small, and stars remain within the main-sequence band on the H\u2013R diagram for most of their lifetimes Star Temperature and Rate of Fusion If a star\u2019s temperature were to double, by what factor would its rate of fusion increase? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 769 Solution Since the rate of fusion (like temperature) goes up to the fourth power, it would increase by a factor of 24, or 16 times. Check Your Learning If the rate of fusion of a star increased 256 times, by what factor would the temperature increase? Answer: The temperature would increase by a factor of 2560.25 (that is, the 4th root of 256), or 4 times. Lifetimes on the Main Sequence How many years a star remains in the main-sequence band depends on its mass. You might think that a more massive star, having more fuel, would last longer, but it\u2019s not that simple. The lifetime of a star in a particular stage of evolution depends on how much nuclear fuel it has and on how quickly it uses up that fuel. (In the same way, how long people can keep spending money depends not only on how much money they have but also on how quickly they spend it. This is why many lottery winners who go on spending sprees quickly wind up poor again.) In the case of stars, more massive ones use up their fuel much more quickly than stars of low mass. The reason massive stars are such spendthrifts is that, as we saw above, the rate of fusion depends very strongly on the star\u2019s core temperature. And what determines how hot a star\u2019s central regions get? It is the mass of the star\u2014the weight of the overlying layers determines how high the pressure in the core must be: higher mass requires higher pressure to balance it. Higher pressure, in turn, is produced by higher temperature. The higher the temperature in the central regions, the faster the star races through its storehouse of central hydrogen. Although massive stars have more fuel, they burn it so prodigiously that their lifetimes are much shorter than those of their low-mass counterparts. You can also understand now why the most massive main-sequence stars are also the most luminous. Like new rock stars with their first platinum album, they spend their resources at an astounding rate. The main-sequence lifetimes of stars of", " different masses are listed in Table 22.1. This table shows that the most massive stars spend only a few million years on the main sequence. A star of 1 solar mass remains there for roughly 10 billion years, while a star of about 0.4 solar mass has a main-sequence lifetime of some 200 billion years, which is longer than the current age of the universe. (Bear in mind, however, that every star spends most of its total lifetime on the main sequence. Stars devote an average of 90% of their lives to peacefully fusing hydrogen into helium.) Lifetimes of Main-Sequence Stars Spectral Type Surface Temperature (K) Mass (Mass of Sun = 1) Lifetime on Main Sequence (years) 54,000 29,200 40 16 1 million 10 million O5 B0 Table 22.1 770 Chapter 22 Stars from Adolescence to Old Age Lifetimes of Main-Sequence Stars Spectral Type Surface Temperature (K) Mass (Mass of Sun = 1) Lifetime on Main Sequence (years) 9600 7350 6050 5240 3750 A0 F0 G0 K0 M0 Table 22.1 3.3 1.7 1.1 0.8 0.4 500 million 2.7 billion 9 billion 14 billion 200 billion These results are not merely of academic interest. Human beings developed on a planet around a G-type star. This means that the Sun\u2019s stable main-sequence lifetime is so long that it afforded life on Earth plenty of time to evolve. When searching for intelligent life like our own on planets around other stars, it would be a pretty big waste of time to search around O- or B-type stars. These stars remain stable for such a short time that the development of creatures complicated enough to take astronomy courses is very unlikely. From Main-Sequence Star to Red Giant Eventually, all the hydrogen in a star\u2019s core, where it is hot enough for fusion reactions, is used up. The core then contains only helium, \u201ccontaminated\u201d by whatever small percentage of heavier elements the star had to begin with. The helium in the core can be thought of as the accumulated \u201cash\u201d from the nuclear \u201cburning\u201d of hydrogen during the main-sequence stage. Energy can no longer be generated by hydrogen fusion in the stellar core because the hydrogen is all gone and, as we will see, the fusion of helium requires much higher temperatures. Since the central temperature is not yet high enough to fuse helium, there is no nuclear energy source to supply heat to the central region of the star. The long period of stability now ends, gravity again takes over, and the core begins to contract. Once more, the star\u2019s energy is partially supplied by gravitational energy, in the way described by Kelvin and Helmholtz (see Sources of Sunshine: Thermal and Gravitational Energy). As the star\u2019s core shrinks, the energy of the inward- falling material is converted to heat. The heat generated in this way, like all heat, flows outward to where it is a bit cooler. In the process, the heat raises the temperature of a layer of hydrogen that spent the whole long main-sequence time just outside the core. Like an understudy waiting in the wings of a hit Broadway show for a chance at fame and glory, this hydrogen was almost (but not quite) hot enough to undergo fusion and take part in the main action that sustains the star. Now, the additional heat produced by the shrinking core puts this hydrogen \u201cover the limit,\u201d and a shell of hydrogen nuclei just outside the core becomes hot enough for hydrogen fusion to begin. New energy produced by fusion of this hydrogen now pours outward from this shell and begins to heat up layers of the star farther out, causing them to expand. Meanwhile, the helium core continues to contract, producing more heat right around it. This leads to more fusion in the shell of fresh hydrogen outside the core (Figure 22.2). The additional fusion produces still more energy, which also flows out into the upper layer of the star. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 771 Figure 22.2. Star Layers during and after the Main Sequence. (a) During the main sequence, a star has a core where fusion takes place and a much larger envelope that is too cold for fusion. (b) When the hydrogen in the core is exhausted (made of helium, not hydrogen), the core is compressed by gravity and heats up. The additional heat starts hydrogen fusion in a layer just outside the core. Note that these parts of the Sun are not drawn to scale. Most stars actually generate more energy each second when they are fusing hydrogen in the shell surrounding the helium core than they did when hydrogen fusion was confined to the central part of the star; thus, they increase in luminosity. With all", " the new energy pouring outward, the outer layers of the star begin to expand, and the star eventually grows and grows until it reaches enormous proportions (Figure 22.3). Figure 22.3. Relative Sizes of Stars. This image compares the size of the Sun to that of Delta Bo\u00f6tis, a giant star, and Xi Cygni, a supergiant. Note that Xi Cygni is so large in comparison to the other two stars that only a small portion of it is visible at the top of the frame. When you take the lid off a pot of boiling water, the steam can expand and it cools down. In the same way, the expansion of a star\u2019s outer layers causes the temperature at the surface to decrease. As it cools, the star\u2019s overall color becomes redder. (We saw in Radiation and Spectra that a red color corresponds to cooler temperature.) So the star becomes simultaneously more luminous and cooler. On the H\u2013R diagram, the star therefore leaves the main-sequence band and moves upward (brighter) and to the right (cooler surface temperature). Over time, massive stars become red supergiants, and lower-mass stars like the Sun become red giants. (We first discussed such giant stars in The Stars: A Celestial Census; here we see how such \u201cswollen\u201d stars originate.) You might also say that these stars have \u201csplit personalities\u201d: their cores are contracting while their outer layers are expanding. (Note that red giant stars do not actually look deep red; their colors are more like orange or orange-red.) Just how different are these red giants and supergiants from a main-sequence star? Table 22.2 compares the Sun with the red supergiant Betelgeuse, which is visible above Orion\u2019s belt as the bright red star that marks the 772 Chapter 22 Stars from Adolescence to Old Age hunter\u2019s armpit. Relative to the Sun, this supergiant has a much larger radius, a much lower average density, a cooler surface, and a much hotter core. Comparing a Supergiant with the Sun Property Sun Betelgeuse Mass (2 \u00d7 1033 g) Radius (km) Surface temperature (K) 1 700,000 5,800 16 500,000,000 3,600 Core temperature (K) 15,000,000 160,000,000 Luminosity (4 \u00d7 1026 W) Average density (g/cm3) Age (millions of years) Table 22.2 1 1.4 4,500 46,000 1.3 \u00d7 10\u20137 10 Red giants can become so large that if we were to replace the Sun with one of them, its outer atmosphere would extend to the orbit of Mars or even beyond (Figure 22.4). This is the next stage in the life of a star as it moves (to continue our analogy to human lives) from its long period of \u201cyouth\u201d and \u201cadulthood\u201d to \u201cold age.\u201d (After all, many human beings today also see their outer layers expand a bit as they get older.) By considering the relative ages of the Sun and Betelgeuse, we can also see that the idea that \u201cbigger stars die faster\u201d is indeed true here. Betelgeuse is a mere 10 million years old, which is relatively young compared with our Sun\u2019s 4.5 billion years, but it is already nearing its death throes as a red supergiant. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 773 Figure 22.4. Betelgeuse. Betelgeuse is in the constellation Orion, the hunter; in the right image, it is marked with a yellow \u201cX\u201d near the top left. In the left image, we see it in ultraviolet with the Hubble Space Telescope, in the first direct image ever made of the surface of another star. As shown by the scale at the bottom, Betelgeuse has an extended atmosphere so large that, if it were at the center of our solar system, it would stretch past the orbit of Jupiter. (credit: Modification of work by Andrea Dupree (Harvard-Smithsonian CfA), Ronald Gilliland (STScI), NASA and ESA) Models for Evolution to the Giant Stage As we discussed earlier, astronomers can construct computer models of stars with different masses and compositions to see how stars change throughout their lives. Figure 22.5, which is based on theoretical calculations by University of Illinois astronomer Icko Iben, shows an H\u2013R diagram with several tracks of evolution from the main sequence to the giant stage. Tracks are shown for stars with different masses (from 0.5 to 15 times the mass of our Sun) and with chemical compositions similar to", " that of the Sun. The red line is the initial or zero-age main sequence. The numbers along the tracks indicate the time, in years, required for each star to reach those points in their evolution after leaving the main sequence. Once again, you can see that the more massive a star is, the more quickly it goes through each stage in its life. 774 Chapter 22 Stars from Adolescence to Old Age Figure 22.5. Evolutionary Tracks of Stars of Different Masses. The solid black lines show the predicted evolution from the main sequence through the red giant or supergiant stage on the H\u2013R diagram. Each track is labeled with the mass of the star it is describing. The numbers show how many years each star takes to become a giant after leaving the main sequence. The red line is the zero-age main sequence. Note that the most massive star in this diagram has a mass similar to that of Betelgeuse, and so its evolutionary track shows approximately the history of Betelgeuse. The track for a 1-solar-mass star shows that the Sun is still in the main-sequence phase of evolution, since it is only about 4.5 billion years old. It will be billions of years before the Sun begins its own \u201cclimb\u201d away from the main sequence\u2014the expansion of its outer layers that will make it a red giant. 22.2 STAR CLUSTERS Learning Objectives By the end of this section, you will be able to: Explain how star clusters help us understand the stages of stellar evolution List the different types of star clusters and describe how they differ in number of stars, structure, and age Explain why the chemical composition of globular clusters is different from that of open clusters The preceding description of stellar evolution is based on calculations. However, no star completes its main- sequence lifetime or its evolution to a red giant quickly enough for us to observe these structural changes as they happen. Fortunately, nature has provided us with an indirect way to test our calculations. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 775 Instead of observing the evolution of a single star, we can look at a group or cluster of stars. We look for a group of stars that is very close together in space, held together by gravity, often moving around a common center. Then it is reasonable to assume that the individual stars in the group all formed at nearly the same time, from the same cloud, and with the same composition. We expect that these stars will differ only in mass. And their masses determine how quickly they go through each stage of their lives. Since stars with higher masses evolve more quickly, we can find clusters in which massive stars have already completed their main-sequence phase of evolution and become red giants, while stars of lower mass in the same cluster are still on the main sequence, or even\u2014if the cluster is very young\u2014undergoing pre-main- sequence gravitational contraction. We can see many stages of stellar evolution among the members of a single cluster, and we can see whether our models can explain why the H\u2013R diagrams of clusters of different ages look the way they do. The three basic types of clusters astronomers have discovered are globular clusters, open clusters, and stellar associations. Their properties are summarized in Table 22.3. As we will see in the next section of this chapter, globular clusters contain only very old stars, whereas open clusters and associations contain young stars. Characteristics of Star Clusters Characteristic Globular Clusters Open Clusters Associations Number in the Galaxy Location in the Galaxy 150 Thousands Halo and central bulge Disk (and spiral arms) Diameter (in light-years) 50\u2013450 <30 Mass MSun 104\u2013106 102\u2013103 104\u2013106 50\u20131000 Red Red or blue 104\u2013106 102\u2013106 Number of stars Color of brightest stars Luminosity of cluster (LSun) Typical ages Thousands Spiral arms 100\u2013500 102\u2013103 102\u2013104 Blue 104\u2013107 Billions of years A few hundred million years to, in the case of unusually large clusters, more than a billion years Up to about 107 years Table 22.3 Globular Clusters Globular clusters were given this name because they are nearly symmetrical round systems of, typically, hundreds of thousands of stars. The most massive globular cluster in our own Galaxy is Omega Centauri, which 776 Chapter 22 Stars from Adolescence to Old Age is about 16,000 light-years away and contains several million stars (Figure 22.6). Note that the brightest stars in this cluster, which are red giants that have already completed the main-sequence phase of their evolution, are red-orange in color. These stars have typical surface temperatures around 4000 K. As we will see, globular clusters are among the oldest parts of our Milky Way Galaxy. Figure 22.6. Omega Centauri. (a) Located at", " about 16,000 light-years away, Omega Centauri is the most massive globular cluster in our Galaxy. It contains several million stars. (b) This image, taken with the Hubble Space Telescope, zooms in near the center of Omega Centauri. The image is about 6.3 light-years wide. The most numerous stars in the image, which are yellow-white in color, are main-sequence stars similar to our Sun. The brightest stars are red giants that have begun to exhaust their hydrogen fuel and have expanded to about 100 times the diameter of our Sun. The blue stars have started helium fusion. (credit a: modification of work by NASA, ESA and the Hubble Heritage Team (STScI/AURA); credit b: modification of work by NASA, ESA, and the Hubble SM4 ERO Team) What would it be like to live inside a globular cluster? In the dense central regions, the stars would be roughly a million times closer together than in our own neighborhood. If Earth orbited one of the inner stars in a globular cluster, the nearest stars would be light-months, not light-years, away. They would still appear as points of light, but would be brighter than any of the stars we see in our own sky. The Milky Way would probably be difficult to see through the bright haze of starlight produced by the cluster. About 150 globular clusters are known in our Galaxy. Most of them are in a spherical halo (or cloud) surrounding the flat disk formed by the majority of our Galaxy\u2019s stars. All the globular clusters are very far from the Sun, and some are found at distances of 60,000 light-years or more from the main disk of the Milky Way. The diameters of globular star clusters range from 50 light-years to more than 450 light-years. Open Clusters Open clusters are found in the disk of the Galaxy. They have a range of ages, some as old as, or even older than, our Sun. The youngest open clusters are still associated with the interstellar matter from which they formed. Open clusters are smaller than globular clusters, usually having diameters of less than 30 light-years, and they typically contain only several dozen to several hundreds of stars (Figure 22.7). The stars in open clusters usually appear well separated from one another, even in the central regions, which explains why they are called \u201copen.\u201d Our Galaxy contains thousands of open clusters, but we can see only a small fraction of them. Interstellar dust, which is also concentrated in the disk, dims the light of more distant clusters so much that they are undetectable. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 777 Figure 22.7. Jewel Box (NGC 4755). This open cluster of young, bright stars is about 6400 light-years away from the Sun. Note the contrast in color between the bright yellow supergiant and the hot blue main-sequence stars. The name comes from John Herschel\u2019s nineteenth-century description of it as \u201ca casket of variously colored precious stones.\u201d (credit: ESO/Y. Beletsky) Although the individual stars in an open cluster can survive for billions of years, they typically remain together as a cluster for only a few million years, or at most, a few hundred million years. There are several reasons for this. In small open clusters, the average speed of the member stars within the cluster may be higher than the cluster\u2019s escape velocity,[1] and the stars will gradually \u201cevaporate\u201d from the cluster. Close encounters of member stars may also increase the velocity of one of the members beyond the escape velocity. Every few hundred million years or so, the cluster may have a close encounter with a giant molecular cloud, and the gravitational force exerted by the cloud may tear the cluster apart. Several open clusters are visible to the unaided eye. Most famous among them is the Pleiades (Figure 20.13), which appears as a tiny group of six stars (some people can see even more than six, and the Pleiades is sometimes called the Seven Sisters). This cluster is arranged like a small dipping spoon and is seen in the constellation of Taurus, the bull. A good pair of binoculars shows dozens of stars in the cluster, and a telescope reveals hundreds. (A car company, Subaru, takes its name from the Japanese term for this cluster; you can see the star group on the Subaru logo.) The Hyades is another famous open cluster in Taurus. To the naked eye, it appears as a V-shaped group of faint stars marking the face of the bull. Telescopes show that Hyades actually contains more than 200 stars. Stellar Associations An association is a group of extremely young stars, typically containing 5 to 50 hot,", " bright O and B stars scattered over a region of space some 100\u2013500 light-years in diameter. As an example, most of the stars in the constellation Orion form one of the nearest stellar associations. Associations also contain hundreds to thousands of low-mass stars, but these are much fainter and less conspicuous. The presence of really hot, luminous stars indicates that star formation in the association has occurred in the last million years or so. Since O stars go through their entire lives in only about a million years, they would not still be around unless star 1 Escape velocity is the speed needed to overcome the gravity of some object or group of objects. The rockets we send up from Earth, for example, must travel faster than the escape velocity of our planet to be able to get to other worlds. 778 Chapter 22 Stars from Adolescence to Old Age formation has occurred recently. It is therefore not surprising that associations are found in regions rich in the gas and dust required to form new stars. It\u2019s like a brand new building still surrounded by some of the construction materials used to build it and with the landscape still showing signs of construction. On the other hand, because associations, like ordinary open clusters, lie in regions occupied by dusty interstellar matter, many are hidden from our view. 22.3 CHECKING OUT THE THEORY Learning Objectives By the end of this section, you will be able to: Explain how the H\u2013R diagram of a star cluster can be related to the cluster\u2019s age and the stages of evolution of its stellar members Describe how the main-sequence turnoff of a cluster reveals its age In the previous section, we indicated that that open clusters are younger than globular clusters, and associations are typically even younger. In this section, we will show how we determine the ages of these star clusters. The key observation is that the stars in these different types of clusters are found in different places in the H\u2013R diagram, and we can use their locations in the diagram in combination with theoretical calculations to estimate how long they have lived. H\u2013R Diagrams of Young Clusters What does theory predict for the H\u2013R diagram of a cluster whose stars have recently condensed from an interstellar cloud? Remember that at every stage of evolution, massive stars evolve more quickly than their lower-mass counterparts. After a few million years (\u201crecently\u201d for astronomers), the most massive stars should have completed their contraction phase and be on the main sequence, while the less massive ones should be off to the right, still on their way to the main sequence. These ideas are illustrated in Figure 22.8, which shows the H\u2013R diagram calculated by R. Kippenhahn and his associates at Munich University for a hypothetical cluster with an age of 3 million years. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 779 Figure 22.8. Young Cluster H\u2013R Diagram. We see an H\u2013R diagram for a hypothetical young cluster with an age of 3 million years. Note that the high-mass (high-luminosity) stars have already arrived at the main-sequence stage of their lives, while the lower-mass (lower-luminosity) stars are still contracting toward the zero-age main sequence (the red line) and are not yet hot enough to derive all of their energy from the fusion of hydrogen. There are real star clusters that fit this description. The first to be studied (in about 1950) was NGC 2264, which is still associated with the region of gas and dust from which it was born (Figure 22.9). Figure 22.9. Young Cluster NGC 2264. Located about 2600 light-years from us, this region of newly formed stars, known as the Christmas Tree Cluster, is a complex mixture of hydrogen gas (which is ionized by hot embedded stars and shown in red), dark obscuring dust lanes, and brilliant young stars. The image shows a scene about 30 light-years across. (credit: ESO) 780 Chapter 22 Stars from Adolescence to Old Age The NGC 2264 cluster\u2019s H\u2013R diagram is shown in Figure 22.10. The cluster in the middle of the Orion Nebula (shown in Figure 21.4 and Figure 21.5) is in a similar stage of evolution. Figure 22.10. NGC 2264 H\u2013R Diagram. Compare this H\u2013R diagram to that in Figure 22.8; although the points scatter a bit more here, the theoretical and observational diagrams are remarkably, and satisfyingly, similar. As clusters get older, their H\u2013R diagrams begin to change. After a short time (less than a million years after they reach the main sequence), the most massive stars use up the hydrogen in their cores and evolve off the main sequence to become red giants and supergiants. As more time", " passes, stars of lower mass begin to leave the main sequence and make their way to the upper right of the H\u2013R diagram To see the evolution of a star cluster in a dwarf galaxy, you can watch this brief animation (https://openstax.org/l/30StarCluster) of how its H\u2013R diagram changes. Figure 22.11 is a photograph of NGC 3293, a cluster that is about 10 million years old. The dense clouds of gas and dust are gone. One massive star has evolved to become a red giant and stands out as an especially bright orange member of the cluster. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 781 Figure 22.11. NGC 3293. All the stars in an open star cluster like NGC 3293 form at about the same time. The most massive stars, however, exhaust their nuclear fuel more rapidly and hence evolve more quickly than stars of low mass. As stars evolve, they become redder. The bright orange star in NGC 3293 is the member of the cluster that has evolved most rapidly. (credit: ESO/G. Beccari) Figure 22.12 shows the H\u2013R diagram of the open cluster M41, which is roughly 100 million years old; by this time, a significant number of stars have moved off to the right and become red giants. Note the gap that appears in this H\u2013R diagram between the stars near the main sequence and the red giants. A gap does not necessarily imply that stars avoid a region of certain temperatures and luminosities. In this case, it simply represents a domain of temperature and luminosity through which stars evolve very quickly. We see a gap for M41 because at this particular moment, we have not caught a star in the process of scurrying across this part of the diagram. 782 Chapter 22 Stars from Adolescence to Old Age Figure 22.12. Cluster M41. (a) Cluster M41 is older than NGC 2264 (see Figure 22.10) and contains several red giants. Some of its more massive stars are no longer close to the zero-age main sequence (red line). (b) This ground-based photograph shows the open cluster M41. Note that it contains several orange-color stars. These are stars that have exhausted hydrogen in their centers, and have swelled up to become red giants. (credit b: modification of work by NOAO/AURA/NSF) H\u2013R Diagrams of Older Clusters After 4 billion years have passed, many more stars, including stars that are only a few times more massive than the Sun, have left the main sequence (Figure 22.13). This means that no stars are left near the top of the main sequence; only the low-mass stars near the bottom remain. The older the cluster, the lower the point on the main sequence (and the lower the mass of the stars) where stars begin to move toward the red giant region. The location in the H\u2013R diagram where the stars have begun to leave the main sequence is called the main- sequence turnoff. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 783 Figure 22.13. H\u2013R Diagram for an Older Cluster. We see the H\u2013R diagram for a hypothetical older cluster at an age of 4.24 billion years. Note that most of the stars on the upper part of the main sequence have turned off toward the red-giant region. And the most massive stars in the cluster have already died and are no longer on the diagram. The oldest clusters of all are the globular clusters. Figure 22.14 shows the H\u2013R diagram of globular cluster 47 Tucanae. Notice that the luminosity and temperature scales are different from those of the other H\u2013R diagrams in this chapter. In Figure 22.13, for example, the luminosity scale on the left side of the diagram goes from 0.1 to 100,000 times the Sun\u2019s luminosity. But in Figure 22.14, the luminosity scale has been significantly reduced in extent. So many stars in this old cluster have had time to turn off the main sequence that only the very bottom of the main sequence remains. 784 Chapter 22 Stars from Adolescence to Old Age Figure 22.14. Cluster 47 Tucanae. This H\u2013R diagram is for the globular cluster 47. Note that the scale of luminosity differs from that of the other H\u2013R diagrams in this chapter. We are only focusing on the lower portion of the main sequence, the only part where stars still remain in this old cluster Check out this brief NASA video with a 3-D visualization (https://openstax.org", "/l/30HRDiagram) of how an H\u2013R diagram is created for the globular cluster Omega Centauri. Just how old are the different clusters we have been discussing? To get their actual ages (in years), we must compare the appearances of our calculated H\u2013R diagrams of different ages to observed H\u2013R diagrams of real clusters. In practice, astronomers use the position at the top of the main sequence (that is, the luminosity at which stars begin to move off the main sequence to become red giants) as a measure of the age of a cluster (the main-sequence turnoff we discussed previously). For example, we can compare the luminosities of the brightest stars that are still on the main sequence in Figure 22.10 and Figure 22.13. Using this method, some associations and open clusters turn out to be as young as 1 million years old, while others are several hundred million years old. Once all of the interstellar matter surrounding a cluster has been used to form stars or has dispersed and moved away from the cluster, star formation ceases, and stars of progressively lower mass move off the main sequence, as shown in Figure 22.10, Figure 22.12, and Figure 22.13. To our surprise, even the youngest of the globular clusters in our Galaxy are found to be older than the oldest open cluster. All of the globular clusters have main sequences that turn off at a luminosity less than that of the Sun. Star formation in these crowded systems ceased billions of years ago, and no new stars are coming on to the main sequence to replace the ones that have turned off (see Figure 22.15). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 785 Figure 22.15. H\u2013R Diagrams for Clusters of Different Ages. This sketch shows how the turn-off point from the main sequence gets lower as we make H\u2013R diagrams for clusters that are older and older. Indeed, the globular clusters are the oldest structures in our Galaxy (and in other galaxies as well). The youngest have ages of about 11 billion years and some appear to be even older. Since these are the oldest objects we know of, this estimate is one of the best limits we have on the age of the universe itself\u2014it must be at least 11 billion years old. We will return to the fascinating question of determining the age of the entire universe in the chapter on The Big Bang. 22.4 FURTHER EVOLUTION OF STARS Learning Objectives By the end of this section, you will be able to: Explain what happens in a star\u2019s core when all of the hydrogen has been used up Define \u201cplanetary nebulae\u201d and discuss their origin Discuss the creation of new chemical elements during the late stages of stellar evolution The \u201clife story\u201d we have related so far applies to almost all stars: each starts as a contracting protostar, then lives most of its life as a stable main-sequence star, and eventually moves off the main sequence toward the red-giant region. As we have seen, the pace at which each star goes through these stages depends on its mass, with more massive stars evolving more quickly. But after this point, the life stories of stars of different masses diverge, with a wider range of possible behavior according to their masses, their compositions, and the presence of any nearby companion stars. Because we have written this book for students taking their first astronomy course, we will recount a simplified version of what happens to stars as they move toward the final stages in their lives. We will (perhaps to your heartfelt relief) not delve into all the possible ways aging stars can behave and the strange things that happen when a star is orbited by a second star in a binary system. Instead, we will focus only on the key stages in the evolution of single stars and show how the evolution of high-mass stars differs from that of low-mass stars (such as our Sun). 786 Chapter 22 Stars from Adolescence to Old Age Helium Fusion Let\u2019s begin by considering stars with composition like that of the Sun and whose initial masses are comparatively low\u2014no more than about twice the mass of our Sun. (Such mass may not seem too low, but stars with masses less than this all behave in a fairly similar fashion. We will see what happens to more massive stars in the next section.) Because there are much more low-mass stars than high-mass stars in the Milky Way, the vast majority of stars\u2014including our Sun\u2014follow the scenario we are about to relate. By the way, we carefully used the term initial masses of stars because, as we will see, stars can lose quite a bit of mass in the process of aging and dying. Remember that red giants start out with a helium core where no energy generation is taking place, surrounded by a shell where hydrogen", " is undergoing fusion. The core, having no source of energy to oppose the inward pull of gravity, is shrinking and growing hotter. As time goes on, the temperature in the core can rise to much hotter values than it had in its main-sequence days. Once it reaches a temperature of 100 million K (but not before such point), three helium atoms can begin to fuse to form a single carbon nucleus. This process is called the triple-alpha process, so named because physicists call the nucleus of the helium atom an alpha particle. When the triple-alpha process begins in low-mass (about 0.8 to 2.0 solar masses) stars, calculations show that the entire core is ignited in a quick burst of fusion called a helium flash. (More massive stars also ignite helium but more gradually and not with a flash.) As soon as the temperature at the center of the star becomes high enough to start the triple-alpha process, the extra energy released is transmitted quickly through the entire helium core, producing very rapid heating. The heating speeds up the nuclear reactions, which provide more heating, and which accelerates the nuclear reactions even more. We have runaway generation of energy, which reignites the entire helium core in a flash. You might wonder why the next major step in nuclear fusion in stars involves three helium nuclei and not just two. Although it is a lot easier to get two helium nuclei to collide, the product of this collision is not stable and falls apart very quickly. It takes three helium nuclei coming together simultaneously to make a stable nuclear structure. Given that each helium nucleus has two positive protons and that such protons repel one another, you can begin to see the problem. It takes a temperature of 100 million K to slam three helium nuclei (six protons) together and make them stick. But when that happens, the star produces a carbon nucleus Stars in Your Little Finger Stop reading for a moment and look at your little finger. It\u2019s full of carbon atoms because carbon is a fundamental chemical building block for life on Earth. Each of those carbon atoms was once inside a red giant star and was fused from helium nuclei in the triple-alpha process. All the carbon on Earth\u2014in you, in the charcoal you use for barbecuing, and in the diamonds you might exchange with a loved one\u2014was \u201ccooked up\u201d by previous generations of stars. How the carbon atoms (and other elements) made their way from inside some of those stars to become part of Earth is something we will discuss in the next chapter. For now, we want to emphasize that our description of stellar evolution is, in a very real sense, the story of our own cosmic \u201croots\u201d\u2014the history of how our own atoms originated among the stars. We are made of \u201cstar-stuff.\u201d This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 787 Becoming a Giant Again After the helium flash, the star, having survived the \u201cenergy crisis\u201d that followed the end of the main-sequence stage and the exhaustion of the hydrogen fuel at its center, finds its balance again. As the star readjusts to the release of energy from the triple-alpha process in its core, its internal structure changes once more: its surface temperature increases and its overall luminosity decreases. The point that represents the star on the H\u2013R diagram thus moves to a new position to the left of and somewhat below its place as a red giant (Figure 22.16). The star then continues to fuse the helium in its core for a while, returning to the kind of equilibrium between pressure and gravity that characterized the main-sequence stage. During this time, a newly formed carbon nucleus at the center of the star can sometimes be joined by another helium nucleus to produce a nucleus of oxygen\u2014another building block of life. Figure 22.16. Evolution of a Star Like the Sun on an H\u2013R Diagram. Each stage in the star\u2019s life is labeled. (a) The star evolves from the main sequence to be a red giant, decreasing in surface temperature and increasing in luminosity. (b) A helium flash occurs, leading to a readjustment of the star\u2019s internal structure and to (c) a brief period of stability during which helium is fused to carbon and oxygen in the core (in the process the star becomes hotter and less luminous than it was as a red giant). (d) After the central helium is exhausted, the star becomes a giant again and moves to higher luminosity and lower temperature. By this time, however, the star has exhausted its inner resources and will soon begin to die. Where the evolutionary track becomes a dashed line, the changes are so rapid that they are difficult to model. However, at a temperature of 100 million K, the inner core is converting its helium fuel or carbon (and a", " bit of oxygen) at a rapid rate. Thus, the new period of stability cannot last very long: it is far shorter than the main- sequence stage. Soon, all the helium hot enough for fusion will be used up, just like the hot hydrogen that was used up earlier in the star\u2019s evolution. Once again, the inner core will not be able to generate energy via fusion. Once more, gravity will take over, and the core will start to shrink again. We can think of stellar evolution as a story of a constant struggle against gravitational collapse. A star can avoid collapsing as long as it can tap energy sources, but once any particular fuel is used up, it starts to collapse again. The star\u2019s situation is analogous to the end of the main-sequence stage (when the central hydrogen got used up), but the star now has a somewhat more complicated structure. Again, the star\u2019s core begins to collapse under its own weight. Heat released by the shrinking of the carbon and oxygen core flows into a shell of helium just above the core. This helium, which had not been hot enough for fusion into carbon earlier, is heated just enough for fusion to begin and to generate a new flow of energy. 788 Chapter 22 Stars from Adolescence to Old Age Farther out in the star, there is also a shell where fresh hydrogen has been heated enough to fuse helium. The star now has a multi-layered structure like an onion: a carbon-oxygen core, surrounded by a shell of helium fusion, a layer of helium, a shell of hydrogen fusion, and finally, the extended outer layers of the star (see Figure 22.17). As energy flows outward from the two fusion shells, once again the outer regions of the star begin to expand. Its brief period of stability is over; the star moves back to the red-giant domain on the H\u2013R diagram for a short time (see Figure 22.16). But this is a brief and final burst of glory. Figure 22.17. Layers inside a Low-Mass Star before Death. Here we see the layers inside a star with an initial mass that is less than twice the mass of the Sun. These include, from the center outward, the carbon-oxygen core, a layer of helium hot enough to fuse, a layer of cooler helium, a layer of hydrogen hot enough to fuse, and then cooler hydrogen beyond. Recall that the last time the star was in this predicament, helium fusion came to its rescue. The temperature at the star\u2019s center eventually became hot enough for the product of the previous step of fusion (helium) to become the fuel for the next step (helium fusing into carbon). But the step after the fusion of helium nuclei requires a temperature so hot that the kinds of lower-mass stars (less than 2 solar masses) we are discussing simply cannot compress their cores to reach it. No further types of fusion are possible for such a star. In a star with a mass similar to that of the Sun, the formation of a carbon-oxygen core thus marks the end of the generation of nuclear energy at the center of the star. The star must now confront the fact that its death is near. We will discuss how stars like this end their lives in The Death of Stars, but in the meantime, Table 22.4 summarizes the stages discussed so far in the life of a star with the same mass as that of the Sun. One thing that gives us confidence in our calculations of stellar evolution is that when we make H\u2013R diagrams of older clusters, we actually see stars in each of the stages that we have been discussing. The Evolution of a Star with the Sun\u2019s Mass Stage Time in This Stage (years) Surface Temperature (K) Luminosity (LSun) Diameter (Sun = 1) Main sequence 11 billion 6000 1 Becomes red giant Table 22.4 1.3 billion 3100 at minimum 2300 at maximum 1 165 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 789 The Evolution of a Star with the Sun\u2019s Mass Stage Time in This Stage (years) Surface Temperature (K) Luminosity (LSun) Diameter (Sun = 1) Helium fusion 100 million Giant again 20 million 4800 3100 50 5200 10 180 Table 22.4 Mass Loss from Red-Giant Stars and the Formation of Planetary Nebulae When stars swell up to become red giants, they have very large radii and therefore a low escape velocity.[2] Radiation pressure, stellar pulsations, and violent events like the helium flash can all drive atoms in the outer atmosphere away from the star, and cause it to lose a substantial fraction of its mass into space. Astronomers estimate that by the time a star like the Sun reaches the point of the helium flash, for example, it will have lost as", " much as 25% of its mass. And it can lose still more mass when it ascends the red-giant branch for the second time. As a result, aging stars are surrounded by one or more expanding shells of gas, each containing as much as 10\u201320% of the Sun\u2019s mass (or 0.1\u20130.2 MSun). When nuclear energy generation in the carbon-oxygen core ceases, the star\u2019s core begins to shrink again and to heat up as it gets more and more compressed. (Remember that this compression will not be halted by another type of fusion in these low-mass stars.) The whole star follows along, shrinking and also becoming very hot\u2014reaching surface temperatures as high as 100,000 K. Such hot stars are very strong sources of stellar winds and ultraviolet radiation, which sweep outward into the shells of material ejected when the star was a red giant. The winds and the ultraviolet radiation heat the shells, ionize them, and set them aglow (just as ultraviolet radiation from hot, young stars produces H II regions; see Between the Stars: Gas and Dust in Space). The result is the creation of some of the most beautiful objects in the cosmos (see the gallery in Figure 22.18 and Figure 22.1). These objects were given an extremely misleading name when first found in the eighteenth century: planetary nebulae. The name is derived from the fact that a few planetary nebulae, when viewed through a small telescope, have a round shape bearing a superficial resemblance to planets. Actually, they have nothing to do with planets, but once names are put into regular use in astronomy, it is extremely difficult to change them. There are tens of thousands of planetary nebulae in our own Galaxy, although many are hidden from view because their light is absorbed by interstellar dust. 2 Recall that the force of gravity depends not only on the mass doing the pulling, but also on our distance from the center of gravity. As a red giant star gets a lot bigger, a point on the surface of the star is now farther from the center, and thus has less gravity. That\u2019s why the speed needed to escape the star goes down. 790 Chapter 22 Stars from Adolescence to Old Age Figure 22.18. Gallery of Planetary Nebulae. This series of beautiful images depicting some intriguing planetary nebulae highlights the capabilities of the Hubble Space Telescope. (a) Perhaps the best known planetary nebula is the Ring Nebula (M57), located about 2000 lightyears away in the constellation of Lyra. The ring is about 1 light-year in diameter, and the central star has a temperature of about 120,000 \u00b0C. Careful study of this image has shown scientists that, instead of looking at a spherical shell around this dying star, we may be looking down the barrel of a tube or cone. The blue region shows emission from very hot helium, which is located very close to the star; the red region isolates emission from ionized nitrogen, which is radiated by the coolest gas farthest from the star; and the green region represents oxygen emission, which is produced at intermediate temperatures and is at an intermediate distance from the star. (b) This planetary nebula, M2-9, is an example of a butterfly nebula. The central star (which is part of a binary system) has ejected mass preferentially in two opposite directions. In other images, a disk, perpendicular to the two long streams of gas, can be seen around the two stars in the middle. The stellar outburst that resulted in the expulsion of matter occurred about 1200 years ago. Neutral oxygen is shown in red, once-ionized nitrogen in green, and twice-ionized oxygen in blue. The planetary nebula is about 2100 light-years away in the constellation of Ophiuchus. (c) In this image of the planetary nebula NGC 6751, the blue regions mark the hottest gas, which forms a ring around the central star. The orange and red regions show the locations of cooler gas. The origin of these cool streamers is not known, but their shapes indicate that they are affected by radiation and stellar winds from the hot star at the center. The temperature of the star is about 140,000 \u00b0C. The diameter of the nebula is about 600 times larger than the diameter of our solar system. The nebula is about 6500 light-years away in the constellation of Aquila. (d) This image of the planetary nebula NGC 7027 shows several stages of mass loss. The faint blue concentric shells surrounding the central region identify the mass that was shed slowly from the surface of the star when it became a red giant. Somewhat later, the remaining outer layers were ejected but not in a spherically symmetric way. The dense clouds formed by this late ejection produce the bright inner regions. The hot central star can be seen faintly near the center of the nebulosity. NGC 7027 is", " about 3000 light-years away in the direction of the constellation of Cygnus. (credit a: modification of work by NASA, ESA, and the Hubble Heritage (STScI/AURA)-ESA/Hubble Collaboration; credit b: modification of work by Bruce Balick (University of Washington), Vincent Icke (Leiden University, The Netherlands), Garrelt Mellema (Stockholm University), and NASA; credit c: modification of work by NASA, The Hubble Heritage Team (STScI/AURA); credit d: modification of work by H. Bond (STScI) and NASA) As Figure 22.18 shows, sometimes a planetary nebula appears to be a simple ring. Others have faint shells surrounding the bright ring, which is evidence that there were multiple episodes of mass loss when the star was a red giant (see image (d) in Figure 22.18). In a few cases, we see two lobes of matter flowing in opposite directions. Many astronomers think that a considerable number of planetary nebulae basically consist of the same structure, but that the shape we see depends on the viewing angle (Figure 22.19). According to this idea, the dying star is surrounded by a very dense, doughnut-shaped disk of gas. (Theorists do not yet have a definite explanation for why the dying star should produce this ring, but many believe that binary stars, which are common, are involved.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 791 Figure 22.19. Model to Explain the Different Shapes of Planetary Nebulae. The range of different shapes that we see among planetary nebulae may, in many cases, arise from the same geometric shape, but seen from a variety of viewing directions. The basic shape is a hot central star surrounded by a thick torus (or doughnut-shaped disk) of gas. The star\u2019s wind cannot flow out into space very easily in the direction of the torus, but can escape more freely in the two directions perpendicular to it. If we view the nebula along the direction of the flow (Helix Nebula), it will appear nearly circular (like looking directly down into an empty ice-cream cone). If we look along the equator of the torus, we see both outflows and a very elongated shape (Hubble 5). Current research on planetary nebulae focuses on the reasons for having a torus around the star in the first place. Many astronomers suggest that the basic cause may be that many of the central stars are actually close binary stars, rather than single stars. (credit \u201cHubble 5\u201d: modification of work by Bruce Balick (University of Washington), Vincent Icke (Leiden University, The Netherlands), Garrelt Mellema (Stockholm University), and NASA/ESA; credit \u201cHelix\u201d: modification of work by NASA, ESA, C.R. O\u2019Dell (Vanderbilt University), and M. Meixner, P. McCullough) As the star continues to lose mass, any less dense gas that leaves the star cannot penetrate the torus, but the gas can flow outward in directions perpendicular to the disk. If we look perpendicular to the direction of outflow, we see the disk and both of the outward flows. If we look \u201cdown the barrel\u201d and into the flows, we see a ring. At intermediate angles, we may see wonderfully complex structures. Compare the viewpoints in Figure 22.19 with the images in Figure 22.18. Planetary nebula shells usually expand at speeds of 20\u201330 km/s, and a typical planetary nebula has a diameter of about 1 light-year. If we assume that the gas shell has expanded at a constant speed, we can calculate that the shells of all the planetary nebulae visible to us were ejected within the past 50,000 years at most. After this amount of time, the shells have expanded so much that they are too thin and tenuous to be seen. That\u2019s a pretty short time that each planetary nebula can be observed (when compared to the whole lifetime of the star). Given the number of such nebulae we nevertheless see, we must conclude that a large fraction of all stars evolve through the planetary nebula phase. Since we saw that low-mass stars are much more common than high-mass stars, this confirms our view of planetary nebulae as sort of \u201clast gasp\u201d of low-mass star evolution. 792 Chapter 22 Stars from Adolescence to Old Age Cosmic Recycling The loss of mass by dying stars is a key step in the gigantic cosmic recycling scheme we discussed in Between the Stars: Gas and Dust in Space. Remember that stars form from vast clouds of gas and dust. As they end their lives, stars return part of their", " gas to the galactic reservoirs of raw material. Eventually, some of the expelled material from aging stars will participate in the formation of new star systems. However, the atoms returned to the Galaxy by an aging star are not necessarily the same ones it received initially. The star, after all, has fused hydrogen and helium to form new elements over the course of its life. And during the red-giant stage, material from the star\u2019s central regions is dredged up and mixed with its outer layers, which can cause further nuclear reactions and the creation of still more new elements. As a result, the winds that blow outward from such stars include atoms that were \u201cnewly minted\u201d inside the stars\u2019 cores. (As we will see, this mechanism is even more effective for high-mass stars, but it does work for stars with masses like that of the Sun.) In this way, the raw material of the Galaxy is not only resupplied but also receives infusions of new elements. You might say this cosmic recycling plan allows the universe to get more \u201cinteresting\u201d all the time The Red Giant Sun and the Fate of Earth How will the evolution of the Sun affect conditions on Earth in the future? Although the Sun has appeared reasonably steady in size and luminosity over recorded human history, that brief span means nothing compared with the timescales we have been discussing. Let\u2019s examine the long-term prospects for our planet. The Sun took its place on the zero-age main sequence approximately 4.5 billion years ago. At that time, it emitted only about 70% of the energy that it radiates today. One might expect that Earth would have been a lot colder than it is now, with the oceans frozen solid. But if this were the case, it would be hard to explain why simple life forms existed when Earth was less than a billion years old. Scientists now think that the explanation may be that much more carbon dioxide was present in Earth\u2019s atmosphere when it was young, and that a much stronger greenhouse effect kept Earth warm. (In the greenhouse effect, gases like carbon dioxide or water vapor allow the Sun\u2019s light to come in but do not allow the infrared radiation from the ground to escape back into space, so the temperature near Earth\u2019s surface increases.) Carbon dioxide in Earth\u2019s atmosphere has steadily declined as the Sun has increased in luminosity. As the brighter Sun increases the temperature of Earth, rocks weather faster and react with carbon dioxide, removing it from the atmosphere. The warmer Sun and the weaker greenhouse effect have kept Earth at a nearly constant temperature for most of its life. This remarkable coincidence, which has resulted in fairly stable climatic conditions, has been the key in the development of complex life-forms on our planet. As a result of changes caused by the buildup of helium in its core, the Sun will continue to increase in luminosity as it grows older, and more and more radiation will reach Earth. For a while, the amount of carbon dioxide will continue to decrease. (Note that this effect counteracts increases in carbon dioxide from human activities, but on a much-too-slow timescale to undo the changes in climate that are likely to occur in the next 100 years.) Eventually, the heating of Earth will melt the polar caps and increase the evaporation of the oceans. Water vapor is also an efficient greenhouse gas and will more than compensate for the decrease in This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 793 carbon dioxide. Sooner or later (atmospheric models are not yet good enough to say exactly when, but estimates range from 500 million to 2 billion years), the increased water vapor will cause a runaway greenhouse effect. About 1 billion years from now, Earth will lose its water vapor. In the upper atmosphere, sunlight will break down water vapor into hydrogen, and the fast-moving hydrogen atoms will escape into outer space. Like Humpty Dumpty, the water molecules cannot be put back together again. Earth will start to resemble the Venus of today, and temperatures will become much too high for life as we know it. All of this will happen before the Sun even becomes a red giant. Then the bad news really starts. The Sun, as it expands, will swallow Mercury and Venus, and friction with our star\u2019s outer atmosphere will make these planets spiral inward until they are completely vaporized. It is not completely clear whether Earth will escape a similar fate. As described in this chapter, the Sun will lose some of its mass as it becomes a red giant. The gravitational pull of the Sun decreases when it loses mass. The result would be that the diameter of Earth\u2019s orbit would increase (remember Kepler\u2019s third law). However, recent calculations also show that forces due to the tides raised on the Sun by Earth will act in the opposite direction, causing Earth\u2019s", " orbit to shrink. Thus, many astrophysicists conclude that Earth will be vaporized along with Mercury and Venus. Whether or not this dire prediction is true, there is little doubt that all life on Earth will surely be incinerated. But don\u2019t lose any sleep over this\u2014we are talking about events that will occur billions of years from now. What then are the prospects for preserving Earth life as we know it? The first strategy you might think of would be to move humanity to a more distant and cooler planet. However, calculations indicate that there are long periods of time (several hundred million years) when no planet is habitable. For example, Earth becomes far too warm for life long before Mars warms up enough. A better alternative may be to move the entire Earth progressively farther from the Sun. The idea is to use gravity in the same way NASA has used it to send spacecraft to distant planets. When a spacecraft flies near a planet, the planet\u2019s motion can be used to speed up the spacecraft, slow it down, or redirect it. Calculations show that if we were to redirect an asteroid so that it follows just the right orbit between Earth and Jupiter, it could transfer orbital energy from Jupiter to Earth and move Earth slowly outward, pulling us away from the expanding Sun on each flyby. Since we have hundreds of millions of years to change Earth\u2019s orbit, the effect of each flyby need not be large. (Of course, the people directing the asteroid had better get the orbit exactly right and not cause the asteroid to hit Earth.) It may seem crazy to think about projects to move an entire planet to a different orbit. But remember that we are talking about the distant future. If, by some miracle, human beings are able to get along for all that time and don\u2019t blow ourselves to bits, our technology is likely to be far more sophisticated than it is today. It may also be that if humans survive for hundreds of millions of years, we may spread to planets or habitats around other stars. Indeed, Earth, by then, might be a museum world to which youngsters from other planets return to learn about the origin of our species. It is also possible that evolution will by then have changed us in ways that allow us to survive in very different environments. Wouldn\u2019t it be exciting to see how the story of the story of the human race turns out after all those billions of years? 794 Chapter 22 Stars from Adolescence to Old Age 22.5 THE EVOLUTION OF MORE MASSIVE STARS Learning Objectives By the end of this section, you will be able to: Explain how and why massive stars evolve much more rapidly than lower-mass stars like our Sun Discuss the origin of the elements heavier than carbon within stars If what we have described so far were the whole story of the evolution of stars and elements, we would have a big problem on our hands. We will see in later chapters that in our best models of the first few minutes of the universe, everything starts with the two simplest elements\u2014hydrogen and helium (plus a tiny bit of lithium). All the predictions of the models imply that no heavier elements were produced at the beginning of the universe. Yet when we look around us on Earth, we see lots of other elements besides hydrogen and helium. These elements must have been made (fused) somewhere in the universe, and the only place hot enough to make them is inside stars. One of the fundamental discoveries of twentieth-century astronomy is that the stars are the source of most of the chemical richness that characterizes our world and our lives. We have already seen that carbon and some oxygen are manufactured inside the lower-mass stars that become red giants. But where do the heavier elements we know and love (such as the silicon and iron inside Earth, and the gold and silver in our jewelry) come from? The kinds of stars we have been discussing so far never get hot enough at their centers to make these elements. It turns out that such heavier elements can be formed only late in the lives of more massive stars. Making New Elements in Massive Stars Massive stars evolve in much the same way that the Sun does (but always more quickly)\u2014up to the formation of a carbon-oxygen core. One difference is that for stars with more than about twice the mass of the Sun, helium begins fusion more gradually, rather than with a sudden flash. Also, when more massive stars become red giants, they become so bright and large that we call them supergiants. Such stars can expand until their outer regions become as large as the orbit of Jupiter, which is precisely what the Hubble Space Telescope has shown for the star Betelgeuse (see Figure 22.4). They also lose mass very effectively, producing dramatic winds and outbursts as they age. Figure 22.20 shows a wonderful image of the very massive star Eta Carinae, with a great deal of ejected material clearly visible. This OpenStax book is available for free at http://cn", "x.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 795 Figure 22.20. Eta Carinae. With a mass at least 100 times that of the Sun, the hot supergiant Eta Carinae is one of the most massive stars known. This Hubble Space Telescope image records the two giant lobes and equatorial disk of material it has ejected in the course of its evolution. The pink outer region is material ejected in an outburst seen in 1843, the largest of such mass loss event that any star is known to have survived. Moving away from the star at a speed of about 1000 km/s, the material is rich in nitrogen and other elements formed in the interior of the star. The inner blue-white region is the material ejected at lower speeds and is thus still closer to the star. It appears blue-white because it contains dust and reflects the light of Eta Carinae, whose luminosity is 4 million times that of our Sun. (credit: modification of work by Jon Morse (University of Colorado) & NASA) But the crucial way that massive stars diverge from the story we have outlined is that they can start additional kinds of fusion in their centers and in the shells surrounding their central regions. The outer layers of a star with a mass greater than about 8 solar masses have a weight that is enough to compress the carbon-oxygen core until it becomes hot enough to ignite fusion of carbon nuclei. Carbon can fuse into still more oxygen, and at still higher temperatures, oxygen and then neon, magnesium, and finally silicon can build even heavier elements. Iron is, however, the endpoint of this process. The fusion of iron atoms produces products that are more massive than the nuclei that are being fused and therefore the process requires energy, as opposed to releasing energy, which all fusion reactions up to this point have done. This required energy comes at the expense of the star itself, which is now on the brink of death (Figure 22.21). What happens next will be described in the chapter on The Death of Stars. 796 Chapter 22 Stars from Adolescence to Old Age Figure 22.21. Interior Structure of a Massive Star Just before It Exhausts Its Nuclear Fuel. High-mass stars can fuse elements heavier than carbon. As a massive star nears the end of its evolution, its interior resembles an onion. Hydrogen fusion is taking place in an outer shell, and progressively heavier elements are undergoing fusion in the higher-temperature layers closer to the center. All of these fusion reactions generate energy and enable the star to continue shining. Iron is different. The fusion of iron requires energy, and when iron is finally created in the core, the star has only minutes to live. Physicists have now found nuclear pathways whereby virtually all chemical elements of atomic weights up to that of iron can be built up by this nucleosynthesis (the making of new atomic nuclei) in the centers of the more massive red giant stars. This still leaves the question of where elements heavier than iron come from. We will see in the next chapter that when massive stars finally exhaust their nuclear fuel, they most often die in a spectacular explosion\u2014a supernova. Heavier elements can be synthesized in the stunning violence of such explosions. Not only can we explain in this way where the elements that make up our world and others come from, but our theories of nucleosynthesis inside stars are even able to predict the relative abundances with which the elements occur in nature. The way stars build up elements during various nuclear reactions really can explain why some elements (oxygen, carbon, and iron) are common and others are quite rare (gold, silver, and uranium). Elements in Globular Clusters and Open Clusters Are Not the Same The fact that the elements are made in stars over time explains an important difference between globular and open clusters. Hydrogen and helium, which are the most abundant elements in stars in the solar neighborhood, are also the most abundant constituents of stars in both kinds of clusters. However, the abundances of the elements heavier than helium are very different. In the Sun and most of its neighboring stars, the combined abundance (by mass) of the elements heavier than hydrogen and helium is 1\u20134% of the star\u2019s mass. Spectra show that most open-cluster stars also have 1\u20134% of their matter in the form of heavy elements. Globular clusters, however, are a different story. The heavy-element abundance of stars in typical globular clusters is found to be only 1/10 to 1/100 that of the Sun. A few very old stars not in clusters have been discovered with even lower abundances of heavy elements. The differences in chemical composition are a direct consequence of the formation of a cluster of stars. The very first generation of stars initially contained only hydrogen and helium. We have seen that these stars, in order to generate energy, created heavier elements in their interiors", ". In the last stages of their lives, they ejected matter, now enriched in heavy elements, into the reservoirs of raw material between the stars. Such matter was then incorporated into a new generation of stars. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 797 This means that the relative abundance of the heavy elements must be less and less as we look further into the past. We saw that the globular clusters are much older than the open clusters. Since globular-cluster stars formed much earlier (that is, they are an earlier generation of stars) than those in open clusters, they have only a relatively small abundance of elements heavier than hydrogen and helium. As time passes, the proportion of heavier elements in the \u201craw material\u201d that makes new stars and planets increases. This means that the first generation of stars that formed in our Galaxy would not have been accompanied by a planet like Earth, full of silicon, iron, and many other heavy elements. Earth (and the astronomy students who live on it) was possible only after generations of stars had a chance to make and recycle their heavier elements. Now the search is on for true first-generation stars, made only of hydrogen and helium. Theories predict that such stars should be very massive, live fast, and die quickly. They should have lived and died long ago. The place to look for them is in very distant galaxies that formed when the universe was only a few hundred million years old, but whose light is only arriving at Earth now. Approaching Death Compared with the main-sequence lifetimes of stars, the events that characterize the last stages of stellar evolution pass very quickly (especially for massive stars). As the star\u2019s luminosity increases, its rate of nuclear fuel consumption goes up rapidly\u2014just at that point in its life when its fuel supply is beginning to run down. After the prime fuel\u2014hydrogen\u2014is exhausted in a star\u2019s core, we saw that other sources of nuclear energy are available to the star in the fusion of, first, helium, and then of other more complex elements. But the energy yield of these reactions is much less than that of the fusion of hydrogen to helium. And to trigger these reactions, the central temperature must be higher than that required for the fusion of hydrogen to helium, leading to even more rapid consumption of fuel. Clearly this is a losing game, and very quickly the star reaches its end. As it does so, however, some remarkable things can happen, as we will see in The Death of Stars. 798 Chapter 22 Stars from Adolescence to Old Age CHAPTER 22 REVIEW KEY TERMS association a loose group of young stars whose spectral types, motions, and positions in the sky indicate a common origin globular cluster one of about 150 large, spherical star clusters (each with hundreds of thousands of stars) that form a system of clusters in the center of our Galaxy helium flash a nearly explosive ignition of helium in the triple-alpha process in the dense core of a red giant star main-sequence turnoff location in the H\u2013R diagram where stars begin to leave the main sequence nucleosynthesis the building up of heavy elements from lighter ones by nuclear fusion open cluster a comparatively loose cluster of stars, containing from a few dozen to a few thousand members, located in the spiral arms or disk of our Galaxy; sometimes referred to as a galactic cluster planetary nebula a shell of gas ejected by and expanding away from an extremely hot low-mass star that is nearing the end of its life (the nebulae glow because of the ultra-violet energy of the central star) triple-alpha process a nuclear reaction by which three helium nuclei are built up (fused) into one carbon nucleus zero-age main sequence a line denoting the main sequence on the H\u2013R diagram for a system of stars that have completed their contraction from interstellar matter and are now deriving all their energy from nuclear reactions, but whose chemical composition has not yet been altered substantially by nuclear reactions SUMMARY 22.1 Evolution from the Main Sequence to Red Giants When stars first begin to fuse hydrogen to helium, they lie on the zero-age main sequence. The amount of time a star spends in the main-sequence stage depends on its mass. More massive stars complete each stage of evolution more quickly than lower-mass stars. The fusion of hydrogen to form helium changes the interior composition of a star, which in turn results in changes in its temperature, luminosity, and radius. Eventually, as stars age, they evolve away from the main sequence to become red giants or supergiants. The core of a red giant is contracting, but the outer layers are expanding as a result of hydrogen fusion in a shell outside the core. The star gets larger, redder, and more luminous as it expands and cools. 22.2 Star Clusters Star clusters provide one of the best tests of our calculations", " of what happens as stars age. The stars in a given cluster were formed at about the same time and have the same composition, so they differ mainly in mass, and thus, in their life stage. There are three types of star clusters: globular, open, and associations. Globular clusters have diameters of 50\u2013450 light-years, contain hundreds of thousands of stars, and are distributed in a halo around the Galaxy. Open clusters typically contain hundreds of stars, are located in the plane of the Galaxy, and have diameters less than 30 light-years. Associations are found in regions of gas and dust and contain extremely young stars. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 799 22.3 Checking Out the Theory The H\u2013R diagram of stars in a cluster changes systematically as the cluster grows older. The most massive stars evolve most rapidly. In the youngest clusters and associations, highly luminous blue stars are on the main sequence; the stars with the lowest masses lie to the right of the main sequence and are still contracting toward it. With passing time, stars of progressively lower masses evolve away from (or turn off) the main sequence. In globular clusters, which are all at least 11 billion years old, there are no luminous blue stars at all. Astronomers can use the turnoff point from the main sequence to determine the age of a cluster. 22.4 Further Evolution of Stars After stars become red giants, their cores eventually become hot enough to produce energy by fusing helium to form carbon (and sometimes a bit of oxygen.) The fusion of three helium nuclei produces carbon through the triple-alpha process. The rapid onset of helium fusion in the core of a low-mass star is called the helium flash. After this, the star becomes stable and reduces its luminosity and size briefly. In stars with masses about twice the mass of the Sun or less, fusion stops after the helium in the core has been exhausted. Fusion of hydrogen and helium in shells around the contracting core makes the star a bright red giant again, but only temporarily. When the star is a red giant, it can shed its outer layers and thereby expose hot inner layers. Planetary nebulae (which have nothing to do with planets) are shells of gas ejected by such stars, set glowing by the ultraviolet radiation of the dying central star. 22.5 The Evolution of More Massive Stars In stars with masses higher than about 8 solar masses, nuclear reactions involving carbon, oxygen, and still heavier elements can build up nuclei as heavy as iron. The creation of new chemical elements is called nucleosynthesis. The late stages of evolution occur very quickly. Ultimately, all stars must use up all of their available energy supplies. In the process of dying, most stars eject some matter, enriched in heavy elements, into interstellar space where it can be used to form new stars. Each succeeding generation of stars therefore contains a larger proportion of elements heavier than hydrogen and helium. This progressive enrichment explains why the stars in open clusters (which formed more recently) contain more heavy elements than do those in ancient globular clusters, and it tells us where most of the atoms on Earth and in our bodies come from. FOR FURTHER EXPLORATION Articles Balick, B. & Frank, A. \u201cThe Extraordinary Deaths of Ordinary Stars.\u201d Scientific American (July 2004): 50. About planetary nebulae, the last gasps of low-mass stars, and the future of our own Sun. Djorgovsky, G. \u201cThe Dynamic Lives of Globular Clusters.\u201d Sky & Telescope (October 1998): 38. Cluster evolution and blue straggler stars. Frank, A. \u201cAngry Giants of the Universe.\u201d Astronomy (October 1997): 32. On luminous blue variables like Eta Carinae. Garlick, M. \u201cThe Fate of the Earth.\u201d Sky & Telescope (October 2002): 30. What will happen when our Sun becomes a red giant. Harris, W. & Webb, J. \u201cLife Inside a Globular Cluster.\u201d Astronomy (July 2014): 18. What would night sky be like there? 800 Chapter 22 Stars from Adolescence to Old Age Iben, I. & Tutokov, A. \u201cThe Lives of the Stars: From Birth to Death and Beyond.\u201d Sky & Telescope (December 1997): 36. Kaler, J. \u201cThe Largest Stars in the Galaxy.\u201d Astronomy (October 1990): 30. On red supergiants. Kalirai, J. \u201cNew Light on Our Sun\u2019s Fate.\u201d Astronomy (February 2014): 44. What will happen to stars like our Sun between the main sequence and the white dwarf stages. Kwok, S. \u201cWhat Is the Real Shape of", " the Ring Nebula?\u201d Sky & Telescope (July 2000): 33. On seeing planetary nebulae from different angles. Kwok, S. \u201cStellar Metamorphosis.\u201d Sky & Telescope (October 1998): 30. How planetary nebulae form. Stahler, S. \u201cThe Inner Life of Star Clusters.\u201d Scientific American (March 2013): 44\u201349. How all stars are born in clusters, but different clusters evolve differently. Subinsky, R. \u201cAll About 47 Tucanae.\u201d Astronomy (September 2014): 66. What we know about this globular cluster and how to see it. Websites BBC Page on Giant Stars: http://www.bbc.co.uk/science/space/universe/sights/giant_stars. Includes basic information and links to brief video excerpts. Encylopedia Brittanica Article on Star Clusters: http://www.britannica.com/topic/star-cluster. Written by astronomer Helen Sawyer Hogg-Priestley. Hubble Image Gallery: Planetary Nebulae: http://hubblesite.org/gallery/album/nebula/planetary/. Click on each image to go to a page with more information available. (See also a similar gallery at the National Optical Astronomy Observatories: https://www.noao.edu/image_gallery/planetary_nebulae.html). Hubble Image Gallery: Star Clusters: http://hubblesite.org/gallery/album/star/star_cluster/. Each image comes with an explanatory caption when you click on it. (See also a similar European Southern Observatory Gallery at: https://www.eso.org/public/images/archive/category/starclusters/). Measuring the Age of a Star Cluster: https://www.e-education.psu.edu/astro801/content/l7_p6.html. From Penn State. Videos Life Cycle of Stars: https://www.youtube.com/watch?v=PM9CQDlQI0A. Short summary of stellar evolution from the Institute of Physics in Great Britain, with astronomer Tim O\u2019Brien (4:58). Missions Take an Unparalleled Look into Superstar Eta Carinae: https://www.youtube.com/ watch?v=0rJQi6oaZf0. NASA Goddard video about observations in 2014 and what we know about the pair of stars in this complicated system (4:00). Star Clusters: Open and Globular Clusters: https://www.youtube.com/watch?v=rGPRLxrYbYA. Three Short Hubblecast Videos from 2007\u20132008 on discoveries involving star clusters (12:24). Tour of Planetary Nebula NGC 5189: https://www.youtube.com/watch?v=1D2cwiZld0o. Brief Hubblecast episode with Joe Liske, explaining planetary nebulae in general and one example in particular (5:22). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 801 COLLABORATIVE GROUP ACTIVITIES A. Have your group take a look at the list of the brightest stars in the sky in Appendix J. What fraction of them are past the main-sequence phase of evolution? The text says that stars spend 90% of their lifetimes in the main-sequence phase of evolution. This suggests that if we have a fair (or representative) sample of stars, 90% of them should be main-sequence stars. Your group should brainstorm why 90% of the brightest stars are not in the main-sequence phase of evolution. B. Reading an H\u2013R diagram can be tricky. Suppose your group is given the H\u2013R diagram of a star cluster. Stars above and to the right of the main sequence could be either red giants that had evolved away from the main sequence or very young stars that are still evolving toward the main sequence. Discuss how you would decide which they are. C. In the chapter on Life in the Universe, we discuss some of the efforts now underway to search for radio signals from possible intelligent civilizations around other stars. Our present resources for carrying out such searches are very limited and there are many stars in our Galaxy. Your group is a committee set up by the International Astronomical Union to come up with a list of the best possible stars with which such a search should begin. Make a list of criteria for choosing the stars on the list, and explain the reasons behind each entry (keeping in mind some of the ideas about the life story of stars and timescales that we discuss in the present chapter.) D. Have your group make a list of the reasons why a star that formed at the very beginning of the universe (soon", " after the Big Bang) could not have a planet with astronomy students reading astronomy textbooks (even if the star has the same mass as that of our Sun). E. Since we are pretty sure that when the Sun becomes a giant star, all life on Earth will be wiped out, does your group think that we should start making preparations of any kind? Let\u2019s suppose that a political leader who fell asleep during large parts of his astronomy class suddenly hears about this problem from a large donor and appoints your group as a task force to make suggestions on how to prepare for the end of Earth. Make a list of arguments for why such a task force is not really necessary. F. Use star charts to identify at least one open cluster visible at this time of the year. (Such charts can be found in Sky & Telescope and Astronomy magazines each month and their websites; see Appendix B.) The Pleiades and Hyades are good autumn subjects, and Praesepe is good for springtime viewing. Go out and look at these clusters with binoculars and describe what you see. G. Many astronomers think that planetary nebulae are among the most attractive and interesting objects we can see in the Galaxy. In this chapter, we could only show you a few examples of the pictures of these objects taken with the Hubble or large telescopes on the ground. Have members of your group search further for planetary nebula images online, and make a \u201ctop ten\u201d list of your favorite ones (do not include more than three that were featured in this chapter.) Make a report (with images) for the whole class and explain why you found your top five especially interesting. (You may want to check Figure 22.19 in the process.) EXERCISES 802 Chapter 22 Stars from Adolescence to Old Age Review Questions 1. Compare the following stages in the lives of a human being and a star: prenatal, birth, adolescence/ adulthood, middle age, old age, and death. What does a star with the mass of our Sun do in each of these stages? 2. What is the first event that happens to a star with roughly the mass of our Sun that exhausts the hydrogen in its core and stops the generation of energy by the nuclear fusion of hydrogen to helium? Describe the sequence of events that the star undergoes. 3. Astronomers find that 90% of the stars observed in the sky are on the main sequence of an H\u2013R diagram; why does this make sense? Why are there far fewer stars in the giant and supergiant region? 4. Describe the evolution of a star with a mass similar to that of the Sun, from the protostar stage to the time it first becomes a red giant. Give the description in words and then sketch the evolution on an H\u2013R diagram. 5. Describe the evolution of a star with a mass similar to that of the Sun, from just after it first becomes a red giant to the time it exhausts the last type of fuel its core is capable of fusing. 6. A star is often described as \u201cmoving\u201d on an H\u2013R diagram; why is this description used and what is actually happening with the star? 7. On which edge of the main sequence band on an H\u2013R diagram would the zero-age main sequence be? 8. How do stars typically \u201cmove\u201d through the main sequence band on an H\u2013R diagram? Why? 9. Certain stars, like Betelgeuse, have a lower surface temperature than the Sun and yet are more luminous. How do these stars produce so much more energy than the Sun? 10. Gravity always tries to collapse the mass of a star toward its center. What mechanism can oppose this gravitational collapse for a star? During what stages of a star\u2019s life would there be a \u201cbalance\u201d between them? 11. Why are star clusters so useful for astronomers who want to study the evolution of stars? 12. Would the Sun more likely have been a member of a globular cluster or open cluster in the past? 13. Suppose you were handed two H\u2013R diagrams for two different clusters: diagram A has a majority of its stars plotted on the upper left part of the main sequence with the rest of the stars off the main sequence; and diagram B has a majority of its stars plotted on the lower right part of the main sequence with the rest of the stars off the main sequence. Which diagram would be for the older cluster? Why? 14. Referring to the H\u2013R diagrams in Exercise 22.13, which diagram would more likely be the H\u2013R diagram for an association? 15. The nuclear process for fusing helium into carbon is often called the \u201ctriple-alpha process.\u201d Why is it called as such, and why must it occur at a much higher temperature than the nuclear process for fusing hydrogen into helium? 16. Pictures of various planetary nebulae show a variety of shapes, but astronomers believe", " a majority of planetary nebulae have the same basic shape. How can this paradox be explained? 17. Describe the two \u201crecycling\u201d mechanisms that are associated with stars (one during each star\u2019s life and the other connecting generations of stars). 18. In which of these star groups would you mostly likely find the least heavy-element abundance for the stars within them: open clusters, globular clusters, or associations? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 22 Stars from Adolescence to Old Age 803 19. Explain how an H\u2013R diagram of the stars in a cluster can be used to determine the age of the cluster. 20. Where did the carbon atoms in the trunk of a tree on your college campus come from originally? Where did the neon in the fabled \u201cneon lights of Broadway\u201d come from originally? 21. What is a planetary nebula? Will we have one around the Sun? Thought Questions 22. Is the Sun on the zero-age main sequence? Explain your answer. 23. How are planetary nebulae comparable to a fluorescent light bulb in your classroom? 24. Which of the planets in our solar system have orbits that are smaller than the photospheric radius of Betelgeuse listed in in Table 22.2? 25. Would you expect to find an earthlike planet (with a solid surface) around a very low-mass star that formed right at the beginning of a globular cluster\u2019s life? Explain. 26. In the H\u2013R diagrams for some young clusters, stars of both very low and very high luminosity are off to the right of the main sequence, whereas those of intermediate luminosity are on the main sequence. Can you offer an explanation for that? Sketch an H\u2013R diagram for such a cluster. 27. If the Sun were a member of the cluster NGC 2264, would it be on the main sequence yet? Why or why not? 28. If all the stars in a cluster have nearly the same age, why are clusters useful in studying evolutionary effects (different stages in the lives of stars)? 29. Suppose a star cluster were at such a large distance that it appeared as an unresolved spot of light through the telescope. What would you expect the overall color of the spot to be if it were the image of the cluster immediately after it was formed? How would the color differ after 1010 years? Why? 30. Suppose an astronomer known for joking around told you she had found a type-O main-sequence star in our Milky Way Galaxy that contained no elements heavier than helium. Would you believe her? Why? 31. Stars that have masses approximately 0.8 times the mass of the Sun take about 18 billion years to turn into red giants. How does this compare to the current age of the universe? Would you expect to find a globular cluster with a main-sequence turnoff for stars of 0.8 solar mass or less? Why or why not? 32. Automobiles are often used as an analogy to help people better understand how more massive stars have much shorter main-sequence lifetimes compared to less massive stars. Can you explain such an analogy using automobiles? Figuring For Yourself 33. The text says a star does not change its mass very much during the course of its main-sequence lifetime. While it is on the main sequence, a star converts about 10% of the hydrogen initially present into helium (remember it\u2019s only the core of the star that is hot enough for fusion). Look in earlier chapters to find out what percentage of the hydrogen mass involved in fusion is lost because it is converted to energy. By how much does the mass of the whole star change as a result of fusion? Were we correct to say that the mass of a star does not change significantly while it is on the main sequence? 804 Chapter 22 Stars from Adolescence to Old Age 34. The text explains that massive stars have shorter lifetimes than low-mass stars. Even though massive stars have more fuel to burn, they use it up faster than low-mass stars. You can check and see whether this statement is true. The lifetime of a star is directly proportional to the amount of mass (fuel) it contains and inversely proportional to the rate at which it uses up that fuel (i.e., to its luminosity). Since the lifetime of the Sun is about 1010 y, we have the following relationship: T = 1010 M L y where T is the lifetime of a main-sequence star, M is its mass measured in terms of the mass of the Sun, and L is its luminosity measured in terms of the Sun\u2019s luminosity. A. Explain in words why this equation works. B. Use the data in Table 18.3 to calculate the ages of the main-sequence stars listed. C. Do low-mass stars have longer main-sequence", " lifetimes? D. Do you get the same answers as those in Table 22.1? 35. You can use the equation in Exercise 22.34 to estimate the approximate ages of the clusters in Figure 22.10, Figure 22.12, and Figure 22.13. Use the information in the figures to determine the luminosity of the most massive star still on the main sequence. Now use the data in Table 18.3 to estimate the mass of this star. Then calculate the age of the cluster. This method is similar to the procedure used by astronomers to obtain the ages of clusters, except that they use actual data and model calculations rather than simply making estimates from a drawing. How do your ages compare with the ages in the text? 36. You can estimate the age of the planetary nebula in image (c) in Figure 22.18. The diameter of the nebula is 600 times the diameter of our own solar system, or about 0.8 light-year. The gas is expanding away from the star at a rate of about 25 mi/s. Considering that distance = velocity \u00d7 time, calculate how long ago the gas left the star if its speed has been constant the whole time. Make sure you use consistent units for time, speed, and distance. 37. If star A has a core temperature T, and star B has a core temperature 3T, how does the rate of fusion of star A compare to the rate of fusion of star B? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 805 23 THE DEATH OF STARS Figure 23.1. Stellar Life Cycle. This remarkable picture of NGC 3603, a nebula in the Milky Way Galaxy, was taken with the Hubble Space Telescope. This image illustrates the life cycle of stars. In the bottom half of the image, we see clouds of dust and gas, where it is likely that star formation will take place in the near future. Near the center, there is a cluster of massive, hot young stars that are only a few million years old. Above and to the right of the cluster, there is an isolated star surrounded by a ring of gas. Perpendicular to the ring and on either side of it, there are two bluish blobs of gas. The ring and the blobs were ejected by the star, which is nearing the end of its life. (credit: modification of work by NASA, Wolfgang Brandner (JPL/IPAC), Eva K. Grebel (University of Washington), You-Hua Chu (University of Illinois Urbana-Champaign)) Chapter Outline 23.1 The Death of Low-Mass Stars 23.2 Evolution of Massive Stars: An Explosive Finish 23.3 Supernova Observations 23.4 Pulsars and the Discovery of Neutron Stars 23.5 The Evolution of Binary Star Systems 23.6 The Mystery of the Gamma-Ray Bursts Thinking Ahead Do stars die with a bang or a whimper? In the preceding two chapters, we followed the life story of stars, from the process of birth to the brink of death. Now we are ready to explore the ways that stars end their lives. Sooner or later, each star exhausts its store of nuclear energy. Without a source of internal pressure to balance the weight of the overlying layers, every star eventually gives way to the inexorable pull of gravity and collapses under its own weight. Following the rough distinction made in the last chapter, we will discuss the end-of-life evolution of stars of lower and higher mass separately. What determines the outcome\u2014bang or whimper\u2014is the mass of the star when it is ready to die, not the mass it was born with. As we noted in the last chapter, stars can lose a significant amount of mass in their middle and old age. 806 Chapter 23 The Death of Stars 23.1 THE DEATH OF LOW-MASS STARS Learning Objectives By the end of this section, you will be able to: Describe the physical characteristics of degenerate matter and explain how the mass and radius of degenerate stars are related Plot the future evolution of a white dwarf and show how its observable features will change over time Distinguish which stars will become white dwarfs Let\u2019s begin with those stars whose final mass just before death is less than about 1.4 times the mass of the Sun (MSun). (We will explain why this mass is the crucial dividing line in a moment.) Note that most stars in the universe fall into this category. The number of stars decreases as mass increases; really massive stars are rare (see The Stars: A Celestial Census). This is similar to the music business where only a few musicians ever become superstars. Furthermore, many stars with an initial mass much greater than 1.4 MSun will be reduced to that level by the time they die. For example, we now know that stars that start out", " with masses of at least 8.0 MSun (and possibly as much as 10 MSun) manage to lose enough mass during their lives to fit into this category (an accomplishment anyone who has ever attempted to lose weight would surely envy). A Star in Crisis In the last chapter, we left the life story of a star with a mass like the Sun\u2019s just after it had climbed up to the red-giant region of the H\u2013R diagram for a second time and had shed some of its outer layers to form a planetary nebula. Recall that during this time, the core of the star was undergoing an \u201cenergy crisis.\u201d Earlier in its life, during a brief stable period, helium in the core had gotten hot enough to fuse into carbon (and oxygen). But after this helium was exhausted, the star\u2019s core had once more found itself without a source of pressure to balance gravity and so had begun to contract. This collapse is the final event in the life of the core. Because the star\u2019s mass is relatively low, it cannot push its core temperature high enough to begin another round of fusion (in the same way larger-mass stars can). The core continues to shrink until it reaches a density equal to nearly a million times the density of water! That is 200,000 times greater than the average density of Earth. At this extreme density, a new and different way for matter to behave kicks in and helps the star achieve a final state of equilibrium. In the process, what remains of the star becomes one of the strange white dwarfs that we met in The Stars: A Celestial Census. Degenerate Stars Because white dwarfs are far denser than any substance on Earth, the matter inside them behaves in a very unusual way\u2014unlike anything we know from everyday experience. At this high density, gravity is incredibly strong and tries to shrink the star still further, but all the electrons resist being pushed closer together and set up a powerful pressure inside the core. This pressure is the result of the fundamental rules that govern the behavior of electrons (the quantum physics you were introduced to in The Sun: A Nuclear Powerhouse). According to these rules (known to physicists as the Pauli exclusion principle), which have been verified in studies of atoms in the laboratory, no two electrons can be in the same place at the same time doing the same thing. We specify the place of an electron by its position in space, and we specify what it is doing by its motion and the way it is spinning. The temperature in the interior of a star is always so high that the atoms are stripped of virtually all their electrons. For most of a star\u2019s life, the density of matter is also relatively low, and the electrons in the star are This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 807 moving rapidly. This means that no two of them will be in the same place moving in exactly the same way at the same time. But this all changes when a star exhausts its store of nuclear energy and begins its final collapse. As the star\u2019s core contracts, electrons are squeezed closer and closer together. Eventually, a star like the Sun becomes so dense that further contraction would in fact require two or more electrons to violate the rule against occupying the same place and moving in the same way. Such a dense gas is said to be degenerate (a term coined by physicists and not related to the electron\u2019s moral character). The electrons in a degenerate gas resist further crowding with tremendous pressure. (It\u2019s as if the electrons said, \u201cYou can press inward all you want, but there is simply no room for any other electrons to squeeze in here without violating the rules of our existence.\u201d) The degenerate electrons do not require an input of heat to maintain the pressure they exert, and so a star with this kind of structure, if nothing disturbs it, can last essentially forever. (Note that the repulsive force between degenerate electrons is different from, and much stronger than, the normal electrical repulsion between charges that have the same sign.) The electrons in a degenerate gas do move about, as do particles in any gas, but not with a lot of freedom. A particular electron cannot change position or momentum until another electron in an adjacent stage gets out of the way. The situation is much like that in the parking lot after a big football game. Vehicles are closely packed, and a given car cannot move until the one in front of it moves, leaving an empty space to be filled. Of course, the dying star also has atomic nuclei in it, not just electrons, but it turns out that the nuclei must be squeezed to much higher densities before their quantum nature becomes apparent. As a result, in white dwarfs, the nuclei do not exhibit degeneracy pressure. Hence, in the white dwarf stage of stellar evolution, it is", " the degeneracy pressure of the electrons, and not of the nuclei, that halts the collapse of the core. White Dwarfs White dwarfs, then, are stable, compact objects with electron-degenerate cores that cannot contract any further. Calculations showing that white dwarfs are the likely end state of low-mass stars were first carried out by the Indian-American astrophysicist Subrahmanyan Chandrasekhar. He was able to show how much a star will shrink before the degenerate electrons halt its further contraction and hence what its final diameter will be (Figure 23.2). When Chandrasekhar made his calculation about white dwarfs, he found something very surprising: the radius of a white dwarf shrinks as the mass in the star increases (the larger the mass, the more tightly packed the electrons can become, resulting in a smaller radius). According to the best theoretical models, a white dwarf with a mass of about 1.4 MSun or larger would have a radius of zero. What the calculations are telling us is that even the force of degenerate electrons cannot stop the collapse of a star with more mass than this. The maximum mass that a star can end its life with and still become a white dwarf\u20141.4 MSun\u2014is called the Chandrasekhar limit. Stars with end-of-life masses that exceed this limit have a different kind of end in store\u2014one that we will explore in the next section. 808 Chapter 23 The Death of Stars Figure 23.2. Relating Masses and Radii of White Dwarfs. Models of white-dwarf structure predict that as the mass of the star increases (toward the right), its radius gets smaller and smaller Subrahmanyan Chandrasekhar Born in 1910 in Lahore, India, Subrahmanyan Chandrasekhar (known as Chandra to his friends and colleagues) grew up in a home that encouraged scholarship and an interest in science (Figure 23.3). His uncle, C. V. Raman, was a physicist who won the 1930 Nobel Prize. A precocious student, Chandra tried to read as much as he could about the latest ideas in physics and astronomy, although obtaining technical books was not easy in India at the time. He finished college at age 19 and won a scholarship to study in England. It was during the long boat voyage to get to graduate school that he first began doing calculations about the structure of white dwarf stars. Chandra developed his ideas during and after his studies as a graduate student, showing\u2014as we have discussed\u2014that white dwarfs with masses greater than 1.4 times the mass of the Sun cannot exist and that the theory predicts the existence of other kinds of stellar corpses. He wrote later that he felt very shy and lonely during this period, isolated from students, afraid to assert himself, and sometimes waiting for hours to speak with some of the famous professors he had read about in India. His calculations soon brought him into conflict with certain distinguished astronomers, including Sir Arthur Eddington, who publicly ridiculed Chandra\u2019s ideas. At a number of meetings of astronomers, such leaders in the field as Henry Norris Russell refused to give Chandra the opportunity to defend his ideas, while allowing his more senior critics lots of time to criticize them. Yet Chandra persevered, writing books and articles elucidating his theories, which turned out not only to be correct, but to lay the foundation for much of our modern understanding of the death of stars. In 1983, he received the Nobel Prize in physics for this early work. In 1937, Chandra came to the United States and joined the faculty at the University of Chicago, where he remained for the rest of his life. There he devoted himself to research and teaching, making major contributions to many fields of astronomy, from our understanding of the motions of stars through the This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 809 Galaxy to the behavior of the bizarre objects called black holes (see Black Holes and Curved Spacetime). In 1999, NASA named its sophisticated orbiting X-ray telescope (designed in part to explore such stellar corpses) the Chandra X-ray Observatory. Figure 23.3. S. Chandrasekhar (1910\u20131995). Chandra\u2019s research provided the basis for much of what we now know about stellar corpses. (credit: modification of work by American Institute of Physics) Chandra spent a great deal of time with his graduate students, supervising the research of more than 50 PhDs during his life. He took his teaching responsibilities very seriously: during the 1940s, while based at the Yerkes Observatory, he willingly drove the more than 100-mile trip to the university each week to teach a class of only a few students. Chandra also had a deep devotion to music, art, and philosophy, writing articles and books about the relationship between the humanities and science. He once wrote that \u201cone", " can learn science the way one enjoys music or art.... Heisenberg had a marvelous phrase \u2018shuddering before the beautiful\u2019... that is the kind of feeling I have.\u201d Using the Hubble Space Telescope, astronomers were able to detect images (https://openstaxcollege.org/l/30hubimgwhidwa) of faint white dwarf stars and other \u201cstellar corpses\u201d in the M4 star cluster, located about 7200 light-years away. The Ultimate Fate of White Dwarfs If the birth of a main-sequence star is defined by the onset of fusion reactions, then we must consider the end of all fusion reactions to be the time of a star\u2019s death. As the core is stabilized by degeneracy pressure, a last shudder of fusion passes through the outside of the star, consuming the little hydrogen still remaining. Now the star is a true white dwarf: nuclear fusion in its interior has ceased. Figure 23.4 shows the path of a star like the Sun on the H\u2013R diagram during its final stages. 810 Chapter 23 The Death of Stars Figure 23.4. Evolutionary Track for a Star Like the Sun. This diagram shows the changes in luminosity and surface temperature for a star with a mass like the Sun\u2019s as it nears the end of its life. After the star becomes a giant again (point A on the diagram), it will lose more and more mass as its core begins to collapse. The mass loss will expose the hot inner core, which will appear at the center of a planetary nebula. In this stage, the star moves across the diagram to the left as it becomes hotter and hotter during its collapse (point B). At first, the luminosity remains nearly constant, but as the star begins to cool off, it becomes less and less bright (point C). It is now a white dwarf and will continue to cool slowly for billions of years until all of its remaining store of energy is radiated away. (This assumes the Sun will lose between 46\u201350% of its mass during the giant stages, based upon various theoretical models). Since a stable white dwarf can no longer contract or produce energy through fusion, its only energy source is the heat represented by the motions of the atomic nuclei in its interior. The light it emits comes from this internal stored heat, which is substantial. Gradually, however, the white dwarf radiates away all its heat into space. After many billions of years, the nuclei will be moving much more slowly, and the white dwarf will no longer shine (Figure 23.5). It will then be a black dwarf\u2014a cold stellar corpse with the mass of a star and the size of a planet. It will be composed mostly of carbon, oxygen, and neon, the products of the most advanced fusion reactions of which the star was capable. Figure 23.5. Visible Light and X-Ray Images of the Sirius Star System. (a) This image taken by the Hubble Space Telescope shows Sirius A (the large bright star), and its companion star, the white dwarf known as Sirius B (the tiny, faint star at the lower left). Sirius A and B are 8.6 light-years from Earth and are our fifth-closest star system. Note that the image has intentionally been overexposed to allow us to see Sirius B. (b) The same system is shown in X-ray taken with the Chandra Space Telescope. Note that Sirius A is fainter in X-rays than the hot white dwarf that is Sirius B. (credit a: modification of work by NASA, ESA, H. Bond, M. Barstow(University of Leicester); credit b: modification of work by NASA/SAO/CXC) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 811 We have one final surprise as we leave our low-mass star in the stellar graveyard. Calculations show that as a degenerate star cools, the atoms inside it in essence \u201csolidify\u201d into a giant, highly compact lattice (organized rows of atoms, just like in a crystal). When carbon is compressed and crystallized in this way, it becomes a giant diamond-like star. A white dwarf star might make the most impressive engagement present you could ever see, although any attempt to mine the diamond-like material inside would crush an ardent lover instantly Learn about a recent \u201cdiamond star\u201d find, a cold, white dwarf star (https://openstaxcollege.org/l/ 30diamondstar) detected in 2014, which is considered the coldest and dimmest found to date, at the website of the National Radio Astronomy Observatory. Evidence That Stars Can Shed a Lot of Mass as They Evolve Whether or not a star will become a white dwarf depends on how much mass is lost in the red", "-giant and earlier phases of evolution. All stars that have masses below the Chandrasekhar limit when they run out of fuel will become white dwarfs, no matter what mass they were born with. But which stars shed enough mass to reach this limit? One strategy for answering this question is to look in young, open clusters (which were discussed in Star Clusters). The basic idea is to search for young clusters that contain one or more white dwarf stars. Remember that more massive stars go through all stages of their evolution more rapidly than less massive ones. Suppose we find a cluster that has a white dwarf member and also contains stars on the main sequence that have 6 times the mass of the Sun. This means that only those stars with masses greater than 6 MSun have had time to exhaust their supply of nuclear energy and complete their evolution to the white dwarf stage. The star that turned into the white dwarf must therefore have had a main-sequence mass of more than 6 MSun, since stars with lower masses have not yet had time to use up their stores of nuclear energy. The star that became the white dwarf must, therefore, have gotten rid of at least 4.6 MSun so that its mass at the time nuclear energy generation ceased could be less than 1.4 MSun. Astronomers continue to search for suitable clusters to make this test, and the evidence so far suggests that stars with masses up to about 8 MSun can shed enough mass to end their lives as white dwarfs. Stars like the Sun will probably lose about 45% of their initial mass and become white dwarfs with masses less than 1.4 MSun. 23.2 EVOLUTION OF MASSIVE STARS: AN EXPLOSIVE FINISH Learning Objectives By the end of this section, you will be able to: Describe the interior of a massive star before a supernova Explain the steps of a core collapse and explosion List the hazards associated with nearby supernovae 812 Chapter 23 The Death of Stars Thanks to mass loss, then, stars with starting masses up to at least 8 MSun (and perhaps even more) probably end their lives as white dwarfs. But we know stars can have masses as large as 150 (or more) MSun. They have a different kind of death in store for them. As we will see, these stars die with a bang. Nuclear Fusion of Heavy Elements After the helium in its core is exhausted (see The Evolution of More Massive Stars), the evolution of a massive star takes a significantly different course from that of lower-mass stars. In a massive star, the weight of the outer layers is sufficient to force the carbon core to contract until it becomes hot enough to fuse carbon into oxygen, neon, and magnesium. This cycle of contraction, heating, and the ignition of another nuclear fuel repeats several more times. After each of the possible nuclear fuels is exhausted, the core contracts again until it reaches a new temperature high enough to fuse still-heavier nuclei. The products of carbon fusion can be further converted into silicon, sulfur, calcium, and argon. And these elements, when heated to a still-higher temperature, can combine to produce iron. Massive stars go through these stages very, very quickly. In really massive stars, some fusion stages toward the very end can take only months or even days! This is a far cry from the millions of years they spend in the main-sequence stage. At this stage of its evolution, a massive star resembles an onion with an iron core. As we get farther from the center, we find shells of decreasing temperature in which nuclear reactions involve nuclei of progressively lower mass\u2014silicon and sulfur, oxygen, neon, carbon, helium, and finally, hydrogen (Figure 23.6). Figure 23.6. Structure of an Old Massive Star. Just before its final gravitational collapse, the core of a massive star resembles an onion. The iron core is surrounded by layers of silicon and sulfur, oxygen, neon, carbon mixed with some oxygen, helium, and finally hydrogen. Outside the core, the composition is mainly hydrogen and helium. (Note that this diagram is not precisely to scale but is just meant to convey the general idea of what such a star would be like.) (credit: modification of work by ESO, Digitized Sky Survey) But there is a limit to how long this process of building up elements by fusion can go on. The fusion of silicon into iron turns out to be the last step in the sequence of nonexplosive element production. Up to this point, each fusion reaction has produced energy because the nucleus of each fusion product has been a bit more stable than the nuclei that formed it. As discussed in The Sun: A Nuclear Powerhouse, light nuclei give up some of their binding energy in the process of fusing into more tightly bound, heavier nuclei. It is this released energy that maintains the outward pressure in the core so that the star does not collapse. But of all the nuclei known, iron is the most tightly bound", " and thus the most stable. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 813 You might think of the situation like this: all smaller nuclei want to \u201cgrow up\u201d to be like iron, and they are willing to pay (produce energy) to move toward that goal. But iron is a mature nucleus with good self-esteem, perfectly content being iron; it requires payment (must absorb energy) to change its stable nuclear structure. This is the exact opposite of what has happened in each nuclear reaction so far: instead of providing energy to balance the inward pull of gravity, any nuclear reactions involving iron would remove some energy from the core of the star. Unable to generate energy, the star now faces catastrophe. Collapse into a Ball of Neutrons When nuclear reactions stop, the core of a massive star is supported by degenerate electrons, just as a white dwarf is. For stars that begin their evolution with masses of at least 10 MSun, this core is likely made mainly of iron. (For stars with initial masses in the range 8 to 10 MSun, the core is likely made of oxygen, neon, and magnesium, because the star never gets hot enough to form elements as heavy as iron. The exact composition of the cores of stars in this mass range is very difficult to determine because of the complex physical characteristics in the cores, particularly at the very high densities and temperatures involved.) We will focus on the more massive iron cores in our discussion. While no energy is being generated within the white dwarf core of the star, fusion still occurs in the shells that surround the core. As the shells finish their fusion reactions and stop producing energy, the ashes of the last reaction fall onto the white dwarf core, increasing its mass. As Figure 23.2 shows, a higher mass means a smaller core. The core can contract because even a degenerate gas is still mostly empty space. Electrons and atomic nuclei are, after all, extremely small. The electrons and nuclei in a stellar core may be crowded compared to the air in your room, but there is still lots of space between them. The electrons at first resist being crowded closer together, and so the core shrinks only a small amount. Ultimately, however, the iron core reaches a mass so large that even degenerate electrons can no longer support it. When the density reaches 4 \u00d7 1011 g/cm3 (400 billion times the density of water), some electrons are actually squeezed into the atomic nuclei, where they combine with protons to form neutrons and neutrinos. This transformation is not something that is familiar from everyday life, but becomes very important as such a massive star core collapses. Some of the electrons are now gone, so the core can no longer resist the crushing mass of the star\u2019s overlying layers. The core begins to shrink rapidly. More and more electrons are now pushed into the atomic nuclei, which ultimately become so saturated with neutrons that they cannot hold onto them. At this point, the neutrons are squeezed out of the nuclei and can exert a new force. As is true for electrons, it turns out that the neutrons strongly resist being in the same place and moving in the same way. The force that can be exerted by such degenerate neutrons is much greater than that produced by degenerate electrons, so unless the core is too massive, they can ultimately stop the collapse. This means the collapsing core can reach a stable state as a crushed ball made mainly of neutrons, which astronomers call a neutron star. We don\u2019t have an exact number (a \u201cChandrasekhar limit\u201d) for the maximum mass of a neutron star, but calculations tell us that the upper mass limit of a body made of neutrons might only be about 3 MSun. So if the mass of the core were greater than this, then even neutron degeneracy would not be able to stop the core from collapsing further. The dying star must end up as something even more extremely compressed, which until recently was believed to be only one possible type of object\u2014the state of ultimate compaction known as a black hole (which is the subject of our next chapter). This is because no force was believed to exist that could stop a collapse beyond the neutron star stage. 814 Chapter 23 The Death of Stars Collapse and Explosion When the collapse of a high-mass star\u2019s core is stopped by degenerate neutrons, the core is saved from further destruction, but it turns out that the rest of the star is literally blown apart. Here\u2019s how it happens. The collapse that takes place when electrons are absorbed into the nuclei is very rapid. In less than a second, a core with a mass of about 1 MSun, which originally was approximately the size of Earth, collapses to a diameter of less than 20 kilometers. The speed with which material falls inward reaches one-", "fourth the speed of light. The collapse halts only when the density of the core exceeds the density of an atomic nucleus (which is the densest form of matter we know). A typical neutron star is so compressed that to duplicate its density, we would have to squeeze all the people in the world into a single sugar cube! This would give us one sugar cube\u2019s worth (one cubic centimeter\u2019s worth) of a neutron star. The neutron degenerate core strongly resists further compression, abruptly halting the collapse. The shock of the sudden jolt initiates a shock wave that starts to propagate outward. However, this shock alone is not enough to create a star explosion. The energy produced by the outflowing matter is quickly absorbed by atomic nuclei in the dense, overlying layers of gas, where it breaks up the nuclei into individual neutrons and protons. Our understanding of nuclear processes indicates (as we mentioned above) that each time an electron and a proton in the star\u2019s core merge to make a neutron, the merger releases a neutrino. These ghostly subatomic particles, introduced in The Sun: A Nuclear Powerhouse, carry away some of the nuclear energy. It is their presence that launches the final disastrous explosion of the star. The total energy contained in the neutrinos is huge. In the initial second of the star\u2019s explosion, the power carried by the neutrinos (1046 watts) is greater than the power put out by all the stars in over a billion galaxies. While neutrinos ordinarily do not interact very much with ordinary matter (we earlier accused them of being downright antisocial), matter near the center of a collapsing star is so dense that the neutrinos do interact with it to some degree. They deposit some of this energy in the layers of the star just outside the core. This huge, sudden input of energy reverses the infall of these layers and drives them explosively outward. Most of the mass of the star (apart from that which went into the neutron star in the core) is then ejected outward into space. As we saw earlier, such an explosion requires a star of at least 8 MSun, and the neutron star can have a mass of at most 3 MSun. Consequently, at least five times the mass of our Sun is ejected into space in each such explosive event! The resulting explosion is called a supernova (Figure 23.7). When these explosions happen close by, they can be among the most spectacular celestial events, as we will discuss in the next section. (Actually, there are at least two different types of supernova explosions: the kind we have been describing, which is the collapse of a massive star, is called, for historical reasons, a type II supernova. We will describe how the types differ later in this chapter). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 815 Figure 23.7. Five Supernova Explosions in Other Galaxies. The arrows in the top row of images point to the supernovae. The bottom row shows the host galaxies before or after the stars exploded. Each of these supernovae exploded between 3.5 and 10 billion years ago. Note that the supernovae when they first explode can be as bright as an entire galaxy. (credit: modification of work by NASA, ESA, and A. Riess (STScI)) Table 23.1 summarizes the discussion so far about what happens to stars and substellar objects of different initial masses at the ends of their lives. Like so much of our scientific understanding, this list represents a progress report: it is the best we can do with our present models and observations. The mass limits corresponding to various outcomes may change somewhat as models are improved. There is much we do not yet understand about the details of what happens when stars die. The Ultimate Fate of Stars and Substellar Objects with Different Masses Initial Mass (Mass of Sun = 1)[1] Final State at the End of Its Life < 0.01 0.01 to 0.08 0.08 to 0.25 0.25 to 8 8 to 10 10 to 40 > 40 Table 23.1 Planet Brown dwarf White dwarf made mostly of helium White dwarf made mostly of carbon and oxygen White dwarf made of oxygen, neon, and magnesium Supernova explosion that leaves a neutron star Supernova explosion that leaves a black hole 1 Stars in the mass ranges 0.25\u20138 and 8\u201310 may later produce a type of supernova different from the one we have discussed so far. These are discussed in The Evolution of Binary Star Systems. 816 Chapter 23 The Death of Stars The Supernova Giveth and the Supernova Taketh Away After the supernova explosion, the life of a massive star comes to an end. But the death of each massive star is an important event in the history of its galaxy. The elements built up by fusion during the", " star\u2019s life are now \u201crecycled\u201d into space by the explosion, making them available to enrich the gas and dust that form new stars and planets. Because these heavy elements ejected by supernovae are critical for the formation of planets and the origin of life, it\u2019s fair to say that without mass loss from supernovae and planetary nebulae, neither the authors nor the readers of this book would exist. But the supernova explosion has one more creative contribution to make, one we alluded to in Stars from Adolescence to Old Age when we asked where the atoms in your jewelry came from. The supernova explosion produces a flood of energetic neutrons that barrel through the expanding material. These neutrons can be absorbed by iron and other nuclei where they can turn into protons. Thus, they build up elements that are more massive than iron, including such terrestrial favorites as gold and silver. This is the only place we know where such heavier atoms as lead or uranium can be made. Next time you wear some gold jewelry (or give some to your sweetheart), bear in mind that those gold atoms were once part of an exploding star! When supernovae explode, these elements (as well as the ones the star made during more stable times) are ejected into the existing gas between the stars and mixed with it. Thus, supernovae play a crucial role in enriching their galaxy with heavier elements, allowing, among other things, the chemical elements that make up earthlike planets and the building blocks of life to become more common as time goes on (Figure 23.8). Figure 23.8. Kepler Supernova Remant. This image shows the expanding remains of a supernova explosion, which was first seen about 400 years ago by sky watchers, including the famous astronomer Johannes Kepler. The bubble-shaped shroud of gas and dust is now 14 light-years wide and is expanding at 2,000 kilometers per second (4 million miles per hour). The remnant emits energy at wavelengths from X-rays (shown in blue and green) to visible light (yellow) and into the infrared (red). The expanding shell is rich in iron, which was produced in the star that exploded. The main image combines the individual single-color images seen at the bottom into one multi-wavelength picture. (credit: modification of work by NASA, ESA, R. Sankrit and W. Blair (Johns Hopkins University)) Supernovae are also thought to be the source of many of the high-energy cosmic ray particles discussed in Cosmic Rays. Trapped by the magnetic field of the Galaxy, the particles from exploded stars continue to circulate around the vast spiral of the Milky Way. Scientists speculate that high-speed cosmic rays hitting the genetic material of Earth organisms over billions of years may have contributed to the steady mutations\u2014subtle changes in the genetic code\u2014that drive the evolution of life on our planet. In all the ways we have mentioned, supernovae have played a part in the development of new generations of stars, planets, and life. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 817 But supernovae also have a dark side. Suppose a life form has the misfortune to develop around a star that happens to lie near a massive star destined to become a supernova. Such life forms may find themselves snuffed out when the harsh radiation and high-energy particles from the neighboring star\u2019s explosion reach their world. If, as some astronomers speculate, life can develop on many planets around long-lived (lower- mass) stars, then the suitability of that life\u2019s own star and planet may not be all that matters for its long-term evolution and survival. Life may well have formed around a number of pleasantly stable stars only to be wiped out because a massive nearby star suddenly went supernova. Just as children born in a war zone may find themselves the unjust victims of their violent neighborhood, life too close to a star that goes supernova may fall prey to having been born in the wrong place at the wrong time. What is a safe distance to be from a supernova explosion? A lot depends on the violence of the particular explosion, what type of supernova it is (see The Evolution of Binary Star Systems), and what level of destruction we are willing to accept. Calculations suggest that a supernova less than 50 light-years away from us would certainly end all life on Earth, and that even one 100 light-years away would have drastic consequences for the radiation levels here. One minor extinction of sea creatures about 2 million years ago on Earth may actually have been caused by a supernova at a distance of about 120 light-years. The good news is that there are at present no massive stars that promise to become supernovae within 50 light- years of the Sun. (This is in part because the kinds of massive stars that become supernovae", " are overall quite rare.) The massive star closest to us, Spica (in the constellation of Virgo), is about 260 light-years away, probably a safe distance, even if it were to explode as a supernova in the near future Extreme Gravity In this section, you were introduced to some very dense objects. How would those objects\u2019 gravity affect you? Recall that the force of gravity, F, between two bodies is calculated as F = GM1 M2 R2 where G is the gravitational constant, 6.67 \u00d7 10\u201311 Nm2/kg2, M1 and M2 are the masses of the two bodies, and R is their separation. Also, from Newton\u2019s second law, where a is the acceleration of a body with mass M. F = M \u00d7 a So let\u2019s consider the situation of a mass\u2014say, you\u2014standing on a body, such as Earth or a white dwarf (where we assume you will be wearing a heat-proof space suit). You are M1 and the body you are standing on is M2. The distance between you and the center of gravity of the body on which you stand is its radius, R. The force exerted on you is Solving for a, the acceleration of gravity on that world, we get F = M 1 \u00d7 a = GM1 M 2/ R2 g = (G \u00d7 M) R2 818 Chapter 23 The Death of Stars Note that we have replaced the general symbol for acceleration, a, with the symbol scientists use for the acceleration of gravity, g. Say that a particular white dwarf has the mass of the Sun (2 \u00d7 1030 kg) but the radius of Earth (6.4 \u00d7 106 m). What is the acceleration of gravity at the surface of the white dwarf? Solution The acceleration of gravity at the surface of the white dwarf is g(white dwarf) = \u239e \u23a0 \u239b \u239dG \u00d7 MSun REarth 2 = \u239b \u239d6.67 \u00d7 10\u221211 m2 /kg s2 \u00d7 2 \u00d7 1030 kg \u239e \u23a0 \u239e \u239b \u239d6.4 \u00d7 106 m \u23a0 2 = 3.26 \u00d7 106 m/s2 Compare this to g on the surface of Earth, which is 9.8 m/s2. Check Your Learning What is the acceleration of gravity at the surface if the white dwarf has the twice the mass of the Sun and is only half the radius of Earth? Answer: g(white dwarf) = \u239e \u23a0 \u239b \u239dG \u00d7 2MSun \u239d0.5REarth \u239e \u23a0 \u239b 2 = \u239b \u239d6.67 \u00d7 10\u221211 m2 /kg s2 \u00d7 4 \u00d7 1030 kg \u239d3.2 \u00d7 106\u239e \u239b \u23a0 2 \u239e \u23a0 = 2.61 \u00d7 107 m/s2 23.3 SUPERNOVA OBSERVATIONS Learning Objectives By the end of this section, you will be able to: Describe the observed features of SN 1987A both before and after the supernova Explain how observations of various parts of the SN 1987A event helped confirm theories about supernovae Supernovae were discovered long before astronomers realized that these spectacular cataclysms mark the death of stars (see Making Connections: Supernovae in History). The word nova means \u201cnew\u201d in Latin; before telescopes, when a star too dim to be seen with the unaided eye suddenly flared up in a brilliant explosion, observers concluded it must be a brand-new star. Twentieth-century astronomers reclassified the explosions with the greatest luminosity as supernovae. From historical records of such explosions, from studies of the remnants of supernovae in our Galaxy, and from analyses of supernovae in other galaxies, we estimate that, on average, one supernova explosion occurs somewhere in the Milky Way Galaxy every 25 to 100 years. Unfortunately, however, no supernova explosion has been observable in our Galaxy since the invention of the telescope. Either we have been exceptionally unlucky or, more likely, recent explosions have taken place in parts of the Galaxy where interstellar dust blocks light from reaching us. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 819 Supernovae in History Although many supernova explosions in our own Galaxy have gone unnoticed, a few were so spectacular that they were clearly seen and recorded by sky watchers and historians at the time. We can use these records, going back two millennia, to help us pinpoint where the exploding stars were and thus where to look for their remnants today. The most dramatic supernova was observed in the year 1006. It appeared in May as a brilliant point of light visible during the", " daytime, perhaps 100 times brighter than the planet Venus. It was bright enough to cast shadows on the ground during the night and was recorded with awe and fear by observers all over Europe and Asia. No one had seen anything like it before; Chinese astronomers, noting that it was a temporary spectacle, called it a \u201cguest star.\u201d Astronomers David Clark and Richard Stephenson have scoured records from around the world to find more than 20 reports of the 1006 supernova ( SN 1006) (Figure 23.9). This has allowed them to determine with some accuracy where in the sky the explosion occurred. They place it in the modern constellation of Lupus; at roughly the position they have determined, we find a supernova remnant, now quite faint. From the way its filaments are expanding, it indeed appears to be about 1000 years old. Figure 23.9. Supernova 1006 Remnant. This composite view of SN 1006 from the Chandra X-Ray Observatory shows the X-rays coming from the remnant in blue, visible light in white-yellow, and radio emission in red. (credit: modification of work by NASA, ESA, Zolt Levay(STScI)) Another guest star, now known as SN 1054, was clearly recorded in Chinese records in July 1054. The remnant of that star is one of the most famous and best-studied objects in the sky, called the Crab Nebula (Figure 23.14). It is a marvelously complex object, which has been key to understanding the death of massive stars. When its explosion was first seen, we estimate that it was about as bright as the planet Jupiter: nowhere near as dazzling as the 1006 event but still quite dramatic to anyone who kept track of objects in the sky. Another fainter supernova was seen in 1181. The next supernova became visible in November 1572 and, being brighter than the planet Venus, was quickly spotted by a number of observers, including the young Tycho Brahe (see Orbits and Gravity). His careful measurements of the star over a year and a half showed that it was not a comet or something in Earth\u2019s atmosphere since it did not move relative to the stars. He correctly deduced that it must be a 820 Chapter 23 The Death of Stars phenomenon belonging to the realm of the stars, not of the solar system. The remnant of Tycho\u2019s Supernova (as it is now called) can still be detected in many different bands of the electromagnetic spectrum. Not to be outdone, Johannes Kepler, Tycho Brahe\u2019s scientific heir, found his own supernova in 1604, now known as Kepler\u2019s Supernova (Figure 23.8). Fainter than Tycho\u2019s, it nevertheless remained visible for about a year. Kepler wrote a book about his observations that was read by many with an interest in the heavens, including Galileo. No supernova has been spotted in our Galaxy for the past 300 years. Since the explosion of a visible supernova is a chance event, there is no way to say when the next one might occur. Around the world, dozens of professional and amateur astronomers keep a sharp lookout for \u201cnew\u201d stars that appear overnight, hoping to be the first to spot the next guest star in our sky and make a little history themselves. At their maximum brightness, the most luminous supernovae have about 10 billion times the luminosity of the Sun. For a brief time, a supernova may outshine the entire galaxy in which it appears. After maximum brightness, the star\u2019s light fades and disappears from telescopic visibility within a few months or years. At the time of their outbursts, supernovae eject material at typical velocities of 10,000 kilometers per second (and speeds twice that have been observed). A speed of 20,000 kilometers per second corresponds to about 45 million miles per hour, truly an indication of great cosmic violence. Supernovae are classified according to the appearance of their spectra, but in this chapter, we will focus on the two main causes of supernovae. Type Ia supernovae are ignited when a lot of material is dumped on degenerate white dwarfs (Figure 23.10); these supernovae will be discussed later in this chapter. For now, we will continue our story about the death of massive stars and focus on type II supernovae, which are produced when the core of a massive star collapses. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 821 Figure 23.10. Supernova 2014J. This image of supernova 2014J, located in Messier 82 (M82), which is also known as the Cigar galaxy, was taken by the Hubble Space Telescope and is superposed on a mosaic image of the galaxy also taken with Hubble. The supernova event is indicated by the box", " and the inset. This explosion was produced by a type Ia supernova, which is theorized to be triggered in binary systems consisting of a white dwarf and another star\u2014and could be a second white dwarf, a star like our Sun, or a giant star. This type of supernova will be discussed later in this chapter. At a distance of approximately 11.5 million light-years from Earth, this is the closest supernova of type Ia discovered in the past few decades. In the image, you can see reddish plumes of hydrogen coming from the central region of the galaxy, where a considerable number of young stars are being born. (credit: modification of work by NASA, ESA, A. Goobar (Stockholm University), and the Hubble Heritage Team (STScI/AURA)) Supernova 1987A Our most detailed information about what happens when a type II supernova occurs comes from an event that was observed in 1987. Before dawn on February 24, Ian Shelton, a Canadian astronomer working at an observatory in Chile, pulled a photographic plate from the developer. Two nights earlier, he had begun a survey of the Large Magellanic Cloud, a small galaxy that is one of the Milky Way\u2019s nearest neighbors in space. Where he expected to see only faint stars, he saw a large bright spot. Concerned that his photograph was flawed, Shelton went outside to look at the Large Magellanic Cloud... and saw that a new object had indeed appeared in the sky (see Figure 23.11). He soon realized that he had discovered a supernova, one that could be seen with the unaided eye even though it was about 160,000 light-years away. 822 Chapter 23 The Death of Stars Figure 23.11. Hubble Space Telescope Image of SN 1987A. The supernova remnant with its inner and outer red rings of material is located in the Large Magellanic Cloud. This image is a composite of several images taken in 1994, 1996, and 1997\u2014about a decade after supernova 1987A was first observed. (credit: modification of work by the Hubble Heritage Team (AURA/STScI/NASA/ESA)) Now known as SN 1987A, since it was the first supernova discovered in 1987, this brilliant newcomer to the southern sky gave astronomers their first opportunity to study the death of a relatively nearby star with modern instruments. It was also the first time astronomers had observed a star before it became a supernova. The star that blew up had been included in earlier surveys of the Large Magellanic Cloud, and as a result, we know the star was a blue supergiant just before the explosion. By combining theory and observations at many different wavelengths, astronomers have reconstructed the life story of the star that became SN 1987A. Formed about 10 million years ago, it originally had a mass of about 20 MSun. For 90% of its life, it lived quietly on the main sequence, converting hydrogen into helium. At this time, its luminosity was about 60,000 times that of the Sun (LSun), and its spectral type was O. When the hydrogen in the center of the star was exhausted, the core contracted and ultimately became hot enough to fuse helium. By this time, the star was a red supergiant, emitting about 100,000 times more energy than the Sun. While in this stage, the star lost some of its mass. This lost material has actually been detected by observations with the Hubble Space Telescope (Figure 23.12). The gas driven out into space by the subsequent supernova explosion is currently colliding with the material the star left behind when it was a red giant. As the two collide, we see a glowing ring. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 823 Figure 23.12. Ring around Supernova 1987A. These two images show a ring of gas expelled about 30,000 years ago when the star that exploded in 1987 was a red giant. The supernova, which has been artificially dimmed, is located at the center of the ring. The left-hand image was taken in 1997 and the right-hand image in 2003. Note that the number of bright spots has increased from 1 to more than 15 over this time interval. These spots occur where high-speed gas ejected by the supernova and moving at millions of miles per hour has reached the ring and blasted into it. The collision has heated the gas in the ring and caused it to glow more brightly. The fact that we see individual spots suggests that material ejected by the supernova is first hitting narrow, inward-projecting columns of gas in the clumpy ring. The hot spots are the first signs of a dramatic and violent collision between the new and old material that will continue over the next few years. By studying these bright spots, astronomers can determine the composition of the ring and hence learn about the nuclear", " processes that build heavy elements inside massive stars. (credit: modification of work by NASA, P. Challis, R. Kirshner (Harvard-Smithsonian Center for Astrophysics) and B. Sugerman (STScI)) Helium fusion lasted only about 1 million years. When the helium was exhausted at the center of the star, the core contracted again, the radius of the surface also decreased, and the star became a blue supergiant with a luminosity still about equal to 100,000 LSun. This is what it still looked like on the outside when, after brief periods of further fusion, it reached the iron crisis we discussed earlier and exploded. Some key stages of evolution of the star that became SN 1987A, including the ones following helium exhaustion, are listed in Table 23.2. While we don\u2019t expect you to remember these numbers, note the patterns in the table: each stage of evolution happens more quickly than the preceding one, the temperature and pressure in the core increase, and progressively heavier elements are the source of fusion energy. Once iron was created, the collapse began. It was a catastrophic collapse, lasting only a few tenths of a second; the speed of infall in the outer portion of the iron core reached 70,000 kilometers per second, about one-fourth the speed of light. Evolution of the Star That Exploded as SN 1987A Phase Central Temperature (K) Central Density (g/cm3) Time Spent in This Phase Hydrogen fusion 40 \u00d7 106 Helium fusion 190 \u00d7 106 Carbon fusion 870 \u00d7 106 Neon fusion 1.6 \u00d7 109 Oxygen fusion 2.0 \u00d7 109 Silicon fusion 3.3 \u00d7 109 Core collapse 200 \u00d7 109 Table 23.2 5 970 170,000 3.0 \u00d7 106 5.6 \u00d7 106 4.3 \u00d7 107 2 \u00d7 1014 8 \u00d7 106 years 106 years 2000 years 6 months 1 year Days Tenths of a second 824 Chapter 23 The Death of Stars In the meantime, as the core was experiencing its last catastrophe, the outer shells of neon, oxygen, carbon, helium, and hydrogen in the star did not yet know about the collapse. Information about the physical movement of different layers travels through a star at the speed of sound and cannot reach the surface in the few tenths of a second required for the core collapse to occur. Thus, the surface layers of our star hung briefly suspended, much like a cartoon character who dashes off the edge of a cliff and hangs momentarily in space before realizing that he is no longer held up by anything. The collapse of the core continued until the densities rose to several times that of an atomic nucleus. The resistance to further collapse then became so great that the core rebounded. Infalling material ran into the \u201cbrick wall\u201d of the rebounding core and was thrown outward with a great shock wave. Neutrinos poured out of the core, helping the shock wave blow the star apart. The shock reached the surface of the star a few hours later, and the star began to brighten into the supernova Ian Shelton observed in 1987. The Synthesis of Heavy Elements The variations in the brightness of SN 1987A in the days and months after its discovery, which are shown in Figure 23.13, helped confirm our ideas about heavy element production. In a single day, the star soared in brightness by a factor of about 1000 and became just visible without a telescope. The star then continued to increase slowly in brightness until it was about the same apparent magnitude as the stars in the Little Dipper. Up until about day 40 after the outburst, the energy being radiated away was produced by the explosion itself. But then SN 1987A did not continue to fade away, as we might have expected the light from the explosion to do. Instead, SN 1987A remained bright as energy from newly created radioactive elements came into play. Figure 23.13. Change in the Brightness of SN 1987A over Time. Note how the rate of decline of the supernova\u2019s light slowed between days 40 and 500. During this time, the brightness was mainly due to the energy emitted by newly formed (and quickly decaying) radioactive elements. Remember that magnitudes are a backward measure of brightness: the larger the magnitude, the dimmer the object looks. One of the elements formed in a supernova explosion is radioactive nickel, with an atomic mass of 56 (that is, the total number of protons plus neutrons in its nucleus is 56). Nickel-56 is unstable and changes spontaneously (with a half-life of about 6 days) to cobalt-56. (Recall that a half-life is the time it takes for half the nuclei in a sample to undergo radioactive decay.) Cobalt-56 in turn decays with a half-life of about 77 days to iron-56, which is stable. Energetic gamma rays are emitted when these radioactive nuclei decay. Those gamma rays then serve This OpenStax book is available for free at", " http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 825 as a new source of energy for the expanding layers of the supernova. The gamma rays are absorbed in the overlying gas and re-emitted at visible wavelengths, keeping the remains of the star bright. As you can see in Figure 23.13, astronomers did observe brightening due to radioactive nuclei in the first few months following the supernova\u2019s outburst and then saw the extra light die away as more and more of the radioactive nuclei decayed to stable iron. The gamma-ray heating was responsible for virtually all of the radiation detected from SN 1987A after day 40. Some gamma rays also escaped directly without being absorbed. These were detected by Earth-orbiting telescopes at the wavelengths expected for the decay of radioactive nickel and cobalt, clearly confirming our understanding that new elements were indeed formed in the crucible of the supernova. Neutrinos from SN 1987A If there had been any human observers in the Large Magellanic Cloud about 160,000 years ago, the explosion we call SN 1987A would have been a brilliant spectacle in their skies. Yet we know that less than 1/10 of 1% of the energy of the explosion appeared as visible light. About 1% of the energy was required to destroy the star, and the rest was carried away by neutrinos. The overall energy in these neutrinos was truly astounding. In the initial second of the event, as we noted earlier in our general discussion of supernovae, their total luminosity exceeded the luminosity of all the stars in over a billion galaxies. And the supernova generated this energy in a volume less than 50 kilometers in diameter! Supernovae are one of the most violent events in the universe, and their light turns out to be only the tip of the iceberg in revealing how much energy they produce. In 1987, the neutrinos from SN 1987A were detected by two instruments\u2014which might be called \u201cneutrino telescopes\u201d\u2014almost a full day before Shelton\u2019s observations. (This is because the neutrinos get out of the exploding star more easily than light does, and also because you don\u2019t need to wait until nightfall to catch a \u201cglimpse\u201d of them.) Both neutrino telescopes, one in a deep mine in Japan and the other under Lake Erie, consist of several thousand tons of purified water surrounded by several hundred light-sensitive detectors. Incoming neutrinos interact with the water to produce positrons and electrons, which move rapidly through the water and emit deep blue light. Altogether, 19 neutrinos were detected. Since the neutrino telescopes were in the Northern Hemisphere and the supernova occurred in the Southern Hemisphere, the detected neutrinos had already passed through Earth and were on their way back out into space when they were captured. Only a few neutrinos were detected because the probability that they will interact with ordinary matter is very, very low. It is estimated that the supernova actually released 1058 neutrinos. A tiny fraction of these, about 30 billion, eventually passed through each square centimeter of Earth\u2019s surface. About a million people actually experienced a neutrino interaction within their bodies as a result of the supernova. This interaction happened to only a single nucleus in each person and thus had absolutely no biological effect; it went completely unnoticed by everyone concerned. Since the neutrinos come directly from the heart of the supernova, their energies provided a measure of the temperature of the core as the star was exploding. The central temperature was about 200 billion K, a stunning figure to which no earthly analog can bring much meaning. With neutrino telescopes, we are peering into the final moment in the life stories of massive stars and observing conditions beyond all human experience. Yet we are also seeing the unmistakable hints of our own origins. 826 Chapter 23 The Death of Stars 23.4 PULSARS AND THE DISCOVERY OF NEUTRON STARS Learning Objectives By the end of this section, you will be able to: Explain the research method that led to the discovery of neutron stars, located hundreds or thousands of light-years away Describe the features of a neutron star that allow it to be detected as a pulsar List the observational evidence that links pulsars and neutron stars to supernovae After a type II supernova explosion fades away, all that is left behind is either a neutron star or something even more strange, a black hole. We will describe the properties of black holes in Black Holes and Curved Spacetime, but for now, we want to examine how the neutron stars we discussed earlier might become observable. Neutron stars are the densest objects in the universe; the force of gravity at their surface is 1011 times greater than what we experience at Earth\u2019s surface. The interior of a neutron star is composed of", " about 95% neutrons, with a small number of protons and electrons mixed in. In effect, a neutron star is a giant atomic nucleus, with a mass about 1057 times the mass of a proton. Its diameter is more like the size of a small town or an asteroid than a star. (Table 23.3 compares the properties of neutron stars and white dwarfs.) Because it is so small, a neutron star probably strikes you as the object least likely to be observed from thousands of light-years away. Yet neutron stars do manage to signal their presence across vast gulfs of space. Properties of a Typical White Dwarf and a Neutron Star Property White Dwarf Neutron Star Mass (Sun = 1) 0.6 (always <1.4) Always >1.4 and <3 Radius Density Table 23.3 7000 km 8 \u00d7 105 g/cm3 10 km 1014 g/cm3 The Discovery of Neutron Stars In 1967, Jocelyn Bell, a research student at Cambridge University, was studying distant radio sources with a special detector that had been designed and built by her advisor Antony Hewish to find rapid variations in radio signals. The project computers spewed out reams of paper showing where the telescope had surveyed the sky, and it was the job of Hewish\u2019s graduate students to go through it all, searching for interesting phenomena. In September 1967, Bell discovered what she called \u201ca bit of scruff\u201d\u2014a strange radio signal unlike anything seen before. What Bell had found, in the constellation of Vulpecula, was a source of rapid, sharp, intense, and extremely regular pulses of radio radiation. Like the regular ticking of a clock, the pulses arrived precisely every 1.33728 seconds. Such exactness first led the scientists to speculate that perhaps they had found signals from an This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 827 intelligent civilization. Radio astronomers even half-jokingly dubbed the source \u201cLGM\u201d for \u201clittle green men.\u201d Soon, however, three similar sources were discovered in widely separated directions in the sky. When it became apparent that this type of radio source was fairly common, astronomers concluded that they were highly unlikely to be signals from other civilizations. By today, more than 2500 such sources have been discovered; they are now called pulsars, short for \u201cpulsating radio sources.\u201d The pulse periods of different pulsars range from a little longer than 1/1000 of a second to nearly 10 seconds. At first, the pulsars seemed particularly mysterious because nothing could be seen at their location on visible-light photographs. But then a pulsar was discovered right in the center of the Crab Nebula, a cloud of gas produced by SN 1054, a supernova that was recorded by the Chinese in 1054 (Figure 23.14). The energy from the Crab Nebula pulsar arrives in sharp bursts that occur 30 times each second\u2014with a regularity that would be the envy of a Swiss watchmaker. In addition to pulses of radio energy, we can observe pulses of visible light and X-rays from the Crab Nebula. The fact that the pulsar was just in the region of the supernova remnant where we expect the leftover neutron star to be immediately alerted astronomers that pulsars might be connected with these elusive \u201ccorpses\u201d of massive stars. Figure 23.14. Crab Nebula. This image shows X-ray emmisions from the Crab Nebula, which is about 6500 light-years away. The pulsar is the bright spot at the center of the concentric rings. Data taken over about a year show that particles stream away from the inner ring at about half the speed of light. The jet that is perpendicular to this ring is a stream of matter and antimatter electrons also moving at half the speed of light. (credit: modification of work by NASA/CXC/SAO) The Crab Nebula is a fascinating object. The whole nebula glows with radiation at many wavelengths, and its overall energy output is more than 100,000 times that of the Sun\u2014not a bad trick for the remnant of a supernova that exploded almost a thousand years ago. Astronomers soon began to look for a connection between the pulsar and the large energy output of the surrounding nebula View an interesting interview (https://openstaxcollege.org/l/30jocbellint) with Jocelyn Bell (Burnell) to learn about her life and work (this is part of a project at the American Institute of Physics to record interviews with pathbreaking scientists while they are still alive). 828 Chapter 23 The Death of Stars A Spinning Lighthouse Model By applying a combination of theory and observation, astronomers eventually concluded that pulsars must be spinning neutron stars. According to this model, a neutron star is something like a lighthouse on a rocky coast", " (Figure 23.15). To warn ships in all directions and yet not cost too much to operate, the light in a modern lighthouse turns, sweeping its beam across the dark sea. From the vantage point of a ship, you see a pulse of light each time the beam points in your direction. In the same way, radiation from a small region on a neutron star sweeps across the oceans of space, giving us a pulse of radiation each time the beam points toward Earth. Figure 23.15. Lighthouse. A lighthouse in California warns ships on the ocean not to approach too close to the dangerous shoreline. The lighted section at the top rotates so that its beam can cover all directions. (credit: Anita Ritenour) Neutron stars are ideal candidates for such a job because the collapse has made them so small that they can turn very rapidly. Recall the principle of the conservation of angular momentum from Newton\u2019s Great Synthesis: if an object gets smaller, it can spin more rapidly. Even if the parent star was rotating very slowly when it was on the main sequence, its rotation had to speed up as it collapsed to form a neutron star. With a diameter of only 10 to 20 kilometers, a neutron star can complete one full spin in only a fraction of a second. This is just the sort of time period we observe between pulsar pulses. Any magnetic field that existed in the original star will be highly compressed when the core collapses to a neutron star. At the surface of the neutron star, in the outer layer consisting of ordinary matter (and not just pure neutrons), protons and electrons are caught up in this spinning field and accelerated nearly to the speed of light. In only two places\u2014the north and south magnetic poles\u2014can the trapped particles escape the strong hold of the magnetic field (Figure 23.16). The same effect can be seen (in reverse) on Earth, where charged particles from space are kept out by our planet\u2019s magnetic field everywhere except near the poles. As a result, Earth\u2019s auroras (caused when charged particles hit the atmosphere at high speed) are seen mainly near the poles. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 829 Figure 23.16. Model of a Pulsar. A diagram showing how beams of radiation at the magnetic poles of a neutron star can give rise to pulses of emission as the star rotates. As each beam sweeps over Earth, like a lighthouse beam sweeping over a distant ship, we see a short pulse of radiation. This model requires that the magnetic poles be located in different places from the rotation poles. (credit \u201cstars\u201d: modification of work by Tony Hisgett) Note that in a neutron star, the magnetic north and south poles do not have to be anywhere close to the north and south poles defined by the star\u2019s rotation. In the same way, we discussed in the chapter on The Giant Planets that the magnetic poles on the planets Uranus and Neptune are not lined up with the poles of the planet\u2019s spin. Figure 23.16 shows the poles of the magnetic field perpendicular to the poles of rotation, but the two kinds of poles could make any angle. In fact, the misalignment of the rotational axis with the magnetic axis plays a crucial role in the generation of the observed pulses in this model. At the two magnetic poles, the particles from the neutron star are focused into a narrow beam and come streaming out of the whirling magnetic region at enormous speeds. They emit energy over a broad range of the electromagnetic spectrum. The radiation itself is also confined to a narrow beam, which explains why the pulsar acts like a lighthouse. As the rotation carries first one and then the other magnetic pole of the star into our view, we see a pulse of radiation each time. Tests of the Model This explanation of pulsars in terms of beams of radiation from highly magnetic and rapidly spinning neutron stars is a very clever idea. But what evidence do we have that it is the correct model? First, we can measure the masses of some pulsars, and they do turn out be in the range of 1.4 to 1.8 times that of the Sun\u2014just what theorists predict for neutron stars. The masses are found using Kepler\u2019s law for those few pulsars that are members of binary star systems. But there is an even-better confirming argument, which brings us back to the Crab Nebula and its vast energy output. When the high-energy charged particles from the neutron star pulsar hit the slower-moving material from the supernova, they energize this material and cause it to \u201cglow\u201d at many different wavelengths\u2014just what we observe from the Crab Nebula. The pulsar beams are a power source that \u201clight up\u201d the nebula long after the initial explosion of the star that made it. Who \u201cp", "ays the bills\u201d for all the energy we see coming out of a remnant like the Crab Nebula? After all, when energy emerges from one place, it must be depleted in another. The ultimate energy source in our model is the rotation of the neutron star, which propels charged particles outward and spins its magnetic field at enormous 830 Chapter 23 The Death of Stars speeds. As its rotational energy is used to excite the Crab Nebula year after year, the pulsar inside the nebula slows down. As it slows, the pulses come a little less often; more time elapses before the slower neutron star brings its beam back around. Several decades of careful observations have now shown that the Crab Nebula pulsar is not a perfectly regular clock as we originally thought: instead, it is gradually slowing down. Having measured how much the pulsar is slowing down, we can calculate how much rotation energy the neutron star is losing. Remember that it is very densely packed and spins amazingly quickly. Even a tiny slowing down can mean an immense loss of energy. To the satisfaction of astronomers, the rotational energy lost by the pulsar turns out to be the same as the amount of energy emerging from the nebula surrounding it. In other words, the slowing down of a rotating neutron star can explain precisely why the Crab Nebula is glowing with the amount of energy we observe. The Evolution of Pulsars From observations of the pulsars discovered so far, astronomers have concluded that one new pulsar is born somewhere in the Galaxy every 25 to 100 years, the same rate at which supernovae are estimated to occur. Calculations suggest that the typical lifetime of a pulsar is about 10 million years; after that, the neutron star no longer rotates fast enough to produce significant beams of particles and energy, and is no longer observable. We estimate that there are about 100 million neutron stars in our Galaxy, most of them rotating too slowly to come to our notice. The Crab pulsar is rather young (only about 960 years old) and has a short period, whereas other, older pulsars have already slowed to longer periods. Pulsars thousands of years old have lost too much energy to emit appreciably in the visible and X-ray wavelengths, and they are observed only as radio pulsars; their periods are a second or longer. There is one other reason we can see only a fraction of the pulsars in the Galaxy. Consider our lighthouse model again. On Earth, all ships approach on the same plane\u2014the surface of the ocean\u2014so the lighthouse can be built to sweep its beam over that surface. But in space, objects can be anywhere in three dimensions. As a given pulsar\u2019s beam sweeps over a circle in space, there is absolutely no guarantee that this circle will include the direction of Earth. In fact, if you think about it, many more circles in space will not include Earth than will include it. Thus, we estimate that we are unable to observe a large number of neutron stars because their pulsar beams miss us entirely. At the same time, it turns out that only a few of the pulsars discovered so far are embedded in the visible clouds of gas that mark the remnant of a supernova. This might at first seem mysterious, since we know that supernovae give rise to neutron stars and we should expect each pulsar to have begun its life in a supernova explosion. But the lifetime of a pulsar turns out to be about 100 times longer than the length of time required for the expanding gas of a supernova remnant to disperse into interstellar space. Thus, most pulsars are found with no other trace left of the explosion that produced them. In addition, some pulsars are ejected by a supernova explosion that is not the same in all directions. If the supernova explosion is stronger on one side, it can kick the pulsar entirely out of the supernova remnant (some astronomers call this \u201cgetting a birth kick\u201d). We know such kicks happen because we see a number of young supernova remnants in nearby galaxies where the pulsar is to one side of the remnant and racing away at several hundred miles per second (Figure 23.17). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 831 Figure 23.17. Speeding Pulsar. This intriguing image (which combines X-ray, visible, and radio observations) shows the jet trailing behind a pulsar (at bottom right, lined up between the two bright stars). With a length of 37 light-years, the jet trail (seen in purple) is the longest ever observed from an object in the Milky Way. (There is also a mysterious shorter, comet-like tail that is almost perpendicular to the purple jet.) Moving at a speed between 2.5 and 5 million miles per hour, the pulsar is traveling away from the core of the supernova remnant where it originated. (credit: X-ray: NASA", "/CXC/ISDC/L.Pavan et al, Radio: CSIRO/ATNF/ATCA Optical: 2MASS/UMass/IPAC-Caltech/NASA/NSF Touched by a Neutron Star On December 27, 2004, Earth was bathed with a stream of X-ray and gamma-ray radiation from a neutron star known as SGR 1806-20. What made this event so remarkable was that, despite the distance of the source, its tidal wave of radiation had measurable effects on Earth\u2019s atmosphere. The apparent brightness of this gamma-ray flare was greater than any historical star explosion. The primary effect of the radiation was on a layer high in Earth\u2019s atmosphere called the ionosphere. At night, the ionosphere is normally at a height of about 85 kilometers, but during the day, energy from the Sun ionizes more molecules and lowers the boundary of the ionosphere to a height of about 60 kilometers. The pulse of X-ray and gamma-ray radiation produced about the same level of ionization as the daytime Sun. It also caused some sensitive satellites above the atmosphere to shut down their electronics. Measurements by telescopes in space indicate that SGR 1806-20 was a special type of fast-spinning neutron star called a magnetar. Astronomers Robert Duncan and Christopher Thomson gave them this name because their magnetic fields are stronger than that of any other type of astronomical source\u2014in this case, about 800 trillion times stronger than the magnetic field of Earth. A magnetar is thought to consist of a superdense core of neutrons surrounded by a rigid crust of atoms about a mile deep with a surface made of iron. The magnetar\u2019s field is so strong that it creates huge stresses inside that can sometimes crack open the hard crust, causing a starquarke. The vibrating crust produces an enormous blast of radiation. An astronaut 0.1 light-year from this particular magnetar would have received a fatal does from the blast in less than a second. 832 Chapter 23 The Death of Stars Fortunately, we were far enough away from magnetar SGR 1806-20 to be safe. Could a magnetar ever present a real danger to Earth? To produce enough energy to disrupt the ozone layer, a magnetar would have to be located within the cloud of comets that surround the solar system, and we know no magnetars are that close. Nevertheless, it is a fascinating discovery that events on distant star corpses can have measurable effects on Earth. 23.5 THE EVOLUTION OF BINARY STAR SYSTEMS Learning Objectives By the end of this section, you will be able to: Describe the kind of binary star system that leads to a nova event Describe the type of binary star system that leads to a type Ia supernovae event Indicate how type Ia supernovae differ from type II supernovae The discussion of the life stories of stars presented so far has suffered from a bias\u2014what we might call \u201csingle- star chauvinism.\u201d Because the human race developed around a star that goes through life alone, we tend to think of most stars in isolation. But as we saw in The Stars: A Celestial Census, it now appears that as many as half of all stars may develop in binary systems\u2014those in which two stars are born in each other\u2019s gravitational embrace and go through life orbiting a common center of mass. For these stars, the presence of a close-by companion can have a profound influence on their evolution. Under the right circumstances, stars can exchange material, especially during the stages when one of them swells up into a giant or supergiant, or has a strong wind. When this happens and the companion stars are sufficiently close, material can flow from one star to another, decreasing the mass of the donor and increasing the mass of the recipient. Such mass transfer can be especially dramatic when the recipient is a stellar remnant such as a white dwarf or a neutron star. While the detailed story of how such binary stars evolve is beyond the scope of our book, we do want to mention a few examples of how the stages of evolution described in this chapter may change when there are two stars in a system. White Dwarf Explosions: The Mild Kind Let\u2019s consider the following system of two stars: one has become a white dwarf and the other is gradually transferring material onto it. As fresh hydrogen from the outer layers of its companion accumulates on the surface of the hot white dwarf, it begins to build up a layer of hydrogen. As more and more hydrogen accumulates and heats up on the surface of the degenerate star, the new layer eventually reaches a temperature that causes fusion to begin in a sudden, explosive way, blasting much of the new material away. In this way, the white dwarf quickly (but only briefly) becomes quite bright, hundreds or thousands of times its previous luminosity. To observers before the invention of the telescope,", " it seemed that a new star suddenly appeared, and they called it a nova.[2] Novae fade away in a few months to a few years. 2 We now know that this historical terminology is quite misleading since novae do not originate from new stars. In fact, quite to the contrary, novae originate from white dwarfs, which are actually the endpoint of stellar evolution for low-mass stars. But since the system of two stars was too faint to be visible to the naked eye, it did seem to people, before telescopes were invented, that a star had appeared where nothing had been visible. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 833 Hundreds of novae have been observed, each occurring in a binary star system and each later showing a shell of expelled material. A number of stars have more than one nova episode, as more material from its neighboring star accumulates on the white dwarf and the whole process repeats. As long as the episodes do not increase the mass of the white dwarf beyond the Chandrasekhar limit (by transferring too much mass too quickly), the dense white dwarf itself remains pretty much unaffected by the explosions on its surface. White Dwarf Explosions: The Violent Kind If a white dwarf accumulates matter from a companion star at a much faster rate, it can be pushed over the Chandrasekhar limit. The evolution of such a binary system is shown in Figure 23.18. When its mass approaches the Chandrasekhar mass limit (exceeds 1.4 MSun), such an object can no longer support itself as a white dwarf, and it begins to contract. As it does so, it heats up, and new nuclear reactions can begin in the degenerate core. The star \u201csimmers\u201d for the next century or so, building up internal temperature. This simmering phase ends in less than a second, when an enormous amount of fusion (especially of carbon) takes place all at once, resulting in an explosion. The fusion energy produced during the final explosion is so great that it completely destroys the white dwarf. Gases are blown out into space at velocities of about 10,000 kilometers per second, and afterward, no trace of the white dwarf remains. Figure 23.18. Evolution of a Binary System. The more massive star evolves first to become a red giant and then a white dwarf. The white dwarf then begins to attract material from its companion, which in turn evolves to become a red giant. Eventually, the white dwarf acquires so much mass that it is pushed over the Chandrasekhar limit and becomes a type Ia supernova. Such an explosion is also called a supernova, since, like the destruction of a high-mass star, it produces a huge amount of energy in a very short time. However, unlike the explosion of a high-mass star, which can leave behind a neutron star or black hole remnant, the white dwarf is completely destroyed in the process, leaving behind no remnant. We call these white dwarf explosions type Ia supernovae. We distinguish type I supernovae from those of supernovae of type II originating from the death of massive stars discussed earlier by the absence of hydrogen in their observed spectra. Hydrogen is the most common element in the universe and is a major component of massive, evolved stars. However, as we learned earlier, hydrogen is absent from the white dwarf remnant, which is primarily composed of carbon and oxygen for masses comparable to the Chandrasekhar mass limit. The \u201ca\u201d subdesignation of type Ia supernovae further refers to the presence of strong silicon absorption lines, which are absent from supernovae originating from the collapse of massive stars. Silicon is one of the products 834 Chapter 23 The Death of Stars that results from the fusion of carbon and oxygen, which bears out the scenario we described above\u2014that there is a sudden onset of the fusion of the carbon (and oxygen) of which the white dwarf was made. Observational evidence now strongly indicates that SN 1006, Tycho\u2019s Supernova, and Kepler\u2019s Supernova (see Supernovae in History) were all type Ia supernovae. For instance, in contrast to the case of SN 1054, which yielded the spinning pulsar in the Crab Nebula, none of these historical supernovae shows any evidence of stellar remnants that have survived their explosions. Perhaps even more puzzling is that, so far, astronomers have not been able to identify the companion star feeding the white dwarf in any of these historical supernovae. Consequently, in order to address the mystery of the absent companion stars and other outstanding puzzles, astronomers have recently begun to investigate alternative mechanisms of generating type Ia supernovae. All proposed mechanisms rely upon white dwarfs composed of carbon and oxygen, which are needed to meet the observed absence of hydrogen in the type Ia spectrum. And because any isolated", " white dwarf below the Chandrasekhar mass is stable, all proposed mechanisms invoke a binary companion to explode the white dwarf. The leading alternative mechanism scientists believe creates a type Ia supernova is the merger of two white dwarf stars in a binary system. The two white dwarfs may have unstable orbits, such that over time, they would slowly move closer together until they merge. If their combined mass is greater than the Chandrasekhar limit, the result could also be a type Ia supernova explosion You can watch a short video (https://openstaxcollege.org/l/30supernovavid) about Supernova SN 2014J, a type Ia supernova discovered in the Messier 82 (M82) galaxy on January 21, 2014, as well as see brief animations of the two mechanisms by which such a supernova could form. Type Ia supernovae are of great interest to astronomers in other areas of research. This type of supernova is brighter than supernovae produced by the collapse of a massive star. Thus, type Ia supernovae can be seen at very large distances, and they are found in all types of galaxies. The energy output from most type Ia supernovae is consistent, with little variation in their maximum luminosities, or in how their light output initially increases and then slowly decreases over time. These properties make type Ia supernovae extremely valuable \u201cstandard bulbs\u201d for astronomers looking out at great distances\u2014well beyond the limits of our own Galaxy. You\u2019ll learn more about their use in measuring distances to other galaxies in The Big Bang. In contrast, type II supernovae are about 5 times less luminous than type Ia supernovae and are only seen in galaxies that have recent, massive star formation. Type II supernovae are also less consistent in their energy output during the explosion and can have a range a peak luminosity values. Neutron Stars with Companions Now let\u2019s look at an even-more mismatched pair of stars in action. It is possible that, under the right circumstances, a binary system can even survive the explosion of one of its members as a type II supernova. In that case, an ordinary star can eventually share a system with a neutron star. If material is then transferred from the \u201cliving\u201d star to its \u201cdead\u201d (and highly compressed) companion, this material will be pulled in by the strong gravity of the neutron star. Such infalling gas will be compressed and heated to incredible temperatures. It will quickly become so hot that it will experience an explosive burst of fusion. The energies involved are so great that we would expect much of the radiation from the burst to emerge as X-rays. And indeed, high-energy This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 835 observatories above Earth\u2019s atmosphere (see Astronomical Instruments) have recorded many objects that undergo just these types of X-ray bursts. If the neutron star and its companion are positioned the right way, a significant amount of material can be transferred to the neutron star and can set it spinning faster (as spin energy is also transferred). The radius of the neutron star would also decrease as more mass was added. Astronomers have found pulsars in binary systems that are spinning at a rate of more than 500 times per second! (These are sometimes called millisecond pulsars since the pulses are separated by a few thousandths of a second.) Such a rapid spin could not have come from the birth of the neutron star; it must have been externally caused. (Recall that the Crab Nebula pulsar, one of the youngest pulsars known, was spinning \u201conly\u201d 30 times per second.) Indeed, some of the fast pulsars are observed to be part of binary systems, while others may be alone only because they have \u201cfully consumed\u201d their former partner stars through the mass transfer process. (These have sometimes been called \u201c black widow pulsars.\u201d) View this short video (https://openstaxcollege.org/l/30scotronvid) to see Dr. Scott Ransom, of the National Radio Astronomy Observatory, explain how millisecond pulsars come about, with some nice animations. And if you thought that a neutron star interacting with a \u201cnormal\u201d star was unusual, there are also binary systems that consist of two neutron stars. One such system has the stars in very close orbits to one another, so much that they continually alter each other\u2019s orbit. Another binary neutron star system includes two pulsars that are orbiting each other every 2 hours and 25 minutes. As we discussed earlier, pulsars radiate away their energy, and these two pulsars are slowly moving toward one another, such that in about 85 million years, they will actually merge. We have now reached the end of our description of", " the final stages of stars, yet one piece of the story remains to be filled in. We saw that stars whose core masses are less than 1.4 MSun at the time they run out of fuel end their lives as white dwarfs. Dying stars with core masses between 1.4 and about 3 MSun become neutron stars. But there are stars whose core masses are greater than 3 MSun when they exhaust their fuel supplies. What becomes of them? The truly bizarre result of the death of such massive stellar cores (called a black hole) is the subject of our next chapter. But first, we will look at an astronomical mystery that turned out to be related to the deaths of stars and was solved through clever sleuthing and a combination of observation and theory. 23.6 THE MYSTERY OF THE GAMMA-RAY BURSTS Learning Objectives By the end of this section, you will be able to: Give a brief history of how gamma-ray bursts were discovered and what instruments made the discovery possible Explain why astronomers think that gamma-ray bursts beam their energy rather than it radiating uniformly in all directions Describe how the radiation from a gamma-ray burst and its afterglow is produced 836 Chapter 23 The Death of Stars Explain how short-duration gamma-ray bursts differ from longer ones, and describe the process that makes short-duration gamma-ray bursts Explain why gamma-ray bursts may help us understand the early universe Everybody loves a good mystery, and astronomers are no exception. The mystery we will discuss in this section was first discovered in the mid-1960s, not via astronomical research, but as a result of a search for the tell-tale signs of nuclear weapon explosions. The US Defense Department launched a series of Vela satellites to make sure that no country was violating a treaty that banned the detonation of nuclear weapons in space. Since nuclear explosions produce the most energetic form of electromagnetic waves called gamma rays (see Radiation and Spectra), the Vela satellites contained detectors to search for this type of radiation. The satellites did not detect any confirmed events from human activities, but they did\u2014to everyone\u2019s surprise\u2014detect short bursts of gamma rays coming from random directions in the sky. News of the discovery was first published in 1973; however, the origin of the bursts remained a mystery. No one knew what produced the brief flashes of gamma rays or how far away the sources were. From a Few Bursts to Thousands With the launch of the Compton Gamma-Ray Observatory by NASA in 1991, astronomers began to identify many more bursts and to learn more about them (Figure 23.19). Approximately once per day, the NASA satellite detected a flash of gamma rays somewhere in the sky that lasted from a fraction of a second to several hundred seconds. Before the Compton measurements, astronomers had expected that the most likely place for the bursts to come from was the main disk of our own (pancake-shaped) Galaxy. If this had been the case, however, more bursts would have been seen in the crowded plane of the Milky Way than above or below it. Instead, the sources of the bursts were distributed isotropically; that is, they could appear anywhere in the sky with no preference for one region over another. Almost never did a second burst come from the same location. Figure 23.19. Compton Detects Gamma-Ray Bursts. (a) In 1991, the Compton Gamma-Ray Observatory was deployed by the Space Shuttle Atlantis. Weighing more than 16 tons, it was one of the largest scientific payloads ever launched into space. (b) This map of gamma-ray burst positions measured by the Compton Gamma-Ray Observatory shows the isotropic (same in all directions), uniform distribution of bursts on the sky. The map is oriented so that the disk of the Milky Way would stretch across the center line (or equator) of the oval. Note that the bursts show no preference at all for the plane of the Milky Way, as many other types of objects in the sky do. Colors indicate the total energy in the burst: red dots indicate long-duration, bright bursts; blue and purple dots show short, weaker bursts. (credit a: modification of work by NASA; credit b: modification of work by NASA/GSFC) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 837 To get a good visual sense of the degree to which the bursts come from all over the sky, watch this short animated NASA video (https://openstaxcollege.org/l/30swiftnasavid) showing the location of the first 500 bursts found by the later Swift satellite. For several years, astronomers actively debated whether the burst sources were relatively nearby or very far away\u2014the two possibilities for bursts that are isotropically distributed. Nearby locations might include the cloud of comets that surrounds the solar system or the halo of our Galaxy, which is large and spherical", ", and also surrounds us in all directions. If, on the other hand, the bursts occurred at very large distances, they could come from faraway galaxies, which are also distributed uniformly in all directions. Both the very local and the very distant hypotheses required something strange to be going on. If the bursts were coming from the cold outer reaches of our own solar system or from the halo of our Galaxy, then astronomers had to hypothesize some new kind of physical process that could produce unpredictable flashes of high-energy gamma rays in these otherwise-quiet regions of space. And if the bursts came from galaxies millions or billions of light-years away, then they must be extremely powerful to be observable at such large distances; indeed they had to be the among the biggest explosions in the universe. The First Afterglows The problem with trying to figure out the source of the gamma-ray bursts was that our instruments for detecting gamma rays could not pinpoint the exact place in the sky where the burst was happening. Early gamma-ray telescopes did not have sufficient resolution. This was frustrating because astronomers suspected that if they could pinpoint the exact position of one of these rapid bursts, then they would be able to identify a counterpart (such as a star or galaxy) at other wavelengths and learn much more about the burst, including where it came from. This would, however, require either major improvements in gamma-ray detector technology to provide better resolution or detection of the burst at some other wavelength. In the end, both techniques played a role. The breakthrough came with the launch of the Italian Dutch BeppoSAX satellite in 1996. BeppoSAX included a new type of gamma-ray telescope capable of identifying the position of a source much more accurately than previous instruments, to within a few minutes of arc on the sky. By itself, however, it was still not sophisticated enough to determine the exact source of the gamma-ray burst. After all, a box a few minutes of arc on a side could still contain many stars or other celestial objects. However, the angular resolution of BeppoSAX was good enough to tell astronomers where to point other, more precise telescopes in the hopes of detecting longer-lived electromagnetic emission from the bursts at other wavelengths. Detection of a burst at visible-light or radio wavelengths could provide a position accurate to a few seconds of arc and allow the position to be pinpointed to an individual star or galaxy. BeppoSAX carried its own X-ray telescope onboard the spacecraft to look for such a counterpart, and astronomers using visible-light and radio facilities on the ground were eager to search those wavelengths as well. Two crucial BeppoSAX burst observations in 1997 helped to resolve the mystery of the gamma-ray bursts. The first burst came in February from the direction of the constellation Orion. Within 8 hours, astronomers working with the satellite had identified the position of the burst, and reoriented the spacecraft to focus BeppoSAX\u2019s X- ray detector on the source. To their excitement, they detected a slowly fading X-ray source 8 hours after the event\u2014the first successful detection of an afterglow from a gamma-ray burst. This provided an even-better 838 Chapter 23 The Death of Stars location of the burst (accurate to about 40 seconds of arc), which was then distributed to astronomers across the world to try to detect it at even longer wavelengths. That very night, the 4.2-meter William Herschel Telescope on the Canary Islands found a fading visible-light source at the same position as the X-ray afterglow, confirming that such an afterglow could be detected in visible light as well. Eventually, the afterglow faded away, but left behind at the location of the original gamma- ray burst was a faint, fuzzy source right where the fading point of light had been\u2014a distant galaxy (Figure 23.20). This was the first piece of evidence that gamma-ray bursts were indeed very energetic objects from very far away. However, it also remained possible that the burst source was much closer to us and just happened to align with a more distant galaxy, so this one observation alone was not a conclusive demonstration of the extragalactic origin of gamma-ray bursts. Figure 23.20. Gamma-Ray Burst. This false-color Hubble Space Telescope image, taken in September 1997, shows the fading afterglow of the gamma-ray burst of February 28, 1997 and the host galaxy in which the burst originated. The left view shows the region of the burst. The enlargement shows the burst source and what appears to be its host galaxy. Note that the gamma-ray source is not in the center of the galaxy. (credit: modification of work by Andrew Fruchter (STScI), Elena Pian (ITSRE-CNR), and NASA, ESA) On May 8 of the same year, a burst came from the direction of the constellation Camelopardalis. In a coordinated international effort, BeppoSAX again fixed a reasonably precise position, and", " almost immediately a telescope on Kitt Peak in Arizona was able to catch the visible-light afterglow. Within 2 days, the largest telescope in the world (the Keck in Hawaii) collected enough light to record a spectrum of the burst. The May gamma-ray burst afterglow spectrum showed absorption features from a fuzzy object that was 4 billion light- years from the Sun, meaning that the location of the burst had to be at least this far away\u2014and possibly even farther. (How astronomers can get the distance of such an object from the Doppler shift in the spectrum is something we will discuss in Galaxies.) What that spectrum showed was clear evidence that the gamma-ray burst had taken place in a distant galaxy. Networking to Catch More Bursts After initial observations showed that the precise locations and afterglows of gamma-ray bursts could be found, astronomers set up a system to catch and pinpoint bursts on a regular basis. But to respond as quickly as needed to obtain usable results, astronomers realized that they needed to rely on automated systems rather than human observers happening to be in the right place at the right time. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 839 Now, when an orbiting high-energy telescope discovers a burst, its rough location is immediately transmitted to a Gamma-Ray Coordinates Network based at NASA\u2019s Goddard Space Flight Center, alerting observers on the ground within a few seconds to look for the visible-light afterglow. The first major success with this system was achieved by a team of astronomers from the University of Michigan, Lawrence Livermore National Laboratory, and Los Alamos National Laboratories, who designed an automated device they called the Robotic Optical Transient Search Experiment ( ROTSE), which detected a very bright visible-light counterpart in 1999. At peak, the burst was almost as bright as Neptune\u2014despite a distance (measured later by spectra from larger telescopes) of 9 billion light-years. More recently, astronomers have been able to take this a step further, using wide-field-of-view telescopes to stare at large fractions of the sky in the hope that a gamma-ray burst will occur at the right place and time, and be recorded by the telescope\u2019s camera. These wide-field telescopes are not sensitive to faint sources, but ROTSE showed that gamma-ray burst afterglows could sometimes be very bright. Astronomers\u2019 hopes were vindicated in March 2008, when an extremely bright gamma-ray burst occurred and its light was captured by two wide-field camera systems in Chile: the Polish \u201cPi of the Sky\u201d and the Russian-Italian TORTORA [Telescopio Ottimizzato per la Ricerca dei Transienti Ottici Rapidi (Italian for Telescope Optimized for the Research of Rapid Optical Transients)] (see Figure 23.21). According to the data taken by these telescopes, for a period of about 30 seconds, the light from the gamma-ray burst was bright enough that it could have been seen by the unaided eye had a person been looking in the right place at the right time. Adding to our amazement, later observations by larger telescopes demonstrated that the burst occurred at a distance of 8 billion light-years from Earth! Figure 23.21. Gamma-Ray Burst Observed in March 2008. The extremely luminous afterglow of GRB 080319B was imaged by the Swift Observatory in X-rays (left) and visible light/ultraviolet (right). (credit: modification of work by NASA/Swift/Stefan Immler, et al.) To Beam or Not to Beam The enormous distances to these events meant they had to have been astoundingly energetic to appear as bright as they were across such an enormous distance. In fact, they required so much energy that it posed a problem for gamma-ray burst models: if the source was radiating energy in all directions, then the energy released in gamma rays alone during a bright burst (such as the 1999 or 2008 events) would have been equivalent to the energy produced if the entire mass of a Sun-like star were suddenly converted into pure radiation. For a source to produce this much energy this quickly (in a burst) is a real challenge. Even if the star producing the gamma-ray burst was much more massive than the Sun (as is probably the case), there is no known means 840 Chapter 23 The Death of Stars of converting so much mass into radiation within a matter of seconds. However, there is one way to reduce the power required of the \u201cmechanism\u201d that makes gamma-ray bursts. So far, our discussion has assumed that the source of the gamma rays gives off the same amount of energy in all directions, like an incandescent light bulb. But as we discuss in Pulsars and the Discovery of Neutron Stars,", " not all sources of radiation in the universe are like this. Some produce thin beams of radiation that are concentrated into only one or two directions. A laser pointer and a lighthouse on the ocean are examples of such beamed sources on Earth (Figure 23.22). If, when a burst occurs, the gamma rays come out in only one or two narrow beams, then our estimates of the luminosity of the source can be reduced, and the bursts may be easier to explain. In that case, however, the beam has to point toward Earth for us to be able to see the burst. This, in turn, would imply that for every burst we see from Earth, there are probably many others that we never detect because their beams point in other directions. Figure 23.22. Burst That Is Beamed. This artist\u2019s conception shows an illustration of one kind of gamma-ray burst. The collapse of the core of a massive star into a black hole has produced two bright beams of light originating from the star\u2019s poles, which an observer pointed along one of these axes would see as a gamma-ray burst. The hot blue stars and gas clouds in the vicinity are meant to show that the event happened in an active star-forming region. (credit: NASA/Swift/Mary Pat Hrybyk-Keith and John Jones) Long-Duration Gamma-Ray Bursts: Exploding Stars After identifying and following large numbers of gamma-ray bursts, astronomers began to piece together clues about what kind of event is thought to be responsible for producing the gamma-ray burst. Or, rather, what kind of events, because there are at least two distinct types of gamma-ray bursts. The two\u2014like the different types of supernovae\u2014are produced in completely different ways. Observationally, the crucial distinction is how long the burst lasts. Astronomers now divide gamma-ray bursts into two categories: short-duration ones (defined as lasting less than 2 seconds, but typically a fraction of a second) and long-duration ones (defined as lasting more than 2 seconds, but typically about a minute). All of the examples we have discussed so far concern the long-duration gamma-ray bursts. These constitute most of the gamma-ray bursts that our satellites detect, and they are also brighter and easier to pinpoint. Many hundreds of long-duration gamma-ray bursts, and the properties of the galaxies in which they occurred, have now been studied in detail. Long-duration gamma-ray bursts are universally observed to come from distant galaxies that are still actively making stars. They are usually found to be located in regions of the galaxy with strong star-formation activity (such as spiral arms). Recall that the more massive a star is, the less time it spends This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 841 in each stage of its life. This suggests that the bursts come from a young and short-lived, and therefore massive type of star. Furthermore, in several cases when a burst has occurred in a galaxy relatively close to Earth (within a few billion light-years), it has been possible to search for a supernova at the same position\u2014and in nearly all of these cases, astronomers have found evidence of a supernova of type Ic going off. A type Ic is a particular type of supernova, which we did not discuss in the earlier parts of this chapter; these are produced by a massive star that has been stripped of its outer hydrogen layer. However, only a tiny fraction of type Ic supernovae produce gamma-ray bursts. Why would a massive star with its outer layers missing sometimes produce a gamma-ray burst at the same time that it explodes as a supernova? The explanation astronomers have in mind for the extra energy is the collapse of the star\u2019s core to form a spinning, magnetic black hole or neutron star. Because the star corpse is both magnetic and spinning rapidly, its sudden collapse is complex and can produce swirling jets of particles and powerful beams of radiation\u2014just like in a quasar or active galactic nucleus (objects you will learn about Active Galaxies, Quasars, and Supermassive Black Holes), but on a much faster timescale. A small amount of the infalling mass is ejected in a narrow beam, moving at speeds close to that of light. Collisions among the particles in the beam can produce intense bursts of energy that we see as a gamma-ray burst. Within a few minutes, the expanding blast from the fireball plows into the interstellar matter in the dying star\u2019s neighborhood. This matter might have been ejected from the star itself at earlier stages in its evolution. Alternatively, it could be the gas out of which the massive star and its neighbors formed. As the high-speed particles from the blast are slowed, they transfer their energy to the surrounding matter in the form of a shock wave. That shocked material emits radiation at longer wavelengths.", " This accounts for the afterglow of X-rays, visible light, and radio waves\u2014the glow comes at longer and longer wavelengths as the blast continues to lose energy. Short-Duration Gamma-Ray Bursts: Colliding Stellar Corpses What about the shorter gamma-ray bursts? The gamma-ray emission from these events lasts less than 2 seconds, and in some cases may last only milliseconds\u2014an amazingly short time. Such a timescale is difficult to achieve if they are produced in the same way as long-duration gamma-ray bursts, since the collapse of the stellar interior onto the black hole should take at least a few seconds. Astronomers looked fruitlessly for afterglows from short-duration gamma-ray bursts found by BeppoSAX and other satellites. Evidently, the afterglows fade away too quickly. Fast-responding visible-light telescopes like ROTSE were not helpful either: no matter how fast these telescopes responded, the bursts were not bright enough at visible wavelengths to be detected by these small telescopes. Once again, it took a new satellite to clear up the mystery. In this case, it was the Swift Gamma-Ray Burst Satellite, launched in 2004 by a collaboration between NASA and the Italian and UK space agencies (Figure 23.23). The design of Swift is similar to that of BeppoSAX. However, Swift is much more agile and flexible: after a gamma- ray burst occurs, the X-ray and UV telescopes can be repointed automatically within a few minutes (rather than a few hours). Thus, astronomers can observe the afterglow much earlier, when it is expected to be much brighter. Furthermore, the X-ray telescope is far more sensitive and can provide positions that are 30 times more precise than those provided by BeppoSAX, allowing bursts to be identified even without visible-light or radio observations. 842 Chapter 23 The Death of Stars Figure 23.23. Artist\u2019s Illustration of Swift. The US/UK/Italian spacecraft Swift contains on-board gamma-ray, X-ray, and ultraviolet detectors, and has the ability to automatically reorient itself to a gamma-ray burst detected by the gamma-ray instrument. Since its launch in 2005, Swift has detected and observed over a thousand bursts, including dozens of short-duration bursts. (credit: NASA, Spectrum Astro) On May 9, 2005, Swift detected a flash of gamma rays lasting 0.13 seconds in duration, originating from the constellation Coma Berenices. Remarkably, the galaxy at the X-ray position looked completely different from any galaxy in which a long-duration burst had been seen to occur. The afterglow originated from the halo of a giant elliptical galaxy 2.7 billion light-years away, with no signs of any young, massive stars in its spectrum. Furthermore, no supernova was ever detected after the burst, despite extensive searching. What could produce a burst less than a second long, originating from a region with no star formation? The leading model involves the merger of two compact stellar corpses: two neutron stars, or perhaps a neutron star and a black hole. Since many stars come in binary or multiple systems, it\u2019s possible to have systems where two such star corpses orbit one another. According to general relativity (which will be discussed in Black Holes and Curved Spacetime), the orbits of a binary star system composed of such objects should slowly decay with time, eventually (after millions or billions of years) causing the two objects to slam together in a violent but brief explosion. Because the decay of the binary orbit is so slow, we would expect more of these mergers to occur in old galaxies in which star formation has long since stopped To learn more about the merger of two neutron stars and how they can produce a burst that lasts less than a second, check out this computer simulation (https://openstaxcollege.org/l/30comsimneustr) by NASA. While it was impossible to be sure of this model based on only a single event (it is possible this burst actually came from a background galaxy and lined up with the giant elliptical only by chance), several dozen more short- This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 843 duration gamma-ray bursts have since been located by Swift, many of which also originate from galaxies with very low star-formation rates. This has given astronomers greater confidence that this model is the correct one. Still, to be fully convinced, astronomers are searching for a \u201csmoking gun\u201d signature for the merger of two ultra-dense stellar remnants. There are two examples we can think of that would provide more direct evidence. One is a very special kind of explosion, produced when neutrons stripped from the neutron stars during the violent final phase of the merger fuse together into heavy elements and then release heat due to radioactivity, producing a short-lived but", " red supernova sometimes called a kilonova. (The term is used because it is about a thousand times brighter than an ordinary nova, but not quite as \u201csuper\u201d as a traditional supernova.) Hubble observations of one short- duration gamma-ray burst in 2013 show suggestive evidence of such a signature, but need to be confirmed by future observations. The second \u201csmoking gun\u201d has been even more exciting to see: the detection of gravitational waves. As will be discussed in Black Holes and Curved Spacetime, gravitational waves are ripples in the fabric of spacetime that general relativity predicts should be produced by the acceleration of extremely massive and dense objects\u2014such as two neutron stars or black holes spiraling toward each other and colliding. The first example of gravitational waves has been observed recently from the merger of two large black holes. If a gravitational wave is observed one day to be coincident in time and space with a gamma-ray burst, this will not only confirm our theories of the origin of short gamma-ray bursts but would also be among the most spectacular demonstrations yet of Einstein\u2019s theory of general relativity. Probing the Universe with Gamma-Ray Bursts The story of how astronomers came to explain the origin of the different kinds of bursts is a good example of how the scientific process sometimes resembles good detective work. While the mystery of short-duration gamma-ray bursts is still being unraveled, the focus of studies for long-duration gamma-ray bursts has begun to change from understanding the origin of the bursts themselves (which is now fairly well-established) to using them as tools to understand the broader universe. The reason that long-duration gamma-ray bursts are useful has to do with their extreme luminosities, if only for a short time. In fact, long-duration gamma-ray bursts are so bright that they could easily be seen at distances that correspond to a few hundred million years after the expansion of the universe began, which is when theorists think that the first generation of stars formed. Some theories predict that the first stars are likely to be massive and complete their evolution in only a million years or so. If this turns out to be the case, then gamma- ray bursts (which signal the death of some of these stars) may provide us with the best way of probing the universe when stars and galaxies first began to form. So far, the most distant gamma-ray burst found (on April 29, 2009) originated a remarkable 13.2 billion light- years away\u2014meaning it happened only 600 million years after the Big Bang itself. This is comparable to the earliest and most distant galaxies found by the Hubble Space Telescope. It is not quite old enough to expect that it formed from the first generation of stars, but its appearance at this distance still gives us useful information about the production of stars in the early universe. Astronomers continue to scan the skies, looking for even more distant events signaling the deaths of stars from even further back in time. 844 Chapter 23 The Death of Stars CHAPTER 23 REVIEW KEY TERMS Chandrasekhar limit the upper limit to the mass of a white dwarf (equals 1.4 times the mass of the Sun) degenerate gas a gas that resists further compression because no two electrons can be in the same place at the same time doing the same thing (Pauli exclusion principle) millisecond pulsar a pulsar that rotates so quickly that it can give off hundreds of pulses per second (and its period is therefore measured in milliseconds) neutron star a compact object of extremely high density composed almost entirely of neutrons nova the cataclysmic explosion produced in a binary system, temporarily increasing its luminosity by hundreds to thousands of times pulsar a variable radio source of small physical size that emits very rapid radio pulses in very regular periods that range from fractions of a second to several seconds; now understood to be a rotating, magnetic neutron star that is energetic enough to produce a detectable beam of radiation and particles type II supernova a stellar explosion produced at the endpoint of the evolution of stars whose mass exceeds roughly 10 times the mass of the Sun SUMMARY 23.1 The Death of Low-Mass Stars During the course of their evolution, stars shed their outer layers and lose a significant fraction of their initial mass. Stars with masses of 8 MSun or less can lose enough mass to become white dwarfs, which have masses less than the Chandrasekhar limit (about 1.4 MSun). The pressure exerted by degenerate electrons keeps white dwarfs from contracting to still-smaller diameters. Eventually, white dwarfs cool off to become black dwarfs, stellar remnants made mainly of carbon, oxygen, and neon. 23.2 Evolution of Massive Stars: An Explosive Finish In a massive star, hydrogen fusion in the core is followed by several other fusion reactions involving heavier elements. Just before it exhausts all sources of energy, a massive star has an iron core surrounded by shells of silicon, sulfur, oxygen, neon, carbon, helium,", " and hydrogen. The fusion of iron requires energy (rather than releasing it). If the mass of a star\u2019s iron core exceeds the Chandrasekhar limit (but is less than 3 MSun), the core collapses until its density exceeds that of an atomic nucleus, forming a neutron star with a typical diameter of 20 kilometers. The core rebounds and transfers energy outward, blowing off the outer layers of the star in a type II supernova explosion. 23.3 Supernova Observations A supernova occurs on average once every 25 to 100 years in the Milky Way Galaxy. Despite the odds, no supernova in our Galaxy has been observed from Earth since the invention of the telescope. However, one nearby supernova (SN 1987A) has been observed in a neighboring galaxy, the Large Magellanic Cloud. The star that evolved to become SN 1987A began its life as a blue supergiant, evolved to become a red supergiant, and returned to being a blue supergiant at the time it exploded. Studies of SN 1987A have detected neutrinos from the core collapse and confirmed theoretical calculations of what happens during such explosions, including This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 845 the formation of elements beyond iron. Supernovae are a main source of high-energy cosmic rays and can be dangerous for any living organisms in nearby star systems. 23.4 Pulsars and the Discovery of Neutron Stars At least some supernovae leave behind a highly magnetic, rapidly rotating neutron star, which can be observed as a pulsar if its beam of escaping particles and focused radiation is pointing toward us. Pulsars emit rapid pulses of radiation at regular intervals; their periods are in the range of 0.001 to 10 seconds. The rotating neutron star acts like a lighthouse, sweeping its beam in a circle and giving us a pulse of radiation when the beam sweeps over Earth. As pulsars age, they lose energy, their rotations slow, and their periods increase. 23.5 The Evolution of Binary Star Systems When a white dwarf or neutron star is a member of a close binary star system, its companion star can transfer mass to it. Material falling gradually onto a white dwarf can explode in a sudden burst of fusion and make a nova. If material falls rapidly onto a white dwarf, it can push it over the Chandrasekhar limit and cause it to explode completely as a type Ia supernova. Another possible mechanism for a type Ia supernova is the merger of two white dwarfs. Material falling onto a neutron star can cause powerful bursts of X-ray radiation. Transfer of material and angular momentum can speed up the rotation of pulsars until their periods are just a few thousandths of a second. 23.6 The Mystery of the Gamma-Ray Bursts Gamma-ray bursts last from a fraction of a second to a few minutes. They come from all directions and are now known to be associated with very distant objects. The energy is most likely beamed, and, for the ones we can detect, Earth lies in the direction of the beam. Long-duration bursts (lasting more than a few seconds) come from massive stars with their outer hydrogen layers missing that explode as supernovae. Short-duration bursts are believed to be mergers of stellar corpses (neutron stars or black holes). FOR FURTHER EXPLORATION Articles Death of Stars Hillebrandt, W., et al. \u201cHow To Blow Up a Star.\u201d Scientific American (October 2006): 42. On supernova mechanisms. Irion, R. \u201cPursuing the Most Extreme Stars.\u201d Astronomy (January 1999): 48. On pulsars. Kalirai, J. \u201cNew Light on Our Sun\u2019s Fate.\u201d Astronomy (February 2014): 44. What will happen to stars like our Sun between the main sequence and the white dwarf stages. Kirshner, R. \u201cSupernova 1987A: The First Ten Years.\u201d Sky & Telescope (February 1997): 35. Maurer, S. \u201cTaking the Pulse of Neutron Stars.\u201d Sky & Telescope (August 2001): 32. Review of recent ideas and observations of pulsars. Zimmerman, R. \u201cInto the Maelstrom.\u201d Astronomy (November 1998): 44. About the Crab Nebula. Gamma-Ray Bursts Fox, D. & Racusin, J. \u201cThe Brightest Burst.\u201d Sky & Telescope (January 2009): 34. Nice summary of the brightest burst observed so far, and what we have learned from it. 846 Chapter 23 The Death of Stars Nadis, S. \u201cDo Cosmic Flashes Reveal Secrets of the Infant Universe?\u201d Astronomy (June 2008): 34. On different types of gamma-ray bursts and what we can learn from them. Na", "eye, R. \u201cDissecting the Bursts of Doom.\u201d Sky & Telescope (August 2006): 30. Excellent review of gamma-ray bursts\u2014how we discovered them, what they might be, and what they can be used for in probing the universe. Zimmerman, R. \u201cSpeed Matters.\u201d Astronomy (May 2000): 36. On the quick-alert networks for finding afterglows. Zimmerman, R. \u201cWitness to Cosmic Collisions.\u201d Astronomy (July 2006): 44. On the Swift mission and what it is teaching astronomers about gamma-ray bursts. Websites Death of Stars Crab Nebula: http://chandra.harvard.edu/xray_sources/crab/crab.html. A short, colorfully written introduction to the history and science involving the best-known supernova remant. Introduction to Neutron Stars: https://www.astro.umd.edu/~miller/nstar.html. Coleman Miller of the University of Maryland maintains this site, which goes from easy to hard as you get into it, but it has lots of good information about corpses of massive stars. Introduction to Pulsars (by Maryam Hobbs at the Australia National Telescope Facility): http://www.atnf.csiro.au/outreach/education/everyone/pulsars/index.html. Magnetars, Soft Gamma Repeaters, and Very Strong Magnetic Fields: http://solomon.as.utexas.edu/ magnetar.html. Robert Duncan, one of the originators of the idea of magnetars, assembled this site some years ago. Gamma-Ray Bursts Brief Intro to Gamma-Ray Bursts (from PBS\u2019 Seeing in the Dark): http://www.pbs.org/seeinginthedark/ astronomy-topics/gamma-ray-bursts.html. Discovery of Gamma-ray Bursts: http://science.nasa.gov/science-news/science-at-nasa/1997/ast19sep97_2/. Gamma-Ray Bursts: Introduction to a Mystery (at NASA\u2019s Imagine the Universe site): http://imagine.gsfc.nasa.gov/docs/science/know_l1/bursts.html. Introduction from the Swift Satellite Site: http://swift.sonoma.edu/about_swift/grbs.html. Missions to Detect and Learn More about Gamma-ray Bursts: \u2022 Fermi Space Telescope: http://fermi.gsfc.nasa.gov/public/. \u2022 INTEGRAL Spacecraft: http://www.esa.int/science/integral. \u2022 SWIFT Spacecraft: http://swift.sonoma.edu/. Videos Death of Stars BBC interview with Antony Hewish: http://www.bbc.co.uk/archive/scientists/10608.shtml. (40:54). Black Widow Pulsars: The Vengeful Corpses of Stars: https://www.youtube.com/watch?v=Fn-3G_N0hy4. A public talk in the Silicon Valley Astronomy Lecture Series by Dr. Roger Romani (Stanford University) (1:01:47). Hubblecast 64: It all ends with a bang!: http://www.spacetelescope.org/videos/hubblecast64a/. HubbleCast Program introducing Supernovae with Dr. Joe Liske (9:48). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 847 Space Movie Reveals Shocking Secrets of the Crab Pulsar: http://hubblesite.org/newscenter/archive/releases/ 2002/24/video/c/. A sequence of Hubble and Chandra Space Telescope images of the central regions of the Crab Nebula have been assembled into a very brief movie accompanied by animation showing how the pulsar affects its environment; it comes with some useful background material (40:06). Gamma-Ray Bursts Gamma-Ray Bursts: The Biggest Explosions Since the Big Bang!: https://www.youtube.com/ watch?v=ePo_EdgV764. Edo Berge in a popular-level lecture at Harvard (58:50). Gamma-Ray Bursts: Flashes in the Sky: https://www.youtube.com/watch?v=23EhcAP3O8Q. American Museum of Natural History Science Bulletin on the Swift satellite (5:59). Overview Animation of Gamma-Ray Burst: http://news.psu.edu/video/296729/2013/11/27/overview-animation- gamma-ray-burst. Brief Animation of what causes a long-duration gamma-ray burst (0:", "55). COLLABORATIVE GROUP ACTIVITIES A. Someone in your group uses a large telescope to observe an expanding shell of gas. Discuss what measurements you could make to determine whether you have discovered a planetary nebula or the remnant of a supernova explosion. B. The star Sirius (the brightest star in our northern skies) has a white-dwarf companion. Sirius has a mass of about 2 MSun and is still on the main sequence, while its companion is already a star corpse. Remember that a white dwarf can\u2019t have a mass greater than 1.4 MSun. Assuming that the two stars formed at the same time, your group should discuss how Sirius could have a white-dwarf companion. Hint: Was the initial mass of the white-dwarf star larger or smaller than that of Sirius? C. Discuss with your group what people today would do if a brilliant star suddenly became visible during the daytime? What kind of fear and superstition might result from a supernova that was really bright in our skies? Have your group invent some headlines that the tabloid newspapers and the less responsible web news outlets would feature. D. Suppose a supernova exploded only 40 light-years from Earth. Have your group discuss what effects there may be on Earth when the radiation reaches us and later when the particles reach us. Would there be any way to protect people from the supernova effects? E. When pulsars were discovered, the astronomers involved with the discovery talked about finding \u201clittle green men.\u201d If you had been in their shoes, what tests would you have performed to see whether such a pulsating source of radio waves was natural or the result of an alien intelligence? Today, several groups around the world are actively searching for possible radio signals from intelligent civilizations. How might you expect such signals to differ from pulsar signals? F. Your little brother, who has not had the benefit of an astronomy course, reads about white dwarfs and neutron stars in a magazine and decides it would be fun to go near them or even try to land on them. Is this a good idea for future tourism? Have your group make a list of reasons it would not be safe for children (or adults) to go near a white dwarf and a neutron star. G. A lot of astronomers\u2019 time and many instruments have been devoted to figuring out the nature of gamma- ray bursts. Does your group share the excitement that astronomers feel about these mysterious high- 848 Chapter 23 The Death of Stars energy events? What are some reasons that people outside of astronomy might care about learning about gamma-ray bursts? EXERCISES Review Questions 1. How does a white dwarf differ from a neutron star? How does each form? What keeps each from collapsing under its own weight? 2. Describe the evolution of a star with a mass like that of the Sun, from the main-sequence phase of its evolution until it becomes a white dwarf. 3. Describe the evolution of a massive star (say, 20 times the mass of the Sun) up to the point at which it becomes a supernova. How does the evolution of a massive star differ from that of the Sun? Why? 4. How do the two types of supernovae discussed in this chapter differ? What kind of star gives rise to each type? 5. A star begins its life with a mass of 5 MSun but ends its life as a white dwarf with a mass of 0.8 MSun. List the stages in the star\u2019s life during which it most likely lost some of the mass it started with. How did mass loss occur in each stage? 6. If the formation of a neutron star leads to a supernova explosion, explain why only three of the hundreds of known pulsars are found in supernova remnants. 7. How can the Crab Nebula shine with the energy of something like 100,000 Suns when the star that formed the nebula exploded almost 1000 years ago? Who \u201cpays the bills\u201d for much of the radiation we see coming from the nebula? 8. How is a nova different from a type Ia supernova? How does it differ from a type II supernova? 9. Apart from the masses, how are binary systems with a neutron star different from binary systems with a white dwarf? 10. What observations from SN 1987A helped confirm theories about supernovae? 11. Describe the evolution of a white dwarf over time, in particular how the luminosity, temperature, and radius change. 12. Describe the evolution of a pulsar over time, in particular how the rotation and pulse signal changes over time. 13. How would a white dwarf that formed from a star that had an initial mass of 1 MSun be different from a white dwarf that formed from a star that had an initial mass of 9 MSun? 14. What do astronomers think are the causes of longer-duration gamma-ray bursts and shorter-duration gamma-ray bursts? 15. How did", " astronomers finally solve the mystery of what gamma-ray bursts were? What instruments were required to find the solution? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 849 Thought Questions 16. Arrange the following stars in order of their evolution: A. A star with no nuclear reactions going on in the core, which is made primarily of carbon and oxygen. B. A star of uniform composition from center to surface; it contains hydrogen but has no nuclear reactions going on in the core. C. A star that is fusing hydrogen to form helium in its core. D. A star that is fusing helium to carbon in the core and hydrogen to helium in a shell around the core. E. A star that has no nuclear reactions going on in the core but is fusing hydrogen to form helium in a shell around the core. 17. Would you expect to find any white dwarfs in the Orion Nebula? (See The Birth of Stars and the Discovery of Planets outside the Solar System to remind yourself of its characteristics.) Why or why not? 18. Suppose no stars more massive than about 2 MSun had ever formed. Would life as we know it have been able to develop? Why or why not? 19. Would you be more likely to observe a type II supernova (the explosion of a massive star) in a globular cluster or in an open cluster? Why? 20. Astronomers believe there are something like 100 million neutron stars in the Galaxy, yet we have only found about 2000 pulsars in the Milky Way. Give several reasons these numbers are so different. Explain each reason. 21. Would you expect to observe every supernova in our own Galaxy? Why or why not? 22. The Large Magellanic Cloud has about one-tenth the number of stars found in our own Galaxy. Suppose the mix of high- and low-mass stars is exactly the same in both galaxies. Approximately how often does a supernova occur in the Large Magellanic Cloud? 23. Look at the list of the nearest stars in Appendix I. Would you expect any of these to become supernovae? Why or why not? 24. If most stars become white dwarfs at the ends of their lives and the formation of white dwarfs is accompanied by the production of a planetary nebula, why are there more white dwarfs than planetary nebulae in the Galaxy? 25. If a 3 and 8 MSun star formed together in a binary system, which star would: A. Evolve off the main sequence first? B. Form a carbon- and oxygen-rich white dwarf? C. Be the location for a nova explosion? 26. You have discovered two star clusters. The first cluster contains mainly main-sequence stars, along with some red giant stars and a few white dwarfs. The second cluster also contains mainly main-sequence stars, along with some red giant stars, and a few neutron stars\u2014but no white dwarf stars. What are the relative ages of the clusters? How did you determine your answer? 27. A supernova remnant was recently discovered and found to be approximately 150 years old. Provide possible reasons that this supernova explosion escaped detection. 850 Chapter 23 The Death of Stars 28. Based upon the evolution of stars, place the following elements in order of least to most common in the Galaxy: gold, carbon, neon. What aspects of stellar evolution formed the basis for how you ordered the elements? 29. What observations or types of telescopes would you use to distinguish a binary system that includes a main-sequence star and a white dwarf star from one containing a main-sequence star and a neutron star? 30. How would the spectra of a type II supernova be different from a type Ia supernova? Hint: Consider the characteristics of the objects that are their source. Figuring For Yourself 31. The ring around SN 1987A (Figure 23.12) initially became illuminated when energetic photons from the supernova interacted with the material in the ring. The radius of the ring is approximately 0.75 light-year from the supernova location. How long after the supernova did the ring become illuminated? 32. What is the acceleration of gravity (g) at the surface of the Sun? (See Appendix E for the Sun\u2019s key characteristics.) How much greater is this than g at the surface of Earth? Calculate what you would weigh on the surface of the Sun. Your weight would be your Earth weight multiplied by the ratio of the acceleration of gravity on the Sun to the acceleration of gravity on Earth. (Okay, we know that the Sun does not have a solid surface to stand on and that you would be vaporized if you were at the Sun\u2019s photosphere. Humor us for the sake of doing these calculations.) 33. What is the escape velocity from the Sun? How much greater is it than the escape velocity from Earth? 34. What is the", " average density of the Sun? How does it compare to the average density of Earth? 35. Say that a particular white dwarf has the mass of the Sun but the radius of Earth. What is the acceleration of gravity at the surface of the white dwarf? How much greater is this than g at the surface of Earth? What would you weigh at the surface of the white dwarf (again granting us the dubious notion that you could survive there)? 36. What is the escape velocity from the white dwarf in Exercise 23.35? How much greater is it than the escape velocity from Earth? 37. What is the average density of the white dwarf in Exercise 23.35? How does it compare to the average density of Earth? 38. Now take a neutron star that has twice the mass of the Sun but a radius of 10 km. What is the acceleration of gravity at the surface of the neutron star? How much greater is this than g at the surface of Earth? What would you weigh at the surface of the neutron star (provided you could somehow not become a puddle of protoplasm)? 39. What is the escape velocity from the neutron star in Exercise 23.38? How much greater is it than the escape velocity from Earth? 40. What is the average density of the neutron star in Exercise 23.38? How does it compare to the average density of Earth? 41. One way to calculate the radius of a star is to use its luminosity and temperature and assume that the star radiates approximately like a blackbody. Astronomers have measured the characteristics of central stars of planetary nebulae and have found that a typical central star is 16 times as luminous and 20 times as hot (about 110,000 K) as the Sun. Find the radius in terms of the Sun\u2019s. How does this radius compare with that of a typical white dwarf? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 23 The Death of Stars 851 42. According to a model described in the text, a neutron star has a radius of about 10 km. Assume that the pulses occur once per rotation. According to Einstein\u2019s theory of relatively, nothing can move faster than the speed of light. Check to make sure that this pulsar model does not violate relativity. Calculate the rotation speed of the Crab Nebula pulsar at its equator, given its period of 0.033 s. (Remember that distance equals velocity \u00d7 time and that the circumference of a circle is given by 2\u03c0R). 43. Do the same calculations as in Exercise 23.42 but for a pulsar that rotates 1000 times per second. 44. If the Sun were replaced by a white dwarf with a surface temperature of 10,000 K and a radius equal to Earth\u2019s, how would its luminosity compare to that of the Sun? 45. A supernova can eject material at a velocity of 10,000 km/s. How long would it take a supernova remnant to expand to a radius of 1 AU? How long would it take to expand to a radius of 1 light-years? Assume that the expansion velocity remains constant and use the relationship: expansion time = distance expansion velocity. 46. A supernova remnant was observed in 2007 to be expanding at a velocity of 14,000 km/s and had a radius of 6.5 light-years. Assuming a constant expansion velocity, in what year did this supernova occur? 47. The ring around SN 1987A (Figure 23.12) started interacting with material propelled by the shockwave from the supernova beginning in 1997 (10 years after the explosion). The radius of the ring is approximately 0.75 light-year from the supernova location. How fast is the supernova material moving, assume a constant rate of motion in km/s? 48. Before the star that became SN 1987A exploded, it evolved from a red supergiant to a blue supergiant while remaining at the same luminosity. As a red supergiant, its surface temperature would have been approximately 4000 K, while as a blue supergiant, its surface temperature was 16,000 K. How much did the radius change as it evolved from a red to a blue supergiant? 49. What is the radius of the progenitor star that became SN 1987A? Its luminosity was 100,000 times that of the Sun, and it had a surface temperature of 16,000 K. 50. What is the acceleration of gravity at the surface of the star that became SN 1987A? How does this g compare to that at the surface of Earth? The mass was 20 times that of the Sun and the radius was 41 times that of the Sun. 51. What was the escape velocity from the surface of the SN 1987A progenitor star? How much greater is it than the escape velocity from Earth? The mass was 20 times that of the Sun and the radius was 41 times that of", " the Sun. 52. What was the average density of the star that became SN 1987A? How does it compare to the average density of Earth? The mass was 20 times that of the Sun and the radius was 41 times that of the Sun. 53. If the pulsar shown in Figure 23.16 is rotating 100 times per second, how many pulses would be detected in one minute? The two beams are located along the pulsar\u2019s equator, which is aligned with Earth. 852 Chapter 23 The Death of Stars This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 853 24 BLACK HOLES AND CURVED SPACETIME Figure 24.1. Stellar Mass Black Hole. On the left, a visible-light image shows a region of the sky in the constellation of Cygnus; the red box marks the position of the X-ray source Cygnus X-1. It is an example of a black hole created when a massive star collapses at the end of its life. Cygnus X-1 is in a binary star system, and the artist\u2019s illustration on the right shows the black hole pulling material away from a massive blue companion star. This material forms a disk (shown in red and orange) that rotates around the black hole before falling into it or being redirected away from the black hole in the form of powerful jets. The material in the disk (before it falls into the black hole) is so hot that it glows with X-rays, explaining why this object is an X-ray source. (credit left: modification of work by DSS; credit right: modification of work by NASA/CXC/M.Weiss) Chapter Outline 24.1 Introducing General Relativity 24.2 Spacetime and Gravity 24.3 Tests of General Relativity 24.4 Time in General Relativity 24.5 Black Holes 24.6 Evidence for Black Holes 24.7 Gravitational Wave Astronomy Thinking Ahead For most of the twentieth century, black holes seemed the stuff of science fiction, portrayed either as monster vacuum cleaners consuming all the matter around them or as tunnels from one universe to another. But the truth about black holes is almost stranger than fiction. As we continue our voyage into the universe, we will discover that black holes are the key to explaining many mysterious and remarkable objects\u2014including collapsed stars and the active centers of giant galaxies. 24.1 INTRODUCING GENERAL RELATIVITY Learning Objectives By the end of this section, you will be able to: 854 Chapter 24 Black Holes and Curved Spacetime Discuss some of the key ideas of the theory of general relativity Recognize that one\u2019s experiences of gravity and acceleration are interchangeable and indistinguishable Distinguish between Newtonian ideas of gravity and Einsteinian ideas of gravity Recognize why the theory of general relativity is necessary for understanding the nature of black holes Most stars end their lives as white dwarfs or neutron stars. When a very massive star collapses at the end of its life, however, not even the mutual repulsion between densely packed neutrons can support the core against its own weight. If the remaining mass of the star\u2019s core is more than about three times that of the Sun (MSun), our theories predict that no known force can stop it from collapsing forever! Gravity simply overwhelms all other forces and crushes the core until it occupies an infinitely small volume. A star in which this occurs may become one of the strangest objects ever predicted by theory\u2014a black hole. To understand what a black hole is like and how it influences its surroundings, we need a theory that can describe the action of gravity under such extreme circumstances. To date, our best theory of gravity is the general theory of relativity, which was put forward in 1916 by Albert Einstein. General relativity was one of the major intellectual achievements of the twentieth century; if it were music, we would compare it to the great symphonies of Beethoven or Mahler. Until recently, however, scientists had little need for a better theory of gravity; Isaac Newton\u2019s ideas that led to his law of universal gravitation (see Orbits and Gravity) are perfectly sufficient for most of the objects we deal with in everyday life. In the past half century, however, general relativity has become more than just a beautiful idea; it is now essential in understanding pulsars, quasars (which will be discussed in Active Galaxies, Quasars, and Supermassive Black Holes), and many other astronomical objects and events, including the black holes we will discuss here. We should perhaps mention that this is the point in an astronomy course when many students start to feel a little nervous (and perhaps wish they had taken botany or some other earthbound course to satisfy the science requirement). This is because in popular culture, Einstein has become a symbol for mathematical brilliance that is simply", " beyond the reach of most people (Figure 24.2). Figure 24.2. Albert Einstein (1879\u20131955). This famous scientist, seen here younger than in the usual photos, has become a symbol for high intellect in popular culture. (credit: NASA) So, when we wrote that the theory of general relativity was Einstein\u2019s work, you may have worried just a bit, convinced that anything Einstein did must be beyond your understanding. This popular view is unfortunate and mistaken. Although the detailed calculations of general relativity do involve a good deal of higher mathematics, the basic ideas are not difficult to understand (and are, in fact, almost poetic in the way they give us a new perspective on the world). Moreover, general relativity goes beyond Newton\u2019s famous \u201cinverse-square\u201d law of This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 855 gravity; it helps explain how matter interacts with other matter in space and time. This explanatory power is one of the requirements that any successful scientific theory must meet. The Principle of Equivalence The fundamental insight that led to the formulation of the general theory of relativity starts with a very simple thought: if you were able to jump off a high building and fall freely, you would not feel your own weight. In this chapter, we will describe how Einstein built on this idea to reach sweeping conclusions about the very fabric of space and time itself. He called it the \u201chappiest thought of my life.\u201d Einstein himself pointed out an everyday example that illustrates this effect (see Figure 24.3). Notice how your weight seems to be reduced in a high-speed elevator when it accelerates from a stop to a rapid descent. Similarly, your weight seems to increase in an elevator that starts to move quickly upward. This effect is not just a feeling you have: if you stood on a scale in such an elevator, you could measure your weight changing (you can actually perform this experiment in some science museums). Figure 24.3. Your Weight in an Elevator. In an elevator at rest, you feel your normal weight. In an elevator that accelerates as it descends, you would feel lighter than normal. In an elevator that accelerates as it ascends, you would feel heavier than normal. If an evil villain cut the elevator cable, you would feel weightless as you fell to your doom. In a freely falling elevator, with no air friction, you would lose your weight altogether. We generally don\u2019t like to cut the cables holding elevators to try this experiment, but near-weightlessness can be achieved by taking an airplane to high altitude and then dropping rapidly for a while. This is how NASA trains its astronauts for the experience of free fall in space; the scenes of weightlessness in the 1995 movie Apollo 13 were filmed in 856 Chapter 24 Black Holes and Curved Spacetime the same way. (Moviemakers have since devised other methods using underwater filming, wire stunts, and computer graphics to create the appearance of weightlessness seen in such movies as Gravity and The Martian.) Watch how NASA uses a \u201cweightless\u201d environment (https://openstax.org/l/30NASAweightra) to help train astronauts. Another way to state Einstein\u2019s idea is this: suppose we have a spaceship that contains a windowless laboratory equipped with all the tools needed to perform scientific experiments. Now, imagine that an astronomer wakes up after a long night celebrating some scientific breakthrough and finds herself sealed into this laboratory. She has no idea how it happened but notices that she is weightless. This could be because she and the laboratory are far away from any source of gravity, and both are either at rest or moving at some steady speed through space (in which case she has plenty of time to wake up). But it could also be because she and the laboratory are falling freely toward a planet like Earth (in which case she might first want to check her distance from the surface before making coffee). What Einstein postulated is that there is no experiment she can perform inside the sealed laboratory to determine whether she is floating in space or falling freely in a gravitational field.[1] As far as she is concerned, the two situations are completely equivalent. This idea that free fall is indistinguishable from, and hence equivalent to, zero gravity is called the equivalence principle. Gravity or Acceleration? Einstein\u2019s simple idea has big consequences. Let\u2019s begin by considering what happens if two foolhardy people jump from opposite banks into a bottomless chasm (Figure 24.4). If we ignore air friction, then we can say that while they freely fall, they both accelerate downward at the same rate and feel no external force acting on them. They can throw a ball back and forth, always aiming it straight at each other, as if there were no gravity. The ball falls at the same", " rate that they do, so it always remains in a line between them. 1 Strictly speaking, this is true only if the laboratory is infinitesimally small. Different locations in a real laboratory that is falling freely due to gravity cannot all be at identical distances from the object(s) responsible for producing the gravitational force. In this case, objects in different locations will experience slightly different accelerations. But this point does not invalidate the principle of equivalence that Einstein derived from this line of thinking. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 857 Figure 24.4. Free Fall. Two people play catch as they descend into a bottomless abyss. Since the people and ball all fall at the same speed, it appears to them that they can play catch by throwing the ball in a straight line between them. Within their frame of reference, there appears to be no gravity. Such a game of catch is very different on the surface of Earth. Everyone who grows up feeling gravity knows that a ball, once thrown, falls to the ground. Thus, in order to play catch with someone, you must aim the ball upward so that it follows an arc\u2014rising and then falling as it moves forward\u2014until it is caught at the other end. Now suppose we isolate our falling people and ball inside a large box that is falling with them. No one inside the box is aware of any gravitational force. If they let go of the ball, it doesn\u2019t fall to the bottom of the box or anywhere else but merely stays there or moves in a straight line, depending on whether it is given any motion. Astronauts in the International Space Station (ISS) that is orbiting Earth live in an environment just like that of the people sealed in a freely falling box (Figure 24.5). The orbiting ISS is actually \u201cfalling\u201d freely around Earth. While in free fall, the astronauts live in a strange world where there seems to be no gravitational force. One can give a wrench a shove, and it moves at constant speed across the orbiting laboratory. A pencil set in midair remains there as if no force were acting on it. 858 Chapter 24 Black Holes and Curved Spacetime Figure 24.5. Astronauts aboard the Space Shuttle. Shane Kimbrough and Sandra Magnus are shown aboard the Endeavour in 2008 with various fruit floating freely. Because the shuttle is in free fall as it orbits Earth, everything\u2014including astronauts\u2014stays put or moves uniformly relative to the walls of the spacecraft. This free-falling state produces a lack of apparent gravity inside the spacecraft. (credit: NASA In the \u201cweightless\u201d environment of the International Space Station, moving takes very little effort. Watch astronaut Karen Nyberg (https://openstax.org/l/30ISSzerogravid) demonstrate how she can propel herself with the force of a single human hair. Appearances are misleading, however. There is a force in this situation. Both the ISS and the astronauts continually fall around Earth, pulled by its gravity. But since all fall together\u2014shuttle, astronauts, wrench, and pencil\u2014inside the ISS all gravitational forces appear to be absent. Thus, the orbiting ISS provides an excellent example of the principle of equivalence\u2014how local effects of gravity can be completely compensated by the right acceleration. To the astronauts, falling around Earth creates the same effects as being far off in space, remote from all gravitational influences. The Paths of Light and Matter Einstein postulated that the equivalence principle is a fundamental fact of nature, and that there is no experiment inside any spacecraft by which an astronaut can ever distinguish between being weightless in remote space and being in free fall near a planet like Earth. This would apply to experiments done with beams of light as well. But the minute we use light in our experiments, we are led to some very disturbing conclusions\u2014and it is these conclusions that lead us to general relativity and a new view of gravity. It seems apparent to us, from everyday observations, that beams of light travel in straight lines. Imagine that a spaceship is moving through empty space far from any gravity. Send a laser beam from the back of the ship to the front, and it will travel in a nice straight line and land on the front wall exactly opposite the point from which it left the rear wall. If the equivalence principle really applies universally, then this same experiment performed in free fall around Earth should give us the same result. Now imagine that the astronauts again shine a beam of light along the length of their ship. But, as shown in Figure 24.6, this time the orbiting space station falls a bit between the time the light leaves the back wall and the time it hits the front wall. (The amount of the fall is grossly exaggerated in Figure 24.6 to illustrate the effect.) This OpenStax book is available for free at http", "://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 859 Therefore, if the beam of light follows a straight line but the ship\u2019s path curves downward, then the light should strike the front wall at a point higher than the point from which it left. Figure 24.6. Curved Light Path. In a spaceship moving to the left (in this figure) in its orbit about a planet, light is beamed from the rear, A, toward the front, B. Meanwhile, the ship is falling out of its straight path (exaggerated here). We might therefore expect the light to strike at B\u2032, above the target in the ship. Instead, the light follows a curved path and strikes at C. In order for the principle of equivalence to be correct, gravity must be able to curve the path of a light beam just as it curves the path of the spaceship. However, this would violate the principle of equivalence\u2014the two experiments would give different results. We are thus faced with giving up one of our two assumptions. Either the principle of equivalence is not correct, or light does not always travel in straight lines. Instead of dropping what probably seemed at the time like a ridiculous idea, Einstein worked out what happens if light sometimes does not follow a straight path. Let\u2019s suppose the principle of equivalence is right. Then the light beam must arrive directly opposite the point from which it started in the ship. The light, like the ball thrown back and forth, must fall with the ship that is in orbit around Earth (see Figure 24.6). This would make its path curve downward, like the path of the ball, and thus the light would hit the front wall exactly opposite the spot from which it came. Thinking this over, you might well conclude that it doesn\u2019t seem like such a big problem: why can\u2019t light fall the way balls do? But, as discussed in Radiation and Spectra, light is profoundly different from balls. Balls have mass, while light does not. Here is where Einstein\u2019s intuition and genius allowed him to make a profound leap. He gave physical meaning to the strange result of our thought experiment. Einstein suggested that the light curves down to meet the front of the shuttle because Earth\u2019s gravity actually bends the fabric of space and time. This radical idea\u2014which we will explain next\u2014keeps the behavior of light the same in both empty space and free fall, but it changes some of our most basic and cherished ideas about space and time. The reason we take Einstein\u2019s suggestion seriously is that, as we will see, experiments now clearly show his intuitive leap was correct. 24.2 SPACETIME AND GRAVITY Learning Objectives By the end of this section, you will be able to: Describe Einstein\u2019s view of gravity as the warping of spacetime in the presence of massive objects Understand that Newton\u2019s concept of the gravitational force between two massive objects and Einstein\u2019s concept of warped spacetime are different explanations for the same observed accelerations of one massive object in the presence of another massive object Is light actually bent from its straight-line path by the mass of Earth? How can light, which has no mass, be affected by gravity? Einstein preferred to think that it is space and time that are affected by the presence of a large mass; light beams, and everything else that travels through space and time, then find their paths affected. 860 Chapter 24 Black Holes and Curved Spacetime Light always follows the shortest path\u2014but that path may not always be straight. This idea is true for human travel on the curved surface of planet Earth, as well. Say you want to fly from Chicago to Rome. Since an airplane can\u2019t go through the solid body of the Earth, the shortest distance is not a straight line but the arc of a great circle. Linkages: Mass, Space, and Time To show what Einstein\u2019s insight really means, let\u2019s first consider how we locate an event in space and time. For example, imagine you have to describe to worried school officials the fire that broke out in your room when your roommate tried cooking shish kebabs in the fireplace. You explain that your dorm is at 6400 College Avenue, a street that runs in the left-right direction on a map of your town; you are on the fifth floor, which tells where you are in the up-down direction; and you are the sixth room back from the elevator, which tells where you are in the forward-backward direction. Then you explain that the fire broke out at 6:23 p.m. (but was soon brought under control), which specifies the event in time. Any event in the universe, whether nearby or far away, can be pinpointed using the three dimensions of space and the one dimension of time. Newton considered space and time to be", " completely independent, and that continued to be the accepted view until the beginning of the twentieth century. But Einstein showed that there is an intimate connection between space and time, and that only by considering the two together\u2014in what we call spacetime\u2014can we build up a correct picture of the physical world. We examine spacetime a bit more closely in the next subsection. The gist of Einstein\u2019s general theory is that the presence of matter curves or warps the fabric of spacetime. This curving of spacetime is identified with gravity. When something else\u2014a beam of light, an electron, or the starship Enterprise\u2014enters such a region of distorted spacetime, its path will be different from what it would have been in the absence of the matter. As American physicist John Wheeler summarized it: \u201cMatter tells spacetime how to curve; spacetime tells matter how to move.\u201d The amount of distortion in spacetime depends on the mass of material that is involved and on how concentrated and compact it is. Terrestrial objects, such as the book you are reading, have far too little mass to introduce any significant distortion. Newton\u2019s view of gravity is just fine for building bridges, skyscrapers, or amusement park rides. General relativity does, however, have some practical applications. The GPS (Global Positioning System) in every smartphone can tell you where you are within 5 to 10 meters only because the effects of general and special relativity on the GPS satellites in orbit around the Earth are taken into account. Unlike a book or your roommate, stars produce measurable distortions in spacetime. A white dwarf, with its stronger surface gravity, produces more distortion just above its surface than does a red giant with the same mass. So, you see, we are eventually going to talk about collapsing stars again, but not before discussing Einstein\u2019s ideas (and the evidence for them) in more detail. Spacetime Examples How can we understand the distortion of spacetime by the presence of some (significant) amount of mass? Let\u2019s try the following analogy. You may have seen maps of New York City that squeeze the full three dimensions of this towering metropolis onto a flat sheet of paper and still have enough information so tourists will not get lost. Let\u2019s do something similar with diagrams of spacetime. Figure 24.7, for example, shows the progress of a motorist driving east on a stretch of road in Kansas where the countryside is absolutely flat. Since our motorist is traveling only in the east-west direction and the terrain is flat, we can ignore the other two dimensions of space. The amount of time elapsed since he left home is shown on the y-axis, and the distance traveled eastward is shown on the x-axis. From A to B he drove at a uniform speed; unfortunately, it was too fast a uniform speed and a police car spotted him. From B to C he stopped to This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 861 receive his ticket and made no progress through space, only through time. From C to D he drove more slowly because the police car was behind him. Figure 24.7. Spacetime Diagram. This diagram shows the progress of a motorist traveling east across the flat Kansas landscape. Distance traveled is plotted along the horizontal axis. The time elapsed since the motorist left the starting point is plotted along the vertical axis. Now let\u2019s try illustrating the distortions of spacetime in two dimensions. In this case, we will (in our imaginations) use a rubber sheet that can stretch or warp if we put objects on it. Let\u2019s imagine stretching our rubber sheet taut on four posts. To complete the analogy, we need something that normally travels in a straight line (as light does). Suppose we have an extremely intelligent ant\u2014a friend of the comic book superhero Ant-Man, perhaps\u2014that has been trained to walk in a straight line. We begin with just the rubber sheet and the ant, simulating empty space with no mass in it. We put the ant on one side of the sheet and it walks in a beautiful straight line over to the other side (Figure 24.8). We next put a small grain of sand on the rubber sheet. The sand does distort the sheet a tiny bit, but this is not a distortion that we or the ant can measure. If we send the ant so it goes close to, but not on top of, the sand grain, it has little trouble continuing to walk in a straight line. Now we grab something with a little more mass\u2014say, a small pebble. It bends or distorts the sheet just a bit around its position. If we send the ant into this region, it finds its path slightly altered by the distortion of the sheet. The distortion is not large, but if we follow the ant\u2019s path carefully,", " we notice it deviating slightly from a straight line. The effect gets more noticeable as we increase the mass of the object that we put on the sheet. Let\u2019s say we now use a massive paperweight. Such a heavy object distorts or warps the rubber sheet very effectively, putting a good sag in it. From our point of view, we can see that the sheet near the paperweight is no longer straight. 862 Chapter 24 Black Holes and Curved Spacetime Figure 24.8. Three-Dimensional Analogy for Spacetime. On a flat rubber sheet, a trained ant has no trouble walking in a straight line. When a massive object creates a big depression in the sheet, the ant, which must walk where the sheet takes it, finds its path changed (warped) dramatically. Now let\u2019s again send the ant on a journey that takes it close to, but not on top of, the paperweight. Far away from the paperweight, the ant has no trouble doing its walk, which looks straight to us. As it nears the paperweight, however, the ant is forced down into the sag. It must then climb up the other side before it can return to walking on an undistorted part of the sheet. All this while, the ant is following the shortest path it can, but through no fault of its own (after all, ants can\u2019t fly, so it has to stay on the sheet) this path is curved by the distortion of the sheet itself. In the same way, according to Einstein\u2019s theory, light always follows the shortest path through spacetime. But the mass associated with large concentrations of matter distorts spacetime, and the shortest, most direct paths are no longer straight lines, but curves. How large does a mass have to be before we can measure a change in the path followed by light? In 1916, when Einstein first proposed his theory, no distortion had been detected at the surface of Earth (so Earth might have played the role of the grain of sand in our analogy). Something with a mass like our Sun\u2019s was necessary to detect the effect Einstein was describing (we will discuss how this effect was measured using the Sun in the next section). The paperweight in our analogy might be a white dwarf or a neutron star. The distortion of spacetime is greater near the surfaces of these compact, massive objects than near the surface of the Sun. And when, to return to the situation described at the beginning of the chapter, a star core with more than three times the mass of the Sun collapses forever, the distortions of spacetime very close to it can become truly mind-boggling. 24.3 TESTS OF GENERAL RELATIVITY Learning Objectives By the end of this section, you will be able to: Describe unusual motion of Mercury around the Sun and explain how general relativity explains the observed behavior Provide examples of evidence for light rays being bent by massive objects, as predicted by general relativity\u2019s theory about the warping of spacetime What Einstein proposed was nothing less than a major revolution in our understanding of space and time. It was a new theory of gravity, in which mass determines the curvature of spacetime and that curvature, in This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 863 turn, controls how objects move. Like all new ideas in science, no matter who advances them, Einstein\u2019s theory had to be tested by comparing its predictions against the experimental evidence. This was quite a challenge because the effects of the new theory were apparent only when the mass was quite large. (For smaller masses, it required measuring techniques that would not become available until decades later.) When the distorting mass is small, the predictions of general relativity must agree with those resulting from Newton\u2019s law of universal gravitation, which, after all, has served us admirably in our technology and in guiding space probes to the other planets. In familiar territory, therefore, the differences between the predictions of the two models are subtle and difficult to detect. Nevertheless, Einstein was able to demonstrate one proof of his theory that could be found in existing data and to suggest another one that would be tested just a few years later. The Motion of Mercury Of the planets in our solar system, Mercury orbits closest to the Sun and is thus most affected by the distortion of spacetime produced by the Sun\u2019s mass. Einstein wondered if the distortion might produce a noticeable difference in the motion of Mercury that was not predicted by Newton\u2019s law. It turned out that the difference was subtle, but it was definitely there. Most importantly, it had already been measured. Mercury has a highly elliptical orbit, so that it is only about two-thirds as far from the Sun at perihelion as it is at aphelion. (These terms were defined in", " the chapter on Orbits and Gravity.) The gravitational effects (perturbations) of the other planets on Mercury produce a calculable advance of Mercury\u2019s perihelion. What this means is that each successive perihelion occurs in a slightly different direction as seen from the Sun (Figure 24.9). Figure 24.9. Mercury\u2019s Wobble. The major axis of the orbit of a planet, such as Mercury, rotates in space slightly because of various perturbations. In Mercury\u2019s case, the amount of rotation (or orbital precession) is a bit larger than can be accounted for by the gravitational forces exerted by other planets; this difference is precisely explained by the general theory of relativity. Mercury, being the planet closest to the Sun, has its orbit most affected by the warping of spacetime near the Sun. The change from orbit to orbit has been significantly exaggerated on this diagram. According to Newtonian gravitation, the gravitational forces exerted by the planets will cause Mercury\u2019s perihelion to advance by about 531 seconds of arc (arcsec) per century. In the nineteenth century, however, it was observed that the actual advance is 574 arcsec per century. The discrepancy was first pointed out in 1859 by Urbain Le Verrier, the codiscoverer of Neptune. Just as discrepancies in the motion of Uranus allowed astronomers to discover the presence of Neptune, so it was thought that the discrepancy in the motion of Mercury could mean the presence of an undiscovered inner planet. Astronomers searched for this planet near the Sun, even giving it a name: Vulcan, after the Roman god of fire. (The name would later be used for the home planet of a famous character on a popular television show about future space travel.) 864 Chapter 24 Black Holes and Curved Spacetime But no planet has ever been found nearer to the Sun than Mercury, and the discrepancy was still bothering astronomers when Einstein was doing his calculations. General relativity, however, predicts that due to the curvature of spacetime around the Sun, the perihelion of Mercury should advance slightly more than is predicted by Newtonian gravity. The result is to make the major axis of Mercury\u2019s orbit rotate slowly in space because of the Sun\u2019s gravity alone. The prediction of general relativity is that the direction of perihelion should change by an additional 43 arcsec per century. This is remarkably close to the observed discrepancy, and it gave Einstein a lot of confidence as he advanced his theory. The relativistic advance of perihelion was later also observed in the orbits of several asteroids that come close to the Sun. Deflection of Starlight Einstein\u2019s second test was something that had not been observed before and would thus provide an excellent confirmation of his theory. Since spacetime is more curved in regions where the gravitational field is strong, we would expect light passing very near the Sun to appear to follow a curved path (Figure 24.10), just like that of the ant in our analogy. Einstein calculated from general relativity theory that starlight just grazing the Sun\u2019s surface should be deflected by an angle of 1.75 arcsec. Could such a deflection be observed? Figure 24.10. Curvature of Light Paths near the Sun. Starlight passing near the Sun is deflected slightly by the \u201cwarping\u201d of spacetime. (This deflection of starlight is one small example of a phenomenon called gravitational lensing, which we\u2019ll discuss in more detail in The Evolution and Distribution of Galaxies.) Before passing by the Sun, the light from the star was traveling parallel to the bottom edge of the figure. When it passed near the Sun, the path was altered slightly. When we see the light, we assume the light beam has been traveling in a straight path throughout its journey, and so we measure the position of the star to be slightly different from its true position. If we were to observe the star at another time, when the Sun is not in the way, we would measure its true position. We encounter a small \u201ctechnical problem\u201d when we try to photograph starlight coming very close to the Sun: the Sun is an outrageously bright source of starlight itself. But during a total solar eclipse, much of the Sun\u2019s light is blocked out, allowing the stars near the Sun to be photographed. In a paper published during World War I, Einstein (writing in a German journal) suggested that photographic observations during an eclipse could reveal the deflection of light passing near the Sun. The technique involves taking a photograph of the stars six months prior to the eclipse and measuring the position of all the stars accurately. Then the same stars are photographed during the eclipse. This is when the starlight has to travel to us by skirting the Sun and moving through measurably warped spacetime. As seen from Earth, the stars closest to the Sun will seem to be \ufffd", "\ufffdout of place\u201d\u2014slightly away from their regular positions as measured when the Sun is not nearby. A single copy of that paper, passed through neutral Holland, reached the British astronomer Arthur S. Eddington, who noted that the next suitable eclipse was on May 29, 1919. The British organized two expeditions to observe it: one on the island of Pr\u00edncipe, off the coast of West Africa, and the other in Sobral, in northern Brazil. Despite some problems with the weather, both expeditions obtained successful photographs. The stars seen near the Sun were indeed displaced, and to the accuracy of the measurements, which was about 20%, the shifts were consistent with the predictions of general relativity. More modern experiments with radio waves This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 865 traveling close to the Sun have confirmed that the actual displacements are within 1% of what general relativity predicts. The confirmation of the theory by the eclipse expeditions in 1919 was a triumph that made Einstein a world celebrity. 24.4 TIME IN GENERAL RELATIVITY Learning Objectives By the end of this section, you will be able to: Describe how Einsteinian gravity slows clocks and can decrease a light wave\u2019s frequency of oscillation Recognize that the gravitational decrease in a light wave\u2019s frequency is compensated by an increase in the light wave\u2019s wavelength\u2014the so-called gravitational redshift\u2014so that the light continues to travel at constant speed General relativity theory makes various predictions about the behavior of space and time. One of these predictions, put in everyday terms, is that the stronger the gravity, the slower the pace of time. Such a statement goes very much counter to our intuitive sense of time as a flow that we all share. Time has always seemed the most democratic of concepts: all of us, regardless of wealth or status, appear to move together from the cradle to the grave in the great current of time. But Einstein argued that it only seems this way to us because all humans so far have lived and died in the gravitational environment of Earth. We have had no chance to test the idea that the pace of time might depend on the strength of gravity, because we have not experienced radically different gravities. Moreover, the differences in the flow of time are extremely small until truly large masses are involved. Nevertheless, Einstein\u2019s prediction has now been tested, both on Earth and in space. The Tests of Time An ingenious experiment in 1959 used the most accurate atomic clock known to compare time measurements on the ground floor and the top floor of the physics building at Harvard University. For a clock, the experimenters used the frequency (the number of cycles per second) of gamma rays emitted by radioactive cobalt. Einstein\u2019s theory predicts that such a cobalt clock on the ground floor, being a bit closer to Earth\u2019s center of gravity, should run very slightly slower than the same clock on the top floor. This is precisely what the experiments observed. Later, atomic clocks were taken up in high-flying aircraft and even on one of the Gemini space flights. In each case, the clocks farther from Earth ran a bit faster. While in 1959 it didn\u2019t matter much if the clock at the top of the building ran faster than the clock in the basement, today that effect is highly relevant. Every smartphone or device that synchronizes with a GPS must correct for this (as we will see in the next section) since the clocks on satellites will run faster than clocks on Earth. The effect is more pronounced if the gravity involved is the Sun\u2019s and not Earth\u2019s. If stronger gravity slows the pace of time, then it will take longer for a light or radio wave that passes very near the edge of the Sun to reach Earth than we would expect on the basis of Newton\u2019s law of gravity. (It takes longer because spacetime is curved in the vicinity of the Sun.) The smaller the distance between the ray of light and the edge of the Sun at closest approach, the longer will be the delay in the arrival time. In November 1976, when the two Viking spacecraft were operating on the surface of Mars, the planet went behind the Sun as seen from Earth (Figure 24.11). Scientists had preprogrammed Viking to send a radio wave toward Earth that would go extremely close to the outer regions of the Sun. According to general relativity, 866 Chapter 24 Black Holes and Curved Spacetime there would be a delay because the radio wave would be passing through a region where time ran more slowly. The experiment was able to confirm Einstein\u2019s theory to within 0.1%. Figure 24.11. Time Delays for Radio Waves near the Sun. Radio signals from the Viking lander on Mars were delayed when they passed near the Sun, where spacetime is curved relatively strongly. In this picture, spac", "etime is pictured as a two-dimensional rubber sheet. Gravitational Redshift What does it mean to say that time runs more slowly? When light emerges from a region of strong gravity where time slows down, the light experiences a change in its frequency and wavelength. To understand what happens, let\u2019s recall that a wave of light is a repeating phenomenon\u2014crest follows crest with great regularity. In this sense, each light wave is a little clock, keeping time with its wave cycle. If stronger gravity slows down the pace of time (relative to an outside observer), then the rate at which crest follows crest must be correspondingly slower\u2014that is, the waves become less frequent. To maintain constant light speed (the key postulate in Einstein\u2019s theories of special and general relativity), the lower frequency must be compensated by a longer wavelength. This kind of increase in wavelength (when caused by the motion of the source) is what we called a redshift in Radiation and Spectra. Here, because it is gravity and not motion that produces the longer wavelengths, we call the effect a gravitational redshift. The advent of space-age technology made it possible to measure gravitational redshift with very high accuracy. In the mid-1970s, a hydrogen maser, a device akin to a laser that produces a microwave radio signal at a particular wavelength, was carried by a rocket to an altitude of 10,000 kilometers. Instruments on the ground were used to compare the frequency of the signal emitted by the rocket-borne maser with that from a similar maser on Earth. The experiment showed that the stronger gravitational field at Earth\u2019s surface really did slow the flow of time relative to that measured by the maser in the rocket. The observed effect matched the predictions of general relativity to within a few parts in 100,000. These are only a few examples of tests that have confirmed the predictions of general relativity. Today, general relativity is accepted as our best description of gravity and is used by astronomers and physicists to understand the behavior of the centers of galaxies, the beginning of the universe, and the subject with which we began this chapter\u2014the death of truly massive stars. Relativity: A Practical Application By now you may be asking: why should I be bothered with relativity? Can\u2019t I live my life perfectly well without it? The answer is you can\u2019t. Every time a pilot lands an airplane or you use a GPS to determine where you are on a drive or hike in the back country, you (or at least your GPS-enabled device) must take the effects of both general and special relativity into account. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 867 GPS relies on an array of 24 satellites orbiting the Earth, and at least 4 of them are visible from any spot on Earth. Each satellite carries a precise atomic clock. Your GPS receiver detects the signals from those satellites that are overhead and calculates your position based on the time that it has taken those signals to reach you. Suppose you want to know where you are within 50 feet (GPS devices can actually do much better than this). Since it takes only 50 billionths of a second for light to travel 50 feet, the clocks on the satellites must be synchronized to at least this accuracy\u2014and relativistic effects must therefore be taken into account. The clocks on the satellites are orbiting Earth at a speed of 14,000 kilometers per hour and are moving much faster than clocks on the surface of Earth. According to Einstein\u2019s theory of relativity, the clocks on the satellites are ticking more slowly than Earth-based clocks by about 7 millionths of a second per day. (We have not discussed the special theory of relativity, which deals with changes when objects move very fast, so you\u2019ll have to take our word for this part.) The orbits of the satellites are 20,000 kilometers above Earth, where gravity is about four times weaker than at Earth\u2019s surface. General relativity says that the orbiting clocks should tick about 45 millionths of a second faster than they would on Earth. The net effect is that the time on a satellite clock advances by about 38 microseconds per day. If these relativistic effects were not taken into account, navigational errors would start to add up and positions would be off by about 7 miles in only a single day. 24.5 BLACK HOLES Learning Objectives By the end of this section, you will be able to: Explain the event horizon surrounding a black hole Discuss why the popular notion of black holes as great sucking monsters that can ingest material at great distances from them is erroneous Use the concept of warped spacetime near a black hole to track what happens to any object that might fall into a black hole Recognize why the concept of a singularity\u2014with its infinite density and zero volume\u2014presents major challenges to our understanding of matter Let\u2019s now apply what we", " have learned about gravity and spacetime curvature to the issue we started with: the collapsing core in a very massive star. We saw that if the core\u2019s mass is greater than about 3 MSun, theory says that nothing can stop the core from collapsing forever. We will examine this situation from two perspectives: first from a pre-Einstein point of view, and then with the aid of general relativity. Classical Collapse Let\u2019s begin with a thought experiment. We want to know what speeds are required to escape from the gravitational pull of different objects. A rocket must be launched from the surface of Earth at a very high speed if it is to escape the pull of Earth\u2019s gravity. In fact, any object\u2014rocket, ball, astronomy book\u2014that is thrown into the air with a velocity less than 11 kilometers per second will soon fall back to Earth\u2019s surface. Only those objects launched with a speed greater than this escape velocity can get away from Earth. The escape velocity from the surface of the Sun is higher yet\u2014618 kilometers per second. Now imagine that we begin to compress the Sun, forcing it to shrink in diameter. Recall that the pull of gravity depends on both the mass that is pulling you and your distance from the center of gravity of that mass. If the Sun is compressed, its mass will remain the same, but the distance between a point on the Sun\u2019s surface and the center will get smaller 868 Chapter 24 Black Holes and Curved Spacetime and smaller. Thus, as we compress the star, the pull of gravity for an object on the shrinking surface will get stronger and stronger (Figure 24.12). Figure 24.12. Formation of a Black Hole. At left, an imaginary astronaut floats near the surface of a massive star-core about to collapse. As the same mass falls into a smaller sphere, the gravity at its surface goes up, making it harder for anything to escape from the stellar surface. Eventually the mass collapses into so small a sphere that the escape velocity exceeds the speed of light and nothing can get away. Note that the size of the astronaut has been exaggerated. In the last picture, the astronaut is just outside the sphere we will call the event horizon and is stretched and squeezed by the strong gravity. When the shrinking Sun reaches the diameter of a neutron star (about 20 kilometers), the velocity required to escape its gravitational pull will be about half the speed of light. Suppose we continue to compress the Sun to a smaller and smaller diameter. (We saw this can\u2019t happen to a star like our Sun in the real world because of electron degeneracy, i.e., the mutual repulsion between tightly packed electrons; this is just a quick \u201cthought experiment\u201d to get our bearings). Ultimately, as the Sun shrinks, the escape velocity near the surface would exceed the speed of light. If the speed you need to get away is faster than the fastest possible speed in the universe, then nothing, not even light, is able to escape. An object with such large escape velocity emits no light, and anything that falls into it can never return. In modern terminology, we call an object from which light cannot escape a black hole, a name popularized by the America scientist John Wheeler starting in the late 1960s (Figure 24.13). The idea that such objects might exist is, however, not a new one. Cambridge professor and amateur astronomer John Michell wrote a paper in 1783 about the possibility that stars with escape velocities exceeding that of light might exist. And in 1796, the French mathematician Pierre-Simon, marquis de Laplace, made similar calculations using Newton\u2019s theory of gravity; he called the resulting objects \u201cdark bodies.\u201d This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 869 Figure 24.13. John Wheeler (1911\u20132008). This brilliant physicist did much pioneering work in general relativity theory and popularized the term black hole starting in the late 1960s. (credit: modification of work by Roy Bishop) While these early calculations provided strong hints that something strange should be expected if very massive objects collapse under their own gravity, we really need general relativity theory to give an adequate description of what happens in such a situation. Collapse with Relativity General relativity tells us that gravity is really a curvature of spacetime. As gravity increases (as in the collapsing Sun of our thought experiment), the curvature gets larger and larger. Eventually, if the Sun could shrink down to a diameter of about 6 kilometers, only light beams sent out perpendicular to the surface would escape. All others would fall back onto the star (Figure 24.14). If the Sun could then shrink just a little more, even that one remaining light beam would no longer be able to escape. Figure 24.14. Light Paths near a Massive Object. Suppose a person could", " stand on the surface of a normal star with a flashlight. The light leaving the flashlight travels in a straight line no matter where the flashlight is pointed. Now consider what happens if the star collapses so that it is just a little larger than a black hole. All the light paths, except the one straight up, curve back to the surface. When the star shrinks inside the event horizon and becomes a black hole, even a beam directed straight up returns. Keep in mind that gravity is not pulling on the light. The concentration of matter has curved spacetime, and light (like the trained ant of our earlier example) is \u201cdoing its best\u201d to go in a straight line, yet is now confronted with a world in which straight lines that used to go outward have become curved paths that lead back in. The 870 Chapter 24 Black Holes and Curved Spacetime collapsing star is a black hole in this view, because the very concept of \u201cout\u201d has no geometrical meaning. The star has become trapped in its own little pocket of spacetime, from which there is no escape. The star\u2019s geometry cuts off communication with the rest of the universe at precisely the moment when, in our earlier picture, the escape velocity becomes equal to the speed of light. The size of the star at this moment defines a surface that we call the event horizon. It\u2019s a wonderfully descriptive name: just as objects that sink below our horizon cannot be seen on Earth, so anything happening inside the event horizon can no longer interact with the rest of the universe. Imagine a future spacecraft foolish enough to land on the surface of a massive star just as it begins to collapse in the way we have been describing. Perhaps the captain is asleep at the gravity meter, and before the crew can say \u201cAlbert Einstein,\u201d they have collapsed with the star inside the event horizon. Frantically, they send an escape pod straight outward. But paths outward twist around to become paths inward, and the pod turns around and falls toward the center of the black hole. They send a radio message to their loved ones, bidding good-bye. But radio waves, like light, must travel through spacetime, and curved spacetime allows nothing to get out. Their final message remains unheard. Events inside the event horizon can never again affect events outside it. The characteristics of an event horizon were first worked out by astronomer and mathematician Karl Schwarzschild (Figure 24.15). A member of the German army in World War I, he died in 1916 of an illness he contracted while doing artillery shell calculations on the Russian front. His paper on the theory of event horizons was among the last things he finished as he was dying; it was the first exact solution to Einstein\u2019s equations of general relativity. The radius of the event horizon is called the Schwarzschild radius in his memory. Figure 24.15. Karl Schwarzschild (1873\u20131916). This German scientist was the first to demonstrate mathematically that a black hole is possible and to determine the size of a nonrotating black hole\u2019s event horizon. The event horizon is the boundary of the black hole; calculations show that it does not get smaller once the whole star has collapsed inside it. It is the region that separates the things trapped inside it from the rest of the universe. Anything coming from the outside is also trapped once it comes inside the event horizon. The horizon\u2019s size turns out to depend only on the mass inside it. If the Sun, with its mass of 1 MSun, were to become a black hole (fortunately, it can\u2019t\u2014this is just a thought experiment), the Schwarzschild radius would be about 3 kilometers; thus, the entire black hole would be about one-third the size of a neutron star of that same mass. Feed the black hole some mass, and the horizon will grow\u2014but not very much. Doubling the mass will make the black hole 6 kilometers in radius, still very tiny on the cosmic scale. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 871 The event horizons of more massive black holes have larger radii. For example, if a globular cluster of 100,000 stars (solar masses) could collapse to a black hole, it would be 300,000 kilometers in radius, a little less than half the radius of the Sun. If the entire Galaxy could collapse to a black hole, it would be only about 1012 kilometers in radius\u2014about a tenth of a light year. Smaller masses have correspondingly smaller horizons: for Earth to become a black hole, it would have to be compressed to a radius of only 1 centimeter\u2014less than the size of a grape. A typical asteroid, if crushed to a small enough size to be a black hole, would have the dimensions of an atomic nucleus The Milky Way\u2019s Black Hole The size of", " the event horizon of a black hole depends on the mass of the black hole. The greater the mass, the larger the radius of the event horizon. General relativity calculations show that the formula for the Schwarzschild radius (RS) of the event horizon is RS = 2GM c2 where c is the speed of light, G is the gravitational constant, and M is the mass of the black hole. Note that in this formula, 2, G, and c are all constant; only the mass changes from black hole to black hole. As we will see in the chapter on The Milky Way Galaxy, astronomers have traced the paths of several stars near the center of our Galaxy and found that they seem to be orbiting an unseen object\u2014dubbed Sgr A* (pronounced \u201cSagittarius A-star\u201d)\u2014with a mass of about 4 million solar masses. What is the size of its Schwarzschild radius? Solution We can substitute data for G, M, and c (from Appendix E) directly into the equation: RS = 2GM c2 = = 1.18 \u00d7 1010 m 2(6.67 \u00d7 10\u221211 N \u00b7 m2 /kg2)(4 \u00d7 106)(1.99 \u00d7 1030 kg) (3.00 \u00d7 108 m/s)2 This distance is about one-fifth of the radius of Mercury\u2019s orbit around the Sun, yet the object contains 4 million solar masses and cannot be seen with our largest telescopes. You can see why astronomers are convinced this object is a black hole. Check Your Learning What would be the size of a black hole that contained only as much mass as a typical pickup truck (about 3000 kg)? (Note that something with so little mass could never actually form a black hole, but it\u2019s interesting to think about the result.) Answer: Substituting the data into our equation gives RS = 2GM 2(6.67 \u00d7 10\u221211 N \u00b7 m2 /kg2)(3000 kg) (3.00 \u00d7 108 m/s)2 c2 = = 1.33 \u00d7 10\u221223 m. 872 Chapter 24 Black Holes and Curved Spacetime For comparison, the size of a proton is usually considered to be about 8 \u00d7 10\u221216 m, which would be about ten million times larger. A Black Hole Myth Much of the modern folklore about black holes is misleading. One idea you may have heard is that black holes go about sucking things up with their gravity. Actually, it is only very close to a black hole that the strange effects we have been discussing come into play. The gravitational attraction far away from a black hole is the same as that of the star that collapsed to form it. Remember that the gravity of any star some distance away acts as if all its mass were concentrated at a point in the center, which we call the center of gravity. For real stars, we merely imagine that all mass is concentrated there; for black holes, all the mass really is concentrated at a point in the center. So, if you are a star or distant planet orbiting around a star that becomes a black hole, your orbit may not be significantly affected by the collapse of the star (although it may be affected by any mass loss that precedes the collapse). If, on the other hand, you venture close to the event horizon, it would be very hard for you to resist the \u201cpull\u201d of the warped spacetime near the black hole. You have to get really close to the black hole to experience any significant effect. If another star or a spaceship were to pass one or two solar radii from a black hole, Newton\u2019s laws would be adequate to describe what would happen to it. Only very near the event horizon of a black hole is the gravitation so strong that Newton\u2019s laws break down. The black hole remnant of a massive star coming into our neighborhood would be far, far safer to us than its earlier incarnation as a brilliant, hot star Gravity and Time Machines Time machines are one of the favorite devices of science fiction. Such a device would allow you to move through time at a different pace or in a different direction from everyone else. General relativity suggests that it is possible, in theory, to construct a time machine using gravity that could take you into the future. Let\u2019s imagine a place where gravity is terribly strong, such as near a black hole. General relativity predicts that the stronger the gravity, the slower the pace of time (as seen by a distant observer). So, imagine a future astronaut, with a fast and strongly built spaceship, who volunteers to go on a mission to such a high-gravity environment. The astronaut leaves in the year 2222, just after graduating from college at age 22. She takes, let\u2019s say, exactly 10 years to get to the black hole. Once there, she orbits some distance from it, taking care not to get pulled in. She is now in a high-gravity realm where time passes much more slowly than", " it does on Earth. This isn\u2019t just an effect on the mechanism of her clocks\u2014time itself is running slowly. That means that every way she has of measuring time will give the same slowed-down reading when compared to time passing on Earth. Her heart will beat more slowly, her hair will grow more slowly, her antique wristwatch will tick more slowly, and so on. She is not aware of this slowing down because all her readings of time, whether made by her own bodily functions or with mechanical equipment, are measuring the same\u2014slower\u2014time. Meanwhile, back on Earth, time passes as it always does. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 873 Our astronaut now emerges from the region of the black hole, her mission of exploration finished, and returns to Earth. Before leaving, she carefully notes that (according to her timepieces) she spent about 2 weeks around the black hole. She then takes exactly 10 years to return to Earth. Her calculations tell her that since she was 22 when she left the Earth, she will be 42 plus 2 weeks when she returns. So, the year on Earth, she figures, should be 2242, and her classmates should now be approaching their midlife crises. But our astronaut should have paid more attention in her astronomy class! Because time slowed down near the black hole, much less time passed for her than for the people on Earth. While her clocks measured 2 weeks spent near the black hole, more than 2000 weeks (depending on how close she got) could well have passed on Earth. That\u2019s equal to 40 years, meaning her classmates will be senior citizens in their 80s when she (a mere 42-year-old) returns. On Earth it will be not 2242, but 2282\u2014and she will say that she has arrived in the future. Is this scenario real? Well, it has a few practical challenges: we don\u2019t think any black holes are close enough for us to reach in 10 years, and we don\u2019t think any spaceship or human can survive near a black hole. But the key point about the slowing down of time is a natural consequence of Einstein\u2019s general theory of relativity, and we saw that its predictions have been confirmed by experiment after experiment. Such developments in the understanding of science also become inspiration for science fiction writers. Recently, the film Interstellar featured the protagonist traveling close to a massive black hole; the resulting delay in his aging relative to his earthbound family is a key part of the plot. Science fiction novels, such as Gateway by Frederik Pohl and A World out of Time by Larry Niven, also make use of the slowing down of time near black holes as major turning points in the story. For a list of science fiction stories based on good astronomy, you can go to www.astrosociety.org/scifi. A Trip into a Black Hole The fact that scientists cannot see inside black holes has not kept them from trying to calculate what they are like. One of the first things these calculations showed was that the formation of a black hole obliterates nearly all information about the star that collapsed to form it. Physicists like to say \u201cblack holes have no hair,\u201d meaning that nothing sticks out of a black hole to give us clues about what kind of star produced it or what material has fallen inside. The only information a black hole can reveal about itself is its mass, its spin (rotation), and whether it has any electrical charge. What happens to the collapsing star-core that made the black hole? Our best calculations predict that the material will continue to collapse under its own weight, forming an infinitely squozen point\u2014a place of zero volume and infinite density\u2014to which we give the name singularity. At the singularity, spacetime ceases to exist. The laws of physics as we know them break down. We do not yet have the physical understanding or the mathematical tools to describe the singularity itself, or even if singularities actually occur. From the outside, however, the entire structure of a basic black hole (one that is not rotating) can be described as a singularity surrounded by an event horizon. Compared to humans, black holes are really very simple objects. Scientists have also calculated what would happen if an astronaut were to fall into a black hole. Let\u2019s take up an observing position a long, safe distance away from the event horizon and watch this astronaut fall toward it. At first he falls away from us, moving ever faster, just as though he were approaching any massive star. However, as he nears the event horizon of the black hole, things change. The strong gravitational field around the black hole will make his clocks run more slowly, when seen from our outside perspective. 874 Chapter 24 Black Holes and Curved Spacetime If, as he approaches the event horizon, he", " sends out a signal once per second according to his clock, we will see the spacing between his signals grow longer and longer until it becomes infinitely long when he reaches the event horizon. (Recalling our discussion of gravitational redshift, we could say that if the infalling astronaut uses a blue light to send his signals every second, we will see the light get redder and redder until its wavelength is nearly infinite.) As the spacing between clock ticks approaches infinity, it will appear to us that the astronaut is slowly coming to a stop, frozen in time at the event horizon. In the same way, all matter falling into a black hole will also appear to an outside observer to stop at the event horizon, frozen in place and taking an infinite time to fall through it. But don\u2019t think that matter falling into a black hole will therefore be easily visible at the event horizon. The tremendous redshift will make it very difficult to observe any radiation from the \u201cfrozen\u201d victims of the black hole. This, however, is only how we, located far away from the black hole, see things. To the astronaut, his time goes at its normal rate and he falls right on through the event horizon into the black hole. (Remember, this horizon is not a physical barrier, but only a region in space where the curvature of spacetime makes escape impossible.) You may have trouble with the idea that you (watching from far away) and the astronaut (falling in) have such different ideas about what has happened. This is the reason Einstein\u2019s ideas about space and time are called theories of relativity. What each observer measures about the world depends on (is relative to) his or her frame of reference. The observer in strong gravity measures time and space differently from the one sitting in weaker gravity. When Einstein proposed these ideas, many scientists also had difficulty with the idea that two such different views of the same event could be correct, each in its own \u201cworld,\u201d and they tried to find a mistake in the calculations. There were no mistakes: we and the astronaut really would see him fall into a black hole very differently. For the astronaut, there is no turning back. Once inside the event horizon, the astronaut, along with any signals from his radio transmitter, will remain hidden forever from the universe outside. He will, however, not have a long time (from his perspective) to feel sorry for himself as he approaches the black hole. Suppose he is falling feet first. The force of gravity that the singularity exerts on his feet is greater than on his head, so he will be stretched slightly. Because the singularity is a point, the left side of his body will be pulled slightly toward the right, and the right slightly toward the left, bringing each side closer to the singularity. The astronaut will therefore be slightly squeezed in one direction and stretched in the other. Some scientists like to call this process of stretching and narrowing spaghettification. The point at which the astronaut becomes so stretched that he perishes depends on the size of the black hole. For black holes with masses billions of times the mass of the Sun, such as those found at the centers of galaxies, the spaghettification becomes significant only after the astronaut passes through the event horizon. For black holes with masses of a few solar masses, the astronaut will be stretched and ripped apart even before he reaches the event horizon. Earth exerts similar tidal forces on an astronaut performing a spacewalk. In the case of Earth, the tidal forces are so small that they pose no threat to the health and safety of the astronaut. Not so in the case of a black hole. Sooner or later, as the astronaut approaches the black hole, the tidal forces will become so great that the astronaut will be ripped apart, eventually reduced to a collection of individual atoms that will continue their inexorable fall into the singularity From the previous discussion, you will probably agree that jumping into a black hole is definitely a once- in-a-lifetime experience! You can see an engaging explanation (https://openstax.org/l/ This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 875 30ndegtystidfor) of death by black hole by Neil deGrasse Tyson, where he explains the effect of tidal forces on the human body until it dies by spaghettification. A similar explanation can be seen in this Discovery Channel video (https://openstax.org/l/ 30dischatidfor) excerpt. 24.6 EVIDENCE FOR BLACK HOLES Learning Objectives By the end of this section, you will be able to: Describe what to look for when seeking and confirming the presence of a stellar black hole Explain how a black hole is inherently black yet can be associated with luminous matter Differentiate between stellar black holes and the black holes in the centers of", " galaxies Theory tells us what black holes are like. But do they actually exist? And how do we go about looking for something that is many light years away, only about a few dozen kilometers across (if a stellar black hole), and completely black? It turns out that the trick is not to look for the black hole itself but instead to look for what it does to a nearby companion star. As we saw, when very massive stars collapse, they leave behind their gravitational influence. What if a member of a double-star system becomes a black hole, and its companion manages to survive the death of the massive star? While the black hole disappears from our view, we may be able to deduce its presence from the things it does to its companion. Requirements for a Black Hole So, here is a prescription for finding a black hole: start by looking for a star whose motion (determined from the Doppler shift of its spectral lines) shows it to be a member of a binary star system. If both stars are visible, neither can be a black hole, so focus your attention on just those systems where only one star of the pair is visible, even with our most sensitive telescopes. Being invisible is not enough, however, because a relatively faint star might be hard to see next to the glare of a brilliant companion or if it is shrouded by dust. And even if the star really is invisible, it could be a neutron star. Therefore, we must also have evidence that the unseen star has a mass too high to be a neutron star and that it is a collapsed object\u2014an extremely small stellar remnant. We can use Kepler\u2019s law (see Orbits and Gravity) and our knowledge of the visible star to measure the mass of the invisible member of the pair. If the mass is greater than about 3 MSun, then we are likely seeing (or, more precisely, not seeing) a black hole\u2014as long as we can make sure the object really is a collapsed star. If matter falls toward a compact object of high gravity, the material is accelerated to high speed. Near the event horizon of a black hole, matter is moving at velocities that approach the speed of light. As the atoms whirl chaotically toward the event horizon, they rub against each other; internal friction can heat them to temperatures of 100 million K or more. Such hot matter emits radiation in the form of flickering X-rays. The last part of our prescription, then, is to look for a source of X-rays associated with the binary system. Since X-rays do not penetrate Earth\u2019s atmosphere, such sources must be found using X-ray telescopes in space. 876 Chapter 24 Black Holes and Curved Spacetime In our example, the infalling gas that produces the X-ray emission comes from the black hole\u2019s companion star. As we saw in The Death of Stars, stars in close binary systems can exchange mass, especially as one of the members expands into a red giant. Suppose that one star in a double-star system has evolved to a black hole and that the second star begins to expand. If the two stars are not too far apart, the outer layers of the expanding star may reach the point where the black hole exerts more gravitational force on them than do the inner layers of the red giant to which the atmosphere belongs. The outer atmosphere then passes through the point of no return between the stars and falls toward the black hole. The mutual revolution of the giant star and the black hole causes the material falling toward the black hole to spiral around it rather than flow directly into it. The infalling gas whirls around the black hole in a pancake of matter called an accretion disk. It is within the inner part of this disk that matter is revolving about the black hole so fast that internal friction heats it up to X-ray\u2013emitting temperatures (see Figure 24.1). Another way to form an accretion disk in a binary star system is to have a powerful stellar wind come from the black hole\u2019s companion. Such winds are a characteristic of several stages in a star\u2019s life. Some of the ejected gas in the wind will then flow close enough to the black hole to be captured by it into the disk (Figure 24.16). Figure 24.16. Binary Black Hole. This artist\u2019s rendition shows a black hole and star (red). As matter streams from the star, it forms a disk around the black hole. Some of the swirling material close to the black hole is pushed outward perpendicular to the disk in two narrow jets. (credit: modification of work by ESO/L. Cal\u00e7ada) We should point out that, as often happens, the measurements we have been discussing are not quite as simple as they are described in introductory textbooks. In real life, Kepler\u2019s law allows us to calculate only the combined mass of the two stars in the binary system. We must learn more about the visible star of the pair and its history to ascertain", " the distance to the binary pair, the true size of the visible star\u2019s orbit, and how the orbit of the two stars is tilted toward Earth, something we can rarely measure. And neutron stars can also have accretion disks that produce X-rays, so astronomers must study the properties of these X-rays carefully when trying to determine what kind of object is at the center of the disk. Nevertheless, a number of systems that clearly contain black holes have now been found. The Discovery of Stellar-Mass Black Holes Because X-rays are such important tracers of black holes that are having some of their stellar companions for lunch, the search for black holes had to await the launch of sophisticated X-ray telescopes into space. These instruments must have the resolution to locate the X-ray sources accurately and thereby enable us to match them to the positions of binary star systems. The first black hole binary system to be discovered is called Cygnus X-1 (see Figure 24.1). The visible star in this binary system is spectral type O. Measurements of the Doppler shifts of the O star\u2019s spectral lines show that it This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 877 has an unseen companion. The X-rays flickering from it strongly indicate that the companion is a small collapsed object. The mass of the invisible collapsed companion is about 15 times that of the Sun. The companion is therefore too massive to be either a white dwarf or a neutron star. A number of other binary systems also meet all the conditions for containing a black hole. Table 24.1 lists the characteristics of some of the best examples. Some Black Hole Candidates in Binary Star Systems Name/Catalog Designation[2] Companion Star Spectral Type Orbital Period (days) Black Hole Mass Estimates (MSun) LMC X-1 Cygnus X-1 O giant O supergiant XTE J1819.3-254 (V4641 Sgr) B giant LMC X-3 B main sequence 4U1543-475 (IL Lup) A main sequence GRO J1655-40 (V1033 Sco) F subgiant GRS 1915+105 GS202+1338 (V404 Cyg) XTE J1550-564 K giant K giant K giant A0620-00 (V616 Mon) K main sequence H1705-250 (Nova Oph 1977) K main sequence GRS1124-683 (Nova Mus 1991) K main sequence GS2000+25 (QZ Vul) K main sequence GRS1009-45 (Nova Vel 1993) XTE J1118+480 XTE J1859+226 Table 24.1 K dwarf K dwarf K dwarf 3.9 5.6 2.8 1.7 1.1 2.6 33.5 6.5 1.5 0.33 0.52 0.43 0.35 0.29 0.17 0.38 10.9 15 6\u20137 7 9 7 14 12 11 9\u201313 5\u20137 7 5\u201310 8\u20139 7 5.4 2 As you can tell, there is no standard way of naming these candidates. The chain of numbers is the location of the source in right ascension and declination (the longitude and latitude system of the sky); some of the letters preceding the numbers refer to objects (e.g., LMC) and constellations (e.g., Cygnus), while other letters refer to the satellite that discovered the candidate\u2014A for Ariel, G for Ginga, and so on. The notations in parentheses are those used by astronomers who study binary star system or novae. 878 Chapter 24 Black Holes and Curved Spacetime Some Black Hole Candidates in Binary Star Systems Name/Catalog Designation Companion Star Spectral Type Orbital Period (days) Black Hole Mass Estimates (MSun) GRO J0422+32 M dwarf 0.21 4 Table 24.1 Feeding a Black Hole After an isolated star, or even one in a binary star system, becomes a black hole, it probably won\u2019t be able to grow much larger. Out in the suburban regions of the Milky Way Galaxy where we live (see The Milky Way Galaxy), stars and star systems are much too far apart for other stars to provide \u201cfood\u201d to a hungry black hole. After all, material must approach very close to the event horizon before the gravity is any different from that of the star before it became the black hole. But, as will see, the central regions of galaxies are quite different from their outer parts. Here, stars and raw material can be quite crowded together, and they can interact much more frequently with each other. Therefore, black holes in the centers of galaxies may have a much better opportunity to find mass close enough to their event", " horizons to pull in. Black holes are not particular about what they \u201ceat\u201d: they are happy to consume other stars, asteroids, gas, dust, and even other black holes. (If two black holes merge, you just get a black hole with more mass and a larger event horizon.) As a result, black holes in crowded regions can grow, eventually swallowing thousands or even millions of times the mass of the Sun. Ground-based observations have provided compelling evidence that there is a black hole in the center of our own Galaxy with a mass of about 4 million times the mass of the Sun (we\u2019ll discuss this further in the chapter on The Milky Way Galaxy). Observations with the Hubble Space Telescope have shown dramatic evidence for the existence of black holes in the centers of many other galaxies. These black holes can contain more than a billion solar masses. The feeding frenzy of such supermassive black holes may be responsible for some of the most energetic phenomena in the universe (see Active Galaxies, Quasars, and Supermassive Black Holes). And evidence from more recent X-ray observations is also starting to indicate the existence of \u201cmiddle-weight\u201d black holes, whose masses are dozens to thousands of times the mass of the Sun. The crowded inner regions of the globular clusters we described in Stars from Adolescence to Old Age may be just the right breeding grounds for such intermediate-mass black holes. Over the past decades, many observations, especially with the Hubble Space Telescope and with X-ray satellites, have been made that can be explained only if black holes really do exist. Furthermore, the observational tests of Einstein\u2019s general theory of relativity have convinced even the most skeptical scientists that his picture of warped or curved spacetime is indeed our best description of the effects of gravity near these black holes. 24.7 GRAVITATIONAL WAVE ASTRONOMY Learning Objectives By the end of this section, you will be able to: Describe what a gravitational wave is, what can produce it, and how fast it propagates Understand the basic mechanisms used to detect gravitational waves This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 879 Another part of Einstein\u2019s ideas about gravity can be tested as a way of checking the theory that underlies black holes. According to general relativity, the geometry of spacetime depends on where matter is located. Any rearrangement of matter\u2014say, from a sphere to a sausage shape\u2014creates a disturbance in spacetime. This disturbance is called a gravitational wave, and relativity predicts that it should spread outward at the speed of light. The big problem with trying to study such waves is that they are tremendously weaker than electromagnetic waves and correspondingly difficult to detect. Proof from a Pulsar We\u2019ve had indirect evidence for some time that gravitational waves exist. In 1974, astronomers Joseph Taylor and Russell Hulse discovered a pulsar (with the designation PSR1913+16) orbiting another neutron star. Pulled by the powerful gravity of its companion, the pulsar is moving at about one-tenth the speed of light in its orbit. According to general relativity, this system of stellar corpses should be radiating energy in the form of gravitational waves at a high enough rate to cause the pulsar and its companion to spiral closer together. If this is correct, then the orbital period should decrease (according to Kepler\u2019s third law) by one ten-millionth of a second per orbit. Continuing observations showed that the period is decreasing by precisely this amount. Such a loss of energy in the system can be due only to the radiation of gravitational waves, thus confirming their existence. Taylor and Hulse shared the 1993 Nobel Prize in physics for this work. Direct Observations Although such an indirect proof convinced physicists that gravitational waves exist, it is even more satisfying to detect the waves directly. What we need are phenomena that are powerful enough to produce gravitational waves with amplitudes large enough that we can measure them. Theoretical calculations suggest some of the most likely events that would give a burst of gravitational waves strong enough that our equipment on Earth could measure it: \u2022 \u2022 \u2022 \u2022 \u2022 the coalescence of two neutron stars in a binary system that spiral together until they merge the swallowing of a neutron star by a black hole the coalescence (merger) of two black holes the implosion of a really massive star to form a neutron star or a black hole the first \u201cshudder\u201d when space and time came into existence and the universe began For the last four decades, scientists have been developing an audacious experiment to try to detect gravitational waves from a source on this list. The US experiment, which was built with collaborators from the UK, Germany, Australia and other countries, is named LIGO (Laser Interferometer Gravitational-Wave Observatory). LIGO currently has two observing stations, one in Louisiana", " and the other in the state of Washington. The effects of gravitational waves are so small that confirmation of their detection will require simultaneous measurements by two widely separated facilities. Local events that might cause small motions within the observing stations and mimic gravitational waves\u2014such as small earthquakes, ocean tides, and even traffic\u2014should affect the two sites differently. Each of the LIGO stations consists of two 4-kilometer-long, 1.2-meter-diameter vacuum pipes arranged in an L-shape. A test mass with a mirror on it is suspended by wire at each of the four ends of the pipes. Ultra-stable laser light is reflected from the mirrors and travels back and forth along the vacuum pipes (Figure 24.17). If gravitational waves pass through the LIGO instrument, then, according to Einstein\u2019s theory, the waves will affect local spacetime\u2014they will alternately stretch and shrink the distance the laser light must travel between the mirrors ever so slightly. When one arm of the instrument gets longer, the other will get shorter, and vice versa. 880 Chapter 24 Black Holes and Curved Spacetime Figure 24.17. Gravitational Wave Telescope. An aerial view of the LIGO facility at Livingston, Louisiana. Extending to the upper left and far right of the image are the 4-kilometer-long detectors. (credit: modification of work by Caltech/MIT/LIGO Laboratory) The challenge of this experiment lies in that phrase \u201cever so slightly.\u201d In fact, to detect a gravitational wave, the change in the distance to the mirror must be measured with an accuracy of one ten-thousandth the diameter of a proton. In 1972, Rainer Weiss of MIT wrote a paper suggesting how this seemingly impossible task might be accomplished. A great deal of new technology had to be developed, and work on the laboratory, with funding from the National Science Foundation, began in 1979. A full-scale prototype to demonstrate the technology was built and operated from 2002 to 2010, but the prototype was not expected to have the sensitivity required to actually detect gravitational waves from an astronomical source. Advanced LIGO, built to be more precise with the improved technology developed in the prototype, went into operation in 2015\u2014and almost immediately detected gravitational waves. What LIGO found was gravitational waves produced in the final fraction of a second of the merger of two black holes (Figure 24.18). The black holes had masses of 20 and 36 times the mass of the Sun, and the merger took place 1.3 billion years ago\u2014the gravitational waves occurred so far away that it has taken that long for them, traveling at the speed of light, to reach us. In the cataclysm of the merger, about three times the mass of the Sun was converted to energy (recall E = mc2). During the tiny fraction of a second for the merger to take place, this event produced power about 50 times the power produced by all the stars in the entire visible universe\u2014but the power was all in the form of gravitational waves and hence was invisible to our instruments, except to LIGO. The event was recorded in Louisiana about 7 milliseconds before the detection in Washington\u2014just the right distance given the speed at which gravitational waves travel\u2014and indicates that the source was located somewhere in the southern hemisphere sky. Unfortunately, the merger of two black holes is not expected to produce any light, so this is the only observation we have of the event. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 881 Figure 24.18. Signal Produced by a Gravitational Wave. (a) The top panel shows the signal measured at Hanford, Washington; the middle panel shows the signal measured at Livingston, Louisiana. The smoother thin curve in each panel shows the predicted signal, based on Einstein\u2019s general theory of relativity, produced by the merger of two black holes. The bottom panel shows a superposition of the waves detected at the two LIGO observatories. Note the remarkable agreement of the two independent observations and of the observations with theory. (b) The painting shows an artist\u2019s impression of two massive black holes spiraling inward toward an eventual merger. (credit a, b: modification of work by SXS) This detection by LIGO (and another one of a different black hole merger a few months later) opens a whole new window on the universe. One of the experimenters compared the beginning of gravitational wave astronomy to the era when silent films were replaced by movies with sound (comparing the vibration of spacetime during the passing of a gravitational wave to the vibrations that sound makes). We can now learn about events, such as the merger of black holes, that can be studied in no other way. For example, this first detected merger involved black holes with masses greater than previously observed for stellar-mass black holes. Such a discovery suggests that", " we may need to make changes to existing models of the evolution of massive stars. Observing the merger of black holes via gravitational waves also means that we can now make tests of Einstein\u2019s general theory of relativity where its effects are very strong\u2014close to black holes\u2014and not weak, as they are near Earth. One remarkable result from this first detection is that the signal measured matched so closely the theoretical predictions made using Einstein\u2019s theory. Once again, Einstein\u2019s revolutionary idea is found to be the correct description of nature. Several facilities similar to LIGO are under construction in other countries to contribute to gravitational wave astronomy and help us pinpoint more precisely where in the sky the signals we detect come from. The European Space Agency (ESA) is also exploring the possibility of building an even larger detector for gravitational waves in space. The goal is to launch a facility called eLISA sometime in the mid 2030s. The design calls for three arms 882 Chapter 24 Black Holes and Curved Spacetime or paths, each a million kilometers in length, for the laser light to travel. This facility could detect the distant merger of supermassive black holes such as might have occurred when the first generation of stars formed only a few hundred million years after the Big Bang. In December 2015, ESA launched LISA Pathfinder to test the technology required to hold two gold-platinum cubes in a state of weightless, perfect rest relative one another. While LISA Pathfinder cannot itself detect gravitational waves, such stability will be required if eLISA is to be able to detect the small changes in path length produced by passing gravitational waves. We should end by acknowledging that the ideas discussed in this chapter may seem strange and overwhelming, especially the first time you read them. The consequences of the general theory of relatively take some getting used to. But you have to admit that they make the universe more interesting and bizarre than you probably thought before you took this course. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 883 CHAPTER 24 REVIEW KEY TERMS accretion disk the disk of gas and dust found orbiting newborn stars, as well as compact stellar remnants such as white dwarfs, neutron stars, and black holes when they are in binary systems and are sufficiently close to their binary companions to draw off material black hole a region in spacetime where gravity is so strong that nothing\u2014not even light\u2014can escape equivalence principle concept that a gravitational force and a suitable acceleration are indistinguishable within a sufficiently local environment event horizon a boundary in spacetime such that events inside the boundary can have no effect on the world outside it\u2014that is, the boundary of the region around a black hole where the curvature of spacetime no longer provides any way out general theory of relativity Einstein\u2019s theory relating gravity and the structure (geometry) of space and time gravitational redshift an increase in wavelength of an electromagnetic wave (light) when propagating from or near a massive object gravitational wave a disturbance in the curvature of spacetime caused by changes in how matter is distributed; gravitational waves propagate at (or near) the speed of light. singularity the point of zero volume and infinite density to which any object that becomes a black hole must collapse, according to the theory of general relativity spacetime system of one time and three space coordinates, with respect to which the time and place of an event can be specified SUMMARY 24.1 Introducing General Relativity Einstein proposed the equivalence principle as the foundation of the theory of general relativity. According to this principle, there is no way that anyone or any experiment in a sealed environment can distinguish between free fall and the absence of gravity. 24.2 Spacetime and Gravity By considering the consequences of the equivalence principle, Einstein concluded that we live in a curved spacetime. The distribution of matter determines the curvature of spacetime; other objects (and even light) entering a region of spacetime must follow its curvature. Light must change its path near a massive object not because light is bent by gravity, but because spacetime is. 24.3 Tests of General Relativity In weak gravitational fields, the predictions of general relativity agree with the predictions of Newton\u2019s law of gravity. However, in the stronger gravity of the Sun, general relativity makes predictions that differ from Newtonian physics and can be tested. For example, general relativity predicts that light or radio waves will be deflected when they pass near the Sun, and that the position where Mercury is at perihelion would change by 43 884 Chapter 24 Black Holes and Curved Spacetime arcsec per century even if there were no other planets in the solar system to perturb its orbit. These predictions have been verified by observation. 24.4 Time in General Relativity General relativity predicts that the stronger the gravity, the more slowly time must run. Experiments on Earth and with spacecraft have confirmed this prediction with remarkable accuracy. When", " light or other radiation emerges from a compact smaller remnant, such as a white dwarf or neutron star, it shows a gravitational redshift due to the slowing of time. 24.5 Black Holes Theory suggests that stars with stellar cores more massive than three times the mass of the Sun at the time they exhaust their nuclear fuel will collapse to become black holes. The surface surrounding a black hole, where the escape velocity equals the speed of light, is called the event horizon, and the radius of the surface is called the Schwarzschild radius. Nothing, not even light, can escape through the event horizon from the black hole. At its center, each black hole is thought to have a singularity, a point of infinite density and zero volume. Matter falling into a black hole appears, as viewed by an outside observer, to freeze in position at the event horizon. However, if we were riding on the infalling matter, we would pass through the event horizon. As we approach the singularity, the tidal forces would tear our bodies apart even before we reach the singularity. 24.6 Evidence for Black Holes The best evidence of stellar-mass black holes comes from binary star systems in which (1) one star of the pair is not visible, (2) the flickering X-ray emission is characteristic of an accretion disk around a compact object, and (3) the orbit and characteristics of the visible star indicate that the mass of its invisible companion is greater than 3 MSun. A number of systems with these characteristics have been found. Black holes with masses of millions to billions of solar masses are found in the centers of large galaxies. 24.7 Gravitational Wave Astronomy General relativity predicts that the rearrangement of matter in space should produce gravitational waves. The existence of such waves was first confirmed in observations of a pulsar in orbit around another neutron star whose orbits were spiraling closer and losing energy in the form of gravitational waves. In 2015, LIGO found gravitational waves directly by detecting the signal produced by the merger of two stellar-mass black holes, opening a new window on the universe. FOR FURTHER EXPLORATION Articles Black Holes Charles, P. & Wagner, R. \u201cBlack Holes in Binary Stars: Weighing the Evidence.\u201d Sky & Telescope (May 1996): 38. Excellent review of how we find stellar-mass black holes. Gezari, S. \u201cStar-Shredding Black Holes.\u201d Sky & Telescope (June 2013): 16. When black holes and stars collide. Jayawardhana, R. \u201cBeyond Black.\u201d Astronomy (June 2002): 28. On finding evidence of the existence of event horizons and thus black holes. Nadis, S. \u201cBlack Holes: Seeing the Unseeable.\u201d Astronomy (April 2007): 26. A brief history of the black hole idea and an introduction to potential new ways to observe them. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 885 Psallis, D. & Sheperd, D. \u201cThe Black Hole Test.\u201d Scientific American (September 2015): 74\u201379. The Event Horizon Telescope (a network of radio telescopes) will test some of the stranger predictions of general relativity for the regions near black holes. The September 2015 issue of Scientific American was devoted to a celebration of the 100th anniversary of the general theory of relativity. Rees, M. \u201cTo the Edge of Space and Time.\u201d Astronomy (July 1998): 48. Good, quick overview. Talcott, R. \u201cBlack Holes in our Backyard.\u201d Astronomy (September 2012): 44. Discussion of different kinds of black holes in the Milky Way and the 19 objects known to be black holes. Gravitational Waves Bartusiak, M. \u201cCatch a Gravity Wave.\u201d Astronomy (October 2000): 54. Gibbs, W. \u201cRipples in Spacetime.\u201d Scientific American (April 2002): 62. Haynes, K., & Betz, E. \u201cA Wrinkle in Spacetime Confirms Einstein\u2019s Gravitation.\u201d Astronomy (May 2016): 22. On the direct detection of gravity waves. Sanders, G., and Beckett, D. \u201cLIGO: An Antenna Tuned to the Songs of Gravity.\u201d Sky & Telescope (October 2000): 41. Websites Black Holes Black Hole Encyclopedia: http://blackholes.stardate.org. From StarDate at the University of Texas McDonald Observatory. Black Holes: http://science.nasa.gov/astrophysics/focus-areas/black-holes. NASA overview of black holes, along with links to the most recent news and discoveries. Black Holes FAQ: http://cfpa.berkeley.edu", "/Education/BHfaq.html. Frequently asked questions about black holes, answered by Ted Bunn of UC\u2013Berkeley\u2019s Center for Particle Astrophysics. Black Holes: Gravity\u2019s Relentless Pull: http://hubblesite.org/explore_astronomy/black_holes/home.html. The Hubble Space Telescope\u2019s Journey to a Black Hole and Black Hole Encyclopedia (a good introduction for beginners). Introduction to Black Holes: http://www.damtp.cam.ac.uk/research/gr/public/bh_intro.html. The Cambridge University Relativity Group\u2019s pages on black holes and related calculations. March 1918: Testing Einstein: http://www.nature.com/nature/podcast/index-pastcast-2014-03-20.html. Nature Podcast about the 1919 eclipse expedition that proved Einstein\u2019s General Theory of Relativity. Movies from the Edge of Spacetime: http://archive.ncsa.illinois.edu/Cyberia/NumRel/MoviesEdge.html. Physicists simulate the behavior of various black holes. Virtual Trips into Black Holes and Neutron Stars: http://antwrp.gsfc.nasa.gov/htmltest/rjn_bht.html. By Robert Nemiroff at Michigan Technological University. Gravitational Waves Advanced LIGO: https://www.advancedligo.mit.edu. The full story on this gravitational wave observatory. eLISA: https://www.elisascience.org. Gravitational Waves Detected, Confirming Einstein\u2019s Theory: http://www.nytimes.com/2016/02/12/science/ ligo-gravitational-waves-black-holes-einstein.html. New York Times article and videos on the discovery of gravitational waves. 886 Chapter 24 Black Holes and Curved Spacetime Gravitational Waves Discovered from Colliding Black Holes: http://www.scientificamerican.com/article/ gravitational-waves-discovered-from-colliding-black-holes1. Scientific American coverage of the discovery of gravitational waves (note the additional materials available in the menu at the right). LIGO Caltech: https://www.ligo.caltech.edu. Videos Black Holes Black Holes: The End of Time or a New Beginning?: https://www.youtube.com/watch?v=mgtJRsdKe6Q. 2012 Silicon Valley Astronomy Lecture by Roger Blandford (1:29:52). Death by Black Hole: http://www.openculture.com/2009/02/death_by_black_hole_and_its_kind_of_funny.htm. Neil deGrasse Tyson explains spaghettification with only his hands (5:34). Hearts of Darkness: Black Holes in Space: https://www.youtube.com/watch?v=4tiAOldypLk. 2010 Silicon Valley Astronomy Lecture by Alex Filippenko (1:56:11). Gravitational Waves Journey of a Gravitational Wave: https://www.youtube.com/watch?v=FlDtXIBrAYE. Introduction from LIGO Caltech (2:55). LIGO\u2019s First Detection of Gravitational Waves: https://www.youtube.com/watch?v=gw-i_VKd6Wo. Explanation and animations from PBS Digital Studio (9:31). Two Black Holes Merge into One: https://www.youtube.com/watch?v=I_88S8DWbcU. Simulation from LIGO Caltech (0:35). What the Discovery of Gravitational Waves Means: https://www.youtube.com/watch?v=jMVAgCPYYHY. TED Talk by Allan Adams (10:58). COLLABORATIVE GROUP ACTIVITIES A. A computer science major takes an astronomy course like the one you are taking and becomes fascinated with black holes. Later in life, he founds his own internet company and becomes very wealthy when it goes public. He sets up a foundation to support the search for black holes in our Galaxy. Your group is the allocation committee of this foundation. How would you distribute money each year to increase the chances that more black holes will be found? B. Suppose for a minute that stars evolve without losing any mass at any stage of their lives. Your group is given a list of binary star systems. Each binary contains one main-sequence star and one invisible companion. The spectral types of the main-sequence stars range from spectral type O to M. Your job is to determine whether any of the invisible companions might be black holes. Which ones are worth observing? Why? (Hint: Remember that in a binary star system, the two stars", " form at the same time, but the pace of their evolution depends on the mass of each star.) C. You live in the far future, and the members of your group have been convicted (falsely) of high treason. The method of execution is to send everyone into a black hole, but you get to pick which one. Since you are doomed to die, you would at least like to see what the inside of a black hole is like\u2014even if you can\u2019t tell anyone outside about it. Would you choose a black hole with a mass equal to that of Jupiter or one with This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 887 a mass equal to that of an entire galaxy? Why? What would happen to you as you approached the event horizon in each case? (Hint: Consider the difference in force on your feet and your head as you cross over the event horizon.) D. General relativity is one of the areas of modern astrophysics where we can clearly see the frontiers of human knowledge. We have begun to learn about black holes and warped spacetime recently and are humbled by how much we still don\u2019t know. Research in this field is supported mostly by grants from government agencies. Have your group discuss what reasons there are for our tax dollars to support such \u201cfar out\u201d (seemingly impractical) work. Can you make a list of \u201cfar out\u201d areas of research in past centuries that later led to practical applications? What if general relativity does not have many practical applications? Do you think a small part of society\u2019s funds should still go to exploring theories about the nature of space and time? E. Once you all have read this chapter, work with your group to come up with a plot for a science fiction story that uses the properties of black holes. F. Black holes seem to be fascinating not just to astronomers but to the public, and they have become part of popular culture. Searching online, have group members research examples of black holes in music, advertising, cartoons, and the movies, and then make a presentation to share the examples you found with the whole class. G. As mentioned in the Gravity and Time Machines feature box in this chapter, the film Interstellar has a lot of black hole science in its plot and scenery. That\u2019s because astrophysicist Kip Thorne at Caltech had a big hand in writing the initial treatment for the movie, and later producing it. Get your group members together (be sure you have popcorn) for a viewing of the movie and then try to use your knowledge of black holes from this chapter to explain the plot. (Note that the film also uses the concept of a wormhole, which we don\u2019t discuss in this chapter. A wormhole is a theoretically possible way to use a large, spinning black hole to find a way to travel from one place in the universe to another without having to go through regular spacetime to get there.) EXERCISES Review Questions 1. How does the equivalence principle lead us to suspect that spacetime might be curved? 2. If general relativity offers the best description of what happens in the presence of gravity, why do physicists still make use of Newton\u2019s equations in describing gravitational forces on Earth (when building a bridge, for example)? 3. Einstein\u2019s general theory of relativity made or allowed us to make predictions about the outcome of several experiments that had not yet been carried out at the time the theory was first published. Describe three experiments that verified the predictions of the theory after Einstein proposed it. 4. If a black hole itself emits no radiation, what evidence do astronomers and physicists today have that the theory of black holes is correct? 5. What characteristics must a binary star have to be a good candidate for a black hole? Why is each of these characteristics important? 888 Chapter 24 Black Holes and Curved Spacetime 6. A student becomes so excited by the whole idea of black holes that he decides to jump into one. It has a mass 10 times the mass of our Sun. What is the trip like for him? What is it like for the rest of the class, watching from afar? 7. What is an event horizon? Does our Sun have an event horizon around it? 8. What is a gravitational wave and why was it so hard to detect? 9. What are some strong sources of gravitational waves that astronomers hope to detect in the future? 10. Suppose the amount of mass in a black hole doubles. Does the event horizon change? If so, how does it change? Thought Questions 11. Imagine that you have built a large room around the people in Figure 24.4 and that this room is falling at exactly the same rate as they are. Galileo showed that if there is no air friction, light and heavy objects that are dropping due to gravity will fall at the same", " rate. Suppose that this were not true and that instead heavy objects fall faster. Also suppose that the man in Figure 24.4 is twice as massive as the woman. What would happen? Would this violate the equivalence principle? 12. A monkey hanging from a tree branch sees a hunter aiming a rifle directly at him. The monkey then sees a flash and knows that the rifle has been fired. Reacting quickly, the monkey lets go of the branch and drops so that the bullet can pass harmlessly over his head. Does this act save the monkey\u2019s life? Why or why not? (Hint: Consider the similarities between this situation and that of Exercise 24.11.) 13. Why would we not expect to detect X-rays from a disk of matter about an ordinary star? 14. Look elsewhere in this book for necessary data, and indicate what the final stage of evolution\u2014white dwarf, neutron star, or black hole\u2014will be for each of these kinds of stars. A. Spectral type-O main-sequence star B. Spectral type-B main-sequence star C. Spectral type-A main-sequence star D. Spectral type-G main-sequence star E. Spectral type-M main-sequence star 15. Which is likely to be more common in our Galaxy: white dwarfs or black holes? Why? 16. If the Sun could suddenly collapse to a black hole, how would the period of Earth\u2019s revolution about it differ from what it is now? 17. Suppose the people in Figure 24.4 are in an elevator moving upward with an acceleration equal to g, but in the opposite direction. The woman throws the ball to the man with a horizontal force. What happens to the ball? 18. You arrange to meet a friend at 5:00 p.m. on Valentine\u2019s Day on the observation deck of the Empire State Building in New York City. You arrive right on time, but your friend is not there. She arrives 5 minutes late and says the reason is that time runs faster at the top of a tall building, so she is on time but you were early. Is your friend right? Does time run slower or faster at the top of a building, as compared with its base? Is this a reasonable excuse for your friend arriving 5 minutes late? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 24 Black Holes and Curved Spacetime 889 19. You are standing on a scale in an elevator when the cable snaps, sending the elevator car into free fall. Before the automatic brakes stop your fall, you glance at the scale reading. Does the scale show your real weight? An apparent weight? Something else? Figuring For Yourself 20. Look up G, c, and the mass of the Sun in Appendix E and calculate the radius of a black hole that has the same mass as the Sun. (Note that this is only a theoretical calculation. The Sun does not have enough mass to become a black hole.) 21. Suppose you wanted to know the size of black holes with masses that are larger or smaller than the Sun. You could go through all the steps in Exercise 24.20, wrestling with a lot of large numbers with large exponents. You could be clever, however, and evaluate all the constants in the equation once and then simply vary the mass. You could even express the mass in terms of the Sun\u2019s mass and make future calculations really easy. Show that the event horizon equation is equivalent to saying that the radius of the event horizon is equal to 3 km times the mass of the black hole in units of the Sun\u2019s mass. 22. Use the result from Exercise 24.21 to calculate the radius of a black hole with a mass equal to: the Earth, a B0-type main-sequence star, a globular cluster, and the Milky Way Galaxy. Look elsewhere in this text and the appendixes for tables that provide data on the mass of these four objects. 23. Since the force of gravity a significant distance away from the event horizon of a black hole is the same as that of an ordinary object of the same mass, Kepler\u2019s third law is valid. Suppose that Earth collapsed to the size of a golf ball. What would be the period of revolution of the Moon, orbiting at its current distance of 400,000 km? Use Kepler\u2019s third law to calculate the period of revolution of a spacecraft orbiting at a distance of 6000 km. 890 Chapter 24 Black Holes and Curved Spacetime This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 891 25 THE MILKY WAY GALAXY Figure 25.1. Milky Way Galaxy. The Milky Way rises over Square Tower, an ancestral pueblo building at Hovenweep National Monument in Utah. Many stars", " and dark clouds of dust combine to make a spectacular celestial sight of our home Galaxy. The location has been designated an International Dark Sky Park by the International Dark Sky Association. Chapter Outline 25.1 The Architecture of the Galaxy 25.2 Spiral Structure 25.3 The Mass of the Galaxy 25.4 The Center of the Galaxy 25.5 Stellar Populations in the Galaxy 25.6 The Formation of the Galaxy Thinking Ahead Today, we know that our Sun is just one of the many billions of stars that make up the huge cosmic island we call the Milky Way Galaxy. How can we \u201cweigh\u201d such an enormous system of stars and measure its total mass? One of the most striking features you can see in a truly dark sky\u2014one without light pollution\u2014is the band of faint white light called the Milky Way, which stretches from one horizon to the other. The name comes from an ancient Greek legend that compared its faint white splash of light to a stream of spilled milk. But folktales differ from culture to culture: one East African tribe thought of the hazy band as the smoke of ancient campfires, several Native American stories tell of a path across the sky traveled by sacred animals, and in Siberia, the diffuse arc was known as the seam of the tent of the sky. In 1610, Galileo made the first telescopic survey of the Milky Way and discovered that it is composed of a multitude of individual stars. Today, we know that the Milky Way comprises our view inward of the huge cosmic pinwheel that we call the Milky Way Galaxy and that is our home. Moreover, our Galaxy is now recognized as just one galaxy among many billions of other galaxies in the cosmos. 892 Chapter 25 The Milky Way Galaxy 25.1 THE ARCHITECTURE OF THE GALAXY Learning Objectives By the end of this section, you will be able to: Explain why William and Caroline Herschel concluded that the Milky Way has a flattened structure centered on the Sun and solar system Describe the challenges of determining the Galaxy\u2019s structure from our vantage point within it Identify the main components of the Galaxy The Milky Way Galaxy surrounds us, and you might think it is easy to study because it is so close. However, the very fact that we are embedded within it presents a difficult challenge. Suppose you were given the task of mapping New York City. You could do a much better job from a helicopter flying over the city than you could if you were standing in Times Square. Similarly, it would be easier to map our Galaxy if we could only get a little way outside it, but instead we are trapped inside and way out in its suburbs\u2014far from the galactic equivalent of Times Square. Herschel Measures the Galaxy In 1785, William Herschel (Figure 25.2) made the first important discovery about the architecture of the Milky Way Galaxy. Using a large reflecting telescope that he had built, William and his sister Caroline counted stars in different directions of the sky. They found that most of the stars they could see lay in a flattened structure encircling the sky, and that the numbers of stars were about the same in any direction around this structure. Herschel therefore concluded that the stellar system to which the Sun belongs has the shape of a disk or wheel (he might have called it a Frisbee except Frisbees hadn\u2019t been invented yet), and that the Sun must be near the hub of the wheel (Figure 25.3). Figure 25.2. William Herschel (1738\u20131822) and Caroline Herschel (1750\u20131848). William Herschel was a German musician who emigrated to England and took up astronomy in his spare time. He discovered the planet Uranus, built several large telescopes, and made measurements of the Sun\u2019s place in the Galaxy, the Sun\u2019s motion through space, and the comparative brightnesses of stars. This painting shows William and his sister Caroline polishing a telescope lens. (credit: modification of work by the Wellcome Library) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 893 To understand why Herschel reached this conclusion, imagine that you are a member of a band standing in formation during halftime at a football game. If you count the band members you see in different directions and get about the same number each time, you can conclude that the band has arranged itself in a circular pattern with you at the center. Since you see no band members above you or underground, you know that the circle made by the band is much flatter than it is wide. Figure 25.3. Herschel\u2019s Diagram of the Milky Way. Herschel constructed this cross section of the Galaxy by counting stars in various directions. We now know that Herschel was right about the shape of our system, but wrong about where the Sun lies within the disk. As we saw in Between the Stars: Gas and Dust in Space,", " we live in a dusty Galaxy. Because interstellar dust absorbs the light from stars, Herschel could see only those stars within about 6000 light-years of the Sun. Today we know that this is a very small section of the entire 100,000-light-year-diameter disk of stars that makes up the Galaxy Harlow Shapley: Mapmaker to the Stars Until the early 1900s, astronomers generally accepted Herschel\u2019s conclusion that the Sun is near the center of the Galaxy. The discovery of the Galaxy\u2019s true size and our actual location came about largely through the efforts of Harlow Shapley. In 1917, he was studying RR Lyrae variable stars in globular clusters. By comparing the known intrinsic luminosity of these stars to how bright they appeared, Shapley could calculate how far away they are. (Recall that it is distance that makes the stars look dimmer than they would be \u201cup close,\u201d and that the brightness fades as the distance squared.) Knowing the distance to any star in a cluster then tells us the distance to the cluster itself. Globular clusters can be found in regions that are free of interstellar dust and so can be seen at very large distances. When Shapley used the distances and directions of 93 globular clusters to map out their positions in space, he found that the clusters are distributed in a spherical volume, which has its center not at the Sun but at a distant point along the Milky Way in the direction of Sagittarius. Shapley then made the bold assumption, verified by many other observations since then, that the point on which the system of globular clusters is centered is also the center of the entire Galaxy (Figure 25.4). 894 Chapter 25 The Milky Way Galaxy Figure 25.4. Harlow Shapley and His Diagram of the Milky Way. (a) Shapley poses for a formal portrait. (b) His diagram shows the location of globular clusters, with the position of the Sun also marked. The black area shows Herschel\u2019s old diagram, centered on the Sun, approximately to scale. Shapley\u2019s work showed once and for all that our star has no special place in the Galaxy. We are in a nondescript region of the Milky Way, only one of 200 to 400 billion stars that circle the distant center of our Galaxy. Born in 1885 on a farm in Missouri, Harlow Shapley at first dropped out of school with the equivalent of only a fifth-grade education. He studied at home and at age 16 got a job as a newspaper reporter covering crime stories. Frustrated by the lack of opportunities for someone who had not finished high school, Shapley went back and completed a six-year high-school program in only two years, graduating as class valedictorian. In 1907, at age 22, he went to the University of Missouri, intent on studying journalism, but found that the school of journalism would not open for a year. Leafing through the college catalog (or so he told the story later), he chanced to see \u201cAstronomy\u201d among the subjects beginning with \u201cA.\u201d Recalling his boyhood interest in the stars, he decided to study astronomy for the next year (and the rest, as the saying goes, is history). Upon graduation Shapley received a fellowship for graduate study at Princeton and began to work with the brilliant Henry Norris Russell (see the Henry Norris Russell feature box). For his PhD thesis, Shapley made major contributions to the methods of analyzing the behavior of eclipsing binary stars. He was also able to show that cepheid variable stars are not binary systems, as some people thought at the time, but individual stars that pulsate with striking regularity. Impressed with Shapley\u2019s work, George Ellery Hale offered him a position at the Mount Wilson Observatory, where the young man took advantage of the clear mountain air and the 60-inch reflector to do his pioneering study of variable stars in globular clusters. Shapley subsequently accepted the directorship of the Harvard College Observatory, and over the next 30 years, he and his collaborators made contributions to many fields of astronomy, including the study of This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 895 neighboring galaxies, the discovery of dwarf galaxies, a survey of the distribution of galaxies in the universe, and much more. He wrote a series of nontechnical books and articles and became known as one of the most effective popularizers of astronomy. Shapley enjoyed giving lectures around the country, including at many smaller colleges where students and faculty rarely got to interact with scientists of his caliber. During World War II, Shapley helped rescue many scientists and their families from Eastern Europe; later, he helped found UNESCO, the United Nations Educational, Scientific, and Cultural Organization. He wrote a pamphlet called Science from Shipboard for men and women in the armed services who", " had to spend many weeks on board transport ships to Europe. And during the difficult period of the 1950s, when congressional committees began their \u201cwitch hunts\u201d for communist sympathizers (including such liberal leaders as Shapley), he spoke out forcefully and fearlessly in defense of the freedom of thought and expression. A man of many interests, he was fascinated by the behavior of ants, and wrote scientific papers about them as well as about galaxies. By the time he died in 1972, Shapley was acknowledged as one of the pivotal figures of modern astronomy, a \u201ctwentieth-century Copernicus\u201d who mapped the Milky Way and showed us our place in the Galaxy To find more information about Shapley\u2019s life and work (https://openstaxcollege.org/l/ 30shapbrumed), see the entry for him on the Bruce Medalists website. (This site features the winners of the Bruce Medal of the Astronomical Society of the Pacific, one of the highest honors in astronomy; the list is a who\u2019s who of some of the greatest astronomers of the last twelve decades.) Disks and Haloes With modern instruments, astronomers can now penetrate the \u201csmog\u201d of the Milky Way by studying radio and infrared emissions from distant parts of the Galaxy. Measurements at these wavelengths (as well as observations of other galaxies like ours) have given us a good idea of what the Milky Way would look like if we could observe it from a distance. Figure 25.5 sketches what we would see if we could view the Galaxy face-on and edge-on. The brightest part of the Galaxy consists of a thin, circular, rotating disk of stars distributed across a region about 100,000 light-years in diameter and about 1000 light-years thick. (Given how thin the disk is, perhaps a CD is a more appropriate analogy than a wheel.) In addition to stars, the dust and gas from which stars form are also found mostly in the thin disk of the Galaxy. The mass of the interstellar matter is about 15% of the mass of the stars in this disk. 896 Chapter 25 The Milky Way Galaxy Figure 25.5. Schematic Representation of the Galaxy. The left image shows the face-on view of the spiral disk; the right image shows the view looking edge-on along the disk. The major spiral arms are labeled. The Sun is located on the inside edge of the short Orion spur. As the diagram in Figure 25.5 shows, the stars, gas, and dust are not spread evenly throughout the disk but are concentrated into a central bar and a series of spiral arms. Recent infrared observations have confirmed that the central bar is composed mostly of old yellow-red stars. The two main spiral arms appear to connect with the ends of the bar. They are highlighted by the blue light from young hot stars. We know many other spiral galaxies that also have bar-shaped concentrations of stars in their central regions; for that reason they are called barred spirals. Figure 25.6 shows two other galaxies\u2014one without a bar and one with a strong bar\u2014to give you a basis for comparison to our own. We will describe our spiral structure in more detail shortly. The Sun is located about halfway between the center of the Galaxy and the edge of the disk and only about 70 light- years above its central plane. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 897 Figure 25.6. Unbarred and Barred Spiral Galaxies. (a) This image shows the unbarred spiral galaxy M74. It contains a small central bulge of mostly old yellow-red stars, along with spiral arms that are highlighted with the blue light from young hot stars. (b) This image shows the strongly barred spiral galaxy NGC 1365. The bulge and the fainter bar both appear yellowish because the brightest stars in them are mostly old yellow and red giants. Two main spiral arms project from the ends of the bar. As in M74, these spiral arms are populated with blue stars and red patches of glowing gas\u2014hallmarks of recent star formation. The Milky Way Galaxy is thought to have a barred spiral structure that is intermediate between these two examples. (credit a: modification of work by ESO/PESSTO/S. Smartt; credit b: modification of work by ESO) Our thin disk of young stars, gas, and dust is embedded in a thicker but more diffuse disk of older stars; this thicker disk extends about 3000 light-years above and below the midplane of the thin disk and contains only about 5% as much mass as the thin disk. Close in to the galactic center (within about 10,000 light-years), the stars are no longer confined to the disk but form a central bulge (or nuclear bulge). When we observe with visible light, we can glimpse the stars in", " the bulge only in those rare directions where there happens to be relatively little interstellar dust. The first picture that actually succeeded in showing the bulge as a whole was taken at infrared wavelengths (Figure 25.7). Figure 25.7. Inner Part of the Milky Way Galaxy. This beautiful infrared map, showing half a billion stars, was obtained as part of the Two Micron All Sky Survey (2MASS). Because interstellar dust does not absorb infrared as strongly as visible light, this view reveals the previously hidden bulge of old stars that surrounds the center of our Galaxy, along with the Galaxy\u2019s thin disk component. (credit: modification of work by 2MASS/J. Carpenter, T. H. Jarrett, and R. Hurt) The fact that much of the bulge is obscured by dust makes its shape difficult to determine. For a long time, astronomers assumed it was spherical. However, infrared images and other data indicate that the bulge is about two times longer than it is wide, and shaped rather like a peanut. The relationship between this elongated inner bulge and the larger bar of stars remains uncertain. At the very center of the nuclear bulge is a tremendous concentration of matter, which we will discuss later in this chapter. 898 Chapter 25 The Milky Way Galaxy In our Galaxy, the thin and thick disks and the nuclear bulge are embedded in a spherical halo of very old, faint stars that extends to a distance of at least 150,000 light-years from the galactic center. Most of the globular clusters are also found in this halo. The mass in the Milky Way extends even farther out, well beyond the boundary of the luminous stars to a distance of at least 200,000 light-years from the center of the Galaxy. This invisible mass has been give the name dark matter because it emits no light and cannot be seen with any telescope. Its composition is unknown, and it can be detected only because of its gravitational effects on the motions of luminous matter that we can see. We know that this extensive dark matter halo exists because of its effects on the orbits of distant star clusters and other dwarf galaxies that are associated with the Galaxy. This mysterious halo will be a subject of the section on The Mass of the Galaxy, and the properties of dark matter will be discussed more in the chapter on The Big Bang. Some vital statistics of the thin and thick disks and the stellar halo are given in Table 25.1, with an illustration in Figure 25.8. Note particularly how the ages of stars correlate with where they are found. As we shall see, this information holds important clues to how the Milky Way Galaxy formed. Characteristics of the Milky Way Galaxy Property Thin Disk Thick Disk Stellar Halo (Excludes Dark Matter) Stellar mass 4 \u00d7 1010 MSun A few percent of the thin disk mass 1010 MSun Luminosity 3 \u00d7 1010 LSun A few percent of the thin disk luminosity 8 \u00d7 108 LSun Typical age of stars 1 million to 10 billion years 11 billion years 13 billion years Heavier-element abundance Rotation Table 25.1 High High Intermediate Very low Intermediate Very low This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 899 Figure 25.8. Major Parts of the Milky Way Galaxy. This schematic shows the major components of our Galaxy. Establishing this overall picture of the Galaxy from our dust-shrouded viewpoint inside the thin disk has been one of the great achievements of modern astronomy (and one that took decades of effort by astronomers working with a wide range of telescopes). One thing that helped enormously was the discovery that our Glaxy is not unique in its characteristics. There are many other flat, spiral-shaped islands of stars, gas, and dust in the universe. For example, the Milky Way somewhat resembles the Andromeda galaxy, which, at a distance of about 2.3 million light-years, is our nearest neighboring giant spiral galaxy. Just as you can get a much better picture of yourself if someone else takes the photo from a distance away, pictures and other diagnostic observations of nearby galaxies that resemble ours have been vital to our understanding of the properties of the Milky Way The Milky Way Galaxy in Myth and Legend To most of us living in the twenty-first century, the Milky Way Galaxy is an elusive sight. We must make an effort to leave our well-lit homes and streets and venture beyond our cities and suburbs into less populated environments. Once the light pollution subsides to negligible levels, the Milky Way can be readily spotted arching over the sky on clear, moonless nights. The Milky Way is especially bright in late summer and early fall in the Northern Hemisphere. Some of the best places to view the Milky Way are in our national and state parks, where residential and industrial developments have been kept to a minimum. Some of these parks host special sky-gazing events that are definitely worth checking out\u2014especially during the two weeks", " surrounding the new moon, when the faint stars and Milky Way don\u2019t have to compete with the Moon\u2019s brilliance. Go back a few centuries, and these starlit sights would have been the norm rather than the exception. Before the advent of electric or even gas lighting, people relied on short-lived fires to illuminate their homes and byways. Consequently, their night skies were typically much darker. Confronted by myriad stellar patterns and the Milky Way\u2019s gauzy band of diffuse light, people of all cultures developed myths to make sense of it all. 900 Chapter 25 The Milky Way Galaxy Some of the oldest myths relating to the Milky Way are maintained by the aboriginal Australians through their rock painting and storytelling. These legacies are thought to go back tens of thousands of years, to when the aboriginal people were being \u201cdreamed\u201d along with the rest of the cosmos. The Milky Way played a central role as an arbiter of the Creation. Taking the form of a great serpent, it joined with the Earth serpent to dream and thus create all the creatures on Earth. The ancient Greeks viewed the Milky Way as a spray of milk that spilled from the breast of the goddess Hera. In this legend, Zeus had secretly placed his infant son Heracles at Hera\u2019s breast while she was asleep in order to give his half-human son immortal powers. When Hera awoke and found Heracles suckling, she pushed him away, causing her milk to spray forth into the cosmos (Figure 25.9). The dynastic Chinese regarded the Milky Way as a \u201csilver river\u201d that was made to separate two star- crossed lovers. To the east of the Milky Way, Zhi Nu, the weaving maiden, was identified with the bright star Vega in the constellation of Lyra the Harp. To the west of the Milky Way, her lover Niu Lang, the cowherd, was associated with the star Altair in the constellation of Aquila the Eagle. They had been exiled on opposite sides of the Milky Way by Zhi Nu\u2019s mother, the Queen of Heaven, after she heard of their secret marriage and the birth of their two children. However, once a year, they are permitted to reunite. On the seventh day of the seventh lunar month (which typically occurs in our month of August), they would meet on a bridge over the Milky Way that thousands of magpies had made (Figure 25.9). This romantic time continues to be celebrated today as Qi Xi, meaning \u201cDouble Seventh,\u201d with couples reenacting the cosmic reunion of Zhi Nu and Niu Lang. Figure 25.9. The Milky Way in Myth. (a) Origin of the Milky Way by Jacopo Tintoretto (circa 1575) illustrates the Greek myth that explains the formation of the Milky Way. (b) The Moon of the Milky Way by Japanese painter Tsukioka Yoshitoshi depicts the Chinese legend of Zhi Nu and Niu Lang. To the Quechua Indians of Andean Peru, the Milky Way was seen as the celestial abode for all sorts of cosmic creatures. Arrayed along the Milky Way are myriad dark patches that they identified with partridges, llamas, a toad, a snake, a fox, and other animals. The Quechua\u2019s orientation toward the dark This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 901 regions rather than the glowing band of starlight appears to be unique among all the myth makers. Likely, their access to the richly structured southern Milky Way had something to do with it. Among Finns, Estonians, and related northern European cultures, the Milky Way is regarded as the \u201cpathway of birds\u201d across the night sky. Having noted that birds seasonally migrate along a north-south route, they identified this byway with the Milky Way. Recent scientific studies have shown that this myth is rooted in fact: the birds of this region use the Milky Way as a guide for their annual migrations. Today, we regard the Milky Way as our galactic abode, where the foment of star birth and star death plays out on a grand stage, and where sundry planets have been found to be orbiting all sorts of stars. Although our perspective on the Milky Way is based on scientific investigations, we share with our forebears an affinity for telling stories of origin and transformation. In these regards, the Milky Way continues to fascinate and inspire us. 25.2 SPIRAL STRUCTURE Learning Objectives By the end of this section, you will be able to: Describe the structure of the Milky Way Galaxy and how astronomers discovered it Compare theoretical models for the formation of spiral arms in disk galaxies Astronomers were able to make tremendous progress in mapping the spiral structure of the Milky Way after the discovery of the 21-cm line that comes from cool hydrogen (see", " Between the Stars: Gas and Dust in Space). Remember that the obscuring effect of interstellar dust prevents us from seeing stars at large distances in the disk at visible wavelengths. However, radio waves of 21-cm wavelength pass right through the dust, enabling astronomers to detect hydrogen atoms throughout the Galaxy. More recent surveys of the infrared emission from stars in the disk have provided a similar dust-free perspective of our Galaxy\u2019s stellar distribution. Despite all this progress over the past fifty years, we are still just beginning to pin down the precise structure of our Galaxy. The Arms of the Milky Way Our radio observations of the disk\u2019s gaseous component indicate that the Galaxy has two major spiral arms that emerge from the bar and several fainter arms and shorter spurs. You can see a recently assembled map of our Galaxy\u2019s arm structure\u2014derived from studies in the infrared\u2014in Figure 25.10. 902 Chapter 25 The Milky Way Galaxy Figure 25.10. Milky Way Bar and Arms. Here, we see the Milky Way Galaxy as it would look from above. This image, assembled from data from NASA\u2019s WISE mission, shows that the Milky Way Galaxy has a modest bar in its central regions. Two spiral arms, Scutum-Centaurus and Perseus, emerge from the ends of the bar and wrap around the bulge. The Sagittarius and Outer arms have fewer stars than the other two arms. (credit: modification of work by NASA/JPL-Caltech/R. Hurt (SSC/Caltech)) The Sun is near the inner edge of a short arm called the Orion Spur, which is about 10,000 light-years long and contains such conspicuous features as the Cygnus Rift (the great dark nebula in the summer Milky Way) and the bright Orion Nebula. Figure 25.11 shows a few other objects that share this small section of the Galaxy with us and are easy to see. Remember, the farther away we try to look from our own arm, the more the dust in the Galaxy builds up and makes it hard to see with visible light. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 903 Figure 25.11. Orion Spur. The Sun is located in the Orion Spur, which is a minor spiral arm located between two other arms. In this diagram, the white lines point to some other noteworthy objects that share this feature of the Milky Way Galaxy with the Sun. (credit: modification of work by NASA/JPL-Caltech) Formation of Spiral Structure At the Sun\u2019s distance from its center, the Galaxy does not rotate like a solid wheel or a CD inside your player. Instead, the way individual objects turn around the center of the Galaxy is more like the solar system. Stars, as well as the clouds of gas and dust, obey Kepler\u2019s third law. Objects farther from the center take longer to complete an orbit around the Galaxy than do those closer to the center. In other words, stars (and interstellar matter) in larger orbits in the Galaxy trail behind those in smaller ones. This effect is called differential galactic rotation. Differential rotation would appear to explain why so much of the material in the disk of the Milky Way is concentrated into elongated features that resemble spiral arms. No matter what the original distribution of the material might be, the differential rotation of the Galaxy can stretch it out into spiral features. Figure 25.12 shows the development of spiral arms from two irregular blobs of interstellar matter. Notice that as the portions of the blobs closest to the galactic center move faster, those farther out trail behind. 904 Chapter 25 The Milky Way Galaxy Figure 25.12. Simplified Model for the Formation of Spiral Arms. This sketch shows how spiral arms might form from irregular clouds of interstellar material stretched out by the different rotation rates throughout the Galaxy. The regions farthest from the galactic center take longer to complete their orbits and thus lag behind the inner regions. If this were the only mechanism for creating spiral arms, then over time the spiral arms would completely wind up and disappear. Since many galaxies have spiral arms, they must be long-lived, and there must be other processes at work to maintain them. But this picture of spiral arms presents astronomers with an immediate problem. If that\u2019s all there were to the story, differential rotation\u2014over the roughly 13-billion-year history of the Galaxy\u2014would have wound the Galaxy\u2019s arms tighter and tighter until all semblance of spiral structure had disappeared. But did the Milky Way actually have spiral arms when it formed 13 billion years ago? And do spiral arms, once formed, last for that long a time? With the advent of the Hubble Space Telescope, it has become possible to observe the structure of very distant galaxies and to see what they were like shortly after they began to form more than 13 billion years ago. What the observations show is that galaxies", " in their infancy had bright, clumpy star-forming regions, but no regular spiral structure. Over the next few billion years, the galaxies began to \u201csettle down.\u201d The galaxies that were to become spirals lost their massive clumps and developed a central bulge. The turbulence in these galaxies decreased, rotation began to dominate the motions of the stars and gas, and stars began to form in a much quieter disk. Smaller star-forming clumps began to form fuzzy, not-very-distinct spiral arms. Bright, well-defined spiral arms began to appear only when the galaxies were about 3.6 billion years old. Initially, there were two well-defined arms. Multi-armed structures in galaxies like we see in the Milky Way appeared only when the universe was about 8 billion years old. We will discuss the history of galaxies in more detail in The Evolution and Distribution of Galaxies. But, even from our brief discussion, you can get the sense that the spiral structures we now observe in mature galaxies have come along later in the full story of how things develop in the universe. Scientists have used supercomputer calculations to model the formation and evolution of the arms. These calculations follow the motions of up to 100 million \u201cstar particles\u201d to see whether gravitational forces can cause them to form spiral structure. What these calculations show is that giant molecular clouds (which we discussed in Between the Stars: Gas and Dust in Space) have enough gravitational influence over their surroundings to initiate the formation of structures that look like spiral arms. These arms then become self-perpetuating and can survive for at least several billion years. The arms may change their brightness over time as star formation comes and goes, but they are not temporary features. The concentration of matter in the arms exerts sufficient gravitational force to keep the arms together over long periods of time. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 905 25.3 THE MASS OF THE GALAXY Learning Objectives By the end of this section, you will be able to: Describe historical attempts to determine the mass of the Galaxy Interpret the observed rotation curve of our Galaxy to suggest the presence of dark matter whose distribution extends well beyond the Sun\u2019s orbit When we described the sections of the Milky Way, we said that the stars are now known to be surrounded by a much larger halo of invisible matter. Let\u2019s see how this surprising discovery was made. Kepler Helps Weigh the Galaxy The Sun, like all the other stars in the Galaxy, orbits the center of the Milky Way. Our star\u2019s orbit is nearly circular and lies in the Galaxy\u2019s disk. The speed of the Sun in its orbit is about 200 kilometers per second, which means it takes us approximately 225 million years to go once around the center of the Galaxy. We call the period of the Sun\u2019s revolution the galactic year. It is a long time compared to human time scales; during the entire lifetime of Earth, only about 20 galactic years have passed. This means that we have gone only a tiny fraction of the way around the Galaxy in all the time that humans have gazed into the sky. We can use the information about the Sun\u2019s orbit to estimate the mass of the Galaxy (just as we could \u201cweigh\u201d the Sun by monitoring the orbit of a planet around it\u2014see Orbits and Gravity). Let\u2019s assume that the Sun\u2019s orbit is circular and that the Galaxy is roughly spherical, (we know the Galaxy is shaped more like a disk, but to simplify the calculation we will make this assumption, which illustrates the basic approach). Long ago, Newton showed that if you have matter distributed in the shape of a sphere, then it is simple to calculate the pull of gravity on some object just outside that sphere: you can assume that gravity acts as if all the matter were concentrated at a point in the center of the sphere. For our calculation, then, we can assume that all the mass that lies inward of the Sun\u2019s position is concentrated at the center of the Galaxy, and that the Sun orbits that point from a distance of about 26,000 light-years. This is the sort of situation to which Kepler\u2019s third law (as modified by Newton) can be directly applied. Plugging numbers into Kepler\u2019s formula, we can calculate the sum of the masses of the Galaxy and the Sun. However, the mass of the Sun is completely trivial compared to the mass of the Galaxy. Thus, for all practical purposes, the result (about 100 billion times the mass of the Sun) is the mass of the Milky Way. More sophisticated calculations based on more sophisticated models give a similar result. Our estimate tells us how much mass is contained in the volume inside the Sun\u2019s orbit. This is a good estimate for the total mass of the Galaxy only if hardly any mass", " lies outside the Sun\u2019s orbit. For many years astronomers thought this assumption was reasonable. The number of bright stars and the amount of luminous matter (meaning any material from which we can detect electromagnetic radiation) both drop off dramatically at distances of more than about 30,000 light-years from the galactic center. Little did we suspect how wrong our assumption was. A Galaxy of Mostly Invisible Matter In science, what seems to be a reasonable assumption can later turn out to be wrong (which is why we continue to do observations and experiments every chance we get). There is a lot more to the Milky Way than meets the eye (or our instruments). While there is relatively little luminous matter beyond 30,000 light-years, we now know that a lot of invisible matter exists at great distances from the galactic center. 906 Chapter 25 The Milky Way Galaxy We can understand how astronomers detected this invisible matter by remembering that according to Kepler\u2019s third law, objects orbiting at large distances from a massive object will move more slowly than objects that are closer to that central mass. In the case of the solar system, for example, the outer planets move more slowly in their orbits than the planets close to the Sun. There are a few objects, including globular clusters and some nearby small satellite galaxies, that lie well outside the luminous boundary of the Milky Way. If most of the mass of our Galaxy were concentrated within the luminous region, then these very distant objects should travel around their galactic orbits at lower speeds than, for example, the Sun does. It turns out, however, that the few objects seen at large distances from the luminous boundary of the Milky Way Galaxy are not moving more slowly than the Sun. There are some globular clusters and RR Lyrae stars between 30,000 and 150,000 light-years from the center of the Galaxy, and their orbital velocities are even greater than the Sun\u2019s (Figure 25.13). Figure 25.13. Rotation Curve of the Galaxy. The orbital speed of carbon monoxide (CO) and hydrogen (H) gas at different distances from the center of the Milky Way Galaxy is shown in red. The blue curve shows what the rotation curve would look like if all the matter in the Galaxy were located inside a radius of 30,000 light-years. Instead of going down, the speed of gas clouds farther out remains high, indicating a great deal of mass beyond the Sun\u2019s orbit. The horizontal axis shows the distance from the galactic center in kiloparsecs (where a kiloparsec equals 3,260 light-years). What do these higher speeds mean? Kepler\u2019s third law tells us how fast objects must orbit a source of gravity if they are neither to fall in (because they move too slowly) nor to escape (because they move too fast). If the Galaxy had only the mass calculated by Kepler, then the high-speed outer objects should long ago have escaped the grip of the Milky Way. The fact that they have not done so means that our Galaxy must have more gravity than can be supplied by the luminous matter\u2014in fact, a lot more gravity. The high speed of these outer objects tells us that the source of this extra gravity must extend outward from the center far beyond the Sun\u2019s orbit. If the gravity were supplied by stars or by something else that gives off radiation, we should have spotted this additional outer material long ago. We are therefore forced to the reluctant conclusion that this matter is invisible and has, except for its gravitational pull, gone entirely undetected. Studies of the motions of the most remote globular clusters and the small galaxies that orbit our own show that the total mass of the Galaxy is at least 2 \u00d7 1012 MSun, which is about twenty times greater than the amount of luminous matter. Moreover, the dark matter (as astronomers have come to call the invisible material) extends to a distance of at least 200,000 light-years from the center of the Galaxy. Observations indicate that this dark matter halo is almost but not quite spherical. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 907 The obvious question is: what is the dark matter made of? Let\u2019s look at a list of \u201csuspects\u201d taken from our study of astronomy so far. Since this matter is invisible, it clearly cannot be in the form of ordinary stars. And it cannot be gas in any form (remember that there has to be a lot of it). If it were neutral hydrogen gas, its 21-cm wavelength spectral-line emission would have been detected as radio waves. If it were ionized hydrogen, it should be hot enough to emit visible radiation. If a lot of hydrogen atoms out there had combined into hydrogen molecules, these should produce dark features in the ultraviolet spectra of objects lying beyond the Galaxy, but such features have not been seen.", " Nor can the dark matter consist of interstellar dust, since in the required quantities, the dust would significantly obscure the light from distant galaxies. What are our other possibilities? The dark matter cannot be a huge number of black holes (of stellar mass) or old neutron stars, since interstellar matter falling onto such objects would produce more X-rays than are observed. Also, recall that the formation of black holes and neutron stars is preceded by a substantial amount of mass loss, which scatters heavy elements into space to be incorporated into subsequent generations of stars. If the dark matter consisted of an enormous number of any of those objects, they would have blown off and recycled a lot of heavier elements over the history of the Galaxy. In that case, the young stars we observe in our Galaxy today would contain much greater abundances of heavy elements than they actually do. Brown dwarfs and lone Jupiter-like planets have also been ruled out. First of all, there would have to be an awful lot of them to make up so much dark matter. But we have a more direct test of whether so many low-mass objects could actually be lurking out there. As we learned in Black Holes and Curved Spacetime, the general theory of relativity predicts that the path traveled by light is changed when it passes near a concentration of mass. It turns out that when the two objects appear close enough together in the sky, the mass closer to us can bend the light from farther away. With just the right alignment, the image of the more distant object also becomes significantly brighter. By looking for the temporary brightening that occurs when a dark matter object in our own Galaxy moves across the path traveled by light from stars in the Magellanic Clouds, astronomers have now shown that the dark matter cannot be made up of a lot of small objects with masses between one- millionth and one-tenth the mass of the Sun. What\u2019s left? One possibility is that the dark matter is composed of exotic subatomic particles of a type not yet detected on Earth. Very sophisticated (and difficult) experiments are now under way to look for such particles. Stay tuned to see whether anything like that turns up. We should add that the problem of dark matter is by no means confined to the Milky Way. Observations show that dark matter must also be present in other galaxies (whose outer regions also orbit too fast \u201cfor their own good\u201d\u2014they also have flat rotation curves). As we will see, dark matter even exists in great clusters of galaxies whose members are now known to move around under the influence of far more gravity than can be accounted for by luminous matter alone. Stop a moment and consider how astounding the conclusion we have reached really is. Perhaps as much as 95% of the mass in our Galaxy (and many other galaxies) is not only invisible, but we do not even know what it is made of. The stars and raw material we can observe may be merely the tip of the cosmic iceberg; underlying it all may be other matter, perhaps familiar, perhaps startlingly new. Understanding the nature of this dark matter is one of the great challenges of astronomy today; you will learn more about this in A Universe of (Mostly) Dark Matter and Dark Energy. 25.4 THE CENTER OF THE GALAXY Learning Objectives By the end of this section, you will be able to: 908 Chapter 25 The Milky Way Galaxy Describe the radio and X-ray observations that indicate energetic phenomena are occurring at the galactic center Explain what has been revealed by high-resolution near-infrared imaging of the galactic center Discuss how these near-infrared images, when combined with Kepler\u2019s third law of motion, can be used to derive the mass of the central gravitating object At the beginning of this chapter, we hinted that the core of our Galaxy contains a large concentration of mass. In fact, we now have evidence that the very center contains a black hole with a mass equivalent to 4.6 million Suns and that all this mass fits within a sphere that has less than the diameter of Mercury\u2019s orbit. Such monster black holes are called supermassive black holes by astronomers, to indicate that the mass they contain is far greater than that of the typical black hole created by the death of a single star. It is amazing that we have very convincing evidence that this black hole really does exist. After all, recall from the chapter on Black Holes and Curved Spacetime that we cannot see a black hole directly because by definition it radiates no energy. And we cannot even see into the center of the Galaxy in visible light because of absorption by the interstellar dust that lies between us and the galactic center. Light from the central region of the Galaxy is dimmed by a factor of a trillion (1012) by all this dust. Fortunately, we are not so blind at other wavelengths. Infrared and radio radiation, which have long wavelengths compared to the sizes of the interstellar dust grains, flow unimpeded past the dust particles and so reach our telescopes with hardly", " any dimming. In fact, the very bright radio source in the nucleus of the Galaxy, now known as Sagittarius A* (pronounced \u201cSagittarius A-star\u201d and abbreviated Sgr A*), was the first cosmic radio source astronomers discovered. A Journey toward the Center Let\u2019s take a voyage to the mysterious heart of our Galaxy and see what\u2019s there. Figure 25.14 is a radio image of a region about 1500 light-years across, centered on Sagittarius A, a bright radio source that contains the smaller Sagittarius A*. Much of the radio emission comes from hot gas heated either by clusters of hot stars (the stars themselves do not produce radio emission and can\u2019t be seen in the image) or by supernova blast waves. Most of the hollow circles visible on the radio image are supernova remnants. The other main source of radio emission is from electrons moving at high speed in regions with strong magnetic fields. The bright thin arcs and \u201cthreads\u201d on the figure show us where this type of emission is produced. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 909 Figure 25.14. Radio Image of Galactic Center Region. This radio map of the center of the Galaxy (at a wavelength of 90 centimeters) was constructed from data obtained with the Very Large Array (VLA) of radio telescopes in Socorro, New Mexico. Brighter regions are more intense in radio waves. The galactic center is inside the region labeled Sagittarius A. Sagittarius B1 and B2 are regions of active star formation. Many filaments or threadlike features are seen, as well as a number of shells (labeled SNR), which are supernova remnants. The scale bar at the bottom left is about 240 light-years long. Notice that radio astronomers also give fanciful animal names to some of the structures, much as visible-light nebulae are sometimes given the names of animals they resemble. (credit: modification of work by N. E. Kassim, D. S. Briggs, T. J. W. Lazio, T. N. LaRosa, and J. Imamura (NRL/RSD)) Now let\u2019s focus in on the central region using a more energetic form of electromagnetic radiation. Figure 25.15 shows the X-ray emission from a smaller region 400 light-years wide and 900 light-years across centered in Sagittarius A*. Seen in this picture are hundreds of hot white dwarfs, neutron stars, and stellar black holes with accretion disks glowing with X-rays. The diffuse haze in the picture is emission from gas that lies among the stars and is at a temperature of 10 million K. 910 Chapter 25 The Milky Way Galaxy Figure 25.15. Galactic Center in X-Rays. This artificial-color mosaic of 30 images taken with the Chandra X-ray satellite shows a region 400 \u00d7 900 light-years in extent and centered on Sagittarius A*, the bright white source in the center of the picture. The X-ray-emitting point sources are white dwarfs, neutron stars, and stellar black holes. The diffuse \u201chaze\u201d is emission from gas at a temperature of 10 million K. This hot gas is flowing away from the center out into the rest of the Galaxy. The colors indicate X-ray energy bands: red (low energy), green (medium energy), and blue (high energy). (credit: modification of work by NASA/CXC/ UMass/D. Wang et al.) As we approach the center of the Galaxy, we find the supermassive black hole Sagittarius A*. There are also thousands of stars within a parsec of Sagittarius A*. Most of these are old, reddish main-sequence stars. But there are also about a hundred hot OB stars that must have formed within the last few million years. There is as yet no good explanation for how stars could have formed recently so close to a supermassive black hole. Perhaps they formed in a dense cluster of stars that was originally at a larger distance from the black hole and subsequently migrated closer. There is currently no star formation at the galactic center, but there is lots of dust and molecular gas that is revolving around the black hole, along with some ionized gas streamers that are heated by the hot stars. Figure 25.16 is a radio map that shows these gas streamers. Figure 25.16. Sagittarius A. This image, taken with the Very Large Array of radio telescopes, shows the radio emission from hot, ionized gas in the center of the Milky Way. The lines slanting across the top of the image are gas streamers. Sagittarius A* is the bright spot in the lower right. (credit: modification of work by Farhad Zadeh et al.", " (Northwestern), VLA, NRAO) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 911 Finding the Heart of the Galaxy Just what is Sagittarius A*, which lies right at the center our Galaxy? To establish that there really is a black hole there, we must show that there is a very large amount of mass crammed into a very tiny volume. As we saw in Black Holes and Curved Spacetime, proving that a black hole exists is a challenge because the black hole itself emits no radiation. What astronomers must do is prove that a black hole is the only possible explanation for our observations\u2014that a small region contains far more mass than could be accounted for by a very dense cluster of stars or something else made of ordinary matter. To put some numbers with this discussion, the radius of the event horizon of a galactic black hole with a mass of about 4 million MSun would be only about 17 times the size of the Sun\u2014the equivalent of a single red giant star. The corresponding density within this region of space would be much higher than that of any star cluster or any other ordinary astronomical object. Therefore, we must measure both the diameter of Sagittarius A* and its mass. Both radio and infrared observations are required to give us the necessary evidence. First, let\u2019s look at how the mass can be measured. If we zero in on the inner few light-days of the Galaxy with an infrared telescope equipped with adaptive optics, we see a region crowded with individual stars (Figure 25.17). These stars have now been observed for almost two decades, and astronomers have detected their rapid orbital motions around the very center of the Galaxy. Figure 25.17. Near-Infrared View of the Galactic Center. This image shows the inner 1 arcsecond, or 0.13 light-year, at the center of the Galaxy, as observed with the giant Keck Telescope. Tracks of the orbiting stars measured from 1995 to 2014 have been added to this \u201csnapshot.\u201d The stars are moving around the center very fast, and their tracks are all consistent with a single massive \u201cgravitator\u201d that resides in the very center of this image. (credit: modification of work by Andrea Ghez, UCLA Galactic Center Group, W.M. Keck Observatory Laser Team Check out an animated version (https://openstaxcollege.org/l/30anifiginfgal) of Figure 25.17, showing the motion of the stars over the years. If we combine observations of their periods and the size of their orbits with Kepler\u2019s third law, we can estimate the mass of the object that keeps them in their orbits. One of the stars has been observed for its full orbit of 15.6 years. Its closest approach takes it to a distance of only 124 AU or about 17 light-hours from the black hole. 912 Chapter 25 The Milky Way Galaxy This orbit, when combined with observations of other stars close to the galactic center, indicates that a mass of 4.6 million MSun must be concentrated inside the orbit\u2014that is, within 17 light-hours of the center of the Galaxy. Even tighter limits on the size of the concentration of mass at the center of the Galaxy come from radio astronomy, which provided the first clue that a black hole might lie at the center of the Galaxy. As matter spirals inward toward the event horizon of a black hole, it is heated in a whirling accretion disk and produces radio radiation. (Such accretion disks were explained in Black Holes and Curved Spacetime.) Measurements of the size of the accretion disk with the Very Long Baseline Array, which provides very high spatial resolution, show that the diameter of the radio source Sagittarius A* is no larger than about 0.3 AU, or about the size of Mercury\u2019s orbit. (In light units, that\u2019s only 2.5 light-minutes!) The observations thus show that 4.6 million solar masses are crammed into a volume that has a diameter that is no larger than the orbit of Mercury. If this were anything other than a supermassive black hole\u2014low-mass stars that emit very little light or neutron stars or a very large number of small black holes\u2014 calculations show that these objects would be so densely packed that they would collapse to a single black hole within a hundred thousand years. That is a very short time compared with the age of the Galaxy, which probably began forming more than 13 billion years ago. Since it seems very unlikely that that we would have caught such a complex cluster of objects just before it collapsed, the evidence for a supermassive black hole at the center of the Galaxy is convincing indeed. Finding the Source Where did our galactic black hole come from? The origin of supermassive black holes in galaxies like ours is currently an active field of research. One possibility is that a large cloud of gas", " near the center of the Milky Way collapsed directly to form a black hole. Since we find large black holes at the centers of most other large galaxies (see Active Galaxies, Quasars, and Supermassive Black Holes)\u2014even ones that are very young\u2014this collapse probably would have taken place when the Milky Way was just beginning to take shape. The initial mass of this black hole might have been only a few tens of solar masses. Another way it could have started is that a massive star might have exploded to leave behind a seed black hole, or a dense cluster of stars might have collapsed into a black hole. Once a black hole exists at the center of a galaxy, it can grow over the next several billion years by devouring nearby stars and gas clouds in the crowded central regions. It can also grow by merging with other black holes. It appears that the monster black hole at the center of our Galaxy is not finished \u201ceating.\u201d At the present time, we observe clouds of gas and dust falling into the galactic center at the rate of about 1 MSun per thousand years. Stars are also on the black hole\u2019s menu. The density of stars near the galactic center is high enough that we would expect a star to pass near the black hole and be swallowed by it every ten thousand years or so. As this happens, some of the energy of infall is released as radiation. As a result, the center of the Galaxy might flare up and even briefly outshine all the stars in the Milky Way. Other objects might also venture too close to the black hole and be pulled in. How great a flare we observe would depend on the mass of the object falling in. In 2013, the Chandra X-ray satellite detected a flare from the center of our Galaxy that was 400 times brighter than the usual output from Sagittarius A*. A year later, a second flare, only half as bright, was also detected. This is much less energy than swallowing a whole star would produce. There are two theories to account for the flares. First, an asteroid might have ventured too close to the black hole and been heated to a very high temperature before being swallowed up. Alternatively, the flares might have involved interactions of the magnetic fields near the galactic center in a process similar to the one described for solar flares (see The Sun: A Garden-Variety Star). Astronomers continue to monitor the galactic center area for flares or other activity. Although the monster in the center of the Galaxy is not close enough to us to represent any danger, we still want to keep our eyes on it. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 913 Andrea Ghez A lover of puzzles, Andrea Ghez has been pursuing one of the greatest mysteries in astronomy: what strange entity lurks within the center of our Milky Way Galaxy? Figure 25.18. Andrea Ghez. Research by Ghez and her team has helped shape our understanding of supermassive black holes. (credit: modification of work by John D. and Catherine T. MacArthur Foundation) As a child living in Chicago during the late 1960s, Andrea Ghez (Figure 25.18) was fascinated by the Apollo Moon landings. But she was also drawn to ballet and to solving all sorts of puzzles. By high school, she had lost the ballet bug in favor of competing in field hockey, playing the flute, and digging deeper into academics. Her undergraduate years at MIT were punctuated by a number of changes in her major\u2014from mathematics to chemistry, mechanical engineering, aerospace engineering, and finally physics\u2014where she felt her options were most open. As a physics major, she became involved in astronomical research under the guidance of one of her instructors. Once she got to do some actual observing at Kitt Peak National Observatory in Arizona, and later at Cerro Tololo Inter-American Observatory in Chile, Ghez had found her calling. Pursuing her graduate studies at Caltech, she stuck with physics but oriented her efforts toward observational astrophysics, an area where Caltech had access to cutting-edge facilities. Though initially attracted to studying the black holes that were suspected of dwelling inside most massive galaxies, Ghez ended up spending most of her graduate study and later postdoctoral research at the University of Arizona studying stars in formation. By taking very high-resolution (detailed) imaging of regions where new stars are born, she discovered that most stars form as members of binary systems. As technologies advanced, she was able to track the orbits danced by these stellar pairings and thereby could ascertain their respective masses. Now an astronomy professor at UCLA, Ghez has since used similar high-resolution imaging techniques to study the orbits of stars in the innermost core of the Milky Way. These orbits take years to delineate, so Ghez and her science team have logged more than 20 years of taking super-resolution infrared images with the giant Keck telescopes", " in Hawaii. Based on the resulting stellar orbits, the UCLA Galactic Center Group has settled (as we saw) on a gravitational solution that requires the presence of a supermassive black hole with a mass equivalent to 4.6 million Suns\u2014all nestled within a space smaller than that 914 Chapter 25 The Milky Way Galaxy occupied by our solar system. Ghez\u2019s achievements have been recognized with one of the \u201cgenius\u201d awards given by the MacArthur Foundation. More recently, her team discovered glowing clouds of warm ionized gas that co-orbit with the stars but may be more vulnerable to the disruptive effects of the central black hole. By monitoring these clouds, the team hopes to better understand the evolution of supermassive black holes and their immediate environs. They also hope to test Einstein\u2019s theory of general relativity by carefully scrutinizing the orbits of stars that careen closest to the intensely gravitating black hole. Besides her pioneering work as an astronomer, Ghez competes as a master swimmer, enjoys family life as a mother of two children, and actively encourages other women to pursue scientific careers. 25.5 STELLAR POPULATIONS IN THE GALAXY Learning Objectives By the end of this section, you will be able to: Distinguish between population I and population II stars according to their locations, motions, heavy- element abundances, and ages Explain why the oldest stars in the Galaxy are poor in elements heavier than hydrogen and helium, while stars like the Sun and even younger stars are typically richer in these heavy elements In the first section of his chapter, we described the thin disk, thick disk, and stellar halo. Look back at Table 25.1 and note some of the patterns. Young stars lie in the thin disk, are rich in metals, and orbit the Galaxy\u2019s center at high speed. The stars in the halo are old, have low abundances of elements heavier than hydrogen and helium, and have highly elliptical orbits randomly oriented in direction (see Figure 25.19). Halo stars can plunge through the disk and central bulge, but they spend most of their time far above or below the plane of the Galaxy. The stars in the thick disk are intermediate between these two extremes. Let\u2019s first see why age and heavier-element abundance are correlated and then see what these correlations tell us about the origin of our Galaxy. Figure 25.19. How Objects Orbit the Galaxy. (a) In this image, you see stars in the thin disk of our Galaxy in nearly circular orbits. (b) In this image, you see the motion of stars in the Galaxy\u2019s halo in randomly oriented and elliptical orbits. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 915 Two Kinds of Stars The discovery that there are two different kinds of stars was first made by Walter Baade during World War II. As a German national, Baade was not allowed to do war research as many other U.S.-based scientists were doing, so he was able to make regular use of the Mount Wilson telescopes in southern California. His observations were aided by the darker skies that resulted from the wartime blackout of Los Angeles. Among the things a large telescope and dark skies enabled Baade to examine carefully were other galaxies\u2014neighbors of our Milky Way Galaxy. We will discuss other galaxies in the next chapter (Galaxies), but for now we will just mention that the nearest Galaxy that resembles our own (with a similar disk and spiral structure) is often called the Andromeda galaxy, after the constellation in which we find it. Baade was impressed by the similarity of the mainly reddish stars in the Andromeda galaxy\u2019s nuclear bulge to those in our Galaxy\u2019s globular clusters and the halo. He also noted the difference in color between all these and the bluer stars found in the spiral arms near the Sun (Figure 25.20). On this basis, he called the bright blue stars in the spiral arms population I and all the stars in the halo and globular clusters population II. Figure 25.20. Andromeda Galaxy (M31). This neighboring spiral looks similar to our own Galaxy in that it is a disk galaxy with a central bulge. Note the bulge of older, yellowish stars in the center, the bluer and younger stars in the outer regions, and the dust in the disk that blocks some of the light from the bulge. (credit: Adam Evans) We now know that the populations differ not only in their locations in the Galaxy, but also in their chemical composition, age, and orbital motions around the center of the Galaxy. Population I stars are found only in the disk and follow nearly circular orbits around the galactic center. Examples are bright supergiant stars, main- sequence stars of high luminosity (spectral classes O and B), which are concentrated in the spiral arms, and members", " of young open star clusters. Interstellar matter and molecular clouds are found in the same places as population I stars. Population II stars show no correlation with the location of the spiral arms. These objects are found throughout the Galaxy. Some are in the disk, but many others follow eccentric elliptical orbits that carry them high above the galactic disk into the halo. Examples include stars surrounded by planetary nebulae and RR Lyrae variable stars. The stars in globular clusters, found almost entirely in the Galaxy\u2019s halo, are also classified as population II. 916 Chapter 25 The Milky Way Galaxy Today, we know much more about stellar evolution than astronomers did in the 1940s, and we can determine the ages of stars. Population I includes stars with a wide range of ages. While some are as old as 10 billion years, others are still forming today. For example, the Sun, which is about 5 billion years old, is a population I star. But so are the massive young stars in the Orion Nebula that have formed in the last few million years. Population II, on the other hand, consists entirely of old stars that formed very early in the history of the Galaxy; typical ages are 11 to 13 billion years. We also now have good determinations of the compositions of stars. These are based on analyses of the stars\u2019 detailed spectra. Nearly all stars appear to be composed mostly of hydrogen and helium, but their abundances of the heavier elements differ. In the Sun and other population I stars, the heavy elements (those heavier than hydrogen and helium) account for 1\u20134% of the total stellar mass. Population II stars in the outer galactic halo and in globular clusters have much lower abundances of the heavy elements\u2014often less than one-hundredth the concentrations found in the Sun and in rare cases even lower. The oldest population II star discovered to date has less than one ten-millionth as much iron as the Sun, for example. As we discussed in earlier chapters, heavy elements are created deep within the interiors of stars. They are added to the Galaxy\u2019s reserves of raw material when stars die, and their material is recycled into new generations of stars. Thus, as time goes on, stars are born with larger and larger supplies of heavy elements. Population II stars formed when the abundance of elements heavier than hydrogen and helium was low. Population I stars formed later, after mass lost by dying members of the first generations of stars had seeded the interstellar medium with elements heavier than hydrogen and helium. Some are still forming now, when further generations have added to the supply of heavier elements available to new stars. The Real World With rare exceptions, we should never trust any theory that divides the world into just two categories. While they can provide a starting point for hypotheses and experiments, they are often oversimplifications that need refinement a research continue. The idea of two populations helped organize our initial thoughts about the Galaxy, but we now know it cannot explain everything we observe. Even the different structures of the Galaxy\u2014disk, halo, central bulge\u2014are not so cleanly separated in terms of their locations, ages, and the heavy element content of the stars within them. The exact definition of the Galaxy\u2019s disk depends on what objects we use to define it, and it has no sharp boundary. The hottest young stars and their associated gas and dust clouds are mostly in a region about 300 light-years thick. Older stars define a thicker disk that is about 3000 light-years thick. Halo stars spend most of their time high above or below the disk but pass through it on their highly elliptical orbits and so are sometimes found relatively near the Sun. The highest density of stars is found in the central bulge, that bar-shaped inner region of the Galaxy. There are a few hot, young stars in the bulge, but most of the bulge stars are more than 10 billion years old. Yet unlike the halo stars of similar age, the abundance of heavy elements in the bulge stars is about the same as in the Sun. Why would that be? Astronomers think that star formation in the crowded nuclear bulge occurred very rapidly just after the Milky Way Galaxy formed. After a few million years, the first generation of massive and short-lived stars then expelled heavy elements in supernova explosions and thereby enriched subsequent generations of stars. Thus, even stars that formed in the bulge more than 10 billion years ago started with a good supply of heavy elements. Exactly the opposite occurred in the Small Magellanic Cloud, a small galaxy near the Milky Way, visible from Earth\u2019s Southern Hemisphere. Even the youngest stars in this galaxy are deficient in heavy elements. We think this is because the little galaxy is not especially crowded, and star formation has occurred quite slowly. As a result there have been, so far, relatively few supernova explosions. Smaller galaxies also have more trouble This OpenStax book is available for free at http://cnx.org/content/col119", "92/1.8 Chapter 25 The Milky Way Galaxy 917 holding onto the gas expelled by supernova explosions in order to recycle it. Low-mass galaxies exert only a modest gravitational force, and the high-speed gas ejected by supernovae can easily escape from them. Which elements a star is endowed with thus depends not only on when the star formed in the history of its galaxy, but also on how many stars in its part of the galaxy had already completed their lives by the time the star is ready to form. 25.6 THE FORMATION OF THE GALAXY Learning Objectives By the end of this section, you will be able to: Describe the roles played by the collapse of a single cloud and mergers with other galaxies in building the Milky Way Galaxy we see today Provide examples of globular clusters and satellite galaxies affected by the Milky Way\u2019s strong gravity. Information about stellar populations holds vital clues to how our Galaxy was built up over time. The flattened disk shape of the Galaxy suggests that it formed through a process similar to the one that leads to the formation of a protostar (see The Birth of Stars and the Discovery of Planets outside the Solar System). Building on this idea, astronomers first developed models that assumed the Galaxy formed from a single rotating cloud. But, as we shall see, this turns out to be only part of the story. The Protogalactic Cloud and the Monolithic Collapse Model Because the oldest stars\u2014those in the halo and in globular clusters\u2014are distributed in a sphere centered on the nucleus of the Galaxy, it makes sense to assume that the protogalactic cloud that gave birth to our Galaxy was roughly spherical. The oldest stars in the halo have ages of 12 to 13 billion years, so we estimate that the formation of the Galaxy began about that long ago. (See the chapter on The Big Bang for other evidence that galaxies in general began forming a little more than 13 billion years ago.) Then, just as in the case of star formation, the protogalactic cloud collapsed and formed a thin rotating disk. Stars born before the cloud collapsed did not participate in the collapse, but have continued to orbit in the halo to the present day (Figure 25.21). 918 Chapter 25 The Milky Way Galaxy Figure 25.21. Monolithic Collapse Model for the Formation of the Galaxy. According to this model, the Milky Way Galaxy initially formed from a rotating cloud of gas that collapsed due to gravity. Halo stars and globular clusters either formed prior to the collapse or were formed elsewhere. Stars in the disk formed later, when the gas from which they were made was already \u201ccontaminated\u201d with heavy elements produced in earlier generations of stars. Gravitational forces caused the gas in the thin disk to fragment into clouds or clumps with masses like those of star clusters. These individual clouds then fragmented further to form stars. Since the oldest stars in the disk are nearly as old as the youngest stars in the halo, the collapse must have been rapid (astronomically speaking), requiring perhaps no more than a few hundred million years. Collision Victims and the Multiple Merger Model In past decades, astronomers have learned that the evolution of the Galaxy has not been quite as peaceful as this monolithic collapse model suggests. In 1994, astronomers discovered a small new galaxy in the direction of the constellation of Sagittarius. The Sagittarius dwarf galaxy is currently about 70,000 light-years away from Earth and 50,000 light-years from the center of the Galaxy. It is the closest galaxy known (Figure 25.22). It is very elongated, and its shape indicates that it is being torn apart by our Galaxy\u2019s gravitational tides\u2014just as Comet Shoemaker-Levy 9 was torn apart when it passed too close to Jupiter in 1992. The Sagittarius galaxy is much smaller than the Milky Way, with only about 150,000 stars, all of which seem destined to end up in the bulge and halo of our own Galaxy. But don\u2019t sound the funeral bells for the little galaxy quite yet; the ingestion of the Sagittarius dwarf will take another 100 million years or so, and the stars themselves will survive. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 919 Figure 25.22. Sagittarius Dwarf. In 1994, British astronomers discovered a galaxy in the constellation of Sagittarius, located only about 50,000 light-years from the center of the Milky Way and falling into our Galaxy. This image covers a region approximately 70\u00b0 \u00d7 50\u00b0 and combines a black-and-white view of the disk of our Galaxy with a red contour map showing the brightness of the dwarf galaxy. The dwarf galaxy lies on the other side of the galactic center from us. The white stars in the red region mark the locations of several globular clusters contained within the Sagittarius", " dwarf galaxy. The cross marks the galactic center. The horizontal line corresponds to the galactic plane. The blue outline on either side of the galactic plane corresponds to the infrared image in Figure 25.7. The boxes mark regions where detailed studies of individual stars led to the discovery of this galaxy. (credit: modification of work by R. Ibata (UBC), R. Wyse (JHU), R. Sword (IoA)) Since that discovery, evidence has been found for many more close encounters between our Galaxy and other neighbor galaxies. When a small galaxy ventures too close, the force of gravity exerted by our Galaxy tugs harder on the near side than on the far side. The net effect is that the stars that originally belonged to the small galaxy are spread out into a long stream that orbits through the halo of the Milky Way (Figure 25.23). 920 Chapter 25 The Milky Way Galaxy Figure 25.23. Streams in the Galactic Halo. When a small galaxy is swallowed by the Milky Way, its member stars are stripped away and form streams of stars in the galactic halo. This image is based on calculations of what some of these tidal streams might look like if the Milky Way swallowed 50 dwarf galaxies over the past 10 billion years. (credit: modification of work by NASA/JPL-Caltech/R. Hurt (SSC/Caltech)) Such a tidal stream can maintain its identity for billions of years. To date, astronomers have now identified streams originating from 12 small galaxies that ventured too close to the much larger Milky Way. Six more streams are associated with globular clusters. It has been suggested that large globular clusters, like Omega Centauri, are actually dense nuclei of cannibalized dwarf galaxies. The globular cluster M54 is now thought to be the nucleus of the Sagittarius dwarf we discussed earlier, which is currently merging with the Milky Way (Figure 25.24). The stars in the outer regions of such galaxies are stripped off by the gravitational pull of the Milky Way, but the central dense regions may survive. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 921 Figure 25.24. Globular Cluster M54. This beautiful Hubble Space Telescope image shows the globular cluster that is now believed to be the nucleus of the Sagittarius Dwarf Galaxy. (credit: ESA/Hubble & NASA) Calculations indicate that the Galaxy\u2019s thick disk may be a product of one or more such collisions with other galaxies. Accretion of a satellite galaxy would stir up the orbits of the stars and gas clouds originally in the thin disk and cause them to move higher above and below the mid-plane of the Galaxy. Meanwhile, the Galaxy\u2019s stars would add to the fluffed-up mix. If such a collision happened about 10 billion years ago, then any gas in the two galaxies that had not yet formed into stars would have had plenty of time to settle back down into the thin disk. The gas could then have begun forming subsequent generations of population I stars. This timing is also consistent with the typical ages of stars in the thick disk. The Milky Way has more collisions in store. An example is the Canis Major dwarf galaxy, which has a mass of about 1% of the mass of the Milky Way. Already long tidal tails have been stripped from this galaxy, which have wrapped themselves around the Milky Way three times. Several of the globular clusters found in the Milky Way may also have come from the Canis Major dwarf, which is expected to merge gradually with the Milky Way over about the next billion years. In about 3 billion years, the Milky Way itself will be swallowed up, since it and the Andromeda galaxy are on a collision course. Our computer models show that after a complex interaction, the two will merge to form a larger, more rounded galaxy (Figure 25.25). 922 Chapter 25 The Milky Way Galaxy Figure 25.25. Collision of the Milky Way with Andromeda. In about 3 billion years, the Milky Way Galaxy and Andromeda Galaxy will begin a long process of colliding, separating, and then coming back together to form an elliptical galaxy. The whole interaction will take 3 to 4 billion years. These images show the following sequence: (1) In 3.75 billion years, Andromeda has approached the Milky Way. (2) New star formation fills the sky 3.85 billion years from now. (3) Star formation continues at 3.9 billion years. (4) The galaxy shapes change as they interact, with Andromeda being stretched and our Galaxy becoming warped, about 4 billion years from now. (5) In 5.1 billion years, the cores of the two galaxies are bright lobes. (6) In 7 billion years, the merged galaxies form a huge elliptical galaxy whose brightness fills the night sky. This artist\u2019s illustrations show events from a vantage point 25,000 light-years from the center of the Milky", " Way. However, we should mention that the Sun may not be at that distance throughout the sequence of events, as the collision readjusts the orbits of many stars within each galaxy. (credit: NASA; ESA; Z. Levay, R. van der Marel, STScl; T. Hallas, and A. Mellinger) We are thus coming to realize that \u201cenvironmental influences\u201d (and not just a galaxy\u2019s original characteristics) play an important role in determining the properties and development of our Galaxy. In future chapters we will see that collisions and mergers are a major factor in the evolution of many other galaxies as well. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 923 CHAPTER 25 REVIEW KEY TERMS central bulge (or nuclear bulge) the central (round) part of the Milky Way or a similar galaxy dark matter nonluminous mass, whose presence can be inferred only because of its gravitational influence on luminous matter; the composition of the dark matter is not known dark matter halo the mass in the Milky Way that extends well beyond the boundary of the luminous stars to a distance of at least 200,000 light-years from the center of the Galaxy; although we deduce its existence from its gravity, the composition of this matter remains a mystery differential galactic rotation the idea that different parts of the Galaxy turn at different rates, since the parts of the Galaxy follow Kepler\u2019s third law: more distant objects take longer to complete one full orbit around the center of the Galaxy halo the outermost extent of our Galaxy (or another galaxy), containing a sparse distribution of stars and globular clusters in a more or less spherical distribution Milky Way Galaxy the band of light encircling the sky, which is due to the many stars and diffuse nebulae lying near the plane of the Milky Way Galaxy population I star a star containing heavy elements; typically young and found in the disk population II star a star with very low abundance of heavy elements; found throughout the Galaxy spiral arm a spiral-shaped region, characterized by relatively dense interstellar material and young stars, that is observed in the disks of spiral galaxies supermassive black hole the object in the center of most large galaxies that is so massive and compact that light cannot escape from it; the Milky Way\u2019s supermassive black hole contains 4.6 millions of Suns\u2019 worth of mass SUMMARY 25.1 The Architecture of the Galaxy The Milky Way Galaxy consists of a thin disk containing dust, gas, and young and old stars; a spherical halo containing populations of very old stars, including RR Lyrae variable stars and globular star clusters; a thick, more diffuse disk with stars that have properties intermediate between those in the thin disk and the halo; a peanut-shaped nuclear bulge of mostly old stars around the center; and a supermassive black hole at the very center. The Sun is located roughly halfway out of the Milky Way, about 26,000 light-years from the center. 25.2 Spiral Structure The gaseous distribution in the Galaxy\u2019s disk has two main spiral arms that emerge from the ends of the central bar, along with several fainter arms and short spurs; the Sun is located in one of those spurs. Measurements show that the Galaxy does not rotate as a solid body, but instead its stars and gas follow differential rotation, such that the material closer to the galactic center completes its orbit more quickly. Observations show that galaxies like the Milky Way take several billion years after they began to form to develop spiral structure. 924 Chapter 25 The Milky Way Galaxy 25.3 The Mass of the Galaxy The Sun revolves completely around the galactic center in about 225 million years (a galactic year). The mass of the Galaxy can be determined by measuring the orbital velocities of stars and interstellar matter. The total mass of the Galaxy is about 2 \u00d7 1012 MSun. As much as 95% of this mass consists of dark matter that emits no electromagnetic radiation and can be detected only because of the gravitational force it exerts on visible stars and interstellar matter. This dark matter is located mostly in the Galaxy\u2019s halo; its nature is not well understood at present. 25.4 The Center of the Galaxy A supermassive black hole is located at the center of the Galaxy. Measurements of the velocities of stars located within a few light-days of the center show that the mass inside their orbits around the center is about 4.6 million MSun. Radio observations show that this mass is concentrated in a volume with a diameter similar to that of Mercury\u2019s orbit. The density of this matter concentration exceeds that of the densest known star clusters by a factor of nearly a million. The only known object with such a high density and total mass is a black hole. 25.5 Stellar Populations in the Galaxy We can roughly", " divide the stars in the Galaxy into two categories. Old stars with few heavy elements are referred to as population II stars and are found in the halo and in globular clusters. Population I stars contain more heavy elements than globular cluster and halo stars, are typically younger and found in the disk, and are especially concentrated in the spiral arms. The Sun is a member of population I. Population I stars formed after previous generations of stars had produced heavy elements and ejected them into the interstellar medium. The bulge stars, most of which are more than 10 billion years old, have unusually high amounts of heavy elements, presumably because there were many massive first-generation stars in this dense region, and these quickly seeded the next generations of stars with heavier elements. 25.6 The Formation of the Galaxy The Galaxy began forming a little more than 13 billion years ago. Models suggest that the stars in the halo and globular clusters formed first, while the Galaxy was spherical. The gas, somewhat enriched in heavy elements by the first generation of stars, then collapsed from a spherical distribution to a rotating disk-shaped distribution. Stars are still forming today from the gas and dust that remain in the disk. Star formation occurs most rapidly in the spiral arms, where the density of interstellar matter is highest. The Galaxy captured (and still is capturing) additional stars and globular clusters from small galaxies that ventured too close to the Milky Way. In 3 to 4 billion years, the Galaxy will begin to collide with the Andromeda galaxy, and after about 7 billion years, the two galaxies will merge to form a giant elliptical galaxy. FOR FURTHER EXPLORATION Articles Blitz, L. \u201cThe Dark Side of the Milky Way.\u201d Scientific American (October 2011): 36\u201343. How we find dark matter and what it tells us about our Galaxy, its warped disk, and its satellite galaxies. Dvorak, J. \u201cJourney to the Heart of the Milky Way.\u201d Astronomy (February 2008): 28. Measuring nearby stars to determine the properties of the black hole at the center. Gallagher, J., Wyse, R., & Benjamin, R. \u201cThe New Milky Way.\u201d Astronomy (September 2011): 26. Highlights all aspects of the Milky Way based on recent observations. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 925 Goldstein, A. \u201cFinding our Place in the Milky Way.\u201d Astronomy (August 2015): 50. On the history of observations that pinpointed the Sun\u2019s location in the Galaxy. Haggard, D., & Bower, G. \u201cIn the Heart of the Milky Way.\u201d Sky & Telescope (February 2016): 16. On observations of the Galaxy\u2019s nucleus and the supermassive black hole and magnetar there. Ibata, R., & Gibson, B. \u201cThe Ghosts of Galaxies Past.\u201d Scientific American (April 2007): 40. About star streams in the Galaxy that are evidence of past mergers and collisions. Irion, R. \u201cA Crushing End for Our Galaxy.\u201d Science (January 7, 2000): 62. On the role of mergers in the evolution of the Milky Way. Irion, R. \u201cHoming in on Black Holes.\u201d Smithsonian (April 2008). On how astronomers probe the large black hole at the center of the Milky Way Galaxy. Kruesi, L. \u201cHow We Mapped the Milky Way.\u201d Astronomy (October 2009): 28. Kruesi, L. \u201cWhat Lurks in the Monstrous Heart of the Milky Way?\u201d Astronomy (October 2015): 30. On the center of the Galaxy and the black hole there. Laughlin, G., & Adams, F. \u201cCelebrating the Galactic Millennium.\u201d Astronomy (November 2001): 39. The long-term future of the Milky Way in the next 90 billion years. Loeb, A., & Cox, T.J. \u201cOur Galaxy\u2019s Date with Destruction.\u201d Astronomy (June 2008): 28. Describes the upcoming merger of Milky Way and Andromeda. Szpir, M. \u201cPassing the Bar Exam.\u201d Astronomy (March 1999): 46. On evidence that our Galaxy is a barred spiral. Tanner, A. \u201cA Trip to the Galactic Center.\u201d Sky & Telescope (April 2003): 44. Nice introduction, with observations pointing to the presence of a black hole. Trimble, V., & Parker, S. \u201cMeet the Milky Way.\u201d Sky & Telescope (January 1995): 26. Overview of our Galaxy. Wakker, B., & Richter, P. \u201cOur Growing, Breathing Galaxy.\u201d Scientific American (January 2004): 38. Evidence that our Galaxy is still being built up by the addition", " of gas and smaller neighbors. Waller, W. \u201cRedesigning the Milky Way.\u201d Sky & Telescope (September 2004): 50. On recent multi-wavelength surveys of the Galaxy. Whitt, K. \u201cThe Milky Way from the Inside.\u201d Astronomy (November 2001): 58. Fantastic panorama image of the Galaxy, with finder charts and explanations. Websites International Dark Sky Sanctuaries: http://darksky.org/idsp/sanctuaries/. A listing of dark-sky sanctuaries, parks, and reserves. Multiwavelength Milky Way: http://mwmw.gsfc.nasa.gov/mmw_sci.html. This NASA site shows the plane of our Galaxy in a variety of wavelength bands, and includes background material and other resources. Shapley-Curtis Debate in 1920: http://apod.nasa.gov/diamond_jubilee/debate_1920.html. In 1920, astronomers Harlow Shapley and Heber Curtis engaged in a historic debate about how large our Galaxy was and whether other galaxies existed. Here you can find historical and educational material about the debate. UCLA Galactic Center Group: http://www.galacticcenter.astro.ucla.edu/. Learn more about the work of Andrea Ghez and colleagues on the central region of the Milky Way Galaxy. 926 Videos Chapter 25 The Milky Way Galaxy Crash of the Titans: http://www.spacetelescope.org/videos/hubblecast55a/. This Hubblecast from 2012 features Jay Anderson and Roeland van der Marel explaining how Andromeda will collide with the Milky Way in the distant future (5:07). Diner at the Center of the Galaxy: https://www.youtube.com/watch?v=UP7ig8Gxftw. A short discussion from NASA ScienceCast of NuSTAR observations of flares from our Galaxy\u2019s central black hole (3:23). Hunt for a Supermassive Black Hole: https://www.ted.com/talks/ andrea_ghez_the_hunt_for_a_supermassive_black_hole. 2009 TED talk by Andrea Ghez on searching for supermassive black holes, particularly the one at the center of the Milky Way (16:19). Journey to the Galactic Center: https://www.youtube.com/watch?v=36xZsgZ0oSo. A brief silent trip into the cluster of stars near the galactic center showing their motions around the center (3:00). COLLABORATIVE GROUP ACTIVITIES A. You are captured by space aliens, who take you inside a complex cloud of interstellar gas, dust, and a few newly formed stars. To escape, you need to make a map of the cloud. Luckily, the aliens have a complete astronomical observatory with equipment for measuring all the bands of the electromagnetic spectrum. Using what you have learned in this chapter, have your group discuss what kinds of maps you would make of the cloud to plot your most effective escape route. B. The diagram that Herschel made of the Milky Way has a very irregular outer boundary (see Figure 25.3). Can your group think of a reason for this? How did Herschel construct his map? C. Suppose that for your final exam in this course, your group is assigned telescope time to observe a star selected for you by your professor. The professor tells you the position of the star in the sky (its right ascension and declination) but nothing else. You can make any observations you wish. How would you go about determining whether the star is a member of population I or population II? D. The existence of dark matter comes as a great surprise, and its nature remains a mystery today. Someday astronomers will know a lot more about it (you can learn more about current findings in The Evolution and Distribution of Galaxies). Can your group make a list of earlier astronomical observations that began as a surprise and mystery, but wound up (with more observations) as well-understood parts of introductory textbooks? E. Physicist Gregory Benford has written a series of science fiction novels that take place near the center of the Milky Way Galaxy in the far future. Suppose your group were writing such a story. Make a list of ways that the environment near the galactic center differs from the environment in the \u201cgalactic suburbs,\u201d where the Sun is located. Would life as we know it have an easier or harder time surviving on planets that orbit stars near the center (and why)? F. These days, in most urban areas, city lights completely swamp the faint light of the Milky Way in our skies. Have each member of your group survey 5 to 10 friends or relatives (you could spread out on campus to investigate or use social media or the phone), explaining what the Milky Way is and then asking if they have seen it. Also ask their age", ". Report back to your group and discuss your reactions to the survey. Is there any This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 927 relationship between a person\u2019s age and whether they have seen the Milky Way? How important is it that many kids growing up on Earth today never (or rarely) get to see our home Galaxy in the sky? EXERCISES Review Questions 1. Explain why we see the Milky Way as a faint band of light stretching across the sky. 2. Explain where in a spiral galaxy you would expect to find globular clusters, molecular clouds, and atomic hydrogen. 3. Describe several characteristics that distinguish population I stars from population II stars. 4. Briefly describe the main parts of our Galaxy. 5. Describe the evidence indicating that a black hole may be at the center of our Galaxy. 6. Explain why the abundances of heavy elements in stars correlate with their positions in the Galaxy. 7. What will be the long-term future of our Galaxy? Thought Questions 8. Suppose the Milky Way was a band of light extending only halfway around the sky (that is, in a semicircle). What, then, would you conclude about the Sun\u2019s location in the Galaxy? Give your reasoning. 9. Suppose somebody proposed that rather than invoking dark matter to explain the increased orbital velocities of stars beyond the Sun\u2019s orbit, the problem could be solved by assuming that the Milky Way\u2019s central black hole was much more massive. Does simply increasing the assumed mass of the Milky Way\u2019s central supermassive black hole correctly resolve the issue of unexpectedly high orbital velocities in the Galaxy? Why or why not? 10. The globular clusters revolve around the Galaxy in highly elliptical orbits. Where would you expect the clusters to spend most of their time? (Think of Kepler\u2019s laws.) At any given time, would you expect most globular clusters to be moving at high or low speeds with respect to the center of the Galaxy? Why? 11. Shapley used the positions of globular clusters to determine the location of the galactic center. Could he have used open clusters? Why or why not? 12. Consider the following five kinds of objects: open cluster, giant molecular cloud, globular cluster, group of O and B stars, and planetary nebulae. A. Which occur only in spiral arms? B. Which occur only in the parts of the Galaxy other than the spiral arms? C. Which are thought to be very young? D. Which are thought to be very old? E. Which have the hottest stars? 13. The dwarf galaxy in Sagittarius is the one closest to the Milky Way, yet it was discovered only in 1994. Can you think of a reason it was not discovered earlier? (Hint: Think about what else is in its constellation.) 928 Chapter 25 The Milky Way Galaxy 14. Suppose three stars lie in the disk of the Galaxy at distances of 20,000 light-years, 25,000 light-years, and 30,000 light-years from the galactic center, and suppose that right now all three are lined up in such a way that it is possible to draw a straight line through them and on to the center of the Galaxy. How will the relative positions of these three stars change with time? Assume that their orbits are all circular and lie in the plane of the disk. 15. Why does star formation occur primarily in the disk of the Galaxy? 16. Where in the Galaxy would you expect to find Type II supernovae, which are the explosions of massive stars that go through their lives very quickly? Where would you expect to find Type I supernovae, which involve the explosions of white dwarfs? 17. Suppose that stars evolved without losing mass\u2014that once matter was incorporated into a star, it remained there forever. How would the appearance of the Galaxy be different from what it is now? Would there be population I and population II stars? What other differences would there be? Figuring For Yourself 18. Assume that the Sun orbits the center of the Galaxy at a speed of 220 km/s and a distance of 26,000 light- years from the center. A. Calculate the circumference of the Sun\u2019s orbit, assuming it to be approximately circular. (Remember that the circumference of a circle is given by 2\u03c0R, where R is the radius of the circle. Be sure to use consistent units. The conversion from light-years to km/s can be found in an online calculator or appendix, or you can calculate it for yourself: the speed of light is 300,000 km/s, and you can determine the number of seconds in a year.) B. Calculate the Sun\u2019s period, the \u201cgalactic year.\u201d Again, be careful with the units. Does it agree with the number we", " gave above? 19. The Sun orbits the center of the Galaxy in 225 million years at a distance of 26,000 light-years. Given that \u23a0 \u00d7 P2, where a is the semimajor axis and P is the orbital period, what is the mass of the a3 = \u239b \u239e \u239dM1 + M2 Galaxy within the Sun\u2019s orbit? 20. Suppose the Sun orbited a little farther out, but the mass of the Galaxy inside its orbit remained the same as we calculated in Exercise 25.19. What would be its period at a distance of 30,000 light-years? 21. We have said that the Galaxy rotates differentially; that is, stars in the inner parts complete a full 360\u00b0 orbit around the center of the Galaxy more rapidly than stars farther out. Use Kepler\u2019s third law and the mass we derived in Exercise 25.19 to calculate the period of a star that is only 5000 light-years from the center. Now do the same calculation for a globular cluster at a distance of 50,000 light-years. Suppose the Sun, this star, and the globular cluster all fall on a straight line through the center of the Galaxy. Where will they be relative to each other after the Sun completes one full journey around the center of the Galaxy? (Assume that all the mass in the Galaxy is concentrated at its center.) 22. If our solar system is 4.6 billion years old, how many galactic years has planet Earth been around? 23. Suppose the average mass of a star in the Galaxy is one-third of a solar mass. Use the value for the mass of the Galaxy that we calculated in Exercise 25.19, and estimate how many stars are in the Milky Way. Give some reasons it is reasonable to assume that the mass of an average star is less than the mass of the Sun. 24. The first clue that the Galaxy contains a lot of dark matter was the observation that the orbital velocities of stars did not decreases with increasing distance from the center of the Galaxy. Construct a rotation curve for the solar system by using the orbital velocities of the planets, which can be found in Appendix F. How does this curve differ from the rotation curve for the Galaxy? What does it tell you about where most of the mass in the solar system is concentrated? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 25 The Milky Way Galaxy 929 25. The best evidence for a black hole at the center of the Galaxy also comes from the application of Kepler\u2019s third law. Suppose a star at a distance of 20 light-hours from the center of the Galaxy has an orbital speed of 6200 km/s. How much mass must be located inside its orbit? 26. The next step in deciding whether the object in Exercise 25.25 is a black hole is to estimate the density of this mass. Assume that all of the mass is spread uniformly throughout a sphere with a radius of 20 lighthours. What is the density in kg/km3? (Remember that the volume of a sphere is given by V = 4 \u03c0R3.) 3 Explain why the density might be even higher than the value you have calculated. How does this density compare with that of the Sun or other objects we have talked about in this book? 27. Suppose the Sagittarius dwarf galaxy merges completely with the Milky Way and adds 150,000 stars to it. Estimate the percentage change in the mass of the Milky Way. Will this be enough mass to affect the orbit of the Sun around the galactic center? Assume that all of the Sagittarius galaxy\u2019s stars end up in the nuclear bulge of the Milky Way Galaxy and explain your answer. 930 Chapter 25 The Milky Way Galaxy This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 931 26 GALAXIES Figure 26.1. Spiral Galaxy. NGC 6946 is a spiral galaxy also known as the \u201cFireworks galaxy.\u201d It is at a distance of about 18 million light-years, in the direction of the constellations Cepheus and Cygnus. It was discovered by William Herschel in 1798. This galaxy is about one-third the size of the Milky Way. Note on the left how the colors of the galaxy change from the yellowish light of old stars in the center to the blue color of hot, young stars and the reddish glow of hydrogen clouds in the spiral arms. As the image shows, this galaxy is rich in dust and gas, and new stars are still being born here. (credit left: modification of work by NASA, ESA, STScI, R. Gendler, and the Subaru Telescope (NAOJ); credit right: modification of work by", " X-ray: NASA/CXC/MSSL/R.Soria et al, Optical: AURA/Gemini OBs) Chapter Outline 26.1 The Discovery of Galaxies 26.2 Types of Galaxies 26.3 Properties of Galaxies 26.4 The Extragalactic Distance Scale 26.5 The Expanding Universe Thinking Ahead In the last chapter, we explored our own Galaxy. But is it the only one? If there are others, are they like the Milky Way? How far away are they? Can we see them? As we shall learn, some galaxies turn out to be so far away that it has taken billions of years for their light to reach us. These remote galaxies can tell us what the universe was like when it was young. In this chapter, we start our exploration of the vast realm of galaxies. Like tourists from a small town making their first visit to the great cities of the world, we will be awed by the beauty and variety of the galaxies. And yet, we will recognize that much of what we see is not so different from our experiences at home, and we will be impressed by how much we can learn by looking at structures built long ago. We begin our voyage with a guide to the properties of galaxies, much as a tourist begins with a guidebook to the main features of the cities on the itinerary. In later chapters, we will look more carefully at the past history of galaxies, how they have changed over time, and how they acquired their many different forms. First, we\u2019ll begin our voyage through the galaxies with the question: is our Galaxy the only one? 932 Chapter 26 Galaxies 26.1 THE DISCOVERY OF GALAXIES Learning Objectives By the end of this section, you will be able to: Describe the discoveries that confirmed the existence of galaxies that lie far beyond the Milky Way Galaxy Explain why galaxies used to be called nebulae and why we don\u2019t include them in that category any more Growing up at a time when the Hubble Space Telescope orbits above our heads and giant telescopes are springing up on the great mountaintops of the world, you may be surprised to learn that we were not sure about the existence of other galaxies for a very long time. The very idea that other galaxies exist used to be controversial. Even into the 1920s, many astronomers thought the Milky Way encompassed all that exists in the universe. The evidence found in 1924 that meant our Galaxy is not alone was one of the great scientific discoveries of the twentieth century. It was not that scientists weren\u2019t asking questions. They questioned the composition and structure of the universe as early as the eighteenth century. However, with the telescopes available in earlier centuries, galaxies looked like small fuzzy patches of light that were difficult to distinguish from the star clusters and gas-and-dust clouds that are part of our own Galaxy. All objects that were not sharp points of light were given the same name, nebulae, the Latin word for \u201cclouds.\u201d Because their precise shapes were often hard to make out and no techniques had yet been devised for measuring their distances, the nature of the nebulae was the subject of much debate. As early as the eighteenth century, the philosopher Immanuel Kant (1724\u20131804) suggested that some of the nebulae might be distant systems of stars (other Milky Ways), but the evidence to support this suggestion was beyond the capabilities of the telescopes of that time. Other Galaxies By the early twentieth century, some nebulae had been correctly identified as star clusters, and others (such as the Orion Nebula) as gaseous nebulae. Most nebulae, however, looked faint and indistinct, even with the best telescopes, and their distances remained unknown. (For more on how such nebulae are named, by the way, see the feature box on Naming the Nebulae in the chapter on interstellar matter.) If these nebulae were nearby, with distances comparable to those of observable stars, they were most likely clouds of gas or groups of stars within our Galaxy. If, on the other hand, they were remote, far beyond the edge of the Galaxy, they could be other star systems containing billions of stars. To determine what the nebulae are, astronomers had to find a way of measuring the distances to at least some of them. When the 2.5-meter (100-inch) telescope on Mount Wilson in Southern California went into operation, astronomers finally had the large telescope they needed to settle the controversy. Working with the 2.5-meter telescope, Edwin Hubble was able to resolve individual stars in several of the brighter spiral-shaped nebulae, including M31, the great spiral in Andromeda (Figure 26.2). Among these stars, he discovered some faint variable stars that\u2014when he analyzed their light curves\u2014turned out to be cepheids. Here were reliable indicators that Hubble could use to measure the distances", " to the nebulae using the technique pioneered by Henrietta Leavitt (see the chapter on Celestial Distances). After painstaking work, he estimated that the Andromeda galaxy was about 900,000 light-years away from us. At that enormous distance, it had to be a separate galaxy of stars located well outside the boundaries of the Milky Way. Today, we know the This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 933 Andromeda galaxy is actually slightly more than twice as distant as Hubble\u2019s first estimate, but his conclusion about its true nature remains unchanged. Figure 26.2. Andromeda Galaxy. Also known by its catalog number M31, the Andromeda galaxy is a large spiral galaxy very similar in appearance to, and slightly larger than, our own Galaxy. At a distance of about 2.5 million light-years, Andromeda is the spiral galaxy that is nearest to our own in space. Here, it is seen with two of its satellite galaxies, M32 (top) and M110 (bottom). (credit: Adam Evans) No one in human history had ever measured a distance so great. When Hubble\u2019s paper on the distances to nebulae was read before a meeting of the American Astronomical Society on the first day of 1925, the entire room erupted in a standing ovation. A new era had begun in the study of the universe, and a new scientific field\u2014extragalactic astronomy\u2014had just been born Edwin Hubble: Expanding the Universe The son of a Missouri insurance agent, Edwin Hubble (Figure 26.3) graduated from high school at age 16. He excelled in sports, winning letters in track and basketball at the University of Chicago, where he studied both science and languages. Both his father and grandfather wanted him to study law, however, and he gave in to family pressure. He received a prestigious Rhodes scholarship to Oxford University in England, where he studied law with only middling enthusiasm. Returning to be the United States, he spent a year teaching high school physics and Spanish as well as coaching basketball, while trying to determine his life\u2019s direction. 934 Chapter 26 Galaxies Figure 26.3. Edwin Hubble (1889\u20131953). Edwin Hubble established some of the most important ideas in the study of galaxies. The pull of astronomy eventually proved too strong to resist, and so Hubble went back to the University of Chicago for graduate work. Just as he was about to finish his degree and accept an offer to work at the soon-to be completed 2.5-meter telescope, the United States entered World War I, and Hubble enlisted as an officer. Although the war had ended by the time he arrived in Europe, he received more officer\u2019s training abroad and enjoyed a brief time of further astronomical study at Cambridge before being sent home. In 1919, at age 30, he joined the staff at Mount Wilson and began working with the world\u2019s largest telescope. Ripened by experience, energetic, disciplined, and a skillful observer, Hubble soon established some of the most important ideas in modern astronomy. He showed that other galaxies existed, classified them on the basis of their shapes, found a pattern to their motion (and thus put the notion of an expanding universe on a firm observational footing), and began a lifelong program to study the distribution of galaxies in the universe. Although a few others had glimpsed pieces of the puzzle, it was Hubble who put it all together and showed that an understanding of the large-scale structure of the universe was feasible. His work brought Hubble much renown and many medals, awards, and honorary degrees. As he became better known (he was the first astronomer to appear on the cover of Time magazine), he and his wife enjoyed and cultivated friendships with movie stars and writers in Southern California. Hubble was instrumental (if you\u2019ll pardon the pun) in the planning and building of the 2.5-meter telescope on Palomar Mountain, and he had begun to use it for studying galaxies when he passed away from a stroke in 1953. When astronomers built a space telescope that would allow them to extend Hubble\u2019s work to distances he could only dream about, it seemed natural to name it in his honor. It was fitting that observations with the Hubble Space Telescope (and his foundational work on expansion of the universe) contributed to the 2011 Nobel Prize in Physics, given for the discovery that the expansion of the universe is accelerating (a topic we will expand upon in the chapter on The Big Bang). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 935 26.2 TYPES OF GALAXIES Learning Objectives By the end of this section, you will be able to: Describe the properties and features of elliptical, spiral, and irregular galaxies Explain what may cause a galaxy\u2019s appearance to change over time Having established", " the existence of other galaxies, Hubble and others began to observe them more closely\u2014noting their shapes, their contents, and as many other properties as they could measure. This was a daunting task in the 1920s when obtaining a single photograph or spectrum of a galaxy could take a full night of tireless observing. Today, larger telescopes and electronic detectors have made this task less difficult, although observing the most distant galaxies (those that show us the universe in its earliest phases) still requires enormous effort. The first step in trying to understand a new type of object is often simply to describe it. Remember, the first step in understanding stellar spectra was simply to sort them according to appearance (see Analyzing Starlight). As it turns out, the biggest and most luminous galaxies come in one of two basic shapes: either they are flatter and have spiral arms, like our own Galaxy, or they appear to be elliptical (blimp- or cigar-shaped). Many smaller galaxies, in contrast, have an irregular shape. Spiral Galaxies Our own Galaxy and the Andromeda galaxy are typical, large spiral galaxies (see Figure 26.2). They consist of a central bulge, a halo, a disk, and spiral arms. Interstellar material is usually spread throughout the disks of spiral galaxies. Bright emission nebulae and hot, young stars are present, especially in the spiral arms, showing that new star formation is still occurring. The disks are often dusty, which is especially noticeable in those systems that we view almost edge on (Figure 26.4). Figure 26.4. Spiral Galaxies. (a) The spiral arms of M100, shown here, are bluer than the rest of the galaxy, indicating young, high-mass stars and star-forming regions. (b) We view this spiral galaxy, NGC 4565, almost exactly edge on, and from this angle, we can see the dust in the plane of the galaxy; it appears dark because it absorbs the light from the stars in the galaxy. (credit a: modification of work by Hubble Legacy Archive, NASA, ESA, and Judy Schmidt; credit b: modification of work by \u201cJschulman555\u201d/ Wikimedia) 936 Chapter 26 Galaxies In galaxies that we see face on, the bright stars and emission nebulae make the arms of spirals stand out like those of a pinwheel on the fourth of July. Open star clusters can be seen in the arms of nearer spirals, and globular clusters are often visible in their halos. Spiral galaxies contain a mixture of young and old stars, just as the Milky Way does. All spirals rotate, and the direction of their spin is such that the arms appear to trail much like the wake of a boat. About two-thirds of the nearby spiral galaxies have boxy or peanut-shaped bars of stars running through their centers (Figure 26.5). Showing great originality, astronomers call these galaxies barred spirals. Figure 26.5. Barred Spiral Galaxy. NGC 1300, shown here, is a barred spiral galaxy. Note that the spiral arms begin at the ends of the bar. (credit: NASA, ESA, and the Hubble Heritage Team(STScI/AURA)) As we noted in The Milky Way Galaxy chapter, our Galaxy has a modest bar too (see Figure 25.10). The spiral arms usually begin from the ends of the bar. The fact that bars are so common suggests that they are long lived; it may be that most spiral galaxies form a bar at some point during their evolution. In both barred and unbarred spiral galaxies, we observe a range of different shapes. At one extreme, the central bulge is large and luminous, the arms are faint and tightly coiled, and bright emission nebulae and supergiant stars are inconspicuous. Hubble, who developed a system of classifying galaxies by shape, gave these galaxies the designation Sa. Galaxies at this extreme may have no clear spiral arm structure, resulting in a lens-like appearance (they are sometimes referred to as lenticular galaxies). These galaxies seem to share as many properties with elliptical galaxies as they do with spiral galaxies At the other extreme, the central bulge is small and the arms are loosely wound. In these Sc galaxies, luminous stars and emission nebulae are very prominent. Our Galaxy and the Andromeda galaxy are both intermediate between the two extremes. Photographs of spiral galaxies, illustrating the different types, are shown in Figure 26.6, along with elliptical galaxies for comparison. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 937 Figure 26.6. Hubble Classification of Galaxies. This figure shows Edwin Hubble\u2019s original classification of galaxies. Elliptical galaxies are on the left. On the right, you can see the basic spiral shapes illustrated, alongside images of actual barred and unbarred spirals. (credit:", " modification of work by NASA, ESA) The luminous parts of spiral galaxies appear to range in diameter from about 20,000 to more than 100,000 light- years. Recent studies have found that there is probably a large amount of galactic material that extends well beyond the apparent edge of galaxies. This material appears to be thin, cold gas that is difficult to detect in most observations. From the observational data available, the masses of the visible portions of spiral galaxies are estimated to range from 1 billion to 1 trillion Suns (109 to 1012 MSun). The total luminosities of most spirals fall in the range of 100 million to 100 billion times the luminosity of our Sun (108 to 1011 LSun). Our Galaxy and M31 are relatively large and massive, as spirals go. There is also considerable dark matter in and around the galaxies, just as there is in the Milky Way; we deduce its presence from how fast stars in the outer parts of the Galaxy are moving in their orbits. Elliptical Galaxies Elliptical galaxies consist almost entirely of old stars and have shapes that are spheres or ellipsoids (somewhat squashed spheres) (Figure 26.7). They contain no trace of spiral arms. Their light is dominated by older reddish stars (the population II stars discussed in The Milky Way Galaxy). In the larger nearby ellipticals, many globular clusters can be identified. Dust and emission nebulae are not conspicuous in elliptical galaxies, but many do contain a small amount of interstellar matter. 938 Chapter 26 Galaxies Figure 26.7. Elliptical Galaxies. (a) ESO 325-G004 is a giant elliptical galaxy. Other elliptical galaxies can be seen around the edges of this image. (b) This elliptical galaxy probably originated from the collision of two spiral galaxies. (credit a: modification of work by NASA, ESA, and The Hubble Heritage Team (STScI/AURA); credit b: modification of work by ESA/Hubble, NASA) Elliptical galaxies show various degrees of flattening, ranging from systems that are approximately spherical to those that approach the flatness of spirals. The rare giant ellipticals (for example, ESO 325-G004 in Figure 26.7) reach luminosities of 1011 LSun. The mass in a giant elliptical can be as large as 1013 MSun. The diameters of these large galaxies extend over several hundred thousand light-years and are considerably larger than the largest spirals. Although individual stars orbit the center of an elliptical galaxy, the orbits are not all in the same direction, as occurs in spirals. Therefore, ellipticals don\u2019t appear to rotate in a systematic way, making it difficult to estimate how much dark matter they contain. We find that elliptical galaxies range all the way from the giants, just described, to dwarfs, which may be the most common kind of galaxy. Dwarf ellipticals (sometimes called dwarf spheroidals) escaped our notice for a long time because they are very faint and difficult to see. An example of a dwarf elliptical is the Leo I Dwarf Spheroidal galaxy shown in Figure 26.8. The luminosity of this typical dwarf is about equal to that of the brightest globular clusters. Intermediate between the giant and dwarf elliptical galaxies are systems such as M32 and M110, the two companions of the Andromeda galaxy. While they are often referred to as dwarf ellipticals, these galaxies are significantly larger than galaxies such as Leo I. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 939 Figure 26.8. Dwarf Elliptical Galaxy. M32, a dwarf elliptical galaxy and one of the companions to the giant Andromeda galaxy M31. M32 is a dwarf by galactic standards, as it is only 2400 light-years across. (credit: NOAO/AURA/NSF) Irregular Galaxies Hubble classified galaxies that do not have the regular shapes associated with the categories we just described into the catchall bin of an irregular galaxy, and we continue to use his term. Typically, irregular galaxies have lower masses and luminosities than spiral galaxies. Irregular galaxies often appear disorganized, and many are undergoing relatively intense star formation activity. They contain both young population I stars and old population II stars. The two best-known irregular galaxies are the Large Magellanic Cloud and Small Magellanic Cloud (Figure 26.9), which are at a distance of a little more than 160,000 light-years away and are among our nearest extragalactic neighbors. Their names reflect the fact that Ferdinand Magellan and his crew, making their round-the-world journey, were the first European travelers to notice them. Although not visible from the United States and Europe, these two systems are prominent from the Southern Hemisphere, where they look like wispy clouds in the", " night sky. Since they are only about one-tenth as distant as the Andromeda galaxy, they present an excellent opportunity for astronomers to study nebulae, star clusters, variable stars, and other key objects in the setting of another galaxy. For example, the Large Magellanic Cloud contains the 30 Doradus complex (also known as the Tarantula Nebula), one of the largest and most luminous groups of supergiant stars known in any galaxy. 940 Chapter 26 Galaxies Figure 26.9. 4-Meter Telescope at Cerro Tololo Inter-American Observatory Silhouetted against the Southern Sky. The Milky Way is seen to the right of the dome, and the Large and Small Magellanic Clouds are seen to the left. (credit: Roger Smith/NOAO/AURA/NSF) The Small Magellanic Cloud is considerably less massive than the Large Magellanic Cloud, and it is six times longer than it is wide. This narrow wisp of material points directly toward our Galaxy like an arrow. The Small Magellanic Cloud was most likely contorted into its current shape through gravitational interactions with the Milky Way. A large trail of debris from this interaction between the Milky Way and the Small Magellanic Cloud has been strewn across the sky and is seen as a series of gas clouds moving at abnormally high velocity, known as the Magellanic Stream. We will see that this kind of interaction between galaxies will help explain the irregular shapes of this whole category of small galaxies View this beautiful album showcasing the different types of galaxies (https://openstaxcollege.org/ l/30galaxphohubb) that have been photographed by the Hubble Space Telescope. Galaxy Evolution Encouraged by the success of the H-R diagram for stars (see Analyzing Starlight), astronomers studying galaxies hoped to find some sort of comparable scheme, where differences in appearance could be tied to different evolutionary stages in the life of galaxies. Wouldn\u2019t it be nice if every elliptical galaxy evolved into a This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 941 spiral, for example, just as every main-sequence star evolves into a red giant? Several simple ideas of this kind were tried, some by Hubble himself, but none stood the test of time (and observation). Because no simple scheme for evolving one type of galaxy into another could be found, astronomers then tended to the opposite point of view. For a while, most astronomers thought that all galaxies formed very early in the history of the universe and that the differences between them had to do with the rate of star formation. Ellipticals were those galaxies in which all the interstellar matter was converted rapidly into stars. Spirals were galaxies in which star formation occurred slowly over the entire lifetime of the galaxy. This idea turned out to be too simple as well. Today, we understand that at least some galaxies have changed types over the billions of years since the universe began. As we shall see in later chapters, collisions and mergers between galaxies may dramatically change spiral galaxies into elliptical galaxies. Even isolated spirals (with no neighbor galaxies in sight) can change their appearance over time. As they consume their gas, the rate of star formation will slow down, and the spiral arms will gradually become less conspicuous. Over long periods, spirals therefore begin to look more like the galaxies at the middle of Figure 26.6 (which astronomers refer to as S0 types). Over the past several decades, the study of how galaxies evolve over the lifetime of the universe has become one of the most active fields of astronomical research. We will discuss the evolution of galaxies in more detail in The Evolution and Distribution of Galaxies, but let\u2019s first see in a little more detail just what different galaxies are like. 26.3 PROPERTIES OF GALAXIES Learning Objectives By the end of this section, you will be able to: Describe the methods through which astronomers can estimate the mass of a galaxy Characterize each type of galaxy by its mass-to-light ratio The technique for deriving the masses of galaxies is basically the same as that used to estimate the mass of the Sun, the stars, and our own Galaxy. We measure how fast objects in the outer regions of the galaxy are orbiting the center, and then we use this information along with Kepler\u2019s third law to calculate how much mass is inside that orbit. Masses of Galaxies Astronomers can measure the rotation speed in spiral galaxies by obtaining spectra of either stars or gas, and looking for wavelength shifts produced by the Doppler effect. Remember that the faster something is moving toward or away from us, the greater the shift of the lines in its spectrum. Kepler\u2019s law, together with such observations of the part of the Andromeda galaxy that is bright in visible light, for example, show it to have a galactic mass of about 4 \u00d7 1011 MSun (", "enough material to make 400 billion stars like the Sun). The total mass of the Andromeda galaxy is greater than this, however, because we have not included the mass of the material that lies beyond its visible edge. Fortunately, there is a handful of objects\u2014such as isolated stars, star clusters, and satellite galaxies\u2014beyond the visible edge that allows astronomers to estimate how much additional matter is hidden out there. Recent studies show that the amount of dark matter beyond the visible edge of Andromeda may be as large as the mass of the bright portion of the galaxy. Indeed, using Kepler\u2019s third law and the velocities of its satellite galaxies, the Andromeda galaxy is estimated to have a mass closer to 1.4 \u00d7 942 Chapter 26 Galaxies 1012 MSun. The mass of the Milky Way Galaxy is estimated to be 8.5 \u00d7 1011 MSun, and so our Milky Way is turning out to be somewhat smaller than Andromeda. Elliptical galaxies do not rotate in a systematic way, so we cannot determine a rotational velocity; therefore, we must use a slightly different technique to measure their mass. Their stars are still orbiting the galactic center, but not in the organized way that characterizes spirals. Since elliptical galaxies contain stars that are billions of years old, we can assume that the galaxies themselves are not flying apart. Therefore, if we can measure the various speeds with which the stars are moving in their orbits around the center of the galaxy, we can calculate how much mass the galaxy must contain in order to hold the stars within it. In practice, the spectrum of a galaxy is a composite of the spectra of its many stars, whose different motions produce different Doppler shifts (some red, some blue). The result is that the lines we observe from the entire galaxy contain the combination of many Doppler shifts. When some stars provide blueshifts and others provide redshifts, they create a wider or broader absorption or emission feature than would the same lines in a hypothetical galaxy in which the stars had no orbital motion. Astronomers call this phenomenon line broadening. The amount by which each line broadens indicates the range of speeds at which the stars are moving with respect to the center of the galaxy. The range of speeds depends, in turn, on the force of gravity that holds the stars within the galaxies. With information about the speeds, it is possible to calculate the mass of an elliptical galaxy. Table 26.1 summarizes the range of masses (and other properties) of the various types of galaxies. Interestingly enough, the most and least massive galaxies are ellipticals. On average, irregular galaxies have less mass than spirals. Characteristics of the Different Types of Galaxies Characteristic Spirals Ellipticals Irregulars Mass (MSun) 109 to 1012 105 to 1013 108 to 1011 Diameter (thousands of light-years) 15 to 150 3 to >700 3 to 30 Luminosity (LSun) 108 to 1011 106 to 1011 107 to 2 \u00d7 109 Populations of stars Old and young Old Old and young Interstellar matter Gas and dust Almost no dust; little gas Much gas; some have little dust, some much dust Mass-to-light ratio in the visible part Mass-to-light ratio for total galaxy Table 26.1 2 to 10 10 to 20 1 to 10 100 100? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 943 Mass-to-Light Ratio A useful way of characterizing a galaxy is by noting the ratio of its mass (in units of the Sun\u2019s mass) to its light output (in units of the Sun\u2019s luminosity). This single number tells us roughly what kind of stars make up most of the luminous population of the galaxy, and it also tells us whether a lot of dark matter is present. For stars like the Sun, the mass-to-light ratio is 1 by our definition. Galaxies are not, of course, composed entirely of stars that are identical to the Sun. The overwhelming majority of stars are less massive and less luminous than the Sun, and usually these stars contribute most of the mass of a system without accounting for very much light. The mass-to-light ratio for low-mass stars is greater than 1 (you can verify this using the data in Table 18.3). Therefore, a galaxy\u2019s mass-to-light ratio is also generally greater than 1, with the exact value depending on the ratio of high-mass stars to low-mass stars. Galaxies in which star formation is still occurring have many massive stars, and their mass-to-light ratios are usually in the range of 1 to 10. Galaxies consisting mostly of an older stellar population, such as ellipticals, in which the massive stars have already completed their evolution and have ceased to shine, have mass-to-light ratios of 10 to 20. But these figures refer", " only to the inner, conspicuous parts of galaxies (Figure 26.10). In The Milky Way Galaxy and above, we discussed the evidence for dark matter in the outer regions of our own Galaxy, extending much farther from the galactic center than do the bright stars and gas. Recent measurements of the rotation speeds of the outer parts of nearby galaxies, such as the Andromeda galaxy we discussed earlier, suggest that they too have extended distributions of dark matter around the visible disk of stars and dust. This largely invisible matter adds to the mass of the galaxy while contributing nothing to its luminosity, thus increasing the mass-to- light ratio. If dark invisible matter is present in a galaxy, its mass-to-light ratio can be as high as 100. The two different mass-to-light ratios measured for various types of galaxies are given in Table 26.1. 944 Chapter 26 Galaxies Figure 26.10. M101, the Pinwheel Galaxy. This galaxy is a face-on spiral at a distance of 21 million light-years. M101 is almost twice the diameter of the Milky Way, and it contains at least 1 trillion stars. (credit: NASA, ESA, K. Kuntz (Johns Hopkins University), F. Bresolin (University of Hawaii), J. Trauger (Jet Propulsion Lab), J. Mould (NOAO), Y.-H. Chu (University of Illinois, Urbana), and STScI) These measurements of other galaxies support the conclusion already reached from studies of the rotation of our own Galaxy\u2014namely, that most of the material in the universe cannot at present be observed directly in any part of the electromagnetic spectrum. An understanding of the properties and distribution of this invisible matter is crucial to our understanding of galaxies. It\u2019s becoming clearer and clearer that, through the gravitational force it exerts, dark matter plays a dominant role in galaxy formation and early evolution. There is an interesting parallel here between our time and the time during which Edwin Hubble was receiving his training in astronomy. By 1920, many scientists were aware that astronomy stood on the brink of important breakthroughs\u2014if only the nature and behavior of the nebulae could be settled with better observations. In the same way, many astronomers today feel we may be closing in on a far more sophisticated understanding of the large-scale structure of the universe\u2014if only we can learn more about the nature and properties of dark matter. If you follow astronomy articles in the news (as we hope you will), you should be hearing more about dark matter in the years to come. 26.4 THE EXTRAGALACTIC DISTANCE SCALE Learning Objectives By the end of this section, you will be able to: This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 945 Describe the use of variable stars to estimate distances to galaxies Explain how standard bulbs and the Tully-Fisher relation can be used to estimate distances to galaxies To determine many of the properties of a galaxy, such as its luminosity or size, we must first know how far away it is. If we know the distance to a galaxy, we can convert how bright the galaxy appears to us in the sky into its true luminosity because we know the precise way light is dimmed by distance. (The same galaxy 10 times farther away, for example, would look 100 times dimmer.) But the measurement of galaxy distances is one of the most difficult problems in modern astronomy: all galaxies are far away, and most are so distant that we cannot even make out individual stars in them. For decades after Hubble\u2019s initial work, the techniques used to measure galaxy distances were relatively inaccurate, and different astronomers derived distances that differed by as much as a factor of two. (Imagine if the distance between your home or dorm and your astronomy class were this uncertain; it would be difficult to make sure you got to class on time.) In the past few decades, however, astronomers have devised new techniques for measuring distances to galaxies; most importantly, all of them give the same answer to within an accuracy of about 10%. As we will see, this means we may finally be able to make reliable estimates of the size of the universe. Variable Stars Before astronomers could measure distances to other galaxies, they first had to establish the scale of cosmic distances using objects in our own Galaxy. We described the chain of these distance methods in Celestial Distances (and we recommend that you review that chapter if it has been a while since you\u2019ve read it). Astronomers were especially delighted when they discovered that they could measure distances using certain kinds of intrinsically luminous variable stars, such as cepheids, which can be seen at very large distances (Figure 26.11). After the variables in nearby galaxies had been used to make distance measurements for a few decades, Walter Baade showed that there were actually two kinds of cepheids and that astronomers had been unwittingly mixing them up", ". As a result, in the early 1950s, the distances to all of the galaxies had to be increased by about a factor of two. We mention this because we want you to bear in mind, as you read on, that science is always a study in progress. Our first tentative steps in such difficult investigations are always subject to future revision as our techniques become more reliable. The amount of work involved in finding cepheids and measuring their periods can be enormous. Hubble, for example, obtained 350 long-exposure photographs of the Andromeda galaxy over a period of 18 years and was able to identify only 40 cepheids. Even though cepheids are fairly luminous stars, they can be detected in only about 30 of the nearest galaxies with the world\u2019s largest ground-based telescopes. As mentioned in Celestial Distances, one of the main projects carried out during the first years of operation of the Hubble Space Telescope was the measurement of cepheids in more distant galaxies to improve the accuracy of the extragalactic distance scale. Recently, astronomers working with the Hubble Space Telescope have extended such measurements out to 108 million light-years\u2014a triumph of technology and determination. 946 Chapter 26 Galaxies Figure 26.11. Cepheid Variable Star. In 1994, using the Hubble Space Telescope, astronomers were able to make out an individual cepheid variable star in the galaxy M100 and measure its distance to be 56 million light-years. The insets show the star on three different nights; you can see that its brightness is indeed variable. (credit: modification of work by Wendy L. Freedman, Observatories of the Carnegie Institution of Washington, and NASA/ESA) Nevertheless, we can only use cepheids to measure distances within a small fraction of the universe of galaxies. After all, to use this method, we must be able to resolve single stars and follow their subtle variations. Beyond a certain distance, even our finest space telescopes cannot help us do this. Fortunately, there are other ways to measure the distances to galaxies. Standard Bulbs We discussed in Celestial Distances the great frustration that astronomers felt when they realized that the stars in general were not standard bulbs. If every light bulb in a huge auditorium is a standard 100-watt bulb, then bulbs that look brighter to us must be closer, whereas those that look dimmer must be farther away. If every star were a standard luminosity (or wattage), then we could similarly \u201cread off\u201d their distances based on how bright they appear to us. Alas, as we have learned, neither stars nor galaxies come in one standard-issue luminosity. Nonetheless, astronomers have been searching for objects out there that do act in some way like a standard bulb\u2014that have the same intrinsic (built-in) brightness wherever they are. A number of suggestions have been made for what sorts of objects might be effective standard bulbs, including the brightest supergiant stars, planetary nebulae (which give off a lot of ultraviolet radiation), and the average globular cluster in a galaxy. One object turns out to be particularly useful: the type Ia supernova. These supernovae involve the explosion of a white dwarf in a binary system (see The Evolution of Binary Star Systems) Observations show that supernovae of this type all reach nearly the same luminosity (about 4.5 \u00d7 109 LSun) at maximum light. With such tremendous luminosities, these supernovae have been detected out to This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 947 a distance of more than 8 billion light-years and are therefore especially attractive to astronomers as a way of determining distances on a large scale (Figure 26.12). Figure 26.12. Type Ia Supernova. The bright object at the bottom left of center is a type Ia supernova near its peak intensity. The supernova easily outshines its host galaxy. This extreme increase and luminosity help astronomers use Ia supernova as standard bulbs. (credit: NASA, ESA, A. Riess (STScI)) Several other kinds of standard bulbs visible over great distances have also been suggested, including the overall brightness of, for example, giant ellipticals and the brightest member of a galaxy cluster. Type Ia supernovae, however, have proved to be the most accurate standard bulbs, and they can be seen in more distant galaxies than the other types of calibrators. As we will see in the chapter on The Big Bang, observations of this type of supernova have profoundly changed our understanding of the evolution of the universe. Other Measuring Techniques Another technique for measuring galactic distances makes use of an interesting relationship noticed in the late 1970s by Brent Tully of the University of Hawaii and Richard Fisher of the National Radio Astronomy Observatory. They discovered that the luminosity of a spiral galaxy is related to its rotational velocity (how", " fast it spins). Why would this be true? The more mass a galaxy has, the faster the objects in its outer regions must orbit. A more massive galaxy has more stars in it and is thus more luminous (ignoring dark matter for a moment). Thinking back to our discussion from the previous section, we can say that if the mass-to-light ratios for various spiral galaxies are pretty similar, then we can estimate the luminosity of a spiral galaxy by measuring its mass, and we can estimate its mass by measuring its rotational velocity. Tully and Fisher used the 21-cm line of cold hydrogen gas to determine how rapidly material in spiral galaxies is orbiting their centers (you can review our discussion of the 21-cm line in Between the Stars: Gas and Dust in Space). Since 21-cm radiation from stationary atoms comes in a nice narrow line, the width of the 21-cm line produced by a whole rotating galaxy tells us the range of orbital velocities of the galaxy\u2019s hydrogen gas. The broader the line, the faster the gas is orbiting in the galaxy, and the more massive and luminous the galaxy turns out to be. 948 Chapter 26 Galaxies It is somewhat surprising that this technique works, since much of the mass associated with galaxies is dark matter, which does not contribute at all to the luminosity but does affect the rotation speed. There is also no obvious reason why the mass-to-light ratio should be similar for all spiral galaxies. Nevertheless, observations of nearer galaxies (where we have other ways of measuring distance) show that measuring the rotational velocity of a galaxy provides an accurate estimate of its intrinsic luminosity. Once we know how luminous the galaxy really is, we can compare the luminosity to the apparent brightness and use the difference to calculate its distance. While the Tully-Fisher relation works well, it is limited\u2014we can only use it to determine the distance to a spiral galaxy. There are other methods that can be used to estimate the distance to an elliptical galaxy; however, those methods are beyond the scope of our introductory astronomy course. Table 26.2 lists the type of galaxy for which each of the distance techniques is useful, and the range of distances over which the technique can be applied. Some Methods for Estimating Distance to Galaxies Method Galaxy Type Approximate Distance Range (millions of light-years) Planetary nebulae All 0\u201370 Cepheid variables Spiral, irregulars 0\u2013110 Tully-Fisher relation Spiral 0\u2013300 Type Ia supernovae All 0\u201311,000 Redshifts (Hubble\u2019s law) All 300\u201313,000 Table 26.2 26.5 THE EXPANDING UNIVERSE Learning Objectives By the end of this section, you will be able to: Describe the discovery that galaxies getting farther apart as the universe evolves Explain how to use Hubble\u2019s law to determine distances to remote galaxies Describe models for the nature of an expanding universe Explain the variation in Hubble\u2019s constant We now come to one of the most important discoveries ever made in astronomy\u2014the fact that the universe is expanding. Before we describe how the discovery was made, we should point out that the first steps in the study of galaxies came at a time when the techniques of spectroscopy were also making great strides. Astronomers using large telescopes could record the spectrum of a faint star or galaxy on photographic plates, guiding their telescopes so they remained pointed to the same object for many hours and collected more light. The resulting spectra of galaxies contained a wealth of information about the composition of the galaxy and the velocities of these great star systems. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 949 Slipher\u2019s Pioneering Observations Curiously, the discovery of the expansion of the universe began with the search for Martians and other solar systems. In 1894, the controversial (and wealthy) astronomer Percival Lowell established an observatory in Flagstaff, Arizona, to study the planets and search for life in the universe. Lowell thought that the spiral nebulae might be solar systems in the process of formation. He therefore asked one of the observatory\u2019s young astronomers, Vesto M. Slipher (Figure 26.13), to photograph the spectra of some of the spiral nebulae to see if their spectral lines might show chemical compositions like those expected for newly forming planets. Figure 26.13. Vesto M. Slipher (1875\u20131969). Slipher spent his entire career at the Lowell Observatory, where he discovered the large radial velocities of galaxies. (credit: Lowell Observatory) The Lowell Observatory\u2019s major instrument was a 24-inch refracting telescope, which was not at all well suited to observations of faint spiral nebulae. With the technology available in those days, photographic plates had to be exposed for 20 to 40 hours", " to produce a good spectrum (in which the positions of the lines could reveal a galaxy\u2019s motion). This often meant continuing to expose the same photograph over several nights. Beginning in 1912, and making heroic efforts over a period of about 20 years, Slipher managed to photograph the spectra of more than 40 of the spiral nebulae (which would all turn out to be galaxies). To his surprise, the spectral lines of most galaxies showed an astounding redshift. By \u201credshift\u201d we mean that the lines in the spectra are displaced toward longer wavelengths (toward the red end of the visible spectrum). Recall from the chapter on Radiation and Spectra that a redshift is seen when the source of the waves is moving away from us. Slipher\u2019s observations showed that most spirals are racing away at huge speeds; the highest velocity he measured was 1800 kilometers per second. Only a few spirals\u2014such as the Andromeda and Triangulum Galaxies and M81\u2014all of which are now known to be our close neighbors, turned out to be approaching us. All the other galaxies were moving away. Slipher first announced this discovery in 1914, years before Hubble showed that these objects were other galaxies and before anyone knew how far away they were. No one at the time quite knew what to make of this discovery. Hubble\u2019s Law The profound implications of Slipher\u2019s work became apparent only during the 1920s. Georges Lema\u00eetre was a Belgian priest and a trained astronomer. In 1927, he published a paper in French in an obscure Belgian journal in which he suggested that we live in an expanding universe. The title of the paper (translated into English) is \u201cA Homogenous Universe of Constant Mass and Growing Radius Accounting for the Radial Velocity of Extragalactic 950 Chapter 26 Galaxies Nebulae.\u201d Lema\u00eetre had discovered that Einstein\u2019s equations of relativity were consistent with an expanding universe (as had the Russian scientist Alexander Friedmann independently in 1922). Lema\u00eetre then went on to use Slipher\u2019s data to support the hypothesis that the universe actually is expanding and to estimate the rate of expansion. Initially, scientists paid little attention to this paper, perhaps because the Belgian journal was not widely available. In the meantime, Hubble was making observations of galaxies with the 2.5-meter telescope on Mt. Wilson, which was then the world\u2019s largest. Hubble carried out the key observations in collaboration with a remarkable man, Milton Humason, who dropped out of school in the eighth grade and began his astronomical career by driving a mule train up the trail on Mount Wilson to the observatory (Figure 26.14). In those early days, supplies had to be brought up that way; even astronomers hiked up to the mountaintop for their turns at the telescope. Humason became interested in the work of the astronomers and, after marrying the daughter of the observatory\u2019s electrician, took a job as janitor there. After a time, he became a night assistant, helping the astronomers run the telescope and record data. Eventually, he made such a mark that he became a full astronomer at the observatory. Figure 26.14. Milton Humason (1891\u20131972). Humason was Hubble\u2019s collaborator on the great task of observing, measuring, and classifying the characteristics of many galaxies. (credit: Caltech Archives) By the late 1920s, Humason was collaborating with Hubble by photographing the spectra of faint galaxies with the 2.5-meter telescope. (By then, there was no question that the spiral nebulae were in fact galaxies.) Hubble had found ways to improve the accuracy of the estimates of distances to spiral galaxies, and he was able to measure much fainter and more distant galaxies than Slipher could observe with his much-smaller telescope. When Hubble laid his own distance estimates next to measurements of the recession velocities (the speed with which the galaxies were moving away), he found something stunning: there was a relationship between distance and velocity for galaxies. The more distant the galaxy, the faster it was receding from us. In 1931, Hubble and Humason jointly published the seminal paper where they compared distances and velocities of remote galaxies moving away from us at speeds as high as 20,000 kilometers per second and were able to show that the recession velocities of galaxies are directly proportional to their distances from us (Figure 26.15), just as Lema\u00eetre had suggested. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 951 Figure 26.15. Hubble\u2019s Law. (a) These data show Hubble\u2019s original velocity-distance relation, adapted from his 1929 paper in the Proceedings of the National Academy of Sciences. (b) These data show Hubble and Humason\u2019s velocity-distance relation, adapted", " from their 1931 paper in The Astrophysical Journal. The red dots at the lower left are the points in the diagram in the 1929 paper. Comparison of the two graphs shows how rapidly the determination of galactic distances and redshifts progressed in the 2 years between these publications. We now know that this relationship holds for every galaxy except a few of the nearest ones. Nearly all of the galaxies that are approaching us turn out to be part of the Milky Way\u2019s own group of galaxies, which have their own individual motions, just as birds flying in a group may fly in slightly different directions at slightly different speeds even though the entire flock travels through space together. Written as a formula, the relationship between velocity and distance is V = H \u00d7 d where v is the recession speed, d is the distance, and H is a number called the Hubble constant. This equation is now known as Hubble\u2019s law Constants of Proportionality Mathematical relationships such as Hubble\u2019s law are pretty common in life. To take a simple example, suppose your college or university hires you to call rich alumni and ask for donations. You are paid $2.50 for each call; the more calls you can squeeze in between studying astronomy and other courses, the more money you take home. We can set up a formula that connects p, your pay, and n, the number of calls p = A \u00d7 n where A is the alumni constant, with a value of $2.50. If you make 20 calls, you will earn $2.50 times 20, or $50. Suppose your boss forgets to tell you what you will get paid for each call. You can calculate the alumni constant that governs your pay by keeping track of how many calls you make and noting your gross pay each week. If you make 100 calls the first week and are paid $250, you can deduce that the constant is 952 Chapter 26 Galaxies $2.50 (in units of dollars per call). Hubble, of course, had no \u201cboss\u201d to tell him what his constant would be\u2014he had to calculate its value from the measurements of distance and velocity. Astronomers express the value of Hubble\u2019s constant in units that relate to how they measure speed and velocity for galaxies. In this book, we will use kilometers per second per million light-years as that unit. For many years, estimates of the value of the Hubble constant have been in the range of 15 to 30 kilometers per second per million light-years The most recent work appears to be converging on a value near 22 kilometers per second per million light-years If H is 22 kilometers per second per million light-years, a galaxy moves away from us at a speed of 22 kilometers per second for every million light-years of its distance. As an example, a galaxy 100 million light-years away is moving away from us at a speed of 2200 kilometers per second. Hubble\u2019s law tells us something fundamental about the universe. Since all but the nearest galaxies appear to be in motion away from us, with the most distant ones moving the fastest, we must be living in an expanding universe. We will explore the implications of this idea shortly, as well as in the final chapters of this text. For now, we will just say that Hubble\u2019s observation underlies all our theories about the origin and evolution of the universe. Hubble\u2019s Law and Distances The regularity expressed in Hubble\u2019s law has a built-in bonus: it gives us a new way to determine the distances to remote galaxies. First, we must reliably establish Hubble\u2019s constant by measuring both the distance and the velocity of many galaxies in many directions to be sure Hubble\u2019s law is truly a universal property of galaxies. But once we have calculated the value of this constant and are satisfied that it applies everywhere, much more of the universe opens up for distance determination. Basically, if we can obtain a spectrum of a galaxy, we can immediately tell how far away it is. The procedure works like this. We use the spectrum to measure the speed with which the galaxy is moving away from us. If we then put this speed and the Hubble constant into Hubble\u2019s law equation, we can solve for the distance Hubble\u2019s Law Hubble\u2019s law (v = H \u00d7 d) allows us to calculate the distance to any galaxy. Here is how we use it in practice. We have measured Hubble\u2019s constant to be 22 km/s per million light-years. This means that if a galaxy is 1 million light-years farther away, it will move away 22 km/s faster. So, if we find a galaxy that is moving away at 18,000 km/s, what does Hubble\u2019s law tells us about the distance to the galaxy? Solution d = v H = 18,000 km/s 22 km/s 1 million light-years = 18,000 22 \u00d7 1 million light-years 1 = 818", " million light-years This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 953 Note how we handled the units here: the km/s in the numerator and denominator cancel, and the factor of million light-years in the denominator of the constant must be divided correctly before we get our distance of 818 million light-years. Check Your Learning Using 22 km/s/million light-years for Hubble\u2019s constant, what recessional velocity do we expect to find if we observe a galaxy at 500 million light-years? Answer: v = d \u00d7 H = 500 million light-years \u00d7 22 km/s 1 million light-years = 11,000 km/s Variation of Hubble\u2019s Constant The use of redshift is potentially a very important technique for determining distances because as we have seen, most of our methods for determining galaxy distances are limited to approximately the nearest few hundred million light-years (and they have large uncertainties at these distances). The use of Hubble\u2019s law as a distance indicator requires only a spectrum of a galaxy and a measurement of the Doppler shift, and with large telescopes and modern spectrographs, spectra can be taken of extremely faint galaxies. But, as is often the case in science, things are not so simple. This technique works if, and only if, the Hubble constant has been truly constant throughout the entire life of the universe. When we observe galaxies billions of light-years away, we are seeing them as they were billions of years ago. What if the Hubble \u201cconstant\u201d was different billions of years ago? Before 1998, astronomers thought that, although the universe is expanding, the expansion should be slowing down, or decelerating, because the overall gravitational pull of all matter in the universe would have a dominant, measureable effect. If the expansion is decelerating, then the Hubble constant should be decreasing over time. The discovery that type Ia supernovae are standard bulbs gave astronomers the tool they needed to observe extremely distant galaxies and measure the rate of expansion billions of years ago. The results were completely unexpected. It turns out that the expansion of the universe is accelerating over time! What makes this result so astounding is that there is no way that existing physical theories can account for this observation. While a decelerating universe could easily be explained by gravity, there was no force or property in the universe known to astronomers that could account for the acceleration. In The Big Bang chapter, we will look in more detail at the observations that led to this totally unexpected result and explore its implications for the ultimate fate of the universe. In any case, if the Hubble constant is not really a constant when we look over large spans of space and time, then the calculation of galaxy distances using the Hubble constant won\u2019t be accurate. As we shall see in the chapter on The Big Bang, the accurate calculation of distances requires a model for how the Hubble constant has changed over time. The farther away a galaxy is (and the longer ago we are seeing it), the more important it is to include the effects of the change in the Hubble constant. For galaxies within a few billion light-years, however, the assumption that the Hubble constant is indeed constant gives good estimates of distance. Models for an Expanding Universe At first, thinking about Hubble\u2019s law and being a fan of the work of Copernicus and Harlow Shapley, you might be shocked. Are all the galaxies really moving away from us? Is there, after all, something special about our 954 Chapter 26 Galaxies position in the universe? Worry not; the fact that galaxies are receding from us and that more distant galaxies are moving away more rapidly than nearby ones shows only that the universe is expanding uniformly. A uniformly expanding universe is one that is expanding at the same rate everywhere. In such a universe, we and all other observers, no matter where they are located, must observe a proportionality between the velocities and distances of equivalently remote galaxies. (Here, we are ignoring the fact that the Hubble constant is not constant over all time, but if at any given time in the evolution of the universe the Hubble constant has the same value everywhere, this argument still works.) To see why, first imagine a ruler made of stretchable rubber, with the usual lines marked off at each centimeter. Now suppose someone with strong arms grabs each end of the ruler and slowly stretches it so that, say, it doubles in length in 1 minute (Figure 26.16). Consider an intelligent ant sitting on the mark at 2 centimeters\u2014a point that is not at either end nor in the middle of the ruler. He measures how fast other ants, sitting at the 4-, 7-, and 12-centimeter marks, move away from him as the ruler stretches. Figure 26.16. Stretching a Ruler. Ants on a stretching ruler see other ants move away", " from them. The speed with which another ant moves away is proportional to its distance. The ant at 4 centimeters, originally 2 centimeters away from our ant, has doubled its distance in 1 minute; it therefore moved away at a speed of 2 centimeters per minute. The ant at the 7-centimeters mark, which was originally 5 centimeters away from our ant, is now 10 centimeters away; it thus had to move at 5 centimeters per minute. The one that started at the 12-centimeters mark, which was 10 centimeters away from the ant doing the counting, is now 20 centimeters away, meaning it must have raced away at a speed of 10 centimeters per minute. Ants at different distances move away at different speeds, and their speeds are proportional to their distances (just as Hubble\u2019s law indicates for galaxies). Yet, notice in our example that all the ruler was doing was stretching uniformly. Also, notice that none of the ants were actually moving of their own accord, it was the stretching of the ruler that moved them apart. Now let\u2019s repeat the analysis, but put the intelligent ant on some other mark\u2014say, on 7 or 12 centimeters. We discover that, as long as the ruler stretches uniformly, this ant also finds every other ant moving away at a speed proportional to its distance. In other words, the kind of relationship expressed by Hubble\u2019s law can be explained by a uniform stretching of the \u201cworld\u201d of the ants. And all the ants in our simple diagram will see the other ants moving away from them as the ruler stretches. For a three-dimensional analogy, let\u2019s look at the loaf of raisin bread in Figure 26.17. The chef has accidentally put too much yeast in the dough, and when she sets the bread out to rise, it doubles in size during the next hour, causing all the raisins to move farther apart. On the figure, we again pick a representative raisin (that is not at the edge or the center of the loaf) and show the distances from it to several others in the figure (before and after the loaf expands). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 955 Figure 26.17. Expanding Raisin Bread. As the raisin bread rises, the raisins \u201csee\u201d other raisins moving away. More distant raisins move away faster in a uniformly expanding bread. Measure the increases in distance and calculate the speeds for yourself on the raisin bread, just like we did for the ruler. You will see that, since each distance doubles during the hour, each raisin moves away from our selected raisin at a speed proportional to its distance. The same is true no matter which raisin you start with. Our two analogies are useful for clarifying our thinking, but you must not take them literally. On both the ruler and the raisin bread, there are points that are at the end or edge. You can use these to pinpoint the middle of the ruler and the loaf. While our models of the universe have some resemblance to the properties of the ruler and the loaf, the universe has no boundaries, no edges, and no center (all mind-boggling ideas that we will discuss in a later chapter). What is useful to notice about both the ants and the raisins is that they themselves did not \u201ccause\u201d their motion. It isn\u2019t as if the raisins decided to take a trip away from each other and then hopped on a hoverboard to get away. No, in both our analogies, it was the stretching of the medium (the ruler or the bread) that moved the ants or the raisins farther apart. In the same way, we will see in The Big Bang chapter that the galaxies don\u2019t have rocket motors propelling them away from each other. Instead, they are passive participants in the expansion of space. As space stretches, the galaxies are carried farther and farther apart much as the ants and the raisins were. (If this notion of the \u201cstretching\u201d of space surprises or bothers you, now would be a good time to review the information about spacetime in Black Holes and Curved Spacetime. We will discuss these ideas further as our discussion broadens from galaxies to the whole universe.) The expansion of the universe, by the way, does not imply that the individual galaxies and clusters of galaxies themselves are expanding. Neither raisins nor the ants in our analogy grow in size as the loaf expands. Similarly, gravity holds galaxies and clusters of galaxies together, and they get farther away from each other\u2014without themselves changing in size\u2014as the universe expands. 956 Chapter 26 Galaxies CHAPTER 26 REVIEW KEY TERMS elliptical galaxy a galaxy whose shape is an ellipse and that contains no conspicuous interstellar material Hubble constant a constant of proportionality in the", " law relating the velocities of remote galaxies to their distances Hubble\u2019s law a rule that the radial velocities of remove galaxies are proportional to their distances from us irregular galaxy a galaxy without any clear symmetry or pattern; neither a spiral nor an elliptical galaxy mass-to-light ratio the ratio of the total mass of a galaxy to its total luminosity, usually expressed in units of solar mass and solar luminosity; the mass-to-light ratio gives a rough indication of the types of stars contained within a galaxy and whether or not substantial quantities of dark matter are present redshift when lines in the spectra are displaced toward longer wavelengths (toward the red end of the visible spectrum) spiral galaxy a flattened, rotating galaxy with pinwheel-like arms of interstellar material and young stars, winding out from its central bulge type Ia supernova luminosity of about 4.5 \u00d7 109 LSun; can be used to determine distances to galaxies on a large scale a supernova formed by the explosion of a white dwarf in a binary system and reach a SUMMARY 26.1 The Discovery of Galaxies Faint star clusters, clouds of glowing gas, and galaxies all appeared as faint patches of light (or nebulae) in the telescopes available at the beginning of the twentieth century. It was only when Hubble measured the distance to the Andromeda galaxy using cepheid variables with the giant 2.5-meter reflector on Mount Wilson in 1924 that the existence of other galaxies similar to the Milky Way in size and content was established. 26.2 Types of Galaxies The majority of bright galaxies are either spirals or ellipticals. Spiral galaxies contain both old and young stars, as well as interstellar matter, and have typical masses in the range of 109 to 1012 MSun. Our own Galaxy is a large spiral. Ellipticals are spheroidal or slightly elongated systems that consist almost entirely of old stars, with very little interstellar matter. Elliptical galaxies range in size from giants, more massive than any spiral, down to dwarfs, with masses of only about 106 MSun. Dwarf ellipticals are probably the most common type of galaxy in the nearby universe. A small percentage of galaxies with more disorganized shapes are classified as irregulars. Galaxies may change their appearance over time due to collisions with other galaxies or by a change in the rate of star formation. 26.3 Properties of Galaxies The masses of spiral galaxies are determined from measurements of their rates of rotation. The masses of elliptical galaxies are estimated from analyses of the motions of the stars within them. Galaxies can be characterized by their mass-to-light ratios. The luminous parts of galaxies with active star formation typically have mass-to-light ratios in the range of 1 to 10; the luminous parts of elliptical galaxies, which contain only old This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 957 stars, typically have mass-to-light ratios of 10 to 20. The mass-to-light ratios of whole galaxies, including their outer regions, are as high as 100, indicating the presence of a great deal of dark matter. 26.4 The Extragalactic Distance Scale Astronomers determine the distances to galaxies using a variety of methods, including the period-luminosity relationship for cepheid variables; objects such as type Ia supernovae, which appear to be standard bulbs; and the Tully-Fisher relation, which connects the line broadening of 21-cm radiation to the luminosity of spiral galaxies. Each method has limitations in terms of its precision, the kinds of galaxies with which it can be used, and the range of distances over which it can be applied. 26.5 The Expanding Universe The universe is expanding. Observations show that the spectral lines of distant galaxies are redshifted, and that their recession velocities are proportional to their distances from us, a relationship known as Hubble\u2019s law. The rate of recession, called the Hubble constant, is approximately 22 kilometers per second per million light-years. We are not at the center of this expansion: an observer in any other galaxy would see the same pattern of expansion that we do. The expansion described by Hubble\u2019s law is best understood as a stretching of space. FOR FURTHER EXPLORATION Articles Andrews, B. \u201cWhat Are Galaxies Trying to Tell Us?\u201d Astronomy (February 2011): 24. Introduction to our understanding of the shapes and evolution of different types of galaxies. Bothun, G. \u201cBeyond the Hubble Sequence.\u201d Sky & Telescope (May 2000): 36. History and updating of Hubble\u2019s classification scheme. Christianson, G. \u201cMastering the Universe.\u201d Astronomy (February 1999): 60. Brief introduction to Hubble\u2019s life and work. Dalcanton, J. \u201cThe Overlooked Gal", "axies.\u201d Sky & Telescope (April 1998): 28. On low-brightness galaxies, which have been easy to miss. Freedman, W. \u201cThe Expansion Rate and Size of the Universe.\u201d Scientific American (November 1992): 76. Hodge, P. \u201cThe Extragalactic Distance Scale: Agreement at Last?\u201d Sky & Telescope (October 1993): 16. Jones, B. \u201cThe Legacy of Edwin Hubble.\u201d Astronomy (December 1989): 38. Kaufmann, G. and van den Bosch, F. \u201cThe Life Cycle of Galaxies.\u201d Scientific American (June 2002): 46. On galaxy evolution and how it leads to the different types of galaxies. Martin, P. and Friedli, D. \u201cAt the Hearts of Barred Galaxies.\u201d Sky & Telescope (March 1999): 32. On barred spirals. Osterbrock, D. \u201cEdwin Hubble and the Expanding Universe.\u201d Scientific American (July 1993): 84. Russell, D. \u201cIsland Universes from Wright to Hubble.\u201d Sky & Telescope (January 1999) 56. A history of our discovery of galaxies. Smith, R. \u201cThe Great Debate Revisited.\u201d Sky & Telescope (January 1983): 28. On the Shapley-Curtis debate concerning the extent of the Milky Way and the existence of other galaxies. 958 Websites Chapter 26 Galaxies ABC\u2019s of Distance: http://www.astro.ucla.edu/~wright/distance.htm. A concise summary by astronomer Ned Wright of all the different methods we use to get distances in astronomy. Cosmic Times 1929: http://cosmictimes.gsfc.nasa.gov/online_edition/1929Cosmic/index.html. NASA project explaining Hubble\u2019s work and surrounding discoveries as if you were reading newspaper articles. Edwin Hubble: The Man Behind the Name: https://www.spacetelescope.org/about/history/ the_man_behind_the_name/. Concise biography from the people at the Hubble Space Telescope. Edwin Hubble: http://apod.nasa.gov/diamond_jubilee/d_1996/sandage_hubble.html. An article on the life and work of Hubble by his student and successor, Allan Sandage. A bit technical in places, but giving a real picture of the man and the science. NASA Science: Introduction to Galaxies: http://science.nasa.gov/astrophysics/focus-areas/what-are-galaxies/. A brief overview with links to other pages, and recent Hubble Space Telescope discoveries. National Optical Astronomy Observatories Gallery of Galaxies: https://www.noao.edu/image_gallery/ galaxies.html. A collection of images and information about galaxies and galaxy groups of different types. Another impressive archive can be found at the European Southern Observatory site: https://www.eso.org/ public/images/archive/category/galaxies/. Sloan Digital Sky Survey: Introduction to Galaxies: http://skyserver.sdss.org/dr1/en/astro/galaxies/ galaxies.asp. Another brief overview. Universe Expansion: http://hubblesite.org/newscenter/archive/releases/1999/19. The background material here provides a nice chronology of how we discovered and measured the expansion of the universe. Videos Edwin Hubble (Hubblecast Episode 89): http://www.spacetelescope.org/videos/hubblecast89a/. (5:59). Galaxies: An Introduction: https://www.youtube.com/watch?v=HYYgangrkZg. A compilation of several short European videos that first describe galaxies in general and then focus on galaxies in Hubble telescope images (12:48). Hubble\u2019s Views of the Deep Universe: https://www.youtube.com/watch?v=argR2U15w-M. A 2015 public talk by Brandon Lawton of the Space Telescope Science Institute about galaxies and beyond (1:26:20). COLLABORATIVE GROUP ACTIVITIES A. Throughout much of the last century, the 100-inch telescope on Mt. Wilson (completed in 1917) and the 200-inch telescope on Palomar Mountain (completed in 1948) were the only ones large enough to obtain spectra of faint galaxies. Only a handful of astronomers (all male\u2014since, until the 1960s, women were not given time on these two telescopes) were allowed to use these facilities, and in general the observers did not compete with each other but worked on different problems. Now there are many other telescopes, and several different groups do often work on the same problem. For example, two different groups have independently developed the techniques for using supernovae to determine", " the distances to galaxies at high redshifts. Which approach do you think is better for the field of astronomy? Which is more cost effective? Why? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 959 B. A distant relative, whom you invite to dinner so you can share all the exciting things you have learned in your astronomy class, says he does not believe that other galaxies are made up of stars. You come back to your group and ask them to help you respond. What kinds of measurements would you make to show that other galaxies are composed of stars? C. Look at Figure 26.1 with your group. What does the difference in color between the spiral arms and the bulge of Andromeda tell you about the difference in the types of stars that populate these two regions of the galaxy? Which side of the galaxy is closer to us? Why? D. What is your reaction to reading about the discovery of the expanding universe? Discuss how the members of the group feel about a universe \u201cin motion.\u201d Einstein was not comfortable with the notion of a universe that had some overall movement to it, instead of being at rest. He put a kind of \u201cfudge factor\u201d into his equations of general relativity for the universe as a whole to keep it from moving (although later, hearing about Hubble and Humason\u2019s work, he called it \u201cthe greatest blunder\u201d he ever made). Do you share Einstein\u2019s original sense that this is not the kind of universe you feel comfortable with? What do you think could have caused space to be expanding? E. In science fiction, characters sometimes talk about visiting other galaxies. Discuss with your group how realistic this idea is. Even if we had fast spaceships (traveling close to the speed of light, the speed limit of the universe) how likely are we to be able to reach another galaxy? Why? F. Despite his son\u2019s fascination with astronomy in college, Edwin Hubble\u2019s father did not want him to go into astronomy as a profession. He really wanted his son to be a lawyer and pushed him hard to learn the law when he won a fellowship to study abroad. Hubble eventually defied his father and went into astronomy, becoming, as you learned in this chapter, one of the most important astronomers of all time. His dad didn\u2019t live to see his son\u2019s remarkable achievements. Do you think he would have reconciled himself to his son\u2019s career choice if he had? Do you or does anyone in your group or among your friends have to face a choice between the passion in your heart and what others want you to do? Discuss how people in college today are dealing with such choices. EXERCISES Review Questions 1. Describe the main distinguishing features of spiral, elliptical, and irregular galaxies. 2. Why did it take so long for the existence of other galaxies to be established? 3. Explain what the mass-to-light ratio is and why it is smaller in spiral galaxies with regions of star formation than in elliptical galaxies. 4. If we now realize dwarf ellipticals are the most common type of galaxy, why did they escape our notice for so long? 5. What are the two best ways to measure the distance to a nearby spiral galaxy, and how would it be measured? 6. What are the two best ways to measure the distance to a distant, isolated spiral galaxy, and how would it be measured? 7. Why is Hubble\u2019s law considered one of the most important discoveries in the history of astronomy? 960 Chapter 26 Galaxies 8. What does it mean to say that the universe is expanding? What is expanding? For example, is your astronomy classroom expanding? Is the solar system? Why or why not? 9. Was Hubble\u2019s original estimate of the distance to the Andromeda galaxy correct? Explain. 10. Does an elliptical galaxy rotate like a spiral galaxy? Explain. 11. Why does the disk of a spiral galaxy appear dark when viewed edge on? 12. What causes the largest mass-to-light ratio: gas and dust, dark matter, or stars that have burnt out? 13. What is the most useful standard bulb method for determining distances to galaxies? 14. When comparing two isolated spiral galaxies that have the same apparent brightness, but rotate at different rates, what can you say about their relative luminosity? 15. If all distant galaxies are expanding away from us, does this mean we\u2019re at the center of the universe? 16. Is the Hubble constant actually constant? Thought Questions 17. Where might the gas and dust (if any) in an elliptical galaxy come from? 18. Why can we not determine distances to galaxies by the same method used to measure the parallaxes of stars? 19. Which is redder\u2014a spiral galaxy or an elliptical galaxy? 20. Suppose the stars in an ellipt", "ical galaxy all formed within a few million years shortly after the universe began. Suppose these stars have a range of masses, just as the stars in our own galaxy do. How would the color of the elliptical change over the next several billion years? How would its luminosity change? Why? 21. Starting with the determination of the size of Earth, outline a sequence of steps necessary to obtain the distance to a remote cluster of galaxies. (Hint: Review the chapter on Celestial Distances.) 22. Suppose the Milky Way Galaxy were truly isolated and that no other galaxies existed within 100 million light-years. Suppose that galaxies were observed in larger numbers at distances greater than 100 million light-years. Why would it be more difficult to determine accurate distances to those galaxies than if there were also galaxies relatively close by? 23. Suppose you were Hubble and Humason, working on the distances and Doppler shifts of the galaxies. What sorts of things would you have to do to convince yourself (and others) that the relationship you were seeing between the two quantities was a real feature of the behavior of the universe? (For example, would data from two galaxies be enough to demonstrate Hubble\u2019s law? Would data from just the nearest galaxies\u2014in what astronomers call \u201cthe Local Group\u201d\u2014suffice?) 24. What does it mean if one elliptical galaxy has broader spectrum lines than another elliptical galaxy? 25. Based on your analysis of galaxies in Table 26.1, is there a correlation between the population of stars and the quantity of gas or dust? Explain why this might be. 26. Can a higher mass-to-light ratio mean that there is gas and dust present in the system that is being analyzed? Figuring For Yourself 27. According to Hubble\u2019s law, what is the recessional velocity of a galaxy that is 108 light-years away from us? (Assume a Hubble constant of 22 km/s per million light-years.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 26 Galaxies 961 28. A cluster of galaxies is observed to have a recessional velocity of 60,000 km/s. Find the distance to the cluster. (Assume a Hubble constant of 22 km/s per million light-years.) 29. Suppose we could measure the distance to a galaxy using one of the distance techniques listed in Table 26.2 and it turns out to be 200 million light-years. The galaxy\u2019s redshift tells us its recessional velocity is 5000 km/s. What is the Hubble constant? 30. Calculate the mass-to-light ratio for a globular cluster with a luminosity of 106 LSun and 105 stars. (Assume that the average mass of a star in such a cluster is 1 MSun.) 31. Calculate the mass-to-light ratio for a luminous star of 100 MSun having the luminosity of 106 LSun. 962 Chapter 26 Galaxies This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 963 27 ACTIVE GALAXIES, QUASARS, AND SUPERMASSIVE BLACK HOLES Figure 27.1. Hubble Ultra-Deep Field. The deepest picture of the sky in visible light (left) shows huge numbers of galaxies in a tiny patch of sky, only 1/100 the area of the full Moon. In contrast, the deepest picture of the sky taken in X-rays (right) shows large numbers of point-like quasars, which astronomers have shown are supermassive black holes at the very centers of galaxies. (credit left: modification of work by NASA, ESA, H. Teplitz and M. Rafelski (IPAC/Caltech), A. Koekemoer (STScI), R. Windhorst (Arizona State University), and Z. Levay (STScI); credit right: modification of work by ESO/Mario Nonino, Piero Rosati, ESO GOODS Team) Chapter Outline 27.1 Quasars 27.2 Supermassive Black Holes: What Quasars Really Are 27.3 Quasars as Probes of Evolution in the Universe Thinking Ahead During the first half of the twentieth century, astronomers viewed the universe of galaxies as a mostly peaceful place. They assumed that galaxies formed billions of years ago and then evolved slowly as the populations of stars within them formed, aged, and died. That placid picture completely changed in the last few decades of the twentieth century. Today, astronomers can see that the universe is often shaped by violent events, including cataclysmic explosions of supernovae, collisions of whole galaxies, and the tremendous outpouring of energy as matter interacts in the environment surrounding very massive black holes. The key event that began to change", " our view of the universe was the discovery of a new class of objects: quasars. 27.1 QUASARS Learning Objectives By the end of this section, you will be able to: Describe how quasars were discovered Explain how astronomers determined that quasars are at the distances implied by their redshifts 964 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes Justify the statement that the enormous amount of energy produced by quasars is generated in a very small volume of space The name \u201c quasars\u201d started out as short for \u201cquasi-stellar radio sources\u201d (here \u201cquasi-stellar\u201d means \u201csort of like stars\u201d). The discovery of radio sources that appeared point-like, just like stars, came with the use of surplus World War II radar equipment in the 1950s. Although few astronomers would have predicted it, the sky turned out to be full of strong sources of radio waves. As they improved the images that their new radio telescopes could make, scientists discovered that some radio sources were in the same location as faint blue \u201cstars.\u201d No known type of star in our Galaxy emits such powerful radio radiation. What then were these \u201cquasi-stellar radio sources\u201d? Redshifts: The Key to Quasars The answer came when astronomers obtained visible-light spectra of two of those faint \u201cblue stars\u201d that were strong sources of radio waves (Figure 27.2). Spectra of these radio \u201cstars\u201d only deepened the mystery: they had emission lines, but astronomers at first could not identify them with any known substance. By the 1960s, astronomers had a century of experience in identifying elements and compounds in the spectra of stars. Elaborate tables had been published showing the lines that each element would produce under a wide range of conditions. A \u201cstar\u201d with unidentifiable lines in the ordinary visible light spectrum had to be something completely new. Figure 27.2. Typical Quasar. The arrow in this image marks the quasar known by its catalog number, PKS 1117-248. Note that nothing in this image distinguishes the quasar from an ordinary star. Its spectrum, however, shows that it is moving away from us at a speed of 36% the speed of light, or 67,000 miles per second. In contrast, the maximum speed observed for any star is only a few hundred miles per second. (credit: modification of work by WIYN Telescope, Kitt Peak National Observatory, NOAO) In 1963 at Caltech\u2019s Palomar Observatory, Maarten Schmidt (Figure 27.3) was puzzling over the spectrum of one of the radio stars, which was named 3C 273 because it was the 273rd entry in the third Cambridge catalog of radio sources (part (b) of Figure 27.3). There were strong emission lines in the spectrum, and Schmidt recognized that they had the same spacing between them as the Balmer lines of hydrogen (see Radiation and Spectra). But the lines in 3C 273 were shifted far to the red of the wavelengths at which the Balmer lines are normally located. Indeed, these lines were at such long wavelengths that if the redshifts were attributed to the Doppler effect, 3C 273 was receding from us at a speed of 45,000 kilometers per second, or about 15% the speed of light! Since stars don\u2019t show Doppler shifts this large, no one had thought of considering high redshifts to be the cause of the strange spectra. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 965 Figure 27.3. Quasar Pioneers and Quasar 3C 273. (a) Maarten Schmidt (left), who solved the puzzle of the quasar spectra in 1963, shares a joke in this 1987 photo with Allan Sandage, who took the first spectrum of a quasar. Sandage was also instrumental in measuring the value of Hubble\u2019s constant. (b) This is the first quasar for which a redshift was measured. The redshift showed that the light from it took about 2.5 billion years to reach us. Despite this great distance, it is still one of the quasars closest to the Milky Way Galaxy. Note also the faint streak going toward the upper left from the quasar. Some quasars, like 3C 273, eject super-fast jets of material. The jet from 3C 273 is about 200,000 lightyears long. (credit a: modification of work by Andrew Fraknoi; credit b: modification of work by ESA/Hubble/NASA) The puzzling emission lines in other star-like radio sources were then reexamined to see if they, too,", " might be well-known lines with large redshifts. This proved to be the case, but the other objects were found to be receding from us at even greater speeds. Their astounding speeds showed that the radio \u201cstars\u201d could not possibly be stars in our own Galaxy. Any true star moving at more than a few hundred kilometers per second would be able to overcome the gravitational pull of the Galaxy and completely escape from it. (As we shall see later in this chapter, astronomers eventually discovered that there was also more to these \u201cstars\u201d than just a point of light.) It turns out that these high-velocity objects only look like stars because they are compact and very far away. Later, astronomers discovered objects with large redshifts that appear star-like but have no radio emission. Observations also showed that quasars were bright in the infrared and X-ray bands too, and not all these X-ray or infrared-bright quasars could be seen in either the radio or the visible-light bands of the spectrum. Today, all these objects are referred to as quasi-stellar objects (QSOs), or, as they are more popularly known, quasars. (The name was also soon appropriated by a manufacturer of home electronics.) Read an interview (https://openstax.org/l/30SchmidtIntv) with Maarten Schmidt on the fiftieth anniversary of his insight about the spectrum of quasars and their redshifts. Over a million quasars have now been discovered, and spectra are available for over a hundred thousand. All these spectra show redshifts, none show blueshifts, and their redshifts can be very large. Yet in a photo they look just like stars (Figure 27.4). 966 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes Figure 27.4. Typical Quasar Imaged by the Hubble Space Telescope. One of these two bright \u201cstars\u201d in the middle is in our Galaxy, while the other is a quasar 9 billion light-years away. From this picture alone, there\u2019s no way to say which is which. (The quasar is the one in the center of the picture.) (credit: Charles Steidel (CIT)/NASA/ESA) In the record-holding quasars, the first Lyman series line of hydrogen, with a laboratory wavelength of 121.5 nanometers in the ultraviolet portion of the spectrum, is shifted all the way through the visible region to the infrared. At such high redshifts, the simple formula for converting a Doppler shift to speed (Radiation and Spectra) must be modified to take into account the effects of the theory of relativity. If we apply the relativistic form of the Doppler shift formula, we find that these redshifts correspond to velocities of about 96% of the speed of light Recession Speed of a Quasar The formula for the Doppler shift, which astronomers denote by the letters z, is z = \u0394\u03bb \u03bb = v c where \u03bb is the wavelength emitted by a source of radiation that is not moving, \u0394\u03bb is the difference between that wavelength and the wavelength we measure, v is the speed with which the source moves away, and c (as usual) is the speed of light. A line in the spectrum of a galaxy is at a wavelength of 393 nanometers (nm, or 10\u20139 m) when the source is at rest. Let\u2019s say the line is measured to be longer than this value (redshifted) by 7.86 nm. Then its redshift z = 7.86 nm 393 nm = 0.02, so its speed away from us is 2% of the speed of light \u239b v c = 0.02\u239e \u23a0. \u239d This formula is fine for galaxies that are relatively nearby and are moving away from us slowly in the expansion of the universe. But the quasars and distant galaxies we discuss in this chapter are moving This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 967 away at speeds close to the speed of light. In that case, converting a Doppler shift (redshift) to a distance must include the effects of the special theory of relativity, which explains how measurements of space and time change when we see things moving at high speeds. The details of how this is done are way beyond the level of this text, but we can share with you the relativistic formula for the Doppler shift: v c = (z + 1)2 \u2212 1 (z + 1)2 + 1 Let\u2019s do an example. Suppose a distant quasar has a redshift of 5. At what fraction of the speed", " of light is the quasar moving away? Solution We calculate the following: v c = (5 + 1)2 \u2212 1 (5 + 1)2 + 1 = 36 \u2212 1 36 + 1 = 35 37 = 0.946 The quasar is thus receding from us at about 95% the speed of light. Check Your Learning Several lines of hydrogen absorption in the visible spectrum have rest wavelengths of 410 nm, 434 nm, 486 nm, and 656 nm. In a spectrum of a distant galaxy, these same lines are observed to have wavelengths of 492 nm, 521 nm, 583 nm, and 787 nm respectively. What is the redshift of this galaxy? What is the recession speed of this galaxy? Answer: Because this is the same galaxy, we could pick any one of the four wavelengths and calculate how much it has shifted. If we use a rest wavelength of 410 nm and compare it to the shifted wavelength of 492 nm, we see that z = \u0394\u03bb \u03bb = (492 nm \u2212 410 nm) 410 nm = 82 nm 410 nm = 0.20 In the classical view, this galaxy is receding at 20% of the speed of light; however, at 20% of the speed of light, relativistic effects are starting to become important. So, using the relativistic Doppler equation, we compute the true recession rate as v c = (z + 1)2 \u2212 1 (z + 1)2 + 1 = (0.2 + 1)2 \u2212 1 (0.2 + 1)2 + 1 = 1.44 \u2212 1 1.44 + 1 = 0.44 2.44 = 0.18 Therefore, the actual recession speed is only 18% of the speed of light. While this may not initially seem like a big difference from the classical measurement, there is already an 11% deviation between the classical and the relativistic solutions; and at greater recession speeds, the divergence between the classical and relativistic speeds increases rapidly! Quasars Obey the Hubble Law The first question astronomers asked was whether quasars obeyed the Hubble law and were really at the large distances implied by their redshifts. If they did not obey the rule that large redshift means large distance, then they could be much closer, and their luminosity could be a lot less. One straightforward way to show that 968 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes quasars had to obey the Hubble law was to demonstrate that they were actually part of galaxies, and that their redshift was the same as the galaxy that hosted them. Since ordinary galaxies do obey the Hubble law, anything within them would be subject to the same rules. Observations with the Hubble Space Telescope provided the strongest evidence showing that quasars are located at the centers of galaxies. Hints that this is true had been obtained with ground-based telescopes, but space observations were required to make a convincing case. The reason is that quasars can outshine their entire galaxies by factors of 10 to 100 or even more. When this light passes through Earth\u2019s atmosphere, it is blurred by turbulence and drowns out the faint light from the surrounding galaxy\u2014much as the bright headlights from an oncoming car at night make it difficult to see anything close by. The Hubble Space Telescope, however, is not affected by atmospheric turbulence and can detect the faint glow from some of the galaxies that host quasars (Figure 27.5). Quasars have been found in the cores of both spiral and elliptical galaxies, and each quasar has the same redshift as its host galaxy. A wide range of studies with the Hubble Space Telescope now clearly demonstrate that quasars are indeed far away. If so, they must be producing a truly impressive amount of energy to be detectable as points of light that are much brighter than their galaxy. Interestingly, many quasar host galaxies are found to be involved in a collision with a second galaxy, providing, as we shall see, an important clue to the source of their prodigious energy output. Figure 27.5. Quasar Host Galaxies. The Hubble Space Telescope reveals the much fainter \u201chost\u201d galaxies around quasars. The top left image shows a quasar that lies at the heart of a spiral galaxy 1.4 billion light-years from Earth. The bottom left image shows a quasar that lies at the center of an elliptical galaxy some 1.5 billion light-years from us. The middle images show remote pairs of interacting galaxies, one of which harbors a quasar. Each of the right images shows long tails of gas and dust streaming away from a galaxy that contains a quasar. Such tails are produced when one galaxy collides with another. (credit: modification of work by John Bahcall, Mike Disney, NASA) The Size of the Energy Source Given their large distances, quasars have to be extremely luminous to be visible to", " us at all\u2014far brighter than any normal galaxy. In visible light alone, most are far more energetic than the brightest elliptical galaxies. But, as we saw, quasars also emit energy at X-ray and ultraviolet wavelengths, and some are radio sources as well. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 969 When all their radiation is added together, some QSOs have total luminosities as large as a hundred trillion Suns (1014 LSun), which is 10 to 100 times the brightness of luminous elliptical galaxies. Finding a mechanism to produce the large amount of energy emitted by a quasar would be difficult under any circumstances. But there is an additional problem. When astronomers began monitoring quasars carefully, they found that some vary in luminosity on time scales of months, weeks, or even, in some cases, days. This variation is irregular and can change the brightness of a quasar by a few tens of percent in both its visible light and radio output. Think about what such a change in luminosity means. A quasar at its dimmest is still more brilliant than any normal galaxy. Now imagine that the brightness increases by 30% in a few weeks. Whatever mechanism is responsible must be able to release new energy at rates that stagger our imaginations. The most dramatic changes in quasar brightness are equivalent to the energy released by 100,000 billion Suns. To produce this much energy we would have to convert the total mass of about ten Earths into energy every minute. Moreover, because the fluctuations occur in such short times, the part of a quasar that is varying must be smaller than the distance light travels in the time it takes the variation to occur\u2014typically a few months. To see why this must be so, let\u2019s consider a cluster of stars 10 light-years in diameter at a very large distance from Earth (see Figure 27.6, in which Earth is off to the right). Suppose every star in this cluster somehow brightens simultaneously and remains bright. When the light from this event arrives at Earth, we would first see the brighter light from stars on the near side; 5 years later we would see increased light from stars at the center. Ten years would pass before we detected more light from stars on the far side. Figure 27.6. How the Size of a Source Affects the Timescale of Its Variability. This diagram shows why light variations from a large region in space appear to last for an extended period of time as viewed from Earth. Suppose all the stars in this cluster, which is 10 light-years across, brighten simultaneously and instantaneously. From Earth, star A will appear to brighten 5 years before star B, which in turn will appear to brighten 5 years earlier than star C. It will take 10 years for an Earth observer to get the full effect of the brightening. Even though all stars in the cluster brightened at the same time, the fact that the cluster is 10 light-years wide means that 10 years must elapse before the increased light from every part of the cluster reaches us. From Earth we would see the cluster get brighter and brighter, as light from more and more stars began to reach us. Not until 10 years after the brightening began would we see the cluster reach maximum brightness. In other words, if an extended object suddenly flares up, it will seem to brighten over a period of time equal to the time it takes light to travel across the object from its far side. We can apply this idea to brightness changes in quasars to estimate their diameters. Because quasars typically vary (get brighter and dimmer) over periods of a few months, the region where the energy is generated can be 970 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes no larger than a few light-months across. If it were larger, it would take longer than a few months for the light from the far side to reach us. How large is a region of a few light-months? Pluto, usually the outermost (dwarf) planet in our solar system, is about 5.5 light-hours from us, while the nearest star is 4 light-years away. Clearly a region a few light months across is tiny relative to the size of the entire Galaxy. And some quasars vary even more rapidly, which means their energy is generated in an even smaller region. Whatever mechanism powers the quasars must be able to generate more energy than that produced by an entire galaxy in a volume of space that, in some cases, is not much larger than our solar system. Earlier Evidence Even before the discovery of quasars, there had been hints that something very strange was going on in the centers of at least some galaxies. Back in 1918, American astronomer Heber Curtis used the large L", "ick Observatory telescope to photograph the galaxy Messier 87 in the constellation Virgo. On that photograph, he saw what we now call a jet coming from the center, or nucleus, of the galaxy (Figure 27.7). This jet literally and figuratively pointed to some strange activity going on in that galaxy nucleus. But he had no idea what it was. No one else knew what to do with this space oddity either. The random factoid that such a central jet existed lay around for a quarter century, until Carl Seyfert, a young astronomer at Mount Wilson Observatory, also in California, found half a dozen galaxies with extremely bright nuclei that were almost stellar, rather than fuzzy in appearance like most galaxy nuclei. Using spectroscopy, he found that these nuclei contain gas moving at up to two percent the speed of light. That may not sound like much, but it is 6 million miles per hour, and more than 10 times faster than the typical motions of stars in galaxies. Figure 27.7. M87 Jet. Streaming out like a cosmic searchlight from the center of the galaxy, M87 is one of nature\u2019s most amazing phenomena, a huge jet of electrons and other particles traveling at nearly the speed of light. In this Hubble Space Telescope image, the blue of the jet contrasts with the yellow glow from the combined light of billions of unseen stars and yellow, point-like globular clusters that make up the galaxy (at the upper left). As we shall see later in this chapter, the jet, which is several thousand light-years long, originates in a disk of superheated gas swirling around a giant black hole at the center of M87. The light that we see is produced by electrons twisting along magnetic field lines in the jet, a process known as synchrotron radiation, which gives the jet its bluish tint. The jet in M87 can be observed in X-ray, radio, and visible light, as shown in the bottom three images. At the extreme left of each bottom image, we see the bright galactic nucleus harboring a supermassive black hole. (credit top: modification of work by NASA, The Hubble Heritage Team(STScI/AURA); credit bottom: modification of work by X-ray: H. Marshall (MIT), et al., CXC, NASA; Radio: F. Zhou, F. Owen (NRAO), J. Biretta (STScI); Optical: E. Perlman (UMBC), et al.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 971 After decades of study, astronomers identified many other strange objects beyond our Milky Way Galaxy; they populate a whole \u201czoo\u201d of what are now called active galaxies or active galactic nuclei (AGN). Astronomers first called them by many different names, depending on what sorts of observations discovered each category, but now we know that we are always looking at the same basic mechanism. What all these galaxies have in common is some activity in their nuclei that produces an enormous amount of energy in a very small volume of space. In the next section, we describe a model that explains all these galaxies with strong central activity\u2014both the AGNs and the QSOs To see a jet for yourself, check out a time-lapse video (https://openstax.org/l/30timelapsejet) of the jet ejected from NGC 3862. 27.2 SUPERMASSIVE BLACK HOLES: WHAT QUASARS REALLY ARE Learning Objectives By the end of this section, you will be able to: Describe the characteristics common to all quasars Justify the claim that supermassive black holes are the source of the energy emitted by quasars (and AGNs) Explain how a quasar\u2019s energy is produced In order to find a common model for quasars (and their cousins, the AGNs), let\u2019s first list the common characteristics we have been describing\u2014and add some new ones: \u2022 Quasars are hugely powerful, emitting more power in radiated light than all the stars in our Galaxy combined. \u2022 Quasars are tiny, about the size of our solar system (to astronomers, that is really small!). \u2022 Some quasars are observed to be shooting out pairs of straight jets at close to the speed of light, in a tight beam, to distances far beyond the galaxies they live in. These jets are themselves powerful sources of radio and gamma-ray radiation. \u2022 Because quasars put out so much power from such a small region, they can\u2019t be powered by nuclear fusion the way stars are; they must use some process that is far more efficient. \u2022 As we shall see later in this chapter, quasars were much more common when the universe was young than they are", " today. That means they must have been able to form in the first billion years or so after the universe began to expand. The readers of this text are in a much better position than the astronomers who discovered quasars in the 1960s to guess what powers the quasars. That\u2019s because the key idea in solving the puzzle came from observations of the black holes. The discovery of the first stellar mass black hole in the binary system Cygnus X-1 was announced in 1971, several years after the discovery of quasars. Proof that there is a black hole at the center of our own Galaxy came even later. Back when astronomers first began trying to figure out what powered quasars, 972 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes black holes were simply one of the more exotic predictions of the general theory of relativity that still waited to be connected to the real world. It was only as proof of the existence of black holes accumulated over several decades that it became clearer that only supermassive black holes could account for all the observed properties of quasars and AGNs. As we saw in The Milky Way Galaxy, our own Galaxy has a black hole in its center, and the energy is emitted from a small central region. While our black hole doesn\u2019t have the mass or energy of the quasar black holes, the mechanism that powers them is similar. The evidence now shows that most\u2014and probably all\u2014elliptical galaxies and all spirals with nuclear bulges have black holes at their centers. The amount of energy emitted by material near the black hole depends on two things: the mass of the black hole and the amount of matter that is falling into it. If a black hole with a billion Suns\u2019 worth of mass inside (109 MSun) accretes (gathers) even a relatively modest amount of additional material\u2014say, about 10 MSun per year\u2014then (as we shall see) it can, in the process, produce as much energy as a thousand normal galaxies. This is enough to account for the total energy of a quasar. If the mass of the black hole is smaller than a billion solar masses or the accretion rate is low, then the amount of energy emitted can be much smaller, as it is in the case of the Milky Way Watch a video (https://openstax.org/l/30mataccrsupblh) of an artist\u2019s impression of matter accreting around a supermassive black hole. Observational Evidence for Black Holes In order to prove that a black hole is present at the center of a galaxy, we must demonstrate that so much mass is crammed into so small a volume that no normal objects\u2014massive stars or clusters of stars\u2014could possibly account for it (just as we did for the black hole in the Milky Way). We already know from observations (discussed in Black Holes and Curved Spacetime) that an accreting black hole is surrounded by a hot accretion disk with gas and dust that swirl around the black hole before it falls in. If we assume that the energy emitted by quasars is also produced by a hot accretion disk, then, as we saw in the previous section, the size of the disk must be given by the time the quasar energy takes to vary. For quasars, the emission in visible light varies on typical time scales of 5 to 2000 days, limiting the size of the disk to that many light-days. In the X-ray band, quasars vary even more rapidly, so the light travel time argument tells us that this more energetic radiation is generated in an even smaller region. Therefore, the mass around which the accretion disk is swirling must be confined to a space that is even smaller. If the quasar mechanism involves a great deal of mass, then the only astronomical object that can confine a lot of mass into a very small space is a black hole. In a few cases, it turns out that the X-rays are emitted from a region just a few times the size of the black hole event horizon. The next challenge, then, is to \u201cweigh\u201d this central mass in a quasar. In the case of our own Galaxy, we used observations of the orbits of stars very close to the galactic center, along with Kepler\u2019s third law, to estimate the mass of the central black hole (The Milky Way Galaxy). In the case of distant galaxies, we cannot measure the orbits of individual stars, but we can measure the orbital speed of the gas in the rotating accretion disk. The Hubble Space Telescope is especially well suited to this task because it is above the blurring of Earth\u2019s This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 973 atmosphere and can obtain spectra very close to the", " bright central regions of active galaxies. The Doppler effect is then used to measure radial velocities of the orbiting material and so derive the speed with which it moves around. One of the first galaxies to be studied with the Hubble Space Telescope is our old favorite, the giant elliptical M87. Hubble Space Telescope images showed that there is a disk of hot (10,000 K) gas swirling around the center of M87 (Figure 27.8). It was surprising to find hot gas in an elliptical galaxy because this type of galaxy is usually devoid of gas and dust. But the discovery was extremely useful for pinning down the existence of the black hole. Astronomers measured the Doppler shift of spectral lines emitted by this gas, found its speed of rotation, and then used the speed to derive the amount of mass inside the disk\u2014applying Kepler\u2019s third law. Figure 27.8. Evidence for a Black Hole at the Center of M87. The disk of whirling gas at right was discovered at the center of the giant elliptical galaxy M87 with the Hubble Space Telescope. Observations made on opposite sides of the disk show that one side is approaching us (the spectral lines are blueshifted by the Doppler effect) while the other is receding (lines redshifted), a clear indication that the disk is rotating. The rotation speed is about 550 kilometers per second or 1.2 million miles per hour. Such a high rotation speed is evidence that there is a very massive black hole at the center of M87. (credit: modification of work by Holland Ford, STScI/JHU; Richard Harms, Linda Dressel, Ajay K. Kochhar, Applied Research Corp.; Zlatan Tsvetanov, Arthur Davidsen, Gerard Kriss, Johns Hopkins; Ralph Bohlin, George Hartig, STScI; Bruce Margon, University of Washington in Seattle; NASA) Modern estimates show that there is a mass of at least 3.5 billion MSun concentrated in a tiny region at the very center of M87. So much mass in such a small volume of space must be a black hole. Let\u2019s stop for a moment and take in this figure: a single black hole that has swallowed enough material to make 3.5 billion stars like the Sun. Few astronomical measurements have ever led to so mind-boggling a result. What a strange environment the neighborhood of such a supermassive black hole must be. Another example is shown in Figure 27.9. Here, we see a disk of dust and gas that surrounds a 300-million-MSun black hole in the center of an elliptical galaxy. (The bright spot in the center is produced by the combined light of stars that have been pulled close together by the gravitational force of the black hole.) The mass of the black hole was again derived from measurements of the rotational speed of the disk. The gas in the disk is moving around at 155 kilometers per second at a distance of only 186 light-years from its center. Given the pull of the mass at the center, we expect that the whole dust disk should be swallowed by the black hole in several billion years. 974 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes Figure 27.9. Another Galaxy with a Black-Hole Disk. The ground-based image shows an elliptical galaxy called NGC 7052 located in the constellation of Vulpecula, almost 200 million light-years from Earth. At the galaxy\u2019s center (right) is a dust disk roughly 3700 light-years in diameter. The disk rotates like a giant merry-go-round: gas in the inner part (186 light-years from the center) whirls around at a speed of 155 kilometers per second (341,000 miles per hour). From these measurements and Kepler\u2019s third law, it is possible to estimate that the disk is orbiting around a central black hole with a mass of 300 million Suns. (credit: modification of work by Roeland P. van der Marel (STScI), Frank C. van den Bosch (University of Washington), NASA) But do we have to accept black holes as the only explanation of what lies at the center of these galaxies? What else could we put in such a small space other than a giant black hole? The alternative is stars. But to explain the masses in the centers of galaxies without a black hole we need to put at least a million stars in a region the size of the solar system. To fit, they would have be only 2 star diameters apart. Collisions between stars would happen all the time. And these collisions would lead to mergers of stars, and very soon the one giant star that they form would collapse into a black hole. So there is really no escape: only a black hole can fit so much mass into so small a space. As we saw earlier, observations now show that all the", " galaxies with a spherical concentration of stars\u2014either elliptical galaxies or spiral galaxies with nuclear bulges (see the chapter on Galaxies)\u2014harbor one of these giant black holes at their centers. Among them is our neighbor spiral galaxy, the Andromeda galaxy, M31. The masses of these central black holes range from a just under a million up to at least 30 billion times the mass of the Sun. Several black holes may be even more massive, but the mass estimates have large uncertainties and need verification. We call these black holes \u201csupermassive\u201d to distinguish them from the much smaller black holes that form when some stars die (see The Death of Stars). So far, the most massive black holes from stars\u2014those detected through gravitational waves detected by LIGO\u2014have masses only a little over 30 solar masses. Energy Production around a Black Hole By now, you may be willing to entertain the idea that huge black holes lurk at the centers of active galaxies. But we still need to answer the question of how such a black hole can account for one of the most powerful sources of energy in the universe. As we saw in Black Holes and Curved Spacetime, a black hole itself can radiate no energy. Any energy we detect from it must come from material very close to the black hole, but not inside its event horizon. In a galaxy, a central black hole (with its strong gravity) attracts matter\u2014stars, dust, and gas\u2014orbiting in the dense nuclear regions. This matter spirals in toward the spinning black hole and forms an accretion disk of material around it. As the material spirals ever closer to the black hole, it accelerates and becomes compressed, This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 975 heating up to temperatures of millions of degrees. Such hot matter can radiate prodigious amounts of energy as it falls in toward the black hole. To convince yourself that falling into a region with strong gravity can release a great deal of energy, imagine dropping a printed version of your astronomy textbook out the window of the ground floor of the library. It will land with a thud, and maybe give a surprised pigeon a nasty bump, but the energy released by its fall will not be very great. Now take the same book up to the fifteenth floor of a tall building and drop it from there. For anyone below, astronomy could suddenly become a deadly subject; when the book hits, it does so with a great deal of energy. Dropping things from far away into the much stronger gravity of a black hole is much more effective in turning the energy released by infall into other forms of energy. Just as the falling book can heat up the air, shake the ground, or produce sound energy that can be heard some distance away, so the energy of material falling toward a black hole can be converted to significant amounts of electromagnetic radiation. What a black hole has to work with is not textbooks but streams of infalling gas. If a dense blob of gas moves through a thin gas at high speed, it heats up as it slows by friction. As it slows down, kinetic (motion) energy is turned into heat energy. Just like a spaceship reentering the atmosphere (Figure 27.10), gas approaching a black hole heats up and glows where it meets other gas. But this gas, as it approaches the event horizon, reaches speeds of 10% the speed of light and more. It therefore gets far, far hotter than a spaceship, which reaches no more than about 1500 K. Indeed, gas near a supermassive black hole reaches a temperature of about 150,000 K, about 100 times hotter than a spaceship returning to Earth. It can even get so hot\u2014millions of degrees\u2014that it radiates X-rays. Figure 27.10. Friction in Earth\u2019s Atmosphere. In this artist\u2019s impression, the rapid motion of a spacecraft (the Apollo mission reentry capsule) through the atmosphere compresses and heats the air ahead of it, which heats the spacecraft in turn until it glows red hot. Pushing on the air slows down the spacecraft, turning the kinetic energy of the spacecraft into heat. Fast-moving gas falling into a quasar heats up in a similar way. (credit: modification of work by NASA) The amount of energy that can be liberated this way is enormous. Einstein showed that mass and energy are interchangeable with his famous formula E = mc2 (see The Sun: A Nuclear Powerhouse). A hydrogen bomb releases just 1% of that energy, as does a star. Quasars are much more efficient than that. The energy released falling to the event horizon of a black hole can easily reach 10% or, in the extreme theoretical limit, 32%, of that energy. (Unlike the hydrogen atoms in a bomb or a star, the gas falling into the black hole is not", " actually losing mass from its atoms to free up the energy; the energy is produced just because the gas is falling closer and closer to the black hole.) This huge energy release explains how a tiny volume like the region around a black hole can release as much power as a whole galaxy. But to radiate all that energy, instead of just falling inside 976 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes the event horizon with barely a peep, the hot gas must take the time to swirl around the star in the accretion disk and emit some of its energy. Most black holes don\u2019t show any signs of quasar emission. We call them \u201cquiescent.\u201d But, like sleeping dragons, they can be woken up by being roused with a fresh supply of gas. Our own Milky Way black hole is currently quiescent, but it may have been a quasar just a few million years ago (Figure 27.11). Two giant bubbles that extend 25,000 light-years above and below the galactic center are emitting gamma rays. Were these produced a few million years ago when a significant amount of matter fell into the black hole at the center of the galaxy? Astronomers are still working to understand what remarkable event might have formed these enormous bubbles. Figure 27.11. Fermi Bubbles in the Galaxy. Giant bubbles shining in gamma-ray light lie above and below the center of the Milky Way Galaxy, as seen by the Fermi satellite. (The gamma-ray and X-ray image is superimposed on a visible-light image of the inner parts of our Galaxy.) The bubbles may be evidence that the supermassive black hole at the center of our Galaxy was a quasar a few million years ago. (credit: modification of work by NASA\u2019s Goddard Space Flight Center) The physics required to account for the exact way in which the energy of infalling material is converted to radiation near a black hole is far more complicated than our simple discussion suggests. To understand what happens in the \u201crough and tumble\u201d region around a massive black hole, astronomers and physicists must resort to computer simulations (and they require supercomputers, fast machines capable of awesome numbers of calculations per second). The details of these models are beyond the scope of our book, but they support the basic description presented here. Radio Jets So far, our model seems to explain the central energy source in quasars and active galaxies. But, as we have seen, there is more to quasars and other active galaxies than the point-like energy source. They can also have long jets that glow with radio waves, light, and sometimes even X-rays, and that extend far beyond the limits of the parent galaxy. Can we find a way for our black hole and its accretion disk to produce these jets of energetic particles as well? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 977 Many different observations have now traced these jets to within 3 to 30 light-years of the parent quasar or galactic nucleus. While the black hole and accretion disk are typically smaller than 1 light-year, we nevertheless presume that if the jets come this close, they probably originate in the vicinity of the black hole. Another characteristic of the jets we need to explain is that they contain matter moving close to the speed of light. Why are energetic electrons and other particles near a supermassive black hole ejected into jets, and often into two oppositely directed jets, rather than in all directions? Again, we must use theoretical models and supercomputer simulations of what happens when a lot of material whirls inward in a crowded black hole accretion disk. Although there is no agreement on exactly how jets form, it has become clear that any material escaping from the neighborhood of the black hole has an easier time doing so perpendicular to the disk. In some ways, the inner regions of black hole accretion disks resemble a baby that is just learning to eat by herself. As much food as goes into the baby\u2019s mouth can sometimes wind up being spit out in various directions. In the same way, some of the material whirling inward toward a black hole finds itself under tremendous pressure and orbiting with tremendous speed. Under such conditions, simulations show that a significant amount of material can be flung outward\u2014not back along the disk, where more material is crowding in, but above and below the disk. If the disk is thick (as it tends to be when a lot of material falls in quickly), it can channel the outrushing material into narrow beams perpendicular to the disk (Figure 27.12). Figure 27.12. Models of Accretion Disks. These schematic drawings show what accretion disks might look like around large black holes for (a) a thin accretion disk and (b) a \u201cfat\u201d disk\u2014the", " type needed to account for channeling the outflow of hot material into narrow jets oriented perpendicular to the disk. Figure 27.13 shows observations of an elliptical galaxy that behaves in exactly this way. At the center of this active galaxy, there is a ring of dust and gas about 400 light-years in diameter, surrounding a 1.2-billion-MSun black hole. Radio observations show that two jets emerge in a direction perpendicular to the ring, just as the model predicts. 978 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes Figure 27.13. Jets and Disk in an Active Galaxy. The picture on the left shows the active elliptical galaxy NGC 4261, which is located in the Virgo Cluster at a distance of about 100 million light-years. The galaxy itself\u2014the white circular region in the center\u2014is shown the way it looks in visible light, while the jets are seen at radio wavelengths. A Hubble Space Telescope image of the central portion of the galaxy is shown on the right. It contains a ring of dust and gas about 800 light-years in diameter, surrounding a supermassive black hole. Note that the jets emerge from the galaxy in a direction perpendicular to the plane of the ring. (credit: modification of work by ESA/HST Quasars and the Attitudes of Astronomers The discovery of quasars in the early 1960s was the first in a series of surprises astronomers had in store. Within another decade they would find neutron stars (in the form of pulsars), the first hints of black holes (in binary X-ray sources), and even the radio echo of the Big Bang itself. Many more new discoveries lay ahead. As Maarten Schmidt reminisced in 1988, \u201cThis had, I believe, a profound impact on the conduct of those practicing astronomy. Before the 1960s, there was much authoritarianism in the field. New ideas expressed at meetings would be instantly judged by senior astronomers and rejected if too far out.\u201d We saw a good example of this in the trouble Chandrasekhar had in finding acceptance for his ideas about the death of stars with cores greater than 1.4 MSun (see the feature box on Subrahmanyan Chandrasekhar). \u201cThe discoveries of the 1960s,\u201d Schmidt continued, \u201cwere an embarrassment, in the sense that they were totally unexpected and could not be evaluated immediately. In reaction to these developments, an attitude has evolved where even outlandish ideas in astronomy are taken seriously. Given our lack of solid knowledge in extragalactic astronomy, this is probably to be preferred over authoritarianism.\u201d[1] That is not to say that astronomers (being human) don\u2019t continue to have prejudices and preferences. For example, a small group of astronomers who thought that the redshifts of quasars were not connected with their distances (which was definitely a minority opinion) often felt excluded from meetings or from access to telescopes in the 1960s and 1970s. It\u2019s not so clear that they actually were This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 979 excluded, as much as that they felt the very difficult pressure of knowing that most of their colleagues strongly disagreed with them. As it turned out, the evidence\u2014which must ultimately decide all scientific questions\u2014was not on their side either. But today, as better instruments bring solutions to some problems and starkly illuminate our ignorance about others, the entire field of astronomy seems more open to discussing unusual ideas. Of course, before any hypotheses become accepted, they must be tested\u2014again and again\u2014against the evidence that nature itself reveals. Still, the many strange proposals published about what dark matter might be (see The Evolution and Distribution of Galaxies) attest to the new openness that Schmidt described. With this black hole model, we have come a long way toward understanding the quasars and active galaxies that seemed very mysterious only a few decades ago. As often happens in astronomy, a combination of better instruments (making better observations) and improved theoretical models enabled us to make significant progress on a puzzling aspect of the cosmos. 27.3 QUASARS AS PROBES OF EVOLUTION IN THE UNIVERSE Learning Objectives By the end of this section, you will be able to: Trace the rise and fall of quasars over cosmic time Describe some of the ways in which galaxies and black holes influence each other\u2019s growth Describe some ways the first black holes may have formed Explain why some black holes are not producing quasar emission but rather are quiescent The quasars\u2019 brilliance and large distance make them ideal probes of the far reaches of the universe and its remote past. Recall that when first introducing quasars, we mentioned that they generally tend to be far away. When we see extremely distant objects, we are seeing", " them as they were long ago. Radiation from a quasar 8 billion light-years away is telling us what that quasar and its environment were like 8 billion years ago, much closer to the time that the galaxy that surrounds it first formed. Astronomers have now detected light emitted from quasars that were already formed only a few hundred million years after the universe began its expansion 13.8 billion years ago. Thus, they give us a remarkable opportunity to learn about the time when large structures were first assembling in the cosmos. The Evolution of Quasars Quasars provide compelling evidence that we live in an evolving universe\u2014one that changes with time. They tell us that astronomers living billions of years ago would have seen a universe that is very different from the universe of today. Counts of the number of quasars at different redshifts (and thus at different times in the evolution of the universe) show us how dramatic these changes are (Figure 27.14). We now know that the number of quasars was greatest at the time when the universe was only 20% of its present age. 1 M. Schmidt, \u201cThe Discovery of Quasars,\u201d in Modern Cosmology in Retrospect, ed. B. Bertotti et al. (Cambridge University Press, 1990). 980 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes Figure 27.14. Relative Number of Quasars and Rate at Which Stars Formed as a Function of the Age of the Universe. An age of 0 on the plots corresponds to the beginning of the universe; an age of 13.8 corresponds to the present time. Both the number of quasars and the rate of star formation were at a peak when the universe was about 20% as old as it is now. As you can see, the drop-off in the numbers of quasars as time gets nearer to the present day is quite abrupt. Observations also show that the emission from the accretion disks around the most massive black holes peaks early and then fades. The most powerful quasars are seen only at early times. In order to explain this result, we make use of our model of the energy source of the quasars\u2014namely that quasars are black holes with enough fuel to make a brilliant accretion disk right around them. The fact that there were more quasars long ago (far away) than there are today (nearby) could be explained if there was more material available to be accreted by black holes early in the history of the universe. You might say that the quasars were more active when their black holes had fuel for their \u201cenergy-producing engines.\u201d If that fuel was mostly consumed in the first few billion years after the universe began its expansion, then later in its life, a \u201chungry\u201d black hole would have very little left with which to light up the galaxy\u2019s central regions. In other words, if matter in the accretion disk is continually being depleted by falling into the black hole or being blown out from the galaxy in the form of jets, then a quasar can continue to radiate only as long as new gas is available to replenish the accretion disk. In fact, there was more gas around to be accreted early in the history of the universe. Back then, most gas had not yet collapsed to form stars, so there was more fuel available for both the feeding of black holes and the forming of new stars. Much of that fuel was subsequently consumed in the formation of stars during the first few billion years after the universe began its expansion. Later in its life, a galaxy would have little left to feed a hungry black hole or to form more new stars. As we see from Figure 27.14, both star formation and black hole growth peaked together when the universe was about 2 billion years old. Ever since, both have been in sharp decline. We are late to the party of the galaxies and have missed some of the early excitement. Observations of nearer galaxies (seen later in time) indicate that there is another source of fuel for the central black holes\u2014the collision of galaxies. If two galaxies of similar mass collide and merge, or if a smaller galaxy is pulled into a larger one, then gas and dust from one may come close enough to the black hole in the other to be devoured by it and so provide the necessary fuel. Astronomers have found that collisions were also much more common early in the history of the universe than they are today. There were more small galaxies in those early times because over time, as we shall see (in The Evolution and Distribution of Galaxies), small galaxies tend to combine into larger ones. Again, this means that we would expect to see more quasars long ago (far away) than we do today (nearby)\u2014as we in fact do. This OpenStax book is available for free at http://cnx.org/content/col119", "92/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 981 Codependence of Black Holes and Galaxies Once black hole masses began to be measured reliably in the late 1990s, they posed an enigma. It looked as though the mass of the central black hole depended on the mass of the galaxy. The black holes in galaxies always seem to be just 1/200 the mass of the galaxy they live in. This result is shown schematically in Figure 27.15, and some of the observations are plotted in Figure 27.16. Figure 27.15. Relationship between Black Hole Mass and the Mass of the Host Galaxy. Observations show that there is a close correlation between the mass of the black hole at the center of a galaxy and the mass of the spherical distribution of stars that surrounds the black hole. That spherical distribution may be in the form of either an elliptical galaxy or the central bulge of a spiral galaxy. (credit: modification of work by K. Cordes, S. Brown (STScI)) Figure 27.16. Correlation between the Mass of the Central Black Hole and the Mass Contained within the Bulge of Stars Surrounding the Black Hole, Using Data from Real Galaxies. The black hole always turns out to be about 1/200 the mass of the stars surrounding it. The horizontal and vertical bars surrounding each point show the uncertainty of the measurement. (credit: modification of work by Nicholas J. McConnell, Chung-Pei Ma, \u201cRevisiting the Scaling Relations of Black Hole Masses and Host Galaxy Properties,\u201d The Astrophysical Journal, 764:184 (14 pp.), February 20, 2013.) 982 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes Somehow black hole mass and the mass of the surrounding bulge of stars are connected. But why does this correlation exist? Unfortunately, astronomers do not yet know the answer to this question. We do know, however, that the black hole can influence the rate of star formation in the galaxy, and that the properties of the surrounding galaxy can influence how fast the black hole grows. Let\u2019s see how these processes work. How a Galaxy Can Influence a Black Hole in Its Center Let\u2019s look first at how the surrounding galaxy might influence the growth and size of the black hole. Without large quantities of fresh \u201cfood,\u201d the surroundings of black holes glow only weakly as bits of local material spiral inward toward the black hole. So somehow large amounts of gas have to find their way to the black hole from the galaxy in order to feed the quasar and make it grow and give off the energy to be noticed. Where does this \u201cfood\u201d for the black hole come from originally and how might it be replenished? The jury is still out, but the options are pretty clear. One obvious source of fuel for the black hole is matter from the host galaxy itself. Galaxies start out with large amounts of interstellar gas and dust, and at least some of this interstellar matter is gradually converted into stars as the galaxy evolves. On the other hand, as stars go through their lives and die, they lose mass all the time into the space between them, thereby returning some of the gas and dust to the interstellar medium. We expect to find more gas and dust in the central regions early in a galaxy\u2019s life than later on, when much of it has been converted into stars. Any of the interstellar matter that ventures too close to the black hole may be accreted by it. This means that we would expect that the number and luminosity of quasars powered in this way would decline with time. And as we have seen, that is just what we find. Today both elliptical galaxies and the nuclear bulges of spiral galaxies have very little raw material left to serve as a source of fuel for the black hole. And most of the giant black holes in nearby galaxies, including the one in our own Milky Way, are now dark and relatively quiet\u2014mere shadows of their former selves. So that fits with our observations. We should note that even if you have a quiescent supermassive black hole, a star in the area could occasionally get close to it. Then the powerful tidal forces of the black hole can pull the whole star apart into a stream of gas. This stream quickly forms an accretion disk that gives off energy in the normal way and makes the black hole region into a temporary quasar. However, the material will fall into the black hole after only a few weeks or months. The black hole then goes back into its lurking, quiescent state, until another victim wanders by. This sort of \u201ccannibal\u201d event happens only once every 100,000 years or so in a typical galaxy. But we can monitor millions of galaxies in the sky, so a few of these \u201c tidal disruption events\u201d are found each year (Figure 27.17). However", ", these individual events, dramatic as they are, are too rare to account for the huge masses of the central black holes. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 983 Figure 27.17. A Black Hole Snacks on a Star. This artist\u2019s impression shows three stages of a star (red) swinging too close to a giant black hole (black circle). The star starts off (top left) in its normal spherical shape, then begins to be pulled into a long football shape by tides raised by the black hole (center). When the star gets closer still, the tides become stronger than the gravity holding the star together, and it breaks up into a streamer (right). Much of the star\u2019s matter forms a temporary accretion disk that lights up as a quasar for a few weeks or months. (credit: modification of work by NASA/CXC/M. Weiss) Another source of fuel for the black hole is the collision of its host galaxy with another galaxy. Some of the brightest galaxies turn out, when a detailed picture is taken, to be pairs of colliding galaxies. And most of them have quasars inside them, not easily visible to us because they are buried by enormous amounts of dust and gas. A collision between two cars creates quite a mess, pushing parts out of their regular place. In the same way, if two galaxies collide and merge, then gas and dust (though not so much the stars) can get pushed out of their regular orbits. Some may veer close enough to the black hole in one galaxy or the other to be devoured by it and so provide the necessary fuel to power a quasar. As we saw, galaxy collisions and mergers happened most frequently when the universe was young and probably help account for the fact that quasars were most common when the universe was only about 20% of its current age. Collisions in today\u2019s universe are less frequent, but they do happen. Once a galaxy reaches the size of the Milky Way, most of the galaxies it merges with will be much smaller galaxies\u2014dwarf galaxies (see the chapter on Galaxies). These don\u2019t disrupt the big galaxy much, but they can supply some additional gas to its black hole. By the way, if two galaxies, each of which contains a black hole, collide, then the two black holes may merge and form an even larger black hole (Figure 27.18). In this process they will emit a burst of gravitational waves. One of the main goals of the European Space Agency\u2019s planned LISA (Laser Interferometer Space Antenna) mission is to detect the gravitational wave signals from the merging of supermassive black holes. 984 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes Figure 27.18. Colliding Galaxies with Two Black Holes. We compare Hubble Space Telescope visible-light (left) and Chandra X-ray (right) images of the central regions of NGC 6240, a galaxy about 400 million light-years away. It is a prime example of a galaxy in which stars are forming, evolving, and exploding at an exceptionally rapid rate due to a relatively recent merger (30 million years ago). The Chandra image shows two bright X-ray sources, each produced by hot gas surrounding a black hole. Over the course of the next few hundred million years, the two supermassive black holes, which are about 3000 light-years apart, will drift toward each other and merge to form an even larger black hole. This detection of a binary black hole supports the idea that black holes can grow to enormous masses in the centers of galaxies by merging with nearby galaxies. (credit left: modification of work by NASA/CXC/MPE/S.Komossa et al; credit right: NASA/STScI/R. P. van der Marel, J. Gerssen Watch two galaxies collide (https://openstax.org/l/30galcolsuperma) to form a supermassive black hole. How Does the Black Hole Influence the Formation of Stars in the Galaxy? We have seen that the material in galaxies can influence the growth of the black hole. The black hole in turn can also influence the galaxy in which it resides. It can do so in three ways: through its jets, through winds of particles that manage to stream away from the accretion disk, and through radiation from the accretion disk. As they stream away from the black hole, all three can either promote star formation by compressing the surrounding gas and dust\u2014or instead suppress star formation by heating the surrounding gas and shredding molecular clouds, thereby inhibiting or preventing star formation. The outflowing energy can even be enough to halt the accretion of new material and starve the black hole of fuel", ". Astronomers are still trying to evaluate the relative importance of these effects in determining the overall evolution of galactic bulges and the rates of star formation. In summary, we have seen how galaxies and supermassive black holes can each influence the evolution of the other: the galaxy supplies fuel to the black hole, and the quasar can either support or suppress star formation. The balance of these processes probably helps account for the correlation between black hole and bulge masses, but there are as yet no theories that explain quantitatively and in detail why the correlation between black hole and bulge masses is as tight as it is or why the black hole mass is always about 1/200 times the mass of the bulge. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 985 The Birth of Black Holes and Galaxies While the connection between quasars and galaxies is increasingly clear, the biggest puzzle of all\u2014namely, how the supermassive black holes in galaxies got started\u2014remains unsolved. Observations show that they existed when the universe was very young. One dramatic example is the discovery of a quasar that was already shining when the universe was only 700 million years old. What does it take to create a large black hole so quickly? A related problem is that in order to eventually build black holes containing more than 2 billion solar masses, it is necessary to have giant \u201cseed\u201d black holes with masses at least 2000 times the mass of the Sun\u2014and they must somehow have been created shortly after the expansion of the universe began. Astronomers are now working actively to develop models for how these seed black holes might have formed. Theories suggest that galaxies formed from collapsing clouds of dark matter and gas. Some of the gas formed stars, but perhaps some of the gas settled to the center where it became so concentrated that it formed a black hole. If this happened, the black hole could form right away\u2014although this requires that the gas should not be rotating very much initially. A more likely scenario is that the gas will have some angular momentum (rotation) that will prevent direct collapse to a black hole. In that case, the very first generation of stars will form, and some of them, according to calculations, will have masses hundreds of times that of the Sun. When these stars finish burning hydrogen, just a few million years later, the supernovae they end with will create black holes a hundred or so times the mass of the Sun. These can then merge with others or accrete the rich gas supply available at these early times. The challenge is growing these smaller black holes quickly enough to make the much larger black holes we see a few hundred million years later. It turns out to be difficult because there are limits on how fast they can accrete matter. These should make sense to you from what we discussed earlier in the chapter. If the rate of accretion becomes too high, then the energy streaming outward from the black hole\u2019s accretion disk will become so strong as to blow away the infalling matter. What if, instead, a collapsing gas cloud doesn\u2019t form a black hole directly or break up and form a group of regular stars, but stays together and makes one fairly massive star embedded within a dense cluster of thousands of lower mass stars and large quantities of dense gas? The massive star will have a short lifetime and will soon collapse to become a black hole. It can then begin to attract the dense gas surrounding it. But calculations show that the gravitational attraction of the many nearby stars will cause the black hole to zigzag randomly within the cluster and will prevent the formation of an accretion disk. If there is no accretion disk, then matter can fall freely into the black hole from all directions. Calculations suggest that under these conditions, a black hole even as small as 10 times the mass of the Sun could grow to more than 10 billion times the mass of the Sun by the time the universe is a billion years old. Scientists are exploring other ideas for how to form the seeds of supermassive black holes, and this remains a very active field of research. Whatever mechanism caused the rapid formation of these supermassive black holes, they do give us a way to observe the youthful universe when it was only about five percent as old as it is now Take a look at some new results (https://openstax.org/l/30chanxrayobser) from the Chandra X-ray Observatory about the formation of supermassive black holes in the early universe. 986 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes CHAPTER 27 REVIEW KEY TERMS active galactic nuclei (AGN) galaxies that are almost as luminous as quasars and share many of their properties, although to a less spectacular degree; abnormal amounts of energy are produced in their centers active galaxies galaxies that house active galactic nuclei quasar an", " object of very high redshift that looks like a star but is extragalactic and highly luminous; also called a quasi-stellar object, or QSO SUMMARY 27.1 Quasars The first quasars discovered looked like stars but had strong radio emission. Their visible-light spectra at first seemed confusing, but then astronomers realized that they had much larger redshifts than stars. The quasar spectra obtained so far show redshifts ranging from 15% to more than 96% the speed of light. Observations with the Hubble Space Telescope show that quasars lie at the centers of galaxies and that both spirals and ellipticals can harbor quasars. The redshifts of the underlying galaxies match the redshifts of the quasars embedded in their centers, thereby proving that quasars obey the Hubble law and are at the great distances implied by their redshifts. To be noticeable at such great distances, quasars must have 10 to 100 times the luminosity of the brighter normal galaxies. Their variations show that this tremendous energy output is generated in a small volume\u2014in some cases, in a region not much larger than our own solar system. A number of galaxies closer to us also show strong activity at their centers\u2014activity now known to be caused by the same mechanism as the quasars. 27.2 Supermassive Black Holes: What Quasars Really Are Both active galactic nuclei and quasars derive their energy from material falling toward, and forming a hot accretion disk around, a massive black hole. This model can account for the large amount of energy emitted and for the fact that the energy is produced in a relatively small volume of space. It can also explain why jets coming from these objects are seen in two directions: those directions are perpendicular to the accretion disk. 27.3 Quasars as Probes of Evolution in the Universe Quasars and galaxies affect each other: the galaxy supplies fuel to the black hole, and the quasar heats and disrupts the gas clouds in the galaxy. The balance between these two processes probably helps explain why the black hole seems always to be about 1/200 the mass of the spherical bulge of stars that surrounds the black hole. Quasars were much more common billions of years ago than they are now, and astronomers speculate that they mark an early stage in the formation of galaxies. Quasars were more likely to be active when the universe was young and fuel for their accretion disk was more available. Quasar activity can be re-triggered by a collision between two galaxies, which provides a new source of fuel to feed the black hole. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 987 FOR FURTHER EXPLORATION Articles Bartusiak, M. \u201cA Beast in the Core.\u201d Astronomy (July 1998): 42. On supermassive black holes at the centers of galaxies. Disney, M. \u201cA New Look at Quasars.\u201d Scientific American (June 1998): 52. Djorgovski, S. \u201cFires at Cosmic Dawn.\u201d Astronomy (September 1995): 36. On quasars and what we can learn from them. Ford, H., & Tsvetanov, Z. \u201cMassive Black Holes at the Hearts of Galaxies.\u201d Sky & Telescope (June 1996): 28. Nice overview. Irion, R. \u201cA Quasar in Every Galaxy?\u201d Sky & Telescope (July 2006): 40. Discusses how supermassive black holes powering the centers of galaxies may be more common than thought. Kormendy, J. \u201cWhy Are There so Many Black Holes?\u201d Astronomy (August 2016): 26. Discussion of why supermassive black holes are so common in the universe. Kruesi, L. \u201cSecrets of the Brightest Objects in the Universe.\u201d Astronomy (July 2013): 24. Review of our current understanding of quasars and how they help us learn about black holes. Miller, M., et al. \u201cSupermassive Black Holes: Shaping their Surroundings.\u201d Sky & Telescope (April 2005): 42. Jets from black hole disks. Nadis, S. \u201cExploring the Galaxy\u2013Black Hole Connection.\u201d Astronomy (May 2010): 28. Overview. Nadis, S. \u201cHere, There, and Everywhere.\u201d Astronomy (February 2001): 34. On Hubble observations showing how common supermassive black holes are in galaxies. Nadis, S. \u201cPeering inside a Monster Galaxy.\u201d Astronomy (May 2014): 24. What X-ray observations tell us about the mechanism that powers the active galaxy M87. Olson, S. \u201cBlack Hole Hunters.", "\u201d Astronomy (May 1999): 48. Profiles four astronomers who search for \u201chungry\u201d black holes at the centers of active galaxies. Peterson, B. \u201cSolving the Quasar Puzzle.\u201d Sky & Telescope (September 2013): 24. A review article on how we figured out that black holes were the power source for quasars, and how we view them today. Tucker, W., et al. \u201cBlack Hole Blowback.\u201d Scientific American (March 2007): 42. How supermassive black holes create giant bubbles in the intergalactic medium. Voit, G. \u201cThe Rise and Fall of Quasars.\u201d Sky & Telescope (May 1999): 40. Good overview of how quasars fit into cosmic history. Wanjek, C. \u201cHow Black Holes Helped Build the Universe.\u201d Sky & Telescope (January 2007): 42. On the energy and outflow from disks around supermassive black holes; nice introduction. Websites Monsters in Galactic Nuclei: http://chandra.as.utexas.edu/stardate.html. An article on supermassive black holes by John Kormendy, from StarDate magazine. Quasar Astronomy Forty Years On: http://www.astr.ua.edu/keel/agn/quasar40.html. A 2003 popular article by William Keel. 988 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes Quasars and Active Galactic Nuclei: www.astr.ua.edu/keel/agn/. An annotated gallery of images showing the wide range of activity in galaxies. There is also an introduction, a glossary, and background information. Also by William Keel. Quasars: \u201cThe Light Fantastic\u201d: http://hubblesite.org/newscenter/archive/releases/1996/35/background/. This brief \u201cbackgrounder\u201d from the public information office at the HubbleSite gives a bit of the history of the discovery and understanding of quasars. Videos Active Galaxies: https://www.youtube.com/watch?v=Y_HgsFmwCeg. Part of the Astronomy: Observations and Theories series; half-hour introduction to quasars and related objects (27:28). Black Hole Chaos: The Environments of the Most Supermassive Black Holes in the Universe: https://www.youtube.com/watch?v=hzSgU-3d8QY. May 2013 lecture by Dr. Belinda Wilkes and Dr. Francesca Civano of the Center for Astrophysics in the CfA Observatory Nights Lecture Series (50:14). Hubble and Black Holes: http://www.spacetelescope.org/videos/hubblecast43a/. Hubblecast on black holes and active galactic nuclei (9:10). Monster Black Holes: https://www.youtube.com/watch?v=LN9oYjNKBm8. May 2013 lecture by Professor Chung-Pei Ma of the University of California, Berkeley; part of the Silicon Valley Astronomy Lecture Series (1:18:03). COLLABORATIVE GROUP ACTIVITIES A. When quasars were first discovered and the source of their great energy was unknown, some astronomers searched for evidence that quasars are much nearer to us than their redshifts imply. (That way, they would not have to produce so much energy to look as bright as they do.) One way was to find a \u201cmismatched pair\u201d\u2014a quasar and a galaxy with different redshifts that lie in very nearly the same direction in the sky. Suppose you do find one and only one galaxy with a quasar very close by, and the redshift of the quasar is six times larger than that of the galaxy. Have your group discuss whether you could then conclude that the two objects are at the same distance and that redshift is not a reliable indicator of distance. Why? Suppose you found three such pairs, each with different mismatched redshifts? Suppose every galaxy has a nearby quasar with a different redshift. How would your answer change and why? B. Large ground-based telescopes typically can grant time to only one out of every four astronomers who apply for observing time. One prominent astronomer tried for several years to establish that the redshifts of quasars do not indicate their distances. At first, he was given time on the world\u2019s largest telescope, but eventually it became clearer that quasars were just the centers of active galaxies and that their redshifts really did indicate distance. At that point, he was denied observing time by the committee of astronomers who reviewed such proposals. Suppose your group had been the committee. What decision would you have made? Why? (In general,", " what criteria should astronomers have for allowing astronomers whose views completely disagree with the prevailing opinion to be able to pursue their research?) C. Based on the information in this chapter and in Black Holes and Curved Spacetime, have your group discuss what it would be like near the event horizon of a supermassive black hole in a quasar or active galaxy. Make a list of all the reasons a trip to that region would not be good for your health. Be specific. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 989 D. Before we understood that the energy of quasars comes from supermassive black holes, astronomers were baffled by how such small regions could give off so much energy. A variety of models were suggested, some involving new physics or pretty \u201cfar out\u201d ideas from current physics. Can your group come up with some areas of astronomy that you have studied in this course where we don\u2019t yet have an explanation for something happening in the cosmos? EXERCISES Review Questions 1. Describe some differences between quasars and normal galaxies. 2. Describe the arguments supporting the idea that quasars are at the distances indicated by their redshifts. 3. In what ways are active galaxies like quasars but different from normal galaxies? 4. Why could the concentration of matter at the center of an active galaxy like M87 not be made of stars? 5. Describe the process by which the action of a black hole can explain the energy radiated by quasars. 6. Describe the observations that convinced astronomers that M87 is an active galaxy. 7. Why do astronomers believe that quasars represent an early stage in the evolution of galaxies? 8. Why were quasars and active galaxies not initially recognized as being \u201cspecial\u201d in some way? 9. What do we now understand to be the primary difference between normal galaxies and active galaxies? 10. What is the typical structure we observe in a quasar at radio frequencies? 11. What evidence do we have that the luminous central region of a quasar is small and compact? Thought Questions 12. Suppose you observe a star-like object in the sky. How can you determine whether it is actually a star or a quasar? 13. Why don\u2019t any of the methods for establishing distances to galaxies, described in Galaxies (other than Hubble\u2019s law itself), work for quasars? 14. One of the early hypotheses to explain the high redshifts of quasars was that these objects had been ejected at very high speeds from other galaxies. This idea was rejected, because no quasars with large blueshifts have been found. Explain why we would expect to see quasars with both blueshifted and redshifted lines if they were ejected from nearby galaxies. 15. A friend of yours who has watched many Star Trek episodes and movies says, \u201cI thought that black holes pulled everything into them. Why then do astronomers think that black holes can explain the great outpouring of energy from quasars?\u201d How would you respond? 16. Could the Milky Way ever become an active galaxy? Is it likely to ever be as luminous as a quasar? 17. Why are quasars generally so much more luminous (why do they put out so much more energy) than active galaxies? 18. Suppose we detect a powerful radio source with a radio telescope. How could we determine whether or not this was a newly discovered quasar and not some nearby radio transmission? 990 Chapter 27 Active Galaxies, Quasars, and Supermassive Black Holes 19. A friend tries to convince you that she can easily see a quasar in her backyard telescope. Would you believe her claim? Figuring For Yourself 20. Show that no matter how big a redshift (z) we measure, v/c will never be greater than 1. (In other words, no galaxy we observe can be moving away faster than the speed of light.) 21. If a quasar has a redshift of 3.3, at what fraction of the speed of light is it moving away from us? 22. If a quasar is moving away from us at v/c = 0.8, what is the measured redshift? 23. In the chapter, we discussed that the largest redshifts found so far are greater than 6. Suppose we find a quasar with a redshift of 6.1. With what fraction of the speed of light is it moving away from us? 24. Rapid variability in quasars indicates that the region in which the energy is generated must be small. You can show why this is true. Suppose, for example, that the region in which the energy is generated is a transparent sphere 1 light-year in diameter. Suppose that in 1 s this region bright", "ens by a factor of 10 and remains bright for two years, after which it returns to its original luminosity. Draw its light curve (a graph of its brightness over time) as viewed from Earth. 25. Large redshifts move the positions of spectral lines to longer wavelengths and change what can be observed from the ground. For example, suppose a quasar has a redshift of \u0394\u03bb \u03bb = 4.1. At what wavelength would you make observations in order to detect its Lyman line of hydrogen, which has a laboratory or rest wavelength of 121.6 nm? Would this line be observable with a ground-based telescope in a quasar with zero redshift? Would it be observable from the ground in a quasar with a redshift of \u0394\u03bb \u03bb = 4.1? 26. Once again in this chapter, we see the use of Kepler\u2019s third law to estimate the mass of supermassive black holes. In the case of NGC 4261, this chapter supplied the result of the calculation of the mass of the black hole in NGC 4261. In order to get this answer, astronomers had to measure the velocity of particles in the ring of dust and gas that surrounds the black hole. How high were these velocities? Turn Kepler\u2019s third law around and use the information given in this chapter about the galaxy NGC 4261\u2014the mass of the black hole at its center and the diameter of the surrounding ring of dust and gas\u2014to calculate how long it would take a dust particle in the ring to complete a single orbit around the black hole. Assume that the only force acting on the dust particle is the gravitational force exerted by the black hole. Calculate the velocity of the dust particle in km/s. 27. In the Check Your Learning section of Example 27.1, you were told that several lines of hydrogen absorption in the visible spectrum have rest wavelengths of 410 nm, 434 nm, 486 nm, and 656 nm. In a spectrum of a distant galaxy, these same lines are observed to have wavelengths of 492 nm, 521 nm, 583 nm, and 787 nm, respectively. The example demonstrated that z = 0.20 for the 410 nm line. Show that you will obtain the same redshift regardless of which absorption line you measure. 28. In the Check Your Learning section of Example 27.1, the author commented that even at z = 0.2, there is already an 11% deviation between the relativistic and the classical solution. What is the percentage difference between the classical and relativistic results at z = 0.1? What is it for z = 0.5? What is it for z = 1? 29. The quasar that appears the brightest in our sky, 3C 273, is located at a distance of 2.4 billion lightyears. The Sun would have to be viewed from a distance of 1300 light-years to have the same apparent magnitude as 3C 273. Using the inverse square law for light, estimate the luminosity of 3C 273 in solar units. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 991 28 THE EVOLUTION AND DISTRIBUTION OF GALAXIES Figure 28.1. Colliding Galaxies. Collisions and mergers of galaxies strongly influence their evolution. On the left is a ground-based image of two colliding galaxies (NCG 4038 and 4039), sometimes nicknamed the Antennae galaxies. The long, luminous tails are material torn out of the galaxies by tidal forces during the collision. The right image shows the inner regions of these two galaxies, as taken by the Hubble Space Telescope. The cores of the twin galaxies are the orange blobs to the lower left and upper right of the center of the image. Note the dark lanes of dust crossing in front of the bright regions. The bright pink and blue star clusters are the result of a burst of star formation stimulated by the collision. (credit left: modification of work by Bob and Bill Twardy/Adam Block/NOAO/AURA/NSF; credit right: modification of work by NASA, ESA, and the Hubble Heritage Team (STScI/AURA)-ESA/Hubble Collaboration) Chapter Outline 28.1 Observations of Distant Galaxies 28.2 Galaxy Mergers and Active Galactic Nuclei 28.3 The Distribution of Galaxies in Space 28.4 The Challenge of Dark Matter 28.5 The Formation and Evolution of Galaxies and Structure in the Universe Thinking Ahead How and when did galaxies like our Milky Way form? Which formed first: stars or galaxies? Can we see direct evidence of the changes galaxies undergo over their lifetimes? If so, what determines whether a galaxy will \u201cgrow up\u201d to be spiral or elliptical? And what is the role of \u201cnature versus nurture\ufffd", "\ufffd? That is to say, how much of a galaxy\u2019s development is determined by what it looks like when it is born and how much is influenced by its environment? Astronomers today have the tools needed to explore the universe almost back to the time it began. The huge new telescopes and sensitive detectors built in the last decades make it possible to obtain both images and spectra of galaxies so distant that their light has traveled to reach us for more than 13 billion years\u2014more than 90% of the way back to the Big Bang: we can use the finite speed of light and the vast size of the universe as a cosmic time machine to peer back and observe how galaxies formed and evolved over time. Studying galaxies so far away in any detail is always a major challenge, largely because their distance makes them appear very faint. However, today\u2019s large telescopes on the ground and in space are finally making such a task possible. 992 Chapter 28 The Evolution and Distribution of Galaxies 28.1 OBSERVATIONS OF DISTANT GALAXIES Learning Objectives By the end of this section, you will be able to: Explain how astronomers use light to learn about distant galaxies long ago Discuss the evidence showing that the first stars formed when the universe was less than 10% of its current age Describe the major differences observed between galaxies seen in the distant, early universe and galaxies seen in the nearby universe today Let\u2019s begin by exploring some techniques astronomers use to study how galaxies are born and change over cosmic time. Suppose you wanted to understand how adult humans got to be the way they are. If you were very dedicated and patient, you could actually observe a sample of babies from birth, following them through childhood, adolescence, and into adulthood, and making basic measurements such as their heights, weights, and the proportional sizes of different parts of their bodies to understand how they change over time. Unfortunately, we have no such possibility for understanding how galaxies grow and change over time: in a human lifetime\u2014or even over the entire history of human civilization\u2014individual galaxies change hardly at all. We need other tools than just patiently observing single galaxies in order to study and understand those long, slow changes. We do, however, have one remarkable asset in studying galactic evolution. As we have seen, the universe itself is a kind of time machine that permits us to observe remote galaxies as they were long ago. For the closest galaxies, like the Andromeda galaxy, the time the light takes to reach us is on the order of a few hundred thousand to a few million years. Typically not much changes over times that short\u2014individual stars in the galaxy may be born or die, but the overall structure and appearance of the galaxy will remain the same. But we have observed galaxies so far away that we are seeing them as they were when the light left them more than 10 billion years ago. By observing more distant objects, we look further back toward a time when both galaxies and the universe were young (Figure 28.2). This is a bit like getting letters in the mail from several distant friends: the farther the friend was when she mailed the letter to you, the longer the letter must have been in transit, and so the older the news is when it arrives in your mailbox; you are learning something about her life at an earlier time than when you read the letter. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 993 Figure 28.2. Astronomical Time Travel. This true-color, long-exposure image, made during 70 orbits of Earth with the Hubble Space Telescope, shows a small area in the direction of the constellation Sculptor. The massive cluster of galaxies named Abell 2744 appears in the foreground of this image. It contains several hundred galaxies, and we are seeing them as they looked 3.5 billion years ago. The immense gravity in Abell 2744 acts as a gravitational lens (see the Astronomy Basics feature box on Gravitational Lensing later in this chapter) to warp space and brighten and magnify images of nearly 3000 distant background galaxies. The more distant galaxies (many of them quite blue) appear as they did more than 12 billion years ago, not long after the Big Bang. Blue galaxies were much more common in that earlier time than they are today. These galaxies appear blue because they are undergoing active star formation and making hot, bright blue stars. (credit: NASA, ESA, STScI) If we can\u2019t directly detect the changes over time in individual galaxies because they happen too slowly, how then can we ever understand those changes and the origins of galaxies? The solution is to observe many galaxies at many different cosmic distances and, therefore, look-back times (how far back in time we are seeing the galaxy). If we can study a thousand very distant \u201cbaby\u201d galaxies when the universe was 1 billion years old, and another thousand slightly closer \u201ct", "oddler\u201d galaxies when it was 2 billion years old, and so on until the present 13.8-billion-year-old universe of mature \u201cadult\u201d galaxies near us today, then maybe we can piece together a coherent picture of how the whole ensemble of galaxies evolves over time. This allows us to reconstruct the \u201clife story\u201d of galaxies since the universe began, even though we can\u2019t follow a single galaxy from infancy to old age. Fortunately, there is no shortage of galaxies to study. Hold up your pinky at arm\u2019s length: the part of the sky blocked by your fingernail contains about one million galaxies, layered farther and farther back in space and time. In fact, the sky is filled with galaxies, all of them, except for Andromeda and the Magellanic Clouds, too faint to see with the naked eye\u2014more than 100 billion galaxies in the observable universe, each one with about 100 billion stars. This cosmic time machine, then, lets us peer into the past to answer fundamental questions about where galaxies come from and how they got to be the way they are today. Astronomers call those galactic changes over cosmic time evolution, a word that recalls the work of Darwin and others on the development of life on Earth. But note that galaxy evolution refers to the changes in individual galaxies over time, while the kind of evolution biologists study is changes in successive generations of living organisms over time. 994 Chapter 28 The Evolution and Distribution of Galaxies Spectra, Colors, and Shapes Astronomy is one of the few sciences in which all measurements must be made at a distance. Geologists can take samples of the objects they are studying; chemists can conduct experiments in their laboratories to determine what a substance is made of; archeologists can use carbon dating to determine how old something is. But astronomers can\u2019t pick up and play with a star or galaxy. As we have seen throughout this book, if they want to know what galaxies are made of and how they have changed over the lifetime of the universe, they must decode the messages carried by the small number of photons that reach Earth. Fortunately (as you have learned) electromagnetic radiation is a rich source of information. The distance to a galaxy is derived from its redshift (how much the lines in its spectrum are shifted to the red because of the expansion of the universe). The conversion of redshift to a distance depends on certain properties of the universe, including the value of the Hubble constant and how much mass it contains. We will describe the currently accepted model of the universe in The Big Bang. For the purposes of this chapter, it is enough to know that the current best estimate for the age of the universe is 13.8 billion years. In that case, if we see an object that emitted its light 6 billion light-years ago, we are seeing it as it was when the universe was almost 8 billion years old. If we see something that emitted its light 13 billion years ago, we are seeing it as it was when the universe was less than a billion years old. So astronomers measure a galaxy\u2019s redshift from its spectrum, use the Hubble constant plus a model of the universe to turn the redshift into a distance, and use the distance and the constant speed of light to infer how far back in time they are seeing the galaxy\u2014the look-back time. In addition to distance and look-back time, studies of the Doppler shifts of a galaxy\u2019s spectral lines can tell us how fast the galaxy is rotating and hence how massive it is (as explained in Galaxies). Detailed analysis of such lines can also indicate the types of stars that inhabit a galaxy and whether it contains large amounts of interstellar matter. Unfortunately, many galaxies are so faint that collecting enough light to produce a detailed spectrum is currently impossible. Astronomers thus have to use a much rougher guide to estimate what kinds of stars inhabit the faintest galaxies\u2014their overall colors. Look again at Figure 28.2 and notice that some of the galaxies are very blue and others are reddish-orange. Now remember that hot, luminous blue stars are very massive and have lifetimes of only a few million years. If we see a galaxy where blue colors dominate, we know that it must have many hot, luminous blue stars, and that star formation must have taken place in the few million years before the light left the galaxy. In a yellow or red galaxy, on the other hand, the young, luminous blue stars that surely were made in the galaxy\u2019s early bursts of star formation must have died already; it must contain mostly old yellow and red stars that last a long time in their main-sequence stages and thus typically formed billions of years before the light that we now see was emitted. Another important clue to the nature of a galaxy is its shape. Spiral galaxies can be distinguished from elliptical galaxies by shape. Observations show that spiral galaxies contain young stars and large amounts of interstellar matter, while ellipt", "ical galaxies have mostly old stars and very little or no star formation. Elliptical galaxies turned most of their interstellar matter into stars many billions of years ago, while star formation has continued until the present day in spiral galaxies. If we can count the number of galaxies of each type during each epoch of the universe, it will help us understand how the pace of star formation changes with time. As we will see later in this chapter, galaxies in the distant universe\u2014that is, young galaxies\u2014look very different from the older galaxies that we see nearby in the present-day universe. The First Generation of Stars In addition to looking at the most distant galaxies we can find, astronomers look at the oldest stars (what we might call the fossil record) of our own Galaxy to probe what happened in the early universe. Since stars are the This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 995 source of nearly all the light emitted by galaxies, we can learn a lot about the evolution of galaxies by studying the stars within them. What we find is that nearly all galaxies contain at least some very old stars. For example, our own Galaxy contains globular clusters with stars that are at least 13 billion years old, and some may be even older than that. Therefore, if we count the age of the Milky Way as the age of its oldest constituents, the Milky Way must have been born at least 13 billion years ago. As we will discuss in The Big Bang, astronomers have discovered that the universe is expanding, and have traced the expansion backward in time. In this way, they have discovered that the universe itself is only about 13.8 billion years old. Thus, it appears that at least some of the globular-cluster stars in the Milky Way must have formed less than a billion years after the expansion began. Several other observations also establish that star formation in the cosmos began very early. Astronomers have used spectra to determine the composition of some elliptical galaxies that are so far away that the light we see left them when the universe was only half as old as it is now. Yet these ellipticals contain old red stars, which must have formed billions of years earlier still. When we make computer models of how such galaxies evolve with time, they tell us that star formation in elliptical galaxies began less than a billion years or so after the universe started its expansion, and new stars continued to form for a few billion years. But then star formation apparently stopped. When we compare distant elliptical galaxies with ones nearby, we find that ellipticals have not changed very much since the universe reached about half its current age. We\u2019ll return to this idea later in the chapter. Observations of the most luminous galaxies take us even further back in time. Recently, as we have already noted, astronomers have discovered a few galaxies that are so far away that the light we see now left them less than a billion years or so after the beginning (Figure 28.3). Yet the spectra of some of these galaxies already contain lines of heavy elements, including carbon, silicon, aluminum, and sulfur. These elements were not present when the universe began but had to be manufactured in the interiors of stars. This means that when the light from these galaxies was emitted, an entire generation of stars had already been born, lived out their lives, and died\u2014spewing out the new elements made in their interiors through supernova explosions\u2014even before the universe was a billion years old. And it wasn\u2019t just a few stars in each galaxy that got started this way. Enough had to live and die to affect the overall composition of the galaxy, in a way that we can still measure in the spectrum from far away. 996 Chapter 28 The Evolution and Distribution of Galaxies Figure 28.3. Very Distant Galaxy. This image was made with the Hubble Space Telescope and shows the field around a luminous galaxy at a redshift z = 8.68, corresponding to a distance of about 13.2 billion light-years at the time when the light was emitted (indicated by the arrow and shown in the upper inset). Long exposures in the far-red and infrared wavelengths were combined to make the image, and additional infrared exposures with the Spitzer Space Telescope, which has lower spatial resolution than the Hubble (lower inset), show the redshifted light of normal stars. The very distant galaxy was detected because it has a strong emission line of hydrogen. This line is produced in regions where the formation of hot, young stars is taking place. (credit: modification of work by I. Labb\u00e9 (Leiden University), NASA/ESA/JPL-Caltech) Observations of quasars (galaxies whose centers contain a supermassive black hole) support this conclusion. We can measure the abundances of heavy elements in the gas near quasar black holes (explained in Active Galaxies, Quas", "ars, and Supermassive Black Holes). The composition of this gas in quasars that emitted their light 12.5 billion light-years ago is very similar to that of the Sun. This means that a large portion of the gas surrounding the black holes must have already been cycled through stars during the first 1.3 billion years after the expansion of the universe began. If we allow time for this cycling, then their first stars must have formed when the universe was only a few hundred million years old. A Changing Universe of Galaxies Back in the middle decades of the twentieth century, the observation that all galaxies contain some old stars led astronomers to the hypothesis that galaxies were born fully formed near the time when the universe began its expansion. This hypothesis was similar to suggesting that human beings were born as adults and did not have to pass through the various stages of development from infancy through the teens. If this hypothesis were correct, the most distant galaxies should have shapes and sizes very much like the galaxies we see nearby. According to this old view, galaxies, after they formed, should then change only slowly, as successive generations of stars within them formed, evolved, and died. As the interstellar matter was slowly used up and fewer new stars formed, the galaxies would gradually become dominated by fainter, older stars and look dimmer and dimmer. Thanks to the new generation of large ground- and space-based telescopes, we now know that this picture of galaxies evolving peacefully and in isolation from one another is completely wrong. As we will see later in this chapter, galaxies in the distant universe do not look like the Milky Way and nearby galaxies such as Andromeda, and the story of their development is more complex and involves far more interaction with their neighbors. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 997 Why were astronomers so wrong? Up until the early 1990s, the most distant normal galaxy that had been observed emitted its light 8 billion years ago. Since that time, many galaxies\u2014and particularly the giant ellipticals, which are the most luminous and therefore the easiest to see at large distances\u2014did evolve peacefully and slowly. But the Hubble, Spitzer, Herschel, Keck, and other powerful new telescopes that have come on line since the 1990s make it possible to pierce the 8-billion-light-year barrier. We now have detailed views of many thousands of galaxies that emitted their light much earlier (some more than 13 billion years ago\u2014see Figure 28.3). Much of the recent work on the evolution of galaxies has progressed by studying a few specific small regions of the sky where the Hubble, Spitzer, and ground-based telescopes have taken extremely long exposure images. This allowed astronomers to detect very faint, very distant, and therefore very young galaxies (Figure 28.4). Our deep space telescope images show some galaxies that are 100 times fainter than the faintest objects that can be observed spectroscopically with today\u2019s giant ground-based telescopes. This turns out to mean that we can obtain the spectra needed to determine redshifts for only the very brightest five percent of the galaxies in these images. Figure 28.4. Hubble Ultra-Deep Field. This image is the result of an 11-day-long observation with the Hubble Space Telescope of a tiny region of sky, located toward the constellation Fornax near the south celestial pole. This is an area that has only a handful of Milky Way stars. (Since the Hubble orbits Earth every 96 minutes, the telescope returned to view the same tiny piece of sky over and over again until enough light was collected and added together to make this very long exposure.) There are about 10,000 objects in this single image, nearly all of them galaxies, each with tens or hundreds of billions of stars. We can see some pinwheel-shaped spiral galaxies, which are like the Milky Way. But we also find a large variety of peculiar-shaped galaxies that are in collision with companion galaxies. Elliptical galaxies, which contain mostly old stars, appear as reddish blobs. (credit: modification of work by NASA, ESA, H. Teplitz and M. Rafelski (IPAC/Caltech), A. Koekemoer (STScI), R. Windhorst (Arizona State University), and Z. Levay (STScI)) Although we do not have spectra for most of the faint galaxies, the Hubble Space Telescope is especially well suited to studying their shapes because the images taken in space are not blurred by Earth\u2019s atmosphere. To the surprise of astronomers, the distant galaxies did not fit Hubble\u2019s classification scheme at all. Remember that Hubble found that nearly all nearby galaxies could be classified into a few categories, depending on whether they were ellipticals or spirals. The distant galaxies observed by the Hubble Space Telescope look very different from present-day galaxies, without identifiable", " spiral arms, disks, and bulges (Figure 28.5). They also tend to be much clumpier than most galaxies today. In other words, it\u2019s becoming clear that the shapes of galaxies have changed significantly over time. In fact, we now know that the Hubble scheme works well for only the last half of the age of the universe. Before then, galaxies were much more chaotic. 998 Chapter 28 The Evolution and Distribution of Galaxies Figure 28.5. Early Galaxies. This Hubble Space Telescope image shows what are probably \u201cgalaxies under construction\u201d in the early universe. The boxes in this color image show enlargements of 18 groups of stars smaller than galaxies as we know them. All these objects emitted their light about 11 billion years ago. They are typically only about 2,000 light-years across, which is much smaller than the Milky Way, with its diameter of 100,000 light-years. These 18 objects are found in a region only 2 million light-years across and are close enough together that they will probably collide and merge to build one or more normal galaxies. (credit: modification of work by Rogier Windhorst (Arizona State University) and NASA) It\u2019s not just the shapes that are different. Nearly all the galaxies at distances greater than 11 billion light- years\u2014that is, galaxies that we are seeing when they were less than 3 billion years old\u2014are extremely blue, indicating that they contain a lot of young stars and that star formation in them is occurring at a higher rate than in nearby galaxies. Observations also show that very distant galaxies are systematically smaller on average than nearby galaxies. Relatively few galaxies present before the universe was about 8 billion years old have masses greater than 1011 MSun. That\u2019s 1/20 the mass of the Milky Way if we include its dark matter halo. Eleven billion years ago, there were only a few galaxies with masses greater than 1010 MSun. What we see instead seem to be small pieces or fragments of galactic material (Figure 28.6). When we look at galaxies that emitted their light 11 to 12 billion years ago, we now believe we are seeing the seeds of elliptical galaxies and of the central bulges of spirals. Over time, these smaller galaxies collided and merged to build up today\u2019s large galaxies. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 999 Bear in mind that stars that formed more than 11 billion years ago will be very old stars today. Indeed when we look nearby (at galaxies we see closer to our time), we find mostly old stars in the nuclear bulges of nearby spirals and in elliptical galaxies. Figure 28.6. One of the Farthest, Faintest, and Smallest Galaxies Ever Seen. The small white boxes, labeled a, b, and c, mark the positions of three images of the same galaxy. These multiple images were produced by the massive cluster of galaxies known as Abell 2744, which is located between us and the galaxy and acts as a gravitational lens. The arrows in the enlarged insets at right point to the galaxy. Each magnified image makes the galaxy appear as much as 10 times larger and brighter than it would look without the intervening lens. This galaxy emitted the light we observe today when the universe was only about 500 million years old. When the light was emitted the galaxy was tiny\u2014only 850 light-years across, or 500 times smaller than the Milky, and its mass was only 40 million times the mass of the Sun. Star formation is going on in this galaxy, but it appears red in the image because of its large redshift. (credit: modification of work by NASA, ESA, A. Zitrin (California Institute of Technology), and J. Lotz, M. Mountain, A. Koekemoer, and the HFF Team (STScI)) What such observations are showing us is that galaxies have grown in size as the universe has aged. Not only were galaxies smaller several billion years ago, but there were more of them; gas-rich galaxies, particularly the less luminous ones, were much more numerous then than they are today. Those are some of the basic observations we can make of individual galaxies (and their evolution) looking back in cosmic time. Now we want to turn to the larger context. If stars are grouped into galaxies, are the galaxies also grouped in some way? In the third section of this chapter, we\u2019ll explore the largest structures known in the universe. 28.2 GALAXY MERGERS AND ACTIVE GALACTIC NUCLEI Learning Objectives By the end of this section, you will be able to: 1000 Chapter 28 The Evolution and Distribution of Galaxies Explain how galaxies grow by merging with other galaxies and by consuming smaller galaxies (for lunch) Describe the effects that supermassive black", " holes in the centers of most galaxies have on the fate of their host galaxies One of the conclusions astronomers have reached from studying distant galaxies is that collisions and mergers of whole galaxies play a crucial role in determining how galaxies acquired the shapes and sizes we see today. Only a few of the nearby galaxies are currently involved in collisions, but detailed studies of those tell us what to look for when we seek evidence of mergers in very distant and very faint galaxies. These in turn give us important clues about the different evolutionary paths galaxies have taken over cosmic time. Let\u2019s examine in more detail what happens when two galaxies collide. Mergers and Cannibalism Figure 28.1 shows a dynamic view of two galaxies that are colliding. The stars themselves in this pair of galaxies will not be affected much by this cataclysmic event. (See the Astronomy Basics feature box Why Galaxies Collide but Stars Rarely Do.) Since there is a lot of space between the stars, a direct collision between two stars is very unlikely. However, the orbits of many of the stars will be changed as the two galaxies move through each other, and the change in orbits can totally alter the appearance of the interacting galaxies. A gallery of interesting colliding galaxies is shown in Figure 28.7. Great rings, huge tendrils of stars and gas, and other complex structures can form in such cosmic collisions. Indeed, these strange shapes are the signposts that astronomers use to identify colliding galaxies. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1001 Figure 28.7. Gallery of Interacting Galaxies. (a and b) M82 (smaller galaxy at top) and M83 (spiral) are seen (a) in a black-and-white visible light image and (b) in radio waves given off by cold hydrogen gas. The hydrogen image shows that the two galaxies are wrapped in a common shroud of gas that is being tugged and stretched by the gravity of the two galaxies. (c) This close-up view by the Hubble Space Telescope shows some of the effects of this interaction on galaxy M82, including gas streaming outward (red tendrils) powered by supernovae explosions of massive stars formed in the burst of star formation that was a result of the collision. (d) Galaxy UGC 10214 (\u201cThe Tadpole\u201d) is a barred spiral galaxy 420 million light-years from the Milky Way that has been disrupted by the passage of a smaller galaxy. The interloper\u2019s gravity pulled out the long tidal tail, which is about 280,000 light-years long, and triggered bursts of star formation seen as blue clumps along the tail. (e) Galaxies NGC 4676 A and B are nicknamed \u201cThe Mice.\u201d In this Hubble Space Telescope image, you can see the long, narrow tails of stars pulled away from the galaxies by the interactions of the two spirals. (e) Arp 148 is a pair of galaxies that are caught in the act of merging to become one new galaxy. The two appear to have already passed through each other once, causing a shockwave that reformed one into a bright blue ring of star formation, like the ripples from a stone tossed into a pond. (credit a, b: modification of work by NRAO/AUI; credit c: modification of work by NASA, ESA, and The Hubble Heritage Team (STScI/AURA); credit d, e: modification of work by NASA, H. Ford (JHU), G. Illingworth (UCSC/LO), M.Clampin (STScI), G. Hartig (STScI), the ACS Science Team, and ESA; credit f: modification of work by NASA, ESA, the Hubble Heritage (STScI/ AURA)-ESA/Hubble Collaboration, and A. Evans (University of Virginia, Charlottesville/NRAO/Stony Brook University)) Why Galaxies Collide but Stars Rarely Do Throughout this book we have emphasized the large distances between objects in space. You might therefore have been surprised to hear about collisions between galaxies. Yet (except at the very cores of galaxies) we have not worried at all about stars inside a galaxy colliding with each other. Let\u2019s see why there is a difference. The reason is that stars are pitifully small compared to the distances between them. Let\u2019s use our Sun as an example. The Sun is about 1.4 million kilometers wide, but is separated from the closest other star by about 4 light-years, or about 38 trillion kilometers. In other words, the Sun is 27 million of its own 1002 Chapter 28 The Evolution and Distribution of Galaxies diameters from its nearest neighbor. If the Sun were a grapefruit in New York City, the nearest star would be another grapefruit in San Francisco", ". This is typical of stars that are not in the nuclear bulge of a galaxy or inside star clusters. Let\u2019s contrast this with the separation of galaxies. The visible disk of the Milky Way is about 100,000 light-years in diameter. We have three satellite galaxies that are just one or two Milky Way diameters away from us (and will probably someday collide with us). The closest major spiral is the Andromeda Galaxy (M31), about 2.4 million light-years away. If the Milky Way were a pancake at one end of a big breakfast table, M31 would be another pancake at the other end of the same table. Our nearest large galaxy neighbor is only 24 of our Galaxy\u2019s diameters from us, and it will begin to crash into the Milky Way in about 3 billion years. Galaxies in rich clusters are even closer together than those in our neighborhood (see The Distribution of Galaxies in Space). Thus, the chances of galaxies colliding are far greater than the chances of stars in the disk of a galaxy colliding. And we should note that the difference between the separation of galaxies and stars also means that when galaxies do collide, their stars almost always pass right by each other like smoke passing through a screen door. The details of galaxy collisions are complex, and the process can take hundreds of millions of years. Thus, collisions are best simulated on a computer (Figure 28.8), where astronomers can calculate the slow interactions of stars, and clouds of gas and dust, via gravity. These calculations show that if the collision is slow, the colliding galaxies may coalesce to form a single galaxy. Figure 28.8. Computer Simulation of a Galaxy Collision. This computer simulation starts with two spiral galaxies merging and ends with a single elliptical galaxy. The colors show the colors of stars in the system; note the bursts of blue color as copious star formation gets triggered by the interaction. The timescale from start to finish in this sequence is about a billion years. (credit: modification of work by P. Jonsson (Harvard-Smithsonian Center for Astrophysics), G. Novak (Princeton University), and T. J. Cox (Carnegie Observatories)) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1003 When two galaxies of equal size are involved in a collision, we call such an interaction a merger (the term applied in the business world to two equal companies that join forces). But small galaxies can also be swallowed by larger ones\u2014a process astronomers have called, with some relish, galactic cannibalism (Figure 28.9). Modern personal computers are more than powerful enough to compute what happens when galaxies collide. Here\u2019s a website and Java applet (https://openstax.org/l/30whgalaxcoll) that will let you try your own hand at crashing two spiral galaxies together from the comfort of your own home or dorm room. By changing a few basic controls such as the relative masses, their separation, and the orientation of each galaxy\u2019s disk, you can create a wide range of resulting merger results. (You can also download a similar app (https://openstax.org/l/30iphoneapp) for your iPhone or iPad.) Figure 28.9. Galactic Cannibalism. (a) This Hubble image shows the eerie silhouette of dark dust clouds against the glowing nucleus of the elliptical galaxy NGC 1316. Elliptical galaxies normally contain very little dust. These clouds are probably the remnant of a small companion galaxy that was cannibalized (eaten) by NGC 1316 about 100 million years ago. (b) The highly disturbed galaxy NGC 6240, imaged by Hubble Space Telescope (background image) and Chandra X-ray Telescope (both insets) is apparently the product of a merger between two gas-rich spiral galaxies. The X-ray images show that there is not one but two nuclei, both glowing brightly in X-rays and separated by only 4000 lightyears. These are likely the locations of two supermassive black holes that inhabited the cores of the two galaxies pre-merger; here they are participating in a kind of \u201cdeath spiral,\u201d in which the two black holes themselves will merge to become one. (credit a: modification of work by NASA, ESA, and The Hubble Heritage Team (STScI/AURA); credit b: X-ray: NASA/CXC/MPE/S.Komossa et al.; Optical: NASA/STScI/R.P.van der Marel & J.Gerssen) The very large elliptical galaxies we discussed in Galaxies probably form by cannibalizing a variety of smaller galaxies in their clusters. These \u201cmonster\u201d galaxies frequently possess more than one nucleus and have probably acquired their unusually high luminosities by", " swallowing nearby galaxies. The multiple nuclei are the remnants of their victims (Figure 28.9). Many of the large, peculiar galaxies that we observe also owe their chaotic shapes to past interactions. Slow collisions and mergers can even transform two or more spiral galaxies into a single elliptical galaxy. 1004 Chapter 28 The Evolution and Distribution of Galaxies A change in shape is not all that happens when galaxies collide. If either galaxy contains interstellar matter, the collision can compress the gas and trigger an increase in the rate at which stars are being formed\u2014by as much as a factor of 100. Astronomers call this abrupt increase in the number of stars being formed a starburst, and the galaxies in which the increase occurs are termed starburst galaxies (Figure 28.10). In some interacting galaxies, star formation is so intense that all the available gas is exhausted in only a few million years; the burst of star formation is clearly only a temporary phenomenon. While a starburst is going on, however, the galaxy where it is taking place becomes much brighter and much easier to detect at large distances. Figure 28.10. Starburst Associated with Colliding Galaxies. (a) Three of the galaxies in the small group known as Stephan\u2019s Quintet are interacting gravitationally with each other (the galaxy at upper left is actually much closer than the other three and is not part of this interaction), resulting in the distorted shapes seen here. Long strings of young, massive blue stars and hundreds of star formation regions glowing in the pink light of excited hydrogen gas are also results of the interaction. The ages of the star clusters range from 2 million to 1 billion years old, suggesting that there have been several different collisions within this group of galaxies, each leading to bursts of star formation. The three interacting members of Stephan\u2019s Quintet are located at a distance of 270 million light-years. (b) Most galaxies form new stars at a fairly slow rate, but members of a rare class known as starburst galaxies blaze with extremely active star formation. The galaxy II Zw 096 is one such starburst galaxy, and this combined image using both Hubble and Spitzer Space Telescope data shows that it is forming bright clusters of new stars at a prodigious rate. The blue colors show the merging galaxies in visible light, while the red colors show infrared radiation from the dusty region where star formation is happening. This galaxy is at a distance of 500 million light-years and has a diameter of about 50,000 light-years, about half the size of the Milky Way. (credit a: modification of work by NASA, ESA, and the Hubble SM4 ERO Team; credit b: modification of work by NASA/JPL-Caltech/STScI) When astronomers finally had the tools to examine a significant number of galaxies that emitted their light 11 to 12 billion years ago, they found that these very young galaxies often resemble nearby starburst galaxies that are involved in mergers: they also have multiple nuclei and peculiar shapes, they are usually clumpier than normal galaxies today, with multiple intense knots and lumps of bright starlight, and they have higher rates of star formation than isolated galaxies. They also contain lots of blue, young, type O and B stars, as do nearby merging galaxies. Galaxy mergers in today\u2019s universe are rare. Only about five percent of nearby galaxies are currently involved in interactions. Interactions were much more common billions of years ago (Figure 28.11) and helped build up the \u201cmore mature\u201d galaxies we see in our time. Clearly, interactions of galaxies have played a crucial role in their evolution. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1005 Figure 28.11. Collisions of Galaxies in a Distant Cluster. The large picture on the left shows the Hubble Space Telescope image of a cluster of galaxies at a distance of about 8 billion light-years. Among the 81 galaxies in the cluster that have been examined in some detail, 13 are the result of recent collisions of pairs of galaxies. The eight smaller images on the right are close-ups of some of the colliding galaxies. The merger process typically takes a billion years or so. (credit: modification of work by Pieter van Dokkum, Marijn Franx (University of Groningen/Leiden), ESA and NASA) Active Galactic Nuclei and Galaxy Evolution While galaxy mergers are huge, splashy events that completely reshape entire galaxies on scales of hundreds of thousands of light-years and can spark massive bursts of star formation, accreting black holes inside galaxies can also disturb and alter the evolution of their host galaxies. You learned in Active Galaxies, Quasars, and Supermassive Black Holes about a family of objects known as active galactic nuclei (AGN), all of them powered by supermassive black holes. If the black hole is surrounded", " by enough gas, some of the gas can fall into the black hole, getting swept up on the way into an accretion disk, a compact, swirling maelstrom perhaps only 100 AU across (the size of our solar system). Within the disk the gas heats up until it shines brilliantly even in X-rays, often outshining the rest of the host galaxy with its billions of stars. Supermassive black holes and their accretion disks can be violent and powerful places, with some material getting sucked into the black hole but even more getting shot out along huge jets perpendicular to the disk. These powerful jets can extend far outside the starry edge of the galaxy. AGN were much more common in the early universe, in part because frequent mergers provided a fresh gas supply for the black hole accretion disks. Examples of AGN in the nearby universe today include the one in galaxy M87 (see Figure 27.7), which sports a jet of material shooting out from its nucleus at speeds close to the speed of light, and the one in the bright galaxy NGC 5128, also known as Centaurus A (see Figure 28.12). 1006 Chapter 28 The Evolution and Distribution of Galaxies Figure 28.12. Composite View of the Galaxy Centaurus A. This artificially colored image was made using data from three different telescopes: submillimeter radiation with a wavelength of 870 microns is shown in orange; X-rays are seen in blue; and visible light from stars is shown in its natural color. Centaurus A has an active galactic nucleus that is powering two jets, seen in blue and orange, reaching in opposite directions far outside the galaxy\u2019s stellar disk, and inflating two huge lobes, or clouds, of hot X-ray-emitting gas. Centaurus is at a distance of 13 million lightyears, making it one of the closest active galaxies we know. (credit: modification of work by ESO/WFI (Optical); MPIfR/ESO/APEX/A. Weiss et al. (Submillimeter); NASA/CXC/CfA/R.Kraft et al. (X-ray)) Many highly accelerated particles move with the jets in such galaxies. Along the way, the particles in the jets can plow into gas clouds in the interstellar medium, breaking them apart and scattering them. Since denser clouds of gas and dust are required for material to clump together to make stars, the disruption of the clouds can halt star formation in the host galaxy or cut it off before it even begins. In this way, quasars and other kinds of AGN can play a crucial role in the evolution of their galaxies. For example, there is growing evidence that the merger of two gas-rich galaxies not only produces a huge burst of star formation, but also triggers AGN activity in the core of the new galaxy. That activity, in turn, could then slow down or shut off the burst of star formation\u2014which could have significant implications for the apparent shape, brightness, chemical content, and stellar components of the entire galaxy. Astronomers refer to that process as AGN feedback, and it is apparently an important factor in the evolution of most galaxies. 28.3 THE DISTRIBUTION OF GALAXIES IN SPACE Learning Objectives By the end of this section, you will be able to: Explain the cosmological principle and summarize the evidence that it applies on the largest scales of the known universe Describe the contents of the Local Group of galaxies Distinguish among groups, clusters, and superclusters of galaxies Describe the largest structures seen in the universe, including voids In the preceding section, we emphasized the role of mergers in shaping the evolution of galaxies. In order to collide, galaxies must be fairly close together. To estimate how often collisions occur and how they affect galaxy evolution, astronomers need to know how galaxies are distributed in space and over cosmic time. Are most of This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1007 them isolated from one another or do they congregate in groups? If they congregate, how large are the groups and how and when did they form? And how, in general, are galaxies and their groups arranged in the cosmos? Are there as many in one direction of the sky as in any other, for example? How did galaxies get to be arranged the way we find them today? Edwin Hubble found answers to some of these questions only a few years after he first showed that the spiral nebulae were galaxies and not part of our Milky Way. As he examined galaxies all over the sky, Hubble made two discoveries that turned out to be crucial for studies of the evolution of the universe. The Cosmological Principle Hubble made his observations with what were then the world\u2019s largest telescopes\u2014the 100-inch and 60-inch reflectors on Mount Wilson. These telescopes have small", " fields of view: they can see only a small part of the heavens at a time. To photograph the entire sky with the 100-inch telescope, for example, would have taken longer than a human lifetime. So instead, Hubble sampled the sky in many regions, much as Herschel did with his star gauging (see The Architecture of the Galaxy). In the 1930s, Hubble photographed 1283 sample areas, and on each print, he carefully counted the numbers of galaxy images ((Figure 28.13).). The first discovery Hubble made from his survey was that the number of galaxies visible in each area of the sky is about the same. (Strictly speaking, this is true only if the light from distant galaxies is not absorbed by dust in our own Galaxy, but Hubble made corrections for this absorption.) He also found that the numbers of galaxies increase with faintness, as we would expect if the density of galaxies is about the same at all distances from us. To understand what we mean, imagine you are taking snapshots in a crowded stadium during a sold-out concert. The people sitting near you look big, so only a few of them will fit into a photo. But if you focus on the people sitting in seats way on the other side of the stadium, they look so small that many more will fit into your picture. If all parts of the stadium have the same seat arrangements, then as you look farther and farther away, your photo will get more and more crowded with people. In the same way, as Hubble looked at fainter and fainter galaxies, he saw more and more of them. Figure 28.13. Hubble at Work. Edwin Hubble at the 100-inch telescope on Mount Wilson. (credit: NASA) Hubble\u2019s findings are enormously important, for they indicate that the universe is both isotropic and homogeneous\u2014it looks the same in all directions, and a large volume of space at any given redshift or distance is much like any other volume at that redshift. If that is so, it does not matter what section of the universe we observe (as long as it\u2019s a sizable portion): any section will look the same as any other. Hubble\u2019s results\u2014and many more that have followed in the nearly 100 years since then\u2014imply not only that the universe is about the same everywhere (apart from changes with time) but also that aside from small-scale local differences, the part we can see around us is representative of the whole. The idea that the universe is 1008 Chapter 28 The Evolution and Distribution of Galaxies the same everywhere is called the cosmological principle and is the starting assumption for nearly all theories that describe the entire universe (see The Big Bang). Without the cosmological principle, we could make no progress at all in studying the universe. Suppose our own local neighborhood were unusual in some way. Then we could no more understand what the universe is like than if we were marooned on a warm south-sea island without outside communication and were trying to understand the geography of Earth. From our limited island vantage point, we could not know that some parts of the planet are covered with snow and ice, or that large continents exist with a much greater variety of terrain than that found on our island. Hubble merely counted the numbers of galaxies in various directions without knowing how far away most of them were. With modern instruments, astronomers have measured the velocities and distances of hundreds of thousands of galaxies, and so built up a meaningful picture of the large-scale structure of the universe. In the rest of this section, we describe what we know about the distribution of galaxies, beginning with those that are nearby. The Local Group The region of the universe for which we have the most detailed information is, as you would expect, our own local neighborhood. It turns out that the Milky Way Galaxy is a member of a small group of galaxies called, not too imaginatively, the Local Group. It is spread over about 3 million light-years and contains more than 54 members. There are three large spiral galaxies (our own, the Andromeda galaxy, and M33), two intermediate ellipticals, and many dwarf ellipticals and irregular galaxies. New members of the Local Group are still being discovered. We mentioned in The Milky Way Galaxy a dwarf galaxy only about 80,000 light-years from Earth and about 50,000 light-years from the center of the galaxy that was discovered in 1994 in the constellation of Sagittarius. (This dwarf is actually venturing too close to the much larger Milky Way and will eventually be consumed by it.) Many of the recent discoveries have been made possible by the new generation of automated, sensitive, wide- field surveys, such as the Sloan Digital Sky Survey, that map the positions of millions of stars across most of the visible sky. By digging into the data with sophisticated computer programs, astronomers have turned up numerous tiny, faint dwarf galaxies that are all but invisible to the eye even in those deep telescopic images. These new", " findings may help solve a long-standing problem: the prevailing theories of how galaxies form predicted that there should be more dwarf galaxies around big galaxies like the Milky Way than had been observed\u2014and only now do we have the tools to find these faint and tiny galaxies and begin to compare the numbers of them with theoretical predictions You can read more about the Sloan survey (https://openstax.org/l/30sloansurvey) and its dramatic results. And check out this brief animation (https://openstax.org/l/30anifliarrgal) of a flight through the arrangement of the galaxies as revealed by the survey. Several new dwarf galaxies have also been found near the Andromeda galaxy. Such dwarf galaxies are difficult to find because they typically contain relatively few stars, and it is hard to distinguish them from the foreground stars in our own Milky Way. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1009 Figure 28.14 is a rough sketch showing where the brighter members of the Local Group are located. The average of the motions of all the galaxies in the Local Group indicates that its total mass is about 4 \u00d7 1012 MSun, and at least half of this mass is contained in the two giant spirals\u2014the Andromeda galaxy and the Milky Way Galaxy. And bear in mind that a substantial amount of the mass in the Local Group is in the form of dark matter. Figure 28.14. Local Group. This illustration shows some members of the Local Group of galaxies, with our Milky Way at the center. The exploded view at the top shows the region closest to the Milky Way and fits into the bigger view at the bottom as shown by the dashed lines. The three largest galaxies among the three dozen or so members of the Local Group are all spirals; the others are small irregular galaxies and dwarf ellipticals. A number of new members of the group have been found since this map was made. Neighboring Groups and Clusters Small galaxy groups like ours are hard to notice at larger distances. However, there are much more substantial groups called galaxy clusters that are easier to spot even many millions of light-years away. Such clusters are described as poor or rich depending on how many galaxies they contain. Rich clusters have thousands or even tens of thousands of galaxies, although many of the galaxies are quite faint and hard to detect. The nearest moderately rich galaxy cluster is called the Virgo Cluster, after the constellation in which it is seen. It is about 50 million light-years away and contains thousands of members, of which a few are shown in Figure 28.15. The giant elliptical (and very active) galaxy M87, which you came to know and love in the chapter on Active Galaxies, Quasars, and Supermassive Black Holes, belongs to the Virgo Cluster. 1010 Chapter 28 The Evolution and Distribution of Galaxies Figure 28.15. Central Region of the Virgo Cluster. Virgo is the nearest rich cluster and is at a distance of about 50 million light-years. It contains hundreds of bright galaxies. In this picture you can see only the central part of the cluster, including the giant elliptical galaxy M87, just below center. Other spirals and ellipticals are visible; the two galaxies to the top right are known as \u201cThe Eyes.\u201d (credit: modification of work by Chris Mihos (Case Western Reserve University)/ESO) A good example of a cluster that is much larger than the Virgo complex is the Coma cluster, with a diameter of at least 10 million light-years (Figure 28.16). Some 250 to 300 million light-years distant, this cluster is centered on two giant ellipticals whose luminosities equal about 400 billion Suns each. Thousands of galaxies have been observed in Coma, but the galaxies we see are almost certainly only part of what is really there. Dwarf galaxies are too faint to be seen at the distance of Coma, but we expect they are part of this cluster just as they are part of nearer ones. If so, then Coma likely contains tens of thousands of galaxies. The total mass of this cluster is about 4 \u00d7 1015 MSun (enough mass to make 4 million billion stars like the Sun). Let\u2019s pause here for a moment of perspective. We are now discussing numbers by which even astronomers sometimes feel overwhelmed. The Coma cluster may have 10, 20, or 30 thousand galaxies, and each galaxy has billions and billions of stars. If you were traveling at the speed of light, it would still take you more than 10 million years (longer than the history of the human species) to cross this giant swarm of galaxies. And if you lived on a planet on the outskirts of one of these galaxies, many other members of the cluster would be close enough to be noteworthy sights in your nighttime sky. This OpenStax book is available for free at http", "://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1011 Figure 28.16. Central Region of the Coma Cluster. This combined visible-light (from the Sloan Digital Sky Survey) and infrared (from the Spitzer Space Telescope) image has been color coded so that faint dwarf galaxies are seen as green. Note the number of little green smudges on the image. The cluster is roughly 320 million light-years away from us. (credit: modification of work by NASA/JPL-Caltech/L. Jenkins (GSFC)) Really rich clusters such as Coma usually have a high concentration of galaxies near the center. We can see giant elliptical galaxies in these central regions but few, if any, spiral galaxies. The spirals that do exist generally occur on the outskirts of clusters. We might say that ellipticals are highly \u201csocial\u201d: they are often found in groups and very much enjoy \u201changing out\u201d with other ellipticals in crowded situations. It is precisely in such crowds that collisions are most likely and, as we discussed earlier, we think that most large ellipticals are built through mergers of smaller galaxies. Spirals, on the other hand, are more \u201cshy\u201d: they are more likely to be found in poor clusters or on the edges of rich clusters where collisions are less likely to disrupt the spiral arms or strip out the gas needed for continued star formation Gravitational Lensing As we saw in Black Holes and Curved Spacetime, spacetime is more strongly curved in regions where the gravitational field is strong. Light passing very near a concentration of matter appears to follow a curved path. In the case of starlight passing close to the Sun, we measure the position of the distant star to be slightly different from its true position. Now let\u2019s consider the case of light from a distant galaxy or quasar that passes near a concentration of matter such as a cluster of galaxies on its journey to our telescopes. According to general relativity, the 1012 Chapter 28 The Evolution and Distribution of Galaxies light path may be bent in a variety of ways; as a result we can observe distorted and even multiple images (Figure 28.17). Figure 28.17. Gravitational Lensing. This drawing shows how a gravitational lens can make two images. Two light rays from a distant quasar are shown being bent while passing a foreground galaxy; they then arrive together at Earth. Although the two beams of light contain the same information, they now appear to come from two different points on the sky. This sketch is oversimplified and not to scale, but it gives a rough idea of the lensing phenomenon. Gravitational lenses can produce not only double images, as shown in Figure 28.17, but also multiple images, arcs, or rings. The first gravitational lens discovered, in 1979, showed two images of the same distant object. Eventually, astronomers used the Hubble Space Telescope to capture remarkable images of the effects of gravitational lenses. One example is shown in Figure 28.18. Figure 28.18. Multiple Images of a Gravitationally Lensed Supernova. Light from a supernova at a distance of 9 billion light-years passed near a galaxy in a cluster at a distance of about 5 billion light-years. In the enlarged inset view of the galaxy, the arrows point to the multiple images of the exploding star. The images are arranged around the galaxy in a cross-shaped pattern called an Einstein Cross. The blue streaks wrapping around the galaxy are the stretched images of the supernova\u2019s host spiral galaxy, which has been distorted by the warping of space. (credit: modification of work by NASA, ESA, and S. Rodney (JHU) and the FrontierSN team; T. Treu (UCLA), P. Kelly (UC Berkeley), and the GLASS team; J. Lotz (STScI) and the Frontier Fields team; M. Postman (STScI) and the CLASH team; and Z. Levay (STScI)) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1013 General relativity predicts that the light from a distant object may also be amplified by the lensing effect, thereby making otherwise invisible objects bright enough to detect. This is particularly useful for probing the earliest stages of galaxy formation, when the universe was young. Figure 28.19 shows an example of a very distant faint galaxy that we can study in detail only because its light path passes through a large concentration of massive galaxies and we now see a brighter image of it. Figure 28.19. Distorted Images of a Distant Galaxy Produced by Gravitational Lensing in a Galaxy Cluster. The rounded outlines show the location of distinct, distorted images of the background galaxy resulting from lensing by the mass in the cluster. The image in", " the box at lower left is a reconstruction of what the lensed galaxy would look like in the absence of the cluster, based on a model of the cluster\u2019s mass distribution, which can be derived from studying the distorted galaxy images. The reconstruction shows far more detail about the galaxy than could have been seen in the absence of lensing. As the image shows, this galaxy contains regions of star formation glowing like bright Christmas tree bulbs. These are much brighter than any star-formation regions in our Milky Way Galaxy. (credit: modification of work by NASA, ESA, and Z. Levay (STScI)) We should note that the visible mass in a galaxy is not the only possible gravitational lens. Dark matter can also reveal itself by producing this effect. Astronomers are using lensed images from all over the sky to learn more about where dark matter is located and how much of it exists. Superclusters and Voids After astronomers discovered clusters of galaxies, they naturally wondered whether there were still larger structures in the universe. Do clusters of galaxies gather together? To answer this question, we must be able to 1014 Chapter 28 The Evolution and Distribution of Galaxies map large parts of the universe in three dimensions. We must know not only the position of each galaxy on the sky (that\u2019s two dimensions) but also its distance from us (the third dimension). This means we must be able to measure the redshift of each galaxy in our map. Taking a spectrum of each individual galaxy to do this is a much more time-consuming task than simply counting galaxies seen in different directions on the sky, as Hubble did. Today, astronomers have found ways to get the spectra of many galaxies in the same field of view (sometimes hundreds or even thousands at a time) to cut down the time it takes to finish their three-dimensional maps. Larger telescopes are also able to measure the redshifts\u2014and therefore the distances\u2014of much more distant galaxies and (again) to do so much more quickly than previously possible. Another challenge astronomers faced in deciding how to go about constructing a map of the universe is similar to that confronted by the first team of explorers in a huge, uncharted territory on Earth. Since there is only one band of explorers and an enormous amount of land, they have to make choices about where to go first. One strategy might be to strike out in a straight line in order to get a sense of the terrain. They might, for example, cross some mostly empty prairies and then hit a dense forest. As they make their way through the forest, they learn how thick it is in the direction they are traveling, but not its width to their left or right. Then a river crosses their path; as they wade across, they can measure its width but learn nothing about its length. Still, as they go on in their straight line, they begin to get some sense of what the landscape is like and can make at least part of a map. Other explorers, striking out in other directions, will someday help fill in the remaining parts of that map. Astronomers have traditionally had to make the same sort of choices. We cannot explore the universe in every direction to infinite \u201cdepth\u201d or sensitivity: there are far too many galaxies and far too few telescopes to do the job. But we can pick a single direction or a small slice of the sky and start mapping the galaxies. Margaret Geller, the late John Huchra, and their students at the Harvard-Smithsonian Center for Astrophysics pioneered this technique, and several other groups have extended their work to cover larger volumes of space Margaret Geller: Cosmic Surveyor Born in 1947, Margaret Geller is the daughter of a chemist who encouraged her interest in science and helped her visualize the three-dimensional structure of molecules as a child. (It was a skill that would later come in very handy for visualizing the three-dimensional structure of the universe.) She remembers being bored in elementary school, but she was encouraged to read on her own by her parents. Her recollections also include subtle messages from teachers that mathematics (her strong early interest) was not a field for girls, but she did not allow herself to be deterred. Geller obtained a BA in physics from the University of California at Berkeley and became the second woman to receive a PhD in physics from Princeton. There, while working with James Peebles, one of the world\u2019s leading cosmologists, she became interested in problems relating to the large-scale structure of the universe. In 1980, she accepted a research position at the Harvard-Smithsonian Center for Astrophysics, one of the nation\u2019s most dynamic institutions for astronomy research. She saw that to make progress in understanding how galaxies and clusters are organized, a far more intensive series of surveys was required. Although it would not bear fruit for many years, Geller and her collaborators began the long, arduous task of mapping the galaxies ((Figure 28.20).). This OpenStax book is available for", " free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1015 Figure 28.20. Margaret Geller. Geller\u2019s work mapping and researching galaxies has helped us to better understand the structure of the universe. (credit: modification of work by Massimo Ramella) Her team was fortunate to be given access to a telescope that could be dedicated to their project, the 60-inch reflector on Mount Hopkins, near Tucson, Arizona, where they and their assistants took spectra to determine galaxy distances. To get a slice of the universe, they pointed their telescope at a predetermined position in the sky and then let the rotation of Earth bring new galaxies into their field of view. In this way, they measured the positions and redshifts of over 18,000 galaxies and made a wide range of interesting maps to display their data. Their surveys now include \u201cslices\u201d in both the Northern and Southern Hemispheres. As news of her important work spread beyond the community of astronomers, Geller received a MacArthur Foundation Fellowship in 1990. These fellowships, popularly called \u201cgenius awards,\u201d are designed to recognize truly creative work in a wide range of fields. Geller continues to have a strong interest in visualization and has (with filmmaker Boyd Estus) made several award-winning videos explaining her work to nonscientists (one is titled So Many Galaxies... So Little Time). She has appeared on a variety of national news and documentary programs, including the MacNeil/Lehrer NewsHour, The Astronomers, and The Infinite Voyage. Energetic and outspoken, she has given talks on her work to many audiences around the country, and works hard to find ways to explain the significance of her pioneering surveys to the public. \u201cIt\u2019s exciting to discover something that nobody\u2019s seen before. [To be] one of the first three people to ever see that slice of the universe [was] sort of being like Columbus.... Nobody expected such a striking pattern!\u201d\u2014Margaret Geller Find out more about Geller and Huchra\u2019s work (including interviews with Geller) in this 4-minute NOVA (https://openstax.org/l/30gellhucwork) video. You can also learn more about their conclusions (https://openstax.org/l/30gellhucconc) and additional research it led to. 1016 Chapter 28 The Evolution and Distribution of Galaxies The largest universe mapping project to date is the Sloan Digital Sky Survey (see the Making Connections feature box Astronomy and Technology: The Sloan Digital Sky Survey at the end of this section). A plot of the distribution of galaxies mapped by the Sloan survey is shown in Figure 28.21. To the surprise of astronomers, maps like the one in the figure showed that clusters of galaxies are not arranged uniformly throughout the universe, but are found in huge filamentary superclusters that look like great arcs of inkblots splattered across a page. The superclusters resemble an irregularly torn sheet of paper or a pancake in shape\u2014they can extend for hundreds of millions of light-years in two dimensions, but are only 10 to 20 million light-years thick in the third dimension. Detailed study of some of these structures shows that their masses are a few times 1016 MSun, which is 10,000 times more massive than the Milky Way Galaxy Check out this animated visualization (https://openstax.org/l/30anivisslosur) of large-scale structure from the Sloan survey. Figure 28.21. Sloan Digital Sky Survey Map of the Large-Scale Structure of the Universe. This image shows slices from the SDSS map. The point at the center corresponds to the Milky Way and might say \u201cYou Are Here!\u201d Points on the map moving outward from the center are farther away. The distance to the galaxies is indicated by their redshifts (following Hubble\u2019s law), shown on the horizontal line going right from the center. The redshift z = \u0394\u03bb/\u03bb, where \u0394\u03bb is the difference between the observed wavelength and the wavelength \u03bb emitted by a nonmoving source in the laboratory. Hour angle on the sky is shown around the circumference of the circular graph. The colors of the galaxies indicate the ages of their stars, with the redder color showing galaxies that are made of older stars. The outer circle is at a distance of two billion light-years from us. Note that red (older stars) galaxies are more strongly clustered than blue galaxies (young stars). The unmapped areas are where our view of the universe is obstructed by dust in our own Galaxy. (credit: modification of work by M. Blanton and the Sloan Digital Sky Survey) Separating the filaments and sheets in a supercluster are voids", ", which look like huge empty bubbles walled in by the great arcs of galaxies. They have typical diameters of 150 million light-years, with the clusters of galaxies concentrated along their walls. The whole arrangement of filaments and voids reminds us of a sponge, the inside of a honeycomb, or a hunk of Swiss cheese with very large holes. If you take a good slice or cross-section through any of these, you will see something that looks roughly like Figure 28.21. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1017 Before these voids were discovered, most astronomers would probably have predicted that the regions between giant clusters of galaxies were filled with many small groups of galaxies, or even with isolated individual galaxies. Careful searches within these voids have found few galaxies of any kind. Apparently, 90 percent of the galaxies occupy less than 10 percent of the volume of space Galaxy Distribution To determine the distribution of galaxies in three-dimensional space, astronomers have to measure their positions and their redshifts. The larger the volume of space surveyed, the more likely the measurement is a fair sample of the universe as a whole. However, the work involved increases very rapidly as you increase the volume covered by the survey. Let\u2019s do a quick calculation to see why this is so. Suppose that you have completed a survey of all the galaxies within 30 million light-years and you now want to survey to 60 million light-years. What volume of space is covered by your second survey? How much larger is this volume than the volume of your first survey? Remember that the volume of a sphere, V, is given by the formula V = 4/3\u03c0R3, where R is the radius of the sphere. Solution Since the volume of a sphere depends on R3 and the second survey reaches twice as far in distance, it will cover a volume that is 23 = 8 times larger. The total volume covered by the second survey will be (4/3)\u03c0 \u00d7 (60 million light-years)3 = 9 \u00d7 1023 light-years3. Check Your Learning Suppose you now want to expand your survey to 90 million light-years. What volume of space is covered, and how much larger is it than the volume of the second survey? Answer: The total volume covered is (4/3)\u03c0 \u00d7 (90 million light-years)3 = 3.05 \u00d7 1024 light-years3. The survey reaches 3 times as far in distance, so it will cover a volume that is 33 = 27 times larger. Even larger, more sensitive telescopes and surveys are currently being designed and built to peer farther and farther out in space and back in time. The new 50-meter Large Millimeter Telescope in Mexico and the Atacama Large Millimeter Array in Chile can detect far-infrared and millimeter-wave radiation from massive starbursting galaxies at redshifts and thus distances more than 90% of the way back to the Big Bang. These cannot be observed with visible light because their star formation regions are wrapped in clouds of thick dust. And in 2018, the 6.5-meter-diameter James Webb Space Telescope is scheduled to launch. It will be the first new major visible light and near-infrared telescope in space since Hubble was launched more than 25 years earlier. One of the major goals of this telescope is to observe directly the light of the first galaxies and even the first stars to shine, less than half a billion years after the Big Bang. At this point, if you have been thinking about our discussions of the expanding universe in Galaxies, you may be wondering what exactly in Figure 28.21 is expanding. We know that the galaxies and clusters of galaxies are 1018 Chapter 28 The Evolution and Distribution of Galaxies held together by their gravity and do not expand as the universe does. However, the voids do grow larger and the filaments move farther apart as space stretches (see The Big Bang). Astronomy and Technology: The Sloan Digital Sky Survey In Edwin Hubble\u2019s day, spectra of galaxies had to be taken one at a time. The faint light of a distant galaxy gathered by a large telescope was put through a slit, and then a spectrometer (also called a spectrograph) was used to separate the colors and record the spectrum. This was a laborious process, ill suited to the demands of making large-scale maps that require the redshifts of many thousands of galaxies. But new technology has come to the rescue of astronomers who seek three-dimensional maps of the universe of galaxies. One ambitious survey of the sky was produced using a special telescope, camera, and spectrograph atop the Sacramento Mountains of New Mexico. Called the Sloan Digital Sky Survey (SDSS), after the foundation that provided a large part of the funding, the program used a 2.5-meter telescope (about the same aperture as the", " Hubble) as a wide-angle astronomical camera. During a mapping program lasting more than ten years, astronomers used the SDSS\u2019s 30 charge-coupled devices (CCDs)\u2014sensitive electronic light detectors similar to those used in many digital cameras and cell phones\u2014to take images of over 500 million objects and spectra of over 3 million, covering more than one-quarter of the celestial sphere. Like many large projects in modern science, the Sloan Survey involved scientists and engineers from many different institutions, ranging from universities to national laboratories. Every clear night for more than a decade, astronomers used the instrument to make images recording the position and brightness of celestial objects in long strips of the sky. The information in each strip was digitally recorded and preserved for future generations. When the seeing (recall this term from Astronomical Instruments) was only adequate, the telescope was used for taking spectra of galaxies and quasars\u2014but it did so for up to 640 objects at a time. The key to the success of the project was a series of optical fibers, thin tubes of flexible glass that can transmit light from a source to the CCD that then records the spectrum. After taking images of a part of the sky and identifying which objects are galaxies, project scientists drilled an aluminum plate with holes for attaching fibers at the location of each galaxy. The telescope was then pointed at the right section of the sky, and the fibers led the light of each galaxy to the spectrometer for individual recording (Figure 28.22). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1019 Figure 28.22. Sloan Digital Sky Survey. (a) The Sloan Digital Sky Survey telescope is seen here in front of the Sacramento Mountains in New Mexico. (b) Astronomer Richard Kron inserts some of the optical fibers into the pre-drilled plate to enable the instruments to make many spectra of galaxies at the same time. (credit a, b: modification of work by the Sloan Digital Sky Survey) About an hour was sufficient for each set of spectra, and the pre-drilled aluminum plates could be switched quickly. Thus, it was possible to take as many as 5000 spectra in one night (provided the weather was good enough). The galaxy survey led to a more comprehensive map of the sky than has ever before been possible, allowing astronomers to test their ideas about large-scale structure and the evolution of galaxies against an impressive array of real data. The information recorded by the Sloan Survey staggers the imagination. The data came in at 8 megabytes per second (this means 8 million individual numbers or characters every second). Over the course of the project, scientists recorded over 15 terabytes, or 15 thousand billion bytes, which they estimate is comparable to the information contained in the Library of Congress. Organizing and sorting this volume of data and extracting the useful scientific results it contains is a formidable challenge, even in our information age. Like many other fields, astronomy has now entered an era of \u201cBig Data,\u201d requiring supercomputers and advanced computer algorithms to sift through all those terabytes of data efficiently. One very successful solution to the challenge of dealing with such large datasets is to turn to \u201ccitizen science,\u201d or crowd-sourcing, an approach the SDSS helped pioneer. The human eye is very good at recognizing subtle differences among shapes, such as between two different spiral galaxies, while computers often fail at such tasks. When Sloan project astronomers wanted to catalog the shapes of some of the millions of galaxies in their new images, they launched the \u201cGalaxy Zoo\u201d project: volunteers around the world were given a short training course online, then were provided with a few dozen galaxy images to classify by eye. The project was wildly successful, resulting in over 40 million galaxy classifications by more than 100,000 volunteers and the discovery of whole new types of galaxies. 1020 Chapter 28 The Evolution and Distribution of Galaxies Learn more about how you can be part of the project of classifying galaxies (https://openstax.org/l/ 30classgalax) in this citizen science effort. This program is part of a whole series of \u201ccitizen science\u201d projects (https://openstax.org/l/30citizscien) that enable people in all walks of life to be part of the research that professional astronomers (and scholars in a growing number of fields) need help with. 28.4 THE CHALLENGE OF DARK MATTER Learning Objectives By the end of this section, you will be able to: Explain how astronomers know that the solar system contains very little dark matter Summarize the evidence for dark matter in most galaxies Explain how we know that galaxy clusters are dominated by dark matter Relate the presence of dark matter to the average mass-to-light ratio of huge volumes of space containing many galaxies So far this chapter has focused almost entirely", " on matter that radiates electromagnetic energy\u2014stars, planets, gas, and dust. But, as we have pointed out in several earlier chapters (especially The Milky Way Galaxy), it is now clear that galaxies contain large amounts of dark matter as well. There is much more dark matter, in fact, than matter we can see\u2014which means it would be foolish to ignore the effect of this unseen material in our theories about the structure of the universe. (As many a ship captain in the polar seas found out too late, the part of the iceberg visible above the ocean\u2019s surface was not necessarily the only part he needed to pay attention to.) Dark matter turns out to be extremely important in determining the evolution of galaxies and of the universe as a whole. The idea that much of the universe is filled with dark matter may seem like a bizarre concept, but we can cite a historical example of \u201cdark matter\u201d much closer to home. In the mid-nineteenth century, measurements showed that the planet Uranus did not follow exactly the orbit predicted from Newton\u2019s laws if one added up the gravitational forces of all the known objects in the solar system. Some people worried that Newton\u2019s laws may simply not work so far out in our solar system. But the more straightforward interpretation was to attribute Uranus\u2019 orbital deviations to the gravitational effects of a new planet that had not yet been seen. Calculations showed where that planet had to be, and Neptune was discovered just about in the predicted location. In the same way, astronomers now routinely determine the location and amount of dark matter in galaxies by measuring its gravitational effects on objects we can see. And, by measuring the way that galaxies move in clusters, scientists have discovered that dark matter is also distributed among the galaxies in the clusters. Since the environment surrounding a galaxy is important in its development, dark matter must play a central role in galaxy evolution as well. Indeed, it appears that dark matter makes up most of the matter in the universe. But what is dark matter? What is it made of? We\u2019ll look next at the search for dark matter and the quest to determine its nature. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1021 Dark Matter in the Local Neighborhood Is there dark matter in our own solar system? Astronomers have examined the orbits of the known planets and of spacecraft as they journey to the outer planets and beyond. No deviations have been found from the orbits predicted on the basis of the masses of objects already discovered in our solar system and the theory of gravity. We therefore conclude that there is no evidence that there are large amounts of dark matter nearby. Astronomers have also looked for evidence of dark matter in the region of the Milky Way Galaxy that lies within a few hundred light-years of the Sun. In this vicinity, most of the stars are restricted to a thin disk. It is possible to calculate how much mass the disk must contain in order to keep the stars from wandering far above or below it. The total matter that must be in the disk is less than twice the amount of luminous matter. This means that no more than half of the mass in the region near the Sun can be dark matter. Dark Matter in and around Galaxies In contrast to our local neighborhood near the Sun and solar system, there is (as we saw in The Milky Way Galaxy) ample evidence strongly suggesting that about 90% of the mass in the entire galaxy is in the form of a halo of dark matter. In other words, there is apparently about nine times more dark matter than visible matter. Astronomers have found some stars in the outer regions of the Milky Way beyond its bright disk, and these stars are revolving very rapidly around its center. The mass contained in all the stars and all the interstellar matter we can detect in the galaxy does not exert enough gravitational force to explain how those fast-moving stars remain in their orbits and do not fly away. Only by having large amounts of unseen matter could the galaxy be holding on to those fast-moving outer stars. The same result is found for other spiral galaxies as well. Figure 28.23 is an example of the kinds of observations astronomers are making, for the Andromeda galaxy, a member of our Local Group. The observed rotation of spiral galaxies like Andromeda is usually seen in plots, known as rotation curves, that show velocity versus distance from the galaxy center. Such plots suggest that the dark matter is found in a large halo surrounding the luminous parts of each galaxy. The radius of the halos around the Milky Way and Andromeda may be as large as 300,000 light-years, much larger than the visible size of these galaxies. 1022 Chapter 28 The Evolution and Distribution of Galaxies Figure 28.23. Rotation Indicates Dark Matter. We see the Milky Way\u2019s sister, the spiral Andromeda galaxy, with a graph that shows the velocity at which stars and clouds of gas orbit", " the galaxy at different distances from the center (red line). As is true of the Milky Way, the rotational velocity (or orbital speed) does not decrease with distance from the center, which is what you would expect if an assembly of objects rotates around a common center. A calculation (blue line) based on the total mass visible as stars, gas, and dust predicts that the velocity should be much lower at larger distances from the center. The discrepancy between the two curves implies the presence of a halo of massive dark matter extending outside the boundary of the luminous matter. This dark matter causes everything in the galaxy to orbit faster than the observed matter alone could explain. (credit background: modification of work by ESO) Dark Matter in Clusters of Galaxies Galaxies in clusters also move around: they orbit the cluster\u2019s center of mass. It is not possible for us to follow a galaxy around its entire orbit because that typically takes about a billion years. It is possible, however, to measure the velocities with which galaxies in a cluster are moving, and then estimate what the total mass in the cluster must be to keep the individual galaxies from flying out of the cluster. The observations indicate that the mass of the galaxies alone cannot keep the cluster together\u2014some other gravity must again be present. The total amount of dark matter in clusters exceeds by more than ten times the luminous mass contained within the galaxies themselves, indicating that dark matter exists between galaxies as well as inside them. There is another approach we can take to measuring the amount of dark matter in clusters of galaxies. As we saw, the universe is expanding, but this expansion is not perfectly uniform, thanks to the interfering hand of gravity. Suppose, for example, that a galaxy lies outside but relatively close to a rich cluster of galaxies. The gravitational force of the cluster will tug on that neighboring galaxy and slow down the rate at which it moves away from the cluster due to the expansion of the universe. Consider the Local Group of galaxies, lying on the outskirts of the Virgo Supercluster. The mass concentrated at the center of the Virgo Cluster exerts a gravitational force on the Local Group. As a result, the Local Group is moving away from the center of the Virgo Cluster at a velocity a few hundred kilometers per second slower than the Hubble law predicts. By measuring such deviations from a smooth expansion, astronomers can estimate the total amount of mass contained in large clusters. There are two other very useful methods for measuring the amount of dark matter in galaxy clusters, and both of them have produced results in general agreement with the method of measuring galaxy velocities: gravitational lensing and X-ray emission. Let\u2019s take a look at both. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1023 As Albert Einstein showed in his theory of general relativity, the presence of mass bends the surrounding fabric of spacetime. Light follows those bends, so very massive objects can bend light significantly. You saw examples of this in the Astronomy Basics feature box Gravitational Lensing in the previous section. Visible galaxies are not the only possible gravitational lenses. Dark matter can also reveal its presence by producing this effect. Figure 28.24 shows a galaxy cluster that is acting like a gravitational lens; the streaks and arcs you see on the picture are lensed images of more distant galaxies. Gravitational lensing is well enough understood that astronomers can use the many ovals and arcs seen in this image to calculate detailed maps of how much matter there is in the cluster and how that mass is distributed. The result from studies of many such gravitational lens clusters shows that, like individual galaxies, galaxy clusters contain more than ten times as much dark matter as luminous matter. Figure 28.24. Cluster Abell 2218. This view from the Hubble Space Telescope shows the massive galaxy cluster Abell 2218 at a distance of about 2 billion light-years. Most of the yellowish objects are galaxies belonging to the cluster. But notice the numerous long, thin streaks, many of them blue; those are the distorted and magnified images of even more distant background galaxies, gravitationally lensed by the enormous mass of the intervening cluster. By carefully analyzing the lensed images, astronomers can construct a map of the dark matter that dominates the mass of the cluster. (credit: modification of work by NASA, ESA, and Johan Richard (Caltech)) The third method astronomers use to detect and measure dark matter in galaxy clusters is to image them in the light of X-rays. When the first sensitive X-ray telescopes were launched into orbit around Earth in the 1970s and trained on massive galaxy clusters, it was quickly discovered that the clusters emit copious X-ray radiation (see Figure 28.25). Most stars do not emit much X-ray radiation, and neither does most of the gas or dust between the stars inside galaxies. What could be emitting the X", "-rays seen from virtually all massive galaxy clusters? It turns out that just as galaxies have gas distributed between their stars, clusters of galaxies have gas distributed between their galaxies. The particles in these huge reservoirs of gas are not just sitting still; rather, they are constantly moving, zooming around under the influence of the cluster\u2019s immense gravity like mini planets around a giant sun. As they move and bump against each other, the gas heats up hotter and hotter until, at temperatures as high as 100 million K, it shines brightly at X-ray wavelengths. The more mass the cluster has, the faster the motions, the hotter the gas, and the brighter the X-rays. Astronomers calculate that the mass present to induce those motions must be about ten times the mass they can see in the clusters, including all the galaxies and all the gas. Once again, this is evidence that the galaxy clusters are seen to be dominated by dark matter. 1024 Chapter 28 The Evolution and Distribution of Galaxies Figure 28.25. X-Ray Image of a Galaxy Cluster. This composite image shows the galaxy cluster Abell 1689 at a distance of 2.3 billion lightyears. The finely detailed views of the galaxies, most of them yellow, are in visible and near-infrared light from the Hubble Space Telescope, while the diffuse purple haze shows X-rays as seen by Chandra X-ray Observatory. The abundant X-rays, the gravitationally lensed images (thin curving arcs) of background galaxies, and the measured velocities of galaxies in the clusters all show that the total mass of Abell 1689\u2014most of it dark matter\u2014is about 1015 solar masses. (credit: modification of work by NASA/ESA/JPL-Caltech/Yale/CNRS) Mass-to-Light Ratio We described the use of the mass-to-light ratio to characterize the matter in galaxies or clusters of galaxies in Properties of Galaxies. For systems containing mostly old stars, the mass-to-light ratio is typically 10 to 20, where mass and light are measured in units of the Sun\u2019s mass and luminosity. A mass-to-light ratio of 100 or more is a signal that a substantial amount of dark matter is present. Table 28.1 summarizes the results of measurements of mass-to-light ratios for various classes of objects. Very large mass-to-light ratios are found for all systems of galaxy size and larger, indicating that dark matter is present in all these types of objects. This is why we say that dark matter apparently makes up most of the total mass of the universe. Mass-To-Light Ratios Type of Object Mass-to-Light Ratio Sun Matter in vicinity of Sun Total mass in Milky Way Small groups of galaxies Rich clusters of galaxies Table 28.1 1 2 10 50\u2013150 250\u2013300 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1025 The clustering of galaxies can be used to derive the total amount of mass in a given region of space, while visible radiation is a good indicator of where the luminous mass is. Studies show that the dark matter and luminous matter are very closely associated. The dark matter halos do extend beyond the luminous boundaries of the galaxies that they surround. However, where there are large clusters of galaxies, you will also find large amounts of dark matter. Voids in the galaxy distribution are also voids in the distribution of dark matter. What Is the Dark Matter? How do we go about figuring out what the dark matter consists of? The technique we might use depends on its composition. Let\u2019s consider the possibility that some of the dark matter is made up of normal particles: protons, neutrons, and electrons. Suppose these particles were assembled into black holes, brown dwarfs, or white dwarfs. If the black holes had no accretion disks, they would be invisible to us. White and brown dwarfs do emit some radiation but have such low luminosities that they cannot be seen at distances greater than a few thousand light-years. We can, however, look for such compact objects because they can act as gravitational lenses. (See the Astronomy Basics feature box Gravitational Lensing.) Suppose the dark matter in the halo of the Milky Way were made up of black holes, brown dwarfs, and white dwarfs. These objects have been whimsically dubbed MACHOs (MAssive Compact Halo Objects). If an invisible MACHO passes directly between a distant star and Earth, it acts as a gravitational lens, focusing the light from the distant star. This causes the star to appear to brighten over a time interval of a few hours to several days before returning to its normal brightness. Since we can\u2019t predict when any given star might brighten this way, we have to monitor huge numbers of stars to catch one in the act. There are", " not enough astronomers to keep monitoring so many stars, but today\u2019s automated telescopes and computer systems can do it for us. Research teams making observations of millions of stars in the nearby galaxy called the Large Magellanic Cloud have reported several examples of the type of brightening expected if MACHOs are present in the halo of the Milky Way (Figure 28.26). However, there are not enough MACHOs in the halo of the Milky Way to account for the mass of the dark matter in the halo. 1026 Chapter 28 The Evolution and Distribution of Galaxies Figure 28.26. Large and Small Magellanic Clouds. Here, the two small galaxies we call the Large Magellanic Cloud and Small Magellanic Cloud can be seen above the auxiliary telescopes for the Very Large Telescope Array on Cerro Paranal in Chile. You can see from the number of stars that are visible that this is a very dark site for doing astronomy. (credit: ESO/J. Colosimo) This result, along with a variety of other experiments, leads us to conclude that the types of matter we are familiar with can make up only a tiny portion of the dark matter. Another possibility is that dark matter is composed of some new type of particle\u2014one that researchers are now trying to detect in laboratories here on Earth (see The Big Bang). The kinds of dark matter particles that astronomers and physicists have proposed generally fall into two main categories: hot and cold dark matter. The terms hot and cold don\u2019t refer to true temperatures, but rather to the average velocities of the particles, analogous to how we might think of particles of air moving in your room right now. In a cold room, the air particles move more slowly on average than in a warm room. In the early universe, if dark matter particles easily moved fast and far compared to the lumps and bumps of ordinary matter that eventually became galaxies and larger structures, we call those particles hot dark matter. In that case, smaller lumps and bumps would be smeared out by the particle motions, meaning fewer small galaxies would get made. On the other hand, if the dark matter particles moved slowly and covered only small distances compared to the sizes of the lumps in the early universe, we call that cold dark matter. Their slow speeds and energy would mean that even the smaller lumps of ordinary matter would survive to grow into small galaxies. By looking at when galaxies formed and how they evolve, we can use observations to distinguish between the two kinds of dark matter. So far, observations seem most consistent with models based on cold dark matter. Solving the dark matter problem is one of the biggest challenges facing astronomers. After all, we can hardly understand the evolution of galaxies and the long-term history of the universe without understanding what its most massive component is made of. For example, we need to know just what role dark matter played in starting the higher-density \u201cseeds\u201d that led to the formation of galaxies. And since many galaxies have large This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1027 halos made of dark matter, how does this affect their interactions with one another and the shapes and types of galaxies that their collisions create? Astronomers armed with various theories are working hard to produce models of galaxy structure and evolution that take dark matter into account in just the right way. Even though we don\u2019t know what the dark matter is, we do have some clues about how it affected the formation of the very first galaxies. As we will see in The Big Bang, careful measurements of the microwave radiation left over after the Big Bang have allowed astronomers to set very tight limits on the actual sizes of those early seeds that led to the formation of the large galaxies that we see in today\u2019s universe. Astronomers have also measured the relative numbers and distances between galaxies and clusters of different sizes in the universe today. So far, most of the evidence seems to weigh heavily in favor of cold dark matter, and most current models of galaxy and large-scale structure formation use cold dark matter as their main ingredient. As if the presence of dark matter\u2014a mysterious substance that exerts gravity and outweighs all the known stars and galaxies in the universe but does not emit or absorb light\u2014were not enough, there is an even more baffling and equally important constituent of the universe that has only recently been discovered: we have called it dark energy in parallel with dark matter. We will say more about it and explore its effects on the evolution of the universe in The Big Bang. For now, we can complete our inventory of the contents of the universe by noting that it appears that the entire universe contains some mysterious energy that pushes spacetime apart, taking galaxies and the larger structures made of galaxies along with it. Observations show that dark energy becomes more and more important relative to gravity as the universe ages. As a result, the expansion", " of the universe is accelerating, and this acceleration seems to be happening mostly since the universe was about half its current age. What we see when we peer out into the universe\u2014the light from trillions of stars in hundreds of billions of galaxies wrapped in intricate veils of gas and dust\u2014is therefore actually only a sprinkling of icing on top of the cake: as we will see in The Big Bang, when we look outside galaxies and clusters of galaxies at the universe as a whole, astronomers find that for every gram of luminous normal matter, such as protons, neutrons, electrons, and atoms in the universe, there are about 4 grams of nonluminous normal matter, mainly intergalactic hydrogen and helium. There are about 27 grams of dark matter, and the energy equivalent (remember Einstein\u2019s famous E = mc2) of about 68 grams of dark energy. Dark matter, and (as we will see) even more so dark energy, are dramatic demonstrations of what we have tried to emphasize throughout this book: science is always a \u201cprogress report,\u201d and we often encounter areas where we have more questions than answers. Let\u2019s next put together all these clues to trace the life history of galaxies and large-scale structure in the universe. What follows is the current consensus, but research in this field is moving rapidly, and some of these ideas will probably be modified as new observations are made. 28.5 THE FORMATION AND EVOLUTION OF GALAXIES AND STRUCTURE IN THE UNIVERSE Learning Objectives By the end of this section, you will be able to: Summarize the main theories attempting to explain how individual galaxies formed Explain how tiny \u201cseeds\u201d of dark matter in the early universe grew by gravitational attraction over billions of years into the largest structures observed in the universe: galaxy clusters and superclusters, filaments, and voids 1028 Chapter 28 The Evolution and Distribution of Galaxies As with most branches of natural science, astronomers and cosmologists always want to know the answer to the question, \u201cHow did it get that way?\u201d What made galaxies and galaxy clusters, superclusters, voids, and filaments look the way they do? The existence of such large filaments of galaxies and voids is an interesting puzzle because we have evidence (to be discussed in The Big Bang) that the universe was extremely smooth even a few hundred thousand years after forming. The challenge for theoreticians is to understand how a nearly featureless universe changed into the complex and lumpy one that we see today. Armed with our observations and current understanding of galaxy evolution over cosmic time, dark matter, and large-scale structure, we are now prepared to try to answer that question on some of the largest possible scales in the universe. As we will see, the short answer to how the universe got this way is \u201cdark matter + gravity + time.\u201d How Galaxies Form and Grow We\u2019ve already seen that galaxies were more numerous, but smaller, bluer, and clumpier, in the distant past than they are today, and that galaxy mergers play a significant role in their evolution. At the same time, we have observed quasars and galaxies that emitted their light when the universe was less than a billion years old\u2014so we know that large condensations of matter had begun to form at least that early. We also saw in Active Galaxies, Quasars, and Supermassive Black Holes that many quasars are found in the centers of elliptical galaxies. This means that some of the first large concentrations of matter must have evolved into the elliptical galaxies that we see in today\u2019s universe. It seems likely that the supermassive black holes in the centers of galaxies and the spherical distribution of ordinary matter around them formed at the same time and through related physical processes. Dramatic confirmation of that picture arrived only in the last decade, when astronomers discovered a curious empirical relationship: as we saw in Active Galaxies, Quasars, and Supermassive Black Holes, the more massive a galaxy is, the more massive its central black hole is. Somehow, the black hole and the galaxy \u201cknow\u201d enough about each other to match their growth rates. There have been two main types of galaxy formation models to explain all those observations. The first asserts that massive elliptical galaxies formed in a single, rapid collapse of gas and dark matter, during which virtually all the gas was turned quickly into stars. Afterward the galaxies changed only slowly as the stars evolved. This is what astronomers call a \u201ctop-down\u201d scenario. The second model suggests that today\u2019s giant ellipticals were formed mostly through mergers of smaller galaxies that had already converted at least some of their gas into stars\u2014a \u201cbottom-up\u201d scenario. In other words, astronomers have debated whether giant ellipticals formed most of their stars in the large galaxy that we see today or in separate small galaxies that subsequently merged", ". Since we see some luminous quasars from when the universe was less than a billion years old, it is likely that at least some giant ellipticals began their evolution very early through the collapse of a single cloud. However, the best evidence also seems to show that mature giant elliptical galaxies like the ones we see nearby were rare before the universe was about 6 billion years old and that they are much more common today than they were when the universe was young. Observations also indicate that most of the gas in elliptical galaxies was converted to stars by the time the universe was about 3 billion years old, so it appears that elliptical galaxies have not formed many new stars since then. They are often said to be \u201cred and dead\u201d\u2014that is, they mostly contain old, cool, red stars, and there is little or no new star formation going on. These observations (when considered together) suggest that the giant elliptical galaxies that we see nearby formed from a combination of both top-down and bottom-up mechanisms, with the most massive galaxies forming in the densest clusters where both processes happened very early and quickly in the history of the universe. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1029 The situation with spiral galaxies is apparently very different. The bulges of these galaxies formed early, like the elliptical galaxies (Figure 28.27). However, the disks formed later (remember that the stars in the disk of the Milky Way are younger than the stars in the bulge and the halo) and still contain gas and dust. However, the rate of star formation in spirals today is about ten times lower than it was 8 billion years ago. The number of stars being formed drops as the gas is used up. So spirals seem to form mostly \u201cbottom up\u201d but over a longer time than ellipticals and in a more complex way, with at least two distinct phases. Figure 28.27. Growth of Spiral Bulges. The nuclear bulges of some spiral galaxies formed through the collapse of a single protogalactic cloud (top row). Others grew over time through mergers with other smaller galaxies (bottom row). Hubble originally thought that elliptical galaxies were young and would eventually turn into spirals, an idea we now know is not true. In fact, as we saw above, it\u2019s more likely the other way around: two spirals that crash together under their mutual gravity can turn into an elliptical. Despite these advances in our understanding of how galaxies form and evolve, many questions remain. For example, it\u2019s even possible, given current evidence, that spiral galaxies could lose their spiral arms and disks in a merger event, making them look more like an elliptical or irregular galaxy, and then regain the disk and arms again later if enough gas remains available. The story of how galaxies assume their final shapes is still being written as we learn more about galaxies and their environment. 1030 Chapter 28 The Evolution and Distribution of Galaxies Forming Galaxy Clusters, Superclusters, Voids, and Filaments If individual galaxies seem to grow mostly by assembling smaller pieces together gravitationally over cosmic time, what about the clusters of galaxies and larger structures such as those seen in Figure 28.21? How do we explain the large-scale maps that show galaxies distributed on the walls of huge sponge- or bubble-like structures spanning hundreds of millions of light-years? As we saw, observations have found increasing evidence for concentrations, filaments, clusters, and superclusters of galaxies when the universe was less than 3 billion years old (Figure 28.28). This means that large concentrations of galaxies had already come together when the universe was less than a quarter as old as it is now. Figure 28.28. Merging Galaxies in a Distant Cluster. This Hubble image shows the core of one of the most distant galaxy clusters yet discovered, SpARCS 1049+56; we are seeing it as it was nearly 10 billion years ago. The surprise delivered by the image was the \u201ctrain wreck\u201d of chaotic galaxy shapes and blue tidal tails: apparently there are several galaxies right in the core that are merging together, the probable cause of a massive burst of star formation and bright infrared emission from the cluster. (credit: modification of work by NASA/STScI/ESA/JPL-Caltech/ McGill) Almost all the currently favored models of how large-scale structure formed in the universe tell a story similar to that for individual galaxies: tiny dark matter \u201cseeds\u201d in the hot cosmic soup after the Big Bang grew by gravity into larger and larger structures as cosmic time ticked on (Figure 28.29). The final models we construct will need to be able to explain the size, shape, age, number, and spatial distribution of galaxies, clusters, and filaments\u2014not only today, but also far", " back in time. Therefore, astronomers are working hard to measure and then to model those features of large-scale structure as accurately as possible. So far, a mixture of 5% normal atoms, 27% cold dark matter, and 68% dark energy seems to be the best way to explain all the evidence currently available (see The Big Bang). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1031 Figure 28.29. Growth of Large-Scale Structure as Calculated by Supercomputers. The boxes show how filaments and superclusters of galaxies grow over time, from a relatively smooth distribution of dark matter and gas, with few galaxies formed in the first 2 billion years after the Big Bang, to the very clumpy strings of galaxies with large voids today. Compare the last image in this sequence with the actual distribution of nearby galaxies shown in Figure 28.21. (credit: modification of work by CXC/MPE/V.Springel) The box at left is labeled \u201cBig Bang,\u201d the box at center is unlabeled and the box at right is labeled \u201cPresent\u201d. A white arrow points from left to right representing the direction of time. Scientists even have a model to explain how a nearly uniform, hot \u201csoup\u201d of particles and energy at the beginning of time acquired the Swiss-cheese-like structure that we now see on the largest scales. As we will see in The Big Bang, when the universe was only a few hundred thousand years old, everything was at a temperature of a few thousand degrees. Theorists suggest that at that early time, all the hot gas was vibrating, much as sound waves vibrate the air of a nightclub with an especially loud band. This vibrating could have concentrated matter into high-density peaks and created emptier spaces between them. When the universe cooled, the concentrations of matter were \u201cfrozen in,\u201d and galaxies ultimately formed from the matter in these high-density regions. The Big Picture To finish this chapter, let\u2019s put all these ideas together to tell a coherent story of how the universe came to look the way it does. Initially, as we said, the distribution of matter (both luminous and dark) was nearly, but not quite exactly, smooth and uniform. That \u201cnot quite\u201d is the key to everything. Here and there were lumps where the density of matter (both luminous and dark) was ever so slightly higher than average. Initially, each individual lump expanded because the whole universe was expanding. However, as the universe continued to expand, the regions of higher density acquired still more mass because they exerted a slightly larger than average gravitational force on surrounding material. If the inward pull of gravity was high enough, the denser individual regions ultimately stopped expanding. They then began to collapse into irregularly shaped blobs (that\u2019s the technical term astronomers use!). In many regions the collapse was more rapid in one 1032 Chapter 28 The Evolution and Distribution of Galaxies direction, so the concentrations of matter were not spherical but came to resemble giant clumps, pancakes, and rope-like filaments\u2014each much larger than individual galaxies. These elongated clumps existed throughout the early universe, oriented in different directions and collapsing at different rates. The clumps provided the framework for the large-scale filamentary and bubble-like structures that we see preserved in the universe today. The universe then proceeded to \u201cbuild itself\u201d from the bottom up. Within the clumps, smaller structures formed first, then merged to build larger ones, like Lego pieces being put together one by one to create a giant Lego metropolis. The first dense concentrations of matter that collapsed were the size of small dwarf galaxies or globular clusters\u2014which helps explain why globular clusters are the oldest things in the Milky Way and most other galaxies. These fragments then gradually assembled to build galaxies, galaxy clusters, and, ultimately, superclusters of galaxies. According to this picture, small galaxies and large star clusters first formed in the highest density regions of all\u2014the filaments and nodes where the pancakes intersect\u2014when the universe was about two percent of its current age. Some stars may have formed even before the first star clusters and galaxies came into existence. Some galaxy-galaxy collisions triggered massive bursts of star formation, and some of these led to the formation of black holes. In that rich, crowded environment, black holes found constant food and grew in mass. The development of massive black holes then triggered quasars and other active galactic nuclei whose powerful outflows of energy and matter shut off the star formation in their host galaxies. The early universe must have been an exciting place! Clusters of galaxies then formed as individual galaxies congregated, drawn together by their mutual gravitational attraction (Figure 28.30). First, a few galaxies came together to form groups, much like our", " own Local Group. Then the groups began combining to form clusters and, eventually, superclusters. This model predicts that clusters and superclusters should still be in the process of gathering together, and observations do in fact suggest that clusters are still gathering up their flocks of galaxies and collecting more gas as it flows in along filaments. In some instances we even see entire clusters of galaxies merging together. Figure 28.30. Formation of Cluster of Galaxies. This schematic diagram shows how galaxies might have formed if small clouds formed first and then congregated to form galaxies and then clusters of galaxies. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1033 Most giant elliptical galaxies formed through the collision and merger of many smaller fragments. Some spiral galaxies may have formed in relatively isolated regions from a single cloud of gas that collapsed to make a flattened disk, but others acquired additional stars, gas, and dark matter through collisions, and the stars acquired through these collisions now populate their halos and bulges. As we have seen, our Milky Way is still capturing small galaxies and adding them to its halo, and probably also pulling fresh gas from these galaxies into its disk. 1034 Chapter 28 The Evolution and Distribution of Galaxies CHAPTER 28 REVIEW KEY TERMS cold dark matter slow-moving massive particles, not yet identified, that don\u2019t absorb, emit, or reflect light or other electromagnetic radiation, and that make up most of the mass of galaxies and galaxy clusters cosmological principle the assumption that, on the large scale, the universe at any given time is the same everywhere\u2014isotropic and homogeneous dark energy an energy that is causing the expansion of the universe to accelerate; the source of this energy is not yet understood evolution (of galaxies) changes in individual galaxies over cosmic time, inferred by observing snapshots of many different galaxies at different times in their lives galactic cannibalism a process by which a larger galaxy strips material from or completely swallows a smaller one homogeneous having a consistent and even distribution of matter that is the same everywhere hot dark matter massive particles, not yet identified, that don\u2019t absorb, emit, or reflect light or other electromagnetic radiation, and that make up most of the mass of galaxies and galaxy clusters; hot dark matter is faster-moving material than cold dark matter isotropic the same in all directions Local Group a small cluster of galaxies to which our Galaxy belongs merger a collision between galaxies (of roughly comparable size) that combine to form a single new structure starburst a galaxy or merger of multiple galaxies that turns gas into stars much faster than usual supercluster a large region of space (more than 100 million light-years across) where groups and clusters of galaxies are more concentrated; a cluster of clusters of galaxies void a region between clusters and superclusters of galaxies that appears relatively empty of galaxies SUMMARY 28.1 Observations of Distant Galaxies When we look at distant galaxies, we are looking back in time. We have now seen galaxies as they were when the universe was about 500 million years old\u2014only about five percent as old as it is now. The universe now is 13.8 billion years old. The color of a galaxy is an indicator of the age of the stars that populate it. Blue galaxies must contain a lot of hot, massive, young stars. Galaxies that contain only old stars tend to be yellowish red. The first generation of stars formed when the universe was only a few hundred million years old. Galaxies observed when the universe was only a few billion years old tend to be smaller than today\u2019s galaxies, to have more irregular shapes, and to have more rapid star formation than the galaxies we see nearby in today\u2019s universe. This shows that the smaller galaxy fragments assembled themselves into the larger galaxies we see today. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1035 28.2 Galaxy Mergers and Active Galactic Nuclei When galaxies of comparable size collide and coalesce we call it a merger, but when a small galaxy is swallowed by a much larger one, we use the term galactic cannibalism. Collisions play an important role in the evolution of galaxies. If the collision involves at least one galaxy rich in interstellar matter, the resulting compression of the gas will result in a burst of star formation, leading to a starburst galaxy. Mergers were much more common when the universe was young, and many of the most distant galaxies that we see are starburst galaxies that are involved in collisions. Active galactic nuclei powered by supermassive black holes in the centers of most galaxies can have major effects on the host galaxy, including shutting off star formation. 28.3 The Distribution of Galaxies in Space Counts of galaxies in various directions establish that the universe on the large scale is homogeneous", " and isotropic (the same everywhere and the same in all directions, apart from evolutionary changes with time). The sameness of the universe everywhere is referred to as the cosmological principle. Galaxies are grouped together in clusters. The Milky Way Galaxy is a member of the Local Group, which contains at least 54 member galaxies. Rich clusters (such as Virgo and Coma) contain thousands or tens of thousands of galaxies. Galaxy clusters often group together with other clusters to form large-scale structures called superclusters, which can extend over distances of several hundred million light-years. Clusters and superclusters are found in filamentary structures that are huge but fill only a small fraction of space. Most of space consists of large voids between superclusters, with nearly all galaxies confined to less than 10% of the total volume. 28.4 The Challenge of Dark Matter Stars move much faster in their orbits around the centers of galaxies, and galaxies around centers of galaxy clusters, than they should according to the gravity of all the luminous matter (stars, gas, and dust) astronomers can detect. This discrepancy implies that galaxies and galaxy clusters are dominated by dark matter rather than normal luminous matter. Gravitational lensing and X-ray radiation from massive galaxy clusters confirm the presence of dark matter. Galaxies and clusters of galaxies contain about 10 times more dark matter than luminous matter. While some of the dark matter may be made up of ordinary matter (protons, neutrons, and electrons), perhaps in the form of very faint stars or black holes, most of it probably consists of some totally new type of particle not yet detected on Earth. Observations of gravitational lensing effects on distant objects have been used to look in the outer region of our Galaxy for any dark matter in the form of compact, dim stars or star remnants, but not enough such objects have been found to account for all the dark matter. 28.5 The Formation and Evolution of Galaxies and Structure in the Universe Initially, luminous and dark matter in the universe was distributed almost\u2014but not quite\u2014uniformly. The challenge for galaxy formation theories is to show how this \u201cnot quite\u201d smooth distribution of matter developed the structures\u2014galaxies and galaxy clusters\u2014that we see today. It is likely that the filamentary distribution of galaxies and voids was built in near the beginning, before stars and galaxies began to form. The first condensations of matter were about the mass of a large star cluster or a small galaxy. These smaller structures then merged over cosmic time to form large galaxies, clusters of galaxies, and superclusters of galaxies. Superclusters today are still gathering up more galaxies, gas, and dark matter. And spiral galaxies like the Milky Way are still acquiring material by capturing small galaxies near them. 1036 Chapter 28 The Evolution and Distribution of Galaxies FOR FURTHER EXPLORATION Articles Andrews, B. \u201cWhat Are Galaxies Trying to Tell Us?\u201d Astronomy (February 2011): 24. Introduction to our understanding of the shapes and evolution of different types of galaxies. Barger, A. \u201cThe Midlife Crisis of the Cosmos.\u201d Scientific American (January 2005): 46. On how our time differs from the early universe in terms of what galaxies are doing, and what role supermassive black holes play. Berman, B. \u201cThe Missing Universe.\u201d Astronomy (April 2014): 24. Brief review of dark matter, what it could be, and modified theories of gravity that can also explain it. Faber, S., et al. \u201cStaring Back to Cosmic Dawn.\u201d Sky & Telescope (June 2014): 18. Program to see the most distant and earliest galaxies with the Hubble. Geller, M., & Huchra, J. \u201cMapping the Universe.\u201d Sky & Telescope (August 1991): 134. On their project mapping the location of galaxies in three dimensions. Hooper, D. \u201cDark Matter in the Discovery Age.\u201d Sky & Telescope (January 2013): 26. On experiments looking for the nature of dark matter. James, C. R. \u201cThe Hubble Deep Field: The Picture Worth a Trillion Stars.\u201d Astronomy (November 2015): 44. Detailed history and results, plus the Hubble Ultra-Deep Field. Kaufmann, G., & van den Bosch, F. \u201cThe Life Cycle of Galaxies.\u201d Scientific American (June 2002): 46. On the evolution of galaxies and how the different shapes of galaxies develop. Knapp, G. \u201cMining the Heavens: The Sloan Digital Sky Survey.\u201d Sky & Telescope (August 1997): 40. Kron, R., & Butler, S. \u201cStars and Strips Forever.\u201d Astronomy (February 1999): 48. On the Sloan Digital Survey. Kruesi, L. \u201cWhat Do We Really Know about Dark Matter?\u201d Astronomy (November 2009): 28.", " Focuses on what dark matter could be and experiments to find out. Larson, R., & Bromm, V. \u201cThe First Stars in the Universe.\u201d Scientific American (December 2001): 64. On the dark ages and the birth of the first stars. Nadis, S. \u201cExploring the Galaxy-Black Hole Connection.\u201d Astronomy (May 2010): 28. About the role of massive black holes in the evolution of galaxies. Nadis, S. \u201cAstronomers Reveal the Universe\u2019s Hidden Structure.\u201d Astronomy (September 2013): 44. How dark matter is the scaffolding on which the visible universe rests. Schilling, G. \u201cHubble Goes the Distance.\u201d Sky & Telescope (January 2015): 20. Using gravitational lensing with HST to see the most distant galaxies. Strauss, M. \u201cReading the Blueprints of Creation.\u201d Scientific American (February 2004): 54. On large-scale surveys of galaxies and what they tell us about the organization of the early universe. Tytell, D. \u201cA Wide Deep Field: Getting the Big Picture.\u201d Sky & Telescope (September 2001): 42. On the NOAO survey of deep sky objects. Villard, R. \u201cHow Gravity\u2019s Grand Illusion Reveals the Universe.\u201d Astronomy (January 2013): 44. On gravitational lensing and what it teaches us. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1037 Websites Assembly of Galaxies: http://jwst.nasa.gov/galaxies.html. Introductory background information about galaxies: what we know and what we want to learn. Brief History of Gravitational Lensing: http://www.einstein-online.info/spotlights/grav_lensing_history. From Einstein OnLine. Cosmic Structures: http://skyserver.sdss.org/dr1/en/astro/structures/structures.asp. Brief review page on how galaxies are organized, from the Sloan Survey. Discovery of the First Gravitational Lens: http://astrosociety.org/wp-content/uploads/2013/02/ab2009-33.pdf. By Ray Weymann, 2009. Gravitational Lensing Discoveries from the Hubble Space Telescope: http://hubblesite.org/newscenter/archive/ releases/exotic/gravitational-lens/. A chronological list of news releases and images. Local Group of Galaxies: http://www.atlasoftheuniverse.com/localgr.html. Clickable map from the Atlas of the Universe project. See also their Virgo Cluster page: http://www.atlasoftheuniverse.com/galgrps/vir.html. RotCurve: http://burro.astr.cwru.edu/JavaLab/RotcurveWeb/main.html. Try your hand at using real galaxy rotation curve data to measure dark matter halos using this Java applet simulation. Sloan Digital Sky Survey Website: http://classic.sdss.org/. Includes nontechnical and technical parts. Spyglasses into the Universe: http://www.spacetelescope.org/science/gravitational_lensing/. Hubble page on gravitational lensing; includes links to videos. Virgo Cluster of Galaxies: http://messier.seds.org/more/virgo.html. A page with brief information and links to maps, images, etc. Videos Cosmic Simulations: http://www.tapir.caltech.edu/~phopkins/Site/Movies_cosmo.html. Beautiful videos with computer simulations of how galaxies form, from the FIRE group. Cosmology of the Local Universe: http://irfu.cea.fr/cosmography. Narrated flythrough of maps of galaxies showing the closer regions of the universe (17:35). Gravitational Lensing: https://www.youtube.com/watch?v=4Z71RtwoOas. Video from Fermilab, with Dr. Don Lincoln (7:14). How Galaxies Were Cooked from the Primordial Soup: https://www.youtube.com/watch?v=wqNNCm7SNyw. A 2013 public talk by Dr. Sandra Faber of Lick Observatory about the evolution of galaxies; part of the Silicon Valley Astronomy Lecture Series (1:19:33). Hubble Extreme Deep Field Pushes Back Frontiers of Time and Space: https://www.youtube.com/ watch?v=gu_VhzhlqGw. Brief 2012 video (2:42). Looking Deeply into the Universe in 3-D: https://www.eso.org/", "tion: http://wwwmpa.mpa-garching.mpg.de/galform/virgo/millennium. A supercomputer in Germany follows the evolution of a representative large box as the universe evolves. Movies of flying through the large-scale local structure: http://www.ifa.hawaii.edu/~tully/. By Brent Tully. 1038 Chapter 28 The Evolution and Distribution of Galaxies Shedding Light on Dark Matter: https://www.youtube.com/watch?v=bZW_B9CC-gI. 2008 TED talk on galaxies and dark matter by physicist Patricia Burchat (17:08). Sloan Digital Sky Survey overview movies: http://astro.uchicago.edu/cosmus/projects/sloanmovie/. Virtual Universe: https://www.youtube.com/watch?v=SY0bKE10ZDM. An MIT model of a section of universe evolving, with dark matter included (4:11). When Two Galaxies Collide: http://www.openculture.com/2009/04/when_galaxies_collide.html. Computer simulation, which stops at various points and shows a Hubble image of just such a system in nature (1:37). COLLABORATIVE GROUP ACTIVITIES A. Suppose you developed a theory to account for the evolution of New York City. Have your group discuss whether it would resemble the development of structure in the universe (as we have described it in this chapter). What elements of your model for NYC resemble the astronomers\u2019 model for the growth of structure in the universe? Which elements do not match? B. Most astronomers believe that dark matter exists and is a large fraction of the total matter in the universe. At the same time, most astronomers do not believe that UFOs are evidence that we are being visited by aliens from another world. Yet astronomers have never actually seen either dark matter or a UFO. Why do you think one idea is widely accepted by scientists and the other is not? Which idea do you think is more believable? Give your reasoning. C. Someone in your group describes the redshift surveys of galaxies to a friend, who says he\u2019s never heard of a bigger waste of effort. Who cares, he asks, about the large-scale structure of the universe? What is your group\u2019s reaction, and what reasons could you come up with for putting money into figuring out how the universe is organized? D. The leader of a small but very wealthy country is obsessed by maps. She has put together a fabulous collection of Earth maps, purchased all the maps of other planets that astronomers have assembled, and now wants to commission the best possible map of the entire universe. Your group is selected to advise her. What sort of instruments and surveys should she invest in to produce a good map of the cosmos? Be as specific as you can. E. Download a high-resolution image of a rich galaxy cluster from the Hubble Space Telescope (see the list of gravitational lens news stories in the \u201cFor Further Exploration\u201d section). See if your group can work together to identify gravitational arcs, the images of distant background galaxies distorted by the mass of the cluster. How many can you find? Can you identify any multiple images of the same background galaxy? (If anyone in the group gets really interested, there is a Citizen Science project called Spacewarps, where you can help astronomers identify gravitational lenses on their images: https://spacewarps.org.) F. You get so excited about gravitational lensing that you begin to talk about it with an intelligent friend who has not yet taken an astronomy course. After hearing you out, this friend starts to worry. He says, \u201cIf gravitational lenses can distort quasar images, sometimes creating multiple, or ghost, images of the same object, then how can we trust any point of light in the sky to be real? Maybe many of the stars we see are just ghost images or lensed images too!\u201d Have your group discuss how to respond. (Hint: Think about the path that the light of a quasar took on its way to us and the path the light of a typical star takes.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1039 G. The 8.4-meter Large Synoptic Survey Telescope (LSST), currently under construction atop Cerro Pach\u00f3n, a mountain in northern Chile, will survey the entire sky with its 3.2-gigapixel camera every few days, looking for transient, or temporary, objects that make a brief appearance in the sky before fading from view, including asteroids and Kuiper belt objects in our solar system, and supernovae and other explosive high- energy events in the distant universe. When it\u2019s fully operating sometime after 2021, the LSST will produce up to 30", " terabytes of data every night. (A terabyte is 1000 gigabytes, which is the unit you probably use to rate your computer or memory stick capacity.) With your group, consider what you think might be some challenges of dealing with that quantity of data every night in a scientifically productive but efficient way. Can you propose any solutions to those challenges? H. Quasars are rare now but were much more numerous when the universe was about one-quarter of its current age. The total star formation taking place in galaxies across the universe peaked at about the same redshift. Does your group think this is a coincidence? Why or why not? I. One way to see how well the ideas in astronomy (like those in this chapter) have penetrated popular culture is to see whether you can find astronomical words in the marketplace. A short web search for the term \u201cdark matter\u201d turns up both a brand of coffee and a brand of \u201cmuscle growth accelerator\u201d with that name. How many other terms used in this chapter can your group find in the world of products? (What\u2019s a really popular type of Android cell phone, for example?) J. What\u2019s your complete address in the universe? Group members should write out their full address, based on the information in this chapter (and the rest of the book). After your postal code and country, you may want to add continent, planet, planetary system, galaxy, etc. Then each group member should explain this address to a family member or student not taking astronomy. EXERCISES Review Questions 1. How are distant (young) galaxies different from the galaxies that we see in the universe today? 2. What is the evidence that star formation began when the universe was only a few hundred million years old? 3. Describe the evolution of an elliptical galaxy. How does the evolution of a spiral galaxy differ from that of an elliptical? 4. Explain what we mean when we call the universe homogeneous and isotropic. Would you say that the distribution of elephants on Earth is homogeneous and isotropic? Why? 5. Describe the organization of galaxies into groupings, from the Local Group to superclusters. 6. What is the evidence that a large fraction of the matter in the universe is invisible? 7. When astronomers make maps of the structure of the universe on the largest scales, how do they find the superclusters of galaxies to be arranged? 8. How does the presence of an active galactic nucleus in a starburst galaxy affect the starburst process? 1040 Chapter 28 The Evolution and Distribution of Galaxies Thought Questions 9. Describe how you might use the color of a galaxy to determine something about what kinds of stars it contains. 10. Suppose a galaxy formed stars for a few million years and then stopped (and no other galaxy merged or collided with it). What would be the most massive stars on the main sequence after 500 million years? After 10 billion years? How would the color of the galaxy change over this time span? (Refer to Evolution from the Main Sequence to Red Giants.) 11. Given the ideas presented here about how galaxies form, would you expect to find a giant elliptical galaxy in the Local Group? Why or why not? Is there in fact a giant elliptical in the Local Group? 12. Can an elliptical galaxy evolve into a spiral? Explain your answer. Can a spiral turn into an elliptical? How? 13. If we see a double image of a quasar produced by a gravitational lens and can obtain a spectrum of the galaxy that is acting as the gravitational lens, we can then put limits on the distance to the quasar. Explain how. 14. The left panel of Figure 27.1 shows a cluster of yellow galaxies that produces several images of blue galaxies through gravitational lensing. Which are more distant\u2014the blue galaxies or the yellow galaxies? The light in the galaxies comes from stars. How do the temperatures of the stars that dominate the light of the cluster galaxies differ from the temperatures of the stars that dominate the light of the blue-lensed galaxy? Which galaxy\u2019s light is dominated by young stars? 15. Suppose you are standing in the center of a large, densely populated city that is exactly circular, surrounded by a ring of suburbs with lower-density population, surrounded in turn by a ring of farmland. From this specific location, would you say the population distribution is isotropic? Homogeneous? 16. Astronomers have been making maps by observing a slice of the universe and seeing where the galaxies lie within that slice. If the universe is isotropic and homogeneous, why do they need more than one slice? Suppose they now want to make each slice extend farther into the universe. What do they need to do? 17. Human civilization is about 10,000 years old as measured by the development of agriculture. If your telescope collects starlight tonight that has been traveling for 10,000 years, is that star inside or outside our Milky Way Galaxy? Is", " it likely that the star has changed much during that time? 18. Given that only about 5% of the galaxies visible in the Hubble Deep Field are bright enough for astronomers to study spectroscopically, they need to make the most of the other 95%. One technique is to use their colors and apparent brightnesses to try to roughly estimate their redshift. How do you think the inaccuracy of this redshift estimation technique (compared to actually measuring the redshift from a spectrum) might affect our ability to make maps of large-scale structures such as the filaments and voids shown in Figure 28.21? Figuring For Yourself 19. Using the information from Example 28.1, how much fainter an object will you have to be able to measure in order to include the same kinds of galaxies in your second survey? Remember that the brightness of an object varies as the inverse square of the distance. 20. Using the information from Example 28.1, if galaxies are distributed homogeneously, how many times more of them would you expect to count on your second survey? 21. Using the information from Example 28.1, how much longer will it take you to do your second survey? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 28 The Evolution and Distribution of Galaxies 1041 22. Galaxies are found in the \u201cwalls\u201d of huge voids; very few galaxies are found in the voids themselves. The text says that the structure of filaments and voids has been present in the universe since shortly after the expansion began 13.8 billion years ago. In science, we always have to check to see whether some conclusion is contradicted by any other information we have. In this case, we can ask whether the voids would have filled up with galaxies in roughly 14 billion years. Observations show that in addition to the motion associated with the expansion of the universe, the galaxies in the walls of the voids are moving in random directions at typical speeds of 300 km/s. At least some of them will be moving into the voids. How far into the void will a galaxy move in 14 billion years? Is it a reasonable hypothesis that the voids have existed for 14 billion years? 23. Calculate the velocity, the distance, and the look-back time of the most distant galaxies in Figure 28.21 using the Hubble constant given in this text and the redshift given in the diagram. Remember the Doppler formula for velocity \u239b \u239dv = c \u00d7 \u0394\u03bb \u03bb these low velocities, you can neglect relativistic effects. \u239e \u23a0 and the Hubble law (v = H \u00d7 d, where d is the distance to a galaxy). For 24. Assume that dark matter is uniformly distributed throughout the Milky Way, not just in the outer halo but also throughout the bulge and in the disk, where the solar system lives. How much dark matter would you expect there to be inside the solar system? Would you expect that to be easily detectable? Hint: For the radius of the Milky Way\u2019s dark matter halo, use R = 300,000 light-years; for the solar system\u2019s radius, use 100 AU; and start by calculating the ratio of the two volumes. 25. The simulated box of galaxy filaments and superclusters shown in Figure 28.29 stretches across 1 billion light-years. If you were to make a scale model where that box covered the core of a university campus, say 1 km, then how big would the Milky Way Galaxy be? How far away would the Andromeda galaxy be in the scale model? 26. The first objects to collapse gravitationally after the Big Bang might have been globular cluster-size galaxy pieces, with masses around 106 solar masses. Suppose you merge two of those together, then merge two larger pieces together, and so on, Lego-style, until you reach a Milky Way mass, about 1012 solar masses. How many merger generations would that take, and how many original pieces? (Hint: Think in powers of 2.) 1042 Chapter 28 The Evolution and Distribution of Galaxies This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1043 29 THE BIG BANG Figure 29.1. Space Telescope of the Future. This drawing shows the James Webb Space Telescope, which is currently planned for launch in 2018. The silver sunshade shadows the primary mirror and science instruments. The primary mirror is 6.5 meters (21 feet) in diameter. Before and during launch, the mirror will be folded up. After the telescope is placed in its orbit, ground controllers will command it to unfold the mirror petals. To see distant galaxies whose light has been shifted to long wavelengths, the telescope will carry several instruments for taking infrared images and spectra", ". (credit: modification of work by NASA) Chapter Outline 29.1 The Age of the Universe 29.2 A Model of the Universe 29.3 The Beginning of the Universe 29.4 The Cosmic Microwave Background 29.5 What Is the Universe Really Made Of? 29.6 The Inflationary Universe 29.7 The Anthropic Principle Thinking Ahead In previous chapters, we explored the contents of the universe\u2014planets, stars, and galaxies\u2014and learned about how these objects change with time. But what about the universe as a whole? How old is it? What did it look like in the beginning? How has it changed since then? What will be its fate? Cosmology is the study of the universe as a whole and is the subject of this chapter. The story of observational cosmology really begins in 1929 when Edwin Hubble published observations of redshifts and distances for a small sample of galaxies and showed the then-revolutionary result that we live in an expanding universe\u2014one which in the past was denser, hotter, and smoother. From this early discovery, astronomers developed many predictions about the origin and evolution of the universe and then tested those predictions with observations. In this chapter, we will describe what we already know about the history of our dynamic universe and highlight some of the mysteries that remain. 1044 Chapter 29 The Big Bang 29.1 THE AGE OF THE UNIVERSE Learning Objectives By the end of this section, you will be able to: Describe how we estimate the age of the universe Explain how changes in the rate of expansion over time affect estimates of the age of the universe Describe the evidence that dark energy exists and that the rate of expansion is currently accelerating Describe some independent evidence for the age of the universe that is consistent with the age estimate based on the rate of expansion To explore the history of the universe, we will follow the same path that astronomers followed historically\u2014beginning with studies of the nearby universe and then probing ever-more-distant objects and looking further back in time. The realization that the universe changes with time came in the 1920s and 1930s when measurements of the redshifts of a large sample of galaxies became available. With hindsight, it is surprising that scientists were so shocked to discover that the universe is expanding. In fact, our theories of gravity demand that the universe must be either expanding or contracting. To show what we mean, let\u2019s begin with a universe of finite size\u2014say a giant ball of a thousand galaxies. All these galaxies attract each other because of their gravity. If they were initially stationary, they would inevitably begin to move closer together and eventually collide. They could avoid this collapse only if for some reason they happened to be moving away from each other at high speeds. In just the same way, only if a rocket is launched at high enough speed can it avoid falling back to Earth. The problem of what happens in an infinite universe is harder to solve, but Einstein (and others) used his theory of general relativity (which we described in Black Holes and Curved Spacetime) to show that even infinite universes cannot be static. Since astronomers at that time did not yet know the universe was expanding (and Einstein himself was philosophically unwilling to accept a universe in motion), he changed his equations by introducing an arbitrary new term (we might call it a fudge factor) called the cosmological constant. This constant represented a hypothetical force of repulsion that could balance gravitational attraction on the largest scales and permit galaxies to remain at fixed distances from one another. That way, the universe could remain still. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1045 Figure 29.2. Einstein and Hubble. (a) Albert Einstein is shown in a 1921 photograph. (b) Edwin Hubble at work in the Mt. Wilson Observatory. About a decade later, Hubble, and his coworkers reported that the universe is expanding, so that no mysterious balancing force is needed. (We discussed this in the chapter on Galaxies.) Einstein is reported to have said that the introduction of the cosmological constant was \u201cthe biggest blunder of my life.\u201d As we shall see later in this chapter, however, relatively recent observations indicate that the expansion is accelerating. Observations are now being carried out to determine whether this acceleration is consistent with a cosmological constant. In a way, it may turn out that Einstein was right after all View this web exhibit (https://openstax.org/l/30exhcosmAIPCHP) on the history of our thinking about cosmology, with images and biographies, from the American Institute of Physics Center for the History of Physics. The Hubble Time If we had a movie of the expanding universe and ran the film backward, what would we see? The galaxies, instead of moving apart, would move together in our movie\u2014getting closer and closer all the time. Eventually, we would find", " that all the matter we can see today was once concentrated in an infinitesimally small volume. Astronomers identify this time with the beginning of the universe. The explosion of that concentrated universe at the beginning of time is called the Big Bang (not a bad term, since you can\u2019t have a bigger bang than one that creates the entire universe). But when did this bang occur? We can make a reasonable estimate of the time since the universal expansion began. To see how astronomers do this, let\u2019s begin with an analogy. Suppose your astronomy class decides to have a party (a kind of \u201cBig Bang\u201d) at someone\u2019s home to celebrate the end of the semester. Unfortunately, everyone is celebrating with so much enthusiasm that the neighbors call the police, who arrive and send everyone away at the same moment. You 1046 Chapter 29 The Big Bang get home at 2 a.m., still somewhat upset about the way the party ended, and realize you forgot to look at your watch to see what time the police got there. But you use a map to measure that the distance between the party and your house is 40 kilometers. And you also remember that you drove the whole trip at a steady speed of 80 kilometers/hour (since you were worried about the police cars following you). Therefore, the trip must have taken: time = distance velocity = 40 kilometers 80 kilometers/hour = 0.5 hours So the party must have broken up at 1:30 a.m. No humans were around to look at their watches when the universe began, but we can use the same technique to estimate when the galaxies began moving away from each other. (Remember that, in reality, it is space that is expanding, not the galaxies that are moving through static space.) If we can measure how far apart the galaxies are now, and how fast they are moving, we can figure out how long a trip it\u2019s been. Let\u2019s call the age of the universe measured in this way T0. Let\u2019s first do a simple case by assuming that the expansion has been at a constant rate ever since the expansion of the universe began. In this case, the time it has taken a galaxy to move a distance, d, away from the Milky Way (remember that at the beginning the galaxies were all together in a very tiny volume) is (as in our example) T0 = d/v where v is the velocity of the galaxy. If we can measure the speed with which galaxies are moving away, and also the distances between them, we can establish how long ago the expansion began. Making such measurements should sound very familiar. This is just what Hubble and many astronomers after him needed to do in order to establish the Hubble law and the Hubble constant. We learned in Galaxies that a galaxy\u2019s distance and its velocity in the expanding universe are related by where H is the Hubble constant. Combining these two expressions gives us V = H \u00d7 d T0 = d v = d (H \u00d7 d) = 1 H We see, then, that the work of calculating this time was already done for us when astronomers measured the Hubble constant. The age of the universe estimated in this way turns out to be just the reciprocal of the Hubble constant (that is, 1/H). This age estimate is sometimes called the Hubble time. For a Hubble constant of 20 kilometers/second per million light-years, the Hubble time is about 15 billion years. The unit used by astronomers for the Hubble constant is kilometers/second per million parsecs. In these units, the Hubble constant is equal to about 70 kilometers/second per million parsecs, again with an uncertainty of about 5%. To make numbers easier to remember, we have done some rounding here. Estimates for the Hubble constant are actually closer to 21 or 22 kilometers/second per million light-years, which would make the age closer to 14 billion years. But there is still about a 5% uncertainty in the Hubble constant, which means the age of the universe estimated in this way is also uncertain by about 5%. To put these uncertainties in perspective, however, you should know that 50 years ago, the uncertainty was a factor of 2. Remarkable progress toward pinning down the Hubble constant has been made in the last couple of decades. The Role of Deceleration The Hubble time is the right age for the universe only if the expansion rate has been constant throughout the time since the expansion of the universe began. Continuing with our end-of-the-semester-party analogy, this is This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1047 equivalent to assuming that you traveled home from the party at a constant rate, when in fact this may not have been the case. At first, mad about having to leave, you may have driven fast, but then as you calmed down\u2014and thought about police cars on the highway\u2014", "you may have begun to slow down until you were driving at a more socially acceptable speed (such as 80 kilometers/hour). In this case, given that you were driving faster at the beginning, the trip home would have taken less than a half-hour. In the same way, in calculating the Hubble time, we have assumed that H has been constant throughout all of time. It turns out that this is not a good assumption. Earlier in their thinking about this, astronomers expected that the rate of expansion should be slowing down. We know that matter creates gravity, whereby all objects pull on all other objects. The mutual attraction between galaxies was expected to slow the expansion as time passed. This means that, if gravity were the only force acting (a big if, as we shall see in the next section), then the rate of expansion must have been faster in the past than it is today. In this case, we would say the universe has been decelerating since the beginning. How much it has decelerated depends on the importance of gravity in slowing the expansion. If the universe were nearly empty, the role of gravity would be minor. Then the deceleration would be close to zero, and the universe would have been expanding at a constant rate. But in a universe with any significant density of matter, the pull of gravity means that the rate of expansion should be slower now than it used to be. If we use the current rate of expansion to estimate how long it took the galaxies to reach their current separations, we will overestimate the age of the universe\u2014just as we may have overestimated the time it took for you to get home from the party. A Universal Acceleration Astronomers spent several decades looking for evidence that the expansion was decelerating, but they were not successful. What they needed were 1) larger telescopes so that they could measure the redshifts of more distant galaxies and 2) a very luminous standard bulb (or standard candle), that is, some astronomical object with known luminosity that produces an enormous amount of energy and can be observed at distances of a billion light-years or more. Recall that we discussed standard bulbs in the chapter on Galaxies. If we compare how luminous a standard bulb is supposed to be and how dim it actually looks in our telescopes, the difference allows us to calculate its distance. The redshift of the galaxy such a bulb is in can tell us how fast it is moving in the universe. So we can measure its distance and motion independently. These two requirements were finally met in the 1990s. Astronomers showed that supernovae of type Ia (see The Death of Stars), with some corrections based on the shapes of their light curves, are standard bulbs. This type of supernova occurs when a white dwarf accretes enough material from a companion star to exceed the Chandrasekhar limit and then collapses and explodes. At the time of maximum brightness, these dramatic supernovae can briefly outshine the galaxies that host them, and hence, they can be observed at very large distances. Large 8- to 10-meter telescopes can be used to obtain the spectra needed to measure the redshifts of the host galaxies (Figure 29.3). 1048 Chapter 29 The Big Bang Figure 29.3. Five Supernovae and Their Host Galaxies. The top row shows each galaxy and its supernova (arrow). The bottom row shows the same galaxies either before or after the supernovae exploded. (credit: modification of work by NASA, ESA, and A. Riess (STScI)) The result of painstaking, careful study of these supernovae in a range of galaxies, carried out by two groups of researchers, was published in 1998. It was shocking\u2014and so revolutionary that their discovery received the 2011 Nobel Prize in Physics. What the researchers found was that these type Ia supernovae in distant galaxies were fainter than expected from Hubble\u2019s law, given the measured redshifts of their host galaxies. In other words, distances estimated from the supernovae used as standard bulbs disagreed with the distances measured from the redshifts. If the universe were decelerating, we would expect the far-away supernovae to be brighter than expected. The slowing down would have kept them closer to us. Instead, they were fainter, which at first seemed to make no sense. Before accepting this shocking development, astronomers first explored the possibility that the supernovae might not really be as useful as standard bulbs as they thought. Perhaps the supernovae appeared too faint because dust along our line of sight to them absorbed some of their light. Or perhaps the supernovae at large distances were for some reason intrinsically less luminous than nearby supernovae of type Ia. A host of more detailed observations ruled out these possibilities. Scientists then had to consider the alternative that the distance estimated from the redshift was incorrect. Distances derived from redshifts assume that the Hubble constant has been truly constant for all time. We saw", " that one way it might not be constant is that the expansion is slowing down. But suppose neither assumption is right (steady speed or slowing down.) Suppose, instead, that the universe is accelerating. If the universe is expanding faster now than it was billions of years ago, our motion away from the distant supernovae has sped up since the explosion occurred, sweeping us farther away from them. The light of the explosion has to travel a greater distance to reach us than if the expansion rate were constant. The farther the light travels, the fainter it appears. This conclusion would explain the supernova observations in a natural way, and this has now been substantiated by many additional observations over the last couple of decades. It really seems that the expansion of the universe is accelerating, a notion so unexpected that astronomers at first resisted considering it. How can the expansion of the universe be speeding up? If you want to accelerate your car, you must supply energy by stepping on the gas. Similarly, energy must be supplied to accelerate the expansion of the universe. The discovery of the acceleration was shocking because scientists still have no idea what the source of the energy is. Scientists call whatever it is dark energy, which is a clear sign of how little we understand it. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1049 Note that this new component of the universe is not the dark matter we talked about in earlier chapters. Dark energy is something else that we have also not yet detected in our laboratories on Earth. What is dark energy? One possibility is that it is the cosmological constant, which is an energy associated with the vacuum of \u201cempty\u201d space itself. Quantum mechanics (the intriguing theory of how things behave at the atomic and subatomic levels) tells us that the source of this vacuum energy might be tiny elementary particles that flicker in and out of existence everywhere throughout the universe. Various attempts have been made to calculate how big the effects of this vacuum energy should be, but so far these attempts have been unsuccessful. In fact, the order of magnitude of theoretical estimates of the vacuum energy based on the quantum mechanics of matter and the value required to account for the acceleration of the expansion of the universe differ by an incredible factor of at least 10120 (that is a 1 followed by 120 zeros)! Various other theories have been suggested, but the bottom line is that, although there is compelling evidence that dark energy exists, we do not yet know the source of that energy. Whatever the dark energy turns out to be, we should note that the discovery that the rate of expansion has not been constant since the beginning of the universe complicates the calculation of the age of the universe. Interestingly, the acceleration seems not to have started with the Big Bang. During the first several billion years after the Big Bang, when galaxies were close together, gravity was strong enough to slow the expansion. As galaxies moved farther apart, the effect of gravity weakened. Several billion years after the Big Bang, dark energy took over, and the expansion began to accelerate (Figure 29.4). Figure 29.4. Changes in the Rate of Expansion of the Universe Since Its Beginning 13.8 Billion Years Ago. The more the diagram spreads out horizontally, the faster the change in the velocity of expansion. After a period of very rapid expansion at the beginning, which scientists call inflation and which we will discuss later in this chapter, the expansion began to decelerate. Galaxies were then close together, and their mutual gravitational attraction slowed the expansion. After a few billion years, when galaxies were farther apart, the influence of gravity began to weaken. Dark energy then took over and caused the expansion to accelerate. (credit: modification of work by Ann Feild (STScI)) Deceleration works to make the age of the universe estimated by the simple relation T0 = 1/H seem older than it really is, whereas acceleration works to make it seem younger. By happy coincidence, our best estimates of how much deceleration and acceleration occurred lead to an answer for the age very close to T0 = 1/H. The best current estimate is that the universe is 13.8 billion years old with an uncertainty of only about 100 million years. Throughout this chapter, we have referred to the Hubble constant. We now know that the Hubble constant does change with time. It is, however, constant everywhere in the universe at any given time. When we say the 1050 Chapter 29 The Big Bang Hubble constant is about 70 kilometers/second/million parsecs, we mean that this is the value of the Hubble constant at the current time. Comparing Ages We now have one estimate for the age of the universe from its expansion. Is this estimate consistent with other observations? For example, are the oldest stars or other astronomical objects younger than 13.8 billion years? After all, the universe has to be at least as old as the oldest objects in it. In our Galaxy", " and others, the oldest stars are found in the globular clusters (Figure 29.5), which can be dated using the models of stellar evolution described in the chapter Stars from Adolescence to Old Age. Figure 29.5. Globular Cluster 47 Tucanae. This NASA/ESA Hubble Space Telescope image shows a globular cluster known as 47 Tucanae, since it is in the constellation of Tucana (The Toucan) in the southern sky. The second-brightest globular cluster in the night sky, it includes hundreds of thousands of stars. Globular clusters are among the oldest objects in our Galaxy and can be used to estimate its age. (credit: NASA, ESA, and the Hubble Heritage (STScI/AURA)-ESA/Hubble Collaboration) The accuracy of the age estimates of the globular clusters has improved markedly in recent years for two reasons. First, models of interiors of globular cluster stars have been improved, mainly through better information about how atoms absorb radiation as they make their way from the center of a star out into space. Second, observations from satellites have improved the accuracy of our measurements of the distances to these clusters. The conclusion is that the oldest stars formed about 12\u201313 billion years ago. This age estimate has recently been confirmed by the study of the spectrum of uranium in the stars. The isotope uranium-238 is radioactive and decays (changes into another element) over time. (Uranium-238 gets its designation because it has 92 protons and 146 neutrons.) We know (from how stars and supernovae make elements) how much uranium-238 is generally made compared to other elements. Suppose we measure the amount of uranium relative to nonradioactive elements in a very old star and in our own Sun, and compare the abundances. With those pieces of information, we can estimate how much longer the uranium has been decaying in the very old star because we know from our own Sun how much uranium decays in 4.5 billion years. The line of uranium is very weak and hard to make out even in the Sun, but it has now been measured in one extremely old star using the European Very Large Telescope (Figure 29.6). Comparing the abundance with that in the solar system, whose age we know, astronomers estimate the star is 12.5 billion years old, with an uncertainty of about 3 billion years. While the uncertainty is large, this work is important confirmation of This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1051 the ages estimated by studies of the globular cluster stars. Note that the uranium age estimate is completely independent; it does not depend on either the measurement of distances or on models of the interiors of stars. Figure 29.6. European Extremely Large Telescope, European Very Large Telescope, and the Colosseum. The European Extremely Large Telescope (E-ELT) is currently under construction in Chile. This image compares the size of the E-ELT (left) with the four 8-meter telescopes of the European Very Large Telescope (center) and with the Colosseum in Rome (right). The mirror of the E-ELT will be 39 meters in diameter. Astronomers are building a new generation of giant telescopes in order to observe very distant galaxies and understand what they were like when they were newly formed and the universe was young. (credit: modification of work by ESO) As we shall see later in this chapter, the globular cluster stars probably did not form until the expansion of the universe had been underway for at least a few hundred million years. Accordingly, their ages are consistent with the 13.8 billion-year age estimated from the expansion rate. 29.2 A MODEL OF THE UNIVERSE Learning Objectives By the end of this section, you will be able to: Explain how the rate of expansion of the universe affects its evolution Describe four possibilities for the evolution of the universe Explain what is expanding when we say that the universe is expanding Define critical density and the evidence that matter alone in the universe is much smaller than the critical density Describe what the observations say about the likely long-term future of the universe Let\u2019s now use the results about the expansion of the universe to look at how these ideas might be applied to develop a model for the evolution of the universe as a whole. With this model, astronomers can make predictions about how the universe has evolved so far and what will happen to it in the future. The Expanding Universe Every model of the universe must include the expansion we observe. Another key element of the models is that the cosmological principle (which we discussed in The Evolution and Distribution of Galaxies) is valid: on the large scale, the universe at any given time is the same everywhere (homogeneous and isotropic). As a result, the expansion rate must be the same everywhere during any epoch of cosmic time. If", " so, we don\u2019t need to think about the entire universe when we think about the expansion, we can just look at any sufficiently large portion 1052 Chapter 29 The Big Bang of it. (Some models for dark energy would allow the expansion rate to be different in different directions, and scientists are designing experiments to test this idea. However, until such evidence is found, we will assume that the cosmological principle applies throughout the universe.) In Galaxies, we hinted that when we think of the expansion of the universe, it is more correct to think of space itself stretching rather than of galaxies moving through static space. Nevertheless, we have since been discussing the redshifts of galaxies as if they resulted from the motion of the galaxies themselves. Now, however, it is time to finally put such simplistic notions behind us and take a more sophisticated look at the cosmic expansion. Recall from our discussion of Einstein\u2019s theory of general relativity (in the chapter on Black Holes and Curved Spacetime) that space\u2014or, more precisely, spacetime\u2014is not a mere backdrop to the action of the universe, as Newton thought. Rather, it is an active participant\u2014affected by and in turn affecting the matter and energy in the universe. Since the expansion of the universe is the stretching of all spacetime, all points in the universe are stretching together. Thus, the expansion began everywhere at once. Unfortunately for tourist agencies of the future, there is no location you can visit where the stretching of space began or where we can say that the Big Bang happened. To describe just how space stretches, we say the cosmic expansion causes the universe to undergo a uniform change in scale over time. By scale we mean, for example, the distance between two clusters of galaxies. It is customary to represent the scale by the factor R; if R doubles, then the distance between the clusters has doubled. Since the universe is expanding at the same rate everywhere, the change in R tells us how much it has expanded (or contracted) at any given time. For a static universe, R would be constant as time passes. In an expanding universe, R increases with time. If it is space that is stretching rather than galaxies moving through space, then why do the galaxies show redshifts in their spectra? When you were young and na\u00efve\u2014a few chapters ago\u2014it was fine to discuss the redshifts of distant galaxies as if they resulted from their motion away from us. But now that you are an older and wiser student of cosmology, this view will simply not do. A more accurate view of the redshifts of galaxies is that the light waves are stretched by the stretching of the space they travel through. Think about the light from a remote galaxy. As it moves away from its source, the light has to travel through space. If space is stretching during all the time the light is traveling, the light waves will be stretched as well. A redshift is a stretching of waves\u2014the wavelength of each wave increases (Figure 29.7). Light from more distant galaxies travels for more time than light from closer ones. This means that the light has stretched more than light from closer ones and thus shows a greater redshift. Figure 29.7. Expansion and Redshift. As an elastic surface expands, a wave on its surface stretches. For light waves, the increase in wavelength would be seen as a redshift. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1053 Thus, what the measured redshift of light from an object is telling us is how much the universe has expanded since the light left the object. If the universe has expanded by a factor of 2, then the wavelength of the light (and all electromagnetic waves from the same source) will have doubled. Models of the Expansion Before astronomers knew about dark energy or had a good measurement of how much matter exists in the universe, they made speculative models about how the universe might evolve over time. The four possible scenarios are shown in Figure 29.8. In this diagram, time moves forward from the bottom upward, and the scale of space increases by the horizontal circles becoming wider. Figure 29.8. Four Possible Models of the Universe. The yellow square marks the present in all four cases, and for all four, the Hubble constant is equal to the same value at the present time. Time is measured in the vertical direction. The first two universes on the left are ones in which the rate of expansion slows over time. The one on the left will eventually slow, come to a stop and reverse, ending up in a \u201cbig crunch,\u201d while the one next to it will continue to expand forever, but ever-more slowly as time passes. The \u201ccoasting\u201d universe is one that expands at a constant rate given by the Hubble constant throughout all of cosmic time. The accelerating universe on the right will continue to expand faster and faster forever. (credit:", " modification of work by NASA/ESA) The simplest scenario of an expanding universe would be one in which R increases with time at a constant rate. But you already know that life is not so simple. The universe contains a great deal of mass and its gravity decelerates the expansion\u2014by a large amount if the universe contains a lot of matter, or by a negligible amount if the universe is nearly empty. Then there is the observed acceleration, which astronomers blame on a kind of dark energy. Let\u2019s first explore the range of possibilities with models for different amounts of mass in the universe and for different contributions by dark energy. In some models\u2014as we shall see\u2014the universe expands forever. In others, it stops expanding and starts to contract. After looking at the extreme possibilities, we will look at recent observations that allow us to choose the most likely scenario. We should perhaps pause for a minute to note how remarkable it is that we can do this at all. Our understanding of the principles that underlie how the universe works on the large scale and our observations of how the objects in the universe change with time allow us to model the evolution of the entire cosmos these days. It is one of the loftiest achievements of the human mind. What astronomers look at in practice, to determine the kind of universe we live in, is the average density of the universe. This is the mass of matter (including the equivalent mass of energy)[1] that would be contained 1 By equivalent mass we mean that which would result if the energy were turned into mass using Einstein\u2019s formula, E = mc2. 1054 Chapter 29 The Big Bang in each unit of volume (say, 1 cubic centimeter) if all the stars, galaxies, and other objects were taken apart, atom by atom, and if all those particles, along with the light and other energy, were distributed throughout all of space with absolute uniformity. If the average density is low, there is less mass and less gravity, and the universe will not decelerate very much. It can therefore expand forever. Higher average density, on the other hand, means there is more mass and more gravity and that the stretching of space might slow down enough that the expansion will eventually stop. An extremely high density might even cause the universe to collapse again. For a given rate of expansion, there is a critical density\u2014the mass per unit volume that will be just enough to slow the expansion to zero at some time infinitely far in the future. If the actual density is higher than this critical density, then the expansion will ultimately reverse and the universe will begin to contract. If the actual density is lower, then the universe will expand forever. These various possibilities are illustrated in Figure 29.9. In this graph, one of the most comprehensive in all of science, we chart the development of the scale of space in the cosmos against the passage of time. Time increases to the right, and the scale of the universe, R, increases upward in the figure. Today, at the point marked \u201cpresent\u201d along the time axis, R is increasing in each model. We know that the galaxies are currently expanding away from each other, no matter which model is right. (The same situation holds for a baseball thrown high into the air. While it may eventually fall back down, near the beginning of the throw it moves upward most rapidly.) The various lines moving across the graph correspond to different models of the universe. The straight dashed line corresponds to the empty universe with no deceleration; it intercepts the time axis at a time, T0 (the Hubble time), in the past. This is not a realistic model but gives us a measure to compare other models to. The curves below the dashed line represent models with no dark energy and with varying amounts of deceleration, starting from the Big Bang at shorter times in the past. The curve above the dashed line shows what happens if the expansion is accelerating. Let\u2019s take a closer look at the future according to the different models. Figure 29.9. Models of the Universe. This graph plots R, the scale of the universe, against time for various cosmological models. Curve 1 represents a universe where the density is greater than the critical value; this model predicts that the universe will eventually collapse. Curve 2 represents a universe with a density lower than critical; the universe will continue to expand but at an ever-slower rate. Curve 3 is a criticaldensity universe; in this universe, the expansion will gradually slow to a stop infinitely far in the future. Curve 4 represents a universe that is accelerating because of the effects of dark energy. The dashed line is for an empty universe, one in which the expansion is not slowed by gravity or accelerated by dark energy. Time is very compressed on this graph. Let\u2019s start with curve 1 in Figure 29.9. In this case, the actual density of the universe is higher than the critical density and there is no dark energy. This universe will stop expanding at some time in the future", " and begin This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1055 contracting. This model is called a closed universe and corresponds to the universe on the left in Figure 29.8. Eventually, the scale drops to zero, which means that space will have shrunk to an infinitely small size. The noted physicist John Wheeler called this the \u201c big crunch,\u201d because matter, energy, space, and time would all be crushed out of existence. Note that the \u201cbig crunch\u201d is the opposite of the Big Bang\u2014it is an implosion. The universe is not expanding but rather collapsing in upon itself. Some scientists speculated that another Big Bang might follow the crunch, giving rise to a new expansion phase, and then another contraction\u2014perhaps oscillating between successive Big Bangs and big crunches indefinitely in the past and future. Such speculation was sometimes referred to as the oscillating theory of the universe. The challenge for theorists was how to describe the transition from collapse (when space and time themselves disappear into the big crunch) to expansion. With the discovery of dark energy, however, it does not appear that the universe will experience a big crunch, so we can put worrying about it on the back burner. If the density of the universe is less than the critical density (curve 2 in Figure 29.9 and the universe second from the left in Figure 29.8), gravity is never important enough to stop the expansion, and so the universe expands forever. Such a universe is infinite and this model is called an open universe. Time and space begin with the Big Bang, but they have no end; the universe simply continues expanding, always a bit more slowly as time goes on. Groups of galaxies eventually get so far apart that it would be difficult for observers in any of them to see the others. (See the feature box on What Might the Universe Be Like in the Distant Future? for more about the distant future in the closed and open universe models.) At the critical density (curve 3), the universe can just barely expand forever. The critical-density universe has an age of exactly two-thirds T0, where T0 is the age of the empty universe. Universes that will someday begin to contract have ages less than two-thirds T0. In an empty universe (the dashed line Figure 29.9 and the coasting universe in Figure 29.8), neither gravity nor dark energy is important enough to affect the expansion rate, which is therefore constant throughout all time. In a universe with dark energy, the rate of the expansion will increase with time, and the expansion will continue at an ever-faster rate. Curve 4 in Figure 29.9, which represents this universe, has a complicated shape. In the beginning, when the matter is all very close together, the rate of expansion is most influenced by gravity. Dark energy appears to act only over large scales and thus becomes more important as the universe grows larger and the matter begins to thin out. In this model, at first the universe slows down, but as space stretches, the acceleration plays a greater role and the expansion speeds up. The Cosmic Tug of War We might summarize our discussion so far by saying that a \u201ctug of war\u201d is going on in the universe between the forces that push everything apart and the gravitational attraction of matter, which pulls everything together. If we can determine who will win this tug of war, we will learn the ultimate fate of the universe. The first thing we need to know is the density of the universe. Is it greater than, less than, or equal to the critical density? The critical density today depends on the value of the expansion rate today, H0. If the Hubble constant is around 20 kilometers/second per million light-years, the critical density is about 10\u201326 kg/m3. Let\u2019s see how this value compares with the actual density of the universe 1056 Chapter 29 The Big Bang Critical Density of the Universe As we discussed, the critical density is that combination of matter and energy that brings the universe coasting to a stop at time infinity. Einstein\u2019s equations lead to the following expression for the critical density (\u03c1crit): \u03c1crit = 3H 2 8\u03c0G where H is the Hubble constant and G is the universal constant of gravity (6.67 \u00d7 10\u201311 Nm2/kg2). Solution Let\u2019s substitute our values and see what we get. Take an H = 22 km/s per million light-years. We need to convert both km and light-years into meters for consistency. A million light-years = 106 \u00d7 9.5 \u00d7 1015 m = 9.5 \u00d7 1021 m. And 22 km/s = 2.2 \u00d7 104 m/s. That makes H = 2.3 \u00d710\u201318 /s and H2 = 5.36 \u00d7 10\u201336 /s2. So", ", \u03c1crit = 3 \u00d7 5.36 \u00d7 10\u201336 8 \u00d7 3.14 \u00d7 6.67 \u00d7 10\u201311 = 9.6 \u00d7 10\u201327 kg/m3 which we can round off to the 10\u201326 kg/m3. (To make the units work out, you have to know that N, the unit of force, is the same as kg \u00d7 m/s2.) Now we can compare densities we measure in the universe to this critical value. Note that density is mass per unit volume, but energy has an equivalent mass of m = E/c2 (from Einstein\u2019s equation E = mc2). Check Your Learning a. A single grain of dust has a mass of about 1.1 \u00d7 10\u201313 kg. If the average mass-energy density of space is equal to the critical density on average, how much space would be required to produce a total mass-energy equal to a dust grain? b. If the Hubble constant were twice what it actually is, how much would the critical density be? Answer: a. In this case, the average mass-energy in a volume V of space is E = \u03c1critV. Thus, for space with critical density, we require that V = Egrain \u03c1crit = 1.1 \u00d7 10\u201313 kg 9.6 \u00d7 10\u201326 kg/m3 = 1.15 \u00d7 1012 m3 = (10,500 m)3 \u2245 (10.5 km)3 Thus, the sides of a cube of space with mass-energy density averaging that of the critical density would need to be slightly greater than 10 km to contain the total energy equal to a single grain of dust! b. Since the critical density goes as the square of the Hubble constant, by doubling the Hubble parameter, the critical density would increase by a factor a four. So if the Hubble constant was 44 km/s per million light-years instead of 22 km/s per million light-years, the critical density would be \u03c1crit = 4 \u00d7 9.6 \u00d7 10\u201327 kg/m3 = 3.8 \u00d7 10\u201326 kg/m3. We can start our survey of how dense the cosmos is by ignoring the dark energy and just estimating the density of all matter in the universe, including ordinary matter and dark matter. Here is where the cosmological principle really comes in handy. Since the universe is the same all over (at least on large scales), we only need to measure how much matter exists in a (large) representative sample of it. This is similar to the way a This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1057 representative survey of a few thousand people can tell us whom the millions of residents of the US prefer for president. There are several methods by which we can try to determine the average density of matter in space. One way is to count all the galaxies out to a given distance and use estimates of their masses, including dark matter, to calculate the average density. Such estimates indicate a density of about 1 to 2 \u00d7 10\u201327 kg/m3 (10 to 20% of critical), which by itself is too small to stop the expansion. A lot of the dark matter lies outside the boundaries of galaxies, so this inventory is not yet complete. But even if we add an estimate of the dark matter outside galaxies, our total won\u2019t rise beyond about 30% of the critical density. We\u2019ll pin these numbers down more precisely later in this chapter, where we will also include the effects of dark energy. In any case, even if we ignore dark energy, the evidence is that the universe will continue to expand forever. The discovery of dark energy that is causing the rate of expansion to speed up only strengthens this conclusion. Things definitely do not look good for fans of the closed universe (big crunch) model What Might the Universe Be Like in the Distant Future? Some say the world will end in fire, Some say in ice. From what I\u2019ve tasted of desire I hold with those who favor fire. \u2014From the poem \u201cFire and Ice\u201d by Robert Frost (1923) Given the destructive power of impacting asteroids, expanding red giants, and nearby supernovae, our species may not be around in the remote future. Nevertheless, you might enjoy speculating about what it would be like to live in a much, much older universe. The observed acceleration makes it likely that we will have continued expansion into the indefinite future. If the universe expands forever (R increases without limit), the clusters of galaxies will spread ever farther apart with time. As eons pass, the universe will get thinner, colder, and darker. Within each galaxy, stars will continue to go through their lives, eventually becoming white dwarfs, neutron stars, and black holes. Low-mass stars might take a long time to finish their evolution, but in this", " model, we would literally have all the time in the world. Ultimately, even the white dwarfs will cool down to be black dwarfs, any neutron stars that reveal themselves as pulsars will slowly stop spinning, and black holes with accretion disks will one day complete their \u201cmeals.\u201d The remains of stars will all be dark and difficult to observe. This means that the light that now reveals galaxies to us will eventually go out. Even if a small pocket of raw material were left in one unsung corner of a galaxy, ready to be turned into a fresh cluster of stars, we will only have to wait until the time that their evolution is also complete. And time is one thing this model of the universe has plenty of. There will surely come a time when all the stars are out, galaxies are as dark as space, and no source of heat remains to help living things survive. Then the lifeless galaxies will just continue to move apart in their lightless realm. 1058 Chapter 29 The Big Bang If this view of the future seems discouraging (from a human perspective), keep in mind that we fundamentally do not understand why the expansion rate is currently accelerating. Thus, our speculations about the future are just that: speculations. You might take heart in the knowledge that science is always a progress report. The most advanced ideas about the universe from a hundred years ago now strike us as rather primitive. It may well be that our best models of today will in a hundred or a thousand years also seem rather simplistic and that there are other factors determining the ultimate fate of the universe of which we are still completely unaware. Ages of Distant Galaxies In the chapter on Galaxies, we discussed how we can use Hubble\u2019s law to measure the distance to a galaxy. But that simple method only works with galaxies that are not too far away. Once we get to large distances, we are looking so far into the past that we must take into account changes in the rate of the expansion of the universe. Since we cannot measure these changes directly, we must assume one of the models of the universe to be able to convert large redshifts into distances. This is why astronomers squirm when reporters and students ask them exactly how far away some newly discovered distant quasar or galaxy is. We really can\u2019t give an answer without first explaining the model of the universe we are assuming in calculating it (by which time a reporter or student is long gone or asleep). Specifically, we must use a model that includes the change in the expansion rate with time. The key ingredients of the model are the amounts of matter, including dark matter, and the equivalent mass (according to E = mc2) of the dark energy along with the Hubble constant. Elsewhere in this book, we have estimated the mass density of ordinary matter plus dark matter as roughly 0.3 times the critical density, and the mass equivalent of dark energy as roughly 0.7 times the critical density. We will refer to these values as the \u201cstandard model of the universe.\u201d The latest (slightly improved) estimates for these values and the evidence for them will be given later in this chapter. Calculations also require the current value of the Hubble constant. For Table 29.1, we have adopted a Hubble constant of 67.3 kilometers/second/ million parsecs (rather than rounding it to 70 kilometers/second/million parsecs), which is consistent with the 13.8 billion-year age of the universe estimated by the latest observations. Once we assume a model, we can use it to calculate the age of the universe at the time an object emitted the light we see. As an example, Table 29.1 lists the times that light was emitted by objects at different redshifts as fractions of the current age of the universe. The times are given for two very different models so you can get a feeling for the fact that the calculated ages are fairly similar. The first model assumes that the universe has a critical density of matter and no dark energy. The second model is the standard model described in the preceding paragraph. The first column in the table is the redshift, which is given by the equation z = \u0394\u03bb/\u03bb0 and is a measure of how much the wavelength of light has been stretched by the expansion of the universe on its long journey to us. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1059 Ages of the Universe at Different Redshifts Redshift Percent of Current Age of Universe When the Light Was Emitted (mass = critical density) Percent of Current Age of Universe When the Light Was Emitted (mass = 0.3 critical density; dark energy = 0.7 critical density) 100 (now) 100 (now) 0 0.5 1.0 2.0 3.0 4.0 5.0 8.0 54 35 19 13 9 7 4 11.9 0.02 Infinite 0", " Table 29.1 63 43 24 16 11 9 5 0.027 0 Notice that as we find objects with higher and higher redshifts, we are looking back to smaller and smaller fractions of the age of the universe. The highest observed redshifts as this book is being written are close to 12 (Figure 29.10). As Table 29.1 shows, we are seeing these galaxies as they were when the universe was only about 3% as old as it is now. They were already formed only about 700 million years after the Big Bang. 1060 Chapter 29 The Big Bang Figure 29.10. Hubble Ultra-Deep Field. This image, called the Hubble Ultra Deep Field, shows faint galaxies, seen very far away and therefore very far back in time. The colored squares in the main image outline the locations of the galaxies. Enlarged views of each galaxy are shown in the black-and-white images. The red lines mark each galaxy\u2019s location. The \u201credshift\u201d of each galaxy is indicated below each box, denoted by the symbol \u201cz.\u201d The redshift measures how much a galaxy\u2019s ultraviolet and visible light has been stretched to infrared wavelengths by the universe\u2019s expansion. The larger the redshift, the more distant the galaxy, and therefore the further astronomers are seeing back in time. One of the seven galaxies may be a distance breaker, observed at a redshift of 11.9. If this redshift is confirmed by additional measurements, the galaxy is seen as it appeared only 380 million years after the Big Bang, when the universe was less than 3% of its present age. (credit: modification of work by NASA, ESA, R. Ellis (Caltech), and the UDF 2012 Team) 29.3 THE BEGINNING OF THE UNIVERSE Learning Objectives By the end of this section, you will be able to: Describe what the universe was like during the first few minutes after it began to expand Explain how the first new elements were formed during the first few minutes after the Big Bang Describe how the contents of the universe change as the temperature of the universe decreases The best evidence we have today indicates that the first galaxies did not begin to form until a few hundred million years after the Big Bang. What were things like before there were galaxies and space had not yet stretched very significantly? Amazingly, scientists have been able to calculate in some detail what was happening in the universe in the first few minutes after the Big Bang. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1061 The History of the Idea It is one thing to say the universe had a beginning (as the equations of general relativity imply) and quite another to describe that beginning. The Belgian priest and cosmologist Georges Lema\u00eetre was probably the first to propose a specific model for the Big Bang itself (Figure 29.11). He envisioned all the matter of the universe starting in one great bulk he called the primeval atom, which then broke into tremendous numbers of pieces. Each of these pieces continued to fragment further until they became the present atoms of the universe, created in a vast nuclear fission. In a popular account of his theory, Lema\u00eetre wrote, \u201cThe evolution of the world could be compared to a display of fireworks just ended\u2014some few red wisps, ashes, and smoke. Standing on a well-cooled cinder, we see the slow fading of the suns and we try to recall the vanished brilliance of the origin of the worlds.\u201d Figure 29.11. Abb\u00e9 Georges Lema\u00eetre (1894\u20131966). This Belgian cosmologist studied theology at Mechelen and mathematics and physics at the University of Leuven. It was there that he began to explore the expansion of the universe and postulated its explosive beginning. He actually predicted Hubble\u2019s law 2 years before its verification, and he was the first to consider seriously the physical processes by which the universe began View a short video (https://openstax.org/l/30Lemaitrevid) about the work of Lema\u00eetre, considered by some to be the father of the Big Bang theory. Physicists today know much more about nuclear physics than was known in the 1920s, and they have shown that the primeval fission model cannot be correct. Yet Lema\u00eetre\u2019s vision was in some respects quite prophetic. We still believe that everything was together at the beginning; it was just not in the form of matter we now know. Basic physical principles tell us that when the universe was much denser, it was also much hotter, and that it cools as it expands, much as gas cools when sprayed from an aerosol can. By the 1940s, scientists knew that fusion of hydrogen into helium was the source of the Sun\u2019s energy. Fusion", " requires high temperatures, and the early universe must have been hot. Based on these ideas, American physicist George Gamow (Figure 29.12) suggested a universe with a different kind of beginning that involved nuclear fusion instead of fission. Ralph Alpher worked out the details for his PhD thesis, and the results were 1062 Chapter 29 The Big Bang published in 1948. (Gamow, who had a quirky sense of humor, decided at the last minute to add the name of physicist Hans Bethe to their paper, so that the coauthors on this paper about the beginning of things would be Alpher, Bethe, and Gamow, a pun on the first three letters of the Greek alphabet: alpha, beta, and gamma.) Gamow\u2019s universe started with fundamental particles that built up the heavy elements by fusion in the Big Bang. Figure 29.12. George Gamow and Collaborators. This composite image shows George Gamow emerging like a genie from a bottle of ylem, a Greek term for the original substance from which the world formed. Gamow revived the term to describe the material of the hot Big Bang. Flanking him are Robert Herman (left) and Ralph Alpher (right), with whom he collaborated in working out the physics of the Big Bang. (The modern composer Karlheinz Stockhausen was inspired by Gamow\u2019s ideas to write a piece of music called Ylem, in which the players actually move away from the stage as they perform, simulating the expansion of the universe.) Gamow\u2019s ideas were close to our modern view, except we now know that the early universe remained hot enough for fusion for only a short while. Thus, only the three lightest elements\u2014hydrogen, helium, and a small amount of lithium\u2014were formed in appreciable abundances at the beginning. The heavier elements formed later in stars. Since the 1940s, many astronomers and physicists have worked on a detailed theory of what happened in the early stages of the universe. The First Few Minutes Let\u2019s start with the first few minutes following the Big Bang. Three basic ideas hold the key to tracing the changes that occurred during the time just after the universe began. The first, as we have already mentioned, is that the universe cools as it expands. Figure 29.13 shows how the temperature changes with the passage of time. Note that a huge span of time, from a tiny fraction of a second to billions of years, is summarized in this diagram. In the first fraction of a second, the universe was unimaginably hot. By the time 0.01 second had elapsed, the temperature had dropped to 100 billion (1011) K. After about 3 minutes, it had fallen to about 1 billion (109) K, still some 70 times hotter than the interior of the Sun. After a few hundred thousand years, the temperature was down to a mere 3000 K, and the universe has continued to cool since that time. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1063 Figure 29.13. Temperature of the Universe. This graph shows how the temperature of the universe varies with time as predicted by the standard model of the Big Bang. Note that both the temperature (vertical axis) and the time in seconds (horizontal axis) change over vast scales on this compressed diagram. All of these temperatures but the last are derived from theoretical calculations since (obviously) no one was there to measure them directly. As we shall see in the next section, however, we have actually detected the feeble glow of radiation emitted at a time when the universe was a few hundred thousand years old. We can measure the characteristics of that radiation to learn what things were like long ago. Indeed, the fact that we have found this ancient glow is one of the strongest arguments in favor of the Big Bang model. The second step in understanding the evolution of the universe is to realize that at very early times, it was so hot that it contained mostly radiation (and not the matter that we see today). The photons that filled the universe could collide and produce material particles; that is, under the conditions just after the Big Bang, energy could turn into matter (and matter could turn into energy). We can calculate how much mass is produced from a given amount of energy by using Einstein\u2019s formula E = mc2 (see the chapter on The Sun: A Nuclear Powerhouse). The idea that energy could turn into matter in the universe at large is a new one for many students, since it is not part of our everyday experience. That\u2019s because, when we compare the universe today to what it was like right after the Big Bang, we live in cold, hard times. The photons in the universe today typically have far-less energy than the amount required to make new matter. In the discussion on the source of the Sun\u2019s energy in The Sun: A Nuclear Power", "house, we briefly mentioned that when subatomic particles of matter and antimatter collide, they turn into pure energy. But the reverse, energy turning into matter and antimatter, is equally possible. This process has been observed in particle accelerators around the world. If we have enough energy, under the right circumstances, new particles of matter (and antimatter) are indeed created \u2014and the conditions were right during the first few minutes after the expansion of the universe began. Our third key point is that the hotter the universe was, the more energetic were the photons available to make matter and antimatter (see Figure 29.13). To take a specific example, at a temperature of 6 billion (6 \u00d7 109) K, the collision of two typical photons can create an electron and its antimatter counterpart, a positron. If the temperature exceeds 1014 K, much more massive protons and antiprotons can be created. The Evolution of the Early Universe Keeping these three ideas in mind, we can trace the evolution of the universe from the time it was about 0.01 second old and had a temperature of about 100 billion K. Why not begin at the very beginning? There are as yet 1064 Chapter 29 The Big Bang no theories that allow us penetrate to a time before about 10\u201343 second (this number is a decimal point followed by 42 zeros and then a one). It is so small that we cannot relate it to anything in our everyday experience. When the universe was that young, its density was so high that the theory of general relativity is not adequate to describe it, and even the concept of time breaks down. Scientists, by the way, have been somewhat more successful in describing the universe when it was older than 10\u201343 second but still less than about 0.01 second old. We will take a look at some of these ideas later in this chapter, but for now, we want to start with somewhat more familiar situations. By the time the universe was 0.01 second old, it consisted of a soup of matter and radiation; the matter included protons and neutrons, leftovers from an even younger and hotter universe. Each particle collided rapidly with other particles. The temperature was no longer high enough to allow colliding photons to produce neutrons or protons, but it was sufficient for the production of electrons and positrons (Figure 29.14). There was probably also a sea of exotic subatomic particles that would later play a role as dark matter. All the particles jiggled about on their own; it was still much too hot for protons and neutrons to combine to form the nuclei of atoms. Figure 29.14. Particle Interactions in the Early Universe. (a) In the first fractions of a second, when the universe was very hot, energy was converted into particles and antiparticles. The reverse reaction also happened: a particle and antiparticle could collide and produce energy. (b) As the temperature of the universe decreased, the energy of typical photons became too low to create matter. Instead, existing particles fused to create such nuclei as deuterium and helium. (c) Later, it became cool enough for electrons to settle down with nuclei and make neutral atoms. Most of the universe was still hydrogen. Think of the universe at this time as a seething cauldron, with photons colliding and interchanging energy, and sometimes being destroyed to create a pair of particles. The particles also collided with one another. Frequently, a matter particle and an antimatter particle met and turned each other into a burst of gamma-ray radiation. Among the particles created in the early phases of the universe was the ghostly neutrino (see The Sun: A Nuclear Powerhouse), which today interacts only very rarely with ordinary matter. In the crowded conditions of the very early universe, however, neutrinos ran into so many electrons and positrons that they experienced frequent interactions despite their \u201cantisocial\u201d natures. By the time the universe was a little more than 1 second old, the density had dropped to the point where neutrinos no longer interacted with matter but simply traveled freely through space. In fact, these neutrinos should now be all around us. Since they have been traveling through space unimpeded (and hence unchanged) since the universe was 1 second old, measurements of their properties would offer one of the best tests of the Big Bang model. Unfortunately, the very characteristic that makes them so useful\u2014the fact that they interact This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1065 so weakly with matter that they have survived unaltered for all but the first second of time\u2014also renders them unable to be measured, at least with present techniques. Perhaps someday someone will devise a way to capture these elusive messengers from the past. Atomic Nuclei Form When the universe was about 3 minutes old and its temperature was down to about 900 million K,", " protons and neutrons could combine. At higher temperatures, these atomic nuclei had immediately been blasted apart by interactions with high-energy photons and thus could not survive. But at the temperatures and densities reached between 3 and 4 minutes after the beginning, deuterium (a proton and neutron) lasted long enough that collisions could convert some of it into helium, (Figure 29.14). In essence, the entire universe was acting the way centers of stars do today\u2014fusing new elements from simpler components. In addition, a little bit of element 3, lithium, could also form. This burst of cosmic fusion was only a brief interlude, however. By 4 minutes after the Big Bang, more helium was having trouble forming. The universe was still expanding and cooling down. After the formation of helium and some lithium, the temperature had dropped so low that the fusion of helium nuclei into still-heavier elements could not occur. No elements beyond lithium could form in the first few minutes. That 4-minute period was the end of the time when the entire universe was a fusion factory. In the cool universe we know today, the fusion of new elements is limited to the centers of stars and the explosions of supernovae. Still, the fact that the Big Bang model allows the creation of a good deal of helium is the answer to a long- standing mystery in astronomy. Put simply, there is just too much helium in the universe to be explained by what happens inside stars. All the generations of stars that have produced helium since the Big Bang cannot account for the quantity of helium we observe. Furthermore, even the oldest stars and the most distant galaxies show significant amounts of helium. These observations find a natural explanation in the synthesis of helium by the Big Bang itself during the first few minutes of time. We estimate that 10 times more helium was manufactured in the first 4 minutes of the universe than in all the generations of stars during the succeeding 10 to 15 billion years These nice animations (https://openstax.org/l/30origelemani) that explain the way in which different elements formed in the history of the universe are from the University of Chicago\u2019s Origins of the Elements site. Learning from Deuterium We can learn many things from the way the early universe made atomic nuclei. It turns out that all of the deuterium (a hydrogen nucleus with a neutron in it) in the universe was formed during the first 4 minutes. In stars, any region hot enough to fuse two protons to form a deuterium nucleus is also hot enough to change it further\u2014either by destroying it through a collision with an energetic photon or by converting it into helium through nuclear reactions. The amount of deuterium that can be produced in the first 4 minutes of creation depends on the density of the universe at the time deuterium was formed. If the density were relatively high, nearly all the deuterium would have been converted into helium through interactions with protons, just as it is in stars. If the density were relatively low, then the universe would have expanded and thinned out rapidly enough that some deuterium 1066 Chapter 29 The Big Bang would have survived. The amount of deuterium we see today thus gives us a clue to the density of the universe when it was about 4 minutes old. Theoretical models can relate the density then to the density now; thus, measurements of the abundance of deuterium today can give us an estimate of the current density of the universe. The measurements of deuterium indicate that the present-day density of ordinary matter\u2014protons and neutrons\u2014is about 5 \u00d7 10\u201328 kg/m3. Deuterium can only provide an estimate of the density of ordinary matter because the abundance of deuterium is determined by the particles that interact to form it, namely protons and neutrons alone. From the abundance of deuterium, we know that not enough protons and neutrons are present, by a factor of about 20, to produce a critical-density universe. We do know, however, that there are dark matter particles that add to the overall matter density of the universe, which is then higher than what is calculated for ordinary matter alone. Because dark matter particles do not affect the production of deuterium, measurement of the deuterium abundance cannot tell us how much dark matter exists. Dark matter is made of some exotic kind of particle, not yet detected in any earthbound laboratory. It is definitely not made of protons and neutrons like the readers of this book. 29.4 THE COSMIC MICROWAVE BACKGROUND Learning Objectives By the end of this section, you will be able to: Explain why we can observe the afterglow of the hot, early universe Discuss the properties of this afterglow as we see it today, including its average temperature and the size of its temperature fluctuations Describe open, flat, and curved universes and explain which type of universe is supported by observations Sum", "marize our current knowledge of the basic properties of the universe including its age and contents The description of the first few minutes of the universe is based on theoretical calculations. It is crucial, however, that a scientific theory should be testable. What predictions does it make? And do observations show those predictions to be accurate? One success of the theory of the first few minutes of the universe is the correct prediction of the amount of helium in the universe. Another prediction is that a significant milestone in the history of the universe occurred about 380,000 years after the Big Bang. Scientists have directly observed what the universe was like at this early stage, and these observations offer some of the strongest support for the Big Bang theory. To find out what this milestone was, let\u2019s look at what theory tells us about what happened during the first few hundred thousand years after the Big Bang. The fusion of helium and lithium was completed when the universe was about 4 minutes old. The universe then continued to resemble the interior of a star in some ways for a few hundred thousand years more. It remained hot and opaque, with radiation being scattered from one particle to another. It was still too hot for electrons to \u201csettle down\u201d and become associated with a particular nucleus; such free electrons are especially effective at scattering photons, thus ensuring that no radiation ever got very far in the early universe without having its path changed. In a way, the universe was like an enormous crowd right after a popular concert; if you get separated from a friend, even if he is wearing a flashing button, it is impossible to see through the dense crowd to spot him. Only after the crowd clears is there a path for the light from his button to reach you. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1067 The Universe Becomes Transparent Not until a few hundred thousand years after the Big Bang, when the temperature had dropped to about 3000 K and the density of atomic nuclei to about 1000 per cubic centimeter, did the electrons and nuclei manage to combine to form stable atoms of hydrogen and helium (Figure 29.14). With no free electrons to scatter photons, the universe became transparent for the first time in cosmic history. From this point on, matter and radiation interacted much less frequently; we say that they decoupled from each other and evolved separately. Suddenly, electromagnetic radiation could really travel, and it has been traveling through the universe ever since. Discovery of the Cosmic Background Radiation If the model of the universe described in the previous section is correct, then\u2014as we look far outward in the universe and thus far back in time\u2014the first \u201cafterglow\u201d of the hot, early universe should still be detectable. Observations of it would be very strong evidence that our theoretical calculations about how the universe evolved are correct. As we shall see, we have indeed detected the radiation emitted at this photon decoupling time, when radiation began to stream freely through the universe without interacting with matter (Figure 29.15). Figure 29.15. Cosmic Microwave Background and Clouds Compared. (a) Early in the universe, photons (electromagnetic energy) were scattering off the crowded, hot, charged particles and could not get very far without colliding with another particle. But after electrons and photons settled into neutral atoms, there was far less scattering, and photons could travel over vast distances. The universe became transparent. As we look out in space and back in time, we can\u2019t see back beyond this time. (b) This is similar to what happens when we see clouds in Earth\u2019s atmosphere. Water droplets in a cloud scatter light very efficiently, but clear air lets light travel over long distances. So as we look up into the atmosphere, our vision is blocked by the cloud layers and we can\u2019t see beyond them. (credit: modification of work by NASA) The detection of this afterglow was initially an accident. In the late 1940s, Ralph Alpher and Robert Herman, working with George Gamow, realized that just before the universe became transparent, it must have been radiating like a blackbody at a temperature of about 3000 K\u2014the temperature at which hydrogen atoms could begin to form. If we could have seen that radiation just after neutral atoms formed, it would have resembled radiation from a reddish star. It was as if a giant fireball filled the whole universe. 1068 Chapter 29 The Big Bang But that was nearly 14 billion years ago, and, in the meantime, the scale of the universe has increased a thousand fold. This expansion has increased the wavelength of the radiation by a factor of 1000 (see Figure 29.7). According to Wien\u2019s law, which relates wavelength and temperature, the expansion has correspondingly lowered the temperature by a factor of 1000 (see the chapter on Radiation and Spectra). Alpher and Herman predicted that the glow from the fireball should now be at radio", " wavelengths and should resemble the radiation from a blackbody at a temperature only a few degrees above absolute zero. Since the fireball was everywhere throughout the universe, the radiation left over from it should also be everywhere. If our eyes were sensitive to radio wavelengths, the whole sky would appear to glow very faintly. However, our eyes can\u2019t see at these wavelengths, and at the time Alpher and Herman made their prediction, there were no instruments that could detect the glow. Over the years, their prediction was forgotten. In the mid-1960s, in Holmdel, New Jersey, Arno Penzias and Robert Wilson of AT&T\u2019s Bell Laboratories had built a delicate microwave antenna (Figure 29.16) to measure astronomical sources, including supernova remnants like Cassiopeia A (see the chapter on The Death of Stars). They were plagued with some unexpected background noise, just like faint static on a radio, which they could not get rid of. The puzzling thing about this radiation was that it seemed to be coming from all directions at once. This is very unusual in astronomy: after all, most radiation has a specific direction where it is strongest\u2014the direction of the Sun, or a supernova remnant, or the disk of the Milky Way, for example. Figure 29.16. Robert Wilson (left) and Arno Penzias (right). These two scientists are standing in front of the horn-shaped antenna with which they discovered the cosmic background radiation. The photo was taken in 1978, just after they received the Nobel Prize in physics. Penzias and Wilson at first thought that any radiation appearing to come from all directions must originate from inside their telescope, so they took everything apart to look for the source of the noise. They even found that some pigeons had roosted inside the big horn-shaped antenna and had left (as Penzias delicately put it) \u201ca layer of white, sticky, dielectric substance coating the inside of the antenna.\u201d However, nothing the scientists did could reduce the background radiation to zero, and they reluctantly came to accept that it must be real, and it must be coming from space. Penzias and Wilson were not cosmologists, but as they began to discuss their puzzling discovery with other scientists, they were quickly put in touch with a group of astronomers and physicists at Princeton University (a short drive away). These astronomers had\u2014as it happened\u2014been redoing the calculations of Alpher and Herman from the 1940s and also realized that the radiation from the decoupling time should be detectable as a faint afterglow of radio waves. The different calculations of what the observed temperature would be for this cosmic microwave background (CMB)[2] were uncertain, but all predicted less than 40 K. 2 Recall that microwaves are in the radio region of the electromagnetic spectrum. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1069 Penzias and Wilson found the distribution of intensity at different radio wavelengths to correspond to a temperature of 3.5 K. This is very cold\u2014closer to absolute zero than most other astronomical measurements\u2014and a testament to how much space (and the waves within it) has stretched. Their measurements have been repeated with better instruments, which give us a reading of 2.73 K. So Penzias and Wilson came very close. Rounding this value, scientists often refer to \u201cthe 3-degree microwave background.\u201d Many other experiments on Earth and in space soon confirmed the discovery by Penzias and Wilson: The radiation was indeed coming from all directions (it was isotropic) and matched the predictions of the Big Bang theory with remarkable precision. Penzias and Wilson had inadvertently observed the glow from the primeval fireball. They received the Nobel Prize for their work in 1978. And just before his death in 1966, Lema\u00eetre learned that his \u201cvanished brilliance\u201d had been discovered and confirmed You may enjoy watching Three Degrees, a 26-minute video (https://openstax.org/l/30threedegvid) from Bell Labs about Penzias and Wilson\u2019s discovery of the cosmic background radiation (with interesting historical footage). Properties of the Cosmic Microwave Background One issue that worried astronomers is that Penzias and Wilson were measuring the background radiation filling space through Earth\u2019s atmosphere. What if that atmosphere is a source of radio waves or somehow affected their measurements? It would be better to measure something this important from space. The first accurate measurements of the CMB were made with a satellite orbiting Earth. Named the Cosmic Background Explorer ( COBE), it was launched by NASA in November 1989. The data it received quickly showed that the CMB closely matches that expected from a blackbody with a temperature of 2.73 K (Figure 29.17). This is exactly the result expected if the CMB was indeed redshifted", " radiation emitted by a hot gas that filled all of space shortly after the universe began. 1070 Chapter 29 The Big Bang Figure 29.17. Cosmic Background Radiation. The solid line shows how the intensity of radiation should change with wavelength for a blackbody with a temperature of 2.73 K. The boxes show the intensity of the cosmic background radiation as measured at various wavelengths by COBE\u2019s instruments. The fit is perfect. When this graph was first shown at a meeting of astronomers, they gave it a standing ovation. The first important conclusion from measurements of the CMB, therefore, is that the universe we have today has indeed evolved from a hot, uniform state. This observation also provides direct support for the general idea that we live in an evolving universe, since the universe is cooler today than it was in the beginning. Small Differences in the CMB It was known even before the launch of COBE that the CMB is extremely isotropic. In fact, its uniformity in every direction is one of the best confirmations of the cosmological principle\u2014 that the universe is homogenous and isotropic. According to our theories, however, the temperature could not have been perfectly uniform when the CMB was emitted. After all, the CMB is radiation that was scattered from the particles in the universe at the time of decoupling. If the radiation were completely smooth, then all those particles must have been distributed through space absolutely evenly. Yet it is those particles that have become all the galaxies and stars (and astronomy students) that now inhabit the cosmos. Had the particles been completely smoothly distributed, they could not have formed all the large-scale structures now present in the universe\u2014the clusters and superclusters of galaxies discussed in the last few chapters. The early universe must have had tiny density fluctuations from which such structures could evolve. Regions of higher-than-average density would have attracted additional matter and eventually grown into the galaxies and clusters that we see today. It turned out that these denser regions would appear to us to be colder spots, that is, they would have lower-than-average temperatures. The reason that temperature and density are related can be explained this way. At the time of decoupling, photons in a slightly denser portion of space had to expend some of their energy to escape the gravitational force exerted by the surrounding gas. In losing energy, the photons became slightly colder than the overall average temperature at the time of decoupling. Vice versa, photons that were located in a slightly less dense portion of space lost less energy upon leaving it than other photons, thus appearing slightly hotter than average. Therefore, if the seeds of present-day galaxies existed at the time that the CMB was emitted, we should see some slight variations in the CMB temperature as we look in different directions in the sky. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1071 Scientists working with the data from the COBE satellite did indeed detect very subtle temperature differences\u2014about 1 part in 100,000\u2014in the CMB. The regions of lower-than-average temperature come in a variety of sizes, but even the smallest of the colder areas detected by COBE is far too large to be the precursor of an individual galaxy, or even a supercluster of galaxies. This is because the COBE instrument had \u201cblurry vision\u201d (poor resolution) and could only measure large patches of the sky. We needed instruments with \u201csharper vision.\u201d The most detailed measurements of the CMB have been obtained by two satellites launched more recently than COBE. The results from the first of these satellites, the Wilkinson Microwave Anisotropy Probe ( WMAP) spacecraft, were published in 2003. In 2015, measurements from the Planck satellite extended the WMAP measurements to even-higher spatial resolution and lower noise (Figure 29.18). Figure 29.18. CMB Observations. This comparison shows how much detail can be seen in the observations of three satellites used to measure the CMB. The CMB is a snapshot of the oldest light in our universe, imprinted on the sky when the universe was just about 380,000 years old. The first spacecraft, launched in 1989, is NASA\u2019s Cosmic Background Explorer, or COBE. WMAP was launched in 2001, and Planck was launched in 2009. The three panels show 10-square-degree patches of all-sky maps. This cosmic background radiation image (bottom) is an all-sky map of the CMB as observed by the Planck mission. The colors in the map represent different temperatures: red for warmer and blue for cooler. These tiny temperature fluctuations correspond to regions of slightly different densities, representing the seeds of all future structures: the stars, galaxies, and galaxy clusters of today. (credit top: modification of work by NASA/JPL-Caltech/ESA; credit bottom: modification", " of work by ESA and the Planck Collaboration) Theoretical calculations show that the sizes of the hot and cold spots in the CMB depend on the geometry of the universe and hence on its total density. (It\u2019s not at all obvious that it should do so, and it takes some pretty 1072 Chapter 29 The Big Bang fancy calculations\u2014way beyond the level of our text\u2014to make the connection, but having such a dependence is very useful.) The total density we are discussing here includes both the amount of mass in the universe and the mass equivalent of the dark energy. That is, we must add together mass and energy: ordinary matter, dark matter, and the dark energy that is speeding up the expansion. To see why this works, remember (from the chapter on Black Holes and Curved Spacetime) that with his theory of general relativity, Einstein showed that matter can curve space and that the amount of curvature depends on the amount of matter present. Therefore, the total amount of matter in the universe (including dark matter and the equivalent matter contribution by dark energy), determines the overall geometry of space. Just like the geometry of space around a black hole has a curvature to it, so the entire universe may have a curvature. Let\u2019s take a look at the possibilities (Figure 29.19). If the density of matter is higher than the critical density, the universe will eventually collapse. In such a closed universe, two initially parallel rays of light will eventually meet. This kind of geometry is referred to as spherical geometry. If the density of matter is less than critical, the universe will expand forever. Two initially parallel rays of light will diverge, and this is referred to as hyperbolic geometry. In a critical-density universe, two parallel light rays never meet, and the expansion comes to a halt only at some time infinitely far in the future. We refer to this as a flat universe, and the kind of Euclidean geometry you learned in high school applies in this type of universe. Figure 29.19. Picturing Space Curvature for the Entire Universe. The density of matter and energy determines the overall geometry of space. If the density of the universe is greater than the critical density, then the universe will ultimately collapse and space is said to be closed like the surface of a sphere. If the density exactly equals the critical density, then space is flat like a sheet of paper; the universe will expand forever, with the rate of expansion coming to a halt infinitely far in the future. If the density is less than critical, then the expansion will continue forever and space is said to be open and negatively curved like the surface of a saddle (where more space than you expect opens up as you move farther away). Note that the red lines in each diagram show what happens in each kind of space\u2014they are initially parallel but follow different paths depending on the curvature of space. Remember that these drawings are trying to show how space for the entire universe is \u201cwarped\u201d\u2014this can\u2019t be seen locally in the small amount of space that we humans occupy. If the density of the universe is equal to the critical density, then the hot and cold spots in the CMB should typically be about a degree in size. If the density is greater than critical, then the typical sizes will be larger than one degree. If the universe has a density less than critical, then the structures will appear smaller. In Figure 29.20, you can see the differences easily. WMAP and Planck observations of the CMB confirmed earlier experiments that we do indeed live in a flat, critical-density universe. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1073 Figure 29.20. Comparison of CMB Observations with Possible Models of the Universe. Cosmological simulations predict that if our universe has critical density, then the CMB images will be dominated by hot and cold spots of around one degree in size (bottom center). If, on the other hand, the density is higher than critical (and the universe will ultimately collapse), then the images\u2019 hot and cold spots will appear larger than one degree (bottom left). If the density of the universe is less than critical (and the expansion will continue forever), then the structures will appear smaller (bottom right). As the measurements show, the universe is at critical density. The measurements shown were made by a balloon-borne instrument called BOOMERanG (Balloon Observations of Millimetric Extragalactic Radiation and Geophysics), which was flown in Antarctica. Subsequent satellite observations by WMAP and Planck confirm the BOOMERanG result. (credit: modification of work by NASA) Key numbers from an analysis of the Planck data give us the best values currently available for some of the basic properties of the universe: \u2022 Age of universe: 13.799 \u00b1 0.038 billion", " years (Note: That means we know the age of the universe to within 38 million years. Amazing!) \u2022 Hubble constant: 67.31 \u00b1 0.96 kilometers/second/million parsecs \u2022 Fraction of universe\u2019s content that is \u201cdark energy\u201d: 68.5% \u00b1 1.3% \u2022 Fraction of the universe\u2019s content that is matter: 31.5% \u00b1 1.3% Note that this value for the Hubble constant is slightly smaller than the value of 70 kilometers/second/million parsecs that we have adopted in this book. In fact, the value derived from measurements of redshifts is 73 kilometers/second/million parsecs. So precise is modern cosmology these days that scientists are working hard to resolve this discrepancy. The fact that the difference between these two independent measurements is so small is actually a remarkable achievement. Only a few decades ago, astronomers were arguing about whether the Hubble constant was around 50 kilometers/second/million parsecs or 100 kilometers/second/ million parsecs. 1074 Chapter 29 The Big Bang Analysis of Planck data also shows that ordinary matter (mainly protons and neutrons) makes up 4.9% of the total density. Dark matter plus normal matter add up to 31.5% of the total density. Dark energy contributes the remaining 68.5%. The age of the universe at decoupling\u2014that is, when the CMB was emitted\u2014was 380,000 years. Perhaps the most surprising result from the high-precision measurements by WMAP and the even higher- precision measurements from Planck is that there were no surprises. The model of cosmology with ordinary matter at about 5%, dark matter at about 25%, and dark energy about 70% has survived since the late 1990s when cosmologists were forced in that direction by the supernovae data. In other words, the very strange universe that we have been describing, with only about 5% of its contents being made up of the kinds of matter we are familiar with here on Earth, really seems to be the universe we live in. After the CMB was emitted, the universe continued to expand and cool off. By 400 to 500 million years after the Big Bang, the very first stars and galaxies had already formed. Deep in the interiors of stars, matter was reheated, nuclear reactions were ignited, and the more gradual synthesis of the heavier elements that we have discussed throughout this book began. We conclude this quick tour of our model of the early universe with a reminder. You must not think of the Big Bang as a localized explosion in space, like an exploding superstar. There were no boundaries and there was no single site where the explosion happened. It was an explosion of space (and time and matter and energy) that happened everywhere in the universe. All matter and energy that exist today, including the particles of which you are made, came from the Big Bang. We were, and still are, in the midst of a Big Bang; it is all around us. 29.5 WHAT IS THE UNIVERSE REALLY MADE OF? Learning Objectives By the end of this section, you will be able to: Specify what fraction of the density of the universe is contributed by stars and galaxies and how much ordinary matter (such as hydrogen, helium, and other elements we are familiar with here on Earth) makes up the overall density Describe how ideas about the contents of the universe have changed over the last 50 years Explain why it is so difficult to determine what dark matter really is Explain why dark matter helped galaxies form quickly in the early universe Summarize the evolution of the universe from the time the CMB was emitted to the present day The model of the universe we described in the previous section is the simplest model that explains the observations. It assumes that general relativity is the correct theory of gravity throughout the universe. With this assumption, the model then accounts for the existence and structure of the CMB; the abundances of the light elements deuterium, helium, and lithium; and the acceleration of the expansion of the universe. All of the observations to date support the validity of the model, which is referred to as the standard (or concordance) model of cosmology. Figure 29.21 and Table 29.2 summarize the current best estimates of the contents of the universe. Luminous matter in stars and galaxies and neutrinos contributes about 1% of the mass required to reach critical density. Another 4% is mainly in the form of hydrogen and helium in the space between stars and in intergalactic space. Dark matter accounts for about an additional 27% of the critical density. The mass equivalent of dark energy (according to E = mc2) then supplies the remaining 68% of the critical density. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1075 Figure 29.21. Composition of the Universe. Only about 5", "% of all the mass and energy in the universe is matter with which we are familiar here on Earth. Most ordinary matter consists of hydrogen and helium located in interstellar and intergalactic space. Only about one-half of 1% of the critical density of the universe is found in stars. Dark matter and dark energy, which have not yet been detected in earthbound laboratories, account for 95% of the contents of the universe. What Different Kinds of Objects Contribute to the Density of the Universe Object Density as a Percent of Critical Density Luminous matter (stars, etc.) <1 Hydrogen and helium in interstellar and intergalactic space 4 Dark matter Equivalent mass density of the dark energy 27 68 Table 29.2 This table should shock you. What we are saying is that 95% of the stuff of the universe is either dark matter or dark energy\u2014neither of which has ever been detected in a laboratory here on Earth. This whole textbook, which has focused on objects that emit electromagnetic radiation, has generally been ignoring 95% of what is out there. Who says there aren\u2019t big mysteries yet to solve in science! Figure 29.22 shows how our ideas of the composition of the universe have changed over just the past three decades. The fraction of the universe that we think is made of the same particles as astronomy students has been decreasing steadily. 1076 Chapter 29 The Big Bang Figure 29.22. Changing Estimates of the Content of the Universe. This diagram shows the changes in our understanding of the contents of the universe over the past three decades. In the 1970s, we suspected that most of the matter in the universe was invisible, but we thought that this matter might be ordinary matter (protons, neutrons, etc.) that was simply not producing electromagnetic radiation. By the 1980s, it was becoming likely that most of the dark matter was made of something we had not yet detected on Earth. By the late 1990s, a variety of experiments had shown that we live in a critical -density universe and that dark energy contributes about 70% of what is required to reach critical density. Note how the estimate of the relative importance of ordinary luminous matter (shown in yellow) has diminished over time. What Is Dark Matter? Many astronomers find the situation we have described very satisfying. Several independent experiments now agree on the type of universe we live in and on the inventory of what it contains. We seem to be very close to having a cosmological model that explains nearly everything. Others are not yet ready to jump on the bandwagon. They say, \u201cshow me the 96% of the universe we can\u2019t detect directly\u2014for example, find me some dark matter!\u201d At first, astronomers thought that dark matter might be hidden in objects that appear dark because they emit no light (e.g., black holes) or that are too faint to be observed at large distances (e.g., planets or white dwarfs). However, these objects would be made of ordinary matter, and the deuterium abundance tells us that no more than 5% of the critical density consists of ordinary matter. Another possible form that dark matter can take is some type of elementary particle that we have not yet detected here on Earth\u2014a particle that has mass and exists in sufficient abundance to contribute 23% of the critical density. Some physics theories predict the existence of such particles. One class of these particles has been given the name WIMPs, which stands for weakly interacting massive particles. Since these particles do not participate in nuclear reactions leading to the production of deuterium, the deuterium abundance puts no limits on how many WIMPs might be in the universe. (A number of other exotic particles have also been suggested as prime constituents of dark matter, but we will confine our discussion to WIMPs as a useful example.) If large numbers of WIMPs do exist, then some of them should be passing through our physics laboratories right now. The trick is to catch them. Since by definition they interact only weakly (infrequently) with other matter, the chances that they will have a measurable effect are small. We don\u2019t know the mass of these particles, but various theories suggest that it might be a few to a few hundred times the mass of a proton. If WIMPs are 60 times the mass of a proton, there would be about 10 million of them passing through your outstretched hand every second\u2014with absolutely no effect on you. If that seems too mind-boggling, bear in mind that neutrinos interact weakly with ordinary matter, and yet we were able to \u201ccatch\u201d them eventually. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1077 Despite the challenges, more than 30 experiments designed to detect WIMPS are in operation or in the planning stages. Predictions of how many times WIMPs might actually collide with the", " nucleus of an atom in the instrument designed to detect them are in the range of 1 event per year to 1 event per 1000 years per kilogram of detector. The detector must therefore be large. It must be shielded from radioactivity or other types of particles, such as neutrons, passing through it, and hence these detectors are placed in deep mines. The energy imparted to an atomic nucleus in the detector by collision with a WIMP will be small, and so the detector must be cooled to a very low temperature. The WIMP detectors are made out of crystals of germanium, silicon, or xenon. The detectors are cooled to a few thousandths of a degree\u2014very close to absolute zero. That means that the atoms in the detector are so cold that they are scarcely vibrating at all. If a dark matter particle collides with one of the atoms, it will cause the whole crystal to vibrate and the temperature therefore to increase ever so slightly. Some other interactions may generate a detectable flash of light. A different kind of search for WIMPs is being conducted at the Large Hadron Collider (LHC) at CERN, Europe\u2019s particle physics lab near Geneva, Switzerland. In this experiment, protons collide with enough energy potentially to produce WIMPs. The LHC detectors cannot detect the WIMPs directly, but if WIMPs are produced, they will pass through the detectors, carrying energy away with them. Experimenters will then add up all the energy that they detect as a result of the collisions of protons to determine if any energy is missing. So far, none of these experiments has detected WIMPs. Will the newer experiments pay off? Or will scientists have to search for some other explanation for dark matter? Only time will tell (Figure 29.23). Figure 29.23. Dark Matter. This cartoon from NASA takes a humorous look at how little we yet understand about dark matter. (credit: NASA) Dark Matter and the Formation of Galaxies As elusive as dark matter may be in the current-day universe, galaxies could not have formed quickly without it. Galaxies grew from density fluctuations in the early universe, and some had already formed only about 1078 Chapter 29 The Big Bang 400\u2013500 million years after the Big Bang. The observations with WMAP, Planck, and other experiments give us information on the size of those density fluctuations. It turns out that the density variations we observe are too small to have formed galaxies so soon after the Big Bang. In the hot, early universe, energetic photons collided with hydrogen and helium, and kept them moving so rapidly that gravity was still not strong enough to cause the atoms to come together to form galaxies. How can we reconcile this with the fact that galaxies did form and are all around us? Our instruments that measure the CMB give us information about density fluctuations only for ordinary matter, which interacts with radiation. Dark matter, as its name indicates, does not interact with photons at all. Dark matter could have had much greater variations in density and been able to come together to form gravitational \u201ctraps\u201d that could then have begun to attract ordinary matter immediately after the universe became transparent. As ordinary matter became increasingly concentrated, it could have turned into galaxies quickly thanks to these dark matter traps. For an analogy, imagine a boulevard with traffic lights every half mile or so. Suppose you are part of a motorcade of cars accompanied by police who lead you past each light, even if it is red. So, too, when the early universe was opaque, radiation interacted with ordinary matter, imparting energy to it and carrying it along, sweeping past the concentrations of dark matter. Now suppose the police leave the motorcade, which then encounters some red lights. The lights act as traffic traps; approaching cars now have to stop, and so they bunch up. Likewise, after the early universe became transparent, ordinary matter interacted with radiation only occasionally and so could fall into the dark matter traps. The Universe in a Nutshell In the previous sections of this chapter, we traced the evolution of the universe progressively further back in time. Astronomical discovery has followed this path historically, as new instruments and new techniques have allowed us to probe ever closer to the beginning of time. The rate of expansion of the universe was determined from measurements of nearby galaxies. Determinations of the abundances of deuterium, helium, and lithium based on nearby stars and galaxies were used to put limits on how much ordinary matter is in the universe. The motions of stars in galaxies and of galaxies within clusters of galaxies could only be explained if there were large quantities of dark matter. Measurements of supernovae that exploded when the universe was about half as old as it is now indicated that the rate of expansion of the universe has sped up since those explosions occurred. Observations of extremely faint galaxies show that galaxies had begun to form when the universe was only 400\u2013500 million years old. And observations of the CMB confirmed early theories that the universe was initially very hot", ". But all this moving further and further backward in time might have left you a bit dizzy. So now let\u2019s instead show how the universe evolves as time moves forward. Figure 29.24 summarizes the entire history of the observable universe from the beginning in a single diagram. The universe was very hot when it began to expand. We have fossil remnants of the very early universe in the form of neutrons, protons, electrons, and neutrinos, and the atomic nuclei that formed when the universe was 3\u20134 minutes old: deuterium, helium, and a small amount of lithium. Dark matter also remains, but we do not yet know what form it is in. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1079 Figure 29.24. History of the Universe. This image summarizes the changes that have occurred in the universe during the last 13.8 billion years. Protons, deuterium, helium, and some lithium were produced in the initial fireball. About 380,000 years after the Big Bang, the universe became transparent to electromagnetic radiation for the first time. COBE, WMAP, Planck, and other instruments have been used to study the radiation that was emitted at that time and that is still visible today (the CMB). The universe was then dark (except for this background radiation) until the first stars and galaxies began to form only a few hundred million years after the Big Bang. Existing space and ground-based telescopes have made substantial progress in studying the subsequent evolution of galaxies. (credit: modification of work by NASA/WMAP Science Team) The universe gradually cooled; when it was about 380,000 years old, and at a temperature of about 3000 K, electrons combined with protons to form hydrogen atoms. At this point, as we saw, the universe became transparent to light, and astronomers have detected the CMB emitted at this time. The universe still contained no stars or galaxies, and so it entered what astronomers call \u201cthe dark ages\u201d (since stars were not lighting up the darkness). During the next several hundred million years, small fluctuations in the density of the dark matter grew, forming gravitational traps that concentrated the ordinary matter, which began to form galaxies about 400\u2013500 million years after the Big Bang. By the time the universe was about a billion years old, it had entered its own renaissance: it was again blazing with radiation, but this time from newly formed stars, star clusters, and small galaxies. Over the next several billion years, small galaxies merged to form the giants we see today. Clusters and superclusters of galaxies began to grow, and the universe eventually began to resemble what we see nearby. During the next 20 years, astronomers plan to build giant new telescopes both in space and on the ground to explore even further back in time. In 2018, the James Webb Space Telescope, a 6.5-meter telescope that is the successor to the Hubble Space Telescope, will be launched and assembled in space. The predictions are that with this powerful instrument (see Figure 29.1) we should be able to look back far enough to analyze in detail the formation of the first galaxies. 1080 Chapter 29 The Big Bang 29.6 THE INFLATIONARY UNIVERSE Learning Objectives By the end of this section, you will be able to: Describe two important properties of the universe that the simple Big Bang model cannot explain Explain why these two characteristics of the universe can be accounted for if there was a period of rapid expansion (inflation) of the universe just after the Big Bang Name the four forces that control all physical processes in the universe The hot Big Bang model that we have been describing is remarkably successful. It accounts for the expansion of the universe, explains the observations of the CMB, and correctly predicts the abundances of the light elements. As it turns out, this model also predicts that there should be exactly three types of neutrinos in nature, and this prediction has been confirmed by experiments with high-energy accelerators. We can\u2019t relax just yet, however. This standard model of the universe doesn\u2019t explain all the observations we have made about the universe as a whole. Problems with the Standard Big Bang Model There are a number of characteristics of the universe that can only be explained by considering further what might have happened before the emission of the CMB. One problem with the standard Big Bang model is that it does not explain why the density of the universe is equal to the critical density. The mass density could have been, after all, so low and the effects of dark energy so high that the expansion would have been too rapid to form any galaxies at all. Alternatively, there could have been so much matter that the universe would have already begun to contract long before now. Why is the universe balanced so precisely on the knife edge of the critical density? Another puzzle is the remarkable uniformity of the universe. The temperature of", " the CMB is the same to about 1 part in 100,000 everywhere we look. This sameness might be expected if all the parts of the visible universe were in contact at some point in time and had the time to come to the same temperature. In the same way, if we put some ice into a glass of lukewarm water and wait a while, the ice will melt and the water will cool down until they are the same temperature. However, if we accept the standard Big Bang model, all parts of the visible universe were not in contact at any time. The fastest that information can go from one point to another is the speed of light. There is a maximum distance that light can have traveled from any point since the time the universe began\u2014that\u2019s the distance light could have covered since then. This distance is called that point\u2019s horizon distance because anything farther away is \u201cbelow its horizon\u201d\u2014unable to make contact with it. One region of space separated by more than the horizon distance from another has been completely isolated from it through the entire history of the universe. If we measure the CMB in two opposite directions in the sky, we are observing regions that were significantly beyond each other\u2019s horizon distance at the time the CMB was emitted. We can see both regions, but they can never have seen each other. Why, then, are their temperatures so precisely the same? According to the standard Big Bang model, they have never been able to exchange information, and there is no reason they should have identical temperatures. (It\u2019s a little like seeing the clothes that all the students wear at two schools in different parts of the world become identical, without the students ever having been in contact.) The only explanation we could suggest was simply that the universe somehow started out being absolutely uniform (which is like saying all students were born liking the same clothes). Scientists are always uncomfortable when they must appeal to a special set of initial conditions to account for what they see. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1081 The Inflationary Hypothesis Some physicists suggested that these fundamental characteristics of the cosmos\u2014its flatness and uniformity\u2014can be explained if shortly after the Big Bang (and before the emission of the CMB), the universe experienced a sudden increase in size. A model universe in which this rapid, early expansion occurs is called an inflationary universe. The inflationary universe is identical to the Big Bang universe for all time after the first 10\u201330 second. Prior to that, the model suggests that there was a brief period of extraordinarily rapid expansion or inflation, during which the scale of the universe increased by a factor of about 1050 times more than predicted by standard Big Bang models (Figure 29.25). Figure 29.25. Expansion of the Universe. This graph shows how the scale factor of the observable universe changes with time for the standard Big Bang model (red line) and for the inflationary model (blue line). (Note that the time scale at the bottom is extremely compressed.) During inflation, regions that were very small and in contact with each other are suddenly blown up to be much larger and outside each other\u2019s horizon distance. The two models are the same for all times after 10\u201330 second. Prior to (and during) inflation, all the parts of the universe that we can now see were so small and close to each other that they could exchange information, that is, the horizon distance included all of the universe that we can now observe. Before (and during) inflation, there was adequate time for the observable universe to homogenize itself and come to the same temperature. Then, inflation expanded those regions tremendously, so that many parts of the universe are now beyond each other\u2019s horizon. Another appeal of the inflationary model is its prediction that the density of the universe should be exactly equal to the critical density. To see why this is so, remember that curvature of spacetime is intimately linked to the density of matter. If the universe began with some curvature of its spacetime, one analogy for it might be the skin of a balloon. The period of inflation was equivalent to blowing up the balloon to a tremendous size. The universe became so big that from our vantage point, no curvature should be visible (Figure 29.26). In the same way, Earth\u2019s surface is so big that it looks flat to us no matter where we are. Calculations show that a universe with no curvature is one that is at critical density. Universes with densities either higher or lower than the critical density would show marked curvature. But we saw that the observations of the CMB in Figure 29.18, which show that the universe has critical density, rule out the possibility that space is significantly curved. 1082 Chapter 29 The Big Bang Figure 29.26. Analogy for Inflation. During a period of rapid inflation,", " a curved balloon grows so large that to any local observer it looks flat. The inset shows the geometry from the ant\u2019s point of view. Grand Unified Theories While inflation is an intriguing idea and widely accepted by researchers, we cannot directly observe events so early in the universe. The conditions at the time of inflation were so extreme that we cannot reproduce them in our laboratories or high-energy accelerators, but scientists have some ideas about what the universe might have been like. These ideas are called grand unified theories or GUTs. In GUT models, the forces that we are familiar with here on Earth, including gravity and electromagnetism, behaved very differently in the extreme conditions of the early universe than they do today. In physical science, the term force is used to describe anything that can change the motion of a particle or body. One of the remarkable discoveries of modern science is that all known physical processes can be described through the action of just four forces: gravity, electromagnetism, the strong nuclear force, and the weak nuclear force (Table 29.3). The Forces of Nature Force Relative Strength Today Range of Action Important Applications Gravity 1 Electromagnetism 1036 Weak nuclear force Strong nuclear force 1033 1038 Table 29.3 Whole universe Whole universe Motions of planets, stars, galaxies Atoms, molecules, electricity, magnetic fields 10\u201317 meters Radioactive decay 10\u201315 meters The existence of atomic nuclei Gravity is perhaps the most familiar force, and certainly appears strong if you jump off a tall building. However, the force of gravity between two elementary particles\u2014say two protons\u2014is by far the weakest of the four This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1083 forces. Electromagnetism\u2014which includes both magnetic and electrical forces, holds atoms together, and produces the electromagnetic radiation that we use to study the universe\u2014is much stronger, as you can see in Table 29.3. The weak nuclear force is only weak in comparison to its strong \u201ccousin,\u201d but it is in fact much stronger than gravity. Both the weak and strong nuclear forces differ from the first two forces in that they act only over very small distances\u2014those comparable to the size of an atomic nucleus or less. The weak force is involved in radioactive decay and in reactions that result in the production of neutrinos. The strong force holds protons and neutrons together in an atomic nucleus. Physicists have wondered why there are four forces in the universe\u2014why not 300 or, preferably, just one? An important hint comes from the name electromagnetic force. For a long time, scientists thought that the forces of electricity and magnetism were separate, but James Clerk Maxwell (see the chapter on Radiation and Spectra) was able to unify these forces\u2014to show that they are aspects of the same phenomenon. In the same way, many scientists (including Einstein) have wondered if the four forces we now know could also be unified. Physicists have actually developed GUTs that unify three of the four forces (but not gravity). In these theories, the strong, weak, and electromagnetic forces are not three independent forces but instead are different manifestations or aspects of what is, in fact, a single force. The theories predict that at high enough temperatures, there would be only one force. At lower temperatures (like the ones in the universe today), however, this single force has changed into three different forces (Figure 29.27). Just as different gases or liquids freeze at different temperatures, we can say that the different forces \u201cfroze out\u201d of the unified force at different temperatures. Unfortunately, the temperatures at which the three forces acted as one force are so high that they cannot be reached in any laboratory on Earth. Only the early universe, at times prior to 10\u201335 second, was hot enough to unify these forces. Figure 29.27. Four Forces That Govern the Universe. The behavior of the four forces depends on the temperature of the universe. This diagram (inspired by some grand unified theories) shows that at very early times when the temperature of the universe was very high, all four forces resembled one another and were indistinguishable. As the universe cooled, the forces took on separate and distinctive characteristics. Many physicists think that gravity was also unified with the three other forces at still higher temperatures, and scientists have tried to develop a theory that combines all four forces. For example, in string theory, the point- like particles of matter that we have discussed in this book are replaced by one-dimensional objects called 1084 Chapter 29 The Big Bang strings. In this theory, infinitesimal strings, which have length but not height or width, are the building blocks used to construct all the forms of matter and energy in the universe. These strings exist in 11-dimensional space (not the 4-dimensional spacetime with which we are familiar). The strings", " vibrate in the various dimensions, and depending on how they vibrate, they are seen in our world as matter or gravity or light. As you can imagine, the mathematics of string theory is very complex, and the theory remains untested by experiments. Even the largest particle accelerators on Earth do not achieve high enough energy to show whether string theory applies to the real world. String theory is interesting to scientists because it is currently the only approach that seems to have the potential of combining all four forces to produce what physicists have termed the Theory of Everything.[3] Theories of the earliest phases of the universe must take both quantum mechanics and gravity into account, but at the simplest level, gravity and quantum mechanics are incompatible. General relativity, our best theory of gravity, says that the motions of objects can be predicted exactly. Quantum mechanics says you can only calculate the probability (chance) that an object will do something. String theory is an attempt to resolve this paradox. The mathematics that underpins string theory is elegant and beautiful, but it remains to be seen whether it will make predictions that can be tested by observations in yet-to-be-developed, high-energy accelerators on Earth or by observations of the early universe. The earliest period in the history of the universe from time zero to 10\u201343 second is called the Planck time. The universe was unimaginably hot and dense, and theorists believe that at this time, quantum effects of gravity dominated physical interactions\u2014and, as we have just discussed, we have no tested theory of quantum gravity. Inflation is hypothesized to have occurred somewhat later, when the universe was between perhaps 10\u201335 and 10\u201333 second old and the temperature was 1027 to 1028 K. This rapid expansion took place when three forces (electromagnetic, strong, and weak) are thought to have been unified, and this is when GUTs are applicable. After inflation, the universe continued to expand (but more slowly) and to cool. An important milestone was reached when the temperature was down to 1015 K and the universe was 10\u201310 second old. Under these conditions, all four forces were separate and distinct. High-energy particle accelerators can achieve similar conditions, and so theories of the history of the universe from this point on have a sound basis in experiments. As yet, we have no direct evidence of what the conditions were during the inflationary epoch, and the ideas presented here are speculative. Researchers are trying to devise some experimental tests. For example, the quantum fluctuations in the very early universe would have caused variations in density and produced gravitational waves that may have left a detectable imprint on the CMB. Detection of such an imprint will require observations with equipment whose sensitivity is improved from what we have today. Ultimately, however, it may provide confirmation that we live in a universe that once experienced an epoch of rapid inflation. If you are typical of the students who read this book, you may have found this brief discussion of dark matter, inflation, and cosmology a bit frustrating. We have offered glimpses of theories and observations, but have raised more questions than we have answered. What is dark matter? What is dark energy? Inflation explains the observations of flatness and uniformity of the university, but did it actually happen? These ideas are at the forefront of modern science, where progress almost always leads to new puzzles, and much more work is needed before we can see clearly. Bear in mind that less than a century has passed since Hubble demonstrated the existence of other galaxies. The quest to understand just how the universe of galaxies came to be will keep astronomers busy for a long time to come. 3 This name became the title of a film about physicist Stephen Hawking in 2014. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1085 29.7 THE ANTHROPIC PRINCIPLE Learning Objectives By the end of this section, you will be able to: Name some properties of the universe that, if different, would have precluded the development of humans Despite our uncertainties, we must admit that the picture we have developed about the evolution of our universe is a remarkable one. With new telescopes, we have begun to collect enough observational evidence that we can describe how the universe evolved from a mere fraction of a second after the expansion began. Although this is an impressive achievement, there are still some characteristics of the universe that we cannot explain. And yet, it turns out that if these characteristics were any different, we would not be here to ask about them. Let\u2019s look at some of these \u201clucky accidents,\u201d beginning with the observations of the cosmic microwave background (CMB). Lucky Accidents As we described in this chapter, the CMB is radiation that was emitted when the universe was a few hundred thousand years old. Observations show that the temperature of the radiation varies from one region to another, typically by about 10 millionths of a degree, and these temperature differences", " signal small differences in density. But suppose the tiny, early fluctuations in density had been much smaller. Then calculations show that the pull of gravity near them would have been so small that no galaxies would ever have formed. What if the fluctuations in density had been much larger? Then it is possible that very dense regions would have condensed, and these would simply have collapsed directly to black holes without ever forming galaxies and stars. Even if galaxies had been able to form in such a universe, space would have been filled with intense X-rays and gamma rays, and it would have been difficult for life forms to develop and survive. The density of stars within galaxies would be so high that interactions and collisions among them would be frequent. In such a universe, any planetary systems could rarely survive long enough for life to develop. So for us to be here, the density fluctuations need to be \u201cjust right\u201d\u2014not too big and not too small. Another lucky accident is that the universe is finely balanced between expansion and contraction. It is expanding, but very slowly. If the expansion had been at a much higher rate, all of the matter would have thinned out before galaxies could form. If everything were expanding at a much slower rate, then gravity would have \u201cwon.\u201d The expansion would have reversed and all of the matter would have recollapsed, probably into a black hole\u2014again, no stars, no planets, no life. The development of life on Earth depends on still-luckier coincidences. Had matter and antimatter been present initially in exactly equal proportions, then all matter would have been annihilated and turned into pure energy. We owe our existence to the fact that there was slightly more matter than antimatter. (After most of the matter made contact with an equal amount of antimatter, turning into energy, a small amount of additional matter must have been present. We are all descendants of that bit of \u201cunbalanced\u201d matter.) If nuclear fusion reactions occurred at a somewhat faster rate than they actually do, then at the time of the initial fireball, all of the matter would have been converted from hydrogen into helium into carbon and all the way into iron (the most stable nucleus). That would mean that no stars would have formed, since the existence of stars depends on there being light elements that can undergo fusion in the main-sequence stage and make the stars shine. In addition, the structure of atomic nuclei had to be just right to make it possible for three 1086 Chapter 29 The Big Bang helium atoms to come together easily to fuse carbon, which is the basis of life. If the triple-alpha process we discussed in the chapter on Stars from Adolescence to Old Age were too unlikely, not enough carbon would have formed to lead to biology as we know it. At the same time, it had to be hard enough to fuse carbon into oxygen that a large amount of carbon survived for billions of years. There are additional factors that have contributed to life like us being possible. Neutrinos have to interact with matter at just the right, albeit infrequent, rate. Supernova explosions occur when neutrinos escape from the cores of collapsing stars, deposit some of their energy in the surrounding stellar envelope, and cause it to blow out and away into space. The heavy elements that are ejected in such explosions are essential ingredients of life here on Earth. If neutrinos did not interact with matter at all, they would escape from the cores of collapsing stars without causing the explosion. If neutrinos interacted strongly with matter, they would remain trapped in the stellar core. In either case, the heavy elements would remain locked up inside the collapsing star. If gravity were a much stronger force than it is, stars could form with much smaller masses, and their lifetimes would be measured in years rather than billions of years. Chemical processes, on the other hand, would not be sped up if gravity were a stronger force, and so there would be no time for life to develop while stars were so short-lived. Even if life did develop in a stronger-gravity universe, life forms would have to be tiny or they could not stand up or move around. What Had to Be, Had to Be In summary, we see that a specific set of rules and conditions in the universe has allowed complexity and life on Earth to develop. As yet, we have no theory to explain why this \u201cright\u201d set of conditions occurred. For this reason, many scientists are beginning to accept an idea we call the anthropic principle\u2014namely, that the physical laws we observe must be what they are precisely because these are the only laws that allow for the existence of humans. Some scientists speculate that our universe is but one of countless universes, each with a different set of physical laws\u2014an idea that is sometimes referred to as the multiverse. Some of those universes might be stillborn, collapsing before any structure forms. Others may expand so quickly that they remain essentially featureless with no stars and galaxies. In other words, there may", " be a much larger multiverse that contains our own universe and many others. This multiverse (existing perhaps in more dimensions that we can become aware of) is infinite and eternal; it generates many, many inflating regions, each of which evolves into a separate universe, which may be completely unlike any of the other separate universes. Our universe is then the way it is because it is the only way it could be and have humans like ourselves in it to discover its properties and ask such questions View the 2011 introductory talk (https://openstax.org/l/30mulcosinfla) on the Multiverse and Cosmic Inflation by Dr. Anthony Aguirre of the University of California, part of the Silicon Valley Astronomy Lecture Series. It is difficult to know how to test these ideas since we can never make contact with any other universe. For most scientists, our discussion in this section borders on the philosophical and metaphysical. Perhaps in the future our understanding of physics will develop to the point that we can know why the gravitational constant is as strong as it is, why the universe is expanding at exactly the rate it is, and why all of the other \u201clucky accidents\u201d This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1087 happened\u2014why they were inevitable and could be no other way. Then this anthropic idea would no longer be necessary. No one knows, however, whether we will ever have an explanation for why this universe works the way it does. We have come a long way in our voyage through the universe. We have learned a remarkable amount about how and when the cosmos came to be, but the question of why the universe is the way it is remains as elusive as ever. 1088 Chapter 29 The Big Bang CHAPTER 29 REVIEW KEY TERMS anthropic principle idea that physical laws must be the way they are because otherwise we could not be here to measure them Big Bang the theory of cosmology in which the expansion of the universe began with a primeval explosion (of space, time, matter, and energy) closed universe a model in which the universe expands from a Big Bang, stops, and then contracts to a big crunch cosmic microwave background (CMB) microwave radiation coming from all directions that is the redshifted afterglow of the Big Bang cosmological constant the term in the equations of general relativity that represents a repulsive force in the universe cosmology the study of the organization and evolution of the universe critical density in cosmology, the density that is just sufficient to bring the expansion of the universe to a stop after infinite time dark energy the energy that is causing the expansion of the universe to accelerate; its existence is inferred from observations of distant supernovae dark matter nonluminous material, whose nature we don\u2019t yet understand, but whose presence can be inferred because of its gravitational influence on luminous matter deuterium a form of hydrogen in which the nucleus of each atom consists of one proton and one neutron flat universe a model of the universe that has a critical density and in which the geometry of the universe is flat, like a sheet of paper fusion the building of heavier atomic nuclei from lighter ones grand unified theories (GUTs) physical theories that attempt to describe the four forces of nature as different manifestations of a single force inflationary universe a theory of cosmology in which the universe is assumed to have undergone a phase of very rapid expansion when the universe was about 10\u201335 second old; after this period of rapid expansion, the standard Big Bang and inflationary models are identical lithium the third element in the periodic table; lithium nuclei with three protons and four neutrons were manufactured during the first few minutes of the expansion of the universe multiverse the speculative idea that our universe is just one of many universes, each with its own set of physical laws open universe a model in which the density of the universe is not high enough to bring the expansion of the universe to a halt This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1089 photon decoupling time when radiation began to stream freely through the universe without interacting with matter weakly interacting massive particles (WIMPs) weakly interacting massive particles are one of the candidates for the composition of dark matter SUMMARY 29.1 The Age of the Universe Cosmology is the study of the organization and evolution of the universe. The universe is expanding, and this is one of the key observational starting points for modern cosmological theories. Modern observations show that the rate of expansion has not been constant throughout the life of the universe. Initially, when galaxies were close together, the effects of gravity were stronger than the effects of dark energy, and the expansion rate gradually slowed. As galaxies moved farther apart, the influence of gravity on the expansion rate weakened. Measurements of distant supernovae show that when the", " universe was about half its current age, dark energy began to dominate the rate of expansion and caused it to speed up. In order to estimate the age of the universe, we must allow for changes in the rate of expansion. After allowing for these effects, astronomers estimate that all of the matter within the observable universe was concentrated in an extremely small volume 13.8 billion years ago, a time we call the Big Bang. 29.2 A Model of the Universe For describing the large-scale properties of the universe, a model that is isotropic and homogeneous (same everywhere) is a pretty good approximation of reality. The universe is expanding, which means that the universe undergoes a change in scale with time; space stretches and distances grow larger by the same factor everywhere at a given time. Observations show that the mass density of the universe is less than the critical density. In other words, there is not enough matter in the universe to stop the expansion. With the discovery of dark energy, which is accelerating the rate of expansion, the observational evidence is strong that the universe will expand forever. Observations tell us that the expansion started about 13.8 billion years ago. 29.3 The Beginning of the Universe Lema\u00eetre, Alpher, and Gamow first worked out the ideas that are today called the Big Bang theory. The universe cools as it expands. The energy of photons is determined by their temperature, and calculations show that in the hot, early universe, photons had so much energy that when they collided with one another, they could produce material particles. As the universe expanded and cooled, protons and neutrons formed first, then came electrons and positrons. Next, fusion reactions produced deuterium, helium, and lithium nuclei. Measurements of the deuterium abundance in today\u2019s universe show that the total amount of ordinary matter in the universe is only about 5% of the critical density. 29.4 The Cosmic Microwave Background When the universe became cool enough to form neutral hydrogen atoms, the universe became transparent to radiation. Scientists have detected the cosmic microwave background (CMB) radiation from this time during the hot, early universe. Measurements with the COBE satellite show that the CMB acts like a blackbody with a temperature of 2.73 K. Tiny fluctuations in the CMB show us the seeds of large-scale structures in the universe. Detailed measurements of these fluctuations show that we live in a critical-density universe and that the critical density is composed of 31% matter, including dark matter, and 69% dark energy. Ordinary matter\u2014the kinds of elementary particles we find on Earth\u2014make up only about 5% of the critical density. CMB measurements also indicate that the universe is 13.8 billion years old. 1090 Chapter 29 The Big Bang 29.5 What Is the Universe Really Made Of? Twenty-seven percent of the critical density of the universe is composed of dark matter. To explain so much dark matter, some physics theories predict that additional types of particles should exist. One type has been given the name of WIMPs (weakly interacting massive particles), and scientists are now conducting experiments to try to detect them in the laboratory. Dark matter plays an essential role in forming galaxies. Since, by definition, these particles interact only very weakly (if at all) with radiation, they could have congregated while the universe was still very hot and filled with radiation. They would thus have formed gravitational traps that quickly attracted and concentrated ordinary matter after the universe became transparent, and matter and radiation decoupled. This rapid concentration of matter enabled galaxies to form by the time the universe was only 400\u2013500 million years old. 29.6 The Inflationary Universe The Big Bang model does not explain why the CMB has the same temperature in all directions. Neither does it explain why the density of the universe is so close to critical density. These observations can be explained if the universe experienced a period of rapid expansion, which scientists call inflation, about 10\u201335 second after the Big Bang. New grand unified theories (GUTs) are being developed to describe physical processes in the universe before and at the time that inflation occurred. 29.7 The Anthropic Principle Recently, many cosmologists have noted that the existence of humans depends on the fact that many properties of the universe\u2014the size of density fluctuations in the early universe, the strength of gravity, the structure of atoms\u2014were just right. The idea that physical laws must be the way they are because otherwise we could not be here to measure them is called the anthropic principle. Some scientists speculate that there may be a multiverse of universes, in which ours is just one. FOR FURTHER EXPLORATION Articles Kruesi, L. \u201cCosmology: 5 Things You Need to Know.\u201d Astronomy (May 2007): 28. Five questions students often ask, and how modern cosmologists answer them. Kruesi, L. \u201cHow Planck Has Redefined the Universe", ".\u201d Astronomy (October 2013): 28. Good review of what this space mission has told us about the CMB and the universe. Lineweaver, C. & Davis, T. \u201cMisconceptions about the Big Bang.\u201d Scientific American (March 2005): 36. Some basic ideas about modern cosmology clarified, using general relativity. Nadis, S. \u201cSizing Up Inflation.\u201d Sky & Telescope (November 2005): 32. Nice review of the origin and modern variants on the inflationary idea. Nadis, S. \u201cHow We Could See Another Universe.\u201d Astronomy (June 2009): 24. On modern ideas about multiverses and how such bubbles of space-time might collide. Nadis, S. \u201cDark Energy\u2019s New Face: How Exploding Stars Are Changing our View.\u201d Astronomy (July 2012): 45. About our improving understanding of the complexities of type Ia supernovae. Naze, Y. \u201cThe Priest, the Universe, and the Big Bang.\u201d Astronomy (November 2007): 40. On the life and work of Georges Lema\u00eetre. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1091 Panek, R. \u201cGoing Over to the Dark Side.\u201d Sky & Telescope (February 2009): 22. A history of the observations and theories about dark energy. Pendrick, D. \u201cIs the Big Bang in Trouble?\u201d Astronomy (April 2009): 48. This sensationally titled article is really more of a quick review of how modern ideas and observations are fleshing out the Big Bang hypothesis (and raising questions.) Reddy, F. \u201cHow the Universe Will End.\u201d Astronomy (September 2014): 38. Brief discussion of local and general future scenarios. Riess, A. and Turner, M. \u201cThe Expanding Universe: From Slowdown to Speedup.\u201d Scientific American (September 2008): 62. Turner, M. \u201cThe Origin of the Universe.\u201d Scientific American (September 2009): 36. An introduction to modern cosmology. Websites Cosmology Primer: https://preposterousuniverse.com/cosmologyprimer/. Caltech Astrophysicist Sean Carroll offers a non-technical site with brief overviews of many key topics in modern cosmology. Everyday Cosmology: http://cosmology.carnegiescience.edu/. An educational website from the Carnegie Observatories with a timeline of cosmological discovery, background materials, and activities. How Big Is the Universe?: http://www.pbs.org/wgbh/nova/space/how-big-universe.html. A clear essay by a noted astronomer Brent Tully summarizes some key ideas in cosmology and introduces the notion of the acceleration of the universe. Universe 101: WMAP Mission Introduction to the Universe: http://map.gsfc.nasa.gov/universe/. Concise NASA primer on cosmological ideas from the WMAP mission team. Cosmic Times Project: http://cosmictimes.gsfc.nasa.gov/. James Lochner and Barbara Mattson have compiled a rich resource of twentieth-century cosmology history in the form of news reports on key events, from NASA\u2019s Goddard Space Flight Center. Videos The Day We Found the Universe: http://www.cfa.harvard.edu/events/mon_video_archive09.html. Distinguished science writer Marcia Bartusiak discusses Hubble\u2019s work and the discovery of the expansion of the cosmos\u2014one of the Observatory Night lectures at the Harvard-Smithsonian Center for Astrophysics (53:46). Images of the Infant Universe: https://www.youtube.com/watch?v=x0AqCwElyUk. Lloyd Knox\u2019s public talk on the latest discoveries about the CMB and what they mean for cosmology (1:16:00). Runaway Universe: https://www.youtube.com/watch?v=kNYVFrnmcOU. Roger Blandford (Stanford Linear Accelerator Center) public lecture on the discovery and meaning of cosmic acceleration and dark energy (1:08:08). From the Big Bang to the Nobel Prize and on to the James Webb Space Telescope and the Discovery of Alien Life: http://svs.gsfc.nasa.gov/vis/a010000/a010300/a010370/index.html. John Mather, NASA Goddard (1:01:02). His Nobel Prize talk from Dec. 8, 2006 can be found at http://www.nobelprize.org/mediaplayer/ index.php?id=74&view=1. Dark Energy and the Fate of the Universe: https", "://webcast.stsci.edu/webcast/ detail.xhtml?talkid=1961&parent=1. Adam Reiss (STScI), at the Space Telescope Science Institute (1:00:00). 1092 Chapter 29 The Big Bang COLLABORATIVE GROUP ACTIVITIES A. This chapter deals with some pretty big questions and ideas. Some belief systems teach us that there are questions to which \u201cwe were not meant to know\u201d the answers. Other people feel that if our minds and instruments are capable of exploring a question, then it becomes part of our birthright as thinking human beings. Have your group discuss your personal reactions to discussing questions like the beginning of time and space, and the ultimate fate of the universe. Does it make you nervous to hear about scientists discussing these issues? Or is it exciting to know that we can now gather scientific evidence about the origin and fate of the cosmos? (In discussing this, you may find that members of your group strongly disagree; try to be respectful of others\u2019 points of view.) B. A popular model of the universe in the 1950s and 1960s was the so-called steady-state cosmology. In this model, the universe was not only the same everywhere and in all directions (homogeneous and isotropic), but also the same at all times. We know the universe is expanding and the galaxies are thinning out, and so this model hypothesized that new matter was continually coming into existence to fill in the space between galaxies as they moved farther apart. If so, the infinite universe did not have to have a sudden beginning, but could simply exist forever in a steady state. Have your group discuss your reaction to this model. Do you find it more appealing philosophically than the Big Bang model? Can you cite some evidence that indicates that the universe was not the same billions of years ago as it is now\u2014that it is not in a steady state? C. One of the lucky accidents that characterizes our universe is the fact that the time scale for the development of intelligent life on Earth and the lifetime of the Sun are comparable. Have your group discuss what would happen if the two time scales were very different. Suppose, for example, that the time for intelligent life to evolve was 10 times greater than the main-sequence lifetime of the Sun. Would our civilization have ever developed? Now suppose the time for intelligent life to evolve is ten times shorter than the main-sequence lifetime of the Sun. Would we be around? (This latter discussion requires considerable thought, including such ideas as what the early stages in the Sun\u2019s life were like and how much the early Earth was bombarded by asteroids and comets.) D. The grand ideas discussed in this chapter have a powerful effect on the human imagination, not just for scientists, but also for artists, composers, dramatists, and writers. Here we list just a few of these responses to cosmology. Each member of your group can select one of these, learn more about it, and then report back, either to the group or to the whole class. \u25e6 The California poet Robinson Jeffers was the brother of an astronomer who worked at the Lick Observatory. His poem \u201cMargrave\u201d is a meditation on cosmology and on the kidnap and murder of a child: http://www.poemhunter.com/best-poems/robinson-jeffers/margrave/. \u25e6 In the science fiction story \u201cThe Gravity Mine\u201d by Stephen Baxter, the energy of evaporating supermassive black holes is the last hope of living beings in the far future in an ever-expanding universe. The story has poetic description of the ultimate fate of matter and life and is available online at: http://www.infinityplus.co.uk/stories/gravitymine.htm. \u25e6 The musical piece YLEM by Karlheinz Stockhausen takes its title from the ancient Greek term for primeval material revived by George Gamow. It tries to portray the oscillating universe in musical terms. Players actually expand through the concert hall, just as the universe does, and then return and expand again. See: http://www.karlheinzstockhausen.org/ylem_english.htm. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1093 \u25e6 The musical piece Supernova Sonata http://www.astro.uvic.ca/~alexhp/new/supernova_sonata.html by Alex Parker and Melissa Graham is based on the characteristics of 241 type Ia supernova explosions, the ones that have helped astronomers discover the acceleration of the expanding universe. \u25e6 Gregory Benford\u2019s short story \u201cThe Final Now\u201d envisions the end of an accelerating open universe, and blends religious and scientific imagery in a very poetic way. Available free", " online at: http://www.tor.com/stories/2010/03/the-final-now. E. When Einstein learned about Hubble\u2019s work showing that the universe of galaxies is expanding, he called his introduction of the cosmological constant into his general theory of relativity his \u201cbiggest blunder.\u201d Can your group think of other \u201cbig blunders\u201d from the history of astronomy, where the thinking of astronomers was too conservative and the universe turned out to be more complicated or required more \u201coutside-the- box\u201d thinking? EXERCISES Review Questions 1. What are the basic observations about the universe that any theory of cosmology must explain? 2. Describe some possible futures for the universe that scientists have come up with. What property of the universe determines which of these possibilities is the correct one? 3. What does the term Hubble time mean in cosmology, and what is the current best calculation for the Hubble time? 4. Which formed first: hydrogen nuclei or hydrogen atoms? Explain the sequence of events that led to each. 5. Describe at least two characteristics of the universe that are explained by the standard Big Bang model. 6. Describe two properties of the universe that are not explained by the standard Big Bang model (without inflation). How does inflation explain these two properties? 7. Why do astronomers believe there must be dark matter that is not in the form of atoms with protons and neutrons? 8. What is dark energy and what evidence do astronomers have that it is an important component of the universe? 9. Thinking about the ideas of space and time in Einstein\u2019s general theory of relativity, how do we explain the fact that all galaxies outside our Local Group show a redshift? 10. Astronomers have found that there is more helium in the universe than stars could have made in the 13.8 billion years that the universe has been in existence. How does the Big Bang scenario solve this problem? 11. Describe the anthropic principle. What are some properties of the universe that make it \u201cready\u201d to have life forms like you in it? 12. Describe the evidence that the expansion of the universe is accelerating. Thought Questions 13. What is the most useful probe of the early evolution of the universe: a giant elliptical galaxy or an irregular galaxy such as the Large Magellanic Cloud? Why? 1094 Chapter 29 The Big Bang 14. What are the advantages and disadvantages of using quasars to probe the early history of the universe? 15. Would acceleration of the universe occur if it were composed entirely of matter (that is, if there were no dark energy)? 16. Suppose the universe expands forever. Describe what will become of the radiation from the primeval fireball. What will the future evolution of galaxies be like? Could life as we know it survive forever in such a universe? Why? 17. Some theorists expected that observations would show that the density of matter in the universe is just equal to the critical density. Do the current observations support this hypothesis? 18. There are a variety of ways of estimating the ages of various objects in the universe. Describe two of these ways, and indicate how well they agree with one another and with the age of the universe itself as estimated by its expansion. 19. Since the time of Copernicus, each revolution in astronomy has moved humans farther from the center of the universe. Now it appears that we may not even be made of the most common form of matter. Trace the changes in scientific thought about the central nature of Earth, the Sun, and our Galaxy on a cosmic scale. Explain how the notion that most of the universe is made of dark matter continues this \u201cCopernican tradition.\u201d 20. The anthropic principle suggests that in some sense we are observing a special kind of universe; if the universe were different, we could never have come to exist. Comment on how this fits with the Copernican tradition described in Exercise 29.19. 21. Penzias and Wilson\u2019s discovery of the Cosmic Microwave Background (CMB) is a nice example of scientific serendipity\u2014something that is found by chance but turns out to have a positive outcome. What were they looking for and what did they discover? 22. Construct a timeline for the universe and indicate when various significant events occurred, from the beginning of the expansion to the formation of the Sun to the appearance of humans on Earth. Figuring For Yourself 23. Suppose the Hubble constant were not 22 but 33 km/s per million light-years. Then what would the critical density be? 24. Assume that the average galaxy contains 1011 MSun and that the average distance between galaxies is 10 million light-years. Calculate the average density of matter (mass per unit volume) in galaxies. What fraction is this of the critical density we calculated in the chapter? 25. The CMB contains roughly 400 million photons per m3. The energy", " of each photon depends on its wavelength. Calculate the typical wavelength of a CMB photon. Hint: The CMB is blackbody radiation at a temperature of 2.73 K. According to Wien\u2019s law, the peak wave length in nanometers is given by \u03bbmax = 3 \u00d7 106 consistent, convert this wavelength from nanometers to meters.. Calculate the wavelength at which the CMB is a maximum and, to make the units T 26. Following up on Exercise 29.27 calculate the energy of a typical photon. Assume for this approximate calculation that each photon has the wavelength calculated in Exercise 29.25. The energy of a photon is, where h is Planck\u2019s constant and is equal to 6.626 \u00d7 10\u201334 J \u00d7 s, c is the speed of light in given by E = hc \u03bb m/s, and \u03bb is the wavelength in m. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 29 The Big Bang 1095 27. Continuing the thinking in Exercise 29.27 and Exercise 29.28, calculate the energy in a cubic meter of space, multiply the energy per photon calculated in Exercise 29.26 by the number of photons per cubic meter given above. 28. Continuing the thinking in the last three exercises, convert this energy to an equivalent in mass, use Einstein\u2019s equation E = mc2. Hint: Divide the energy per m3 calculated in Exercise 29.27 by the speed of light squared. Check your units; you should have an answer in kg/m3. Now compare this answer with the critical density. Your answer should be several powers of 10 smaller than the critical density. In other words, you have found for yourself that the contribution of the CMB photons to the overall density of the universe is much, much smaller than the contribution made by stars and galaxies. 29. There is still some uncertainty in the Hubble constant. (a) Current estimates range from about 19.9 km/ s per million light-years to 23 km/s per million light-years. Assume that the Hubble constant has been constant since the Big Bang. What is the possible range in the ages of the universe? Use the equation in the text, T0 = 1 H, and make sure you use consistent units. (b) Twenty years ago, estimates for the Hubble constant ranged from 50 to 100 km/s per Mps. What are the possible ages for the universe from those values? Can you rule out some of these possibilities on the basis of other evidence? 30. It is possible to derive the age of the universe given the value of the Hubble constant and the distance to a galaxy, again with the assumption that the value of the Hubble constant has not changed since the Big Bang. Consider a galaxy at a distance of 400 million light-years receding from us at a velocity, v. If the Hubble constant is 20 km/s per million light-years, what is its velocity? How long ago was that galaxy right next door to our own Galaxy if it has always been receding at its present rate? Express your answer in years. Since the universe began when all galaxies were very close together, this number is a rough estimate for the age of the universe. 1096 Chapter 29 The Big Bang This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1097 30 LIFE IN THE UNIVERSE Figure 30.1. Astrobiology: The Road to Life in the Universe. In this fanciful montage produced by a NASA artist, we see one roadmap for discovering life in the universe. Learning more about the origin, evolution, and properties of life on Earth aids us in searching for evidence of life beyond our planet. Our neighbor world, Mars, had warmer, wetter conditions billions of years ago that might have helped life there begin. Farther out, Jupiter\u2019s moon Europa represents the icy moons of the outer solar system. Beneath their shells of solid ice may lie vast oceans of liquid water that could support biology. Beyond our solar system are stars that host their own planets, some of which might be similar to Earth in the ability to support liquid water\u2014and a thriving biosphere\u2014at the planet\u2019s surface. Research is pushing actively in all these directions with the goal of proving a scientific answer to the question, \u201cAre we alone?\u201d (credit: modification of work by NASA) Chapter Outline 30.1 The Cosmic Context for Life 30.2 Astrobiology 30.3 Searching for Life beyond Earth 30.4 The Search for Extraterrestrial Intelligence Thinking Ahead As we have learned more about the universe, we have naturally wondered whether there might be other forms of life out there. The ancient question, \u201cAre we alone in the universe?\u201d connects us to generations of humans before us", ". While in the past, this question was in the realm of philosophy or science fiction, today we have the means to seek an answer through scientific inquiry. In this chapter, we will consider how life began on Earth, whether the same processes could have led to life on other worlds, and how we might seek evidence of life elsewhere. This is the science of astrobiology. The search for life on other planets is not the same as the search for intelligent life, which (if it exists) is surely much rarer. Learning more about the origin, evolution, and properties of life on Earth aids us in searching for evidence of all kinds of life beyond that on our planet. 1098 Chapter 30 Life in the Universe 30.1 THE COSMIC CONTEXT FOR LIFE Learning Objectives By the end of this section, you will be able to: Describe the chemical and environmental conditions that make Earth hospitable to life Discuss the assumption underlying the Copernican principle and outline its implications for modern-day astronomers Understand the questions underlying the Fermi paradox We saw that the universe was born in the Big Bang about 14 billion years ago. After the initial hot, dense fireball of creation cooled sufficiently for atoms to exist, all matter consisted of hydrogen and helium (with a very small amount of lithium). As the universe aged, processes within stars created the other elements, including those that make up Earth (such as iron, silicon, magnesium, and oxygen) and those required for life as we know it, such as carbon, oxygen, and nitrogen. These and other elements combined in space to produce a wide variety of compounds that form the basis of life on Earth. In particular, life on Earth is based on the presence of a key unit known as an organic molecule, a molecule that contains carbon. Especially important are the hydrocarbons, chemical compounds made up entirely of hydrogen and carbon, which serve as the basis for our biological chemistry, or biochemistry. While we do not understand the details of how life on Earth began, it is clear that to make creatures like us possible, events like the ones we described must have occurred, resulting in what is called the chemical evolution of the universe. What Made Earth Hospitable to Life? About 5 billion years ago, a cloud of gas and dust in this cosmic neighborhood began to collapse under its own weight. Out of this cloud formed the Sun and its planets, together with all the smaller bodies, such as comets, that also orbit the Sun (Figure 30.2). The third planet from the Sun, as it cooled, eventually allowed the formation of large quantities of liquid water on its surface. Figure 30.2. Comet Hyakutake. This image was captured in 1996 by NASA photographer Bill Ingalls. Comet impacts can deliver both water and a variety of interesting chemicals, including some organic chemicals, to Earth. (credit: NASA/Bill Ingalls) The chemical variety and moderate conditions on Earth eventually led to the formation of molecules that could make copies of themselves (reproduce), which is essential for beginning life. Over the billions of years of Earth history, life evolved and became more complex. The course of evolution was punctuated by occasional planet- wide changes caused by collisions with some of the smaller bodies that did not make it into the Sun or one of its This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1099 accompanying worlds. As we saw in the chapter on Earth as a Planet, mammals may owe their domination of Earth\u2019s surface to just such a collision 65 million years ago, which led to the extinction of the dinosaurs (along with the majority of other living things). The details of such mass extinctions are currently the focus of a great deal of scientific interest. Through many twisting turns, the course of evolution on Earth produced a creature with self-consciousness, able to ask questions about its own origins and place in the cosmos (Figure 30.3). Like most of Earth, this creature is composed of atoms that were forged in earlier generations of stars\u2014in this case, assembled into both its body and brain. We might say that through the thoughts of human beings, the matter in the universe can become aware of itself. Figure 30.3. Young Human. Human beings have the intellect to wonder about their planet and what lies beyond it. Through them (and perhaps other intelligent life), the universe becomes aware of itself. (credit: Andrew Fraknoi) Think about those atoms in your body for a minute. They are merely on loan to you from the lending library of atoms that make up our local corner of the universe. Atoms of many kinds circulate through your body and then leave it\u2014with each breath you inhale and exhale and the food you eat and excrete. Even the atoms that take up more permanent residence in your tissues will not be part of you much longer than you are alive. Ultimately, you will return your atoms to the vast reservoir of Earth", ", where they will be incorporated into other structures and even other living things in the millennia to come. This picture of cosmic evolution, of our descent from the stars, has been obtained through the efforts of scientists in many fields over many decades. Some of its details are still tentative and incomplete, but we feel reasonably confident in its broad outlines. It is remarkable how much we have been able to learn in the short time we have had the instruments to probe the physical nature of the universe. The Copernican Principle Our study of astronomy has taught us that we have always been wrong in the past whenever we have claimed that Earth is somehow unique. Galileo, using the newly invented technology of the telescope, showed us that Earth is not the center of the solar system, but merely one of a number of objects orbiting the Sun. Our study of the stars has demonstrated that the Sun itself is a rather undistinguished star, halfway through its long main- sequence stage like so many billions of others. There seems nothing special about our position in the Milky Way Galaxy either, and nothing surprising about our Galaxy\u2019s position in either its own group or its supercluster. The discovery of planets around other stars confirms our idea that the formation of planets is a natural consequence of the formation of stars. We have identified thousands of exoplanets\u2014planets orbiting around 1100 Chapter 30 Life in the Universe other stars, from huge ones orbiting close to their stars (informally called \u201chot Jupiters\u201d) down to planets smaller than Earth. A steady stream of exoplanet discoveries is leading to the conclusion that earthlike planets occur frequently\u2014enough that there are likely many billions of \u201cexo-Earths\u201d in our own Milky Way Galaxy alone. From a planetary perspective, smaller planets are not unique. Philosophers of science sometimes call the idea that there is nothing special about our place in the universe the Copernican principle. Given all of the above, most scientists would be surprised if life were limited to our planet and had started nowhere else. There are billions of stars in our Galaxy old enough for life to have developed on a planet around them, and there are billions of other galaxies as well. Astronomers and biologists have long conjectured that a series of events similar to those on the early Earth probably led to living organisms on many planets around other stars, and possibly even on other planets in our solar system, such as Mars. The real scientific issue (which we do not currently know the answer to) is whether organic biochemistry is likely or unlikely in the universe at large. Are we a fortunate and exceedingly rare outcome of chemical evolution, or is organic biochemistry a regular part of the chemical evolution of the cosmos? We do not yet know the answer to this question, but data, even an exceedingly small amount (like finding \u201cunrelated to us\u201d living systems on a world like Europa), will help us arrive at it. So Where Are They? If the Copernican principle is applied to life, then biology may be rather common among planets. Taken to its logical limit, the Copernican principle also suggests that intelligent life like us might be common. Intelligence like ours has some very special properties, including an ability to make progress through the application of technology. Organic life around other (older) stars may have started a billion years earlier than we did on Earth, so they may have had a lot more time to develop advanced technology such as sending information, probes, or even life-forms between stars. Faced with such a prospect, physicist Enrico Fermi asked a question several decades ago that is now called the Fermi paradox: where are they? If life and intelligence are common and have such tremendous capacity for growth, why is there not a network of galactic civilizations whose presence extends even into a \u201clatecomer\u201d planetary system like ours? Several solutions have been suggested to the Fermi paradox. Perhaps life is common but intelligence (or at least technological civilization) is rare. Perhaps such a network will come about in the future but has not yet had the time to develop. Maybe there are invisible streams of data flowing past us all the time that we are not advanced enough or sensitive enough to detect. Maybe advanced species make it a practice not to interfere with immature, developing consciousness such as our own. Or perhaps civilizations that reach a certain level of technology then self-destruct, meaning there are no other civilizations now existing in our Galaxy. We do not yet know whether any advanced life is out there and, if it is, why we are not aware of it. Still, you might want to keep these issues in mind as you read the rest of this chapter Is there a network of galactic civilizations beyond our solar system? If so, why can\u2019t we see them? Explore the possibilities in the cartoon video (https://openstax.org/l/30fermparadox) \u201cThe Fermi Paradox\u2014Where Are All the Aliens?\u201d This OpenStax", " book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1101 30.2 ASTROBIOLOGY Learning Objectives By the end of this section, you will be able to: Describe the chemical building blocks required for life Describe the molecular systems and processes driving the origin and evolution of life Describe the characteristics of a habitable environment Describe some of the extreme conditions on Earth, and explain how certain organisms have adapted to these conditions Scientists today take a multidisciplinary approach to studying the origin, evolution, distribution, and ultimate fate of life in the universe; this field of study is known as astrobiology. You may also sometimes hear this field referred to as exobiology or bioastronomy. Astrobiology brings together astronomers, planetary scientists, chemists, geologists, and biologists (among others) to work on the same problems from their various perspectives. Among the issues that astrobiologists explore are the conditions in which life arose on Earth and the reasons for the extraordinary adaptability of life on our planet. They are also involved in identifying habitable worlds beyond Earth and in trying to understand in practical terms how to look for life on those worlds. Let\u2019s look at some of these issues in more detail. The Building Blocks of Life While no unambiguous evidence for life has yet been found anywhere beyond Earth, life\u2019s chemical building blocks have been detected in a wide range of extraterrestrial environments. Meteorites (which you learned about in Cosmic Samples and the Origin of the Solar System) have been found to contain two kinds of substances whose chemical structures mark them as having an extraterrestrial origin\u2014amino acids and sugars. Amino acids are organic compounds that are the molecular building blocks of proteins. Proteins are key biological molecules that provide the structure and function of the body\u2019s tissues and organs and essentially carry out the \u201cwork\u201d of the cell. When we examine the gas and dust around comets, we also find a number of organic molecules\u2014compounds that on Earth are associated with the chemistry of life. Expanding beyond our solar system, one of the most interesting results of modern radio astronomy has been the discovery of organic molecules in giant clouds of gas and dust between stars. More than 100 different molecules have been identified in these reservoirs of cosmic raw material, including formaldehyde, alcohol, and others we know as important stepping stones in the development of life on Earth. Using radio telescopes and radio spectrometers, astronomers can measure the abundances of various chemicals in these clouds. We find organic molecules most readily in regions where the interstellar dust is most abundant, and it turns out these are precisely the regions where star formation (and probably planet formation) happen most easily (Figure 30.4). 1102 Chapter 30 Life in the Universe Figure 30.4. Cloud of Gas and Dust. This cloud of gas and dust in the constellation of Scorpius is the sort of region where complex molecules are found. It is also the sort of cloud where new stars form from the reservoir of gas and dust in the cloud. Radiation from a group of hot stars (off the picture to the bottom left) called the Scorpius OB Association is \u201ceating into\u201d the cloud, sweeping it into an elongated shape and causing the reddish glow seen at its tip. (credit: Dr. Robert Gendler) Clearly the early Earth itself produced some of the molecular building blocks of life. Since the early 1950s, scientists have tried to duplicate in their laboratories the chemical pathways that led to life on our planet. In a series of experiments known as the Miller-Urey experiments, pioneered by Stanley Miller and Harold Urey at the University of Chicago, biochemists have simulated conditions on early Earth and have been able to produce some of the fundamental building blocks of life, including those that form proteins and other large biological molecules known as nucleic acids (which we will discuss shortly). Although these experiments produced encouraging results, there are some problems with them. The most interesting chemistry from a biological perspective takes place with hydrogen-rich or reducing gases, such as ammonia and methane. However, the early atmosphere of Earth was probably dominated by carbon dioxide (as Venus\u2019 and Mars\u2019 atmospheres still are today) and may not have contained an abundance of reducing gases comparable to that used in Miller-Urey type experiments. Hydrothermal vents\u2014seafloor systems in which ocean water is superheated and circulated through crustal or mantle rocks before reemerging into the ocean\u2014have also been suggested as potential contributors of organic compounds on the early Earth, and such sources would not require Earth to have an early reducing atmosphere. Both earthly and extraterrestrial sources may have contributed to Earth\u2019s early supply of organic molecules, although we have more direct evidence for the latter. It is even conceivable that life itself originated elsewhere and was seeded onto our planet\u2014although this, of course, does not solve the problem of how that life originated", " to begin with Hydrothermal vents are beginning to seem more likely as early contributors to the organic compounds found on Earth. Read about hydrothermal vents, watch videos and slideshows on these and other deep- This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1103 sea wonders, and try an interactive simulation of hydrothermal circulation at the Woods Hole Oceanographic Institution (https://openstax.org/l/30wohooceins) website. The Origin and Early Evolution of Life The carbon compounds that form the chemical basis of life may be common in the universe, but it is still a giant step from these building blocks to a living cell. Even the simplest molecules of the genes (the basic functional units that carry the genetic, or hereditary, material in a cell) contain millions of molecular units, each arranged in a precise sequence. Furthermore, even the most primitive life required two special capabilities: a means of extracting energy from its environment, and a means of encoding and replicating information in order to make faithful copies of itself. Biologists today can see ways that either of these capabilities might have formed in a natural environment, but we are still a long way from knowing how the two came together in the first life-forms. We have no solid evidence for the pathway that led to the origin of life on our planet except for whatever early history may be retained in the biochemistry of modern life. Indeed, we have very little direct evidence of what Earth itself was like during its earliest history\u2014our planet is so effective at resurfacing itself through plate tectonics (see the chapter on Earth as a Planet) that very few rocks remain from this early period. In the earlier chapter on Cratered Worlds, you learned that Earth was subjected to a heavy bombardment\u2014a period of large impact events\u2014some 3.8 to 4.1 billion years ago. Large impacts would have been energetic enough to heat- sterilize the surface layers of Earth, so that even if life had begun by this time, it might well have been wiped out. When the large impacts ceased, the scene was set for a more peaceful environment on our planet. If the oceans of Earth contained accumulated organic material from any of the sources already mentioned, the ingredients were available to make living organisms. We do not understand in any detail the sequence of events that led from molecules to biology, but there is fossil evidence of microbial life in 3.5-billion-year-old rocks, and possible (debated) evidence for life as far back as 3.8 billion years. Life as we know it employs two main molecular systems: the functional molecules known as proteins, which carry out the chemical work of the cell, and information-containing molecules of DNA (deoxyribonucleic acid) that store information about how to create the cell and its chemical and structural components. The origin of life is sometimes considered a \u201cchicken and egg problem\u201d because, in modern biology, neither of these systems works without the other. It is our proteins that assemble DNA strands in the precise order required to store information, but the proteins are created based on information stored in DNA. Which came first? Some origin of life researchers believe that prebiotic chemistry was based on molecules that could both store information and do the chemical work of the cell. It has been suggested that RNA (ribonucleic acid), a molecule that aids in the flow of genetic information from DNA to proteins, might have served such a purpose. The idea of an early \u201cRNA world\u201d has become increasingly accepted, but a great deal remains to be understood about the origin of life. Perhaps the most important innovation in the history of biology, apart from the origin of life itself, was the discovery of the process of photosynthesis, the complex sequence of chemical reactions through which some living things can use sunlight to manufacture products that store energy (such as carbohydrates), releasing oxygen as one by-product. Previously, life had to make do with sources of chemical energy available on Earth or delivered from space. But the abundant energy available in sunlight could support a larger and more productive biosphere, as well as some biochemical reactions not previously possible for life. One of these was the production of oxygen (as a waste product) from carbon dioxide, and the increase in atmospheric levels of oxygen about 2.4 billion years ago means that oxygen-producing photosynthesis must have emerged and 1104 Chapter 30 Life in the Universe become globally important by this time. In fact, it is likely that oxygen-producing photosynthesis emerged considerably earlier. Some forms of chemical evidence contained in ancient rocks, such as the solid, layered rock formations known as stromatolites, are thought to be the fossils of oxygen-producing photosynthetic bacteria in rocks that are almost 3.5 billion years old (Figure 30.5). It is generally thought that a simpler form of photosynthesis that does not produce oxygen (and is still used by", " some bacteria today) probably preceded oxygen-producing photosynthesis, and there is strong fossil evidence that one or the other type of photosynthesis was functioning on Earth at least as far back as 3.4 billion years ago. Figure 30.5. Stromatolites Preserve the Earliest Physical Representation of Life on Earth. In their reach for sunlight, the single-celled microbes formed mats that trapped sediments in the water above them. Such trapped sediments fell and formed layers on top of the mats. The microbes then climbed atop the sediment layers and trapped more sediment. What is found in the rock record are (a) the solidified, curved sedimentary layers that are signatures of biological activity. The earliest known stromatolite is 3.47 billion years old and is found in Western Australia. (b) This more recent example is in Lake Thetis, also in Western Australia. (credit a: modification of work by James St. John; credit b: modification of work by Ruth Ellison) The free oxygen produced by photosynthesis began accumulating in our atmosphere about 2.4 billion years ago. The interaction of sunlight with oxygen can produce ozone (which has three atoms of oxygen per molecule, as compared to the two atoms per molecule in the oxygen we breathe), which accumulated in a layer high in Earth\u2019s atmosphere. As it does on Earth today, this ozone provided protection from the Sun\u2019s damaging ultraviolet radiation. This allowed life to colonize the landmasses of our planet instead of remaining only in the ocean. The rise in oxygen levels was deadly to some microbes because, as a highly reactive chemical, it can irreversibly damage some of the biomolecules that early life had developed in the absence of oxygen. For other microbes, it was a boon: combining oxygen with organic matter or other reduced chemicals generates a lot of energy\u2014you can see this when a log burns, for example\u2014and many forms of life adopted this way of living. This new energy source made possible a great proliferation of organisms, which continued to evolve in an oxygen-rich environment. The details of that evolution are properly the subject of biology courses, but the process of evolution by natural selection (survival of the fittest) provides a clear explanation for the development of Earth\u2019s remarkable variety of life-forms. It does not, however, directly solve the mystery of life\u2019s earliest beginnings. We hypothesize that life will arise whenever conditions are appropriate, but this hypothesis is just another form of the Copernican principle. We now have the potential to address this hypothesis with observations. If a second example of life is found in our solar system or a nearby star, it would imply that life emerges commonly enough that the universe is likely filled with biology. To make such observations, however, we must first decide where to focus our search. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1105 Just how did life arise in the first place? And could it have happened with a different type of chemistry? Watch the 15-minute video Making Matter Come Alive (https://openstax.org/l/30makmattcomali) in which a chemistry expert explores some answers to these questions, from a 2011 TED Talk. Habitable Environments Among the staggering number of objects in our solar system, Galaxy, and universe, some may have conditions suitable for life, while others do not. Understanding what conditions and features make a habitable environment\u2014an environment capable of hosting life\u2014is important both for understanding how widespread habitable environments may be in the universe and for focusing a search for life beyond Earth. Here, we discuss habitability from the perspective of the life we know. We will explore the basic requirements of life and, in the following section, consider the full range of environmental conditions on Earth where life is found. While we can\u2019t entirely rule out the possibility that other life-forms might have biochemistry based on alternatives to carbon and liquid water, such life \u201cas we don\u2019t know it\u201d is still completely speculative. In our discussion here, we are focusing on habitability for life that is chemically similar to that on Earth. Life requires a solvent (a liquid in which chemicals can dissolve) that enables the construction of biomolecules and the interactions between them. For life as we know it, that solvent is water, which has a variety of properties that are critical to how our biochemistry works. Water is abundant in the universe, but life requires that water be in liquid form (rather than ice or gas) in order to properly fill its role in biochemistry. That is the case only within a certain range of temperatures and pressures\u2014too high or too low in either variable, and water takes the form of a solid or a gas. Identifying environments where water is present within the appropriate range of temperature and pressure is thus an important first step in identifying habitable environments.", " Indeed, a \u201cfollow the water\u201d strategy has been, and continues to be, a key driver in the exploration of planets both within and beyond our solar system. Our biochemistry is based on molecules made of carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur. Carbon is at the core of organic chemistry. Its ability to form four bonds, both with itself and with the other elements of life, allows for the formation of a vast number of potential molecules on which to base biochemistry. The remaining elements contribute structure and chemical reactivity to our biomolecules, and form the basis of many of the interactions among them. These \u201cbiogenic elements,\u201d sometimes referred to with the acronym CHNOPS (carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur), are the raw materials from which life is assembled, and an accessible supply of them is a second requirement of habitability. As we learned in previous chapters on nuclear fusion and the life story of the stars, carbon, nitrogen, oxygen, phosphorus, and sulfur are all formed by fusion within stars and then distributed out into their galaxy as those stars die. But how they are distributed among the planets that form within a new star system, in what form, and how chemical, physical, and geological processes on those planets cycle the elements into structures that are accessible to biology, can have significant impacts on the distribution of life. In Earth\u2019s oceans, for example, the abundance of phytoplankton (simple organisms that are the base of the ocean food chain) in surface waters can vary by a thousand-fold because the supply of nitrogen differs from place to place (Figure 30.6). Understanding what processes control the accessibility of elements at all scales is thus a critical part of identifying habitable environments. 1106 Chapter 30 Life in the Universe Figure 30.6. Chlorophyll Abundance. The abundance of chlorophyll (an indicator of photosynthetic bacteria and algae) varies by almost a thousand-fold across the ocean basins. That variation is almost entirely due to the availability of nitrogen\u2014one of the major \u201cbiogenic elements\u201d in forms that can be used by life. (credit: modification of work by NASA, Gene C. Feldman) With these first two requirements, we have the elemental raw materials of life and a solvent in which to assemble them into the complicated molecules that drive our biochemistry. But carrying out that assembly and maintaining the complicated biochemical machinery of life takes energy. You fulfill your own requirement for energy every time you eat food or take a breath, and you would not live for long if you failed to do either on a regular basis. Life on Earth makes use of two main types of energy: for you, these are the oxygen in the air you breathe and the organic molecules in your food. But life overall can use a much wider array of chemicals and, while all animals require oxygen, many bacteria do not. One of the earliest known life processes, which still operates in some modern microorganisms, combines hydrogen and carbon dioxide to make methane, releasing energy in the process. There are microorganisms that \u201cbreathe\u201d metals that would be toxic to us, and even some that breathe in sulfur and breathe out sulfuric acid. Plants and photosynthetic microorganisms have also evolved mechanisms to use the energy in light directly. Water in the liquid phase, the biogenic elements, and energy are the fundamental requirements for habitability. But are there additional environmental constraints? We consider this in the next section. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1107 Figure 30.7. Grand Prismatic Spring in Yellowstone National Park. This hot spring, where water emerges from the bluish center at temperatures near the local boiling point (about 92 \u00baC), supports a thriving array of microbial life. The green, yellow, and orange colors around the edges come from thick \u201cmats\u201d of photosynthetic bacteria. In fact, their coloration in part demonstrates their use of light energy\u2014some wavelengths of incoming sunlight are selectively captured for energy; the rest are reflected back. Since it lacks the captured wavelengths, this light is now different in color than the sunlight that illuminates it. The blue part of the spring has temperatures too high to allow photosynthetic life (hence the lack of color except that supplied by water itself), but life is still present. Here, at nearly boiling temperatures, bacteria use the chemical energy supplied by the combination of hydrogen and other chemicals with oxygen. (credit: modification of work by Domenico Salvagnin) Life in Extreme Conditions At a chemical level, life consists of many types of molecules that interact with one another to carry out the processes of life. In addition to water, elemental raw materials, and energy, life also needs an environment in which those complicated molecules are stable (don\u2019t break down before they can do their jobs", ") and their interactions are possible. Your own biochemistry works properly only within a very narrow range of about 10 \u00b0C in body temperature and two-tenths of a unit in blood pH (pH is a numerical measure of acidity, or the amount of free hydrogen ions). Beyond those limits, you are in serious danger. Life overall must also have limits to the conditions in which it can properly work but, as we will see, they are much broader than human limits. The resources that fuel life are distributed across a very wide range of conditions. For example, there is abundant chemical energy to be had in hot springs that are essentially boiling acid (see Figure 30.7). This provides ample incentive for evolution to fill as much of that range with life as is biochemically possible. An organism (usually a microbe) that tolerates or even thrives under conditions that most of the life around us would consider hostile, such as very high or low temperature or acidity, is known as an extremophile (where the suffix -phile means \u201clover of\u201d). Let\u2019s have a look at some of the conditions that can challenge life and the organisms that have managed to carve out a niche at the far reaches of possibility. Both high and low temperatures can cause a problem for life. As a large organism, you are able to maintain an almost constant body temperature whether it is colder or warmer in the environment around you. But this is not possible at the tiny size of microorganisms; whatever the temperature in the outside world is also the temperature of the microbe, and its biochemistry must be able to function at that temperature. High temperatures are the enemy of complexity\u2014increasing thermal energy tends to break apart big molecules into smaller and smaller bits, and life needs to stabilize the molecules with stronger bonds and special proteins. But this approach has its limits. Nevertheless, as noted earlier, high-temperature environments like hot springs and hydrothermal vents often offer abundant sources of chemical energy and therefore drive the evolution of organisms that can tolerate high temperatures (see Figure 30.8); such an organism is called a thermophile. Currently, the high temperature record holder is a methane-producing microorganism that can grow at 122 \u00b0C, where the pressure 1108 Chapter 30 Life in the Universe also is so high that water still does not boil. That\u2019s amazing when you think about it. We cook our food\u2014meaning, we alter the chemistry and structure of its biomolecules\u2014by boiling it at a temperature of 100 \u00b0C. In fact, food begins to cook at much lower temperatures than this. And yet, there are organisms whose biochemistry remains intact and operates just fine at temperatures 20 degrees higher. Figure 30.8. Hydrothermal Vent on the Sea Floor. What appears to be black smoke is actually superheated water filled with minerals of metal sulfide. Hydrothermal vent fluid can represent a rich source of chemical energy, and therefore a driver for the evolution of microorganisms that can tolerate high temperatures. Bacteria feeding on this chemical energy form the base of a food chain that can support thriving communities of animals\u2014in this case, a dense patch of red and white tubeworms growing around the base of the vent. (credit: modification of work by the University of Washington; NOAA/OAR/OER) Cold can also be a problem, in part because it slows down metabolism to very low levels, but also because it can cause physical changes in biomolecules. Cell membranes\u2014the molecular envelopes that surround cells and allow their exchange of chemicals with the world outside\u2014are basically made of fatlike molecules. And just as fat congeals when it cools, membranes crystallize, changing how they function in the exchange of materials in and out of the cell. Some cold-adapted cells (called psychrophiles) have changed the chemical composition of their membranes in order to cope with this problem; but again, there are limits. Thus far, the coldest temperature at which any microbe has been shown to reproduce is about \u201325 \u00baC. Conditions that are very acidic or alkaline can also be problematic for life because many of our important molecules, like proteins and DNA, are broken down under such conditions. For example, household drain cleaner, which does its job by breaking down the chemical structure of things like hair clogs, is a very alkaline solution. The most acid-tolerant organisms (acidophiles) are capable of living at pH values near zero\u2014about ten million times more acidic than your blood (Figure 30.9). At the other extreme, some alkaliphiles can grow at pH levels of about 13, which is comparable to the pH of household bleach and almost a million times more alkaline than your blood. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1109 Figure 30.9. Spain\u2019", "s Rio Tinto. With a pH close to 2, Rio Tinto is literally a river of acid. Acid-loving microorganisms (acidophiles) not only thrive in these waters, their metabolic activities help generate the acid in the first place. The rusty red color that gives the river its name comes from high levels of iron dissolved in the waters. High levels of salts in the environment can also cause a problem for life because the salt blocks some cellular functions. Humans recognized this centuries ago and began to salt-cure food to keep it from spoiling\u2014meaning, to keep it from being colonized by microorganisms. Yet some microbes have evolved to grow in water that is saturated in sodium chloride (table salt)\u2014about ten times as salty as seawater (Figure 30.10). Figure 30.10. Salt Ponds. The waters of an evaporative salt works near San Francisco are colored pink by thriving communities of photosynthetic organisms. These waters are about ten times as salty as seawater\u2014enough for sodium chloride to begin to crystallize out\u2014yet some organisms can survive and thrive in these conditions. (credit: modification of work by NASA) Very high pressures can literally squeeze life\u2019s biomolecules, causing them to adopt more compact forms that do not work very well. But we still find life\u2014not just microbial, but even animal life\u2014at the bottoms of our ocean trenches, where pressures are more than 1000 times atmospheric pressure. Many other adaptions 1110 Chapter 30 Life in the Universe to environmental \u201cextremes\u201d are also known. There is even an organism, Deinococcus radiodurans, that can tolerate ionizing radiation (such as that released by radioactive elements) a thousand times more intense than you would be able to withstand. It is also very good at surviving extreme desiccation (drying out) and a variety of metals that would be toxic to humans. From many such examples, we can conclude that life is capable of tolerating a wide range of environmental extremes\u2014so much so that we have to work hard to identify places where life can\u2019t exist. A few such places are known\u2014for example, the waters of hydrothermal vents at over 300 \u00b0C appear too hot to support any life\u2014and finding these places helps define the possibility for life elsewhere. The study of extremophiles over the last few decades has expanded our sense of the range of conditions life can survive and, in doing so, has made many scientists more optimistic about the possibility that life might exist beyond Earth. 30.3 SEARCHING FOR LIFE BEYOND EARTH Learning Objectives By the end of this section, you will be able to: Outline what we have learned from exploration of the environment on Mars Identify where in the solar system life is most likely sustainable and why Describe some key missions and their findings in our search for life beyond our solar system Explain the use of biomarkers in the search for evidence of life beyond our solar system Astronomers and planetary scientists continue to search for life in the solar system and the universe at large. In this section, we discuss two kinds of searches. First is the direct exploration of planets within our own solar system, especially Mars and some of the icy moons of the outer solar system. Second is the even more difficult task of searching for evidence of life\u2014a biomarker\u2014on planets circling other stars. In the next section, we will examine SETI, the search for extraterrestrial intelligence. As you will see, the approaches taken in these three cases are very different, even though the goal of each is the same: to determine if life on Earth is unique in the universe. Life on Mars The possibility that Mars hosts, or has hosted, life has a rich history dating back to the \u201ccanals\u201d that some people claimed to see on the martian surface toward the end of the nineteenth century and the beginning of the twentieth. With the dawn of the space age came the possibility to address this question up close through a progression of missions to Mars that began with the first successful flyby of a robotic spacecraft in 1964 and have led to the deployment of NASA\u2019s Curiosity rover, which landed on Mars\u2019 surface in 2012. The earliest missions to Mars provided some hints that liquid water\u2014one of life\u2019s primary requirements\u2014may once have flowed on the surface, and later missions have strengthened this conclusion. The NASA Viking landers, whose purpose was to search directly for evidence of life on Mars, arrived on Mars in 1976. Viking\u2019s onboard instruments found no organic molecules (the stuff of which life is made), and no evidence of biological activity in the martian soils it analyzed. This result is not particularly surprising because, despite the evidence of flowing liquid water in the past, liquid water on the surface of Mars is generally not stable today. Over much of Mars, temperatures and pressures at the surface are so low that pure water would either freeze or boil away (under very low pressures, water will boil", " at a much lower temperature than usual). To make matters worse, unlike Earth, Mars does not have a magnetic field and ozone layer to protect the surface from harmful solar ultraviolet radiation and energetic particles. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1111 However, Viking\u2019s analyses of the soil said nothing about whether life may have existed in Mars\u2019 distant past, when liquid water was more abundant. We do know that water in the form of ice exists in abundance on Mars, not so deep beneath its surface. Water vapor is also a constituent of the atmosphere of Mars. Since the visit of Viking, our understanding of Mars has deepened spectacularly. Orbiting spacecraft have provided ever-more detailed images of the surface and detected the presence of minerals that could have formed only in the presence of liquid water. Two bold surface missions, the Mars Exploration Rovers Spirit and Opportunity (2004), followed by the much larger Curiosity Rover (2012), confirmed these remote-sensing data. All three rovers found abundant evidence for a past history of liquid water, revealed not only from the mineralogy of rocks they analyzed, but also from the unique layering of rock formations. Curiosity has gone a step beyond evidence for water and confirmed the existence of habitable environments on ancient Mars. \u201cHabitable\u201d means not only that liquid water was present, but that life\u2019s requirements for energy and elemental raw materials could also have been met. The strongest evidence of an ancient habitable environment came from analyzing a very fine-grained rock called a mudstone\u2014a rock type that is widespread on Earth but was unknown on Mars until Curiosity found it (see Figure 30.11). The mudstone can tell us a great deal about the wet environments in which they formed. Figure 30.11. Mudstone. Shown are the first holes drilled by NASA\u2019s Curiosity Mars rover into a mudstone, with \u201cfresh\u201d drill-pilings around the holes. Notice the difference in color between the red ancient martian surface and the gray newly exposed rock powder that came from the drill holes. Each drill hole is about 0.6 inch (1.6 cm) in diameter. (credit: modification of work by NASA/JPL-Caltech/MSSS) Five decades of robotic exploration have allowed us to develop a picture of how Mars evolved through time. Early Mars had epochs of warmer and wetter conditions that would have been conducive to life at the surface. However, Mars eventually lost much of its early atmosphere and the surface water began to dry up. As that happened, the ever-shrinking reservoirs of liquid water on the martian surface became saltier and more acidic, until the surface finally had no significant liquid water and was bathed in harsh solar radiation. The surface thus became uninhabitable, but this might not be the case for the planet overall. Reservoirs of ice and liquid water could still exist underground, where pressure and temperature conditions make it stable. There is recent evidence to suggest that liquid water (probably very salty water) can occasionally (and briefly) flow on the surface even today. Thus, Mars might even have habitable conditions in the present day, but of a much different sort than we normally think of on Earth. Our study of Mars reveals a planet with a fascinating history\u2014one that saw its ability to host surface life dwindle billions of years ago, but perhaps allowing life to adapt and survive in favorable environmental niches. Even if life did not survive, we expect that we might find evidence of life if it ever took hold on Mars. If it is there, it is hidden in the crust, and we are still learning how best to decipher that evidence. 1112 Chapter 30 Life in the Universe Life in the Outer Solar System The massive gas and ice giant planets of the outer solar system\u2014Jupiter, Saturn, Uranus, and Neptune\u2014are almost certainly not habitable for life as we know it, but some of their moons might be (see Figure 30.12). Although these worlds in the outer solar system contain abundant water, they receive so little warming sunlight in their distant orbits that it was long believed they would be \u201cgeologically dead\u201d balls of hard-frozen ice and rock. But, as we saw in the chapter on Rings, Moons, and Pluto, missions to the outer solar system have found something much more interesting. Jupiter\u2019s moon Europa revealed itself to the Voyager and Galileo missions as an active world whose icy surface apparently conceals an ocean with a depth of tens to perhaps a hundred kilometers. As the moon orbits Jupiter, the planet\u2019s massive gravity creates tides on Europa\u2014just as our own Moon\u2019s gravity creates our ocean tides\u2014and the friction of all that pushing and pulling generates enough heat to keep the water in liquid form (Figure 30.13). Similar tides act upon other moons if they orbit close to the planet.", " Scientists now think that six or more of the outer solar system\u2019s icy moons may harbor liquid water oceans for the same reason. Among these, Europa and Enceladus, a moon of Saturn, have thus far been of greatest interest to astrobiologists. Figure 30.12. Jupiter\u2019s Moons. The Galilean moons of Jupiter are shown to relative scale and arranged in order of their orbital distance from Jupiter. At far left, Io orbits closest to Jupiter and so experiences the strongest tidal heating by Jupiter\u2019s massive gravity. This effect is so strong that Io is thought to be the most volcanically active body in our solar system. At far right, Callisto shows a surface scarred by billions of years\u2019 worth of craters\u2014an indication that the moon\u2019s surface is old and that Callisto may be far less active than its sibling moons. Between these hot and cold extremes, Europa, second from left, orbits at a distance where Jupiter\u2019s tidal heating may be \u201cjust right\u201d to sustain a liquid water ocean beneath its icy crust. (credit: modification of work by NASA/JPL/DLR) Europa has probably had an ocean for most or all of its history, but habitability requires more than just liquid water. Life also requires energy, and because sunlight does not penetrate below the kilometers-thick ice crust of Europa, this would have to be chemical energy. One of Europa\u2019s key attributes from an astrobiology perspective is that its ocean is most likely in direct contact with an underlying rocky mantle, and the interaction of water and rocks\u2014especially at high temperatures, as within Earth\u2019s hydrothermal vent systems\u2014yields a reducing chemistry (where molecules tend to give up electrons readily) that is like one half of a chemical battery. To complete the battery and provide energy that could be used by life requires that an oxidizing chemistry (where molecules tend to accept electrons readily) also be available. On Earth, when chemically reducing vent fluids meet oxygen-containing seawater, the energy that becomes available often supports thriving communities of microorganisms and animals on the sea floor, far from the light of the Sun. The Galileo mission found that Europa\u2019s icy surface does contain an abundance of oxidizing chemicals. This means that availability of energy to support life depends very much on whether the chemistry of the surface and the ocean can mix, despite the kilometers of ice in between. That Europa\u2019s ice crust appears geologically \u201cyoung\u201d (only tens of millions of years old, on average) and that it is active makes it tantalizing to think that such mixing might indeed occur. Understanding whether and how much exchange occurs between the surface This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1113 and ocean of Europa will be a key science objective of future missions to Europa, and a major step forward in understanding whether this moon could be a cradle of life. Figure 30.13. Jupiter\u2019s Moon Europa, as Imaged by NASA\u2019s Galileo Mission. The relative scarcity of craters on Europa suggests a surface that is \u201cgeologically young,\u201d and the network of colored ridges and cracks suggests constant activity and motion. Galileo\u2019s instruments also strongly suggested the presence of a massive ocean of salty liquid water beneath the icy crust. (credit: modification of work by NASA/JPL-Caltech/SETI Institute) In 2005, the Cassini mission performed a close flyby of a small (500-kilometer diameter) moon of Saturn, Enceladus (Figure 30.14), and made a remarkable discovery. Plumes of gas and icy material were venting from the moon\u2019s south polar region at a collective rate of about 250 kilograms of material per second. Several observations, including the discovery of salts associated with the icy material, suggest that their source is a liquid water ocean beneath tens of kilometers of ice. Although it remains to be shown definitively whether the ocean is local or global, transient or long-lived, it does appear to be in contact, and to have reacted, with a rocky interior. As on Europa, this is probably a necessary\u2014though not sufficient\u2014condition for habitability. What makes Enceladus so enticing to planetary scientists, though, are those plumes of material that seem to come directly from its ocean: samples of the interior are there for the taking by any spacecraft sent flying through. For a future mission, such samples could yield evidence not only of whether Enceladus is habitable but, indeed, of whether it is home to life. Figure 30.14. Image of Saturn\u2019s Moon Enceladus from NASA\u2019s Cassini Mission. The south polar region was found to have multiple plumes of ice and gas that, combined, are venting about 250 kilograms of material per second into space. Such", " features suggest that Enceladus, like Europa, has a sub-ice ocean. (credit: NASA/JPL/ SSI) Saturn\u2019s big moon Titan is very different from both Enceladus and Europa (see Figure 30.15). Although it may host a liquid water layer deep within its interior, it is the surface of Titan and its unusual chemistry that 1114 Chapter 30 Life in the Universe makes this moon such an interesting place. Titan\u2019s thick atmosphere\u2014the only one among moons in the solar system\u2014is composed mostly of nitrogen but also of about 5% methane. In the upper atmosphere, the Sun\u2019s ultraviolet light breaks apart and recombines these molecules into more complex organic compounds that are collectively known as tholins. The tholins shroud Titan in an orange haze, and imagery from Cassini and from the Huygens probe that descended to Titan\u2019s surface show that heavier particles appear to accumulate on the surface, even forming \u201cdunes\u201d that are cut and sculpted by flows of liquid hydrocarbons (such as liquid methane). Some scientists see this organic chemical factory as a natural laboratory that may yield some clues about the solar system\u2019s early chemistry\u2014perhaps even chemistry that could support the origin of life. Figure 30.15. Image of Saturn\u2019s Moon Titan from NASA\u2019s Cassini Mission. (a) The hazy orange glow comes from Titan\u2019s thick atmosphere (the only one known among the moons of the solar system). That atmosphere is mostly nitrogen but also contains methane and potentially a variety of complex organic compounds. The bright spot near the top of the image is sunlight reflected from a very flat surface\u2014almost certainly a liquid. We see this effect, called \u201cglint,\u201d when sunlight reflects off the surface of a lake or ocean. (b) Cassini radar imagery shows what look very much like landforms and lakes on the surface of Titan. But the surface lakes and oceans of Titan are not water; they are probably made of liquid hydrocarbons like methane and ethane. (credit a: modification of work by NASA/JPL/University of Arizona/DLR; credit b: modification of work by NASA/JPL-Caltech/ASI In January 2005, the Huygens probe descended to the surface of Titan and relayed data, including imagery of the landing site, for about 90 minutes. You can watch a video (https://openstax.org/l/ 30huytatsurf) about the descent of Huygens to Titan\u2019s surface. Habitable Planets Orbiting Other Stars One of the most exciting developments in astronomy during the last two decades is the ability to detect exoplanets\u2014planets orbiting other stars. As we saw in the chapter on the formation of stars and planets, since the discovery of the first exoplanet in 1995, there have been thousands of confirmed detections, and many more candidates that are not yet confirmed. These include several dozen possibly habitable exoplanets. Such numbers finally allow us to make some predictions about exoplanets and their life-hosting potential. The This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1115 majority of stars with mass similar to the Sun appear to host at least one planet, with multi-planet systems like our own not unusual. How many of these planets might be habitable, and how could we search for life there The NASA Exoplanet Archive (https://openstax.org/l/30NASAexoarc) is an up-to-date searchable online source of data and tools on everything to do with exoplanets. Explore stellar and exoplanet parameters and characteristics, find the latest news on exoplanet discoveries, plot your own data interactively, and link to other related resources. In evaluating the prospect for life in distant planetary systems, astrobiologists have developed the idea of a habitable zone\u2014a region around a star where suitable conditions might exist for life. This concept focuses on life\u2019s requirement for liquid water, and the habitable zone is generally thought of as the range of distances from the central star in which water could be present in liquid form at a planet\u2019s surface. In our own solar system, for example, Venus has surface temperatures far above the boiling point of water and Mars has surface temperatures that are almost always below the freezing point of water. Earth, which orbits between the two, has a surface temperature that is \u201cjust right\u201d to keep much of our surface water in liquid form. Whether surface temperatures are suitable for maintaining liquid water depends on a planet\u2019s \u201cradiation budget\u201d \u2014how much starlight energy it absorbs and retains\u2014and whether or how processes like winds and ocean circulation distribute that energy around the planet. How much stellar energy a planet receives,", " in turn, depends on how much and what sort of light the star emits and how far the planet is from that star,[1] how much it reflects back to space, and how effectively the planet\u2019s atmosphere can retain heat through the greenhouse effect (see Earth as a Planet). All of these can vary substantially, and all matter a lot. For example, Venus receives about twice as much starlight per square meter as Earth but, because of its dense cloud cover, also reflects about twice as much of that light back to space as Earth does. Mars receives only about half as much starlight as Earth, but also reflects only about half as much. Thus, despite their differing orbital distances, the three planets actually absorb comparable amounts of sunlight energy. Why, then, are they so dramatically different? As we learned in several chapters about the planets, some of the gases that make up planetary atmospheres are very effective at trapping infrared light\u2014the very range of wavelengths at which planets radiate thermal energy back out to space\u2014and this can raise the planet\u2019s surface temperature quite a bit more than would otherwise be the case. This is the same \u201cgreenhouse effect\u201d that is of such concern for global warming on our planet. Earth\u2019s natural greenhouse effect, which comes mostly from water vapor and carbon dioxide in the atmosphere, raises our average surface temperature by about 33 \u00b0C over the value it would have if there were no greenhouse gases in the atmosphere. Mars has a very thin atmosphere and thus very little greenhouse warming (about 2 \u00b0C worth), while Venus has a massive carbon dioxide atmosphere that creates very strong greenhouse warming (about 510 \u00b0C worth). These worlds are much colder and much hotter, respectively, than Earth would be if moved into their orbits. Thus, we must consider the nature of any atmosphere as well as the distance from the star in evaluating the range of habitability. 1 The amount of starlight received per unit area of a planet\u2019s surface (per square meter, for example) decreases with the square of the distance from the star. Thus, when the orbital distance doubles, the illumination decreases by 4 times (22), and when the orbital distance increases tenfold, the illumination decreases by 100 times (102). Venus and Mars orbit the sun at about 72% and 152% of Earth\u2019s orbital distance, respectively, so Venus receives about 1/(0.72)2 = 1.92 (about twice) and Mars about 1/(1.52)2 = 0.43 (about half) as much light per square meter of planet surface as Earth does. 1116 Chapter 30 Life in the Universe Of course, as we have learned, stars also vary widely in the intensity and spectrum (the wavelengths of light) they emit. Some are much brighter and hotter (bluer), while others are significantly dimmer and cooler (redder), and the distance of the habitable zone varies accordingly. For example, the habitable zone around M-dwarf stars is 3 to 30 times closer in than for G-type (Sun-like) stars. There is a lot of interest in whether such systems could be habitable because\u2014although they have some potential downsides for supporting life\u2014M-dwarf stars are by far the most numerous and long-lived in our Galaxy. The luminosity of stars like the Sun also increases over their main-sequence lifetime, and this means that the habitable zone migrates outward as a star system ages. Calculations indicate that the power output of the Sun, for example, has increased by at least 30% over the past 4 billion years. Thus, Venus was once within the habitable zone, while Earth received a level of solar energy insufficient to keep the modern Earth (with its present atmosphere) from freezing over. In spite of this, there is plenty of geological evidence that liquid water was present on Earth\u2019s surface billions of years ago. The phenomenon of increasing stellar output and an outwardly migrating habitable zone has led to another concept: the continuously habitable zone is defined by the range of orbits that would remain within the habitable zone during the entire lifetime of the star system. As you might imagine, the continuously habitable zone is quite a bit narrower than the habitable zone is at any one time in a star\u2019s history. The nearest star to the Sun, Proxima Centauri, is an M star that has a planet with a mass of at least 1.3 Earth masses, taking about 11 days to orbit. At the distance for such a quick orbit (0.05 AU), the planet may be in the habitable zone of its star, although whether conditions on such a planet near such a star are hospitable for life is a matter of great scientific debate. Even when planets orbit within the habitable zone of their star, it is no guarantee that they are habitable. For example, Venus today has virtually no water, so even if it were suddenly moved to a \u201cjust right\u201d orbit within the habitable zone, a critical requirement for life would still be", " lacking. Scientists are working to understand all the factors that define the habitable zone and the habitability of planets orbiting within that zone because this will be our primary guide in targeting exoplanets on which to seek evidence of life. As technology for detecting exoplanets has advanced, so too has our potential to find Earth-size worlds within the habitable zones of their parent stars. Of the confirmed or candidate exoplanets known at the time of writing, nearly 300 are considered to be orbiting within the habitable zone and more than 10% of those are roughly Earth-size Explore the habitable universe at the online Planetary Habitability Laboratory (https://openstax.org/ l/30planhabitlab) created by the University of Puerto Rico at Arecibo. See the potentially habitable exoplanets and other interesting places in the universe, watch video clips, and link to numerous related resources on astrobiology. Biomarkers Our observations suggest increasingly that Earth-size planets orbiting within the habitable zone may be common in the Galaxy\u2014current estimates suggest that more than 40% of stars have at least one. But are any of them inhabited? With no ability to send probes there to sample, we will have to derive the answer from the light and other radiation that come to us from these faraway systems (Figure 30.16). What types of observations might constitute good evidence for life? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1117 Figure 30.16. Earth, as Seen by NASA\u2019s Voyager 1. In this image, taken from 4 billion miles away, Earth appears as a \u201cpale blue dot\u201d representing less than a pixel\u2019s worth of light. Would this light reveal Earth as a habitable and inhabited world? Our search for life on exoplanets will depend on an ability to extract information about life from the faint light of faraway worlds. (credit: modification of work by NASA/JPL-Caltech) To be sure, we need to look for robust biospheres (atmospheres, surfaces, and/or oceans) capable of creating planet-scale change. Earth hosts such a biosphere: the composition of our atmosphere and the spectrum of light reflected from our planet differ considerably from what would be expected in the absence of life. Presently, Earth is the only body in our solar system for which this is true, despite the possibility that habitable conditions might prevail in the subsurface of Mars or inside the icy moons of the outer solar system. Even if life exists on these worlds, it is very unlikely that it could yield planet-scale changes that are both telescopically observable and clearly biological in origin. What makes Earth \u201cspecial\u201d among the potentially habitable worlds in our solar system is that it has a photosynthetic biosphere. This requires the presence of liquid water at the planet\u2019s surface, where organisms have direct access to sunlight. The habitable zone concept focuses on this requirement for surface liquid water\u2014even though we know that subsurface habitable conditions could prevail at more distant orbits\u2014exactly because these worlds would have biospheres detectable at a distance. Indeed, plants and photosynthetic microorganisms are so abundant at Earth\u2019s surface that they affect the color of the light that our planet reflects out into space\u2014we appear greener in visible wavelengths and reflect more near-infrared light than we otherwise would. Moreover, photosynthesis has changed Earth\u2019s atmosphere at a large scale\u2014more than 20% of our atmosphere comes from the photosynthetic waste product, oxygen. Such high levels would be very difficult to explain in the absence of life. Other gases, such as nitrous oxide and methane, when found simultaneously with oxygen, have also been suggested as possible indicators of life. When sufficiently abundant in an atmosphere, such gases could be detected by their effect on the spectrum of light that a planet emits or reflects. (As we saw in the chapter on exoplanets, astronomers today are beginning to have the capability of detecting the spectrum of the atmospheres of some planets orbiting other stars.) Astronomers have thus concluded that, at least initially, a search for life outside our solar system should focus on exoplanets that are as much like Earth as possible\u2014roughly Earth-size planets orbiting in the habitable zone\u2014and look for the presence of gases in the atmosphere or colors in the visible spectrum that are hard to 1118 Chapter 30 Life in the Universe explain except by the presence of biology. Simple, right? In reality, the search for exoplanet life poses many challenges. As you might imagine, this task is more challenging for planetary systems that are farther away and, in practical terms, this will limit our search to the habitable worlds closest to our own. Should we become limited to a very small number of nearby targets, it will also become important to consider the habitability of planets orbiting the M-dwarfs we discussed above", ". If we manage to separate out a clean signal from the planet and find some features in the light spectrum that might be indicative of life, we will need to work hard to think of any nonbiological process that might account for them. \u201cLife is the hypothesis of last resort,\u201d noted astronomer Carl Sagan\u2014meaning that we must exhaust all other explanations for what we see before claiming to have found evidence of extraterrestrial biology. This requires some understanding of what processes might operate on worlds that we will know relatively little about; what we find on Earth can serve as a guide but also has potential to lead us astray (Figure 30.17). Recall, for example, that it would be extremely difficult to account for the abundance of oxygen in Earth\u2019s atmosphere except by the presence of biology. But it has been hypothesized that oxygen could build up to substantial levels on planets orbiting M-dwarf stars through the action of ultraviolet radiation on the atmosphere\u2014with no need for biology. It will be critical to understand where such \u201cfalse positives\u201d might exist in carrying out our search. We need to understand that we might not be able to detect biospheres even if they exist. Life has flourished on Earth for perhaps 3.5 billion years, but the atmospheric \u201cbiosignatures\u201d that, today, would supply good evidence for life to distant astronomers have not been present for all of that time. Oxygen, for example, accumulated to detectable levels in our atmosphere only a little over 2 billion years ago. Could life on Earth have been detected before that time? Scientists are working actively to understand what additional features might have provided evidence of life on Earth during that early history, and thereby help our chances of finding life beyond. Figure 30.17. Spectrum of Light Transmitted through Earth\u2019s Atmosphere. This graph shows wavelengths ranging from ultraviolet (far left) to infrared. The many downward \u201cspikes\u201d come from absorption of particular wavelengths by molecules in Earth\u2019s atmosphere. Some of these compounds, like water and the combination oxygen/ozone and methane, might reveal Earth as both habitable and inhabited. We will have to rely on this sort of information to seek life on exoplanets, but our spectra will be of much poorer quality than this one, in part because we will receive so little light from the planet. (credit: modification of work by NASA) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1119 30.4 THE SEARCH FOR EXTRATERRESTRIAL INTELLIGENCE Learning Objectives By the end of this section, you will be able to: Explain why spaceships from extraterrestrial civilizations are unlikely to have visited us List efforts by humankind to communicate with other civilizations via messages on spacecraft Understand the various SETI programs scientists are undertaking Given all the developments discussed in this chapter, it seems likely that life could have developed on many planets around other stars. Even if that life is microbial, we saw that we may soon have ways to search for chemical biosignatures. This search is of fundamental importance for understanding biology, but it does not answer the question, \u201cAre we alone?\u201d that we raised at the beginning of this chapter. When we ask this question, many people think of other intelligent creatures, perhaps beings that have developed technology similar to our own. If any intelligent, technical civilizations have arisen, as has happened on Earth in the most recent blink of cosmic time, how could we make contact with them? This problem is similar to making contact with people who live in a remote part of Earth. If students in the United States want to converse with students in Australia, for example, they have two choices. Either one group gets on an airplane and travels to meet the other, or they communicate by sending a message remotely. Given how expensive airline tickets are, most students would probably select the message route. In the same way, if we want to get in touch with intelligent life around other stars, we can travel, or we can try to exchange messages. Because of the great distances involved, interstellar space travel would be very slow and prohibitively expensive. The fastest spacecraft the human species has built so far would take almost 80,000 years to get to the nearest star. While we could certainly design a faster craft, the more quickly we require it to travel, the greater the energy cost involved. To reach neighboring stars in less than a human life span, we would have to travel close to the speed of light. In that case, however, the expense would become truly astronomical. Interstellar Travel Bernard Oliver, an engineer with an abiding interest in life elsewhere, made a revealing calculation about the costs of rapid interstellar space travel. Since we do not know what sort of technology we (or other civilizations) might someday develop, Oliver considered a trip to the nearest star (and back again) in a spaceship with a \u201cperfect engine\u201d\u2014", "one that would convert its fuel into energy with 100% efficiency. Even with a perfect engine, the energy cost of a single round-trip journey at 70% the speed of light turns out to be equivalent to several hundred thousand years\u2019 worth of total U.S. electrical energy consumption. The cost of such travel is literally out of this world. This is one reason astronomers are so skeptical about claims that UFOs are spaceships from extraterrestrial civilizations. Given the distance and energy expense involved, it seems unlikely that the dozens of UFOs (and even UFO abductions) claimed each year could be visitors from other stars so fascinated by Earth civilization that they are willing to expend fantastically large amounts of energy or time to reach us. Nor does it seem credible that these visitors have made this long and expensive journey and then systematically avoided contacting our governments or political and intellectual leaders. Not every UFO report has been explained (in many cases, the observations are sketchy or contradictory). But investigation almost always converts them to IFOs (identified flying objects) or NFOs (not-at-all flying objects). While some are hoaxes, others are natural phenomena, such as bright planets, ball lightning, fireballs (bright 1120 Chapter 30 Life in the Universe meteors), or even flocks of birds that landed in an oil slick to make their bellies reflective. Still others are human craft, such as private planes with some lights missing, or secret military aircraft. It is also interesting that the group of people who most avidly look at the night sky, the amateur astronomers, have never reported UFO sightings. Further, not a single UFO has ever left behind any physical evidence that can be tested in a laboratory and shown to be of nonterrestrial origin. Another common aspect of belief that aliens are visiting Earth comes from people who have difficulty accepting human accomplishments. There are many books and TV shows, for example, that assert that humans could not have built the great pyramids of Egypt, and therefore they must have been built by aliens. The huge statues (called Moai) on Easter Island are also sometimes claimed to have been built by aliens. Some people even think that the accomplishments of space exploration today are based on alien technology. However, the evidence from archaeology and history is clear: ancient monuments were built by ancient people, whose brains and ingenuity were every bit as capable as ours are today, even if they didn\u2019t have electronic textbooks like you do. Messages on Spacecraft While space travel by living creatures seems very difficult, robot probes can travel over long distances and over long periods of time. Five spacecraft\u2014two Pioneers, two Voyagers, and New Horizons\u2014are now leaving the solar system. At their coasting speeds, they will take hundreds of thousands or millions of years to get anywhere close to another star. On the other hand, they were the first products of human technology to go beyond our home system, so we wanted to put messages on board to show where they came from. Each Pioneer carries a plaque with a pictorial message engraved on a gold-anodized aluminum plate (Figure 30.18). The Voyagers, launched in 1977, have audio and video records attached, which allowed the inclusion of over 100 photographs and a selection of music from around the world. Given the enormous space between stars in our section of the Galaxy, it is very unlikely that these messages will ever be received by anyone. They are more like a note in a bottle thrown into the sea by a shipwrecked sailor, with no realistic expectation of its being found soon but a slim hope that perhaps someday, somehow, someone will know of the sender\u2019s fate. Figure 30.18. Interstellar Messages. (a) This is the image engraved on the plaques aboard the Pioneer 10 and 11 spacecraft. The human figures are drawn in proportion to the spacecraft, which is shown behind them. The Sun and planets in the solar system can be seen at the bottom, with the trajectory that the spacecraft followed. The lines and markings in the left center show the positions and pulse periods for a number of pulsars, which might help locate the spacecraft\u2019s origins in space and time. (b) Encoded onto a gold-coated copper disk, the Voyager record contains 118 photographs, 90 minutes of music from around the world, greetings in almost 60 languages, and other audio material. It is a summary of the sights and sounds of Earth. (credit a, b: modification of work by NASA) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1121 The Voyager Message An Excerpt from the Voyager Record: \u201cWe cast this message into the cosmos. It is likely to survive a billion years into our future, when our civilization is profoundly altered.... If [another] civilization intercepts Voyager and can understand these recorded contents, here is our message: This is a present from a small, distant world, a", " token of our sounds, our science, our images, our music, our thoughts, and our feelings. We are attempting to survive our time so we may live into yours. We hope, someday, having solved the problems we face, to join a community of galactic civilizations. This record represents our hope and our determination, and our goodwill in a vast and awesome universe.\u201d \u2014Jimmy Carter, President of the United States of America, June 16, 1977 Communicating with the Stars If direct visits among stars are unlikely, we must turn to the alternative for making contact: exchanging messages. Here the news is a lot better. We already use a messenger\u2014light or, more generally, electromagnetic waves\u2014that moves through space at the fastest speed in the universe. Traveling at 300,000 kilometers per second, light reaches the nearest star in only 4 years and does so at a tiny fraction of the cost of sending material objects. These advantages are so clear and obvious that we assume they will occur to any other species of intelligent beings that develop technology. However, we have access to a wide spectrum of electromagnetic radiation, ranging from the longest- wavelength radio waves to the shortest-wavelength gamma rays. Which would be the best for interstellar communication? It would not be smart to select a wavelength that is easily absorbed by interstellar gas and dust, or one that is unlikely to penetrate the atmosphere of a planet like ours. Nor would we want to pick a wavelength that has lots of competition for attention in our neighborhood. One final criterion makes the selection easier: we want the radiation to be inexpensive enough to produce in large quantities. When we consider all these requirements, radio waves turn out to be the best answer. Being the lowest-frequency (and lowest-energy) band of the spectrum, they are not very expensive to produce, and we already use them extensively for communications on Earth. They are not significantly absorbed by interstellar dust and gas. With some exceptions, they easily pass through Earth\u2019s atmosphere and through the atmospheres of the other planets we are acquainted with. The Cosmic Haystack Having made the decision that radio is the most likely means of communication among intelligent civilizations, we still have many questions and a daunting task ahead of us. Shall we send a message, or try to receive one? Obviously, if every civilization decides to receive only, then no one will be sending, and everyone will be disappointed. On the other hand, it may be appropriate for us to begin by listening, since we are likely to be among the most primitive civilizations in the Galaxy who are interested in exchanging messages. We do not make this statement to insult the human species (which, with certain exceptions, we are rather fond of). Instead, we base it on the fact that humans have had the ability to receive (or send) a radio message across interstellar distances for only a few decades. Compared to the ages of the stars and the Galaxy, this is a mere 1122 Chapter 30 Life in the Universe instant. If there are civilizations out there that are ahead of us in development by even a short time (in the cosmic sense), they are likely to have a technology head start of many, many years. In other words, we, who have just started, may well be the \u201cyoungest\u201d species in the Galaxy with this capability (see the discussion in Example 30.1). Just as the youngest members of a community are often told to be quiet and listen to their elders for a while before they say something foolish, so may we want to begin our exercise in extraterrestrial communication by listening. Even restricting our activities to listening, however, leaves us with an array of challenging questions. For example, if an extraterrestrial civilization\u2019s signal is too weak to be detected by our present-day radio telescopes, we will not detect them. In addition, it would be very expensive for an extraterrestrial civilization to broadcast on a huge number of channels. Most likely, they select one or a few channels for their particular message. Communicating on a narrow band of channels also helps distinguish an artificial message from the radio static that comes from natural cosmic processes. But the radio band contains an astronomically large number of possible channels. How can we know in advance which one they have selected, and how they have coded their message into the signal? Table 30.1 summarizes these and other factors that scientists must grapple with when trying to tune in to radio messages from distant civilizations. Because their success depends on either guessing right about so many factors or searching through all the possibilities for each factor, some scientists have compared their quest to looking for a needle in a haystack. Thus, they like to say that the list of factors in Table 30.1 defines the cosmic haystack problem. Factors From which direction (which star) is the message coming? On what channels (or frequencies) is the message being broadcast? How wide in frequency is the channel? How strong is the signal (can our radio telescopes detect it)? Is the signal continuous, or does it shut off at times (as, for example", ", a lighthouse beam does when it turns away from us)? Does the signal drift (change) in frequency because of the changing relative motion of the source and the receiver? How is the message encoded in the signal (how do we decipher it)? Can we even recognize a message from a completely alien species? Might it take a form we don\u2019t at all expect? Table 30.1 The Cosmic Haystack Problem: Some Questions about an Extraterrestrial Message Radio Searches Although the cosmic haystack problem seems daunting, many other research problems in astronomy also require a large investment of time, equipment, and patient effort. And, of course, if we don\u2019t search, we\u2019re sure not to find anything. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1123 The very first search was conducted by astronomer Frank Drake in 1960, using the 85-foot antenna at the National Radio Astronomy Observatory (Figure 30.19). Called Project Ozma, after the queen of the exotic Land of Oz in the children\u2019s stories of L. Frank Baum, his experiment involved looking at about 7200 channels and two nearby stars over a period of 200 hours. Although he found nothing, Drake demonstrated that we had the technology to do such a search, and set the stage for the more sophisticated projects that followed. Figure 30.19. Project Ozma and the Allen Telescope Array. (a) This 25th anniversary photo shows some members of the Project Ozma team standing in front of the 85-foot radio telescope with which the 1960 search for extraterrestrial messages was performed. Frank Drake is in the back row, second from the right. (b) The Allen Telescope Array in California is made up of 42 small antennas linked together. This system allows simultaneous observations of multiple sources with millions of separate frequency channels. (credit a: modification of work by NRAO; credit b: modification of work by Colby Gutierrez-Kraybill) Receivers are constantly improving, and the sensitivity of SETI programs\u2014 SETI stands for the search for extraterrestrial life\u2014is advancing rapidly. Equally important, modern electronics and software allow simultaneous searches on millions of frequencies (channels). If we can thus cover a broad frequency range, the cosmic haystack problem of guessing the right frequency largely goes away. One powerful telescope array (funded with an initial contribution from Microsoft founder Paul Allen) that is built for SETI searches is the Allen Telescope in Northern California. Other radio telescopes being used for such searches include the giant Arecibo radio dish in Puerto Rico and the Green Bank Telescope in West Virginia, which is the largest steerable radio telescope in the world. What kind of signals do we hope to pick up? We on Earth are inadvertently sending out a flood of radio signals, dominated by military radar systems. This is a kind of leakage signal, similar to the wasted light energy that is beamed upward by poorly designed streetlights and advertising signs. Could we detect a similar leakage of radio signals from another civilization? The answer is just barely, but only for the nearest stars. For the most part, therefore, current radio SETI searches are looking for beacons, assuming that civilizations might be intentionally drawing attention to themselves or perhaps sending a message to another world or outpost that lies in our direction. Our prospects for success depend on how often civilizations arise, how long they last, and how patient they are about broadcasting their locations to the cosmos. 1124 Chapter 30 Life in the Universe Jill Tarter: Trying to Make Contact 1997 was quite a year for Jill Cornell Tarter (Figure 30.20), one of the world\u2019s leading scientists in the SETI field. The SETI Institute announced that she would be the recipient of its first endowed chair (the equivalent of an endowed research professorship) named in honor of Bernard Oliver. The National Science Foundation approved a proposal by a group of scientists and educators she headed to develop an innovative hands-on high school curriculum based on the ideas of cosmic evolution (the topics of this chapter). And, at roughly the same time, she was being besieged with requests for media interviews as news reports identified her as the model for Ellie Arroway, the protagonist of Contact, Carl Sagan\u2019s best- selling novel about SETI. The book had been made into a high-budget science fiction film, starring Jodie Foster, who had talked with Tarter before taking the role. Figure 30.20. Jill Tarter (credit: Christian Schidlowski) Tarter is quick to point out, \u201cCarl Sagan wrote a book about a woman who does what I do, not about me.\u201d Still, as the only woman in such a senior position in the small field of SETI, she was the center of a great deal of public attention. (However, colleagues and reporters pointed out that this was nothing compared to what would happen if her search for radio signals from other civilizations", " recorded a success.) Being the only woman in a group is not a new situation to Tarter, who often found herself the only woman in her advanced science or math classes. Her father had encouraged her, both in her interest in science and her \u201ctinkering.\u201d As an undergraduate at Cornell University, she majored in engineering physics. That training became key to putting together and maintaining the complex systems that automatically scan for signals from other civilizations. Switching to astrophysics for her graduate studies, she wrote a PhD thesis that, among other topics, considered the formation of failed stars\u2014those whose mass was not sufficient to ignite the nuclear reactions that power more massive stars like our own Sun. Tarter coined the term \u201cbrown dwarf\u201d for these small, dim objects, and it has remained the name astronomers use ever since. It was while she was still in graduate school that Stuart Bowyer, one of her professors at the University of California, Berkeley, asked her if she wanted to be involved in a small experiment to siphon off a bit of radiation from a radio telescope as astronomers used it year in and year out and see if there was any hint This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1125 of an intelligently coded radio message buried in the radio noise. Her engineering and computer programming skills became essential to the project, and soon she was hooked on the search for life elsewhere. Thus began an illustrious career working full time searching for extraterrestrial civilizations, leading Jill Tarter to receive many awards, including being elected fellow of the American Association for the Advancement of Science in 2002, the Adler Planetarium Women in Space Science Award in 2003, and a 2009 TED Prize, among others Watch the TED talk (https://openstax.org/l/30TarterSETI) Jill Tarter gave on the fascination of the search for intelligence The Drake Equation At the first scientific meeting devoted to SETI, Frank Drake wrote an equation on the blackboard that took the difficult question of estimating the number of civilizations in the Galaxy and broke it down into a series of smaller, more manageable questions. Ever since then, both astronomers and students have used this Drake equation as a means of approaching the most challenging question: How likely is it that we are alone? Since this is at present an unanswerable question, astronomer Jill Tarter has called the Drake equation a \u201cway of organizing our ignorance.\u201d (See Figure 30.21.) Figure 30.21. Drake Equation. A plaque at the National Radio Astronomy Observatory commemorates the conference where the equation was first discussed. (credit: NRAO/NSF/AUI) The form of the Drake equation is very simple. To estimate the number of communicating civilizations that currently exist in the Galaxy (we will define these terms more carefully in a moment), we multiply the rate of formation of such civilizations (number per year) by their average lifetime (in years). In symbols, 1126 Chapter 30 Life in the Universe N = Rtotal \u00d7 L To make this formula easier to use (and more interesting), however, Drake separated the rate of formation Rtotal into a series of probabilities: Rtotal = Rstar \u00d7 fp \u00d7 fe \u00d7 fl \u00d7 fi \u00d7 fc Rstar is the rate of formation of stars like the Sun in our Galaxy, which is about 10 stars per year. Each of the other terms is a fraction or probability (less than or equal to 1.0), and the product of all these probabilities is itself the total probability that each star will have an intelligent, technological, communicating civilization that we might want to talk to. We have: \u2022 \u2022 \u2022 \u2022 \u2022 fp = the fraction of these stars with planets fe = the fraction of the planetary systems that include habitable planets fl = the fraction of habitable planets that actually support life fi = the fraction of inhabited planets that develop advanced intelligence fc = the fraction of these intelligent civilizations that develop science and the technology to build radio telescopes and transmitters Each of these factors can be discussed and perhaps evaluated, but we must guess at many of the values. In particular, we don\u2019t know how to calculate the probability of something that happened once on Earth but has not been observed elsewhere\u2014and these include the development of life, of intelligent life, and of technological life (the last three factors in the equation). One important advance in estimating the terms of the Drake equation comes from the recent discovery of exoplanets. When the Drake equation was first written, no one had any idea whether planets and planetary systems were common. Now we know they are\u2014another example of the Copernican principle. Solution Even if we don\u2019t know the answers, we can make some guesses and calculate the resulting number N. Let\u2019s start with the optimism implicit in the Copernican principle and set the last three terms equal to 1.0. If R is 10 stars/year and if", " we measure the average lifetime of a technological civilization in years, the units of years cancel. If we also assume that fp is 0.1, and fe is 1.0, the equation becomes N = Rtotal \u00d7 L = L Now we see the importance of the term L, the lifetime of a communicating civilization (measured in years). We have had this capability (to communicate at the distances of the stars) for only a few decades. Check Your Learning Suppose we assume that this stage in our history lasts only one century. Answer: With our optimistic assumptions about the other factors, L = 100 years and N = 100 such civilizations in the entire Galaxy. In that case, there are so few other civilizations like ours that we are unlikely to detect any signals in a SETI search. But suppose the average lifetime is a million years; in that case, there are a million such civilizations in the Galaxy, and some of them may be within range for radio communication. The most important conclusion from this calculation is that even if we are extremely optimistic about the This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1127 probabilities, the only way we can expect success from SETI is if other civilizations are much older (and hence probably much more advanced) than ours Read Frank Drake\u2019s own account (https://openstax.org/l/30drakeequat) of how he came up with his \u201cequation.\u201d And here is a recent interview (https://openstax.org/l/30frandrakinter) with Frank Drake by one of the authors of this textbook. SETI outside the Radio Realm For the reasons discussed above, most SETI programs search for signals at radio wavelengths. But in science, if there are other approaches to answering an unsolved question, we don\u2019t want to neglect them. So astronomers have been thinking about other ways we could pick up evidence for the existence of technologically advanced civilizations. Recently, technology has allowed astronomers to expand the search into the domain of visible light. You might think that it would be hopeless to try to detect a flash of visible light from a planet given the brilliance of the star it orbits. This is why we usually cannot measure the reflected light of planets around other stars. The feeble light of the planet is simply swamped by the \u201cbig light\u201d in the neighborhood. So another civilization would need a mighty strong beacon to compete with their star. However, in recent years, human engineers have learned how to make flashes of light brighter than the Sun. The trick is to \u201cturn on\u201d the light for a very brief time, so that the costs are manageable. But ultra-bright, ultra- short laser pulses (operating for periods of a billionth of a second) can pack a lot of energy and can be coded to carry a message. We also have the technology to detect such short pulses\u2014not with human senses, but with special detectors that can be \u201ctuned\u201d to hunt automatically for such short bursts of light from nearby stars. Why would any civilization try to outshine its own star in this way? It turns out that the cost of sending an ultra- short laser pulse in the direction of a few promising stars can be less than the cost of sweeping a continuous radio message across the whole sky. Or perhaps they, too, have a special fondness for light messages because one of their senses evolved using light. Several programs are now experimenting with \u201c optical SETI\u201d searches, which can be done with only a modest telescope. (The term optical here means using visible light.) If we let our imaginations expand, we might think of other possibilities. What if a truly advanced civilization should decide to (or need to) renovate its planetary system to maximize the area for life? It could do so by breaking apart some planets or moons and building a ring of solid material that surrounds or encloses the star and intercepts some or all of its light. This huge artificial ring or sphere might glow very brightly at infrared wavelengths, as the starlight it receives is eventually converted to heat and re-radiated into space. That infrared radiation could be detected by our instruments, and searches for such infrared sources are also underway (Figure 30.22). 1128 Chapter 30 Life in the Universe Figure 30.22. Wide-Field Infrared Survey Explorer (WISE). Astronomers have used this infrared satellite to search for infrared signatures of enormous construction projects by very advanced civilizations, but their first survey did not reveal any. (credit: modification of work by NASA/ JPL-Caltech) Should We Transmit in Addition to Listening? Our planet has some leakage of radio waves into space, from FM radio, television, military radars, and communication between Earth and our orbiting spacecraft. However, such leakage radiation is still quite weak, and therefore difficult to detect at the distances of the stars,", " at least with the radio technology we have. So at the present time our attempts to communicate with other civilizations that may be out there mostly involve trying to receive messages, but not sending any ourselves. Some scientists, however, think that it is inconsistent to search for beacons from other civilizations without announcing our presence in a similar way. (We discussed earlier the problem that if every other civilization confined itself to listening, no one would ever get in touch.) So, should we be making regular attempts at sending easily decoded messages into space? Some scientists warn that our civilization is too immature and defenseless to announce ourselves at this early point in our development. The decision whether to transmit or not turns out to be an interesting reflection of how we feel about ourselves and our place in the universe. Discussions of transmission raise the question of who should speak for planet Earth. Today, anyone and everyone can broadcast radio signals, and many businesses, religious groups, and governments do. It would be a modest step for the same organizations to use or build large radio telescopes and begin intentional transmissions that are much stronger than the signals that leak from Earth today. And if we intercept a signal from an alien civilization, then the issue arises whether to reply. Who should make the decision about whether, when, and how humanity announces itself to the cosmos? Is there freedom of speech when it comes to sending radio messages to other civilizations? Do all the nations of Earth have to agree before we send a signal strong enough that it has a serious chance of being received at the distances of the stars? How our species reaches a decision about these kinds of questions may well be a test of whether or not there is intelligent life on Earth. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe Conclusion 1129 Whether or not we ultimately turn out to be the only intelligent species in our part of the Galaxy, our exploration of the cosmos will surely continue. An important part of that exploration will still be the search for biomarkers from inhabited planets that have not produced technological creatures that send out radio signals. After all, creatures like butterflies and dolphins may never build radio antennas, but we are happy to share our planet with them and would be delighted to find their counterparts on other worlds. Whether or not life exists elsewhere is just one of the unsolved problems in astronomy that we have discussed in this book. A humble acknowledgment of how much we have left to learn about the universe is one of the fundamental hallmarks of science. This should not, however, prevent us from feeling exhilarated about how much we have already managed to discover, and feeling curious about what else we might find out in the years to come. Our progress report on the ideas of astronomy ends here, but we hope that your interest in the universe does not. We hope you will keep up with developments in astronomy through media and online, or by going to an occasional public lecture by a local scientist. Who, after all, can even guess all the amazing things that future research projects will reveal about both the universe and our connection with it? 1130 Chapter 30 Life in the Universe CHAPTER 30 REVIEW KEY TERMS amino acids organic compounds that are the molecular building blocks of proteins astrobiology the multidisciplinary study of life in the universe: its origin, evolution, distribution, and fate; similar terms are exobiology and bioastronomy biomarker evidence of the presence of life, especially a global indication of life on a planet that could be detected remotely (such as an unusual atmospheric composition) DNA (deoxyribonucleic acid) a molecule that stores information about how to replicate a cell and its chemical and structural components Drake equation a formula for estimating the number of intelligent, technological civilizations in our Galaxy, first suggested by Frank Drake extremophile an organism (usually a microbe) that tolerates or even thrives under conditions that most of the life around us would consider hostile, such as very high or low temperature or acidity gene the basic functional unit that carries the genetic (hereditary) material contained in a cell habitable environment an environment capable of hosting life habitable zone the region around a star in which liquid water could exist on the surface of terrestrial-sized planets, hence the most probable place to look for life in a star\u2019s planetary system organic compound a compound containing carbon, especially a complex carbon compound; not necessarily produced by life organic molecule a combination of carbon and other atoms\u2014primarily hydrogen, oxygen, nitrogen, phosphorus, and sulfur\u2014some of which serve as the basis for our biochemistry photosynthesis a complex sequence of chemical reactions through which some living things can use sunlight to manufacture products that store energy (such as carbohydrates), releasing oxygen as one by-product protein a key biological molecule that provides the structure and function of the body\u2019s tissues and organs, and essentially carries out the chemical work of the cell RNA (ribonucleic acid) a molecule that aids in the flow of genetic information from DNA to proteins SETI", " the search for extraterrestrial intelligence; usually applied to searches for radio signals from other civilizations stromatolites solid, layered rock formations that are thought to be the fossils of oxygen-producing photosynthetic bacteria in rocks that are 3.5 billion years old thermophile an organism that can tolerate high temperatures This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1131 SUMMARY 30.1 The Cosmic Context for Life Life on Earth is based on the presence of a key unit known as an organic molecule, a molecule that contains carbon, especially complex hydrocarbons. Our solar system formed about 5 billion years ago from a cloud of gas and dust enriched by several generations of heavier element production in stars. Life is made up of chemical combinations of these elements made by stars. The Copernican principle, which suggests that there is nothing special about our place in the universe, implies that if life could develop on Earth, it should be able to develop in other places as well. The Fermi paradox asks why, if life is common, more advanced life-forms have not contacted us. 30.2 Astrobiology The study of life in the universe, including its origin on Earth, is called astrobiology. Life as we know it requires water, certain elemental raw materials (carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur), energy, and an environment in which the complex chemistry of life is stable. Carbon-based (or organic) molecules are abundant in space and may also have been produced by processes on Earth. Life appears to have spread around our planet within 400 million years after the end of heavy bombardment, if not sooner. The actual origin of life\u2014the processes leading from chemistry to biology\u2014is not completely understood. Once life took hold, it evolved to use many energy sources, including first a range of different chemistries and later light, and diversified across a range of environmental conditions that humans consider \u201cextreme.\u201d This proliferation of life into so many environmental niches, so relatively soon after our planet became habitable, has served to make many scientists optimistic about the chances that life could exist elsewhere. 30.3 Searching for Life beyond Earth The search for life beyond Earth offers several intriguing targets. Mars appears to have been more similar to Earth during its early history than it is now, with evidence for liquid water on its ancient surface and perhaps even now below ground. The accessibility of the martian surface to our spacecraft offers the exciting potential to directly examine ancient and modern samples for evidence of life. In the outer solar system, the moons Europa and Enceladus likely host vast sub-ice oceans that may directly contact the underlying rocks\u2014a good start in providing habitable conditions\u2014while Titan offers a fascinating laboratory for understanding the sorts of organic chemistry that might ultimately provide materials for life. And the last decade of research on exoplanets leads us to believe that there may be billions of habitable planets in the Milky Way Galaxy. Study of these worlds offers the potential to find biomarkers indicating the presence of life. 30.4 The Search for Extraterrestrial Intelligence Some astronomers are engaged in the search for extraterrestrial intelligent life (SETI). Because other planetary systems are so far away, traveling to the stars is either very slow or extremely expensive (in terms of energy required). Despite many UFO reports and tremendous media publicity, there is no evidence that any of these are related to extraterrestrial visits. Scientists have determined that the best way to communicate with any intelligent civilizations out there is by using electromagnetic waves, and radio waves seem best suited to the task. So far, they have only begun to comb the many different possible stars, frequencies, signal types, and other factors that make up what we call the cosmic haystack problem. Some astronomers are also undertaking searches for brief, bright pulses of visible light and infrared signatures of huge construction projects by advanced civilizations. If we do find a signal someday, deciding whether to answer and what to answer may be two of the greatest challenges humanity will face. 1132 Chapter 30 Life in the Universe FOR FURTHER EXPLORATION Articles Astrobiology Chyba, C. \u201cThe New Search for Life in the Universe.\u201d Astronomy (May 2010): 34. An overview of astrobiology and the search for life out there in general, with a brief discussion of the search for intelligence. Dorminey, B. \u201cA New Way to Search for Life in Space.\u201d Astronomy (June 2014): 44. Finding evidence of photosynthesis on other worlds. McKay, C., & Garcia, V. \u201cHow to Search for Life on Mars.\u201d Scientific American (June 2014): 44\u201349. Experiments future probes could perform. Reed, N. \u201cWhy We Haven\u2019t Found Another Earth Yet.\u201d Astronomy (February 2016): 25. On the search for smaller earthlike planets in their star\u2019s habitable zones, and", " where we stand. Shapiro, R. \u201cA Simpler Origin of Life.\u201d Scientific American (June 2007): 46. New ideas about what kind of molecules formed first so life could begin. Simpson, S. \u201cQuestioning the Oldest Signs of Life.\u201d Scientific American (April 2003): 70. On the difficulty of interpreting biosignatures in rocks and the implications for the search for life on other worlds. SETI Chandler, D. \u201cThe New Search for Alien Intelligence.\u201d Astronomy (September 2013): 28. Review of various ways of finding other civilizations out there, not just radio wave searches. Crawford, I. \u201cWhere Are They?\u201d Scientific American (July 2000): 38. On the Fermi paradox and its resolutions, and on galactic colonization models. Folger, T. \u201cContact: The Day After.\u201d Scientific American (January 2011): 40\u201345. Journalist reports on efforts to prepare for ET signals; protocols and plans for interpreting messages; and discussions of active SETI. Kuhn, J., et al. \u201cHow to Find ET with Infrared Light.\u201d Astronomy (June 2013): 30. On tracking alien civilizations by the heat they put out. Lubick, N. \u201cAn Ear to the Stars.\u201d Scientific American (November 2002): 42. Profile of SETI researcher Jill Tarter. Nadis, S. \u201cHow Many Civilizations Lurk in the Cosmos?\u201d Astronomy (April 2010): 24. New estimates for the terms in the Drake equation. Shostak, S. \u201cClosing in on E.T.\u201d Sky & Telescope (November 2010): 22. Nice summary of current and proposed efforts to search for intelligent life out there. Websites Astrobiology Astrobiology Web: http://astrobiology.com/. A news site with good information and lots of material. Exploring Life\u2019s Origins: http://exploringorigins.org/index.html. A website for the Exploring Origins Project, part of the multimedia exhibit of the Boston Museum of Science. Explore the origin of life on Earth with an interactive timeline, gain a deeper knowledge of the role of RNA, \u201cbuild\u201d a cell, and explore links to learn more about astrobiology and other related information. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1133 History of Astrobiology: https://astrobiology.nasa.gov/about/history-of-astrobiology/. By Marc Kaufman, on the NASA Astrobiology site. Life, Here and Beyond: https://astrobiology.nasa.gov/about/. By Marc Kaufman, on the NASA Astrobiology site. SETI Berkeley SETI Research Center: https://seti.berkeley.edu/. The University of California group recently received a $100 million grant from a Russian billionaire to begin the Breakthrough: Listen project. Fermi Paradox: http://www.seti.org/seti-institute/project/details/fermi-paradox. Could we be alone in our part of the Galaxy or, more dramatic still, could we be the only technological society in the universe? A useful discussion. Planetary Society: http://www.planetary.org/explore/projects/seti/. This advocacy group for exploration has several pages devoted to the search for life. SETI Institute: http://www.seti.org. A key organization in the search for life in the universe; the institute\u2019s website is full of information and videos about both astrobiology and SETI. SETI: http://www.skyandtelescope.com/tag/seti/. Sky & Telescope magazine offers good articles on this topic. Videos Astrobiology Copernicus Complex: Are We Special in the Cosmos?: https://www.youtube.com/watch?v=ERp0AHYRm_Q. A video of a popular-level talk by Caleb Scharf of Columbia University (1:18:54). Life at the Edge: Life in Extreme Environments on Earth and the Search for Life in the Universe: https://www.youtube.com/watch?v=91JQmTn0SF0. A video of a 2009 nontechnical lecture by Lynn Rothschild of NASA Ames Research Center (1:31:21). Saturn\u2019s Moon Titan: A World with Rivers, Lakes, and Possibly Even Life: https://www.youtube.com/ watch?v=bbkTJeHoOKY. A video of a 2011 talk by Chris McKay of NASA Ames Research Center (1:23:33). SETI Allen Telescope Array: The Newest Pitchfork for Exploring the Cosmic Haystack: https://www.youtube.com/ watch?v=aqsI1HZCgUM", ". A 2013 popular-level lecture by Jill Tarter of the SETI Institute (1:45:55). Confessions of an Alien Hunter: http://fora.tv/2009/03/31/Seth_Shostak_Confessions_of_an_Alien_Hunter. 2009 interview with Seth Shostak on FORA TV (36:27). Search for Extra-Terrestrial Intelligence: Necessarily a Long-Term Strategy: http://www.longnow.org/seminars/ 02004/jul/09/the-search-for-extra-terrestrial-intelligence-necessarily-a-long-term-strategy/. 2004 talk by Jill Tarter at the Long Now Foundation (1:21:13). Search for Intelligent Life Among the Stars: New Strategies: https://www.youtube.com/ watch?v=m9WxW2ktcKU. A 2010 nontechnical talk by Seth Shostak of the SETI Institute (1:29:58). COLLABORATIVE GROUP ACTIVITIES A. If one of the rocks from Mars examined by a future mission to the red planet does turn out to have unambiguous signs of ancient life that formed on Mars, what does your group think would be the 1134 Chapter 30 Life in the Universe implications of such a discovery for science and for our view of life elsewhere? Would such a discovery have any long-term effects on your own thinking? B. Suppose we receive a message from an intelligent civilization around another star. What does your group think the implications of this discovery would be? How would your own thinking or personal philsophy be affected by such a discovery? C. A radio message has been received from a civilization around a star 40 light-years away, which contains (in pictures) quite a bit of information about the beings that sent the message. The president of the United States has appointed your group a high-level commission to advise whether humanity should answer the message (which was not particularly directed at us, but comes from a beacon that, like a lighthouse, sweeps out a circle in space). How would you advise the president? Does your group agree on your answer or do you also have a minority view to present? D. If there is no evidence that UFOs are extraterrestrial visitors, why does your group think that television shows, newspapers, and movies spend so much time and effort publicizing the point of view that UFOs are craft from other worlds? Make a list of reasons. Who stands to gain by exaggerating stories of unknown lights in the sky or simply fabricating stories that alien visitors are already here? E. Does your group think scientists should simply ignore all the media publicity about UFOs or should they try to respond? If so, how should they respond? Does everyone in the group agree? F. Suppose your group is the team planning to select the most important sights and sounds of Earth to record and put on board the next interstellar spacecraft. What pictures (or videos) and sounds would you include to represent our planet to another civilization? G. Let\u2019s suppose Earth civilization has decided to broadcast a message announcing our existence to other possible civilizations among the stars. Your group is part of a large task force of scientists, communications specialists, and people from the humanities charged with deciding the form and content of our message. What would you recommend? Make a list of ideas. H. Think of examples of contact with aliens you have seen in movies and on TV. Discuss with your group how realistic these have been, given what you have learned in this class. Was the contact in person (through traveling) or using messages? Why do you think Hollywood does so many shows and films that are not based on our scientific understanding of the universe? I. Go through the Drake equation with your group and decide on values for each factor in the estimate. (If you disagree on what a factor should be within the group, you can have a \u201cminority report.\u201d) Based on the factors, how many intelligent, communicating civilizations do you estimate to be thriving in our Galaxy right now? EXERCISES Review Questions 1. What is the Copernican principle? Make a list of scientific discoveries that confirm it. 2. Where in the solar system (and beyond) have scientists found evidence of organic molecules? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Chapter 30 Life in the Universe 1135 3. Give a short history of the atoms that are now in your little finger, going back to the beginning of the universe. 4. What is a biomarker? Give some possible examples of biomarkers we might look for beyond the solar system. 5. Why are Mars and Europa the top targets for the study of astrobiology? 6. Why is traveling between the stars (by creatures like us) difficult? 7. What are the advantages to using radio waves for communication between civilizations that live around different", " stars? List as many as you can. 8. What is the \u201ccosmic haystack problem\u201d? List as many of its components as you can think of. 9. What is a habitable zone? 10. Why is the simultaneous detection of methane and oxygen in an atmosphere a good indication of the existence of a biosphere on that planet? 11. What are two characteristic properties of life that distinguish it from nonliving things? 12. What are the three requirements that scientists believe an environment needs to supply life with in order to be considered habitable? 13. Can you name five environmental conditions that, in their extremes, microbial life has been challenged by and has learned to survive on Earth? Thought Questions 14. Would a human have been possible during the first generation of stars that formed right after the Big Bang? Why or why not? 15. If we do find life on Mars, what might be some ways to check whether it formed separately from Earth life, or whether exchanges of material between the two planets meant that the two forms of life have a common origin? 16. What kind of evidence do you think would convince astronomers that an extraterrestrial spacecraft has landed on Earth? 17. What are some reasons that more advanced civilizations might want to send out messages to other star systems? 18. What are some answers to the Fermi paradox? Can you think of some that are not discussed in this chapter? 19. Why is there so little evidence of Earth\u2019s earliest history and therefore the period when life first began on our planet? 20. Why was the development of photosynthesis a major milestone in the evolution of life? 21. Does all life on Earth require sunshine? 22. Why is life unlikely to be found on the surface of Mars today? 23. In this chapter, we identify these characteristic properties of life: life extracts energy from its environment, and has a means of encoding and replicating information in order to make faithful copies of itself. Does this definition fully capture what we think of as \u201clife\u201d? How might our definition be biased by our terrestrial environment? 1136 Chapter 30 Life in the Universe 24. Given that no sunlight can penetrate Europa\u2019s ice shell, what would be the type of energy that could make some form of europan life possible? 25. Why is Saturn\u2019s moon Enceladus such an exciting place to send a mission? 26. In addition to an atmosphere dominated by nitrogen, how else is Saturn\u2019s moon Titan similar to Earth? 27. How can a planet\u2019s atmosphere affect the width of the habitable zone in its planetary system? 28. Why are we limited to finding life on planets orbiting other stars to situations where the biosphere has created planet-scale changes? Figuring For Yourself 29. Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 lightyears away. Their message encourages us to send a radio answer, which we decide to do. Suppose our governing bodies take 2 years to decide whether and how to answer. When our answer arrives there, their governing bodies also take two of our years to frame an answer to us. How long after we get their first message can we hope to get their reply to ours? (A question for further thinking: Once communication gets going, should we continue to wait for a reply before we send the next message?) 30. The light a planet receives from the Sun (per square meter of planet surface) decreases with the square of the distance from the Sun. So a planet that is twice as far from the Sun as Earth receives (1/2)2 = 0.25 times (25%) as much light and a planet that is three times as far from the Sun receives (1/3)2 = 0.11 times (11%) as much light. How much light is received by the moons of Jupiter and Saturn (compared to Earth), worlds which orbit 5.2 and 9.5 times farther from the Sun than Earth? 31. Think of our Milky Way Galaxy as a flat disk of diameter 100,000 light-years. Suppose we are one of 1000 civilizations, randomly distributed through the disk, interested in communicating via radio waves. How far away would the nearest such civilization be from us (on average)? This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix A 1137 A HOW TO STUDY FOR AN INTRODUCTORY ASTRONOMY CLASS In this brief appendix, we want to give you some hints for the effective study of astronomy. These suggestions are based on ideas from good teachers and good students around the United States. Your professor will probably have other, more specific suggestions for doing well in your class. Astronomy, the study of the universe beyond the borders of our planet, is one of the most exciting and rapidly changing branches of science. Even scientists from other fields often confess to having had a lifelong interest in astronomy, though they may now be doing something earthbound\u2014like biology,", " chemistry, engineering, or writing software. But some of the things that make astronomy so interesting also make it a challenge for the beginning student. The universe is a big place, full of objects and processes that do not have familiar counterparts here on Earth. Like a visitor to a new country, it will take you a while to feel familiar with the territory or the local customs. Astronomy, like other sciences, also has its own special vocabulary, some of which you will have to learn to communicate well with your professor and classmates. Still, hundreds of thousands of non-science majors take an introductory astronomy course every year, and surveys show that students from a wide range of backgrounds have succeeded in (and even enjoyed) these classes. Astronomy is for everyone, not just those who are \u201cscience oriented.\u201d So, here are some suggestions to help you increase your chances of doing well in your astronomy class. 1. The best advice we can give you is to be sure to leave enough time in your schedule to study the material in this class regularly. It sounds obvious, but it is not very easy to catch up with a subject like astronomy by trying to do everything just before an exam. (As astronomers like to put it, you can\u2019t learn the whole universe in one night!) Try to put aside some part of each day, or every other day, when you can have uninterrupted time for reading and studying astronomy. 2. In class, put your phone away and focus on the class activities. If you have to use a laptop or tablet in class, make a pact with yourself that you will not check email, get on social media, or play games during class. A number of careful studies of student behavior and grades have shown that students are not as good at such multi-tasking as they think they are, and that students who do not use screens during class get significantly better grades in the end. 3. Try to take careful notes during class. Many students start college without good note-taking habits. If you are not a good note-taker, try to get some help. Many colleges and universities have student learning centers that offer short courses, workbooks, tutors, or videos on developing good study habits. Good note- taking skills will also be useful for many jobs or activities you are likely get involved with after college. 4. Try to read each assignment in the textbook twice, once before it is discussed in class, and once afterward. Take notes as you read or use a highlighter to outline ideas that you may want to review later. 5. Form a small astronomy study group with people in your class. Get together with them regularly and discuss what you have been learning. Also, focus on the topics that may be giving group members trouble. Make up sample exam questions and make sure everyone in the group can answer them confidently. If you have always studied alone, you may at first resist this idea, but don\u2019t be too hasty to say no. Study groups are a very effective way to digest a large amount of new information. 6. Before each exam, create a concise outline of the main ideas discussed in class and presented in your text. Compare your outline with those of other students as a check on your own study habits. 7. If your professor suggests doing web-based sample quizzes, or looking at online apps, animations, or study guides, take advantage of these resources to enhance your studying. 1138 Appendix A 8. At the end of each chapter in this textbook you, will find four kinds of questions. The Collaborative Group Activities are designed to encourage you follow up on the material in the chapter as a group, rather than individually. Review Questions help you see if you have learned the material in the chapter. Thought Questions test deeper understanding by asking you to apply your knowledge to new situations. And Figuring for Yourself exercises test and extend some of the mathematical examples in the chapter. (Not all professors will use the math sections; if they don\u2019t, you may not have homework from this section.) 9. If you find a topic in the text or in class especially difficult or interesting, talk to your professor or teaching assistant. Many students are scared to show their ignorance in front of their teacher, but we can assure you that most professors and TA\u2019s like it when students come to office hours and show that they care enough about the course to ask for help. 10. Don\u2019t stay up all night before a test and then expect your mind to respond well. For the same reason, don\u2019t eat a big meal just before a test, since we all get a little sleepy and don\u2019t think as clearly after a big meal. Take many deep breaths and try to relax during the test itself. 11. Don\u2019t be too hard on yourself! If astronomy is new to you, many of the ideas and terms in this book may be unfamiliar. Astronomy is like any new language: it may take a while to become a good conversational", "ist. Practice as much as you can, but also realize that it is natural to feel overwhelmed by the vastness of the universe and the variety of things that are going on in it. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix B 1139 B ASTRONOMY WEBSITES, IMAGES, AND APPS Throughout the textbook, we suggest useful resources for students on the specific topics in a given chapter. Here, we offer some websites for exploring astronomy in general, plus good sites for viewing and downloading the best astronomy images, and guides to astronomical apps for smartphones and tablets. This is not an exhaustive listing, but merely a series of suggestions to whet the appetite of those wanting to go beyond the textbook. Websites For Exploring Astronomy In General Astronomical Organizations Amateur Astronomy Clubs. In most large cities and a number of rural areas, there are amateur astronomy clubs, where those interested in the hobby of astronomy gather to observe the sky, share telescopes, hear speakers, and help educate the public about the night sky. To find an astronomy club near you, you can try the following sites: Night Sky Network club finder: http://nightsky.jpl.nasa.gov/club-map.cfm. Sky & Telescope Magazine astronomy clubs and organizations: http://www.skyandtelescope.com/ community/organizations. Astronomy Magazine club finder: http://www.astronomy.com/groups.aspx. Astronomical League astronomy clubs and societies: http://www.astroleague.org/societies/all. Go-Astronomy club search: http://www.go-astronomy.com/astro-club-search.htm. American Astronomical Society: http://www.aas.org. Composed mainly of professional astronomers. They have an active education office and various materials for students and the public on the education pages of their website. Astronomical League: http://www.astroleague.org. The league is the umbrella organization of American astronomy clubs. They offer a newsletter, national observing programs, and support for how to form and support a club. Astronomical Society of the Pacific: http://www.astrosociety.org. Founded in 1889, this international society is devoted to astronomy education and outreach. They have programs, publications, and materials for families, teachers, amateur astronomers, museum guides, and anyone interested in astronomy. European Space Agency (ESA): http://www.esa.int/. Information on European space missions with an excellent gallery of images. International Astronomical Union (IAU): http://www.iau.org/. International organization for professional astronomers; see the menu choice \u201cIAU for the Public\u201d for information on naming astronomical objects and other topics of interest to students. International Dark-Sky Association: http://www.darksky.org. Dedicated to combating light pollution, the encroachment of stray light that wastes energy and washes out the glories of the night sky. NASA: http://www.nasa.gov. NASA has a wide range of information on its many websites; the trick is to find what you need. Most space missions and NASA centers have their own sites. Planetary Society: http://www.planetary.org. Founded by the late Carl Sagan and others, this group works to encourage planetary exploration and the search for life elsewhere. While much of their work is advocacy, they have some educational outreach too. 1140 Appendix B Royal Astronomical Society of Canada: http://www.rasc.ca/. Unites professional and amateur astronomers around Canada; has 28 centers with local activities, plus national magazines and meetings. Some Astronomical Publications Students Can Read Astronomy Now: http://www.astronomynow.com/. A colorful British monthly, with excellent articles about astronomy, the history of astronomy, and stargazing. Astronomy: http://www.astronomy.com. Has the largest circulation of any magazine devoted to the universe and is designed especially for astronomy hobbyists and armchair astronomers. Free Astronomy: http://www.astropublishing.com/. A new web-based publication, with European roots. Scientific American: http://www.sciam.com. Offers one astronomy article about every other issue. These articles, a number of which are reproduced on their website, are at a slightly higher level, but\u2014often being written by the astronomers who have done the work\u2014are authoritative and current. Sky & Telescope: http://skyandtelescope.com. An older and somewhat higher-level magazine for astronomy hobbyists. Many noted astronomers write for this publication. Sky News: http://www.skynews.ca/. A Canadian publication, featuring both astronomy and stargazing information. It also lists Canadian events for hobbyists. StarDate: https://stardate.org/. Magazine", " that accompanies the brief radio program, with a useful website for beginners. Sites that Cover Astronomy News Exploring the Universe: http://fraknoi.blogspot.com. An astronomy news blog by one of the original authors of this textbook. Portal to the Universe: http://www.portaltotheuniverse.org/. A site that gathers online astronomy and space news items, blogs, and pictures. Science@NASA news stories and newscasts: http://science.nasa.gov/science-news/. Well-written stories with, of course, a NASA focus. Space.com: http://www.space.com/news/. A commercial site, but with wide coverage of space and astronomy news. Universe Today: http://www.universetoday.com/. Another commercial site, with good articles by science journalists, but a lot of ads. Sites for Answering Astronomical Questions Ask an Astrobiologist: http://astrobiology.nasa.gov/ask-an-astrobiologist/. On this site from the National Astrobiology Institute at NASA, astronomer David Morrison answered questions about the search for life on other planets, the origin of life on Earth, and many other topics. Ask an Astronomer at Lick Observatory: http://www.ucolick.org/~mountain/AAA/. Graduate students and staff members at this California observatory answered selected astronomy questions, particularly from high school students. Ask an Astrophysicist: http://imagine.gsfc.nasa.gov/docs/ask_astro/ask_an_astronomer.html. Questions and answers at NASA\u2019s Laboratory for High-Energy Astrophysics focus on X-ray and gamma-ray astronomy, and such objects as black holes, quasars, and supernovae. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix B 1141 Ask an Infrared Astronomer: http://coolcosmos.ipac.caltech.edu/cosmic_classroom/ask_astronomer/faq/ index.shtml. A site from the California Institute of Technology, with an archive focusing on infrared (heat-ray) astronomy and the discoveries it makes about cool objects in the universe. No longer taking new questions. Ask the Astronomer: http://www.astronomycafe.net/qadir/qanda.html. This site, run by astronomer Sten Odenwald, is no longer active, but lists 3001 answers to questions asked in the mid-1990s. They are nicely organized by topic. Ask the Experts at PhysLink: http://www.physlink.com/Education/AskExperts/index.cfm. Lots of physics questions answered, with some astronomy as well, at this physics education site. Most answers are by physics teachers, not astronomers. Still taking new questions. Ask the Space Scientist: http://image.gsfc.nasa.gov/poetry/ask/askmag.html. An archive of questions about the Sun and its interactions with Earth, answered by astronomer Sten Odenwald. Not accepting new questions. Curious about Astronomy?: http://curious.astro.cornell.edu. An ask-an-astronomer site run by graduate students and professors of astronomy at Cornell University. Has searchable archives and is still answering new questions. Miscellaneous Sites of Interest A Guide to Careers in Astronomy: http://aas.org/files/resources/Careers-in-Astronomy.pdf. From the American Astronomical Society. Astronomical Pseudo-Science: A Skeptic\u2019s Resource List: http://www.astrosociety.org/pseudo. Readings and websites that analyze such claims as astrology, UFOs, moon-landing denial, creationism, human faces on other worlds, astronomical disasters, and more. Astronomy for Beginners: http://www.skyandtelescope.com/astronomy-information/. A page to find resources for getting into amateur astronomy. Science Fiction Stories with Good Astronomy and Physics: http://www.astrosociety.org/scifi. Space Calendar: http://www2.jpl.nasa.gov/calendar/. Ron Baalke at the Jet Propulsion Laboratory keeps a listing of what space events happened on each day of the year; great if you need a reason to have a space-theme party. Unheard Voices: The Astronomy of Many Cultures: http://multiverse.ssl.berkeley.edu/multicultural. A guide to resources about the astronomy of native, African, Asian, and other non-Western groups. Selected Websites For Viewing And Downloading Astronomical Images The Top Image Sites Astronomy Picture of the Day: http://antwrp.gsfc.nasa.gov", "/apod/lib/aptree.html. Two space scientists scour the internet and feature one interesting astronomy image each day. European Southern Observatory Photo Gallery: http://www.eso.org/public/images/. Magnificent color images from ESO\u2019s largest telescopes. See the topical menu at the top. Hubble Space Telescope Images: http://hubblesite.org/newscenter/archive/browse/images/. Starting at this page, you can select from among many hundreds of Hubble pictures by subject. Other ways to approach these images are through the more public-oriented Hubble Gallery (http://hubblesite.org/gallery/) or the European ESO site (http://www.spacetelescope.org/images/). 1142 Appendix B National Optical Astronomy Observatories Image Gallery: http://www.noao.edu/image_gallery/. Growing archive of images from the many telescopes that are at the United States\u2019 National Observatories. Planetary Photojournal: http://photojournal.jpl.nasa.gov/index.html. Features thousands of images from NASA\u2019s extensive set of planetary exploration missions with a good search menu. Does not include most of the missions from other countries. The World at Night: http://www.twanight.org/newTWAN/index.asp. Dramatic night-sky images by professional photographers who are amateur astronomers. Note that while many of the astronomy sites allow free use of their images, these are copyrighted by photographers who make their living selling them. Other Useful General Galleries Anglo-Australian Observatory: http://203.15.109.22/images/. Soon at https://www.aao.gov.au/public/images. Great copyrighted color images by leading astro-photographer David Malin and others. Canada-France-Hawaii Telescope: http://www.cfht.hawaii.edu/HawaiianStarlight/images.html. Remarkable color images from a major telescope on top of the Mauna Kea peak in Hawaii. European Space Agency Gallery: http://www.esa.int/spaceinimages/Images. Access images from such missions as Mars Express, Rosetta, and Herschel. Gemini Observatory Images: http://www.gemini.edu/index.php?option=com_gallery. Images from a pair of large telescopes in the northern and the southern hemispheres. Isaac Newton Group of Telescopes Image Gallery: http://www.ing.iac.es/PR/images_index.html. Beautiful images from the Herschel, Newton, and Kapteyn telescopes on La Palma. National Radio Astronomy Observatory Image Gallery: http://images.nrao.edu/. Organized by topic, the images show objects and processes that give off radio waves. Our Infrared World Gallery: http://coolcosmos.ipac.caltech.edu/image_galleries/missions_gallery.html. Images from a variety of infrared astronomy telescopes and missions. See also their \u201cCool Cosmos\u201d site for the public: http://coolcosmos.ipac.caltech.edu/. Some Galleries on Specific Subjects Astronaut Photography of Earth: http://eol.jsc.nasa.gov/. Chandra X-Ray Observatory Images: http://chandra.harvard.edu/photo/category.html. NASA Human Spaceflight Gallery: https://www.flickr.com/photos/nasa2explore or http://spaceflight1.nasa.gov/gallery/index.html. Astronaut images. Robert Gendler: http://www.robgendlerastropics.com/. One of the amateur astro-photographers who comes closest to being professional. Sloan Digital Sky Survey Images: http://www.sdss.org/gallery/. Solar Dynamics Observatory Gallery: http://sdo.gsfc.nasa.gov/gallery/main. Sun images. Spitzer Infrared Telescope Images: http://www.spitzer.caltech.edu/images. Astronomy Apps For Smartphones And Tablets A pretty comprehensive listing of such apps with brief descriptions and links to their websites can be found at: http://dx.doi.org/10.3847/AER2011036. The list is now a few years old, but most of the apps are still available. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix B 1143 Listings and Reviews of Apps 11 Best Astronomy Apps for Amateur Star Gazers: http://www.businessinsider.com/11-best-astronomy-apps- for-amateurs-2013-10. From Kelly Dickerson (2013). 14 Best Astronomy Apps for Stargazers and Space Lovers: http://nerdsmagazine.com/best-astronomy-apps-", " for-android/. Viney Dhiman\u2019s recommendations, part of Nerd\u2019s Magazine (2014). 15 Best Astronomy Applications for iPhone: http://www.iphoneness.com/iphone-apps/top-astronomy- applications-for-iphone/. From iPhoneness. Apps for Stargazing: http://appadvice.com/appguides/show/astronomy-apps. App Advice site\u2019s reviews. NASA Apps for Smartphones and Tablets: https://www.nasa.gov/connect/apps.html. Phone/Tablet Apps and the Practical Astronomer: http://www.cloudynights.com/page/articles/cat/user- reviews/phonetablet-apps-and-the-practical-astronomer-r2925. Active amateur astronomer Tom Fowler reviews 22 apps (2014). Sky & Telescope Mobile Apps: http://www.skyandtelescope.com/sky-and-stargazing-apps/. Apps from Sky & Telescope Magazine. Smartphone Apps Can Make Astronomy as Easy as Point and Gaze: http://www.heraldnet.com/article/ 20140511/LIVING/140519988.\u201d Mike Lynch for HeraldNet (2014). 1144 Appendix B This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix C 1145 C SCIENTIFIC NOTATION In astronomy (and other sciences), it is often necessary to deal with very large or very small numbers. In fact, when numbers become truly large in everyday life, such as the national debt in the United States, we call them astronomical. Among the ideas astronomers must routinely deal with is that the Earth is 150,000,000,000 meters from the Sun, and the mass of the hydrogen atom is 0.00000000000000000000000000167 kilograms. No one in his or her right mind would want to continue writing so many zeros! Instead, scientists have agreed on a kind of shorthand notation, which is not only easier to write, but (as we shall see) makes multiplication and division of large and small numbers much less difficult. If you have never used this powers-of-ten notation or scientific notation, it may take a bit of time to get used to it, but you will soon find it much easier than keeping track of all those zeros. Writing Large Numbers In scientific notation, we generally agree to have only one number to the left of the decimal point. If a number is not in this format, it must be changed. The number 6 is already in the right format, because for integers, we understand there to be a decimal point to the right of them. So 6 is really 6., and there is indeed only one number to the left of the decimal point. But the number 965 (which is 965.) has three numbers to the left of the decimal point, and is thus ripe for conversion. To change 965 to proper form, we must make it 9.65 and then keep track of the change we have made. (Think of the number as a weekly salary and suddenly it makes a lot of difference whether we have $965 or $9.65.) We keep track of the number of places we moved the decimal point by expressing it as a power of ten. So 965 becomes 9.65 \u00d7 102 or 9.65 multiplied by ten to the second power. The small raised 2 is called an exponent, and it tells us how many times we moved the decimal point to the left. Note that 102 also designates 10 squared, or 10 \u00d7 10, which equals 100. And 9.65 \u00d7 100 is just 965, the number we started with. Another way to look at scientific notation is that we separate out the messy numbers out front, and leave the smooth units of ten for the exponent to denote. So a number like 1,372,568 becomes 1.372568 times a million (106) or 1.372568 times 10 multiplied by itself 6 times. We had to move the decimal point six places to the left (from its place after the 8) to get the number into the form where there is only one digit to the left of the decimal point. The reason we call this powers-of-ten notation is that our counting system is based on increases of ten; each place in our numbering system is ten times greater than the place to the right of it. As you have probably learned, this got started because human beings have ten fingers and we started counting with them. (It is interesting to speculate that if we ever meet intelligent life-forms with only eight fingers, their counting system would probably be a powers-of-eight notation!) So, in the example we started with, the number of meters from Earth to the Sun is 1.5 \u00d7 1011. Elsewhere in the book, we mention that a string 1 light-", "year long would fit around Earth\u2019s equator 236 million or 236,000,000 times. In scientific notation, this would become 2.36 \u00d7 108. Now if you like expressing things in millions, as the annual reports of successful companies do, you might like to write this number as 236 \u00d7 106. However, the usual convention is to have only one number to the left of the decimal point. Writing Small Numbers Now take a number like 0.00347, which is also not in the standard (agreed-to) form for scientific notation. To put it into that format, we must make the first part of it 3.47 by moving the decimal point three places to the right. 1146 Appendix C Note that this motion to the right is the opposite of the motion to the left that we discussed above. To keep track, we call this change negative and put a minus sign in the exponent. Thus 0.00347 becomes 3.47 \u00d7 10\u22123. In the example we gave at the beginning, the mass of the hydrogen atom would then be written as 1.67 \u00d7 10\u221227 kg. In this system, one is written as 100, a tenth as 10\u22121, a hundredth as 10\u22122, and so on. Note that any number, no matter how large or how small, can be expressed in scientific notation. Multiplication And Division Scientific notation is not only compact and convenient, it also simplifies arithmetic. To multiply two numbers expressed as powers of ten, you need only multiply the numbers out front and then add the exponents. If there are no numbers out front, as in 100 \u00d7 100,000, then you just add the exponents (in our notation, 102 \u00d7 105 = 107). When there are numbers out front, you have to multiply them, but they are much easier to deal with than numbers with many zeros in them. Here\u2019s an example: And here\u2019s another example: \u239d3 \u00d7 105\u239e \u239b \u23a0 \u00d7 \u239d2 \u00d7 109\u239e \u239b \u23a0 = 6 \u00d7 1014 0.04 \u00d7 6,000,000 = \u23a0 \u00d7 \u239d6 \u00d7 106\u239e \u239b \u23a0 \u239b \u239d4 \u00d7 10\u22122\u239e = 24 \u00d7 104 = 2.4 \u00d7 105 Note in the second example that when we added the exponents, we treated negative exponents as we do in regular arithmetic (\u22122 plus 6 equals 4). Also, notice that our first result had a 24 in it, which was not in the acceptable form, having two places to the left of the decimal point, and we therefore changed it to 2.4 and changed the exponent accordingly. To divide, you divide the numbers out front and subtract the exponents. Here are several examples: = 106 (6 \u2212 3) = 103 1,000,000 103 = 10 1000 9 \u00d7 1012 2 \u00d7 103 = 4.5 \u00d7 109 2.8 \u00d7 102 6.2 \u00d7 105 = 0.452 \u00d7 10\u22123 = 4.52 \u00d7 10\u22124 In the last example, our first result was not in the standard form, so we had to change 0.452 into 4.52, and change the exponent accordingly. If this is the first time that you have met scientific notation, we urge you to practice many examples using it. You might start by solving the exercises below. Like any new language, the notation looks complicated at first but gets easier as you practice it. Exercises 1. At the end of September, 2015, the New Horizons spacecraft (which encountered Pluto for the first time in July 2015) was 4.898 billion km from Earth. Convert this number to scientific notation. How many astronomical units is this? (An astronomical unit is the distance from Earth to the Sun, or about 150 million km.) This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix C 1147 2. During the first six years of its operation, the Hubble Space Telescope circled Earth 37,000 times, for a total of 1,280,000,000 km. Use scientific notation to find the number of km in one orbit. 3. In a large university cafeteria, a soybean-vegetable burger is offered as an alternative to regular hamburgers. If 889,875 burgers were eaten during the course of a school year, and 997 of them were veggie-burgers, what fraction and what percent of the burgers does this represent? 4. In a 2012 Kelton Research poll, 36 percent of adult Americans thought that alien beings have actually landed on Earth. The number of adults in the United States in 2012 was about 222,000,000. Use scientific notation to determine how many adults believe aliens have visited Earth. 5. In the school year 2009\u20132010, American colleges and universities awarded 2,", "354,678 degrees. Among these were 48,069 PhD degrees. What fraction of the degrees were PhDs? Express this number as a percent. (Now go and find a job for all those PhDs!) 6. A star 60 light-years away has been found to have a large planet orbiting it. Your uncle wants to know the distance to this planet in old-fashioned miles. Assume light travels 186,000 miles per second, and there are 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, and 365 days in a year. How many miles away is that star? Answers 1. 4.898 billion is 4.898 \u00d7 109 km. One astronomical unit (AU) is 150 million km = 1.5 \u00d7 108 km. Dividing the 2. 3. first number by the second, we get 3.27 \u00d7 10(9 \u2013 8) = 3.27 \u00d7 101 AU. 1.28 \u00d7 109 km 3.7 \u00d7 104 orbits = 0.346 \u00d7 10 (9 \u2212 4) = 0.346 \u00d7 105 = 3.46 \u00d7 104 km per orbit. 9.97 \u00d7 102 veggie burgers 8.90 \u00d7 105 total burgers = 1.12 \u00d7 10 (2 \u2212 5) = 1.12 \u00d7 10 (2 \u2212 5) = 1.12 \u00d7 10\u22123 (or roughly about one thousandth) 1.12 \u00d7 10\u22123 10\u22122 of the burgers were vegetarian. Percent means per hundred. So = 1.12 \u00d7 10 (\u22123 \u2212 (\u22122)) = 1.12 \u00d7 10\u22121 percent (which is roughly one tenth of one percent). 4. 36% is 36 hundredths or 0.36 or 3.6 \u00d7 10\u22121. Multiply that by 2.22 \u00d7 108 and you get about 7.99 \u00d7 10(\u22121 + 8) = 7.99 \u00d7 107 or almost 80 million people who believe that aliens have landed on our planet. We need more astronomy courses to educate all those people. 5. 4.81 \u00d7 104 2.35 \u00d7 106 = 2.05 \u00d7 10 (4 \u2212 6) = 2.05 \u00d7 10\u22122 = about 2%. (Note that in these examples we are rounding off some of the numbers so that we don\u2019t have more than 2 places after the decimal point.) 6. One light-year is the distance that light travels in one year. (Usually, we use metric units and not the old British system that the United States is still using, but we are going to humor your uncle and stick with miles.) If light travels 186,000 miles every second, then it will travel 60 times that in a minute, and 60 times that in an hour, and 24 times that in a day, and 365 times that in a year. So we have 1.86 \u00d7 105 \u00d7 6.0 \u00d7 101 \u00d7 6.0 \u00d7 101 \u00d7 2.4 \u00d7 101 \u00d7 3.65 \u00d7 102. So we multiply all the numbers out front together and add all the exponents. We get 586.57 \u00d7 1010 = 5.86 \u00d7 1012 miles in a light year (which is roughly 6 trillion miles\u2014a heck of a lot of miles). So if the star is 60 light-years away, its distance in miles is 6 \u00d7 101 \u00d7 5.86 \u00d7 1012 = 35.16 \u00d7 1013 = 3.516 \u00d7 1014 miles. 1148 Appendix C This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix D 1149 D UNITS USED IN SCIENCE In the American system of measurement (originally developed in England), the fundamental units of length, weight, and time are the foot, pound, and second, respectively. There are also larger and smaller units, which include the ton (2240 lb), the mile (5280 ft), the rod (16 1/2 ft), the yard (3 ft), the inch (1/12 ft), the ounce (1/16 lb), and so on. Such units, whose origins in decisions by British royalty have been forgotten by most people, are quite inconvenient for conversion or doing calculations. In science, therefore, it is more usual to use the metric system, which has been adopted in virtually all countries except the United States. Its great advantage is that every unit increases by a factor of ten, instead of the strange factors in the American system. The fundamental units of the metric system are: length: 1 meter (m) mass: 1 kilogram (kg) time: 1 second (s) A meter was originally intended to be 1 ten-millionth of the distance from the equator to the North Pole along the surface of Earth. It is about 1.1 yd. A kilogram is the mass that on Earth results in a weight of about 2.2 lb. The second is the same in metric and American units. Length", " The most commonly used quantities of length of the metric system are the following. Conversions 1 kilometer (km) = 1000 meters = 0.6214 mile 1 meter (m) = 0.001 km = 1.094 yards = 39.37 inches 1 centimeter (cm) = 0.01 meter = 0.3937 inch 1 millimeter (mm) = 0.001 meter = 0.1 cm 1 micrometer (\u00b5m) = 0.000001 meter = 0.0001 cm 1 nanometer (nm) = 10\u22129 meter = 10\u22127 cm Table D1 Length To convert from the American system, here are a few helpful factors: 1 mile = 1.61 km 1 inch = 2.54 cm Mass Although we don\u2019t make the distinction very carefully in everyday life on Earth, strictly speaking the kilogram is a unit of mass (measuring the quantity of matter in a body, roughly how many atoms it has,) while the pound is a unit of weight (measuring how strongly Earth\u2019s gravity pulls on a body). 1150 Appendix D The most commonly used quantities of mass of the metric system are the following. Conversions 1 metric ton = 106 grams = 1000 kg (and it produces a weight of 2.205 \u00d7 103 lb on Earth) 1 kg = 1000 grams (and it produces a weight of 2.2046 lb on Earth) 1 gram (g) = 0.0353 oz (and the equivalent weight is 0.002205 lb) 1 milligram (mg) = 0.001 g Table D2 Mass A weight of 1 lb is equivalent on Earth to a mass of 0.4536 kg, while a weight of 1 oz is produced by a mass of 28.35 g. Temperature Three temperature scales are in general use: Fahrenheit (F); water freezes at 32 \u00b0F and boils at 212 \u00b0F. Celsius or centigrade[1] (C); water freezes at 0 \u00b0C and boils at 100 \u00b0C. Kelvin or absolute (K); water freezes at 273 K and boils at 373 K. All molecular motion ceases at about \u2212459 \u00b0F = \u2212273 \u00b0C = 0 K, a temperature called absolute zero. Kelvin temperature is measured from this lowest possible temperature, and it is the temperature scale most often used in astronomy. Kelvins have the same value as centigrade or Celsius degrees, since the difference between the freezing and boiling points of water is 100 degrees in each. (Note that we just say \u201ckelvins,\u201d not kelvin degrees.) On the Fahrenheit scale, the difference between the freezing and boiling points of water is 180 degrees. Thus, to convert Celsius degrees or kelvins to Fahrenheit degrees, it is necessary to multiply by 180/100 = 9/5. To convert from Fahrenheit degrees to Celsius degrees or kelvins, it is necessary to multiply by 100/180 = 5/9. The full conversion formulas are: K = \u00b0C + 273 \u00b0C = 0.555 \u00d7 (\u00b0F \u2013 32) \u00b0F = (1.8 \u00d7 \u00b0C) + 32 1 Celsius is now the name used for centigrade temperature; it has a more modern standardization but differs from the old centigrade scale by less than 0.1\u00b0. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix E 1151 E SOME USEFUL CONSTANTS FOR ASTRONOMY Physical Constants Name Value speed of light (c) 2.9979 \u00d7 108 m/s gravitational constant (G) 6.674 \u00d7 10\u221211 m3/(kg s2) Planck\u2019s constant (h) mass of a hydrogen atom (MH) mass of an electron (Me) Rydberg constant ( R\u221e ) 6.626 \u00d7 10\u221234 J-s 1.673 \u00d7 10\u221227 kg 9.109 \u00d7 10\u221231 kg 1.0974 \u00d7 107 m\u22121 Stefan-Boltzmann constant (\u03c3) 5.670 \u00d7 10\u22128 J/(s\u00b7m2 deg4)[1] Wien\u2019s law constant (\u03bbmaxT) electron volt (energy) (eV) energy equivalent of 1 ton TNT Table E1 Astronomical Constants 2.898 \u00d7 10\u22123 m K 1.602 \u00d7 10\u221219 J 4.2 \u00d7 109 J Name Value astronomical unit (AU) Light-year (ly) parsec (pc) sidereal year (y) mass of Earth (MEarth) 1.496 \u00d7 1011 m 9.461 \u00d7 1015 m 3.086 \u00d7 1016 m = 3.262 light-years 3.156 \u00d7 107 s 5.974 \u00d7 1024 kg equatorial radius of Earth (REarth) 6.378 \u00d7 106 m obliquity of ecliptic 23.4\u00b0 26\u2019", " Table E2 1 deg stands for degrees Celsius or kelvins 1152 Appendix E Astronomical Constants Name Value surface gravity of Earth (g) 9.807 m/s2 escape velocity of Earth (vEarth) 1.119 \u00d7 104 m/s mass of Sun (MSun) 1.989 \u00d7 1030 kg equatorial radius of Sun (RSun) 6.960 \u00d7 108 m luminosity of Sun (LSun) 3.85 \u00d7 1026 W solar constant (flux of energy received at Earth) (S) 1.368 \u00d7 103 W/m2 Hubble constant (H0) approximately 20 km/s per million light-years, or approximately 70 km/s per megaparsec Table E2 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix F 1153 F PHYSICAL AND ORBITAL DATA FOR THE PLANETS Physical Data for the Major Planets Major Planet Mean Diameter (km) Mean Diameter (Earth = 1) Mass (Earth = 1) Mean Density (g/cm3) Rotation Period (d) Inclination of Equator to Orbit (\u00b0) Mercury 4879 0.38 0.055 5.43 58. Venus 12,104 0.95 0.815 5.24 \u2212243. Earth 12,756 1.00 1.00 5.51 1.000 Mars 6779 0.53 0.11 3.93 1.026 0.0 177 23.4 25.2 Jupiter 140,000 10.9 318 1.33 0.414 3.1 Saturn 117,000 9.13 95.2 0.69 0.440 26.7 Uranus 50,700 3.98 14.5 1.27 \u22120.718 97.9 Neptune 49,200 3.86 17.2 1.64 0.671 29.6 Table F1 Physical Data for Well-Studied Dwarf Planets WellStudied Dwarf Planet Diameter (km) Diameter (Earth = 1) Mass (Earth = 1) Mean Density (g/cm3) Rotation Period (d) Inclination of Equator to Orbit (\u00b0) Ceres 950 Pluto 2470 Haumea 1700 Makemake 1400 Eris 2326 Table F2 0.07 0.18 0.13 0.11 0.18 0.0002 2.2 0.378 3 0.0024 1.9 \u22126.387 122 0.0007 0.0005 3 2 0.0028 2.5 0.163 0.321 1.25[1] \u2014 \u2014 \u2014 Surface Gravity (Earth = 1[g]) Velocity of Escape (km/s) 0.38 4.3 0.90 10.4 1.00 11.2 0.38 2.53 1.07 0.89 1.14 5.0 60. 36. 21. 23. Surface Gravity (Earth = 1[g]) Velocity of Escape (km/s) 0.03 0.06 \u2014 \u2014 \u2014 0.5 1.3 0.8 0.8 1.4 1 This measurement is quite uncertain. 1154 Appendix F Orbital Data for the Major Planets Major Planet Semimajor Axis (AU) Semimajor Axis (106 km) Sidereal Period (y) Sidereal Period (d) Mean Orbital Speed (km/s) Orbital Eccentricity Inclination of Orbit to Ecliptic (\u00b0) Mercury 0.39 Venus 0.72 Earth 1.00 Mars 1.52 Jupiter 5.20 Saturn 9.54 Uranus 19.19 Neptune 30.06 Table F3 58 108 149 228 778 1427 2871 4497 0.24 88.0 47.9 0.206 0.6 1.00 1.88 11.86 29.46 84.01 164.82 224.7 35.0 0.007 365.2 29.8 0.017 687.0 24.1 0.093 \u2014 \u2014 \u2014 \u2014 13.1 0.048 9.6 6.8 5.4 0.056 0.046 0.010 7.0 3.4 0.0 1.9 1.3 2.5 0.8 1.8 Orbital Data for Well-Studied Dwarf Planets Semimajor Axis (AU) Semimajor Axis (106 km) Sidereal Period (y) Mean Orbital Speed (km/s) Orbital Eccentricity Inclination of Orbit to Ecliptic (\u00b0) 414.0 4.6 5915 6452 6850 248.6 283.3 309.9 18 4.7 4.5 4.4 3.4 0.08 0.25 0.19 0.16 0.44 11 17 28 29 44 WellStudied Dwarf Planet Ceres Pluto Haumea 2.77 39.5 43.1 Makemake 45.8 Eris 68.0 10", ",120 560.9 Table F4 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix G 1155 G SELECTED MOONS OF THE PLANETS Note: As this book goes to press, nearly two hundred moons are now known in the solar system and more are being discovered on a regular basis. Of the major planets, only Mercury and Venus do not have moons. In addition to moons of the planets, there are many moons of asteroids. In this appendix, we list only the largest and most interesting objects that orbit each planet (including dwarf planets). The number given for each planet is discoveries through 2015. For further information see https://solarsystem.nasa.gov/planets/solarsystem/ moons and https://en.wikipedia.org/wiki/List_of_natural_satellites. Selected Moons of the Planets Planet (moons) Satellite Name Discovery Semimajor Axis (km \u00d7 1000) Period (d) Diameter (km) Mass (1020 kg) Density (g/cm3) Earth (1) Moon \u2014 384 27.32 3476 735 3.3 Mars (2) Phobos Hall (1877) 9.4 Deimos Hall (1877) 23.5 0.32 1.26 23 13 0.50 200 0.67 90 1 \u00d7 10\u22124 2.0 2 \u00d7 10\u22125 1.7 \u2014 \u2014 \u2014 \u2014 1.77 3630 894 3.6 3.55 3138 480 3.0 181 222 422 671 1070 7.16 5262 1482 1.9 1883 16.69 4800 1077 1.9 11,460 251 170 \u2014 \u2014 133.6 0.58 20 3 \u00d7 10\u22125 \u2014 137.7 0.60 40 \u2014 \u2014 Jupiter (67) Amalthea Thebe Io Europa Ganymede Callisto Himalia Pan Atlas Saturn (62) Table G1 Barnard (1892) Voyager (1979) Galileo (1610) Galileo (1610) Galileo (1610) Galileo (1610) Perrine (1904) Voyager (1985) Voyager (1980) 1156 Appendix G Selected Moons of the Planets Planet (moons) Satellite Name Discovery Semimajor Axis (km \u00d7 1000) Period (d) Diameter (km) Mass (1020 kg) Density (g/cm3) Prometheus Pandora Janus Epimetheus Mimas Enceladus Tethys Dione Rhea Titan Hyperion Iapetus Phoebe Uranus (27) Puck Table G1 Voyager (1980) Voyager (1980) Dollfus (1966) Fountain, Larson (1980) Herschel (1789) Herschel (1789) Cassini (1684) Cassini (1684) Cassini (1672) Huygens (1655) Bond, Lassell (1848) Cassini (1671) Pickering (1898) Voyager (1985) 139.4 0.61 80 141.7 0.63 100 151.4 0.69 190 151.4 0.69 120 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 186 238 295 377 527 0.94 394 0.4 1.2 1.37 502 0.8 1.2 1.89 1048 7.5 1.3 2.74 1120 4.52 1530 11 25 1.3 1.3 1222 15.95 5150 1346 1.9 1481 21.3 270 \u2014 \u2014 3561 79.3 1435 12,950 550 (R)[1] 220 86.0 0.76 170 19 \u2014 \u2014 1.2 \u2014 \u2014 1 R stands for retrograde rotation (backward from the direction that most objects in the solar system revolve and rotate). This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix G 1157 Selected Moons of the Planets Planet (moons) Satellite Name Discovery Semimajor Axis (km \u00d7 1000) Period (d) Diameter (km) Mass (1020 kg) Density (g/cm3) Miranda Ariel Umbriel Titania Oberon Neptune (14) Despina Galatea Larissa Triton Nereid Pluto (5) Charon Kuiper (1948) Lassell (1851) Lassell (1851) Herschel (1787) Herschel (1787) Voyager (1989) Voyager (1989) Voyager (1989) Lassell (1846) Kuiper (1949) Christy (1978) 130 191 266 436 583 53 62 118 355 1.41 485 0.8 1.3 2.52 1160 4.14 1190 8.71 1610 13.5 1550 0.33 150 0.40 150 1.12 400 13 13 35 29 \u2014", " \u2014 \u2014 1.6 1.4 1.6 1.5 \u2014 \u2014 \u2014 5.88 (R)[2] 2720 220 2.1 5511 360 340 19.7 6.39 1200 Styx Nix Kerberos 42 48 58 Showalter et al (2012) Weaver et al (2005) Showalter et al (2011) 20 20 24 24 46 28 Table G1 2 R stands for retrograde rotation (backward from the direction that most objects in the solar system revolve and rotate). \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 1.7 \u2014 2.1 1.4 1158 Appendix G Selected Moons of the Planets Planet (moons) Satellite Name Discovery Semimajor Axis (km \u00d7 1000) Period (d) Diameter (km) Mass (1020 kg) Density (g/cm3) Hydra Weaver et al (2005) Eris (1) Dysnomea Makemake (1) (MK2) Haumea (2) Hi\u2019iaka Namaka Brown et al (2005) Parker et al (2016) Brown et al (2005) Brown et al (2005) 65 38 \u2014 50 39 Table G1 38 16 \u2014 49 35 61 684 160 400 200 \u2014 \u2014 \u2014 \u2014 \u2014 0.8 \u2014 \u2014 \u2014 \u2014 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix H 1159 H FUTURE TOTAL ECLIPSES Future Total Solar Eclipses We also include eclipses that are annular\u2014where the Moon is directly in front of the Sun, but doesn\u2019t fully cover it\u2014leaving a ring of light around the dark Moon\u2019s edges) Future Total Solar Eclipses Date Type of Eclipse Location on Earth[1] September 1, 2016 Annular S Atlantic Ocean, C Africa, Madagascar, Indian Ocean February 26, 2017 Annular SW Africa, S tip of South America August 21, 2017 Total U.S. and oceans on either side July 2, 2019 Total SW South America, Pacific Ocean December 26, 2019 Annular Saudi Arabia, S India, Malaysia June 21, 2020 Annular (very short) C Africa, Pakistan, India, China December 14, 2020 Total Chile, Argentina, and oceans on either side June 10, 2021 Annular N Canada, Greenland December 4, 2021 Total Only in Antarctica April 20, 2023 Total[2] Mostly in Indian and Pacific oceans, Indonesia October 14, 2023 Annular OR, NV, UT, NM, TX, C America, Colombia, Brazil April 8, 2024 Total N Mexico, U.S. (TX to ME), SE Canada and oceans on either side October 2, 2024 Annular S Chile, S Argentina, and oceans on either side February 17, 2026 Annular Only in Antarctica August 12, 2026 Total Greenland, Iceland, Spain February 6, 2027 Annular S Pacific, Argentina, Chile, Uruguay, S Atlantic August 2, 2027 Total Spain, Morocco, Egypt, Saudi Arabia, Yemen, Arabian Sea Table H1 1 Remember that a total or annular eclipse is only visible on a narrow track. The same eclipse will be partial over a much larger area, but partial eclipses are not as spectacular as total ones. 2 This is a so-called hybrid eclipse, which is total in some places and annular in others. 1160 Appendix H Future Total Solar Eclipses Date Type of Eclipse Location on Earth January 26, 2028 Annular Ecuador, Peru, Brazil, North Atlantic Ocean, Portugal, Spain July 22, 2028 Total Indian Ocean, Australia, New Zealand, South Pacific Ocean Table H1 Future Total Lunar Eclipses Future Total Lunar Eclipses Date Location on Earth January 31, 2018 Asia, Australia, W North America July 27, 2018 S America, Asia, Africa, Australia, Indian Ocean January 21, 2019 N America, S America, W Africa, W Europe May 26, 2021 E Asia, Australia, Pacific Ocean, W North America, W South America May 16, 2022 N America, S America, Europe, Africa November 8, 2022 Asia, Australia, Pacific Ocean, N America, S America March 14, 2025 Pacific Ocean, N America, S America, Atlantic Ocean, W Europe, W Africa September 7, 2025 Europe, Africa, Asia, Australia, Indian Ocean March 3, 2026 E Asia, Australia, Pacific Ocean, N America, C America June 26, 2029 E North America, S America, Atlantic Ocean, W Europe, W Africa December 20, 2029 E North America, E South America, Atlantic Ocean, Europe, Africa, Asia Table H2 Additional Resources For more information and detailed maps about eclipses, see these resources. NASA\u2019s Eclipse Site: http://eclipse.gsfc.nasa.gov/ Mr. Eclipse site for beginners by Dr. Fred Espenak: http://www.mreclipse.com/ Eclipse Weather and Maps by Meteorologist Jay Anderson", ": http://home.cc.umanitoba.ca/~jander/ This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix H 1161 Eclipse Maps by Michael Zeiler: http://www.eclipse-maps.com/Eclipse-Maps/Welcome.html Eclipse Information and Maps by Xavier Jubier: http://xjubier.free.fr/en/site_pages/eclipses.html 1162 Appendix H This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix I 1163 I THE NEAREST STARS, BROWN DWARFS, AND WHITE DWARFS The Nearest Stars, Brown Dwarfs, and White Dwarfs Star System Discovery Name Distance (lightyear) Spectral Type Location: RA[1] Location: Dec[2] Luminosity (Sun = 1 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 Sun \u2014 G2 V \u2014 \u2014 1 Proxima Centauri 4.2 M5.5 V 14 29 \u221262 40 5 \u00d7 10\u22125 Alpha Centauri A 4.4 G2 V 14 39 \u221260 50 Alpha Centauri B 4.4 K2 IV 14 39 \u221260 50 1.5 0.5 Barnard\u2019s Star Wolf 359 Lalande 21 185 Sirius A Sirius B Luyten 726-8 A Luyten 726-8 B (UV Ceti) Ross 154 Ross 248 (HH Andromedae) 6.0 7.8 8.3 8.6 8.6 8.7 8.7 9.7 10.3 M4 V 17 57 +04 42 4.4 \u00d7 10\u22124 M6 V 10 56 +07 00 2 \u00d7 10\u22125 M2 V 11 03 +35 58 5.7 \u00d7 10\u22123 A1 V 06 45 \u221216 42 23.1 DA2[3] 06 45 \u221216 43 2.5 \u00d7 10\u22123 M5.5 V 01 39 \u221217 57 M6 V 01 39 \u221217 57 6 \u00d7 10\u22125 4 \u00d7 10\u22125 M.05 V 18 49 \u221223 50 5 \u00d7 10\u22124 M5.5 V 23 41 +44 10 1.0 \u00d7 10\u22124 Epsilon Eridani 10.5 K2 V 03 32 \u221209 27 0.29 Lacaille 9352 Ross 128 (FI Virginis) 10.7 10.9 M0.5 V 23 05 \u221235 51 0.011 M4 V 11 47 +00 48 3.4 \u00d7 10\u22124 16 13 Luyten 789-6 A (EZ Aquarii A) 11.3 M5 V 22 38 \u221215 17 5 \u00d7 10\u22125 Table I1 1 2 Location (right ascension) given for Epoch 2000.0 Location (declination) given for Epoch 2000.0 3 White dwarf stellar remnant 1164 Appendix I The Nearest Stars, Brown Dwarfs, and White Dwarfs Star System Discovery Name Distance (lightyear) Spectral Type Location: RA Location: Dec Luminosity (Sun = 1) 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 Luyten 789-6 B (EZ Aquarii B) Luyten 789-6 C (EZ Aquarii C) 14 61 Cygni A 61 Cygni B 15 Procyon A Procyon B 16 Sigma 2398 A Sigma 2398 B 17 Groombridge 34 A (GX Andromedae) Groombridge 34 B (GQ Andromedae) 11.3 M5.5 V 22 38 \u221215 15 5 \u00d7 10\u22125 11.3 M6.5 V 22 38 \u221215 17 2 \u00d7 10\u22125 11.4 11.4 11.4 11.4 11.5 11.5 11.6 K5 V 21 06 +38 44 0.086 K7 V 21 06 +38 44 0.041 F51V wd[4] 07 39 +05 13 7.38 07 39 +05 13 5.5 \u00d7 10\u22124 M3 V 18 42 +59 37 0.003 M3.5 V 18 42 +59 37 1.4 \u00d7 10\u22123 M1.5 V 00 18 +44 01 6.4 \u00d7 10\u22123 11.6 M3.5 V 00 18 +44 01 4.1 \u00d7 10\u22124 18 Epsilon Indi A 11.8 Epsilon Indi Ba 11.7 Epsilon Indi Bb 11.7 K5 V T1[5] T6[6] 22 03 \u221256 46 0.150 22 04 \u221256 46 22 04 \u221256 46 \u2014 \u2014 19 20 21 22 G 51-15 (DX Cancri) Tau Ceti Luyten 372", "-58 Luyten 725-32 (YZ Ceti) 11.8 M6.5 V 08 29 +26 46 1 \u00d7 10\u22125 11.9 12.0 12.1 G8.5 V 01 44 \u221215 56 0.458 M5 V 03 35 \u221244 30 7 \u00d7 10\u22125 M4.5 V 01 12 \u221216 59 1.8 \u00d7 10\u22124 Table I1 4 White dwarf stellar remnant 5 6 Brown dwarf Brown dwarf This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix I 1165 The Nearest Stars, Brown Dwarfs, and White Dwarfs Star System Discovery Name Distance (lightyear) Spectral Type Location: RA Location: Dec Luminosity (Sun = 1) Luyten\u2019s Star 12.4 M3.5 V 07 27 +05 13 SCR J184-6357 A 12.6 M8.5 V 18 45 \u221263 57 1.4 \u00d7 10\u22123 1 \u00d7 10\u22126 SCR J184-6357 B 12.7 T6[7] 18 45 \u221263 57 \u2014 Teegarden\u2019s Star 12.5 M6 V 02 53 +16 52 1 \u00d7 10\u22125 Kapteyn\u2019s Star Lacaille 8760 (AX Microscopium) 12.8 12.9 M1 V 05 11 \u221245 01 3.8 \u00d7 10\u22123 K7 V 21 17 \u221238 52 0.029 23 24 25 26 27 34 35 36 37 38 39 Table I1 7 Brown dwarf 1166 Appendix I This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix J 1167 J THE BRIGHTEST TWENTY STARS Note: These are the stars that appear the brightest visually, as seen from our vantage point on Earth. They are not necessarily the stars that are intrinsically the most luminous. Figure J1 The brightest stars typically have names from antiquity. Next to each star\u2019s ancient name, we have added a column with its name in the system originated by Bayer (see the Naming Stars feature box.) The distances of the more remote stars are estimated from their spectral types and apparent brightnesses and are only approximate. The luminosities for those stars are approximate to the same degree. Right ascension and declination is given for Epoch 2000.0. 1168 Appendix J This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix K 1169 K THE CHEMICAL ELEMENTS The Chemical Elements Element Symbol Atomic Number Atomic Weight[1] Percentage of Naturally Occurring Elements in the Universe Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium H He Li Be B C N O F Ne Na Magnesium Mg Al Si P S Cl Ar K Aluminum Silicon Phosphorus Sulfur Chlorine Argon Potassium Table K1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1.008 4.003 6.94 9.012 10.821 12.011 14.007 15.999 18.998 75 23 6 \u00d7 10\u22127 1 \u00d7 10\u22127 1 \u00d7 10\u22127 0.5 0.1 1 4 \u00d7 10\u22125 20.180 0.13 22.990 0.002 24.305 0.06 26.982 0.005 28.085 0.07 30.974 7 \u00d7 10\u22124 32.06 35.45 0.05 1 \u00d7 10\u22124 39.948 0.02 39.098 3 \u00d7 10\u22124 1 Where mean atomic weights have not been well determined, the atomic mass numbers of the most stable isotopes are given in parentheses. 1170 Appendix K The Chemical Elements Element Symbol Atomic Number Atomic Weight Percentage of Naturally Occurring Elements in the Universe Calcium Scandium Titanium Vanadium Chromium Ca Sc Ti V Cr Manganese Mn Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton Rubidium \u2019Strontium Yttrium Zirconium Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Niobium Nb Table K1 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 40.078 0.007 44.956 47.867 50.942 3 \u00d7 10\u22126 3 \u00d7 10\u22124 3 \u00d7 10\u22124 51.996 0.0015 54.938 8 \u00d7 10\u22124 55.845 0.11 58.933 3 \u00d7 10\u22124 58.693 0.006 63.546 6 \u00d7 10\u22126 65.38 69.723 72.630 74.922 78.", "971 79.904 83.798 85.468 87.62 88.906 91.224 92.906 3 \u00d7 10\u22125 1 \u00d7 10\u22126 2 \u00d7 10\u22125 8 \u00d7 10\u22127 3 \u00d7 10\u22126 7 \u00d7 10\u22127 4 \u00d7 10\u22126 1 \u00d7 10\u22126 4 \u00d7 10\u22126 7 \u00d7 10\u22127 5 \u00d7 10\u22126 2 \u00d7 10\u22127 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix K 1171 The Chemical Elements Element Symbol Atomic Number Atomic Weight Percentage of Naturally Occurring Elements in the Universe Molybdenum Mo Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon Cesium Barium Lanthanum Cerium Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Ce Praseodymium Pr Neodymium Nd Promethium Pm Samarium Europium Sm Eu Table K1 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 95.95 (98) 5 \u00d7 10\u22127 \u2014 101.07 4 \u00d7 10\u22127 102.906 6 \u00d7 10\u22128 106.42 2 \u00d7 10\u22127 107.868 6 \u00d7 10\u22128 112.414 2 \u00d7 10\u22127 114.818 3 \u00d7 10\u22128 118.710 4 \u00d7 10\u22127 121.760 4 \u00d7 10\u22128 127.60 9 \u00d7 10\u22127 126.904 1 \u00d7 10\u22127 131.293 1 \u00d7 10\u22126 132.905 8 \u00d7 10\u22128 137.327 1 \u00d7 10\u22126 138.905 2 \u00d7 10\u22127 140.116 1 \u00d7 10\u22126 140.907 2 \u00d7 10\u22127 144.242 1 \u00d7 10\u22126 (145) \u2014 150.36 5 \u00d7 10\u22127 151.964 5 \u00d7 10\u22128 1172 Appendix K The Chemical Elements Element Symbol Atomic Number Atomic Weight Percentage of Naturally Occurring Elements in the Universe Gd Tb Dy Ho Er Tm Yb Lu Hf Ta W Re Os Ir Pt Au Hg TI Pb Bi Po At Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium Lead Bismuth Polonium Astatine Table K1 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 157.25 2 \u00d7 10\u22127 158.925 5 \u00d7 10\u22128 162.500 2 \u00d7 10\u22127 164.930 5 \u00d7 10\u22128 167.259 2 \u00d7 10\u22127 168.934 1 \u00d7 10\u22128 173.054 2 \u00d7 10\u22127 174.967 1 \u00d7 10\u22128 178.49 7 \u00d7 10\u22128 180.948 8 \u00d7 10\u22129 183.84 5 \u00d7 10\u22128 186.207 2 \u00d7 10\u22128 190.23 3 \u00d7 10\u22127 192.217 2 \u00d7 10\u22127 195.084 5 \u00d7 10\u22127 196.967 6 \u00d7 10\u22128 200.592 1 \u00d7 10\u22127 204.38 5 \u00d7 10\u22128 207.2 1 \u00d7 10\u22126 208.980 7 \u00d7 10\u22128 (209) (210) \u2014 \u2014 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix K 1173 The Chemical Elements Element Symbol Atomic Number Atomic Weight Percentage of Naturally Occurring Elements in the Universe Radon Francium Radium Actinium Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Rn Fr Ra Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Mendelevium Md Nobelium Lawrencium No Lr Rutherfordium Rf Dubnium Seaborgium Bohrium Db Sg Bh Table K1 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 (222) (223) (226) (227) \u2014 \u2014 \u2014 \u2014 232.038 4 \u00d7 10\u22128 231.036 \u2014 238.029 2 \u00d7 10\u22128 (237) (244) (243) (247) (247) (251) (252) (257) (258) (259) (262) (267) (268) (271) (272) \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 1174 Appendix K The Chemical Elements Element Symbol Atomic Number Atomic Weight Percentage of Naturally Occurring Elements in the Universe Hassium Meitnerium Hs Mt Darm", "stadtium Ds Roentgenium Copernicium Rg Cn Ununtrium Uut Flerovium Fl Ununpentium Uup Livermorium Lv Ununseptium Uus Ununoctium Uuo Table K1 108 109 110 111 112 113 114 115 116 117 118 (270) (276) (281) (280) (285) (284) (289) (288) (293) (294) (294) \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 Note: Some of the newest elements near the bottom of the table have suggested names that are still under review, and those names are not yet listed here. For example, Tennessine is suggested for element 117, after the state where the Oak Ridge National Laboratory is located. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix L 1175 L THE CONSTELLATIONS The Constellations Constellation (Latin name) Genitive Case Ending English Name or Description Abbreviation Approximate Position: \u03b1 (h) Approximate Position: \u03b4 (\u00b0) Andromeda Andromedae Princess of Ethiopia And Antila Apus Antilae Air pump Ant Apodis Bird of Paradise Aps Aquarius Aquarii Water bearer Aquila Aquilae Eagle Ara Aries Arae Arietis Altar Ram Auriga Aurigae Charioteer Bo\u00f6tes Bo\u00f6tis Herdsman Caelum Cael Graving tool Camelopardus Camelopardis Giraffe Cancer Cancri Crab Aqr Aql Ara Ari Aur Boo Cae Cam Cnc Canes Venatici Canum Venaticorum Hunting dogs CVn Canis Major Canis Majoris Big dog CMa Canis Minor Canis Minoris Little dog Capricornus Capricorni Sea goat Carina[1] Carinae Cassiopeia Cassiopeiae Keel of Argonauts\u2019 ship Queen of Ethiopia CMi Cap Car Cas Table L1 1 10 16 23 20 17 3 6 15 5 6 9 13 7 8 21 9 1 1 The four constellations Carina, Puppis, Pyxis, and Vela originally formed the single constellation Argo Navis. +40 \u221235 \u221275 \u221215 +5 \u221255 +20 +40 +30 \u221240 +70 +20 +40 \u221220 +5 \u221220 \u221260 +60 1176 Appendix L The Constellations Constellation (Latin name) Genitive Case Ending English Name or Description Abbreviation Approximate Position: \u03b1 (h) Approximate Position: \u03b4 (\u00b0) Centaurus Centauri Centaur Cen Cepheus Cephei King of Ethiopia Cep Cetus Ceti Sea monster (whale) Cet Chamaeleon Chamaeleontis Chameleon Cha Circinus Circini Compasses Columba Columbae Dove Cir Col Coma Berenices Comae Berenices Berenice\u2019s hair Com Corona Australis Corona Borealis Coronae Australis Coronae Borealis Southern crown CrA Northern crown CrB Corvus Corvi Crow Crater Crateris Cup Crv Crt Crux Crucis Cross (southern) Cru Cygnus Cygni Swan Delphinus Delphini Porpoise Dorado Doradus Swordfish Draco Draconis Dragon Equuleus Equulei Little horse Eridanus Eridani River Fornax Fornacis Furnace Cyg Del Dor Dra Equ Eri For Gemini Geminorum Twins Gem Table L1 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 13 22 2 11 15 6 13 19 16 12 11 12 21 21 5 17 21 3 3 7 \u221250 +70 \u221210 \u221280 \u221260 \u221235 +20 \u221240 +30 \u221220 \u221215 \u221260 +40 +10 \u221265 +65 +10 \u221220 \u221230 +20 Appendix L 1177 The Constellations Constellation (Latin name) Genitive Case Ending English Name or Description Abbreviation Approximate Position: \u03b1 (h) Approximate Position: \u03b4 (\u00b0) Grus Gruis Crane Hercules Herculis Hercules, son of Zeus Horologium Horologii Clock Hydra Hydrae Sea serpent Hydrus Hydri Water snake Indus Indi Indian Lacerta Lacertae Lizard Leo Leonis Lion Leo Minor Leonis Minoris Little lion Lepus Libra Lupus Lynx Lyra Leporis Hare Librae Balance Lupi Lyncis Lyrae Wolf Lynx Lyre or harp Gru Her Hor Hya Hyi Ind Lac Leo LMi Lep Lib Lup Lyn Lyr Mensa Mensae Table Mountain Men Microscopium Microscopii Microscope Monoceros Monocerotis Unicorn Musca Muscae Fly Mic Mon Mus Norma Normae Carpenter\ufffd", "\ufffds level Nor Octans Octantis Octant Ophiuchus Ophiuchi Holder of serpent Oct Oph Table L1 22 17 3 10 2 21 22 11 10 6 15 15 8 19 5 21 7 12 16 22 17 \u221245 +30 \u221260 \u221220 \u221275 \u221255 +45 +15 +35 \u221220 \u221215 \u221245 +45 +40 \u221280 \u221235 \u22125 \u221270 \u221250 \u221285 0 1178 Appendix L The Constellations Constellation (Latin name) Genitive Case Ending English Name or Description Abbreviation Approximate Position: \u03b1 (h) Approximate Position: \u03b4 (\u00b0) Orion Orionis Orion, the hunter Pavo Pavonis Peacock Pegasus Pegasi Perseus Persei Pegasus, the winged horse Perseus, hero who saved Andromeda Phoenix Phoenicis Phoenix Pictoris Easel Piscium Fishes Pictor Pisces Piscis Austrinus Ori Pav Peg Per Phe Pic Psc 5 20 22 3 1 6 1 Piscis Austrini Southern fish PsA 22 Puppis[2] Puppis Pyxis[3] (=Malus) Pyxidus Stern of the Argonauts\u2019 ship Compass of the Argonauts\u2019 ship Reticulum Reticuli Net Sagitta Sagittae Arrow Sagittarius Sagittarii Archer Scorpius Scorpii Scorpion Pup Pyx Ret Sge Sgr Sco Sculptor Sculptoris Sculptor\u2019s tools Scl Scutum Scuti Shield Serpens Serpentis Serpent Sct Ser Table L1 8 9 4 20 19 17 0 19 17 2 3 The four constellations Carina, Puppis, Pyxis, and Vela originally formed the single constellation Argo Navis. The four constellations Carina, Puppis, Pyxis, and Vela originally formed the single constellation Argo Navis. This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 +5 \u221265 +20 +45 \u221250 \u221255 +15 \u221230 \u221240 \u221230 \u221260 +10 \u221225 \u221240 \u221230 \u221210 0 Appendix L 1179 The Constellations Constellation (Latin name) Genitive Case Ending English Name or Description Abbreviation Approximate Position: \u03b1 (h) Approximate Position: \u03b4 (\u00b0) Sextans Sextantis Sextant Taurus Tauri Bull Telescopium Telescopii Telescope Triangulum Trianguli Triangle Triangulum Australe Trianguli Australis Southern triangle Tucana Tucanae Toucan Ursa Major Ursae Majoris Big bear Ursa Minor Ursae Minoris Little bear Vela[4] Velorum Sail of the Argonauts\u2019 ship Virgo Virginis Virgin Volans Volantis Flying fish Vulpecula Vulpeculae Fox Table L1 Sex Tau Tel Tri TrA Tuc UMa UMi Vel Vir Vol Vul 10 4 19 2 16 0 11 15 9 13 8 20 0 +15 \u221250 +30 \u221265 \u221265 +50 +70 \u221250 0 \u221270 +25 4 The four constellations Carina, Puppis, Pyxis, and Vela originally formed the single constellation Argo Navis. 1180 Appendix L This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Appendix M 1181 M STAR CHART AND SKY EVENT RESOURCES Star Charts To obtain graphic charts of the sky over your head tonight, there are a number of free online resources. One of the easiest is at the Skymaps website: http://www.skymaps.com/downloads.html. Here you can print out a free PDF version of the northern sky (for roughly latitude 40\u00b0, which is reasonable for much of the United States). Maps of the sky from the equator and the southern hemisphere can also be printed. Sky charts and other summaries of astronomical information can also be found at http://www.heavens- above.com/. A free, open-source computer application that shows the sky at any time from any place is called Stellarium. You can find it at http://www.stellarium.org/. Appendix B provides a section of information for finding astronomy apps for cell phones and tablets. Many of these also provide star charts. If you have a smartphone, you can find a variety of inexpensive apps that allow you to simply hold your phone upward to see what is in the sky behind your phone. A planisphere is a sky chart that turns inside a round frame and can show you the night sky at your latitude on any date and time of the year. You can buy them at science supply and telescope stores or online. Or you can construct your own from templates at these two websites: Dennis Schatz\u2019s AstroAdventures Star Finder: http://dennisschatz.org/activities/Star", "%20Finder.pdf Uncle Al\u2019s Star Wheel: http://www.lawrencehallofscience.org/do_science_now/ science_apps_and_activities/starwheels Calendars Of Night Sky Events The following resources offer calendars of night sky events. Sea and Sky: http://www.seasky.org/astronomy/astronomy.html (click on the fourth menu button) Sky & Telescope\u2019s This Week\u2019s Sky at a Glance: http://www.skyandtelescope.com/observing/sky-at-a- glance/ Astronomy Magazine\u2019s The Sky This Week: http://www.astronomy.com/observing/sky-this-week NASA Sky Cal: http://eclipse.gsfc.nasa.gov/SKYCAL/SKYCAL.html Night Sky Network Sky Planner: https://nightsky.jpl.nasa.gov/planner.cfm 1182 Appendix M This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Index INDEX Symbols 21-cm line, 695, 901, 947 3C 273, 964, 965 A absorption spectrum, 164, 181 accelerate, 58, 62 accretion, 504, 517 accretion disk, 876, 883, 972, 977, 977, 980 active galactic nuclei (AGN), 971, 986 active galaxies, 971, 986 active region, 544, 551 Adams, 92 adaptive optics, 205, 225 afterglow, 837 Alcor, 626 Algol, 633, 634 Allen Telescope Array, 1123 Alpha Centauri, 152, 624, 661, 663, 665 Alpher, 1062, 1067 Amino acids, 1101 amino acids, 1130 1183 asteroid, 258, 482 Bohr, 169 asteroid belt, 86, 96, 382, 451, 482 Bohr model, 172 asteroids, 241, 425, 450, 450, 451, Borexino experiment, 583 452, 453, 460, 499, 501, 515 Brahe, 70, 819 astrobiology, 1101, 1130 Brookhaven National Laboratory, astrology, 50, 62 581 astronomical unit, 658 brown dwarf, 614, 647 astronomical unit (AU), 74, 96 brown dwarfs, 603, 604, 630 astronomy, 13 B\u2013V index, 598 Atacama Large Millimeter/ submillimeter Array, 213 atmosphere, 391, 398 aurora, 549, 551 auroras, 534 Australia Telescope Compact Array, 213 B Baade, 915, 945 Balmer lines, 602 bar, 278, 295 Barnard, 701 barred spiral galaxy, 897 baryon cycle, 711, 715 basalt, 268, 295 C C-type asteroids, 453 calendar, 117 Callisto, 409, 410, 410, 418 Callisto\u2019s surface, 411 Canberra Deep Space Communication Complex, 212 Cannon, 599, 602 Carrington Event, 545 Cassini, 384 celestial equator, 34, 62 celestial poles, 34, 62 celestial sphere, 33, 62 centaurs, 474 central bulge, 897, 923 cepheid, 670, 679 Andromeda galaxy, 24, 25, 915, Bayer, 664 921, 935, 941, 974, 992, 1009 Bayer letter, 665 Ceres, 87, 87, 236, 237, 425, 450, Andromeda Galaxy, 922 Bell, 826 451, 453, 459, 459 angular momentum, 80, 96 BeppoSAX, 837, 838 Chandra X-Ray Observatory, 222 Ant Nebula, 767 Bessel, 661 Chandrasekhar, 808 anthropic principle, 1086, 1088 Betelgeuse, 41, 597, 601, 637, 664, Chandrasekhar limit, 807, 833, 844 antimatter, 565, 1063, 1085 664, 726, 771, 773, 774 CHARA, 215 antiparticle, 564 aperture, 225 apertures, 192 aphelion, 85, 96 apogee, 85, 96 Apollo program, 305 apparent brightness, 592, 597, 614 apparent magnitude, 62 apparent magnitudes, 46 Bethe, 1062 charge-coupled device (CCD), 225 Big Bang, 1045, 1060, 1080, 1088 charge-coupled devices (CCDs), big crunch, 1055 207 Big Telescope Altazimuth, 198 Charon, 423, 424, 430, 431 binary star system, 832 Che", "lyabinsk, 290 binary stars, 625, 647 binding energy, 566 biomarker, 1110, 1130 black dwarf, 810, 844 Cherenkov Telescope Array, 223 Chicxulub, 292 Chiron, 473 CHNOPS, 1105 black hole, 813, 826, 835, 841, 842, chromatic aberration, 194, 225 apparent solar time, 115, 136 867, 868, 870, 872, 873, 875, 877, chromosphere, 530, 551 Arecibo Observatory, 212 878, 883, 907, 908, 911, 911, 972, circumpolar zone, 35, 62 Aristarchus, 43 Aristotle, 43 Armstrong, 314 association, 777, 798 974, 981, 984 circumstellar disk, 507 black widow pulsar, 835 climate, 281 blackbody, 157, 181 blueshift, 178 closed universe, 1055, 1088 CNO cycle, 570 1184 Index COBE, 1069 crater, 315, 316 Earth-approaching asteroids, 461, cold dark matter, 1026, 1034 crater counts, 315 462, 463 collisions, 921 color index, 597, 614 Coma cluster, 1010 comet, 258, 464, 482 crust, 268, 295 Curtis, 970 craters, 251, 288 Earth\u2019s atmosphere, 278, 280 critical density, 1054, 1088 Earth\u2019s crust, 268, 270, 271 Comet Churyumov-Gerasimenko, Cygnus X-1, 853, 876 471, 472 comet dust, 468 Comet Hale-Bopp, 449, 466, 471 Comet Halley, 464, 465, 466, 469, 469, 481, 492 Comet LINEAR, 479 Comet Shoemaker-Levy 9, 477, 479, 480, 481 Comet Shoemaker\u2013Levy 9, 249, 250 comets, 241, 466, 468, 469, 491, 493, 501, 506, 691, 701 Compton Gamma-Ray Observatory, 836 conduction, 574, 584 conservation of angular momentum, 80 constellation, 37, 41 continuous spectrum, 164, 181 convection, 271, 295, 575, 577, 584 convection currents, 397 convective zone, 528 coordinates, 104 Copernican principle, 1100 Copernicus, 54, 55, 56, 60, 1094 core, 269, 295 corona, 531, 532, 551 coronal hole, 545, 551 coronal holes, 533 coronal mass ejection, 546, 581 coronal mass ejection (CME), 543, 551 cosmic haystack problem, 1122 cosmic microwave background (CMB), 1068, 1088 Cosmic rays, 708 cosmic rays, 715, 816 cosmological constant, 1044, 1088 cosmological principle, 1008, D Dactyl, 455, 456 dark energy, 1027, 1034, 1048, 1088 dark matter, 906, 907, 923, 937, 941, 943, 1020, 1021, 1022, 1025, 1027, 1076, 1088 dark matter halo, 898, 923 Darwin, 128 Davis, 581, 583 declination, 106, 136 degenerate gas, 807, 844 Deimos, 456, 456 Delta Cephei, 670, 674, 677 density, 79, 96, 239 detector, 190, 225 deuterium, 1065, 1088 differential galactic rotation, 903, 923 differential rotation, 536, 551 Differentiation, 247 differentiation, 258 disk, 895, 898 dispersion, 162, 181 distance, 656 DNA (deoxyribonucleic acid), 1103, 1130 Doppler, 176, 607 Doppler effect, 177, 179, 181, 509, 629, 741, 941 Doppler shift, 875, 966 Drake, 1125 Drake equation, 1125, 1130 Draper, 612 Dreyer, 691 dwarf planet, 425 E Eagle Nebula, 725 Earth\u2019s interior, 267 Earth\u2019s magnetic field, 269 Earth\u2019s magnetosphere, 270 Earth\u2019s surface, 268 eccentricity, 72, 96 eclipsing binary, 633, 647 ecliptic, 36, 38, 62 Eddington, 646 Effelsberg 100-m Telescope, 212 Einstein, 562, 563, 854, 859, 862, 1023, 1044 electromagnetic radiation, 148, 181 electromagnetic spectrum, 153, 181 ellip", "se, 71, 96 Elliptical galaxies, 937 elliptical galaxy, 956 emission spectrum, 164, 181 Enceladus, 409, 435, 436, 437, 516, 1113 energy, 562 energy flux, 160, 181 energy level, 170, 181 epicycle, 49, 62 equivalence principle, 856, 883 Eratosthenes, 44, 45 Eris, 236, 237, 425, 474 Eros, 457 escape speed, 90, 96 escape velocity, 867 Eta Carinae, 794, 795 Europa, 409, 410, 412, 413, 414, 418, 451, 1112, 1113 European Extremely Large Telescope, 197, 223 event horizon, 870, 870, 883, 974 evolution, 993 evolution (of galaxies), 1034 Ewen, 695 excitation, 173, 181 exoplanet, 509, 517, 740, 759 1034, 1051, 1056 Earth, 19, 87, 104, 129, 239, 247, exoplanets, 508 cosmology, 42, 62, 1088 265, 266, 270, 278, 283, 288, 371, expanding universe, 954 Crab Nebula, 12, 819, 827, 827, 385, 513, 515 829 extremophile, 1107, 1130 eyepiece, 193, 225 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Index 1185 F F Ring, 435, 440, 441 fault, 295 faults, 275 Fermi, 1100 Fermi Gamma-ray Space Telescope, 222 Fermi paradox, 1100 Fisher, 947 fission, 566, 584 Five-hundred-meter Aperture Spherical Telescope, 216 Flamsteed, 664 Flamsteed number, 665 flat universe, 1072, 1088 Fleming, 599 focus, 72, 96, 193, 225 formaldehyde, 698 Foucault, 106 Fraunhofer, 162, 598 frequency, 149, 181 fusion, 566, 567, 584, 786, 812, 1061, 1088 G Gaia, 222 Ghez, 913 Giacconi, 221 giant, 614 heat transfer, 576 heliocentric, 55, 62 helioseismology, 580, 584 giant impact hypothesis, 321 helium flash, 786, 798 Giant Magellan Telescope, 197 Henderson, 661 giant molecular clouds, 725, 759 Herbig, 732 giant planet, 258 Herbig-Haro (HH) object, 731, 759 giant planets, 238, 380, 385 Herman, 1062, 1067 giant stars, 637 giants, 605 Gilbert, 315 Herschel, 92, 625, 691, 892 Hertz, 151 Hertzsprung, 639 globular cluster, 775, 796, 798 Hewish, 826 Globular clusters, 775 globular clusters, 1050 Goldstone Deep Space highlands, 311, 329 Hipparchus, 46, 593 HL Tau, 508, 739 Communications Complex, 212 Hobby\u2013Eberly Telescope, 198 Goodricke, 633, 673 Holden, 701 Gran Telescopio Canarias, 198 homogeneous, 1007, 1034 grand unified theories, 1082, 1088 horizon, 32 granite, 268, 295 granulation, 529, 551 horizon (astronomical), 62 horizon distance, 1080 gravitational energy, 561 horoscope, 50, 62 gravitational lens, 1012, 1025 hot dark matter, 1026, 1034 gravitational redshift, 866, 883 hot Jupiter, 511 gravitational wave, 879, 879, 880, hot Jupiters, 743, 756 883 Hubble, 202, 673, 932, 933, 950, galactic cannibalism, 1003, 1034 gravitational waves, 843 1043 galactic mass, 941 galaxies, 932, 935, 941, 992 galaxy clusters, 1009 galaxy collisions, 1002 gravity, 81, 96, 865, 872 Hubble constant, 951, 956, 1046, great circle, 104, 136 Great Red Spot, 400 1055, 1073 Hubble law, 967, 1046 Green Bank Telescope, 212 Hubble Space Telescope, 205, 219, Galileo, 57, 58, 59, 60, 70, 103, 191, greenhouse effect, 286, 295, 349 221, 968, 972 383, 536, 625 Galileo probe, 384 Galle, 92 gamma rays, 154, 181 gamma-ray burst, 835, 840, 841 gamma-ray bursts, 840 Gamow, 1067 Ganymede, 248, 409, 410, 411, 412, 418 Gaspra,", " 455 Geller, 1014 Gemini North, 198 Gemini South, 198 gene, 1130 general theory of relativity, 854, 883 genes, 1103 geocentric, 32, 62 greenhouse gas, 295 greenhouse gases, 287 Gregorian calendar, 119 ground state, 173, 181 H H II region, 693, 715 habitable environment, 1105, 1130 habitable zone, 1115, 1130 Hale, 201, 613 Hale Telescope, 199 half-life, 252, 258 Halley, 76, 464, 465 halo, 898, 898, 923 Haro, 732 Hubble time, 1046 Hubble\u2019s law, 951, 952, 956, 1058 Huggins, 598, 608 Humason, 950 hydrostatic equilibrium, 574, 584 H\u2013R diagram, 641, 647, 675, 733, 768, 778, 809 I Ida, 455, 456, 515 igneous rock, 270, 295 impact craters, 315, 319 infalling material, 976 inflationary universe, 1081, 1088 infrared, 155, 181 Interface Region Imaging Hat Creek Radio Observatory, 213 Spectrograph, 531 Haumea, 237, 474 Interference, 214 1186 Index interference, 225 Keck, 613 interferometer, 214, 225 Keck I and II, 198, 216 interferometer array, 214, 225 Kelvin, 561 International Astronomical Union, Kepler, 71, 72, 73, 222, 740, 820, 245 905 International Date Line, 116, 136 Kepler mission, 750 International Gamma-Ray Kepler\u2019s first law, 74, 96 Astrophysics Laboratory, 222 Kepler\u2019s second law, 75, 96 International Thermonuclear Kepler\u2019s Supernova, 820, 834 Experimental Reactor (ITER), 571 Kepler\u2019s third law, 75, 75, 96 interstellar dust, 688, 703, 706, Koshiba, 583 715 Kuiper belt, 474, 477, 482 M M-type asteroids, 453 M87, 970, 973, 973, 1005 MACHO, 1025 Magellan Telescopes, 198 magnetic field, 390 magnetism, 147 magnetogram, 539 magnetosphere, 269, 295, 545, 549 magnitude, 593, 594, 614 main sequence, 641, 643, 647, 768, 769 main-sequence stars, 643 main-sequence turnoff, 782, 798 interstellar extinction, 704, 715 interstellar gas, 692, 694, 696 interstellar mass, 690 interstellar medium, 710, 710, 711 interstellar medium (ISM), 688, 715 interstellar molecules, 698 inverse square law, 152, 181 Io, 270, 391, 409, 410, 414, 415, 416, 417, 417 ion, 175, 181 ionization, 175, 181 IRAM, 213 iron meteorite, 517 irons, 498 irregular galaxy, 939, 956 isotope, 181 isotopes, 168 isotropic, 1007, 1034 Itokawa, 457 J James Clerk Maxwell Telescope, 213 James Webb Space Telescope, 222 Jansky, 210 Janssen, 191 jets, 976 L Large Binocular Telescope, 198, 216 Large Magellanic Cloud, 24, 31, 670, 672, 821, 825, 939, 1026 major axis, 72, 96 Makemake, 237, 425, 474 Large Synoptic Survey Telescope, mantle, 269, 295 198, 223 mare, 329 law of conservation of energy, 560 Le Verrier, 92 Leavitt, 670, 932 Lema\u00eetre, 949 Levy, 477 Lick, 612 light, 151 light curve, 669, 679 light-year, 663 LIGO, 879, 879 line broadening, 942 Lippershey, 191 lithium, 1065, 1088 Little Ice Age, 549 Local Bubble, 712, 715 Local Fluff, 713, 715 Local Group, 26, 1008, 1009, 1022, 1034 Mars, 12, 48, 72, 87, 233, 239, 247, 280, 285, 336, 337, 340, 341, 350, 353, 354, 357, 359, 359, 363, 369, 371, 413, 456, 513, 515, 1110 martian meteorites, 353 mass, 79 mass extinction, 291, 295 mass loss, 790 mass-luminosity relation, 630, 647 mass-to-light ratio, 943, 956, 1024 Maunder, 549 Maunder Minimum, 549, 551 Maxwell, 146, 147, 1083 Maxwell Mountains, 514 Mayor, 742 McDonald, 583", " mean solar time, 116, 136 Mercury, 87, 238, 239, 247, 248, 304, 322, 324, 326, 328, 862, 863 long-duration gamma-ray burst, Mercury\u2019s composition, 322 Jupiter, 60, 87, 205, 236, 238, 239, 840 246, 247, 250, 250, 379, 380, 381, Lovell Telescope, 212 381, 382, 384, 385, 386, 386, 387, Lowell, 337, 338, 340, 423 388, 389, 390, 393, 395, 398, 399, luminosity, 592, 614 400, 407, 409, 410, 432, 477, 479, luminosity class, 679 480, 481, 510, 604, 630 Jupiter\u2019s storms, 400 K Kajita, 583 Kant, 932 luminosity classes, 676 lunar eclipse, 129, 136 lunar exploration, 304 lunar history, 311 lunar missions, 308 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 Mercury\u2019s orbit, 322 Mercury\u2019s rotation, 324 Mercury\u2019s surface, 326 merger, 1003, 1034 meridian, 104, 136 MESSENGER, 326 Messier, 691 metamorphic rock, 295 Metamorphic rocks, 271 meteor, 242, 258, 490, 517 Index 1187 Meteor Crater, 291, 319 neutrino, 565, 568, 581, 582, 584, perigee, 85, 97 meteor shower, 491, 494, 517 814, 825, 1064, 1064, 1086 perihelion, 85, 97 meteorite, 242, 258, 495, 517 neutron star, 813, 826, 828, 831, period-luminosity relation, 670, meteorite fall, 496 meteorites, 499 meter, 656 Metius, 191 Michell, 868 834, 841, 842, 844 New Horizons, 428 672, 679 perturbation, 97 Newton, 76, 81, 161, 465, 854, 905 perturbations, 92 Newton\u2019s first law, 77, 96 phases, 120 Newton\u2019s second law, 77, 96 phases of the Moon, 136 microwave, 155, 181 Newton\u2019s third law, 77, 96 Phobos, 456, 456 Milky Way Galaxy, 21, 22, 31, 210, Nobeyama Radio Observatory, photochemistry, 395, 402 667, 689, 696, 703, 709, 710, 756, 213 photon, 151, 171, 182 892, 898, 899, 922, 923, 942, 976, nova, 832, 844 photon decoupling time, 1067, nuclear bulge, 897 1089 nucleosynthesis, 796, 798 photosphere, 528, 551 1009 Miller, 1102 Miller-Urey, 1102 millisecond pulsar, 844 millisecond pulsars, 835 mini-Neptune, 759 mini-Neptunes, 750 Mizar, 626 molecular cloud, 715 molecular clouds, 697, 724 momentum, 77, 96 Moon, 19, 120, 123, 124, 125, 129, 234, 237, 248, 252, 303, 304, 309, 310, 315, 320, 410, 513, 632, 660 moons, 408, 410 Moon\u2019s highlands, 311 Moon\u2019s interior, 309 Moon\u2019s origin, 321 Moon\u2019s properties, 304 Moon\u2019s surface, 310 Multi-Element Radio Linked Interferometer Network, 213 Multi-Mirror Telescope, 198 multiverse, 1086, 1088 N near-Earth asteroid (NEA), 482 nucleus, 167, 467 nucleus (of a comet), 482 nucleus (of an atom), 182 O Oliver, 1119 Olympus Mons, 354, 354, 354, 355, 514, 514 Omega Centauri, 775, 776 Oort, 475 Oort cloud, 475, 482 open cluster, 775, 798, 811 Open clusters, 776 open universe, 1055, 1088 optical SETI, 1127 orbit, 71, 97 orbital period, 73 orbital period (P), 97 orbital speed, 73, 97 organic compound, 1130 organic compounds, 1101 organic molecule, 1098, 1130 Orion Nebula, 16, 256, 507, 630, 692, 726, 727, 902 oscillating theory of the universe, near-Earth asteroids (NEAs), 461 1055 near-Earth object (NEO), 482 Osterbrock, 702 near-Earth objects (N", "EOs), 460 ozone, 279, 295 nebula, 688, 691, 715 Neptune, 88, 92, 238, 239, 247, 379, 381, 382, 383, 385, 386, 387, 389, 390, 391, 397, 397, 399, 401, 409, 421, 423, 432, 474 Neptune\u2019s Great Dark Spot, 401 Neptune\u2019s rings, 438, 439 P Pallas, 450 parallax, 43, 62, 659, 662, 667, 679 Parkes Observatory, 212 parsec, 662, 679 partial eclipse, 133 Pauli, 565 photosynthesis, 285, 295, 1103, 1130 Piazzi, 450 Pickering, 626 Pioneer, 1120 Pioneer 10, 382 Pioneer 11, 382 plage, 551 plages, 541 Planck, 170 Planck time, 1084 planet, 62 planetary evolution, 371 planetary nebula, 798 planetary nebulae, 789 planetary system, 506 planetesimal, 255, 504 planetesimals, 256, 258 planets, 39 plasma, 527, 551, 573 Plate tectonics, 271 plate tectonics, 295 Pleiades, 704, 777 Pluto, 236, 237, 237, 238, 248, 423, 424, 425, 427, 427, 428, 429, 430, 474 Polaris, 35, 47, 106, 595, 670 population I, 915 population I star, 923 population II, 915 population II star, 923 positron, 568, 584 precession, 46 precession (of Earth), 62 prime focus, 194, 225 Pauli exclusion principle, 806 primitive meteorites, 499 1188 Index primitive rock, 271, 295 right ascension, 106, 136 Slipher, 949 prominence, 551 prominences, 541 ring system, 431 ring systems, 433 Sloan Digital Sky Survey, 1016, 1018 proper motion, 608, 614 RNA (ribonucleic acid), 1103, 1130 Small Magellanic Cloud, 24, 31, protein, 1130 Proteins, 1101 ROTSE, 839 621, 670, 916, 939, 1026 RR Lyrae, 670, 674, 679 SN 1006, 819, 834 protogalactic cloud, 917 runaway greenhouse effect, 349, SN 1054, 819, 827, 834 proton-proton chain, 568, 570, 372 584 protoplanet, 505 protostar, 730, 759 Proxima Centauri, 20, 623, 624, 665, 690, 758, 1116 Ptolemy, 47, 49 pulsar, 830, 844 pulsars, 827 pulsating variable star, 679 pulsating variable stars, 670 Purcell, 695 Pythagoras, 42 Q quantum mechanics, 1049 quasar, 964, 971, 986, 996 quasars, 965, 979 Queloz, 742 quiescent, 976 R Radar, 216 radar, 225, 657 radial velocity, 178, 182, 607, 614 radiation, 146, 575, 584 radiative zone, 528 radio astronomy, 912 radio observations, 901 radio waves, 156, 182 radioactive decay, 824 radioactivity, 252, 258 Reber, 210 reddening, 704 reddening (interstellar), 715 redshift, 178, 949, 956, 1052 reflecting telescope, 194, 225 refracting telescope, 193, 225 Resolution, 204 resolution, 225 resonance, 440, 443 retrograde motion, 48, 62 rift zone, 296 rift zones, 274 Russell, 639, 640, 808 Rutherford, 166 S S-type asteroids, 453 Sagan, 242 Sagittarius A, 908, 909 Sagittarius A*, 908, 910, 911 satellite, 85, 88, 97 Saturn, 87, 238, 239, 241, 241, 246, 379, 381, 381, 384, 385, 386, 387, 388, 389, 394, 395, 396, 398, 409, 418, 432, 505 Saturn\u2019s rings, 385, 394, 433, 434, 435, 440, 505 scarps, 327 Schiaparelli, 337 Schmidt, 978 Schmitt, 306 Schwabe, 536 Schwarzschild, 870 science, 13 Scott, 312 seasons, 107, 112 sedimentary rock, 296 Sedimentary rocks, 271 seeing, 203, 225 seismic wave, 296 seismic waves, 267 selection effect, 624, 647 semimajor axis, 72, 97 SETI, 1123, 1124, 1130 Seyfert, 970 Shapley, 893 short-", "duration gamma-ray burst, 841 sidereal day, 114, 136 sidereal month, 123, 136 singularity, 873, 883 Sirius, 39, 115, 592, 593, 594, 601, 608, 609, 624, 629, 632, 636, 645, 646, 646, 664, 665, 810 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8 solar atmosphere, 525 solar cycle, 535, 539, 540, 544, 550 solar day, 114, 136 solar dynamo, 539 solar eclipse, 129, 136 solar flare, 542, 546, 551, 581 solar interior, 572, 579 solar maximum, 539, 546, 546, 548 solar month, 123, 136 solar nebula, 255, 258, 502, 503 solar neutrino, 581 solar wind, 533, 551 solstice, 110 Southern African Large Telescope, 198 space velocity, 609, 614 space weather, 545, 548 spacetime, 860, 860, 861, 883 spectra, 164 spectral class, 599, 600, 614 spectral line, 537 spectrometer, 162, 182 spectroscopic binary, 626, 647 spiral arm, 923 spiral arms, 903 spiral galaxies, 935, 936 spiral galaxy, 956 Spitzer Space Telescope, 222 Sp\u04e7rer, 549 Square Kilometre Array, 213 standard bulb, 945, 946, 1047 star cluster, 774 star diameter, 632 star formation, 729 star mass, 625 starburst, 1004, 1034 Stefan-Boltzmann law, 160, 182 stellar association, 775, 777 stellar color, 596 stellar evolution, 774, 785, 797 stellar spectra, 599 stellar temperature, 596 stellar wind, 731, 759 Index 1189 stones, 498 stony meteorite, 517 stony-iron meteorite, 517 stony-irons, 498 stratosphere, 279, 296 string theory, 1083 stromatolites, 1104, 1130 strong nuclear force, 566 Struve, 661 Subaru Telescope, 198 subduction, 275, 296 tidal heating, 412, 417, 443 tides, 125, 136 time zones, 114 Venus, 60, 87, 237, 239, 247, 280, 336, 336, 340, 341, 342, 343, 344, 345, 347, 349, 371, 513, 515, Titan, 248, 384, 409, 410, 418, 420, 593 421, 516, 1113 Very Large Array, 213 Titan\u2019s atmosphere, 419 Very Large Telescope, 198, 216 Titan\u2019s surface, 419 Tombaugh, 424, 426 total eclipse, 130 Very Long Baseline Array, 213 Vesta, 87, 248, 252, 450, 451, 454, 454, 459 trans-Neptunian object, 236 Virgo Cluster, 978, 1009, 1022 Tully-Fisher relation, 948 Voyager record, 1120 transit, 744, 759 Sudbury Neutrino Observatory, transit depth, 745 582 transit technique, 509 Sun, 20, 108, 129, 524, 535, 541, transition region, 531, 551 545, 560, 567, 572, 579, 596, 601, tree of life, 284 657, 713, 771 sunspot, 551, 580 triple-alpha process, 786, 798 Triton, 409, 421, 422, 423, 477 sunspot cycle, 536, 538, 551 Triton\u2019s surface, 422 sunspot maxima, 537 sunspot minima, 537 sunspots, 535 super-Earth, 759 super-Earths, 750 supercluster, 1034 superclusters, 1016 supergiant stars, 637 troposphere, 278, 296 Tully, 947 Tunguska, 290 Tycho\u2019s Supernova, 820, 834 type Ia supernova, 834, 946, 947, 953, 956 type Ia supernovae, 833 supermassive black hole, 910, type Ic supernova, 841 912, 923, 972, 985, 985 type II supernova, 814, 834, 844 supermassive black holes, 908 Tyson, 244 Virgo Supercluster, 26 visible light, 155, 182 visual binary, 625, 647 Vogel, 633 void, 1034 voids, 1016 volcano, 296 Volcanoes, 277 volume, 79 Voyager 2, 383 W", " watt, 560 wave equation, 150 wavelength, 149, 182 waves, 148 weakly interacting massive particles, 1076, 1089 weather, 280 Wegener, 272 Westerbork Synthesis Radio supernova, 709, 814, 815, 818 supernova 1987A, 822 Swift, 841 synchronous rotation, 123, 136 synchrotron radiation, 390, 402 T T Tauri star, 730 tail, 464, 482 Tarter, 1124 tectonic, 346, 372 telescope, 59, 190, 225 terrestrial planet, 258 terrestrial planets, 237 theory of general relativity, 1044, 1052, 1064, 1072 theory of relativity, 562 thermophile, 1107, 1130 Thirty-Meter Telescope, 197 Thomson, 166 tidal disruption event, 982 U ultraviolet, 155, 182 unbarred spiral galaxy, 897 Telescope, 213 universal law of gravitation, 81 Wheeler, 868, 1055 Uranus, 87, 92, 238, 239, 247, 379, Whipple, 468 381, 382, 385, 386, 387, 388, 389, white dwarf, 645, 647, 832 390, 397, 399, 409, 432, 438, 441 white dwarfs, 641, 806, 807, 826 Uranus\u2019 rings, 437, 438 Wide-field Infrared Survey Urey, 1102 V Valles Marineris, 12, 355, 356, 357 Van Allen, 392 Van Allen belt, 393 variable star, 669 variable stars, 945 Vega, 47, 596, 597, 601, 609, 611, 624, 661, 663, 900 Vela, 836 velocity, 77, 97 Explorer, 222 Wien\u2019s law, 159, 182 WMAP, 1071 Wolf, 450 X X-rays, 154, 182 XMM-Newton, 222 Y year, 36, 62 Yerkes, 201, 613 1190 Index Z Zeeman effect, 537, 538 zenith, 32, 63 zero-age main sequence, 768, 798 zodiac, 40, 63 This OpenStax book is available for free at http://cnx.org/content/col11992/1.8\u03bd1 \u03bdl. \u00b7\u00b7\u00b7 (1.51) Thus, each upper index gets transformed like a vector, and each lower index gets transformed like a dual vector. Although we have de\ufb01ned tensors as linear maps from sets of vectors and tangent vectors to R, there is nothing that forces us to act on a full collection of arguments. Thus, a (1, 1) tensor also acts as a map from vectors to vectors: T \u00b5 \u03bd : V \u03bd T \u00b5 \u03bdV \u03bd. \u2192 (1.52) You can check for yourself that T \u00b5 \u03bdV \u03bd is a vector (i.e. obeys the vector transformation law). Similarly, we can act one tensor on (all or part of) another tensor to obtain a third tensor. For example, U \u00b5 \u03bd = T \u00b5\u03c1 \u03c3S\u03c3 \u03c1\u03bd (1.53) is a perfectly good (1, 1) tensor. You may be concerned that this introduction to tensors has been somewhat too brief, given the esoteric nature of the material. In fact, the notion of tensors does not require a great deal of e\ufb00ort to master; it\u2019s just a matter of keeping the indices straight, and the rules for manipulating them are very natural. Indeed, a number of books like to de\ufb01ne tensors as 1 SPECIAL RELATIVITY AND FLAT SPACETIME 15 collections of numbers transforming according to (1.51). While this is operationally useful, it tends to obscure the deeper meaning of tensors as geometrical entities with a life independent of any chosen coordinate system. There is, however, one subtlety which we have glossed over. The notions of dual vectors and tensors and bases and linear maps belong to the realm of linear algebra, and are appropriate whenever we have an abstract vector space at hand. In the case of interest to us we have not just a vector space, but a vector space at each point in spacetime. More often than not we are interested in tensor \ufb01elds, which can be thought of as tensor-valued functions on spacetime. Fortunately, none of the manipulations we de\ufb01ned above really care whether we are dealing with a single vector space or a collection of vector spaces, one for each event. We will be able to get away with simply calling things functions of x\u00b5 when appropriate. However, you should keep straight the logical independence of the notions we have introduced and their speci\ufb01c application to spacetime and relativity.", " Now let\u2019s turn to some examples of tensors. First we consider the previous example of column vectors and their duals, row vectors. In this system a (1, 1) tensor is simply a matrix, M i j. Its action on a pair (\u03c9, V ) is given by usual matrix multiplication: M(\u03c9, V ) = (\u03c91 \u03c92 \u00b7 \u00b7 \u00b7 \u03c9nV n \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8eb \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed = \u03c9iM i jV j. (1.54) If you like, feel free to think of tensors as \u201cmatrices with an arbitrary number of indices.\u201d In spacetime, we have already seen some examples of tensors without calling them that. The most familiar example of a (0, 2) tensor is the metric, \u03b7\u00b5\u03bd. The action of the metric on two vectors is so useful that it gets its own name, the inner product (or dot product): \u03b7(V, W ) = \u03b7\u00b5\u03bdV \u00b5W \u03bd = V W. \u00b7 (1.55) Just as with the conventional Euclidean dot product, we will refer to two vectors whose dot product vanishes as orthogonal. Since the dot product is a scalar, it is left invariant under Lorentz transformations; therefore the basis vectors of any Cartesian inertial frame, which are chosen to be orthogonal by de\ufb01nition, are still orthogonal after a Lorentz transformation (despite the \u201cscissoring together\u201d we noticed earlier). The norm of a vector is de\ufb01ned to be inner product of the vector with itself; unlike in Euclidean space, this number is not positive de\ufb01nite: < 0, V \u00b5 is timelike = 0, V \u00b5 is lightlike or null > 0, V \u00b5 is spacelike. if \u03b7\u00b5\u03bdV \u00b5V \u03bd is \uf8f1 \uf8f4\uf8f2 \uf8f4\uf8f3 1 SPECIAL RELATIVITY AND FLAT SPACETIME 16 (A vector can have zero norm without being the zero vector.) You will notice that the terminology is the same as that which we earlier used to classify the relationship between two points in spacetime; it\u2019s no accident, of course, and we will go into more detail later. Another tensor is the Kronecker delta \u03b4\u00b5 \u03bd, of type (1, 1), which you already know the components of. Related to this and the metric is the inverse metric \u03b7\u00b5\u03bd, a type (2, 0) tensor de\ufb01ned as the inverse of the metric: \u03b7\u00b5\u03bd\u03b7\u03bd\u03c1 = \u03b7\u03c1\u03bd\u03b7\u03bd\u00b5 = \u03b4\u03c1 \u00b5. (1.56) In fact, as you can check, the inverse metric has exactly the same components as the metric itself. (This is only true in \ufb02at space in Cartesian coordinates, and will fail to hold in more general situations.) There is also the Levi-Civita tensor, a (0, 4) tensor: \u01eb\u00b5\u03bd\u03c1\u03c3 = \uf8f1 \uf8f4\uf8f2 +1 if \u00b5\u03bd\u03c1\u03c3 is an even permutation of 0123 1 if \u00b5\u03bd\u03c1\u03c3 is an odd permutation of 0123 \u2212 0 otherwise. (1.57) \uf8f4\uf8f3 \u2212 Here, a \u201cpermutation of 0123\u201d is an ordering of the numbers 0, 1, 2, 3 which can be obtained by starting with 0123 and exchanging two of the digits; an even permutation is obtained by an even number of such exchanges, and an odd permutation is obtained by an odd number. Thus, for example, \u01eb0321 = 1. It is a remarkable property of the above tensors \u2013 the metric, the inverse metric, the Kronecker delta, and the Levi-Civita tensor \u2013 that, even though they all transform according to the tensor transformation law (1.51), their components remain unchanged in any Cartesian coordinate system in \ufb02at spacetime. In some sense this makes them bad examples of tensors, since most tensors do not have this property. In fact, even these tensors do not have this property once we go to more general coordinate systems, with the single exception of the Kronecker delta. This tensor has exactly the same", " components in any coordinate system in any spacetime. This makes sense from the de\ufb01nition of a tensor as a linear map; the Kronecker tensor can be thought of as the identity map from vectors to vectors (or from dual vectors to dual vectors), which clearly must have the same components regardless of coordinate system. The other tensors (the metric, its inverse, and the Levi-Civita tensor) characterize the structure of spacetime, and all depend on the metric. We shall therefore have to treat them more carefully when we drop our assumption of \ufb02at spacetime. A more typical example of a tensor is the electromagnetic \ufb01eld strength tensor. We all know that the electromagnetic \ufb01elds are made up of the electric \ufb01eld vector Ei and the magnetic \ufb01eld vector Bi. (Remember that we use Latin indices for spacelike components 1,2,3.) Actually these are only \u201cvectors\u201d under rotations in space, not under the full Lorentz 1 SPECIAL RELATIVITY AND FLAT SPACETIME 17 group. In fact they are components of a (0, 2) tensor F\u00b5\u03bd, de\ufb01ned by E3 E2 \u2212 E1 \u2212 0 \u2212 E1 B2 B3 \u2212 0 B1 E2 \u2212 B3 0 B1 E3 B2 \u2212 0 \uf8f7 \uf8f7 \uf8f7 \uf8f8 F\u00b5\u03bd = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ed \uf8f6 = F\u03bd\u00b5. \u2212 (1.58) From this point of view it is easy to transform the electromagnetic \ufb01elds in one reference frame to those in another, by application of (1.51). The unifying power of the tensor formalism is evident: rather than a collection of two vectors whose relationship and transformation properties are rather mysterious, we have a single tensor \ufb01eld to describe all of electromagnetism. (On the other hand, don\u2019t get carried away; sometimes it\u2019s more convenient to work in a single coordinate system using the electric and magnetic \ufb01eld vectors.) With some examples in hand we can now be a little more systematic about some properties of tensors. First consider the operation of contraction, which turns a (k, l) tensor into a (k 1) tensor. Contraction proceeds by summing over one upper and one lower index: 1, l \u2212 \u2212 S\u00b5\u03c1 \u03c3 = T \u00b5\u03bd\u03c1 \u03c3\u03bd. (1.59) You can check that the result is a well-de\ufb01ned tensor. Of course it is only permissible to contract an upper index with a lower index (as opposed to two indices of the same type). Note also that the order of the indices matters, so that you can get di\ufb00erent tensors by contracting in di\ufb00erent ways; thus, T \u00b5\u03bd\u03c1 \u03c3\u03bd 6 = T \u00b5\u03c1\u03bd \u03c3\u03bd (1.60) in general. The metric and inverse metric can be used to raise and lower indices on tensors. That \u03b3\u03b4, we can use the metric to de\ufb01ne new tensors which we choose to is, given a tensor T \u03b1\u03b2 denote by the same letter T : T \u03b1\u03b2\u00b5 \u03b2 T\u00b5 T\u00b5\u03bd \u03b4 = \u03b7\u00b5\u03b3T \u03b1\u03b2 \u03b3\u03b4 = \u03b7\u00b5\u03b1T \u03b1\u03b2 \u03c1\u03c3 = \u03b7\u00b5\u03b1\u03b7\u03bd\u03b2\u03b7\u03c1\u03b3\u03b7\u03c3\u03b4T \u03b1\u03b2 \u03b3\u03b4, \u03b3\u03b4, \u03b3\u03b4, (1.61) and so forth. Notice that raising and lowering does not change the position of an index relative to other indices, and also that \u201cfree\u201d indices (which are not summed over) must be the same on both sides of an equation, while \u201cdummy\u201d indices (which are summed over) only appear on one side. As an example, we can turn vectors and dual vectors into each other by raising and lowering indices: V\u00b5 = \u03b7\u00b5\u03bdV \u03bd \u03c9\u00b5 = \u03b7\u00b5\u03bd\u03c9\u03bd. (1.62) 1 SPECIAL RELATIVITY AND FLAT SPACETIME 18 This explains why the gradient in three-dimensional \ufb02at Euclidean space is usually thought of as an ordinary vector, even though we have seen that it arises as a dual vector; in Euclidean space (where the metric is diagonal with all entries +1) a dual vector is turned into a vector with precisely the same components when we raise its index. You may then wonder why we have belabored the distinction at all. One simple reason,", " of course, is that in a Lorentzian spacetime the components are not equal: \u03c9\u00b5 = ( \u03c90, \u03c91, \u03c92, \u03c93). \u2212 (1.63) In a curved spacetime, where the form of the metric is generally more complicated, the difference is rather more dramatic. But there is a deeper reason, namely that tensors generally have a \u201cnatural\u201d de\ufb01nition which is independent of the metric. Even though we will always have a metric available, it is helpful to be aware of the logical status of each mathematical object we introduce. The gradient, and its action on vectors, is perfectly well de\ufb01ned regardless of any metric, whereas the \u201cgradient with upper indices\u201d is not. (As an example, we will eventually want to take variations of functionals with respect to the metric, and will therefore have to know exactly how the functional depends on the metric, something that is easily obscured by the index notation.) Continuing our compilation of tensor jargon, we refer to a tensor as symmetric in any of its indices if it is unchanged under exchange of those indices. Thus, if S\u00b5\u03bd\u03c1 = S\u03bd\u00b5\u03c1, (1.64) we say that S\u00b5\u03bd\u03c1 is symmetric in its \ufb01rst two indices, while if S\u00b5\u03bd\u03c1 = S\u00b5\u03c1\u03bd = S\u03c1\u00b5\u03bd = S\u03bd\u00b5\u03c1 = S\u03bd\u03c1\u00b5 = S\u03c1\u03bd\u00b5, (1.65) we say that S\u00b5\u03bd\u03c1 is symmetric in all three of its indices. Similarly, a tensor is antisymmetric (or \u201cskew-symmetric\u201d) in any of its indices if it changes sign when those indices are exchanged; thus, A\u00b5\u03bd\u03c1 = A\u03c1\u03bd\u00b5 \u2212 (1.66) means that A\u00b5\u03bd\u03c1 is antisymmetric in its \ufb01rst and third indices (or just \u201cantisymmetric in \u00b5 and \u03c1\u201d). If a tensor is (anti-) symmetric in all of its indices, we refer to it as simply (anti-) symmetric (sometimes with the redundant modi\ufb01er \u201ccompletely\u201d). As examples, the metric \u03b7\u00b5\u03bd and the inverse metric \u03b7\u00b5\u03bd are symmetric, while the Levi-Civita tensor \u01eb\u00b5\u03bd\u03c1\u03c3 and the electromagnetic \ufb01eld strength tensor F\u00b5\u03bd are antisymmetric. (Check for yourself that if you raise or lower a set of indices which are symmetric or antisymmetric, they remain that way.) Notice that it makes no sense to exchange upper and lower indices with each other, so don\u2019t succumb to the temptation to think of the Kronecker delta \u03b4\u03b1 \u03b2 as symmetric. On the other hand, the fact that lowering an index on \u03b4\u03b1 \u03b2 gives a symmetric tensor (in fact, the metric) 1 SPECIAL RELATIVITY AND FLAT SPACETIME 19 means that the order of indices doesn\u2019t really matter, which is why we don\u2019t keep track index placement for this one tensor. Given any tensor, we can symmetrize (or antisymmetrize) any number of its upper or lower indices. To symmetrize, we take the sum of all permutations of the relevant indices and divide by the number of terms: T(\u00b51\u00b52 \u00b5n)\u03c1 \u00b7\u00b7\u00b7 \u03c3 = 1 n! (T\u00b51\u00b52 \u00b5n\u03c1 \u00b7\u00b7\u00b7 \u03c3 + sum over permutations of indices \u00b51 \u00b7 \u00b7 \u00b7 \u00b5n), (1.67) while antisymmetrization comes from the alternating sum: 1 n! \u00b7\u00b7\u00b7 \u00b5n]\u03c1 \u03c3 = T[\u00b51\u00b52 (T\u00b51\u00b52 \u00b5n). (1.68) By \u201calternating sum\u201d we mean that permutations which are the result of an odd number of exchanges are given a minus sign, thus: \u03c3 + alternating sum over permutations of indices \u00b51 \u00b7 \u00b7 \u00b7 \u00b5n\u03c1 \u00b7\u00b7\u00b7 T[\u00b5\u03bd\u03c1]\u03c3 = 1 6 (T\u00b5\u03bd\u03c1\u03c3 \u2212 T\u00b5\u03c1\u03bd\u03c3 + T\u03c1\u00b5\u03bd\u03c3 \u2212 T\u03bd\u00b5\u03c1\u03c3 + T\u03bd\u03c1\u00b5\u03c3 \u2212 T\u03c1\u03bd\u00b5\u03c3). (1.69) Notice that round/square brackets denote symmetrization/antisymmetrization. Furthermore, we may sometimes want to (anti-) symmetrize indices which are not next to each other, in which case we use vertical bars to denote indices not included in the sum: T", "(\u00b5 | \u03bd | \u03c1) = 1 2 (T\u00b5\u03bd\u03c1 + T\u03c1\u03bd\u00b5). (1.70) Finally, some people use a convention in which the factor of 1/n! is omitted. The one used here is a good one, since (for example) a symmetric tensor satis\ufb01es S\u00b51 \u00b5n = S(\u00b51 \u00b7\u00b7\u00b7 \u00b5n), \u00b7\u00b7\u00b7 (1.71) and likewise for antisymmetric tensors. We have been very careful so far to distinguish clearly between things that are always true (on a manifold with arbitrary metric) and things which are only true in Minkowski space in Cartesian coordinates. One of the most important distinctions arises with partial derivatives. If we are working in \ufb02at spacetime with Cartesian coordinates, then the partial derivative of a (k, l) tensor is a (k, l + 1) tensor; that is, T\u03b1 \u00b5 \u03bd = \u2202\u03b1R\u00b5 \u03bd (1.72) transforms properly under Lorentz transformations. However, this will no longer be true in more general spacetimes, and we will have to de\ufb01ne a \u201ccovariant derivative\u201d to take the place of the partial derivative. Nevertheless, we can still use the fact that partial derivatives 1 SPECIAL RELATIVITY AND FLAT SPACETIME 20 give us tensor in this special case, as long as we keep our wits about us. (The one exception to this warning is the partial derivative of a scalar, \u2202\u03b1\u03c6, which is a perfectly good tensor [the gradient] in any spacetime.) We have now accumulated enough tensor know-how to illustrate some of these concepts using actual physics. Speci\ufb01cally, we will examine Maxwell\u2019s equations of electrodynamics. In 19th-century notation, these are B \u2212 \u2207 \u00d7 \u2202tE = 4\u03c0J E = 4\u03c0\u03c1 \u2207 \u00b7 E + \u2202tB = 1.73) Here, E and B are the electric and magnetic \ufb01eld 3-vectors, J is the current, \u03c1 is the charge density, and are the conventional curl and divergence. These equations are invariant under Lorentz transformations, of course; that\u2019s how the whole business got started. But they don\u2019t look obviously invariant; our tensor notation can \ufb01x that. Let\u2019s begin by writing these equations in just a slightly di\ufb00erent notation, \u2207\u00d7 and \u2207\u00b7 \u01ebijk\u2202jBk \u2212 \u22020Ei = 4\u03c0J i \u2202iEi = 4\u03c0J 0 \u01ebijk\u2202jEk + \u22020Bi = 0 \u2202iBi = 0. (1.74) In these expressions, spatial indices have been raised and lowered with abandon, without any attempt to keep straight where the metric appears. This is because \u03b4ij is the metric on \ufb02at 3-space, with \u03b4ij its inverse (they are equal as matrices). We can therefore raise and lower indices at will, since the components don\u2019t change. Meanwhile, the three-dimensional Levi-Civita tensor \u01ebijk is de\ufb01ned just as the four-dimensional one, although with one fewer index. We have replaced the charge density by J 0; this is legitimate because the density and current together form the current 4-vector, J \u00b5 = (\u03c1, J 1, J 2, J 3). From these expressions, and the de\ufb01nition (1.58) of the \ufb01eld strength tensor F\u00b5\u03bd, it is easy to get a completely tensorial 20th-century version of Maxwell\u2019s equations. Begin by noting that we can express the \ufb01eld strength with upper indices as F 0i = Ei F ij = \u01ebijkBk. (1.75) (To check this, note for example that F 01 = \u03b700\u03b711F01 and F 12 = \u01eb123B3.) Then the \ufb01rst two equations in (1.74) become \u2202jF ij \u2212 \u22020F 0i = 4\u03c0J i 1 SPECIAL RELATIVITY AND FLAT SPACETIME \u2202iF 0i = 4\u03c0J 0. 21 (1.76) Using the antisymmetry of F \u00b5\u03bd, we see that these may be combined into the single tensor equation \u2202\u00b5F \u03bd\u00b5 = 4\u03c0J \u03bd. (1.77) A similar line of reasoning, which is left as an exercise to you, reveals that the third and", " fourth equations in (1.74) can be written \u2202[\u00b5F\u03bd\u03bb] = 0. (1.78) The four traditional Maxwell equations are thus replaced by two, thus demonstrating the economy of tensor notation. More importantly, however, both sides of equations (1.77) and (1.78) manifestly transform as tensors; therefore, if they are true in one inertial frame, they must be true in any Lorentz-transformed frame. This is why tensors are so useful in relativity \u2014 we often want to express relationships without recourse to any reference frame, and it is necessary that the quantities on each side of an equation transform in the same way under change of coordinates. As a matter of jargon, we will sometimes refer to quantities which are written in terms of tensors as covariant (which has nothing to do with \u201ccovariant\u201d as opposed to \u201ccontravariant\u201d). Thus, we say that (1.77) and (1.78) together serve as the covariant form of Maxwell\u2019s equations, while (1.73) or (1.74) are non-covariant. Let us now introduce a special class of tensors, known as di\ufb00erential forms (or just \u201cforms\u201d). A di\ufb00erential p-form is a (0, p) tensor which is completely antisymmetric. Thus, scalars are automatically 0-forms, and dual vectors are automatically one-forms (thus explaining this terminology from a while back). We also have the 2-form F\u00b5\u03bd and the 4-form \u01eb\u00b5\u03bd\u03c1\u03c3. The space of all p-forms is denoted \u039bp, and the space of all p-form \ufb01elds over a manifold M is denoted \u039bp(M). A semi-straightforward exercise in combinatorics reveals that the number of linearly independent p-forms on an n-dimensional vector space is n!/(p!(n p)!). So at a point on a 4-dimensional spacetime there is one linearly independent 0-form, four 1-forms, six 2-forms, four 3-forms, and one 4-form. There are no p-forms for p > n, since all of the components will automatically be zero by antisymmetry. \u2212 Why should we care about di\ufb00erential forms? This is a hard question to answer without some more work, but the basic idea is that forms can be both di\ufb00erentiated and integrated, without the help of any additional geometric structure. We will delay integration theory until later, but see how to di\ufb00erentiate forms shortly. Given a p-form A and a q-form B, we can form a (p + q)-form known as the wedge product A \u2227 B by taking the antisymmetrized tensor product: (A \u2227 B)\u00b51 \u00b5p+q = \u00b7\u00b7\u00b7 (p + q)! p! q! A[\u00b51 \u00b5pB\u00b5p+1 \u00b7\u00b7\u00b7 \u00b5p+q]. \u00b7\u00b7\u00b7 (1.79) 1 SPECIAL RELATIVITY AND FLAT SPACETIME Thus, for example, the wedge product of two 1-forms is (A B)\u00b5\u03bd = 2A[\u00b5B\u03bd] = A\u00b5B\u03bd \u2212 \u2227 A\u03bdB\u00b5. Note that \u2227 so you can alter the order of a wedge product if you are careful with signs. \u2212 \u2227 A B = ( 1)pqB A, 22 (1.80) (1.81) The exterior derivative \u201cd\u201d allows us to di\ufb00erentiate p-form \ufb01elds to obtain (p+1)-form \ufb01elds. It is de\ufb01ned as an appropriately normalized antisymmetric partial derivative: The simplest example is the gradient, which is the exterior derivative of a 1-form: (dA)\u00b51 \u00b5p+1 = (p + 1)\u2202[\u00b51A\u00b52 \u00b7\u00b7\u00b7 \u00b5p+1]. \u00b7\u00b7\u00b7 (d\u03c6)\u00b5 = \u2202\u00b5\u03c6. (1.82) (1.83) The reason why the exterior derivative deserves special attention is that it is a tensor, even in curved spacetimes, unlike its cousin the partial derivative. Since we haven\u2019t studied curved spaces yet, we cannot prove this, but (1.82) de\ufb01nes an honest tensor no matter what the metric and coordinates are. Another interesting fact about exterior di\ufb00erentiation is that, for any form A, d(dA) = 0, (1.84) which is", " often written d2 = 0. This identity is a consequence of the de\ufb01nition of d and the fact that partial derivatives commute, \u2202\u03b1\u2202\u03b2 = \u2202\u03b2\u2202\u03b1 (acting on anything). This leads us to the following mathematical aside, just for fun. We de\ufb01ne a p-form A to be closed if dA = 0, and exact if A = dB for some (p 1)-form B. Obviously, all exact forms are closed, but the converse is not necessarily true. On a manifold M, closed p-forms comprise a vector space Z p(M), and exact forms comprise a vector space Bp(M). De\ufb01ne a new vector space as the closed forms modulo the exact forms: \u2212 H p(M) = Z p(M) Bp(M). (1.85) This is known as the pth de Rham cohomology vector space, and depends only on the topology of the manifold M. (Minkowski space is topologically equivalent to R4, which is uninteresting, so that all of the H p(M) vanish for p > 0; for p = 0 we have H 0(M) = R. Therefore in Minkowski space all closed forms are exact except for zero-forms; zero-forms can\u2019t be exact since there are no 1-forms for them to be the exterior derivative of.) It is striking that information about the topology can be extracted in this way, which essentially involves the solutions to di\ufb00erential equations. The dimension bp of the space H p(M) is \u2212 1 SPECIAL RELATIVITY AND FLAT SPACETIME 23 called the pth Betti number of M, and the Euler characteristic is given by the alternating sum n \u03c7(M) = 1)pbp. \u2212 ( Xp=0 (1.86) Cohomology theory is the basis for much of modern di\ufb00erential topology. Moving back to reality, the \ufb01nal operation on di\ufb00erential forms we will introduce is Hodge duality. We de\ufb01ne the \u201cHodge star operator\u201d on an n-dimensional manifold as a map from p-forms to (n p)-forms, \u2212 A)\u00b51 ( \u2217 \u00b5n\u2212p = \u00b7\u00b7\u00b7 1 p! \u01eb\u03bd1 \u03bdp \u00b7\u00b7\u00b7 \u00b51 \u00b5n\u2212pA\u03bd1 \u00b7\u00b7\u00b7 \u03bdp, \u00b7\u00b7\u00b7 (1.87) mapping A to \u201cA dual\u201d. Unlike our other operations on forms, the Hodge dual does depend on the metric of the manifold (which should be obvious, since we had to raise some indices on the Levi-Civita tensor in order to de\ufb01ne (1.87)). Applying the Hodge star twice returns either plus or minus the original form: A = ( 1)s+p(n p)A, \u2212 (1.88) \u2212 \u2217 \u2217 where s is the number of minus signs in the eigenvalues of the metric (for Minkowski space, s = 1). Two facts on the Hodge dual: First, \u201cduality\u201d in the sense of Hodge is di\ufb00erent than the relationship between vectors and dual vectors, although both can be thought of as the space of linear maps from the original space to R. Notice that the dimensionality of the space of p)-forms is equal to that of the space of p-forms, so this has at least a chance of being (n p)-form acting on a p-form is true. In the case of forms, the linear map de\ufb01ned by an (n p) is an (n given by the dual of the wedge product of the two forms. Thus, if A(n p)-form and B(p) is a p-form at some point in spacetime, we have \u2212 \u2212 \u2212 \u2212 (A(n \u2212 p) B(p)) R. \u2208 \u2227 \u2217 (1.89) The second fact concerns di\ufb00erential forms in 3-dimensional Euclidean space. The Hodge dual of the wedge product of two 1-forms gives another 1-form: (U \u2217 \u2227 V )i = \u01ebi jkUjVk. (1.90) (All of the prefactors cancel.) Since 1-forms in Euclidean space are just like vectors, we have a map from two vectors to a single vector. You should convince yourself that this is just the conventional cross product, and that the appearance of the Levi-Civita tensor explains why the cross product changes sign under parity (interchange of two coordinates, or equivalently basis vectors). This is why the cross product only exists", " in three dimensions \u2014 because only 1 SPECIAL RELATIVITY AND FLAT SPACETIME 24 in three dimensions do we have an interesting map from two dual vectors to a third dual vector. If you wanted to you could de\ufb01ne a map from n 1 one-forms to a single one-form, but I\u2019m not sure it would be of any use. \u2212 Electrodynamics provides an especially compelling example of the use of di\ufb00erential forms. From the de\ufb01nition of the exterior derivative, it is clear that equation (1.78) can be concisely expressed as closure of the two-form F\u00b5\u03bd: dF = 0. (1.91) Does this mean that F is also exact? Yes; as we\u2019ve noted, Minkowski space is topologically trivial, so all closed forms are exact. There must therefore be a one-form A\u00b5 such that F = dA. (1.92) This one-form is the familiar vector potential of electromagnetism, with the 0 component given by the scalar potential, A0 = \u03c6. If one starts from the view that the A\u00b5 is the fundamental \ufb01eld of electromagnetism, then (1.91) follows as an identity (as opposed to a dynamical law, an equation of motion). Gauge invariance is expressed by the observation that the theory is invariant under A A + d\u03bb for some scalar (zero-form) \u03bb, and this is also immediate from the relation (1.92). The other one of Maxwell\u2019s equations, (1.77), can be expressed as an equation between three-forms: \u2192 d( \u2217 F ) = 4\u03c0( J), \u2217 (1.93) where the current one-form J is just the current four-vector with index lowered. Filling in the details is left for you to do. As an intriguing aside, Hodge duality is the basis for one of the hottest topics in theoretical physics today. It\u2019s hard not to notice that the equations (1.91) and (1.93) look very similar. Indeed, if we set J\u00b5 = 0, the equations are invariant under the \u201cduality transformations1.94) We therefore say that the vacuum Maxwell\u2019s equations are duality invariant, while the invariance is spoiled in the presence of charges. We might imagine that magnetic as well as electric monopoles existed in nature; then we could add a magnetic current term 4\u03c0( JM ) to the right hand side of (1.91), and the equations would be invariant under duality transformations plus the additional replacement J JM. (Of course a nonzero right hand side to (1.91) is inconsistent with F = dA, so this idea only works if A\u00b5 is not a fundamental variable.) Long ago Dirac considered the idea of magnetic monopoles and showed that a necessary condition for their existence is that the fundamental monopole charge be inversely proportional to \u2194 \u2217 1 SPECIAL RELATIVITY AND FLAT SPACETIME 25 the fundamental electric charge. Now, the fundamental electric charge is a small number; electrodynamics is \u201cweakly coupled\u201d, which is why perturbation theory is so remarkably successful in quantum electrodynamics (QED). But Dirac\u2019s condition on magnetic charges implies that a duality transformation takes a theory of weakly coupled electric charges to a theory of strongly coupled magnetic monopoles (and vice-versa). Unfortunately monopoles don\u2019t exist (as far as we know), so these ideas aren\u2019t directly applicable to electromagnetism; but there are some theories (such as supersymmetric non-abelian gauge theories) for which it has been long conjectured that some sort of duality symmetry may exist. If it did, we would have the opportunity to analyze a theory which looked strongly coupled (and therefore hard to solve) by looking at the weakly coupled dual version. Recently work by Seiberg and Witten and others has provided very strong evidence that this is exactly what happens in certain theories. The hope is that these techniques will allow us to explore various phenomena which we know exist in strongly coupled quantum \ufb01eld theories, such as con\ufb01nement of quarks in hadrons. We\u2019ve now gone over essentially everything there is to know about the care and feeding of tensors. In the next section we will look more carefully at the rigorous de\ufb01nitions of manifolds and tensors, but the basic mechanics have been pretty well covered. Before jumping to more abstract mathematics, let\u2019s review how physics works in Minkowski spacetime. Start with the worldline of a single particle. This is speci\ufb01ed by a map R M", ", where M is the manifold representing spacetime; we usually think of the path as a parameterized curve x\u00b5(\u03bb). As mentioned earlier, the tangent vector to this path is dx\u00b5/d\u03bb (note that it depends on the parameterization). An object of primary interest is the norm of the tangent vector, which serves to characterize the path; if the tangent vector is timelike/null/spacelike at some parameter value \u03bb, we say that the path is timelike/null/spacelike at that point. This explains why the same words are used to classify vectors in the tangent space and intervals between two points \u2014 because a straight line connecting, say, two timelike separated points will itself be timelike at every point along the path. \u2192 Nevertheless, it\u2019s important to be aware of the sleight of hand which is being pulled here. The metric, as a (0, 2) tensor, is a machine which acts on two vectors (or two copies of the same vector) to produce a number. It is therefore very natural to classify tangent vectors according to the sign of their norm. But the interval between two points isn\u2019t something quite so natural; it depends on a speci\ufb01c choice of path (a \u201cstraight line\u201d) which connects the points, and this choice in turn depends on the fact that spacetime is \ufb02at (which allows a unique choice of straight line between the points). A more natural object is the line element, or in\ufb01nitesimal interval: ds2 = \u03b7\u00b5\u03bddx\u00b5dx\u03bd. (1.95) From this de\ufb01nition it is tempting to take the square root and integrate along a path to obtain a \ufb01nite interval. But since ds2 need not be positive, we de\ufb01ne di\ufb00erent procedures 1 SPECIAL RELATIVITY AND FLAT SPACETIME 26 t \u00b5 \u03bb x ( ) timelike \u00b5 dx -d \u03bb null spacelike x for di\ufb00erent cases. For spacelike paths we de\ufb01ne the path length \u2206s = Z \u03b7\u00b5\u03bd s dx\u00b5 d\u03bb dx\u03bd d\u03bb d\u03bb, (1.96) where the integral is taken over the path. For null paths the interval is zero, so no extra formula is required. For timelike paths we de\ufb01ne the proper time \u2206\u03c4 = Z \u03b7\u00b5\u03bd s\u2212 dx\u00b5 d\u03bb dx\u03bd d\u03bb d\u03bb, (1.97) which will be positive. Of course we may consider paths that are timelike in some places and spacelike in others, but fortunately it is seldom necessary since the paths of physical particles never change their character (massive particles move on timelike paths, massless particles move on null paths). Furthermore, the phrase \u201cproper time\u201d is especially appropriate, since \u03c4 actually measures the time elapsed on a physical clock carried along the path. This point of view makes the \u201ctwin paradox\u201d and similar puzzles very clear; two worldlines, not necessarily straight, which intersect at two di\ufb00erent events in spacetime will have proper times measured by the integral (1.97) along the appropriate paths, and these two numbers will in general be di\ufb00erent even if the people travelling along them were born at the same time. Let\u2019s move from the consideration of paths in general to the paths of massive particles (which will always be timelike). Since the proper time is measured by a clock travelling on a timelike worldline, it is convenient to use \u03c4 as the parameter along the path. That is, we use (1.97) to compute \u03c4 (\u03bb), which (if \u03bb is a good parameter in the \ufb01rst place) we can invert to obtain \u03bb(\u03c4 ), after which we can think of the path as x\u00b5(\u03c4 ). The tangent vector in this 1 SPECIAL RELATIVITY AND FLAT SPACETIME parameterization is known as the four-velocity, U \u00b5: U \u00b5 = dx\u00b5 d\u03c4. Since d\u03c4 2 = \u03b7\u00b5\u03bddx\u00b5dx\u03bd, the four-velocity is automatically normalized: \u2212 \u03b7\u00b5\u03bdU \u00b5U \u03bd = 1. \u2212 27 (1.98) (1.99) (It will always be negative, since we are only de\ufb01ning it for timelike trajectories. You could de\ufb01ne an analogous vector for spacelike paths as well; null paths give some extra problems since the norm is zero.) In the rest frame of a particle, its four-velocity has components U \u00b5 = (1,", " 0, 0, 0). A related vector is the energy-momentum four-vector, de\ufb01ned by p\u00b5 = mU \u00b5, (1.100) where m is the mass of the particle. The mass is a \ufb01xed quantity independent of inertial frame; what you may be used to thinking of as the \u201crest mass.\u201d It turns out to be much more convenient to take this as the mass once and for all, rather than thinking of mass as depending on velocity. The energy of a particle is simply p0, the timelike component of its energy-momentum vector. Since it\u2019s only one component of a four-vector, it is not invariant under Lorentz transformations; that\u2019s to be expected, however, since the energy of a particle at rest is not the same as that of the same particle in motion. In the particle\u2019s rest frame we have p0 = m; recalling that we have set c = 1, we \ufb01nd that we have found the equation that made Einstein a celebrity, E = mc2. (The \ufb01eld equations of general relativity are actually 1 2Rg\u00b5\u03bd = 8\u03c0GT\u00b5\u03bd\u201d doesn\u2019t elicit the visceral much more important than this one, but \u201cR\u00b5\u03bd \u2212 reaction that you get from \u201cE = mc2\u201d.) In a moving frame we can \ufb01nd the components of p\u00b5 by performing a Lorentz transformation; for a particle moving with (three-) velocity v along the x axis we have (1.101) where \u03b3 = 1/\u221a1 2 mv2 (what we usually think of as rest energy plus kinetic energy) and p1 = mv (what we usually think of as [Newtonian] momentum). So the energy-momentum vector lives up to its name. v2. For small v, this gives p0 = m + 1 p\u00b5 = (\u03b3m, v\u03b3m, 0, 0), \u2212 The centerpiece of pre-relativity physics is Newton\u2019s 2nd Law, or f = ma = dp/dt. An analogous equation should hold in SR, and the requirement that it be tensorial leads us directly to introduce a force four-vector f \u00b5 satisfying d2 d\u03c4 2 x\u00b5(\u03c4 ) = The simplest example of a force in Newtonian physics is the force due to gravity. In relativity, however, gravity is not described by a force, but rather by the curvature of spacetime itself. f \u00b5 = m p\u00b5(\u03c4 ). (1.102) d d\u03c4 1 SPECIAL RELATIVITY AND FLAT SPACETIME 28 Instead, let us consider electromagnetism. The three-dimensional Lorentz force is given by f = q(E + v B), where q is the charge on the particle. We would like a tensorial generalization of this equation. There turns out to be a unique answer: \u00d7 f \u00b5 = qU \u03bbF\u03bb \u00b5. (1.103) You can check for yourself that this reduces to the Newtonian version in the limit of small velocities. Notice how the requirement that the equation be tensorial, which is one way of guaranteeing Lorentz invariance, severely restricted the possible expressions we could get. This is an example of a very general phenomenon, in which a small number of an apparently endless variety of possible physical laws are picked out by the demands of symmetry. Although p\u00b5 provides a complete description of the energy and momentum of a particle, for extended systems it is necessary to go further and de\ufb01ne the energy-momentum tensor (sometimes called the stress-energy tensor), T \u00b5\u03bd. This is a symmetric (2, 0) tensor which tells us all we need to know about the energy-like aspects of a system: energy density, pressure, stress, and so forth. A general de\ufb01nition of T \u00b5\u03bd is \u201cthe \ufb02ux of four-momentum p\u00b5 across a surface of constant x\u03bd\u201d. To make this more concrete, let\u2019s consider the very general category of matter which may be characterized as a \ufb02uid \u2014 a continuum of matter described by macroscopic quantities such as temperature, pressure, entropy, viscosity, etc. In fact this de\ufb01nition is so general that it is of little use. In general relativity essentially all interesting types of matter can be thought of as perfect \ufb02uids, from stars to electromagnetic \ufb01elds to the entire universe. Schutz de\ufb01nes a perfect \ufb02uid to be one with no heat conduction and no viscosity, while Weinberg de\ufb01nes it as", " a \ufb02uid which looks isotropic in its rest frame; these two viewpoints turn out to be equivalent. Operationally, you should think of a perfect \ufb02uid as one which may be completely characterized by its pressure and density. To understand perfect \ufb02uids, let\u2019s start with the even simpler example of dust. Dust is de\ufb01ned as a collection of particles at rest with respect to each other, or alternatively as a perfect \ufb02uid with zero pressure. Since the particles all have an equal velocity in any \ufb01xed inertial frame, we can imagine a \u201cfour-velocity \ufb01eld\u201d U \u00b5(x) de\ufb01ned all over spacetime. (Indeed, its components are the same at each point.) De\ufb01ne the number-\ufb02ux four-vector to be N \u00b5 = nU \u00b5, (1.104) where n is the number density of the particles as measured in their rest frame. Then N 0 is the number density of particles as measured in any other frame, while N i is the \ufb02ux of particles in the xi direction. Let\u2019s now imagine that each of the particles have the same mass m. Then in the rest frame the energy density of the dust is given by \u03c1 = nm. (1.105) 1 SPECIAL RELATIVITY AND FLAT SPACETIME 29 By de\ufb01nition, the energy density completely speci\ufb01es the dust. But \u03c1 only measures the energy density in the rest frame; what about other frames? We notice that both n and m are 0-components of four-vectors in their rest frame; speci\ufb01cally, N \u00b5 = (n, 0, 0, 0) and p\u00b5 = (m, 0, 0, 0). Therefore \u03c1 is the \u00b5 = 0, \u03bd = 0 component of the tensor p N as measured in its rest frame. We are therefore led to de\ufb01ne the energy-momentum tensor for dust: \u2297 T \u00b5\u03bd dust = p\u00b5N \u03bd = nmU \u00b5U \u03bd = \u03c1U \u00b5U \u03bd, (1.106) where \u03c1 is de\ufb01ned as the energy density in the rest frame. Having mastered dust, more general perfect \ufb02uids are not much more complicated. Remember that \u201cperfect\u201d can be taken to mean \u201cisotropic in its rest frame.\u201d This in turn means that T \u00b5\u03bd is diagonal \u2014 there is no net \ufb02ux of any component of momentum in an orthogonal direction. Furthermore, the nonzero spacelike components must all be equal, T 11 = T 22 = T 33. The only two independent numbers are therefore T 00 and one of the T ii; we can choose to call the \ufb01rst of these the energy density \u03c1, and the second the pressure p. (Sorry that it\u2019s the same letter as the momentum.) The energy-momentum tensor of a perfect \ufb02uid therefore takes the following form in its rest frame \u00b5\u03bd = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ed \uf8f6. (1.107) We would like, of course, a formula which is good in any frame. For dust we had T \u00b5\u03bd = \u03c1U \u00b5U \u03bd, so we might begin by guessing (\u03c1 + p)U \u00b5U \u03bd, which gives \uf8eb \uf8ec \uf8ec \uf8ec \uf8ed To get the answer we want we must therefore add. 1.1081.109) Fortunately, this has an obvious covariant generalization, namely p\u03b7\u00b5\u03bd. Thus, the general form of the energy-momentum tensor for a perfect \ufb02uid is T \u00b5\u03bd = (\u03c1 + p)U \u00b5U \u03bd + p\u03b7\u00b5\u03bd. (1.110) This is an important formula for applications such as stellar structure and cosmology. 1 SPECIAL RELATIVITY AND FLAT SPACETIME 30 As further examples, let\u2019s consider the energy-momentum tensors of electromagnetism and scalar \ufb01eld theory. Without any explanation at all, these are given by 1 T \u00b5\u03bd e+m = \u2212 4\u03c0 (F \u00b5\u03bbF \u03bd 1 4 \u03bb \u2212 \u03b7\u00b5\u03bdF \u03bb\u03c3F\u03bb\u03c3), (1.111) and 1 2 You can check for yourself that, for example, T 00 in each case is equal to what you would expect the energy density to be. T \u00b5\u03bd scalar = \u03b7\u00b5\u03bb\u03b7\u03bd\u03c3\u2202\u03bb\u03c6\u2202\u03c3\u03c6 \u03b7\ufffd", "\ufffd\u03bd(\u03b7\u03bb\u03c3\u2202\u03bb\u03c6\u2202\u03c3\u03c6 + m2\u03c62). (1.112) \u2212 Besides being symmetric, T \u00b5\u03bd has the even more important property of being conserved. In this context, conservation is expressed as the vanishing of the \u201cdivergence\u201d: \u2202\u00b5T \u00b5\u03bd = 0. (1.113) This is a set of four equations, one for each value of \u03bd. The \u03bd = 0 equation corresponds to conservation of energy, while \u2202\u00b5T \u00b5k = 0 expresses conservation of the kth component of the momentum. We are not going to prove this in general; the proof follows for any individual source of matter from the equations of motion obeyed by that kind of matter. In fact, one way to de\ufb01ne T \u00b5\u03bd would be \u201ca (2, 0) tensor with units of energy per volume, which is conserved.\u201d You can prove conservation of the energy-momentum tensor for electromagnetism, for example, by taking the divergence of (1.111) and using Maxwell\u2019s equations as previously discussed. A \ufb01nal aside: we have already mentioned that in general relativity gravitation does not count as a \u201cforce.\u201d As a related point, the gravitational \ufb01eld also does not have an energymomentum tensor. In fact it is very hard to come up with a sensible local expression for the energy of a gravitational \ufb01eld; a number of suggestions have been made, but they all have their drawbacks. Although there is no \u201ccorrect\u201d answer, it is an important issue from the point of view of asking seemingly reasonable questions such as \u201cWhat is the energy emitted per second from a binary pulsar as the result of gravitational radiation?\u201d December 1997 Lecture Notes on General Relativity Sean M. Carroll 2 Manifolds After the invention of special relativity, Einstein tried for a number of years to invent a Lorentz-invariant theory of gravity, without success. His eventual breakthrough was to replace Minkowski spacetime with a curved spacetime, where the curvature was created by (and reacted back on) energy and momentum. Before we explore how this happens, we have to learn a bit about the mathematics of curved spaces. First we will take a look at manifolds in general, and then in the next section study curvature. In the interest of generality we will usually work in n dimensions, although you are permitted to take n = 4 if you like. A manifold (or sometimes \u201cdi\ufb00erentiable manifold\u201d) is one of the most fundamental concepts in mathematics and physics. We are all aware of the properties of n-dimensional Euclidean space, Rn, the set of n-tuples (x1,..., xn). The notion of a manifold captures the idea of a space which may be curved and have a complicated topology, but in local regions looks just like Rn. (Here by \u201clooks like\u201d we do not mean that the metric is the same, but only basic notions of analysis like open sets, functions, and coordinates.) The entire manifold is constructed by smoothly sewing together these local regions. Examples of manifolds include: \u2022 \u2022 \u2022 Rn itself, including the line (R), the plane (R2), and so on. This should be obvious, since Rn looks like Rn not only locally but globally. The n-sphere, Sn. This can be de\ufb01ned as the locus of all points some \ufb01xed distance from the origin in Rn+1. The circle is of course S1, and the two-sphere S2 will be one of our favorite examples of a manifold. The n-torus T n results from taking an n-dimensional cube and identifying opposite sides. Thus T 2 is the traditional surface of a doughnut. identify opposite sides 31 2 MANIFOLDS 32 \u2022 A Riemann surface of genus g is essentially a two-torus with g holes instead of just one. S2 may be thought of as a Riemann surface of genus zero. For those of you who know what the words mean, every \u201ccompact orientable boundaryless\u201d two-dimensional manifold is a Riemann surface of some genus. genus 0 genus 1 genus 2 \u2022 \u2022 More abstractly, a set of continuous transformations such as rotations in Rn forms a manifold. Lie groups are manifolds which also have a group structure. The direct product of two manifolds is a manifold. That is, given manifolds M and M \u2032 of dimension n and n\u2032, we can construct a manifold M M \u2032, of dimension n + n\u2032, consisting of ordered pairs (p, p\u2032) for all p", " M and p\u2032 \u2208 With all of these examples, the notion of a manifold may seem vacuous; what isn\u2019t a manifold? There are plenty of things which are not manifolds, because somewhere they do not look locally like Rn. Examples include a one-dimensional line running into a twodimensional plane, and two cones stuck together at their vertices. (A single cone is okay; you can imagine smoothing out the vertex.) \u00d7 M \u2032. \u2208 We will now approach the rigorous de\ufb01nition of this simple idea, which requires a number of preliminary de\ufb01nitions. Many of them are pretty clear anyway, but it\u2019s nice to be complete. 2 MANIFOLDS 33 The most elementary notion is that of a map between two sets. (We assume you know what a set is.) Given two sets M and N, a map \u03c6 : M N is a relationship which assigns, to each element of M, exactly one element of N. A map is therefore just a simple generalization of a function. The canonical picture of a map looks like this: \u2192 M \u03d5 N Given two maps \u03c6 : A C \u2192 by the operation (\u03c8 C. The \u2208 order in which the maps are written makes sense, since the one on the right acts \ufb01rst. In pictures: \u2192 \u03c6)(a) = \u03c8(\u03c6(a)). So a C, we de\ufb01ne the composition \u03c8 B, and thus (\u03c8 B and \u03c8 : B \u25e6 \u03c6)(a) A, \u03c6(a map \u03c6 is called one-to-one (or \u201cinjective\u201d) if each element of N has at most one element of M mapped into it, and onto (or \u201csurjective\u201d) if each element of N has at least one element of M mapped into it. (If you think about it, a better name for \u201cone-to-one\u201d R. Then \u03c6(x) = ex is one-to-one, but would be \u201ctwo-to-two\u201d.) Consider a function \u03c6 : R x is onto, but not one-to-one; \u03c6(x) = x3 is both; and \u03c6(x) = x2 is not onto; \u03c6(x) = x3 neither. \u2192 \u2212 The set M is known as the domain of the map \u03c6, and the set of points in N which M N, the set of elements of 1(U). A map which is gets mapped into is called the image of \u03c6. For some subset U M which get mapped to U is called the preimage of U under \u03c6, or \u03c6\u2212 \u2282 2 MANIFOLDS 34 e x 3 x x one-to-one, not onto 3 x - x 2 x x onto, not one-to-one x both x neither both one-to-one and onto is known as invertible (or \u201cbijective\u201d). In this case we can de\ufb01ne 1 is the inverse map \u03c6\u2212 used for both the preimage and the inverse map, even though the former is always de\ufb01ned and the latter is only de\ufb01ned in some special cases.) Thus: \u03c6)(a) = a. (Note that the same symbol \u03c6\u2212 M by (\u03c61 N The notion of continuity of a map between topological spaces (and thus manifolds) is actually a very subtle one, the precise formulation of which we won\u2019t really need. However Rn between the intuitive notions of continuity and di\ufb00erentiability of maps \u03c6 : Rm Euclidean spaces are useful. A map from Rm to Rn takes an m-tuple (x1, x2,..., xm) to an n-tuple (y1, y2,..., yn), and can therefore be thought of as a collection of n functions \u03c6i of \u2192 2 MANIFOLDS m variables: y1 = \u03c61(x1, x2,..., xm) y2 = \u03c62(x1, x2,..., xm) 35 \u00b7 \u00b7 \u00b7 yn = \u03c6n(x1, x2,..., xm). We will refer to any one of these functions as C p if it is continuous and p-times di\ufb00erentiable, and refer to the entire map \u03c6 : Rm Rn as C p if each of its component functions are at least C p. Thus a C 0 map is continuous but not necessarily di\ufb00erentiable, while a C \u221e map is continuous and can be di\ufb00erentiated as many", " times as you like. C \u221e maps are sometimes called smooth. We will call two sets M and N di\ufb00eomorphic if there exists a C \u221e map \u03c6 : M M; the map \u03c6 is then called a di\ufb00eomorphism. N with a C \u221e inverse \u03c6\u2212 1 : N \u2192 (2.1) Aside: The notion of two spaces being di\ufb00eomorphic only applies to manifolds, where a notion of di\ufb00erentiability is inherited from the fact that the space resembles Rn locally. But \u201ccontinuity\u201d of maps between topological spaces (not necessarily manifolds) can be de\ufb01ned, and we say that two such spaces are \u201chomeomorphic,\u201d which means \u201ctopologically equivalent to,\u201d if there is a continuous map between them with a continuous inverse. It is therefore conceivable that spaces exist which are homeomorphic but not di\ufb00eomorphic; topologically the same but with distinct \u201cdi\ufb00erentiable structures.\u201d In 1964 Milnor showed that S7 had 28 di\ufb00erent di\ufb00erentiable structures; it turns out that for n < 7 there is only one di\ufb00erentiable structure on Sn, while for n > 7 the number grows very large. R4 has in\ufb01nitely many di\ufb00erentiable structures. \u2192 \u2192 One piece of conventional calculus that we will need later is the chain rule. Let us Rl, and therefore the composition Rn and g : Rn imagine that we have maps f : Rm Rl. (g f ) : Rm \u2192 \u25e6 \u2192 \u2192 We can represent each space in terms of coordinates: xa on Rm, yb on Rn, and zc on Rl, where the indices range over the appropriate values. The chain rule relates the partial 36 (2.2) (2.3) 2 MANIFOLDS derivatives of the composition to the partial derivatives of the individual maps: \u2202 \u2202xa (g \u25e6 f )c = \u2202f b \u2202xa \u2202gc \u2202yb. Xb This is usually abbreviated to \u2202 \u2202xa = \u2202yb \u2202xa \u2202 \u2202yb. Xb There is nothing illegal or immoral about using this form of the chain rule, but you should be able to visualize the maps that underlie the construction. Recall that when m = n the determinant of the matrix \u2202yb/\u2202xa is called the Jacobian of the map, and the map is invertible whenever the Jacobian is nonzero. These basic de\ufb01nitions were presumably familiar to you, even if only vaguely remembered. We will now put them to use in the rigorous de\ufb01nition of a manifold. Unfortunately, a somewhat baroque procedure is required to formalize this relatively intuitive notion. We will \ufb01rst have to de\ufb01ne the notion of an open set, on which we can put coordinate systems, and then sew the open sets together in an appropriate way. y < r for some \ufb01xed y Start with the notion of an open ball, which is the set of all points x in Rn such that yi)2]1/2. Note that x R, where | this is a strict inequality \u2014 the open ball is the interior of an n-sphere of radius r centered at y. Rn and r i(xi = [ \u2212 \u2212 \u2212 open ball An open set in Rn is a set constructed from an arbitrary (maybe in\ufb01nite) union of open V, there is an open ball centered balls. In other words, V at y which is completely inside V. Roughly speaking, an open set is the interior of some (n 1)-dimensional closed surface (or the union of several such interiors). By de\ufb01ning a notion of open sets, we have equipped Rn with a topology \u2014 in this case, the \u201cstandard metric topology.\u201d Rn is open if, for any y \u2282 \u2212 \u2208 2 MANIFOLDS 37 A chart or coordinate system consists of a subset U of a set M, along with a one-toRn, such that the image \u03c6(U) is open in R. (Any map is onto its image, \u03c6(U) is invertible.) We then can say that U is an open set in M. (We one map \u03c6 : U \u2192 so the map \u03c6 : U have thus induced a topology on M, although we will not explore this.) \u2192 M U \u03d5 n R \u03d5( ) U A C \u221e atlas is an indexed collection of charts (U\u03b1, \u03c6\u03b1) } { which satis\ufb01", "es two conditions: 1. The union of the U\u03b1 is equal to M; that is, the U\u03b1 cover M. 2. The charts are smoothly sewn together. More precisely, if two charts overlap, U\u03b1\u2229 U\u03b2 6 = Rn, U\u03b2) \u03b2 ) takes points in \u03c6\u03b2(U\u03b1 \u2229 \u03c6\u2212, then the map (\u03c6\u03b1 \u25e6 \u2205 \u2282 and all of these maps must be C \u221e where they are de\ufb01ned. This should be clearer in pictures: Rn onto \u03c6\u03b1(U\u03b1 \u2229 U\u03b2) \u2282 1 M U\u03b2 U\u03b1 \u03d5 \u03b1 \u03d5 \u03b2 \u03d5 \u03d5 \u03b1 -1 \u03b2 n R \u03d5 ( ) \u03b2 U\u03b2 \u03d5 \u03d5 -1 \u03b1 \u03b2 n R \u03d5 ( ) U\u03b1 \u03b1 these maps are only defined on the shaded regions, and must be smooth there. 2 MANIFOLDS 38 So a chart is what we normally think of as a coordinate system on some open set, and an atlas is a system of charts which are smoothly related on their overlaps. At long last, then: a C \u221e n-dimensional manifold (or n-manifold for short) is simply a set M along with a \u201cmaximal atlas\u201d, one that contains every possible compatible chart. (We can also replace C \u221e by C p in all the above de\ufb01nitions. For our purposes the degree of di\ufb00erentiability of a manifold is not crucial; we will always assume that any manifold is as di\ufb00erentiable as necessary for the application under consideration.) The requirement that the atlas be maximal is so that two equivalent spaces equipped with di\ufb00erent atlases don\u2019t count as di\ufb00erent manifolds. This de\ufb01nition captures in formal terms our notion of a set that looks locally like Rn. Of course we will rarely have to make use of the full power of the de\ufb01nition, but precision is its own reward. One thing that is nice about our de\ufb01nition is that it does not rely on an embedding of the manifold in some higher-dimensional Euclidean space. In fact any n-dimensional manifold can be embedded in R2n (\u201cWhitney\u2019s embedding theorem\u201d), and sometimes we will make use of this fact (such as in our de\ufb01nition of the sphere above). But it\u2019s important to recognize that the manifold has an individual existence independent of any embedding. We have no reason to believe, for example, that four-dimensional spacetime is stuck in some larger space. (Actually a number of people, string theorists and so forth, believe that our four-dimensional world is part of a ten- or eleven-dimensional spacetime, but as far as GR is concerned the 4-dimensional view is perfectly adequate.) Why was it necessary to be so \ufb01nicky about charts and their overlaps, rather than just covering every manifold with a single chart? Because most manifolds cannot be covered with just one chart. Consider the simplest example, S1. There is a conventional coordinate system, \u03b8 : S1 R, where \u03b8 = 0 at the top of the circle and wraps around to 2\u03c0. However, in the de\ufb01nition of a chart we have required that the image \u03b8(S1) be open in R. If we include either \u03b8 = 0 or \u03b8 = 2\u03c0, we have a closed interval rather than an open one; if we exclude both points, we haven\u2019t covered the whole circle. So we need at least two charts, as shown. \u2192 S1 U 2 U1 A somewhat more complicated example is provided by S2, where once again no single 2 MANIFOLDS 39 chart will cover the manifold. A Mercator projection, traditionally used for world maps, misses both the North and South poles (as well as the International Date Line, which involves the same problem with \u03b8 that we found for S1.) Let\u2019s take S2 to be the set of points in R3 de\ufb01ned by (x1)2 + (x2)2 + (x3)2 = 1. We can construct a chart from an open set U1, de\ufb01ned to be the sphere minus the north pole, via \u201cstereographic projection\u201d: 3 x x1 2 x 2 3 (y, y ) 3 x = -1 Thus, we draw a straight line from the north pole to the plane de\ufb01ned by x3 = 1, and assign to the point on S2 intercepted by the line the Cartesian coordinates (y1, y2) of the appropriate point on the plane. Explicitly, the map", " is given by \u2212 \u03c61(x1, x2, x3) (y1, y2) = \u2261 2x1 1 \u2212 x3, 2x2 x3! 1 \u2212. (2.4) You are encouraged to check this for yourself. Another chart (U2, \u03c62) is obtained by projecting from the south pole to the plane de\ufb01ned by x3 = +1. The resulting coordinates cover the sphere minus the south pole, and are given by \u03c62(x1, x2, x3) (z1, z2) = \u2261 2x1 1 + x3, 2x2 1 + x3!. Together, these two charts cover the entire manifold, and they overlap in the region x3 < +1. Another thing you can check is that the composition \u03c62 \u25e6 is given by \u03c6\u2212 1 1 zi = 4yi [(y1)2 + (y2)2], (2.5) 1 < \u2212 (2.6) and is C \u221e in the region of overlap. As long as we restrict our attention to this region, (2.6) is just what we normally think of as a change of coordinates. We therefore see the necessity of charts and atlases: many manifolds cannot be covered with a single coordinate system. (Although some can, even ones with nontrivial topology. Can you think of a single good coordinate system that covers the cylinder, S1 R?) Nevertheless, it is very often most convenient to work with a single chart, and just keep track of the set of points which aren\u2019t included. \u00d7 2 MANIFOLDS 40 The fact that manifolds look locally like Rn, which is manifested by the construction of coordinate charts, introduces the possibility of analysis on manifolds, including operations such as di\ufb00erentiation and integration. Consider two manifolds M and N of dimensions m N, and n, with coordinate charts \u03c6 on M and \u03c8 on N. Imagine we have a function f : M \u2192 M f N \u03d5-1 \u03d5 -1\u03c8 \u03c8 m R \u03c8 f \u03d5-1 n R f \u03c6\u2212 Just thinking of M and N as sets, we cannot nonchalantly di\ufb00erentiate the map f, since we don\u2019t know what such an operation means. But the coordinate charts allow us to construct Rn. (Feel free to insert the words \u201cwhere the maps are the map (\u03c8 de\ufb01ned\u201d wherever appropriate, here and later on.) This is just a map between Euclidean spaces, and all of the concepts of advanced calculus apply. For example f, thought of as an N-valued function on M, can be di\ufb00erentiated to obtain \u2202f /\u2202x\u00b5, where the x\u00b5 represent Rm. The point is that this notation is a shortcut, and what is really going on is 1) : Rm \u2192 \u25e6 \u25e6 \u2202f \u2202x\u00b5 \u2261 \u2202 \u2202x\u00b5 (\u03c8 f \u25e6 \u25e6 \u03c6\u2212 1)(x\u00b5). (2.7) It would be far too unwieldy (not to mention pedantic) to write out the coordinate maps explicitly in every case. The shorthand notation of the left-hand-side will be su\ufb03cient for most purposes. Having constructed this groundwork, we can now proceed to introduce various kinds of structure on manifolds. We begin with vectors and tangent spaces. In our discussion of special relativity we were intentionally vague about the de\ufb01nition of vectors and their relationship to the spacetime. One point that was stressed was the notion of a tangent space \u2014 the set of all vectors at a single point in spacetime. The reason for this emphasis was to remove from your minds the idea that a vector stretches from one point on the manifold to another, but instead is just an object associated with a single point. What is temporarily lost by adopting this view is a way to make sense of statements like \u201cthe vector points in 2 MANIFOLDS 41 the x direction\u201d \u2014 if the tangent space is merely an abstract vector space associated with each point, it\u2019s hard to know what this should mean. Now it\u2019s time to \ufb01x the problem. Let\u2019s imagine that we wanted to construct the tangent space at a point p in a manifold M, using only things that are intrinsic to M (no embeddings in higher-dimensional spaces etc.). One \ufb01rst guess might be to use our intuitive knowledge that there are objects called \u201ctangent vectors to curves\u201d which belong in the tangent space. We might therefore consider the set of all parameterized curves through p \u2014 that is, the space of all (n", "ondegenerate) maps \u03b3 : R M such that p is in the image of \u03b3. The temptation is to de\ufb01ne the tangent space as simply the space of all tangent vectors to these curves at the point p. But this is obviously cheating; the tangent space Tp is supposed to be the space of vectors at p, and before we have de\ufb01ned this we don\u2019t have an independent notion of what \u201cthe tangent vector to a curve\u201d is supposed to mean. In some coordinate system x\u00b5 any curve through p de\ufb01nes an element of Rn speci\ufb01ed by the n real numbers dx\u00b5/d\u03bb (where \u03bb is the parameter along the curve), but this map is clearly coordinate-dependent, which is not what we want. \u2192 F \u2192 Nevertheless we are on the right track, we just have to make things independent of to be the space of all smooth functions on M (that coordinates. To this end we de\ufb01ne R). Then we notice that each curve through p de\ufb01nes an operator is, C \u221e maps f : M on this space, the directional derivative, which maps f df /d\u03bb (at p). We will make the following claim: the tangent space Tp can be identi\ufb01ed with the space of directional derivative operators along curves through p. To establish this idea we must demonstrate two things: \ufb01rst, that the space of directional derivatives is a vector space, and second that it is the vector space we want (it has the same dimensionality as M, yields a natural idea of a vector pointing along a certain direction, and so on). \u2192 d\u03bb and d The \ufb01rst claim, that directional derivatives form a vector space, seems straightforward enough. Imagine two operators d d\u03b7 representing derivatives along two curves through p. There is no problem adding these and scaling by real numbers, to obtain a new operator a d d\u03bb + b d It is not immediately obvious, however, that the space closes; i.e., that the d\u03b7. resulting operator is itself a derivative operator. A good derivative operator is one that acts linearly on functions, and obeys the conventional Leibniz (product) rule on products of functions. Our new operator is manifestly linear, so we need to verify that it obeys the Leibniz rule. We have a d d\u03bb + b d d\u03b7! (f g) = af = a dg d\u03bb df d\u03bb + ag + b df d\u03bb df d\u03b7! + bf dg d\u03b7 + bg g + a dg d\u03bb df d\u03b7 dg d\u03b7! + b f. (2.8) As we had hoped, the product rule is satis\ufb01ed, and the set of directional derivatives is therefore a vector space. 2 MANIFOLDS 42 Is it the vector space that we would like to identify with the tangent space? The easiest way to become convinced is to \ufb01nd a basis for the space. Consider again a coordinate chart with coordinates x\u00b5. Then there is an obvious set of n directional derivatives at p, namely the partial derivatives \u2202\u00b5 at p at p form a basis for We are now going to claim that the partial derivative operators the tangent space Tp. (It follows immediately that Tp is n-dimensional, since that is the number of basis vectors.) To see this we will show that any directional derivative can be decomposed into a sum of real numbers times partial derivatives. This is in fact just the familiar expression for the components of a tangent vector, but it\u2019s nice to see it from the Rn, a big-machinery approach. Consider an n-manifold M, a coordinate chart \u03c6 : M curve \u03b3 : R R. This leads to the following tangle of maps: M, and a function f : M \u2202\u00b5-1 \u00b5 x If \u03bb is the parameter along \u03b3, we want to expand the vector/operator d d\u03bb in terms of the 2 MANIFOLDS 43 partials \u2202\u00b5. Using the chain rule (2.2), we have d d\u03bb f = = = = (f \u03b3) \u25e6 d d\u03bb d d\u03bb d(\u03c6 dx\u00b5 d\u03bb [(f \u25e6 d\u03bb \u03c6\u2212 1) \u25e6 \u2202(f \u25e6 \u03b3)\u00b5 (\u03c6 \u03b3)] \u25e6 1) \u03c6\u2212 \u25e6 \u2202x\u00b5 \u2202\u00b5f. (2.9) The \ufb01rst line simply takes the informal expression on the left hand side and rewrites it as an honest derivative of the function (f R. The second line just comes from the 1", " (and associativity of the operation of composition). The de\ufb01nition of the inverse map \u03c6\u2212 third line is the formal chain rule (2.2), and the last line is a return to the informal notation of the start. Since the function f was arbitrary, we have \u03b3) : R \u2192 \u25e6 d d\u03bb = dx\u00b5 d\u03bb \u2202\u00b5. (2.10) { do indeed represent a good basis for the vector space of directional Thus, the partials derivatives, which we can therefore safely identify with the tangent space. \u2202\u00b5} Of course, the vector represented by d d\u03bb is one we already know; it\u2019s the tangent vector to the curve with parameter \u03bb. Thus (2.10) can be thought of as a restatement of (1.24), where we claimed the that components of the tangent vector were simply dx\u00b5/d\u03bb. The only di\ufb00erence is that we are working on an arbitrary manifold, and we have speci\ufb01ed our basis vectors to be \u02c6e(\u00b5) = \u2202\u00b5. This particular basis (\u02c6e(\u00b5) = \u2202\u00b5) is known as a coordinate basis for Tp; it is the formalization of the notion of setting up the basis vectors to point along the coordinate axes. There is no reason why we are limited to coordinate bases when we consider tangent vectors; it is sometimes more convenient, for example, to use orthonormal bases of some sort. However, the coordinate basis is very simple and natural, and we will use it almost exclusively throughout the course. One of the advantages of the rather abstract point of view we have taken toward vectors is that the transformation law is immediate. Since the basis vectors are \u02c6e(\u00b5) = \u2202\u00b5, the basis vectors in some new coordinate system x\u00b5\u2032 are given by the chain rule (2.3) as We can get the transformation law for vector components by the same technique used in \ufb02at space, demanding the the vector V = V \u00b5\u2202\u00b5 be unchanged by a change of basis. We have \u2202\u00b5\u2032 = \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202\u00b5. (2.11) V \u00b5\u2202\u00b5 = V \u00b5\u2032 \u2202\u00b5\u2032 = V \u00b5\u2032 \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202\u00b5, (2.12) 2 MANIFOLDS and hence (since the matrix \u2202x\u00b5\u2032 /\u2202x\u00b5 is the inverse of the matrix \u2202x\u00b5/\u2202x\u00b5\u2032 ), V \u00b5\u2032 = \u2202x\u00b5\u2032 \u2202x\u00b5 V \u00b5. 44 (2.13) = \u039b\u00b5\u2032 Since the basis vectors are usually not written explicitly, the rule (2.13) for transforming components is what we call the \u201cvector transformation law.\u201d We notice that it is compatible with the transformation of vector components in special relativity under Lorentz transformations, V \u00b5\u2032 \u00b5V \u00b5, since a Lorentz transformation is a special kind of coordinate transformation, with x\u00b5\u2032 \u00b5x\u00b5. But (2.13) is much more general, as it encompasses the behavior of vectors under arbitrary changes of coordinates (and therefore bases), not just linear transformations. As usual, we are trying to emphasize a somewhat subtle ontological distinction \u2014 tensor components do not change when we change coordinates, they change when we change the basis in the tangent space, but we have decided to use the coordinates to de\ufb01ne our basis. Therefore a change of coordinates induces a change of basis: = \u039b\u00b5\u2032 \u00b5 x \u03c1 2 \u03c1 1 \u00b5\u2019 x \u03c1 1 \u2019 \u03c1 2 \u2019 Having explored the world of vectors, we continue to retrace the steps we took in \ufb02at space, and now consider dual vectors (one-forms). Once again the cotangent space T \u2217p is the set of linear maps \u03c9 : Tp \u2192 R. The canonical example of a one-form is the gradient of a function f, denoted df. Its action on a vector d d\u03bb is exactly the directional derivative of the function: df d d\u03bb! = df d\u03bb. (2.14) It\u2019s tempting to think, \u201cwhy shouldn\u2019t the function f itself be considered the one-form, and df /d\u03bb its action?\u201d The point is that a one-form, like a vector, exists only at the point it is de\ufb01ned, and does not depend on information at other points on M. If you know a function in some neighborhood of a point you can take its derivative, but not just from knowing its value at", " the point; the gradient, on the other hand, encodes precisely the information 2 MANIFOLDS 45 necessary to take the directional derivative along any curve through p, ful\ufb01lling its role as a dual vector. Just as the partial derivatives along coordinate axes provide a natural basis for the tangent space, the gradients of the coordinate functions x\u00b5 provide a natural basis for the cotangent space. Recall that in \ufb02at space we constructed a basis for T \u2217p by demanding that \u02c6\u03b8(\u00b5)(\u02c6e(\u03bd)) = \u03b4\u00b5 \u03bd. Continuing the same philosophy on an arbitrary manifold, we \ufb01nd that (2.14) leads to dx\u00b5(\u2202\u03bd) = \u2202x\u00b5 \u2202x\u03bd = \u03b4\u00b5 \u03bd. (2.15) Therefore the gradients { form is expanded into components as \u03c9 = \u03c9\u00b5 dx\u00b5. } dx\u00b5 are an appropriate set of basis one-forms; an arbitrary one- The transformation properties of basis dual vectors and components follow from what is by now the usual procedure. We obtain, for basis one-forms, and for components, dx\u00b5\u2032 = \u2202x\u00b5\u2032 \u2202x\u00b5 dx\u00b5, \u03c9\u00b5\u2032 = \u2202x\u00b5 \u2202x\u00b5\u2032 \u03c9\u00b5. (2.16) (2.17) We will usually write the components \u03c9\u00b5 when we speak about a one-form \u03c9. The transformation law for general tensors follows this same pattern of replacing the Lorentz transformation matrix used in \ufb02at space with a matrix representing more general coordinate transformations. A (k, l) tensor T can be expanded T = T \u00b51 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u03bdl\u2202\u00b51 \u2297 \u00b7 \u00b7 \u00b7 \u2297 \u00b7\u00b7\u00b7 \u2202\u00b5k \u2297 dx\u03bd1 \u2297 \u00b7 \u00b7 \u00b7 \u2297 dx\u03bdl, and under a coordinate transformation the components change according to T \u00b5\u2032 1\u00b7\u00b7\u00b7 \u00b5\u2032 k \u03bd\u2032 \u03bd\u2032 l 1\u00b7\u00b7\u00b7 = 1 \u2202x\u00b5\u2032 \u2202x\u00b51 \u00b7 \u00b7 \u00b7 k \u2202x\u00b5\u2032 \u2202x\u00b5k \u2202x\u03bd1 \u2202x\u03bd\u2032 1 \u00b7 \u00b7 \u00b7 \u2202x\u03bdl \u2202x\u03bd\u2032 l T \u00b51 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u03bdl. \u00b7\u00b7\u00b7 (2.18) (2.19) This tensor transformation law is straightforward to remember, since there really isn\u2019t anything else it could be, given the placement of indices. However, it is often easier to transform a tensor by taking the identity of basis vectors and one-forms as partial derivatives and gradients at face value, and simply substituting in the coordinate transformation. As an example consider a symmetric (0, 2) tensor S on a 2-dimensional manifold, whose components in a coordinate system (x1 = x, x2 = y) are given by S\u00b5\u03bd = x 0 0 1. (2.20) 2 MANIFOLDS This can be written equivalently as S = S\u00b5\u03bd(dx\u00b5 dx\u03bd) = x(dx)2 + (dy)2, \u2297 46 (2.21) where in the last line the tensor product symbols are suppressed for brevity. Now consider new coordinates This leads directly to x\u2032 = x1/3 y\u2032 = ex+y. x = (x\u2032)3 (x\u2032)3 y = ln(y\u2032) \u2212 dx = 3(x\u2032)2 dx\u2032 1 y\u2032 dy = dy\u2032 \u2212 3(x\u2032)2 dx\u2032. (2.22) (2.23) We need only plug these expressions directly into (2.21) to obtain (remembering that tensor products don\u2019t commute, so dx\u2032 dy\u2032 S = 9(x\u2032)4[1 + (x\u2032)3](dx\u2032)2 (dx\u2032 dy\u2032 + dy\u2032 dx\u2032) + 1 (y\u2032)2 (dy\u2032)2, (2.24) or 9(x\u2032)4[1 + (x\u2032)3] 3 (x\u2032)2 y\u2032 Notice that it is still symmetric. We did not use the transformation law (2.19) directly, but doing so would have yielded the same result, as you can check. S\u00b5\u2032\u03bd\u2032 = (2.25) 1 (y\u2032)x\u2032)2 y\u2032 For the most part the various tensor operations we de\ufb01ned in \ufb02at space are unaltered in a more general setting: contraction, symmetrization, etc. There are three important exceptions: partial derivatives", ", the metric, and the Levi-Civita tensor. Let\u2019s look at the partial derivative \ufb01rst. The unfortunate fact is that the partial derivative of a tensor is not, in general, a new tensor. The gradient, which is the partial derivative of a scalar, is an honest (0, 1) tensor, as we have seen. But the partial derivative of higher-rank tensors is not tensorial, as we can see by considering the partial derivative of a one-form, \u2202\u00b5W\u03bd, and changing to a new coordinate system: \u2202 \u2202x\u00b5\u2032 W\u03bd\u2032 = = \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202 \u2202x\u00b5 \u2202x\u03bd \u2202x\u03bd\u2032 \u2202x\u03bd \u2202x\u03bd\u2032 W\u03bd \u2202 \u2202x\u00b5 W\u03bd!! + W\u03bd \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202 \u2202x\u00b5 \u2202x\u03bd \u2202x\u03bd\u2032. (2.26) = dy\u2032 dx\u2032): (x\u2032)2 y\u2032 \u2212 3 6 2 MANIFOLDS 47 The second term in the last line should not be there if \u2202\u00b5W\u03bd were to transform as a (0, 2) tensor. As you can see, it arises because the derivative of the transformation matrix does not vanish, as it did for Lorentz transformations in \ufb02at space. On the other hand, the exterior derivative operator d does form an antisymmetric (0, p+1) tensor when acted on a p-form. For p = 1 we can see this from (2.26); the o\ufb00ending nontensorial term can be written W\u03bd \u2202 \u2202x\u00b5 \u2202x\u00b5 \u2202x\u00b5\u2032 \u22022x\u03bd \u2202x\u00b5\u2032 \u2202x\u03bd\u2032. \u2202x\u03bd \u2202x\u03bd\u2032 = W\u03bd This expression is symmetric in \u00b5\u2032 and \u03bd\u2032, since partial derivatives commute. But the exterior derivative is de\ufb01ned to be the antisymmetrized partial derivative, so this term vanishes (the antisymmetric part of a symmetric expression is zero). We are then left with the correct tensor transformation law; extension to arbitrary p is straightforward. So the exterior derivative is a legitimate tensor operator; it is not, however, an adequate substitute for the partial derivative, since it is only de\ufb01ned on forms. In the next section we will de\ufb01ne a covariant derivative, which can be thought of as the extension of the partial derivative to arbitrary manifolds. (2.27) The metric tensor is such an important object in curved space that it is given a new symbol, g\u00b5\u03bd (while \u03b7\u00b5\u03bd is reserved speci\ufb01cally for the Minkowski metric). There are few restrictions on the components of g\u00b5\u03bd, other than that it be a symmetric (0, 2) tensor. It is usually taken to be non-degenerate, meaning that the determinant g = doesn\u2019t vanish. This allows us to de\ufb01ne the inverse metric g\u00b5\u03bd via g\u00b5\u03bd| | g\u00b5\u03bdg\u03bd\u03c3 = \u03b4\u00b5 \u03c3. (2.28) The symmetry of g\u00b5\u03bd implies that g\u00b5\u03bd is also symmetric. Just as in special relativity, the metric and its inverse may be used to raise and lower indices on tensors. It will take several weeks to fully appreciate the role of the metric in all of its glory, but for purposes of inspiration we can list the various uses to which g\u00b5\u03bd will be put: (1) the metric supplies a notion of \u201cpast\u201d and \u201cfuture\u201d; (2) the metric allows the computation of path length and proper time; (3) the metric determines the \u201cshortest distance\u201d between two points (and therefore the motion of test particles); (4) the metric replaces the Newtonian gravitational \ufb01eld \u03c6; (5) the metric provides a notion of locally inertial frames and therefore a sense of \u201cno rotation\u201d; (6) the metric determines causality, by de\ufb01ning the speed of light faster than which no signal can travel; (7) the metric replaces the traditional Euclidean three-dimensional dot product of Newtonian mechanics; and so on. Obviously these ideas are not all completely independent, but we get some sense of the importance of this tensor. In our discussion of path lengths in special relativity we (somewhat handwavingly) introduced the line element as ds2 = \u03b7\u00b5\u03bddx", "\u00b5dx\u03bd, which was used to get the length of a path. 2 MANIFOLDS 48 Of course now that we know that dx\u00b5 is really a basis dual vector, it becomes natural to use the terms \u201cmetric\u201d and \u201cline element\u201d interchangeably, and write ds2 = g\u00b5\u03bd dx\u00b5 dx\u03bd. (2.29) (To be perfectly consistent we should write this as \u201cg\u201d, and sometimes will, but more often than not g is used for the determinant.) For example, we know that the Euclidean line element in a three-dimensional space with Cartesian coordinates is g\u00b5\u03bd| | ds2 = (dx)2 + (dy)2 + (dz)2. (2.30) We can now change to any coordinate system we choose. For example, in spherical coordinates we have x = r sin \u03b8 cos \u03c6 y = r sin \u03b8 sin \u03c6 z = r cos \u03b8, which leads directly to ds2 = dr2 + r2 d\u03b82 + r2 sin2 \u03b8 d\u03c62. (2.31) (2.32) Obviously the components of the metric look di\ufb00erent than those in Cartesian coordinates, but all of the properties of the space remain unaltered. Perhaps this is a good time to note that most references are not su\ufb03ciently picky to distinguish between \u201cdx\u201d, the informal notion of an in\ufb01nitesimal displacement, and \u201cdx\u201d, the rigorous notion of a basis one-form given by the gradient of a coordinate function. In fact our notation \u201cds2\u201d does not refer to the exterior derivative of anything, or the square of anything; it\u2019s just conventional shorthand for the metric tensor. On the other hand, \u201c(dx)2\u201d refers speci\ufb01cally to the (0, 2) tensor dx dx. A good example of a space with curvature is the two-sphere, which can be thought of as the locus of points in R3 at distance 1 from the origin. The metric in the (\u03b8, \u03c6) coordinate system comes from setting r = 1 and dr = 0 in (2.32): \u2297 ds2 = d\u03b82 + sin2 \u03b8 d\u03c62. (2.33) This is completely consistent with the interpretation of ds as an in\ufb01nitesimal length, as illustrated in the \ufb01gure. As we shall see, the metric tensor contains all the information we need to describe the curvature of the manifold (at least in Riemannian geometry; we will actually indicate somewhat more general approaches). In Minkowski space we can choose coordinates in which the components of the metric are constant; but it should be clear that the existence of curvature 2 MANIFOLDS 49 2 S \u03b8 d ds sin \u03b8 d\u03c6 is more subtle than having the metric depend on the coordinates, since in the example above we showed how the metric in \ufb02at Euclidean space in spherical coordinates is a function of r and \u03b8. Later, we shall see that constancy of the metric components is su\ufb03cient for a space to be \ufb02at, and in fact there always exists a coordinate system on any \ufb02at space in which the metric is constant. But we might not want to work in such a coordinate system, and we might not even know how to \ufb01nd it; therefore we will want a more precise characterization of the curvature, which will be introduced down the road. A useful characterization of the metric is obtained by putting g\u00b5\u03bd into its canonical form. In this form the metric components become g\u00b5\u03bd = diag ( 1, 1,..., \u2212 1, +1, +1,..., +1, 0, 0,..., 0), \u2212 (2.34) \u2212 \u2212 If n is the dimension of where \u201cdiag\u201d means a diagonal matrix with the given elements. the manifold, s is the number of +1\u2019s in the canonical form, and t is the number of 1\u2019s, then s t is the signature of the metric (the di\ufb00erence in the number of minus and plus signs), and s + t is the rank of the metric (the number of nonzero eigenvalues). If a metric is continuous, the rank and signature of the metric tensor \ufb01eld are the same at every point, and if the metric is nondegenerate the rank is equal to the dimension n. We will always deal with continuous, nondegenerate metrics. If all of the signs are positive (t", " = 0) the metric is called Euclidean or Riemannian (or just \u201cpositive de\ufb01nite\u201d), while if there is a single minus (t = 1) it is called Lorentzian or pseudo-Riemannian, and any metric with some +1\u2019s and some 1\u2019s is called \u201cinde\ufb01nite.\u201d (So the word \u201cEuclidean\u201d sometimes means that the space is \ufb02at, and sometimes doesn\u2019t, but always means that the canonical form is strictly positive; the terminology is unfortunate but standard.) The spacetimes of interest in general relativity have Lorentzian metrics. \u2212 \u2212 We haven\u2019t yet demonstrated that it is always possible to but the metric into canonical M, but in general it will In fact it is always possible to do so at some point p form. \u2208 2 MANIFOLDS 50 only be possible at that single point, not in any neighborhood of p. Actually we can do slightly better than this; it turns out that at any point p there exists a coordinate system in which g\u00b5\u03bd takes its canonical form and the \ufb01rst derivatives \u2202\u03c3g\u00b5\u03bd all vanish (while the second derivatives \u2202\u03c1\u2202\u03c3g\u00b5\u03bd cannot be made to all vanish). Such coordinates are known as Riemann normal coordinates, and the associated basis vectors constitute a local Lorentz frame. Notice that in Riemann normal coordinates (or RNC\u2019s) the metric at p looks like that of \ufb02at space \u201cto \ufb01rst order.\u201d This is the rigorous notion of the idea that \u201csmall enough regions of spacetime look like \ufb02at (Minkowski) space.\u201d (Also, there is no di\ufb03culty in simultaneously constructing sets of basis vectors at every point in M such that the metric takes its canonical form; the problem is that in general this will not be a coordinate basis, and there will be no way to make it into one.) We won\u2019t consider the detailed proof of this statement; it can be found in Schutz, pp. 158160, where it goes by the name of the \u201clocal \ufb02atness theorem.\u201d (He also calls local Lorentz frames \u201cmomentarily comoving reference frames,\u201d or MCRF\u2019s.) It is useful to see a sketch of the proof, however, for the speci\ufb01c case of a Lorentzian metric in four dimensions. The idea is to consider the transformation law for the metric g\u00b5\u2032\u03bd\u2032 = \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u03bd \u2202x\u03bd\u2032 g\u00b5\u03bd, (2.35) and expand both sides in Taylor series in the sought-after coordinates x\u00b5\u2032 of the old coordinates x\u00b5 looks like. The expansion x\u00b5 = \u2202x\u00b5 \u2202x\u00b5\u2032!p x\u00b5\u2032 + 1 2 \u22022x\u00b5 1 \u2202x\u00b5\u2032 \u2202x\u00b5\u2032 2!p x\u00b5\u2032 1x\u00b5\u2032 2 + 1 6 \u22023x\u00b5 1\u2202x\u00b5\u2032 \u2202x\u00b5\u2032 2 \u2202x\u00b5\u2032 3!p x\u00b5\u2032 1x\u00b5\u2032 2x\u00b5\u2032 3 +, (2.36) \u00b7 \u00b7 \u00b7 with the other expansions proceeding along the same lines. x\u00b5(p) = x\u00b5\u2032 (2.35) to second order is (For simplicity we have set (p) = 0.) Then, using some extremely schematic notation, the expansion of (g\u2032)p + (\u2202\u2032g\u2032)p x\u2032 + (\u2202\u2032\u2202\u2032g\u2032)p x\u2032x\u2032 \u22022x \u2202x\u2032\u2202x\u2032 \u2202x \u2202x\u2032 \u2202x \u2202x\u2032 \u2202x \u2202x\u2032 + = g!p + \u2202x \u2202x\u2032 \u22023x \u2202x\u2032\u2202x\u2032\u2202x\u2032 g + \u22022x \u2202x\u2032\u2202x\u2032 g + \u2202\u2032g x\u2032 \u2202x \u2202x\u2032 \u2202x \u2202x\u2032 \u22022x \u2202x\u2032\u2202x\u2032 g +!p \u2202x \u2202x\u2032 \u22022x \u2202x\u2032\u2202x\u2032 \u2202\u2032g + \u2202x \u2202x\u2032 \u2202x \u2202x\u2032 \u2202\u2032\u2202\u2032g!p x\u2032x\u2032.(2.37) We can set terms of equal order in x\u2032 on each side equal to each other. Therefore, the components g\u00b5\u2032\u03bd", "\u2032(p), 10 numbers in all (to describe a symmetric two-index tensor), are determined by the matrix (\u2202x\u00b5/\u2202x\u00b5\u2032 4 matrix with no constraints; thus, )p. This is a 4 16 numbers we are free to choose. Clearly this is enough freedom to put the 10 numbers of g\u00b5\u2032\u03bd\u2032(p) into canonical form, at least as far as having enough degrees of freedom is concerned. \u00d7 2 MANIFOLDS 51 (In fact there are some limitations \u2014 if you go through the procedure carefully, you \ufb01nd for example that you cannot change the signature and rank.) The six remaining degrees of freedom can be interpreted as exactly the six parameters of the Lorentz group; we know that these leave the canonical form unchanged. At \ufb01rst order we have the derivatives \u2202\u03c3\u2032g\u00b5\u2032\u03bd\u2032(p), four derivatives of ten components for a total of 40 numbers. But looking at the right hand side of (2.37) we see that we now have the additional freedom to choose (\u22022x\u00b5/\u2202x\u00b5\u2032 2)p. In this set of numbers there are 10 independent choices of the indices \u00b5\u20321 and \u00b5\u20322 (it\u2019s symmetric, since partial derivatives commute) and four choices of \u00b5, for a total of 40 degrees of freedom. This is precisely the amount of choice we need to determine all of the \ufb01rst derivatives of the metric, which we can therefore set to zero. At second order, however, we are concerned with \u2202\u03c1\u2032\u2202\u03c3\u2032g\u00b5\u2032\u03bd\u2032(p); this is symmetric in \u03c1\u2032 and \u03c3\u2032 as well as \u00b5\u2032 and \u03bd\u2032, for a total of 10 10 = 100 numbers. Our ability to make additional choices is contained in (\u22023x\u00b5/\u2202x\u00b5\u2032 2 \u2202x\u00b5\u2032 3)p. This is symmetric in the three lower indices, which gives 20 possibilities, times four for the upper index gives us 80 degrees of freedom \u2014 20 fewer than we require to set the second derivatives of the metric to zero. So in fact we cannot make the second derivatives vanish; the deviation from \ufb02atness must therefore be measured by the 20 coordinate-independent degrees of freedom representing the second derivatives of the metric tensor \ufb01eld. We will see later how this comes about, when we characterize curvature using the Riemann tensor, which will turn out to have 20 independent components. \u00d7 1\u2202x\u00b5\u2032 1 \u2202x\u00b5\u2032 The \ufb01nal change we have to make to our tensor knowledge now that we have dropped \u00b5n. Remember that \u00b7\u00b7\u00b7 \u00b5n, was de\ufb01ned as the assumption of \ufb02at space has to do with the Levi-Civita tensor, \u01eb\u00b51\u00b52 the \ufb02at-space version of this object, which we will now denote by \u02dc\u01eb\u00b51\u00b52 \u00b7\u00b7\u00b7 \u02dc\u01eb\u00b51\u00b52 \u00b5n = \uf8f1 \u00b7\u00b7\u00b7 \uf8f4\uf8f2 +1 if \u00b51\u00b52 \u00b7 \u00b7 \u00b7 1 if \u00b51\u00b52 \u00b7 \u00b7 \u00b7 \u2212 0 otherwise. \u00b5n is an even permutation of 01 \u00b5n is an odd permutation of 01 (n (), 1), (2.38) \uf8f4\uf8f3 We will now de\ufb01ne the Levi-Civita symbol to be exactly this \u02dc\u01eb\u00b51\u00b52 \u00b5n \u2014 that is, an object with n indices which has the components speci\ufb01ed above in any coordinate system. This is called a \u201csymbol,\u201d of course, because it is not a tensor; it is de\ufb01ned not to change under coordinate transformations. We can relate its behavior to that of an ordinary tensor by \ufb01rst noting that, given some n n matrix M \u00b5 \u00b5\u2032, the determinant obeys M \u00b7\u00b7\u00b7 \u00d7 \u02dc\u01eb\u00b5\u2032 1\u00b5\u2032 2\u00b7\u00b7\u00b7 \u00b5\u2032 n| M | = \u02dc\u01eb\u00b51\u00b52 \u00b5nM \u00b51 \u00b7\u00b7\u00b7 \u00b5\u2032 1 M \u00b52 \u00b5\u2032 2 \u00b7 \u00b7 \u00b7 n. \u00b5\u2032 (2.39) | | M \u00b5n This is just a true fact about the determinant which you can \ufb01nd in a su\ufb03ciently enlightened linear algebra book. If follows that, setting M \u00b5 \u00b5\u2032 = \u2202x\u00b5/\u2202x\u00b5\u2032, we have \u02dc\u01eb\u00b5\u2032 1\u00b5\u2032 2\u00b7\u00b7\u00b7 n = \u00b5\u2032 \u02dc\u01eb\u00b51\u00b52 \u00b5n \u00b7\u00b7\u00b7", " \u2202x\u00b51 \u2202x\u00b5\u2032 1 \u2202x\u00b52 \u2202x\u00b5\u2032 2 \u00b7 \u00b7 \u00b7 \u2202x\u00b5n \u2202x\u00b5\u2032 n. (2.40) \u2202x\u00b5\u2032 \u2202x\u00b5 2 MANIFOLDS 52 This is close to the tensor transformation law, except for the determinant out front. Objects which transform in this way are known as tensor densities. Another example is given by the determinant of the metric, g =. It\u2019s easy to check (by taking the determinant of both sides of (2.35)) that under a coordinate transformation we get g\u00b5\u03bd| | g(x\u00b5\u2032 ) = 2 \u2212 g(x\u00b5). (2.41) \u2202x\u00b5\u2032 \u2202x\u00b5 Therefore g is also not a tensor; it transforms in a way similar to the Levi-Civita symbol, 2 power. The power to which the Jacobian is except that the Jacobian is raised to the raised is known as the weight of the tensor density; the Levi-Civita symbol is a density of weight 1, while g is a (scalar) density of weight \u2212 2. However, we don\u2019t like tensor densities, we like tensors. There is a simple way to convert w/2, where w is the weight of the density a density into an honest tensor \u2014 multiply by (the absolute value signs are there because g < 0 for Lorentz metrics). The result will transform according to the tensor transformation law. Therefore, for example, we can de\ufb01ne the Levi-Civita tensor as g | | \u2212 \u01eb\u00b51\u00b52 \u00b5n = g \u02dc\u01eb\u00b51\u00b52 \u00b5n. (2.42) \u00b7\u00b7\u00b7 | q It is this tensor which is used in the de\ufb01nition of the Hodge dual, (1.87), which is otherwise unchanged when generalized to arbitrary manifolds. Since this is a real tensor, we can raise indices, etc. Sometimes people de\ufb01ne a version of the Levi-Civita symbol with upper indices, \u00b5n, whose components are numerically equal to the symbol with lower indices. This \u02dc\u01eb\u00b51\u00b52 turns out to be a density of weight 1, and is related to the tensor with upper indices by \u00b7\u00b7\u00b7 \u00b7\u00b7\u00b7 | \u2212 \u01eb\u00b51\u00b52 \u00b5n = sgn(g) \u00b7\u00b7\u00b7 \u02dc\u01eb\u00b51\u00b52 \u00b5n. \u00b7\u00b7\u00b7 1 g | | q (2.43) As an aside, we should come clean and admit that, even with the factor of, the | Levi-Civita tensor is in some sense not a true tensor, because on some manifolds it cannot be globally de\ufb01ned. Those on which it can be de\ufb01ned are called orientable, and we will deal exclusively with orientable manifolds in this course. An example of a non-orientable manifold is the M\u00a8obius strip; see Schutz\u2019s Geometrical Methods in Mathematical Physics (or a similar text) for a discussion. | q g One \ufb01nal appearance of tensor densities is in integration on manifolds. We will not do this subject justice, but at least a casual glance is necessary. You have probably been exposed to the fact that in ordinary calculus on Rn the volume element dnx picks up a factor of the Jacobian under change of coordinates: dnx\u2032 = dnx. \u2202x\u00b5\u2032 \u2202x\u00b5 (2.44) 2 MANIFOLDS 53 There is actually a beautiful explanation of this formula from the point of view of di\ufb00erential forms, which arises from the following fact: on an n-dimensional manifold, the integrand is properly understood as an n-form. The naive volume element dnx is itself a density rather than an n-form, but there is no di\ufb03culty in using it to construct a real n-form. To see how this works, we should make the identi\ufb01cation dnx dx0 \u2194 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dxn 1. \u2212 (2.45) The expression on the right hand side can be misleading, because it looks like a tensor (an n-form, actually) but is really a density. Certainly if we have two functions f and g on M, then df and dg are one-forms, and df dg is a two-form. But we would like to interpret the right hand side of (2.45) as a coordinate-dependent object which, in the x\u00b5 coordinate 1", ". This sounds tricky, but in fact it\u2019s just an ambiguity of system, acts like dx0 notation, and in practice we will just use the shorthand notation \u201cdnx\u201d. \u2227 \u00b7 \u00b7 \u00b7 \u2227 dxn \u2227 \u2212 To justify this song and dance, let\u2019s see how (2.45) changes under coordinate transfor- mations. First notice that the de\ufb01nition of the wedge product allows us to write dx0 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dxn 1 = \u2212 1 n! \u02dc\u01eb\u00b51 \u00b5n dx\u00b51 \u00b7\u00b7\u00b7 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dx\u00b5n, (2.46) since both the wedge product and the Levi-Civita symbol are completely antisymmetric. Under a coordinate transformation \u02dc\u01eb\u00b51 \u00b5n stays the same while the one-forms change according to (2.16), leading to \u00b7\u00b7\u00b7 \u02dc\u01eb\u00b51 \u00b5n dx\u00b51 \u00b7\u00b7\u00b7 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dx\u00b5n = \u02dc\u01eb\u00b51 \u00b5n \u2202x\u00b5n \u2202x\u00b5\u2032 n dx\u00b5\u2032 1 \u2202x\u00b51 \u2202x\u00b5\u2032 \u02dc\u01eb\u00b5\u2032 1\u00b7\u00b7\u00b7 1 \u00b7 \u00b7 \u00b7 n dx\u00b5\u2032 \u00b5\u2032 1 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dx\u00b5\u2032 n. dx\u00b5\u2032 n \u00b7\u00b7\u00b7 \u2202x\u00b5 \u2202x\u00b5\u2032 = (2.47) It is clear that the naive volume element dnx transforms as a density, not a tensor, but Multiplying by the Jacobian on both sides recovers (2.44). \u2227 \u00b7 \u00b7 \u00b7 \u2227 it is straightforward to construct an invariant volume element by multiplying by g : | | q dx0\u2032 g\u2032| dx(n \u2212 1)\u2032 = g dx0 dxn 1, \u2212 (2.48) | | q \u2227 \u00b7 \u00b7 \u00b7 \u2227 \u00b5n dx\u00b51 \u00b7\u00b7\u00b7 \u2227 \u00b7 \u00b7 \u00b7 \u2227 1\u01eb\u00b51 which is of course just (n!)\u2212 usually write the volume element as dxn | q dx\u00b5n. In the interest of simplicity we will dnx, rather than as the explicit wedge product 1; it will be enough to keep in mind that it\u2019s supposed to be an n-form. g | | As a \ufb01nal aside to \ufb01nish this section, let\u2019s consider one of the most elegant and powerful q theorems of di\ufb00erential geometry: Stokes\u2019s theorem. This theorem is the generalization of b. Imagine that we have an n-manifold the fundamental theorem of calculus dx0 \u2212 a b dx = a R \u2212 2 MANIFOLDS 54 1)-form \u03c9 on M. (We haven\u2019t discussed manifolds with M with boundary \u2202M, and an (n boundaries, but the idea is obvious; M could for instance be the interior of an (n 1)dimensional closed surface \u2202M.) Then d\u03c9 is an n-form, which can be integrated over M, while \u03c9 itself can be integrated over \u2202M. Stokes\u2019s theorem is then \u2212 \u2212 d\u03c9 = \u03c9. Z\u2202M ZM (2.49) You can convince yourself that di\ufb00erent special cases of this theorem include not only the fundamental theorem of calculus, but also the theorems of Green, Gauss, and Stokes, familiar from vector calculus in three dimensions. December 1997 Lecture Notes on General Relativity Sean M. Carroll 3 Curvature In our discussion of manifolds, it became clear that there were various notions we could talk about as soon as the manifold was de\ufb01ned; we could de\ufb01ne functions, take their derivatives, consider parameterized paths, set up tensors, and so on. Other concepts, such as the volume of a region or the length of a path, required some additional piece of structure, namely the introduction of a metric. It would be natural to think of the notion of \u201ccurvature\u201d, which we have already used informally, is something that depends on the metric. Actually this turns out to be not quite true, or at least incomplete. In fact there is one additional structure we need to introduce \u2014 a \u201cconnection\u201d \u2014 which is characterized by the curvature. We will show how the existence of a metric implies a certain connection, whose curvature may be thought of as that of the metric. The connection becomes necessary when we attempt to address the problem of the partial derivative not being a good tensor operator. What we would like is a covariant derivative; that is, an operator which reduces to the partial derivative in \ufb02at space with Cart", "esian coordinates, but transforms as a tensor on an arbitrary manifold. It is conventional to spend a certain amount of time motivating the introduction of a covariant derivative, but in fact the need is obvious; equations such as \u2202\u00b5T \u00b5\u03bd = 0 are going to have to be generalized to curved space somehow. So let\u2019s agree that a covariant derivative would be a good thing to have, and go about setting it up. In \ufb02at space in Cartesian coordinates, the partial derivative operator \u2202\u00b5 is a map from (k, l) tensor \ufb01elds to (k, l+1) tensor \ufb01elds, which acts linearly on its arguments and obeys the Leibniz rule on tensor products. All of this continues to be true in the more general situation we would now like to consider, but the map provided by the partial derivative depends on the coordinate system used. We would therefore like to de\ufb01ne a covariant derivative operator to perform the functions of the partial derivative, but in a way independent of coordinates. be a map from (k, l) tensor \ufb01elds to (k, l + 1) tensor \ufb01elds which \u2207 We therefore require that has these two properties: \u2207 1. linearity: (T + S. Leibniz (product) rule: (T S). If \u2297 \u2207 \u2207 \u2297 is going to obey the Leibniz rule, it can always be written as the partial derivative plus some linear transformation. That is, to take the covariant derivative we \ufb01rst take the partial derivative, and then apply a correction to make the result covariant. (We aren\u2019t going to prove this reasonable-sounding statement, but Wald goes into detail if you are interested.) \u2207 \u2207 \u2297 55 3 CURVATURE 56 Let\u2019s consider what this means for the covariant derivative of a vector V \u03bd. It means that, for each direction \u00b5, the covariant derivative \u2207\u00b5 will be given by the partial derivative \u2202\u00b5 plus a correction speci\ufb01ed by a matrix (\u0393\u00b5)\u03c1 n matrix, where n is the dimensionality of the manifold, for each \u00b5). In fact the parentheses are usually dropped and we write these matrices, known as the connection coe\ufb03cients, with haphazard index placement as \u0393\u03c1 \u00b5\u03c3. We therefore have \u03c3 (an n \u00d7 \u2207\u00b5V \u03bd = \u2202\u00b5V \u03bd + \u0393\u03bd \u00b5\u03bbV \u03bb. (3.1) Notice that in the second term the index originally on V has moved to the \u0393, and a new index is summed over. If this is the expression for the covariant derivative of a vector in terms of the partial derivative, we should be able to determine the transformation properties of \u0393\u03bd \u00b5\u03bb by demanding that the left hand side be a (1, 1) tensor. That is, we want the transformation law to be \u2202x\u00b5 \u2202x\u00b5\u2032 Let\u2019s look at the left side \ufb01rst; we can expand it using (3.1) and then transform the parts that we understand: \u2207\u00b5\u2032V \u03bd\u2032 \u2202x\u03bd\u2032 \u2202x\u03bd \u2207\u00b5V \u03bd. (3.2) = \u2207\u00b5\u2032V \u03bd\u2032 = \u2202\u00b5\u2032V \u03bd\u2032 \u2202x\u00b5 \u2202x\u00b5\u2032 \u00b5\u2032\u03bb\u2032V \u03bb\u2032 + \u0393\u03bd\u2032 \u2202x\u03bd\u2032 \u2202x\u03bd \u2202\u00b5V \u03bd + = \u2202x\u00b5 \u2202x\u00b5\u2032 V \u03bd \u2202 \u2202x\u00b5 \u2202x\u03bd\u2032 \u2202x\u03bd + \u0393\u03bd\u2032 \u00b5\u2032\u03bb\u2032 \u2202x\u03bb\u2032 \u2202x\u03bb V \u03bb. The right side, meanwhile, can likewise be expanded: \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u03bd\u2032 \u2202x\u03bd \u2207\u00b5V \u03bd = \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u03bd\u2032 \u2202x\u03bd \u2202\u00b5V \u03bd + \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u03bd\u2032 \u2202x\u03bd \u0393\u03bd \u00b5\u03bbV \u03bb. (3.3) (3.4) These last two expressions are to be equated; the \ufb01rst terms in each are identical and therefore cancel, so we have \u0393\u03bd\u2032 \u00b5\u2032\u03bb\u2032 \u2202x\u03bb\u2032 \u2202x\u03bb V \u03bb + \u2202x\u00b5 \u2202x\u00b5\u2032 V \u03bb \u2202 \u2202x\ufffd", "\ufffd \u2202x\u03bd\u2032 \u2202x\u03bb = \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u03bd\u2032 \u2202x\u03bd \u0393\u03bd \u00b5\u03bbV \u03bb, (3.5) where we have changed a dummy index from \u03bd to \u03bb. This equation must be true for any vector V \u03bb, so we can eliminate that on both sides. Then the connection coe\ufb03cients in the primed coordinates may be isolated by multiplying by \u2202x\u03bb/\u2202x\u03bb\u2032. The result is \u0393\u03bd\u2032 \u00b5\u2032\u03bb\u2032 = \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u03bb \u2202x\u03bb\u2032 \u2202x\u03bd\u2032 \u2202x\u03bd \u0393\u03bd \u00b5\u03bb \u2212 \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u03bb \u2202x\u03bb\u2032 \u22022x\u03bd\u2032 \u2202x\u00b5\u2202x\u03bb. (3.6) This is not, of course, the tensor transformation law; the second term on the right spoils it. That\u2019s okay, because the connection coe\ufb03cients are not the components of a tensor. They are purposefully constructed to be non-tensorial, but in such a way that the combination (3.1) transforms as a tensor \u2014 the extra terms in the transformation of the partials and 3 CURVATURE 57 the \u0393\u2019s exactly cancel. This is why we are not so careful about index placement on the connection coe\ufb03cients; they are not a tensor, and therefore you should try not to raise and lower their indices. What about the covariant derivatives of other sorts of tensors? By similar reasoning to that used for vectors, the covariant derivative of a one-form can also be expressed as a partial derivative plus some linear transformation. But there is no reason as yet that the matrices representing this transformation should be related to the coe\ufb03cients \u0393\u03bd In general we could write something like \u00b5\u03bb. \u2207\u00b5\u03c9\u03bd = \u2202\u00b5\u03c9\u03bd + \u0393\u03bb \u00b5\u03bd\u03c9\u03bb, (3.7) \u0393\u03bb e \u00b5\u03bd is a new set of matrices for each \u00b5. (Pay attention to where all of the various where e indices go.) It is straightforward to derive that the transformation properties of \u0393 must be the same as those of \u0393, but otherwise no relationship has been established. To do so, we need to introduce two new properties that we would like our covariant derivative to have (in addition to the two above): e 3. commutes with contractions: \u2207\u00b5(T \u03bb \u03bb\u03c1) = ( 4. reduces to the partial derivative on scalars: T )\u00b5 \u03bb \u03bb\u03c1, \u2207 \u2207\u00b5\u03c6 = \u2202\u00b5\u03c6. There is no way to \u201cderive\u201d these properties; we are simply demanding that they be true as part of the de\ufb01nition of a covariant derivative. Let\u2019s see what these new properties imply. Given some one-form \ufb01eld \u03c9\u00b5 and vector \ufb01eld V \u00b5, we can take the covariant derivative of the scalar de\ufb01ned by \u03c9\u03bbV \u03bb to get \u2207\u00b5(\u03c9\u03bbV \u03bb) = ( \u2207\u00b5\u03c9\u03bb)V \u03bb + \u03c9\u03bb( \u00b5\u03bb\u03c9\u03c3V \u03bb + \u03c9\u03bb(\u2202\u00b5V \u03bb) + \u03c9\u03bb\u0393\u03bb \u0393\u03c3 \u2207\u00b5V \u03bb) = (\u2202\u00b5\u03c9\u03bb)V \u03bb + \u00b5\u03c1V \u03c1. But since \u03c9\u03bbV \u03bb is a scalar, this must also be given by the partial derivative: e \u2207\u00b5(\u03c9\u03bbV \u03bb) = \u2202\u00b5(\u03c9\u03bbV \u03bb) = (\u2202\u00b5\u03c9\u03bb)V \u03bb + \u03c9\u03bb(\u2202\u00b5V \u03bb). (3.8) (3.9) This can only be true if the terms in (3.8) with connection coe\ufb03cients cancel each other; that is, rearranging dummy indices, we must have But both \u03c9\u03c3 and V \u03bb are completely arbitrary, so e 0 = \u00b5\u03bb\u03c9\u03c3V \u03bb + \u0393\u03c3 \u0393\u03c3 \u00b5\u03bb\u03c9\u03c3V \u03bb. \u0393\u03c3 \u00b5\u03bb. \u2212 \u0393\u03c3 \u00b5\u03bb = e (3.10) (3.11) 3 CURVATURE 58 The two extra conditions we have imposed therefore allow us to express the covariant derivative of a one-form using the same connection coe\ufb03cients as were used for the vector, but now with a minus sign (and indices matched up", " somewhat di\ufb00erently): \u2207\u00b5\u03c9\u03bd = \u2202\u00b5\u03c9\u03bd \u2212 \u0393\u03bb \u00b5\u03bd\u03c9\u03bb. (3.12) It should come as no surprise that the connection coe\ufb03cients encode all of the information necessary to take the covariant derivative of a tensor of arbitrary rank. The formula is quite straightforward; for each upper index you introduce a term with a single +\u0393, and for each lower index a term with a single \u0393: \u2207\u03c3T \u00b51\u00b52 \u00b5k \u00b7\u00b7\u00b7 \u03bd1\u03bd2 \u2212 \u03bdl = \u2202\u03c3T \u00b51\u00b52 \u00b7\u00b7\u00b7 \u00b5k \u00b7\u00b7\u00b7 +\u0393\u00b51 \u03c3\u03bb T \u03bb\u00b52 \u0393\u03bb \u03c3\u03bd1T \u00b51\u00b52 \u00b7\u00b7\u00b7 \u03bd1\u03bd2 \u00b5k \u03bdl \u00b7\u00b7\u00b7 \u03bd1\u03bd2 \u00b5k \u00b7\u00b7\u00b7 \u03bb\u03bd2 \u2212 \u00b7\u00b7\u00b7 \u03bdl + \u0393\u00b52 \u03c3\u03bb T \u00b51\u03bb \u00b5k \u00b7\u00b7\u00b7 \u0393\u03bb \u03c3\u03bd2T \u00b51\u00b52 \u03bdl \u2212 \u00b7\u00b7\u00b7 \u00b7\u00b7\u00b7 \u03bd1\u03bd2 \u00b5k \u00b7\u00b7\u00b7 \u03bd1\u03bb \u03bdl + \u00b7 \u00b7 \u00b7 \u03bdl \u2212 \u00b7 \u00b7 \u00b7 \u00b7\u00b7\u00b7. (3.13) This is the general expression for the covariant derivative. You can check it yourself; it comes from the set of axioms we have established, and the usual requirements that tensors of various sorts be coordinate-independent entities. Sometimes an alternative notation is used; just as commas are used for partial derivatives, semicolons are used for covariant ones: \u03bdl \u2261 \u00b7\u00b7\u00b7 Once again, I\u2019m not a big fan of this notation. \u03bd1\u03bd2 \u2207\u03c3T \u00b51\u00b52 \u00b5k \u00b7\u00b7\u00b7 T \u00b51\u00b52 \u00b5k \u00b7\u00b7\u00b7 \u03bd1\u03bd2 \u03bdl;\u03c3. \u00b7\u00b7\u00b7 (3.14) To de\ufb01ne a covariant derivative, then, we need to put a \u201cconnection\u201d on our manifold, which is speci\ufb01ed in some coordinate system by a set of coe\ufb03cients \u0393\u03bb \u00b5\u03bd (n3 = 64 independent components in n = 4 dimensions) which transform according to (3.6). (The name \u201cconnection\u201d comes from the fact that it is used to transport vectors from one tangent space to another, as we will soon see.) There are evidently a large number of connections we could de\ufb01ne on any manifold, and each of them implies a distinct notion of covariant di\ufb00erentiation. In general relativity this freedom is not a big concern, because it turns out that every metric de\ufb01nes a unique connection, which is the one used in GR. Let\u2019s see how that works. The \ufb01rst thing to notice is that the di\ufb00erence of two connections is a (1, 2) tensor. If \u0393\u03bb \u00b5\u03bd \u00b5\u03bd, their di\ufb00erence S\u00b5\u03bd \u00b5\u03bd and \u03bb = \u0393\u03bb we have two sets of connection coe\ufb03cients, \u0393\u03bb (notice index placement) transforms as \u00b5\u03bd \u2212 \u0393\u03bb b b S\u00b5\u2032\u03bd\u2032 \u03bb\u2032 = = \u0393\u03bb\u2032 \u00b5\u2032\u03bd\u2032 \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u00b5 \u2202x\u00b5\u2032 = = \u0393\u03bb\u2032 \u00b5\u2032\u03bd\u2032 \u2212 \u2202x\u03bb\u2032 \u2202x\u03bd \u2202x\u03bb \u0393\u03bb b \u2202x\u03bd\u2032 \u2202x\u03bb\u2032 \u2202x\u03bd \u2202x\u03bb (\u0393\u03bb \u2202x\u03bd\u2032 \u2202x\u03bb\u2032 \u2202x\u03bd \u2202x\u03bb S\u00b5\u03bd \u2202x\u03bd\u2032 \u00b5\u03bd \u2212 \u00b5\u03bd \u2212 \u03bb. \u2202x\u00b5 \u2202x\u00b5\u2032 \u0393\u03bb \u00b5\u03bd) \u2202x\u03bd \u2202x\u03bd\u2032 \u22022x\u03bb\u2032 \u2202x\u00b5\u2202x\u03bd \u2212 \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u03bd \u2202x\u03bd\u2032 \u2202x\u03bb\u2032 \u2202x\u03bb \u2202x\u00b5 \u2202x\u00b5\u2032 \u2202x\u03bd \u2202x\u03bd\u2032 \u22022x\u03bb\u2032 \u2202x\u00b5\u2202x\u03bd \u0393\u03bb \u00b5\u03bd + b b (3.15) 3 CURVATURE 59 Next notice that, given a connection speci\ufb01ed by \u0393\u03bb \u03bb is indeed a tensor. This implies that any This is just the tensor transormation law, so S\u00b5\u03bd set of connections can be expressed as some \ufb01ducial connection plus a tensorial correction.", " \u00b5\u03bd, we can immediately form another connection simply by permuting the lower indices. That is, the set of coe\ufb03cients \u0393\u03bb \u03bd\u00b5 will also transform according to (3.6) (since the partial derivatives appearing in the last term can be commuted), so they determine a distinct connection. There is thus a tensor we can associate with any given connection, known as the torsion tensor, de\ufb01ned by T\u00b5\u03bd \u03bb = \u0393\u03bb \u00b5\u03bd \u2212 \u03bd\u00b5 = 2\u0393\u03bb \u0393\u03bb [\u00b5\u03bd]. (3.16) It is clear that the torsion is antisymmetric its lower indices, and a connection which is symmetric in its lower indices is known as \u201ctorsion-free.\u201d We can now de\ufb01ne a unique connection on a manifold with a metric g\u00b5\u03bd by introducing two additional properties: torsion-free: \u0393\u03bb \u00b5\u03bd = \u0393\u03bb (\u00b5\u03bd). metric compatibility: \u2207\u03c1g\u00b5\u03bd = 0. \u2022 \u2022 A connection is metric compatible if the covariant derivative of the metric with respect to that connection is everywhere zero. This implies a couple of nice properties. First, it\u2019s easy to show that the inverse metric also has zero covariant derivative, \u2207\u03c1g\u00b5\u03bd = 0. Second, a metric-compatible covariant derivative commutes with raising and lowering of indices. Thus, for some vector \ufb01eld V \u03bb, (3.17) g\u00b5\u03bb\u2207\u03c1V \u03bb = \u2207\u03c1(g\u00b5\u03bbV \u03bb) = \u2207\u03c1V\u00b5. (3.18) With non-metric-compatible connections one must be very careful about index placement when taking a covariant derivative. Our claim is therefore that there is exactly one torsion-free connection on a given manifold which is compatible with some given metric on that manifold. We do not want to make these two requirements part of the de\ufb01nition of a covariant derivative; they simply single out one of the many possible ones. We can demonstrate both existence and uniqueness by deriving a manifestly unique expression for the connection coe\ufb03cients in terms of the metric. To accomplish this, we expand out the equation of metric compatibility for three di\ufb00erent permutations of the indices: \u2207\u03c1g\u00b5\u03bd = \u2202\u03c1g\u00b5\u03bd \u2212 \u0393\u03bb \u03c1\u00b5g\u03bb\u03bd \u2212 \u0393\u03bb \u03c1\u03bdg\u00b5\u03bb = 0 3 CURVATURE \u2207\u00b5g\u03bd\u03c1 = \u2202\u00b5g\u03bd\u03c1 \u2212 \u2207\u03bdg\u03c1\u00b5 = \u2202\u03bdg\u03c1\u00b5 \u2212 \u0393\u03bb \u00b5\u03bdg\u03bb\u03c1 \u2212 \u0393\u03bb \u03bd\u03c1g\u03bb\u00b5 \u2212 \u0393\u03bb \u0393\u03bb \u00b5\u03c1g\u03bd\u03bb = 0 \u03bd\u00b5g\u03c1\u03bb = 0. 60 (3.19) We subtract the second and third of these from the \ufb01rst, and use the symmetry of the connection to obtain It is straightforward to solve this for the connection by multiplying by g\u03c3\u03c1. The result is \u2202\u03c1g\u00b5\u03bd \u2212 \u2202\u00b5g\u03bd\u03c1 \u2212 \u2202\u03bdg\u03c1\u00b5 + 2\u0393\u03bb \u00b5\u03bdg\u03bb\u03c1 = 0. (3.20) 1 2 \u0393\u03c3 \u00b5\u03bd = g\u03c3\u03c1(\u2202\u00b5g\u03bd\u03c1 + \u2202\u03bdg\u03c1\u00b5 \u2212 This is one of the most important formulas in this subject; commit it to memory. Of course, we have only proved that if a metric-compatible and torsion-free connection exists, it must be of the form (3.21); you can check for yourself (for those of you without enough tedious computation in your lives) that the right hand side of (3.21) transforms like a connection. \u2202\u03c1g\u00b5\u03bd). (3.21) This connection we have derived from the metric is the one on which conventional general It is known relativity is based (although we will keep an open mind for a while longer). by di\ufb00erent names: sometimes the Christo\ufb00el connection, sometimes the Levi-Civita connection, sometimes the Riemannian connection. The associated connection coe\ufb03cients are sometimes called Christo\ufb00el symbols and written as ; we will sometimes call them Christo\ufb00el symbols, but we won\u2019t use the funny notation. The study of manifolds with metrics and their associated connections is called \u201cRiemannian geometry.\u201d As far as I can tell the study of more general connections can be traced back to Cartan, but I\u2019ve never heard it called \u201cCartanian geometry.\ufffd", "\ufffd \u03c3 \u00b5\u03bd o n Before putting our covariant derivatives to work, we should mention some miscellaneous properties. First, let\u2019s emphasize again that the connection does not have to be constructed from the metric. In ordinary \ufb02at space there is an implicit connection we use all the time \u2014 the Christo\ufb00el connection constructed from the \ufb02at metric. But we could, if we chose, use a di\ufb00erent connection, while keeping the metric \ufb02at. Also notice that the coe\ufb03cients of the Christo\ufb00el connection in \ufb02at space will vanish in Cartesian coordinates, but not in curvilinear coordinate systems. Consider for example the plane in polar coordinates, with metric ds2 = dr2 + r2d\u03b82. (3.22) The nonzero components of the inverse metric are readily found to be grr = 1 and g\u03b8\u03b8 = r\u2212 2. (Notice that we use r and \u03b8 as indices in an obvious notation.) We can compute a typical connection coe\ufb03cient: \u0393r rr = = 1 2 1 2 gr\u03c1(\u2202rgr\u03c1 + \u2202rg\u03c1r \u2212 grr(\u2202rgrr + \u2202rgrr \u2212 \u2202\u03c1grr) \u2202rgrr) 3 CURVATURE 1 gr\u03b8(\u2202rgr\u03b8 + \u2202rg\u03b8r \u2212 2 (1)(0 + 0 0) + \u2212 \u2202\u03b8grr) 1 2 (0)(0 + 0 0) \u2212 + 1 2 = 0. = Sadly, it vanishes. But not all of them do: \u0393r \u03b8\u03b8 = = = = 1 2 1 2 1 2 gr\u03c1(\u2202\u03b8g\u03b8\u03c1 + \u2202\u03b8g\u03c1\u03b8 \u2212 grr(\u2202\u03b8g\u03b8r + \u2202\u03b8gr\u03b8 \u2212 (1)(0 + 0 2r) \u2212 r. \u2212 \u2202\u03c1g\u03b8\u03b8) \u2202rg\u03b8\u03b8) Continuing to turn the crank, we eventually \ufb01nd \u03b8r = \u0393r \u0393r \u0393\u03b8 rr = 0 r\u03b8 = \u0393\u03b8 \u0393\u03b8 \u0393\u03b8 \u03b8\u03b8 = 0. r\u03b8 = 0 \u03b8r = 1 r 61 (3.23) (3.24) (3.25) The existence of nonvanishing connection coe\ufb03cients in curvilinear coordinate systems is the ultimate cause of the formulas for the divergence and so on that you \ufb01nd in books on electricity and magnetism. Contrariwise, even in a curved space it is still possible to make the Christo\ufb00el symbols vanish at any one point. This is just because, as we saw in the last section, we can always make the \ufb01rst derivative of the metric vanish at a point; so by (3.21) the connection coe\ufb03cients derived from this metric will also vanish. Of course this can only be established at a point, not in some neighborhood of the point. Another useful property is that the formula for the divergence of a vector (with respect to the Christo\ufb00el connection) has a simpli\ufb01ed form. The covariant divergence of V \u00b5 is given by \u2207\u00b5V \u00b5 = \u2202\u00b5V \u00b5 + \u0393\u00b5 It\u2019s easy to show (see pp. 106-108 of Weinberg) that the Christo\ufb00el connection satis\ufb01es \u00b5\u03bbV \u03bb. (3.26) and we therefore obtain \u0393\u00b5 \u00b5\u03bb = 1 g | | q \u2202\u03bb, g | | q \u2207\u00b5V \u00b5 = 1 g | | q \u2202\u00b5( | q g V \u00b5). | (3.27) (3.28) 3 CURVATURE 62 There are also formulas for the divergences of higher-rank tensors, but they are generally not such a great simpli\ufb01cation. As the last factoid we should mention about connections, let us emphasize (once more) that the exterior derivative is a well-de\ufb01ned tensor in the absence of any connection. The reason this needs to be emphasized is that, if you happen to be using a symmetric (torsionfree) connection, the exterior derivative (de\ufb01ned to be the antisymmetrized partial derivative) happens to be equal to the antisymmetrized covariant derivative: \u2207[\u00b5\u03c9\u03bd] = \u2202[\u00b5\u03c9\u03bd] \u2212 = \u2202[\u00b5\u03c9\u03bd]. \u0393\u03bb [\u00b5", "\u03bd]\u03c9\u03bb (3.29) This has led some misfortunate souls to fret about the \u201cambiguity\u201d of the exterior derivative in spaces with torsion, where the above simpli\ufb01cation does not occur. There is no ambiguity: the exterior derivative does not involve the connection, no matter what connection you happen to be using, and therefore the torsion never enters the formula for the exterior derivative of anything. Before moving on, let\u2019s review the process by which we have been adding structures to our mathematical constructs. We started with the basic notion of a set, which you were presumed to know (informally, if not rigorously). We introduced the concept of open subsets of our set; this is equivalent to introducing a topology, and promoted the set to a topological space. Then by demanding that each open set look like a region of Rn (with n the same for each set) and that the coordinate charts be smoothly sewn together, the topological space became a manifold. A manifold is simultaneously a very \ufb02exible and powerful structure, and comes equipped naturally with a tangent bundle, tensor bundles of various ranks, the ability to take exterior derivatives, and so forth. We then proceeded to put a metric on the manifold, resulting in a manifold with metric (or sometimes \u201cRiemannian manifold\u201d). Independently of the metric we found we could introduce a connection, allowing us to take covariant derivatives. Once we have a metric, however, there is automatically a unique (In principle there is nothing to stop us from torsion-free metric-compatible connection. introducing more than one connection, or more than one metric, on any given manifold.) The situation is thus as portrayed in the diagram on the next page. 3 CURVATURE 63 introduce a topology (open sets) n R locally like introduce a connection set topological space manifold manifold with connection introduce a metric Riemannian manifold (automatically has a connection) Having set up the machinery of connections, the \ufb01rst thing we will do is discuss parallel transport. Recall that in \ufb02at space it was unnecessary to be very careful about the fact that vectors were elements of tangent spaces de\ufb01ned at individual points; it is actually very natural to compare vectors at di\ufb00erent points (where by \u201ccompare\u201d we mean add, subtract, take the dot product, etc.). The reason why it is natural is because it makes sense, in \ufb02at space, to \u201cmove a vector from one point to another while keeping it constant.\u201d Then once we get the vector from one point to another we can do the usual operations allowed in a vector space. q keep vector constant p The concept of moving a vector along a path, keeping constant all the while, is known as parallel transport. As we shall see, parallel transport is de\ufb01ned whenever we have a 3 CURVATURE 64 connection; the intuitive manipulation of vectors in \ufb02at space makes implicit use of the Christo\ufb00el connection on this space. The crucial di\ufb00erence between \ufb02at and curved spaces is that, in a curved space, the result of parallel transporting a vector from one point to another will depend on the path taken between the points. Without yet assembling the complete mechanism of parallel transport, we can use our intuition about the two-sphere to see that this is the case. Start with a vector on the equator, pointing along a line of constant longitude. Parallel transport it up to the north pole along a line of longitude in the obvious way. Then take the original vector, parallel transport it along the equator by an angle \u03b8, and then move it up to the north pole as before. It is clear that the vector, parallel transported along two paths, arrived at the same destination with two di\ufb00erent values (rotated by \u03b8). It therefore appears as if there is no natural way to uniquely move a vector from one tangent space to another; we can always parallel transport it, but the result depends on the path, and there is no natural choice of which path to take. Unlike some of the problems we have encountered, there is no solution to this one \u2014 we simply must learn to live with the fact that two vectors can only be compared in a natural way if they are elements of the same tangent space. For example, two particles passing by each other have a well-de\ufb01ned relative velocity (which cannot be greater than the speed of light). But two particles at di\ufb00erent points on a curved manifold do not have any well-de\ufb01ned notion of relative velocity \u2014 the concept simply makes no sense. Of course, in certain special situations it is still useful to talk as if it did make sense, but it is necessary to understand that occasional usefulness", " is not a substitute for rigorous de\ufb01nition. In cosmology, for example, the light from distant galaxies is redshifted with respect to the frequencies we would observe from a nearby stationary source. Since this phenomenon bears such a close resemblance to the conventional Doppler e\ufb00ect due to relative motion, it is very tempting to say that the galaxies are \u201creceding away from us\u201d at a speed de\ufb01ned by their redshift. At a rigorous level this is nonsense, what Wittgenstein would call a \u201cgrammatical mistake\u201d \u2014 the galaxies are not receding, since the notion of their velocity with respect to us is not well-de\ufb01ned. What is actually happening is that the metric of spacetime between us and the galaxies has changed (the universe has 3 CURVATURE 65 expanded) along the path of the photon from here to there, leading to an increase in the wavelength of the light. As an example of how you can go wrong, naive application of the Doppler formula to the redshift of galaxies implies that some of them are receding faster than light, in apparent contradiction with relativity. The resolution of this apparent paradox is simply that the very notion of their recession should not be taken literally. Enough about what we cannot do; let\u2019s see what we can. Parallel transport is supposed to be the curved-space generalization of the concept of \u201ckeeping the vector constant\u201d as we move it along a path; similarly for a tensor of arbitrary rank. Given a curve x\u00b5(\u03bb), the requirement \u2202T of constancy of a tensor T along this curve in \ufb02at space is simply dT \u2202x\u00b5 = 0. We therefore de\ufb01ne the covariant derivative along the path to be given by an operator d\u03bb = dx\u00b5 d\u03bb D d\u03bb = dx\u00b5 d\u03bb \u2207\u00b5. (3.30) We then de\ufb01ne parallel transport of the tensor T along the path x\u00b5(\u03bb) to be the requirement that, along the path, D d\u03bb T \u00b51\u00b52 \u00b5k \u00b7\u00b7\u00b7 \u03bd1\u03bd2 \u03bdl \u2261 \u00b7\u00b7\u00b7 dx\u03c3 d\u03bb \u2207\u03c3T \u00b51\u00b52 \u00b5k \u00b7\u00b7\u00b7 \u03bd1\u03bd2 \u03bdl = 0. \u00b7\u00b7\u00b7 (3.31) This is a well-de\ufb01ned tensor equation, since both the tangent vector dx\u00b5/d\u03bb and the covariant T are tensors. This is known as the equation of parallel transport. For a derivative \u2207 vector it takes the form d d\u03bb V \u00b5 + \u0393\u00b5 \u03c3\u03c1 dx\u03c3 d\u03bb V \u03c1 = 0. (3.32) We can look at the parallel transport equation as a \ufb01rst-order di\ufb00erential equation de\ufb01ning an initial-value problem: given a tensor at some point along the path, there will be a unique continuation of the tensor to other points along the path such that the continuation solves (3.31). We say that such a tensor is parallel transported. The notion of parallel transport is obviously dependent on the connection, and di\ufb00erent connections lead to di\ufb00erent answers. If the connection is metric-compatible, the metric is always parallel transported with respect to it: D d\u03bb g\u00b5\u03bd = dx\u03c3 d\u03bb \u2207\u03c3g\u00b5\u03bd = 0. (3.33) It follows that the inner product of two parallel-transported vectors is preserved. That is, if V \u00b5 and W \u03bd are parallel-transported along a curve x\u03c3(\u03bb), we have D d\u03bb (g\u00b5\u03bdV \u00b5W \u03bd) = D d\u03bb g\u00b5\u03bd V \u00b5W \u03bd + g\u00b5\u03bd D d\u03bb V \u00b5 W \u03bd + g\u00b5\u03bdV \u00b5 D d\u03bb W \u03bd (3.34) = 0. This means that parallel transport with respect to a metric-compatible connection preserves the norm of vectors, the sense of orthogonality, and so on. 3 CURVATURE 66 One thing they don\u2019t usually tell you in GR books is that you can write down an explicit and general solution to the parallel transport equation, although it\u2019s somewhat formal. First x\u03c3(\u03bb), solving the parallel transport equation for a vector notice that for some path \u03b3 : \u03bb \u2192 V \u00b5 amounts to \ufb01nding a matrix P \u00b5 \u03c1(\u03bb, \u03bb0) which relates the vector at its initial value V \u00b5(\u03bb0) to its value somewhere later down the path: V \u00b5(\u03bb) = P \u00b5 \u03c1(\u03bb, \u03bb0)V \u03c1(\u03bb0). (3.35) Of course the", " matrix P \u00b5 \u03c1(\u03bb, \u03bb0), known as the parallel propagator, depends on the path \u03b3 (although it\u2019s hard to \ufb01nd a notation which indicates this without making \u03b3 look like an index). If we de\ufb01ne \u2212 where the quantities on the right hand side are evaluated at x\u03bd(\u03bb), then the parallel transport equation becomes (3.36) A\u00b5 \u03c1(\u03bb) = \u0393\u00b5 \u03c3\u03c1 dx\u03c3 d\u03bb, d d\u03bb V \u00b5 = A\u00b5 \u03c1V \u03c1. (3.37) Since the parallel propagator must work for any vector, substituting (3.35) into (3.37) shows that P \u00b5 \u03c1(\u03bb, \u03bb0) also obeys this equation: d d\u03bb P \u00b5 \u03c1(\u03bb, \u03bb0) = A\u00b5 \u03c3(\u03bb)P \u03c3 \u03c1(\u03bb, \u03bb0). (3.38) To solve this equation, \ufb01rst integrate both sides: P \u00b5 \u03c1(\u03bb, \u03bb0) = \u03b4\u00b5 \u03c1 + \u03bb \u03bb0 Z A\u00b5 \u03c3(\u03b7)P \u03c3 \u03c1(\u03b7, \u03bb0) d\u03b7. (3.39) The Kronecker delta, it is easy to see, provides the correct normalization for \u03bb = \u03bb0. We can solve (3.39) by iteration, taking the right hand side and plugging it into itself repeatedly, giving P \u00b5 \u03c1(\u03bb, \u03bb0) = \u03b4\u00b5 \u03c1 + \u03bb \u03bb0 Z A\u00b5 \u03c1(\u03b7) d\u03b7 + \u03bb \u03b7 \u03bb0 Z Z \u03bb0 A\u00b5 \u03c3(\u03b7)A\u03c3 \u03c1(\u03b7\u2032) d\u03b7\u2032d\u03b7 +. \u00b7 \u00b7 \u00b7 (3.40) The nth term in this series is an integral over an n-dimensional right triangle, or n-simplex. 3 CURVATURE 67 \u03bb \u03bb0 Z A(\u03b71) d\u03b71 \u03bb \u03b72 \u03bb0 Z \u03bb0 Z A(\u03b72)A(\u03b71) d\u03b71d\u03b72 \u03bb \u03b73 \u03b72 \u03bb0 Z Z \u03bb0 Z \u03bb0 A(\u03b73)A(\u03b72)A(\u03b71) d3\u03b7 It would simplify things if we could consider such an integral to be over an n-cube instead of an n-simplex; is there some way to do this? There are n! such simplices in each cube, so we would have to multiply by 1/n! to compensate for this extra volume. But we also want to get the integrand right; using matrix notation, the integrand at nth order \u03b71. We A(\u03b71), but with the special property that \u03b7n \u2265 is A(\u03b7n)A(\u03b7n therefore de\ufb01ne the path-ordering symbol, In other words, the expression, to ensure that this condition holds. 1 \u2265 \u00b7 \u00b7 \u00b7 \u2265 \u00b7 \u00b7 \u00b7 \u03b7n 1) P \u2212 \u2212 [A(\u03b7n)A(\u03b7n 1) A(\u03b71)] (3.41) \u2212 stands for the product of the n matrices A(\u03b7i), ordered in such a way that the largest value of \u03b7i is on the left, and each subsequent value of \u03b7i is less than or equal to the previous one. We then can express the nth-order term in (3.40) as \u00b7 \u00b7 \u00b7 P \u03bb \u03b7n \u03b72 A(\u03b7n)A(\u03b7n \u00b7 \u00b7 \u00b7 Z \u03bb \u03bb0 \u03bb \u03bb 1) \u2212 \u00b7 \u00b7 \u00b7 A(\u03b71) dn\u03b7 \u03bb0 Z \u03bb0 Z = 1 n! Z \u03bb0 Z \u03bb0 \u00b7 \u00b7 \u00b7 Z \u03bb0 P [A(\u03b7n)A(\u03b7n 1) \u2212 \u00b7 \u00b7 \u00b7 A(\u03b71)] dn\u03b7. (3.42) This expression contains no substantive statement about the matrices A(\u03b7i); it is just notation. But we can now write (3.40) in matrix form as P (\u03bb, \u03bb0) = 1 + \u221e Xn=1 \u03bb 1 n! Z [A(\u03b7n)A(\u03b7n 1) \u2212 \u00b7 \u00b7 \u00b7 A(\u03b71)] dn\u03b7. (3.43) \u03bb0 P This formula is just the series expression for an exponential; we therefore say that the parallel propagator is", " given by the path-ordered exponential P (\u03bb, \u03bb0) = exp P \u03bb Z \u03bb0 A(\u03b7) d\u03b7,! (3.44) 3 CURVATURE 68 where once again this is just notation; the path-ordered exponential is de\ufb01ned to be the right hand side of (3.43). We can write it more explicitly as P \u00b5 \u03bd(\u03bb, \u03bb0) = exp P \u03bb \u2212 Z \u03bb0 \u0393\u00b5 \u03c3\u03bd dx\u03c3 d\u03b7. d\u03b7! (3.45) It\u2019s nice to have an explicit formula, even if it is rather abstract. The same kind of expression appears in quantum \ufb01eld theory as \u201cDyson\u2019s Formula,\u201d where it arises because the Schr\u00a8odinger equation for the time-evolution operator has the same form as (3.38). As an aside, an especially interesting example of the parallel propagator occurs when the path is a loop, starting and ending at the same point. Then if the connection is metriccompatible, the resulting matrix will just be a Lorentz transformation on the tangent space at the point. This transformation is known as the \u201cholonomy\u201d of the loop. If you know the holonomy of every possible loop, that turns out to be equivalent to knowing the metric. This fact has let Ashtekar and his collaborators to examine general relativity in the \u201cloop representation,\u201d where the fundamental variables are holonomies rather than the explicit metric. They have made some progress towards quantizing the theory in this approach, although the jury is still out about how much further progress can be made. With parallel transport understood, the next logical step is to discuss geodesics. A geodesic is the curved-space generalization of the notion of a \u201cstraight line\u201d in Euclidean space. We all know what a straight line is: it\u2019s the path of shortest distance between two points. But there is an equally good de\ufb01nition \u2014 a straight line is a path which parallel transports its own tangent vector. On a manifold with an arbitrary (not necessarily Christo\ufb00el) connection, these two concepts do not quite coincide, and we should discuss them separately. We\u2019ll take the second de\ufb01nition \ufb01rst, since it is computationally much more straightforward. The tangent vector to a path x\u00b5(\u03bb) is dx\u00b5/d\u03bb. The condition that it be parallel transported is thus D d\u03bb dx\u00b5 d\u03bb = 0, (3.46) or alternatively d2x\u00b5 d\u03bb2 + \u0393\u00b5 dx\u03c1 d\u03bb This is the geodesic equation, another one which you should memorize. We can easily see that it reproduces the usual notion of straight lines if the connection coe\ufb03cients are the Christo\ufb00el symbols in Euclidean space; in that case we can choose Cartesian coordinates in which \u0393\u00b5 \u03c1\u03c3 = 0, and the geodesic equation is just d2x\u00b5/d\u03bb2 = 0, which is the equation for a straight line. dx\u03c3 d\u03bb (3.47) = 0. \u03c1\u03c3 That was embarrassingly simple; let\u2019s turn to the more nontrivial case of the shortest distance de\ufb01nition. As we know, there are various subtleties involved in the de\ufb01nition of 3 CURVATURE 69 distance in a Lorentzian spacetime; for null paths the distance is zero, for timelike paths it\u2019s more convenient to use the proper time, etc. So in the name of simplicity let\u2019s do the calculation just for a timelike path \u2014 the resulting equation will turn out to be good for any path, so we are not losing any generality. We therefore consider the proper time functional, \u03c4 = g\u00b5\u03bd Z \u2212 dx\u00b5 d\u03bb 1/2 dx\u03bd d\u03bb! d\u03bb, (3.48) where the integral is over the path. To search for shortest-distance paths, we will do the usual calculus of variations treatment to seek extrema of this functional. (In fact they will turn out to be curves of maximum proper time.) We want to consider the change in the proper time under in\ufb01nitesimal variations of the path, x\u00b5 \u2192 g\u00b5\u03bd \u2192 x\u00b5 + \u03b4x\u00b5 g\u00b5\u03bd + \u03b4x\u03c3\u2202\u03c3g\u00b5\u03bd. (3.49) (The second line comes from Taylor expansion in curved spacetime, which as you can see uses the partial derivative, not the covariant derivative.) Plugging this into (3.48), we get \u03c4 +", " \u03b4\u03c4 = g\u00b5\u03bd Z \u2212 = g\u00b5\u03bd Z \u2212 dx\u00b5 d\u03bb dx\u00b5 d\u03bb dx\u03bd d\u03bb \u2212 dx\u03bd d\u03bb! 1/2 \u2202\u03c3g\u00b5\u03bd dx\u00b5 d\u03bb dx\u03bd d\u03bb 1 + \uf8ee g\u00b5\u03bd \u2212 d(\u03b4x\u03bd) 1/2 d\u03bb! d\u03bb dx\u00b5 d\u03bb 1 \u03b4x\u03c3 dx\u00b5 d\u03bb \u2212 2g\u00b5\u03bd \u2212 dx\u03bd d\u03bb! dx\u00b5 d\u03bb \u2202\u03c3g\u00b5\u03bd \u00d7 \u2212 \uf8f0 dx\u00b5 d\u03bb dx\u03bd d\u03bb \u03b4x\u03c3 \u2212 2g\u00b5\u03bd d(\u03b4x\u03bd) 1/2 d\u03bb!# d\u03bb. (3.50) Since \u03b4x\u03c3 is assumed to be small, we can expand the square root of the expression in square brackets to \ufb01nd \u03b4\u03c4 = g\u00b5\u03bd Z \u2212 1/2 dx\u00b5 d\u03bb \u2212 dx\u03bd d\u03bb! 1 2 \u2212 \u2202\u03c3g\u00b5\u03bd dx\u00b5 d\u03bb dx\u03bd d\u03bb \u03b4x\u03c3 g\u00b5\u03bd \u2212 dx\u00b5 d\u03bb d(\u03b4x\u03bd) d\u03bb! d\u03bb. (3.51) It is helpful at this point to change the parameterization of our curve from \u03bb, which was arbitrary, to the proper time \u03c4 itself, using d\u03bb = g\u00b5\u03bd \u2212 1/2 \u2212 dx\u00b5 d\u03bb dx\u03bd d\u03bb! d\u03c4. (3.52) We plug this into (3.51) (note: we plug it in for every appearance of d\u03bb) to obtain \u03b4\u03c4 = Z \"\u2212 1 2 \u2202\u03c3g\u00b5\u03bd dx\u00b5 d\u03c4 dx\u03bd d\u03c4 \u03b4x\u03c3 g\u00b5\u03bd \u2212 dx\u00b5 d\u03c4 d(\u03b4x\u03bd) d\u03c4 # d\u03c4 3 CURVATURE = Z \"\u2212 1 2 \u2202\u03c3g\u00b5\u03bd dx\u00b5 d\u03c4 dx\u03bd d\u03c4 + d d\u03c4 g\u00b5\u03c3 dx\u00b5 d\u03c4!# \u03b4x\u03c3 d\u03c4, 70 (3.53) where in the last line we have integrated by parts, avoiding possible boundary contributions by demanding that the variation \u03b4x\u03c3 vanish at the endpoints of the path. Since we are searching for stationary points, we want \u03b4\u03c4 to vanish for any variation; this implies dx\u00b5 d\u03c4 where we have used dg\u00b5\u03c3/d\u03c4 = (dx\u03bd/d\u03c4 )\u2202\u03bdg\u00b5\u03c3. Some shu\ufb04ing of dummy indices reveals d2x\u00b5 d\u03c4 2 = 0, \u2202\u03bdg\u00b5\u03c3 + g\u00b5\u03c3 dx\u00b5 d\u03c4 dx\u03bd d\u03c4 dx\u03bd d\u03c4 \u2202\u03c3g\u00b5\u03bd 1 2 \u2212 + (3.54) d2x\u00b5 d\u03c4 2 + and multiplying by the inverse metric \ufb01nally leads to \u2202\u03c3g\u00b5\u03bd + \u2202\u03bdg\u00b5\u03c3 + \u2202\u00b5g\u03bd\u03c3) ( \u2212 g\u00b5\u03c3 1 2 dx\u00b5 d\u03c4 dx\u03bd d\u03c4 = 0, d2x\u03c1 d\u03c4 2 + 1 2 g\u03c1\u03c3 (\u2202\u00b5g\u03bd\u03c3 + \u2202\u03bdg\u03c3\u00b5 \u2212 \u2202\u03c3g\u00b5\u03bd) dx\u00b5 d\u03c4 dx\u03bd d\u03c4 = 0. (3.55) (3.56) We see that this is precisely the geodesic equation (3.32), but with the speci\ufb01c choice of Christo\ufb00el connection (3.21). Thus, on a manifold with metric, extremals of the length functional are curves which parallel transport their tangent vector with respect to the Christo\ufb00el connection associated with that metric. It doesn\u2019t matter if there is any other connection de\ufb01ned on the same manifold. Of course, in GR the Christo\ufb00el connection is the only one which is used, so the two notions are the same. The primary usefulness of geodesics in general relativity is that they are the paths followed by unaccelerated particles. In fact, the geodesic equation can be thought of as the generalization of Newton\u2019s law f = ma for the case f = 0. It is also possible to introduce forces by adding terms to the right hand side; in fact, looking back to the expression (1.103) for the Lorentz force in special relativity, it is tempting to guess that the equation of motion for a particle of mass m and charge q in general relativity should be d2x\u00b5 d\u03c4 2 + \u0393\u00b5 \u03c1\u03c3 dx\u03c1 d\u03c4 dx\u03c3 d\u03c4 = q m F \u00b5 \u03bd dx\u03bd d\u03c4. (3.", "57) We will talk about this more later, but in fact your guess would be correct. Having boldly derived these expressions, we should say some more careful words about the parameterization of a geodesic path. When we presented the geodesic equation as the requirement that the tangent vector be parallel transported, (3.47), we parameterized our path with some parameter \u03bb, whereas when we found the formula (3.56) for the extremal of the spacetime interval we wound up with a very speci\ufb01c parameterization, the proper time. Of course from the form of (3.56) it is clear that a transformation \u03bb = a\u03c4 + b, \u03c4 \u2192 (3.58) 3 CURVATURE 71 for some constants a and b, leaves the equation invariant. Any parameter related to the proper time in this way is called an a\ufb03ne parameter, and is just as good as the proper time for parameterizing a geodesic. What was hidden in our derivation of (3.47) was that the demand that the tangent vector be parallel transported actually constrains the parameterization of the curve, speci\ufb01cally to one related to the proper time by (3.58). In other words, if you start at some point and with some initial direction, and then construct a curve by beginning to walk in that direction and keeping your tangent vector parallel transported, you will not only de\ufb01ne a path in the manifold but also (up to linear transformations) de\ufb01ne the parameter along the path. Of course, there is nothing to stop you from using any other parameterization you like, but then (3.47) will not be satis\ufb01ed. More generally you will satisfy an equation of the form d2x\u00b5 d\u03b12 + \u0393\u00b5 \u03c1\u03c3 dx\u03c1 d\u03b1 dx\u03c3 d\u03b1 = f (\u03b1) dx\u00b5 d\u03b1, (3.59) for some parameter \u03b1 and some function f (\u03b1). Conversely, if (3.59) is satis\ufb01ed along a curve you can always \ufb01nd an a\ufb03ne parameter \u03bb(\u03b1) for which the geodesic equation (3.47) will be satis\ufb01ed. An important property of geodesics in a spacetime with Lorentzian metric is that the character (timelike/null/spacelike) of the geodesic (relative to a metric-compatible connection) never changes. This is simply because parallel transport preserves inner products, and the character is determined by the inner product of the tangent vector with itself. This is why we were consistent to consider purely timelike paths when we derived (3.56); for spacelike paths we would have derived the same equation, since the only di\ufb00erence is an overall minus sign in the \ufb01nal answer. There are also null geodesics, which satisfy the same equation, except that the proper time cannot be used as a parameter (some set of allowed parameters will exist, related to each other by linear transformations). You can derive this fact either from the simple requirement that the tangent vector be parallel transported, or by extending the variation of (3.48) to include all non-spacelike paths. Let\u2019s now explain the earlier remark that timelike geodesics are maxima of the proper time. The reason we know this is true is that, given any timelike curve (geodesic or not), we can approximate it to arbitrary accuracy by a null curve. To do this all we have to do is to consider \u201cjagged\u201d null curves which follow the timelike one: 3 CURVATURE 72 timelike null As we increase the number of sharp corners, the null curve comes closer and closer to the timelike curve while still having zero path length. Timelike geodesics cannot therefore be curves of minimum proper time, since they are always in\ufb01nitesimally close to curves of zero proper time; in fact they maximize the proper time. (This is how you can remember which twin in the twin paradox ages more \u2014 the one who stays home is basically on a geodesic, and therefore experiences more proper time.) Of course even this is being a little cavalier; actually every time we say \u201cmaximize\u201d or \u201cminimize\u201d we should add the modi\ufb01er \u201clocally.\u201d It is often the case that between two points on a manifold there is more than one geodesic. For instance, on S2 we can draw a great circle through any two points, and imagine travelling between them either the short way or the long way around. One of these is obviously longer than the other, although both are stationary points of the length", " functional. The \ufb01nal fact about geodesics before we move on to curvature proper is their use in mapping the tangent space at a point p to a local neighborhood of p. To do this we notice that any geodesic x\u00b5(\u03bb) which passes through p can be speci\ufb01ed by its behavior at p; let us choose the parameter value to be \u03bb(p) = 0, and the tangent vector at p to be dx\u00b5 d\u03bb (\u03bb = 0) = k\u00b5, (3.60) for k\u00b5 some vector at p (some element of Tp). Then there will be a unique point on the manifold M which lies on this geodesic where the parameter has the value \u03bb = 1. We de\ufb01ne M, via the exponential map at p, expp : Tp \u2192 expp(k\u00b5) = x\u03bd(\u03bb = 1), (3.61) where x\u03bd(\u03bb) solves the geodesic equation subject to (3.60). For some set of tangent vectors k\u00b5 near the zero vector, this map will be well-de\ufb01ned, and in fact invertible. Thus in the 3 CURVATURE 73 T p \u00b5 k p M \u03bb=1 \u03bd \u03bb x ( ) neighborhood of p given by the range of the map on this set of tangent vectors, the the tangent vectors themselves de\ufb01ne a coordinate system on the manifold. In this coordinate system, any geodesic through p is expressed trivially as x\u00b5(\u03bb) = \u03bbk\u00b5, (3.62) for some appropriate vector k\u00b5. We won\u2019t go into detail about the properties of the exponential map, since in fact we won\u2019t be using it much, but it\u2019s important to emphasize that the range of the map is not necessarily the whole manifold, and the domain is not necessarily the whole tangent space. The range can fail to be all of M simply because there can be two points which are not connected by any geodesic. (In a Euclidean signature metric this is impossible, but not in a Lorentzian spacetime.) The domain can fail to be all of Tp because a geodesic may run into a singularity, which we think of as \u201cthe edge of the manifold.\u201d Manifolds which have such singularities are known as geodesically incomplete. This is not merely a problem for careful mathematicians; in fact the \u201csingularity theorems\u201d of Hawking and Penrose state that, for reasonable matter content (no negative energies), spacetimes in general relativity are almost guaranteed to be geodesically incomplete. As examples, the two most useful spacetimes in GR \u2014 the Schwarzschild solution describing black holes and the FriedmannRobertson-Walker solutions describing homogeneous, isotropic cosmologies \u2014 both feature important singularities. Having set up the machinery of parallel transport and covariant derivatives, we are at last prepared to discuss curvature proper. The curvature is quanti\ufb01ed by the Riemann tensor, which is derived from the connection. The idea behind this measure of curvature is that we know what we mean by \u201c\ufb02atness\u201d of a connection \u2014 the conventional (and usually implicit) Christo\ufb00el connection associated with a Euclidean or Minkowskian metric has a number of properties which can be thought of as di\ufb00erent manifestations of \ufb02atness. These include the fact that parallel transport around a closed loop leaves a vector unchanged, that covariant derivatives of tensors commute, and that initially parallel geodesics remain parallel. As we 3 CURVATURE 74 shall see, the Riemann tensor arises when we study how any of these properties are altered in more general contexts. We have already argued, using the two-sphere as an example, that parallel transport of a vector around a closed loop in a curved space will lead to a transformation of the vector. The resulting transformation depends on the total curvature enclosed by the loop; it would be more useful to have a local description of the curvature at each point, which is what the Riemann tensor is supposed to provide. One conventional way to introduce the Riemann tensor, therefore, is to consider parallel transport around an in\ufb01nitesimal loop. We are not going to do that here, but take a more direct route. (Most of the presentations in the literature are either sloppy, or correct but very di\ufb03cult to follow.) Nevertheless, even without working through the details, it is possible to see what form the answer should take. Imagine that we parallel transport a vector V \u03c3 around a closed loop de\ufb01ned by two vectors", " A\u03bd and B\u00b5: \u00b5 A \u03b4 \u03b4 ( a, b) (0, b)\u03b4 \u03bd B (0, 0) \u03bd B \u00b5 A \u03b4 ( a, 0) The (in\ufb01nitesimal) lengths of the sides of the loop are \u03b4a and \u03b4b, respectively. Now, we know the action of parallel transport is independent of coordinates, so there should be some tensor which tells us how the vector changes when it comes back to its starting point; it will be a linear transformation on a vector, and therefore involve one upper and one lower index. But it will also depend on the two vectors A and B which de\ufb01ne the loop; therefore there should be two additional lower indices to contract with A\u03bd and B\u00b5. Furthermore, the tensor should be antisymmetric in these two indices, since interchanging the vectors corresponds to traversing the loop in the opposite direction, and should give the inverse of the original answer. (This is consistent with the fact that the transformation should vanish if A and B are the same vector.) We therefore expect that the expression for the change \u03b4V \u03c1 experienced by this vector when parallel transported around the loop should be of the form \u03b4V \u03c1 = (\u03b4a)(\u03b4b)A\u03bdB\u00b5R\u03c1 \u03c3\u00b5\u03bd V \u03c3, (3.63) where R\u03c1 \u03c3\u00b5\u03bd is a (1, 3) tensor known as the Riemann tensor (or simply \u201ccurvature tensor\u201d). 3 CURVATURE 75 It is antisymmetric in the last two indices: R\u03c1 \u03c3\u00b5\u03bd = R\u03c1 \u03c3\u03bd\u00b5. (3.64) \u2212 (Of course, if (3.63) is taken as a de\ufb01nition of the Riemann tensor, there is a convention that needs to be chosen for the ordering of the indices. There is no agreement at all on what this convention should be, so be careful.) Knowing what we do about parallel transport, we could very carefully perform the necessary manipulations to see what happens to the vector under this operation, and the result would be a formula for the curvature tensor in terms of the connection coe\ufb03cients. It is much quicker, however, to consider a related operation, the commutator of two covariant derivatives. The relationship between this and parallel transport around a loop should be evident; the covariant derivative of a tensor in a certain direction measures how much the tensor changes relative to what it would have been if it had been parallel transported (since the covariant derivative of a tensor in a direction along which it is parallel transported is zero). The commutator of two covariant derivatives, then, measures the di\ufb00erence between parallel transporting the tensor \ufb01rst one way and then the other, versus the opposite ordering. \u2206 \u03bd \u2206 \u00b5 \u00b5\u2206 \u2206 \u03bd The actual computation is very straightforward. Considering a vector \ufb01eld V \u03c1, we take \u2207\u00b5, [ \u2207\u03bd]V \u03c1 = \u2212 \u2207\u03bd\u2207\u00b5V \u03c1 \u2207\u00b5\u2207\u03bdV \u03c1 \u0393\u03bb \u2207\u03bdV \u03c1) \u00b5\u03bd\u2207\u03bbV \u03c1 + \u0393\u03c1 = \u2202\u00b5( \u2212 = \u2202\u00b5\u2202\u03bdV \u03c1 + (\u2202\u00b5\u0393\u03c1 \u03bd\u03c3)V \u03c3 + \u0393\u03c1 \u03bd\u03bbV \u03bb \u00b5\u03c3\u0393\u03c3 +\u0393\u03c1 \u00b5\u03bb\u0393\u03bb \u00b5\u03c3\u2202\u03bdV \u03c3 + \u0393\u03c1 \u00b5\u03c3 + \u0393\u03c1 = (\u2202\u00b5\u0393\u03c1 \u2202\u03bd\u0393\u03c1 \u03bd) \u00b5\u03c3\u2207\u03bdV \u03c3 (\u00b5 \u2194 \u2212 \u03bd\u03c3\u2202\u00b5V \u03c3 \u0393\u03bb \u00b5\u03bd\u2202\u03bbV \u03c1 \u2212 \u03bd) (\u00b5 \u2194 \u2212 \u0393\u03c1 \u00b5\u03c3)V \u03c3 \u03bd\u03bb\u0393\u03bb 2\u0393\u03bb \u03bd\u03c3 \u2212 \u2212 \u03bd\u03c3 \u2212 \u00b5\u03bd\u0393\u03c1 \u0393\u03bb \u03bb\u03c3V \u03c3 \u2212 [\u00b5\u03bd]\u2207\u03bbV \u03c1. (3.65) In the last step we have relabeled some dummy indices and eliminated some terms that cancel when antisymmetrized. We recognize that the last term is simply the torsion tensor, and that the left hand side is manifestly a tensor; therefore the expression in parentheses must be a tensor itself. We write \u2207\u00b5, [ \u2207\u03bd]V \u03c1 = R\u03c1 \u03c3\u00b5\u03bdV \u03c3 \u03bb T\u00b5\u03bd \u2207\u03bbV \u03c1, \u2212 (3.66", ") 3 CURVATURE where the Riemann tensor is identi\ufb01ed as \u03bd\u03c3 \u2212 There are a number of things to notice about the derivation of this expression: \u03bd\u03c3 \u2212 R\u03c1 \u03c3\u00b5\u03bd = \u2202\u00b5\u0393\u03c1 \u2202\u03bd\u0393\u03c1 \u00b5\u03c3 + \u0393\u03c1 \u00b5\u03bb\u0393\u03bb \u0393\u03c1 \u03bd\u03bb\u0393\u03bb \u00b5\u03c3. 76 (3.67) \u2022 \u2022 \u2022 \u2022 \u2022 Of course we have not demonstrated that (3.67) is actually the same tensor that appeared in (3.63), but in fact it\u2019s true (see Wald for a believable if tortuous demonstration). It is perhaps surprising that the commutator [ \u2207\u03bd], which appears to be a di\ufb00erential operator, has an action on vector \ufb01elds which (in the absence of torsion, at any rate) is a simple multiplicative transformation. The Riemann tensor measures that part of the commutator of covariant derivatives which is proportional to the vector \ufb01eld, while the torsion tensor measures the part which is proportional to the covariant derivative of the vector \ufb01eld; the second derivative doesn\u2019t enter at all. \u2207\u00b5, Notice that the expression (3.67) is constructed from non-tensorial elements; you can check that the transformation laws all work out to make this particular combination a legitimate tensor. The antisymmetry of R\u03c1 its derivation. \u03c3\u00b5\u03bd in its last two indices is immediate from this formula and We constructed the curvature tensor completely from the connection (no mention of the metric was made). We were su\ufb03ciently careful that the above expression is true for any connection, whether or not it is metric compatible or torsion free. \u2022 Using what are by now our usual methods, the action of [ a tensor of arbitrary rank. The answer is \u2207\u03c1, \u2207\u03c3] can be computed on \u2207\u03c1, [ \u2207\u03c3]X \u00b51 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u03bdl = \u00b7\u00b7\u00b7 \u2212 \u03bb T\u03c1\u03c3 +R\u00b51 R\u03bb \u2207\u03bbX \u00b51 \u00b7\u00b7\u00b7 \u03bb\u03c1\u03c3X \u03bb\u00b52 \u03bd1\u03c1\u03c3X \u00b51 \u00b7\u00b7\u00b7 \u00b5k \u00b7\u00b7\u00b7 \u00b5k \u03bdl \u03bd1 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u03bb\u03bd2 \u2212 \u03bdl + R\u00b52 \u00b7\u00b7\u00b7 R\u03bb \u03bdl \u2212 \u00b7\u00b7\u00b7 \u00b5k \u03bb\u03c1\u03c3X \u00b51\u03bb \u00b7\u00b7\u00b7 \u00b5k \u03bd2\u03c1\u03c3X \u00b51 \u00b7\u00b7\u00b7 \u03bd1\u03bb \u03bd1 \u03bdl + \u00b7 \u00b7 \u00b7 \u00b7\u00b7\u00b7 \u03bdl \u2212 \u00b7 \u00b7 \u00b7 \u00b7\u00b7\u00b7.(3.68) A useful notion is that of the commutator of two vector \ufb01elds X and Y, which is a third vector \ufb01eld with components [X, Y ]\u00b5 = X \u03bb\u2202\u03bbY \u00b5 Y \u03bb\u2202\u03bbX \u00b5. \u2212 (3.69) Both the torsion tensor and the Riemann tensor, thought of as multilinear maps, have elegant expressions in terms of the commutator. Thinking of the torsion as a map from two vector \ufb01elds to a third vector \ufb01eld, we have T (X, Y ) = \u2207XY \u2212 \u2207Y X \u2212 [X, Y ], (3.70) 3 CURVATURE 77 and thinking of the Riemann tensor as a map from three vector \ufb01elds to a fourth one, we have R(X, Y )Z = \u2207X\u2207Y Z \u2207X = X \u00b5 \u2212 \u2207Y \u2207XZ \u2212 \u2207[X,Y ]Z. (3.71) \u2207X refers to the covariant derivative along the vector \ufb01eld In these expressions, the notation X; in components, \u2207\u00b5. Note that the two vectors X and Y in (3.71) correspond to the two antisymmetric indices in the component form of the Riemann tensor. The last term in (3.71), involving the commutator [X, Y ], vanishes when X and Y are taken to be the coordinate basis vector \ufb01elds (since [\u2202\u00b5, \u2202\u03bd] = 0), which is why this term did not arise when we originally took the commutator of two covariant derivatives. We will not use this notation extensively, but you might see it in the literature, so you should be able to decode it. Having de\ufb01ned the curvature tensor as something which characterizes the connection, let us now", " admit that in GR we are most concerned with the Christo\ufb00el connection. In this case the connection is derived from the metric, and the associated curvature may be thought of as that of the metric itself. This identi\ufb01cation allows us to \ufb01nally make sense of our informal notion that spaces for which the metric looks Euclidean or Minkowskian are \ufb02at. In fact it works both ways: if the components of the metric are constant in some coordinate system, the Riemann tensor will vanish, while if the Riemann tensor vanishes we can always construct a coordinate system in which the metric components are constant. The \ufb01rst of these is easy to show. If we are in some coordinate system such that \u2202\u03c3g\u00b5\u03bd = 0 (everywhere, not just at a point), then \u0393\u03c1 \u03c3\u00b5\u03bd = 0 by (3.67). But this is a tensor equation, and if it is true in one coordinate system it must be true in any coordinate system. Therefore, the statement that the Riemann tensor vanishes is a necessary condition for it to be possible to \ufb01nd coordinates in which the components of g\u00b5\u03bd are constant everywhere. \u00b5\u03bd = 0 and \u2202\u03c3\u0393\u03c1 \u00b5\u03bd = 0; thus R\u03c1 It is also a su\ufb03cient condition, although we have to work harder to show it. Start by choosing Riemann normal coordinates at some point p, so that g\u00b5\u03bd = \u03b7\u00b5\u03bd at p. (Here we are using \u03b7\u00b5\u03bd in a generalized sense, as a matrix with either +1 or 1 for each diagonal element and zeroes elsewhere. The actual arrangement of the +1\u2019s and 1\u2019s depends on the canonical form of the metric, but is irrelevant for the present argument.) Denote the basis vectors at p by \u02c6e(\u00b5), with components \u02c6e\u03c3 \u2212 \u2212 (\u00b5). Then by construction we have (\u00b5)\u02c6e\u03c1 (\u03bd)(p) = \u03b7\u00b5\u03bd. g\u03c3\u03c1\u02c6e\u03c3 (3.72) Now let us parallel transport the entire set of basis vectors from p to another point q; the vanishing of the Riemann tensor ensures that the result will be independent of the path taken between p and q. Since parallel transport with respect to a metric compatible connection preserves inner products, we must have g\u03c3\u03c1\u02c6e\u03c3 (\u00b5)\u02c6e\u03c1 (\u03bd)(q) = \u03b7\u00b5\u03bd. (3.73) 3 CURVATURE 78 We therefore have speci\ufb01ed a set of vector \ufb01elds which everywhere de\ufb01ne a basis in which the metric components are constant. This is completely unimpressive; it can be done on any manifold, regardless of what the curvature is. What we would like to show is that this is a coordinate basis (which can only be true if the curvature vanishes). We know that if the \u02c6e(\u00b5)\u2019s are a coordinate basis, their commutator will vanish: [\u02c6e(\u00b5), \u02c6e(\u03bd)] = 0. (3.74) What we would really like is the converse: that if the commutator vanishes we can \ufb01nd coordinates y\u00b5 such that \u02c6e(\u00b5) = \u2202 \u2202y\u00b5. In fact this is a true result, known as Frobenius\u2019s Theorem. It\u2019s something of a mess to prove, involving a good deal more mathematical apparatus than we have bothered to set up. Let\u2019s just take it for granted (skeptics can consult Schutz\u2019s Geometrical Methods book). Thus, we would like to demonstrate (3.74) for the vector \ufb01elds we have set up. Let\u2019s use the expression (3.70) for the torsion: [\u02c6e(\u00b5), \u02c6e(\u03bd)] = T (\u02c6e(\u00b5), \u02c6e(\u03bd)). \u2207\u02c6e(\u00b5) \u02c6e(\u03bd) \u2212 \u2207\u02c6e(\u03bd) \u02c6e(\u00b5) \u2212 The torsion vanishes by hypothesis. The covariant derivatives will also vanish, given the method by which we constructed our vector \ufb01elds; they were made by parallel transporting along arbitrary paths. If the \ufb01elds are parallel transported along arbitrary paths, they are certainly parallel transported along the vectors \u02c6e(\u00b5), and therefore their covariant derivatives in the direction of these vectors will vanish. Thus (3.", "70) implies that the commutator vanishes, and therefore that we can \ufb01nd a coordinate system y\u00b5 for which these vector \ufb01elds are the partial derivatives. In this coordinate system the metric will have components \u03b7\u00b5\u03bd, as desired. (3.75) The Riemann tensor, with four indices, naively has n4 independent components in an n-dimensional space. In fact the antisymmetry property (3.64) means that there are only 1)/2 independent values these last two indices can take on, leaving us with n3(n n(n 1)/2 independent components. When we consider the Christo\ufb00el connection, however, there are a number of other symmetries that reduce the independent components further. Let\u2019s consider these now. \u2212 \u2212 The simplest way to derive these additional symmetries is to examine the Riemann tensor with all lower indices, R\u03c1\u03c3\u00b5\u03bd = g\u03c1\u03bbR\u03bb \u03c3\u00b5\u03bd. (3.76) Let us further consider the components of this tensor in Riemann normal coordinates established at a point p. Then the Christo\ufb00el symbols themselves will vanish, although their derivatives will not. We therefore have R\u03c1\u03c3\u00b5\u03bd = g\u03c1\u03bb(\u2202\u00b5\u0393\u03bb \u03bd\u03c3 \u2212 \u2202\u03bd\u0393\u03bb \u00b5\u03c3) 3 CURVATURE 79 = = 1 2 1 2 g\u03c1\u03bbg\u03bb\u03c4 (\u2202\u00b5\u2202\u03bdg\u03c3\u03c4 + \u2202\u00b5\u2202\u03c3g\u03c4 \u03bd \u2212 (\u2202\u00b5\u2202\u03c3g\u03c1\u03bd \u2212 \u2202\u00b5\u2202\u03c1g\u03bd\u03c3 \u2212 \u2202\u00b5\u2202\u03c4 g\u03bd\u03c3 \u2212 \u2202\u03bd\u2202\u03c3g\u03c1\u00b5 + \u2202\u03bd\u2202\u03c1g\u00b5\u03c3). \u2202\u03bd\u2202\u00b5g\u03c3\u03c4 \u2212 \u2202\u03bd\u2202\u03c3g\u03c4 \u00b5 + \u2202\u03bd\u2202\u03c4 g\u00b5\u03c3) (3.77) In the second line we have used \u2202\u00b5g\u03bb\u03c4 = 0 in RNC\u2019s, and in the third line the fact that partials commute. From this expression we can notice immediately two properties of R\u03c1\u03c3\u00b5\u03bd ; it is antisymmetric in its \ufb01rst two indices, and it is invariant under interchange of the \ufb01rst pair of indices with the second: R\u03c1\u03c3\u00b5\u03bd = R\u03c3\u03c1\u00b5\u03bd, \u2212 R\u03c1\u03c3\u00b5\u03bd = R\u00b5\u03bd\u03c1\u03c3. (3.78) (3.79) With a little more work, which we leave to your imagination, we can see that the sum of cyclic permutations of the last three indices vanishes: R\u03c1\u03c3\u00b5\u03bd + R\u03c1\u00b5\u03bd\u03c3 + R\u03c1\u03bd\u03c3\u00b5 = 0. (3.80) This last property is equivalent to the vanishing of the antisymmetric part of the last three indices: R\u03c1[\u03c3\u00b5\u03bd] = 0. (3.81) All of these properties have been derived in a special coordinate system, but they are all tensor equations; therefore they will be true in any coordinates. Not all of them are independent; with some e\ufb00ort, you can show that (3.64), (3.78) and (3.81) together imply (3.79). The logical interdependence of the equations is usually less important than the simple fact that they are true. Given these relationships between the di\ufb00erent components of the Riemann tensor, how many independent quantities remain? Let\u2019s begin with the facts that R\u03c1\u03c3\u00b5\u03bd is antisymmetric in the \ufb01rst two indices, antisymmetric in the last two indices, and symmetric under interchange of these two pairs. This means that we can think of it as a symmetric matrix R[\u03c1\u03c3][\u00b5\u03bd], where the pairs \u03c1\u03c3 and \u00b5\u03bd are thought of as individual indices. An m m symmetric matrix has m(m + 1)/2 independent components, while an n n antisymmetric matrix has n(n 1)/2 independent components. We therefore have \u00d7 \u00d7 \u2212 1 2 1 2 n(n \u2212 1) 1 2 n(n \u2212 1) + 1 = 1 8 (n4 \u2212 2n3 + 3n2 2n) \u2212 (3.82) independent components. We still have to deal with the additional symmetry (3.81). An immediate consequence of (3.81) is that the totally antisymmetric part of the Riemann tensor vanishes, R[\u03c1\u03c3\u00b5\u03bd] = 0. (3.83) 3 CURVATURE 80 In fact, this equation plus the", " other symmetries (3.64), (3.78) and (3.79) are enough to imply (3.81), as can be easily shown by expanding (3.83) and messing with the resulting terms. Therefore imposing the additional constraint of (3.83) is equivalent to imposing (3.81), once the other symmetries have been accounted for. How many independent restrictions does this represent? Let us imagine decomposing R\u03c1\u03c3\u00b5\u03bd = X\u03c1\u03c3\u00b5\u03bd + R[\u03c1\u03c3\u00b5\u03bd]. (3.84) It is easy to see that any totally antisymmetric 4-index tensor is automatically antisymmetric in its \ufb01rst and last indices, and symmetric under interchange of the two pairs. Therefore these properties are independent restrictions on X\u03c1\u03c3\u00b5\u03bd, unrelated to the requirement (3.83). Now a totally antisymmetric 4-index tensor has n(n 3)/4! terms, and therefore (3.83) reduces the number of independent components by this amount. We are left with 2)(n 1)(n \u2212 \u2212 \u2212 1 8 (n4 \u2212 2n3 + 3n2 2n) \u2212 1 24 \u2212 n(n \u2212 1)(n \u2212 2)(n \u2212 3) = 1 12 n2(n2 1) \u2212 (3.85) independent components of the Riemann tensor. In four dimensions, therefore, the Riemann tensor has 20 independent components. (In one dimension it has none.) These twenty functions are precisely the 20 degrees of freedom in the second derivatives of the metric which we could not set to zero by a clever choice of coordinates. This should reinforce your con\ufb01dence that the Riemann tensor is an appropriate measure of curvature. In addition to the algebraic symmetries of the Riemann tensor (which constrain the number of independent components at any point), there is a di\ufb00erential identity which it obeys (which constrains its relative values at di\ufb00erent points). Consider the covariant derivative of the Riemann tensor, evaluated in Riemann normal coordinates: \u2207\u03bbR\u03c1\u03c3\u00b5\u03bd = \u2202\u03bbR\u03c1\u03c3\u00b5\u03bd = 1 2 \u2202\u03bb(\u2202\u00b5\u2202\u03c3g\u03c1\u03bd \u2212 \u2202\u00b5\u2202\u03c1g\u03bd\u03c3 \u2212 \u2202\u03bd\u2202\u03c3g\u03c1\u00b5 + \u2202\u03bd\u2202\u03c1g\u00b5\u03c3). (3.86) We would like to consider the sum of cyclic permutations of the \ufb01rst three indices: \u2207\u03bbR\u03c1\u03c3\u00b5\u03bd + \u2207\u03c1R\u03c3\u03bb\u00b5\u03bd + 1 (\u2202\u03bb\u2202\u00b5\u2202\u03c3g\u03c1\u03bd \u2212 = 2 +\u2202\u03c1\u2202\u00b5\u2202\u03bbg\u03c3\u03bd \u2212 +\u2202\u03c3\u2202\u00b5\u2202\u03c1g\u03bb\u03bd \u2212 \u2207\u03c3R\u03bb\u03c1\u00b5\u03bd \u2202\u03bb\u2202\u00b5\u2202\u03c1g\u03bd\u03c3 \u2212 \u2202\u03c1\u2202\u00b5\u2202\u03c3g\u03bd\u03bb \u2212 \u2202\u03c3\u2202\u00b5\u2202\u03bbg\u03bd\u03c1 \u2212 = 0. \u2202\u03bb\u2202\u03bd\u2202\u03c3g\u03c1\u00b5 + \u2202\u03bb\u2202\u03bd\u2202\u03c1g\u00b5\u03c3 \u2202\u03c1\u2202\u03bd\u2202\u03bbg\u03c3\u00b5 + \u2202\u03c1\u2202\u03bd\u2202\u03c3g\u00b5\u03bb \u2202\u03c3\u2202\u03bd\u2202\u03c1g\u03bb\u00b5 + \u2202\u03c3\u2202\u03bd\u2202\u03bbg\u00b5\u03c1) (3.87) Once again, since this is an equation between tensors it is true in any coordinate system, even though we derived it in a particular one. We recognize by now that the antisymmetry 3 CURVATURE 81 R\u03c1\u03c3\u00b5\u03bd = R\u03c3\u03c1\u00b5\u03bd allows us to write this result as \u2212 \u2207[\u03bbR\u03c1\u03c3]\u00b5\u03bd = 0. This is known as the Bianchi identity. (Notice that for a general connection there would be additional terms involving the torsion tensor.) It is closely related to the Jacobi identity, since (as you can show) it basically expresses (3.88) \u2207\u03bb, [[ \u2207\u03c1], \u2207\u03c3] + [[ \u2207\u03c1, \u2207\u03c3], \u2207\u03bb] + [[ \u2207\u03c3, \u2207\u03bb], \u2207\u03c1] = 0. (3.89) It is frequently useful to consider contractions of the Riemann tensor. Even without the metric, we can form a contraction known as the Ricci tensor: R\u00b5\u03bd = R\u03bb \u00b5\u03bb\u03bd. (3.90) Notice that, for the curvature tensor formed from an arbitrary (not necessarily Christo\ufb00el", ") connection, there are a number of independent contractions to take. Our primary concern is with the Christo\ufb00el connection, for which (3.90) is the only independent contraction (modulo conventions for the sign, which of course change from place to place). The Ricci tensor associated with the Christo\ufb00el connection is symmetric, R\u00b5\u03bd = R\u03bd\u00b5, (3.91) as a consequence of the various symmetries of the Riemann tensor. Using the metric, we can take a further contraction to form the Ricci scalar: R = R\u00b5 \u00b5 = g\u00b5\u03bdR\u00b5\u03bd. (3.92) An especially useful form of the Bianchi identity comes from contracting twice on (3.87): = \u2207 or 0 = g\u03bd\u03c3g\u00b5\u03bb( \u2207\u03c1R\u03c3\u03bb\u00b5\u03bd + \u2207\u03bbR\u03c1\u03c3\u00b5\u03bd + \u03bdR\u03c1\u03bd, \u00b5R\u03c1\u00b5 \u2212 \u2207\u03c1R + \u2207 \u00b5R\u03c1\u00b5 = \u2207 1 2\u2207\u03c1R. \u2207\u03c3R\u03bb\u03c1\u00b5\u03bd) (3.93) (3.94) (Notice that, unlike the partial derivative, it makes sense to raise an index on the covariant derivative, due to metric compatibility.) If we de\ufb01ne the Einstein tensor as then we see that the twice-contracted Bianchi identity (3.94) is equivalent to G\u00b5\u03bd = R\u00b5\u03bd \u2212 1 2 Rg\u00b5\u03bd, \u00b5G\u00b5\u03bd = 0. \u2207 (3.95) (3.96) 3 CURVATURE 82 The Einstein tensor, which is symmetric due to the symmetry of the Ricci tensor and the metric, will be of great importance in general relativity. The Ricci tensor and the Ricci scalar contain information about \u201ctraces\u201d of the Riemann tensor. It is sometimes useful to consider separately those pieces of the Riemann tensor which the Ricci tensor doesn\u2019t tell us about. We therefore invent the Weyl tensor, which is basically the Riemann tensor with all of its contractions removed. It is given in n dimensions by C\u03c1\u03c3\u00b5\u03bd = R\u03c1\u03c3\u00b5\u03bd \u2212 (n 2 \u2212 2) g\u03c1[\u00b5R\u03bd]\u03c3 \u2212 g\u03c3[\u00b5R\u03bd]\u03c1 + 2 1)(n 2) \u2212 (n \u2212 Rg\u03c1[\u00b5g\u03bd]\u03c3. (3.97) This messy formula is designed so that all possible contractions of C\u03c1\u03c3\u00b5\u03bd vanish, while it retains the symmetries of the Riemann tensor: C\u03c1\u03c3\u00b5\u03bd = C[\u03c1\u03c3][\u00b5\u03bd], C\u03c1\u03c3\u00b5\u03bd = C\u00b5\u03bd\u03c1\u03c3, C\u03c1[\u03c3\u00b5\u03bd] = 0. (3.98) The Weyl tensor is only de\ufb01ned in three or more dimensions, and in three dimensions it vanishes identically. For n 4 it satis\ufb01es a version of the Bianchi identity, \u2265 \u03c1C\u03c1\u03c3\u00b5\u03bd = \u2207 (n (n 2 \u2212 \u2212 \u2212 3) 2) \u2207[\u00b5R\u03bd]\u03c3 + 2(n 1 \u2212 g\u03c3[\u03bd\u2207\u00b5]R! 1). (3.99) One of the most important properties of the Weyl tensor is that it is invariant under conformal transformations. This means that if you compute C\u03c1\u03c3\u00b5\u03bd for some metric g\u00b5\u03bd, and then compute it again for a metric given by \u21262(x)g\u00b5\u03bd, where \u2126(x) is an arbitrary nonvanishing function of spacetime, you get the same answer. For this reason it is often known as the \u201cconformal tensor.\u201d After this large amount of formalism, it might be time to step back and think about what curvature means for some simple examples. First notice that, according to (3.85), in 1, 2, 3 and 4 dimensions there are 0, 1, 6 and 20 components of the curvature tensor, respectively. (Everything we say about the curvature in these examples refers to the curvature associated with the Christo\ufb00el connection, and therefore the metric.) This means that one-dimensional manifolds (such as S1) are never curved; the intuition you have that tells you that a circle is curved comes from thinking of it embedded in a certain \ufb02at two-dimensional plane. (There is something called \u201cextrinsic curvature,\u201d which characterizes the way something is embedded in a higher dimensional space. Our notion of curvature is \ufffd", "\ufffdintrinsic,\u201d and has nothing to do with such embeddings.) The distinction between intrinsic and extrinsic curvature is also important in two dimensions, where the curvature has one independent component. (In fact, all of the information 3 CURVATURE 83 identify \u00d7 about the curvature is contained in the single component of the Ricci scalar.) Consider a S1. Although this looks curved from our point of view, it should be clear cylinder, R that we can put a metric on the cylinder whose components are constant in an appropriate coordinate system \u2014 simply unroll it and use the induced metric from the plane. In this metric, the cylinder is \ufb02at. (There is also nothing to stop us from introducing a di\ufb00erent metric in which the cylinder is not \ufb02at, but the point we are trying to emphasize is that it can be made \ufb02at in some metric.) The same story holds for the torus: identify We can think of the torus as a square region of the plane with opposite sides identi\ufb01ed (in other words, S1 S1), from which it is clear that it can have a \ufb02at metric even though it looks curved from the embedded point of view. \u00d7 A cone is an example of a two-dimensional manifold with nonzero curvature at exactly one point. We can see this also by unrolling it; the cone is equivalent to the plane with a \u201cde\ufb01cit angle\u201d removed and opposite sides identi\ufb01ed: 3 CURVATURE 84 In the metric inherited from this description as part of the \ufb02at plane, the cone is \ufb02at everywhere but at its vertex. This can be seen by considering parallel transport of a vector around various loops; if a loop does not enclose the vertex, there will be no overall transformation, whereas a loop that does enclose the vertex (say, just one time) will lead to a rotation by an angle which is just the de\ufb01cit angle. Our favorite example is of course the two-sphere, with metric ds2 = a2(d\u03b82 + sin2 \u03b8 d\u03c62), (3.100) where a is the radius of the sphere (thought of as embedded in R3). Without going through the details, the nonzero connection coe\ufb03cients are \u0393\u03b8 \u03b8\u03c6 = \u0393\u03c6 \u0393\u03c6 \u03c6\u03c6 = \u03c6\u03b8 = cot \u03b8. \u2212 sin \u03b8 cos \u03b8 (3.101) Let\u2019s compute a promising component of the Riemann tensor: R\u03b8 \u03c6\u03b8\u03c6 = \u2202\u03b8\u0393\u03b8 \u03c6\u03c6 \u2212 \u2202\u03c6\u0393\u03b8 \u03b8\u03c6 + \u0393\u03b8 \u03b8\u03bb\u0393\u03bb \u03c6\u03c6 \u2212 \u03c6\u03bb\u0393\u03bb \u0393\u03b8 \u03b8\u03c6 3 CURVATURE = (sin2 \u03b8 = sin2 \u03b8. \u2212 cos2 \u03b8) (0) + (0) ( \u2212 \u2212 \u2212 sin \u03b8 cos \u03b8)(cot \u03b8) 85 (3.102) (The notation is obviously imperfect, since the Greek letter \u03bb is a dummy index which is summed over, while the Greek letters \u03b8 and \u03c6 represent speci\ufb01c coordinates.) Lowering an index, we have R\u03b8\u03c6\u03b8\u03c6 = g\u03b8\u03bbR\u03bb = g\u03b8\u03b8R\u03b8 \u03c6\u03b8\u03c6 = a2 sin2 \u03b8. \u03c6\u03b8\u03c6 (3.103) It is easy to check that all of the components of the Riemann tensor either vanish or are related to this one by symmetry. We can go on to compute the Ricci tensor via R\u00b5\u03bd = g\u03b1\u03b2R\u03b1\u00b5\u03b2\u03bd. We obtain R\u03b8\u03b8 = g\u03c6\u03c6R\u03c6\u03b8\u03c6\u03b8 = 1 R\u03b8\u03c6 = R\u03c6\u03b8 = 0 R\u03c6\u03c6 = g\u03b8\u03b8R\u03b8\u03c6\u03b8\u03c6 = sin2 \u03b8. The Ricci scalar is similarly straightforward: R = g\u03b8\u03b8R\u03b8\u03b8 + g\u03c6\u03c6R\u03c6\u03c6 = 2 a2. (3.104) (3.105) Therefore the Ricci scalar, which for a two-dimensional manifold completely characterizes the curvature, is a constant over this two-sphere. This is a re\ufb02ection of the fact that the manifold is \u201cmaximally symmetric,\u201d a concept we will de\ufb01ne more precisely later (although it means what you think it should). In any number", " of dimensions the curvature of a maximally symmetric space satis\ufb01es (for some constant a) R\u03c1\u03c3\u00b5\u03bd = a\u2212 2(g\u03c1\u00b5g\u03c3\u03bd \u2212 g\u03c1\u03bdg\u03c3\u00b5), (3.106) which you may check is satis\ufb01ed by this example. Notice that the Ricci scalar is not only constant for the two-sphere, it is manifestly positive. We say that the sphere is \u201cpositively curved\u201d (of course a convention or two came into play, but fortunately our conventions conspired so that spaces which everyone agrees to call positively curved actually have a positive Ricci scalar). From the point of view of someone living on a manifold which is embedded in a higher-dimensional Euclidean space, if they are sitting at a point of positive curvature the space curves away from them in the same way in any direction, while in a negatively curved space it curves away in opposite directions. Negatively curved spaces are therefore saddle-like. Enough fun with examples. There is one more topic we have to cover before introducing general relativity itself: geodesic deviation. You have undoubtedly heard that the de\ufb01ning 3 CURVATURE 86 positive curvature negative curvature property of Euclidean (\ufb02at) geometry is the parallel postulate: initially parallel lines remain parallel forever. Of course in a curved space this is not true; on a sphere, certainly, initially parallel geodesics will eventually cross. We would like to quantify this behavior for an arbitrary curved space. The problem is that the notion of \u201cparallel\u201d does not extend naturally from \ufb02at to curved spaces. Instead what we will do is to construct a one-parameter family of geodesics, \u03b3s(t). R, \u03b3s is a geodesic parameterized by the a\ufb03ne parameter t. The That is, for each s collection of these curves de\ufb01nes a smooth two-dimensional surface (embedded in a manifold M of arbitrary dimensionality). The coordinates on this surface may be chosen to be s and t, provided we have chosen a family of geodesics which do not cross. The entire surface is the set of points x\u00b5(s, t) M. We have two natural vector \ufb01elds: the tangent vectors to the geodesics, \u2208 \u2208 and the \u201cdeviation vectors\u201d S\u00b5 = \u2202x\u00b5 \u2202s T \u00b5 = \u2202x\u00b5 \u2202t,. (3.107) (3.108) This name derives from the informal notion that S\u00b5 points from one geodesic towards the neighboring ones. The idea that S\u00b5 points from one geodesic to the next inspires us to de\ufb01ne the \u201crelative velocity of geodesics,\u201d V \u00b5 = ( \u2207T S)\u00b5 = T \u03c1 \u2207\u03c1S\u00b5, (3.109) and the \u201crelative acceleration of geodesics,\u201d \u2207T V )\u00b5 = T \u03c1 You should take the names with a grain of salt, but these vectors are certainly well-de\ufb01ned. \u2207\u03c1V \u00b5. a\u00b5 = ( (3.110) 3 CURVATURE 87 Since S and T are basis vectors adapted to a coordinate system, their commutator van- ishes: [S, T ] = 0. We would like to consider the conventional case where the torsion vanishes, so from (3.70) we then have \u2207\u03c1T \u00b5 = T \u03c1 With this in mind, let\u2019s compute the acceleration: S\u03c1 \u2207\u03c1S\u00b5. a\u00b5 = T \u03c1 = T \u03c1 = (T \u03c1 = (S\u03c1 = (S\u03c1 = R\u00b5 \u2207\u03c1(T \u03c3 \u2207\u03c1(S\u03c3 \u2207\u03c1S\u03c3)( \u2207\u03c1T \u03c3)( \u2207\u03c1T \u03c3)( \u03bd\u03c1\u03c3T \u03bdT \u03c1S\u03c3. \u2207\u03c3S\u00b5) \u2207\u03c3T \u00b5) \u2207\u03c3T \u00b5) + T \u03c1S\u03c3 \u2207\u03c3T \u00b5) + T \u03c1S\u03c3( \u2207\u03c3T \u00b5) + S\u03c3 \u2207\u03c1\u2207\u03c3T \u00b5 \u2207\u03c3\u2207\u03c1T \u00b5 + R\u00b5 (S\u03c3 \u2207\u03c1T \u00b5) \u2212 \u2207\u03c3(T \u03c1 \u03bd\u03c1\u03c3T \u03bd) \u2207\u03c3T \u03c1) (3.111) \u2207\u03c1T \u00b5 + R\u00b5 \u03bd\u03c1\u03c3T \u03bdT \u03c1S\u03c3 (3.112) Let\u2019s think about this line by line. The \ufb01rst line is the de\ufb01", "nition of a\u00b5, and the second line comes directly from (3.111). The third line is simply the Leibniz rule. The fourth line replaces a double covariant derivative by the derivatives in the opposite order plus the Riemann tensor. In the \ufb01fth line we use Leibniz again (in the opposite order from usual), and then we cancel two identical terms and notice that the term involving T \u03c1 \u2207\u03c1T \u00b5 vanishes because T \u00b5 is the tangent vector to a geodesic. The result, a\u00b5 = D2 dt2 S\u00b5 = R\u00b5 \u03bd\u03c1\u03c3T \u03bdT \u03c1S\u03c3, (3.113) 3 CURVATURE 88 is known as the geodesic deviation equation. It expresses something that we might have expected: the relative acceleration between two neighboring geodesics is proportional to the curvature. Physically, of course, the acceleration of neighboring geodesics is interpreted as a manifestation of gravitational tidal forces. This reminds us that we are very close to doing physics by now. There is one last piece of formalism which it would be nice to cover before we move on to gravitation proper. What we will do is to consider once again (although much more concisely) the formalism of connections and curvature, but this time we will use sets of basis vectors in the tangent space which are not derived from any coordinate system. It will turn out that this slight change in emphasis reveals a di\ufb00erent point of view on the connection and curvature, one in which the relationship to gauge theories in particle physics is much more transparent. In fact the concepts to be introduced are very straightforward, but the subject is a notational nightmare, so it looks more di\ufb03cult than it really is. Up until now we have been taking advantage of the fact that a natural basis for the tangent space Tp at a point p is given by the partial derivatives with respect to the coordinates at that point, \u02c6e(\u00b5) = \u2202\u00b5. Similarly, a basis for the cotangent space T \u2217p is given by the gradients of the coordinate functions, \u02c6\u03b8(\u00b5) = dx\u00b5. There is nothing to stop us, however, from setting up any bases we like. Let us therefore imagine that at each point in the manifold we introduce a set of basis vectors \u02c6e(a) (indexed by a Latin letter rather than Greek, to remind us that they are not related to any coordinate system). We will choose these basis vectors to be \u201corthonormal\u201d, in a sense which is appropriate to the signature of the manifold we are working on. That is, if the canonical form of the metric is written \u03b7ab, we demand that the inner product of our basis vectors be g(\u02c6e(a), \u02c6e(b)) = \u03b7ab, (3.114) where g(, ) is the usual metric tensor. Thus, in a Lorentzian spacetime \u03b7ab represents the Minkowski metric, while in a space with positive-de\ufb01nite metric it would represent the Euclidean metric. The set of vectors comprising an orthonormal basis is sometimes known as a tetrad (from Greek tetras, \u201ca group of four\u201d) or vielbein (from the German for \u201cmany legs\u201d). In di\ufb00erent numbers of dimensions it occasionally becomes a vierbein (four), dreibein (three), zweibein (two), and so on. (Just as we cannot in general \ufb01nd coordinate charts which cover the entire manifold, we will often not be able to \ufb01nd a single set of smooth basis vector \ufb01elds which are de\ufb01ned everywhere. As usual, we can overcome this problem by working in di\ufb00erent patches and making sure things are well-behaved on the overlaps.) The point of having a basis is that any vector can be expressed as a linear combination of basis vectors. Speci\ufb01cally, we can express our old basis vectors \u02c6e(\u00b5) = \u2202\u00b5 in terms of the 3 CURVATURE new ones: \u02c6e(\u00b5) = ea \u00b5\u02c6e(a). 89 (3.115) \u00b5 form an n The components ea n invertible matrix. (In accord with our usual practice of blurring the distinction between objects and their components, we will refer to the ea \u00b5 as the tetrad or vielbein, and often in the plural as \u201cvielbeins.\u201d) We denote their inverse by switching indices to obtain e\u00b5 a, which satisfy \u00d7", " aea e\u00b5 \u03bd = \u03b4\u00b5 \u03bd, \u00b5e\u00b5 ea b = \u03b4a b. These serve as the components of the vectors \u02c6e(a) in the coordinate basis: In terms of the inverse vielbeins, (3.114) becomes \u02c6e(a) = e\u00b5 a \u02c6e(\u00b5). g\u00b5\u03bde\u00b5 ae\u03bd b = \u03b7ab, (3.116) (3.117) (3.118) or equivalently \u00b5eb This last equation sometimes leads people to say that the vielbeins are the \u201csquare root\u201d of the metric. g\u00b5\u03bd = ea (3.119) \u03bd\u03b7ab. We can similarly set up an orthonormal basis of one-forms in T \u2217p, which we denote \u02c6\u03b8(a). They may be chosen to be compatible with the basis vectors, in the sense that \u02c6\u03b8(a)(\u02c6e(b)) = \u03b4a b. (3.120) It is an immediate consequence of this that the orthonormal one-forms are related to their coordinate-based cousins \u02c6\u03b8(\u00b5) = dx\u00b5 by and \u02c6\u03b8(\u00b5) = e\u00b5 a \u02c6\u03b8(a) \u02c6\u03b8(a) = ea \u00b5 \u02c6\u03b8(\u00b5). (3.121) (3.122) The vielbeins ea \u00b5 thus serve double duty as the components of the coordinate basis vectors in terms of the orthonormal basis vectors, and as components of the orthonormal basis one-forms in terms of the coordinate basis one-forms; while the inverse vielbeins serve as the components of the orthonormal basis vectors in terms of the coordinate basis, and as components of the coordinate basis one-forms in terms of the orthonormal basis. Any other vector can be expressed in terms of its components in the orthonormal basis. If a vector V is written in the coordinate basis as V \u00b5\u02c6e(\u00b5) and in the orthonormal basis as V a\u02c6e(a), the sets of components will be related by V a = ea \u00b5V \u00b5. (3.123) 3 CURVATURE 90 So the vielbeins allow us to \u201cswitch from Latin to Greek indices and back.\u201d The nice property of tensors, that there is usually only one sensible thing to do based on index placement, is of great help here. We can go on to refer to multi-index tensors in either basis, or even in terms of mixed components: V a b = ea \u00b5V \u00b5 b = e\u03bd b V a \u03bd = ea \u00b5e\u03bd b V \u00b5 \u03bd. (3.124) Looking back at (3.118), we see that the components of the metric tensor in the orthonormal basis are just those of the \ufb02at metric, \u03b7ab. (For this reason the Greek indices are sometimes referred to as \u201ccurved\u201d and the Latin ones as \u201c\ufb02at.\u201d) In fact we can go so far as to raise and lower the Latin indices using the \ufb02at metric and its inverse \u03b7ab. You can check for yourself that everything works okay (e.g., that the lowering an index with the metric commutes with changing from orthonormal to coordinate bases). By introducing a new set of basis vectors and one-forms, we necessitate a return to our favorite topic of transformation properties. We\u2019ve been careful all along to emphasize that the tensor transformation law was only an indirect outcome of a coordinate transformation; the real issue was a change of basis. Now that we have non-coordinate bases, these bases can be changed independently of the coordinates. The only restriction is that the orthonormality property (3.114) be preserved. But we know what kind of transformations preserve the \ufb02at metric \u2014 in a Euclidean signature metric they are orthogonal transformations, while in a Lorentzian signature metric they are Lorentz transformations. We therefore consider changes of basis of the form where the matrices \u039ba\u2032 a(x) represent position-dependent transformations which (at each point) leave the canonical form of the metric unaltered: \u02c6e(a) \u2192 \u02c6e(a\u2032) = \u039ba\u2032 a(x)\u02c6e(a), (3.125) \u039ba\u2032 a\u039bb\u2032 b\u03b7ab = \u03b7a\u2032b\u2032. (3.126) In fact these matrices correspond to what in \ufb02at space we called the inverse Lorentz", " transformations (which operate on basis vectors); as before we also have ordinary Lorentz transformations \u039ba\u2032 a, which transform the basis one-forms. As far as components are concerned, as before we transform upper indices with \u039ba\u2032 a and lower indices with \u039ba\u2032 a. So we now have the freedom to perform a Lorentz transformation (or an ordinary Euclidean rotation, depending on the signature) at every point in space. These transformations are therefore called local Lorentz transformations, or LLT\u2019s. We still have our usual freedom to make changes in coordinates, which are called general coordinate transformations, or GCT\u2019s. Both can happen at the same time, resulting in a mixed tensor transformation law: T a\u2032\u00b5\u2032 b\u2032\u03bd\u2032 = \u039ba\u2032 a \u2202x\u00b5\u2032 \u2202x\u00b5 \u039bb\u2032 b \u2202x\u03bd \u2202x\u03bd\u2032 T a\u00b5 b\u03bd. (3.127) 3 CURVATURE 91 Translating what we know about tensors into non-coordinate bases is for the most part merely a matter of sticking vielbeins in the right places. The crucial exception comes when In our ordinary formalism, the covariant derivative of a we begin to di\ufb00erentiate things. tensor is given by its partial derivative plus correction terms, one for each index, involving the tensor and the connection coe\ufb03cients. The same procedure will continue to be true for the non-coordinate basis, but we replace the ordinary connection coe\ufb03cients \u0393\u03bb \u00b5\u03bd by the spin connection, denoted \u03c9\u00b5 b. Each Latin index gets a factor of the spin connection in the usual way: a \u2207\u00b5X a b = \u2202\u00b5X a b + \u03c9\u00b5 a cX c \u03c9\u00b5 c bX a c. b \u2212 (3.128) (The name \u201cspin connection\u201d comes from the fact that this can be used to take covariant derivatives of spinors, which is actually impossible using the conventional connection coe\ufb03cients.) In the presence of mixed Latin and Greek indices we get terms of both kinds. The usual demand that a tensor be independent of the way it is written allows us to \u00b5\u03bb\u2019s. Consider the derive a relationship between the spin connection, the vielbeins, and the \u0393\u03bd covariant derivative of a vector X, \ufb01rst in a purely coordinate basis: \u2207 X = ( \u2207\u00b5X \u03bd)dx\u00b5 = (\u2202\u00b5X \u03bd + \u0393\u03bd \u2202\u03bd \u00b5\u03bbX \u03bb)dx\u00b5 \u2297 \u2202\u03bd. \u2297 (3.129) Now \ufb01nd the same object in a mixed basis, and convert into the coordinate basis: X = ( \u2207 \u02c6e(a) \u2297 bX b)dx\u00b5 a beb a \u2207\u00b5X a)dx\u00b5 = (\u2202\u00b5X a + \u03c9\u00b5 = (\u2202\u00b5(ea = e\u03c3 a(ea = (\u2202\u00b5X \u03bd + e\u03bd \u03bdX \u03bd) + \u03c9\u00b5 \u03bd\u2202\u00b5X \u03bd + X \u03bd\u2202\u00b5ea a\u2202\u00b5ea \u02c6e(a) \u2297 \u03bbX \u03bb)dx\u00b5 a \u03bd + \u03c9\u00b5 aeb \u03bbX \u03bb + e\u03bd \u03bb\u03c9\u00b5 (e\u03c3 a\u2202\u03c3) \u2297 \u03bbX \u03bb)dx\u00b5 beb bX \u03bb)dx\u00b5 a Comparison with (3.129) reveals \u00b5\u03bb = e\u03bd \u0393\u03bd a\u2202\u00b5ea \u03bb + e\u03bd aeb \u03bb\u03c9\u00b5 a b, or equivalently a \u03c9\u00b5 b = ea \u03bde\u03bb b \u0393\u03bd \u00b5\u03bb \u2212 b \u2202\u00b5ea e\u03bb \u03bb. \u2202\u03c3 \u2202\u03bd. \u2297 \u2297 (3.130) (3.131) (3.132) A bit of manipulation allows us to write this relation as the vanishing of the covariant derivative of the vielbein, \u2207\u00b5ea \u03bd = 0, (3.133) which is sometimes known as the \u201ctetrad postulate.\u201d Note that this is always true; we did not need to assume anything about the connection in order to derive it. Speci\ufb01cally, we did not need to assume that the connection was metric compatible or torsion free. 3 CURVATURE 92 Since the connection may be thought of as something we need to \ufb01x up the transformation law of the", " covariant derivative, it should come as no surprise that the spin connection does not itself obey the tensor transformation law. Actually, under GCT\u2019s the one lower Greek index does transform in the right way, as a one-form. But under LLT\u2019s the spin connection transforms inhomogeneously, as a\u2032 b\u2032 = \u039ba\u2032 \u03c9\u00b5 a\u039bb\u2032 b\u03c9\u00b5 a \u039bb\u2032 c\u2202\u00b5\u039ba\u2032 c. b \u2212 (3.134) You are encouraged to check for yourself that this results in the proper transformation of the covariant derivative. So far we have done nothing but empty formalism, translating things we already knew into a new notation. But the work we are doing does buy us two things. The \ufb01rst, which we already alluded to, is the ability to describe spinor \ufb01elds on spacetime and take their covariant derivatives; we won\u2019t explore this further right now. The second is a change in viewpoint, in which we can think of various tensors as tensor-valued di\ufb00erential forms. For a, which we think of as a (1, 1) tensor written with mixed indices, example, an object like X\u00b5 can also be thought of as a \u201cvector-valued one-form.\u201d It has one lower Greek index, so we think of it as a one-form, but for each value of the lower index it is a vector. Similarly a b, antisymmetric in \u00b5 and \u03bd, can be thought of as a \u201c(1, 1)-tensor-valued twotensor A\u00b5\u03bd form.\u201d Thus, any tensor with some number of antisymmetric lower Greek indices and some number of Latin indices can be thought of as a di\ufb00erential form, but taking values in the tensor bundle. (Ordinary di\ufb00erential forms are simply scalar-valued forms.) The usefulness a of this viewpoint comes when we consider exterior derivatives. If we want to think of X\u00b5 as a vector-valued one-form, we are tempted to take its exterior derivative: a (dX)\u00b5\u03bd a = \u2202\u00b5X\u03bd a \u2202\u03bdX\u00b5 a. \u2212 (3.135) It is easy to check that this object transforms like a two-form (that is, according to the transformation law for (0, 2) tensors) under GCT\u2019s, but not as a vector under LLT\u2019s (the Lorentz transformations depend on position, which introduces an inhomogeneous term into the transformation law). But we can \ufb01x this by judicious use of the spin connection, which can be thought of as a one-form. (Not a tensor-valued one-form, due to the nontensorial transformation law (3.134).) Thus, the object (dX)\u00b5\u03bd a + (\u03c9 \u2227 X)\u00b5\u03bd a = \u2202\u00b5X\u03bd a \u2212 \u2202\u03bdX\u00b5 a + \u03c9\u00b5 a bX\u03bd b \u03c9\u03bd a bX\u00b5 b, \u2212 (3.136) as you can verify at home, transforms as a proper tensor. An immediate application of this formalism is to the expressions for the torsion and curvature, the two tensors which characterize any given connection. The torsion, with two a. The antisymmetric lower indices, can be thought of as a vector-valued two-form T\u00b5\u03bd 3 CURVATURE 93 curvature, which is always antisymmetric in its last two indices, is a (1, 1)-tensor-valued two-form, Ra b\u00b5\u03bd. Using our freedom to suppress indices on di\ufb00erential forms, we can write the de\ufb01ning relations for these two tensors as and T a = dea + \u03c9a eb b \u2227 Ra b = d\u03c9a b + \u03c9a \u03c9c b. c \u2227 (3.137) (3.138) These are known as the Maurer-Cartan structure equations. They are equivalent to the usual de\ufb01nitions; let\u2019s go through the exercise of showing this for the torsion, and you can check the curvature for yourself. We have T\u00b5\u03bd a aT\u00b5\u03bd a(\u2202\u00b5e\u03bd \u03bb = e\u03bb = e\u03bb = \u0393\u03bb a \u2212 \u03bd\u00b5, \u0393\u03bb \u00b5\u03bd \u2212 \u2202\u03bde\u00b5 a + \u03c9\u00b5 a be\u03bd b \u03c9\u03bd a be\u00b5 b) \u2212 (3.139) which is just the original de\ufb01nition we gave. Here we have used", " (3.131), the expression for the \u0393\u03bb \u00b5\u03bd\u2019s in terms of the vielbeins and spin connection. We can also express identities obeyed by these tensors as dT a + \u03c9a T b = Ra eb b \u2227 b \u2227 (3.140) and dRa b + \u03c9a c \u2227 The \ufb01rst of these is the generalization of R\u03c1 Rc Ra \u03c9c b = 0. c \u2227 b \u2212 [\u03c3\u00b5\u03bd] = 0, while the second is the Bianchi identity (3.141) | \u03c3 \u00b5\u03bd] = 0. (Sometimes both equations are called Bianchi identities.) R\u03c1 \u2207[\u03bb | The form of these expressions leads to an almost irresistible temptation to de\ufb01ne a \u201ccovariant-exterior derivative\u201d, which acts on a tensor-valued form by taking the ordinary exterior derivative and then adding appropriate terms with the spin connection, one for each Latin index. Although we won\u2019t do that here, it is okay to give in to this temptation, and in fact the right hand side of (3.137) and the left hand sides of (3.140) and (3.141) can be thought of as just such covariant-exterior derivatives. But be careful, since (3.138) cannot; you can\u2019t take any sort of covariant derivative of the spin connection, since it\u2019s not a tensor. So far our equations have been true for general connections; let\u2019s see what we get for the Christo\ufb00el connection. The torsion-free requirement is just that (3.137) vanish; this does not lead immediately to any simple statement about the coe\ufb03cients of the spin connection. Metric compatibility is expressed as the vanishing of the covariant derivative of the metric: g = 0. We can see what this leads to when we express the metric in the orthonormal basis, \u2207 where its components are simply \u03b7ab: \u2207\u00b5\u03b7ab = \u2202\u00b5\u03b7ab \u2212 c \u03c9\u00b5 a\u03b7cb \u2212 c \u03c9\u00b5 b\u03b7ac 3 CURVATURE Then setting this equal to zero implies = \u03c9\u00b5ab \u2212 \u2212 \u03c9\u00b5ba. 94 (3.142) \u03c9\u00b5ab = \u03c9\u00b5ba. (3.143) \u2212 Thus, metric compatibility is equivalent to the antisymmetry of the spin connection in its Latin indices. (As before, such a statement is only sensible if both indices are either upstairs or downstairs.) These two conditions together allow us to express the spin connection in terms of the vielbeins. There is an explicit formula which expresses this solution, but in practice it is easier to simply solve the torsion-free condition \u03c9ab \u2227 eb = dea, \u2212 (3.144) using the asymmetry of the spin connection, to \ufb01nd the individual components. We now have the means to compare the formalism of connections and curvature in Riemannian geometry to that of gauge theories in particle physics. (This is an aside, which is hopefully comprehensible to everybody, but not an essential ingredient of the course.) In both situations, the \ufb01elds of interest live in vector spaces which are assigned to each point in spacetime. In Riemannian geometry the vector spaces include the tangent space, the cotangent space, and the higher tensor spaces constructed from these. In gauge theories, on the other hand, we are concerned with \u201cinternal\u201d vector spaces. The distinction is that the tangent space and its relatives are intimately associated with the manifold itself, and were naturally de\ufb01ned once the manifold was set up; an internal vector space can be of any dimension we like, and has to be de\ufb01ned as an independent addition to the manifold. In math lingo, the union of the base manifold with the internal vector spaces (de\ufb01ned at each point) is a \ufb01ber bundle, and each copy of the vector space is called the \u201c\ufb01ber\u201d (in perfect accord with our de\ufb01nition of the tangent bundle). Besides the base manifold (for us, spacetime) and the \ufb01bers, the other important ingredient in the de\ufb01nition of a \ufb01ber bundle is the \u201cstructure group,\u201d a Lie group which acts on the \ufb01bers to describe how they are sewn together on overlapping coordinate patches. Without going into details, the structure group for the tangent bundle in a four-dimensional spacetime is generally GL(4, R), the group of real invertible 4 4 matrices; if we have a", " Lorentzian metric, this may be reduced to the Lorentz group SO(3, 1). Now imagine that we introduce an internal three-dimensional vector space, and sew the \ufb01bers together with ordinary rotations; the structure group of this new bundle is then SO(3). A \ufb01eld that lives in this bundle might be denoted \u03c6A(x\u00b5), where A runs from one to three; it is a three-vector (an internal one, unrelated to spacetime) for each point on the manifold. We have freedom to choose the basis in the \ufb01bers in any way we wish; this means that \u201cphysical quantities\u201d should be left invariant under local SO(3) transformations such as \u00d7 \u03c6A(x\u00b5) \u2192 \u03c6A\u2032 (x\u00b5) = OA\u2032 A(x\u00b5)\u03c6A(x\u00b5), (3.145) 3 CURVATURE 95 where OA\u2032 A(x\u00b5) is a matrix in SO(3) which depends on spacetime. Such transformations are known as gauge transformations, and theories invariant under them are called \u201cgauge theories.\u201d For the most part it is not hard to arrange things such that physical quantities are invariant under gauge transformations. The one di\ufb03culty arises when we consider partial derivatives, \u2202\u00b5\u03c6A. Because the matrix OA\u2032 A(x\u00b5) depends on spacetime, it will contribute an unwanted term to the transformation of the partial derivative. By now you should be able to guess the solution: introduce a connection to correct for the inhomogeneous term in the transformation law. We therefore de\ufb01ne a connection on the \ufb01ber bundle to be an object A\u00b5 B, with two \u201cgroup indices\u201d and one spacetime index. Under GCT\u2019s it transforms as a one-form, while under gauge transformations it transforms as A A\u2032 B\u2032 = OA\u2032 A\u00b5 AOB\u2032 BA\u00b5 A OB\u2032 C\u2202\u00b5OA\u2032 C. B \u2212 (3.146) (Beware: our conventions are so drastically di\ufb00erent from those in the particle physics literature that I won\u2019t even try to get them straight.) With this transformation law, the \u201cgauge covariant derivative\u201d D\u00b5\u03c6A = \u2202\u00b5\u03c6A + A\u00b5 A B\u03c6B (3.147) (In transforms \u201ctensorially\u201d under gauge transformations, as you are welcome to check. ordinary electromagnetism the connection is just the conventional vector potential. No indices are necessary, because the structure group U(1) is one-dimensional.) It is clear that this notion of a connection on an internal \ufb01ber bundle is very closely related to the connection on the tangent bundle, especially in the orthonormal-frame picture we have been discussing. The transformation law (3.146), for example, is exactly the same as the transformation law (3.134) for the spin connection. We can also de\ufb01ne a curvature or \u201c\ufb01eld strength\u201d tensor which is a two-form, F A B = dAA B + AA AC B, C \u2227 (3.148) in exact correspondence with (3.138). We can parallel transport things along paths, and there is a construction analogous to the parallel propagator; the trace of the matrix obtained by parallel transporting a vector around a closed curve is called a \u201cWilson loop.\u201d We could go on in the development of the relationship between the tangent bundle and internal vector bundles, but time is short and we have other \ufb01sh to fry. Let us instead \ufb01nish by emphasizing the important di\ufb00erence between the two constructions. The di\ufb00erence stems from the fact that the tangent bundle is closely related to the base manifold, while other \ufb01ber bundles are tacked on after the fact. It makes sense to say that a vector in the tangent space at p \u201cpoints along a path\u201d through p; but this makes no sense for an internal vector bundle. There is therefore no analogue of the coordinate basis for an internal space \u2014 3 CURVATURE 96 partial derivatives along curves have nothing to do with internal vectors. It follows in turn that there is nothing like the vielbeins, which relate orthonormal bases to coordinate bases. The torsion tensor, in particular, is only de\ufb01ned for a connection on the tangent bundle, not for any gauge theory connections; it can be thought of as the covariant exterior derivative of the vielbein, and no such construction is available on an internal bundle. You should appreciate the relationship between the", " di\ufb00erent uses of the notion of a connection, without getting carried away. December 1997 Lecture Notes on General Relativity Sean M. Carroll 4 Gravitation Having paid our mathematical dues, we are now prepared to examine the physics of gravitation as described by general relativity. This subject falls naturally into two pieces: how the curvature of spacetime acts on matter to manifest itself as \u201cgravity\u201d, and how energy and momentum in\ufb02uence spacetime to create curvature. In either case it would be legitimate to start at the top, by stating outright the laws governing physics in curved spacetime and working out their consequences. Instead, we will try to be a little more motivational, starting with basic physical principles and attempting to argue that these lead naturally to an almost unique physical theory. The most basic of these physical principles is the Principle of Equivalence, which comes in a variety of forms. The earliest form dates from Galileo and Newton, and is known as the Weak Equivalence Principle, or WEP. The WEP states that the \u201cinertial mass\u201d and \u201cgravitational mass\u201d of any object are equal. To see what this means, think about Newton\u2019s Second Law. This relates the force exerted on an object to the acceleration it undergoes, setting them proportional to each other with the constant of proportionality being the inertial mass mi: f = mia. (4.1) The inertial mass clearly has a universal character, related to the resistance you feel when you try to push on the object; it is the same constant no matter what kind of force is being exerted. We also have the law of gravitation, which states that the gravitational force exerted on an object is proportional to the gradient of a scalar \ufb01eld \u03a6, known as the gravitational potential. The constant of proportionality in this case is called the gravitational mass mg: \u03a6. fg = mg\u2207 \u2212 On the face of it, mg has a very di\ufb00erent character than mi; it is a quantity speci\ufb01c to the gravitational force. If you like, it is the \u201cgravitational charge\u201d of the body. Nevertheless, Galileo long ago showed (apocryphally by dropping weights o\ufb00 of the Leaning Tower of Pisa, actually by rolling balls down inclined planes) that the response of matter to gravitation was universal \u2014 every object falls at the same rate in a gravitational \ufb01eld, independent of the composition of the object. In Newtonian mechanics this translates into the WEP, which is simply (4.2) for any object. An immediate consequence is that the behavior of freely-falling test particles is universal, independent of their mass (or any other qualities they may have); in fact we mi = mg (4.3) 97 4 GRAVITATION have 98 a = \u03a6. (4.4) \u2212\u2207 The universality of gravitation, as implied by the WEP, can be stated in another, more popular, form. Imagine that we consider a physicist in a tightly sealed box, unable to observe the outside world, who is doing experiments involving the motion of test particles, for example to measure the local gravitational \ufb01eld. Of course she would obtain di\ufb00erent answers if the box were sitting on the moon or on Jupiter than she would on the Earth. But the answers would also be di\ufb00erent if the box were accelerating at a constant velocity; this would change the acceleration of the freely-falling particles with respect to the box. The WEP implies that there is no way to disentangle the e\ufb00ects of a gravitational \ufb01eld from those of being in a uniformly accelerating frame, simply by observing the behavior of freely-falling particles. This follows from the universality of gravitation; it would be possible to distinguish between uniform acceleration and an electromagnetic \ufb01eld, by observing the behavior of particles with di\ufb00erent charges. But with gravity it is impossible, since the \u201ccharge\u201d is necessarily proportional to the (inertial) mass. To be careful, we should limit our claims about the impossibility of distinguishing gravity from uniform acceleration by restricting our attention to \u201csmall enough regions of spacetime.\u201d If the sealed box were su\ufb03ciently big, the gravitational \ufb01eld would change from place to place in an observable way, while the e\ufb00ect of acceleration is always in the same direction. In a rocket ship or elevator, the particles always fall straight down: In a very big box in a gravitational \ufb01eld, however, the particles will move toward the center of the Earth (for example), which might be a di\ufb00erent direction in di\ufb00erent regions: 4 GRAVITATION 99 Earth The W", "EP can therefore be stated as \u201cthe laws of freely-falling particles are the same in a gravitational \ufb01eld and a uniformly accelerated frame, in small enough regions of spacetime.\u201d In larger regions of spacetime there will be inhomogeneities in the gravitational \ufb01eld, which will lead to tidal forces which can be detected. After the advent of special relativity, the concept of mass lost some of its uniqueness, as it became clear that mass was simply a manifestation of energy and momentum (E = mc2 and all that). It was therefore natural for Einstein to think about generalizing the WEP to something more inclusive. His idea was simply that there should be no way whatsoever for the physicist in the box to distinguish between uniform acceleration and an external gravitational \ufb01eld, no matter what experiments she did (not only by dropping test particles). This reasonable extrapolation became what is now known as the Einstein Equivalence Principle, or EEP: \u201cIn small enough regions of spacetime, the laws of physics reduce to those of special relativity; it is impossible to detect the existence of a gravitational \ufb01eld.\u201d In fact, it is hard to imagine theories which respect the WEP but violate the EEP. Consider a hydrogen atom, a bound state of a proton and an electron. Its mass is actually less than the sum of the masses of the proton and electron considered individually, because there is a negative binding energy \u2014 you have to put energy into the atom to separate the proton and electron. According to the WEP, the gravitational mass of the hydrogen atom is therefore less than the sum of the masses of its constituents; the gravitational \ufb01eld couples to electromagnetism (which holds the atom together) in exactly the right way to make the gravitational mass come out right. This means that not only must gravity couple to rest mass universally, but to all forms of energy and momentum \u2014 which is practically the claim of the EEP. It is possible to come up with counterexamples, however; for example, we could imagine a theory of gravity in which freely falling particles began to rotate as they moved through a gravitational \ufb01eld. Then they could fall along the same paths as they would in an accelerated frame (thereby satisfying the WEP), but you could nevertheless detect the 4 GRAVITATION 100 existence of the gravitational \ufb01eld (in violation of the EEP). Such theories seem contrived, but there is no law of nature which forbids them. Sometimes a distinction is drawn between \u201cgravitational laws of physics\u201d and \u201cnongravitational laws of physics,\u201d and the EEP is de\ufb01ned to apply only to the latter. Then one de\ufb01nes the \u201cStrong Equivalence Principle\u201d (SEP) to include all of the laws of physics, I don\u2019t \ufb01nd this a particularly useful distinction, and won\u2019t gravitational and otherwise. belabor it. For our purposes, the EEP (or simply \u201cthe principle of equivalence\u201d) includes all of the laws of physics. It is the EEP which implies (or at least suggests) that we should attribute the action of gravity to the curvature of spacetime. Remember that in special relativity a prominent role is played by inertial frames \u2014 while it was not possible to single out some frame of reference as uniquely \u201cat rest\u201d, it was possible to single out a family of frames which were \u201cunaccelerated\u201d (inertial). The acceleration of a charged particle in an electromagnetic \ufb01eld was therefore uniquely de\ufb01ned with respect to these frames. The EEP, on the other hand, implies that gravity is inescapable \u2014 there is no such thing as a \u201cgravitationally neutral object\u201d with respect to which we can measure the acceleration due to gravity. It follows that \u201cthe acceleration due to gravity\u201d is not something which can be reliably de\ufb01ned, and therefore is of little use. Instead, it makes more sense to de\ufb01ne \u201cunaccelerated\u201d as \u201cfreely falling,\u201d and that is what we shall do. This point of view is the origin of the idea that gravity is not a \u201cforce\u201d \u2014 a force is something which leads to acceleration, and our de\ufb01nition of zero acceleration is \u201cmoving freely in the presence of whatever gravitational \ufb01eld happens to be around.\u201d This seemingly innocuous step has profound implications for the nature of spacetime. In SR, we had a procedure for starting at some point and constructing an inertial frame which stretched throughout spacetime, by joining together rigid rods and attaching clocks to them. But, again due to inhomogeneities in the gravitational \ufb01e", "ld, this is no longer possible. If we start in some freely-falling state and build a large structure out of rigid rods, at some distance away freely-falling objects will look like they are \u201caccelerating\u201d with respect to this reference frame, as shown in the \ufb01gure on the next page. 4 GRAVITATION 101 The solution is to retain the notion of inertial frames, but to discard the hope that they can be uniquely extended throughout space and time. Instead we can de\ufb01ne locally inertial frames, those which follow the motion of freely falling particles in small enough regions of spacetime. (Every time we say \u201csmall enough regions\u201d, purists should imagine a limiting procedure in which we take the appropriate spacetime volume to zero.) This is the best we can do, but it forces us to give up a good deal. For example, we can no longer speak with con\ufb01dence about the relative velocity of far away objects, since the inertial reference frames appropriate to those objects are independent of those appropriate to us. So far we have been talking strictly about physics, without jumping to the conclusion that spacetime should be described as a curved manifold. It should be clear, however, why such a conclusion is appropriate. The idea that the laws of special relativity should be obeyed in su\ufb03ciently small regions of spacetime, and further that local inertial frames can be established in such regions, corresponds to our ability to construct Riemann normal coordinates at any one point on a manifold \u2014 coordinates in which the metric takes its canonical form and the Christo\ufb00el symbols vanish. The impossibility of comparing velocities (vectors) at widely separated regions corresponds to the path-dependence of parallel transport on a curved manifold. These considerations were enough to give Einstein the idea that gravity was a manifestation of spacetime curvature. But in fact we can be even more persuasive. (It is impossible to \u201cprove\u201d that gravity should be thought of as spacetime curvature, since scienti\ufb01c hypotheses can only be falsi\ufb01ed, never veri\ufb01ed [and not even really falsi\ufb01ed, as Thomas Kuhn has famously argued]. But there is nothing to be dissatis\ufb01ed with about convincing plausibility arguments, if they lead to empirically successful theories.) Let\u2019s consider one of the celebrated predictions of the EEP, the gravitational redshift. Consider two boxes, a distance z apart, moving (far away from any matter, so we assume in the absence of any gravitational \ufb01eld) with some constant acceleration a. At time t0 the trailing box emits a photon of wavelength \u03bb0. 4 GRAVITATION 102 The boxes remain a constant distance apart, so the photon reaches the leading box after a time \u2206t = z/c in the reference frame of the boxes. In this time the boxes will have picked up an additional velocity \u2206v = a\u2206t = az/c. Therefore, the photon reaching the lead box will be redshifted by the conventional Doppler e\ufb00ect by an amount \u2206\u03bb \u03bb0 = \u2206v c = az c2. (4.5) (We assume \u2206v/c is small, so we only work to \ufb01rst order.) According to the EEP, the same thing should happen in a uniform gravitational \ufb01eld. So we imagine a tower of height z sitting on the surface of a planet, with ag the strength of the gravitational \ufb01eld (what Newton would have called the \u201cacceleration due to gravity\u201d). z \u03bb 0 This situation is supposed to be indistinguishable from the previous one, from the point of view of an observer in a box at the top of the tower (able to detect the emitted photon, but 4 GRAVITATION 103 otherwise unable to look outside the box). Therefore, a photon emitted from the ground with wavelength \u03bb0 should be redshifted by an amount \u2206\u03bb \u03bb0 = agz c2. (4.6) This is the famous gravitational redshift. Notice that it is a direct consequence of the EEP, not of the details of general relativity. It has been veri\ufb01ed experimentally, \ufb01rst by Pound and Rebka in 1960. They used the M\u00a8ossbauer e\ufb00ect to measure the change in frequency in \u03b3-rays as they traveled from the ground to the top of Je\ufb00erson Labs at Harvard. \u2207 The formula for the redshift is more often stated in terms of the Newtonian potential \u03a6, where ag = \u03a6. (The sign is changed with respect to the usual convention, since we are thinking of ag as the acceleration of the reference frame", ", not of a particle with respect to this reference frame.) A non-constant gradient of \u03a6 is like a time-varying acceleration, and the equivalent net velocity is given by integrating over the time between emission and absorption of the photon. We then have \u2206\u03bb \u03bb0 = 1 c Z \u2207 1 c2 Z = \u2206\u03a6, = \u03a6 dt \u2202z\u03a6 dz (4.7) where \u2206\u03a6 is the total change in the gravitational potential, and we have once again set c = 1. This simple formula for the gravitational redshift continues to be true in more general circumstances. Of course, by using the Newtonian potential at all, we are restricting our domain of validity to weak gravitational \ufb01elds, but that is usually completely justi\ufb01ed for observable e\ufb00ects. The gravitational redshift leads to another argument that we should consider spacetime as curved. Consider the same experimental setup that we had before, now portrayed on the spacetime diagram on the next page. The physicist on the ground emits a beam of light with wavelength \u03bb0 from a height z0, which travels to the top of the tower at height z1. The time between when the beginning of any single wavelength of the light is emitted and the end of that same wavelength is emitted is \u2206t0 = \u03bb0/c, and the same time interval for the absorption is \u2206t1 = \u03bb1/c. Since we imagine that the gravitational \ufb01eld is not varying with time, the paths through spacetime followed by the leading and trailing edge of the single wave must be precisely congruent. (They are represented by some generic curved paths, since we do not pretend that we know just what the paths will be.) Simple geometry tells us that the times \u2206t0 and \u2206t1 must be the same. But of course they are not; the gravitational redshift implies that \u2206t1 > \u2206t0. (Which we can interpret as \u201cthe clock on the tower appears to run more quickly.\u201d) The fault lies with 4 GRAVITATION t 104 tsimple geometry\u201d; a better description of what happens is to imagine that spacetime is curved. All of this should constitute more than enough motivation for our claim that, in the presence of gravity, spacetime should be thought of as a curved manifold. Let us now take this to be true and begin to set up how physics works in a curved spacetime. The principle of equivalence tells us that the laws of physics, in small enough regions of spacetime, look like those of special relativity. We interpret this in the language of manifolds as the statement that these laws, when written in Riemannian normal coordinates x\u00b5 based at some point p, are described by equations which take the same form as they would in \ufb02at space. The simplest example is that of freely-falling (unaccelerated) particles. In \ufb02at space such particles move in straight lines; in equations, this is expressed as the vanishing of the second derivative of the parameterized path x\u00b5(\u03bb): d2x\u00b5 d\u03bb2 = 0. According to the EEP, exactly this equation should hold in curved space, as long as the coordinates x\u00b5 are RNC\u2019s. What about some other coordinate system? As it stands, (4.8) is not an equation between tensors. However, there is a unique tensorial equation which reduces to (4.8) when the Christo\ufb00el symbols vanish; it is (4.8) d2x\u00b5 d\u03bb2 + \u0393\u00b5 \u03c1\u03c3 dx\u03c1 d\u03bb dx\u03c3 d\u03bb = 0. (4.9) Of course, this is simply the geodesic equation. In general relativity, therefore, free particles move along geodesics; we have mentioned this before, but now you know why it is true. As far as free particles go, we have argued that curvature of spacetime is necessary to describe gravity; we have not yet shown that it is su\ufb03cient. To do so, we can show how the usual results of Newtonian gravity \ufb01t into the picture. We de\ufb01ne the \u201cNewtonian limit\u201d by three requirements: the particles are moving slowly (with respect to the speed of light), the 4 GRAVITATION 105 gravitational \ufb01eld is weak (can be considered a perturbation of \ufb02at space), and the \ufb01eld is also static (unchanging with time). Let us see what these assumptions do to the geodesic equation, taking the proper time \u03c4 as an a\ufb03ne parameter. \u201cMoving slowly\u201d means that so the geodesic equation becomes dxi d\u03c4", " << dt d\u03c4 d2x\u00b5 d\u03c4 2 + \u0393\u00b5 00 dt d\u03c4!, 2 = 0. Since the \ufb01eld is static, the relevant Christo\ufb00el symbols \u0393\u00b5 00 simplify: \u0393\u00b5 00 = = 1 2 \u2212 g\u00b5\u03bb(\u22020g\u03bb0 + \u22020g0\u03bb \u2212 1 g\u00b5\u03bb\u2202\u03bbg00. 2 \u2202\u03bbg00) (4.10) (4.11) (4.12) Finally, the weakness of the gravitational \ufb01eld allows us to decompose the metric into the Minkowski form plus a small perturbation: g\u00b5\u03bd = \u03b7\u00b5\u03bd + h\u00b5\u03bd, h\u00b5\u03bd| | << 1. (4.13) (We are working in Cartesian coordinates, so \u03b7\u00b5\u03bd is the canonical form of the metric. The \u201csmallness condition\u201d on the metric perturbation h\u00b5\u03bd doesn\u2019t really make sense in other coordinates.) From the de\ufb01nition of the inverse metric, g\u00b5\u03bdg\u03bd\u03c3 = \u03b4\u00b5 \u03c3, we \ufb01nd that to \ufb01rst order in h, g\u00b5\u03bd = \u03b7\u00b5\u03bd h\u00b5\u03bd, (4.14) \u2212 where h\u00b5\u03bd = \u03b7\u00b5\u03c1\u03b7\u03bd\u03c3h\u03c1\u03c3. In fact, we can use the Minkowski metric to raise and lower indices on an object of any de\ufb01nite order in h, since the corrections would only contribute at higher orders. Putting it all together, we \ufb01nd The geodesic equation (4.11) is therefore \u0393\u00b5 00 = 1 2 \u2212 \u03b7\u00b5\u03bb\u2202\u03bbh00. d2x\u00b5 d\u03c4 2 = 1 2 \u03b7\u00b5\u03bb\u2202\u03bbh00 2 dt d\u03c4!. Using \u22020h00 = 0, the \u00b5 = 0 component of this is just d2t d\u03c4 2 = 0. (4.15) (4.16) (4.17) 4 GRAVITATION 106 That is, dt spacelike components of \u03b7\u00b5\u03bd are just those of a 3 d\u03c4 is constant. To examine the spacelike components of (4.16), recall that the 3 identity matrix. We therefore have \u00d7 2 d2xi d\u03c4 2 = 1 2 dt d\u03c4! \u2202ih00. (4.18) Dividing both sides by from \u03c4 to t, leaving us with dt d\u03c4 2 has the e\ufb00ect of converting the derivative on the left-hand side 1 2 This begins to look a great deal like Newton\u2019s theory of gravitation. In fact, if we compare this equation to (4.4), we \ufb01nd that they are the same once we identify \u2202ih00. (4.19) d2xi dt2 = or in other words h00 = 2\u03a6, \u2212 g00 = (1 + 2\u03a6). \u2212 (4.20) (4.21) Therefore, we have shown that the curvature of spacetime is indeed su\ufb03cient to describe gravity in the Newtonian limit, as long as the metric takes the form (4.21). It remains, of course, to \ufb01nd \ufb01eld equations for the metric which imply that this is the form taken, and that for a single gravitating body we recover the Newtonian formula \u03a6 = GM r \u2212, (4.22) but that will come soon enough. Our next task is to show how the remaining laws of physics, beyond those governing freelyfalling particles, adapt to the curvature of spacetime. The procedure essentially follows the paradigm established in arguing that free particles move along geodesics. Take a law of physics in \ufb02at space, traditionally written in terms of partial derivatives and the \ufb02at metric. According to the equivalence principle this law will hold in the presence of gravity, as long as we are in Riemannian normal coordinates. Translate the law into a relationship between tensors; for example, change partial derivatives to covariant ones. In RNC\u2019s this version of the law will reduce to the \ufb02at-space one, but tensors are coordinate-independent objects, so the tensorial version must hold in any coordinate system. This procedure is sometimes given a name, the Principle of Covariance. I\u2019m not sure that it deserves its own name, since it\u2019s really a consequence of the EEP plus the requirement that the laws of physics be independent of coordinates. (The requirement that laws of physics be independent of coordinates is essentially impossible to", " even imagine being untrue. Given some experiment, if one person uses one coordinate system to predict a result and another one uses a di\ufb00erent coordinate system, they had better agree.) Another name 4 GRAVITATION 107 is the \u201ccomma-goes-to-semicolon rule\u201d, since at a typographical level the thing you have to do is replace partial derivatives (commas) with covariant ones (semicolons). We have already implicitly used the principle of covariance (or whatever you want to call it) in deriving the statement that free particles move along geodesics. For the most part, it is very simple to apply it to interesting cases. Consider for example the formula for conservation of energy in \ufb02at spacetime, \u2202\u00b5T \u00b5\u03bd = 0. The adaptation to curved spacetime is immediate: \u2207\u00b5T \u00b5\u03bd = 0. This equation expresses the conservation of energy in the presence of a gravitational \ufb01eld. (4.23) Unfortunately, life is not always so easy. Consider Maxwell\u2019s equations in special relativity, where it would seem that the principle of covariance can be applied in a straightforward way. The inhomogeneous equation \u2202\u00b5F \u03bd\u00b5 = 4\u03c0J \u03bd becomes \u2207\u00b5F \u03bd\u00b5 = 4\u03c0J \u03bd, (4.24) and the homogeneous one \u2202[\u00b5F\u03bd\u03bb] = 0 becomes \u2207[\u00b5F\u03bd\u03bb] = 0. On the other hand, we could also write Maxwell\u2019s equations in \ufb02at space in terms of di\ufb00erential forms as (4.25) and d( \u2217 F ) = 4\u03c0( J), \u2217 dF = 0. (4.26) (4.27) These are already in perfectly tensorial form, since we have shown that the exterior derivative is a well-de\ufb01ned tensor operator regardless of what the connection is. We therefore begin to worry a little bit; what is the guarantee that the process of writing a law of physics in tensorial form gives a unique answer? In fact, as we have mentioned earlier, the di\ufb00erential forms versions of Maxwell\u2019s equations should be taken as fundamental. Nevertheless, in this case it happens to make no di\ufb00erence, since in the absence of torsion (4.26) is identical to (4.24), and (4.27) is identical to (4.25); the symmetric part of the connection doesn\u2019t contribute. Similarly, the de\ufb01nition of the \ufb01eld strength tensor in terms of the potential A\u00b5 can be written either as or equally well as F\u00b5\u03bd = \u2207\u00b5A\u03bd \u2212 \u2207\u03bdA\u00b5, F = dA. (4.28) (4.29) The worry about uniqueness is a real one, however. Imagine that two vector \ufb01elds X \u00b5 and Y \u03bd obey a law in \ufb02at space given by Y \u00b5\u2202\u00b5\u2202\u03bdX \u03bd = 0. (4.30) 4 GRAVITATION 108 The problem in writing this as a tensor equation should be clear: the partial derivatives can be commuted, but covariant derivatives cannot. If we simply replace the partials in (4.30) by covariant derivatives, we get a di\ufb00erent answer than we would if we had \ufb01rst exchanged the order of the derivatives (leaving the equation in \ufb02at space invariant) and then replaced them. The di\ufb00erence is given by \u2212 Y \u00b5 Y \u00b5 \u2207\u00b5\u2207\u03bdX \u03bd \u2207\u03bd\u2207\u00b5X \u03bd = The prescription for generalizing laws from \ufb02at to curved spacetimes does not guide us in choosing the order of the derivatives, and therefore is ambiguous about whether a term such as that in (4.31) should appear in the presence of gravity. (The problem of ordering covariant derivatives is similar to the problem of operator-ordering ambiguities in quantum mechanics.) R\u00b5\u03bd Y \u00b5X \u03bd. (4.31) \u2212 In the literature you can \ufb01nd various prescriptions for dealing with ambiguities such as this, most of which are sensible pieces of advice such as remembering to preserve gauge invariance for electromagnetism. But deep down the real answer is that there is no way to resolve these problems by pure thought alone; the fact is that there may be more than one way to adapt a law of physics to curved space, and ultimately only experiment can decide between the alternatives. In fact, let us be honest about the principle of equivalence: it serves as a useful guideline, but it does not", " deserve to be treated as a fundamental principle of nature. From the modern point of view, we do not expect the EEP to be rigorously true. Consider the following alternative version of (4.24): \u2207\u00b5[(1 + \u03b1R)F \u03bd\u00b5] = 4\u03c0J \u03bd, (4.32) where R is the Ricci scalar and \u03b1 is some coupling constant. If this equation correctly described electrodynamics in curved spacetime, it would be possible to measure R even in an arbitrarily small region, by doing experiments with charged particles. The equivalence principle therefore demands that \u03b1 = 0. But otherwise this is a perfectly respectable equation, consistent with charge conservation and other desirable features of electromagnetism, which reduces to the usual equation in \ufb02at space. Indeed, in a world governed by quantum mechanics we expect all possible couplings between di\ufb00erent \ufb01elds (such as gravity and electromagnetism) that are consistent with the symmetries of the theory (in this case, gauge invariance). So why is it reasonable to set \u03b1 = 0? The real reason is one of scales. Notice that the Ricci tensor involves second derivatives of the metric, which is dimensionless, so R has 2 (with c = 1). Therefore \u03b1 must have dimensions of (length)2. But dimensions of (length)\u2212 since the coupling represented by \u03b1 is of gravitational origin, the only reasonable expectation for the relevant length scale is l2 P, \u03b1 \u223c (4.33) 4 GRAVITATION where lP is the Planck length lP = 1/2 G\u00afh c3! = 1.6 \u00d7 10\u2212 33 cm, 109 (4.34) where \u00afh is of course Planck\u2019s constant. So the length scale corresponding to this coupling is extremely small, and for any conceivable experiment we expect the typical scale of variation for the gravitational \ufb01eld to be much larger. Therefore the reason why this equivalenceprinciple-violating term can be safely ignored is simply because \u03b1R is probably a fantastically small number, far out of the reach of any experiment. On the other hand, we might as well keep an open mind, since our expectations are not always borne out by observation. Having established how physical laws govern the behavior of \ufb01elds and objects in a curved spacetime, we can complete the establishment of general relativity proper by introducing Einstein\u2019s \ufb01eld equations, which govern how the metric responds to energy and momentum. We will actually do this in two ways: \ufb01rst by an informal argument close to what Einstein himself was thinking, and then by starting with an action and deriving the corresponding equations of motion. The informal argument begins with the realization that we would like to \ufb01nd an equation which supersedes the Poisson equation for the Newtonian potential: 2\u03a6 = 4\u03c0G\u03c1, \u2207 (4.35) \u2207 2 = \u03b4ij\u2202i\u2202j is the Laplacian in space and \u03c1 is the mass density. (The explicit form of where \u03a6 given in (4.22) is one solution of (4.35), for the case of a pointlike mass distribution.) What characteristics should our sought-after equation possess? On the left-hand side of (4.35) we have a second-order di\ufb00erential operator acting on the gravitational potential, and on the right-hand side a measure of the mass distribution. A relativistic generalization should take the form of an equation between tensors. We know what the tensor generalization of the mass density is; it\u2019s the energy-momentum tensor T\u00b5\u03bd. The gravitational potential, meanwhile, should get replaced by the metric tensor. We might therefore guess that our new equation will have T\u00b5\u03bd set proportional to some tensor which is second-order in derivatives of the metric. In fact, using (4.21) for the metric in the Newtonian limit and T00 = \u03c1, we see that in this limit we are looking for an equation that predicts but of course we want it to be completely tensorial. 2h00 = \u2207 8\u03c0GT00, \u2212 (4.36) The left-hand side of (4.36) does not obviously generalize to a tensor. The \ufb01rst choice might be to act the D\u2019Alembertian 2 = \u2207\u00b5 on the metric g\u00b5\u03bd, but this is automatically zero by metric compatibility. Fortunately, there is an obvious quantity which is not zero \u2207 \u00b5 4 GRAVITATION 110 and is constructed from second derivatives (and \ufb01rst derivatives) of the metric: the Riemann tensor R\u03c1 \u03c3\u00b5\u03bd. It", " doesn\u2019t have the right number of indices, but we can contract it to form the Ricci tensor R\u00b5\u03bd, which does (and is symmetric to boot). It is therefore reasonable to guess that the gravitational \ufb01eld equations are R\u00b5\u03bd = \u03baT\u00b5\u03bd, (4.37) for some constant \u03ba. In fact, Einstein did suggest this equation at one point. There is a problem, unfortunately, with conservation of energy. According to the Principle of Equivalence, the statement of energy-momentum conservation in curved spacetime should be which would then imply \u00b5T\u00b5\u03bd = 0, \u2207 \u00b5R\u00b5\u03bd = 0. \u2207 (4.38) (4.39) This is certainly not true in an arbitrary geometry; we have seen from the Bianchi identity (3.94) that \u00b5R\u00b5\u03bd = \u2207 1 2\u2207\u03bdR. (4.40) But our proposed \ufb01eld equation implies that R = \u03bag\u00b5\u03bdT\u00b5\u03bd = \u03baT, so taking these together we have \u2207\u00b5T = 0. The covariant derivative of a scalar is just the partial derivative, so (4.41) is telling us that T is constant throughout spacetime. This is highly implausible, since T = 0 in vacuum while T > 0 in matter. We have to try harder. (4.41) (Actually we are cheating slightly, in taking the equation \u00b5T\u00b5\u03bd = 0 so seriously. If as we said, the equivalence principle is only an approximate guide, we could imagine that there are nonzero terms on the right-hand side involving the curvature tensor. Later we will be more precise and argue that they are strictly zero.) \u2207 Of course we don\u2019t have to try much harder, since we already know of a symmetric (0, 2) tensor, constructed from the Ricci tensor, which is automatically conserved: the Einstein tensor G\u00b5\u03bd = R\u00b5\u03bd \u2212 Rg\u00b5\u03bd, 1 2 which always obeys \u2207 \u00b5G\u00b5\u03bd = 0. We are therefore led to propose G\u00b5\u03bd = \u03baT\u00b5\u03bd (4.42) (4.43) as a \ufb01eld equation for the metric. This equation satis\ufb01es all of the obvious requirements; the right-hand side is a covariant expression of the energy and momentum density in the 4 GRAVITATION 111 form of a symmetric and conserved (0, 2) tensor, while the left-hand side is a symmetric and conserved (0, 2) tensor constructed from the metric and its \ufb01rst and second derivatives. It only remains to see whether it actually reproduces gravity as we know it. To answer this, note that contracting both sides of (4.43) yields (in four dimensions) and using this we can rewrite (4.43) as R = \u03baT, \u2212 R\u00b5\u03bd = \u03ba(T\u00b5\u03bd \u2212 1 2 T g\u00b5\u03bd). (4.44) (4.45) This is the same equation, just written slightly di\ufb00erently. We would like to see if it predicts In Newtonian gravity in the weak-\ufb01eld, time-independent, slowly-moving-particles limit. this limit the rest energy \u03c1 = T00 will be much larger than the other terms in T\u00b5\u03bd, so we want to focus on the \u00b5 = 0, \u03bd = 0 component of (4.45). In the weak-\ufb01eld limit, we write (in accordance with (4.13) and (4.14)) g00 = g00 = 1 + h00, \u2212 h00. 1 \u2212 \u2212 The trace of the energy-momentum tensor, to lowest nontrivial order, is T = g00T00 = T00. \u2212 (4.46) (4.47) Plugging this into (4.45), we get 1 2 This is an equation relating derivatives of the metric to the energy density. To \ufb01nd the explicit expression in terms of the metric, we need to evaluate R00 = R\u03bb 0\u03bb0. In fact we only need Ri 000 = 0. We have 0i0, since R0 R00 = \u03baT00. (4.48) Ri 0j0 = \u2202j\u0393i 00 \u2212 \u22020\u0393i j0 + \u0393i j\u03bb\u0393\u03bb 00 \u2212 0\u03bb\u0393\u03bb \u0393i j0. (4.49) The second term here is a time derivative, which vanishes for static \ufb01elds. The third and fourth terms are of the form", " (\u0393)2, and since \u0393 is \ufb01rst-order in the metric perturbation these contribute only at second order, and can be neglected. We are left with Ri 00. From this we get 0j0 = \u2202j\u0393i R00 = Ri = \u2202i 0i0 1 gi\u03bb(\u22020g\u03bb0 + \u22020g0\u03bb \u2212 2 1 \u03b7ij\u2202i\u2202jh00 2 = \u2212 \u2202\u03bbg00) 4 GRAVITATION = 1 2 \u2207 \u2212 2h00. 112 (4.50) Comparing to (4.48), we see that the 00 component of (4.43) in the Newtonian limit predicts 2h00 = \u2207 \u03baT00. \u2212 (4.51) But this is exactly (4.36), if we set \u03ba = 8\u03c0G. So our guess seems to have worked out. With the normalization \ufb01xed by comparison with the Newtonian limit, we can present Einstein\u2019s equations for general relativity: R\u00b5\u03bd \u2212 1 2 Rg\u00b5\u03bd = 8\u03c0GT\u00b5\u03bd. (4.52) These tell us how the curvature of spacetime reacts to the presence of energy-momentum. Einstein, you may have heard, thought that the left-hand side was nice and geometrical, while the right-hand side was somewhat less compelling. Einstein\u2019s equations may be thought of as second-order di\ufb00erential equations for the metric tensor \ufb01eld g\u00b5\u03bd. There are ten independent equations (since both sides are symmetric two-index tensors), which seems to be exactly right for the ten unknown functions of the \u00b5G\u00b5\u03bd = 0 represents four constraints on metric components. However, the Bianchi identity the functions R\u00b5\u03bd, so there are only six truly independent equations in (4.52). In fact this is appropriate, since if a metric is a solution to Einstein\u2019s equation in one coordinate system x\u00b5 it should also be a solution in any other coordinate system x\u00b5\u2032. This means that there are four unphysical degrees of freedom in g\u00b5\u03bd (represented by the four functions x\u00b5\u2032 (x\u00b5)), and we should expect that Einstein\u2019s equations only constrain the six coordinate-independent degrees of freedom. \u2207 As di\ufb00erential equations, these are extremely complicated; the Ricci scalar and tensor are contractions of the Riemann tensor, which involves derivatives and products of the Christo\ufb00el symbols, which in turn involve the inverse metric and derivatives of the metric. Furthermore, the energy-momentum tensor T\u00b5\u03bd will generally involve the metric as well. The equations are also nonlinear, so that two known solutions cannot be superposed to \ufb01nd a third. It is therefore very di\ufb03cult to solve Einstein\u2019s equations in any sort of generality, and it is usually necessary to make some simplifying assumptions. Even in vacuum, where we set the energy-momentum tensor to zero, the resulting equations (from (4.45)) R\u00b5\u03bd = 0 (4.53) can be very di\ufb03cult to solve. The most popular sort of simplifying assumption is that the metric has a signi\ufb01cant degree of symmetry, and we will talk later on about how symmetries of the metric make life easier. The nonlinearity of general relativity is worth remarking on. In Newtonian gravity the potential due to two point masses is simply the sum of the potentials for each mass, but 4 GRAVITATION 113 clearly this does not carry over to general relativity (outside the weak-\ufb01eld limit). There is a physical reason for this, namely that in GR the gravitational \ufb01eld couples to itself. This can be thought of as a consequence of the equivalence principle \u2014 if gravitation did not couple to itself, a \u201cgravitational atom\u201d (two particles bound by their mutual gravitational attraction) would have a di\ufb00erent inertial mass (due to the negative binding energy) than gravitational mass. From a particle physics point of view this can be expressed in terms of Feynman diagrams. The electromagnetic interaction between two electrons can be thought of as due to exchange of a virtual photon: - e e - photon But there is no diagram in which two photons exchange another photon between themselves; electromagnetism is linear. The gravitational interaction, meanwhile, can be thought of as due to exchange of a virtual graviton (a quantized perturbation of the metric). The nonlinearity manifests itself as the fact that both electrons and gravitons (and anything else) can exchange virtual gravitons, and", " therefore exert a gravitational force: - e e - graviton gravitons There is nothing profound about this feature of gravity; it is shared by most gauge theories, such as quantum chromodynamics, the theory of the strong interactions. (Electromagnetism is actually the exception; the linearity can be traced to the fact that the relevant gauge group, U(1), is abelian.) But it does represent a departure from the Newtonian theory. (Of course this quantum mechanical language of Feynman diagrams is somewhat inappropriate for GR, which has not [yet] been successfully quantized, but the diagrams are just a convenient shorthand for remembering what interactions exist in the theory.) 4 GRAVITATION 114 To increase your con\ufb01dence that Einstein\u2019s equations as we have derived them are indeed the correct \ufb01eld equations for the metric, let\u2019s see how they can be derived from a more modern viewpoint, starting from an action principle. (In fact the equations were \ufb01rst derived by Hilbert, not Einstein, and Hilbert did it using the action principle. But he had been inspired by Einstein\u2019s previous papers on the subject, and Einstein himself derived the equations independently, so they are rightly named after Einstein. The action, however, is rightly called the Hilbert action.) The action should be the integral over spacetime of a Lagrange density (\u201cLagrangian\u201d for short, although strictly speaking the Lagrangian is the integral over space of the Lagrange density): SH = dnx LH. Z (4.54) \u2212 g times a scalar. What The Lagrange density is a tensor density, which can be written as \u221a scalars can we make out of the metric? Since we know that the metric can be set equal to its canonical form and its \ufb01rst derivatives set to zero at any one point, any nontrivial scalar must involve at least second derivatives of the metric. The Riemann tensor is of course made from second derivatives of the metric, and we argued earlier that the only independent scalar we could construct from the Riemann tensor was the Ricci scalar R. What we did not show, but is nevertheless true, is that any nontrivial tensor made from the metric and its \ufb01rst and second derivatives can be expressed in terms of the metric and the Riemann tensor. Therefore, the only independent scalar constructed from the metric, which is no higher than second order in its derivatives, is the Ricci scalar. Hilbert \ufb01gured that this was therefore the simplest possible choice for a Lagrangian, and proposed LH = \u221a gR. \u2212 (4.55) The equations of motion should come from varying the action with respect to the metric. In fact let us consider variations with respect to the inverse metric g\u00b5\u03bd, which are slightly easier but give an equivalent set of equations. Using R = g\u00b5\u03bdR\u00b5\u03bd, in general we will have \u03b4S = dnx \u221a h \u2212 = (\u03b4S)1 + (\u03b4S)2 + (\u03b4S)3. Z gg\u00b5\u03bd\u03b4R\u00b5\u03bd + \u221a gR\u00b5\u03bd\u03b4g\u00b5\u03bd + R\u03b4\u221a \u2212 g \u2212 i (4.56) The second term (\u03b4S)2 is already in the form of some expression times \u03b4g\u00b5\u03bd; let\u2019s examine the others more closely. Recall that the Ricci tensor is the contraction of the Riemann tensor, which is given by R\u03c1 \u00b5\u03bb\u03bd = \u2202\u03bb\u0393\u03bb \u03bd\u00b5 + \u0393\u03c1 \u03bb\u03c3\u0393\u03c3 \u03bd\u00b5 \u2212 (\u03bb \u2194 \u03bd). (4.57) The variation of this with respect the metric can be found \ufb01rst varying the connection with respect to the metric, and then substituting into this expression. Let us however consider 4 GRAVITATION arbitrary variations of the connection, by replacing \u0393\u03c1 \u03bd\u00b5 \u2192 \u03bd\u00b5 + \u03b4\u0393\u03c1 \u0393\u03c1 \u03bd\u00b5. 115 (4.58) The variation \u03b4\u0393\u03c1 can thus take its covariant derivative, \u03bd\u00b5 is the di\ufb00erence of two connections, and therefore is itself a tensor. We \u2207\u03bb(\u03b4\u0393\u03c1 \u03bd\u00b5) = \u2202\u03bb(\u03b4\u0393\u03c1 \u03bd\u00b5) + \u0393\u03c1 \u03bb\u03c3\u03b4\u0393\u03c3 \u03bd\u00b5 \u2212 \u03bb\u03bd\u03b4\u0393\u03c1 \u0393\u03c3 \u03c3\u00b5 \u2212 \u03bb\u00b5\u03b4\u0393\u03c1 \u0393\u03c3 \u03bd\u03c3. Given this expression (and a small amount of labor) it is easy to", " show that \u03b4R\u03c1 \u00b5\u03bb\u03bd = \u2207\u03bb(\u03b4\u0393\u03c1 \u03bd\u00b5) \u2212 \u2207\u03bd(\u03b4\u0393\u03c1 \u03bb\u00b5). (4.59) (4.60) You can check this yourself. Therefore, the contribution of the \ufb01rst term in (4.56) to \u03b4S can be written Z Z = \u2207\u03c3 g g\u00b5\u03bd dnx\u221a dnx\u221a \u2212 g \u2212 (\u03b4S)1 = \u2212 \u2207\u03bd(\u03b4\u0393\u03bb \u03bb\u00b5) g\u00b5\u03bd(\u03b4\u0393\u03c3 \u2207\u03bb(\u03b4\u0393\u03bb \u03bd\u00b5) h g\u00b5\u03c3(\u03b4\u0393\u03bb \u03bb\u00b5) h where we have used metric compatibility and relabeled some dummy indices. But now we have the integral with respect to the natural volume element of the covariant divergence of a vector; by Stokes\u2019s theorem, this is equal to a boundary contribution at in\ufb01nity which we can set to zero by making the variation vanish at in\ufb01nity. (We haven\u2019t actually shown that Stokes\u2019s theorem, as mentioned earlier in terms of di\ufb00erential forms, can be thought of this way, but you can easily convince yourself it\u2019s true.) Therefore this term contributes nothing to the total variation. i \u00b5\u03bd) (4.61) \u2212 i, To make sense of the (\u03b4S)3 term we need to use the following fact, true for any matrix M: Tr(ln M) = ln(det M). (4.62) Here, ln M is de\ufb01ned by exp(ln M) = M. (For numbers this is obvious, for matrices it\u2019s a little less straightforward.) The variation of this identity yields Tr(M \u2212 1\u03b4M) = 1 det M \u03b4(det M). (4.63) Here we have used the cyclic property of the trace to allow us to ignore the fact that M \u2212 and \u03b4M may not commute. Now we would like to apply this to the inverse metric, M = g\u00b5\u03bd. Then det M = g\u2212 1 (where g = det g\u00b5\u03bd), and 1 \u03b4(g\u2212 1) = 1 g g\u00b5\u03bd\u03b4g\u00b5\u03bd. (4.64) 4 GRAVITATION Now we can just plug in: \u03b4\u221a \u2212 = g = \u03b4[( \u2212 1 ( 2 1 2 \u2212 = \u2212 g\u2212 1)\u2212 1/2] g\u2212 1) g\u2212 1)\u2212 3/2\u03b4( \u2212 gg\u00b5\u03bd\u03b4g\u00b5\u03bd. \u2212 \u221a \u2212 Hearkening back to (4.56), and remembering that (\u03b4S)1 does not contribute, we \ufb01nd \u03b4S = dnx\u221a Z g \u2212 R\u00b5\u03bd \u2212 1 2 Rg\u00b5\u03bd \u03b4g\u00b5\u03bd. 116 (4.65) (4.66) This should vanish for arbitrary variations, so we are led to Einstein\u2019s equations in vacuum: 1 \u221a g \u2212 \u03b4S \u03b4g\u00b5\u03bd = R\u00b5\u03bd \u2212 1 2 Rg\u00b5\u03bd = 0. (4.67) The fact that this simple action leads to the same vacuum \ufb01eld equations as we had previously arrived at by more informal arguments certainly reassures us that we are doing something right. What we would really like, however, is to get the non-vacuum \ufb01eld equations as well. That means we consider an action of the form S = 1 8\u03c0G SH + SM, (4.68) where SM is the action for matter, and we have presciently normalized the gravitational action (although the proper normalization is somewhat convention-dependent). Following through the same procedure as above leads to 1 \u221a g \u2212 \u03b4S \u03b4g\u00b5\u03bd = 1 8\u03c0G R\u00b5\u03bd \u2212 1 2 Rg\u00b5\u03bd + 1 \u221a g \u2212 \u03b4SM \u03b4g\u00b5\u03bd = 0, and we recover Einstein\u2019s equations if we can set T\u00b5\u03bd = 1 \u221a g \u2212 \u2212 \u03b4SM \u03b4g\u00b5\u03bd. (4.69) (4.70) What makes us think that we can make such an identi\ufb01cation? In fact (4.70) turns out to be the best way to de\ufb01ne a symmetric energy-momentum tensor. The tricky part is to show that it is conserved, which is in fact automatically true, but which we will not justify until the next section. We say that (", "4.70) provides the \u201cbest\u201d de\ufb01nition of the energy-momentum tensor because it is not the only one you will \ufb01nd. In \ufb02at Minkowski space, there is an alternative de\ufb01nition which is sometimes given in books on electromagnetism or \ufb01eld theory. In this context 4 GRAVITATION 117 energy-momentum conservation arises as a consequence of symmetry of the Lagrangian under spacetime translations. Noether\u2019s theorem states that every symmetry of a Lagrangian implies the existence of a conservation law; invariance under the four spacetime translations leads to a tensor S\u00b5\u03bd which obeys \u2202\u00b5S\u00b5\u03bd = 0 (four relations, one for each value of \u03bd). The details can be found in Wald or in any number of \ufb01eld theory books. Applying Noether\u2019s procedure to a Lagrangian which depends on some \ufb01elds \u03c8i and their \ufb01rst derivatives \u2202\u00b5\u03c8i, we obtain S\u00b5\u03bd = \u03b4 L \u03b4(\u2202\u00b5\u03c8i) \u2202\u03bd\u03c8i \u03b7\u00b5\u03bd, L \u2212 (4.71) where a sum over i is implied. You can check that this tensor is conserved by virtue of the equations of motion of the matter \ufb01elds. S\u00b5\u03bd often goes by the name \u201ccanonical energymomentum tensor\u201d; however, there are a number of reasons why it is more convenient for us to use (4.70). First and foremost, (4.70) is in fact what appears on the right hand side of Einstein\u2019s equations when they are derived from an action, and it is not always possible to generalize (4.71) to curved spacetime. But even in \ufb02at space (4.70) has its advantages; it is manifestly symmetric, and also guaranteed to be gauge invariant, neither of which is true for (4.71). We will therefore stick with (4.70) as the de\ufb01nition of the energy-momentum tensor. Sometimes it is useful to think about Einstein\u2019s equations without specifying the theory of matter from which T\u00b5\u03bd is derived. This leaves us with a great deal of arbitrariness; consider for example the question \u201cWhat metrics obey Einstein\u2019s equations?\u201d In the absence of some constraints on T\u00b5\u03bd, the answer is \u201cany metric at all\u201d; simply take the metric of your choice, compute the Einstein tensor G\u00b5\u03bd for this metric, and then demand that T\u00b5\u03bd be equal to G\u00b5\u03bd. (It will automatically be conserved, by the Bianchi identity.) Our real concern is with the existence of solutions to Einstein\u2019s equations in the presence of \u201crealistic\u201d sources of energy and momentum, whatever that means. The most common property that is demanded of T\u00b5\u03bd is that it represent positive energy densities \u2014 no negative masses are allowed. In a locally inertial frame this requirement can be stated as \u03c1 = T00 \u2265 0. To turn this into a coordinate-independent statement, we ask that T\u00b5\u03bdV \u00b5V \u03bd 0, \u2265 for all timelike vectors V \u00b5. (4.72) This is known as the Weak Energy Condition, or WEC. It seems like a fairly reasonable requirement, and many of the important theorems about solutions to general relativity (such as the singularity theorems of Hawking and Penrose) rely on this condition or something very close to it. Unfortunately it is not set in stone; indeed, it is straightforward to invent otherwise respectable classical \ufb01eld theories which violate the WEC, and almost impossible to invent a quantum \ufb01eld theory which obeys it. Nevertheless, it is legitimate to assume that the WEC holds in all but the most extreme conditions. (There are also stronger energy conditions, but they are even less true than the WEC, and we won\u2019t dwell on them.) 4 GRAVITATION 118 We have now justi\ufb01ed Einstein\u2019s equations in two di\ufb00erent ways: as the natural covariant generalization of Poisson\u2019s equation for the Newtonian gravitational potential, and as the result of varying the simplest possible action we could invent for the metric. The rest of the course will be an exploration of the consequences of these equations, but before we start on that road let us brie\ufb02y explore ways in which the equations could be modi\ufb01ed. There are an uncountable number of such ways, but we will consider four di\ufffd", "\ufffd\ufffderent possibilities: the introduction of a cosmological constant, higher-order terms in the action, gravitational scalar \ufb01elds, and a nonvanishing torsion tensor. The \ufb01rst possibility is the cosmological constant; George Gamow has quoted Einstein as calling this the biggest mistake of his life. Recall that in our search for the simplest possible action for gravity we noted that any nontrivial scalar had to be of at least second order in derivatives of the metric; at lower order all we can create is a constant. Although a constant does not by itself lead to very interesting dynamics, it has an important e\ufb00ect if we add it to the conventional Hilbert action. We therefore consider an action given by where \u039b is some constant. The resulting \ufb01eld equations are S = dnx\u221a Z g(R \u2212 \u2212 2\u039b), R\u00b5\u03bd \u2212 1 2 Rg\u00b5\u03bd + \u039bg\u00b5\u03bd = 0, (4.73) (4.74) and of course there would be an energy-momentum tensor on the right hand side if we had included an action for matter. \u039b is the cosmological constant; it was originally introduced by Einstein after it became clear that there were no solutions to his equations representing a static cosmology (a universe unchanging with time on large scales) with a nonzero matter content. If the cosmological constant is tuned just right, it is possible to \ufb01nd a static solution, but it is unstable to small perturbations. Furthermore, once Hubble demonstrated that the universe is expanding, it became less important to \ufb01nd static solutions, and Einstein rejected his suggestion. Like Rasputin, however, the cosmological constant has proven di\ufb03cult to kill o\ufb00. If we like we can move the additional term in (4.74) to the right hand side, and think of it as a kind of energy-momentum tensor, with T\u00b5\u03bd = \u039bg\u00b5\u03bd (it is automatically conserved by metric compatibility). Then \u039b can be interpreted as the \u201cenergy density of the vacuum,\u201d a source of energy and momentum that is present even in the absence of matter \ufb01elds. This interpretation is important because quantum \ufb01eld theory predicts that the vacuum should In ordinary quantum mechanics, an harmonic have some sort of energy and momentum. oscillator with frequency \u03c9 and minimum classical energy E0 = 0 upon quantization has a ground state with energy E0 = 1 2\u00afh\u03c9. A quantized \ufb01eld can be thought of as a collection of an in\ufb01nite number of harmonic oscillators, and each mode contributes to the ground state energy. The result is of course in\ufb01nite, and must be appropriately regularized, for example \u2212 4 GRAVITATION 119 by introducing a cuto\ufb00 at high frequencies. The \ufb01nal vacuum energy, which is the regularized sum of the energies of the ground state oscillations of all the \ufb01elds of the theory, has no good reason to be zero and in fact would be expected to have a natural scale m4 P, \u039b \u223c (4.75) 5 grams. Observations of the where the Planck mass mP is approximately 1019 GeV, or 10\u2212 universe on large scales allow us to constrain the actual value of \u039b, which turns out to be smaller than (4.75) by at least a factor of 10120. This is the largest known discrepancy between theoretical estimate and observational constraint in physics, and convinces many people that the \u201ccosmological constant problem\u201d is one of the most important unsolved problems today. On the other hand the observations do not tell us that \u039b is strictly zero, and in fact allow values that can have important consequences for the evolution of the universe. This mistake of Einstein\u2019s therefore continues to bedevil both physicists, who would like to understand why it is so small, and astronomers, who would like to determine whether it is really small enough to be ignored. A somewhat less intriguing generalization of the Hilbert action would be to include scalars of more than second order in derivatives of the metric. We could imagine an action of the form S = dnx\u221a g(R + \u03b11R2 + \u03b12R\u00b5\u03bdR\u00b5\u03bd + \u03b13g\u00b5\u03bd \u2212 \u2207\u00b5R \u2207\u03bdR + ), \u00b7 \u00b7 \u00b7 (4.76) Z where the \u03b1\u2019s are coupling constants and the dots represent every other scalar we can make from the curvature tensor, its contractions, and its derivatives. Traditionally, such terms have been neglected on the reasonable grounds that they merely complicate", " a theory which is already both aesthetically pleasing and empirically successful. However, there are at least three more substantive reasons for this neglect. First, as we shall see below, Einstein\u2019s equations lead to a well-posed initial value problem for the metric, in which \u201ccoordinates\u201d and \u201cmomenta\u201d speci\ufb01ed at an initial time can be used to predict future evolution. With higherderivative terms, we would require not only those data, but also some number of derivatives of the momenta. Second, the main source of dissatisfaction with general relativity on the part of particle physicists is that it cannot be renormalized (as far as we know), and Lagrangians with higher derivatives tend generally to make theories less renormalizable rather than more. Third, by the same arguments we used above when speaking about the limitations of the principle of equivalence, the extra terms in (4.76) should be suppressed (by powers of the Planck mass to some power) relative to the usual Hilbert term, and therefore would not be expected to be of any practical importance to the low-energy world. None of these reasons are completely persuasive, and indeed people continue to consider such theories, but for the most part these models do not attract a great deal of attention. A set of models which does attract attention are known as scalar-tensor theories of gravity, since they involve both the metric tensor g\u00b5\u03bd and a fundamental scalar \ufb01eld, \u03bb. The 4 GRAVITATION action can be written S = dnx\u221a Z f (\u03bb)R + g \u2212 1 2 g\u00b5\u03bd(\u2202\u00b5\u03bb)(\u2202\u03bd\u03bb) V (\u03bb), \u2212 120 (4.77) where f (\u03bb) and V (\u03bb) are functions which de\ufb01ne the theory. Recall from (4.68) that the coe\ufb03cient of the Ricci scalar in conventional GR is proportional to the inverse of Newton\u2019s constant G. In scalar-tensor theories, then, where this coe\ufb03cient is replaced by some function of a \ufb01eld which can vary throughout spacetime, the \u201cstrength\u201d of gravity (as measured by the local value of Newton\u2019s constant) will be di\ufb00erent from place to place and time to time. In fact the most famous scalar-tensor theory, invented by Brans and Dicke and now named after them, was inspired by a suggestion of Dirac\u2019s that the gravitational constant varies with time. Dirac had noticed that there were some interesting numerical coincidences one could discover by taking combinations of cosmological numbers such as the Hubble constant H0 (a measure of the expansion rate of the universe) and typical particle-physics parameters such as the mass of the pion, m\u03c0. For example, m3 \u03c0 H0 \u223c cG \u00afh2. (4.78) If we assume for the moment that this relation is not simply an accident, we are faced with the problem that the Hubble \u201cconstant\u201d actually changes with time (in most cosmological models), while the other quantities conventionally do not. Dirac therefore proposed that in fact G varied with time, in such a way as to maintain (4.78); satisfying this proposal was the motivation of Brans and Dicke. These days, experimental test of general relativity are su\ufb03ciently precise that we can state with con\ufb01dence that, if Brans-Dicke theory is correct, the predicted change in G over space and time must be very small, much slower than that (See Weinberg for details on Brans-Dicke theory necessary to satisfy Dirac\u2019s hypothesis. and experimental tests.) Nevertheless there is still a great deal of work being done on other kinds of scalar-tensor theories, which turn out to be vital in superstring theory and may have important consequences in the very early universe. As a \ufb01nal alternative to general relativity, we should mention the possibility that the connection really is not derived from the metric, but in fact has an independent existence as a fundamental \ufb01eld. We will leave it as an exercise for you to show that it is possible to consider the conventional action for general relativity but treat it as a function of both the metric g\u00b5\u03bd and a torsion-free connection \u0393\u03bb \u03c1\u03c3, and the equations of motion derived from varying such an action with respect to the connection imply that \u0393\u03bb \u03c1\u03c3 is actually the Christo\ufb00el connection associated with g\u00b5\u03bd. We could drop the demand that the connection be torsionfree, in which case the torsion tensor could lead to additional propagating degrees of freedom. Without going into details,", " the basic reason why such theories do not receive much attention is simply because the torsion is itself a tensor; there is nothing to distinguish it from other, 4 GRAVITATION 121 \u201cnon-gravitational\u201d tensor \ufb01elds. Thus, we do not really lose any generality by considering theories of torsion-free connections (which lead to GR) plus any number of tensor \ufb01elds, which we can name what we like. With the possibility in mind that one of these alternatives (or, more likely, something we have not yet thought of) is actually realized in nature, for the rest of the course we will work under the assumption that general relativity as based on Einstein\u2019s equations or the Hilbert action is the correct theory, and work out its consequences. These consequences, of course, are constituted by the solutions to Einstein\u2019s equations for various sources of energy and momentum, and the behavior of test particles in these solutions. Before considering speci\ufb01c solutions in detail, lets look more abstractly at the initial-value problem in general relativity. In classical Newtonian mechanics, the behavior of a single particle is of course governed by f = ma. If the particle is moving under the in\ufb02uence of some potential energy \ufb01eld \u03a6(x), then the force is f = \u03a6, and the particle obeys \u2212\u2207 m d2xi dt2 = \u2202i\u03a6. \u2212 (4.79) This is a second-order di\ufb00erential equation for xi(t), which we can recast as a system of two coupled \ufb01rst-order equations by introducing the momentum p: dpi dt dxi dt = = \u2202i\u03a6 \u2212 1 m pi. (4.80) The initial-value problem is simply the procedure of specifying a \u201cstate\u201d (xi, pi) which serves as a boundary condition with which (4.80) can be uniquely solved. You may think of (4.80) as allowing you, once you are given the coordinates and momenta at some time t, to evolve them forward an in\ufb01nitesimal amount to a time t + \u03b4t, and iterate this procedure to obtain the entire solution. We would like to formulate the analogous problem in general relativity. Einstein\u2019s equations G\u00b5\u03bd = 8\u03c0GT\u00b5\u03bd are of course covariant; they don\u2019t single out a preferred notion of \u201ctime\u201d through which a state can evolve. Nevertheless, we can by hand pick a spacelike hypersurface (or \u201cslice\u201d) \u03a3, specify initial data on that hypersurface, and see if we can evolve uniquely from it to a hypersurface in the future. (\u201cHyper\u201d because a constant-time slice in four dimensions will be three-dimensional, whereas \u201csurfaces\u201d are conventionally two-dimensional.) This process does violence to the manifest covariance of the theory, but if we are careful we should wind up with a formulation that is equivalent to solving Einstein\u2019s equations all at once throughout spacetime. Since the metric is the fundamental variable, our \ufb01rst guess is that we should consider the values g\u00b5\u03bd|\u03a3 of the metric on our hypersurface to be the \u201ccoordinates\u201d and the time 4 GRAVITATION 122 t \u03a3 Initial Data derivatives \u2202tg\u00b5\u03bd|\u03a3 (with respect to some speci\ufb01ed time coordinate) to be the \u201cmomenta\u201d, which together specify the state. (There will also be coordinates and momenta for the matter \ufb01elds, which we will not consider explicitly.) In fact the equations G\u00b5\u03bd = 8\u03c0GT\u00b5\u03bd do involve second derivatives of the metric with respect to time (since the connection involves \ufb01rst derivatives of the metric and the Einstein tensor involves \ufb01rst derivatives of the connection), \u2207\u00b5G\u00b5\u03bd = 0. so we seem to be on the right track. However, the Bianchi identity tells us that We can rewrite this equation as \u22020G0\u03bd = \u2202iGi\u03bd \u2212 \u2212 \u0393\u00b5 \u00b5\u03bbG\u03bb\u03bd \u2212 \u00b5\u03bbG\u00b5\u03bb. \u0393\u03bd (4.81) A close look at the right hand side reveals that there are no third-order time derivatives; therefore there cannot be any on the left hand side. Thus, although G\u00b5\u03bd as a whole involves second-order time derivatives of the metric, the speci\ufb01c components G0\u03bd do not. Of the ten independent components in Einstein\u2019s equations, the four represented by G0\u03bd = 8\u03c0GT 0\u03bd (4", ".82) cannot be used to evolve the initial data (g\u00b5\u03bd, \u2202tg\u00b5\u03bd)\u03a3. Rather, they serve as constraints on this initial data; we are not free to specify any combination of the metric and its time derivatives on the hypersurface \u03a3, since they must obey the relations (4.82). The remaining equations, Gij = 8\u03c0GT ij (4.83) are the dynamical evolution equations for the metric. Of course, these are only six equations for the ten unknown functions g\u00b5\u03bd(x\u03c3), so the solution will inevitably involve a fourfold ambiguity. This is simply the freedom that we have already mentioned, to choose the four coordinate functions throughout spacetime. It is a straightforward but unenlightening exercise to sift through (4.83) to \ufb01nd that t gij appears in t g0\u03bd. Therefore a \u201cstate\u201d in general relativity will consist of a speci\ufb01cation not all second time derivatives of the metric appear. In fact we \ufb01nd that \u22022 (4.83), but not \u22022 4 GRAVITATION 123 of the spacelike components of the metric gij|\u03a3 and their \ufb01rst time derivatives \u2202tgij|\u03a3 on the hypersurface \u03a3, from which we can determine the future evolution using (4.83), up to an unavoidable ambiguity in \ufb01xing the remaining components g0\u03bd. The situation is precisely analogous to that in electromagnetism, where we know that no amount of initial data can su\ufb03ce to determine the evolution uniquely since there will always be the freedom to perform a A\u00b5 +\u2202\u00b5\u03bb. In general relativity, then, coordinate transformations gauge transformation A\u00b5 \u2192 play a role reminiscent of gauge transformations in electromagnetism, in that they introduce ambiguity into the time evolution. One way to cope with this problem is to simply \u201cchoose a gauge.\u201d In electromagnetism this means to place a condition on the vector potential A\u00b5, which will restrict our freedom to perform gauge transformations. For example we can choose Lorentz gauge, in which \u2207\u00b5A\u00b5 = 0, or temporal gauge, in which A0 = 0. We can do a similar thing in general relativity, by \ufb01xing our coordinate system. A popular choice is harmonic gauge (also known as Lorentz gauge and a host of other names), in which 2x\u00b5 = 0. (4.84) \u00b5 Here 2 = \u2207\u00b5 is the covariant D\u2019Alembertian, and it is crucial to realize when we take the covariant derivative that the four functions x\u00b5 are just functions, not components of a vector. This condition is therefore simply \u2207 0 = 2x\u00b5 = g\u03c1\u03c3\u2202\u03c1\u2202\u03c3x\u00b5 \u03c1\u03c3. = g\u03c1\u03c3\u0393\u03bb \u2212 g\u03c1\u03c3\u0393\u03bb \u03c1\u03c3\u2202\u03bbx\u00b5 \u2212 (4.85) In \ufb02at space, of course, Cartesian coordinates (in which \u0393\u03bb \u03c1\u03c3 = 0) are harmonic coordinates. (As a general principle, any function f which satis\ufb01es 2f = 0 is called an \u201charmonic function.\u201d) To see that this choice of coordinates successfully \ufb01xes our gauge freedom, let\u2019s rewrite the condition (4.84) in a somewhat simpler form. We have g\u03c1\u03c3\u0393\u00b5 \u03c1\u03c3 = 1 2 g\u03c1\u03c3g\u00b5\u03bd(\u2202\u03c1g\u03c3\u03bd + \u2202\u03c3g\u03bd\u03c1 \u2212 \u2202\u03bdg\u03c1\u03c3), (4.86) from the de\ufb01nition of the Christo\ufb00el symbols. Meanwhile, from \u2202\u03c1(g\u00b5\u03bdg\u03c3\u03bd) = \u2202\u03c1\u03b4\u00b5 have \u03c3 = 0 we \u2212 Also, from our previous exploration of the variation of the determinant of the metric (4.65), we have g\u00b5\u03bd\u2202\u03c1g\u03c3\u03bd = g\u03c3\u03bd\u2202\u03c1g\u00b5\u03bd. (4.87) 1 2 g\u03c1\u03c3\u2202\u03bdg\u03c1\u03c3 = \u2212 1 \u221a g \u2212 \u2202\u03bd\u221a g. \u2212 (4.88) 4 GRAVITATION Putting it all together, we \ufb01nd that (in general), g\u03c1\u03c3\u0393\u00b5 \u03c1\u03c3 = 1 \u221a g \u2212 \u2202\u03bb(\u221a gg\u03bb\u00b5). \u2212 The harmonic gauge condition (4.85) therefore is equivalent to Taking the partial derivative of this with respect to t = x0 yields \u2202\u03bb(\u221a gg\u03bb\u00b5) = 0. \u2212 \u22022 \u2202", "t2 (\u221a \u2212 gg0\u03bd) = \u2202 \u2202xi \" \u2202 \u2202t \u2212 (\u221a ggi\u03bd) \u2212. # 124 (4.89) (4.90) (4.91) This condition represents a second-order di\ufb00erential equation for the previously unconstrained metric components g0\u03bd, in terms of the given initial data. We have therefore succeeded in \ufb01xing our gauge freedom, in that we can now solve for the evolution of the entire metric in harmonic coordinates. (At least locally; we have been glossing over the fact our gauge choice may not be well-de\ufb01ned globally, and we would have to resort to working in patches as usual. The same problem appears in gauge theories in particle physics.) Note that we still have some freedom remaining; our gauge condition (4.84) restricts how the coordinates stretch from our initial hypersurface \u03a3 throughout spacetime, but we can still choose coordinates xi on \u03a3 however we like. This corresponds to the fact that making a x\u00b5 + \u03b4\u00b5, with 2\u03b4\u00b5 = 0, does not violate the harmonic gauge coordinate transformation x\u00b5 condition. \u2192 We therefore have a well-de\ufb01ned initial value problem for general relativity; a state is speci\ufb01ed by the spacelike components of the metric and their time derivatives on a spacelike hypersurface \u03a3; given these, the spacelike components (4.83) of Einstein\u2019s equations allow us to evolve the metric forward in time, up to an ambiguity in coordinate choice which may be resolved by choice of gauge. We must keep in mind that the initial data are not arbitrary, but must obey the constraints (4.82). (Once we impose the constraints on some spacelike hypersurface, the equations of motion guarantee that they remain satis\ufb01ed, as you can check.) The constraints serve a useful purpose, of guaranteeing that the result remains spacetime covariant after we have split our manifold into \u201cspace\u201d and \u201ctime.\u201d Speci\ufb01cally, the Gi0 = 8\u03c0GT i0 constraint implies that the evolution is independent of our choice of coordinates on \u03a3, while G00 = 8\u03c0GT 00 enforces invariance under di\ufb00erent ways of slicing spacetime into spacelike hypersurfaces. Once we have seen how to cast Einstein\u2019s equations as an initial value problem, one issue of crucial importance is the existence of solutions to the problem. That is, once we have speci\ufb01ed a spacelike hypersurface with initial data, to what extent can we be guaranteed that a unique spacetime will be determined? Although one can do a great deal of hard work 4 GRAVITATION 125 \u03a3 to answer this question with some precision, it is fairly simple to get a handle on the ways in which a well-de\ufb01ned solution can fail to exist, which we now consider. It is simplest to \ufb01rst consider the problem of evolving matter \ufb01elds on a \ufb01xed background spacetime, rather than the evolution of the metric itself. We therefore consider a spacelike hypersurface \u03a3 in some manifold M with \ufb01xed metric g\u00b5\u03bd, and furthermore look at some connected subset S in \u03a3. Our guiding principle will be that no signals can travel faster than the speed of light; therefore \u201cinformation\u201d will only \ufb02ow along timelike or null trajectories (not necessarily geodesics). We de\ufb01ne the future domain of dependence of S, denoted D+(S), as the set of all points p such that every past-moving, timelike or null, inextendible curve through p must intersect S. (\u201cInextendible\u201d just means that the curve goes on forever, not ending at some \ufb01nite point.) We interpret this de\ufb01nition in such a way that S itself is a subset of D+(S). (Of course a rigorous formulation does not require additional interpretation over and above the de\ufb01nitions, but we are not being as rigorous as we could be right now.) Similarly, we de\ufb01ne the past domain of dependence D\u2212(S) in the same way, but with \u201cpastmoving\u201d replaced by \u201cfuture-moving.\u201d Generally speaking, some points in M will be in one of the domains of dependence, and some will be outside; we de\ufb01ne the boundary of D+(S) to be the future Cauchy horizon H +(S), and likewise the boundary of D\u2212(S) to be the past Cauchy horizon H \u2212(S). You can convince yourself that they", " are both null surfaces. + H (S) + D (S) \u03a3 S H (S) D (S) 4 GRAVITATION 126 The usefulness of these de\ufb01nitions should be apparent; if nothing moves faster than light, than signals cannot propagate outside the light cone of any point p. Therefore, if every curve which remains inside this light cone must intersect S, then information speci\ufb01ed on S should be su\ufb03cient to predict what the situation is at p. (That is, initial data for matter \ufb01elds given on S can be used to solve for the value of the \ufb01elds at p.) The set of all points for which we can predict what happens by knowing what happens on S is simply the union D+(S) D\u2212(S). We can easily extend these ideas from the subset S to the entire hypersurface \u03a3. The important point is that D+(\u03a3) D\u2212(\u03a3) might fail to be all of M, even if \u03a3 itself seems like a perfectly respectable hypersurface that extends throughout space. There are a number of ways in which this can happen. One possibility is that we have just chosen a \u201cbad\u201d hypersurface (although it is hard to give a general prescription for when a hypersurface is bad in this sense). Consider Minkowski space, and a spacelike hypersurface \u03a3 which remains to the past of the light cone of some point. \u222a \u222a \u03a3 D ( )\u03a3+ In this case \u03a3 is a nice spacelike surface, but it is clear that D+(\u03a3) ends at the light cone, and we cannot use information on \u03a3 to predict what happens throughout Minkowski space. Of course, there are other surfaces we could have picked for which the domain of dependence would have been the entire manifold, so this doesn\u2019t worry us too much. A somewhat more nontrivial example is known as Misner space. This is a twodimensional spacetime with the topology of R1 S1, and a metric for which the light cones progressively tilt as you go forward in time. Past a certain point, it is possible to travel on a timelike trajectory which wraps around the S1 and comes back to itself; this is known as a closed timelike curve. If we had speci\ufb01ed a surface \u03a3 to this past of this point, then none of the points in the region containing closed timelike curves are in the domain of dependence of \u03a3, since the closed timelike curves themselves do not intersect \u03a3. This is obviously a worse problem than the previous one, since a well-de\ufb01ned initial value problem does not seem to \u00d7 4 GRAVITATION 127 identify closed timelike curve Misner space \u03a3 exist in this spacetime. (Actually problems like this are the subject of some current research interest, so I won\u2019t claim that the issue is settled.) A \ufb01nal example is provided by the existence of singularities, points which are not in the manifold even though they can be reached by travelling along a geodesic for a \ufb01nite distance. Typically these occur when the curvature becomes in\ufb01nite at some point; if this happens, the point can no longer be said to be part of the spacetime. Such an occurrence can lead to the emergence of a Cauchy horizon \u2014 a point p which is in the future of a singularity cannot be in the domain of dependence of a hypersurface to the past of the singularity, because there will be curves from p which simply end at the singularity. + \u03a3 H ( ) \u03a3+ D ( ) \u03a3 All of these obstacles can also arise in the initial value problem for GR, when we try to evolve the metric itself from initial data. However, they are of di\ufb00erent degrees of trouble- 4 GRAVITATION 128 someness. The possibility of picking a \u201cbad\u201d initial hypersurface does not arise very often, especially since most solutions are found globally (by solving Einstein\u2019s equations throughout spacetime). The one situation in which you have to be careful is in numerical solution of Einstein\u2019s equations, where a bad choice of hypersurface can lead to numerical di\ufb03culties even if in principle a complete solution exists. Closed timelike curves seem to be something that GR works hard to avoid \u2014 there are certainly solutions which contain them, but evolution from generic initial data does not usually produce them. Singularities, on the other hand, are practically unavoidable. The simple fact that the gravitational force is always attractive tends to pull matter together, increasing the curvature, and generally leading to some sort of singularity. This is something which we apparently must learn to live with", ", although there is some hope that a well-de\ufb01ned theory of quantum gravity will eliminate the singularities of classical GR. December 1997 Lecture Notes on General Relativity Sean M. Carroll 5 More Geometry With an understanding of how the laws of physics adapt to curved spacetime, it is undeniably tempting to start in on applications. However, a few extra mathematical techniques will simplify our task a great deal, so we will pause brie\ufb02y to explore the geometry of manifolds some more. When we discussed manifolds in section 2, we introduced maps between two di\ufb00erent manifolds and how maps could be composed. We now turn to the use of such maps in carrying along tensor \ufb01elds from one manifold to another. We therefore consider two manifolds M and N, possibly of di\ufb00erent dimension, with coordinate systems x\u00b5 and y\u03b1, respectively. We imagine that we have a map \u03c6 : M N and a function f : N R y\u03b1 n R It is obvious that we can compose \u03c6 with f to construct a map (f R, which is simply a function on M. Such a construction is su\ufb03ciently useful that it gets its own name; we de\ufb01ne the pullback of f by \u03c6, denoted \u03c6 \u03c6) : M f, by \u2192 \u25e6 \u2217 f = (f \u03c6 \u2217 \u25e6 \u03c6). (5.1) The name makes sense, since we think of \u03c6 as \u201cpulling back\u201d the function f from N to M. We can pull functions back, but we cannot push them forward. If we have a function g : M R, there is no way we can compose g with \u03c6 to create a function on N; the arrows don\u2019t \ufb01t together correctly. But recall that a vector can be thought of as a derivative operator that maps smooth functions to real numbers. This allows us to de\ufb01ne the pushforward of \u2192 \u2217 129 5 MORE GEOMETRY 130 a vector; if V (p) is a vector at a point p on M, we de\ufb01ne the pushforward vector \u03c6\u2217V at the point \u03c6(p) on N by giving its action on functions on N: (\u03c6\u2217V )(f ) = V (\u03c6 f ). \u2217 (5.2) So to push forward a vector \ufb01eld we say \u201cthe action of \u03c6\u2217V on any function is simply the action of V on the pullback of that function.\u201d This is a little abstract, and it would be nice to have a more concrete description. We know that a basis for vectors on M is given by the set of partial derivatives \u2202\u00b5 = \u2202 \u2202x\u00b5, and a basis on N is given by the set of partial derivatives \u2202\u03b1 = \u2202 \u2202y\u03b1. Therefore we would like to relate the components of V = V \u00b5\u2202\u00b5 to those of (\u03c6\u2217V ) = (\u03c6\u2217V )\u03b1\u2202\u03b1. We can \ufb01nd the sought-after relation by applying the pushed-forward vector to a test function and using the chain rule (2.3): (\u03c6\u2217V )\u03b1\u2202\u03b1f = V \u00b5\u2202\u00b5(\u03c6 = V \u00b5\u2202\u00b5(f = V \u00b5 \u2202y\u03b1 f ) \u2217 \u03c6) \u25e6 \u2202x\u00b5 \u2202\u03b1f. (5.3) This simple formula makes it irresistible to think of the pushforward operation \u03c6\u2217 as a matrix operator, (\u03c6\u2217V )\u03b1 = (\u03c6\u2217)\u03b1 \u00b5V \u00b5, with the matrix being given by (\u03c6\u2217)\u03b1 \u00b5 = \u2202y\u03b1 \u2202x\u00b5. (5.4) The behavior of a vector under a pushforward thus bears an unmistakable resemblance to the vector transformation law under change of coordinates. In fact it is a generalization, since when M and N are the same manifold the constructions are (as we shall discuss) identical; but don\u2019t be fooled, since in general \u00b5 and \u03b1 have di\ufb00erent allowed values, and there is no reason for the matrix \u2202y\u03b1/\u2202x\u00b5 to be invertible. \u2192 It is a rewarding exercise to convince yourself that, although you can push vectors forward from M to N (given a map \u03c6 : M N), you cannot in general pull them back \u2014 just keep trying to invent an appropriate construction until the futility of the attempt becomes clear. Since one-forms are dual to vectors, you should not be surprised to hear that one-forms can be pulled back (but not in general pushed forward). To", " do this, remember that one-forms \u03c9 of a one-form \u03c9 on N are linear maps from vectors to the real numbers. The pullback \u03c6 can therefore be de\ufb01ned by its action on a vector V on M, by equating it with the action of \u03c9 itself on the pushforward of V : \u2217 \u03c9)(V ) = \u03c9(\u03c6\u2217V ). (\u03c6 \u2217 Once again, there is a simple matrix description of the pullback operator on forms, (\u03c6 (\u03c6 \u03b1\u03c9\u03b1, which we can derive using the chain rule. It is given by )\u00b5 \u2217 )\u00b5 \u03b1 = (\u03c6 \u2217 \u2202y\u03b1 \u2202x\u00b5. (5.5) \u03c9)\u00b5 = \u2217 (5.6) 5 MORE GEOMETRY 131 That is, it is the same matrix as the pushforward (5.4), but of course a di\ufb00erent index is contracted when the matrix acts to pull back one-forms. There is a way of thinking about why pullbacks and pushforwards work on some objects but not others, which may or may not be helpful. If we denote the set of smooth functions (M), then a vector V (p) at a point p on M (i.e., an element of the tangent space on M by (M) to R. But we already know that the TpM) can be thought of as an operator from (M) (just as \u03c6 itself maps M to N, but pullback operator on functions maps in the opposite direction). Therefore we can de\ufb01ne the pushforward \u03c6 acting on vectors \u2217 simply by composing maps, as we \ufb01rst de\ufb01ned the pullback of functions: (N) to F F F F R V(p) \u03c6*(V(p)) = V(p) \u03c6 * F (M) \u03c6 * F (N) Similarly, if TqN is the tangent space at a point q on N, then a one-form \u03c9 at q (i.e., an element of the cotangent space T \u2217q N) can be thought of as an operator from TqN to R. Since the pushforward \u03c6\u2217 maps TpM to T\u03c6(p)N, the pullback \u03c6 of a one-form can also be thought of as mere composition of maps: \u2217 (\u03c9) = \u03c9 \u03c6* \u03c6 * R \u03c9 \u03c6* T Mp \u03c6(p)T N If this is not helpful, don\u2019t worry about it. But do keep straight what exists and what doesn\u2019t; the actual concepts are simple, it\u2019s just remembering which map goes what way that leads to confusion. You will recall further that a (0, l) tensor \u2014 one with l lower indices and no upper ones \u2014 is a linear map from the direct product of l vectors to R. We can therefore pull back not only one-forms, but tensors with an arbitrary number of lower indices. The de\ufb01nition is simply the action of the original tensor on the pushed-forward vectors: T )(V (1), V (2),..., V (l)) = T (\u03c6\u2217V (1), \u03c6\u2217V (2),..., \u03c6\u2217V (l)), (5.7) (\u03c6 \u2217 5 MORE GEOMETRY 132 where T\u03b11 by acting it on pulled-back one-forms: \u03b1l is a (0, l) tensor on N. We can similarly push forward any (k, 0) tensor S\u00b51 \u00b7\u00b7\u00b7 \u00b5k \u00b7\u00b7\u00b7 (\u03c6\u2217S)(\u03c9(1), \u03c9(2),..., \u03c9(k)) = S(\u03c6 \u03c9(1), \u03c6 \u2217 \u2217 \u03c9(2),..., \u03c6 \u03c9(k)). \u2217 (5.8) Fortunately, the matrix representations of the pushforward (5.4) and pullback (5.6) extend to the higher-rank tensors simply by assigning one matrix to each index; thus, for the pullback of a (0, l) tensor, we have (\u03c6 \u2217 T )\u00b51 \u00b5l = \u00b7\u00b7\u00b7 \u2202y\u03b11 \u2202x\u00b51 \u00b7 \u00b7 \u00b7 \u2202y\u03b1l \u2202x\u00b5l T\u03b11 \u03b1l, \u00b7\u00b7\u00b7 while for the pushforward of a (k, 0) tensor we have (\u03c6\u2217S)\u03b11 \u03b1k = \u00b7\u00b7\u00b7 \u2202y\u03b11 \u2202x\u00b51 \u00b7 \u00b7 \u00b7 \ufffd", "\ufffdy\u03b1k \u2202x\u00b5k S\u00b51 \u00b5k. \u00b7\u00b7\u00b7 Our complete picture is therefore: (5.9) (5.10) M ( )k 0 ( ) 0 l \u03c6* \u03c6 \u03c6 * ( )k 0 ( )0 l N Note that tensors with both upper and lower indices can generally be neither pushed forward nor pulled back. This machinery becomes somewhat less imposing once we see it at work in a simple example. One common occurrence of a map between two manifolds is when M is actually a submanifold of N; then there is an obvious map from M to N which just takes an element of M to the \u201csame\u201d element of N. Consider our usual example, the two-sphere embedded in R3, as the locus of points a unit distance from the origin. If we put coordinates x\u00b5 = (\u03b8, \u03c6) on M = S2 and y\u03b1 = (x, y, z) on N = R3, the map \u03c6 : M N is given by \u2192 \u03c6(\u03b8, \u03c6) = (sin \u03b8 cos \u03c6, sin \u03b8 sin \u03c6, cos \u03b8). (5.11) In the past we have considered the metric ds2 = dx2 + dy2 + dz2 on R3, and said that it induces a metric d\u03b82 + sin2 \u03b8 d\u03c62 on S2, just by substituting (5.11) into this \ufb02at metric on 5 MORE GEOMETRY 133 R3. We didn\u2019t really justify such a statement at the time, but now we can do so. (Of course it would be easier if we worked in spherical coordinates on R3, but doing it the hard way is more illustrative.) The matrix of partial derivatives is given by \u2202y\u03b1 \u2202x\u00b5 = cos \u03b8 cos \u03c6 cos \u03b8 sin \u03c6 sin \u03b8 sin \u03c6 sin \u03b8 cos \u03c6 \u2212 \u2212 sin \u03b8 0. The metric on S2 is obtained by simply pulling back the metric from R3, (\u03c6\u2217g)\u00b5\u03bd = = \u2202y\u03b1 \u2202y\u03b2 \u2202x\u03bd g\u03b1\u03b2 \u2202x\u00b5 1 0 0 sin2 \u03b8, (5.12) (5.13) as you can easily check. Once again, the answer is the same as you would get by naive substitution, but now we know why. We have been careful to emphasize that a map \u03c6 : M N can be used to push certain things forward and pull other things back. The reason why it generally doesn\u2019t work both ways can be traced to the fact that \u03c6 might not be invertible. If \u03c6 is invertible (and both \u03c6 1 are smooth, which we always implicitly assume), then it de\ufb01nes a di\ufb00eomorphism and \u03c6\u2212 between M and N. In this case M and N are the same abstract manifold. The beauty of 1 to move tensors from M to N; this will di\ufb00eomorphisms is that we can use both \u03c6 and \u03c6\u2212 allow us to de\ufb01ne the pushforward and pullback of arbitrary tensors. Speci\ufb01cally, for a (k, l) tensor \ufb01eld T \u00b51 \u00b5l on M, we de\ufb01ne the pushforward by \u2192 \u00b5k \u03bd1 \u00b7\u00b7\u00b7 \u00b7\u00b7\u00b7 (\u03c6\u2217T )(\u03c9(1),..., \u03c9(k), V (1),..., V (l)) = T (\u03c6 \u03c9(1),..., \u03c6 \u03c9(k), [\u03c6\u2212 1]\u2217V (1),..., [\u03c6\u2212 1]\u2217V (l)), \u2217 \u2217 (\u03c6\u2217T )\u03b11 (5.14) where the \u03c9(i)\u2019s are one-forms on N and the V (i)\u2019s are vectors on N. In components this becomes \u2202y\u03b1k \u2202x\u00b5k The appearance of the inverse matrix \u2202x\u03bd /\u2202y\u03b2 is legitimate because \u03c6 is invertible. Note that we could also de\ufb01ne the pullback in the obvious way, but there is no need to write separate equations because the pullback \u03c6 is the same as the pushforward via the inverse map, [\u03c6\u2212 \u2202y\u03b11 \u2202x\u00b51 \u00b7 \u00b7 \u00b7 \u2202x\u03bd1 \u2202y\u03b21 \u00b7 \u00b7 \u00b7 \u2202x\u03bdl \u2202y\u03b2l (5.15) \u03b2l = T \u00b51 1]\u2217. \u03bdl.", " \u03b1k \u00b5k \u03b21 \u03bd1 \u00b7\u00b7\u00b7 \u00b7\u00b7\u00b7 \u00b7\u00b7\u00b7 \u00b7\u00b7\u00b7 \u2217 We are now in a position to explain the relationship between di\ufb00eomorphisms and coordinate transformations. The relationship is that they are two di\ufb00erent ways of doing precisely the same thing. If you like, di\ufb00eomorphisms are \u201cactive coordinate transformations\u201d, while traditional coordinate transformations are \u201cpassive.\u201d Consider an n-dimensional manifold Rn. To change coordinates we can either simply M with coordinate functions x\u00b5 : M \u2192 Rn (\u201ckeep the manifold \ufb01xed, change the coordinate introduce new functions y\u00b5 : M \u2192 5 MORE GEOMETRY 134 M, after which the maps\u201d), or we could just as well introduce a di\ufb00eomorphism \u03c6 : M Rn (\u201cmove the points on the mancoordinates would just be the pullbacks (\u03c6 ifold, and then evaluate the coordinates of the new points\u201d). In this sense, (5.15) really is the tensor transformation law, just thought of from a di\ufb00erent point of view. x)) * \u2192 M and a tensor \ufb01eld T \u00b51 Since a di\ufb00eomorphism allows us to pull back and push forward arbitrary tensors, it provides another way of comparing tensors at di\ufb00erent points on a manifold. Given a di\ufb00eo\u00b5k \u00b5l(x), we can form the di\ufb00erence between morphism \u03c6 : M \u00b7\u00b7\u00b7 \u00b5l(\u03c6(p))], its value at \u03c6(p) pulled the value of the tensor at some point p and \u03c6 back to p. This suggests that we could de\ufb01ne another kind of derivative operator on tensor \ufb01elds, one which categorizes the rate of change of the tensor as it changes under the di\ufb00eomorphism. For that, however, a single discrete di\ufb00eomorphism is insu\ufb03cient; we require a one-parameter family of di\ufb00eomorphisms, \u03c6t. This family can be thought of as a smooth \u03c6t = \u03c6s+t. Note map R that this last condition implies that \u03c60 is the identity map. R \u03c6t is a di\ufb00eomorphism and \u03c6s \u25e6 M, such that for each t \u03bd1 \u00b7\u00b7\u00b7 [T \u00b51 \u2192 M \u00d7 \u2208 \u00b5k \u03bd1 \u00b7\u00b7\u00b7 \u00b7\u00b7\u00b7 \u2217 One-parameter families of di\ufb00eomorphisms can be thought of as arising from vector \ufb01elds (and vice-versa). If we consider what happens to the point p under the entire family \u03c6t, it is clear that it describes a curve in M; since the same thing will be true of every point on M, these curves \ufb01ll the manifold (although there can be degeneracies where the di\ufb00eomorphisms have \ufb01xed points). We can de\ufb01ne a vector \ufb01eld V \u00b5(x) to be the set of tangent vectors to each of these curves at every point, evaluated at t = 0. An example on S2 is provided by the di\ufb00eomorphism \u03c6t(\u03b8, \u03c6) = (\u03b8, \u03c6 + t). We can reverse the construction to de\ufb01ne a one-parameter family of di\ufb00eomorphisms from any vector \ufb01eld. Given a vector \ufb01eld V \u00b5(x), we de\ufb01ne the integral curves of the vector \ufb01eld to be those curves x\u00b5(t) which solve Note that this familiar-looking equation is now to be interpreted in the opposite sense from our usual way \u2014 we are given the vectors, from which we de\ufb01ne the curves. Solutions to dx\u00b5 dt = V \u00b5. (5.16) 5 MORE GEOMETRY 135 \u03c6 (5.16) are guaranteed to exist as long as we don\u2019t do anything silly like run into the edge of our manifold; any standard di\ufb00erential geometry text will have the proof, which amounts to \ufb01nding a clever coordinate system in which the problem reduces to the fundamental theorem of ordinary di\ufb00erential equations. Our di\ufb00eomorphisms \u03c6t represent \u201c\ufb02ow down the integral curves,\u201d and the associated vector \ufb01eld is referred to as the generator of the di\ufb00eomorphism. (Integral curves are used all the time in elementary physics, just not given the name. The \ufffd", "\ufffdlines of magnetic \ufb02ux\u201d traced out by iron \ufb01lings in the presence of a magnet are simply the integral curves of the magnetic \ufb01eld vector B.) Given a vector \ufb01eld V \u00b5(x), then, we have a family of di\ufb00eomorphisms parameterized by t, and we can ask how fast a tensor changes as we travel down the integral curves. For each t we can de\ufb01ne this change as \u2206tT \u00b51 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u00b5l(p) = \u03c6t \u2217 \u00b7\u00b7\u00b7 [T \u00b51 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u00b5l(\u03c6t(p))] \u00b7\u00b7\u00b7 \u2212 T \u00b51 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u00b5l(p). \u00b7\u00b7\u00b7 (5.17) Note that both terms on the right hand side are tensors at p. M \u03c6 t * \u03c6 [T( (p))] t T(p) p \u03c6 T[ (p)] t \u00b5 x (t) \u03c6 t (p) We then de\ufb01ne the Lie derivative of the tensor along the vector \ufb01eld as \u00a3V T \u00b51 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u00b5l = limt 0 \u00b7\u00b7\u00b7 \u2192 \u2206t T \u00b51 \u00b5k \u00b7\u00b7\u00b7 t \u03bd1 \u00b5l \u00b7\u00b7\u00b7. (5.18) 5 MORE GEOMETRY 136 The Lie derivative is a map from (k, l) tensor \ufb01elds to (k, l) tensor \ufb01elds, which is manifestly independent of coordinates. Since the de\ufb01nition essentially amounts to the conventional de\ufb01nition of an ordinary derivative applied to the component functions of the tensor, it should be clear that it is linear, \u00a3V (aT + bS ) = a\u00a3V T + b\u00a3V S, and obeys the Leibniz rule, \u00a3V (T \u2297 S ) = (\u00a3V T ) S + T \u2297 \u2297 (\u00a3V S ), (5.19) (5.20) where S and T are tensors and a and b are constants. The Lie derivative is in fact a more primitive notion than the covariant derivative, since it does not require speci\ufb01cation of a connection (although it does require a vector \ufb01eld, of course). A moment\u2019s re\ufb02ection shows that it reduces to the ordinary derivative on functions, \u00a3V f = V (f ) = V \u00b5\u2202\u00b5f. (5.21) To discuss the action of the Lie derivative on tensors in terms of other operations we know, it is convenient to choose a coordinate system adapted to our problem. Speci\ufb01cally, we will work in coordinates x\u00b5 for which x1 is the parameter along the integral curves (and the other coordinates are chosen any way we like). Then the vector \ufb01eld takes the form V = \u2202/\u2202x1; that is, it has components V \u00b5 = (1, 0, 0,..., 0). The magic of this coordinate system is that a di\ufb00eomorphism by t amounts to a coordinate transformation from x\u00b5 to y\u00b5 = (x1 + t, x2,..., xn). Thus, from (5.6) the pullback matrix is simply (\u03c6t \u2217 )\u00b5 \u03bd = \u03b4\u03bd \u00b5, and the components of the tensor pulled back from \u03c6t(p) to p are simply \u03c6t \u2217 [T \u00b51 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u00b5l(\u03c6t(p))] = T \u00b51 \u00b7\u00b7\u00b7 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u00b5l(x1 + t, x2,..., xn). \u00b7\u00b7\u00b7 In this coordinate system, then, the Lie derivative becomes \u2202 \u2202x 1 T \u00b51 and speci\ufb01cally the derivative of a vector \ufb01eld U \u00b5(x) is \u00a3V T \u00b51 \u00b5l = \u00b5k \u03bd1 \u00b7\u00b7\u00b7 \u00b7\u00b7\u00b7 \u00b7\u00b7\u00b7 \u00b5k \u03bd1 \u00b5l, \u00b7\u00b7\u00b7 \u00a3V U \u00b5 = \u2202U \u00b5 \u2202x 1. (5.22) (5.23) (5.24) (5.25) Although this expression is clearly not covariant, we know that the commutator [V, U] is a well-de\ufb01ned tensor, and in this coordinate system [V, U]\u00b5 = V \u03bd\u2202\u03bdU \u00b5 U \u03bd\u2202\u03bdV \u00b5 \u2212 5 MORE GEOMETRY 137 \u2202U \u00b5 \u2202x1. Therefore the Lie", " derivative of U with respect to V has the same components in this coordinate system as the commutator of V and U; but since both are vectors, they must be equal in any coordinate system: (5.26) = \u00a3V U \u00b5 = [V, U ]\u00b5. (5.27) As an immediate consequence, we have \u00a3V S = \u2212 commutator is sometimes called the \u201cLie bracket.\u201d \u00a3W V. It is because of (5.27) that the To derive the action of \u00a3V on a one-form \u03c9\u00b5, begin by considering the action on the scalar \u03c9\u00b5U \u00b5 for an arbitrary vector \ufb01eld U \u00b5. First use the fact that the Lie derivative with respect to a vector \ufb01eld reduces to the action of the vector itself when applied to a scalar: \u00a3V (\u03c9\u00b5U \u00b5) = V (\u03c9\u00b5U \u00b5) = V \u03bd\u2202\u03bd(\u03c9\u00b5U \u00b5) = V \u03bd(\u2202\u03bd\u03c9\u00b5)U \u00b5 + V \u03bd\u03c9\u00b5(\u2202\u03bdU \u00b5). (5.28) Then use the Leibniz rule on the original scalar: \u00a3V (\u03c9\u00b5U \u00b5) = (\u00a3V \u03c9)\u00b5U \u00b5 + \u03c9\u00b5(\u00a3V U )\u00b5 = (\u00a3V \u03c9)\u00b5U \u00b5 + \u03c9\u00b5V \u03bd\u2202\u03bdU \u00b5 \u03c9\u00b5U \u03bd\u2202\u03bdV \u00b5. \u2212 (5.29) Setting these expressions equal to each other and requiring that equality hold for arbitrary U \u00b5, we see that \u00a3V \u03c9\u00b5 = V \u03bd\u2202\u03bd\u03c9\u00b5 + (\u2202\u00b5V \u03bd)\u03c9\u03bd, (5.30) which (like the de\ufb01nition of the commutator) is completely covariant, although not manifestly so. By a similar procedure we can de\ufb01ne the Lie derivative of an arbitrary tensor \ufb01eld. The answer can be written \u00a3V T \u00b51 \u00b52 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u03bd2 \u03bdl = V \u03c3\u2202\u03c3T \u00b51\u00b52 \u00b7\u00b7\u00b7 \u00b5k \u00b7\u00b7\u00b7 \u03bd1\u03bd2 (\u2202\u03bbV \u00b51)T \u03bb\u00b52 \u00b7\u00b7\u00b7 \u2212 +(\u2202\u03bd1V \u03bb)T \u00b51\u00b52 \u00b7\u00b7\u00b7 \u03bdl \u00b7\u00b7\u00b7 \u00b5k \u00b5k \u03bd1\u03bd2 \u03bb\u03bd2 (\u2202\u03bbV \u00b52)T \u00b51\u03bb \u00b7\u00b7\u00b7 \u03bdl \u2212 \u00b7\u00b7\u00b7 \u03bdl + (\u2202\u03bd2V \u03bb)T \u00b51\u00b52 \u00b7\u00b7\u00b7 \u00b7\u00b7\u00b7 \u00b5k \u00b5k \u03bd1\u03bd2 \u03bd1\u03bb \u03bdl \u2212 \u00b7 \u00b7 \u00b7 \u00b7\u00b7\u00b7 \u03bdl + \u00b7 \u00b7 \u00b7 \u00b7\u00b7\u00b7.(5.31) Once again, this expression is covariant, despite appearances. It would undoubtedly be comforting, however, to have an equivalent expression that looked manifestly tensorial. In fact it turns out that we can write \u00a3V T \u00b51 \u00b52 \u00b5k \u00b7\u00b7\u00b7 \u03bd1 \u03bd2 \u03bdl = V \u03c3 \u00b7\u00b7\u00b7 \u00b7\u00b7\u00b7 \u00b5k \u03bd1\u03bd2 \u2207\u03c3T \u00b51\u00b52 \u2207\u03bbV \u00b51)T \u03bb\u00b52 ( \u2212 \u2207\u03bd1V \u03bb)T \u00b51\u00b52 +( \u03bdl \u00b7\u00b7\u00b7 \u00b5k \u00b7\u00b7\u00b7 \u03bd1\u03bd2 \u00b5k \u00b7\u00b7\u00b7 \u03bb\u03bd2 \u2207\u03bbV \u00b52)T \u00b51\u03bb ( \u03bdl \u2212 \u00b7\u00b7\u00b7 \u2207\u03bd2V \u03bb)T \u00b51\u00b52 \u03bdl + ( \u00b7\u00b7\u00b7 \u00b5k \u00b7\u00b7\u00b7 \u00b5k \u00b7\u00b7\u00b7 \u03bd1\u03bd2 \u03bd1\u03bb \u03bdl \u2212 \u00b7 \u00b7 \u00b7 \u00b7\u00b7\u00b7 \u03bdl + \u00b7 \u00b7 \u00b7 \u00b7\u00b7\u00b7,(5.32) 5 MORE GEOMETRY 138 \u2207\u00b5 represents any symmetric (torsion-free) covariant derivative (including, of course, where one derived from a metric). You can check that all of the terms which would involve connection coe\ufb03cients if we were to expand (5.32) would cancel, leaving only (5.31). Both versions of the formula for a Lie derivative are useful at di\ufb00erent times. A particularly useful formula is for the Lie derivative of the metric: \u00a3V g\u00b5\u03bd = V \u03c3 \u2207\u03c3g\u00b5\u03bd + ( \u2207\u03bdV\u00b5 \u2207\u00b5V\u03bd + \u2207(\u00b5V\u03bd), = = 2 \u2207\u00b5V \u03bb)g\u03bb\u03bd + ( \u2207\u03bdV \u03bb)g\u00b5\u03bb (5.33) where", " \u2207\u00b5 is the covariant derivative derived from g\u00b5\u03bd. \u2217 \u2217 g\u00b5\u03bd, \u03c6 Let\u2019s put some of these ideas into the context of general relativity. You will often hear it proclaimed that GR is a \u201cdi\ufb00eomorphism invariant\u201d theory. What this means is that, if the universe is represented by a manifold M with metric g\u00b5\u03bd and matter \ufb01elds \u03c8, and \u03c6 : M \u2192 M is a di\ufb00eomorphism, then the sets (M, g\u00b5\u03bd, \u03c8) and (M, \u03c6 \u03c8) represent the same physical situation. Since di\ufb00eomorphisms are just active coordinate transformations, this is a highbrow way of saying that the theory is coordinate invariant. Although such a statement is true, it is a source of great misunderstanding, for the simple fact that it conveys very little information. Any semi-respectable theory of physics is coordinate invariant, including those based on special relativity or Newtonian mechanics; GR is not unique in this regard. When people say that GR is di\ufb00eomorphism invariant, more likely than not they have one of two (closely related) concepts in mind: the theory is free of \u201cprior geometry\u201d, and there is no preferred coordinate system for spacetime. The \ufb01rst of these stems from the fact that the metric is a dynamical variable, and along with it the connection and volume element and so forth. Nothing is given to us ahead of time, unlike in classical mechanics or SR. As a consequence, there is no way to simplify life by sticking to a speci\ufb01c coordinate system adapted to some absolute elements of the geometry. This state of a\ufb00airs forces us to be very careful; it is possible that two purportedly distinct con\ufb01gurations (of matter and metric) in GR are actually \u201cthe same\u201d, related by a di\ufb00eomorphism. In a path integral approach to quantum gravity, where we would like to sum over all possible con\ufb01gurations, special care must be taken not to overcount by allowing physically indistinguishable con\ufb01gurations to contribute more than once. In SR or Newtonian mechanics, meanwhile, the existence of a preferred set of coordinates saves us from such ambiguities. The fact that GR has no preferred coordinate system is often garbled into the statement that it is coordinate invariant (or \u201cgenerally covariant\u201d); both things are true, but one has more content than the other. On the other hand, the fact of di\ufb00eomorphism invariance can be put to good use. Recall that the complete action for gravity coupled to a set of matter \ufb01elds \u03c8i is given by a sum of the Hilbert action for GR plus the matter action, S = 1 8\u03c0G SH[g\u00b5\u03bd] + SM [g\u00b5\u03bd, \u03c8i]. (5.34) 5 MORE GEOMETRY 139 The Hilbert action SH is di\ufb00eomorphism invariant when considered in isolation, so the matter action SM must also be if the action as a whole is to be invariant. We can write the variation in SM under a di\ufb00eomorphism as \u03b4SM = dnx Z \u03b4SM \u03b4g\u00b5\u03bd \u03b4g\u00b5\u03bd + dnx Z \u03b4SM \u03b4\u03c8i \u03b4\u03c8i. (5.35) We are not considering arbitrary variations of the \ufb01elds, only those which result from a di\ufb00eomorphism. Nevertheless, the matter equations of motion tell us that the variation of SM with respect to \u03c8i will vanish for any variation (since the gravitational part of the action doesn\u2019t involve the matter \ufb01elds). Hence, for a di\ufb00eomorphism invariant theory the \ufb01rst term on the right hand side of (5.35) must vanish. If the di\ufb00eomorphism in generated by a vector \ufb01eld V \u00b5(x), the in\ufb01nitesimal change in the metric is simply given by its Lie derivative along V \u00b5; by (5.33) we have \u03b4g\u00b5\u03bd = \u00a3V g\u00b5\u03bd = 2 \u2207(\u00b5V\u03bd). Setting \u03b4SM = 0 then implies 0 = = dnx \u03b4SM \u03b4g\u00b5\u03bd \u2207\u00b5V\u03bd dnx\u221a gV\u03bd\u2207\u00b5 \u2212 Z \u2212 Z 1 \u221a g \u2212 \u03b4SM \u03b4g\u00b5\u03bd!, (5.36) (5.37) \u2207(", "\u00b5V\u03bd) since \u03b4SM /\u03b4g\u00b5\u03bd is already symmetric. where we are able to drop the symmetrization of Demanding that (5.37) hold for di\ufb00eomorphisms generated by arbitrary vector \ufb01elds V \u00b5, and using the de\ufb01nition (4.70) of the energy-momentum tensor, we obtain precisely the law of energy-momentum conservation, \u2207\u00b5T \u00b5\u03bd = 0. This is why we claimed earlier that the conservation of T\u00b5\u03bd was more than simply a consequence of the Principle of Equivalence; it is much more secure than that, resting only on the di\ufb00eomorphism invariance of the theory. (5.38) There is one more use to which we will put the machinery we have set up in this section: symmetries of tensors. We say that a di\ufb00eomorphism \u03c6 is a symmetry of some tensor T if the tensor is invariant after being pulled back under \u03c6: Although symmetries may be discrete, it is more common to have a one-parameter family of symmetries \u03c6t. If the family is generated by a vector \ufb01eld V \u00b5(x), then (5.39) amounts to T = T. \u03c6 \u2217 (5.39) \u00a3V T = 0. (5.40) 5 MORE GEOMETRY 140 By (5.25), one implication of a symmetry is that, if T is symmetric under some one-parameter family of di\ufb00eomorphisms, we can always \ufb01nd a coordinate system in which the components of T are all independent of one of the coordinates (the integral curve coordinate of the vector \ufb01eld). The converse is also true; if all of the components are independent of one of the coordinates, then the partial derivative vector \ufb01eld associated with that coordinate generates a symmetry of the tensor. The most important symmetries are those of the metric, for which \u03c6 g\u00b5\u03bd = g\u00b5\u03bd. A di\ufb00eomorphism of this type is called an isometry. If a one-parameter family of isometries is generated by a vector \ufb01eld V \u00b5(x), then V \u00b5 is known as a Killing vector \ufb01eld. The condition that V \u00b5 be a Killing vector is thus \u2217 \u00a3V g\u00b5\u03bd = 0, (5.41) or from (5.33), \u2207(\u00b5V\u03bd) = 0. This last version is Killing\u2019s equation. If a spacetime has a Killing vector, then we know we can \ufb01nd a coordinate system in which the metric is independent of one of the coordinates. By far the most useful fact about Killing vectors is that Killing vectors imply conserved quantities associated with the motion of free particles. If x\u00b5(\u03bb) is a geodesic with tangent vector U \u00b5 = dx\u00b5/d\u03bb, and V \u00b5 is a Killing vector, then (5.42) U \u03bd \u2207\u03bd(V\u00b5U \u00b5) = U \u03bdU \u00b5 = 0, \u2207\u03bdV\u00b5 + V\u00b5U \u03bd \u2207\u03bdU \u00b5 (5.43) where the \ufb01rst term vanishes from Killing\u2019s equation and the second from the fact that x\u00b5(\u03bb) is a geodesic. Thus, the quantity V\u00b5U \u00b5 is conserved along the particle\u2019s worldline. This can be understood physically: by de\ufb01nition the metric is unchanging along the direction of the Killing vector. Loosely speaking, therefore, a free particle will not feel any \u201cforces\u201d in this direction, and the component of its momentum in that direction will consequently be conserved. Long ago we referred to the concept of a space with maximal symmetry, without o\ufb00ering a rigorous de\ufb01nition. The rigorous de\ufb01nition is that a maximally symmetric space is one which possesses the largest possible number of Killing vectors, which on an n-dimensional manifold is n(n + 1)/2. We will not prove this statement, but it is easy to understand at an informal level. Consider the Euclidean space Rn, where the isometries are well known to us: translations and rotations. In general there will be n translations, one for each direction we can move. There will also be n(n 1 directions in which we can rotate it, but we must divide by two to prevent overcounting (rotating x into y and rotating y into x are two versions of the same thing). We therefore 1)/2 rotations; for each of n", " dimensions there are n \u2212 \u2212 5 MORE GEOMETRY have n + n(n 1) = n(n + 1) 2 \u2212 2 141 (5.44) independent Killing vectors. The same kind of counting argument applies to maximally symmetric spaces with curvature (such as spheres) or a non-Euclidean signature (such as Minkowski space), although the details are marginally di\ufb00erent. Although it may or may not be simple to actually solve Killing\u2019s equation in any given spacetime, it is frequently possible to write down some Killing vectors by inspection. (Of course a \u201cgeneric\u201d metric has no Killing vectors at all, but to keep things simple we often deal with metrics with high degrees of symmetry.) For example in R2 with metric ds2 = dx2+dy2, independence of the metric components with respect to x and y immediately yields two Killing vectors: X \u00b5 = (1, 0), Y \u00b5 = (0, 1). (5.45) These clearly represent the two translations. The one rotation would correspond to the vector R = \u2202/\u2202\u03b8 if we were in polar coordinates; in Cartesian coordinates this becomes R\u00b5 = ( y, x). \u2212 (5.46) You can check for yourself that this actually does solve Killing\u2019s equation. \u2265 Note that in n 2 dimensions, there can be more Killing vectors than dimensions. This is because a set of Killing vector \ufb01elds can be linearly independent, even though at any one point on the manifold the vectors at that point are linearly dependent. It is trivial to show (so you should do it yourself) that a linear combination of Killing vectors with constant coe\ufb03cients is still a Killing vector (in which case the linear combination does not count as an independent Killing vector), but this is not necessarily true with coe\ufb03cients which vary over the manifold. You will also show that the commutator of two Killing vector \ufb01elds is a Killing vector \ufb01eld; this is very useful to know, but it may be the case that the commutator gives you a vector \ufb01eld which is not linearly independent (or it may simply vanish). The problem of \ufb01nding all of the Killing vectors of a metric is therefore somewhat tricky, as it is sometimes not clear when to stop looking. December 1997 Lecture Notes on General Relativity Sean M. Carroll 6 Weak Fields and Gravitational Radiation When we \ufb01rst derived Einstein\u2019s equations, we checked that we were on the right track by considering the Newtonian limit. This amounted to the requirements that the gravitational \ufb01eld be weak, that it be static (no time derivatives), and that test particles be moving slowly. In this section we will consider a less restrictive situation, in which the \ufb01eld is still weak but it can vary with time, and there are no restrictions on the motion of test particles. This will allow us to discuss phenomena which are absent or ambiguous in the Newtonian theory, such as gravitational radiation (where the \ufb01eld varies with time) and the de\ufb02ection of light (which involves fast-moving particles). The weakness of the gravitational \ufb01eld is once again expressed as our ability to decompose the metric into the \ufb02at Minkowski metric plus a small perturbation, g\u00b5\u03bd = \u03b7\u00b5\u03bd + h\u00b5\u03bd, h\u00b5\u03bd| | << 1. (6.1) We will restrict ourselves to coordinates in which \u03b7\u00b5\u03bd takes its canonical form, \u03b7\u00b5\u03bd = 1, +1, +1, +1). The assumption that h\u00b5\u03bd is small allows us to ignore anything that is diag( higher than \ufb01rst order in this quantity, from which we immediately obtain \u2212 g\u00b5\u03bd = \u03b7\u00b5\u03bd h\u00b5\u03bd, \u2212 (6.2) where h\u00b5\u03bd = \u03b7\u00b5\u03c1\u03b7\u03bd\u03c3h\u03c1\u03c3. As before, we can raise and lower indices using \u03b7\u00b5\u03bd and \u03b7\u00b5\u03bd, since the corrections would be of higher order in the perturbation. In fact, we can think of the linearized version of general relativity (where e\ufb00ects of higher than \ufb01rst order in h\u00b5\u03bd are neglected) as describing a theory of a symmetric tensor \ufb01eld h\u00b5\u03bd propagating on a \ufb02at background spacetime. This theory is Lorentz invariant in the sense of special relativity; under a Lorentz transformation x\u00b5\u2032 \u00b5x\u00b5, the \ufb02at metric \u03b7\u00b5\u03bd is invariant, while the perturbation", " transforms as = \u039b\u00b5\u2032 h\u00b5\u2032\u03bd\u2032 = \u039b\u00b5\u2032 \u00b5\u039b\u03bd\u2032 \u03bdh\u00b5\u03bd. (6.3) (Note that we could have considered small perturbations about some other background spacetime besides Minkowski space. In that case the metric would have been written g\u00b5\u03bd = g(0) \u00b5\u03bd + h\u00b5\u03bd, and we would have derived a theory of a symmetric tensor propagating on the curved space with metric g(0) \u00b5\u03bd. Such an approach is necessary, for example, in cosmology.) We want to \ufb01nd the equation of motion obeyed by the perturbations h\u00b5\u03bd, which come by examining Einstein\u2019s equations to \ufb01rst order. We begin with the Christo\ufb00el symbols, which are given by \u0393\u03c1 \u00b5\u03bd = 1 2 g\u03c1\u03bb(\u2202\u00b5g\u03bd\u03bb + \u2202\u03bdg\u03bb\u00b5 \u2212 \u2202\u03bbg\u00b5\u03bd) 142 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 143 = \u03b7\u03c1\u03bb(\u2202\u00b5h\u03bd\u03bb + \u2202\u03bdh\u03bb\u00b5 \u2212 Since the connection coe\ufb03cients are \ufb01rst order quantities, the only contribution to the Riemann tensor will come from the derivatives of the \u0393\u2019s, not the \u03932 terms. Lowering an index for convenience, we obtain \u2202\u03bbh\u00b5\u03bd). (6.4) 1 2 R\u00b5\u03bd\u03c1\u03c3 = \u03b7\u00b5\u03bb\u2202\u03c1\u0393\u03bb \u03b7\u00b5\u03bb\u2202\u03c3\u0393\u03bb \u03bd\u03c1 \u03bd\u03c3 \u2212 \u2202\u03c3\u2202\u03bdh\u00b5\u03c1 \u2212 The Ricci tensor comes from contracting over \u00b5 and \u03c1, giving (\u2202\u03c1\u2202\u03bdh\u00b5\u03c3 + \u2202\u03c3\u2202\u00b5h\u03bd\u03c1 \u2212 = 1 2 \u2202\u03c1\u2202\u00b5h\u03bd\u03c3). (6.5) R\u00b5\u03bd = 1 2 (\u2202\u03c3\u2202\u03bdh\u03c3 \u00b5 + \u2202\u03c3\u2202\u00b5h\u03c3 \u03bd \u2212 \u2202\u00b5\u2202\u03bdh \u2212 2h\u00b5\u03bd), (6.6) which is manifestly symmetric in \u00b5 and \u03bd. In this expression we have de\ufb01ned the trace of the perturbation as h = \u03b7\u00b5\u03bdh\u00b5\u03bd = h\u00b5 \u00b5, and the D\u2019Alembertian is simply the one from \ufb02at space, 2 = t + \u22022 \u22022 z. Contracting again to obtain the Ricci scalar yields x + \u22022 y + \u22022 \u2212 Putting it all together we obtain the Einstein tensor: R = \u2202\u00b5\u2202\u03bdh\u00b5\u03bd 2h. \u2212 (6.7) G\u00b5\u03bd = R\u00b5\u03bd \u2212 = 1 2 (\u2202\u03c3\u2202\u03bdh\u03c3 1 2 \u03b7\u00b5\u03bdR \u00b5 + \u2202\u03c3\u2202\u00b5h\u03c3 \u2202\u00b5\u2202\u03bdh \u03bd \u2212 2h\u00b5\u03bd \u2212 \u2212 \u03b7\u00b5\u03bd\u2202\u00b5\u2202\u03bdh\u00b5\u03bd + \u03b7\u00b5\u03bd2h). (6.8) Consistent with our interpretation of the linearized theory as one describing a symmetric tensor on a \ufb02at background, the linearized Einstein tensor (6.8) can be derived by varying the following Lagrangian with respect to h\u00b5\u03bd: = L 1 2 (\u2202\u00b5h\u00b5\u03bd)(\u2202\u03bdh) \u2212 (\u2202\u00b5h\u03c1\u03c3)(\u2202\u03c1h\u00b5 \u03c3) + 1 2 \u03b7\u00b5\u03bd(\u2202\u00b5h\u03c1\u03c3)(\u2202\u03bdh\u03c1\u03c3) 1 2 \u2212 \u03b7\u00b5\u03bd(\u2202\u00b5h)(\u2202\u03bdh). (6.9) I will spare you the details. The linearized \ufb01eld equation is of course G\u00b5\u03bd = 8\u03c0GT\u00b5\u03bd, where G\u00b5\u03bd is given by (6.8) and T\u00b5\u03bd is the energy-momentum tensor, calculated to zeroth order in h\u00b5\u03bd. We do not include higher-order corrections to the energy-momentum tensor because the amount of energy and momentum must itself be small for the weak-\ufb01eld limit to apply. In other words, the lowest nonvanishing order in T\u00b5\u03bd is automatically of the same order of magnitude as the perturbation. Notice that the conservation law to lowest order is simply \u2202\u00b5T \u00b5\u03bd = 0. We will most often be concerned with the", " vacuum equations, which as usual are just R\u00b5\u03bd = 0, where R\u00b5\u03bd is given by (6.6). 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 144 With the linearized \ufb01eld equations in hand, we are almost prepared to set about solving them. First, however, we should deal with the thorny issue of gauge invariance. This issue arises because the demand that g\u00b5\u03bd = \u03b7\u00b5\u03bd + h\u00b5\u03bd does not completely specify the coordinate system on spacetime; there may be other coordinate systems in which the metric can still be written as the Minkowski metric plus a small perturbation, but the perturbation will be di\ufb00erent. Thus, the decomposition of the metric into a \ufb02at background plus a perturbation is not unique. We can think about this from a highbrow point of view. The notion that the linearized theory can be thought of as one governing the behavior of tensor \ufb01elds on a \ufb02at background can be formalized in terms of a \u201cbackground spacetime\u201d Mb, a \u201cphysical spacetime\u201d Mp, Mp. As manifolds Mb and Mp are \u201cthe same\u201d (since and a di\ufb00eomorphism \u03c6 : Mb \u2192 they are di\ufb00eomorphic), but we imagine that they possess some di\ufb00erent tensor \ufb01elds; on Mb we have de\ufb01ned the \ufb02at Minkowski metric \u03b7\u00b5\u03bd, while on Mp we have some metric g\u03b1\u03b2 which obeys Einstein\u2019s equations. (We imagine that Mb is equipped with coordinates x\u00b5 and Mp is equipped with coordinates y\u03b1, although these will not play a prominent role.) The di\ufb00eomorphism \u03c6 allows us to move tensors back and forth between the background and physical spacetimes. Since we would like to construct our linearized theory as one taking place on the \ufb02at background spacetime, we are interested in the pullback (\u03c6 g)\u00b5\u03bd of the physical metric. We can de\ufb01ne the perturbation as the di\ufb00erence between the pulled-back physical metric and the \ufb02at one: \u2217 h\u00b5\u03bd = (\u03c6 g)\u00b5\u03bd \u2212 \u2217 \u03b7\u00b5\u03bd. (6.10) From this de\ufb01nition, there is no reason for the components of h\u00b5\u03bd to be small; however, if the gravitational \ufb01elds on Mp are weak, then for some di\ufb00eomorphisms \u03c6 we will have << 1. We therefore limit our attention only to those di\ufb00eomorphisms for which this is true. Then the fact that g\u03b1\u03b2 obeys Einstein\u2019s equations on the physical spacetime means that h\u00b5\u03bd will obey the linearized equations on the background spacetime (since \u03c6, as a di\ufb00eomorphism, can be used to pull back Einstein\u2019s equations themselves). h\u00b5\u03bd| | M b \u03b7 \u00b5\u03bd \u03c6 ( g)\u00b5\u03bd * \u03c6 \u03c6 * M p g\u03b1\u03b2 In this language, the issue of gauge invariance is simply the fact that there are a large number of permissible di\ufb00eomorphisms between Mb and Mp (where \u201cpermissible\u201d means 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 145 that the perturbation is small). Consider a vector \ufb01eld \u03be\u00b5(x) on the background spacetime. Mb. For This vector \ufb01eld generates a one-parameter family of di\ufb00eomorphisms \u03c8\u01eb : Mb \u2192 \u01eb su\ufb03ciently small, if \u03c6 is a di\ufb00eomorphism for which the perturbation de\ufb01ned by (6.10) is small than so will (\u03c6 \u03c8\u01eb) be, although the perturbation will have a di\ufb00erent value * Speci\ufb01cally, we can de\ufb01ne a family of perturbations parameterized by \u01eb: h(\u01eb) \u00b5\u03bd = [(\u03c6 = [\u03c8\u01eb \u2217 g]\u00b5\u03bd \u2212 \u03c8\u01eb) g)]\u00b5\u03bd \u2212 \u2217 \u25e6 (\u03c6 \u2217 \u03b7\u00b5\u03bd \u03b7\u00b5\u03bd. M p (6.11) The second equality is based on the fact that the pullback under a composition is given by the composition of the pullbacks in the opposite order, which follows from the fact that the pullback itself moves", " things in the opposite direction from the original map. Plugging in the relation (6.10), we \ufb01nd h(\u01eb) \u00b5\u03bd = \u03c8\u01eb = \u03c8\u01eb \u2217 \u2217 (h + \u03b7)\u00b5\u03bd \u2212 (h\u00b5\u03bd) + \u03c8\u01eb \u03b7\u00b5\u03bd (\u03b7\u00b5\u03bd) \u2217 \u03b7\u00b5\u03bd \u2212 (6.12) (since the pullback of the sum of two tensors is the sum of the pullbacks). Now we use our assumption that \u01eb is small; in this case \u03c8\u01eb (h\u00b5\u03bd) will be equal to h\u00b5\u03bd to lowest order, while the other two terms give us a Lie derivative: \u2217 h(\u01eb) \u00b5\u03bd = \u03c8\u01eb (h\u00b5\u03bd) + \u01eb \u2217 \" = h\u00b5\u03bd + \u01eb\u00a3\u03be\u03b7\u00b5\u03bd = h\u00b5\u03bd + 2\u01eb\u2202(\u00b5\u03be\u03bd). \u03c8\u01eb \u03b7\u00b5\u03bd \u2217 (\u03b7\u00b5\u03bd) \u01eb \u2212 # (6.13) The last equality follows from our previous computation of the Lie derivative of the metric, (5.33), plus the fact that covariant derivatives are simply partial derivatives to lowest order. The in\ufb01nitesimal di\ufb00eomorphisms \u03c6\u01eb provide a di\ufb00erent representation of the same physical situation, while maintaining our requirement that the perturbation be small. Therefore, the result (6.12) tells us what kind of metric perturbations denote physically equivalent spacetimes \u2014 those related to each other by 2\u01eb\u2202(\u00b5\u03be\u03bd), for some vector \u03be\u00b5. The invariance of 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 146 our theory under such transformations is analogous to traditional gauge invariance of elecA\u00b5 + \u2202\u00b5\u03bb. (The analogy is di\ufb00erent from the previous analogy tromagnetism under A\u00b5 \u2192 we drew with electromagnetism, relating local Lorentz transformations in the orthonormalframe formalism to changes of basis in an internal vector bundle.) In electromagnetism the \u2202\u03bdA\u00b5 is left unchanged invariance comes about because the \ufb01eld strength F\u00b5\u03bd = \u2202\u00b5A\u03bd \u2212 by gauge transformations; similarly, we \ufb01nd that the transformation (6.13) changes the linearized Riemann tensor by \u03b4R\u00b5\u03bd\u03c1\u03c3 = 1 2 = 0. (\u2202\u03c1\u2202\u03bd\u2202\u00b5\u03be\u03c3 + \u2202\u03c1\u2202\u03bd\u2202\u03c3\u03be\u00b5 + \u2202\u03c3\u2202\u00b5\u2202\u03bd\u03be\u03c1 + \u2202\u03c3\u2202\u00b5\u2202\u03c1\u03be\u03bd \u2202\u03c3\u2202\u03bd\u2202\u03c1\u03be\u00b5 \u2212 \u2202\u03c3\u2202\u03bd\u2202\u00b5\u03be\u03c1 \u2212 \u2212 \u2202\u03c1\u2202\u00b5\u2202\u03bd\u03be\u03c3 \u2212 \u2202\u03c1\u2202\u00b5\u2202\u03c3\u03be\u03bd) (6.14) Our abstract derivation of the appropriate gauge transformation for the metric perturbation is veri\ufb01ed by the fact that it leaves the curvature (and hence the physical spacetime) unchanged. Gauge invariance can also be understood from the slightly more lowbrow but considerably more direct route of in\ufb01nitesimal coordinate transformations. Our di\ufb00eomorphism \u03c8\u01eb can be thought of as changing coordinates from x\u00b5 to x\u00b5 (The minus sign, which is unconventional, comes from the fact that the \u201cnew\u201d metric is pulled back from a small distance forward along the integral curves, which is equivalent to replacing the coordinates by those a small distance backward along the curves.) Following through the usual rules for transforming tensors under coordinate transformations, you can derive precisely (6.13) \u2014 although you have to cheat somewhat by equating components of tensors in two di\ufb00erent coordinate systems. See Schutz or Weinberg for an example. \u01eb\u03be\u00b5. \u2212 When faced with a system that is invariant under some kind of gauge transformations, our \ufb01rst instinct is to \ufb01x a gauge. We have already discussed the harmonic coordinate system, and will return to it now in the context of the weak \ufb01eld limit. Recall that this gauge was speci\ufb01ed by 2x\u00b5 = 0, which we showed was equivalent to In the weak \ufb01eld limit this becomes g\u00b5\u03bd\u0393\u03c1 \u00b5\u03bd = 0. 1 2 \u03b7\u00b5\u03bd\u03b7\u03bb\u03c1(\u2202\u00b5h", "\u03bd\u03bb + \u2202\u03bdh\u03bb\u00b5 \u2212 \u2202\u03bbh\u00b5\u03bd) = 0, (6.15) (6.16) or 1 2 This condition is also known as Lorentz gauge (or Einstein gauge or Hilbert gauge or de Donder gauge or Fock gauge). As before, we still have some gauge freedom remaining, since we can change our coordinates by (in\ufb01nitesimal) harmonic functions. \u2202\u03bbh = 0. (6.17) \u2202\u00b5h\u00b5 \u03bb \u2212 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 147 In this gauge, the linearized Einstein equations G\u00b5\u03bd = 8\u03c0GT\u00b5\u03bd simplify somewhat, to 2h\u00b5\u03bd \u2212 1 2 \u03b7\u00b5\u03bd2h = 16\u03c0GT\u00b5\u03bd, \u2212 while the vacuum equations R\u00b5\u03bd = 0 take on the elegant form 2h\u00b5\u03bd = 0, (6.18) (6.19) which is simply the conventional relativistic wave equation. Together, (6.19) and (6.17) determine the evolution of a disturbance in the gravitational \ufb01eld in vacuum in the harmonic gauge. It is often convenient to work with a slightly di\ufb00erent description of the metric pertur- bation. We de\ufb01ne the \u201ctrace-reversed\u201d perturbation \u00afh\u00b5\u03bd by 1 2 \u03b7\u00b5\u03bdh. (6.20) \u00afh\u00b5\u03bd = h\u00b5\u03bd \u2212 h\u00b5 The name makes sense, since \u00afh\u00b5 \u00b5 = Ricci tensor.) In terms of \u00afh\u00b5\u03bd the harmonic gauge condition becomes \u2212 \u00b5. (The Einstein tensor is simply the trace-reversed The full \ufb01eld equations are \u2202\u00b5\u00afh\u00b5 \u03bb = 0. 2\u00afh\u00b5\u03bd = 16\u03c0GT\u00b5\u03bd, \u2212 from which it follows immediately that the vacuum equations are 2\u00afh\u00b5\u03bd = 0. (6.21) (6.22) (6.23) From (6.22) and our previous exploration of the Newtonian limit, it is straightforward to derive the weak-\ufb01eld metric for a stationary spherical source such as a planet or star. Recall that previously we found that Einstein\u2019s equations predicted that h00 obeyed the Poisson equation (4.51) in the weak-\ufb01eld limit, which implied h00 = 2\u03a6, \u2212 (6.24) GM/r. Let us now assume that where \u03a6 is the conventional Newtonian potential, \u03a6 = the energy-momentum tensor of our source is dominated by its rest energy density \u03c1 = T00. (Such an assumption is not generally necessary in the weak-\ufb01eld limit, but will certainly hold for a planet or star, which is what we would like to consider for the moment.) Then the other components of T\u00b5\u03bd will be much smaller than T00, and from (6.22) the same must hold for \u00afh\u00b5\u03bd. If \u00afh00 is much larger than \u00afhij, we will have \u2212 h = \u00afh = \u2212 \u03b7\u00b5\u03bd\u00afh\u00b5\u03bd = \u00afh00, \u2212 (6.25) 6 WEAK FIELDS AND GRAVITATIONAL RADIATION and then from (6.20) we immediately obtain The other components of \u00afh\u00b5\u03bd are negligible, from which we can derive \u00afh00 = 2h00 = 4\u03a6. \u2212 and hi0 = \u00afhi0 \u2212 1 2 \u03b7i0\u00afh = 0, hij = \u00afhij \u2212 1 2 \u03b7ij\u00afh = 2\u03a6\u03b4ij. \u2212 The metric for a star or planet in the weak-\ufb01eld limit is therefore ds2 = (1 + 2\u03a6)dt2 + (1 \u2212 \u2212 2\u03a6)(dx2 + dy2 + dz2). 148 (6.26) (6.27) (6.28) (6.29) A somewhat less simplistic application of the weak-\ufb01eld limit is to gravitational radiation. Those of you familiar with the analogous problem in electromagnetism will notice that the procedure is almost precisely the same. We begin by considering the linearized equations in vacuum (6.23). Since the \ufb02at-space D\u2019Alembertian has the form 2 = 2, the \ufb01eld equation is in the form of a wave equation for \u00afh\u00b5\u03bd. As all good physicists know, the thing to do when faced with such an equation is to write down complex-valued solutions, and then take the real part at the very end of the day. So", " we recognize that a particularly useful set of solutions to this wave equation are the plane waves, given by \u22022 t + \u2207 \u2212 \u00afh\u00b5\u03bd = C\u00b5\u03bdeik\u03c3x\u03c3, (6.30) where C\u00b5\u03bd is a constant, symmetric, (0, 2) tensor, and k\u03c3 is a constant vector known as the wave vector. To check that it is a solution, we plug in: 0 = 2\u00afh\u00b5\u03bd = \u03b7\u03c1\u03c3\u2202\u03c1\u2202\u03c3\u00afh\u00b5\u03bd = \u03b7\u03c1\u03c3\u2202\u03c1(ik\u03c3\u00afh\u00b5\u03bd) \u03b7\u03c1\u03c3k\u03c1k\u03c3\u00afh\u00b5\u03bd = k\u03c3k\u03c3\u00afh\u00b5\u03bd. = \u2212 \u2212 (6.31) Since (for an interesting solution) not all of the components of h\u00b5\u03bd will be zero everywhere, we must have k\u03c3k\u03c3 = 0. (6.32) The plane wave (6.30) is therefore a solution to the linearized equations if the wavevector is null; this is loosely translated into the statement that gravitational waves propagate at the speed of light. The timelike component of the wave vector is often referred to as the 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 149 frequency of the wave, and we write k\u03c3 = (\u03c9, k1, k2, k3). moving with four-velocity U \u00b5 would observe the wave to have a frequency \u03c9 = Then the condition that the wave vector be null becomes (More generally, an observer k\u00b5U \u00b5.) \u2212 \u03c92 = \u03b4ijkikj. (6.33) Of course our wave is far from the most general solution; any (possibly in\ufb01nite) number of distinct plane waves can be added together and will still solve the linear equation (6.23). Indeed, any solution can be written as such a superposition. There are a number of free parameters to specify the wave: ten numbers for the coe\ufb03cients C\u00b5\u03bd and three for the null vector k\u03c3. Much of these are the result of coordinate freedom and gauge freedom, which we now set about eliminating. We begin by imposing the harmonic gauge condition, (6.21). This implies that 0 = \u2202\u00b5 \u00afh\u00b5\u03bd = \u2202\u00b5(C \u00b5\u03bdeik\u03c3x\u03c3 = iC \u00b5\u03bdk\u00b5eik\u03c3x\u03c3 ), (6.34) which is only true if k\u00b5C \u00b5\u03bd = 0. We say that the wave vector is orthogonal to C \u00b5\u03bd. These are four equations, which reduce the number of independent components of C\u00b5\u03bd from ten to six. (6.35) Although we have now imposed the harmonic gauge condition, there is still some coor- dinate freedom left. Remember that any coordinate transformation of the form \u2192 will leave the harmonic coordinate condition x\u00b5 x\u00b5 + \u03b6 \u00b5 satis\ufb01ed as long as we have 2x\u00b5 = 0 2\u03b6 \u00b5 = 0. (6.36) (6.37) (6.38) Of course, (6.38) is itself a wave equation for \u03b6 \u00b5; once we choose a solution, we will have used up all of our gauge freedom. Let\u2019s choose the following solution: \u03b6\u00b5 = B\u00b5eik\u03c3x\u03c3, (6.39) where k\u03c3 is the wave vector for our gravitational wave and the B\u00b5 are constant coe\ufb03cients. We now claim that this remaining freedom allows us to convert from whatever coe\ufb03cients C (old) \u00b5\u03bd that characterize our gravitational wave to a new set C (new) \u00b5\u03bd, such that C (new)\u00b5 \u00b5 = 0 (6.40) 6 WEAK FIELDS AND GRAVITATIONAL RADIATION and C (new) 0\u03bd = 0. 150 (6.41) (Actually this last condition is both a choice of gauge and a choice of Lorentz frame. The choice of gauge sets U \u00b5C (new) \u00b5\u03bd = 0 for some constant timelike vector U \u00b5, while the choice of frame makes U \u00b5 point along the time axis.) Let\u2019s see how this is possible by solving explicitly for the necessary coe\ufb03cients B\u00b5. Under the transformation (6.36), the resulting change in our metric perturbation can be written which induces a change in the trace-reversed perturbation, \u00b5\u03bd = h(old) h(new) \u00b5\u03bd \u2212 \u2202\u00b5\u03b6\u03bd \u2212 \u2202\u03bd\u03b6\u00b5, \u00afh(new) \u00b5\u03bd \u03b7\u00b5\u03bdh(new) = h(new) \u00b5\u03bd \u2212 = h(", "old) = \u00afh(old) \u00b5\u03bd \u2212 \u00b5\u03bd \u2212 1 2 \u2202\u00b5\u03b6\u03bd \u2212 \u2202\u00b5\u03b6\u03bd \u2212 1 \u2202\u03bd\u03b6\u00b5 \u2212 2 \u2202\u03bd\u03b6\u00b5 + \u03b7\u00b5\u03bd\u2202\u03bb\u03b6 \u03bb. \u03b7\u00b5\u03bd(h(old) \u2212 2\u2202\u03bb\u03b6 \u03bb) (6.42) (6.43) Using the speci\ufb01c forms (6.30) for the solution and (6.39) for the transformation, we obtain C (new) \u00b5\u03bd = C (old) \u00b5\u03bd \u2212 ik\u00b5B\u03bd \u2212 ik\u03bdB\u00b5 + i\u03b7\u00b5\u03bdk\u03bbB\u03bb. Imposing (6.40) therefore means or 0 = C (old)\u00b5 \u00b5 + 2ik\u03bbB\u03bb, k\u03bbB\u03bb = i 2 C (old)\u00b5 \u00b5. Then we can impose (6.41), \ufb01rst for \u03bd = 0: 0 = C (old) = C (old) 00 \u2212 00 \u2212 2ik0B0 \u2212 2ik0B0 + ik\u03bbB\u03bb 1 2 C (old)\u00b5 \u00b5, or Then impose (6.41) for \u03bd = j: B0 = \u2212 i 2k0 C (old) 00 + 1 2 C (old)\u00b5 \u00b5. 0 = C (old) = C (old) 0j \u2212 0j \u2212 ik0Bj \u2212 ik0Bj \u2212 ikjB0 i \u2212 2k0 C (old) 00 + ikj 1 2 C (old)\u00b5 \u00b5, (6.44) (6.45) (6.46) (6.47) (6.48) (6.49) 6 WEAK FIELDS AND GRAVITATIONAL RADIATION or Bj = i 2(k0)2 2k0C (old) 0j + kj \u2212 00 + C (old) 1 2 C (old)\u00b5 \u00b5. 151 (6.50) To check that these choices are mutually consistent, we should plug (6.48) and (6.50) back into (6.40), which I will leave to you. Let us assume that we have performed this transformation, and refer to the new components C (new) simply as C\u00b5\u03bd. \u00b5\u03bd Thus, we began with the ten independent numbers in the symmetric matrix C\u00b5\u03bd. Choosing harmonic gauge implied the four conditions (6.35), which brought the number of independent components down to six. Using our remaining gauge freedom led to the one condition (6.40) and the four conditions (6.41); but when \u03bd = 0 (6.41) implies (6.35), so we have a total of four additional constraints, which brings us to two independent components. We\u2019ve used up all of our possible freedom, so these two numbers represent the physical information characterizing our plane wave in this gauge. This can be seen more explicitly by choosing our spatial coordinates such that the wave is travelling in the x3 direction; that is, k\u00b5 = (\u03c9, 0, 0, k3) = (\u03c9, 0, 0, \u03c9), (6.51) where we know that k3 = \u03c9 because the wave vector is null. In this case, k\u00b5C\u00b5\u03bd = 0 and C0\u03bd = 0 together imply C3\u03bd = 0. (6.52) The only nonzero components of C\u00b5\u03bd are therefore C11, C12, C21, and C22. But C\u00b5\u03bd is traceless and symmetric, so in general we can write 0 0 0 C11 0 C12 \u2212 0 0 0 C12 C11 0. 0 0 0 0 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 C\u00b5\u03bd = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ed (6.53) Thus, for a plane wave in this gauge travelling in the x3 direction, the two components C11 and C12 (along with the frequency \u03c9) completely characterize the wave. In using up all of our gauge freedom, we have gone to a subgauge of the harmonic gauge known as the transverse traceless gauge (or sometimes \u201cradiation gauge\u201d). The name comes from the fact that the metric perturbation is traceless and perpendicular to the wave vector. Of course, we have been working with the trace-reversed perturbation \u00afh\u00b5\u03bd rather than the perturbation h\u00b5\u03bd itself; but since \u00afh\u00b5\u03bd is traceless (because C\u00b5\u03bd is), and is equal to the trace-reverse of h", "\u00b5\u03bd, in this gauge we have \u00afhTT \u00b5\u03bd = hTT \u00b5\u03bd (transverse traceless gauge). (6.54) Therefore we can drop the bars over h\u00b5\u03bd, as long as we are in this gauge. 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 152 One nice feature of the transverse traceless gauge is that if you are given the components of a plane wave in some arbitrary gauge, you can easily convert them into the transverse traceless components. We \ufb01rst de\ufb01ne a tensor P\u00b5\u03bd which acts as a projection operator: P\u00b5\u03bd = \u03b7\u00b5\u03bd \u2212 n\u00b5n\u03bd. (6.55) You can check that this projects vectors onto hyperplanes orthogonal to the unit vector n\u00b5. Here we take n\u00b5 to be a spacelike unit vector, which we choose to lie along the direction of propagation of the wave: Then the transverse part of some perturbation h\u00b5\u03bd is simply the projection P\u00b5 the transverse traceless part is obtained by subtracting o\ufb00 the trace: n0 = 0, nj = kj/\u03c9. (6.56) \u03c1P\u03bd \u03c3h\u03c1\u03c3, and \u03c3h\u03c1\u03c3 \u2212 For details appropriate to more general cases, see the discussion in Misner, Thorne and Wheeler. P\u00b5\u03bdP \u03c1\u03c3h\u03c1\u03c3. hTT \u00b5\u03bd = P\u00b5 (6.57) \u03c1P\u03bd 1 2 To get a feeling for the physical e\ufb00ects due to gravitational waves, it is useful to consider the motion of test particles in the presence of a wave. It is certainly insu\ufb03cient to solve for the trajectory of a single particle, since that would only tell us about the values of the coordinates along the world line. (In fact, for any single particle we can \ufb01nd transverse traceless coordinates in which the particle appears stationary to \ufb01rst order in h\u00b5\u03bd.) To obtain a coordinate-independent measure of the wave\u2019s e\ufb00ects, we consider the relative motion of nearby particles, as described by the geodesic deviation equation. If we consider some nearby particles with four-velocities described by a single vector \ufb01eld U \u00b5(x) and separation vector S\u00b5, we have D2 d\u03c4 2 S\u00b5 = R\u00b5 \u03bd\u03c1\u03c3U \u03bdU \u03c1S\u03c3. (6.58) We would like to compute the left-hand side to \ufb01rst order in h\u00b5\u03bd. If we take our test particles to be moving slowly then we can express the four-velocity as a unit vector in the time direction plus corrections of order h\u00b5\u03bd and higher; but we know that the Riemann tensor is already \ufb01rst order, so the corrections to U \u03bd may be ignored, and we write U \u03bd = (1, 0, 0, 0). Therefore we only need to compute R\u00b5 00\u03c3, or equivalently R\u00b500\u03c3. From (6.5) we have R\u00b500\u03c3 = 1 2 (\u22020\u22020h\u00b5\u03c3 + \u2202\u03c3\u2202\u00b5h00 \u2212 \u2202\u03c3\u22020h\u00b50 \u2212 \u2202\u00b5\u22020h\u03c30). But h\u00b50 = 0, so R\u00b500\u03c3 = 1 2 \u22020\u22020h\u00b5\u03c3. (6.59) (6.60) (6.61) 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 153 Meanwhile, for our slowly-moving particles we have \u03c4 = x0 = t to lowest order, so the geodesic deviation equation becomes \u22022 \u2202t2 S\u00b5 = 1 2 For our wave travelling in the x3 direction, this implies that only S1 and S2 will be a\ufb00ected \u2014 the test particles are only disturbed in directions perpendicular to the wave vector. This is of course familiar from electromagnetism, where the electric and magnetic \ufb01elds in a plane wave are perpendicular to the wave vector. \u2202t2 h\u00b5 S\u03c3 \u22022 (6.62) \u03c3. Our wave is characterized by the two numbers, which for future convenience we will = C12. Let\u2019s consider their e\ufb00ects separately, beginning with rename as C+ = C11 and C the case C = 0. Then we have \u00d7 \u00d7 \u22022 \u2202t2 S1 = 1 2 S1 \u22022 \u2202t2 (C+eik\u03c3x\u03c3 ) and \u2202", "t2 (C+eik\u03c3x\u03c3 These can be immediately solved to yield, to lowest order, \u2212 ). \u22022 \u2202t2 S2 = S2 \u22022 1 2 S1 = 1 + 1 2 C+eik\u03c3x\u03c3 S1(0) and S2 = 1 1 2 \u2212 C+eik\u03c3x\u03c3 S2(0). (6.63) (6.64) (6.65) (6.66) Thus, particles initially separated in the x1 direction will oscillate back and forth in the x1 direction, and likewise for those with an initial x2 separation. That is, if we start with a ring of stationary particles in the x-y plane, as the wave passes they will bounce back and forth in the shape of a \u201c+\u201d: y x On the other hand, the equivalent analysis for the case where C+ = 0 but C yield the solution S1 = S1(0) + 1 2 eik\u03c3x\u03c3 S2(0) C \u00d7 = 0 would \u00d7 6 (6.67) 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 154 and S2 = S2(0) + 1 2 C \u00d7 eik\u03c3x\u03c3 S1(0). In this case the circle of particles would bounce back and forth in the shape of a \u201c (6.68) \u201d: \u00d7 y x The notation C+ and C independent modes of linear polarization of the gravitational wave. consider right- and left-handed circularly polarized modes by de\ufb01ning should therefore be clear. These two quantities measure the two If we liked we could \u00d7 CR = CL = 1 \u221a2 1 \u221a2 (C+ + iC ), \u00d7 (C+ \u2212 iC \u00d7 ). (6.69) The e\ufb00ect of a pure CR wave would be to rotate the particles in a right-handed sense, y x and similarly for the left-handed mode CL. (Note that the individual particles do not travel around the ring; they just move in little epicycles.) We can relate the polarization states of classical gravitational waves to the kinds of particles we would expect to \ufb01nd upon quantization. The electromagnetic \ufb01eld has two independent polarization states which are described by vectors in the x-y plane; equivalently, a single polarization mode is invariant under a rotation by 360\u25e6 in this plane. Upon quantization this theory yields the photon, a massless spin-one particle. The neutrino, on the other hand, is also a massless particle, described by a \ufb01eld which picks up a minus sign under rotations by 360\u25e6; it is invariant under rotations of 720\u25e6, and we say it has spin- 1 2. 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 155 The general rule is that the spin S is related to the angle \u03b8 under which the polarization modes are invariant by S = 360\u25e6/\u03b8. The gravitational \ufb01eld, whose waves propagate at the speed of light, should lead to massless particles in the quantum theory. Noticing that the polarization modes we have described are invariant under rotations of 180\u25e6 in the x-y plane, we expect the associated particles \u2014 \u201cgravitons\u201d \u2014 to be spin-2. We are a long way from detecting such particles (and it would not be a surprise if we never detected them directly), but any respectable quantum theory of gravity should predict their existence. With plane-wave solutions to the linearized vacuum equations in our possession, it remains to discuss the generation of gravitational radiation by sources. For this purpose it is necessary to consider the equations coupled to matter, 2\u00afh\u00b5\u03bd = 16\u03c0GT\u00b5\u03bd. \u2212 (6.70) The solution to such an equation can be obtained using a Green\u2019s function, in precisely the same way as the analogous problem in electromagnetism. Here we will review the outline of the method. The Green\u2019s function G(x\u03c3 y\u03c3) for the D\u2019Alembertian operator 2 is the solution of the wave equation in the presence of a delta-function source: \u2212 2xG(x\u03c3 \u2212 y\u03c3) = \u03b4(4)(x\u03c3 y\u03c3), \u2212 (6.71) where 2x denotes the D\u2019Alembertian with respect to the coordinates x\u03c3. The usefulness of such a function resides in the fact that the general solution to an equation such as (6.70) can be written \u00afh\u00b5\u03bd(x\u03c3) = 16\u03c0G G(x\u03c3 y\u03c3)T\u00b5\u03bd(y\u03c3) d4y, \u2212 (Notice that no factors of \u221a \u2212 Z g are necessary, since", " our as can be veri\ufb01ed immediately. background is simply \ufb02at spacetime.) The solutions to (6.71) have of course been worked out long ago, and they can be thought of as either \u201cretarded\u201d or \u201cadvanced,\u201d depending on whether they represent waves travelling forward or backward in time. Our interest is in the retarded Green\u2019s function, which represents the accumulated e\ufb00ects of signals to the past of the point under consideration. It is given by \u2212 (6.72) G(x\u03c3 \u2212 y\u03c3) = 1 x \u2212 y | \u2212 4\u03c0 | x \u03b4[ | \u2212 y | \u2212 (x0 \u2212 y0)] \u03b8(x0 y0). \u2212 (6.73) Here we have used boldface to denote the spatial vectors x = (x1, x2, x3) and y = (y1, y2, y3), y0) equals 1 when with norm x0 > y0, and zero otherwise. The derivation of (6.73) would take us too far a\ufb01eld, but it can be found in any standard text on electrodynamics or partial di\ufb00erential equations in physics. yj)]1/2. The theta function \u03b8(x0 = [\u03b4ij(xi yi)(xj \u2212 \u2212 \u2212 \u2212 x y | | 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 156 Upon plugging (6.73) into (6.72), we can use the delta function to perform the integral over y0, leaving us with \u00afh\u00b5\u03bd(t, x) = 4G 1 Z x | \u2212 y | T\u00b5\u03bd(t x y | \u2212 \u2212 |, y) d3y, where t = x0. The term \u201cretarded time\u201d is used to refer to the quantity tr = t x. y | \u2212 \u2212 | (6.74) (6.75) The interpretation of (6.74) should be clear: the disturbance in the gravitational \ufb01eld at (t, x) is a sum of the in\ufb02uences from the energy and momentum sources at the point (tr, x y) on the past light cone. \u2212 t xi yi (t, y )i r Let us take this general solution and consider the case where the gravitational radiation is emitted by an isolated source, fairly far away, comprised of nonrelativistic matter; these approximations will be made more precise as we go on. First we need to set up some conventions for Fourier transforms, which always make life easier when dealing with oscillatory phenomena. Given a function of spacetime \u03c6(t, x), we are interested in its Fourier transform (and inverse) with respect to time alone, \u03c6(\u03c9, x) = e \u03c6(t, x) = 1 \u221a2\u03c0 Z 1 \u221a2\u03c0 Z dt ei\u03c9t\u03c6(t, x), d\u03c9 e\u2212 i\u03c9t \u03c6(\u03c9, x). (6.76) Taking the transform of the metric perturbation, we obtain e \u00afh\u00b5\u03bd(\u03c9, x) = e 1 \u221a2\u03c0 Z dt ei\u03c9t\u00afh\u00b5\u03bd(t, x) 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 157 = = 4G \u221a2\u03c0 Z 4G \u221a2\u03c0 Z dt d3y ei\u03c9t T\u00b5\u03bd(t dtr d3y ei\u03c9tr ei\u03c9 | x x y, y) | \u2212 y \u2212 | T\u00b5\u03bd(tr, y) \u2212 | x | y \u2212 | = 4G d3y ei\u03c9 | x y | \u2212 Z T\u00b5\u03bd(\u03c9, y6.77) In this sequence, the \ufb01rst equation is simply the de\ufb01nition of the Fourier transform, the second line comes from the solution (6.74), the third line is a change of variables from t to tr, and the fourth line is once again the de\ufb01nition of the Fourier transform. We now make the approximations that our source is isolated, far away, and slowly moving. This means that we can consider the source to be centered at a (spatial) distance R, with the di\ufb00erent parts of the source at distances R + \u03b4R such that \u03b4R << R. Since it is slowly moving, most of the radiation emitted will be at frequencies \u03c9 su\ufb03ciently low that 1. (Essentially, light traverses the source much faster than the components of the \u03b4R << \u03c9\u2212 source itself do.) observer", " R R\u03b4 source Under these approximations, the term ei\u03c9 outside the integral. This leaves us with y x | \u2212 |/ x | y | \u2212 can be replaced by ei\u03c9R/R and brought \u00afh\u00b5\u03bd(\u03c9, x) = 4G e ei\u03c9R R Z e d3y T\u00b5\u03bd(\u03c9, y). (6.78) \u00afh\u00b5\u03bd(\u03c9, x), since the harmonic e In fact there is no need to compute all of the components of gauge condition \u2202\u00b5 \u00afh\u00b5\u03bd(t, x) = 0 in Fourier space implies \u00afh0\u03bd = i \u03c9 \u00afhi\u03bd. \u2202i e e We therefore only need to concern ourselves with the spacelike components of From (6.78) we therefore want to take the integral of the spacelike components of (6.79) \u00afh\u00b5\u03bd(\u03c9, x). T\u00b5\u03bd(\u03c9, y). e e 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 158 We begin by integrating by parts in reverse: d3y T ij(\u03c9, y) = \u2202k(yi T kj) d3y Z Z \u2212 Z yi(\u2202k T kj) d3y. (6.80) e The \ufb01rst term is a surface integral which will vanish since the source is isolated, while the second can be related to T 0j by the Fourier-space version of \u2202\u00b5T \u00b5\u03bd = 0: e e Thus, d3y Z e \u2202k T k\u00b5 = i\u03c9 T 0\u00b5. \u2212 e e (6.81) yi T 0j d3y (yi T 0j + yj e T 0i) d3y e Z = T ij(\u03c9, y) = i\u03c9 i\u03c9 2 Z i\u03c9 2 Z h \u03c92 2 Z \u2212 = = \u2202l(yiyj e yiyj(\u2202l T 0l) e \u2212 T 00 d3y. e yiyj T 0l) d3y i e (6.82) The second line is justi\ufb01ed since we know that the left hand side is symmetric in i and j, while the third and fourth lines are simply repetitions of reverse integration by parts and conservation of T \u00b5\u03bd. It is conventional to de\ufb01ne the quadrupole moment tensor of the energy density of the source, e qij(t) = 3 yiyjT 00(t, y) d3y, Z (6.83) a constant tensor on each surface of constant time. In terms of the Fourier transform of the quadrupole moment, our solution takes on the compact form \u00afhij(\u03c9, x) = 2G\u03c92 3 ei\u03c9R R \u2212 qij(\u03c9), or, transforming back to t, e e \u00afhij(t, x) = d\u03c9 e\u2212 i\u03c9(t \u2212 R)\u03c92 qij(\u03c9) \u2212 1 \u221a2\u03c0 1 \u221a2\u03c0 2G 3R 2G 3R Z d2 2G dt2 3R d2qij dt2 (tr), Z d\u03c9 e\u2212 i\u03c9tr e qij(\u03c9) e = = R. where as before tr = t \u2212 The gravitational wave produced by an isolated nonrelativistic object is therefore proportional to the second derivative of the quadrupole moment of the energy density at the (6.84) (6.85) 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 159 point where the past light cone of the observer intersects the source. In contrast, the leading contribution to electromagnetic radiation comes from the changing dipole moment of the charge density. The di\ufb00erence can be traced back to the universal nature of gravitation. A changing dipole moment corresponds to motion of the center of density \u2014 charge density in the case of electromagnetism, energy density in the case of gravitation. While there is nothing to stop the center of charge of an object from oscillating, oscillation of the center of mass of an isolated system violates conservation of momentum. (You can shake a body up and down, but you and the earth shake ever so slightly in the opposite direction to compensate.) The quadrupole moment, which measures the shape of the system, is generally smaller than the dipole moment, and for this reason (as well as the weak coupling of matter to gravity) gravitational radiation is typically much weaker than electromagnetic radiation. It is always educational to take a general solution and apply it to a speci\ufb01c case of interest. One case of genuine interest is the gravitational radiation emitted by a binary star (two stars in orbit", " around each other). For simplicity let us consider two stars of mass M in a circular orbit in the x1-x2 plane, at distance r from their common center of mass We will treat the motion of the stars in the Newtonian approximation, where we can discuss their orbit just as Kepler would have. Circular orbits are most easily characterized by equating the force due to gravity to the outward \u201ccentrifugal\u201d force: which gives us GM 2 (2r)2 = Mv2 r, v = 1/2 GM 4r. The time it takes to complete a single orbit is simply T = 2\u03c0r v, (6.86) (6.87) (6.88) 6 WEAK FIELDS AND GRAVITATIONAL RADIATION but more useful to us is the angular frequency of the orbit, \u2126 = 2\u03c0 T = GM 4r3 1/2. In terms of \u2126 we can write down the explicit path of star a, x1 a = r cos \u2126t, x2 a = r sin \u2126t, and star b, The corresponding energy density is x1 b = r cos \u2126t, \u2212 x2 b = r sin \u2126t. \u2212 160 (6.89) (6.90) (6.91) T 00(t, x) = M\u03b4(x3). (6.92) i The profusion of delta functions allows us to integrate this straightforwardly to obtain the quadrupole moment from (6.83): r sin \u2126t) + \u03b4(x1 + r cos \u2126t)\u03b4(x2 + r sin \u2126t) r cos \u2126t)\u03b4(x2 \u03b4(x1 h \u2212 \u2212 q11 = 6Mr2 cos2 \u2126t = 3Mr2(1 + cos 2\u2126t) q22 = 6Mr2 sin2 \u2126t = 3Mr2(1 cos 2\u2126t) q12 = q21 = 6Mr2(cos \u2126t)(sin \u2126t) = 3Mr2 sin 2\u2126t \u2212 qi3 = 0. (6.93) From this in turn it is easy to get the components of the metric perturbation from (6.85): \u00afhij(t, x) = 8GM R \u21262r2 \uf8eb \u2212 \u2212 cos 2\u2126tr \u2212 sin 2\u2126tr 0 sin 2\u2126tr cos 2\u2126tr 6.94) The remaining components of \u00afh\u00b5\u03bd could be derived from demanding that the harmonic gauge condition be satis\ufb01ed. (We have not imposed a subsidiary gauge condition, so we are still free to do so.) It is natural at this point to talk about the energy emitted via gravitational radiation. Such a discussion, however, is immediately beset by problems, both technical and philosophical. As we have mentioned before, there is no true local measure of the energy in the gravitational \ufb01eld. Of course, in the weak \ufb01eld limit, where we think of gravitation as being described by a symmetric tensor propagating on a \ufb01xed background metric, we might hope to derive an energy-momentum tensor for the \ufb02uctuations h\u00b5\u03bd, just as we would for electromagnetism or any other \ufb01eld theory. To some extent this is possible, but there are still di\ufb03culties. As a result of these di\ufb03culties there are a number of di\ufb00erent proposals in the literature for what we should use as the energy-momentum tensor for gravitation in the 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 161 weak \ufb01eld limit; all of them are di\ufb00erent, but for the most part they give the same answers for physically well-posed questions such as the rate of energy emitted by a binary system. At a technical level, the di\ufb03culties begin to arise when we consider what form the energymomentum tensor should take. We have previously mentioned the energy-momentum tensors for electromagnetism and scalar \ufb01eld theory, and they both shared an important feature \u2014 they were quadratic in the relevant \ufb01elds. By hypothesis our approach to the weak \ufb01eld limit has been to only keep terms which are linear in the metric perturbation. Hence, in order to keep track of the energy carried by the gravitational waves, we will have to extend our calculations to at least second order in h\u00b5\u03bd. In fact we have been cheating slightly all along. In discussing the e\ufffd", "\ufffd\ufffdects of gravitational waves on test particles, and the generation of waves by a binary system, we have been using the fact that test particles move along geodesics. But \u2207\u00b5T \u00b5\u03bd = 0. as we know, this is derived from the covariant conservation of energy-momentum, In the order to which we have been working, however, we actually have \u2202\u00b5T \u00b5\u03bd = 0, which would imply that test particles move on straight lines in the \ufb02at background metric. This is a symptom of the fundamental inconsistency of the weak \ufb01eld limit. In practice, the best that can be done is to solve the weak \ufb01eld equations to some appropriate order, and then justify after the fact the validity of the solution. Keeping these issues in mind, let us consider Einstein\u2019s equations (in vacuum) to second order, and see how the result can be interpreted in terms of an energy-momentum tensor for the gravitational \ufb01eld. If we write the metric as g\u00b5\u03bd = \u03b7\u00b5\u03bd + h\u00b5\u03bd, then at \ufb01rst order we have G(1) \u00b5\u03bd [\u03b7 + h] = 0, (6.95) where G(1) \u00b5\u03bd is Einstein\u2019s tensor expanded to \ufb01rst order in h\u00b5\u03bd. These equations determine h\u00b5\u03bd up to (unavoidable) gauge transformations, so in order to satisfy the equations at second order we have to add a higher-order perturbation, and write g\u00b5\u03bd = \u03b7\u00b5\u03bd + h\u00b5\u03bd + h(2) \u00b5\u03bd. (6.96) The second-order version of Einstein\u2019s equations consists of all terms either quadratic in h\u00b5\u03bd or linear in h(2) \u00b5\u03bd. Since any cross terms would be of at least third order, we have G(1) \u00b5\u03bd [\u03b7 + h(2)] + G(2) \u00b5\u03bd [\u03b7 + h] = 0. (6.97) Here, G(2) tion. It can be computed from the second-order Ricci tensor, which is given by \u00b5\u03bd is the part of the Einstein tensor which is of second order in the metric perturba- R(2) \u00b5\u03bd = 1 2 h\u03c1\u03c3\u2202\u00b5\u2202\u03bdh\u03c1\u03c3 \u2212 + \u2202\u03c3(h\u03c1\u03c3\u2202\u03c1h\u00b5\u03bd) h\u03c1\u03c3\u2202\u03c1\u2202(\u00b5h\u03bd)\u03c3 + 1 4 \u2212 1 2 (\u2202\u03c1h\u00b5\u03bd)\u2202\u03c1h 1 4 (\u2202\u00b5h\u03c1\u03c3)\u2202\u03bdh\u03c1\u03c3 + (\u2202\u03c3h\u03c1 \u03bd)\u2202[\u03c3h\u03c1]\u00b5 (\u2202\u03c3h\u03c1\u03c3 \u2212 1 2 \u2212 \u2202\u03c1h)\u2202(\u00b5h\u03bd)\u03c1. (6.98) 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 162 We can cast (6.97) into the suggestive form simply by de\ufb01ning G(1) \u00b5\u03bd [\u03b7 + h(2)] = 8\u03c0Gt\u00b5\u03bd, t\u00b5\u03bd = 1 8\u03c0G \u2212 G(2) \u00b5\u03bd [\u03b7 + h]. (6.99) (6.100) The notation is of course meant to suggest that we think of t\u00b5\u03bd as an energy-momentum tensor, speci\ufb01cally that of the gravitational \ufb01eld (at least in the weak \ufb01eld regime). To make this claim seem plausible, note that the Bianchi identity for G(1) \u00b5\u03bd [\u03b7 + h(2)] implies that t\u00b5\u03bd is conserved in the \ufb02at-space sense, \u2202\u00b5t\u00b5\u03bd = 0. (6.101) Unfortunately there are some limitations on our interpretation of t\u00b5\u03bd as an energymomentum tensor. Of course it is not a tensor at all in the full theory, but we are leaving that aside by hypothesis. More importantly, it is not invariant under gauge transformations (in\ufb01nitesimal di\ufb00eomorphisms), as you could check by direct calculation. However, we can construct global quantities which are invariant under certain special kinds of gauge transformations (basically, those that vanish su\ufb03ciently rapidly at in\ufb01nity; see Wald). These include the total energy on a surface \u03a3 of constant time, and the total energy radiated through to in\ufb01nity, t00 d3x, E = Z\u03a3 \u2206E = ZS t0\u00b5n\u00b5 d2", "x dt. (6.102) (6.103) Here, the integral is taken over a timelike surface S made of a spacelike two-sphere at in\ufb01nity and some interval in time, and n\u00b5 is a unit spacelike vector normal to S. Evaluating these formulas in terms of the quadrupole moment of a radiating source involves a lengthy calculation which we will not reproduce here. Without further ado, the amount of radiated energy can be written where the power P is given by \u2206E = P dt, Z P = G 45 \" d3Qij dt3 d3Qij dt3 #tr, and here Qij is the traceless part of the quadrupole moment, Qij = qij \u2212 1 3 \u03b4ij\u03b4klqkl. (6.104) (6.105) (6.106) 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 163 For the binary system represented by (6.93), the traceless part of the quadrupole is (1 + 3 cos 2\u2126t) 3 sin 2\u2126t 0 3 sin 2\u2126t (1 \u2212 3 cos 2\u2126t) 0 Qij = Mr2 \uf8eb \uf8ec \uf8ed and its third time derivative is therefore d3Qij dt3 = 24Mr2\u21263 \uf8eb \uf8ec \uf8ed The power radiated by the binary is thus sin 2\u2126t cos 2\u2126t 0 \u2212 \u2212 \u2212 cos 2\u2126t sin 2\u2126t \uf8f6 \uf8f7 \uf8f8 or, using expression (6.89) for the frequency, P = 27 5 GM 2r4\u21266, P = 2 5 G4M 5 r5. (6.107) (6.108) (6.109) (6.110) Of course, this has actually been observed. In 1974 Hulse and Taylor discovered a binary system, PSR1913+16, in which both stars are very small (so classical e\ufb00ects are negligible, or at least under control) and one is a pulsar. The period of the orbit is eight hours, extremely small by astrophysical standards. The fact that one of the stars is a pulsar provides a very accurate clock, with respect to which the change in the period as the system loses energy can be measured. The result is consistent with the prediction of general relativity for energy loss through gravitational radiation. Hulse and Taylor were awarded the Nobel Prize in 1993 for their e\ufb00orts. December 1997 Lecture Notes on General Relativity Sean M. Carroll 7 The Schwarzschild Solution and Black Holes We now move from the domain of the weak-\ufb01eld limit to solutions of the full nonlinear Einstein\u2019s equations. With the possible exception of Minkowski space, by far the most important such solution is that discovered by Schwarzschild, which describes spherically symmetric vacuum spacetimes. Since we are in vacuum, Einstein\u2019s equations become R\u00b5\u03bd = 0. Of course, if we have a proposed solution to a set of di\ufb00erential equations such as this, it would su\ufb03ce to plug in the proposed solution in order to verify it; we would like to do better, however. In fact, we will sketch a proof of Birkho\ufb00\u2019s theorem, which states that the Schwarzschild solution is the unique spherically symmetric solution to Einstein\u2019s equations in vacuum. The procedure will be to \ufb01rst present some non-rigorous arguments that any spherically symmetric metric (whether or not it solves Einstein\u2019s equations) must take on a certain form, and then work from there to more carefully derive the actual solution in such a case. \u201cSpherically symmetric\u201d means \u201chaving the same symmetries as a sphere.\u201d (In this section the word \u201csphere\u201d means S2, not spheres of higher dimension.) Since the object of interest to us is the metric on a di\ufb00erentiable manifold, we are concerned with those metrics that have such symmetries. We know how to characterize symmetries of the metric \u2014 they are given by the existence of Killing vectors. Furthermore, we know what the Killing vectors of S2 are, and that there are three of them. Therefore, a spherically symmetric manifold is one that has three Killing vector \ufb01elds which are just like those on S2. By \u201cjust like\u201d we mean that the commutator of the Killing vectors is the same in either case \u2014 in fancier language, that the algebra generated by the vectors is the same. Something", " that we didn\u2019t show, but is true, is that we can choose our three Killing vectors on S2 to be (V (1), V (2), V (3)), such that [V (1), V (2)] = V (3) [V (2), V (3)] = V (1) [V (3), V (1)] = V (2). (7.1) The commutation relations are exactly those of SO(3), the group of rotations in three dimensions. This is no coincidence, of course, but we won\u2019t pursue this here. All we need is that a spherically symmetric manifold is one which possesses three Killing vector \ufb01elds with the above commutation relations. Back in section three we mentioned Frobenius\u2019s Theorem, which states that if you have a set of commuting vector \ufb01elds then there exists a set of coordinate functions such that the vector \ufb01elds are the partial derivatives with respect to these functions. In fact the theorem 164 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 165 does not stop there, but goes on to say that if we have some vector \ufb01elds which do not commute, but whose commutator closes \u2014 the commutator of any two \ufb01elds in the set is a linear combination of other \ufb01elds in the set \u2014 then the integral curves of these vector \ufb01elds \u201c\ufb01t together\u201d to describe submanifolds of the manifold on which they are all de\ufb01ned. The dimensionality of the submanifold may be smaller than the number of vectors, or it could be equal, but obviously not larger. Vector \ufb01elds which obey (7.1) will of course form 2-spheres. Since the vector \ufb01elds stretch throughout the space, every point will be on exactly one of these spheres. (Actually, it\u2019s almost every point \u2014 we will show below how it can fail to be absolutely every point.) Thus, we say that a spherically symmetric manifold can be foliated into spheres. Let\u2019s consider some examples to bring this down to earth. The simplest example is \ufb02at three-dimensional Euclidean space. If we pick an origin, then R3 is clearly spherically symmetric with respect to rotations around this origin. Under such rotations (i.e., under the \ufb02ow of the Killing vector \ufb01elds) points move into each other, but each point stays on an S2 at a \ufb01xed distance from the origin. z R 3 x y It is these spheres which foliate R3. Of course, they don\u2019t really foliate all of the space, since the origin itself just stays put under rotations \u2014 it doesn\u2019t move around on some two-sphere. But it should be clear that almost all of the space is properly foliated, and this will turn out to be enough for us. We can also have spherical symmetry without an \u201corigin\u201d to rotate things around. An S2. If we suppress a dimension example is provided by a \u201cwormhole\u201d, with topology R and draw our two-spheres as circles, such a space might look like this: \u00d7 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 166 In this case the entire manifold can be foliated by two-spheres. This foliated structure suggests that we put coordinates on our manifold in a way which is adapted to the foliation. By this we mean that, if we have an n-dimensional manifold foliated by m-dimensional submanifolds, we can use a set of m coordinate functions ui on m coordinate functions vI to tell us which submanifold we the submanifolds and a set of n are on. (So i runs from 1 to m, while I runs from 1 to n m.) Then the collection of v\u2019s and u\u2019s coordinatize the entire space. If the submanifolds are maximally symmetric spaces (as two-spheres are), then there is the following powerful theorem: it is always possible to choose the u-coordinates such that the metric on the entire manifold is of the form \u2212 \u2212 ds2 = g\u00b5\u03bddx\u00b5dx\u03bd = gIJ(v)dvIdvJ + f (v)\u03b3ij(u)duiduj. (7.2) Here \u03b3ij(u) is the metric on the submanifold. This theorem is saying two things at once: that there are no cross terms dvIduj, and that both gIJ(v) and f (v", ") are functions of the vI alone, independent of the ui. Proving the theorem is a mess, but you are encouraged to look in chapter 13 of Weinberg. Nevertheless, it is a perfectly sensible result. Roughly speaking, if gIJ or f depended on the ui then the metric would change as we moved in a single submanifold, which violates the assumption of symmetry. The unwanted cross terms, meanwhile, can be eliminated by making sure that the tangent vectors \u2202/\u2202vI are orthogonal to the submanifolds \u2014 in other words, that we line up our submanifolds in the same way throughout the space. We are now through with handwaving, and can commence some honest calculation. For the case at hand, our submanifolds are two-spheres, on which we typically choose coordinates (\u03b8, \u03c6) in which the metric takes the form d\u21262 = d\u03b82 + sin2 \u03b8 d\u03c62. (7.3) Since we are interested in a four-dimensional spacetime, we need two more coordinates, which we can call a and b. The theorem (7.2) is then telling us that the metric on a spherically 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 167 symmetric spacetime can be put in the form ds2 = gaa(a, b)da2 + gab(a, b)(dadb + dbda) + gbb(a, b)db2 + r2(a, b)d\u21262. (7.4) Here r(a, b) is some as-yet-undetermined function, to which we have merely given a suggestive label. There is nothing to stop us, however, from changing coordinates from (a, b) to (a, r), by inverting r(a, b). (The one thing that could possibly stop us would be if r were a function of a alone; in this case we could just as easily switch to (b, r), so we will not consider this situation separately.) The metric is then ds2 = gaa(a, r)da2 + gar(a, r)(dadr + drda) + grr(a, r)dr2 + r2d\u21262. (7.5) Our next step is to \ufb01nd a function t(a, r) such that, in the (t, r) coordinate system, there are no cross terms dtdr + drdt in the metric. Notice that dt = \u2202t \u2202a da + \u2202t \u2202r dr, so dt2 = 2 \u2202t \u2202a! da2 + \u2202t \u2202a! \u2202t \u2202r! (dadr + drda) + 2 \u2202t \u2202r! dr2. We would like to replace the \ufb01rst three terms in the metric (7.5) by for some functions m and n. This is equivalent to the requirements mdt2 + ndr2, and m 2 \u2202t \u2202a! = gaa, n + m 2 \u2202t \u2202r! = grr, m \u2202t \u2202a! \u2202t \u2202r! = gar. (7.6) (7.7) (7.8) (7.9) (7.10) (7.11) We therefore have three equations for the three unknowns t(a, r), m(a, r), and n(a, r), just enough to determine them precisely (up to initial conditions for t). (Of course, they are \u201cdetermined\u201d in terms of the unknown functions gaa, gar, and grr, so in this sense they are still undetermined.) We can therefore put our metric in the form ds2 = m(t, r)dt2 + n(t, r)dr2 + r2d\u21262. (7.12) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 168 \u2212 To this point the only di\ufb00erence between the two coordinates t and r is that we have chosen r to be the one which multiplies the metric for the two-sphere. This choice was motivated by what we know about the metric for \ufb02at Minkowski space, which can be written ds2 = dt2 + dr2 + r2d\u21262. We know that the spacetime under consideration is Lorentzian, so either m or n will have to be negative. Let us choose m, the coe\ufb03cient of dt2, to be negative. This is not a choice we are simply allowed to make, and in fact we will see", " later that it can go wrong, but we will assume it for now. The assumption is not completely unreasonable, since we know that Minkowski space is itself spherically symmetric, and will therefore be described by (7.12). With this choice we can trade in the functions m and n for new functions \u03b1 and \u03b2, such that ds2 = e2\u03b1(t,r)dt2 + e2\u03b2(t,r)dr2 + r2d\u21262. \u2212 (7.13) This is the best we can do for a general metric in a spherically symmetric spacetime. The next step is to actually solve Einstein\u2019s equations, which will allow us to determine explicitly the functions \u03b1(t, r) and \u03b2(t, r). It is unfortunately necessary to compute the Christo\ufb00el symbols for (7.13), from which we can get the curvature tensor and thus the Ricci tensor. If we use labels (0, 1, 2, 3) for (t, r, \u03b8, \u03c6) in the usual way, the Christo\ufb00el symbols are given by \u03930 \u03b2)\u22021\u03b1 \u03930 00 = \u22020\u03b1 00 = e2(\u03b1 \u03931 \u2212 12 = 1 \u03932 r 2\u03b2 sin2 \u03b8 re\u2212 01 = \u22021\u03b1 \u03931 22 = \u03932 \u2212 33 = \u03931 01 = \u22020\u03b2 2\u03b2 re\u2212 \u03b1)\u22020\u03b2 \u2212 11 = \u22021\u03b2 11 = e2(\u03b2 \u03930 \u03931 13 = 1 \u03933 r 23 = cos \u03b8 \u03933 sin \u03b8 cos \u03b8 sin \u03b8. \u2212 \u03931 33 = \u2212 (7.14) (Anything not written down explicitly is meant to be zero, or related to what is written by symmetries.) From these we get the following nonvanishing components of the Riemann tensor: R0 R0 R0 R0 R0 R1 R1 R2 \u2212 re\u2212 re\u2212 re\u2212 re\u2212 101 = e2(\u03b2 202 = 303 = 212 = 313 = 212 = re\u2212 313 = re\u2212 323 = (1 \u2212 \u2212 \u2212 \u2212 \u2212 0\u03b2 + (\u22020\u03b2)2 \u03b1)[\u22022 2\u03b2\u22021\u03b1 2\u03b2 sin2 \u03b8 \u22021\u03b1 2\u03b1\u22020\u03b2 2\u03b1 sin2 \u03b8 \u22020\u03b2 2\u03b2\u22021\u03b2 2\u03b2 sin2 \u03b8 \u22021\u03b2 e\u2212 2\u03b2) sin2 \u03b8. \u22020\u03b1\u22020\u03b2] + [\u22021\u03b1\u22021\u03b2 \u2212 \u22022 1\u03b1 \u2212 \u2212 (\u22021\u03b1)2] (7.15) Taking the contraction as usual yields the Ricci tensor: R00 = [\u22022 0 \u03b2 + (\u22020\u03b2)2 \u2212 \u22020\u03b1\u22020\u03b2] + e2(\u03b1 \u2212 \u03b2)[\u22022 1 \u03b1 + (\u22021\u03b1)2 \u22021\u03b1\u22021\u03b2 + \u2212 2 r \u22021\u03b1] 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 169 \u22021\u03b1\u22021\u03b2 \u2212 2 r \u2212 \u22021\u03b2] + e2(\u03b2 \u2212 \u03b1)[\u22022 0\u03b2 + (\u22020\u03b2)2 \u22020\u03b1\u22020\u03b2] \u2212 [\u22022 1 \u03b1 + (\u22021\u03b1)2 R11 = R01 = \u2212 2 \u22020\u03b2 r 2\u03b2[r(\u22021\u03b2 R22 = e\u2212 R33 = R22 sin2 \u03b8. \u22021\u03b1) \u2212 \u2212 1] + 1 Our job is to set R\u00b5\u03bd = 0. From R01 = 0 we get \u22020\u03b2 = 0. If we consider taking the time derivative of R22 = 0 and using \u22020\u03b2 = 0, we get We can therefore write \u22020\u22021\u03b1 = 0. \u03b2 = \u03b2(r) \u03b1 = f (r) + g(t). (7.16) (7.17) (7.18) (7.19) The \ufb01rst term in the metric (7.13) is therefore rede\ufb01ne our time coordinate by replacing dt t such that g(t) = 0, whence \u03b1(t, r) = f (r). We therefore have \u2212 e\u2212 \u2192 e2f (r)e2g(t)dt2. But we could always simply g(t)dt; in other words, we are free to choose ds2 = e2\u03b1(r)dt2 + e\u03b2(r)dr2 + r2d\u21262. \u2212 (7.20) All of", " the metric components are independent of the coordinate t. We have therefore proven a crucial result: any spherically symmetric vacuum metric possesses a timelike Killing vector. This property is so interesting that it gets its own name: a metric which possesses a timelike Killing vector is called stationary. There is also a more restrictive property: a metric is called static if it possesses a timelike Killing vector which is orthogonal to a family of hypersurfaces. (A hypersurface in an n-dimensional manifold is simply an (n 1)dimensional submanifold.) The metric (7.20) is not only stationary, but also static; the Killing vector \ufb01eld \u22020 is orthogonal to the surfaces t = const (since there are no cross terms such as dtdr and so on). Roughly speaking, a static metric is one in which nothing is moving, while a stationary metric allows things to move but only in a symmetric way. For example, the static spherically symmetric metric (7.20) will describe non-rotating stars or black holes, while rotating systems (which keep rotating in the same way at all times) will be described by stationary metrics. It\u2019s hard to remember which word goes with which concept, but the distinction between the two concepts should be understandable. \u2212 Let\u2019s keep going with \ufb01nding the solution. Since both R00 and R11 vanish, we can write 0 = e2(\u03b2 \u03b1)R00 + R11 = \u2212 2 r (\u22021\u03b1 + \u22021\u03b2), (7.21) (7.22) (7.23) (7.24) (7.25) (7.26) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 170 which implies \u03b1 = our coordinates, so we have \u2212 \u03b2 + constant. Once again, we can get rid of the constant by scaling Next let us turn to R22 = 0, which now reads \u03b1 = \u03b2. \u2212 This is completely equivalent to We can solve this to obtain e2\u03b1(2r\u22021\u03b1 + 1) = 1. \u22021(re2\u03b1) = 1. e2\u03b1 = 1 + \u00b5 r, where \u00b5 is some undetermined constant. With (7.22) and (7.25), our metric becomes ds2 = 1 + dt2 + 1 + \u00b5 r \u2212 1 dr2 + r2d\u21262. \u00b5 \u2212 r We now have no freedom left except for the single constant \u00b5, so this form better solve the remaining equations R00 = 0 and R11 = 0; it is straightforward to check that it does, for any value of \u00b5. The only thing left to do is to interpret the constant \u00b5 in terms of some physical parameter. The most important use of a spherically symmetric vacuum solution is to represent the spacetime outside a star or planet or whatnot. In that case we would expect to recover the weak \ufb01eld limit as r. In this limit, (7.26) implies \u2192 \u221e g00(r grr(r ) = ) = \u2192 \u221e \u2192 \u221e The weak \ufb01eld limit, on the other hand, has g00 = \u2212 grr = (1 (1 + 2\u03a6), 2\u03a6), \u2212, (7.27) (7.28) with the potential \u03a6 = \u00b5 = 2GM. GM/r. Therefore the metrics do agree in this limit, if we set \u2212 \u2212 Our \ufb01nal result is the celebrated Schwarzschild metric, ds2 = 1 This is true for any spherically symmetric vacuum solution to Einstein\u2019s equations; M functions as a parameter, which we happen to know can be interpreted as the conventional 1 \u2212 r (7.29) \u2212 \u2212 dr2 + r2d\u21262. dt2 + 1 2GM \u2212 r 2GM 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 171 Newtonian mass that we would measure by studying orbits at large distances from the gravitating source. Note that as M 0 we recover Minkowski space, which is to be expected. Note also that the metric becomes progressively Minkowskian as we go to r ; this property is known as asymptotic \ufb02atness. \u2192 \u221e \u2192 The fact that the Schwarzschild metric is not just a good solution, but is the unique spherically symmetric vacuum solution, is known as Birkho\ufb00\u2019s theorem. It is interesting to note that the result is a static metric. We did not say anything about the source except that it be spherically symmetric. Speci\ufb01cally, we did not demand that the source itself be static; it could be a collapsing star, as long as the collapse", " were symmetric. Therefore a process such as a supernova explosion, which is basically spherical, would be expected to generate very little gravitational radiation (in comparison to the amount of energy released through other channels). This is the same result we would have obtained in electromagnetism, where the electromagnetic \ufb01elds around a spherical charge distribution do not depend on the radial distribution of the charges. Before exploring the behavior of test particles in the Schwarzschild geometry, we should say something about singularities. From the form of (7.29), the metric coe\ufb03cients become in\ufb01nite at r = 0 and r = 2GM \u2014 an apparent sign that something is going wrong. The metric coe\ufb03cients, of course, are coordinate-dependent quantities, and as such we should not make too much of their values; it is certainly possible to have a \u201ccoordinate singularity\u201d which results from a breakdown of a speci\ufb01c coordinate system rather than the underlying manifold. An example occurs at the origin of polar coordinates in the plane, where the 2 of the inverse metric ds2 = dr2 + r2d\u03b82 becomes degenerate and the component g\u03b8\u03b8 = r\u2212 metric blows up, even though that point of the manifold is no di\ufb00erent from any other. What kind of coordinate-independent signal should we look for as a warning that something about the geometry is out of control? This turns out to be a di\ufb03cult question to answer, and entire books have been written about the nature of singularities in general relativity. We won\u2019t go into this issue in detail, but rather turn to one simple criterion for when something has gone wrong \u2014 when the curvature becomes in\ufb01nite. The curvature is measured by the Riemann tensor, and it is hard to say when a tensor becomes in\ufb01nite, since its components are coordinate-dependent. But from the curvature we can construct various scalar quantities, and since scalars are coordinate-independent it will be meaningful to say that they become in\ufb01nite. This simplest such scalar is the Ricci scalar R = g\u00b5\u03bdR\u00b5\u03bd, but we can also construct higher-order scalars such as R\u00b5\u03bdR\u00b5\u03bd, R\u00b5\u03bd\u03c1\u03c3R\u00b5\u03bd\u03c1\u03c3, R\u00b5\u03bd\u03c1\u03c3R\u03c1\u03c3\u03bb\u03c4 R\u03bb\u03c4 \u00b5\u03bd, and so on. If any of these scalars (not necessarily all of them) go to in\ufb01nity as we approach some point, we will regard that point as a singularity of the curvature. We should also check that the point is not \u201cin\ufb01nitely far away\u201d; that is, that it can be reached by travelling a \ufb01nite distance along a curve. We therefore have a su\ufb03cient condition for a point to be considered a singularity. It is 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 172 not a necessary condition, however, and it is generally harder to show that a given point is nonsingular; for our purposes we will simply test to see if geodesics are well-behaved at the point in question, and if so then we will consider the point nonsingular. In the case of the Schwarzschild metric (7.29), direct calculation reveals that R\u00b5\u03bd\u03c1\u03c3R\u00b5\u03bd\u03c1\u03c3 = 12G2M 2 r6. (7.30) This is enough to convince us that r = 0 represents an honest singularity. At the other trouble spot, r = 2GM, you could check and see that none of the curvature invariants blows up. We therefore begin to think that it is actually not singular, and we have simply chosen a bad coordinate system. The best thing to do is to transform to more appropriate coordinates if possible. We will soon see that in this case it is in fact possible, and the surface r = 2GM is very well-behaved (although interesting) in the Schwarzschild metric. Having worried a little about singularities, we should point out that the behavior of Schwarzschild at r 2GM is of little day-to-day consequence. The solution we derived is valid only in vacuum, and we expect it to hold outside a spherical body such as a star. However, in the case of the Sun we are dealing with a body which extends to a radius of \u2264 R \u2299 = 106GM \u2299. (7.31) Thus, r = 2GM metric to imply. In fact, realistic stellar interior solutions are of the form is far inside the solar interior, where we do not expect the Schwarzschild \u2299 ds2 = 1 \u2212 \u2212 2Gm(r) r! dt2 + 1 1 \u2212", " 2Gm(r) r! \u2212 dr2 + r2d\u21262. (7.32) See Schutz for details. Here m(r) is a function of r which goes to zero faster than r itself, so there are no singularities to deal with at all. Nevertheless, there are objects for which the full Schwarzschild metric is required \u2014 black holes \u2014 and therefore we will let our imaginations roam far outside the solar system in this section. The \ufb01rst step we will take to understand this metric more fully is to consider the behavior of geodesics. We need the nonzero Christo\ufb00el symbols for Schwarzschild: 00 = GM \u03931 2GM) r3 (r \u2212 12 = 1 \u03932 r \u03931 33 = (r \u2212 \u2212 2GM) sin2 \u03b8 \u03931 \u03931 22 = 11 = \u2212 r(r \u2212 (r \u2212 \u2212 \u03932 33 = \u2212 GM 2GM ) 2GM) sin \u03b8 cos \u03b8 01 = GM \u03930 r(r \u2212 13 = 1 \u03933 r 23 = cos \u03b8 \u03933 2GM ) sin \u03b8. (7.33) The geodesic equation therefore turns into the following four equations, where \u03bb is an a\ufb03ne parameter: d2t d\u03bb2 + 2GM r(r \u2212 2GM) dr d\u03bb dt d\u03bb = 0, (7.34) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 173 d2r d\u03bb2 + GM r3 (r \u2212 2GM) (r \u2212 \u2212 2GM) \uf8ee d\u03b8 d\u03bb! 2 dt d\u03bb! 2 \u2212 + sin2 \u03b8 GM \u2212 d\u03c6 d\u03bb! r(r 2GM) 2 2 dr d\u03bb! = 0, \uf8f9 \uf8fb d2\u03b8 d\u03bb2 + 2 r d\u03b8 d\u03bb \uf8f0 dr d\u03bb \u2212 sin \u03b8 cos \u03b8 2 d\u03c6 d\u03bb! = 0, and d2\u03c6 d\u03bb2 + 2 r d\u03c6 d\u03bb dr d\u03bb + 2 cos \u03b8 sin \u03b8 d\u03b8 d\u03bb d\u03c6 d\u03bb = 0. (7.35) (7.36) (7.37) There does not seem to be much hope for simply solving this set of coupled equations by inspection. Fortunately our task is greatly simpli\ufb01ed by the high degree of symmetry of the Schwarzschild metric. We know that there are four Killing vectors: three for the spherical symmetry, and one for time translations. Each of these will lead to a constant of the motion for a free particle; if K \u00b5 is a Killing vector, we know that K\u00b5 dx\u00b5 d\u03bb = constant. (7.38) In addition, there is another constant of the motion that we always have for geodesics; metric compatibility implies that along the path the quantity \u01eb = g\u00b5\u03bd \u2212 dx\u00b5 d\u03bb dx\u03bd d\u03bb (7.39) \u2212 is constant. Of course, for a massive particle we typically choose \u03bb = \u03c4, and this relation g\u00b5\u03bdU \u00b5U \u03bd = +1. For a massless particle we always have \u01eb = 0. We will simply becomes \u01eb = also be concerned with spacelike geodesics (even though they do not correspond to paths of particles), for which we will choose \u01eb = \u2212 1. Rather than immediately writing out explicit expressions for the four conserved quantities associated with Killing vectors, let\u2019s think about what they are telling us. Notice that the symmetries they represent are also present in \ufb02at spacetime, where the conserved quantities they lead to are very familiar. Invariance under time translations leads to conservation of energy, while invariance under spatial rotations leads to conservation of the three components of angular momentum. Essentially the same applies to the Schwarzschild metric. We can think of the angular momentum as a three-vector with a magnitude (one component) and direction (two components). Conservation of the direction of angular momentum means that the particle will move in a plane. We can choose this to be the equatorial plane of our coordinate system; if the particle is not in this plane, we can rotate coordinates until it is. Thus, the two Killing vectors which lead to conservation of the direction of angular momentum imply \u03b8 =. (7.40) \u03c0 2 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 174 The two remaining Killing vectors correspond to energy and the magnitude of angular momentum. The energy arises from the timelike Killing vector K = \u2202t, or K\u00b5 = 1 \u2212 \u2212 2GM r, 0, 0, 0. (7.41) The Killing vector whose conserved quantity is the magnitude of the", " angular momentum is L = \u2202\u03c6, or L\u00b5 = 0, 0, 0, r2 sin2 \u03b8. (7.42) Since (7.40) implies that sin \u03b8 = 1 along the geodesics of interest to us, the two conserved quantities are and 1 \u2212 2GM r dt d\u03bb r2 d\u03c6 d\u03bb = L. = E, (7.43) (7.44) For massless particles these can be thought of as the energy and angular momentum; for massive particles they are the energy and angular momentum per unit mass of the particle. Note that the constancy of (7.44) is the GR equivalent of Kepler\u2019s second law (equal areas are swept out in equal times). Together these conserved quantities provide a convenient way to understand the orbits of particles in the Schwarzschild geometry. Let us expand the expression (7.39) for \u01eb to obtain 1 \u2212 \u2212 2GM r 2 dt d\u03bb! + 1 \u2212 1 2GM \u2212 r 2 dr d\u03bb! + r2 2 d\u03c6 d\u03bb! = \u01eb. \u2212 If we multiply this by (1 \u2212 2GM/r) and use our expressions for E and L, we obtain E2 + \u2212 2 dr d\u03bb! + 1 \u2212 2GM r L2 r2 + \u01eb! = 0. (7.45) (7.46) This is certainly progress, since we have taken a messy system of coupled equations and obtained a single equation for r(\u03bb). It looks even nicer if we rewrite it as where 2 1 2 dr d\u03bb! + V (r) = 1 2 E2, V (r) = 1 2 \u01eb \u2212 \u01eb GM r + L2 2r2 \u2212 GML2 r3. (7.47) (7.48) In (7.47) we have precisely the equation for a classical particle of unit mass and \u201cenergy\u201d 1 2E2 moving in a one-dimensional potential given by V (r). (The true energy per unit mass is E, but the e\ufb00ective potential for the coordinate r responds to 1 2E2.) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 175 Of course, our physical situation is quite di\ufb00erent from a classical particle moving in one dimension. The trajectories under consideration are orbits around a star or other object: \u03bbr( ) \u03bb r( ) The quantities of interest to us are not only r(\u03bb), but also t(\u03bb) and \u03c6(\u03bb). Nevertheless, we can go a long way toward understanding all of the orbits by understanding their radial behavior, and it is a great help to reduce this behavior to a problem we know how to solve. A similar analysis of orbits in Newtonian gravity would have produced a similar result; the general equation (7.47) would have been the same, but the e\ufb00ective potential (7.48) would not have had the last term. (Note that this equation is not a power series in 1/r, it is exact.) In the potential (7.48) the \ufb01rst term is just a constant, the second term corresponds exactly to the Newtonian gravitational potential, and the third term is a contribution from angular momentum which takes the same form in Newtonian gravity and general relativity. The last term, the GR contribution, will turn out to make a great deal of di\ufb00erence, especially at small r. Let us examine the kinds of possible orbits, as illustrated in the \ufb01gures. There are di\ufb00erent curves V (r) for di\ufb00erent values of L; for any one of these curves, the behavior of the orbit can be judged by comparing the 1 2E2 to V (r). The general behavior of the particle will be to move in the potential until it reaches a \u201cturning point\u201d where V (r) = 1 2E2, where it will begin moving in the other direction. Sometimes there may be no turning point to hit, in which case the particle just keeps going. In other cases the particle may simply move in a circular orbit at radius rc = const; this can happen if the potential is \ufb02at, dV /dr = 0. Di\ufb00erentiating (7.48), we \ufb01nd that the circular orbits occur when \u01ebGMr2 c \u2212 L2rc + 3GML2\u03b3 = 0, (7.49) where \u03b3 = 0 in Newtonian gravity and \u03b3 = 1 in general relativity. Circular orbits will be stable if they correspond to a minimum of the potential, and unstable if they correspond to a maximum. Bound orbits which are not circular will oscillate around the radius of the stable circular orbit. Turning to Newtonian gravity, we", " \ufb01nd that circular orbits appear at rc = L2 \u01ebGM. (7.50) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 176 0.8 0.6 0.4 0.2 0 0 0.8 0.6 0.4 0.2 0 0 Newtonian Gravity massive particles 5 4 3 2 L=1 10 20 30 r Newtonian Gravity massless particles L=5 4 3 2 1 10 20 30 r 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 177 \u221e For massless particles \u01eb = 0, and there are no circular orbits; this is consistent with the \ufb01gure, which illustrates that there are no bound orbits of any sort. Although it is somewhat obscured in this coordinate system, massless particles actually move in a straight line, since the Newtonian gravitational force on a massless particle is zero. (Of course the standing of massless particles in Newtonian theory is somewhat problematic, but we will ignore that for now.) In terms of the e\ufb00ective potential, a photon with a given energy E will come in from and gradually \u201cslow down\u201d (actually dr/d\u03bb will decrease, but the speed of light isn\u2019t r = changing) until it reaches the turning point, when it will start moving away back to r =. \u221e The lower values of L, for which the photon will come closer before it starts moving away, are simply those trajectories which are initially aimed closer to the gravitating body. For massive particles there will be stable circular orbits at the radius (7.50), as well as bound orbits which oscillate around this radius. If the energy is greater than the asymptotic value E = 1, the orbits will be unbound, describing a particle that approaches the star and then recedes. We know that the orbits in Newton\u2019s theory are conic sections \u2014 bound orbits are either circles or ellipses, while unbound ones are either parabolas or hyperbolas \u2014 although we won\u2019t show that here. \u2212 \u2192 \u2192 \u221e \u2212\u221e GML2/r3, as r 0 the potential goes to In general relativity the situation is di\ufb00erent, but only for r su\ufb03ciently small. Since the the behaviors are identical in the two di\ufb00erence resides in the term theories. But as r as in the Newtonian rather than + case. At r = 2GM the potential is always zero; inside this radius is the black hole, which we will discuss more thoroughly later. For massless particles there is always a barrier (except for L = 0, for which the potential vanishes identically), but a su\ufb03ciently energetic photon will nevertheless go over the barrier and be dragged inexorably down to the center. (Note that \u201csu\ufb03ciently energetic\u201d means \u201cin comparison to its angular momentum\u201d \u2014 in fact the frequency of the photon is immaterial, only the direction in which it is pointing.) At the top of the barrier there are unstable circular orbits. For \u01eb = 0, \u03b3 = 1, we can easily solve (7.49) to obtain \u221e rc = 3GM. (7.51) This is borne out by the \ufb01gure, which shows a maximum of V (r) at r = 3GM for every L. This means that a photon can orbit forever in a circle at this radius, but any perturbation will cause it to \ufb02y away either to r = 0 or r = \u221e For massive particles there are once again di\ufb00erent regimes depending on the angular. momentum. The circular orbits are at L2 \u00b1 rc = \u221aL4 12G2M 2L2 \u2212 2GM. For large L there will be two circular orbits, one stable and one unstable. In the L (7.52) \u2192 \u221e 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 178 General Relativity massive particles 10 20 30 r General Relativity massless particles 0.8 0.6 0.4 0.2 0 0 0.8 0.6 0.4 0.2 5 4 3 2 L=1 L=5 4 3 2 1 0 0 10 20 30 r 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 179 limit their radii are given by L2 rc = \u00b1 L2(1 6G2M 2/L2) \u2212 2GM = L2 GM, 3GM.! (7.53) In this limit the stable circular orbit becomes farther and farther away, while the unstable one approaches 3GM, behavior which parallels the massless case. As we decrease L the two circular orbits come closer together; they coincide when the discriminant in (7.52) vanishes, at for which L = \ufffd", "\ufffd12GM, rc = 6GM, (7.54) (7.55) and disappear entirely for smaller L. Thus 6GM is the smallest possible radius of a stable circular orbit in the Schwarzschild metric. There are also unbound orbits, which come in from in\ufb01nity and turn around, and bound but noncircular ones, which oscillate around the stable circular radius. Note that such orbits, which would describe exact conic sections in Newtonian gravity, will not do so in GR, although we would have to solve the equation for d\u03c6/dt to demonstrate it. Finally, there are orbits which come in from in\ufb01nity and continue all the way in to r = 0; this can happen either if the energy is higher than the barrier, or for L < \u221a12GM, when the barrier goes away entirely. We have therefore found that the Schwarzschild solution possesses stable circular orbits for r > 6GM and unstable circular orbits for 3GM < r < 6GM. It\u2019s important to remember that these are only the geodesics; there is nothing to stop an accelerating particle from dipping below r = 3GM and emerging, as long as it stays beyond r = 2GM. Most experimental tests of general relativity involve the motion of test particles in the solar system, and hence geodesics of the Schwarzschild metric; this is therefore a good place to pause and consider these tests. Einstein suggested three tests: the de\ufb02ection of light, the precession of perihelia, and gravitational redshift. The de\ufb02ection of light is observable in the weak-\ufb01eld limit, and therefore is not really a good test of the exact form of the Schwarzschild geometry. Observations of this de\ufb02ection have been performed during eclipses of the Sun, with results which agree with the GR prediction (although it\u2019s not an especially clean experiment). The precession of perihelia re\ufb02ects the fact that noncircular orbits are not closed ellipses; to a good approximation they are ellipses which precess, describing a \ufb02ower pattern. Using our geodesic equations, we could solve for d\u03c6/d\u03bb as a power series in the eccentricity e of the orbit, and from that obtain the apsidal frequency \u03c9a, de\ufb01ned as 2\u03c0 divided by the time it takes for the ellipse to precess once around. For details you can look in Weinberg; the answer is \u03c9a = 3(GM)3/2 e2)r5/2, c2(1 \u2212 (7.56) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 180 where we have restored the c to make it easier to compare with observation. (It is a good exercise to derive this yourself to lowest nonvanishing order, in which case the e2 is missing.) Historically the precession of Mercury was the \ufb01rst test of GR. For Mercury the relevant numbers are GM c2 \u2299 = 1.48 a = 5.55 105 cm, 1012 cm, \u00d7 \u00d7 (7.57) \u00d7 1010 cm/sec. This gives \u03c9a = 2.35 1. In other words, and of course c = 3.00 the major axis of Mercury\u2019s orbit precesses at a rate of 42.9 arcsecs every 100 years. The observed value is 5601 arcsecs/100 yrs. However, much of that is due to the precession of equinoxes in our geocentric coordinate system; 5025 arcsecs/100 yrs, to be precise. The gravitational perturbations of the other planets contribute an additional 532 arcsecs/100 yrs, leaving 43 arcsecs/100 yrs to be explained by GR, which it does quite well. 14 sec\u2212 10\u2212 \u00d7 The gravitational redshift, as we have seen, is another e\ufb00ect which is present in the weak \ufb01eld limit, and in fact will be predicted by any theory of gravity which obeys the Principle of Equivalence. However, this only applies to small enough regions of spacetime; over larger distances, the exact amount of redshift will depend on the metric, and thus on the theory under question. It is therefore worth computing the redshift in the Schwarzschild geometry. We consider two observers who are not moving on geodesics, but are stuck at \ufb01xed spatial coordinate values (r1, \u03b81, \u03c61) and (r2, \u03b82, \u03c62). According to (7.45), the proper time of observer i will be related to the coordinate time t by d\u03c4i dt = 1 2GM 1/2. ri O1 emits a light pulse which travels", " to the observer O2, such that Suppose that the observer O1 measures the time between two successive crests of the light wave to be \u2206\u03c41. Each crest follows the same path to O2, except that they are separated by a coordinate time \u2212 (7.58) \u2206t = 1/2 1 \u2212 2GM \u2212 r1 \u2206\u03c41. (7.59) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 181 This separation in coordinate time does not change along the photon trajectories, but the second observer measures a time between successive crests given by \u2206\u03c42 = = 1 1 1 \u2212 \u2212 \u2212 \u2206t 2GM 1/2 r2 2GM/r2 2GM/r1! 1/2 \u2206\u03c41. (7.60) Since these intervals \u2206\u03c4i measure the proper time between two crests of an electromagnetic wave, the observed frequencies will be related by \u03c92 \u03c91 = = \u2206\u03c41 \u2206\u03c42 1 1 \u2212 \u2212 2GM/r1 2GM/r2! 1/2. (7.61) This is an exact result for the frequency shift; in the limit r >> 2GM we have \u03c92 \u03c91 GM = 1 r1 \u2212 = 1 + \u03a61 \u2212 This tells us that the frequency goes down as \u03a6 increases, which happens as we climb out of a gravitational \ufb01eld; thus, a redshift. You can check that it agrees with our previous calculation based on the equivalence principle. GM r2 (7.62) \u03a62. + Since Einstein\u2019s proposal of the three classic tests, further tests of GR have been proposed. The most famous is of course the binary pulsar, discussed in the previous section. Another is the gravitational time delay, discovered by (and observed by) Shapiro. This is just the fact that the time elapsed along two di\ufb00erent trajectories between two events need not be the same. It has been measured by re\ufb02ecting radar signals o\ufb00 of Venus and Mars, and once again is consistent with the GR prediction. One e\ufb00ect which has not yet been observed is the Lense-Thirring, or frame-dragging e\ufb00ect. There has been a long-term e\ufb00ort devoted to a proposed satellite, dubbed Gravity Probe B, which would involve extraordinarily precise gyroscopes whose precession could be measured and the contribution from GR sorted out. It has a ways to go before being launched, however, and the survival of such projects is always year-to-year. We now know something about the behavior of geodesics outside the troublesome radius r = 2GM, which is the regime of interest for the solar system and most other astrophysical situations. We will next turn to the study of objects which are described by the Schwarzschild solution even at radii smaller than 2GM \u2014 black holes. (We\u2019ll use the term \u201cblack hole\u201d for the moment, even though we haven\u2019t introduced a precise meaning for such an object.) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 182 One way of understanding a geometry is to explore its causal structure, as de\ufb01ned by the light cones. We therefore consider radial null curves, those for which \u03b8 and \u03c6 are constant and ds2 = 0: ds2 = 0 = 1 \u2212 \u2212 dt2 + 1 \u2212 r 2GM 1 2GM \u2212 r dr2, (7.63) from which we see that dt dr = 1 1 2GM \u2212 r. \u2212 This of course measures the slope of the light cones on a spacetime diagram of the t-r plane. 1, as it would be in \ufb02at space, while as we approach r = 2GM we For large r the slope is get dt/dr, and the light cones \u201cclose up\u201d: \u00b1 \u00b1 \u2192 \u00b1\u221e (7.64) t 2GM r Thus a light ray which approaches r = 2GM never seems to get there, at least in this coordinate system; instead it seems to asymptote to this radius. As we will see, this is an illusion, and the light ray (or a massive particle) actually has no trouble reaching r = 2GM. But an observer far away would never be able to tell. If we stayed outside while an intrepid observational general relativist dove into the black hole, sending back signals all the time, we would simply see the signals reach us more and more slowly. This should be clear from the pictures, and is con\ufb01rmed by our computation of \u2206\u03c41/\u2206\u03c42 when we discussed the gravitational redshift (7.61). As infalling astronauts approach r = 2GM, any \ufb01xed interval \u2206", "\u03c41 of their proper time corresponds to a longer and longer interval \u2206\u03c42 from our point of view. This continues forever; we would never see the astronaut cross r = 2GM, we would just see them move more and more slowly (and become redder and redder, almost as if they were embarrassed to have done something as stupid as diving into a black hole). The fact that we never see the infalling astronauts reach r = 2GM is a meaningful statement, but the fact that their trajectory in the t-r plane never reaches there is not. It is highly dependent on our coordinate system, and we would like to ask a more coordinateindependent question (such as, do the astronauts reach this radius in a \ufb01nite amount of their proper time?). The best way to do this is to change coordinates to a system which is better 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 183 t \u2206\u03c4 1 \u2206\u03c4 1 2GM \u2206\u03c4 > \u2206\u03c4 \u2019 2 2 \u2206\u03c4 2 r behaved at r = 2GM. There does exist a set of such coordinates, which we now set out to \ufb01nd. There is no way to \u201cderive\u201d a coordinate transformation, of course, we just say what the new coordinates are and plug in the formulas. But we will develop these coordinates in several steps, in hopes of making the choices seem somewhat motivated. The problem with our current coordinates is that dt/dr along radial null geodesics which approach r = 2GM; progress in the r direction becomes slower and slower with respect to the coordinate time t. We can try to \ufb01x this problem by replacing t with a coordinate which \u201cmoves more slowly\u201d along null geodesics. First notice that we can explicitly solve the condition (7.64) characterizing radial null curves to obtain \u2192 \u221e where the tortoise coordinate r\u2217 is de\ufb01ned by t = r\u2217 + constant, \u00b1 (7.65) (7.66) 1. r\u2217 = r + 2GM ln r 2GM \u2212 (The tortoise coordinate is only sensibly related to r when r 2GM, but beyond there our coordinates aren\u2019t very good anyway.) In terms of the tortoise coordinate the Schwarzschild metric becomes \u2265 2GM ds2 = 1 dt2 + dr\u2217 2 + r2d\u21262, (7.67) \u2212 \u2212 where r is thought of as a function of r\u2217. This represents some progress, since the light cones now don\u2019t seem to close up; furthermore, none of the metric coe\ufb03cients becomes in\ufb01nite at r = 2GM (although both gtt and gr\u2217r\u2217 become zero). The price we pay, however, is that the surface of interest at r = 2GM has just been pushed to in\ufb01nity. r Our next move is to de\ufb01ne coordinates which are naturally adapted to the null geodesics. If we let \u02dcu = t + r\u2217 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 184 t r = 2GM r* = - 8 r* \u02dcv = t r\u2217, \u2212 (7.68) then infalling radial null geodesics are characterized by \u02dcu = constant, while the outgoing ones satisfy \u02dcv = constant. Now consider going back to the original radial coordinate r, but replacing the timelike coordinate t with the new coordinate \u02dcu. These are known as Eddington-Finkelstein coordinates. In terms of them the metric is ds2 = 1 \u2212 \u2212 2GM r d\u02dcu2 + (d\u02dcudr + drd\u02dcu) + r2d\u21262. (7.69) Here we see our \ufb01rst sign of real progress. Even though the metric coe\ufb03cient g\u02dcu\u02dcu vanishes at r = 2GM, there is no real degeneracy; the determinant of the metric is g = r4 sin2 \u03b8, \u2212 (7.70) which is perfectly regular at r = 2GM. Therefore the metric is invertible, and we see once and for all that r = 2GM is simply a coordinate singularity in our original (t, r, \u03b8, \u03c6) system. In the Eddington-Finkelstein coordinates the condition for radial null curves is solved by d\u02dcu dr = ( 0, 2 2GM r 1 \u2212. (infalling) (outgoing) (7.71) 1 \u2212 We can therefore see what has happened: in this coordinate system the light cones remain well-behaved at r = 2GM, and this surface is at a \ufb01nite coordinate value. There is no problem in", " tracing the paths of null or timelike particles past the surface. On the other hand, something interesting is certainly going on. Although the light cones don\u2019t close up, they do tilt over, such that for r < 2GM all future-directed paths are in the direction of decreasing r. The surface r = 2GM, while being locally perfectly regular, globally functions as a point of no return \u2014 once a test particle dips below it, it can never come back. For this reason 2GM can in\ufb02uence any other r = 2GM is known as the event horizon; no event at r \u2264 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 185 ~ u ~ u = const r r = 0 r = 2GM event at r > 2GM. Notice that the event horizon is a null surface, not a timelike one. Notice also that since nothing can escape the event horizon, it is impossible for us to \u201csee inside\u201d \u2014 thus the name black hole. Let\u2019s consider what we have done. Acting under the suspicion that our coordinates may not have been good for the entire manifold, we have changed from our original coordinate t to the new one \u02dcu, which has the nice property that if we decrease r along a radial curve null curve \u02dcu = constant, we go right through the event horizon without any problems. (Indeed, a local observer actually making the trip would not necessarily know when the event horizon had been crossed \u2014 the local geometry is no di\ufb00erent than anywhere else.) We therefore conclude that our suspicion was correct and our initial coordinate system didn\u2019t do a good job of covering the entire manifold. The region r 2GM should certainly be included in our spacetime, since physical particles can easily reach there and pass through. However, there is no guarantee that we are \ufb01nished; perhaps there are other directions in which we can extend our manifold. \u2264 In fact there are. Notice that in the (\u02dcu, r) coordinate system we can cross the event horizon on future-directed paths, but not on past-directed ones. This seems unreasonable, since we started with a time-independent solution. But we could have chosen \u02dcv instead of \u02dcu, in which case the metric would have been ds2 = 1 \u2212 \u2212 2GM r d\u02dcv2 \u2212 (d\u02dcvdr + drd\u02dcv) + r2d\u21262. (7.72) Now we can once again pass through the event horizon, but this time only along past-directed curves. This is perhaps a surprise: we can consistently follow either future-directed or pastdirected curves through r = 2GM, but we arrive at di\ufb00erent places. It was actually to be expected, since from the de\ufb01nitions (7.68), if we keep \u02dcu constant and decrease r we must have t, while if we keep \u02dcv constant and decrease r we must have t. (The 2GM.) So we have extended spacetime in two tortoise coordinate r\u2217 goes to di\ufb00erent directions, one to the future and one to the past. \u2192 \u2212\u221e \u2212\u221e as r \u2192 \u2192 \u221e + 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 186 ~ v ~ v = const r r = 0 r = 2GM The next step would be to follow spacelike geodesics to see if we would uncover still more regions. The answer is yes, we would reach yet another piece of the spacetime, but let\u2019s shortcut the process by de\ufb01ning coordinates that are good all over. A \ufb01rst guess might be to use both \u02dcu and \u02dcv at once (in place of t and r), which leads to ds2 = 1 1 2 \u2212 2GM r (d\u02dcud\u02dcv + d\u02dcvd\u02dcu) + r2d\u21262, (7.73) with r de\ufb01ned implicitly in terms of \u02dcu and \u02dcv by 1 2 (\u02dcu \u2212 \u02dcv) = r + 2GM ln r 2GM \u2212 1. (7.74) We have actually re-introduced the degeneracy with which we started out; in these coordinates r = 2GM is \u201cin\ufb01nitely far away\u201d (at either \u02dcu = ). The thing to do is to change to coordinates which pull these points into \ufb01nite coordinate values; a good choice is or \u02dcv = + \u2212\u221e \u221e u\u2032 = e\u02dcu/4GM v\u2032 = e\u2212 \u02dcv/4GM, which in terms of our original (t, r) system is u\u2032 = v\u2032 = r 2GM \u2212 r 2GM \u2212 1/2 1/2 1 1 e(r", "+t)/4GM e(r t)/4GM. \u2212 (7.75) (7.76) In the (u\u2032, v\u2032, \u03b8, \u03c6) system the Schwarzschild metric is ds2 = 16G3M 3 r \u2212 e\u2212 r/2GM (du\u2032dv\u2032 + dv\u2032du\u2032) + r2d\u21262. (7.77) Finally the nonsingular nature of r = 2GM becomes completely manifest; in this form none of the metric coe\ufb03cients behave in any special way at the event horizon. 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 187 Both u\u2032 and v\u2032 are null coordinates, in the sense that their partial derivatives \u2202/\u2202u\u2032 and \u2202/\u2202v\u2032 are null vectors. There is nothing wrong with this, since the collection of four partial derivative vectors (two null and two spacelike) in this system serve as a perfectly good basis for the tangent space. Nevertheless, we are somewhat more comfortable working in a system where one coordinate is timelike and the rest are spacelike. We therefore de\ufb01ne and u = = 1 2 v\u2032) \u2212 (u\u2032 r 2GM \u2212 1/2 1 er/4GM cosh(t/4GM) (7.78) v = = 1 2 (u\u2032 + v\u2032) r 2GM \u2212 1/2 1 er/4GM sinh(t/4GM), (7.79) in terms of which the metric becomes ds2 = 32G3M 3 r e\u2212 r/2GM ( dv2 + du2) + r2d\u21262, \u2212 where r is de\ufb01ned implicitly from (u2 \u2212 v2) = r 2GM \u2212 1 er/2GM. (7.80) (7.81) The coordinates (v, u, \u03b8, \u03c6) are known as Kruskal coordinates, or sometimes KruskalSzekres coordinates. Note that v is the timelike coordinate. The Kruskal coordinates have a number of miraculous properties. Like the (t, r\u2217) coor- dinates, the radial null curves look like they do in \ufb02at space: Unlike the (t, r\u2217) coordinates, however, the event horizon r = 2GM is not in\ufb01nitely far away; in fact it is de\ufb01ned by v = u + constant. \u00b1 (7.82) \u00b1 consistent with it being a null surface. More generally, we can consider the surfaces r = constant. From (7.81) these satisfy v = u, (7.83) u2 \u2212 v2 = constant. (7.84) Thus, they appear as hyperbolae in the u-v plane. Furthermore, the surfaces of constant t are given by = tanh(t/4GM), (7.85) v u 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 188 which de\ufb01nes straight lines through the origin with slope tanh(t/4GM). Note that as t this becomes the same as (7.83); therefore these surfaces are the same as r = 2GM. Now, our coordinates (v, u) should be allowed to range over every value they can take \u00b1\u221e \u2192 without hitting the real singularity at r = 2GM; the allowed region is therefore \u2212\u221e \u2264 and v2 < u2 + 1. We can now draw a spacetime diagram in the v-u plane (with u \u03b8 and \u03c6 suppressed), known as a \u201cKruskal diagram\u201d, which represents the entire spacetime corresponding to the Schwarzschild metric. \u2264 \u221e r = 2GM t = - 8 r = const r = 2GM = 2GM t = + 8 u t = const r = 2GM t = - 8 Each point on the diagram is a two-sphere. 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 189 Our original coordinates (t, r) were only good for r > 2GM, which is only a part of the manifold portrayed on the Kruskal diagram. It is convenient to divide the diagram into four regions: IV II III I The region in which we started was region I; by following future-directed null rays we reached region II, and by following past-directed null rays we reached region III. If we had explored spacelike geodesics, we would have been led to region IV. The de\ufb01nitions (7.78) and (7.79) which relate (u, v) to (t, r) are really only good in region I; in the other regions it is necessary to introduce appropriate minus signs to prevent the coordinates from", " becoming imaginary. Having extended the Schwarzschild geometry as far as it will go, we have described a remarkable spacetime. Region II, of course, is what we think of as the black hole. Once anything travels from region I into II, it can never return. In fact, every future-directed path in region II ends up hitting the singularity at r = 0; once you enter the event horizon, you are utterly doomed. This is worth stressing; not only can you not escape back to region I, you cannot even stop yourself from moving in the direction of decreasing r, since this is simply the timelike direction. (This could have been seen in our original coordinate system; for r < 2GM, t becomes spacelike and r becomes timelike.) Thus you can no more stop moving toward the singularity than you can stop getting older. Since proper time is maximized along a geodesic, you will live the longest if you don\u2019t struggle, but just relax as you approach the singularity. Not that you will have long to relax. (Nor that the voyage will be very relaxing; as you approach the singularity the tidal forces become in\ufb01nite. As you fall toward the singularity your feet and head will be pulled apart from each other, while your torso is squeezed to in\ufb01nitesimal thinness. The grisly demise of an astrophysicist falling into a black hole is detailed in Misner, Thorne, and Wheeler, section 32.6. Note that they use orthonormal frames [not that it makes the trip any more enjoyable].) Regions III and IV might be somewhat unexpected. Region III is simply the time-reverse of region II, a part of spacetime from which things can escape to us, while we can never get there. It can be thought of as a \u201cwhite hole.\u201d There is a singularity in the past, out of which the universe appears to spring. The boundary of region III is sometimes called the past 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 190 event horizon, while the boundary of region II is called the future event horizon. Region IV, meanwhile, cannot be reached from our region I either forward or backward in time (nor can anybody from over there reach us). It is another asymptotically \ufb02at region of spacetime, a mirror image of ours. It can be thought of as being connected to region I by a \u201cwormhole,\u201d a neck-like con\ufb01guration joining two distinct regions. Consider slicing up the Kruskal diagram into spacelike surfaces of constant v: E D C B A Now we can draw pictures of each slice, restoring one of the angular coordinates for clarity: A B C D E r = 2GM v So the Schwarzschild geometry really describes two asymptotically \ufb02at regions which reach toward each other, join together via a wormhole for a while, and then disconnect. But the wormhole closes up too quickly for any timelike observer to cross it from one region into the next. It might seem somewhat implausible, this story about two separate spacetimes reaching toward each other for a while and then letting go. In fact, it is not expected to happen in the real world, since the Schwarzschild metric does not accurately model the entire universe. 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 191 Remember that it is only valid in vacuum, for example outside a star. If the star has a radius larger than 2GM, we need never worry about any event horizons at all. But we believe that there are stars which collapse under their own gravitational pull, shrinking down to below r = 2GM and further into a singularity, resulting in a black hole. There is no need for a white hole, however, because the past of such a spacetime looks nothing like that of the full Schwarzschild solution. Roughly, a Kruskal-like diagram for stellar collapse would look like the following: r = 0 r = 2GM interior of star vacuum (Schwarzschild) The shaded region is not described by Schwarzschild, so there is no need to fret about white holes and wormholes. While we are on the subject, we can say something about the formation of astrophysical black holes from massive stars. The life of a star is a constant struggle between the inward pull of gravity and the outward push of pressure. When the star is burning nuclear fuel at its core, the pressure comes from the heat produced by this burning. (We should put \u201cburning\u201d in quotes, since nuclear fusion is unrelated to oxidation.) When the fuel is used up, the temperature declines and the star begins to shrink as gravity starts winning the struggle. Eventually this process is stopped when the electrons are pushed so close together that they resist further compression simply on the basis of the Pauli exclusion principle (no two ferm", "ions can be in the same state). The resulting object is called a white dwarf. If the mass is su\ufb03ciently high, however, even the electron degeneracy pressure is not enough, and the electrons will combine with the protons in a dramatic phase transition. The result is a neutron star, which consists of almost entirely neutrons (although the insides of neutron stars are not understood terribly well). Since the conditions at the center of a neutron star are very di\ufb00erent from those on earth, we do not have a perfect understanding of the equation of state. Nevertheless, we believe that a su\ufb03ciently massive neutron star will itself 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 192 be unable to resist the pull of gravity, and will continue to collapse. Since a \ufb02uid of neutrons is the densest material of which we can presently conceive, it is believed that the inevitable outcome of such a collapse is a black hole. The process is summarized in the following diagram of radius vs. mass: M/M D 1.5 1.0 0.5 neutron stars B C white dwarfs A 1 2 3 4 log R 10 (km) The point of the diagram is that, for any given mass M, the star will decrease in radius until it hits the line. White dwarfs are found between points A and B, and neutron stars between points C and D. Point B is at a height of somewhat less than 1.4 solar masses; the height of D is less certain, but probably less than 2 solar masses. The process of collapse is complicated, and during the evolution the star can lose or gain mass, so the endpoint of any given star is hard to predict. Nevertheless white dwarfs are all over the place, neutron stars are not uncommon, and there are a number of systems which are strongly believed to contain black holes. (Of course, you can\u2019t directly see the black hole. What you can see is radiation from matter accreting onto the hole, which heats up as it gets closer and emits radiation.) We have seen that the Kruskal coordinate system provides a very useful representation of the Schwarzschild geometry. Before moving on to other types of black holes, we will introduce one more way of thinking about this spacetime, the Penrose (or Carter-Penrose, or conformal) diagram. The idea is to do a conformal transformation which brings the entire manifold onto a compact region such that we can \ufb01t the spacetime on a piece of paper. Let\u2019s begin with Minkowski space, to see how the technique works. The metric in polar coordinates is Nothing unusual will happen to the \u03b8, \u03c6 coordinates, but we will want to keep careful track ds2 = dt2 + dr2 + r2d\u21262. \u2212 (7.86) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 193 of the ranges of the other two coordinates. In this case of course we have < t < + r < + \u221e. \u221e \u2212\u221e 0 \u2264 (7.87) Technically the worldline r = 0 represents a coordinate singularity and should be covered by a di\ufb00erent patch, but we all know what is going on so we\u2019ll just act like r = 0 is well-behaved. Our task is made somewhat easier if we switch to null coordinatest + r) r), (t \u2212 with corresponding ranges given by \u2212\u221e \u2212\u221e < 7.88) (7.89) These ranges are as portrayed in the \ufb01gure, on which each point represents a 2-sphere of t v = const r u = const radius r = u \u2212 v. The metric in these coordinates is given by ds2 = 2(dudv + dvdu) + (u \u2212 v)2d\u21262. \u2212 (7.90) We now want to change to coordinates in which \u201cin\ufb01nity\u201d takes on a \ufb01nite coordinate value. A good choice is U = arctan u 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 194 U = arctan u \u03c0 /2 \u03c0 - /2 u V = arctan v. (7.91) The ranges are now To get the metric, use and \u03c0/2 < U < +\u03c0/2 \u03c0/2 < V < +\u03c0/2 \u2212 \u2212 U. V \u2264 dU = du 1 + u2, cos(arctan u) = 1 \u221a1 + u2, (7.92) (7.93) (7.94) and likewise for v. We are led to dudv + dvdu = 1 cos2 U cos2 V (dUdV + dV dU). (7", ".95) Meanwhile, (u \u2212 tan V )2 = v)2 = (tan U \u2212 1 cos2 U cos2 V 1 cos2 U cos2 V = sin2(U (sin U cos V cos U sin V )2 (7.96) \u2212 V ). \u2212 Therefore, the Minkowski metric in these coordinates is ds2 = 1 cos2 U cos2 V \u2212 h 2(dUdV + dV dU) + sin2(U V )d\u21262. (7.97) \u2212 i This has a certain appeal, since the metric appears as a fairly simple expression multiplied by an overall factor. We can make it even better by transforming back to a timelike coordinate \u03b7 and a spacelike (radial) coordinate \u03c7, via \u03b7 = U + V 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES \u03c7 = U V, \u2212 with ranges Now the metric is where \u03c0 < \u03b7 < +\u03c0 \u03c7 < +\u03c0. \u2212 0 \u2264 ds2 = \u03c9\u2212 2 d\u03b72 + d\u03c72 + sin2 \u03c7 d\u21262 \u2212, \u03c9 = cos U cos V 1 2 = (cos \u03b7 + cos \u03c7). 195 (7.98) (7.99) (7.100) (7.101) The Minkowski metric may therefore be thought of as related by a conformal transfor- mation to the \u201cunphysical\u201d metric d\u00afs2 = \u03c92ds2 = \u2212 d\u03b72 + d\u03c72 + sin2 \u03c7 d\u21262. (7.102) \u00d7 S3, where the 3-sphere is maximally symmetric and static. This describes the manifold R There is curvature in this metric, and it is not a solution to the vacuum Einstein\u2019s equations. This shouldn\u2019t bother us, since it is unphysical; the true physical metric, obtained by a conformal transformation, is simply \ufb02at spacetime. In fact this metric is that of the \u201cEinstein static universe,\u201d a static (but unstable) solution to Einstein\u2019s equations with a perfect \ufb02uid S3 would and a cosmological constant. Of course, the full range of coordinates on R \u03c0, while Minkowski space is mapped into the subspace usually be \u2264 \u2264 S3 can be drawn as a cylinder, in which each circle is a de\ufb01ned by (7.99). The entire R three-sphere, as shown on the next page. < \u03b7 < + \u2212\u221e, 0 \u221e \u00d7 \u00d7 \u03c7 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 196 \u03c0 The shaded region represents Minkowski space. Note that each point (\u03b7, \u03c7) on this cylinder is half of a two-sphere, where the other half is the point (\u03b7, \u03c7). We can unroll the shaded region to portray Minkowski space as a triangle, as shown in the \ufb01gure. The is the Penrose \u2212 \u03b7, t \u03c7, r \u03c7=0 t = const r = const + i i - + I - I 0 i diagram. Each point represents a two-sphere. In fact Minkowski space is only the interior of the above diagram (including \u03c7 = 0); the boundaries are not part of the original spacetime. Together they are referred to as conformal in\ufb01nity. The structure of the Penrose diagram allows us to subdivide conformal in\ufb01nity 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 197 into a few di\ufb00erent regions: i+ = future timelike in\ufb01nity (\u03b7 = \u03c0, \u03c7 = 0) i0 = spatial in\ufb01nity (\u03b7 = 0, \u03c7 = \u03c0) i\u2212 = past timelike in\ufb01nity (\u03b7 = + = future null in\ufb01nity (\u03b7 = \u03c0 I \u03c0, \u03c7 = 0) \u2212 \u03c7, 0 < \u03c7 < \u03c0) \u2212 \u2212 = past null in\ufb01nity () I I I I + and + and \u2212 \u2212 are pronounced as \u201cscri-plus\u201d and \u201cscri-minus\u201d, respectively.) Note that i+, ( I i0, and i\u2212 are actually points, since \u03c7 = 0 and \u03c7 = \u03c0 are the north and south poles of S3. \u2212 are actually null surfaces, with the topology of R Meanwhile There are a number of important features of the Penrose diagram for Minkowski spacetime. The points i+, and i\u2212 can be", " thought of as the limits of spacelike surfaces whose normals are timelike; conversely, i0 can be thought of as the limit of timelike surfaces whose 45\u25e6 in the diagram. All timelike normals are spacelike. Radial null geodesics are at geodesics begin at i\u2212 and end at i+; all null geodesics begin at +; all spacelike geodesics both begin and end at i0. On the other hand, there can be non-geodesic timelike curves that end at null in\ufb01nity (if they become \u201casymptotically null\u201d). \u2212 and end at S2. \u00d7 \u00b1 I I It is nice to be able to \ufb01t all of Minkowski space on a small piece of paper, but we don\u2019t really learn much that we didn\u2019t already know. Penrose diagrams are more useful when we want to represent slightly more interesting spacetimes, such as those for black holes. The original use of Penrose diagrams was to compare spacetimes to Minkowski space \u201cat in\ufb01nity\u201d \u2014 the rigorous de\ufb01nition of \u201casymptotically \ufb02at\u201d is basically that a spacetime has a conformal in\ufb01nity just like Minkowski space. We will not pursue these issues in detail, but instead turn directly to analysis of the Penrose diagram for a Schwarzschild black hole. We will not go through the necessary manipulations in detail, since they parallel the Minkowski case with considerable additional algebraic complexity. We would start with the null version of the Kruskal coordinates, in which the metric takes the form 16G3M 3 r where r is de\ufb01ned implicitly via ds2 = \u2212 e\u2212 r/2GM (du\u2032dv\u2032 + dv\u2032du\u2032) + r2d\u21262, (7.103) u\u2032v\u2032 = r 2GM \u2212 1 er/2GM. (7.104) Then essentially the same transformation as was used in \ufb02at spacetime su\ufb03ces to bring in\ufb01nity into \ufb01nite coordinate values: u\u2032\u2032 = arctan u\u2032 \u221a2GM! 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES with ranges v\u2032\u2032 = arctan v\u2032 \u221a2GM!, \u03c0/2 < u\u2032\u2032 < +\u03c0/2 \u2212 \u03c0/2 < v\u2032\u2032 < +\u03c0/2 \u2212 \u03c0 < u\u2032\u2032 + v\u2032\u2032 < \u03c0. \u2212 198 (7.105) The (u\u2032\u2032, v\u2032\u2032) part of the metric (that is, at constant angular coordinates) is now conformally related to Minkowski space. In the new coordinates the singularities at r = 0 are straight lines that stretch from timelike in\ufb01nity in one asymptotic region to timelike in\ufb01nity in the other. The Penrose diagram for the maximally extended Schwarzschild solution thus looks like this = const i 0 I - r = const i - r = 0 i - The only real subtlety about this diagram is the necessity to understand that i+ and i\u2212 are distinct from r = 0 (there are plenty of timelike paths that do not hit the singularity). Notice also that the structure of conformal in\ufb01nity is just like that of Minkowski space, consistent with the claim that Schwarzschild is asymptotically \ufb02at. Also, the Penrose diagram for a collapsing star that forms a black hole is what you might expect, as shown on the next page. Once again the Penrose diagrams for these spacetimes don\u2019t really tell us anything we didn\u2019t already know; their usefulness will become evident when we consider more general black holes. In principle there could be a wide variety of types of black holes, depending on the process by which they were formed. Surprisingly, however, this turns out not to be the case; no matter how a black hole is formed, it settles down (fairly quickly) into a state which is characterized only by the mass, charge, and angular momentum. This property, which must be demonstrated individually for the various types of \ufb01elds which one might imagine go into the construction of the hole, is often stated as \u201cblack holes have no hair.\u201d You 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 199 i can demonstrate, for example, that a hole which is formed from an initially inhomogeneous collapse \u201cshakes o\ufb00\u201d any lumpiness by emitting gravitational radiation. This is an example of a \u201cno-hair theorem.", "\u201d If we are interested in the form of the black hole after it has settled down, we thus need only to concern ourselves with charged and rotating holes. In both cases there exist exact solutions for the metric, which we can examine closely. But \ufb01rst let\u2019s take a brief detour to the world of black hole evaporation. It is strange to think of a black hole \u201cevaporating,\u201d but in the real world black holes aren\u2019t truly black \u2014 they radiate energy as if they were a blackbody of temperature T = \u00afh/8\u03c0kGM, where M is the mass of the hole and k is Boltzmann\u2019s constant. The derivation of this e\ufb00ect, known as Hawking radiation, involves the use of quantum \ufb01eld theory in curved spacetime and is way outside our scope right now. The informal idea is nevertheless understandable. In quantum \ufb01eld theory there are \u201cvacuum \ufb02uctuations\u201d \u2014 the spontaneous creation and annihilation of particle/antiparticle pairs in empty space. These \ufb02uctuations are precisely analogous to the zero-point \ufb02uctuations of a simple harmonic oscillator. Normally such \ufb02uctuations are t +e e- e- + e r r = 2GM 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 200 impossible to detect, since they average out to give zero total energy (although nobody knows why; that\u2019s the cosmological constant problem). In the presence of an event horizon, though, occasionally one member of a virtual pair will fall into the black hole while its partner escapes to in\ufb01nity. The particle that reaches in\ufb01nity will have to have a positive energy, but the total energy is conserved; therefore the black hole has to lose mass. (If you like you can think of the particle that falls in as having a negative mass.) We see the escaping particles as Hawking radiation. It\u2019s not a very big e\ufb00ect, and the temperature goes down as the mass goes up, so for black holes of mass comparable to the sun it is completely negligible. Still, in principle the black hole could lose all of its mass to Hawking radiation, and shrink to nothing in the process. The relevant Penrose diagram might look like this radiation I - r = 0 i - On the other hand, it might not. The problem with this diagram is that \u201cinformation is lost\u201d \u2014 if we draw a spacelike surface to the past of the singularity and evolve it into the future, part of it ends up crashing into the singularity and being destroyed. As a result the radiation itself contains less information than the information that was originally in the spacetime. (This is the worse than a lack of hair on the black hole. It\u2019s one thing to think that information has been trapped inside the event horizon, but it is more worrisome to think that it has disappeared entirely.) But such a process violates the conservation of information that is implicit in both general relativity and quantum \ufb01eld theory, the two theories that led to the prediction. This paradox is considered a big deal these days, and there are a number of e\ufb00orts to understand how the information can somehow be retrieved. A currently popular explanation relies on string theory, and basically says that black holes have a lot of hair, in the form of virtual stringy states living near the event horizon. I hope you will not be disappointed to hear that we won\u2019t look at this very closely; but you should know what the problem is and that it is an area of active research these days. 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 201 With that out of our system, we now turn to electrically charged black holes. These seem at \ufb01rst like reasonable enough objects, since there is certainly nothing to stop us from throwing some net charge into a previously uncharged black hole. In an astrophysical situation, however, the total amount of charge is expected to be very small, especially when compared with the mass (in terms of the relative gravitational e\ufb00ects). Nevertheless, charged black holes provide a useful testing ground for various thought experiments, so they are worth our consideration. In this case the full spherical symmetry of the problem is still present; we know therefore that we can write the metric as ds2 = e2\u03b1(r,t)dt2 + e2\u03b2(r,t)dr2 + r2d\u21262. \u2212 (7.106) Now, however, we are no longer in vacuum, since the hole will have a nonzero electromagnetic \ufb01eld, which in turn acts as a source of energy-momentum. The energy-momentum tensor for electromagnetism is given by T\ufffd", "\ufffd\u03bd = 1 4\u03c0 (F\u00b5\u03c1F\u03bd \u03c1 1 4 \u2212 g\u00b5\u03bdF\u03c1\u03c3F \u03c1\u03c3), (7.107) where F\u00b5\u03bd is the electromagnetic \ufb01eld strength tensor. Since we have spherical symmetry, the most general \ufb01eld strength tensor will have components Ftr = f (r, t) = Frt F\u03b8\u03c6 = g(r, t) sin \u03b8 = \u2212 F\u03c6\u03b8, \u2212 (7.108) where f (r, t) and g(r, t) are some functions to be determined by the \ufb01eld equations, and components not written are zero. Ftr corresponds to a radial electric \ufb01eld, while F\u03b8\u03c6 corresponds to a radial magnetic \ufb01eld. (For those of you wondering about the sin \u03b8, recall that the thing which should be independent of \u03b8 and \u03c6 is the radial component of the magnetic \ufb01eld, Br = \u01eb01\u00b5\u03bd F\u00b5\u03bd. For a spherically symmetric metric, \u01eb\u03c1\u03c3\u00b5\u03bd = 1 g \u02dc\u01eb\u03c1\u03c3\u00b5\u03bd is proportional \u221a \u2212 1, so we want a factor of sin \u03b8 in F\u03b8\u03c6.) The \ufb01eld equations in this case are both to (sin \u03b8)\u2212 Einstein\u2019s equations and Maxwell\u2019s equations: g\u00b5\u03bd \u2207\u00b5F\u03bd\u03c3 = 0 \u2207[\u00b5F\u03bd\u03c1] = 0. The two sets are coupled together, since the electromagnetic \ufb01eld strength tensor enters Einstein\u2019s equations through the energy-momentum tensor, while the metric enters explicitly into Maxwell\u2019s equations. (7.109) The di\ufb03culties are not insurmountable, however, and a procedure similar to the one we followed for the vacuum case leads to a solution for the charged case as well. We will not 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 202 go through the steps explicitly, but merely quote the \ufb01nal answer. The solution is known as the Reissner-Nordstr\u00f8m metric, and is given by where ds2 = \u2206dt2 + \u2206\u2212 \u2212 1dr2 + r2d\u21262, \u2206 = 1 2GM r \u2212 + G(p2 + q2) r2. (7.110) (7.111) In this expression, M is once again interpreted as the mass of the hole; q is the total electric charge, and p is the total magnetic charge. Isolated magnetic charges (monopoles) have never been observed in nature, but that doesn\u2019t stop us from writing down the metric that they would produce if they did exist. There are good theoretical reasons to think that monopoles (Of course, there is also the possibility that a black hole exist, but are extremely rare. could have magnetic charge even if there aren\u2019t any monopoles.) In fact the electric and magnetic charges enter the metric in the same way, so we are not introducing any additional complications by keeping p in our expressions. The electromagnetic \ufb01elds associated with this solution are given by q r2 Ftr = F\u03b8\u03c6 = p sin \u03b8. \u2212 (7.112) Conservatives are welcome to set p = 0 if they like. The structure of singularities and event horizons is more complicated in this metric than it was in Schwarzschild, due to the extra term in the function \u2206(r) (which can be thought of as measuring \u201chow much the light cones tip over\u201d). One thing remains the same: at r = 0 there is a true curvature singularity (as could be checked by computing the curvature scalar R\u00b5\u03bd\u03c1\u03c3R\u00b5\u03bd\u03c1\u03c3). Meanwhile, the equivalent of r = 2GM will be the radius where \u2206 vanishes. This will occur at This might constitute two, one, or zero solutions, depending on the relative values of GM 2 and p2 + q2. We therefore consider each case separately. = GM r \u00b1 \u00b1 G2M 2 G(p2 + q2). \u2212 (7.113) q Case One \u2014 GM 2 < p2 + q2 In this case the coe\ufb03cient \u2206 is always positive (never zero), and the metric is completely regular in the (t, r, \u03b8, \u03c6) coordinates all the way down to r = 0. The coordinate t is always timelike, and r is always spacelike. But there still is the singularity at r = 0, which is now a timelike line. Since there is no event horizon, there is", " no obstruction to an observer travelling to the singularity and returning to report on what was observed. This is known as a naked singularity, one which is not shielded by an horizon. A careful analysis of the geodesics 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 203 \u2206 (r) (1) 2 GM < p + q 2 2 (3) 2 GM = p + q 2 2 r- GM r+ 2GM r (2) 2 2 GM > Schwarzschild) reveals, however, that the singularity is \u201crepulsive\u201d \u2014 timelike geodesics never intersect r = 0, instead they approach and then reverse course and move away. (Null geodesics can reach the singularity, as can non-geodesic timelike curves.) As r the solution approaches \ufb02at spacetime, and as we have just seen the causal structure is \u201cnormal\u201d everywhere. The Penrose diagram will therefore be just like that of Minkowski space, except that now r = 0 is a singularity. \u2192 \u221e r = 0 (singularity The nakedness of the singularity o\ufb00ends our sense of decency, as well as the cosmic censorship conjecture, which roughly states that the gravitational collapse of physical matter con\ufb01gurations will never produce a naked singularity. (Of course, it\u2019s just a conjecture, and it may not be right; there are some claims from numerical simulations that collapse of spindlelike con\ufb01gurations can lead to naked singularities.) In fact, we should not ever expect to \ufb01nd 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 204 a black hole with GM 2 < p2 + q2 as the result of gravitational collapse. Roughly speaking, this condition states that the total energy of the hole is less than the contribution to the energy from the electromagnetic \ufb01elds alone \u2014 that is, the mass of the matter which carried the charge would have had to be negative. This solution is therefore generally considered to be unphysical. Notice also that there are not good Cauchy surfaces (spacelike slices for which every inextendible timelike line intersects them) in this spacetime, since timelike lines can begin and end at the singularity. Case Two \u2014 GM 2 > p2 + q2 q \u00b1 \u2212 G2M 2 The surfaces de\ufb01ned by r = r This is the situation which we expect to apply in real gravitational collapse; the energy in the electromagnetic \ufb01eld is less than the total energy. In this case the metric coe\ufb03cient \u2206(r) is positive at large r and small r, and negative inside the two vanishing points r = \u00b1 G(p2 + q2). The metric has coordinate singularities at both r+ and r GM ; in both cases these could be removed by a change of coordinates as we did with Schwarzschild. are both null, and in fact they are event horizons (in a sense we will make precise in a moment). The singularity at r = 0 is a timelike line (not a spacelike surface as in Schwarzschild). If you are an observer falling into the black hole from far away, r+ is just like 2GM in the Schwarzschild metric; at this radius r switches from being a spacelike coordinate to a timelike coordinate, and you necessarily move in the direction of decreasing r. Witnesses outside the black hole also see the same phenomena that they would outside an uncharged hole \u2014 the infalling observer is seen to move more and more slowly, and is increasingly redshifted. \u00b1 \u2212 \u2212 But the inevitable fall from r+ to ever-decreasing radii only lasts until you reach the null surface r = r, where r switches back to being a spacelike coordinate and the motion in the direction of decreasing r can be arrested. Therefore you do not have to hit the singularity at r = 0; this is to be expected, since r = 0 is a timelike line (and therefore not necessarily in your future). In fact you can choose either to continue on to r = 0, or begin to move in the direction of increasing r back through the null surface at r = r. Then r will once again be a timelike coordinate, but with reversed orientation; you are forced to move in the direction of increasing r. You will eventually be spit out past r = r+ once more, which is like emerging from a white hole into the rest of the universe. From here you can choose to go back into the black hole \u2014 this time, a di\ufb00erent hole than the one you entered in the \ufb01rst place \u2014 and repeat the voyage as many times as you like. This little story corresponds to the accompanying Penrose diagram, which of course can be derived more rigorously by", " choosing appropriate coordinates and analytically extending the Reissner-Nordstr\u00f8m metric as far as it will go. \u2212 How much of this is science, as opposed to science \ufb01ction? Probably not much. If you think about the world as seen from an observer inside the black hole who is about to cross the event horizon at r, you will notice that they can look back in time to see the entire history \u2212 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 205 Reissner-Nordstrom: 2 GM > p + q 2 i 0 i 0 const r = surfaces - +r r- +r +r +r r- r- +r +r r- r- +r + timelike trajectories i 0 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 206 of the external (asymptotically \ufb02at) universe, at least as seen from the black hole. But they see this (in\ufb01nitely long) history in a \ufb01nite amount of their proper time \u2014 thus, any signal that gets to them as they approach r is in\ufb01nitely blueshifted. Therefore it is reasonable to believe (although I know of no proof) that any non-spherically symmetric perturbation that comes into a Reissner-Nordstr\u00f8m black hole will violently disturb the geometry we have It\u2019s hard to say what the actual geometry will look like, but there is no very described. good reason to believe that it must contain an in\ufb01nite number of asymptotically \ufb02at regions connecting to each other via various wormholes. Case Three \u2014 GM 2 = p2 + q2 \u2212 This case is known as the extreme Reissner-Nordstr\u00f8m solution (or simply \u201cextremal black hole\u201d). The mass is exactly balanced in some sense by the charge \u2014 you can construct exact solutions consisting of several extremal black holes which remain stationary with respect to each other for all time. On the one hand the extremal hole is an amusing theoretical toy; these solutions are often examined in studies of the information loss paradox, and the role of black holes in quantum gravity. On the other hand it appears very unstable, since adding just a little bit of matter will bring it to Case Two The extremal black holes have \u2206(r) = 0 at a single radius, r = GM. This does represent an event horizon, but the r coordinate is never timelike; it becomes null at r = GM, but is spacelike on either side. The singularity at r = 0 is a timelike line, as in the other cases. So 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 207 for this black hole you can again avoid the singularity and continue to move to the future to extra copies of the asymptotically \ufb02at region, but the singularity is always \u201cto the left.\u201d The Penrose diagram is as shown. We could of course go into a good deal more detail about the charged solutions, but let\u2019s instead move on to spinning black holes. It is much more di\ufb03cult to \ufb01nd the exact solution for the metric in this case, since we have given up on spherical symmetry. To begin with all that is present is axial symmetry (around the axis of rotation), but we can also ask for stationary solutions (a timelike Killing vector). Although the Schwarzschild and ReissnerNordstr\u00f8m solutions were discovered soon after general relativity was invented, the solution for a rotating black hole was found by Kerr only in 1963. His result, the Kerr metric, is given by the following mess: ds2 = dt2 + \u2212 \u03c12 \u2206 dr2 + \u03c12d\u03b82 + (r2 + a2) sin2 \u03b8 d\u03c62 + 2GMr \u03c12 (a sin2 \u03b8 d\u03c6 \u2212 dt)2, (7.114) where and \u2206(r) = r2 \u2212 2GMr + a2, \u03c12(r, \u03b8) = r2 + a2 cos2 \u03b8. (7.115) (7.116) Here a measures the rotation of the hole and M is the mass. It is straightforward to include (q2 + p2)/G; electric and magnetic charges q and p, simply by replacing 2GMr with 2GMr the result is the Kerr-Newman metric. All of the interesting phenomena persist in the absence of charges, so we will set q = p = 0 from now on. \u2212 The coordinates (t, r, \u03b8, \u03c6) are known as Boyer-Lindquist coordinates. It is straight0 they reduce to Schwarzschild coordinates. If we", " keep a \ufb01xed 0, however, we recover \ufb02at spacetime but not in ordinary polar coordinates. \u2192 forward to check that as a and let M The metric becomes \u2192 ds2 = dt2 + \u2212 (r2 + a2 cos2 \u03b8)2 (r2 + a2) dr2 + (r2 + a2 cos2 \u03b8)2d\u03b82 + (r2 + a2) sin2 \u03b8 d\u03c62, (7.117) and we recognize the spatial part of this as \ufb02at space in ellipsoidal coordinates. They are related to Cartesian coordinates in Euclidean 3-space by x = (r2 + a2)1/2 sin \u03b8 cos(\u03c6) y = (r2 + a2)1/2 sin \u03b8 sin(\u03c6) z = r cos \u03b8. (7.118) There are two Killing vectors of the metric (7.114), both of which are manifest; since the metric coe\ufb03cients are independent of t and \u03c6, both \u03b6 \u00b5 = \u2202t and \u03b7\u00b5 = \u2202\u03c6 are Killing vectors. 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 208 \u03b8 = const r = const a r = 0 Of course \u03b7\u00b5 expresses the axial symmetry of the solution. The vector \u03b6 \u00b5 is not orthogonal to t = constant hypersurfaces, and in fact is not orthogonal to any hypersurfaces at all; hence this metric is stationary, but not static. (It\u2019s not changing with time, but it is spinning.) What is more, the Kerr metric also possesses something called a Killing tensor. This is any symmetric (0, n) tensor \u03be\u00b51 \u00b5n which satis\ufb01es \u00b7\u00b7\u00b7 \u2207(\u03c3\u03be\u00b51 \u00b5n) = 0. \u00b7\u00b7\u00b7 (7.119) Simple examples of Killing tensors are the metric itself, and symmetrized tensor products of Killing vectors. Just as a Killing vector implies a constant of geodesic motion, if there exists a Killing tensor then along a geodesic we will have \u03be\u00b51 \u00b5n \u00b7\u00b7\u00b7 dx\u00b51 d\u03bb \u00b7 \u00b7 \u00b7 dx\u00b5n d\u03bb = constant. (7.120) (Unlike Killing vectors, higher-rank Killing tensors do not correspond to symmetries of the metric.) In the Kerr geometry we can de\ufb01ne the (0, 2) tensor \u03be\u00b5\u03bd = 2\u03c12l(\u00b5n\u03bd) + r2g\u00b5\u03bd. (7.121) In this expression the two vectors l and n are given (with indices raised) by l\u00b5 = n\u00b5 = Both vectors are null and satisfy 1 r2 + a2, \u2206, 0, a \u2206 1 \u2206, 0, a 2\u03c12 r2 + a2, \u2212. (7.122) l\u00b5l\u00b5 = 0, n\u00b5n\u00b5 = 0, l\u00b5n\u00b5 = 1. \u2212 (7.123) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 209 (For what it is worth, they are the \u201cspecial null vectors\u201d of the Petrov classi\ufb01cation for this spacetime.) With these de\ufb01nitions, you can check for yourself that \u03be\u00b5\u03bd is a Killing tensor. Let\u2019s think about the structure of the full Kerr solution. Singularities seem to appear at both \u2206 = 0 and \u03c1 = 0; let\u2019s turn our attention \ufb01rst to \u2206 = 0. As in the Reissner-Nordstr\u00f8m solution there are three possibilities: G2M 2 > a2, G2M 2 = a2, and G2M 2 < a2. The last case features a naked singularity, and the extremal case G2M 2 = a2 is unstable, just as in Reissner-Nordstr\u00f8m. Since these cases are of less physical interest, and time is short, we will concentrate on G2M 2 > a2. Then there are two radii at which \u2206 vanishes, given by = GM r \u00b1 \u00b1 \u221aG2M 2 a2. \u2212 (7.124) Both radii are null surfaces which will turn out to be event horizons. The analysis of these surfaces proceeds in close analogy with the Reissner-Nordstr\u00f8m case; it is straightforward to \ufb01nd coordinates which extend through the horizons. Besides the event horizons at r, the Kerr solution also features an additional surface of interest. Recall that in the spherically", " symmetric solutions, the \u201ctimelike\u201d Killing vector \u03b6 \u00b5 = \u2202t actually became null on the (outer) event horizon, and spacelike inside. Checking to see where the analogous thing happens for Kerr, we compute \u00b1 \u03b6 \u00b5\u03b6\u00b5 = 1 \u03c12 (\u2206 \u2212 \u2212 a2 sin2 \u03b8). (7.125) This does not vanish at the outer event horizon; in fact, at r = r+ (where \u2206 = 0), we have \u03b6 \u00b5\u03b6\u00b5 = a2 \u03c12 sin2 \u03b8 0. \u2265 (7.126) So the Killing vector is already spacelike at the outer horizon, except at the north and south poles (\u03b8 = 0) where it is null. The locus of points where \u03b6 \u00b5\u03b6\u00b5 = 0 is known as the Killing horizon, and is given by while the outer event horizon is given by (r \u2212 GM)2 = G2M 2 a2 cos2 \u03b8, \u2212 (r+ \u2212 GM)2 = G2M 2 a2. \u2212 (7.127) (7.128) There is thus a region in between these two surfaces, known as the ergosphere. Inside the ergosphere, you must move in the direction of the rotation of the black hole (the \u03c6 direction); however, you can still towards or away from the event horizon (and there is no trouble exiting the ergosphere). It is evidently a place where interesting things can happen even before you cross the horizon; more details on this later. 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 210 outer event horizon r + Killing horizon ergosphere r - r=0 inner event horizon Before rushing to draw Penrose diagrams, we need to understand the nature of the true curvature singularity; this does not occur at r = 0 in this spacetime, but rather at \u03c1 = 0. Since \u03c12 = r2 + a2 cos2 \u03b8 is the sum of two manifestly nonnegative quantities, it can only vanish when both quantities are zero, or 7.129) This seems like a funny result, but remember that r = 0 is not a point in space, but a disk; the set of points r = 0, \u03b8 = \u03c0/2 is actually the ring at the edge of this disk. The rotation has \u201csoftened\u201d the Schwarzschild singularity, spreading it out over a ring. What happens if you go inside the ring? A careful analytic continuation (which we will not perform) would reveal that you exit to another asymptotically \ufb02at spacetime, but not an identical copy of the one you came from. The new spacetime is described by the Kerr metric with r < 0. As a result, \u2206 never vanishes and there are no horizons. The Penrose diagram is much like that for Reissner-Nordstr\u00f8m, except now you can pass through the singularity. Not only do we have the usual strangeness of these distinct asymptotically \ufb02at regions connected to ours through the black hole, but the region near the ring singularity has additional pathologies: closed timelike curves. If you consider trajectories which wind around in \u03c6 while keeping \u03b8 and t constant and r a small negative value, the line element along such a path is ds2 = a2 1 + 2GM r d\u03c62, (7.130) which is negative for small negative r. Since these paths are closed, they are obviously CTC\u2019s. You can therefore meet yourself in the past, with all that entails. Of course, everything we say about the analytic extension of Kerr is subject to the same caveats we mentioned for Schwarzschild and Reissner-Nordstr\u00f8m; it is unlikely that realistic gravitational collapse leads to these bizarre spacetimes. It is nevertheless always useful to have exact solutions. Furthermore, for the Kerr metric there are strange things happening even if we stay outside the event horizon, to which we now turn. 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 211 Kerr: 2 2 G M > a 2 (r = + )8 I + i 0 (r = + )8 (r = - )8 I - I + r = 0 r- +r r- +r I + (r = + )8 +r r- r- r = 0 (r = - )8 (r = + )8 I I + i 0 (r = + )8 I - +r r + r- r - +r +r = + )8 (r = - ) 8 timelike trajectories (r = - ) 8 (r = + )8 i 0 +r I - (r = + )8 r = 0 r = 0 7 THE SCHWARZSCHILD", " SOLUTION AND BLACK HOLES 212 We begin by considering more carefully the angular velocity of the hole. Obviously the conventional de\ufb01nition of angular velocity will have to be modi\ufb01ed somewhat before we can apply it to something as abstract as the metric of spacetime. Let us consider the fate of a photon which is emitted in the \u03c6 direction at some radius r in the equatorial plane (\u03b8 = \u03c0/2) of a Kerr black hole. The instant it is emitted its momentum has no components in the r or \u03b8 direction, and therefore the condition that it be null is ds2 = 0 = gttdt2 + gt\u03c6(dtd\u03c6 + d\u03c6dt) + g\u03c6\u03c6d\u03c62. (7.131) This can be immediately solved to obtain gt\u03c6 g\u03c6\u03c6 \u00b1 v u u t If we evaluate this quantity on the Killing horizon of the Kerr metric, we have gtt = 0, and the two solutions are gt\u03c6 g\u03c6\u03c6! gtt g\u03c6\u03c6 (7.132) d\u03c6 dt \u2212 \u2212 =. 2 d\u03c6 dt = 0, d\u03c6 dt = 2a (2GM)2 + a2. (7.133) The nonzero solution has the same sign as a; we interpret this as the photon moving around the hole in the same direction as the hole\u2019s rotation. The zero solution means that the photon directed against the hole\u2019s rotation doesn\u2019t move at all in this coordinate system. (This isn\u2019t a full solution to the photon\u2019s trajectory, just the statement that its instantaneous velocity is zero.) This is an example of the \u201cdragging of inertial frames\u201d mentioned earlier. The point of this exercise is to note that massive particles, which must move more slowly than photons, are necessarily dragged along with the hole\u2019s rotation once they are inside the Killing horizon. This dragging continues as we approach the outer event horizon at r+; we can de\ufb01ne the angular velocity of the event horizon itself, \u2126H, to be the minimum angular velocity of a particle at the horizon. Directly from (7.132) we \ufb01nd that \u2126H = d\u03c6 dt! \u2212 (r+) = a + + a2. r2 (7.134) Now let\u2019s turn to geodesic motion, which we know will be simpli\ufb01ed by considering the conserved quantities associated with the Killing vectors \u03b6 \u00b5 = \u2202t and \u03b7\u00b5 = \u2202\u03c6. For the purposes at hand we can restrict our attention to massive particles, for which we can work with the four-momentum p\u00b5 = m dx\u00b5 d\u03c4, (7.135) where m is the rest mass of the particle. Then we can take as our two conserved quantities the actual energy and angular momentum of the particle, E = \u03b6\u00b5p\u00b5 = m \u2212 2GMr \u03c12! dt d\u03c4 + 2mGMar \u03c12 1 \u2212 sin2 \u03b8 d\u03c6 d\u03c4 (7.136) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 213 and L = \u03b7\u00b5p\u00b5 = 2mGMar \u03c12 \u2212 sin2 \u03b8 dt d\u03c4 + m(r2 + a2)2 \u2212 \u03c12 m\u2206a2 sin2 \u03b8 sin2 \u03b8 d\u03c6 d\u03c4. (7.137) (These di\ufb00er from our previous de\ufb01nitions for the conserved quantities, where E and L were taken to be the energy and angular momentum per unit mass. They are conserved either way, of course.) The minus sign in the de\ufb01nition of E is there because at in\ufb01nity both \u03b6 \u00b5 and p\u00b5 are timelike, so their inner product is negative, but we want the energy to be positive. Inside the ergosphere, however, \u03b6 \u00b5 becomes spacelike; we can therefore imagine particles for which E = \u03b6\u00b5p\u00b5 < 0. \u2212 (7.138) The extent to which this bothers us is ameliorated somewhat by the realization that all particles outside the Killing horizon must have positive energies; therefore a particle inside the ergosphere with negative energy must either remain on a geodesic inside the Killing horizon, or be accelerated until its energy is positive if it is to escape. Still, this realization leads to a way to extract energy from a rotating black hole; the method is known as the Penrose process. The idea is simple; starting from outside the ergosphere, you arm yourself with a large rock and leap toward the black hole.", " If we call the four-momentum of the (you + rock) system p(0)\u00b5, then the energy E(0) = \u03b6\u00b5p(0)\u00b5 is certainly positive, and conserved as you move along your geodesic. Once you enter the ergosphere, you hurl the rock with all your might, in a very speci\ufb01c way. If we call your momentum p(1)\u00b5 and that of the rock p(2)\u00b5, then at the instant you throw it we have conservation of momentum just as in special relativity: \u2212 p(0)\u00b5 = p(1)\u00b5 + p(2)\u00b5. Contracting with the Killing vector \u03b6\u00b5 gives E(0) = E(1) + E(2). (7.139) (7.140) But, if we imagine that you are arbitrarily strong (and accurate), you can arrange your throw such that E(2) < 0, as per (7.158). Furthermore, Penrose was able to show that you can arrange the initial trajectory and the throw such that afterwards you follow a geodesic trajectory back outside the Killing horizon into the external universe. Since your energy is conserved along the way, at the end we will have E(1) > E(0). (7.141) Thus, you have emerged with more energy than you entered with. 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 214 (top view) ergosphere Killing horizon \u00b5 (2) p (1) \u00b5 p (0) \u00b5 p There is no such thing as a free lunch; the energy you gained came from somewhere, and that somewhere is the black hole. In fact, the Penrose process extracts energy from the rotating black hole by decreasing its angular momentum; you have to throw the rock against the hole\u2019s rotation to get the trick to work. To see this more precisely, de\ufb01ne a new Killing vector \u03c7\u00b5 = \u03b6 \u00b5 + \u2126H \u03b7\u00b5. (7.142) On the outer horizon \u03c7\u00b5 is null and tangent to the horizon. (This can be seen from \u03b6 \u00b5 = \u2202t, \u03b7\u00b5 = \u2202\u03c6, and the de\ufb01nition (7.134) of \u2126H.) The statement that the particle with momentum p(2)\u00b5 crosses the event horizon \u201cmoving forwards in time\u201d is simply Plugging in the de\ufb01nitions of E and L, we see that this condition is equivalent to p(2)\u00b5\u03c7\u00b5 < 0. L(2) < E(2) \u2126H. (7.143) (7.144) Since we have arranged E(2) to be negative, and \u2126H is positive, we see that the particle must have a negative angular momentum \u2014 it is moving against the hole\u2019s rotation. Once you have escaped the ergosphere and the rock has fallen inside the event horizon, the mass and angular momentum of the hole are what they used to be plus the negative contributions of the rock: \u03b4M = E(2) \u03b4J = L(2). (7.145) Here we have introduced the notation J for the angular momentum of the black hole; it is given by J = Ma. (7.146) 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 215 We won\u2019t justify this, but you can look in Wald for an explanation. Then (7.144) becomes a limit on how much you can decrease the angular momentum: \u03b4J < \u03b4M \u2126H. (7.147) If we exactly reach this limit, as the rock we throw in becomes more and more null, we have the \u201cideal\u201d process, in which \u03b4J = \u03b4M/\u2126H. We will now use these ideas to prove a powerful result: although you can use the Penrose process to extract energy from the black hole, you can never decrease the area of the event horizon. For a Kerr metric, one can go through a straightforward computation (projecting the metric and volume element and so on) to compute the area of the event horizon: A = 4\u03c0(r2 + + a2). (7.148) To show that this doesn\u2019t decrease, it is most convenient to work instead in terms of the irreducible mass of the black hole, de\ufb01ned by M 2 irr = = = = + + a2) A 16\u03c0G2 1 4G2 (r2 Ma/G)2 (J/G)2. We can di\ufb00erentiate to obtain, after a bit of work, \u03b4Mir", "r = a 4G\u221aG2M 2 (\u2126\u2212 1 H \u03b4M \u03b4J). \u2212 a2Mirr \u2212 (7.149) (7.150) (I think I have the factors of G right, but it wouldn\u2019t hurt to check.) Then our limit (7.147) becomes \u03b4Mirr > 0. (7.151) The irreducible mass can never be reduced; hence the name. It follows that the maximum amount of energy we can extract from a black hole before we slow its rotation to zero is M \u2212 Mirr = M \u2212 1 \u221a2 M 2 + M 4 q \u2212 (J/G)2 1/2. (7.152) The result of this complete extraction is a Schwarzschild black hole of mass Mirr. It turns out that the best we can do is to start with an extreme Kerr black hole; then we can get out approximately 29% of its total energy. 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 216 The irreducibility of Mirr leads immediately to the fact that the area A can never decrease. From (7.149) and (7.150) we have \u03b4A = 8\u03c0G a \u2126H\u221aG2M 2 a2 \u2212 (\u03b4M \u2212 \u2126H\u03b4J ), which can be recast as where we have introduced \u03b4M = \u03ba 8\u03c0G \u03b4A + \u2126H \u03b4J, \u03ba = \u221aG2M 2 a2 2GM(GM + \u221aG2M 2 \u2212. a2) \u2212 (7.153) (7.154) (7.155) The quantity \u03ba is known as the surface gravity of the black hole. It was equations like (7.154) that \ufb01rst started people thinking about the relationship between black holes and thermodynamics. Consider the \ufb01rst law of thermodynamics, dU = T dS + work terms. (7.156) It is natural to think of the term \u2126H \u03b4J as \u201cwork\u201d that we do on the black hole by throwing rocks into it. Then the thermodynamic analogy begins to take shape if we think of identifying the area A as the entropy S, and the surface gravity \u03ba as 8\u03c0G times the temperature T. In fact, in the context of classical general relativity the analogy is essentially perfect. The \u201czeroth\u201d law of thermodynamics states that in thermal equilibrium the temperature is constant throughout the system; the analogous statement for black holes is that stationary black holes have constant surface gravity on the entire horizon (true). As we have seen, the \ufb01rst law (7.156) is equivalent to (7.154). The second law, that entropy never decreases, is simply the statement that the area of the horizon never decreases. Finally, the third law is that it is impossible to achieve T = 0 in any physical process, which should imply that it is impossible to achieve \u03ba = 0 in any physical process. It turns out that \u03ba = 0 corresponds to the extremal black holes (either in Kerr or Reissner-Nordstr\u00f8m) \u2014 where the naked singularities would appear. Somehow, then, the third law is related to cosmic censorship. The missing piece is that real thermodynamic bodies don\u2019t just sit there; they give o\ufb00 blackbody radiation with a spectrum that depends on their temperature. Black holes, it was thought before Hawking discovered his radiation, don\u2019t do that, since they\u2019re truly black. Historically, Bekenstein came up with the idea that black holes should really be honest black bodies, including the radiation at the appropriate temperature. This annoyed Hawking, who set out to prove him wrong, and ended up proving that there would be radiation after all. So the thermodynamic analogy is even better than we had any right to expect \u2014 although it is safe to say that nobody really knows why. December 1997 Lecture Notes on General Relativity Sean M. Carroll 8 Cosmology Contemporary cosmological models are based on the idea that the universe is pretty much the same everywhere \u2014 a stance sometimes known as the Copernican principle. On the face of it, such a claim seems preposterous; the center of the sun, for example, bears little resemblance to the desolate cold of interstellar space. But we take the Copernican principle to only apply on the very largest scales, where local variations in density are averaged over. Its validity on such scales is manifested in a number of di\ufb00erent observations, such as number counts of galaxies and observations of di\ufb00use X-ray and \u03b3-ray backgrounds, but is most clear in the 3\u25e6 microwave background radiation. Although we now know that the microwave background is not perfectly smooth (and nobody ever", " expected that it was), the 5 or less, certainly an adequate basis for deviations from regularity are on the order of 10\u2212 an approximate description of spacetime on large scales. The Copernican principle is related to two more mathematically precise properties that a manifold might have: isotropy and homogeneity. Isotropy applies at some speci\ufb01c point in the space, and states that the space looks the same no matter what direction you look in. More formally, a manifold M is isotropic around a point p if, for any two vectors V and W in TpM, there is an isometry of M such that the pushforward of W under the isometry is parallel with V (not pushed forward). It is isotropy which is indicated by the observations of the microwave background. Homogeneity is the statement that the metric is the same throughout the space. In other words, given any two points p and q in M, there is an isometry which takes p into q. Note that there is no necessary relationship between homogeneity and isotropy; a manifold S2 in the usual metric), or it can can be homogeneous but nowhere isotropic (such as R be isotropic around a point without being homogeneous (such as a cone, which is isotropic around its vertex but certainly not homogeneous). On the other hand, if a space is isotropic everywhere then it is homogeneous. (Likewise if it is isotropic around one point and also homogeneous, it will be isotropic around every point.) Since there is ample observational evidence for isotropy, and the Copernican principle would have us believe that we are not the center of the universe and therefore observers elsewhere should also observe isotropy, we will henceforth assume both homogeneity and isotropy. \u00d7 There is one catch. When we look at distant galaxies, they appear to be receding from us; the universe is apparently not static, but changing with time. Therefore we begin construction of cosmological models with the idea that the universe is homogeneous and isotropic in space, but not in time. In general relativity this translates into the statement that the universe can be foliated into spacelike slices such that each slice is homogeneous and isotropic. 217 8 COSMOLOGY 218 We therefore consider our spacetime to be R \u03a3, where R represents the time direction and \u03a3 is a homogeneous and isotropic three-manifold. The usefulness of homogeneity and isotropy is that they imply that \u03a3 must be a maximally symmetric space. (Think of isotropy as invariance under rotations, and homogeneity as invariance under translations. Then homogeneity and isotropy together imply that a space has its maximum possible number of Killing vectors.) Therefore we can take our metric to be of the form \u00d7 ds2 = dt2 + a2(t)\u03b3ij(u)duiduj. \u2212 (8.1) Here t is the timelike coordinate, and (u1, u2, u3) are the coordinates on \u03a3; \u03b3ij is the maximally symmetric metric on \u03a3. This formula is a special case of (7.2), which we used to derive the Schwarzschild metric, except we have scaled t such that gtt = 1. The function a(t) is known as the scale factor, and it tells us \u201chow big\u201d the spacelike slice \u03a3 is at the moment t. The coordinates used here, in which the metric is free of cross terms dt dui and the spacelike components are proportional to a single function of t, are known as comoving coordinates, and an observer who stays at constant ui is also called \u201ccomoving\u201d. Only a comoving observer will think that the universe looks isotropic; in fact on Earth we are not quite comoving, and as a result we see a dipole anisotropy in the cosmic microwave background as a result of the conventional Doppler e\ufb00ect. \u2212 Our interest is therefore in maximally symmetric Euclidean three-metrics \u03b3ij. We know that maximally symmetric metrics obey (3)Rijkl = k(\u03b3ik\u03b3jl \u2212 \u03b3il\u03b3jk), (8.2) where k is some constant, and we put a superscript (3) on the Riemann tensor to remind us that it is associated with the three-metric \u03b3ij, not the metric of the entire spacetime. The Ricci tensor is then (3)Rjl = 2k\u03b3jl. (8.3) If the space is to be maximally symmetric, then it will certainly be spherically symmetric. We already know something about spherically symmetric spaces from our exploration of the Schwarzschild solution; the metric can be put in the form d\u03c3", "2 = \u03b3ijdui duj = e2\u03b2(r)dr2 + r2(d\u03b82 + sin2 \u03b8 d\u03c62). (8.4) The components of the Ricci tensor for such a metric can be obtained from (7.16), the Ricci tensor for a spherically symmetric spacetime, by setting \u03b1 = 0 and \u22020\u03b2 = 0, which gives 2 (3)R11 = r (3)R22 = e\u2212 (3)R33 = [e\u2212 \u22021\u03b2 2\u03b2(r\u22021\u03b2 2\u03b2(r\u22021\u03b2 1) + 1 1) + 1] sin2 \u03b8. \u2212 \u2212 (8.5) 8 COSMOLOGY We set these proportional to the metric using (8.3), and can solve for \u03b2(r): This gives us the following metric on spacetime: \u03b2 = 1 2 \u2212 ln(1 \u2212 kr2). ds2 = dt2 + a2(t) \u2212 1 \" dr2 kr2 + r2(d\u03b82 + sin2 \u03b8 d\u03c62) \u2212. # 219 (8.6) (8.7) This is the Robertson-Walker metric. We have not yet made use of Einstein\u2019s equations; those will determine the behavior of the scale factor a(t). Note that the substitutions 8.8) leave (8.7) invariant. Therefore the only relevant parameter is k/, and there are three cases of interest: k = 1 case corresponds to constant negative curvature on \u03a3, and is called open; the k = 0 case corresponds to no curvature on \u03a3, and is called \ufb02at; the k = +1 case corresponds to positive curvature on \u03a3, and is called closed. 1, k = 0, and k = +1. The k = \u2212 \u2212 k | | Let us examine each of these possibilities. For the \ufb02at case k = 0 the metric on \u03a3 is d\u03c32 = dr2 + r2d\u21262 = dx2 + dy2 + dz2, (8.9) which is simply \ufb02at Euclidean space. Globally, it could describe R3 or a more complicated manifold, such as the three-torus S1 S1. For the closed case k = +1 we can de\ufb01ne r = sin \u03c7 to write the metric on \u03a3 as S1 \u00d7 \u00d7 d\u03c32 = d\u03c72 + sin2 \u03c7 d\u21262, (8.10) In this case the only possible global structure is which is the metric of a three-sphere. actually the three-sphere (except for the non-orientable manifold RP3). Finally in the open k = 1 case we can set r = sinh \u03c8 to obtain \u2212 d\u03c32 = d\u03c82 + sinh2 \u03c8 d\u21262. (8.11) This is the metric for a three-dimensional space of constant negative curvature; it is hard to visualize, but think of the saddle example we spoke of in Section Three. Globally such a space could extend forever (which is the origin of the word \u201copen\u201d), but it could also 8 COSMOLOGY 220 describe a non-simply-connected compact space (so \u201copen\u201d is really not the most accurate description). With the metric in hand, we can set about computing the connection coe\ufb03cients and curvature tensor. Setting \u02d9a \u2261 da/dt, the Christo\ufb00el symbols are given by a \u02d9a \u03930 11 = kr2 1 \u2212 \u03931 01 = \u03931 10 = \u03932 \u03931 r(1 22 = \u2212 \u2212 21 = \u03933 12 = \u03932 \u03932 \u03932 sin \u03b8 cos \u03b8 33 = 02 = \u03932 kr2) 13 = \u03933 \u2212 22 = a \u02d9ar2 \u03930 33 = a \u02d9ar2 sin2 \u03b8 \u03930 \u02d9a a kr2) sin2 \u03b8 30 = \u2212 03 = \u03933 r(1 20 = \u03933 \u03931 33 = 1 31 = r 23 = \u03933 \u03933 \u2212 32 = cot \u03b8. The nonzero components of the Ricci tensor are R00 = \u00a8a a 3 \u2212 a\u00a8a + 2 \u02d9a2 + 2k R11 = 1 R22 = r2(a\u00a8a + 2 \u02d9a2 + 2k) R33 = r2(a\u00a8a + 2 \u02d9a2 + 2k) sin2 \u03b8, k", "r2 \u2212 (8.12) (8.13) and the Ricci scalar is then R = 6 a2 (a\u00a8a + \u02d9a2 + k). The universe is not empty, so we are not interested in vacuum solutions to Einstein\u2019s equations. We will choose to model the matter and energy in the universe by a perfect \ufb02uid. We discussed perfect \ufb02uids in Section One, where they were de\ufb01ned as \ufb02uids which are isotropic in their rest frame. The energy-momentum tensor for a perfect \ufb02uid can be written (8.14) T\u00b5\u03bd = (p + \u03c1)U\u00b5U\u03bd + pg\u00b5\u03bd, (8.15) where \u03c1 and p are the energy density and pressure (respectively) as measured in the rest frame, and U \u00b5 is the four-velocity of the \ufb02uid. It is clear that, if a \ufb02uid which is isotropic in some frame leads to a metric which is isotropic in some frame, the two frames will coincide; that is, the \ufb02uid will be at rest in comoving coordinates. The four-velocity is then and the energy-momentum tensor is U \u00b5 = (1, 0, 0, 0), 0 0 gijp \u03c1 0 0 0 0. \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 T\u00b5\u03bd = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ed (8.16) (8.17) 8 COSMOLOGY With one index raised this takes the more convenient form T \u00b5 Note that the trace is given by \u03bd = diag( \u03c1, p, p, p + 3p. \u2212 221 (8.18) (8.19) Before plugging in to Einstein\u2019s equations, it is educational to consider the zero compo- nent of the conservation of energy equation: 0 = \u2207\u00b5T \u00b5 = \u2202\u00b5T \u00b5 = \u22020\u03c1 \u2212 0 0 + \u0393\u00b5 \u00b50T 0 \u02d9a a \u2212 3 0 \u2212 (\u03c1 + p). \u00b50T \u00b5 \u0393\u03bb \u03bb (8.20) To make progress it is necessary to choose an equation of state, a relationship between \u03c1 and p. Essentially all of the perfect \ufb02uids relevant to cosmology obey the simple equation of state p = w\u03c1, (8.21) where w is a constant independent of time. The conservation of energy equation becomes which can be integrated to obtain \u02d9\u03c1 \u03c1 = 3(1 + w) \u2212 \u02d9a a, a\u2212 3(1+w). \u03c1 \u221d (8.22) (8.23) The two most popular examples of cosmological \ufb02uids are known as dust and radiation. Dust is collisionless, nonrelativistic matter, which obeys w = 0. Examples include ordinary stars and galaxies, for which the pressure is negligible in comparison with the energy density. Dust is also known as \u201cmatter\u201d, and universes whose energy density is mostly due to dust are known as matter-dominated. The energy density in matter falls o\ufb00 as a\u2212 3. \u03c1 \u221d (8.24) This is simply interpreted as the decrease in the number density of particles as the universe expands. (For dust the energy density is dominated by the rest energy, which is proportional to the number density.) \u201cRadiation\u201d may be used to describe either actual electromagnetic radiation, or massive particles moving at relative velocities su\ufb03ciently close to the speed of light that they become indistinguishable from photons (at least as far as their equation of state is concerned). Although radiation is a perfect \ufb02uid and thus has an energy-momentum 8 COSMOLOGY 222 tensor given by (8.15), we also know that T\u00b5\u03bd can be expressed in terms of the \ufb01eld strength as T \u00b5\u03bd = The trace of this is given by 1 4\u03c0 (F \u00b5\u03bbF \u03bd \u03bb \u2212 1 4 g\u00b5\u03bdF \u03bb\u03c3F\u03bb\u03c3). T \u00b5 \u00b5 = 1 4\u03c0 F \u00b5\u03bbF\u00b5\u03bb \u2212 1 4 (4)F \u03bb\u03c3F\u03bb\u03c3 = 0. But this must also equal (8.19), so the equation of state is p = 1 3 \u03c1. (8.25) (8.26) (8.27) A universe in which most of the energy density is in the form of radiation is known as radiation-dominated. The energy density in radiation falls o\ufb00 as a\u2212 4. \u03c1 \u221d (8.28) Thus", ", the energy density in radiation falls o\ufb00 slightly faster than that in matter; this is because the number density of photons decreases in the same way as the number density of 1 as they redshift, as nonrelativistic particles, but individual photons also lose energy as a\u2212 we will see later. (Likewise, massive but relativistic particles will lose energy as they \u201cslow down\u201d in comoving coordinates.) We believe that today the energy density of the universe 106. However, in the past the universe was much is dominated by matter, with \u03c1mat/\u03c1rad \u223c smaller, and the energy density in radiation would have dominated at very early times. There is one other form of energy-momentum that is sometimes considered, namely that Introducing energy into the vacuum is equivalent to introducing a of the vacuum itself. cosmological constant. Einstein\u2019s equations with a cosmological constant are G\u00b5\u03bd = 8\u03c0GT\u00b5\u03bd \u2212 \u039bg\u00b5\u03bd, (8.29) which is clearly the same form as the equations with no cosmological constant but an energymomentum tensor for the vacuum, This has the form of a perfect \ufb02uid with T (vac) \u00b5\u03bd = \u039b 8\u03c0G \u2212 g\u00b5\u03bd. \u03c1 = p = \u2212 \u039b 8\u03c0G. (8.30) (8.31) 1, and the energy density is independent of a, which is what we We therefore have w = would expect for the energy density of the vacuum. Since the energy density in matter and \u2212 8 COSMOLOGY 223 radiation decreases as the universe expands, if there is a nonzero vacuum energy it tends to win out over the long term (as long as the universe doesn\u2019t start contracting). If this happens, we say that the universe becomes vacuum-dominated. We now turn to Einstein\u2019s equations. Recall that they can be written in the form (4.45): The \u00b5\u03bd = 00 equation is and the \u00b5\u03bd = ij equations give R\u00b5\u03bd = 8\u03c0G T\u00b5\u03bd \u2212 1 2 g\u00b5\u03bdT. 3 \u00a8a a \u2212 = 4\u03c0G(\u03c1 + 3p), \u00a8a a + 2 2 \u02d9a a + 2 k a2 = 4\u03c0G(\u03c1 \u2212 p). (8.32) (8.33) (8.34) (There is only one distinct equation from \u00b5\u03bd = ij, due to isotropy.) We can use (8.33) to eliminate second derivatives in (8.34), and do a little cleaning up to obtain \u00a8a a = 4\u03c0G 3 \u2212 (\u03c1 + 3p), (8.35) and 8\u03c0G 3 Together these are known as the Friedmann equations, and metrics of the form (8.7) which obey these equations de\ufb01ne Friedmann-Robertson-Walker (FRW) universes. k a2. \u02d9a a (8.36) \u2212 = \u03c1 2 There is a bunch of terminology which is associated with the cosmological parameters, and we will just introduce the basics here. The rate of expansion is characterized by the Hubble parameter, H =. (8.37) \u02d9a a The value of the Hubble parameter at the present epoch is the Hubble constant, H0. There is currently a great deal of controversy about what its actual value is, with measurements falling in the range of 40 to 90 km/sec/Mpc. (\u201cMpc\u201d stands for \u201cmegaparsec\u201d, which is 1024 cm.) Note that we have to divide \u02d9a by a to get a measurable quantity, since the 3 overall scale of a is irrelevant. There is also the deceleration parameter, \u00d7 q = a\u00a8a \u02d9a2, \u2212 (8.38) which measures the rate of change of the rate of expansion. Another useful quantity is the density parameter, \u2126 = 8\u03c0G 3H 2 \u03c1 8 COSMOLOGY = \u03c1 \u03c1crit, where the critical density is de\ufb01ned by \u03c1crit = 3H 2 8\u03c0G. 224 (8.39) (8.40) This quantity (which will generally change with time) is called the \u201ccritical\u201d density because the Friedmann equation (8.36) can be written 1 = \u2126 \u2212 k H 2a2. (8.41) The sign of k is therefore determined by whether \u2126 is greater than, equal to, or less than one. We have \u2194 \u2194 \u2194 The density parameter, then, tells us which of the three Robertson-Walker geometries describes our universe. Determining it observationally is an area of intense investigation. \u2194 \u2194 \u2194 \u03c1 < \ufffd", "\ufffdcrit \u2194 \u03c1 = \u03c1crit \u2194 \u03c1 > \u03c1crit \u2194 open \ufb02at closed. It is possible to solve the Friedmann equations exactly in various simple cases, but it is often more useful to know the qualitative behavior of various possibilities. Let us for the moment set \u039b = 0, and consider the behavior of universes \ufb01lled with \ufb02uids of positive energy (\u03c1 > 0) and nonnegative pressure (p 0). Then by (8.35) we must have \u00a8a < 0. Since we know from observations of distant galaxies that the universe is expanding ( \u02d9a > 0), this means that the universe is \u201cdecelerating.\u201d This is what we should expect, since the gravitational attraction of the matter in the universe works against the expansion. The fact that the universe can only decelerate means that it must have been expanding even faster in the past; if we trace the evolution backwards in time, we necessarily reach a singularity at a = 0. Notice that if \u00a8a were exactly zero, a(t) would be a straight line, and the age of the universe would be H \u2212 0. Since \u00a8a is actually negative, the universe must be somewhat younger than that. \u2265 1 This singularity at a = 0 is the Big Bang. It represents the creation of the universe It might be from a singular state, not explosion of matter into a pre-existing spacetime. hoped that the perfect symmetry of our FRW universes was responsible for this singularity, but in fact it\u2019s not true; the singularity theorems predict that any universe with \u03c1 > 0 and p 0 must have begun at a singularity. Of course the energy density becomes arbitrarily high as a 0, and we don\u2019t expect classical general relativity to be an accurate description of nature in this regime; hopefully a consistent theory of quantum gravity will be able to \ufb01x things up. \u2192 \u2265 8 COSMOLOGY a(t) 225 Big Bang -1H 0 now t The future evolution is di\ufb00erent for di\ufb00erent values of k. For the open and \ufb02at cases, 0, (8.36) implies k \u2264 \u02d9a2 = 8\u03c0G 3 \u03c1a2 +. k | | (8.42) The right hand side is strictly positive (since we are assuming \u03c1 > 0), so \u02d9a never passes through zero. Since we know that today \u02d9a > 0, it must be positive for all time. Thus, the open and \ufb02at universes expand forever \u2014 they are temporally as well as spatially open. (Please keep in mind what assumptions go into this \u2014 namely, that there is a nonzero positive energy density. Negative energy density universes do not have to expand forever, even if they are \u201copen\u201d.) How fast do these universes keep expanding? Consider the quantity \u03c1a3 (which is constant in matter-dominated universes). By the conservation of energy equation (8.20) we have d dt (\u03c1a3) = a3 \u02d9\u03c1 + 3\u03c1 = 3pa2 \u02d9a. \u2212 \u02d9a a The right hand side is either zero or negative; therefore d dt (\u03c1a3) 0. \u2264 This implies in turn that \u03c1a2 must go to zero in an ever-expanding universe, where a Thus (8.42) tells us that \u02d9a2. k | \u2192 | (8.43) (8.44). \u2192 \u221e (8.45) (Remember that this is true for k limiting value \u02d9a slowly. \u2192 1 the expansion approaches the 1, while for k = 0 the universe keeps expanding, but more and more 0.) Thus, for k = \u2264 \u2212 8 COSMOLOGY For the closed universes (k = +1), (8.36) becomes \u02d9a2 = 8\u03c0G 3 \u03c1a2 1. \u2212 226 (8.46) The argument that \u03c1a2 still applies; but in that case (8.46) would become negative, which can\u2019t happen. Therefore the universe does not expand inde\ufb01nitely; a possesses an upper bound amax. As a approaches amax, (8.35) implies 0 as a \u2192 \u221e \u2192 \u00a8a \u2192 \u2212 4\u03c0G 3 (\u03c1 + 3p)amax < 0. (8.47) Thus \u00a8a is \ufb01nite and negative at this point, so a reaches amax and starts decreasing, whereupon (since \u00a8a < 0) it will inevitably continue to contract to zero \u2014 the Big Crunch. Thus, the closed universes (again, under our assumptions of positive \u03c1 and nonnegative p) are closed in time as well", " as space. a(t) k = -1 k = 0 k = +1 bang now crunch t We will now list some of the exact solutions corresponding to only one type of energy density. For dust-only universes (p = 0), it is convenient to de\ufb01ne a development angle \u03c6(t), rather than using t as a parameter directly. The solutions are then, for open universes, for \ufb02at universescosh \u03c6 2 (sinh \u03c6 1) \u03c6) \u2212 \u2212 (k = 1), \u2212 and for closed universes, a = 1/3 t2/3 9C 4 (k = 01 2 (\u03c6 ( cos \u03c6) sin \u03c6) \u2212 \u2212 (k = +1), (8.48) (8.49) (8.50) 8 COSMOLOGY where we have de\ufb01ned C = 8\u03c0G 3 \u03c1a3 = constant. 227 (8.51) For universes \ufb01lled with nothing but radiation, p = 1 3\u03c1, we have once again open universes, a = \u221aC \u2032 \uf8ee 1 + \uf8f0 2 t \u221aC \u2032! 1/2 1 \u2212 \uf8f9 \uf8fb (k = 1), \u2212 a = (4C \u2032)1/4t1/2 (k = 0), (8.52) (8.53) \ufb02at universes, and closed universes, a = \u221aC \u2032 \uf8ee 1 1 \u2212 \u2212 where this time we de\ufb01ned \uf8f0 C \u2032 = 8\u03c0G 3 2 t \u221aC \u2032! 1/2 \uf8f9 \uf8fb (k = +1), (8.54) \u03c1a4 = constant. (8.55) You can check for yourselves that these exact solutions have the properties we argued would hold in general. For universes which are empty save for the cosmological constant, either \u03c1 or p will be negative, in violation of the assumptions we used earlier to derive the general behavior of a(t). In this case the connection between open/closed and expands forever/recollapses is lost. We begin by considering \u039b < 0. In this case \u2126 is negative, and from (8.41) this can only happen if k = 1. The solution in this case is \u2212 a = 3 \u2212 \u039b s sin \uf8eb s \u039b \u2212 3 t \uf8f6. \uf8ed \uf8f8 There is also an open (k = 1) solution for \u039b > 0, given by \u2212 a = 3 \u039b s sinh \uf8eb s \u039b 3 t \uf8f6. A \ufb02at vacuum-dominated universe must have \u039b > 0, and the solution is \uf8ed \uf8f8 exp a \u221d \uf8eb \u00b1s \u039b 3 t \uf8f6, while the closed universe must also have \u039b > 0, and satis\ufb01es \uf8ed \uf8f8 a = 3 \u039b s cosh 8.56) (8.57) (8.58) (8.59) 8 COSMOLOGY 228 These solutions are a little misleading. In fact the three solutions for \u039b > 0 \u2014 (8.57), (8.58), and (8.59) \u2014 all represent the same spacetime, just in di\ufb00erent coordinates. This spacetime, known as de Sitter space, is actually maximally symmetric as a spacetime. (See Hawking and Ellis for details.) The \u039b < 0 solution (8.56) is also maximally symmetric, and is known as anti-de Sitter space. It is clear that we would like to observationally determine a number of quantities to decide which of the FRW models corresponds to our universe. Obviously we would like to determine H0, since that is related to the age of the universe. (For a purely matter-dominated, k = 0 universe, (8.49) implies that the age is 2/(3H0). Other possibilities would predict similar relations.) We would also like to know \u2126, which determines k through (8.41). Given the de\ufb01nition (8.39) of \u2126, this means we want to know both H0 and \u03c10. Unfortunately both quantities are hard to measure accurately, especially \u03c1. But notice that the deceleration parameter q can be related to \u2126 using (8.35): q = = = = = \u2212 a\u00a8a \u02d9a2 2 \u00a8a H \u2212 a \u2212 4\u03c0G 3H 2 (\u03c1 + 3p) 4\u03c0G 3H 2 \u03c1(1 + 3w) 1 + 3w \u2126. 2 (8.60) Therefore, if we think we know", " what w is (i.e., what kind of stu\ufb00 the universe is made of), we can determine \u2126 by measuring q. (Unfortunately we are not completely con\ufb01dent that we know w, and q is itself hard to measure. But people are trying.) To understand how these quantities might conceivably be measured, let\u2019s consider geodesic motion in an FRW universe. There are a number of spacelike Killing vectors, but no timelike Killing vector to give us a notion of conserved energy. There is, however, a Killing tensor. If U \u00b5 = (1, 0, 0, 0) is the four-velocity of comoving observers, then the tensor K\u00b5\u03bd = a2(g\u00b5\u03bd + U\u00b5U\u03bd) (8.61) satis\ufb01es if a particle has four-velocity V \u00b5 = dx\u00b5/d\u03bb, the quantity \u2207(\u03c3K\u00b5\u03bd) = 0 (as you can check), and is therefore a Killing tensor. This means that K 2 = K\u00b5\u03bdV \u00b5V \u03bd = a2[V\u00b5V \u00b5 + (U\u00b5V \u00b5)2] (8.62) will be a constant along geodesics. Let\u2019s think about this, \ufb01rst for massive particles. Then we will have V\u00b5V \u00b5 = 1, or \u2212 (V 0)2 = 1 + 2, ~V | | (8.63) 8 COSMOLOGY where ~V | | 2 = gijV iV j. So (8.61) implies ~V | | = K a. 229 (8.64) The particle therefore \u201cslows down\u201d with respect to the comoving coordinates as the universe expands. In fact this is an actual slowing down, in the sense that a gas of particles with initially high relative velocities will cool down as the universe expands. A similar thing happens to null geodesics. In this case V\u00b5V \u00b5 = 0, and (8.62) implies U\u00b5V \u00b5 = K a. (8.65) U\u00b5V \u00b5. The But the frequency of the photon as measured by a comoving observer is \u03c9 = frequency of the photon emitted with frequency \u03c91 will therefore be observed with a lower frequency \u03c90 as the universe expands: \u2212 \u03c90 \u03c91 = a1 a0. (8.66) Cosmologists like to speak of this in terms of the redshift z between the two events, de\ufb01ned by the fractional change in wavelength: z = = \u03bb1 1. \u03bb0 \u2212 \u03bb1 a0 a1 \u2212 (8.67) Notice that this redshift is not the same as the conventional Doppler e\ufb00ect; it is the expansion of space, not the relative velocities of the observer and emitter, which leads to the redshift. The redshift is something we can measure; we know the rest-frame wavelengths of various spectral lines in the radiation from distant galaxies, so we can tell how much their wavelengths have changed along the path from time t1 when they were emitted to time t0 when they were observed. We therefore know the ratio of the scale factors at these two times. But we don\u2019t know the times themselves; the photons are not clever enough to tell us how much coordinate time has elapsed on their journey. We have to work harder to extract this information. Roughly speaking, since a photon moves at the speed of light its travel time should simply be its distance. But what is the \u201cdistance\u201d of a far away galaxy in an expanding universe? The comoving distance is not especially useful, since it is not measurable, and furthermore because the galaxies need not be comoving in general. Instead we can de\ufb01ne the luminosity distance as d2 L = L 4\u03c0F, (8.68) where L is the absolute luminosity of the source and F is the \ufb02ux measured by the observer (the energy per unit time per unit area of some detector). The de\ufb01nition comes from the 8 COSMOLOGY 230 fact that in \ufb02at space, for a source at distance d the \ufb02ux over the luminosity is just one over the area of a sphere centered around the source, F/L = 1/A(d) = 1/4\u03c0d2. In an FRW universe, however, the \ufb02ux will be diluted. Conservation of photons tells us that the total number of photons emitted by the source will eventually pass through a sphere at comoving distance r from the emitter. Such a sphere is at a physical distance d = a0r, where a0 is the scale factor when the photons", " are observed. But the \ufb02ux is diluted by two additional e\ufb00ects: the individual photons redshift by a factor (1 + z), and the photons hit the sphere less frequently, since two photons emitted a time \u03b4t apart will be measured at a time (1+z)\u03b4t apart. Therefore we will have F L 1 = 4\u03c0a2 0r2(1 + z)2, or dL = a0r(1 + z). The luminosity distance dL is something we might hope to measure, since there are some astrophysical sources whose absolute luminosities are known (\u201cstandard candles\u201d). But r is not observable, so we have to remove that from our equation. On a null geodesic (chosen to be radial for convenience) we have For galaxies not too far away, we can expand the scale factor in a Taylor series about its present value: or 0 = ds2 = dt2 + \u2212 1 a2 kr2 dr2, \u2212 t0 t1 Z dt a(t) = r (1 0 Z dr kr2)1/2. \u2212 1 2 a(t1) = a0 + ( \u02d9a)0(t1 \u2212 t0) + We can then expand both sides of (8.72) to \ufb01nd (\u00a8a)0(t1 \u2212 t0)2 +.... r = a\u2212 0 1 t1) + (t0 \u2212 1 2 H0(t0 \u2212 t1)2 +.... Now remembering (8.67), the expansion (8.73) is the same as 1 1 + z = 1 + H0(t1 \u2212 t0) \u2212 1 2 q0H 2 0 (t1 \u2212 t0)2 +.... For small H0(t1 \u2212 t0) this can be inverted to yield t1 = H \u2212 0 1 t0 \u2212 1 + z \u2212 q0 2 z2 +.... (8.69) (8.70) (8.71) (8.72) (8.73) (8.74) (8.75) (8.76) 8 COSMOLOGY Substituting this back again into (8.74) gives r = 1 a0H0 z 1 2 \u2212 (1 + q0) z2 +.... Finally, using this in (8.70) yields Hubble\u2019s Law: dL = H \u2212 0 1 z + 1 2 (1 q0)z2 +.... \u2212 231 (8.77) (8.78) Therefore, measurement of the luminosity distances and redshifts of a su\ufb03cient number of galaxies allows us to determine H0 and q0, and therefore takes us a long way to deciding what kind of FRW universe we live in. The observations themselves are extremely di\ufb03cult, and the values of these parameters in the real world are still hotly contested. Over the next decade or so a variety of new strategies and more precise application of old strategies could very well answer these questions once and for all.l to propagate from one to the other. This particular di\ufb03culty may be resolved by introducing the concept of the electromagnetic \ufb01eld. Then we may suppose that one charge does not act directly on the other, but on the \ufb01eld in its immediate vicinity; this in turn a\ufb00ects the \ufb01eld further out, and so on. By supposing that the \ufb01eld itself can carry energy and momentum, we can reinstate the conservation laws, which are among the most important consequences of Newton\u2019s laws. However, this does not completely remove the di\ufb03culty, for there is still an apparent contradiction between this classical electromagnetic theory 10 Classical Mechanics \u00a7 1.1. This arises from the fact and the principle of relativity discussed in that if the speed of light is constant with respect to one inertial frame \u2014 as it should be according to electromagnetic theory \u2014 then the usual rules for combining velocities would lead to the conclusion that it is not constant with respect to a relatively moving frame, in contradiction with the statement that all inertial frames are equivalent. This paradox can only be resolved by the introduction of Einstein\u2019s theory of relativity (i.e., \u2018special\u2019 relativity). Classical electromagnetic theory and classical mechanics can be incorporated into a single self-consistent theory, but only by ignoring the relativity principle and sticking to one \u2018preferred\u2019 inertial frame. 1.3 The Concepts of Mass and Force It is an important general principle of physics (though not universally applied!) that no quantity should be introduced into the theory which cannot, at least in principle, be measured. Now, Newton\u2019s", " laws involve not only the concepts of velocity and acceleration, which can be measured by measuring distances and times, but also the new concepts of mass and force. To give the laws a physical meaning we have, therefore, to show that these are measurable quantities. This is not quite as trivial as it might seem, because any experiment designed to measure these quantities must necessarily involve Newton\u2019s laws themselves in its interpretation. Thus the operational de\ufb01nitions of mass and force \u2014 the prescriptions of how they may be measured \u2014 which are required to make the laws physically signi\ufb01cant, are actually contained in the laws themselves. This is by no means an unusual or logically objectionable situation, but it may clarify the status of these concepts to reformulate the laws in such a way as to isolate their de\ufb01nitional element. Let us consider \ufb01rst the measurement of mass. Since the units of mass are arbitrary, we have to specify a way of comparing the masses of two given bodies. It is important to realize that we are discussing here the inertial mass, which appears in Newton\u2019s second law, (1.1) and not the gravitational mass, which appears in (1.5). The two are of course proportional, but this equivalence principle is a physical law derived from experimental observation (in particular from Galileo\u2019s observations of falling bodies, from which he deduced that in a vacuum all bodies would fall equally fast) rather than an a priori assumption. To verify the law, we must be able to measure each kind of mass separately. This rules out, for example, the use of a balance, Introduction 11 which compares gravitational masses. Clearly, we can compare the inertial masses of two bodies by subjecting them to equal forces and comparing their accelerations, but this does not help unless we have some way of knowing that the forces are equal. However there is one case in which we do know this, because of Newton\u2019s third law. If we isolate the two bodies from all other matter, and compare their mutually induced accelerations, then according to (1.1) and (1.3), m1a1 = m2a2, \u2212 (1.7) so that the accelerations are oppositely directed, and inversely proportional to the masses. If we allow two small bodies to collide, then during the collision the e\ufb00ects of more remote bodies are generally negligible in comparison with their e\ufb00ect on each other, and we may treat them approximately as an isolated system. (Such collisions will be discussed in detail in Chapters 2 and 7.) The mass ratio can then be determined from measurements of their velocities before and after the collision, by using (1.7) or its immediate consequence, the law of conservation of momentum, m1v1 + m2v2 = constant. (1.8) If we wish to separate the de\ufb01nition of mass from the physical content of equation (1.7), we may adopt as a fundamental axiom the following: In an isolated two-body system, the accelerations always satisfy the relation a1 = k21a2, where the scalar k21 \u2212 is, for two given bodies, a constant independent of their positions, velocities and internal states. If we choose the \ufb01rst body to be a standard body, and conventionally assign it unit mass (say m1 = 1 kg), then we may de\ufb01ne the mass of the second to be k21 in units of this standard mass (here m2 = k21 kg). Note that for consistency, we must have k12 = 1/k21. We must also assume of course that if we compare the masses of three bodies in this way, we obtain consistent results: For any three bodies, the constants kij satisfy k31 = k32k21. It then follows that for any two bodies, k32 is the mass ratio: k32 = m3/m2. To complete the list of fundamental axioms, we need one which deals with systems containing more than two bodies, analogous to the law of composition of forces, (1.2). This may be stated as follows: 12 Classical Mechanics The acceleration induced in one body by another is some de\ufb01nite function of their positions, velocities and internal structure, and is una\ufb00ected by the presence of other bodies. In a many-body system, the acceleration of any given body is equal to the sum of the accelerations induced in it by each of the other bodies individually. These laws, which appear in a rather unfamiliar form, are actually completely equivalent to Newton\u2019s laws, as stated in the previous section. In view of the apparently fundamental role played by the concept of force in Newtonian mechanics, it is remarkable that we have been able to reformulate the basic laws without mentioning this concept. It can of", " course be introduced, by de\ufb01ning it through Newton\u2019s second law, (1.1). The utility of this de\ufb01nition arises from the fact that forces satisfy Newton\u2019s third law, (1.3), while accelerations satisfy only the more complicated law, (1.7). Since the mutually induced accelerations of two given bodies are always proportional, they are essentially determined by a single function, and it is useful to introduce the more symmetric concept of force, for which this becomes obvious. It is interesting to note, \ufb01nally, that one consequence of our basic laws is the additive nature of mass. Let us take a three-body system. Then, returning to the notation of the previous section, the equations of motion for the three bodies are m1a1 = F 12 + F 13, m2a2 = F 21 + F 23, m3a3 = F 31 + F 32. (1.9) If we add these equations, then, in view of (1.3), the terms on the right cancel in pairs, and we are left with m1a1 + m2a2 + m3a3 = 0, (1.10) which is the generalization of (1.7). Now, if we suppose that the force between the second and third is such that they are rigidly bound together to form a composite body, their accelerations must be equal: a2 = a3. In that case, we get m1a1 = \u2212 (m2 + m3)a2, which shows that the mass of the composite body is just m23 = m2 + m3. Introduction 13 1.4 External Forces To \ufb01nd the motion of the various bodies in any dynamical system, we have to solve two closely interrelated problems. First, given the positions and velocities at any one instant of time, we have to determine the forces acting on each body. Second, given the forces acting, we have to compute the new positions and velocities after a short interval of time has elapsed. In a general case, these two problems are inextricably bound up with each other, and must be solved simultaneously. If, however, we are concerned with the motions of a small body, or group of small bodies, then we can often neglect its e\ufb00ect on other bodies, and in that case the two problems can be separated. For example, in discussing the motion of an arti\ufb01cial satellite, we can clearly ignore its e\ufb00ect on the Earth. Since the motion of the Earth is already known, we can calculate the force on the satellite as a function of its position and (if atmospheric resistance is included) its velocity. Then, taking the force as known, we can solve separately the problem of its motion. In the latter problem, we are really concerned with the satellite alone. The Earth enters simply as a known external in\ufb02uence. In many cases, therefore, it is useful to concentrate our attention on a small part of a dynamical system, and to represent the e\ufb00ect of everything outside this by external forces, which we suppose to be known in advance, as functions of position, velocity and time. This is the kind of problem with which we shall be mainly concerned in the next few chapters. Typically, we shall consider the motion of a particle under a known external force. In Chapter 6, we consider, for the gravitational and electrostatic cases, the complementary problem of determining the force from a knowledge of the positions of other bodies. Later, in Chapter 7, we return to the more complex type of problem in which the system of immediate interest cannot be taken to be merely a single particle. 1.5 Summary To some extent, the selection of a group of basic concepts, in terms of which others are to be de\ufb01ned, is a matter of choice. We have chosen to regard position and time (relative to some frame of reference) as basic. From this point of view, Newton\u2019s laws must be regarded as containing de\ufb01nitions in addition to physical laws. The \ufb01rst law contains the de\ufb01nition of an inertial 14 Classical Mechanics frame, together with the physical assertion that such frames exist, while the second and third laws contain de\ufb01nitions of mass and force. These laws, supplemented by the laws of force, such as the law of universal gravitation, provide the equations from which we can determine the motion of any dynamical system. Problems Note. Here and in later chapters, starred problems are somewhat harder. 1. An object A moving with velocity v collides with a stationary object B. After the collision, A is moving with velocity 1 2 v and B with velocity 3 2 v. Find the ratio of their masses. If, instead of bouncing apart, the two bodies stuck together", " after the collision, with what velocity would they then move? 2. The two components of a double star are observed to move in circles of radii r1 and r2. What is the ratio of their masses? (Hint: Write down their accelerations in terms of the angular velocity of rotation, \u03c9.) 3. Consider a system of three particles, each of mass m, whose motion is described by (1.9). If particles 2 and 3, even though not rigidly bound together, are regarded as forming a composite body of mass 2m located at the mid-point r = 1 2 (r2+r3), \ufb01nd the equations describing the motion of the two-body system comprising particle 1 and the composite body (2+3). What is the force on the composite body due to particle 1? Show that the equations agree with (1.7). When the masses are unequal, what is the correct de\ufb01nition of the position of the composite (2 + 3) that will make (1.7) still hold? \u00d7 4. Find the distance r between two protons at which the electrostatic repulsion between them will equal the gravitational attraction of the 19 C, proton mass = Earth on one of them. (Proton charge = 1.6 1.7 27 kg.) 10\u2212 \u00d7 10\u2212 5. Consider a transformation to a relatively uniformly moving frame of vt. reference, where each position vector ri is replaced by ri = ri \u2212 (Here v is a constant, the relative velocity of the two frames.) How does a relative position vector rij transform? How do momenta and forces transform? Show explicitly that if equations (1.1) to (1.4) hold in the original frame, then they also hold in the new one. 6. A body of mass 50 kg is suspended by two light, inextensible cables of Introduction 15 lengths 15 m and 20 m from rigid supports placed 25 m apart on the same level. Find the tensions in the cables. (Note that by convention \u2018light\u2019 2. This and the following means \u2018of negligible mass\u2019. Take g = 10 m s\u2212 two problems are applications of vector addition.) 7. *An aircraft is to \ufb02y to a destination 800 km due north of its starting 1. The wind is from the east at a speed point. Its airspeed is 800 km h\u2212 1. On what compass heading should the pilot \ufb02y? How long of 30 m s\u2212 1, and the will the \ufb02ight take? If the wind speed increases to 50 m s\u2212 wind backs to the north-east, but no allowance is made for this change, how far from its destination will the aircraft be at its expected arrival time, and in what direction? 8. *The two front legs of a tripod are each 1.4 m long, with feet 0.8 m apart. The third leg is 1.5 m long, and its foot is 1.5 m directly behind the midpoint of the line joining the other two. Find the height of the tripod, and the vectors representing the positions of its three feet relative to the top. (Hint: Choose a convenient origin and axes and write down the lengths of the legs in terms of the position vector of the top.) Given that the tripod carries a weight of mass 2 kg, \ufb01nd the forces in the legs, assuming they are purely compressional (i.e., along the direction of the leg) and that the legs themselves have negligible weight. (Take g = 10 m s\u2212 9. *Discuss the possibility of using force rather than mass as the basic quantity, taking for example a standard weight (at a given latitude) as the unit of force. How should one then de\ufb01ne and measure the mass of a body? 2.) 10. The \ufb01rst estimate of Newton\u2019s constant was made by the astronomer Nevil Maskelyne in 1774 by measuring the angle between the directions of the apparent plumb-line vertical on opposite sides of the Scottish mountain Schiehallion (height 1081 m, chosen for its regular conical shape). Find a rough estimate of the angle through which a plumb line is deviated by the gravitational attraction of the mountain, by modelling 3, the mountain as a sphere of radius 500 m and density 2.7 and assuming that its gravitational e\ufb00ect is the same as though the total mass were concentrated at its centre. (This latter assumption will be justi\ufb01ed, for a spherical object, in Chapter 6.) 103 kg m\u2212 \u00d7 16 Classical Mechanics Chapter 2 Linear Motion In this chapter we discuss the motion of a body which is free to move only in one dimension. The problems discussed are chosen to illustrate the concepts and techniques which will be of use in the more general case of three-dimensional motion. 2.1 Conservative Forces;", " Conservation of Energy We consider \ufb01rst a particle moving along a line, under a force which is given as a function of its position, F (x). Then the equation of motion (1.1) is m\u00a8x = F (x). (2.1) Since this equation is of second order in the time derivatives, we shall have to integrate twice to \ufb01nd x as a function of t. Thus the solution will contain two arbitrary constants that may be \ufb01xed by specifying the initial values of x and \u02d9x. When the force depends only on x, we can always \ufb01nd a \u2018\ufb01rst integral\u2019, a function of x and \u02d9x whose value is constant in time. Let us consider the kinetic energy, T = 1 2 m \u02d9x2. Di\ufb00erentiating, we \ufb01nd for its rate of change \u02d9T = m \u02d9x\u00a8x = F (x) \u02d9x, using (2.1). Integrating with respect to time, we \ufb01nd T = F (x) \u02d9x dt = F (x) dx. 17 (2.2) (2.3) (2.4) 18 Classical Mechanics If we de\ufb01ne the potential energy, V (x) = x \u2212 x0 F (x) dx, we can write (2.4) in the form T + V = E = constant. (2.5) (2.6) Here x0 is an arbitrary constant, corresponding to the arbitrary additive constant of integration in (2.4). There is a corresponding arbitrariness up to an additive constant in the total energy E. Note that (2.5) can be inverted to give the force in terms of the potential energy: F (x) = dV dx \u2261 \u2212 \u2212 V (x). (2.7) The equation (2.6) is the law of conservation of energy. A force of this type, depending only on x, is called a conservative force. A great deal of information about the motion can be obtained from this conservation law, even without integrating again to \ufb01nd x explicitly as a function of t. If the initial position and velocity are given, we can calculate the value of the constant E. Then (2.6) in the form 1 2 m \u02d9x2 = E V (x) \u2212 (2.8) gives the velocity of the particle (except for sign) when it is at any given position x. Since the kinetic energy is obviously positive, the motion is con\ufb01ned to the region where V (x) E. \u2264 An example may help to explain this. Example: The simple pendulum A simple pendulum comprises a bob of mass m supported by a light rigid rod of length l. (We choose a rod rather than a string so that we may consider motion above the point of support. By \u2018light\u2019 we mean \u2018of negligible mass\u2019.) The bob starts with velocity v from the equilibrium position. What kinds of motion are possible for di\ufb00erent values of v? Linear Motion 19 The distance moved by the bob is x = l\u03b8, where \u03b8 is the angular displacement from the downward vertical. Corresponding to the restoring force F = mg sin(x/l), the potential energy function is mg sin \u03b8 = \u2212 \u2212 V = mgl[1 \u2212 cos(x/l)] = mgl(1 cos \u03b8). \u2212 It is plotted in Fig. 2.1. (Note that the points \u03b8 = \u03c0 and \u03b8 = be identi\ufb01ed.) \u2212 (2.9) \u03c0 may V E (2) E (1) 2 1 \u2212mv\u2032 2 \u2212\u03c0 \u2212\u03b8 0 \u03b8 0 \u03c0 \u03b8 Fig. 2.1 Initially \u03b8 = 0, and the bob is moving with velocity v. Since we have chosen the arbitrary constant in V so that V (0) = 0, the total energy is E = 1 2 mv2. Now if E < 2mgl (E(1) in the \ufb01gure), the motion will be con\ufb01ned between the two angles \u03b80 where V (\u03b80) = E, i.e. \u00b1 cos \u03b80 = v2/2gl. 1 \u2212 These are the points where the kinetic energy vanishes, so that the pendulum bob is instantaneously at rest. The motion is an oscillation of amplitude \u03b80. On the other hand, if the bob is pushed so hard that E > 2mgl (E(2) in the \ufb01gure), then the kinetic energy will never vanish. When the bob reaches the upward vertical, it still has positive energy, namely 1 2 mv 2", " = E \u2212 2mgl = 1 2 mv2 2mgl. \u2212 In this case, the motion is a continuous rotation rather than an oscillation. 20 Classical Mechanics 2.2 Motion near Equilibrium; the Harmonic Oscillator A particle can be in equilibrium only if the total force acting on it is zero. For a conservative force this means, by (2.7), that the potential energy curve is horizontal at the position of the particle. Let us now consider the motion of a particle near a position of equilibrium. Without loss of generality, we may choose the equilibrium point to be the origin, x = 0, and choose the arbitrary constant in V so that V (0) = 0. If we are interested only in small displacements, we may expand V (x) in a Maclaurin\u2013Taylor series, V (x) = V (0) + xV (0) + 1 2 x2V (0) +, \u00b7 \u00b7 \u00b7 where the primes denote di\ufb00erentiation with respect to x. Since we have chosen V (0) = 0, the constant term is absent, and since the equilibrium condition is V (0) = 0, the linear term is absent also. Thus near x = 0 we can write, approximately, V (x) = 1 2 kx2, k = V (0). (2.10) (We assume for simplicity that V (0) does not also vanish.) Because motion near almost any point of equilibrium is described approximately by this potential energy function, it is remarkably ubiquitous; it plays as important a role in quantum mechanics as in classical mechanics. It will therefore be useful to analyze it in some detail. The potential energy curve here is a parabola (see Fig. 2.2). Thus for k > 0 and any energy E > 0 (Fig. 2.2(a)), there are two points where V (x) = E, namely x = a, \u00b1 a = 2E/k. (2.11) The motion is an oscillation between these two points. On the other hand, if k < 0, the curve is an inverted parabola (see Fig. 2.2(b)). In this case, two kinds of motion are possible. If E < 0 the particle may approach to some minimum distance, where it comes momentarily to rest before reversing direction (E(2) in \ufb01gure). But if E > 0 (E(1) in \ufb01gure), it has enough energy to surmount the barrier and will never come to rest: it will continue moving in the same direction for ever, with decreasing speed as it approaches x = 0 and then increasing speed. The force corresponding to the potential energy function (2.10) is, by (2.7), F (x) = kx. \u2212 (2.12) Linear Motion 21 V E (a) \u2212a a x Fig. 2.2 V E (1) E (2) (b) x It is an attractive or repulsive force, according as k > 0 or k < 0, proportional to displacement from the point of equilibrium. The equation of motion may be written m\u00a8x + kx = 0. (2.13) This equation is very easy to solve. We could proceed directly from the energy conservation equation in the form (2.8) and integrate again to obtain 2E m \u2212 k m 1/2 \u2212 x2 dx = dt. (2.14) However, since we shall encounter a number of similar equations later, it will be useful to discuss an alternative method of solution that can be adapted later to other examples. Solution of the harmonic oscillator equation Equation (2.13) is a linear di\ufb00erential equation; that is, one involving only linear terms in x and its derivatives. Such equations have the important property that their solutions satisfy the superposition principle: if x1(t) and x2(t) are solutions, then so is any linear combination x(t) = a1x1(t) + a2x2(t), (2.15) where a1 and a2 are constants; for, clearly, m\u00a8x + kx = a1(m\u00a8x1 + kx1) + a2(m\u00a8x2 + kx2) = 0. 22 Classical Mechanics Moreover, if x1 and x2 are linearly independent solutions (that is, unless one is simply a constant multiple of the other), then (2.15) is actually the general solution. For, since (2.13) is of second order, we could obtain its solution by integrating twice, and the general solution must therefore contain just two arbitrary constants of integration. So to \ufb01nd the general solution, all we have to do is to \ufb01nd any two independent solutions x1(t) and x2(t).", " Let us consider \ufb01rst the case where k < 0, so that V (x) has a maximum at x = 0. Then (2.13) can be written p2x = 0, p = \u00a8x \u2212 k/m. \u2212 (2.16) It is easy to verify that this equation is satis\ufb01ed by the functions x = ept and x = e\u2212 pt. Thus the general solution is x = 1 2 Aept + 1 2 Be\u2212 pt. (2.17) (We introduce the factor of 1 2 for later convenience; it is a matter of convention whether we call the arbitrary constants A and B or 1 2 B.) Clearly, a small displacement will in general lead to an exponential increase of x with time, which continues until the approximation involved in (2.10) ceases to be valid. Thus the equilibrium is unstable, as we should expect when V has a maximum. 2 A and 1 We now turn to the case where k > 0 and V has a minimum at x = 0. Then the potential energy function (2.10) is that of a simple harmonic oscillator. The equation of motion (2.13) becomes \u00a8x + \u03c92x = 0, \u03c9 = k/m. (2.18) It is again very easy to check that the functions x = cos \u03c9t and x = sin \u03c9t are solutions, and the general solution is therefore x = c cos \u03c9t + d sin \u03c9t. (2.19) The arbitrary constants c and d are to be determined by the initial conditions. If at t = 0 the particle is at x0 with velocity v0, then we easily \ufb01nd c = x0, d = v0/\u03c9. An alternative form of (2.19) is x = a cos(\u03c9t \u03b8), \u2212 (2.20) Linear Motion 23 where the constants a, \u03b8 are related to c, d by c = a cos \u03b8, d = a sin \u03b8. The constant a is called the amplitude. It is identical to the constant introduced in (2.11), and de\ufb01nes the limits between which the particle oscillates, x = a. The motion is a periodic oscillation, of period \u03c4 given by \u00b1 \u03c4 = 2\u03c0/\u03c9. (2.21) (See Fig. 2.3.) The frequency f is the number of oscillations per unit time, namely f = 1 \u03c4 = \u03c9 2\u03c0. (2.22) \u03c4 x a \u2212a t Fig. 2.3 Notice that this discussion applies to the motion of a particle near a stable equilibrium point of any potential energy function. For su\ufb03ciently small displacements, any system of this kind behaves like a simple harmonic oscillator. In particular, the period or frequency of small oscillations may always be found from the second derivative of the potential energy function at the equilibrium position. Example: Charged particle between two \ufb01xed charges A particle of charge q is constrained to move along a line between a. What is the period of two other equal charges q, \ufb01xed at x = small oscillations? \u00b1 24 Classical Mechanics Here, the force (in the region < a) is x | | F = q2 4\u03c00(a + x)2 \u2212 q2 4\u03c00(a \u2212 x)2, corresponding to the potential energy function V = q2 4\u03c00 1 a + x + a 1 = x q2 4\u03c00 2a a2 x2. \u2212 Clearly, the position of equilibrium, where V = 0, is x = 0, and we easily \ufb01nd \u2212 k = V (0) = q2a \u03c00 a2 + 3x2 x2)3 (a2 \u2212 = q2 \u03c00a3. x=0 Hence, \u03c92 = q2/\u03c00a3m, and \u03c4 = (2\u03c0/q)\u221a\u03c00a3m. 2.3 Complex Representation It is often very convenient, in discussing periodic phenomena, to use complex numbers. In particular, this allows us to treat both cases, k > 0 and k < 0, of the previous section together. If we allow p to be complex, then the solution in both cases is given by (2.16) and (2.17). When k > 0, p becomes purely imaginary, p = i\u03c9, and the solution (2.17) is x = 1 2 Aei\u03c9t + 1 2 Be\u2212 i\u03c9t. (2.23) Of course, x must be a real number, so that A and B must be complex conjugates. If we write and use the relation A = c id, \u2212 B = c + id, ei\u03c9t = cos \u03c9t + i sin \u03c9t,", " we recover the solution (2.19). Similarly, if we use the polar form of a complex number, A = ae\u2212 i\u03b8, B = aei\u03b8, we obtain the solution in the form (2.20). Linear Motion 25 It is also useful to note that if z = x + iy is a complex solution of (2.13), m\u00a8z + kz = (m\u00a8x + kx) + i(m\u00a8y + ky) = 0, then its real and imaginary parts, x = Re(z) and y = Im(z), must separately each satisfy the equation. This method of obtaining solutions will be useful later. If we plot the complex solution, z = x + iy = Aei\u03c9t, in the xy-plane, we see that z moves in a circle around the origin with angular velocity \u03c9 (see Fig. 2.4). For this reason, \u03c9 is usually called the y c \u03b8 a d x z Fig. 2.4 angular frequency. The constant A is a complex amplitude, whose absolute value (the real amplitude a) is the radius of the circle, and whose phase \u03b8 de\ufb01nes the initial direction of the vector from the origin to z. The real part of the solution, x, may be regarded as the projection of this circular motion on the real axis. 2.4 The Law of Conservation of Energy This law was originally derived from Newton\u2019s laws for the case where the force is a function only of x. It has, however, a much wider application. 26 Classical Mechanics By introducing additional forms of energy (heat, chemical, electromagnetic and so on), it has been extended far beyond the \ufb01eld of mechanics, to the point where it is now recognized as one of the most fundamental of all physical laws. The existence of such a law, and of the laws of conservation of momentum and angular momentum, is in fact closely related to the relativity principle, as we shall see later in Chapter 12. Many non-conservative (or dissipative) forces may be regarded as macroscopic e\ufb00ects of forces which are really conservative on a small scale. For example, when a particle penetrates a retarding medium, such as the atmosphere, it experiences a force which is velocity-dependent, and therefore non-conservative. However, if we look at the situation on a sub-microscopic scale, we see that what happens is that the particle makes a series of collisions with the molecules of the medium. In each collision, energy is conserved, and some of the kinetic energy of the incoming particle is transferred to the molecules with which it collides. (Such collisions will be discussed in detail in Chapter 7.) By means of further collisions, this energy is gradually distributed among the surrounding molecules. The net result is to retard the incoming particle, and to increase the average energy of the molecules in the medium. This increased energy appears macroscopically as heat, and results in a rise in the temperature of the medium. For an arbitrary force F, the rate of change of the kinetic energy T is \u02d9T = F \u02d9x. Thus the increase in kinetic energy in a time given by (2.3): interval dt, during which the particle moves a distance dx, is where dT = dW, dW = F dx (2.24) (2.25) is called the work done by the force F in the in\ufb01nitesimal displacement dx. (Thus for a conservative force, the potential energy V (x) is equal to minus the work done by the force in the displacement from the \ufb01xed point x0 to x.) The work is therefore a measure of the amount of energy converted to kinetic energy from other forms, and the rate P of doing work (the power ) is de\ufb01ned as F \u02d9x. In a real mechanical system, there is usually some loss of mechanical (kinetic or potential) energy to heat or other forms. Correspondingly, there will be dissipative forces acting on the system. (Of course, it is also possible to convert heat to mechanical energy, as in an internal combustion engine.) Linear Motion 27 2.5 The Damped Oscillator \u00a7 We saw in 2.2 that a particle near a position of stable equilibrium under a conservative force may always be treated approximately as a simple harmonic oscillator. If there is energy loss, we must include in the equation of motion a force depending on the velocity. So long as we are concerned only with small displacements from the equilibrium position, we may treat both x and \u02d9x as small quantities, and neglect x2, x \u02d9x and \u02d9x2 in comparison. Thus we are led to consider the damped harmonic oscillator for which F = kx \u2212 \u2212 \u03bb \u02d9x, where \u03bb is another", " constant. The equation of motion now becomes m\u00a8x + \u03bb \u02d9x + kx = 0. (2.26) Equations of this form turn up in many branches of physics. For example, the oscillations of an electrical circuit containing an inductance L, resistance R and capacitance C in series are described by the equation L\u00a8q + R \u02d9q + 1 C q = 0, in which the variable q represents the charge on one plate of the condenser. The equation (2.26) may be solved as before by looking for solutions of the form x = ept. Substituting in (2.26), we obtain for p the equation whose roots are where mp2 + \u03bbp + k = 0, p = \u03b3 \u2212 \u00b1 \u03c92 0, \u2212 \u03b32 \u03b3 = \u03bb 2m, and \u03c90 = k m. (2.27) (2.28) Note that \u03c90 is the frequency of the undamped oscillator, as in (2.18). The \u03bb \u02d9x2. If \u03bb were negative, rate at which work is done by the force the particle would therefore be gaining, rather than losing, energy. So we shall assume that \u03bb is positive. \u03bb \u02d9x is \u2212 \u2212 28 Classical Mechanics Large damping If \u03bb is so large that \u03b3 > \u03c90, then both roots for p are real and negative: \u00b1 The general solution is then p = \u03b3 \u2212, = \u03b3 \u03b3 \u00b1 \u00b1 \u03c92 0. \u2212 \u03b32 x = 1 2 Ae\u2212 \u03b3+t + 1 2 Be\u2212 \u03b3\u2212t, (2.29) where A and B are arbitrary constants. Hence the displacement tends exponentially to zero. For large times, the dominant term is that containing. Thus the characteristic time in in the exponent the smaller coe\ufb03cient, \u03b3 which the displacement x is reduced by a factor 1/e is of order 1/\u03b3 \u2212. \u2212 Small damping Now let us consider the case where \u03bb is small, so that \u03b3 < \u03c90. Then the two roots for p are complex conjugates, \u00b1 The general solution may be written in the alternative forms \u2212 p = \u03b3 i\u03c9, \u03c9 = \u03c92 0 \u2212 \u03b32. (2.30) 2 Be\u2212 i\u03c9t \u2212 \u03b3t x = 1 2 Aei\u03c9t \u2212 = Re(Aei\u03c9t \u03b3t + 1 \u03b3t) \u03b3t cos(\u03c9t \u2212 \u2212 \u03b8), (2.31) = ae\u2212 where A = ae\u2212 i\u03b8, B = aei\u03b8. From the last form of (2.31) we see that this solution represents an oscilla\u03b3t, and angular frequency tion with exponentially decreasing amplitude ae\u2212 \u03c9. (See Fig. 2.5.) Note that \u03c9 is always less than the frequency \u03c90 of the undamped oscillator. The time in which the amplitude is reduced by a factor 1/e, called the relaxation time of the oscillator, is 1/\u03b3 = 2m/\u03bb. It is often convenient to introduce the quality factor, or simply \u2018Q\u2019, of the oscillator, de\ufb01ned as the dimensionless number Q = m\u03c90 \u03bb = \u03c90 2\u03b3. (2.32) Linear Motion 29 x t Fig. 2.5 If the damping is small then Q is large. In a single oscillation period, the \u03c0/Q. The amplitude is reduced by the factor e\u2212 number of periods in a relaxation time is roughly Q/\u03c0. 2\u03c0\u03b3/\u03c9, or approximately e\u2212 Critical damping The limiting case, \u03b3 = \u03c90, is the case of critical damping, in which \u03c9 = 0 and the two roots for p coincide. Then the solution (2.29) or (2.31) involves only one arbitrary constant, A + B, and cannot be the general solution. We need to \ufb01nd a second independent solution in this case. In fact, it is easy to verify by direct substitution that the equation (2.26) in this special case is also satis\ufb01ed by the function te\u2212 pt. Thus the general solution is x = (a + bt)e\u2212 \u03b3t. (2.33) Critical damping is often the ideal. For example, in a measuring instrument, we want to damp out the oscillations of the pointer about its correct position as quickly as possible, but too much damping would lead to a very slow response. Let us assume that k is \ufb01xed and the amount of damping varied. When the damping is less than critical (\u03b3 < \u03c90), the characteristic time of response is the", " relaxation time 1/\u03b3, which of course decreases as \u03b3 is increased. However, when \u03b3 > \u03c90, the characteristic time is 1/\u03b3, as \u2212 decreases, we noted above. It is not hard to verify that as \u03b3 increases, \u03b3 increases (see Fig. 2.6). Thus the shortest so that the response time 1/\u03b3 possible response time is obtained by choosing \u03b3 = \u03c90, that is for critical damping. \u2212 \u2212 30 Classical Mechanics Fig. 2.6 2.6 Oscillator under Simple Periodic Force In an isolated system the forces are functions of position and velocity, but not explicitly of the time. However, we are often interested in the response of an oscillatory system to an applied external force, which is given as a function of the time, F (t). Then we have to deal with the equation m\u00a8x + \u03bb \u02d9x + kx = F (t). (2.34) Now, if x1(t) is any solution of this equation, and x0(t) is a solution of the corresponding homogeneous equation (2.26) for the unforced oscillator, then, as one can easily check by direct substitution, x1(t) + x0(t) will be another solution of (2.34). Hence we only need to \ufb01nd one particular solution of this equation. Its general solution is then obtained by adding to this particular solution the general solution of the homogeneous equation (2.26). (This will be the general solution because it contains two arbitrary constants.) Let us consider \ufb01rst the case where the applied force is periodic in time, with the simple form F (t) = F1 cos \u03c91t, (2.35) where F1 and \u03c91 are real constants. It is convenient to write this in the form F (t) = Re(F1ei\u03c91t), Linear Motion 31 and to solve \ufb01rst the equation with a complex force, m\u00a8z + \u03bb \u02d9z + kz = F1ei\u03c91t. (2.36) Then the real part x of this solution will be a solution of the equation (2.34) with the force (2.35). We now look for a particular solution of (2.36) which is periodic in time with the same period as the applied force, z = A1ei\u03c91t = a1ei(\u03c91t \u2212 \u03b81), where A1 = a1e\u2212 an equation for A1: i\u03b81 is a complex constant. Substituting in (2.36), we obtain m\u03c92 1 + i\u03bb\u03c91 + k)A1 = F1. ( \u2212 (2.37) Dividing by me\u2212 i\u03b81 and rearranging the terms gives (\u03c92 0 + 2i\u03b3\u03c91 \u2212 \u03c92 1)a1 = F1 m ei\u03b81, where \u03b3 and \u03c90 are de\ufb01ned as before by (2.28). Hence, equating real and imaginary parts, (\u03c92 0 \u2212 \u03c92 1)a1 = (F1/m) cos \u03b81, 2\u03b3\u03c91a1 = (F1/m) sin \u03b81. The amplitude is found by squaring these equations and adding, to give a1 = F1/m 1)2 + 4\u03b32\u03c92 \u03c92 1. (\u03c92 0 \u2212 Dividing one equation by the other yields the phase: tan \u03b81 = 2\u03b3\u03c91 \u03c92 0 \u2212 \u03c92 1. (2.38) (2.39) If F1 > 0, the correct choice between the two solutions for \u03b81 di\ufb00ering by \u03c0 is the one lying in the range 0 < \u03b81 < \u03c0. We have found a particular solution of (2.36). Its real part, the corre- sponding particular solution of our original equation (2.34), is simply x = a1 cos(\u03c91t \u03b81). \u2212 (2.40) 32 Classical Mechanics The general solution, as we discussed above, is obtained by adding to this particular solution the general solution of the corresponding homogeneous equation, namely (2.29) or (2.31). In the case where the damping is less than critical (\u03b3 < \u03c90), we obtain x = a1 cos(\u03c91t \u2212 \u03b81) + ae\u2212 \u03b3t cos(\u03c9t \u03b8). \u2212 (2.41) Here, of course, a1 and \u03b81 are given by (2.38) and (2.39), but a and \u03b8 are arbitrary constants to be \ufb01xed by the initial conditions. The second term in", " the general solution (2.41), which represents a free oscillation, dies away exponentially with time. It is therefore called the transient. After a long time, the displacement x will be given by the \ufb01rst term of (2.41), that is, by (2.40), alone. Thus, no matter what initial conditions we choose, the oscillations are ultimately governed solely by the external force. Note that their period is the period of the applied force, not the period of the unforced oscillator. Resonance The amplitude a1 and phase \u03b81 of the forced oscillations are strongly dependent on the angular frequencies \u03c90 and \u03c91. In particular, if the damping is small, the amplitude can become very large when the frequencies are almost equal. If we \ufb01x \u03b3 and the forcing frequency \u03c91, and vary the oscillator frequency \u03c90, the amplitude is a maximum when \u03c90 = \u03c91. In this case, we say that the system is in resonance. At resonance, the amplitude is a1 = F1 2m\u03b3\u03c91 = F1 \u03bb\u03c91, (2.42) which can be very large if the damping constant \u03bb is small. If, on the other hand, we \ufb01x the parameters \u03b3 and \u03c90 of the oscillator, and vary the forcing frequency \u03c91, the maximum amplitude actually occurs for a frequency slightly lower than \u03c90, namely \u03c91 = \u03c92 0 \u2212 2\u03b32. However, if \u03b3 is small, this does not di\ufb00er much from \u03c90. Note that the natural frequency \u03c9 of the oscillator lies between this resonant frequency and the frequency \u03c90 of the undamped oscillator. Near resonance, the dependence of the amplitude a1 on the forcing frequency \u03c91 is of the form illustrated in Fig. 2.7(a). The width of the resonance, that is the range of frequencies over which the amplitude is large, Linear Motion 33 a 1 large Q F 1 m\u03c9 2 0 small Q \u03b3 \u03c9 0 (a) \u03b8 1 \u03c0 \u03c0 2 \u03c9 1 Fig. 2.7 large Q small Q \u03c9 0 (b) \u03c9 1 is determined by \u03b3. For, the amplitude is reduced to 1/\u221a2 of its peak value when the two terms in the denominator of (2.38) become roughly equal, \u03b3. Therefore, \u03b3 is called the and for small \u03b3 this occurs when \u03c91 = \u03c90 \u00b1 half-width of the resonance. Notice the inverse relationship between the width and peak amplitude: the narrower the resonance, the higher is its peak. The quality factor Q = \u03c90/2\u03b3 provides a quantitative measure of the sharpness of the resonance peak. Indeed, the ratio of the amplitude at resonance, given by (2.42), to the amplitude at \u03c91 = 0 (which describes the response to a time-independent force) is precisely Q. The phenomenon of resonance occurs with any oscillatory system, and is of great practical importance. Since quite small forces can set up large oscillations if the frequencies are in resonance, great care must be taken in the design of any mechanical structure to avoid this possibility. This led to problems for example with the original design of the Millennium Bridge in London. The constant \u03b81 speci\ufb01es the phase relation between the applied force and the induced oscillations. If the force is slowly oscillating, \u03c91 is small, 0, so that the induced oscillations are in phase with the force. In and \u03b81 \u2248 that case, the amplitude (2.38) is a1 \u2248 F1 m\u03c92 0 = F1 k. Thus the position x at any time t, given by (2.40), is approximately the 34 Classical Mechanics \u2212 kx. At resonance, the equilibrium position under the force F1 cos \u03c91t phase shift is \u03b81 = 1 2 \u03c0, and the induced oscillations lag behind by a quarter period. (The variation of \u03b81 with \u03c91 is shown in Fig. 2.7(b).) For very \u03c0, and the oscillations are almost exactly rapidly oscillating forces, \u03b81 \u2248 1 and the oscillations out of phase. correspond to those of a free particle under the applied oscillatory force. Note that the value of the damping constant is important only in the region near the resonance. In this limiting case, a1 \u2248 F1/m\u03c92 2.7 General Periodic Force The solution we have obtained can immediately be generalized to the case where the applied force is a sum of periodic terms, F (t) = F1ei\u03c91t + F2ei\u03c92t + Frei\u03c9rt. (2.43) = \u00b7 \u00b7 \u00b7", " r We can always ensure that this force is real by including with each term its complex conjugate, so it is not really necessary to write Re in front of the sum. (In particular, a simple periodic force may be written as a sum of two terms proportional to ei\u03c9t and e\u2212 i\u03c9t.) Because of the linearity of (2.34), the corresponding solution is x = Arei\u03c9rt + transient, (2.44) r where each Ar is related to the corresponding Fr by (2.37). This may easily be veri\ufb01ed by direct substitution in (2.34). In particular, let us take F (t) = \u221e Fnein\u03c9t, (2.45) n= \u2212\u221e where n runs over all integers, positive, negative and zero. Here \u03c9 represents a fundamental angular frequency and the frequencies n\u03c9 are its harmonics. Each term in the sum is unchanged by the replacement t t + 2\u03c0/\u03c9, so F (t) is clearly periodic in the sense that \u2192 F (t + \u03c4 ) = F (t), where \u03c4 = 2\u03c0/\u03c9. (2.46) In fact, any periodic force with period \u03c4 can be expressed in the form (2.45), which is called a Fourier series. If we are given F (t) it is easy to \ufb01nd the Linear Motion 35 corresponding coe\ufb03cients Fn. For one easily sees that 1 \u03c4 \u03c4 0 ei(n \u2212 m)tdt = \u03b4nm, (2.47) where \u03b4nm is Kronecker\u2019s delta symbol de\ufb01ned by \u03b4nm = 1 if n = m, 0 if n = m. (2.48) Thus if we multiply (2.45) by e\u2212 term in the sum which survives is the one with n = m. Hence we obtain im\u03c9t and integrate over a period, the only Fm = \u03c4 1 \u03c4 0 F (t)e\u2212 im\u03c9tdt. (2.49) A simple example involves a force with \u2018square-wave\u2019 form. Example: Square-wave force Find the position after a long time of an oscillator subjected to a force with the square-wave form shown in Fig. 2.8, and given by F(t) F 0 \u2212F \u03c4 t Fig. 2.8 F (t) = F, F, \u2212 for r = 0, 1, \u00b1 \u00b1 2,.... (r + 1 r\u03c4 < t \u2264 1 2 )\u03c4 < t (r \u2212 \u2264 2 )\u03c4, r\u03c4, 36 Classical Mechanics Since \u03c9 = 2\u03c0/\u03c4, we \ufb01nd \u03c4 /2 e\u2212 2\u03c0int/\u03c4 dt F \u03c4 \u2212 \u03c4 \u03c4 /2 e\u2212 2\u03c0int/\u03c4 dt Fn = = 0 F \u03c4 F i\u03c0n [1 ( \u2212 \u2212 1)n]. Note that Fn = 0 for all even values of n, because the expression in square brackets vanishes. This happens because in addition to repeating after the period \u03c4, F (t) repeats with opposite sign after half this period: F (t + 1 F (t). For odd values of n, we \ufb01nd from (2.37) that the 2 \u03c4 ) = amplitudes An are \u2212 An = i\u03c0nm(\u03c92 0 \u2212 2F n2\u03c92 + 2in\u03b3\u03c9) (n odd). The position of the oscillator is then given by \u221e x = r= \u2212\u221e A2r+1ei(2r+1)\u03c9t + transient. After a long time, the transient can be ignored, and the position is given by the sum alone. Note that if the damping is small and \u03c90 is close to one of the values n\u03c9, then one pair of amplitudes, An and A n will be much larger than the rest, and the oscillation will be almost simple harmonic motion at this frequency. In general, for large n, the amplitudes An decrease rapidly in magnitude as n increases, so that we can obtain a good approximation to x by keeping only a few terms of the series. \u2212 In this way, one can obtain the solution to (2.34) for an arbitrary periodic force. As a matter of fact, this kind of treatment can be extended even further, by replacing the sum in (2.45) by an integral over all possible angular frequencies \u03c9. Any force whatever, subject to some very general mathematical requirements, can be written in this form. (This is the Fourier integral theorem.) However, we shall not pursue this method, because we shall \ufb01nd in the next section an alternative method of solving the problem, which is often simpler to use. Linear Motion 37 2.8 Impulsive Forces; the Green\u2019s Function Method There are many physical situations", ", for example collisions, in which very large forces act for very short times. Let us consider a force F acting on a particle during the time interval \u2206t. The resulting change of momentum of the particles is \u2206p = p(t + \u2206t) p(t) = \u2212 t t+\u2206t F dt. (2.50) The quantity on the right is called the impulse I delivered to the particle. It is natural to consider an idealized situation in which the time interval \u2206t tends to zero, and the whole of the impulse I is delivered to the particle instantaneously. in such a way that the impulse I remains \ufb01nite. At the instant when the impulse is delivered the momentum of the particle changes discontinuously. Although in reality the force must always remain \ufb01nite, this is a good description of, for example, the e\ufb00ect of a sudden blow. In other words, we let \u2206t 0 and F \u2192 \u221e \u2192 In the next section, we shall discuss some simple collision problems. Here, we shall consider the e\ufb00ect of an impulsive force on an oscillator. Let us suppose \ufb01rst that the oscillator is at rest at its equilibrium position x = 0, and that at time t = 0 it experiences a blow of impulse I. Immediately after the blow, its position is still x = 0 (since the velocity remains \ufb01nite, the position does not change discontinuously), while its velocity is v0 = I/m. Inserting these initial conditions in the general solution (2.31) for the oscillatory case \u03b3 < \u03c90, we \ufb01nd that the position is given by x = 0, I m\u03c9 \u23a7 \u23a8 t < 0, e\u2212 \u03b3t sin \u03c9t, t > 0. (2.51) This solution is illustrated in Fig. 2.9. \u23a9 Just as in the case of periodic forces, we can immediately generalize this result to a sum of impulsive forces. If the oscillator is subjected to a series of blows, of impulse Ir at time tr, its position may be found by adding together the corresponding set of solutions of the form (2.51). We obtain in this way x(t) = G(t r tr)Ir + transient, \u2212 (2.52) 38 Classical Mechanics G(t) t Fig. 2.9 where G(t \u2212 t) = 0, 1 m\u03c9 \u23a7 \u23a8 t < t, e\u2212 \u03b3(t \u2212 t ) sin \u03c9(t t), t > t. \u2212 (2.53) \u2212 \u23a9 t) is called the Green\u2019s function of the oscillator. It The function G(t represents the response to a blow of unit impulse delivered at the time t. We can now use this result to obtain a solution to (2.34) for an arbitrary force F (t). The method is, in a sense, complementary to Fourier\u2019s. Instead of expressing F (t) as a sum of periodic forces, we express it as a sum of impulsive forces. If we divide the time scale into small intervals \u2206t, then the impulse delivered to the particle between t and t+\u2206t will be approximately F (t)\u2206t. Moreover, if \u2206t is short enough, it will be a good approximation to suppose that the whole of this impulse is delivered instantaneously at the time t. Thus the solution should be given approximately by (2.52), in which the interval between tr and tr+1 is \u2206t and Ir = F (tr)\u2206t. The approximation will improve as \u2206t is reduced, and become exact in the limit 0. In this limit, the sum in (2.52) goes over into an integral, and we \u2206t \ufb01nally obtain \u2192 t x(t) = G(t t0 \u2212 t)F (t) dt + transient. (2.54) The lower limit t0 is the initial time at which the initial conditions determining the arbitrary constants in the transient term are to be imposed. t) vanishes for t > t. The upper limit may be taken to be t because G(t (In other words, blows subsequent to t do not a\ufb00ect the position at t.) \u2212 The solution (2.54) is very useful in practice, because it is an explicit It is particularly solution requiring the evaluation of only one integral. Linear Motion 39 well adapted to numerical solution of the problem when F (t) is known numerically. Example: Step-function force An oscillator is initially at rest and is subject to a step-function force, F (t) = 0, t < 0, F, t > 0, with F constant. What is", " the position at time t > 0? To perform the integration in (2.54), it is convenient to write the sine function as the imaginary part of an exponential: F m\u03c9 t 0 Im e( \u2212 \u03b3+i\u03c9)(t \u2212 t ) dt t x(t) = G(t \u2212 1 0 F m\u03c9 Im t)F dt = \u03b3t+i\u03c9t e\u2212 \u2212 \u03b3 i\u03c9 F m\u03c9(\u03b32 + \u03c92) \u2212 Im (\u03b3 + i\u03c9) = = = \u03b3t+i\u03c9t e\u2212 1 \u2212 F m\u03c9\u03c92 0 [\u03c9(1 \u2212 \u03b3t cos \u03c9t) e\u2212 \u2212 \u03b3t sin \u03c9t], \u03b3e\u2212 using \u03b32 + \u03c92 = \u03c92 displacement for a constant force F, namely x = F/m\u03c92 0. 0. Note that for large t, this tends as it must to the 2.9 Collision Problems So far, we have been discussing the motion of a single particle under a known external force. We now turn to the problem of an isolated system of two bodies moving in one dimension under a mutual force F which is given in terms of their positions and velocities. The equations of motion are then m1 \u00a8x1 = F, m2 \u00a8x2 = F. \u2212 (2.55) As we have already seen in Chapter 1, the sum of these two equations leads to the law of conservation of momentum, p1 + p2 = P = constant. (2.56) 40 Classical Mechanics According to the relativity principle, F must be a function only of the relative distance and relative velocity, x = x1 \u2212 x2 and \u02d9x = \u02d9x1 \u2212 \u02d9x2. When it is a function only of x the force is conservative, and we can introduce a potential energy function V (x) as before. The law of conservation of energy then takes the form T + V = E = constant, (2.57) with T = 1 2 m1 \u02d9x2 1 + 1 2 m2 \u02d9x2 2. This may easily be veri\ufb01ed by di\ufb00erentiating (2.57) and using (2.55) and the relation between V and F, which is still (2.7). We are particularly interested in collision problems, in which the force is generally small except when the bodies are very close together. The potential energy function is then a constant (say zero) for large values of x, and rises very sharply for small values. (See Fig. 2.10(a).) An ideal impulsive conservative force corresponds to a potential energy function with V V x (a) x (b) Fig. 2.10 a discontinuity or step (Fig. 2.10(b)). So long as the initial kinetic energy is less than the height of the step, the bodies will bounce o\ufb00 one another. (If the kinetic energy exceeds this value, one body will pass through the other. For completely impenetrable bodies, the potential step would have to be in\ufb01nitely high.) From the law of conservation of energy, we know that the \ufb01nal value of the kinetic energy, when the bodies are again far apart, is the same as the Linear Motion 41 initial value well before the collision. If we denote the initial velocities by u1, u2 and the \ufb01nal velocities by v1, v2, then (see Fig. 2.11 initial final Fig. 2.11 1 2 m1v2 1 + 1 2 m2v2 2 = 1 2 m1u2 1 + 1 2 m2u2 2. (2.58) Collisions of this type, in which there is no loss of kinetic energy, are known as elastic collisions. They are typical of very hard bodies, like billiard balls. We saw earlier that a particle which bounces back from a potential barrier emerges with velocity equal (except for sign) to its initial velocity. We can easily derive a similar result for the case of two-body collisions. To do this we write the energy and momentum conservation equations in the forms m1v2 1 \u2212 m1v1 \u2212 1 = m2u2 m1u2 2 \u2212 m1u1 = m2u2 \u2212 m2v2 2, m2v2. We can then divide the \ufb01rst equation by the second, to obtain or, equivalently, v1 + u1 = u2 + v2, v2 \u2212 v1 = u1 \u2212 u2. (2.59) This shows that the relative velocity is just reversed by the collision. Equation (2.59) and the momentum conservation equation (2.56) may be solved for", " the \ufb01nal velocities in terms of the initial ones. In particular, if the second body is initially at rest (u2 = 0), the result is v1 = m2 m1 \u2212 m1 + m2 u1, v2 = 2m1 m1 + m2 u1. (2.60) Note that if the masses are equal, then the \ufb01rst body is brought to rest by the collision, and its velocity transferred to the second body. If m1 > m2, the \ufb01rst body continues in the same direction, with reduced velocity, 42 Classical Mechanics whereas if m1 < m2, it rebounds in the opposite direction. In the limit where m2 is much larger than m1, we obtain v1 = u1, which agrees with the previous result for a particle rebounding from a \ufb01xed potential barrier. \u2212 Inelastic collisions In practice there is generally some loss of energy in a collision, for instance in the form of heat. In that case, the relative velocity will be reduced in magnitude by the collision. We de\ufb01ne the coe\ufb03cient of restitution e in a particular collision by v2 \u2212 v1 = e(u1 \u2212 The usefulness of this quantity derives from the experimental fact that, for any two given bodies, e is approximately a constant for a wide range of velocities. (2.61) u2). The \ufb01nal velocities may again be found from (2.61) and the momentum conservation equation (2.56). When u2 = 0, we get em2 m1 \u2212 m1 + m2 (1 + e)m1 m1 + m2 v2 = v1 = u1, u1. (2.62) For e = 1 these reduce to the relations (2.60) for the case of an elastic collision. At the other extreme, we have the case of very soft bodies that stick together on impact. Then e = 0, and the collision is called perfectly inelastic (or, more accurately perhaps, perfectly plastic). The energy loss in an inelastic collision is easily evaluated. The initial 1 + 2. Substituting for the \ufb01nal velocities from (2.62) we \ufb01nd, after a kinetic energy is T = 1 1 2 m2v2 simple calculation, that the fractional loss of kinetic energy is 1, while the \ufb01nal kinetic energy is T = 1 2 m1u2 2 m1v2 T = T \u2212 T e2)m2 (1 m1 + m2 \u2212. (2.63) This shows, incidentally, that e must always be less than one, unless some energy is released, as for example by an explosion. 2.10 Summary The conservation laws for energy and momentum are among the most important consequences of Newton\u2019s laws. They are valid quite generally for an isolated system, as we shall see in later chapters. For a system which is Linear Motion 43 only part of a larger system, and subject to external forces, there may be some transfer of energy and momentum from the system to its surroundings (or vice versa). For a single particle moving on a straight line, the total (kinetic plus potential) energy is conserved if the force is a function only of x. If the force is velocity-dependent, however, there will be some loss of energy to the surroundings. The importance of the harmonic oscillator, treated in detail in this chapter, lies in the fact that any system with one degree of freedom behaves like a harmonic oscillator when su\ufb03ciently near to a point of stable equilibrium. The methods we have discussed for solving the oscillator equation can therefore be applied to a great variety of problems. In particular, the phenomenon of resonance can occur in any system subjected to periodic forces. It occurs when there is a natural frequency of oscillation close to the forcing frequency, and leads, if the damping is small, to oscillations of very large amplitude. Problems 3 m, with \u02d9x = 8 m s\u2212 1. A harmonic oscillator of angular frequency 2 s\u2212 1 is initially at x = 1. Write the solution in each of the three forms \u2212 (2.19), (2.20) and (2.23). Find the \ufb01rst time at which x = 0, and the \ufb01rst at which \u02d9x = 0. (A sketch of the solution may help.) 2. A weight of mass m is hung from the end of a spring which provides a restoring force equal to k times its extension. The weight is released from rest with the spring unextended. Find its position as a function of time, assuming negligible damping. 3. When a mass is suspended from a spring, the equilibrium length is increased by 50 mm. The mass is", " then given a blow which starts it 1. Find the period and amplitude of the moving vertically at 200 mm s\u2212 resulting oscillations, assuming negligible damping. 4. A pendulum of period 2 s is released from rest at an inclination of 5\u25e6 to the (downward) vertical. What is its angular velocity when it reaches the vertical? When it \ufb01rst returns to its starting point, it is given an impulsive blow towards the vertical that increases the amplitude of swing to 10\u25e6. Find its subsequent angular position as a function of time. 5. Write down the potential energy function corresponding to the force GM m/x2 experienced by a particle of mass m at a distance x from a \u2212 44 Classical Mechanics m). The particle is released from rest at a distance planet of mass M ( a from the centre of the planet, whose radius is R, and falls under gravity. What is its speed when it strikes the surface? Evaluate this speed for a particle falling to the Earth from a = 2R. (Use R = 6400 km, and GM/R2 = g = 10 m s\u2212 2.) 6. A particle of mass m moves under a force F = cx3, where c is a positive constant. Find the potential energy function. If the particle starts from rest at x = a, what is its velocity when it reaches x = 0? Where in the subsequent motion does it instantaneously come to rest? 7. A particle of mass m moves (in the region x > 0) under a force F = kx+c/x, where k and c are positive constants. Find the corresponding \u2212 potential energy function. Determine the position of equilibrium, and the frequency of small oscillations about it. \u2212 \u2212 8. A particle of mass m has the potential energy function V (x) = mk, x | | where k is a positive constant. What is the force when x > 0, and when x < 0? Sketch the function V and describe the motion. If the particle starts from rest at x = a, \ufb01nd the time it takes to reach x = a. 9. A particle of mass m moves under a conservative force with potential \u2212 energy function given by V (x) = 1 2 k(a2 0 x2) \u2212 for for x | | x | < a, a, | \u2265 where a and k are constants, and a > 0. What is the force on the particle? Sketch the function V, for both cases k > 0 and k < 0, and describe the possible types of motion. \u2212 \u2212 10. The particle of Problem 9 with k = m\u03c92 < 0 is initially in the region x < a, moving to the right with velocity v. When it emerges into the region x > a, will it do so earlier or later than if it were moving freely under no force? Find an expression for the time di\ufb00erence. (To do the sin \u03b8.) required integral, use a substitution of the form x = constant \u00d7 1 2 c(x2 \u2212 a2)2, where c and a are positive constants. Sketch this function, and describe the possible types of motion in the three cases (a) E > 0, (b) E < 11. The potential energy function of a particle of mass m is V = \u2212 1 1 2 ca4 < E < 0. 2 ca4, and (c) 12. *The potential energy function of a particle of mass m is V = cx/(x2 + a2), where c and a are positive constants. Sketch V as a function of x. Find the position of stable equilibrium, and the period of small oscillations about it. Given that the particle starts from this point \u2212 \u2212 Linear Motion 45 \u2212\u221e. \u221e with velocity v, \ufb01nd the ranges of values of v for which it (a) oscillates, (b) escapes to, and (c) escapes to + 13. *A particle falling under gravity is subject to a retarding force proportional to its velocity. Find its position as a function of time, if it starts from rest, and show that it will eventually reach a terminal velocity. [The equation of motion can be integrated once to give, with a suitable choice of origin and de\ufb01nition of \u03b3 (di\ufb00ering from (2.28) by a factor of gt. To integrate again, use an integrating factor, i.e. 2), a function f (t) such that when the equation is multiplied by f (t) the left-hand side becomes an exact derivative, in fact the derivative of zf. The \ufb01nal stage requires an integration by parts.] \u02d9z + \u03b3z = \u2212 1. Find the 1, and the distance it has fallen 14. The terminal speed of the particle in Problem 13 is 50 m s\u2212 time it takes to reach a speed of 40 m", " s\u2212 in that time. (Take g = 10 m s\u2212 2.) 15. *A particle moves vertically under gravity and a retarding force proportional to the square of its velocity. (This is more appropriate than a linear relation for larger particles \u2014 see Problem 8.10.) If v is its upkv2, respectively, where k ward or downward speed, show that \u02d9v = is a constant. If the particle is moving upwards, show that its position at time t is given by z = z0 + (1/k) ln cos[\u221agk (t0 \u2212 t)], where z0 and t0 are integration constants. If its initial velocity at t = 0 is u, \ufb01nd the time at which it comes instantaneously to rest, and its height then. loge x. You [Note that ln always denotes the natural logarithm: ln x \u2261 2 ln(1 + tan2 x) useful.] may \ufb01nd the identity ln cos x = \u2212 16. *Show that if the particle of the previous question falls from rest its g/k tanh(\u221agk t). What is its speed after a time t is given by v = limiting speed? How long does it take to hit the ground if dropped from height h? \u2213 \u2212 g 1 17. The pendulum described in 2.1 is released from rest at an angle \u03b80 to the downward vertical. Find its angular velocity as a function of \u03b8, and the period of small oscillations about the position of stable equilibrium. Write down the solution for \u03b8 as a function of time, assuming that \u03b80 is small. \u00a7 18. *For the pendulum described in 2.1, \ufb01nd the equation of motion for small displacements from the position of unstable equilibrium, \u03b8 = \u03c0. Show that if it is released from rest at a small angle to the upward vertical, then the time taken for the angular displacement to increase by l/g ln 20. Evaluate this time for a factor of 10 will be approximately \u00a7 46 Classical Mechanics a pendulum of period 2 s, and \ufb01nd the angular velocity of the pendulum when it reaches the downward vertical. 19. *The particle of Problem 11 starts from rest at x = a, and is given a small push to start it moving to the right. What is its velocity when it reaches the point x? Given that t = 0 is the instant when it reaches x = 0, \ufb01nd its position as a function of time. (Assume that the push has negligible e\ufb00ect on its energy.) \u2212 \u2212 \u2212 m\u03c92(x 20. *A particle of mass m moves in the region x > 0 under the force a4/x3), where \u03c9 and a are constants. Sketch the F = potential energy function. Find the position of equilibrium, and the period of small oscillations about it. The particle starts from this point with velocity v. Find the limiting values of x in the subsequent motion. (To do the Show that the period of oscillation is independent of v. integration, transform to the variable y = x2, then add a constant to \u2018complete the square\u2019, and \ufb01nally use a trigonometric substitution.) 21. Repeat the calculation of Problem 2 assuming that the system is critically damped. Given that the \ufb01nal position of equilibrium is 0.4 m below the point of release, \ufb01nd how close to the equilibrium position the particle is after 1 s. 22. *A pendulum whose period in a vacuum is 1 s is placed in a resistive medium. Its amplitude on each swing is observed to be half that on the previous swing. What is its new period? Given that the pendulum bob is of mass 0.1 kg, and is subjected to a periodic force of amplitude 0.02 N and period 1 s, \ufb01nd the angular amplitude of the resulting forced oscillation. Compare your answer with the angular deviation that would be induced by a constant force of this magnitude in a pendulum at rest. 23. Write down the solution to the oscillator equation for the case \u03c90 > \u03b3 if the oscillator starts from x = 0 with velocity v. Show that, as \u03c90 is reduced to the critical value \u03b3, the solution tends to the corresponding solution for the critically damped oscillator. 24. *Solve the problem of an oscillator under a simple periodic force (turned on at t = 0) by the Green\u2019s function method, and verify that it repro[Assume that the damping is less than duces the solution of \u00a7 critical. To do the integral, write sin \u03c9t as (ei\u03c9t i\u03c9t)/2i.] 2.6. e\u2212 25. *For an oscillator under a periodic force", " F (t) = F1 cos \u03c91t, calculate the power, the rate at which the force does work. (Neglect the transient.) Show that the average power is P = m\u03b3\u03c92 1, and hence verify that it is equal to the average rate at which energy is dissipated against the resistive force. Show that the power P is a maximum, as a function 1a2 \u2212 Linear Motion 47 of \u03c91, at \u03c91 = \u03c90, and \ufb01nd the values of \u03c91 for which it has half its maximum value. 26. *Find the average value \u00afE of the total energy of an oscillator under a periodic force. If W is the work done against friction in one period, show that when \u03c91 = \u03c90 the ratio W/ \u00afE is related to the quality factor Q by W/ \u00afE = 2\u03c0/Q. 27. Three perfectly elastic bodies of masses 5 kg, 1 kg, 5 kg are arranged in that order on a straight line, and are free to move along it. Initially, 1, and the others are at the middle one is moving with velocity 27 m s\u2212 rest. Find how many collisions take place in the subsequent motion, and verify that the \ufb01nal value of the kinetic energy is equal to the initial value. 28. A ball is dropped from height h and bounces. The coe\ufb03cient of restitution at each bounce is e. Find the velocity immediately after the \ufb01rst bounce, and immediately after the nth bounce. Show that the ball \ufb01nally comes to rest after a time 1 + e 1 \u2212 e 2h g. 29. *An oscillator with free period \u03c4 is critically damped and subjected to a force with the \u2018saw-tooth\u2019 form F (t) = c(t \u2212 n\u03c4 ) for (n \u2212 1 2 )\u03c4 < t \u2264 (n + 1 2 )\u03c4, for each integer n (c is a constant). Find the ratios of the amplitudes an of oscillation at the angular frequencies 2\u03c0n/\u03c4. 30. *A particle moving under a conservative force oscillates between x1 and x2. Show that the period of oscillation is x2 \u03c4 = 2 x1 0(x2 2 m\u03c92 \u2212 m 2[V (x2) V (x)] \u2212 dx. bx4), show that the period for oscilla- In particular, if V = 1 tions of amplitude a is \u03c4 = 2 \u03c90 a dx \u221aa2 a x2 1 \u2212 \u2212 \u2212. b(a2 + x2) Using the binomial theorem to expand in powers of b, and the substitution x = a sin \u03b8, show that for small amplitude the period is approximately \u03c4 2\u03c0(1 + 3 4 ba2)/\u03c90. \u2248 48 Classical Mechanics 31. Use the result of the preceding question to obtain an estimate of the correction factor to the period of a simple pendulum when its angular amplitude is 30\u25e6. (Write cos \u03b8 1 32. *Find the Green\u2019s function of an oscillator in the case \u03b3 > \u03c90. Use it to solve the problem of an oscillator that is initially in equilibrium, and is subjected from t = 0 to a force increasing linearly with time, F = ct. 2 \u03b82 + 1 24 \u03b84.) \u2248 \u2212 1 Chapter 3 Energy and Angular Momentum In this chapter, we generalize the discussion of Chapter 2 to the case of motion in two or three dimensions. Throughout this chapter, we shall be concerned with the problem of a particle moving under a known external force F. 3.1 Energy; Conservative Forces The kinetic energy of a particle of mass m free to move in three dimensions is de\ufb01ned to be T = 1 2 m \u02d9r2 = 1 2 m( \u02d9x2 + \u02d9y2 + \u02d9z2). The rate of change of the kinetic energy is, therefore, \u02d9T = m( \u02d9x\u00a8x + \u02d9y\u00a8y + \u02d9z \u00a8z) = m \u02d9r \u00a8r = \u02d9r F, \u00b7 \u00b7 (3.1) (3.2) by the equation of motion (1.1). The change in kinetic energy in a time interval dt during which the particle moves a (vector) distance dr is then dT = dW, (3.3) with dW = F \u00b7 dr = Fxdx + Fydy + Fzdz. (3.4) This is the three-dimensional expression for the work done by the force F in the displacement dr. Note that it is equal to the distance travelled, dr, | | multiplied by the component of F in the direction of the displacement.", " 49 50 Classical Mechanics One might think at \ufb01rst sight that a conservative force in three dimensions should be de\ufb01ned to be a force F (r) depending only on the position r of the particle. However, this is not su\ufb03cient to ensure the existence of a law of conservation of energy, which is the essential feature of a conservative force. We require that T + V = E = constant, (3.5) where T is given by (3.1), and the potential energy V is a function of position, V (r). Now, the rate of change of the potential energy is \u02d9V (r) = \u2202V \u2202x \u02d9x + \u2202V \u2202y \u02d9y + \u2202V \u2202z \u02d9z, or, in terms of the gradient of V, \u2207V = i \u2202V \u2202x + j \u2202V \u2202y + k \u2202V \u2202z, (see (A.21)), by \u00b7 Thus, di\ufb00erentiating (3.5), and using (3.2) and (3.6) for \u02d9T and \u02d9V, we obtain \u02d9V = \u02d9r \u2207V. (3.6) (F + \u2207V ) = 0. \u02d9r \u00b7 Since this must hold for any value of the velocity \u02d9r of the particle, the quantity in parentheses must vanish, i.e. F (r) = \u2207V (r). \u2212 (3.7) This is the analogue for the three-dimensional case of (2.7). In terms of components, it reads Fx = \u2202V \u2202x, \u2212 Fy = \u2202V \u2202y, \u2212 Fz = \u2202V \u2202z. \u2212 For example, suppose that V has the form V = 1 2 kr2 = 1 2 k(x2 + y2 + z2), (3.8) (3.9) Energy and Angular Momentum 51 which describes a three-dimensional harmonic oscillator (discussed in more detail in Chapter 4). Then (3.8) yields F = ( \u2212 kx, ky, \u2212 \u2212 kz) = kr. \u2212 (3.10) It is not hard to \ufb01nd the necessary and su\ufb03cient condition on the force F (r) to ensure the existence of a potential energy function V (r) satisfying (3.7). Any vector function F (r) of the form (3.7) obeys the relation F = 0, \u2207 \u2227 (3.11) that is, its curl vanishes (see (A.26)). For instance, the z component of (3.11) is \u2202Fy \u2202x \u2212 \u2202Fx \u2202y = 0, and this is true because of (3.8) and the symmetry of the second partial derivative: \u22022V \u2202x \u2202y = \u22022V \u2202y \u2202x. Thus (3.11) is a necessary condition for the force F (r) to be conservative. It can be shown (see Appendix A) that it is also a su\ufb03cient condition. More speci\ufb01cally, if the force F (r) satis\ufb01es this condition, then the work done by the force in a displacement from r0 to r is independent of the path chosen between these points. Therefore, we can de\ufb01ne the potential energy V (r) in terms of the force by a relation analogous to (2.5). 3.2 Projectiles As a \ufb01rst example of motion in two dimensions let us consider a projectile moving under gravity. The equation of motion, assuming that atmospheric friction is negligible, is m\u00a8r = mg, (3.12) where g is the acceleration due to gravity, a vector of constant magnitude 2 directed in the downward vertical direction. (We ignore the g = 9.81 m s\u2212 small e\ufb00ects due to the Earth\u2019s rotation, which will be discussed in Chapter 52 Classical Mechanics 5, as well as the variation of g with height \u2014 see Chapter 6.) If we choose gk, so that (3.12) takes the form the z-axis vertically upwards, then g = \u2212 \u00a8x = 0, \u00a8y = 0, \u00a8z = g. \u2212 (3.13) If we choose the x- and y-axes so that initially the particle is moving in the xz-plane, then it will do so always. Thus the motion is e\ufb00ectively two-dimensional. The solutions of (3.13) are x = a + bt, z = c + dt 1 2 g", "t2, \u2212 where a, b, c, d are constants to be \ufb01xed by the initial conditions. Example: Projectile range A projectile is launched from the surface of the Earth (assumed \ufb02at) with velocity v at an angle \u03b1 to the horizontal. What is its range? We may choose the point of launch as the origin, and the time of launch as t = 0. At that instant, \u02d9x = v cos \u03b1, \u02d9z = v sin \u03b1. Thus the solution with the required initial conditions is x = vt cos \u03b1, z = vt sin \u03b1 1 2 gt2. \u2212 (3.14) (The trajectory in the xz-plane is obtained by eliminating t between these equations, and is easily seen to be a parabola. However, the explicit equation is not needed here.) The projectile will hit the ground when z = 0 again, i.e., at the time t = 2(v sin \u03b1)/g. Substituting into the \ufb01rst of Eqs. (3.14) yields for the range x = 2(v2 cos \u03b1 sin \u03b1)/g. (3.15) It is interesting to note that to maximize the range we should choose \u03b1 = \u03c0/4; the maximum value is xmax = v2/g. Finally, let us consider the much less trivial problem of a projectile subject also to a resistive force, such as atmospheric drag. As in Chapter 2, we assume that the retarding force is proportional to the velocity, so that Energy and Angular Momentum 53 the equation of motion becomes m\u00a8r = \u2212 \u03bb \u02d9r + mg. (3.16) The actual dependence of atmospheric drag on velocity is de\ufb01nitely nonlinear, but this equation nevertheless gives a reasonable qualitative picture of the motion. De\ufb01ning \u03b3 = \u03bb/m, we may write (3.16) as \u00a8x = \u03b3 \u02d9x, \u2212 \u00a8z = \u03b3 \u02d9z \u2212 \u2212 g. These equations may be integrated by the use of integrating factors (see Chapter 2, Problem 13). If the initial velocity is v = (u, 0, w), the required solution is x = (1 e\u2212 \u03b3t), z = u \u03b3 \u2212 w \u03b3 + g \u03b32 e\u2212 \u03b3t) (1 \u2212 gt \u03b3. \u2212 Eliminating t between these two equations yields the equation of the trajectory, z = \u03b3w + g \u03b3u x + g \u03b32 ln \u03b3x u 1 \u2212. (3.17) A distinctive feature of this equation is that, even if the launch point is far above the ground, there is a maximum value of x. The projectile can never reach beyond x = u/\u03b3, and as it approaches that value, z. Thus the trajectory ends in a near-vertical drop, when all the forward momentum has been spent. \u2192 \u2212\u221e 3.3 Moments; Angular Momentum The moment about the origin of a force F acting on a particle at position r is de\ufb01ned to be the vector product G = r F. \u2227 (3.18) The components of the vector G are the moments about the x-, y- and z-axes, Gx = yFz \u2212 zFy, Gy = zFx \u2212 xFz, Gz = xFy \u2212 yFx. The direction of the vector G is that of the normal to the plane of r and F. It may be regarded as de\ufb01ning the axis about which the force tends to 54 Classical Mechanics rotate the particle. The magnitude of G is G = rF sin \u03b8 = bF, where \u03b8 is the angle between r and F, and b is the perpendicular distance from the origin to the line of the force. (See Fig. 3.1.) Moments of forces G F r b \u03b8 Fig. 3.1 play a particularly important role in the dynamics of rigid bodies, as we shall see in Chapters 8 and 9. Correspondingly, we de\ufb01ne the vector angular momentum (sometimes called moment of momentum) about the origin of a particle at position r, and moving with momentum p, to be \u2227 Its components, the angular momenta about the x-, y- and z-axes, are \u2227 J = r p = mr \u02d9r. (3.19) Jx = m(y \u02d9z z \u02d9y), \u2212 Jy = m(z \u02d9x x \u02d9z), \u2212 Jz = m(x \u02d9y y \u02d9x). \u2212 The momentum p is often called linear momentum when it is important to emphasize the distinction between p and J. The rate of change of angular momentum,", " obtained by di\ufb00erentiating (3.19) is \u02d9J = m d dt (r \u2227 \u02d9r) = m( \u02d9r \u02d9r + r \u00a8r). \u2227 \u2227 (3.20) (Recall that the usual rule for di\ufb00erentiating a product applies also to vector products, provided that the order of the two factors is preserved.) Now, the \ufb01rst term in (3.20) is zero, because it is the vector product of a vector F = G. Thus we obtain with itself. By (1.1), the second term is simply r \u2227 Energy and Angular Momentum 55 the important result that the rate of change of the angular momentum is equal to the moment of the applied force: \u02d9J = G. (3.21) This should be compared with the equation \u02d9p = F for the rate of change of the linear momentum. Since the de\ufb01nition of the vector product (see Appendix, A.3) depends on the choice of a right-hand screw convention, the directions of the vectors G and J also depend on this convention. A vector of this type is known as an axial vector. It is to be contrasted with an ordinary, or polar, vector. Axial vectors are often associated with rotation about an axis. What is speci\ufb01ed physically is not the direction along the axis, but the sense of rotation about it (see Fig. 3.1). \u00a7 3.4 Central Forces; Conservation of Angular Momentum An external force is said to be central if it is always directed towards or away from a \ufb01xed point, called the centre of force. If we choose the origin to be this centre, this means that F is always parallel to the position vector r. Since the vector product of two parallel vectors is zero, the condition for a force F to be central is that its moment about the centre should vanish: G = r \u2227 F = 0. (3.22) From (3.21) it follows that if the force is central, the angular momentum is a constant: J = constant. (3.23) This is the law of conservation of angular momentum in its simplest form. It really contains two statements: that the direction of J is constant, and that its magnitude is constant. Let us look at these in turn. The direction of J is that of the normal to the plane of r and v = \u02d9r. Hence, the statement that this direction is \ufb01xed implies that r and v must always lie in a \ufb01xed plane. In other words, the motion of the particle is con\ufb01ned to the plane containing the initial position vector and velocity vector (see Fig. 3.2). This is obvious physically; for, since the force is central, it has no component perpendicular to this plane, and, since the normal component of velocity is initially zero, it must always remain zero. 56 Classical Mechanics J r v Fig. 3.2 To understand the meaning of the second part of the law, the constancy of the magnitude J of J, it is convenient to introduce polar co-ordinates r and \u03b8 in the plane of the motion. In a short time interval, in which the co-ordinates change by amounts dr and d\u03b8, the distances travelled in the radial and transverse directions are dr and rd\u03b8, respectively (see Fig. 3.3). Thus the radial and transverse components of the velocity are dr rd\u03b8 d\u03b8 r Fig. 3.3 vr = \u02d9r, v\u03b8 = r \u02d9\u03b8. The magnitude of the angular momentum is then J = mrv\u03b8 = mr2 \u02d9\u03b8. (3.24) (3.25) It is now easy to \ufb01nd a geometrical interpretation of the statement that J is a constant. We note that when the angle \u03b8 changes by an amount d\u03b8, the radius vector sweeps out an area dA = 1 2 r2d\u03b8. Energy and Angular Momentum 57 Thus the rate of sweeping out area is dA dt = 1 2 r2 \u02d9\u03b8 = J 2m = constant. (3.26) One of the most important applications is to planetary motion, and in that context, the constancy of the rate of sweeping out area is generally referred to as Kepler\u2019s second law, although as we have seen it applies more generally to motion under any central force. It applies even if the force is non-conservative \u2014 for example, to a particle attached to the end of a string which is gradually being wound in towards the centre. An equivalent statement of the law, which follows at once from (3.25), is that the transverse component of", " the velocity, v\u03b8, varies inversely with the radial distance r. When the force is both central and conservative, the two conservation laws, (3.5) and (3.23), together give us a great deal of information about the motion of the particle. This important special case is the subject of the next chapter. 3.5 Polar Co-ordinates In problems with particular symmetries, it is often convenient to use nonCartesian co-ordinates. In particular, in the case of axial or spherical symmetry, we may use cylindrical polar co-ordinates \u03c1, \u03d5, z, or spherical polar co-ordinates r, \u03b8, \u03d5. These are related to Cartesian co-ordinates by Fig. 3.4 58 Classical Mechanics x = \u03c1 cos \u03d5 = r sin \u03b8 cos \u03d5, y = \u03c1 sin \u03d5 = r sin \u03b8 sin \u03d5, z = z = r cos \u03b8. (See Fig. 3.4.) The relation between the two is clearly z = r cos \u03b8 \u03c1 = r sin \u03b8, \u03d5 = \u03d5. (3.27) (3.28) It is easy to \ufb01nd the components of velocity in the various co-ordinate directions, analogous to (3.24). In the cylindrical polar case, the co-ordinates \u03c1 and \u03d5 are just plane polar co-ordinates in the xy-plane, and we obviously have v\u03c1 = \u02d9\u03c1, v\u03d5 = \u03c1 \u02d9\u03d5 vz = \u02d9z. (3.29) In the spherical polar case, a curve along which \u03b8 varies, while r and \u03d5 are \ufb01xed, is a circle of radius r (corresponding on the Earth to a meridian of longitude). A curve along which only \u03d5 varies is a circle of radius \u03c1 = r sin \u03b8 (corresponding to a parallel of latitude). Thus the elements of length in the three co-ordinate directions are dr, r d\u03b8 and r sin \u03b8 d\u03d5, and we get vr = \u02d9r, v\u03b8 = r \u02d9\u03b8, v\u03d5 = r sin \u03b8 \u02d9\u03d5. (3.30) For use later in this chapter, it will be very useful to have expressions for the kinetic energy in terms of polar co-ordinates. These co-ordinates, though curvilinear, are still orthogonal in the sense that the three coordinate directions at each point are mutually perpendicular. Thus we can write, for example, v2 = v2 \u03c1 + v2 \u03d5 + v2 z. (This would not be true if the co-ordinates were oblique.) Hence from (3.29) and (3.30) we \ufb01nd and T = 1 2 m( \u02d9\u03c12 + \u03c12 \u02d9\u03d52 + \u02d9z2), T = 1 2 m( \u02d9r2 + r2 \u02d9\u03b82 + r2 sin2 \u03b8 \u02d9\u03d52). (3.31) (3.32) Energy and Angular Momentum 59 Note the characteristic di\ufb00erence between these expressions and the corresponding one for the case of Cartesian co-ordinates: here T depends not only on the time derivatives of the co-ordinates, but also on the co-ordinates themselves. We could now go on and \ufb01nd expressions for the acceleration in terms of these co-ordinates, and hence write down equations of motion. While it is not particularly hard to do this directly in these simple cases, it is very convenient to have a general prescription for writing down the equations of motion in arbitrary co-ordinates. A method of doing this that we shall use extensively later is due to the eighteenth-century mathematician Joseph\u2013 Louis Lagrange. It can be done as soon as we have found the expression for the kinetic energy, like (3.31) or (3.32). The simplest \u2014 though not the only \u2014 way of deriving Lagrange\u2019s equations is to use what is known as a variational principle. This is a principle which states that some quantity has a minimum value, or, more generally, a stationary value. Principles of this kind are used in many branches of physics, particularly in quantum mechanics and in optics. It may therefore be useful, before discussing the particular principle we shall require, to consider the general kind of problem in which such methods are used. This we shall do in the next section. 3.6 The Calculus of Variations Let us begin with what may seem a ridiculously simple example: what is the shortest path between two points in a plane? Of course we know the answer already, but the method we shall use", " to derive it can also be applied to less trivial examples \u2014 for instance, to \ufb01nd the shortest path between two points on a curved surface. Suppose the two points are (x0, y0) and (x1, y1). Any curve joining them is represented by an equation y = y(x) such that the function satis\ufb01es the boundary conditions y(x0) = y0, y(x1) = y1. (3.33) Consider two neighbouring points on this curve. The distance dl between them is given by dl = dx2 + dy2 = 1 + y 2 dx, 60 Classical Mechanics where y = dy/dx. Thus the total length of the curve is x1 l = 1 + y 2 dx. (3.34) The problem, therefore, is to \ufb01nd that function y(x), subject to the boundary conditions (3.33), that will make this integral a minimum. x0 This problem di\ufb00ers from the more familiar kind of minimum-value problem in that what we have to vary is not just a single variable or set of variables, but a function y(x). However, we can still apply the same criterion: when the integral has a minimum value, it must be unchanged to \ufb01rst order by making a small variation in the function y(x). (We shall not be concerned with the problem of distinguishing maxima from minima. All we shall do is to \ufb01nd the stationary values.) More generally, we may be interested in \ufb01nding the stationary values of an integral of the form x1 I = f (y, y) dx, x0 (3.35) where f (y, y) is a speci\ufb01ed function of y and its \ufb01rst derivative. We shall solve this general problem, and then apply the result to the integral (3.34). Consider a small variation \u03b4y(x) in the function y(x), subject to the condition that the values of y at the end-points are unchanged (see Fig. 3.5): y (x, y ) 1 1 \u03b4y(x) y(x) (x, y ) 0 0 x x Fig. 3.5 \u03b4y(x0) = 0, \u03b4y(x1) = 0. (3.36) Energy and Angular Momentum 61 To \ufb01rst order, the variation in f (y, y) is \u03b4f = \u2202f \u2202y \u03b4y + \u2202f \u2202y \u03b4y, where \u03b4y = d(\u03b4y) dx. Thus the variation of the integral I is \u03b4I = x1 x0 \u2202f \u2202y \u03b4y + \u2202f \u2202y d(\u03b4y) dx dx. In the second term, we may integrate by parts. The integrated term, namely \u2202f \u2202y x1 \u03b4y, x0 vanishes because of the conditions (3.36). Hence we obtain \u03b4I = x1 x0 \u2202f \u2202y \u2212 d dx \u2202f \u2202y \u03b4y(x) dx. (3.37) Now, in order that I be stationary, this variation \u03b4I must vanish for an arbitrary small variation \u03b4y(x) (subject only to the boundary conditions (3.36)). This is only possible if the integrand vanishes identically. Thus we require \u2202f \u2202y \u2212 d dx \u2202f \u2202y = 0. (3.38) This is known as the Euler\u2013Lagrange equation. It is in general a secondorder di\ufb00erential equation for the unknown function y(x), whose solution contains two arbitrary constants that may be determined from the boundary conditions. We can now return to solving the problem we started with, of \ufb01nding the curves of shortest length (called geodesics) in a plane, pretending for the moment that we don\u2019t know the answer! Example: Geodesics on a plane What is the shortest path between two given points in a plane? In this case, comparing (3.34) with (3.35), we have to choose f = 1 + y 2 and, therefore, \u2202f \u2202y = 0, \u2202f \u2202y = y 1 + y. 2 62 Classical Mechanics Thus the Euler\u2013Lagrange equation (3.38) reads d dx y 1 + y 2 = 0. This equation states that the expression inside the parentheses is a constant, and hence that y is a constant. Its solutions are therefore the straight lines y = ax + b (a, b = constants). Thus we have proved that the shortest path between two points in a plane is a straight line! The", " constants a and b are of course \ufb01xed by the conditions (3.33). So far, we have used x as the independent variable, but in the applications we consider later we shall be concerned instead with functions of the time t. It is easy to generalize the discussion to the case of a function f of n variables q1, q2,..., qn and their time derivatives \u02d9q1, \u02d9q2,..., \u02d9qn. In order that the integral t1 I = f (q1,..., qn, \u02d9q1,..., \u02d9qn)dt t0 be stationary, it must be unchanged to \ufb01rst order by a variation in any one of the functions qi(t) (i = 1, 2,..., n), subject to the conditions \u03b4qi(t0) = \u03b4qi(t1) = 0. Thus we require the n Euler\u2013Lagrange equations \u2202f \u2202qi \u2212 d dt \u2202f \u2202 \u02d9qi = 0, i = 1, 2,..., n. (3.39) These n second-order partial di\ufb00erential equations determine the n functions qi(t) to within 2n arbitrary constants of integration. 3.7 Hamilton\u2019s Principle; Lagrange\u2019s Equations We shall now show that the equations of motion for a particle moving under a conservative force can be written as the Euler\u2013Lagrange equations corresponding to a suitable integral. We de\ufb01ne the Lagrangian function L = T \u2212 V = 1 2 m( \u02d9x2 + \u02d9y2 + \u02d9z2) \u2212 V (x, y, z). (3.40) Energy and Angular Momentum 63 (Note the minus sign \u2014 this is not the total energy.) Its derivatives are \u2202L \u2202 \u02d9x = m \u02d9x = px, \u2202L \u2202x = \u2202V \u2202x \u2212 = Fx, (3.41) with similar equations for the y and z components. Thus the equation of motion \u02d9px = Fx can be written in the form d dt \u2202L \u2202 \u02d9x = \u2202L \u2202x. (3.42) But this has precisely the form of an Euler\u2013Lagrange equation for the integral t1 I = L dt. t0 (3.43) This is generally known as the action integral. Strictly speaking, this terminology does not accord with the original, in which the term \u2018action\u2019 was used for a related but di\ufb00erent quantity (the integral of twice the kinetic energy), which is relatively little used. However, it is more convenient, and has now become standard practice, to use this simple name for the commonly occurring function (3.43). Thus we have arrived at Hamilton\u2019s principle (named after the nineteenth-century Irish mathematician and physicist Sir William Rowan Hamilton): the action integral I is stationary under arbitrary variations \u03b4x, \u03b4y, \u03b4z which vanish at the limits of integration t0 and t1. The importance of this principle lies in the fact that it can immediately be applied to any set of co-ordinates. If, instead of using Cartesian coordinates x, y, z, we employ a set of curvilinear co-ordinates q1, q2, q3, then V in terms of q1, q2, q3 we can express the Lagrangian function L = T and their time derivatives, \u02d9q1, \u02d9q2, \u02d9q3. The action integral must then be stationary with respect to arbitrary variations \u03b4q1, \u03b4q2, \u03b4q3, subject to the conditions that the variations are zero at the limits t0 and t1. Thus we obtain Lagrange\u2019s equations: \u2212 d dt \u2202L \u2202 \u02d9qi = \u2202L \u2202qi, i = 1, 2, 3. (3.44) These are the equations of motion in terms of the co-ordinates q1, q2, q3. 64 Classical Mechanics By analogy with (3.41), we may de\ufb01ne the generalized momenta and generalized forces, pi = \u2202L \u2202 \u02d9qi and Fi = \u2202L \u2202qi, (3.45) so that the equations (3.44) read \u02d9pi = Fi. (3.46) It must be emphasized, however, that in general the pi and Fi are not components of the momentum vector p and the force vector F \u2014 we shall see this explicitly in the examples below", ". To see how this works in a simple case, let us return to the simple pendulum discussed in 2.1. \u00a7 Example: Simple pendulum revisited Find the equation of motion for a simple pendulum. Using the potential energy function (2.9), we \ufb01nd L = T \u2212 V = 1 2 ml2 \u02d9\u03b82 mgl(1 \u2212 \u2212 cos \u03b8). Thus \u2202L \u2202 \u02d9\u03b8 = ml2 \u02d9\u03b8, \u2202L \u2202\u03b8 = \u2212 mgl sin \u03b8, and Lagrange\u2019s equation yields ml2 \u00a8\u03b8 = mgl sin \u03b8. \u2212 Note that to \ufb01nd this equation of motion, we have not needed expressions for the polar components of acceleration, as we would to derive it directly from Newton\u2019s laws; nor have we needed to introduce the tension in the rod. We can also apply these equations to the polar coordinates discussed 3.5. In the cylindrical polar case, the derivatives of the kinetic energy \u00a7 in Energy and Angular Momentum 65 function (3.31) are \u2202T \u2202 \u02d9\u03c1 \u2202T \u2202\u03c1 = m \u02d9\u03c1, = m\u03c1 \u02d9\u03d52, \u2202T \u2202 \u02d9\u03d5 \u2202T \u2202\u03d5 = m\u03c12 \u02d9\u03d5, = 0, \u2202T \u2202 \u02d9z \u2202T \u2202z = m \u02d9z, = 0. Hence Lagrange\u2019s equations (3.44) take the form d dt (m \u02d9\u03c1) = m\u03c1 \u02d9\u03d52 \u2202V \u2202\u03c1, \u2212, (3.47) d dt (m\u03c12 \u02d9\u03d5) = \u2212 d dt (m \u02d9z) = \u2202V \u2202\u03d5 \u2202V \u2202z. \u2212 The generalized momenta m \u02d9\u03c1 and m \u02d9z corresponding to the co-ordinates \u03c1 and z are the components of the momentum vector p in the \u03c1 and z directions. However, the generalized momentum m\u03c12 \u02d9\u03d5 is an angular momentum; in fact, the angular momentum Jz about the z-axis. It is easy to see that, whenever qi is an angle, the corresponding generalized momentum pi has the dimensions of angular, rather than linear, momentum. Note that when the potential energy is independent of \u03d5, then the second of Eqs. (3.47) expresses the fact that the component Jz of angular momentum is conserved. One must be careful, when using the notation p\u03d5 to specify whether this is the \u03d5 component of the momentum vector p or the \u03d5 generalized momentum, Jz. Proceeding in a similar way, we can write down the equations of motion in spherical polars, using the derivatives of the kinetic energy function (3.32). We \ufb01nd (m \u02d9r) = mr( \u02d9\u03b82 + sin2 \u03b8 \u02d9\u03d52) d dt (mr2 \u02d9\u03b8) = mr2 sin \u03b8 cos \u03b8 \u02d9\u03d52 \u2202V \u2202r, \u2212 \u2202V \u2202\u03b8, \u2212 (3.48) d dt d dt (mr2 sin2 \u03b8 \u02d9\u03d5) = \u2202V \u2202\u03d5. \u2212 Once again, the generalized momentum mr2 sin2 \u03b8 \u02d9\u03d5 is the component of angular momentum Jz about the z-axis, and is a constant if V is independent of \u03d5. The generalized momentum mr2 \u02d9\u03b8 also has the dimensions of angular momentum. Since it is equal to the radial distance r multiplied by the \u03b8 component of momentum, mr \u02d9\u03b8, it is in fact the \u03d5 component of 66 Classical Mechanics angular momentum, J\u03d5. Note, however, that even when V is independent of both angular variables \u03b8 and \u03d5, as is the case for a central force, J\u03d5 is not in general a constant. This is because the \u03d5 direction is not a \ufb01xed direction in space, like the z-axis, but changes according to the position of the particle. Thus, even when J is a \ufb01xed vector, the component J\u03d5 may vary. Before concluding this section, it will be useful to discuss brie\ufb02y the special case of a central conservative force, which is treated in detail in the following chapter. In spherical polars, the components of the force F (not the generalized forces) are given in terms of the potential by Fr = \u2202V \u2202r, \u2212 F\u03b8 = 1 r \u2202V \ufffd", "\ufffd\u03b8, \u2212 F\u03d5 = 1 r sin \u03b8 \u2202V \u2202\u03d5. \u2212 (See (A.50).) Thus the force will be purely radial if, and only if, V is independent of both angular variables \u03b8 and \u03d5. The potential energy function (3.9) of the three-dimensional harmonic oscillator is an example. Then we have F = dV (r) dr. \u02c6r \u2212 (3.49) This shows incidentally that the magnitude of a central, conservative force depends only on r, not on \u03b8 or \u03d5. We know from the discussion of 3.4 that the angular momentum vector J in this case is a constant, and that the motion is con\ufb01ned to a plane. The equations of motion are considerably simpli\ufb01ed by an appropriate choice of the polar axis \u03b8 = 0. If it is chosen so that \u02d9\u03d5 = 0 initially, then by the third of Eqs. (3.48), \u02d9\u03d5 will always be zero. The motion is then in a plane \u03d5 = constant, through the polar axis, in which r and \u03b8 play the role of plane polar co-ordinates. The angular momentum J = J\u03d5 = mr2 \u02d9\u03b8 is then a constant, by the second of Eqs. (3.48). \u00a7 Similarly, if the particle is originally moving in the \u2018equatorial\u2019 plane, 2 with \u02d9\u03b8 = 0, then from the second equation it follows that \u03b8 is always \u03b8 = \u03c0 \u03c0 2. In this case, the plane polar co-ordinates are r and \u03d5, and the angular momentum is J = Jz = mr2 \u02d9\u03d5. 3.8 Summary The rate of change of kinetic energy of a particle is the rate at which the F. The force is conservative if it has the form force does work, \u02d9T = \u02d9r \u00b7 Energy and Angular Momentum 67 \u2227 \u02d9J = r \u2212 \u2207V (r), and in that case the total energy T + V is a constant. The F = condition for the existence of such a function is that F should be a function of position such that \u2207 F = 0. The rate of change of angular momentum of a particle is equal to the F. When the force is central, the angular moment of the force, momentum is conserved. Then the motion is con\ufb01ned to a plane, and the rate of sweeping out area in this plane is a constant. \u2227 The use of these conservation laws greatly simpli\ufb01es the treatment of any problem involving central or conservative forces. When the force is both central and conservative, they provide all the information we need to determine the motion of the particle, as we shall see in the following chapter. Lagrange\u2019s equations are of great importance in advanced treatments of classical mechanics (and also in quantum mechanics). We have seen that they can be used to write down equations of motion in any system of coordinates, as soon as we have found the expressions for the kinetic energy and potential energy. In later chapters, we shall see that the method can readily and usefully be extended to more complicated systems than a single particle. Problems 1. Find which of the following forces are conservative, and for those that are \ufb01nd the corresponding potential energy function (a and b are constants, and a is a constant vector): (a) Fx = ax + by2, Fy = az + 2bxy, Fz = ay + bz2; (b) Fx = ay, Fy = az, Fz = ax; (c) Fr = 2ar sin \u03b8 sin \u03d5, F\u03b8 = ar cos \u03b8 sin \u03d5, F\u03d5 = ar cos \u03d5; (d) F = a \u2227 (e) F = ra; (f) F = a(a r). r; 2. Given that the force is as in Problem 1(a), evaluate the work done in taking a particle from the origin to the point (1, 1, 0): (i) by moving \ufb01rst along the x-axis and then parallel to the y-axis, and (ii) by going in a straight line. Verify that the result in each case is equal to minus the change in the potential energy function. \u00b7 3. Repeat the calculations of Problem 2 for the force in 1(b). 4. Compute the work done in taking a particle around the circle x2 + y2 = 68 Classical Mechanics a2, z = 0 if the force is (a) F = yi, and (b) F = xi. What do you conclude about these forces? (Use the parametrization x = a cos \u03d5, y = a", " sin \u03d5, z = 0.) 5. Evaluate the force corresponding to the potential energy function V (r) = cz/r3, where c is a constant. Write your answer in vector notation, and also in spherical polars, and verify that it satis\ufb01es \u2207 F = 0. 1 at 60\u25e6 to the horizontal. Atmospheric drag is negligible. Find the maximum height attained and the range. What other angle of launch would give the same range? Find the time of \ufb02ight in each of the two cases. 6. A projectile is launched with velocity 100 m s\u2212 \u2227 7. *Find the equation for the trajectory of a projectile launched with velocity v at an angle \u03b1 to the horizontal, assuming negligible atmospheric resistance. Given that the ground slopes at an angle \u03b2, show that the range of the projectile (measured horizontally) is x = 2v2 g sin(\u03b1 \u03b2) cos \u03b1 \u2212 cos \u03b2. At what angle should the projectile be launched to achieve the maximum range? 8. *By expanding the logarithm in (3.17), \ufb01nd the approximate equation for the trajectory of a projectile subject to small atmospheric drag to \ufb01rst order in \u03b3. (Note that this requires terms up to order \u03b3 3 in the logarithm.) Show that to this order the range (on level ground) is x = 2uw g \u2212 8\u03b3uw2 3g2, and hence that to maximize the range for given launch speed v the angle of launch should be chosen to satisfy cos 2\u03b1 = \u221a2\u03b3v/3g. (Hint : In the term containing \u03b3, you may use the zeroth-order approximation for the angle.) For a projectile whose terminal speed if dropped from rest (see 1, estimate the optimal angle Chapter 2, Problem 13) would be 500 m s\u2212 and the range if the launch speed is 100 m s\u2212 1. 9. *Show that in the limit of strong damping (large \u03b3) the time of \ufb02ight 1 (w/g + 1/\u03b3)(1 \u03b3w/g). Show that to the same order of accuracy the range is x \u03b3w/g). For a projectile launched at 800 m s\u2212 of a projectile (on level ground) is approximately t e\u2212 \u2212 (u/\u03b3)(1 0.1 s\u2212 1, estimate the range for launch angles of 30\u25e6, 20\u25e6 and 10\u25e6. \u2212 \u2248 1 with \u03b3 = e\u2212 \u2212 \u2248 \u2212 1 10. A particle of mass m is attached to the end of a light string of length l. The other end of the string is passed through a small hole and is Energy and Angular Momentum 69 slowly pulled through it. Gravity is negligible. The particle is originally spinning round the hole with angular velocity \u03c9. Find the angular velocity when the string length has been reduced to 1 2 l. Find also the tension in the string when its length is r, and verify that the increase in kinetic energy is equal to the work done by the force pulling the string through the hole. 11. A particle of mass m is attached to the end of a light spring of equilibrium length a, whose other end is \ufb01xed, so that the spring is free to rotate in a horizontal plane. The tension in the spring is k times its extension. Initially the system is at rest and the particle is given an impulse that starts it moving at right angles to the spring with velocity v. Write down the equations of motion in polar co-ordinates. Given that the maximum radial distance attained is 2a, use the energy and angular momentum conservation laws to determine the velocity at that point, and to \ufb01nd v in terms of the various parameters of the system. Find also the values of \u00a8r when r = a and when r = 2a. 12. *A light rigid cylinder of radius 2a is able to rotate freely about its axis, which is horizontal. A particle of mass m is \ufb01xed to the cylinder at a distance a from the axis and is initially at rest at its lowest point. A light string is wound on the cylinder, and a steady tension F applied to it. Find the angular acceleration and angular velocity of the cylinder after it has turned through an angle \u03b8. Show that there is a limiting tension F0 such that if F < F0 the motion is oscillatory, but if F > F0 it continues to accelerate. Estimate the value of F0 by numerical approximation. 13. In the system of Problem 12, instead of a \ufb01xed tension applied to the string, a weight of mass m/2 is hung on it. Use the energy conservation equation to \ufb01nd the angular velocity as a function of \u03b8. Find also the (Compare your angular acceleration and the tension in the string. results with those in Problem", " 12.) Show that there is a point at which the tension falls to zero, and \ufb01nd the angle at which this occurs. What happens immediately beyond this point? 14. A wedge-shaped block of mass M rests on a smooth horizontal table. A small block of mass m is placed on its upper face, which is also smooth and inclined at an angle \u03b1 to the horizontal. The system is released from rest. Write down the horizontal component of momentum, and the kinetic energy of the system, in terms of the velocity v of the wedge and the velocity u of the small block relative to it. Using conservation of momentum and the equation for the rate of change of kinetic energy, 70 Classical Mechanics \ufb01nd the accelerations of the blocks. Given that M = 1 kg and m = 250 g, \ufb01nd the angle \u03b1 that will maximize the acceleration of the wedge. 15. *A particle starts from rest and slides down a smooth curve under gravity. Find the shape of the curve that will minimize the time taken [Take the origin as the starting point and between two given points. the z axis downwards. Show that the time taken is z1 0 1 + (dx/dz)2 2gz 1/2 dz, and hence that for a minimum dx dz 2 = c2z c2z 1 \u2212, where c is an integration constant. To complete the integration, use 2 sin2 \u03b8. This famous curve is known as the the substitution z = c\u2212 brachistochrone. It is in fact an example of a cycloid, the locus of a point on the rim of a circle of radius 1/2c2 being rolled beneath the x-axis. This problem was \ufb01rst posed by Johann Bernoulli on New Year\u2019s Day 1697 as an open challenge. Newton\u2019s brilliant solution method initiated the calculus of variations. Bernoulli had an equally brilliant idea, using an optical analogy with refraction of a light ray through a sequence of plates and Fermat\u2019s principle of least time.] 16. *The position on the surface of a cone of semi-vertical angle \u03b1 is speci\ufb01ed by the distance r from the vertex and the azimuth angle \u03d5 about the axis. Show that the shortest path (or geodesic) along the surface between two given points is speci\ufb01ed by a function r(\u03d5) obeying the equation r d2r d\u03d52 \u2212 2 dr d\u03d5 2 \u2212 r2 sin2 \u03b1 = 0. Show that the solution is r = r0 sec[(\u03d5 \u03d50) sin \u03b1], where r0 and \u03d50 are constants. [The equation may be solved by the standard technique of introducing a new dependent variable u = dr/d\u03d5, and writing \u2212 d2r d\u03d52 = du d\u03d5 = dr d\u03d5 du dr = u du dr. The substitution u2 = v then reduces the equation to a linear form that may be solved by using an integrating factor (see Chapter 2, Problem Energy and Angular Momentum 71 13). Finally, to calculate \u03d5 = r = 1/x.] (1/u)dr, one may use the substitution 17. *Find the geodesics on a sphere of unit radius. [Hint : Use \u03b8 as independent variable, and look for the path \u03d5 = \u03d5(\u03b8). To perform the integration, use the substitution x = cot \u03b8.] 18. *Parabolic co-ordinates (\u03be, \u03b7) in a plane are de\ufb01ned by \u03be = r + x, \u03b7 = x. Find x and y in terms of \u03be and \u03b7. Show that the kinetic energy r of a particle of mass m is \u2212 T = m 8 (\u03be + \u03b7) \u02d9\u03be2 \u03be + \u02d9\u03b72 \u03b7. Hence \ufb01nd the equations of motion. 19. Write down the equations of motion in polar co-ordinates for a particle of unit mass moving in a plane under a force with potential energy function V = k ln r + cr + gr cos \u03b8, where k, c and g are positive constants. Find the positions of equilibrium (a) if c > g, and (b) if c < g. By considering the equations of motion near these points, determine whether the equilibrium is stable (i.e., will the particle, if given a small displacement, tend to return repeatedly?). \u2212 20. If q1, q2, q3 are orthogonal curvilinear co-ordinates, and the element of length in the qi direction is hidqi, write down (a) the kinetic energy T in terms of the generalized velocities \u02d9qi, (b) the generalized momentum p of the momentum vector p in", " the qi pi and (c) the component ei \u00b7 direction. (Here ei is a unit vector in the direction of increasing qi.) 21. *By comparing the Euler\u2013Lagrange equations with the corresponding \u2207V, show that the components of the equation of motion m\u00a8r = component of the acceleration vector in the qi direction is \u2212 \u00a8r = ei \u00b7 1 mhi d dt \u2202T \u2202 \u02d9qi \u2212 \u2202T \u2202qi. Use this result to identify the components of the acceleration in cylindrical and spherical polars. 22. For the case of plane polar co-ordinates r, \u03b8, write the unit vectors er (= \u02c6r) and e\u03b8 in terms of i and j. Hence show that \u2202er/\u2202\u03b8 = e\u03b8 and er. By starting with r = rer and di\ufb00erentiating, rederive \u2202e\u03b8/\u2202\u03b8 = the expressions for the components of the velocity and acceleration vectors. \u2212 23. *Find the corresponding formulae for \u2202ei/\u2202qj for spherical polar co- ordinates, and hence verify the results obtained in Problem 21. 72 Classical Mechanics 24. *The motion of a particle in a plane may be described in terms of elliptic co-ordinates \u03bb, \u03b8 de\ufb01ned by x = c cosh \u03bb cos \u03b8, y = c sinh \u03bb sin \u03b8, (\u03bb 0, 0 \u03b8 \u2264 \u2264 \u2265 2\u03c0), where c is a positive constant. Show that the kinetic energy function may be written T = 1 2 mc2(cosh2 \u03bb \u2212 cos2 \u03b8)( \u02d9\u03bb2 + \u02d9\u03b82). Hence write down the equations of motion. 25. *The method of Lagrange multipliers (see Appendix A, Problem 11) can be extended to the calculus of variations. To \ufb01nd the maxima and minima of an integral I, subject to the condition that another integral \u03bbJ J = 0, we have to \ufb01nd the stationary points of the integral I under variations of the function y(x) and of the parameter \u03bb. Apply this method to \ufb01nd the catenary, the shape in which a uniform heavy chain hangs between two \ufb01xed supports. [The required shape is the one that minimizes the total potential energy, subject to the condition that the total length is \ufb01xed. Show that this leads to a variational problem with \u2212 and hence to the equation f (y, y) = (y \u03bb) 1 + y 2, \u2212 \u03bb)y = 1 + y 2. (y \u2212 Solve this equation by introducing the new variable u = y and solving for u(y).] 26. *A curve of given total length is drawn in a plane between the points a, 0). Using the method of Problem 25, \ufb01nd the shape that will ( \u00b1 enclose the largest possible area between the curve and the x-axis. Chapter 4 Central Conservative Forces The most important forces of this type are the gravitational and electrostatic forces, which obey the \u2018inverse square law\u2019. Much of this chapter will therefore be devoted to this special case. We begin, however, by discussing a problem for which it is particularly easy to solve the equations of motion \u2014 the three-dimensional analogue of the harmonic oscillator discussed in 2.2. \u00a7 4.1 The Isotropic Harmonic Oscillator We consider a particle moving under the action of a central restoring force proportional to its distance from the origin, with k constant. The corresponding equation of motion is F = kr, \u2212 (4.1) m\u00a8r + kr = 0, or, in terms of components, m\u00a8x + kx = 0, m\u00a8y + ky = 0, m\u00a8z + kz = 0. The equation of motion for each co-ordinate is thus identical with the equation (2.13) for the simple harmonic oscillator. This oscillator is called isotropic because all directions (not only those along the axes!) are equivalent. The anisotropic oscillator (which we shall not discuss) is described by similar equations, but with di\ufb00erent constants in the three equations. 73 74 Classical Mechanics The general solution is again given by (2.19): x = cx cos \u03c9t + dx sin \u03c9t, y = cy cos \u03c9t + dy sin \u03c9t, z = cz cos \u03c9t + dz sin \u03c9t, where \u03c92 = k/m. In vector notation, the solution is r = c cos \u03c9t + d sin \u03c9t. (4.2) Clearly, the motion is periodic, with period \u03c4 = 2\u03c0/", "\u03c9. (Other forms of the solution can be given, analogous to (2.20) or (2.23).) As in the one-dimensional case, the arbitrary constant vectors c and d can be \ufb01xed by the initial conditions. If at t = 0 the particle is at r0 and moving with velocity v0, then c = r0, d = v0/\u03c9. (4.3) We can easily verify the conservation laws for angular momentum and energy. From (4.2) it is clear that r always lies in the plane of c and d, so that the direction of J is \ufb01xed. Using \u02d9r = \u2212 \u03c9c sin \u03c9t + \u03c9d cos \u03c9t, we \ufb01nd explicitly J = mr \u02d9r = m\u03c9c \u2227 which is obviously a constant. d = mr0 \u2227 \u2227 v0, (4.4) The potential energy function corresponding to the force (4.1) \u2014 or (3.10) \u2014 is just (3.9), namely V = 1 2 kr2 = 1 2 k(x2 + y2 + z2). Thus, evaluating the energy, we \ufb01nd E = 1 2 m \u02d9r2 + 1 2 kr2 = 1 2 k(c2 + d2) = 1 2 mv2 0 + 1 2 kr2 0, (4.5) (4.6) which is again a constant. To \ufb01nd the shape of the particle orbit, it is convenient to rewrite (4.2) in a slightly di\ufb00erent, but equivalent, form. If \u03b8 is any \ufb01xed angle, we can write r = a cos(\u03c9t \u03b8) + b sin(\u03c9t \u03b8), \u2212 \u2212 (4.7) Central Conservative Forces 75 c = a cos \u03b8 b sin \u03b8, d = a sin \u03b8 + b cos \u03b8. \u2212 (4.8) where or, equivalently, a = c cos \u03b8 + d sin \u03b8, b = \u2212 c sin \u03b8 + d cos \u03b8. We can now choose \u03b8 so that a and b are perpendicular. This requires 0 = a b = \u00b7 (c2 \u2212 \u2212 d2) sin \u03b8 cos \u03b8 + c d(cos2 \u03b8 \u00b7 \u2212 sin2 \u03b8), or tan 2\u03b8 = d d2. 2c c2 \u00b7 \u2212 We are now free to choose our axes so that the x-axis is in the direction of a and the y-axis in the direction of b. Equation (4.7) then becomes x = a cos(\u03c9t \u03b8), \u2212 y = b sin(\u03c9t \u03b8), \u2212 z = 0. The equation of the orbit is obtained by eliminating the time from these equations. It is x2 a2 + y2 b2 = 1, z = 0. (4.9) This is the well-known equation of an ellipse with centre at the origin, and semi-axes a and b. (See Appendix B, Eq. (B.2), and Fig. 4.1.) The vectors c and d are what is known in geometry as a pair of conjugate semi-diameters of the ellipse. b O d c a Fig. 4.1 v 0 \u03b1 76 Classical Mechanics The simplest way to determine the magnitudes of the semi-axes a and b from the initial conditions (that is, from c and d) is to use the constancy of E and J. Equating the expressions for these quantities, (4.6) and (4.4), to the corresponding expressions in terms of a and b, we obtain a2 + b2 = c2 + d2, b = c d. a \u2227 \u2227 If \u03b1 is the (These equations may also be obtained directly from (4.8).) angle between c and d, the second equation yields ab = cd sin \u03b1. Hence eliminating b (or a) between these two equations, we obtain for a2 (or b2) the quadratic equation a4 \u2212 (c2 + d2)a2 + c2d2 sin2 \u03b1 = 0. (4.10) Conventionally, a is taken to be the larger root, and called the semi-major axis. 4.2 The Conservation Laws We now consider the general case of a central, conservative force. It corresponds, as we saw in 3.7, to a potential energy function V (r) depending only on the radial co-ordinate r. There are two conservation laws, one for energy, \u00a7 1 2 m \u02d9r2 + V (r) = E = constant, and one for angular momentum mr \u2227 \u02d9r = J", " = constant. According to the discussion of 3.4, the second of these laws implies that the motion is con\ufb01ned to a plane, so that the problem is e\ufb00ectively twodimensional. Introducing polar co-ordinates r, \u03b8 in this plane, we may write the two conservation laws in the form \u00a7 1 2 m( \u02d9r2 + r2 \u02d9\u03b82) + V (r) = E, mr2 \u02d9\u03b8 = J. (4.11) A great deal of information about the motion can be obtained directly from these equations, without actually having to solve them to \ufb01nd r and Central Conservative Forces 77 \u03b8 as functions of the time. We note that \u02d9\u03b8 may be eliminated to yield an equation involving only r and \u02d9r: 1 2 m \u02d9r2 + J 2 2mr2 + V (r) = E. (4.12) We shall call this the radial energy equation. For a given value of J, it has precisely the same form as the one-dimensional energy equation with an e\ufb00ective potential energy function U (r) = J 2 2mr2 + V (r). (4.13) It is easy to understand the physical signi\ufb01cance of the extra term J 2/2mr2 in this \u2018potential energy\u2019 function. It corresponds to a \u2018force\u2019 J 2/mr3. This is precisely the \u2018centrifugal force\u2019 mr \u02d9\u03b82 (the \ufb01rst term on the right hand side of the radial equation in (3.48)), expressed in terms of the constant J rather than the variable \u02d9\u03b8. We can use the radial energy equation just as we did in is positive, the motion is limited to the range of values of r for which 2.1. Since \u02d9r2 \u00a7 J 2 2mr2 + V (r) The maximum and minimum radial distances are given by the values of r for which the equality holds. U (r) = (4.14) E. \u2264 U E b a r Fig. 4.2 As an example, let us consider again the case of the isotropic harmonic oscillator, for which V (r) = 1 2 kr2. The corresponding e\ufb00ective potential energy function U (r) is shown in Fig. 4.2. It has a minimum (corresponding 78 Classical Mechanics to a position of stable equilibrium in the one-dimensional case) at r = J 2 mk 1/4. (4.15) When the value of E is equal to the minimum value of U, \u02d9r must always be zero, and r is \ufb01xed at the minimum position. In this case, the particle must move in a circle around the origin. It is interesting to note that we could also obtain (4.15) by equating the attractive force kr to the centrifugal force in the circular orbit, J 2/mr3. For any larger value of E, the motion is con\ufb01ned to the region b a between the two limiting values of r, given by the solutions of the equality in (4.14). If the particle is initially at a distance r0 from the origin, and moving with velocity v0 in a direction making an angle \u03b1 with the radius vector (as in Fig. 4.1), then the values of E and J are \u2264 \u2264 r E = 1 2 mv2 0 + 1 2 kr2 0, J = mr0v0 sin \u03b1. Thus the equation whose roots are a and b becomes (on multiplying by 2r2/k) r4 \u2212 m k r2 0 + v2 0 r2 + 0v2 r2 0 sin2 \u03b1 = 0. m k By (4.3), this is identical with the equation (4.10) found previously. 4.3 The Inverse Square Law We now consider a force F = k r2 \u02c6r, where k is a constant. The corresponding potential energy function is V (r) = k/r. The constant k may be either positive or negative; in the \ufb01rst case, the force is repulsive, and in the second, attractive. The radial energy equation for this case is 1 2 m \u02d9r2 + J 2 2mr2 + k r = E. (4.16) It corresponds to the \u2018e\ufb00ective potential energy\u2019 function U (r) = J 2 2mr2 + k r. Central Conservative Forces 79 Repulsive case We suppose \ufb01rst that k > 0. Then U (r) decreases monotonically from + at r = 0 to 0 at r = \u221e (see Fig.", " 4.3). Thus it has no minima, and circular \u221e U E r 1 r Fig. 4.3 motion is impossible, as is physically obvious. For any positive value of E, there is a minimum value of r, r1 say, which is the unique positive root of U (r) = E, but no maximum value. If the radial velocity is initially inward, the particle must follow an orbit on which r decreases to r1 (at which point the velocity is purely transverse), and then increases again without limit. As is well known, and as we shall show in the next section, the orbit is actually a hyperbola. Let us consider an example. Example: Distance of closest approach A charged particle of charge q moves in the \ufb01eld of a \ufb01xed point charge q, with qq > 0. Initially it is approaching the centre of force with velocity v (at a large distance) along a path which, if continued in a straight line, would pass the centre at a distance b. Find the distance of closest approach. This distance b is known as the impact parameter, and will appear frequently in our future work. (See Fig. 4.4.) Since the particle starts from a great distance, its initial potential energy is negligible. (Note that we have chosen the arbitrary additive constant in V so that V ( ) = 0.) Thus the total energy is simply \u221e E = 1 2 mv2. (4.17) 80 Classical Mechanics m v \u0398 b r 1 Fig. 4.4 Moreover, since the component of r perpendicular to v is b, the angular momentum is J = mvb. (4.18) The distance of closest approach, r1, is obtained by substituting these values, and k = qq/4\u03c00, into the radial energy equation, and then setting \u02d9r = 0. This yields r2 1 \u2212 2ar1 \u2212 b2 = 0, where a = qq 4\u03c00mv2. The required solution is the positive root r1 = a + a2 + b2. (4.19) (4.20) Attractive case We now suppose that k < 0, as in the gravitational case. It will be useful to de\ufb01ne a quantity l, with the dimensions of length, by 4.21) Then the e\ufb00ective potential energy function is U (r) = k | | l 2r2 \u2212 1 r. Central Conservative Forces 81 It is plotted in Fig. 4.5. Evidently, U ( 1 2 l) = 0, and U (r) has a minimum at /2l. k r = l, with minimum value U (l) = | \u2212| U E\u2032 E |k| 2l Fig. 4.5 Here, di\ufb00erent types of motion are possible according to the value of E. We may distinguish four cases: (a) E = k | | 2l. \u2212 This is the minimum value of U. Hence, \u02d9r must always be zero, and the particle must move in a circle of radius l. It is easy to \ufb01nd the orbital V. Since the potential energy velocity v from the kinetic energy T = E /l, we must have is V = | \u2212| \u2212 k T = | k | 2l, whence v = k | | ml. (4.22) This can also be found by equating the attractive force trifugal force\u2019 mv2/l. k | /l2 to the \u2018cen| Note the interesting result that for a circular orbit the potential energy is always twice as large in magnitude as the kinetic energy (see Problem 21). (b) k | | 2l \u2212 < E < 0. This case is illustrated by the lower line (E) in Fig. 4.5. The radial distance is limited between a minimum distance r1 and a maximum distance r2. As we shall see, the orbit is in fact an ellipse. 82 Classical Mechanics (c) E = 0. In this case, there is a minimum distance, r1 = 1 2 l, but the maximum distance r2 is in\ufb01nite. Thus the particle has just enough energy to escape to in\ufb01nity, with kinetic energy tending to zero at large distances. The orbit will be shown to be a parabola. (d) E > 0. This corresponds to the upper line (E) in Fig. 4.5. Again, there is a minimum distance but no maximum distance. Now, however, the particle can escape to in\ufb01nity with non-zero limiting velocity. The orbit is in fact a hyperbola. As a simple example of the use of these equations, we compute the escape velocity from the Earth. Example: Escape velocity What is the minimum velocity with which a projectile", " launched from the surface of the Earth (taken to be a sphere of mass M and radius R) can escape, and how does this depend on the angle of launch? Suppose the projectile is launched with velocity v at an angle \u03b1 to the vertical. The energy and angular momentum are E = 1 2 mv2 GM m R, \u2212 J = mRv sin \u03b1, (4.23) where G is Newton\u2019s gravitational constant. To express the energy in terms of more familiar quantities, we note that the gravitational force on a particle at the Earth\u2019s surface is mg = GM m R2, from which we obtain the useful result GM = R2g. (4.24) Thus E = 1 that E 2 mv2 Rgm. The projectile will escape to in\ufb01nity provided 0; or equivalently that its velocity exceeds the escape velocity, \u2212 \u2265 ve = 2Rg. Central Conservative Forces 83 Note that this condition is independent of the angle of launch \u03b1 (so long as we neglect atmospheric drag). Using the values R = 6370 km, g = 9.81 m s\u2212 2, we \ufb01nd for the escape velocity from the Earth ve = 11.2 km s\u2212 1. If the projectile is launched with a speed less than escape velocity, it will reach a maximum height and then fall back. (This is a generalization 3.2.) The problem of \ufb01nding this distance of the problem discussed in provides another useful example. \u00a7 Example: Maximum height of projectile If the launch speed v is equal to the circular orbit speed vc in an orbit just above the Earth\u2019s surface, how high will the projectile reach for a given angle of launch \u03b1? The maximum distance r2 from the centre is the larger root of the equation U (r) = E, or, with the values (4.23) for E and J, (2Rg \u2212 v2)r2 \u2212 2R2gr + R2v2 sin2 \u03b1 = 0. The circular orbit velocity, by (4.22) and (4.24), is vc = Rg = 7.9 km s\u2212 1. Thus the equation reduces to r2 2Rr + R2 sin2 \u03b1 = 0, so that \u2212 r2 = R(1 + cos \u03b1). For a vertical launch, the maximum distance is 2R. For any other angle, it is less, and in the limit of an almost horizontal launch, the orbit is nearly a circle of radius R. This example illustrates the kind of problem which may readily be solved by using the radial energy equation. It is particularly useful when we are interested only in r, and not in the polar angle \u03b8 or the time t. Energy levels of the hydrogen atom This is not a problem that can be solved by the methods of classical mechanics alone. However, the energy levels derived from quantum mechanics can 84 Classical Mechanics also be obtained by imposing on the classical orbits an ad hoc \u2018quantization rule\u2019. Indeed, historically, this is how they were \ufb01rst obtained. According to Bohr\u2019s \u2018old quantum theory\u2019, the electron in an atom cannot occupy any orbit, but only a certain discrete set of orbits. In the case of circular orbits, the quantization rule he imposed was that the angular momentum J be an integer multiple of (Planck\u2019s constant divided by 2\u03c0). The constant k in e2/4\u03c00, where e is the electric charge on the proton, or this case is k = minus the charge on the electron. Thus for J = n the radius an of the orbit is, by (4.21), \u2212 an = 4\u03c00J 2 me2 = n2a1, where a1, the radius of the \ufb01rst Bohr orbit (usually denoted by a0), is a1 = 4\u03c002 me2 = 5.29 \u00d7 10\u2212 11 m. The corresponding energy levels are En = e2 8\u03c00an \u2212 = 1 n2 e2 8\u03c00a1. \u2212 These values agree well with the energies of atomic transitions, as determined from the spectrum of hydrogen. 4.4 Orbits We now turn to the problem of determining the orbit of a particle moving under a central conservative force. This can be done by eliminating the time from the two conservation equations, (4.11), to obtain an equation relating r and \u03b8. The simplest way of doing this is to work not with r itself, but with the variable u = 1/r, and look for an equation determining u as a function of \u03b8. Now whence du d\u03b8 = 1 r2 dr d\u03b8, \u2212 \u02d9r = dr d\u03b8 \u02d9\u03b8 = r2 \u02d9\u03b8 \u2212 du d\u03b8 = J m du d\u03b8. \u2212 Central Conservative Forces 85 Thus substituting for", " \u02d9r in the radial energy equation (4.12), we obtain J 2 2m du d\u03b8 2 + J 2 2m u2 + V = E, (4.25) in which of course V is to be regarded as a function of 1/u. This equation can be integrated (numerically, if not analytically) to give the equation of the orbit. We shall consider explicitly only the case of the inverse square law, for which V = ku. We shall treat both cases k > 0 and k < 0 together, by writing V = u. It will be useful, as in the preceding section, to de\ufb01ne | the length parameter \u00b1| k Then, multiplying (4.25) by 2/ l = J 2 k m |, we obtain | | k | 2 du d\u03b8 l + lu2 2u = \u00b1 (4.26), 2E k | | where the upper sign refers to the repulsive case, k > 0, and the lower to the attractive case, k < 0. To solve this equation, we multiply both sides by l and add 1 to \u2018com- plete the square\u2019. We now introduce the new variable z = lu 1, \u00b1 so that dz d\u03b8 = l du d\u03b8. Then we may write the equation as 2 dz d\u03b8 + z2 = 2El k | | + 1 = e2, (4.27) say, de\ufb01ning a new dimensionless constant e. Note that since the left hand side is a sum of squares, it can have a solution only when the the right hand side is positive, in agreement with our earlier result that the minimum value of E is k /2l. | \u2212| The general solution of this equation is z = lu \u00b1 1 = e cos(\u03b8 \u03b80), \u2212 where \u03b80 is an arbitrary constant of integration. Thus \ufb01nally, replacing u by 1/r and multiplying by r, we \ufb01nd that the orbit equation is, in the 86 Classical Mechanics repulsive case r[e cos(\u03b8 \u03b80) \u2212 \u2212 1] = l, (k > 0) (4.28) and in the attractive case r[e cos(\u03b8 \u2212 \u03b80) + 1] = l. (k < 0) (4.29) These are the polar equations of conic sections, referred to a focus as origin (see Appendix B). The constant e, the eccentricity, determines the shape of the orbit; l, called the semi-latus rectum, determines its scale; and \u03b80 its orientation relative to the coordinate axes. In the repulsive case, e must be greater than unity, since otherwise the square bracket in (4.28) is always negative. Therefore, by (4.27), E must be positive. In the attractive case, e = 0 when E has its minimum value, /2l; | the orbit is then the circle r = l. So long as e < 1, or E < 0, the square bracket in (4.29) is always positive, and there is a value of r for every \u03b8. Thus the orbit is closed. When e 0, r can become in\ufb01nite \u2265 when the square bracket vanishes. This is in agreement with our previous conclusion that the particle will escape to in\ufb01nity if and only if its total energy is positive. 1, or E \u2212| \u2265 k Note that r takes its minimum value when \u03b8 = \u03b80. Thus \u03b80 speci\ufb01es the direction of the point of closest approach. In the attractive case, the constant l also has a simple geometrical meaning. It is the radial distance at right angles to this direction; that is, r = l when \u03b8 = \u03b80 \u00b1 1 2 \u03c0. Elliptic orbits (E < 0, 0 \u2264 e < 1) For most applications, it is better to use the orbit equation directly in polar form. It is, however, straightforward to rewrite it in the possibly more familiar Cartesian form (see Appendix B). If we choose the axes so that \u03b80 = 0, we obtain after some algebra (x + ae)2 a2 + y2 b2 = 1, where a = l 1 \u2212 k e2 = | | E 2 | | and b2 = al = J 2 E 2m | | (4.30) Central Conservative Forces 87 This is the equation of an ellipse with centre at ( ae, 0), and semi-axes a and b. (See Fig. 4.6) It is useful to note that the semi-major axis a is \u2212 b l a ae Fig. 4.6 determined by the value of the energy E, while the semi-", "latus rectum l is \ufb01xed by the angular momentum J. The time taken by the particle to traverse any part of its orbit may be found from the relation (3.26) between angular momentum and rate of sweeping out area, dA dt = J 2m. All we have to do is to evaluate the area swept out by the radius vector, and multiply by 2m/J. In particular, since the area of the ellipse is A = \u03c0ab, the orbital period is \u03c4 = 2m\u03c0ab/J. Thus by (4.26) and (4.30), and thus \u03c4 2\u03c0 2 = m2a2b2 J 2 = m2a3l l k m | |, \u03c4 2\u03c0 2 = a3. m k | | (4.31) (It is easy to verify directly from (4.22) that this gives the correct period for a circular orbit of radius a.) For a planet or satellite orbiting a central body of mass M, \u03c4 2\u03c0 2 = a3 GM. (4.32) This yields Kepler\u2019s third law of planetary motion: the square of the orbital period is proportional to the cube of the semi-major axis. 88 Classical Mechanics Example: Time spent in the two halves of Earth\u2019s orbit If the Earth\u2019s orbit is divided in two by its minor axis, how much longer does it spend in one half than the other? The eccentricity of Earth\u2019s orbit is e = 0.0167. It is clear from Fig. 4.6 that the areas swept out by the position vector of the Earth in the two halves are equal to half the area of the ellipse the area of a triangle with base 2b and height ae, namely 1 aeb. Thus the times taken are ( 1 e/\u03c0) years. The di\ufb00erence between the two is 2e/\u03c0 years, or 3.88 days. 2 \u03c0ab 2 \u00b1 \u00b1 \u00b1 Hyperbolic orbits (E > 0, e > 1) For both the repulsive and attractive cases, the Cartesian equation of the orbit is (x ae)2 \u2212 a2 y2 b2 = 1, \u2212 where a = e2 l \u2212 = | k | 2E 1 and b2 = al = J 2 2mE. (4.33) This is the equation of a hyperbola with centre at (+ae, 0) and semi-axes a and b. (See Fig. 4.7.) One branch of the hyperbola (on the left in the \ufb01gure) corresponds to the orbit in the attractive case and the other to the orbit in the repulsive case. As before, a is determined by the energy E, and l by ae \u0398 a l b Fig. 4.7 Central Conservative Forces 89 the angular momentum J. Note that by (4.17) and (4.18), the semi-minor axis b is identical with the impact parameter introduced earlier. arccos(1/e), and, in the attractive case, \u03b8 = The directions in which r becomes in\ufb01nite are, in the repulsive case, \u03b8 = arccos(1/e)]. In both cases, therefore, the angle through which the particle is de\ufb02ected from its original direction of motion is [\u03c0 \u00b1 \u00b1 \u2212 \u0398 = \u03c0 \u2212 2 arccos(1/e). (4.34) This angle \u0398 is called the scattering angle. For later use, it will be helpful to \ufb01nd the relation between the impact parameter b, the scattering angle \u0398, and the limiting velocity v at in\ufb01nity. From (4.34), we have e = sec 1 \u0398). Substituting in (4.33), this gives 2 (\u03c0 \u2212 b2 = a2(e2 1) = a2[sec2 1 2 (\u03c0 \u0398) \u2212 \u2212 1] = a2 cot2 1 2 \u0398. \u2212 But also from (4.33), using E = 1 obtain 2 mv2, we have a = k | /mv2. Thus we | b = | k mv2 cot 1 | 2 \u0398. (4.35) Let us complete this section by considering a rather special orbit. Example: Orbit of a comet The minimum distance of a comet from the Sun is half the radius of the Earth\u2019s orbit (assumed circular), and its velocity at that point is twice the orbital velocity of the Earth. Find its velocity when it crosses the Earth\u2019s orbit, and the angle at which the orbits cross. Will the comet subsequently escape from the solar system? What kind of orbit does it follow? At perihelion, r = aE/2 and v = 2vc,E. Hence", " the energy is E = 2mv2 c,E \u2212 2GMSm aE = 0, using (4.22). It follows that the comet\u2019s velocity v when it crosses the Earth\u2019s orbit is given by 1 2 mv2 = GMSm aE = mv2 c,E, 90 Classical Mechanics whence v = \u221a2vc,E. The angular momentum is J = maEvc,E, whence by (4.26), l = aE. Thus the angle \u03b1 at which the orbits cross is given by maEv cos \u03b1 = J = maEvc,E, which yields \u03b1 = 45\u25e6. Since E = 0, the comet has just enough energy to escape from the solar system. Its orbit is a parabola. 4.5 Scattering Cross-sections One of the most important ways of obtaining information about the structure of small bodies (for example atomic nuclei) is to bombard them with particles and measure the number of particles scattered in various directions. The angular distribution of scattered particles will depend on the shape of the target and on the nature of the forces between the particles and the target. To be able to interpret the results of such an experiment, we must know how to calculate the expected angular distribution when the forces are known. We shall consider \ufb01rst a particularly simple case. We suppose that the target is a \ufb01xed, hard (that is, perfectly elastic) sphere of radius R, and that a uniform, parallel beam of particles impinges on it. Let the particle \ufb02ux in the beam, that is, the number of particles crossing unit area normal to the beam direction per unit time, be f. Then the number of particles which strike the target in unit time is w = f \u03c3, where \u03c3 is the cross-sectional area presented by the target, namely \u03c3 = \u03c0R2. (4.36) (4.37) Now let us consider one of these particles. We suppose that it impinges on the target with velocity v and impact parameter b. Then, as is clear from Fig. 4.8, it will hit the target at an angle \u03b1 to the normal given by b = R sin \u03b1. The force on the particle is an impulsive central conservative force, corresponding to a potential V (r) which is zero for r > R, and rises very sharply in the neighbourhood of r = R. Thus the kinetic energy and angular momentum must be the same before and after the collision. Let us take the Central Conservative Forces 91 \u03b1 \u0398 \u03b1 b R z Fig. 4.8 positive z direction (\u03b8 = 0) to be the direction of motion of the incoming particles. Then, by the axial symmetry of the problem, the particle must move in a plane \u03d5 = constant. From energy conservation, its velocity must be the same in magnitude before and after the collision. Then, from angular momentum conservation, it follows that the particle will bounce o\ufb00 the sphere at an angle to the normal equal to the incident angle \u03b1. Thus the particle is de\ufb02ected through an angle \u03b8 = \u03c0 2\u03b1, related to the impact parameter by \u2212 b = R cos 1 2 \u03b8. (4.38) We can now calculate the number of particles scattered in a direction speci\ufb01ed by the polar angles \u03b8, \u03d5, within angular ranges d\u03b8, d\u03d5. The particles scattered through angles between \u03b8 and \u03b8 + d\u03b8 are those that came in with impact parameters between b and b + db, where db = \u2212 1 2 R sin 1 2 \u03b8 d\u03b8. (Note that in our case db is actually negative.) Consider now a cross-section of the incoming beam. The particles we are interested in are those which cross a small element of area d\u03c3 = b d\u03d5. db | | (4.39) 92 Classical Mechanics L d\u03b8 dA L sin \u03b8 d\u03d5 b d\u03d5 db d\u03c3 b d\u03d5 \u0398 Fig. 4.9 (See Fig. 4.9.) Inserting the values of b and db, we \ufb01nd d\u03c3 = 1 4 R2 sin \u03b8 d\u03b8 d\u03d5. (4.40) The rate at which particles cross this area, and therefore the rate at which they emerge in the given angular range, is dw = f d\u03c3. (4.41) In order to measure this rate, we may place a small detector at a large distance from the target in the speci\ufb01ed direction. We therefore want to express our result in terms of the cross-sectional area dA of the detector, and its distance L from the target (which we assume to be much greater Central Conservative Forces 93 than R). Now from the discussion of (3.30)), we see that the element of area on a sphere of radius", " L is 3.5 (see the paragraph preceding \u00a7 We de\ufb01ne the solid angle subtended at the origin by the area dA to be dA = L d\u03b8 L sin \u03b8 d\u03d5. \u00d7 so that d\u2126 = sin \u03b8 d\u03b8 d\u03d5, dA = L2d\u2126. (4.42) (4.43) The solid angle is measured in steradians (sr). It plays the same role for a sphere as does the angle in radians for a circle; Eq. (4.43) is the analogue of the equation ds = L d\u03b8 for a circle of radius L. Just as the total angle subtended by an entire circle is 2\u03c0, so the total solid angle subtended by an entire sphere is d\u2126 = 1 L2 \u03c0 2\u03c0 dA = sin \u03b8 d\u03b8 d\u03d5 = 4\u03c0. 0 The important quantity is not the cross-sectional area d\u03c3 itself, but rather the ratio d\u03c3/d\u2126, which is called the di\ufb00erential cross-section. By (4.41) and (4.43), the rate dw at which particles enter the detector is 0 dw = f d\u03c3 d\u2126 dA L2. (4.44) It is useful to note an alternative de\ufb01nition of the di\ufb00erential crosssection, which is applicable even if we cannot follow the trajectory of each individual particle, and therefore cannot say just which of the incoming particles are those that emerge in a particular direction. We may de\ufb01ne d\u03c3/d\u2126 to be the ratio of scattered particles per unit solid angle to the number of incoming particles per unit area. Then the rate at which particles are detected is obtained by multiplying the di\ufb00erential cross-section by the \ufb02ux of incoming particles, and by the solid angle subtended at the target by the detector, as in (4.44). Note that d\u03c3/d\u2126 has the dimensions of area (solid angle, like angle, is dimensionless); it is measured in square metres per steradian (m2 sr\u2212 1). In the particular case of scattering from a hard sphere, the di\ufb00erential cross-section is, by (4.40) and (4.42), d\u03c3 d\u2126 = 1 4 R2. (4.45) 94 Classical Mechanics It has the special feature of being isotropic, or independent of the scattering angle. Thus the rate at which particles enter the detector is, in this case, independent of the direction in which it is placed. We note that the total cross-section (4.37) is correctly given by integrating (4.45) over all solid angles; in this case, we have merely to multiply by the total solid angle, 4\u03c0. 4.6 Mean Free Path The total cross-section \u03c3 is useful in discussing the attenuation of a beam of particles passing through matter. Let us consider \ufb01rst a particle moving through a material containing n atoms per unit volume, and suppose that the total cross-section for scattering by a single atom is \u03c3. Consider a cylinder whose axis is the line of motion of the particle, and whose crosssectional area is \u03c3. Then the particle will collide with any atoms whose centres lie within the cylinder. Now, the number of such atoms in a length x of the cylinder is n\u03c3x. This is therefore the average number of collisions made by the particle when it travels a distance x. Thus the mean distance travelled between collisions \u2014 the mean free path \u03bb \u2014 is x/n\u03c3x, i.e. \u03bb = 1 n\u03c3. (4.46) Now consider a beam of particles, with \ufb02ux f, impinging normally on a wall. How far will they penetrate? We need to calculate the \ufb02ux of particles that penetrate to a depth x without su\ufb00ering a collision. Let this \ufb02ux be f (x), and consider a thin slice of wall of thickness dx and area A at the depth x (see Fig. 4.10). The rate at which unscattered particles enter the slice is Af (x), and the rate at which they emerge on the other side is Af (x + dx). The di\ufb00erence between the two must be the rate at which collisions occur within the slice. Now the total number of atoms in the slice is nAdx, and the total cross-sectional area presented by all of them is \u03c3nA dx (assuming dx is small enough that none of them overlap). Thus the rate at which collisions occur in the slice is f (x)\u03c3nA dx. Equating the two quantities, we have Af (x) \u2212 Af (x + dx) = f (x)\u03c3", "nA dx = f (x) A \u03bb dx, by (4.46). Equivalently, df (x) dx = 1 \u03bb \u2212 f (x). Central Conservative Forces 95 A f f(x + dx) dx f(x) x Fig. 4.10 This equation may be integrated to give f (x) = Ce\u2212 x/\u03bb, where C is an arbitrary constant. But f (0) = f, so \ufb01nally f (x) = f e\u2212 x/\u03bb. (4.47) Thus the number of particles in the beam decreases by a factor of 1/e in one mean free path. Very often, when we wish to study the structure of a small object, like an atom, it would be quite impractical to use a target consisting of only a single atom. Instead, we have to use a target containing a large number N of atoms. If the thickness of the target is x, the rate at which collisions occur within the target will be Af (0) \u2212 Af (x) = A(1 e\u2212 x/\u03bb)f. \u2212 If x is small compared to the mean free path, then we may retain only the linear term in the expansion of the exponential. In this case, the rate is approximately Axf /\u03bb = Axn\u03c3f. Since N = nAx, this shows that for a thin target the number of particles scattered will be just N times the number scattered by a single atom. For a thick target, it would be less, because the atoms would e\ufb00ectively screen each other. When the target is thin, the probability of multiple collisions, in which a particle strikes several target atoms in succession, will be small. If we 96 Classical Mechanics assume that it is negligible, then we can conclude that the angular distribution of scattered particles will also be the same as that from a single target atom. This will be the case if all the dimensions of the target are small in comparison with \u03bb. If we use a detector whose distance from the target, L, is large compared to the target size (so that the scattering angle does not depend appreciably on which of the atoms in the target is struck), then the rate at which particles enter the detector will be dw = N f d\u03c3 d\u2126 dA L2, (4.48) that is, just N times the rate for a single target atom. 4.7 Rutherford Scattering We discuss in this section a problem which was of crucial importance in obtaining an understanding of the structure of the atom. In a classic experiment, performed in 1911, Rutherford bombarded atoms with \u03b1-particles (helium nuclei). Because these particles are much heavier than electrons, they are de\ufb02ected only very slightly by the electrons in the atom (see Chapter 7), and can therefore be used to study the heavy atomic nucleus. From observations of the angular distribution of the scattered \u03b1-particles, Rutherford was able to show that the law of force between \u03b1-particle and nucleus is the inverse square law down to very small distances. Thus he concluded that the positive charge is concentrated in a very small nuclear volume rather than being spread out over the whole volume of the atom. Let us now calculate the di\ufb00erential cross-section for the scattering of a particle of charge q and mass m by a \ufb01xed point charge q. The impact parameter b is related to the scattering angle \u03b8 by (4.35), i.e. b = a cot 1 2 \u03b8, a = qq 4\u03c00mv2. (4.49) Thus db = a d\u03b8 2 sin2 1 2 \u03b8, \u2212 so that, substituting in (4.39), we obtain d\u03c3 = a2 cos 1 2 \u03b8 d\u03b8 d\u03d5 2 sin3 1 2 \u03b8. Central Conservative Forces 97 Dividing by the solid angle (4.42), we \ufb01nd for the di\ufb00erential cross-section d\u03c3 d\u2126 = a2 4 sin4 1 2 \u03b8. (4.50) This is the Rutherford scattering cross-section. \u00a7 We note that, in contrast to the di\ufb00erential cross-section for hard-sphere scattering, this cross-section is strongly dependent both on the velocity of the incoming particle and on the scattering angle. It also increases rapidly with increasing charge. For scattering of an \u03b1-particle on a nucleus of atomic number Z, qq = 2Ze2 (where e is the electronic charge). Thus we \u2212 expect the number of particles scattered to increase like Z 2 with increasing atomic number. We saw in 4.3 that the minimum distance of approach is given by (4.20). Thus to investigate the structure of the atom at small distances, we must use high-velocity particles, for which a is small, and examine largeangle scattering, corresponding to particles with small", " impact parameter b. The cross-section (4.50) is large for small values of the scattering angle, but physically it is the large-angle scattering which is of most interest. For, the fact that particles can be scattered through large angles is an indication that there are very strong forces acting at short distances. If the positive nuclear charge were spread out over a large volume, the force would be inverse-square-law only down to the radius of the charge distribution. Beyond that point, it would decrease as we go to even smaller distances. Consequently, the particles that penetrate to within this distance would experience a weaker force than the inverse square law predicts, and would be scattered through smaller angles. A peculiar feature of the di\ufb00erential cross-section (4.50) is that the corresponding total cross-section is in\ufb01nite. This is a consequence of the in\ufb01nite range of the Coulomb force. However far away from the nucleus a particle may be, it still experiences some force, and is scattered through a non-zero (though small) angle. Thus the total number of particles scattered through any angle, however small, is indeed in\ufb01nite. We can easily calculate the number of particles scattered through any angle greater than some small lower limit \u03b80. These are the particles which had impact parameters b less than b0 = a cot 1 2 \u03b80. The corresponding cross-section is, therefore, \u03c3(\u03b8 > \u03b80) = \u03c0b2 0 = \u03c0a2 cot2 1 2 \u03b80. (4.51) (This may also be veri\ufb01ed by direct integration of (4.50).) 98 Classical Mechanics 4.8 Summary For a particle moving under any central, conservative force, information about the radial motion may be obtained from the radial energy equation, which results from eliminating \u02d9\u03b8 between the conservation equations for energy and angular momentum. The values of E and J can be determined from the initial conditions, and this equation then tells us the radial velocity at any value of r. When information about the angle \u03b8 is needed, we must \ufb01nd the equation of the orbit. For the inverse square law, the orbit is an ellipse or a hyperbola, according as E < 0 or E > 0. The semi-major axis is \ufb01xed by E, and the semi-latus rectum by J. If we are concerned with \ufb01nding the time taken to traverse part of the orbit, we can use the relation between the angular momentum and the rate of sweeping out area. When a beam of particles strikes a target, the angular distribution of scattered particles may be found from the di\ufb00erential cross-section d\u03c3/d\u2126. This may be calculated from a knowledge of the relation between the scattering angle and the impact parameter. The attenuation of the beam is related to the total cross-section \u03c3, obtained by integrating d\u03c3/d\u2126 over all solid angles. Problems 1. The orbits of synchronous communications satellites have been chosen so that viewed from the Earth they appear to be stationary. Find the radius of the orbits. 2. Find the radii of synchronous orbits about Jupiter and about the Sun. [Their mean rotation periods are 10 hours and 27 days, respectively. The mass of Jupiter is 318 times that of the Earth. The semi-major axis of the Earth\u2019s orbit, or astronomical unit (au) is 1.50 108 km.] 3. The semi-major axis of Jupiter\u2019s orbit is 5.20 au. Find its orbital period in years, and its mean (time-averaged) orbital speed. (Mean orbital speed of Earth = 29.8 km s\u2212 1.) \u00d7 4. The orbit of an asteroid extends from the Earth\u2019s to Jupiter\u2019s, just touching both. Find its orbital period. (Treat the planetary orbits as circular and coplanar.) Central Conservative Forces 99 5. Find the maximum and minimum orbital speeds of the asteroid in Prob- lem 4. 6. The Moon\u2019s mass and radius are 0.0123 ME and 0.273 RE (E = Earth). For Jupiter the corresponding \ufb01gures are 318 ME and 11.0 RE. Find in each case the gravitational acceleration at the surface, and the escape velocity. 7. Calculate the period of a satellite in an orbit just above the Earth\u2019s atmosphere (whose thickness may be neglected). Find also the periods for close orbits around the Moon and Jupiter. 8. The Sun has an orbital speed of about 220 km s\u2212 1 around the centre of the Galaxy, whose distance is 28 000 light years. Estimate the total mass of the Galaxy in solar masses. 9. A particle of mass m moves under the action of a harmonic oscillator force with potential energy 1", " 2 kr2. Initially, it is moving in a circle of radius a. Find the orbital speed v. It is then given a blow of impulse mv in a direction making an angle \u03b1 with its original velocity. Use the conservation laws to determine the minimum and maximum distances from the origin during the subsequent motion. Explain your results physically for the two limiting cases \u03b1 = 0 and \u03b1 = \u03c0. 10. Write down the e\ufb00ective potential energy function U (r) for the system described in Chapter 3, Problem 11. Initially, the particle is moving in a circular orbit of radius 2a. Find the orbital angular velocity \u03c9 in terms of the natural angular frequency \u03c90 of the oscillator when not rotating. If the motion is lightly disturbed, the particle will execute small oscillations about the circular orbit. By considering the e\ufb00ective potential energy function U (r) near its minimum, \ufb01nd the angular frequency \u03c9 of small oscillations. Hence describe the disturbed orbit qualitatively. 11. Show that the comet discussed at the end of 4.4 crosses the Earth\u2019s orbit at opposite ends of a diameter. Find the time it spends inside the Earth\u2019s orbit. (To evaluate the area required, write the equation of the orbit in Cartesian co-ordinates. See Appendix B.) \u00a7 12. A star of mass M and radius R is moving with velocity v through a cloud of particles of density \u03c1. If all the particles that collide with the star are trapped by it, show that the mass of the star will increase at a rate dM dt = \u03c0\u03c1v R2 + 2GM R v2. 100 Classical Mechanics Given that M = 1031 kg and R = 108 km, \ufb01nd how the e\ufb00ective crosssectional area compares with the geometric cross-section \u03c0R2 for velocities of 1000 km s\u2212 1 and 10 km s\u2212 1, 100 km s\u2212 1. 13. *Find the polar equation of the orbit of an isotropic harmonic oscillator by solving the di\ufb00erential equation (4.25), and verify that it is an ellipse with centre at the origin. (Hint : Change to the variable v = u2.) Check also that the period is given correctly by \u03c4 = 2mA/J. 14. Discuss qualitatively the orbits of a particle under a repulsive force with potential energy function V = 1 2 kr2 where k is negative, using the e\ufb00ective potential energy function U. How would the orbit equation, as found in Problem 13, di\ufb00er in this case? What shape is the orbit? 15. *If the Earth\u2019s orbit is divided in two by the latus rectum, show that the di\ufb00erence in time spent in the two halves, in years, is 2 \u03c0 e 1 \u2212 e2 + arcsin e, and hence for small e about twice as large as the di\ufb00erence computed 4.4. (Hint : Use Cartesian co-ordinates to evaluate in the example in e2 = arcsin e may be the required area. The identity \u03c0/2 useful.) arcsin \u221a1 \u2212 \u2212 \u00a7 16. A spacecraft is to travel from Earth to Jupiter along an elliptical orbit that just touches each of the planetary orbits (i.e., the orbit of the asteroid in Problem 4). Use the results of Problems 3, 4 and 5 to \ufb01nd the relative velocity of the spacecraft with respect to the Earth just after launching and that with respect to Jupiter when it nears that planet, neglecting in each case the gravitational attraction of the planet. Where in its orbit must Jupiter be at the time of launch, relative to the Earth? Where will the Earth be when it arrives? [This semi-elliptical trajectory is known as a Hohmann transfer and it is energy-e\ufb03cient for interplanetary travel using high-thrust rockets in that a discrete boost is required at the beginning and at the end of the journey in order respectively to leave Earth and to arrive at the target planet. A similarly careful choice of timing for initiating a return to Earth is necessary, so that there is an inevitable minimum time that must be spent in the region of (here) Jupiter before this can be done. It is an interesting and important exercise to compare timings for round trips of planetary exploration using this form of transfer. For Mars, an obvious target in the short term, the time required is about 32 Earth months with about 15 months spent at Mars. There are of course other Central Conservative Forces 101 ways of e\ufb00ecting transfer, but at rather greater cost, either in fuel or in time spent at the destination!] 17. *Suppose that the asteroid of Problems 4 and 5 approaches the Earth with an impact parameter of 5RE, where RE = Earth\u2019s radius,", " moving (This is an improbably close in the same plane and overtaking it. encounter for a large asteroid; however the spectacular impact of comet Shoemaker\u2013Levy 9 with Jupiter in July 1994 should prevent us from being too complacent about this threat!) Find the distance of closest approach and the angle through which the asteroid is scattered, in the frame of reference in which the Earth is at rest. (Assume that the asteroid is small enough to have negligible e\ufb00ect on the Earth\u2019s orbit.) What is its new velocity v relative to the Sun? Show that the semiv2), where aE and vE are major axis of its new orbit is aEv2 the Earth\u2019s orbital radius and orbital velocity. Find the asteroid\u2019s new orbital period. E/(2v2 E \u2212 18. *Show that the position of a planet in its elliptical orbit can be expressed, using a frame with x-axis in the direction of perihelion (point of closest approach to the Sun), in terms of an angular parameter \u03c8 by x = a(cos \u03c8 e), y = b sin \u03c8. (See Problem B.1. In the literature, \u03c8 is sometimes called the eccentric anomaly, while the polar angle \u03b8 is the true anomaly.) Show that r = a(1 e cos \u03c8), and that the time from e sin \u03c8) (Kepler\u2019s equation). perihelion is given by t = (\u03c4 /2\u03c0)(\u03c8 \u2212 19. Use the parametrization of Problem 18 to calculate the time-averaged values of the kinetic and potential energies T and V over a complete period. Hence verify the virial theorem, 2 \u2212 \u2212 V av = \u2212 T av. 20. *Find a parametrization similar to that of Problem 18 for a hyperbolic orbit, using hyperbolic functions. 21. On reaching the vicinity of Jupiter, the spacecraft in Problem 16 is swung around the planet by its gravitational attraction \u2014 a \u2018slingshot\u2019 manoeuvre. Consider this encounter in the frame of reference in which Jupiter is at rest. What is the magnitude and direction of the spacecraft\u2019s velocity before scattering? What is its magnitude after scattering? If the scattering angle in this frame is 90\u25e6, what must be the impact parameter? What is the distance of closest approach to the planet, in terms of Jupiter radii? (MJ = 318ME, RJ = 11.0RE.) 22. *If the manoeuvre in Problem 21 is in the orbital plane, so that the \ufb01nal velocity of the spacecraft relative to Jupiter is radially away from the Sun, what is its velocity in magnitude and direction relative to the Sun? Use the radial energy equation to determine the spacecraft\u2019s 102 Classical Mechanics farthest distance from the Sun (its aphelion distance) in astronomical units. Find also its new orbital period. When it returns, what will be its perihelion distance? 23. An alternative to the manoeuvre described in Problem 22 is for the spacecraft to be scattered out of the orbital plane. Assume that relative to Jupiter its velocity after scattering is directed normal to the orbital plane. What is its velocity relative to the Sun? What will be its aphelion distance and orbital period? How far from the orbital plane will it reach? (Hint : Immediately after scattering, the radial component of its velocity is zero. This is therefore the perihelion point of the new orbit. The farthest point from the original orbital plane will occur when it is at one end of the semi-minor axis of the orbit.) 24. *A ballistic rocket (one that moves freely under gravity after its initial launch) is \ufb01red from the surface of the Earth with velocity v < \u221aRg at an angle \u03b1 to the vertical. (Ignore the Earth\u2019s rotation.) Find the equation of its orbit. Express the range 2R\u03b8 (measured along the Earth\u2019s surface) in terms of the parameters l and a, and hence show that to maximize the range, we should choose \u03b1 so that l = 2a R. (Hint : A sketch may help.) Deduce that the maximum range is 2R\u03b8 where sin \u03b8 = v2/(2Rg v2). Given that the maximum range is 3600 nautical miles, \ufb01nd the launch velocity and the angle at which the rocket should be launched. (Note: 1 nautical mile = 1 minute of arc over the Earth\u2019s surface.) \u2212 \u2212 \u2212 25. Discuss the possible types of orbit for a particle moving under a central inverse-cube-law force, described by the potential energy function V = k/2r2. For the repulsive case (k > 0), show that the orbit equation is \u03b80) = b, where n, b and \u03b80 are constants. Show", " that for the r cos n(\u03b8 attractive case (k < 0), the nature of the orbit depends on the signs of J 2 + mk and of E. Find the equation of the orbit for each possible type. (Include the cases where one of these parameters vanishes.) 26. Show that the scattering angle for particles of mass m and initial ve\u03c0/n (see locity v scattered by a repulsive inverse-cube-law force is \u03c0 Problem 25). Hence \ufb01nd the di\ufb00erential cross-section. \u2212 27. *The potential energy of a particle of mass m is V (r) = k/r + c/3r3, where k < 0 and c is a small constant. (The gravitational potential energy in the equatorial plane of the Earth has approximately this form, because of its \ufb02attened shape \u2014 see Chapter 6.) Find the angular velocity \u03c9 in a circular orbit of radius a, and the angular frequency \u03c9 of small radial oscillations about this circular orbit. Hence show that a Central Conservative Forces 103 k | a2)\u03c9. | \u2248 nearly circular orbit is approximately an ellipse whose axes precess at an angular velocity \u2126 (c/ 1029 atoms per m3. 28. A beam of particles strikes a wall containing 2 15 m. Find the thickEach atom behaves like a sphere of radius 3 ness of wall that exactly half the particles will penetrate without scattering. What thickness would be needed to stop all but one particle in 106? \u00d7 10\u2212 \u00d7 29. An \u03b1-particle of energy 4 keV (1 eV = 1.6 \u00d7 10\u2212 19 J) is scattered by an aluminium atom through an angle of 90\u25e6. Calculate the distance of closest approach to the nucleus. (Atomic number of \u03b1-particle = 2, 19 C.) A beam of such partiatomic number of Al = 13, e = 1.6 1 strikes a target containing 50 mg of cles with a \ufb02ux of 3 aluminium. A detector of cross-sectional area 400 mm2 is placed 0.6 m from the target in a direction at right angles to the beam direction. Find the rate of detection of \u03b1-particles. (Atomic mass of Al = 27 u; 1 u = 1.66 108 m\u2212 27 kg.) 2 s\u2212 10\u2212 \u00d7 \u00d7 \u00a7 3.4 that Kepler\u2019s second law of planetary motion implies that the force is central. Show that his \ufb01rst law \u2014 that the orbit is an ellipse with the Sun at a focus \u2014 implies the inverse square law. (Hint : By di\ufb00erentiating the orbit equation l/r = 1 + e cos \u03b8, and using (3.26), \ufb01nd \u02d9r and \u00a8r in terms of r and \u03b8. Hence calculate the radial acceleration.) 31. Show that Kepler\u2019s third law, \u03c4 a3/2, implies that the force on a \u221d planet is proportional to its mass. \u00afr3/2, where \u00afr is a [This law was originally expressed by Kepler as \u03c4 \u2018mean value\u2019 of r. For an ellipse, the mean over angle \u03b8 is in fact b; the mean over time is actually a(1 + 1 2 e2); it is the mean over arc length \u2014 or the median \u2014 which is given by a! Of course, for most planets in our Solar System these values are almost equal.] \u221d 10\u2212 30. *It was shown in \u00d7 104 Classical Mechanics Chapter 5 Rotating Frames Hitherto, we have always used inertial frames, in which the laws of motion take on the simple form expressed in Newton\u2019s laws. There are, however, a number of problems that can most easily be solved by using a non-inertial frame. For example, when discussing the motion of a particle near the Earth\u2019s surface, it is often convenient to use a frame which is rigidly \ufb01xed to the Earth, and rotates with it. In this chapter, we shall \ufb01nd the equations of motion with respect to such a frame, and discuss some applications of them. 5.1 Angular Velocity; Rate of Change of a Vector Let us consider a solid body which is rotating with constant angular velocity \u03c9 about a \ufb01xed axis. Let n be a unit vector along the axis, whose direction is de\ufb01ned by the right-hand rule: it is the direction in which a right-handthread screw would move when turned in the direction of the rotation. Then we de\ufb01ne the vector angular velocity \u03c9 to be a vector of magnitude \u03c9 in the direction of n: \u03c9 = \u03c9n. Clearly, angular velocity, like angular momentum, is an axial vector (see 3.3). For example,", " for the Earth, \u03c9 is a vector pointing along the polar axis, towards the north pole. Its magnitude is equal to 2\u03c0 divided by the length of the sidereal day (the rotation period with respect to the \ufb01xed stars, which is less than that with respect to the Sun by one part in 365), that is \u00a7 \u03c9 = 2\u03c0 86164 s\u2212 1 = 7.292 10\u2212 5 s\u2212 1. \u00d7 (5.1) If we take the origin to lie on the axis of rotation, then the velocity of 105 106 Classical Mechanics a point of the body at position r is given by the simple formula v = \u03c9 r. \u2227 (5.2) To prove this, we note that the point moves with angular velocity \u03c9 around a circle of radius \u03c1 = r sin \u03b8 (see Fig. 5.1). Thus its speed is v \u03c9\u03c9\u03c9\u03c9 \u03c1 \u03b8 r Fig. 5.1 v = \u03c9\u03c1 = \u03c9r sin \u03b8 = \u03c9 | Moreover, the direction of v is that of \u03c9 r; for clearly, v is perpendicular to the plane containing \u03c9 and r, and it is easy to see from the \ufb01gure that its sense is given correctly by the right-hand rule. Thus (5.2) is correct both as regards magnitude and direction.. | \u2227 \u2227 r It is not necessary that r should be the position vector of a point of the rotating body. If a is any vector \ufb01xed in the rotating body, then by the same argument, da dt = \u03c9 a. \u2227 (5.3) In particular, if i, j, k are orthogonal unit vectors \ufb01xed in the body, then di dt = \u03c9 i, \u2227 dj dt = \u03c9 j, \u2227 dk dt = \u03c9 k. \u2227 (5.4) For example, if \u03c9 is in the k direction, then di dt = \u03c9j, dj dt = \u03c9i, \u2212 dk dt = 0. (Compare Fig. 5.2.) Rotating Frames 107 k dj j i di Fig. 5.2 Now consider a vector a speci\ufb01ed with respect to the rotating axes i, j, k by the components ax, ay, az, so that a = axi + ayj + azk. (Our discussion applies quite generally, but it may be helpful to think of a as the position vector of a particle with respect to the Earth.) We must distinguish two kinds of \u2018rates of change\u2019, and it will be convenient to suspend for the moment the convention whereby da/dt and \u02d9a mean the same thing. We shall denote by da/dt the rate of change of a as measured by an inertial observer at rest relative to the origin, and by \u02d9a the rate of change as measured by an observer rotating with the solid body. We wish to \ufb01nd the relation between these two rates of change. Now, although our two observers di\ufb00er about the rate of change of a vector, they will always agree about the rate of change of any scalar quantity, and in particular about the rates of change of the three components ax, ay, az. (Note that for the inertial observer, these are not the components of a with respect to his own set of axes. An observer outside the Earth will always agree with one on the Earth about the latitude and longitude of a small body moving on the surface, though these are not the co-ordinates he would naturally use himself to describe its position.) Hence we can write dax dt = \u02d9ax, day dt = \u02d9ay, daz dt = \u02d9az. (5.5) According to the observer on the rotating body, the rate of change of a is fully described by the rates of change of its three components, so that \u02d9a = \u02d9axi + \u02d9ayj + \u02d9azk. (5.6) 108 Classical Mechanics However, to the inertial observer, the axes i, j, k are themselves changing with time according to (5.4), so that da dt = dax dt i + day dt j + daz dt k = ( \u02d9axi + \u02d9ayj + \u02d9azk) + \u03c9 + ax + ay di dt (axi + ayj + azk), dj dt + az dk dt \u2227 by (5.5) and (5.4). Thus, \ufb01nally, we obtain da dt = \u02d9a + \u03c9 a. \u2227 (5.7) Applied to the position vector r, this result is almost obvious,", " for it states that the velocity with respect to the inertial observer is the sum of the r of a velocity \u02d9r with respect to the rotating frame and the velocity \u03c9 particle at r rotating with the body. We have given a detailed proof because of the central importance of this result, and because its application to other vectors is much less obvious. \u2227 \u2227 a. We shall frequently encounter in our later work equations similar to (5.3): da/dt = \u03c9 It is important to realize that one can reverse the argument that led up to it. If a satis\ufb01es this equation, then it must be a vector of constant length rotating with angular velocity \u03c9 about the direction of \u03c9. For, if we introduce a frame of reference rotating with angular velocity \u03c9, then according to (5.7), \u02d9a = 0, so that a is \ufb01xed in the rotating frame. 5.2 Particle in a Uniform Magnetic Field A particle of charge q moving with velocity v in a magnetic \ufb01eld B(r) experiences a force proportional to its velocity, and perpendicular to it, The equation of motion is then F = qv B. \u2227 m dv dt = qv B. \u2227 (5.8) (5.9) Now let us suppose that the magnetic \ufb01eld is uniform (independent of position) and constant in time. Then (5.9) has precisely the form of (5.3), dv dt = \u03c9 v, \u2227 where \u03c9 = q m \u2212 B. (5.10) Rotating Frames 109 Hence, by the argument of the preceding section, the velocity vector v rotates about the direction of B with a constant angular velocity. It follows that the path of the particle is a helix whose axis points along the \ufb01eld lines (see Fig. 5.3). If we take the direction of B to be the z-direction, then v has B Fig. 5.3 a constant z-component, and a component in the xy-plane which rotates around a circle with angular velocity \u03c9. In particular, if v is initially parallel to B, the particle will continue to move with uniform velocity along a \ufb01eld line. The other limiting case, in which v is perpendicular to B, is more interesting. In that case, the particle describes a circle of radius r = v \u03c9 = mv qB. (5.11) This e\ufb00ect is used in accelerating particles in the cyclotron. In order to prevent the particles from escaping, a strong magnetic \ufb01eld is employed to constrain them to move in circles. To accelerate them, an alternating voltage is applied between two D-shaped pole pieces enclosing two halves of the circles (see Fig. 5.4). The angular frequency of this alternating voltage is chosen to coincide with the cyclotron frequency \u03c9c = qB m. (5.12) Thus, by the time the particles have made a half-revolution, the direction of the electric \ufb01eld has reversed, and they experience an accelerating \ufb01eld each time they cross the gap. Each semicircle is therefore slightly larger than the preceding one, and the particles spiral outwards with increasing energy. Finally, one obtains a beam of high-velocity particles at the outer edge of the cyclotron. This method can be used only so long as the speed remains small compared to the speed of light; for, according to relativity theory, the angular frequency \u03c9c is not precisely constant when the speed is large, but decreases with increasing speed. Quite similar methods can be used, however, even at the highest energies. 110 Classical Mechanics E E ~ Fig. 5.4 If there are both electric and magnetic \ufb01elds present, a charged particle B). An interesting case is experiences the Lorentz force, F = q(E + v that of crossed \ufb01elds. \u2227 Example: Crossed electric and magnetic \ufb01elds A particle of mass m and charge q moves in uniform perpendicular electric and magnetic \ufb01elds, E = Ej and B = Bk. Show that if it is initially moving in the xy-plane, it will continue to do so, and that it has a circular motion with a superimposed uniform drift velocity in a direction perpendicular to both E and B. Find the solution for the case when the particle starts from rest at the origin. The equation of motion m\u00a8r = qE+q \u02d9r B is, in terms of components, m\u00a8x = qB \u02d9y, m\u00a8y = qE \u2212 \u2227 qB \u02d9x, m\u00a8z = 0. Clearly, if z = \u02d9z = 0 initially", ", this will be true always. We can eliminate the term in E by going over to a frame moving with relative velocity (E/B)i (see Chapter 1, Problem 5), in which the co-ordinates are given by x = x \u2212 (E/B)t, y = y, z = z. Rotating Frames 111 Then \u02d9x = \u02d9x E/B, \u02d9y = \u02d9y, \u02d9z = \u02d9z. Thus the equations of motion become \u2212 m\u00a8x = qB \u02d9y, m\u00a8y = \u2212 qB \u02d9x, m\u00a8z = 0. So in the moving frame, the particle moves in a circle, as in the discussion above. In the original frame, this means circular motion with a uniform velocity superimposed. The initial conditions yield \u02d9x = E/B, so the radius of the circular motion is E/B\u03c9c. If the particle starts from the origin, the complete solution is \u2212 x = E B t \u2212 sin \u03c9ct \u03c9c, y = E B\u03c9c (1 \u2212 cos \u03c9ct), z = 0. 5.3 Acceleration; Apparent Gravity We can use the formula (5.7) twice to obtain the relation between the absolute acceleration d2r/dt2 of a particle and its acceleration \u00a8r relative to a rotating frame. The velocity of the particle relative to an inertial frame is v = dr dt = \u02d9r + \u03c9 r. \u2227 Applying the same formula to the rate of change of v, we have d2r dt2 = But, from (5.13), assuming that \u03c9 is constant, we \ufb01nd = \u02d9v + \u03c9 dv dt v. \u2227 (5.13) (5.14) \u02d9v = \u00a8r + \u03c9 \u02d9r, \u2227 and v = \u03c9 \u03c9 \u2227 \u02d9r + \u03c9 (\u03c9 \u2227 \u2227 r). \u2227 Hence, substituting in (5.14), we obtain d2r dt2 = \u00a8r + 2\u03c9 \u2227 \u02d9r + \u03c9 (\u03c9 \u2227 \u2227 r). (5.15) The second term on the right is called the Coriolis acceleration, and the third term is the centripetal acceleration. The latter is directed inwards 112 Classical Mechanics towards the axis of rotation, and perpendicular to it, as may be seen by writing it in the form (see (A.16)) (\u03c9 \u03c9 \u2227 \u2227 r) = (\u03c9 r)\u03c9 \u00b7 \u2212 \u03c92r. The most important application of (5.15) is to a particle moving near the surface of the Earth. For a particle moving under gravity, and under an additional, mechanical force F, the equation of motion is d2r dt2 = mg + F, where g is a vector of magnitude g pointing downward (towards the centre of the Earth). Using (5.15), and moving all the terms except the relative acceleration to the right side of the equation, we can write it as m m\u00a8r = mg + F 2m\u03c9 \u02d9r \u2227 \u2212 m\u03c9 \u2227 (\u03c9 \u2227 r). \u2212 (5.16) The last two terms are apparent (or \ufb01ctitious) forces, which arise because of the non-inertial nature of the frame of reference. We shall postpone discussion of the e\ufb00ects of the third term \u2014 the Coriolis force \u2014 to the next section, and consider here the last term, which is the familiar centrifugal force. This force is a slowly varying function of position, proportional, like gravity, to the mass of the particle. When we make a laboratory measurement of the acceleration due to gravity, what we actually measure is not g but g\u2217 = g \u03c9 \u2212 \u2227 (\u03c9 \u2227 r). (5.17) In particular, a plumb line does not point directly towards the centre of the Earth, but is swung outwards through a small angle by the centrifugal force. (See Fig. 5.5, in which this e\ufb00ect is greatly exaggerated.) Let us consider a point in colatitude (\u03c0/2 latitude) \u03b8. Then \u03c9 | Thus the horizontal and vertical components of the e\ufb00ective gravitational acceleration g\u2217 (using \u2018vertical\u2019 to mean towards the Earth\u2019s centre) are r) | \u03c9 | = \u03c92r sin \u03b8. (\u03c9 \u2227 \u2227 \u2227 r | g\u2217h = \u03c92r sin \u03b8 cos \u03b8, g\u2217v = g \u2212 \u03c92r sin2 \u03b8. (There are further corrections, to", " be discussed later, due to the nonspherical shape of the Earth.) \u2212 = \u03c9 Rotating Frames 113 \u03c9 \u03c9 \u2212\u03c9\u2227(\u03c9\u2227r) g \u03b1 g* Fig. 5.5 The magnitude of the centrifugal force may be found by inserting the value (5.1) for \u03c9, and for r the mean radius of the Earth, 6371 km. We \ufb01nd \u03c92r = 34 mm s\u2212 2. (5.18) Since \u03c92r approximately g, the angle \u03b1 between the apparent and true verticals is \u03b1 \u2248 g\u2217h g\u2217v \u2248 \u03c92r g sin \u03b8 cos \u03b8. The maximum value occurs at \u03b8 = 45\u25e6, and is about 0\u25e66. At the pole, there is no centrifugal force, and g\u2217 = g. On the equator, \u03c92r. Thus we might expect the measured value of the acceleration 2. The actual measured g\u2217 = g due to gravity to be larger at the pole by 34 mm s\u2212 di\ufb00erence is somewhat larger than this, \u2212 \u2206g\u2217 = g\u2217pole \u2212 g\u2217equator = 52 mm s\u2212 2. (5.19) This discrepancy arises from the fact that the Earth is not a perfect sphere, but more nearly spheroidal in shape, \ufb02attened at the poles. Thus the gravitational acceleration g, even excluding the centrifugal term, is itself larger at the pole than on the equator. These two e\ufb00ects are not really independent, for the \ufb02attening of the Earth is a consequence of its rotation. We shall discuss this point in the next chapter. As another example of the e\ufb00ect of the centrifugal force, we shall con- sider the problem of a liquid in a rotating vessel. 114 Classical Mechanics Example: Surface of a rotating liquid Liquid in a vessel rotating with angular velocity \u03c9 is rotating with the same angular velocity. Find the equation for its surface. We use a frame rotating with the vessel, so that in the rotating frame the liquid is at rest, but subject to the centrifugal force. Now this force is conservative, and corresponds to a potential energy Vcent = 1 2 m\u03c92\u03c12 = \u2212 \u2212 1 2 m\u03c92(x2 + y2). (5.20) Consider a particle on the surface of the liquid. Unless the surface is one on which potential energy is constant, the particle will tend to move towards regions of lower potential energy. Hence if the liquid is in equilibrium under the gravitational and centrifugal forces, its surface must be an equipotential surface, 2 \u03c92(x2 + y2) = constant. This is a paraboloid of revolution about the z-axis. gz \u2212 1 This e\ufb00ect is used in constructing some high-powered optical telescopes. Because a paraboloid is the ideal shape to focus light from directly above it, the mirror is made by spinning a thin \ufb01lm of liquid, for example mercury. This is very much cheaper than grinding glass mirrors to a precise shape \u2014 though it has the disadvantage that the telescope can only look at objects near the zenith. 5.4 Coriolis Force \u2227 \u2212 2m\u03c9 The Coriolis force, \u02d9r, is an apparent, velocity-dependent force arising from the Earth\u2019s rotation. To understand its physical origin, it may be helpful to consider a \ufb02at, rotating disc. Suppose that a particle moves across the disc under no forces, so that an inertial observer sees it moving diametrically across in a straight line. (See Fig. 5.6(a).) Then, because the disc is rotating, an observer on the disc will see the particle crossing successive radii, following a curved track as in Fig. 5.6(b). If he is unaware that the disc is rotating, he will ascribe this curvature to a force acting on the particle at right angles to its velocity. This is the Coriolis force. We may regard Fig. 5.6 as representing the Earth viewed from the north pole. Thus we see that the e\ufb00ect of the Coriolis force is to make a particle Rotating Frames 115 \u03c9 (a) (b) Fig. 5.6 deviate to the right in the northern hemisphere, and to the left in the southern. (It also has a vertical component, in general.) It is responsible for some well-known meteorological phenomena, which we shall discuss below. First, however, we examine some e\ufb00ects of the Coriolis force which are observable in the laboratory. Let us consider a particle moving near the surface of the Earth in colatitude \u03b8. We shall suppose that the distance through which it moves is", " su\ufb03ciently small for both the gravitational and centrifugal forces to be e\ufb00ectively constant. Then we may combine them in a constant e\ufb00ective gravitational acceleration g\u2217. For convenience, we shall drop the star, and write this simply as g. The equation of motion is then m\u00a8r = mg + F 2m\u03c9 \u02d9r. \u2227 \u2212 (5.21) As an example, let us take a small body falling freely under gravity towards the Earth\u2019s surface. Example: Freely falling body A body is dropped from rest at a height h above the ground. How far from the point directly below its starting point does it land? To analyze this problem, it is convenient to choose our axes so that i is east, j is north, and k up. (See Fig. 5.7.) Here \u2018up\u2019 means opposite to g\u2217, and the colatitude \u03b8 is the angle between that direction and the Earth\u2019s axis. The angular velocity vector then has the components \u03c9 = (0, \u03c9 sin \u03b8, \u03c9 cos \u03b8). 116 Classical Mechanics \u03c9\u03c9 j k \u03b8 North Fig. 5.7 Hence the Coriolis force is 2m\u03c9 \u2212 \u2227 \u02d9r = 2m\u03c9( \u02d9y cos \u03b8 \u02d9z sin \u03b8, \u2212 \u2212 \u02d9x cos \u03b8, \u02d9x sin \u03b8). (5.22) Neglecting the Coriolis force, the motion is described by x = 0, y = 0, z = h 1 2 gt2. \u2212 We shall calculate the e\ufb00ect of the Coriolis force only to \ufb01rst order; that is, we neglect terms of order \u03c92. Since the Coriolis force (5.22) contains an explicit factor of \u03c9, this means that we may substitute for \u02d9r the zeroth-order value \u02d9r = (0, 0, gt). The equations of motion (5.21) then read \u2212 m\u00a8x = 2m\u03c9gt sin \u03b8, m\u00a8y = 0, m\u00a8z = mg. \u2212 The solution with appropriate initial conditions is, therefore, x = 1 3 \u03c9gt3 sin \u03b8, y = 0, z = h 1 2 gt2. \u2212 Thus the body will hit the ground, z = 0, at a point east of that vertically below its point of release, at a distance x = 8h3 g \u03c9 3 1/2 sin \u03b8. (5.23) For example, if a particle is dropped from a height of 100 m in latitude 45\u25e6, the deviation is about 16 mm. It is instructive to consider how an inertial observer would describe this experiment. Since the particle is dropped from rest relative to the Earth, it has a component of velocity towards the east relative to the inertial Rotating Frames 117 h x (b) x h (a) Fig. 5.8 observer. As it falls, the angular momentum about the Earth\u2019s axis remains constant, and therefore its angular velocity increases, so that it gets ahead of the ground beneath it. Figs. 5.8(a) and 5.8(b) show the experiment as it appears (but much exaggerated) to an observer on the Earth and to an inertial observer, respectively. Foucault\u2019s pendulum Another way of observing the e\ufb00ect of the Coriolis force is to use Foucault\u2019s pendulum. This is simply an ordinary pendulum free to swing in any direction, and carefully arranged to be perfectly symmetric, so that its periods of oscillation in all directions are precisely equal. (It must also be long, and fairly heavy, so that it will go on swinging freely for several hours at least, despite the resistance of the air.) Examples of the Foucault pendulum are to be found in many science museums. If the amplitude is small, the pendulum equation is simply the equation of two-dimensional simple harmonic motion. The vertical component of the Coriolis force is negligible, for it is merely a small correction to g, whose sign alternates on each half-period. (Even for a velocity \u02d9x as large g.) The important components as 10 m s\u2212 are the horizontal ones. For small amplitude, the velocity of the pendulum bob is almost horizontal, so that \u02d9z 0. Thus the equations of motion for the x and y co-ordinates are 1, we have 2\u03c9 \u02d9x = 1.5 mm s\u2212 \u2248 2 \u00a8x = g l \u2212 x + 2\u03c9 \u02d9y cos \u03b8, \u00a8y = g l y \u2212 \u2212 2\u03c9 \u02d9x cos \u03b8, (", "5.24) or, in vector notation, \u00a8r = g l r \u2212 \u2212 2\u03c9 cos \u03b8 k \u02d9r. \u2227 (5.25) 118 Classical Mechanics Let us \ufb01rst suppose that the pendulum is at the north pole (\u03b8 = 0). Then it is clear from the equation of motion in a non-rotating frame that it must swing in a \ufb01xed direction in space, while the Earth rotates beneath it. Thus, relative to the Earth, its oscillation plane must rotate around \u03c9. Now, at any other latitude, the the vertical with angular velocity \u2212 only di\ufb00erence in Eq. (5.25) is that in place of the angular velocity \u03c9 we have only its vertical component, \u03c9 cos \u03b8 k. Hence, we should expect that the oscillation plane of the pendulum rotates with angular velocity \u03c9 cos \u03b8 around the vertical. In e\ufb00ect, we may regard the Earth\u2019s \u2212 surface in colatitude \u03b8 as rotating about the vertical with angular velocity \u03c9 cos \u03b8 (and also with angular velocity \u03c9 sin \u03b8 about a horizontal northsouth axis, but this component leads only to a vertical Coriolis force, which does not appreciably a\ufb00ect the motion.) \u2126 = \u2212 We can verify this conclusion by obtaining an explicit solution of Eqs. (5.24). A neat way of doing this is to combine the two equations by introducing a complex variable z = x + iy (not of course to be confused i \u02d9z, so that, adding i times with the vertical co-ordinate). Then \u02d9y the second equation to the \ufb01rst, we get i \u02d9x = \u2212 \u2212 \u00a8z + 2i\u2126 \u02d9z + \u03c92 0z = 0, where \u2126 = \u03c9 cos \u03b8 and \u03c92 z = Aept and, as in \u00a7 0 = g/l. We now look for solutions of the form 2.5, \ufb01nd for p the equation p2 + 2i\u2126p + \u03c92 0 = 0. The roots of this equation are p = the general solution of the equation for z is i\u2126 \u00b1 \u2212 i\u03c91, where \u03c92 1 = \u03c92 0 + \u21262. Hence, z = Ae\u2212 i(\u2126 \u2212 \u03c91)t + Be\u2212 i(\u2126+\u03c91)t. In particular, if the pendulum is released from rest at position (a, 0) at t = 0, we must choose A = B = a/2, so that the solution is z = ae\u2212 i\u2126t cos \u03c91t, or, in terms of x and y, x = a cos \u2126t cos \u03c91t, y = \u2212 a sin \u2126t cos \u03c91t. Rotating Frames 119 \u03c90 and \u03c90 \u00b1 So at time t, since \u2126 between the extreme points oscillation with amplitude a in a plane rotating with angular velocity \u03c91, the pendulum bob oscillates a(cos \u2126t, sin \u2126t). The solution represents an \u2126. It should be noted that in deriving (5.24) we neglected terms of order \u21262 (in particular, the variation in the centrifugal force is of this order). Thus the di\ufb00erence between the angular frequency \u03c91 in our solution and \u03c90 is not signi\ufb01cant. The period of the pendulum will not be substantially a\ufb00ected. \u2212 At the pole, the plane of oscillation makes a complete revolution in just 24 hours. At any other latitude, the period is greater than this, and is in fact 2\u03c0/\u03c9 cos \u03b8. In latitude 45\u25e6, it is about 34 hours, while on the equator (where \u03c9 is purely horizontal), it is in\ufb01nite. Cyclones and trade winds There are also some important large-scale e\ufb00ects of the Coriolis force. Suppose that for some reason a region of low pressure develops in the northern hemisphere. The air around it will be pushed inwards by the force of the pressure gradient. As it starts to move, however, the Coriolis force cause it to curve to the right. Thus an anticlockwise rotation is set up around the low-pressure zone. The process will continue until an approximate equilibrium is established between the pressure force, acting inwards, and the Coriolis force (plus the centrifugal force of the rotation) acting outwards. This con\ufb01guration is a cyclone, or depression, familiar to those who live in temperate", " latitudes. (As in the case of Foucault\u2019s pendulum, there is no e\ufb00ect of this kind on the equator.) More generally, in these latitudes, the wind velocity is not directly from regions of high to low pressure, but more nearly along the isobars, keeping the low pressure on the left in the northern hemisphere. The same e\ufb00ect is responsible, on a yet larger scale, for the trade winds. The heating of the Earth\u2019s surface near the equator causes the air to rise, and be replaced by cooler air \ufb02owing in from higher latitudes. However, because of the Coriolis force, it does not \ufb02ow directly north or south, but is made to deviate towards the west. Thus we have the north-east trade winds in the northern hemisphere, and the south-east trade winds in the southern. For reasons that are too complex to discuss in detail here, this pattern does not extend to high latitudes. Further from the equator there are belts of high pressure, which, like the equatorial region itself, are characterized 120 Classical Mechanics by light and variable winds. Beyond these, the direction of the pressure gradient is reversed. Thus around 40\u25e6 to 50\u25e6 N or S the prevailing wind direction is westerly rather than easterly. Near the poles, there is a further reversal, and the circumpolar winds blow from the east. 5.5 Larmor E\ufb00ect As a rather di\ufb00erent example of the use of rotating frames, we shall consider the e\ufb00ect of a magnetic \ufb01eld on a particle of charge q moving in an orbit q. The equation of motion, including the around a \ufb01xed point charge magnetic force (5.8), is \u2212 k r2 \u02c6r + q where k = qq/4\u03c00. Let us rewrite this equation in terms of a rotating frame of reference, using (5.7) and (5.15). We obtain dr dt \u2227 d2r dt2 = B, m \u2212 \u00a8r + 2\u03c9 \u02d9r + \u03c9 (\u03c9 \u2227 \u2227 r) = \u2227 k mr2 \u02c6r + q m \u2212 ( \u02d9r + \u03c9 r) \u2227 \u2227 B. Now, if we choose \u03c9 = (q/2m)B, then the terms in \u02d9r drop out. The last term on the left only cancels half the last term on the right, however, so we obtain \u2212 \u00a8r = k mr2 \u02c6r + \u2212 q 2m 2 B (B r). \u2227 \u2227 We now assume that the magnetic \ufb01eld is su\ufb03ciently weak for the quadratic term in B to be negligible in comparison to the electrostatic force term. The necessary condition for this is that \u03c92 = qB 2m 2 k mr3 = qq 4\u03c00mr3 \u2248 \u03c92 0, (5.26) where we have used (4.31) to express the right side in terms of the mean angular velocity \u03c90 of the particle in its orbit. When this condition is satis\ufb01ed, we obtain the approximate equation \u00a8r = k mr2 \u02c6r. \u2212 But this is just the usual equation of motion for a particle in an inversesquare-law \ufb01eld. Consequently, a bounded orbit in the rotating frame is an ellipse. In the original, non-rotating frame, it is a slowly precessing ellipse, Rotating Frames 121 precessing with angular velocity \u03c9. (See Fig. 5.9, which is drawn for the special case in which the orbit lies in the plane normal to B. In general, it B Fig. 5.9 will be inclined, and the plane of the orbit will precess about the direction of B.) Note that, in view of the restriction (5.26), the axes of the ellipse will precess only through a small angle on each revolution. This e\ufb00ect is known as the Larmor e\ufb00ect, and the angular velocity of precession, \u03c9L = qB 2m, is called the Larmor frequency. Note that it is just half the cyclotron frequency (5.12). The di\ufb00erence arises from the factor of 2 in the Coriolis acceleration term of (5.15). This e\ufb00ect leads to observable changes in the spectra emitted by atoms in the presence of a magnetic \ufb01eld, since the Bohr energy level corresponding to a given angular momentum is slightly shifted in the presence of a magnetic \ufb01eld. (The shift in spectral", " lines is known as the Zeeman e\ufb00ect.) When the weak magnetic \ufb01eld approximation is not applicable then the resulting motion can be very complex \u2014 see 14.4. \u00a7 5.6 Angular Momentum and the Larmor E\ufb00ect The Larmor e\ufb00ect is a particular example of a more general, and very important, phenomenon. As we shall see later in a number of di\ufb00erent examples, the e\ufb00ect of a small force on a rotating system is often to make the axis of rotation precess \u2014 that is, revolve about some \ufb01xed direction. 122 Classical Mechanics It will therefore be useful to consider here an alternative treatment of the problem which is of more general applicability. If the magnetic \ufb01eld is weak, then to a \ufb01rst approximation the particle must move in an inverse-square-law orbit whose characteristics change slowly with time. Since the magnetic force is perpendicular to the velocity, it does no work. Hence the energy is constant, and therefore so is the semimajor axis of the orbit. However the angular momentum J does change with time, according to the equation dJ dt = r \u2227 F = qr (v \u2227 \u2227 B) = q[(r B)v (r \u00b7 \u2212 \u00b7 v)B], (5.27) using (A.16). To be speci\ufb01c, let us suppose that the magnetic \ufb01eld is in the z-direction, B = Bk, and that the particle is moving in a circular orbit of radius r in a plane inclined to the xy-plane at an angle \u03b1. (See Fig. 5.10.) It will be convenient to introduce three orthogonal unit vectors, n normal to the k n \u03b1 b r \u03c8 a v v\u2227B Fig. 5.10 plane of the orbit, a in the direction of k angular momentum is \u2227 n and b = n a. Then the \u2227 J = mr \u2227 v = mrvn. (5.28) If we specify the position of the particle in its orbit by the angle \u03c8 between a and r, then in terms of the axes a, b, n, the components of r, v and B Rotating Frames 123 are r = (r cos \u03c8, r sin \u03c8, 0), v = ( v sin \u03c8, v cos \u03c8, 0), \u2212 B = (0, B sin \u03b1, B cos \u03b1). (There is also a small contribution to v from the fact that the plane of the orbit is changing, but for small magnetic \ufb01elds this will be negligible.) Now, in a circular orbit, r of (5.27) is \u00b7 v is always zero. Hence the right hand side qBr sin \u03b1 sin \u03c8 v = qBrv sin \u03b1( \u2212 sin2 \u03c8, sin \u03c8 cos \u03c8, 0). (5.29) Since the magnetic \ufb01eld is weak, the change in J in a single orbit will be small. Thus the oscillatory term in (5.29) is unimportant: it leads only to a very small periodic oscillation in J. The important term is the one which leads to a secular variation of J (a steady change in one direction). It follows that we can replace (5.29) by its average value over a complete oscillation period, i.e., over all values of \u03c8 from 0 to 2\u03c0. Thus we replace sin2 \u03c8 = 1 2, and sin \u03c8 cos \u03c8 by 0. This yields the equation cos 2\u03c8) by 1 2 (1 \u2212 dJ dt = r F av = \u2227 \u2212 1 2 qBrv sin \u03b1 a. (5.30) a. Since dJ /dt is perpenThus the vector J will move in the direction of dicular to J, its magnitude remains unchanged, while its direction precesses around the direction of B. In fact, (5.30) can be seen to be another example of an equation of the form (5.3). From (5.28), we have \u2212 J = mrvk k \u2227 \u2227 n = mrv sin \u03b1 a. Thus (5.30) may be written dJ dt = qB 2m k \u2212 \u2227 J. (5.31) This is the equation of a vector J rotating with angular velocity \u03c9 = qB 2m. \u2212 124 Classical Mechanics Since the direction of J is that of the normal to the orbital plane, this agrees with our previous conclusion that the orbital plane precesses around the direction of the magnetic \ufb01eld, with a precessional angular velocity equal to the angular Larmor frequency. Comparison with a current loop It is instructive to contrast this behaviour of a", " charged particle with the super\ufb01cially similar case of a current loop. A circular current loop in the same position as the particle orbit would experience a similar force, produca-axis. The e\ufb00ect of the force, however, would ing a moment about the be quite di\ufb00erent. Since the loop itself (that is, its mass) is not rotating, the e\ufb00ect would be simply to rotate it about this axis, so that its normal tends to become aligned with the direction of B. (This is the familiar fact that a current loop behaves like a magnetic dipole.) \u2212 To understand why a particle moving in an orbit behaves so di\ufb00erently, it may be helpful to consider, in place of a continuously acting force, a small impulsive blow delivered once every orbit, say at the farthest point above the xy-plane. The e\ufb00ect of such a blow is to give the particle a small component of velocity perpendicular to its original orbit. The next orbit will be in a plane slightly tilted with respect to the original one, but reaching just as far from the xy-plane. The net e\ufb00ect is to make the orbit swing around the z-axis. The essential point here is that it is the velocity of the particle, rather than its position, that is changed instantaneously. Thus the e\ufb00ect of a small blow is not seen in a shift of the orbit at the point of the blow, but rather at a point 90\u25e6 later. We shall see in Chapter 9 that a rigid body displays a very similar type of behaviour. If one applies a small force downwards on the rim of a rapidly rotating wheel, it is not that point that moves down, but a point 90\u25e6 later. (This is easy to verify with a freely spinning bicycle wheel.) 5.7 Summary In problems involving a rotating body \u2014 particularly the Earth \u2014 it is often convenient to use a rotating frame of reference. The equations of motion in such a frame contain additional terms representing apparent forces which arise because the frame is non-inertial. These are the centrifugal force, directed outwards from the axis of rotation, and the velocity-dependent Rotating Frames 125 2m\u03c9 \u02d9r. Because the magnetic force on a charged particle Coriolis force, tends to produce rotation about the direction of the magnetic \ufb01eld, rotating frames are also useful in many problems involving a magnetic \ufb01eld. \u2212 \u2227 Problems \u00d7 \u00d7 1. Find the centrifugal acceleration at the equator of the planet Jupiter and of the Sun. In each case, express your answer also as a fraction of the surface gravity. (The rotation periods are 10 hours and 27 days, 105 km, and the masses 104 km and 7.0 respectively, the radii 7.1 1027 kg and 2.0 1.9 \u00d7 1030 kg.) \u00d7 2. Water in a rotating container of radius 50 mm is 30 mm lower in the centre than at the edge. Find the angular velocity of the container. 3. The water in a circular lake of radius 1 km in latitude 60\u25e6 is at rest relative to the Earth. Find the depth by which the centre is depressed relative to the shore by the centrifugal force. For comparison, \ufb01nd the height by which the centre is raised by the curvature of the Earth\u2019s surface. (Earth radius = 6400 km.) 4. Find the velocity relative to an inertial frame (in which the centre of the Earth is at rest) of a point on the Earth\u2019s equator. An aircraft is \ufb02ying 1. Assuming that it \ufb02ies straight and above the equator at 1000 km h\u2212 level (i.e., at a constant altitude above the surface) what is its velocity relative to the inertial frame (a) if it \ufb02ies north, (b) if it \ufb02ies west, and (c) if it \ufb02ies east? 5. The apparent weight of the aircraft in Problem 4 when on the ground at the equator is 100 t weight. What is its apparent weight in each of the three cases (a)\u2013(c)? 6. A bird of mass 2 kg is \ufb02ying at 10 m s\u2212 1 in latitude 60\u25e6N, heading due east. Find the horizontal and vertical components of the Coriolis force acting on it. 7. The wind speed in colatitude \u03b8 is v. By considering the forces on a small volume of air, show that the pressure gradient required to balance the horizontal component of the Coriolis force, and thus to maintain a constant wind direction, is dp/dx = 2\u03c9\u03c1v cos \u03b8, where \u03c1 is the 1 for a wind density of the air. Evaluate this gradient in mbar km\u2212 1 in latitude 30\u25e6N. (1 bar", " = 105 Pa; density of air = speed of 50 km h\u2212 1.3 kg m\u2212 3.) 126 Classical Mechanics 8. An aircraft is \ufb02ying at 800 km h\u2212 1 in latitude 55\u25e6N. Find the angle through which it must tilt its wings to compensate for the horizontal component of the Coriolis force. 9. An orbiting space station may be made to rotate to provide an arti\ufb01cial gravity. Given that the radius is 25 m, \ufb01nd the rotation period required to produce an apparent gravity equal to 0.7g. A man whose normal weight is 75 kg weight runs around the station in one direction and then the other (i.e., on a circle on the inside of the cylindrical wall) 1. Find his apparent weight in each case. What e\ufb00ects will at 5 m s\u2212 he experience if he climbs a ladder to a higher level (i.e., closer to the axis), climbing at 1 m s\u2212 1? 10. A beam of particles of charge q and velocity v is emitted from a point source, roughly parallel with a magnetic \ufb01eld B, but with a small angular dispersion. Show that the e\ufb00ect of the \ufb01eld is to focus the beam to a point at a distance z = 2\u03c0mv/ B from the source. Calculate the q | | focal distance for electrons of kinetic energy 500 eV in a magnetic \ufb01eld 31 kg, 1.6 of 0.01 T. (Charge on electron = 10\u2212 1 eV = 1.6 19 C, mass = 9.1 19 J.) 10\u2212 10\u2212 \u00d7 \u2212 \u00d7 11. *Write down the equation of motion for a charged particle in uniform, parallel electric and magnetic \ufb01elds, both in the z-direction, and solve it, given that the particle starts from the origin with velocity (v, 0, 0). A screen is placed at x = a, where a mv/qB. Show that the locus of points of arrival of particles with given m and q, but di\ufb00erent speeds v, is approximately a parabola. How does this locus depend on m and q? \u00d7 12. A beam of particles with velocity (v, 0, 0) enters a region containing crossed electric and magnetic \ufb01elds, as in the example at the end of 5.2. Show that if the ratio E/B is correctly chosen the particles are \u00a7 undeviated, while particles with other speeds follow curved trajectories. Suppose the particles have velocities equal to v in magnitude, but with a small angular dispersion. Show that if the path length l is correctly chosen, all such particles are focussed onto a line parallel to the z-axis. (Thus a slit at that point can be used to select particles with a given speed.) For electrons of velocity 108 m s\u2212 1 in a magnetic \ufb01eld of 0.02 T, \ufb01nd the required electric \ufb01eld, and the correct (smallest possible) choice for l. 13. The angular velocity of the electron in the lowest Bohr orbit of the 1. What is the largest hydrogen atom is approximately 4 1016 s\u2212 \u00d7 Rotating Frames 127 \u00a7 magnetic \ufb01eld which may be regarded as small in this case, in the sense of 5.5? Determine the Larmor frequency in a \ufb01eld of 2 T. 14. *The orbit of an electron (charge e) around a nucleus (charge Ze) \u2212 is a circle of radius a in a plane perpendicular to a uniform magnetic \ufb01eld B. By writing the equation of motion in a frame rotating with the electron, show that the angular velocity \u03c9 is given by one of the roots of the equation m\u03c92 eB\u03c9 \u2212 \u2212 Ze2/4\u03c00a3 = 0. Verify that for small values of B, this agrees with two roots if B = 105 T, Z = 1 and a = 5.3 10\u2212 that in reality a would be changed by the \ufb01eld.) \u00d7 5.5. Evaluate the 11 m. (Note, however, \u00a7 15. *A projectile is launched due north from a point in colatitude \u03b8 at an angle \u03c0/4 to the horizontal, and aimed at a target whose distance is y (small compared to Earth\u2019s radius R). Show that if no allowance is made for the e\ufb00ects of the Coriolis force, the projectile will miss its target by a distance x = \u03c9 2y3 g 1/2 (cos \u03b8 \u2212 1 3 sin \u03b8). Evaluate this distance if \u03b8 = 45\u25e6 and y = 40 km. Why is it that the deviation is to the east near the north pole, but", " to the west both on the equator and near the south pole? (Neglect atmospheric resistance.) 16. *Solve the problem of a particle falling from height h above the equator by using an inertial frame, and verify that the answer agrees with that (Hint : Use equations (3.48). Recall found using a rotating frame. Fig. 5.8.) 17. Find the equations of motion for a particle in a frame rotating with variable angular velocity \u03c9, and show that there is another apparent force of the form r. Discuss the physical origin of this force. m \u02d9\u03c9 18. Find the equation of motion for a particle in a uniformly accelerated frame, with acceleration a. Show that for a particle moving in a uniform gravitational \ufb01eld, and subject to other forces, the gravitational \ufb01eld may be eliminated by a suitable choice of a. \u2212 \u2227 19. *The co-ordinates (x, y, z) of a particle with respect to a uniformly rotating frame may be related to those with respect to a \ufb01xed inertial 128 Classical Mechanics frame, (x\u2217, y\u2217, z\u2217), by the transformation x y z \u23a1 \u23a3 \u23a4 \u23a6 = \u23a1 \u2212 cos \u03c9t sin \u03c9t 0 sin \u03c9t cos \u03c9t 0 1 0 0 x\u2217 y\u2217 zHere, we use matrix notation: this stands for three separate equations, x = cos \u03c9t x\u2217 + sin \u03c9t y\u2217, \u00b7 \u00b7 etc.) Write down the inverse relation giving (x\u2217, y\u2217, z\u2217) in terms of (x, y, z). By di\ufb00erentiating with respect to t, rederive the relation (5.15) between d2r/dt2 and \u00a8r. [Hint : Note that \u00a8r = (\u00a8x, \u00a8y, \u00a8z), while d2r/dt2 is the vector obtained by applying the above transformation to (\u00a8x\u2217, \u00a8y\u2217, \u00a8z\u2217).] 20. Another way of deriving the equation of motion (5.16) is to use Lagrange\u2019s equations. Express the kinetic energy 1 2 m(dr/dt)2 in terms of (x, y, z), and show that Lagrange\u2019s equations (3.44) reproduce (5.16) for the case where the force is conservative. Chapter 6 Potential Theory This chapter is complementary to the preceding ones. In it we shall discuss not the problem of determining the motion of a particle under known forces, but the problem of \ufb01nding the forces from a knowledge of the positions of other bodies. We deal speci\ufb01cally with gravitational and electrostatic forces obeying the inverse square law, which are determined by the positions of other masses and charges. We return here to the convention whereby \u02d9r = dr/dt. We shall not use rotating frames again until Chapter 9. 6.1 Gravitational and Electrostatic Potentials The gravitational potential energy of a particle of mass m moving in the \ufb01eld of a \ufb01xed mass m at r is. If we have several masses Gmm/ | \u2212 mj, located at the points rj, then the potential energy is the sum r | r \u2212 V (r) = \u2212 Gmmj rj| r | \u2212. j (The fact that the potential energies add follows from the additive property of forces.) Since the mass m appears only as an overall factor, we may de\ufb01ne the gravitational potential \u03a6(r) to be the potential energy per unit mass so that V (r) = m\u03a6(r), \u03a6(r) = \u2212 Gmj rj| r | \u2212. j 129 (6.1) (6.2) 130 Classical Mechanics Note that \u03a6 is always negative. (The potential has sometimes been de\ufb01ned to be minus the potential energy per unit mass, and therefore positive. We prefer, however, to retain the direct correspondence between potential and potential energy, so that particles tend to move towards regions of lower potential.) The acceleration of a particle moving under gravitational forces is given by m\u00a8r = \u2212 \u2207V (r) = \u2212 m\u2207\u03a6(r). Since this acceleration is independent of the mass m, we may de\ufb01ne the gravitational acceleration or gravitational \ufb01eld g(r) by g(r) = \u2207\u03a6(r), \u2212 (6.3) Thus (6.2) and (6.3) are all that is needed to calculate the acceleration induced in a particle by a given distribution of masses. The electrostatic case is very similar. We de\ufb01ne the electrostatic", " poten- tial \u03c6(r) to be the electrostatic potential energy per unit charge, V (r) = q\u03c6(r). The potential due to charges qj at rj is then \u03c6(r) =. rj| The acceleration of a charged particle is given by j \u2212 qj r 4\u03c00| where the electric \ufb01eld E(r) is de\ufb01ned by m\u00a8r = qE, E(r) = \u2207\u03c6(r), \u2212 (6.4) (6.5) (6.6) Note that the acceleration depends on the charge-to-mass ratio q/m. (The corresponding ratio in the gravitational case is the ratio of gravitational to inertial mass, which is of course a constant, set equal to unity in conventional units. This proportionality is the content of the equivalence principle.) Unlike the gravitational potential, the electrostatic potential \u03c6 may have either sign, because both positive and negative charges exist. Potential Theory 131 We can now forget about the particle at r, and concentrate on the problem of calculating the potential from given information about the positions of the masses or charges. 6.2 The Dipole and Quadrupole The electric dipole consists of two equal and opposite charges, q and q, placed close together, say at a and at the origin, respectively. (See Fig. 6.1.) The potential is \u2212 +q a \u03b8 \u2212q r\u2212a r Fig. 6.1 \u03c6(r) = q r 4\u03c00| | We shall assume that a r, so that we may expand by the binomial theorem. Since we shall need the result later, we evaluate the terms up to order a2, but neglect a3. If \u03b8 is the angle between a and r, then q r 4\u03c00| (6.7 = r2 \u2212 2ar cos \u03b8 + a2. Hence /2 \u2212 a r cos \u03b8 + a2 r2 2 a r \u2212 cos \u03b8 + a2 r2 a r2 cos \u03b8 + a2 r3 ( 3 2 cos2 ) +. \u00b7 \u00b7 \u00b7 \u2212 cos \u03b8 + a2 r2 2 \u2212 \u00b7 \u00b7 \u00b7 (6.8) (The general term in this series is of the form (al/rl+1)Pl(cos \u03b8), where Pl is a polynomial, known as the Legendre polynomial of degree l. Here, 132 Classical Mechanics however, we shall only need the terms up to l = 2.) In vector notation, we may write (6.8) as r3 + \u00b7 3(a r)2 2r5 \u2212 \u00b7 a2r2 +. \u00b7 \u00b7 \u00b7 (6.9) We now return to (6.7). Keeping only the linear term in a and neglecting a2, we can write it as d r 4\u03c00r3 = \u00b7 where d is the electric dipole moment, d = qa. The corresponding electric \ufb01eld, given by (6.6), has the spherical polar components (see (A.50)) d cos \u03b8 4\u03c00r2, \u03c6(r) = (6.10) Er = E\u03b8 = E\u03d5 = \u2202\u03c6 \u2202r 1 r \u2202\u03c6 \u2202\u03b8 1 r sin \u03b8 \u2212 \u2212 \u2212 = = 2d cos \u03b8 4\u03c00r3, d sin \u03b8 4\u03c00r3, \u2202\u03c6 \u2202\u03d5 = 0. (6.11) (To avoid any possible risk of confusion, we denote the potential and the azimuth angle by the distinct symbols \u03c6 and \u03d5.) This \ufb01eld is illustrated in Fig. 6.2, in which the solid lines are \ufb01eld lines, drawn in the direction of E, and the dashed lines are the equipotential surfaces, \u03c6 = constant. Note that the two always intersect at right angles. We can repeat the process of putting two charges (or monopoles) together to form a dipole, by putting two dipoles together to form a quadrupole. If we place oppositely oriented dipoles, with dipole moments d and d, at a and at the origin, then the potential is \u2212 \u03c6(r) = (r d \u00b7 r 4\u03c00| \u2212 \u2212 a) a | 3 \u2212 d r 4\u03c00r3. \u00b7 Now, neglecting terms of order a2, we have r | Hence, to this approximation a | 3 \u2248 \u2212 1 1 r3 1 2 a r \u00b7 r2 \u2212 3/2 \u2212 1 r3 + 3 a r r5. \u00b7 \u2248 \u03c6(r) = 3(d \u00b7 r)(a r) \u00b7 \u2212 4\u03c00r5 (d \u00b7 a)r2. (6.12) Potential", " Theory 133 Fig. 6.2 In the special case, where the dipoles are placed end-on, so that d is parallel to a, we may take this common direction to be the z-axis, and obtain \u03c6(r) = Q 16\u03c00r3 (3 cos2 \u03b8 1), \u2212 (6.13) where Q is the electric quadrupole moment, Q = 4da. (The factor of 4 is purely conventional, and serves to simplify some of the later formulae.) The corresponding electric \ufb01eld is Er = E\u03b8 = 3Q 16\u03c00r4 (3 cos2 \u03b8 3Q 8\u03c00r4 cos \u03b8 sin \u03b8, \u2212 1), (6.14) E\u03d5 = 0. It is illustrated in Fig. 6.3. 134 Classical Mechanics Fig. 6.3 6.3 Spherical Charge Distributions When we have a continuous distribution of charge (or mass) we can replace the sum in (6.5) by an integral: \u03c6(r) = \u03c1(r) r 4\u03c00| where \u03c1(r) is the charge density, and d3r denotes the element of volume. In spherical polars, the volume element is the product of the elements of length in the three co-ordinate directions, namely (see d3r, (6.15) r| 3.5) \u2212 \u00a7 d3r = dx dy dz = r 2dr sin \u03b8d\u03b8 d\u03d5. Note that \u03b8 ranges from 0 \u03c0, and \u03d5 from 0 We shall consider in this section spherically symmetric distributions of charge, for which \u03c1 is a function only of the radial co-ordinate r. Let us \ufb01rst take a uniform thin spherical shell of charge density \u03c1, radius a and thickness da. Choosing the direction of r to be the z-axis, we can write 2\u03c0. \u2192 \u2192 Potential Theory 135 (6.15) as \u03c6(r) = \u03c1a2da 4\u03c00 (r2 \u2212 sin \u03b8 d\u03b8 d\u03d5 2ar cos \u03b8 + a2)1/2. The \u03d5 integration gives a factor of 2\u03c0. The \u03b8 integration can easily be performed by the substitution u = cos \u03b8, and yields \u03c6(r) = \u03c1a2da (r + a) r \u2212 | 20ar a |. \u2212 We now have to consider separately the cases r > a and r < a. In terms of the total charge dq = 4\u03c0\u03c1a2da of the spherical shell, we obtain \u03c6(r) = \u23a7 \u23aa\u23a8 dq 4\u03c00r dq 4\u03c00a,, r > a, r < a. (6.16) Outside the shell, the potential is the same as that of a charge dq concentrated at the origin, and the electric \ufb01eld is E = \u02c6r dq/4\u03c00r2. Inside, the potential is a constant, and the electric \ufb01eld vanishes. \u23aa\u23a9 It is now easy to \ufb01nd the potential of any spherically symmetric distribution, by summing over all the spherical shells. It is clear that the electric \ufb01eld at a distance r from the centre is equal to that of a point charge located at the origin, whose magnitude is the total charge enclosed within a sphere of radius r. (A similar result holds in the gravitational case.) A simple example is a uniformly charged sphere. Example: Uniformly charged sphere What is the potential and electric \ufb01eld for a sphere of radius a, with constant charge density \u03c1 inside it? For r > a the potential is clearly just \u03c6(r) = q 4\u03c00r, with q = \u03c0a3\u03c1. 4 3 Inside the sphere, we have to separate the contributions from the regions r < r and r > r, and obtain \u03c6(r) = r 2 \u03c1r 0r 0 dr + a \u03c1r 0 dr. r 136 Classical Mechanics Performing the integrations, we \ufb01nd \u03c6(r) = \u23a7 \u23aa\u23a8, q 4\u03c00r q 4\u03c00 3 2a \u2212 r2 2a3, r > a, r < a. (6.17) This potential is illustrated in Fig. 6.4. The corresponding electric \ufb01eld is \u23aa\u23a9 \u03c6 q 4\u03c0\u03b5 a 0 a Fig. 6.4 r \u02c6r r > a, q 4\u03c00r2, qr 4\u03c00a3, Thus, outside the sphere, the \ufb01eld obeys the inverse square law. Inside, it increases linearly from zero", " at the centre to the surface value q/4\u03c00a2. E(r) = \u23a7 \u23aa\u23a8 r < a. (6.18) \u23aa\u23a9 \u02c6r So far as its e\ufb00ect on outside bodies is concerned, any spherically symmetric distribution of charge or mass may be replaced by a point charge or mass located at the centre. This is the justi\ufb01cation for treating the Sun, for example, as a point mass in discussing the motion of the planets. Of course, any deviation from spherical symmetry will lead to a modi\ufb01cation in the inverse square law. We shall discuss the nature of this correction later. Potential Theory 137 6.4 Expansion of Potential at Large Distances It is only for simple cases that we can calculate the potential exactly. In general, we must use approximation methods. In particular, we are often interested in the form of the potential at distances from the charge or mass distribution which are large compared to its dimensions. Then in (6.5) or (6.15), r is much larger than rj or r, and we may use the series expansion (6.8) to obtain an expansion of the potential in powers of 1/r: \u03c6(r) = \u03c60(r) + \u03c61(r) + \u03c62(r) +. \u00b7 \u00b7 \u00b7 The leading (\u2018monopole\u2019) term is \u03c60(r) = q 4\u03c00r, (6.19) (6.20) where q is the total charge q = qj or q = \u03c1(r) d3r. (6.21) j Thus at very great distances, only the total charge is important, not the shape of the distribution. This is the potential we should get for an exactly spherically symmetric distribution. Thus the further terms in the series (6.19) may be regarded as measuring deviations from spherical symmetry. The next term in (6.19) is \u03c61(r) = d r 4\u03c00r3, \u00b7 (6.22) where d is the dipole moment, de\ufb01ned by d = qjrj or d = \u03c1(r)r d3r. (6.23) j If we shift the origin through a distance R, the total charge is clearly una\ufb00ected, but the dipole moment changes to d = j qj(rj \u2212 R) = d qR. \u2212 (6.24) Thus, if the total charge is non-zero, we can always make the dipole moment vanish by choosing the origin to lie at the centre of charge R = = d q. qjrj qj \" \" (6.25) 138 Classical Mechanics In this case, the dipole moment gives us information about the position of the centre of charge. However, if the total charge is zero, then by (6.24) the dipole moment is independent of the choice of origin \u2014 the electric dipole and quadrupole discussed in the preceding section are examples of this. Note that in the gravitational case, the total mass can never vanish, so we can always ensure that the gravitational \u2018dipole moment\u2019 is zero by choosing our origin at the centre of mass. The \u2018quadrupole\u2019 term in the potential (6.19), obtained by substituting the third term of (6.9) into (6.15), is \u03c62(r) = 3(r \u03c1(r) r2r 2 \u00b7 r)2 \u2212 8\u03c00r5 d3r. (6.26) One can write this expression in a form similar to (6.12). However, we shall not consider the general case here (see Appendix A, Problem 17), but restrict the discussion to the special case of axial symmetry, where \u03c1 depends on r and \u03b8 (or \u03c1 and z) but not on \u03d5. This case is of particular interest in connection with the gravitational potential of the Earth, which is to a good approximation axially symmetric, though \ufb02attened at the poles. We shall see that in the case of axial symmetry, (6.26) can be written in a form similar to (6.13) with a suitably de\ufb01ned quadrupole moment Q. To do this, we examine the numerator of the integrand in (6.26). Written out in terms of components, it reads x2(2x 2 2 y \u2212 \u2212 z 2) + y2(2y 2 \u2212 2 x \u2212 z 2) + z2(2z 2 2 x y 2) \u2212 \u2212 + 6xyxy + 6xzxz + 6yzyz. (6.27) Now, because of ax", "ial symmetry, any integral involving an odd power of x or y will vanish, for example \u03c1(r)xz d3r = 0, because the contributions from (x, y, z) and ( Moreover, axial symmetry also implies that \u2212 x, y, z) exactly cancel. \u03c1(r)x 2 d3r = \u03c1(r)y 2 d3r, since the x- and y-axes are in no way distinguished from one another. Hence 2 without we can replace 2x a\ufb00ecting the value of the integral. When we do this, we see that the three 2 in the integrand by 1 2 + y 2 (x 2) y z z \u2212 \u2212 \u2212 2 2 Potential Theory 139 remaining terms of (6.26) all involve the same integral, namely Q = \u03c1(r)(2z 2 2 x \u2212 \u2212 y 2) d3r. (6.28) Thus (6.26) can be written \u03c62(r) = 2z2 x2 y2 16\u03c00r5 Q = \u2212 \u2212 Q 16\u03c00r3 (3 cos2 \u03b8 1). \u2212 (6.29) This is identical with the potential (6.12) of an axially symmetric quadrupole. The quantity Q de\ufb01ned by (6.28) is the quadrupole moment of the distribution. Example: Linear quadrupole Two equal charges q are positioned at the points (0, 0, a charge write down the potential and electric \ufb01eld at large distances. a), and 2q at the origin. Find the quadrupole moment, and \u2212 \u00b1 Here the central particle contributes nothing to Q, and the other two each give the same contribution, so Q = 2q(2a2) = 4qa2. The potential and electric \ufb01eld at large distances are given by (6.13) and (6.14) with this value of Q. Thus we see that a better approximation than treating a charge distribution as a single point charge is to treat it as a point charge plus an electric quadrupole of moment given by (6.28). For a spherically symmetric charge 2 all yield the distribution, Q = 0, because the integrals over x same value. The value of Q may be regarded as a measure of the \ufb02attening of the distribution. For a distribution with uniform positive charge density, Q is positive if the shape is prolate (egg-shaped, with the z-axis longer than the others), and negative if it is oblate, like the Earth. In particular, for a spheroid (ellipsoid of revolution), with uniform charge density and semiaxes a, a, c, we can evaluate the integral in (6.28) explicitly. We shall have occasion to evaluate a very similar integral in connection with moments of inertia in Chapter 9, and therefore omit the details here. The result is 2 and z 2, y Q = 2 5 q(c2 a2). \u2212 (6.30) 140 Classical Mechanics 6.5 The Shape of the Earth The Earth is approximately an oblate spheroid whose equatorial radius a exceeds its polar radius c by about 21.4 km. Its oblateness is de\ufb01ned to be = a c \u2212 a \u2248 1 297. (6.31) Its gravitational potential is therefore not precisely the inverse-square-law GM/r. The most important correction is the \u2018quadrupole\u2019 potential, term, \u2212 \u03a62(r) = GQ 4r3 (3 cos2 \u03b8 \u2212 \u2212 1). Since the Earth is oblate, Q here is actually negative. If the density of the Earth were uniform, the quadrupole moment would be given simply by (6.30): Q = 2 5 M (a2 c2) 4 5 M a2. \u2212 \u2248 \u2212 \u2212 ), and dropped terms of order 2. In fact, Here we have used c = a(1 the density is considerably greater near the centre than at the surface, so that large values of r contribute proportionately less to the integral (6.28). Thus we must expect Q to be rather smaller in magnitude than this value (see Problem 15). It will be convenient to write \u2212 Q = \u2212 2M a2J2, where J2 is a dimensionless parameter, somewhat less than 2 5, de\ufb01ning the magnitude of the quadrupole moment. (In general, Jl parametrizes the term in \u03a6 proportional to Pl(cos \u03b8) \u2014 see the note following Eq. (6.8).) Then the gravitational potential becomes \u03a6(r) GM r + GM a2J2 2r3 \u2248 \u2212 (3 cos2 \u03b8 1). \u2212 (6.32) We shall see that", " both and J2 can be determined from measurements of the acceleration due to gravity and the use of an equilibrium principle. The \ufb02attening of the Earth is in fact a consequence of its rotation. Over long periods of time, the Earth is capable of plastic deformation, and behaves more nearly like a liquid than a solid \u2014 and of course the larger part of the surface is covered by a liquid. Although its crust does have some rigidity, it could not preserve its shape for long unless it were at Potential Theory 141 least approximately in equilibrium under the combined gravitational and centrifugal forces. Now, as in the problem discussed at the end of 5.3, the surface of a liquid in equilibrium under conservative forces must be an equipotential surface, for otherwise there would be a tendency for the liquid to \ufb02ow towards regions of lower potential. We can include the centrifugal force by adding to the potential a term \u00a7 \u03a6cent(r) = 1 2 \u03c92(x2 + y2) = 1 2 \u03c92r2 sin2 \u03b8. \u2212 \u2212 The equipotential equation is then \u03a6(r) + \u03a6cent(r) = constant, (6.33) with \u03a6 given by (6.32). Thus equating the values at the pole (\u03b8 = 0) and on the equator (\u03b8 = \u03c0/2), we obtain the equation GM c \u2212 + GM a2J2 c3 = GM GM J2 \u2212 a \u2212 2a \u2212 1 2 \u03c92a2. Since and J2 are both small quantities, we may neglect 2 and J2. Thus, in the quadrupole term, we may ignore the di\ufb00erence between c and a. In the inverse-square-law term, we can write 1/c (1/a)(1 + ). Hence to \ufb01rst order we \ufb01nd \u2248 GM a \u2212 + 3GM J2 2a = \u2212 1 2 \u03c92a2. Writing g0 for the gravitational acceleration without the centrifugal term, g0 = GM/a2, and multiplying by 2/ag0, we obtain \u03c92a g0 = 2 3J2. \u2212 (6.34) If we made the approximation of treating the Earth as of uniform den5, then this equation would determine the 1/230, which is sity, and therefore set J2 = 2 oblateness in terms of the angular velocity. This yields appreciably larger than the observed value. \u2248 On the other hand, if we regard and J2 as independent, then we need another relation to \ufb01x them both. Such a relation can be found from the \u2207\u03a6 measured values of g. (See 5.3.) Now, the gravitational \ufb01eld g = \u00a7 \u2212 142 Classical Mechanics corresponding to the potential (6.32) is gr = g\u03b8 = GM r2 + \u2212 3GM a2J2 r4 3GM a2J2 2r4 cos \u03b8 sin \u03b8. (3 cos2 \u03b8 1), \u2212 (6.35) Note that g\u03b8 is directed away from the poles, and towards the equator, as one might expect from thinking of it as due to the attraction of the gr is decreased at the equatorial bulge. The inward radial acceleration poles, and increased at the equator, by the quadrupole term, so that the di\ufb00erence between the two values is not as large as the inverse square law would predict. \u2212 At the poles, and on the equator, g\u03b8 = 0, and the magnitude of g is equal to \u2212 gr. Thus we \ufb01nd gpole = geq = GM c2 \u2212 GM a2 + 3GM a2J2 c4 3GM J2 2a2 \u2248 \u2248 g0(1 + 2 3J2) \u2212 g0(1 + 3 2 J2), making the same approximations as before. Thus, for the di\ufb00erence, we \ufb01nd the expression \u2206g g0 = gpole \u2212 g0 geq = 2 9 2 \u2212 J2. (6.36) For the value including the centrifugal term, \u2206g\u2217 = \u2206g + \u03c92a, we have to add (6.34) to (6.36). This yields \u2206g\u2217 g0 = 4 15 2 \u2212 J2. (6.37) Using the known values quoted in (5.18) and (5.19), we may solve the simultaneous equations (6.34) and (6.37) for and J2. This yields = 0.0034, J2 = 0.0011. The value of is in good agreement with the the oblateness deduced from direct measurements of the polar and equatorial radii.", " As expected, the value of J2 is somewhat less than 2 5. The relation between the two is determined by the density distribution within the Earth. (See Problem 15.) One consequence of the fact that the Earth\u2019s surface is approximately an equipotential surface should be noted. Because of the general property Potential Theory 143 that the force \ufb01eld is always perpendicular to the equipotentials, it means that the e\ufb00ective g\u2217, which is the \ufb01eld derived from \u03a6 + \u03a6cent, is always perpendicular to the Earth\u2019s surface. Thus, although a plumb line does not point towards the Earth\u2019s centre, it is perpendicular to the surface at that point \u2014 apart, of course, from minor deviations caused by mountains and the like. (Compare Chapter 1, Problem 10.) Satellite orbits The \u2018quadrupole\u2019 term in the Earth\u2019s gravitational \ufb01eld has two important e\ufb00ects on the orbit of a close arti\ufb01cial satellite. Indeed, observations of such orbits provide the most reliable means of measuring its magnitude (and those of the even smaller higher-order correction terms). They have revealed quite substantial deviations from spheroidal shape and from hydrostatic equilibrium: the Earth\u2019s surface is only rather approximately an equipotential. The \ufb01rst e\ufb00ect of the quadrupole term, arising mainly from the deviation of the radial component of g from the inverse square law, is a precession of the major axis of the orbit within the orbital plane. (Compare Chapter 4, Problem 27.) The major axis precesses in the forward direction for orbits of small inclination to the equator, and in the retrograde direction for orbits (The di\ufb00erence is a with inclination greater than arcsin re\ufb02ection of the di\ufb00ering sign of the quadrupole contribution to gr on the equator and near the poles.) 4/5 = 63.4\u25e6. The second e\ufb00ect occurs because the force is no longer precisely central, so that the angular momentum changes with time according to dJ dt = mr g. \u2227 (6.38) Thus, as in the case of the Larmor e\ufb00ect discussed in 5.6, the orbital plane precesses around the direction of the Earth\u2019s axis. The rate of precession may be calculated by a method very similar to the one used there (see Problem 25). The precession is in fact greater for orbits of small inclination, and is zero for an orbit passing over the poles. For a close satellite with small inclination, it can be nearly 10\u25e6 per day. It is always in a retrograde sense (opposite to the direction of revolution of the satellite in its orbit). \u00a7 Both e\ufb00ects are strongly dependent on the radius of the satellite orbit. 7/2. At the radius of the In fact, the rate of precession decreases like r\u2212 Moon\u2019s orbit, the precessional angular velocity is only a few seconds of arc 144 Classical Mechanics per year. (The Moon\u2019s orbit does precess, at about 19\u25e6 per year, but this is a consequence of the non-uniformity of the Sun\u2019s gravitational \ufb01eld, not of the shape of the Earth. See Problem 26.) Planetary precession It should be noted that a small oblateness of the Sun produces a slow precession of the Kepler orbits around it. The major part of the precession of planetary orbits is due to the gravitational action of the planets on each other, leading for example in the case of Mercury to a predicted precession of the axes of the ellipse of about 531 seconds of arc every century. These e\ufb00ects can be modelled quite well for each planet by smearing the mass of each of the other planets into an appropriate coplanar uniform circular ring. The modi\ufb01cations to the inverse square law potential depend, of course, on whether the planetary distance being considered is inside or outside the ring. For Mercury (alone) there is an observed residual precession of 43 per century, not explainable by the Newtonian theory. In fact, general relativity does explain this precession \u2014 indeed that it does so is one of the celebrated tests of the theory. 6.6 The Tides Tidal forces arise because the gravitational attraction of the Moon, and to a lesser extent of the Sun, is not uniform over the surface of the Earth. The attraction is stronger than average on the side of the Earth facing the Moon, and weaker than average on the far side, so there is a tendency for the Earth to be elongated along the line of centres. Let a be the position of the Moon relative to the Earth\u2019s centre, and consider a point r on the Earth", ". The potential at this point is (see Fig. 6.5) \u03a6(r) = Gm a r | \u2212 |, \u2212 (6.39) a, we may expand in where m is the mass of the Moon. Now, since r powers of r/a. (This is the reverse of the situation encountered in 6.2, where r was much larger than a.) Taking the direction of the Moon to be \u00a7 Potential Theory 145 h(\u03b8) E r \u03b8 a M Fig. 6.5 the z-direction, and using (6.8), we obtain \u03a6(r) = Gm \u2212 + r a2 cos \u03b8 + r2 a3 ( 3 2 cos2 \u03b8 1 a 1 2 ) + \u2212. \u00b7 \u00b7 \u00b7 The \ufb01rst term is a constant and does not yield any force. The second may be written \u03a61(r) = Gm a2 z. \u2212 It yields a uniform gravitational acceleration Gm/a2 directed towards the Moon. This term therefore describes the major e\ufb00ect of the Moon\u2019s gravitational force, which is to accelerate the Earth as a whole. It is irrelevant in discussing the phenomenon of the tides, since we are interested in motion relative to the centre of the Earth. In fact, we can eliminate it by going over to an accelerated frame, the frame in which the Earth\u2019s centre of mass is at rest. (See Chapter 5, Problem 18.) The important term for our purposes is the quadratic term, which leads to a gravitational \ufb01eld gr = g\u03b8 = Gmr a3 (3 cos2 \u03b8 3Gmr a3 \u2212 cos \u03b8 sin \u03b8. 1), \u2212 (6.40) This \ufb01eld is directed outwards along the z-axis, towards and away from the Moon, and inwards in the xy-plane, as expected. It should be noted that the \ufb01eld described by (6.40) is much weaker even than the corrections to the Earth\u2019s \ufb01eld discussed in the preceding section. (If it were not, it would show up in measurements of g.) It is easy to compute its magnitude. It is smaller than g0 = GM/r2 by the factor 146 Classical Mechanics mr3/M a3. For the Moon, m/M = 1/81.3 and r/a = 1/60.3, so that this factor is mr3 M a3 = 5.60 \u00d7 10\u2212 8. (6.41) If we denote the mass of the Sun by m and its mean distance, the semimajor axis of the Earth\u2019s orbit, by a, then the corresponding values for 5. Thus the factor the Sun are m/M = 3.33 in that case is 105 and r/a = 4.26 10\u2212 \u00d7 \u00d7 mr3 M a 3 = 2.57 10\u2212 8. \u00d7 (6.42) By a remarkable coincidence (unique in the solar system), these two \ufb01elds are of roughly the same order of magnitude: the e\ufb00ect of the Sun is rather less than half that of the Moon. The only reason why these very small \ufb01elds can lead to signi\ufb01cant e\ufb00ects is that they change with time. As the Earth rotates, the value of \u03b8 at any point on its surface varies. Thus the \ufb01eld described by (6.40) oscillates periodically with time. Because of the symmetry of (6.40) in the central plane, the main term has an oscillation period of 12 hours in the case of the Sun, and slightly more for the Moon, because of its changing position. This explains the most noticeable feature of the tides \u2014 their twice-daily periodicity. Unless the Moon is directly over the equator, however, a point on the Earth that passes directly beneath it will not also pass directly opposite, so that the two daily tides may be of unequal height. In other words, there is an additional term in the \ufb01eld with a period of 24 rather than 12 hours. At new moon or full moon, the Sun and Moon are acting in the same direction, and the tides are unusually high; these are the spring tides. On the other hand, at the \ufb01rst and third quarters, when the Sun and Moon are at right angles to each other, their e\ufb00ects partially cancel, and we have the relatively low neap tides. (See Fig. 6.6.) Measurements of the relative heights of the tides at these times provided one of the earliest methods (A more accurate method will be of estimating the mass of the Moon. discussed in the next chapter.) To get some idea of the order of magnitude of the e", "\ufb00ects produced by the tidal \ufb01eld, (6.40), let us make the simplest possible assumptions. We consider a perfectly rigid solid Earth, completely covered by ocean, and suppose that the natural periods of oscillation are short compared to the In that case, the rotation is slow enough for the water rotation period. Potential Theory 147 S S M M (a) Spring tides (b) Neap tides Fig. 6.6 to reach equilibrium under the combined forces of Earth and Moon. The problem is then one in hydrostatics rather than hydrodynamics. In equilibrium, the surface of the water must be an equipotential surface. We can thus calculate the height h(\u03b8) through which it is raised, as a function of the angle \u03b8 to the Moon\u2019s direction (see Fig. 6.5). Since h(\u03b8) is certainly small, the change in the Earth\u2019s gravitational potential is approximately g0h(\u03b8). This change must be balanced by the potential due to the Moon, so that g0h(\u03b8) = Gmr2 a3 2 cos2 \u03b8 ( 3 1 2 ). \u2212 Thus, using g0 = GM/r2, we \ufb01nd h(\u03b8) = h0( 3 2 cos2 \u03b8 1 2 ), \u2212 where h0 = mr4 M a3. (6.43) This h0 is just (6.41) times an extra factor of r. Thus, using r = 6370 km, we \ufb01nd, for the Moon, h0 = 0.36 m, and for the Sun, h0 = 0.16 m. These \ufb01gures must not be taken as more than rough order-of-magnitude estimates of the height of the tides. Even on a completely ocean-covered Earth, there would be important modifying e\ufb00ects. First, we have neglected the gravitational attraction of the ocean itself. The tidal bulges exert an attraction that tends to increase h0 slightly, in fact by about 12 per cent (see Problem 19). More important, however, is the fact the Earth is not perfectly rigid, and is itself distorted by tidal forces. Because the observed tidal range refers not to absolute height but to the relative heights of sea surface and sea \ufb02oor, this reduces the e\ufb00ective height by a substantial factor. Indeed, in the extreme case of a \ufb02uid Earth enclosed by a completely \ufb02exible crust, there would be essentially no observable tide at all. 148 Classical Mechanics The values obtained from (6.43) may seem rather small, especially when further reduced by the e\ufb00ect just mentioned. However, it is important to remember that they refer to the unrealistic case of an Earth without continents. The observed tidal range in mid-ocean (which can be measured by ranging from satellites) is in fact quite small, normally less than a metre. Large tides are a feature of continental shelf areas, and are strongly dependent on the local topography. Note too that the Moon\u2019s attraction at any point on the Earth\u2019s surface is a periodic force, whose e\ufb00ect may 2.6. be greatly enhanced by the phenomenon of resonance, discussed in The natural periods of oscillation of a body of water depend on a variety of factors, including its size, shape and depth, and there are critical values for which one of the periods is close to 12 hours. Then, because the damping produced by tidal friction is normally quite small except in very shallow water, very large tides can be set up. \u00a7 Another e\ufb00ect of resonance is to delay the oscillations, so that they are not precisely in phase with the applied force. At resonance, the phase lag is a quarter period (about three hours in this case). Thus we should not normally expect the times of high tide to coincide with the times when the Moon is overhead, or directly opposite. In practice, the times are determined in a very complicated way by the actual shapes of the oceans. 6.7 The Field Equations It is often convenient to obtain the electric or gravitational potential by solving a di\ufb00erential equation rather than performing an integration. Although we shall not in fact need to use this technique in this book, we include a short discussion of it because of its very important role in more advanced treatments of mechanics and electromagnetic theory. To \ufb01nd the relevant equations, we consider \ufb01rst a single charge q located at the origin. The electric \ufb01eld is then E = q 4\u03c00r2 \u02c6r. Now let us consider a closed surface S surrounding the charge, and evaluate the surface integral of the component of E normal to the surface, n dS = E \u00b7 S q 4\u03c00 S \u02c6r n r2 dS, \u00b7 Potential Theory 149", " where dS is the element of surface area, and n is a unit vector normal to the surface, directed outwards (see Fig. 6.7). If \u03b1 is the angle between \u02c6r n \u03b1 E dS q d\u2126 Fig. 6.7 \u00b7 and n, then \u02c6r n dS = dS cos \u03b1 is the projection of the area dS on a plane normal to the radius vector r. Hence it follows from the de\ufb01nition of solid n dS/r2 is equal to the solid angle d\u2126 subtended angle (see (4.42)) that \u02c6r at the origin by the element of area dS. Hence \u00b7 n dS = E \u00b7 S q 4\u03c00 d\u2126 = q 0, (6.44) since the total solid angle subtended by a closed surface around the origin is 4\u03c0. (It is to ensure that the factor of 4\u03c0 here cancels that Coulomb\u2019s law in SI units is written with an explicit factor of 1/4\u03c0.) We have implicitly assumed that each radial line cuts the surface only once. However, we can easily remove this restriction. Evidently, a radial line must always cut it in an odd number of points. (See Fig. 6.8.) Moreover, the contributions to the surface integral from points where it goes into, rather n = cos \u03b1 is than out of, the surface are all of opposite sign (because \u02c6r negative at these points) and of equal magnitude. Thus all the terms but one will cancel, and we obtain the same answer (6.44). \u00b7 Similarly, if we consider a surface S which does not enclose the origin, we can see that the surface integral will vanish. For each radial line cuts S in an even number of points, half in each direction, so that all contributions will cancel. 150 Classical Mechanics n n n q Fig. 6.8 It is now clear how we may generalize this result to an arbitrary distribution of charges. Any charge q located anywhere within a closed surface S will contribute to the surface integral an amount q/0, while charges outside contribute nothing. Therefore, the value of the integral is 1/0 times the total charge enclosed within the surface. If we have a continuous distribution of charge, with charge density \u03c1(r), then n dS = E \u00b7 S 1 0 V \u03c1(r) d3r, (6.45) where V is the volume enclosed by the surface S. This is often called Gauss\u2019s law. Now, by the theorem of Gauss (see (A.36)), the surface integral is equal to the volume integral of the divergence of E: n dS = E \u00b7 S V E d3r. \u2207 \u00b7 Hence (6.45) may be written (\u2207 E \u00b7 \u2212 1 0 \u03c1) d3r = 0. \u2212 V This equation must be true for an arbitrary volume V, and therefore the integrand itself must vanish: E = \u2212 1 0 \u03c1. \u2207 \u00b7 (6.46) Potential Theory 151 This is the di\ufb00erential form of Gauss\u2019s law. The electric \ufb01eld is in fact completely determined (except for a possible additive uniform \ufb01eld extending over the whole of space) by this equation together with the condition that it be a conservative \ufb01eld, E = 0. \u2207 \u2227 (6.47) This latter equation guarantees the existence of the potential \u03c6 related to 6.6.) Substituting in (6.46), we obtain Poisson\u2019s E by E = \u00a7 \u2212 equation for the potential, \u2207\u03c6. (See \u22072\u03c6 = 1 0 \u03c1. \u2212 \u2212 (6.48) The main importance of these equations lies in the fact that they provide local relations between the potential or \ufb01eld and the charge density. The expression (6.15) for \u03c6 in terms of \u03c1 is non-local, in the sense that it expresses the potential \u03c6 at one point as an integral involving the charge density at all points of space. To use this equation, we must know \u03c1 everywhere. Frequently, however, we are interested in the \ufb01eld in some restricted region of space, where we do know the value of \u03c1, while, instead of having information about \u03c1 outside this region, we have some conditions on the \ufb01elds at the boundary. This type of problem may be solved by looking for solutions of (6.48) with appropriate boundary conditions. A particularly important special case is that in which \u03c1 vanishes inside the region of interest. Then (6.48) becomes Laplace\u2019s equation, \u22072\u03c6 = 0. (6.49) The gravitational case is of course entirely similar.", " The \ufb01eld equations are where \u03c1 is now the mass density, and g = \u2207 \u00b7 \u2212 4\u03c0G\u03c1, g = 0. \u2207 \u2227 (6.50) (6.51) Note the explicit appearance of the factor of 4\u03c0 on the right hand side of (6.50), which arises because the gravitational force is conventionally expressed in terms of unrationalized units, i.e., there is no factor of 1/4\u03c0 in (1.5). Note also the di\ufb00ering signs in (6.46) and (6.50), which again re\ufb02ect the di\ufb00erence between (1.5) and (1.6). 152 Classical Mechanics Poisson\u2019s equation for the gravitational potential is \u22072\u03a6 = 4\u03c0G\u03c1. (6.52) 6.8 Summary The great advantage of calculating the potential, rather than the \ufb01eld itself directly, is that it is much easier to add scalar quantities than vector ones. Even in simple cases, like that of the uniformly charged sphere, a direct integration of the forces produced by each element of the sphere is di\ufb03cult to perform. In fact, the introduction of the potential was one of the major advances in mechanics, which allowed many previously intractable problems to be handled relatively simply. This chapter, in which we have discussed the method of determining the force on a particle from a knowledge of the positions of other masses or charges, is complementary to the previous chapters, where we discussed the motion of a particle under known forces. Together, they provide a method of solving most of the problems in which the object of immediate interest is (or may be taken to be) a single particle, moving under the action of conservative forces. Problems? \u2192 \u221e 1. Find the potential and electric \ufb01eld at points on the axis of symmetry of a uniformly charged \ufb02at circular disc of charge q and radius a. What happens to the \ufb01eld if we keep the charge per unit area \u03c3 = q/\u03c0a2 \ufb01xed, and let a 2. Calculate the quadrupole moment of the disc of Problem 1, and deduce the form of the \ufb01eld at large distance in any direction. Verify that on the axis it agrees with the exact value when r/a is su\ufb03ciently large. 3. Write down the potential energy of a pair of charges, q at a and q \u2212 at the origin, in a \ufb01eld with potential \u03c6(r). By considering the limit 0, show that the potential energy of a dipole of moment d is a \u2192 V = If the electric \ufb01eld is uniform, when is this potential energy a minimum? Show that the dipole experiences a net moment, or couple, G = d E, and that in a non-uniform \ufb01eld there is also a \u2227 \u2207)E. (Take d in the z-direction, and show that net force, F = (d \u00b7 F = d \u2202E/\u2202z.) E. \u2212 d \u00b7 Potential Theory 153 4. *Write the electric \ufb01eld of a dipole in vector notation. Using the result of Problem 3, \ufb01nd the potential energy of a dipole of moment d in the \ufb01eld of another dipole of moment d. (Take d at the origin and d at position r.) Find the forces and couples acting between the dipoles if they are placed on the z-axis and (a) both are pointing in the zdirection, (b) both are pointing in the x-direction, (c) d is in the zdirection, and d in the x-direction, and (d) d is in the x-direction and d in the y-direction. 5. Show that the work done in bringing two charges q1 and q2, initially far apart, to a separation r12 is q1q2/4\u03c00r12. Write down the corresponding expression for a system of many charges. Show that the energy stored in the charge distribution is V = 1 2 qj\u03c6j (rj), j where \u03c6j(rj) is the potential at rj due to all the other charges. Why does a factor of 1 2 appear here, but not in the corresponding expression for the potential energy in an external potential \u03c6(r)? 6. Find the energy stored in a sphere of charge q and radius a with uniform charge density, and show that in\ufb01nite energy is required to compress the sphere to a point. Find also the stored energy in the case where the charge is uniformly spread over the surface of the sphere. 7. Near the Earth\u2019s surface, there is normally a vertical electric \ufb01eld, typ1", ". Find the corresponding total charge on the ically of order 100 V m\u2212 Earth, and the stored energy per unit surface area. 8. Find the quadrupole moment of a distribution of charge on the surface 1 2 ). of a sphere of radius a with surface charge density \u03c3 = \u03c30( 3 Find the total energy stored in this distribution. 9. *Two equal charges q are located at the points ( a, 0, 0), and two a, 0). Find the leading term in the potential at charges large distances, and the corresponding electric \ufb01eld. 2 cos2 \u03b8 q at (0, \u2212 \u00b1 \u00b1 \u2212 10. Find the gravitational potential at large distances of a thin circular 3. Find also the loop of radius a and mass m, up to terms of order r\u2212 potential, and the leading term in the gravitational \ufb01eld near the origin, at distances r a. 11. *Six equal point masses m are located at the points aj and ak. Show that the quadrupole term in the potential vanishes, and 6Gm/r. (Note: This \u00b1 \ufb01nd the leading correction to the monopole term requires expansion of the potential up to terms of order a4/r5.) ai, \u00b1 \u2212 \u00b1 154 Classical Mechanics 12. A di\ufb00use spherical cloud of gas of density \u03c1 is initially at rest, and starts to collapse under its own gravitational attraction. Find the radial velocity of a particle which starts at a distance a from the centre when it reaches the distance r. Hence, neglecting other forces, show that every particle will reach the centre at the same instant, and that the time 3. 3\u03c0/32\u03c1G. Evaluate this time in years if \u03c1 = 10\u2212 taken is (Hint : Assume that particles do not overtake those that start nearer the centre. Verify that your solution is consistent with this assumption. The substitution r = a sin2 \u03b8 may be used to perform the integration.) Estimate the collapse time for the Earth and for the Sun, taking a suitable value of \u03c1 in each case. 19 kg m\u2212 13. Given that the mass of the gas cloud of Problem 12 is 1030 kg, and that its contraction is halted by the build-up of pressure when a star of radius 106 km has been formed, use the result of Problem 6 to \ufb01nd the total energy released, assuming that the density of the star is still uniform. 14. The rotation period of Jupiter is approximately 10 hours. Its mass and radius are 318ME and 11.0RE, respectively (E = Earth). Calculate approximately its oblateness, neglecting the variation in density. (The observed value is about 1/15.) 15. *Assume that the Earth consists of a core of uniform density \u03c1c, surrounded by a mantle of uniform density \u03c1m, and that the boundary between the two is of similar shape to the outer surface, but with a radius only three-\ufb01fths as large. Find what ratio of densities \u03c1c/\u03c1m is required to explain the observed quadrupole moment. (Hint : Treat the \u03c1m. Earth as a superposition of two ellipsoids of densities \u03c1m and \u03c1c \u2212 Note that in reality neither core nor mantle is of uniform density.) 16. The distance between the Earth and the Moon is gradually increasing 8.4). Estimate the height of the tides (because of tidal friction \u2014 see when the Moon was 10 Earth radii away. How far away will it be when the lunar and solar tides are equal in magnitude? \u00a7 17. Suppose that at some time the Moon had been ocean-covered and rotating relative to the Earth. Find the ratio between the heights of the tides raised on the Moon by the Earth and on the Earth by the Moon. (The Moon\u2019s radius is RM = 0.27RE.) Estimate how high the tides on the Moon would have been when the Earth\u2013Moon distance was 10RE. 18. Jupiter has a satellite Io only a little heavier than the Moon (mass 1022 kg). The semi-major axis of 105 km for the Moon. 8.93 its orbit is 4.22 1022 kg, as compared to 7.35 105 km, compared with 3.84 \u00d7 \u00d7 \u00d7 \u00d7 Potential Theory 155 Estimate the height of the tides raised on Jupiter by Io. (See Problem 14 for Jovian data.) 19. *The tidal bulge in the ocean may be regarded as adding a surface mass distribution of the same form as the surface charge distribution of Problem 8, positive near \u03b8 = 0 and \u03c0 and negative near \u03b8 = \u03c0/2. Find the quadrupole moment of this distribution, and show that when it is included in (6.43) the e\ufb00ective height of the tides is increased by 1, where \u03c1", "o is the mean density of the oceans a factor (1 and \u03c1E that of the Earth as a whole. Evaluate this factor, given that 3. (Note, however, \u03c1E = 5.52 that the e\ufb00ective height is decreased by a larger factor due to the nonrigidity of the Earth.) 3 and \u03c1o = 1.03 103 kg m\u2212 103 kg m\u2212 3\u03c1o/5\u03c1E)\u2212 \u2212 \u00d7 \u00d7 20. *Two small identical uniform spheres of density \u03c1 and radius r are orbiting the Earth in a circular orbit of radius a. Given that the spheres are just touching, with their centres in line with the Earth\u2019s centre, and that the only force between them is gravitational, show that they will be pulled apart by the Earth\u2019s tidal force if a is less than ac = 2(\u03c1E/\u03c1)1/3RE, where \u03c1E is the mean density of the Earth and RE its radius. (This is an illustration of the existence of the Roche limit, within which small planetoids would be torn apart by tidal forces. The actual limit is larger than the one found here, because the spheres themselves would be distorted by the tidal force, thus enhancing the e\ufb00ect. It is ac = 2.45(\u03c1E/\u03c1)1/3RE. For the mean density of the Moon, for example, 3, this gives ac = 2.89RE.) \u03c1 = 3.34 21. Verify that the potential of a uniformly charged sphere satis\ufb01es Pois(You will need son\u2019s equation, both inside and outside the sphere. (A.59) for this and subsequent problems.) 103 kg m\u2212 \u00d7 22. The potential \u03c6(r) = (q/4\u03c00r)e\u2212 \u00b5r may be regarded as representing the e\ufb00ect of screening of a charge q at the origin by mobile charges in a plasma. Calculate the charge density \u03c1 (at points where r = 0) and \ufb01nd the total charge throughout space, excluding the origin. 23. By considering the equilibrium of a small volume element, show that in a \ufb02uid in equilibrium under pressure and gravitational forces, \u2207p = \u03c1g, where \u03c1 is the density and p the pressure (the equation of hydrostatic equilibrium). Deduce that, for an incompressible \ufb02uid of uniform density \u03c1, p + \u03c1\u03a6 is a constant. Use this result to obtain a rough estimate (Mean density of Earth of the pressure at the centre of the Earth. 3. Note that the pressure at the surface is essentially = 5.5 zero. The actual pressure at the centre is more than this estimate \u2014 103 kg m\u2212 \u00d7 156 Classical Mechanics \u00d7 1011 Pa \u2014 because of the non-uniformity of \u03c1. It in fact about 3.6 should be noted that \u2014 as indicated in Problem 12 \u2014 even very small departures from hydrostatic equilibrium would result in collapse on a short time scale, only checked by an increase in internal pressure via other processes.) 24. *Assume that the pressure p in a star with spherical symmetry is related to the density \u03c1 by the (distinctly unrealistic) equation of state p = 1 2 k\u03c12, where k is a constant. Use the \ufb02uid equilibrium equation obtained in Problem 23 to \ufb01nd a relation between \u03c1 and \u03a6. Hence show that Poisson\u2019s equation yields d2[r\u03c1(r)] dr2 = 4\u03c0G k \u2212 r\u03c1(r). Solve this equation with the boundary conditions that \u03c1 is \ufb01nite at r = 0 and vanishes at the surface of the star. Hence show that the radius a of the star is determined solely by k and is independent of its mass M. Show also that M = (4/\u03c0)a3\u03c1(0). 25. *Show that the moment of the Earth\u2019s gravitational force may be writ- ten in the form mr \u2227 g = 3GM ma2J2 r5 (k \u00b7 r)(k r). \u2227 Consider a satellite in a circular orbit of radius r in a plane inclined to the equator at an angle \u03b1. By introducing a pair of axes in the plane 5.6 (see Fig. 5.10), show that the average value of of the orbit, as in this moment is \u00a7 mr g av = \u2227 \u2212 3GM ma2J2 2r3 cos \u03b1 (k n), \u2227 where n is the normal to the orbital plane. Hence show that the orbit precesses around the direction k of the Earth\u2019s axis at a rate (3J2a2/2r2)\u03c9 cos \u03b1, where \u03c9 is the orbital angular velocity. Eval\u2126 = uate this rate for an orbit 400", " km above the Earth\u2019s surface, with an inclination of 30\u25e6. Find also the precessional period. \u2212 26. *Show that the moment of the solar tidal force mgS on an Earth satel- lite is mr \u2227 gS = 3GMSm a5 S (r \u00b7 aS)(r aS), \u2227 Potential Theory 157 where aS is the position vector of the Sun relative to the Earth. Consider the e\ufb00ect on the Moon, whose orbit is inclined at \u03b1 = 5\u25e6 to the ecliptic (the Earth\u2019s orbital plane). Introduce axes i and j in the plane of the ecliptic, with i in the direction where the Moon\u2019s orbit crosses it. Write down the positions of the Sun and Moon, relative to Earth, as functions of time. By averaging both over a month and over a year (i.e., over the positions in their orbits of both bodies), show that the moment leads to a precession of the Moon\u2019s orbital plane, at a precessional angular velocity \u2126 = 32 4\u03c9 where is the Earth\u2019s orbital angular velocity. Compute the precessional period. (Note that this calculation is only approximately correct.) cos \u03b1 k, \u2212 158 Classical Mechanics Chapter 7 The Two-Body Problem We shall be mainly concerned in this chapter with an isolated system of two particles, subject only to the force between the two. However, as it is no harder to solve, and considerably extends the range of applicability of the results, we shall also allow the presence of a uniform gravitational \ufb01eld. 7.1 Centre-of-mass and Relative Co-ordinates We denote the positions and masses of the two particles by r1, r2 and m1, m2. If the force on the \ufb01rst particle due to the second is F, then, by F. Thus, in a Newton\u2019s third law, that on the second due to the \ufb01rst is uniform gravitational \ufb01eld g, the equations of motion are \u2212 m1\u00a8r1 = m1g + F, m2\u00a8r2 = m2g F. \u2212 (7.1) It is convenient to introduce new variables in place of r1 and r2. We de\ufb01ne the position of the centre of mass, R = m1r1 + m2r2 m1 + m2, (7.2) and the relative position, r = r1 \u2212 (See Fig. 7.1.) The centre of mass is often denoted by \u00afr. This symbol is, however, typographically inconvenient when dots are used to denote time derivatives. From (7.1), we get the equation of motion for R by adding: (7.3) r2. M \u00a8R = M g, where M = m1 + m2. (7.4) 159 160 Classical Mechanics m 1 r 1 r R r 2 Fig. 7.1 m 2 One virtue of the notation R for the centre of mass is that it serves to emphasize the fact that R is associated with the total mass, M. The equation for r is obtained by dividing by the masses and then subtracting: \u00b5\u00a8r = F, where \u00b5 = m1m2 m1 + m2. (7.5) The mass \u00b5 is called the reduced mass, because it is always less than either m1 or m2. These two equations are now completely separate. Equation (7.4) shows that the centre of mass moves with uniform acceleration g. In the case g = 0 it is equivalent to the law of conservation of momentum, M \u02d9R = m1 \u02d9r1 + m2 \u02d9r2 = P = constant. (7.6) The equation of motion (7.5) for the relative position is identical with that for a single particle of mass \u00b5 moving under the force F. If we can solve this one-particle problem, then we can also solve the two-particle problem. When we have found R and r as functions of time, we can obtain the positions of the particles by solving the simultaneous equation (7.2) and (7.3): r1 = R + m2 M r, r2 = R m1 M r. \u2212 (7.7) The separation between centre-of-mass and relative motion extends to the expressions for the total angular momentum and kinetic energy. We have J = m1r1 \u2227 R + = m1 \u02d9r1 + m2r2 \u2227 m2 r M \u2227 \u02d9r2 \u02d9R + m2 M \u02d9r + m2 R m1 M r \u2212 \u02d9R \u2212 m1 M \u02d9r, \u2227 The Two-Body Problem 161 by", " (7.7). It is easy to see that the cross terms between R and r cancel, and we are left with J = M R \u02d9R + \u00b5r \u02d9r. \u2227 \u2227 Similarly, substituting (7.7) into T = 1 2 m1 \u02d9r2 1 + 1 2 m2 \u02d9r2 2, we \ufb01nd, after a little algebra, T = 1 2 M \u02d9R 2 + 1 2 \u00b5 \u02d9r2. (7.8) (7.9) \u00a7 It is instructive to examine this problem also from the Lagrangian point 3.7. Let us suppose that the internal, two-body of view introduced in force is conservative, and depends only on the relative co-ordinate r. It then corresponds to a potential energy function Vint(r). The total potential energy V includes not only Vint but also the potential energy of the external gravitational forces. Now the potential energy of a particle of mass m in a uniform gravitational \ufb01eld g is z-direction, this is \u2212 the familiar expression mgz.) Thus the Lagrangian function is r. (If g is in the mg \u2212 \u00b7 L = T = 1 V \u2212 2 m1 \u02d9r2 1 + 1 2 m2 \u02d9r2 2 + m1g r1 + m2g r2 \u2212 Vint(r1 \u2212 \u00b7 r2). \u00b7 (7.10) It is easy to verify that Lagrange\u2019s equations (3.44) yield the correct equations of motion (7.1). Now, just as in 3.7, we may express L in terms of any six independent co-ordinates in place of x1, y1, z1, x2, y2, z2. In particular, we may choose the coordinates of R and r. Then, from (7.9) and (7.10), we \ufb01nd \u00a7 L = 1 2 M \u02d9r2 \u00b7 \u2212 Vint(r). (7.11) The complete separation of the terms involving R from those involving r is now obvious. It is easy to check that Lagrange\u2019s equations for these co-ordinates are just (7.4) and (7.5). In this form, it is clear that the crucial point is the separation of the potential energy. The kinetic energy may always be separated according to (7.9), but only for a uniform gravitational \ufb01eld does the potential energy separate in a similar way. 162 Classical Mechanics 7.2 The Centre-of-mass Frame It is often convenient to describe the motion of the system in terms of a frame of reference in which the centre of mass is at rest at the origin. (In a gravitational \ufb01eld, this is an accelerated, non-inertial frame, but it is still useful.) This is the centre-of-mass (CM ) frame. We shall denote quantities referred to it by an asterisk. The relative position r is of course independent of the choice of origin, so, setting R\u2217 = 0 in (7.7), we \ufb01nd r\u22171 = m2 M r, r\u22172 = m1 M \u2212 r. (7.12) In this frame, the momenta of the two particles are equal and opposite: m1 \u02d9r\u22171 = \u2212 m2 \u02d9r\u22172 = \u00b5 \u02d9r = p\u2217, (7.13) say. As we shall see explicitly later, it is often convenient to solve a problem \ufb01rst in the CM frame. To \ufb01nd the solution in some other frame, we then need the relations between the momenta in the two frames. Let us consider a frame in which the centre of mass is moving with velocity \u02d9R. Then the velocities of the two particles are \u02d9r1 = \u02d9R + \u02d9r\u22171, \u02d9r2 = \u02d9R + \u02d9r\u22172. Hence by (7.13), their momenta are p1 = m1 \u02d9r1 = m1 \u02d9R + p\u2217, p2 = m2 \u02d9r2 = m2 \u02d9R p\u2217. \u2212 (7.14) From (7.8) and (7.9), it follows that the total momentum, angular momentum and kinetic energy in the CM frame are P \u2217 = 0, J \u2217 = \u00b5r \u02d9r = r \u2227 2 \u00b5 \u02d9r2 = T \u2217 = 1 p\u2217,. \u2227 2 p\u2217 2\u00b5 Thus in any other frame we can write P = M \u02d9RR T = 1", " \u02d9R + J \u2217, + T \u2217. (7.15) (7.16) The Two-Body Problem 163 To obtain the values in any frame from those in the CM frame, we have only to add the contribution of a particle of mass M located at the centre of mass R. We shall see in the next chapter that this is a general conclusion, not restricted to two-particle systems. As an example, let us consider two bodies moving under their mutual gravitational attraction, for instance the Moon and the Earth. Example: Motion of a satellite How does the period of the Moon\u2019s orbit relate to the Earth\u2013 Moon distance? How far does the apparent position of the Sun oscillate because of the in\ufb02uence of the Moon on the Earth\u2019s position? Equation (7.5) becomes \u00b5\u00a8r = Gm1m2 r2 \u02c6r \u2212 = \u02c6r \u2212 GM \u00b5 r2. This is identical with the equation for a particle moving around a \ufb01xed point mass M. In particular, for an elliptic orbit, the period is given by (4.31), \u03c4 2\u03c0 2 = a3 GM. Note that a here is the semi-major axis of the relative orbit (the median distance between the bodies), and that M is the sum of the masses, rather than the mass of the heavier body. Thus Kepler\u2019s third law is only approximately correct: the orbital period depends not only on the semi-major axis, but also on the mass. (See also Chapter 4, Problem 31.) The only cases in the solar system for which the lighter mass is an appreciable fraction of the total are those of the Earth\u2013Moon system, for which m1/m2 = 1/81.3, and the Pluto\u2013Charon system, for which m1/m2 = 1/8.5. If we were to compute the period of the Moon\u2019s orbit from Kepler\u2019s third law by comparing it with the period of a small Earth satellite, we should obtain the value given above but with m2 in place of M. This would yield a period about 4 hours too long. In the CM frame of the Earth\u2013Moon system, we see from (7.12) that both bodies move in ellipses around the centre of mass, with semi-major 164 axes Classical Mechanics a1 = m2 M a, a2 = m1 M a. \u00d7 105 km. The Earth therefore For the Earth\u2013Moon system, a = 3.84 moves around the centre of mass in a small ellipse with semi-major axis (Note that the centre of mass is inside the a2 = a/82.3 = 4670 km. Earth.) This leads to a small oscillation in the direction of the Sun (see a2/A, Fig. 7.2 \u2014 which is not drawn to scale), of angular amplitude \u03b1 \u2248 108 km, \u03b1 = 6.4 where A is the distance to the Sun; since A = 1.5 seconds of arc. \u00d7 Moon a 1 a 2 Earth A \u03b1 Sun Fig. 7.2 A larger e\ufb00ect can be obtained by looking at the direction of some object closer to us than the Sun, such as an asteroid. This e\ufb00ect provides a method of determining the lunar mass. In the frame in which the Sun is at rest, the centre of mass of the Earth\u2013Moon system moves in an ellipse around the Sun, and the motion of the two bodies around the centre of mass is superimposed on this larger orbital motion. In the approximation in which the Sun\u2019s gravitational \ufb01eld is uniform over the dimension of the Earth\u2013Moon system, the two types of motion are completely separate, and no transfer of energy or angular momentum from one to the other can occur. However, as we shall see later, there are small e\ufb00ects due to the non-uniformity of the Sun\u2019s \ufb01eld. The Two-Body Problem 165 7.3 Elastic Collisions A collision between two particles is called elastic if there is no loss of kinetic energy in the collision \u2014 that is, if the total kinetic energy after the collision is the same as that before it. Such collisions are typical of very hard bodies, like billiard balls. (They are also important in atomic and nuclear collision problems.) It is easy to describe such a collision in the CM frame. The particles p\u2217, must approach each other with equal and opposite momenta, p\u2217 and and recede after the collision again with equal and opposite momenta, q\u2217 q\u2217. Thus each particle is scattered through the same angle \u03b8\u2217 (see and Fig. 7.3). Since the collision is elastic, we have, from (7.15), \u2212 \u2212 1 \u2217 p \u2217 q \ufffd", "\ufffd \u2217 \u2217 \u2212p 2 \u2217 \u2212q Fig. 7.3 T \u2217 = 2 p\u2217 2\u00b5 = 2 q\u2217 2\u00b5. (7.17) Thus the magnitudes of the momenta before and after the collision are the same, p\u2217 = q\u2217. (7.18) In practice, most experiments are performed with one particle initially at rest (or nearly so) in the laboratory. To interpret such an experiment, we therefore need to use the laboratory (Lab) frame, in which the momentum of particle 2, the target, before the collision is zero, so that p2 = 0. We shall denote the Lab momentum of the incoming particle by p1, and the momenta after the collision by q1 and q2, and the angles of scattering and recoil by \u03b8 and \u03b1 (see Fig. 7.4). We could work out the relations 166 Classical Mechanics between these quantities by using the conservation laws of energy and mo Fig. 7.4 mentum directly in the Lab frame, but it is actually simpler to relate them to the CM quantities, using the formulae obtained in 7.2. \u00a7 Since p2 = 0, we have from (7.14), 1 m2 \u02d9R = p\u2217, and also p1 = m1 m2 p\u2217 + p\u2217 = M m2 p\u2217. (7.19) (7.20) The momenta after the collision are again given by (7.14), but with q\u2217 in place of p\u2217. They are then, using also (7.19), q1 = m1 m2 p\u2217 + q\u2217, q2 = p\u2217 q\u2217. \u2212 All these relations may be conveniently summarized in a vector diagram, drawn in Fig. 7.5, which also incorporates the momentum conservation p \u2217 q \u03b8\u2217 p 1 Fig. 7.5 The Two-Body Problem 167 equation in the Lab, p1 = q1 + q2. Any desired relation between momenta, energies or angles may be extracted from this diagram without di\ufb03culty. We consider explicitly only a few important relations. By (7.18), the vectors p\u2217, q\u2217, q2 form an isosceles triangle. Thus the recoil angle and the magnitude of the recoil momentum q2 are given in terms of CM quantities by \u03b1 = 1 2 (\u03c0 \u03b8\u2217), \u2212 q2 = 2p\u2217 sin 1 2 \u03b8\u2217. (7.21) The Lab kinetic energy transferred to the target particle is therefore T2 = q2 2 2m2 = 2 2p\u2217 m2 sin2 1 2 \u03b8\u2217. On the other hand, the total kinetic energy in the Lab is just the kinetic energy of the incoming particle, T = p2 1 2m1 = 2 M 2p\u2217 2m1m2 2, by (7.20). The interesting quantity is the fraction of the total kinetic energy which is transferred. This is T2 T = 4m1m2 M 2 sin2 1 2 \u03b8\u2217. (7.22) The maximum possible kinetic energy transfer occurs for a head-on collision (\u03b8\u2217 = \u03c0), and is T2/T = 4m1m2/(m1 + m2)2. Clearly this can be close to unity only if m1 and m2 are comparable in magnitude. If the incoming particle is very light, it bounces o\ufb00 the target with little loss of energy; if it is very heavy, it is hardly de\ufb02ected at all from its original trajectory, and again loses little of its energy. For example, in a proton\u2013\u03b1-particle collision (m1/m2 = 4 or 1 4 ), the maximum fractional energy transfer is 64 per cent. In an electron\u2013proton collision (m1/m2 = 1/1836) it is about 0.2 per cent. (These conclusions are changed by relativistic e\ufb00ects for speeds approaching that of light.) Another important relation is that between the Lab and CM scattering angles. It is easy to prove by elementary trigonometry (most simply by dropping a perpendicular from the upper vertex in Fig. 7.5) that tan \u03b8 = sin \u03b8\u2217 (m1/m2) + cos \u03b8\u2217. (7.23) 168 Classical Mechanics Note that this relation is independent of the momenta of the two particles, and depends only on their mass ratio. As \u03b8\u2217 varies from 0 to \u03c0, the vector q\u2217 sweeps out a semicircle of radius p\u2217. If the target is the heavier particle (m1/m2 < 1), the left-hand vertex of Fig. 7.5 lies inside", " this semicircle, and \u03b8 also varies from 0 to \u03c0. However, if the target is the lighter particle (m1/m2 > 1), the semicircle excludes this vertex. In that case, \u03b8 = 0 both when \u03b8\u2217 = 0 and when \u03b8\u2217 = \u03c0, and there is an intermediate value of \u03b8\u2217 for which \u03b8 is a maximum. The maximum scattering angle occurs when q1 is tangent to the circle (see Fig. 7.6). It is given by * q q 1 \u03b8 max m 1 m 2 * p Fig. 7.6 sin \u03b8max = m2/m1. (7.24) For instance, an \u03b1-particle can only be scattered by a proton through an angle of almost 14.5\u25e6, and a proton can only be scattered by an electron through an angle less than 0.031\u25e6. In the special case of equal-mass particles, the right side of (7.23) sim- pli\ufb01es to tan 1 2 \u03b8\u2217, so that we obtain the simple relation \u03b8 = 1 2 \u03b8\u2217, (m1 = m2). (7.25) In this case, the maximum scattering angle is \u03c0/2. 7.4 CM and Lab Cross-sections Let us now consider the scattering of a beam of particles by a target. We shall suppose that the target contains a large number N of identical target particles, but, as in 4.5, is still small enough so that the particles do not screen each other, and so that the geometry of the scattering event is not \u00a7 The Two-Body Problem 169 seriously a\ufb00ected by which target particle is struck. We also suppose that the beam is a parallel beam of particles, each with the same mass and velocity, with a particle \ufb02ux of f particles crossing unit area per unit time. The incident particles may or may not be the same as those in the target, but for the moment we shall assume that they can be distinguished, so that it is possible to set up a detector to count the number of scattered particles emerging in some direction, without including also the recoiling target particles. We can now introduce the concept of di\ufb00erential cross-section, just as we did for the case of a \ufb01xed target in 4.5. Indeed the fact that the particles \u00a7 that are struck recoil out of the target makes no essential di\ufb00erence. In any one collision, the scattering angle \u03b8 will be determined in some way (depending on the shape of the particles) by the impact parameter b. Thus the particles scattered through angles between \u03b8 and \u03b8 + d\u03b8 will be those of the incident particles which strike any one of the target particles with impact parameters between the corresponding values b and b + db. To \ufb01nd the number emerging within a solid angle in some speci\ufb01ed direction, we have to calculate the corresponding cross-sectional area of the incident beam, db | (compare (4.39) and Fig. 4.9), and multiply by the number of target particles, N, and by the \ufb02ux, f. d\u03c3 = b (7.26) d\u03d5. | If we set up a detector of cross-sectional area dA at a large distance L from the target, the rate of detection will be dw = N f d\u03c3 d\u2126 dA L2, (7.27) exactly as in (4.44). The ratio d\u03c3/d\u2126 is the Lab di\ufb00erential cross-section. We shall return to the question of calculating the Lab di\ufb00erential crosssection later. First, however, we wish to discuss a slightly di\ufb00erent type of experiment. Let us imagine two beams of particles approaching from opposite directions \u2014 a \u2018colliding-beam\u2019 experiment. In particular, we shall be interested in the case where the momenta of the particles in the two beams are equal and opposite, so that we are directly concerned with the CM frame. To be speci\ufb01c, let us suppose that the particles in one beam are hard spheres of radius a1 and those in the other beam are hard spheres of radius a2. Evidently, a particular pair of particles will collide if the distance b between 170 Classical Mechanics the lines of motion of their centres is less than a = a1 + a2 (see Fig. 7.7). \u03b1 \u03b1 \u03b8\u2217 a 1 b a 2 \u03b8\u2217 Fig. 7.7 Let us select one of the particles in the second beam. It will collide with any particles in the \ufb01rst beam whose centres cross an area \u03c3 = \u03c0a2. This is the total cross-section for the collision, and is the same whether", " the target is at rest or moving towards the beam. Note that it is the sum of the radii which enters here. This serves to emphasize the fact that the cross-section is a property of the pair of particles involved in the collision, not of either particle individually. To compute the number of collisions occurring in a short time interval dt, we may imagine the cross-sectional area \u03c3 to be attached to our selected particle and moving with it. Then the probability that the centre of one of the particles in the \ufb01rst beam crosses this area in the time dt will be n1v\u03c3 dt, where n1 is the number of particles per unit volume in the beam, and v is the relative velocity (= v1 + v2). If the number of particles in the second beam per unit volume is n2, then the total number of collisions occurring in volume V is n1n2v\u03c3V dt. Now, if we are interested in the number of particles emerging in a given solid angle, we have to \ufb01nd the relation between b and the CM scattering angle \u03b8\u2217, and evaluate the corresponding cross-sectional area (7.26). In our case, it is clear from Fig. 7.7 that b = a sin \u03b1 = a cos 1 2 \u03b8\u2217. (7.28) The Two-Body Problem 171 Hence, introducing the solid angle d\u2126\u2217 = sin \u03b8\u2217 d\u03b8\u2217 d\u03d5, as in (4.42), we \ufb01nd that the CM di\ufb00erential cross-section is d\u03c3 d\u2126\u2217 = 1 4 a2. (7.29) If we set up a detector arranged to record particles scattered in a particular direction from a small volume V, then the detection rate of particles will be w = n1n2vV d\u03c3 d\u2126\u2217 dA L2. (7.30) Note that N and f in (7.27) correspond to n2V and n1v in this formula. In our particular case, of hard-sphere scattering, the di\ufb00erential crosssection (7.29) is independent of \u03b8\u2217, and the scattering is therefore isotropic in the CM frame. In other words, the number of particles detected is independent of the direction in which we place the detector (so long as it is outside the unscattered beam). This is a very special property of hard spheres, and is not true in general. Nor is it true in the Lab frame. Indeed, the reason for calculating the CM di\ufb00erential cross-section \ufb01rst is that the relation between b and \u03b8\u2217 is much simpler than that between b and \u03b8, so that the cross-section takes a simpler form in the CM frame than in the Lab frame. The easiest way to \ufb01nd the Lab di\ufb00erential cross-section is often to compute the CM di\ufb00erential cross-section \ufb01rst, and then use the relation d\u2126\u2217 d\u2126 = sin \u03b8\u2217 d\u03b8\u2217 sin \u03b8 d\u03b8 = dz\u2217 dz, with z = cos \u03b8, z\u2217 = cos \u03b8\u2217. Let us assume (for a reason to be given later) that m1 < m2. Then it follows that d\u03c3 d\u2126 = d\u03c3 d\u2126\u2217 dz\u2217 dz. (7.31) The value of dz\u2217/dz may be found from (7.23), which may be written z2 = [(m1/m2) + z\u2217]2 1 + 2(m1/m2)z\u2217 + (m1/m2)2. It is straightforward, if tedious, to solve for z \u2217 and di\ufb00erentiate, but the resulting expression for the general case is not particularly illuminating, and we shall omit it. 172 Classical Mechanics If m1 > m2, there is a complication which arises from the fact that for each value of \u03b8 less than the maximum scattering angle \u03b8max there are two possible values for \u03b8\u2217, as is clear from Fig. 7.6. If we measure only the direction of the scattered particles, we cannot distinguish these two, and must add the contributions to d\u03c3/d\u2126 from d\u03c3/d\u2126\u2217 at these two separate values of \u03b8\u2217. (Moreover, for one of these two, dz\u2217/dz is negative, and must be replaced by its absolute value.) The two values of \u03b8\u2217 can be distinguished if we can measure also the energy or momentum of the scattered particles. We shall consider explicitly only the case of equal mass particles, m1 =", " m2. Example: Scattering of equal hard spheres Find the Lab-frame di\ufb00erential cross-section for the scattering of equal-mass hard spheres. Here the relation between the scattering angles, (7.23), simpli\ufb01es to 1. Consequently, dz\u2217/dz = 4z, and the Lab \u03b8\u2217 = 2\u03b8, or z\u2217 = 2z2 \u2212 di\ufb00erential cross-section corresponding to (7.29) is d\u03c3 d\u2126 = a2 cos \u03b8, (\u03b8 < 1 2 \u03c0). (7.32) Of course, no particles are scattered through angles greater than \u03c0/2, so d\u03c3 d\u2126 = 0, (\u03b8 > 1 2 \u03c0). (7.33) In this case, the cross-section is peaked towards the forward direction, and more particles will enter a detector placed at a small angle to the beam direction than one placed nearly at right angles. As a check of this result, we may evaluate the total cross-section \u03c3, 1 4 a2 = \u03c0a2 obtained in the which must agree with the value \u03c3 = 4\u03c0 CM frame. We \ufb01nd \u00d7 \u03c3 = d\u03c3 d\u2126 2\u03c0 \u03c0/2 d\u2126 = d\u03d5 a2 cos \u03b8 sin \u03b8 d\u03b8 = \u03c0a2, 0 as expected. (Note the upper limit of \u03c0/2 on the \u03b8 integration, which arises because of (7.33).) 0 The Two-Body Problem 173 We assumed at the beginning of this section that the target particles and beam particles were distinguishable, and that the aim of the experiment was to count scattered beam particles only. However, it is easy to relax this condition. We shall now calculate the rate at which recoiling target particles enter the detector. In a collision in which the incoming particle has impact parameter b, and is moving in a plane speci\ufb01ed by the angle \u03d5, the target particle will emerge in a direction speci\ufb01ed by the polar angles \u03b1, \u03c8, where \u03b1 is related to \u03b8\u2217 or b by (7.21) and \u03c8 = \u03c0 + \u03d5. The number of target particles emerging within the angular range d\u03b1, d\u03c8 is therefore equal to the number of incident particles in the corresponding range db, d\u03d5. To determine the number of recoiling particles entering the detector, we have to relate d\u03c3 to the solid angle d\u21262 = sin \u03b1 d\u03b1 d\u03c8. By (7.21), we have cos \u03b8\u2217 = 1 2 cos2 \u03b1, whence \u2212 d\u03c3 d\u21262 = d\u03c3 d\u2126\u2217 4 cos \u03b1 (\u03b1 < 1 2 \u03c0). (7.34) In the particular case of equal-mass hard spheres, this di\ufb00erential crosssection has exactly the same form as (7.32). This shows that in the scattering of identical hard spheres the numbers of scattered particles and recoiling target particles entering the detector are precisely equal, and the total detection rate is obtained by doubling that found earlier. 7.5 Summary For our later work, the most important result of this chapter is that the total momentum, angular momentum and kinetic energy in an arbitrary frame di\ufb00er from those in the CM frame by an amount equal to the contribution of a particle of mass M moving with the centre of mass. We have seen that for a two-particle system the use of the CM frame is often a considerable simpli\ufb01cation, and this is true also for more complicated systems. When we need results in some other frame, it is often best to solve the problem \ufb01rst in the CM frame, and then transform to the required frame. This is true for example in the calculation of scattering cross-sections in two-particle collisions. 174 Classical Mechanics Problems 1. A double star is formed of two components, each with mass equal to that of the Sun. The distance between them is 1 AU (see Chapter 4, Problem 2). What is the orbital period? 2. Where is the centre of mass of the Sun\u2013Jupiter system? (The mass ratio is MS/MJ = 1047. See Chapter 4, Problems 2 and 3.) Through what angle does the Sun\u2019s position as seen from the Earth oscillate because of the gravitational attraction of Jupiter? 3. The parallax of a star (the angle subtended at the star by the radius of the Earth\u2019s orbit) is. The star\u2019s position is observed to oscillate with angular amplitude \u03b1 and period \u03c4. If the oscillation is interpreted as being due to the existence of a planet moving in a circular orbit around the star, show that its mass m1 is", " given by m1 MS = \u03b1 M \u03c4E MS\u03c4 2/3, where M is the total mass of star plus planet, MS is the Sun\u2019s mass, and \u03c4E = 1 year. Evaluate the mass m1 if M = 0.25MS, \u03c4 = 16 years, = 0.5 and \u03b1 = 0.01. What conclusion can be drawn without making the assumption that the orbit is circular? 4. Two particles of masses m1 and m2 are attached to the ends of a light spring. The natural length of the spring is l, and its tension is k times its extension. Initially, the particles are at rest, with m1 at a height l above m2. At t = 0, m1 is projected vertically upward with velocity v. Find the positions of the particles at any subsequent time (assuming that v is not so large that the spring is expanded or compressed beyond its elastic limit). 5. *Prove that in an elastic scattering process the angle \u03b8 + \u03b1 between the emerging particles is related to the recoil angle \u03b1 by tan(\u03b8 + \u03b1) tan \u03b1 = m1 + m2 m2 m1 \u2212. (Hint : Express both tangents in terms of tan 1 ratio if the particles emerge at right angles to each other? 2 \u03b8\u2217.) What is the mass 6. A proton is elastically scattered through an angle of 56\u25e6 by a nucleus, which recoils at an angle \u03b1 = 60\u25e6. Find the atomic mass of the nucleus, and the fraction of the kinetic energy transferred to it. The Two-Body Problem 175 7. An experiment is to be designed to measure the di\ufb00erential cross-section for elastic pion\u2013proton scattering at a CM scattering angle of 70\u25e6 and a pion CM kinetic energy of 490 keV. (The electron-volt (eV) is the atomic unit of energy.) Find the angles in the Lab at which the scattered pions, and the recoiling protons, should be detected, and the required Lab kinetic energy of the pion beam. (The ratio of pion to proton mass is 1/7.) 8. *An unstable particle of mass M = m1 + m2 decays into two particles of masses m1 and m2, releasing an amount of energy Q. Determine the kinetic energies of the two particles in the CM frame. Given that m1/m2 = 4, Q = 1 MeV, and that the unstable particle is moving in the Lab with kinetic energy 2.25 MeV, \ufb01nd the maximum and minimum Lab kinetic energies of the particle of mass m1. 9. The molecules in a gas may be treated as identical hard spheres. Find the average loss of kinetic energy of a molecule with kinetic energy T in a collision with a stationary molecule. (Hint : Use the fact that the collisions are isotropic in the CM frame, so that all values of cos \u03b8\u2217 between 1 are equally probable.) How many collisions are required, on average, to reduce the velocity of an exceptionally fast molecule by a factor of 1000? \u00b1 10. Two identical charged particles, each of mass m and charge e, are initially far apart. One of the particles is at rest at the origin, and the other is approaching it with velocity v along the line x = b, y = 0, where b = e2/2\u03c00mv2. Find the scattering angle in the CM frame, and the directions in which the two particles emerge in the Lab. (See 4.7.) 11. *Find the distance of closest approach for the particles in Problem 10, \u00a7 and the velocity of each at the moment of closest approach. 12. Obtain the relation between the total kinetic energy in the CM and Lab frames. Discuss the limiting cases of very large and very small mass for the target. \u00d7 13. *Suppose that the asteroid of Chapter 4, Problem 17, has a mass of 1020 kg. Find the proportional change in the kinetic energy of the 6 Earth in this encounter. What is the resulting change in the semi-major axis of the Earth\u2019s orbit? By how much is its orbital period lengthened? (Note that the postulated event is exceedingly improbable.) 14. Calculate the di\ufb00erential cross-section for the scattering of identical hard spheres directly in the Lab frame. 176 Classical Mechanics 15. Find the Lab di\ufb00erential cross-section for the scattering of identical particles of charge e and mass m, if the incident velocity is v. (See (4.50).) 16. At low energies, protons and neutrons behave roughly like hard spheres 14 m. A parallel beam of of equal mass and radius about 1.3 1, strikes a target neutrons, with a \ufb02ux of 3 1022 protons. A circular detector of radius 20 mm is containing 4 placed 0.7 m", " from the target, in a direction making an angle of 30\u25e6 to the beam direction. Calculate the rate of detection of neutrons, and of protons. \u00d7 1010 neutrons m\u2212 2 s\u2212 10\u2212 \u00d7 \u00d7 17. *Write down the equations of motion for a pair of charged particles of q, in a uniform electric \ufb01eld E. equal masses m, and of charges q and Show that the \ufb01eld does not a\ufb00ect the motion of the centre of mass. Suppose that the particles are moving in circular orbits with angular velocity \u03c9 in planes parallel to the xy-plane, with E in the z-direction. Write the equations in a frame rotating with angular velocity \u03c9, and hence \ufb01nd the separation of the planes. \u2212 Chapter 8 Many-Body Systems Any material object may be regarded as composed of a large number of small particles, small enough to be treated as essentially point-like, but still large enough to obey the laws of classical rather than quantum mechanics. These particles interact in complicated ways with each other and with the environment. However, as we shall see, if we are interested only in the motion of the object as a whole, many of these details are irrelevant. We shall consider in this chapter a general system of N particles labelled by an index i = 1, 2,..., N, interacting through two-body forces, and also subjected to external forces due to bodies outside the system. We denote the force on the ith particle due to the jth particle by F ij, and the external force on the ith particle by F i. Then the equations of motion are mi\u00a8ri = F i1 + F i2 + + F iN + F i \u2261 \u00b7 \u00b7 \u00b7 F ij + F i. (8.1) j Here, and throughout this chapter, the sum is over all particles of the system, j = 1, 2,..., N. Of course, there is no force on the ith particle due to itself, and so F ii = 0. The sum in this case is really over the other N 1 particles. \u2212 8.1 Momentum; Centre-of-mass Motion The position R of the centre of mass is de\ufb01ned, as in (7.2), by R = 1 M i miri, where M = mi. (8.2) i 177 178 Classical Mechanics The total momentum is P = mi \u02d9ri = M \u02d9R. (8.3) i It is equal to the momentum of a particle of mass M located at the centre of mass. The rate of change of momentum, by (8.1), is \u02d9P = F ij + F i. i j i (8.4) Now the two-body forces must satisfy Newton\u2019s third law F ji = F ij. (8.5) \u2212 Thus, for every term F ij in the double sum in (8.4), there is an equal and opposite term F ji. The terms therefore cancel in pairs and the double sum is zero. (Compare the discussion at the end of 1.3 for the three-particle case.) \u00a7 Hence we obtain the important result that the rate of change of mo- mentum is equal to the sum of the external forces alone: \u02d9P = M \u00a8R = F i. i (8.6) In the special case of an isolated system of particles, acted on by no external forces, this yields the law of conservation of momentum, P = M \u02d9R = constant. (8.7) Let us now regard our system of particles as forming a composite body. If the body is isolated, then according to Newton\u2019s \ufb01rst law it moves with uniform velocity. Thus we see that to maintain this law for composite bodies, we should de\ufb01ne the position of such a body to mean the position of its centre of mass. Moreover, with this de\ufb01nition, Newton\u2019s second law is just (8.6), provided that we interpret the force on the body in the obvious way to mean the sum of the external forces on all of its constituent particles, and the mass as the sum of their masses. It is also clear that if Newton\u2019s third law applies to each pair of particles from two composite bodies, then it will apply to the bodies as a whole. Thus, with a suitable (and very natural) interpretation of the concepts involved, Newton\u2019s three basic laws may be applied to composite bodies as well as to point particles. (Though we have phrased the discussion in terms Many-Body Systems 179 of collections of particles, we could also include the case of a continuous distribution of matter, by dividing it up into in\ufb01nitesimal particles.) It follows that, so long as we", " are interested only in the motion of a body as a whole, we may replace it by a particle of mass M located at the centre of mass. This is a result of the greatest importance, for it allows us to apply our earlier discussion of particle motion to real physical bodies. We have of course implicitly assumed it in many of our applications; for example, we treated the planets as point particles in discussing their orbital motion. There is one point about which one must be careful. We may still need some information about the actual shape of the body to calculate the total force acting on it. This force is not necessarily equal to that on a particle of mass M at the centre of mass. In the gravitational case, it happens to be so if the body is spherical, or if the gravitational \ufb01eld is uniform, but not in general otherwise. As an example of the use of the momentum conservation equation for an isolated system, we consider the motion of a rocket. Example: Rockets A rocket, initially at rest, emits matter with a constant velocity u relative to the rocket (though not necessarily at a uniform rate). Gravity is negligible. If the initial total mass of the rocket is M0, \ufb01nd its velocity v when the mass has fallen to M. Consider the emission of a small mass dm. After the emission, the mass of the rocket will be reduced by this amount, dM = dm, \u2212 (8.8) and its velocity will be increased to some value v + dv. (See Fig. 8.1.) This velocity can be found from the momentum conservation equation (M \u2212 dm)(v + dv) + dm(v u) = M v. \u2212 M v before dm M\u2212dm u\u2212v v+dv after Fig. 8.1 180 Classical Mechanics Neglecting second-order in\ufb01nitesimals, we \ufb01nd By (8.8), we can write this in the form M dv = u dm. (8.9) \u2212 whence, integrating, and using the initial condition M = M0 when v = 0, we have dv u = dM M, v u = ln \u2212 M M0. This relation may be solved to give the mass as a function of the velocity attained: M = M0e\u2212 v/u. (8.10) This shows that to accelerate a rocket to a velocity equal to its ejection velocity u, we must eject all but a fraction 1/e of its original mass. Note that the velocity of the rocket depends only on the ejection velocity and the fraction of the initial mass which has been ejected, and not on the rate of ejection. It makes no di\ufb00erence whether the acceleration is brief and intense, or prolonged and gentle. (This assumes of course that the rocket is not subjected to other forces, such as gravity, during its acceleration. It would clearly be useless to try to escape from the Earth with a rocket providing a long, slow acceleration, because it would be constantly retarded by gravity.) For interplanetary \ufb02ights, rockets are normally used only for brief spells, between which the spacecraft moves in a free orbit \u2014 as, for example, in Chapter 4, Problem 16, where these brief spells are at or near planetary departures and arrivals. If the duration of the rocket bursts is su\ufb03ciently short for the change in position of the rocket during each burst to be negligible, then it is a good approximation to assume that on each occasion the velocity is changed instantaneously, by an amount known as the velocity impulse. The relevant quantity for determining the mass of the rocket required to deliver a given payload, using a given ejection velocity, via a given orbital manoeuvre, is then the sum of the velocity impulses (in the ordinary, not the vector, sense). For example, the minimum velocity im1. For the return trip, pulse needed to escape from the Earth is 11 km s\u2212 Many-Body Systems 181 it is 22 km s\u2212 produced by the rockets rather than atmospheric friction. 1, if the deceleration on re-entry into the atmosphere is to be 8.2 Angular Momentum; Central Internal Forces The total angular momentum of our system of particles is J = i miri \u2227 \u02d9ri. The rate of change of J is \u02d9J = miri \u2227 \u00a8ri = ri \u2227 F ij + ri \u2227 F i. i i j i (8.11) (8.12) Now let us examine the contribution to (8.12) from the internal force between a particular pair of particles, say 1 and 2. (See Fig. 8.2.) For simF. plicity, let us write r = r1 \u2212 r2, and F = F 12, so that also F 21 = \u2212 1 F r 1 2 \u2212F r r 2 Fig. 8.2 The contribution", " consists of two terms, r1 \u2227 F 12 + r2 \u2227 F 21 = r1 \u2227 F r2 \u2227 \u2212 F = r F. \u2227 (8.13) This contribution will be zero if F is a central internal force, parallel to r. Let us for the moment simply assume that all the internal forces are central. (We shall discuss the possible justi\ufb01cation for this assumption shortly.) Then all the terms in the double sum in (8.12) will cancel in pairs, just as they did in the evaluation of the rate of change of total linear momentum. The total moment of all the internal forces will be 182 Classical Mechanics zero. So, when the internal forces are central, the rate of change of angular momentum is equal to the sum of the moments of the external forces only: \u02d9J = ri \u2227 F i. i (8.14) In particular, for an isolated system, with all F i = 0, we have the law of conservation of angular momentum, J = constant. (8.15) More generally, this is true if all the external forces are directed towards, or away from, the origin (which may be true for example if they are due to the gravitational attraction of a spherical body). Although we have only shown that (8.14) and (8.15) hold in the special case of central internal forces, they are actually of much more general validity, for reasons that should become clearer in Chapter 12. It is certainly not true that all the internal forces in real solids, liquids or gases are central. Many are, but there are obvious exceptions, notably the electromagnetic force between moving charges. It is better to regard (8.14) as a basic postulate of the dynamics of composite bodies, justi\ufb01ed by the fact that the predictions derived from it agree with observations. It is often convenient, as we did in Chapter 7, to separate the contributions to J from the centre-of-mass motion and the relative motion. We de\ufb01ne the positions r\u2217i of the particles relative to the centre of mass by ri = R + r\u2217i. (8.16) Clearly, the position of the centre of mass relative to itself is zero, so that M R\u2217 = mir\u2217i = 0. (8.17) Now, substituting (8.16) into (8.11), we obtain i \u02d9R + mir\u2217i \u2227 \u02d9R + R mi \u02d9r\u2217i \u2227 i i J = i + \u2227 mi R mir\u2217i \u2227 \u02d9r\u2217i. i The second and third terms vanish by virtue of (8.17). Thus we can write J = M R \u02d9R + J \u2217, \u2227 (8.18) Many-Body Systems 183 where J \u2217 is the angular momentum about the centre of mass, namely J \u2217 = i mir\u2217i \u2227 \u02d9r\u2217i. It is easy to \ufb01nd the rate of change of J \u2217. For, by (8.6), d dt (M R \u2227 \u02d9R) = M R \u00a8R = R \u2227 F i. \u2227 i Hence, subtracting from (8.14), and using (8.16), we obtain \u02d9J \u2217 = r\u2217i \u2227 F i. i (8.19) (8.20) (8.21) Thus the rate of change of the angular momentum J \u2217 about the centre of mass is equal to the sum of the moments of the external forces about the centre of mass. This is a remarkable result. For, it must be remembered that the centre of mass is not in general moving uniformly. In general, we may take moments about the origin of any inertial frame, but it would be quite wrong to take moments about an accelerated point. Only in very special cases, such as where the point is the centre of mass, are we allowed to do this. This result means that in discussing the rotational motion of a body we can ignore the motion of the centre of mass, and treat it as though it were \ufb01xed. It is particularly important in the case of rigid bodies, which we discuss in Chapter 9. For an isolated system, it follows from (8.21) that J \u2217 as well as J is a constant. More generally, J \u2217 is a constant if the external forces have zero total moment about the centre of mass. For example, for a system of particles in a uniform gravitational \ufb01eld, the resultant force acts at the centre of mass, and J \u2217 is therefore a constant. 8.3 The Earth\u2013Moon System As an interesting example of the use of the angular momentum conservation law, we consider the system comprising", " the Earth and the Moon. (We ignore the other planets, and treat the Sun as \ufb01xed.) The angular momentum of this system is J = M R \u02d9R + J \u2217, \u2227 (8.22) 184 Classical Mechanics where M = m + m is the total mass of Earth plus Moon, and J \u2217 is the angular momentum about the centre of mass. This angular momentum J \u2217 can again be separated into an orbital term, due to the motion of the Earth and Moon around their common centre of mass, and two rotational terms, J \u2217E and J \u2217M, due to the rotation of each body about its own centre. The orbital angular momentum, according to (7.15), is \u00b5r \u02d9r, where \u00b5 = mm/M. Thus \u2227 J \u2217 = \u00b5r \u2227 \u02d9r + J \u2217E + J \u2217M. (8.23) Let us \ufb01rst neglect the non-uniformity of the Sun\u2019s gravitational \ufb01eld over the dimensions of the Earth\u2013Moon system. Then J \u2217 as well as J will be constant. Indeed, to a good approximation, the individual terms of (8.23) are separately constant. However, over long periods of time they do change, for at least two reasons. One is the non-spherical shape of the Earth, which, as we saw in 6.5, leads to a precession of the Moon\u2019s orbital plane. This is too small to be of much importance in the present context (the non-uniformity of the Sun\u2019s \ufb01eld produces a much bigger e\ufb00ect), but because of angular momentum conservation there is a corresponding precession of the Earth\u2019s angular momentum J \u2217E which is important. We shall discuss this e\ufb00ect further in 10.1. \u00a7 The second reason is the e\ufb00ect of tidal friction. The dissipation of energy by the tides has the e\ufb00ect of gradually slowing the Earth\u2019s rotation. We can picture this e\ufb00ect as follows: as the Earth rotates, it tries to carry with it the tidal \u2018bulges\u2019, while the Moon\u2019s attraction is pulling them back into line. (See Fig. 8.3.) Thus there is a couple acting to slow the Earth\u2019s \u00a7 Moon Fig. 8.3 rotation, and a corresponding equal and opposite couple tending to increase the orbital angular momentum. To investigate this e\ufb00ect, we shall make the simplest possible approximations. We neglect the rotational angular momentum J \u2217M of the Moon Many-Body Systems 185 (which is in fact very small compared to the other terms in (8.23)), and assume that the Earth\u2019s axis is perpendicular to the Moon\u2019s orbital plane, so that the remaining terms of (8.23) have the same direction. The Earth\u2019s rotational angular momentum, as we shall see in the next chapter, has the form J \u2217E = I\u03c9, where I is the so-called moment of inertia, and \u03c9 the rotational angular velocity. For a sphere of uniform density, mass m and radius r, I has the value I = 2 In fact, the Earth\u2019s density increases towards its centre, and I is therefore somewhat less than this, approximately I = 0.33mr2. Thus the total angular momentum about the centre of mass is 5 mr2. J \u2217 = \u00b5a2\u2126 + 0.33mr2\u03c9, (8.24) where \u2126 is the Moon\u2019s orbital angular velocity. It is convenient to express the angular velocities in terms of the present value \u03c90 of \u03c9, given by (5.1). Using m/\u00b5 = 82.3, we may write the conservation law (8.24) (dividing by \u00b5r2\u03c90) as a r 2 \u2126 \u03c90 + 27.2 \u03c9 \u03c90 = 160, (8.25) where the value of the constant on the right is obtained by using the present values a/r = 60.3 and \u2126/\u03c90 = 0.0365. Note that at present the \ufb01rst term of (8.25) is almost \ufb01ve times as large as the second. Now the orbital angular velocity \u2126 is related to a by (4.32), i.e., or, equivalently \u21262a3 = GM, \u2126 \u03c90 = 17.1 r a 3/2. (8.26) Inserting this into (8.25) yields for the Earth\u2019s rotational angular velocity \u03c9 \u03c90 = 5", ".88 0.629 \u2212 a r 1/2. (8.27) As \u03c9 decreases due to tidal friction, a must increase, i.e., the Moon moves further away from the Earth. Indeed this e\ufb00ect can now be measured: the 1. Hence \u2126 Moon is currently receding from the Earth at about 3.8 cm yr\u2212 decreases too. If we use (8.26) and (8.27) to plot \u2126 and \u03c9 as functions of a we obtain curves like those of Fig. 8.4 (which are not drawn to scale). 186 Classical Mechanics \u03c9,\u2126 \u03c9 0 \u21260 \u03c9 \u2126 Fig. 8.4 a It is important to note that if at any time \u2126 were actually larger than \u03c9, so that the Moon revolved around the Earth in less than a day, then the tidal forces would act to increase the angular velocities and decrease the separation. On this simple picture, the smallest distance there can have been between the bodies corresponds to the \ufb01rst point where the curves cross (about 2.3 Earth radii, with a common rotation period of under 5 hours). However, it is in fact impossible that the Moon should ever have been so close, because it would then have been inside the Roche limit (about 2.9 Earth radii) where the tidal forces become so strong that they would break up the Moon (see Chapter 6, Problem 20). It is now generally believed that the material that later formed the Moon was ejected from a proto-Earth in a violent collision with another planet early in its history, and then gradually condensed into the present Moon in an orbit with a radius of a few Earth radii. The \ufb01nal state of the system in the very distant future would correspond to the second point where the curves cross. Then \u03c9 and \u2126 would again be equal, and tidal forces would play no further role. In this \ufb01nal state, the Earth would always present the same face to the Moon, as the Moon already does to the Earth (except for a small wobble). The distance between Earth and Moon would then be about half as large again as it now is, and the common rotation period (the \u2018day\u2019 or the \u2018month\u2019) would be nearly 48 days. There are, however, several factors which might modify this picture. It was at one time believed that Mercury and the Sun had reached the \ufb01nal state described above, and that Mercury always presents the same face to the Sun. However, it is now known that the ratio of the orbital Many-Body Systems 187 and rotation periods is not unity but 3/2. There are several other cases in the solar system of such rational ratios. The stability of this con\ufb01guration depends on the unusually large eccentricity of Mercury\u2019s orbit. It is unlikely that such a state could be reached by the Earth\u2013Moon system. The \u2018\ufb01nal\u2019 state would in any case not be the end of the evolution of the system, because the solar tides would still be active. They would tend to slow the Earth\u2019s rotation even more, and would therefore gradually pull the Moon back towards the Earth. In the very remote future, the two bodies would collide, although it is probable that the Sun will have reached the end of its life, and engulfed the whole system, long before that happens. A signi\ufb01cant over-simpli\ufb01cation in our argument is the assumption that the Moon\u2019s orbit is a circle in a plane normal to the Earth\u2019s axis. In reality, both its eccentricity and its inclination can change with time, and of course the Earth\u2019s axis also precesses is a complicated way. The e\ufb00ect of tidal dissipation is not only to make the Moon\u2019s orbit expand, but also to make it more eccentric and less inclined. In the distant past, it may have been considerably more steeply inclined than it now is. The Sun has another e\ufb00ect which has been omitted from our discussion. The solar ocean tides are indeed small compared to the lunar ones. However, there are quite substantial semi-diurnal tides in the Earth\u2019s atmosphere, noticeable as \ufb02uctuations in the barometric pressure at sea level. They are predominantly of thermal, rather than gravitational, origin, due to heating during the hours of daylight, and are very di\ufb00erent in character from the ocean tides. The most important component has a period of half a solar, rather than lunar, day. It is a little surprising that the dominant period is 12 hours, rather than 24. This may be an example of the phenomenon of resonance (see 2.6): the amplitude of the 12-hour oscillation may be enhanced by being closer to", " a natural oscillation period of the atmosphere than is 24 hours. As on the oceans, the Sun exerts a gravitational couple on the resulting bulges in the atmosphere. (So does the Moon of course, but its e\ufb00ect averages out to zero over a lunar month.) It happens that the maxima in this case occur before, not after, 12 noon and 12 midnight. Thus the couple acts to speed the Earth\u2019s rotation rather than to slow it. This process has the e\ufb00ect of transferring angular momentum from the Earth\u2019s orbital motion to its rotation. It therefore tends to decrease the Earth\u2019s orbital radius, pulling it (immeasurably slowly) closer to the Sun. A purely gravitational process could not have this e\ufb00ect, for the energy (kinetic plus potential) is actually increasing. This energy comes from solar heating, with the Earth acting as a kind of giant heat engine \u00a7 188 Classical Mechanics converting heat to mechanical work. However, the net e\ufb00ect of the atmospheric tides in speeding the Earth\u2019s rotation is not nearly large enough to counterbalance that of the ocean tides in slowing it. 8.4 Energy; Conservative Forces The kinetic energy of our system of particles is T = 1 2 mi \u02d9r2 i. i (8.28) As in the case of angular momentum, T can be separated into centre-ofmass and relative terms. Substituting (8.16) into (8.28), we again \ufb01nd that the cross terms drop out because of (8.17). Thus we obtain T = 1 2 M \u02d9R 2 + T \u2217, where T \u2217, the kinetic energy relative to the centre of mass, is T \u2217 = 1 2 mi \u02d9r\u2217 i 2. i (8.29) (8.30) The relations (8.3), (8.18) and (8.29) show that, exactly as in the twoparticle case (see (7.16)), the momentum, angular momentum and kinetic energy in any frame are obtained from those in the CM frame by adding the contribution of a particle of mass M located at the centre of mass R. The rate of change of kinetic energy is \u02d9T = mi \u02d9ri \u00b7 \u00a8ri = F ij + \u02d9ri \u00b7 F i. \u02d9ri \u00b7 i i j i (8.31) This equation may be compared with (8.12). Let us again examine the F 21 = F between particles 1 contribution from the internal force F 12 = and 2 (see Fig. 8.2). As in (8.13), it is \u2212 F \u2212 F. \u02d9r1 \u00b7 F = \u02d9r \u02d9r2 \u00b7 This is of course the rate at which the force F does work. However, in contrast to the previous situation, there is in general no reason for this to vanish, and \u02d9T cannot therefore be written in terms of the external forces alone. This is not at all surprising, because in general we must expect there to be some change in the internal potential energy of the system. \u00b7 We may note that there are certain special cases in which a force does no work, and may be omitted from the energy change equation. This will Many-Body Systems 189 be the case, for example, if \u02d9r = 0 (e.g., the reaction at a \ufb01xed pivot), or if F is always perpendicular to \u02d9r (e.g., the magnetic force on a charged particle). An important special case is that of a rigid body, in which the distances between all the particles are \ufb01xed. Then r is constant, so that \u02d9r = 0. If the force is central, then it has the same direction as r, and r is always perpendicular to \u02d9r. Thus a central force between particles whose distance apart is \ufb01xed does no work. \u00b7 Now, however, let us make the less restrictive assumption that all the internal forces are conservative. Then the force F must correspond to a potential energy function of particles 1 and 2, and the rate of working \u02d9r F will be equal to minus the rate of change of this potential energy. Let us denote the internal potential energy \u2014 the sum of the potential energies for all the 1 1) pairs of particles \u2014 by Vint. Then clearly the rate of working of all the internal forces will be minus the rate of change of Vint. Thus we obtain 2 N (N \u2212 \u00b7 d dt (T + Vint) = F i. \u02d9ri \u00b7 i (8.32) The rate of change of the kinetic plus internal potential energy is equal to the rate at which the external forces do work. We can also \ufb01nd the rate of change of energy relative to", " the centre of mass. From (8.6), we have d dt 2 2 M \u02d9R ( 1 ) = M \u02d9R \u00a8R = \u02d9R \u00b7 F i. \u00b7 i Hence, subtracting from (8.32) and using (8.29) and (8.16), we \ufb01nd d dt (T \u2217 + Vint) = F i. \u02d9r\u2217i \u00b7 i (8.33) This equation is the analogue of (8.21). Note that Vint is a function only of rj, and is therefore una\ufb00ected by the choice of reference the di\ufb00erences ri \u2212 frame. Thus T \u2217 + Vint is the total energy in the CM frame. Now let us suppose that the external forces are also conservative. Then there must exist a corresponding external potential energy function Vext, whose rate of change is minus the rate of working of the external forces. From (8.32) we then \ufb01nd the law of conservation of energy T + V = E = constant, where V = Vext + Vint. (8.34) 190 Classical Mechanics It is important to notice, however, that Vext depends on the co-ordinates themselves, not merely the co-ordinate di\ufb00erences, and is therefore dependent on the choice of frame, unlike Vint. Thus in general it is not possible to write down a similar equation in which only the kinetic energy T \u2217 relative to the centre of mass appears. 8.5 Lagrange\u2019s Equations \u00a7 We close this chapter with a brief discussion of the extension of the Lagrangian method of 3.7 to a many-particle system. We consider only the case where all the forces are conservative, so that there exists a total potential energy function V. We saw in 7.1 (see the discussion following r2) (7.9)) that the derivatives of the potential energy function Vint(r1 \u2212 corresponding to a two-body force with respect to the components of r1 or r2 give correctly minus the forces on the corresponding particles. Thus the components of the total force on the ith particle will be minus the derivatives of V with respect to the components of ri. The equations of motion may thus be written \u00a7 mi \u00a8xi = \u2202V \u2202xi, \u2212 (i = 1, 2,..., N ), (8.35) together with similar equations for the y and z components. Now, clearly, \u2202T \u2202 \u02d9xi = mi \u02d9xi, while \u2202T /\u2202xi and \u2202V /\u2202 \u02d9xi both vanish. Thus (8.35) are identical with Lagrange\u2019s equations d dt \u2202L \u2202 \u02d9xi = \u2202L \u2202xi, (i = 1, 2,..., N ), (8.36) with L = T V. Exactly as in \u2212 3.7, it follows that the action integral, \u00a7 t1 I = L dt, t0 is stationary under arbitrary small variations of the 3N co-ordinates ri = (xi, yi, zi) which vanish at the limits of integration. Then, again as in 3.7, we may re-express L in terms of any other set of 3N co-ordinates, \u00a7 Many-Body Systems 191 q1, q2,..., q3N. The corresponding equations of motion will be Lagrange\u2019s equations d dt \u2202L \u2202 \u02d9q\u03b1 = \u2202L \u2202q\u03b1, (\u03b1 = 1, 2,..., 3N ). (8.37) As an example, let us suppose that external forces are provided by a uniform gravitational \ufb01eld g. Example: Uniform gravitational \ufb01eld Show that the centre-of-mass and internal motions can be completely separated in the case of a uniform gravitational \ufb01eld, with two separate energy conservation laws. The potential energy function for the uniform \ufb01eld is Vext = \u2212 mig ri = M g R. \u00b7 \u2212 \u00b7 i Now, let us write the Lagrangian in terms of the three co-ordinates R and the co-ordinates r\u2217i (of which only 3N 3 are independent, because of (8.17)). Then, using (8.29), we \ufb01nd \u2212 L = 1 2 M \u02d9R 2 + M g R + \u00b7 mi \u02d9r\u2217 2 i \u2212 Vint. (8.38) i rj = Since Vint is a function only of the co-ordinate di\ufb00erences ri \u2212 r\u2217j, this Lagrangian function exhibits a complete separation between r", "\u2217i \u2212 the terms involving R and those involving the r\u2217i, just as in the two7.1. For the case of a uniform gravitational \ufb01eld, the particle case of motion of the centre of mass and the relative motion are uncoupled. In particular, there are two separate conservation laws for energy, \u00a7 2 1 2 M \u02d9R M g \u00b7 \u2212 R = constant, T \u2217 + Vint = constant. 8.2, there is also in this case a separate conservation law and As we saw in \u00a7 for angular momentum relative to the centre of mass. There is an important physical consequence of this separation: no observation of the internal motion of a system can ever reveal the presence of 192 Classical Mechanics a uniform gravitational \ufb01eld, since such a \ufb01eld in no way a\ufb00ects the equations of motion for the relative co-ordinates r\u2217i. Note that a laboratory is subjected to other external forces besides the Earth\u2019s gravitational \ufb01eld, since it is supported by the ground. Indeed if the Earth were removed, but the supporting forces somehow retained, there would be no observable difference inside the laboratory, which would of course be accelerated upwards with acceleration g! This was an important consideration in the argument which led to Einstein\u2019s general theory of relativity. 8.6 Summary The centre of mass of any system moves like a particle of mass M acted on by a force equal to the total force on the system. The contribution of this motion to the angular momentum or kinetic energy may be completely separated from the contributions of the relative motion, and J or T may (Of course, the only be written as a sum of two corresponding terms. contribution to P comes from the centre-of-mass motion.) When the internal forces are central, the rate of change of angular momentum is equal to the sum of the moments of the external forces. When they are conservative, the rate of change of the kinetic energy plus the internal potential energy is equal to the rate of working of the external forces. In both cases, the same thing is true for the motion relative to the centre of mass. If the external forces are also central, or conservative, then the total angular momentum, or total energy (including external potential energy), respectively, are conserved. In particular, for an isolated system, P, J and T + Vint are all constants. Problems 1. A rocket is launched from the surface of the Earth, to reach a height of 50 km. Find the required velocity impulse, neglecting the variation of g with height (and the Earth\u2019s rotation). Given that the mass of the payload and rocket without fuel is 100 kg, and the ejection velocity is 2 km s\u2212 1, \ufb01nd the required initial mass. 2. A satellite is orbiting the Earth in a circular orbit 230 km above the equator. Calculate the total velocity impulse needed to place it in a Many-Body Systems 193 synchronous orbit (see Chapter 4, Problem 1), using an intermediate (Hint : semi-elliptical transfer orbit which just touches both circles. From the orbit parameters a and ae, \ufb01nd l, and hence the velocities.) Given that the \ufb01nal mass to be placed in orbit is 30 kg, and the ejection velocity of the rocket is 2.5 km s\u2212 1, \ufb01nd the necessary initial mass. 3. Assume that the residual mass of a rocket, without payload or fuel, is a given fraction \u03bb of the initial mass including fuel (but still without payload). Show that the total take-o\ufb00 mass required to accelerate a payload m to velocity v is M0 = m 1 \u2212 v/u e\u2212 \u03bb \u2212. \u03bb If \u03bb = 0.15, what is the upper limit to the velocity attainable with an ejection velocity of 2.5 km s\u2212 1? 4. Find a formula analogous to that of Problem 3 for a two-stage rocket, in which each stage produces the same velocity impulse. (The \ufb01rststage rocket is discarded when its fuel is burnt out.) With the \ufb01gures of Problem 3, what is the minimum number of stages required to reach 1)? With that number, what take-o\ufb00 mass escape velocity (11.2 km s\u2212 is required, if the payload mass is 100 kg? 5. Find the velocity impulse needed to launch the spacecraft on its trip to Jupiter, described in Chapter 4, Problem 16. (Hint : Use energy conservation to \ufb01nd the velocity at the surface of the Earth needed to give the appropriate relative velocity for the spacecraft once it has escaped.) If a three-stage rocket is used, and the parameter \u03bb of Problem 3 is 0.1, what is the minimum required ejection velocity? Given that 1, and that the mass of the payload is", " 500 kg, \ufb01nd the u = 2.5 km s\u2212 total take-o\ufb00 mass. 6. Find the gain in kinetic energy when a rocket emits a small amount of matter. Hence calculate the total energy which must be supplied from chemical or other sources to accelerate the rocket to a given velocity. Show that this is equal to the energy required if an equal amount of matter is ejected while the rocket is held \ufb01xed on a test-bed. 7. *A rocket with take-o\ufb00 mass M0 is launched vertically upward, as in Problem 1. Consider the e\ufb00ect of a \ufb01nite burn-up time. Show that, if the rocket ejects matter at a constant rate a, then its height at time t is z = ut uM a ln M0 M \u2212 \u2212 1 2 gt2, with M = M0 \u2212 at. 194 Classical Mechanics Hence, show that if it burns out after a time t1, leaving a \ufb01nal mass M1, then [provided that t1 < (u/g) ln(M0/M1)], the maximum height reached is z = ln u2 2g 2 M0 M1 ut1 \u2212 M0 M0 \u2212 M1 ln M0 M1 \u2212 1. Using the same values of u, M0 and M1 as in Problem 1, \ufb01nd the maximum height reached if the burn-up time is (a) 10 s, and (b) 30 s. 8. *A satellite is to be launched into a synchronous orbit directly from the surface of the Earth, using a rocket launched vertically from a point on the equator. Find the required launch speed to achieve the desired apogee, and the required velocity increment. With the same \ufb01gures as in Problem 2, \ufb01nd the required take-o\ufb00 mass. 9. Two billiard balls are resting on a smooth table, and just touching. A third identical ball moving along the table with velocity v perpendicular to their line of centres strikes both balls simultaneously. Find the velocities of the three balls immediately after impact, assuming that the collision is elastic. 10. A spherical satellite of radius r is moving with velocity v through a uniform tenuous atmosphere of density \u03c1. Find the retarding force on the satellite if each particle which strikes it (a) adheres to the surface, and (b) bounces o\ufb00 it elastically. Can you explain why the two answers are equal, in terms of the scattering cross-section of a hard sphere? 11. *If the orbit of the satellite of Problem 10 is highly elliptical, the retarding force is concentrated almost entirely in the lowest part of the orbit. Replace it by an impulsive force of impulse I delivered once every orbit, at perigee. By considering changes in energy and angular momentum, \ufb01nd the changes in the parameters a and l. Show that e)2, and hence that the e\ufb00ect is to decrease the period and \u03b4l = \u03b4a(1 apogee distance, while leaving the perigee distance una\ufb00ected. (The orbit therefore becomes more and more circular with time.) Show that the velocity at apogee increases, while that at perigee decreases. 12. *Suppose that the satellite of Problems 10 and 11 has achieved a circular orbit of radius a. Find the rates of change of energy and angular momentum, and hence show that the rates of change of a and l are equal, so that the orbit remains approximately circular. Show also that the velocity of the satellite must increase. \u2212 13. If the satellite orbit of Problem 12 is 500 km above the Earth\u2019s surface, the mass and radius of the satellite are 30 kg and 0.7 m, and Many-Body Systems 195 13 kg m\u2212 \u03c1 = 10\u2212 year. If the height is 200 km and \u03c1 = 10\u2212 a single orbit. 3, \ufb01nd the changes in orbital period and height in a 3, \ufb01nd the changes in 10 kg m\u2212 14. Find the lengths of the \u2018day\u2019 and the \u2018month\u2019 (a) when the Moon was 10 Earth radii away, and (b) when the solar and lunar tides become equal in magnitude. (See Chapter 6, Problem 16.) 15. Show that in a conservative N -body system, a state of minimal total energy for a given total z-component of angular momentum is necessarily one in which the system is rotating as a rigid body about the z-axis. [Use the method of Lagrange multipliers (see Appendix A, Problem 11), and treat the components of the positions ri and velocities \u02d9ri as independent variables.] 16. *A planet of mass M is surrounded by a cloud of small", " particles in orbits around it. Their mutual gravitational attraction is negligible. Due to collisions between the particles, the energy will gradually decrease from its initial value, but the angular momentum will remain \ufb01xed, J = J 0, say. The system will thus evolve towards a state of minimum energy, subject to this constraint. Show that the particles will tend to form a ring around the planet. [As in Problem 15, the constraint may be imposed by the method of Lagrange multipliers. In this case, because there are three components of the constraint equation, we need three Lagrange multipliers, say \u03c9x, \u03c9y, \u03c9z. We have to minimize the function J 0) with respect to variations of the positions ri and E velocities \u02d9ri, and with respect to \u03c9. Show by minimizing with respect to \u02d9ri that once equilibrium has been reached the cloud rotates as a rigid body, and by minimizing with respect to ri that all particles occupy the same orbit.] What happens to the energy lost? Why does the argument not necessarily apply to a cloud of particles around a hot star? (J \u2212 \u2212 \u03c9 \u00b7 17. *An N -body system is interacting only through the gravitational forces between the bodies. Show that the potential energy function V satis\ufb01es the equation \u2207iV = ri \u00b7 V, \u2212 i where \u2207i = (\u2202/\u2202xi, \u2202/\u2202yi, \u2202/\u2202zi). (Hint : Show that each two-body term Vij satis\ufb01es this equation. This condition expresses the fact that V is a homogeneous function of the co-ordinates of degree 1.) \u2212 196 Classical Mechanics 18. *Under the conditions of Problem 17, show that the total kinetic and potential energies T and V satisfy the virial equation, 2T + V = d2K dt2, where K = 1 2 mir2 i. i (Note that K relates to the overall scale of the system. We may de\ufb01ne a root-mean-square radius r by K = 1 2 M r2.) Deduce that, if the scale of the system as measured by K is, on average, neither growing nor shrinking, then the time-averaged value of the total energy is equal to minus the time-averaged value of the kinetic energy \u2014 the virial theorem. (Compare Chapter 4, Problem 19.) Chapter 9 Rigid Bodies The principal characteristic of a solid body is its rigidity. Under normal circumstances, its size and shape vary only slightly under stress, changes in temperature, and the like. Thus it is natural to consider the idealization of a perfectly rigid body, whose size and shape are permanently \ufb01xed. Such a body may be characterized by the requirement that the distance between any two points of the body remains \ufb01xed. In this chapter, we shall be concerned with the mechanics of rigid bodies. 9.1 Basic Principles It will be convenient to simplify the notation of the previous chapter by omitting the particle label i from sums over all particles in the rigid body. Thus, for example, we shall write P = m \u02d9r, and J = in place of (8.3) and (8.11). mr \u02d9r, \u2227 The motion of the centre of mass of the body is completely speci\ufb01ed by (8.6): \u02d9P = M \u00a8R = F. (9.1) Our main interest in this chapter will be centred on the rotational motion of the rigid body. For the moment, let us assume, as we originally did in 8.3, that the internal forces are central. Then, according to (8.14), \u00a7 \u02d9J = F. r \u2227 197 (9.2) 198 Classical Mechanics We shall see later that these two equations are su\ufb03cient to determine the motion completely. One very important application of (9.1) and (9.2) should be noted. For \u02d9r = 0 for every particle, and thus both P and J a rigid body at rest, vanish. Clearly, the body can remain at rest only if the right hand sides of both (9.1) and (9.2) vanish, i.e., if the sum of the forces and the sum of their moments are both zero. In fact, as we shall see, this is not only a necessary, but also a su\ufb03cient, condition for equilibrium. Under the same assumption of central internal forces, we saw in that the internal forces do no work, so that \u02d9T = F. \u02d9r \u00b7 8.5 \u00a7 (9.3) This might at \ufb01rst sight appear to be a third independent equation. However, we shall see later that it is actually a consequence", " of the other two. It is of course particularly useful in the case when the external forces are conservative, since it then leads to the conservation law T + V = E = constant, (9.4) where V is the external potential energy, previously denoted by Vext. (Note that there is no mention here of internal potential energy; in a rigid body, this does not change.) \u00a7 The assumption that the internal forces are central is much stronger than it need be. Indeed, as we discussed in 8.3, it could not be justi\ufb01ed from our knowledge of the internal forces in real solids, which are certainly not exclusively central, and in any case cannot be adequately described by classical mechanics. All we actually require is the validity of the basic equations (9.1) and (9.2), and it is better to regard these as basic assumptions of rigid body dynamics, whose justi\ufb01cation lies in the fact that their consequences agree with observation. To some extent, the success of these postulates, particularly (9.2), may seem rather fortuitous. However, we shall see in Chapter 12 that it is closely related to the relativity principle discussed in Chapter 1. 9.2 Rotation about an Axis Let us now apply these basic equations to a rigid body which is free to rotate only about a \ufb01xed axis, which for simplicity we take to be the z-axis. Rigid Bodies 199 We also choose the origin on this axis, positioned so that the z co-ordinate of the centre of mass is zero. In cylindrical polars, the z and \u03c1 co-ordinates of every point are \ufb01xed, while the \u03d5 co-ordinate varies according to \u02d9\u03d5 = \u03c9, the angular velocity of the body (not necessarily constant). We examine \ufb01rst the component of angular momentum about the axis of rotation. By (3.29), it is where Jz = m\u03c1v\u03d5 = m\u03c12\u03c9 = I\u03c9, I = m\u03c12 (9.5) (9.6) is the moment of inertia about the axis. Since I is obviously constant, the z component of (9.2) yields \u02d9Jz = I \u02d9\u03c9 = \u03c1F\u03d5. (9.7) This equation determines the rate of change of angular velocity, and may be called the equation of motion of the rotating body. The condition for equilibrium here is simply that the right hand side of (9.7) vanishes, i.e., that the forces have no net moment about the z-axis. For example, consider a rectangular lamina (i.e., a plane, two-dimensional object) of size a b and negligible mass, pivoted at one corner, carrying a weight M g and supported by a horizontal force F, as shown in Fig. 9.1. The aM g, so for equilibrium we require F = (a/b)M g. total moment here is bF \u00d7 \u2212 F b Q a Mg Fig. 9.1 The kinetic energy may also be expressed in terms of I. For 1 2 m(\u03c1 \u02d9\u03d5)2 = 1 2 I\u03c92. T = (9.8) 200 Classical Mechanics The equation (9.3) is, therefore, \u02d9T = I\u03c9 \u02d9\u03c9 = (\u03c1 \u02d9\u03d5)F\u03d5 = \u03c9 \u03c1F\u03d5. This is clearly a consequence of (9.7), and gives us no additional information. The momentum equation (9.1) serves to determine the reaction at the axis, which has of course no moment about the axis, and so does not appear in (9.7). Let us denote the force on the body at the axis by Q, and separate this from all the other forces on the body. Then from (9.1) \u02d9P = M \u00a8R = Q + F. (9.9) For example, in the case illustrated in Fig. 9.1, the equilibrium condition \u02d9P = 0 immediately yields Q = (F, M g, 0). (Note that the z-axis here is horizontal.) The centre of mass is of course \ufb01xed in the body, so that, by (5.2), R. Di\ufb00erentiating again to \ufb01nd the acceleration of the centre of \u02d9R = \u03c9 \u2227 mass, we obtain \u00a8R = \u02d9\u03c9 R + \u03c9 \u2227 \u2227 \u02d9R = \u02d9\u03c9 R + \u03c9 (\u03c9 \u2227 \u2227 R). \u2227 (9.10) The \ufb01rst term is the tangential acceleration, R \u02d9\u03c9 in the \u03d5 direction, and the \u03c92R in the \u03c1 direction. (", "Compare (3.47).) second the radial acceleration, Together, equations (9.9) and (9.10) serve to determine Q. \u2212 As a simple example, let us consider a compound pendulum \u2014 a rigid body pivoted about a horizontal axis, and moving under gravity. Example: Compound pendulum Find the period of small oscillations of a compound pendulum, and the reaction at the pivot as a function of angle. To be consistent, we shall still take the z-axis to be the axis of rotation, and choose the x-axis vertically downward. (See Fig. 9.2.) Then the force (the weight) acts at the centre of mass, and has the components F = (M g, 0, 0). Thus the equation of motion (9.7) is I \u00a8\u03d5 = \u2212 M gR sin \u03d5. (9.11) Rigid Bodies 201 y R \u03d5 F x Fig. 9.2 This is identical with the equation of motion of a simple pendulum of length l = I/M R. (This is the length of the equivalent simple pendulum) Thus the period of small oscillations is 2\u03c0 I/M gR. As we shall see, l is always larger than the distance R from the pivot to the centre of mass. l/g = 2\u03c0 The energy conservation equation is T + V = 1 2 I \u02d9\u03d52 \u2212 M gR cos \u03d5 = E = constant. (9.12) It may be obtained from (9.11) by multiplying by \u02d9\u03d5 and integrating. As usual, E is determined by the initial conditions, and (9.12) then serves to \ufb01x the angular velocity \u02d9\u03d5 for any given inclination \u03d5. The components of the reaction Q, obtained from (9.9), are Qz = 0 Q\u03c1 = Q\u03d5 = M g sin \u03d5 + M R \u00a8\u03d5. M g cos \u03d5 \u2212 \u2212 M R \u02d9\u03d52, Thus, substituting for \u00a8\u03d5 in terms of \u03d5 using (9.11), and for \u02d9\u03d5 in terms of \u03d5 by (9.12), we \ufb01nd Qz = 0 Q\u03c1 = \u2212 Q\u03d5 = M g(1 M g(1 + 2R/l) cos \u03d5 2E/l, \u2212 R/l) sin \u03d5. \u2212 202 Classical Mechanics Note, however, that this result is not the same as the reaction at the pivot of a simple pendulum of length l. As a rather di\ufb00erent illustration of the use of these equations, let us consider the e\ufb00ect of a sudden blow. Example: E\ufb00ect of sudden blow Suppose that the pendulum is initially in equilibrium and is given a sharp horizontal blow of impulse K at a distance d below the axis (see Fig. 9.3). For what value of d does the impulsive reaction at the pivot vanish? S d K Fig. 9.3 From (9.11) integrated over a short time interval, we learn that the angular velocity \u03c9 just after the blow is given by I\u03c9 = dK. As before, (9.9) serves to \ufb01x the reaction at the axis, in this case the impulsive reaction, S say. Since the velocity of the centre of mass immediately after the blow is \u03c9R, the integrated form of (9.9) is whence M \u03c9R = S + K, \u2212 S = 1 \u2212 M dR I K. The interesting thing about this result is that it is possible to make S vanish by choosing d equal to the length of the equivalent simple pendulum, d = I/M R = l. This is the ideal point at which to hit a ball with a bat (sometimes called the \u2018sweet spot\u2019 in sporting applications \u2014 cricket, tennis, baseball, etc.). Rigid Bodies 203 9.3 Perpendicular Components of Angular Momentum We now consider the remaining components of the angular momentum vector. We shall see that they give us further information about the reaction on the axis. In Cartesian co-ordinates, the velocity of a point r of the body is given by Thus we \ufb01nd \u02d9x = \u03c9y, \u2212 \u02d9y = \u03c9x, \u02d9z = 0. (9.13) Jx = Jy = m( z \u02d9y) = m(z \u02d9x) = \u2212 \u2212 \u2212 mxz\u03c9, myz\u03c9. \" \" \" \" We can write all three components of J in the form Jx = Ixz\u03c9, Jy = Iyz\u03c9, Jz = Izz\u03c9, (9.14) where Ixz = \u2212 mxz, Iyz = \u2212", " myz, Izz = m(x2 + y2). (9.15) Here Izz is the moment of inertia about the z-axis, previously denoted by I. The quantities Ixz and Iyz are called products of inertia. At \ufb01rst sight it may seem surprising that J has components in directions perpendicular to \u03c9. A simple example may help to clarify the reason for this. Consider a light rigid rod with equal masses m at its two ends, rigidly \ufb01xed at an angle \u03b8 to an axis through its mid-point. (See Fig. 9.4.) If the positions of the masses are r and r, the total angular momentum is \u2212 J = mr \u02d9r + m( \u2212 r) ( \u2212 \u2227 \u2227 \u02d9r) = 2mr (\u03c9 \u2227 \u2227 r). Clearly, J is perpendicular to r, as shown in the \ufb01gure. When the rod is in the xz-plane, the masses are moving in the y directions, and there is a component of angular momentum about the x-axis as well as one about the z-axis. \u00b1 In this example, the centre of mass lies on the axis. Thus if no external force is applied, the total reaction on the axis is zero. There is, however, a resultant couple on the axis (here represented by the forces F ), which is required to balance the couple produced by the centrifugal forces. The magnitude of this couple may be determined from the remaining pair of the 204 Classical Mechanics \u03c9\u03c9 J \u2212r \u2212F \u2032 F \u2032 d/2 \u03b8 d/2 r Fig. 9.4 equations (9.2). From (9.15), we see that, unlike Izz the products of inertia are not constants, in general. Using (9.13), we \ufb01nd \u02d9Ixz = \u03c9Iyz, \u2212 \u02d9Iyz = \u03c9Ixz, \u02d9Izz = 0. Thus if G is the couple on the axis, we have from (9.2) \u02d9Jx = Ixz \u02d9\u03c9 \u2212 Iyz\u03c92 = Gx + \u02d9Jy = Iyz \u02d9\u03c9 + Ixz\u03c92 = Gy + (yFz \u2212 (zFx \u2212 zFy), xFz). If there are no other external forces, then \u03c9 is a constant, and G precisely balances the centrifugal couple. When the rod is in the xz-plane, the only non-vanishing component of G is the moment about the y-axis, Gy = dF x = Ixz\u03c92 = \u2212 2mr2\u03c92 sin \u03b8 cos \u03b8. This determines F. \u2212 \u2212 Rigid Bodies 205 9.4 Principal Axes of Inertia We have seen that, in general, the angular momentum vector J is in a di\ufb00erent direction from the angular velocity vector \u03c9. There are special cases, however, in which the products of inertia Ixz and Iyz vanish. Then J is also in the z direction. In that case, the z-axis is called a principal axis of inertia. When a body is rotating freely about a principal axis through its centre of mass, there is no resultant force or couple on the axis. The z-axis is a principal axis, in particular, if the xy-plane is a plane of re\ufb02ection symmetry; for then the contributions to the products of inertia Ixz and Iyz from any point (x, y, z) is exactly cancelled by that from (x, y, z). Similarly, it is a principal axis if it is an axis of rotational symmetry, for then the contribution from (x, y, z) is cancelled by that from ( y, z). x, \u2212 For bodies with three symmetry axes \u2014 such as a rectangular parallelepiped or an ellipsoid \u2014 there are obviously three perpendicular principal axes, and we shall see later that this is true more generally. In this case, it is clearly an advantage to choose these as our co-ordinate axes. So we shall no longer assume that the axis of rotation is the z-axis, but take it to have an arbitrary inclination. Then \u03c9 has three components (\u03c9x, \u03c9y, \u03c9z). The angular momentum J may be written, using (A.16), in the form J = mr (\u03c9 \u2227 \u2227 r) = m[r2\u03c9 (r \u00b7 \u2212 \u03c9)r]. (9.16) Its components are therefore linear functions of the components of \u03c9, which we may write in matrix notation as Jx Jy Jz \u23a1 = \u23a4 Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz", " \u23a1 \u03c9x \u03c9y \u03c9z, \u23a4 \u23a4 \u23a1 (9.17) \u23a3 \u23a6 \u23a3 \u23a6 \u23a3 \u23a6 that is, Jx = Ixx\u03c9x + Ixy\u03c9y + Ixz\u03c9z, and two similar equations. The nine elements Ixx, Ixy,..., Izz of the 3 3 matrix may be regarded as the components of a single entity I, in much the same way as the three quantities Jx, Jy, Jz are regarded as the components of the vector J. The entity I is called a tensor, in this case the inertia tensor. The elementary properties of tensors are described in A.9. However, we shall need to use only one result from this general theory. \u00d7 \u00a7 It is obvious from the de\ufb01nition (9.15) that the products of inertia sat3 matrix in (9.17) is isfy symmetry relations like Ixy = Iyx. The 3 therefore unchanged by re\ufb02ection in the leading diagonal. A tensor I with this property is called symmetric. \u00d7 206 Classical Mechanics If the three co-ordinate axes are all axes of symmetry, then all the products of inertia vanish, and I has the diagonal form Ixx 0 0 0 Iyy 0 0 Izz 0 \u23a1. \u23a4 \u23a6 In this case, the relations (9.17) simplify to \u23a3 Jx = Ixx\u03c9x, Jy = Iyy\u03c9y, Jz = Izz\u03c9z. (9.18) Thus J is parallel to \u03c9 if \u03c9 is along any one of the three symmetry axes, but not in general otherwise. \u00a7 It is shown in A.10 that for any given symmetric tensor one can always \ufb01nd a set of axes with respect to which it is diagonal. Thus for any rigid body, we can \ufb01nd three perpendicular axes through any given point which are principal axes of inertia. It will be convenient to introduce three unit vectors e1, e2, e3 along these axes. Then, if we write \u03c9 = \u03c91e1 + \u03c92e2 + \u03c93e3, (9.19) the components of J in these three directions will be obtained by multiplying the components of \u03c9 by the appropriate moments of inertia, as in (9.18). Thus we obtain J = I1\u03c91e1 + I2\u03c92e2 + I3\u03c93e3. (9.20) The three diagonal elements of the inertia tensor, I1, I2, I3 are called principal moments of inertia. We shall always use a single subscript for the principal moments, to distinguish them from moments of inertia about arbitrary axes. It is important to realize that the principal axes are \ufb01xed in the body, not in space, and therefore rotate with it. It is often convenient to use these axes to de\ufb01ne our frame of reference, particularly since the principal moments I1, I2, I3 are constants. This is, however, a rotating frame, not an inertial one. The kinetic energy may also be expressed in terms of the angular velocity and the inertia tensor. We have T = 1 2 m \u02d9r2 = 1 2 m(\u03c9 \u2227 r)2 = 1 2 m[\u03c92r2 (\u03c9 \u00b7 \u2212 r)2], Rigid Bodies 207 by a standard formula of vector algebra (see Appendix A, Problem 7). Comparing with (9.16), we see that T = 1 2 \u03c9 J. \u00b7 (9.21) (Equations (9.5) and (9.8) provide a special case of this general result.) Thus from (9.19) and (9.20), we \ufb01nd T = 1 2 I1\u03c92 1 + 1 2 I2\u03c92 2 + 1 2 I3\u03c92 3. (9.22) These equations for rotational motion may be compared with the corresponding ones for translational motion with velocity v, namely T = 2 M v2 = 1 1 P. The principal di\ufb00erence is that the mass, unlike the inertia 2 v tensor, has no directional properties, so that the coe\ufb03cients of v2 y and v2 z are all equal. x, v2 \u00b7 Symmetric bodies The term symmetric applied to a rigid body has a special, technical significance. It means that two of the principal moments of inertia coincide, say I1 = I2. This is true in particular if the third axis e3 is a symmetry axis of appropriate type. It may be an axis of cylindrical symmetry \u2014 for example, the axis of a spheroid or a circular cylinder or cone. More generally,", " I1 = I2 if e3 is an axis of more than two-fold rotational symmetry \u2014 for example the axis of an equilateral triangular prism, or a square pyramid. (In the case of the prism, this is not entirely obvious. It follows from the fact that one cannot \ufb01nd a pair of perpendicular axes e1 and e2 which are in any way distinguished from other possible pairs.) For a symmetric body, (9.20) becomes J = I1(\u03c91e1 + \u03c92e2) + I3\u03c93e3. (9.23) Note that if \u03c93 = 0, then J = I1\u03c9, so J is parallel to \u03c9. In other words, any axis in the plane of e1 and e2 is a principal axis. Assuming that = I1, the symmetry axis e3 is of course uniquely determined, but the I3 axes e1 and e2 are not. We may choose them to be any pair of orthogonal axes in the plane normal to e3. Indeed, they need not even be \ufb01xed in the body, so long as they always remain perpendicular to e3 and to each other. Equation (9.23) will still hold so long as this condition is satis\ufb01ed. This freedom of choice will prove to be very useful later. It can happen that all three principal moments of inertia are equal, I1 = I2 = I3. This is the case for a sphere (with origin at the centre), a cube, 208 Classical Mechanics or a regular tetrahedron (or, indeed, any of the \ufb01ve regular solids). It may of course also happen by accidental coincidence of the three values. We may call such bodies totally symmetric. They have the property that J = I1\u03c9 whatever direction \u03c9 may take, so that every axis is a principal axis, and the choice of the orthonormal triad e1, e2, e3 is completely arbitrary. 9.5 Calculation of Moments of Inertia The moments and products of inertia of any body with respect to a given origin may be calculated from the de\ufb01nitions (9.15). For continuous distributions of matter, we must replace sums by integrals: Ixx = \u03c1(r)(y2 + z2) d3r, Ixy = etc., where \u03c1(r) is the density. \u03c1(r)( \u2212 xy) d3r, (9.24) Shift of origin When the rigid body is pivoted so that one point is \ufb01xed, it is convenient to choose that point to be the origin. If there is no \ufb01xed point, we generally choose the origin to be the centre of mass. Thus it is useful to be able to relate the moments and products of inertia about an arbitrary point to those about the centre of mass. As usual, we shall distinguish quantities referred to the centre of mass as origin by an asterisk. To \ufb01nd the desired relations, we substitute in (9.15) the relation (8.16), r = R + r\u2217, and use the conditions (8.17), namely mx\u2217 = my\u2217 = mz\u2217 = 0. Because of these relations, the cross terms between R and r\u2217 drop out, exactly as they did in (8.18) or (8.29). For example, Ixy = \u2212 m(X + x\u2217)(Y + y\u2217) = M XY \u2212 \u2212 mx\u2217y\u2217. The last term here is just the product of inertia I \u2217xy referred to the centre of mass as origin. Thus we obtain relations of the form Ixx = M (Y 2 + Z 2) + I \u2217xx, Ixy = \u2212 M XY + I \u2217xy, (9.25) etc. Note that, as in previous cases, the components of the inertia tensor with respect to an arbitrary origin are obtained from those with respect to Rigid Bodies 209 the centre of mass by adding the contribution of a particle of mass M at R. (For the moments of inertia, this is known as the parallel axes theorem.) Because of this result, it is only necessary, for any given body, to compute the moments and products of inertia with respect to the centre of mass; those with respect to any other origin are then given by (9.25). It is important to realize that the principal axes at a given origin are not necessarily parallel to those at the centre of mass. If we choose the axes at the centre of mass to be principal axes, then the products of inertia I \u2217xy, I \u2217xz, I \u2217yz will all be zero, but it is clear from (9.25) that this does not necessarily imply that Ixy, I", "xz, Iyz are zero. In fact, this will be true if, and only if, the chosen origin lies on one of the principal axes through the centre of mass, so that two of the three co-ordinates X, Y, Z are zero. Routh\u2019s rule There is a simple and convenient formula that allows us to write down at once the moments of inertia of a large class of bodies. We consider bodies with uniform density \u03c1 and three perpendicular symmetry planes. The principal axes are then obvious, and we shall take them to be the co-ordinate axes. From (9.24), we see that the principal moments of inertia may be written in the form I \u22171 = Ky + Kz, I \u22172 = Kx + Kz, I \u22173 = Kx + Ky, where, for example, Kz = \u03c1z2 dx dy dz, V integrated over the volume V of the body. The mass of the body is of course M = \u03c1 dx dy dz. V Now let us denote the lengths of the three co-ordinate axes by 2a, 2b, 2c, respectively, and consider together all bodies of the same type (e.g., ellipsoids, or rectangular parallelepipeds) but with di\ufb00erent values of a, b, c. It is then easy to \ufb01nd how M and Kz depend on these three lengths. Making the substitution x = a\u03be, y = b\u03b7, z = c\u03b6, we see that M = \u03c1abc d\u03be d\u03b7 d\u03b6, V0 210 Classical Mechanics where V0 is a standard body of this type with \u03c1 = 1 and a = b = c = 1. It follows that Kz = Hence M \u03bbzM c2, where \u03bbz is a dimensionless number, the same for all bodies of this type. Hence we obtain Routh\u2019s rule, which asserts that In a similar way, Kz \u221d \u03c1abc3. \u03c1abc. \u221d I \u22171 = M (\u03bbyb2 + \u03bbzc2), I \u22172 = M (\u03bbxa2 + \u03bbzc2), I \u22173 = M (\u03bbxa2 + \u03bbyb2). (9.26) The values of the constants \u03bbx, \u03bby, \u03bbz may be found by examining the standard volume V0. For ellipsoids, for example, V0 is a sphere of unit radius, for which Kz0 = \u03b62 d\u03be d\u03b7 d\u03b6 = V0 1 1 \u2212 \u03b62\u03c0(1 \u2212 \u03b6 2) d\u03b6 = 4\u03c0 15. Since also M0 = 4\u03c0/3, we have \u03bbz = Kz0/M0 = 1 5. Thus for an ellipsoid \u03bbx = \u03bby = \u03bbz = 1 5. (ellipsoid) A similar calculation shows that for a rectangular parallelepiped \u03bbx = \u03bby = \u03bbz = 1 3. (parallelepiped) As an example of a body for which the three constants are not all equal, we may quote the elliptic cylinder of length 2c, for which \u03bbx = \u03bby = 1 4 \u03bbz = 1 3. (cylinder) All three examples may be summarized by saying that \u03bb is 1/3 for \u2018rectangular\u2019 axes, 1/4 for \u2018elliptic\u2019 ones, and 1/5 for \u2018ellipsoidal\u2019 ones. This formula covers several special cases of particular interest. A sphere is of course an ellipsoid with a = b = c, and each of its three principal moments of inertia is 2 5 M a2. Similarly a cube is a parallelepiped with a = b = c, and each of its principal moments is 2 3 M a2. A parallelepiped with c = 0 is a \ufb02at rectangular plate. Its principal moments are I \u22171 = 1 3 M b2, I \u22172 = 1 3 M a2, I \u22173 = 1 3 M (a2 + b2). Rigid Bodies 211 Similarly, a \ufb02at circular plate is a cylinder with a = b and c = 0. In that case, I \u22171 = I \u22172 = 1 4 M a2, I \u22173 = 1 2 M a2. Finally, a thin rod is a limiting case of either a parallelepiped or a cylinder with a = b = 0. Its principal moments are I \u22171 = I \u22172 = 1 3 M c2, I \u22173 = 0. 9.6 E\ufb00ect", " of a Small Force on the Axis So far we have assumed that the axis about which the rigid body is rotating is \ufb01xed. We now go on to consider how it moves if only one point on the axis is \ufb01xed. As we shall see, rapidly rotating bodies have great stability, which forms the basis, for example, of the gyroscope. We suppose that the rigid body is free to rotate about a \ufb01xed smooth pivot, and that initially it is rotating freely about a principal axis, say e3. Then, if the angular velocity is \u03c9 = \u03c9e3, the angular momentum will be J = I3\u03c9e3. So long as no external force acts on the body the angular momentum equation (9.2) tells us that \u02d9J = I3 \u02d9\u03c9 = 0, so the axis will remain \ufb01xed in space and the angular velocity will be constant. (This would not be true if the axis of rotation were not a principal axis. We shall see what happens in that case later.) Now suppose that a small force F is applied to the axis at a point r (see Fig. 9.5). Then the equation of motion becomes \u02d9J = r F. \u2227 (9.27) This force will cause the axis to change direction, and the body will therefore acquire a small component of angular velocity perpendicular to its axis e3. However, provided the force is small enough, the angular velocity with which the axis moves will be small in comparison to the angular velocity of rotation about the axis. In that case, we may neglect the angular 212 Classical Mechanics F r\u2227F \u03c9\u03c9 r Fig. 9.5 momentum components normal to the axis, and again write \u02d9J = I3 \u02d9\u03c9 = r F. \u2227 (9.28) Since the moment r velocity does not change. (Recall that d(\u03c92)/dt = 2\u03c9 direction does change, and the axis will move in the direction of r is perpendicular to the applied force. F is perpendicular to \u03c9, the magnitude of the angular \u02d9\u03c9.) However, its F, that \u2227 \u2227 \u00b7 The e\ufb00ect may be rather surprising at \ufb01rst sight. However, it is easy to see it in action in, for example, a freely rolling bicycle wheel (see Fig. 9.6). If the wheel leans to the right, it will not necessarily fall over, as it would if it were not rotating, but will turn to the right. The force F in this case is the force of gravity, which produces a moment about a horizontal axis, as shown in Fig. 9.6. Thus, in a small time interval dt, the angular velocity \u03c9 of a wheel acquires a small additional horizontal component, d\u03c9. This shows that the axis of the wheel must change direction. When turning a corner on a bicycle, this e\ufb00ect plays a role, but there are other more important e\ufb00ects, related to the way the handlebars and front fork are pivoted. (See Jones, Physics Today, 23, 34\u201340, 1970 and Fajans, American Journal of Physics, 68, 654\u2013659, 2000.) Another familiar example is the spinning top, or toy gyroscope (which we shall discuss in more detail in Chapters 10 and 12). The force in this case is again the force of gravity, F = M gk, acting at the centre of mass position, R = Re3 (see Fig. 9.7). Thus (9.28) yields \u2212 Rigid Bodies 213 \u03c9\u03c9 \u03c9\u03c9 r \u2227 Mg Mg r Fig. 9.6 \u03c9\u03c9 \u2126 k R Mg Fig. 9.7 I3\u03c9 \u02d9e3 = M gRe3 \u2227 \u2212 k. This equation is of a form we have encountered several times already. It may be written \u02d9e3 = \u2126 e3, \u2227 214 where Classical Mechanics \u2126 = M gR I3\u03c9 k. (9.29) \u00a7 5.1) it describes a vector e3 rotating with constant angular Thus (see velocity \u2126 about the vertical. The e\ufb00ect is similar to the Larmor precession (see 6.5). Note that the angular velocity \u2126 is independent of the inclination of the axis. 5.6) or the precession of a satellite orbit (see \u00a7 \u00a7 Our treatment here is valid only if \u2126 \u03c9, or, equivalently, if M gR I3\u03c92. In other words, we require the rotational kinetic energy (T = 1 2 I3\u03c92) to be much greater than the possible change in gravitational potential energy (of order M gR). We shall discuss what happens when this condition is", " not satis\ufb01ed later (see 12.4). Note that the precessional angular velocity (9.29) is inversely proportional both to the moment of inertia and the rotational angular velocity \u03c9. Thus, to minimize the e\ufb00ect of a given force, we should use a fat, rapidly spinning body. 10.3 and \u00a7 \u00a7 The great stability of rapidly rotating bodies is the basis of the gyroscope. Essentially, this consists of a spinning body suspended in such a way that its axis is free to rotate relative to its support. The bearings are made as nearly frictionless as possible, to minimize the torques on the gyroscope. Then, no matter how we turn the support, the axis of the gyroscope will remain pointing very closely to the same direction in space. The gyroscope is particularly useful for navigational instruments in aircraft. It may be employed, for example, to provide an \u2018arti\ufb01cial horizon\u2019 which allows the pilot to \ufb02y a level course even in cloud (where our normal sense of balance is notoriously unreliable). A similar instrument, designed to show the orientation in a horizontal plane, serves as a direction indicator. Precession of the equinoxes The non-spherical shape of the Earth leads to another example of the phenomenon of precession. The e\ufb00ect is closely related to the precession of a satellite orbit discussed in 6.5. Since the Earth exerts a moment on the satellite, the satellite exerts an equal and opposite moment on the Earth. For a small satellite, this is of course quite negligible, but a similar moment is exerted by both the Sun and the Moon, and, although the e\ufb00ect is still quite small, it leads to observable changes over long periods of time. \u00a7 Rigid Bodies 215 Let us consider for example the e\ufb00ect of the Sun. Because it attracts the nearer equatorial bulge more strongly that the farther one, it exerts a couple on the Earth\u2019s axis, in a direction which would tend (if the Earth were not rotating) to align it with the normal to the plane of the Earth\u2019s orbit (called the ecliptic plane, because it is the plane in which eclipses can occur.) (See Fig. 9.8.) However, because of the Earth\u2019s rotation, the e 3 n \u03b1 Sun Fig. 9.8 actual e\ufb00ect is to make the axis precess around the normal to the ecliptic plane. Since the equator is inclined to the ecliptic at about 23.45\u25e6, the axis therefore describes a cone in space with semi-vertical angle 23.45\u25e6. It turns out that this e\ufb00ect depends on the same parameter, (6.41) or (6.42), that appeared in the discussion of the tides. Thus, as in that case, the e\ufb00ect of the Moon is rather more than twice as large as that of the Sun (though somewhat complicated by the fact that the Moon\u2019s orbital plane is not \ufb01xed in space). The two bodies together lead to a precessional angular velocity of about 50 per year, which corresponds to a precessional period of 26 000 years. The e\ufb00ect means that the pole of the Earth\u2019s axis moves relative to the \ufb01xed stars, so that 13 000 years ago our present pole star was some 47\u25e6 away from the pole. (The irregularity of the Moon\u2019s motion, which is due to the moment exerted by the Sun on the Earth\u2013Moon system, leads to a superimposed wobble of the axis with an amplitude of 9 and a period of 18.6 years.) The equinoxes occur when the apparent direction of the Sun crosses the equator. The points on the ecliptic where this happens also precess with the 26 000-year period, which is why this phenomenon is known as \u2018precession of the equinoxes\u2019. 216 Classical Mechanics 9.7 Instantaneous Angular Velocity We shall now turn to the problem of the general motion of a rigid body, without assuming that it is spinning about a principal axis, or that the In this section, we shall discuss the concept of forces acting are small. angular velocity, and show that, no matter how the body is moving, it is always possible to de\ufb01ne an instantaneous angular velocity vector \u03c9 (which in general is constant in neither direction nor magnitude). We consider \ufb01rst a rigid body free to rotate about a \ufb01xed pivot. Then the position of every point of the body is \ufb01xed if we specify the directions of the three principal axes e1, e2, e3. (This would be true for any three axes \ufb01", "xed in the body, but it will be convenient later if we choose them to be principal axes.) Since the position of any point of the body is \ufb01xed relative to these axes, its velocity is determined by the three velocities \u02d9e1, \u02d9e2, \u02d9e3. In fact, if the point in question is r = r1e1 + r2e2 + r3e3, then r1, r2, r3 are constants, and \u02d9r = r1 \u02d9e1 + r2 \u02d9e2 + r3 \u02d9e3. (9.30) (9.31) We now wish to show that, no matter how the body is moving, we can always de\ufb01ne an instantaneous angular velocity \u03c9. Let us de\ufb01ne the nine quantities Now, since e1 is a unit vector, e2 \u02d9ej. aij = ei \u00b7 1 = 1. Hence, di\ufb00erentiating, we obtain 0 = d dt (e2 1) = 2e1 \u00b7 \u02d9e1 = 2a11. It follows that \u02d9e1 has the form \u02d9e1 = a21e2 + a31e3. (9.32) Moreover, since e1 and e2 are always perpendicular, we have 0 = d dt (e1 \u00b7 e2) = \u02d9e1 \u00b7 e2 + e1 \u00b7 \u02d9e2 = a21 + a12. Rigid Bodies 217 Thus we can de\ufb01ne a vector \u03c9 by setting \u03c91 = \u03c92 = \u03c93 = \u2212 \u2212 \u2212 a23 = a32, a31 = a13, a12 = a21. With this de\ufb01nition, (9.32) becomes simply where of course \u02d9e1 = \u03c93e2 \u2212 \u03c92e3 = \u03c9 e1, \u2227 \u03c9 = \u03c91e1 + \u03c92e2 + \u03c93e3 (9.33) (9.34) is the instantaneous angular velocity. Substituting this, and two similar equations, into (9.31), we \ufb01nd that the velocity of any point in the body is given by \u02d9r = \u03c9 r. \u2227 (9.35) Thus we have shown that an instantaneous angular velocity vector \u03c9 always exists. We note that the direction of \u03c9 is the instantaneous axis of rotation: for points r lying on this axis, the instantaneous velocity \u02d9r is zero. Now let us consider the general case, in which no point of the body is permanently \ufb01xed. Then we may specify the position of the body by the position R of its centre of mass (or of any other designated point), and by the orientations of the principal axes at that point. We can now de\ufb01ne, as \u02d9R before, an angular velocity vector \u03c9 relative to the centre of mass. If is the velocity of the centre of mass, then the instantaneous velocity of a point whose position relative to the centre of mass is r\u2217 is \u02d9r = \u02d9R + \u02d9r\u2217. But, since r\u2217 is \ufb01xed in the body, we can apply the same argument as above to de\ufb01ne an instantaneous angular velocity vector \u03c9 and write \u02d9r = \u02d9R + \u03c9 r\u2217. \u2227 (9.36) We have not attached an asterisk to \u03c9, because the angular velocity is in fact independent of the choice of origin. For example, if the body is rotating with angular velocity \u03c9 about a \ufb01xed pivot, then \u02d9\u2217. \u2227 \u2227 \u2227 218 Classical Mechanics But the instantaneous velocity of the centre of mass is of course \u02d9R = \u03c9 whence subtracting from (9.36), we \ufb01nd \u2227 R, \u02d9r\u2217 = \u03c9 r\u2217, \u2227 which shows that \u03c9 is also the angular velocity relative to the centre of mass. The equation of motion of a rigid body rotating about a \ufb01xed pivot is the angular momentum equation (9.2), \u02d9J = F, r \u2227 (9.37) where J is related to \u03c9 by (9.20): J = I1\u03c91e1 + I2\u03c92e2 + I3\u03c93e3. 9.2, the momentum equation (9.1) serves to determine the reaction As in \u00a7 at the pivot. In the general case, we have to determine both the velocity of the centre of mass and the angular velocity. The motion of the", " centre of mass is described by the momentum equation (9.1): \u02d9P = M \u00a8R = F, and the rotational motion by \u02d9J \u2217 = r\u2217 F. \u2227 (9.38) (9.39) The angular momentum J \u2217 about the centre of mass is of course related to the angular velocity by J \u2217 = I \u22171 \u03c91e\u22171 + I \u22172 \u03c92e\u22172 + I \u22173 \u03c93e\u22173, where e\u22171, e\u22172, e\u22173 are the principal axes at the centre of mass, and I \u22171, I \u22172, I \u22173 are the corresponding principal moments. 9.8 Rotation about a Principal Axis The principal axes e1, e2, e3 rotate with the body. Thus if we wish to use directly the expression for J in terms of its components with respect to these axes, we have to remember that they constitute a rotating frame. In this section, we shall return to the notation of Chapter 5, and distinguish the absolute rate of change dJ /dt from the relative rate of change \u02d9J. The Rigid Bodies 219 equation of motion (9.37) refers of course to the absolute rate of change, so we must write it as dJ dt = F = G, r \u2227 (9.40) say. (The argument here may be applied equally to a rigid body rotating about a \ufb01xed pivot, or to the rotation about the centre of mass.) The relative rate of change is \u02d9J = I1 \u02d9\u03c91e1 + I2 \u02d9\u03c92e2 + I3 \u02d9\u03c93e3, since the principal moments of inertia are constants. The two rates of change are related by dJ dt = \u02d9J + \u03c9 J. \u2227 Substituting in (9.40), we obtain or, in terms of components, \u02d9J + \u03c9 \u2227 J = G, I1 \u02d9\u03c91 + (I3 \u2212 I2 \u02d9\u03c92 + (I1 \u2212 I3 \u02d9\u03c93 + (I2 \u2212 I2)\u03c92\u03c93 = G1, I3)\u03c93\u03c91 = G2, I1)\u03c91\u03c92 = G3. (9.41) These equations may be solved, in principle, to give the angular velocity components as functions of time. However, when there are external forces, they are not often particularly useful, because these forces are usually speci\ufb01ed in terms of their components with respect to a given \ufb01xed set of axes. Even if the external force F is a constant, its components F1, F2, F3 are variable, and depend on the unknown orientation of the body. For this kind of problem, an alternative method of solution, using Lagrange\u2019s equations (to be described in the next chapter), is usually preferable. We shall con\ufb01ne our discussion here to the case where there are no external forces, so that the right hand sides of (9.41) vanish. If the body is initially rotating about the principal axis e3, so that \u03c91 = \u03c92 = 0, then we see from (9.41) that they remain zero, and \u03c93 is a constant. Thus we have veri\ufb01ed our earlier assertion that a body rotating freely about a principal axis will continue to rotate with constant angular velocity. Now, however, we wish to investigate the stability of this motion. 220 Classical Mechanics Example: Stability of free rotation about a principal axis A rigid body is rotating freely about a principal axis, say e3, and is given a small displacement. Will its rotation axis remain close to e3? If the displacement is small enough, we may treat \u03c91 and \u03c92 (at least initially) as small, and neglect the product \u03c91\u03c92. Thus the third equation, with G3 = 0, tells us that \u03c93 remains almost constant. We can solve the remaining two equations (with G1 = G2 = 0) by looking for solutions of the form \u03c91 = a1ept, \u03c92 = a2ept, where a1, a2 and p are constants. (This is essentially the method we used to treat the the problem of stability of equilibrium in one dimension in 2.2, 2.3. We shall discuss the general problem of stability by a similar \u00a7\u00a7 method in the next chapter.) Substituting, and dividing by ept, we obtain I1pa1 + (I3 \u2212 I2pa2 + (I1 \u2212 Eliminating the ratio a1/a2, we \ufb01nd I2)\u03c93a2", " = 0, I3)\u03c93a1 = 0. p2 = (I3 \u2212 I2)(I1 \u2212 I1I2 I3) \u03c92 3. The denominator is obviously positive, as is \u03c92 3. Thus if I3 > I1 and I3 > I2, or if I3 < I1 and I3 < I2, the two roots for p are pure imaginary, and we have an oscillatory solution. On the other hand, if I1 > I3 > I2 or I1 < I3 < I2, then the roots are real, and the values of \u03c91 and \u03c92 will in general increase exponentially with time, until the approximation of treating them as small is no longer valid. Therefore, we may conclude that the rotation about the axis e3 is stable if I3 is either the largest or the smallest of the three principal moments, but not if it is the middle one. This interesting result is easy to verify. For example, if one throws a matchbox in the air, it is not hard to get it to spin about its longest or its shortest axis, but it will not spin stably about the other one. The result applies to any rigid body. It is often called the tennis racquet theorem, and we will examine this problem from a di\ufb00erent perspective in 13.6. \u00a7 Rigid Bodies 221 9.9 Euler\u2019s Angles The orientation of a rigid body about a \ufb01xed point \u2014 or about its centre of mass \u2014 must be speci\ufb01ed by three angles. These may be chosen in various ways, but one convenient choice is the set of angles known as Euler\u2019s angles. Two of these are required to \ufb01x the direction of one of the axes, say e3 \u2014 they are just the polar angles \u03b8, \u03d5 \u2014 and the third speci\ufb01es the angle through which the body has been rotated from a standard position about this axis. We can reach an arbitrary orientation by starting with the body in a standard position, and making three successive rotations. Let us suppose that initially the axes e1, e2, e3 of the body coincide with the \ufb01xed axes i, j, k. (See Fig. 9.9.) The \ufb01rst two rotations are designed to bring the \u2032\u2032 1 e \u2032 1 Fig. 9.9 axis e3 to its required position, speci\ufb01ed by the polar angles \u03b8, \u03d5. First, we make a rotation through an angle \u03d5 about the axis k. This brings the three axes into the positions e1, e2, k. Next, we make a rotation through an angle \u03b8 about the axis e2, bringing the axes to the positions e1, e2, e3. 222 Classical Mechanics Finally, we make a rotation through an angle \u03c8 about e3. This brings all three axes to their required positions e1, e2, e3. Since the three Euler angles, \u03d5, \u03b8, \u03c8, \ufb01x the orientations of the three axes, they specify completely the orientation of the rigid body. The angular velocity of the body is clearly determined by the rates of change of these three angles. A small change in \u03d5 corresponds to a rotation about k. Hence if \u03d5 is changing at a rate \u02d9\u03d5, and the other angles are constant, the angular velocity is \u02d9\u03d5k. Similarly, if only \u03b8 or \u03c8 is changing, the angular velocity is \u02d9\u03b8e2 or \u02d9\u03c8e3. When all three angles are changing, the angular velocity is the sum of these three contributions, \u03c9 = \u02d9\u03d5k + \u02d9\u03b8e2 + \u02d9\u03c8e3. (9.42) In order to \ufb01nd the angular momentum or kinetic energy, using (9.20) or (9.22), we should have to \ufb01nd the components of \u03c9 in the directions of the three principal axes, e1, e2, e3. This is not di\ufb03cult, but we shall not go through the derivation in the general case. Instead, we limit the discussion to the case of a symmetric body, with symmetry axis e3, and I1 = I2. Then, according to the discussion of 9.4, any orthogonal pair of axes in the plane of e1 and e2 will serve as principal axes, together with e3. It is not necessary that the axes we choose be \ufb01xed in the body, so long as they are always principal axes. In particular, we may use the axes e1, e2. This choice has the advantage that two of the three terms in (9", ".44) are already expressed in terms of these axes e1, e2, e3. \u00a7 The components of k in the directions of e1, e2, e3 are clearly sin \u03b8, 0, cos \u03b8), so that ( \u2212 k = \u2212 sin \u03b8 e1 + cos \u03b8 e3. (9.43) Thus from (9.42), we \ufb01nd \u03c9 = \u2212 \u02d9\u03d5 sin \u03b8 e1 + \u02d9\u03b8e2 + ( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8)e3. (9.44) We can now \ufb01nd the angular momentum from (9.20). All we have to do is to multiply the three components of \u03c9 by the principal moments I1, I1, I3. We obtain J = \u2212 I1 \u02d9\u03d5 sin \u03b8 e1 + I1 \u02d9\u03b8e2 + I3( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8)e3. (9.45) To \ufb01nd the equations of motion, we could either re-express J in terms of the \ufb01xed axes i, j, k, or write down equations directly in the rotating frame Rigid Bodies 223 de\ufb01ned by e1, e2, e3. We shall, however, \ufb01nd a simpler method in the next chapter. The kinetic energy is obtained similarly from (9.22). It is T = 1 2 I1 \u02d9\u03d52 sin2 \u03b8 + 1 2 I1 \u02d9\u03b82 + 1 2 I3( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8)2. (9.46) Free motion of a symmetric rigid body As an example of the use of the Euler angles, let us consider a symmetric body moving under no forces. Its centre of mass moves of course with uniform velocity. The interesting part of the motion is the rotation about the centre of mass. The same formalism applies to a body rotating freely about a \ufb01xed pivot. Since there are no external forces, the angular momentum equation is simply dJ /dt = 0. The angular momentum vector therefore points in a \ufb01xed direction in space. Though we could solve the problem in terms of arbitrary axes, it will be convenient to choose the axis k to be in the direction of the constant vector J. Then, by (9.43), J = Jk = \u2212 J sin \u03b8 e1 + J cos \u03b8 e3. For consistency, this must be identically equal to the expression (9.45). Hence, equating the three components, we obtain I1 \u02d9\u03d5 sin \u03b8 = J sin \u03b8, I1 \u02d9\u03b8 = 0, I3( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8) = J cos \u03b8. (9.47) From the second equation, we learn that \u03b8 is a constant, and from the other = 0 or \u03c0) \u02d9\u03d5 and \u02d9\u03c8 are constants. Hence the axis of the two that (provided \u03b8 body, e3 rotates around the direction of J at a constant rate \u02d9\u03d5, maintaining a constant angle \u03b8 to it, and in addition the body spins about its axis with constant angular velocity \u02d9\u03c8 (see Fig 9.10). The angular velocity vector \u03c9 = \u2212 \u02d9\u03d5 sin \u03b8 e1 + ( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8)e3 (9.48) has \ufb01xed components with respect to e1, e2, e3. It too therefore maintains a constant angle with e3, and also with J. In space, the angular velocity vector \u03c9 thus describes a cone around the direction of J, precessing at the 224 Classical Mechanics e 3 \u03c8 J \u03d5 \u03c9\u03c9 \u03b2 \u03b8 Fig. 9.10 \u2212e \u20321 rate \u02d9\u03d5. This is known as the space cone. In the body, since \u03c9 maintains a constant angle to e3, it must describe a cone around this axis \u2014 the body cone. The rate at which it describes this body cone is \u02d9\u03c8. (For, \u03c9 is \ufb01xed with respect to the axes e1, e2, e3, and the body rotates relative to them with angular velocity \u02d9\u03c8.) Since the direction of \u03c9 is the instantaneous axis of rotation, the body cone is instantaneously at rest along the line in the direction of \u03c9, which is where it touches the space cone. (See Fig. 9.10.) Thus the motion may be described by saying that the", " body cone rolls round the \ufb01xed space cone. From (9.48) we see that the semi-vertical angle \u03b2 of the body cone is given by tan \u03b2 = \u03c91 \u03c93 \u2212 = \u02d9\u03d5 sin \u03b8 \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8 = I3 I1 tan \u03b8, (9.49) I3| by (9.47). In particular, if the body is nearly totally symmetric, so that I1 \u2212 I3, then the angles \u03b2 and \u03b8 are nearly equal, and the space cone | is much narrower than the body cone. In this case, it is easy to see that \u02d9\u03c8 \u02d9\u03d5, so that the period taken to describe the body cone is much greater than that for the space cone. The Earth\u2019s axis has a very small oscillation of this kind, called the Chandler wobble. It precesses around a \ufb01xed direction in space roughly once every day, and moves round a cone in the Earth with semi-vertical angle of order 0.1 (a circle of about 3 m radius at the pole) in a period which Rigid Bodies 225 should be about 300 days for a perfectly rigid Earth, but is actually around 14 months, with considerable variation. In this case I3 > I1, and therefore \u03b2 > \u03b8, so the body cone actually rolls around with the space cone inside it, and \u02d9\u03c8 < 0 (see Fig. 9.11). (Note that this precession is quite independent e 3 \u2212\u03c8 \u03d5 \u03c9\u03c9 J \u03b2 \u03b8 Fig. 9.11 \u2212e \u20321 of the much larger e\ufb00ect, the precession of the equinoxes, discussed at the end of 9.6). \u00a7 9.10 Summary Determining the motion of a rigid body under known forces can be divided into two quite separate problems. The motion of the centre of mass is determined, via the momentum equation (9.1), by the total force acting on the body \u2014 or, if one point is \ufb01xed, the force is determined by the motion of the centre of mass. The rotational motion about the centre of mass is determined, via the angular momentum equation (9.2), by the total moment of the forces acting. The relation between angular momentum and angular velocity is described by the inertia tensor. It is always possible to \ufb01nd a set of principal axes (\ufb01xed in the body) for which all the products of inertia vanish. Then J is obtained from \u03c9 simply by multiplying each component by the appropriate principal moment of inertia. Similarly the kinetic energy T is the sum of the three terms corresponding to rotation about each of the three principal axes. 226 Classical Mechanics In general, it is convenient to describe the rotational motion in terms of a set of principal axes. Normally these rotate with the body, though in the special case of a symmetric rigid body (with I1 = I2) they may be chosen as in 9.9. In that case, the orientation of the body is conveniently described by Euler\u2019s angles. We shall see in the following chapter that the Lagrangian method is very useful for obtaining equations of motion in terms of Euler\u2019s angles. \u00a7 Problems 1. A uniform solid cube of edge length 2a is suspended from a horizontal axis along one edge. Find the length of the equivalent simple pendulum. Given that the cube is released from rest with its centre of mass level with the axis, \ufb01nd its angular velocity when it reaches the lowest point. 2. An insect of mass 100 mg is resting on the edge of a \ufb02at uniform disc of mass 3 g and radius 50 mm, which is rotating at 60 r.p.m. about a smooth pivot. The insect crawls in towards the centre of the disc. Find the angular velocity when it reaches it, and the gain in kinetic energy. Where does this kinetic energy come from, and what happens to it when the insect crawls back out to the edge? 3. A uniform solid cube of edge 2a is sliding with velocity v on a smooth horizontal table when its leading edge is suddenly brought to rest by a small ridge on the table. Which dynamical variables are conserved (a) before impact, (b) during impact, and (c) after impact? Find the angular velocity immediately after impact, and the fractional loss of kinetic energy. Determine the minimum value of v for which the cube topples over rather than falling back. 4. A pendulum consists of a light rigid rod of length 250 mm, with two identical uniform solid spheres of radius 50 mm attached one on either side of its lower end. Find the period of small oscillations (a) perpendicular to the line of centres, and (b) along it", ". 5. A uniform rod of mass M and length 2a hangs from a smooth hinge at one end. Find the length of the equivalent simple pendulum. It is struck sharply, with impulse X, at a point a distance b below the hinge. Use the angular momentum equation to \ufb01nd the initial value of the angular velocity. Find also the initial momentum. Determine the point at which the rod may be struck without producing any impulsive reaction at the hinge. Show that, if the rod is struck elsewhere, the Rigid Bodies 227 direction of the impulsive reaction depends on whether the point of impact is above or below this point. 6. *(a) A simple pendulum supported by a light rigid rod of length l is released from rest with the rod horizontal. Find the reaction at the pivot as a function of the angle of inclination. (b) For the cube of Problem 1, \ufb01nd the horizontal and vertical components of the reaction on the axis as a function of its angular position. Compare your answer with the corresponding expressions for the equivalent simple pendulum. \u00d7 7. *Find the principal moments of inertia of a \ufb02at rectangular plate of 60 mm. Given that the plate is romass 30 g and dimensions 80 mm 1, \ufb01nd the comtating about a diagonal with angular velocity 15 rad s\u2212 ponents of the angular momentum parallel to the edges. Given that the axis is of total length 120 mm, and is held vertical by bearings at its ends, \ufb01nd the horizontal component of the force on each bearing. 8. Find the principal moments of inertia of a uniform solid cube of mass m and edge length 2a (a) with respect to the mid-point of an edge, and (b) with respect to a vertex. 9. Find the moment of inertia about an axis through its centre of a uniform hollow sphere of mass M and outer and inner radii a and b. (Hint : Think of it as a sphere of density \u03c1 and radius a, with a sphere of density \u03c1 and radius b removed.) 10. A spaceship of mass 3 t has the form of a hollow sphere, with inner radius 2.5 m and outer radius 3 m. Its orientation in space is controlled by a uniform circular \ufb02ywheel of mass 10 kg and radius 0.1 m. Given that the \ufb02ywheel is set spinning at 2000 r.p.m., \ufb01nd how long it takes the spaceship to rotate through 1\u25e6. Find also the energy dissipated in this manoeuvre. 11. A long, thin hollow cylinder of radius a is balanced on a horizontal knife edge, with its axis parallel to it. It is given a small displacement. Calculate the angular displacement at the moment when the cylinder ceases to touch the knife edge. (Hint : This is the moment when the radial component of the reaction falls to zero.) 12. *Calculate the principal moments of inertia of a uniform, solid cone of vertical height h, and base radius a about its vertex. For what value of the ratio h/a is every axis through the vertex a principal axis? For this case, \ufb01nd the position of the centre of mass and the principal moments of inertia about it. 228 Classical Mechanics 13. A top consists of a uniform, solid cone of height 50 mm and base radius It is spinning with its vertex \ufb01xed at 7200 r.p.m. Find the 20 mm. precessional period of the axis about the vertical. 14. A gyroscope consisting of a uniform solid sphere of radius 0.1 m is spinning at 3000 r.p.m. about a horizontal axis. Due to faulty construction, the \ufb01xed point is not precisely at the centre, but 20 \u00b5m away from it along the axis. Find the time taken for the axis to move through 1\u25e6. 15. A gyroscope consisting of a uniform circular disc of mass 100 g and radius 40 mm is pivoted so that its centre of mass is \ufb01xed, and is spinning about its axis at 2400 r.p.m. A 5 g mass is attached to the axis at a distance of 100 mm from the centre. Find the angular velocity of precession of the axis. 16. *A uniformly charged sphere is spinning freely with angular velocity \u03c9 in a uniform magnetic \ufb01eld B. Taking the z axis in the direction of \u03c9, and B in the xz-plane, write down the moment about the centre of the magnetic force on a particle at r. Evaluate the total moment of the B, magnetic force on the sphere, and show that it is equal to (q/2M )J where q and M are the total charge and mass, respectively. Hence show that the axis will precess around the direction of the magnetic \ufb01eld with precessional angular velocity equal to the Larmor", " frequency 5.5. What di\ufb00erence would it make if the charge distribution were of spherically symmetric, but non-uniform? \u2227 \u00a7 17. *A wheel of radius a, with its mass concentrated on the rim, is rolling with velocity v round a circle of radius R ( a), maintaining a constant inclination \u03b1 to the vertical. Show that v = a\u03c9 = R\u2126, where \u03c9 is the angular velocity of the wheel about its axis, and \u2126 ( \u03c9) is the precessional angular velocity of the axis. Use the momentum equation to \ufb01nd the horizontal and vertical components of the force at the point of contact. Then show from the angular momentum equation about the centre of mass that R = 2v2/g tan \u03b1. Evaluate R for v = 5 m s\u2212 1 and \u03b1 = 30\u25e6. 18. A solid rectangular box, of dimensions 100 mm 20 mm, is spinning freely with angular velocity 240 r.p.m. Determine the frequency of small oscillations of the axis, if the axis of rotation is (a) the longest, and (b) the shortest, axis. 60 mm \u00d7 \u00d7 19. *A rigid body of spheroidal shape, spinning rapidly about its axis of symmetry, is placed on a smooth \ufb02at table. Show by considering the moment of the force at the point of contact that its axis will precess in one direction if it is oblate (c < a = b, e.g. a discus) and in the Rigid Bodies 229 opposite direction if it is prolate (c > a = b, e.g. a rugby ball). Show also that if there is a small frictional force, the axis will become more nearly vertical, so that if the body is oblate its centre of mass will fall, but if it is prolate it will rise. 20. *The average moment exerted by the Sun on the Earth is, except for sign, identical with the expression found in Chapter 6, Problem 26, provided we interpret m as the mass of the Sun, and r as the distance I1) and hence that the precessional to the Sun. Show that Q = angular velocity produced by this moment is 2(I3 \u2212 \u2212 \u2126 = I1 3 2 I3 \u2212 I3 \u2212 2 \u03c9 cos \u03b1 n, where is the Earth\u2019s orbital angular velocity, and \u03b1 = 23.45\u25e6 is the tilt between the Earth\u2019s axis and the normal to the orbital plane, the oblateness of the (the ecliptic). Show also that (I3 \u2212 Earth, and hence evaluate \u2126. Why is this e\ufb00ect less sensitive to the distribution of density within the Earth than the complementary one discussed in I1)/I3 \u2248 6.5? \u00a7 21. *The axis of a gyroscope is free to rotate within a smooth horizontal circle in colatitude \u03bb. Due to the Coriolis force, there is a couple on the gyroscope. To \ufb01nd the e\ufb00ect of this couple, use the equation for the rate of change of angular momentum in a frame rotating with the Earth (e.g., that of Fig. 5.7), \u02d9J +\u2126 J = G, where G is the couple restraining the axis from leaving the horizontal plane, and \u2126 is the Earth\u2019s angular velocity. (Neglect terms of order \u21262, in particular the contribution of \u2126 to J.) From the component along the axis, show that the angular velocity \u03c9 about the axis is constant; from the vertical component show that the angle \u03d5 between the axis and east obeys the equation \u2227 I1 \u00a8\u03d5 \u2212 I3\u03c9\u2126 sin \u03bb cos \u03d5 = 0. Show that the stable position is with the axis pointing north. Determine the period of small oscillations about this direction if the gyroscope is a \ufb02at circular disc spinning at 6000 r.p.m. in latitude 30\u25e6 N. Explain why this system is sensitive to the horizontal component of \u2126, and describe the e\ufb00ect qualitatively from the point of view of an inertial observer. 230 Classical Mechanics Chapter 10 Lagrangian Mechanics We have already seen that the equations of motion for a system of N particles moving under conservative forces may be obtained from the Lagrangian function in terms of any set of 3N independent co-ordinates. (See 8.5.) In this chapter, we shall give a more systematic account of the Lagrangian method, and apply it in particular to the case of rigid bodies. \u00a7 10.1 Generalized Co-ordinates; Holonomic Systems Let us consider a rigid body, composed of a large number N of particles. The positions of all the particles", " may be speci\ufb01ed by 3N co-ordinates. However, these 3N co-ordinates cannot all vary independently, but are subject to constraints \u2014 the rigidity conditions. In fact, the position of every particle may be \ufb01xed by specifying the values of just six quantities \u2014 for instance, the three co-ordinates X, Y, Z of the centre of mass and the three Euler angles \u03d5, \u03b8, \u03c8 which determine the orientation. These six constitute a set of generalized co-ordinates for the rigid body. In particular problems, these co-ordinates may be subject to further constraints, which may be of two kinds. First, we might for example \ufb01x the position of one point of the body, say the centre of mass. Such constraints are represented by algebraic conditions on the co-ordinates (e.g., X = Y = Z = 0), which may be used to eliminate some of the co-ordinates. In this particular case, the three Euler angles alone su\ufb03ce to \ufb01x the position of every particle. The second type of constraint is represented by conditions on the velocities rather than the co-ordinates. For example, we might constrain the centre of mass to move with uniform velocity, or to move round a circle with uniform angular velocity. Then, in place of algebraic equations, we 231 232 Classical Mechanics \u02d9X = u). In simple cases, these equations have di\ufb00erential equations (e.g., can be solved to \ufb01nd some of the co-ordinates as explicit functions of time (e.g., X = X0 + ut). Then the position of every particle will be determined by the values of the remaining generalized co-ordinates and the time t. In general, we say that q1, q2,..., qn is a set of generalized co-ordinates for a given system if the position of every particle in the system is a function of these variables, and perhaps also explicitly of time: ri = ri(q1, q2,..., qn, t). (10.1) The number of co-ordinates which can vary independently is called the number of degrees of freedom of the system. If it is possible to solve the constraint equations, and eliminate some of the co-ordinates, leaving a set equal in number to the number of degrees of freedom, the system is called holonomic. If this elimination introduces explicit functions of time, the system is said to be forced ; on the other hand, if all the constraints are purely algebraic, so that t does not appear explicitly in (10.1), the system is natural. There do exist non-holonomic systems, for which the constraint equations cannot be solved to eliminate some of the co-ordinates. Consider, for example, a sphere rolling on a rough plane. Its position may be speci\ufb01ed by \ufb01ve generalized co-ordinates \u2014 X, Y, \u03d5, \u03b8, \u03c8. (Z is a constant and may be omitted.) However, the sphere can roll in only two directions, and the number of degrees of freedom is two (three if we allow it also to spin about the vertical). The constraint equations \u2014 the rolling conditions \u2014 serve to determine the angular velocity in terms of the velocity of the centre of mass. Using the fact that the instantaneous axis of rotation must be a horizontal axis through the point of contact, it is not hard to show that \u02d9R/a, where a is the radius of the sphere. (This follows from the \u03c9 = k \u2227 fact that \u02d9R = \u03c9 \u03c9 = 0.) However, these equations cannot be integrated to \ufb01nd the orientation in terms of the position of the centre of mass; for, we could roll the sphere round a circle so that it returns to its starting point but with a di\ufb00erent orientation. ak, together with k \u2227 \u00b7 For the present, we shall assume that the constraint equations can be solved, and the number of generalized coordinates taken equal to the number of degrees of freedom. The distinction between a natural and a forced system can be expressed in another way, which will be useful later. Di\ufb00erentiating (10.1), we \ufb01nd that the velocity of the ith particle is a linear function of \u02d9q1,..., \u02d9qn, though Lagrangian Mechanics 233 in general it depends in a more complicated way on the co-ordinates themselves: \u02d9ri = n \u03b1=1 \u2202ri \u2202q\u03b1 \u02d9q\u03b1 + \u2202ri \u2202t. The last term arises from the explicit dependence on t in (10.1), and is 2 m \u02d9r2, we", " absent for a natural system. When we substitute in T = obtain a quadratic function of the time derivatives \u02d9q1,..., \u02d9qn. For a natural system, it is a homogeneous quadratic function; but for a forced system there are also linear terms and terms independent of the velocities. \" 1 For example, according to (8.29) and (9.46), the kinetic energy of a symmetric rigid body is T = 1 2 M ( \u02d9X 2 + \u02d9Y 2 + \u02d9Z 2) + 1 2 I \u22171 ( \u02d9\u03d52 sin2 \u03b8 + \u02d9\u03b82) + 1 2 I \u22173 ( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8)2. If we impose further algebraic constraints, such as X = 0, the corresponding terms drop out, and we are still left with a homogeneous quadratic function of the remaining time derivatives. On the other hand, if we impose a di\ufb00erential constraint, such as \u02d9X = u or \u02d9\u03d5 = \u03c9, we obtain a function with constant or linear terms. 10.2 Lagrange\u2019s Equations To obtain a more general form of Lagrange\u2019s equations than the one found 3.7 \u2014 in particular, one that is not restricted to conservative forces \u2014 in let us return for the moment to the case of a single particle, moving under an arbitrary force F. We consider variations of the integral \u00a7 t1 I = T dt, where t0 T = 1 2 m \u02d9r2. (10.2) Now let us make a small variation \u03b4x(t) in the x co-ordinate, subject to the boundary conditions \u03b4x(t0) = \u03b4x(t1) = 0. Then clearly \u03b4T = m \u02d9x \u03b4 \u02d9x. Substituting in (10.2), and performing an integration by parts, in which the integrated term vanishes by virtue of the boundary conditions, we \ufb01nd t1 m\u00a8x \u03b4x dt. \u03b4I = \u2212 t0 \u03b4W, where \u03b4W is the work done by the Now, the integrand is force F in the displacement \u03b4x. Similarly, if we make variations of all three Fx \u03b4x = \u2212 \u2212 234 Classical Mechanics coordinates, we \ufb01nd \u03b4I = t1 \u2212 t0 \u03b4W dt, (10.3) \u00b7 where \u03b4W = F The advantage of (10.3) is that, like Hamilton\u2019s principle (see \u03b4r is the work done by the force in the displacement \u03b4r. 3.7), it makes no explicit reference to any particular set of co-ordinates. Hence we can use it to \ufb01nd equations of motion in terms of an arbitrary set of co-ordinates q1, q2, q3. Let us de\ufb01ne the generalized forces F1, F2, F3 corresponding to F by \u00a7 \u03b4W = F1\u03b4q1 + F2\u03b4q2 + F3\u03b4q3. (10.4) Then we may equate \u03b4I as given by (10.3) to the general expression (3.37). Consider, for example, a variation \u03b4q1 of q1. Then, according to (3.37), t1 \u03b4I = \u2202T \u2202q1 \u2212 d dt \u2202T \u2202 \u02d9q1 \u03b4q1(t) dt. On the other hand, by (10.3) and (10.4), t0 \u03b4I = t1 \u2212 t0 F1 \u03b4q1(t) dt. These two expressions must be equal for arbitrary variations subject only to the boundary conditions that \u03b4q1(t0) = \u03b4q1(t1) = 0. Hence the integrands must be equal, and we obtain Lagrange\u2019s equations in the form d dt \u2202T \u2202 \u02d9q\u03b1 = \u2202T \u2202q\u03b1 + F\u03b1. (10.5) For example, for a particle moving in a plane, and described by polar co-ordinates, T = 1 2 m( \u02d9r2 + r2 \u02d9\u03b82). Hence Lagrange\u2019s equations are d dt (m \u02d9r) = mr \u02d9\u03b82 + Fr, d dt (mr2 \u02d9\u03b8) = F\u03b8. (10.6) It is important to remember that F\u03b8 here does not mean the component of the vector F in the direction of", " increasing \u03b8. (Compare the discussion following (3.47).) It is de\ufb01ned by (10.4), or \u03b4W = Fr\u03b4r + F\u03b8\u03b4\u03b8. Lagrangian Mechanics 235 The work done in producing a small angular displacement \u03b4\u03b8 is equal to \u03b4\u03b8 multiplied by the moment of the force F about the origin. Thus F\u03b8 here is in fact the moment of F about the origin. (Recall that the generalized momentum p\u03b8 = mr2 \u02d9\u03b8 is the angular momentum. Thus the second of Eqs. (10.6) expresses the fact that the rate of change of angular momentum is equal to the moment of the force.) If the force is conservative, then \u03b4W = energy. It then follows from (10.4) that \u2212 \u03b4V, where V is the potential F\u03b1 = \u2202V \u2202q\u03b1. \u2212 (10.7) Thus, de\ufb01ning the Lagrangian function L = T the form (3.44) obtained earlier, \u2212 V, we can write (10.5) in = d dt \u2202L \u2202 \u02d9q\u03b1 Note that the \u2018generalized force\u2019 \u2202L/\u2202q\u03b1 on the right hand side of this equation includes not only the force F\u03b1 determined by F but also the \u2018apparent\u2019 force \u2202T /\u2202q\u03b1 arising from the curvilinear nature of the co-ordinates (for example, the centrifugal term mr \u02d9\u03b82 in (10.6)). \u2202L \u2202q\u03b1 (10.8). More generally, even if F is non-conservative, it may be possible to \ufb01nd an e\ufb00ective \u2018potential energy\u2019 function V depending on the velocities as well as the co-ordinates, and such that F\u03b1 = \u2202V \u2202q\u03b1 \u2212 + d dt \u2202V \u2202 \u02d9q\u03b1. (10.9) In this case too, we can de\ufb01ne L = T V, and write Lagrange\u2019s equations in the form (10.8). The most important example of a force of this type in the electromagnetic force on a charged particle, which we shall discuss in 10.5. Note that in this case the \u2018potential energy\u2019 V contributes not only to the \u2018generalized force\u2019 on the right side of (10.8) but also to the \u2018generalized momentum\u2019 on the left. \u2212 \u00a7 There is nothing in the discussion leading to Lagrange\u2019s equations (10.5) or (10.8) that restricts their applicability to the case of a single particle. The general principle embodied in (10.3) applies to any system whatsoever, If the system is holonomic, with n containing any number of particles. degrees of freedom, then we can make independent arbitrary variations of the generalized co-ordinates q1, q2,..., qn, and obtain Lagrange\u2019s equations 236 Classical Mechanics (10.5), in which the generalized forces are de\ufb01ned, as in (10.4), by \u03b4W = F1\u03b4q1 + F2\u03b4q2 + \u00b7 \u00b7 \u00b7 n + Fn\u03b4qn = F\u03b1\u03b4q\u03b1. (10.10) \u03b1=1 Finally, if the forces are conservative, or more generally if they can be written in the form (10.9), then Lagrange\u2019s equations may be expressed in the form (10.8). 10.3 Precession of a Symmetric Top As a \ufb01rst example of the use of Lagrange\u2019s equations, we consider in more detail the problem of the symmetric top, discussed in the previous chapter. This is a symmetric rigid body, pivoted at a point on its axis of symmetry, and moving under gravity. (Compare Fig. 9.7.) Example: Steady precession of a top Find the conditions under which steady precession at a given constant angle of inclination of the axis, \u03b8, can occur. Show that for any \u03b8 < \u03c0/2, there is a minimum value of the rotational angular velocity about the axis for which this is possible. This system has three degrees of freedom, and we use the three Euler angles as generalized co-ordinates. The kinetic energy is given by (9.46), and the potential energy is V = M gR cos \u03b8. Thus the Lagrangian function is L = 1 2 I1 \u02d9\u03d52 sin2 \u03b8 + 1 2 I1 \u02d9\u03b82 + 1 2 I3( \u02d9\u03c8 + \u02d9", "\u03d5 cos \u03b8)2 M gR cos \u03b8. (10.11) \u2212 Lagrange\u2019s equation (10.8) for \u03b8 is therefore d dt (I1 \u02d9\u03b8) = I1 \u02d9\u03d52 sin \u03b8 cos \u03b8 \u2212 I3( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8) \u02d9\u03d5 sin \u03b8 + M gR sin \u03b8. (10.12) The Lagrangian function does not involve the other two Euler angles, but only their time derivatives, so the corresponding equations express the constancy of the generalized momenta p\u03d5 and p\u03c8: d dt [I1 \u02d9\u03d5 sin2 \u03b8 + I3( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8) cos \u03b8] = 0, d dt [I3( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8)] = 0. (10.13) Lagrangian Mechanics 237 Note that this last equation tells us that the component of angular velocity \u03c93 about the axis of symmetry is constant \u03c93 = \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8 = constant. (10.14) From the equations (10.13), given that \u03b8 remains constant, we learn that both \u02d9\u03d5 and \u02d9\u03c8 must be constant. Thus the axis of the top precesses around the vertical with constant angular velocity \u02d9\u03d5 = \u2126, say. To \ufb01nd the relation between the angular velocities \u2126 and \u03c93, we examine the \u03b8 = equation, (10.12). Since the left side must vanish, we obtain (for sin \u03b8 0, that is unless the axis is vertical), I1\u21262 cos \u03b8 \u2212 I3\u03c93\u2126 + M gR = 0. (10.15) The minimum value of \u03c93 for which real roots for \u2126 exist is given by 3 \u03c92 I 2 3 = 4I1M gR cos \u03b8. If the top is spinning more slowly than this, it will start to wobble. For any larger value of \u03c93 there are two possible values of \u2126, the roots of (10.15). It is interesting to examine the special case where \u03c93 is large. Then there M gR/I3\u03c93, and is one root for \u2126 which is much less than \u03c93, given by \u2126 \u2248 I3\u03c93/I1 cos \u03b8. The another which is comparable in magnitude to \u03c93, \u2126 \ufb01rst of these is just the approximate result we obtained in 9.6 on the assumption that \u2126 was much smaller than \u03c93 (see (9.29)). The second represents a rapid precessional motion in which the gravitational force is negligible. It corresponds precisely to the free precessional motion of a rigid body, discussed in 9.9. \u2248 \u00a7 Note that if we allow \u03b8 to be greater than \u03c0/2 \u2014 that is, if we consider a top suspended below its point of support \u2014 then steady precession is possible for any value of \u03c93. In particular, for \u03c93 = 0, we \ufb01nd the possible angular velocities of a compound pendulum swinging in a circle, \u00a7 \u2126 = \u00b1 M gR cos \u03b8 I1| |. We shall discuss the more general motion of this system in Chapter 12, using the Hamiltonian methods developed there. 238 Classical Mechanics 10.4 Pendulum Constrained to Rotate about an Axis Next, let us consider the system illustrated in Fig. 10.1. It consists of a light rigid rod, of length l, carrying a mass m at one end, and hinged at \u03c9 \u03b8 l m Fig. 10.1 the other to a vertical axis, so that it can swing freely in a vertical plane. We suppose initially that, in addition to gravity, a known torque G(t) is applied to rotate it about the axis. This system has two degrees of freedom. Its position may be described by the two polar angles \u03b8, \u03d5. (For convenience, we take the z-axis vertically downward, so that \u03b8 = 0 is the equilibrium position.) The Lagrangian function is L = 1 2 ml2( \u02d9\u03b82 + \u02d9\u03d52 sin2 \u03b8) mgl(1 \u2212 \u2212 cos \u03b8). (10.16) Since the work done by the torque G is \u03b4W = G \u03b4\u03d5, Lagrange\u2019s equations are d dt \u2202L \u2202 \u02d9\u03b8 = \u2202L \u2202\ufffd", "\ufffd, d dt \u2202L \u2202 \u02d9\u03d5 = \u2202L \u2202\u03d5 + G, or, explicitly, ml2 \u00a8\u03b8 = ml2 \u02d9\u03d52 sin \u03b8 cos \u03b8 mgl sin \u03b8, \u2212 (ml2 \u02d9\u03d5 sin2 \u03b8) = G. d dt (10.17) (10.18) Now let us suppose that the torque G is adjusted to constrain the system to rotate with constant angular velocity \u03c9 about the vertical. This imposes Lagrangian Mechanics 239 the constraint \u02d9\u03d5 = \u03c9, and the system may be regarded as a system with one degree of freedom, described by the co-ordinate \u03b8. Substituting this constraint into the Lagrangian function (10.16), we \ufb01nd L = 1 2 ml2( \u02d9\u03b82 + \u03c92 sin2 \u03b8) mgl(1 \u2212 \u2212 cos \u03b8). (10.19) Note the appearance in the kinetic energy part of L of a term independent of \u02d9\u03b8, which is characteristic of a forced system. It is often useful in problems of this kind to separate the Lagrangian function into a \u2018kinetic\u2019 V, term T, quadratic in the time derivatives, and a \u2018potential\u2019 term independent of them. In this case, we write L = T V, where \u2212 T = 1 2 ml2 \u02d9\u03b82, V = mgl(1 cos \u03b8) \u2212 \u2212 \u2212 2 ml2\u03c92 sin2 \u03b8. 1 Physically, this corresponds to using a rotating frame, rotating with angular velocity \u03c9. Here T is the kinetic energy of the motion relative to this frame, and the extra term in V is the potential energy corresponding to the centrifugal force. For a forced system, the constraining forces can do work on the system, so the total energy T + V is not in general a constant. However, in our case, there is still a conservation law. Multiplying (10.17) by \u02d9\u03b8 and integrating, we \ufb01nd T + V = E = constant. (10.20) This is not the total energy T + V, because the centrifugal term appears with the opposite sign. It is easy to verify the consistency of this by calculating the work done on the system. The rate at which the torque does work is G\u03c9. Thus, from (10.18), d dt (T + V ) = G\u03c9 = d dt (ml2\u03c92 sin2 \u03b8). Since T + V = T + V ml2\u03c92 sin2 \u03b8, this is equivalent to (10.20). We can use this conservation law, just as we did in \u2212 2.1, to \ufb01nd the \u00a7 qualitative features of the motion. Example: Types of motion of the constrained pendulum Describe the possible types of motion of the constrained pendulum for a given value of \u03c9. 240 Classical Mechanics To do this, we draw the \u2018e\ufb00ective potential energy\u2019 diagram for the function V. There are two distinct cases, depending on the value of If \u03c92 < g/l (that is, if the rotation period is longer than the free \u03c9. oscillation period of the pendulum), then V has a minimum at \u03b8 = 0, and a maximum at \u03b8 = \u03c0, as shown in the upper curve in Fig. 10.2. The motion is qualitatively just like that of the ordinary pendulum, though the period is longer. V \u2032 2mgl 0 \u03b8 0 \u03c0 \u03b8 Fig. 10.2 If \u03c92 > g/l, then V has maxima at both 0 and \u03c0, and an intermediate minimum at \u03b8 = \u03b80 = arccos(g/l\u03c92), as shown in the lower curve in Fig. 10.2. The equilibrium position at \u03b8 = 0 is then unstable. The stable equilibrium position at \u03b80 is the one where the transverse components of the gravitational and centrifugal forces are in equilibrium. For this case, three types of motion are possible. If E < 0, the pendulum oscillates around this stable position, without ever reaching the vertical. For 0 < E < 2mgl, it swings from one side to the other as before, though \u03b8 = 0 is no longer the position of maximum velocity. Finally, if E > 2mgl, we have a continuous circular motion, passing through both the upward and downward verticals. There is a more general way of handling constraints (holonomic or not), using the method of Lagrange multipliers (see Appendix A, Problem 10). To impose the constraint \u02d9\u03d5 = \u03c9 on our system, we introduce a new variable \ufffd", "\ufffd, the Lagrange multiplier, and subtract \u03bb times the constraint from the Lagrangian function. From (10.16), we obtain in this way L = 1 2 ml2( \u02d9\u03b82 + \u02d9\u03d52 sin2 \u03b8) mgl(1 cos \u03b8) \u03bb( \u02d9\u03d5 \u03c9). \u2212 \u2212 \u2212 \u2212 (10.21) Lagrangian Mechanics 241 The Euler\u2013Lagrange equation for \u03b8, (10.17) is unchanged. That for \u03d5 becomes d dt (ml2 \u02d9\u03d5 sin2 \u03b8 \u03bb) = 0. \u2212 (10.22) In addition, we have the Euler\u2013Lagrange equation for \u03bb, which simply reproduces the constraint, 0 = \u02d9\u03d5 \u03c9. \u2212 Comparing (10.22) with (10.18), we see that \u02d9\u03bb is physically the torque G required to impose the constraint. Equivalently, we could have integrated the constraint, and subtracted \u03bb(\u03d5 In that case, \u03bb itself In general, the physical signi\ufb01cance of \u03bb may be would be the torque. found by considering a small virtual change in the constraint equation. This method is particularly useful when we want to \ufb01nd the constraining forces or torques. \u03c9t) from L. \u2212 10.5 Charged Particle in an Electromagnetic Field We now wish to consider one of the most important examples of a nonconservative force. We consider a particle of charge q moving in an electric \ufb01eld E and a magnetic \ufb01eld B. The force on the particle is then or, in terms of components, F = q(E + v B), \u2227 Fx = qEx + q( \u02d9yBz \u2212 \u02d9zBy), (10.23) (10.24) together with two similar equations obtained by cyclic permutation of x, y, z. Now, we wish to show that this force may be written in the form (10.9) with a suitably chosen function V. To do this, we have to make use of a standard result of electromagnetic theory (see A.7), according to which it is always possible to \ufb01nd a scalar potential \u03c6 and a vector potential A, functions of r and t, such that \u00a7 E = \u2207\u03c6 \u2212 \u2212 \u2202A \u2202t, B = \u2207 A. \u2227 (10.25) For time-independent \ufb01elds, \u03c6 is simply the electrostatic potential of Chapter 6. 242 Classical Mechanics Now, let us consider the function V = q\u03c6(r, t) = q\u03c6 \u2212 q \u02d9r A(r, t) \u2212 \u00b7 q( \u02d9xAx + \u02d9yAy + \u02d9zAz). (10.26) Clearly, Also, \u2202V \u2202x \u2212 = q \u2212 \u2202\u03c6 \u2202x + q \u02d9x \u2202Ax \u2202x + \u02d9y \u2202Ay \u2202x + \u02d9z \u2202Az \u2202x. d dt \u2202V \u2202 \u02d9x = q \u2212 dAx dt, so, bearing in mind that Ax varies with time both because of its explicit time dependence, and because of its dependence on the particle position r(t), we \ufb01nd d dt \u2202V \u2202 \u02d9x = q \u2212 \u2202Ax \u2202t + \u2202Ax \u2202x \u02d9x + \u2202Ax \u2202y \u02d9y + \u2202Ax \u2202z \u02d9z. Adding, we \ufb01nd that the terms in \u02d9x cancel, whence \u2202V \u2202x \u2212 + d dt \u2202V \u2202 \u02d9x = q \u2202\u03c6 \u2202x \u2212 \u2212 \u2202Ax \u2202t q \u02d9y \u2202Ay \u2202x \u2212 = qEx + q( \u02d9yBz \u2212 + \u2202Ax \u2202y \u02d9zBy), \u2202Az \u2202x \u2212 \u2202Ax \u2202z + \u02d9z by (10.25). But this is just the expression for Fx given by (10.24). Hence we have veri\ufb01ed that F is given by (10.9) with V equal to (10.26). It follows that the equations of motion for a particle in an electromag- netic \ufb01eld may be obtained from the Lagrangian function L = 1 2 m \u02d9r2 + q \u02d9r A(r, t) q\u03c6(r, t). \u2212 \u00b7 Note the appearance in L of terms", " linear in the time derivatives. This function cannot be separated into two parts, T V, one quadratic in \u02d9r and one independent of it. Another consequence of the appearance of these terms is that the generalized momentum px is no longer equal to the familiar mechanical momentum, m \u02d9x. Instead, \u2212 (10.27) px = \u2202L \u2202 \u02d9x = m \u02d9x + qAx, Lagrangian Mechanics 243 or, more generally, p = m \u02d9r + qA. (10.28) We can now obtain the equations of motion in terms of arbitrary coordinates from the Lagrangian function (10.27). For example, in terms of cylindrical polars, it reads L = 1 2 m( \u02d9\u03c12 + \u03c12 \u02d9\u03d52 + \u02d9z2) + q( \u02d9\u03c1A\u03c1 + \u03c1 \u02d9\u03d5A\u03d5 + \u02d9zAz) q\u03c6. \u2212 (10.29) Example: Uniform magnetic \ufb01eld A charged particle moves in a uniform static magnetic \ufb01eld B. Find the solutions of the equations of motion in which \u03c1 is constant. For a uniform magnetic \ufb01eld, we may take \u2227 or, if we choose the z-axis in the direction of B, \u03c6 = 0 and A = 1 2 B r, \u03c6 = 0, A\u03c1 = 0, A\u03d5 = 1 2 B\u03c1, Az = 0. (It is easy to verify the relation B = \u2207 Lagrangian function is \u2227 A using (A.55).) Thus the L = 1 2 m( \u02d9\u03c12 + \u03c12 \u02d9\u03d52 + \u02d9z2) + 1 2 qB\u03c12 \u02d9\u03d5. Lagrange\u2019s equations are therefore m\u00a8\u03c1 = m\u03c1 \u02d9\u03d52 + qB\u03c1 \u02d9\u03d5, d dt (m\u03c12 \u02d9\u03d5 + 1 2 qB\u03c12) = 0, m\u00a8z = 0. (10.30) (10.31) In particular, in the case where \u03c1 is a constant, we learn from the last two equations that \u02d9\u03d5 and \u02d9z are also constants, and from the \ufb01rst equation that either \u02d9\u03d5 = 0 (particle moving parallel to the z-axis), or \u02d9\u03d5 = qB m. \u2212 This is of course precisely the solution we obtained in 5.2. \u00a7 244 Classical Mechanics The second of the three equations (10.31) is particularly interesting. It shows that, although in general the z component of the particle angular momentum is not a constant, there is still a corresponding conservation law for the quantity p\u03d5 = m\u03c12 \u02d9\u03d5 + 1 2 qB\u03c12. The reason for the existence of such a conservation law will be discussed in Chapter 12. 10.6 The Stretched String As a \ufb01nal example of the use of the Lagrangian method, we consider a rather di\ufb00erent kind of problem. This is an example of a system with an in\ufb01nite number of degrees of freedom \u2014 a string of length l, and mass \u00b5 per unit length, with \ufb01xed ends, and stretched to a tension F. We shall consider small transverse oscillations of the string. In place of a \ufb01nite set of generalized co-ordinates, we now have a continuous function, the displacement y(x, t) of the string from its equilibrium position. For the partial derivatives of y we use the notations \u2202y \u2202t The kinetic energy of a small element of string of length dx is 1 Thus the total kinetic energy is \u2202y \u2202x y = \u02d9y =,. 2 (\u00b5 dx) \u02d9y2. T = l 0 1 2 \u00b5 \u02d9y2 dx. When the string is in equilibrium, its length is l. However, when it is displaced, its length is increased, say to l + \u2206l, and is given by (3.34), namely l l + \u2206l = 1 + y 2 dx. The work done against the tension in increasing the length by \u2206l is F \u2206l. This is the potential energy of the string. For small displacements, we can replace 2. Thus we may take 2 by dx. (10.32) Lagrangian Mechanics 245 Our Lagrangian function is, therefore, l 2 \u00b5 \u02d9y2 ( ) dx. (10.33) As before, we can use Hamilton\u2019s principle to derive the equation of motion. To do this, it will", " be useful to consider a more general problem, in which the Lagrangian function has the form l L = (y, \u02d9y, y) dx, 0 L (10.34) where may be called the Lagrangian density. (In our particular case,, given by the integrand of (10.33), is actually independent of y, and a L L function only of its derivatives.) The action integral is t1 l I = (y, \u02d9y, y) dx dt. 0 L t0 Let us consider a small variation of the displacement \u03b4y(x, t) which vanishes at t0 and t1, and also (because the ends of the string are \ufb01xed) at x = 0 and x = l. The variation of the action integral is \u03b4I = t1 l t0 0 \u2202 L \u2202y \u03b4y + \u2202 L \u2202 \u02d9y \u2202(\u03b4y) \u2202t + \u2202 L \u2202y \u2202(\u03b4y) \u2202x dx dt. Now, as in and with respect to x in the third. vanish because \u03b4y is zero at the limits of integration. Thus we obtain 3.6, we integrate by parts, with respect to t in the second term, In both cases, the integrated terms \u00a7 t1 l \u03b4I = \u2202 L \u2202y \u2212 \u2202 \u2202t \u2202 L \u2202 \u02d9y \u2202 \u2202x \u2212 \u2202 L \u2202y \u03b4y(x, t) dx dt. t0 0 Hamilton\u2019s principle requires that this expression should be zero for arbitrary variations \u03b4y(x, t) vanishing at the limits. This is possible only if the integrand vanishes identically. Hence we obtain Lagrange\u2019s equation \u2202 L \u2202y \u2212 \u2202 \u2202t \u2202 L \u2202 \u02d9y \u2212 \u2202 \u2202x \u2202 L \u2202y = 0. (10.35) In the particular case of the stretched string, the derivatives of the Lagrangian density are \u2202 L \u2202y = 0, \u2202 L \u2202 \u02d9y = \u00b5 \u02d9y, \u2202 L \u2202y F y. = \u2212 246 Classical Mechanics Thus Lagrange\u2019s equation becomes \u00a8y = c2y, where c2 = F/\u00b5. (10.36) In place of a discrete set of ordinary di\ufb00erential equations for the generalized co-ordinates q\u03b1(t), we now have a partial di\ufb00erential equation for the function y(x, t) of two independent variables. Note that c2 has the dimensions of (velocity)2. In fact, as we show below, c is the velocity of propagation of waves along the string. Equation (10.36) is the one-dimensional wave equation. Similar equations turn up in many branches of physics, whenever wave phenomena are encountered. Example: Solution of the stretched-string equation Given the initial displacement y(x, 0) and velocity \u02d9y(x, 0) of the string \ufb01nd its displacement at a later time t. Ignoring for the moment the boundary conditions at x = 0 and x = l, the general solution of the wave equation (10.36) involves two arbitrary functions, which may be determined by the initial values of y and \u02d9y. It is easy to verify by direct substitution that, for any function f, y(x, t) = f (x ct) \u2212 is a solution. It represents a wave travelling along the string with velocity c to the right; for, the shape of the displacement y is the same at time t as at time 0, but shifted to the right by a distance ct (see Fig. 10.3). Similarly, y(x, t) = f (x + ct) is a solution, representing a wave travelling to the left along the string. The general solution is y = f (x + ct) + g(x ct), \u2212 where f and g are arbitrary functions. To satisfy the boundary conditions at x = 0 and x = l both terms in the solution must be present. For a string of \ufb01nite length, a wave cannot travel inde\ufb01nitely in one direction. When it gets to the end of the string it must be re\ufb02ected back. Lagrangian Mechanics 247 y time 0 time t ct x Fig. 10.3 The condition y = 0 at x = 0 requires that f (ct) + g( \u2212 ct) = 0 for all t. Thus the functions f and g di\ufb00er only in sign, and in the sign of the argument, and we can write the solution as The condition y = 0", " at x = l requires that y = f (x + ct) f (ct x). \u2212 \u2212 (10.37) f (ct + l) f (ct \u2212 \u2212 l) = 0 for all t. In other words, f must be a periodic function, of period 2l, satisfying f (x + 2l) = f (x). (10.38) To establish the connection between the periodic function f and the initial conditions, let us write f as the sum of an even function and an odd function: f (x) = 1 2 [h(x) + k(x)], (10.39) where h(x) = f (x) + f ( f ( k(x) = f (x) \u2212 x), x). \u2212 \u2212 To \ufb01x f, we have to specify its values in the interval from l to l, or, equivalently, to specify both functions h and k in the interval from 0 to l. Now from (10.37), the initial value of y is \u2212 y(x, 0) = f (x) f ( \u2212 \u2212 x) = k(x). (10.40) 248 Classical Mechanics This therefore determines the odd part of f. Moreover, the initial value of \u02d9y is \u02d9y(x, 0) = cf (x) cf ( x) = ch(x). (10.41) \u2212 x).) Thus the initial value of \u02d9y deter(Remember that [f ( mines the even part of f (up to an irrelevant additive constant, which cancels in (10.37)). x)] = f ( \u2212 \u2212 \u2212 \u2212 10.7 Summary The position of every part of a system may be \ufb01xed by specifying the values of a set of generalized co-ordinates. If these co-ordinates can all vary independently, the system is holonomic. This is the case in all the examples we have considered. The system is natural if the functions specifying the positions of particles in terms of the generalized co-ordinates do not involve the time explicitly. In that case, the kinetic energy is a homogeneous quadratic function of the \u02d9q\u03b1. For a forced system, on the other hand, T may contain linear and constant terms. In either case, the equations of motion are given by Lagrange\u2019s equations. If the forces are conservative (and sometimes in V. In other cases too), all we need is the Lagrangian function L = T general, for dissipative forces, the generalized forces F\u03b1 corresponding to the generalized co-ordinates q\u03b1 must be found by evaluating the work done in a small displacement. \u2212 Problems 1. Masses m and 2m are joined by a light inextensible string which runs without slipping over a uniform circular pulley of mass 2m and radius a. Using the angular position of the pulley as generalized co-ordinate, write down the Lagrangian function, and Lagrange\u2019s equation. Find the acceleration of the masses. 2. A uniform cylindrical drum of mass M and radius a is free to rotate about its axis, which is horizontal. A cable of negligible mass and length l is wound on the drum, and carries on its free end a mass m. Write down the Lagrangian function in terms of an appropriate generalized co-ordinate, assuming no slipping or stretching of the cable. If the cable Lagrangian Mechanics 249 is initially fully wound up, and the system is released from rest, \ufb01nd the angular velocity of the drum when it is fully unwound. 3. Treat the system of Problem 2 as one with two generalized co-ordinates, the angular position of the drum and the free length of cable, with an appropriate constraint. Hence \ufb01nd the tension in the cable. (Show that it is equal to the Lagrange multiplier.) 4. *Find the Lagrangian function for the system of Problem 2 if the cable is elastic, with elastic potential energy 1 2 kx2, where x is the extension of the cable. Show that the motion of the mass m is a uniform acceleration at the same rate as before, with a superimposed oscillation of angular frequency given by \u03c92 = k(M + 2m)/M m. Find the amplitude of this oscillation if the system is released from rest with the cable unextended. 5. Write down the kinetic energy of a particle in cylindrical polar coordinates in a frame rotating with angular velocity \u03c9 about the z-axis. Show that the terms proportional to \u03c9 and \u03c92 reproduce the Coriolis force and centrifugal force respectively. 6. A light inextensible string passes over a light smooth pulley, and carries a mass 4m on one end. The other end supports a second pulley with a string over it carrying masses 3m and m on the two ends. Using a suitable pair", " of generalized co-ordinates, write down the Lagrangian function for the system, and Lagrange\u2019s equations. Find the downward accelerations of the three masses. 7. Find the tensions in the strings in Problem 6. Explain why the \ufb01rst pulley turns, although the total mass on each side is the same. 8. Evaluate accurately the two possible precessional angular velocities of the top described in Chapter 9, Problem 13, if the axis makes an angle of 30\u25e6 with the vertical. Compare the slower value with the approximate result found earlier. Find also the minimum angular velocity \u03c93 for which steady precession at this angle is possible. 9. A simple pendulum of mass m and length l hangs from a trolley of mass M running on smooth horizontal rails. The pendulum swings in a plane parallel to the rails. Using the position x of the trolley and the angle of inclination \u03b8 of the pendulum as generalized co-ordinates, write down the Lagrangian function, and Lagrange\u2019s equations. Obtain an equation of motion for \u03b8 alone. If the system is released from rest with the pendulum inclined at 30\u25e6 to the vertical, use energy conservation to \ufb01nd its angular velocity when it reaches the vertical, given that M = 2 kg, m = 1 kg, and l = 2 m. 250 Classical Mechanics 10. *Show that the kinetic energy of the gyroscope described in Chapter 9, Problem 21, is T = 1 2 I1(\u2126 sin \u03bb cos \u03d5)2 + 1 2 I1( \u02d9\u03d5 + \u2126 cos \u03bb)2 + 1 2 I3( \u02d9\u03c8 + \u2126 sin \u03bb sin \u03d5)2. From Lagrange\u2019s equations, show that the angular velocity \u03c93 about the axis is constant, and obtain the equation for \u03d5 without neglecting \u21262. Show that motion with the axis pointing north becomes unstable for very small values of \u03c93, and \ufb01nd the smallest value for which it is stable. What are the stable positions when \u03c93 = 0? Interpret this result in terms of a non-rotating frame. 11. *Find the Lagrangian function for a symmetric top whose pivot is free to slide on a smooth horizontal table, in terms of the generalized coordinates X, Y, \u03d5, \u03b8, \u03c8, and the principal moments I \u22171, I \u22171, I \u22173 about the centre of mass. (Note that Z is related to \u03b8.) Show that the horizontal motion of the centre of mass may be completely separated from the rotational motion. What di\ufb00erence is there in the equation (10.15) for steady precession? Are the precessional angular velocities greater or less than in the case of a \ufb01xed pivot? Show that steady precession at a given value of \u03b8 can occur for a smaller value of \u03c93 than in the case of a \ufb01xed pivot. 12. *A uniform plank of length 2a is placed with one end on a smooth horizontal \ufb02oor and the other against a smooth vertical wall. Write down the Lagrangian function, using two generalized co-ordinates, the distance x of the foot of the plank from the wall, and its angle \u03b8 of inclination to the horizontal, with a suitable constraint between the two. Given that the plank is initially at rest at an inclination of 60\u25e6, \ufb01nd the angle at which it loses contact with the wall. (Hint : First write the co-ordinates of the centre of mass in terms of x and \u03b8. Note that the reaction at the wall is related to the Lagrange multiplier.) 13. Use Hamilton\u2019s principle to show that if F is any function of the generalized co-ordinates, then the Lagrangian functions L and L + dF/dt must yield the same equations of motion. Hence show that the equations of motion of a charged particle in an electromagnetic \ufb01eld are una\ufb00ected by the \u2018gauge transformation\u2019 (A.42). (Hint : Take F = q\u039b.) 14. The stretched string of 10.6 is released from rest with its mid-point displaced a distance a, and each half of the string straight. Find the function f (x). Describe the shape of the string after (a) a short time, (b) a time l/2c, and (c) a time l/c. \u00a7 \u2212 Lagrangian Mechanics 251 15. *Two bodies of masses M1 and M2 are moving in circular orbits of radii a1 and a2 about their centre of mass. The restricted three-body problem concerns the motion of a third", " small body of mass m ( M1 or M2) in their gravitational \ufb01eld (e.g., a spacecraft in the vicinity of the Earth\u2013Moon system). Assuming that the third body is moving in the plane of the \ufb01rst two, write down the Lagrangian function of the system, using a rotating frame in which M1 and M2 are \ufb01xed. Find the equations of motion. (Hint : The identities GM1 = \u03c92a2a2 and GM2 = \u03c92a2a1 may be useful, with a = a1 + a2 and \u03c92 = GM/a3.) 16. *For the system of Problem 15, \ufb01nd the equations that must be satis\ufb01ed for \u2018equilibrium\u2019 in the rotating frame (i.e., circular motion with the same angular velocity as M1 and M2). Consider \u2018equilibrium\u2019 positions on the line of centres of M1 and M2. By roughly sketching the e\ufb00ective potential energy curve, show that there are three such positions, but that all three are unstable. (Note: The positions are actually the solutions of a quintic equation.) Show also that there are two \u2018equilibrium\u2019 positions o\ufb00 the line of centres, in each of which the three bodies form an equilateral triangle. (The stability of these so-called Lagrangian points is the subject of Problem 12, Chapter 12. There is further consideration of this important problem in 14.4.) \u00a7 252 Classical Mechanics Chapter 11 Small Oscillations and Normal Modes \u00a7 In this chapter, we shall discuss a generalization of the harmonic oscillator problem treated in 2.2 \u2014 the oscillations of a system of several degrees of freedom near a position of equilibrium. We consider only conservative, holonomic systems, described by n generalized co-ordinates q1, q2,..., qn. Without loss of generality, we may choose the position of equilibrium to be q1 = q2 = = qn = 0. We shall begin by investigating the form of the kinetic and potential energy functions near this point. \u00b7 \u00b7 \u00b7 11.1 Orthogonal Co-ordinates We shall restrict our attention to natural systems, for which the kinetic energy is a homogeneous quadratic function of \u02d9q1, \u02d9q2,..., \u02d9qn. (More gener10.4 \u2014 for ally, we could include also those forced systems \u2014 like that of which L can be written as a sum of a quadratic term T and a term V independent of the time derivatives. The only essential restriction is that there should be no linear terms.) \u2212 \u00a7 For example, for n = 2, we have T = 1 2 a11 \u02d9q2 1 + a12 \u02d9q1 \u02d9q2 + 1 2 a22 \u02d9q2 2. (11.1) In general, the coe\ufb03cients a11, a12, a22 will be functions of q1 and q2. However, if we are interested only in small values of q1 and q2, we may neglect this dependence, and treat them as constants (equal to their values at q1 = q2 = 0). For a particle described by curvilinear co-ordinates, the co-ordinates are called orthogonal if the co-ordinate curves always intersect at right angles 2, \u02d9q2 3, (see but no cross products like \u02d9q1 \u02d9q2. By an extension of this terminology, the In that case, the kinetic energy contains terms in \u02d9q2 1, \u02d9q2 3.5). \u00a7 253 254 Classical Mechanics generalized co-ordinates q1, q2,..., qn are called orthogonal if T is a sum of squares, with no cross products \u2014 for example, if a12 above is zero. It is a considerable simpli\ufb01cation to choose the co-ordinates to be or- thogonal, and this can always be done. For instance, we may set q1 = q1 + a12 a11 q2, so that, in terms of q1 and q2, (11.1) becomes T = 1 2 a11 \u02d9q1 2 + 1 2 a22 \u02d9q2 2, with a22 = a22 \u2212 a2 12 a11. (11.2) We can even go further. Since T is necessarily positive for all possible \u02d9q1, \u02d9q2, the coe\ufb03cients in (11.2) must be positive numbers. Hence we can de\ufb01ne new co-ordinates q1 = \u221aa11q1, q", "2 = a22q2, so that T is reduced to a simple sum of squares: T = 1 2 \u02d9q1 2 + 1 2 \u02d9q2 2. (11.3) A similar procedure can be used in the general case. (It is called the Gram\u2013Schmidt orthogonalization procedure.) We may \ufb01rst eliminate the cross products involving \u02d9q1 by means of the transformation to q1 = q1 + a12 a11 q2 + + \u00b7 \u00b7 \u00b7 a1n a11 qn, then those involving \u02d9q2, and so on. Thus we can always reduce T to the standard form T = 1 2 \u02d9q2 \u03b1, n \u03b1=1 (11.4) where we now drop any primes. As an illustration of these ideas, let us consider the double pendulum illustrated in Fig. 11.1. Small Oscillations and Normal Modes 255 Example: Kinetic energy of the double pendulum \u03b8 L M L\u03b8 \u03d5 l l\u03d5 L\u03b8 m Fig. 11.1 A double pendulum consists of a simple pendulum of mass M and length L, with a second simple pendulum of mass m and length l suspended from it. Consider only motion in a vertical plane, so that the system has two degrees of freedom. Find a standard set of co-ordinates in terms of which the kinetic energy takes the form (11.4). T = 1 As generalized co-ordinates, we may initially choose the inclinations \u03b8 and \u03d5 to the downward vertical. The velocity of the upper pendulum bob is L \u02d9\u03b8. That of the lower bob has two components \u2014 the velocity L \u02d9\u03b8 of its point of support, and the velocity l \u02d9\u03d5 relative to that point. The angle between these is \u03d5 \u03b8. Hence the kinetic energy is \u2212 2 m[L2 \u02d9\u03b82 + l2 \u02d9\u03d52 + 2Ll \u02d9\u03b8 \u02d9\u03d5 cos(\u03d5 2 M L2 \u02d9\u03b82 + 1 \u2212 For small values of \u03b8 and \u03d5, we may approximate cos(\u03d5 \u03b8) by 1. Since there is a term in \u02d9\u03b8 \u02d9\u03d5, these co-ordinates are not orthogonal, but we can make them so by adding an appropriate multiple of \u03d5 to \u03b8, or, equally well, of \u03b8 to \u03d5. In fact, it is easy to see physically that a pair of orthogonal co-ordinates is provided by the displacements of the two bobs, which for small angles are \u03b8)]. \u2212 x = L\u03b8, y = L\u03b8 + l\u03d5. (11.5) 256 Classical Mechanics In terms of these, T becomes T = 1 2 M \u02d9x2 + 1 2 m \u02d9y2. (11.6) To complete the reduction to the standard form (11.4), we may de\ufb01ne the new co-ordinates q1 = \u221aM x, q2 = \u221amy. (11.7) In practice, this is not always essential, though it makes the discussion of the general case much easier. 11.2 Equations of Motion for Small Oscillations Now let us consider the potential energy function V. With T given by (11.4), the equations of motion are simply \u00a8q\u03b1 = \u2202V \u2202q\u03b1, \u2212 for \u03b1 = 1, 2,..., n. (11.8) Thus the condition for equilibrium is that all n partial derivatives of V should vanish at the equilibrium position. For small values of the co-ordinates, we can expand V as a series, just as we did for a single co-ordinate in 2.2. For example, for n = 2, V = V0 + (b1q1 + b2q2) + ( 1 1 + k12q1q2 + 1 2 k22q2 2) +. \u00b7 \u00b7 \u00b7 \u00a7 2 k11q2 The equilibrium conditions require that the linear terms should be zero, b1 = b2 = 0, just as in 2.2. Moreover, the constant term V0 is arbitrary, and may be set equal to zero without changing the equations of motion. Thus the leading terms are the quadratic ones, and for small values of q1 and q2, we may approximate V by \u00a7 V = 1 2 k11q2 1 + k12q1q2 + 1 2 k22q2 2. (11.9) Then the equations of motion (11.8) become k11q1 \u2212 k21q1 \u2212 where for the sake of symmetry we have written k21 = k12 in the second", " equation. k12q2, k22q2, \u00a8q1 = \u00a8q2 = (11.10) \u2212 \u2212 Small Oscillations and Normal Modes 257 In the general case, V may be taken to be a homogeneous quadratic function of the co-ordinates, which can be written V = n n \u03b1=1 \u03b2=1 1 2 k\u03b1\u03b2q\u03b1q\u03b2, (11.11) with k\u03b2\u03b1 = k\u03b1\u03b2. (Notice that each term with \u03b1 example 1 2 k12q1q2 and 1 equations of motion are = \u03b2 appears twice, for 2 k21q2q1, which are of course equal.) Then the \u00a8q\u03b1 = n \u2212 \u03b2=1 or, in matrix notation, k\u03b1\u03b2q\u03b2, for \u03b1 = 1, 2,..., n, (11.12) \u00a8q1 \u00a8q2... \u00a8qn \u23a1 \u23a2 \u23a2 \u23a2 \u23a3 k11 k12... k1n k21 k22... k2n......... kn1 kn2... knn \u23a1 \u23a2 \u23a2 \u23a2 \u23a3 q1 q2... qn \u23a4. \u23a5 \u23a5 \u23a5 \u23a6 \u23a4 \u23a1 \u23a5 \u23a5 \u23a5 \u23a6 \u23a2 \u23a2 \u23a2 \u23a3 (11.13) = \u2212 \u23a4 \u23a5 \u23a5 \u23a5 \u23a6 Example: Equations of motion of the double pendulum Find the equations of motion of the double pendulum, in terms of the orthogonal co-ordinates x and y. The potential energy of the double pendulum is easily seen to be V = (M + m)gL(1 cos \u03b8) + mgl(1 cos \u03d5). \u2212 \u2212 For small angles, we may approximate (1 of x and y, we have \u2212 cos \u03b8) by 1 2 \u03b82. Hence in terms V = 1 2 (M + m)gL\u03b82 + 1 2 mgl\u03d52 = (M + m)g 2L x2 + mg 2l (y \u2212 x)2. Thus the equations of motion are M \u00a8x \u23a1 \u23a3 m\u00a8y \u23a4 \u23a6 \u2212 = \u23a1 \u23a2 \u23a3 (M + m)g L mg l \u2212 mg l mg l mg l \u2212 x \u23a4 \u23a1 \u23a5 \u23a6 \u23a3 y\u23a4 \u23a6. (11.14) Note the appearance of the masses on the left hand side, because we have not gone over to the normalized variables of (11.7). 258 Classical Mechanics 11.3 Normal Modes The general solution of a pair of second-order di\ufb00erential equations like (11.14) must involve four arbitrary constants, which may be \ufb01xed by the initial values of q1, q2, \u02d9q1, \u02d9q2. Similarly, the general solution of (11.13) must involve 2n arbitrary constants. To \ufb01nd the general solution, we adopt a generalization of the method used for the damped harmonic oscillator in 2.5: we look \ufb01rst for solutions in which all the co-ordinates are oscillating \u00a7 with the same frequency \u03c9, of the form q\u03b1 = A\u03b1ei\u03c9t, (11.15) where the A\u03b1 are complex constants. (As in 2.5, the physical solution may be taken to be the real part of (11.15).) Such solutions are called normal modes of the system. 2.3 and \u00a7 \u00a7 Substituting (11.15) into (11.13), we obtain a set of n simultaneous equations for the n amplitudes An, \u03c92A\u03b1 = \u2212 n \u2212 \u03b2=1 k\u03b1\u03b2A\u03b2. (11.16) Let us consider \ufb01rst the case n = 2. Then these equations are = \u03c92 k11 k12 k21 k22 A1 A2 This is what is known as an eigenvalue equation. The values of \u03c92 for which non-zero solutions exist are called the eigenvalues of the 2 2 matrix with elements k\u03b1\u03b2. The column vector formed by the A\u03b1 is an eigenvector of the matrix. A1 A2 (11.17) \u00d7. The equations (11.17) can alternatively be written as \u03c92 k11 \u2212 k21 k12 k22 \u2212 A1 A2 =. 0 0 \u03c92 These equations have a non-zero solution if and only if the determinant of the coe\ufb03cient matrix vanishes,", " i.e., if and only if \u03c92 k11 \u2212 k21 k12 k22 \u2212 \u03c92 = (k11 \u2212 \u03c92)(k22 \u2212 \u03c92) \u2212 k2 12 = 0. (11.18) Small Oscillations and Normal Modes 259 This is called the characteristic equation for the system. It determines the frequencies \u03c9 of the normal modes, which are the square roots of the eigenvalues \u03c92. Equation (11.18) is a quadratic equation for \u03c92. Its discriminant may 12, which is clearly positive. Hence it always be written (k11 \u2212 has two real roots. The condition for stability is that both roots should be positive. A negative root, say \u03b3 2, would yield a solution of the form k22)2 + 4k2 \u2212 q\u03b1 = A\u03b1e\u03b3t + B\u03b1e\u2212 \u03b3t, \u00d7 2 matrix, corresponding to the eigenvalue where both the A\u03b1 and the B\u03b1 coe\ufb03cients constitute eigenvectors of the \u03b3 2. Except in the degenerate 2 case of two equal eigenvalues, this means that the A\u03b1 and B\u03b1 coe\ufb03cients must be proportional, since the eigenvector is unique up to an overall factor. In general, therefore, the solution yields an exponential increase in the displacements with time. \u2212 This stability condition, that all the eigenvalues be positive, is a natural generalization of the requirement, in the one-dimensional case, that the second derivative of the potential energy function be positive (see 2.2). If \u03c92 is chosen equal to one of the two roots of (11.18), then either of the two equations in (11.17) \ufb01xes the ratio A1/A2. Since the coe\ufb03cients are real numbers, the ratio is obviously real. (This is a special case of a general theorem about symmetric matrices \u2014 those satisfying k\u03b2\u03b1 = k\u03b1\u03b2. A.10.) This means that A1 and A2 have the same phase (or phases See di\ufb00ering by \u03c0), so that q1 and q2 not only oscillate with the same frequency, but actually in (or directly out of) phase. The ratio of q1 to q2 remains \ufb01xed throughout the motion. \u00a7 \u00a7 There remains in A1 and A2 a common arbitrary complex factor, which serves to \ufb01x the overall amplitude and phase of the normal mode solution. Thus each normal mode solution contains two arbitrary real constants. Since the equations of motion (11.10) are linear, any superposition of solutions is again a solution. Hence the general solution is simply a super2 are the roots of position of the two normal mode solutions. If \u03c92 and \u03c9 (11.18), it may be written as the real part of q1 = A1ei\u03c9t + A1ei\u03c9 q2 = A2ei\u03c9t + A2ei\u03c9 t, t, (11.19) in which the ratios A1/A2 and A1/A2 are \ufb01xed by (11.17). (11.20) (11.21) 260 Classical Mechanics Example: Normal modes of the double pendulum Find the normal mode frequencies of the double pendulum. Here the equations (11.17) may be written (M + m)g M L + mg M l \u2212 g l \u2212 mg M l g l \u23a1 \u23a2 \u23a3 Ax Ay \u23a1 \u23a3 \u23a4 \u23a6 = \u03c92 Ax Ay \u23a1 \u23a3. \u23a4 \u23a6 \u23a4 \u23a5 \u23a6 The characteristic equation (11.18) simpli\ufb01es to \u03c94 \u03c92 + M + m M g2 Ll = 0. The roots of this equation determine the frequencies of the two normal modes. It is interesting to examine special limiting cases. First, let us suppose that the upper pendulum is very heavy (M m). Then, provided that l is not too close to L, the two roots, with the corresponding ratios determined by (11.20), are, approximately and \u03c92 g l, \u2248 Ax Ay \u2248 m M L l \u2212, L \u03c92 g L, \u2248 Ax Ay \u2248 L l. \u2212 L In the \ufb01rst mode, the upper pendulum is practically stationary, while the lower is swinging with its natural frequency. In the second mode, whose frequency is that of the upper pendulum, the amplitudes are comparable. At the other extreme, if M m, the two normal modes are and \u03c92 m M \u2248 g L + l \u03c92 \u2248,, g L + g l Ax Ay \u2248 L L + l, Ax", " Ay \u2248 \u2212 m M L + l L. In the \ufb01rst mode, the pendulums swing almost like a single rigid pendulum of length L + l. In the second, the lower bob remains almost stationary, while the upper one executes a very rapid oscillation. Small Oscillations and Normal Modes 261 The normal modes of a system with n degrees of freedom may be found by a very similar method. The condition for consistency of the simultaneous equations (11.16) is again that the determinant of the coe\ufb03cients should vanish. For example, for n = 3, we require \u03c92 k11 \u2212 k21 k31 k12 k22 \u2212 k32 \u03c92 k13 k23 k33 \u2212 \u03c92 = 0. (11.22) This is a cubic equation for \u03c92. It can be proved that its three roots are all real (see A.10). As before, the condition for stability is that all three roots should be positive. The roots then determine the frequencies of the three normal modes. \u00a7 For each normal mode, the ratios of the amplitudes are \ufb01xed by the equations (11.16). As in the case n = 2, these ratios are all real, so that in a normal mode all the co-ordinates oscillate in phase (or 180\u25e6 out of phase). Each normal mode solution involves just two arbitrary real constants, and the general solution is a superposition of all the normal modes. 11.4 Coupled Oscillators One often encounters examples of physical systems that may be described as two (or more) harmonic oscillators, which are approximately independent, but with some kind of relatively weak coupling between the two. (As a speci\ufb01c example, we shall consider below the system shown in Fig. 11.2, which consists of a pair of identical pendulums coupled by a spring.) 1 If the co-ordinate q of a harmonic oscillator is normalized so that T = 2 \u03c92q2, where \u03c9 is the angular frequency. Hence for a pair 2 \u02d9q2, then V = 1 of uncoupled oscillators, the coe\ufb03cients in (11.10) are k11 = \u03c92 1, k12 = 0, k22 = \u03c92 2. When the oscillators are weakly coupled, these equalities will still be approximately true, so that in particular k12 is small in comparison to k11 and k22. Thus from (11.18) it is clear that the characteristic frequencies of the system are given by \u03c92 k22; as one might expect, they are close to the frequencies of the uncoupled oscillators. Then from (11.17) we see that, in the \ufb01rst normal mode, the ratio A2/A1 is approxk22). Thus, unless the frequencies of the two normal imately k12/(k11 \u2212 modes are nearly equal, the normal modes di\ufb00er very little from those of k11 and \u03c92 \u2248 \u2248 262 Classical Mechanics the uncoupled system, and the coupling is not of great importance. The interesting case, in which even a weak coupling can be important, is that in which the two frequencies are equal, or nearly so. As a speci\ufb01c example of this case, we consider a pair of pendulums, each of mass m and length l, coupled by a weak spring (see Fig. 11.2). We shall l m y l m x Fig. 11.2 use the displacements x and y as generalized co-ordinates. Then, in the absence of coupling, the potential energy is approximately 1 0(x2 + y2), where \u03c92 0 = g/l gives the free oscillation frequency. For small values of x and y, the potential energy of the spring has the form 1 y)2. It will \u2212 be convenient to introduce another frequency, \u03c9s de\ufb01ned by \u03c92 s = k/m. In fact, \u03c9s is the angular frequency of the spring if one end is held \ufb01xed and the other attached to a mass m. Thus we take 2 m\u03c92 2 k(x T = 1 2 m( \u02d9x2 + \u02d9y2), V = 1 2 m(\u03c92 0 + \u03c92 s )(x2 + y2) m\u03c92 s xy. \u2212 (11.23) The normal mode equations now read 0 + \u03c92 \u03c92 \u03c92 s \u03c92 s \u2212 s 0 + \u03c92 \u03c92 \u2212 Ax Ay = \u03c92 Ax Ay. (11.24) s The two solutions of the characteristic equation are easily seen to be \u03c92 = \u03c92 0 0 + 2\u03c92 \u03c92 = \ufffd", "\ufffd2 s with with Ax/Ay = 1, Ax/Ay = 1. \u2212 Small Oscillations and Normal Modes 263 In the \ufb01rst normal mode, the two pendulums oscillate together in phase, with equal amplitude (see Fig. 11.3(a)). Since the spring is neither expanded nor compressed in this motion, it is not surprising that the frequency is just that of the uncoupled pendulums. In the second normal mode, which has a somewhat higher frequency, the pendulums swing in opposite directions, alternately expanding and compressing the spring (see Fig. 11.3(b)). (a) (b) Fig. 11.3 The general solution is a superposition of these two normal modes, and may be written as the real part of x = Aei\u03c90t + Aei\u03c9 Aei\u03c9 y = Aei\u03c90t \u2212 t, t, (11.25) 2 = \u03c92 where \u03c9 initial conditions. 0 + 2\u03c92 s. The constants A and A may be determined by the Example: Motion of coupled pendulums Find the solution for the positions of the bobs if the system is released from rest with one bob displaced a distance a from its equilibrium position. Describe the motion. 264 Classical Mechanics Here, at t = 0, we have x = a, y = 0, \u02d9x = 0, \u02d9y = 0, whence we \ufb01nd A = A = a/2, so the solution is x = 1 y = 1 2 a cos \u03c90t + 1 2 a cos \u03c90t 2 a cos \u03c9t = a cos \u03c9 1 2 a cos \u03c9t = a sin \u03c9 t cos \u03c9+t, t sin \u03c9+t, \u2212 \u2212 \u2212 where \u03c9 tities. \u00b1 = 1 2 (\u03c9 \u00b1 \u03c90), and we have used standard trigonometric iden- \u2212 Since the spring is weak, \u03c9 is only slightly greater than \u03c90, and \u03c9+. We have beats between two nearly equal frequentherefore \u03c9 cies. Thus we may describe the motion as follows. Initially, the \ufb01rst spring swings with angular frequency \u03c9+ and gradually decreasing amt. Meanwhile, the second pendulum starts to swing with plitude, a cos \u03c9 the same angular frequency, but 90\u25e6 out of phase, and with gradually increasing amplitude a sin \u03c9, the \ufb01rst pendulum has t. After a time \u03c0/2\u03c9 come momentarily to rest, and the second is oscillating with amplitude a. Then its amplitude starts to decrease, while the \ufb01rst increases again. The whole process is then repeated inde\ufb01nitely (though in practice there will of course be some damping). \u2212 \u2212 \u2212 This behaviour should be contrasted with that of a pair of coupled oscillators of very di\ufb00erent frequencies. In such a case, if one is started oscillating, one of the two normal modes will have a much larger amplitude than the other. Thus only a very small oscillation will be set up in the second oscillator, and the amplitude of the \ufb01rst will be practically constant. Normal co-ordinates The two normal modes of this system (or the n normal modes in the general case) are completely independent. We can make this fact explicit by introducing new \u2018normal\u2019 co-ordinates. In the case of the coupled pendulums, we introduce, in place of x and y, the new co-ordinates q1 = m 2 (x + y), q2 = m 2 (x \u2212 y). In terms of these co-ordinates, the solution (11.25) is q1 = A1ei\u03c90t, t, q2 = A2ei\u03c9 A1 = \u221a2mA, A2 = \u221a2mA. (11.26) (11.27) Small Oscillations and Normal Modes 265 in each normal mode, one co-ordinate only is oscillating. Co- Thus, ordinates with this property are called normal co-ordinates. The independence of the two normal co-ordinates may also be seen by examining the Lagrangian function. From (11.23) and (11.26), we \ufb01nd T = 1 2 \u02d9q2 1 + 1 2 \u02d9q2 2, whence the Lagrangian function is V = 1 2 \u03c92 0(q2 1 + q2 2) + \u03c92 s q2 2, L = 1 2 ( \u02d9q2 1 \u2212 0q2 \u03c92 1) + 1 2 ( \u02d9q2 2 \u2212 \u03c9 2q2 2). (11.28) In e\ufb00ect, we have reduced the Lagrangian to that for", " a pair of uncoupled oscillators, with angular frequencies \u03c90 and \u03c9. The normal co-ordinates are very useful in studying the e\ufb00ect on the system of a prescribed external force. Example: Forced oscillation of coupled pendulums Find the amplitudes of forced oscillations of the coupled pendulums if one of them is subjected to a periodic force F (t) = F1 cos \u03c91t. We write the force as the real part of F1ei\u03c91t. To \ufb01nd the equations of motion in the presence of this force, we must evaluate the work done 10.2). This is in a small displacement (see \u00a7 F (t)\u03b4x = F (t) \u221a2m (\u03b4q1 + \u03b4q2). Hence the equations of motion are \u00a8q1 = \u00a8q2 = \u2212 \u2212 \u03c92 0q1 + \u03c9 2q2 +, F1ei\u03c91t \u221a2m F1ei\u03c91t \u221a2m. (11.29) These independent oscillator equations may be solved exactly as in 2.6. In particular, the amplitudes of the forced oscillations are given \u00a7 by A1 = F1/\u221a2m \u03c92 \u03c92 1 0 \u2212, A2 = F1/\u221a2m \u03c92 \u03c9 1 2 \u2212. 266 Classical Mechanics The corresponding forced oscillation amplitudes for x and y are found by solving (11.26), and are Ax = A1 + A2 \u221a2m, Ay = A2 A1 \u2212 \u221a2m. Note that if the forcing frequency is very close to \u03c90, the \ufb01rst normal mode will predominate, and the pendulums will swing in the same direction; while if it is close to \u03c9 the second will be more important. (Of course, we should really include the e\ufb00ects of damping, so that the amplitudes do not become in\ufb01nite at resonance, and so that transient e\ufb00ects disappear in time.) 11.5 Oscillations of Particles on a String Consider a light string of length (n + 1)l, stretched to a tension F, with n equal masses m placed along it at regular intervals l. We shall consider transverse oscillations of the particles, and use as our generalized co-ordinates the displacements y1, y2,..., yn. (See Fig. 11.4, drawn for the case n = 3.) Since the kinetic energy is Fig. 11.4 T = 1 2 m( \u02d9y2 1 + \u02d9y2 2 + + \u02d9y2 n), \u00b7 \u00b7 \u00b7 (11.30) these co-ordinates are orthogonal. Next, we must calculate the potential energy. Let us consider the length of string between the jth and (j + 1)th particles. In equilibrium, its length is l, but when the particles are displaced it is l + \u03b4l = l2 + (yj+1 \u2212 yj)2 \u2248 l 1 + yj)2 (yj+1 \u2212 2l2, assuming the displacements are small. This calculation also applies to the segments of string at each end, provided we set y0 = yn+1 = 0. The work done against the tension in increasing the length of the string to this extent Small Oscillations and Normal Modes 267 is F \u03b4l. Hence, adding the contributions from each segment of string, we \ufb01nd that the potential energy is V = F 2l y2 1 + (y2 \u2212 y1)2 + + (yn \u2212 yn \u2212 \u00b7 \u00b7 \u00b7 1)2 + y2 n. (11.31) It is worth noting that the potential energy of a continuous string may be yj)2/l2 obtained as a limiting case, as n is approximately y and l 2, so we recover the expression (10.32). 0. For small l, (yj+1\u2212 \u2192 \u221e \u2192 From (11.30) and (11.31), we \ufb01nd that Lagrange\u2019s equations are \u00a8y1 = \u00a8y2 =... \u00a8yn = F ml F ml F ml 2y1 + y2), ( \u2212 (y1 \u2212 2y2 + y3), (11.32) (yn \u2212 1 \u2212 2yn). It will be convenient to write \u03c92 0 = F/ml. Then, substituting the normal mode solution yj = Ajei\u03c9t, we obtain the equations \u23a1 \u03c92 2\u03c92 0... 0 0 0 \u2212 \u03c92 0 2\u03c92 \u03c92 0... 0 0 \u2212 \u2212 0", " 2\u03c92 \u03c92 0... 0 0............ 0... 2\u03c92 0 0 0 \u2212 \u23a4 \u23a1 \u23a4 A1 A2 A3... An \u23a4. A1 A2 A3... An = \u03c92 \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 \u23a5 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 Let us look at the \ufb01rst few values of n. For n = 1, there is of course just 0. For n = 2, the characteristic equation is one normal mode, with \u03c92 = 2\u03c92 \u23a5 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 \u23a5 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 (11.33) 0 \u2212 and we obtain two normal modes: (2\u03c92 \u03c92)2 \u03c94 0 = 0, \u2212 \u03c92 = \u03c92 0, \u03c92 = 3\u03c92 0, A1/A2 = 1, A1/A2 = 1. \u2212 Now consider n = 3. The characteristic equation here is 2\u03c92 \u03c92 0 \u2212 \u03c92 0 \u2212 0 \u03c92 0 \u2212 \u03c92 2\u03c92 0 \u2212 \u03c92 0 \u2212 0 \u03c92 0 \u2212 2\u03c92 0 \u2212 \u03c92 = 0. (11.34) 268 Classical Mechanics Expanding this determinant by the usual rules, we obtain a cubic equation for \u03c92: (2\u03c92 0 \u2212 \u03c92)3 \u2212 2\u03c94 0(2\u03c92 0 \u2212 \u03c92) = 0. The roots of this equation are 2\u03c92 three normal modes 0 and (2 \u00b1 \u221a2)\u03c92 0. Hence we obtain the \u221a2)\u03c92 0, \u03c92 = (2 \u2212 \u03c92 = 2\u03c92 0, \u03c92 = (2 + \u221a2)\u03c92 0, A1 : A2 : A3 = 1 : \u221a2 : 1; A1 : A2 : A3 = 1 : 0 : 1; A1 : A2 : A3 = 1 : \u2212 \u221a2 : 1. \u2212 These normal modes are illustrated in Fig. 11.5. Fig. 11.5 Higher values of n may be treated similarly. For n = 4, the character4 determinant, which may be istic equation requires the vanishing of a 4 expanded by similar rules to yield \u00d7 (2\u03c92 0 \u2212 \u03c92)4 \u2212 3\u03c94 0(2\u03c92 0 \u2212 \u03c92)2 + \u03c98 0 = 0. The roots of this equation are given by (2\u03c92 0 \u2212 \u03c92)2 = 3 \u221a5 \u00b1 2 \u03c94 0 = 1 \u221a5 \u00b1 2 \u03c92 0 2. Small Oscillations and Normal Modes 269 Thus we obtain the four normal modes: \u03c92 = 0.38 \u03c92 0, \u03c92 = 1.38 \u03c92 0, \u03c92 = 2.62 \u03c92 0, \u03c92 = 3.62 \u03c92 0, (See Fig. 11.6.) A1 : A2 : A3 : A4 = 1 : 1.62 : 1.62 : 1; A1 : A2 : A3 : A4 = 1.62 : 1 : A1 : A2 : A3 : A4 = 1.62 : A1 : A2 : A3 : A4 = 1 : \u2212 1.62 : 1.62 : \u2212 1 : 1 : \u2212 \u2212 1.62; 1 : 1.62; 1. \u2212 \u2212 Fig. 11.6 For every value of n, the slowest mode is the one in which all the masses are oscillating in the same direction, while the fastest is one in which alternate masses oscillate in opposite directions. For large values of n, the normal modes approach those of a continuous stretched string, which we discuss in the following section. 11.6 Normal Modes of a Stretched String We now wish to discuss the problem treated in of normal modes. We start from the equations of motion, (10.36), 10.6 from the point of view \u00a7 \u00a8y = c2y, c2 = F/\u00b5, (11.35) 270 Classical Mechanics and look for normal mode solutions of the form y(x, t) = A(x)ei\u03c9t. (11.36) Substituting in (11.35), we obtain A(x) + k2A(x) = 0,", " k = \u03c9/c. Thus, in place of a set of simultaneous equations for the amplitudes Aj, we obtain a di\ufb00erential equation for the amplitude function A(x). The general solution of this equation is A(x) = a cos kx + b sin kx. However, because the ends of the string are \ufb01xed, we must impose the boundary conditions A(0) = A(l) = 0. (Note that l here is the full length of the string, denoted by (n + 1)l in the preceding section.) Thus a = 0 and moreover sin kl = 0. The possible values of k are k = n\u03c0 l, n = 1, 2, 3,.... (11.37) Each of these values corresponds to a normal mode of the string. The corresponding angular frequencies are \u03c9 = n\u03c0c l, n = 1, 2, 3,.... (11.38) Note that they are all multiples of the fundamental frequency \u03c91 = \u03c0c l = \u03c0 F M l, where M is the total mass of the string. The solution for the nth normal mode can be written as y(x, t) = Re Anein\u03c0ct/l sin n\u03c0x l, (11.39) where An is an arbitrary complex constant. It represents a \u2018standing wave\u2019 1 nodes, or points where y = 0. The \ufb01rst few of wavelength 2l/n, with n normal modes are illustrated in Fig. 11.7. \u2212 The general solution for the stretched string is a superposition of all the normal modes (11.39). It is easy to establish the connection with the general solution (10.37) obtained in the previous chapter. According to Small Oscillations and Normal Modes 271 Fig. 11.7 (10.38), f (x) is a periodic function of x, with period 2l. Hence it may be expanded in a Fourier series (see Eq. (2.45)), + \u221e f (x) = fnein\u03c0x/l. n= \u2212\u221e Thus the solution (10.37) is + \u221e y(x, t) = fn ein\u03c0(ct+x)/l ein\u03c0(ct \u2212 x)/l \u2212 n= \u2212\u221e + \u221e n= \u2212\u221e = 2ifnein\u03c0ct/l sin n\u03c0x l. \u2212 Now recall that fn and f n must be complex conjugates (so that the two terms add to give a real contribution to y). Thus we may restrict the sum to positive values of n. (Note that there is no contribution from n = 0 because the sine function then vanishes.) If we de\ufb01ne An = 2ifn, we can write the solution as y(x, t) = 2Re Anein\u03c0ct/l sin n\u03c0x l. + \u221e n=1 272 Classical Mechanics 11.7 Summary Near a position of equilibrium of any natural, conservative system, the kinetic energy may be taken to be a homogeneous quadratic function of the \u02d9q\u03b1, with constant coe\ufb03cients, and the potential energy to be a homogeneous quadratic function of the q\u03b1. We can always \ufb01nd a set of orthogonal co-ordinates, in terms of which T is reduced to a sum of squares. Lagrange\u2019s equations then take on a simple form. To \ufb01nd the normal modes of oscillation, we substitute solutions of the form q\u03b1 = A\u03b1ei\u03c9t, and obtain a set of simultaneous linear equations for the coe\ufb03cients. The condition for consistency of these equations is the characteristic equation, which determines the frequencies of the normal modes. The stability condition is that all the roots of this equation for \u03c92 should be positive. The problem of \ufb01nding the normal modes is equivalent to that of \ufb01nding normal co-ordinates, which reduce not only T but also V to a sum of squares. In terms of normal co-ordinates, the system is reduced to a set of uncoupled harmonic oscillators, whose frequencies are the characteristic frequencies of the system. The general solution to the equations of motion is a superposition of all the normal modes. In it, each normal co-ordinate is oscillating at its own frequency, and with amplitude and phase determined by the initial conditions. The linearized analysis of small amplitude oscillations near to a position of stable equilibrium in the form of normal modes is a technique which is applicable generally. For some systems, which are special but important, the idea of a normal mode may be generalized. Such systems may then be analyzed as a combination of \u2018nonlinear\u2019 normal modes, where no small amplitude approximation needs to", " be made \u2014 see 14.1. \u00a7 Problems 1. A double pendulum, consisting of a pair, each of mass m and length l, is released from rest with the pendulums displaced but in a straight line. Find the displacements of the pendulums as functions of time. 2. Find the normal modes of a pair of coupled pendulums (like those of Fig. 11.2) if the two are of di\ufb00erent masses M and m, but still the same length l. Given that the pendulum of mass M is started oscillating with amplitude a, \ufb01nd the maximum amplitude of the other pendulum in Small Oscillations and Normal Modes 273 the subsequent motion. Does the amplitude of the \ufb01rst pendulum ever fall to zero? 3. A spring of negligible mass, and spring constant (force/extension) k, supports a mass m, and beneath it a second, identical spring, carrying a second, identical mass. Using the vertical displacements x and y of the masses from their positions with the springs unextended as generalized co-ordinates, write down the Lagrangian function. Find the position of equilibrium, and the normal modes and frequencies of vertical oscillations. 4. Three identical pendulums are coupled, as in Fig. 11.2, with springs between the \ufb01rst and second and between the second and third. Find the frequencies of the normal modes, and the ratios of the amplitudes. 5. The \ufb01rst of the three pendulums of Problem 4 is initially displaced a distance a, while the other two are vertical. The system is released from rest. Find the maximum amplitudes of the second and third pendulums in the subsequent motion. 6. Three identical springs, of negligible mass, spring constant k, and natural length a are attached end-to-end, and a pair of particles, each of mass m, are \ufb01xed to the points where they meet. The system is stretched between \ufb01xed points a distance 3l apart (l > a). Find the frequencies of normal modes of (a) longitudinal, and (b) transverse oscillations. 7. *A bead of mass m slides on a smooth circular hoop of mass M and radius a, which is pivoted at a point on its rim so that it can swing freely in its plane. Write down the Lagrangian in terms of the angle of inclination \u03b8 of the diameter through the pivot and the angular position \u03d5 of the bead relative to a \ufb01xed point on the hoop. Find the frequencies of the normal modes, and sketch the con\ufb01guration of hoop and bead at the extreme point of each. 8. *The system of Problem 7 is released from rest with the centre of the hoop vertically below the pivot and the bead displaced by a small angle \u03d50. Given that M = 8m and that 2a is the length of a simple pendulum of period 1 s, \ufb01nd the angular displacement \u03b8 of the hoop as a function of time. Determine the maximum value of \u03b8 in the subsequent motion, and the time at which it \ufb01rst occurs. 9. A simple pendulum of mass m, whose period when suspended from a rigid support is 1 s, hangs from a supporting block of mass 2m which can move along a horizontal line (in the plane of the pendulum), and is restricted by a harmonic-oscillator restoring force. The period of the 274 Classical Mechanics oscillator (with the pendulum removed) is 0.1 s. Find the periods of the two normal modes. When the pendulum bob is swinging in the slower mode with amplitude 100 mm, what is the amplitude of the motion of the supporting block? 10. *The system of Problem 9 is initially at rest, and the pendulum bob is 1. given an impulsive blow which starts it moving with velocity 0.5 m s\u2212 Find the position of the support as a function of time in the subsequent motion. 11. *A particle of charge q and mass m is free to slide on a smooth horaj, and two \ufb01xed 2ai. Find the electrostatic potential near the origin 6.2). Show that this is a position of stable equilibrium, and \ufb01nd izontal table. Two \ufb01xed charges q are placed at charges 12q at (see the frequencies of the normal modes of oscillation near it. \u00b1 \u00b1 \u00a7 12. *A rigid rod of length 2a is suspended by two light, inextensible strings of length l joining its ends to supports also a distance 2a apart and level with each other. Using the longitudinal displacement x of the centre of the rod, and the transverse displacements y1, y2 of its ends, as generalized co-ordinates, \ufb01nd the Lagrangian function (for small x", ", y1, y2). Determine the normal modes and frequencies. (Hint : First \ufb01nd the height by which each end is raised, the co-ordinates of the centre of mass and the angle through which the rod is turned.) 13. *Each of the pendulums in Fig 11.2 is subjected to a damping force, of magnitude \u03b1 \u02d9x and \u03b1 \u02d9y respectively, while there is a damping force \u03b2( \u02d9x \u2212 \u02d9y) in the spring. Show that the equations for the normal co-ordinates q1 and q2 are still uncoupled. Find the amplitudes of the forced oscillations obtained by applying a periodic force to one pendulum. Given that the forcing frequency is that of the uncoupled pendulums, and that \u03b2 is negligible, \ufb01nd the range of values of \u03b1 for which the amplitude of the second pendulum is less than half that of the \ufb01rst. 14. *Show that a stretched string is equivalent mathematically to an in\ufb01nite number of uncoupled oscillators, described by the co-ordinates qn(t) = 2 l l 0 y(x, t) sin n\u03c0x l dx. Determine the amplitudes of the various normal modes in the motion described in Chapter 10, Problem 14. Why, physically, are the modes for even values of n not excited? 15. Show that a typical equation of the set (11.33) may be satis\ufb01ed by setting A\u03b1 = sin \u03b1k (\u03b1 = 1, 2,..., n), provided that \u03c9 = 2\u03c90 sin 1 2 k. Small Oscillations and Normal Modes 275 Hence show by considering the required condition when \u03b1 = n + 1 that the frequencies of the normal modes are \u03c9r = 2\u03c90 sin[r\u03c0/2(n + 1)], with r = 1, 2,..., n. Why may we ignore values of r greater than n + 1? Show that, in the limit of large n, the frequency of the rth normal mode tends to the corresponding frequency of the continuous string with the same total length and mass. 16. A particle moves under a conservative force with potential energy V (r). The point r = 0 is a position of equilibrium, and the axes are so chosen that x, y, z are normal co-ordinates. Show that, if V satis\ufb01es Laplace\u2019s equation, \u22072V = 0 (see 6.7), then the equilibrium is necessarily unstable, and hence that stable equilibrium under purely gravitational and electrostatic forces is impossible. (Of course, dynamic equilibrium \u2014 stable periodic motion \u2014 can occur. Note also that the two-dimensional stable equilibrium of Problem 11 does not contradict this result because there is another force imposed, con\ufb01ning the charge to the horizontal plane.) \u00a7 276 Classical Mechanics Chapter 12 Hamiltonian Mechanics We have already seen, in several examples, the value of the Lagrangian method, which allows us to \ufb01nd equations of motion for any system in terms of an arbitrary set of generalized co-ordinates. In this chapter, we shall discuss an extension of the method, due to Hamilton. Its principal feature is the use of the generalized momenta p1, p2,..., pn in place of the generalized velocities \u02d9q1, \u02d9q2,..., \u02d9qn. It is particularly valuable when, as often happens, some of the generalized momenta are constants of the motion. More generally, it is well suited to \ufb01nding conserved quantities, and making use of them. 12.1 Hamilton\u2019s Equations The Lagrangian function L for a natural system is a function of q1, q2,..., qn and \u02d9q1, \u02d9q2,..., \u02d9qn. For brevity, we shall indicate this dependence by writing L(q, \u02d9q), where q stands for all the generalized coordinates, and \u02d9q for all their time derivatives. Lagrange\u2019s equations may be written in the form \u02d9p\u03b1 = \u2202L \u2202q\u03b1, where the generalized momenta are de\ufb01ned by p\u03b1 = \u2202L \u2202 \u02d9q\u03b1. (12.1) (12.2) Here and in the following equations, \u03b1 runs over 1, 2,..., n. The instantaneous position and velocity of every part of our system may be speci\ufb01ed by the values of the 2n variables q and \u02d9q. However, we 277 278 Classical Mechanics can alternatively solve the equations (12.2) for the \u02d9q", " in terms of q and p, obtaining, say, \u02d9q\u03b1 = \u02d9q\u03b1(q, p), (12.3) so the 2n variables q and p also serve equally well to specify the instantaneous position and velocity of every particle. For example, for a particle moving in a plane, and described by polar coordinates, the generalized momenta are given by pr = m \u02d9r and p\u03b8 = mr2 \u02d9\u03b8. In this case, Eqs. (12.3) read \u02d9r = pr m, \u02d9\u03b8 = p\u03b8 mr2. (12.4) The instantaneous position and velocity of the particle may be \ufb01xed by the values of r, \u03b8, pr and p\u03b8. We now de\ufb01ne a function of q and p, the Hamiltonian function, by n H(q, p) = p\u03b2 \u02d9q\u03b2(q, p) \u03b2=1 L(q, \u02d9q(q, p)). \u2212 (12.5) Next, we compute the derivatives of H with respect to its 2n independent variables. We di\ufb00erentiate \ufb01rst with respect to p\u03b1. One term in this derivative is the coe\ufb03cient of p\u03b1 in the sum p \u02d9q, namely \u02d9q\u03b1. Other terms arise from the dependence of each \u02d9q\u03b2, either in the sum or as an argument of L, on p\u03b1. Altogether, we obtain \" \u2202H \u2202p\u03b1 = \u02d9q\u03b1 + p\u03b2 \u2202 \u02d9q\u03b2 \u2202p\u03b1 \u2212 n \u03b2=1 n \u03b2=1 \u2202L \u2202 \u02d9q\u03b2 \u2202 \u02d9q\u03b2 \u2202p\u03b1. Now, by (12.2), the second and third terms cancel. Hence we are left with \u2202H \u2202p\u03b1 = \u02d9q\u03b1. (12.6) We now examine the derivative with respect to q\u03b1. Again, there are two kinds of terms, the term coming from the explicit dependence of L on q\u03b1, and those from the dependence of each \u02d9q\u03b2 on q\u03b1. We \ufb01nd \u2202H \u2202q\u03b1 = \u2202L \u2202q\u03b1 \u2212 + p\u03b2 \u2202 \u02d9q\u03b2 \u2202q\u03b1 \u2212 n \u03b2=1 n \u03b2=1 \u2202L \u2202 \u02d9q\u03b2 \u2202 \u02d9q\u03b2 \u2202q\u03b1. Hamiltonian Mechanics 279 As before, the second and third terms cancel. Thus, using Lagrange\u2019s equations (12.1), we obtain \u2202H \u2202q\u03b1 = \u02d9p\u03b1. \u2212 (12.7) The equations (12.6) and (12.7) together constitute Hamilton\u2019s equations. Note that whereas Lagrange\u2019s equations are a set of n second-order di\ufb00erential equations, Hamilton\u2019s constitute a set of 2n \ufb01rst-order equations. Let us consider, for example, a particle moving in a plane under a cen- tral, conservative force. Example: Central conservative force Obtain Hamilton\u2019s equations for a particle moving in a plane with potential energy function V (r). Here the Lagrangian is L = 1 2 m \u02d9r2 + 1 2 mr2 \u02d9\u03b82 V (r). \u2212 Thus the Hamiltonian function is H = (pr \u02d9r + p\u03b8 \u02d9\u03b8) 2 m \u02d9r2 + 1 [ 1 2 mr2 \u02d9\u03b82 \u2212 V (r)], \u2212 or, using (12.4) to eliminate the velocities, H = p2 r 2m + p2 \u03b8 2mr2 + V (r). (12.8) Note that this is the expression for the total energy, T + V. This is no accident, but a general property of natural systems, as we shall see below. The \ufb01rst pair of Hamilton\u2019s equation, (12.6), are \u02d9r = \u2202H \u2202pr = pr m, \u02d9\u03b8 = \u2202H \u2202p\u03b8 = p\u03b8 mr2. (12.9) They simply reproduce the relations (12.4) between velocities and momenta. The second pair, (12.7), are \u02d9pr = \u2212 \u2202H \u2202r p2 \u03b8 mr3 + dV dr, = \u2212 \u02d9p\u03b8 = \u2212 \u2202H \u2202\u03b8 = 0. (12.10) 280 Classical Mechanics The second of these two equations yields the law of conservation of", " angular momentum, p\u03b8 = J = constant. (12.11) The \ufb01rst gives the radial equation of motion, \u02d9pr = m\u00a8r = J 2 mr3 \u2212 dV dr. It may be integrated to give what we termed the \u2018radial energy equation\u2019 (4.12) in Chapter 4. 12.2 Conservation of Energy \u00a7 We saw in 10.1 that a natural system is characterized by the fact that the kinetic energy contains no explicit dependence on the time, and is a homogeneous quadratic function of the time derivatives \u02d9q. This latter condition may be expressed algebraically by the equation \u2202T \u2202 \u02d9q\u03b1 n \u03b1=1 \u02d9q\u03b1 = 2T. For example, for n = 2, T has the form (11.1). Thus \u2202T \u2202 \u02d9q1 \u02d9q1 + \u2202T \u2202 \u02d9q2 \u02d9q2 = (a11 \u02d9q1 + a12 \u02d9q2) \u02d9q1 + (a21 \u02d9q1 + a22 \u02d9q2) \u02d9q2 = 2T. Since p\u03b1 = \u2202T /\u2202 \u02d9q\u03b1, we therefore have H = n \u03b2=1 p\u03b2 \u02d9q\u03b2 \u2212 n L = \u03b2=1 = 2T \u2202T \u2202 \u02d9q\u03b2 \u02d9q\u03b2 \u2212 (T \u2212 V ) (T \u2212 V ) = T + V. \u2212 Thus, for a natural system, the value of the Hamiltonian function is equal to the total energy of the system. \u00a7 For a forced system, the Lagrangian can sometimes be written, as we saw V, where T is a homogeneous quadratic 10.4, in the form L = T in in the variables \u02d9q, and V is independent of them. In such a case, the Hamiltonian function is equal to T + V, which is in general not the total energy. \u2212 Now let us examine the time derivative of H. We shall now allow for the possibility that H may contain an explicit time dependence (as it does for Hamiltonian Mechanics 281 some forced systems), and write H = H(q, p, t). Then the value of H varies with time for two reasons: \ufb01rstly, because of its explicit time dependence, and, secondly, because the variables q and p are themselves functions of time. Thus the total time derivative is dH dt = \u2202H \u2202t + n \u03b1=1 \u2202H \u2202q\u03b1 \u02d9q\u03b1 + \u2202H \u2202p\u03b1 \u02d9p\u03b1. n \u03b1=1 Now, if we express \u02d9q and \u02d9p in terms of derivatives of H, using Hamilton\u2019s equations (12.6) and (12.7), we obtain dH dt = \u2202H \u2202t + n \u03b1=1 \u2202H \u2202q\u03b1 \u2202H \u2202p\u03b1 \u2212 \u2202H \u2202p\u03b1 \u2202H \u2202q\u03b1. Obviously, the terms in parentheses cancel, whence dH dt = \u2202H \u2202t. (12.12) This equation asserts that the value of H changes in time only because of its explicit time dependence. The net change induced by the fact that q and p vary with time is zero. In particular, for a natural, conservative system, neither T nor V contains any explicit dependence on the time. Thus \u2202H/\u2202t = 0, and it follows that dH dt = 0. (12.13) Thus there is a law of conservation of energy, H = T + V = E = constant. (12.14) In a forced system, if H = T + V, and is time-independent, we again have a conservation law \u2014 like (10.20) \u2014 though not for T + V. Even when H is not of the form T + V, an energy conservation law may exist. The prime example of this is that of a charged particle moving in static electric and magnetic \ufb01elds E and B. Since the magnetic force is perpendicular to \u02d9r, it does no work, and the sum of the kinetic and electrostatic potential energies is a constant. It is interesting to see how this emerges from the Hamiltonian formalism. 282 Classical Mechanics Example: Particle in electric and magnetic \ufb01elds Find the Hamiltonian for a charged particle in electric and magnetic \ufb01elds. Show that, if the \ufb01elds are time-independent, there is an energy conservation law. We begin with the Lagrangian (10.27): L = 1 2 m \u02d9r2 + q \u02d9", "r A(r, t) q\u03c6(r, t), \u00b7 from which, as in (10.28), it follows that the generalized momentum is p = m \u02d9r + qA. Thus we \ufb01nd \u2212 H = p \u02d9r \u00b7 \u2212 L = (p qA)2 \u2212 2m + q\u03c6. (12.15) Now, if E and B are time-independent, it is possible to choose the scalar and vector potentials \u03c6 and A also to be so (see A.7). Thus H has no explicit time-dependence, and therefore is conserved: H = E = constant. \u00a7 Note that in this case, L does have a term linear in \u02d9r: it has the form L = L2 + L1 + L0, where Lk is of degree k in \u02d9r. In such a case, it L0. The \u2018magnetic\u2019 term L1 apparently does turns out that H = L2 \u2212 not appear. In fact, B appears in H only via the relation between \u02d9r and p. The Hamiltonian formalism is particularly well suited to \ufb01nding conservation laws, or constants of the motion. The conservation law for energy is the \ufb01rst of a large class of conservation laws which we shall discuss in the following sections. 12.3 Ignorable Co-ordinates It sometimes happens that one of the generalized co-ordinates, say q\u03b1, does not appear in the Hamiltonian function (though the corresponding momentum, p\u03b1, does). In that case, the co-ordinate q\u03b1 is said to be ignorable \u2014 for a reason we shall explain in a moment. For an ignorable co-ordinate, Hamilton\u2019s equation (12.7) yields \u02d9p\u03b1 = \u2212 \u2202H \u2202q\u03b1 = 0. (12.16) Hamiltonian Mechanics 283 It leads immediately to a conservation law for the corresponding generalized momentum, p\u03b1 = constant. (12.17) For example, for a particle moving in a plane under a central, conservative force, H is independent of the angular co-ordinate \u03b8 and we therefore have the law of conservation of angular momentum, (12.11). The term \u2018ignorable co-ordinate\u2019 means just what it says: that for many purposes we can ignore the co-ordinate q\u03b1, and treat the corresponding p\u03b1 simply as a constant appearing in the Hamiltonian function. This is, in e\ufb00ect, what we did for the central force problem in Chapter 4. Because of the conservation law for angular momentum, we were able to deal with an e\ufb00ectively one-dimensional problem involving only the radial co-ordinate r. The generalized momentum p\u03b8 = J was simply a constant appearing in the equation of motion or the energy conservation equation. Let us re-examine this problem from the Hamiltonian point of view. Since the Hamiltonian (12.8) is independent of \u03b8, so \u03b8 is ignorable. Thus, we may regard (12.8) as the Hamiltonian for a system with one degree of freedom, described by the co-ordinate r and its corresponding momentum pr, in which a constant p\u03b8 appears. It is identical with the Hamiltonian for a particle moving in one dimension under a conservative force with potential energy function U (r) = p2 \u03b8 2mr2 + V (r). (12.18) This is precisely the \u2018e\ufb00ective potential energy function\u2019 of (4.13). Hamilton\u2019s equations for r and pr are \u02d9r = \u2202H \u2202pr = pr m, \u02d9pr = \u2212 \u2202H \u2202r = dU dr. To solve the central force problem, we solve \ufb01rst this one-dimensional problem (for example, using its energy conservation equation, the \u2018radial energy equation\u2019). Our solution gives us complete information about the radial motion \u2014 it gives \u02d9r as a function of r, and therefore r as a function of t, by integrating. Any required information about the angular part of the motion can then be found from the remaining pair of Hamilton\u2019s equations, one of which is the angular momentum conservation equation, \u02d9p\u03b8 = 0, while the other gives \u02d9\u03b8 in terms of p\u03b8, in the form \u02d9\u03b8 = p\u03b8/mr2. Clearly, though we did not then 284 Classical Mechanics introduce the Hamiltonian, this is essentially just the method we used in Chapter 4. We shall use the same method in the following section to discuss the general motion of a symmetric top. Example: The symmetric top Show that for the symmetric top two of the three Euler angles are ignorable co-ordinates, and", " \ufb01nd the \u2018e\ufb00ective potential energy function\u2019 for the remaining co-ordinate. We start from the Lagrangian function (10.11): 2 I3( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8)2 2 I1 \u02d9\u03d52 sin2 \u03b8 + 1 2 I1 \u02d9\u03b82 + 1 L = 1 M gR cos \u03b8. \u2212 The corresponding generalized momenta are p\u03d5 = I1 \u02d9\u03d5 sin2 \u03b8 + I3( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8) cos \u03b8, p\u03b8 = I1 \u02d9\u03b8, p\u03c8 = I3( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8). Solving these equations for \u02d9\u03d5, \u02d9\u03b8, \u02d9\u03c8, we obtain, (12.19) p\u03c8 cos \u03b8 I1 sin2 \u03b8, \u02d9\u03d5 = \u02d9\u03b8 = \u02d9\u03c8 = p\u03d5 \u2212 p\u03b8 I1 p\u03c8 I3 \u2212 p\u03d5 \u2212 p\u03c8 cos \u03b8 cos \u03b8. I1 sin2 \u03b8 The simplest way to construct the Hamiltonian function is to use the fact that H = T + V, and express T in terms of the generalized momenta using (12.19). In this way, we \ufb01nd H = (p\u03d5 \u2212 p\u03c8 cos \u03b8)2 2I1 sin2 \u03b8 + p2 \u03b8 2I1 + p2 \u03c8 2I3 + M gR cos \u03b8. (12.20) It is easy to verify that the \ufb01rst set of Hamilton\u2019s equations, (12.6), correctly reproduce (12.19). It is clear that the two co-ordinates \u03d5 and \u03c8 here are ignorable, and there are two corresponding conservation laws, p\u03d5 = constant and p\u03c8 = constant. Thus the problem can be reduced to that of a system with one degree of freedom only, described by the co-ordinate \u03b8. The Hamiltonian Hamiltonian Mechanics 285 function (12.20) may be written H = p2 \u03b8 2I1 + U (\u03b8), where the e\ufb00ective potential energy function U (\u03b8) is U (\u03b8) = (p\u03d5 \u2212 p\u03c8 cos \u03b8)2 2I1 sin2 \u03b8 + p2 \u03c8 2I3 + M gR cos \u03b8. (12.21) 12.4 General Motion of the Symmetric Top Hamilton\u2019s equations for \u03b8 and p\u03b8 give I1 \u00a8\u03b8 = \u2212 \u02d9p\u03b8 = \u2212 \u2202H \u2202\u03b8 = dU d\u03b8. (12.22) This is obviously a rather complicated equation to solve. However, the qualitative features of the motion can be found from the energy conservation equation, p2 \u03b8 2I1 + U (\u03b8) = E = constant. (12.23) \u2264 In particular, the angles \u03b8 at which \u02d9\u03b8 = 0 are given by the equation U (\u03b8) = E, and the motion is con\ufb01ned to the region where U (\u03b8) E. \u00b1 + Now let us examine the function U (\u03b8). We exclude for the moment the special case where p\u03d5 = p\u03c8. (We return to this important special case in the next section.) Then it is clear from (12.21) that as \u03b8 approaches either. Hence it has roughly the form shown in Fig. 12.1, 0 or \u03c0, U (\u03b8) with a minimum at some value of \u03b8, say \u03b80, between 0 and \u03c0. It can be shown that there is only one minimum (see Problem 7). When E is equal to this minimum value, we have an \u2018equilibrium\u2019 situation, and \u03b8 remains \ufb01xed at \u03b80. This corresponds to steady precession. For any larger value of E, the angle \u03b8 oscillates between a minimum \u03b81 and a maximum \u03b82. \u2192 \u221e It is not hard to describe the motion of the top. We note that, according to (12.19), the angular velocity of the axis about the vertical, \u02d9\u03d5, is zero when cos \u03b8 = p\u03d5/p\u03c8. If this angle lies outside the range between \u03b81 and \u03b82 (or > 1), then \u02d9\u03d5 never vanishes and the axis precesses round the if vertical in a \ufb01xed direction, and wobbles up and down between \u03b81 and \u03b8", "2. p\u03d5/p\u03c8| | 286 Classical Mechanics Fig. 12.1 This motion is illustrated in Fig. 12.2, which shows the track of the end of the axis on a sphere. \u03b8 1 \u03b8 2 Fig. 12.2 On the other hand, if \u03b81 < arccos(p\u03d5/p\u03c8) < \u03b82, the axis moves in loops, as shown in Fig. 12.3. The angular velocity has one sign near the top of the loop, and the opposite near the bottom. The limiting case between the two kinds of motion occurs when arccos(p\u03d5/p\u03c8) = \u03b81. Then the loops shrink to cusps, as shown in Fig. 12.4. The axis of the top comes instantaneously to rest at the top of each loop. Hamiltonian Mechanics 287 \u03b8 1 \u03b8 2 \u03b8 1 \u03b8 2 Fig. 12.3 Fig. 12.4 This kind of motion will occur if the top is set spinning with its axis initially at rest. (It is impossible to have cusped motion with the cusps at the bottom, for they correspond to points of minimum kinetic energy, and the motion must always be below such points. A top set spinning with its axis stationary cannot rise without increasing its energy.) 288 Classical Mechanics It is easy to observe these kinds of motion using a small gyroscope. In practice, because of frictional e\ufb00ects, the type of motion will change slowly with time. Example: Stability of a vertical top A top is set spinning with angular velocity \u03c93 with its axis vertical. How will it move? If the axis of the top passes through the vertical, \u03b8 = 0, then it is clear that U (0) must be \ufb01nite. This is possible only if p\u03d5 = p\u03c8, and both generalized momenta must be equal to I3\u03c93. (We could also consider types of motion for which the axis passes through the downward vertical, I3\u03c93. The treatment is \u03b8 = \u03c0. In that case, we require p\u03d5 = entirely similar \u2014 but less interesting because a downward-pointing top is always stable.) p\u03c8 = \u2212 \u2212 If we set p\u03d5 = p\u03c8 = I3\u03c93, the e\ufb00ective potential energy function (12.21) becomes U (\u03b8) = I 2 3 \u03c92 3 2I1 tan2 1 2 \u03b8 + 1 2 I3\u03c92 3 + M gR cos \u03b8, (12.24) where we have used the identity (1 cos \u03b8)/ sin \u03b8 = tan 1 2 \u03b8. \u2212 For small values of \u03b8, we may expand U (\u03b8), and retain only the terms up to order \u03b82, obtaining U (\u03b8) \u2248 2 I3\u03c92 ( 1 3 + M gR) + 3 \u03c92 I 2 3 4I1 \u2212 1 2 M gR \u03b82. (12.25) Since there is no linear term, \u03b8 = 0 is always a position of equilibrium. It is a position of stable equilibrium if U (\u03b8) has a minimum at \u03b8 = 0, that is, if the coe\ufb03cient of \u03b82 is positive. Thus there is a minimum value of \u03c93 for which the vertical top is stable, given by \u03c92 3 = 4I1M gR I 2 3 = \u03c92 0, say. (12.26) If the top is set spinning with angular velocity greater than this critical value, it will remain vertical. When the angular velocity falls below the critical value (as it eventually will, because of friction), the top will begin to wobble. The energy of the vertical top is E = 1 2 I3\u03c92 3 + M gR. Hamiltonian Mechanics 289 Thus the angles at which \u02d9\u03b8 = 0 are given by U (\u03b8) = E (see Fig. 12.5), or, from (12.24), by I 2 3 \u03c92 3 2I1 tan2 1 2 \u03b8 = M gR(1 \u2212 cos \u03b8) = 2M gR sin2 1 2 \u03b8. U E 0 \u03b8 2 \u03c0 \u03b8 Fig. 12.5 They are easily seen to be \u03b81 = 0 and \u03b82 = 2 arccos(\u03c93/\u03c90). Thus if the top is set spinning with its axis vertical and almost stationary, with angular velocity less than the critical value \u03c90, it will oscillate in the subsequent motion between the vertical and the angle \u03b82. Note that \u03b82 increases as \u03c93 is decreased, and tends to \u03c0 as \u03c93 approaches zero. When \u03c93 = 0, the top behaves", " like a compound pendulum, and swings in a circle through both the upward and downward verticals. 12.5 Liouville\u2019s Theorem This theorem really belongs to statistical mechanics, but it is interesting to consider it here because it is a very direct consequence of Hamilton\u2019s equations. The instantaneous position and velocity of every particle in our system is speci\ufb01ed by the 2n variables (q1,..., qn, p1,..., pn). It is convenient to think of these as co-ordinates in a 2n-dimensional space, called the phase space of the system. Symbolically, we may write r = (q1,..., qn, p1,..., pn). 290 Classical Mechanics As time progresses, the changing state of the system can be described by a curve r(t) in the phase space. Hamilton\u2019s equations, (12.6) and (12.7), prescribe the rates of change ( \u02d9q1,..., \u02d9qn, \u02d9p1,..., \u02d9pn). These may be regarded as the components of a 2n-dimensional velocity vector, v = \u02d9r = ( \u02d9q1,..., \u02d9qn, \u02d9p1,..., \u02d9pn). Now suppose that we have a large number of copies of our system, starting out with slightly di\ufb00erent initial values of the co-ordinates and momenta. For example, we may repeat many times an experiment on the same system, but with small random variations in the initial conditions. Each copy of the system is represented by a point in the phase space, moving according to Hamilton\u2019s equations. We thus have a swarm of points, occupying some volume in phase space, rather like the particles in a \ufb02uid. Liouville\u2019s theorem concerns how this swarm moves. What it says is a very simple, but very remarkable, result, namely that the representative points in phase space move as though they formed an incompressible \ufb02uid. The 2n-dimensional volume occupied by the swarm does not change with time, though of course its shape may, and usually does, change in very complicated ways. To prove this, we need to apply a generalization of the divergence operation. In three dimensions, it is shown in Appendix A (see (A.24)) that the \ufb02uid velocity in an incompressible \ufb02uid satis\ufb01es the condition \u2207 v = 0. It is easy to see that the argument generalizes to any number of dimensions. The condition that the phase-space volume does not change with time in the \ufb02ow described by the velocity \ufb01eld v is simply that the 2n-dimensional divergence of v is zero, i.e., \u00b7 v = \u2207 \u00b7 \u2202 \u02d9q1 \u2202q1 + + \u00b7 \u00b7 \u00b7 \u2202 \u02d9qn \u2202qn + \u2202 \u02d9p1 \u2202p1 \u00b7 \u00b7 \u00b7 + \u2202 \u02d9pn \u2202pn = 0. (12.27) But, by (12.6) and (12.7), the 1st and (n+1)st terms are + \u2202 \u02d9q1 \u2202q1 \u2202H \u2202p1 which is indeed zero. Similarly, all the other terms in (12.27) cancel in pairs. Thus the theorem is proved. \u2202 \u2202p1 \u2212 \u2202 \u2202q1 \u2202H \u2202q1 \u2202 \u02d9p1 \u2202p1 = +, In general, except for special cases, the motion in phase space is complicated. The phase-space volume containing the swarm of representative points maintains its volume, but becomes extremely distorted, rather like a drop of immiscible coloured liquid in a glass of water which is then stirred. Hamiltonian Mechanics 291 The points cannot just go anywhere in phase space, because of the energy conservation equation. They must remain on the same constant-energy surface, H(q, p) = E, and of course there might be other conservation laws that also restrict the accessible region of phase space. But one might expect that in time the phase-space volume would become thinly distributed throughout almost all the accessible parts of phase space. Such behaviour is called ergodic, and is commonly assumed in statistical mechanics. Averaged properties of the system over a long time can then be estimated by averaging over the accessible phase space. Remarkably, however, many quite complicated systems do not behave in this way, but show surprising almost-periodic behaviour, to be discussed in Chapter 14. The study of which systems do and do not behave ergodically, and particularly the transition between", " one type and another as the parameters of the system are varied, is now one of the most active \ufb01elds of mathematical physics. It has revealed an astonishing range of possibilities. 12.6 Symmetries and Conservation Laws \u00a7\u00a7 In 12.2 and 12.3, we found some examples of conserved quantities, but so far we have not discussed the physical reasons for their existence. In fact, they are expressions of symmetry properties possessed by the system. For example, the conservation law of angular momentum for the central force problem, (12.11), arises from the fact that the Hamiltonian is independent of \u03b8. This is an expression of the rotational symmetry of the system \u2014 in other words, of the fact that there is no preferred orientation in the plane. Explicitly, the equation \u2202H/\u2202\u03b8 = 0 means that the energy of the system is unchanged if we rotate it to a new position, replacing \u03b8 by \u03b8 + \u03b4\u03b8, without changing r, pr or p\u03b8. Thus angular momentum is conserved (p\u03b8 = constant) for systems possessing this rotational symmetry. Of course, if the force is non-central, it does determine a preferred orientation is space, and angular momentum is not conserved. In a three-dimensional problem, every component of the angular momentum J is conserved if the force is purely central. If the force is noncentral, but still possesses axial symmetry \u2014 so that H depends on \u03b8 but not on \u03d5 \u2014 then only the component of J along the axis of symmetry, namely p\u03d5, is conserved. Similarly, for the symmetric top, the equation \u2202H/\u2202\u03d5 = 0 is an expression of the rotational symmetry of the system about the vertical. The 292 Classical Mechanics corresponding conserved quantity p\u03d5 is the vertical component of J ; for, by (9.43) and (9.45), Jz = k \u00b7 J = I1 \u02d9\u03d5 sin2 \u03b8 + I3( \u02d9\u03c8 + \u02d9\u03d5 cos \u03b8) cos \u03b8 = p\u03d5. The equation \u2202H/\u2202\u03c8 = 0 expresses the rotational symmetry of the top itself about its own axis. The energy is clearly unchanged by rotating the top about its axis. In this case, we see from (9.45) that the conserved quantity p\u03c8 is the component of J along the axis of the top, p\u03c8 = J3 = e3 \u00b7 Now of course not all symmetries are expressible simply by saying that H is independent of some particular co-ordinate. For example, we might consider the central force problem in terms of the Cartesian co-ordinates x and y. Then the Hamiltonian is J. H = x + p2 p2 y 2m + V x2 + y2. Since it depends on both x and y, neither co-ordinate is ignorable. It does, however, possess a symmetry under rotations. If we make a small rotation through an angle \u03b4\u03b8, the changes in the co-ordinates and momenta are (see Fig. A.4) \u03b4x = \u03b4px = y \u03b4\u03b8, py \u03b4\u03b8, \u2212 \u2212 \u03b4y = x \u03b4\u03b8, \u03b4py = px \u03b4\u03b8. (12.28) Under this transformation, \u03b4(x2 + y2) = 0 and \u03b4(p2 \u03b4H = 0. x + p2 y) = 0, so clearly Now we know from our earlier discussion in terms of polar co-ordinates that this symmetry is related to the conservation of angular momentum, J = xpy \u2212 ypx = constant. The problem is to understand the relationship between the transformation (12.28) and the conserved quantity J. Let us consider a general function of the co-ordinates, momenta, and time, G(q, p, t). We de\ufb01ne the transformation generated by G to be \u03b4q\u03b1 = \u2202G \u2202p\u03b1 \u03b4\u03bb, \u03b4p\u03b1 = \u2202G \u2202q\u03b1 \u2212 \u03b4\u03bb, (12.29) where \u03b4\u03bb is an in\ufb01nitesimal parameter. For example, the function G = p1 generates the transformation in which \u03b4q1 = \u03b4\u03bb, while all the remaining coordinates and momenta are unchanged. Using Hamilton\u2019s equations (12.6) Hamiltonian Mechanics 293 and (12.7), we see that the transformation generated by the Hamiltonian is \u03b4q\u03b1 = \u02d9q\u03b1 \u03b4\u03bb, \u03b4p\u03b1 = \u02d9p\u03b1 \u03b4\u03bb. (12.30) If \u03b4\u03bb is interpreted as a small time interval, this represents the time development of the", " system. We can now return to the function J. The transformation it generates is given by \u03b4x = \u03b4px = \u2202J \u2202px \u2202J \u2202x \u2212 \u03b4\u03bb = \u03b4\u03bb = y \u03b4\u03bb, \u03b4y = py \u03b4\u03bb, \u03b4py = \u2212 \u2212 \u2202J \u2202py \u2202J \u2202y \u2212 \u03b4\u03bb = x \u03b4\u03bb, \u03b4\u03bb = px \u03b4\u03bb. This is clearly identical to the in\ufb01nitesimal rotation (12.28). Thus we have established a connection between J and this transformation (12.28). The next problem is to understand why the fact that this transformation represents a symmetry property of the system should lead to a conservation law. To this end, we return to a general function G, and consider the e\ufb00ect of the transformation (12.29) on some other function F (q, p, t). The change in F is n \u03b4F = \u03b1=1 \u2202F \u2202q\u03b1 \u03b4q\u03b1 + \u2202F \u2202p\u03b1 n \u03b4p\u03b1 = \u03b1=1 \u2202F \u2202q\u03b1 \u2202G \u2202p\u03b1 \u2212 \u2202F \u2202p\u03b1 \u2202G \u2202q\u03b1 \u03b4\u03bb. This kind of sum, involving the derivatives of two functions, appears quite frequently, and it is convenient to introduce an abbreviated notation. We de\ufb01ne the Poisson bracket of F and G to be [F, G] = n \u03b1=1 \u2202F \u2202q\u03b1 \u2202G \u2202p\u03b1 \u2212 \u2202F \u2202p\u03b1 \u2202G \u2202q\u03b1. (12.31) Then we can write the change in F under the transformation generated by G in the form \u03b4F = [F, G] \u03b4\u03bb. (12.32) A particular example is provided by the transformation (12.30) gener- ated by H. The rate of change of F is dF dt = \u2202F \u2202t + n \u03b1=1 \u2202F \u2202q\u03b1 \u02d9q\u03b1 + \u2202F \u2202p\u03b1 \u2202F \u2202t \u02d9p\u03b1 = + [F, H]. (12.33) 294 Classical Mechanics The extra term here arises from the fact that we have now allowed F to have an explicit dependence on the parameter t, in addition to the dependence via q and p. Now an obvious property of the Poisson bracket is its antisymmetry. If we interchange F and G, we merely change the sign: [G, F ] = [F, G]. \u2212 (12.34) This has the important consequence that, if F is unchanged by the transformation generated by G, then reciprocally G is unchanged by the transformation generated by F. We are now \ufb01nally in a position to apply this discussion to the case of a symmetry property of the system. Let us suppose that there exists a transformation of the co-ordinates and momenta which leaves the Hamiltonian una\ufb00ected, and which is generated by a function G. From (12.29) we see that the generator G is unique, apart from an arbitrary additive function of t, independent of q and p. In particular, if the transformation does not involve the time explicitly, then G may be chosen to contain no explicit t dependence. The condition that H should be unchanged is \u03b4H = [H, G] \u03b4\u03bb = 0. (12.35) It then follows from the reciprocity relation (12.34) that [G, H] = 0 also. Hence if \u2202G/\u2202t = 0, we \ufb01nd from (12.33) that dG dt = [G, H] = 0. (12.36) Thus we have shown that if H is unaltered by a t-independent transformation of this type, then the corresponding generator is conserved. The number of independent symmetries possessed by a system, and hence the number of conserved quantities, has a profound e\ufb00ect on the way that the system may behave. By exploiting the conservation laws, the complexity of problems may be reduced progressively. If the number of independent conserved quantities for a system, i.e., constants or integrals of the motion, is at least equal to the number of degrees of freedom (the number of independent co-ordinates), then the reduction may be complete, and the system is termed integrable (in the sense of Liouville \u2014 see 14.1). The motion is then ordered and \u2018regular\u2019. If there are insu\ufb03cient conserved quantities to bring this about, then the motion may exhibit disorder or \u2018chaos\u2019. \u00a7 Hamiltonian Mechanics 295 The central, conservative force problems considered in Chapter 4", ", and the symmetric top considered in 12.3, 12.4, are examples of 10.3 and integrable systems, since they possess respectively two and three conserved quantities, equal in each case to the number of degrees of freedom. The restricted three-body problem, considered in Problems 15 and 16 of Chapter 10, and in Problems 12 and 13 at the end of this chapter, does not possess such a complete set of conserved quantities, and so is not integrable. \u00a7\u00a7 \u00a7 The formal description of the connection between symmetry properties and invariance is contained in a famous theorem due to Emmy Noether (1918). 12.7 Galilean Transformations \u00a7 To illustrate the ideas of the preceding section, we shall consider a general, isolated system of N particles, and investigate the symmetry properties implied by the relativity principle of 1.1. The system has 3N degrees of freedom, and may be described by the particle positions ri and momenta pi (i = 1, 2,..., N ). We shall consider four distinct symmetry properties, associated with the requirements that there should be no preferred zero of the time scale, origin in space, orientation of axes, or standard of rest. The corresponding symmetry transformations are translations in time, spatial translations, rotations, and transformations between frames moving with uniform relative velocity (sometimes called boosts). A combination of these four types of transformation is the most general transformation which takes one inertial frame into another. They are known collectively as Galilean transformations. We consider them in turn. Time translations The changes in r and p in an in\ufb01nitesimal time \u03b4t are generated by the Hamiltonian function H. The condition for invariance of H under this transformation is [H, H] = 0, which is certainly true because of (12.34). Thus, as we showed in 12.2, if H contains no explicit time dependence, then it is in fact conserved, \u00a7 dH dt = 0. (12.37) 296 Classical Mechanics Spatial translations An in\ufb01nitesimal translation of the system through a distance \u03b4x in the x direction is represented by the transformation \u03b4xi = \u03b4x, \u03b4pxi = 0, \u03b4yi = 0, \u03b4pyi = 0, \u03b4zi = 0, \u03b4pzi = 0. (12.38) The corresponding generator is easily seen to be the total x-component of momentum, N Px = pxi. i=1 The condition for H to be invariant under this transformation is 0 = [H, Px] \u03b4x = \u2202H \u2202xi \u03b4x. N i=1 It is satis\ufb01ed if H depends only on the co-ordinate di\ufb00erences xi \u2212 xj; for then, changing each xi by the same amount cannot a\ufb00ect H. When this condition holds, we obtain a conservation law for the x-component of momentum, dPx/dt = 0. More generally, a translation in the direction of the unit vector n is generated by the component of the total momentum P in this direction, n P. When the system possesses translational invariance in all directions, then all components of P are conserved: \u00b7 dP dt = 0. (12.39) This is physically very reasonable. We know from our earlier work that momentum is conserved for an isolated system, but not for a system subjected to external forces which determine a preferred origin (for example, a centre of force). Rotations An in\ufb01nitesimal rotation through an angle \u03b4\u03d5 about the z-axis yields \u03b4xi = \u03b4pxi = yi \u03b4\u03d5, pyi \u03b4\u03d5, \u2212 \u2212 \u03b4yi = xi \u03b4\u03d5, \u03b4pyi = pxi \u03b4\u03d5, \u03b4zi = 0, \u03b4pzi = 0. (12.40) Hamiltonian Mechanics 297 The corresponding generator is the z-component of the total angular momentum, Jz = N i=1 (xipyi \u2212 yipxi). The condition for H to be rotationally symmetric, [H, Jz] = 0, is satis\ufb01ed provided that H involves the x and y co-ordinates and momenta only through invariant combinations like xixj + yiyj. In general, a rotation through an angle \u03b4\u03d5 about an axis in the direction of the unit vector n may be written in the form \u03b4ri = n ri \u03b4\u03d5, \u2227 \u03b4pi = n pi \u03b4\u03d5. \u2227 (12.41) \u00b7 It is generated by the appropriate component of the angular momentum, n J. The Hamiltonian function is invariant under this transformation provided that it is a scalar function of ri and pi, involving only the squares", " and scalar products of vectors. It is easy to verify that scalar quantities are indeed unchanged by (12.41). For example, \u03b4(ri \u00b7 rj) = (\u03b4ri) rj + ri \u00b7 \u00b7 (\u03b4rj) = [(n ri) rj + ri \u00b7 \u00b7 (n \u2227 \u2227 rj)]\u03b4\u03d5 = 0, by the symmetry of the scalar triple product (see (A.13)). When the Hamiltonian possesses complete rotational symmetry, then every component of J is conserved, and so dJ dt = 0. (12.42) So far, we have shown that the conservations laws of energy, momentum, and angular momentum are expressions of symmetry properties required by the relativity principle. They have, therefore, a much more general validity than the speci\ufb01c assumptions used in their original derivation. For example, we have not assumed that the forces in our system are all two-body forces, or that they are central or conservative. All we have assumed is the existence of the Hamiltonian function, and the relativity principle. There remains one type of Galilean transformation, which is in some ways rather di\ufb00erent from the others. 298 Classical Mechanics Transformations to moving frames Let us consider the e\ufb00ect of giving our system a small overall velocity \u03b4v in the x direction \u2014 a small \u2018boost\u2019. The corresponding transformations are \u03b4xi = t \u03b4v, \u03b4pxi = mi \u03b4v, \u03b4yi = 0, \u03b4pyi = 0, \u03b4zi = 0, \u03b4pzi = 0. (12.43) This transformation di\ufb00ers from the others we have considered in that t appears explicitly in (12.43). The generator of this transformation is N Gx = (pxit i=1 mixi) = Pxt M X, \u2212 \u2212 where M is the total mass, and X is the x co-ordinate of the centre of mass. This generator is also explicitly time-dependent. It is clearly the x component of the vector G = P t M R. \u2212 (12.44) Transformations to frames moving in other directions are generated by appropriate components of this vector. We must now be careful. For, because of the explicit time-dependence, it is no longer true that H must be invariant under (12.43). Indeed, we know that the energy of a system does depend on the choice of reference frame, though not on the choice of origin or axes. What the relativity principle actually requires is that the equations of motion should be unchanged by the transformation. It can be shown (see Problem 17) that the condition for this is still dG dt = 0. (12.45) However, this no longer implies the invariance of H. In fact, using (12.33) and (12.34), we \ufb01nd \u03b4H = [H, G] \u03b4\u03bb = [G, H] \u03b4\u03bb = \u2212 \u2202G \u2202t \u03b4\u03bb. (12.46) Thus the change in H is related to the explicit time-dependence of G. The fourth of the basic conservation laws, for the quantity (12.44), is dG dt = d dt (P t \u2212 M R) = 0. (12.47) Hamiltonian Mechanics 299 Though this is an unfamiliar form, the equation is actually quite familiar. For, since dP /dt = 0, it may be written P M \u2212 dR dt = 0, which is simply the relation (8.7) between the total momentum and the centre-of-mass velocity. This relation too is therefore a consequence of the relativity principle. Let us now turn to the question of what this symmetry implies for the form of the Hamiltonian function. For a transformation in the x direction, the change in H, given by (12.46), is \u03b4H = \u2202Gx \u2202t \u03b4v = Px \u03b4v. (12.48) Now, if we write H = T + V, where T = p2 i 2mi, N i=1 then we \ufb01nd directly from (12.43) that N \u03b4T = pxi \u03b4v = Px \u03b4v. i=1 Thus the change in kinetic energy is exactly what is demanded by (12.48), and this relation reduces to \u03b4V = 0. (12.49) It is interesting to examine the conditions imposed on V by this requirement. We have already seen that V must be a scalar function, and must rj. involve the particle positions only through the di\ufb00erences rij = ri \u2212 Since these are una\ufb00ected by (12.43), the requirement (12.49) imposes no further restrictions", " on the r dependence. Moreover, V must contain no explicit dependence on time. However, none of the conditions imposed so far requires it to be independent of the momenta. To satisfy (12.49), it must contain them only through the combinations vij = pi mi \u2212 pj mj, 300 Classical Mechanics which are easily seen to be invariant under (12.43), but this is the only new requirement. Thus the most general form of interaction in an N particle system which is invariant under all Galilean transformations is one described by a \u2018potential energy function\u2019 which is an arbitrary scalar function of the relative position vectors and the relative velocity vectors. Re\ufb02ections; parity There is one other type of co-ordinate transformation which might be mentioned. All those we have considered so far have the property that if we start with a right-handed set of axes, then the transformation takes us to another right-handed set. However, we could also consider transformar) tions like re\ufb02ections (say x which lead from a right-handed set of axes to a left-handed set. These are called improper co-ordinate transformations. They di\ufb00er from the proper transformations, such as rotations, in being discrete rather than continuous \u2014 no continuous change can ever take a right-handed set of axes into a left-handed set. z) or inversions (r \u2192 \u2212 \u2192 \u2212 x, y y, z \u2192 \u2192 The condition for the Hamiltonian to be unchanged also under improper co-ordinate transformations is that it should be a true scalar function, like rk, which changes sign ri \u00b7 under inversion. If this condition is ful\ufb01lled, the equations of motion will have the same form in right-handed and left-handed frames of reference. rj, rather than a pseudoscalar, like (ri \u2227 rj) \u00b7 Because of the discontinuous nature of these transformations, this symmetry does not lead to a conservation law for some continuous variable. In fact, in classical mechanics, it does not lead to a conservation law at all. However, in quantum mechanics, it yields a conservation law for a quantity known as the parity, which has only two possible values 1. Until 1957, it was believed that all physical laws were unchanged by re\ufb02ections, but it was then discovered by Wu et al, following a theoretical prediction by Lee and Yang, that parity is in fact not conserved in the process of radioactive decay of atomic nuclei. The laws describing such processes do not have the same form in right-handed and left-handed frames of reference. \u00b1 12.8 Summary The Hamiltonian method is an extremely powerful tool in dealing with complex problems. In particular, when the Hamiltonian function is independent Hamiltonian Mechanics 301 of some particular co-ordinate q\u03b1, then the corresponding generalized momentum p\u03b1 is conserved. In such a case, the number of degrees of freedom is e\ufb00ectively reduced by one. More generally, we have seen that any symmetry property of the system leads to a corresponding conservation law. This can be of great importance in practice, since the amount of labour involved in solving a complicated problem can be greatly reduced by making full use of all the available symmetries. If there is a su\ufb03cient number of symmetries, then the system is \u2018integrable\u2019 (in the sense of Liouville) and the conservation laws may then be exploited to produce (in principle) the complete solution to the problem. The Hamiltonian function is also of great importance in quantum mechanics, and many of the features of our discussion carry over to that case. We have seen that the variables appear in pairs. To each co-ordinate q\u03b1 there corresponds a momentum p\u03b1. Such pairs are called canonically conjugate. This relationship between pairs of variables is of central importance in quantum mechanics, where there is an \u2018uncertainty principle\u2019 according to which it is impossible to measure both members of such a pair simultaneously with arbitrary accuracy. The relationship between symmetries and conservation laws also applies to quantum mechanics. In relativity, the transformations we consider are slightly di\ufb00erent (Lorentz transformations rather than Galilean), but the same principles apply, and lead to very similar conservation laws. The relationship between the relativity principle and the familiar conservation laws (including the \u2018conservation law\u2019 P = M \u02d9R) is of the greatest importance for the whole of physics. It is the basic reason for the universal character of these laws, which were originally derived as rather special consequences of Newton\u2019s laws, but can now be seen as having a far more fundamental role. Problems 1. A particle of mass m slides on the inside of a smooth cone of semivertical angle \u03b1, whose axis points vertically upwards. Obtain the Hamiltonian function, using the distance r from the vertex, and the az", "imuth angle \u03d5 as generalized co-ordinates. Show that stable circular motion is possible for any value of r, and determine the corresponding angular velocity, \u03c9. Find the angle \u03b1 if the frequency of small oscillations about this circular motion is also \u03c9. 302 Classical Mechanics 2. Find the Hamiltonian function for the forced pendulum considered in 10.4, and verify that it is equal to T + V. Determine the frequency of \u00a7 small oscillations about the stable \u2018equilibrium\u2019 position when \u03c92 > g/l. 3. A light, inextensible string passes over a small pulley and carries a mass 2m on one end. On the other end is a mass m, and beneath it, supported by a spring with spring constant k, a second mass m. Find the Hamiltonian function, using the distance x of the \ufb01rst mass beneath the pulley, and the extension y in the spring, as generalized coordinates. Show that x is ignorable. To what symmetry property does this correspond? (In other words, what operation can be performed on the system without changing its energy?) If the system is released from rest with the spring unextended, \ufb01nd the positions of the particles at any later time. 4. A particle of mass m moves in three dimensions under the action of a central, conservative force with potential energy V (r). Find the Hamiltonian function in terms of spherical polar co-ordinates, and show that \u03d5, but not \u03b8, is ignorable. Express the quantity J 2 = m2r4( \u02d9\u03b82 + sin2 \u03b8 \u02d9\u03d52) in terms of the generalized momenta, and show that it is a second constant of the motion. 5. *Find the Hamiltonian for the pendulum hanging from a trolley described in Chapter 10, Problem 9. Show that x is ignorable. To what symmetry does this correspond? 6. *Obtain the Hamiltonian function for the top with freely sliding pivot described in Chapter 10, Problem 11. Find whether the minimum angular velocity required for stable vertical rotation is greater or less than in the case of a \ufb01xed pivot. Can you explain this result physically? 7. *To prove that the e\ufb00ective potential energy function U (\u03b8) of the symmetric top (see 12.4) has only a single minimum, show that the equation U (\u03b8) = E can be written as a cubic equation in the variable z = cos \u03b8, with three roots in general. Show, however, that f (z) has the same sign at both z = 1, and hence that there are either two roots or none between these points: for every E there are at most two values of \u03b8 for which U (\u03b8) = E. \u00b1 \u00a7 8. Find the Hamiltonian for a charged particle in electric and magnetic \ufb01elds in cylindrical polars, starting from the Lagrangian function (10.29). Show that in the case of an axially symmetric, static magnetic \ufb01eld, described by the single component A\u03d5(\u03c1, z) of the vector Hamiltonian Mechanics 303 potential, it takes the form H = 1 2m z + p2 p2 \u03c1 + (p\u03d5 \u2212 q\u03c1A\u03d5)2 \u03c12. (Note: Remember that the subscripts \u03d5 on the generalized momentum p\u03d5 and on the component A\u03d5 mean di\ufb00erent things.) \u2212 9. A particle of mass m and charge q is moving around a \ufb01xed point q (qq > 0), and in a uniform magnetic \ufb01eld B. The motion charge is con\ufb01ned to the plane perpendicular to B. Write down the Lagrangian function in polar co-ordinates rotating with the Larmor angular velocity \u03c9L = 5.5). Hence \ufb01nd the Hamiltonian function. Show that \u03d5 is ignorable, and interpret the conservation law. (Note that Jz is not a constant of the motion.) qB/2m (see \u2212 \u00a7 10. Consider a system like that of Problem 9, but with a charge +q at the origin. By examining the e\ufb00ective radial potential energy function, \ufb01nd the radius of a stable circular orbit with angular velocity \u03c9L, and determine the angular frequency of small oscillations about it. 11. *A particle of mass m and charge q is moving in the equatorial plane z = 0 of a magnetic dipole of moment \u00b5, described (see Appendix A, Problem 12) by a vector potential with the single non-zero component A\u03d5 = \u00b50\u00b5 sin \u03b8/4\u03c0r2. Show that it will continue to move in this plane. Initially, it is approaching from a great distance with velocity", " v and impact parameter b, whose sign is de\ufb01ned to be that of p\u03d5. Show that v and p\u03d5 are constants of the motion, and that the distance of closest approach to the dipole is 1 a2 b), according as b > a or b < a, where a2 = \u00b50q\u00b5/\u03c0mv. (Here q\u00b5 is assumed positive.) Find also the range of values of b for which the velocity can become purely radial, and the distances at which it does so. Describe qualitatively the appearance of the orbits for di\ufb00erent values of b. (Hint : It may be useful to sketch the e\ufb00ective radial potential energy function.) 2 (\u221ab2 \u2213 \u00b1 12. *Find the Hamiltonian for the restricted three-body problem described in Chapter 10, Problems 15 and 16. Investigate the stability of one of the Lagrangian \u2018equilibrium\u2019 positions o\ufb00 the line of centres by assumm\u03c9x0 are all ing a solution where x small quantities proportional to ept, with p constant. Show that the possible values for p are given by y0, px + m\u03c9y0 and py \u2212 x0, y \u2212 \u2212 p4 + \u03c92p2 + 27M1M2\u03c94 4(M1 + M2)2 = 0, 304 Classical Mechanics and hence that the points are stable provided that the masses M1 and M2 are su\ufb03ciently di\ufb00erent. Speci\ufb01cally, given that M1 > M2 show that the minimum possible ratio for stability is slightly less than 25. 13. The stability condition of Problem 12 is well satis\ufb01ed for the case of the Sun and Jupiter, for which M1/M2 = 1047. Indeed, in that case these positions are occupied by the so-called Trojan asteroids, whose orbital periods are the same as Jupiter\u2019s, 11.86 years. Find for this case the periods of small oscillations about the \u2018equilibrium\u2019 points (in the plane of the orbit). 14. *The magnetic \ufb01eld in a particle accelerator is axially symmetric (as in Problem 8), and in the plane z = 0 has only a z component. De\ufb01ning q\u03c1Bz, q\u03c1A\u03d5, show, using (A.40) and (A.55), that \u2202J/\u2202\u03c1 = J = p\u03d5 \u2212 and \u2202J/\u2202z = q\u03c1B\u03c1. What is the relation between \u02d9\u03d5 and J? Treat the third term of the Hamiltonian in Problem 8 as an e\ufb00ective potential energy function U (\u03c1, z) = J 2/2m\u03c12, compute its derivatives, and write down the \u2018equilibrium\u2019 conditions \u2202U/\u2202\u03c1 = \u2202U/\u2202z = 0. Hence show that a particle of mass m and charge q can move in a circle of any given radius a in the plane z = 0 with angular velocity equal to the cyclotron frequency for the \ufb01eld at that radius (see 5.2). \u2212 15. *To investigate the stability of the motion described in the preceding question, evaluate the second derivatives of U at \u03c1 = a, z = 0, and show that they may be written \u00a7 \u22022U \u2202\u03c12 = q2 m Bz Bz + \u03c1 \u2202Bz \u2202\u03c1, \u03c1=a,z=0 \u22022U \u2202\u03c1 \u2202z = 0, \u22022U \u2202z2 = q2 m \u2212 \u2202Bz \u2202\u03c1 Bz\u03c1. \u03c1=a,z=0 (Hint : You will need to use the \u03d5 component of the equation \u2207 B = 0, and the fact that, since B\u03c1 = 0 for all \u03c1, \u2202B\u03c1/\u2202\u03c1 = 0 also.) Given that the dependence of Bz on \u03c1 near the equilibrium orbit is described by Bz \u221d (a/\u03c1)n, show that the orbit is stable if 0 < n < 1. 16. Show that the Poisson brackets of the components of angular momen- \u2227 tum are [Jx, Jy] = Jz (together with two other relations obtained by cyclic permutation of x, y, z). Interpret this result in terms of the transformation of one component generated by another. Hamiltonian Mechanics 305 17. *Show that the condition that Hamilton\u2019s equations remain unchanged under the transformation generated by G is dG/dt = 0 even in the case when G has an explicit time-dependence, in addition to its dependence via q(t) and p(t). Proceed as follows. The \ufffd", "\ufffd\ufffdrst set of Hamilton\u2019s equations, (12.6), will be unchanged provided that d dt (\u03b4q\u03b1) = \u03b4 \u2202H \u2202p\u03b1. Write both sides of this equation in terms of G and use (12.33) applied both to \u2202G/\u2202p\u03b1 and to G itself to show that it is equivalent to the condition \u2202 \u2202p\u03b1 dG dt = 0. Thus dG/dt is independent of each p\u03b1. Similarly, by using the other set of Hamilton\u2019s equations, (12.7), show that it is independent of each q\u03b1. Thus dG/dt must be a function of t alone. But since we can always add to G any function of t alone without a\ufb00ecting the transformation it generates, this means we can choose it so that dG/dt = 0. 306 Classical Mechanics Chapter 13 Dynamical Systems and Their Geometry In this chapter we will begin to look at continuous dynamical problems in a new, powerful and largely geometrical way, which had its origins in the work of Poincar\u00b4e. This approach enables us to move on from the solution of the di\ufb00erential equations for a system having one set of parameters and initial conditions to the consideration of the geometry of all the possible solutions. Not only does this new perspective bring out important features of systems which are already familiar, but it also enables us to gain qualitative (and some quantitative) information about systems in which there is complex interaction, which may be highly nonlinear and accessible by other methods only with di\ufb03culty, if at all. The ideas are of very wide applicability in many \ufb01elds, not only in mechanics, other branches of physics, engineering and applied mathematics, but also in, for example, biology, chemistry and economics. [For some systems it is appropriate and useful to observe at discrete time intervals (not necessarily equal). Di\ufb00erential equations are then replaced by maps \u2014 see 14.2 and appendix D.] \u00a7 13.1 Phase Space and Phase Portraits Many systems have a \ufb01nite number n of distinct dynamic elements, which may be represented by functions xi(t) of time t where i = 1, 2,..., n. It should be noted that n is not necessarily the same as the number of degrees of freedom if, for example, the system is mechanical. Here we are dealing with the number of elements which describe the dynamic state and these might therefore include, for example, positions and velocities or momenta (as in 12.1). \u00a7 307 308 Classical Mechanics We may denote the state of the system by the vector x(t) and the n equations of motion can be written in vector form dx dt describing the evolution of the dynamic elements with \u2018velocity\u2019 F. = F (x, t), \u2261 \u02d9x (13.1) In practice F usually contains a set c of control parameters which are characteristic of the physical problem, like masses, spring constants and so on. The idea is that we \ufb01x c and consider the solutions of (13.1) for various di\ufb00erent prescribed values of the xi at a particular initial time, so determining the evolution of these di\ufb00erent initial states. It is extremely useful to view (13.1) as the prescription for the way in which the point x evolves with time along trajectories (or orbits) in an n-dimensional space \u2014 the phase space of the system. The function F in (13.1) is then called the phase velocity. The set of all trajectories in this space for various initial prescribed values of x, together with the velocity directions upon them, constitutes the phase portrait of the system. The evolutions of a continuous dynamical system pictured in this manner are often described as a \ufb02ow, by analogy with \ufb02uid motion (see Fig. 13.1). The phase space is usually taken to be Euclidean in its geometry. While this is not absolutely necessary, it is useful to do so, although there are obvious di\ufb03culties in graphical representation when n > 3. Fig. 13.1 Dynamical Systems and Their Geometry 309 It is important to note that the \ufb01rst-order system (13.1) is not as special as it appears, since any system having higher-order derivatives can be put into this standard form by a suitable change of variables. If we have dnx dtn = G dx dt,..., t, x, dn \u2212 dtn 1x 1 \u2212, (13.2) then we may write xi = di 1 for i = 1, 2,..., n, and recover the \u2212 form (13.1). Simple examples of this technique (for n = 2", ") will be given below. 1x/dti \u2212 In (13.1) the functions Fi contained within the phase velocity F may not depend explicitly on time, and the system is then said to be autonomous. Of course, it is immediately apparent that any non-autonomous system may be made autonomous through the substitution xn+1 = t, but at the expense of raising by one the dimension of the phase space. We will not be concerned further with this matter here, but will concentrate almost exclusively on autonomous systems [cf. also \u2018natural systems\u2019 of 10.1]. Evidently the phase portrait for a particular dynamical system may look very di\ufb00erent for di\ufb00erent values of the control parameters c in (13.1) and, for many such systems, there is considerable interest in the dynamical transitions which occur in their behaviour at particular parameter values. The importance of this lies in what is called the structural stability of systems; abrupt changes at particular c are bifurcations or catastrophes. \u00a7 13.2 First-order Systems \u2014 the Phase Line (n = 1) Let us consider the logistic di\ufb00erential equation for the dynamic state variable x(t): dx dt \u02d9x \u2261 = kx \u03c3x2, \u2212 with x(0) = x0. (13.3) Here k, \u03c3 are positive control parameters. The equation, which is due to Verhulst, is a model of population growth (among many other applications) and the general motivation is that the growth rate \u02d9x/x decreases as the population x(t) increases on account of overcrowding or lack of resources. Naturally the interest for this biological application is in x0 \u2265 0. 310 Classical Mechanics Now this equation (13.3) is \ufb01rst-order separable and it may be solved explicitly to give x(t) = x(t) \u2261 [\u03c3x0 + (k 0, \u2212 kx0 \u03c3x0) exp( \u2212 kt)], for x0 > 0, for x0 = 0. (13.4) This solution may be plotted as a function of t as in Fig. 13.2. \u2192 \u221e It is evident from (13.4) and from Fig. 13.2 that, for all x0 > 0, the k/\u03c3. For 0 < x0 < k/\u03c3 the \u2018sigmoid\u2019 evolution of x with t is such that x \u2192 curves obtained represent the way in which the exponential growth (which would be present if \u03c3 = 0) is inhibited by overcrowding. For x0 > k/\u03c3 there k/\u03c3 is steady decay of the population towards the value k/\u03c3. Since x \u2192 as t k/\u03c3 is in each case, then the equilibrium population x(t) said to be asymptotically stable and it is an attractor. The equilibrium 0 is unstable, in that it is only realizable if x0 = 0. Any population x(t) positive change in x0, however small, results in explosive growth until the overcrowding takes e\ufb00ect and x is attracted towards k/\u03c3, as above. So x(t) \u2261 Now here we have been fortunate to be able to arrive at the complete solution of (13.3) in the form (13.4). The comments detailed above result from interpretation of (13.4) and of Fig. 13.2 derived from it. However, the important qualitative behaviour of the solutions is obtainable more directly from (13.3) by considering the phase space and the phase portrait. 0 is a repeller. \u2261 \u2261 Fig. 13.2 Dynamical Systems and Their Geometry 311 Example: Phase portrait of the logistic system Find the qualitative features of the logistic system by considering the phase portrait. The phase space is here one-dimensional (a phase line). (See Fig. 13.3.) In constructing the phase portrait for this example the equilibria x = 0, x = k/\u03c3 are found by putting \u02d9x 0 in (13.3). It is important to note that these equilibria, or critical points, are the only values of x at which \u02d9x can change sign. The arrow directions are those of the phase velocity and, here, the complete portrait can be found by testing just a single x value away from the equilibria. From this we see immediately that x = k/\u03c3 is an attractor and x = 0 is a repeller. \u2261 Fig. 13.3 The above is su\ufb03cient argument, but with an eye to developments for higher-order systems we can consider a small-amplitude approximation near to each critical point, very similar to the analysis carried out in, for example, 2.2. If we write x(t) = k/\u03c3 +", " \u03be(t), where \u03be(t) is small enough for quadratic terms involving it to be neglected, then we \ufb01nd that \u00a7 \u02d9\u03be = k\u03be \u2212 (13.5) 0 exponentially as t to \ufb01rst order in \u03be and \u03be(t), since k is positive, con\ufb01rming of course that x = k/\u03c3 is an attractor. Note that this solution of (13.5) is consistent with our assumption that quadratic terms in \u03be could be neglected. The other critical point x = 0 may be treated similarly and con\ufb01rmed as a repeller. \u2192 \u221e \u2192 An important point to note here is that for this system and for other such \ufb01rst-order systems the qualitative behaviour is completely determined by the position of the critical points, by the local behaviour on the phase line near to them and through the strong geometrical requirement that \u02d9x cannot change sign at any other value of x. We might consider the equation \u02d9x = kx \u03c3x2 f, \u2212 \u2212 (13.6) 312 Classical Mechanics where f represents a positive constant extraction rate. For example, x(t) could be the population of \ufb01sh in a lake and f the rate of \ufb01shing this population. The exact solution is now rather more awkward to obtain and the use of the one-dimensional phase space provides clear advantage. In particular, the extra control parameter f which has been introduced can be varied and predictions made about the evolution of x from resulting changes in the phase portrait. (See Problem 1.) The analysis given above will be generalized in following sections to deal with higher-order systems. Naturally for many biological systems a discrete, rather than continuous, model is more appropriate [see Appendix D]. 13.3 Second-order Systems \u2014 the Phase Plane (n = 2) Again let us proceed by considering a particular and familiar example. Example: Simple harmonic oscillator Determine the phase-space trajectories of the simple harmonic oscillator. The oscillator equation (2.13) may be written \u00a8x = \u03c92x, \u2212 \u03c9 = k/m, (13.7) which is in the form (13.2) and which may be converted to the form (13.1) as \u02d9x = y, \u02d9y = \u03c92x. (13.8) \u2212 The trajectories of this autonomous system in the (x, y) phase plane are evidently given by the solutions of dy dx \u2261 \u02d9y \u02d9x = \u03c92x y, \u2212 (13.9) which may be solved easily to give \u03c92x2 + y2 = A2, with A2 constant. (13.10) The phase portrait is a family of ellipses centred at the origin. (See Fig. 13.4.) Here, along each trajectory the energy E of the oscillator is Dynamical Systems and Their Geometry 313 Fig. 13.4 constant and given by E = 1 2 mA2. Evidently (x, y) = (0, 0) is a critical point of the system (13.8) and any small perturbation leads to the system traversing an elliptical trajectory around this equilibrium point. Each trajectory is closed and is therefore a periodic orbit. In this case, we were able to \ufb01nd the trajectories by direct integration of (13.9). In the case of a general system this may well not be easy, or possible, except by numerical computation. The general second-order autonomous system with which we are pre- sented is \u02d9x = F (x, y), \u02d9y = G(x, y), where F, G are suitable functions. It is apparent that the equation dy dx = G F (13.11) (13.12) certainly determines the slope of the unique trajectory in the phase plane at each point except at the critical points (x0, y0), for which F (x0, y0) = 0 = G(x0, y0). Each such (x0, y0) is an equilibrium solution of the system and trajectories can only intersect at such points and nowhere else. The interesting question to ask, from the physical point of view, concerns the fate of the system when it is displaced slightly from such a critical point. The system could, for general small displacements: 314 Classical Mechanics 1. Move so as to tend towards (x0, y0) as t system is asymptotically stable., in which case we say the \u2192 \u221e 2. Move increasingly away from (x0, y0), in which case we say it is unstable. 3. Move in an arbitrarily small neighbourhood of (x0, y0) without neces, in", " which case we say it is sarily tending towards (x0, y0) as t stable. \u2192 \u221e The simple harmonic oscillator is therefore stable at x = 0, y hardly a surprising conclusion! \u2261 \u02d9x = 0 \u2014 The behaviour of the trajectories near to each critical point (x0, y0) in the phase plane can be found by an analysis similar to that introduced in 13.2 using x = x0 + \u03be, y = y0 + \u03b7, in which \u03be, \u03b7 are small and we expand \u00a7 functions F (x, y), G(x, y) locally. We obtain \u02d9\u03be = \u03be \u02d9\u03b7 = \u03be \u2202F \u2202x \u2202G \u2202x 0 0 + \u03b7 + \u03b7 \u2202F \u2202y \u2202G \u2202y 0 0 + F1(\u03be, \u03b7), + G1(\u03be, \u03b7). (13.13) Here the partial derivatives are evaluated at the critical point (x0, y0) and F1, G1 are generally quadratic in \u03be, \u03b7. In matrix notation, these equations are \u02d9\u03be \u02d9\u03b7 \u2261 d dt \u03be \u03b7 = M \u03be \u03b7 + higher-order terms, (13.14) \u00d7 and the 2 2 Jacobian matrix M has constant entries which are special to each critical point. In nearly all cases the correct behaviour of the almost linear system near \u03be = 0 = \u03b7 is given by the behaviour of the linear system which is obtained by neglecting the higher-order terms F1, G1 in the above. The behaviour near (x0, y0) of the linear system may be classi\ufb01ed using the eigenvalues \u03bb of M. We seek a solution of the linear system which is a linear combination of \u2018modes\u2019, in each of which \u03be, \u03b7 are constant multiples of e\u03bbt. The analysis appears similar to that involved in the calculation of the frequencies of the normal modes in 11.3, but we are dealing with matrices M which are not necessarily symmetric, so that the roots of the quadratic characteristic equation for \u03bb need not be real. Evidently, the signs of the real parts of these roots are crucial in determining whether trajectories near to (x0, y0) go inwards towards this critical point or go outwards from it. \u00a7 Dynamical Systems and Their Geometry 315 Details of the classi\ufb01cation of critical points in the various cases are given in Appendix C. The curves which go directly in and directly out of a critical point are called the stable and unstable manifolds respectively and, in the linear and almost linear systems, they correspond directly and are tangent to one another. We note that the critical point (x0, y0) = (0, 0) for the simple harmonic oscillator is a centre, i.e. nearby trajectories form closed curves around it. (See Appendix C, Fig. C.6.) The importance of the local behaviours near critical points of system (13.11) is that we can usually obtain a qualitative picture of the full phase portrait of the system by this means. The trajectories are constrained in that they cannot intersect elsewhere and they are locally parallel to the x, y axes respectively when crossing the curves G(x, y) = 0, F (x, y) = 0. The fate of particular initial states becomes apparent from the phase portrait. Detailed quantitative information can, if required, be obtained from exact computational plots. Apart from systems for which (13.12) may be integrated in straightforward fashion to give the curve family f (x, y) = constant, there are some special cases of particular interest in mechanics. Conservative systems with one degree of freedom For the systems considered in \u02d9x = y, \u02d9y = \u00a7 F m 2.1 we have, with F = dV (x) dx, \u2212 (13.15) and we \ufb01nd the trajectories are given by 1 2 my2 + V (x) = E, y = \u00b1 2 m [E \u2212 V (x)]. (13.16) When the potential V (x) is known the phase portrait may be sketched immediately. (See Fig. 13.5.) Since M = 0 V (x0) m 1 0\u23a4 \u23a1 \u2212 at equilibria (x0, 0), the eigenvalues of M when V (x) is a local minimum are pure imaginary, since V (x0) is positive, and (x0, 0) is a centre, which is stable. Similarly, local maxima of V (x) lead to saddles in the phase \u23a3 \u23a6 316 Classical Mechanics Fig. 13.5", " portrait and these are unstable, i.e. most trajectories approach the critical point, but then turn away. (See Appendix C, Fig. C.2.) An important physical example is provided by the simple pendulum of 2.1. \u00a7 Example: Simple pendulum Determine the possible types of motion of a simple pendulum by considering the phase portrait. From the 2\u03c0-periodic potential function V (\u03b8) sketched in Fig. 2.1 we can sketch the phase portrait. (See Fig. 13.6.) The phase portrait is also 2\u03c0-periodic. The simple oscillatory motions corresponding to closed orbits around (0, 0) are called librations cos \u03b8), it is necessary that E < 2mgl for these and, since V (\u03b8) = mgl(1 \u2212 Dynamical Systems and Their Geometry 317 Fig. 13.6 to occur (E(1) in Fig. 2.1). If E > 2mgl (E(2) in Fig. 2.1) the pendulum makes complete revolutions or rotations, either in the positive or negative sense. In the special case E = 2mgl the trajectory is a separatrix connecting the saddles ( \u03c0, 0), (\u03c0, 0), and corresponds to motions of the pendulum in which it is vertically above the point of support at t. The period of libration motion of the pendulum increases with increasing amplitude and becomes in\ufb01nite as the phase-plane trajectory corresponds more and more closely to the separatrices from the inside. \u2192 \u00b1\u221e \u2212 It should be noted that (13.15) in general, and hence the simple harmonic oscillator and simple pendulum in particular, are Hamiltonian systems with one degree of freedom. For such systems the critical points in the phase plane are either centres or saddles \u2014 spirals and nodes do not occur! Gradient systems These systems have equations of the form \u02d9x = \u2202U (x, y) \u2202x, \u2212 \u02d9y = \u2202U (x, y) \u2202y, \u2212 (13.17) \u2207U. The trajectories in the phase so that in (13.1) F takes the form \u2212 plane are everywhere perpendicular to the curves U (x, y) = constant, i.e. they are orthogonal trajectories of the level curves (contours) of U. Here 318 Classical Mechanics almost all trajectories tend towards local minima of the function U (x, y) and away from local maxima. For these systems the critical points in the phase plane are nodes and saddles \u2014 spirals and centres do not occur. A physical example of a gradient system is the \u2018runo\ufb00\u2019 problem of water \ufb02owing downhill through surface soil. (See Fig. 13.7 and Problem 5.) Fig. 13.7 13.4 Prey\u2013Predator, Competing-species Systems and War The general method of analysis in the phase plane introduced in very wide application. \u00a7 13.3 \ufb01nds An interesting example, which is deceptively simple, leads to the equa- tions \u02d9x = ax \u02d9y = bxy, cy + dxy, \u2212 \u2212 (13.18) where a, b, c, d are positive constants. These equations have been used by Volterra (1926) to model biological populations and by Lotka (1920) to model chemical reactions. They are usually known as the Lotka\u2013Volterra system. Dynamical Systems and Their Geometry 319 The biological application is to a simple ecological model of population dynamics, where x, y are prey, predators respectively (e.g. rabbits, foxes) and the terms proportional to xy model the e\ufb00ect of interaction between these species through encounters \u2014 advantageous to the predators, disadvantageous to the prey. Without the interactions the prey would proliferate exponentially in time, while the predators would die out. The parameters a, b, c, d are therefore rates of growth, decay and competitive e\ufb03ciency. The ecology, even of such isolated sets of species, is, of course, not this simple, but the consequences of idealized models prove useful as a basis for more extensive investigations. Example: Lotka\u2013Volterra system Find the critical points and their nature for the Lotka\u2013Volterra system. In the local linear analysis there are two critical points for the system (13.18): a 0 c 0 \u2212 (0, 0) with M =, so that the eigenvalues are \u03bb1 = a, \u03bb2 = c. \u2212 We have a saddle with eigenvectors parallel to the x, y co-ordinate axes. \u2022 \u2022 bc d \u2212 0 with M = \u23a1 0 ad b i\u221aac. We have a centre \u03bb1", ",2 = \u00b1, so that the eigenvalues are imaginary: The question arises as to whether the centre is a true centre for the full system in this case (see Appendix C and Problem 6). That it is such is demonstrated by the realization that (13.18) leads to dy dx = y( \u2212 x(a c + dx) by) \u2212, (13.19) which is separable, so that it may be integrated to give f (x, y) c ln x | | \u2261 \u2212 + dx a ln y | | \u2212 + by = constant. (13.20) The phase portrait may be sketched as the set of contours of f (x, y). (See Fig. 13.8.) Naturally, the biological interest is con\ufb01ned to the \ufb01rst 320 Classical Mechanics Fig. 13.8 0, y quadrant (x 0) and the trajectories are closed curves around the critical point at (c/d, a/b), so con\ufb01rming the local linear analysis that this is indeed a centre. \u2265 \u2265 Small oscillations around the point of stable equilibrium have frequency \u221aac and period T = 2\u03c0/\u221aac, which increases with the size of the orbit, although an explicit analytic demonstration of this is not easy and will not be given here. However, since T 0 we \ufb01nd that dt = \u02d9x x T (a \u2212 0 by) dt = ln x(t) T 0 = 0, (13.21) \u00afy = 1 T T 0 y dt = a b, so that the mean value of y around any periodic orbit is given by the equilibrium value a/b. Similarly \u00afx = c/d. Dynamical Systems and Their Geometry 321 There are ecological consequences of this model in that there are cyclic variations in x, y, which are not in phase, and changes in parameter values can be investigated. The e\ufb00ect here of a proportional extraction rate is to reduce a and increase c, so that the mean populations of prey/predators respectively increase/decrease \u2014 the paradoxical advantage to the prey of this extraction is due to the removal of some natural predators, and is known as Volterra\u2019s principle. The model has one further interest for us in that it is, in e\ufb00ect, a Hamiltonian system, as detailed in Chapter 12. For the \ufb01rst quadrant we may substitute x = ep, y = eq in order to obtain the Hamilton canonical form \u02d9q = \u2212 c + d ep = \u02d9p = a \u2212 b eq = \u2212 \u2202H \u2202p \u2202H \u2202q,, (13.22) cp + d ep aq + b eq. For this autonomous system, H is a where H(q, p) = \u2212 conserved quantity. (See 12.2.) In the original variables x, y the function H(q, p) becomes equivalent to f (x, y) in (13.20) and it is an \u2018energy\u2019, but not mechanical energy. \u2212 \u00a7 More complicated systems involving interactions are very common and the outcomes are even less intuitively transparent. For example \u02d9x = k1x \u02d9y = k2y \u03c31x2 \u03c32y2 \u2212 \u2212 \u2212 \u2212 \u03b11xy, \u03b12xy, (13.23) where k1, k2, \u03c31, \u03c32, \u03b11, \u03b12 are positive control parameters, models logistic behaviour (see (13.3)) for each of x, y with similarly destructive interactions. This system is certainly not usually Hamiltonian, but the phase portrait is accessible by the local linear analysis near the critical points, of which there may be four. The in\ufb02uence of the parameters is crucial. An example of such a portrait is given in Fig. 13.9, for which the parameters have been chosen to be k1 = \u03c31 = \u03b11 = 1, k2 = 1 4. (See Problem 7 for the details.) 4, \u03b12 = 3 2, \u03c32 = 1 Here there are two asymptotically stable nodes at (1, 0), (0, 2), an unstable node at (0, 0) and a saddle (unstable) at ( 1 2 ). Nearly all initial conditions lead to extinction for one or other species and this is an example of the \u2018Principle of Competitive Exclusion\u2019 [Charles Darwin, The Origin of Species (1859)]. 2, 1 322 Classical Mechanics Fig. 13.9 In the general case (13.23) and where there is a critical point in the \ufb01rst quadrant, an extinction is inevitable when k1\u03c32 < k2\u03b11, k2\u03c31 < k", "1\u03b12 (so that \u03c31\u03c32 < \u03b11\u03b12). On the other hand, peaceful coexistence via an asymptotically stable node in the \ufb01rst quadrant is guaranteed when these inequalities are all reversed (again, see Problem 7). In human con\ufb02ict there are mathematical theories of war. The simplest such model involves forces x(t), y(t) which are e\ufb00ectively isolated apart from the confrontation between them. Example: Combat model The simple system \u02d9x = \u02d9y = ay, bx, \u2212 \u2212 where a, b are positive constants, models the attrition rates of each force on the other. Determine how the \ufb01nal outcome of a con\ufb02ict depends on the initial conditions in this model. (13.24) Dynamical Systems and Their Geometry 323 Evidently (0, 0) is the only critical point and it is a saddle. Indeed the trajectories are given by ay2 \u2212 bx2 = ay2 0 \u2212 bx2 0 \u2261 K, (13.25) where x0, y0 are the initial strengths at t = 0. These are hyperbolic arcs in general (see Fig. 13.10). Fig. 13.10 Of course x \u2018wins\u2019 if K < 0 and y \u2018wins\u2019 if K > 0. One interesting feature is that the trajectories diverge \u2014 the discrepancy between the two sides becomes more marked as time progresses, so the likely outcome becomes clearer and the weaker party may be induced to surrender. The most damaging con\ufb02icts are those that start near the line K = 0 which leads to a \u2018draw\u2019 in which both sides are completely annihilated, since in this case the forces were initially perfectly balanced. The measure of \ufb01ghting e\ufb00ectiveness of e.g. force y as ay2 0 at t = 0, i.e. quadratic in numerical strength, is Lanchester\u2019s Law of Conventional Combat (1916). Through the ages, the general success of the empirical rule \u2018Divide and conquer!\u2019 is con\ufb01rmation of the e\ufb00ect of this simple qualitative law. 324 Classical Mechanics One may also model not the con\ufb02ict itself but the arms race leading up to it. Richardson (1939) introduced such a model in the form \u02d9x = a2y \u02d9y = a1x \u2212 c1x + g1, c2y + g2. (13.26) \u2212 Here x(t), y(t) are war potentials/armaments of two nations/coalitions, (a1, a2) are response parameters to the armament of the opponent (note that in contrast to the example above, these coe\ufb03cients enter with pos(See itive sign), (c1, c2) are cost deterrents and g1, g2 basic grievances. Problem 8.) 13.5 Limit Cycles So far we have looked at phase portraits in which the attractors and repellers have been equilibria which are isolated critical points. For some systems there are other possibilities. Rayleigh\u2019s equation (1883) \u02d9x(1 \u00a8x \u2212 \u2212 \u02d9x2) + x = 0, (13.27) where is a positive control parameter, has been used to model various selfexcited oscillations, for example the bowing of a violin string. By reference 2.5 it is evident that (13.27) has positive or to the damped oscillator of negative damping respectively according as < 1. The sort > 1 or of physical system modelled by this equation is illustrated schematically in Fig. 13.11, where a block is acted upon by a spring and by a dry frictional force at contact with a belt, which is itself moving uniformly. We may write \u02d9x | \u02d9x | \u00a7 | | Fig. 13.11 (13.27) in the form \u02d9x = y, \u02d9y = \u2212 x + y(1 y2). \u2212 (13.28) Dynamical Systems and Their Geometry 325 The only critical point is at (0, 0) and is always unstable (spiral or node). However trajectories come inwards from large values of x and y. The phase portraits for progressive values are sketched in Fig. 13.12. Fig. 13.12 In each case there is a periodic orbit attractor, which is an asymptoti- cally stable limit cycle. Other trajectories tend towards this cycle as t \u2192 \u221e and the form of the cycle depends crucially on the parameter. As increases, the variation of x with time t changes from essentially sinusoidal in Fig. 13", ".12(a) to a markedly non-symmetrical oscillation in Fig. 13.12(c). There is then a sharp rise and fall in x as t varies (see Fig. 13.13(a)). For this type of oscillation, which is called a relaxation oscillation, there is a mechanical analogy in the seesaw mechanism (see Fig. 13.13(b)). Here liquid is added at a constant rate to the reservoir on the right-hand side and periodically tips the balance, so that the reservoir is emptied suddenly and the cycle can be repeated. 326 Classical Mechanics Fig. 13.13 There are many mechanical and electrical examples of Rayleigh\u2019s equation and of another which may be derived from it by the substitution \u02d9x = v/\u221a3. The equation for v is then (see Problem 10) \u02d9v(1 \u00a8v \u2212 \u2212 v2) + v = 0. (13.29) This equation was studied by Van der Pol (1926) in order to model various electrical circuits and in his study of the heartbeat. A spectacular example of a limit cycle was in the behaviour of the Tacoma Narrows bridge near Seattle. The cycle of torsional oscillations, up to 1.5 m in the bridge deck, drew its energy from a constant wind \ufb02ow across it. The bridge became known as \u2018Galloping Gertie\u2019, at least up to its spectacular collapse on 7 November 1940, fortunately and memorably recorded on \ufb01lm. Relaxation oscillations have also been used to model natural phenom- ena, like geysers and earthquakes. Analytic demonstration of the existence of limit cycles, both attractors and repellers, is beyond our scope here, but it is important to note that some systems exhibit bifurcation phenomena as a control parameter is varied. Hopf bifurcation \u00a7 At the end of 13.1 we referred to the way in which critical points and phase portraits may have very di\ufb00erent character when control parameters are varied. Consider now that the stability of a focus can change as a control parameter c is varied, in such a way that the system dynamics \u2018gives birth\u2019 to a limit cycle. This is usually known as a Hopf bifurcation and it is Dynamical Systems and Their Geometry 327 Fig. 13.14 termed supercritical or subcritical respectively according as the resulting limit cycle is an attractor or repeller. For the supercritical bifurcation an asymptotically stable focus becomes unstable at a critical value c0 at which an attracting limit cycle appears. (See Fig. 13.14(a).) For the subcritical bifurcation an unstable focus becomes asymptotically stable at the critical value of c, at which a repelling limit cycle appears. (See Fig. 13.14(b).) The e\ufb00ect of a supercritical Hopf bifurcation is relatively benign (\u2018soft\u2019) in that the change results in the system remaining very much in the same neighbourhood of the phase plane when it is subject to a perturbation. However, near a subcritical Hopf bifurcation even a small perturbation can take the system outside the repelling limit cycle and then far away in the phase plane, despite the asymptotic stability of the critical point in the linear analysis \u2014 the e\ufb00ect is \u2018hard\u2019. The \u2018Brusselator\u2019 is a simple model of chemical oscillation proposed by Prigogine and Lefever (1968). 328 Classical Mechanics Example: The \u2018Brusselator\u2019 The chemical concentrations x(t), y(t) of two particular reactants are related by the equations: \u02d9x = a \u02d9y = bx \u2212 \u2212 (1 + b)x + x2y, x2y. (13.30) Here a, b are concentrations of other reactants which, once chosen, are kept constant, with a, b both positive. Find the nature of the attractors for this model. There is only one critical point, at (a, b/a), with Jacobian matrix b 1 a2 \u2212 a2 b \u2212. \u2212 In consequence the critical point is asymptotically stable when b < 1+a2. When b > 1 + a2 the critical point is unstable (a spiral or node according as b < or > (1 + a)2). At b = 1 + a2 there is a supercritical Hopf bifurcation, so that the unstable critical point is accompanied by a limitcycle attractor (see Fig. 13.15, where a = 1, b = 3). Fig. 13.15 Brusselator limit cycle attractor (a = 1, b = 3). Dynamical Systems and Their Geometry 329", " 13.6 Systems of Third (and Higher) Order We have seen that, for \ufb01rst-order systems, there are point attractors and repellers and, for second-order systems, there may also be limit cycles. Many higher-order systems are special, in that their dynamical behaviour is e\ufb00ectively lower-dimensional, through, for example, the existence of constants of the motion. Some important speci\ufb01c examples of this will be given in Chapter 14 for systems of fourth order and beyond, but in this section we will consider third-order systems which exhibit certain properties typical of systems of order higher than second. Rigid-body rotation Consider the equations of rotational motion of a rigid body, which were introduced in 9.8. When the rotation is about the centre of mass and external forces have zero moment we may, for convenience, choose to express the equations (9.41) in terms of angular momentum components (J1, J2, J3) (I1\u03c91, I2\u03c92, I3\u03c93) to obtain \u00a7 \u2261 \u02d9J1 \u2212 I2 \u2212 I3 I2I3 and two similar equations obtained by cyclic interchange of 1, 2, 3. In the (J1, J2, J3) phase space each trajectory is then the intersection of two surfaces (see Problem 12) J2J3 = 0 (13.31) sphere: ellipsoid, J 2 1 I1 + J 2 2 I2 + J 2 3 I3 = 2T, (13.32) with J, T constant. The \ufb01rst of these relations expresses the constancy of the magnitude of the angular momentum vector J and the second expresses the constancy of the kinetic energy T. (See Fig. 13.16, with principal moments of inertia such that I1 < I2 < I3.) \u00a7 If we take J to be \ufb01xed and look at the critical points on the corresponding phase sphere, it is evident that Fig 13.16 con\ufb01rms the results obtained in 9.8, since the saddle-point geometry leading to instability is at the axis of the intermediate principal moment of inertia, while the centres are at the axes of the largest and the smallest moment. The result holds, of course, for all rigid bodies (from a match-box to a space station), but it is often known as the tennis racquet theorem. 330 Classical Mechanics Fig. 13.16 When there is axial symmetry of the rigid body, as in 9.9, there is similar symmetry in the phase space. (See Fig. 13.17, with principal moments of inertia such that I1 = I2 < I3.) \u00a7 In this case J3 is constant so that J3 = I3\u2126 (say), while the components J1, J2 of angular momentum are simple harmonic with frequency [(I3 \u2212 I1)/I1]\u2126. Relative to axes \ufb01xed in space J is itself \ufb01xed and the angular velocity vector \u03c9 precesses around J in the way described in In \u2126 then this precession rate is given approximately by the case \u03c91, \u03c92 (I3/I1)\u2126. Hence the precession rate for the Chandler wobble of the Earth is about \u2126 (i.e. once per day), while the precession rate is about 1 2 \u2126 for a spun pass with an American football and about 2\u2126 for a spun coin, discus or dinner plate! 9.9. \u00a7 The problem of rotational motion, when the external force moment G is not special, can be very complicated, although it is of course an important one for many feedback control problems in, for example, the aerospace industry. Some of the complexity which may result is indicated by the following examples of other physical systems. Dynamical Systems and Their Geometry 331 Fig. 13.17 The Lorenz system The system of equations \u02d9x = \u03c3(y \u02d9y = \u03c1x x), \u2212 y \u2212 \u2212 \u02d9z = \u2212 \u03b2z + xy, xz, (13.33) where \u03c3, \u03c1, \u03b2 are positive control parameters, is an important system which was originally put forward by Lorenz in 1963 to provide a very simple and idealized model of convection in a slab of \ufb02uid. The aim was to gain insight into the dynamics of weather systems. (See Fig. 13.18.) For a general discussion of the origins of this system, see Lorenz, The Essence of Chaos, UCL Press (1993). The parameters have the following physical meaning, after scaling: \u03c3 quanti\ufb01es the ratio of di\ufb00usion rates of momentum and heat in the \ufb02uid (\u03c3 \u03c1 quant", "i\ufb01es the temperature di\ufb00erence applied across the slab of \ufb02uid. \u03b2 quanti\ufb01es the slab geometry through an aspect ratio. kinematic viscosity/thermal conductivity). \u2261 \u2022 \u2022 \u2022 332 Classical Mechanics Fig. 13.18 Despite its geophysical origin the system (13.33) and its wide range of behaviours have been the subject of extensive study, particularly because the system \ufb01nds application in other physical contexts, for example in the study of lasers. It is instructive (from both a practical and a theoretical point of view) to look at the (x, y, z) phase space, say with \u03c3, \u03b2 \ufb01xed and with a sequence of di\ufb00erent values of \u03c1. We can \ufb01nd the critical points of the system and 13.3 and Appendix examine their local character through the methods of C, suitably extended. However the characteristic equation is now a cubic, so that there are three eigenvalues, although the general principles, with regard to stable and unstable manifolds according to the sign of real eigenvalues, still apply. \u00a7 When \u03c3, \u03b2 are indeed \ufb01xed the behaviour of the system (13.33) in the (x, y, z) phase space can be summarized as follows (see Problem 14): \u2022 \u2022 (0, 0, 0) is the only critical point of the 0 < \u03c1 < 1: The origin P1 \u2261 system, and it is a point attractor. This corresponds physically to the case of no motion and steady heat conduction across the slab (see Fig. 13.19(a)). 1 < \u03c1: The origin P1 is now unstable and two new critical points appear, P2, P3 \u2261 It can be shown that when: \u00b1 \u03b2(\u03c1 1), \u00b1 \u2212 \u03b2(\u03c1 \u2212 1), (\u03c1 \u2212 1). * 1 < \u03c1 < \u03c1crit = \u03c3(\u03c3 + \u03b2 + 3) 1) \u03b2 (\u03c3 \u2212 \u2212 in the practical case \u03c3 > \u03b2 + 1, Dynamical Systems and Their Geometry 333 all the eigenvalues in the analysis of P2, P3 are negative real or have negative real part, so that P2, P3 are point attractors. This corresponds physically to steady convection rolls in either of the two senses (see Fig. 13.19(b)). * \u03c1crit < \u03c1, the points P1, P2, P3 are all unstable, so that a single point attractor does not exist. In fact the loss of asymptotic stability of P2, P3 can be shown to be via a subcritical Hopf bifurcation. Fig. 13.19 We can show that trajectories are attracted towards a bounded region containing the points P1, P2, P3 and there should be an attractor, since \u02d9x = \u2207 \u00b7 \u2202 \u02d9x \u2202x + \u2202 \u02d9y \u2202y + \u2202 \u02d9z \u2202z = \u2212 (\u03c3 + 1 + \u03b2) < 0 (13.34) everywhere, which means that the phase-space volume elements are reduced progressively in volume by the \ufb02ow \u02d9x. (See 12.5 and relation (A.24).) In fact, it turns out that there is a value \u00af\u03c1, such that for \u00af\u03c1 < \u03c1 there is a highly disordered motion, which corresponds physically to turbulent convection. Trajectories in the phase space tend asymptotically towards a structure which is multi-leaved on a hierarchy of scales (a \u2018fractal \u2019) and \u00a7 334 Classical Mechanics looped around the unstable critical points P2, P3. This is called a chaotic or strange attractor, on account of its peculiar properties. (See for example, Fig 13.20 in which \u03c3 = 10, \u03b2 = 8 3 and \u03c1 = 28, so that \u03c1crit = 24.74). Fig. 13.20 Lorenz attractor [Lanford, (Lecture Notes in Mathematics, 615, 114, Springer 1977)] For the particular values of \u03c3, \u03b2 chosen here a strange attractor of this general type is found to occur when \u03c1 > \u00af\u03c1 = 24.06. Evidently in a (here) small range \u00af\u03c1 < \u03c1 < \u03c1crit there are actually three attractors \u2014 two points (steady convection) and one strange (turbulent convection). In practice each of these will have its own basin of attraction of states attracted to it, so that the attractor which is observed to occur depends on the initial state of the system. The strange attractor has zero phase-space volume on account of (13.34) above, and a fractal geometric structure. However, states which", " are close initially are stretched out along the attractor exponentially with time. The latter property is characteristic of chaotic systems and there is more discussion of this property in the following section ( \u00a7 13.7). The Rikitake two-disc dynamo In order to model irregular reversals of the Earth\u2019s magnetic \ufb01eld Rikitake put forward (in Proceedings of the Cambridge Philosophical Society, 54, 89\u2013 Dynamical Systems and Their Geometry 335 105, 1958) a system containing two identical coupled disc dynamos which does mimic this type of behaviour. (See Fig. 13.21.) Fig. 13.21 If X1, X2 are scaled currents and Y1, Y2 are scaled rotation rates of the discs, which are driven by a constant external torque, the current in each disc feeds the coil of the other. Then the equations may be written \u02d9X1 = Y X2 \u2212 \u00b5X1, \u02d9X2 = (Y A)X1 \u2212 \u2212 \u02d9Y = 1 X1X2, \u2212 \u00b5X2, (13.35) i \u2212 \u00b1 \u00b1 \u2261 k, 2\u00b5, 1/k2). where Y 1/k, \u00b5k2), where A = \u00b5(k2 Y1 = Y2 + A and A, \u00b5 are control parameters, with \u00b5 positive. For this system there are two critical points in the (X1, X2, Y ) phase (See Problem space, at ( 15.) Although the three eigenvalues in the local linear approximation k2 + 1/k2, the apparent centre behaviour associated with the are pure imaginary pair of eigenvalues is here not carried through for the full system. The critical points correspond to senses of North/South magnetic polarity and the irregular reversals which are associated with this model (see Fig. 13.22) mimic the \ufb01eld reversals of the Earth\u2019s magnetic \ufb01eld which are known, from the interpretation of geological records, to have occurred at irregular intervals during more than the last 160 million years. \u00b1 \u2212 There is, of course, no suggestion that the two-disc dynamo provides other than an interesting analogy for features of the magnetohydrodynamic 336 Classical Mechanics Fig. 13.22 Projection on the (X2, Y ) plane of a trajectory in phase space for \u00b5 = 1, K = 2 [Cook and Roberts (Proceedings of the Cambridge Philosophical Society, 68, 547\u2013569, 1970)] motions in the Earth\u2019s core \u2014 it is a \u2018toy\u2019 system. The forced damped pendulum If the simple pendulum of \u00a7 forcing we may write the equation of motion in the form 2.1 and \u00a7 13.3 is subjected to damping and ml\u00a8\u03b8 + \u03bb \u02d9\u03b8 + mg sin \u03b8 = F cos \u2126t, (13.36) where m, l, \u03bb, g, F, \u2126 are control parameters. This equation may be scaled suitably, so that we study the equivalent third-order autonomous system \u02d9\u03b8 = \u03c9, \u02d9\u03c9 = \u2212 \u02d9\u03c6 = \u2126, A\u03c9 \u2212 sin \u03b8 + B cos \u03c6, (13.37) where A, B, \u2126 are control parameters. The nonlinearity of this system, through the sin \u03b8, cos \u03c6 terms, makes for a very rich variety of behaviour as A, B, \u2126 are changed. Dynamical Systems and Their Geometry 337 The variety of behaviours instanced in this whole section is common for third- and higher-order continuous systems, but systems have to be \u2018special\u2019 in order to be e\ufb00ectively reducible to lower order via, for example, constants of the motion. Systems need to be of at least third order if they are to exhibit chaotic behaviour. 13.7 Sensitivity to Initial Conditions and Predictability An important consequence of the e\ufb00ective order of a dynamical system is in the way that the resulting dimension of the phase space can restrict the evolution along trajectories of states which may be initially close together. In the line and plane, typically, states which are initially close, whose trajectories remain in a bounded region and do not intersect one another, move apart at a rate which is at most essentially uniform, on average. If we examine a dynamical system with closed trajectories in phase space, we can examine the evolution of two neighbouring initial states P \u2261 (x0, y0), Q \u2261 (x0 + \u2206x0, y0 + \u2206y0). (See Fig. 13.23.) Fig. 13.23 Now, if the system is the simple harmonic oscillator (see 13.3) then, in this special case, the period along each closed orbit is the same and it can be", " shown that the (\u2018Pythagorean\u2019) distance in phase space between P and Q remains bounded in time. This behaviour is not typical and, if Fig. 13.23 \u00a7 338 Classical Mechanics is taken to represent, for example, libration motions of a simple pendulum, it should be evident that di\ufb00erent orbits have di\ufb00erent periods. For this, the more typical situation, the distance between P and Q increases broadly linearly with time, as measured along the orbit. A useful analogy is to consider runners travelling at slightly di\ufb00erent rates in neighbouring lanes of a running track (see Problems 16, 17). In three- and higher-dimensional phase spaces the neighbouring trajectories can diverge exponentially from one another within a bounded region, without crossing one another. They can do this by wrapping over and under one another in a complex tangle, which may still have a layered structure. (See Fig. 13.20 and Fig. 13.22.) This geometrical behaviour is central to the \u2018strangeness\u2019 of a strange attractor to which we referred in 13.6. The states which are initially close move apart, in such a way that the distance d in phase space between them increases with time, on average, like e\u03bbt, with \u03bb positive. (See Fig. 13.24.) \u00a7 Fig. 13.24 The parameter \u03bb in the exponent is called a Lyapunov exponent. When a typical small volume of initial states in phase space is considered, it is evident that dynamical evolution will result, in general, in stretching and squeezing of this volume. There will be a number of principal deformations, which is equal to the dimension of the phase space, leading to a set of Lyapunov exponents. Each of the positive exponents leads to a corresponding strong divergence of initial states. We can evidently test \ufb02ows in phase space for this divergence property. The strong divergence is called sensitivity to initial conditions and it is the prime property which identi\ufb01es the irregular behaviour now called chaos. Dynamical Systems and Their Geometry 339 The e\ufb00ect on the global behaviour of dynamical systems produced by sensitivity to initial conditions is often called the butter\ufb02y e\ufb00ect, which was \ufb01rst put forward by Lorenz in an address to the American Association for the Advancement of Science in December 1972 \u2014 \u2018Predictability: Does the \ufb02ap of a butter\ufb02y\u2019s wings in Brazil set o\ufb00 a tornado in Texas?\u2019 \u2014 printed in his book The Essence of Chaos referred to above in the description of the Lorenz system (13.33). The point he was making is that chaotic regimes in weather systems can render predictions of them utterly useless, except on a very short time scale. The term \u2018butter\ufb02y e\ufb00ect\u2019 has become a much used metaphor for describing chaos, which has experienced an explosive growth in application. The consequences for all dynamical systems and for the real situations they model are immense. A direct attack is made on the long-standing philosophical idea of determinism and on the concept of a \u2018clockwork universe\u2019. The Laplacian point of view identi\ufb01es the initial state of a system as the cause, from which (if it is known precisely) the e\ufb00ects may be calculated with in\ufb01nite precision inde\ufb01nitely far into the future, or into the past. There is, in this world view, no room for chance or free will. The sensitivity to initial conditions, which is typical of many, if not most, systems and is made very apparent by the geometrical approach pioneered by Poincar\u00b4e, leads to a loss of predictability. This comes about because the inevitable imprecision in the initial description of even a classical system, however small this may be, leads later to a vastly greater imprecision in its predicted fate. It is in this way that deterministic systems, where the mechanisms may be known completely and precisely, can have an essentially stochastic (\u2018random\u2019) output, so that outcomes can be predicted only as a matter of probability. A good illustration of this apparent paradox is provided by the Galton board (see Fig. 13.25) where small spheres of lead shot fall through a regular array of scattering pins. The path of an individual piece of lead shot, which falls through such an array of pins, is impossible to predict, because of the sensitivity of the collisions at individual pins. However the typical outcome for the resulting distribution of shot along the channels at the base is, in this case, very like the normal probability distribution. In this way predictability of the fate of an individual piece of lead shot gives way to a collective predictability for many pieces in the form", " of chance. (See Problem 18.) 340 Classical Mechanics Fig. 13.25 13.8 Summary The analysis of the geometry of dynamical systems in phase space is a very powerful way of investigating their behaviour for all possible initial conditions. The dynamics is dependent crucially on attractors and on their stability. For systems which are dynamically one-dimensional or two-dimensional the only attractors are critical points (equilibria) in the phase line and the phase plane and, for the latter, limit cycles. As control parameters are varied, the character of critical points and limit cycles may change, so that there is a corresponding change in the phase portraits and in the behaviour of the system. In systems of third (or higher) order additional attractors may arise which can have a fractal and sensitive structure \u2014 strange (chaotic) attractors. In the phase plane the system trajectories are constrained such that the divergence of initial states takes place essentially at a uniform rate, at most. For a strange attractor, in systems of third or higher order, the divergence will be essentially exponential and this leads to a breakdown of predictability. The consequence is that deterministic systems can produce e\ufb00ectively stochastic output. This has profound implications for our view Dynamical Systems and Their Geometry 341 of physical and other dynamical systems; it can also be expected to a\ufb00ect the relationship between classical physics and its quantum counterpart. Problems 1. The population x(t) of \ufb01sh in a pond is a function of time t and when the \ufb01sh are removed at a constant rate f the function x(t) satis\ufb01es equation (13.6). (a) Show that there is a critical \ufb01shing rate \u00aff such that for f < \u00aff there are two equilibrium \ufb01sh populations, of which only one is asymptotically stable. (b) If the \ufb01sh population is initially in stable equilibrium (with f = 0 0 and in a and \ufb01shing commences at a rate f > \u00aff, show that x \ufb01nite time t0 (which need not be found explicitly!). \u2192 Why would it not be prudent to \ufb01sh at a rate f which is only just less than \u00aff? 2. For the pendulum equation \u00a8\u03b8 + (g/l) sin \u03b8 = 0 \ufb01nd the equations of the trajectories in the phase plane and sketch the phase portrait of the system. Show that on the separatrices \u03b8(t) = (2n + 1)\u03c0 \u00b1 4 arctan[exp(\u03c9t + \u03b1)], g/l. [Hint : Use the where n is an integer, \u03b1 = constant and \u03c9 = substitution u = tan 1 \u03b8 (2n + 1)\u03c0.] 4 { \u2212 } 3. The relativistic equivalent of the simple harmonic oscillator equation for a spring with constant k and a rest mass m0 attached is d dt m0y y2/c2 1 \u2212 + kx = 0 with \u02d9x = y, where c is the speed of light. Show that the phase trajectories are given by m0c2/ 1 \u2212 y2/c2 + 1 2 kx2 = constant and sketch the phase portrait for this system. 4. Draw the phase portrait of the damped linear oscillator, whose displace0x = 0, in the phase plane (x, y), where ment x(t) satis\ufb01es \u00a8x + \u00b5 \u02d9x + \u03c92 342 Classical Mechanics y = \u02d9x. Distinguish the cases (a) under- (or light) damping (b) over-damping (c) critical damping 0 < \u00b5 < 2\u03c90, \u00b5 > 2\u03c90, \u00b5 = 2\u03c90. 5. Consider the gradient system (13.17) in the case U (x, y) = x2(x 1)2+y2. Find the critical points and their character. Sketch the phase portrait for the system. \u2212 \u2265 0, y 6. For the Lotka\u2013Volterra system (13.18) show that the trajectories in the phase plane are given by f (x, y) = constant as in (13.20). In the \ufb01rst 0, the intersections of a line y = constant with a quadrant x c ln x + dx = constant. Hence show that there trajectory are given by are 0, 1 or 2 such intersections, so that the equilibrium point (c/d, a/b) fore this system is a true centre (i.e. it cannot be a spiral point). Using the substitution x = ep, y = eq, show that the system takes on the Hamiltonian canonical form (13.22). \u2265 \u2212 7. Consider the", " \u2018competing species\u2019 system (13.23). For the case k1 = 4, \u03b12 = 3 \u03c31 = \u03b11 = 1, k2 = 1 4 \ufb01nd the critical points and their character in order to con\ufb01rm all the features of Fig. 13.9. What happens in the case k1 = 1, \u03c31 = 2, \u03b11 = 1 2, k2 = 3, \u03c32 = 2, \u03b12 = 4? 2, \u03c32 = 1 8. For the Arms-Race model system (13.26) with all parameters positive show that there is an asymptotically stable coexistence or a runaway escalation according as c1c2 > a1a2 or c1c2 < a1a2. 9. *A simple model for the dynamics of malaria due to Ross (1911) and Macdonald (1952) is abM \u02d9x = N \u02d9y = ax(1 y(1 y) x) \u00b5y, \u2212 \u2212 \u2212 rx, \u2212 where: x, y are the infected proportions of the human host, female mosquito populations, N, M are the numerical sizes of the human, female mosquito populations, a is the biting rate by a single mosquito, b is the proportion of infected bites that result in infection, r, \u00b5 are per capita rates of recovery, mortality for humans, mosquitoes, respectively. Show that the disease can maintain itself within these populations or Dynamical Systems and Their Geometry 343 must die out according as R = M N a2b \u00b5r > 1 or < 1. 10. Show that Rayleigh\u2019s equation in the form (13.28) has a single critical point at (0, 0) and that this is always unstable. Making use of substitution \u02d9x = v/\u221a3 show that v satis\ufb01es the Van der Pol equation (13.29). 11. Show that the origin is the only critical point for the system y + \u03b1x(\u03b2 \u02d9x = \u02d9y = x + \u03b1y(\u03b2 \u2212 \u2212 \u2212 x2 x2 y2), \u2212 y2), \u2212 where \u03b1, \u03b2 are real parameters, with \u03b1 \ufb01xed and positive and \u03b2 allowed to take di\ufb00erent values. Show that the character of the critical point and the existence of a limit cycle depend on the parameter \u03b2, so that the system undergoes a supercritical Hopf bifurcation at \u03b2 = 0. (Hint : Make the change from Cartesian co-ordinates to plane polars.) 12. *For the rotation of a rigid body about its centre of mass with zero torque the equations for the angular momentum components J1, J2, J3 are given by (13.31). (a) Show that constant), so that the angular mo- mentum J must lie on a sphere in (J1, J2, J3) phase space. (b) When I1 < I2 < I3 show that there are six critical points on this phase sphere and show that, in local expansion, four of these are centres and two are saddles. (Hence the tennis racquet theorem of 13.6.) \u00a7 (c) Show that when I1 = I2 J1, J2 are simple harmonic (with frequency [ = I3 then J3 is constant ( \u2261 I1| I3 \u2212 | I3\u2126) and that /I1]\u2126). (d) A space station with I1 < I2 < I3 is executing a tumbling motion with \u03c91, \u03c92, \u03c93 nonzero. It is to be stabilized by reducing \u03c9 to 0 \u03c9, with \u00b5 constant, so that the rightwith an applied torque \u00b5 | \u03c91, etc. About which of its axes \u00b5 hand side of (13.31) becomes | does the space station tend to be spinning as \u03c9 \u2212| \u2212| 0? \u2192 13. *The simple SIR model equations for the transmission of a disease are aSI, \u02d9S = \u2212 \u02d9I = aSI \u02d9R = bI, bI, \u2212 344 Classical Mechanics where S(t), I(t), R(t) are respectively susceptibles, moved/recovered and a, b are positive constants. infectives, re- (a) Show that the overall population N = S +I +R remains constant, so that we may consider (S, I) in a projected phase plane. Hence show that a trajectory with initial values (S0, I0) has equation I(S) = I0 + S0 \u2212 (b) Using the function I(S) show that an epidemic can occur only if the number of susceptibles S0 in the population exceeds the threshold level b/a", " and that the disease stops spreading through lack of infectives rather than through lack of susceptibles. S + (b/a) ln(S/S0). (c) For the trajectory which corresponds to S0 = (b/a) + \u03b4, I0 = with \u03b4, small and positive, show that, to a good approximation, \u03b4 susceptibles who escape infection [the Kermack\u2013 there are (b/a) McKendrick theorem of epidemiology (1926/27)]. \u2212 14. *Consider the Lorenz system (13.33). (a) Show that the origin P1 (0, 0, 0) is a critical point and that its sta- bility depends on eigenvalues \u03bb satisfying the cubic (\u03bb + \u03b2)[\u03bb2 + (\u03c3 + 1)\u03bb + \u03c3(1 \u03c1)] = 0. \u2212 Hence show that P1 is asymptotically stable only when 0 < \u03c1 < 1. (b) Show that there are two further critical points P2, P3 \u2261 [ \u00b1 \u03b2(\u03c1 1), \u00b1 \u2212 \u03b2(\u03c1 1), (\u03c1 1)], \u2212 \u2212 when \u03c1 > 1, and that their stability depends on eigenvalues \u03bb satisfying the cubic \u03bb3 + \u03bb2(\u03c3 + \u03b2 + 1) + \u03bb\u03b2(\u03c3 + \u03c1) + 2\u03c3\u03b2(\u03c1 1) = 0. \u2212 (c) Show that when \u03c1 = 1 the roots of the cubic in (b) are 0, and that in order for the roots to have the form real) we must have \u00b5, \u2212 \u00b1 \u03b2, (1+\u03c3) \u2212 \u2212 i\u03bd (with \u00b5, \u03bd \u03c1 = \u03c1crit = \u03c3(\u03c3 + \u03b2 + 3) 1) \u03b2 (\u03c3 \u2212 \u2212 > 0. (d) By considering how the roots of the cubic in (b) change continuously with \u03c1 (with (\u03c3, \u03b2 kept constant), show that P2, P3 are asymptotically stable for 1 < \u03c1 < \u03c1crit and unstable for \u03c1 > \u03c1crit. Dynamical Systems and Their Geometry 345 (e) Show that if \u00afz = z \u03c1 \u2212 \u2212 \u03c3, then 1 2 d dt (x2 + y2 + \u00afz2) = \u03c3x2 y2 \u2212 \u2212 \u2212 \u03b2[\u00afz + 1 2 (\u03c1 + \u03c3)]2 + 1 4 \u03b2(\u03c1 + \u03c3)2, so that (x2 + y2 + \u00afz2)1/2 decreases for all states outside any sphere which contains a particular ellipsoid (implying the existence of an attractor). 15. *For the Rikitake dynamo system (13.35): (a) Show that there are two real critical points at ( k, (X1, X2, Y ) phase space, where k is given by A = \u00b5(k2 \u00b1 \u00b1 1/k, \u00b5k2) in the 1/k2). \u2212 (b) Show that the stability of these critical points is determined by eigen- values \u03bb satisfying the cubic (\u03bb + 2\u00b5) \u03bb2 + k2 + 1 k2 = 0, so that the points are not asymptotically stable in this approximation. (For the full system they are actually unstable.) (c) Show that the divergence of the phase-space \ufb02ow velocity is negative, so that the \ufb02ow causes volume to contract. (d) Given that \u00afY = \u221a2(Y A/2), show that \u2212 2 + \u00afY 2) = (X 2 1 + X 2 1 2 d dt \u00b5(2 \u00afY \u2212 and use this result to determine in which region of the space the trajectories all have a positive inward component towards X1 = X2 = \u00afY = 0 on surfaces X 2 1 + X 2 16. For the simple harmonic oscillator system (13.8) we may solve for the equation of the trajectory which passes through x(x0, y0, t), y(x0, y0, t). If we now consider perturbations \u2206x0, \u2206y0 in the initial data (x0, y0) at t = 0, \ufb01nd the resulting changes \u2206x, \u2206y in x, y and show that they remain bounded in time, so that the Pythagorean distance d 2 + \u00afY 2 = constant. \u2261 (\u2206x)2 + (\u2206y)2 then remains bounded. 17. *In contrast to Problem 16, consider the perfectly elastic bouncing of a ball vertically under gravity above the plane x = 0. We have \u02d9x = y, \u02d9y = g and we can solve for x(x0, y0, t), y(", "x0, y0, t) in terms of the initial \u2212 data (x0, y0) at t = 0. Show that in this case the resulting perturbations \u2206x, \u2206y essentially grow linearly with time t along the trajectory when we make perturbations \u2206x0, \u2206y0 in the initial data. That is to say the 346 Classical Mechanics distance along the trajectory d and \u03ba is a suitable constant. \u2261 (\u2206x)2 + (\u2206y)2 \u223c \u03bat when t is large 18. For the Galton board of Fig. 13.25 we may arrange things so that each piece of lead shot has an equal chance of rebounding just to the left or to the right at each direct encounter with a scattering pin at each level. Show that the probabilities of each piece of shot passing between the pins ( 1 2 )n where the binomial along a particular row n are then given by r)!r!] and r = 0, 1,..., n. Use the result coe\ufb03cient n n+1 to generate the probability distribution for row n = r r+1 16. (For large numbers of pieces of shot and large n the distribution of shot in the collection compartments approximates the standard normal error curve y = k exp( x2/2s2) where k, s are constants.) = n!/[(n n r+1 n r + \u2212 = n r \u2212 Chapter 14 Order and Chaos in Hamiltonian Systems In this chapter we consider the geometrical aspects of the particular class of dynamical systems which have the Hamiltonian structure described in Chapter 12. The dynamics of these systems depends crucially on the number of symmetries which they possess, leading to conserved quantities. It will become apparent that the larger the number of symmetries the greater is the restriction in the freedom of a trajectory in the corresponding phase space and the more ordered is the motion. In contrast a de\ufb01cit in the number of such symmetries allows the development of irregularity, or chaos, in the evolution of states along the system trajectories. 14.1 Integrability \u00a7 12.6 we discussed the direct relation between symmetries and conservaIn tion laws, indicating that conserved quantities (constants of motion) may be used to reduce progressively the e\ufb00ective order of a system. If there are enough symmetries and independent conserved quantities, then the reduction may be complete, in that it leads to the full solution of the system of equations (\u2018integrability\u2019). For a system with n degrees of freedom the Hamiltonian H may depend on the co-ordinates q1,..., qn, momenta p1,..., pn and on time t. While there is no unique de\ufb01nition of integrability, it turns out that, for the complete solution to be obtainable via a sequence of integrations, we need n independent functions Fi (i = 1,..., n) of the same variables, which are themselves constants of the motion. That is to say, we require the same number of symmetries as there are degrees of freedom in the system and the system is then said to be \u2018integrable (in the sense of Liouville)\u2019. If there is not such a full set of n conserved quantities then the system is said to be non-integrable. 347 348 Classical Mechanics Recalling (12.33), dFi dt = \u2202Fi \u2202t + [Fi, H] = 0. (14.1) We will con\ufb01ne our discussion here to autonomous systems, for which H and other Fi have no explicit dependence on t, and we then have n Poisson-bracket relations [Fi, H] = 0 from (14.1). Of course H itself can be taken as one of the functions Fi and the invariance of H corresponds to symmetry with respect to time translation. An additional technical condition required of the Fi is that for each pair of them we require [Fi, Fj ] = 0 for i, j = 1,... n. (14.2) The functions Fi are then said to be in involution. \u2212 Each of the n functions Fi = constant con\ufb01nes a particular trajectory to a (2n 1)-dimensional subspace of the full 2n-dimensional phase space of the system. So the complete set of these restrictions con\ufb01nes a trajectory to the n-dimensional intersection M of these subspaces. That this is a strong dimensional restriction on the way that a trajectory can move in the full phase space is indicated by Table 14.1. Table 14.1 Number of degrees of freedom Dimension of: phase space surface Fi = constant 2n 2n \u2212 1 n Naturally the values of co-ord", "inates and momenta at a given initial time determine the n values of the Fi constants and hence pick out the particular M to which the trajectory having these initial values is then con\ufb01ned for all time. Now at each point on a particular (2n 1)-dimensional surface Fi = constant, the normal to the surface is in the direction given by the 2ncomponent gradient vector (\u2207q Fi, \u2207p Fi). (See A.5.) \u2212 The Poisson-bracket relation (14.2) is a scalar-product relation and it expresses the orthogonality conditions between members of a set of n vector \u2207q Fj), j = 1,..., n and the above gradient vector. It \ufb01elds vj = (\u2207p Fj, is apparent therefore that the vector \ufb01elds vj at a particular point in the phase space are all parallel to the surface M which contains that point. The \u2212 \u00a7 Order and Chaos in Hamiltonian Systems 349 \ufb01elds vj allow us to de\ufb01ne a co-ordinate grid continuously over the surface M and without any singularity. The topology of M is then necessarily that of a torus \u2014 an n-dimensional \u2018ring doughnut\u2019 embedded within the full 2n-dimensional phase space. For a system of one degree of freedom each M is a 1-torus F1 \u2261 H = constant, i.e. a one dimensional loop in the two-dimensional phase space. (See Fig. 14.1(a).) Evidently v1 here is tangent to the curve at each point of M. For di\ufb00erent values of F1 we obtain di\ufb00erent loops M, members of a set of \u2018nested\u2019 tori, as for example in Fig. 13.4. For a system of two degrees of freedom each M is a 2-torus, i.e. a two-dimensional \u2018ring doughnut\u2019 surface embedded in the four-dimensional phase space. (See Fig. 14.1(b).) In this case we can take v1, v2 everywhere parallel to distinct loops which can parametrize the surface. It is the case that M cannot have, for example, the topology of a sphere, because all v1, v2 parametrizations of such a surface are necessarily singular, at least somewhere. In Fig. 14.1(c) the attempt to use a grid of lines of latitude and longitude is singular at the poles. That all such attempts for a sphere are bound to be singular is a consequence of the \u2018hairy ball theorem\u2019 of topology; this theorem also implies, among other things, the existence of at least one crown parting in the combing of a head of hair, and that a spherical magnetic bottle must leak! Fig. 14.1 The n-tori for an integrable system of n degrees of freedom in its 2ndimensional phase space are often called invariant tori because a trajectory (orbit) which is initially on one of them (M ) remains on it for ever. Different initial conditions then lead to tori which are nested within the phase space. 350 Classical Mechanics On a particular torus M the separate vector \ufb01elds correspond to n independent types of linking circuit (see Fig. 14.1(a),(b)), and the integrable motion is then exactly equivalent to the combination of n separate corresponding periodic motions, one for each type of linking circuit. Each of these periodic motions of an integrable system is a nonlinear normal mode, similar to those detailed in 11.3, but now with an analysis which is exact \u2014 no approximation has been made for small oscillations. This makes integrability a very special property. \u00a7 The separate normal modes have associated natural frequencies \u03c9i, i = 1,..., n on a particular torus M and any trajectory on M has n associated natural periods 2\u03c0/\u03c9i. The trajectory is then said to be multiply periodic. If the trajectory on M is closed (see Fig. 14.2(a)) then it does not \ufb01ll M and is exactly periodic. For closure it is required that the frequencies of the the modes are rationally related, i.e. \u03c9i/\u03c9j is rational for all i, j. This degeneracy is the exception rather than the usual case, so that closure If the trajectory on M is not closed, because of is a special property. irrational \u03c9i/\u03c9j, then typically the trajectory covers M densely eventually (see Fig. 14.2(b)) \u2014 we have quasiperiodicity and the trajectory is said to be ergodic on M. Fig. 14.2 Two important examples are given by: Example: Double harmonic oscillator and central", " force motion Find suitable independent constants F1, F2 for (a) the double harmonic oscillator, (b) central force motion. Order and Chaos in Hamiltonian Systems 351 Are trajectories closed? (a) For the double harmonic oscillator we have H = 1 2m (p2 1 + p2 2) + 1 2 m(\u03c92 1q2 1 + \u03c92 2q2 2) H1 + H2 \u2261 (14.3) with H1, H2 the Hamiltonians for the two separate oscillators. Here we may take F1, F2 to be any pair of H, H1, H2, since then [Fi, H] = 0. The frequencies are \u03c91, \u03c92 and this evidently leads to closed trajectories (periodicity) only when \u03c91/\u03c92 is rational, leading to the familiar Lissajous \ufb01gures. (b) Central-force motion was discussed in Chapter 4. The Hamiltonian takes the form (12.8) and we can take F1 = H, F2 = p\u03b8, so that [Fi, H] = 0. For the isotropic harmonic oscillator of 4.1 we can take \u03c91 = \u03c92 in (14.3) above, so that all trajectories are closed. The oscillator describes an ellipse with centre at the origin and the frequencies of radial and transverse motions are related by \u03c9r = 2\u03c9\u03b8. For the inverse square law attractive force of 4.3 all bounded trajectories are closed. They are ellipses in general, with the centre of force at a focus and \u03c9r = \u03c9\u03b8. That the isotropic harmonic oscillator and the inverse square law of attraction are the only such force laws for which all bounded orbits are closed is known as Bertrand\u2019s theorem. \u00a7 \u00a7 14.2 Surfaces of Section If we consider two dynamical states which are initially close in the phase space of an integrable system, they do, in general, have di\ufb00erent values of F1, F2 and so they reside on di\ufb00erent tori. The evolution of these states along their respective trajectories takes place with slightly di\ufb00erent frequencies, so that the distance between the states increases essentially linearly with time t. So, just as for the discussion of the libration motions of a simple pendulum in 13.7, there is for an integrable system no exponential divergence of states with time and no sensitivity to initial conditions \u2014 the motion is ordered. \u00a7 When a full complement of constants of the motion does not exist, so that we have a system which is not integrable, then trajectories have much less restriction on their room to manoeuvre and sensitivity to initial 352 Classical Mechanics conditions is certain in at least some regions of the phase space \u2014 there is then inevitably chaotic behaviour, as indicated in 13.7. A method of detecting these ordered and chaotic behaviours was put forward by Poincar\u00b4e and involves the concept of a surface of section. If we consider an autonomous system with two degrees of freedom, so that we have a four-dimensional phase space, with three-dimensional surfaces of constant \u2018energy\u2019 (H = constant), then we can consider taking a \u2018slice\u2019 through the space, and this may or may not be planar according to convenience. \u00a7 If we look at a particular trajectory of the system and identify the points at which this intersects the section (in the same sense) then the dynamics induces a map of the section to itself. In Fig. 14.3 we have H(q1, q2, p1, p2) = constant and we have chosen a section q2 = 0 (with p2 > 0) to demonstrate how a trajectory maps points in the (q1, p1) plane successively. Fig. 14.3 When we map di\ufb00erent (q1, p1) points using di\ufb00erent trajectories we develop a Poincar\u00b4e return map of the section to itself. \u00a7 For integrable systems we may take a section through the nested tori 14.1. For two degrees of freedom the section through a 2-torus (ring of doughnut) is a closed curve C (see Fig. 14.4) and a particular point P0 on this curve is mapped successively by the trajectory through it to P1, then P2, and so on. If the trajectory is closed (see 14.1) then we have a periodic motion and the map has the property that after N steps, for some integer N, it carries P0 back to P0. The successive points P are twisted around the curve C. If the motion is only quasiperiodic, however, there is no such \u00a7 Order and Chaos in Hamiltonian Systems 353 Fig.", " 14.4 exact repetition and the successive points P eventually \ufb01ll the whole curve C. Integrable systems are characterized by a \ufb01nite set of discrete points or by curve-\ufb01lling in the surface of section. For systems which are not integrable the full torus structure does not exist. For two degrees of freedom we expect the system to explore the full three-dimensional \u2018energy\u2019 surface. The section return map is then generally two-dimensional and is not con\ufb01ned to points and lines. Systems which are non-integrable are characterized by area-\ufb01lling in the surface of section. \u00a7 It should be noted that the Poincar\u00b4e return map for a Hamiltonian system can be shown to be area-preserving. This property is related to 12.5, although it is not derivable from it. Liouville\u2019s theorem, treated in Despite the equal-area property, the map induced by a particular dynamical system typically involves stretching and squeezing of \ufb01nite and in\ufb01nitesimal area elements, associated with, respectively, positive and negative Lya13.7) whose e\ufb00ects balance. When these e\ufb00ects are punov exponents (see combined with a folding of trajectories, necessitated by con\ufb01nement to a bounded region of phase space, it is clear that the divergence of neighbouring trajectories leading to the sensitivity to initial conditions of chaos can then typically occur. \u00a7 For systems which are not Hamiltonian the concept of a Poincar\u00b4e return map for a surface of section is still useful. When the dynamics involves dissipation the equal-area property is replaced by area reduction at each iteration of the map \u2014 this would be the case e.g. for the Lorenz system of 13.6 \u2014 and here the e\ufb00ect of a negative Lyapunov exponent overcomes the \u00a7 e\ufb00ect of one which is positive. (There is some reference to the properties of discrete maps in Appendix D.) 354 Classical Mechanics 14.3 Action/Angle Variables Quite often the initial choice of variables qi, pi (i = 1,..., n) used to describe a system of n degrees of freedom may not be the best. For integrable systems there is a natural set of co-ordinates and momenta which is particularly convenient and useful. We recall that these systems have n distinct constants of the motion (Fi = constant) and we can transform to a new set of co-ordinates \u03c6i and momenta Ii, in such a way that the Hamiltonian form of the equations of motion is preserved, the new momenta Ii are functions of the Fi, the new co-ordinates \u03c6i are all ignorable (as in 12.3). \u00a7 \u2022 \u2022 \u2022 For a suitable choice the \u03c6i are angle variables and the Ii are the corresponding canonically conjugate action variables (see Table 14.2). Table 14.2 Old description New description \u02d9qi = (qi, pi) (i = 1,..., n) H(qi, pi) \u2202H \u2202pi \u02d9pi = \u2212 \u2202H \u2202qi (\u03c6j, Ij) (j = 1,..., n) K(Ij) \u2261 H \u02d9\u03c6j = \u2202K \u2202Ij \u02d9Ij = \u2212 \u2202K \u2202\u03c6j \u2261 \u03c9j(Ii) = 0 It is, of course, apparent that the new momenta Ij are all constants of the motion and that the evolution of each \u03c6j is given by \u03c6j = \u03c9jt+\u03b2j, where \u03c9j, \u03b2j are constants. Evidently the angle variables evolve at a uniform rate. Although these are general results, let us consider how the change of variables may be e\ufb00ected for a system of one degree of freedom (n = 1). For such an autonomous system we have H = constant (\u2018energy\u2019) closed trajectories in the phase plane, representing libration motions. (See Fig. 14.5(a).) We need to choose the action I to be a suitable function of H, such that the phase plane in the new description is as given in Fig. 14.5(b). For Hamilton\u2019s equations to be preserved in form it is necessary that the change of variables (q, p) (\u03c6, I) preserves areas, so that a suitable choice for I is derived from the area within the corresponding trajectory in \u2192 Order and Chaos in Hamiltonian Systems 355 Fig. 14.5 the (q, p) phase plane by I = 1 2\u03c0 p dq, (14.4) % as indicated by the equality of the", " shaded areas //// in Fig. 14.5(a),(b). Naturally I depends directly on H, and the natural frequency of the system for the corresponding trajectory is then given by (see Table 14.2) \u03c9(I) = \u2202K \u2202I \u2261 \u2202H \u2202I. (14.5) It also follows from the preservation of area that the de\ufb01ning relation for \u03c6 is given by \u03c6 = \u2202 \u2202I 0 q p dq \u03c9 \u2202 \u2202H \u2261 q p dq, (14.6) 0 the shaded areas indicated by the equality of again as in Fig. 14.5(a),(b) for small changes \u2206H, \u2206I in H, I respectively. The \u2018angle\u2019 nature of the \u03c6 variable can be emphasized by a further change of variables to polars (R, \u03c6), where R = \u221a2I in order to maintain the preservation of area property. (See Fig. 14.5(c).) For the oscillator problems of 13.3 we have 2.1 and \\\\\\\\ \u00a7 \u00a7 H(q, p) = p2 2m + V (q) = E, (14.7) so that p = 2m[H \u00b1 I = 1 2\u03c0 \u2212 % V (q)] and p dq = 2\u221a2m 2\u03c0 q2 q1 H \u2212 V (q) dq. (14.8) 356 Classical Mechanics Example: Simple harmonic oscillator Find action/angle variables for the simple harmonic oscillator. Here V (q) = 1 2 kq2, so that I = H/\u03c90, where \u03c90 = k/m, (14.9) p dq is the area within an ellipse with semi-axes since \u221a2mH. (See Fig. 14.6.) & 2H/k and Fig. 14.6 For the simple harmonic oscillator we have action ergy/frequency. Evidently and action \u03c9 = \u2202H \u2202I = \u03c90 (here) I = p2 + kmq2 2m\u03c90, with the angle variable \u03c6 = \u2202 \u2202I q 0 p dq = \u2202 \u2202I q 2m\u03c90I 0 q kmq2 dq \u2212 = m\u03c90 0 dq 2m\u03c90I \u2212 kmq2 = arcsin q \u03c90t + \u03b2. \u2261 m\u03c90 2I en- \u2261 (14.10) (14.11) (14.12) Order and Chaos in Hamiltonian Systems 357 Of course, the original variables q, p can be found in terms of \u03c6, I in the form q = 2I m\u03c90 sin \u03c6, p = 2m\u03c90I cos \u03c6. (14.13) These relations, together with the linear evolution of \u03c6 with time give the evolution of the original q, p variables. The elastic bouncer For a perfectly elastic ball bouncing normally between two walls (see E = 1 2 mv2. Here the acFig. 14.7) the Hamiltonian is H = p2/2m tion I = \u221a2mH L/\u03c0 so that \u03c9 = \u03c02I/mL2 \u03c0v/L and the period of this \u2018oscillation\u2019 is \u03c4 = 2\u03c0/\u03c9 = 2L/v (of course). \u2261 \u2261 Fig. 14.7 The results are hardly very surprising for these two cases, but the method provides a powerful means of obtaining the frequency (and so the period) of a general nonlinear libration oscillation. The method may be extended to deal with rotations, which we will not consider further here, and to deal with systems which have two or more degrees of freedom. So we have the following: Central force motion (See Chapter 4.) 358 Classical Mechanics We may write the Hamiltonian in plane polars in the form H = p2 r 2m + p2 \u03b8 2mr2 + V (r) E, \u2261 (14.14) with constants of the motion F1 = H, F2 = p\u03b8. We can then calculate the action variables as I2 = I1 = 1 2\u03c0 1 2\u03c0 % % p\u03b8 d\u03b8 = p\u03b8, pr dr = \u221a2m 2\u03c0 H % p2 \u03b8 2mr2 \u2212 \u2212 V (r) dr. (14.15) Example: Inverse square law For the inverse square attractive force law V (r) = and Problem 9) we obtain k \u2212| /r (see | \u00a7 4.3 I1 = k I2 + | | 2 \u2212 2m H \u2212. (14.16) Show that in this case the nonlinear normal mode frequencies are the same, so that the orbits are", " closed. Evidently H = E = mk2 2(I1 + I2)2. \u2212 (14.17) The Hamiltonian depends only on the action variables, and the corresponding angle variables are ignorable, by design. The natural frequencies for this system are then given by \u03c91 = \u03c92 = mk2 (I1 + I2)3 \u2261 m k | \u2212 | 3/2 2E m 2\u03c0 \u03c4, \u2261 (14.18) where \u03c4 is the period. In fact this is just Kepler\u2019s third law of planetary motion (4.32), since the semi-major axis a of the elliptical orbit of negative energy E is given by a = \u2212| Note that important results for this system have been obtained without /2E (as in (4.30)). | k explicit calculation of the form of the angle variables: \u2022 The equality of the natural frequencies of the radial and transverse modes of oscillation, which leads to orbital closure. (See 14.1.) \u00a7 Order and Chaos in Hamiltonian Systems 359 \u2022 The relation between the period of oscillation and the energy E of a bound orbit. A further important property of the action variables will be indicated later in 14.5. \u00a7 14.4 Some Hamiltonian Systems which Exhibit Chaos We have seen in Chapter 13 and in previous sections of this current chapter that a system must have at least two degrees of freedom in order to have a phase space of large enough dimension to exhibit any chaotic behaviour. Again the system must also then have fewer symmetries (leading to conservation laws) than are required to give integrability and order. Orbital motion \u2014 n-body problem \u00a7\u00a7 14.1, 14.2 we indicated that the central force problem of Chapter In 4, with a \ufb01xed centre of force and two degrees of freedom, is integrable, so that it cannot exhibit the property of sensitivity to initial conditions characteristic of chaos. This is also true for the two-body problem in the centre-of-mass frame. (See 7.2.) Less intuitively apparent is that the motion of a particle in a plane under the inverse square law of attraction to two \ufb01xed centres may also be integrated. \u00a7 However the three-body problem is, in general, not integrable, even when it is restricted (as in Chapter 10, Problem 15); this problem, together with the extension to more bodies, was the strong motivation for Poincar\u00b4e in his general considerations for dynamical systems. The three-body problem has nine degrees of freedom and so gives rise to a system of di\ufb00erential equations of order eighteen. We may reduce this to order twelve by making use of the conservation of linear momentum, i.e. by going to centre-of-mass co-ordinates. By making use of conservation of energy and conservation of angular momentum we can reduce the order to eight. By further device and restricting our consideration to motions which are planar, the order can be reduced to four. This is the best that can be done in general. Even after all these reductions the problem is still extremely complicated and has kept mathematicians busy for several hundred years. As an indication we can consider the equations of the planar circular restricted three-body problem (Chapter 10, Problem 15). In the frame of 360 Classical Mechanics reference rotating with the primary masses M1, M2 about their centre of mass (see Fig. 14.8), the equations for the third (small mass m) body can be written in the form Fig. 14.8 2\u03c9 \u02d9y = \u00a8x \u2212 \u00a8y + 2\u03c9 \u02d9x = \u2202V \u2202x \u2202V \u2202y \u2212 \u2212. with and V = \u2212 1 2 \u03c92(x2 + y2) GM1 \u2212 r1 \u2212 GM2 r2, r1 = (x + a1)2 + y2, r2 = (x a2)2 + y2. \u2212 (14.19) This system can be expressed in Hamiltonian form with two degrees of freedom and there is only a single conserved quantity, i.e. the Hamiltonian itself, so that 1 2 ( \u02d9x2 + \u02d9y2) + V is a constant, which is known as the Jacobi integral. While there are some stable equilibria for this system (depending on the mass ratio M1/M2, see Chapter 12, Problem 12) there are also motions which are highly chaotic. However, the motion with given initial conditions is restricted by the curves of zero velocity at which \u02d9x = 0 = \u02d9y. (See Problem 12 of this Chapter.) Interest in the restricted problem re\ufb02ects the hierarchical nature of our Solar System and an increasing interest in space missions. The primaries Order and Chaos in Hamiltonian Systems 361 are variously Sun\u2013Jupiter (Chapter 12, Problem 13", "), Earth\u2013Moon (Problem 12) and so on. The \ufb01ve relative equilibria referred to in Chapter 10, Problem 16 are apparent in these applications \u2014 as sites for orbiting observatories and as staging points for more extended missions by NASA and ESA. The Lagrangian equilateral triangle rotating in its plane about its centre of mass is actually a relative equilibrium for the three-body problem with general masses. Example: Three-body problem Show that the three-body problem with three general masses m1, m2, m3 has a relative equilibrium (with the masses not in a straight line) if and only if the triangle formed by the masses is equilateral and \ufb01nd the rotation rate \u2126. Fig. 14.9 Note that the origin in this \ufb01gure is at the centre of mass. It is apparent that the masses remain in a plane. Then we have m1r1 + m2r2 + m3r3 = 0, (14.20) together with Gm1m2(r2 \u2212 r3 12 r1) + Gm1m3(r3 \u2212 r3 31 r1) = \u2212 m1\u21262r1, (14.21) 362 Classical Mechanics and two similar equations for m2, m3. From (14.20) and (14.21) we obtain a relation \u03b1r2 + \u03b2r3 = 0 where \u03b1, \u03b2 are scalars. Since r2, r3 are by hypothesis not parallel, we must have \u03b1 = \u03b2 = 0, which then leads to r12 = r31 = a (say). Of course the similar equations give r23 = a also, and by substitution we obtain \u21262 = G(m1 + m2 + m3)/a3. Naturally, stability of this con\ufb01guration is quite another matter \u2014 in general it is surely unstable and, typically, the system will separate. The search for exact periodic solutions of the three-body problem has resulted in an astonishing solution for the case of three equal masses \u2014 the masses follow one another round a \u2018\ufb01gure of eight\u2019 in a plane, visiting in turn each of the Euler three-body straight-line equilibria with one mass at the mid-point of the line joining the other two (see Chenciner and Montgomery, Annals of Mathematics, 152, 881\u2013901, 2000). While this solution is extremely special, it is exact and extensive searches have uncovered further exotic n-body solutions. In general the vast majority of n-body systems are chaotic, particularly on account of close two-body encounters in their evolution. Charged particle in a magnetic \ufb01eld \u00a7 In 5.5 as an example of rotating frames of reference we considered the e\ufb00ect of a magnetic \ufb01eld B on a charged particle moving in an orbit around a \ufb01xed charge. We assumed there that the magnetic \ufb01eld was su\ufb03ciently weak for the quadratic term in B to be neglected, leading to the Larmor precession of the Kepler elliptical orbit around the magnetic \ufb01eld vector. If the weak magnetic \ufb01eld approximation is not made, then the equation of motion of the moving charge can be written relative to the frame which rotates about the magnetic \ufb01eld with the Larmor frequency in terms of the Hamiltonian H = p2 2m \u2212 k1| | r + 1 k2| 2 | \u03c12. (14.22) Here the constant B \ufb01eld de\ufb01nes the z axis and \u03c12 = r2 z2, so that the Kepler inverse square law force of attraction towards the origin (having full rotational symmetry about all three axes) is supplemented by a Hooke attraction towards the B \ufb01eld axis (having rotational symmetry only about this axis). As a result the Hamiltonian system is not integrable \u2014 it is, in e\ufb00ect, a \u2018many-body\u2019 Hamiltonian and exhibits chaos. This broken symme- \u2212 Order and Chaos in Hamiltonian Systems 363 try of e.g. the classical hydrogen atom is complex, but very important in its implications for the quantum mechanical treatment of the same problem (see e.g. Friedrich and Wintgen, Physics Reports, 183, 37\u201379, 1989). Tops \u00a7 \u00a7\u00a7 10.3 and in In 12.3,12.4 we indicated the Lagrangian and Hamiltonian descriptions of the symmetric top, pivoted at a point on its axis of symmetry. The problem has three degrees of freedom and there are three independent conserved quantities H, p\u03d5, p\u03c8, so that the system is integrable and the motions detailed are regular and ordered. More complicated systems abound when the equations of", " rotational motion of an asymmetric body are considered, or where the system involves rolling motion, so that it is non-holonomic. For example, various asymmetric \u2018tops\u2019 appear to exhibit ordered and chaotic motion regimes and this is common for systems which are modelled by the Euler equations of 9.8 with general torques G. \u00a7 Billiard systems After Berry (European Journal of Physics, 2, 91\u2013102, 1981) we can consider the motion of a point \u2018ball\u2019 moving within a closed boundary (see Fig. 14.10) and bouncing perfectly on the wall. We assume motion in a straight line between collisions with the wall, simple re\ufb02ection at each bounce and no dissipation, i.e. we assume that the Hamiltonian is H = p2/2m within the enclosure. The ball follows a path just like that of a light ray with a boundary wall which is a perfect mirror. Fig. 14.10 364 Classical Mechanics Despite its apparent simplicity, this system turns out to be very rich and instructive, with dynamics depending particularly on the shape of the enclosure. The dynamics can be examined using natural canonical variables, s, p, where s is the distance measured along the boundary and p = cos \u03b1, where \u03b1 is the angle of incidence (and re\ufb02ection) at the wall. (Note that p here is not the magnitude of the linear momentum,.) This leads to a successive | bounce map (sn, pn) (sn+1, pn+1) which preserves area and leads to a Poincar\u00b4e section for the dynamics \u2014 the slice through the overall dynamical space is the (non-planar) boundary shape. p | \u2192 For motion in a planar region of this kind the system has two degrees of freedom and H is a constant of the motion. For some special enclosure shapes e.g. circular (see Fig. 14.11), elliptical (see Fig. 14.12) there is in each case another conserved quantity leading to integrability and order. Example: Circular billiard Fig. 14.11 Billiards in a circle: (a) basic orbit geometry; (b) typical orbit (never closing); (c) two closed orbits; (d) phase-space trajectories for orbits in (b) and (c). Show that there is a second conserved quantity (independent of H) for the circular billiard. Order and Chaos in Hamiltonian Systems 365 Evidently the billiard trajectory for a circular enclosure is a succession of chords, each at the same perpendicular distance from the centre of the circle. So the angular momentum of the ball about the centre (= a\u221a2mH p, where a is the radius and p = cos \u03b1) remains constant through successive bounces at the boundary. The angular momentum is the second conserved quantity and is evidently independent of the Hamiltonian H. Note that when a chord meets the circular boundary the angle \u03b1 between it and the tangent to the circle is the same at each bounce \u2014 the bounce map for a particular trajectory therefore consists of a \ufb01nite number of points or a line (each at p = cos \u03b1) according as the trajectory is closed or is not closed. For the elliptical enclosure the second conserved quantity is the product of the angular momenta of the ball measured about the two foci of the ellipse. (See Problem 10.) Fig. 14.12 Orbits in an ellipse: (a) repeatedly touching a confocal ellipse; (b) repeatedly touching confocal hyperbolae; (c) ellipse billiard mapping. 366 Classical Mechanics In these cases the bounce map (Poincar\u00b4e section) is made up of points and lines characteristic of integrability. When the shape of the boundary is not so special, the bounce map has regions in which there is area \ufb01lling, leading to chaos, as well as regions where ordered motion predominates. (See Fig. 14.13.) Billiard systems with various di\ufb00erent properties have been used to model other types of physical behaviour. Fig. 14.13 Oval billiard mapping: 25 orbits followed through 200 bounces. Dice and coins The motion when dice and coins are thrown is regular when there are no external torques, i.e. when moving and rotating through space, without any air resistance. However, there are impulsive forces which act at each bounce, so that for typical shapes the motion can be highly chaotic. Whereas billiards (see above) involve a regular object bouncing on an irregular boundary, dice (and coins) are a sort of \u2018dual\u2019 in that they are irregular objects bouncing on a regular surface. For each of these types of system the e\ufb00ects of small changes in physical geometry can be large in the statistical properties of the outcomes (see 13.7), whether any dissipation which may be", " present is very small or if it is such as to lead to rapid energy decay. \u00a7 Order and Chaos in Hamiltonian Systems 367 For example, when a regular cubical die is subject to a small change \u2206 in aspect ratio (away from 1 for the perfect cube), each 1% change in this geometric measure leads to a 3.4% change, approximately, in the probabilities associated with the faces displayed, a change modelled locally by 3) (Berkshire, British Association for the Advancement \u22062\u221a2/\u03c0(4 2/3 \u2212 of Science, 1985). The swinging Atwood\u2019s machine A simple physical system which exhibits a variety of dynamical behaviours is the swinging Atwood\u2019s machine, which consists of two masses M, m connected by a light inextensible string which passes over a small pulley and through a small hole in a smooth vertical plane, so that the mass m can rotate in this plane. The string is such that collisions of masses with pulley or hole can be discounted. (See Fig. 14.14.) Fig. 14.14 This system was introduced \ufb01rst by Tu\ufb01llaro, Abbott and Gri\ufb03ths (American Journal of Physics, 52, 895\u2013903, 1984) and has been the subject of various subsequent papers. The Hamiltonian for the system is H = p2 r 2m(1 + \u00b5) + p2 \u03b8 2mr2 + mgr(\u00b5 \u2212 cos \u03b8), with \u00b5 = M/m, (14.23) and the motion is not, in general, integrable, since H is usually the only constant of the motion. In the case \u00b5 > 1 we can show that the motion of m is always bounded by a curve of zero velocity (p = 0), which is an ellipse whose shape depends on the mass ratio \u00b5 and on the energy H. (See 368 Classical Mechanics Problem 11.) When \u00b5 energy and eventually the mass M passes over the pulley. 1 the motion is not, in general, bounded for any \u2264 The system is integrable in the case \u00b5 = 3 and apparently only for this particular value of the mass ratio. In that special case, there is a second conserved quantity given by J = p\u03b8 4m pr cos \u03b8 2 \u2212 2p\u03b8 r sin \u03b8 2 + mgr2 sin \u03b8 2 cos2 \u03b8 2. (14.24) \u02d9pr = (p2 From Hamilton\u2019s equations, cos \u03b8), \u02d9\u03b8 = p\u03b8/mr2, \u02d9p\u03b8 = mgr sin \u03b8, one can show by brute force that \u02d9J = 0. We note that while J is also quadratic in the momenta pr, p\u03b8, it is independent of the Hamiltonian H. \u02d9r = pr/4m, \u03b8/mr3) mg(3 \u2212 \u2212 \u2212 It is certainly not yet clear what \u2018physical\u2019 property is represented by this \u2018hidden symmetry\u2019. When \u00b5 = 3 the motion is completely ordered. For all other mass ratio values there are at least pockets of highly irregular and chaotic motion. The H\u00b4enon\u2013Heiles system From the previous examples it should be apparent that a paradigm of chaotic dynamics is provided by a Hamiltonian system with two degrees of freedom which has only one conserved quantity, e.g. the Hamiltonian itself. Under these circumstances the system dynamics can show all the features of sensitivity to initial conditions and loss of predictability indi13.7. As a \ufb01nal example consider the motion of a particle of unit cated in mass in a two-dimensional asymmetric potential well (see Fig. 14.15) so that its Hamiltonian is \u00a7 H = 1 2m where (p2 x + p2 y) + V (x, y), (14.25) V = 1 1 2 (x2 + y2) + x2y 3 r3 sin 3\u03b8 2 r2 + 1 \u2261 \u2212 1 3 y3 in plane polars. This system, with the analogy of a small ball rolling under gravity in an asymmetric bowl, was introduced by H\u00b4enon and Heiles (Astronomical Journal, 69, 73\u201379, 1964) in order to model the motion of a star in a galaxy which has a simple smoothed-out gravitational potential to represent the attraction of the other stars. Order and Chaos in Hamiltonian Systems 369 Fig. 14.15 H\u00b4enon\u2013Heiles potential contours of V near the local minimum at (0, 0). For small energies the cubic terms in V are insigni\ufb01cant when compared with the quadratic terms and in that limit the motion is integrable, consisting as it does of a combination of normal modes as obtained by the methods of", " Chapter 11. Of course the motion for small energies is just that of the isotropic oscillator of 4.1. For larger energies corresponding to larger displacements away from the potential minimum at the origin the motion becomes highly irregular. \u00a7 There are very similar consequences for many other such systems, e.g. 11.1, when we now allow \u03b8, \u03d5 not to be the double pendulum system of restricted to be small. \u00a7 14.5 Slow Change of Parameters \u2014 Adiabatic Invariance Let us suppose that in a cyclic system (librating or rotating) we allow some of the control parameters to change on a very long time scale, i.e. on a scale of many basic oscillation periods. Then it emerges that the change in an action I is then also very slow, in the sense that we can change the parameters by as large an amount as we may want, while I stays as near constant as we want, provided that the change of parameters is made slowly enough. The change of parameters is then said to be carried out adiabatically and the action variables are then called adiabatic invariants. Now a general proof of this property is beyond our scope here, so let us 2.2, but consider by way of illustration the simple harmonic oscillator of \u00a7 370 Classical Mechanics now with a frequency which is slowly-varying. That is to say \u00a8x + [\u03c9(t)]2x = 0, (14.26) k/m, where \u02d9\u03c9/\u03c9 with \u03c9 = \u03c9. The variation in \u03c9 could, for example, be brought about by a spring constant k which varies on a time scale which is very long compared with the natural period \u03c4 = 2\u03c0/\u03c9. Over a single period (during which \u03c9 does not change appreciably) we may write x(t) = a cos[\u03c9(t)t \u03b8], \u2212 as in (2.20). The energy of the oscillation is E = 1 2 m( \u02d9x2 + \u03c92x2) 1 2 m\u03c92a2, \u2261 essentially, and this now changes with time at the slow rate \u02d9E = m( \u02d9x\u00a8x + \u03c92x \u02d9x + \u03c9 \u02d9\u03c9x2) = m\u03c9 \u02d9\u03c9x2. By averaging over a period of the oscillation, so that e.g. E (above) we obtain E = 1 \u03c4 d dt E = m\u03c9 \u02d9\u03c9 x2 = 1 2 m\u03c9 \u02d9\u03c9a2 \u02d9\u03c9 \u03c9. E We obtain immediately \u2261 (14.27) (14.28) (14.29) \u03c4 0 E d\u03c4 = (14.30) d dt E \u03c9 = 1 \u03c9 d dt E \u2212 \u02d9\u03c9 \u03c92 E = 0, (14.31) so that action I of this oscillator. (See (14.9).) /\u03c9 is essentially constant. This is, of course, our measure of the E For more general systems the demonstration is carried out via the definition (14.4) of the action variable. Particular physical applications are very common and valuable conclusions can be drawn about the variations of other variables on the assumption of the applicability of the principle of adiabatic invariance. Example: Pendulum with slowly varying length Find how the amplitude of (small) oscillation of a simple pendulum changes as the length is varied slowly. Order and Chaos in Hamiltonian Systems 371 We can consider the simple pendulum of 2.1, but now with length l(t) changing slowly with time, and we can simplify considerably by con\ufb01ning ourselves to small oscillations. We then need only put x = l\u03b8 in the above analysis of the simple harmonic oscillator, together with \u03c9 = g/l. In this case the action I is given by \u00a7 I = E/\u03c9 = 1 = 1 2 m\u03c9a2 2 m gl3(\u03b8max)2, (14.32) since a = l\u03b8max here. The assumption that I is invariant leads to \u03b8max \u221d 3/4 (see Problem 13) and we can \ufb01nd from this how the energy, displacement, maximum acceleration of the pendulum bob change as l is changed, possibly by orders of magnitude, on the assumption that the change in l is e\ufb00ected su\ufb03ciently slowly. l\u2212 A practical problem to contemplate when considering this matter is the consumption of a piece of spaghetti, although this is not usually a simple pendulum and politeness may well rule out the necessary long time scale for consumption! However, the general principle is the same, particularly with regard to the dangers of spaghetti sauce! The elastic bouncer \u00a7 In 14.3 we considered a", " perfectly elastic ball bouncing normally between two walls and the action I for that case was found to be \u221a2mH L/\u03c0 where m, H are, respectively, the mass and energy of the ball and L is the distance between the walls. If the walls are now moved either towards or away from one another very slowly (\u2018adiabatically\u2019), then the principle of adiabatic invariance tells us that L\u221aH should be essentially constant, so that v 1/L 1/L2 (see Problem 14). Again it should be noted that v, H and energy H can change by orders of magnitude, while the action I remains essentially constant. We can note here that we can model a monatomic gas with elastic molecules bouncing in a cubical box and, under reasonable assumptions, (density)5/3 for adiabatic change. obtain the relationship that pressure \u221d \u221d \u221d Attractive central force problem with varying strength Evidently the force of attraction, e.g. of the Sun, would diminish as its mass decreases over what is, for us, a long time. It has also been suggested 372 Classical Mechanics \u00a7 by Dirac that the gravitational \u2018constant\u2019 may, in fact, change on a long time scale. In either case we can look to the action/angle formulation 14.3, where I2 is an exact invariant and I1 can be taken now as an in adiabatic invariant. As a consequence we can show, in particular, that, as of the attracting force decreases, the period and the semithe strength major axis of the Kepler elliptical orbit of a planet both increase (as we might expect), but that (rather unexpectedly) the eccentricity of the orbit remains constant and the orbit retains its shape. (See Problem 16.) k | | The adiabatic invariance of the action variables has enormous importance in the transition to quantum mechanics. The pendulum with slow variation in length was a discussion problem for Lorentz and Einstein in 1911, when the foundations of quantum mechanics were being laid. The e\ufb00ective constancy of the action variables by the principle of adiabatic invariance, in the classical formulation as control parameters are changed very slowly, makes the action variables the prime candidates for quantization. 14.6 Near-integrable Systems When we have a system which is Hamiltonian with H = H0(q, p) and integrable, so that its dynamics are regular and ordered, we can ask what happens for a system which is not itself integrable, but which is close by. That is to say this latter system has Hamiltonian H = H0 + H1 where is very small. When there is only one degree of freedom (n = 1) there is a welldeveloped perturbation theory, which allows us to \ufb01nd out how the action/angle variables for the H0 system are changed by the perturbation H1. We won\u2019t detail this here, but it turns out that the only real di\ufb03culty with the perturbation expansions (as power series in ) is associated with attempting to expand near where the basic frequency \u03c90 of the unperturbed system H0 is zero. An example of this is provided by the simple pendulum and the libration/rotation orbits which are situated close to the separatrices (at which \u03c90 = 0). The perturbation expansions break down in this neighbourhood, although they generally work well elsewhere. A simple remedy would seem to suggest itself strongly \u2014 avoid separatrices! For systems with at least two degrees of freedom (n 2) the perturbation analysis seems to be plagued by a similar di\ufb03culty, but much worse in implication. The expansions for the n = 1 case contain denominators \u03c90, \u2265 Order and Chaos in Hamiltonian Systems 373 \u2265 2 case we \ufb01nd in the expansions denominators like \u03c90 \u00b7 so that the expansions become non-uniformly valid near \u03c90 \u223c= 0. For the N where the n lattice vector N has integer components and \u03c90 is the vector of normal frequencies of the particular torus (see 14.1) about which we are trying to expand our series solution. \u00a7 Now when this denominator is actually zero we have a degeneracy (or resonance), in that some of the normal frequencies have rational ratios. This is bad enough, but the trouble is that, even for \u03c90 not exactly resonant in this manner, we can apparently make choices of N components in N as close as we might (not) like to zero \u2014 thus deorder to make \u03c90 \u00b7 stroying the usefulness of our expansion. This is usually called the problem of small denominators and it seems to imply that, only if the perturbation H1 to our Hamiltonian H0 is such that H = H0 + H1 is still integ", "rable through the preservation of a su\ufb03cient number of conserved quantities, is the phase space torus structure itself preserved, although distorted, and with convergent perturbation expansions. Very elaborate methods have been developed in an attempt to patch up the expansions in the general case. Especially noteworthy is a tour de force by Delaunay for the Sun\u2013Earth\u2013Moon system (\u2018lunar theory\u2019) undertaken in the middle of the nineteenth century, which involved about twenty years of endeavour, but all such e\ufb00orts seemed doomed to failure. It seemed that the slightest perturbation, however small, would cause the torus structure of an integrable system to collapse irretrievably. That this is not in fact the case was demonstrated by Kolmogorov, Arnol\u2019d and Moser (KAM) in work originating between 1954 and 1963. This work has led to an explosion of research, results and applications. KAM showed that, as long as the perturbation introduced to a smooth system is itself smooth enough, then the torus destruction with a perturbed system is progressive as the size of the perturbation increases. Key roles are played in the theory, and its consequences, by the following: \u2022 \u2022 Rational approximation. This means the closeness by which real numbers can be approximated by rationals. This is important, because the torus breakdown grows with the perturbation around the resonant (\u03c90 with rational ratios) tori. Fractals. These are \u2018self-similar\u2019 structures in maps, i.e. in Poincar\u00b4e sections (like e.g. Fig. 14.13) and in full phase space. Properties of maps are explored brie\ufb02y in Appendix D. 374 Classical Mechanics \u2022 \u2022 Divergence of trajectories. Breakdown of the torus structure gives freedom to system trajectories to diverge strongly from their neighbours, leading to \u2018extreme sensitivity to initial conditions\u2019 (see 13.7) and to chaotic behaviour. Lack of predictability. We have e\ufb00ectively stochastic output resulting even from strictly deterministic systems, so that we lose knowledge of the connection between cause and e\ufb00ect. \u00a7 To the big question \u2018Is the Solar System stable?\u2019, which has motivated a lot of work in Mechanics, the theory provides the answer \u2018a (de\ufb01nite) maybe\u2019. The Solar System is an n-body problem, which is markedly hierarchical, so that the e\ufb00ect of e.g. planet/planet interactions is of lower order than the planet/Sun interactions which produce Kepler ellipses. For this reason the broad structure of the system is robust (fortunately for the human race!). For large con\ufb01gurations, stable and unstable regimes may well be closely interwoven, so that in an idealized system we might not be able to tell from (necessarily) imprecise observations quite what regime we are in! However we should \ufb01nd the hierarchical nature of our system, with a strong central Sun, to be reassuring apart from the possibility of asteroid impact \u2014 in practice the fate of the whole Solar System is actually rather likely to be determined largely by dissipative and other non-gravitational forces. 14.7 Summary In this chapter the importance of the number of symmetries of a Hamiltonian system (leading to conservation laws) has been shown in the restrictions they impose on the freedom of the system trajectories in the appropriate phase space. Integrability of the system leads to a natural formulation in terms of action/angle variables, which allow us to detail the nonlinear normal modes of such a system, i.e. modes which are exact and involve no approximations for small-amplitude motions. Examples have been given of simple/important Hamiltonian systems which have regimes which exhibit chaos and a resulting breakdown of predictability. The action variables are e\ufb00ectively constant when system control parameters are changed very slowly. This principle of adiabatic invariance is a general result, which has non-intuitive and useful consequences for the way in which other quantities of physical interest may change, possibly by orders of magnitude on these long time scales. Order and Chaos in Hamiltonian Systems 375 The adiabatic invariance of action variables identi\ufb01es them as prime candidates to be quantized in the transition from classical to quantum mechanics. There are other routes to chaos. Indeed e.g. the limit cycles of Chapter 13, each with a single-point Poincar\u00b4e section, can go through a perioddoubling cascade to chaos (see Appendix D) as parameters are changed. When trajectories are stretched, squeezed and folded then chaos and \u2018mixing\u2019 typically result. It is \ufb01tting that this book should end with the mention of results (K", "AM) which point the way to a whole host of future developments, by no means the least of which is the question of how to quantize a system which is classically chaotic! As we said in the Preface, Classical Mechanics is a very old subject. However it is very much alive! Problems 1. A particle of mass m is projected outward radially from the surface (q = R) of a spherical planet. Show that the Hamiltonian is given by H = p2/2m /q (with k constant), so that H is a constant of the | motion ( energy E). Sketch the phase portrait in the (q, p) phase plane \u2261 for q R, distinguishing between trajectories which correspond to the particle returning and not returning to the planet\u2019s surface. When the particle does return show that the time taken to do this is \u2212 | \u2265 k t0 = \u221a2m h R dq E + k | /q |. k, where h = | | E | | /m. (Hint : the | 0 k | Evaluate this integral to \ufb01nd t0 in terms of h, R, substitution q = h sin2 \u03b8 is helpful!) By considering the limit R/h show that the result is in accord with Kepler\u2019s third law (4.32). \u2192 2. For an autonomous Hamiltonian system with two degrees of freedom the Hamiltonian H is a constant of the motion and such a system is integrable if another constant F (independent of H) can be found \u2014 i.e. such that [F, H] = 0. (a) For the double harmonic oscillator (14.3) show that we may take F = p2 (b) For central force motion (14.14) show that we may take F = p\u03b8 as the second constant. 1 as the second constant. 1/2m + 1 2 m\u03c92 1q2 376 Classical Mechanics (c) *When the potential V (r) = the vector k \u2212| /r, show that the components of | | \u2212 r r A = p J \u2227 k m | with p, J respectively linear, angular momentum) are constants of the motion. (A is the Laplace\u2013Runge\u2013Lenz vector. It can be shown using the results 4.4 that it has magnitude given by the eccentricity e of the conicin section orbit and is directed from the centre of force along the major axis towards the position of closest approach (\u2018perihelion\u2019).) \u00a7 3. A string of length l with a mass m at each end passes through a hole in a frictionless horizontal plane. One mass moves horizontally on the plane and the other mass hangs vertically downwards. Show that a suitable Hamiltonian for the system is H = p2 r 4m + p2 \u03b8 2mr2 \u2212 mg(l r), \u2212 where (r, \u03b8) are the polar co-ordinates of the particle on the plane and pr, p\u03b8 are the corresponding momenta. Identify two constants of the motion. Show that a steady motion with r = r0 is possible (for any r0 > 0), if p\u03b8 is chosen suitably, and that the period of small oscillations about this motion is 2\u03c0 2r0/3g. 4. Consider again the particle sliding on the inside of a smooth cone, as in Chapter 12, Problem 1. Show that when a stable circular motion is disturbed the resulting small oscillations are only closed if \u221a3 sin \u03b1 is a rational number. 5. For central force motion with an inverse square law force of attraction the Hamiltonian is (14.14), i.e. | + k | r H = ( \u2261 energy E). p2 \u03b8 2mr2 \u2212 p2 r 2m If we \ufb01x p\u03b8, show that r is bounded (r1 \u2264 r2) only when k2m/2p2 E < 0 and that the energy minimum corresponds to \u2212 an orbit in physical space which is a circle. Sketch the curves H = constant in the (r, pr) projection of the full four-dimensional phase space for this system. Consider this projection in the light of the discussion of surfaces of section in 14.2. \u03b8 \u2264 \u2264 r 6. A particle of mass m is constrained to move under the action of gravity in the vertical (x, z) plane on a smooth cycloid curve given paramet- \u00a7 Order and Chaos in Hamiltonian Systems 377 rically by x = l(\u03b8 + sin \u03b8), z = l(1 Hamiltonian is \u2212 cos \u03b8). Show that a suitable H = p2 \u03b8 4ml2(1 + cos \u03b8) + mgl(1 cos \u03b8). \u2212 Use action/angle variables to show that the frequency of oscillation of the particle is independent of its amplitude, i.", "e. it is the same for all initial conditions with < \u03c0. (The substitution s = sin 1 2 \u03b8 is useful. This tautochrone property of the cycloid was known to Huygens in the seventeenth century and, in principle at least, it leads to some quite accurate clock mechanisms. Contrast the tautochrone property with the brachistochrone property of Chapter 3, Problem 15.) \u03b8 | | l)2 for x 2 k( | 7. *A particle of mass m is attached to the origin by a light elastic string of natural length l, so that it is able to move freely along the x-axis if its distance from the origin is less than l, but otherwise moves in a potential V (x) = 1 > l. If the particle always moves x | | in a straight line, sketch the potential and the phase-plane trajectories \u03b2)2, I\u2126 + \u03b22 for di\ufb00erent values of the energy E. Show that E = ( where I is the action, \u03b2 = \u221a2k l/\u03c0, \u2126 = k/m. Explain brie\ufb02y (without detailed calculation) how the angle variable \u03c6 conjugate to the action may be found in the form \u03c6(x) and how x may be found as a function of the time t. What happens in the (separate) limits of small and large energies E? | \u2212 \u2212 1 8. *A particle of mass m moves in a one-dimensional potential V (q) = 2 (kq2 + \u03bb/q2), where k, \u03bb, q > 0. Sketch the potential and the phase portrait. Show that the energy E and action I are related by E = \u221ak\u03bb + 2I k/m and that the period is then independent of amplitude. Discuss how the dependence of q on the angle variable \u03c6 may be found and then the dependence of q on the time t. (This one-dimensional problem models the purely radial part of the motion of the isotropic harmonic oscillator of 4.1. The integral \u00a7 x2 x1 (x2 \u2212 x)(x \u2212 x1) dx x = 1 2 \u03c0(x1 + x2) \u03c0\u221ax1x2 \u2212 where x2 > x1 > 0 and x = q2 will prove useful!) 9. *For central force motion with an inverse square law force of attraction (see Problem 5) con\ufb01rm the result of (14.16) that the radial action I1 = 378 Classical Mechanics | I2 + 1 k 2 | 2m/( E), where I2 = p\u03b8, so that the natural frequencies \u2212 of radial and transverse motions of the system are the same. (In order to evaluate the integral in (14.15) the result given at the end of Problem 8 will prove similarly useful!) \u2212 10. For the billiard in an elliptical enclosure (Fig. 14.12) use the result of Appendix B, Problem 2 to show that the product \u039b of the angular momenta of the ball measured about the two foci of the ellipse is preserved through each bounce, so that it is conserved and therefore a constant of the motion. Using elliptical co-ordinates (see Chapter 3, Problem 24), show that \u039b = (cosh2 \u03bb 1(sinh2 \u03bb p2 \u03bb) and that H = [2mc2(cosh2 \u03bb \u03b8). Hence show that the re\ufb02ected trajectory from the boundary x2/a2 + y2/b2 = 1 is necessarily tangent (when \u039b > 0) to the ellipse \u03bb = arcsinh [Closure for such a trajectory implies closure for all trajectories tangent to the same inner ellipse \u2014 an example of a general result due to Poncelet (1822).] cos2 \u03b8)\u2212 1(p2 \u2212 cos2 \u03b8)]\u2212 \u039b/2mc2H. sin2 \u03b8 p2 \u03bb + p2 \u03b8 \u2212 \u2212 11. *Consider the motion of the swinging Atwood\u2019s machine (see Fig. 14.14) for various di\ufb00erent values of the mass ratio \u00b5 = M/m. (a) Show that when \u00b5 > 1 the motion of m is always bounded by a zero-velocity curve which is an ellipse, whose shape depends on \u00b5 and on the constant energy E. (b) Show that when \u00b5 1 the motion is not in general bounded for any E by sketching the zero-velocity curves for mass m in the r, \u03b8 plane (in various cases). \u2264 (Hint : The results of velocity curves.) \u00a7 B.2 will prove helpful in identifying the zero- \u2212 12. *Show that the equations of", " motion (14.19) for the restricted threebody problem may be put into Hamiltonian form using the substitution px = \u02d9x \u03c9y, py = \u02d9y + \u03c9x. Show that, since the system is autonomous, 1 2 ( \u02d9x2 + \u02d9y2) + V = C, where C is a constant of the motion (Jacobi). C of possible motions in the x, y plane, Hence sketch the regions V for various values of C and of the mass ratio M1/M2, considering particularly those values of C corresponding to the three \u2018equilibria\u2019 on the x axis and to the two Lagrangian equilateral triangle \u2018equilibria\u2019 \u2014 see Chapter 10, Problem 16. (Note that for the Earth/Moon system with M1/M2 = 81.3 there is in consequence a value of C below which an Earth\u2013Moon transfer is not possible.) \u2264 Order and Chaos in Hamiltonian Systems 379 13. For the small oscillations of a pendulum whose length l is varying very 3/4. slowly show that the maximum angular displacement \u03b8max \u221d Hence show that the maximum sideways displacement from the vertical 3/4. (Note that this is last result implies, as l decreases slowly, an increasing risk for spaghetti eaters from sauce detachment!) l1/4 and that the maximum acceleration is l\u2212 l\u2212 \u221d \u221d \u221d 14. For the elastic bouncer (see Fig. 14.7) between two walls whose separation is being changed slowly (\u2018adiabatically\u2019) show that the bouncer 1/L. We can model (crudely) a monatomic gas, con\ufb01ned speed v within a cubical box of side L, by assuming that each molecule has the same speed v and that one third of the molecules move normally to each pair of faces of the box! If intermolecular collisions do not a\ufb00ect these assumptions then L can be changed very slowly. Assuming that the kinetic energy of the molecules gives a measure of temperature and that (volume) the gas is \u2018ideal\u2019, so satisfying the ideal gas law: (pressure) (volume)5/3 for (temperature), obtain the relationship (pressure) \u00d7 \u221d adiabatic change of a monatomic gas. \u221d \u2265 15. A particle of mass m moves smoothly up and down a smooth inclined plane (inclined at an angle \u03b1 to the horizontal). The particle Hamiltonian is H = p2/2m + mgq sin \u03b1 (= E) where p = m \u02d9q, the co-ordinate 0, being measured upwards along the plane from a \ufb01xed point q q = 0 at which the particle is perfectly elastically re\ufb02ected at each impact. Show that the energy E and action I for this oscillator are related by E = [(9\u03c02/8)g2m sin2 \u03b1]1/3I 2/3 and \ufb01nd the frequency \u03c9 of small oscillation in terms of g, \u03b1 and q0 (the amplitude of the motion). Given that the angle \u03b1 now decreases very slowly use the principle of adiabatic invariance to show that during the long time in which \u03b1 decreases from \u03c0/3 to \u03c0/6 the energy of the system decreases by about 31% and the amplitude and period increase by about 20% and 44% respectively. 16. *For central force motions with an inverse square law force of attraction the relation between energy E and actions I1, I2 is (14.17), demonstrated in Problem 9. If the strength of the force (i.e. ) decreases | slowly (\u2018tired sun\u2019), use the principle of adiabatic invariance to show that the period \u03c4 of a bounded orbit (an ellipse, see Problem 5) varies 2 (i.e. \u03c4 increases). Find how the semi-major axis of so that \u03c4 the ellipse varies with k and show that the eccentricity of the ellipse remains constant. k | k\u2212 \u221d 380 Classical Mechanics Appendix A Vectors In this appendix, we give a summary of the properties of vectors which are used in the text. A.1 De\ufb01nitions and Elementary Properties A vector a is an entity speci\ufb01ed by a magnitude, written a or, and a | direction in space. It is to be contrasted with a scalar, which is speci\ufb01ed by a magnitude alone. The vector a may be represented geometrically by an arrow of length a drawn from any point in the appropriate direction. In particular, the position of a point P with respect to a given origin O may be speci\ufb01ed by the position vector r drawn from O to P as in Fig. A.", "1 Fig. A.1 381 382 Classical Mechanics Any vector can be speci\ufb01ed, with respect to a given set of Cartesian axes, by three components. If x, y, z are the Cartesian co-ordinates of P, with O as origin, then we write r = (x, y, z), and say that x, y, z are the components of r. (See Fig. A.1.) We often speak of P as \u2018the point r\u2019. When P coincides with O, its position vector is the zero vector 0 = (0, 0, 0) of length 0 and indeterminate direction. For a general vector, we write a = (ax, ay, az), where ax, ay, az are its components. The product of a vector a and a scalar c is ca = (cax, cay, caz). If c > 0, it is a vector in the same direction as a, and of length ca; if c < 0, it is in the opposite direction, and of length a. In particular, if c = 1/a, we | obtain the unit vector in the direction of a, \u02c6a = a/a. c | Addition of two vectors a and b may be de\ufb01ned geometrically by drawing one vector from the head of the other, as in Fig. A.2. (This is the \u2018parallelogram law\u2019 for addition of forces \u2014 or vectors in general.) Subtraction b \u03b8 a + b a a Fig. A.2 b is de\ufb01ned similarly by Fig. A.3. In terms of components, a a \u2212 b b Fig. A.3 a + b = (ax + bx, ay + by, az + bz), a bz). bx, ay \u2212 b = (ax \u2212 by, az \u2212 \u2212 Vectors 383 It is often useful to introduce three unit vectors i, j, k, pointing in the directions of the x-, y-, z-axes, respectively. They form what is known as an orthonormal triad \u2014 a set of three mutually perpendicular vectors of unit length. It is clear from Fig. A.1 that any vector r can be written as a sum of three vectors along the three axes, r = xi + yj + zk. (A.1) Mathematically, any set of three quantities may be grouped together and regarded as the components of a vector. It is important to realize, however, that when we say that some physical quantity is a vector we mean more than just that it needs three numbers to specify it. What we mean is that these three numbers must transform in the correct way under a change of axes. For example, consider a new set of axes i, j, k related to i, j, k by a rotation through an angle \u03d5 about the z-axis (see \ufb01g. A.4): j\u2032 j \u03d5 y x\u2032 r \u03d5 x y\u2032 i\u2032 i Fig. A.4 i = i cos \u03d5 + j sin \u03d5, i sin \u03d5 + j cos \u03d5, j = \u2212 k = k. (A.2) The co-ordinates x, y, z of P with respect to the new axes are de\ufb01ned by r = xi + yj + zk. Substituting (A.2) and comparing with (A.1), we see that x = x cos \u03d5 \u2212 y sin \u03d5, etc, or equivalently x = x cos \u03d5 + y sin \u03d5, etc. Physically, then, a vector a is an object represented with respect to any set of axes by three 384 Classical Mechanics components (ax, ay, az) which transform under rotations in the same way as (x, y, z), i.e., in matrix notation, = ax ay az \u23a1 \u23a3 \u23a4 \u23a6 cos \u03d5 sin \u03d5 0 sin \u03d5 cos \u03d5 0 1 0 0 \u23a1 \u2212 \u23a3 ax ay az \u23a1 \u23a4 \u23a6 \u23a3. \u23a4 \u23a6 (A.3) A.2 The Scalar Product If \u03b8 is the angle between the vectors a and b, then by elementary trigonometry the length of their sum is given by a + b | | 2 = a2 + b2 + 2ab cos \u03b8. It is useful to de\ufb01ne their scalar product a b (\u2018a dot b\u2019) as \u00b7 b = ab cos \u03b8. a \u00b7 (A.4) (A.5) Note that this is equal to the length of a multiplied by the projection of b on a, or vice versa. (See Fig. A.5.) b \ufffd", "\ufffd a \u22c5 b a Fig. A.5 In particular, the square of a is a2 = a \u00b7 a = a2. Thus we can write (A.4) as (a + b)2 = a2 + b2 + 2a and, similarly, the square of the di\ufb00erence is b)2 = a2 + b2 (a \u2212 2a \u2212 b, b. \u00b7 \u00b7 Vectors 385 All the ordinary rules of algebra are valid for the sums and scalar products of vectors, save one. (For example, the commutative law of addition, a + b = b + a is obvious from Fig. A.2, and the other laws can be deduced from appropriate \ufb01gures.) The one exception is the following: for two scalars, ab = 0 implies that either a = 0 or b = 0 (or, of course, both), but we can \ufb01nd two non-zero vectors a and b for which a b = 0. In fact, this is the case if \u03b8 = \u03c0/2, that is, if the vectors are orthogonal: \u00b7 b = 0 a \u00b7 if b. a \u22a5 The scalar products of the unit vectors i, j, k are i2 = j2 = k2 = 1. Thus, taking the scalar product of each in turn with (A.1), we \ufb01nd r = x, i \u00b7 r = y, j \u00b7 r = z. k \u00b7 These relations express the fact that the components of r are equal to its projections on the three co-ordinate axes. More generally, if we take the scalar product of two vectors a and b, we \ufb01nd and in particular, b = axbx + ayby + azbz, a \u00b7 r2 = r2 = x2 + y2 + z2. (A.6) (A.7) A.3 The Vector Product Any two non-parallel vectors a and b drawn from O de\ufb01ne a unique axis through O perpendicular to the plane containing a and b. It is useful to de\ufb01ne the vector product a b) to be a vector along this axis whose magnitude is the area of the parallelogram with edges a and b, b (\u2018a cross b\u2019, sometimes also written a \u00d7 \u2227 a | \u2227 b | = ab sin \u03b8. (A.8) (See Fig. A.6.) To distinguish between the two opposite directions along the axis, we introduce a convention: the direction of a b is that in which a right-hand screw would move when turned from a to b. \u2227 386 Classical Mechanics a\u2227b b A \u03b8 a Fig. A.6 \u00a7 A vector whose sense is merely conventional, and would be reversed by changing from a right-hand to a left-hand convention, is called an axial vector, as opposed to an ordinary, or polar, vector. For example, velocity and force are polar vectors, but angular velocity is an axial vector (see 5.1). The vector product of two polar vectors is thus an axial vector. The vector product has one very important, but unfamiliar, property. If we interchange a and b, we reverse the sign of the vector product: a = b \u2227 a \u2212 \u2227 b. (A.9) It is essential to remember this fact when manipulating any expression involving vector products. In particular, the vector product of a vector with itself is the zero vector, a = 0. a \u2227 More generally, a \u2227 b vanishes if \u03b8 = 0 or \u03c0: b = 0 if a \u2227 b. a If we choose our co-ordinate axes to be right-handed, then the vector products of i, j, k are k = 0, \u2227 i = j j = k, k = i, i = j, i, j. \u2212 \u2212 \u2212 (A.10) Vectors 387 Thus, when we form the vector product of two arbitrary vectors a and b, we obtain a b = i(aybz \u2212 azby) + j(azbx \u2212 axbz) + k(axby \u2212 \u2227 aybx). This relation may conveniently be expressed in the form of a determinant b = a \u2227 i j k ax ay az bx by bz. (A.11) From any three vectors a, b, c, we can form the scalar triple product (a c. Geometrically, it represents the volume V of the parallelepiped with adjacent edges a, b, c (see Fig. A.7). For, if \u03d5 is the angle between c b) \u2227 \u00b7 a\u2227b c \u03d5 h b A a Fig. A.7 and a \u2227 b, then \u00b7 b b", ") (a a | c = \u2227 c cos \u03d5 = Ah = V, | where A is the area of the base, and h = c cos \u03d5 is the height. The volume is reckoned positive if a, b, c form a right-handed triad, and negative if they form a left-handed triad. For example, (i 1. In terms of components, we can evaluate the scalar triple product by k = 1, but (i j = k) j) \u2212 \u2227 \u2227 \u2227 \u00b7 \u00b7 taking the scalar product of c with (A.11). We \ufb01nd (a b) c = (A.12) Either from this formula, or from its geometric interpretation, we see that the scalar triple product is unchanged by any cyclic permutation of \u2227 \u00b7 ax ay az bx by bz cx cy cz. 388 Classical Mechanics a, b, c, but changes sign if any pair is interchanged: (a b) \u00b7 \u2227 c = (b = \u2227 (b \u2212 c) a = (c c = \u00b7 a) \u2227 \u00b7 a) \u2227 (c \u2212 \u00b7 \u2227 b b) a = (a c) b. \u00b7 \u2227 \u2212 \u00b7 Moreover, we may interchange the dot and the cross: (a b) \u00b7 \u2227 c = a (b \u00b7 \u2227 c). (A.13) (A.14) (For this reason, a more symmetrical notation, [a, b, c], is sometimes used.) Note that the scalar triple product vanishes if any two vectors are equal, or parallel. More generally, it vanishes if a, b, c are coplanar. From three vectors we can also form the vector triple product (a c. b, it must lie in the plane of a and Since this vector is perpendicular to a b, and must therefore be a linear combination of these two vectors. It is not hard to show by writing out the components, that b) \u2227 \u2227 \u2227 Similarly, (a b) \u2227 \u2227 c = (a (b a \u2227 \u2227 c) = (a c)b (b \u00b7 \u2212 c)a. c)b (a \u00b7 \u2212 b)c. \u00b7 \u00b7 (A.15) (A.16) Note that these vectors are unequal, so that we cannot omit the brackets in a vector triple product. It is useful to note that in both of these formulae the term with the positive sign is the middle vector times the scalar product of the other two. A.4 Di\ufb00erentiation and Integration of Vectors We are often concerned with vectors which are functions of some scalar parameter, for example the position vector of a particle as a function of time, r(t). The vector distance travelled by the particle in a short time interval \u2206t is \u2206r = r(t + \u2206t) r(t). \u2212 (See Fig. A.8.) The velocity, or derivative of r with respect to t, is de\ufb01ned just as for scalars, as the limit of a ratio, \u02d9r = dr dt = lim 0 \u2206t \u2192 \u2206r \u2206t. (A.17) Vectors 389 r(t+\u2206t) v \u2206r r(t) Fig. A.8 In the limit, the direction of this vector is that of the tangent to the path of the particle, and its magnitude is the speed in the usual sense. In terms of co-ordinates, \u02d9r = ( \u02d9x, \u02d9y, \u02d9z). Derivatives of other vectors are de\ufb01ned similarly. di\ufb00erentiate again to form the acceleration vector \u00a8r = d2r/dt2. In particular, we can It is easy to show that all the usual rules for di\ufb00erentiating sums and products apply also to vectors. For example, d dt (a \u2227 b) = da dt \u2227 b + a db dt, \u2227 though in this particular case one must be careful to preserve the order of the two factors, because of the antisymmetry of the vector product. Note that the derivative of the magnitude of r, dr/dt, is not the same dr/dt. For example, if the thing as the magnitude of the derivative, | | particle is moving in a circle, r is constant, so that \u02d9r = 0, but clearly is not zero. In fact, applying the rule for di\ufb00erentiating a scalar product to r2, we obtain \u02d9r | | 2r \u02d9r = d dt (r2) = d dt (r2) = 2r \u02d9r, \u00b7 which may also be", " written \u02d9r = \u02c6r \u02d9r. \u00b7 (A.18) Thus the rate of change of the distance r from the origin is equal to the radial component of the velocity vector. 390 Classical Mechanics We can also de\ufb01ne the integral of a vector. If v = dr/dt, then we also write r = v dt, and say that r is the integral of v. If we are given v(t) as a function of time, and the initial value of r, r(t0), then the position at any later time is given by the de\ufb01nite integral r(t1) = r(t0) + v(t) dt. (A.19) t1 t0 This is equivalent to three scalar equations for the components, for example, x(t1) = x(t0) + vx(t) dt. t1 t0 One can show, exactly as for scalars, that the integral in (A.19) may be expressed as the limit of a sum. A.5 Gradient, Divergence and Curl There are many quantities in physics which are functions of position in space; for example, temperature, gravitational potential, or electric \ufb01eld. Such quantities are known as \ufb01elds. A scalar \ufb01eld is a scalar function \u03c6(x, y, z) of position in space; a vector \ufb01eld is a vector function A(x, y, z). We can also indicate the position in space by the position vector r and write \u03c6(r) or A(r). Now let us consider the three partial derivatives of a scalar \ufb01eld, \u2202\u03c6/\u2202x, \u2202\u03c6/\u2202y, \u2202\u03c6/\u2202z. They form the components of a vector \ufb01eld, known as the gradient of \u03c6, and written grad \u03c6, or \u2207\u03c6 (\u2018del \u03c6\u2019, or occasionally \u2018nabla \u03c6\u2019). To show that they really are the components of a vector, we have to show that it is de\ufb01ned in a manner which is independent of the choice of axes. We note that if r and r + dr are two neighbouring points, then the di\ufb00erence between the values of \u03c6 at these points is d\u03c6 = \u03c6(r + dr) \u03c6(r) = \u2212 \u2202\u03c6 \u2202x dx + \u2202\u03c6 \u2202y dy + \u2202\u03c6 \u2202z dz = dr \u00b7 \u2207\u03c6. (A.20) is \ufb01xed, then this scalar product takes on its Now, if the distance maximum value when dr is in the direction of \u2207\u03c6. Hence we conclude that the direction of \u2207\u03c6 is the direction in which \u03c6 increases most rapidly. dr | | Vectors 391 Moreover, its magnitude is the rate of increase of \u03c6 with distance in this direction. (This is the reason for the name \u2018gradient\u2019.) Clearly, therefore, we could de\ufb01ne \u2207\u03c6 by these properties, which are independent of any choice of axes. We are often interested in the value of the scalar \ufb01eld \u03c6 evaluated at the position of a moving particle, \u03c6(r(t)). From (A.20) it follows that the rate of change of \u03c6 is d\u03c6(r(t)) dt = \u02d9r \u00b7 \u2207\u03c6. (A.21) The symbol \u2207 may be regarded as a vector which is also a di\ufb00erential operator (like d/dx), given by \u2207 = i \u2202 \u2202x + j \u2202 \u2202y + k \u2202 \u2202z. We can also apply it to a vector \ufb01eld A. The divergence of A is de\ufb01ned to be the scalar \ufb01eld div A = \u2207 A = \u00b7 \u2202Ax \u2202x + \u2202Ay \u2202y + \u2202Az \u2202z, (A.22) and the curl of A to be the vector \ufb01eld i j k curl A = \u2207 \u2227. (A.23) This latter expression is an abbreviation for the expanded form A = i \u2207 \u2227 \u2202Az \u2202y \u2212 \u2202Ay \u2202z + j \u2202Az \u2202x + k \u2202Ay \u2202x \u2212 \u2202Ax \u2202y. (Instead of curl A, the alternative notation rot A is sometimes used, particularly in non-English-speaking countries.) To understand the physical signi\ufb01cance of these operations, it is", " helpful to think of the velocity \ufb01eld in a \ufb02uid: v(r) is the \ufb02uid velocity at the point r. Let us consider a small volume of \ufb02uid, \u03b4V = \u03b4x \u03b4y \u03b4z, and try to \ufb01nd its rate of change as it moves with the \ufb02uid. Consider \ufb01rst the length \u03b4x. To a \ufb01rst approximation, over a short time interval dt, the velocity components in the y and z directions are irrelevant; the length \u03b4x changes because the x components of velocity, vx, at its two ends are slightly di\ufb00erent, by \u2202 \u2202 \u2202 \u2202x \u2202y \u2202z Ax Ay Az A = \u2202Ax \u2202z \u2212 392 Classical Mechanics an amount (\u2202vx/\u2202x) \u03b4x. Thus in a time dt, the change in \u03b4x is d \u03b4x = (\u2202vx/\u2202x) \u03b4x dt, whence d \u03b4x dt = \u2202vx \u2202x \u03b4x. Taking account of similar changes in \u03b4y and \u03b4z, we have d dt (\u03b4x \u03b4y \u03b4z) = \u2202vx \u2202x + \u2202vy \u2202y + \u2202vz \u2202z \u03b4x \u03b4y \u03b4z, or, equivalently, d \u03b4V dt = (\u2207 \u00b7 v) \u03b4V. (A.24) \u00b7 \u00b7 v represents the proportional rate of increase of volume: positive v compression. In particular, if the v means expansion, negative \u2207 Thus \u2207 \u2207 \ufb02uid is incompressible, then \u2207 \u00b7 It is possible to show in a similar way that a non-zero \u2207 v means that locally the \ufb02uid is rotating. This vector, called the vorticity, represents the local angular velocity of rotation (times 2; see Problem 10). \u00b7 v = 0. \u2227 The rule for di\ufb00erentiating products can also be applied to expressions involving \u2207. For example, \u2207 B) is a sum of two terms, in one of \u2227 which \u2207 acts on A only and in the other on B only. The gradient of a product of scalar \ufb01elds can be written (A \u00b7 and similarly \u2207(\u03c6\u03c8) = \u03c8\u2207\u03c6 + \u03c6\u2207\u03c8, \u2207 \u00b7 (\u03c6A) = A \u2207\u03c6 + \u03c6\u2207 A. \u00b7 \u00b7 But, when vector products are involved, we have to remember that the order of the factors as a product of vectors cannot be changed without a\ufb00ecting the signs. Thus we have (A \u2207 \u00b7 \u2227 B) = B (\u2207 A) A \u00b7 \u2212 \u2227 \u00b7 (\u2207 \u2227 B), and, similarly, (\u03c6A) = \u03c6(\u2207 \u2207 \u2227 A) A \u2227 \u2212 \u2227 (\u2207\u03c6). Vectors 393 We may apply the vector di\ufb00erential operator \u2207 twice. The divergence of the gradient of a scalar \ufb01eld \u03c6 is called the Laplacian of \u03c6, \u22072\u03c6 = \u2207 \u2207\u03c6 = \u00b7 \u22022\u03c6 \u2202x2 + \u22022\u03c6 \u2202y2 + \u22022\u03c6 \u2202z2. (A.25) Some operations always give zero. Just as a curl of a gradient vanishes, a = 0, we \ufb01nd that the \u2227 For example, its z component is \u2207\u03c6 = 0. \u2207 \u2227 (A.26) \u2202 \u2202x \u2202\u03c6 \u2202y \u2212 \u2202 \u2202y \u2202\u03c6 \u2202x = 0. Similarly, one can show that the divergence of a curl vanishes: (\u2207 \u2207 \u00b7 \u2227 A) = 0. (A.27) An important identity, analogous to the expansion of the vector triple product (A.16), gives the curl of a curl, (\u2207 \u2207 \u2227 \u2227 A) = \u2207(\u2207 A) \u00b7 \u2212 \u22072A, (A.28) where of course \u22072A = \u22022A \u2202x2 + \u22022A \u2202y2 + \u22022A \u2202z2. It may easily be proved by inserting the expressions in terms of components. A.6 Integral Theorems There are three important theorems for vectors which are generalizations of the fundamental theorem of the calculus, x1 x0 df dx dx = f (x1", ") f (x0). \u2212 First, consider a curve C in space, running from r0 to r1 (see Fig. A.9). Let the directed element of length along C be dr. If \u03c6 is a scalar \ufb01eld, then according to (A.20), the change in \u03c6 along this element of length is 394 Classical Mechanics r + dr r 1 dr r r 0 Fig. A.9 d\u03c6 = dr \u00b7 integral theorems, \u2207\u03c6. Thus, integrating from r0 to r1, we obtain the \ufb01rst of the r 1 r 0 dr \u00b7 \u2207\u03c6 = \u03c6(r1) \u03c6(r0). \u2212 (A.29) The integral on the left is called the line integral of \u2207\u03c6 along C. (Note that, as here, it is often more convenient to place the di\ufb00erential symbol dr to the left of the integrand.) This theorem may be used to relate the potential energy function V (r) for a conservative force to the work done in going from some \ufb01xed point r0, where V is chosen to vanish, to r. Thus, if F = \u2207V, then \u2212 r dr \u2212 r 0 \u00b7 V (r) = F. (A.30) When F is conservative, this integral depends only on its end-points, and not on the path C chosen between them. Conversely, if this condition is satis\ufb01ed, we can de\ufb01ne V by (A.30), and the force must be conservative. The condition that two line integrals of the form (A.30) should be equal whenever their end-points coincide may be restated by saying the the line integral round any closed path should vanish. Physically, this means that no work is done in taking a particle round a loop which returns to its starting point. The integral round a closed loop is usually denoted by the symbol C. Thus we require & F = 0, dr \u00b7 %C (A.31) for all closed loops C. This condition may be simpli\ufb01ed by using the second of the integral theorems \u2014 Stokes\u2019 theorem. Consider a curved surface S, bounded by the closed curve C. If one side of S is chosen to be the \u2018positive\u2019 side, then the Vectors 395 positive direction round C may be de\ufb01ned by the right-hand-screw convention (see Fig. A.10). Take a small element of the surface, of area dS, and let n C dS S Fig. A.10 n be a unit vector normal to the element, and directed towards its positive side. Then the directed element of area is de\ufb01ned to be dS = n dS. Stokes\u2019 theorem states that if A is any vector \ufb01eld, then dS (\u2207 \u00b7 \u2227 A) = A. dr \u00b7 %C S (A.32) The application of this theorem to (A.31) is immediate. If the line integral round C is required to vanish for all closed curves C, then the surface integral must vanish for all surfaces S. But this is only possible if the integrand vanishes identically. So the condition for a force to be conservative is F = 0. \u2207 \u2227 (A.33) We shall not prove Stokes\u2019 theorem. However, it is easy to verify for a small rectangular surface. (The proof proceeds by splitting up the surface into small sub-regions.) Suppose S is a rectangle in the xy-plane, of area dx dy. Then dS = k dx dy, so the surface integral is k (\u2207 A) dx dy = \u00b7 \u2227 \u2202Ay \u2202x \u2212 \u2202Ax \u2202y dx dy. (A.34) The line integral consists of four terms, one from each edge. The two terms arising from the edges parallel to the x-axis involve the x component of A evaluated for di\ufb00erent values of y. They therefore contribute Ax(y) dx \u2212 Ax(y + dy) dx = \u2202Ax \u2202y \u2212 dx dy. 396 Classical Mechanics Similarly, the other pair of edges yields the \ufb01rst term of (A.34). One can also \ufb01nd a necessary and su\ufb03cient condition for a \ufb01eld B(r) to have the form of a curl, By (A.27), such a \ufb01eld must satisfy B = \u2207 A. \u2227 B = 0. \u2207 \u00b7 (A.35) The proof that this is also a su\ufb03cient condition (which we shall not give in detail) follows much the same lines as before. One can show that it is su\ufb03cient", " that the surface integral of B over any closed surface should vanish: B = 0, (S closed) dS \u00b7 S and then use the third of the integral theorems, Gauss\u2019 theorem. This states that if V is a volume in space bounded by the closed surface S, then for any vector \ufb01eld B, dV \u2207 B = \u00b7 dS B, \u00b7 S V (A.36) where dV denotes the volume element dV = dx dy dz, and the positive side of S is taken to be the outside. It is again easy to verify Gauss\u2019 theorem for a small rectangular volume, dV = dx dy dz. The volume integral is \u2202Bx \u2202x + \u2202By \u2202y + \u2202Bz \u2202z dx dy dz. (A.37) The surface integral consists of six terms, one for each face. Consider the faces parallel to the xy-plane, with directed surface elements k dx dy and B = Bz, evaluated for di\ufb00erent k dx dy. Their contributions involve k \u2212 values of z. Thus they contribute \u00b7 Bz(z + dz) dx dy Bz(z) dx dy = \u2212 \u2202Bz \u2202z dx dy dz. Similarly, the other terms of (A.37) come from the other pairs of faces. Vectors 397 A.7 Electromagnetic Potentials An important application of these theorems is to the electromagnetic \ufb01eld. The basic equations of electromagnetic theory are Maxwell\u2019s equations. In the absence of dielectric or magnetic media, they may be expressed in terms of two \ufb01elds, the electric \ufb01eld E and the magnetic \ufb01eld B. There is one pair of homogeneous equations, E + \u2207 \u2227 \u2202B \u2202t = 0, B = 0, \u2207 \u00b7 (A.38) and a second pair involving also the electric charge density \u03c1 and current density j, 1 0 \u2207 \u00b5\u2212 B 0 \u2212 \u2227 \u2202E \u2202t = j, 0\u2207 \u00b7 E = \u03c1, (A.39) in which \u00b50 and 0 are universal constants. The second equation in (A.38) is just the condition (A.35). It follows that there must exist a vector potential A such that B = \u2207 A. \u2227 (A.40) Then, substituting in the \ufb01rst of the equations (A.38), we \ufb01nd that \u2207 \u2227 (E + \u2202A/\u2202t) = 0. It follows that there must exist a scalar potential \u03c6 such that E = \u2207\u03c6 \u2212 \u2212 \u2202A \u2202t. (A.41) These potentials are not unique. If \u039b is any scalar \ufb01eld, then the po- tentials \u03c6 = \u03c6 + \u2202\u039b \u2202t, A = A \u2207\u039b \u2212 (A.42) de\ufb01ne the same \ufb01elds E and B as do \u03c6 and A. This is called a gauge transformation. We may eliminate this arbitrariness by imposing an extra condition, for example the radiation gauge (or Coulomb gauge) condition A = 0. \u2207 \u00b7 (A.43) In the static case, where all the \ufb01elds are time-independent, Maxwell\u2019s equations separate into a pair of electrostatic equations, and a magnetostatic pair. Then \u03c6 becomes the ordinary electrostatic potential, satisfying 398 Classical Mechanics Poisson\u2019s equation (6.48). The vector potential, by (A.39) and (A.40) satis\ufb01es \u2227 Using (A.28), and imposing the radiation gauge condition (A.43), we \ufb01nd \u2227 \u2207 (\u2207 A) = \u00b50j. \u22072A = \u00b50j. \u2212 (A.44) This is the analogue of Poisson\u2019s equation. The solution is of the same form as (6.15), namely A(r) = \u00b50 4\u03c0 j(r) r| r | \u2212 d3r. (A.45) Thus, given any static distribution of charges and currents, we may calculate the potentials \u03c6 and A, and hence the \ufb01elds E and B. A.8 Curvilinear Co-ordinates Another use of the integral theorems is to provide expressions for the gradient, divergence and curl in terms of curvilinear co-ordinates. Consider a set of orthogonal curvilinear co-ordinates (see 3.5) q1, q2, q3. Let us denote the elements of length along", " the three co-ordinate curves by h1 dq1, h2 dq2, h3 dq3. For example, in cylindrical polars \u00a7 h\u03c1 = 1, h\u03d5 = \u03c1, hz = 1, (A.46) while in spherical polars hr = 1, h\u03b8 = r, h\u03d5 = r sin \u03b8. (A.47) Now consider a scalar \ufb01eld \u03c8, and two neighbouring points (q1, q2, q3) and (q1 +dq1, q2, q3). Then the di\ufb00erence between the values of \u03c8 at these points is \u2202\u03c8 \u2202q1 dq1 = d\u03c8 = dr \u00b7 \u2207\u03c8 = h1 dq1 (\u2207\u03c8)1, where (\u2207\u03c8)1 is the component of \u2207\u03c8 in the direction of increasing q1. Hence we \ufb01nd (\u2207\u03c8)1 = 1 h1 \u2202\u03c8 \u2202q1, (A.48) Vectors 399 with similar expressions for the other components. Thus in cylindrical and spherical polars, we have and \u2207\u03c8 = \u2202\u03c8 \u2202\u03c1, 1 \u03c1 \u2202\u03c8 \u2202\u03d5, \u2202\u03c8 \u2202z, \u2207\u03c8 = \u2202\u03c8 \u2202r, 1 r \u2202\u03c8 \u2202\u03b8, 1 r sin \u03b8 \u2202\u03c8 \u2202\u03d5. (A.49) (A.50) To \ufb01nd an expression for the divergence, we use Gauss\u2019 theorem, applied to a small volume bounded by the co-ordinate surfaces. The volume integral is (\u2207 \u00b7 A) h1 dq1 h2 dq2 h3 dq3. In the surface integral, the terms arising from the faces which are surfaces of constant q3 are of the form A3h1 dq1 h2 dq2, evaluated for two di\ufb00erent values of q3. They therefore contribute \u2202 \u2202q3 (h1h2A3) dq1 dq2 dq3. Adding the terms from all three pairs of faces, and comparing with the volume integral, we obtain A = \u2207 \u00b7 1 h1h2h3 \u2202(h2h3A1) \u2202q1 + \u2202(h3h1A2) \u2202q2 + \u2202(h1h2A3) \u2202q3 (A.51). In particular, in cylindrical and spherical polars, A = \u2207 \u00b7 1 \u03c1 \u2202(\u03c1A\u03c1) \u2202\u03c1 + 1 \u03c1 \u2202A\u03d5 \u2202\u03d5 + \u2202Az \u2202z, (A.52) and A = \u2207 \u00b7 1 r2 \u2202(r2Ar) \u2202r + 1 r sin \u03b8 \u2202(sin \u03b8 A\u03b8) \u2202\u03b8 + 1 r sin \u03b8 \u2202A\u03d5 \u2202\u03d5. (A.53) To \ufb01nd the curl, we use Stokes\u2019 theorem in a similar way. Let us consider a small element of a surface q3 = constant, bounded by curves of constant q1 and of q2. Then the surface integral is (\u2207 \u2227 A)3 h1 dq1 h2 dq2. 400 Classical Mechanics In the line integral around the boundary, the two edges of constant q2 involve A1h1 dq1 evaluated for di\ufb00erent values of q2, and so contribute \u2202 \u2202q2 \u2212 (h1A1) dq1 dq2. Hence, adding the contribution from the other pair of edges, we obtain (\u2207 \u2227 A)3 = 1 h1h2 \u2202(h2A2) \u2202(h1A1) \u2202q1 \u2212 \u2202q2, (A.54) with similar expressions for the other components. Thus, in particular, in cylindrical and spherical polars A = \u2207 \u2227 1 \u03c1 \u2202Az \u2202\u03d5 \u2212 \u2202A\u03d5 \u2202z, \u2202A\u03c1 \u2202z \u2212 \u2202Az \u2202\u03c1, 1 \u03c1 \u2202(\u03c1A\u03d5) \u2202\u03c1 \u2212 \u2202A\u03c1 \u2202\u03d5, (A.55) and A = \u2207 \u2227 1 r sin \u03b8 \u2202(sin \u03b8 A\u03d5) \u2202\u03b8 \u2202Ar \u2202\u03d5 \u2212 1 r sin \u03b8 \u2202A\u03b8 \u2202\u03d5 \u2212 \u2202(rA\u03d5) 1 \u2202r r,, \u2202(rA\u03b8", ") \u2202r \u2212 1 r \u2202Ar \u2202\u03b8. (A.56) Finally, combining the expressions for the divergence and gradient, we can \ufb01nd the Laplacian of a scalar \ufb01eld. It is \u22072\u03c8 = 1 h1h2h3 \u2202 \u2202q1 h2h3 h1 \u2202\u03c8 \u2202q1 + \u2202 \u2202q2 + h3h1 h2 \u2202 \u2202q3 \u2202\u03c8 \u2202q2 h1h2 h3 \u2202\u03c8 \u2202q3. In cylindrical polars, \u22072\u03c8 = 1 \u03c1 \u2202 \u2202\u03c1 \u2202\u03c8 \u2202\u03c1 \u03c1 + 1 \u03c12 \u22022\u03c8 \u2202\u03d52 + \u22022\u03c8 \u2202z2, and, in spherical polars, (A.57) (A.58) \u22072\u03c8 = 1 r2 \u2202 \u2202r r2 \u2202\u03c8 \u2202r + 1 r2 sin \u03b8 \u2202 \u2202\u03b8 sin \u03b8 \u2202\u03c8 \u2202\u03b8 + 1 r2 sin2 \u03b8 \u22022\u03c8 \u2202\u03d52. (A.59) Vectors 401 A.9 Tensors Scalars and vectors are the \ufb01rst two members of a family of objects known collectively as tensors, and described by 1, 3, 9, 27,... components. Scalars and vectors are called tensors of valence 0 and valence 1, respectively. (Sometimes the word rank is used instead of \u2018valence\u2019, but there is then a possibility of confusion with a di\ufb00erent usage of the same word in matrix theory.) In this section, we shall be concerned with the next member of the family, the tensors of valence 2, often called dyadics. We shall use the word \u2018tensor\u2019 in this restricted sense, to mean a tensor of valence 2. Tensors occur most frequently when one vector b is given as a linear function of another vector a, according to the matrix equation bx by bz \u23a1 = \u23a1 \u23a4 Txx Txy Txz Tyx Tyy Tyz Tzx Tzy Tzz ax ay az. \u23a4 \u23a4 \u23a1 (A.60) \u23a3 \u23a6 \u23a3 \u23a6 \u23a3 \u23a6 An example is the relation (9.17) between the angular momentum J and angular velocity \u03c9 of a rigid body. The nine elements of the 3 3 matrix in (A.60) are the components of a tensor, which we shall denote by the sans-serif capital T. By an obvious extension of the dot product notation for the scalar product of two vectors, we may write (A.60) as \u00d7 b = T a. \u00b7 (A.61) For example, (9.17) may be written J = I \u03c9, where I is the inertia tensor. We can go on to form the scalar product of (A.61) with another vector, a. Note that in general this is not the same as T \u00b7 c, obtaining a scalar, c c. In factT \u00b7 a, \u00b7 (A.62) where \u02dcT is the transposed tensor of T, obtained by re\ufb02ecting in the leading diagonal, e.g., \u02dcTxy = Tyx. The tensor T is called symmetric if \u02dcT = T, i.e., if Tji = Tij for all i, j. It is antisymmetric if \u02dcT = T, or Tji = Tij for all i, j. \u2212 \u2212 An interesting example of an antisymmetric tensor is provided by the relation (5.2) giving the velocity v as a function of position r in a body rotating with angular velocity \u03c9. It is a linear relation and so may be 402 Classical Mechanics written in the form (A.60), speci\ufb01cally as vx vy vz \u23a1 = \u23a4 \u23a1 \u03c9z \u03c9y 0 \u2212 \u03c9z \u03c9x 0 \u2212 0 \u03c9y \u03c9x \u2212 x y z \u23a1 \u23a4 \u23a6 \u23a3. \u23a4 \u23a6 \u23a3 There is an important special tensor \u23a1, \u23a4 called the unit tensor or identity tensor, with the property that 1 for all vectors a(b From any two vectors a and b, we can form a tensor T by multiplying their elements together (without adding), i.e., Tij = aibj. This is the tensor product (or dy", "adic product or outer product ) of a and b, written T = ab, with no dot or cross. Note that T c), so the brackets are c = (ab) in fact unnecessary. In matrix notation, ab is the product of the column vector a and the row vector b, while the scalar product (or inner product ) a b is the row a times the column b. We can deduce the correct transformation law of a tensor under a rotation of axes: its components transform just like the products of components of two vectors. If we symbolize (A.2) formally as a = R a, then the correct \u02dcR. (This denotes a product of transformation law of a tensor is T = R three 3 3 matrices.) T \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 The use of the tensor product allows us to write some old results in a new way. For example, for any vector a, 1 \u00b7 a = a = i(i a) + j(j \u00b7 \u00b7 a) + k(k \u00b7 a) = (ii + jj + kk) a, \u00b7 whence ii + jj + kk = 1, (A.63) as may easily be veri\ufb01ed by writing out the components. Similarly, we may write the relation (9.16) between angular momentum and angular velocity in the form J = m(r2\u03c9 rr \u00b7 \u2212 \u03c9) = I \u03c9, \u00b7 where the inertia tensor I is given explicitly by I = m(r21 rr). \u2212 \u00d7 Vectors 403 Note the di\ufb00erence between the unit tensor 1 and the inertia tensor I. It is easy to check that the nine components of this equation reproduce the relations (9.15). Note that if T = ab, then \u02dcT = ba, whence in particular the inertia tensor I is symmetric. A.10 Eigenvalues; Diagonalization of a Symmetric Tensor In this section, we discuss a theorem that has very wide applicability. Let T be a symmetric tensor. A vector a is called an eigenvector of T, with eigenvalue \u03bb, if a = \u03bba, T \u00b7 (A.64) or, equivalently (T a = 0. (Compare (11.17), which is also an eigenvalue equation.) The condition for the existence of a non-trivial solution is that the determinant of the coe\ufb03cients vanishes, \u03bb1) \u2212 \u00b7 det(T \u03bb1) = \u2212 \u03bb Txx \u2212 Tyx Tzx Txy Tyy \u2212 Tzy \u03bb Txz Tyz Tzz \u2212 = 0. \u03bb This is a cubic equation for \u03bb. Its three roots are either all real, or else one real and one complex conjugate pair. However, for a symmetric tensor T with real elements the latter possibility can be ruled out. To see this, suppose that \u03bb is a complex eigenvalue, and let a be the corresponding eigenvector, whose components may also be complex. Now, taking the complex conjugate of T a\u2217 = \u03bb\u2217a\u2217, where \u03bb\u2217 denotes the complex conjugate of \u03bb, and a\u2217 = (a\u2217x, a\u2217y, a\u2217z). Multiplying these two equations by a\u2217 and a respectively, we obtain a = \u03bba, we obtain T \u00b7 \u00b7 a\u2217 T \u00b7 \u00b7 a = \u03bba\u2217 a, \u00b7 and T a \u00b7 \u00b7 a\u2217 = \u03bb\u2217a a\u2217. \u00b7 2 + ay| | But since T is symmetric, the left-hand sides of these equations are equal, a = by (A.62). Hence the right-hand sides must be equal too. Since a\u2217 2 + a\u2217, this means that \u03bb\u2217 = \u03bb, i.e., \u03bb must be real. ax| | Thus we have shown that there are three real eigenvalues, say \u03bb1, \u03bb2, \u03bb3, and three corresponding real eigenvectors, a1, a2, a3. (We consider the case where two eigenvalues are equal below.) Next, we show that the eigenvectors are orthogonal. For, if 2 = a az| | \u00b7 \u00b7 a1 = \u03bb1a1, T \u00b7 a2 = \u03bb2a2, T \u00b7 404 Classical Mechanics then, multiplying the \ufb01rst equation by a2 and the second by a1, and again using the symmetry of T, we obtain a1 = \u03bb2a1 \u00b7 a2. \u03bb1a2 \u00b7 a2 = 0. = \u03bb2, then a1 \u00b7 Thus if \u03bb1 If", " all three eigenvalues are distinct, then the three eigenvectors are orthogonal. Moreover, it is clear that if a is an eigenvector, then so is any multiple of a, so that we may choose to normalize it, de\ufb01ning e1 = a1/a1. Then the three normalized eigenvectors form an orthonormal triad, e1, e2, e3. If we choose these as axes, then T must take the diagonal form T = \u03bb1 0 0 0 \u03bb2 0 0 0 \u03bb3 \u23a1. \u23a4 (A.65) \u23a3 \u23a6 For, since e1 = (1, 0, 0), T \u00b7 be \u03bb1e1 = (\u03bb1, 0, 0). Similarly for the other columns. e1 is simply the \ufb01rst column of T, and this must This relationship between T and the eigenvectors may also be expressed, using the tensor-product notation, in a co-ordinate-independent form, namely T = \u03bb1e1e1 + \u03bb2e2e2 + \u03bb3e3e3. (A.66) Finally, we have to show that these results still hold if two or three eigenvalues coincide. The simplest way to do this is to add a small quantity to one of the diagonal components of T, to make the eigenvalues slightly di\ufb00er= 0, the tensor must have three orthonormal eigenvectors. ent. So long as By continuity, this must still be true in the limit 0. (The symmetry of T is important here, because without the consequent orthogonality of eigenvectors we could not exclude the possibility that two eigenvectors that Indeed, this does are distinct for happen for non-symmetric tensors, as will be seen in a di\ufb00erent context in Appendix C.) = 0 have the same limit as \u2192 \u2192 0. We have shown, therefore, that any symmetric tensor may be diagonalized by a suitable choice of axes. This was the result we used for the inertia tensor in Chapter 9. In that case, the eigenvectors are the principal axes, and the eigenvalues the principal moments of inertia. The procedure for 11.2, \ufb01nding normal co-ordinates for an oscillating system, discussed in is essentially the same. In that case, it is the potential energy function that is brought to \u2018diagonal\u2019 form. Eigenvalue equations also appear in the \u00a7 Vectors 405 analysis of dynamical systems in Chapter 13 and in many other branches of physics, in particular playing a big role in quantum mechanics. Problems 1. Given a = (3, 1, 2), b = (0, 1, 1) and c = (2, 2, 1), \ufb01nd: \u2212 (a) a b, a \u00b7 \u00b7 (b) a b, a \u2227 (c) (a b) \u2227 (d) (a b) \u2227 c)b (e) (a \u00b7 \u00b7 \u2227 \u2212 \u2212 c and a (b + c); \u00b7 c and a (b + c); \u2227 \u2227 c) c and (a b; \u2227 \u00b7 c and (a c) \u2227 \u2227 c)a and (a (b b; b)c \u00b7 2. Find the angles between the vectors a \u2227 and b, where a, b, c are as in Problem 1. \u2212 \u00b7 \u00b7 (b c)a. b and c, and between a c \u2227 3. Show that c = (ab + ba)/(a + b) bisects the angle between a and b, where a and b are any two vectors. \u2207\u03c6 = 0, and evaluate \u22072\u03c6. 4. Find \u2207\u03c6 if \u03c6 = x3 5. (a) Find the gradients of u = x + y2/x and v = y + x2/y, and show xyz. Verify that \u2207 \u2212 \u2227 that they are always orthogonal. (b) Describe the contour curves of u and v in the xy-plane. What does (a) tell you about these curves? 6. Draw appropriate \ufb01gures to give geometric proofs for the following laws of vector algebra: (a + b) + c = a + (b + c); \u03bb(a + b) = \u03bba + \u03bbb; a \u00b7 (b + c) = a b + a c. \u00b7 \u00b7 c. Hence show that \u00b7 (Note that a, b, c need not be coplanar.) d (c 7. Show that (a d) = a c b b) (a \u2227 b)2", " = a2b2 (a \u2227 8. Express \u2207 9. If the vector \ufb01eld v(r) is de\ufb01ned by v = \u03c9k \u00b7 (a b) in terms of scalar products. \u2227 b)2, verify that \u2207 v = 0, \u00b7 \u2227 \u2227 and evaluate the vorticity \u2207 v. 10. *Show that, if u and v are scalar \ufb01elds, the maxima and minima of u on the surface v = 0 are points where \u2207u = \u03bb\u2207v for some value of \u03bb. Interpret this equation geometrically. (Hint : On v = 0 only two coordinates can vary independently. Thus \u03b4z for example can usually be expressed in terms of \u03b4x and \u03b4y. We require that \u03b4u should vanish for 406 Classical Mechanics all in\ufb01nitesimal variations satisfying this constraint.) Show that this problem is equivalent to \ufb01nding the unrestricted maxima and minima \u03bbv as a function of the four independent of the function w(r, \u03bb) = u variables x, y, z and \u03bb. Here \u03bb is called a Lagrange multiplier. What is the role of the equation \u2202w/\u2202\u03bb = 0? \u2212 11. *Evaluate the components of \u22072A in cylindrical polar co-ordinates by using the identity (A.28). Show that they are not the same as the scalar Laplacians of the components of A. 12. *Find the radiation-gauge vector potential at large distances from a circular loop of radius a carrying an electric current I. [Hint : Consider \ufb01rst a point (x, 0, z), and expand the integrand in powers of a/r, keeping only the linear term. Then express your answer in spherical polars.] Hence \ufb01nd the magnetic \ufb01eld \u2014 the \ufb01eld of a magnetic dipole. Express the results in terms of the magnetic moment \u00b5, a vector normal to the loop, of magnitude \u00b5 = \u03c0a2I. 13. *Calculate the vector potential due to a short segment of wire of directed length ds, carrying a current I, placed at the origin. Evaluate the corresponding magnetic \ufb01eld. Find the force on another segment, of length ds, carrying current I, at r. (To compute the force, treat the current element as a collection of moving charges.) Show that this force does not satisfy Newton\u2019s third law. (To preserve the law of conservation of momentum, one must assume that, while this force is acting, some momentum is transferred to the electromagnetic \ufb01eld.) 14. *Given u = cos \u03b8 and v = ln r, evaluate A = u\u2207v divergence and curl of A, and verify that \u2207 that \u2207 A = 2\u2207u \u2207v. \u00b7 \u2227 \u2227 r = R 15. *Show that the rotation which takes the axes i, j, k into i, j, k may r, where the tensor R is R = ii+ jj + kk. be speci\ufb01ed by r Write down the matrix of components of R if the rotation is through an angle \u03b8 about the y-axis. What is the tensor corresponding to the rotation which takes i, j, k back into i, j, k? Show that \u02dcR R = 1. (Such tensors are said to be orthogonal.) \u2192 \u00b7 \u00b7 16. *The trace of a tensor T is the sum of its diagonal elements, tr(T) = i Tii. Show that the trace is equal to the sum of the eigenvalues, and that the determinant det(T) is equal to the product of the eigenvalues. \" T) = 17. *The double dot product of two tensors is de\ufb01ned as S:T = tr(S j Sij Tji. Evaluate 1 : 1 and 1 : rr. Show that \u00b7 i \" \" (3rr \u2212 r 21) : (rr 1 3 r21) = 3(r r)2 \u00b7 \u2212 r 2r2. \u2212 \u2212 A = u\u22072v v\u2207u. Find the v\u22072u and \u2212 Vectors 407 Hence show that the expansion (6.19) of the potential may be written 1 \u00b7 2 rr 1 r Q : + d \u03c6(r) = 3 r21 \u2212 r5 1 q 4\u03c00 r r3 + 1 and write down an expression for the quadrupole tensor Q. Show that tr(Q) = 0, and that in the axially symmetric case it has diagonal el1 2 Q,", " Q, where Q is the quadrupole moment de\ufb01ned in ements Chapter 6. Show also that the gravitational quadrupole tensor is related to the inertia tensor I by Q = tr(I)1 - 3I. 1 2 Q, \u00b7 \u00b7 \u00b7 + \u2212 \u2212, 18. *In an elastic solid in equilibrium, the force across a small area may have both a normal component (of compression or tension) and transverse components (shearing stress). Denote the ith component of force per unit area across an area with normal in the jth direction by Tij. These are the components of the stress tensor T. By considering the equilibrium of a small volume, show that the force across area A with normal in the direction of the unit vector n is F = T nA. Show also by considering the equilibrium of a small rectangular volume that T is symmetric. What physical signi\ufb01cance attaches to its eigenvectors? \u00b7 408 Classical Mechanics Appendix B Conics Conic sections, or conics for short, are most simply de\ufb01ned as curves in a plane whose equation in Cartesian co-ordinates is quadratic in x and y. The name derives from the fact that they can be obtained by making a plane section through a circular cone. They turn up in several physical applications, particularly in the theory of orbits under an inverse square law force. It may be useful to gather together the relevant mathematical information. B.1 Cartesian Form The most general conic would have an equation of the form Ax2 + 2Bxy + Cy2 + 2Dx + 2Ey + F = 0, where A, B,..., F are real constants, but by choosing the axes appropriately we can reduce this to a simpler form. First, we look at the quadratic part, Ax2 + 2Bxy + Cy2. It is always possible by rotating the axes to eliminate the constant B. This is another A.10. The example of the diagonalization process described in 2 + quadratic part of the equation is then reduced to a sum of squares, Ax 2. We then forget about the original co-ordinates, and drop the primes. Cy The nature of the curve is largely determined by the ratio A/C of the new constants. 11.3 and \u00a7 \u00a7 Let us assume for the moment that A and C are both non-zero (we will come back later to the special case where that isn\u2019t true). Then we can choose to shift the origin (adding constants to x and y, e.g., x = x + D/A) so as to remove D and E. If F is also non-zero, we can move it to the other F, to get the standard form of the side of the equation, and divide by \u2212 409 410 Classical Mechanics equation, Ax2 + Cy2 = 1. (B.1) F = 0 is a degenerate case: if A and C have the same sign, the only solution is x = y = 0; if they are of opposite sign, the equation factorizes, and so represents a pair of straight lines, y = \u00b1 We cannot allow both A and C in (B.1) to be negative; the equation A/C x. \u2212 would then have no solutions at all. So we can distinguish two cases: 1. Both A and C are positive. Without loss of generality we can assume that A C. (If A > C, we simply interchange the x and y axes.) De\ufb01ning new positive constants a and b by A = 1/a2 and C = 1/b2, we \ufb01nally arrive at the canonical form of the equation, \u2264 x2 a2 + This is the equation of an ellipse (see Fig. B.1). Here a b; a is the semi-major axis and b is the semi-minor axis. (In the special case y2 b2 = 1. (B.2) \u2265 y b a x Fig. B.1 a = b, we have a circle of radius a.) 2. A and C have opposite signs. Again, without loss of generality, we can 1/b2, assume that A > 0 and C < 0. So, de\ufb01ning A = 1/a2 and C = we get \u2212 x2 a2 \u2212 y2 b2 = 1, (B.3) the equation of a hyperbola (see Fig. B.2); a and b are still called the semi-major axis and the semi-minor axis respectively, although it is no longer necessarily true that a is the larger. Note that this curve has two 411 Conics y b a x Fig. B.2 separate branches on opposite sides of the origin. At large distances, (b/a)x shown it asymptot", "ically approaches the two straight lines y = on the \ufb01gure. \u00b1 We still have to consider the special case where one of the constants A and C vanishes. (They cannot both vanish, otherwise we have simply a linear equation, representing a straight line.) Without loss of generality, we may assume that A = 0 and C = 0. As before, we can shift the origin in the y direction to eliminate E. On the other hand, D cannot be zero (otherwise x doesn\u2019t appear at all in the equation). This time, we can choose the origin y l l/2 x Fig. B.3 in the x direction to make F = 0 (by setting x = x + F/2D). Finally, de\ufb01ning l = D/C, we arrive at the canonical form \u2212 which is the equation of a parabola (see Fig. B.3). y2 = 2lx, (B.4) 412 Classical Mechanics Areas are easy to compute using the Cartesian form of the equation. For example, we can solve (B.2) for y and integrate to \ufb01nd the area of the ellipse; the result, which generalizes the familiar \u03c0r2 for a circle, is \u03c0ab. (In fact, the ellipse may be thought of as a circle of radius a which has been squashed uniformly in the y direction in the ratio b/a.) B.2 Polar Form When we are looking for orbits under the action of a central force, it is usually convenient to use polar co-ordinates. The form of the equation that emerged from the discussion in 4.4 was \u00a7 r(e cos \u03b8 1) = l, \u00b1 (B.5) where e and l are constants satisfying e signs refer to the attractive and repulsive cases, respectively). \u2265 0, l > 0 (the upper and lower It is interesting to see how this form is related to the Cartesian form above. If we rearrange (B.5) and square it, we obtain for both signs the equation x2 + y2 = (l ex)2. \u2212 (B.6) This can easily be put into one of the canonical forms above; which one depends on the value of e. I. If e < 1, we can \u2018complete the square\u2019 in (B.6) and write it as e2)x2 + 2elx + (1 \u2212 e2l2 1 \u2212 e2 + y2 = l2 1 \u2212 e2. Dividing by l2/(1 \u2212 e2), this reduces almost to the form (B.2), where a = l 1 \u2212 e2, b = \u221a1 l \u2212. e2 (B.7) (B.8) The only di\ufb00erence is that the origin is not at the centre of the ellipse: (B.7) is equivalent to (x + ae)2 a2 + y2 b2 = 1, (B.9) an ellipse with centre at ( \u2212 ae, 0). Conics 413 II. If e > 1, we complete the square in the same way and divide by l2/(e2 \u2212 1), obtaining where now (x ae)2 \u2212 a2 y2 b2 = 1, \u2212 a = e2 l \u2212, 1 b = l \u221ae2. 1 \u2212 (B.10) (B.11) This is a hyperbola with centre at (ae, 0). The left-hand branch, intersecting the x-axis at (ae a, 0), corresponds to an orbit under an attractive inverse square law force, while the right-hand one, meeting it at (ae + a, 0), corresponds to the repulsive case. \u2212 III. Finally, if e = 1, the equation can be written y2 = l2 2lx, \u2212 (B.12) which is a parabola with its apex at (l/2, 0), and oriented in the opposite direction to (B.4). In all these cases, the position of the origin is one focus of the conic. In cases I and II there is a second focus symmetrically placed on the other side of the centre; for the parabola, the second focus is at in\ufb01nity. (The plural of focus is foci.) The reason for the name is an intriguing geometric property (see Problem 2): if we have a perfect mirror in the shape of an ellipse light from a source at one focus will converge to the second focus. Similarly, a source at the focus of a parabolic mirror generates a parallel beam, which makes parabolic mirrors ideal for certain applications. For a hyperbolic mirror with a source at one focus, the re\ufb02ected light will appear to come from a virtual image", " at the second focus. Problems 1. The equation (B.2) of an ellipse can be written in parametric form as e2)a2)] that x = a cos \u03c8, y = b sin \u03c8. Show [using the identity b2 = (1 \u2212 ae, 0), are the distances between the point labelled \u03c8 and the two foci, ( a(1 e cos \u03c8), and hence that the sum of the two distances is a constant. (This result provides a commonly used method of drawing an ellipse, by tying a string between two pegs at the foci, stretching it round a pencil, and drawing a curve while keeping the string taut.) \u2213 \u00b1 414 Classical Mechanics 2. *Using the parametrization of the previous question, show that the slope of the curve is given by dy/dx = (b/a) cot \u03c8. Hence show that the \u2212 angles between the curve and the two lines joining it to the foci are equal. (One way is to \ufb01nd the scalar products between the unit vector tangent to the curve and the unit vectors from the two foci. This result provides a proof of the focussing property: light from one focus converges to the other.) [The results stated in Problems 1, 2 imply that all radiation originating at one focus of an ellipse at a particular time is then re\ufb02ected to the other focus with the same time of arrival \u2014 a consequence with many applications, both peaceful and otherwise.] Appendix C Phase Plane Analysis near Critical Points In this appendix we give a summary of the types of behaviour exhibited by a general autonomous dynamical system near critical points in the phase plane (n = 2), as indicated in 13.3. \u00a7 C.1 Linear Systems and their Classi\ufb01cation 13.3 that, in the local expansion near a critical point (x0, y0), We saw in the key to the local behaviour and to the stability of the equilibrium at the critical point is, normally, the behaviour of the linear system \u00a7 \u02d9\u03be \u02d9\u03b7 d dt \u03be \u03b7 \u2261 \u03be \u03b7 = M, (C.1) which is obtained from (13.14) by neglecting higher-order terms in the expansion. The 2 A.10)] has constant entries, which are found as derivatives of the functions F (x, y), G(x, y) evaluated at the critical points (x0, y0), as in (13.11), (13.13). 2 Jacobian matrix M [a tensor of valence 2 ( \u00a7 \u00d7 Consider M = a b c d, where a, b, c, d are real constants. For the critical point itself, at which \u03be = 0, \u03b7 = 0, to be an isolated critical point it is necessary that the determinant of M is non-zero. That is to say ad = 0 and M then has an inverse. If \u2212 this condition is not satis\ufb01ed, so that M is singular, then there is at least a critical line through \u03be = 0, \u03b7 = 0, rather than just the single point; we do not consider this case further here. bc 415 416 Classical Mechanics If we seek a solution to (C.1) in the form then we require \u03be(t) \u03be \u03b7 \u2261 = \u03be0 \u03b70 e\u03bbt \u2261 \u03be0e\u03bbt, M \u03be0 = \u03bb\u03be0 (C.2) (C.3) and this is an eigenvalue/eigenvector problem. may not now be symmetric.) (See \u00a7 A.10, although M Here the eigenvalues \u03bb1, \u03bb2 satisfy the quadratic equation \u03bb2 \u2212 (a + d)\u03bb + (ad bc) = 0, \u2212 that is \u03bb2 \u2212 (trM )\u03bb + (det M ) (\u03bb \u2212 \u2261 \u03bb1)(\u03bb \u2212 \u03bb2) = 0, (C.4) so that \u03bb1 + \u03bb2 = trM, the trace of M and \u03bb1\u03bb2 = det M, the determinant of M (see Appendix A, Problem 16). The eigenvalues \u03bb1, \u03bb2 lead to corresponding eigenvectors \u03be01, \u03be02 respectively, in principle, but, since the matrix M is not necessarily symmetric we have eigenvalues which may not be real and a set of eigenvectors which may not be orthogonal or even complete. There are various cases depending on the nature of the eigenvalues and we can consider separately the cases \u03bb1 1. \u03bb1 write = \u03bb2 and \u03bb1 = \u03bb2. = \u03bb2. In this case, because of the linearity of the system", ", we can \u03be(t) = c1\u03be01e\u03bb1t + c2\u03be02e\u03bb2t, (C.5) with c1, c2 constants. The vectors \u03be01, \u03be02 are independent in this case and any vector can be expressed as a linear combination of them. In \u03be(0) leads to the unique values of c1, c2 corresponding particular \u03be0 \u2261 to given initial conditions. In this situation we can carry out a linear change of variables = S \u03be \u03b7 \u00af\u03be \u00af\u03b7, Phase Plane Analysis near Critical Points 417 similar to the change to normal co-ordinates in d dt \u00af\u03be \u00af\u03b7 = \u03bb1 0 0 \u03bb2 \u00af\u03be \u00af\u03b7 11.4, in such a way that (C.6) \u00a7. \u00d7 2 matrix S = [\u03be01... \u03be02] and the In this similarity transformation the 2 1M S. It should diagonal matrix in (C.6) above then takes the form S \u2212 be noted here that \u03bbi, \u03be0i, ci (i = 1, 2) could be complex, but even then (C.5) is the formal expression of the solution for \u03be. In the case when M is symmetric then the eigenvalues \u03bb1, \u03bb2 are real and the eigenvectors \u03be01, \u03be02 are orthogonal. If the eigenvectors are normalized to have unit length then S\u2212 \u02dcS, i.e. S is a rotation matrix. 1 \u2261 2. \u03bb1 = \u03bb2 ( \u03bb). In this case, \u03bb is necessarily real and we may \ufb01nd that the matrix reduction to diagonal form indicated above may, or may not, be possible: \u2261 (a) If we can \ufb01nd two distinct eigenvectors corresponding to \u03bb then the above machinery will go through trivially, since the matrix M = \u03bbI in this case, where I is the unit matrix, and all non-zero vectors are eigenvectors! (b) If there are not two distinct eigenvectors corresponding to \u03bb then the best that can be done by a linear transformation is to reduce the system to d dt \u00af\u03be \u00af\u03b7 = \u03bb 0 1 \u03bb \u00af\u03be \u00af\u03b7, (C.7) since the diagonal form is not now achievable. The system (C.7) has the solution \u00af\u03be \u00af\u03b7 = c1 0 1 e\u03bbt + c2 e\u03bbt, 1 t (C.8) with c1, c2 constants. Depending on the eigenvalues \u03bb1, \u03bb2 there are then various possible cases to consider. These are listed below together with sketches of typical patterns of local trajectories. The orientation and sense of rotation in these patterns depends on the system concerned. However, in each case the directions of the arrows indicate evolution with time t along the trajectories. 418 Case 1 Classical Mechanics \u03bb1, \u03bb2 real, unequal, same sign = (Fig. C.1). \u21d2 (improper ) node, e.g. negative sign Fig. C.1 All trajectories except for one pair approach the critical point tangent to the same line. The critical point is asymptotically stable. If the sign of \u03bb1, \u03bb2 is positive then the local structure is similar to that above, but with the sense of the arrows reversed. The critical point is then unstable. Case 2 \u03bb1, \u03bb2 real, equal or unequal magnitude, opposite sign = saddle (or hyperbolic point ) (Fig. C.2). This type of critical point is always unstable. \u21d2 Fig. C.2 Phase Plane Analysis near Critical Points 419 Case 3 \u03bb1 = \u03bb2 = \u03bb (necessarily real). 1. When M = \u03bbI we have a (proper ) node, e.g. \u03bb negative (Fig. C.3). This critical point is asymptotically stable. Fig. C.3 If \u03bb is positive then the local structure is similar to that above, but with the sense of the arrows reversed. The critical point is then unstable. (A proper node is sometimes called a star, focus, source or sink as appropriate.) 2. When M may not be diagonalized, so that there is only a single eigenvector corresponding to \u03bb, we have an improper (or in\ufb02ected ) node, e.g. \u03bb negative (Fig. C.4). This critical point is asymptotically stable. Fig. C.4 420 Classical Mechanics If \u03bb is positive then the local structure is similar to that above, but with the sense of the arrows reversed. The critical point is then unstable", ". Case 4 \u03bb1,2 a complex conjugate pair \u00b5 \u00b5 negative (Fig. C.5). \u00b1 i\u03bd, with \u00b5 = 0 = \u21d2 spiral (point ), e.g. Fig. C.5 This critical point is asymptotically stable. If \u00b5 is positive then the local structure is similar to that above, but with the sense of the arrows reversed. The critical point is then unstable. is sometimes called a spiral source or spiral sink as (A spiral appropriate.) Case 5 \u03bb1,2 a pure imaginary conjugate pair (Fig. C.6). \u00b1 i\u03bd = \u21d2 centre (or elliptic point) The sense of the arrows may be di\ufb00erent, but this type of critical point is always stable. We can solve equation (C.4) for the eigenvalues \u03bb1, \u03bb2 in terms of trM and det M obtaining \u03bb1,2 = 1 2 (trM \u221a\u2206), \u00b1 (C.9) where the discriminant \u2206 = (trM )2 4(det M ). We may then represent the \u2212 types of behaviour in the phase plane near a critical point schematically. Phase Plane Analysis near Critical Points 421 Fig. C.6 (See Fig. C.7.) Note that along the line det M = 0 the critical point is not isolated. Fig. C.7 C.2 Almost Linear Systems We have seen that the local analysis near a typical critical point in the phase plane leads to (13.14) and this equation di\ufb00ers from the linear system (C.1) in that it includes some higher-order terms. For the almost linear system (13.14) the classi\ufb01cation of the corresponding linear system (C.1) 422 Classical Mechanics determines the local phase portrait and the type of stability in almost every case. Small changes produced by the higher-order terms are evidently going to be crucial, if at all, only in the particular cases: Case 3 (node) or i (spiral ), where is small. However, the stability of the critical point The equal real eigenvalues \u03bb, \u03bb (node) could split to give \u03bb \u03bb would still be just that predicted by the linear system analysis. \u00b1 \u00b1 Case 5 i(\u03bd + ) (centre) or The pure imaginary conjugate pair of eigenvalues i\u03bd (centre) could become \u00b1 i\u03bd (spiral ), where is small. Naturally a centre \u00b1 would still indicate that the critical point is stable. However, the spiral would be crucially dependent for its stability on the sign of the new real part of the eigenvalue pair. If > 0 then the critical point is unstable, whereas if < 0 then the critical point is asymptotically stable. \u00b1 So, for the systems we are considering, it is only when the exactly linear C.1 predicts that a critical point is a centre that we need to be analysis of suspicious of the predictions of the exactly linear analysis. Whether (13.14) has a true centre or an unstable or asymptotically stable spiral has to be resolved by a closer scrutiny of the particular system in hand. \u00a7 It is the case, in fact, that the trajectories near a critical point in the phase plane have a topological equivalence in the linear and almost linear systems except when there is a zero eigenvalue (i.e. the critical point is not isolated) or when the eigenvalues are pure imaginary (i.e. a centre) \u2014 this is guaranteed by a theorem due to Hartman and Grobman. For example, we can examine the system dx dt dy dt = x, y + x2, = \u2212 (C.10) which has only one critical point (at the origin x = 0, y = 0). For the linear system, in the expansion about the origin we have M = 1 0 1 0 \u2212, Phase Plane Analysis near Critical Points 423 so that the eigenvalues are \u03bb1 = 1, \u03bb2 =, 1 0 1 with eigenvectors \u2212 0 1 respectively. The trajectories near the origin, which is a saddle, are indicated in Fig. C.8(a). For the exact nonlinear system (C.10) we can write dy dx = y x \u2212 + x, (C.11) so that y = x2 3 + c x, with c constant, together with a second solution x = 0 (for all y). The exact family of trajectories near the origin is indicated in Fig. C.8(b). It should be noted that the trajectories which go directly into and directly out of the critical point O (respectively the stable and unstable manifolds) correspond directly at and near O for the exactly linear and almost linear systems \u2014 a general result usually known as the stable manifold theorem. Fig. C.8 C.", "3 Systems of Third (and Higher) Order 13.6, higher-order systems can be analyzed in a similar As we indicated in fashion to that carried out in 13.3 and earlier in this appendix. That is to say the critical points are found and local analysis e\ufb00ected about each of \u00a7 \u00a7 424 Classical Mechanics them by linearization. The resulting eigenvalue/eigenvector problems determine the local stability of the critical points and the local phase portrait structures normally determine the global phase portrait for the complete system. \u00a7 \u00d7 For a third-order system, e.g. the Lorenz system of 3 [again a tensor of valence 2 ( \u00a7 13.6, the matrix M for a particular critical point is 3 A.9,10)], so that the three eigenvalues satisfy a cubic equation. This implies that at least one of the eigenvalues must be real, with the others either both real or a complex conjugate pair. We only note here that, if all the eigenvalues are negative real or have negative real part, then the critical point is asymptotically stable. Even if only one of the eigenvalues is positive or if the complex conjugate pair has positive real part, then the critical point is unstable. Problems 1. Find the critical points of the following systems and classify them ac- cording to their local linear approximations: 3x + y, \u2212 (a) \u02d9x = (b) \u02d9x = 3x + y, (c) \u02d9x = \u2212 (d) \u02d9x = \u2212 (e) \u02d9x = 4 (f) \u02d9x = sin y, \u2212 4x2 \u2212 6x + 2xy 2x \u2212 2y; \u02d9y = 4x \u02d9y = 2x + 2y; 8, \u02d9y = xy; \u02d9y = 3xy; \u02d9y = y2 \u2212 y + 2, y2, \u02d9y = x + x3; y2 \u03b1 \u2212 \u2212 1+x2 \u2212 \u03c92 (g) \u02d9x = y, \u02d9y = x2; \u2212 x in the cases \u03c92 < \u03b1 and \u03c92 > \u03b1. 2. For the nonlinear oscillator equation \u00a8x + x = x3, write \u02d9x = y and show that there are two saddle points and one centre in the linear approximation about the critical points in the (x, y) phase plane. Integrate the equations of the system to obtain an \u2018energy\u2019 equation and use this to show that (a) the centre is a true centre for the full system; (b) the equation of the separatrices through the saddles is 2y2 = x2(x2 2) + 1. \u2212 Appendix D Discrete Dynamical Systems \u2014 Maps In this appendix we consider discrete dynamical systems in which a space is e\ufb00ectively mapped onto itself repeatedly. We recognized in Chapter 13 that for some systems it is appropriate and useful to observe at discrete time intervals, which are not necessarily equal \u2014 this is, for example, often the case for biological systems. \u00a7 Also we saw in 14.2 the concept of a Poincar\u00b4e return map, where the evolution of a system through its dynamics induces a map of a Poincar\u00b4e section onto itself. Examining properties of maps, in their own right, will give insight into mechanisms of chaotic breakdown in continuous systems as well. D.1 One-dimensional Maps We consider a map given by xn+1 = F (xn), (D.1) for n = 0, 1, 2,... and with F a known function, and consider possible behaviours of xn for suitable initial values x0, as we iterate to \ufb01nd successively F (2)(x0), etc. We can expect to x1 = F (x0), x2 = F (x1) F (x0) \ufb01nd any \ufb01xed points X as solutions of X = F (X). (D.2) \u2261 \u2261 F To examine the stability of the \ufb01xed point X, we may write xn + n = X, for each n, and, when n is small, we can expand F (xn) in (D.1) in the form F (xn) = F (X n) = F (X) \u2212 \u2212 nF (X) + 1 2 2 nF (X) +..., (D.3) 425 426 Classical Mechanics where F (X) = [dF (x)/dx]x=X, etc. A \ufb01xed point X is asymptotically stable (and therefore an attractor) if \u2022 < 1. We may", " consider di\ufb00erent cases (where n \u2192 F (X) | | 0 < F (X) n+1 < 1 = | \u21d2 | \ufb01rst-order convergence. F (X) = 0, F (X) have second-order convergence. n+1 \u2212 F (X)n as n 2 F (X)2 = 0 = \u21d2 n as n \u2192 \u221e \u2022 1 0):, and we have, and we \u2192 \u221e While this sequence may be continued, the key criterion is that stated above > 1 leads to instability of X, for and that the case = 1 depends more speci\ufb01cally on the function F (X).. We note that the case F (X) | | F (X) | | F (X) | | A familiar example of what is normally second-order convergence is the Newton\u2013Raphson iteration process to \ufb01nd roots of a single equation f (xn)/f (xn) and each root has a basin of f (x) = 0. Here xn+1 = xn \u2212 attraction, so that we can \ufb01nd all the roots by judicious choices of x0. The very simplest map is the linear map: xn+1 = rxn, (D.4) with r constant (and, say, non-negative), and it is evident that xn = rnx0 in this case. Here there are various behaviours depending on r: \u2264 r < 1: xn \u2192 0 for all x0 [asymptotic stability of X = 0]. 0 r = 1: xn = x0 for all x0 [steady state]. r > 1: xn = x0 exp(n ln r) [exponential growth]. \u2022 \u2022 \u2022 If this were a biological model, of e.g. a seasonal breeding population xn, then the rate constant r is crucial in determining the fate of any initial population x0. The logistic map A simple nonlinear map, derived from (D.4), is xn+1 = rxn \u2212 sx2 n, (D.5) which is called the logistic map and has apparent similarity with the logistic di\ufb00erential equation (13.3). In a biological context the rate constant r, quantifying the ability of the population to reproduce, is balanced by the parameter s, which quanti\ufb01es the e\ufb00ect of overcrowding. This model formed the centrepiece of what has become a very in\ufb02uential paper \u2014 \u2018Simple Discrete Dynamical Systems \u2014 Maps 427 mathematical models with very complicated dynamics\u2019, May, Nature, 261, 459\u2013467, 1976. A simple scaling \u00afxn = sxn/r leads to \u00afxn+1 = r\u00afxn(1 \u00afxn), and it is evident that the overbar may then be dropped, in order to \ufb01nd the map \u2212 xn+1 = rxn(1 xn), \u2212 (D.6) which is the logistic map in standard form, with r the single key parameter. Naturally the primary physical/biological interest is in the case where 4 the x interval [0, 1] is mapped by (D.6) onto [0, 1], which requires 0 \u2014 for other applications this r restriction might well be absent. \u2264 \u2264 r Despite the apparent similarity with the continuous system (13.3) the maps (D.5), (D.6) have some very di\ufb00erent and complex properties. We see immediately that there are two \ufb01xed points of (D.6) \u2014 at X = 0, X = 1 1/r \u2014 and, in each case, F (X) = r(1 2X). \u2212 \u2212 It is helpful for maps to consider [see Fig. D.1(a)] a plot of y = x and y = F (x) so that by tracing the vertical and horizontal lines between the..., and thus see whether two we can follow the sequence x0 \u2192 or not it might converge. x2 \u2192 x1 \u2192 Fig. D.1 The behaviours of the map (D.6) for di\ufb00erent values of r can be sum- marized as follows: \u2022 \u2022 r < 1: X = 0 is asymptotically stable and X = 1 1/r is unsta[X = 0 is a point attractor \u2014 if the corresponding linear model 0 \u2264 ble. population cannot sustain itself then overcrowding makes it worse!] 1 < r < 3: X = 0 is unstable and X = 1 [X = 1 \u2014 see Fig. D.1(a) for example. 1/r is asymptotically stable 1/r is a point attractor \u2014 \u2212 \u2212 \u2212 428 \u2022", " Classical Mechanics exponential growth is stabilized by overcrowding, in very similar fashion to the behaviour of the logistic di\ufb00erential equation (13.3).] 3 < r 1/r are now both unstable. As r \u2212 increases successive \u2018period-doubling bifurcations\u2019 occur as asymptotic stability is exchanged between lower- and higher-order cycles (termed a supercritical \ufb02ip bifurcation): 4: X = 0 and 6 = 3.449 48 an asymptotically stable : xn \u2192 If we examine xn+2 = F (xn+1) = 2-cycle (or period-2 solution). F (2)(xn), then we obtain an equation of degree 4 for the \ufb01xed points of this iterated mapping. Of course X = 0, X = 1 1/r are two of the roots of this equation. The nontrivial solutions X1, X2, such that F (X1) = X2, F (X2) = X1 [see Fig. D.1(b)], are roots of the quadratic r2X 2 r(r + 1)X + (r + 1) = 0. For asymptotic stability \u2212 < 1 or equivalently, it is necessary (see Problem 1) that for positive r, 3 < r < 1 + \u221a6. 4 + 2r | r2 \u2212 \u2212 | * 3.449 48 \u00b7 \u00b7 \u00b7 < r < 3.544 09 an asymptotically stable 4: xn \u2192 cycle found from an equation of degree 16; four more roots are the 1/r, X1, X2 (the other roots, when real, give trivial solutions 0, 1 \u2212 4-cycles arising through a di\ufb00erent process). \u00b7 \u00b7 \u00b7 * 3.544 09 < r < 3.564 40 an asymptotically stable 8- : xn \u2192 \u00b7 \u00b7 \u00b7 cycle, and so on (each time in a shorter interval in r)... until \u00b7 \u00b7 \u00b7 * r = 3.569 94 * 3.569 94 \u00b7 \u00b7 \u00b7 < r : Accumulation point of 2\u221e-cycle. \u2264 \u00b7 \u00b7 \u00b7 4: For some values of r there are asymptotically stable cycles of di\ufb00erent lengths, but for others the xn values range seemingly over a whole continuous interval, in an apparently random fashion. An intriguing fact is that odd-period cycles only appear for r > 3.678 57. \u00b7 \u00b7 \u00b7 The logistic map attractors are shown in Fig. D.2. The numbers along the top are the cycle periods. It should be noted that this \ufb01gure has an approximate self-similarity at higher magni\ufb01cation, in that the period-doubling cascade is broadly repeated as other-period asymptotically stable cycles become unstable \u2014 e.g. the period-3 cycle in Fig. D.2. That the period-doubling proceeds broadly in geometric fashion in the limit was discovered by Feigenbaum in 1975. Here the r-intervals \u2206i between bifurcations and the measures di of width in X of successive cycles Discrete Dynamical Systems \u2014 Maps 429 Fig. D.2 are such that (see Figs. D.2, D.3) \u2206i/\u2206i+1 \u2192 di/di+1 \u2192 \u03b4 = 4.669 201 6..., \u03b1 = 2.502 907 8..., (D.7) in each case in the limit as i. \u2192 \u221e Feigenbaum noted that essentially all \u2018humped\u2019 mapping functions F (x) lead to intersections of y = x with y = F (x), y = F (2)(x), y = F (4)(x),... which are similar up to a rescaling. In a process of \u2018renormalization\u2019 the precise form of F (x) is lost and in 1976 Feigenbaum discovered a universal function g(x), which is self-reproducing under such rescaling and iteration, so describing this universal property: g(x) = \u2212 \u03b1g g x \u2212 \u03b1 (D.8) (see Feigenbaum, Journal of Statistical Physics, 21, 669\u2013706, 1979). The universality leads to \u03b4, \u03b1 as universal constants and \u03b4 is usually identi\ufb01ed as the Feigenbaum number. 430 Classical Mechanics X 1 2 Schematic attractor diagram \u221e 2 accumulation \u2018Chaos\u2019 and stable windows d1 d3 d2 1\u2206 2\u2206 3\u2206 0 1 3 3.44949 3.54409 3.56995 r Fig. D.3 As to the onset of cycles with other periods", ", there is a theorem, due to Sarkovskii (1964), which is as follows: If a mapping function F (x) has a point xp which is cyclic of order p, then it must also have a point xq of period q, for every q which precedes p in the sequence: 1 4 2 \u21d0 \u21d0 \u21d0 \u00b7 \u00b7 \u00b7 \u21d0 2n \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 2m.9 2m.7 2m.5 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 22.9 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 22.7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 22.5 2.9 9 2.7 7 2.5 5 \u21d0 \u21d0 \u21d0 \u21d0 \u21d0 \u21d0 \u21d0 \u21d0 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u21d0 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u21d0 \u00b7 \u00b7 \u00b7 \u21d0 \u00b7 \u00b7 \u00b7 \u21d0 \u21d0 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 2m.3 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 22.3 2.3 3. \u21d0 \u21d0 \u21d0 \u21d0 For example, the existence of a 3-cycle implies the existence of cycles of all the other periods! Naturally the odd periods cannot arise from period-doubling, but do so via a rather di\ufb00erent process which may be examined by similar methods to those employed above. This theorem can be proved using a continuity/intermediate value theorem argument. However it says nothing about stability of these cycles, or the ranges of r (in our logistic example) for which they may be observed. The vast majority of these cycles are unstable when all are present, and it is this which leads to a brief summary statement in the form \u2018Period 3 Discrete Dynamical Systems \u2014 Maps 431 implies chaos\u2019, as essentially random behaviour of iterates xn occurs (Li and Yorke, American Mathematical Monthly, 82, 985\u2013992, 1975, in which, incidentally, the term chaos was \ufb01rst introduced!). What can be said about values of r in the logistic map which lead to distributions of iterates, rather than to asymptotically stable cycles? Analytically this is a tough problem. However, it happens that there are two positive values of r for which a formal exact solution of (D.6) is known \u2014 r = 2 and r = 4. The former (r = 2) allows (1 n 2xn+1) = (1 ]. This is not especially interesting, since xn \u2192 2xn)2 leading to xn = 1 2 as we \u2212 \u2212 1 2 [1 (1 should expect. \u2212 \u2212 2x0)2 \u2261 \u2212 cos(2\u03c0\u03b8n)] However the latter possibility (r = 4) allows us to substitute xn = sin2(\u03c0\u03b8n) and this leads us to \u03b8n+1 = 2\u03b8n and then 1 2 [1 \u03b8n = 2n\u03b80. Here \u03b8 is evidently periodic, with period 1, in that the same x is generated by \u03b8 and by \u03b8 + 1. Thus we may write any \u03b80 we choose in a binary representation just using negative powers of 2 \u2014 for example, \u03b80 = 1 = 0.101 101.... Then \u03b81 = 0.011 01..., \u03b82 = 0.110 1..., etc, since the integer part may be cancelled at each stage on account of the periodic property. 16 + 1 8 + 1 2 + 1 64 + \u00b7 \u00b7 \u00b7 For almost all choices of \u03b80 then the \u03b8n will be uniformly distributed on the interval [0, 1], since each digit in the binary expansion of \u03b80 could be chosen with equal likelihood to be a 0 or a 1. As a consequence of \u03b8 being uniformly distributed, then x is not. Indeed x has a probability distribution given by P (x) = 1/[\u03c0 x)] (see Figs. D.4, D.2). x(1 \u2212 P(x) = 1/ [\u03c0\u221ax(1\u2212x)] 0 1 x Fig. D.4 432 Classical Mechanics We note that P (x) itself is the attractor here for r = 4, in the sense that for almost all choices of x0 the distribution of xn will approach P (x) for large n. It is called an invariant probability distribution. For other values of r there are theoretical results for the corresponding probability distribution attractors (via an equation for the distribution function \u2014 Perron\u2013Frobenius), but no slick result like that above for r = 4. The distribution attractors are also characterized by an exponential divergence of iterates, leading to the sensitivity to initial conditions characteristic of chaos. If we choose to examine x0 and x0 + 0", ", with 0 very small, 0 exp(\u03bbn) on average, and we have divergence or convergence of then n iterates according as \u03bb > 0 or \u03bb < 0. Here \u03bb is a Lyapunov exponent (see 13.7). Since we have F (n)(x0 + 0) F (n)(x0) \u2212 0e\u03bbn, then we have \u03bb lim n \u2192\u221e \u2019 = lim n \u2192\u221e 1 n 1 n d dx F (n)(x). |) F (xi) | ln n 1 \u2212 ln i=0 ( (D.9) \u00a7 Values of \u03bb can be found by numerical computation. In measuring \u03bb empirically in a particular practical case, we allow n to become large enough so that our estimate of \u03bb can settle down to a steady value. We also average over various di\ufb00erent x0, in order to avoid an atypical result through a single unfortunate choice. For the logistic map Fig. D.5 shows a plot of \u03bb as r varies. Here F (x) = r(1 2x), so that we expect: \u2212 0 < r < 1: \u03bb = ln r, r 2 1 < r < 3: \u03bb = ln | \u2212 | 3 < r < 1 + \u221a6: \u03bb = 1 2 ln r(1 2 ln | in\ufb01nite spike when X1 = 1 2 (requiring r = 1 + \u221a5 = 3.236)], x(1 r = 4: since P (x) = 1/[\u03c0 x)] so we have 2X1) | 2X2) | r(1 | + 1 (leading to an in\ufb01nite spike when r = 2), \u2212 \u2212 [leading to an \u2212 \u2022 \u2022 \u2022 \u2022 \u03bb = 1 0 (ln )P (x) dx = F (x) | | 1 ln 0 4(1 | \u03c0 \u2212 x(1 2x) | x) dx \u2212 = ln 2 = 0.693 147.... In any event, the regions of r for which there are positive values of \u03bb indicate the sensitivity to initial conditions at these r values. While there are other measures of the complexity/disorder of the iterates in these cases \u2014 Discrete Dynamical Systems \u2014 Maps 433 Fig. D.5 1996)] [Baker and Gollub (Chaotic Dynamics, 2nd ed., Cambridge University Press, e.g. the system \u2018entropy\u2019 \u2014 these will not be pursued further here. However, see Problem 4 for another (and simpler) example of a chaotic map. The logistic map (D.6), together with other similar maps, involves stretching and folding, so that states diverging exponentially are still broadly con\ufb01ned to a bounded region. The loss of information about initial conditions, as the iteration process proceeds in a chaotic regime, is associated with the non-invertibility of the mapping function F (x), i.e. while xn+1 is uniquely determined from xn, each xn can come from 2 possible 2, etc, and eventually from 2n possible x0. Hence xn system memory of initial conditions becomes blurred! 1, 4 possible xn \u2212 \u2212 Many continuous systems \u2014 for example the Lorenz system of exhibit a similar period-doubling in their dynamics. 13.6 \u2014 \u00a7 D.2 Two-dimensional Maps For two-dimensional maps any new universality has proved harder to \ufb01nd! However, there are interesting phenomena. We may write our generic map in the form xn+1 = F (xn, yn), yn+1 = G(xn, yn), (D.10) 434 Classical Mechanics for integer n and with F, G known functions. Again we may seek \ufb01xed points (X, Y ) as solutions of X = F (X, Y ), Y = G(X, Y ). (D.11) By a Taylor expansion, near X, Y and similar to that carried out in and in Appendix C, we obtain \u00a7 13.3 M X Y X Y, with M = xn+1 \u2212 yn+1 \u2212 xn \u2212 yn \u2212 It now follows that (xn, yn) matrix M all have modulus less than 1; this is needed to force M n zero matrix as n (X, Y ) if and only if the eigenvalues of the the \u2202F \u2202y \u2202G \u2202y (x,y)=(X,Y ) \u2202F \u2202x \u2202G \u2202x (D.12) \u2192 \u2192. When instability sets in as parameters are changed, we then typically have a 2-cycle to examine with (X1, Y1) (X2", ", Y2) and where these points are the solutions of. \u2192 \u221e X = F [F (X, Y ), G(X, Y )], Y = G[F (X, Y ), G(X, Y )], (D.13) other than the \ufb01xed points of (D.10) found earlier and which satisfy (D.11). The asymptotic stability of this 2-cycle depends on the eigenvalues of the matrix product M1M2 = \u2202F \u2202x \u2202G \u2202x \u2202F \u2202y \u2202G \u2202y (x,y)=(X1,Y1) \u2202F \u2202x \u2202G \u2202x \u2202F \u2202y \u2202G \u2202y (x,y)=(X2,Y2), (D.14) and so on. The H\u00b4enon map Probably the most celebrated example is the H\u00b4enon map (H\u00b4enon, Communications in Mathematical Physics, 50, 69\u201377, 1976): xn+1 = 1 \u2212 yn+1 = bxn, ax2 n + yn, (D.15) with a, b real parameters, and where the normal physical interest is in | \u2264 1. This map was constructed to exhibit some behaviours similar to those 13.6. Geometrically the map may be considered of the Lorenz system of b | \u00a7 Discrete Dynamical Systems \u2014 Maps 435 to be a composition of three separate simple maps \u2014 an area-preserving fold, a contraction in the x direction, an area-preserving rotation. b | | There are two \ufb01xed points given by 1 [ \u2212 bX \u00b1 \u2212, b)2 + 4a]/2a, b) \u00b1 (1 \u2212 (D.16) b)2. One and these are real and distinct if and only if a > a0 = of the \ufb01xed points is then always unstable and the other is asymptotically stable if a < a1 = 3 b)2 (see Problem 7). For a > a1, both \ufb01xed points are unstable and we get period-doubling to a 2-cycle, 4-cycle, etc. The two-cycle stability is determined through (D.14) by the eigenvalues of the matrix product 4 (1 1 4 (1 \u2212 \u2212 \u2212 M1M2 = 4a2X1X2 + b 2abX2 \u2212 \u2212 2aX1 b. This leads to asymptotic stability of the 2-cycle only when a1 < a < a2 4 (1 + b)2. The period-doubling cascade then continues with a2 = (1 \u2212 (see Fig. D.6). b)2 + 1 b 1 0.3 Henon chaotic attractor * \u2212 = \u2212 \u2212(1\u2212b) 4 0 one stable, one unstable fixed point a 2 stable tw \u22121 2 3 a = \u2212(1\u2212b) 4 1 o-cycle b 0 0.3 a a 1 0.75 0 \u22120.25 \u22120.1225 0.3675 \u221e a(2 ) \u22121.40 \u22121.06 a( ) * 1.40 b=0 corresponds to the one-dimensional logistic map (see Problem 2) Fig. D.6 436 Classical Mechanics Beyond the cascade, and embedded among other-period cycles, there are (a, b) values where the attractor is very complex (see, e.g., Fig. D.7, where a = 1.4, b = 0.3). Fig. D.7 Here there is stretching along the strands of the attractor and squeezing across them \u2014 with associated Lyapunov exponents which are respectively positive (along) and negative (across) (see Problem 8). This chaotic attractor \u2014 termed \u2018strange\u2019 \u2014 has some similar features to the Lorenz attractor of Fig. 13.20. It also has a fractal character in that it has a broad similarity in features and relative scale at all magni\ufb01cations. The repetition of stretching, squeezing and folding onto the original region is characteristic of a technical construct \u2014 the Smale \u2018horseshoe\u2019 (1960) \u2014 which is now known to be a trademark of chaotic systems. Smale \u2018stretched\u2019 the unit square in one co-ordinate direction and \u2018squeezed\u2019 it in the other direction, then placing the resulting strip (with one \u2018fold\u2019) over the original square \u2014 with inevitable overlap. In\ufb01nite repetition of this sequence of operations leads to the identi\ufb01cation of an attractor consisting of all the points which remain within the original square inde\ufb01nitely. This attractor has", " great topological complexity and, while it guarantees the \u2018sensitivity to initial conditions\u2019 of chaos, proving its existence for a particular system may be very tough. Discrete Dynamical Systems \u2014 Maps 437 A formal demonstration of chaotic dynamics for the H\u00b4enon map is contained in a paper by Benedicks and Carleson (Annals of Mathematics, 133, 73\u2013169, 1991). | \u00a7 det M In passing (and for reference in D.3) we note that a map which pre= 1 in (D.12) and then the stretching and serves area would have squeezing exactly compensate each other (see Problem 11). For the H\u00b4enon map this is the case only when = 1. Equal-area maps are of special interest since the Poincar\u00b4e return map for a section through a Hamiltonian 14.2) has the equal-area property. Some consequences are system (see explored brie\ufb02y in D.3. b | \u00a7 | | \u00a7 D.3 Twist Maps and Torus Breakdown \u00a7 In 14.1 we noted that each trajectory of an integrable Hamiltonian system with n degrees of freedom is con\ufb01ned to the surface of an n-torus in the 2ndimensional phase space. The n-tori corresponding to the range of initial conditions are \u2018nested\u2019 in the phase space. \u00a7 In 14.2 we introduced the concept of a Poincar\u00b4e surface of section, as a slice through the dynamical structure. For n = 2 degrees of freedom this section of the nested torus structure is a continuum of closed curves, each one of which is intersected by one of its own torus trajectories in a sequence of points (see Fig. 14.4). If we make use of action/angle variables, as described in 14.3 and particularly in the polar form of Fig. 14.5(c), then the closed curves of the Poincar\u00b4e section can be taken as concentric circles. The intersection points of a trajectory with its own particular one of these circles (of radius r) will necessarily be twisted successively around the origin (the centre of the circle) through an angle 2\u03c0\u03b1. Here \u03b1 is the rotation number, which is the ratio of normal frequencies characterizing the particular torus concerned \u2014 it therefore depends on r. \u00a7 We can then introduce the notion of a twist map, T [Moser (1973)]; in polars: rn+1 \u03b8n+1 = rn \u03b8n + 2\u03c0\u03b1(rn) T \u2261 rn \u03b8n. (D.17) When a perturbation Hamiltonian is introduced, as in 14.6, we can \u00a7 438 Classical Mechanics model the modi\ufb01ed situation using a perturbed twist map, T: rn+1 \u03b8n+1 = rn + f (rn, \u03b8n) \u03b8n + 2\u03c0\u03b1(rn) + g(rn, \u03b8n) T rn \u03b8n, \u2261 (D.18) with small and positive and with f, g known smooth functions. We can now ask what happens to a circle of a particular radius, which is mapped to itself (with a twist) by T, as the perturbation quanti\ufb01ed by in T grows from zero \u2014 i.e. as the integrable system T becomes a near-integrable system T, in such a way that area is still preserved. In fact the answer to this question depends on the rotation number \u03b1 corresponding to our particular circle chosen. If \u03b1 is an irrational number then, for small enough, the circle undergoes some distortion (perturbation) but is certainly not destroyed. This is also true for the corresponding torus in the full phase space. This result is in accord with the KAM theory referred to in 14.6. As we shall see, it is circles for (D.17), and their tori, corresponding to rational \u03b1, which break down under perturbation, leading to sensitivity to initial conditions and chaos. \u00a7 The rational \u03b1 are \u2018scanty, but dense\u2019 among the real numbers (see Problem 12) and these \u03b1 correspond to resonances in the system. As in the 14.6, the sensitivity to discussion of the problem of small denominators in initial conditions is strongest for the rational \u03b1 = k/s with small values of s. As the perturbation grows with, the breakdown associated with each such rational \u03b1 broadens, so that progressive overlap occurs, leading eventually to complete breakdown of the torus structure. \u00a7 Let us examine the twist map (D.17) for three neighbouring circles < k/s, \u03b1 = k/s, \u03b1+ > k/s, C, C+ corresponding respectively to \u03b1 C, \u03b1+ irrational. (We have here (say), with k, s positive integers and with \u03b1", " chosen to take \u03b1 to be an increasing function of radius r.) \u2212 \u2212 \u2212 \u2212 Then applying the twist map T successively s times we \ufb01nd that the e\ufb00ect of T (s) (see Fig. D.8) is to map the circles to themselves, with C invariant and with C, C+ twisted \u2018rigidly\u2019 clockwise, anticlockwise respec< 2\u03c0k and 2\u03c0s\u03b1+ > 2\u03c0k. All points of C tively \u2014 this is so since 2\u03c0s\u03b1 are \ufb01xed under the iterated mapping T (s)., \u03b1+ are irrational the circles C, C+ are only mildly distorted when we apply the map T successively s times instead \u2014 i.e. when we iterate to consider the e\ufb00ect of T (s). However, the inner and outer cir, C+ are still twisted clockwise, anticlockwise respectively \u2014 see cles C Fig. D.9. Since \u03b1 \u2212 \u2212 \u2212 \u2212 Discrete Dynamical Systems \u2014 Maps 439 C + C O C \u2212 Invariant circles under the twist map Fig. D.8 E T C(s(s) + \u03b5 C H C E The action of the perturbed twist map Fig. D.9 Moving outwards radially from O in any direction we can, by continuity, \ufb01nd a point on that radius arm which undergoes no net twist when the iterated map T (s) is applied \u2014 P is mapped only radially (to Q) by this process. By considering all possible radius arms we construct the closed curve consisting of all such points P which are mapped only radially by C T (s) \u2014 to \u00af. C For the map T to model a section of a Hamiltonian dynamical system \u2014 must be one (as, of course, does T itself) the map T \u2014 and hence T (s) of equal-area, as re\ufb02ected in 14.2. \u00a7 440 Classical Mechanics Since the areas contained within the curves must be equal, there is in general an even number of intersections of these curves, which correspond, of course, to \ufb01xed points of the iterated map T (s). These points are all that remains \ufb01xed from the original invariant circle C of the unperturbed iterated map T (s). We now note that the \ufb01xed points are of alternating type as we move around ) \u2014 see Fig. D.10, where the iterated mapping sense of \ufb02ow is indicated. (or \u00af C C C and \u00af C (a) Elliptic point (E) stable (b) Hyperbolic point (H) unstable Fig. D.10 There are evidently in all 2ns such \ufb01xed points of the iterated map T (s), where n is a positive integer (usually 1). The statement of existence of these \ufb01xed points, of their multiplicity and their alternating stability is a result known as the Poincar\u00b4e\u2013Birkho\ufb00 Theorem (1927). It turns out that the elliptic \ufb01xed points E are themselves surrounded (at higher scales) by elliptic and hyperbolic \ufb01xed points corresponding to even higher-order frequency resonances. \u00a7 For the hyperbolic \ufb01xed points H the unstable and stable manifolds (q.v. also in C.2) for neighbouring such points in the same family cross to form what are called homoclinic intersections. The resulting instability at all scales leads inevitably to sensitivity to initial conditions, in that trajectories of the system have to twist and turn, forming a homoclinic tangle, in order not to self -intersect, while maintaining the equal-area property of the return map. This results in the stretching, squeezing and folding associated with the Smale horseshoe referred to in D.2 and this forces a dense interweaving of ordered and chaotic motions within Hamiltonian systems (Fig. D.11). Recognition by Poincar\u00b4e of the existence of such tangles was the \ufb01rst mathematical realization of the presence of what we now call chaos. \u00a7 Discrete Dynamical Systems \u2014 Maps 441 KAM torus KAM torus Schematic chaotic breakdown Hyperbolic point H Elliptic point E Fig. D.11 An example of a map in which some of the breakdown to chaos in a Poincar\u00b4e section is apparent is that for the oval billiard map (Fig. 14.13). For further exploration of breakdown to chaos under parameter change, see Problem 13. Problems 1. For the logistic map (D.6) with r non-negative: \u2264 r < 3. with parameter s ( (a) Show that there is a point attractor for 0 (", "b) Show that there is a two-cycle attractor for 3 < r < 1 + \u221a6. (c) Show that a, b, s can be found such that yn satis\ufb01es the logistic map = r) and xn = a + byn. (d) Hence determine the principal period-doubling bifurcation points for 2 the logistic map on the range \u2212 ay2 n is just the logistic map (D.6) for xn with xn, yn related linearly by yn = \u03b1 + \u03b2xn and a = 1 2). [This is an example of the fact that all quadratic maps yn+1 = A + Byn + Cy2 n are essentially just the logistic map, in that (D.6) can be obtained by a suitably chosen linear relation between yn and xn.] 4 r(r 4. \u2264 \u2212 \u2264 \u2212 r 2. Show that the map yn+1 = 1 3. *Allowing r and xn to be complex in the logistic map (D.6), \ufb01nd regions of the complex r plane for which the map has (a) a point attractor, (b) a 2-cycle attractor. (c) Sketch the corresponding regions when these results are expressed in terms of the complex a plane for the yn map of Problem 2. 442 Classical Mechanics 4. The tent map xn+1 = 2xn, 2(1 \u2212 xn), (0 xn \u2264 \u2264 ( 1 2 < xn \u2264 1 2 ) 1) has unstable \ufb01xed points. Show that this map exhibits extreme sensitivity to initial conditions, in that an uncertainty 0 in x0 is rapidly magni\ufb01ed. Estimate the number of iterations after which the range of uncertainty in the iterates is the complete interval [0, 1]. 5. For the cubic map xn+1 = axn \u2212 x3 n, where a is real, show that, when < 1, there is an asymptotically stable \ufb01xed point X = 0 and that, 1. What a | when 1 < a < 2 there are two such \ufb01xed points at X = happens when a becomes > 2? \u221aa \u2212 \u00b1 | 6. Explore the one-dimensional maps of Problems 1\u20135 using a programmable calculator or (better) a computer. Further interesting examples are xn+1 = exp[a(1 xn)], xn+1 = a sin xn. 7. For the H\u00b4enon map (D.15) show that: \u2212 (a) when 1 4 (1 \u2212 \u2212 b)2 < a < 3 4 (1 \u2212 one of which is asymptotically stable, b)2 < a < (1 b)2 + 1 (b) *when 3 4 (1 \u2212 stable 2-cycle. \u2212 b)2 there are two real \ufb01xed points, 4 (1 + b)2 there is an asymptotically 2aX 1 0 b [Hint : Since M = \u2212 the eigenvalues of M are \u03bb1, \u03bb2 with \u03bbi| | \u2212 b, \u03bb1 + \u03bb2 = \u03bb1\u03bb2 = 2aX. To determine stability, it is useful to \u03bb and look for points consider a sketch of the function f (\u03bb) where f (\u03bb) = 2aX in order to \ufb01nd (a, b) such that both the eigenvalues satisfy < 1.] b/\u03bb \u2261 \u2212 \u2212 8. *Show that a small circle of radius centred at any (X, Y ) becomes a small ellipse under a single iteration of the H\u00b4enon map (D.15). Explain how the semi-axes of the ellipse are related to Lyapunov exponents \u03bb1, \u03bb2 and show that \u03bb1 + \u03bb2 = ln with \u03bb1 > 0 > \u03bb2, implying simultaneous \u2018stretch\u2019 and \u2018squeeze\u2019. b | | 9. *The Lozi map is (D.10) with F (x, y) = 1 + y a, b are real parameters. a, G(x, y) = bx, where x | | \u2212 b | (a) When < 1 and | stable \ufb01xed point. (b) Find the 2-cycle when a | | < 1 stability. b, show that there is one asymptotically \u2212 < 1 and a > 1 b | | \u2212 b and determine its 10. Explore the two-dimensional maps of Problems 7\u20139 using a computer. Discrete Dynamical Systems \u2014 Maps 443 11. By calculating Lyapunov exponents", " examine sensitivity to initial condi- tions of the equal-area maps of the unit square (0 x, y 1): \u2264 \u2264 (a) Arnold\u2019s cat map xn+1 = xn + yn, yn+1 = xn + 2yn (each modulo 1). (b) The baker\u2019s transformation (xn+1, yn+1) = (2xn, 1 (2xn \u2212 12. Consider a system of two degrees of freedom with two natural frequencies \u03c91, \u03c92 in the light of the discussion of periodicity and degeneracy in 14.1,14.6 and the real ratio \u03c91/\u03c92. Show that for numbers on the real xn < 1 xn \u2264 (0 \u2264 ( 1 2 \u2264 2 (yn + 1)) 2 yn) 1, 1 2 ), 1). \u00a7\u00a7 line: (a) between any two irrationals we can certainly \ufb01nd a rational; (b) between any two rationals we can certainly \ufb01nd an irrational. (Note that, despite (a), there are vastly more irrationals than rationals \u2014 unlike the former the latter are \u2018countable\u2019, so that the rationals are \u2018scanty, but dense\u2019. This emphasizes that for most systems periodicity (closure) is relatively rare, although it is still the case that for any irrational \u03c91/\u03c92 there are rationals arbitrarily close by.) 13. *As an exercise on near-integrable systems explore the \u2018standard map\u2019 (Chirikov\u2013Taylor) analytically/computationally: In+1 = In + K sin \u03c6n, \u03c6n+1 = \u03c6n + In+1. 14.2 and in more detail in (Note: This equal-area map models the twist around a Poincar\u00b4e section by the dynamics (as in D.3), when the integrable case (K = 0) is expressed in action/angle variables (I, \u03c6). As K increases there appear resonance zones, periodic orbits and bands of chaos as tori I = constant undergo progressive breakdown. See Chirikov, Physics Reports, 52, 263\u2013379, 1979.) \u00a7 \u00a7 444 Classical Mechanics Answers to Problems CHAPTER 1 1. mA/mB = 3; 3v/4. 2. m1/m2 = r2/r1. 3. 2m\u00a8r = F = F 21 + F 31; r = (m2r2 + m3r3)/(m2 + m3). 4. 0.12 m. 5. rij = rij, pi = pi \u2212 6. 400 N, 300 N. 7. arcsin 0.135 = 7.76\u25e6 E of N; 60.6 min; 130 km, 8.62\u25e6 W of S. 8. 1.2 m; ( miv, F ij = F ij. 1.2); 7 N, 7 N, 10 N. 1.2), (0, 0.9, \u2212 \u2212 0.4, 0.6, 3)\u25e6 = 7.9. \u2212 \u00b1 10\u2212 10. (2.2 \u00d7 CHAPTER 2 1. (x in m, t in s) x = 3 cos 2t + 4 sin 2t = 5 cos(2t 2.214) = \u2212 \u2212 k/m. \u2212 \u2212 \u2212 3 (mg/k)(1 4i)e2it]; t = 0.322 s, 1.107 s. cos \u03c9t), \u03c9 = Re[( 2. z = \u2212 3. 0.447 s, 14.2 mm. 1, \u03b8 = 5\u25e6 cos \u03c0t 4. 15.7\u25e6 s\u2212 5. V = GM m/x; \u2212 6. V = 1 4 cx4; a. 7. V = 1 2 kx2 c/k; \u03c9 = 8. F = mk, F = mk; oscillation; 2 9. F = kx, c/2m a2; x = c ln x; x = \u2212 2GM (R\u2212 < a; F = 0, \u2212 a\u2212 \u2212 \u00b1 \u2212 1 x | | x | | 2k/m. 2a/k. 8.66\u25e6 sin \u03c0t = 10\u25e6 cos(\u03c0t + \u03c0/3). 1); 8 km s\u2212 1. > a; oscillation between two turning points if k < 0 and E < 0, 1 turning point if k > 0 and E < 1 otherwise no turning points. 2 ka2, 10. Earlier by (2a/v) 11. (a) no turning points, (b) 1 turning point, (c) 1 or 2 turning points. (2/\u03c9", ") arctan(\u03c9a/v). \u2212 445 c/ma, (b) v < 2c/ma. g/\u03b3. c/ma or \u2212 g/l t). \u2212 l/g; \u03b8 = \u03b80 cos( 1. k/m; 16 mm. 446 Classical Mechanics k/g u), (1/2k) ln(1 + ku2/g). \u221ae2kh | \u2212 \u2212 \u2212 < \u2212 \u00b1 1). v | e\u2212 gt/\u03b3; \u02d9z 12. x = 2ma3/c; (a) 2(g/l)(cos \u03b8 \u2192 \u2212 1/2 ln(ekh \u2212 cos \u03b80); 2\u03c0 2c/ma, (c) v > \u03b3t) \u03b1, \u00a8\u03b1 = (g/l)\u03b1; 0.95 s; 2\u03c0 s\u2212 x2); x = a tanh( c/m at). a2 + v2/4\u03c92 v/2\u03c9. \u00b1 \u03b3t], \u03b3 = 1 + (1 + \u03b3t)e\u2212 a; 2\u03c0 c/ma < v < 13. z = (g/\u03b32)(1 14. 8.05 s, 202 m. 15. \u221agk arctan( g/k, (gk)\u2212 16. 17. 18. For \u03b8 = \u03c0 \u2212 c/m(a2 19. \u2212 20. x = a; \u03c0/\u03c9; 21. z = (mg/k)[ 22. 1.006 s; 5.33\u25e6, 1.17\u25e6. \u03b3t sin \u03c9t 23. x = (v/\u03c9)e\u2212 \u03c92 0 + \u03b32 \u03b3. 25. \u03c91 = \u00b1 26. \u00afE = 1 0), W = 2\u03c0m\u03b3\u03c91a2 1 + \u03c92 1(\u03c92 4 ma2 1. 27. 3; \ufb01nal velocities: 28. vn = en\u221a2gh. 29. an = c/m\u03c92n(1 + n2). 31. \u03c4 = 1.017 2\u03c0 \u00d7 \u03b3\u2212t 32. G(t) = (e\u2212 x = c m l/g. \u03b3+t)/m(\u03b3+ \u2212 \u03b3 e\u2212 \u2212 1 \u03b3+t e\u2212 e\u2212 \u03b32 \u2212 \u2212 1, 7 m s\u2212 6 m s\u2212 vte\u2212 1 \u03b32 + \u2212 \u2212 \u03b3t. \u2192 \u03b3\u2212t \u2212 \u2212 \u03b3\u2212 \u03b3+ 1 \u2212 CHAPTER 3 1, 10 m s\u2212 1; T = 364.5 J. ), t > 0; 2\u03b3 \u03c94 0 + t \u03c92 0. r2k]/r5; Fr = 2c cos \u03b8/r3, F\u03b8 = c sin \u03b8/r3, F\u03d5 = 0. ar2 sin \u03b8 sin \u03d5, 1 1 \u00b7 \u2212 \u2212 \u2212 ayz 2 a. \u2212 \u2212 bxy2 \u2212 r)2. 2 ax2 1 2 (a 3 bz3, (c) V = 1. (a) V = (f) V = 2. 1 2 a + b. 3. (i) 0, (ii) 1 4. (a) \u03c0a2, F not conservative; (b) 0, F may be conservative. 5. F = c[3(k r)r 6. 382 m, 883 m; 30\u25e6; 17.7 s, 10.2 s. 7. z = x tan \u03b1 8. z = wx/u \u2212 9. 6.89 km, 7.35 km, 7.18 km. 10. 4\u03c9; m\u03c93l4/r3; \u2206T = 3 2 m\u03c92l2. 11. v/2, v2 = 4ka2/3m; 4ka/3m, 12. \u00a8\u03b8 = 2F/ma 5ka/6m. (g/a) sin \u03b8, \u02d9\u03b82 = 4F \u03b8/ma gx2/2v2 cos2 \u03b1; \u03b1 = \u03c0/4 + \u03b2/2. \u03b3gx3/3u3; 42.3\u25e6, 823 m. \u2212 gx2/2u2 (2g/a)(1 \u2212 \u2212 \u2212 \u00b7 cos \u03b8); \u2212 \u2212 F0 = 0.362 mg. \u2212 Answers to Problems 447 13. \u00a8\u03b8 = (1 sin \u03b8)g/3a, \u02d9\u03b82 = (\u03b8 1 + cos \u03b8)2g/3a; F = mg(1 + 2 sin \u03b8)/6; \u2212", " \u2212 \u03b8 = 7\u03c0/6; thereafter the two bodies move independently until string tautens. \u02d9u = (M + m)g sin \u03b1/(M + m sin2 \u03b1), \u02d9v = mg sin \u03b1 cos \u03b1/(M + m sin2 \u03b1); \u03b1 = arcsin(2/3) = 41.8\u25e6. 14. 2 sin2 \u03b8, x = c\u2212 15. z = c\u2212 17. cot \u03b8 = cot \u03b80 cos(\u03d5 \u00a8\u03be 18. m 4 \u03be+\u03b7 \u03be 1 2 \u03b7 \u2212 \u02d9\u03b7 \u03b7 \u02d9\u03be \u03be \u2212 \u2212 1 2(\u03b8 2 sin 2\u03b8). \u03d50), (\u03b80, \u03d50 constants). \u2212 2 = F\u03be, m 4 \u03be+\u03b7 \u03b7 \u00a8\u03b7 1 2 \u03be \u2212 \u02d9\u03be \u03be \u2212 \u02d9\u03b7 \u03b7 2 = F\u03b7. g), \u2212 19. (a) and (b): r = k/(c + g), \u03b8 = 0, unstable; (a) only: r = k/(c \u03b8 = \u03c0, stable. i h2 2 m 20. T = 1 21. \u00a8r = (\u00a8\u03c1 i \u02d9q2 i, pi = mh2 i \u02d9qi, ei \u00b7 p = mhi \u02d9qi. \u03c1 \u02d9\u03d52)e\u03c1 + (\u03c1 \u00a8\u03d5 + 2 \u02d9\u03c1 \u02d9\u03d5)e\u03d5 + \u00a8zk = (\u00a8r \" \u2212 23. \u2202er/\u2202\u03b8 = e\u03b8, \u2202e\u03b8/\u2202\u03b8 = \u2212 (r \u00a8\u03b8 + 2 \u02d9r \u02d9\u03b8 (er sin \u03b8 + e\u03b8 cos \u03b8), others zero. \u2212 r sin \u03b8 cos \u03b8 \u02d9\u03d52)e\u03b8 + (r sin \u03b8 \u00a8\u03d5 + 2r cos \u03b8 \u02d9\u03b8 \u02d9\u03d5 + 2 \u02d9r sin \u03b8 \u02d9\u03d5)e\u03d5. r \u02d9\u03b82 \u2212 \u2212 r sin2 \u03b8 \u02d9\u03d52)er + er, \u2202er/\u2202\u03d5 = e\u03d5 sin \u03b8, \u2202e\u03b8/\u2202\u03d5 = e\u03d5 cos \u03b8, \u2202e\u03d5/\u2202\u03d5 = \u2212 24. mc2[(cosh2 \u03bb mc2[(cosh2 \u03bb 25. y = \u03bb + a cosh[(x 26. A circle, x2 + y2 \u2212 \u2212 cos2 \u03b8)\u00a8\u03bb + 1 cos2 \u03b8)\u00a8\u03b8 1 2 sinh 2\u03bb( \u02d9\u03bb2 2 sin 2\u03b8( \u02d9\u03bb2 \u2212 b)/a], a, b, \u03bb constants. 2by = a2, b constant. \u2212 \u2212 \u2212 \u02d9\u03b82) + sin 2\u03b8 \u02d9\u03bb \u02d9\u03b8] = F\u03bb, \u2212 \u02d9\u03b82) + sinh 2\u03bb \u02d9\u03bb \u02d9\u03b8] = F\u03b8. CHAPTER 4 107 km. \u00d7 \u00d7 \u00d7 104 km. 105 km, 0.176 au = 2.64 1. 1. 4.22 2. 1.61 3. 11.9 yrs, 13.1 km s\u2212 4. 5.46 yrs. 5. 38.6 km s\u2212 6. 1.62 m s\u2212 7. 84.4 min, 108 min, 173 min. 8. 1.0 1, 7.4 km s\u2212 2, 2.38 km s\u2212 1. 1; 25.8 m s\u2212 \u00d7 2, 60.2 km s\u2212 1. 1011 MS (assuming that the mass distribution is spherical \u2014 this is the mass inside the radius of the Sun\u2019s orbit). 2 a2(3 + 2 cos \u03b1 k/m a; r2 = 1 \u221a5 + 4 cos \u03b1); r = 2a, a; r = a, 0. \u00b1 a)2; \u03c9 = 1/2 \u03c90; \u03c9 = 5/2 \u03c90; 2.24 radial 9. 10. U = J 2/2mr2 + 1 2 k(r oscillations per orbit. 11. (2/3\u03c0) yrs = 77.5 days. 12. Ratio is 1.013, 2.33, 134. 13. 1/r2 = mE/J 2 + \u2212 (mE/J 2)2 mk/J 2 cos 2(\u03b8 \u03b8", "0). \u2212 \u2212 448 Classical Mechanics 14. Hyperbola with origin at the centre. 15. 7.77 days. 16. 8.8 km s\u2212 17. 4.26RE, 18.4\u25e6; 38.3 km s\u2212 19. GM m/2a, 20. x = a(e 1, 5.7 km s\u2212 1, 4.87 yrs. GM m/a. \u2212 cosh \u03c8), y = b sinh \u03c8; r = a(e cosh \u03c8 1), \u2212 1; 97\u25e6 ahead of Earth; 82\u25e6 ahead of Jupiter. 1, opposite to Jupiter\u2019s orbital motion; 5.7 km s\u2212 1; \u2212 t = (abm/J)(e sinh \u03c8 \u03c8). \u2212 106 km = 56RJ, 23RJ. 21. 5.7 km s\u2212 3.9 \u00d7 22. 14.3 km s\u2212 3.6 au. 23. 14.3 km s\u2212 1 at 23.5\u25e6 to Jupiter\u2019s orbital direction; 9.2 au, 16.2 yrs; 1, in plane normal to Jupiter\u2019s orbit, at 23.5\u25e6 to orbital direction; 7.8 au, 16.2 yrs; 2.5 au. l/a; 60\u25e6, 6.45 km s\u2212 1 l/R)/ 24. cos \u03b8 = (1 \u2212 \u2212, b2 = J 2/2m 1 + mk/J 2 25. With n2 = | | 1. \u2212 \u03b80) = J 2 + mk > 0, E > 0 : r cos n(\u03b8 J 2 + mk = 0, E > 0 : r(\u03b8 J 2 + mk < 0, E > 0 : r sinh n(\u03b8 J 2 + mk < 0, E = 0 : re\u00b1 J 2 + mk < 0, E < 0 : r cosh n(\u03b8 \u03b8)/mv2\u03b82(2\u03c0 \u2212 \u2212 n\u03b8 = r0; \u2212 c)/ma5, \u03c9 = : E | | \u03b80) = b; b; \u00b1 \u03b80) = b; \u00b1 \u03b80) = b. \u03b8)2 sin \u03b8. ka2 + c)/ma5. \u2212 \u2212 ( \u2212 103 s\u2212 \u02d9r = eJ sin \u03b8/ml, \u00a8r = eJ 2 cos \u03b8/m2lr2, rad. accel. = 11 m, 8.1 1. \u00d7 ka2 ( \u2212 26. d\u03c3/d\u2126 = k\u03c02(\u03c0 27. \u03c9 = \u2212 28. 0.123 m, 2.44 m. 29. 1.13 \u00d7 30. 10\u2212 J 2/m2lr2. \u2212 CHAPTER 5 \u00d7 \u00d7 10\u2212 10\u2212 5 gS. 3 m s\u2212 2 = 1.9 1; (a) 542 m s\u2212 2 = 0.086 gJ, 5.1 1. 2.2 m s\u2212 1. 2. 15.3 s\u2212 3. 0.20 mm, 78 mm. 4. 465 m s\u2212 5. (a) 99.88 t wt, (b) 100.29 t wt, (c) 99.46 t wt. 3 N to south, 1.46 6. 2.53 \u00d7 7. 0.013 mbar km\u2212 8. 0.155\u25e6. 9. 12.0 s; 100 kg wt, 20 kg wt; decreasing weight and a Coriolis force of 1, (b) 187 m s\u2212 1, (c) 743 m s\u2212 3 N up. 10\u2212 10\u2212 1. 1. \u00d7 79 N. 10. 47.4 mm. Answers to Problems 449 B) with E = Ek, B = Bk; 11. F = m\u00a8r = q(E + v \u2227 x = (mv/qB) sin(qBt/m), y = (mv/qB)[cos(qBt/m) z = qEt2/2m; z = (2mE/qa2B2)y2; depends only on m/q. 1], \u2212 1, l = 0.089 m. 106 V m\u2212 1. 1016 s\u2212 1. \u00d7 1011 s\u2212 3.3 \u00d7 \u223c \u2212 12. l = \u03c0mv/qB; E = 2 105 T, 1.76 13. \u00d7 1, 1016 s\u2212 14. 5.1 \u00d7 15. 124 m. 18. m\u00a8r = F x\u2217 y\u2217 z\u2217 ma. cos \u03c9t sin \u03c9t 0 20. T = 1 \u23a3 \u23a6 \u23a3 2 m", "( \u02d9r + \u03c9 19. = \u2212 \u23a4 \u23a1 \u23a1 \u2227 CHAPTER 6 sin \u03c9t 0 0 1 \u2212 cos \u03c9t 0 r)2 | ). | z \u2212 | ), E = (qk/2\u03c00a2) | (q/4\u03c00)(1/z 1. \u03c6 = (q/2\u03c00a2)(\u221aa2 + z2 E k(\u03c3/20)(z/ \u2192 2 qa2; for \u03b8 = 0, \u03c6 2. Q = 3. When d is in same direction as E. 4. E = (3d (a) F = (b) F = (c) F = (d) F = F = 0, G = 3d 6k(dd/4\u03c00r4), G = G = 0; rr \u2212 F = F = 3k(dd/4\u03c00r4), G = G = 0; F = 3i(dd/4\u03c00r4), G = r2d)/4\u03c00r5, V = (r2d \u2212 k(dd/4\u03c00r3). \u00b7 \u2212 \u2212 \u2212 G = d \u2248 \u2212 \u2212 \u2212 \u00b7 a2/4z3). \u2212 \u2212 z z | | \u2212 z \u221aa2+z2 ; r d \u00b7 \u00b7 r)/4\u03c00r5; j(dd/4\u03c00r3), G = 2G; =j(qiqj/4\u03c00rij ). i 2 (q2/4\u03c00a). 5 (q2/4\u03c00a); 1 \" 105 C, 0.28 J m\u2212 5. V = 1 2 6. 3 7. 4.5 8. 8\u03c0\u03c30a4/5, 2\u03c0\u03c32 9. 3qa2(x2 2. 0a3/250. y2)/4\u03c00r5, \u00d7 \u2212 10. (3x2 \u2212 \u2212 \u2212 \u2212 \u2212 E = (3qa2/4\u03c00r7) 7y2 Gm/r + (Gma2/r3)(3 cos2 \u03b8 Gm/a + (Gm/4a3)(2z2 x2 \u2212 6Gm/r + (7Gma4/4r5)[3 a\u2212 11. 12. 13. 4.0 \u00d7 14. 1/9.5. 15. 3.0. 16. 79 m; 78 RE. 17. (ME/MM)2(RM/RE)4 = 35; 2.8 km. (8\u03c0G\u03c10a3/3)(r\u2212 \u2212 1); 6.7 1040 J. \u2212 \u00d7 1 3y2 + 2z2)y, 5(x2 \u2212 2z2)x, (7x2 1); y2), g = (Gm/2a3)(x, y, \u2212 \u2212 \u2212 5(x4 + y4 + z4)/r4]. 2z). \u2212 y2)z. \u2212 106 yrs; 14.9 mins, 29.5 mins. 450 Classical Mechanics \u00b5r/4\u03c00r; 18. 12.5 m. 19. (8\u03c0/5)\u03c10d0r4; 1.13. 22. q\u00b52e\u2212 q. \u2212 23. 2\u03c0G\u03c12R2/3 = 1.7 24. \u03a6 = 25. 26. 17.9 yrs (should be 18.6 yrs). \u03c0k/4G. 1; 51 days. k\u03c1; a = 10\u2212 6 s\u2212 \u2212 1.43 \u00d7 \u00d7 \u2212 1011 Pa =1.7 Mbar. CHAPTER 7 105 km from centre of Sun, i.e., just outside the Sun; 0.28\u25e6. \u00d7 1. 258 days. 2. 7.4 3. 0.00125 M0; m1 \u2265 1 2 gt2 + (m2v/M \u03c9) sin \u03c9t, 4. z1 = l + m1vt/M \u2212 1 2 gt2 z2 = m1vt/M 0.00125 M0. \u2212 \u2212 (m1v/M \u03c9) sin \u03c9t, with \u03c9 = k/\u00b5 and v < l\u03c9. 5. m1/m2 = 1. 6. 12; 0.071. 7. 62.7\u25e6, 55.0\u25e6, 640 keV. 8. T \u22171 = m2Q/M, T \u22172 = m1Q/M ; 3.2 MeV, 0.8 MeV. 9. ln 106/ ln 2 \u2248 10. 90\u25e6; 45\u25e6, 45\u25e6. 11. 2.41b; (", "0.65v, 0.15v, 0), (0.35v, 12. T \u2217/T = m2/M ; 13. 3 10\u2212 15. a2 cos \u03b8(1/ sin4 \u03b8 + 1/ cos4 \u03b8), where a = e2/2\u03c00mv2. (The second \u2192 6, +450 km, +2.4 min. 0.15v, 0). 1 or 0. 20. \u2212 \u00d7 term comes from recoiling target particles.) 103 s\u2212 16. 1.8 \u00d7 17. 2m \u00a8R = 0, 1 1, same for both. 2 m\u00a8r = qE \u2212 (q2/4\u03c00r3)r; z = 2qE/m\u03c92. CHAPTER 8 1, 164 kg. 1. 0.99 km s\u2212 2. (2.44 + 1.48 =) 3.91 km s\u2212 1. 3. 4.74 km s\u2212 4. 3 stages, 1.48 5. 14.2 km s\u2212 6. 1 2 M0u2(1 7. 44.6 km, 33.9 km. 1, (3.07 8. 10.3 km s\u2212 9. If u1 = (v, 0): ( 105 kg. 1, 2.06 km s\u2212 v/u). e\u2212 \u2212 \u00d7 1, 143 kg. 1, 2.8 106 kg. \u00d7 0.07 =) 3.0 km s\u2212 \u2212 1, 0)v/5, (3, 1; 6.13 t. v2| \u221a3)v/5, ( | \u00b1 \u2212 = 2\u221a3v/5). Answers to Problems 451 2(I/m) \u03c1Av2, A = \u03c0r2; because scattering is isotropic. (1 + e)a3/GM (1 2\u03c1Ava/m. 3.37 s, 10. \u2212 11. \u03b4a = \u2212 12. da/dt = dl/dt = \u2212 13. 16.9 km; 14. (a) 6.2 h, 1.85 d; (b) 3.15 d, 40.4 d. 2.78 km. 20.9 s, e). \u2212 \u2212 \u2212 \u2212 \u2212 CHAPTER 9 1. 4\u221a2a/3, (3\u221a2g/4a)1/2. 2. 64 r.p.m., 5.3 3. (a) E, P ; (b) J about leading edge; (c) E; 3v/8a, 5/8; 10\u2212 6 J from work done by insect; dissipated to heat. \u00d7 [16(\u221a2 1)ga/3]1/2. \u2212 4. (a) 1.011 s, (b) 1.031 s. 5. 4a/3; 3bX/4M a2, 3bX/4a; b = 4a/3. 6. (a) 3M g cos \u03d5; (b) (M g/8)( \u2212 sin 2\u03d5, 1 + cos 2\u03d5). 9 sin 2\u03d5, 11 + 9 cos 2\u03d5), (3M g/2)( 10\u2212 (1.08, 1.44, 0) \u00d7 \u2212 \u00d7 10\u2212 7. 9 6 kg m2, 25 1; 6.3 6 kg m2, 16 10\u2212 \u00d7 4 kg m2 s\u2212 10\u2212 10\u2212 8. (a) (8, 8, 2)M a2/3; (b) (11, 11, 2)M a2/3. 9. 2M (a5 10. 25.6 s, 1.097 11. 60\u25e6. 12. I1 = I2 = 3M (a2 + 4h2)/20, I3 = 3M a2/10; 1/2; Z = 3h/4, 6 kg m2; 3 N. b5)/5(a3 \u2212 103 J. \u00d7 \u00d7 b3). \u2212 \u00d7 I \u22171 = I \u22172 = 51M a2/320, I \u22173 = I3. 1. 13. 1.55 s. 14. 112 s. 15. 0.244 s\u2212 17. 8.83 m. 18. (a) 2.64 Hz (\u2126 = 16.6 s\u2212 20. 2.50 \u00d7 21. 22.3 s. 12 s\u2212 10\u2212 1 = 16.3 yr\u2212 1. 1); (b) 3.44 Hz (\u2126 = 21.6 s\u2212 1). CHAPTER 10 \u00b1 g/4. 4mgl/(M + 2m)a2. 1. 2. 3. M mg/(M + 2", "m). 4. M 2mg/k(M + 2m)2. 6. g/7, 3g/7, 5g/7. 7. 24mg/7, 12mg/7. \u2212 452 Classical Mechanics 1, 4.347 s\u2212 8. 62.62 s\u2212 9. (M + m sin2 \u03b8)l \u00a8\u03b8 + ml \u02d9\u03b82 cos \u03b8 sin \u03b8 + (M + m)g sin \u03b8 = 0; 1.40 s\u2212 10. I1 \u00a8\u03d5 = I3\u03c93\u2126 sin \u03bb cos \u03d5 1. I1\u21262 sin2 \u03bb sin \u03d5 cos \u03d5; (I1/I3)\u2126 sin \u03bb; east 1 (3548 r.p.m.). 1 (cf. 4.065 s\u2212 1); 371.5 s\u2212 and west. \u2212 11. I1, I3 are replaced by I \u22171 < I1, I \u22173 = I3; large \u2126 is bigger, small \u2126 is slightly smaller. 12. arcsin(1/\u221a3) = 35.3\u25e6. 14. (a) as at t = 0 except that for l/2 y(x, 0). (b) y = 0; (c) y(x, l/c) = 15. \u00a8x \u2212 \u03c92x = \u03c92y = 2\u03c9 \u02d9y \u00a8y + 2\u03c9 \u02d9x 1 = (x + a1)2 + y2, r2 r2 \u2212 \u2212 \u2212 \u2212 \u2212 GM1(x + a1)/r3 GM1y/r3 2 = (x GM2(x 1 \u2212 \u2212 GM2y/r3 2, with a2)2 + y2, \u03c92 = GM/a3. a2)/r3 2, 1 \u2212 \u2212 CHAPTER 11 ct < x < l/2 + ct, y = a 2act/l; \u2212 \u2212 1. x = a(cos \u03c91t cos \u03c92t + 1 2 \u221a2 sin \u03c91t sin \u03c92t), 2 \u221a2 sin \u03c91t sin \u03c92t), = ) and \u03c9 y = a(2 cos \u03c91t cos \u03c92t + 3 where \u03c91,2 = 1 \u2212 \u03c9 2 (\u03c9+ \u00b1 (2 2. \u03c92 = g/l, g/l + k/M + k/m; AX /Ax = 1, 3. x0 = 2mg/k, y0 = 3mg/k; \u03c92 = (3 4. \u03c92 = \u03c92 0 + \u03c92 Ax : Ay : Az = 1 : 1 : 1, 1 : 0 : 0 + 3\u03c92 s ; s, \u03c92 0, \u03c92 1, 1 : \u00b1 \u00b1 \u00b1 \u2212 \u221a5)k/2m. \u2212 2 : 1. \u2212 \u221a2)g/l. m/M ; 2M a/(M + m); no. 5. 2a/3, a. 6. (a) \u03c92 = k/m, 3k/m; (b) \u03c92 = (1 7. \u03c92 = (M + m)g/ma, g/2a. 8. \u03b8 = (\u03d50/10)(cos 2\u03c0t 9. 1.0025 s, 0.099 75 s; 0.5025 mm. \u2212 a/l)k/m, 3(1 a/l)k/m. \u2212 \u2212 cos 3\u03c0t) (t in s); \u03d50/5, t = 1 s. 1 = q2/4\u03c00a3m, \u03c92 2 = 4\u03c92 1. 1 = \u03c92 0, \u03c92 2 = \u03c92 0 + 2\u03c92 s, 0.0399 sin 63.0t) mm, (t in s). 10. (0.401 sin 6.27t 11. \u03c6 = (q/4\u03c00)[14/a + (4x2 + y2)/2a3]; \u03c92 12. \u03c92 = g/l, g/l, 3g/l. 13. A1,2 = (F/\u221a2m)(\u03c92 \u2212 \u03c92 + 2i\u03b31,2\u03c9), with \u03c92 1,2 \u2212 \u03b31 = \u03b1/2m, \u03b32 = (\u03b1 + 2\u03b2)/2m; \u03b1 > \u221a3k/\u03c90. 14. q2r = 0, q2r+1 = ( \u2212 1)r4\ufffd", "\ufffd2la/\u03c02(2r + 1)2. CHAPTER 12 1. \u03c92 = g cos \u03b1/r sin2 \u03b1; arcsin(1/\u221a3) = 35.3\u25e6. 2. \u21262 = \u03c92 \u2212 3. H = (px \u2212 y = mg(1 \u2212 g2/l2\u03c92. y/2m + 1 py)2/6m + p2 cos \u03c9t)/k, x = x0 \u2212 2 ky2 mgy; y/4, \u03c92 = 4k/3m. \u2212 5. H = 4. J 2 = p2 \u03b8 + p2 p2 \u03b8 2ml2 + X + p2 p2 Y 2M 6. H = + p2 \u03c8 2I \u22173 7. (p\u03d5 \u2212 8. H = (p \u2212 p2 \u03c1 2m 9. H = p\u03c8z)2 Answers to Problems 453 \u03d5/ sin2 \u03b8. p\u03b8 cos \u03b8)2 (lpx \u2212 2(M + m sin2 \u03b8)l2 + p2 \u03b8 2(I \u22171 + M R2 sin2 \u03b8) + + mgl(1 cos \u03b8). \u2212 (p\u03d5 \u2212 p\u03c8 cos \u03b8)2 2I \u22171 sin2 \u03b8 + M gR cos \u03b8; \u03c92 3,min = 4I \u22171 M gR/I 2 3, reduced by a factor I \u22171 /I1. p2 z2)(E \u03c8/2I3) = 0. M grz \u2212 \u2212 \u2212 \u2212 2I1(1 qA)2/2m + q\u03c6. qq 4\u03c00\u03c1 p2 \u03d5 2m\u03c12 \u2212 + + q2B2\u03c12 2m. 2 (25 + \u221a621) = 24.96. qm/\u03c00qB2; \u221a3\u03c9L. 10. 11. 0 < b < a, a2/4b. 12. M1/M2 > 1 13. \u21262 = p2 = 1 2 \u03c92 1 { \u2212 147.4 yrs. J m\u03c12 ; \u02d9\u03d5 = 14. \u2212 \u00b1 J 2 m\u03c13 \u2212 \u2202U \u2202\u03c1 = \u2212 1 27M1M2/(M1 + M2)2 ; 11.90 yrs, } qJBz m\u03c1, \u2202U \u2202z = qJB\u03c1 m\u03c1. CHAPTER 13 1. (a) \u00aff = k2/4\u03c3; equilibria x \u00b1 asymptotically stable. (b) \u02d9x \u21d2 practice, \u00aff is not precisely known or constant. k2 \u00b1 \u2212 f < 0 = = (k \u00aff \u2212 \u2264 4f \u03c3)/2\u03c3, x+ is k/\u03c3(f t0 \u2264 \u2212 \u00aff ). In \u02d9\u03b82 2. Trajectories 1 2 \u03c92 0x 4. (g/l) cos \u03b8 = constant (= g/l on separatrices). \u00b5y. Critical point (0, 0) which is (i) asymptotically \u02d9x = y, stable spiral, (ii) (improper ) asymptotically stable node, (iii) (in\ufb02ected ) asymptotically stable node. \u2212 \u2212 \u02d9y = \u2212 5. (0, 0) and (1, 0) are asymptotically stable nodes; ( 1 [Note that the nodes are local minima of U (x, y).] 2, 0) is a saddle. \u2212 \u2192 \u221e 0+ \u2192 6. f (x) = c ln x + dx has a single minimum and f + when x + \u2192 and when x. \u221e 7. (0, 0) has \u03bb1,2 = 1, 1 2 with eigenvectors (1, 0), (0, 1): unstable node; 1, (0, 2) has \u03bb1,2 = (1, 0) has \u03bb1,2 = 1, node; ( 1 2, 1 Nearly all initial conditions lead to extinction for one or other species in Fig. 13.9 = = 1 2 with (1, 3), (0, 1): asymptotically stable node; 1 4 with (1, 0), (1, 3 4 ): asymptotically stable 2 + \u03bbi): saddle. no stable coexistence. Second set of parameters \u2212 \u2212 2 ) has \u03bb1,2 = ( (0, 0): unstable node; (0, 3 \u221a57)/16 with ( 2, 1 2, 0): saddles;", " ( 1 2 ) and ( 1 4, 1): \u2212 \u2212 \u21d2 \u00b1 \u2212 \u2212 \u2212 5 1 \u21d2 454 Classical Mechanics asymptotically stable node; in this case there is asymptotically stable coexistence. 8. M = = c1 a2 \u2212 c2 a1 \u2212 c1c2 > a1a2 = \u21d2 coexistence in \ufb01rst quadrant; and c1c2 < a1a2 = opposite sign: saddle in third quadrant and trajectories run away \u21d2 \u03bb1, \u03bb2 real, negative: asymptotically stable node, \u03bb2 + (c1 + c2)\u03bb + (c1c2 \u2212 a1a2) = 0. Then \u03bb1, \u03bb2 real, \u21d2 +. \u221e \u2192 9. Critical points at (0, 0), 1) \u00b5(R (\u00b5R+a), a(R 1) \u2212 R(a+\u00b5) \u2212. R < 1: asymptotically stable node, saddle. R > 1: saddle, asymptotically stable node. Disease maintains itself only when R > 1. 10. At (0, 0), \u03bb1,2 = ( 2). ( \u2265 \u221a2 \u00b1 \u2212 4)/2; unstable spiral ( < 2) or node \u2192 11. At (0, 0), \u03bb1,2 = \u03b1\u03b2 i, so we have an asymptotically stable spiral (\u03b2 < 0), unstable spiral (\u03b2 > 0) or centre (\u03b2 = 0). In polars, \u02d9\u03b8 = 1 and \u02d9r = \u03b1r(\u03b2 0. r r2); for \u03b2 > 0, r \u221a\u03b2 as t and \u03b8 \u2192 \u221e \u2192 \u2212 \u00b1 ; when \u03b2 < 0, J, 0, 0), (0, 0, 12. (a) J1 \u02d9J1 + J2 \u02d9J2 + J3 \u02d9J3 = 0; then integrate to get J) are centres on the sphere; (0, \u00b1 J3 = I3\u2126. J1,2 satisfy \u00a8Ji + [(I1 \u2212 ( \u00b1 (c) \u02d9J3 = 0 = (d) Critical point (0, 0, 0), with \u03bb1,2,3 = (1, 0, 0), (0, 1, 0), (0, 0, 1). As \u03c9 largest moment of inertia, I3. J = constant. (b) | | J, 0) are saddles. \u00b1 I3)/I1]2\u21262Ji = 0. /I1,2,3 and eigenvectors \u00b5 | rotation about axis with the 0, \u03c9 \u2212| \u2192 \u21d2 \u2192 13. (a) dN/dt 0 and dI/dS = \u02d9I/ \u02d9S = 1 + b/aS. (b) dI/dS = 0 when b/aS2 < 0, so maximum. I = 0 at, S+. (c) S0 = b/a + \u03b4, I0 =. Write S = b/a + \u03be and expand \u2261 S = b/a and d2I/dS2 = S = S = arbitrarily close to S susceptibles escape infection. \u21d2 \u2212 \u2212 \u2212 trajectory passes through (b/a \u2212 \u2212 \u03b4, ). Since S0 can be made, so S+ is arbitrarily close to b/a \u03b4 and these \u2212 I t (/a S \u2212 S Answers to Problems 455 14. For a critical point (X, Y, Z), the matrix )\u03c3, so that \u2212 \u2212 \u23a4 \u2212 \u2212 1 2 (\u03c3 + 1) \u03b2 < 0, \u03bb2,3 = 4(1 (a) \u03bb1 = \u03bb2,3 are real and are both negative only when 0 < \u03c1 < 1. (c) \u03c1 = 1 = has the form (\u03bb + \u00b5)(\u03bb2 + \u03bd2) = 0 then \u00b5 = \u03c3 + \u03b2 + 1, \u03bd2 = \u03b2(\u03c3 + \u03c1), \u00b5\u03bd2 = 2\u03c3\u03b2(\u03c1 result for \u03c1crit. (d) Write \u03c1 = 1 + (with \u03c3, \u03b2 cubic becomes \u03bb(\u03bb + \u03b2)(\u03bb + \u03c3 + 1) = 0. If the cubic 1) = (\u03c3 + 1)2 \u21d2 \u00b1 \u2212 \u2212 1 2 \u23a3 \u2212 \u21d2 constant, small) and \ufb01nd changes in eigenvalues by putting \u03bb = \u03bbi + \u03be() and performing a linear analysis of the cubic in (b). (e) Evidently RHS = 0 is an ellipsoid and the distance of (x,", " y, z) from \u00afO (0, 0, \u03c1 + \u03c3) decreases where RHS < 0, i.e., outside the ellipsoid and so a fortiori outside any sphere centred at \u00afO which contains it. \u2261 15. (b) For a critical point (X1, X2, Y ), the matrix M = Y \u23a1 \u00b5 Y X2 \u2212 \u00b5 X1 A \u2212 \u2212 0 X1 X2 \u2212 2\u00b5 < 0. (d) X 2 \u2212. \u23a4 \u2212 ( \u02d9x) = \u23a3 \u00b7 (c) \u2207 ellipsoid (oblate). Trajectories move towards smaller C values whenever (X1, X2, \u00afY ) is below the paraboloid \u00afY = \u00b5(X 2 1 + X 2 2 + \u00afY 2 = C, where C > 0, is an 1 + X 2 \u23a6 2 )/\u221a2. \u03c9 \u2206x0 sin \u03c9t + \u2206y0 cos \u03c9t. 16. \u2206x = \u2206x0 cos \u03c9t + (\u2206y0/\u03c9) sin \u03c9t, \u2206y = Each is bounded and so is (\u2206x)2 + (\u2206y)2. \u2212 17. Trajectories in the xy-plane are parabolic arcs in x 0. The higher \u2265 the energy the longer the period of oscillation (between successive bounces). Between bounces, \u2206x = \u2206x0 + \u2206y0t, \u2206y = \u2206y0, so that \u03bat for t increasing, and bounces don\u2019t a\ufb00ect this (\u2206x)2 + (\u2206y)2 \u223c 18. Evidently, we have (1), ( 1 4, 1 (row of Pascal\u2019s triangle). When n = 16, we have 4 ), etc. in successive rows, i.e., 2 ), ( 1 2, 1 2, 1 result. ( 1 2 )n \u00d7 456 Classical Mechanics ( 1 2 )16[1,16,120,560,1820,4368,8008,11440, 12870,11440,8008,4368, 1820,560,120,16,1]. CHAPTER 14 1. Return trajectories have 2h3m/ \u03c0/2 t0 = k /R \u2212| | arcsin \u2264 R/h E < 0. k | | \u03c0 as R/h \u2212 \u2192 h3m/2 k | | major axis h in this limit. + (R/h)(1 R/h) \u2212 \u2192 0, i.e., Kepler\u2019s third law for a \ufb02at ellipse, 2. Explicitly, [F, H] = 0 in each case. In (c), the vector A may be considered using Cartesians or polars. (Then the orientation of the orbit may be speci\ufb01ed using \u03b1 = arctan(Ay/Ax) and 2 = 1 A | | 3. H ( /mk2), consistent with (4.30).) | E) and p\u03b8 are constants of the motion. Steady r = r0 when g/r0. Small oscillations r = r0 + \u2206 = (2J 2 E | \u2212 \u2261 p\u03b8 = mr2 SHM for \u2206 with frequency 0\u21260 with \u21260 = \u00b1 3g/2r0. g cos \u03b1/r0 sin2 \u03b1, rotation rate at r0. Small oscillations 4. \u03c9 = \u21d2 r = r0 + \u2206 = for rational \u221a3 sin \u03b1. 5. E\ufb00ective potential p2 \u21d2 SHM for \u2206 with frequency \u221a3(\u03c9 sin \u03b1). Closure \u2261 \u03b8/2mr2 k \u2212 | k | \u03b8/m minimum /r = | \u03b8 when, circle; (r, pr) curves closed (around r = r0) when | r = r0 = p2 E < 0, corresponding to (r, \u03b8) ellipses; E r system \u2014 given E, Poincar\u00b4e section is a single point on an (r, pr) curve, a di\ufb00erent point for each choice of \u03b8 section., (r, \u03b8) hyperbolae, parabola. We always have closure for this 0 curves stretch to \u2192 \u221e \u21d2 \u2265 \u2212 k2m/2p2 l/g = \u21d2 E = 1 2 g/l I, and frequency 6. H E here. Action I = 2E \u2261 \u03c9 = \u2202H/\u2202I = 1 2 g/l. 7. Oscillation between x = [l + \u00b1 2E", "/k]. Action I\u2126 + \u03b22 I = (2l/\u03c0)\u221a2mE + E/\u2126 = \u03c6(x) = (\u2202/\u2202I) \u03c9 = \u2126\u221aE/(\u221aE + \u03b2), T = 2\u03c0/\u03c9. Small E = 2\u03c0/\u2126. large E = \u221aE = \u2212 x p dx with \u03c6 = \u03c9(I)t + constant, T \u21d2 \u21d2 \u03b2. as E\u2212 1/2; \u2192 \u221e 8. I = 1 2 T \u21d2 \u221amk[E/k with \u03c9 = \u2202E/\u2202I = 2 \u03bbk \u2192 \u2212 E \u03bb/k] = E(I); \u03c6(q) = (\u2202/\u2202I) k/m. Evaluating \u03c6(q) gives \u21d2 1/2 q p dq \u2261 \u03c9t + \u03b2, k + \u221aE2 q(t) = oscillator there are two radial oscillations for each complete elliptical trajectory.). (Note. For the isotropic sin(\u03c9t + \u03b2) \u2212 k Answers to Problems 457 9. Action I1 = (\u221a r2 2mE/\u03c0) r1 \u2212 r1 + r2 = /E and r1r2 = | \u03c91 = \u2202E/\u2202I1 \u2261 (r2 \u2212 r)(r \u2212 I 2 2 /2mE = \u21d2 \u2212 \u03c92 (see (14.18)). \u2212| k r1)(dr/r), where result. So E(I1, I2) and 10. Consider the lines from a bounce point to the foci and the angles they make with the trajectory just before and just after the bounce. ae)py \u2212 \u039b = [(x + ae)py \u2212 \u2212 from Chapter 3, Problem 24 using p\u03bb = mc2(cosh2 \u03bb p\u03b8 = mc2(cosh2 \u03bb \u2212 tangency condition. cos2 \u03b8) \u02d9\u03b8. Turning value for \u03bb is when p\u03bb = 0 = pxy]. Change to \u03bb, \u03b8 variables; H cos2 \u03b8) \u02d9\u03bb, pxy][(x \u21d2 \u2212 \u2212 11. (a) Motion within curve (E/mg\u00b5r) = 1 (1/\u00b5) cos \u03b8 (ellipse). (b) \u00b5 = 1: bounding curve E = 2mgr sin2(\u03b8/2); \u00b5 < 1: curve as in (a) = 0, two straight lines for E = 0. \u2014 hyperbola for E \u02d9x = px + \u03c9y, \u02d9px = \u03c9py \u2212 12. Hamilton\u2019s equations: \u02d9y = py \u2212 U. System autonomous, H = constant; U 2 (py \u2212 possible motion and for the Earth/Moon system the critical C is that corresponding to the \u2018equilibrium\u2019 point between Earth and Moon. 2 (px + \u03c9y)2+ C for \u03c92y + \u2202U/\u2202y and H = 1 \u03c92x + \u2202U/\u2202x, \u03c9px \u2212 \u03c9x, \u02d9py = \u03c9x)2 \u2212 \u2212 \u2265 1 1/2. Maximum sideways displacement \u221d = l\u2212 13. Action I = E/\u03c9, so E \u03c9 \u221d l1/4. Maximum acceleration L\u221aH and H 1/L5 = = l\u03b8max \u221d 14. Action I \u221d pressure 15. Action I = (2m/3\u03c0) Frequency \u03c9 = 2\u03c0 1/3, period 2\u03c0/\u03c9 (sin \u03b1)2/3, q0 \u221d E E2/E1 = 0.69, q02/q01 = 1.20, \u03c91/\u03c92 = 1.44. v2 = \u21d2 pressure \u221d \u21d2 2gq3 \u21d2 \u221d \u221d \u221d v \u221d 0 sin \u03b1 and E = mgq0 sin \u03b1 = g sin \u03b1/8q0. Evidently, I = constant = (sin \u03b1)\u2212 (sin \u03b1)\u2212 \u21d2 \u21d2 2/3. So result. l\u2212 3/4. \u03c92l\u03b8max| \u221d | 1/L. Temperature \u221d (density)5/3. 1/L2 \u221d 16. Evidently E = k | 1 + 2EI 2 a \u221d e = \u221d \u21d2 1/ | k\u2212 2. Since \u03c4 k2 and \u03c4 k decreases, a increases. Since eccentricity 2 /mk2 (see (4.30)), e remains constant. a3/ k | \u221d \u221d", " | (see (4.31)), APPENDIX A 1. (a) 1, 2, 3; (b) ( 3, 3, 0), ( (d) ( 3, \u2212 1, 3, 3, 3), ( 3, 7, 8), ( 3); (e) as (d). \u2212 6, 4, 11); (c) 15, 15; \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 xz, xy); 6x. \u2212 2. 164.2\u25e6, 16.2\u25e6. 4. (3x2 \u2212 5. (a) (1 yz, \u2212 \u2212 y2/x2, 2y/x, 0), (2x/y, 1 x2/y2, 0); (b) circles passing through the origin with centres on x- or y-axes, respectively, intersecting at right angles. \u2212 458 Classical Mechanics \u00b7 \u2207)a b) + (b 8. a(\u2207 9. 2\u03c9k. 11. (\u22072A)\u03c1 = \u22072(A\u03c1) \u00b7 (\u22072A)\u03d5 = \u22072(A\u03d5) \u2212 b(\u2207 a) \u00b7 \u2212 \u2212 (a \u2207)b. \u2212 1 \u03c12 1 \u03c12 A\u03c1 + 2 2 A\u03d5 \u2212 \u00b7 \u2202A\u03d5 \u2202\u03d5 \u2202A\u03c1 \u2202\u03d5, 12. Ar = A\u03b8 = 0, A\u03d5 = \u00b50\u00b5 sin \u03b8/4\u03c0r2; Br = \u00b50\u00b5 cos \u03b8/2\u03c0r3, B\u03b8 = \u00b50\u00b5 sin \u03b8/4\u03c0r3, B\u03d5 = 0; r2\u00b5). A = (\u00b50/4\u03c0r3)\u00b5 \u2227 13. B = (\u00b50I/4\u03c0r3)ds (ds rr r; F = (\u00b50II /4\u03c0r3)ds r; B = (\u00b50/4\u03c0r5)(3\u00b5 \u00b7 \u2212 \u2227 r); \u2227, (\u22072A)z = \u22072(Az). (ds ds) \u2227 1 ln r sin \u03b8, A\u03d5 = 0; r\u2212 = 0. \u2227 2(1 + 2 ln r) cos \u03b8; \u2227 F + F = (\u00b50II /4\u03c0r3)r 1 cos \u03b8, A\u03b8 = r\u2212 14. Ar = r\u2212 2 sin \u03b8). (0, 0, 2r\u2212 0 cos \u03b8 1 0 sin \u03b8 0 sin \u03b8 0 cos \u03b8 \u23a6 \u03c1(r)(3rr \u2212 17. Q = \u23a3. \u23a4 15. \u23a1 r 21)d3r. \u2212 APPENDIX B 1. Distances r1,2 are given by r2 1,2 = (a cos \u03c8 2. Tangent vector is t = ( \u2213 ae)2 + (b sin \u03c8)2 = a2(1 e cos \u03c8)2. a sin \u03c8, b cos \u03c8), unit vectors from two foci are e, (b/a) sin \u03c8), scalar products are \u2213 \u2212 1(cos \u03c8 n1,2 = (1 n1,2 = t \u00b7 \u2213 \u00b1 e cos \u03c8)\u2212 ae sin \u03c8. \u2213 APPENDIX C 1. (a) (0, 0), \u03bb1,2 = stable node. 5 2 \u00b1 1 2 \u2212 \u221a17, eigenvectors (1, 3 + \u03bbi), asymptotically (b) (0, 0), \u03bb1,2 = 4, 1, eigenvectors (1, \u03bbi \u2212 (c) ( \u2212 \u221a5, eigenvectors ( 1), \u03bb1,2 = 1, \u00b1 5 \u2212 3), unstable node. 2, \u03bbi + 8), asymptotically \u2212 \u2212 stable node; (4, 4), \u03bb1,2 = 5 (d) (0, 2), \u03bb1,2 = \u2212 (1, 0) \u03bb1,2 = 1, (e) (0, 2), \u03bb1,2 = \u00b1 2), \u03bb1,2 = (0, \u2212 \u00b1 1 \u00b1 i\u221a55, unstable spiral. i, asymptotically stable spiral ; \u00b1 2, eigenvectors ( \u2212 i2\u221a6, centre (\u2234 stable); \u2212 i2\u221a6, centre (\u2234 stable); 1, \u03bbi + 2), saddle (\u2234 unstable). Answers to Problems 459 8, 3, eigenvectors (1, 0), (0, 1", "), resp., saddle \u2212 3, eigenvectors (1, 0), (0, 1), resp., saddle \u2212 (1, 0), \u03bb1,2 = (\u2234 unstable); ( \u2212 (\u2234 unstable). 1, 0), \u03bb1,2 = 8, (f) (0, n\u03c0), n odd: \u03bb1,2 = i, centres (\u2234 stable); \u00b1 1, eigenvectors (1, 1), (1, n even: \u03bb1,2 = (\u2234 unstable). (g) (0, 0), \u03bb1,2 = \u00b1 \u03c92 < \u03b1, centre (\u2234 stable). \u00b1 \u2212 \u221a\u03c92 \u03b1, so \u03c92 > \u03b1, saddle (\u2234 unstable) and 1), resp., saddles \u2212 (Note: In each case, consider local sketches near the critical points and how these build into the global phase portrait.) \u02d9x = y, (0, 0), centre, ( 1, 0), saddles; \u02d9y = x3 x = 2. \u21d2 \u2212 (a) dy/dx = (x3 1 2 x4 (b) separatrices given by c = 1 4. x)/ x4 2 y2 = 1 1 2 x2 + c, i.e., \u2212 x2 + 2c; symmetry about x axis rules out spirals; APPENDIX D 1. (b) d asymptotically stable when dx F (2)(x) = F (x)F (F (x)) = < 1 = \u21d2 r2 + 2r + 4 \u2265 \u21d2 | \u2212 F (X1)F (X2) | | r 0). (c) a = 1 1, s = 2 (d) Applying (c) to (a), (b), etc, yields asymptotically stable points/cycles: 1/r, b = 2/r < 1 = \u21d2 \u2212 \u2212 \u2212 r. | 3 < r < 1 + \u221a6 (if 6 3 \u2192 Y 0 1/s 1 \u2212 Y1(s), Y2(s) 3.57 4 2\u221e accum. range limit \u2212 r X 1 1 \u221a.57 1 1/r \u2212 0 X1(r), X2(r) 2\u221e accum. 2 range limit \u2212 \u2212 2. r = 1 3. (a) X = 0 asymptotically stable in circle \u221a1 + 4a, \u03b2 = 2\u03b1 = r/a. \u00b1 \u2212 asymptotically stable in circle (b) asymptotically stable 2-cycle when r | 1 + \u221a5 r. | | \u221atwo disjoint ovals). r | < 1. | \u2212 < 1, X = 1 1/r \u2212 | r2 + 2r + 4 < 1, or | 460 Classical Mechanics Note: If we continue in this way, adding the regions of the complex r plane in which there are cycles of any \ufb01nite length we arrive at the \ufb01gure shown: (c) In the complex a plane, the circles become a heart-shaped region 1 4 ; the 2-cycles yield a circle with a cusp at a = entire set yields the Mandelbrot \u2018signature\u2019 (see Peitgen and Richter, The Beauty of Fractals, Springer, 1986): 4. The = 1 a | \u2212 \u2212 1 | 4. Fixed points at X = 0, X = 2 Uncertainty n = 02n = \u2265 Lyapunov exponent \u03bb = ln 2 > 0. 3 have 1 when n = 2 = F (X) | \u21d2 | ln(1/0)/ ln 2. Note: n = 02n \u2265 instability. 5. X = 0 asymptotically stable if 3 | < 1. X = < 1, or 1 < a < 2. At a = 2, there is a | stable when period-doubling on each branch, followed by a period-doubling cascade. 1 asymptotically 2a \u2212 \u00b1 \u2212 | | \u221aa \u21d2 Answers to Problems 461 ) is, Y \u2212 b)2. b); real \u2212 7. (a) Both | a(1 < 1 \u03bb2| | < 1 =, \u03bb1| | 2aX | < 1 and also b | \u21d2 | always unstable; (X+, Y+) is asymptotically stable if a < 3 (b) Nontrivial 2-cycle = a2X 2 \u2212 \u2212 roots if a > 3 4 (1 (4a2X1X2 + 2b)\u03bb + b2 = 0 = \u03bb2 \u2212 4a + 2b and 4(1 | an argument similar to that in the Hint]. \u2212 b", ")2. Eigenvalues \u03bb of M1M2 satisfy a = 0, with Y = b(1 \u21d2 < 1 + b2 = Xi roots of b)X + (1 b)2 + 1 a < (1 \u21d2 b)2 b)2 \u21d2 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 | for asymptotic stability aX 2)/(1 b. (X 4 (1 + b)2 [using aX 2 + Y + \u00af\u03be, yn+1 = bX + \u00af\u03b7, \u2212 \u2212 \u21d2 8. xn = X + \u03be, yn = Y + \u03b7 = xn+1 = 1 with \u00af\u03be = 2aX\u03be + \u03b7, \u00af\u03b7 = b\u03be. Circle \u03be2 + \u03b72 = 2 becomes [ \u00af\u03be + (2aX/b)\u00af\u03b7]2 + \u00af\u03b72/b2 = 2, ellipse with semi-axes /\u221a\u00b51, /\u221a\u00b52 where \u00b51\u00b52 = 1/b2, \u00b51 + \u00b52 = 1 + (1 + 4a2X 2)/b2. Area of ellipse = \u03c0 \u21d2 0 < \u00b51 < 1 < \u00b52 (say) = 9. (a) Fixed points P1 = (1 + a Lyapunov exponents \u03bb1 > 0 > \u03bb2. area reduction. (1 \u00b52) < 0 = 2 = | \u00b51)(1 b | \u21d2 \u21d2 \u2212 \u2212 \u2212 1(1, b) if a > 1 b)\u2212 1(1, b) if a > b 1, a b b. If | | P1 asymptotically stable. \u2212 < 1 and | | \u2212 \u2212 \u21d2 < 1 a = \u2212 \u2212 b only P1 a b)\u2212 \u2212 b/\u03bb = P2 = (1 exists, \u03bb (b) For nontrivial 2-cycle Q1 = (X1, Y1) Q2 = (X2, Y2), X1, X2 must be of opposite sign. Hence X1,2 = (1 Y1,2 = bX2,1. x y 1 1 1 2 Lyapunov exponents \u03bb1,2 = ln \u00b51,2 = \u03ben+1 \u03b7n+5 = a)/[a2 + (1 \u03ben \u03b7n eigenvalues b)2], \u21d2 \u21d2 \u2212 = = = \u2213 \u2212 \u2212 = b \u00b51,2 = 3 2 \u00b1 ln 2.618 = \u00b1 \u21d2 0.9624 \u2014 stretch and squeeze. See diagram: 11. (a) \u00b1 (b) eigenvalues \u00b51 = 2, \u00b52 = 1 diagram: 2 = \u21d2 \u03bb1,2 = ln 2 = \u00b1 \u00b1 0.6931. See 462 Classical Mechanics 12. (a) \u03b1, \u03b2 distinct irrationals with \u03b2 \u03b1 > > 0, then choose integer N such that > 1/N ; mesh integer multiples of 1/N, at least one is between \u03b1 and \u03b2. (b) Since \u221a2 is irrational, then e.g., (1 rationals p1/q1 and p2/q2. 1/\u221a2)(p1/q1) + (1/\u221a2)(p2/q2) is irrational and lies between the \u2212 13. We may take both \u03b8 and I to be 2\u03c0-periodic here. See diagram: \u2212 Bibliography The following is a short list of selected books which may be found helpful for further reading on classical mechanics and related subjects. Abraham, R. and Marsden, J.E.: Foundations of Mechanics, 2nd ed., Benjamin/Cummings, 1994. Abraham, R.H. and Shaw, C.D.: Dynamics \u2014 the Geometry of Behavior, 2nd ed., Addison\u2013Wesley, 1992 [originally published by Aerial Press]. Arnol\u2019d, V.I.: Mathematical Methods of Classical Mechanics, 2nd ed., translated by K. Vogtmann and A. Weinstein, Springer\u2013Verlag, 1995. Arnol\u2019d, V.I., Kozlov, V.V. and Neishtadt, A.I.: Mathematical Aspects of Classical and Celestial Mechanics, 2nd ed., Springer, 1996. Baker, G.L. and Gollub, J.P.: Chaotic Dynamics \u2014 an Introduction, 2nd ed., Cambridge University Press, 1996. Chandrasekhar, S.: Newton\u2019s \u201cPrincipia\u201d for the Common Reader, Clarendon, 1995. Cvitanovic, P.: Universality in Chaos \u2014 a reprint selection, 2nd ed., Adam Hil", "ger/IOP Publishing, 1989. Fetter, A.L. and Walecka, J.D.: Theoretical Mechanics of Particles and Continua, McGraw\u2013Hill, 1980. Feynman, R.P.: Lectures on Physics, vol. I, Addison\u2013Wesley, 1963. Goldstein, H., Poole, C.P. and Safko, J.L.: Classical Mechanics, 3rd ed., Addison\u2013Wesley, 2002. Guckenheimer, J. and Holmes, P.: Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields, Springer\u2013Verlag, 1983. Gutzwiller, M.C.: Chaos in Classical and Quantum Mechanics, Springer\u2013Verlag, 1990. Hofbauer, J. and Sigmund, K.: Evolutionary Games and Population Dynamics, Cambridge University Press, 1998. 463 464 Classical Mechanics Hunter, S.C.: Mechanics of Continuous Media, 2nd ed., Ellis Horwood, 1983. Jackson, E.A.: Perspectives of Nonlinear Dynamics, Cambridge Univer- sity Press, 1989. Jammer, M.: Concepts of Force: A Study in the Foundations of Dy- namics, Dover, 1999. Jammer, M.: Concepts of Mass in Classical Mechanics, Dover, 1997. Landau, L.D. and Lifshitz, E.M.: Mechanics (vol. 1 of Course of Theo- retical Physics), 3rd ed., Butterworth\u2013Heinemann, 1982. Lichtenberg, A.J. and Lieberman, M.A.: Regular and Stochastic Motion, Springer\u2013Verlag, 1983. Lorenz, E.N.: The Essence of Chaos, UCL Press/University of Wash- ington Press, 1993. May, R.M.: Stability and Complexity in Model Ecosystems, 2nd ed., Princeton University Press, 1974 and 2001. Murray, C.D. and Dermott, S.F.: Solar System Dynamics, Cambridge University Press, 1999. Ott, E.: Chaos in Dynamical Systems, 2nd ed., Cambridge University Press, 2002. Park, D.: Classical Dynamics and its Quantum Analogues, 2nd ed., Springer, 1990. Pars, L.A.: A Treatise on Analytical Dynamics, Heinemann, 1965 [reprinted by Oxbow Press]. Peitgen, H.-O., J\u00a8urgens, H. and Saupe, D.: Chaos and Fractals, Springer\u2013Verlag, 1992. Percival, I. and Richards, D.: Introduction to Dynamics, Cambridge University Press, 1982. Index Note: Page numbers in italics refer to items in the problems. accelerated frame, 127, 145 acceleration, 6 absolute nature of, 3 gravitational, xv, 51, 99, 112, 130 in rotating frame, 111\u2013112 polar components of, 71 accelerator, 304 action and reaction, 7 action integral, 63, 190 action variable, 354, 369 action/angle variables, 354\u2013359, 437 addition of vectors, 382 additivity of mass, 12 adiabatic change, 379 adiabatic invariance, 369\u2013372 principle of, 370 adiabatic invariant, 369 aircraft, 3, 15, 126 algebra of vectors, 385 almost linear system, 314, 421\u2013423 alpha particle, 96, 103, 167 amplitude, 23 at resonance, 32 complex, 25, 258 of forced oscillation, 31, 265 angle variable, 354 angular frequency, 25 of damped oscillation, 28 angular momentum, 54 about an axis, 54, 65 about centre of mass, 183 and Euler angles, 222 and Larmor e\ufb00ect, 121 as generator of rotation, 293, 296\u2013297 conservation of, 55, 65, 66, 74, 76, 182, 280, 283, 292, 296\u2013297 of Earth\u2013Moon system, 183\u2013188 of many-body system, 181\u2013183 of rigid body, 197, 199, 205\u2013208, 218 of two-body system, 161, 162 perpendicular components, 203\u2013204 rate of change, 54, 181\u2013183 total, 161, 181, 182 angular velocity, 105\u2013106 and Euler angles, 222 instantaneous, 216\u2013218 of Earth, xv, 105, 185\u2013186 of precession, 121, 123, 144, 156, 214, 237 vector, 105 anomaly, 101 antisymmetric tensor, 401 aphelion, 102 apparent force, 235 apparent gravity, 112\u2013113, 142 area of ellipse, 87, 412 area, directed element of, 395 area-\ufb01lling map, 353 465 466 Classical Mechanics area-preserving map, 353 Aristotle, 3 arms race, 324 Arnold\u2019s cat map, 443 asteroid, 98, 101, 164, 175 Trojan, 304 astronomical unit, 98 asymptotically stable point, 310,", " 314, 418, 419, 426 atmospheric drag, 45, 52, 194 atmospheric pressure, xv atmospheric tides, 187 atomic spectrum, 84, 121 attenuation, 94 attractor, 310, 426, 436 strange, 334, 338 Atwood\u2019s machine, 367\u2013368, 378 autonomous system, 309, 415 general second-order, 313 axes of orbit, 75, 76, 87, 88 rotation of, 383 unit vectors along, 383 axial vector, 55, 105, 386 axioms of Euclid, 1 axioms of mechanics, 11 axis of rotation, 105 instantaneous, 217, 224 axis of symmetry, 207 baker\u2019s transformation, 443 basin of attraction, 334, 426 beam, colliding, 169 Bernoulli, 70 Bertrand\u2019s theorem, 351 bicycle wheel, 124 bifurcation, 309, 428 billiard balls, 41 billiard systems, 363\u2013366, 378 circular, 364 elliptical, 365, 378 oval, 366, 441 body cone, 224 Bohr orbit, 84, 121, 126 boost, 295, 298\u2013300 bounce map, 364, 366 bouncing ball, 345 boundary conditions, 61 brachistochrone, 70 broken symmetry, 363 Brusselator, 328 butter\ufb02y e\ufb00ect, 339 calculus of variations, 59\u201362, 233 canonically conjugate pairs, 301 Cartesian co-ordinates, 4, 5, 382 catastrophe, 309 catenary, 72 cause and e\ufb00ect, 339 central conservative force, 8, 66, 73\u201398, 279, 283\u2013284, 350, 357\u2013359, 375 Hamiltonian for, 279 impulsive, 90 of varying strength, 371 potential energy of, 66 central force, 55\u201357 and angular momentum conservation, 291 internal, 181, 189, 198 centre, 315, 420, 422 true, 422, 424 centre of charge, 137 centre of mass, 6, 138, 159, 177 equation of motion of, 159, 178, 218 centre-of-mass frame, 162\u2013164 centrifugal force, 77, 112, 235, 249 potential energy of, 114, 141, 239 centripetal acceleration, 111 Chandler wobble, 224, 330 chaos, 294, 334, 337, 338, 347 routes to, 375 chaotic attractor, 334 characteristic equation, 259, 314 charge, 8 centre of, 137 total, 137 charge density, 134, 150, 397 charge-to-mass ratio, 130 charged particle in electromagnetic \ufb01eld, 241\u2013244, 302 Index 467 chemical oscillation, 327 circular billiard, 364 circular orbit, 78, 81, 410 velocity in, 83 clockwork universe, 339 closest approach distance, 79\u201380 co-ordinates Cartesian, 4, 5, 382 curvilinear, 58, 398\u2013400 elliptic, 72, 378 ignorable, 282\u2013285 normal, 264\u2013266 orthogonal, 58, 253\u2013256, 398 parabolic, 71 polar, 57\u201359, 398\u2013400 coe\ufb03cient of restitution, 42 coins, 366 collision, 26, 37 elastic, 41, 165\u2013173 hard-sphere, 90\u201394, 169, 172 inelastic, 42 multiple, 95 two-body, 11, 14, 39\u201342, 165\u2013173 combat model, 322\u2013323 comet, 89, 99 competing species, 318\u2013322, 342 competitive exclusion principle, 321 complex amplitude, 25, 258 of forced oscillation, 31, 265 components, of vector, 382 composite body, 14, 178 force on, 178 mass of, 12, 178 position of, 5, 14, 178 compound pendulum, 200\u2013202 cone, 207 moments of inertia, 227 particle on, 301 conics, 86, 409\u2013413 Cartesian form, 409\u2013412 polar form, 412\u2013413 conservation law in constrained system, 239 of angular momentum, 55, 65, 66, 74, 76, 182, 280, 283, 291\u2013293, 296\u2013297 of energy, 9, 18, 25\u201326, 40, 50, 74, 76, 188\u2013190, 198, 280\u2013282, 295 of momentum, 11, 39, 160, 166, 178, 296 conservation laws, 2, 76\u201378 and symmetries, 291\u2013301, 347 for symmetric top, 284 in polar co-ordinates, 76 conservative force, 17\u201318, 49\u201351, 188\u2013190 condition for, 50, 395 impulsive, 40 internal, 189 potential energy for, 18, 394 conservative system, 315\u2013317 constraint equation, 231\u2013232, 239 and Lagrange multiplier, 240 continuous dynamics, 307, 312 control parameters, 308 convection, 331 Coriolis acceleration, 111 Coriolis force, 112, 114\u2013120, 125, 229, 249 Coulomb gauge, 397 Coulomb scattering, 79\u201380 Coulomb\u2019s law, 8 coupled oscillators,", " 261\u2013266 critical damping, 29, 46 critical line, 415 critical point, 311, 313, 415\u2013424 isolated, 415 cross-section, 90\u201397, 168\u2013173 di\ufb00erential, 93, 97, 169, 171, 172 total, 94, 170 cube, moments of inertia, 210 cubic map, 442 curl, 391 in polar co-ordinates, 400 of curl, 393 of gradient, 393 vanishing, 51, 394 current density, 397 current loop, 124 curvilinear co-ordinates, 58, 63, 398\u2013400 468 Classical Mechanics unit vectors for, 71 cycloid, 70, 376 cyclone, 119 cyclotron, 109 cyclotron frequency, 109, 304 cylinder, 207 moments of inertia, 210 cylindrical polars, 57\u201359, 65, 71, 398\u2013400 kinetic energy in, 58 Lagrange\u2019s equations in, 65 damped oscillator, 27\u201329, 274, 341 day, 105, 186 degeneracy, 350, 373 degrees of freedom, 232, 294 Delaunay, 373 denominators, small, 373 depression, 119 determinant, 258, 416 determinism, 339 diagonalization, 403\u2013404 dice, 366 di\ufb00erential cross-section, 93 for Rutherford scattering, 97 hard-sphere, 93, 172 in CM frame, 171 in Lab frame, 169, 171 di\ufb00erentiation of vector, 388 dipole electric, 131 magnetic, 124, 406 dipole moment, 132, 137 discrete dynamics, 307, 312 discrete map, 353 discrete systems, 425\u2013441 disease, 343 divergence, 391 in polar co-ordinates, 399 of curl, 393 divergence of trajectories, 374 dot product, 384 of tensors, 401 double dot product, 406 double pendulum, 255\u2013257, 260, 369 double star, 14, 174 dyadic, 401 dynamo, 334\u2013336, 345 Earth, xv, 140\u2013144, 163, 214 angular velocity of, xv, 105, 185\u2013186 core of, 154, 336 escape velocity, 82 magnetic \ufb01eld of, 334 motion near rotating, 112\u2013113 orbit of, xv, 88, 98, 100 precession of axis of, 214, 224 shape of, 113, 140\u2013144, 214 Earth\u2013Moon distance, xvi, 154, 163 Earth\u2013Moon system, 163, 183\u2013188, 215 eccentric anomaly, 101 eccentricity, 86 of Earth\u2019s orbit, xv, 88 ecliptic, 215 e\ufb00ective potential energy, 77, 235 for central force, 283 for constrained system, 239 for inverse square law, 78, 80 for symmetric top, 285, 288, 302 eigenvalue equation, 258, 314, 403, 416 eigenvalues, of Jacobian matrix, 314 eigenvector, 258, 403 Einstein, 1, 3, 10, 192 elastic bouncer, 357, 371, 379 elastic collision, 41, 165\u2013173 electric charge, 8 electric dipole, 131 electric \ufb01eld, 130, 150, 241, 397 electric quadrupole, 132 electromagnetic \ufb01eld, 9, 397\u2013398 particle in, 110, 241\u2013244, 282, 302 electromagnetic force, 9, 110, 235, 241 electron, xv, 84, 167 electrostatic force, 8, 14, 130 electrostatic potential, 129\u2013131 elements, 3 ellipse, 75, 81, 87, 410, 412 ellipsoid, moments of inertia of, 210 elliptic co-ordinates, 72, 378 Index 469 elliptic orbit, 75, 86\u201387 elliptic point, 420, 440 elliptical billiard, 365, 378 energy, 25\u201326 and time translation symmetry, 295 conservation of, 9, 18, 25\u201326, 40, 50, 74, 76, 188\u2013190, 198, 280\u2013282, 291 kinetic, 17, 49, 161, 162, 188 lost in inelastic collision, 42 potential, 18, 50, 161, 190, 394 total, 18, 50, 280 transferred in elastic collision, 166 energy levels of hydrogen atom, 84 entropy, 433 equal-area map, 437 equations of motion for relative motion, 160 in electromagnetic \ufb01eld, 242, 250 of centre of mass, 159, 178, 218 of rigid body, 199, 218 of small oscillations, 256\u2013257 equilibrium, 253, 311 condition for, 256 hydrostatic, 155 motion near, 20\u201325, 253\u2013271 of rigid body, 198 solution, 313 stability of, 22, 259 equipotential surface, 132, 141, 142, 147 equivalence principle, 10, 130 equivalent simple pendulum, 201, 202, 226 ergodic behaviour, 291 ergodic trajectory, 350 escape velocity, 82\u201383, 99 from Earth, xv, 82 Euclid\u2019s", " axioms, 1 Euclidean geometry, 2 Euler angles, 221\u2013223, 231 Euler\u2013Lagrange equations, 61, 62 external forces, 13, 177 external potential energy, 189, 191, 198 extinction, 321 Feigenbaum, 428 Feigenbaum number, 429 \ufb01eld, 390 electric, 130, 150, 241, 397 electromagnetic, 9, 241, 397\u2013398 gravitational, 130, 151 \ufb01eld equations, 148\u2013152, 397\u2013398 \ufb01gure of eight, 362 \ufb01rst-order convergence, 426 \ufb01sh population, 312, 341 \ufb01xed point, 425, 434, 440 \ufb02ow, 308 \ufb02uid, 114, 146 incompressible, 290, 392 velocity \ufb01eld in, 391 \ufb02ux of particles, 90, 94, 169 \ufb02ywheel, 227 focus, 86, 413, 419 of charged-particle beam, 126 of conic, 413, 414 folding, 353, 433, 436, 440 force, 6, 12 addition of, 6, 382 apparent, 112, 235 as basic quantity, 15 central, 55\u201357, 181, 189, 198 central conservative, 8, 66, 73\u201398, 279, 283\u2013284, 357\u2013359, 375 centrifugal, 77, 112, 114, 235 conservative, 17\u201318, 49\u201351, 188\u2013190, 394, 395 Coriolis, 112, 114\u2013120 de\ufb01nition of, 12 dissipative, 9, 26 electromagnetic, 9, 110, 235, 241 electrostatic, 8, 14, 130 external, 13, 177 generalized, 64, 234 gravitational, 8, 14, 130 impulsive, 37, 40, 202, 226 inverse square law, 78\u201384, 129, 358 Lorentz, 110, 241 magnetic, 108 on composite body, 178 periodic, 30\u201336, 265 saw-tooth, 47 470 Classical Mechanics square-wave, 35 step-function, 39 three-body, 7 tidal, 147, 184 total, 6 two-body, 6, 177 work done by, 26, 49, 188, 234 forced oscillation, 30\u201339, 265\u2013266 forced system, 232\u2013233, 239, 280 Foucault\u2019s pendulum, 117 four-cycle, 435 Fourier integral theorem, 36 Fourier series, 34, 271 fractal, 333, 373, 436 frame, 4 accelerated, 127, 145 centre-of-mass, 162\u2013164 inertial, 3\u20134, 295 laboratory, 165 rotating, 105\u2013125, 218, 239 transformation to moving, 14, 298\u2013300 free motion of rigid body, 219, 223\u2013225 freely falling body, 115\u2013117, 127 frequency, 23 angular, 25 cyclotron, 109 fundamental, 270 Larmor, 121 of normal mode, 350 friction, 9 Galaxy, 99 Galilean transformation, 295\u2013301 Galileo, 2, 10 Galton board, 339, 346 gauge transformation, 250, 397 Gauss law, 150 Gauss\u2019 theorem, 396, 399 general relativity, 1, 3, 144, 192 generalized co-ordinates, 231\u2013233 generalized force, 64, 234 for angular co-ordinate, 234 generalized momentum, 64, 65, 235, 277 in electromagnetic \ufb01eld, 242 generating function, 292 geodesic, 61, 70 gradient, 390, 398 in polar coordinates, 398 gradient system, 317, 342 gravitational acceleration, 51, 99, 130 apparent, 112\u2013113, 142 of Earth, xv gravitational constant, xv, 8, 15, 82 gravitational \ufb01eld, 130, 151 uniform, 159\u2013161, 183, 191 gravitational force, 8, 14, 130 gravitational mass, 10 gravitational potential, 129\u2013131, 152 Green\u2019s function, 37 gyroscope, 212, 214, 228, 229, 250 hairy ball theorem, 349 half-width of resonance, 33 Hamilton\u2019s equations, 277\u2013280 Hamilton\u2019s principle, 63, 234 for stretched string, 245 Hamiltonian function, 278 and total energy, 280 for central conservative force, 279 for charged particle, 302 for symmetric top, 284 rate of change of, 280 Hamiltonian system, 321 hard-sphere collision, 90\u201394, 169, 172 harmonic oscillator, 20\u201326, 73\u201376, 312\u2013313, 341, 345 action/angle variables, 356\u2013357 anisotropic, 73 complex representation of, 24\u201325 damped, 27\u201329, 341 double, 350, 375 forced, 30\u201339 isotropic, 51, 73\u201376, 100, 377 orbit of, 75 with varying frequency, 369 Hartman\u2013Grobman theorem, 422 heat, 26 heat conduction, 332 helium nucleus, 96 Index 471 helix, 109 H\u00b4enon map,", " 434, 437, 442 H\u00b4enon\u2013Heiles system, 368\u2013369 hidden symmetry, 368 holonomic system, 232, 235 homoclinic intersection, 440 Hopf bifurcation, 326\u2013328, 343 Huygens, 377 hydrogen atom, xv, 83\u201384, 126 hyperbola, 79, 82, 88, 410 in polar form, 413 hyperbolic point, 418, 440 ignorable co-ordinate, 282\u2013285 impact parameter, 79, 89 improper node, 418 improper transformations, 300 impulse, 37 velocity, 180 impulsive force, 37, 40, 202, 226 incompressible \ufb02uid, 290, 392 inelastic collision, 42 inertia tensor, 205, 401 inertial frame, 3\u20134 transformation of, 295, 298\u2013300 inertial mass, 10 in\ufb01nite precision, 339 inner product of vectors, 402 insect, 226 integrability, 347\u2013351 integrable system, 294, 347, 353, 437, 438 integral theorems, 393\u2013396 integration of vector, 390 internal force, 177 central, 181, 189, 198 conservative, 189 invariant circle, 440 invariant probability distribution, 432 invariant torus, 349 inverse cube law, 102 inverse square law, 78\u201384, 358 orbit, 85\u201389 inversion symmetry, 300 involution, 348 isobars, 119 isolated critical point, 415 isolated system, 6, 178 isotropic harmonic oscillator, 51, 73\u201376, 100 iteration, 425 Jacobi integral, 360 Jacobian matrix, 314, 415 Jupiter, 98, 100, 125, 154, 193, 304 e\ufb00ect on Sun, 174 KAM, 373, 438 Kepler\u2019s \ufb01rst law, 103 Kepler\u2019s second law, 57, 103 Kepler\u2019s third law, 87, 103, 163, 358, 375 Kermack\u2013McKendrick theorem, 344 kinetic energy, 17, 49 and Euler angles, 223 in CM frame, 188 in orthogonal co-ordinates, 58, 254 near equilibrium, 253 of forced system, 233, 239 of many-body system, 188 of natural system, 233, 280 of particles on string, 266 of rotating body, 199, 207 of stretched string, 244 of two-body system, 161, 162, 166 rate of change, 17, 26, 49, 188, 198 transferred, 166 Kronecker\u2019s symbol, 35 laboratory frame, 165 di\ufb00erential cross-section in, 169, 171, 172 Lagrange multiplier, 72, 195, 240, 249, 406 Lagrange\u2019s equations, 62\u201366, 128, 231\u2013248 for conservative system, 235 for many-body system, 190\u2013192 for non-conservative force, 234 for two-body system, 161 in polar co-ordinates, 64\u201366, 234 Lagrangian density, 245 472 Classical Mechanics Lagrangian function, 62, 235 for charged particle, 242 for coupled oscillators, 265 for many-body system, 191 for pendulum, 64, 238 for stretched string, 245 for symmetric top, 236 for two-body system, 161 in uniform gravitational \ufb01eld, 191 Lagrangian points, 251, 303, 378 lamina, 199 Lanchester\u2019s law, 323 Laplace\u2019s equation, 151, 275 Laplace\u2013Runge\u2013Lenz vector, 376 Laplacian, 393 in polar co-ordinates, 400 of vector \ufb01eld, 393, 406 Larmor e\ufb00ect, 120\u2013124, 126, 362 Larmor frequency, 121, 228, 303 laws of mechanics, 1\u201312, 14 laws of nature, 1 Legendre polynomial, 131 libration, 316, 351, 354 limit cycle, 324\u2013328, 375 line integral, 394 linear independence, 22 linear map, 426 linear system, 314, 415\u2013421 Liouville\u2019s theorem, 289\u2013291 liquid surface, rotating, 114, 125 Lissajous \ufb01gures, 351 logistic equation, 309\u2013311, 426 logistic map, 426, 432, 441 attractors, 428 in standard form, 427 Lorentz force, 110, 241 Lorentz transformation, 301 Lorenz, 331, 339 Lorenz attractor, 436 Lorenz system, 331\u2013334, 344, 353, 424, 433, 434 Lotka\u2013Volterra system, 318\u2013321, 342 Lozi map, 442 lunar theory, 373 Lyapunov exponent, 338, 353, 432, 442 Macdonald, 342 magnetic bottle, 349 magnetic dipole, 124, 406 magnetic \ufb01eld, 241, 397 of Earth, 334 particle in, 108\u2013109, 120\u2013124 uniform, 243 magnitude of vector, 381 malaria", ", 342 Mandelbrot set, 460 many-body system, 177\u2013192 angular momentum of, 181\u2013183 energy conservation in, 188\u2013190 Lagrange\u2019s equation for, 190\u2013192 maps, 307, 373, 425\u2013441 area-\ufb01lling, 353 area-preserving, 353 Arnold\u2019s cat, 443 bounce, 364, 366 cubic, 442 discrete, 353 equal-area, 437 H\u00b4enon, 434, 437, 442 linear, 426 logistic, 426, 432, 441 Lozi, 442 nonlinear, 426 Poincar\u00b4e return, 352, 353 standard, 443 tent, 442 twist, 437, 438 Maskelyne, 15 mass, 6 additive nature of, 12 gravitational and inertial, 10 measurement of, 10 of composite body, 12, 178 reduced, 160 matchbox, 220 maximum scattering angle, 168 Maxwell\u2019s equations, 397 mean free path, 94\u201396 Mercury, 144, 186 moment of force, 53 moment of inertia, 199, 203 calculation of, 208\u2013211 Index 473 principal, 206 shift of origin, 208 momentum, 6 as generator of translation, 296 conservation of, 11, 39, 160, 178, 296 generalized, 64, 65, 235, 242, 277 in CM frame, 162 of many-body system, 178 of rigid body, 197 rate of change, 6, 178, 197 monopole, 132, 137 Moon, xvi, 99, 163, 183\u2013188 precession of orbit, 157 tides due to, 144\u2013148, 154 moving frame, 14, 298\u2013300 multiple collisions, 95 multiply periodic trajectory, 350 n-body problem, 359\u2013362, 374 n-torus, 437 natural system, 232\u2013233 condition for, 280 nautical mile, 102 neap tides, 146 near-integrable system, 372\u2013374, 438 nested tori, 349, 437 Newton, 2, 70 Newton\u2019s \ufb01rst law, 4, 178 Newton\u2019s gravitational constant, xv, 8, 15, 82 for stretched string, 274 normal modes, 258\u2013271 nonlinear, 350 of coupled oscillators, 261\u2013266 of double pendulum, 260 of particles on string, 266\u2013269 of stretched string, 269\u2013271 nucleus, 96 oblateness, 139 of Earth, 140, 141 of Jupiter, 154 one-dimensional maps, 425\u2013433 orbit, 84\u201389 Bohr, 84 circular, 81 elliptical, 86\u201387, 101 hyperbolic, 88\u201389 in phase space, 308 in rotating frame, 120 inverse cube law, 102 inverse square law, 85\u201389 of comet, 89, 99 of Earth, 88, 100 of Jupiter, 98 of oscillator, 75 of satellite, 98, 143, 156 precession of, 103, 120, 143 transfer, 100, 193 orientation of rigid body, 221 orthogonal co-ordinates, 58, 253\u2013256, Newton\u2019s law of gravity, 8 Newton\u2019s laws, 5\u201310 Newton\u2019s second law, 6, 12, 178 Newton\u2019s third law, 7, 9, 11, 159, 178, 398 orthogonal tensor, 406 orthogonality, 348 of eigenvectors, 403 406 Newton\u2013Raphson iteration, 426 node, 422 improper, 418 proper, 419 nodes on stretched string, 270 non-holonomic system, 232 non-integrable system, 347, 353 non-invertibility, 433 nonlinear map, 426 normal co-ordinates, 264\u2013266 orthonormal triad, 383, 404 oscillation, forced, 30\u201339, 265\u2013266 oscillator, 20\u201326, 73\u201376, 312\u2013313, 341, 345 action/angle variables, 356\u2013357 complex representation of, 24\u201325 damped, 27\u201329, 274 forced, 30\u201339, 46 isotropic, 51, 73\u201376, 100, 377 with varying frequency, 369 oscillators, coupled, 261\u2013266 474 Classical Mechanics outer product of vectors, 402 oval billiard, 366, 441 parabola, 52, 82, 90, 126, 411 in polar form, 413 parabolic co-ordinates, 71 paraboloid, 114 parallax, 174 parallel axes theorem, 209 parallelepiped, 387 moments of inertia, 210 parallelogram law, 6, 382 parity, 300 particle, 5 path, shortest, 59, 61, 70 pendulum, 18\u201319, 43, 45, 46, 64, 226, 227, 316\u2013317, 341 compound, 200\u2013202 constrained, 238\u2013241, 302 coupled, 261\u2013266 double, 255\u2013257, 260, 369 equivalent simple, 201, 226 forced damped, 336 Foucault\u2019s, 117 hanging from trolley, 249, 302 of varying length, 370", ", 379 period of, 48 perihelion, 101 period of orbit, xv, 74, 87, 163 of oscillation, 23, 200 of rotation, 105 period-doubling, 375 period-doubling bifurcation, 428 period-doubling cascade, 428, 435 periodic force, 30\u201336, 265 periodic trajectory, 350 Perron\u2013Frobenius equation, 432 perturbed twist map, 437 phase line, 309\u2013312 phase of forced oscillation, 31, 33 at resonance, 34 phase plane, 312\u2013318, 415\u2013424 phase portraits, 307\u2013309 local or global, 424 phase space, 289, 307\u2013309 phase velocity, 308 pion, 175 planet, 44, 101, 103, 195 plumb line, 15, 112, 143 Pluto\u2013Charon system, 163 Poincar\u00b4e, 307, 339 Poincar\u00b4e return map, 352, 353 Poincar\u00b4e section, 364, 366, 437, 441, 443 Poincar\u00b4e\u2013Birkho\ufb00 theorem, 440 point particle, 5 Poisson bracket, 293, 304, 348 Poisson\u2019s equation, 151, 397 polar co-ordinates, 57\u201359, 398\u2013400 acceleration in, 71 Lagrange\u2019s equation in, 64\u201366 unit vectors for, 71 velocity in, 56 polar vector, 55, 386 position vector, 5, 6, 381 of composite body, 5, 178 relative, 7, 159, 162, 181, 182, 299 potential, 129\u2013152 electrostatic, 129\u2013131, 151 expansion at large distances, 137 gravitational, 129\u2013131, 152 of centrifugal force, 141 of dipole, 132 of Earth, 140 of quadrupole, 132 of spherical distribution, 134 of spheroid, 139 scalar and vector, 241, 397 tidal, 144 potential energy, 18, 50, 394 e\ufb00ective, 77, 78, 80, 235, 239, 283, 285, 288 external, 189, 191, 198 in many-body system, 189 internal, 161, 189, 191 inverse square law, 78 near equilibrium, 20, 256 of central force, 66 of centrifugal force, 114, 141, 239 of harmonic oscillator, 20, 50 of impulsive force, 40 Index 475 of particles on string, 267 of simple pendulum, 19 of stretched string, 244 rate of change, 50 precession, 103, 108, 211\u2013215, 330 Larmor, 120, 362 of Earth\u2019s axis, 224, 330 of equinoxes, 214\u2013215, 229 of orbit, 143, 156, 157 of rolling wheel, 228 of symmetric top, 213, 228, 236\u2013237, 249 planetary, 144 predictability, 337\u2013339 loss of, 339, 374 pressure, in \ufb02uid, 155 prey\u2013predator system, 318\u2013321 Prigogine and Lefever, 327 principal axes of inertia, 205\u2013211 rotation about, 218\u2013220 principal moment of inertia, 206, 329 Principia, 2 probability distribution attractors, 432 product of inertia, 203 shift of origin, 208 projectile, 51\u201353, 68, 82, 127 range of, 52, 68, 102 trajectory of, 52, 53, 68 projection of vector, 384 prolate spheroid, 139 proper node, 419 proton, xv, 14, 167 pseudoscalar, 300 pulley, 248 quadrupole, 132 quadrupole moment, 133, 139 of Earth, 140, 154, 155 quadrupole tensor, 407 quality factor, 28, 33, 47 quantum mechanics, 2, 9, 20, 67, 83, 301 quasiperiodic trajectory, 350 radial energy equation, 77, 280 for inverse square law, 78 radiation gauge, 397 range of projectile, 52, 68, 102 range of tides, 147 rate constant, 426 rate of particle detection, 92, 171 rational approximation, 373 Rayleigh\u2019s equation, 324\u2013326, 343 reaction on axis, 200\u2013203, 218, 226 rectangular plate moments of inertia, 210 reduced mass, 160 re\ufb02ection symmetry, 205, 300 relative position, 7, 159, 162, 181, 182, 299 relative velocity, 7, 299 in elastic collision, 41 relativity, 3 general, 1, 3, 144, 192 principle of, 3\u20134, 10, 26, 198, 295\u2013301 special, 2, 10, 109, 301 relaxation oscillation, 325 relaxation time, 28 renormalization, 429 repeller, 310 resonance, 32\u201334, 148, 187, 373 half-width of, 33 Richardson, 324 right-hand rule, 105, 385 rigid body, 197\u2013225 and Euler angles, 221 angular momentum of, 197, 199, 203\u2013208, 218 angular velocity of, 216\u2013218 energy of, 198 free motion of, 223\u2013", "225, 228 generalized co-ordinates for, 231 kinetic energy of, 199, 207, 233 precession of, 211\u2013215 symmetric, 207\u2013208, 222 rigid-body rotation, 329\u2013330, 343 Rikitake dynamo, 334\u2013336, 345 Roche limit, 155, 186 rocket, 102, 179\u2013181, 192 Ross, 342 476 Classical Mechanics rotating frame, 4, 105\u2013125, 218, 239 acceleration in, 111\u2013112 rate of change of vector in, 106\u2013108 rotation, 317 rotation about an axis, 198\u2013202 stability of, 218\u2013220 rotation matrix, 417 rotation number, 437, 438 rotation of axes, 383 rotational symmetry, 205 and angular momentum conservation, 291, 292, 296\u2013297 Routh\u2019s rule, 209\u2013211 runo\ufb00 problem, 318 Rutherford scattering, 96\u201397 saddle, 418 Sarkovskii, 430 satellite, 13, 163 orbit of, 143, 156, 194 synchronous, 98, 192 scalar, 5, 300, 381 scalar \ufb01eld, 390 scalar potential, 241, 397 scalar product, 384 scalar triple product, 387 scattering, 90\u201394, 96\u201397, 165\u2013173 by multiple targets, 95 Coulomb, 79\u201380 cross-section, 90\u201397, 168\u2013173 hard-sphere, 90\u201394, 172 Rutherford, 96\u201397 scattering angle, 89, 91 maximum, 168 Schiehallion, 15 second-order convergence, 426 semi-axes, 75, 87, 88, 410 semi-latus rectum, 86 semi-major axis, 76, 87, 410 of Earth\u2019s orbit, xv, 98 semi-minor axis, 89, 410 sensitivity to initial conditions, 337\u2013339, 353, 432, 440 separatrix, 317, 341, 372 sidereal day, 105 sigmoid curve, 310 similarity transformation, 417 SIR model, 343 slingshot, 101 Smale horseshoe, 436, 440 small denominators, 373 small oscillations, 253\u2013271 of particles on string, 266\u2013269 of pendulum, 200 of rotating body, 220 of stretched string, 244\u2013248, 269\u2013271 Solar System, 360, 374 solid angle, 93 source or sink, 419 space, 2\u20134 space cone, 224 space translation, 296 spacecraft, 100, 180, 193, 227, 251 sphere moments of inertia, 210, 227 potential of, 135 rolling on plane, 232 spherical charge distribution, 134\u2013136 spherical polars, 57\u201359, 65, 71, 398\u2013400 kinetic energy in, 58 Lagrange\u2019s equations in, 65 volume element in, 134 spherical shell, 134 spheroid, 139, 207, 228 spiral point, 419, 422 spiral source or sink, 419 spring, 69 spring tides, 146 square of vector, 384 squeezing, 353, 436, 440 stability and Laplace\u2019s equation, 275 asymptotic, 310 of equilibrium, 22, 259 of Lagrangian points, 251, 303 of rotation, 218\u2013220, 239 of vertical top, 288\u2013289 structural, 309 stable manifold, 315 Index 477 theorem, 423 stable point, 314 standard map, 443 standing wave, 270 star, 419 steady convection, 333 steradian, 93 Stokes\u2019 theorem, 394 strange attractor, 334, 338, 436 stress tensor, 407 stretched string, 244\u2013248 normal modes of, 269\u2013271 particles on, 266\u2013269 stretching, 353, 433, 436, 440 structural stability, 309 subcritical bifurcation, 327 subtraction of vectors, 382 Sun, xvi, 98, 125, 304 angular position of, 174 e\ufb00ect on Earth\u2019s rotation, 215 potential of, 136 tides due to, 144\u2013148, 187 supercritical bifurcation, 327, 343 supercritical \ufb02ip bifurcation, 428 superposition principle, 21, 259 surface of rotating liquid, 114, 141 surface of section, 351\u2013353 surface, closed, 148, 396 symmetric rigid body, 207\u2013208, 222 angular momentum of, 207, 222 free motion of, 223\u2013225 kinetic energy of, 223, 233 symmetric tensor, 205, 401 symmetric top, 236\u2013237, 284\u2013289, 363 conservation laws for, 284, 291 free to slide, 250, 302 Hamiltonian for, 284 Lagrangian for, 236 precession of, 213, 228, 236\u2013237 vertical, 288\u2013289 symmetries and conservation laws, 291\u2013301, 347 tautochrone, 377 tennis racquet theorem, 220, 329 tension in string, 244 tensor, 400\u2013405, 415, 424 inertia, 205, 401 symmetric, 205, 401, 403\u2013404 tensor product of vectors, 402 tent map, 442 terminal speed, 45 three-body problem, 359\u2013362 restricted", ", 251, 303 three-body system, 12, 14 tides, 144\u2013148, 154, 184, 195 atmospheric, 187 due to Earth, 154, 155 due to oceans, 155 spring and neap, 146 time, 2\u20134 time translation, 295 top, 212, 236\u2013237, 284\u2013289, 363 torus, 349, 352, 373 torus breakdown, 437 total cross-section, 94 and mean free path, 94 for hard-sphere scattering, 94 for Rutherford scattering, 97 trace, 406, 416 trade winds, 119 trajectory closed, 350 in phase space, 308 of projectile, 52, 53, 68 transient, 32 translation symmetry, 295\u2013296 transposed tensor, 401 triple product, 387\u2013388 Trojan asteroids, 304 true anomaly, 101 true centre, 422, 424 turbulent convection, 333 twist map, 437, 438 two-body collision, 11, 14, 39\u201342, synchronous satellite, 98, 192 165\u2013173 Tacoma Narrows bridge, 326 target particle, 165 two-body system, 159\u2013173 Lagrangian for, 161 two-cycle, 434, 435 478 Classical Mechanics two-dimensional maps, 433\u2013437 unaccelerated observer, 3 uncertainty principle, 301 uniform gravitational \ufb01eld, 159\u2013161, 183 uniform magnetic \ufb01eld, 108\u2013109, 243 unit vector, 5, 382 units, xvi universal function, 429 unstable manifold, 315 unstable point, 310, 314, 418, 419 valence, of tensor, 400 Van der Pol equation, 326, 343 variation, of integral, 61, 233 vector, 4\u20135, 381 angular velocity, 105 axial, 55, 105, 386 components of, 382 di\ufb00erentiation of, 388 integration of, 390 polar, 55, 386 position, 5, 381 unit, 5, 382 zero, 382 vector \ufb01eld, 390 Laplacian of, 393, 406 line integral of, 394 vector potential, 241, 397 vector product, 385 vector triple product, 388 velocity, 6 circular orbital, 81 escape, 82\u201383, 99 in rotating body, 106\u2013108 in rotating frame, 111 of centre of mass, 160, 178, 299 of light, xv of propagation, 246 polar components of, 56, 58 relative, 7, 40, 299 velocity impulse, 180 Verhulst, 309 vertical top, 288\u2013289 violin string, 324 virial theorem, 101, 196 Volterra\u2019s principle, 321 vorticity, 392 war, 322\u2013324 wave equation, 246 wave, electromagnetic, 9 wavelength, of normal mode of string, 270 width, of resonance, 32 work done by force, 26, 49, 188, 234 Zeeman e\ufb00ect, 121 zero vector, 382 zero-velocity curve, 360, 367, 378-order contributions to the Taylor expansion are immaterial in the long-range continuum limit. 6 From particles to \ufb01elds T \u03c6 M S S[\u03c6] Figure 1.3 Schematic visualization of a \ufb01eld: a mapping \u03c6 from a base manifold M into a target space T (in this case, T are the real numbers; but, in general, T can be more complicated). A functional assigns to each \u03c6 a real number S[\u03c6]. The grid embedded into M indicates that \ufb01elds in condensed matter physics arise as continuum limits of discrete mappings. where the Lagrangian density L has dimensionality [energy]/[length] and we have designated the particle mass by the more common symbol m \u2261 M. Similarly, the classical action assumes the continuum form L S[\u03c6] = dt L[\u03c6] = dt 0 dx L(\u03c6, \u2202x\u03c6, \u02d9\u03c6). (1.5) We have thus succeeded in abandoning the N -point particle description in favor of one involving continuous degrees of freedom, a (classical) \ufb01eld. The dynamics of the latter are speci\ufb01ed by the functionals L and S, which represent the continuum generalizations of the discrete classical Lagrangian and action, respectively. INFO The continuum variable \u03c6 is our \ufb01rst encounter with a \ufb01eld. Before proceeding with our example, let us pause to make some preliminary remarks on the general de\ufb01nition of these objects. This will help to place the subsequent discussion of the atomic chain into a broader context. Formally, a \ufb01eld is a smooth mapping \u03c6 : M \u2192 T, z \u2192 \u03c6(z), from a certain manifold M,4 often called the \u201cbase manifold,\u201d into a \u201ctarget\u201d or \u201c\ufb01eld manifold\u201d T (see Fig. 1.3).", "5 In our present example, M = [0, L] \u00d7 [0, t] \u2282 R2 is the product of intervals in space and time, and T = R. In fact, the factorization M \u2282 R \u00d7 T into a space-like manifold R multiplied by a one-dimensional time-like manifold T is inherent in most applications of condensed matter physics.6 4 If you are unfamiliar with the notion of manifolds (for a crash course, see page 537), think of M and T as subsets of some vector space. For the moment, this limitation won\u2019t do any harm. 5 In some (rare) cases it becomes necessary to de\ufb01ne \ufb01elds in a more general sense (e.g. as sections of mathematical objects known as \ufb01ber bundles). However, in practically all condensed matter applications the more restrictive de\ufb01nition above will su\ufb03ce. 6 By contrast, the condition of Lorentz invariance implies the absence of such factorizations in relativistic \ufb01eld theory. In classical statistical \ufb01eld theories, i.e. theories probing the thermodynamic behavior of large systems, M is just space-like. 1.1 Classical harmonic chain: phonons 7 However, the individual factors R and T may, of course, be more complex than in our prototypical problem above. As to the target manifold, not much can be said in general; depending on the application, the realizations of T range from real or complex numbers over vector spaces and groups to the \u201cfanciest objects\u201d of mathematical physics. In applied \ufb01eld theory, \ufb01elds appear not as \ufb01nal objects but rather as input to functionals (see Fig. 1.3). Mathematically, a functional S : \u03c6 \u2192 S[\u03c6] \u2208 R is a mapping that takes a \ufb01eld as its argument and maps it into the real numbers. The functional pro\ufb01le S[\u03c6] essentially determines the character of a \ufb01eld theory. Notice that the argument of a functional is commonly indicated in square brackets [ ]. While these formulations may appear unnecessarily abstract, remembering the mathematical backbone of the theory often helps to avoid confusion. At any rate, it takes some time and practice to get used to the concept of \ufb01elds and functionals. Conceptual di\ufb03culties in handling these objects can be overcome by remembering that any \ufb01eld in condensed matter physics arises as the limit of a discrete mapping. In the present example, the \ufb01eld \u03c6(x) is obtained as a continuum approximation of the discrete vector {\u03c6I } \u2208 RN ; the functional L[\u03c6] is the continuum limit of the function L : RN \u2192 R, etc. While in practical calculations \ufb01elds are usually easier to handle than their discrete analogs, it is sometimes helpful to think about problems of \ufb01eld theory in a discrete language. Within the discrete picture, the mathematical apparatus of \ufb01eld theory reduces to \ufb01nite-dimensional calculus. Although Eq. (1.4) contains the full information about the model, we have not yet learned much about its actual behavior. To extract concrete physical information from Eq. (1.4) we need to derive equations of motion. At \ufb01rst sight, it may not be entirely clear what is meant by the term \u201cequations of motion\u201d in the context of an in\ufb01nite-dimensional model: the equations of motion relevant for the present problem are obtained as the generalization of the conventional Lagrange equations of N -particle classical mechanics to a model with in\ufb01nitely many degrees of freedom. To derive these equations we need to generalize Hamilton\u2019s extremal principle (i.e. the route from an action to the associated equations of motion) to in\ufb01nite dimensions. As a warm-up, let us brie\ufb02y recapitulate how the extremal principle works for a system with one degree of freedom. Suppose the dynamics of a classical point particle with coordinate x(t) is described by the dt L(x, \u02d9x). Hamilton\u2019s extremal princlassical Lagrangian L(x, \u02d9x), and action S[x] = ciple states that the con\ufb01gurations x(t) that areactually realized are those that extremize the action, \u03b4S[x] = 0. This means (for a substantiated discussion, see Section 1.2 below) that, for any smooth curve t \u2192 y(t), lim \u21920 1 (S[x + y] \u2212 S[x]) = 0.", " (1.6) To \ufb01rst order in, the action has to remain invariant. Applying this condition, one \ufb01nds that it is ful\ufb01lled if and only if x satis\ufb01es Lagrange\u2019s equation of motion d dt (\u2202 \u02d9xL) \u2212 \u2202xL = 0. (1.7) EXERCISE Recapitulate the derivation of (1.7) from the classical action. 8 From particles to \ufb01elds \u03c6 \u03c6 (x,t) \u03b5\u03b7 (x,t) x T t L Figure 1.4 Schematic showing the variation of the \ufb01eld associated with the action functional. Notice that the variation \u03b7 is supposed to vanish on the boundaries of the base M = [0, L] \u00d7 [0, t]. In Eq. (1.5) we are dealing with a system of in\ufb01nitely many degrees of freedom, \u03c6(x, t). Yet Hamilton\u2019s principle is general and we may see what happens if Eq. (1.5) is subjected to an extremal principle analogous to Eq. (1.6). To do so, we substitute \u03c6(x, t) \u2192 \u03c6(x, t) +\u03b7 (x, t) into Eq. (1.5) and require vanishing of the \ufb01rst-order contribution to an expansion in (see Fig. 1.4). When applied to the speci\ufb01c Lagrangian (1.4), substituting the \u201cvaried\u201d \ufb01eld leads to S[\u03c6 + \u03b7] = S[\u03c6] + dt dx m \u02d9\u03c6 \u02d9\u03b7 \u2212 ksa2\u2202x\u03c6\u2202x\u03b7 + O(2). L 0 Integrating by parts and requiring that the contribution linear in vanishes, one obtains lim \u21920 1 (S[\u03c6 + \u03b7] \u2212 S[\u03c6]) = \u2212 dt 0 L dx m \u00a8\u03c6 \u2212 ksa2\u22022 x\u03c6 \u03b7!= 0. (Notice that the boundary terms vanish identically.) Now, since \u03b7 was de\ufb01ned to be an arbitrary smooth function, the integral above can vanish only if the factor in parentheses is globally vanishing. Thus the equation of motion takes the form of a wave equation m\u22022 t \u2212 ksa2\u22022 x \u03c6 = 0. (1.8) x = \u2013 vt The solutions of Eq. (1.8) have the general form \u03c6+(x \u2212 vt) +\u03c6 \u2212(x + vt) where v = a ks/m, and \u03c6\u00b1 are arbitrary smooth functions of the argument. From this we can deduce that the basic low-energy elementary excitations of our model are lattice vibrations propagating as sound waves to the left or right at a constant velocity v (see the \ufb01gure).7 The trivial behavior of our model is of course a direct consequence of its simplistic de\ufb01nition \u2013 no dissipation, dispersion, or other non-trivial ingredients. Adding these re\ufb01nements leads to the general classical theory of lattice vibrations (such as that described in the text by Ashcroft and Mermin8). Finally, notice that x = vt \u03c6 \u2013 \u03c6 + 7 Strictly speaking, the modeling of our system enforces a periodicity constraint \u03c6\u00b1(x + L) =\u03c6 \u00b1(x). However, in the limit of a large system, this aspect becomes inessential. 8 N. W. Ashcroft and N. D. Mermin, Solid State Physics (Holt\u2013Saunders International, 1983). 1.1 Classical harmonic chain: phonons 9 the elementary excitations of the chain have little in common with its \u201cmicroscopic\u201d constituents (the atomic oscillators). Rather they are collective excitations, i.e. elementary excitations comprising a macroscopically large number of microscopic degrees of freedom. INFO The \u201crelevant\u201d excitations of a condensed matter system can, but need not, be of collective type. For example, the interacting electron gas (a system to be discussed in detail below) supports microscopic excitations \u2013 charged quasi-particles standing in 1:1 correspondence with the electrons of the original microscopic system \u2013 while the collective excitations are plasmon modes of large wavelength and engaging many electrons. Typically, the nature of the fundamental excitations cannot be straightforwardly inferred from the microscopic de\ufb01nition of a model. Indeed, the mere identi\ufb01cation of the relevant excitations often represents the most important step in the solution of a condensed matter", " problem. Hamiltonian formulation An important characteristic of any excitation is its energy. How much energy is stored in the sound waves of the harmonic chain? To address this question, we need to switch back to a Hamiltonian formulation. Once again, this is achieved by generalizing standard manipulations from point mechanics to the continuum. Remembering that, for a Lagrangian of a point particle, p \u2261 \u2202 \u02d9xL is the momentum conjugate to the coordinate x, let us consider the Lagrangian density and de\ufb01ne9 Sir William Rowan Hamilton 1805\u201365 A mathematician credited with the discovery of quaternions, the \ufb01rst non-commutative algebra to be studied. He also invented important new methods in mechanics. (Image from W. R. Hamilton, Collected Papers, vol. II, Cambridge University Press, 1940. \u03c0(x) \u2261 \u2202L(\u03c6, \u2202x\u03c6, \u02d9\u03c6) \u2202 \u02d9\u03c6(x), (1.9) as the canonical momentum associated with \u03c6 (at the point x). In common with \u03c6, the momentum \u03c0 is a continuum degree of freedom. At each space point it may take an independent value. Notice that \u03c0(x) is nothing but the continuum generalization of the lattice momentum PI of Eq. (1.2). (Applied to PI, a continuum approximation like \u03c6I \u2192 \u03c6(x) would produce \u03c0(x).) The Hamiltonian density is then de\ufb01ned as usual through the Legendre transformation, H(\u03c6, \u2202x\u03c6, \u03c0) = \u03c0 \u02d9\u03c6 \u2212 L(\u03c6, \u2202x\u03c6, \u02d9\u03c6) \u02d9\u03c6= \u02d9\u03c6(\u03c6,\u03c0), (1.10) 9 In \ufb01eld theory literature it is popular to denote the momentum by a Greek letter. 10 From particles to \ufb01elds from where the full Hamiltonian is obtained as H = L 0 dx H. EXERCISE Verify that the transition L \u2192 H is a straightforward continuum generalization of the Legendre transformation of the N -particle Lagrangian L({\u03c6I }, { \u02d9\u03c6I }). Having introduced a Hamiltonian, we are in a position to determine the energy of the sound waves. Application of Eq. (1.9) and (1.10) to the Lagrangian of the atomic chain yields \u03c0(x, t) = m \u02d9\u03c6(x, t) and H[\u03c0, \u03c6] = dx \u03c02 2m + ksa2 2 (\u2202x\u03c6)2. (1.11) Considering, say, a right-moving sound-wave excitation, \u03c6(x, t) =\u03c6 +(x \u2212 vt), we \ufb01nd that \u03c0(x, t) = \u2212mv\u2202x\u03c6+(x \u2212 vt) and H[\u03c0, \u03c6] = ksa2 dx[\u2202x\u03c6+(x \u2212 vt)]2 = ksa2 dx [\u2202x\u03c6+(x)]2, i.e. a positive de\ufb01nite, time-independent expression, as one would expect. Before proceeding, let us note an interesting feature of the energy functional: in the limit of an in\ufb01nitely shallow excitation, \u2202x\u03c6+ \u2192 0, the energy vanishes. This sets the stage for the last of the principles (4) hitherto unconsidered, symmetry. The Hamiltonian of an atomic chain is invariant under simultaneous translation of all atom coordinates by a \ufb01xed increment: \u03c6I \u2192 \u03c6I + \u03b4, where\u03b4 is constant. This expresses the fact that a global translation of the solid as a whole does not a\ufb00ect the internal energy. Now, the ground state of any speci\ufb01c realization of the solid will be de\ufb01ned through a static array of atoms, each located at a \ufb01xed coordinate RI = Ia \u21d2 \u03c6I = 0. We say that the translational symmetry is \u201cspontaneously broken,\u201d i.e. the solid has to decide where exactly it wants to rest. However, spontaneous breakdown of a symmetry does not imply that the symmetry disappeared. On the contrary, in\ufb01nite-wavelength deviations from the pre-assigned ground state come close to global translations of (macroscopically large portions of) the solid and, therefore, cost a vanishingly small amount of energy. This is the reason for the vanishing of the sound wave energy in the limit \u2202x\u03c6 \u2192 0", ". It is also our \ufb01rst encounter with the aforementioned phenomenon that symmetries lead to the formation of soft, i.e. low-energy, excitations. A much more systematic exposition of these connections will be given in Chapter 6. Ludwig Boltzmann 1844\u20131906 A physicist whose greatest achievement was in the development of statistical mechanics, which explains and predicts how the properties of atoms (such as mass, charge, and structure) determine the visible properties of matter (such as viscosity, thermal conductivity, and di\ufb00usion). To conclude our discussion of the classical harmonic chain, let us consider the speci\ufb01c heat, a quantity directly accessible in experiment. A rough estimate of this quantity can be readily obtained from the microscopic harmonic Hamiltonian (1.2). According to the principles of statis- tical mechanics, the thermodynamic energy density is given by d\u0393 e\u2212\u03b2H H d\u0393 e\u2212\u03b2H = \u2212 1 L \u2202\u03b2 ln u = 1 L d\u0393 e\u2212\u03b2H, where \u03b2 = 1/kBT, Z \u2261 space volume element d\u0393 = d\u0393e\u2212\u03b2H is the Boltzmann partition function and the phase N I=1 dRI dPI. We will set kB = 1 throughout. The speci\ufb01c heat 1.2 Functional analysis and variational principles 11 is then obtained as c = \u2202T u. To determine the temperature dependence of this quantity, we make use of the fact that, upon rescaling of integration variables, RI \u2192 \u03b2\u22121/2XI, PI \u2192 \u03b2\u22121/2YI, the exponent \u03b2H(R, P ) \u2192 H(X, Y ) becomes independent of temperature (a property that relies on the quadratic dependence of H on both R and P ). The integration measure transforms as d\u0393 \u2192 \u03b2\u2212N I=1 dXI dYI \u2261 \u03b2\u2212N d\u0393. Expressed in terms of the rescaled variables, one obtains the energy density N u = \u2212 1 L \u03b2\u2212N K \u2202\u03b2 ln = \u03c1T, where \u03c1 = N/L is the density of atoms, and we have made use of the fact that the constant d\u0393 e\u2212H(X,Y ) is independent of temperature. We thus \ufb01nd a temperature indepenK \u2261 dent speci\ufb01c heat c = \u03c1. Notice that c is fully universal, i.e. independent of the material constants M and ks determining H. (In fact, we could have anticipated this result from the equipartition theorem of classical mechanics, i.e. the law that in a system with N degrees of freedom, the energy scales as U = N T.) How do these \ufb01ndings compare with experiment? Figure 1.5 shows the speci\ufb01c heat of the insulating compound EuCoO3.10 For large temperatures, the speci\ufb01c heat approaches a constant value, in accord with our analysis. However, for lower temperatures, substantial deviations from c = const. appear. Yet, this strong temperature dependence does not re\ufb02ect a failure of the simplistic microscopic modeling. Rather, the deviation is indicative of a quantum phenomenon. Indeed, we have so far totally neglected the quantum nature of the atomic oscillators. In the next chapter we will rectify this de\ufb01ciency and discuss how the e\ufb00ective low-energy theory of the harmonic chain can be promoted to a quantum \ufb01eld theory. However, before proceeding with the development of the theory let us pause to introduce a number of mathematical concepts that surfaced above, in a way that survives generalization to richer problems. 1.2 Functional analysis and variational principles Let us revisit the derivation of the equations of motion associated with the harmonic chain, Eq. (1.8). Although straightforward, neither was the calculation e\ufb03cient, nor did it reveal general structures. In fact, what we did \u2013 expanding explicitly to \ufb01rst order in the variational parameter \u2013 had the same status as evaluating derivatives by explicitly taking limits: f (x) = lim\u21920 ((f (x + ) \u2212 f (x))/). Moreover, the derivation made explicit use of the particular form of the Lagrangian, thereby being of limited use with regard to a general understanding of the construction scheme. Given the importance attached to extremal principles in all of \ufb01eld theory, it is worthwhile investing some e\ufb00ort in constructing a more e\ufb03cient scheme for the general variational analysis of continuum theories. In order to carry out this program we \ufb01rst need to introduce a mathematical tool of functional analysis, namely the concept of functional di\ufb00erentiation. 10 In metals, the speci", "\ufb01c heat due to lattice vibrations exceeds the speci\ufb01c heat of the free conduction electrons for temperatures larger than a few degrees kelvin. 12 From particles to \ufb01elds 100 50 ) EuCoO3 T3 100 200 300 T (K) Figure 1.5 Speci\ufb01c heat cp of the insulator EuCoO3. At large temperatures, the speci\ufb01c heat starts to approach a constant value, as predicted by the analysis of the classical harmonic chain. However, for small temperatures, deviations from cp = const. are substantial. Such deviations can be ascribed to quantum mechanical e\ufb00ects. (Courtesy of M. Kriener, A. Reichl, T. Lorenz, and A. Freimuth.) In working with functionals, one is often concerned with how a given functional behaves under (small) variations of its argument function. In particular, given a certain function f, suspected to make a functional F [f ] stationary, one would like to \ufb01nd out whether, indeed, the functional remains invariant under variations f \u2192 f +h, where h is an \u201cin\ufb01nitely small\u201d increment function. In ordinary analysis, questions of this type are commonly addressed by exploring derivatives, i.e. we need to generalize the concept of a derivative to functionals. This is achieved by the following de\ufb01nition: a functional F is called di\ufb00erentiable if F [f + g] \u2212 F [f ] = \u00b7 DFf [g] +O (2), where the di\ufb00erential DFf is a linear functional (i.e. one with DFf [g1 + g2] = DFf [g1] + DFf [g2]), is a small parameter, and g is an arbitrary function. The subscript indicates that the di\ufb00erential generally depends on the \u201cbase argument\u201d f. A functional F is said to be stationary on f, if and only if DFf = 0. In principle, the de\ufb01nition above answers our question concerning a stationarity condition. However, to make use of the de\ufb01nition, we still need to know how to compute the di\ufb00erential DF and how to relate the di\ufb00erentiability criterion to the concepts of ordinary calculus. In order to understand how answers to these questions can be systematically found, it is helpful to return temporarily to a discrete way of thinking, i.e. to interpret the argument f of a functional F [f ] as the limit N \u2192 \u221e of a discrete vector f = {fn \u2261 f (xn), n = 1,..., N }, where {xn} denotes a discretization of the support of f (cf. the harmonic chain, \u03c6 \u2194 f ). Prior to taking the continuum limit, N \u2192 \u221e, f has the status of an N -dimensional vector and F (f ) is a function de\ufb01ned over N -dimensional space. After the continuum limit, f \u2192 f becomes a function itself and F (f ) \u2192 F [f ] becomes a functional. Now, within the discrete picture, it is clear how the variational behavior of functions is to be analyzed. For example, the condition that, for all and all vectors g, the linear expansion of F (f + g) ought to vanish is simply to say that the ordinary di\ufb00erential, dFf, de\ufb01ned 1.2 Functional analysis and variational principles 13 through F (f + g) \u2212 F (f ) = \u00b7 dFf (g) +O (2), must be zero. In practice, one often expresses conditions of this type in terms of a certain basis. In a Cartesian basis of N unit vectors, en, n = 1,..., N, dFf (g) \u2261 \u2207Ff, g, where f, g \u2261 n=1 fngn denotes the standard scalar product, \u2207Ff = {\u2202fn F } represents the gradient with the partial derivative de\ufb01ned as N \u2202fn F (f ) \u2261 lim \u21920 1 [F (f + en) \u2212 F (f )]. From these identities, the di\ufb00erential is identi\ufb01ed as dFf (g) = \u2202fn F (f )gn. n (1.12) (1.13) The vanishing of the di\ufb00erential amounts to vanishing of all partial derivatives \u2202fn F = 0. Equations (1.12) and (1.13) can now be straightforwardly generalized to the continuum limit whereupon the summation de\ufb01ning the \ufb01nite-dimensional scalar product translates", " to an integral, f, g = N n=1 fngn \u2192 f, g = dx f (x)g(x). The analog of the nth unit vector is a \u03b4-distribution, en \u2192 \u03b4x, where \u03b4x(x) \u2261 \u03b4(x \u2212 x), as can be seen from the following correspondence: fn!=f, en = m fm(en)m \u2192 f (x)!=f, \u03b4x = dx f (x)\u03b4x(x). Here (en)m = \u03b4nm denotes the mth component of the nth unit vector. The correspondence (unit vector \u2194 \u03b4-distribution) is easy to memorize: while the components of en vanish, save for the nth component, that equals unity, \u03b4x is a function that vanishes everywhere, save for x where it is in\ufb01nite. That a unit component is replaced by \u201cin\ufb01nity\u201d re\ufb02ects the fact that the support of the \u03b4-distribution is in\ufb01nitely narrow; to obtain a unit-normalized integral \u03b4x, the function must be singular. As a consequence of these identities, Eq. (1.13) translates to the continuum di\ufb00erential, DFf [g] = dx \u03b4F [f ] \u03b4f (x) g(x), where the generalization of the partial derivative, \u03b4F [f ] \u03b4f (x) \u2261 lim \u21920 1 (F [f + \u03b4x] \u2212 F [f ]), (1.14) (1.15) is commonly denoted by \u201c\u03b4\u201d instead of \u201c\u2202.\u201d Equations (1.14) and (1.15) establish the conceptual connection between ordinary and functional di\ufb00erentiation. Notice that we have not yet learned how to calculate the di\ufb00erential practically, i.e. to evaluate expressions like Eq. (1.15) for concrete functionals. Nevertheless, the identities above are very useful, enabling us to generalize more-complex derivative operations of ordinary calculus by 14 From particles to \ufb01elds Table 1.1 Summary of basic de\ufb01nitions of discrete and continuum calculus. Entity Discrete Argument Function(al) Di\ufb00erential Cartesian basis Scalar product, \u201cPartial derivative\u201d vector f multidimensional function F (f ) dFf (g) en n fngn \u2202fn F (f ) Continuum function f functional F [f ] DFf [g] \u03b4x dx f (x)g(x) \u03b4F [f ] \u03b4f (x) straightforward transcription. For example, the generalization of the standard chain rule, \u2202fn F (g(f )) = m \u2202gm F (g) g=g(f ) \u2202fn gm(f ) reads dy = \u03b4F [g] \u03b4g(y) \u03b4F [g[f ]] \u03b4f (x) \u03b4g(y)[f ] \u03b4f (x). g=g[f ] Here g[f ] is the continuum generalization of an Rm-valued function, g : Rn \u2192 Rm, a function whose components g(y)[f ] are functionals by themselves. Furthermore, given some functional F [f ], we can construct its Taylor expansion as F [f ] = F [0] + dx1 \u03b4F [f ] \u03b4f (x1) f =0 f (x1) + 1 2 dx1 dx2 \u03b42F [f ] \u03b4f (x2)\u03b4f (x1) f =0 f (x1)f (x2) +\u00b7 \u00b7 \u00b7, where (exercise) \u03b42F [f ] \u03b4f (x2)\u03b4f (x1) = lim 1,2\u21920 1 12 (F [f + 1\u03b4x1 + 2\u03b4x2 ] \u2212 F [f + 1\u03b4x1 ] \u2212 F [f + 2\u03b4x2 ] +F [f ]) generalizes a two-fold partial derivative. The validity of these identities can be made plausible by applying the transcription, Table 1.1, to the corresponding relations of standard calculus. To actually verify the formulae, one has to take the continuum limit of each step taken in the discrete variant of the corresponding proofs. At any rate, experience shows that it takes some time to get used to the concept of functional di\ufb00erentiation. However, after some practice it will become clear that this operation is not only extremely useful but also as easy to handle as conventional partial di\ufb00erentiation.", " We \ufb01nally address the question how to compute functional derivatives in practice. In doing so, we will make use of the fact that, in all but a few cases, the functionals encountered in \ufb01eld theory are of the structure S[\u03c6] = dmx L(\u03c6i, \u2202\u03bc\u03c6i). M (1.16) Here, we assume the base manifold M to be parameterized by an m-dimensional coordinate vector x = {x\u03bc}. (In most practical applications, m = d + 1, and x = (x0, x1,..., xd) 1.3 Maxwell\u2019s equations as a variational principle 15 contains one time-like component x0 = t and d space-like components xk, k = 1,..., d.11) We further assume that the \ufb01eld manifold has dimensionality n and that \u03c6i, i = 1,..., n, are the coordinates of the \ufb01eld. What makes the functional S[\u03c6] easy to handle is that all of its information is stored in the function L. Owing to this simpli\ufb01cation, the functional derivative can be related to an ordinary derivative of L. To see this, all that we have to do is to evaluate the general de\ufb01nition Eq. (1.14) on the functional S: S[\u03c6 + \u03b8] \u2212 S[\u03c6] = dmx [L(\u03c6 + \u03b8, \u2202\u03bc\u03c6 + \u2202\u03bc\u03b8) \u2212 L(\u03c6, \u2202\u03bc\u03c6)] = dmx M M \u2202L \u2202\u03c6i \u03b8i + \u2202L \u2202\u2202\u03bc\u03c6i \u2202\u03bc\u03b8i + O(2) = dmx M \u2202L \u2202\u03c6i \u2212 \u2202\u03bc \u2202L \u2202\u2202\u03bc\u03c6i \u03b8i + O(2), where in the last line we have assumed that the \ufb01eld variation vanishes on the boundary of the base manifold, \u03b8 |\u2202M = 0. Comparison with Eq. (1.14) identi\ufb01es the functional derivative as \u03b4S[\u03c6] \u03b4\u03c6i(x) = \u2202L \u2202\u03c6i(x) \u2212 \u2202\u03bc \u2202L \u2202(\u2202\u03bc\u03c6i(x)). We conclude that stationarity of the functional (1.16) is equivalent to the condition \u2200x, i : \u2202L \u2202\u03c6i(x) \u2212 \u2202\u03bc \u2202L \u2202\u2202\u03bc\u03c6i(x) = 0. (1.17) Equation (1.17) is known as the Euler\u2013Lagrange equation of \ufb01eld theory. In fact, for d = 0 andx 0 = t, Eq. (1.17) reduces to the familiar Euler\u2013Lagrange equation of a point particle in n-dimensional space. For d = 1 and (x0, x1) = (t, x) we get back to the stationarity equations discussed in the previous section. In the next section we will apply the formalism to a higher-dimensional problem. 1.3 Maxwell\u2019s equations as a variational principle As a second example, let us consider the archetype of classical \ufb01eld theory, classical electrodynamics. Indeed, as well as exemplifying the application of continuum variational principles on a problem with which we are all acquainted, this example illustrates the unifying potential of the approach: that problems as di\ufb00erent as the low-lying vibrational modes of a crystalline solid, and electrodynamics can be described by almost identical language indicates that we are dealing with a useful formalism. Speci\ufb01cally, our aim will be to explore how the equations of motion of electrodynamics, the inhomogeneous Maxwell equations, \u2207 \u00b7 E = \u03c1, \u2207 \u00d7B \u2212 \u2202tE = j, (1.18) 11 Following standard conventions we denote space-like components by small Latin indices k = 1,..., d. In con- trast, space-time indices are denoted by Greek indices \u03bc = 0,..., d. 16 From particles to \ufb01elds can be obtained from variational principles. (For simplicity, we restrict ourselves to a vacuum theory, i.e. E = D and B = H. Further, we have set the velocity of light to unity,c = 1. Within the framework of the variational principle, the homogeneous equations, \u2207 \u00d7 E + \u2202tB = 0, \u2207 \u00b7 B = 0,", " (1.19) are regarded as ab initio constraints imposed on the \u201cdegrees of freedom\u201d E and B.) James Clerk Maxwell 1831\u201379 Scottish theoretical physicist and mathematician. Amongst many other achievements, he is credited with the formulation of the theory of electromagnetism, synthesizing all previous unrelated experiments and equations of electricity, magnetism and optics into a consistent theory. (He is also known for creating the \ufb01rst true color photograph in 1861.) To formulate Maxwell\u2019s theory as a variational principle we require (1) a \ufb01eld formulated in a set of suitable \u201cgeneralized coordinates,\u201d and (2) its action. As to coordinates, the natural choice will be the coe\ufb03cients of the electromagnetic (EM) 4-potential, A\u03bc = (\u03c6, \u2212A), where \u03c6 is the scalar and A is the vector potential. The potential A is unconstrained and uniquely determines the \ufb01elds E and B through the standard equations E = \u2212\u2207\u03c6 \u2212 \u2202tA and B = \u2207 \u00d7A. (In fact, the set of coordinates A\u03bc is \u201coverly free\u201d in the sense that gauge transformations A\u03bc \u2192 A\u03bc + \u2202\u03bc\u0393, where \u0393 is an arbitrary function, leave the physical \ufb01elds invariant. Later we will comment explicitly on this point.) The connection between A and the physical \ufb01elds can be expressed in a more symmetric way by introducing the EM \ufb01eld tensor, F = {F\u03bc\u03bd} = \u23a1 \u23a2 \u23a2 \u23a3 E1 E3 E2 0 \u2212B3 B2 0 \u2212B1 0 \u2212E1 \u2212E2 B3 \u2212E3 \u2212B2 B1 0 \u23a4 \u23a5 \u23a5 \u23a6. (1.20) The relation between \ufb01elds and potential now reads F\u03bc\u03bd = \u2202\u03bcA\u03bd \u2212\u2202\u03bdA\u03bc where x\u03bc = (t, \u2212x) and \u2202\u03bc = (\u2202t, \u2207). EXERCISE Con\ufb01rm that this connection follows from the de\ufb01nition of the vector potential. To verify that the constraint (1.19) is automatically included in the de\ufb01nition (1.20), compute the construct \u2202\u03bbF\u03bc\u03bd + \u2202\u03bcF\u03bd\u03bb + \u2202\u03bd F\u03bb\u03bc, where (\u03bb\u03bd\u03bc) represent arbitrary but di\ufb00erent indices. This produces four di\ufb00erent terms, identi\ufb01ed as the left-hand side of Eq. (1.19). Evaluation of the same construct on F\u03bc\u03bd \u2261 \u2202\u03bcA\u03bd \u2212 \u2202\u03bd A\u03bc produces zero by the symmetry of the right-hand side. 1.3 Maxwell\u2019s equations as a variational principle 17 As to the structure of the action S[A] we can proceed in di\ufb00erent ways. One option would be to regard Maxwell\u2019s equations as fundamental, i.e. to construct an action that produces these equations upon variation (by analogy with the situation in classical mechanics where the action functional was designed so as to reproduce Newton\u2019s equations). However, we can also be a little bit more ambitious and ask whether the structure of the action can be motivated independently of Maxwell\u2019s equations. In fact, there is just one principle in electrodynamics as \u201cfundamental\u201d as Maxwell\u2019s equations: symmetry. A theory of electromagnetism must be Lorentz invariant, i.e. invariant under relativistic coordinate transformations. Hendrik Antoon Lorentz 1853\u2013 1928 1902 Nobel Laureate in Physics (with Pieter Zeeman) in recognition of the extraordinary service their rendered researches into the in\ufb02uence of magnetism upon radiation phenomena. Lorentz derived the transformation equations subsequently used by Albert Einstein to describe space and time. they by INFO Let us brie\ufb02y recapitulate the notion of Lorentz invariance. Suppose we are given a 4-vector X\u03bc. A linear coordinate transformation X\u03bc \u2192 X \u03bc \u2261 T\u03bc\u03bd X\u03bd is a Lorentz transformation if it leaves the 4-metric g = {g\u03bc\u03bd } = 1 \u23a1 \u23a2 \u23a3 \u22121 \u22121 \u23a4 \u23a5 \u23a6, \u22121 (1.21) invariant: T T gT = g. To concisely formulate the invariance properties of relativistic theories, it is common to introduce the notion of raised and lowered indices. De\ufb01ning X \u03bc \u2261 g\u03bc\u03bd X\u03bd, Lorentz invariance is expressed as X \u03bc", "X\u03bc = X \u03bcX \u03bc. Aided by the symmetry criterion, we can attempt to conjecture the structure of the action from three basic assumptions, all independent of Maxwell\u2019s equations: the action should be invariant under (a) Lorentz transformations and (b) gauge transformations, and (c) it should be simple! The most elementary choice compatible with these conditions is S[A] = d4x (c1 F\u03bc\u03bdF \u03bc\u03bd + c2 A\u03bcj\u03bc), (1.22) \u03bc dx\u03bc = dt dx1 dx2 dx3 denotes the measure, j\u03bc = (\u03c1, \u2212j) the 4-current, where d4x = and c1,2 are undetermined constants. Up to quadratic order in A, Eq. (1.22) in fact de\ufb01nes the only possible structure consistent with gauge and Lorentz invariance. EXERCISE Using the continuity equation \u2202\u03bcj\u03bc = 0, verify that the Aj-coupling is gauge invariA\u03bcA\u03bc would not be gauge ant. (Hint: Integrate by parts.) Verify that a contribution like invariant. 18 From particles to \ufb01elds Having de\ufb01ned a trial action, we can apply the variational principle Eq. (1.17) to compute equations of motion. In the present context, the role of the \ufb01eld \u03c6 is taken by the four components of A. Variation of the action with respect to A\u03bc gives four equations of motion, \u2202L \u2202A\u03bc \u2212 \u2202\u03bd \u2202L \u2202(\u2202\u03bdA\u03bc) = 0, \u03bc = 0,..., 3, (1.23) d4x L. With the speci\ufb01c form of L, it is where the Lagrangian density is de\ufb01ned by S = L = \u22124c1F \u03bc\u03bd. Substitution of these L = c2j\u03bc and \u2202\u2202\u03bd A\u03bc straightforward to verify that \u2202A\u03bc building blocks into the equations of motion \ufb01nally yields 4c1\u2202\u03bdF\u03bd\u03bc = c2j\u03bc. Comparing this with the de\ufb01nition of the \ufb01eld tensor (1.20), and setting c1/c2 = 1/4, we arrive at Maxwell\u2019s equations (1.18). We \ufb01nally \ufb01x the overall multiplicative constant c1 (= c2/4) by requiring that the Hamiltonian density associated with the Lagrangian density L reproduce the known energy density of the EM \ufb01eld (see Problem 1.8). This leads to c1 = \u22121/4, so that we have identi\ufb01ed L(A\u03bc, \u2202\u03bdA\u03bc) = \u2212 1 4 F\u03bc\u03bdF \u03bc\u03bd + A\u03bcj\u03bc, (1.24) as the Lagrangian density of the electromagnetic \ufb01eld. The corresponding action is given by S[A] = d4x L(A\u03bc, \u2202\u03bdA\u03bc). At \ufb01rst sight, this result does not look particularly surprising. After all, Maxwell\u2019s equations can be found on the \ufb01rst page of most textbooks on electrodynamics. However, further re\ufb02ection will show that our achievement is actually quite remarkable. By invoking only symmetry, the algebraic structure of Maxwell\u2019s equations has been established unambiguously. We have thus proven that Maxwell\u2019s equations are relativistically invariant, a fact not obvious from the equations themselves. Further, we have shown that Eq. (1.18) are the only equations of motion linear in the current\u2013density distribution and consistent with the invariance principle. One might object that, in addition to symmetry, we have also imposed an ad hoc \u201csimplicity\u201d criterion on the action S[A]. However, later we will see that this was motivated by more than mere aesthetic principles. Finally, we note that the symmetry-oriented modeling that led to Eq. (1.22) is illustrative of a popular construction scheme in modern \ufb01eld theory. The symmetry-oriented approach stands complementary to the \u201cmicroscopic\u201d formulation exempli\ufb01ed in Section 1.1. Crudely speaking, these are the two principal approaches to constructing e\ufb00ective low-energy \ufb01eld theories: Microscopic analysis: Starting from a microscopically de\ufb01ned system, one projects onto those degrees of freedom that one believes are relevant for the low-energy dynamics. Ideally, this \u201cbelief\u201d is backed up by a small expansion parameter stabilizing the mathematical parts of the analysis. Advant", "ages: The method is rigorous and \ufb01xes the resulting \ufb01eld theory completely. Disadvantages: The method is time-consuming and, for su\ufb03ciently complex systems, not even viable. Symmetry considerations: One infers an e\ufb00ective low-energy theory on the basis of only fundamental symmetries of the physical system. Advantages: The method is fast and 1.4 Quantum chain 19 elegant. Disadvantages: It is less explicit than the microscopic approach. Most importantly, it does not \ufb01x the coe\ufb03cients of the di\ufb00erent contributions to the action. Thus far, we have introduced some basic concepts of \ufb01eld theoretical modeling in condensed matter physics. Starting from a microscopic model Hamiltonian, we have illustrated how principles of universality and symmetry can be applied to distill e\ufb00ective continuum \ufb01eld theories capturing the low-energy content of the system. We have formulated such theories in the language of Lagrangian and Hamiltonian continuum mechanics, respectively, and shown how variational principles can be applied to extract concrete physical information. Finally, we have seen that \ufb01eld theory provides a unifying framework whereby analogies between seemingly di\ufb00erent physical systems can be uncovered. In the next section we discuss how the formalism of classical \ufb01eld theory can be elevated to the quantum level. 1.4 Quantum chain Earlier we saw that, at low temperatures, the excitation pro\ufb01le of the classical atomic chain di\ufb00ers drastically from that observed in experiment. Generally, in condensed matter physics, low-energy phenomena with pronounced temperature sensitivity are indicative of a quantum mechanism at work. To introduce and exemplify a general procedure whereby quantum mechanics can be incorporated into continuum models, we next consider the low-energy physics of the quantum mechanical atomic chain. The \ufb01rst question to ask is conceptual: how can a model like Eq. (1.4) be quantized in general? Indeed, there exists a standard procedure for quantizing continuum theories, which closely resembles the quantization of Hamiltonian point mechanics. Consider the de\ufb01ning Eq. (1.9) and (1.10) for the canonical momentum and the Hamiltonian, respectively. Classically, the momentum \u03c0(x) and the coordinate \u03c6(x) are canonically conjugate variables: {\u03c0(x), \u03c6(x)} = \u2212\u03b4(x\u2212x) where{, } is the Poisson bracket and the \u03b4-function arises through continuum generalization of the discrete identity {PI, RI } = \u2212\u03b4II, I, I = 1,..., N. The theory is quantized by generalization of the canonical quantization procedure for the discrete pair of conjugate coordinates (RI, PI ) to the continuum: (i) promote \u03c6(x) and\u03c0 (x) to operators, \u03c6 \u2192 \u02c6\u03c6, \u03c0 \u2192 \u02c6\u03c0, and (ii) generalize the canonical commutation relation [PI, RI ] = \u2212i\u03b4II to12 [\u02c6\u03c0(x), \u02c6\u03c6(x)] = \u2212i\u03b4(x \u2212 x). (1.25) Operator-valued functions like \u02c6\u03c6 and \u02c6\u03c0 are generally referred to as quantum \ufb01elds. For clarity, the relevant relations between canonically conjugate classical and quantum \ufb01elds are summarized in Table 1.2. INFO By introducing quantum \ufb01elds, we have departed from the conceptual framework laid out on page 6: being operator-valued, the quantized \ufb01eld no longer represents a mapping into an 12 Note that the dimensionality of both the quantum and the classical continuum \ufb01eld is compatible with the dimensionality of the Dirac \u03b4-function, [\u03b4(x \u2212 x)] = [length]\u22121, i.e. [\u03c6(x)] = [\u03c6I ] \u00b7 [length]\u22121/2 and similarly for \u03c0. 20 From particles to \ufb01elds Table 1.2 Relations between discrete and continuum canonically conjugate variables/operators. Classical Quantum Discrete Continuum {PI, RI } = \u2212\u03b4II {\u03c0(x), \u03c6(x)} = \u2212\u03b4(x \u2212 x) [ \u02c6PI, \u02c6RI ] = \u2212i\u03b4II [\u02c6\u03c0(x), \u02c6\u03c6(x)] = \u2212i\u03b4(x \u2212 x) ordinary di\ufb00erentiable manifold.13 It is thus legitimate to ask why we bothered to give a lengthy exposition of \ufb01elds as \u201cordinary\u201d functions. The reason", " is that, in the not too distant future, after the framework of functional \ufb01eld integration has been introduced, we will return to the comfortable ground of the de\ufb01nition of page 6. Employing these de\ufb01nitions, the classical Hamiltonian density (1.10) becomes the quantum operator \u02c6H( \u02c6\u03c6, \u02c6\u03c0) = 1 2m \u02c6\u03c02 + ksa2 2 (\u2202x \u02c6\u03c6)2. (1.26) The Hamiltonian above represents a quantum \ufb01eld theoretical formulation of the problem but not yet a solution. In fact, the development of a spectrum of methods for the analysis of quantum \ufb01eld theoretical models will represent a major part of this text. At this point the objective is merely to exemplify the way physical information can be extracted from models like Eq. (1.26). As a word of caution, let us mention that the following manipulations, while mathematically straightforward, are conceptually deep. To disentangle di\ufb00erent aspects of the problem, we will \ufb01rst concentrate on plain operational aspects. Later, in Section 1.4, we will re\ufb02ect on \u201cwhat has really happened.\u201d As with any function, operator-valued functions can be represented in a variety of di\ufb00erent ways. In particular, they can be subjected to Fourier transformation, \u02c6\u03c6k \u02c6\u03c0k \u2261 1 L1/2 L 0 dx e{\u2213ikx \u02c6\u03c6(x) \u02c6\u03c0(x), \u02c6\u03c6(x) \u02c6\u03c0(x) = 1 L1/2 k e{\u00b1ikx \u02c6\u03c6k \u02c6\u03c0k, (1.27) where k represents the sum over all Fourier coe\ufb03cients indexed by quantized momenta k = 2\u03c0m/L, m \u2208 Z (not to be confused with the \u201coperator momentum\u201d \u02c6\u03c0!). Note that the real classical \ufb01eld \u03c6(x) quantizes to a Hermitian quantum \ufb01eld \u02c6\u03c6(x), implying that \u2020 \u02c6\u03c6k = \u02c6\u03c6 \u2212k (and similarly for \u02c6\u03c0k). The corresponding Fourier representation of the canonical commutation relations reads (exercise) [\u02c6\u03c0k, \u02c6\u03c6k ] = \u2212i\u03b4kk. (1.28) 13 At least if we ignore the mathematical subtlety that a linear operator can also be interpreted as an element of a certain manifold. 1.4 Quantum chain 21 When expressed in the Fourier representation, making use of the identity dx (\u2202x \u02c6\u03c6)2 = (\u2212ik \u02c6\u03c6k)(\u2212ik \u02c6\u03c6k ) k,k \u03b4k+k,0! dx e\u2212i(k+k)x= 1 L k k2 \u02c6\u03c6k \u02c6\u03c6\u2212k = k k2| \u02c6\u03c6k|2 together with a similar relation for dx H( \u02c6\u03c6, \u02c6\u03c0) assumes the near diagonal form dx \u02c6\u03c02, the Hamiltonian \u02c6H = \u02c6H = k 1 2m \u02c6\u03c0k \u02c6\u03c0\u2212k + \u02c6\u03c6k \u02c6\u03c6\u2212k, m\u03c92 k 2 (1.29) where \u03c9k = v|k| and v = a ks/m denotes the classical sound wave velocity. In this form, the Hamiltonian can be identi\ufb01ed as nothing but a superposition of independent harmonic oscillators.14 This result is actually not di\ufb03cult to understand (see \ufb01gure): Classically, the system supports a discrete set of wave excitations, each indexed by a wave number k = 2\u03c0m/L. (In fact, we could have performed a Fourier transformation of the classical \ufb01elds \u03c6(x) and\u03c0 (x) to represent the Hamiltonian function as a superposition of classical harmonic oscillators.) Within the quantum picture, each of these excitations is described by an oscillator Hamiltonian operator with a k-dependent frequency. However, it is important not to confuse the atomic constituents, also oscillators (albeit coupled), with the independent collective oscillator modes described by \u02c6H. The description above, albeit perfectly valid, still su\ufb00ers from a de\ufb01ciency: our analysis amounts to explicitly describing the e\ufb00ective low-energy excitations of the system (the waves) in terms", " of their microscopic constituents (the atoms). Indeed the di\ufb00erent contributions to \u02c6H keep track of details of the microscopic oscillator dynamics of individual k-modes. However, it would be much more desirable to develop a picture where the relevant excitations of the system, the waves, appear as fundamental units, without explicit account of underlying microscopic details. (As with hydrodynamics, information is encoded in terms of collective density variables rather than through individual molecules.) As preparation for the construction of this improved formulation of the system, let us temporarily focus on a single oscillator mode. Revision of the quantum harmonic oscillator Consider a standard harmonic oscillator (HO) Hamiltonian \u02c6H = \u02c6p2 2m + m\u03c92 2 \u02c6x2. 14 The only di\ufb00erence between Eq. (1.29) and the canonical form of an oscillator Hamiltonian \u02c6H = \u02c6p2/(2m) + k = \u02c6\u03c6\u2212k). As we will show shortly, \u2020 m\u03c92 \u02c6x2/2 is the presence of the sub-indices k and \u2212k (a consequence of \u02c6\u03c6 this di\ufb00erence is inessential. 22 From particles to \ufb01elds \u03c9 Figure 1.6 Low-lying energy levels/states of the harmonic oscillator. The \ufb01rst few energy levels n = \u03c9 and the associated Hermite polynomial eigenfunctions are displayed schematically in Fig. 1.6. (To simplify the notation we henceforth set = 1.) n + 1 2 The HO has, of course, the status of a single-particle problem. However, the equidistance of its energy levels suggests an alternative interpretation. One can think of a given energy state n as an accumulation of n elementary entities, or quasi-particles, each having energy \u03c9. What can be said about the features of these new objects? First, they are structureless, i.e. the only \u201cquantum number\u201d identifying the quasi-particles is their energy \u03c9 (otherwise n-particle states formed of the quasi-particles would not be equidistant). This implies that the quasi-particles must be bosons. (The same state \u03c9 can be occupied by more than one particle, see Fig. 1.6.) This idea can be formulated in quantitative terms by employing the formalism of ladder operators in which the operators \u02c6p and \u02c6x are traded for the pair of Hermitian adjoint 2 (\u02c6x\u2212 i operators \u02c6a \u2261 m\u03c9 \u02c6p). Up to a factor of i, the transformation (\u02c6x, \u02c6p) \u2192 (\u02c6a, \u02c6a\u2020) is canonical, i.e. the new operators obey the canonical commutation relation m\u03c9 \u02c6p), \u02c6a\u2020 \u2261 2 (\u02c6x+ i m\u03c9 m\u03c9 [\u02c6a, \u02c6a\u2020] = 1. More importantly, the a-representation of the Hamiltonian is very simple, namely \u02c6H = \u03c9 \u02c6a\u2020\u02c6a +, 1 2 (1.30) (1.31) as can be checked by direct substitution. Suppose, now, we had been given a zero eigenvalue state |0 of the operator \u02c6a: \u02c6a|0 = 0. As a direct consequence, \u02c6H|0 = (\u03c9/2)|0, i.e. |0 is identi\ufb01ed as the ground state of the oscillator.15 The complete hierarchy of higher energy states can now be generated by setting |n \u2261 (n!)\u22121/2 (\u02c6a\u2020)n|0. EXERCISE Using the canonical commutation relation, verify that \u02c6H|n = \u03c9(n + 1/2)|n and n|n = 1. Formally, the construction above represents yet another way of constructing eigenstates of the quantum HO. However, its \u201creal\u201d advantage is that it naturally a\ufb00ords a many-particle interpretation. To this end, let us declare |0 to represent a \u201cvacuum\u201d state, i.e. a state with zero particles present. Next, imagine that \u02c6a\u2020|0 is a state with a single featureless particle 15 This can be veri\ufb01ed by explicit construction. Switching to a real-space representation, the solution of the m\u03c9/2\u03c0e\u2212m\u03c9x2 /2. equation [x + \u2202x/(m\u03c9)]x|0 = 0 obtains the familiar ground state wavefunction x|0 = 1.4 Quantum chain 23 \ufffd", "\ufffd \u03c9 1 k1 \u03c9 2 k2 \u03c9 3 k3 k Figure 1.7 Diagram visualizing an excited state of the chain. Here, the number of quasi-particles decreases with increasing energy \u03c9k. (the operator \u02c6a\u2020 does not carry any quantum number labels) of energy \u03c9. Similarly, (\u02c6a\u2020)n|0 is considered as a many-body state with n particles, i.e. within the new picture, \u02c6a\u2020 is an operator that creates particles. The total energy of these states is given by \u03c9 \u00d7 (occupation number). Indeed, it is straightforward to verify (see exercise above) that \u02c6a\u2020\u02c6a|n = n|n, i.e. the Hamiltonian basically counts the number of particles. While, at \ufb01rst sight, this may look unfamiliar, the new interpretation is internally consistent. Moreover, it achieves what we had asked for above, i.e. it allows an interpretation of the HO states as a superposition of independent structureless entities. INFO The representation above illustrates the capacity to think about individual quantum problems in complementary pictures. This principle \ufb01nds innumerable applications in modern condensed matter physics. The existence of di\ufb00erent interpretations of a given system is by no means heretical but, rather, re\ufb02ects a principle of quantum mechanics: there is no \u201cabsolute\u201d system that underpins the phenomenology. The only thing that matters is observable phenomena. For example, we will see later that the \u201c\ufb01ctitious\u201d quasi-particle states of oscillator systems behave as \u201creal\u201d particles, i.e. they have dynamics, can interact, be detected experimentally, etc. From a quantum point of view these object are, then, real particles. Quasi-particle interpretation of the quantum chain Returning to the oscillator chain, one can transform the Hamiltonian (1.29) to a form analogous to (1.31) by de\ufb01ning the ladder operators16 \" \" \u02c6ak \u2261 m\u03c9k 2 \u02c6\u03c6k + i m\u03c9k \u02c6\u03c0\u2212k, \u2020 \u02c6a k \u2261 m\u03c9k 2 \u02c6\u03c6\u2212k \u2212 i m\u03c9k \u02c6\u03c0k. (1.32) With this de\ufb01nition, applying the commutation relations Eq, (1.28), one \ufb01nds that the ladder operators obey commutation relations generalizing Eq. (1.30): # $ \u02c6ak, \u02c6a \u2020 k # $ \u2020 k, \u02c6a \u02c6a \u2020 k = \u03b4kk, [\u02c6ak, \u02c6ak ] = = 0. (1.33) \u2020 16 As to the consistency of these de\ufb01nitions, recall that \u02c6\u03c6 k = \u02c6\u03c6\u2212k and \u02c6\u03c0 \u2020 k = \u02c6\u03c0\u2212k. Under these conditions the second of the de\ufb01nitions following in the text follows from the \ufb01rst upon taking the Hermitian adjoint. 24 \u03c9 From particles to \ufb01elds 0 1 1 1 1 0.5 0.5 0.5 0.5 1 1 q Figure 1.8 Phonon spectra of the transition metal oxide Sr2RuO4 measured along di\ufb00erent axes in momentum space. Notice the approximate linearity of the low-energy branches (acoustic phonons) at small momenta. Superimposed at high frequencies are various branches of optical phonons. (Source: Courtesy of M. Braden, II. Physikalisches Institut, Universit\u00a8at zu K\u00a8oln.) Expressing the operators ( \u02c6\u03c6k, \u02c6\u03c0k) in terms of (\u02c6ak, \u02c6a Hamiltonian into the quasi-particle oscillator form (exercise) \u2020 k), it is now straightforward to bring the \u02c6H = k \u03c9k \u02c6a \u2020 k\u02c6ak + 1 2. (1.34) Equations (1.34) and (1.33) represent the \ufb01nal result of our analysis. The Hamiltonian \u02c6H takes the form of a sum of harmonic oscillators with characteristic frequencies \u03c9k. In the limit k \u2192 0 (i.e. long wavelength), one \ufb01nds \u03c9k \u2192 0; excitations with this property are said to be massless. An excited state of the system is indexed by a set {nk} = (n1, n2,... ) of quasi-particles with energy {\u03c9k} (see Fig", ". 1.7). Physically, the quasi-particles of the harmonic chain are identi\ufb01ed with the phonon modes of the solid. A comparison with measured phonon spectra (Fig. 1.8) reveals that, at low momenta, \u03c9k \u223c |k| in agreement with our simplistic model (even in spite of the fact that the spectrum was recorded for a three-dimensional solid with non-trivial unit cell \u2013 universality!). While the linear dispersion was already a feature of the classical sound wave spectrum, the low-temperature speci\ufb01c heat re\ufb02ected non-classical behavior. It is left as an exercise (problem 1.8) to verify that the quantum nature of the phonons resolves the problem with the low-temperature speci\ufb01c heat discussed in Section 1.1. (For further discussion of phonon modes in atomic lattices we refer to Chapter 2 of the text by Kittel.17) 1.5 Quantum electrodynamics The generality of the procedure outlined above suggests that the quantization of the EM \ufb01eld Eq. (1.24) proceeds in a manner analogous to the phonon system. However, there are a number of practical di\ufb00erences that make quantization of the EM \ufb01eld a harder (but also more interesting!) enterprise. Firstly, the vectorial character of the vector potential, in combination with the condition of relativistic covariance, gives the problem a non-trivial internal geometry. Closely related, the gauge freedom of the vector potential introduces redundant degrees of freedom whose removal on the quantum level is not easily achieved. For 17 C. Kittel, Quantum Theory of Solids, 2nd edition (Wiley, 1987). 1.5 Quantum electrodynamics 25 example, quantization in a setting where only physical degrees of freedom are kept \u2013 i.e. the two polarization directions of the transverse photon \ufb01eld \u2013 is technically cumbersome, the reason being that the relevant gauge condition is not relativistically covariant. In contrast, a manifestly covariant scheme, while technically more convenient, introduces spurious \u201cghost degrees of freedom\u201d that are di\ufb03cult to remove. To circumvent a discussion of these issues, we will not discuss the problem of EM \ufb01eld quantization in full detail.18 On the other hand, the photon \ufb01eld plays a much too important role in condensed matter physics for us to drop the problem altogether. We will therefore aim at an intermediate exposition, largely insensitive to the problems outlined above but su\ufb03ciently general to illustrate the main principles. Field quantization & Consider the Lagrangian of the matter-free EM \ufb01eld, L = \u2212 1 d3x F\u03bc\u03bdF \u03bc\u03bd. As a \ufb01rst step 4 towards quantization of this system we \ufb01x a gauge. In the absence of charge, a particularly convenient choice is the Coulomb gauge, \u2207 \u00b7A = 0, with the scalar component \u03c6 = 0. (Keep in mind that, once a gauge has been set, we cannot expect further results to display \u201cgauge invariance.\u201d) Using the gauge conditions, one may verify that the Lagrangian assumes the form % L = 1 2 d3x (\u2202tA)2 \u2212 (\u2207 \u00d7 A)2. (1.35) By analogy with our discussion of the atomic chain, we would now proceed to \u201cdecouple\u201d the theory by expanding the action in terms of eigenfunctions of the Laplace operator. The di\ufb00erence with our previous discussion is that we are dealing (i) with the full threedimensional Laplacian (instead of a simple second derivative) acting on (ii) the vectorial quantity A that is (iii) subject to the constraint \u2207 \u00b7A = 0. It is these aspects which lead to the complications outlined above. We can circumvent these di\ufb03culties by considering cases where the geometry of the system reduces the complexity of the eigenvalue problem. This restriction is less arti\ufb01cial than it might appear. For example, in anisotropic electromagnetic waveguides, the solutions of the eigenvalue equation can be formulated as19 \u2212\u22072Rk(x) = \u03bbkRk(x), (1.36) where k \u2208 R is a one-dimensional index and the vector-valued functions Rk are real and Rk \u00b7 Rk = \u03b4kk. The dependence of the eigenvalues \u03bbk on k depends on orthonormalized, details of the geometry (see Eq. (1.38) below) and need not be speci\ufb01ed for the moment. INFO An electrodynamic", " waveguide is a quasi-one-dimensional cavity with metallic boundaries (see Fig. 1.9). The practical advantage of waveguides is that they are good at con\ufb01ning EM 18 Readers interested in learning more about EM \ufb01eld quantization are referred to, e.g., L. H. Ryder, Quantum Field Theory (Cambridge University Press, 1996). 19 More precisely, one should say that Eq. (1.36) de\ufb01nes the set of eigenfunctions relevant for the low-energy dynamics of the waveguide. More-complex eigenfunctions of the Laplace operator exist but they carry much higher energy. 26 From particles to \ufb01elds z y x Ly Lz Figure 1.9 EM waveguide with rectangular cross-section. The structure of the eigenmodes of the EM \ufb01eld is determined by boundary conditions at the walls of the cavity. waves. At large frequencies, where the wavelengths are of order meters or less, radiation loss in conventional conductors is high. In these frequency domains, hollow conductors provide the only practical way of transmitting radiation. EM \ufb01eld propagation inside a waveguide is constrained by boundary conditions. Assuming the walls of the system to be perfectly conducting, E(xb) = 0, B\u22a5(xb) = 0, (1.37) where xb is a point at the system boundary and E (B\u22a5) is the parallel (perpendicular) component of the electric (magnetic) \ufb01eld. For concreteness, returning to the problem of \ufb01eld quantization, let us consider a cavity with uniform rectangular cross-section Ly \u00d7 Lz. To conveniently represent the Lagrangian of the system, we wish to express the vector potential in terms of eigenfunctions Rk that are consistent with the boundary conditions (1.37). A complete set of functions ful\ufb01lling this condition is given by \u239b \u239e Rk = Nk \u239d c1 cos(kxx) sin(kyy) sin(kzz) c2 sin(kxx) cos(kyy) sin(kzz) c3 sin(kxx) sin(kyy) cos(kzz) \u23a0. Here, ki = ni\u03c0/Li, ni \u2208 N, i = x, y, z, Nk is a factor normalizing Rk to unit modulus, and the coe\ufb03cients ci are subject to the condition c1kx + c2ky + c3kz = 0. Indeed, it is straightforward to verify that a general superposition of the type A(x, t) \u2261 k \u03b1k(t)Rk(x), \u03b1k(t) \u2208 R, is divergenceless, and generates an EM \ufb01eld compatible with (1.37). Substitution of Rk into Eq. (1.36) identi\ufb01es the eigenvalues as \u03bbk = k2 z. In the physics and electronic engineering literature, eigenfunctions of the Laplace operator in a quasi-one-dimensional geometry are commonly described as modes. As we will see shortly, the energy of a mode (i.e. the Hamiltonian evaluated on a speci\ufb01c mode con\ufb01guration) grows with |\u03bbk|. In cases where one is interested in the lowenergy dynamics of the EM \ufb01eld, only con\ufb01gurations with small |\u03bbk| are relevant. For example, let us consider a massively anisotropic waveguide with Lz < Ly Lx. In this case the modes with smallest |\u03bbk| are those with kz = 0, ky = \u03c0/Ly, and kx \u2261 k L\u22121 z,y. (Why is it not possible to set both ky and kz to zero?) With this choice, x + k2 y + k2 \u03bbk = k2 + 2, \u03c0 Ly (1.38) and a scalar index k su\ufb03ces to label both eigenvalues and eigenfunctions Rk. A caricature of the spatial structure of the functions Rk is shown in Fig. 1.9. The dynamical properties of these con\ufb01gurations will be discussed in the text. 1.5 Quantum electrodynamics 27 Returning to the problem posed by Eq. (1.35) and (1.36), one can expand the vector potential in terms of eigenfunctions Rk as A(x, t) = k \u03b1k(t)Rk(x), where the sum runs over all allowed values of the index parameter k. (In a waveguide,k = \u03c0n/L, n \u2208 N", ", where L is the length of the guide.) Substituting this expansion into Eq. (1.35) and using the normalization properties of Rk, we obtain L = 1, i.e. a decoupled repre2 sentation where the system is described in terms of independent dynamical systems with coordinates \u03b1k. From this point on, quantization proceeds along the lines of the standard algorithm. Firstly, de\ufb01ne momenta through the relation \u03c0k = \u2202 \u02d9\u03b1k L = \u02d9\u03b1k. This produces the Hamiltonian H = 1 k (\u03c0k\u03c0k + \u03bbk\u03b1k\u03b1k). Next quantize the theory by promoting \ufb01elds 2 to operators \u03b1k \u2192 \u02c6\u03b1k and \u03c0k \u2192 \u02c6\u03c0k, and declaring [\u02c6\u03c0k, \u02c6\u03b1k ] = \u2212i\u03b4kk. The quantum Hamiltonian operator, again of harmonic oscillator type, then reads \u2212 \u03bbk\u03b12 k \u02d9\u03b12 k k \u02c6H = 1 2 k \u02c6\u03c0k \u02c6\u03c0k + \u03c92 k \u02c6\u03b1k \u02c6\u03b1k, where \u03c92 operators ak \u2261 \u03c9k 2 (\u02c6\u03b1k + i \u03c9k assumes the now familiar form k = \u03bbk. Following the same logic as marshaled in Section 1.4, we then de\ufb01ne ladder 2 (\u02c6\u03b1k \u2212 i \u03c9k \u02c6\u03c0k), whereupon the Hamiltonian \u03c9k \u02c6\u03c0k), a \u2261 \u2020 k \u02c6H = \u03c9k a \u2020 kak + 1 2 k. (1.39) &1/2 For the speci\ufb01c problem of the \ufb01rst excited mode in a waveguide of width Ly, \u03c9k = % k2 + (\u03c0/Ly)2. Equation (1.39) represents our \ufb01nal result for the quantum Hamiltonian of the EM waveguide. Before concluding this section let us make a few comments on the structure of the result: Firstly, notice that the construction above almost completely paralleled our previous discussion of the harmonic chain.20 The structural similarity between the two systems \ufb01nds its origin in the fact that the free \ufb01eld Lagrangian (1.35) is quadratic and, therefore, bound to map onto an oscillator-type Hamiltonian. That we obtained a simple one-dimensional superposition of oscillators is due to the boundary conditions speci\ufb01c to a narrow waveguide. For less restrictive geometries, e.g. free space, a more complex superposition of vectorial degrees of freedom in three-dimensional space would have been obtained. However, the principal mapping of the free EM \ufb01eld onto a superposition of oscillators is independent of geometry. Physically, the quantum excitations described by Eq. (1.39) are, of course, the photons of the EM \ufb01eld. The unfamiliar appearance of the dispersion relation \u03c9k is again a peculiarity of the waveguide. However, in the limit of large longitudinal wave numbers k L\u22121 y, the dispersion approaches \u03c9k \u223c |k|, i.e. the relativistic dispersion of the photon \ufb01eld. Also notice that, due to the equality of the Hamiltonians (1.34) and (1.39), all that has been said about the behavior of the phonon modes of the atomic chain carries over to the photon modes of the waveguide. 20 Technically, the only di\ufb00erence is that instead of index pairs (k, \u2212k) all indices (k, k) are equal and positive. This can be traced back to the fact that we have expanded in terms of the real eigenfunctions of the closed waveguide instead of the complex eigenfunctions of the circular oscillator chain. 28 From particles to \ufb01elds As with their phononic analogue, the oscillators described by Eq. (1.39) exhibit zero-point \ufb02uctuations. It is a fascinating aspect of quantum electrodynamics that these oscillations, caused by quantization of the most relativistic \ufb01eld, surface at various points of non-relativistic physics. In the section below two prominent manifestations of zero-point \ufb02uctuations in condensed matter physics will be brie\ufb02y discussed. Vacuum \ufb02uctuations in matter The quantum oscillation of the electromagnetic \ufb01eld manifests itself in a multitude of physical phenomena. In the following, we shall brie\ufb02y discuss some of the more prominent e\ufb00ects caused", " by the dynamics of the quantum vacuum. One of the most important phenomena induced by vacuum \ufb02uctuations is the Casimir e\ufb00ect:21 two parallel conducting plates embedded into the vacuum exert an attractive force on each other. This phenomenon is not only of conceptual importance \u2013 better than anything else it demonstrates that the vacuum is \u201calive\u201d \u2013 but also of applied relevance. For example, the force balance of hydrophobic suspensions of particles of size 0.1\u20131 \u03bcm in electrolytes is believed to be strongly in\ufb02uenced by Casimir forces. At least it is applications in colloidal chemistry which prompted Casimir to his famous analysis of the idealized vacuum problem. Qualitatively, the origin of the Casimir force is readily understood. Like their classical analog, quantum photons exert a certain radiation pressure on macroscopic media. The di\ufb00erence with the classical case is that, due to zero-point oscillations, even the quantum vacuum is capable of creating radiation pressure. For a single conducting body embedded into the in\ufb01nite vacuum, the net pressure vanishes by symmetry. However, for two parallel plates, the situation is di\ufb00erent. Mode quantization arguments similar to the ones used in the previous section show that the density of quantum modes between the plates is lower than in the semi-in\ufb01nite outer spaces. Hence, the force (density) created by outer space exceeds the counter-pressure from the inside; the plates \u201cattract\u201d each other. The unambiguous measurement of the Casimir force required advanced nano-technological instrumentation and has succeeded only recently.22 A second important phenomenon where vacuum \ufb02uctuations play a role is van der Waals forces: atoms or molecules attract each other by a potential that, at small separation r, scales as r\u22126. Early attempts to explain classically the phenomenon produced results for the force that, in con\ufb02ict with experimental \ufb01ndings, were strongly temperature dependent. It was considered a major breakthrough of the new quantum mechanics when London proposed a model whereby a temperature-independent r\u22126 law was obtained. The essence of London\u2019s 21 H. B. G. Casimir and D. Polder, The in\ufb02uence of retardation on the London\u2013van der Waals forces, Phys. Rev. 73, 360\u201372 (1948); H. B. G. Casimir, On the attraction between two perfectly conducting plates, Proc. Kon. Nederland. Akad. Wetensch. 51, 793\u20136 (1948). 22 S. K. Lamoreaux, Demonstration of the Casimir force in the 0.6 to 6 m Range, Phys. Rev. Lett. 78, 5\u20139 (1997); G. Bressi et al., Measurement of the Casimir force between parallel metallic surfaces, Phys. Rev. Lett. 88, 41804\u20139 (2002). For a review on recent developments see M. Bordag et al., New developments in the Casimir e\ufb00ect, Phys. Rep. 353, 1\u2013205 (2001). 1.5 Quantum electrodynamics 29 dB dA r Figure 1.10 On the creation of van der Waals forces by EM vacuum \ufb02uctuations. An atom or molecule B is exposed to \ufb02uctuations of the EM quantum vacuum (depicted by wavy lines). This leads to the creation of a dipole element dB. A nearby partner atom A sees both the vacuum amplitude and the dipole \ufb01eld created by dB. The cooperation of the two \ufb01eld strengths leads to a net lowering of the energy of A. idea is easily explained: imagine each of the atoms/molecules as an oscillator. The zeropoint motion of these oscillators, measured in terms of some coordinate x, creates a dipole moment of strength \u223c x. At close distances, the systems interact through a dipole\u2013dipole interaction.23 Writing down coupled oscillator equations (see Problem 1.8), one readily obtains a lowering of the ground state energy by a contribution V (r) \u223c r\u22126. However, there is a subtle point about the seemingly innocuous coupling of the quantum oscillator to the classical Coulomb potential. Close inspection (see Milonni\u2019s text for details24) shows that, to maintain the quantum commutation relations of the oscillator degrees of freedom, the EM \ufb01eld must not be assumed entirely classical. In fact, once the quantum nature of the EM \ufb01eld has entered the stage, it becomes possible to explain the force without need for phenomenological introduction of atomic oscillator degrees of freedom. Johannes Diderik van der Waals 1837\u20131923 1910 Nobel Laureate in physics in recognition of his work on the", " state equation of non-ideal gases. Van der Waals was one of the \ufb01rst to postulate inter-molecular forces, at a time when the existence of atoms and molecules was still disputed. Other important work includes the derivation of \ufb01rst variants of thermodynamic scaling relations. Consider an atom A exposed to an electric \ufb01eld E (see Fig. 1.10). The \ufb01eld gives rise to the atomic Stark e\ufb00ect, i.e. a level shift of order W \u223c \u03b1E2(xA), where \u03b1 is the atomic polarizability, E = \u2202tA is the operator representing the EM \ufb01eld, and... a quantum expectation value. For an atom in empty space, E(xA) \u2261 E0(xA) is an operator measuring the vacuum \ufb02uctuations of the free \ufb01eld. However, in the neighborhood of a second atom B, an induced dipole contribution adds to E0. The point is that the zero-point amplitude E0 induces a dipole element dB \u221d \u03b1E0(xB) which, in turn, creates a dipole \ufb01eld EB 23 In passing, we mention that various animal species bene\ufb01t from the short-distance e\ufb03ciency of the van der Waals force. For instance, geckos and spiders owe their ability to climb up planar surfaces of basically any material to the presence of bushels of ultra-\ufb01ne hair (about three orders of magnitude thinner than human hair) on their feet. The tips of these hairs come close enough to the atoms of the substrate material to make the van der Waals force sizeable. Impressively, this mechanism provides a force of about two orders of magnitude larger than that required to support a spider\u2019s full body weight. Both spiders and geckos have to \u201croll\u201d their feet o\ufb00 the surface to prevent getting stuck by the enormous power of the forces acting on their many body hairs! 24 P. W. Milonni, The Quantum Vacuum (Academic Press, 1994). 30 From particles to \ufb01elds contributing to the total \ufb01eld amplitude felt by atom A: E(xA) = (E0 + EB)(xA) (see Fig. 1.10). The dominant contribution to the level shift due to the proximity of B is then given by V = W \u2212 \u03b1E2 0(xA) \u223c \u03b1E0(xA) \u00b7 EB(xA) \u223c \u03b12ET 0 (xA)K(xA \u2212 xB)E(xB), (1.40) where K is a matrix kernel describing the geometric details of the dipole interaction. (Here, we have subtracted the vacuum level shift because it represents an undetectable o\ufb00set.) Notice that Eq. (1.40) predicts the van der Waals interaction to be proportional to the square of the zero-point \ufb01eld amplitude. Unfortunately, the quantitative evaluation of the matrix element, involving a geometric average over the all polarization directions of is somequantum \ufb01eld amplitude, what involved. As a \ufb01nal result of a calculation25 one obtains V (r) = \u2212 3\u03c90\u03b12 4r6, (1.41) Johannes Stark 1874\u20131957 Nobel laureate 1919 for his discovery of the Stark e\ufb00ect in 1913. Later, Stark became infamous for the role he played during the third Reich. He attacked theoretical physics as \u201cJewish\u201d and stressed that scienti\ufb01c positions in Nazi Germany should only be held by pure-blooded Germans. Image cThe Nobel Foundation where r = |xA \u2212 xB| is the distance between the atoms and \u03c90 is the transition frequency between the ground state and the \ufb01rst excited state of the atom. (This parameter enters the result through the dependence of the microscopic polarizability on the transition frequencies.) 1.6 Noether\u2019s theorem According to a basic paradigm of physics every continuous symmetry of a system entails a conservation law.26 Conservation laws, in turn, greatly simplify the solution of any problem which is why one gets acquainted with the correspondence (symmetry \u2194 conservation law) at a very early stage of the physics curriculum, e.g. the connection between rotational symmetry and the conservation of angular momentum. However, it is not all that trivial to see (at least within the framework of Newtonian mechanics) that the former entails the latter. One needs to know what to look for (angular momentum) to identify the conserved quantity corresponding to rotational invariance. It is one of the major advantages of Lagrangian over Newtonian mechanics that it provides one with a tool \u2013 No", "ether\u2019s theorem \u2013 to automatically identify the conservation laws generated by the symmetries of classical mechanics. 25 Detailed, e.g., in Milonni, The Quantum Vacuum, Section 3.11. 26 Before exploring the rami\ufb01cations of symmetries and conservation laws for \ufb01elds, it may be instructive to recapitulate Noether\u2019s theorem in the context of classical point-particle mechanics \u2013 see, e.g., L. D. Landau and E. M. Lifshitz, Classical Mechanics (Pergamon, 1960). 1.6 Noether\u2019s theorem 31 Amalie E. Noether 1882\u20131935 German mathematician known for her groundbreaking contributions to abstract algebra and theoretical physics. Alive at a time when women were not supposed to attend college preparatory schools, she was often forbidden from lecturing under her own name. Despite these obstacles, Noether became one of the greatest algebraists of the century. Described by Albert Einstein as the most signi\ufb01cant creative mathematical genius thus far produced since the higher education of women began, she revolutionized the theories of rings, \ufb01elds, and algebras. In physics, Noether\u2019s theorem explains connection between symmetry and conservation laws. In 1933, she lost her teaching position due to being a Jew and a woman, and was forced out of Germany by the Nazis. What happens when one advances from point to continuum mechanics?27 Clearly, multidimensional continuum theories leave more room for the emergence of complex symmetries; even more so than in classical mechanics, we are in need of a tool identifying the corresponding conservation laws. Fortunately, it turns out that Noether\u2019s theorem of point mechanics a\ufb00ords a more or less straightforward to higher dimensions. Starting from the general form of the action of a continuum system, Eq. (1.16), the continuum version of Noether\u2019s theorem will be derived below. In not referring to a speci\ufb01c physical problem, our discussion will be somewhat dry. This lack of physical context is, however, more than outweighed by the general applicability of the result: the generalized form of Noether\u2019s theorem can be \u2013 without much further thought \u2013 applied to generate the conservation laws of practically any physical symmetry. In this section, we will illustrate the application of the formalism on the simple (yet important) example of space\u2013time translational invariance. A much more intriguing case study will be presented in Section 4.3 after some further background of quantum \ufb01eld theory has been introduced. generalization fundamental the Symmetry transformations The symmetries of a physical system manifest themselves in the invariance of its action under certain transformations. Mathematically, symmetry transformations are described by two pieces of input data: \ufb01rstly, a mapping M \u2192 M, x \u2192 x(x) that assigns to any point of the base manifold some \u201ctransformed\u201d point; secondly, the \ufb01eld con\ufb01gurations themselves may undergo some change, i.e. there may be a mapping (\u03c6 : M \u2192 T ) \u2192 (\u03c6 : M \u2192 T ) that de\ufb01nes transformed values \u03c6(x) = F [{\u03c6(x)}] in terms of the \u201cold\u201d \ufb01eld. It is important to understand that these two operations may, in general, be independent of each other. 27 Later on we shall see that all we are discussing here carries over to the quantum level. 32 From particles to \ufb01elds For example, suppose we want to explore the invariance properties of a theory under translations in space\u2013time. In this case, we can consider a mapping x = x+a, a \u2208 Rm, \u03c6(x) = \u03c6(x). This describes the translation of a \ufb01eld by a \ufb01xed o\ufb00set a in space\u2013time (see \ufb01gure). The system is translationally invariant if and only if S[\u03c6] = S[\u03c6]. As a second example, let us probe rotational symmetry: x = Rx, where R \u2208 O(m) is a rotation of Euclidean space\u2013time. In this case it would, in general, be unphysical to de\ufb01ne \u03c6(x) = \u03c6(x). To illustrate this point, consider the example of a vector \ufb01eld in two dimensions n = m = 2 (see the \ufb01gure.) A properly rotated \ufb01eld con\ufb01guration is de\ufb01ned by \u03c6(x) =R\u03c6 (x), i.e. the \ufb01eld amplitude actively participates in the", " operation. In fact, one often considers symmetry operations where only the \ufb01elds are transformed while the base manifold is left untouched.28 For example, the intrinsic29 rotational invariance of a magnet is probed by x = x, m(x) =R \u00b7 m(x), where the vector \ufb01eld m describes the local magnetization. To understand the impact of a symmetry transformation, it is fully su\ufb03cient to consider its in\ufb01nitesimal version. (Any \ufb01nite transformation can be generated by successive application of in\ufb01nitesimal ones.) Consider, thus, the two mappings x\u03bc \u2192 x \u03bc = x\u03bc + \u2202x\u03bc \u2202\u03c9a \u03c9a(x), \u03c9=0 \u03c6i(x) \u2192 \u03c6i(x) = \u03c6i(x) +\u03c9 a(x)F i a[\u03c6], (1.42) expressing the change of both \ufb01elds and coordinates to linear order in a set of parameter functions {\u03c9a} characterizing the transformation. (For a three-dimensional rotation, (\u03c91, \u03c92, \u03c93) = (\u03c6, \u03b8, \u03c8) would be the rotation angles, etc.) The functionals {F i } \u2013 which a need not depend linearly on the \ufb01eld \u03c6, and may explicitly depend on the coordinate x \u2013 de\ufb01ne the incremental change \u03c6(x) \u2212 \u03c6(x). We now ask how the action Eq. (1.16) changes under the transformation Eq. (1.42), i.e. we wish to compute the di\ufb00erence \u0394S = dmx L(\u03c6i(x), \u2202x \u03bc \u03c6i(x)) \u2212 dmx L(\u03c6i(x), \u2202x\u03bc \u03c6i(x)). 28 For example, the standard symmetry transformations of classical mechanics, q(t) \u2192 q(t), belong to this class: the coordinate vector of a point particle, q (a \u201c\ufb01eld\u201d in 0 + 1 space\u2013time dimensions) changes while the \u201cbase\u201d (time t) does not transform. 29 \u201cIntrinsic\u201d means that we rotate just the spins but not the entire magnet (as in our second example). 1.6 Noether\u2019s theorem 33 Inserting Eq. (1.42), using the fact that \u2202x with30 1 + (\u2202/\u2202x\u03bc) (\u03c9a\u2202x\u03bc/\u2202\u03c9a), one obtains \u2202x \u03bc/\u2202x\u03bd = \u03b4\u03bc\u03bd + (\u2202/\u2202x\u03bd) (\u03c9a\u2202x\u03bc/\u2202\u03c9a), together \u03bc/\u2202x\u03bd \u0394S dmx (1 + \u2202x\u03bc (\u03c9a\u2202\u03c9a x\u03bc))L(\u03c6i + F i a\u03c9a, (\u03b4\u03bc\u03bd \u2212 \u2202x\u03bc (\u03c9a\u2202\u03c9a x\u03bd))\u2202x\u03bd (\u03c6i + F i a\u03c9a)) \u2212 dmx L(\u03c6i(x), \u2202x\u03bc \u03c6i(x)). So far, we did not use the fact that the transformation was actually meant to be a symmetry transformation. By de\ufb01nition, we are dealing with a symmetry if, for constant parameters \u03c9a \u2013 a uniform rotation, global translation, etc. \u2013 the action di\ufb00erence \u0394S vanishes. In other words, the leading contribution to the action di\ufb00erence of a symmetry transformation must be linear in the derivatives \u2202\u03bc\u03c9a. A straightforward expansion of the formula above for \u0394S shows that these terms are given by \u0394S sym. = \u2212 dmx ja \u03bc(x)\u2202\u03bc\u03c9a, where the so-called Noether current is given by \u2202L \u2202(\u2202\u03bc\u03c6i) ja \u03bc = \u2202\u03bd\u03c6i \u2212 L\u03b4\u03bc\u03bd \u2202x\u03bd \u2202\u03c9a \u03c9=0 \u2212 \u2202L \u2202(\u2202\u03bc\u03c6i) F i a. (1.43) For a completely general \ufb01eld con\ufb01guration, there is not much can be said about the Noether current (whether or not the theory possesses the symmetry). However, if the \ufb01eld \u03c6 obeys the classical equations of motion and the theory is symmetric, the Noether current is locally conserved, \u2202\u03bcja \u03bc = 0. This follows from the fact that, for a solution \u03c6 of the Euler\u2013Lagrange equations, the linear variation of the action in any parameter must vanish. Speci\ufffd", "\ufffdcally, the vanishing of \u0394S[\u03c6] for arbitrary symmetry transformations {\u03c9a} enforces the condition \u2202\u03bcja \u03bc = 0. (But keep in mind the fact that the conservation law holds only on the classical level!) Therefore, in summary, we have Noether\u2019s theorem: A continuous symmetry entails a classically conserved current. The local conservation of the current entails the existence of a globally conserved \u201ccharge.\u201d For example, for a base manifold x = (x0, x1,..., xd) in Euclidean (1 + d)-dimensional space\u2013time, integration over the space\u2013like directions, and application of Stokes\u2019 theorem, gives \u22020Qa = 0, where31 Qa \u2261 ddx ja 0, (1.44) is the conserved charge and we have assumed that the current density vanishes at spatial in\ufb01nity. 30 This follows from \u2202x \u03bc \u2202x\u03bd = det \u2202x \u03bc \u2202x\u03bd = exp tr ln \u2202x \u03bc \u2202x\u03bd exp \u2202 \u2202x\u03bc \u03c9a \u2202x\u03bc \u2202\u03c9a \u2202x\u03bc (Exercise: Show that det A = exp tr ln A, whereA is a linear operator. Hint: Use the eigenbasis.) 1 + \u2202 31 Notice that the integral involved in the de\ufb01nition of Q runs only over spatial coordinates.. \u03c9a \u2202x\u03bc \u2202\u03c9a 34 From particles to \ufb01elds Notice that nowhere in the discussion above have we made assumptions about the internal structure of the Lagrangian. In particular, all results apply equally to the Minkowskian and the Euclidean formulation of the theory. Example: translational invariance As an example that is as elementary as it is important, we consider the case of space\u2013time translational invariance. The corresponding symmetry transformation is de\ufb01ned by x \u03bc = x\u03bc+a\u03bc, \u03c6(x) = \u03c6(x). The in\ufb01nitesimal version of this transformation reads as x \u03bc = x\u03bc+\u03c9\u03bc where we have identi\ufb01ed the parameter indexa with the space\u2013time index \u03bc. Noether\u2019s current, which in the case of translational invariance is called the energy\u2013momentum tensor, is given byT \u03bd \u03bc : T \u03bd \u03bc (x) = \u2202L \u2202(\u2202\u03bc\u03c6i) \u2202\u03bd\u03c6i \u2212 \u03b4\u03bc\u03bdL. The conserved \u201ccharges\u201d corresponding to this quantity are P \u03bd \u2261 ddx \u2202L \u2202(\u22020\u03c6i) \u2202\u03bd\u03c6i \u2212 \u03b4\u03bd0L. Speci\ufb01cally, P 0 is the energy and P l, l = 1,..., d the total momentum carried by the system. EXERCISE Consider the one-dimensional elastic chain discussed in Section 1.1. Convince yourdx\u03c1(x, t) = \u2202t(\u03c6(x2, t) \u2212 self that the continuity equation of the system assumes the form \u2202t \u03c6(x1, t)), where \u03c1(x, t) is the local density of the medium. Use this result to show that \u03c1 = \u2202x\u03c6. Show that the momentum dx\u03c1\u2202t\u03c6 carried by the system coincides with the Noether momentum. dx (particle density) \u00d7 (velocity) = x2 x1 EXERCISE Compute the energy-momentum tensor for the Lagrangian of the free electromagnetic \ufb01eld. Derive the corresponding energy\u2013momentum vector and convince yourself that you obtain results familiar from electrodynamics. Admittedly, translational invariance is not a particularly exciting symmetry and, if this were the only example of relevance, the general discussion above would not have been worth the e\ufb00ort. However, later in the text, we shall exemplify (see, for example, Section 4.3) how concepts hinging on Noether\u2019s theorem can be applied to obtain far-reaching results that are not readily accessible by di\ufb00erent means. 1.7 Summary and outlook In this chapter we have introduced the general procedure whereby classical continuum theories are quantized. Employing the elementary harmonic oscillator as a example, we have seen that the Hilbert spaces of these theories a\ufb00ord di\ufb00erent interpretations. Of particular use was a quasi-particle picture whereby the collective excitations of the continuum theories acquired the status of elementary particles. 1.8 Problems 35 Suspiciously, both examples discussed in this text, the harmonic quantum chain and free quantum electrodynamics, led", " to exactly solvable free \ufb01eld theories. However, it takes only little imagination to foresee that only few continuum theories will be as simple. Indeed, the exact solvability of the atomic chain would have been lost had we included higher-order contributions in the expansion in powers of the lattice displacement. Such terms would hinder the free wave-like propagation of the phonon modes. Put di\ufb00erently, phonons would begin to scatter o\ufb00 each other (i.e. interact). Similarly, the free status of electrodynamics is lost once the EM \ufb01eld is allowed to interact with a matter \ufb01eld. Needless to say, interacting \ufb01eld theories are in\ufb01nitely more complex, but also more interesting, than the systems considered so far. Technically, we have seen that the phonon/photon interpretation of the \ufb01eld theories discussed in this chapter could conveniently be formulated in terms of ladder operators. However, the applications discussed so far provide only a glimpse of the advantages of this language. In fact, the formalism of ladder operators, commonly described as \u201csecond quantization,\u201d represents a central, and historically the oldest, element of quantum \ufb01eld theory. The next chapter is devoted to a more comprehensive discussion of both the formal aspects and applications of this formulation. 1.8 Problems Electrodynamics from a variational principle Choosing the Lorentz-gauged components of the vector potential as generalized coordinates, the aim of this problem is to show how the wave equations of electrodynamics can be obtained as a variational principle. \u22022 t \u2212 \u22072 Electrodynamics can be described by Maxwell\u2019s equations or, equivalently, by wave-like equations for the vector potential. Working in the Lorentz gauge, \u2202t\u03c6 = \u2207 \u00b7A, these A = j. Using relativistically covariant notation, equations read the form of the equations can be compressed further to \u2202\u03bc\u2202\u03bcA\u03bd = j\u03bd. Starting from the Lagrangian action, S[A] = \u2212 4 F\u03bc\u03bdF \u03bc\u03bd + A\u03bcj\u03bc, obtain these equations by applying the variational principle. Compare the Lorentz gauge representation of the action of the \ufb01eld with the action of the elastic chain. What are the di\ufb00erences/parallels? \u22022 t \u03c6 = \u03c1, \u2212 \u22072 d4x 1 Answer: Using the de\ufb01nition of the electromagnetic \ufb01eld tensor F\u03bc\u03bd = \u2202\u03bcA\u03bd \u2212 \u2202\u03bdA\u03bc, and integrating by parts, the action assumes the form S[A] = \u2212 d4x \u2212 1 2 A\u03bd [\u2202\u03bc\u2202\u03bcA\u03bd \u2212 \u2202\u03bc\u2202\u03bdA\u03bc] +j \u03bcA\u03bc. Owing to the Lorentz gauge condition, the second contribution in the square brackets vanishes and we obtain S[A] = \u2212, where we have again integrated by parts. Applying the general variational Eq. (1.17) one \ufb01nally obtains the wave equation. 2 \u2202\u03bcA\u03bd\u2202\u03bcA\u03bd + j\u03bcA\u03bc d4x 1 36 From particles to \ufb01elds Hamiltonian of electromagnetic \ufb01eld Here it is shown that the Hamiltonian canonically conjugate to the Lagrangian of the electromagnetic \ufb01eld indeed coincides with the energy density familiar from elementary electrodynamics. Consider the electromagnetic \ufb01eld in the absence of matter, j = 0. Verify that the total d3x H(x) whereH (x) = E2(x) +B 2(x) is the energy stored in the \ufb01eld is given by H \u2261 familiar expression for the EM energy density. (Hint: Use the vacuum form of Maxwell\u2019s equations and the fact that, for an in\ufb01nite system, the energy is de\ufb01ned only up to surface terms.) Answer: Following the canonical prescription, let us \ufb01rst consider the Lagrangian density, L = \u2212 1 4 F\u03bc\u03bdF \u03bc\u03bd = 1 2 3 i=1 (\u22020Ai \u2212 \u2202iA0)(\u22020Ai \u2212 \u2202iA0) \u2212 1 4 3 i,j=1 (\u2202iAj \u2212 \u2202jAi)(\u2202iAj \u2212 \u2202jAi). We next determine the components of the canonical momentum through the relation \u03c0\u03bc = L : \u03c00 = 0, \u03c0i = \u22020Ai \u2212 \u2202iA0 = Ei. Using the fact that \u2202iAj \u2212 \ufffd", "\ufffdjAi is a component \u2202\u22020A\u03bc of the magnetic \ufb01eld, the Hamiltonian density can now be written as H = \u03c0\u03bc\u22020A\u03bc \u2212 L = 1 2 (2E \u00b7 \u22020A \u2212 E2 + B2) (1) = 1 2 (2E \u00b7 \u2207\u03c6 + E2 + B2) (2) = 1 2 (2\u2207 \u00b7 (E\u03c6) +E 2 + B2), where (1) is based on addition and subtraction of a term 2E \u00b7 \u2207\u03c6, and (2) on \u2207 \u00b7E = 0 combined with the identity \u2207 \u00b7 (af ) = \u2207 \u00b7 af + a \u00b7 \u2207f (valid for general vector [scalar] functions a(f )). Substitution of this expression into the de\ufb01nition of the Hamiltonian yields H = 1 2 d3x 2\u2207 \u00b7 (E\u03c6) +E 2 + B2 = 1 2 d3x E2 + B2, where we have used the fact that the contribution \u2207 \u00b7( E\u03c6) is a surface term that vanishes upon integration by parts. Phonon speci\ufb01c heat In the text, we have stated that the mode quantization of elastic media manifests itself in lowtemperature anomalies of the speci\ufb01c heat. In this problem, concepts of elementary quantum statistical mechanics are applied to determine the temperature pro\ufb01le of the speci\ufb01c heat. Compute the energy density u = \u2212(1/L)\u2202\u03b2 ln Z of one-dimensional longitudinal phonons with dispersion \u03c9k = v|k|, where Z = tr e\u2212\u03b2 \u02c6H denotes the quantum partition function. As an intermediate result, show that the thermal expectation value of the energy density of the system can be represented as u = 1 L k \u03c9k 2 + \u03c9knB(\u03c9k), (1.45) 1.8 Problems 37 e\u03b2 \u2212 1 \u22121 where nB() = is the Bose\u2013Einstein distribution function. Approximate the sum over k by an integral and show that cv \u223c T. At what temperatureT cl does the speci\ufb01c heat cross over to the classical result cv = const? (Remember that the linear dispersion \u03c9k = v|k| is based on a quadratic approximation to the Hamiltonian of the system and, therefore, holds only for |k| < \u039b, where \u039b is some cuto\ufb00 momentum.) Reiterate the discussion of Section 1.4 for a d-dimensional isotropic solid of volume Ld (i.e. assume that the atomic exchange constants remain the same in all directions). Show that the dispersion generalizes to \u03c9k = v|k|, where k = (n1,..., nd)2\u03c0/L and ni \u2208 Z. Show that the speci\ufb01c heat shows the temperature dependence cv \u223c T d. Answer: As discussed in the text, the quantum eigenstates of the system are given by |n1, n2,..., where nm is the number of phonons of wavenumber km = 2\u03c0m/L, E|n1,n2,... = m the eigenenergy, and \u03c9m = v|km|. In the energy represen- m nm tation, the quantum partition function then takes the form m \u03c9km (nm + 1/2) \u2261 Z = tr e\u2212\u03b2 \u02c6H = states e\u2212\u03b2Estate ='\u221e e\u2212\u03b2\u03c9m(nm+1/2) ='m=1,2,... nm=0 m=1,2,... e\u2212\u03b2\u03c9m/2 1 \u2212 e\u2212\u03b2\u03c9m, where nm is the occupation number of the state with wavenumber km. Hence, # ln Z = \u2212 \u03b2 m \u03c9m 2 + ln 1 \u2212 e\u2212\u03b2\u03c9m. $ Di\ufb00erentiation with respect to \u03b2 yields Eq. (1.45) and, replacing arrive at u = C1 + 1 2\u03c0 dk |k|<\u039b v|k| e\u03b2v|k| \u2212 1 = C1 + \u03b2\u22122C2. \u2192 (L/2\u03c0) m dk, we Here, C1 is the temperature-independent constant accounting for the \u201czero-point energies\u201d \u03c9m/2. In the second equality, we have scaled k \u2192 \u03b2k. This produces a prefactor \u03b2\u22122 multiplied by a temperature-independent (up to the temperature dependence of the boundaries \u039b \u2192 \u03b2\u039b) integral that we denoted by C2", ". Di\ufb00erentiation with respect to T then leads to the relation cv = \u2202T u \u223c T. However, for temperatures T > v\u039b higher than the highest frequencies stored in the phonon modes, the procedure above no longer makes sense (formally, due to the now non-negligible temperature dependence of the boundaries). Yet in this regime we may expand e\u03b2v|k| \u2212 1 \u03b2v|k|, which brings us back to the classical result cv = const. d \u2212 i=1(\u03c6R+ei \u03c6R)2 (ei: unit vector in the i-direction). Taking a continuum limit leads to a contribution (ksa2/2)(\u2207\u03c6(x))2. Proceeding along the lines of the text, we \ufb01nd that the relevant excitations are now waves with wave vector k = 2\u03c0(n1,..., nd)/L and energy \u03c9k = |k|v. ddk. Scaling Accordingly, the sum in our formula for the speci\ufb01c heat becomes ki \u2192 \u03b2ki then generates a prefactor \u03b2\u2212d\u22121 and we arrive at the relation cv \u223c T d. Consider now a d-dimensional solid with \u201cisotropic\u201d exchange coupling ks 2 \u223c k 38 From particles to \ufb01elds Van der Waals force We here discuss a semi-phenomenological approach to the van der Waals force: modelling the constituents participating in the interaction as quantum oscillators, we \ufb01nd that the ground state energy of the ensuing two-body system is lower than that of the isolated compounds. London\u2019s phenomenological approach to explaining the van der Waals force starts out from the Hamiltonian \u02c6H = \u02c6p2 1 2m + \u02c6p2 2 2m + m\u03c92 o 2 1 + \u02c6x2 \u02c6x2 2 + mK \u02c6x1 \u02c6x2, where \u02c6xi is the coordinate of an oscillator phenomenologically representing one of two atoms i = 1, 2. The atoms are coupled through a dipole\u2013dipole interaction, where the dipole operators are proportional to \u02c6xi and K(r) = qe2/(mr3) encapsulates the details of the interaction. Here r denotes the distance between the atoms and q is a dipole\u2013dipole orientation factor. Compute the spectrum of the two-particle system described by \u02c6H and verify that the dipole coupling leads to a lowering of the ground state energy by an amount V (r) = K 2/8\u03c93 o. Using the fact that the classical polarizability of an oscillator is given by \u03b1 = e2/m\u03c92 o this becomes V = q2\u03c9o\u03b12/8r6. Finally, using the fact that the directional average of the geometric factor q2 equals 2, and multiplying by three to take account of the three-dimensional character of a \u201creal\u201d atomic oscillator, we recover the result Eq. (1.41). Answer: We begin by formulating the potential \u02c6U of the oscillator system in a matrix notation, \u02c6U = \u02c6xT A\u02c6x, where \u02c6xT = (\u02c6x1, \u02c6x2)T and A = m 2 \u03c92 K K \u03c92. Diagonalizing the matrix, A = U DU T, and transforming coordinates, x \u2192 U T x = x, the system decouples into two independent oscillators, \u02c6H = i=1,2 \u02c6p2 i 2m + m\u03c92 i 2 \u02c6x2 i, are determined by the eigenwhere the new characteristic frequencies \u03c91/2 = values of the matrix A. The ground state of this system is given by E0 = (\u03c91 + \u03c92)/2 \u2248 \u03c9o \u2212 K 2/8\u03c93 lower than the energy of the isolated atoms. o, which lies by an amountV = \u2212K 2/ 8\u03c93 o \u03c92 o \u00b1 K 1/2 2 Second quantization The purpose of this chapter is to introduce and apply the method of second quantization, a technique that underpins the formulation of quantum many-particle theories. The \ufb01rst part of the chapter focuses on methodology and notation, while the remainder is devoted to the development of applications designed to engender familiarity with and \ufb02uency in the technique. Speci\ufb01cally, we will investigate the physics of the interacting electron gas, charge density wave propagation in one-dimensional quantum wires, and spin waves in a quantum Heisenberg (anti)ferromagnet. Indeed, many of these examples and", " their descendants will reappear as applications in our discussion of methods of quantum \ufb01eld theory in subsequent chapters. In the previous chapter we encountered two \ufb01eld theories that could conveniently be represented in the language of \u201csecond quantization,\u201d i.e. a formulation based on the algebra of certain ladder operators \u02c6ak.1 There were two remarkable facts about this formulation: \ufb01rstly, second quantization provides a compact way of representing the many-body space of excitations; secondly, the properties of the ladder operators were encoded in a simple set of commutation relations (cf. Eq. (1.33)) rather than in some explicit Hilbert space representation. Apart from a certain aesthetic appeal, these observations would not be of much relevance if it were not for the fact that the formulation can be generalized to a comprehensive and highly e\ufb03cient formulation of many-body quantum mechanics in general. In fact, second quantization can be considered the \ufb01rst major cornerstone on which the theoretical framework of quantum \ufb01eld theory was built. This being so, extensive introductions to the concept can be found throughout the literature. We will therefore not develop the formalism in full mathematical rigor but rather proceed pragmatically by \ufb01rst motivating and introducing its basic elements, followed by a discussion of the \u201csecond quantized\u201d version of standard operations of quantum mechanics (taking matrix elements, changing bases, representing operators, etc.). The second part of the chapter is concerned with developing \ufb02uency in the method by addressing a number of applications. Readers familiar with the formalism may therefore proceed directly to these sections. 1 The term \u201csecond quantization\u201d is unfortunate. Historically, this terminology was motivated by the observation that the ladder operator algebra fosters an interpretation of quantum excitations as discrete \u201cquantized\u201d units. Fundamentally, however, there is nothing like \u201ctwo\u201d superimposed quantization steps in single- or many-particle quantum mechanics. Rather, one is dealing with a particular representation of the \u201c\ufb01rst and only quantized\u201d theory tailored to the particular problem at hand. 39 40 Second quantization 2.1 Introduction to second quantization Motivation n\u03bb n\u03bb some \u03b5\u03bb \u03b5 4 \u03b5 3 fundamental notions We begin our discussion by recapitulating of many-body quantum mechanics, as forlanguage of mulated in the traditional symmetrized/anti-symmetrized wavefunctions. Consider the (normalized) set of wavefunctions |\u03bb of some single-particle Hamiltonian \u02c6H : \u02c6H|\u03bb = \u03bb|\u03bb, where \u03bb are the eigenvalues. With this de\ufb01nition, fermions the normalized two-particle wavefunction \u03c8F(\u03c8B) of two fermions (bosons) populating levels \u03bb1 and \u03bb2 is given by the antisymmetrized (symmetrized) product bosons \u03c8F(x1, x2) = \u03c8B(x1, x2) = 1\u221a 2 1\u221a 2 (x1|\u03bb1x2|\u03bb2 \u2212 x1|\u03bb2x2|\u03bb1), (x1|\u03bb1x2|\u03bb2 + x1|\u03bb2x2|\u03bb1). In the Dirac bracket representation, the two-body states |\u03bb1, \u03bb2F(B) corresponding to the wave functions \u03c8F(B)(x1, x2) = (x1| \u2297 x2|) |\u03bb1, \u03bb2F(B) above can be presented as |\u03bb1, \u03bb2F(B) \u2261 1\u221a 2 (|\u03bb1 \u2297 |\u03bb2 + \u03b6|\u03bb2 \u2297 |\u03bb1), where \u03b6 = \u22121 for fermions while \u03b6 = 1 for bosons. Note that the explicit symmetrization of the wavefunctions is necessitated by quantum mechanical indistinguishability: for fermions (bosons) the wave function has to be anti-symmetric (symmetric) under exchange.2 More generparticle ally, an appropriately symmetrized N -particle wavefunction can be expressed in the form Enrico Fermi 1901\u201354 Nobel Laureate in Physics in 1938 for his demonstrations of the existence of new radioactive elements produced by neutron irradiation, and for his related discovery of nuclear reactions brought about by slow neutrons. (Image c The Nobel Foundation.) |\u03bb1, \u03bb2,..., \u03bbN \u2261 1 \u221e \u03bb=0(n\u03bb!) N! P \u03b6 (1\u2212sgn P)/2|\u03bbP1 \u2297", " |\u03bbP2 \u2297 \u00b7 \u00b7 \u00b7 \u2297 |\u03bbPN, (2.1) 2 Notice, however, that in two-dimensional systems the standard doctrine of fully symmetric/anti-symmetric many particle wave functions is too narrow and more general types of exchange statistics can be realized, cf. our discussion on page 41. 2.1 Introduction to second quantization 41 where n\u03bb represents the total number of particles in state \u03bb (for fermions, Pauli exclusion enforces the constraint n\u03bb \u2264 1) \u2013 see the schematic \ufb01gure above. The summation runs over all N! permutations of the set of quantum numbers {\u03bb1,..., \u03bbN }, and sgn P denotes the sign of the permutation P. (sgn P = 1 [\u22121] if the number of transpositions of two elements which brings the permutation (P1, P2,..., PN ) back to its original form (1, 2,..., N ) is \u03bb(n\u03bb!) normalizes the many-body wavefunction. In even [odd].) The prefactor 1/ the fermionic case, wave functions corresponding to the states above are known as Slater determinants. N! Finally, notice that it will be useful to assume that the quantum numbers {\u03bbi} de\ufb01ning the state |\u03bb1, \u03bb2,..., \u03bbN are ordered according to some convention. (For example, for \u03bbi = xi a one-dimensional coordinate representation, we might order according to the rule x1 \u2264 x2 \u2264 \u00b7 \u00b7 \u00b7 \u2264x N.) Once an ordered sequence of states has been \ufb01xed we may \u2013 for notational convenience \u2013 label our quantum states by integers, \u03bbi = 1, 2,.... Any initially non-ordered state (e.g. |2, 1, 3) can be brought into an ordered form (|1, 2, 3) at the cost of, at most, a change of sign. David Hilbert 1862\u20131943 He contributed to many branches of mathematics, including invariants, algebraic number \ufb01elds, functional analysis, integral equations, mathematical physics, and the calculus of variations. His work in geometry had the greatest in\ufb02uence in that area after Euclid. A systematic study of the axioms of Euclidean geometry led Hilbert to propose twenty one such axioms and he analyzed their signi\ufb01cance. INFO For the sake of completeness, let us spell out the connection between the permutation group and many-body quantum mechanics in a more mathematical language. The basic arena wherein N -body quantum mechanics takes place is the product space, HN \u2261 H \u2297 \u00b7 \u00b7 \u00b7 \u2297 H N copies, of N single-particle Hilbert spaces. In this space, we have a linear representation of the permutation group, SN,3 assigning to each P \u2208 SN the permutation (no ordering of the \u03bbs implied at this stage), P : HN \u2192 HN, |\u03bb1 \u2297 \u00b7 \u00b7 \u00b7 \u2297 |\u03bbN \u2192 |\u03bbP1 \u2297 \u00b7 \u00b7 \u00b7 \u2297 |\u03bbPN. 3 Recall that a linear representation of a group G is a mapping that assigns to each g \u2208 G a linear mapping \u03c1g : V \u2192 V of some vector space V. For notational convenience one usually writes g : V \u2192 V instead of \u03c1g : V \u2192 V. Conceptually, however, it is often important to carefully discriminate between the abstract group elements g and the matrices (also g) assigned to them by a given representation. (Consider, for example the symmetry group G = SU(2) of quantum mechanical spin. SU(2) is the two-dimensional group of unitary matrices with determinant one. However, when acting in the Hilbert space of a quantum spin S = 5, say, elements of SU(2) are represented by (2S + 1 = 11)-dimensional matrices.) Two representations \u03c1 and \u03c1 that di\ufb00er only by a unitary transformation, \u2200g \u2208 G : \u03c1g = U \u03c1 gU \u22121, are called unitary equivalent. If a transformation U can be found such that all representation matrices \u03c1g assume a block structure, the representation is called reducible, and otherwise irreducible. Notice that the di\ufb00erent sub-blocks of a reducible representation by themselves form irreducible representation spaces. The identi\ufb01cation of all distinct irreducible representations of a given group is one of the most important objectives of group theory. 42 Second quantization The identi\ufb01cation of all irreducible subspaces of this representation is a", " formidable task that, thanks to a fundamental axiom of quantum mechanics, we need not address in full. All we need to know is that SN has two particularly simple one-dimensional irreducible representations: one wherein each P \u2208 SN acts as the identity transform P(\u03a8) \u2261 \u03a8 and, another, the alternating representation P(\u03a8) = sgn P \u00b7 \u03a8. According to a basic postulate of quantum mechanics, the state vectors \u03a8 \u2208 HN describing bosons/fermions must transform according to the identity/alternating representation. The subset F N \u2282 HN of all states showing this transformation behavior de\ufb01nes the physical N -body Hilbert space. To construct a basis of F N, one may apply the symmetrization operator P s \u2261 P (sgn P)P) to the basis vectors |\u03bb1 \u2297 \u00b7 \u00b7 \u00b7 \u2297 |\u03bbN of HN. Up to normalization, this operation obtains the states (2.1). P P (anti-symmetrization operator P a \u2261 Some readers may wonder why we mention these representation-theoretic aspects. Being pragmatic, all we really need to know is the symmetrization/anti-symmetrization postulate, and its implementation through Eq. (2.1). Notice, however, that one may justly question what we actually mean when we talk about the permutation exchange of quantum numbers. For example, when we compare wavefunctions that di\ufb00er by an exchange of coordinates, we should, at least in principle, be able to tell by what physical operation we e\ufb00ect this exchange (for, otherwise, we cannot really compare them other than in a formal and, in fact, in an ambiguous sense). Oddly enough, decades passed before this crucial issue in quantum mechanics was critically addressed. In a now seminal work by Leinaas and Myrheim4 it was shown that the standard paradigm of permutation exchange is far from innocent. In particular, it turned out that its applicability is tied to the dimensionality of space! Speci\ufb01cally, in two-dimensional spaces (in a sense, also in d = 1) a more elaborate scheme is needed. (Still one may use representation-theoretic concepts to describe particle exchange. However, the relevant group \u2013 the braid group \u2013 now di\ufb00ers from the permutation group.) Physically, these phenomena manifest themselves in the emergence of quantum particles di\ufb00erent from both bosons and fermions. For a further discussion of these \u201canyons\u201d we refer to Chapter 9. While representations like Eq. (2.1) can be used to represent the full Hilbert space of manybody quantum mechanics, a moment\u2019s thought shows that this formulation is not at all convenient: It takes little imagination to anticipate that practical computation in the language of Eq. (2.1) will be cumbersome. For example, to compute the overlap of two wavefunctions one needs to form no less than (N!)2 di\ufb00erent products. The representation is tailor-made for problems with \ufb01xed particle number N. However, we know from statistical mechanics that for N = O(1023) it is much more convenient to work in a grand canonical formulation where N is allowed to \ufb02uctuate. Closely related to the above, in applications one will often ask questions such as, \u201cWhat is the amplitude for injection of a particle into the system at a certain space-time coordinate (x1, t1) followed by annihilation at some later time (x2, t2)?\u201d Ideally, one would work with a representation that supports the intuition a\ufb00orded by thinking in terms of such 4 J. M. Leinaas and J. Myrheim, On the theory of identical particles, Il Nuovo Cimento B 37 (1977), 1\u201323. 2.1 Introduction to second quantization 43 processes: i.e. a representation where the quantum numbers of individual quasi-particles rather than the entangled set of quantum numbers of all constituents are fundamental. The \u201csecond quantized\u201d formulation of many-body quantum mechanics, as introduced in the next subsection, will remove all these di\ufb03culties in an elegant and e\ufb03cient manner. The apparatus of second quantization Occupation number representation and Fock space Some of the disadvantages of the representation (2.1) can be avoided with relatively little e\ufb00ort. In our present notation, quantum states are represented by \u201cN -letter words\u201d of the form |1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 6, 6,.... Obviously, this notation contains a lot of redundancy. A more e\ufb03cient", " encoding of the state above might read |4, 2, 3, 1, 0, 2,..., where theith number signals how many particles occupy state number i; no more information is needed to characterize a symmetrized state. (For fermions, these occupation numbers take a value of either zero or one.) This de\ufb01nes the \u201coccupation number representation.\u201d In the new representation, the basis states of F N are speci\ufb01ed by |n1, n2,..., where i ni = N. Any state |\u03a8 in F N can be obtained by a linear superposition |\u03a8 = n1,n2,..., ni=N cn1,n2,...|n1, n2,.... As pointed out above, eventually we will want to emancipate ourselves from the condition of a \ufb01xed particle number N. A Hilbert space large enough to accommodate a state with an undetermined number of particles is given by F \u2261 \u221e( N =0 F N. (2.2) the direct sum conNotice that tains a curious contribution F 0, the \u201cvacuum space.\u201d This is a onedimensional Hilbert space which describes the sector of the theory with no particles present. Its single normalized basis state, the vacuum is denoted by |0. We will state, soon see why it is convenient to add this strange animal to our family of basis states. The space F is called Fock space and it de\ufb01nes the principal arena of quantum many-body theory. Vladimir Aleksandrovich Fock 1898\u20131974 One of the main participants in the history of the general theory of relativity in Russia. His groundbreaking contributions to manybody theory include the introduction of Fock space and the development of perhaps the most important many-particle approximation scheme, the Hartree-Fock approximation (see Chapter 5). To obtain a basis of F, we need only take the totality of our previous basis states {|n1, n2,...}, and drop the condition i ni = N on the occupation numbers. A general many-body state |\u03a8 can then be represented by a linear superposition |\u03a8 = 44 Second quantization n1,n2,... cn1,n2,...|n1, n2,.... Notice that states of di\ufb00erent particle numbers may contribute to the linear superposition forming |\u03a8. We shall see that such mixtures play an important role, for example in the theory of superconductivity. Foundations of second quantization The occupation number representation introduced above provides a step in the right direction, but it does not yet solve our main problem, the need for explicit symmetrization/antisymmetrization of a large number of states in each quantum operation. As a \ufb01rst step towards the construction of a more powerful representation, let us recall an elementary fact of linear algebra: a linear map A : V \u2192 V of a vector space into itself is fully determined by de\ufb01ning the images wi \u2261 Avi of the action of A on a basis {vi}. Now let us use this scheme to introduce a set of linear operators acting in Fock space. For every i = 1, 2,..., we de\ufb01ne operators a i : F \u2192 F through \u2020 a \u2020 i |n1,..., ni,... \u2261 (ni + 1)1/2\u03b6 si |n1,..., ni + 1,..., (2.3) where si = mod 2. Speci\ufb01cally, (1 + 1) = 0 mod 2, i.e. the application of a annihilates this state. i\u22121 j=1 nj. In the fermionic case, the occupation numbers ni have to be understood \u2020 i to a state with ni = 1 Notice that by virtue of this de\ufb01nition we are able to generate every basis state of F by \u2020 i s to the vacuum state. (From a formal point of view, this fact repeated application of a alone is motivation enough to add the vacuum space to the de\ufb01nition of Fock space.) Indeed, repeated application of Eq. (2.3) leads to the important relation |n1, n2,... ='i 1 (ni!)1/2 (a \u2020 i )ni |0. (2.4) Notice that Eq. (2.4) presents a strong statement: the complicated permutation \u201centanglement\u201d implied in the de\ufb01nition (2.1) of the Fock states can be generated by straightforward application of a set of linear operators to a single reference state. Physically", ", N -fold application of operators a\u2020 to the empty vacuum state generates an N -particle state, which is why the a\u2020s are commonly called creation operators. Of course, the introduction of creation operators might still turn out to be useless, i.e. consistency with the properties of the Fock states (such as the fact that, in the fermionic case, the numbers ni = 0, 1 are de\ufb01ned only \u2020 mod 2), might invalidate the simple relation (2.3) with its (ni-independent!) operator a i. However, as we shall demonstrate below, this is not the case. \u2020 i and a j for i = j. From the de\ufb01nition (2.3), one may readily i )|n1, n2,... = 0. Holding for every basis vector, this relation implies Consider two operators a \u2020 \u2020 i a j \u2020 ja \u2020 \u2020 verify that (a \u2020 that [a i, a \u2212\u03b6a \u2020 j]\u03b6 = 0, where [ \u02c6A, \u02c6B]\u03b6 \u2261 \u02c6A \u02c6B \u2212 \u03b6 \u02c6B \u02c6A, i.e. [, ]\u03b6=1 \u2261 [, ] is the commutator and [, ]\u03b6=\u22121 \u2261 {, } \u2261 [, ]+ the anti-commutator. \u2020 Turning to the case i = j, we note that, for fermions, the two-fold application of a i to any \u20202 i = 0 is nilpotent, a fact that can be formulated as state leads to its annihilation. Thus, a 2.1 Introduction to second quantization 45... F 2 a a+ F 1 a a+ a 0 F 0 Figure 2.1 Visualization of the generation of the Fock subspaces F N by repeated action of creation operators on the vacuum space F 0. \u2020 i, a \u2020 i ]+ = 0. For bosons we have, of course, [a [a Summarizing, we have found that the creation operators obey the commutation relation \u2020 i ] = 0 (identical operators commute!). \u2020 i, a \u2200i. (2.5) Now, quantum mechanics is a unitary theory so, whenever one meets a new operator \u02c6A, one should determine its Hermitian adjoint \u02c6A\u2020. To understand the action of the Hermitian = ai of the creation operators we may take the complex conjugates of all adjoints basis matrix elements of Eq. (2.3): \u2020 \u2020 i a n1,..., ni,... |a 1,..., n |n i,... |ai|n1,..., ni,...\u2217 i,... = (n = n1/2 1,..., n i + 1)1/2 \u03b6 s \u03b6 si \u03b4n \u2020 i i \u21d2 n i \u03b4n1,n 1... \u03b4ni,n i+1... 1,n1... \u03b4n i,ni\u22121... Holding for every bra n 1,..., n i,... |, the last line tells us that ai|n1,..., ni,... = n1/2 i \u03b6 si |n1,..., ni \u2212 1,..., (2.6) a relation that identi\ufb01es ai as an operator that \u201cannihilates\u201d particles. The action of creation and annihilation operators in Fock space is illustrated in Fig. 2.1. Creation operators a\u2020 : F N \u2192 F N +1 increase the particle number by one, while annihilation operators a : F N \u2192 F N \u22121 lower it by one; the application of an annihilation operator to the vacuum state, ai|0 = 0, annihilates it. (Do not confuse the vector |0 with the number zero.) Taking the Hermitian adjoint of Eq. (2.5) we obtain [ai, aj]\u03b6 = 0. Further, a straight\u2020 forward calculation based on the de\ufb01nitions (2.3) and (2.6) shows that [ai, a j]\u03b6 = \u03b4ij. Altogether, we have shown that the creation and annihilation operators satisfy the algebraic closure relation [ai, a \u2020 j]\u03b6 = \u03b4ij, [ai, aj]\u03b6 = 0, [a \u2020 i, a \u2020 j]\u03b6 = 0. (2.7) \u2020 Given that", " the full complexity of Fock space is generated by application of a i s to a single reference state, the simplicity of the relations obeyed by these operators seems remarkable and surprising. INFO Perhaps less surprising is that, behind this phenomenon, there lingers some mathematical structure. Suppose we are given an abstract algebra A of objects ai, a\u2020 i satisfying the relation (2.7). (Recall that an algebra is a vector space whose elements can be multiplied by each other.) Further suppose that A is irreducibly represented in some vector space V, i.e. that there is a mapping assigning to each ai \u2208 A a linear mapping ai : V \u2192 V, such that every vector |v \u2208 V can be reached from any other |w \u2208V by (possibly iterated) application of operators ai and a\u2020 i 46 Second quantization (irreducibility).5 According to the Stone\u2013von Neumann theorem (a) such a representation is unique (up to unitary equivalence), and (b) there is a unique state |0 \u2208 V that is annihilated by every ai. All other states can then be reached by repeated application of a\u2020 i s. The precise formulation of this theorem, and its proof \u2013 a good practical exercise in working with creation/annihilation operators \u2013 are left as Problem 2.4. From the Stone\u2013von Neumann theorem, we can infer that the Fock space basis could have been constructed in reverse. Not knowing the basis {|n1, n2,...}, we could have started from a set of operators obeying the commutation relations (2.7) acting in some a priori unknown space F. Outgoing from the unique state |0, the prescription (2.4) would then have yielded an equally unique basis of the entire space F (up to unitary transformations). In other words, the algebra (2.7) fully characterizes the operator action and provides information equivalent to the de\ufb01nitions (2.3) and (2.6). Practical aspects Our next task will be to promote the characterization of Fock space bases introduced above to a full reformulation of many-body quantum mechanics. To this end, we need to \ufb01nd out how changes from one single-particle basis {|\u03bb} to another {|\u02dc\u03bb} a\ufb00ect the operator algebra {a\u03bb}. (In this section we shall no longer use integers to identify di\ufb00erent elements \u2020 \u03bb creates a particle in of a given single-particle basis. Rather, we use Greek labels \u03bb, i.e. a state \u03bb.) Equally important, we need to understand in what way generic operators acting in many-particle Hilbert spaces can be represented in terms of creation and annihilation operators. Change of basis: Using the resolution of identity id = |\u03bb\u03bb|\u02dc\u03bb, |\u03bb \u2261a \u03bb \u2020 \u03bb |0, and| \u02dc\u03bb \u2261a \u2020 \u02dc\u03bb |\u03bb\u03bb|, the relations |\u02dc\u03bb = |0 immediately give rise to the transformation law \u03bb=0 \u221e \u03bb|\u02dc\u03bba \u2020 \u03bb, a \u2020 \u02dc\u03bb = \u03bb a\u02dc\u03bb = \u03bb \u02dc\u03bb|\u03bba\u03bb. (2.8) In many applications, we will be dealing with continuous sets of quantum numbers (such as position coordinates). In these cases, the quantum numbers are commonly denoted by a bracket notation a\u03bb \u2192 a(x) = x|\u03bba\u03bb and the summations appearing in the dx\u03bb|xa(x). transformation formula above translate to integrals: a\u03bb = \u03bb EXERCISE The transformation from the coordinate to the Fourier momentum representation in a \ufb01nite one-dimensional system of length L would read L ak = dx k|xa(x), a(x) = x|kak, (2.9) where k|x \u2261 x|k\u2217 = e\u2212ikx/ 0 \u221a L. k 5 To appropriately characterize the representation, we need to be a bit more precise. Within A, ai and a \u2020 i are independent objects, i.e. in general there exists no notion of Hermitian adjointness in A. We require, though, \u2020 i the Hermitian adjoint (in V ) of the image of ai. Also, we have to require that the representation assigns to a j ] :V \u2192 V where, in the latter expression, the commutator involves the that [ai, a ordinary product of matrices ai, a j ] \u2208 A be mapped onto [ai.1 Introduction to second quantization 47 Representation of operators (one-body): Single-particle or one-body operators \u02c6O1 N acting in the N -particle Hilbert space F N generally", " take the form \u02c6O1 = n=1 \u02c6on, where \u02c6on is an ordinary single-particle operator acting on the nth particle. A typical example is the kinetic energy operator \u02c6T = n/2m, where \u02c6pn is the momentum operator acting on the nth particle. Other examples include the one-particle potential operator \u02c6Sn, etc. Since \u02c6V = we have seen that, by applying \ufb01eld operators to the vacuum space, we can generate the Fock space in general and any N -particle Hilbert space in particular, it must be possible to represent any operator \u02c6O1 in an a-representation. n V (\u02c6xn), where V (x) is a scalar potential, the total spin operator n \u02c6p2 n Now, although the representation of n-body operators is, after all, quite straightforward, the construction can, at \ufb01rst sight, seem daunting. A convenient way of \ufb01nding such a representation is to express the operator in terms of a basis in which it is diagonal, and only later transform to an arbitrary basis. For this purpose it is useful to de\ufb01ne the occupation number operator \u02c6n\u03bb = a \u2020 \u03bba\u03bb, (2.10) n|0. with the property that, for bosons or fermions (exercise), \u02c6n\u03bb a \u2020 |n\u03bb1, n\u03bb2,... = Since \u02c6n\u03bb commutes with all a \u03bb=\u03bb, Eq. (2.4) readily implies that \u02c6n\u03bbj |n\u03bb1, n\u03bb2,..., i.e., \u02c6n\u03bb simply counts the number of particles in state \u03bb (hence the name n\u03bbj \u201coccupation number operator\u201d). Let us now consider a one-body operator, \u02c6O1, which is diagonal in the basis |\u03bb, with \u02c6o = |\u03bbi\u03bbi|, o\u03bbi = \u03bbi|\u02c6o|\u03bbi. With this de\ufb01nition, one \ufb01nds that n|0 = n i o\u03bbi \u2020 \u03bb \u2020 \u03bb a n \u03bb1, n \u03bb2,...| \u02c6O1|n\u03bb1, n\u03bb2,... = o\u03bbi n\u03bbi n \u03bb1 i = n \u03bb1, n \u03bb2,...|,...| n\u03bb1, n\u03bb2,..., n \u03bb2 o\u03bbi \u02c6n\u03bbi i |n\u03bb1, n\u03bb2,.... Since this equality holds for any set of states, one can infer the second quantized representation of the operator \u02c6O1, \u02c6O1 = \u221e \u03bb=0 o\u03bb \u02c6n\u03bb = \u221e \u03bb=0 \u03bb|\u02c6o|\u03bba \u2020 \u03bba\u03bb. The result is straightforward: a one-body operator engages a single particle at a time \u2013 the others are just spectators. In the diagonal representation, one simply counts the number of particles in a state \u03bb and multiplies by the corresponding eigenvalue of the one-body operator. Finally, by transforming from the diagonal representation to a general basis, one obtains the general result, \u02c6O1 = \u03bc|\u02c6o|\u03bda\u2020 \u03bca\u03bd. \u03bc\u03bd (2.11) 48 Second quantization To cement these ideas, let us consider some speci\ufb01c examples: representing the matrix 2 (\u03c3i)\u03b1\u03b1, where \u03b1, \u03b1 is a two- elements of the single-particle spin operator as (Si)\u03b1\u03b1 = 1 component spin index and \u03c3i are the Pauli spin matrices \u03c31 =, \u03c32 =, \u03c33 = 0 1 1 0 0 \u2212i 0 i, 1 0 0 \u22121 (2.12) the spin operator of a many-body system assumes the form \u02c6S = a \u2020 \u03bb\u03b1 S\u03b1\u03b1a\u03bb\u03b1. (2.13) \u03bb (Here, \u03bb denotes the set of additional quantum numbers, e.g. a lattice site index.) When second quantized in the position representation, one can show that the one-body Hamiltonian for a free particle is given as a sum of kinetic and potential energy as Wolfgang Ernst Pauli 1900\u201358 Nobel Laureate in Physics in 1945 \u201cfor the discovery of the Exclusion Principle, also called the Pauli Principle.\u201d The exclusion principle crystallized the existing knowledge of atomic structure at the time it was postulated", " and it led to the recognition of the two-valued variable required to characterize the state of an electron. Pauli was the \ufb01rst to recognize the existence of the neutrino. (Image c The Nobel Foundation.) \u02c6H = ddr a\u2020(r) \u02c6p2 2m + V (r) a(r), (2.14) where \u02c6p = \u2212i\u2202. EXERCISE Starting with momentum representation (in which the kinetic energy is diagonal), transform to the position representation and thereby establish Eq. (2.14). The local density operator \u02c6\u03c1(r), measuring the particle density at a certain coordinate r, is simply given by \u02c6\u03c1(r) = a\u2020(r)a(r). (2.15) Finally, the total occupation number operator, obtained by integrating over the particle density, is de\ufb01ned by \u02c6N = ddr a\u2020(r)a(r). In a theory with discrete quantum numbers, this operator assumes the form \u02c6N = \u03bb a \u2020 \u03bba\u03bb. Representation of operators (two-body): Two-body operators \u02c6O2 are needed to describe pairwise interactions between particles. Although pair-interaction potentials are straightforwardly included in classical many-body theories, their embedding into conventional many-body quantum mechanics is made cumbersome by particle indistinguishability. The formulation of interaction processes within the language of second quantization is considerably more straightforward: Initially, let us consider particles subject to the symmetric two-body potential V (rm, rn) \u2261 V (rn, rm) between two particles at position rm and rn. Our aim is to \ufb01nd an operator \u02c6V in second quantized form whose action on a many-body state gives 2.1 Introduction to second quantization 49 (presently, it is more convenient to use the original representation Eq. (2.1) rather than the occupation number representation) \u02c6V |r1, r2,...r N = N n<m V (rn, rm)|r1, r2,...r N = 1 2 N n=m V (rn, rm)|r1, r2,...r N. When this is compared with the one-point function, one might immediately guess that ddr ddra\u2020(r)a\u2020(r)V (r, r)a(r)a(r). \u02c6V = 1 2 That this is the correct answer can be con\ufb01rmed by applying the operator to a many-body state: a\u2020(r)a\u2020(r)a(r)a(r)|r1, r2,..., rN = a\u2020(r)a\u2020(r)a(r)a(r)a\u2020(r1) \u00b7 \u00b7 \u00b7a \u2020(rN )|0 \u03b6 n\u22121\u03b4(r \u2212 rn)a\u2020(rn) \u02c6\u03c1(r)! a\u2020(r)a(r) a\u2020(r1) \u00b7 \u00b7 \u00b7a \u2020(rn\u22121)a\u2020(rn+1) \u00b7 \u00b7 \u00b7a \u2020(rN )|0 \u03b6 n\u22121\u03b4(r \u2212 rn) N m(=n) \u03b4(r \u2212 rm)a\u2020(rn)a\u2020(r1) \u00b7 \u00b7 \u00b7a \u2020(rn\u22121)a\u2020(rn+1) \u00b7 \u00b7 \u00b7a \u2020(rN )|0 \u03b4(r \u2212 rn)\u03b4(r \u2212 rm)|r1, r2, \u00b7 \u00b7 \u00b7r N. = = = N n=1 N n=1 N n,m=n Multiplying by V (r, r)/2, and integrating over r and r, one con\ufb01rms the validity of the expression. It is left as an exercise to con\ufb01rm that the naive expression ddrV (r, r)\u02c6\u03c1(r)\u02c6\u03c1(r) does not reproduce the two-body operator. More generally, 1 2 turning to a non-diagonal basis, it is straightforward to con\ufb01rm that a general two-body operator can be expressed in the form ddr \u02c6O2 = \u03bb\u03bb\u03bc\u03bc O\u03bc,\u03bc,\u03bb,\u03bb a\u2020 \u03bca \u2020 \u03bc a\u03bba\u03bb, (2.16) where O\u03bc,\u03bc,\u03bb,\u03bb \u2261 \u03bc, \u03bc| \u02c6O2|\u03bb, \u03bb. As well as the pairwise Coulomb interaction formulated above, another important interaction, frequently encountered in problems of quantum magnetism, is the spin\u2013spin interaction. From our discussion of", " the second-quantized representation of spin \u02c6S above, we can infer that the general spin\u2013spin interaction can be presented in second-quantized form as ddr ddr \u02c6V = 1 2 \u03b1\u03b1\u03b2\u03b2 J(r, r)S\u03b1\u03b2 \u00b7 S\u03b1\u03b2 a\u2020 \u2020 \u03b1(r)a \u03b1 (r)a\u03b2 (r)a\u03b2(r), where J(r, r) denotes the exchange interaction. 50 Second quantization In principle, one may proceed in the same manner and represent general n-body interactions in terms of second-quantized operators. However, as n >2 interactions appear infrequently, we refer to the literature for discussion. This completes our formal introduction to the method of second quantization. To develop \ufb02uency in the operation of the method, we will continue by addressing a number of problems chosen from the realm of condensed matter. In doing so, we will see that second quantization often leads to considerable simpli\ufb01cation of the analysis of many-particle systems. The e\ufb00ective model Hamiltonians that appear below provide the input for subsequent applications considered in this text. Readers not wishing to get distracted from our main focus \u2013 the development of modern methods of quantum \ufb01eld theory in the condensed matter setting \u2013 may safely skip the next sections and turn directly to Chapter 3 below. It is worthwhile keeping in mind, however, that the physical motivation for the study of various prototypical model systems considered later in the text is given in Section 2.2. 2.2 Applications of second quantization Starting from the prototype Hamiltonian (1.1) introduced in Chapter 1, we have already explored generic aspects of lattice dynamics in condensed matter systems. In much of the remaining text we will explore examples from the complementary sector focusing on the electronic degrees of freedom. Drawing on the \ufb01rst of the principles discussed in Chapter 1, we will begin our discussion by reducing the full Hamiltonian to a form that contains the essential elements of the electron dynamics. As well as the pure electron sub-Hamiltonian He, the reduced Hamiltonian will involve the interaction between the electrons and the positively charged ionic background lattice. However, typically, lattice distortions due to both the motion of the ions and the ion\u2013ion interaction couple only indirectly. (Exercise: Try to think of a prominent example where the electron sector is crucially in\ufb02uenced by the dynamics of the host lattice.) To a \ufb01rst approximation, we may, therefore, describe the electron system through the simpli\ufb01ed Hamiltonian, \u02c6H = \u02c6H0 + \u02c6Vee, where \u02c6H0 = \u02c6Vee = 1 2 ddr a\u2020 \u03c3(r) \u02c6p2 2m + V (r) a\u03c3(r) ddr ddr Vee(r \u2212 r)a\u2020 \u2020 \u03c3(r)a \u03c3 (r)a\u03c3 (r)a\u03c3(r), \u23ab \u23aa\u23aa\u23ac (2.17) \u23aa\u23aa\u23ad I Vei(RI \u2212 r) denotes the lattice potential experienced by the electrons, and V (r) = the coordinates of the lattice ions RI are assumed \ufb01xed. For completeness, we have also endowed the electrons with a spin index, \u03c3 =\u2191 / \u2193. The Hamiltonian de\ufb01nes the problem of the interacting electron gas embedded in a solid state system. Despite its seemingly innocuous structure, the interacting electron Hamiltonian (2.17) accommodates a wide variety of electron phases from metals and magnets to insulators. To classify the phase behavior of the model, it is helpful to divide our considerations, focusing \ufb01rst on the properties of the non-interacting single-particle system \u02c6H0 and then, later, restoring the electron interaction Vee. 2.2 Applications of second quantization 51 Electrons in a periodic potential As we know from Bloch\u2019s theorem, eigenstates of a periodic Hamiltonian can be presented in the form of Bloch waves6 \u03c8kn(r) =e ik\u00b7rukn(r), where the components of the crystal momentum k take values inside the Brillouin zone, ki \u2208 [\u2212\u03c0/a, \u03c0/a], and we have assumed that the periodicity of the lattice potential is the same in all directions, i.e. V (r + aei) =V (r). The index n labels the separate energy bands of the solid, and the functions ukn(r + aei) =u kn(r) are periodic on the lattice. Now, depending on the nature of the bonding, there are two complementary classes of materials where the general structure of the Bl", "och functions can be simpli\ufb01ed signi\ufb01cantly. Felix Bloch 1905\u201383 Nobel Laureate in Physics in 1952 \u201cfor the development [with Edward M. Purcell] of new methods for nuclear magnetic precision measurements and discoveries in connection therewith.\u201d (Image c The Nobel Foundation.) Nearly free electron systems For certain materials, notably the elemental metals drawn from groups I\u2013IV of the periodic table, the outermost itinerant conduction electrons behave as if they were \u201cnearly free,\u201d i.e. their dynamic is largely oblivious to both the Coulomb potential created by the positively charged ion background and their mutual interaction. INFO Loosely speaking, Pauli exclusion of the bound state inner core electrons prevents the conduction electrons from exploring the region close to the ion core, thereby e\ufb00ectively screening the nuclear charge. In practice, the conduction electrons experience a renormalized pseudopotential, which accommodates the e\ufb00ect of the lattice ions and core electrons. Moreover, the high mobility of the conduction electrons provides an e\ufb03cient method of screening their own mutual Coulomb interaction. In nearly free electron compounds, complete neglect of the lattice potential is usually a good approximation (as long as one considers crystal momenta remote from the boundaries of the Brillouin zone, ki = \u00b1\u03c0/a). In practice, this means that we are at liberty to set the Bloch function to unity, ukn = 1, and regard the eigenstates of the non-interacting Hamiltonian as plane waves. This in turn motivates the representation of the \ufb01eld operators in momentum space Eq. (2.9), whereupon the non-interacting part of the Hamiltonian assumes the free particle form (once again, we have set = 1) \u02c6H0 = k k2 2m a \u2020 k\u03c3ak\u03c3, (2.18) 6 For a more detailed discussion one may refer to one of the many texts on the basic elements of solid state physics, e.g., Ashcroft and Mermin, Solid State Physics (Holt-Saunders International, 1995). 52 Second quantization where the sum extends over all wavevectors k and, as usual, summation convention of the spin indices is assumed. Turning to the Coulomb potential between conduction electrons, in the Fourier representation, the two-body interaction takes the form \u02c6Vee = 1 2Ld k,k,q Vee(q)a \u2020 k\u2212q\u03c3a \u2020 k+q\u03c3 ak\u03c3 ak\u03c3, (2.19) where (choosing units such that 4\u03c0 = 1), Vee(q) = e2/q2 represents the Fourier transform of the Coulomb interaction potential Vee(r) = e2/|r|. Now, as written, this expression neglects the fact that, in ionized solids, the negative charge density of the electron cloud will be compensated by the charge density of the positively ionized background. The latter can be incorporated into Eq. (2.19) by placing on the sum over q the restriction that q = 0 (exercise). Taken together, the free electron Hamiltonian \u02c6H0 and the Coulomb interaction potential \u02c6Vee are known as the Jellium model. The interaction described by Eq. (2.19) can be illustrated graphically, as shown in the \ufb01gure (for a more elaborate discussion of such diagrams, see Chapter 5): an electron of momentum k is scattered into a new momentum state k\u2212q while another electron is scattered from k \u2192 k + q. In concrete applications of lowcondensed matter temperature physics, one will typically consider low excitation energies. The solution of such problems is naturally organized around the zero-temperature ground state as a reference platform. However, the accurate calculation of the ground state energy of the system is a complicated problem of many-body physics that cannot be solved in closed form. Therefore, assuming that interactions will not substantially alter the ground state of the free particle problem Eq. (2.18) \u2013 which is often not the case! \u2013 one usually uses the ground state of the latter as a reference state. Niels Henrik David Bohr 1885\u2013 1962 Recipient of the 1922 Nobel Prize in Physics \u201cfor his services in the investigation of the structure of atoms and of the radiation emac nating from them.\u201d (Image The Nobel Foundation.) INFO Deferring a more quali\ufb01ed discussion to later, a preliminary justi\ufb01cation for this assumption can be given as follows. Suppose that the density of the electron gas is such that each of its N constituent particles occupies an average volume of O(ad). The average kinetic energy per particle is then estimated to be T \u223c 1/ma2, while", " the Coulomb interaction potential will scale as V \u223c e2/a. Thus, for a much smaller than theBohr radius a0 = 1/e2m, the interaction contribution is much smaller than the average kinetic energy. In other words, for the dense electron gas, the interaction energy can indeed be treated as a perturbation. Unfortunately, for most metals one \ufb01nds that a \u223c a0 and neither high- nor low- density approximations are strictly justi\ufb01ed. 2.2 Applications of second quantization 53 The ground state of the system occupied by N non-interacting particles can be readily inferred from Eq. (2.18). The Pauli principle implies that all energy states k = k2/2m will be uniformly occupied up to a cuto\ufb00 Fermi energy, EF. Speci\ufb01cally, for a system of size L, allowed momentum states k have components ki = 2\u03c0ni/L, ni \u2208 Z. The summation extends up to momenta with |k| \u2264k F, where the Fermi momentum kF is de\ufb01ned through the relation k2 F/2m = EF (see \ufb01gure at right). The relation between the Fermi momentum and the occupation number can be established by dividing the volume of theFermi sphere \u223c kd viz. N = C(kFL)d where C denotes a dimensionless geometry-dependent constant. F by the momentum space volume per mode (2\u03c0/L)d, In the language of second quantization, the ground state can now be represented as |\u03a9 \u2261 N'|k|\u2264kF,\u03c3 a \u2020 k\u03c3 |0, (2.20) where |0 denotes the state with zero electrons present. When the interaction is weak, one may anticipate that low-temperature properties will be governed by energetically lowlying excitations superimposed upon the state |\u03a9. Therefore, remembering the philosophy whereby excitations rather than microscopic constituents play a prime role, one would like to declare the \ufb01lled Fermi sea, |\u03a9 (rather than the empty state |0), to be the \u201cphysical vacuum\u201d of the theory. To make this compatible with the language of second quantization, we \u2020 need to identify a new operator algebra ck\u03c3, c k\u03c3 such that the new operators ck\u03c3 annihilate the Fermi sea. This can be easily engineered by de\ufb01ning c \u2020 k\u03c3 = ) \u2020 a k\u03c3, k > kF, ak\u03c3, k \u2264 kF, * ) ck\u03c3 = * ak\u03c3, k > kF, \u2020 k\u03c3, k \u2264 kF. a (2.21) It is then a straightforward matter to verify that ck\u03c3|\u03a9 = 0, and that the canonical commutation relations are preserved. The Hamiltonian de\ufb01ned through Eq. (2.18) and (2.19), represented in terms of the operator algebra (2.21) and the vacuum (2.20), forms the basis of the theory of interactions in highly mobile electron compounds. The investigation of the role of Coulomb interactions in such systems will provide a useful arena to apply the methods of quantum \ufb01eld theory formulated in subsequent chapters. Following our classi\ufb01cation of electron systems, let us now turn our attention to a complementary class of materials where the lattice potential presents a strong perturbation on the conduction electrons. In such situations, realized, for example, in transition metal oxides, a description based on \u201calmost localized\u201d electron states will be used to represent the Hamiltonian (2.17). 1\u221a N B.Z. k R 54 Second quantization Tight\u2013binding systems Let us consider a \u201crare\ufb01ed\u201d lattice in which the constituent ion cores are separated by a distance in excess of the typical Bohr radius of the valence band electrons. In this \u201catomic limit\u201d the weight of the electron wavefunctions is \u201ctightly bound\u201d to the lattice centers. Here, to formulate a microscopic theory of interactions, it is convenient to expand the Hamiltonian in a local basis that re\ufb02ects the atomic orbital states of the isolated ion. Such a representation is presented by the basis of Wannier states de\ufb01ned by |\u03c8Rn \u2261 1\u221a N B.Z. k e\u2212ik\u00b7R|\u03c8kn, |\u03c8kn = eik\u00b7R|\u03c8Rn, (2.22) where R denote the coordinates of the lattice centers, and represents a summation over all momenta k in the \ufb01rst Brillouin zone. For a system with a vanishingly weak interatomic", " overlap, i.e. a lattice where V approaches a superposition of independent atomic potentials, the Wannier function \u03c8Rn(r) \u2261 r|\u03c8Rn converges on the nth orbital of an isolated atom centered at coordinate R. However, when the interatomic coupling is non-zero, i.e. in a \u201creal\u201d solid, the N formerly degenerate states labeled by n split to form an energy band (see \ufb01gure below). Here, the Wannier functions (which, note, are not eigenfunctions of the Hamiltonian) di\ufb00er from those of the atomic orbitals through residual oscillations in the region of overlap to ensure orthogonality of the basis. Signi\ufb01cantly, in cases where the Fermi energy lies between two energetically separated bands, the system presents insulating behavior. Conversely, when the Fermi energy is located within a band, one may expect metallic behavior. Ignoring the complications that arise when bands begin to overlap, we will henceforth focus on metallic systems where the Fermi energy is located within a de\ufb01nite band n0. E EF n = 1 n = 2 How can the Wannier basis be exploited to obtain a simpli\ufb01ed representation of the general Hamiltonian (2.17)? The \ufb01rst thing to notice is that the Wannier states {|\u03c8Rn} form an orthonormal basis of the singleparticle Hilbert space, i.e. the transformation atomic limit between the real space and the Wannier R(r)|\u03c8R. (Here, since we are |\u03c8R\u03c8R|r = representation is unitary, |r = interested only in contributions arising from the particular \u201cmetallic\u201d band n0 in which the Fermi energy lies, we have dropped the remaining set of bands n = n0 and, with them, reference to the speci\ufb01c band index.) (Exercise: By focusing on just a single band n0, in what sense is the Wannier basis now complete?) As such, it induces a transformation insulator metal R \u03c8\u2217 V \u2013Vatom R a\u2020 \u03c3(r) = R \u03c8\u2217 \u2020 R(r)a R\u03c3 \u2261 i \u03c8\u2217 Ri \u2020 (r)a i\u03c3, (2.23) between the real space and the Wannier space operator basis, respectively. In the second representation, following a convention commonly used in the literature, we have labeled the lattice center coordinates R \u2261 Ri by a counting index i = 1,..., N. Similarly, the 2.2 Applications of second quantization 55 unitary transformation between Bloch and Wannier states Eq. (2.22) induces an operator transformation a \u2020 k\u03c3 = 1\u221a N i eik\u00b7Ri a \u2020 i\u03c3, a \u2020 i\u03c3 = 1\u221a N B.Z. k e\u2212ik\u00b7Ri a \u2020 k\u03c3. (2.24) We can now use the transformation formulae (2.23) and (2.24) to formulate a Wannier representation of the Hamiltonian (2.17). Using the fact that the Bloch states diagonalize the single-particle component \u02c6H0, we obtain \u02c6H0 = ka \u2020 k\u03c3ak\u03c3 (2.24) = eik(Ri\u2212Ri )ka \u2020 i\u03c3ai\u03c3 \u2261 a \u2020 i\u03c3 tii ai\u03c3, k ii k k eik\u00b7(Ri\u2212Ri )k. The new representation of \u02c6H0 describes where we have de\ufb01nedt ii = N \u22121 electrons hopping from one lattice center i to another, i. The strength of the hopping matrix element tii is controlled by the e\ufb00ective overlap of neighboring atoms. In the extreme atomic limit, where the levels k = const. are degenerate, tii \u221d \u03b4ii and no inter-atomic transport is possible. The tight-binding representation becomes useful when ti=i is non-vanishing, but the orbital overlap is so weak that only nearest neighbor hopping e\ufb00ectively contributes. ii 1 N EXERCISE Taking a square lattice geometry, and setting tii = \u2212t for i, i nearest neighbors and zero otherwise, diagonalize the two-dimensional tight-binding Hamiltonian \u02c6H0. Show that the eigenvalues are given by k = \u22122t(cos(kxa) + cos(kya)). Sketch contours of constant energy in the Brillouin zone and note the geometry at half-\ufb01lling. To assess the potency of the tight-binding approximation,", " let us consider its application to two prominent realizations of carbon-based lattice systems, graphene and carbon nanotubes. INFO Graphene is a single layer of graphite: a planar hexagonal lattice of sp2-hybridized7 carbon atoms connected by strong covalent bonds of their three planar \u03c3-orbitals. (See the Fig. 2.2 and \ufb01gure overleaf for a schematic.) The remaining pz orbitals \u2013 oriented vertically to the lattice plane \u2013 overlap weakly to form a band of mobile electrons. For a long time, it was thought that graphene sheets in isolation will inevitably be destabilized by thermal \ufb02uctuations; only layered stacks of graphene would mutually stabilize to form a stable compound \u2013 graphite. It thus came as a surprise when in 2004 a team of researchers8 succeeded in the isolation of large (micron-sized) graphene \ufb02akes on an SiO2 substrate. (Meanwhile, the isolation of even free standing graphene layers has become possible. In fact, our whole conception of the stability of the compound has changed. It is now believed that whenever you draw a pencil line, a trail of graphene \ufb02akes will be left behind.) Immediately after its discovery, it became clear that graphene possesses unconventional conduction properties. Nominally a gapless semiconductor, it has an electron mobility \u223c 2 \u00d7 105cm2/Vs, by far more than that of even the purest silicon based semiconductors; it shows manifestations of the integer quantum Hall e\ufb00ect qualitatively di\ufb00erent from those of conventional two-dimensional electron compounds (cf. Chapter 9 for a general discussion of the quantum Hall 7 Although this will not be necessary to follow the text, readers may \ufb01nd it rewarding to recapitulate the quantum chemistry of sp2-hybridized carbon and its covalent bonds. 8 K. S. Novoselov, et al., Electric \ufb01eld e\ufb00ect in atomically thin carbon \ufb01lms, Science 306 (2004), 666\u20139. 56 Second quantization Figure 2.2 Left: optical microscopy image of graphene \ufb02akes. Regions labeled by \u2018I\u2019 de\ufb01ne monolayer graphene sheets. Right: STM image of the graphene samples shown in the left part. Images taken from E. Stolyarova et al., High-resolution scanning tunneling microscopy imaging of mesoscopic graphene sheets on an insulating surface, PNAS 104 (2007), 9210\u201312. e\ufb00ect), etc. Although an in-depth discussion of graphene is beyond the scope of this text, we note that it owes most of its fascinating properties to its band structure: electrons in graphene show a linear dispersion and behave like two-dimensional Dirac fermions! By way of an illustration of the concepts discussed above, we here derive this unconventional band dispersion from a tight binding formulation of the system. To a \ufb01rst approximation, graphene\u2019s \u03c0-electron system can be modeled as a tight-binding Hamiltonian characterized by a single hopping matrix element between neighboring atoms \u2212t, and the energy o\ufb00set of the p-electron states. To determine the spectrum of the system, one may introduce a system of bi-atomic unit cells (see the ovals in the schematic \ufb01gure right) and two (nonorthogonal) unit vectors of the hexagonal lattice 3, \u22121)a/2, where a1 = ( a = |a1| = |a2| 2.46 \u02daA denotes the lattice constant. The tight-binding Hamiltonian can then be represented in the form \u02c6H = \u2212 1(r)a2(r) + h.c.), where the sum runs over all nearest neighbor basis vectors of the unit cells (the coordinate vectors of the bottom left \u2020 1(2)(r) creates a state in the \ufb01rst (second) atom of the cell at vector r. Switching atom) and a to a Fourier representation, the Hamiltonian assumes the form 3, 1)a/2 anda 2 = ( \u2013a1 a2 \u2013 a1 r,r(ta a1 a2 \u221a \u221a \u2020 \u02c6H = k\u03c3 a \u2020 1\u03c3 a \u2020 2\u03c3 0 \u2212tf \u2217(k) \u2212tf (k) 0, a1\u03c3 a2\u03c3 (2.25) where f (k) = 1 + e\u2212ik1a + ei(\u2212k1+k2)a. \u03b5 (a) 2.2 Applications of second quantization 57 ky kx ky (b) ky Figure 2.3 (a) Spectrum of the tight-binding Hamiltonian (2.24) showing the point-like structure", " of the Fermi surface when EF = 0. (b) A contour plot of the same. EXERCISE (Recall the concept of the reciprocal lattice in solid state theory.) To derive the Fourier representation above, show that a system of two reciprocal lattice vectors conjugate to the \u221a unit vectors above is given by G1/2 = 2\u03c0\u221a 3). Next, show that the Fourier decomposition 3a k e\u2212i a 2\u03c0 (k1G1+k2G2)\u00b7raa,k, where ki \u2208 [0, 2\u03c0/a] is of a \ufb01eld operator reads as aa(r) = 1\u221a N quantized in units 2\u03c0/Li. (Li is the extension of the system in the direction of ai and N its total number of unit cells.) Substitute this decomposition into the real space representation of the Hamiltonian to arrive at the Fourier representation. (1, \u00b1 Diagonalizing the Hamiltonian, one obtains the dispersion relation,9 k = \u00b1t|f (k)| = \u00b1t 3 + 2 cos(k1a) + 2 cos((k1 \u2212 k2)a) + 2 cos(k2a). (2.26) Here, in contrast to the square lattice tight-binding Hamiltonian, the half-\ufb01lled system is characterized by a point-like Fermi surface (see Figure 2.3). When lightly doped away from half-\ufb01lling, the spectrum divides into Dirac-like spectra with a linear dispersion. Fig. 2.3 shows the full spectrum of the tight binding Hamiltonian. Notice that of the six Dirac points displayed in the \ufb01gure only two are independent. The complementary four can be reached from those two points by addition of a reciprocal lattice vector and, therefore, do not represent independent states. EXERCISE Derive an explicit representation of the Dirac-Hamiltonian describing the lowenergy physics of the system. To this end, choose two inequivalent (not connected by reciprocal lattice vectors) zero-energy points k1,2 in the Brillouin zone. Expand the Hamiltonian (2.25) around these two points in small momentum deviations q \u2261 k \u2212 k1,2 up to linear order. Show that in this approximation, \u02c6H reduces to the sum of two two-dimensional Dirac Hamiltonians. Now, against this background, let us consider the carbon nanotube system. A single-wall nanotube describes a one-dimensional structure involving a graphene sheet rolled into a cylinder (see Fig 2.4 for an STM image of a carbon nanotube). Tubes of comparatively simple structure are obtained by rolling the hexagonal pattern of the sheet along one of its two axes of symmetry: along the a1-direction one obtains a zig-zag tube and along the 9 P. R. Wallace, The band theory of graphite, Phys. Rev. 71 (1947), 622\u201334. 58 Second quantization Figure 2.4 STM image of a carbon nanotube contacted with Pt electrodes. (Source: Courtesy of C. Dekker.) (a1 + a2)-direction an armchair tube (Exercise: Where do you see an armchair?). More complex, \u201cchiral\u201d structures involve twists along the circumference of the tube. \u03b5 2 By knowing the band structure of graphene, the dispersion of the nanotube system can be straightforwardly inferred. The situation is most easily visualized for the achiral geometries. Essentially a graphene sheet with periodic boundary conditions and length L in the direction transverse to the length of the system has tube, a spectrum that obtains by projection of the two-dimensional dispersion onto lines indexed by the discrete values k\u22a5 along the compact axis. For example, the dispersion of lowest transverse harmonic, k\u22a5 = 0, of the armchair tube is given by (see the \ufb01gure) the quasi-one-dimensional 6 k a \u20132 \u20136 \u20131 \u20132 \u20134 1 4 2 + k = \u00b1t 3 + 4 cos(ka/2) + 2 cos(ka). Notice the presence of two nodal points in the spectrum. Electrons with longitudinal momentum close to one of these two \u201chot spots\u201d propagate with approximately linear dispersion. The physics of e\ufb00ectively one-dimensional electron systems of this type will be discussed in Section 2.2 below. \u221a \u221a EXERCISE Verify the one-dimensional dispersion relation k above. To this end, notice that the single-electron wave functions of the system must obey periodic boundary conditions \u03c8(r + N (a1 + a2)) = \u03c8(r), where N is the number of cells in the trans", "verse direction (i.e. L\u22a5 \u2261 N |a1 + a2| = N a 3 is the circumference of the tube). Use this condition to obtain the quantization rule k1 + k2 = 2\u03c0 3m/L\u22a5, where m is integer. When evaluated for the lowest harmonic, m = 0, the dispersion relation (2.26) collapses to k (where k = k1 \u2212 k2 is the momentum in the longitudinal direction). Show that the zig-zag tube supports zero-energy excitations only for speci\ufb01c values of the transverse lattice spacings N. In general, it is insulating with a band gap of ca. 0.5 eV. In the case of the chiral tubes, it may be shown that a third of the tubes are semi-metallic while all others are insulating.10 10 For futher details on the electronic structure of carbon nanotubes, we refer to the text Physical Properties of Carbon Nanotubes by R. Saito, G. Dresselhaus, and M. S. Dresselhaus (Imperial College Press, 1998). 2.2 Applications of second quantization 59 Interaction e\ufb00ects in the tight-binding system Although the pseudopotential of the nearly free electron system accommodates the e\ufb00ects of Coulomb interaction between the conduction and valence band electrons, the mutual Coulomb interaction between the conduction electrons themselves may lead to new physical phenomena. These e\ufb00ects can substantially alter the material parameters (e.g. e\ufb00ective masses), however, they neither change the nature of the ground state, nor that of the elementary quasi-particle excitations in any fundamental way \u2013 this is the content of Fermiliquid theory, and a matter to which we will return. By contrast even weak interaction e\ufb00ects signi\ufb01cantly in\ufb02uence the physics of the tight-binding system. Here, commensurability e\ufb00ects combined with interaction may drive the system towards a correlated magnetic state or an insulating phase. To understand why, let us re-express the interaction in \ufb01eld operators associated with the Wannier states. Once again, to keep our discussion simple (yet generic in scope), let us focus attention on a single sub-band and drop reference to the band index. Then, applied to the Coulomb interaction, the transformation (2.23) leads to the expansion \u02c6Vee = iijj Uiijj a \u2020 i\u03c3a \u2020 i\u03c3 aj\u03c3aj\u03c3 where Uiijj = 1 2 ddr ddr \u03c8\u2217 Ri (r)\u03c8Rj (r)V (r \u2212 r)\u03c8\u2217 Ri (r)\u03c8Rj (r). (2.27) Taken together, the combination of the contributions \u02c6H0 and \u02c6Vee, \u02c6H = ii a \u2020 i\u03c3tii ai\u03c3 + iijj Uiijj a \u2020 i\u03c3a \u2020 i\u03c3 aj\u03c3 aj\u03c3, (2.28) where the sum of repeated spin indices is implied, de\ufb01nes the tight-binding representation of the interaction Hamiltonian. Apart from the neglect of the neighboring sub-bands, the Hamiltonian is exact. Yet, to assimilate the e\ufb00ects of the interaction, it is useful to assess the relative importance of the di\ufb00erent matrix elements drawing on the nature of the atomic limit that justi\ufb01ed the tight-binding description. We will thus focus on contributions to Uiijj where the indices are either equal or, at most, nearest neighbors. Focusing on the most relevant of these matrix elements, a number of physically di\ufb00erent contributions can be identi\ufb01ed: \u03c3 a tions and couple density \ufb02uctuations at neighboring sites, The direct terms Uiiii \u2261 Vii involve integrals over square moduli of Wannier funci=i Vii \u02c6ni \u02c6ni, where \u02c6ni = \u2020 i\u03c3ai\u03c3. This contribution accounts for the \u2013 essentially classical \u2013 interaction between charges localized at neighboring sites (see Fig. 2.5). In certain materials, interactions of this type have the capacity to induce global instabilities in the charge distribution known as charge density wave instabilities. A second important contribution derives from the exchange coupling, which induces \u2261 Uijji, and making use of magnetic correlations among the electron spins. Setting J F ij 60 Second quantization (a) (b) (c) (d) Figure 2.5 Di\ufb00erent types of interaction mechanism induced by the tight-binding interaction Vee. The curves symbolically indicate", " wavefunction envelopes. (a) Direct Coulomb interaction between neighboring sites. Taking account of the exchange interaction, parallel alignment of spins (b) is preferred since it enforces anti-symmetry of the spatial wave function. By contrast, for anti-parallel spin con\ufb01gurations (c) the wave function amplitude in the repulsion zone is enhanced. (d) Coulomb interaction between electrons of opposite spin populating the same site. Pauli matrix identities (see below), one obtains i=j Uijjia \u2020 i\u03c3a \u2020 j\u03c3 ai\u03c3 aj\u03c3 = \u22122 J F ij \u02c6Si \u00b7 \u02c6Sj + \u02c6ni \u02c6nj. 1 4 i=j Such contributions tend to induce weak ferromagnetic coupling of neighboring spins (i.e. J F > 0). The fact that an e\ufb00ective magnetic coupling is born out of the electrostatic interaction between quantum particles is easily understood. Consider two electrons inhabiting neighboring sites. The Coulomb repulsion between the particles is minimized if the orbital two-particle wave function is anti-symmetric and, therefore, has low amplitude in the interaction zone between the particles. Since the overall wavefunction must be anti-symmetric, the energetically favored real-space con\ufb01guration enforces a symmetric alignment of the two spins. Such a mechanism is familiar from atomic physics where it is manifested as Hund\u2019s rule. In general, magnetic interactions in solids are usually generated as an indirect manifestation of the much stronger Coulomb interaction. EXERCISE Making use of the Pauli matrix identity \u03c3\u03b1\u03b2 \u00b7 \u03c3\u03b3\u03b4 = 2\u03b4\u03b1\u03b4\u03b4\u03b2\u03b3 \u2212 \u03b4\u03b1\u03b2\u03b4\u03b3\u03b4, show that \u02c6Si \u00b7 \u02c6Sj = \u2212a\u2020 j\u03b2ai\u03b2aj\u03b1/2 \u2212 \u02c6ni \u02c6nj/4 where, as usual, \u02c6Si = a\u2020 i\u03b1\u03c3\u03b1\u03b2ai\u03b2/2 denotes the operator for spin 1/2, and the lattice sites i and j are assumed distinct. i\u03b1a\u2020 Finally, far into the atomic limit, where the atoms are very well separated, and the overlap between neighboring orbitals is weak, the matrix elements tij and J F ij are exponentially small in the interatomic separation. In this limit, the \u201con-site\u201d Coulomb or Hubbard interaction, Uiiii \u2261 U/2, i U \u02c6ni\u2191 \u02c6ni\u2193, generates the dominant interaction mechanism. Taking only the nearest neighbor contribution to the hopping matrix elements, and neglecting the energy o\ufb00set due to the diagonal term, the e\ufb00ective \u2020 i\u03c3 ai\u03c3 ai\u03c3 = i\u03c3\u03c3 Uiiiia \u2020 i\u03c3a 2.2 Applications of second quantization 61 Hamiltonian takes a simpli\ufb01ed form known as the Hubbard model, \u02c6H = \u2212t ij a \u2020 i\u03c3aj\u03c3 + U i \u02c6ni\u2191 \u02c6ni\u2193, (2.29) where ij is a shorthand used to denote neighboring lattice sites. In hindsight, a model of this structure could have been proposed from the outset on purely phenomenological grounds: electrons tunnel between atomic orbitals localized on individual lattice sites while double occupancy of a lattice site incurs an energetic penalty associated with the mutual Coulomb interaction. Mott\u2013Hubbard transition and the magnetic state Deceptive in its simplicity, the Hubbard model is acknowledged as a paradigm of strong electron correlation in condensed matter. Yet, after 40 years of intense investigation, the properties of this seemingly simple model system \u2013 the character of the ground state and nature of the quasi-particle excitations \u2013 are still the subject of controversy (at least in dimensions higher than one \u2013 see below). Nevertheless, given the importance attached to this system, we will close this section with a brief discussion of the remarkable phenomenology that is believed to characterize the Hubbard system. As well as dimensionality, the phase behavior of the Hubbard Hamiltonian is characterized by three dimensionless parameters: the ratio of the Coulomb interaction scale to the bandwidth U/t, the particle density or \ufb01lling fraction n (i.e. the average number of electrons per site), and the (dimensionless) temperature, T /t. The symmetry of the Hamiltonian under particle\u2013hole interchange (exercise) allows one to limit consideration to densities in the range 0 \u2264 n \u2264 1 while densities 1 < n \u2264 2 can be inferred by \u201cre\ufb02ection.\u201d Focusing \ufb01rst on the low temperature system, in the dilute limit n 1, the typical electron wavelength is greatly in excess of the site separation and the dynamics are free. Here", " the local interaction presents only a weak perturbation and one can expect the properties of the Hubbard system to mirror those of the weakly interacting nearly free electron system. While the interaction remains weak one expects a metallic behavior to persist. Sir Neville Francis Mott 1905\u2013 96 Nobel Laureate in Physics in 1977, with Philip W. Anderson and John H. van Vleck, for their \u201cfundamental theoretical investigations of the electronic structure of magnetic and disordered systems.\u201d Amongst his contributions to science, Mott provided a theoretical basis to understand the transition of materials from metallic to nonmetallic states (the Mott transition). (Image c The Nobel Foundation.) By contrast, let us consider the half-\ufb01lled system where the average site occupancy is unity. Here, if the interaction is weak, U/t 1, one may again expect properties reminiscent 62 Second quantization of the weakly interacting electron system.11 If, on the other hand, the interaction is very strong, U/t 1, site double occupancy is inhibited and electrons in the half-\ufb01lled system become \u201cjammed\u201d: migration of an electron to a neighboring lattice site necessitates site double occupancy incurring an energy cost U. Here, in this strongly correlated state, the mutual Coulomb interaction between the electrons drives the system from a metallic to an insulating phase with properties very di\ufb00erent from those of a conventional band insulator. INFO Despite the ubiquity of the experimental phenomenon (\ufb01rst predicted in a celebrated work by Mott) the nature of the Mott\u2013Hubbard transition from the metallic to the insulating phase in the half-\ufb01lled system has been the subject of considerable discussion and debate. In the original formulation, following a suggestion of Peierls, Mott conceived of an insulator characterized by two \u201cHubbard bands\u201d with a bandwidth \u223c t separated by a charge gap U.12 States of the upper band engage site double occupancy while those states that make up the lower band do not. The transition between the metallic and the insulating phase was predicted to occur when the interaction was su\ufb03ciently strong that a charge gap develops between the bands. Later, starting from the weakly interacting Fermi-liquid, Brinkman and Rice13 proposed that the transition was associated with the localization of quasi-particles created by an interaction driven renormalization of the e\ufb00ective mass. Finally, a third school considers the transition to the Mott insulating phase as inexorably linked to the development of magnetic correlations in the weak coupling system \u2013 the Slater instability. In fact, the characterizations of the transition above are not mutually exclusive. Indeed, in the experimental system, one \ufb01nds that all three possibilities are, in a sense, realized. In particular, a transition between the Mott insulating phase and an itinerant electron phase can be realized in two ways. In the \ufb01rst case, one can reduce the interaction strength U/t while, in the second, one can introduce charge carriers (or vacancies) into the half-\ufb01lled system. Experimentally, the characteristic strength of the interaction is usually tuned by changing the bandwidth t through external pressure (see Fig. 2.6), while a system may be doped away from half-\ufb01lling by chemical substitution. Remarkably, by focusing on the scaling behavior close to the critical end-point, researchers have been able to show that the Mott transition in this system belongs to the universality class of the three-dimensional Ising model (see the discussion of critical phenomena in Chapter 8). Lowering the temperature, both the Mott insulating phase and the strongly correlated metallic phase exhibit a transition to a magnetic phase where the local moments order antiferromagnetically (for the explanation of this phenomenon, see below). Experimentally, it is often found that the low-temperature phase of the Mott insulator is accompanied by the anti-ferromagnetic ordering of the local moments. The origin of these magnetic correlations can be traced to a mechanism known as superexchange and can be understood straightforwardly within the framework of the Hubbard model system. 11 In fact, one has to exercise some caution since the commensurability of the Fermi wavelength with the lattice can initiate a transition to an insulating spin density wave state characterized by a small quasi-particle energy gap. In later chapters, we will discuss the nature of this Slater instability (J. C. Slater, Magnetic e\ufb00ects and the Hartree\u2013Fock equation, Phys. Rev. 82 (1951), 538\u201341) within the framework of the quantum \ufb01eld integral. 12 N. F. Mott, The basis of the electron theory of metals with special reference to the transition metals, Proc. Roy. Soc. A 62 (1949), 416\u201322", " \u2013 for a review see, e.g., N. F. Mott, Metal\u2013insulator transition, Rev. Mod. Phys. 40 (1968), 677\u201383, or N. F. Mott, Metal\u2013Insulator Transitions, 2nd ed. (Taylor and Francis, 1990). 13 W. Brinkman and T. M. Rice, Application of Gutzwiller\u2019s variational method to the metal\u2013insulator transition, Phys. Rev. B 2 (1970), 4302\u20134. 2.2 Applications of second quantization 63 A M et a l 250 300 350 T (K) 400 450 500 1800 1500 1200 900 600 300 \u03c3 ( \u03a9 c m ) \u2013 1 6000 5000 I n s u l a t o r 1000 4000 3000 2000 P (bar) Figure 2.6 Conductivity of Cr-doped V2O3 as a function of decreasing pressure and temperature. At temperatures below the Mott\u2013Hubbard transition point (Pc = 3738 bar, Tc = 457.5 K) the conductivity reveals hysteretic behavior characteristic of a \ufb01rst-order transition. (Reprinted from P. Limelette, A. Georges, D. J\u00b4erome, et al., Universality and critical behavior at the Mott transition, Science 302 (2003), 89\u201392. Copyright 2003 AAAS.) \u2020 \u2020 \u2020 \u2020 \u2020 2\u2191a 2\u2193|\u03a9, and four states with Sz \u2020 1\u2191a 1\u2193|\u03a9, |d1 = a \u2020 1\u2193a \u2020 1\u2193|\u03a9, and |d2 = a To this end, one may consider a simple \u201ctwo-site\u201d system from which the characteristics of the in\ufb01nite lattice system can be inferred. For the two-site system, at half-\ufb01lling (i.e. with just two electrons to share between the two sites), one can identify a total of six 2\u2191|\u03a9, a total = 0: basis states: two spin-polarized states a \u2020 \u2020 |s1 = a 2\u2193|\u03a9. Recalling 2\u2193|\u03a9, |s2 = a 1\u2191a 1\u2191a the constraints imposed by the Pauli principle, it is evident that the fully spin polarized states are eigenstates of the Hubbard Hamiltonian with zero energy, while the remaining eigenstates involve superpositions of the basis states |si and |di. In the strong coupling limit U/t 1, the ground state will be composed predominantly of states with no double occupancy, |si. To determine the precise structure of the ground state, we could simply diagonalize the 4 \u00d7 4 Hamiltonian exactly \u2013 a procedure evidently infeasible in the large lattice system. Instead, to gain some intuition for the extended system, we will e\ufb00ect a perturbation theory which projects the insulating system onto a low-energy e\ufb00ective spin Hamiltonian. Speci\ufb01cally, we will treat the hopping part of the Hamiltonian \u02c6Ht as a weak perturbation of the Hubbard interaction \u02c6HU. \u2020 2\u2191a \u2020 To implement the perturbation theory, it is helpful to invoke a canonical transformation of the Hamiltonian, namely \u02c6H \u2192 \u02c6H \u2261 e\u2212t \u02c6O \u02c6Het \u02c6O = e\u2212t[ \u02c6O, ] \u02c6H \u2261 \u02c6H \u2212 t[ \u02c6O, \u02c6H] + t2 2! [ \u02c6O, [ \u02c6O, \u02c6H]] + \u00b7 \u00b7 \u00b7, (2.30) 64 Second quantization where the exponentiated commutator is de\ufb01ned by the series expansion on the right. EXERCISE Prove the second equality. (Hint: Consider the derivative of \u02c6H with respect to t.) By choosing the operator \u02c6O such that \u02c6Ht + t[ \u02c6HU, \u02c6O] = 0, all terms at \ufb01rst order in t can be eliminated from the transformed Hamiltonian. As a result, the e\ufb00ective Hamiltonian is brought to the form \u02c6H = \u02c6HU + t 2 [ \u02c6Ht, \u02c6O] +O (t3). (2.31) Applying the ansatz, t \u02c6O = [ \u02c6Ps \u02c6Ht \u02c6Pd\u2212 \u02c6Pd \u02c6Ht \u02c6Ps]/U, where \u02c6Ps and \u02c6Pd are operators that project onto the singly", " and doubly occupied subspaces respectively, the \ufb01rst-order cancellation is assured. EXERCISE To verify this statement, take the matrix elements of the \ufb01rst-order equation with the basis states. Alternatively, it can be con\ufb01rmed by inspection, noting that \u02c6Ps \u02c6Pd = 0, \u02c6HU \u02c6Ps = 0, and, in the present case, \u02c6Ps \u02c6Ht \u02c6Ps = \u02c6Pd \u02c6Ht \u02c6Pd = 0. Substituting this result into Eq. (2.31) and projecting onto the singly occupied subspace one obtains \u02c6Ps \u02c6H \u02c6Ps = \u2212 1 U \u02c6Ps \u02c6Ht \u02c6Pd \u02c6Ht \u02c6Ps = \u22122 t2 U \u02c6Ps 1 + a \u2020 1\u03c3a \u2020 2\u03c3 a1\u03c3 a2\u03c3 \u02c6Ps = J \u02c6S1 \u00b7 \u02c6S2 \u2212 1 4, where J = 4t2/U denotes the strength of the anti-ferromagnetic exchange interaction that couples the spins on neighboring sites. EXERCISE Remembering the anti-commutation relations of the electron operators, \ufb01nd the matrix elements of the Hubbard Hamiltonian on the four basis states |si and |di. Diagonalizing the 4\u00d74 matrix Hamiltonian, obtain the eigenstates of the system. In the strong coupling system U/t 1, determine the spin and energy dependence of the ground state. The perturbation theory above shows that electrons subject to a strong repulsive Coulomb interaclocal tion have a tendency to adopt an antiparallel or antiferromagnetic spin con\ufb01guration between neighboring sites. This observation has a simple physical interpretation. Anti-parallel spins can take advantage of the hybridization (however small) and reduce their kinetic energy by hopping to a neighboring site (see Fig. 2.7). Parallel spins on the other hand are restricted from participating in this virtual process by the Pauli principle. This mechanism, which involves a two-step process, was \ufb01rst formulated by Anderson14 and is known as superexchange. Philip W. Anderson 1923\u2013 in Nobel Laureate in Physics 1977, with Sir Neville Mott and John H. van Vleck, their \u201cfundamental theoretical investigations of the electronic structure of magnetic and disordered systems.\u201d Anderson made numerous contributions to theoretical physics from theories of localization and antiferromagnetism to superconductivity. (Image c The Nobel Foundation.) for 14 P. W. Anderson, Antiferromagnetism. Theory of superexchange interaction, Phys. Rev. 79 (1950) 350\u20136. 2.2 Applications of second quantization 65 J U\u20131 J Figure 2.7 Top: hybridization of spin polarized states is forbidden by Pauli exclusion. Bottom: superexchange mechanism by which two antiparallel spins can lower their energy by a virtual process in which the upper Hubbard band is occupied. The calculation presented above is easily generalized to an extended lattice system. Once again, projecting onto a basis in which all sites are singly occupied, virtual exchange processes favor an antiferromagnetic arrangement of neighboring spins. Such a correlated magnetic insulator is described by the quantum spin-(1/2) Heisenberg Hamiltonian \u02c6H = J mn \u02c6Sm \u00b7 \u02c6Sn, (2.32) where, as usual, mn denotes a sum of neighboring spins on the lattice and the positive exchange constant J \u223c t2/U. While, in the insulating magnetic phase, the charge degrees of freedom remain \u201cquenched,\u201d spin \ufb02uctuations can freely propagate. When doped away from half-\ufb01lling, the behavior of the Hubbard model is notoriously di\ufb03cult to resolve. The removal of electrons from the half-\ufb01lled system introduces vacancies into the \u201clower Hubbard band\u201d that may propagate through the lattice. For a low concentration of holes, the strong coupling Hubbard system may be described by the e\ufb00ective t\u2013J Hamiltonian, \u02c6Ht\u2212J = \u2212t mn \u02c6Psa\u2020 m\u03c3an\u03c3 \u02c6Ps + J mn \u02c6Sm \u00b7 \u02c6Sn. However, the passage of vacancies is frustrated by the antiferromagnetic spin correlations of the background. Here transport depends sensitively on the competition between the exchange energy of the spins and the kinetic energy of the holes. Oddly, at J = 0 (i.e. U = \u221e), the ground state spin con\ufb01guration is known to be driven ferromagnetic by", " a single hole while, for J > 0, it is generally accepted that a critical concentration of holes is required to destabilize antiferromagnetic order. EXERCISE At U = \u221e, all 2N states of the half-\ufb01lled Hubbard model are degenerate \u2013 each site is occupied by a single electron of arbitrary spin. This degeneracy is lifted by the removal of a single electron from the lower Hubbard band. In such a case, there is a theorem due to 66 Second quantization Nagaoka15 that, on a bipartite lattice (i.e. one in which the neighbors of one sublattice belong to the other sublattice), the ground state is ferromagnetic. For a four-site \u201cplaquette\u201d with three electrons determine the eigenspectrum of the Hubbard system with U = \u221e within the manifold (a) Sz total = 1/2. In each case, determine the total spin of the ground state. (Hint: In (b) there are a total of 12 basis states \u2013 here it useful to arrange these states in the order in which they are generated by application of the Hamiltonian.) total = 3/2, and (b) Sz INFO The rich behavior of the Mott\u2013Hubbard system is nowhere more exempli\ufb01ed than in the ceramic cuprate system \u2013 the basic material class of the high-temperature superconductors. Cuprates are built of layers of CuO2 separated by heavy rare earth ions such as lanthanum. Here, the copper ions adopt a square lattice con\ufb01guration separated by oxygen ions. At half-\ufb01lling, electrons in the outermost occupied shell of the copper sites in the plane adopt a partially \ufb01lled 3 d9 con\ufb01guration, while the oxygen sites are completely \ufb01lled. Elevated in energy by a frozen lattice distortion, the Fermi energy lies in the dx2\u2212y2 orbital of copper. According to a simple band picture, the single band is exactly half-\ufb01lled (one electron per Cu site) and, therefore, according to the standard band picture, should be metallic. However, strong electron interaction drives the cuprate system into an insulating antiferromagnetic Mott\u2013Hubbard phase. When doped away from half-\ufb01lling (by, for example, replacing the rare earth atoms by others with a di\ufb00erent stoichiometry; see the \ufb01gure, where the phase diagram of La2\u2212xSrxCuO4 is shown as a function of the concentration x of Sr atoms replacing La atoms and temperature), charge carriers are introduced into the \u201clower Hubbard band.\u201d However, in this case, the collapse of the Hubbard gap and loss of antiferromagnetic (AF) order is accompanied by the development of a high-temperature unconventional superconducting (SC) phase whose mechanism is believed to be rooted in the exchange of antiferromagnetic spin \ufb02uctuations. Whether the rich phenomenology of the cuprate system is captured by the Hubbard model remains a subject of great interest and speculation. T 300 200 100 AF SC 0.1 0.2 0.3 x This concludes our preliminary survey of the rich phenomenology of the interacting electron system. Notice that, so far, we have merely discussed ways to distill a reduced model from the original microscopic many-body Hamiltonian (2.17). However, save for the two examples of free \ufb01eld theories analyzed in Chapter 1, we have not yet learned how methods of second quantization can be applied to actually solve problems. To this end, in the following section we will illustrate the application of the method on a prominent strongly interacting problem. 15 Y. Nagaoka, Ferromagnetism in a narrow, almost half-\ufb01lled s-band, Phys. Rev. 147 (1966), 392\u2013405. 2.2 Applications of second quantization 67 Interacting fermions in one dimension \u03bc\u03bd a\u2020 Within the context of many-body physics, a theory is termed free if the Hamiltonian is bilinear in creation and annihilation operators, i.e. \u02c6H \u223c \u03bcH\u03bc\u03bda\u03bd, where H may be a \ufb01nite- or in\ufb01nite-dimensional matrix.16 Such models are \u201csolvable\u201d in the sense that the solution of the problem simply amounts to a diagonalization of the matrix H\u03bc\u03bd (subject to the preservation of the commutation relations of the operators a and a\u2020). However, only a few models of interest belong to this category. In general, interaction contributions typically quartic in the \ufb01eld operators are present and complete analytical solutions are out of reach. Yet there are a few precious examples of genuinely interacting systems that are amenable to (nearly", ") exact solution. In this section we will address an important representative of this class, namely the one-dimensional interacting electron gas. Not only is the analysis of this system physically interesting but, in addition, it provides an opportunity to practice working with the second quantized operator formalism on a deeper level. Qualitative discussion Consider the nearly free electron Hamiltonian (2.18) and (2.19) reduced to a one-dimensional environment. Absorbing the chemical potential EF into the de\ufb01nition of the Hamiltonian, and neglecting spin degrees of freedom (e.g. one might consider a fully spin polarized band), \u02c6H = k a \u2020 k k2 2m \u2212 EF ak + 1 2L kk,q=0 V (q)a \u2020 k\u2212qa \u2020 k+qak ak. (2.33) INFO At \ufb01rst sight, the treatment of a one-dimensional electron system may seem an academic exercise. However, e\ufb00ective one-dimensional interacting fermion systems are realized in a surprisingly rich spectrum of materials. We have already met with carbon nanotubes above. A nanotube is surrounded by clouds of mobile electrons (see earlier discussion in section 2.2). With the latter, con\ufb01nement of the circumferential direction divides the system into a series of one-dimensional bands, each classi\ufb01ed by a sub-band index and a wavenumber k. At low temperatures, the Fermi surface typically intersects a single sub-band, allowing attention to be drawn to a strictly one-dimensional system. A similar mechanism renders certain organic molecules (such as the Bechgaard salt (TMTSF)2PF6, where TMTSF stands for the tetramethyl-tetraselenafulvalene) one-dimensional conductors. A third, solid state, realization is presented by arti\ufb01cial low-dimensional structures fabricated from semiconducting devices. Redistribution of electron charge at the interface of a GaAs/AlGaAs heterostructure results in the formation of a two-dimensional electron gas. By applying external gates, it is possible to fabricate quasi-one-dimensional semiconductor quantum wires in which electron motion in the transverse direction is impeded by a large potential gradient (Fig. 2.8 (a)). At su\ufb03ciently low Fermi energies, only the lowest eigenstate of the transverse Schr\u00a8odinger equation (the lowest \u201cquantum mode\u201d) is populated and one is left with a strictly one-dimensional electron system. There are other realizations, such as edge modes in quantum Hall systems, \u201cstripe phases\u201d in high-tempterature superconductors, or certain inorganic crystals, but we shall not discuss these here explicitly. 16 More generally, a free Hamiltonian may also contain contributions \u223c a\u03bca\u03bd and a\u2020 \u03bca\u2020 \u03bd. 68 Second quantization (a) (b) Figure 2.8 Di\ufb00erent realizations of one-dimensional electron systems. (a) Steep potential well (realizable in, e.g., gated two-dimensional electron systems). (b) (Approximately) cylindrical quantum system (carbon nanotubes, quasi-one-dimensional molecules, etc.). In both cases, the single-particle spectrum is subject to mechanisms of size quantization. This leads to the formation of \u201cminibands\u201d (indicated by shaded areas in the \ufb01gure), structureless in the transverse direction and extended in the longitudinal direction. The one-dimensional fermion system exhibits a number of features not shared by higherdimensional systems. The origin of these peculiarities can be easily understood from a simple qualitative picture. Consider an array of interacting fermions con\ufb01ned to a line. To optimize their energy the electrons can merely \u201cpush\u201d each other around, thereby creating density \ufb02uctuations. By contrast, in higher-dimensional systems, electrons are free to avoid contact by moving around each other. A slightly di\ufb00erent formulation of the same picture can be given in momentum space. The Fermi \u201csphere\u201d of the one-dimensional system is de\ufb01ned through the interval [\u2212kF, kF] of \ufb01lled momentum states. The Fermi \u201csurface\u201d consists of two isolated points, {kF, \u2212kF} (see the \ufb01gure below). By contrast, higher-dimensional systems typically exhibit continuous and simply connected Fermi surfaces. It takes little imagination to anticipate that an extended Fermi sphere provides more phase space to two-particle interaction processes than the two isolated Fermi energy sectors of the onedimensional system. The one-dimensional electron system represents a rare", " exception of an interacting system that can be solved under no more than a few, physically weak, simplifying assumptions. This makes it a precious test system on which non-perturbative quantum manifestations of many-body interactions can be explored. Quantitative analysis We now proceed to develop a quantitative picture of the charge density excitations of the one-dimensional electron system. Anticipating that, at low temperatures, the relevant dynamics will take place in the vicinity of the two Fermi points {kF, \u2212kF}, the Hamiltonian (2.33) can be reduced further to an e\ufb00ective model describing the propagation of left and right moving excitations. To this end, we \ufb01rst introduce the notation that the subscripts \u2020 (+/\u2212)kF+q creates an electron that moves to the right/left R/L indicate that an operator a with velocity vF \u2261 kF/m. We next observe (see the \ufb01gure below) that, in the immediate vicinity of the Fermi points, the dispersion relation is approximately linear, implying that the non-interacting part of 2.2 Applications of second quantization 69 the Hamiltonian can be represented as (exercise) \u02c6H0 s=R,L q a\u2020 sq\u03c3svFq asq, (2.34) where \u03c3s = (+/\u2212) for s = R/L and the summation over q is restricted by some momentum E EF q F \u03c5 q cut\u2013o\ufb00 |q| < \u0393 beyond which the linearization of the dispersion is invalid. (Throughout this section, all momentum summations will be subject to this constraint.) Turning to the interacting part of the Hamiltonian, let us \ufb01rst de\ufb01ne the operator \u02c6\u03c1sq = a k \u2020 sk+qask. (2.35) k kF \u2013kF Crucially, the de\ufb01nition of these operators is not just motivated by notational convenience. It is straightforward to verify (exercise) that \u02c6\u03c1s(q) is obtained from the Fourier transform of the local density operator \u02c6\u03c1(x). In other words, \u02c6\u03c1sq measures density \ufb02uctuations of characteristic wavelength q\u22121 supported by electron excitations with characteristic momentum \u00b1kF (see Fig. 2.9 (a)). From our heuristic argument above, suggesting charge density modulations to be the basic excitations of the system, we expect the operators \u02c6\u03c1sq to represent the central degrees of freedom of the theory. Represented in terms of the density operators, the interaction contribution to the Hamil- tonian may be recast as \u02c6Vee = 1 2L kkq Vee(q) a \u2020 k\u2212qa \u2020 k+qak ak \u2261 1 2L qs [g4 \u02c6\u03c1sq \u02c6\u03c1s\u2212q + g2 \u02c6\u03c1sq \u02c6\u03c1\u00afs\u2212q], (2.36) where \u00afs = L/R denotes the complement of s = R/L, and the constants g2 and g4 measure the strength of the interaction in the vicinity of the Fermi points, i.e. where q 0 and q 2kF. (With the notation g2,4 we follow a common convention in the literature.) EXERCISE Explore the relation between the coupling constants g2, g4 and the Fourier transform kkq, not only terms with (k, k, q) (\u00b1kF, \u00b1kF, 0), but of Vee. Show that to the summation also terms with (k, k, q) (\u00b1kF, \u2213kF, 2kF) contribute. When adequately ordered (do it!), these contributions can be arranged into the form of the right-hand side of Eq. (2.36). (For a detailed discussion see, e.g., T. Giamarchi, Quantum Physics in One Dimension (Oxford University Press, 2004) or G. Mahan, Many Particle Physics (Plenum Press, 1981)). At any rate, the only point that matters for our present discussion is that the interaction can be represented through density operators with positive constants g2,4 determined by the interaction strength. INFO Working with second quantized theories, one frequently needs to compute commutators of operators \u02c6A(a, a\u2020) polynomial in the elementary boson/fermion operators of the theory (e.g. \u02c6A = aa\u2020, \u02c6A = aaa\u2020a\u2020, etc. where we have omitted the quantum number subscripts generally carried by a and a\u2020). Such types of operation are made easier by a number of operations of elementary 70 Second quantization \u03c5 F", " q\u20131 EF q (a) (b) Figure 2.9 Two di\ufb00erent interpretations of the excitations created by the density operators \u02c6\u03c1sq. (a) Real space; \u02c6\u03c1sq creates density modulations of characteristic wavelength q\u22121 and characteristic velocity vF. (b) Momentum space; application of \u02c6\u03c1sq to the ground state excites electrons from states k to k + q. This creates particle\u2013hole excitations of energy k+q \u2212 k = vFq independent of the particle/hole momentum k. Both particles and holes forming the excitation travel with the same velocity vF, implying that the excitation does not disperse (i.e. decay). commutator algebra. The most basic identity, from which all sorts of other formulae can be generated recursively, is the following: [ \u02c6A, \u02c6B \u02c6C]\u00b1 = [ \u02c6A, \u02c6B]\u00b1 \u02c6C \u2213 \u02c6B[ \u02c6A, \u02c6C]\u2212. Iteration of this equation for boson operators a, a\u2020 shows that [a\u2020, an] = \u2212nan\u22121. (2.37) (2.38) (Due to the fact that a2 = 0 in the fermionic case, there is no fermion analog of this equation.) Taylor expansion then shows that, for any analytic function F (a), [a\u2020, F (a)] = \u2212F (a). Similarly, another useful formula which follows from the above is the relation a\u2020F (aa\u2020) =F (a\u2020a)a\u2020, which is also veri\ufb01ed by series expansion. So far, we have merely rewritten parts of the Hamiltonian in terms of density operators. Ultimately, however, we wish to arrive at a representation whereby these operators, instead of the original electron operators, represent the fundamental degrees of freedom of the theory. Since the de\ufb01nition of the operators \u03c1 involves the squares of two Fermi operators, we expect the density operators to resemble bosonic excitations. Thus, as a \ufb01rst and essential step towards the construction of the new picture, we explore the commutation relations between the operators \u02c6\u03c1sq. From the de\ufb01nition (2.35) and the auxiliary identity (2.37) it is straightforward to verify \u2020 the commutation relation [\u02c6\u03c1sq, \u02c6\u03c1sq ] = \u03b4ss sk+q+q ask). As it stands, this k(a relation is certainly not of much practical use. To make further progress, we must resort to a (not very restrictive) approximation. Ultimately we will want to compute some observables involving quantum averages taken on the ground state of the theory, \u03a9|... |\u03a9. To simplify the structure of the theory, we may thus replace the right-hand side of the relation by its ground state expectation value: sk+qask\u2212q \u2212a \u2020 [\u02c6\u03c1sq, \u02c6\u03c1sq ] \u2248 \u03b4ss \u03a9|a \u2020 sk+qask\u2212q \u2212 a \u2020 sk+q+q ask|\u03a9 = \u03b4ss \u03b4q,\u2212q \u03a9|(\u02c6nsk+q \u2212 \u02c6nsk)|\u03a9, k k skask |\u03a9 = \u03b4kk. where, as usual, \u02c6nsk = a Although this is an uncontrolled approximation, it is expected to become better the closer we stay to the zero-temperature ground state |\u03a9 of the theory (i.e. at low excitation energies). skask, and we have made use of the fact that \u03a9|a \u2020 \u2020 2.2 Applications of second quantization 71 EXERCISE Try to critically assess the validity of the approximation. (For a comprehensive discussion, see the text by Giamarchi.17) At \ufb01rst glance, it would seem that the right-hand side of our simpli\ufb01ed commutator relation \u03a9|\u02c6nsk|\u03a9 actually vanishes. A simple shift of the summation index, indicates that the two terms contributing to the sum cancel. However, this argument is certainly too naive. It ignores the fact that our summation is limited by a cut-o\ufb00 momentum \u0393. Since the shift k \u2192 k \u2212 q changes the cut-o\ufb00, the interpretation above is invalid. \u03a9|\u02c6nsk+q|\u03a9?= k k To obtain a more accurate result, let us consider the case s = R and q > 0. We know that, in the ground state, all states with momentum k < 0 are occupied while all states with k \u2265 0 are", " empty. This implies that \u239b \u239e \u03a9|(\u02c6nRk+q \u2212 \u02c6nRk)|\u03a9 = \u239d k + + \u2212q<k\u22640 0<k<\u0393 \u23a0 \u03a9|(\u02c6nRk+q \u2212 \u02c6nRk)|\u03a9 \u2212\u0393<k\u2264\u2212q = \u2212q\u2264k\u22640 \u03a9|(\u02c6nRk+q \u2212 \u02c6nRk)|\u03a9 = \u2212 qL 2\u03c0, where, with the last equality, we have used the fact that a momentum interval of size q contains q/(2\u03c0/L) quantized momentum states. Similar reasoning for s = L shows that the e\ufb00ective form of the commutator relation reads [\u02c6\u03c1sq, \u02c6\u03c1sq ] = \u2212\u03b4ss \u03b4q,\u2212q \u03c3s qL 2\u03c0. (2.39) Now, if it were not for the q-dependence of the right-hand side of this relation, we would indeed have found (approximate) bosonic commutation relations. Therefore, to make the connection to bosons explicit, let us de\ufb01ne bq \u2261 nq \u02c6\u03c1Lq, b\u2020 q \u2261 nq \u02c6\u03c1L(\u2212q), b\u2212q \u2261 nq \u02c6\u03c1R(\u2212q), \u2020 b \u2212q \u2261 nq \u02c6\u03c1Rq, \u23ab \u23aa\u23aa\u23ac \u23aa\u23aa\u23ad (2.40) where q > 0 andn q \u2261 (2\u03c0/Lq)1/2. It is easily con\ufb01rmed that the newly de\ufb01ned operators bq obey canonical commutation relations (exercise), i.e. we have indeed found that, apart from the scaling factors nq, the density excitations of the system behave as bosonic \u201cparticles.\u201d Expressed in terms of the operators b, the interaction part of the Hamiltonian takes the form (exercise) Vee = 1 2\u03c0 q>0 q (bq \u2020 \u2212q) b g4 g2 g2 g4 b\u2020 q b\u2212q. Notice that we have succeeded in representing a genuine two-body interaction, a contribution that usually renders a model unsolvable, in terms of a quadratic representation. However, 17 T. Giamarchi, Quantum Physics in One Dimension (Oxford University Press, 2004). 72 Second quantization the free boson representation of the interaction term will be of little use until the kinetic part of the Hamiltonian \u02c6H0 is represented in terms of the b operators. There are various ways of achieving this goal. The most straightforward route, a direct construction of a representation of \u02c6H0 in terms of the Bose operators, is cumbersome in practice. However, there exists a more e\ufb03cient way that is based on indirect reasoning. As follows from the discussion of Section 2.1, the properties of second quantized operators are \ufb01xed by their commutation relations.18 So what we are going to do is search for an operator \u02c6H 0(b, b\u2020) that has the same commutation relations with the boson operators (b, b\u2020) as the original kinetic energy operator \u02c6H0(a, a\u2020). Using Eq. (2.34), the de\ufb01nition (2.35), and the auxiliary identity (2.37), it is straightforward to verify that [ \u02c6H0, \u02c6\u03c1sq] = qvF\u03c3s \u02c6\u03c1sq. On the other hand, using Eq. (2.39) one \ufb01nds that the same commutation relations hold with the operator \u02c6H 0 = 2\u03c0vF L qs \u02c6\u03c1sq \u02c6\u03c1s\u2212q, 0, \u02c6\u03c1sq] = qvF\u03c3s \u02c6\u03c1sq. Following the logic of our argument we thus identify \u02c6H0 = \u02c6H i.e. [ \u02c6H to inessential constants) and substitute \u02c6H 0 for the non-interacting Hamiltonian. 0 (up EXERCISE To gain some con\ufb01dence in the identi\ufb01cation \u02c6H0 = \u02c6H 0 +const., and to show that the undetermined constant actually equals zero, compute the energy expectation value of the state |\u03a8sq \u2261 \u02c6\u03c1sq|\u03a9 both as \u03a8sq| \u02c6H0|\u03a8sq and as \u03a8sq| \u02c6H 0|\u03a8sq.", " Con\ufb01rm that the two expressions coincide. Finally, using Eq. (2.40) and adding the interaction contribution Vee we arrive at the e\ufb00ective Hamiltonian \u02c6H = q>0 q (bq \u2020 \u2212q) b vF + g4 2\u03c0 g2 2\u03c0 g2 2\u03c0 vF + g4 2\u03c0 b\u2020 q b\u2212q. (2.41) We have thus succeeded in mapping the full interacting problem onto a free bosonic theory. The mapping a \u2192 \u02c6\u03c1 \u2192 b is our \ufb01rst example of a technique known as bosonization. Such techniques play an important role in 2(= 1 space + 1 time)-dimensional \ufb01eld theory in general. More sophisticated bosonization schemes will be discussed in Sections 4.3 and 9.4.4. Conversely, it is sometimes useful to represent a boson problem in terms of fermions via fermionization. One may wonder why it is indeed possible to e\ufb00ortlessly represent the lowlying excitations of a gas of fermions in terms of bosons. Fermi\u2013Bose transmutability is indeed a peculiarity of one-dimensional quantum systems. Particles con\ufb01ned to a line cannot pass \u201caround\u201d each other. That means that the whole issue of sign factors arising from the interchange of particle coordinates does not arise, and much of the exclusion-type 18 This argument can be made quantitative by group theoretical reasoning: Eq. (2.4) and (2.7) de\ufb01ne the irre] de\ufb01nes a product in the space of ducible representation of an operator algebra \u2013 an algebra because [, }, arepresentation because the operators act in a vector space (namely Fock space F ), which generators {a\u03bb, a is irreducible because all states |\u03bb1,..., \u03bbN \u2208 F can be reached by iterative application of operators onto a unique reference state (e.g. |\u03a9). Under these conditions, Schur\u2019s lemma \u2013 to be discussed in more detail in Chapter 4 \u2013 states that two operators \u02c6A1 and \u02c6A2 having identical commutation relations with all {a\u03bb, a } are equal up to a constant. \u2020 \u03bb \u2020 \u03bb 2.2 Applications of second quantization 73 characteristics of the Fermi system are inactivated. A more systematic formulation of Fermi \u2194 Bose transformations will be discussed in Chapter 4. \u2020 \u2020 0 1 \u03bb b q b\u2020 \u2212qb\u2020 is given by \u02c6N = \u03bbb\u03bb. Now, if the Hamiltonian has the form \u02c6H = Now, there is one last problem that needs to be overcome to actually solve the interacting qbq as problem. In Chapter 1, we learned how to interpret Hamiltonians of the structure superpositions of harmonic oscillators. However, in our present problem, terms of the type bqb\u2212q and b q appear. To return to familiar terrain, we need to eliminate these terms. However, before doing so, it is instructive to discuss the physical meaning of the problem. Firstly, let us recall that the total number operator of a theory described by operators \u2020 \u03bcH\u03bc\u03bdb\u03bd, \u03bb, b\u03bb b the total number operator commutes with \u02c6H, i.e. [ \u02c6N, \u02c6H] = 0 (exercise). This means that \u02c6H and \u02c6N can be simultaneously diagonalized, or, in more physical terms, that the Hamiltonian enjoys the feature of particle number conservation. More generally, any Hamiltonian in which operators appear as polynomials containing equal numbers of creation and annihilation operators (e.g. b\u2020b\u2020bb, b\u2020b\u2020b\u2020bbb, etc.) has this property. This is because any operator of this structure creates as many particles as it annihilates. In problems where the total number of particles is conserved (e.g. the theory of interacting electrons in an isolated piece of metal), the Hamiltonian is bound to have this structure. Conversely, in situations where the number of excitations is not \ufb01xed (e.g. a theory of photons or phonons) particle number violating terms like bb or b\u2020b\u2020 can appear. Such a situation is realized in our present problem; the number of density excitations in an electron system is certainly not a conserved quantity which explains why contributions like bqb\u2212q appear in \u02c6H. \u03bc\u03bd b\u2020 To eliminate the non-particle-number-conserving contributions we should, somehow, transform the matrix K \u2261 vF + g4 2\u03c0 g2 2\u03c0 g2 2\u03c0", " vF + g4 2\u03c0, q to a di\ufb00erent representation. Speci\ufb01cally, with \u03a8q \u2261 (b\u2020 to a diagonal structure. Transformations of K can be generated by transforming the operators bq and b\u2020 q, b\u2212q)T, we may de\ufb01ne \u03a8 \u2261 T \u22121\u03a8q, whereT is a 2 \u00d7 2 matrix acting on the two components of \u03a8. (Since K does q not depend on q, T can be chosen to have the same property.) After the transformation, the Hamiltonian will have the form H = q \u03a8\u2020 qK\u03a8q \u2192 q>0 q>0 q \u03a8\u2020 q T \u2020KT! K \u03a8 q, (2.42) \u2261 \u2212q T q, b b\u2020 with a new matrix K \u2261 T \u2020KT. We will seek for a transformation T that makes K diagonal. However, an important point to be kept in mind is that not all 2 \u00d7 2 matrices T qualify as transformations. We must ensure that the transformed \u201cvector\u201d again has the structure \u03a8, with a boson creation/annihilation operator in the \ufb01rst/second q component \u2013 i.e. the commutation relations of the operators must be conserved by the transformation. Remembering that the algebraic properties of the operators b are speci\ufb01ed through commutation relations, this condition can be cast in mathematical form by requiring that the commutator be invariant under the transformation. Using the fact that \u03a8 = T \u22121\u03a8, this condition is seen to be equivalent to the pseudo-unitarity condition, T \u2020\u03c33T!= \u03c33. = (\u2212\u03c33)ij \u03a8qi, \u03a8 qi, \u03a8 \u03a8 \u2020 qj \u2020 qj!= & & % % 74 Second quantization With this background, we are now in a position to \ufb01nd a transformation that brings the matrix K to a 2 \u00d7 2 diagonal form. Multiplication of the de\ufb01nition K = T \u2020KT by \u03c33 leads to T \u2020KT = K \u21d4 \u03c33T \u2020\u03c33! T \u22121 This means that the matrix \u03c33K is obtained by a similarity transformation T \u22121(\u00b7 \u00b7 \u00b7) T from the matrix \u03c33K, or, in other words, that the matrix \u03c33K contains the eigenvalues \u00b1u of \u03c33K on its diagonal. (That the eigenvalues sum to 0 follows from the fact that the trace vanishes, tr(\u03c33K) = 0.) However, the eigenvalues of \u03c33K are readily computed as \u03c33KT = \u03c33K. v\u03c1 = % 1 2\u03c0 (2\u03c0vF + g4)2 \u2212 g2 2 &1/2. (2.43) Thus, with \u03c33K = \u03c33v\u03c1 we arrive at K = v\u03c1 \u00b7 id., where \u201cid.\u201d stands for the unit matrix.19 Substitution of this result into Eq. (2.42) \ufb01nally leads to the diagonal Hamiltonian \u02c6H = v\u03c1 q, or equivalently, making use of the identity \u03a8\u2020 \u2020 \u2212qb\u2212q + 1, q = b\u2020 qbq + b q>0 q \u03a8\u2020 q \u03a8 q \u03a8 \u02c6H = v\u03c1 |q|b\u2020 qbq. (2.44) q Nicolai Nikolaevich Bogoliubov 1909\u201392 A theoretical physicist acclaimed for his works in nonlinear mechanics, statistical physics, theory of super\ufb02uidity and superconductivity, quantum \ufb01eld theory, renormalization group theory, proof of dispersion relations, and elementary particle theory. Here we have ignored an overall constant and omitted the prime on our new Bose operators. In the literature, the transformation procedure outlined above is known as a Bogoliubov transformation. Transformations of this type are frequently applied in quantum magnetism (see below), superconductivity, or, more generally, all problems where the particle number is not conserved. Notice that the possibility to transform to a representation \u223c b\u2020b does not imply that miraculously the theory has become particle number conserving. The new \u201cquasi-particle\u201d operators b are related to the original Bose operators through a transformation that mixes b and b\u2020. While the quasi-particle number is conserved, the number of original density excitations is not. Equations (2.43) and (2.44) represent our \ufb01nal solution of the problem of spinless interacting fermions in one dimension. We have suceeded in mapping the problem onto a form analogous to our", " previous results (1.34) and (1.39) for the phonon and the photon system, respectively. Indeed, all that has been said about those Hamiltonians applies equally to Eq. (2.44): the basic elementary excitations of the one-dimensional fermion system are waves, i.e. excitations with linear dispersion \u03c9 = v\u03c1|q|. In the present context, they are 19 Explicit knowledge of the transformation matrix T, i.e. knowledge of the relation between the operators b and b, is not needed for our construction. However, for the sake of completeness, we mention that T = cosh \u03b8k sinh \u03b8k sinh \u03b8k cosh \u03b8k with tanh(2\u03b8) =\u2212 g2/(2\u03c0vF + g4) represents a suitable parameterization. 2.2 Applications of second quantization 75 termed charge density waves (CDW). The Bose creation operators describing these excitations are, up to the Bogoliubov transformation, and a momentum dependent scaling factor (2\u03c0/Lq)1/2, equivalent to the density operators of the electron gas. For a non-interacting system, g2 = g4 = 0, and the CDW propagates with the velocity of the free Fermi particles, vF. A \ufb01ctitious interaction that does not couple particles of opposite Fermi momentum, g2 = 0, g4 = 0, speeds up the CDW. Heuristically, this can be interpreted as an \u201cacceleration process\u201d whereby a CDW pushes its own charge front. By contrast, interactions between left and right movers, g2 = 0, diminish the velocity, i.e. due to the Coulomb interaction it is di\ufb03cult for distortions of opposite velocities to penetrate each other. (Notice that, for a theory with g2 = 0, no Bogoliubov transformation would be needed to diagonalize the Hamiltonian, i.e. in this case, undisturbed left- and right-moving waves would be the basic excitations of the theory.) Our discussion above neglected the spin carried by the conduction electrons. Had we included the electron spin, the following picture would have emerged (see Problem 2.4): the long-range dynamics of the electron gas is governed by two independently propagating wave modes, the charge density wave discussed above, and a spin density wave (SDW).20 The SDW carries a spin current, but is electrically neutral. As with the CDW, its dispersion is linear with an interaction-renormalized velocity, vs (which, however, is generally larger than the velocity v\u03c1 of the CDW). To understand the consequences of this phenomenon, imagine an electron had been thrown into the system (e.g. by attaching a tunnel contact somewhere along the wire). As discussed above, a single electron does not represent a stable excitation of the one-dimensional electron gas. What will happen is that the spectral weight of the particle21 disintegrates into a collective charge excitation and a spin excitation. The newly excited waves then propagate into the bulk of the system at di\ufb00erent velocities \u00b1v\u03c1 and \u00b1vs. In other words, the charge and the spin of the electron e\ufb00ectively \u201cdisintegrate\u201d into two separate excitations, a phenomenon known as spin\u2013charge separation. Spin\u2013 charge separation in one-dimensional metals exempli\ufb01es a mechanism frequently observed in condensed matter systems: the set of quantum numbers carried by elementary particles may get e\ufb00ectively absorbed by di\ufb00erent excitation channels. One of the more spectacular manifestations of this e\ufb00ect is the appearance of fractionally charged excitations in quantum Hall systems, to be discussed in more detail in Chapter 9. The theory of spin and charge density waves in one-dimensional conductors has a long history spanning four decades. However, despite the rigor of the theory its experimental veri\ufb01cation has proved excruciatingly di\ufb03cult! While various experiments are consistent with theory (for a review, see Ref.17), only recently have signatures of spin and charge density wave excitations been experimentally observed. 20 One may think of the charge density of the electron gas \u03c1 = \u03c1\u2191 + \u03c1\u2193 as the sum of the densities of the spin up and spin down populations, respectively. The local spin density is then given by \u03c1s \u2261 \u03c1\u2191 \u2212 \u03c1\u2193. After what has been said above, it is perhaps not too surprising that \ufb02uctuations of these two quantities represent the dominant excitations of the electron gas. What is surprising, though, is that these two excitations do not interact with each other. 21 For a precise de\ufb01nition of this term, see Chapter 7", ". 76 Second quantization Quantum spin chains In the previous section, the emphasis was placed on charging e\ufb00ects generated by Coulomb interaction. However, as we have seen in Section 2.2, Coulomb interaction may also lead to the indirect generation of magnetic interactions. In one dimension, one can account for these mechanisms by adding to our previously structureless electrons a spin degree of freedom. This leads to the Tomonaga\u2013Luttinger liquid, a system governed by the coexistence of collective spin and charge excitations. However, to introduce the phenomena brought about example, in Mott by quantum magnetic correlations, it is best to \ufb01rst consider systems where the charge degrees of freedom are frozen and only spin excitations remain. Such systems are realized, for insulators where interaction between the spins of localized electrons is mediated by virtual exchange processes between neighboring can describe these correlations through models of localized quantum spins \u2013 either in chains or, more generally, in higher-dimensional quantum spin lattices. We begin our discussion with the ferromagnetic spin chain. Werner Heisenberg 1901\u201376 in Nobel Laureate in Physics 1932 \u201cfor the creation of quantum mechanics, the application of which has, inter alia, led to the discovery of the allotropic forms of hydrogen.\u201d In 1927 he published his uncertainty principle, for which he is perhaps best known. He also made important contributions to the theories of hydrodynamics of turbulent \ufb02ows, ferromagnetism, cosmic rays, and subatomic particles, and he was instrumental in planning the \ufb01rst West German nuclear reactor at Karlsruhe, together with a research reactor in Munich, in 1957. (Image c The Nobel Foundation.) electrons. One The quantum Heisenberg ferromagnet is speci\ufb01ed by the Hamiltonian Quantum ferromagnet \u02c6H = \u2212J mn \u02c6Sm \u00b7 \u02c6Sn, (2.45) where J > 0, \u02c6Sm represents the quantum mechanical spin operator at lattice site m, and, as before, mn denotes summation over neighboring sites. In Section 2.1 (see Eq. (2.13)) the quantum mechanical spin was represented through an electron basis. However, one can conceive of situations where the spin sitting at site m is carried by a di\ufb00erent object (e.g. an atom with non-vanishing magnetic moment). At any rate, for the purposes of our present discussion, we need not specify the microscopic origin of the spin. All we need to know is (i) that the lattice operators \u02c6Si m obey the SU(2) commutator algebra, = i\u03b4mnijk \u02c6Sk n, % \u02c6Si m, \u02c6Sj & n (2.46) characteristic of quantum spins, and (ii) the total spin at each lattice site is S.22 22 Remember that the \ufb01nite-dimensional representations of the spin operator are of dimension 2S + 1 where S may be integer or half integer. While a single electron has spin S = 1/2, the total magnetic moment of electrons bound to an atom may be much larger. 2.2 Applications of second quantization 77 Figure 2.10 Showing the spin con\ufb01guration of an elementary spin-wave excitation from the spin polarized ground state. Now, due to the positivity of the coupling constant J, the Hamiltonian favors con\ufb01gurations where the spins at neighboring sites are aligned in the same direction (cf. Fig. 2.10). A 2 |Sm, where |Sm represents a state with ground state of the system is given by |\u03a9 \u2261 |Sm = S|Sm. We have written \u201ca\u201d ground state instead of maximal spin-z component: Sz m \u201cthe\u201d ground state because the system is highly degenerate: a simultaneous change of the orientation of all spins does not change the ground state energy, i.e. the system possesses a global rotation symmetry. m EXERCISE Compute the energy expectation value of the state |\u03a9. De\ufb01ning global spin operators m, consider the state |\u03b1 \u2261 exp(i\u03b1\u00b7\u02c6S)|\u03a9. Verify that the state \u03b1 is degenerate through \u02c6Si \u2261 \u02c6Si with |\u03a9. Explicitly compute the state |(\u03c0/2, 0, 0). Convince yourself that, for general \u03b1, |\u03b1 can be interpreted as a state with rotated quantization axis. m As with our previous examples, we expect that a global continuous symmetry will involve the presence of energetically low-lying excitations. Indeed, it is obvious that, in the limit of long wavelength \u03bb, a weak distortion of a ground state con\ufb01guration (see Fig. 2.10) will cost", " vanishingly small energy. To quantitatively explore the physics of these spin waves, we adopt a \u201csemiclassical\u201d picture, where the spin S 1 is assumed to be large. In this limit, the rotation of the spins around the ground state con\ufb01guration becomes similar to the rotation of a classical magnetic moment. INFO To better understand the mechanism behind the semi-classical approximation, consider the Heisenberg uncertainty relation, \u0394Si \u0394Sj \u223c |[ \u02c6Si, \u02c6Sj]| = ijk| \u02c6Sk|, where \u0394Si is the root mean square of the quantum uncertainty of spin component i. Using the fact that | \u02c6Sk| \u2264 S, we obtain for the relative uncertainty, \u0394Si/S, I.e., for S 1, quantum \ufb02uctuations of the spin become less important. \u0394Si S \u0394Sj S \u223c S S2 S1\u2212\u2192 0. In the limit of large spin S, and at low excitation energies, it is natural to describe the ordered phase in terms of small \ufb02uctuations of the spins around their expectation values (cf. the description of the ordered phase of a crystal in terms of small \ufb02uctuations of the atoms around the ordered lattice sites). These \ufb02uctuations are conveniently represented in 78 Second quantization J2 (K) 0.130 0.125 0.120 0.115 0.110 0.105 0.600 0.610 J1 (K110\u232a \u2329100\u232a \u2329111.2 0.4 0.6 0.8 1.0 1.2 Wave number (\u00c5\u20131) Figure 2.11 Spin-wave spectrum of europium oxide as measured by inelastic neutron scattering at a reference temperature of 5.5 K. Note that, at low values of momenta q, the dispersion is quadratic, in agreement with the low-energy theory. (Exercise: A closer inspection of the data shows the existence of a small gap in the spectrum at q = 0. To what may this gap be attributed?) Figure reprinted with permission from L. Passell, O. W. Dietrich, and J. Als-Nielser, Neutron scattering from the Heisenberg ferromagnets EuO and EuS I: the exchange interaction, Phys. Rev. B 14 (1976), 4897\u2013907. Copyright (1976) by the American Physical Society. terms of spin raising and lowering operators: with \u02c6S\u00b1 m verify that \u2261 S1 m \u00b1 iS2 m, it is straightforward to % m, \u02c6S\u00b1 \u02c6Sz n & = \u00b1\u03b4mnS\u00b1 m, % m, \u02c6S\u2212 \u02c6S+ n & = 2\u03b4mnSz m. (2.47) \u2212(+) m Application of \u02c6S lowers (raises) the z-component of the spin at site m by one. To actually make use of the fact that deviations around |\u03a9 are small, a representation known as the Holstein\u2013Primako\ufb00 transformation23 was introduced in which the spin operators \u02c6S\u00b1, \u02c6S are speci\ufb01ed in terms of bosonic creation and annihilation operators a\u2020 and a: m = a\u2020 \u02c6S\u2212 m 2S \u2212 a\u2020 mam 1/2, \u02c6S+ m = 2S \u2212 a\u2020 mam 1/2 am, m = S \u2212 a\u2020 \u02c6Sz mam. EXERCISE Con\ufb01rm that the spin operators satisfy the commutation relations (2.47). The utility of this representation is clear. When the spin is large, S 1, an expansion in powers of 1/S gives \u02c6Sz (2S)1/2am. In this (2S)1/2a\u2020 m = S \u2212 a\u2020 m, and \u02c6S+ mam, \u02c6S\u2212 m m 23 T. Holstein and H. Primako\ufb00, Field dependence of the intrinsic domain magnetisation of a ferromagnet, Phys. Rev. 58 (1940), 1098\u2013113. 2.2 Applications of second quantization 79 approximation, the one-dimensional Heisenberg Hamiltonian takes the form \u02c6H = \u2212J ) \u02c6Sz m m = \u2212JN S2 \u2212 JS \u02c6Sz m+1 + 0 \u02c6S+ m 1 2 \u02c6S\u2212 m+1 + \u02c6S\u2212 m \u02c6S+ m+1 * \u22122a\u2020 mam + a\u2020 mam+1 + h.c. 1 + O(S0) = \u2212JN", " S2 + JS m m (a \u2020 m+1 \u2212 a\u2020 m)(am+1 \u2212 am) +O (S0). Keeping \ufb02uctuations at leading order in S, the quadratic Hamiltonian can be diagonalized by Fourier transformation. In this case, it is convenient to impose periodic boundary conditions: \u02c6Sm+N = \u02c6Sm, anda m+N = am, where N denotes the total number of lattice sites. De\ufb01ning ak = 1\u221a N N m=1 eikmam, am = 1\u221a N B.Z. k e\u2212ikmak, & % ak, a \u2020 k = \u03b4kk, where the summation over k runs over the Brillouin zone, the Hamiltonian for the onedimensional lattice system takes the form (exercise) \u02c6H = \u2212JN S2 + B.Z. k \u03c9ka \u2020 kak + O(S0). (2.48) Here \u03c9k = 2JS(1 \u2212 cos k) = 4JS sin2(k/2) represents the dispersion relation of the spin excitations. In particular, in the limit k \u2192 0, the energy of the elementary excitations vanishes, \u03c9k \u2192 JSk2. These massless low-energy excitations, known as magnons, describe the elementary spin-wave excitations of the ferromagnet. Taking into account terms at higher order in the parameter 1/S, one \ufb01nds interactions between the magnons. A comparison of these theoretical predictions and experiment is shown in Fig. 2.11. Having explored the elementary excitation spectrum of the ferromagnet, we now turn to the discussion of the spin S Heisenberg antiferromagnetic Hamiltonian Quantum antiferromagnet \u02c6H = J mn \u02c6Sm \u00b7 \u02c6Sn, where, once again, J > 0. As we have seen above, such antiferromagnetic systems occur 80 Second quantization (a) (b) Figure 2.12 (a) Example of a two-dimensional bipartite lattice. (b) Example of a non-bipartite lattice. Notice that, with the latter, no antiferromagnetic arrangement of the spins can be made that recovers the maximum exchange energy from each and every bond. from its Louis N\u00b4eel 1904\u20132000 Nobel Laureate in Physics in 1970, shared with Hannes Olof for his \u201cfundaG\u00a8osta Alfv\u00b4en, discoveries and mental work concerning antiferromagnetism and ferrimagnetism that have led to important applications in solid state physics.\u201d (Image c The Nobel Foundation.) in the arena of strongly correlated electron compounds. Although the Hamiltonian di\ufb00ers ferromagnetic relative \u201conly\u201d by a change of sign, the di\ufb00erences in the physics are drastic. Firstly, the phenomenology displayed by the \u02c6H antiferromagnetic Hamiltonian depend sensitively on the morphology of the underlying lattice. For a bipartite lattice, i.e. one in which the neighbors of one sublattice A belong to the other sublattice B (see Fig. 2.12(a)), the ground states of the Heisenberg antiferromagnet are close24 to a staggered spin con\ufb01guration, known as a N\u00b4eel state, where all neighboring spins are antiparallel (see Fig. 2.12). Again the ground state is degenerate, i.e. a global rotation of all spins by the same amount does not change the energy. By contrast, on nonbipartite lattices such as the triangular lattice shown in Fig. 2.12(b), no spin arrangement can be found wherein each bond recovers the full exchange energy J. Spin models of this kind are said to be frustrated. EXERCISE Engaging only symmetry considerations, try to identify a possible classical ground state of the triangular lattice Heisenberg antiferromagnet. (Hint: Construct the classical ground state of a three-site plaquette and then develop the periodic continuation.) Show that the classical antiferromagnetic ground state of the Kagom\u00b4e lattice \u2013 a periodic array of corner-sharing \u201cstars of David\u201d \u2013 has a continuous spin degeneracy generated by local spin rotations. How might the degeneracy a\ufb00ect the transition to an ordered phase? Returning to the one-dimensional system, we \ufb01rst note that a chain is trivially bipartite. As before, our strategy will be to expand the Hamiltonian in terms of bosonic operators. However, before doing so, it is convenient to apply a canonical transformation to the Hamiltonian in which the spins on one sublattice", ", say B, are rotated through 180\u25e6 about the 24 It is straightforward to verify that the classical ground state \u2013 the N\u00b4eel state \u2013 is now not an exact eigenstate of the quantum Hamiltonian. The true ground state exhibits zero-point \ufb02uctuations reminiscent of the quantum harmonic oscillator or atomic chain. However, when S 1, it serves as a useful reference state from which \ufb02uctuations can be examined. 2.2 Applications of second quantization 81 (a) (b) Figure 2.13 (a) N\u00b4eel state con\ufb01guration of the spin chain. (b) Cartoon of an antiferromagnetic spin wave. \u2192 3Sx B, Sy B = Sx x-axis, i.e. Sx B. That is, when repreB sented in terms of the new operators, the N\u00b4eel ground state looks like a ferromagnetic state, with all spins aligned. We expect that a gradual distortion of this state will produce the antiferromagnetic analog of the spin waves discussed in the previous section (see Fig. 2.11). B, and Sz B B = \u2212Sz B = \u2212Sy B \u2192 3Sy \u2192 3Sz Represented in terms of the transformed operators, the Hamiltonian takes the form \u02c6H = \u2212J m Sz m 3Sz m+1 \u2212 1 2 S+ m 3S+ m+1 + S\u2212 m 3S\u2212 m+1. Once again, applying an expansion of the Holstein\u2013Primako\ufb00 representation, S\u2212 m (2S)1/2a\u2020 m, etc., one obtains the Hamiltonian # \u02c6H = \u2212N JS2 + JS a\u2020 mam + a m+1am+1 + amam+1 + a\u2020 ma \u2020 $ \u2020 m+1 + O(S0). m At \ufb01rst sight the structure of this Hamiltonian, albeit quadratic in the Bose operators, looks awkward. However, after Fourier transformation, am = N \u22121/2 k e\u2212ikmak it assumes the more accessible form \u02c6H = \u2212N JS(S + 1) +JS k a \u2020 k a\u2212k 1 \u03b3k \u03b3k 1 ak \u2020 a \u2212k + O(S0), where \u03b3k = cos k. Apart from the de\ufb01nition of the matrix kernel between the Bose operators, \u02c6H is equivalent to the Hamiltonian (2.41) discussed in connection with the charge density wave. Performing the same steps as before, the non-particle-number-conserving con- 82 Second quantization 350 300 250 200 150 100 50 ) 3/4,1/4) (1/2,1/2) (1/2,0) (3/4,1/4) (1,0) (1/2,0) Figure 2.14 Experimentally obtained spin-wave dispersion of the high-Tc parent compound LaCuO4 \u2013 a prominent spin 1/2 antiferromagnet. Reprinted with permission from R. Coldea S. M. Hayder, G. Aeppli, et al., Spin waves and electronic excitations in La2CuO4, Phys. Rev. Lett. 86 (2001), 5377\u201380. Copyright (2001) by the American Physical Society. tributions a\u2020a\u2020 can be removed by Bogoliubov transformation. As a result, the transformed Hamiltonian assumes the diagonal form \u02c6H = \u2212N JS2 + 2JS | sin k| \u03b1 \u2020 k\u03b1k +. 1 2 k (2.49) Thus, in contrast to the ferromagnet, the spin-wave excitations of the antiferromagnet (Fig. 2.14) display a linear spectrum in the limit k \u2192 0. Surprisingly, although developed in the limit of large spin, experiment shows that even for S = 1/2 spin chains, the integrity of the linear dispersion is maintained (see Fig. 2.14). More generally, it turns out that, for chains of arbitrary half integer spin S = 1/2, 3/2, 5/2,..., the low-energy spectrum is linear, in agreement with the results of the harmonic approximation. In contrast, for chains of integer spin S = 1, 2, 3..., the lowenergy spectrum contains a gap, i.e. these systems do not support long-range excitations. As a rule, the sensitivity of a physical phenomenon to the characteristics of a sequence of numbers \u2013 such as half integer vs. integer \u2013 signals the presence of a mechanism of topological origin.25 At the same time, the formation of a gap (observed for integer chains) represents an interaction", " e\ufb00ect; at orders beyond the harmonic approximation, spin waves begin to interact nonlinearly with each other, a mechanism that may (S integer) but need not (S half integer) destroy the-wave like nature of low-energy excitations. In Section 9.3.3 \u2013 in a chapter devoted to a general discussion of the intriguing condensed matter phenom- 25 Speci\ufb01cally, the topological signature of a spin \ufb01eld con\ufb01guration will turn out to be the number of times the classical analog of a spin (a vector on the unit sphere) will wrap around the sphere in (1 + 1)-dimensional space time. 2.3 Summary and outlook 83 ena generated by the conspiracy of global (topological) structures with local interaction mechanisms \u2013 we will discuss these phenomena on a deeper level. 2.3 Summary and outlook This concludes our preliminary discussion of applications of the second quantization. Additional examples can be found in the problems. In this chapter, we have introduced second quantization as a tool whereby problems of many-body quantum mechanics can be addressed more e\ufb03ciently than by the traditional language of symmetrized many-body wave functions. We have discussed how the two approaches are related to each other and how the standard operations of quantum mechanics can be performed by second quantized methods. One may note that, beyond qualitative discussions, the list of concrete applications encountered in this chapter involved problems that either were non-interacting from the outset, or could be reduced to a quadratic operator structure by a number of suitable manipulations. However, we carefully avoided dealing with interacting problems where no such reductions are possible \u2013 the majority by far of the problems encountered in condensed matter physics. What can be done in situations where interactions, i.e. operator contributions of fourth or higher order, are present and no tricks like bosonization can be played? Generically, either interacting problems of many-body physics are fundamentally inaccessible to perturbation theory, or they necessitate perturbative analyses of in\ufb01nite order in the interaction contribution. Situations where a satisfactory result can be obtained by \ufb01rst- or second-order perturbation theory are exceptional. Within second quantization, large-order perturbative expansions in interaction operators lead to complex polynomials of creation and annihilation operators. Quantum expectation values taken over such structures can be computed by a reductive algorithm, known as Wick\u2019s theorem. However, from a modern perspective, the formulation of perturbation theory in this way is not very e\ufb03cient. More importantly, problems that are principally non-perturbative have emerged as the focus of interest. To understand the language of modern condensed matter physics, we thus need to develop another layer of theory, known as \ufb01eld integration. In essence, the latter is a concept generalizing the e\ufb00ective action approach of Chapter 1 to the quantum level. However, before discussing quantum \ufb01eld theory, we should understand how the concept works in principle, i.e. on the level of single particle quantum mechanics. This will be the subject of the next chapter. 2.4 Problems Stone\u2013von Neumann theorem In the text we introduced creation and annihilation operators in a constructive manner, i.e. by specifying their action on a \ufb01xed Fock space state. We saw that this de\ufb01nition implied remarkably simple algebraic relations between the newly introduced operators \u2013 the Heisenberg algebra (2.7). In this problem we explore the mathematical structure behind this observation. (The problem has been included for the 84 Second quantization bene\ufb01t of the more mathematically inclined. Readers primarily interested in the practical aspects of second quantization may safely skip it.) Let us de\ufb01ne an abstract algebra of objects a\u03bb and \u02dca\u03bb by [a\u03bb, \u02dca\u03bc]\u03b6 = \u03b4\u03bb\u03bc, [a\u03bb, a\u03bc]\u03b6 = [\u02dca\u03bb, \u02dca\u03bc]\u03b6 = 0. Further, let us assume that this algebra is unitarily represented in some vector space F. This means that (i) to every a\u03bb and \u02dca\u03bb we assign a linear map Ta\u03bb : F \u2192 Fsuch that (ii) T[a\u03bb,\u02dca\u03bc]\u03b6 = [Ta\u03bb, T\u02dca\u03bc ]\u03b6, and (iii) T\u02dca\u03bb = T \u2020. To keep the notation simple, we will denote a\u03bb Ta\u03bb by a\u03bb (now regarded as a linear map F \u2192 F) andT \u02dca\u03bb by a \u2020 \u03bb. The Stone\u2013von Neumann theorem states that the representation above is unique, i.e. that, up to unitary transformations of basis, there is only one such representation. The statement is proven by explicit construction of a basis", " on which the operators act in a speci\ufb01c and wellde\ufb01ned way. We will see that this action is given by Eq. (2.6), i.e. the reference basis is but the Fock space basis used in the text. This proves that the Heisenberg algebra encapsulates the full mathematical structure of the formalism of second quantization. (a) We begin by noting that the operators \u02c6n\u03bb \u2261 a \u2020 \u03bba\u03bb are Hermitian and commute with each other, i.e. they can be simultaneously diagonalized. Let |n\u03bb1, n\u03bb2,... be an orthonormalized eigenbasis of the operators {\u02c6n\u03bb}, i.e. \u02c6n\u03bbi |n\u03bb1, n\u03bb2,.... Show that, up to unit-modular factors, this basis is unique. (Hint: Use the irreducibility of the transformation.) |n\u03bb1, n\u03bb2,... = n\u03bbi (b) Show that a\u03bbi \u2212 1. Use this |n\u03bb1, n\u03bb2,... is an eigenstate of \u02c6n\u03bbi with eigenvalue n\u03bbi information to show that all eigenvalues n\u03bbi are positive integers. (Hint: note positivity of the scalar norm.) Show that the explicit representation of the basis is given by |n\u03bb1, n\u03bb2,... ='\u2020n\u03bbi a \u03bbi n\u03bbi! i |0, (2.50) where |0 is the unique state which has eigenvalue 0 for all \u02c6ni. Comparison with Eq. (2.4) shows that the basis constructed above indeed coincides with the Fock space basis considered in the text. Answer: (a) Suppose we had identi\ufb01ed two bases {|n\u03bb1, n\u03bb2,...} and {|n\u03bb1, n\u03bb2,...} on which all operators \u02c6ni assumed equal eigenvalues. The irreducibility of the representation implies the existence of some polynomial P ({a\u03bci, a\u2020 }) such that |n\u03bb1, n\u03bb2,... = \u03bci })|n\u03bb1, n\u03bb2,.... Now, the action of P must not change any of the eigenP ({a\u03bci, a\u2020 \u03bci values of \u02c6ni, which means that P contains the operators a\u03bc and a\u2020 \u03bc in equal numbers. }) = \u02dcP ({\u02c6n\u03bci Reordering operators, we may thus bring P into the form P ({a\u03bci, a\u2020 }). \u03bci However, the action of this latter expression on |n\u03bb1, n\u03bb2,... just produces a number, i.e. the bases are equivalent. 2.4 Problems 85 (b) For a given state |n, (concentrating on a \ufb01xed element of the single-particle basis, we suppress the subscript \u03bbi throughout), let us choose an integer q such that \u02c6naq\u22121|n = (n \u2212 q + 1)aq\u22121|n with n \u2212 q + 1 > 0 while n \u2212 q \u2264 0. We then obtain 0 \u2265 (n \u2212 q)n|a\u2020qaq|n = n|a\u2020q \u02c6naq|n = n|(a\u2020)q+1aq+1|n \u2265 0. The only way to satisfy this sequence of inequalities is to require that n|a\u2020q+1|aq+1|n = 0 andn \u2212 q = 0. The last equation implies the \u201cinteger-valuedness\u201d of n. (In principle, we ought to prove that a zero-eigenvalue state |0 exists. To show this, take any reference state |n\u03bb1, n\u03bb2,... and apply operators a\u03bbi as long as it takes to lower all eigenvalues n\u03bbi to zero.) Using the commutation relations, it is then straightforward to verify that the r.h.s. of Eq. (2.50) (a) is unit normalized and (b) has eigenvalue n\u03bbi for each \u02c6n\u03bbi. Semiclassical spin waves In Chapter 1, the development of a theory of lattice vibrations in the harmonic atom chain was motivated by the quantization of the continuum classical theory. The latter provided insight into the nature of the elementary collective excitations. Here we will employ the semiclassical theory of spin dynamics to explore the nature of elementary spin-wave excitations. i, \u02c6S\u03b2 j ] = i\u03b4ij\u03b1\u03b2", "\u03b3 \u02c6S\u03b3 (a) Making use of the spin commutation relation, [ \u02c6S\u03b1 i, apply the operator \u02d9\u02c6Si = [\u02c6Si, \u02c6H] to express the equation of motion of a spin in a nearest neighbor identity i spin-S one-dimensional Heisenberg ferromagnet as a di\ufb00erence equation (N.B. = 1). (b) Interpreting the spins as classical vectors, and taking the continuum limit, show that the equation of motion of the hydrodynamic modes takes the form \u02d9S = JS \u00d7 \u22022S where we have assumed a unit lattice spacing. (Hint: In taking the continuum limit, apply a Taylor expansion to the spins i.e. Si+1 = Si + \u2202Si + \u00b7 \u00b7 \u00b7.) Find and sketch a wave-like solution describing small angle precession around a globally magnetized state Si = Sez (i.e. a solution as shown in Fig. 2.10). Answer: (a) Making use of the equation of motion, and the commutation relation, substitution of the Heisenberg ferromagnetic Hamiltonian gives the di\ufb00erence equation \u02d9\u02c6Si = J \u02c6Si \u00d7 (\u02c6Si+1 + \u02c6Si\u22121). (b) Interpreting the spins as classical vectors, and applying the Taylor expansion Si+1 \u2192 S(x + 1) = S + \u2202S + \u22022S/2 + \u00b7 \u00b7 \u00b7, one obtains the classical equation of motion shown. Making the ansatz S = (c cos(kx \u2212 \u03c9t), c sin(kx \u2212 \u03c9t), S2 \u2212 c2) one may con\ufb01rm that the equation of motion is satis\ufb01ed if \u03c9 = Jk2. \u221a 86 Second quantization (a) (b) (c) Figure 2.15 (a) An sp2-hybridized polymer chain. (b) One of the con\ufb01gurations of the Peierls distorted chain. The double bonds represent the short links of the lattice. (c) A topological defect separating two domains of the ordered phase. Su\u2013Shrie\ufb00er\u2013Heeger model of a conducting polymer chain Polyacetylene consists of bonded CH groups forming an isomeric long-chain polymer. According to molecular orbital theory, the carbon atoms are expected to be sp2-hybridized suggesting a planar con\ufb01guration of the molecule. An unpaired electron is expected to occupy a single \u03c0-orbital which is oriented perpendicular to the plane. The weak overlap of the \u03c0-orbitals delocalizes the electrons into a narrow conduction band. According to the nearly free electron theory, one might expect the half-\ufb01lled conduction band of a polyacetylene chain to be metallic. However, the energy of a half-\ufb01lled band of a one-dimensional system can always be lowered by imposing a periodic lattice distortion known as a Peierls instability (see Fig. 2.15). The aim of this problem is to explore the instability. (a) At its simplest level, the conduction band of polyacetylene can be modeled by a simple (arguably over-simpli\ufb01ed) microscopic Hamiltonian, due to Su, Shrie\ufb00er, and Heeger,26 in which the hopping matrix elements of the electrons are modulated by the lattice distortion of the atoms. By taking the displacement of the atomic sites to be un, and treating their dynamics as classical, the e\ufb00ective Hamiltonian can be cast in the form \u02c6H = \u2212t N n=1 % (1 + un) c\u2020 n\u03c3cn+1\u03c3 + h.c. N & + n=1 ks 2 (un+1 \u2212 un)2, where, for simplicity, the boundary conditions are taken to be periodic, and summation over the spins \u03c3 is assumed. The \ufb01rst term describes the hopping of electrons between neighboring sites with a matrix element modulated by the periodic distortion of the bond-length, while the last term represents the associated increase in the elastic energy. Taking the lattice distortion to be periodic, un = (\u22121)n\u03b1, and the number of sites to be even, bring the Hamiltonian to diagonal form. (Hint: Note that the lattice distortion lowers the symmetry of the lattice. The Hamiltonian is most easily diagonalized by distinguishing the two sites of the sublattice \u2013 i.e. doubling the size of the elementary unit cell.) Show that the Peierls distortion of the lattice opens a gap in the spectrum at the Fermi level of the half-\ufb01lled system. (", "b) By estimating the total electronic and elastic energy of the half-\ufb01lled band (i.e. an average of one electron per lattice site), show that the one-dimensional system is always unstable towards the Peierls distortion. To complete this calculation, you will need 26 W. P. Su, J. R. Schrie\ufb00er, and A. J. Heeger, Solitons in polyacetylene, Phys. Rev. Lett. 42 (1979), 1698\u2013701. 2.4 Problems 87 1/2 the approximate formula for the (elliptic) integral, 2+(a1 \u2212b1 ln \u03b12)\u03b12 +O(\u03b12 ln \u03b12), where a1 and b1 are (unspeci\ufb01ed) numerical constants. (c) For an even number of sites, the Peierls instability has two degenerate con\ufb01gurations (see Fig. 2.15(a)) \u2013 ABABAB... and BABABA... Comment on the qualitative form of the ground state lattice con\ufb01guration if the number of sites is odd (see Fig. 2.15(b)). Explain why such con\ufb01gurations give rise to mid-gap states. \u03c0/2 \u2212\u03c0/2 dk 1 \u2212 \u03b12 sin2 k 1 \u2212 Answer: (a) Since each unit cell is of twice the dimension of the original lattice, we begin by recasting the Hamiltonian in a sublattice form \u02c6H = \u2212t N/2 0 % (1 + \u03b1) a\u2020 m\u03c3bm\u03c3 + h.c. & % + (1\u2212 \u03b1) b\u2020 m\u03c3am+1\u03c3 + h.c. &1 + 2N ks\u03b12, m=1,\u03c3 where the creation operators a\u2020 m act on the two sites of the elemental unit cell of k e2ikmak (similarly the distorted lattice. Switching to the Fourier basis, am = 2/N bm), where k takes N/2 values uniformly on the interval [\u2212\u03c0/2, \u03c0/2] and the lattice spacing of the undistorted system is taken to be unity, the Hamiltonian takes the form m and b\u2020. ak\u03c3 bk\u03c3 &1/2 \u02c6H = 2N ks\u03b12 \u2212 t \u2020 \u2020 k\u03c3, b k\u03c3 a k 0 (1 + \u03b1) + (1 \u2212 \u03b1)e\u22122ik (1 + \u03b1) + (1 \u2212 \u03b1)e2ik 0 % Diagonalizing the 2 \u00d7 2 matrix, one obtains (k) =\u00b1 2t. Reassuringly, in the limit \u03b1 \u2192 0, one recovers the cosine spectrum characteristic of the undistorted tight-binding problem while, in the limit \u03b1 \u2192 1, pairs of monomers become decoupled and we obtain a massively degenerate bonding and antibonding spectrum. (b) According to the formula given in the text, the total shift in energy is given by \u03b4 = \u22124t(a1 \u2212 b1 ln \u03b12)\u03b12 + 2ks\u03b12. Maximizing the energy gain with respect to \u03b1, one \ufb01nds that the stable con\ufb01guration is found when \u03b12 = exp[ a1 b1 1 + (\u03b12 \u2212 1) sin2 k \u2212 1 \u2212 ks\u03b1 2tb1 (c) If the number of sites is odd, the Peierls distortion is inevitably frustrated. The result is that the polymer chain must accommodate a topological excitation. The excitation is said to be topological because the defect cannot be removed by a smooth continuous deformation. Its e\ufb00ect on the spectrum of the model is to introduce a state that lies within the band gap of the material. The consideration of an odd number of sites forces a topological defect into the system. However, even if the number of sites is even, one can create low energy topological excitations of the system either by doping (see Fig. 2.15(b)), or by the creation of excitons, particle\u2013hole excitations of the system. Indeed, such topological excitations can dominate the transport properties of the system. ]. 88 Second quantization Schwinger boson representation As with the Holstein\u2013Primako\ufb00 representation, the Schwinger boson provides another representation of quantum mechanical spin. The aim of this problem is to con\ufb01rm the validity of the representation. For practical purposes, the value of the particular representation depends on its application. In the Schwinger boson representation, the quantum mechanical spin is expressed in terms of two bosonic operators a and b in the form \u02c6S+ = a\u2020b, \u02c6Sz = 1 2 \ufffd", "\ufffdS\u2212 = ( \u02c6S+)\u2020, a\u2020a \u2212 b\u2020b. Julian Schwinger 1918\u20131994 Nobel Laureate in Physics with Sin-Itiro Tomonaga and Richard P. Feynman, for their fundamental work in quantum electrodynamics, with far-reaching consequences for the physics of elementary particles. (Image c The Nobel Foundation.) (a) Show that this de\ufb01nition is consistent with the commutation relations for spin: [ \u02c6S+, \u02c6S\u2212] = 2 \u02c6Sz. (b) Using the bosonic commutation relations, show that |S, m = (a\u2020)S+m (S + m)! (b\u2020)S\u2212m (S \u2212 m)! |\u03a9, is compatible with the de\ufb01nition of an eigenstate of the total spin operator S2 and Sz. Here |\u03a9 denotes the vacuum of the Schwinger bosons, and the total spin S de\ufb01nes the physical subspace {|na, nb|na + nb = 2S}. Answer: (a) Using the commutation relation for bosons, one \ufb01nds [ \u02c6S+, \u02c6S\u2212] =a \u2020b b\u2020a \u2212 b\u2020a a\u2020b = a\u2020a \u2212 b\u2020b = 2 \u02c6Sz, as required. (b) Using the identity \u02c6S2 = ( \u02c6Sz)2 + 1 \ufb01nds that \u02c6S2|S, m = Similarly, one \ufb01nds \u02c6Sz|S, m = 1 |S, m to be an eigenstate of the operator \u02c6Sz with eigenvalue m. 2 (\u02c6na + \u02c6nb) one |S, m = S(S + 1)|S, m, as required. 2 (na \u2212 nb)|na = S + m, nb = S \u2212 m = m|S, m showing m2 + (S + m)(S \u2212 m) +S 4 (\u02c6na \u2212 \u02c6nb)2 + \u02c6na \u02c6nb + 1 2 ( \u02c6S+ \u02c6S\u2212 + \u02c6S\u2212 \u02c6S+) = 1 & % Jordan\u2013Wigner transformation So far we have shown how the algebra of quantum mechanical spin can be expressed using boson operators \u2013 cf. the Holstein\u2013Primako\ufb00 transformation and the Schwinger boson representation. In this problem we show that a representation for spin-(1/2) can be obtained in terms of Fermion operators. Let us represent an up spin as a particle and a down spin as the vacuum, i.e. |\u2191 \u2261 |1 = f \u2020|0, | \u2193 \u2261 |0 = f |1. In this representation the spin raising and lowering operators are expressed in the forms \u02c6S+ = f \u2020 and \u02c6S\u2212 = f, while \u02c6Sz = f \u2020f \u2212 1/2. (a) With this de\ufb01nition, con\ufb01rm that the spins obey the algebra [ \u02c6S+, \u02c6S\u2212] = 2 \u02c6Sz. 2.4 Problems 89 However, there is a problem: spins on di\ufb00erent sites commute while fermion operators \u2020 \u2020 \u2020 anticommute, e.g. S+ i. To obtain a faithful spin reprei f sentation, it is necessary to cancel this unwanted sign. Although a general procedure is hard to formulate, in one dimension this can be achieved by a nonlinear transformation, namely j = S+ i, but f j = \u2212f i S+ j S+ \u2020 j f \u02c6S+ l = f \u2020 l ei\u03c0 j<l \u02c6nj, l = e\u2212i\u03c0 \u02c6S\u2212 j<l \u02c6nj fl, \u02c6Sz l = f \u2020 l fl \u2212 1 2. Operationally, this seemingly complicated transformation is straightforward: in one dimension, the particles can be ordered on the line. By counting the number of particles \u201cto the left\u201d we can assign an overall sign of +1 or \u22121 to a given con\ufb01guration and thereby transmute the particles into fermions. (Put di\ufb00erently, the exchange of two fermions induces a sign change that is compensated by the factor arising from the phase \u2013 the \u201cJordan\u2013Wigner string.\u201d) (b) Using the Jordan\u2013Wigner representation, show that \u02c6S+ m (c) For the spin-(1/2) anisotropic", " quantum Heisenberg spin chain, the spin Hamiltonian assumes the form \u02c6H = \u2212 \u02c6Sz n+1 + J\u22a5. Turning to the 2 Jordan\u2013Wigner representation, show that the Hamiltonian can be cast in the form mfm+1. $ \u02c6S\u2212 n+1 + \u02c6S\u2212 \u02c6S\u2212 m+1 = f \u2020 \u02c6S+ n+1 Jz \u02c6Sz n \u02c6S+ n # n n \u02c6H = \u2212 n J\u22a5 2 f \u2020 nfn+1 + h.c. + Jz 1 4 \u2212 f \u2020 nfn + f \u2020 nfnf \u2020 n+1fn+1. (d) The mapping above shows that the one-dimensional quantum spin-(1/2) XY-model (i.e. Jz = 0) can be diagonalized as a non-interacting theory of spinless fermions. In this case, show that the spectrum assumes the form (k) = \u2212J\u22a5 cos ka. Answer: (a) Using the fermionic anti-commutation relations, one \ufb01nds [ \u02c6S+, \u02c6S\u2212]\u2212 = [f \u2020, f ]\u2212 = f \u2020f \u2212 f f \u2020 = 2f \u2020f \u2212 1 = 2 \u02c6Sz. mei\u03c0 j<m nj e\u2212i\u03c0 \u02c6S\u2212 m+1 = f \u2020 (b) Using the fact that the number operators on di\ufb00erent sites commute, one \ufb01nds me\u2212i\u03c0nm fm+1 = f \u2020 l<m+1 nl fm+1 = f \u2020 mfm+1, where we me\u2212i\u03c0nm \u2261 f \u2020 m. \u02c6S+ m have made use of the fact that, for fermionic particles, f \u2020 (c) The fermion representation is simply obtained by substitution. (d) With Jz = 0, the spin Hamiltonian assumes the form of a non-interacting tight-binding Hamiltonian \u02c6H = J\u22a5 nfn+1 +h.c.). This Hamiltonian, which has been encountered 2 previously, is diagonalized in the Fourier space, after which one obtains the cosine band dispersion. n(f \u2020 90 Second quantization Spin\u2013charge separation in one-dimension In Section 2.2 a free theory of interacting spinless fermions was developed in one dimension making use of the bosonization formalism. This analysis showed that the low-energy degrees of freedom were described by hydrodynamic charge (i.e. density) \ufb02uctuations that propagated with a linear dispersion. However, as well as charge, the electron degrees of freedom carry spin. The aim of this problem is to explore the fate of the spin degrees of freedom in a one-dimensional environment. \u2020 k a As a \ufb01rst step, we introduce operators (cf. Eq. (2.35)) \u02c6\u03c1sq\u03b1 = s(k+q)\u03b1ask\u03b1, \u03b1 =\u2191, \u2193 generalizing the previously introduced density operators for the presence of spin. Similarly, the bosonic degrees of freedom of the theory (cf. Eq. (2.40)) now carry a spin index, bq \u2192 bq\u03b1. One aspect that makes the problem more di\ufb03cult to tackle than the previously explored spinless case is that the 2kF-momentum transfer interaction |kF + q + q1, \u2191; kF + q \u2212 q1, \u2193 \u2192 | \u2212 kF + q + q2, \u2191; \u2212kF + q \u2212 q2, \u2193 in which a right-moving spin up electron is scattered to a left-moving spin up electron cannot be expressed in terms of slowly \ufb02uctuating density operators. (If you don\u2019t believe this, try!) However, using the renormalisation group methods introduced in Chapter 8, it can be shown that this type of interaction is physically largely irrelevant and can be neglected from the outset. Concentrating on the low-momentum-transfer interaction, the e\ufb00ective bosonic Hamilto- nian assumes the form (exercise) \u02c6H = q>0,s,\u03b1 vFqb\u2020 sq\u03b1bsq\u03b1 + q>0,s,\u03b1,\u03b1 # g2 2\u03c0 |q| b\u2020 sq\u03b1b \u2020 \u00afsq\u03b1 + h.c. + g4 2\u03c0 b\u2020 sq\u03b1bsq\u03b1 $. Introducing operators that create charge (\u03c1) and spin (\u03c3) \ufb02uctuations, bsq\u03c1", " = 1\u221a (bsq\u2191 + 2 (bsq\u2191 \u2212 bsq\u2193), rearrange the Hamiltonian, and thereby show that it assumes bsq\u2193), bsq\u03c3 = 1\u221a 2 a diagonal form with the spin and charge degrees of freedom exhibiting di\ufb00erent velocities. This is a manifestation of spin\u2013charge separation: even without the introduction of spindependent forces, the spin and charge degrees of freedom of the electron in the metallic conductor separate and propagate at di\ufb00erent velocities. In this sense, there is no way to adiabatically continue from non-interacting electrons to the collective charge and spin excitations of the system. Despite the \u201cfragility\u201d of the electron, and the apparent ubiquity of this phenomenon, the observation of spin\u2013charge separation in one-dimensional conductors has presented a signi\ufb01cant challenge to experimentalists. The reason is subtle. The completion of an electrical circuit necessarily requires contact of the quantum wire with bulk leads. The leads involve a reservoir of electrons with conventional Fermi-liquid character. Electrical transport requires the recombination of the collective charge (holon) and spin (spinon) degrees of freedom at the contact to reconstitute physical electrons. It is an exasperating fact that this reconstitution of the physical electron \u201cmasks\u201d the character of spin\u2013charge separation. Instead, the phenomenon of spin\u2013charge separation has been inferred indirectly through spectroscopic techniques. Answer: Motivated by the separation into spin and charge degrees of freedom, a rearrangement of the Hamiltonian gives 4 \u02c6H = vFq q>0,s 2.4 Problems 91 b\u2020 sq\u03c1bsq\u03c1 + b\u2020 sq\u03c3bsq\u03c3 + |q| # g2 \u03c0 b\u2020 sq\u03c1b \u2020 \u00afsq\u03c1 + h.c. + g4 \u03c0 $5. b\u2020 sq\u03c1bsq\u03c1 Once again, applying a Bogoliubov transformation, the Hamiltonian is brought to the diagonal form # $ \u02c6H = |q| (vF + g4/\u03c0)2 \u2212 (g2/\u03c0)2\u03b1\u2020 sq\u03c1\u03b1sq\u03c1 + |q|vF\u03b1\u2020 sq\u03c3\u03b1sq\u03c3 + const.. q>0,s The Kondo problem Historically, the Kondo problem has assumed a place of great signi\ufb01cance in the development of the \ufb01eld of strongly correlated quantum systems. It represents perhaps the simplest example of a phenomenon driven by strong electron interaction and, unusually for this arena of physics, admits a detailed theoretical understanding. Further, in respect of the principles established in Chapter 1, it exempli\ufb01es a number of important ideas from the concept of reducibility \u2013 the collective properties of the system may be captured by a simpli\ufb01ed e\ufb00ective Hamiltonian which includes only the relevant low-energy degrees of freedom \u2013 and the renormalization group. In the following problem, we will seek to develop the low-energy theory of the \u201cKondo impurity system\u201d leaving the discussion of its phenomenology to Problems 5.5 and 8.8.5 in subsequent chapters. The Kondo e\ufb00ect is rooted in the experimental observation that, when small amounts of magnetic ion impurities are embedded in a metallic host (such as manganese in copper, or iron in CuAu alloys), a pronounced minimum develops in the temperature dependence of the resistivity. Although the phenomenon was discovered experimentally in 1934,27 it was not until 1964 that a \ufb01rm understanding of the phenomenon was developed by Kondo.28 Historically, the \ufb01rst step towards the resolution of this phenomenon came with a suggestion by Anderson that the system could be modeled as an itinerant band of electron states interacting with local dilute magnetic moments associated with the ion impurities.29 Anderson proposed that the integrity of the local moment was protected by a large local Coulomb repulsion which inhibited multiple occupancy of the orbital state \u2013 a relative of the Hubbard U -interaction. Such behavior is encoded in the Anderson impurity Hamiltonian # \u02c6H = k\u03c3 kc \u2020 k\u03c3ck\u03c3 + Vkd\u2020 \u03c3ck\u03c3 + h.c. $ + dnd\u03c3 + U nd\u2191nd\u2193, \u03c3 where the operators c while the operators d\u2020 Here we have used nd\u03c3 = d\u2020 \u2020 k\u03c3 create itinerant electrons of spin \u03c3 and k in the metallic host \u03c3 create electrons of spin \u03c3 on the local impurity at position r = 0. \u03c3d\u03c3 to denote the number operator. While electrons in the band 27 de Haas, de Boer, and van den Berg, The electrical resistance of gold, copper, and lead at low temperatures, Physica 1 (1934), 1115", ". 28 J. Kondo, Resistance minimum in dilute magnetic alloys, Prog. Theor. Phys. 32 (1964), 37\u201369. 29 P. W. Anderson, Localized magnetic states in metals, Phys. Rev. 124 (1961), 41\u201353. 92 Second quantization are assumed to be characterized by a Fermi-liquid-like behavior, those associated with the impurity state experience an on-site Coulomb interaction of a strength characterized by a Hubbard energy U. According to the experimental phenomenology, the Fermi level F is assumed to lie somewhere in between the single-particle impurity level d and d + U so that, on average, the site occupancy of the impurity is unity. Nevertheless, the matrix element coupling the local dr V (r)eik\u00b7r admits the existence of moment to the itinerant electron states Vk = L\u2212d/2 virtual processes in which the site occupancy can \ufb02uctuate between zero and two. These virtual \ufb02uctuations allow the spin on the impurity site to \ufb02ip through exchange. The discussion of the half-\ufb01lled Hubbard system in Section 2.2 suggests that it will be helpful to transform the Anderson impurity Hamiltonian to an e\ufb00ective theory which exposes the low-energy content of the system. To this end, let us express the total wavefunction of the many-body Hamiltonian |\u03c8 as the sum of terms |\u03c80, |\u03c81, and| \u03c82, where the subscript denotes the occupancy of the impurity site. With this decomposition, the \u02c6Hmn|\u03c8n = Schr\u00a8odinger equation for the Hamiltonian can be cast in matrix form, E|\u03c8m, where \u02c6Hmn = \u02c6Pm \u02c6H \u02c6Pn, and the operators \u02c6Pm project onto the subspace with m electrons on the impurity (i.e. \u02c6P0 = \u03c3(1 \u2212 nd\u03c3), etc.). 2 n=0 (a) Construct the operators \u02c6Hmn explicitly and explain why \u02c6H20 = \u02c6H02 = 0. (b) Since we are interested in the e\ufb00ect of virtual excitations from the |\u03c81 subspace, we may proceed by formally eliminating |\u03c80 and |\u03c82 from the Schr\u00a8odinger equation. Doing so, show that the equation for |\u03c81 can be written as \u02c6H10 1 E \u2212 \u02c6H00 \u02c6H01 + \u02c6H11 + \u02c6H12 1 E \u2212 \u02c6H22 \u02c6H21 |\u03c81 = E|\u03c81. (c) At this stage, the equation for |\u03c81 is exact. Show that, when substituted into this expression, an expansion to leading order in 1/U and 1/d leads to the expression \u02c6H12 1 E \u2212 \u02c6H22 \u2212 kk\u03c3\u03c3 \u02c6H21 + \u02c6H10 6 VkV \u2217 k 1 E \u2212 \u02c6H00 \u2020 \u2020 c k\u03c3ck\u03c3 d\u03c3d \u03c3 U + d \u2212 k \u02c6H01 + ck\u03c3 c \u2020 \u03c3 d\u03c3 \u2020 k\u03c3d k \u2212 d 7. To obtain the \ufb01rst term in the expression, consider the commutation of (E \u2212 \u02c6H22)\u22121 with \u02c6H21 and make use of the fact that the total operator acts upon the singly occupied subspace. A similar line of reasoning will lead to the second term in the expression. Here U + d \u2212 k and d \u2212 k denote the respective excitation energies of the virtual states. Making use of the Pauli matrix identity, \u03c3\u03b1\u03b2 \u00b7 \u03c3\u03b3\u03b4 = 2\u03b4\u03b1\u03b4\u03b4\u03b2\u03b3 \u2212 \u03b4\u03b1\u03b2\u03b4\u03b3\u03b4, it follows that (exercise) c \u2020 k\u03c3ck\u03c3 d \u2020 \u03c3 d\u03c3 = 2\u02c6skk \u00b7 \u02c6Sd + 1 2 \u03c3\u03c3 c \u2020 k\u03c3ck\u03c3nd\u03c3, \u03c3\u03c3 where \u02c6Sd = \u03b1\u03b2 d\u2020 with the impurity and \u02c6skk = \u03b1 \u03c3\u03b1\u03b2d\u03b2/2 denotes the impurity spin 1/2 degree of freedom associated \u03c3\u03b1\u03b2ck\u03b2/2. Combining this result with that \u03b1\u03b2 c \u2020 k\u03b1 2.4 Problems 93 obtained above, up to an irrelevant constant, the total e\ufb00ective Hamiltonian (including \u02c6H11) acting in the projected subspace, |\u03c81 is given by \u02c6Hsd = k\u03c3", " kc \u2020 k\u03c3ck\u03c3 + kk 2Jk,k \u02c6skk \u00b7 \u02c6Sd + Kk,k \u03c3 c \u2020 k\u03c3ck\u03c3, where Jk,k = V \u2217 k Vk Kk,k = V \u2217 k Vk \u2212,. With both U + d and d greatly in excess of the typical excitation energy scales, one may safely neglect the particular energy dependence of the parameters Jk,k and Kk,k. In this case, the exchange interaction Jk,k can be treated as local, the scattering term Kk,k can be absorbed into a shift of the single-particle energy of the itinerant band, and the positive (i.e. antiferromagnetic) exchange coupling can be accommodated through the e\ufb00ective sd-Hamiltonian \u02c6Hsd = k\u03c3 kc \u2020 k\u03c3ck\u03c3 + 2J \u02c6Sd \u00b7 \u02c6s(r = 0), (2.51) \u2020 where \u02c6s(r = 0) = k\u03c3\u03c3\u03c3\u03c3 ck\u03c3 /2 denotes the spin density of the itinerant electron band at the impurity site. To understand how the magnetic impurity a\ufb00ects the low-temperature transport, we refer to Problem 5.5, where the sd-Hamiltonian is explored in the framework of a diagrammatic perturbation theory in the spin interaction. kk\u03c3\u03c3 c Answer: (a) Since the diagonal elements \u02c6Hmm leave the occupation number \ufb01xed, they may be identi\ufb01ed with the diagonal elements of the microscopic Hamiltonian, i.e. \u02c6H00 = kc \u2020 k\u03c3ck\u03c3, \u02c6H11 = kc \u2020 k\u03c3ck\u03c3 + d, \u02c6H22 = kc \u2020 k\u03c3ck\u03c3 + 2d + U. k k k The o\ufb00-diagonal terms arise from the hybridization between the free electron states and the impurity. Since the coupling involves only the transfer of single electrons, \u02c6H02 = \u02c6H20 = 0 and \u02c6H10 = Vkd\u2020 \u03c3(1 \u2212 nd\u00af\u03c3)ck\u03c3, \u02c6H21 = Vkd\u2020 \u03c3nd\u00af\u03c3ck\u03c3, k\u03c3 k\u03c3 where \u00af\u03c3 =\u2191 for \u03c3 =\u2193 and vice versa, \u02c6H01 = \u02c6H \u2020 10 and \u02c6H12 = \u02c6H \u2020 21. (b) Using the fact that \u02c6H00|\u03c80 + \u02c6H01|\u03c81 = E|\u03c80, one may set|\u03c8 0 = (E \u2212 \u02c6H00)\u22121 \u02c6H01|\u03c81 and, similarly, |\u03c82 = (E \u2212 \u02c6H22)\u22121 \u02c6H21|\u03c81. Then, substituting into the equation for |\u03c81, one obtains the required expression. 94 Second quantization (c) Making use of the expressions from part (a), we have \u02c6H12 \u02c6H10 1 E \u2212 \u02c6H22 1 E \u2212 \u02c6H00 \u02c6H21 = \u02c6H01 = kk\u03c3\u03c3 kk\u03c3\u03c3 VkV \u2217 k c \u2020 k\u03c3nd\u00af\u03c3d\u03c3 1 E \u2212 \u02c6H22 d \u2020 \u03c3 nd\u00af\u03c3 ck\u03c3, VkV \u2217 k d\u2020 \u03c3(1 \u2212 nd\u00af\u03c3)ck\u03c3 1 E \u2212 \u02c6H00 c \u2020 k\u03c3 (1 \u2212 nd\u00af\u03c3 )d\u03c3. Then, substituting for \u02c6H22 and \u02c6H00 from (a), and commuting operators, we have 1 E \u2212 \u02c6H22 d \u2020 \u03c3 nd\u00af\u03c3 ck\u03c3 = \u2212 \u2020 \u03c3 nd\u00af\u03c3 ck\u03c3 \u2212 \u02c6H00 U + d \u2212 k \u22121, 1 E \u2212 \u02c6H00 c \u2020 k\u03c3 (1 \u2212 nd\u00af\u03c3 )d\u03c3 = \u2212 c \u2020 k\u03c3 (1 \u2212 nd\u00af\u03c3 )d\u03c3 k \u2212 d 1 \u2212 E \u2212 d \u2212 \u02c6H00 k \u2212 d \u22121. Then expanding in large U and d, to leading order we obtain \u02c6H12 1 E \u2212 \u02c6H22 \u2212 VkV \u2217 k kk\u03c3\u03c3 \u02c6H21 + \u02c6H10 6 c \u2020 k\u03c3nd,\u00af\u03c3d\u03c3d \u02c6H01 1 E \u2212 \u02c6H00 \u2020 \u03c3 nd", "\u00af\u03c3 ck\u03c3 U + d \u2212 k + d\u2020 \u03c3(1 \u2212 nd,\u00af\u03c3)ck\u03c3c k\u03c3 (1 \u2212 nd\u00af\u03c3 )d\u03c3 \u2020 k \u2212 d 7. Finally, noting that this operator acts upon the singly-occupied subspace spanned by |\u03c81, we see that the factors involving nd\u03c3 are redundant and can be dropped. As a result, swapping the momentum and spin indices in the second part of the expression, we obtain the required expression. 3 Feynman path integral The aim of this chapter is to introduce the concept of the Feynman path integral. As well as developing the general construction scheme, particular emphasis is placed on establishing the interconnections between the quantum mechanical path integral, classical Hamiltonian mechanics, and classical statistical mechanics. The practice of path integration is discussed in the context of several pedagogical applications. As well as the canonical examples of a quantum particle in a single and a double potential well, we discuss the generalization of the path integral scheme to tunneling of extended objects (quantum \ufb01elds), dissipative and thermally assisted quantum tunneling, and the quantum mechanical spin. In this chapter we temporarily leave the arena of many-body physics and second quantization and, at least super\ufb01cially, return to single-particle quantum mechanics. By establishing the path integral approach for ordinary quantum mechanics, we will set the stage for the introduction of \ufb01eld integral methods for many-body theories explored in the next chapter. We will see that the path integral not only represents a gateway to higher-dimensional functional integral methods but, when viewed from an appropriate perspective, already represents a \ufb01eld theoretical approach in its own right. Exploiting this connection, various concepts of \ufb01eld theory, namely stationary phase analysis, the Euclidean formulation of \ufb01eld theory, instanton techniques, and the role of topology in \ufb01eld theory, are introduced in this chapter. 3.1 The path integral: general formalism Broadly speaking, there are two approaches to the formulation of quantum mechanics: the \u201coperator approach\u201d based on the canonical quantization of physical observables and the associated operator algebra, and the Feynman path integral.1 Whereas canonical quantization is usually taught \ufb01rst in elementary courses on quantum mechanics, path integrals seem to have acquired the reputation of being a sophisticated concept that is better reserved for 1 For a more extensive introduction to the Feynman path integral, one can refer to one of the many standard texts including R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals (McGraw-Hill, 1965), J. W. Negele and H. Orland, Quantum Many Particle Systems (Addison-Wesley, 1988), and L. S. Schulman, Techniques and Applications of Path Integration (Wiley, 1981). Alternatively, one may turn to the original paper, R. P. Feynman, Space-time approach to non-relativistic quantum mechanics, Rev. Mod. Phys. 20 (1948), 362\u201387. Historically, Feynman\u2019s development of the path integral was motivated by earlier work by Dirac on the connection between classical and quantum mechanics, P. A. M. Dirac, On the analogy between classical and quantum mechanics, Rev. Mod. Phys. 17 (1945), 195\u20139. 95 96 Feynman path integral advanced courses. Yet this treatment is hardly justi\ufb01ed! In fact, formulation has the path integral many advantages, most of which explicitly support an intuitive understanding of quantum mechanics. even the Moreover, in\ufb01nite-dimensional encountered below \u2013 are hardly more abstract than in\ufb01nite-dimensional linear operators. Further merits of the path integral include the following: \u2013 ones integrals Richard P. Feynman 1918\u201388 Nobel Laureate in Physics in 1965 (with Sin-Itiro Tomonaga, and Julian Schwinger) for \u201cfundamental work in quantum electrodynamics, with far-reaching consequences for the physics of elementary particles.\u201d He was also well known for his unusual life style and for his popular books and lectures on mathematics and physics. (Image c The Nobel Foundation.) Whereas the classical limit is not always easy to retrieve within the canonical formulation of quantum mechanics, it constantly remains visible in the path integral approach. The latter makes explicit use of classical mechanics as a \u201cplatform\u201d on which to construct a theory of quantum \ufb02uctuations. The classical solutions of Hamilton\u2019s equation of motion always remain a central ingredient of the formalism.2 Path integrals allow for an e\ufb03cient formulation of non-perturbative approaches to the solution of quantum mechanical problems. Examples include the \u201cinstanton\u201d formulation", " of quantum tunneling discussed below. The extension of such methods to continuum theories has led to some of the most powerful concepts of quantum \ufb01eld theory. The Feynman path integral represents a prototype of the higher-dimensional \ufb01eld integrals to be introduced in the next chapter. However, even the basic \u201czero-dimensional\u201d path integral is of relevance to applications in many-body physics. Very often, one encounters environments such as the superconductor, super\ufb02uid, or strongly correlated few-electron devices where a macroscopically large number of degrees of freedom \u201clock\u201d to form a single collective variable. (For example, to a \ufb01rst approximation, the phase information carried by the order parameter \ufb01eld in moderately large superconducting grains can often be described in terms of a single phase degree of freedom, i.e. a \u201cquantum particle\u201d living on the unit circle.) Path integral techniques have proven ideally suited to the analysis of such systems. What then is the basic idea of the path integral approach? More than any other formulation of quantum mechanics, the path integral formalism is based on connections to classical mechanics. The variational approach employed in Chapter 1 relied on the fact that classically allowed trajectories in con\ufb01guration space extremize an action functional. A principal constraint to be imposed on any such trajectory is energy conservation. By contrast, quantum particles have a little bit more freedom than their classical counterparts. In particular, by the Uncertainty Principle, energy conservation can be violated by an amount \u0394E over a time \u223c /\u0394E (here, and throughout this chapter, we will use for clarity). The connection 2 For this reason, path integration has turned out to be an indispensable tool in \ufb01elds such as quantum chaos where the quantum manifestations of classically non-trivial behavior are investigated \u2013 for more details, see Section 3.3. 3.2 Construction of the path integral 97 to action principles of classical mechanics becomes particularly apparent in problems of quantum tunneling: a particle of energy E may tunnel through a potential barrier of height barrier dx p/), V > E. However, this process is penalized by a damping factor \u223c exp(i 2m(E \u2212 V ), i.e. the exponent of the (imaginary) action associated with the where p = classically forbidden path. These observations motivate the idea of a new formulation of quantum propagation: could it be that, as in classical mechanics, the quantum amplitude A for propagation between any two points in coordinate space is again controlled by the action functional \u2013 controlled in a relaxed sense where not just a single extremal path xcl(t), but an entire manifold of neighboring paths contribute? More speci\ufb01cally, one might speculate that the quantum x(t) exp(iS[x]/), where amplitude is obtained as A \u223c x(t) symbolically stands for a summation over all paths compatible with the initial conditions of the problem, and S denotes the classical action. Although, at this stage, no formal justi\ufb01cation for the path integral has been presented, with this ansatz some features of quantum mechanics would obviously be borne out correctly. Speci\ufb01cally, in the classical limit ( \u2192 0), the quantum mechanical amplitude would become increasingly dominated by the contribution to the sum from the classical path xcl(t). This is because non-extremal con\ufb01gurations would be weighted by a rapidly oscillating amplitude associated with the large phase S/ and would, therefore, average to zero.3 Secondly, quantum mechanical tunneling would be a natural element of the theory; non-classical paths do contribute to the net amplitude, but at the cost of a damping factor speci\ufb01ed by the imaginary action (as in the traditional formulation). Fortunately, no fundamentally novel \u201cpicture\u201d of quantum mechanics needs to be declared x(t) exp(iS[x]/) to a working theory. As we to promote the idea of the path \u201cintegral\u201d will see in the next section, the new formulation can be developed from the established principles of canonical quantization. 3.2 Construction of the path integral All information about an autonomous4 quantum system is contained in its time evolution operator. A formal integration of the time-dependent Schr\u00a8odinger equation i\u2202t|\u03a8 = \u02c6H|\u03a8 gives the time evolution operator |\u03a8(t) = \u02c6U (t, t)|\u03a8(t), \u02c6U (t, t) = e\u2212 i \u02c6H(t\u2212t)\u0398(t \u2212 t). (3.1) The operator \u02c6U (t, t) describes dynamical evolution under the in\ufb02uence of the Hamiltonian from a time", " t to time t. Causality implies that t > t, as indicated by the step or Heaviside \u0398-function. In the real space representation we can write \u03a8(q, t) = q|\u03a8(t) = q| \u02c6U (t, t)\u03a8(t) = dq U (q, t; q, t)\u03a8(q, t), 3 More precisely, in the limit of small, the path sum can be evaluated by saddle-point methods, as detailed below. 4 A system is classi\ufb01ed as autonomous if its Hamiltonian does not explicitly depend on time. Actually the construction of the path integral can be straightforwardly extended so as to include time-dependent problems. However, in order to keep the introductory discussion as simple as possible, here we assume time independence. 98 Feynman path integral \u02c6H(t\u2212t)|q\u0398(t \u2212 t) de\ufb01nes the (q, q)-component of the time where U (q, t; q, t) = q|e\u2212 i evolution operator. As the matrix element expresses the probability amplitude for a particle to propagate between points q and q in a time t \u2212 t, it is sometimes known as the propagator of the theory. The basic idea behind Feynman\u2019s path integral approach is easy to formulate. Rather than attacking the Schr\u00a8odinger equation governing the time evolution for general times t, one may \ufb01rst attempt to solve the much simpler problem of describing the time evolution for in\ufb01nitesimally small times \u0394t. In order to formulate this idea one must \ufb01rst divide the time evolution into N 1 time steps, e\u2212i \u02c6Ht/ = # e\u2212i \u02c6H\u0394t/ $N, (3.2) where \u0394t = t/N. Albeit nothing more than a formal rewriting of Eq. (3.1), the representation (3.2) has the advantage that the factors e\u2212i \u02c6H\u0394t/ (or, rather, their expectation values) are small. (More precisely, if \u0394t is much smaller than the [reciprocal of the] eigenvalues of the Hamiltonian in the regime of physical interest, the exponents are small in comparison with unity and, as such, can be treated perturbatively.) A \ufb01rst simpli\ufb01cation arising from this fact is that the exponentials can be factorized into two pieces, each of which can be readily diagonalized. To achieve this factorization, we make use of the identity e\u2212i \u02c6H\u0394t/ = e\u2212i \u02c6T \u0394t/e\u2212i \u02c6V \u0394t/ + O(\u0394t2), where the Hamiltonian \u02c6H = \u02c6T + \u02c6V is the sum of a kinetic energy \u02c6T = \u02c6p2/2m, and some potential energy operator \u02c6V.5 (The following analysis, restricted for simplicity to a onedimensional Hamiltonian, is easily generalized to arbitrary spatial dimension.) The advantage of this factorization is that the eigenstates of each factor e\u2212i \u02c6T \u0394t/ and e\u2212i \u02c6V \u0394t/ are known independently. To exploit this fact we consider the time evolution operator factorized as a product, # qf | e\u2212i \u02c6H\u0394t/ $N |qi qf | \u2227e \u2212i \u02c6T \u0394t/e\u2212i \u02c6V \u0394t/ \u2227 \u00b7 \u00b7 \u00b7 \u2227 e\u2212i \u02c6T \u0394t/e\u2212i \u02c6V \u0394t/|qi, (3.3) and insert at each of the positions indicated by the symbol \u201c\u2227\u201d the resolution of identity id = dqn dpn |qnqn|pnpn|. (3.4) Here |qn and |pn represent a complete set of position and momentum eigenstates respectively, and n = 1,..., N serves as an index keeping track of the time steps at which the unit operator is inserted. The rationale behind the particular choice (3.4) is clear. The unit operator is arranged in such a way that both \u02c6T and \u02c6V act on the corresponding 5 Although this ansatz covers a wide class of quantum problems, many applications (e.g. Hamiltonians involving spin or magnetic \ufb01elds) do not \ufb01t into this framework. For a detailed exposition covering its realm of applicability, we refer to the specialist literature such as, e.g., Schulman1. 3.2 Construction of the path integral 99 qf N", " N\u20131 qi (a) tn 0 (b) t Figure 3.1 (a) Visualization of a set of phase space points contributing to the discrete time con\ufb01guration integral (3.5). (b) In the continuum limit, the set of points becomes a smooth curve. eigenstates. Inserting Eq. (3.4) into (3.3), and making use of the identity q|p = p|q\u2217 = eiqp//(2\u03c0)1/2, one obtains qf |e\u2212i \u02c6Ht/|qi N \u22121' n=1 qN =qf,q0=qi dqn N' n=1 dpn 2\u03c0 e N \u22121 n=0 \u2212i \u0394t V (qn)+T (pn+1)\u2212pn+1 qn+1\u2212qn \u0394t. (3.5) Thus, the matrix element of the time evolution operator has been expressed as a (2N \u2212 1)dimensional integral over eigenvalues. Up to corrections of higher order in V \u0394t/ and T \u0394t/, the expression (3.5) is exact. At each \u201ctime step\u201d tn = n\u0394t, n = 1,..., N, we are integrating over a pair of coordinates xn \u2261 (qn, pn) parameterizing the classical phase space. Taken together, the points {xn} form an N -point discretization of a path in this space (see Fig. 3.1). To make further progress, we need to develop some intuition for the behavior of the integral (3.5). We \ufb01rst notice that rapid \ufb02uctuations of the integration arguments xn as a function of the index n are strongly inhibited by the structure of the integrand. When taken together, contributions for which (qn+1 \u2212 qn)pn+1 > O() (i.e. when the phase of the exponential exceeds 2\u03c0) tend to lead to a \u201crandom phase cancellation.\u201d In the language of wave mechanics, the superposition of partial waves of erratically di\ufb00erent phases destructively interferes. The smooth variation of the paths that contribute signi\ufb01cantly motivates the application of a continuum limit analogous to that employed in Chapter 1. To be speci\ufb01c, sending N \u2192 \u221e whilst keeping t = N \u0394t \ufb01xed, the formerly discrete set tn = n\u0394t, n = 1,..., N, becomes dense on the time interval [0, t], and the set of phase space points {xn} becomes a continuous curve x(t). In the same limit, \u0394t N \u22121 n=0 \u2192 t 0 dt, qn+1 \u2212 qn \u0394t \u2192 \u2202t q |t=tn \u2261 \u02d9q|t=tn, while [V (qn) +T (pn+1)] \u2192 [T (p|t=tn ) +V (q|t=tn )] \u2261 H(x|t=tn ) denotes the classical Hamiltonian. In the limit N \u2192 \u221e, the fact that kinetic and potential energies are evaluated 100 Feynman path integral at neighboring time slices, n and n + 1, becomes irrelevant.6 Finally, lim N\u2192\u221e N \u22121' n=1 qN =qf,q0=qi dqn N' n=1 dpn 2\u03c0 \u2261 Dx, q(t)=qf q(0)=qi de\ufb01nes the integration measure of the integral. INFO Integrals extending over in\ufb01nite-dimensional integration measures like D(q, p) are generally called functional integrals (recall our discussion of functionals in Chapter 1). The question of the way functional integration can be rigorously de\ufb01ned is far from innocent and represents a subject of current, and partly controversial, mathematical research. In this book \u2013 as in most applications in physics \u2013 we take a pragmatic point of view and deal with the in\ufb01nite-dimensional integration naively unless mathematical problems arise (which actually will not be the case!). Then, applying these conventions to Eq. (3.5), one \ufb01nally obtains qf |e\u2212i \u02c6Ht/|qi = Dx exp q(t)=qf q(0)=qi i t 0 dt (p \u02d9q \u2212 H(p, q)). (3.6) Equation (3.6) represents the Hamiltonian formulation of the path integral. The integration extends over all possible paths through the classical phase space of the system which begin and end at the same con\ufb01guration points qi", " and qf respectively (cf. Fig. 3.1). The contribution of each path is weighted by its Hamiltonian action. INFO Remembering the connection of the Hamiltonian to the Lagrangian through the Legendre transform, H(p, q) = p \u02d9q \u2212 L(p, q), the classical action of a trajectory t \u2192 q(t) is given by S[p, q] = 0 dt L(q, \u02d9q) = 0 dt [p \u02d9q \u2212 H(p, q)]. t t Before we turn to the discussion of the path integral (3.6), it is useful to recast the integral in an alternative form which will be both convenient in applications and physically instructive. The search for an alternative formulation is motivated by the resemblance of Eq. (3.6) to the Hamiltonian formulation of classical mechanics. Given that, classically, Hamiltonian and Lagrangian mechanics can be equally employed to describe dynamical evolution, it is natural to seek a Lagrangian analog of Eq. (3.6). Focusing on Hamiltonians for which the kinetic energy T (p) is quadratic in p, the Lagrangian form of the path integral can indeed be inferred from Eq. (3.6) by straightforward Gaussian integration. 6 To see this formally, one may Taylor expand T (pn+1) =T (p(t + \u0394t))| t=n\u0394t around p(t). For smooth p(t), all but the zeroth-order contribution T (p(t)) scale with powers of \u0394t, thereby becoming irrelevant. Note, however, that all of these arguments are based on the assertion that the dominant contributions to the path integral are smooth in the sense qn+1 \u2212qn \u223c O(\u0394t). A closer inspection, however, shows that in fact qn+1 \u2212qn \u223c O( \u0394t) (see Schulman1.) In some cases, the most prominent one being the quantum mechanics of a particle in a magnetic \ufb01eld, the lowered power of \u0394t spoils the naive form of the continuity argument above, and more care must be applied in taking the continuum limit. In cases where a \u201cnew\u201d path integral description of a quantum mechanical problem is developed, it is imperative to delay taking the continuum limit until the \ufb02uctuation behavior of the discrete integral across individual time slices has been thoroughly examined. \u221a 3.2 Construction of the path integral 101 To make this point clear, let us rewrite the integral in a way that emphasizes its depen- dence on the momentum variable p: qf |e\u2212i \u02c6Ht/|qi = q(t)=qf q(0)=qi Dq e\u2212 i t 0 dtV (q) t 0 dt \u2212 i p2 2m \u2212p \u02d9q. Dp e (3.7) The exponent is quadratic in p which means that we are dealing with the continuum generalization of a Gaussian integral. Carrying out the integration by means of Eq. (3.13) below, one obtains qf |e\u2212i \u02c6Ht/|qi = Dq exp q(t)=qf q(0)=qi i t 0 dtL(q, \u02d9q), (3.8) N/2 N m it2\u03c0 N \u22121 n=1 dqn denotes the functional measure of the remaining where Dq = limN\u2192\u221e q-integration, and L(q, \u02d9q) = m \u02d9q2/2 \u2212 V (q) represents the classical Lagrangian. Strictly speaking, the (\ufb01nite-dimensional) integral formula (3.13) is not directly applicable to the in\ufb01nite-dimensional Gaussian integral (3.7). This, however, does not represent a substantial problem as we can always rediscretize the integral (3.7), apply Eq. (3.13), and reinstate the continuum limit after integration (exercise). Together Eq. (3.6) and (3.8) represent the central results of this section. A quantum mechanical transition amplitude has been expressed in terms of an in\ufb01nite-dimensional integral extending over paths through phase space, Eq. (3.6), or coordinate space, Eq. (3.8). including number All paths begin (end) at the initial (\ufb01nal) coordinate of the matrix element. Each path is weighted by its classical action. Notice in particular that the quantum transition amplitude has been cast in a form that does not contain quantum mechanical operators. Nonetheless, quantum mechanics is still fully present! The point is that the integration extends over all paths", " and not just the subset of solutions of the classical equations of motion. (The distinguished role classical paths play in the path integral will be discussed below in Section 3.2.) The two forms of the path integral, Eq. (3.6) and Eq. (3.8), represent the formal implementation of the \u201calternative picture\u201d of quantum mechanics proposed heuristically at the beginning of the chapter. Johann Carl Friedrich Gauss 1777\u20131855 Worked in a wide variety of \ufb01elds in both mathematics and physics theory, analysis, di\ufb00erential geometry, geodesy, magnetism, astronomy, and optics. As well as several books, Gauss published a number of memoirs (reports of his experiences), mainly in the journal of the Royal Society of G\u00a8ottingen. However, in general, he was unwilling to publish anything that could be regarded as controversial and, as a result, some of his most brilliant work was found only after his death. INFO Gaussian integration: Apart from a few rare exceptions, all integrals encountered in this book will be of Gaussian form. In most cases the dimension of the integrals will be large if not in\ufb01nite. Yet, after a bit of practice, it will become clear that high-dimensional representatives of Gaussian integrals are no more di\ufb03cult to handle than their one-dimensional counterparts. 102 Feynman path integral Therefore, considering the important role played by Gaussian integration in \ufb01eld theory, we will here derive the principal formulae once and for all. Our starting point is the one-dimensional integral (both real and complex). The ideas underlying the proofs of the one-dimensional formulae will provide the key to the derivation of more complex functional identities that will be used liberally throughout the remainder of the text. One-dimensional Gaussian integral: The basic ancestor of all Gaussian integrals is the identity \u221e \u2212\u221e dx e\u2212 a 2 x2 = 2\u03c0 a, Re a > 0. (3.9) In the following we will need various generalizations of Eq. (3.9). Firstly, we have 2\u03c0/a3, a result established either by substituting a \u2192 a + in Eq. (3.9) and expanding both the left and the right side of the equation to leading order in, or by di\ufb00erentiating Eq. (3.9). Often one encounters integrals where the exponent is not purely quadratic from the outset but rather contains both quadratic and linear pieces. The generalization of Eq. (3.9) to this case reads \u221e \u2212\u221e dx e\u2212 a 2 x2+bx = b2 2a. e 2\u03c0 a (3.10) \u221e \u2212\u221e dx e\u2212ax2/2x2 = To prove this identity, one simply eliminates the linear term by means of the change of variables x \u2192 x + b/a which transforms the exponent \u2212ax2/2 + bx \u2192 \u2212ax2/2 + b2/2a. The constant factor scales out and we are left with Eq. (3.9). Note that Eq. (3.10) holds even for complex b. The reason is that as a result of shifting the integration contour into the complex plane no singularities are encountered, i.e. the integral remains invariant. Later, we will be concerned with the generalization of the Gaussian integral to complex argu- ments. The extension of Eq. (3.9) to this case reads d(\u00afz, z) e\u2212\u00afzwz =, Re w > 0, \u03c0 w \u221e where \u00afz represents the complex conjugate of z. Here, \u2212\u221e dx dy represents the independent integration over the real and imaginary parts of z = x + iy. The identity is easy to prove: owing to the fact that \u00afzz = x2 + y2, the integral factorizes into two pieces, each of which is equivalent to Eq. (3.9) with a = w. Similarly, it may be checked that the complex generalization of Eq. (3.10) is given by d(\u00afz, z) \u2261 d(\u00afz, z) e\u2212\u00afzwz+\u00afuz+\u00afzv = \u00afuv w, Re w > 0. e \u03c0 w (3.11) More importantly \u00afu and v may be independent complex numbers; they need not be related to each other by complex conjugation (exercise). Gaussian integration in more than one dimension: All of the integrals above have higherdimensional counterparts. Although the real and complex versions of the N -dimensional integral formulae can be derived in a perfectly analogous manner, it is better to discuss them separately in order not to", " confuse the notation. (a) Real case: The multi-dimensional generalization of the prototype integral (3.9) reads dv e\u2212 1 2 vT Av = (2\u03c0)N/2det A\u22121/2, (3.12) 3.2 Construction of the path integral 103 where A is a positive de\ufb01nite real symmetric N -dimensional matrix and v is an N -component real vector. The proof makes use of the fact that A (by virtue of being symmetric) can be diagonalized by orthogonal transformation, A = OT DO, where the matrix O is orthogonal, and all elements of the diagonal matrix D are positive. The matrix O can be absorbed into the integration vector by means of the variable transformation, v \u2192 Ov, which has unit Jacobian, det O = 1. As a result, we are left with a Gaussian integral with exponent \u2212vT Dv/2. Due to the diagonality of D, the integral factorizes into N independent Gaussian integrals, each of which contributes a factor 2\u03c0/di, whered i, i = 1,..., N, is thei th entry of the matrix D. Noting that The multi-dimensional generalization of Eq. (3.10) reads N i=1 di = det D = det A, Eq. (3.12) is derived. dv e\u2212 1 2 vT Av+jT \u00b7v = (2\u03c0)N/2det A\u22121/2e 1 2 jT A\u22121j, (3.13) 2 vT Av + 1 where j is an arbitrary N -component vector. Equation (3.13) is proven by analogy with Eq. (3.10), i.e. by shifting the integration vector according to v \u2192 v + A\u22121j, which does not change the value of the integral but removes the linear term from the exponent, \u2212 1 2 vT Av + 2 jT A\u22121j. The resulting integral is of the type (3.12), and we arrive at jT \u00b7 v \u2192 \u2212 1 Eq. (3.13). The integral (3.13) not only is of importance in its own right, but also serves as a \u201cgenerator\u201d |j=0 to the of other useful integral identities. Applying the di\ufb00erentiation operation \u22022 vT Avvmvn = dv e\u2212 1 left- and the right-hand side of Eq. (3.13), one obtains the identity7 (2\u03c0)N/2det A\u22121/2A\u22121 mn. This result can be more compactly formulated as jmjn 2 vmvn = A\u22121 mn, where we have introduced the shorthand notation \u00b7 \u00b7 \u00b7 \u2261 (2\u03c0)\u2212N/2det A1/2 dv e\u2212 1 2 vT Av(\u00b7 \u00b7 \u00b7 ), (3.14) (3.15) suggesting an interpretation of the Gaussian weight as a probability distribution. Indeed, the di\ufb00erentiation operation leading to Eq. (3.14) can be iterated. Di\ufb00erentiating four times, one obtains vmvnvqvp = A\u22121 nq. One way of memorizing the structure of this \u2013 important \u2013 identity is that the Gaussian \u201cexpectation\u201d value vmvnvpvq is given by all \u201cpairings\u201d of type (3.14) that can be formed from the four components vm. This rule generalizes to expectation values of arbitrary order: 2n-fold di\ufb00erentiation of Eq. (3.13) yields qp + A\u22121 np + A\u22121 mnA\u22121 mpA\u22121 mqA\u22121 vi1 vi2... vi2n = pairings of {i1,...,i2n } A\u22121 ik1 ik2... A\u22121 ik2n\u22121. ik2n (3.16) This result is the mathematical identity underlying Wick\u2019s theorem (for real bosonic \ufb01elds), to be discussed in more physical terms below. 7 Note that the notation A\u22121 mn refers to the mn-element of the matrix A\u22121. 104 Feynman path integral (b) Complex case: The results above are straightforwardly extended to multi-dimensional com- plex Gaussian integrals. The complex version of Eq. (3.12) is given by d(v\u2020, v) e\u2212v\u2020Av = \u03c0N det A\u22121, (3.17) where v is a complex N -component vector, d(v\u2020, v) \u2261 N i=1 d", " Re vi d Im vi, andA is a complex matrix with positive de\ufb01nite Hermitian part. (Remember that every matrix can be 2 (A \u2212 decomposed into a Hermitian and an anti-Hermitian component, A = 1 A\u2020).) For Hermitian A, the proof of Eq. (3.17) is analogous to that of Eq. (3.12), i.e. A is unitarily diagonalizable, A = U\u2020AU, the matrices U can be transformed into v, the resulting integral factorizes, etc. For non-Hermitian A the proof is more elaborate, if unedifying, and we refer to the literature for details. The generalization of Eq. (3.17) to exponents with linear contributions reads 2 (A + A\u2020) + 1 d(v\u2020, v) e\u2212v\u2020Av+w\u2020\u00b7v+v\u2020\u00b7w = \u03c0N det A\u22121ew\u2020A\u22121w. (3.18) Note that w and w may be independent complex vectors. The proof of this identity mirrors that of Eq. (3.13), i.e. by e\ufb00ecting the shift v\u2020 \u2192 v\u2020 + w\u2020, v \u2192 v + w.8 As with Eq. (3.13), Eq. (3.18) may also serve as a generator of integral identities. Di\ufb00erentiating the integral twice according to \u22022 w |w=w=0 gives m, \u00afwn \u00afvmvn = A\u22121 nm, where \u00b7 \u00b7 \u00b7 \u2261 \u03c0\u2212N det A gives \u00afvn\u00afvmvpvq = A\u22121 pmA\u22121 d(v\u2020, v) e\u2212v\u2020Av(\u00b7 \u00b7 \u00b7 ). The iteration to more than two derivatives qn + A\u22121 qm and, eventually, pn A\u22121 \u00afvi1 \u00afvi2 \u00b7 \u00b7 \u00b7 \u00afvin vj1 vj2 \u00b7 \u00b7 \u00b7 vjn = P A\u22121 j1iP 1 \u00b7 \u00b7 \u00b7 A\u22121 jniP n, where P represents for the sum over all permutations of n integers. Gaussian functional integration: With this preparation, we are in a position to investigate the main practice of quantum and statistical \ufb01eld theory \u2013 the method of Gaussian functional integration. Turning to Eq. (3.13), let us suppose that the components of the vector v parameterize the weight of a real scalar \ufb01eld on the sites of a one-dimensional lattice. In the continuum limit, the set {vi} translates to a function v(x), and the matrix Aij is replaced by an operator kernel or propagator A(x, x). In this limit, the natural generalization of Eq. (3.13) is Dv(x) exp \u2212 1 2 dx dx v(x)A(x, x)v(x) + dx j(x)v(x) \u221d (det A)\u22121/2exp dx dx j(x)A\u22121(x, x)j(x), (3.19) 1 2 8 For an explanation of why v and v\u2020 may be shifted independently of each other, cf. the analyticity remarks made in connection with Eq. (3.11). 3.2 Construction of the path integral 105 where the inverse kernel A\u22121(x, x) satis\ufb01es the equation dx A(x, x)A\u22121(x, x) = \u03b4(x \u2212 x), (3.20) i.e. A\u22121(x, x) can be interpreted as the Green function of the operator A(x, x). The notation Dv(x) is used to denote the measure of the functional integral. Although the constant of proportionality, (2\u03c0)N, left out of Eq. (3.19) is formally divergent in the thermodynamic limit N \u2192 \u221e, it does not a\ufb00ect averages that are obtained from derivatives of such integrals. For example, for Gaussian distributed functions, Eq. (3.14) has the generalization v(x)v(x) = A\u22121(x, x). Accordingly, Eq. (3.16) assumes the form v(x1)v(x2) \u00b7 \u00b7 \u00b7 v(x2n) = pairings of {x1,...,x2n } A\u22121(xk1, xk2 ) \u00b7 \u00b7 \u00b7 A\u22121(xk2n\u22121, xk2n ). (3.21) The generalization of the other Gaussian averaging formulae discussed above should be obvious. To make sense of Eq. (3", ".19) one must interpret the meaning of the determinant, det A. When the variables entering the Gaussian integral were discrete, the integral simply represented the determinant of the (real symmetric) matrix. In the present case, one must interpret A as a Hermitian operator having an in\ufb01nite set of eigenvalues. The determinant simply represents the product over this in\ufb01nite set (see, e.g., Section 3.3). Before turning to speci\ufb01c applications of the Feynman path integral, let us stay with the general structure of the formalism and identify two fundamental connections of the path integral to classical point mechanics and classical and quantum statistical mechanics. Path integral and statistical mechanics The path integral reveals a connection between quantum mechanics and classical (and quantum) statistical mechanics whose importance to all areas of \ufb01eld theory and statistical physics can hardly be exaggerated. To reveal this link, let us for a moment forget about quantum mechanics and consider, by way of an example, a perfectly classical, onedimensional continuum model describing a \u201c\ufb02exible string.\u201d We assume that our string is held under constant tension, and con\ufb01ned to a \u201cgutter-like potential\u201d (as shown in Fig. 3.2). For simplicity, we also assume that the mass density of the string is pretty high, so that its \ufb02uctuations are \u201casymptotically slow\u201d (the kinetic contribution to its energy is negligible). Transverse \ufb02uctuations of the string are then penalized by its line tension, and by the external potential. Assuming that the transverse displacement of the string u(x) is small, the potential energy stored in the string separates into two parts. The \ufb01rst arises from the line tension stored in the string, and the second comes from the external potential. Starting with the former, a transverse \ufb02uctuation of a line segment of length dx by an amount du leads to a potential 106 Feynman path integral x V(u) u Figure 3.2 A string held under tension and con\ufb01ned to a potential well V. L energy of magnitude \u03b4Vtension = \u03c3[(dx2 +du2)1/2 \u2212dx] \u03c3dx (\u2202xu)2/2, where \u03c3 denotes the tension. Integrating over the length of the string, one obtains Vtension[\u2202xu] \u2261 \u03b4Vtension = 1 0 dx \u03c3(\u2202xu(x))2. The second contribution, arising from the external potential, is given 2 L by Vexternal[u] \u2261 0 dx V (u(x)). Adding the two contributions, we \ufb01nd that the total energy 0 dx [ \u03c3 of the string is given by V = Vtension + Vexternal = 2 (\u2202xu)2 + V (u)]. L EXERCISE Find an expression for the kinetic energy contribution assuming that the string has a mass per unit length of m. How does this model compare to the continuum model of lattice vibrations discussed in Chapter 1? Convince yourself that in the limit m \u2192 \u221e, the kinetic e\u2212\u03b2H contribution to the partition function Z = tr is inessential. According to the general principles of statistical mechanics, the equilibrium properties of a system are encoded in the partition function Z = tr, where \u201ctr\u201d denotes a summation over all possible con\ufb01gurations of the system and V is the total potential energy functional. Applied to the present case, tr \u2192 Du stands for the functional integration over all con\ufb01gurations of the string u(x), x \u2208 [0, L]. Thus, the partition function of the string is given by Du, where e\u2212\u03b2V % & L 0 \u03c3 2 dx Z = Du exp \u2212\u03b2 (\u2202xu)2 + V (u). (3.22) A comparison of this result with Eq. (3.8) shows that the partition function of the classical system coincides with the quantum mechanical amplitude Z = dq q|e\u2212it \u02c6H/|q, =1/\u03b2, t=\u2212iL evaluated at an imaginary \u201ctime\u201d t \u2192 \u2212i\u03c4 \u2261 \u2212iL, where \u02c6H = \u02c6p2/2\u03c3 + V (q), and Planck\u2019s constant is identi\ufb01ed with the \u201ctemperature,\u201d = 1/\u03b2. (Here we have assumed that our string is subject to periodic boundary conditions.) To see this explicitly, let us assume that we had reason to consider quantum propagation in imaginary time, i.e. e\u2212it \u02c6H/ \u2192 e\u2212\u03c4 \u02c6H/, ort \u2192 \u2212i", "\u03c4. Assuming convergence (i.e. positivity of the eigenvalues of \u02c6H), a construction scheme perfectly analogous to the one outlined in Section 3.1 would have led to a path integral formula of the structure (3.8). Formally, the only di\ufb00erences would be (a) that the Lagrangian would be integrated along the imaginary time axis t \u2192 \u2212i\u03c4 \u2208 [0, \u2212i\u03c4 ] and (b) that there would be a change of the sign of the kinetic 3.2 Construction of the path integral 107 energy term, i.e. (\u2202t q)2 \u2192 \u2212(\u2202\u03c4 q)2. After a suitable exchange of variables, \u03c4 \u2192 L, \u2192 1/\u03b2, the coincidence of the resulting expression with the partition function (3.22) is clear. The connection between quantum mechanics and classical statistical mechanics outlined above generalizes to higher dimensions. There are close analogies between quantum \ufb01eld theories in d dimensions and classical statistical mechanics in d + 1. (The equality of the path integral above with the one-dimensional statistical model is merely the d = 0 version of this connection.) In fact, this connection turned out to be one of the major driving forces behind the success of path integral techniques in modern \ufb01eld theory/statistical mechanics. It o\ufb00ered, for the \ufb01rst time, a possibility to draw connections between systems that had seemed unrelated. However, the concept of imaginary times not only provides a bridge between quantum and classical statistical mechanics, but also plays a role within a purely quantum mechanical context. Consider the partition function of a single-particle quantum mechanical system, Z = tr[e\u2212\u03b2 \u02c6H ] = dq q|e\u2212\u03b2 \u02c6H |q. The partition function can be interpreted as a trace over the transition amplitude q|e\u2212i \u02c6Ht/|q evaluated at an imaginary time t = \u2212i\u03b2. Thus, real time dynamics and quantum statistical mechanics can be treated on the same footing, provided that we allow for the appearance of imaginary times. Later we will see that the concept of imaginary or even generalized complex times plays an important role in all of \ufb01eld theory. There is even some nomenclature regarding imaginary times. The transformation t \u2192 \u2212i\u03c4 is described as a Wick rotation (alluding to the fact that a multiplication by the imaginary unit can be interpreted as a (\u03c0/2)-rotation in the complex plane). Imaginary time representations of Lagrangian actions are termed Euclidean, whereas the real time forms are called Minkowski actions. INFO The origin of this terminology can be understood by considering the structure of the action of, say, the phonon model (1.4). Forgetting for a moment about the magnitude of the coupling constants, we see that the action has the bilinear structure \u223c x\u03bcg\u03bc\u03bd x\u03bd, where\u03bc = 0, 1, the vector x\u03bc = \u2202\u03bc\u03c6, and the diagonal matrix g = diag(\u22121, 1) is the two-dimensional version of a Minkowski metric. (In three spatial dimensions, g would take the form of the standard Minkowski metric of special relativity.) On Wick rotation of the time variable, the factor \u22121 in the metric changes sign to +1, and g becomes a positive de\ufb01nite Euclidean metric. The nature of this transformation motivates the notation above. Once one has grown accustomed to the idea that the interpretation of time as an imaginary quantity can be useful, yet more general concepts can be conceived. For example, one may contemplate propagation along temporal contours that are neither purely real nor purely imaginary but rather are generally complex. Indeed, it has turned out that path integrals with curvilinear integration contours in the complex \u201ctime plane\u201d \ufb01nd numerous applications in statistical and quantum \ufb01eld theory. 108 Feynman path integral Semiclassics from the path integral In deriving the two path integral representations (3.6) and (3.8) no approximations were made. Yet the vast majority of quantum mechanical problems are unsolvable in closed form, and it would be hoping for too much to expect that within the path integral approach this situation would be any di\ufb00erent. In fact no more than the path integrals of problems with a quadratic Hamiltonian \u2013 corresponding to the quantum mechanical harmonic oscillator and generalizations thereof \u2013 can be carried out in closed form. Yet what counts more than the (rare) availability of exact solutions is the \ufb02exibility with which approximation schemes can be developed. As to the path integral formulation, it is particularly strong in cases where semiclassical limits of quantum theories are explored. Here, by \u201csemiclassical,\u201d we mean the limit", " \u2192 0, i.e. the case where the theory is expected to be largely governed by classical structures with quantum \ufb02uctuations superimposed. To see more formally how classical structures enter the path integral approach, let us explore Eq. (3.6) and (3.8) in the limit of small. In this case the path integrals are dominated by path con\ufb01gurations with stationary action. (Non-stationary contributions to the integral imply massive phase \ufb02uctuations that largely average to zero.) Now, since the exponents of the two path integrals (3.6) and (3.8) involve the classical action functionals in their Hamiltonian and Lagrangian forms respectively, the extremal path con\ufb01gurations are simply the solutions of the classical equations of motion, namely, Hamiltonian : \u03b4S[x] = 0 \u21d2 dtx = {H(x), x} \u2261\u2202 pH \u2202qx \u2212 \u2202qH \u2202px, Lagrangian : \u03b4S[q] = 0 \u21d2 (dt\u2202 \u02d9q \u2212 \u2202q) L(q, \u02d9q) = 0. 8 (3.23) These equations are to be solved subject to the boundary conditions q(0) = qi and q(t) = qf. (Note that these boundary conditions do not uniquely specify a solution, i.e. in general there are many solutions to the equations (3.23). As an exercise, one may try to invent examples!) Now the very fact that the stationary phase con\ufb01gurations are classical does not imply that quantum mechanics has disappeared completely. As with saddle-point approximations in general, it is not just the saddle-point itself that matters but also the \ufb02uctuations around it. At least it is necessary to integrate out Gaussian (quadratic) \ufb02uctuations around the point of stationary phase. In the case of the path integral, \ufb02uctuations of the action around the stationary phase con\ufb01gurations involve non-classical (in that they do not solve the classical equations of motion) trajectories through phase or coordinate space. Before exploring how this mechanism works in detail, let us consider the stationary phase analysis of functional integrals in general. INFO Stationary phase approximation: Consider a general functional integral Dx e\u2212F [x] N where Dx = limN\u2192\u221e n=1 dxn represents a functional measure resulting from taking the continuum limit of some \ufb01nite-dimensional integration space, and the \u201caction\u201d F [x] may be an arbitrary complex functional of x (leading to convergence of the integral). The function resulting from taking the limit of in\ufb01nitely many discretization points, {xn}, is denoted by x : t \u2192 x(t) (where t plays the role of the formerly discrete index n). Evaluating the integral above within a stationary phase approximation amounts to performing the following steps: 3.2 Construction of the path integral 109 h1/2 q q t Figure 3.3 Quantum \ufb02uctuations around a classical path in coordinate space (here we assume a set of two-dimensional coordinates). Non-classical paths q \ufb02uctuating around a classical solution qcl typically extend a distance O(h1/2). All paths begin and end at qi and qf, respectively. 1. Firstly, \ufb01nd the \u201cpoints\u201d of stationary phase, i.e. con\ufb01gurations \u00afx quali\ufb01ed by the condition of vanishing functional derivative (cf. Section 1.2), DFx = 0 \u21d4 \u2200t : \u03b4F [x] \u03b4x(t) = 0. x=\u00afx Although there may, in principle, be one or many solutions, for clarity we \ufb01rst discuss the case in which the stationary phase con\ufb01guration \u00afx is unique. 2. Secondly, Taylor expand the functional to second order around \u00afx, i.e. F [x] =F [\u00afx + y] =F [\u00afx] + 1 2 dt dt y(t)A(t, t)y(t) +\u00b7 \u00b7 \u00b7, (3.24) where A(t, t) = \u03b42F [x] arity of \u00afx, no \ufb01rst-order contribution can appear. \u03b4x(t) \u03b4x(t) x=\u00afx denotes the second functional derivative. Due to the station- 3. Thirdly, check that the operator \u02c6A \u2261 {A(t, t)} is positive de\ufb01nite. If it is not, there is a problem \u2013 the integration over the Gaussian \ufb02uct", "uations y diverges. (In practice, where the analysis is rooted in a physical context, such eventualities arise only rarely. In situations where problems do occur, the resolution can usually be found in a judicious rotation of the integration contour.) For positive de\ufb01nite \u02c6A, however, the functional integral over y can be 2\u03c0 )\u22121/2, (cf. the discussion of performed, after which one obtains Gaussian integrals above and, in particular, Eq. (3.19)). Dx e\u2212F [x] e\u2212F [\u00afx] det( \u02c6A 4. Finally, if there are many stationary phase con\ufb01gurations, \u00afxi, the individual contributions have to be added: Dx e\u2212F [x] i e\u2212F [\u00afxi] det \u02c6Ai 2\u03c0 \u22121/2. (3.25) Equation (3.25) represents the most general form of the stationary phase evaluation of a (real) functional integral. \u221e 0 dx xze\u2212x, with z complex, show EXERCISE Applied to the Gamma function, \u0393(z + 1) = that the stationary phase approximation is consistent with Stirling\u2019s approximation, i.e. \u0393(s + 1) = 2\u03c0ses(ln s\u22121). \u221a 110 Feynman path integral Applied to the Lagrangian form of the Feynman path integral, this program can be implemented directly. In this case, the extremal \ufb01eld con\ufb01guration \u00afq(t) is identi\ufb01ed as the classical solution associated with the Lagrangian, i.e. \u00afq(t) \u2261 qcl(t). De\ufb01ning r(t) = q(t) \u2212 qcl(t) as the deviation of a general path, q(t), from a nearby classical path, qcl(t) (see Fig. 3.3), and assuming for simplicity that there exists only one classical solution connecting qi with qf in a time t, a stationary phase analysis obtains qf |e\u2212i \u02c6Ht/|qi eiS[qcl]/ Dr exp r(0)=r(t)=0 t 0 i 2 dt dtr(t) \u03b42S[q] \u03b4q(t) \u03b4q(t) q=qcl r(t), (3.26) as the Gaussian approximation to the path integral (cf. Eq. (3.24)). For free Lagrangians of the form L(q, \u02d9q) =m \u02d9q2/2 \u2212 V (q), the second functional derivative of the action can be straightforwardly computed by means of the rules of functional di\ufb00erentiation formulated in Chapter 1. Alternatively, one can obtain this result by simply expanding the action as a Taylor series in the deviation r(t). As a result, one obtains (exercise) t dt 0 1 2 dt r(t) \u03b42S[q] \u03b4q(t) \u03b4q(t) r(t) = \u2212 1 2 q=qcl % dt r(t) m\u22022 t + V (qcl(t)) & r(t), (3.27) where V (qcl(t)) \u2261 \u22022 q V (q)|q=qcl represents an ordinary derivative of the potential function. Thus, the Gaussian integration over r yields the square root of the determinant of the t + V (qcl(t)) \u2013 interpreted as an operator acting in the space of functions r(t) operator m\u22022 with boundary conditions r(0) = r(t) = 0. (Note that, as we are dealing with a di\ufb00erential operator, the issue of boundary conditions is crucial.) INFO More generally, Gaussian integration over \ufb02uctuations around the stationary phase con\ufb01guration obtains the formal expression qf |e\u2212i \u02c6Ht/|qi det i 2\u03c0 \u22022S[qcl] \u2202qi \u2202qf 1/2 i S[qcl], e (3.28) as the \ufb01nal result for the transition amplitude evaluated in the semiclassical approximation. (In cases where there is more than one classical solution, the individual contributions have to be added.) To derive this expression, one shows that the operator controlling the quadratic action (3.27) ful\ufb01ls some di\ufb00erential relations which can again be related back to the classical action. While a detailed formulation of this calculation9 is beyond the scope of the present text, the heuristic interpretation of the result is straightforward, as detailed below", ". According to the rules of quantum mechanics P (qf, qi, t) = |qf |e\u2212i \u02c6Ht/|qi|2 de\ufb01nes the probability density function for a particle injected at coordinate qi to arrive at coordinate qf after a time t. In the semiclassical approximation, the probability density function assumes the form P (qf, qi, t) = | det( 1 )|. We can gain some physical insight into this expression from the 2\u03c0 following consideration: for a \ufb01xed initial coordinate qi, the \ufb01nal coordinate qf (qi, pi) becomes a function of the initial momentum pi. The classical probability density function P (qi, qf ) can \u22022S[qcl] \u2202qi\u2202qf 9 See, e.g., Schulman1. 3.2 Construction of the path integral 111 then be related to the probability density function \u02dcP (qi, pi) for a particle to leave from the initial phase space coordinate (qi, pi) according to P (qi, qf )dqidqf = P (qi, qf (qi, pi)) det \u2202qf \u2202pi dqidpi = \u02dcP (qi, pi)dqidpi. Now, if we say that our particle actually left at the phase space coordinate (qi, pi), \u02dcP must be singular at (qi, pi) while being zero everywhere else. In quantum mechanics, however, all we can say is that our particle was initially con\ufb01ned to a Planck cell centered around (qi, pi) : \u02dcP (qi, pi) = 1/(2\u03c0)d. We thus conclude that P (qi, qf ) = | det(\u2202pi/\u2202qf )|(2\u03c0)\u2212d. Finally, noticing that pi = \u2212\u2202qi S we arrive at the result of the semiclassical analysis above. In deriving Eq. (3.28) we have restricted ourselves to the consideration of quadratic \ufb02uctuations around the classical paths. Under what conditions is this semiclassical approximation justi\ufb01ed? Unfortunately there is no rigorous and generally applicable answer to this question. For \ufb01nite, the quality of the approximation depends largely on the sensitivity of the action to path variations. Whether or not the approximation is legitimate is a question that has to be judged from case to case. However, the asymptotic stability of the semiclassical approximation in the limit \u2192 0 can be deduced simply from power counting. From the structure of Eq. (3.28) it is clear that the typical magnitude of \ufb02uctuations r(t) scales asr \u223c (/\u03b42 q S is a symbolic shorthand for the functional variation of the action. (Variations larger than that lead to phase \ufb02uctuations > 2\u03c0, thereby being negligible.) Non-Gaussian contributions to the action would have the structure \u223c \u22121rn\u03b4n r, this is of the order q S)n/2 \u00d7 n/2\u22121. Since theS -dependent factors are classical (-independent), these \u223c \u03b4n contributions scale to zero as \u2192 0. q S, n >2. For a typical q S)1/2, where\u03b4 2 q S/(\u03b42 This concludes the conceptual part of the chapter. Before turning to the discussion of applications of the path integral, let us \ufb01rst brie\ufb02y summarize the main steps taken in its construction. Construction recipe of the path integral Consider a general quantum transition amplitude \u03c8|e\u2212i \u02c6Ht/|\u03c8, where t may be real, purely imaginary, or generally complex. To construct a functional integral representation of the amplitude: 1. Partition the time interval into N 1 steps, e\u2212i \u02c6Ht/ = # e\u2212i \u02c6H\u0394t/ $N, \u0394t = t/N. 112 Feynman path integral 2. Regroup the operator content appearing in the expansion of each factor e\u2212i \u02c6H\u0394t/ accord- ing to the relation e\u2212i \u02c6H\u0394t/ = 1 + \u0394t cmn \u02c6Am \u02c6Bn + O(\u0394t2), mn where the eigenstates |a, |b of \u02c6A, \u02c6B are known and the coe\ufb03cients cmn are c-numbers. (In the quantum mechanical application above \u02c6A = \u02c6p, \u02c6B = \u02c6q.) This \u201cnormal ordering\u201d procedure emphasizes that distinct quantum mechanical systems may be associated with the same classical action. 3. Insert resolutions", " of identity according to e\u2212i \u02c6H\u0394t/ = |aa| 1 + \u0394t 6 7 cmn \u02c6Am \u02c6Bn + O(\u0394t2) |bb| a,b mn |aa|e\u2212iH(a,b)\u0394t/|bb| + O(\u0394t2), = a,b where H(a, b) is the Hamiltonian evaluated at the eigenvalues of \u02c6A and \u02c6B. 4. Regroup terms in the exponent: due to the \u201cmismatch\u201d of the eigenstates at neighboring time slices n and n + 1, not only the Hamiltonians H(a, b), but also sums over di\ufb00erences of eigenvalues, appear (cf. the last term in the action (3.5)). 5. Take the continuum limit. 3.3 Applications of the Feynman path integral Having introduced the general machinery of path integration we now turn to the discussion of speci\ufb01c applications. Our starting point will be an investigation of a low-energy quantum particle con\ufb01ned to a single potential well, and the phenomenon of tunneling in a double well. With the latter, we become acquainted with instanton techniques and the role of topology in \ufb01eld theory. The ideas developed in this section are generalized further in the investigation of quantum mechanical decay and quantum dissipation. Finally, we turn our attention to the development of the path integral for quantum mechanical spin and, as a case study, explore the semiclassical trace formulae for quantum chaos. The simplest example of a quantum mechanical problem is that of a free particle ( \u02c6H = \u02c6p2/2m). Yet, within the framework of the path integral, this example, which can be dealt with straightforwardly by elementary means, is far from trivial: the Gaussian functional integral engaged in its construction involves divergences which must be regularized by rediscretizing the path integral. Nevertheless, its knowledge will be useful as a means to normalize the path integral in the applications below. Therefore, we leave it as an exercise to show10 Gfree(qf, qi; t) \u2261 qf |e\u2212 i \u02c6p2 2m t|qi\u0398(t) = 1/2 m 2\u03c0it i m 2t (qf \u2212qi)2 e \u0398(t), (3.29) 10 Compare this result with the solution of a classical di\ufb00usion equation. 3.3 Applications of the Feynman path integral 113 where the Heaviside \u0398-function re\ufb02ects causality.11 EXERCISE Derive Eq. (3.29) by the standard methodology of quantum mechanics. (Hint: Insert a resolution of identity and perform a Gaussian integral.) EXERCISE Using the path integral, obtain a perturbative expansion for the scattering amplitude p|U (t \u2192 \u221e, t \u2192 \u2212\u221e)|p of a free particle from a short-range central potential V (r). In particular, show that the \ufb01rst-order term in the expansion recovers the Born scattering amplitude \u2212ie\u2212i(t\u2212t )E(p)/\u03b4(E(p) \u2212 E(p))p|V |p. Quantum particle in a well As a \ufb01rst application of the path integral, let us consider the problem of a quantum particle in a one-dimensional potential well (see \ufb01gure). The discussion of this example illustrates how the semiclassical evaluation scheme discussed above works in practice. For simplicity we assume the potential to be symmetric, V (q) = V (\u2212q) with V (0) = 0. The quantity we wish to compute is the probability amplitude that a particle injected at q = 0 returns after a time t, i.e. with \u02c6H = \u02c6p2/2m + V (\u02c6q), G(0, 0; t) \u2261 qf = 0|e\u2212i \u02c6Ht/|qi = 0\u0398(t). Drawing on our previous discussion, the path integral representation of the transition amplitude is given by G(0, 0; t) = Dq exp q(t)=q(0)=0 i t 0 dt L(q, \u02d9q), where L = m \u02d9q2/2 \u2212 V (q) represents the corresponding Lagrangian. V Now, for a generic potential V (q), the path integral cannot be evaluated exactly. Instead, we wish to invoke the semiclassical analysis outlined above. Accordingly, we must \ufb01rst \ufb01nd solutions to the classical equation of motion. Minimizing the action with respect to variations of q", "(t), one obtains the Euler\u2013 Lagrange equation of motion m\u00a8q = \u2212V (q). According to the Feynman path integral, this equation must q be solved subject to the boundary conditions q(t) = q(0) = 0. One solution is obvious, namely qcl(t) = 0. Assuming that this is in fact the only solution,12 we obtain (cf. Eq. (3.26) \u03c9 11 Motivated by its interpretation as a Green function, in the following we refer to the quantum transition prob- ability amplitude by the symbol G (as opposed to U used above). 12 In general, this assumption is wrong. For smooth potentials V (q), a Taylor expansion of V at small q obtains the harmonic oscillator potential, V (q) =V 0 + m\u03c92q2/2 +\u00b7 \u00b7 \u00b7. For times t that are commensurate with \u03c0/\u03c9, one has periodic solutions, qcl(t) \u221d sin(\u03c9t) that start out from the origin at time t = 0 and revisit it at just the right time t. In the next section we will see why the restriction to just the trivial solution was nonetheless legitimate (for arbitrary times t). 114 and (3.27)) Feynman path integral G(0, 0; t) Dr exp r(0)=r(t)=0 \u2212 i t 0 dtr(t) m 2 t + \u03c92 \u22022 r(t), where, by de\ufb01nition, m\u03c92 \u2261 V (0) is the second derivative of the potential at the origin.13 Note that, in this case, the contribution to the action from the stationary phase \ufb01eld con\ufb01guration vanishes: S[qcl] = 0. Following the discussion of Section 3.2, Gaussian functional integration over r then leads to the semiclassical expansion G(0, 0; t) J det \u2212m(\u22022 t + \u03c92)/2 \u22121/2, (3.30) where the prefactor J absorbs various constant prefactors. Operator determinants are usually most conveniently obtained by presenting them as a product over eigenvalues. In the present case, the eigenvalues n are determined by the equation \u2212 m 2 t + \u03c92 \u22022 rn = nrn, which is to be solved subject to the boundary condition rn(t) = rn(0) = 0. A complete set of solutions to this equation is given by,14 rn(t) = sin(n\u03c0t/t), n = 1, 2,..., with eigenvalues n = m[(n\u03c0/t)2 \u2212 \u03c92]/2. Applying this to the determinant, one \ufb01nds det \u2212m(\u22022 t + \u03c92)/2 \u22121/2 \u221e' = n=1 m 2 2 n\u03c0 t \u22121/2 \u2212 \u03c92. To interpret this result, one must make sense of the in\ufb01nite product (which even seems divergent for times commensurate with \u03c0/\u03c9). Moreover the value of the constant J has yet to be determined. To resolve these di\ufb03culties, one may exploit the facts that (a) we do know the transition amplitude Eq. (3.29) of the free particle system, and (b) the latter coincides with the transition amplitude G in the special case where the potential V \u2261 0. In other words, had we computed Gfree via the path integral, we would have obtained the same constant J and, more importantly, an in\ufb01nite product like the one above, but with \u03c9 = 0. This allows the transition amplitude to be regularized as G(0, 0; t) \u2261 G(0, 0; t) Gfree(0, 0; t) Gfree(0, 0; t) = 2 \u22121/2 \u221e' n=1 \u03c9t n\u03c0 1 \u2212 1/2 m 2\u03c0it \u0398(t). Then, with the help of the identity the result \u221e n=1[1 \u2212 (x/n\u03c0)2]\u22121 = x/ sin x, one \ufb01nally arrives at G(0, 0; t) \" m\u03c9 2\u03c0i sin(\u03c9t) \u0398(t). (3.31) 13 Those who are uncomfortable with functional di\ufb00erentiation can arrive at the same expression simply by sub- stituting q(t) =q cl(t) +r (t) into the action and expanding in r. 14 To \ufb01nd the solutions of this equation, recall the structure of the Schr\u00a8odinger equation of a particle in a", " one- dimensional box of width L = t. 3.3 Applications of the Feynman path integral 115 In the case of the harmonic oscillator, the expansion of the potential necessarily truncates at quadratic order and, in this case, the expression above is exact. (For a more wideranging discussion of the path integral for the quantum harmonic oscillator system, see Problem 3.5.) For a general potential, the semiclassical approximation e\ufb00ectively involves the replacement of V (q) by a quadratic potential with the same curvature. The calculation above also illustrates how coordinate space \ufb02uctuations around a completely static solution may reinstate the zero-point \ufb02uctuations characteristic of quantum mechanical bound states. Double well potential: tunneling and instantons V As a second application of the path integral let us now consider the motion of a particle in a double well potential (see \ufb01gure). Our aim will be to estimate the quantum probability amplitude for a particle either to stay at the bottom of one of the local minima or to go from one minimum to the other. In doing so, it is understood that the energy range accessible to the particle (i.e. \u0394E \u223c /t) is well below the potential barrier height, i.e. quantum mechanical transfer between minima is by tunneling. Here, in contrast to the single well system, it is far from clear what kind of classical stationary phase solutions may serve as a basis for a description of the quantum dynamics; there appear to be no classical paths connecting the two minima. Of course one may think of particles \u201crolling\u201d over the potential hill. Yet, these are singular and, by assumption, energetically inaccessible. q The key to resolving these di\ufb03culties is an observation, already made above, that the time argument appearing in the path integral should be considered as a general complex quantity that can (according to convenience) be set to any value in the complex plane. In the present case, a Wick rotation to imaginary times will reveal a stationary point of the action. At the end of the calculation, the real time amplitudes we seek can be obtained by analytic continuation. INFO The mechanism of quantum double (or multiple) well tunneling plays a role in a number of problems of condensed matter physics. A prominent example is in the physics of amorphous solids such as glasses. A caricature of a glass is shown in the \ufb01gure. The absence of long-range order in the system implies that individual chemical bonds cannot assume their optimal binding lengths. For understretched bonds this leads to the formation of two approximately equal metastable minima around the ideal binding axis (see the inset). The energetically lowest excitations of the system are transitions of individual atoms between nearly degenerate minima of this type, i.e. \ufb02ips of atoms around V d 116 Feynman path integral the binding axis. A prominent phenomenological model15 describes the system by an ensemble of quantum double wells of random center height and width. This model e\ufb00ortlessly explains the existence of a vast system of metastable points in the landscape of low-energy con\ufb01gurations of glassy systems. To be speci\ufb01c, let us consider the imaginary time transition amplitudes exp GE(a, \u00b1a; \u03c4 ) \u2261 = GE(\u2212a, \u2213a; \u03c4 ), \u00b1 a \u02c6H a 9 : \u2212 \u03c4 (3.32) where the coordinates \u00b1a coincide with the two minima of the potential. From Eq. (3.32) the real time amplitudes G(a, \u00b1a; t) =G E(a, \u00b1a; \u03c4 \u2192 it) can be recovered by the analytic continuation \u03c4 \u2192 it. According to Section 3.2, the Euclidean path integral formulation of the transition amplitudes is given by GE(a, \u00b1a; \u03c4 ) = q(0)=\u00b1a,q(\u03c4 )=a Dq exp \u2212 1 \u03c4 0 d\u03c4 m 2 \u02d9q2 + V (q), (3.33) where the function q now depends on imaginary time. From Eq. (3.33) we obtain the stationary phase (or saddle-point) equations \u2212m\u00a8q + V (q) = 0. (3.34) From this result, one can infer that, as a consequence of the Wick rotation, there is an e\ufb00ective inversion of the potential, V \u2192 \u2212V (shown dashed in the \ufb01gure on page 115). The crucial point is that, within the inverted potential landscape, the barrier has become a sink, i.e. within the new formulation, there are classical solutions connecting the two points, \u00b1a. More precisely, there are three di\ufb00erent types of classical solution that ful\ufb01ll the condition to", " be at coordinates \u00b1a at times 0 and/or \u03c4 : (a) the solution wherein the particle rests permanently at a;16 (b) the corresponding solution staying at \u2212a; and, most importantly, (c) the solution in which the particle leaves its initial position at \u00b1a, accelerates through the minimum at 0 and eventually reaches the \ufb01nal position \u2213a at time \u03c4. In computing the transition amplitudes, all three types of path have to be taken into account. As to (a) and (b), by computing quantum \ufb02uctuations around these solutions, one can recover the physics of the zero-point motion described in Section 3.3 for each well individually. (Exercise: Convince yourself that this is true!) Now let us see what happens if the paths connecting the two coordinates are added to this picture. 15 P. W. Anderson, B. I. Halperin, and C. M. Varma, Anomalous low-temperature thermal properties of glasses and spin glasses, Phil. Mag. 25 (1972), 1\u20139. 16 Note that the potential inversion answers a question that arose above, i.e. whether or not the classical solution staying at the bottom of the single well was actually the only one to be considered. As with the double well, we could have treated the single well within an imaginary time representation, whereupon the well would have become a hill. Clearly, the boundary condition requires the particle to start and \ufb01nish at the top of the hill, i.e. the solution that stays there forever. By formulating the semiclassical expansion around that path, we would have obtained Eq. (3.31) with t \u2192 \u2212i\u03c4, which, upon analytic continuation, would have led back to the real time result. 3.3 Applications of the Feynman path integral 117 The instanton gas The classical solution of the Euclidean equation of motion that connects the two potential maxima is called an instanton solution while a solution traversing the same path but in the opposite direction (\u201c\u2212a \u2192 a\u201d \u2192 \u201ca \u2192 \u2212a\u201d) is called an anti-instanton. The name \u201cinstanton\u201d was invented by \u2019t Hooft with the idea that these objects are very similar in their mathematical structure to \u201csolitons,\u201d particle-like solutions of classical \ufb01eld theories. However, unlike solitons, they are structures in time (albeit Euclidean time); thus the \u201cinstant-.\u201d As another etymological remark, note that the syllable \u201c-on\u201d in \u201cinstanton\u201d hints at an interpretation of these states as a kind of particle. The background is that, as a function of the time coordinate, instantons are almost everywhere constant save for a short region of variation (see below). Alluding to the interpretation of time as something akin to a spatial dimension, these states can be interpreted as a well-localized excitation or, according to standard \ufb01eld theoretical practice, a particle.17 Gerardus \u2019t Hooft 1946\u2013 in Nobel Laureate in Physics J. G. 1999, with Martinus elucidating the Veltman, \u201cfor quantum structure elecof troweak interactions in physics.\u201d Together, they were able to identify the properties of the W and Z particles. The \u2019t Hooft\u2013Veltman model allowed scientists to calculate the physical properties of other particles, including the mass of the top quark, which was directly observed in 1995. (Image c The Nobel Foundation.) To proceed, we must \ufb01rst compute the classical action associated with a single-instanton solution. Multiplying Eq. (3.34) by \u02d9qcl, integrating over time (i.e. performing the \ufb01rst integral of the equation of motion), and using the fact that, at qcl = \u00b1a, \u2202\u03c4 qcl = V = 0, one \ufb01nds that m 2 \u02d9q2 cl = V (qcl). (3.35) With this result, one obtains the instanton action Sinst = \u03c4 0 m 2 d\u03c4 m \u02d9q2 cl! \u02d9q2 cl + V (qcl) = d\u03c4 dqcl d\u03c4 (m \u02d9qcl) = a \u2212a dq (2mV (q))1/2. (3.36) Notice that Sinst is determined solely by the functional pro\ufb01le of the potential V (i.e. does not depend on the structure of the solution qcl). Secondly, let us explore the structure of the instanton as a function of time. De\ufb01ning the second derivative of the potential at \u00b1a by V (\u00b1a)", " = m\u03c92, Eq. (3.35) implies that, for large times (where the particle is close to the right maximum), \u02d9qcl = \u2212\u03c9(qcl \u2212 a), which integrates to qcl(\u03c4 ) \u03c4 \u2192\u221e\u2212\u2192 a \u2212 e\u2212\u03c4 \u03c9. Thus the temporal extension of the instanton is set by the oscillator frequencies of the local potential minima (the maxima of the inverted potential) and, in 17 In addition to the original literature, the importance that has been attached to the instanton method has inspired a variety of excellent and pedagogical reviews of the \ufb01eld. Of these, the following are highly recommended: A. M. Polyakov, Quark con\ufb01nement and topology of gauge theories, Nucl. Phys. B120 (1977), 429\u201358 \u2013 see also A. M. Polyakov, Gauge Fields and Strings (Harwood, 1987); S. Coleman, Aspects of Symmetry \u2013 Selected Erice Lectures (Cambridge University Press, 1985), Chapter 7. 118 Feynman path integral cases where tunneling takes place on time scales much larger than that, can be regarded as short (see Fig. 3.4). V \u2013 a a q q a \u2013 a Figure 3.4 Single-instanton con\ufb01guration. \u03c9 \u20131 \u03c4 The con\ufb01nement of the instanton con\ufb01guration to a narrow interval of time has an important implication \u2013 there must exist approximate solutions of the stationary equation involving further anti-instanton/instanton pairs (physically, the particle repeatedly bouncing to and fro in the inverted potential). According to the general philosophy of the saddle-point scheme, the path integral is obtained by summing over all solutions of the saddle-point equations and hence over all instanton con\ufb01gurations. The summation over multi-instanton con\ufb01gurations \u2013 termed the \u201cinstanton gas\u201d \u2013 is substantially simpli\ufb01ed by the fact that individual instantons have short temporal support (events of overlapping con\ufb01gurations are rare) and that not too many instantons can be accommodated in a \ufb01nite time interval (the instanton gas is dilute). The actual density is dictated by the competition between the con\ufb01gurational \u201centropy\u201d (favoring high density), and the \u201cenergetics,\u201d the exponential weight implied by the action (favoring low density) \u2013 see the estimate below. In practice, multi-instanton con\ufb01gurations imply a transition amplitude G(a, \u00b1a; \u03c4 ) n even/odd K n \u03c4 \u03c41 \u03c4n\u22121 d\u03c41 d\u03c42... 0 0 0 d\u03c4n An(\u03c41,..., \u03c4n), (3.37) where An denotes the amplitude associated with n instantons, and we have taken into account the fact that, in order to connect a with \u00b1a, the number of instantons must be even/odd. The n instanton bounces contributing to each An can take place at arbitrary times \u03c4i \u2208 [0, \u03c4 ], i = 1,..., n, and all these possibilities have to be added (i.e. integrated). Here K denotes a (dimensionful) constant absorbing the temporal dimension [time]n introduced by the time integrations, and An(\u03c41,..., \u03c4n) is the transition amplitude, evaluated within the semiclassical approximation around a con\ufb01guration of n instanton bounces at times 0 \u2264 \u03c4n \u2264 \u03c4n\u22121 \u2264 \u00b7 \u00b7 \u00b7 \u2264 \u03c41 \u2264 \u03c4 (see Fig. 3.5). In the following, we \ufb01rst focus on the transition amplitude An, which controls the exponential dependence of the tunneling amplitude, returning later to consider the prefactor K. According to the general semiclassical principle, each amplitude An = An,cl \u00d7 An,qu factorizes into two parts: a classical contribution An,cl accounting for the action of the instanton con\ufb01guration; and a quantum contribution An,qu resulting from quadratic \ufb02uctuations around the classical path. Focusing initially on An,cl we note that, at intermediate times, \u03c4i \u03c4 \u03c4i+1, where the particle rests on top of either of the maxima at \u00b1a, no 3.3 Applications of the Feynman path integral 119 Figure 3.5 Dilute instanton gas con\ufb01guration. action accumulates (cf. the previous section). However, each instanton bounce has a \ufb01nite action Sinst (see Eq. (3.36)) and these contributions add up to give the full classical action, An,cl(\u03c41,.", ".., \u03c4n) = e\u2212nSinst/, (3.38) which is independent of the time coordinates \u03c4i. (The individual instantons \u201cdo not know of each other\u201d; their action is independent of their relative position.) As to the quantum factor An,qu, there are, in principle, two contributions. Whilst the particle rests on either of the hills (the straight segments in Fig. 3.5), quadratic \ufb02uctuations around the classical (i.e. spatially constant) con\ufb01guration play the same role as the quantum \ufb02uctuations considered in the previous section, the only di\ufb00erence being that we are working in a Wick rotated picture. There it was found that quantum \ufb02uctuations around a classical con\ufb01guration which stays for a (real) time t at the bottom of the well result in a factor 1/ sin(\u03c9t) (the remaining constants being absorbed into the prefactor K n). Rotating to imaginary times, t \u2192 \u2212i\u03c4, one can infer that the quantum \ufb02uctuation accumulated during the stationary time \u03c4i+1 \u2212 \u03c4i is given by ; 1 sin(\u2212i\u03c9(\u03c4i+1 \u2212 \u03c4i)) \u223c e\u2212\u03c9(\u03c4i+1\u2212\u03c4i)/2, where we have used the fact that, for the dilute con\ufb01guration, the typical separation times between bounces are much larger than the inverse of the characteristic oscillator scales of each of the minima. (It takes the particle much longer to tunnel through a high barrier than to oscillate in either of the wells of the real potential.) Now, in principle, there are also \ufb02uctuations around the \u201cbouncing\u201d segments of the path. However, due to the fact that a bounce takes a time of O(\u03c9\u22121) \u0394\u03c4, where \u0394\u03c4 represents the typical time between bounces, one can neglect these contributions (which is to say that they can be absorbed into the prefactor K without explicit calculation). Within this approximation, setting \u03c40 \u2261 0, \u03c4n+1 \u2261 \u03c4, the overall quantum \ufb02uctuation correction is given by An,qu(\u03c41,..., \u03c4n) = n' i=0 e\u2212\u03c9(\u03c4i+1\u2212\u03c4i)/2 = e\u2212\u03c9\u03c4 /2, (3.39) 120 Feynman path integral again independent of the particular spacing con\ufb01guration {\u03c4i}. Combining Eq. (3.38) and (3.39), one \ufb01nds that G(a, \u00b1a; \u03c4 ) \u221e K ne\u2212nSinst/e\u2212\u03c9\u03c4 /2 n even/odd = e\u2212\u03c9\u03c4 /2 n even/odd \u03c4 Ke\u2212Sinst/ n. 1 n! \u03c4 \u03c41 \u03c4 n/n! \u03c4n\u22121 d\u03c41 d\u03c42... 0 0 0! d\u03c4n Finally, performing the summation, one obtains the transition amplitude G(a, \u00b1a; \u03c4 ) Ce\u2212\u03c9\u03c4 /2 ) \u03c4 Ke\u2212Sinst/ \u03c4 Ke\u2212Sinst/, cosh sinh * (3.40) (3.41) where C is some factor that depends in a non-exponential way on the transition time. Before we turn to a discussion of the physical content of this result, let us check the self-consistency of our central working hypothesis \u2013 the diluteness of the instanton gas. To this end, consider the representation of G in terms of the partial amplitudes (3.40). To determine the typical number of instantons contributing to the sum, one may make use of the fact that, for a general sum n cn of positive quantities cn > 0, the \u201ctypical\u201d value of the summation index can be estimated as n \u2261 n cn. With the abbreviation X \u2261 \u03c4 Ke\u2212Sinst/ the application of this estimate to our current sum yields n cnn/ n \u2261 n nX n/n! n X n/n! = X, where we have used the fact that, as long as n 1, the even/odd distinction in the sum is irrelevant. Thus, we can infer that the average instanton density, n/\u03c4 = Ke\u2212Sinst/, is both exponentially small in the instanton action Sinst, and independent of \u03c4, con\ufb01rming the validity of our diluteness assumptions above. Finally, let us discuss how the form of the transition amplitude (3.41) can be understood in physical terms. To this end, let us reconsider the basic structure of the problem we are dealing with (see Fig. 3.6). While there is no coupling across the barrier, the Hamiltonian has two independent, oscillator-like sets of low", "-lying eigenstates sitting in the two local minima. Allowing for a weak inter-barrier coupling, the oscillator ground states (like all higher states) split into a doublet of a symmetric and an antisymmetric eigenstate, |S and |A with energies A and S, respectively. Focusing on the low-energy sector formed by the ground state doublet, we can express the transition amplitudes (3.32) as G(a, \u00b1a; \u03c4 ) a| |Se\u2212S \u03c4 /S| + |Ae\u2212A\u03c4 /A| | \u00b1a. Setting A/S = \u03c9/2 \u00b1 \u0394/2, where \u0394 represents the tunnel-splitting, the symmetry properties |a|S|2 = |\u2212a|S|2 = C/2 and a|AA| \u2212a = \u2212|a|A|2 = \u2212C/2 imply that G(a, \u00b1a; \u03c4 ) C 2 e\u2212(\u03c9\u2212\u0394)\u03c4 /2 \u00b1 e\u2212(\u03c9+\u0394)\u03c4 /2 = Ce\u2212\u03c9\u03c4 /2 cosh(\u0394\u03c4 /), sinh(\u0394\u03c4 /). ) 3.3 Applications of the Feynman path integral 121 V \u03c8 A S q q Figure 3.6 Quantum states of the double well. The thick lines indicate energy levels of harmonic oscillator states; the thin and dotted lines indicate exact symmetric (S) and antisymmetric (A) eigenstates. Comparing this expression with Eq. (3.41) the interpretation of the instanton calculation becomes clear: at long times, the transition amplitude engages the two lowest states \u2013 the symmetric and anti-symmetric combinations of the two oscillator ground states. The energy splitting \u0394 accommodates the energy shift due to the tunneling between the two wells. Remarkably, the e\ufb00ect of tunneling was obtained from a purely classical picture (formulated in imaginary time!). The instanton calculation also produced a prediction for the tunnel splitting of the energies, namely \u0394 = K exp(\u2212Sinst/), which, up to the prefactor, agrees with the result of a WKB-type analysis of the tunnel process. Before leaving this section, some general remarks on instantons are in order: In hindsight, was the approximation scheme used above consistent? In particular, terms at second order in were neglected, while terms non-perturbative in (the instanton) were kept. Yet, the former typically give rise to a larger correction to the energy than the latter. However, the large perturbative shift a\ufb00ects the energies of the symmetric and antisymmetric states equally. The instanton contribution gives the leading correction to the splitting of the levels. It is the latter that is likely to be of more physical signi\ufb01cance. Secondly, it may \u2013 legitimately \u2013 appear as though the development of the machinery above was a bit of an \u201coverkill\u201d for describing a simple tunneling process. As a matter of fact, the basic result Eq. (3.41) could have been obtained in a simpler way by more elementary means (using, for example, the WKB method). Why then did we discuss instantons at such length? One reason is that, even within a purely quantum mechanical framework, the instanton formulation of tunneling is much stronger than WKB. The 122 Feynman path integral latter represents, by and large, an uncontrolled approximation. In general it is hard to tell whether WKB results are accurate or not. In contrast, the instanton approximation to the path integral is controlled by a number of well-de\ufb01ned expansion parameters. For example, by going beyond the semiclassical approximation and/or softening the diluteness assumption, the calculation of the transition amplitudes can, in principle, be driven to arbitrary accuracy. A second and, for our purposes, more important motivation is that instanton techniques are of crucial importance within higher-dimensional \ufb01eld theories (here we regard the path integral formulation of quantum mechanics as a (0 space + 1 time) = 1-dimensional \ufb01eld theory). The reason is that instantons are intrinsically non-perturbative objects, which is to say that instanton solutions to stationary phase equations describe a type of physics that cannot be obtained by a perturbative expansion around a non-instanton sector of the theory. (For example, the bouncing orbits in the example above cannot be incorporated into the analysis by doing a kind of perturbative expansion around a trivial orbit.) This non-perturbative nature of instantons can be understood by topological reasoning. Relatedly, one of the features of the instanton analysis above was that the number of instantons involved was a stable quantity; \u201cstable\u201d in the sense that by including perturbative \ufffd", "\ufffductuations around the n-instanton sector, say, one does not connect with the n + 2 sector. Although no rigorous proof of this statement has been given, it should be heuristically clear: a trajectory involving n bounces between the hills of the inverted potential cannot be smoothly connected with one of a di\ufb00erent number. Suppose for instance we should forcibly attempt to interpolate between two paths with di\ufb00erent bounce numbers. Inevitably, some of the intermediate con\ufb01gurations would be charged with actions that are far apart from any stationary phase-like value. Thus, the di\ufb00erent instanton sectors are separated by an energetic barrier that cannot be penetrated by smooth interpolation and, in this sense, they are topologically distinct. INFO Fluctuation determinant: Our analysis above provided a method to extract the tunneling rate between the quantum wells to a level of exponential accuracy. However, in some applications, it is useful to compute the exponential prefactor K. Although such a computation follows the general principles outlined above and implemented explicitly for the single well, there are some idiosyncrasies in the tunneling system that warrant discussion. According to the general principles outlined in Section 3.2, integrating over Gaussian \ufb02uctuations around the saddle-point \ufb01eld con\ufb01gurations, the contribution to the transition amplitude from the n-instanton sector is given by Gn = J det \u2212m\u22022 \u03c4 + V (qcl,n)! e\u2212nSinst, where qcl,n(\u03c4 ) represents an n-instanton con\ufb01guration and J the normalization. Now, in the zeroinstanton sector, the evaluation of the functional determinant recovers the familiar harmonic oscillator result, G(a, a, \u03c4 ) = (m\u03c9/\u03c0) 2 exp[\u2212\u03c9\u03c4 /2]. Let us now consider the one-instanton sector of the theory. To evaluate the functional determinant, one must consider the spectrum \u03c4 + V (qcl,1). Di\ufb00erentiating the de\ufb01ning equation for qcl,1 Eq. (3.34), one of the operator \u2212m\u22022 may con\ufb01rm that 1 \u2212m\u22022 \u03c4 + V (qcl,1) \u2202\u03c4 qcl,1 = 0,! 3.3 Applications of the Feynman path integral 123 i.e. the function \u2202\u03c4 qcl,1 presents a zero mode of the operator! Physically, the origin of the zero mode is elucidated by noting that a translation of the instanton along the time axis, qcl,1(\u03c4 ) \u2192 qcl,1(\u03c4 +\u03b4\u03c4 ), should leave the action approximately invariant. However, for small \u03b4\u03c4, qcl,1(\u03c4 +\u03b4\u03c4 ) qcl,1(\u03c4 ) +\u03b4\u03c4 \u2202 \u03c4 qcl,1, i.e. to \ufb01rst order, the addition of the increment function \u2202\u03c4 qcl,1 leaves the action invariant, and \u03b4\u03c4 is a \u201czero mode coordinate.\u201d With this interpretation, it becomes clear how to repair the formula for the \ufb02uctuation determinant. While the Gaussian integral over \ufb02uctuations is controlled for the non-zero eigenvalues, its execution for the zero mode must be rethought. Indeed, by integrating over the coordinate \u03c4 of the instanton, that is 0 d\u03c40 = \u03c4, one \ufb01nds that the contribution to the transition amplitude in the one-instanton sector is given by det e\u2212Sinst, \u03c4 + V (qcl,1) \u2212m\u22022 \u22121/2 J\u03c4 Sinst 2\u03c0 where the prime indicates the exclusion of the zero mode from the determinant, and the factor Sinst/2\u03c0 re\ufb02ects the Jacobian associated with the change to a new set of integration variables which contains the zero mode coordinate \u03c4 as one of its elements.18 To \ufb01x the, as yet, undetermined coupling constant J, we normalize by the \ufb02uctuation determinant of the (imaginary time) harmonic oscillator, i.e. we use the fact that (cf. Section 3.3), for the harmonic oscillator, \u03c4 + \u03c92)/2)\u22121/2 = (m\u03c9/\u03c0)1/2 e\u2212\u03c9\u03c4 /2, the return amplitude evaluates to G(a, a, \u03c4 ) = J det(m(\u2212\u22022 where the \ufb01rst/second representation is the imaginary time variant of Eq. (3.30)/Eq.(3.31). Using this result, and noting that the zero mode analysis above generalizes to the n-instanton sector, we \ufffd", "\ufffdnd that the pre-exponential constant K used in our analysis of the double-well problem above a\ufb00ords the explicit representation \" Sinst 2\u03c0 K = \u03c9 m\u03c92det \u2212m\u22022 det [\u2212m\u22022 \u03c4 + V (qcl,1) \u03c4 + m\u03c92] #\u22121/2. Naturally, the instanton determinant depends sensitively on the particular nature of the potential V (q). For the quartic potential V (q) = m\u03c92(x2 \u2212 a2)2/8a2, it may be con\ufb01rmed that m\u03c92det \u2212m\u22022 det [\u2212m\u22022 \u03c4 + V (qcl,1) \u03c4 + m\u03c92] = 1 12, while Sinst = (2/3)m\u03c9a2. For further details of the calculation, we refer to, e.g., Zinn-Justin (1993). 18 Escape from a metastable minimum: \u201cbounces\u201d The instanton gas approximation for the double-well system can be easily adapted to explore the problem of quantum mechanical tunneling from a metastable state such as that presented by an unstable nucleus. In particular, suppose one wishes to estimate the \u201csurvival probability\u201d of a particle captured in a metastable minimum of a one-dimensional potential such as that shown in Fig. 3.7. According to the path integral scheme, the survival probability, de\ufb01ned by the probability amplitude of remaining at the potential minimum qm, i.e. the propagator G(qm, qm; t), can 18 See J. Zinn-Justin, Quantum Field Theory and Critical Phenomena (Oxford University Press, 1993) for an explicit calculation of this Jacobian. 124 Feynman path integral V \u2013V q qm q q qm \u03c4 Figure 3.7 E\ufb00ective potential showing a metastable minimum together with the inverted potential and a sketch of a bounce solution. To obtain the tunneling rate it is necessary to sum over a dilute gas of bounce trajectories. be evaluated by making use of the Euclidean time formulation of the Feynman path integral. As with the double well, in the Euclidean time formalism the dominant contribution to the transition probability arises from the classical path minimizing the action corresponding to the inverted potential (see Fig. 3.7). However, in contrast to the double-well potential, the classical solution takes the form of a \u201cbounce\u201d (i.e. the particle spends only a short time away from the potential minimum \u2013 there is only one metastable minimum of the potential). As with the double well, one can expect multiple bounce trajectories to present a signi\ufb01cant contribution. Summing over all bounce trajectories (note that in this case we have an exponential series \u2013 no even/odd parity e\ufb00ect), one obtains the survival probability G(qm, qm; \u03c4 ) = Ce\u2212\u03c9\u03c4 /2 exp # \u03c4 Ke\u2212Sbounce/ $. Applying an analytic continuation to real time, one \ufb01nds G(qm, qm; t) = Ce\u2212i\u03c9t/2 exp[\u2212 \u0393 2 t], where the decay rate is given by \u0393/2 =| K|e\u2212Sbounce/. (Note that on physical grounds we can see that K must be imaginary.19) EXERCISE Consider a heavy nucleus having a \ufb01nite rate of \u03b1-decay. The nuclear forces can be considered very short-range so that the rate of \u03b1 particle emission is controlled by tunneling under a Coulomb barrier. Taking the e\ufb00ective potential to be spherically symmetric with a deep minimum core of radius r0 beyond which it decays as U (r) = 2(Z \u2212 1)e2/r where Z is the nuclear charge, \ufb01nd the temperature of the nuclei above which \u03b1-decay will be thermally assisted if the energy of the emitted particles is E0. Estimate the mean energy of the \u03b1 particles as a function of temperature. EXERCISE A uniform electric \ufb01eld E is applied perpendicular to the surface of a metal with work function W. Assuming that the electrons in the metal describe a Fermi gas of density n, with exponential accuracy, \ufb01nd the tunneling current at zero temperature (\u201ccold emission\u201d). Show that, e\ufb00ectively, only electrons with energy near the Fermi level are tunneling. With the same accuracy, \ufb01nd the current at \ufb01nite temperature (\u201chot emission\u201d). What is the most probable energy of tunneling electrons as a function of temperature? 19 In fact, a more careful analysis shows that this estimate of the", " decay rate is too large by a factor of 2 (for further details see, e.g., the discussion in Coleman.17 3.3 Applications of the Feynman path integral 125 V x \u2013a a \u03c6 Figure 3.8 Snapshot of a \ufb01eld con\ufb01guration \u03c6(x, t = const.) in a potential landscape with two nearly degenerate minima. For further discussion, see the text. Tunneling of quantum \ufb01elds: \u201cfate of the false vacuum\u201d Hitherto we have focused on applications of the Feynman path integral to the quantum mechanics of isolated point-like particles. In this setting, the merit of the path integral scheme over, say, standard perturbative methods or the \u201cWKB\u201d approach is perhaps not compelling. Therefore, by way of motivation, let us present an example that builds on the structures elucidated above and illustrates the power of the path integral method. To this end, let us consider a theory involving a continuous classical \ufb01eld that can adopt two homogeneous equilibrium states with di\ufb00erent energy densities. To be concrete, one may consider a harmonic chain con\ufb01ned to one or other minimum of an asymmetric quasi-onedimensional \u201cgutter-like\u201d double-well potential (see Fig. 3.8). When quantized, the state of higher energy density becomes unstable through barrier penetration \u2013 it is said to be a \u201cfalse vacuum\u201d.20 Speci\ufb01cally, drawing on our discussion of the harmonic chain in Chapter 1, let us consider a quantum system speci\ufb01ed by the Hamiltonian density \u02c6H = \u02c6\u03c02 2m + ksa2 2 (\u2202x \u02c6\u03c6)2 + V ( \u02c6\u03c6), (3.42) where [\u02c6\u03c0(x), \u02c6\u03c6(x)] = \u2212i\u03b4(x \u2212 x). Here we have included a potential V (\u03c6) which, in the present case, assumes the form of a double well. The inclusion of a weak bias \u2212f \u03c6 in V (\u03c6) identi\ufb01es a stable and a metastable potential minimum. Previously, we have seen that, in the absence of the con\ufb01ning potential, the quantum string exhibits low-energy collective wave-like excitations \u2013 phonons. In the con\ufb01ning potential, these harmonic \ufb02uctuations are rendered massive. However, drawing on the quantum mechanical principles established in the single-particle system, one might assume that the string tunnels freely between the two potential minima. To explore the capacity of the system to tunnel, let us suppose that, at 20 For a detailed discussion of the history and rami\ufb01cations of this idea, we refer to the original insightful paper by Sidney Coleman, Fate of the false vacuum: semiclassical theory, Phys. Rev. D 15 (1977), 2929\u201336. In fact, many of the ideas developed in this work were anticipated in an earlier analysis of metastability in the context of classical \ufb01eld theories by J. S. Langer, Theory of the condensation point, Ann. Phys. (NY) 41 (1967), 108\u201357. 126 Feynman path integral some time t = 0, the system adopts a \ufb01eld con\ufb01guration in which the string is located in the (metastable) minimum of the potential at, say, \u03c6 = \u2212a. What is the probability that the entire string of length L will tunnel across the barrier into the potential minimum at \u03c6 = a in a time t? INFO The tunneling of \ufb01elds between nearly degenerate ground states plays a role in numerous physical contexts. By way of example, consider a superheated liquid. In this context, the \u201cfalse\u201d vacuum is the liquid state, the true one the gaseous phase. The role of the \ufb01eld is taken by the local density distribution in the liquid. Thermodynamic \ufb02uctuations trigger the continuous appearance of vapor bubbles in the liquid. For bubbles of too small a diameter, the gain in volume energy is outweighed by the surface energy cost \u2013 the bubble will collapse. However, for bubbles beyond a certain critical size, the energy balance is positive. The bubble will grow and, eventually, swallow the entire mass density of the system; the liquid has vaporized or, more formally, the density \ufb01eld tunneled21 from the false ground state into the true one. More speculative (but also potentially more damaging) manifestations of the phenomenon have been suggested in the context of cosmology:22 what if the big bang released our universe not into its true vacuum con\ufb01guration but into", " a state separated by a huge barrier from a more favorable sector of the energy landscape. In this case, everything would depend on the tunneling rate: If this time scale is of the order of milliseconds, the universe is still hot when the false vacuum decays... if this time is of the order of years, the decay will lead to a sort of secondary big bang with interesting cosmological consequences. If this time is of the order of 109 years, we have occasion for anxiety. (S. Coleman) Previously, for the point-particle system, we have seen that the transition probability between the minima of the double well is most easily accessed by exploring the classical \ufb01eld con\ufb01gurations of the Euclidean time action. In the present case, anticipating to some extent our discussion of the quantum \ufb01eld integral in the next chapter, the Euclidean time action associated with the Hamiltonian density (3.42) assumes the form23 T L S[\u03c6] = d\u03c4 dx 0 0 m 2 (\u2202\u03c4 \u03c6)2 + ksa2 2 (\u2202x\u03c6)2 + V (\u03c6), where the time integral runs over the interval [0, T = it]. Here, for simplicity, let us assume that the string obeys periodic boundary conditions in space, namely \u03c6(x+L, \u03c4 ) \u2261 \u03c6(x, \u03c4 ). To estimate the tunneling amplitude, we will explore the survival probability of the metastable state, imposing the boundary conditions \u03c6(x, \u03c4 = 0) = \u03c6(x, \u03c4 = T ) =\u2212a on the path integral. Once again, when the potential barrier is high, and the time T is long, one may assume that the path integral is dominated by the saddle-point \ufb01eld con\ufb01guration of the 21 At this point, readers should no longer be confused regarding the mentioning of \u201ctunneling\u201d in the context of a classical system. Within the framework of the path integral, the classical partition sum maps onto the path integral of a \ufb01ctitious quantum system. It is this tunneling that we have in mind. 22 See note 25. 23 Those readers who wish to develop a more rigorous formulation of the path integral for the string may either turn to the discussion of the \ufb01eld integral in the next chapter or, alternatively, may satisfy themselves of the validity of the Euclidean action by (re-)discretizing the harmonic chain, presenting the transition amplitude as a series of Feynman path integrals for each element of the string and, \ufb01nally, taking the continuum limit. 3.3 Applications of the Feynman path integral 127 Euclidean action. In this case, varying the action with respect to the \ufb01eld \u03c6(x, \u03c4 ), one obtains the classical equation of motion m\u22022 \u03c4 \u03c6 + ksa2\u22022 x\u03c6 = \u2202\u03c6V (\u03c6), which must be solved subject to the boundary conditions above. Now, motivated by our consideration of the point-particle problem, one might seek a solution in which the string tunnels as a single rigid entity without \u201c\ufb02exing.\u201d However, it is evident from the spatial translational invariance of the system that the instanton action would scale with the system size L. In the in\ufb01nite system L \u2192 \u221e, such a trajectory would therefore not contribute signi\ufb01cantly to the tunneling amplitude. Instead, one must consider a di\ufb00erent type of \ufb01eld con\ufb01guration in which the transfer of the chain is by degree. In this, x elements of the string cross the barrier in a consecutive sequence as two outwardly propagating \u201cdomain walls\u201d (see the \ufb01gure, where the emergence of such a double-kink con\ufb01guration is shown as a function of space and time; notice the circular shape of the resulting space-time droplet \u2013 a consequence of the rotational symmetry of the rescaled problem). Such a \ufb01eld con\ufb01guration can be motivated from symmetry considerations by noting that, after rescaling x \u2192 ksa2/m denotes the classical vsx (where vs = sound wave velocity), the saddle-point equation assumes the isotropic form m\u22022\u03c6 = \u2202\u03c6V (\u03c6), x2 + (\u03c4 \u2212 T /2)2, and sending (T, L) \u2192 \u221e, the where \u22022 = \u22022 space-time rotational symmetry suggests a solution of the form \u03c6 = \u03c6(r), where \u03c6(r) obeys the radial di\ufb00usion equation x. Then, setting r = a \u03c4 +", " \u22022 \u2013a \u03c6 \u03c4 m r with the boundary condition limr\u2192\u221e \u03c6(r) = \u2212a. This equation describes the onedimensional motion of a particle in a potential \u2212V and subject to a strange \u201cfriction force\u201d \u2212mr\u22121\u2202r\u03c6 whose strength is inversely proportional to \u201ctime\u201d r. \u2202r\u03c6 = \u2202\u03c6V, r \u03c6 + m\u22022 To understand the pro\ufb01le of the bounce solution, suppose that at time r = 0 the particle has been released from rest at a position slightly to the left of the (inverted) potential maximum at a. After rolling through the potential minimum it will climb the potential hill at \u2212a. Now, the initial position may be \ufb01ne-tuned in such a way that the viscous damping of the particle compensates for the excess potential energy (which would otherwise make the particle overshoot and disappear to in\ufb01nity): there exists a solution where the particle starts close to \u03c6 = a and eventually winds up at \u03c6 = \u2212a, in accord with the imposed boundary conditions. In general, the analytical solution for the bounce depends sensitively on the form of the con\ufb01ning potential. However, while we assume that the well asymmetry imposed by external potential \u2212f \u03c6 is small, the radial equation may be considerably simpli\ufb01ed. In this limit, one may invoke a \u201cthin-wall\u201d approximation in which one assumes that the bounce 128 Feynman path integral con\ufb01guration is described by a domain wall of thickness \u0394r, at a radius r0 \u0394r separating an inner region where \u03c6(r < r0) = a from the outer region where \u03c6(r > r0) = \u2212a. In this case, and to lowest order in an expansion in f, the action of the friction force is immaterial, i.e. we may set m\u22022 r \u03c6 = \u2202\u03c6V \u2013 the very instanton equation formulated earlier for the point-particle system! Then, substituting back into S, one \ufb01nds that the bounce (or kink-like) solution is char- acterized by the Euclidean action S = vs % 2\u03c0r0Sinst \u2212 \u03c0r2 02af &, where Sinst denotes the action of the instanton associated with the point-particle system Eq. (3.36), and the last term accommodates the e\ufb00ect of the potential bias on the \ufb01eld con\ufb01guration. Crucially, one may note that the instanton contribution to the action scales with the circumference of the domain wall in the space-time, while that of the potential bias scales with the area of the domain. From this scaling dependence, it is evident that, however small the external force f, at large enough r0 the contribution of the second term will always outweigh the \ufb01rst and the string will tunnel from the metastable minimum to the global minimum of the potential. More precisely, the optimal size of domain is found by minimizing the action with respect to r0. In doing so, one \ufb01nds that r0 = Sinst/2af. Then, substituting back into the action, one obtains the tunneling rate \u0393 \u223c exp \u2212 1 \u03c0vsS2 2af inst. From this result, one can conclude that, in the absence of an external force f, the tunneling of the string across the barrier is completely quenched! In the zero-temperature unbiased system, the symmetry of the quantum Hamiltonian is broken: the ground state exhibits a two-fold degeneracy in which the string is con\ufb01ned to one potential minimum or another. The rami\ufb01cations of the tunneling amplitude suppression can be traced to the statistical mechanics of the corresponding classical system. As emphasized in Section 3.2, any Euclidean time path integral of a d-dimensional system can be identi\ufb01ed with the statistical mechanics of a classical system (d + 1)-dimensional problem. In the double-well system, the Euclidean time action of the point-particle quantum system is isomorphic to the onedimensional realization of the classical Ising ferromagnet, namely \u03b2HIsing = L dx 0 t 2 m2 + um4 + K 2 (\u2207m)2. (3.43) Translated into this context, the saddle-point (or mean-\ufb01eld) analysis suggests that the system will exhibit a spontaneous symmetry breaking to an ordered phase (m = 0) when the parameter t (the reduced temperature) becomes negative. However, drawing on our analysis of the quantum point-particle system, in the thermodynamic limit", ", we see that \ufb02uctuations (non-perturbative in temperature) associated with instanton \ufb01eld con\ufb01gurations of the Hamiltonian m(x) may restore the symmetry of the system and destroy long-range order at any \ufb01nite temperature 1/\u03b2. Whether this happens or not depends on the competition between the energy cost of instanton creation and the entropy gained by integrating over the instanton zero-mode coordinates. It turns out that, in d = 1, the latter wins, i.e. 3.3 Applications of the Feynman path integral 129 the system is \u201cdisordered\u201d at any \ufb01nite temperature. In contrast, for d \u2265 2, the creation of instantons is too costly, i.e. the system will remain in its energetically preferred ground state. For further discussion of these issues, we refer to the Info block in Section 8.1.1. Tunneling in a dissipative environment In the condensed matter context it is, of course, infeasible to completely divorce a system from its environment. Indeed, in addition to the dephasing e\ufb00ect of thermal \ufb02uctuations, the realization of quantum mechanical phenomena depends sensitively on the strength and nature of the coupling to the external degrees of freedom. For example, the tunneling of an atom from one interstitial site in a crystal to another is likely to be heavily in\ufb02uenced by its coupling to the phonon degrees of freedom that characterize the crystal lattice. By exchanging energy with the phonons, which act in the system as an external bath, a quantum particle can lose its phase coherence and with it, its quantum mechanical character. Beginning with the seminal work of Caldeira and Leggett,24 there have been numerous theoretical investigations of the e\ufb00ect of an environment on the quantum mechanical properties of a system. Such e\ufb00ects are particularly acute in systems where the quantum mechanical degree of freedom is macroscopic, such as the magnetic \ufb02ux trapped in a superconducting quantum interference device (SQUID). In the following, we show that the Feynman path integral provides a natural (and almost unique) setting in which the e\ufb00ects of the environment on a microscopic or macroscopic quantum mechanical degree of freedom can be explored. For further discussion of the response of quantum wave coherence to environmental coupling, we refer to Chapter 11. Before we begin, let us note that the phenomenon of macroscopic quantum tunneling represents an extensive and still active area of research recently reinvigorated by the burgeoning \ufb01eld of quantum computation. By contrast, our discussion here will be necessarily limited in scope, targeting a particular illustrative application, and highlighting only the guiding principles. For a more thorough and detailed discussion, we refer the reader to one of the many comprehensive reviews.25 Caldeira\u2013Leggett model Previously, we have discussed the ability of the Feynman path integral to describe quantum mechanical tunneling of a particle q across a potential barrier V (q). In the following, we will invoke the path integral to explore the capacity for quantum mechanical tunneling when the particle is coupled to degrees of freedom of an external environment. Following Caldeira and 24 A. O. Caldeira and A. J. Leggett, In\ufb02uence of dissipation on quantum tunneling in macroscopic systems, Phys. Rev. Lett. 46 (1981), 211\u201314. 25 See, e.g., A. J. Leggett et al., Dynamics of the dissipative two-state system, Rev. Mod. Phys 59 (1976), 1\u201385, and U. Weiss, Quantum Dissipative Systems (World Scienti\ufb01c Publishing, 1993). 130 Feynman path integral Leggett\u2019s original formulation, let us represent the environment by a bath of quantum harmonic oscillators characterized by a set of frequencies {\u03c9\u03b1}, \u02c6Hbath[q\u03b1] = N \u03b1 \u02c6p2 \u03b1 2m\u03b1 + m\u03b1 2 \u03b1q2 \u03c92 \u03b1, where N is the number of bath-oscillators. For simplicity, let us suppose that, in the leading approximation, the particle\u2013bath coupling is linear in the bath coordinates and such that N \u02c6Hc[q, q\u03b1] =\u2212 \u03b1 f\u03b1[q]q\u03b1, where f\u03b1[q] represents some function of the particle coordinate q. Expressed as a Feynman path integral, the survival probability of a particle con\ufb01ned to a metastable minimum at a position q = a, and coupled to an external environment, can then be expressed as ( = 1) a|e\u2212i \u02c6Ht/|a = Dq eiSpart.[q] Dq", "\u03b1 eiSbath[q\u03b1]+iSc[q,q\u03b1], q(0)=q(t)=a where \u02c6H = \u02c6Hpart + \u02c6Hbath + \u02c6Hc denotes the total Hamiltonian of the system, $ # Spart[q] = 0 t dt m 2 \u02d9q2 \u2212 V (q), Sbath[q\u03b1] = t dt 0 \u03b1 % m\u03b1 2 \u02d9q2 \u03b1 \u2212 \u03c92 \u03b1q2 \u03b1 &, denote, respectively, the actions of the particle and bath, while Sc[q, q\u03b1] = \u2212 t 0 dt 6 \u03b1 f\u03b1[q]q\u03b1 + 7 a f\u03b1[q]2 2ma\u03c92 a represents their coupling.26 Here we assume that the functional integral over q\u03b1(t) is taken over all \ufb01eld con\ufb01gurations of the bath while, as before, the path integral on q(t) is subject to the boundary conditions q(0) = q(t) = a. To reveal the e\ufb00ect of the bath on the capacity for tunneling of the particle, it is instructive to integrate out \ufb02uctuations q\u03b1 and thereby obtain an e\ufb00ective action for q. Being Gaussian in the coordinates q\u03b1, the integration can be performed straightforwardly. Although not crucial, since we are dealing with quantum mechanical tunneling, it is useful to transfer to the Euclidean time representation. Taking the boundary conditions on the \ufb01elds q\u03b1(\u03c4 ) to be periodic on the interval [0, T \u22121 \u2261 \u03b2], it may be con\ufb01rmed that the Gaussian functional integral over q\u03b1 induces a time non-local interaction of the particle (exercise) a|e\u2212i \u02c6Ht/|a = Dq e\u2212Seff [q] where a constant of integration has been absorbed into the measure and Se\ufb00 [q] = Spart[q] + 1 2T \u03c9n,\u03b1 nf\u03b1[q(\u03c9n)]f\u03b1[q(\u2212\u03c9n)] \u03c92 m\u03b1\u03c92 \u03b1(\u03c92 \u03b1 + \u03c92 n). 26 The second term in the coupling action has been added to keep the e\ufb00ect of the environment minimally invasive (purely dissipative). If it were not present, the coupling to the oscillator degrees of freedom would e\ufb00ectively shift the extremum of the particle potential, i.e. change its potential landscape. (Exercise: Substitute the solutions of the Euler\u2013Lagrange equations \u03b4q\u03b1 S[q, q\u03b1] = 0 [computed for a \ufb01xed realization of q] into the action to obtain the said shift.) 3.3 Applications of the Feynman path integral 131 \u03c9n Here, the sum runs over the discrete set of Fourier frequencies \u03c9n = 2\u03c0nT with n integer.27 By integrating out of the bath degrees of freedom, the particle action acquires an induced contribution. To explore its e\ufb00ect on dissipation and tunneling, it is necessary to specialize our discussion to a particular form of coupling. In the particular case that the coupling to the bath is linear, i.e. f\u03b1[q(\u03c4 )] = c\u03b1q(\u03c4 ), the e\ufb00ective action assumes the form (exercise) Se\ufb00 [q] = Spart[q] \u2212 T \u03b2 0 d\u03c4 d\u03c4 K(\u03c4 \u2212 \u03c4 )q(\u03c4 )q(\u03c4 ), where K(\u03c4 ) = \u221e 0 d\u03c9 \u03c0 J(\u03c9)D\u03c9(\u03c4 ), J(\u03c9) = \u03c0 2 D\u03c9(\u03c4 ) = \u2212 \u03b4(\u03c9 \u2212 \u03c9\u03b1) and \u03b1 c2 \u03b1 m\u03b1\u03c9\u03b1 \u03c92 n \u03c9(\u03c92 + \u03c92 n) \u03c9n ei\u03c9n\u03c4, resembles the Green function of a boson with energy \u03c9. Physically, the non-locality of the action is easily understood. By exchanging \ufb02uctuations with the external bath, a particle can e\ufb00ect a self-interaction, retarded in time. Taken as a whole, the particle and the bath maintain quantum phase coherence. However, when projected onto the particle degree of freedom, the total energy of the system appears to \ufb02uctuate and the phase coherence of the particle transport is diminished. To explore the properties of the dissipative action, it is helpful to separate the non-local interaction according to the identity q(\u03c4 )q(\u03c4 ) = [q2(\u03c4 ) + q2(\u03c4 )]/2 \u2212 [q(\u03c4 ) \u2212 q(\u03c4 )]2/2. The", " \ufb01rst square-bracketed contribution presents an innocuous renormalization of the potential V (q) and, applying equally to the classically allowed motion and quantum tunneling, presents an unobservable perturbation. Therefore, we will suppose that its e\ufb00ect has been absorbed into a rede\ufb01nition of the particle potential V (q). By contrast, the remaining contribution is always positive. Either the particular form of the \u201cspectral function\u201d J(\u03c9) may be obtained from an a priori knowledge of the microscopic interactions of the bath or, phenomenologically, it can be inferred from the structure of the classical damped equations of motion. For example, for a system subject to an \u201cohmic\u201d dissipation (where, in real time, the classical equations of motion obtain a dissipative term \u2212\u03b7 \u02d9q with a \u201cfriction coe\ufb03cient\u201d \u03b7), one has J(\u03c9) = \u03b7|\u03c9| for all frequencies smaller than some characteristic cut-o\ufb00 (at the scale of the inverse Drude relaxation time of the environment). By contrast, for a defect in a three-dimensional crystal, interaction with acoustic phonons presents a frequency dependence of \u03c93 or \u03c95 depending on whether \u03c9 is below or above the Debye frequency. INFO Consider, for example, the coupling of a particle to a continuum of bosonic modes whose spectral density J(\u03c9) = \u03b7\u03c9 grows linearly with frequency. In this case, K(\u03c9n) = \u221e 0 \u03c92 n \u03c0 d\u03c9 J(\u03c9) \u03c9(\u03c92 + \u03c92 n) = \u03b7 2 |\u03c9n|, 27 More precisely, anticipating our discussion of the Matsubara frequency representation, we have de\ufb01ned m qmei\u03c9m\u03c4, qm = the Fourier decomposition on the Euclidean time interval T, namely q(\u03c4 ) = T 0 d\u03c4 q(\u03c4 )e\u2212i\u03c9m \u03c4, where\u03c9 m = 2\u03c0m/\u03b2 with m integer. \u03b2 132 Feynman path integral describes Ohmic dissipation of the particle. Fourier transforming this expression we obtain K(\u03c4 ) = \u03c0T \u03b7 2 1 sin2(\u03c0T \u03c4 ) \u22121 \u03c4T \u03b7 2\u03c0T 1 \u03c4 2, (3.44) i.e. a strongly time non-local \u201cself-interaction\u201d of the particle. Disssipative quantum tunneling To return to the particular problem at hand, previously we have seen that the tunneling rate of a particle from a metastable potential minimum can be inferred from the extremal \ufb01eld con\ufb01gurations of the Euclidean action: the bounce trajectory. To explore the e\ufb00ect of the dissipative coupling, it is necessary to understand how it revises the structure of the bounce solution. Now, in general, the non-local character of the interaction inhibits access to an exact solution of the classical equation of motion. In such cases, the e\ufb00ect of the dissipative coupling can be explored perturbatively or with the assistance of the renormalization group (see the discussion in Section 8.2). However, by tailoring our choice of potential V (q), we can gain some intuition about the more general situation. To this end, let us consider a particle of mass m con\ufb01ned in a metastable minimum by a (semi-in\ufb01nite) harmonic potential trap (see \ufb01gure), V (q) = c q2/2, m\u03c92 \u2212\u221e, 0 < |q| \u2264 a, |q| > a. Further, let us assume that the environment imparts an ohmic dissipation with a damping or viscosity \u03b7. To keep our discussion general, let us consider the combined impact of dissipation and temperature on the rate of tunneling from the potential trap. To do so, following Langer28 it is natural to investigate the \u201cquasi-equilibrium\u201d quantum partition function Z of the combined system. In this case, the tunneling rate appears as an imaginary contribution to the free energy F = \u2212T ln Z, namely \u0393 = \u2212(2/)Im F. By drawing on the path integral, the quantum partition function of the system can be q(\u03b2)=q(0) Dq e\u2212Seff / where, as we have seen above, presented as a functional integral Z = for ohmic coupling, the Euclidean action assumes the form V Se\ufb00 [q] = m 2 d\u03c4 \u03b2 0 + \u03b7 4\u03c0 0 \u02d9q2 + V (q) \u03b2 d\u03c4 d\u03c4 q(\u03c4 ) \u2212 q(\u03c4 ) \u03c4 \u2212 \u03c4 2. a q", " Once again, to estimate the tunneling rate, we will suppose that the barrier is high and the temperature is low, so that the path integral is dominated by stationary con\ufb01gurations of the action. In this case, one may identify three distinct solutions. In the \ufb01rst place, the particle may 28 J. S. Langer, Theory of the condensation point, Ann. Phys. (NY) 41 (1967), 108\u201357. 3.3 Applications of the Feynman path integral 133 remain at q = 0 poised precariously on the maximum of the inverted harmonic potential. Contributions from this solution and the associated harmonic \ufb02uctuations reproduce terms in the quantum partition function associated with states of the closed harmonic potential trap. Secondly, there exists a singular solution in which the particle remains at the minimum of the inverted potential, i.e. perched on the potential barrier. The latter presents a negligible contribution to the quantum partition function and can be neglected. Finally, there exists a bounce solution in which the particle injected at a position q inside the well accelerates down the inverted potential gradient, is re\ufb02ected from the potential barrier, and returns to the initial position q in a time \u03b2. While, in the limit \u03b2 \u2192 \u221e, the path integral singles out the boundary condition q(0) = q(\u03b2) \u2192 0, at \ufb01nite \u03b2 the boundary condition will depart from 0 in a manner that depends non-trivially on the temperature. It is this general bounce solution that governs the decay rate. Since, in the inverted potential, the classical bounce trajectory stays within the interval over which the potential is quadratic, a variation of the Euclidean action with respect to q(\u03c4 ) obtains the classical equation of motion \u2212m\u00a8q + m\u03c92 c q + \u03b2 0 \u03b7 \u03c0 d\u03c4 q(\u03c4 ) \u2212 q(\u03c4 ) (\u03c4 \u2212 \u03c4 )2 = A\u03b4(\u03c4 \u2212 \u03b2/2), where the term on the right-hand side of the equation imparts an impulse that changes discontinuously the velocity of the particle, while the coe\ufb03cient A is chosen to ensure symmetry of the bounce solution on the Euclidean time interval. Turning to the Fourier representation, the solution of the saddle-point equation then assumes the form qn = AT e\u2212i\u03c9n\u03b2/2g(\u03c9n), g(\u03c9n) \u2261 [m(\u03c92 n + \u03c92 c ) +\u03b7 |\u03c9n|]\u22121. Imposing the condition that q(\u03c4 = \u03b2/2) = a, one \ufb01nds that A = a/f where f \u2261 T Finally, the action of the bounce is given by Sbounce = 1 2T n (m(\u03c92 n + \u03c92 c ) +\u03b7 |\u03c9m|)|qn|2 = a2 2f. (3.45) n g(\u03c9n). (3.46) 1. To make sense of these expressions, as a point of reference, let us \ufb01rst determine the zero-temperature tunneling rate in the absence of dissipation, that is \u03b7 \u2192 0 and \u03b2 \u2192 \u221e. In this case, the (Matsubara) frequency summation translates to the continuous \u221e \u2212\u221e(d\u03c9/2\u03c0)g(\u03c9) = (2m\u03c9c)\u22121. Using this result, the bounce action (3.46) integral, f = takes the form Sbounce = m\u03c9ca2. As one would expect, the tunneling rate \u0393 \u223c e\u2212Sbounce is controlled by the ratio of the potential barrier height m\u03c92 c a2/2 to the attempt frequency \u03c9c. Also notice that the bounce trajectory is given by q(\u03c4 ) = a f \u221e \u2212\u221e d\u03c9 2\u03c0 ei\u03c9(\u03c4 \u2212\u03b2/2)g(\u03c9) = a e\u2212\u03c9c|\u03c4 \u2212\u03b2/2|, i.e., as expected from our discussion in section 3.3, the particle spends only a time 1/\u03c9c in the under-barrier region. 2. Now, restricting attention to the zero-temperature limit, let us consider the in\ufb02uence of dissipation on the nature of the bounce solution and the capacity for tunneling. Focusing on the limit in which the dynamics of the particle is overdamped, \u03b7 m\u03c9c, 134 Feynman path integral \u221e \u2212\u221e d\u03c9g(\u03c9) (2/\u03c0\u03b7) ln (\u03b7/m\u03c9c), which implies Sbounce = \u03c0\u03b7a2/(4 ln[\u03b7/(m\u03c9c)]). In f = particular, this result shows that, in the limit \u03b7 \u2192 \u221e, the coupling of the particle", " to the ohmic bath leads to an exponential suppression of the tunneling rate while only a weak dependence on the jump frequency persists. Physically, this result is easy to rationalize: under-barrier tunneling is a feature of the quantum mechanical system. In the transfer of energy to and from the external bath, the phase coherence of the particle is lost. At zero temperature, the tunneling rate becomes suppressed and the particle con\ufb01ned. 3. Let us now consider the in\ufb02uence of temperature on the tunneling rate when the dissipative coupling is inactive, \u03b7 \u2192 0. In this case, the discrete frequency summation takes the form29 f = T n g(\u03c9n) = (coth(\u03b2\u03c9c/2)/2\u03c9cm). Using this result, one obtains the action Sbounce = m\u03c9ca2 tanh(\u03b2\u03c9c/2). In the low temperature limit \u03b2 \u2192 \u221e, Sbounce = m\u03c9ca2 as discussed above. At high temperatures \u03b2 \u2192 0, as expected, one recovers a classical activated dependence of the escape rate, namely Sbounce \u03b2m\u03c92 c a2/2. 4. Finally, let us brie\ufb02y remark on the interplay of thermal activation with ohmic dis\u2212 sipation. Applying the the Euler\u2013Maclaurin formula f (0) 12 + \u00b7 \u00b7 \u00b7 to relate discrete sums over Matsubara frequencies to their zero-temperature integral limits, one \ufb01nds that Sbounce(T ) \u2212 Sbounce(T = 0) \u221d \u03b7T 2. This shows that, in the dissipative regime, an increase in temperature diminishes the tunneling rate with a scale proportion to the damping. \u221e 0 dx f (x) + f (0) m=0 f (m) = \u221e 2 This concludes our cursory discussion of the application of the Feynman path integral to dissipative quantum tunneling. As mentioned above, our brief survey was able only to touch upon the broad \ufb01eld of research. Those interested in learning more about the \ufb01eld of macroscopic quantum tunneling are referred to the wider literature. To close this chapter, we turn now to our penultimate application of the path integral \u2013 quantum mechanical spin. Path integral for spin The quantum mechanics of a spin-(1/2) particle is a standard example in introductory courses. Indeed, there is hardly any other system whose quantum mechanics is as easy to formulate. Given that, it is perhaps surprising that for a long time the spin problem de\ufb01ed all attempts to cast it in path integral form: Feynman, the architect of the path integral, did not succeed in incorporating spin into the new formalism. It took several decades to \ufb01ll this gap (for a review of the early history up to 1980, see Schulman\u2019s text1), and a fully satisfactory formulation of the subject was obtained no earlier than 1988. (The present exposition follows closely the lines of the review by Michael Stone.30) Why then is it so di\ufb03cult to \ufb01nd a path integral of spin? In hindsight it turns out that the spin path integral is in fact no more complex than any other path integral, it merely appears to be a bit unfamiliar. The reason is that, on the one hand, the integrand of the path integral is essentially the exponentiated classical action whilst, on the other, the classical 29 For details on how to implement the discrete frequency summation, see the Info block on page 170 below. 30 M. Stone, Supersymmetry and the quantum mechanics of spin, Nucl. Phys B 314 (1989), 577\u201386. 3.3 Applications of the Feynman path integral 135 mechanics of spin is a subject that is not standard in introductory or even advanced courses. In other words, the path integral approach must, by necessity, lead to an unusual object. The fact that the classical mechanics of spin is hardly ever mentioned is related not only to the common view that spin is something \u201cfundamentally quantum\u201d but also to the fact that the mechanics of a classical spin (see below) cannot be expressed within the standard formulation of Hamiltonian mechanics, i.e. there is no formulation in terms of a set of globally de\ufb01ned coordinates and equally many global momenta. It is therefore inevitable that one must resort to the (less widely applied) symplectic formulation of Hamiltonian mechanics.31 However, as we will see below, the classical mechanics of spin can nevertheless be quite easily understood physically. Besides attempting to elucidate the connections between quantum and classical mechanics of spin, there is yet another motivation for discussing the spin path integral. Pretending that we have forgotten essential quantum mechanics, we will formulate the path integral ignoring the fact that spin quantum numbers are half integer or integer. The quant", "ization of spin will then be derived in hindsight, by way of a geometric consideration. In other words, the path integral formulation demonstrates how quantum mechanical results can be obtained by geometric rather than standard algebraic reasoning. Finally, the path integral of spin will serve as a basic platform on which our analysis of higher-dimensional spin systems below will be based. A reminder of \ufb01nite-dimensional SU(2)-representation theory In order to formulate the spin path integral, it is necessary to recapitulate some facts regarding the role of SU(2) in quantum mechanics. The special unitary group in two dimensions, SU(2), is de\ufb01ned by SU(2) = {g \u2208 Mat(2 \u00d7 2, C)|g\u2020g = 12, det g = 1}, where 12 is the two-dimensional unit matrix. Counting independent components one \ufb01nds that the group has three free real parameters or, equivalently, that its Lie algebra, su(2), is threedimensional. As we have seen, the basis vectors of the algebra \u2013 the group generators \u2013 \u02c6Si, i = x, y, z, satisfy the closure relation [ \u02c6Si, \u02c6Sj] = iijk \u02c6Sk, where ijk is the familiar fully antisymmetric tensor. Another useful basis representation of su(2) is given by the spin raising and lowering operators, \u02c6S\u00b1 = \u02c6Sx \u00b1 i \u02c6Sy. Again, as we have seen earlier, the algebra { \u02c6S+, \u02c6S\u2212, \u02c6Sz} is de\ufb01ned by the commutation relations [ \u02c6S+, \u02c6S\u2212] = 2 \u02c6Sz, [ \u02c6Sz, \u02c6S\u00b1] = \u00b1 \u02c6S\u00b1. Each group element can be uniquely parameterized in terms of the exponentiated algebra. For example, in the Euler angle representation, the group is represented as 5 4 SU(2) = g(\u03c6, \u03b8, \u03c8) = e\u2212i\u03c6 \u02c6S3 e\u2212i\u03b8 \u02c6S2 e\u2212i\u03c8 \u02c6S3 \u03c6, \u03c8 \u2208 [0, 2\u03c0], \u03b8 \u2208 [0, \u03c0]. The Hilbert spaces HS of quantum spin are irreducible representation spaces of SU(2). Within the spaces HS, SU(2) acts in terms of representation matrices (which will be denoted by g) and the matrix representations of its generators \u02c6Si. The index S is the so-called weight 31 Within this formulation, the phase space is regarded as a di\ufb00erential manifold with a symplectic structure (cf. Arnold\u2019s text on classical mechanics: V. I. Arnold, Mathematical Methods of Classical Mechanics [SpringerVerlag, 1978]). In the case of spin, this manifold is the 2-sphere S2. 136 Feynman path integral of the representation (physically: the total spin).32 Within each HS, there is a distinguished state, a state of highest weight | \u2191, which is de\ufb01ned as the (normalized) eigenstate of \u02c6Sz with maximum eigenvalue, S (physically: a spin state polarized in the 3-direction). Owing to the irreducibility of the representation, each (normalized) state of the Hilbert space HS can be obtained by applying the Euler-angle-parameterized elements of the representation to the maximum weight state. Being a compact group, SU(2) can be integrated over, i.e. it makes sense to de\ufb01ne objects SU(2) dg f (g), where f is some function of g and dg is a realization of a group measure.33 like Among the variety of measures that can be de\ufb01ned in principle, the (unique) Haar measure plays a distinguished role. It has the convenient property that it is invariant under left and right multiplication of g by \ufb01xed group elements, i.e. \u2200h \u2208 SU(2) : dg f (gh) = dg f (hg) = where, for notational simplicity, we have omitted the subscript in dg f (g), SU(2). Construction of the path integral With this background, we are now in a position to formulate the Feynman path integral. To be speci\ufb01c, let us consider a particle of spin S subject to the Hamiltonian \u02c6H = B \u00b7 \u02c6S, where B is a magnetic \ufb01eld and \u02c6S \u2261 ( \u02c6S1, \u02c6S2, \u02c6S3)", " is a vector of spin operators in the spinS representation. Our aim is to calculate the imaginary time path integral representation of the partition function Z \u2261 tr e\u2212\u03b2 \u02c6H. In constructing the path integral we will follow the general strategy outlined at the end of Section 3.2, i.e. the \ufb01rst step is to represent Z as Z = tr (e\u2212 \u02c6H )N, where = \u03b2/N. Next, we have \u2013 the most important step in the construction \u2013 to insert a suitably chosen resolution of identity between each of the factors e\u2212 \u02c6H. A representation that will lead us directly to the \ufb01nal form of the path integral is speci\ufb01ed by id = C dg |gg|, (3.47) where \u201cid\u201d represents the unit operator in HS, dg is a group integral over the Haar measure, C is some constant, and |g \u2261 g| \u2191 is the state obtained by letting the representation matrix g act on the maximum weight state | \u2191 (cf. the glossary of SU(2) representation theory above). Of course it remains to be veri\ufb01ed that the integral (3.47) is indeed proportional to the unit operator. That this is so follows from Schur\u2019s lemma, which states that, if, and only if, an operator \u02c6A commutes with all representation matrices of an irreducible group 32 The index S is de\ufb01ned in terms of the eigenvalues of the Casimir operator (physically: the total angular momentum operator) \u02c6S2 \u2261 i according to the relation \u2200|s \u2208 HS : \u02c6S2|s = S(S + 1)|s. \u02c6S2 i 33 To de\ufb01ne group measures in a mathematically clean way, one makes use of the fact that (as a Lie group) SU(2) is a three-dimensional di\ufb00erentiable manifold. Group measures can then be de\ufb01ned in terms of the associated volume form (see the primer in di\ufb00erential geometry on page 537). 3.3 Applications of the Feynman path integral 137 representation (in our case the gs acting in the Hilbert space HS), \u02c6A is proportional to the unit matrix. That the group above integral ful\ufb01lls the global commutativity criterion follows from the properties of the Haar measure: \u2200h \u2208 SU(2): h dg |gg| = dg |hgg| Haar= dg |hh\u22121gh\u22121g| = dg |gg|h. dg |gg| is, indeed, proportional to the unit operator. The proportionality constant Thus, appearing in Eq. (3.47) will not be of any concern to us \u2013 apart from the fact that it is non-zero.34 Substituting the resolution of identity into the time-sliced partition function and making use of the fact that gi+1|e\u2212B\u00b7\u02c6S|gi gi+1|gi \u2212 gi+1|B \u00b7 \u02c6S|gi gi|gi=1 = 1 \u2212 gi|gi + gi+1|gi \u2212 gi+1|B \u00b7 \u02c6S|gi exp gi+1|gi \u2212 gi|gi \u2212 gi+1|B \u00b7 \u02c6S|gi, one obtains Z = lim N\u2192\u221e N' gN =g0 i=0 dgi exp \u2212 N \u22121 i=0 \u2212 gi+1|gi \u2212 gi|gi + gi+1|B \u00b7 \u02c6S|gi. By taking the limit N \u2192 \u221e, this can be cast in path integral form, Z = Dg exp \u2212 \u03b2 0 d\u03c4 \u2212\u2202\u03c4 g|g + g|B \u00b7 \u02c6S|g, (3.48) where the HS-valued function |g(\u03c4 ) is the continuum limit of |gi. Equation (3.48) is our \ufb01nal, albeit somewhat over-compact, representation of the path integral. In order to give this expression physical interpretation, we need to examine more thoroughly the meaning of the states |g. In the literature, the states |g expressed in the Euler-angle representation |3g(\u03c6, \u03b8, \u03c8) \u2261 e\u2212i\u03c6 \u02c6S3 e\u2212i\u03b8 \u02c6S2 e\u2212i\u03c8 \u02c6S3 | \u2191, are referred to as spin coherent states. Before discussing the origin of this terminology, it is useful to explore the algebraic structure of these states. Firstly, note that the maximum weight state | \u2191", " is, by de\ufb01nition, an eigenstate of \u02c6S3 with maximum eigenvalue S. Thus, |3g(\u03c6, \u03b8, \u03c8) \u2261e \u2212i\u03c6 \u02c6S3 e\u2212i\u03b8 \u02c6S2 | \u2191e\u2212i\u03c8S and the angle \u03c8 enters the coherent state merely as a phase or gauge factor. By contrast, the two remaining angles \u03b8 and \u03c6 act through true rotations. Now, the angular variables \u03c6 \u2208 [0, 2\u03c0) and\u03b8 \u2208 [0, \u03c0) de\ufb01ne a standard representation of the 2-sphere. In view of the fact that the states |\u02dcg(\u03c6, \u03b8, \u03c8) cover the entire Hilbert space 34 Actually, the constant C can be straightforwardly computed by taking the trace of Eq. (3.47), which leads to C = (dimension of the representation space)/(volume of the group). 138 Feynman path integral HS, we are led to suspect that the latter bears similarity with a sphere.35 To substantiate this view, let us compute the expectation values ni \u2261 \u02dcg(\u03c6, \u03b8, \u03c8)| \u02c6Si|\u02dcg(\u03c6, \u03b8, \u03c8), i = 1, 2, 3. (3.49) To this end, we \ufb01rst derive an auxiliary identity which will spare us much of the trouble that will arise in expanding the exponentials appearing in the de\ufb01nition of |3g. By making use of the identity (i = j) e\u2212i\u03c6 \u02c6Si \u02c6Sjei\u03c6 \u02c6Si = e\u2212i\u03c6[ \u02c6Si, ] \u02c6Sj = \u02c6Sj cos \u03c6 + ijk \u02c6Sk sin \u03c6, (3.50) where the last equality follows from the fact that cos x (sin x) contain x in even (odd) orders and [ \u02c6Sj, ]2 \u02c6Si = \u02c6Si, it is straightforward to obtain (exercise) n = S(sin \u03b8 cos \u03c6, sin \u03b8 sin \u03c6, cos \u03b8), i.e. n is the product of S and a unit vector parameterized in terms of spherical coordinates. \u03c8 \u03b8 \u03c6 This is the key to understanding the terminology \u201cspin coherent states.\u201d The vectors |3g(\u03c6, \u03b8, \u03c8) represent the closest approximation of a classical angular momentum state one can formulate in Hilbert space (see \ufb01gure). Let us now see what happens if we employ the Euler angle representation in formulating the path integral. A \ufb01rst and important observation is that the path integral is gauge invariant \u2013 in the sense that it does not depend on the U(1)-phase, \u03c8. As to the B-dependent part of the action, the gauge invariance is manifest: Eq. (3.49) implies that SB[\u03c6, \u03b8] \u2261 \u03b2 0 d\u03c4 \u02dcg|B \u00b7 \u02c6S|\u02dcg = \u03b2 0 d\u03c4 g|B \u00b7 \u02c6S|g = S \u03b2 0 d\u03c4 n \u00b7 B = SB \u03b2 0 d\u03c4 cos \u03b8. Here, we have introduced the gauge-independent part |g of the state vector by setting |3g \u2261 |g exp(\u2212iS\u03c8) or, equivalently, |g(\u03c6, \u03b8) \u2261 e\u2212i\u03c6 \u02c6S3 e\u2212i\u03b8 \u02c6S2 | \u2191. Substituting this representation into the \ufb01rst term of the action of Eq. (3.48), one obtains Stop[\u03c6, \u03b8] \u2261 \u2212 \u03b2 \u03b2 d\u03c4 \u2202\u03c4 \u02dcg|\u02dcg = \u2212 0 d\u03c4 \u2202\u03c4 e\u2212iS\u03c8g|ge\u2212iS\u03c8 \u03b2 d\u03c4 (\u2202\u03c4 g|g \u2212iS\u2202 \u03c4 \u03c8g|g) = \u2212 0 d\u03c4 \u2202\u03c4 g|g, (3.51) = \u2212 0 \u03b2 0 where the last equality holds because g|g = 1 is constant and \u03c8 is periodic in \u03b2. We thus conclude that the path integral does not depend on the gauge phase, i.e. it e\ufb00ectively extends over paths living on the 2-sphere (rather than the entire group manifold SU(2)). This \ufb01nding is reassuring in the sense that a degree of freedom living on a sphere comes 35 There is a group theoretical identity behind this observation, namely the isomorphism SU(2) \u223c= S2 \u00d7 U(", "1), where U(1) is the \u201cgauge\u201d subgroup contained in SU(2). 3.3 Applications of the Feynman path integral 139 close to what one might intuitively expect to be the classical counterpart of a quantum particle of de\ufb01nite angular momentum. Let us now proceed by exploring the action of the path integral. By using the auxiliary identity (3.50) it is a straightforward matter to show that Stop[\u03c6, \u03b8] = \u2212 \u03b2 0 d\u03c4 \u2202\u03c4 g|g = \u2212iS \u03b2 0 d\u03c4 \u2202\u03c4 \u03c6 cos \u03b8 = iS \u03b2 0 d\u03c4 \u2202\u03c4 \u03c6(1 \u2212 cos \u03b8). (3.52) Combining this with the B-dependent term discussed above, one obtains S[\u03b8, \u03c6] =S B[\u03c6, \u03b8] +S top[\u03c6, \u03b8] = S \u03b2 0 for the action of the path integral for spin. d\u03c4 [B cos \u03b8 + i(1 \u2212 cos \u03b8)\u2202\u03c4 \u03c6], (3.53) EXERCISE Derive the Euler\u2013Lagrange equations associated with this action. Show that they are equivalent to the Bloch equations i\u2202\u03c4 n = B \u00d7 n of a spin with expectation value S = Sn subject to a magnetic \ufb01eld. Here, n(\u03c6, \u03b8) \u2208 S2 is the unit vector de\ufb01ned by the two angles \u03c6, \u03b8. Analysis of the action To formulate the second term in the action (3.53) in a more illuminating way, we note that the velocity of the point n moving on the unit sphere is given by \u02d9n = \u02d9\u03b8\u02c6e\u03b8 + \u02d9\u03c6 sin \u03b8 \u02c6e\u03c6, where (\u02c6er, \u02c6e\u03b8, \u02c6e\u03c6) form a spherical orthonormal system. We can thus rewrite Eq. (3.52) as where Stop[\u03c6, \u03b8] = iS \u03b2 0 < d\u03c4 \u02d9n \u00b7 A = iS dn \u00b7 A, \u03b3 A = 1 \u2212 cos \u03b8 sin \u03b8 \u02c6e\u03c6. (3.54) (3.55) Notice that, in spite of its compact appearance, Eq. (3.54) does not represent a coordinate invariant formulation of the action Stop. (The \ufb01eld A(\u03c6, \u03b8) explicitly depends on the coordinates (\u03c6, \u03b8).) In fact, we will see in Chapter 9 that the action Stop cannot be expressed in a coordinate invariant manner, for reasons deeply rooted in the topology of the 2-sphere. A second observation is that Eq. (3.54) can be read as the (Euclidean time) action of a particle of charge S moving under the in\ufb02uence of a vector potential A.36 Using standard formulae of vector calculus37 one \ufb01nds Bm \u2261 \u2207 \u00d7 A = er, i.e. our particle moves in a radial magnetic \ufb01eld of constant strength unity. Put di\ufb00erently, the particle experiences the \ufb01eld of a magnetic \u201ccharge\u201d of strength 4\u03c0 centered at the origin of the sphere. INFO If you \ufb01nd this statement di\ufb03cult to reconcile with the Maxwell equation \u2207 \u00b7 B = 0 \u2194 S B \u00b7 dS = 0 for any closed surface S, notice that \u2207 \u00b7B = \u2207 \u00b7(\u2207 \u00d7 A) = 0 holds only if A is non-singular. However, the vector potential Eq. (3.55) is manifestly singular along the line 36 See, e.g., H. Goldstein, Classical Mechanics (Addison-Wesley, 1981). 37 See, e.g., J. D. Jackson,Classical Electrodynamics (Wiley, 1975). 140 Feynman path integral (r, \u03b8 = \u03c0) through the south pole of the sphere. The physical picture behind this singularity is as follows: imagine an in\ufb01nitely thin solenoid running from r = \u221e through the south pole of the sphere to its center. Assuming that the solenoid contains a magnetic \ufb02ux 4\u03c0, the center of the sphere becomes a source of magnetic \ufb02ux, the so-called Dirac monopole. This picture is consistent with the presence of a \ufb01eld B = er. It also explains the singularity of A along the string. (Of course, the solenoidal construction does not lead to the prediction of a genuine monopole potential: somewhere, at", " r = \u221e, our auxiliary magnetic coil has to end, and this is where the \ufb02ux lines emanating from the point r = 0 terminate.) The postulate of a \ufb02ux line at the singularity of A merely helps to reconcile the presence of a radial magnetic \ufb01eld with the principles of electrodynamics. However, as far as our present discussion goes, this extra structure is not essential, i.e. we may simply interpret r = 0 as the position of a magnetic \u201ccharge.\u201d To explore the consequences of this phenomenon, we apply Stokes\u2019 theorem to write Stop[n] =iS < \u03b3 dn \u00b7 A = iS dS \u00b7 (\u2207 \u00d7A) =iS dS \u00b7 er = iSA\u03b3,n. (3.56) A\u03b3,n A\u03b3,n Here, A\u03b3,n is the domain on the 2-sphere which (a) has the curve \u03b3 as its boundary, and Stokes Sir George Gabriel 1819\u20131903 As Lucasian Professor of Mathematics at Cambridge, Stokes etablished the science of hydrodynamics with his law of viscosity (1851), describing the velocity of a small sphere through a viscous \ufb02uid. Furthermore, he investigated the wave theory of light, named and explained the phenomenon of \ufb02uorescence, and theorized an explanation of the Fraunhofer lines in the solar spectrum. (Figure reproduced from Sir George Gabriel Stokes Memoirs presented to the Cambridge Philosophical Society on the occasion of the Jubilee of Stokes volume XVIII of the Transactions of the Cambridge Philosophical Society, Cambridge University Press, 1900.) (b) contains the north pole (see \ufb01gure on next page). The integral produces the area of this surface which we again denote by A\u03b3,n. Curiously, the action Stop is but a measure of the area bounded by the curve \u03b3 : \u03c4 \u2192 n(\u03c4 ). However, simple as it is, this result should raise some suspicion: by assigning a designated role to the northern hemisphere of the sphere some symmetry breaking, not present in the original problem, has been introduced. Indeed, we might have de\ufb01ned our action \u03b3 dn \u00b7 A where by Stop[\u03c6, \u03b8] =iS A = \u2212 1+cos \u03b8 sin \u03b8 \u02c6e\u03c6 = A \u2212 2\u2207\u03c6 di\ufb00ers from A only by a gauge transformation.38 The newly de\ufb01ned vector potential is non-singular in the southern hemisphere, so that application of Stokes\u2019 theorem leads to the conclusion dS \u00b7 Bm = \u2212iSA\u03b3,s. Here, A\u03b3,s is the area of a surface bounded by \u03b3 Stop[n] = \u2212iS but covering the south pole of the sphere. The absolute minus sign is due to the outward orientation of the surface A\u03b3,s. A\u03b3,s = 38 You may, with some justi\ufb01cation, feel uneasy about the fact that \u03c6 is not a true \u201cfunction\u201d on the sphere (or, dn \u00b7 \u2207\u03c6 = \u03c6(\u03b2) \u2212 \u03c6(0) may be a non-vanishing multiple of 2\u03c0). We will alternatively, about the fact that return to the discussion of this ambiguity shortly. (Notice that a similarly hazardous manipulation is performed in the last equality of Eq. (3.52).) 3.3 Applications of the Feynman path integral 141 One has to concede that the result obtained for the action Stop depends on the chosen gauge of the monopole vector potential! The di\ufb00erence between the northern and the southern variant of our analysis is given by iS dS \u00b7 Bm + iS dS \u00b7 Bm = iS dS \u00b7 er = 4\u03c0iS, A\u03b3,n A\u03b3,s S2 where we have made use of the fact that A\u03b3,n \u222a A\u03b3,s = S2 is the full sphere. At \ufb01rst sight, A\u03b3,n it looks as if our analysis has led us to a gauge-dependent, and therefore pathological, result. Let us recall, however, that physical quantities are determined by the exponentiated action exp(\u2212S[n]) and not by the action itself. Now, S is either integer or half integer, which implies the factor exp(4\u03c0iS) = 1 is irrelevant. In the operator representation of the theory, spin quantization follows from the representation theory of the algebra su(2). It is a \u201cnon-local\u201d feature, in the sense that the action of the spin operators on all eigenstates has to be considered to \ufb01x the dimensionality 2S + 1 of HS. In hindsight, it is thus not too surprising", " that the same information is encapsulated in a \u201cglobal\u201d condition (gauge invariance) imposed on the action of the path integral. \u03b3 Summarizing, we have found that the classical dynamics of a spin is that of a massless point particle on a sphere coupled to a monopole \ufb01eld Bm. We have seen that the vector potential of the \ufb01eld cannot be globally continuous on the full sphere. More generally, the phase space S2 cannot be represented in terms of a global system of \u201ccoordinates and momenta\u201d which places it outside the scope of traditional treatments of classical mechanics. This probably explains the failure of early attempts to describe the spin in terms of a path integral or, equivalently, in terms of a Hamiltonian action. In Chapter 9 we will use the path action (3.53) as a building block for our construction of the \ufb01eld theory of higher-dimensional spin systems. However, before concluding this section, let us make some more remarks on the curious properties of the monopole action Stop. Unlike all other Euclidean actions encountered thus far, the action (3.54) is imaginary. In fact, it will stay imaginary upon Wick rotation \u03c4 \u2192 it back to real times. More generally, Stop is invariant under the rescaling \u03c4 \u2192 c\u03c4, and invariant even under arbitrary reparameterizations \u03c4 \u2192 g(\u03c4 ) \u2261 \u03c4. This invariance is a hallmark of a topological term. Loosely speaking (see Chapter 9 for a deeper discussion), a topological term is a contribution to the action of a \ufb01eld theory that depends on the global geometry of a \ufb01eld con\ufb01guration rather than on its local structure. In contrast, \u201cconventional\u201d operators in \ufb01eld theoretical actions measure the energy cost of dynamical or spatial \ufb01eld \ufb02uctuations. In doing so they must relate to a speci\ufb01c spatio-temporal reference frame, i.e. they cannot be invariant under reparameterization. Summarizing our results, we have found that: 142 Feynman path integral 1. The classical action of a spin is one of a massless particle (there is no standard kinetic energy term in Eq. (3.48)) moving on a unit sphere. The particle carries a magnetic moment of magnitude S. It is coupled (a) to a conventional magnetic \ufb01eld via its magnetic moment, and (b) to a monopole \ufb01eld via its orbital motion. Note that we have come, \ufb01nally, to a position which hints at the di\ufb03culties plaguing attempts to formulate a classical mechanics of spin. The vector potential of a monopole, A, cannot be globally de\ufb01ned on the entire sphere. The underlying physical reason is that, by the very nature of the monopole (\ufb02ux going radially outwards everywhere), the associated vector potential must be singular at one point of the surface.39 As a consequence, the classical phase space of the system, the sphere, cannot be covered by a global choice of coordinate system. (Unlike most standard problems of classical mechanics there is no system of globally de\ufb01ned \u201cp\u201ds and \u201cq\u201ds.) This fact largely spoils a description within the standard \u2013 coordinate-oriented \u2013 formulation of Hamiltonian mechanics (cf. the discussion in the article by Stone40). 2. Terms akin to the monopole contribution to the spin action appear quite frequently within path integral formulations of systems with non-trivial topology (like the twosphere above). Depending on the particular context under consideration, one distinguishes between Wess\u2013Zumino\u2013Witten (WZW) terms, \u03b8-terms, Chern\u2013Simons terms and a few other terms of topological origin. What makes these contributions generally important is that the value taken by these terms depends only on the topology of a \ufb01eld con\ufb01guration but not on structural details. For further discussion of phenomena driven by non-trivial topological structures we refer to Chapter 9 below. Trace formulae and quantum chaos As a \ufb01nal application of the path integral, we turn now to the consideration of problems in which the dynamics of the classical system is, itself, non-trivial. Introductory courses on classical mechanics usually convey the impression that dynamical systems behave in a regular and, at least in principle, mathematically predictable way. However, experience shows that the majority of dynamical processes in nature do not conform to this picture. Partly, or even fully, chaotic motion (i.e. motion that depends in a singular and, thereby, in an essentially unpredictable way on initial conditions) is the rule rather than the exception. In view of", " the drastic di\ufb00erences in the observable behavior of classically integrable and chaotic systems, an obvious question arises: in what way does the quantum phenomenology of chaotic systems di\ufb00er from that associated with integrable dynamics? This question de\ufb01nes the \ufb01eld of quantum chaos. 39 To better understand this point, consider the integral of A along an in\ufb01nitesimal closed curve \u03b3 on the sphere. If A were globally continuous, we would have two choices to transform the integral into a surface integral over B: an integral over the \u201clarge\u201d or the \u201csmall\u201d surface area bounded by \u03b3. The monopole nature of B would demand that both integrals are proportional to the respective areas of the integration domain which, by assumption, are di\ufb00erent, \u21d2 contradiction. The resolution of this paradox is that A must be discontinuous at one point on the sphere, i.e. we cannot globally set B = \u2207 \u00d7 A, and the choice of the integration area is prescribed by the condition that it must not encompass the singular point. 40 M. Stone, Supersymmetry and the quantum mechanics of spin, Nucl. Phys. B 314 (1986), 557\u201386. 3.3 Applications of the Feynman path integral 143 Understanding signatures of classically chaotic motion in quantum mechanics is an issue not only of conceptual, but also of great practical relevance, impinging on areas such as quantum electron transport in condensed matter systems. The inevitable presence of impurities and imperfections in any macroscopic solid renders the long-time dynamics of electronic charge carriers chaotic. Relying on a loose interpretation of the Heisenberg principle, \u0394t \u223c /\u0394E, i.e. the relation between long-time dynamical behavior and small-scale structures in energy, one would expect that signatures of chaotic quantum dynamics are especially important in the low-energy response in which one is usually interested. This expectation has been con\ufb01rmed for innumerable observables related to low-temperature electronic transport in solid state systems. Disordered conducting media represent but one example of a wide class of dynamical systems with long-time chaotic dynamics. Indeed, recent experimental advances have made it possible to realize a plethora of e\ufb00ectively non-disordered chaotic dynamical systems in condensed matter devices. For example, by employing modern semiconductor device technology, it has become possible to manufacture small two-dimensional conducting systems, of a size O(< 1 \u03bcm) and of almost any geometric shape. Here, the number of imperfections can be reduced to a negligible minimum, i.e. electrons propagate ballistically along straight trajectories, as with a billiard system. The smallness of the devices further implies that the ratio between Fermi wavelength and system size is of O(10\u22121\u201310\u22123), so, while semiclassical concepts will surely be applicable, the wave aspects of quantum propagation remain visible. In recent years, the experimental and theoretical study of electron transport in such quantum billiards has emerged as a \ufb01eld in its own right. How then can signatures of chaotic dynamics in quantum systems be sought? The most fundamental characteristic of a quantum system is its spectrum. Although not a direct observable, it determines the majority of properties accessible to measurement. On the other hand, it is clear that the manifestations of chaos we are looking for must relate back to the classical dynamical properties of the system. The question then is, how can a link between classical mechanics and quantum spectra be drawn? This problem is tailor-made for analysis by path integral techniques. Semiclassical approximation to the density of states The close connection between the path integral and classical mechanics should be evident from the previous sections. However, to address the problem raised above, we still need to understand how the path integral can be employed to analyze the spectrum of a quantum system. The latter is described by the (single-particle) density of states \u03c1() = tr \u03b4( \u2212 \u02c6H) = a \u03b4( \u2212 a), (3.57) where {a} represents the complete set of energy levels. To compute the sum, one commonly employs a trick based on the Dirac identity, lim \u03b40 1 x + i\u03b4 = \u2212i\u03c0\u03b4(x) +P 1 x, (3.58) 144 Feynman path integral where P(1/x) denotes the principal part of 1/x. By taking the imaginary part of Eq. (3.58), ), where + \u2261 Eq. (3.57) can be represented as \u03c1() = \u2212 1 1 \u03c0 Im +\u2212 \u02c6H \u221e 0 dt eix+t, and + i\u03b4 and the limit lim\u03b40 is implicit. Using the identity 1/x+ = \u2212i representing the", " trace tr \u02c6A = dq q| \u02c6A|q as a real space integral, \u03c0 Im tr ( 1 +\u2212a = \u2212 1 a \u03c1() = 1 \u03c0 \u221e 0 dt Re tr(ei(+\u2212 \u02c6H)t/) = 1 \u03c0 Re \u221e 0 dt ei+t/ dqq|e\u2212i \u02c6Ht/|q, (3.59) we have made the connection between the density of states and the quantum propagation amplitude explicit. Without going into full mathematical detail (see, for example, the work by Haake,41 for a modern discourse) we now outline how this integral is evaluated by path integral techniques within the semiclassical approximation. Although, for brevity, some of the more tricky steps of the calculation are swept under the carpet, the sketch will be accurate enough to make manifest some aesthetic connections between the spectral theory of chaotic quantum systems and classically chaotic dynamics. (For a more formal and thorough discussion, we refer to 41 Haake and Gutzwiller.42) Making use of the semiclassical approximation (3.28) established earlier, and substituting S[qcl], where, following into Eq. (3.59), one obtains \u03c1() 1 \u221e 0 dt ei+t/ \u03c0 Re dq A[qcl]e i our discussion in Section 3.2, we have de\ufb01ned A[qcl] \u2261 det and qcl represents a closed classical path that begins at q at time zero and ends at the same coordinate at time t. Again relying on the semiclassical condition S[qcl], the integrals over q and t can be performed in a stationary phase approximation. Beginning with the time integral, and noticing that \u2202tS[qcl] =\u2212 qcl is the (conserved) energy of the path qcl, we obtain the saddle-point condition!= qcl and i 2\u03c0 \u22022S[qcl] \u2202q(0) \u2202q(t) 1/2 \u03c1() 1 \u03c0 Re dq A[qcl,]e i S[qcl,], where the symbol qcl, indicates that only paths q \u2192 q of energy are taken into account, and the contribution coming from the quadratic integration around the saddle-point has been absorbed into a rede\ufb01nition of A[qcl,]. Turning to the q-integration, making use of the fact that \u2202qi S[qcl] = \u2212pi, \u2202qf S[qcl] = pf, whereq i,f are the initial and \ufb01nal coordinates of a path qcl, andp i,f are the initial and \ufb01nal momentum, the stationary phase condition assumes the form 0!= dqS[qcl,] = (\u2202qi + \u2202qf ) S[qcl,]|qi=qf =q = pf \u2212 pi, i.e. the stationarity of the integrand under the q-integration requires that the initial and \ufb01nal momentum of the path qcl, be identical. We thus \ufb01nd that the paths contributing to the integrated transition amplitude are periodic not only in coordinate space but even in phase space. Such paths are called periodic 41 F. Haake, Quantum Signatures of Chaos (Springer-Verlag, 2001). 42 M. C. Gutzwiller, Chaos in Classical and Quantum Mechanics (Springer-NY, 1991), and Haake, Quantum Signatures of Chaos. 3.3 Applications of the Feynman path integral 145 orbits \u2013\u201cperiodic\u201d because the path comes back to its initial phase space coordinate after a certain revolution time. As such, the orbit will be traversed repeatedly as time goes by (see the \ufb01gure, where a periodic orbit \u03b1 with initial coordinates x = (p, q) is shown). According to our analysis above, each coordinate point q lying on a periodic orbit is a stationary phase point of the q-integral. The stationary phase approximation of the integral can thus be formulated as \u03c1() 1 \u03c0 Re \u221e n=1 \u03b1 \u03b1 dq A\u03b1e i nS\u03b1, \u03b1 stands for a sum over all periodic orbits (of energy ) and S\u03b1 is the action where corresponding to one traversal of the orbit (all at \ufb01xed energy ). The index n accounts for the fact that, owing to its periodicity, the orbit can be traversed repeatedly, with total action nS\u03b1. Furthermore, \u03b1 dq is an integral over all coordinates lying on the orbit and we have again absorbed a contribution coming from the quadratic integration around the stationary phase points in the pre-exponential amplitude A\u03b1. Finally, noting that \u03b1 dq", " \u221d T\u03b1, whereT \u03b1 is the period of one traversal of the orbit \u03b1 (at energy ), we arrive at the result \u03c1() 1 \u03c0 Re \u221e n=1 \u03b1 T\u03b1A\u03b1e i nS\u03b1. (3.60) This is a (simpli\ufb01ed, see info block below) representation of the famous Gutzwiller trace formula. The result is actually quite remarkable: the density of states, an observable of quantum mechanical signi\ufb01cance, has been expressed entirely in terms of classical quantities. EXERCISE Making use of the Feynman path integral, show that the propagator for a particle of mass m con\ufb01ned by a square well potential of in\ufb01nite strength is given by G(qf, qi; t) = m 2\u03c0it $ \u221e exp n=\u2212\u221e im(qf \u2212 qi + 2na)2 2t \u2212 exp im(qf + qi + 2na)2 2t %. INFO Had we carefully kept track of all determinants arising from the stationary phase integrals, the prefactor A\u03b1 would have read A\u03b1 = 1 ei \u03c0 |detM r 2 \u03bd\u03b1 \u03b1 \u2212 1| 1 2, where \u03bd\u03b1 is known as the Maslov index (an integer-valued factor associated with the singular points on the orbit, i.e. the classical turning points). The meaning of this object can be understood, e.g., by applying the path integral to the problem of a quantum particle in a box. To correctly reproduce the spectrum, the contribution of each path must be weighted by (\u2212)n = exp(i\u03c0n), where n is the number of its turning points in the box potential), and M\u03b1 represents the monodromy matrix. To understand the meaning of this object, notice that a phase space point \u00afx on a periodic orbit can be interpreted as a \ufb01xed point of the classical time evolution 146 Feynman path integral operator U (T\u03b1): U (T\u03b1, \u00afx) = \u00afx, which is just to say that the orbit is periodic. As with any other smooth mapping, U can be linearized in the vicinity of its \ufb01xed points,U (T\u03b1, \u00afx + y) = \u00afx + M\u03b1y, where the linear operator M\u03b1 is the monodromy matrix. Evidently, M\u03b1 determines the stability of the orbit under small distortions, which makes it plausible that it appears as a controlling prefactor of the stationary phase approximation to the density of states. 3.4 Summary and outlook In this chapter we have introduced the path integral formulation of quantum mechanics, an approach independent of, yet (modulo certain mathematical imponderabilities related to continuum functional integration) equivalent to, the standard route of canonical operator quantization. While a few precious exactly solvable quantum problems (e.g. the evolution of a free particle, the harmonic oscillator, and, perhaps intriguingly, quantum mechanical spin) are more e\ufb03ciently formulated by the standard approach, a spectrum of unique features makes the path integral an indispensable tool of modern quantum mechanics. The path integral approach is highly intuitive, powerful in the treatment of non-perturbative problems, and tailor-made to formulation of semiclassical limits. Perhaps most importantly, we have seen that it provides a unifying link whereby quantum problems can be related to classical statistical mechanics. Indeed, we have found that the path integral of a quantum point particle is, in many respects, equivalent to the partition function of a classical one-dimensional continuum system. We have hinted at a generalization of this principle, i.e. an equivalence principle relating d-dimensional quantum \ufb01eld theory to (d + 1)-dimensional statistical mechanics. However, before exploring this bridge further, we \ufb01rst need to generalize the concept of path integration to problems involving quantum \ufb01elds. This will be the subject of the next chapter. 3.5 Problems Quantum harmonic oscillator As emphasized in the main text, the quantum harmonic oscillator provides a valuable environment in which to explore the Feynman path integral and methods of functional integration. In this, along with a small number of other precious examples, the path integral may be computed exactly, and the Feynman propagator explored rigorously. (a) Starting with the Feynman path integral, show that the propagator for the onedimensional quantum harmonic oscillator, \u02c6H = \u02c6p2/2m + m\u03c92 \u02c6q2/2, takes the form qf |e\u2212i \u02c6Ht/|qi = m\u03c9 2\u03c0i sin \u03c9t 1/2 exp m\u03c9 i 2 & % i + q2 q2 f cot \u03c9t \u2212 2qiqf sin \u03c9t.", " Suggest why the propagator varies periodically on the time interval t, and explain the origin of the singularities at t = n\u03c0/\u03c9, n = 1, 2,... Taking the frequency \u03c9 \u2192 0, show that the propagator for the free particle is recovered. (b) Show that the wavepacket \u03c8(q, t = 0) = (2\u03c0a)\u22121/4 exp[\u2212q2/4a] remains Gaussian at all 3.5 Problems 147 subsequent times. Obtain the width a(t) as a function of time. (c) Semiclassical limit: Taking the initial wavepacket to be of the form \u03c8(q, t = 0) = (2\u03c0a)\u22121/4 exp i mvq \u2212 1 4a q2, (which corresponds to a wavepacket centered at an initial position q = 0 with a velocity v), \ufb01nd the wavepacket at times t >0, and determine the mean position, mean velocity, and mean width as functions of time. Answer: (a) Making use of the Feynman path integral, the propagator can be expressed as the functional integral, qf |e\u2212i \u02c6Ht/|qi = q(t)=qf q(0)=qi Dq eiS[q]/, S[q] = t 0 dt m 2 \u02d9q2 \u2212 \u03c92q2. The evaluation of the functional integral over \ufb01eld con\ufb01gurations q(t) is facilitated by parameterizing the path in terms of \ufb02uctuations around the classical trajectory. Setting q(t) = qcl(t) + r(t) whereq cl(t) satis\ufb01es the classical equation of motion m\u00a8qcl = \u2212m\u03c92qcl, and applying the boundary conditions, one obtains the solution qcl(t) = A sin(\u03c9t) +B cos(\u03c9t), with the coe\ufb03cients B = qi and A = qf / sin(\u03c9t) \u2212 qi cot(\u03c9t). Being Gaussian in q, the action separates as S[q] = S[qcl] +S [r], where S[qcl] = = m\u03c92 2 m\u03c9 2 (A2 \u2212 B2) sin(2\u03c9t) 2\u03c9 f ) cot(\u03c9t) \u2212 2qiqf sin(\u03c9t) + 2AB. (q2 i + q2 cos(2\u03c9t) \u2212 1 2\u03c9 Finally, integrating over the \ufb02uctuations and applying the identity z/ sin z = \u221e' n=1 (1 \u2212 z2/\u03c02n2)\u22121, one obtains the required result, periodic in t with frequency \u03c9, and singular at t = n\u03c0/\u03c9. In particular, a careful regularization of the expression for the path integral shows that qf |e\u2212i \u02c6Ht/|qi \u2192 \u03b4(qf \u2212 qi), \u03b4(qf + qi), t = 2\u03c0n/\u03c9, t = \u03c0(2n + 1)/\u03c9. Physically, the origin of the singularity is clear. The harmonic oscillator is peculiar in having a spectrum with energies uniformly spaced in units of \u03c9. Noting the eigenfunction expansion qf |e\u2212i \u02c6Ht/|qi = qf |nn|qie\u2212i\u03c9nt, this means that when \u03c9 \u00d7 t/ = 2\u03c0 \u00d7 integer there is a coherent superposition of the states and the initial state is recovered. Furthermore, since the ground state and its even integer descendants are symmetric n 148 Feynman path integral while the odd states are antisymmetric, it is straightforward to prove the identity for the odd periods (exercise). (b) Given the initial condition \u03c8(q, t = 0), the time evolution of the wavepacket can be \u221e \u2212\u221e dq q|e\u2212i \u02c6Ht/|q\u03c8(q, 0), from which determined from the propagator as \u03c8(q, t) = one obtains \u221e \u03c8(q, t) = J(t) \u2212\u221e dq 1 (2\u03c0a)1/4 e\u2212q 2/4ae i m\u03c9 2 [q2+q 2] cot(\u03c9t)\u2212 2qq sin(\u03c9t), where J(t) represents the time-dependent contribution arising from the \ufb02uctuations around the classical trajectory. Being Gaussian in q, the integral can be performed explicitly. Setting \u03b1 = 1/2a \u2212 im\u03c9 cot(\u03c9t)/, \u03b2 = im\u03c9q/( sin(\u03c9", "t)), and performing the Gaussian integral over q, one obtains \" \u03c8(q, t) = J(t) 1 (2\u03c0a)1/4 2\u03c0 \u03b1 e\u03b22/2\u03b1 exp m\u03c9q2 cot(\u03c9t), i 2 where \u03b22/2\u03b1 = \u2212(1 + i\u03ba cot(\u03c9t))q2/4a(t). Rearranging terms, it is straightforward # to show that \u03c8(q, t) = (2\u03c0a(t))\u22121/4 exp ei\u03d5(q,t), where a(t) = a[cos2(\u03c9t) + \u03ba\u22122 sin2(\u03c9t)], \u03ba = 2am\u03c9/, and \u03d5(q, t) represents a pure phase.43 As required, under the action of the propagator the normalization of the wavepacket is preserved. (A graphical representation of the time evolution is shown in Fig. 3.9(a).) Note that, if a = /2m\u03c9 (i.e. \u03ba = 1), a(t) = a for all times \u2013 i.e. it is a pure eigenstate. \u2212 q2 4a(t) $ m\u03c9 sin(\u03c9t) \u00d7 (q \u2212 v (c) Still of a Gaussian form, the integration can again be performed explicitly for the new initial condition. In this case, we obtain an expression of the form above but with \u03b2 = i \u03c9 sin(\u03c9t)). Reading o\ufb00 the coe\ufb03cients, we \ufb01nd that the position and velocity of the wavepacket have the forms q0(t) = (v/\u03c9) sin(\u03c9t), v(t) = v cos(\u03c9t), coinciding with those of classical dynamics. Note that, as above, the width a(t) of the wavepacket oscillates at frequency \u03c9. (A graphical representation of the time evolution is shown in Fig. 3.9(b).) |\u03c8 |2 \u03c0/\u03c9 |\u03c8 |2 t (a) q t (b) q Figure 3.9 (a) Variation of a \u201cstationary\u201d Gaussian wavepacket in the harmonic oscillator taken from the solution, and (b) variation of the moving wavepacket. 43 For completeness, we note that \u03d5(q, t) =\u2212 1 2 tan\u22121( 1 \u03ba cot(\u03c9t)) \u2212 \u03baq2 4a cot(\u03c9t)( a a(t) \u2212 1). 3.5 Problems Density matrix 149 Using the results derived in the previous section as an example, we explore how real-time dynamical information can be converted into quantum statistical information. Using the results of the previous question, obtain the density matrix \u03c1(q, q) = q|e\u2212\u03b2 \u02c6H |q for the harmonic oscillator at \ufb01nite temperature, \u03b2 = 1/T (kB = 1). Obtain and comment on the asymptotics: (i) T \u03c9 and (ii) T \u03c9. (Hint: In the high-temperature case, be sure to carry out the expansion in \u03c9/T to second order.) Answer: The density matrix can be deduced from the general solution of the previous question. Turning to the Euclidean time formulation, \u03c1(q, q) =q = |e\u2212\u03b2H |q = q|e\u2212(i/)H(\u03b2/i)|q 1/2 \u2212 m\u03c9 2 m\u03c9 2\u03c0 sinh(\u03b2\u03c9) exp (q2 + q2) coth(\u03b2\u03c9) \u2212 2qq sinh(\u03b2\u03c9). (i) In the low-temperature limit T \u03c9 (\u03b2\u03c9 1), coth(\u03b2\u03c9) \u2192 1, sinh(\u03b2\u03c9) \u2192 e\u03b2\u03c9/2, and \u03c1(q, q) 1/2 m\u03c9 \u03c0e\u03b2\u03c9 # \u2212 m\u03c9 2 exp $ (q2 + q2) = q|n = 0 e\u2212\u03b2E0 n = 0|q. (ii) Using the relations coth(x) x1= 1/x + x/3 + \u00b7 \u00b7 \u00b7 and 1/ sinh(x) x1= 1/x \u2212 x/6 + \u00b7 \u00b7 \u00b7, the high-temperature expansion (T \u03c9) of the density operator gives \u03c1(q, q) e\u2212m(q\u2212q)2/2\u03b22 1/2 m 2\u03c0\u03b22 \u03b4(q \u2212 q)e\u2212 \u03b2m\u03c92 q2 2, exp \u2212 \u03b2m\u03c92 6 (q2 + q2 + qq) i.e. one recovers the classical Maxwell", "\u2013Boltzmann distribution! Depinning transition and bubble nucleation In Section 3.3 we explored the capacity for a quantum \ufb01eld to tunnel from the metastable minimum of a potential, the \u201cfalse vacuum.\u201d Yet, prior to the early work of Coleman on the quantum mechanical problem, similar ideas had been developed by Langer in the context of classical bubble nucleation. The following problem is an attempt to draw the connections between the classical and the quantum problem. As posed, the quantum formulation describes the depinning of a \ufb02ux line in a superconductor from a columnar defect. 150 Feynman path integral Consider a quantum elastic string embedded in a three-dimensional space and \u201cpinned\u201d by a columnar defect potential V oriented parallel to the z-axis. The corresponding Euclidean time action is given by S[u] = d\u03c4 dz 1 2 \u03c1 \u02d9u2 + 1 2 \u03c3(\u2202zu)2 + V (|u|), u where the two-dimensional vector \ufb01eld u(z, \u03c4 ) denotes the string displacement within the xy-plane, \u03c1 represents the density per unit length, and \u03c3 de\ufb01nes the tension in the string. On this system (See \ufb01gure), let us suppose that an external in-plane \ufb01eld f is imposed along the x-direction, Sext = \u2212f dz u\u00b7ex. Following the steps below, determine the probability (per unit time and per unit length) for the string to detach from the defect: d\u03c4 f y z x (a) Derive a saddle-point equation in the two-dimensional z\u03c4 -space. Rescaling the coordi- nates, transform the equation of motion to a problem with circular symmetry. (b) If the \ufb01eld is weak, one can invoke a \u201cthin-wall\u201d or \u201cbubble\u201d approximation to describe the saddle-point solution u(z, t) by specifying two regions of space-time, where the string is free, or is completely locked to the defect, respectively. In this approximation, \ufb01nd u(z, t). (Hint: Use the fact that, in either case, complete locking or complete freedom, the potential does not exert a net force on the string.) (c) With exponential accuracy, determine the detaching probability. You may assume that, for all values of ux obtained in (b), V (|u|) V0 = const. (Exercise: Think how the quantum model can be related to the classical system.) Answer: (a) Varying the action with respect to ux, the saddle-point equation assumes the form \u03c1\u00a8ux + \u03c3\u22022ux = \u2212f + V (u)(ux/u), where u = |u|. Applying the rescaling \u03c4 = (\u03c1/\u03c3)1/43\u03c4 and \u03c3\u03c1\u22022ux = \u2212f + V (u)(ux/u), z = (\u03c3/\u03c1)1/43z, the equation takes the symmetrized form where \u22022 = \u22022 z and boundary conditions \u03c4 + \u22022 \u221a ux(r) = 0, r > R, g(r), r < R, on the radial coordinate (3\u03c4, 3z) \u2192 (r, \u03c6) are imposed. (b) In the thin-wall approximation, the potential gradient can be neglected. In this case \u03c3\u03c1, with the solution g = the saddle-point equation assumes the form \u22072g = \u2212f / (R2 \u2212 r2)f / 4\u03c3\u03c1. \u221a \u221a (c) With the result, the tunneling rate can be estimated from the saddle-point action R Sbubble = dr \u221a \u03c3\u03c1 2 (3\u2202g)2 + V0 \u2212 f g = \u2212\u03c0R2 3f 2R2 16(\u03c3\u03c1)1/2 \u2212 V0. 3.5 Problems 151 Minimizing over R, one obtains the optimal radius R2 the estimate for the tunneling rate W \u221d e\u2212S(R\u2217) = exp \u2217 = 8V0 # \u221a \u03c3\u03c1 3f 2 \u2212 4\u03c0V 2 0 3f 2 \u03c3\u03c1.. As a result, we obtain \u221a $ Tunneling in a dissipative environment In Section 3.3 we considered the impact of dissipation on the action of a point-particle in a quantum well. There a model was chosen in which the degrees of freedom of the environment were represented phenomenologically by a bath of harmonic oscillators. In the following we will explore a model in which the particle is coupled to the \ufb02uctuations of a quantum mechanical \u201c", "string.\u201d Later, in Section 8.2, we will see that this model provides a description of tunneling through a single impurity in a Luttinger liquid. (a) A quantum particle of mass m is con\ufb01ned by a sinusoidal potential U (q) = 2g sin2(\u03c0q/q0). Employing the Euclidean (imaginary time) Feynman path integral ( = 1), Z = Dq(\u03c4 ) e\u2212Spart[q(\u03c4 )], Spart[q(\u03c4 )] = d\u03c4 \u2212\u221e \u221e $ \u02d9q2 + U (q), # m 2 con\ufb01rm by direct substitution that the extremal contribution to the propagator connecting two neighboring degenerate minima (q(\u03c4 = \u2212\u221e) = 0 and q(\u03c4 = \u221e) = q0) is given by the instanton trajectory qcl(\u03c4 ) = (2q0/\u03c0) arctan (exp [\u03c90\u03c4 ]), where \u03c90 = (2\u03c0/q0) g/m. Show that S[qcl] = (2/\u03c02)mq2 0\u03c90. (Note: Although the equation of motion associated with the minimum of the Euclidean path integral is nonlinear, the solution above is exact. It is known in the literature as a soliton con\ufb01guration.) (b) If the quantum particle is coupled at one point to an in\ufb01nite \u201cstring,\u201d the path integral is given by Z = Du(x, \u03c4 ) Dq(\u03c4 ) \u03b4 (q(\u03c4 ) \u2212 u(\u03c4, x = 0)) e\u2212Sstring[u(x,\u03c4 )]\u2212Spart[q(\u03c4 )], where the classical action of the string is given by (cf. the action functional for phonons discussed in Section 1.1) Sstring[u(x, \u03c4 )] = \u221e \u221e d\u03c4 dx \u2212\u221e \u2212\u221e \u03c1 2 \u02d9u2 + \u03c3 2 2 \u2202u \u2202x. Here \u03b4(q(\u03c4 ) \u2212 u(\u03c4, x = 0)) represents a functional \u03b4-function which enforces the condition q(\u03c4 ) =u( \u03c4, x = 0) for all times \u03c4. Operationally, it can be understood from the n \u03b4(q(\u03c4n) \u2212 u(\u03c4n, x = 0)). By representing the functional \u03b4-function discretized form as the functional integral \u221e \u03b4(q(\u03c4 ) \u2212 u(\u03c4, 0)) = Df (\u03c4 ) exp i d\u03c4 f (\u03c4 )(q(\u03c4 ) \u2212 u(\u03c4, 0)), \u2212\u221e and integrating over the \ufb02uctuations of the string, show that the dynamics of the par(d\u03c9/2\u03c0)|\u03c9q(\u03c9)|2, ticle is governed by the e\ufb00ective action Se\ufb00 [q] =S part[q] + (\u03b7/2) 152 Feynman path integral \u221a where \u03b7 = Section 3.3? \u03c1\u03c3. How does this result compare with the dissipative action discussed in (c) Treating the correction to the particle action as a perturbation, use your result from (a) to show that the e\ufb00ective action for an instanton\u2013anti-instanton pair q(\u03c4 ) = qcl(\u03c4 + \u00af\u03c4 /2) \u2212 qcl(\u03c4 \u2212 \u00af\u03c4 /2), where \u03c90 \u00af\u03c4 1, is given approximately by Se\ufb00 [q] = 2Spart[qcl] \u2212 \u03b7q2 0 \u03c0 ln (\u03c90 \u00af\u03c4 ). (Hint: Note that, in \ufb01nding the Fourier decomposition of qcl(\u03c4 ), a crude estimate is su\ufb03cient.) (d) Using this result, estimate the typical separation of the pair (i.e. interpret the overall action as an e\ufb00ective probability distribution function for \u00af\u03c4, and evaluate \u00af\u03c4 = d\u00af\u03c4 \u00af\u03c4 e\u2212Seff ). Comment on the implications of your result for the nature of the tunneling probability. Answer: (a) Varying the Euclidean time path integral with respect to q(\u03c4 ) one \ufb01nds that the extremal \ufb01eld con\ufb01guration obeys the classical sine\u2013Gordon equation m\u00a8q \u2212 2\u03c0g q0 sin 2\u03c0q q0 = 0. Applying the trial solution, one \ufb01nds that the equation of motion is satis\ufb01ed if \u03c90 = (2\u03c0/q0) g/m. From this result one obtains the classical action \ufffd", "\ufffd S[qcl] = d\u03c4 0 # m 2 $ \u221e \u02d9q2 cl + U (qcl) = d\u03c4 m \u02d9q2 cl = m 0 q0 0 dq \u02d9qcl = 2 mq2 0 \u03c02 \u03c90. (b) In Fourier space, the action of the classical string takes the form \u221e \u221e Sstring = \u2212\u221e d\u03c9 2\u03c0 dk 2\u03c0 \u03c1\u03c92 + \u03c3k2 2 \u2212\u221e |u(\u03c9, k)|2. Representing the functional \u03b4-function as the functional integral \u221e Df exp i \u2212\u221e d\u03c9 2\u03c0 f (\u03c9) q(\u2212\u03c9) \u2212 \u221e \u2212\u221e dk 2\u03c0 u(\u2212\u03c9, \u2212k), and performing the integral over the degrees of freedom of the string, one obtains \u221e \u221e d\u03c4 f (\u03c4 )u(\u03c4,0) \u221d exp \u2212 \u2212\u221e d\u03c9 2\u03c0 dk 2\u03c0 1 2 1 (\u03c1\u03c92 + \u03c3k2) \u2212\u221e |f (\u03c9)|2. Du e\u2212Sstring\u2212i Integrating over k, and performing the Gaussian functional integral over the Lagrange multiplier \ufb01eld f (\u03c9), one obtains the e\ufb00ective action as required. 3.5 Problems 153 (c) Approximating the instanton/anti-instanton pair q(\u03c4 ) = qcl(\u03c4 + \u00af\u03c4 )\u2212qcl(\u03c4 \u2212 \u00af\u03c4 ) by a \u201ctop\u00af\u03c4 /2 \u2212\u00af\u03c4 /2 d\u03c4 q0ei\u03c9\u03c4 = q0 \u00af\u03c4 sin(\u03c9\u00af\u03c4 /2)/(\u03c9\u00af\u03c4 /2). Treating hat\u201d function, one \ufb01nds that q(\u03c9) = the dissipative term as a perturbation, the action then takes the form Se\ufb00 \u2212 2Spart = \u03c90 0 \u03b7 2 d\u03c9 2\u03c0 |\u03c9|(q0 \u00af\u03c4 )2 sin2(\u03c9\u00af\u03c4 /2) (\u03c9\u00af\u03c4 /2)2 q2 0 \u03c0 \u03b7 ln(\u03c90 \u00af\u03c4 ), where \u03c90 serves as a high-frequency cut-o\ufb00. (d) Interpreted as a probability distribution for the instanton separation, one \ufb01nds \u00af\u03c4 = d\u00af\u03c4 \u00af\u03c4 exp \u2212 q2 0 \u03c0 \u221e \u03b7 ln(\u03c90 \u00af\u03c4 ) \u223c d\u00af\u03c4 \u00af\u03c4 1\u2212q2 0 \u03b7/\u03c0. The divergence of the integral shows that, for \u03b7 > 2\u03c0/q2 0, instanton\u2013anti-instanton pairs are con\ufb01ned and particle tunneling is deactivated. Later, in Chapter 8 we revisit the dissipative phase transition from the standpoint of the renormalization group. Winding numbers In the main text, we considered the application of the Feynman path integral to model systems where trajectories could be parameterized in terms of their harmonic (Fourier) expansion. However, very often, one is interested in applications of the path integral to spaces that are not simply connected. In this case, one must include classes of trajectories which cannot be simply continued. Rather, trajectories are classi\ufb01ed by their \u201cwinding number\u201d on the space. To illustrate the point, let us consider the application of the path integral to a particle on a ring. (a) Starting with the Hamiltonian \u02c6H = \u2212(1/2I)(\u22022/\u2202\u03b82), where \u03b8 denotes an angle variable, show from \ufb01rst principles that the quantum partition function Z = tre\u2212\u03b2 \u02c6H is given by Z = \u221e n=\u2212\u221e exp \u2212\u03b2. n2 2I (3.61) (b) Formulated as a Feynman path integral, show that the quantum partition function can be cast in the form Z = 0 2\u03c0 d\u03b8 \u221e m=\u2212\u221e \u03b8(0) = \u03b8 \u03b8(\u03b2) = \u03b8(0) + 2\u03c0m D\u03b8(\u03c4 ) exp \u03b2 0 \u2212 I 2 d\u03c4 \u02d9\u03b82. (c) Varying the Euclidean action with respect to \u03b8, show that the path integral is minimized by the classical trajectories \u00af\u03b8(\u03c4 ) = \u03b8 + 2\u03c0m\u03c4 /\u03b2. Parameterizing a general path as \u03b8(\u03c4 ) = \u00af\u03b8(\u03c4 ) +\u03b7 (\u03c4 ), where \u03b7(\u03c4 ) is a path with no net winding, show that Z = Z0 \u221e m=\u2212\u221e exp \u2212 I 2 (2\u03c0m)2 \u03b2, (3.62)", " where Z0 represents the quantum partition function for a free particle with open boundary conditions. Making use of the free particle propagator, show that Z0 = I/2\u03c0\u03b2. 154 Feynman path integral (d) Finally, making use of Poisson\u2019s summation formula, show that Eq. (3.62) coincides with Eq. (3.61). m h(m) = \u221e \u2212\u221e d\u03c6 h(\u03c6)e2\u03c0in\u03c6, n Answer: (a) Solving the Schr\u00a8odinger equation, the wavefunctions obeying periodic boundary conditions take the form \u03c8n = ein\u03b8/ 2\u03c0, n integer, and the eigenvalues are given by En = n2/2I. Cast in the eigenbasis representation, the partition function assumes the form Eq. (3.61). \u221a (b) Interpreted as a Feynman path integral, the quantum partition function takes the form of a propagator with 2\u03c0 Z = 0 d\u03b8 \u03b8|e\u2212\u03b2 \u02c6H |\u03b8 = 2\u03c0 d\u03b8 0 \u03b8(\u03b2) =\u03b8 (0) = \u03b8 D\u03b8(\u03c4 ) exp \u2212 \u03b2 0 d\u03c4 \u02d9\u03b82. I 2 The trace implies that paths \u03b8(\u03c4 ) must start and \ufb01nish at the same point. However, to accommodate the invariance of the \ufb01eld con\ufb01guration \u03b8 under translation by 2\u03c0 we must impose the boundary conditions shown in the question. (c) Varying the action with respect to \u03b8 we obtain the classical equation I \u00a8\u03b8 = 0. Solving this equation subject to the boundary conditions, we obtain the solution given in the question. Evaluating the Euclidean action, we \ufb01nd that \u03b2 0 (\u2202\u03c4 \u03b8)2d\u03c4 = \u03b2 d\u03c4 0 2\u03c0m \u03b2 2 + \u2202\u03c4 \u03b7 = \u03b2 2 \u03b2 + 0 2\u03c0m \u03b2 d\u03c4 (\u2202\u03c4 \u03b7)2. Thus, we obtain the partition function (3.62), where ; Z0 = D\u03b7(\u03c4 ) exp d\u03c4 (\u2202\u03c4 \u03b7)2 = \u2212 I 2 \u03b2 0 I 2\u03c0\u03b2, denotes the free particle partition function. This can be obtained from direct evaluation of the free particle propagator. (d) Applying the Poisson summation formula with h(x) = exp[\u2212 (2\u03c0)2I 2\u03b2 x2], one \ufb01nds that \u221e m=\u2212\u221e e\u2212 (2\u03c0)2 Im2 2\u03b2 = \u221e \u221e n=\u2212\u221e \u2212\u221e d\u03c6 e\u2212 (2\u03c0)2 I 2\u03b2 \u03c62+2\u03c0in\u03c6 = \" Multiplication by Z0 obtains the result. \u03b2 2\u03c0I \u221e n=\u2212\u221e e\u2212 \u03b2 2I n2. Particle in a periodic potential In Section 3.3 it was shown that the quantum probability amplitude for quantum mechanical tunneling can be expressed as a sum over instanton \ufb01eld con\ufb01gurations of the Euclidean action. By generalizing this approach, the aim of the present problem is to explore quantum mechanical tunneling in a periodic potential. Such an analysis allows us to draw a connection to the problem of the Bloch spectrum. 3.5 Problems 155 (a) A quantum mechanical particle moves in a periodic lattice potential V with period a. Taking the Euclidean action for the instanton connecting two neighboring minima to be Sinst, express the Euclidean time propagator G(ma, na; \u03c4 ), with m and n integer, as a sum over instanton and anti-instanton \ufb01eld con\ufb01gurations. (b) Making use of the identity \u03b4qq = G(ma, na; \u03c4 ) \u223c e\u2212\u03c9\u03c4 /2 2\u03c0 0 d\u03b8 ei(q\u2212q)\u03b8/(2\u03c0) show that \u0394\u03c4 e\u2212i(n\u2212m)\u03b8 exp 2\u03c0 d\u03b8 2\u03c0 0, 2 cos\u03b8 where our notation is taken from Section 3.3. (c) Keeping in mind that, in the periodic system, the eigenfunctions are Bloch states \u03c8p\u03b1(q) = eipqup\u03b1(q) whereu p\u03b1(q + ma) \u2261 up\u03b1(q) denotes the periodic part of the Bloch function, show that the propagator is compatible with a spectrum of the lowest band \u03b1 = 0, p = \u03c9/2 \u2212 2\u0394 cos(pa). Answer: (a) In the double-well potential, the extremal \ufb01eld con\ufb01gurations", " of the Euclidean action involve consecutive sequences of instanton\u2013anti-instanton pairs. However, in the periodic potential, the q instantons and q anti-instantons can appear in any sequence provided only that q \u2212 q = m \u2212 n. In this case, the Feynman amplitude takes the form G(ma, na; \u03c4 ) \u223c \u221e \u221e q=0 q=0 \u03b4q\u2212q,n\u2212m q! q! (\u03c4 Ke\u2212Sinst/)q+q e\u2212\u03c9\u03c4 /2. (b) To evaluate the instanton summation, we can make use of the identity \u03b4q\u2212q,n\u2212m = 2\u03c0 0 d\u03b8 ei(q\u2212q\u2212n+m)\u03b8/(2\u03c0). As a result, we obtain G(ma, na; \u03c4 ) \u223c e\u2212\u03c9\u03c4 /2 e\u2212i(n\u2212m)\u03b8 2\u03c0 d\u03b8 2\u03c0 \u221e q=0 (\u03c4 Kei\u03b8e\u2212Sinst/)q q! \u221e q=0 (\u03c4 Ke\u2212i\u03b8e\u2212Sinst/)q q! d\u03b8 2\u03c0 e\u2212i(n\u2212m)\u03b8 exp \u0394\u03c4 ei\u03b8 exp \u0394\u03c4 e\u2212i\u03b8, \u223c e\u2212\u03c9\u03c4 /2 0 2\u03c0 0 from which can be obtained the required result. (c) Expanded in terms of the Bloch states of the lowest band of the periodic potential \u03b1 = 0, one obtains G(ma, na; \u03c4 ) = p p(ma)\u03c8p(na)e\u2212p\u03c4 / = \u03c8\u2217 p |up(0)|2eip(n\u2212m)ae\u2212p\u03c4 /. Interpreting \u03b8 = pa, and taking |up(0)|2 = const. independent of p, we can draw the correspondence p = \u03c9/2 \u2212 2\u0394 cos(pa). 4 Functional \ufb01eld integral In this chapter, the concept of path integration is generalized to integration over quantum \ufb01elds. Speci\ufb01cally we will develop an approach to quantum \ufb01eld theory that takes as its starting point an integration over all con\ufb01gurations of a given \ufb01eld, weighted by an appropriate action. To emphasize the importance of the formulation that, methodologically, represents the backbone of the remainder of the text, we have pruned the discussion to focus only on the essential elements. This being so, conceptual aspects stand in the foreground and the discussion of applications is postponed to the following chapters. In this chapter, the concept of path integration is extended from quantum mechanics to quantum \ufb01eld theory. Our starting point is a situation very much analogous to that outlined at the beginning of the previous chapter. Just as there are two di\ufb00erent approaches to quantum mechanics, quantum \ufb01eld theory can also be formulated in two di\ufb00erent ways: the formalism of canonically quantized \ufb01eld operators, and functional integration. As to the former, although much of the technology needed to e\ufb03ciently implement this framework \u2013 essentially Feynman diagrams \u2013 originated in high-energy physics, it was with the development of condensed matter physics through the 1950s, 1960s, and 1970s that this approach was driven to unprecedented sophistication. The reason is that, almost as a rule, problems in condensed matter investigated at that time necessitated perturbative summations to in\ufb01nite order in the non-trivial content of the theory (typically interactions). This requirement led to the development of advanced techniques to sum perturbation series in many-body interaction operators to in\ufb01nite order. In the 1970s, however, essentially non-perturbative problems began to attract more and more attention \u2013 a still prevailing trend \u2013 and it turned out that the formalism of canonically quantized operators was not tailored to this type of physics. By contrast, the alternative approach to many-body problems, functional integration, is ideally suited! The situation is similar to the one described in the last chapter, where we saw that the Feynman path integral provided a spectrum of novel routes to approaching quantum mechanical problems (controlled semiclassical limits, analogies to classical mechanics, statistical mechanics, concepts of topology and geometry, etc.). Similarly, the introduction of \ufb01eld integration into many-body physics spawned new theoretical developments, many of which went beyond perturbation theory. In fact, the advantage of the path integral approach in many-body physics is more pronounced than in single-particle quantum mechanics: higher dimensionality introduces more complex \ufb01elds, and the concept of \ufb01eld integration is ideally suited to explore the ensuing structures. Also, the connections to classical statistical mechanics 156 Functional \ufb01", "eld integral 157 play a more important role than in single-particle quantum mechanics. All of these concepts will begin to play a role in subsequent chapters when applications of the \ufb01eld integral are discussed. Degrees of freedom Path integral QM q q \u03c6 \u03c6 QFT x x t t Figure 4.1 The concept of \ufb01eld integration. Upper panels: path integral of quantum mechanics \u2013 integration over all time-dependent con\ufb01gurations of a point particle degree of freedom leads to integrals over curves. Lower panels: \ufb01eld integral \u2013 integration over time-dependent con\ufb01gurations of d-dimensional continuum mappings (\ufb01elds) leads to integrals over generalized (d+1)-dimensional surfaces. Before embarking on the quantitative construction of the \ufb01eld integral \u2013 the subject of the following sections \u2013 let us anticipate the kind of structures that one should expect. In quantum mechanics, we were starting from a single point particle degree of freedom, characterized by some coordinate q (or some other quantum numbers for that matter). Path integration then meant integration over all time-dependent con\ufb01gurations q(t), i.e. a set of curves t \u2192 q(t) (see Fig. 4.1, upper panel). By contrast, the degrees of freedom of \ufb01eld theory are continuous objects \u03c6(x) by themselves, where x parameterizes some ddimensional base manifold and \u03c6 takes values in some target manifold (Fig. 4.1, lower panel). The natural generalization of a \u201cpath\u201d integral then implies integration over a single copy of these objects at each instant of time, i.e. we shall have to integrate over generalized surfaces, mappings from (d+1)-dimensional space-time into the \ufb01eld manifold, (x, t) \u2192 \u03c6(x, t). While this notion may sound worrying, it is important to realize that, conceptually, nothing much changes in comparison with the path integral: instead of a one-dimensional manifold \u2013 a curve \u2013 our object of integration will be a (d + 1)-dimensional manifold. We now proceed to formulate these ideas in quantitative terms. EXERCISE If necessary, recapitulate the general construction scheme of path integrals (Section 3.2) and the connection between quantum \ufb01elds and second quantized operators. 158 Functional \ufb01eld integral 4.1 Construction of the many-body path integral The construction of a path integral for \ufb01eld operators follows the general scheme outlined at the end of Section 3.2. The basic idea is to segment the time evolution of a quantum (many-body) Hamiltonian into in\ufb01nitesimal time slices and to absorb as much as possible of the quantum dynamical phase accumulated during the short-time propagation into a set of suitably chosen eigenstates. But how should these eigenstates be chosen? In the context of single-particle quantum mechanics, the structure of the Hamiltonian suggested a representation in terms of coordinate and momentum eigenstates. Now, given that many-particle Hamiltonians are conveniently expressed in terms of creation/annihilation operators, an obvious idea would be to search for eigenstates of these operators. Such states indeed exist and are called coherent states. Coherent states (bosons) Our goal is, therefore, to \ufb01nd eigenstates of the (non-Hermitian) Fock space operators a\u2020 and a. Although the general form of these states will turn out to be the same for bosons and fermions, there are major di\ufb00erences regarding their algebraic structure. The point is that the anti-commutation relations of fermions require that the eigenvalues of an annihilation operator themselves anti-commute, i.e. they cannot be ordinary numbers. Postponing the introduction of the unfamiliar concept of anti-commuting \u201cnumbers\u201d to the next section, we \ufb01rst concentrate on the bosonic case where problems of this kind do not arise. So what form do the eigenstates |\u03c6 of the bosonic Fock space operators a and a\u2020 take? Being a state of the Fock space, an eigenstate |\u03c6 can be expanded as |\u03c6 = n1,n2,... Cn1,n2,...|n1, n2,..., |n1, n2,... = \u2020 1)n1 (a \u221a n1! \u2020 2)n2 (a \u221a n2!...| 0, \u2020 i creates a boson in state i, Cn1,n2,... represents a set of expansion coe\ufb03cients, where a and |0 represents the vacuum. Here, for reasons of clarity, it", " is convenient to adopt this convention for the vacuum as opposed to the notation |\u03a9 used previously. Furthermore, the many-body state |n1, n2,... is indexed by a set of occupation numbers: n1 in state |1, n2 in state |2, and so on. Importantly, the state |\u03c6 can, in principle (and will in practice) contain a superposition of basis states which have di\ufb00erent numbers of particles. Now, if |\u03c6 must be n0 + 1. the minimum number of particles in state |\u03c6 is n0, the minimum of a Clearly the creation operators a \u2020 i \u2020 i themselves cannot possess eigenstates. However, with annihilation operators this problem does not arise. Indeed, the annihilation operators do possess eigenstates, known as boson coherent states, |\u03c6 \u2261 exp |0, \u03c6ia \u2020 i i (4.1) 4.1 Construction of the many-body path integral 159 where the elements of \u03c6 = {\u03c6i} represent a set of complex numbers. The states |\u03c6 are eigenstates in the sense that, for all i, ai|\u03c6 = \u03c6i|\u03c6, (4.2) \u2020 i.e. they simultaneously diagonalize all annihilation operators. Noting that ai and a j, with j = i, commute, Eq. (4.2) can be veri\ufb01ed by showing that a exp(\u03c6a\u2020)|0 = \u03c6 exp(\u03c6a\u2020)|0.1 Although not crucial to the practice of \ufb01eld integration, in the construction of the path integral it will be useful to assimilate some further properties of coherent states: By taking the Hermitian conjugate of Eq. (4.2), we \ufb01nd that the \u201cbra\u201d associated with the \u201cket\u201d |\u03c6 is a left eigenstate of the set of creation operators, i.e. for all i, \u03c6|a \u2020 i = \u03c6| \u00af\u03c6i, (4.3) where \u00af\u03c6i is the complex conjugate of \u03c6i, and \u03c6| = 0|exp[ i \u00af\u03c6iai]. It is a straightforward matter \u2013 e.g. by a Taylor expansion of Eq. (4.1) \u2013 to show that the action of a creation operator on a coherent state yields the identity a \u2020 i |\u03c6 = \u2202\u03c6i |\u03c6. (4.4) Reassuringly, it may be con\ufb01rmed that Eq. (4.4) and (4.2) are consistent with the commutation relations [ai, a j]|\u03c6 = (\u2202\u03c6j \u03c6i \u2212 \u03c6i\u2202\u03c6j )|\u03c6 = \u03b4ij|\u03c6. \u2020 j] = \u03b4ij: [ai, a \u2020 Making use of the relation \u03b8|\u03c6 = 0|e \u00af\u03b8iai |\u03c6 = e i \u00af\u03b8i\u03c6i 0|\u03c6, one \ufb01nds that the i overlap between two coherent states is given by \u03b8|\u03c6 = exp \u00af\u03b8i\u03c6i. i From this result, one can infer that the norm of a coherent state is given by \u03c6|\u03c6 = exp \u00af\u03c6i\u03c6i. i (4.5) (4.6) Most importantly, the coherent states form a complete \u2013 in fact an overcomplete \u2013 set of states in Fock space:'i d \u00af\u03c6id\u03c6i \u03c0 e\u2212 \u00af\u03c6i\u03c6i |\u03c6\u03c6| = 1F, i (4.7) where d \u00af\u03c6id\u03c6i = d Re \u03c6i d Im \u03c6i, and 1F represents the unit operator or identity in the Fock space. 1 Using the result [a, (a\u2020)n] = n(a\u2020)n\u22121 (cf. Eq. (2.38)) a Taylor expansion shows a exp(\u03c6a\u2020)|0 = \u221e \u03c6n\u22121 (n\u22121)! (a\u2020)n\u22121|0 = n\u03c6n n! (a\u2020)n\u22121|0 = \u03c6 \u03c6n n! [a, (a\u2020)n]|0 = \u221e \u221e n=1 n=1 n=0 [a, exp(\u03c6a\u2020)]|0 = \u03c6 exp(\u03c6a\u2020)|0. 160 Functional \ufb01eld integral INFO The proof of Eq. (4.7) proceeds by application of Schur\u2019s lemma (cf", ". our discussion of the completeness of the spin coherent states in the previous chapter). The operator family {ai}, {a\u2020 } acts irreducibly in Fock space. According to Schur\u2019s lemma, the proportionality of the lefti hand side of Eq. (4.7) to the unit operator is, therefore, equivalent to its commutativity with all creation and annihilation operators. Indeed, this property is easily con\ufb01rmed: ai d( \u00af\u03c6, \u03c6) e\u2212 i \u00af\u03c6i\u03c6i |\u03c6\u03c6| = = by parts = = \u2212 d( \u00af\u03c6, \u03c6) e\u2212 & i d( \u00af\u03c6, \u03c6) d( \u00af\u03c6, \u03c6) e\u2212 \u2202 \u00af\u03c6i \u00af\u03c6i\u03c6i \u03c6i|\u03c6\u03c6|'e\u2212 \u00af\u03c6i\u03c6i i |\u03c6\u03c6| i \u00af\u03c6i\u03c6i |\u03c6! \u2202 \u00af\u03c6i \u03c6| d( \u00af\u03c6, \u03c6) e\u2212 i \u00af\u03c6i\u03c6i |\u03c6\u03c6|ai, (4.8) i d \u00af\u03c6i d\u03c6i/\u03c0. Taking the adjoint of Eq. (4.8), one may where, for brevity, we have set d( \u00af\u03c6, \u03c6) \u2261 further check that the left-hand side of Eq. (4.7) commutes with the set of creation operators, i.e. it must be proportional to the unit operator. To \ufb01x the constant of proportionality, one may simply take the overlap with the vacuum: d( \u00af\u03c6, \u03c6) e\u2212 i \u00af\u03c6i\u03c6i 0|\u03c6\u03c6|0 = d( \u00af\u03c6, \u03c6) e\u2212 i \u00af\u03c6i\u03c6i = 1. (4.9) Taken together, Eq. (4.8) and (4.9) prove (4.7). Note that the coherent states are overcomplete in the sense that they are not pairwise orthogonal (see Eq. (4.5)). The exponential weight \u00af\u03c6i\u03c6i appearing in the resolution of the identity compensates for the overcounting achieved e\u2212 by integrating over the whole set of coherent states. i With these de\ufb01nitions we have all that we need to construct the path integral for the bosonic system. However, before doing so, we will \ufb01rst introduce the fermionic version of the coherent state. This will allow us to construct the path integrals for bosons and fermions simultaneously, thereby emphasizing the similarity of their structure. Coherent states (fermions) Much of the formalism above generalizes to the fermionic case: as before, it is evident that creation operators cannot possess eigenstates. Following the bosonic system, let us suppose that the annihilation operators are characterized by a set of coherent states such that, for all i, ai|\u03b7 = \u03b7i|\u03b7, (4.10) where \u03b7i is the eigenvalue. Although the structure of this equation appears to be equivalent to its bosonic counterpart Eq. (4.2) it has one frustrating feature: anti-commutativity of the fermionic operators, [ai, aj]+ = 0, where i = j, implies that the eigenvalues \u03b7i also have to anti-commute, \u03b7i\u03b7j = \u2212\u03b7j\u03b7i. (4.11) 4.1 Construction of the many-body path integral 161 Clearly, these objects cannot be ordinary numbers. In order to de\ufb01ne a fermionic version of coherent states, we now have two choices: we may (a) accept Eq. (4.11) as a working de\ufb01nition and pragmatically explore its consequences, or (b) \ufb01rst try to remove any mystery from the de\ufb01nitions (4.10) and (4.11). This latter task is tackled in the Info block below where objects {\u03b7i} with the desired properties are de\ufb01ned in a mathematically clean way. Readers wishing to proceed in a maximally streamlined manner may skip this exposition and directly turn to the more praxis-oriented discussion below. now called Hermann Gnther Grassmann 1809\u201377 Credited with inventing what is algebra.(Figure reproduced from Hermann Grassmann, Gesamnelte Mathematische and Physikalische Werke (Druck and Verlag von B. G. Teubner, 1894).) exterior INFO There is a mathematical structure ideally suited to generalize the concept of ordinary number (", "\ufb01elds), namely algebras. An algebraA is a vector space endowed with a multiplication rule A \u00d7 A \u2192 A. So let usconstruct an algebra A by starting out from a set of elements, or generators, \u03b7i \u2208 A, i = 1,..., N, and imposing the rules: (i) The elements \u03b7i can be added and multiplied by complex numbers, i.e. c0 + ci\u03b7i + cj\u03b7j \u2208 A, c0, ci, cj \u2208 C, (4.12) i.e. A is a complex vector space. (ii) The product, A \u00d7 A \u2192 A, (\u03b7i, \u03b7j) \u2192 \u03b7i\u03b7j, is associative and anti-commutative, i.e. it obeys the anti-commutation relation (4.11). Because of the associativity of this operation, there is no ambiguity when it comes to forming products of higher order, i.e. (\u03b7i\u03b7j)\u03b7k = \u03b7i(\u03b7j\u03b7k) \u2261 \u03b7i\u03b7j\u03b7k. The de\ufb01nition requires that products of odd order in the number of generators anticommute, while (even, even) and (even, odd) combinations commute (exercise). By virtue of (i) and (ii), the set A of all linear combinations \u221e N c0 + n=1 i1,...in=1 ci1,...,in \u03b7i1... \u03b7in, c0, ci1,...,in \u2208 C, spans a \ufb01nite-dimensional associative algebra A,2 known as the Grassmann algebra (and sometimes also the exterior algebra). For completeness we mention that Grassmann algebras \ufb01nd a number of realizations in mathematics, the most basic being exterior multiplication in linear algebra. Given an N -dimensional vector space V, letV \u2217 be the dual space, i.e. the space of all linear mappings, or \u201cforms\u201d \u039b : V \u2192 C, v \u2192 \u039b(v), where v \u2208 V. (LikeV, V \u2217 is a vector space of dimension N.) Next, de\ufb01ne exterior multiplication through (\u039b, \u039b) \u2192 \u039b \u2227 \u039b, where \u039b \u2227 \u039b is the mapping v, v) \u2192 \u039b(v)\u039b(v) \u2212 \u039b(v)\u039b(v). 2 Whose dimension can be shown to be 2N (exercise). 162 Functional \ufb01eld integral This operation is manifestly anti-commutative: \u039b \u2227 \u039b = \u2212\u039b \u2227 \u039b. Identifying the N linear basis forms \u039bi \u2194 \u03b7i with generators and \u2227 with the product, we see that the space of exterior forms de\ufb01nes a Grassmann algebra. Apart from their anomalous commutation properties, the generators {\u03b7i}, and their product generalizations {\u03b7i\u03b7j, \u03b7i\u03b7j\u03b7k,...}, resemble ordinary, albeit anti-commutative, numbers. (In practice, the algebraic structure underlying their de\ufb01nition can safely be ignored. All we will need to work with these objects is the basic rule Eq. (4.11) and the property Eq. (4.12).) We emphasize that A contains not only anti-commuting but also commuting elements, i.e. linear combinations of an even number of Grassmann numbers \u03b7i are overall commutative. (This mimics the behavior of the Fock space algebra: products of an even number of annihilation operators aiaj... commute with all other linear combinations of operators ai. In spite of this similarity, the \u201cnumbers\u201d \u03b7i must not be confused with the Fock space operators; there is nothing on which they act.) To make practical use of the new concept, we need to go beyond the level of pure arithmetic. Speci\ufb01cally, we need to introduce functions of anti-commuting numbers, and elements of calculus. Remarkably, not only do most of the concepts of calculus \u2013 di\ufb00erentiation, integration, etc. \u2013 naturally generalize to anti-commuting number \ufb01elds, but, contrary to what one might expect, they turn out to be simpler than in ordinary calculus. Functions of Grassmann numbers are de\ufb01ned via their Taylor expansion: f (\u03be1,..., \u03bek) = \u221e k n=0 i1,...,in=1 1 n! \u2202n", "f \u00b7 \u00b7 \u00b7\u2202\u03be in \u2202\u03bei1 \u03be=0 \u03bein \u00b7 \u00b7 \u00b7\u03be i1, \u03be1,..., \u03bek \u2208 A, (4.13) where f is an analytic function. Note that the anti-commutation properties of the algebra imply that the series terminates after a \ufb01nite number of terms. For example, in the simple case where \u03b7 is \ufb01rst-order in the generators of the algebra, N = 1, and f (\u03b7) = f (0)+f (0)\u03b7 (since \u03b72 = 0) \u2013 functions of Grassmann variables are fully characterized by a \ufb01nite number of Taylor coe\ufb03cients! Di\ufb00erentiation with respect to Grassmann numbers is de\ufb01ned by \u2202\u03b7i \u03b7j = \u03b4ij. (4.14) Note that, in order to be consistent with the commutation relations, the di\ufb00erential operator \u2202\u03b7i must itself be anti-commutative. In particular, \u2202\u03b7i \u03b7j\u03b7i i=j = \u2212\u03b7j. Integration over Grassmann variables is de\ufb01ned by d\u03b7i = 0, d\u03b7i \u03b7i = 1. (4.15) Note that the de\ufb01nitions (4.13), (4.14), and (4.15) imply that the actions of Grassmann di\ufb00erentiation and integration are e\ufb00ectively identical, that is d\u03b7 f (\u03b7) = d\u03b7 (f (0) + f (0)\u03b7) = f (0) = \u2202\u03b7f (\u03b7). 4.1 Construction of the many-body path integral 163 With this background, let us now proceed to apply the Grassmann algebra to the construction of fermion coherent states. To this end we need to enlarge the algebra so as to allow for a multiplication of Grassmann numbers by fermion operators. In order to be consistent with the anti-commutation relations, we need to require that fermion operators and Grassmann generators anti-commute, [\u03b7i, aj]+ = 0. (4.16) It then becomes a straightforward matter to demonstrate that fermionic coherent states are de\ufb01ned by |\u03b7 = exp \u2212 |0, \u03b7ia \u2020 i i (4.17) i.e. by a structure perfectly analogous to the bosonic states (4.1).3 It is a straightforward matter \u2013 and also a good exercise \u2013 to demonstrate that the properties (4.3), (4.4), (4.5), (4.6), and, most importantly, (4.7) carry over to the fermionic case. One merely has to identify ai with a fermionic operator and replace the complex variables \u03c6i by \u03b7i \u2208 A. Apart from a few sign changes and the A-valued arguments, the fermionic coherent states di\ufb00er only in two respects from their bosonic counterpart: \ufb01rstly, the Grassmann variables \u00af\u03b7i appearing in the adjoint of a fermion coherent state, \u03b7| = 0|exp \u2212 i ai \u00af\u03b7i = 0|exp \u00af\u03b7iai, i are not related to the \u03b7is of the state |\u03b7 via some kind of complex conjugation. Rather \u03b7i and \u00af\u03b7i are strictly independent variables.4 Secondly, the Grassmann version of a Gaussian d\u00af\u03b7 d\u03b7 e\u2212\u00af\u03b7\u03b7 = 1 does not contain the factors of \u03c0 characteristic of integral (exercise), standard Gaussian integrals. Thus, the measure of the fermionic analog of Eq. (4.7) does not contain a \u03c0 in the denominator. For the sake of future reference, the most important properties of Fock space coherent states are summarized in Table 4.1. \u2020 3 To prove that the states (4.17) indeed ful\ufb01ll the de\ufb01ning relation (4.10), we note that aiexp(\u2212\u03b7ia i )|0 (4.13) = i )|0 (4.16) i )|0. This, in combination with the ai(1 \u2212 \u03b7ia j (i = j) commute, proves Eq. (4.10). Note that the proof has actually been simpler fact that ai and \u03b7j a than in the bosonic case. The fermionic Taylor series terminates after the \ufb01rst contribution. This observation is representative of a general rule: Grassmann calculus is simpler than", " standard calculus \u2013 all series are \ufb01nite, integrals always converge, etc. |0 = \u03b7i|0 = \u03b7i(1 \u2212 \u03b7ia i )|0 = \u03b7iexp(\u2212\u03b7ia = \u03b7iaia \u2020 \u2020 i \u2020 \u2020 \u2020 4 In the literature, complex conjugation of Grassmann variables is sometimes de\ufb01ned. Although appealing from an aesthetic point of view \u2013 symmetry between bosons and fermions \u2013 this concept is problematic. The di\ufb03culties become apparent when supersymmetric theories are considered, i.e. theories where operator algebras contain both bosons and fermions (the so-called super-algebras). It is not possible to introduce a complex conjugation that leads to compatibility with the commutation relations of a super-algebra. It therefore seems to be better to abandon the concept of Grassmann complex conjugation altogether. (Unlike with the bosonic case where complex conjugation is inevitable in order to de\ufb01ne convergent Gaussian integrals, no such need arises in the fermionic case.) 164 Functional \ufb01eld integral Table 4.1 Basic properties of coherent states for bosons (\u03b6 = 1, \u03c8i \u2208 C) and fermions (\u03b6 = \u22121, \u03c8i \u2208 A). In the last line, the integration measure is de\ufb01ned as i d( \u00af\u03c8, \u03c8) \u2261 De\ufb01nition Action of ai Action of a\u2020 i Overlap Completeness d \u00af\u03c8i d\u03c8i \u03c0(1+\u03b6)/2. &'|\u03c8 = exp \u03b6 i ai|\u03c8 = \u03c8i|\u03c8, a\u2020 |\u03c8 = \u03b6\u2202\u03c8i i \u03c8|\u03c8 = exp d( \u00af\u03c8, \u03c8) e\u2212 i \u03c8ia\u2020 |0 \u03c8|ai = \u2202 \u00af\u03c8i \u03c8| \u03c8|a\u2020 i = \u03c8| \u00af\u03c8i'i\u03c8i \u00af\u03c8 i \u00af\u03c8i\u03c8i |\u03c8\u03c8| = 1F i |\u03c8, & INFO Grassmann Gaussian integration: Finally, before turning to the development of the \ufb01eld integral, it is useful to digress on the generalization of Gaussian integrals for Grassmann variables. The prototype of all Grassmann Gaussian integration formulae is the identity d\u00af\u03b7 d\u03b7 e\u2212\u00af\u03b7a\u03b7 = a, (4.18) where a \u2208 C takes arbitrary values. Equation (4.18) is derived by a \ufb01rst-order Taylor expansion of the exponential and application of Eq. (4.15). The multi-dimensional generalization of Eq. (4.18) is given by d( \u00af\u03c6, \u03c6) e\u2212 \u00af\u03c6T A\u03c6 = det A, (4.19) N where \u00af\u03c6 and \u03c6 are N -component vectors of Grassmann variables, the measure d( \u00af\u03c6, \u03c6) \u2261 i=1 d \u00af\u03c6i d\u03c6i, and A may be an arbitrary complex matrix. For matrices that are unitarily diagonalizable, A = U\u2020DU, with U unitary, and D diagonal, Eq. (4.19) is proven in the same way as its complex counterpart (3.17): one changes variables \u03c6 \u2192 U\u2020\u03c6, \u00af\u03c6 \u2192 UT \u00af\u03c6. Since det U = 1, the transform leaves the measure invariant (see below) and leaves us with N decoupled integrals of the type Eq. (4.18). The resulting product of N eigenvalues is just the determinant of A (cf. the later discussion of the partition function of the non-interacting gas). For general (non-diagonalizable) A, the identity is established by a straightforward expansion of the exponent. The expansion terminates at N th order and, by commuting through integration variables, it may be shown that the resulting N th-order polynomial of matrix elements of A is the determinant.5 5 As with ordinary integrals, Grassmann integrals can also be subjected to variable transforms. Suppose we are given an integral d( \u00af\u03c6, \u03c6) f ( \u00af\u03c6, \u03c6) and wish to change variables according to \u00af\u03bd = M \u00af\u03c6, \u03bd = M \u03c6, (4.20) where, for simplicity, M and M are complex matrices (i.e. we here restrict ourselves to linear transforms). One can show that \u00af\u03bd1 \u00b7 \u00b7 \u00b7 \u00af\u03bdN = (det M) \u00af\ufffd", "\ufffd1 \u00b7 \u00b7 \u00b7 \u00af\u03c6N, \u03bd1 \u00b7 \u00b7 \u00b7\u03bd N = (det M )\u03c61 \u00b7 \u00b7 \u00b7\u03c6 N. (4.21) (There are di\ufb00erent ways to prove this identity. The most straightforward is by explicitly expanding Eq. (4.20) in components and commuting all Grassmann variables to the right. A more elegant way is to argue that the coe\ufb03cient relating the right- and left-hand sides of Eq. (4.21) must be an N th-order polynomial of matrix 4.2 Field integral for the quantum partition function 165 Keeping the analogy with ordinary commuting variables, the Grassmann version of Eq. (3.18) reads d( \u00af\u03c6, \u03c6) e\u2212 \u00af\u03c6T A\u03c6+\u00af\u03bdT \u00b7\u03c6+ \u00af\u03c6T \u00b7\u03bd = det A e\u00af\u03bdT A\u22121\u03bd. (4.22) d\u03b7 f (\u03b7) = To prove the latter, we note that d\u03b7 f (\u03b7 +\u03bd), i.e. in Grassmann integration one can shift variables as in the ordinary case. The proof of the Gaussian relation above thus proceeds in complete analogy to the complex case. As with Eq. (3.18), Eq. (4.22) can also be employed d( \u00af\u03c6, \u03c6) e\u2212 \u00af\u03c6T A\u03c6(\u00b7 \u00b7 \u00b7 ), and to generate further integration formulae. De\ufb01ning \u00b7 \u00b7 \u00b7 \u2261 det A\u22121 expanding both the left-and the right-hand side of Eq. (4.22) to leading order in the \u201cmonomial\u201d \u00af\u03bdj\u03bdi, one obtains \u03c6j \u00af\u03c6i = A\u22121 ji. The N -fold iteration of this procedure gives \u03c6j1 \u03c6j2 \u00b7 \u00b7 \u00b7 \u03c6jn \u00af\u03c6in \u00b7 \u00b7 \u00b7 \u00af\u03c6i2 \u00af\u03c6i1 = P (sgnP )A\u22121 j1iP 1 \u00b7 \u00b7 \u00b7 A\u22121 jniP n, where the sign of the permutation accounts for the sign changes accompanying the interchange of Grassmann variables. Finally, as with Gaussian integration over commuting variables, by taking N \u2192 \u221e the Grassmann integration can be translated to a Gaussian functional integral. 4.2 Field integral for the quantum partition function Having introduced the coherent states, the construction of path integrals for many-body systems no longer presents substantial di\ufb03culties. However, before proceeding, we should address the question: what does the phrase \u201cpath integral for manybody systems\u201d actually mean? In the next chapter we will see that much of the information on quantum is encoded many-particle systems in expectation values of products of creation and annihilation operators, i.e. expressions of the structure a\u2020a \u00b7 \u00b7 \u00b7. By an analogy to be explained then, objects of this type are generally called correlation functions. More important for our present discussion, at any \ufb01nite temperature, the average \u00b7 \u00b7 \u00b7 entering the de\ufb01nition of the Josiah Willard Gibbs 1839\u20131903 Credited with the development thermodynamics, of he introduced concepts of free energy and chemical potential. reproduced from The (Figure Collected Works of J. Willard Gibbs, vol. I (Longmans, Green and Co., 1928).) chemical elements of M. In order to be consistent with the anti-commutation behavior of Grassmann variables, the polynomial must obey commutation relations which uniquely characterize a determinant. Exercise: Check the relation for N = 2.) On the other hand, the integral of the new variables must obey the de\ufb01ning relation, N i=1 d\u00af\u03bdi and the sign on the right-hand side is d\u03bd2 \u03bd2 = \u22121. Together Eq. (4.21) and d\u03bd1 \u03bd1 d\u03bd\u03bd1 \u00b7 \u00b7 \u00b7\u03bd N = (\u2212)N (N \u22121)/2, whered \u00af\u03bd = d\u03bd1 d\u03bd2 \u03bd1\u03bd2 = \u2212 attributed to ordering of the integrand, i.e. (4.20) enforce the identities d\u00af\u03bd = (det M)\u22121d \u00af\u03c6, d\u03bd = (det M)\u22121d\u03c6, which combine to give d\u00af\u03bd \u00af\u03bd1 \u00b7 \u00b7 \u00b7 \u00af\u03bdN = d( \u00af\u03c6, \u03c6) f ( \u00af\u03c6, \u03c6) = det(MM ) d(\u00af\u03bd, \u03bd) f ( \u00af\u03c6(\u00af\u03bd), \u03c6(\u03bd)). 166 Functional \ufb01eld integral correlation function runs over the quantum Gibbs distribution \u02c6\u03c1 \u2261 e\u2212\u03b2( \u02c6H\u2212\u03bc \u02c6", "N )/Z, where, as usual, Z = tr e\u2212\u03b2( \u02c6H\u2212\u03bc \u02c6N ) = n|e\u2212\u03b2( \u02c6H\u2212\u03bc \u02c6N )|n, (4.23) n is the quantum partition function, \u03b2 \u2261 1/T, \u03bc denotes the chemical potential, and the sum extends over a complete set of Fock space states {|n}. (For the time being we specify neither the statistics of the system \u2013 bosonic or fermionic \u2013 nor the structure of the Hamiltonian.) Ultimately, we will want to construct the path integral representations of correlation functions. Later we will see that all of these representations can be derived by a few straightforward manipulations from a prototypical path integral, namely that for Z. Further, the (path integral of the) partition function is of importance in its own right, as it contains much of the information needed to characterize the thermodynamic properties of a manybody quantum system.6 We thus begin our journey into many-body \ufb01eld theory with a construction of the path integral for Z. To prepare the representation of the partition function (4.23) in terms of coherent states, one must insert the resolution of identity Z = d( \u00af\u03c8, \u03c8) e\u2212 \u00af\u03c8i\u03c8i i n|\u03c8\u03c8|e\u2212\u03b2( \u02c6H\u2212\u03bc \u02c6N )|n. (4.24) n n We now wish to get rid of the \u2013 now redundant \u2013 Fock space summation over |n (another |nn| = 1F, one must resolution of identity). To bring the summation to the form commute the factor n|\u03c8 to the right-hand side. However, in performing this operation, we must be careful not to miss a sign change whose presence will have important consequences for the structure of the fermionic path integral. Indeed, it may be checked that, whilst for bosons n|\u03c8\u03c8|n = \u03c8|nn|\u03c8, the fermionic coherent states change sign upon permutation, n|\u03c8\u03c8|n = \u2212\u03c8|nn|\u03c8 (i.e. \u2212\u03c8| \u2261 0|exp ). The presence of the sign is a direct consequence of the anti-commutation relations between Grassmann variables and Fock space operators (exercise). Note that, as both \u02c6H and \u02c6N contain elements even in the creation/annihilation operators, the sign is insensitive to the presence of the Boltzmann factor in Eq. (4.24). Making use of the sign factor \u03b6, the result of the interchange can be formulated as the general expression \u00af\u03c8iai \u2212 i Z = d( \u00af\u03c8, \u03c8) e\u2212 = d( \u00af\u03c8, \u03c8) e\u2212 \u00af\u03c8i\u03c8i i n \u03b6\u03c8|e\u2212\u03b2( \u02c6H\u2212\u03bc \u02c6N )|nn|\u03c8 \u00af\u03c8i\u03c8i \u03b6\u03c8|e\u2212\u03b2( \u02c6H\u2212\u03bc \u02c6N )|\u03c8. i (4.25) 6 In fact, the statement above is not entirely correct. Strictly speaking, thermodynamic properties involve the thermodynamic potential \u03a9 = \u2212T ln Z rather than the partition function itself. At \ufb01rst sight it seems that the di\ufb00erence between the two is arti\ufb01cial \u2013 one might \ufb01rst calculate Z and then take the logarithm. However, typically, one is unable to determine Z in closed form, but rather one has to perform a perturbative expansion, i.e. the result of a calculation of Z will take the form of a series in some small parameter. Now a problem arises when the logarithm of the series is taken. In particular, the Taylor series expansion of Z to a given order in does not automatically determine the expansion of \u03a9 to the same order. Fortunately, the situation is not all that bad. As we will see in the next chapter, the logarithm essentially rearranges the perturbation series in an order known as a cumulant expansion. 4.2 Field integral for the quantum partition function 167 Equation (4.25) can now be directly subjected to the general construction scheme of the path integral. To be concrete, let us assume that the Hamiltonian is limited to a maximum of two-body interactions (see Eq. (2.11) and (2.16)), \u02c6H(a\u2020, a) = hija \u2020 i aj + Vijkla \u2020 i a \u2020 jakal. (4.26) ij ijkl Note that we have arranged for all of the annihilation operators to stand to the right of the", " creation operators. Fock space operators of this structure are said to be normal ordered.7 The reason for emphasizing normal ordering is that such an operator can be readily diagonalized by means of coherent states. Dividing the \u201ctime interval\u201d \u03b2 into N segments and inserting coherent state resolutions of identity (steps 1, 2, and 3 of the general scheme), Eq. (4.25) assumes the form Z = N' \u00af\u03c80=\u03b6 \u00af\u03c8N \u03c80=\u03b6\u03c8N n=1 d( \u00af\u03c8n, \u03c8n) e\u2212\u03b4 N \u22121 n=0 [\u03b4\u22121( \u00af\u03c8n\u2212 \u00af\u03c8n+1)\u00b7\u03c8n+H( \u00af\u03c8n+1,\u03c8n)\u2212\u03bcN ( \u00af\u03c8n+1,\u03c8n)], (4.27) \u03c8| \u02c6H(a\u2020,a)|\u03c8 \u03c8|\u03c8 \u2261 H( \u00af\u03c8, \u03c8) (simiwhere \u03b4 = \u03b2/N and larly N ( \u00af\u03c8, \u03c8)) and we have adopted the shorthand \u03c8n = {\u03c8n }, etc. Finally, sending N \u2192 \u221e i and taking limits analogous to those leading from Eq. (3.5) to (3.6) we obtain the continuum version of the path integral,8 ijkl Vijkl \u00af\u03c8j\u03c8 \u00af\u03c8i\u03c8 ij hij k\u03c8 j + \u00af\u03c8i = l Z = D( \u00af\u03c8, \u03c8) e\u2212S[ \u00af\u03c8,\u03c8], S[ \u00af\u03c8, \u03c8] = \u03b2 0 % d\u03c4 \u00af\u03c8\u2202\u03c4 \u03c8 + H( \u00af\u03c8, \u03c8) \u2212 \u03bcN ( \u00af\u03c8, \u03c8) &, (4.28) where D( \u00af\u03c8, \u03c8) = limN\u2192\u221e N n=1 d( \u00af\u03c8n, \u03c8n), and the \ufb01elds satisfy the boundary condition \u00af\u03c8(0) = \u03b6 \u00af\u03c8(\u03b2), \u03c8(0) = \u03b6\u03c8(\u03b2). (4.29) Written in a more explicit form, the action associated with the general pair-interaction Hamiltonian (4.26) can be cast in the form \u23a1 \u03b2 0 S = \u23a3 d\u03c4 ij \u00af\u03c8i(\u03c4 ) [(\u2202\u03c4 \u2212 \u03bc)\u03b4ij + hij] \u03c8j(\u03c4 ) + ijkl Vijkl \u00af\u03c8i(\u03c4 ) \u00af\u03c8j(\u03c4 )\u03c8k(\u03c4 )\u03c8l(\u03c4 ) \u23a6. (4.30) \u23a4 Notice that the structure of the action \ufb01ts nicely into the general scheme discussed in the previous chapter. By analogy, one would expect that the exponent of the many-body 7 More generally, an operator is de\ufb01ned to be \u201cnormal ordered\u201d with respect to a given vacuum state |0 if, and only if, it annihilates |0. Note that the vacuum need not necessarily be de\ufb01ned as a zero-particle state. If the vacuum contains particles, normal ordering need not lead to a representation where all annihilators stand to the right. If, for whatever reason, one is given a Hamiltonian whose structure di\ufb00ers from Eq. (4.26), one can always e\ufb00ect a normal ordered form at the expense of introducing commutator terms. For example, normal ordering the quartic term leads to the appearance of a quadratic contribution that can be absorbed into h\u03b1\u03b2. 8 Whereas the bosonic continuum limit is indeed perfectly equivalent to the one taken in constructing the quantum mechanical path integral (lim\u03b4\u21920 \u03b4\u22121( \u00af\u03c8n+1 \u2212 \u00af\u03c8n) =\u2202 \u03c4 |\u03c4 =n\u03b4 \u00af\u03c8(\u03c4 ) gives an ordinary derivative, etc.), a novelty arises in the fermionic case. The notion of replacing di\ufb00erences by derivatives is purely symbolic for Grassmann variables. There is no sense in which \u00af\u03c8n+1 \u2212 \u00af\u03c8n is small. The symbol \u2202\u03c4 \u00af\u03c8 rather denotes the formal (and well-de\ufb01ned expression) lim\u03b4\u21920 \u03b4\u22121( \u00af\u03c8n+1 \u2212 \u00af\u03c8n). 168 Functional \ufb01eld integral path integral carries the signi\ufb01cance of the Hamiltonian action, S \u223c d(p \u02d9q \u2212 H), where (q, p) symbolically stands for a set of generalized coordinates and momenta. In the present case, the natural pair of canonically", " conjugate operators is (a, a\u2020). One would then interpret the eigenvalues (\u03c8, \u00af\u03c8) as \u201ccoordinates\u201d (much as (q, p) are the eigenvalues of the operators (\u02c6q, \u02c6p)). Adopting this interpretation, we see that the exponent of the path integral indeed has the canonical form of a Hamiltonian action and, therefore, is easy to memorize. Equations (4.28) and (4.30) de\ufb01ne the functional integral in the time representation (in the sense that the \ufb01elds are functions of a time variable). In practice we shall mostly \ufb01nd it useful to represent the action in an alternative, Fourier conjugate representation. To this end, note that, due to the boundary conditions (4.29), the functions \u03c8(\u03c4 ) can be interpreted as functions on the entire Euclidean time axis that are periodic/antiperiodic on the interval [0, \u03b2]. As such, they can be represented in terms of a Fourier series, \u03c8(\u03c4 ) = 1\u221a \u03b2 \u03c9n \u03c8ne\u2212i\u03c9n\u03c4, \u03c8n = 1\u221a \u03b2 \u03b2 d\u03c4 \u03c8(\u03c4 )ei\u03c9n\u03c4, where ) 2n\u03c0T, (2n + 1)\u03c0T, \u03c9n = bosons, fermions, 0 * n \u2208 Z, (4.31) are known as Matsubara frequencies. Substituting this representation into Eq. (4.28) n d( \u00af\u03c8n, \u03c8n) de\ufb01nes the and (4.30), we obtain Z = measure (for each Matsubara index n we have an integration over a coherent state basis {|\u03c8n}),9 and the action takes the form D( \u00af\u03c8, \u03c8) e\u2212S[ \u00af\u03c8,\u03c8], whereD ( \u00af\u03c8, \u03c8) = S[ \u00af\u03c8, \u03c8] = ij,n \u00af\u03c8in [(\u2212i\u03c9n \u2212 \u03bc) \u03b4ij + hij] \u03c8jn + 1 \u03b2 ijkl,ni Vijkl \u00af\u03c8in1 \u00af\u03c8jn2 \u03c8kn3 \u03c8ln4 \u03b4n1+n2,n3+n4. Here we have used the identity quency representation of the action.10 (4.32) 0 d\u03c4 e\u2212i\u03c9n\u03c4 = \u03b2\u03b4\u03c9n0. Equation (4.32) de\ufb01nes the fre- \u03b2 INFO In performing calculations in the Matsubara representation, one sometimes runs into convergence problems (which will manifest themselves in the form of ill-convergent Matsubara frequency summations). In such cases it will be important to remember that Eq. (4.32) does not actually represent the precise form of the action. What is missing is a convergence generating factor whose presence follows from the way in which the integral was constructed, and which will save us in cases of non-convergent sums (except, of course, in cases where divergences have a physical origin). More precisely, since the \ufb01elds \u00af\u03c8 are evaluated in\ufb01nitesimally later than the \ufb01elds \u03c8 (cf. Eq. (4.27)), the h- and\u03bc-dependent contributions to the action acquire a factor 9 Notice, however, that the \ufb01elds \u03c8n carry dimension [energy]\u22121/2, i.e. the frequency coherent state integral is normalized as d( \u00af\u03c8n, \u03c8n) e\u2212 \u00af\u03c8n \u03c8n = (\u03b2)\u2212\u03b6. 10 As to the signs of the Matsubara indices appearing in Eq. (4.32), note that the Fourier representation of \u00af\u03c8 is 0 d\u03c4 \u00af\u03c8(\u03c4 )e\u2212i\u03c9n \u03c4. In the bosonic case, this sign convention is de\ufb01ned as \u00af\u03c8(\u03c4 ) = 1\u221a \u03b2 motivated by \u00af\u03c8 being the complex conjugate of \u03c8. For reasons of notational symmetry, this convention is also adopted in the fermionic case. \u00af\u03c8ne+i\u03c9n \u03c4, \u00af\u03c8n = 1\u221a \u03b2 n \u03b2 4.2 Field integral for the quantum partition function 169 exp(\u2212i\u03c9n\u03b4), \u03b4 in\ufb01nitesimal. Similarly, the V contribution acquires a factor exp(\u2212i(\u03c9n1 + \u03c9n2)\u03b4). In cases where the convergence is not critical, we will omit these contributions. However, once in a while it is necessary to remember their presence. Partition function of", " non-interacting gas As a \ufb01rst exercise, let us consider the quantum partition function of the non-interacting gas. (Later, this object will prove to be a \u201creference\u201d in the development of weakly interacting theories.) In some sense, the \ufb01eld integral formulation of the non-interacting partition function has a status similar to that of the path integral for the quantum harmonic oscillator: the direct quantum mechanical solution of the problem is straightforward and application of the full artillery of the \ufb01eld integral seems somewhat ludicrous. From a pedagogical point of view, however, the free partition function is a good problem; it provides us with the welcome opportunity to introduce a number of practical concepts of \ufb01eld integration within a comparatively simple setting. Also, the \ufb01eld integral representation of the free partition function will be an important operational building block for our subsequent analysis of interacting problems. Consider, then, the partition function (4.28) with H0( \u00af\u03c8, \u03c8) = \u00af\u03c8iH0,ij\u03c8j. Diagonalizing H0 by a unitary transformation U, H0 = U DU \u2020, and transforming integration variables, \u00af\u03c6an(\u2212i\u03c9n + \u03bea)\u03c6an, where \u03bea \u2261 a \u2212 \u03bc U \u2020\u03c8 \u2261 \u03c6, the action assumes the form S = a and a are the single-particle eigenvalues. Remembering that the \ufb01elds \u03c6a(\u03c4 ) are independent integration variables (Exercise: Why does the transformation \u03c8 \u2192 \u03c6 have a Jacobian of unity?), we \ufb01nd that the partition function decouples, Z = Za, where \u03c9n ij a Za = D( \u00af\u03c6a, \u03c6a) e\u2212 \u00af\u03c6an(\u2212i\u03c9n+\u03bea)\u03c6an = n'n [\u03b2(\u2212i\u03c9n + \u03bea)]\u2212\u03b6, (4.33) and the last equality follows from the fact that the integrals over \u03c6an are one-dimensional complex or Grassmann Gaussian integrals. Here, let us recall our convention de\ufb01ning \u03b6 = 1 (\u22121) for bosonic (fermionic) \ufb01elds. At this stage, we have left all aspects of \ufb01eld integration behind us and reduced the problem to one of computing an in\ufb01nite product over factors i\u03c9n \u2212 \u03bea. Since products are usually more di\ufb03cult to get under control than sums, we take the logarithm of Z to obtain the free energy F = \u2212T ln Z = T \u03b6 (4.34) INFO Before proceeding with this expression, let us take a second look at the intermediate an identity (4.33). Our calculation showed the partition function to be the product over all eigenvalues of the operator \u2212i\u02c6\u03c9 + \u02c6H \u2212 \u03bc \u02c6N de\ufb01ning the action of the non-interacting system (here, \u02c6\u03c9 = {\u03c9n\u03b4nn }). As such, it can be written compactly as: ln[\u03b2(\u2212i\u03c9n + \u03bea)]. ( Z = det \u03b2(\u2212i\u02c6\u03c9 + \u02c6H \u2212 \u03bc \u02c6N ) )\u2212\u03b6. (4.35) This result was derived by \ufb01rst converting to an eigenvalue integration and then performing the one-dimensional integrals over \u201ceigencomponents\u201d \u03c6an. While technically straightforward, that \u2013 explicitly representation-dependent \u2013 procedure is not well suited to generalization to 170 Functional \ufb01eld integral more complex problems. (Keep in mind that later on we will want to embed the free action of the non-interacting problem into the more general framework of an interacting theory.) Indeed, it is not necessary to refer to an eigenbasis at all. In the bosonic case, Eq. (3.17) tells us that Gaussian integration over a bilinear \u223c \u00af\u03c6 \u02c6X\u03c6 generates the inverse determinant of \u02c6X. Similarly, as we have seen, Gaussian integration extends to the Grassmann case with the determinants appearing in the numerator rather than in the denominator (as exempli\ufb01ed by Eq. (4.35)). (As a matter of fact, Eq. (4.33) is already a proof of this relation.) We now have to face up to a technical problem: how do we compute Matsubara sums n ln(i\u03c9n \u2212 x)? Indeed, it takes little imagination to foresee that sums of of the form n1,n2,... X(\u03c9n1,", " \u03c9n2,...), where X symbolically stands for some function, will the type be a recurrent structure in the analysis of functional integrals. A good ansatz would be to argue that, for su\ufb03ciently low temperatures (i.e. temperatures smaller than any other characteristic energy scale in the problem), the sum can be traded for an integral, speci\ufb01cally d\u03c9/(2\u03c0). However, this approximation is too crude to capture much of the T characteristic temperature dependence in which one is usually interested. Yet there exists an alternative, and much more accurate, way of computing sums over Matsubara frequencies (see Info below). \u2192 n INFO Consider a single Matsubara frequency summation, S \u2261 n h(\u03c9n), (4.36) where h is some function and \u03c9n may be either bosonic or fermionic (cf. Eq. (4.31)). The basic idea behind the standard scheme of evaluating sums of this type is to introduce a complex auxiliary function g(z) that has simple poles at z = i\u03c9n. The sumS then emerges as the sum of residues obtained by integrating the product gh along a suitably chosen path in the complex plane. Typical choices of g include g(z) = * \u03b2 exp(\u03b2z)\u22121, \u03b2 exp(\u03b2z)+1, + bosons, fermions, or g(z) = $ \u03b2 2 coth(\u03b2z/2), \u03b2 2 tanh(\u03b2z/2), % bosons, fermions, (4.37) where, in much of this section, we will employ the functions of the \ufb01rst column. (Notice the similarity between these functions and the familiar Fermi and Bose distribution functions.) In practice, the choice of the counting function is mostly a matter of taste, save for some cases where one of the two options is dictated by convergence criteria. Integration over the path shown in the left part of Fig. 4.2 then produces, \u03b6 2\u03c0i dz g(z)h(\u2212iz) = \u03b6 n Res(g(z)h(\u2212iz))|z=i\u03c9n = n h(\u03c9n) = S, where, in the third identity, we have used the fact that the \u201ccounting functions\u201d g are chosen so as to have residue \u03b6 and it is assumed that the integration contour closes at z \u2192 \u00b1i\u221e. (The di\ufb00erence between using the \ufb01rst and the second column of Eq. (4.37) lies in the value of the residue. In the latter case, it is equal to unity rather than \u03b6.) Now, the integral along a contour in the immediate vicinity of the poles of g is usually intractable. However, as long as we are careful not to cross any singularities of g or the function h(\u2212iz) (symbolically indicated 4.2 Field integral for the quantum partition function 171 \u03c9 \u03b3 1 \u03c9 \u03b3 2 (a) (b) Figure 4.2 (a) The integration contour employed in calculating the sum (4.36). (b) The deformed integration contour. by isolated crosses in Fig. 4.211) we are free to distort the integration path, ideally to a contour along which the integral can be done. Finding a suitable contour is not always straightforward. If the product hg decays su\ufb03ciently fast as |z| \u2192 \u221e (i.e. faster than z\u22121), one will usually try to \u201cin\ufb02ate\u201d the original contour to an in\ufb01nitely large circle (Fig. 4.2(b)).12 The integral along the outer perimeter of the contour then vanishes and one is left with the integral around the singularities of the function h. In the simple case where h(\u2212iz) possesses a number of isolated singularities at {zk} (i.e. the situation indicated in the \ufb01gure) we thus obtain, S = \u03b6 2\u03c0i dz h(\u2212iz)g(z) =\u2212\u03b6 k Res h(\u2212iz)g(z) z=zk, (4.38) where the contour integral encircles the singularities of h(\u2212iz) in clockwise direction. The computation of the in\ufb01nite sum S has been now been reduced to the evaluation of a \ufb01nite number of residues \u2013 a task that is always possible! To illustrate the procedure on a simple example, let us consider the function h(\u03c9n) =\u2212 \u03b6T i\u03c9ne\u2212i\u03c9n\u03b4 \u2212 \u03be, where \u03b4 is a positive in\ufb01nitesimal.13 To evaluate", " the sum S = n h(\u03c9n), we \ufb01rst observe that the product h(\u2212iz)g(z) has benign convergence properties. Further, the function h(\u2212iz) has a 11 Remember that a function that is bounded and analytic in the entire complex plane is constant, i.e. every \u201cinteresting\u201d function will have singularities. 12 Notice that the condition lim|z|\u2192\u221e |hg| < z\u22121 is not as restrictive as it may seem. The reason is that the function h will be mostly related to physical observables that approach some limit (or vanish) for large excitation energies. This implies vanishing in at least portions of the complex plane. The convergence properties of g depend on the concrete choice of the counting function. (Exercise: Explore the convergence properties of the functions shown in Eq. (4.37).) 13 In fact, this choice of h is actually not as arti\ufb01cial as it may seem. The expectation value of the number of particles in the grand canonical ensemble is de\ufb01ned through the identity N \u2261 \u2212\u2202F/\u2202\u03bc where F is the free energy. In the non-interacting case, F is given by Eq. (4.34) and, remembering that \u03bea = a \u2212 \u03bc, one obtains. Now, why did we write \u201c\u2248\u201d instead of \u201c=\u201d? The reason is that the right-hand side, N \u2248 \u03b6T obtained by naive di\ufb00erentiation of Eq. (4.34), is ill-convergent. (The sum n+x, x arbitrary, does not exist!) At this point we have to remember the remark made on page 168, i.e., had we carefully treated the discretization of the \ufb01eld integral, both the logarithm of the free energy and \u2202\u03bcF would have acquired in\ufb01nitesimal phases exp(\u2212i\u03c9n\u03b4). As an exercise, try to keep track of the discretization of the \ufb01eld integral from 1 \u2212i\u03c9n+\u03bea \u221e n=\u2212\u221e an 1 172 Functional \ufb01eld integral simple pole that, in the limit \u03b4 \u2192 0, lies on the real axis at z = \u03be. This leads to the result n h(\u03c9n) =\u2212\u03b6 Res g(z)h(\u2212iz)| z=\u03be = 1 e\u03b2\u03be \u2212 \u03b6. We have thus arrived at the important identity \u2212\u03b6T n * 1 i\u03c9n \u2212 \u03bea = nB(a), bosons, nF(a), fermions, where nF() = 1 exp( \u2212 \u03bc) + 1, nB() = 1 exp( \u2212 \u03bc) \u2212 1, (4.39) (4.40) are the Fermi/Bose distribution functions. As a corollary we note that the expectation value for the number of particles in a non-interacting quantum gas assumes the familiar form N = a nF/B(a). Before returning to our discussion of the partition function, let us note that life is not always as simple as the example above. More often than not, the function h contains not only isolated singularities but also cuts or worse singularities. In such circumstances, \ufb01nding a good choice of the integration contour can be far from straightforward! Returning to the problem of computing the sum (4.34), consider for a moment a \ufb01xed \u03be \u03c9 = eigenvalue \u03bea \u2261 a \u2212 \u03bc. In this case, we need to evaluate the sum S \u2261 n h(\u03c9n), where h(\u03c9n) \u2261 \u03b6T ln[\u03b2(\u2212i\u03c9n + \u03be)] = \u03b6T ln[\u03b2(i\u03c9n \u2212 \u03be)] + C and C is an inessential constant. As discussed before, the sum can be represented dz g(z)h(\u2212iz), where g(z) =\u03b2 (e\u03b2z \u2212 \u03b6)\u22121 as S = \u03b6 2\u03c0i is (\u03b2 times) the distribution function and the contour encircles the poles of g as in Fig. 4.2(a). Now, there is an essential di\ufb00erence from the example discussed previously, i.e. the function h(\u2212iz) =\u03b6T ln(z \u2212 \u03be) +C has a branch cut along the real axis, z \u2208 (\u2212\u221e, \u03be) (see the \ufb01gure). To avoid contact with this singularity one must distort the integration contour as shown in the \ufb01gure. Noticing that the (suitably regularized, cf. our previous", " discussion of the particle number N ) integral along the perimeter vanishes, we conclude that S = T 2\u03c0i \u221e \u2212\u221e d g() ln(+ \u2212 \u03be) \u2212 ln(\u2212 \u2212 \u03be), its de\ufb01nition to Eq. (4.34) to show that the accurate expression for N reads N = \u03b6T an 1 \u2212i\u03c9ne\u2212i\u03c9n \u03b4 + \u03bea = a n h(\u03c9n)|, \u03be=\u03bea where h is the function introduced above. (Note that the necessity to keep track of the lifebuoy e\u2212i\u03b4\u03c9n does not arise too often. Most Matsubara sums of physical interest relate to functions f that decay faster than z\u22121.) 4.3 Field theoretical bosonization: a case study 173 where \u00b1 = \u00b1 i\u03b7, \u03b7 is a positive in\ufb01nitesimal, and we have used the fact that g(\u00b1) g() is continuous across the cut. (Also, without changing the value of the integral (exercise: why?), we have enlarged the integration interval from (\u2212\u221e, \u03be] to (\u2212\u221e, \u221e).) To evaluate the integral, we observe that g() = \u03b6\u2202 ln and integrate by parts: 1 \u2212 \u03b6e\u2212\u03b2 d ln 1 \u2212 \u03b6e\u2212\u03b2 S = \u2212 \u03b6T 2\u03c0i 3.58) = \u03b6T ln 1 \u2212 \u03b6e\u2212\u03b2\u03be. Insertion of this result into Eq. (4.34) \ufb01nally gives the familiar expression F = \u03b6T ln 1 \u2212 \u03b6e\u2212\u03b2(a\u2212\u03bc), a (4.41) for the free energy of the non-interacting Fermi/Bose gas. While this result could have been obtained much more straightforwardly by the methods of quantum statistical mechanics, we will shortly see how powerful a tool the methods discussed in this section are when it comes to the analysis of less elementary problems! 4.3 Field theoretical bosonization: a case study The \ufb01eld integral (4.28) provides an exact representation of the quantum partition function; it contains the full information on the microscopic Hamiltonian operator. However, what we are actually interested in is the universal large-scale behavior of a quantum system. To extract this information from the \ufb01eld integral we will need to identify the relevant longrange degrees of freedom and to dispense with the abundance of microscopic data controlling the short-range behavior. In other words, we will have to pass from the microscopic \ufb01eld theory to some e\ufb00ective long-range theory. In Chapter 1 we saw that there are usually two principal strategies to obtain e\ufb00ective long range theories of microscopic systems: explicit construction \u2013 the subject of the next two chapters \u2013 and more phenomenological approaches based on consistency considerations and symmetry arguments. Besides a certain lack of rigor, the principal disadvantage of the second route is the lack of quantitative control of the results (which implies susceptibility to mistakes). On the other hand, the phenomenological approach is far less laborious and involves a minimal amount of technical preparation. Often, the phenomenological deduction of a low-energy \ufb01eld theory precedes its rigorous construction (sometimes by years). In fact, there are cases where phenomenology is the only viable route. Below we will illustrate the power of the phenomenological approach on the example of the interacting one-dimensional electron gas. We will map the microscopic partition function of the system onto a free (and thus exactly solvable) bosonic theory.14 In this section the emphasis is placed on purely methodological aspects, i.e. we will derive an e\ufb00ective theory, but will not do anything with it. (Nonetheless, the derivation is instructive and helps us to understand the essential physics of the system!) In later chapters, the \ufb01eld theory derived 14 A preliminary account of the ideas underlying this mapping has already been given in Section 2.2. 174 Functional \ufb01eld integral below will then serve as the starting point for the discussion of a number of interesting applications. One-dimensional electron gas (fermionic theory) Non-interacting system Let us begin by considering the action of a non-interacting one-dimensional electron gas S0[\u03c8\u2020, \u03c8] = s=\u00b11 dxd\u03c4 \u03c8\u2020 s (\u2212isvF\u2202x + \u2202\u03c4 ) \u03c8s, where \u03c8+/\u2212 are the right-/left-moving fermions and we have denoted the Grassmann \ufb01eld conjugate to \u03c8 by \u03c8\u2020.15 For later reference we recall that the left-/right-moving fermion operators are projections of the global momentum-dependent f", "ermion operator to the vicin\u2020 \u2212kF+q, where |q| kF. +(q) =\u03c8 ity of the left/right Fermi point, i.e. \u03c8 Fourier transforming this expression we obtain the approximate decomposition \u2020 \u2212(q) =\u03c8 \u2020 kF+q, \u03c8 \u2020 \u03c8(x) =e ikFx\u03c8+(x) +e \u2212ikFx\u03c8\u2212(x). (4.42) Before proceeding, let us rewrite the action in a form that emphasizes the symmetries of the problem: S0[\u03c8\u2020, \u03c8] = d2x \u03c8\u2020 (\u03c30\u2202x0 + i\u03c33\u2202x1 ) \u03c8 = d2x \u00af\u03c8 (\u03c31\u2202x0 + \u03c32\u2202x1 ) \u03c8, (4.43) where we have setv F = 1 for notational simplicity. Here, \u03c8 = (\u03c8+, \u03c8\u2212)T is a two-component \ufb01eld comprising left- and right-moving fermions, x = (x0, x1) = (\u03c4, x) parameterizes (1 + 1)dimensional space-time, and \u00af\u03c8 \u2261 \u03c8\u2020\u03c31. The second equality identi\ufb01es the action of the free one-dimensional fermion with that of a (1 + 1)-dimensional Dirac particle.16 We next turn to the discussion of the symmetries of the problem. For one thing, the action is clearly invariant under the transformation, \u03c8 \u2192 ei\u03c6v \u03c8, where \u03c6v = const. What is the resulting conserved current? The in\ufb01nitesimal variant of this transformation is described by \u03c8 \u2192 \u03c8 + (i\u03b4\u03c6v)\u03c8 or, in a notation adapted to Eq. (1.42), \u03c8 \u2194 \u03c6i, \u03c9a \u2194 i\u03c6v, a = 1. Equation (1.43) then gives the conserved current j\u03bd,\u03bc = \u2202L \u2202(\u2202\u03bc\u03c8) \u03c8 = \u00af\u03c8\u03b3\u03bc\u03c8. For later F i reference, we mention that, under a rotation of space-time, x\u03bc \u2192 (R \u00b7 x)\u03bc, the components of jv transform like a vector, j\u03bc \u2192 (R \u00b7 j)\u03bc. In relativistic \ufb01eld theory, jv is therefore usually called a vector current. Notice that the two components of the vector, j0 = \u03c8\u2020\u03c8 = \u03c8 \u2020 \u2212\u03c8\u2212 \u2261 \u03c1 and j1 = \u2261 ij, are the charge density, \u03c1, of the system and (i times) \u2020 +\u03c8+ + \u03c8 \u03c8 \u2020 +\u03c8+ \u2212 \u03c8 \u2020 \u2212\u03c8\u2212 i\u03c8\u2020\u03c33\u03c8 = i 15 Following the remarks earlier, this represents an abuse of notation; there is no Grassmann complex conjugation! However, within the context of relativistic fermions our standard symbol \u00af\u03c8 is reserved for another object (see below). 16 For a review of the theory of the standard four-dimensional massless Dirac equation \u03b3\u03bc\u2202\u03bc\u03c8 = 0, see, e.g., L. H. Ryder, Quantum Field Theory (Cambridge University Press, 1996). The Dirac equation a\ufb00ords a natural generalization to any even-dimensional space. Speci\ufb01cally, for d = 2, the algebra of Dirac \u03b3-matrices, {\u03b3\u03bc, \u03b3\u03bd } = 2\u03b4\u03bc\u03bd, \u03bc, \u03bd = 0, 1, 5, is satis\ufb01ed by \u03b30 \u2261 \u03c31, \u03b31 \u2261 \u03c32, \u03b35 \u2261 \u2212i\u03b30\u03b31 = \u03c33. 4.3 Field theoretical bosonization: a case study 175 the current density, j, respectively.17 Thus, the equation \u2212i\u2202\u03bcj\u03bc = \u2212i\u2202\u03c4 \u03c1 + \u2202xj = 0 simply expresses the conservation of particle current. INFO A particular example of the general fact that the U(1)-symmetry of quantum mechanics (the freedom to multiply wave functions or operators of a second quantized approach by a constant phase ei\u03c6) implies the conservation of particle current. EXERCISE Subject the action of the general \ufb01eld integral (4.28) to the transformation \u03c8 \u2192 ei\u03c6\u03c8, \u00af\u03c8 \u2192 \u00af\u03c8e\u2212i\u03c6 and compute the resulting Noether current. Convince yourself that the components of the current are the coherent", " state representation of the standard density/current operator of quantum mechanics. Now, the action (4.43) possesses a somewhat less obvious second symmetry: it remains invariant under the transformation \u03c8 \u2192 ei\u03c6a\u03c33 \u03c8, \u00af\u03c8 \u2192 \u00af\u03c8ei\u03c6a\u03c33. (Notice that this is not a unitary symmetry, i.e. the matrices transforming \u03c8 and \u00af\u03c8 are not inverse to each other.) Indeed, one may note that ( \u00af\u03c8ei\u03c6a\u03c33 )\u03c3\u03bc\u2202\u03bc(ei\u03c6a\u03c33 \u03c8) \u00af\u03c8\u03c3\u03bc\u2202\u03bc\u03c8. A straightforward application of Noether\u2019s theorem based on the in\ufb01nitesimal variant \u03c8 \u2192 \u03c8 + (i\u03c6a)\u03c33\u03c8 gives the conserved current ja,\u03bc = i \u00af\u03c8\u03c3\u03bc\u03c33\u03c8 = \u03bc\u03bd \u00af\u03c8\u03c3\u03bd\u03c8. Introducing a unit vector e2 pointing into a \ufb01ctitious third dimension perpendicular to the space-time plane, the current can be written as ja = e2 \u00d7 jv. This representation shows that, under rotations, ja transforms like an axial vector (similar to, say, a magnetic \ufb01eld). For this reason, ja is commonly called an axial current. [\u03c33,\u03c3\u03bc]+=0 = INFO The axial symmetry of the relativistic electron gas is a prominent example of symmetry that does not pervade to the quantum level. I.e. the conservation of the axial current breaks down once quantum \ufb02uctuations are taken into account, a phenomenon known as the chiral or axial anomaly.18 Although we will meet with various manifestations of the chiral anomaly, a thorough discussion of all its implications is beyond the scope of this text. (However, most textbooks on particle physics contain an extensive coverage of anomalies.) Finally, notice that the existence of two distinct symmetries, vectorial and axial, a\ufb00ords a simple interpretation in the original representation of the theory (the \ufb01rst equality in Eq. (4.43)). All it means is that the left- and right-moving fermion states do not couple, i.e. that the left- and the right-moving fermion particle currents are separately conserved. (Compute the corresponding Noether currents!) Given that we are dealing with but the simplest one-dimensional theory one can imagine, the formal discussion of symmetries may seem to be a bit of an overkill. However, we shall see in a moment that the e\ufb00ort was well invested: as soon as we switch on interactions, the fermionic theory ceases to be exactly solvable. It turns out, however, that our symmetry discussion above provides the key to a bosonic reformulation of the problem which does enjoy exact solvability. Yet, before turning to the bosonic approach, let us brie\ufb02y recapitulate how interactions couple to the model. 17 Notice that, for a one-dimensional Fermi system with uniform Fermi velocity vF = 1, the current density is equal to the density of right movers minus that of the left movers. 18 In \ufb01eld theory, the quantum violation of a classical conservation law is generally called an anomaly. 176 Functional \ufb01eld integral Interacting case As in Section 2.2 we assume a short-range interaction between the left- and right-moving densities. Quantitatively, this is described by the coherent state representation of the second quantized Hamiltonian (2.36), i.e. Sint[\u03c8\u2020, \u03c8] = 1 2 s dxd\u03c4 (g2 \u02c6\u03c1s \u02c6\u03c1\u00afs + g4 \u02c6\u03c1s \u02c6\u03c1s), (4.44) where \u02c6\u03c1s \u2261 \u03c8\u2020 s\u03c8s. Notice that the interaction term leaves the vectorial/axial symmetry of the system intact (why?). But what else can we say about the interacting system? In fact, we have seen in Section 2.2 that it is di\ufb03cult to understand the physics of the system in the microscopic language of interacting fermion states. Rather, one should turn to a formulation in terms of the e\ufb00ective long-range degrees of freedom of the model \u2013 non-dispersive charge density \ufb02uctuations. In the next section, we will formulate the dynamics of these excitations in a \ufb01eld theoretical language. Remarkably, it will turn out that all we need to extract this formulation from the microscopic model is a minimal investment of phenomenological input plus symmetry considerations. One-dimensional electron gas (bosonic theory) We have seen in Section 2.2 that the one-", "dimensional fermion system supports collective bosonic excitations. In this section, we apply phenomenological and symmetry arguments to construct a \ufb01eld theory of these excitations.19 Consider the electron operators c\u2020(x) of a one-dimensional system. As seen in Chapter 2, the operators c\u2020 a\ufb00ord a representation in terms of bosons \u2013 the Jordan\u2013Wigner transformation. As we are after an e\ufb00ective bosonic theory, it is certainly a good idea to switch to this Bose representation right from the x<x dx b\u2020(x)b(x)b\u2020(x), outset. Expressed in terms of Jordan\u2013Wigner bosons, c\u2020(x) =e i\u03c0 where b\u2020(x) creates a (bosonic) unit-charge excitation at x and the \u201cJordan\u2013Wigner string\u201d x<x dx b\u2020(x)b(x) implements the fermionic exchange statistics of the cs in terms of \u223c Bose operators. Now, we have seen in Section 2.2 that the dominant excitations of the (spinless) system are charge-density \ufb02uctuations. Thus, our next step is to switch to a representation that separately treats density and phase degrees of freedom. To this end, we de\ufb01ne charge and phase operators through 1/2 1/2 b(x) \u2261 + \u02c6\u03c1(x) ei \u02c6\u03c6(x), b\u2020(x) \u2261 e\u2212i \u02c6\u03c6(x) + \u02c6\u03c1(x), (4.45) kF \u03c0 kF \u03c0 where kF/\u03c0 is the average background density of the system, \u02c6\u03c1 describes density \ufb02uctuations, and \u02c6\u03c6 describes the phase of the Bose excitations. INFO Charge-density representations of this type are frequently employed in bosonic theories. Importantly, the transformation (b, b\u2020) \u2192 (\u02c6\u03c1, \u02c6\u03c6) is canonical, i.e. the density and the phase 19 For a detailed account of the rigorous construction (elements of which have been used in Section 2.2) we refer to H. Sch\u00a8oller and J. von Delft, Bosonization for beginners \u2013 refermionization for experts, Ann. Phys. 7 (1998), 225\u2013305. 4.3 Field theoretical bosonization: a case study 177 of a bosonic excitation form a canonically conjugate pair, [\u02c6\u03c1(x), \u02c6\u03c6(x)] = i\u03b4(x \u2212 x).20 To check this assertion, let us temporarily shift the density operator \u02c6\u03c1 \u2192 \u02c6\u03c1+kF /\u03c0, so that \u02c6\u03c1 now describes the total density of the system. (Of course, the shift by a number does not alter the commutation relations.) We thus write b = \u02c6\u03c11/2exp(i \u02c6\u03c6) andb \u2020 = exp(\u2212i \u02c6\u03c6)\u02c6\u03c11/2, where the position index has been dropped for notational transparency, and check that [b, b\u2020] = 1 indeed requires [\u02c6\u03c1, \u02c6\u03c6] = i: [b, b\u2020] = ( \u02c6\u03c11/2ei \u02c6\u03c6, e\u2212i \u02c6\u03c6 \u02c6\u03c11/2 ) = \u02c6\u03c1 \u2212 e\u2212i \u02c6\u03c6 \u02c6\u03c1ei \u02c6\u03c6 = \u02c6\u03c1 \u2212 e\u2212i[ \u02c6\u03c6, ] \u02c6\u03c1 = \u02c6\u03c1 \u2212 \u02c6\u03c1 + i[ \u02c6\u03c6, \u02c6\u03c1] \u2212 1 2 [ \u02c6\u03c6, [ \u02c6\u03c6, \u02c6\u03c1]] + \u00b7 \u00b7 \u00b7 [ \u02c6\u03c1, \u02c6\u03c6]=i = 1, as required. (In the last line we have used the general operator identity e \u02c6A \u02c6Be\u2212 \u02c6A = e[ \u02c6A,.] \u02c6B.) To represent the fermions in terms of bosonic degrees of freedom we might now substitute the charge/phase decomposition directly into our expression for the Jordan\u2013Wigner transformed fermion. It turns out, however, that the resulting representation \u2013 though perfectly valid \u2013 is not quite what we want. To obtain a representation better adjusted to the present problem, let us, rather, construct a fermion \u201cfrom scratch.\u201d To this end, let us think of the fermion as a structureless charge endowed with fermionic exchange statistics. The charge is created by application of ei \ufffd", "\ufffd\u03c6,21 while the fermionic exchange statistics are then generated in a second step through multiplication by a Jordan\u2013Wigner string. In fact, there is a certain ambiguity in the de\ufb01nition of the string variable: de\ufb01ning \u02c6\u03b8(x) \u2261 \u03c0 \u2212\u221e dx \u02c6\u03c1(x), it is straightforward to show that for m, m odd integers x # eim\u02c6\u03b8(x)ei \u02c6\u03c6(x), eim \u02c6\u03b8(x)ei \u02c6\u03c6(x) $ + \u221d ei\u03c0m\u0398(x\u2212x) + ei\u03c0m\u0398(x\u2212x) = 0, which implies that the most general representation of the fermion operator reads as,22 m=\u00b11 eim(kFx+\u02c6\u03b8(x)/\u03c0)ei \u02c6\u03c6(x). Here we have introduced a scalar prefactor \u0393 whose c\u2020(x) = \u0393 presence is needed to regularize the right-hand side of the equation in the limit of vanishing lattice spacing (the short distance cuto\ufb00 of the theory, see the problem set). We have also reinstalled the constant shift, \u02c6\u03c1 \u2192 \u02c6\u03c1 + kF/\u03c0 \u21d2 \u02c6\u03b8 \u2192 \u02c6\u03b8 + kFx. EXERCISE Verify the commutator identity above. (Hint: Use the general identities e \u02c6A \u02c6Be\u2212 \u02c6A = e[ \u02c6A,] \u02c6B and e \u02c6Ae \u02c6Be\u2212 \u02c6A = ee \u02c6A \u02c6Be \u2212 \u02c6A.) For most purposes, it is su\ufb03cient to keep the lowest two contributions to the harmonic s=\u00b11 eis(kFx+\u02c6\u03b8(x)/\u03c0)ei \u02c6\u03c6(x). Comparison with expansion of the fermion operator: c\u2020(x) = \u0393 Eq. (4.42) then leads to the tentative identi\ufb01cation s(x) = \u0393eis\u02c6\u03b8(x)ei \u02c6\u03c6(x). c\u2020 (4.46) 20 This implies, in particular, that the corresponding \ufb01eld integral transformation (\u03c8, \u00af\u03c8) \u2192 (\u03c1, \u03c6) from complex integration variables to the two real variables (\u03c1, \u03c6) has a unit Jacobian (check this!). 21 Notice that \u02c6\u03c6 is the \u201cmomentum\u201d conjugate to \u02c6\u03c1, i.e. the \u201ctranslation operator\u201d ei \u02c6\u03c6(x) increases the charge at x by unity. 22 To obtain the correct normalization of the anti-commutator at coinciding arguments x = x, all expansion coe\ufb03cients have to be equal to each other. 178 Functional \ufb01eld integral Finally, one may notice that the two \ufb01elds \u02c6\u03b8 and \u02c6\u03c6 \u2013 our principal degrees of freedom throughout \u2013 are characterized by the commutation relations [ \u02c6\u03c6(x), \u02c6\u03b8(x)] = i\u03c0\u0398(x \u2212 x). (4.47) At \ufb01rst sight, the results derived above may look strange: we have traded our simple fermion operator for a nonlinear representation in terms of bosonic degrees of freedom \u2013 but why? The point is that we are usually interested not in expressions linear in the fermion operators but in fermion bilinears (currents, densities, etc.). In contrast to the case of a single fermion, fermion bilinears have very simple expressions in terms of bosons. In particular, the Hamiltonian operator of the interacting system becomes quadratic (manifestly solvable) when expressed in terms of bosons. This aspect of the theory \u2013 undoubtedly the prime motivation behind the method of bosonization \u2013 will be explored next. Non-interacting system We now apply a combination of symmetry and dynamical arguments to identify the Hamiltonian of the non-interacting system. The fermionic prototype action is invariant under global rotations of space-time, x \u2192 x = R\u00b7x, \u03c8(x) = \u03c8(x). This symmetry must pertain to the bosonic description of the theory. Turning to the system\u2019s \u201cintrinsic\u201d symmetries, we notice that the vectorial and axial symmetry operations considered in the previous section act on the left-/right-moving fermion states as \u03c8", "s \u2192 ei\u03c6v \u03c8s and \u03c8s \u2192 eis\u03c6a \u03c8s, respectively. Of course, these transformations must continue to be symmetries no matter which representation of the theory (bosonic, fermionic, or whatever) is chosen. A glance at Eq. (4.46) shows that the symmetry transformation acts on the bosonic variables by a simple shift operation, vectorial: (\u03c6, \u03b8) \u2192 (\u03c6 + \u03c6v, \u03b8), axial: (\u03c6, \u03b8) \u2192 (\u03c6, \u03b8 + \u03c6a). For \u03c6a,v = const., these transformations must not change the action, which excludes the presence of non-derivative terms. (For example, a contribution such as \u223c \u03b82 would not be invariant under a uniform axial transformation, etc.) Of course, symmetries alone do not su\ufb03ce to \ufb01x the action of the system. What we need, in addition, is a minimal amount of dynamical input. More speci\ufb01cally, what we shall build on is the fact that the creation of a density distortion \u03c1 = \u2202x\u03b8/\u03c0 in the system costs a certain amount of energy U. Assuming that U \u223c \u03c12 (i.e. that screening has rendered the Coulomb interaction e\ufb00ectively short-range), the Lagrangian action of the charge displacement \ufb01eld \u03b8 will contain a term \u223c (\u2202x\u03b8)2. However, this expression lacks rotational invariance. Its canonical rotationally invariant extension reads (\u2202x\u03b8)2 + (\u2202\u03c4 \u03b8)2. Thus, up to second order in derivatives, the Lagrangian action is given by S0[\u03b8] = c dx d\u03c4 [(\u2202x\u03b8)2 + (\u2202\u03c4 \u03b8)2], where 2 the coupling constant c needs to be speci\ufb01ed. This expression tells us that the \ufb01eld \u03b8 has a linear dispersion and propagates at constant velocity (recall our discussion of, e.g., the phonon action in Chapter 1). Recalling that \u2202x\u03b8 \u223c \u03c1, this con\ufb01rms our results regarding the behavior of density distortions in one-dimensional electron systems derived in Section 2.2. 4.3 Field theoretical bosonization: a case study 179 To \ufb01x the value of the coupling constant c, we must compute the \u201ccorrelation function\u201d23 C(x, \u03c4 ) \u2261 (\u03c8 \u2020 +\u03c8\u2212)(x, \u03c4 )(\u03c8 \u2020 \u2212\u03c8+)(0, 0)\u03c8, (4.48) \ufb01rst in the fermionic, then in the bosonic, language, and require coinciding answers. Referring for a detailed discussion to the problem set, we here merely note that the result obtained for the free fermion action Eq. (4.43) reads C(x, \u03c4 ) = (4\u03c02(x2 + \u03c4 2))\u22121. The bosonic variant C(x, \u03c4 ) = \u03934 \u03b8 leads to the same expression provided that we set c = 1/\u03c0 and \u0393 = 1/(2\u03c0a)1/2. e2i\u03b8(x,\u03c4 )e\u22122i\u03b8(0,0) : 9 Thus, our \ufb01nal result for the Lagrangian of the non-interacting system reads S[\u03b8] = 1 2\u03c0 dx d\u03c4 [(\u2202x\u03b8)2 + (\u2202\u03c4 \u03b8)2]. Before proceeding to include interactions, let us turn from the Lagrangian to the Hamiltonian formulation. Using the fact that the canonical momentum corresponding to \u03b8 is \u03c0\u03b8 = \u2202\u2202\u03c4 \u03b8L = \u2202\u03c4 \u03b8/\u03c0, we obtain the Hamiltonian density 1 and the action 2\u03c0 & % S[\u03b8, \u03c0\u03b8] = 1 2 1 \u03c0 dx d\u03c4 (\u2202x\u03b8)2 + \u03c0\u03c02 \u03b8 + 2i\u2202\u03c4 \u03b8\u03c0\u03b8. (\u2202x\u03b8)2 + \u03c02\u03c02 \u03b8 At the same time, the relation (4.47) identi\ufb01es the canonical momentum as \u03c0\u03b8 = \u2202x\u03c6/\u03c0. Expressed in terms of the two fundamental degrees of freedom of the theory, the Hamiltonian action thus reads S[\u03b8, \u03c6] = 1 2\u03c0 dx d\u03c4 (\u2202x\u03b8)2 + (\u2202x\u03c6)2 + 2i\u2202\u03c4 \u03b8\u2202x", "\u03c6. (4.49) It is instructive to inspect the Noether current corresponding to the vectorial symmetry in the bosonic language. A straightforward application of Noether\u2019s theorem to the transformation \u03c6 \u2192 \u03c6 + \u03c6v obtains \u2202\u03bcjv,\u03bc = 0, where jv,0 = ij = \u2212i\u2202x\u03c6/\u03c0.24 While the \ufb01rst of these relations merely reiterates the de\ufb01nition of the displacement \ufb01eld, the second contains new information. Remembering that \u03c1 = \u03c1+ + \u03c1\u2212 and j = \u03c1+ \u2212 \u03c1\u2212, we obtain the important identi\ufb01cation = \u03c1 = \u2202x\u03b8/\u03c0 and jv,1 p. 174 p. 174 \u03c1\u00b1 = 1 2\u03c0 [\u2202x\u03b8 \u2213 \u2202x\u03c6], (4.50) of the left- and right-moving densities in terms of the Bose \ufb01eld. EXERCISE Compute the bosonic representation of the axial Noether current. 23 The function C describes the correlation of the bilinear \u00af\u03c8+\u03c8\u2212 with itself measured at di\ufb00erent values of space and time. We have more to say on the subject of correlation functions in the next chapter. 24 The (arbitrary) normalization constant \u03c0\u22121 has been chosen to obtain consistency with previous de\ufb01nitions, see below. 180 Functional \ufb01eld integral Interacting system We are now in a position to turn to the interacting case. In fact, all the hard work necessary to get the interacting problem under control has already been done! We simply substitute the bosonic representation Eq. (4.50) into the Coulomb action Eq. (4.44) and obtain s Sint = = 1 8\u03c02 1 4\u03c02 dx d\u03c4 [g2\u2202x(\u03b8 \u2212 s\u03c6)\u2202x(\u03b8 + s\u03c6) +g 4\u2202x(\u03b8 \u2212 s\u03c6)\u2202x(\u03b8 \u2212 s\u03c6)] % (g2 + g4)(\u2202x\u03b8)2 + (g4 \u2212 g2)(\u2202x\u03c6)2 &, dx d\u03c4 i.e. an action that is still quadratic and, thus, exactly solvable. INFO In the \ufb01eld theoretical literature (especially the literature on conformal \ufb01eld theory), interactions of this type are commonly referred to as current\u2013current interactions. This is because Sint can be expressed as a bilinear form in the Noether currents generated by the symmetries of the system. Adding Sint to the action of the non-interacting action Eq. (4.49), we arrive at the \ufb01nal expression S[\u03b8, \u03c6] = 1 2\u03c0 dx d\u03c4 g\u22121v(\u2202x\u03b8)2 + gv(\u2202x\u03c6)2 + 2i\u2202\u03c4 \u03b8\u2202x\u03c6, (4.51) where we have introduced the parameters v \u2261 vF + 1 2 2 \u2212 g2 2\u03c0 g4 2\u03c0, g \u2261 vF + g4\u2212g2 2\u03c0 vF + g4+g2 2\u03c0 1 2, (4.52) and have reinstated vF. Comparison with the results of Section 2.2 identi\ufb01es v as the e\ufb00ective velocity of the charge-density wave excitations of the system.25 Equation (4.51) represents the main result of our analysis. We have succeeded in mapping a non-linear (i.e. interacting) problem onto a linear bosonic \ufb01eld theory. Critical readers may justly object that this result does not contain much new information. After all, a (second quantized) representation of the theory in terms of free bosons has been derived previously in Section 2.2. Nonetheless, the analysis of this section is very valuable from both a methodological and a conceptual point of view. Methodologically, it turns out that the \ufb01eld integral D(\u03b8, \u03c6) exp(\u2212S[\u03b8, \u03c6]) is very convenient for calculation. Many intriguing pheformulation nomena displayed by one-dimensional electron systems (we will meet some examples) can straightforwardly be addressed in this language. From a conceptual point of view, it is quite remarkable that no \u201cmicroscopic\u201d calculations had to be performed to determine the e\ufb00ective \ufb01eld theory above. \u201cAll\u201d we had to invest was symmetry arguments, a minimal amount of phenomenological input, and a number of consistency checks. In Chapter 9", " we will meet a related problem \u2013 the low-energy \ufb01eld theory of quantum spin chains \u2013 where 25 Of course, we do not need to resort to the results of Section 2.2 to draw this conclusion: the structure of Eq. (4.51) by itself determines the interpretation of v as an e\ufb00ective velocity. To see this, one may integrate over \u03c6 to arrive at an e\ufb00ective wave-type Lagrangian for \u03b8 whose characteristic velocity is set by v (exercise). 4.5 Problems 181 a microscopic construction scheme for the e\ufb00ective \ufb01eld theory is not known at all. In this case a (suitably generalized) version of the symmetry analysis above will represent the only viable route towards the solution of the problem. 4.4 Summary and outlook This concludes our preliminary introduction to the \ufb01eld integral. We have learned how to represent the partition function of a quantum many-body system in terms of a generalized path integral. The \ufb01eld integral representation of the partition function will be the basic platform on which all our further developments will be based. In fact, we are now in a position to face up to the main problem addressed in this text: practically none of the \u201cnon-trivial\u201d \ufb01eld integrals in which one might be interested can be executed in closed form. This re\ufb02ects the fact that, save for a few exceptions (such as the one considered in Section 4.3 above), interacting many-body problems do not admit closed solutions. Before employing the \ufb01eld integral to solve serious problems, we need to develop a spectrum of approximation strategies \u2013 perturbation theory, linear response theory, mean-\ufb01eld methods, instanton techniques and the like. The construction and application of such methods will be the subject of the following chapters. 4.5 Problems Exercises on fermion coherent states To practice the coherent state method, we begin with a few simple exercises on the fermionic coherent state which complement the structures discussed in the main text. \u2020 i |\u03b7 = \u2212\u2202\u03b7i \u03b7|\u00af\u03b7i, (b) a Considering a fermionic coherent state |\u03b7, verify the following identities: (a) \u03b7|a \u2020 i =, and (d) i d\u00af\u03b7i d\u03b7i. Finally, (e) show that n|\u03c8\u03c8|n = \u03b6\u03c8|nn|\u03c8, where |n is an n-particle state in Fock space while |\u03c8 is a coherent state. |\u03b7 and \u03b7|ai = \u2202\u00af\u03b7i i \u00af\u03b7i\u03b7i |\u03b7\u03b7| = 1F, where d(\u00af\u03b7, \u03b7) \u2261 \u03b7|, (c) \u03b7|\u03bd = exp d(\u00af\u03b7, \u03b7) d\u03b7i e\u2212 i \u00af\u03b7i\u03bdi Answer: Making use of the rules of Grassmann algebra, (a) \u03b7|a i = 0|exp \u2020 \u2212 j aj \u00af\u03b7j \u00af\u03b7i j=i \u2020 a i = 0| j (1 \u2212 aj \u00af\u03b7j) = 0| (1 \u2212 aj \u00af\u03b7j)a \u2020 i = 0|(1 \u2212 ai \u00af\u03b7i)a \u2020 i j=i (1 \u2212 aj \u00af\u03b7j) (1 \u2212 aj \u00af\u03b7j)\u00af\u03b7i = \u03b7|\u00af\u03b7i. = \u2020 0|aia! i j (1 \u2212 \u03b7ja \u2020 j)|0 = \u2212\u2202\u03b7i j=i (1 \u2212 \u03b7ja \u2020 j)|0 = \u2212\u2202\u03b7i |\u03b7, \u2020 i ) (1 \u2212 ai \u00af\u03b7i)ai! =ai=\u2212\u2202 \u00af\u03b7i ai \u00af\u03b7i=\u2202 \u00af\u03b7i (1\u2212ai \u00af\u03b7i) j j = \u2202\u00af\u03b7i 0| (1 \u2212 aj \u00af\u03b7j) = \u2202\u00af\u03b7i \u03b7|. (b) a \u2020 i |\u03b7 = =a \u03b7|ai = 0| \u2020 i ]+=0| =0|[ai,a \u2020 i (1 \u2212 \u03b7ia a! i =\u2212\u2202\u03b7i (1\u2212\u03b7ia \u2020 i =\u2202\u03b7i \u03b7ia \u2020 i ) \u2020 (1 \u2212 aj \u00af\u03b7j) j=i 182 Functional \ufb01eld integral (c) \u03b7|\u03bd = \u03b7| i (1 \u2212 \u03bdia! \u2020 i ) (", "1+a \u2020 i \u03bdi) |0 = \u03b7| i (1 + \u00af\u03b7i\u03bdi)! i \u00af\u03b7i\u03bdi] exp[ |0 = exp \u00af\u03b7i\u03bdi. i (d) To prove the completeness of fermion coherent states, we apply Schur\u2019s lemma, i.e. we need to show that [a(\u2020) j, d(\u00af\u03b7, \u03b7)e\u2212 i \u00af\u03b7i\u03b7i |\u03b7\u03b7|] = 0. a \u2020 j = d(\u00af\u03b7, \u03b7) \u2202\u03b7j d(\u00af\u03b7, \u03b7) e\u2212 i \u00af\u03b7i\u03b7i |\u03b7\u03b7| = \u2212 e\u2212! i \u00af\u03b7i\u03b7i |\u03b7\u03b7| = =\u00af\u03b7j e\u2212 i \u00af\u03b7i \u03b7i d(\u00af\u03b7, \u03b7) e\u2212 i \u00af\u03b7i\u03b7i |\u03b7\u03b7| = aj = d(\u00af\u03b7, \u03b7) e\u2212 i \u00af\u03b7i\u03b7i |\u03b7\u2202\u00af\u03b7j \u03b7| = d(\u00af\u03b7, \u03b7) e\u2212 i \u00af\u03b7i\u03b7i \u2202\u03b7j |\u03b7\u03b7| d(\u00af\u03b7, \u03b7) e\u2212 i \u00af\u03b7i\u03b7i |\u03b7\u03b7|a \u2020 j, d(\u00af\u03b7, \u03b7) e\u2212 i \u00af\u03b7i\u03b7i \u03b7j! =\u2212\u2202 \u00af\u03b7j (e\u2212 i \u00af\u03b7i \u03b7i) d(\u00af\u03b7, \u03b7) e\u2212 i \u00af\u03b7i\u03b7i |\u03b7\u03b7|aj. |\u03b7\u03b7| The constant of proportionality is \ufb01xed by taking the expectation value with the vacuum. 0| d(\u00af\u03b7, \u03b7)e\u2212 i \u00af\u03b7i\u03b7i |\u03b7\u03b7|0 = d(\u00af\u03b7, \u03b7) e\u2212 i \u00af\u03b7i\u03b7i = 1. \u2020 (e) A general n-particle state is given as |n = a 1... a\u2020 n |0, n| = 0|an... a1, where we neglected a normalization factor. The matrix element n|\u03c8, thus, reads n|\u03c8 = 0|an... a1|\u03c8 = 0|\u03c8n... \u03c81|\u03c8 = \u03c8n... \u03c81. Similarly, we obtain \u03c8|n = \u00af\u03c81... \u00af\u03c8n. Using these results, n|\u03c8\u03c8|n = \u03c8n... \u03c81 \u00af\u03c81... \u00af\u03c8n = \u03c81 \u00af\u03c81... \u03c8n \u00af\u03c8n = (\u03b6 \u00af\u03c81\u03c81)...( \u03b6 \u00af\u03c8n\u03c8n) = (\u03b6 \u00af\u03c81)...( \u03b6 \u00af\u03c8n)\u03c8n... \u03c81 = \u03b6\u03c8|nn|\u03c8. Feynman path integral from the functional \ufb01eld integral The abstraction of the coherent state representation betrays the close similarity between the Feynman and coherent state path integrals. To help elucidate the connection, the goal of the present problem is to con\ufb01rm that the Feynman path integral of the quantum harmonic oscillator follows from the coherent state path integral. Consider the simplest bosonic many-body Hamiltonian, \u02c6H = \u03c9(a\u2020a+ 1 2 ), where a\u2020 creates \u201cstructureless\u201d particles, i.e. states in a one-dimensional Hilbert space. Note that \u02c6H can be interpreted as the Hamiltonian of a single oscillator degree of freedom. Show that the \ufb01eld integral for the partition function Z = tr[exp(\u2212\u03b2 \u02c6H)] can be mapped onto the (imaginarytime) path integral of a harmonic oscillator by a suitable variable transformation. (Hint: Let yourself be guided by the fact that the conjugate operator pair (a, a\u2020) is related to the momentum and coordinate operators (\u02c6p, \u02c6q) through a canonical transformation.) 4.5 Problems 183 Answer: In the coherent state representation, the quantum partition function of the oscillator Hamiltonian is expressed in terms of the path integral ( = 1): Z = D( \u00af\u03c6, \u03c6) exp \u2212 \u03b2 0 d\u03c4 \u00af\u03c6\u2202\u03c4 \u03c6 + \u03c9 \u00af\u03c6\ufffd", "\ufffd, (4.53) where \u03c6(\u03c4 ) denotes a complex scalar \ufb01eld, the constant factor e\u2212\u03b2\u03c9/2 has been absorbed integral D( \u00af\u03c6, \u03c6), and we have set the chemical into the measure of the functional potential \u03bc = 0. The connection between the coherent state and Feynman integral is established by the change of \ufb01eld variables, \u03c6(\u03c4 ) = (m\u03c9/2)1/2 (q(\u03c4 ) +ip( \u03c4 )/m\u03c9), where p(\u03c4 ) andq (\u03c4 ) represent real \ufb01elds. Substituting this representation in Eq. (4.53), and rearranging some terms by integrating by parts, the connection is established: Z =. (Of course, the \u201cabsorption\u201d of the constant \u03c9/2 in the Hamiltonian into the measure has been a bit of cheating. In operator quantum mechanics, the correspondence between the (q, p) and the (a, a\u2020) representation of the Hamiltonian includes that constant. Keeping in mind that the constant re\ufb02ects the non-commutativity of operators, think how it might be recovered within the path integral formalism. Hint: it is best to consider the mapping between representations within a time-slice discretized setting.) 2m + m\u03c92 \u2212ip \u02d9q + p2 D(p, q) exp \u03b2 0 d\u03c4 2 q2 $ \u2212 # Quantum partition function of the harmonic oscillator The following involves a practice exercise on elementary \ufb01eld integral manipulations, and in\ufb01nite products. \u221e Compute the partition function of the harmonic oscillator Hamiltonian in the \ufb01eld integral formulation. To evaluate the resulting in\ufb01nite product over Matsubara frequencies apply \u22121. (Hint: The normalization of the result can the formula x/ sin x = be \ufb01xed by requiring that, in the zero-temperature limit, the oscillator occupy its ground state.) Finally, compute the partition function by elementary means and check your result. As an additional exercise, repeat the same steps for the \u201cfermionic oscillator,\u201d i.e. with a, a\u2020 fermion operators. Here you will need the auxiliary identity cos x =. 1 \u2212 x2/(\u03c0n)2 \u221e 1 \u2212 n=1 x2 (\u03c0(n+1/2))2 n=1 Answer: Making use of the Gaussian functional Eq. (4.53) ( = 1) integral for complex \ufb01elds, one obtains from ZB \u223c det(\u2202\u03c4 + \u03c9)\u22121 \u223c'(\u2212i\u03c9n + \u03c9) \u22121 \u223c \u03c9n \u22121 2 \u221e' 1 + \u223c n=1 \u03b2\u03c9 2\u03c0n \u223c 1 sinh(\u03b2\u03c9/2). \u221e' n=1 \u22121 2 2n\u03c0 \u03b2 + \u03c92 184 Functional \ufb01eld integral Now, in the limit of small temperatures, the partition function is dominated by the ground state, lim\u03b2\u2192\u221e ZB = exp[\u2212\u03b2\u03c9/2], which \ufb01xes the constant of proportionality. Thus, ZB = [2 sinh(\u03b2\u03c9/2)] \u22121. In the fermionic case, the Gaussian integration gives a product over eigenvalues in the numerator and we have to use fermionic Matsubara frequencies, \u03c9n = (2n + 1)\u03c0/\u03b2:'\u03c9n (\u2212i\u03c9n + \u03c9) \u223c 2 ZF \u223c det(\u2202\u03c4 + \u03c9) \u223c \u221e' 1 + \u223c n=1 \u03b2\u03c9 (2n + 1)\u03c0 \u223c cosh(\u03b2\u03c9/2). \u221e' n=1 (2n + 1)\u03c0 \u03b2 2 + \u03c92 Fixing the normalization, one obtains ZF = 2e\u2212\u03b2\u03c9 cosh(\u03b2\u03c9/2). Taken together, these results are easily con\ufb01rmed by direct computation, viz. ZB = e\u2212\u03b2\u03c9/2 ZF = e\u2212\u03b2\u03c9/2 \u221e n=0 1 n=0 e\u2212n\u03b2\u03c9 = e\u2212\u03b2\u03c9/2 1 \u2212 e\u2212\u03b2\u03c9 = 1 2 sinh(\u03b2\u03c9/2), e\u2212n\u03b2\u03c9 = e\u2212\u03b2\u03c9/2(1 + e\u2212\u03b2\u03c9) = 2e\u2212\u03b2\u03c9 cosh(\u03b2\u03c9/2). Boson\u2013fermion duality The equivalence of the bosonic and the fermionic representation of the one-dimensional electron gas is exempli\ufb01ed by computation of the correlation function Eq. (4.48) considered in the main", " text. (a) Employ the free fermion \ufb01eld integral with action (4.43) to compute the zerotemperature limit of the correlation function (4.48) considered in the text (assume x >0). (b) Considering a free scalar bosonic \ufb01eld \u03b8 with action S[\u03b8] = 1 2c dx d\u03c4 [(\u2202\u03c4 \u03b8)2 + (\u2202x\u03b8)2], compute the correlation function K(x, \u03c4 ) \u2261 \u03b8(x, \u03c4 )\u03b8(0, 0) \u2212 \u03b8(0, 0)\u03b8(0, 0) for x >0. (c) Compute the correlation function C(x, \u03c4 ) = \u03b32 exp[2i\u03b8(x, \u03c4 )] exp[\u22122i\u03b8(0, 0)]. Answer: (a) Setting vF = 1 and de\ufb01ning Z\u00b1 \u2261 dx d\u03c4 \u00af\u03c8(\u2202\u03c4 \u2213 i\u2202x)\u03c8, we obtain D( \u00af\u03c8, \u03c8) exp \u2212S\u00b1[ \u00af\u03c8, \u03c8], where S\u00b1[ \u00af\u03c8, \u03c8] = G\u00b1(x, \u03c4 ) = Z \u22121 \u00b1 = \u2212 T L D( \u00af\u03c8, \u03c8) \u00af\u03c8(x, \u03c4 ) \u03c8(0, 0)e\u2212S\u00b1[ \u00af\u03c8,\u03c8] = \u2212(\u2202\u03c4 \u2213 i\u2202x )\u22121 (x,\u03c4 ;0,0) p,\u03c9n 1 \u2212i\u03c9n \u2213 p e\u2212ipx\u2212i\u03c9n\u03c4. 4.5 Problems 185 Assuming for de\ufb01niteness that x >0 and integrating over momenta, we arrive at G\u00b1(x, \u03c4 ) = \u2213iT n \u0398(\u00b1n)e\u03c9n(\u2213x\u2212i\u03c4 ) 1 2\u03c0 1 \u00b1ix \u2212 \u03c4, where in the last equality we have approximated the frequency sum by an integral. Thus, the correlation function (4.48) is given by C(x, \u03c4 ) = G+(x, \u03c4 )G\u2212(\u2212x, \u03c4 ) = 1 1 x2+\u03c4 2. |\u03b8q,n|2(q2+ (b) Expressed in a frequency/momentum Fourier representation, S[\u03b8] = L 2cT (2\u03c0)2 q,n \u03c92 n). Performing the Gaussian integral over \u03b8, we obtain eiqx+i\u03c9n\u03c4 q2 + \u03c92 n \u2212 1 cT 2 K(x, \u03c4 ) = cT L e\u2212|\u03c9n|x+i\u03c9n\u03c4 \u2212 1 |\u03c9n| q,n c 4\u03c0 a\u22121 d\u03c9 0 n e\u2212\u03c9(x\u2212i\u03c4 ) \u2212 1 \u03c9 + c.c. x,\u03c4 a \u2212 c 4\u03c0 ln((x2 + \u03c4 2)/a2), where we have approximated the momentum sum by an integral and the frequency sum by an integral, cut o\ufb00 at large frequencies by EF vFa\u22121 vF=1= a\u22121. (c) Using the results derived in (b), > C(x, \u03c4 ) = \u03b32 e2i(\u03b8(x,\u03c4 )\u2212\u03b8(0,0))? = \u03b32exp \u2212 c \u03c0 ln x2 + \u03c4 2 a2 = \u03b32e\u22122(\u03b8(x,\u03c4 )\u2212\u03b8(0,0))2 c = \u03b32 a2 x2 + \u03c4 2 \u03c0. Setting c = \u03c0 and \u0393 = 1/2\u03c0a, we obtain equivalence to the fermionic representation of the correlation function considered in (a). Frequency summations Using the frequency summation techniques developed in the text, this problem involves the computation of two basic correlation functions central to the theory of the interacting Fermi gas. (a) The pair correlation function \u03c7c n,q is an important building block entering the calculation of the Cooper pair propagator in superconductors (see Section 6.4). It is given by \u03c7c n,q \u2261 \u2212 T Ld m,p G0(p, i\u03c9m)G0(\u2212p+q, \u2212i\u03c9m +i\u03c9n) = 1 Ld p 1 \u2212 nF(\u03bep) \u2212 nF(\u03be\u2212p+q) i\u03c9n \u2212 \u03bep \u2212 \u03be\u2212p+q, where G0(p, i\u03c9m) = 1/(i\u03c9m \u2212 \u03bep). Verify the second equality. (Note that \u03c9m = (2m + 1)\u03c0\u03b2 are f", "ermionic Matsubara frequencies, while \u03c9n = 2\u03c0nT is a bosonic Matsubara frequency.) (b) Another correlation function central to the theory of the interacting Fermi gas (see Section 5.2), the so-called density\u2013density response function, is given by G0(p, i\u03c9m)G0(p + q, i\u03c9m + i\u03c9n) = \u2212 1 Ld \u2261 \u2212 T Ld nF(\u03bep) \u2212 nF(\u03bep+q) i\u03c9n + \u03bep \u2212 \u03bep+q q,\u03c9n \u03c7d. p p,\u03c9m Again verify the second equality. 186 Answer: Functional \ufb01eld integral (a) To evaluate a sum over fermionic frequencies \u03c9m, we employ the Fermi function \u03b2nF(z) =\u03b2 (e\u03b2z + 1)\u22121 de\ufb01ned in the left column of Eq. (4.37). Noting that the function G0(p1, z)G0(p2, z + i\u03c9n) has simple poles at z = \u03bep1 and z = i\u03c9n \u2212 \u03bep2, h and h = G0G0), we and applying Eq. (4.38) (with the identi\ufb01cation S = obtain S =. Using the fact that nF(x + i\u03c9n) = nF(x) and i\u03c9n\u2212\u03bep1 nF(\u2212x) = 1 \u2212 nF(x) we arrive at the result. \u2212nF(\u03bep1 )+nF(\u03be\u2212p2 +i\u03c9n) \u2212\u03bep2 (b) One may proceed as in part (a). Pauli paramagnetism There are several mechanisms whereby a Fermi gas subject to an external magnetic \ufb01eld responds to the perturbation. One of these, the phenomenon of Pauli paramagnetism, is purely quantum mechanical in nature. Its origin lies in the energy balance of spinful fermions rearranging at the Fermi surface in response to the \ufb01eld. We explore the resulting contribution to the magnetic susceptibility of the electron gas. Fermions couple to a magnetic \ufb01eld by their orbital momentum as well as by their spin. Concentrating on the latter mechanism, consider the Hamiltonian \u02c6Hz = \u2212\u03bc0B \u00b7 \u02c6S, \u02c6S = 1 2 a\u2020 \u03b1\u03c3\u03c3\u03c3\u03c3 a\u03b1\u03c3, where \u03c3 = (\u03c3x, \u03c3y, \u03c3z)T is a vector of Pauli matrices, \u03b1 an orbital quantum number, and \u03b5 B =0 \u03b5 B >0 \u03bc p \u03bc p \u03bc0 = e/(2m) the Bohr magneton. It turns out that the presence of \u02c6Hz in the energy balance leads to the generation of a net paramagnetic response of purely quantum mechanical origin. To understand the origin of the e\ufb00ect, consider a two-fold (spin!) degenerate singleparticle band of free electrons states (see the \ufb01gure). Both bands are \ufb01lled up to a certain chemical potential \u03bc. Upon the switching on of an external \ufb01eld, the degeneracy is lifted and the two bands shift in opposite directions by an amount \u223c \u03bc0B. While, deep in the bands, the Pauli principle forbids a rearrangement of spin con\ufb01gurations, up at the Fermi energy, \u2193 states can turn to energetically more favorable \u2191 states. More precisely, for bands shifted by an amount \u223c \u03bc0B, a number \u223c \u03bc0B\u03c1(\u03bc) of states may change their spin direction, which leads to a total energy change of \u0394E \u223c \u2212\u03bc2 0B2\u03c1(\u03bc). Di\ufb00erentiating twice with respect to the magnetic \ufb01eld gives a positive contribution \u03c7 \u223c \u2212\u22022 0\u03c1(\u03bc) to the magnetic susceptibility of the system. B\u0394E \u223c \u03bc2 (a) To convert the qualitative estimate above into a quantitative result, write down the \u03b1\u03c3\u03b1a\u03b1\u03c3 coherent state action of the full Hamiltonian \u02c6H = \u02c6H0 + \u02c6Hz, where \u02c6H0 = \u03b1,\u03c3 a\u2020 4.5 Problems 187 is the non-magnetic part of the Hamiltonian. Integrate out the Grassmann \ufb01elds to obtain the free energy F as a sum over frequencies. (b) Show that, at low temperatures, the spin contribution to the magnetic susceptibility B=0F is given by \u03c7 T \u21920\u2212\u2192 \u03bc2 0 2 \u03c1(\u03bc), (4.54) \ufffd", "\ufffd \u2261 \u2212\u22022 B where \u03c1() = convenient to perform the \ufb01eld derivatives prior to the frequency summation.) \u03b1 \u03b4( \u2212 \u03b1) denotes the single-particle density of states. (Hint: It is Answer: (a) Choosing the quantization axis parallel to the magnetic \ufb01eld the Hamiltonian assumes 2 (\u03c3z)\u03c3\u03c3 a\u03b1\u03c3 and the (frequency representation \u00af\u03c8\u03b1\u03c3n(\u2212i\u03c9n + \u03be\u03b1 \u2212 \u03bc0B 2 (\u03c3z)\u03c3\u03c3)\u03c8\u03b1\u03c3n. Integrating \u03b1;n \u03b22 a diagonal form \u02c6H = \u03b1\u03c3 a\u2020 \u03b1\u03c3 of the) action reads S[ \u00af\u03c8, \u03c8] = over \u03c8, we obtain the partition function Z = (\u2212i\u03c9n + \u03be\u03b1)2 \u2212 1 % \u03b1 \u2212 \u03bc0B 4 (\u03bc0B)2 and \u03b1\u03c3n & F = \u2212T ln Z = \u2212T ln \u03b22 (\u2212i\u03c9n + \u03be\u03b1)2 \u2212 1 4 (\u03bc0B)2. \u03b1;n (b) Di\ufb00erentiating the free energy twice with respect to B, we obtain \u03c7 = \u2212 1 2 \u03bc2 0T 0T (\u2212i\u03c9n + \u03be\u03b1)\u22122, Eq. (4.38) can \u03be\u03b1)\u22122. De\ufb01ning \u03c7 = be applied to perform the frequency sum. Noting that the function h(\u2212iz) has poles of second order at z = \u03be\u03b1, i.e. Res[h(\u2212iz)g(z)]|z=\u03be\u03b1 = g(\u03be\u03b1), we obtain n\u03b1 h\u03b1(\u03c9n), where h\u03b1(\u03c9n) = 1 2 \u03bc2 \u03b1\u03c9n (\u2212i\u03c9n+ \u03c7 = \u2212 \u03bc2 0 2 \u03b1 F(\u03be\u03b1) = \u2212 \u03bc2 n 0 2 \u221e \u2212\u221e d \u03c1()n F( \u2212 \u03bc). At low temperatures, T \u2192 0, the Fermi distribution function approaches a step function, nF() \u2192 \u03b8(\u2212), i.e. n F() = \u2212\u03b4() and our result reduces to Eq. (4.54). Electron\u2013phonon coupling As follows from the structure of our prototypical condensed matter \u201cmaster Hamiltonian\u201d (1.1), mobile electrons in solids are susceptible to the vibrations of the host ions, the phonons. This coupling mechanism generates a net attractive interaction between the electrons. Referring for a qualitative discussion of this interaction mechanism to page 266 below, it is the purpose of this problem to quantitatively explore the pro\ufb01le of the phonon mediated electron\u2013electron interaction. In Section 6.4 we will see that this interaction lies at the root of conventional BCS superconductivity. Consider the three-dimensional variant of the phonon Hamiltonian (1.34), \u02c6Hph = \u03c9qa q,j \u2020 q,jaq,j + const., 188 Functional \ufb01eld integral where \u03c9q is the phonon dispersion (here assumed to depend only on the modulus of the momentum, |q| = q) and the index j = 1, 2, 3 accounts for the fact that the lattice ions can oscillate in three directions in space (i.e. there are three linearly independent oscillator modes26). Electrons in the medium sense the induced charge \u03c1ind \u223c \u2207 \u00b7 P, where P \u223c u is the polarization generated by the local distortion u of the lattice (u(r) is the three-dimensional generalization of the displacement \ufb01eld \u03c6(r) considered in Chapter 1). Expressed in terms of phonon creation and annihilation operators (cf. Eq. (1.32)), \u2020 \u2212q,j)/(2m\u03c9q)1/2, where ej is the unit vector in the j-direction,26 and we uq = ej(aq,j + a conclude that the electron\u2013phonon Hamiltonian reads \u02c6Hel\u2212ph = \u03b3 ddr\u02c6n (r)\u2207 \u00b7u (r) = \u03b3 k,q,j iqj (2m\u03c9q)1/2 \u02c6nq(aq,j + a \u2020 \u2212q,j). Here, \u02c6nq \u2261 and annihilation operators, and the electron spin has been neglected for simplicity. \u2020 k+qck denotes the electronic density expressed in terms of fermion creation k c (a) Formulate the coherent state action of the electron\u2013phonon system. (b) Integrate out the phonon \ufb01elds, and show that an attractive interaction between elec- tr", "ons is generated. Answer: (a) Introducing a Grassmann \ufb01eld \u03c8 (a complex \ufb01eld \u03c6) to represent the electron (phonon) operators, one obtains the coherent state \ufb01eld integral Z = D[ \u00af\u03c8, \u03c8] D[ \u00af\u03c6, \u03c6] e\u2212Sel[ \u00af\u03c8,\u03c8]\u2212Sph[ \u00af\u03c6,\u03c6]\u2212Sel\u2212ph[ \u00af\u03c8,\u03c8, \u00af\u03c6,\u03c6], where Sph[ \u00af\u03c6, \u03c6] = \u00af\u03c6qj(\u2212i\u03c9n + \u03c9q)\u03c6qj, Sel\u2212ph[ \u00af\u03c8, \u03c8, \u00af\u03c6, \u03c6] =\u03b3 q,j q,j iqj (2m\u03c9q)1/2 \u03c1q(\u03c6qj + \u00af\u03c6\u2212qj), \u00af\u03c8k+q\u03c8k, and the electron action need not be speci\ufb01ed explicitly. Here we have \u03c1q = adopted a short-hand convention setting q = (\u03c9n, q).27 k 26 For more details see N. W. Ashcroft and N. D. Mermin, Solid State Physics (Holt-Saunders International, 1983). 27 Do not confuse the 4-momentum q with the modulus |q| = q. 4.5 Problems 189 (b) We next perform the Gaussian integration over the phonon \ufb01elds to obtain the e\ufb00ective electron action Se\ufb00 [ \u00af\u03c8, \u03c8] =S el[ \u00af\u03c8, \u03c8] \u2212 ln D[ \u00af\u03c6, \u03c6] e\u2212(Sph[ \u00af\u03c6,\u03c6]+Sel\u2212ph)[ \u00af\u03c8,\u03c8, \u00af\u03c6,\u03c6] = Sel[ \u00af\u03c8, \u03c8] \u2212 \u03b3 2m q q2 n + \u03c92 q \u03c92 \u03c1q\u03c1\u2212q. Sloppily transforming from Matsubara to real frequencies, \u03c9n \u2192 \u2212i\u03c9, we notice that, for every momentum mode q, the interaction is attractive at low frequencies, \u03c9 < \u03c9q. Disordered quantum wires In this problem, we consider a one-dimensional interacting Fermi system \u2013 a \u201cquantum wire\u201d \u2013 in the presence of impurities. Building on the results obtained in Section 4.3, we derive an e\ufb00ective low-energy action of this system. (The actual analysis of the large-scale behavior of the disordered quantum wire necessitates the application of renormalization group methods and is postponed to Chapter 8.) In Sections 2.2 and 4.3 we discussed the physics of interacting fermions in one dimension. We saw that, unlike in a Fermi liquid, the fundamental excitations of the system are charge (and spin) density waves \u2013 collective excitations describing the wave-like propagation of spin and charge degrees of freedom, respectively. Going beyond the level of an idealized translationally invariant environment, the question we wish to address currently is to what extent the propagation of these modes will be hampered by the presence of spatially localized imperfections. This problem is of considerable practical relevance. All physical realizations of one-dimensional conductive systems \u2013 semiconductor quantum wires, conducting polymers, carbon nanotubes, quantum Hall edges, etc. \u2013 generally contain imperfections. Further, and unlike systems of higher dimensionality, a spin/charge degree of freedom propagating down a one-dimensional channel will inevitably hit any impurity blocking its way. We thus expect that impurity scattering has a stronger impact on the transport coe\ufb03cients than in higher dimensions. However, there is a second and less obvious mechanism behind the strong impact of r\u20131 2kF disorder scattering on the conduction behavior of one-dimensional quantum wires: imagine a wavepacket of characteristic momentum kF colliding with an impurity at position x = 0 (see \ufb01gure). The total wave amplitude to the left of the impurity, \u03c8(x) \u223c exp(ikFx) + r exp(\u2212ikFx) will be a linear superposition of the incoming amplitude \u223c exp(ikFx) and the re\ufb02ected outgoing amplitude \u223c r exp(\u2212ikFx), where r is the re\ufb02ection coe\ufb03cient. Thus, the electronic density pro\ufb01le is given by \u03c1(x) =| \u03c8(x)|2 \u223c 1 +| r|2 + 2Re, which contains an oscillatory contribution known as a Friedel oscillation. Moreover, a closer analysis (see exercise below) shows that, in one dimension, the amplitude of these oscillations decays rather slowly", ", varying as \u223c |x|\u22121. The key point is that, in the presence of electron\u2013electron re\u22122ikFx 0 x 190 Functional \ufb01eld integral interactions, other particles approaching the impurity will notice, not only the impurity itself, but also the charged density pattern of the Friedel oscillation. The additional scattering potential then creates a secondary Friedel oscillation, etc. We thus expect that even a weak imperfection in a Luttinger liquid acts as a \u201ccatalyst\u201d for the recursive accumulation of a strong potential. In this problem, we will derive the e\ufb00ective low-energy action describing the interplay of interaction and impurity scattering. The actual catalytic ampli\ufb01cation mechanism outlined above is then explored in Chapter 8 by renormalization group methods. EXERCISE To explore the Friedel oscillatory response of the one-dimensional electron gas to a local perturbation, consider the connected density\u2013density correlation function \u03a0(x, t) = \u02c6\u03c1(x, t)\u02c6\u03c1(0, 0) \u2212 \u02c6\u03c1(x, t)\u02c6\u03c1(0, 0), q,s vF(pF + sq)a\u2020 where \u00b7 \u00b7 \u00b7 denotes the ground state expectation value, \u02c6\u03c1 = a\u2020a and a(x) =e ikFxa+(x) + e\u2212ikFxa\u2212(x) splits into a left- and a right-moving part as usual. Using the fact that \u02c6H = sqasq and the von Neumann equation \u02d9asq = i[ \u02c6H, asq], show that the timedependence of the annihilation operators is given by asq(t) =e \u2212ivF(pF+sq)tasq. Use this result, the canonical operator commutation relations, and the ground state property a\u00b1,q|\u03a9 = 0 for \u00b1q > 0 to show that \u03a0(x, t) = 1 4\u03c02 1 (x \u2212 vFt)2 + 1 (x + vFt)2 + 2 cos(2pFx) x2 \u2212 (vFt)2. Use this result to argue why the static response to an impurity potential decays as \u223c |x|\u22121. Consider the one-dimensional quantum wire, as described by the actions Eq. (4.43) and (4.44). Further, assume that, at x = 0, the system contains an imperfection or impurity. Within the e\ufb00ective action approach, this is described by # $ Simp[\u03c8\u2020, \u03c8] = d\u03c4 v+\u03c8 \u2020 +\u03c8+ + v\u2212\u03c8 \u2020 +\u03c8+ + v\u03c8 \u2020 +\u03c8\u2212 + \u00afv\u03c8 \u2020 \u2212\u03c8+, where all \ufb01eld amplitudes are evaluated at x = 0 and the constants v\u00b1 \u2208 R and v \u2208 C describe the amplitudes of forward and backward scattering, respectively. (a) Show that the forward scattering contributions can be removed by a gauge transformation. This demonstrates that forward scattering is inessential as long as only gauge invariant observables are considered. What is the reason for the insigni\ufb01cance of forward scattering? We next reformulate the problem in a bosonic language. While the clean system is described by Eq. (4.51), substitution of (4.46) into the impurity action gives Simp[\u03b8] = d\u03c4 cos(2\u03b8(\u03c4 )), where \u03b3 = 2v\u03932 and we have assumed the backward scattering amplitude \u03b3 to be real. (Any phase carried by the scattering amplitude can be removed by a global gauge transformation of the \ufb01elds \u03c8\u00b1. How?) Notice the independence of Simp on the \ufb01eld \u03c6. 4.5 Problems 191 (b) Integrate out the Gaussian \ufb01eld \u03c6 to obtain the Lagrangian formulation of the action, S[\u03b8] = 1 2\u03c0g % dxd\u03c4 v(\u2202x\u03b8)2 + v\u22121(\u2202\u03c4 \u03b8)2 & + Simp[\u03b8]. The formulation of the problem derived in (b) still contains redundancy. The point is that, everywhere except for x = 0, the action is Gaussian. This observation suggests that one may integrate out all \ufb01eld degrees of freedom \u03b8(x = 0), thus reducing the problem to one that is local in space (though, as we shall see, non-local in time). To this end, we reformulate the \ufb01eld integral as Z = D \u02dc\u03b8 exp(\u2212S[\u02dc\u03b8]), where 6 7", " exp(\u2212S[\u02dc\u03b8]) = D\u03b8 \u03b4(\u02dc\u03b8(\u03c4 ) \u2212 \u03b8(0, \u03c4 )) exp(\u2212S[\u03b8]),'\u03c4 \u03c4 \u03b4(\u02dc\u03b8(\u03c4 ) \u2212 \u03b8(0, \u03c4 )) is a is the action integrated over all \ufb01eld amplitudes save for \u03b8(0, \u03c4 ) and product of \u03b4-functions (one for each time slice) imposing the constraints \u03b8(0, \u03c4 ) = \u02dc\u03b8(\u03c4 ). We next represent these \u03b4- functions as \u03b4(\u02dc\u03b8 \u2212 \u03b8(0, \u03c4 )) = 1 dk(\u03c4 ) exp(ik(\u03c4 )(\u02dc\u03b8(\u03c4 ) \u2212 \u03b8(0, \u03c4 ))) to 2\u03c0 obtain exp(\u2212S[\u02dc\u03b8]) = D\u03b8 Dk exp = D\u03b8Dkexp \u2212 dxd\u03c4 \u2212S[\u03b8] +i d\u03c4 k(\u03c4 )(\u02dc\u03b8(\u03c4 ) \u2212 \u03b8(0, \u03c4 )) % 1 2\u03c0g v(\u2202x\u03b8)2 + v\u22121(\u2202\u03c4 \u03b8)2 & + (c cos(2\u02dc\u03b8) +ik (\u02dc\u03b8 \u2212 \u03b8))\u03b4(x). The advantage gained by this representation is that it permits us to replace cos(2\u03b8(0, \u03c4 )) \u2192 cos(2\u02dc\u03b8(\u03c4 )), whereupon the \u03b8-dependence of the action becomes purely quadratic. (c) Integrate out the \ufb01eld \u03b8(x, \u03c4 ) to obtain the representation Z = D\u03b8 e\u2212S[\u03b8], Se\ufb00 [\u03b8] = 1 \u03c0T g n \u03b8n|\u03c9n|\u03b8\u2212n + \u03b3 d\u03c4 cos(2\u03b8(\u03c4 )), (4.55) entirely in terms of a single time-dependent degree of freedom \u03b8(\u03c4 ). Notice that the entire e\ufb00ect of the bulk of the electron gas at x = 0 went into the \ufb01rst, dissipative term. We have, thus, reduced the problem to one involving a single time-dependent degree of freedom subject to a dissipative damping mechanism and a periodic potential (cf. our discussion of this problem in Problem 3.5 above). 0 x INFO To understand the physical origin of the dissipative damping mechanism notice that, in the absence of the impurity, the system is described by a set of harmonic oscillators. We can thus think of the degree of freedom \u03b8(0, \u03c4 ) as the coordinate of a \u201cbead\u201d embedded into an in\ufb01nitely extended harmonic chain. From the point of view of this bead, the neighboring degrees of freedom hamper its free kinematic motion, i.e., in order to move, the bead has to drag an 192 Functional \ufb01eld integral entire \u201cstring\u201d of oscillators behind. In other words, a local excitation of the x = 0 oscillator will lead to the dissipation of kinetic energy into the continuum of neighboring oscillators. Clearly, the rate of dissipation will increase with both the sti\ufb00ness of the oscillator chain (g\u22121) and the frequency of the excitation (\u03c9n), as described by the \ufb01rst term in the last operator in Eq. (4.55). Answer: (a) Consider the gauge transformation \u03c8+(x, \u03c4 ) \u2192 e\u2212iv\u22121 F v+\u03b8(x). While Sint and Simp are gauge invariant and do not change, substitution of the transformed \ufb01eld into the noninteracting action leads to S0[\u03c8\u2020, \u03c8] \u2192 S0[\u03c8\u2020, \u03c8] \u2212 v+ d\u03c4 \u03c8 \u2020 +\u03c8+. The induced term cancels against the v+ contribution to Simp. A similar transformation removes the v\u2212 contribution. The physical reason for the insigni\ufb01cance of the forward scattering operators is that they describe the scattering of states | \u00b1k F into the same states | \u00b1k F. The optional phase shift picked up in these processes is removed by the transformation above. (b) This involves an elementary Gaussian integral. (c) Expressed in momentum space, the e\ufb00ective action assumes the form e\u2212S[\u02dc\u03b8] = D\u03b8 Dk exp # \u2212 T L q,\u03c9n 1 2\u03c0g vq2 + v\u22121\u03c92 n $ |\u03b8q,n|2 + ikn\u03b8q,\u2212n = Dk exp \u2212 \u03c0gT 2L = N Dk exp = exp \u2212 1 \u03c0T g n \u2212ikn ", "\u02dc\u03b8\u2212n \u2212 Simp[\u02dc\u03b8] q,\u03c9n kn(vq2 + v\u22121\u03c92 n)\u22121k\u2212n \u2212 ikn \u02dc\u03b8\u2212n \u2212 Simp[\u02dc\u03b8] \u2212 \u03c0T g 4 \u03c9n kn|\u03c9n|\u22121k\u2212n \u2212 ikn \u02dc\u03b8\u2212n \u2212 Simp[\u02dc\u03b8] \u02dc\u03b8n|\u03c9n|\u02dc\u03b8\u2212n \u2212 Simp[\u02dc\u03b8]. Denoting \u02dc\u03b8(\u03c4 ) by\u03b8 (\u03c4 ), we obtain the e\ufb00ective action Eq. (4.55). 5 Perturbation theory In this chapter, we introduce the analytical machinery to investigate the properties of many-body systems perturbatively. Speci\ufb01cally, employing the \u201c\u03c64-theory\u201d as an example, we learn how to describe systems that are not too far from a known reference state by perturbative means. Diagrammatic methods are introduced as a tool to e\ufb03ciently implement perturbation theory at large orders. The new concepts are then applied to the analysis of various properties of the weakly interacting electron gas. In previous chapters we have emphasized repeatedly that the majority of many-particle problems cannot be solved in closed form. Therefore, in general, one is compelled to think about approximation strategies. One promising ansatz leans on the fact that, when approaching the low-temperature physics of a many-particle system, we often have some idea, however vague, of its preferred states and/or its low-energy excitations. One may then set out to explore the system by using these prospective ground state con\ufb01gurations as a working platform. For example, one might expand the Hamiltonian in the vicinity of the reference state and check that, indeed, the residual \u201cperturbations\u201d acting in the low-energy sector of the Hilbert space are weak and can be dealt with by some kind of approximate expansion. Consider, for example, the quantum Heisenberg magnet. In dimensions higher than one, an exact solution of this system is out of the question. However, we know (or, more conservatively, \u201cexpect\u201d) that, at zero temperature, the spins will be frozen into con\ufb01gurations aligned along some (domain-wise) constant magnetisation axes. Residual \ufb02uctuations around these con\ufb01gurations, described by the Holstein\u2013Primako\ufb00 boson excitations, or spin waves, discussed before can be described in terms of a controlled expansion scheme. Similar programs work for countless other physical systems. These considerations dictate much of our further strategy. We will need to construct methods to identify and describe the lowest-energy con\ufb01gurations of many-particle systems \u2013 often called \u201cmean-\ufb01elds\u201d \u2013 and learn how to implement perturbation theory around them. In essence, the \ufb01rst part of that program amounts to solving a variational problem, a relatively straightforward task. However, the formulation of perturbation strategies requires some preparation and, equally important, a good deal of critical caution (because many systems notoriously defy perturbative assaults \u2013 a fact easily overlooked or misjudged!). We thus turn the logical sequence of the two steps upside down and devote this chapter to an introduction to many-body perturbation theory. This will include a number of applications, i.e. problems where the mean-\ufb01eld is trivial and perturbation theory on its own su\ufb03ces to 193 194 Perturbation theory produce meaningful results. Perturbation theory superimposed on non-trivial mean-\ufb01elds will then be the subject of the next chapter. 5.1 General structures and low-order expansions As with any other perturbative approach, many-body perturbation theory amounts to an expansion of observables in powers of some parameter \u2013 typically the coupling strength of an interaction operator. However, before discussion of how this program is implemented in practice, it is imperative to develop some understanding of the mathematical status of such \u201cseries expansions.\u201d (To motivate the point: it may, and often does, happen that the in\ufb01niteorder expansion in the \u201csmall parameter\u201d of the problem does not exist in a mathematical sense!) This can be achieved by considering the following. Consider the integral An instructive integral I(g) = \u221e \u2212\u221e dx\u221a 2\u03c0 exp \u2212 1 2 x2 \u2212 gx4. (5.1) This can be regarded as a caricature of a particle subject to some harmonic potential (x2) together with an \u201cinteraction\u201d (x4). For small g 1, it seems natural to treat", " the interaction perturbatively, i.e. to develop the expansion I(g) \u2248 n gnIn, where, applying Stirling\u2019s approximation, n! n1\u223c nne\u2212n, gnIn = (\u2212g)n n! \u221e \u2212\u221e dx\u221a 2\u03c0 e\u2212 1 2 x2 x4n = (\u2212g)n (4n \u2212 1)!! n! n1\u223c \u2212 gn e n. This estimate should alarm us: strictly speaking, it states that a series expansion in the \u201csmall parameter\u201d g does not exist. No matter how small g, at roughly the (1/g)th order in the perturbative expansion the series begins to diverge. In fact, it is easy to predict this breakdown on qualitative grounds: for g > 0 (g < 0), the integral (5.1) is convergent (divergent). This implies that the series expansion of the function I(g) around g = 0 must have zero radius of convergence. However, there is also a more \u201cphysical\u201d way of understanding the phenomenon. Consider a one-dimensional version of Eq. (3.16), where the \u2018Gaussian average\u2019 is given by Eq. (3.15): \u221e \u2212\u221e dx\u221a 2\u03c0 e\u2212 1 2 x2 x4n = 1 = (4n \u2212 1)!!. all possible pairings of 4n objects The factor (4n \u2212 1)!! measures the combinatorial freedom to pair up 4n objects. This suggests an interpretation of the breakdown of the perturbative expansion as the result of a competition between the smallness of the expansion parameter g and the combinatorial proliferation of equivalent contributions, or \u201cpairings,\u201d to the Gaussian integral. Physically, the combinatorial factor can be interpreted as the number of di\ufb00erent \u201cpartial amplitudes\u201d contributing to the net result at any given order of perturbation theory. Eventually, the 5.1 General structures and low-order expansions 195 exponential growth of this \ufb01gure overpowers the smallness of the expansion parameter, which is when perturbation theory breaks down. (Oddly the existence of this rather general mechanism is usually not mentioned in textbook treatments of quantum perturbation theory!) Does the ill-convergence of the series imply that perturbative approaches to problems of the structure of Eq. (5.1) are doomed to fail? Fortunately, this is not the case. While nmax n=0 gnIn can yield n=0 gnIn is divergent, a partial resummation the in\ufb01nite series excellent approximations to the exact result I(g). To see this, let us use the fact that |e\u2212gx4 \u2212 to estimate the error \u221e nmax n=0 (\u2212gx4)n n! | \u2264 (gx4)nmax+1 (nmax+1)! gnIn nmax n=0 I(g) \u2212 \u2264 gnmax+1|Inmax+1| nmax1\u223c gnmax e nmax. Variation with respect to nmax shows that the error reaches its minimum when nmax \u223c g\u22121 where it scales like e\u22121/g. (Notice the exponential dependence of the error on the coupling g \u2013 e.g. for a small coupling g \u2248 0.01, the 100th order of the perturbation theory would lead to an approximation of astronomic absolute precision e\u2212100.) By contrast, for g \u2248 0.3, perturbation theory becomes poor after the third order! Summarizing, the moral to be taken from the analysis of the integral (5.1) (and its generalizations to theories of a more complex structure) is that perturbative expansions should not be confused with rigorous Taylor expansions. Rather they represent asymptotic expansions, in the sense that, for weaker and weaker coupling, a partial resummation of the perturbation series leads to an ever more precise approximation to the exact result. For weak enough coupling the distinction between Taylor expansion and asymptotic expansion becomes academic (at least for physicists). However, for intermediate or strong coupling theories, the asymptotic character of perturbation theory must be kept in mind. \u03c64-theory While the ordinary integral discussed in the previous section conveyed something of the general status of perturbation theory, we need to proceed to the level of functional integrals to learn more about the practical implementation of perturbative methods. The simplest interacting \ufb01eld theory displaying all relevant structures is de\ufb01ned through the \ufb01eld integral Z \u2261 D\u03c6 e\u2212S[\u03c6], S[\u03c6] \u2261 ddx 1 2 (\u2202\u03c6)2 + r 2 \u03c62 +", " g\u03c64, (5.2) where \u03c6 is a scalar bosonic \ufb01eld. Owing to the structure of the interaction, this model is often referred to as the \u03c64-theory. The\u03c6 4-model not only provides a prototypical environment in which features of interacting \ufb01eld theories can be explored, but also appears in numerous applications. For example, close to its critical point, the d-dimensional Ising model is described by the \u03c64-action (see Info below). More generally, it can be shown that the long-range behavior of classical statistical systems with a single order parameter (e.g. 196 Perturbation theory the density of a \ufb02uid, uniaxial magnetization, etc.) is described by the \u03c64-action.1 Within the context of statistical mechanics, S[\u03c6] is known as the Ginzburg\u2013Landau free energy functional (and less frequently also as the Landau\u2013Wilson model). INFO The d-dimensional Ising model describes the classical magnetism of a lattice of magnetic moments Si \u2208 {1, \u22121} that can take only two values \u00b11. It is de\ufb01ned through the Hamiltonian HIsing = ij SiCijSj \u2212 H Si, i (5.3) where Cij = C(|i \u2212 j|) is a (translationally invariant) correlation matrix describing the mutual interaction of the spins, H is an external magnetic \ufb01eld, and the sums run over the sites of a ddimensional lattice (assumed hypercubic for simplicity). The Ising model represents the simplest Hamiltonian describing classical magnetism. In low dimensions d = 1, 2, it can be solved exactly, i.e. the partition function and all observables depending on it can be computed rigorously (see our discussion in Chapter 8). However, for higher dimensions, no closed solutions exist and one has to resort to approximation strategies to analyze the properties of the partition function. Below we will show that the long-range physics of the system is described by \u03c64-theory. Notice that (save for the exceptional case d = 1 discussed in Section 8.1) the system is expected to display a magnetic phase transition. As a corollary this implies that the \u03c64-model must exhibit much more interesting behavior than its innocent appearance suggests! Consider the classical partition function Z = e ij SiKij Sj + i hiSi, (5.4) {Si} where K \u2261 \u2212\u03b2C, and hi \u2261 \u03b2Hi, and we have generalized Eq. (??) to the case of a spatially varying magnetic \ufb01eld, Hi. The feature that prevents us from rigorously computing the con\ufb01gurational sum is, of course, the interaction between the spins. However, at a price, the interaction \u22121)ij \u03c8j, whereD\u03c8 \u2261 can be removed: let us consider the \u201cfat unity,\u201d 1 = N det(4\u03c0K) is a factor normalj KijSj, brings the D\u03c8 e\u2212 1 4 i d\u03c8i, K \u22121 is the inverse of the correlation matrix, and N = 1/ izing the integral to unity. A shift of the integration variables, \u03c8i \u2192 \u03c8i \u2212 2 integral into the form ij \u03c8i(K 1 =N D\u03c8 e\u2212 1 4 ij \u03c8i(K \u22121)ij \u03c8j + i Si\u03c8i\u2212 ij SiKij Sj. Incorporating the fat unity under the spin sum in the partition function, one obtains Z = N D\u03c8 e\u2212 1 4 ij \u03c8i(K \u22121)ij \u03c8j + i Si(\u03c8i+hi). (5.5) {Si} Thus, we have removed the interaction between the spin variables at the expense of introducing a new continuous \ufb01eld {\u03c8i}. Why should one do this? A multi-dimensional integral D\u03c8 is usually easier to work with than a multi-dimensional sum {Si} over discrete objects. Moreover, the new representation may provide a more convenient platform for approximation strategies. The 1 Heuristically, this is explained by the fact that S[\u03c6] is the simplest interacting (i.e. non-Gaussian) model action invariant under inversion \u03c6 \u2190\u2192 \u2212\u03c6. (The action of a uniaxial magnet should depend on the value of the local magnetization, but not on its sign.) A purely Gaussian theory might describe wave-like \ufb02uctuations of the magnetization, but not the \u201ccritical\u201d phenomenon of a magnetic transition. One thus needs, at least, a \u03c64-inter", "action term. Later on we will see that more complex monomials of \u03c6, such as\u03c6 6 or (\u2202\u03c6)4, are inessential in the long-range limit. 5.1 General structures and low-order expansions 197 transformation leading from Eq. (5.4) to (5.5) is our \ufb01rst example of a Hubbard\u2013Stratonovich transformation. The interaction of one \ufb01eld is decoupled at the expense of the introduction of another. Notice that, in spite of the somewhat high-minded designation, the transformation is tantamount to a simple shift of a Gaussian integration variable, a feature shared by all Hubbard\u2013 Stratonovich transformations! The summation {Si} = Si can now be trivially performed: i Z = N D\u03c8 e\u2212 1 4 = N D\u03c8 e\u2212 1 4 ij \u03c8i(K \u22121)ij \u03c8j - i (2 cosh(\u03c8i + hi)) ij (\u03c8i\u2212hi)(K \u22121)ij (\u03c8j \u2212hj )+ i ln(cosh \u03c8i), where we have absorbed the inessential factor N. Finally, changing integration variables from \u03c8i to \u03c6i \u2261 1 2 intermediate result i 2 into a rede\ufb01nition of the normalization j[K \u22121]ij\u03c8j, one arrives at the Z = N D\u03c6 e\u2212 ij \u03c6iKij \u03c6j + i \u03c6ihi+ i ln cosh(2 j Kij \u03c6j ). This representation of the problem still does not look very inviting. To bring it into a form amenable to further analytical evaluation, we need to make the simplifying assumption that we are working at low temperatures such that the exponential weight Kij = \u03b2C(|i \u2212 j|) inhibits strong \ufb02uctuations of the \ufb01eld \u03c6. More precisely, we assume that |\u03c6i| 1 and that the spatial pro\ufb01le of the \ufb01eld is smooth. To make use of these conditions, we switch to a Fourier reprek e\u2212ik\u00b7(ri\u2212rj )K(k), and expand ln cosh(x) = sentation, \u03c6i = 1\u221a N 2 k2K (0)\u03c6(k) +O (k4), we 2 x2 \u2212 1 1 conclude that the low-temperature expansion of the action has the general structure 12 x4 + \u00b7 \u00b7 \u00b7. Noting that (K\u03c6)(k) =K (k)\u03c6(k) = K(0)\u03c6(k) + 1 k e\u2212ik\u00b7ri \u03c6(k), Kij = 1 N S[\u03c6] = [\u03c6k(c1 + c2k \u00b7 k)\u03c6\u2212k + c3\u03c6kh\u2212k] k c4 N + k1,...,k4 \u03c6k1 \u03c6k2 \u03c6k3 \u03c6k4 \u03b4k1+k2+k3+k4,0 + O(k4, h2, \u03c66). EXERCISE Show that the coe\ufb03cients ci are given by c1 = K(0)(1 \u2212 2K(0)), c2 = 1 4K(0)), c3 = 1, c4 = 4K(0)4. 3 2 K (0)(1 \u2212 Switching back to a real space representation and taking a continuum limit, S[\u03c6] assumes the form of a prototypical \u03c64-action S[\u03c6] = ddx c2(\u2202\u03c6)2 + c1\u03c62 + c3\u03c6h + c4\u03c64.! A rescaling of variables \u03c6 \u2192 1\u221a cients r = c1/c2 and g = c4/(2c2).2 2c2 \u03c6 \ufb01nally brings the action into the form Eq. (5.2) with coe\ufb03- We have thus succeeded in describing the low-temperature phase of the Ising model in terms of a \u03c64-model. While the structure of the action could have been guessed on symmetry grounds, the \u201cmicroscopic\u201d derivation has the advantage that it yields explicit expressions for the coupling constants. There is actually one interesting aspect of the dependence of these constants on the 2 The only di\ufb00erence is that the magnetic \u03c64-action contains a term linear in \u03c6 and h. The reason is that, in the presence of a \ufb01nite magnetic \ufb01eld, the action is no longer invariant under inversion \u03c6 \u2192 \u2212\u03c6. 198 Perturb", "ation theory parameters of the microscopic model. Consider the constant c1 controlling the k-independent contribution to the Gaussian action: c1 \u221d K(0)(1 \u2212 2K(0)) \u221d (1 \u2212 2\u03b2C(0)). Since C(0) must be positive to ensure the overall stability of the model (exercise: why?) the constant c1 will change sign at a certain \u201ccritical temperature\u201d \u03b2\u2217. For temperatures lower than \u03b2\u2217, the Gaussian action is unstable (i.e. \ufb02uctuations with low wavevector become unbound) and the anharmonic term \u03c64 alone controls the stability of the model. Clearly, the behavior of the system will change drastically at this point. Indeed, the critical temperature c1(\u03b2\u2217) = 0 marks the position of the magnetic phase transition, a point to be discussed in more detail below. Let us begin our primer of perturbation theory by introducing some nomenclature.3 For simplicity, let us \ufb01rst de\ufb01ne the notation \u00b7 \u00b7 \u00b7 \u2261 D\u03c6 e\u2212S[\u03c6]( \u00b7 \u00b7 \u00b7) D\u03c6 e\u2212S[\u03c6], (5.6) for the functional integral, weighted by the action S, of any expression ( \u00b7 \u00b7 \u00b7). Due to the structural similarity to thermal averages of statistical mechanics, \u00b7 \u00b7 \u00b7 is sometimes called a functional average or functional expectation value. Similarly, let us de\ufb01ne \u00b7 \u00b7 \u00b7 0 \u2261 D\u03c6 e\u2212S0[\u03c6]( \u00b7 \u00b7 \u00b7) D\u03c6 e\u2212S0[\u03c6], (5.7) for the functional average over the Gaussian action S0 \u2261 S|g=0. The average over a product of \ufb01eld variables, Cn(x1, x2,..., xn) \u2261 \u03c6(x1)\u03c6(x2) \u00b7 \u00b7 \u00b7\u03c6 (xn), (5.8) four is known as an n-point correlation function or, for brevity, just the n-point function.4 The one-point function C1(x) = \u03c6(x) simply measures the expectation value of the \ufb01eld amplitude. For the particular case of the \u03c64problem above the phase transition and, more generally, the majority of \ufb01eld theories with an action even in the \ufb01eld amplitudes, C1 = 0 and the \ufb01rst non-vanishing correlation function is the two-point function George Green 1793\u20131841 His only schooling consisted of terms in 1801/1802. He owned and worked a Nottingham windmill. Green made major contributions to potential theory although where he learnt his mathematical skills is a mystery. The inventor of Green functions, he used the method of sources and sinks in potential \ufb02ows. He published only ten mathematical works, the \ufb01rst and most important being published at his own expense in 1828, \u201cAn essay on the application of mathematical analysis to the theories of electricity and magnetism.\u201d He left his mill and became an undergraduate at Cambridge in 1833 at the age of 40, then a Fellow of Gonville and Caius College in 1839. (Why does C2 depend only on the di\ufb00erence of its arguments?) The two-point function is sometimes also called the propagator of the theory, the Green function or, especially in the more formal literature, the resolvent operator. G(x1 \u2212 x2) \u2261 C2(x1, x2). (5.9) 3 Needless to say, the jargon introduced below is not restricted to the \u03c64 example! 4 Notice that, depending on the context and/or scienti\ufb01c community, the phrase \u201cn-point function\u201d sometimes refers to C2n instead of Cn. 5.1 General structures and low-order expansions 199 The existence of di\ufb00erent names suggests that we have met with an important object. Indeed, we will shortly see that the Green function not only represents a central building block of the theory but also carries profound physical signi\ufb01cance. INFO To develop some understanding of the physical meaning of the correlation function, let us recall that the average of a linear \ufb01eld amplitude, \u03c6(0), vanishes. (See the \ufb01gure, where a diagram of a few \u201ctypical\u201d \ufb01eld con\ufb01gurations is sketched as functions of a coordinate.) \u03c6 However, the average of the squared amplitude (\u03c6(0))2 is certainly non-vanishing simply because we are integrating over a positive object. Now, what happens if we split our single observation point into two, \u03c62(0)0 \u2192", " \u03c6(0)\u03c6(x)0 = G(x)? For asymptotically large values of x, it is likely that the two amplitudes \ufb02uctuate indepen- 0 |x|\u2192\u221e\u2212\u2192 0. However, this decoupling will not happen locally. The dently of each other, i.e. G(x) reason is that the \ufb01eld amplitudes are correlated over a certain range in space. For example, if \u03c6(0) > 0, the \ufb01eld amplitude will, on average, stay positive in an entire neighborhood of 0 since rapid \ufb02uctuations of the \ufb01eld are energetically costly (i.e. due to the gradient term in the action!). The spatial correlation pro\ufb01le of the \ufb01eld is described by the function G(x). x How does the correlation behavior of the \ufb01eld relate to the basic parameters of the action? A quick answer can be given by dimensional analysis. The action of the theory must be dimensionless (because it appears as the argument of an exponential). Denoting the dimension of = Ld, [\u2202] =L \u22121, inspection of Eq. (5.2) any quantity X by [X], and using the fact that obtains the set of relations ddx Ld\u22122[\u03c6]2 = 1, Ld[r][\u03c6]2 = 1, Ld[g][\u03c6]4 = 1, from which it follows that [\u03c6] =L \u2212(d\u22122)/2, [r] =L \u22122, [g] =L d\u22124. In general, both system parameters, g and r, carry a non-zero length-dimension. However, temporarily concentrating on the non-interacting sector, g = 0, the only parameter of the theory, r, has dimensionality L\u22122. Arguing in reverse, we conclude that any intrinsic length scale produced by the theory (e.g. the range over which the \ufb01elds are correlated), must scale as \u223c r\u22121/2. A more quantitative description can be obtained by considering the free propagator of the theory, G0(x) \u2261 \u03c6(0)\u03c6(x)0. (5.10) Since the momentum representation of the Gaussian action is simply given by S0[\u03c6] = it is convenient to \ufb01rst compute G0 in reciprocal space: G0,p \u2261 1 2 p \u03c6p\u03c6p 0. Using the Gaussian contraction rule Eq. (3.14), the free p \u03c6p(p2 + r)\u03c6\u2212p, ddx eip\u00b7xG0(x) = functional average takes the form \u03c6p\u03c6p 0 = \u03b4p+p,0(p2 + r)\u22121, i.e.5 G0,p = \u03c6p\u03c6\u2212p0 = 1 p2 + r. (5.11) 5 The result G0,p = (p2 + r)\u22121 clari\ufb01es why G is referred to as a \u201cGreen function.\u201d Indeed, G0,p is (the Fourier representation of the) Green function of the di\ufb00erential equation (\u2212\u22022 r + r)G(r, r) =\u03b4 (r \u2212 r). 200 Perturbation theory To obtain G(x), we need to compute the inverse transform G0(x) = 1 Ld p e\u2212ip\u00b7xG0,p \u2248 ddp (2\u03c0)d e\u2212ip\u00b7x p2 + r, (5.12) where we have assumed that the system is large, i.e. the sum over momenta can be exchanged for an integral. For simplicity, let us compute the integral for a one-dimensional system. (For the two- and three-dimensional cases see exercise below.) Setting p2 + r = (p + ir1/2)(p \u2212 ir1/2), we note that the (complex extension of the) p integral has simple poles at \u00b1ir1/2. Forx smaller (larger) than zero, the integrand is analytic in the upper (lower) complex p-plane and closure of the integration contour to a semicircle of in\ufb01nite radius gives G0(x) = dp 2\u03c0 e\u2212ipx (p + ir1/2)(p \u2212 ir1/2) = e\u2212r1/2|x| 2r1/2. (5.13) This result conveys an interesting observation: typically, correlations decay exponentially, at a rate set by thecorrelation length \u03be \u2261 r\u22121/2", ". However, as r approaches 0, the system becomes long-range correlated. The origin of this phenomenon can be understood by inspecting the structure of the Gaussian contribution to the action (5.2). For r \u2192 0 (and still neglecting the \u03c64 contribution) nothing prevents the constant \ufb01eld mode \u03c6(x) = \u03c60 = const. from becoming in\ufb01nitely large, i.e. the \ufb02uctuating contribution to the \ufb01eld becomes relatively less important than the constant o\ufb00set. The increasing \u201csti\ufb00ness\u201d of the \ufb01eld in turn manifests itself in a growth of spatial correlations (cf. the \ufb01gure on page 201). Notice that this dovetails with our previous statement that r = 0 marks the position of a phase transition. Indeed, the build-up of in\ufb01nitely long-range spatial correlations is known to be a hallmark of second-order phase transitions (see Chapter 8). EXERCISE Referring to Eq. (5.12), show that, in dimensions d = 2 and d = 3, G0(x) d=2= G0(x) d=3= d2k (2\u03c0)2 e\u2212ik\u00b7x k2 + r d3k (2\u03c0)3 e\u2212ik\u00b7x k2 + r = = \u221a K0( 1 2\u03c0 r|x|) = \u221a r|x| e\u2212 4\u03c0|x|. * 2\u03c0 ln \u221a \u2212 1 1 2 (2\u03c0 \u221a r|x|, 2 r|x|)\u2212 1 2 e\u2212 |x| 1/ |x| 1/ \u221a \u221a r, r, \u221a r|x|, Notice that, in both cases, the Green function diverges in the limit |x| \u2192 0 and decays exponentially (at a rate \u223c r\u22121/2) for |x| r\u22121/2. Perturbation theory at low orders Having discussed the general structure of the theory and of its free propagator, let us turn our attention to the role of the interaction contribution to the action, Sint[\u03c6] \u2261 g ddx \u03c64. Within the jargon of \ufb01eld theory, an integrated monomial of a \ufb01eld variable (like \u03c64) is commonly called an (interaction) operator or a vertex (operator). Keeping in mind the words of caution given in Section 5.1, we wish to explore perturbatively how the interaction 5.1 General structures and low-order expansions 201 vertex a\ufb00ects the functional expectation value of any given \ufb01eld observable, i.e. we wish to analyze expansions of the type... (\u2212g)n n! X[\u03c6]( X[\u03c6] \u2248 n=0... ddx \u03c64)n0 (\u2212g)n n! ( ddx \u03c64)n0 n=0 \u2248 nmax n=0 X (n), (5.14) where X may be any observable and X (n) denotes the contribution of nth order to the expansion in g. The limits on the summation in the numerator and denominator are symbolic because, as explained above, we will need to terminate the total perturbative expansion at a certain \ufb01nite order nmax. EXERCISE To navigate the following section, it is helpful to recapitulate Section 3.2 on continuum Gaussian integration. To keep the discussion concrete, let us focus on the perturbative expansion of the propagator in the coupling constant, g. (A physical application relating to this expansion will be discussed below.) The zeroth-order contribution G(0) = G0 has been discussed before, so the \ufb01rst non-trivial term we have to explore is G(1): @ @ A A A @ G(1)(x, x) = \u2212g \u03c6(x) ddy \u03c6(y)4\u03c6(x) \u2212 \u03c6(x)\u03c6(x) ddy \u03c6(y)4. (5.15) 0 0 0 Since the functional average is now over a Gaussian action, this expression can be evaluated by Wick\u2019s theorem, Eq. (3.21). For example, the functional average of the \ufb01rst of the two terms leads to (integral signs and constants stripped o\ufb00 for clarity) 0 = 3\u03c6(x)\u03c6(x)0 [\u03c6(y)\u03c6(y)0]2 \u03c6(x)\u03c6(y)4\u03c6(x) 9 : +12\u03c6(x)\u03c6(y)0\u03c6", "(y)\u03c6(y)0\u03c6(y)\u03c6(x)0 = 3G0(x \u2212 x)G0(0)2 + 12G0(x \u2212 y)G0(0)G0(y \u2212 x), (5.16) where we have used the fact that the operator inverse of the Gaussian action is, by de\ufb01nition, the free Green function (cf. Eq. (5.10)). Further, notice that the total number of terms appearing on the right-hand side is equal to 15 = (6\u22121)!! which is just the number of distinct pairings of six objects (cf. Eq. (3.21) and with our discussion of Section 5.1). Similarly, the second contribution to G(1) leads to \u03c6(x)\u03c6(x)0\u03c6(y)40 = 3\u03c6(x)\u03c6(x)0[\u03c6(y)20]2 = 3G0(x \u2212 x)G0(0)2. Before analyzing these structures in more detail, let us make some general observations. The \ufb01rst-order expansion of G contains a number of factors of G0(0), the free Green function evaluated at coinciding points. This bears disturbing consequences. To see this, consider G0(0) evaluated in momentum space: G0(0) = ddp (2\u03c0)d 1 p2 + r. (5.17) For dimensions d >1, the integral is divergent at large momenta or short wavelengths; we have met with an ultraviolet (UV) divergence. Physically, the divergence implies that, 202 Perturbation theory already at \ufb01rst order, our expansion runs into a di\ufb03culty that is obviously related to the short-distance structure of the system. How can this problem be overcome? One way out is to remember that \ufb01eld theories like the \u03c64-model represent e\ufb00ective low-temperature, or long-wavelength, approximations to more microscopic models. The range of applicability of the action must be limited to wavelengths in excess of some microscopic lattice cuto\ufb00 a (e.g. the lattice spacing), or momenta k < a\u22121. It seems that, once that cuto\ufb00 has been built in, the convergence problem is solved. However, there is something unsatisfactory in this argument. All our perturbative corrections, and therefore the \ufb01nal result of the analysis, exhibit sensitivity to the microscopic cuto\ufb00 parameter. But this is not what we expect of a sensible low-energy theory (cf. the discussion of Chapter 1)! The UV problem signals that something more interesting is going on than a naive cuto\ufb00 regularization has the capacity to describe. We discuss this point extensively in Chapter 8. However, even if we temporarily close our eyes to the UV-phenomenon, there is another problem. For dimensions d \u2264 2, and in the limit r \u2192 0, G0(0) also diverges at small momenta, aninfrared (IR) divergence. Being related to structures at large wavelengths, this type of singularity should attract our attention even more than the UV-divergence mentioned above. Indeed, it is intimately related to the accumulation of long-range correlations in the limit r \u2192 0 (cf. the structure of the integral (5.12)). We come back to the discussion of the IR singularities, and their connection to the UV phenomenon, in Chapter 8. The considerations above show that the perturbative analysis of functional integrals will be accompanied by all sorts of divergences. Moreover, there is another, less fundamental, but also important, point: referring to Eq. (5.16), we have to concede that the expression does not look particularly inviting. To emphasize the point, let us consider the core contribution to the expansion at second order in g. EXERCISE Show that the 10th-order contraction leads to the 945=(10-1)!! terms. / 0 = 9G0(x \u2212 x)G0(0)4 + 72G0(x \u2212 x)G0(y \u2212 y)2G0(0)2 \u03c6(x)\u03c6(y)4\u03c6(y)4\u03c6(x) +24G0(x \u2212 x)G0(y \u2212 y)4 + 36G0(x \u2212 y)G0(x \u2212 y)G0(0)3 +144(G0(x \u2212 y)G0(x \u2212 y)G0(y \u2212 y)2G0(0) +G 0(x \u2212 y)G0(x \u2212 y)G0(0)2G0(", "y \u2212 y)) +96G0(x \u2212 y)G0(x \u2212 y)G0(y \u2212 y)3 + (y \u2194 y) (5.18). Note: Our further discussion will not rely on this result. It only serves an illustrative purpose. Clearly Eq. (5.18) is highly opaque. There are eight groups of di\ufb00erent terms, but it is not obvious how to attribute any meaning to these contributions. Further, should we consider the full second-order Green function G(2), i.e. take account of the expansion of both numerator and denominator in Eq. (5.14), we would \ufb01nd that some contributions cancel (see Problem 5.5). Clearly, the situation will not improve at third and higher orders in g. 5.1 General structures and low-order expansions 203 x y x\u2032 3 x x y +12 x y x\u2032 Figure 5.1 Graphical representation of a \ufb01rst-order-in-g contraction contributing to the expansion of the Green function. \u03c6 (y) \u03c6 (x) 0 y x \u2329 \u232a To e\ufb03ciently apply perturbative concepts beyond lowest orders, a more e\ufb03cient formulation of the expansion is needed. The key to the construction of a better language lies in the observation that our previous notation is full of redundancy, i.e. in the full contraction of a perturbative contribution, we represent our \ufb01elds by \u03c6(x). A more compact way of keeping track of the presence of that \ufb01eld is shown in the upper portion of the \ufb01gure to the right. Draw a point (with an optional \u201cx\u201d labeling its position) and attach a little leg to it. The leg indicates that the \ufb01elds are sociable objects, i.e. they need to \ufb01nd a partner with which to pair. After the contraction, a pair \u03c6(x)\u03c6(y) \u2192 G0(x \u2212 y) becomes a free Green function. Graphically, this information can be represented by a pairwise connection of the legs of the \ufb01eld symbols to lines, where each line is identi\ufb01ed with a Green function connecting the two terminating points. The full contraction of a free correlation function \u03c6(x1)\u03c6(x2) \u00b7 \u00b7 \u00b7 \u03c6(x2n)0 is represented by the set of all distinct diagrams formed by pairwise connection of the \ufb01eld vertices. G0 (x, y) x y Figure 5.1 shows the graphical representation of the contraction of Eq. (5.16). (The cross appearing on the left-hand side represents four \ufb01eld operators sitting at the same point y.) According to our rule formulated above, each of the two diagrams on the right-hand side represents the product of three Green functions, taken between the speci\ufb01ed coordinates. Further, each contribution is weighted by a combinatorial factor, i.e. the number of identical diagrams of that structure. Consider, for example, the second contribution on the righthand side. It is formed by connecting the \u201cexternal\u201d \ufb01eld vertex at x to any of the legs of the internal vertex at y: four possibilities. Next, the vertex at x is connected with one of the remaining three unsaturated vertices at y: three possibilities. The last contraction y \u2194 y is \ufb01xed, i.e. we obtain altogether 3 \u00d7 4 = 12 equivalent diagrams \u2013 \u201cequivalent\u201d in that each of these represents the same con\ufb01guration of Green functions. EXERCISE Verify that the graphical representation of the second-order contraction Eq. (5.18) is as shown in Fig. 5.2.6 Associate the diagrams with individual contributions appearing in Eq. (5.18) and try to reproduce the combinatorial factors. (For more details, see Problem 5.5.) The graphical representation of the contractions shown in Fig. 5.1 and 5.2 provides us with su\ufb03cient background to list some general aspects of the diagrammatic approach: 6 In the \ufb01gure, the coordinates carried by the \ufb01eld vertices have been dropped for notational simplicity. To restore the full information carried by any of these \u201cnaked\u201d graphs one attaches coordinates x and x to the external \ufb01eld vertices and integration coordinates yi to each of the i nodes that do not connect to an external \ufb01eld vertex. Since no information is lost, diagrams are often represented without explicit reference to coordinates. 204 Perturbation theory 9 + 72 + 24 + 72 + 288 +", " 192 + 288 Figure 5.2 Graphical representation of the second-order correction to the Green function. In the main text, the seven types of diagram contributing to the contraction will be referred to (in the order they appear above) as diagrams 1 to 7. Firstly, diagrammatic methods help to e\ufb03ciently represent the perturbative expansion. However, we are still left with the problem (see the discussion above) of computing the analytical expressions corresponding to individual diagrams. To go back from an nth-order graph to its analytical representation one (i) attaches coordinates to all \ufb01eld vertices, (ii) identi\ufb01es lines between points with Green functions, (iii) multiplies the graph by the overall constant gn/n!, and (iv) integrates over all of the internal coordinates. When one encounters expressions like G(n) = \u201csum of graphs,\u201d the operations (i)\u2013(iv) are implicit. As should be clear from the formulation of our basic rules, there is no \ufb01xed rule as to how to represent a diagram. As long as no lines are cut, any kind of reshaping, twisting, rotating, etc. of the diagram leaves its content invariant. (At large orders of perturbation theory, it often takes a second look to identify two diagrams as equivalent.) From the assembly of diagrams contributing to any given order, a number of internal structures common to the series expansion become apparent. For example, looking at the diagrams shown in Fig. 5.2, we notice that some are connected, and some are not. Among the set of connected diagrams (nos. 5, 6, 7) there are some whose \u201ccore portion,\u201d i.e. the content of the diagram after the legs connecting to the external vertices have been removed, can be cut into two pieces just by cutting one more line (no. 7). Diagrams of this type are called one-particle reducible while the others are termed one-particle irreducible. More generally, a diagram whose core region can be cut by severing n lines is called n-particle reducible. (For example, no. 6 is three-particle reducible, no. 7 oneparticle reducible, etc.) One can also attach a loop order to a diagram, i.e. the number of inequivalent loops formed by segments of Green functions (for Fig. 5.2: 4, 3, 3, 3, 2, 2, 2, in that order). One (correctly) expects that these structures, which are di\ufb03cult to discern from the equivalent analytical representation, will re\ufb02ect themselves in the mathematics of the perturbative expansion. We return to the discussion of this point below. Then there is the issue of combinatorics. The diagrammatic representation simpli\ufb01es the determination of the combinatorial factors appearing in the expansion. However, the problem of getting the combinatorics right remains non-trivial. (If you are not impressed with the factors entering the second-order expansion, consider the (14 \u2212 1)!! = 135135 5.1 General structures and low-order expansions 205 terms contributing at third order!) In some sub-disciplines of theoretical physics, the art of identifying the full set of combinatorial coe\ufb03cients at large orders of perturbation theory has been developed to a high degree of sophistication. Indeed, one can set up re\ufb01ned sets of diagrammatic construction rules which to a considerable extent automate the combinatorics. Pedagogical discussions of these rules can be found, for example, in the textbooks by Negele and Orland, and Ryder.7 However, as we will see shortly, the need to explicitly carry out a large-order expansion, with account of all diagrammatic sub-processes, rarely arises in modern condensed matter physics; mostly one is interested in subclasses of diagrams, for which the combinatorics is less problematic. For this reason, the present text does not contain a state-of-the-art exposition of all diagrammatic tools and interested readers are referred to the literature. Finally, and perhaps most importantly, the diagrammatic representation of a given contribution to the perturbative expansion often suggests a physical interpretation of the corresponding physical process. (After all, any term contributing to the expansion of a physical observable must correspond to some \u201creal\u201d physical process.) Unfortunately, the \u03c64-theory is not well suited to illustrate this aspect, i.e., being void of any dynamical content, it is a little bit too simple. However, the possibility of \u201creading\u201d individual diagrams will become evident in the next section when we discuss an application to the interacting electron gas. Above we have introduced the diagrammatic approach on the example of \ufb01eld expectation values \u03c6(x)(", "\u03c6(y)4)n\u03c6(x)0. However, to obtain the Green function to any given order in perturbation theory, we need to add to these expressions the contributions emanating from the expansion of the denominator of the functional average (cf. Eq. (5.14) and (5.15)). While, at \ufb01rst sight, the need to keep track of even more terms seems to complicate matters, we will see that, in fact, quite the opposite is true! The combined expansion of numerator and denominator leads to a miraculous \u201ccancellation mechanism\u201d that greatly simpli\ufb01es the analysis. \u2013 0 0 0 + 12 \u2013 3 = 3 = 12 Figure 5.3 Graphical representation of the \ufb01rst-order correction to the Green function: vacuum graphs cancel out. Let us exemplify the mechanism of cancellation on G(1). The three diagrams corresponding to the contractions of Eq. (5.15) are shown in Fig. 5.3, where integral signs and coordinates are dropped for simplicity. On the left-hand side of the equation, the brackets \u00b7 \u00b7 \u00b7 0 7 J. W. Negele and H. Orland, Quantum Many Particle Systems (Addison-Wesley, 1988); L. H. Ryder, Quantum Field Theory (Cambridge University Press, 1996). 206 Perturbation theory indicate that the second contribution comes from the expansion of the denominator. The point to be noticed is that the graph produced by the contraction of that term cancels against a contribution arising from the numerator. One further observes that the canceled graph is of a special type: it contains an interaction vertex that does not connect to any of the external vertices. Diagrams with that property are commonly termed vacuum graphs.8 EXERCISE Construct the diagrammatic representation of G(2) and verify that the expansion of the denominator eliminates all vacuum graphs of the numerator. In particular, show that G(2) is given by the sum of connected diagrams shown in Fig. 5.4. (For more details, see Problem 5.5.) G(2) = 192 + 288 + 288 Figure 5.4 Graphical representation of the second-order contribution to the Green function. Indeed, the cancellation of vacuum graphs pertains to higher-order correlation functions and to all orders of the expansion: The contribution to a correlation function C (2n)(x1,..., x2n) at lth order of perturbation theory is given by the sum of all graphs, excluding vacuum graphs. 1 C(4)(1,2,3,4) = 24 For example, the \ufb01rst-order expansion of the four-point function C (4)(x1,... x4) is shown in the \ufb01gure, where coordinates xi \u2194 i are abbreviated by indices and \u201c+ perm.\u201d stands for the six permutations obtained by interchanging arguments. In the literature, the statement of vacuum graph cancellation is sometimes referred to as the linked cluster theorem. Notice that the linked cluster feature takes care of two problems: \ufb01rstly we are relieved of the burden of a double expansion of numerator and denominator, and secondly only non-vacuum contributions to the expansion of the former need to be kept. + perm. 12 + 4 1 3 4 2 2 3 INFO The proof of the linked cluster theorem is straightforward. Consider a contribution of nth order to the expansion of the numerator of Eq. (5.14): (\u2212g)n. The contraction of this expression will lead to a sum of vacuum graphs of pth-order and non-vacuum graphs of (n \u2212 p)th-order, where p runs from 0 to n. The pth-order contribution is given by x\u03c64)n dd X[\u03c6]( 0 1 3 n! /. 0 2 n.v. n\u2212p p 1 n! n p X[\u03c6] \u03c64 \u03c64, 0 0 where the superscript \u00b7 \u00b7 \u00b7 n.v. indicates that the contraction excludes vacuum graphs and the combinatorial coe\ufb03cient counts the number of possibilities to choose p vertices \u03c64 of a total of n 8 The term \u201cvacuum graph\u201d has its origin in the diagrammatic methods invented in the 1950s in the context of particle theory. Instead of thermal averages \u00b7 \u00b7 \u00b7 0, one considered matrix elements \u03a9| \u00b7 \u00b7 \u00b7 |\u03a9 taken on the ground state or \u201cvacuum\u201d of the \ufb01eld theory. This caused matrix elements \u03a9|(Sint[\u03c6])n|\u03a9 not containing an external \ufb01eld vertex to be dubbed \u201cvacuum graphs.\u201d 5.1 General structures", " and low-order expansions 207 p 1 \u2013p 1 \u2013p 2 \u2013p 3 p p 2 p 3 \u2013p 3 p p 1 +12 p p p 1 p 2 Figure 5.5 Momentum space representation of a \ufb01rst-order contribution to the Green function. Internal momenta pi are integrated over. vertices to form a vacuum graph. Summing over p, we \ufb01nd that the expansion of the numerator, split into vacuum and non-vacuum contributions, reads \u221e n n=0 p=0 (\u2212g)n (n \u2212 p)! p! 0 n\u2212p 1 n.v. 2 p 3 X[\u03c6], \u03c64, \u03c64. 0 0 By a straightforward rearrangement of the summations, this can be rewritten as \u221e n=0 (\u2212g)n n! 2 X[\u03c6], \u03c64 n 3 n.v. \u221e 0 p=0 (\u2212g)p p! 2 p 3, \u03c64. 0 The p-summation recovers exactly the expansion of the denominator, so we are left with the sum over all non-vacuum contractions. Before concluding this section, let us discuss one last technical point. The translational invariance of the \u03c64-action suggests a representation of the theory in reciprocal space. Indeed, the momentum space representation of the propagator Eq. (5.11) is much simpler than the real space form, and the subsequent analytical evaluation of diagrams will be formulated in momentum space anyway (cf. the prototypical expression (5.17)). The diagrammatic formulation of the theory in momentum space is straightforward. All we need to do is to slightly adjust the graphical code. Inspection of Eq. (5.11) shows that the elementary contraction should now be formulated as indicated in the \ufb01gure. Only \ufb01elds with opposite momentum can be contracted; the line carries this momenta as a label. Notice that the momentum representation of the \ufb01eld vertex \u03c64(x) is not given by \u03c64 p. Rather, Fourier transformation of the vertex leads to the threefold convolution \u03c6 p p \u03c6 \u2013p \u2013p \u2329...\u232a 0 p G0( p) ddx \u03c64(x) \u2192 1 Ld p1,...,p4 \u03c6p1 \u03c6p2 \u03c6p3 \u03c6p4 \u03b4p1+p2+p3+p4. The graphical representation of the \ufb01rst-order correction to the Green function (i.e. the momentum space analog of Fig. 5.3) is shown in Fig. 5.5. It is useful to think about the vertices of the momentum-space diagrammatic language in the spirit of \u201cKirchho\ufb00 laws\u201d: the sum of all momenta \ufb02owing into a vertex is equal to zero. Consequently (exercise) the total sum of all momenta \u201c\ufb02owing\u201d into a diagram from external \ufb01eld vertices must also i=1 pi,0(\u00b7 \u00b7 \u00b7 ). This fact expresses the conservation of the equal zero: \u03c6p1 \u03c6p2 total momentum characteristic for theories with global momentum conservation. \u00b7 \u00b7 \u00b7 \u03c6pn \u2192 \u03b4 0 n 208 Perturbation theory EXERCISE Represent the diagrams of the second-order contraction shown in Fig. 5.2 in momentum space. Convince yourself that the \u201cKirchho\ufb00 law\u201d su\ufb03ces to \ufb01x the result. Observe that the number of summations over internal momenta is equal to the number of loops. This concludes the \ufb01rst part of our introduction to the formal elements of perturbation theory. Critical readers will object that, while we undertook some e\ufb00orts to e\ufb03ciently represent the perturbative expansion, we have not in the least addressed the question of how interactions will actually modify the results of the free theory. Indeed, we are not yet in a position to quantitatively address this problem, the reason being that we \ufb01rst need to better understand the origin and remedy of the UV/IR divergences observed above. However, temporarily ignoring the presence of this roadblock, let us try to outline what kind of information can be extracted from perturbative analyses, in principle. One important point to be noted is that, in condensed matter physics,9 low-order perturbation theory is usually not enough to obtain quantitative results. The fact that the \u201cperturbation\u201d couples to a macroscopic number of degrees of freedom10 usually necessitates summation of in\ufb01nite (sub)series of a perturbative expansion or even the application of non-perturb", "ative methods. This, however, does not mean that the tools developed above are useless: given a system subject to unfamiliar interactions, low-order perturbation theory will usually be applied as a \ufb01rst step to qualitatively explore the situation. For example, a malign divergence of the expansion in the interaction operator may signal the presence of an instability towards the formation of a di\ufb00erent phase. Or it may turn out that certain contributions to the expansion are \u201cphysically more relevant\u201d than others. Technically, such contributions usually correspond to diagrams of a regular graphical structure. If so, a summation over all \u201crelevant processes\u201d may be in reach. In either case, low-order expansions provide vital hints as to the appropriate strategy of further analysis. In the following we discuss two examples that may help to make these remarks more transparent. 5.2 Ground state energy of the interacting electron gas In Section 2.2 we began to consider the physics of highly mobile electron compounds. We argued that such a system can be described in terms of the free particle Hamiltonian (2.18) together with the interaction operator (2.19). While we have reviewed the physics of the non-interacting system, nothing has hitherto been said about the role of electron\u2013electron interactions. Yet by now we have developed enough analytical machinery to begin to address this problem. Below we will apply concepts of perturbation theory to estimate the contribution of electronic correlations to the ground state energy of a Fermi system. However, before plunging into the technicalities of this analysis, it is worthwhile discussing some qualitative aspects of the problem. 9 There are subdisciplines of physics where the situation is di\ufb00erent. For example, consider the high-precision scattering experiments of atomic and sub-atomic physics. In these areas, the power of a theory to quantitatively predict the dependence of scattering rates on the strength of the projectile/target interaction (the \u201cperturbation\u201d) is a measure of its quality. Such tests involve large-order expansions in the physical coupling parameters. 10 In contrast, low-order expansions in the external perturbation (e.g. experimentally applied electric or magnetic \ufb01elds) are usually secure; see Chapter 7. 5.2 Ground state energy of the interacting electron gas 209 Qualitative aspects A principal question that we will need to address is under what physical conditions are interactions \u201cweak\u201d (in comparison to the kinetic energy), i.e. when does a perturbative approach with the interacting electron system make sense at all? To estimate the relative magnitude of the two contributions to the energy, let us assume that each electron occupies an average volume r3 0. According to the uncertainty relation, the minimum kinetic energy per particle will be of order O(2/mr2 0). On the other hand, assuming that each particle interacts predominantly with its nearest neighbors, the Coulomb energy is of order O(e2/r0). The ratio of the two energy scales de\ufb01nes the dimensionless density parameter (see \ufb01gure) e2 r0 mr2 0 2 = r0 a0 \u2261 rs, rs where a0 = 2/e2m denotes the Bohr radius.11 Physically, rs is the radius of the spherical volume containing one electron on average; for the Coulomb interaction, the denser the electron gas, the smaller rs. We have thus identi\ufb01ed the electron density as the relevant parameter controlling the relative strength of electron\u2013electron interactions. Below, we will be concerned with the regime of high density, rs 1, or weak Coulomb interaction. In the opposite limit, rs 1, properties become increasingly dominated by electronic correlations. Ultimately, for su\ufb03ciently large rs (low density) it is believed that the electron gas undergoes a (\ufb01rst order) transition to a condensed phase known as a Wigner crystal. Although Wigner crystals have never been unambiguously observed, several experiments performed on low-density electron gases are consistent with a Wigner crystal ground state. Monte Carlo simulation suggests that Wigner crystallization may occur for densities with rs > 35 in the two-dimensional electron gas and rs > 106 in three.12 (Note that this scenario relies on being at low temperature, Eugene P. Wigner 1902\u201395 Nobel Laureate in Physics in 1963 \u201cfor his contributions to the theory of the atomic nucleus and the elementary particles, particularly through the discovery and application of fundamental symmetry principles.\u201d (Image c The Nobel Foundation.) 11 Notice that the estimate of the relative magnitude of energy scales mimics Bohr\u2019s famous qualitative discussion of the average size of the hydrogen atom. 12 With the earliest reference to electron crystallization appearing in E. Wigner, On the interaction of electrons in metals, Phys. Rev. 46 (1934), 100", "2\u201311, a discussion of the results of quantum Monte Carlo simulations in three-dimensions can be found in the following papers: D. M. Ceperley and B. J. Alder, Ground state of the electron gas by a stochastic method, Phys. Rev. Lett. 45 (1980), 566\u2013569 and N. D. Drummond, Z. Radnai, J. R. Train, M. D. Towler, and R. J. Needs, Di\ufb00usion quantum Monte Carlo study of three-dimensional Wigner crystals, Phys. Rev. B 69 (2004), 085116, and in two-dimensions in B. Tanatar and D. M. Ceperley, Ground state of the two-dimensional electron gas, Phys. Rev. B 39 (1989) 5005-5016. For a further discussion of the subtleties of the transition in two-dimensions, we refer to B. Spivak and S. A. Kivelson, Phases intermediate between a two-dimensional electron liquid and Wigner crystal, Phys. Rev. B 70 (2004), 155114. 210 Perturbation theory Table 5.1 Density parameters of a number of metals. Metal rs Metal rs Metal rs Metal rs Li K 3.2 4.9 Be Sn 1.9 2.2 Na Cu 3.9 2.7 Al Pb 2.1 2.3 Source: Data taken from Ashcroft and Mermin, Solid State Physics.a a N. W. Ashcroft and N. D. Mermin, Solid State Physics (Holt- Saunders International, 1983). and on the long-range nature of the Coulomb interaction. In particular, if the Coulomb interaction is subject to some screening mechanism, V (r) \u223c e\u2212r/\u03bb, rs \u223c (r0/a0)e\u2212r0/\u03bb, and the in\ufb02uence of Coulomb interaction at low densities becomes diminished.) For rs \u223c O(1), the potential and kinetic energies are comparable. This regime of intermediate coupling is notoriously di\ufb03cult to describe quantitatively. Yet most metals lie in a regime of intermediate coupling 2 < rs < 6 (see Table 5.1). Fortunately, there is overwhelming evidence to suggest that a weak coupling description holds even well outside the regime over which the microscopic theory can be justi\ufb01ed. The phenomenology of the intermediate coupling regime is the realm of Landau\u2019s Fermi liquid Theory.13 The fundamental principle underliquid theory is lying the Fermi one of \u201cadiabatic continuity\u201d:14 in the absence of an electronic phase transition (such as Wigner crystallization), a non-interacting ground state evolves smoothly or adiabatically into the interacting ground state as the strength of interaction is increased.15 An elementary excitation of the non-interacting system represents an \u201capproximate excitation\u201d of the interacting system (i.e. its \u201clifetime\u201d is long). Excitations are quasi-particles (and quasi-holes) above a sharply de\ufb01ned Fermi surface. Lev D. Landau 1908\u201368 Nobel Laureate in Physics in 1962 \u201cfor his pioneering theories for condensed matter, especially liquid helium.\u201d Landau\u2019s work covers all branches of theoretical physics, ranging from \ufb02uid mechanics to quantum \ufb01eld theory. A large portion of his papers refer to the theory of the condensed state. They started in 1936 with a formulation of a general thermodynamical theory of the phase transitions of the second order. After P.L. Kapitsa\u2019s discovery, in 1938, of the super\ufb02uidity of liquid helium, Landau began extensive research which led him to the construction of the complete theory of the \u201cquantum liquids\u201d at very low temperatures. (Image c The Nobel Foundation.) 13 L. D. Landau, The theory of a Fermi liquid, Sov. Phys. JETP 3 (1956), 920. 14 P. W. Anderson, Basic Notions in Condensed Matter Physics (Benjamin, 1984). 15 As a simple non-interacting example, consider the adiabatic evolution of the bound states of a quantum particle as the con\ufb01ning potential is changed from a box to a harmonic potential well (see \ufb01gure). While the wavefunctions and energies evolve, the topological characteristics of the wavefunctions, i.e. the number of the nodes, and therefore the assignment of the corresponding quantum numbers, remain unchanged. 5.2 Ground state energy of the interacting electron gas 211 The starting point of Fermi liquid theory is a few phenomenological assumptions, all rooted in the adiabaticity principle. For example, it is postulated that", " the density of quasiparticles can be described in terms of a momentum-dependent density distribution n(p) which, in the limit of zero interaction, evolves into the familiar Fermi distribution. From this assumption (and a few more postulates) a broad spectrum of observables can be analyzed without further \u201cmicroscopic\u201d calculation. Its remarkable success (as well as the few notorious failures) has made Landau Fermi liquid theory a powerful tool in the development of modern condensed matter physics but one which we are not going to explore in detail.16 Instead, motivated in part by the phenomenological success of the \u201cadiabatic continuity,\u201d we will continue with the development of a microscopic theory of the weakly interacting three-dimensional electron gas, with rs 1. Perturbative approach The starting point of the perturbative analysis is the functional representation of the free energy F = \u2212T ln Z through the quantum partition function. (Here, as usual, we set kB = 1.) Expressed as a coherent state path integral, Z = D( \u00af\u03c8, \u03c8) e\u2212S[ \u00af\u03c8,\u03c8], where S[\u03c8, \u00af\u03c8] = \u00af\u03c8p\u03c3 \u2212i\u03c9n + p \u2212 \u03bc \u03c8p\u03c3 + p2 2m T 2L3 ppq \u00af\u03c8p+q\u03c3 \u00af\u03c8p\u2212q\u03c3 V (q)\u03c8p\u03c3 \u03c8p\u03c3. Here, for brevity, we have introduced the \u201c4-momentum\u201d p \u2261 (p, \u03c9n) comprising both frequency and momentum.17 As with the Green function discussed in the previous section, the free energy can be expanded in terms of an interaction parameter. To \ufb01x a reference scale against which to compare the correlation energies, let us begin by computing the free energy Eq. (4.41) of the non-interacting electron gas: F (0) = \u2212T ln p\u03c3 1 + e\u2212\u03b2(p2/2m\u2212\u03bc) T \u21920\u2212\u2192 p2/2m<\u03bc,\u03c3 p2 2m \u2212 \u03bc \u2212 2 5 N \u03bc, (5.19) where \u03bc \u2261 p2 F/2m, N = (23/2/3\u03c02)m3/2L3\u03bc3/2 is the number of particles, and the last estimate is obtained by replacing the sum over momenta by an integral. According to Eq. (5.19), the average kinetic energy per particle is equal to 3/5 of the Fermi energy. To relate this scale to the density parameter rs, we choose to measure all energies in units of the Rydberg energy (alias the ionization energy of hydrogen), ERy = me4/22 = 13.6 eV, F (0) ERy \u223c N r2 s. (5.20) 16 Interested readers are referred to one of several excellent reviews, e.g. P. W. Anderson, Basic Notions in Condensed Matter Physics (Benjamin, 1984). 17 Notice that \u201c4-momentum\u201d p and the modulus of the \u201c3-momentum\u201d, p = |p|, are denoted by the same symbol (standard notation!). However, in its context, this should not cause confusion. 212 Perturbation theory If we turn now to the discussion of interactions, a formal expansion of F to \ufb01rst order in V gives F (1) = T 2 2L3 B ppq \u00af\u03c8p+q\u03c3 \u00af\u03c8p\u2212q\u03c3 V (q)\u03c8p\u03c3 \u03c8p\u03c3 C, 0 (5.21) where \u00b7 \u00b7 \u00b7 0 denotes the functional average with respect to the non-interacting action. The two18 diagrams contributing to this expression are shown in the \ufb01gure overleaf. To account for the speci\ufb01cs of the electron gas, we are using a diagrammatic code slightly di\ufb00erent from that of the previous section: The Coulomb interaction is represented by a wavy line labeled by the momentum argu- ment q. A contraction \u03c8p\u03c3 \u00af\u03c8p\u03c30 is indicated by a solid arrow representing the free Green func- tion of the electron gas, Gp \u2261 1 \u2212i\u03c9n + p2 2m \u2212 \u03bc, (5.22) i.e. the inverse of the free action. The labeling of the contraction by an arrow (instead of a unidirectional line as in \u03c64-theory) is motivated by two considerations: \ufb01rstly, the \u00af\u03c8\u03bb0 describing the creation mnemonic has the e\ufb00ect of indicating that a contraction \u03c8\u03b7 of an electron with quantum numbers \u03bb, followed by the annihilation of", " an electron at \u03b7, is a directed process; secondly, there are situations (e.g. when a magnetic \ufb01eld is present) where \u03c8n\u03c3(r) \u00af\u03c8n\u03c3(r)0 = \u03c8n\u03c3(r) \u00af\u03c8n\u03c3(r)0. The sum of all 4-momenta emanating from an interaction vertex formed by a wavy line and two electron \ufb01eld lines is equal to zero (exercise) \u2013 the \u201cKirchho\ufb00 law.\u201d Finally, we have to be careful about sign factors arising when Grassmann variables are interchanged. However, the anti-commutativity of the \ufb01elds merely leads to an overall factor (\u2212)Nl, where Nl is the number of loops of a diagram. (To verify this claim, one may notice that a loop is formed by a \u201cringwise contraction\u201d \u00af\u03c81\u03c82 \u00af\u03c83... \u03c8N 0, (2 \u2192 3)(4 \u2192 5)...((N \u2212 2) \u2192 (N \u2212 1))(N \u2192 1). The last contraction introduces the minus sign.) First-order perturbation theory Turning to the discussion of the speci\ufb01c diagrams shown in the \ufb01gure below, we notice that the \ufb01rst, generally known as a Hartree contribution, vanishes. Technically, this is a consequence of the fact that the interaction line connecting the two loops carries zero momentum. However, as discussed in Section 2.2, V (q = 0) = 0. Physically, the vanishing of the Hartree contribution is a consequence of charge neutrality. Douglas R. Hartree, 1897\u20131958 Originally appointed as a professor of applied mathematics he became one of the \ufb01rst \u201ccomputational physicists.\u201d He developed methods for numerically solving the Schr\u00a8odinger equation, among them the celebrated \u201cHartree approximation.\u201d 18 Remember that, in a theory with complex or Grassmann \ufb01elds, only contractions \u223c \u00af\u03c8\u03c80 exist, i.e. there is a total of n! distinct contributions to a contraction \u00af\u03c8\u03c8... \u03c80 of 2n \ufb01eld operators. 5.2 Ground state energy of the interacting electron gas 213 Indeed, the two Green function loops p Gp measure the local particle density of the electron gas (cf. the discussion on page 171). Global charge neutrality requires that the electronic density cancels against the density of the ionic background. However, notice that this cancellation mechanism relies on our assumption of overall spatial homogeneity. Only in a spatially uniform system does the density of the electron gas locally compensate the positive counter-density. In realistic metals, the inevitable presence of impurities breaks translational invariance and there is no reason for the Hartree contribution to vanish. Indeed, the analysis of Hartree-type contributions to the correlation energy in disordered electronic media is a subject of ongoing research. While the Hartree term describes the classical interaction of charge densities through the Coulomb potential, the second diagram shown in the \ufb01gure, known as a Fock contribution, is quantum. Translating the diagrammatic language back into Green functions, we obtain F (1) = \u2212 T 2 L3 = \u2212 1 L3 T \u21920= \u2212 1 L3 p,p p,p p,p <\u03bc GpGp V (p \u2212 p) nF(p)nF(p ) e2 |p \u2212 p|2 e2 |p \u2212 p|2 = \u2212 e2L3p4 (2\u03c0)5.23) Here, the sign factor in the \ufb01rst equality is due to the odd number of fermion loops, and the evaluation of the last sum, which can be found, for example, in the text by Kittel,19 is left as an exercise. (Up to numerical factors, the result follows from dimensional considerations. We are integrating the inverse square of the distance in momentum space ([momentum]\u22122) over two Fermi spheres ([momentum]6). Since the integral is convergent at low momenta, it must scale as the fourth power of the upper cuto\ufb00 \u223c p4 Division by the Rydberg energy leads to the scaling F.) \u0393(x) 4 F (1) ERy = const. \u00b7 N rs, (5.24) 3 2 1 where the constant is of order unity. This result conforms with our previous estimate of the density dependence of correlation energies. 0.5 1 x 1.5 19 C. Kittel, Quantum Theory of Solids (Wiley, 1963). 214 Perturbation theory INFO A closer analysis shows that other predictions following from E", "q. (5.23) are less sensible. To understand what goes wrong, let us rewrite the sum of F (0) and F (1) as \u239b \u239d p2 2m \u2212 \u03bc \u2212 1 L3 e2 |p \u2212 p|2 F (0) + F (1) = \u23a0. \u239e |p|<pF p <\u03bc This way of writing suggests that the interaction of an electron of momentum p with all partner electrons of momentum p leads to a reduction20 of the single particle energy by an amount L\u22123 p <\u03bc e2/|p \u2212 p|2. Evaluation of the internal sum (exercise) leads to21 F (0) + F (1) = p2 2m \u2212 \u03bc \u2212 e2pF 2\u03c0 \u0393(|p|/pF), |p|<pF 1+x 1\u2212x where the function \u0393(x) = 2 + 1\u2212x2 is shown in the \ufb01gure. The most important characteristic of the scaling function \u0393 is a logarithmically diverging derivative at x = 1 (exercise). To appreciate the consequences of this feature, let us consider the single particle density of states \u03c1() = 2\u03c0 \u0393(p/pF) is the \ufb01rst-order Fock single-particle energy. Approximating the sum by an integral, we obtain p\u03c3 \u03b4( \u2212 p), where p = p2 \u2212 e2pF ln 2m x \u03c1() = \u221e 0 L3 \u03c02 p2dp \u03b4( \u2212 p) = L3p2 \u03c02 dp dp \u22121, where p is de\ufb01ned through p =. The singularity of the derivative of \u0393(p/pF) at p = pF implies that the density of states is predicted to vanish as approaches the Fermi energy, clearly a nonsensical result.22 The origin of this pathological divergence can be traced back to the longrange (|r|\u22121 \u2192 |q|\u22122) decay of the Coulomb potential (cf. Eq. (5.21)). Yet we know that, in systems with mobile charge carriers, electrostatic forces are screened, i.e. they decay exponentially. Microscopically, screening is due to the polarization of a medium in response to an applied electric \ufb01eld. This by itself is an interaction e\ufb00ect, albeit one of \u201chigher order\u201d in the Coulomb interaction, i.e. if we ever want to observe signatures of the screening mechanism, we need to advance to higher orders in the expansion in V. Second-order perturbation theory Let us then consider the second-order contribution F (2) = \u2212 T 2 2 T 2L3 B\u239b \u239d ppq \u00af\u03c8p+q\u03c3 \u00af\u03c8p\u2212q\u03c3 V (q)\u03c8p\u03c3 \u03c8p\u03c3 \u239e 2Cc \u23a0, 0 where the superscript indicates that only connected diagrams contribute. 20 Why a reduction? Charge neutrality implies that the classical correlation energy vanishes. However, being fermionic in nature, electrons tend to avoid each other even more than classical particles, i.e. quantum statistics leads to a lowering of the correlation energy below the classical value of zero. 21 See, e.g., J. Hubbard, The description of collective motion in terms of many-body perturbation theory, Proc. Roy. Soc. (London) A243 (1957), 336. 22 The majority of transport observables in metals are proportional to the density of states at the Fermi energy, i.e. \u03c1(\u03bc) = 0 would entail the vanishing of almost all transport coe\ufb03cients! 5.2 Ground state energy of the interacting electron gas 215 EXERCISE Using the \ufb01eld integral representation of Z, show that second-order expansion of ) F = \u2212T ln Z in the interaction operator Sint leads to F (2) = \u2212(T /2). Convince yourself that the second term cancels disconnected diagrams. Apply arguments similar to those involved in the proof of the linked cluster theorem to verify that the cancellation of disconnected graphs pertains to all orders in the expansion of F. I.e. the free energy can be obtained by expanding the partition function Z (not its logarithm) and keeping only connected diagrams; dropping disconnected contributions is equivalent to taking the logarithm. Sint S2 int (. \u2212 / 2 0 0 Connected contraction of the eight \ufb01eld operators leads to four distinct types of diagram (exercise) of which two are of Hartree type (i.e. contain a zero-momentum interaction line V (q = 0)). The non-van", "ishing diagrams F (2),1 and F (2),2 are shown in the \ufb01gure below.23 Translating these diagrams to momentum summations over Green functions, one obtains (exercise) F (2),1 = \u2212 T 3 L6 F (2),2 = 1 2 T 3 L6 p1,p2,q p,q1,q2 Gp1 Gp1+qGp2 Gp2+qV (q)2, GpGp\u2212q1 Gp\u2212q1\u2212q2 Gp\u2212q2 V (q1)V (q2). (5.25) While, at \ufb01rst sight, these expressions do not look very illuminating, closer inspection reveals some structure. Re\ufb02ecting the fact that electronic transport in solids is carried by excitations at the Fermi energy, the electron Green function (5.22) assumes large values for momenta p pF. This implies that only con\ufb01gurations where all momentum arguments carried by the Green function are close to the Fermi surface contributes signi\ufb01cantly to the sums (5.25). Considering the \ufb01rst sum, we see that, for small |q| and |pi| p F, this condition is met, i.e. there are two unbound summations over momentum shells around the Fermi surface. However, with the second sum, the situation is less favourable. For \ufb01xed |p1| p F, \ufb01netuning of both q1 and q2 is necessary to bring all momenta close to pF, i.e. e\ufb00ectively one momentum summation is frozen out. There is no need to enter into detailed calculations to predict that, due to the relatively larger \u201cphase volume,\u201d F (2),1/F (2),2 1, where the ratio will be proportional to the area of the Fermi surface which, in turn, is proportional to the density of the electron gas. For large densities, the second Fock diagram can be neglected in comparison with the \ufb01rst. 23 In fact, one more non-vanishing contribution is obtained by drawing a single \u201cring\u201d of Green functions containing two non-crossing interaction lines. This diagram, and its obvious generalization to higher order processes containing sequential \u201cself\u2013interactions\u201d of a single Green function do not play an essential role (in the present context.) The reasons why will become clear in section 5.3 when we introduce the notion of self-energies. 216 Perturbation theory (a) (b) Figure 5.6 Diagram of second-order Fock contributions to the free energy of the electron gas. (a) Interaction process involving a single electron\u2013hole pair. An electron is virtually excited above the Fermi energy, and subsequently recombines with a hole state left behind by a second electron excitation which then jumps into the hole created by the initial excitation. (b) Virtual excitation of independent electron\u2013hole pairs. F (2),1 Of course, there must be a \u2018more physical\u2019 way of understanding this observation. The Green function lines in the diagrams F (2),i describe the propagation of quasi-particles and quasi-holes24 on the background of the interacting medium. Now, the diagram F (2),2 contains a simply connected propagator line: a single particle\u2013hole excitation undergoes a second-order interaction process with itself (see Fig. 5.6(a)). By contrast, the \ufb01rst diagram F (2),1 involves two independent electron\u2013hole excitations, as shown in Fig. 5.6(b). Since, in a dense electron gas, a second-order interaction process will more likely involve di\ufb00erent particles, this type of contribution is more important. Notice that the process shown in Fig. 5.6(b) can be interpreted as a \u201cpolarization\u201d of the medium due to the excitation of electron\u2013hole pairs. F (2),2 Higher orders in perturbation theory The picture above readily generalizes to interaction processes of higher order. In the highdensity limit, dominant contributions to the free energy should contain one free integration over the Fermi momentum per interaction process. A moment\u2019s thought shows that only diagrams of \u201cring graph\u201d structure (see the \ufb01gure) meet this condition. Expanding the free energy functional to nth order in the interaction operator, and retaining only diagrams of that structure, one obtains 6 RPA = \u2212 T F (n) 2n q 2T L3 V (q) p 7n GpGp+q. (5.26) (To understand the origin of the", " multiplicative factor 1/n, notice that F (n) results from the connected contraction of an operator \u223c 24 In principle, the system consists of physical electrons immersed into a globally positive background. However (cf. the discussion of Section 2.2), keeping in mind that at low temperatures dynamical processes take place in the immediate vicinity of the Fermi surface, a more problem-oriented way of thinking about states is in terms of quasi-particles and quasi-holes, i.e. electronic states immediately above and below the Fermi surface. 5.2 Ground state energy of the interacting electron gas 217 (n!)\u22121(Sint[ \u00af\u03c8, \u03c8])nc 0. There are (n\u22121)! di\ufb00erent ways of arranging the interaction operators Sint in a ring-shaped structure, i.e. the diagram carries a global factor (n \u2212 1)!/n! = 1/n.) In Eq. (5.26) the subscript \u201cRPA\u201d stands for random phase approximation.25 However, more important than the designation are the facts: (a) we have managed to identify a particularly relevant subclass of diagrams contributing to the plethora of interaction processes; (b) there is a physical parameter controlling the dominance of these diagrams; and (c) we are apparently able to sum up the entire series of nth-order RPA-interaction contributions. Indeed, summation over n leads to the simple result FRPA \u2261 n F (n) RPA = T 2 q ln (1 \u2212 V (q)\u03a0q), (5.27) where we have introduced the polarization operator,26 \u03a0q \u2261 2T L3 p GpGp+q. (5.28) Equation (5.27) represents our \ufb01rst example of an in\ufb01nite-order expansion. However, before turning to the discussion of further aspects of in\ufb01nite-order perturbation theory, let us stay for a moment with the RPA free energy. Speci\ufb01cally, let us investigate whether the RPA repairs the pathology in the Fermi surface density of states observed above. The last unknown we need to compute before turning to the discussion of the RPA free energy is the polarization operator. Drawing on the frequency summation in Problem 4.5, the polarization operator can be written explicitly as \u03a0q = 2T L3 p 1 \u2212i\u03c9n + \u03bep 1 \u2212i\u03c9n+m + \u03bep+q = 2 L3 p nF(p+q) \u2212 nF(p) i\u03c9m + \u03bep+q \u2212 \u03bep. (5.29) The evaluation of the momentum sum is straightforward if a bit tedious. As a result of the calculation outlined in the Info block below one obtains \u03a0q,\u03c9m = \u2212\u03bd0 1 \u2212 i\u03c9m 2vFq ln i\u03c9m + vFq i\u03c9m \u2212 vFq, where \u03bd0 \u2261 1 Ld p,\u03c3 \u03b4(\u03bc \u2212 p) = 2 d3p (2\u03c0)3 \u03b4(p \u2212 \u03bc) = mpF \u03c02, (5.30) (5.31) 25 The attribute \u201crandom phase\u201d seems to allude to the fact that the quantum mechanical phase carried by the particle\u2013hole excitations stirred up by interactions gets lost after each elementary polarization process. This contrasts with more-generic contributions to F where quantum phases may survive more complex interaction processes. Also notice that more than one approximation scheme in statistical physics has been dubbed \u201crandom phase.\u201d 26 The de\ufb01nition (5.28) applies to the speci\ufb01c case of a three-dimensional translationally invariant system. More generally, the polarization operator is de\ufb01ned as the frequency/momentum Fourier transform of the connected average \u00af\u03c8(x, \u03c4 )\u03c8(x, \u03c4 ) \u00af\u03c8(x, \u03c4 )\u03c8(x, \u03c4 )c. 218 Perturbation theory is the density of states per volume of non-interacting electrons at the Fermi surface. INFO For an isotropic three-dimensional system at zero temperature, the polarization operator (5.29) becomes a function of |q| and \u03c9n, known as theLindhard function. The evaluation of the momentum sum is greatly simpli\ufb01ed by noting that, due to the di\ufb00erence of the Fermi functions, only momenta with |p| pF signi\ufb01cantly contribute. Assuming that |q| p F, we may thus e\ufb00ect the linearization \u03bep+q \u2212 \u03bep = (1/m)p \u00b7 q + O(q2), where the", " term of O(q2) is negligibly small. Similarly, nF(p+q) \u2212 nF(p) \u2202p nF(p)(1/m)p \u00b7 q \u2212\u03b4(p \u2212 \u03bc)(1/m)p \u00b7 q, where, in the zero-temperature limit, the last equality becomes exact. Converting the momentum sum into an integral, we thus obtain \u03a0q,\u03c9m = \u22122 d3p (2\u03c0)3 \u03b4(p \u2212 \u03bc) 1 m p \u00b7 q. i\u03c9m + 1 m p \u00b7 q Setting 1/(2\u03c0)3 dimensional unit sphere, and p = pn, with|n| = 1, the integral can be evaluated, thus: d3p = 1/(2\u03c0)3 d\u03a9 is the integral over the three- d\u03a9, where dp p2 \u03a0q,\u03c9m = \u2212 2 (2\u03c0)3 = \u2212 2 (2\u03c0)3 dp p2 dp p2 d\u03a9 \u03b4(p \u2212 \u03bc) d\u03a9 \u03b4(p \u2212 \u03bc) \u03bd0 vFxq i\u03c9m + vFxq = \u2212 \u03bd0 2 1 dx \u22121 = \u2212\u03bd0 1 \u2212 i\u03c9m 2vFq ln. i\u03c9m + vFq i\u03c9m \u2212 vFq vFn \u00b7 q i\u03c9m + vFn \u00b7 q 1 d\u03a9 d\u03a9 vFn \u00b7 q i\u03c9m + vFn \u00b7 q To obtain the free energy, we need to substitute the Lindhard function into Eq. (5.27) and sum over q = {q, \u03c9m}. Yet, due to the complicated dependence of the function \u03a0q,\u03c9m on its arguments, this \ufb01nal part of the calculation is far from trivial. An asymptotic expression for the high-density expansion of the RPA approximation to F was \ufb01rst obtained in a famous study by Gell-Mann and Brueckner.27 Measuring in atomic units, they obtained the free energy per particle of Murray Gell-Mann 1929\u2013 Nobel Laureate in Physics in 1969 \u201cfor his contributions and discoveries concerning the classi\ufb01cation of elementary particles and their interactions.\u201d Among his many accomplishments, he formulated the quark model of hadronic resonances, and identi\ufb01ed the SU(3) \ufb02avor symmetry of the light quarks, extending isospin to include strangeness, which he also discovered. (Image c The Nobel Foundation.) FRPA ERyN = \u22120.142 + 0.0622 ln rs. (5.32) For a derivation of this result we refer the interested reader to the literature. (In the late 1950s, motivated largely by applications to the physics of heavy nuclei, the perturbative calculation of ground state energies had become an industry. A comprehensive discussion of 27 M. Gell-Mann and K. Brueckner, Correlation energy of an electron gas at high density, Phys. Rev. 106 (1957), 364\u20138. To explore this and related literature, it is useful to refer to D. Pines, The Many-Body Problem (Benjamin, 1962) where the important reprints are collected. 5.2 Ground state energy of the interacting electron gas 219 the subject, and details of the asymptotic rs- expansion of FRPA can be found, for example, in Mahan.28) When added to the kinetic energy Eq. (5.20) and the \ufb01rst-order correlation energy Eq. (5.24), the structure of the density expansion of the free energy becomes clear: the sum over all RPA diagrams yields the coe\ufb03cient of O(r0 s ) in the expansion in rs.29 More important for our present discussion is the conceptual meaning of the RPA, notably the connection to screening. To this end, let us temporarily consider the expectation value of the particle number N = \u2212\u2202\u03bcF, rather than the free energy itself. Speci\ufb01cally, we wish to compare the \ufb01rst order correction (with the non-interacting result) N (1) = \u2212\u2202\u03bcF to the RPA, NRPA = \u2212\u2202\u03bcFRPA. Noting that \u2202\u03bcGp = \u2212(Gp)2 we readily \ufb01nd (cf. Eq. (5.23)) N (1) = \u2212 2T 2 L3 p,q (Gp)2Gp+qV (q). After a second di\ufb00erentiation, \u03c1(1) = \u2202\u03bcN (1), this expression would lead to the pathological density of states,", " a consequence of the unscreened interaction line. (The diagrammatic visualization of N (1) is shown in Fig. 5.7(a).) Now, consider the \u03bc-derivative of FRPA, Eq. (5.27): NRPA = T 2 = \u2212 2T 2 L3 q V (q)\u2202\u03bc\u03a0q 1 \u2212 V (q)\u03a0q = \u2212 T 2 L3 V (q) 1 \u2212 V (q)\u03a0q p q q V (q) 1 \u2212 V (q)\u03a0q Gp+q(Gp)2 = \u2212 2T 2 L3 p Gp+q(Gp)2 + (q \u2194 \u2212q) q Ve\ufb00 (q) p Gp+q(Gp)2, (5.33) where we have de\ufb01ned the \u201ce\ufb00ective\u201d interaction Ve\ufb00 (q) \u2261 1 V (q)\u22121 \u2212 \u03a0q \u2261 V (q) (q), with the generalized dielectric function (q) \u2261 1 \u2212 V (q)\u03a0q. (5.34) (5.35) Structurally, the expression for NRPA resembles the \ufb01rst-order expression N (1), only with the \u201cbare\u201d Coulomb interaction replaced by the e\ufb00ective interaction Ve\ufb00. From its de\ufb01nition, it is clear that Ve\ufb00 represents a geometric series over polarization bubbles, augmented by bare interaction lines. This is visualized in Fig. 5.7(b), where the fat line is de\ufb01ned in the bottom part of the \ufb01gure. In fact, we do not need to stare at the analytical expression (5.33) to understand its origin. The \u03bc-di\ufb00erentiation acting on FRPA may pick any of the n rings contributing to F (n) RPA in Eq. (5.26). The \u201cdi\ufb00erentiated ring\u201d becomes the bubble in Fig. 5.7, while all other rings conspire to form the (n \u2212 1)th-order contribution to the e\ufb00ective interaction line. 28 G. Mahan, Many Particle Physics (Plenum Press, 1981). 29 Here we follow a convention (used mostly in the older literature) where the RPA starts from the second-order ring diagram F (2),1. However, henceforth we will refer to the \u201cRPA\u201d as the sum over all ring diagrams, including the \ufb01rst, F (1). 220 Perturbation theory p + q q p (b) p + q q p (a) = = + + + +... Figure 5.7 Diagrammatic visualization of expectation value of the particle number. (a) First-order Fock correction and (b) the RPA approximation, where the de\ufb01nition of the RPA interaction line is shown in the bottom part of the \ufb01gure. At this stage, the connection between RPA and the collective electromagnetic response of the charged system becomes discernible. We remember that the electric \ufb01eld E in a medium is related to the vacuum \ufb01eld D through the relation D(q, \u03c9) = (q, \u03c9)E(q, \u03c9), where the dielectric function (q, \u03c9) = 1+4\u03c0\u03c7(q, \u03c9) is determined by the electromagnetic susceptibility, i.e. a function that measures the tendency of the medium to \u201crespond\u201d or adjust to an external electromagnetic perturbation. Identifying E(D) with the gradient of the \u201cdressed\u201d potential Ve\ufb00 (the \u201cbare\u201d potential V ), we conclude that, on the level of the microscopic theory, 4\u03c0\u03c7(q, \u03c9) =\u2212V (q)\u03a0q,\u03c9n, i.e. the susceptibility is proportional to the polarization operator \u03a0q. These connections motivate the introduction of the dielectric function as in Eq. (5.35) above. The full complexity of the polarization of the dense homogeneous electron gas is encoded by the Lindhard function (5.30). However, there are some interesting limiting cases, where the situation simpli\ufb01es. Inspection of Eq. (5.30) shows that the Lindhard function depends on the dimensionless ratio between two characteristic length scales: the \u201cwavelength\u201d q\u22121, and the distance vF/\u03c9 an excitation propagating with Fermi velocity travels during the time \u03c9\u22121. For small frequencies, vF/\u03c9 q\u22121, the electron gas has enough time to optim", "ally adjust to the spatial variation \u223c q\u22121 of the potential, i.e. to screen out electro-neutrality \u03c9qvF= \u2212\u03bd0 + violating potential \ufb02uctuations. In this static limit, we may expand \u03a0q,\u03c9m O(\u03c9/vFq). In this case, the electron gas interacts through the e\ufb00ective potential, Ve\ufb00 (q) \u03c9qvF= 1 V (q)\u22121 + \u03bd0 \u22121/2 = 4\u03c0e2 q2 + 4\u03c0e2\u03bd0 \u2261 4\u03c0e2 q2 + \u03bb\u22122, where the constant \u03bb \u2261 is known as the Thomas\u2013Fermi screening length. Indeed, it is straightforward to verify that the inverse Fourier transform leads to an e\ufb00ective interaction potential 4\u03c0e2\u03bd0 Ve\ufb00 (r) = e2 e\u2212|r|/\u03bb |r|, (5.36) i.e. the potential is exponentially supressed, or screened, on length scales |r| > \u03bb. 5.2 Ground state energy of the interacting electron gas 221 EXERCISE To verify Eq. (5.36), introduce polar coordinates q = (q, \u03b8, \u03c6) in q-space and perform \ufb01rst the elementary integrations over \u03c6 and \u03b8. It is then straightforward to complete the remaining q-integration over the radial coordinate by analogy to the contour integral (5.13). INFO Let us brie\ufb02y recapitulate the heuristic interpretation of Thomas\u2013Fermi screening. Imagine a test charge e has been immersed in an electron gas. The host system will respond to this perturbation by a local distortion of its density. To compute the distortion, we note that the e\ufb00ective potential Ve\ufb00 (r) created by the test charge changes the electronic energy levels according to p \u2192 p \u2212 Ve\ufb00. (Here, we assume that the external perturbation changes so slowly that the simultaneous usage of momentum and coordinate quantum numbers is not in con\ufb02ict with the uncertainty relation.) The induced charge density thus reads as \u03c1ind(r) =\u2212e d3p (2\u03c0)3 [nF(p \u2212 Ve\ufb00 (r)) \u2212 nF(p)] \u2248 eVe\ufb00 (r) d3p (2\u03c0)3 \u2202nF(p) T \u21920 \u2212eVe\ufb00 (r) d d3p (2\u03c0)3 \u03b4( \u2212 p) \u03bd() \u03b4( \u2212 \u03bc) = \u2212eVe\ufb00 (r)\u03bd0, where in the second equality we assumed weakness of the induced potential and in the third equality used the low-temperature approximation \u2202nF() \u2212\u03b4( \u2212 \u03bc). A substitution of this result into the Fourier transform of the Poisson equation \u22072Ve\ufb00 (r) = \u22124\u03c0e\u03c1(r) = \u22124\u03c0e(e\u03b4(r) + \u03c1ind(r)) gives Ve\ufb00 (q) = 4\u03c0e2 q2 + 4\u03c0e2\u03bd0. We also note that Thomas\u2013Fermi screening repairs our problem with the density of states. The pathology discussed on page 213 found its origin in the (1/q2)-singularity of the unscreened potential. This IR singularity is now cut o\ufb00 at momenta q \u223c \u03bb\u22121, implying benign behavior of the density of states. Using the fact that the density parameter rs scales with the Fermi energy as (exercise) rs \u223c \u03bc\u22121/2, and di\ufb00erentiating \u03c1 = \u22022 \u03bcF, it is indeed straightforward to verify that the RPA expansion of F no longer leads to singular behavior of \u03c1 at the Fermi energy. It is also interesting to consider the limit vF/\u03c9 q\u22121 of essentially dynamic polarization. Expansion of \u03a0q,\u03c9m in Eq. (5.30) to \ufb01rst order in vFq/\u03c9 1 leads to the expression Ve\ufb00 (q, \u03c9m) = 4\u03c0e2 q2 1 1 + 4\u03c0\u03bd0e2v2 F 3\u03c92 m = 4\u03c0e2 q2 1 1 \u2212 \u03c92 p \u03c92, where \u03c9p \u2261 (4\u03c0ne2/m)1/2 denotes the plasma frequency and we have used the fact that (exercise) the particle density n \u2261 N/L3 = 2k3 F /6\u03c02. In fact, the form of the denominator hints at a collective instability of the electron gas at frequencies \u223c \u03c9p. Referring for a more rigorous discussion to", " the next chapter, one can understand the origin of the instability by another heuristic argument. What we need to remember is that the current formalism is 222 Perturbation theory developed for imaginary times \u03c4. Eventually we will be interested in real-time dynamics, i.e. we need to analytically continue back to t \u223c \u2212i\u03c4. In the language of frequencies, this amounts to an analytic continuation process from Matsubara frequencies to real frequencies, i\u03c9m \u2192 \u03c9. Sloppily substituting a real frequency into the potential, we obtain Ve\ufb00 (q, \u03c9m \u2192 \u2212i\u03c9) \u221d, i.e. the system seems to respond singularly to excitations with frequency \u03c9 \u223c \u03c9p. The cause of this instability is the well-known plasmon mode of the electron gas. 1 \u2212 \u03c92 p/\u03c92 \u22122 E x INFO The physics of the plasmon mode can be understood as follows. Imagine the electron gas uniformly displaced by a distance x against the positively charged background (see \ufb01gure). This will lead to the formation of oppositely charged surface layers at the two ends of the system. The surface charge densities \u03c1\u00b1 = \u00b1exn lead to an electric \ufb01eld E = 4\u03c0enx directed opposite to the displacement vector. Mobile charge carriers inside the system are thus subject to a force \u22124\u03c0e2nx. The solution of the equation of motion m\u00a8x = \u22124\u03c0e2nx oscillates at a frequency \u03c9p = (4\u03c0e2n/m)1/2, the plasma frequency. Since the motion of the charge carriers is in turn responsible for the accumulation of the charged surface layers, we conclude that the system performs a collective oscillatory motion, known as the plasmon excitation. At this point we conclude our preliminary discussion of the electron gas. We have seen that large-order perturbation theory can be applied to successfully explain various features of the interacting system: energetic lowering due to quantum correlation, screening, and even collective instabilities. The interacting electron gas is but one example of the many applications of the diagrammatic perturbation theory. After the full potential of the approach had become evident \u2013 in the late 1950s and early 1960s \u2013 diagrammatic techniques of great sophistication were developed, and applied to a plethora of many-body problems. Indeed, more than two decades passed before large-order perturbation theory eventually ceased to be the most important tool of theoretical condensed matter physics. Re\ufb02ecting the great practical relevance of the approach, there is a huge body of textbook literature concentrating on perturbative methods.30 Although it would make little sense to develop the \ufb01eld in its full depth once again, a few generally important concepts of diagrammatic perturbation theory are summarized in the next section. 30 See, e.g., A. A. Abrikosov, L. P. Gorkov, and I. E. Dzyaloshinkii, Methods of Quantum Field Theory in Statistical Physics (Dover Publications, Inc., 1975), A. Fetter and J. D. Walecka, Quantum Theory of ManyParticle Systems (McGraw-Hill, 1971), D. Pines and P. Nozi`eres, The Theory of Quantum Liquids \u2013 Normal Fermi Liquids (Addison-Wesley, 1989), and S. Doniach and E. H. Sondheimer, Green Functions for Solid State Physicists (Benjamin/Cummings, 1974). 5.3 In\ufb01nite-order expansions 223 5.3 In\ufb01nite-order expansions Turning back to the prototypical \u03c64-model, it is the purpose of the present section to introduce a number of general concepts of in\ufb01nite-order perturbative summations. As should be clear from the discussion above, a meaningful summation over an in\ufb01nite set of diagrams necessitates the existence of a class of perturbative corrections that is \u201cmore important\u201d than others. In practice, what we need is a small parameter that discriminates between diagrams of di\ufb00erent structure. In our example above, this parameter was the e\ufb00ective density rs of the electron gas. However, in other settings, the control parameter N may be de\ufb01ned quite di\ufb00erently: large values of a spin, S, the number of colors, Nc, in QCD, the number of spatial dimensions, d, the number of modes of an optical wave guide, etc. Unfortunately, in most physical contexts, these parameters are typically far from large, S = 1/2, d = Nc = 3, etc. So we have to resort to a \u201cpoor man\u2019s\u201d strategy where we develop", " a controlled and self-consistent theory in the limit of asymptotically large control parameters and hope that some fragments of truth survive the limit down to more mundane values of N. Perhaps unexpectedly, this strategy often works astonishingly well down to values N = O(1). So let us, then, begin by introducing a large control parameter into a \u03c64type theory. This can be achieved by promoting \u03c6 from a scalar to an N component vector \ufb01eld \u03c6 = {\u03c6a}, a = 1,..., N. The self-interaction of the \ufb01eld is modeled as g \u03c6-space. The action of our modi\ufb01ed theory is thus given by ddx \u03c6a\u03c6a\u03c6b\u03c6b, i.e. an expression that is \u201crotationally\u201d invariant in S[\u03c6] \u2261 ddx 1 2 \u2202\u03c6 \u00b7 \u2202\u03c6 + r 2 \u03c6 \u00b7 \u03c6 + g 4N (\u03c6 \u00b7 \u03c6)2, (5.37) where the factor 1/N in front of the interaction constant has been introduced for later convenience. As before, we shall concentrate on the Green function Gab(x \u2212 y) = \u03c6a(x)\u03c6b(y) as a \u201ctest observable.\u201d Denoting the Green function by a bold line, the free Green function G0 \u2261 \u03c6a(x)\u03c6b(y)0 \u221d \u03b4ab by a thin line, and the interaction operator by a wavy line,31 the structure of the \ufb01rst- and second-order expansion of the Green function is shown in the upper portion of Fig. 5.8. For simplicity, the combinatorial factors weighting individual diagrams have been omitted. Self-energy operator Even without resorting to the large-N structure of the theory, it is possible to bring some order in the spaghetti of diagrams contributing to the expansion. Indeed, there are two distinct subclasses of diagrams: diagrams that are one-particle reducible (i.e. can be cut into two halves by cutting a single internal line, see the classi\ufb01cation on page 204) and those that are not. This observation motivates the collection of all one-particle irreducible 31 Since the four \ufb01eld vertices entering the interaction are no longer indiscriminate, the interaction \u201cpoint\u201d rep- resentation of Section 5.1 is no longer suitable. 224 Perturbation theory + + + + + +... + + +... = + + + + = = + + = + + + + + + +... Figure 5.8 Expansion of the Green function of \u03c64-theory. Bottom: expansion of the self-energy operator. subportions of the diagrammatic expansion into a structural unit. In Fig. 5.8 this entity, which is commonly called the self-energy operator, and sometimes also the e\ufb00ective mass operator, is denoted by a hatched circle. The \ufb01rst- and second-order expansion of the self-energy is shown in the bottom part of the \ufb01gure. With that de\ufb01nition, the Green function becomes a \u201cchain\u201d of selfenergy operators, separated by free Green function lines, as shown in the second equality of the \ufb01gure. A convenient representation of the expansion is shown in the third equality. An insertion of the full Green function after the \ufb01rst selfenergy correction recursively generates the full series. Let us translate these statements into the language Freeman Dyson, 1923\u2013 a mathematician, Trained as Dyson turned to physics in the 1940s. His work in condensed matter physics, statistical mechanics, and several other areas has had lasting in\ufb02uence on the development of modern theoretical physics. Beyond his professional work in physics, Dyson has written several books on the social implications of modern science. (Photo by Randall Hagadorn. Courtesy of the Institute of Advanced Study.) 5.3 In\ufb01nite-order expansions 225 of formulae. Denoting the set of all self-energy diagrams by \u02c6\u03a3 ={\u03a3 ab(x \u2212 y)},32 the expansion of the Green function assumes the form \u02c6G = \u02c6G0 + \u02c6G0 \u02c6\u03a3 \u02c6G0 + \u02c6G0 \u02c6\u03a3 \u02c6G0 \u02c6\u03a3 \u02c6G0 + \u00b7 \u00b7 \u00b7 = \u02c6G0 + \u02c6G0 \u02c6\u03a3 \u02c6G. (5.38) Here, the operator products involve summation over coordinates and internal indices, i.e (", " \u02c6A \u02c6B)ab(x \u2212 y) = ddz Aac(x \u2212 z)Bcb(z \u2212 y). Recursion relations of this type are commonly referred to as Dyson equations. The Dyson equation states that the problem of calculating \u02c6G is essentially tantamount to that of analyzing the self-energy. To make this point more explicit, let us reformulate the Dyson equation in momentum space: \u02c6Gp = \u02c6G0,p + \u02c6G0,p \u02c6\u03a3p \u02c6Gp \u21d4 [1 \u2212 \u02c6G0,p \u02c6\u03a3p] \u02c6Gp = \u02c6G0,p. Here we have used the convolution theorem or, more physically, the fact that all scattering processes lumped into the self-energy conserve momentum separately. Matrix multiplication of this identity from the right by [1 \u2212 \u02c6G0,p \u02c6\u03a3p]\u22121 leads to the expression \u02c6Gp = [1 \u2212 \u02c6G0,p \u02c6\u03a3p]\u22121 \u02c6G0,p = \u02c6G\u22121 0,p \u2212 \u02c6\u03a3p # $\u22121. Finally, using the fact that $ab # \u02c6G\u22121 0,p = (p2 + r)\u03b4ab, we arrive at the formal solution # Gab p = (p2 + r \u2212 \u02c6\u03a3p)\u22121 $ab. (5.39) This equation provides two lessons: \ufb01rstly, the full information about the Green function is indeed stored in the self-energy; secondly, the self-energy somehow \u201cadds\u201d to the arguments p2 and r entering the quadratic action, a point to be discussed in more detail below. But how then do we compute the self-energy operator? In fact, the construction recipe follows from what has been said above. By de\ufb01nition, the nth-order contributions to the selfenergy are generated by a connected and one-particle irreducible contraction of n interaction operators (weighted with the appropriate combinatorial factor 1/n!). Two \ufb01eld vertices stay uncontracted as connectors to the free Green function lines contacting the self-energy. For example, the \ufb01rst-order contribution is given by (exercise) $ab # \u03a3(1) p = \u2212\u03b4ab g Ld \u239b \u239d 1 N \u239e p G0,p + p G0,p\u2212p \u23a0, where the \ufb01rst (second) contribution corresponds to the \ufb01rst (second) diagram in the selfenergy expansion of Fig. 5.8. EXERCISE Represent the second-order contribution \u03a3(2) in terms of Green functions. Once the self-energy has been computed to any desired order, the result is substituted into Eq. (5.39) and one obtains the Green function. 32 The conservation of global momentum of the theory implies (exercise: think about it!) that, like the Green function, the self-energy depends only on the di\ufb00erence of its coordinate arguments. 226 Perturbation theory INFO The critical reader will note that there are some problems with the line of argument above. Firstly, we have tacitly ignored the issue of combinatorics. (How do we know that once we have plugged the expansion of the self-energy into the Dyson equation we get the same result as a brute-force direct expansion of the Green function would have obtained?) To understand that the two-step program \u2013 \u201c\ufb01rst compute the self-energy, and then substitute into the Dyson series\u201d \u2013 indeed produces correct results, let us consider the nth-order contribution to the expansion of the Green function with its overall combinatorial factor 1/n!. Now imagine that we want to distribute those diagrams that contain, say, one free internal Green function over two self-energy operators according to the expression G0\u03a3G0\u03a3G0. Assuming that the \ufb01rst self-energy operator is of order m < n and the second one of order n \u2212 m, we notice that there are possibilities to distribute the interaction vertices over the two self-energies. That means that we obtain an m! (n\u2212m)!. But 1/m! and 1/(n \u2212 m)! are precisely the overall combinatorial factor of 1 n! combinatorial factors that appear in the de\ufb01nition of an mth- and an (n \u2212 m)th-order selfenergy operator, respectively. Arguing in reverse, we conclude that the prescription above indeed produces the correct comb", "inatorics second objection concerns the consistency of the expansion, i.e. the nth-order expansion of the self-energy is, of course, by no means equivalent to nth-order expansion of the Green function, nor to any speci\ufb01c order of the expansion. Indeed, when working with the concept of a self-energy, structuring the expansion according to its order in the interaction operator does not make much sense. We should rather focus on the summation of speci\ufb01c in\ufb01nite-order diagram classes as exempli\ufb01ed in the previous section and discussed in more general terms below. Large-N expansion So far we have not made reference to the N -component structure of the theory. However, let us now assume that N is very large, i.e. that we may be content with an expansion of the Green function to leading order in 1/N. This condition can be made explicit by sending N \u2192 \u221e and declaring limN\u2192\u221e Gaa p to be our observable of interest. The limit of large N entails a drastic simpli\ufb01cation of the diagrammatic expansion. Each interaction vertex comes with an overall factor of 1/N which must be compensated by a summation over \ufb01eld components to produce a contribution that survives the limit N \u2192 \u221e. This condition removes numerous diagrams contributing to the series. For example, in the Green function expansion of Fig. 5.8 only the \ufb01rst, third, eighth and ninth diagrams survive the limit. In all other contributions, interaction and Green function lines are interwoven in a way that does not leave room for one \ufb01eld-index summation per interaction vertex. Max Born, 1882\u20131970 Worked on the mathematical basis of quantum mechanics. Amongst his major contributions was the interpretation of the absolute square of the wave function as a measure of the probability of \ufb01nding the particle at a given location. Born shared the 1954 Nobel Prize in Physics with Walter Bothe for \u201cfor his fundamental research in quantum mechanics, especially for his statistical interpretation of the wavefunction.\u201d (Image c The Nobel Foundation.) 5.3 In\ufb01nite-order expansions 227 Inspection of the series shows that only diagrams void of crossing interaction lines (cf. the \ufb01gure on the right) survive the limit of large N. The approximation \u2013 indeed in the limit of in\ufb01nite N it becomes exact \u2013 that retains only these contributions is commonly called the non-crossing approximation (NCA). More poetically, the diagrams contributing to the reduced expansion are sometimes called \u201crainbow diagrams.\u201d... = + + = Importantly, the NCA self-energy can be computed in closed form. All one has to realize is that the summation over all rainbow diagrams amounts to substitution of the full NCA Green function under a single interaction line (exercise). Using the fact that in the NCA the self-energy is proportional to unity in the \ufb01eld-index space, we can express this fact through the formula \u02c6\u03a3p NCA= \u2212gL\u2212d Gp = \u2212gL\u2212d p2 + r \u2212 \u03a3p \u22121. (5.40) p p In the literature, this equation goes under the name self-consistent Born approximation. It is a \u201cBorn approximation\u201d because, formally, it resembles a \ufb01rst-order perturbative correction (the overall factor of g). The approximation is \u201cself-consistent\u201d because the selfenergy recursively appears on the right-hand side of the equation again, i.e. the equation is in fact not of \ufb01rst but of in\ufb01nite order. INFO Although the objective of the present section is to expose general structures, let us brie\ufb02y review the solution of a Born equation. To keep things simple, let us assume that we are dealing with the low-energy approximation to some microscopic model, i.e. that the momentum summations must be cut o\ufb00 at some upper limit \u039b. We further make the assumption (to be checked self-consistently) that the solution for the self-energy operator will turn out to be momentum-independent: \u03a3p = \u03a3. This leads to the expression \u03a3 \u2248 \u2212g \u039b ddp (2\u03c0)d 1 p2 + r \u2212 \u03a3. The evaluation of the integral depends on dimensionality and on the analytical structure of the energy denominator. For example, taking d = 2 and assuming that the parameter r is much smaller than the self-energy induced by scattering \u2013 an assumption also to be checked self-consistently later on \u2013 we obtain \u03a3 \u2248 \u2212 g 4\u03c0 \u039b2 0 d(p2) 1 p", "2 \u2212 \u03a3 \u2212 g 4\u03c0 ln. \u2212 \u039b2 \u03a3 A solution \u03a3(g, \u039b) can now be sought by approximate analytical methods, or graphically. (One plots both sides of the equation as functions of \u03a3 and seeks a crossing point.) However, for our present discussion, more important than the detailed dependence of \u03a3 on the input parameters g and \u039b is the principal meaning of the self-energy: apparently, \u03a3 adds to the parameter r of the naked Green function. (Notice that the solution of the equation determining \u03a3 will be negative.) Remembering that r \u223c \u03be\u22121/2, one concludes that the interaction operator lowers the spatial correlation of the system. This is indeed what one should expect intuitively: scattering due to interactions acts as a source of \u201cdisorder\u201d inside the system. 228 Perturbation theory (p,a) (p,a) ( (p\u2032,b) (p+q,a) C(4) = = (p,a) (p\u2032,b) (p+q,a) (p\u2032+q,b) + + = = (p+q,a) (p\u2032+q,b) + + +... + +... = + + +... Figure 5.9 Expansion of the four-point function. Notice that the arrows represent the full Green function, i.e. all diagrams \u201crenormalizing\u201d the two-particle subunits of the diagram are automatically included. At this stage, it is worthwhile to take a step back and see what we have achieved. We have managed to compute the Green function to in\ufb01nite order in an expansion in the set of \u201crelevant\u201d diagrams. How does that \ufb01t together with what has been said in Section 5.1 about the \u201casymptotic\u201d nature of perturbative series? In fact, the exponential proliferation of the number of diagrams, i.e. the mechanism that led to the eventual breakdown of the perturbative expansion, is blocked by the limit N \u2192 \u221e. Only subclasses of diagrams, with far fewer members, contribute and the series remains summable. The large-N principle is actually not limited to the expansion of the Green function. To illustrate the point, let us brie\ufb02y consider the expansion of the four-point correlation function C (4) q = 1 N 1 L2d ab p,p \u03c6a p\u03c6a \u2212p\u2212q\u03c6b \u2212p \u03c6b p+q. (5.41) In the next chapter we will see that objects of this architecture represent the most important information carriers of the theory. Unlike the Green function discussed previously, they relate directly to observable quantities. In the many-body literature, the four-point function is described as a two-particle propagator, indicating that it describes the joint propagation of two particles. Leaving a more substantial discussion of the meaning of this object to the next chapter, we concentrate here on the formal aspects of its perturbative expansion. The structure of the expansion of the four-point function is shown in Fig. 5.9 where, for simplicity, momentum/component-indices are indicated only once. The simplest diagram contributing to the expansion consists of just two Green functions. It encapsulates all dis\u223c \u03b4ab\u03b4pp contributing to C (4). All \u2212p\u2212q\u03c6b \u03c6a connected contractions, i.e. p\u03c6b \u03c6a : 9 9 : p+q \u2212p 5.3 In\ufb01nite-order expansions 229 other contractions simultaneously involve all four \ufb01eld operators, i.e. they contain interaction lines between the Green functions. The sum of all these contributions is represented by the diagram containing the hatched surface. A few low-order contributions to the expansion are explicitly shown in the second line. Notice that all arrows appearing in these diagrams are fat. This means that diagrams \u201cdressing\u201d the two-particle sub-units of the expansion are automatically included. For example, the second contribution, containing just a single interaction line between the two-particle propagators, in fact represents an entire series of diagrams obtained by substituting the expansion of Fig. 5.8 for the full Green function. (In an analytical calculation, one will take account of these contributions simply by substitution of the self-energy renormalized Green function for each arrow.) EXERCISE Write down the analytical expressions contributing to the low-order diagrams shown in the \ufb01rst and second lines of the \ufb01gure. All diagrams involving interactions between the two Green functions have the common factor that they contain four external \u201clegs,\u201d i.e. the Green function connectors to the external \ufb01eld operators. If you", " imagine these legs removed, you end up with a \u201ccore contribution\u201d often called thevertex. (The designation is motivated by the fact that the \ufb01rst contribution to the vertex is, indeed, the contracted interaction vertex of the action; see diagram no. 2.) In Fig. 5.9, the vertex is denoted by a tightly hatched area. As with the Green function, the expansion of the vertex can be given some structure. To this end, notice that some of the diagrams contributing to the vertex (e.g. the second diagram in the second line with external legs removed) can be cut into two just by cutting two internal Green function lines. Vertex diagrams of this type are called two-particle reducible, by analogy with the \u201cone-particle reducible\u201d contributions to the expansion of the Green function. As with the expansion of the latter, one can lump all irreducible contributions to the vertex (e.g. the last diagram in the \ufb01rst line, or the \ufb01rst diagram in the second line, external legs removed) into a structural subunit called the irreducible vertex. In Fig. 5.9, that unit is denoted by a lightly hatched area. The \ufb01rst two diagrams contributing to the irreducible vertex are shown in the fourth line of the \ufb01gure. Here, one can see that the irreducible vertex plays a role similar to the self-energy of the Green function. Expressed in terms of the irreducible vertex, the expansion assumes the regular form shown in the third line. To represent these graphical relations in analytical form, we denote the full vertex by the symbol \u02c6\u0393 = {\u0393aa,bb } and the irreducible verp,p,q tex by \u02c6\u03930 = {\u0393aa,bb } where the 0,p,p,q indices a, a, b, b keep track of the index labels carried by the four Green functions entering the vertex. (Although we have de\ufb01ned our correlation function in such a way that a = a, b = b, a generalization Hans Albrecht Bethe, 1906\u2013 2005 Nobel Laureate in physics in 1967 for his \u201ccontributions to the thereactions, espeory of nuclear cially his discoveries concerning the energy production in stars.\u201d As well as nuclear matter, he has also contributed substantially to atomic and condensed matter physics.\u201d (Image c The Nobel Foundation.) 230 Perturbation theory to a four-fold index label is necessary to formulate the recursion Eq. (5.42).) The three momentum arguments represent the momenta of the Green functions connecting to the vertex operators, as indicated explicitly in the \ufb01rst line of the \ufb01gure. (Remember that the theory has overall momentum conservation, i.e. three momentum arguments su\ufb03ce to unambiguously \ufb01x the momentum dependence of \u0393 and C (4).) The content of the third line of the \ufb01gure can then be expressed in terms of a closed recursion relation: p,p,q = \u0393aa,bb \u0393aa,bb 0,q,p,p + 1 Ld c,c,p \u0393aa,cc 0,p,p,q Gc p Gc p+q \u0393cc,bb p,p,q. (5.42) Equations of this type are (often) called Bethe\u2013Salpeter equations. Comparison with Eq. (5.38) shows that this equation appears to be conceptually similar to the Dyson equation for the one-particle Green function.33 Indeed, the principle behind most recursion relations of perturbation theory is a structure like \u02c6X = \u02c6X0 + \u02c6X0 \u2217 \u02c6Z \u2217 \u02c6X, (5.43) where \u02c6X is our object of interest (e.g. \u02c6\u0393), \u02c6X0 its \u201cfree\u201d version, \u02c6Z a subunit that is, in some sense, \u201cirreducible\u201d ( \u02c6\u03a3 or \u02c6\u03930GG), and \u2217 some generalized matrix convolution. C(4) = + + +... Figure 5.10 Large-N expansion of the two-particle correlation function into a \u201cladder structure\u201d. Owing to the importance of the two-particle propagator, the solution of Bethe\u2013Salpeter type equations is a central issue of many areas of many-body physics. In most cases only approximate solutions can be obtained. With our present example, \u201capproximate\u201d means that one sends", " N to large values and seeks a solution to leading order in N \u22121. In that limit, the only surviving contribution to the irreducible vertex is the \ufb01rst, i.e. a plain interaction line (see Fig. 5.10). As with the self-energy operator discussed in the previous section, all diagrams with \u201centangled\u201d interaction lines are frustrated in the sense that we do not have as many index summations as interaction constants \u223c g/N. Such contributions vanish in the limit of large N. We thus conclude that the Bethe\u2013Salpeter equation assumes the simple form p,p,q = \u2212 g \u0393ab N Ld \u2212 g N 1 Ld c,p Ga p Gc p+q \u0393cb p,p,q, 33 We can make the analogy perfect by de\ufb01ning a \u201cone-particle vertex\u201d \u02c6\u0393(1) = \u02c6G(0)\u22121[ \u02c6G \u2212 \u02c6G(0)] \u02c6G(0)\u22121. Inspection of the second equality of Fig. 5.8 shows that the expansion of \u0393(1) starts and ends with a self-energy operator, i.e. the \ufb01rst free Green function line G0 is removed, and so are the two external G0s connecting to the self-energy operator. In direct analogy to Eq. (5.42), the analytical formula for \u0393(1) then reads \u02c6\u0393(1) = \u03a3 + \u03a3G0 \u02c6\u0393(1). 5.3 In\ufb01nite-order expansions 231 \u03b4bb \u03b4aa where \u0393ab \u2261 \u0393aa,bb.34 This equation can be simpli\ufb01ed even further by making the ansatz \u0393ab p,p,q = \u0393q, where \u0393 is independent of discrete indices and input momenta p and p. (When solving a perturbative recursion relation, it is always a good idea to try an ansatz of maximal simplicity, i.e. one that is no more complex than the constituting elements of the equation.) Then, \u0393q = \u2212 g N Ld \u2212 gPq\u0393q \u21d2 \u0393q = \u2212 g N Ld 1 1 + gPq, (5.44) where we have introduced the abbreviationP q = 1 p GpGp+q. In principle, one may now Ld proceed by substituting the large-N expansion of the Green function (5.39) and computing the function Pq by integration over p. This would produce a closed expression for \u0393 and, by virtue of the relation C (4) q = Pq 1 Ld + N \u0393qPq, our correlation function. Since the emphasis in this section is on conceptual aspects of perturbation theory, we will not pursue the analysis to its very end. (For an analysis of the Bethe\u2013Salpeter equation in a context more interesting than \u03c64-theory, see the discussion of the quantum disorder problem in the next chapter.) Yet there is one aspect of the expression for \u0393q worth noticing. Consider Pq expanded as a Taylor series in q and focus on the zeroth-order contribution, P0. From the de\ufb01nition of the Green function (5.39) we have (5.40) = \u2212g\u22121\u2202\u03a3\u03a3 =\u2212 g\u22121. (5.45) P0 = 1 Ld p G2 p = \u2202\u03a3 1 Ld Gp p When substituted into the formula for \u0393, this shows that for small momenta the expansion of the numerator of the vertex starts with a power of q. (By symmetry, the \ufb01rst nonvanishing contribution will be of O(q2).) This means that both the vertex and the four-point correlation function are long-ranged objects, i.e. unlike the Green function, they decay not exponentially, but as a power law. The long-range character of the four-point function has observable consequences, as discussed in the next chapter. Summarizing, our discussion of the two- and four-point functions, and of the RPA theory of the interacting electron gas has shown that, if a large parameter is present, (a) relevant subclasses of perturbative contributions can be identi\ufb01ed and (b) they can be summed to in\ufb01nite order. As a matter of fact, there are not too many of these summable diagram classes: ring-diagrams, rainbow-diagrams and ladder-diagrams nearly exhaust the set of \u201cfriendly\u201d corrections, amenable to analytical summation. Notice that, so", " far, we have largely restricted ourselves to the discussion of abstract summation schemes; i.e. we still need to learn more about the way intermediate results like Eq. (5.39) or (5.44) can be translated into concrete physical information. These aspects are discussed \u2013 on applications more rewarding than 34 Here we have used the fact that the large-N approximation of the irreducible vertex forces the two input indices a, a to be equal (and the same for the output indices). As a word of caution, strictly speaking, our largeN approximation of the irreducible vertex explicitly uses the fact that the input/output indices entering our de\ufb01nition of the vertex are equal, i.e. should we compute a correlation function where the two input/output indices are di\ufb00erent (a rare occurrence in realistic applications), the large-N approximation of the four-point functions would no longer assume the simple form of a regular ladder. 232 Perturbation theory plain \u03c64-theory \u2013 a little later. However, at this point we leave the discussion of formal perturbation theory. While a state-of-the-art exposition of the subject would require much more space \u2013 for pedagogical and condensed-matter-oriented texts on perturbative methods, see Pines and Nozi`eres, and Schulman35 \u2013 the material introduced in this section su\ufb03ces for nearly all purposes of the present text. Turning back to our prototypical problem of the interacting electron gas, in the next chapter we extend the purely diagrammatic methods discussed above to a more powerful hybrid approach comprising perturbative concepts in combination with functional methods. 5.4 Summary and outlook This concludes our preliminary introduction to the concepts of perturbation theory. We have seen that general perturbative expansions mostly have the status of \u201casymptotic\u201d rather than convergent series. We have learned how to e\ufb03ciently encode perturbative series by graphical methods and how to assess the \u201cimportance\u201d of individual contributions. Further, we have seen how the presence of a large parameter can be utilized to \ufb01rmly establish in\ufb01nite-order expansions. A number of recursive techniques have been introduced to sum up diagram sequences of in\ufb01nite order. However, a second look at the discussion of the previous sections shows that the central tool, the functional integral, did not play much of a role. All it did was to provide the combinatorial framework of the perturbative expansion of correlation functions. However, for that we hardly need the full machinery of functional integration. Indeed, the foundations of the perturbative approach were laid down in the 1950s, long before people even began to think about the conventional path integral. (For a pure operator construction of the perturbative expansion, see the problem set.) More importantly, the analysis so far has a serious methodological weakness: all subclasses of relevant diagrams had the common feature that they contained certain \u201csub-units,\u201d more structured than the elementary propagator or the interaction line. For example, the RPA diagrams were organized in terms of polarization bubbles, the NCA diagrams had their rainbows, and the ladder diagrams their rungs. Within the diagrammatic approach, in each diagram these units are reconstructed \u201cfrom scratch,\u201d i.e. in terms of elementary propagators and interaction lines. However, taking seriously the general philosophy on information reduction declared at the beginning of the text, we should strive to make the \u201cimportant\u201d structural elements of an expansion our elementary degrees of freedom. This program is hardly feasible within purely diagrammatic theory. However, the functional integral is ideally suited to introducing degrees of freedom hierarchically, i.e. trading microscopic objects for entities of higher complexity. The combination of functional integral techniques with perturbation theory presents a powerful theoretical machinery which is the subject of the next chapter. 35 See D. Pines and P. Nozi`eres, The Theory of Quantum Liquids \u2014 Normal Fermi Liquids (Addison-Wesley, 1989), and L. S. Schulman, Techniques and Applications of Path Integration (Wiley, 1981). 5.5 Problems 5.5 Problems 233 Technical aspects of diagrammatic perturbation theory Taking the second-order expansion of the \u03c64 Green function as an example, it is the purpose of this problem to discuss a number of technical aspects relating to the classi\ufb01cation and the combinatorics of diagrams. (a) Show that the full contraction of 9G0(x \u2212 x)G4! 1 0(0) + 72G0(x \u2212 x)G2 0(y \u2212 y)G2! 2 0(0) generates the 945 terms + 24G0(x \u2212 x)G4 0(y \u2212 y)! 3 9 \u03c6(x)", "\u03c64(y)\u03c64(y)\u03c6(x) : +! 4 36G0(x \u2212 y)G0(x \u2212 y)G3 + 144G0(x \u2212 y)G0(x \u2212 y)G2 0(0) 0(y \u2212 y)G0(0)! 5 + 96G0(x \u2212 y)G0(x \u2212 y)G3! 6 + 144G0(x \u2212 y)G0(x \u2212 y)G0(y \u2212 y)G2 0(y \u2212 y) 0(0)! 7 + (y \u2190\u2192 y). (5.46) Try to reproduce the combinatorial prefactors. (b) Check explicitly that the disconnected terms cancel against the \u201cvacuum loops\u201d from the expansion of the denominator. (c) Represent the corresponding diagrams in momentum space. Convince yourself that the \u201cKirchho\ufb00 law\u201d discussed in the main text su\ufb03ces to unambiguously \ufb01x the result. Answer: (a) The seven distinct pairings of the ten \u03c6-\ufb01elds are shown in Fig. 5.2 (and referred to here as diagrams 1\u20137 in the order in which they appear in the \ufb01gure). By way of example, let us consider the combinatorial factor of diagram 2. There are 3 \u00b7 2 = 6 ways to pair two \ufb01elds \u03c6(y) and the same for\u03c6 (y). From these two contractions, we obtain a factor (6G(0))2. Then, there are two possible ways of pairing the two remaining \u03c6(y) with the two remaining \u03c6(y) leading to a factor 2G2 0(y \u2212 y). Finally, the pairing of \u03c6(x) with \u03c6(x) obtains a single factor G0(x \u2212 x). Multiplying all contributions, we obtain term 2. (b) To obtain all terms contributing to G(2), both numerator and denominator of Eq. (5.14) have to be expanded to second order (where the identi\ufb01cation X[\u03c6] =\u03c6 (x)\u03c6(x) is understood). In symbolic notation, N0 + N1 + 1 1 + D1 + 1 2 N2 + \u00b7 \u00b7 \u00b7 2 D2 + \u00b7 \u00b7 \u00b7 N0 + N1 \u2212 N0D1! G(1) + 1 2 N2 \u2212 N1D1 + N0D2 1! G(2) \u2212 1 2. N0D2 234 Perturbation theory The expression for N2 is given by formula (5.46). From the main text, we know that N1 = 3G0(x \u2212 x)G2 0(0). Further, N0 = G0(x \u2212 x). Finally, D2 reads 0(0) + 12G0(x \u2212 y)G0(0)G0(y \u2212 x) and D1 = 3G2 9 \u03c64(y)\u03c64(y) : = 9G4 0(0) + 72G2 0(0)G2 0(y \u2212 y) + 24G4 0(y \u2212 y). Collecting all the terms, we obtain N2 \u2212 18G0(x \u2212 x)G4 0(0)G0(y \u2212 x) 0(0) \u221272G0(x \u2212 y)G3! \u2212(4+y\u2194y) +18G0(x \u2212 x)G4 0(0) \u221272G0(x \u2212 x)G2! \u22122 0(0)G2 0(y \u2212 y) 0(0) \u22129G0(x \u2212 x)G4! \u22121 \u221224G0(x \u2212 x)G4 0(y \u2212 y)! \u22123 = 144G0(x \u2212 y)G0(x \u2212 y)G2! 5 + 144G0(x \u2212 y)G0(x \u2212 y)G0(y\u2212 y)G2 0(y \u2212 y)G0(0) 0(0)! 6 + 96G0(x \u2212 y)G0(x \u2212 y)G3 0(y\u2212 y)! 7 +(y \u2190\u2192 y), i.e. the set of connected diagrams shown in Fig. 5.4. (c) This translates to a straightforward exercise in Fourier transformation. Self-consistent T-matrix approximation As well as scattering from lattice vibrations and other electrons, the low-temperature conductivity of a metal is heavily in\ufb02uenced by defects or impurities. The general role of weak", " impurity scattering is discussed in some detail in Section 6.5. As a warm-up to our more comprehensive discussion below, the aim of the present problem is to determine the scattering rate imposed by a collection of isolated impurities using diagrammatic perturbation theory. Consider a system of spinless electrons subject to a random pattern of Nimp non-magnetic scattering impurities and described by the Hamiltonian \u02c6H = \u02c6H0 + \u02c6Himp, where \u02c6H0 = k kc \u2020 kck, \u02c6Himp = v0ad ddr Nimp i=1 \u03b4(r \u2212 Ri)c\u2020(r)c(r) = v0ad Nimp i=1 c\u2020(Ri)c(Ri), and we have assumed that an impurity at positionR i creates a local potential v0ad\u03b4(r\u2212Ri). (Here a is a constant, of dimensionality [length], identi\ufb01ed roughly with the extension of the impurity.) 5.5 Problems 235 Our aim is to compute the single-particle Green function Gn(r \u2212 r) \u2261 c\u2020 n(r)cn(r)imp where n is a Matsubara frequency index and the con\ufb01gurational average...imp \u2261 L\u2212dNimp ddRi is de\ufb01ned by integration over all impurity coordinates. i (a) To begin, let us consider scattering from a single impurity (i.e. Nimp = 1). By developing a perturbative expansion in the impurity potential, show that the Green function can be written as \u02c6Gn = \u02c6G0,n + \u02c6G0,n \u02c6Tn \u02c6G0,n, where \u02c6Tn = \u02c6Himp + \u02c6Himp \u02c6G0,n \u02c6Himp + \u02c6Himp \u02c6G0,n \u02c6Himp \u02c6G0,n \u02c6Himp + \u00b7 \u00b7 \u00b7 imp, (5.47) denotes the T-matrix,36 and \u02c6Gn = {Gn(r \u2212 r)}. Show that the T-matrix equation is (v0ad)\u22121 \u2212 G0,n(0) solved by Tn(r, r) = \u03b4(r \u2212 r)L\u2212d \u22121. (b) For a collection of random impurities, the capacity for multiple scattering makes the concept of the T-matrix de\ufb01ned for the full operator \u02c6Himp unilluminating. Instead, one would like to \ufb01nd an expression for the con\ufb01gurational average of the Green function involving the T-matrix of a single isolated impurity. Presenting the Green function as a series expansion in the impurity potential v, enumerate and classify the contributions entering the con\ufb01gurational average up to third order in v. Organising the series at this order in a diagrammatic expansion, show that a subset (all but one) of the diagrams can be incorporated into a Dyson series expansion, Gk = G0,k +G0,k\u03a3kGk, whereG k \u2261 Gn,k denotes the Green function in a four-momentum representation, \u03a3k = Nimpk| \u02c6Tn|k is the self-energy and \u02c6Tn represents the T-matrix for a single isolated impurity. The Dyson series expansion can be inverted to give Gk = (i\u03c9n \u2212 \u03bek \u2212 \u03a3k)\u22121. When Fourier transformed back to time/space, one may note that the imaginary part of the self-energy can be associated with (one half of) the scattering rate, Im\u03a3k = \u2212 1 2\u03c4 sgn(n) (cf. the Info block on page 379 for further discussion of this point). To understand the sign factor sgn(n), notice that the imaginary part of Tn is determined by the imaginary part of G0,n. However, it is straightforward to verify that sgn(ImG0,n) \u221d sgnn. (For a comprehensive discussion of the analyticity properties of the self-energy, we refer to Chapter 7.) (c) Sometimes, one is interested in the T-matrix for a \ufb01xed (unaveraged) pattern of dilute impurities. In this case, the single-impurity T-matrix is given by the right hand side of Eq. (5.47), without the averaging brackets; it is no longer diagonal in momentum space, i.e. \u02c6Tn = {k| \u02c6Tn|k", "} = {Tn,kk }. Making use of the operator identity (1\u2212 \u02c6O2)\u22121\u2212(1\u2212 \u02c6O1)\u22121 = (1\u2212 \u02c6O2)\u22121( \u02c6O2\u2212 \u02c6O1)(1\u2212 \u02c6O1)\u22121, and de\ufb01ning the real-frequency analytical continuation \u02c6T \u00b1() \u2261 \u02c6T (i\u03c9n \u2192 \u00b1 i\u03b4) (the same for \u02c6G) show that \u02c6T +() \u2212 \u02c6T \u2212() = \u02c6T \u2212()( \u02c6G+ 0 () \u2212 \u02c6G\u2212 0 ()) \u02c6T +(). (5.48) Using this result, show that the scattering rate is given by 1 2\u03c4 \u2261 Im \u03a3( = k + i0) = \u03c0 k |k| \u02c6T +(k)|k|2 \u03b4(k \u2212 k ). (5.49) 36 For a discussion of the physical content of the T-matrix and its connection to the scattering matrix, we refer to any elementary text on scattering theory. 236 Answer: Perturbation theory (a) Representing the Green function as a coherent state path integral and performing the Gaussian integral over the Grassmann \ufb01eld, we obtain the formal result \u02c6Gn = (i\u03c9n \u2212 \u02c6H0 \u2212 \u02c6Himp)\u22121. We next expand \u02c6Gn in \u02c6Himp and compare with the de\ufb01nition of the Tmatrix to identify the series in Eq. (5.47). Substitution of \u02c6Himp = v0ad\u03b4(r\u2212R), where R is the impurity position, we obtain \u02c6T = \u03b4(r\u2212R)\u03b4(r \u2212R)impv0ad(1\u2212v0adG0,n(0, 0))\u22121. Performing the average and rearranging terms we arrive at the result (see Fig. 5.11). = + + + + + + + +... + Figure 5.11 Diagrammatic series expansion of the impurity averaged Green function (for a discussion, see the main text). (b) Presented as a series expansion in the random potential, \u02c6Gn = \u02c6G0,n + \u02c6G0,n \u02c6Himp \u02c6G0,n + \u02c6G0,n \u02c6Himp \u02c6G0,n \u02c6Himp \u02c6G0,n + \u00b7 \u00b7 \u00b7 imp. Turning to a momentum space representation and using the fact that k| \u02c6Himp|k = v0 (a/L)d i eiRi(k\u2212k), the \ufb01rst term in the series assumes the form a L eiRi(k\u2212k)impG0,k = G0,kv0 Nimp d a L G(1) k,k = G0,kv0 d G0,k\u03b4kk. i At second order, we obtain G(2) k,k = G0,kv2 0 = G0,kv2 0 2d 2d k 6 i,i Nimp a L a L eiRi(k\u2212k)G0,k eiRi (k\u2212k)impG0,k 7 G0,k + N 2 impG0,k G0,k\u03b4kk, k where the \ufb01rst (second) contribution describes the two-fold scattering o\ufb00 a single (onefold scattering o\ufb00 two di\ufb00erent) impurities. Notice that, upon averaging, both types of scattering conserve momentum, i.e. Gkk \u221d \u03b4kk. A graphical representation of the two processes is shown in the \ufb01rst line of Fig. 5.11 where crosses represent individual impurities (averaging over position understood) while the dashed lines denote scattering events. At third order in v, one may identify contributions from where all three impurities are distinct, where only two are distinct, and where the scattering occurs at the same impurity (see the bottom line of the \ufb01gure). Of these terms, one may note that the penultimate diagram is special as it involves a \u201ccrossing\u201d of the impurity (i.e. the dotted) lines. Later, in the discussion of the weakly disordered system, we show that such contributions enter with a suppression factor of 1/kF 1, where = vF\u03c4 is the elastic mean free path, kF the Fermi wavevector,", " and the inequality holds in all but the dirtiest metals. Physically, diagrams with crossed lines represent processes wherein an 5.5 Problems 237 electron alternately scatters o\ufb00 di\ufb00erent impurities. Clearly, such events become highly improbable in the limit of a low impurity concentration. Since impurity averages engaging a single impurity involve zero momentum transfer, one may note that the remaining series can be organised into the form of the required Dyson series expansion. Here, since the blocks entering the series do not correlate di\ufb00erent impurities, the self-energy may be identi\ufb01ed as Nimp times the single-impurity T-matrix; the approximation neglecting crossed lines is called the T-matrix approximation. Now, when pursuing to fourth order and higher, one may note the existence of a second class of \u201crainbow\u201d diagrams which (a) do not involve crossed impurity lines and (b) are not included in the Dyson series (see the \ufb01gure above.) These contributions can be incorporated by replacing the bare Green function \u02c6G0,k by the impurity averaged Green \u02c6Gk in the T-matrix expansion Eq. (5.47), an exercise that can be performed only selfconsistently. Such a scheme is known as the self-consistent T-matrix approximation (SCTA). (c) Noting that \u02c6T \u00b1() = \u02c6Himp(1 \u2212 \u02c6G\u00b1 0 () \u02c6Himp)\u22121, a straightforward application of the given identity gives the required expression. Now, applying this result to the self-energy, substituting = k for the energy arguments of the Green functions, and noting that Im k| \u02c6G+ 0 (k)|k = \u03c0\u03b4(k \u2212 k ), one obtains Eq. (5.49), a manifestation of the \u201cGolden Rule.\u201d Kondo e\ufb00ect: perturbation theory In Problem 2.4, we introduced the Kondo e\ufb00ect, describing the interaction of a local impurity with an itinerant band of carriers. (Those unfamiliar with the physical context and background to the problem are referred back to that section.) There we determined an e\ufb00ective Hamiltonian for the coupled system, describing the spin exchange interaction that acts between the local moment of the impurity state and the itinerant band. In the following, motivated by the seminal work of Kondo, we employ methods of perturbation theory to explore the impact of magnetic \ufb02uctuations on transport. In doing so, we elucidate the mechanism responsible for the observed minimum of electrical resistance found in magnetic impurity systems. The perturbation theory of the Kondo e\ufb00ect is one of those problems where the traditional formalism of second quantized operators is superior to the \ufb01eld integral (the reason being that perturbative manipulations on quantum spins are di\ufb03cult to formulate in a \ufb01eld integral language). More speci\ufb01cally, the method of choice would be second quantized perturbation theory formulated in the interaction picture, a concept discussed in great detail in practically any textbook on many-body perturbation theory, but not in this text. For this reason, the solution scheme discussed below is somewhat awkward and surely not very e\ufb03cient (yet fully su\ufb03cient for understanding the essence of the phenomenon). 238 Perturbation theory Although the phenomenology of the Kondo system can be explored in several di\ufb00erent ways ranging from the exact analytical solution of the Kondo Hamiltonian system37 to a variational analysis of the Anderson impurity Hamiltonian,38 in the following we focus on the perturbative scheme developed in the original work of Kondo. Later, in Problem 8.8.5, we will introduce a more advanced approach based on the renormalization group. The starting point of the analysis is the e\ufb00ective sd\u2013Hamiltonian (2.51) introduced in Problem 2.4. Setting \u02c6Hsd = \u02c6H0 + \u02c6Himp where \u02c6H0 = \u2020 k\u03c3ck\u03c3, kc k\u03c3 \u02c6Himp = 2J \u02c6S \u00b7 \u02c6s(r = 0), kk\u03c3\u03c3 c \u2020 and \u02c6s(0) = 1 k\u03c3\u03c3\u03c3\u03c3 ck\u03c3, our aim here is to develop a perturbative expansion in J 2 to explore the scattering properties of the model. Here we have assumed that the exchange constant J is characterized by a single parameter, positive in sign (i.e. antiferromagnetic). In Problem 5.5, we have seen that the scattering rate associated with a system of impu-", " rities is speci\ufb01ed by the T-matrix: 1 2\u03c4 = \u03c0Nimp 9 k,\u03c3 |k, \u03c3|T +()|k, \u03c3|2 : S \u03b4(k \u2212 k ), where the symbol \u00b7 \u00b7 \u00b7 S \u2261 trS(\u00b7 \u00b7 \u00b7) /trS(1) indicates that the calculation of the electron self-energy implies an average over all con\ufb01gurations of the impurity spin. (Indeed, it is the presence of an internal impurity degree of freedom which makes the problem distinct from conventional disorder scattering.) Via the Drude formula, \u03c1 = m/e2n\u03c4 (n is the electron density), the scattering time can be used to estimate the e\ufb00ect of the impurities on the electric resistivity \u03c1 of the system.39 (a) Show that to leading order in the exchange constant J the scattering rate is given by 1 2\u03c4 (F ) = \u03c0cimp\u03bdJ 2S(S + 1), where \u03bd denotes the density of states at the Fermi level. From this result, one can infer a resistivity \u03c1imp \u223c (m/e2F)cimpJ 2S(S + 1), independent of temperature. Taken at this order, scattering o\ufb00 magnetic impurities clearly does not explain the existence of a resistance minimum. (b) At second order in the expansion in \u02c6Himp, the T-matrix assumes the form \u02c6T (2) = \u02c6Himp(+\u2212 \u02c6H0)\u22121 \u02c6Himp. Assuming for simplicity that we are working at zero temperature, \u2020 |\u03a9 is a con\ufb01guration where a single particle with momentum k, |k| > pF, |k, \u03c3 = c k\u03c3 and spin \u03c3 is superimposed on the \ufb01lled Fermi sphere. Convince yourself that the state \u02c6Himp|k, \u03c3 is a linear combination of a single-particle state and a two-particle\u2013one-hole 37 N. Andrei, Diagonalization of the Kondo Hamiltonian, Phys. Rev. Lett. 45 (1980), 379\u201382, P. B. Wiegmann, Exact solution of the s\u2013d exchange model (Kondo problem), J. Phys. C. 14 (1981), 1463\u201378. 38 C. M. Varma and Y. Yafet, Magnetic susceptibility of mixed-valence rare-earth compounds, Phys. Rev. B 13 (1976), 2950\u20134. 39 Notice that, besides magnetic impurities, the system will typically contain non-magnetic scattering centers, i.e. \u03c1total = \u03c10 + \u03b4\u03c1, where \u03c10 is the resistivity due to conventional disorder and we used the fact that the inverse scattering times corresponding to di\ufb00erent types of disorder add independently, \u03c4 \u22121 total = \u03c4 \u22121 0 + \u03c4 \u22121. 5.5 Problems 239 S k \u2032,\u03c3 \u2032 S\u2033 p,\u03bc S\u2032 k,\u03c3 S S\u2033 k \u2032,\u03c3 \u2032 S\u2032 k,\u03c3 p,\u03bc Figure 5.12 The two spin-scattering processes contributing to the electron T-matrix at secondorder perturbation theory. The second process involves a two-particle\u2013one-hole con\ufb01guration as an intermediate state. state (see Fig. 5.12). Determine the excitation energies of the two con\ufb01gurations. Use \u2212 \u02c6H0)\u22121 \u02c6Himp|k, \u03c3. Then show that the real part of the this result to compute (+ k second-order matrix element is given by Re k, \u03c3| \u02c6T (2)|k, \u03c3 = J 2 p 1 \u03bek \u2212 \u03bep [S(S + 1) \u2212 S \u00b7 \u03c3\u03c3\u03c3 (\u0398(\u03bep) \u2212 \u0398(\u2212\u03bep))]. (5.50) (Hint: (i) In deriving this result, all \u201cvacuum contributions\u201d have to be discarded [for they cancel against the partition function denominator of the properly normalized perturbation series]. The de\ufb01ning property of a vacuum contribution is that it factorizes into two independent ground state matrix elements. For example, \u03a9|ck\u03c3 \u02c6Himp(+ \u2212 k\u03c3 |\u03a9 \u00b7 \u03a9| \u02c6Himp|\u03a9 is of this type. (ii) Note that (\u02c6S \u00b7 \u03c3)2 = \u02c6S2 \u2212 \u03c3 \u00b7 \u02c6S. (iii) \u02c6H0)\u22121c Taking the real part of", " the T- matrix amounts to omitting the in\ufb01nitesimal imaginary increment in the energy denominators.) At \ufb01nite temperature, the Pauli blocking factors \u0398() \u2192 (1\u2212nF()) and \u0398(\u2212) \u2192 nF() generalize to Fermi functions and we obtain \u2020 Rek, \u03c3| \u02c6T (2)|k, \u03c3 = J 2 # p 1 \u03bek \u2212 \u03bep S(S + 1)\u03b4\u03c3\u03c3 + (2nF(\u03bep) \u2212 1)\u02c6S \u00b7 \u03c3\u03c3\u03c3. (5.51) $ Neglecting the \ufb01rst (non-singular, why?) contribution, one may note that the second can be absorbed into a renormalization of the term \ufb01rst-order in J derived above, viz. 1 2\u03c4 = \u03c0\u03bdcimpJ 2 e\ufb00 S(S + 1), where Je\ufb00 = J(1 + 2Jg(, T )) and the function g(, T ) = P depends sensitively on the bandwidth 2D of the itinerant electrons \u03bd 2 and on the energy \u2261 k of the reference state. Noting that \u2212D d\u03be tanh(\u03b2\u03be/2) \u03be\u2212 D lim \u21920 g(, 0) = \u03bd ln, D lim T \u21920 g(0, T ) = \u03bd ln D kBT, the e\ufb00ective exchange constant can be written as 1 + 2\u03bd(F )J ln Je\ufb00 = J. D max(, T ) 240 Perturbation theory On substituting into the expression for the scattering rate, one \ufb01nds that the resistivity diverges logarithmically with temperature, \u03c1(T ) \u03c1(0) 1 \u2212 4\u03bdJ ln. T D Finally, on combining with the e\ufb00ect of electron scattering from phonons, which provides a contribution to the resistivity that scales as T 5, one may con\ufb01rm that the resistance develops a shallow minimum. Although the perturbation theory suggests a divergence of the resistivity with temperature, the result remains valid only up to the characteristic Kondo temperature scale, TK = D exp \u2212 1 2\u03bdJ, i.e. a temperature where the logarithmic correction becomes of the order of the \ufb01rstorder term. Experimentally, it is found that, at temperatures T TK, the resistance saturates. In fact, the low-energy physics can be reconciled through the formation of a resonance in which an electron in the itinerant band combines with the electron on the impurity site to form a singlet bound state e\ufb00ectively screening the magnetic impurity. Formally, one can explore the formation of the resonance state by developing |\u03a9, a variational analysis based on the trial wavefunction |\u03c8 = where |\u03a9 represents a \ufb01lled Fermi sea. In doing so, one may con\ufb01rm that a bound state develops with a binding energy \u0394K = TK set by the Kondo temperature. k\u03c3 \u03b1\u03b1d\u2020 \u03b10 + \u03c3ck\u03c3 % & INFO Although it was originally conceived for the problem of magnetic impurities in metals, lately the e\ufb00ects of Kondo resonance formation have been observed in arti\ufb01cial quantum dot structures.40 Here a microscopic quantum dot (with dimensions of ca. 1 \u03bcm) is sandwiched between two metallic leads. In the so-called Coulomb blockade regime (see Problem 6.7), the charging energy of the dot plays the role of the local Hubbard interaction while the leads act as the Fermi sea. The development of a Kondo resonance below TK appears as a signature in the quantum transport through the dot. In particular, for temperatures T < TK, the di\ufb00erential conductance dI/dV shows a peak corresponding to the suppression of scattering o\ufb00 the impurity state. 40 D. Goldhaber-Gordon, H. Shtrikman, D. Mahalu, et al., Kondo e\ufb00ect in a single-electron transistor, Nature 391 (1998), 156\u20139. Answer: 5.5 Problems 241 (a) To leading order in \u02c6Himp, the T-matrix is given by \u02c6T \u02c6T (1) = \u02c6Himp. Using the fact that k, \u03c3| \u02c6Himp|k, \u03c3 = JL\u2212d\u03c3|S \u00b7 \u03c3|\u03c3 we then obtain? > 1 2\u03c4 \u03c0Nimp k,\u03c3 = \u03c0cimpJ 2L\u2212d |k, \u03c3| ", "\u02c6Himp|k, \u03c3|2 S \u03c3|S \u00b7 \u03c3|\u03c3\u03c3|S \u00b7 \u03c3|\u03c3S \u03b4(k \u2212 k ) \u03c3 = \u03c0cimp\u03bdJ 2\u03c3|S \u00b7 \u03c3S \u00b7 \u03c3|\u03c3S = \u03c0J 2S(S + 1)cimp\u03bd, k \u03b4(k \u2212 k ) where in the last line we have used the fact that \u03c3)S = S(S + 1)\u00b7 1. k \u03b4(k \u2212 k ) =L d\u03bd and (S \u00b7 \u03c3)(S \u00b7 (b) Substituting the explicit form of \u02c6Himp and using the anti-commutation relations of fermions, we obtain (summation convention) \u02c6Himp|k, \u03c3 = J c\u2020 p1\u03bc1 (S \u00b7 \u03c3\u03bc1\u03bc2 )cp2\u03bc2 c p1p2 = J p1 (S \u00b7 \u03c3\u03bc1\u03c3 )c\u2020 p1\u03bc1 |\u03a9 \u2212J \u2020 k,\u03c3 |\u03a9 (S \u00b7 \u03c3\u03bc1\u03bc2 )c\u2020 p1\u03bc1 p1p2 c \u2020 k\u03c3 cp2\u03bc2 |\u03a9, i.e. a linear combination of a one-particle state and a two-particle\u2013one-hole state. Noting that the energies of the two contributions are given by p1 and p1 +k \u2212p2, respectively, multiplying by the \u201cbra\u201d k, \u03c3| \u02c6Himp, and observing that the overlap between a oneparticle state and a two-particle\u2013one-hole state vanishes, we obtain Rek, \u03c3| \u02c6T (2)|k, \u03c3 = J 2(S \u00b7 \u03c3\u03c3\u03bc 1 )(S \u00b7 \u03c3\u03bc1\u03c3 ) p1p 1 \u03a9|cp \u2020 c 1\u03bc p1\u03bc1 k \u2212 p1 1 |\u03a9 +J 2(S \u00b7 \u03c3\u03bc 2\u03bc 1 \u03a9|c )(S \u00b7 \u03c3\u03bc1\u03bc2 ) \u2020 k\u03c3 cp2\u03bc2 |\u03a9 \u2020 2\u03bc p 2 ck\u03c3cp 1\u03bc k \u2212 p1 1 \u2020 c p1\u03bc1 c \u2212 k + p2 \u0398(\u2212\u03bep) k \u2212 p, p p1p2p 1p 2 \u0398(\u03bep) k \u2212 p p \u2192 J 2(S \u00b7 \u03c3\u03c3\u03bc)(S \u00b7 \u03c3\u03bc\u03c3 ) + J 2(S \u00b7 \u03c3\u03bc\u03c3 )(S \u00b7 \u03c3\u03c3\u03bc) where the arrow indicates that vacuum contributions have been discarded. Application of the spin identity then leads directly to Eq. (5.50). 6 Broken symmetry and collective phenomena Previously, we have seen how the \ufb01eld integral method can be deployed to formulate perturbative approximation schemes to explore weakly interacting theories. In this chapter, we will learn how elements of the perturbative approach can be formulated more e\ufb03ciently by staying \ufb01rmly within the framework of the \ufb01eld integral. More importantly, in doing so, we will see how the \ufb01eld integral provides a method for identifying and exploring non-trivial reference ground states \u2013 \u201cmean-\ufb01elds.\u201d A fusion of perturbative and mean-\ufb01eld methods will provide us with analytical machinery powerful enough to address a spectrum of rich applications ranging from super\ufb02uidity and superconductivity to metallic magnetism and the interacting electron gas. As mentioned in Chapter 5, the perturbative machinery is but one part of a larger framework. In fact, the diagrammatic series already contained hints indicating that a straightforward expansion of a theory in the interaction operator might not always be an optimal strategy: all previous examples that contained a large parameter \u201cN \u201d \u2013 and usually it is only problems of this type that are amenable to controlled analytical analysis \u2013 shared the property that the diagrammatic analysis bore structures of \u201chigher complexity.\u201d (For example, series of polarization operators appeared rather than series of the elementary Green functions, etc.) This phenomenon suggests that large-N problems should qualify for a more e\ufb03cient and, indeed, a more physical formulation. While these remarks appear to be largely methodological, the rationale behind searching for an improved theoretical formulation is, in fact, much deeper. With our previous examples, the perturbative expansion was benign. However, we already saw some glimpses indicating that more drastic things may happen. For example, for frequencies approaching the plasma frequency, the polarization operator of the weakly interacting electron gas developed an instability. The appearance of such instabilities usually indicates that one is formulating a theory around the \u201cwrong\u201d reference state", " (in that case, the uniformly \ufb01lled Fermi sphere of the non-interacting electron gas). Thus, what we would like to develop is a theoretical framework (a) that is capable of detecting the \u201cright\u201d reference states, or \u201cmean-\ufb01elds\u201d of a system, and that enables us (b) to e\ufb03ciently apply perturbative methods around these states and, \ufb01nally, (c) to do this in a language that draws upon the \u201cphysical\u201d rather than the plain microscopic degrees of freedom as the fundamental units. To this end, in the following sections we develop a functional-integral-based approach that meets these criteria. In contrast to the previous chapters, the discussion here is decidedly biased towards concrete application to physically motivated problems. After the formulation 242 6.2 Plasma theory of the interacting electron gas 243 of the general strategy of \ufb01eld-integral-based mean-\ufb01eld methods, the next section addresses a problem that we have encountered before, the weakly interacting electron gas. The exempli\ufb01cation of the new concepts on a known problem enables us to better understand the intimate connection between the mean-\ufb01eld approach and straightforward perturbation theory. In subsequent sections we then turn to the discussion of problems that lie \ufb01rmly outside the range of direct perturbative summation, super\ufb02uidity and superconductivity. 6.1 Mean-\ufb01eld theory Roughly speaking, the functional approach to problems with a large parameter proceeds according to the following program: 1. In the \ufb01rst place, one must identify the relevant structural units of the theory. (That part of the program can be e\ufb03ciently carried out by the straightforward methods discussed above.) 2. Secondly, it is necessary to introduce a new \ufb01eld \u2013 let us call it \u03c6 for concreteness \u2013 that encapsulates the relevant degrees of freedom of the low-energy theory. 3. With this in hand, one can then trade integration over the \u201cmicroscopic \ufb01elds\u201d for an integration over \u03c6, a step often e\ufb00ected by an operation known as the Hubbard\u2013Stratonovich transformation. 4. The low-energy content of the theory can often be explored by subjecting the resulting action S[\u03c6] to a stationary phase analysis. (The justi\ufb01cation for applying stationary phase methods is provided by the existence of a large parameter N 1.) Often, at this stage, instabilities in the theory show up \u2013 an indication of a physically interesting problem! 5. Finally, the nature of the elementary (collective) excitations above the ground state can be explored by expanding the functional integral around the solution of the stationary phase equations \u2013 the \u201cmean-\ufb01eld.\u201d From this low-energy e\ufb00ective action, one can compute physical observables. In the next section, we will illustrate how such a program can be implemented on a speci\ufb01c example, studied earlier by diagrammatic means. 6.2 Plasma theory of the interacting electron gas To begin, let us return to the \ufb01eld theory of the interacting electron gas (see the action de\ufb01ned in section 5.2), S[\u03c8, \u00af\u03c8] = \u00af\u03c8p\u03c3 \u2212i\u03c9n + p \u2212 \u03bc \u03c8p\u03c3 + p2 2m T 2L3 ppq \u00af\u03c8p+q\u03c3 \u00af\u03c8p\u2212q\u03c3 V (q)\u03c8p\u03c3 \u03c8p\u03c3, where the pairwise Coulomb interaction assumes the form V (q) = 4\u03c0e2/|q|2. Here the summation runs over the 4-momenta k = (\u03c9n, k) comprising Matsubara and momentum components and summation over repeated spin indices is implied. Being quartic in the \ufb01elds \u03c8\u03c3, the Coulomb interaction prevents an explicit computation of the \u03c8-integral along the 244 Broken symmetry and collective phenomena lines of the free fermion integrals discussed in the previous chapter. However, it is actually a straightforward matter to reduce, or \u201cdecouple,\u201d the interaction operator, bringing it to a form quadratic in the \ufb01elds \u03c8. Let us multiply the functional integral by the \u201cfat unity\u201d 1 \u2261 D\u03c6 exp \u2212 e2\u03b2 2Ld q \u03c6qV \u22121(q)\u03c6\u2212q, where \u03c6 represents a real bosonic \ufb01eld variable, and a normalization constant has been absorbed in the de\ufb01nition of the functional measure D\u03c6. Notice that the summation runs over a 4-mom", "entum q = (\u03c9m, q) comprising a vectorial momentum and a bosonic Matsubara \u00af\u03c8p\u03c3\u03c8p+q\u03c3 frequency. Employing the variable shift \u03c6q \u2192 \u03c6q + ie\u22121V (q)\u03c1q/\u03b2, where\u03c1 q \u2261 (summation on \u03c3 implied), one obtains p 1 = D\u03c6 exp q 1 Ld \u2212 e2\u03b2 2 \u03c6qV \u22121(q)\u03c6\u2212q + ie\u03c1q\u03c6\u2212q + 1 2\u03b2 \u03c1qV (q)\u03c1\u2212q. The rationale behind this exercise can be seen in the last contribution to the exponent: this term is equivalent to the quartic interaction contribution to the fermionic path integral, albeit with opposite sign. Therefore, multiplication of our unity by Z leads to the \ufb01eld integral Z = D\u03c6 S[\u03c6, \u00af\u03c8\u03c3, \u03c8\u03c3] = \u03c6qq2\u03c6\u2212q + \u00af\u03c8p\u03c3 \u2212i\u03c9n + \u2212 \u03bc \u03b4pp + \u03c8p\u03c3, (6.1) D( \u00af\u03c8\u03c3, \u03c8\u03c3) e\u2212S[\u03c6, \u00af\u03c8\u03c3,\u03c8\u03c3], where \u03b2 8\u03c0Ld q pp p2 2m ie Ld \u03c6p\u2212p denotes the action, i.e. an expression that is free of quartic \ufb01eld interactions of \u03c8\u03c3. Before proceeding, to acquire some intuition for the nature of the action it is helpful to rewrite S \u03b2 ddr e\u2212iq\u00b7r+i\u03c9\u03c4 \u03c6(r, \u03c4 ), one may con\ufb01rm in a real space representation. With \u03c6q = T 0 d\u03c4 that (exercise) S[\u03c6, \u00af\u03c8, \u03c8] = d\u03c4 ddr ) 1 8\u03c0 (\u2202\u03c6)2 + \u00af\u03c8\u03c3 \u2202\u03c4 \u2212 \u22022 2m \u2212 \u03bc + ie Ld \u03c6 * \u03c8\u03c3. Physically, \u03c6 couples to the electron degrees of freedom as a space/time-dependent (imaginary) potential, while the \ufb01rst term re\ufb02ects the Lagrangian energy density associated with the electric component of the electromagnetic \ufb01eld. Before proceeding, let us step back and discuss the general philosophy of the manipulations that led from the original partition function to the two-\ufb01eld representation Eq. (6.1). INFO The sequence of manipulations developed above, i.e. the \u201cdecoupling\u201d of a quartic interaction through an auxiliary \ufb01eld, is known more generally as a Hubbard\u2013Stratonovich transformation (see our discussion of the Ising model on page 196). The essence of the transformation is a straightforward manipulation of a Gaussian integral. To make this point more transparent, let us reformulate the Hubbard\u2013Stratonovich transformation in a notation that is not burdened by the presence of model-speci\ufb01c constants. Consider an interaction operator of the form Sint = V\u03b1\u03b2\u03b3\u03b4 \u00af\u03c8\u03b1\u03c8\u03b2 \u00af\u03c8\u03b3\u03c8\u03b4 (summation convention implied), where \u00af\u03c8 and \u03c8 may be either bosonic or fermionic \ufb01eld variables, the indices \u03b1, \u03b2,... refer to an unspeci\ufb01ed set of quantum numbers, Matsubara frequencies, etc., and V\u03b1\u03b2\u03b3\u03b4 is an interaction matrix element. Now, let us introduce composite operators \u02c6\u03c1\u03b1\u03b2 \u2261 \u00af\u03c8\u03b1\u03c8\u03b2 to rewrite the interaction as Sint = V\u03b1\u03b2\u03b3\u03b4 \u02c6\u03c1\u03b1\u03b2 \u02c6\u03c1\u03b3\u03b4. The notation 6.2 Plasma theory of the interacting electron gas 245 can be simpli\ufb01ed still further by introducing composite indices m \u2261 (\u03b1\u03b2), n \u2261 (\u03b3\u03b4), whereupon the action Sint = \u02c6\u03c1mVmn \u02c6\u03c1n acquires the structure of a generalized bilinear form. To reduce the action to a form quadratic in the \u03c8s, one may simply multiply the exponentiated action by unity, i.e. e\u2212 \u02c6\u03c1mVmn \u02c6\u03c1n = D\u03c6 e\u2212 1 4 \u03c6mV 1 \u22121 mn\u03c6n e\u2212 \u02c6\u03c1mVmn \u02c6\u03c1n, where \u03c6 is bosonic. (Notice that here V \u22121 mn represents the matrix elements of the inverse and not the inverse (Vmn)\u22121 of individual matrix elements.) Finally, applying the variable change \u03c6m \u2192 \u03c6m + 2i(V \u02c6\u03c1)m, where the notation (V", " \u02c6\u03c1) is shorthand for Vmn \u02c6\u03c1n, one obtains exp [\u2212\u02c6\u03c1mVmn \u02c6\u03c1n] = D\u03c6 exp \u2212 1 4 \u03c6mV \u22121 mn\u03c6n \u2212 i\u03c6m \u02c6\u03c1m, i.e. the term quadratic in \u02c6\u03c1 is cancelled.1 This completes the formulation of the Hubbard\u2013 Stratonovich transformation. The interaction operator has been traded for (a) an integration over an auxiliary \ufb01eld (b) coupled to a \u03c8-bilinear (the operator \u03c6m \u02c6\u03c1m). In essence, the Hubbard\u2013Stratonovich transformation is tantamount to the Gaussian integral identity Eq. (3.13) but read in reverse. An exponentiated square is removed in exchange for a linear coupling. (In (3.13) we showed how terms linear in the integration variable can be removed.) To make the skeleton outlined above a well-de\ufb01ned prescription, one has to be more speci\ufb01c about the meaning of the Gaussian integration over the kernel \u03c6mV \u22121 mn\u03c6n, i.e. the integration variables can be real or complex, and V must be a positive matrix (which is usually the case on physical grounds). There is some freedom as to the choice of the integration variable. For example, the factor of 1/4 in front of the Gaussian weight \u03c6mV \u22121 mn\u03c6n was introduced for mere convenience (i.e. to generate a coupling \u03c6m \u02c6\u03c1m free of numerical factors). If one does not like to invert the matrix kernel Vmn, one can scale \u03c6m \u2192 (V \u03c6)m, whereupon the key formula reads e\u2212 \u02c6\u03c1mVmn \u02c6\u03c1n = D\u03c6 e\u2212 1 4 \u03c6mVmn\u03c6n\u2212i\u03c6mVmn \u02c6\u03c1n. EXERCISE Show that the passage from the Lagrangian to the Hamiltonian formulation of the Feynman path integral can be interpreted as a Hubbard\u2013Stratonovich transformation. As de\ufb01ned, the Hubbard\u2013Stratonovich transformation is exact. However, to make it a meaningful operation, it must be motivated by some physical considerations. In our discussion above, we split up the interaction by choosing \u02c6\u03c1\u03b1\u03b2 as a composite operator. However, there is clearly some arbitrariness with this choice. Why not, for example, pair the fermion-bilinears according to ( \u00af\u03c8\u03b1\u03c8\u03b2)( \u00af\u03c8\u03b3\u03c8\u03b4), or otherwise? The three inequivalent choices of pairing up operators are shown in Fig. 6.1 where, as usual, the wavy line with attached \ufb01eld vertices represents the interaction, and the dashed ovals indicate how the \ufb01eld operators are paired. 1 Here we have assumed that the matrix V is symmetric. If it is not, we can apply the relation \u02c6\u03c1mVmn \u02c6\u03c1n \u2261 \u02c6\u03c1T V \u02c6\u03c1 = 1 2 \u02c6\u03c1T (V + V T ) \u02c6\u03c1 to symmetrize the interaction. 246 Broken symmetry and collective phenomena a) (b) (c) Figure 6.1 On the di\ufb00erent channels of decoupling an interaction by Hubbard\u2013Stratonovich transformation. (a) Decoupling in the \u201cdensity\u201d channel; (b) decoupling in the \u201cpairing\u201d or \u201cCooper\u201d channel; and (c) decoupling in the \u201cexchange\u201d channel. The version of the transformation discussed above corresponds to Fig. 6.1(a). That type of pairing is sometimes referred to as decoupling in the direct channel. The designation becomes more transparent if we consider the example of the spinful electron\u2013electron interaction, d\u03c4 ddr ddr \u00af\u03c8\u03c3(r, \u03c4 ) \u00af\u03c8\u03c3 (r, \u03c4 )V (r \u2212 r)\u03c8\u03c3 (r, \u03c4 )\u03c8\u03c3(r, \u03c4 ), Sint = 1 2 i.e. here \u03b1 = \u03b2 = (r, \u03c4, \u03c3), \u03b3 = \u03b4 = (r, \u03c4, \u03c3), and V\u03b1\u03b2\u03b3\u03b4 = V (r \u2212 r). The \u201cdirect\u201d decoupling proceeds via the most obvious choice, i.e. the density operator \u02c6\u03c1(r, \u03c4 ) = \u00af\u03c8\u03c3(r, \u03c4 )\u03c8\u03c3(r, \u03c4 ). One speaks about decoupling in a \u201cchannel\u201d because, as will be elucidated below,", " the propagator of the decoupling \ufb01eld can be interpreted in terms of two Green function lines tied together by multiple interactions, a sequential object reminiscent of a \u201cchannel.\u201d However, more important than the terminology is the fact that there are other choices for \u03c1. Decoupling in the exchange channel (Fig. 6.1(c)) is generated by the choice \u02c6\u03c1\u03b1\u03b4 \u223c \u00af\u03c8\u03b1\u03c8\u03b4 where, in the context of the Coulomb interaction, the reversed pairing of \ufb01eld operators is reminiscent of the exchange contraction generating Fock-type contributions. Finally, one may decouple in the Cooper channel (Fig. 6.1(b)) \u02c6\u03c1\u03b1\u03b3 = \u00af\u03c8\u03b1 \u00af\u03c8\u03b3, \u02c6\u03c1\u03b2\u03b3 = \u02c6\u03c1\u2020 \u03b3\u03b2. Here, the pairing \ufb01eld is conjugate to two creation operators. Below we will see that this type of decoupling is tailored to problems involving superconductivity. The remarks above may convey the impression of a certain arbitrariness inherent in the Hubbard\u2013Stratonovich scheme. Indeed, the \u201ccorrect\u201d choice of decoupling can be motivated only by physical reasoning, not by plain mathematics. Put di\ufb00erently, the transformation as such is exact, no matter what channel we choose. However, later on we will want to derive an e\ufb00ective low-energy theory based on the decoupling \ufb01eld. In cases where one has accidentally decoupled in an \u201cunphysical\u201d channel, it will be di\ufb03cult, if not impossible, to distill a meaningful lowenergy theory for the \ufb01eld \u03c6 conjugate to \u03c1. Although the initial model still contains the full microscopic information (by virtue of the exactness of the transformation) it is not amenable to further approximation schemes. In fact, one is frequently confronted with situations where more than one Hubbard\u2013 Stratonovich \ufb01eld is needed to capture the full physics of the problem. To appreciate this point, consider the Coulomb interaction in momentum space. Sint[ \u00af\u03c8, \u03c8] = 1 2 p1,...,p4 \u00af\u03c8\u03c3p1 \u00af\u03c8\u03c3p3 V (p1 \u2212 p2)\u03c8\u03c3p4 \u03c8\u03c3p2 \u03b4p1\u2212p2+p3\u2212p4. (6.2) 6.2 Plasma theory of the interacting electron gas 247 In principle, we can decouple this interaction in any one of the three channels discussed above. However, \u201cinteresting\u201d physics is usually generated by processes where one of the three unbounded momenta entering the interaction vertex is small. Only these interaction processes have a chance to accumulate an overall collective excitation of low energy (see our previous discussion of the RPA, the interacting electron gas, and many of the examples to follow). It may be instructive to imagine the situation geometrically: in the three-dimensional Cartesian space of free momentum coordinates (p1, p2, p3) entering the vertex, there are three thin layers, where one of the momenta is small, (q, p2, p3), (p1, q, p3), (p1, p2, q), |q| |pi|. (Why not make all momenta small? Because that would be in con\ufb02ict with the condition that the Green functions connecting to the vertex be close to the Fermi surface.) One will thus often choose to break down the full momentum summation to a restricted summation over the small-momentum sublayers: Sint[ \u00af\u03c8, \u03c8] 1 2 p,p,q \u00af\u03c8\u03c3p\u03c8\u03c3p+qV (q) \u00af\u03c8\u03c3p \u03c8\u03c3p\u2212q \u2212 \u00af\u03c8\u03c3p\u03c8\u03c3p+qV (p \u2212 p) \u00af\u03c8\u03c3p+q\u03c8\u03c3p \u2212 \u00af\u03c8\u03c3p \u00af\u03c8\u03c3\u2212p+qV (p \u2212 p)\u03c8\u03c3p \u03c8\u03c3\u2212p+q!. Now, each of these three contributions has its own predestined choice of a slow decoupling \ufb01eld. \u00af\u03c8\u03c3p\u03c8\u03c3p+q, the second in The \ufb01rst term should be decoupled in the direct channel \u02c6\u03c1d,q \u223c \u00af\u03c8\u03c3p\u03c8\u03c3p+q, and the third in the Cooper channel \u02c6\u03c1c,\u03c3\u03c3q \u223c the exchange channel \u02c6\u03c1x,\u03c3\u03c3q \u223c \u00af\u03c8\u03c3p \u00af\u03c8\u03c3\u2212p+q. One thus winds up with an e\ufb00ective theory that contains three", " independent slow Hubbard\u2013Stratonovich \ufb01elds. (Notice that the decoupling \ufb01elds in the exchange and in the Cooper channel explicitly carry a spin structure.) p p p In our discussion of the high-density limit of the electron gas above, we e\ufb00ected a decoupling in the direct channel. That choice is made because, drawing on our previous discussion, we already know in advance that relevant contributions to the free energy of the system are generated by \u00af\u03c8\u03c3p\u03c8\u03c3p+q, whereq is small. If we had not known, RPA-type contraction of operators \u02c6\u03c1d,q \u223c a more careful three-fold Hubbard\u2013Stratonovich decoupling, followed (see below) by a careful analysis of the decoupled action, would have identi\ufb01ed the density channel as relevant. More generally, if in doubt, one should decouple in all available channels, and let the mean-\ufb01eld analysis discriminate between the relevant \ufb01elds. p After this digression on the principles of the Hubbard\u2013Stratonovich transformation, let us now return to the discussion of the electron gas. At the expense of introducing a second \ufb01eld, the Hubbard\u2013Stratonovich transformation provides an action quadratic in the fermion \ufb01elds. In this case, the fermion integration can be undertaken exactly. Making use of the Gaussian functional integral Eq. (4.19), one obtains Z = D\u03c6 e\u2212 \u03b2 8\u03c0L3 q \u03c6qq2\u03c6\u2212q det \u2212i\u02c6\u03c9 + \u02c6p2 2m \u2212 \u03bc + ie Ld \u02c6\u03c6, where, as usual, the circum\ufb02exes appearing in the argument of the determinant indicate that symbols have to be interpreted as operators (acting in the space of Matsubara and Hilbert space components). The standard procedure to deal with the determinants generated at intermediate stages of the manipulation of a \ufb01eld integral is to simply re-exponentiate them. This is achieved 248 Broken symmetry and collective phenomena by virtue of the identity ln det \u02c6A = tr ln \u02c6A, (6.3) valid for arbitrary (non-singular) operators \u02c6A.2 Thus, the quantum partition function takes the form Z = D\u03c6 e\u2212S[\u03c6], where S[\u03c6] = \u03b2 8\u03c0Ld q \u03c6qq2\u03c6\u2212q \u2212 tr ln \u2212i\u02c6\u03c9 + \u02c6p2 2m \u2212 \u03bc + \u02c6\u03c6. ie Ld (6.4) This is as far as purely formal exact manipulations can carry us. We have managed to trade the integration over the interacting Grassmann \ufb01eld \u03c8\u03c3 for an integration over an auxiliary \ufb01eld \u03c6; a \ufb01eld that we believe encapsulates the relevant degrees of freedom of the model. This completes steps 1, 2, and 3 of the general program outlined above. The next step in the program is to subject the action to a stationary phase analysis, i.e. to seek solutions of the set of saddle-point equations such that \u2200q = (q = 0, \u03c9) : \u03b4S[\u03c6] \u03b4\u03c6q!= 0. The solution \u03c6(x, t) \u2194 \u03c6q is commonly referred to as the mean-\ufb01eld. This terminology can be understood by inspection of the argument of the \u201ctr ln\u201d above. The structure \u02c6p2/2m \u2212 \u03bc + ie\u03c6/Ld, where\u03c6 is a \ufb01xed con\ufb01guration (to be determined by solving the saddle-point equations), resembles the Hamiltonian operator of particles subject to some background potential, or \u201cmean\u201d \ufb01eld. The notation on the left-hand side of the saddle-point equations indicates that our original interaction V (q) and, therefore, the decoupling \ufb01eld \u03c6 do not possess a zero-momentum mode (a consequence of charge neutrality). The concrete evaluation of the functional derivative \u03b4S/\u03b4\u03c6 leads us to question how one di\ufb00erentiates the trace of the logarithm of the operator, \u02c6G\u22121[\u03c6] = i\u02c6\u03c9 \u2212 \u02c6p2 2m + \u03bc \u2212 ie Ld \u02c6\u03c6, with respect to its argument. (The notation \u02c6G\u22121 is motivated by its structural similarity to an inverse Green function.) Owing to the presence of the trace, the di\ufb00erentiation can be carried", " out as if \u02c6G were a function:3 (\u03b4/\u03b4\u03c6q)tr ln( \u02c6G\u22121) = tr( \u02c6G(\u03b4/\u03b4\u03c6q) \u02c6G\u22121), where we 2 Equation (6.3) is readily established by switching to an eigenbasis, whereupon one obtains ln det \u02c6A = tr ln \u02c6A, where a are the eigenvalues of \u02c6A and we have used the fact that the eigenvalues of ln \u02c6A are ln a. 3 Consider an operator \u02c6A(x) depending on some parameter x. Let f ( \u02c6A) be an arbitrary function (f ( \u02c6A) = ln \u02c6A in a ln a = the present application). Then \u2202xtr(f ( \u02c6A)) = \u2202x = n n n! n f (n)(0) n! tr( \u02c6An) = n f (n)(0) n! tr((\u2202x \u02c6A) \u02c6An\u22121 + \u02c6A(\u2202x \u02c6A) \u02c6An\u22122 + \u00b7 \u00b7 \u00b7 + \u02c6An\u22121\u2202x \u02c6A) f (n)(0)tr( \u02c6An\u22121(\u2202x \u02c6A)) = tr(f ( \u02c6A)\u2202x \u02c6A), where, in the third equality, we have used the cyclic invariance of the trace. 6.2 Plasma theory of the interacting electron gas 249 omitted the argument [\u03c6] for notational clarity. Then, making use of the identity tr \u02c6G \u02c6G\u22121 = 2 \u02c6Gq1q2 \u03b4 \u03b4\u03c6q q1q2 \u03b4 \u03b4\u03c6q \u02c6G\u22121 q2q1 = \u2212 2ie Ld q1q2 \u02c6Gq1q2 \u03b4q1\u2212q2,q, the saddle-point equation assumes the form \u03b4 \u03b4\u03c6q S[\u03c6] = \u03b2 4\u03c0Ld q2\u03c6\u2212q + 2ie Ld q1 \u02c6Gq1,q1\u2212q!= 0. (6.5) Here, the factor of two appearing in front of the double summation accounts for the electron spin. Further, we have used the fact that \u02c6G\u22121 is linear in its argument \u03c6 and that, following \u02c6\u03c6)q1q2 = \u03b4q1\u2212q2,q. In fact (owing to the discreteness the rules of functional di\ufb00erentiation, (\u03b4\u03c6q of the momentum) we are not even di\ufb00erentiating functionals: \u02c6\u03c6 is a matrix in momentum space. Its matrix elements are given by components \u03c6q : ( \u02c6\u03c6)q1,q2 = \u03c6q1\u2212q2. Di\ufb00erentiation of this matrix with respect to \u03c6q then produces another matrix with matrix elements one for q1 \u2212q2 = q and zero otherwise. (Admittedly, it takes some time to get used to di\ufb00erentiation formulae of this type.) Equations of this form are solved not from scratch, but by making a physically motivated ansatz, i.e. by guessing the solution! Naturally, the \ufb01rst guess is a homogeneous solution, \u03c6(r, t) \u2261 \u00af\u03c6 = const., i.e. one relies on the picture that a spatially and temporarily varying \ufb01eld con\ufb01guration is energetically more costly than a constant one, and therefore cannot provide a stable extremum. INFO Be aware that there exist translationally invariant problems with inhomogeneous mean\ufb01elds; or a homogeneous solution exists, but it is energetically inferior to a textured \ufb01eld con\ufb01guration. Indeed, there may be sets of degenerate solutions, etc. Often, when new theories describing an unknown territory have been developed, the search for the \u201ccorrect\u201d mean-\ufb01eld turns out to be a matter of long, and sometimes controversial, research. In the present context, spatial and temporal homogeneity translates to the solution \u03c6q,\u03c9 = 0 if either \u03c9 or q is non-zero. Indeed, such an ansatz solves the mean-\ufb01eld equation: assuming that all \ufb02uctuating components of \u03c6 are switched o\ufb00, the Green function operator becomes diagonal in momentum space", ", ( \u02c6G[\u03c6])q1,q2 \u221d \u03b4q1,q2. We thus see that, for non-vanishing q, both terms in the equation vanish. Moreover, since charge neutrality requires \u03c6q=0 = 0 one may identify \u02c6\u03c6 = 0 as the solution of the mean-\ufb01eld equation, settling step 4 of the general program. We now proceed to expand the functional in \ufb02uctuations around \u03c6 = 0. Since the mean\ufb01eld solution vanishes, it makes no sense to introduce new notation, i.e. we will denote the \ufb02uctuations again by the symbol \u03c6. As regards the \ufb01rst term in the action (6.4), it already has a quadratic form. The logarithmic contribution can be expanded as if we were dealing with a function (again, a consequence of the trace), i.e. tr ln \u02c6G\u22121 = tr ln \u02c6G\u22121 0 + ie Ld tr( \u02c6G0 \u02c6\u03c6) + 1 2 2 e Ld tr( \u02c6G0 \u02c6\u03c6 \u02c6G0 \u02c6\u03c6) +\u00b7 \u00b7 \u00b7, where \u02c6G\u22121 2m + \u03bc is the momentum and frequency diagonal operator whose matrix 0 elements give the free Green function of the electron gas. Now, let us discuss the terms \u2261 i\u02c6\u03c9 \u2212 \u02c6p2 250 Broken symmetry and collective phenomena appearing on the right-hand side in turn. Being \u03c6-independent, the \ufb01rst term generates an overall constant multiplying the functional integral, i.e. a constant that must describe the non-interacting content of the theory. Indeed, one may note that etr ln \u02c6G\u22121 0 = e\u2212tr ln \u02c6G0 = det \u02c6G\u22121 \u2261 Z0 is just the partition function of the non-interacting electron gas. Linear in \u02c6\u03c6, 0 the second term of the expansion must, by virtue of the mean-\ufb01eld analysis, vanish after all. (We are expanding around an extremum! Exercise: Write out the momentum representation of the \ufb01rst-order contribution to con\ufb01rm that it vanishes identically.) The third term is the interesting one. Remembering that \u02c6\u03c6 couples to the theory as a voltage, this term describes the way potential \ufb02uctuations are a\ufb00ected by the presence of the electron gas, i.e. it must encode screening. To resolve this connection, let us make the momentum dependence of the second-order term explicit (exercise): 2 e Ld 1 2 tr( \u02c6G0 \u02c6\u03c6 \u02c6G0 \u02c6\u03c6) = e Ld 2 p,q G0,p\u03c6qG0,p+q\u03c6\u2212q = e2 2T Ld q \u03a0q\u03c6q\u03c6\u2212q, where, once again, we encounter the polarization operator (5.28). Combining with the \ufb01rst term in the action, one \ufb01nally obtains Se\ufb00 [\u03c6] = 1 2T L3 q \u03c6q q2 4\u03c0 \u2212 e2\u03a0q \u03c6\u2212q + O(\u03c64), (6.6) where we note that odd powers of\u03c6 vanish by the symmetry of the problem (exercise). Indeed, the form of the e\ufb00ective action (6.6) is suggestive. Ignoring the coupling to the electron gas, \u03c6 represents a \ufb01eld with bare \u201cpropagator\u201d \u223c q\u22122, i.e. the long-range correlation mediated by the electric \ufb01eld in vacuum. Coupling the \ufb01eld to a medium leads to the appearence of a \u201cself-energy\u201d \u223c \u03a0q. As we saw above, the self-energy converts the long-range power law correlation into something exponential \u2013 screening. The form of the screened propagator exactly coinicides with the e\ufb00ective RPA interaction (5.34) derived diagrammatically above. In passing we note that the structures emerging in the functional integral analysis support a uni\ufb01ed way of \u201creading\u201d the theory: there appear \u201cpropagators,\u201d \u201cself-energies,\u201d etc. irrespective of the concrete context, i.e. irrespective of whether we are dealing with a fermionic \ufb01eld (\u03c8) or a bosonic \ufb01eld (\u03c6). From here it is a one", "-line calculation to reproduce the result for the RPA free energy of the electron gas discussed previously. Gaussian integration over the \ufb01eld \u03c6 (step 5 of the q2 \u03a0q)\u22121/2, where we have noted that program) leads to the expression ZRPA = Z0 the \u03c6-integration was normalized to unity, i.e. for \u03a0 = 0, the integral collapses to unity. Taking the logarithm, we obtain the free energy q(1\u2212 4\u03c0e2 FRPA = \u2212T (ln Z \u2212 ln Z0) = T 2 ln q 1 \u2212 4\u03c0e2 q2 \u03a0q, in agreement with Eq. (5.27). At this point, it is instructive to pause and compare the two approaches to the problem: diagrammatics and \ufb01eld integration. We \ufb01rst note that the functional integral formulation indeed leads to the \u201creduced\u201d description we sought: the minimal degree of freedom (\u03c6) 6.3 Bose\u2013Einstein condensation and super\ufb02uidity 251 directly couples to the physically relevant entities of the theory (\u03a0q). The downside is that, to formulate the approach, we had to go through some preparatory analysis, notably work that required prior physical insight into the problem (i.e. the correct choice of the Hubbard\u2013Stratonovich decoupling). However, this turned out to be an e\ufb00ort well invested. After the identi\ufb01cation of \u03c6 as the \u201cappropriate\u201d \ufb01eld, the further construction of the theory proceeded along the lines of a largely \u201cautomated program\u201d (seeking saddle-points, expanding, etc.). In particular, there was no need to do battle with combinatorial problems. Further, in functional integral approaches, the risk of missing relevant contributions, or diagrams, in the expansion of the theory is far less pronounced than in direct diagrammatic expansions. But undoubtedly the most important advantage of the functional integral route is its extensibility. For example, an expansion of the theory to higher orders in \u03c6 would have generated an interacting theory of voltage \ufb02uctuations. Correlations on the level of that theory correspond to diagrams beyond the RPA level, i.e. diagrams whose direct and error-free summation would require more re\ufb01ned skills, a program beyond the scope of the present text. The mean-\ufb01eld optimizing the problem above was particularly simple with \u03c6 = 0. Moreinteresting situations arise when one encounters non-vanishing mean-\ufb01eld con\ufb01gurations, i.e. the perturbation theory has to be organized around a state di\ufb00erent from the trivial vacuum of the theory. Referring to Problem 6.7, we will discuss how, under certain circumstances, Coulomb interaction can favor an itinerant magnetic state. 6.3 Bose\u2013Einstein condensation and super\ufb02uidity Previously, we have considered the in\ufb02uence of weak Coulomb interaction on the properties of the electron gas. In the following, our goal will be to consider the phases realized by a weakly interacting Bose gas. To this end, let us introduce the quantum partition function Z = D( \u00af\u03c8, \u03c8) e\u2212S[ \u00af\u03c8,\u03c8], where # \u00af\u03c8(r, \u03c4 )(\u2202\u03c4 + \u02c6H0 \u2212 \u03bc)\u03c8(r, \u03c4 ) + S[ \u00af\u03c8, \u03c8] = ddr d\u03c4 ( \u00af\u03c8(r, \u03c4 )\u03c8(r, \u03c4 ))2. (6.7) $ g 2 Here \u03c8 represents a complex \ufb01eld subject to the periodic boundary condition \u03c8(r, \u03b2) =\u03c8 (r, 0). The integral Z describes the functional physics of a system of bosonic particles in d dimensions subject to a repulsive contact interaction of strength g > 0. For the moment the speci\ufb01c structure of the one-body operator \u02c6H0 need not be speci\ufb01ed. The most remarkable phenomena displayed by systems of this type are Bose\u2013Einstein condensation and super\ufb02uidity. However, contrary to a widespread belief, these two e\ufb00ects Satyendranath Bose 1894\u20131974 He undertook important work in quantum theory, in particular on Planck\u2019s black body radiation law. His work was enthusiastically endorsed by Einstein. He also published on statistical mechanics, leading most famously to the concept of Bose-Einstein statistics. Dirac coined the term \u201cboson\u201d for particles obeying such statistics. 252 Broken symmetry and collective phenomena do not depend on each other: super\ufb02u", "idity can arise without condensation and vice versa. We begin our discussion with the more elementary of the two phenomena. Bose\u2013Einstein condensation As may be recalled from elementary statistical mechanics, at su\ufb03ciently low temperatures the ground state of a bosonic system can involve the condensation of a macroscopic fraction of particles into a single state. This phenomenon, predicted in a celebrated work by Einstein, is known as Bose\u2013Einstein condensation. To see how this phenomenon is born out of the functional integral let us temporarily switch o\ufb00 the interaction and turn to the basis in which the one-particle Hamiltonian is diagonal. Expressed in the frequency representation, the partition function of the non-interacting system is given by Albert Einstein 1879\u20131955 Nobel Laureate in Physics in 1921 \u201cfor his services to theoretical physics, and especially for his discovery of the law of the photoelectric e\ufb00ect.\u201d He is perhaps best known for his theory of relativity and speci\ufb01cally mass\u2013 energy equivalence, expressed by the equation E = mc2. His work on the low-temperature behavior of the bosonic quantum gas is published in A. Einstein, Quantentheorie des einatomigen idealen Gases, Sitzungsber. Preuss. Akad. Wiss. (1925), 261-7. (Image c The Nobel Foundation.) formalism, Z0 \u2261 Z|g=0 = D( \u00af\u03c8, \u03c8) exp \u2212 \u00af\u03c8an (\u2212i\u03c9n + a \u2212 \u03bc) \u03c8an. an Without loss of generality, we may assume that the eigenvalues a \u2265 0 are positive with a ground state 0 = 0.4 (In contrast to the fermionic systems discussed above, we should not have in mind low-energy excitations superimposed on high-energy microscopic degrees of freedom. Here, everything will take place in the vicinity of the ground state of the microscopic single-particle Hamiltonian.) Further, we note that, to ensure stability, the chemical potential determining the number of particles in the system must be negative for, otherwise, the Gaussian weight corresponding to the low-lying states a < \u2212\u03bc would change sign, resulting in an ill-de\ufb01ned theory. From our discussion of Section 4.2 we recall that the number of particles in the system is set by the relation N (\u03bc) =\u2212 \u2202\u03bcF = T na 1 i\u03c9n \u2212 a + \u03bc = a nB(a), 4 The chemical potential \u03bc can always be adjusted so as to meet this condition. 6.3 Bose\u2013Einstein condensation and super\ufb02uidity 253 where, as usual, nB() = (e\u03b2(\u2212\u03bc) \u22121)\u22121 denotes the Bose distribution. For a given number of particles, this equation determines the temperature dependence of the chemical potential, \u03bc(T ). As the temperature is reduced, the distribution function controlling the population of individual states decreases. Since the number of particles must be kept constant, this scaling must be counter-balanced by a corresponding increase in the chemical potential. \u03bc 0 Tc T Below a certain critical temperature Tc, even the maximum value of the chemical potential, \u03bc = 0, will not su\ufb03ce to keep the distribution function nB(a=0) large enough to T <Tc\u2261 accommodate all particles in the states of non-vanishing energy, i.e. N1 < N. I.e. below the critical temperature, the chemical potential stays constant at \u03bc = 0 (see the \ufb01gure). As a result, a macroscopic number of particles, N \u2212 N1, must accumulate in the single-particle ground state: Bose\u2013Einstein condensation. a>0 nB(a)|\u03bc=0 EXERCISE For a three-dimensional free particle spectrum, k = 2k2/2m, show that the critical temperature is set by Tc = c02/ma2, wherea = \u03c1\u22121/3 is the average inter-particle spacing, and c0 is a constant of order unity. Show that for temperatures T < Tc, the density of particles in the condensate (k = 0) is given by \u03c10(T ) = \u03c1[1 \u2212 (T /Tc)3/2]. Since its prediction in the early 1920s, the phenomenon of Bose\u2013 Einstein condensation has been a standard component of undergraduate texts. However, took some seven decades before the condensation of bosonic particles was directly5 observed in experiment. The reason for this delay is that the critical condensation temperature of particles (atoms) that are comfortably accessible to experiment is absurdly low. Eric A. Cornell 1961\u2013 (left), WolfKetterle gang 1957\u2013 (center), and Carl E.", " Wieman 1951\u2013 (right) Joint recipients of the 2001 Nobel Prize in Physics \u201cfor the achievement of Bose-Einstein condensation in dilute gases of alkali atoms, and for early fundamental studies of the properties of the condensates.\u201d (Images c The Nobel Foundation.) it INFO In 1995 the groups of Cornell and Wieman at Colorado University and, soon after, Ketterle at MIT succeeded in cooling a system of rubidium atoms down to temperatures of 20 billionths(!) of a kelvin.6 To reach these temperatures, a gas of rubidium atoms was caught in a magnetic 5 Here, by \u201cdirect\u201d we refer to the controlled preparation of a state of condensed massive bosonic particles. There are numerous \u201cindirect\u201d manifestations of condensed states, e.g. the anomalous properties of the helium liquids at low temperatures, or of Cooper-pair condensates in superconductors. 6 M. H. Anderson, J.R. Ensher, M.R. Matthews, C.E. Wieman, and E. A. Cornell, Observation of Bose\u2013Einstein condensation in a dilute atomic vapor, Science 269 (1995), 198-201. 254 Broken symmetry and collective phenomena Figure 6.2 Spectroscopic images of the Bose\u2013Einstein condensation process for three values of temperature (400 nK, 200 nK, and 50 nK from left to right). (Figure courtesy of the JILA Institute, University of Colorado.) trap, i.e. a con\ufb01guration of magnetic \ufb01eld gradients that couple to the magnetic moments of the atoms so as to keep the system spatially localized. The gas of atoms was then brought to a temperature of O(10\u22125) K \u2013 still much too hot to condense \u2013 by \u201claser cooling\u201d: crudely speaking, a technique where atoms, subjected to a suitably adjusted ray of monochromatic light, may transmit more of their kinetic energy to the photons than they get back. To lower the temperature still further, the principle of \u201cevaporative cooling\u201d was applied: by lowering the potential well of the trap, a fraction of the atoms, namely those with large kinetic energy, is allowed to escape. The remaining atoms have a low kinetic energy and, therefore, a low temperature. What sounds like a simple recipe actually represents a most delicate experimental procedure. (For example, if the trap potential is lowered too strongly, all atoms escape and there is nothing left to condense. If, on the other hand, trapping is too strong, the atoms remain too hot, etc.) However, after more than a decade of intensive experimental preparation, the required temperatures have been reached. Spectroscopic images of the Bose\u2013Einstein condensation process are shown in Fig. 6.2 at three temperatures. The peak in the density distribution signals the onset of condensation. On lowering the temperature, one may observe the transition to a condensed phase by monitoring the formation of a peak in the density distribution. The preparation of a Bose-Einstein condensed state of matter was recognized with the award of the 2001 Nobel prize in physics. Since 1995, research on atomic condensates has blossomed into a broad arena of research. Already, it is possible to prepare complex states of Bose condensed matter such as atomic vortices in rotating Bose\u2013Einstein condensates, condensates in di\ufb00erent dimensionalities, or even an arti\ufb01cial crystalline state of matter (see Fig. 6.3). Regrettably, a detailed discussion of these interesting developments is beyond the scope of the present text. Those interested in learning more about this area are referred to the many reviews of the \ufb01eld. With this background, let us now try to understand how the phenomenon of Bose\u2013Einstein condensation can be implemented in the functional integral representation. Evidently, the characteristics of the condensate will be described by the zero \ufb01eld component \u03c80(\u03c4 ). The problem with this zero mode is that, below the condensation transition, its action appears to be unbounded: both the chemical potential and the eigenvalue are zero. This means that the action of the zero Matsubara component \u03c80,0 vanishes. We will deal with this di\ufb03culty 6.3 Bose\u2013Einstein condensation and super\ufb02uidity 255 in a pragmatic way. That is, we will treat \u03c80(\u03c4 ) not as an integration variable but rather as a time-independent Lagrange multiplier to be used to \ufb01x the number of particles below the transition. More precisely, we introduce a reduced action of the form S0[ \u00af\u03c8, \u03c8] = \u2212 \u00af\u03c80\u03b2\u03bc\u03c80 + a=0,n \u00af\u03c8an (\u2212i\u03c9", "n + a \u2212 \u03bc) \u03c8an, where we did not yet set \u03bc = 0 (since we still need \u03bc as a di\ufb00erentiation variable). To understand the rationale behind this simpli\ufb01cation one may note that N = \u2212\u2202\u03bcF0|\u03bc=0\u2212 = T \u2202\u03bc ln Z0|\u03bc=0\u2212 = \u00af\u03c80\u03c80 + T a=0,n 1 i\u03c9n \u2212 a = \u00af\u03c80\u03c80 + N1, (6.8) determines the number of particles. According to this expression, \u00af\u03c80\u03c80 = N0 sets the number of particles in the condensate. Now, what enables us to regard \u03c80 as a time-independent \ufb01eld? Remembering the construction of the path integral, we note that the introduction of time-dependent \ufb01elds, or \u201ctime slicing,\u201d was necessitated by the fact that the operators appearing in the Hamiltonian of a quantum theory do not, in general, commute. (Otherwise we could have decoupled the expression tr(e\u2212\u03b2( \u02c6H\u2212\u03bc \u02c6N )(a\u2020,a)) d( \u00af\u03c8, \u03c8) e\u2212\u03b2( \u02c6H\u2212\u03bc \u02c6N )( \u00af\u03c8,\u03c8) in terms of a single coherent state resolution, i.e. a \u201cstatic\u201d con\ufb01guration.) Reading this observation in reverse, we conclude that the dynamic content of the \ufb01eld integral represents the quantum character of a theory. (Alluding to this fact, the temporal \ufb02uctuations of \ufb01eld variables are often referred to as quantum \ufb02uctuations.) Conversely, Figure 6.3 The creation of an arti\ufb01cial mono-atomic crystal from a Bose\u2013Einstein condensate. A system of bosons in a condensed state \u2013 the state signaled by the pronounced coherent intensity peaks in the left portion of the \ufb01gure \u2013 is exposed to a \u201cgrid\u201d formed by two crossing \ufb01elds of intense laser radiation. For strong enough \ufb01eld amplitudes, the electromagnetic radiation interferes, presenting a periodic lattice of deep potential minima. These minima become occupied by one atom each. The result is an arti\ufb01cial mono-atomic crystal, characterized by the absence of coherent superposition of atomic wavefunction amplitudes, i.e. an equilibrated density pro\ufb01le (middle). If the laser intensity is lowered, the atoms re-arrange to the characteristic density distribution of the condensate (right). (Source: Courtesy of Max-Planck-Institut f\u00a8ur Quantenoptik, Garching, Germany.) 256 Broken symmetry and collective phenomena A static approximation in a \ufb01eld integral \u03c8(\u03c4 ) = \u03c80 = const. amounts to replacing a quantum degree of freedom by its classical approximation. (In order to distinguish them from quantum, \ufb02uctuations in the \u201cclassical\u201d static sector of the theory are called thermal \ufb02uctuations.) To justify the approximation of a0 \u2194 \u03c80 by a \u2020 0a0 will assume \u201cmacroscopically classical object, notice that, upon condensation, N0 = a \u2020 0] = 1 continues to be of O(1). It large\u201d values. On the other hand, the commutator [a0, a thus seems to be legitimate to neglect all commutators of the zero operator a0 in comparison with its expectation value \u2013 a classical approximation.7 Now, we are still left with the problem that the \u03c80-integration appears to be unde\ufb01ned. The way out is to remember that the partition function should extend over those states that contain an average number N of particles. That is, Eq. (6.8) has to be interpreted as a relation that \ufb01xes the modulus \u00af\u03c80\u03c80 so as to adjust the appropriate value of N.8 The weakly interacting Bose gas Now, with this background, let us restore the interactions focusing on a small but nonzero coupling constant g. To keep the discussion concrete, we specialize to the case of a free single-particle system, \u02c6H0 = \u02c6p2/2m. (Notice that the ground state wavefunction of this system is described by a spatially constant zero-momentum state.) By adiabatic continuity we expect that much of the picture developed above will survive generalization to non-zero interaction strengths. In particular, the ground state, which in the case under consideration corresponds to a temporally and spatially constant mode \u03c80,", " will continue to be macroscopically occupied. In these circumstances, the dominant contribution to the action will again come from the classical \u03c80 sector: T S[ \u00af\u03c80, \u03c80] = \u2212\u03bc \u00af\u03c80\u03c80 + g 2Ld ( \u00af\u03c80\u03c80)2. (6.9) (Notice the similarity of the action to the integrand of the toy problem discussed in Section 5.1.) Crucially, the stability of the action is now guaranteed by the interaction vertex, no matter how small is g > 0 (see the schematic plot of the action in the \ufb01gure on page 263). Accordingly, we will treat \u03c80 no longer as a \ufb01xed parameter but rather as an ordinary integration variable. Integration over all \ufb01eld components will produce a partition function Z(\u03bc) that depends parametrically on the chemical potential. As usual in statistical physics, the function can then be employed to \ufb01x the particle number. (Notice that, vis-`a-vis aspects 7 Notice the similarity of that reasoning to the arguments employed in connection with the semi-classical treatment of spin systems in the limit of large S (Section 2.2). Unfortunately, the actual state of a\ufb00airs with the classical treatment of the condensate is somewhat more complex than the simple argument above suggests. (For a good discussion, see A. A. Abrikosov, L. P. Gorkov, and I. E. Dzyaloshinskii, Methods of Quantum Field Theory in Statistical Physics [Dover Publications, 1975]) However, the net result of a more thorough analysis, i.e. an integration over all dynamically \ufb02uctuating components \u03c80,n=0, shows that the treatment of \u03c80 as classical represents a legitimate approximation. 8 For a more rigorous discussion of the choice of the thermodynamic variables in the present context, we again refer to Abrikosov et al., ibid. 6.3 Bose\u2013Einstein condensation and super\ufb02uidity 257 of thermodynamics, the interacting system appears to behave more \u201cnaturally\u201d than its ideal, non-interacting approximation. This re\ufb02ects a general feature of bosonic systems: interactions \u201cregularize\u201d a number of pathological features of the ideal gas.) Returning to the \u03c80-integration, we observe that, for low enough temperatures, the problem is an ideal candidate for saddle-point analysis. Variation of the action with respect to \u03c80 gives g Ld This equation is solved by any constant complex \ufb01eld con\ufb01guration \u03c80 with modulus |\u03c80| = \u03bcLd/g \u2261 \u03b3. In spite of its innocent appearance, this equation reveals much about the \u00af\u03c80\u03c80 \u2212\u03bc + = 0. \u00af\u03c80 nature of the system: For \u03bc <0 (i.e. above the condensation threshold of the non-interacting system), the equation exhibits only the trivial solution \u03c80 = 0. This means that no stable condensate amplitude exists. Below the condensation threshold (i.e. for \u03bc \u2265 0),9 the equation is solved by any con\ufb01g\u03bcLd/g. (Notice that \u00af\u03c80\u03c80 \u221d Ld, re\ufb02ecting the macroscopic uration with |\u03c80| = \u03b3 \u2261 population of the ground state.) The equation couples only to the modulus of \u03c80. That is, the solution of the stationary phase equation is continuously degenerate: each con\ufb01guration \u03c80 = \u03b3 exp(i\u03c6), \u03c6 \u2208 [0, 2\u03c0], is a solution. For our present discussion, the last of the three aspects mentioned above is the most important. It raises the question of which con\ufb01guration \u03c80 = \u03b3exp(i\u03c6) is the \u201cright\u201d one? Without loss of generality, we may choose \u03c80 = \u03b3 \u2208 R as a reference con\ufb01guration for our theory. This choice amounts to selecting a particular minimum lying in the \u201cMexican hat\u201d pro\ufb01le of the action shown above. However, it is clear that an expansion of the action around that minimum will be singular: \ufb02uctuations \u03c80 \u2192 \u03c80 + \u03b4\u03c8 that do not leave the azimuthally symmetric well of degenerate minima do not change the action and, therefore, have vanishing expansion coe\ufb03cients. As a result, in the present situation, we will not be able to implement a simple scheme, i.e. \u201csaddle-point plus quadratic \ufffd", "\ufffd\ufffductuations\u201d; there is nothing that constrains the deviations \u03b4\u03c8 to be small. In the next section, we discuss the general principles behind this phenomenon. In section 6.3, we then continue to explore its rami\ufb01cations in the physics of the Bose system. Spontaneous symmetry breaking The mechanism encountered here is one of spontaneous symmetry breaking. To understand the general principle, consider an action S[\u03c8] with a global continuous symmetry under some transformation g (not to be confused with the aforementioned coupling constant of the Bose gas): speci\ufb01cally, the action remains invariant under a global transformation of 9 Owing to the stabilization of the zero mode integration by the interaction constant, \u03bc \u2264 0 is no longer a strict condition. 258 Broken symmetry and collective phenomena the \ufb01elds such that \u2200i \u2208 M : \u03c8i \u2192 g\u03c8i, where M is the base manifold, i.e. S[\u03c8] = S[g\u03c8]. The transformation is \u201ccontinuous\u201d in the sense that g takes values in some manifold, typically a group G. Examples: The action of a Heisenberg ferromagnet is invariant under rotation of all spins simultaneously by the same amount, Si \u2192 gSi. In this case, g \u2208 G = O(3), the three-dimensional group of rotations (g not to be confused with the coupling constant of the interaction). The action of the displacement \ufb01elds u describing elastic deformations of a solid (phonons) is invariant under simultaneous translation of all displacements ui \u2192 ui + a, i.e. the symmetry manifold is the d-dimensional translation group G \u223c= Rd. In the example above, we encountered a U(1) symmetry under phase multiplication \u03c80 \u2192 ei\u03c6\u03c80. This phase freedom expresses the global gauge symmetry of quantum mechanics under transformation by a phase, a point we discuss in more detail below. S Now, given a theory with globally G invariant action, two scenarios are conceivable: either the ground states share the invariance properties of the action or they do not. The two alternatives are illustrated in the \ufb01gure for the example of the Bose system. For \u03bc < 0, the action S[ \u00af\u03c80, \u03c80] has a single ground state at \u03c80 = 0. This state is trivially symmetric under the action of G = U(1). However, for positive \u03bc, i.e. in the situation discussed above, there is an entire manifold of degenerate ground states, de\ufb01ned through the relation |\u03c80| = \u03b3. These ground states transform into each other under the action of the gauge group. However, none of them is individually invariant. Re\u03c8 Im\u03c8 With the other examples mentioned above, the situation is similar. For symmetry groups more complex than the one-dimensional manifold U(1), the ground states will, in general, be invariant under transformation by the elements of a certain subgroup H \u2286 G (that includes the two extremes H = {1} and H = G). For example, below the transition temperature, the ground state of the Heisenberg magnet will be given by (domain-wise) aligned con\ufb01gurations of spins. Assuming that the spins are oriented along the z-direction, the ground state is invariant under the abelian subgroup H \u2282 O(3) containing all rotations around the z-axis. However, invariance under the full rotation group is manifestly broken. Solids represent states where the translation symmetry is fully broken, i.e. all atoms collectively occupy a \ufb01xed pattern of spatial positions in space, H = {1}, etc. In spite of the undeniable existence of solids, magnets, and Bose-Einstein condensates of de\ufb01nite phase, the notion of a ground state that does not share the full symmetry of the theory may appear paradoxical, or at least \u201cunnatural.\u201d For example, even if any particular ground state of the \u201cMexican hat\u201d potential shown in the \ufb01gure above \u201cbreaks\u201d the rotational symmetry, should not all these states enter the partition sum with equal statistical weight, such that the net outcome of the theory is again fully symmetric? To understand why symmetry breaking is a \u201cnatural\u201d and observable phenomenon, it is instructive to perform a gedankenexperiment. To this end, consider the partition function 6.3 Bose\u2013Einstein condensation and super\ufb02uidity 259 of a classical10 ferromagnet, e\u2212\u03b2(H\u2212h\u00b7 i Si), Z = tr where H is the rotationally invariant part of", " the energy functional and h represents a weak external \ufb01eld. (Alternatively, we can think of h as an internal \ufb01eld, caused by a slight structural imperfection of the system.) In the limit of vanishing \ufb01eld strength, the theory becomes manifestly symmetric. Symbolically, lim N\u2192\u221e lim h\u21920 Z \u2212\u2192 rot. sym., where the limit N \u2192 \u221e serves as a mnemonic indicating that we consider systems of macroscopic size. However, keeping in mind the fact that the model ought to describe a physical magnetic system, the order of limits taken above appears questionable. Since the external perturbation couples to a macroscopic number of spins, a more natural description of an \u201calmost\u201d symmetric situation would be lim h\u21920 lim N\u2192\u221e Z \u2212\u2192? The point is that the two orders of limits lead to di\ufb00erent results. In this case, for any h, the N \u2192 \u221e system is described by an explicitly symmetry broken action. No matter how small the magnetic \ufb01eld, the energetic cost to rotate N \u2192 \u221e spins against the \ufb01eld is too high, i.e. the ground state |S below the transition temperature will be uniquely aligned, Si h. When we then send h \u2192 0 in a subsequent step, that particular state will remain the observable reference state of the system. Although, formally, a spontaneous thermal \ufb02uctuation rotating all spins by the same amount |S \u2192 |gS would not cost energy, that \ufb02uctuation can be excluded by entropic reasoning.11 (By analogy, one rarely observes kettles crashing into the kitchen wall as a consequence of a concerted thermal \ufb02uctuation of the water molecules!) However, the appearance of non-trivial ground states is just one manifestation of spontaneous symmetry breaking. Equally important, residual \ufb02uctuations around the ground state lead to the formation of soft modes (sometimes known as massless modes), i.e. \ufb01eld con\ufb01gurations \u03c6q whose action S[\u03c6] vanishes in the limit of long wavelengths, q \u2192 0. Speci\ufb01cally, the soft modes formed on top of a symmetry broken ground state are called Goldstone modes. As a rule, the presence of soft modes in a continuum theory has important phenomenological consequences. To understand this point, notice that the general structure of a soft mode action is given by S[\u03c6] = q,i % \u03c6q ci 1 |qi| + ci 2q2 i & \u03c6\u2212q + O(\u03c64, q3), (6.10) where ci 1,2 are coe\ufb03cients. As discussed in Section 5.1, the absence of a constant contribution to the action (i.e. a contribution that does not vanish in the limit q \u2192 0) signals the existence of long-range power-law correlations in the system. As we will see shortly, the vanishing of 10 The same argument can be formulated for the quantum magnet. 11 In Chapter 8, we show that this (overly) simple picture in fact breaks down in dimensions d \u2264 2. 260 Broken symmetry and collective phenomena the action in the long-wavelength limit q \u2192 0 further implies that the contribution of the soft modes dominates practically all observable properties of the system. EXERCISE Explore the structure of the propagator G(q) \u2261 \u03c6q\u03c6\u2212q associated with S[\u03c6] and convince yourself that the arguments formulated for the speci\ufb01c case of the \u03c64-theory are of general validity. To this end, notice that, for small q, G(q) \u223c |q|\u2212n, where n denotes the index of the \ufb01rst non-vanishing coe\ufb03cient c1, c2,..., i.e. the propagator is dominated by the smallest q-power appearing in the action. The power-law behavior of the correlation function implies a homogeneity relation G(q/\u03bb) =\u03bb nG(q). Show that this scaling relation implies that the Fourier transform G(r \u2261 |r|) = \u03c6(r)\u03c6(0) obeys the \u201cscaling law\u201d G(\u03bbr) =\u03bb \u2212d+nG(r). This, in turn, implies that the real space correlation function also decays as a power law, namely G(r) \u223c |r|\u2212d+n, i.e. in a \u201clong-range\u201d manner. Explore the breakdown of the argument for an action with a \ufb01nite mass term. Convince yourself that, in this case, the decay would be exponential, i.e.", " \u201cshort-range.\u201d \u03c8 0 0 0 \u03bb g\u03c8 g(x)\u03c8 What, then, are the origin and nature of the soft Goldstone modes caused by the spontaneous breakdown of a symmetry? To address this point let us consider the action of a symmetry group element g on a (symmetry broken) ground state \u03c80 (cf. the middle row of the \ufb01gure). By de\ufb01nition, S[g\u03c80] = S[\u03c80] still assumes its extremal value. Assuming that g is close a \u03c6aTa], where the {Ta} are generators to the group identity, we may express g = exp[ living in the Lie algebra of the group and \u03c6a are some expansion coe\ufb03cients.12 Expressing \ufb02uctuations around \u03c80 in terms of the \u201ccoordinates\u201d \u03c6a, we conclude that the action S[\u03c6] = 0. However, if we promote the global transformation to one with a weakly \ufb02uctuating spatial pro\ufb01le, g \u2192 g(r), \u03c80 \u2192 g(r)\u03c80 (bottom row of the \ufb01gure), some price must be paid. That is, for a spatially \ufb02uctuating coordinate pro\ufb01le {\u03c6a(r)}, S[\u03c6] = 0, where the energy cost depends inversely on the \ufb02uctuation rate \u03bb of the \ufb01eld \u03c6. The expansion of S in terms of gradients of \u03c6 is thus bound to lead to a soft mode action of the type Eq. (6.10). In view of their physical signi\ufb01cance, it is important to ask how many independent soft modes exist. The answer can be straightforwardly given on the basis of the geometric picture developed above. Suppose our symmetry group G has dimension r, i.e. its Lie algebra is spanned by r linearly independent generators Ta, a = 1,..., r. If the subgroup H \u2282 G has dimension s < r, s of these generators can be chosen so as to leave the ground state invariant. On the other hand, the remaining p \u2261 r \u2212 s generators inevitably create Goldstone modes. In the language of group theory, these generators span the coset space G/H. For example, for the ferromagnet, H = O(2) is the one-dimensional subgroup of rotations around the quantization axis (e.g. the z-axis). Since the rotation group has dimension 3, there must be two independent Goldstone modes. These can be generated by the action of the rotation, or angular momentum generators Jx,y acting on the z-aligned ground state. The coset space 12 Here we have used the fact that, for any reasonable model, G will be a Lie group, i.e. a group with the structure of a di\ufb00erentiable manifold. 6.3 Bose\u2013Einstein condensation and super\ufb02uidity 261 O(3)/O(2) can be shown to be isomorphic to the 2-sphere, i.e. the sphere traced out by the spins as they \ufb02uctuate around the ground state. Finally, you may ask why we bothered to formulate these concepts in the abstract language of Lie groups and generators, etc. The reason is that the connection between the coordinates parameterizing the Goldstone modes \u03c6i, i = 1,..., p, residual \u201cmassive modes\u201d \u03baj, j = 1,..., n \u2212 p, and the original coordinates \u03c8i, i = 1,..., n, of the problem, respectively, is usually nonlinear and sometimes not even very transparent. With problems more complex than the three prototypical examples mentioned above, it is usually pro\ufb01table to \ufb01rst develop a good understanding of the geometry of the problem before speci\ufb01c coordinate systems are introduced. After these general considerations it is time to get back to the Bose system and to explore the physical consequences of Goldstone mode formation on the level of a concrete example! Super\ufb02uidity The theory of the weakly interacting super\ufb02uid to be discussed below was originally conceived by Bogoliubov, then in the language of second quantization.13 We will reformulate the theory in the language of the \ufb01eld integrals starting with the action of the weakly interacting Bose gas (6.7). Focussing on temperatures below Tc (\u03bc > 0), let us expand the theory around the particular mean-\ufb01eld ground state \u00af\u03c80 = \u03c80 = (\u03bcLd/g)1", "/2 = \u03b3. (Of course, any other state lying in the \u201cMexican hat\u201d minimum of the action would be just as good.) Notice that the quantum ground state corresponding to the con\ufb01guration \u03c80 is unconventional in the sense that it cannot have a de\ufb01nite particle number. The reason is that, according to the correspondence \u03c8 \u2194 a between coherent states and operators, respectively, a non-vanishing functional expectation value of \u03c80 is equivalent to a non-vanishing quantum expectation value a0. Assuming that, at low temperatures, the thermal average... will project onto the ground state |\u03a9, we conclude that \u03a9|a0|\u03a9 = 0, i.e. |\u03a9 cannot be a state with a de\ufb01nite number of particles.14 The symmetry group U(1) acts on this state by multiplication, \u03c80 \u2192 ei\u03c6\u03c80 and \u00af\u03c80 \u2192 e\u2212i\u03c6 \u00af\u03c80. Knowing that the action of a weakly modulated \ufb01eld \u03c6(r, \u03c4 ) will be massless, let us introduce coordinates \u03c8(r, \u03c4 ) = \u03c11/2(r, \u03c4 )ei\u03c6(r,\u03c4 ), \u00af\u03c8(r, \u03c4 ) = \u03c11/2(r, \u03c4 )e\u2212i\u03c6(r,\u03c4 ), Im \u03c8 S = extr. \u03c6 \u03b4\u03c1 Re \u03c8 where \u03c1(r, \u03c4 ) =\u03c1 0 + \u03b4\u03c1(r, \u03c4 ) and \u03c10 = \u00af\u03c80\u03c80/Ld is the condensate density. Evidently, the variable \u03b4\u03c1 parametrizes deviations of the \ufb01eld \u03c8(r, \u03c4 ) from the extremum. These excursions are energetically costly, i.e. \u03b4\u03c1 will turn out to be a massive mode. Also notice that the transformation of coordinates ( \u00af\u03c8, \u03c8) \u2192 (\u03c1, \u03c6), viewed as a change of integration variables, has a Jacobian of unity. 13 N. N. Bogoliubov, On the theory of super\ufb02uidity, J. Phys. (USSR) 11, 23-32 (1947) (reprinted in D. Pines, The Many-Body Problem, [Benjamin, 1961]). 14 However, as usual with grand canonical descriptions, in the thermodynamic limit the relative uncertainty in the number of particles, ( \u02c6N 2 \u2212 \u02c6N 2)/ \u02c6N 2, will become vanishingly small. 262 Broken symmetry and collective phenomena INFO As we are dealing with a (functional) integral, there is a lot of freedom as to the choice of integration parameters. (I.e., in contrast to the operator formulation, there is no a priori constraint for a transform to be \u201ccanonical.\u201d) However, physically meaningful changes of representation will usually be canonical transformations, in the sense that the corresponding transformations of operators would conserve the commutation relations. Indeed, as we have seen in the info block starting on page 176, the operator transformation a(r) \u2261 \u02c6\u03c1(r)1/2 ei \u02c6\u03c6(r), a\u2020(r) \u2261 e\u2212i \u02c6\u03c6(r) \u02c6\u03c1(r)1/2, ful\ufb01lls this criterion. We next substitute the density\u2013phase relation into the action and expand to second order around the reference mean-\ufb01eld. Ignoring gradients acting on the density \ufb01eld (in comparison with the \u201cpotential\u201d cost of these \ufb02uctuations), we obtain S[\u03c1, \u03c6] \u2248 d\u03c4 ddr i\u03b4\u03c1\u2202\u03c4 \u03c6 + \u03c10 2m (\u2207\u03c6)2 +. g\u03c12 2 (6.11) The \ufb01rst term of the action has the canonical structure \u201cmomentum \u00d7 \u2202\u03c4 (coordinate)\u201d indicative of a canonically conjugate pair. The second term measures the energy cost of spatially varying phase \ufb02uctuations. Notice that \ufb02uctuations with \u03c6(r, \u03c4 ) = const. do not incur an energy cost \u2013 \u03c6 is a Goldstone mode. Finally, the third term records the energy cost of massive \ufb02uctuations from the potential minimum. Equation (6.11) represents the Hamiltonian version of the action, i.e. an action comprising coordinates \u03c6 and momenta \u03c1. Gaussian integration over the \ufb01eld \u03b4\u03c1 leads us to the Lagrangian form of the action (exercise): d\u03c4 ddr S[", "\u03c6] \u2248 1 2 1 g (\u2202\u03c4 \u03c6)2 + \u03c10 m (\u2207\u03c6)2. (6.12) Comparison with Eq. (1.4) identi\ufb01es this action as the familiar d-dimensional oscillator. Drawing on the results of Chapter 1 (see, e.g., Eq. (1.29)), we \ufb01nd that the energy \u03c9k carried by elementary excitations of the system scales linearly with momentum, \u03c9k = |k|(g\u03c10/m)1/2. Let us now discuss the physical rami\ufb01cations of these results. The actions (6.11) and (6.12) describe the phenomenon of super\ufb02uidity. To make the connection between the fundamental degree of freedom of a super\ufb02uid system, the supercurrent, and the phase \ufb01eld explicit, let us consider the quantum mechanical current operator \u02c6j(r, \u03c4 ) = fun. int.\u2212\u2192 % % i 2m i 2m (\u2207a\u2020(r, \u03c4 ))a(r, \u03c4 ) \u2212 a\u2020(r, \u03c4 )\u2207a(r, \u03c4 ) & (\u2207 \u00af\u03c8(r, \u03c4 ))\u03c8(r, \u03c4 ) \u2212 \u00af\u03c8(r, \u03c4 )\u2207\u03c8(r, \u03c4 ) & \u2248 \u03c10 m \u2207\u03c6(r, \u03c4 ), (6.13) where the arrow indicates the functional integral correspondence of the operator description and we have neglected all contributions arising from spatial \ufb02uctuations of the density pro\ufb01le. (Indeed, these \u2013 massive \u2013 \ufb02uctuations describe the \u201cnormal\u201d contribution to the current \ufb02ow.) INFO Super\ufb02uidity is one of the most counterintuitive and fascinating phenomena displayed by condensed matter systems. Experimentally, the most straightforward access to super\ufb02uid states of matter is provided by the helium liquids. Representative of many other e\ufb00ects displayed by 6.3 Bose\u2013Einstein condensation and super\ufb02uidity 263 super\ufb02uid states of helium, we mention the capability of thin \ufb01lms to \ufb02ow up the walls of a vessel (if the reward is that on the outer side of the container a low-lying basin can be reached \u2013 the fountain experiment) or to e\ufb00ortlessly propagate through porous media that no normal \ufb02uid may penetrate. Readers interested in learning more about the phenomenology of super\ufb02uid states of matter may refer to the seminal text by Pines and Nozi`eres.15 The gradient of the phase variable is therefore a measure of the (super)current \ufb02ow in the system. The behavior of that degree of freedom can be understood by inspection of the stationary phase equations \u2013 a.k.a. the Hamilton or Lagrange equations of motion \u2013 associated with the action (6.11) or (6.12). Turning to the Hamiltonian formulation, one obtains (exercise) i\u2202\u03c4 \u03c6 = \u2212g\u03c1 + \u03bc, i\u2202\u03c4 \u03c1 = \u03c10 m \u22072\u03c6 = \u2207 \u00b7 j. The second of these equations represents (the Euclidean time version of) a continuity equation. A current \ufb02ow with non-vanishing divergence is accompanied by dynamical distortions in the density pro\ufb01le. The \ufb01rst equation tells us that the system adjusts to spatial \ufb02uctuations of the density by a dynamical phase \ufb02uctuation. The most remarkable feature of these equations is that they possess steady state solutions with non-vanishing current \ufb02ow. Setting \u2202\u03c4 \u03c6 = \u2202\u03c4 \u03c1 = 0, we obtain the conditions \u03b4\u03c1 = 0 and \u2207 \u00b7j = 0, i.e. below the condensation temperature, a con\ufb01guration with a uniform density pro\ufb01le can support a steady state divergenceless (super)current. Notice that a \u201cmass term\u201d in the \u03c6-action would spoil this property, i.e., within our present approach, the phenomenon of supercurrent \ufb02ow is intimately linked to the Goldstone mode character of the \u03c6 \ufb01eld. EXERCISE Add a \ufb01ctitious mass term to the \u03c6-action and explore its consequences. How do the features discussed above present themselves in the Lagrange picture? It is very instructive to interpret the phenomenology of supercurrent \ufb02ow from a di\ufb00erent, more microscopic perspective. Steady state current \ufb02ow in normal environments is prevented by the mechanism of energy dissipation, i.e. particles constit", "uting the current \ufb02ow scatter o\ufb00 imperfections inside the system, thereby converting part of their energy into the creation of elementary excitations. (Macroscopically, the conversion of kinetic energy into the creation of excitations manifests itself as heat production.) Apparently, this mechanism is inactivated in super\ufb02uid states of matter, i.e. the current \ufb02ow is dissipationless. How can the dissipative loss of energy be avoided? Trivially, no energy can be exchanged if there are no elementary excitations to create. In reality, this means that the excitations of the system are energetically so high-lying that the kinetic energy stored in the current-carrying particles is insu\ufb03cient to create them. But this is not the situation that we encounter in the super\ufb02uid! As we saw above, there is no energy gap separating the quasi-particle excitations of the system from the ground state. Rather, the dispersion \u03c9(k) vanishes linearly as k \u2192 0. However, there is an ingenious argument due to Landau showing that a linear excitation spectrum indeed su\ufb03ces to stabilize dissipationless transport. 15 D. Pines and P. Nozi`eres, The Theory of Quantum Liquids: Super\ufb02uid Bose Liquids (Addison-Wesley, 1989). 264 Broken symmetry and collective phenomena V p (a) (c) \u2013V (b) (d) Figure 6.4 (a) Flow of a \ufb02uid through a rough pipe. (b) The same viewed from the rest frame of the \ufb02uid. (c) Dissipative creation of a (quasi-particle) excitation. (d) The same viewed from the laboratory frame. INFO Consider the \ufb02ow of some \ufb02uid through a pipe (see Fig. 6.4(a)). To be concrete, let us assume that the \ufb02ow occurs at a uniform velocity V. Taking the mass (of a certain portion of the \ufb02uid) to be M, the current carries a total kinetic energy E1 = M V2/2. Now, suppose we view the situation from the point of view of the \ufb02uid, i.e. we perform a Galilean transformation into its own rest frame (Fig. 6.4(b)). From the perspective of the \ufb02uid, the walls of the pipe appear as though they were moving with velocity \u2212V. Now, suppose that frictional forces between \ufb02uid and the wall lead to the creation of an elementary excitation of momentum p and energy (p), i.e. the \ufb02uid is no longer at rest but carries kinetic energy (Fig. 6.4(c)). After a Galilean transformation back to the laboratory frame (Fig. 6.4(d)), one \ufb01nds that the energy of the \ufb02uid after the creation of the excitation is given by (exercise) E2 = M V2 2 + p \u00b7 V + (p). Now, since all of the energy needed to manufacture the excitation must have been provided by the liquid itself, energy conservation requires that E1 = E2, or \u2212p \u00b7 V = (p). Since p \u00b7 V > \u2212|p||V|, this condition can only be met if |p||V| > (p). While systems with a \u201cnormal\u201d gapless dispersion, (p) \u223c p2, are compatible with this energy-balance relation (i.e. no matter how small |V|, quasi-particles of low momentum can always be excited), both gapped dispersions p\u21920\u2212\u2192 const. and linear dispersions are incompatible if V becomes smaller than a certain (p) critical velocity V\u2217. Speci\ufb01cally for a linear dispersion (p) =v |p|, the critical velocity is given by V\u2217 = v. For currents slower than that, the \ufb02ow is necessarily dissipationless. Let us conclude our preliminary discussion of the weakly interacting Bose gas with a very important remark. Super\ufb01cially, Eq. (6.11) and (6.12) suggest that we have managed to describe the long-range behavior of the condensed matter system in terms of a free Gaussian theory. However, one must recall that \u03c6 is a phase \ufb01eld, de\ufb01ned only modulo 2\u03c0. (In Eq. (6.11) and (6.12) this condition is understood implicitly. At this point, it is perhaps worth reiterating that when dealing with Goldstone modes it is important to keep the underlying geometry in mind and not focus too tightly on a speci\ufb01c coordinate representation.) The fact that \u03c6", " is de\ufb01ned only up to integer multiples of 2\u03c0 manifests itself in the 6.4 Superconductivity 265 formation of the most interesting excitations of the super\ufb02uid: vortices, i.e. phase con\ufb01gurations \u03c6(r, \u03c4 ) that change by a multiple of 2\u03c0 as one moves around a certain reference coordinate, the vortex center. Existing in parallel with harmonic phonon-like excitations discussed above, these excitations lead to a wealth of observable phenomena, to be discussed in more detail in Chapter 8. However, for the moment let us turn to the discussion of another prominent super\ufb02uid, the condensate of Cooper pairs, more generally known as the superconductor. 6.4 Superconductivity 1853\u2013 Kammerlingh Onnes 1926 Nobel Laureate in Physics in 1913 \u201cfor his investigations on the properties of matter at low temperatures which led, inter alia to the production of liquid c The Nobel helium.\u201d (Image Foundation.) The electrical resistivity of many metals and alloys drops abruptly to zero when the material is cooled to a su\ufb03ciently low temperature. This phenomenon, which goes by the name ofsuperconductivity, was \ufb01rst observed by Kammerlingh Onnes three in Leiden in 1911, years after he \ufb01rst lique\ufb01ed helium. Equally striking, a superconductor cooled below its transition temperature in a magnetic \ufb01eld expels all magnetic \ufb02ux from its interior. (One of the more spectacular manifestations of the \ufb01eld-aversion of superconductors is exempli\ufb01ed in the \ufb01gure below: a magnet levitated by a superconductor due to the expulsion of magnetic \ufb02ux.) This phenomenon of perfect diamagnetism is known as the Meissner e\ufb00ect and is characteristic of superconductivity. Indeed, the Meissner e\ufb00ect and dissipationless transport are but two of a plethora of phenomena accompanying superconductivity.16 Along with the introduction of more advanced theoretical machinery, a variety of superconducting phenomena is discussed in the remainder of this text. The present, introductory section is devoted to the formulation of the theoretical foundations of the conventional \u201cBCS\u201d theory of superconductivity, cast into the language of the \ufb01eld integral. Although the presentation is self-contained, our focus is on the theoretical aspects. Depending on taste, some readers may \ufb01nd it useful to motivate their encounter with the formalism developed below by \ufb01rst familiarizing themselves with the basic phenomenology of the BCS superconductor. 16 In fact, it is not even appropriate to speak about the phenomenon of \u201csuperconductivity\u201d as deriving from the same microscopic origin: since the discovery of the class of high-temperature cuprate superconductors in 1986, it has become increasingly evident that the physical mechanisms responsible for high-temperature and \u201cconventional\u201d superconductivity are likely to be strikingly di\ufb00erent. 266 Broken symmetry and collective phenomena Basic concepts of BCS theory of Bardeen John 1908\u20131991 (left), Leon N. Cooper 1930\u2013 (center), and John R. Schrie\ufb00er 1931\u2013 (right) Nobel Laureates in Physics in 1972 for their theory of superconductivity. (Bardeen was also awarded the 1956 Nobel Prize in Physics for his research on semiconductors c The Nobel and discovery of the transistor e\ufb00ect.) (Images Foundation.) Superconductivity an ordered involves state conduction electrons in a metal, caused by the presence of a residual attractive interaction at the Fermi surface. The nature and origin of the ordering were elucidated in a seminal work by Bardeen, Cooper, and Schrie\ufb00er \u2013 BCS theory17 \u2013 some 50 years after its discovery! At low temperatures, an attractive pairwise interaction can induce an instability of the electron gas towards the formation of bound pairs of time-reversed states k \u2191 and \u2212k \u2193 in the vicinity of the Fermi surface. From where does an attractive interaction between charged particles appear? In conventional (BCS) superconductors, attractive correlations between electrons are due to the exchange of lattice vibrations, or phonons: The motion of an electron through a metal causes a dynamic local distortion of the ionic crystal. Crucially, this process is governed by two totally di\ufb00erent time scales. For an electron, it takes a time \u223c E\u22121 F to traverse the immediate vicinity of a lattice ion and to trigger a distortion out of its equilibrium position into a con\ufb01guration that both particles \ufb01nd energetically bene\ufb01cial (top right panel of the \ufb01gure). \u20131 E F", " \u03c9\u22121 D E\u22121 F However, once the ion has been excited it needs a time of O to relax back into its equilibrium position (middle left). Here, \u03c9D denotes the Debye frequency, i.e. the characteristic scale for phonon excitations. This means that, long after the \ufb01rst electron has passed, a second electron may bene\ufb01t from the distorted ion potential (middle right). Only after the ion has been left alone for a time > \u03c9\u22121 D does it relax back into its equilibrium con\ufb01guration (bottom left and right). The net e\ufb00ect of this retardation mechanism is an attractive interaction between the two electrons. Since the maximum energy scale of ionic excitations is given by the Debye frequency, the range of the interaction is limited to energies \u223c \u03c9D around the Fermi surface. (For a more quantitative formulation, see Problem 4.5.) As regards the high-temperature cuprate superconductors, the particular mechanism of pair formation remains (at the time of writing) controversial, although the consensus is that its origin is rooted in spin \ufb02uctuations. \u223c\u03c9 D \u20131 17 J. Bardeen, L. N. Cooper, and J. R. Schrie\ufb00er, Microscopic theory of superconductivity, Phys. Rev. 106 (1957), 162-4; Theory of Superconductivity, 108 (1957), 1175-204. 6.4 Superconductivity 267 Comprising two fermions, the electron\u2013electron bound states, known as Cooper pairs, mimic the behavior of bosonic composite particles.18 At low temperatures, these quasibosonic degrees of freedom form a condensate which is responsible for the remarkable properties of superconductors, such as perfect diamagnetism. To appreciate the tendency to pair formation in the electron system, consider the diagram shown in the \ufb01gure below. The region of attractive correlation is indicated as a shaded ring of width \u223c \u03c9D/vF. Now, consider a two-electron state |k \u2191, \u2212k \u2193 formed by two particles of (near) opposite momentum and opposite spin.19 Momentum conserving scattering of the constituent particles may lead to the formation of a new state |(k + p) \u2191, \u2212(k + p) \u2193 \u2261 |k \u2191, \u2212k \u2193 of the same, opposite-momentum structure. Crucially, the momentum transfer p may trace out a large set of values of O(kd\u22121 F \u03c9D/vF) without violating the condition that the \ufb01nal states be close to the Fermi momentum. (By contrast, if the initial state had not been formed by particles of opposite momentum, the phase space for scattering would have been greatly diminished.) Remembering our previous discussion of the RPA approximation, we recognize a familiar mechanism: an a priori weak interaction may amplify its e\ufb00ect by conspiring with a large phase space volume. \u03c9 D \u2044 EF k' \u03c5 k F To explore this mechanism in quantitative terms, we will adopt a simpli\ufb01ed model de\ufb01ned by the Hamiltonian \u2013k' \u02c6H = k\u03c3 k \u02c6nk\u03c3 \u2212 g Ld k,k,q c \u2020 k+q\u2191c \u2020 \u2212k\u2193c\u2212k+q\u2193ck\u2191, (6.14) where g represents a (positive) constant. The Hamiltonian \u02c6H should be interpreted as an e\ufb00ective Hamiltonian describing the physics of a thin shell of states of width O(\u03c9D) centered around the Fermi surface (i.e. the region where a net attractive interaction prevails). Although a more realistic model of attraction would involve a complicated momentumdependent interaction such as the one obtained from the detailed consideration of the electron\u2013phonon interaction (see Problem 4.5), the simple constant pairing interaction captures the essential physics.20 After the trio who \ufb01rst explored its phenomenology, the model Hamiltonian (6.14) is commonly referred to as the BCS Hamiltonian. 18 Strictly speaking, this identi\ufb01cation deserves some quali\ufb01cation. In the superconducting context, Cooper pairs have a length scale (the coherence length to be introduced below) which typically exceeds the average particle spacing of the electron gas (usually by as much as three orders of magnitude). In this sense, it can be misleading to equate a pair with a single composite particle. The interpretation of the BCS state as a Bose\u2013Einstein condensate of Cooper pairs is developed more fully in the problem set in the discussion of the BEC\u2013BCS crossover \u2013 see Problem 6.7. 19 Note that there are a", " minority of superconducting materials \u2013 the spin triplet superconductors \u2013 in which electrons of equal spin are paired. 20 More importantly, to simplify our discussion, we will take the electrons to be otherwise non-interacting. In fact, the presence of a repulsive Coulomb interaction of the electrons plays a crucial role in controlling the properties of the superconductor. For an in-depth discussion of the role played by repulsive interactions, see A. I. Larkin and \u2013k 268 Broken symmetry and collective phenomena Now the preliminary discussion above does not explain why an attractive interaction is so special. Nor does the discussion elucidate the phenomenological consequences of pair scattering at the Fermi surface. In the following we will address these issues from a number of di\ufb00erent angles. The result will be a heuristic picture of the superconductor that will guide us in constructing the more rigorous \ufb01eld integral approach below. Mainly for illustrative purposes, we begin our discussion with a brief perturbative analysis of Cooper pair scattering. Proceeding in close analogy to the previous discussion of the RPA, we discover the dramatic consequences of an attractive interaction on the ground state of the system. (However, this part of the discussion is an optional (if instructive) element of the development of the theory. Readers who did not yet navigate Section 5.3 may choose to skip this part of the discussion and proceed directly to Section 6.4 where the mean-\ufb01eld picture of the superconductor is developed.) Cooper instability To explore the fate of a Cooper pair under multiple scattering, let us consider the four-point correlation function C(q, \u03c4 ) = 1 L2d k,k 9 \u00af\u03c8k+q\u2191(\u03c4 ) \u00af\u03c8\u2212k\u2193(\u03c4 ) \u03c8k+q\u2193(0)\u03c8\u2212k\u2191(0) :. This describes the amplitude of Cooper pair propagation |k + q \u2191, \u2212k \u2193 \u2192 |k + q \u2191, \u2212k \u2193 in an imaginary time \u03c4 and averaged over all initial and \ufb01nal particle momenta.21 As is usual with problems whose solution must depend only on time di\ufb00erences, it is convenient 0 d\u03c4 e\u2212i\u03c9m\u03c4 C(q, \u03c4 ), to switch to a frequency representation. With C(q) \u2261 C(q, \u03c9m) = T where \u03c9m denotes a bosonic Matsubara frequency, it is straightforward to verify that C(q) = : T 2 L2d \u00af\u03c8k+q\u2191 \u00af\u03c8\u2212k\u2193\u03c8k+q\u2193\u03c8\u2212k\u2191 k,k 9 \u03b2. To calculate the correlation function, let us draw on the perturbative methods introduced in Section 5.3. As with the analysis of the RPA, the density of the electron gas will play the role of a large parameter, i.e. one must expand the correlation function in pair interaction vertices g and retain only those terms that appear with one free momentum summation per interaction. Summation over these contributions leads to the ladder diagram series shown in Fig. 6.5, where the momentum labels of the Green functions are hidden for clarity. According to the de\ufb01nition of the correlation function, the two Green functions entering the ladder carry momenta k +q and \u2212k, respectively. Momentum conservation then implies that the Green functions de\ufb01ning each consecutive rung of the ladder also carry near opposite momenta p + q and \u2212p, wherep is a summation variable. A. A. Varlamov, Fluctuation phenomena in superconductors, in Handbook on Superconductivity: Conventional and Unconventional Superconductors, ed. K.-H. Bennemann and J. B. Ketterson (Springer-Verlag, 2002). 21 As with applications of the path integral discussed previously, information about the real time dynamics of the pair can be extracted from the analytical continuation \u03c4 \u2192 it. Further, notice that the \u201ccenter of mass\u201d |k + q \u2191, \u2212k \u2193 can be interpreted as a variable Fourier conjugate to the center momentum q of a pair 1 Ld of the pair. (Exercise: Show this by inverse Fourier transform.) An equivalent interpretation of the correlation function C is that it describes the wandering of the Cooper pair under the in\ufb02uence of scattering. k C = = + + 6.4 Superconductivity 269 + +... = + p + q, \u2013p, Figure 6.5 Two-particle propagator in the presence of an (attractive) interaction. The two Green function lines de\ufb01ning each rung of the ladder carry momenta p + q and \u2212p, respectively, where p is a free summation variable. The vertex of the propagator, de\ufffd", "\ufffd\ufffdned through the second line, obeys the Bethe\u2013Salpeter equation de\ufb01ned on the bottom right. EXERCISE Convince yourself that the ladder diagrams shown in the \ufb01gure are the only diagrams that contain one free momentum summation per interaction vertex. As with the previous discussion in Section 5.3, the central part of the correlation function is described by a vertex \u0393. The diagrammatic de\ufb01nition of that object is shown in the bottom right part of Fig. 6.5. Translating from diagrammatic to an algebraic formulation, one obtains the Cooper version of a Bethe\u2013Salpeter equation \u0393q = g + gT p Gp+qG\u2212p\u0393q, Ld where we have anticipated that a solution independent of the intermediate momenta can be found. Solving this equation for \u0393q, we arrive at an equation structurally similar to Eq. (5.44): \u0393q = 1 \u2212 gT Ld g p Gp+qG\u2212p. (6.15) Drawing on the results of Problem 4.5, the frequency part of the summation over p gives T Ld p Gp+qG\u2212p = = 1 Ld 1 Ld p p 1 \u2212 nF(\u03bep+q) \u2212 nF(\u03be\u2212p) i\u03c9m + \u03bep+q + \u03be\u2212p 1 2 \u2212 nF(\u03bep) 1 i\u03c9m + \u03bep+q + \u03be\u2212p + (q \u2194 \u2212q), where, in the last line, we have made use of the symmetry of the energy arguments, \u03bep = \u03be\u2212p. The summation for non-zero q is left as an instructive exercise in Fermi-surface integration. However, for the sake of our present argument, it will be su\ufb03cient to perform the sum for zero external momentum q = (0, 0) (i.e. we will probe the fate of spatially homogeneous and static pair con\ufb01gurations). Using the identity 1 d \u03bd()F () to replace Ld the momentum sum by an energy integral, and remembering that the pairing interaction is limited to a thin shell around the Fermi surface, we then obtain p F (p) = T Ld p GpG\u2212p = \u03c9D \u2212\u03c9D d \u03bd() 1 \u2212 2nF() 2 \u03bd \u03c9D T d = \u03bd ln, \u03c9D T (6.16) 270 Broken symmetry and collective phenomena where we have used the fact that, at energies \u223c T, the 1/ singularity of the integrand is cut o\ufb00 by the Fermi distribution function. Substitution of this result back into the expression for the vertex leads to the result \u0393(0,0) g 1 \u2212 g\u03bd ln. \u03c9D T From this, one can read o\ufb00 essential elements of the transition to the superconducting phase. We \ufb01rst note that the interaction constant appears in combination with the density of states, i.e. even a weak interaction can lead to sizeable e\ufb00ects if the density of states is large enough. From our previous qualitative discussion it should be clear that the scaling factor \u03bd simply measures the number of \ufb01nal states accessible to the scattering mechanism depicted in the \ufb01gure on page 266. The net strength of the Cooper pair correlation grows upon increasing the energetic range \u03c9D of the attractive force or, equivalently, on lowering the temperature. Obviously, something drastic happens when g\u03bd ln = 1, i.e. when \u03c9D T T = Tc \u2261 \u03c9Dexp \u2212 1 g\u03bd. At this critical temperature, the vertex develops a singularity. Since the vertex and the correlation function are related by multiplication by a number of (non-singular) Green functions, the same is true for the correlation function itself. As we will soon see, Tc marks the transition temperature to the superconducting state. At and below Tc a perturbative approach based on the Fermi sea of the non-interacting system as a reference state breaks down. The Cooper instability signals that we will have to look for an alternative ground state or \u201cmean-\ufb01eld,\u201d i.e. one that accounts for the strong binding of Cooper pairs. In the next section, we explore the nature of the superconducting state from a complementary perspective. Mean-\ufb01eld theory of superconductivity The discussion of the previous section suggests that, at the transition, the system develops an instability towards pair binding, or \u201ccondensation.\u201d In the next section we build on this observation to construct a quantitative approach, based on", " a Hubbard\u2013Stratonovich decoupling in the Cooper channel. However, for the moment, let us stay on a more informal level and assume that the ground state |\u03a9s of the theory is characterized by the presence of a macroscopic number of Cooper pairs. More speci\ufb01cally, let us assume that the operator k c\u2212k\u2193ck\u2191 acquires a non-vanishing ground state expectation value, \u0394 = g Ld k \u03a9s|c\u2212k\u2193ck\u2191|\u03a9s, \u00af\u0394 = g Ld k \u03a9s|c \u2020 k\u2191c \u2020 \u2212k\u2193|\u03a9s, (6.17) 6.4 Superconductivity 271 where we have included the coupling constant of the theory for later convenience. The assumption that \u0394 assumes non-zero values (vanishes) below (above) the transition temperature Tc is tantamount to declaring \u0394 to be the order parameter of the superconducting transition. However, at the present stage, this statement has the status of a mere presumption; we will have to explore its validity below. At any rate, the non-vanishing expectation value of \u0394 looks strange. It clearly implies that the fermion many-body state |\u03a9s cannot have a de\ufb01nite number of particles (cf. the coherent states). However, a better way to think about the problem is to remember the \u2020 \u2020 bosonic nature of the two-fermion pair state |k \u2191, \u2212k \u2193. From this perspective, c k\u2191c \u2212k\u2193 appears as the operator that creates a bosonic excitation. Non-vanishing of its expectation value implies a condensation phenomenon akin to the condensates discussed in Section 6.3. Indeed, much of the remainder of this section is devoted to the (semi-phenomenological) construction of a \u201cbosonic\u201d mean-\ufb01eld picture of the superconductor. To develop this description, let us substitute k c\u2212k+q\u2193ck\u2191 = Ld\u0394 g + k c\u2212k+q\u2193ck\u2191 \u2212 \u0394Ld g! small, into the microscopic Hamiltonian and retain only those terms that appear as bilinears in the electron operators. Adding the chemical potential, and setting \u03bek = k \u2212 \u03bc, the \u201cmean-\ufb01eld\u201d Hamiltonian takes the form \u02c6H \u2212 \u03bc \u02c6N # k \u03bekc \u2020 k\u03c3ck\u03c3 \u2212 \u00af\u0394c\u2212k\u2193ck\u2191 + \u0394c \u2020 k\u2191c \u2020 \u2212k\u2193 $ + Ld|\u0394|2 g, known (in the Russian literature) as the Bogoliubov or Gor\u2019kov Hamiltonian after its authors (while the terminology Bogoliubov\u2013de Gennes Hamiltonian has become more widespread in the Anglo-Saxon literature, re\ufb02ecting the promotion of the mean-\ufb01eld description by de Gennes). Indeed, although perfectly Hermitian, the Gor\u2019kov Hamiltonian does not conserve particle number. Instead, pairs of particles are born out of, and annihilated into, the vacuum. To bring the mean-\ufb01eld Hamiltonian to a diagonal form, we proceed in a manner analogous to that of Section 2.2 (where appeared a Hamiltonian of similar structure, namely a\u2020a + aa + a\u2020a\u2020). Speci\ufb01cally, let us recast the fermion operators in a two-component Nambu spinor representation \u03a8 \u2020 k = c \u2020 k\u2191, c\u2212k\u2193, \u03a8k = 6 7 ck\u2191 \u2020 c \u2212k\u2193, k2 \u03bb k k1 272 Broken symmetry and collective phenomena comprising \u2191-creation and \u2193-annihilation operators in a single object. It is then straightforward to show that the Hamiltonian assumes the bilinear form \u02c6H \u2212 \u03bc \u02c6N = \u2020 k \u03a8 k \u2212\u0394 \u03bek \u2212 \u00af\u0394 \u2212\u03bek \u03a8k + \u03bek + k Ld|\u0394|2 g. Now, being bilinear in the Nambu operators, the mean-\ufb01eld Hamiltonian can be brought to a diagonal form by employing the unitary transformation22 6 7 \u03c7k \u2261 \u03b1k\u2191 \u2020 \u03b1 \u2212k\u2193 = sin \u03b8k cos \u03b8k sin \u03b8k \u2212 cos \u03b8k 6 7 ck\u2191 \u2020 c \u2212k\u2193 \u2261 Uk\u03a8k, (under which the anti-commutation relations of the new electron operators \u03b1k\u03c3 are main\u2020 \u2020 tained \u2013 exercise). Note that the operators \u03b1 k\u2191 and c\u2212k\u2193", ", i.e. k\u2191 involve superpositions of c the quasi-particle states created by these operators contain linear combinations of particle and hole states. Choosing \u0394 to be real23 and setting tan (2\u03b8k) =\u2212 \u0394/\u03bek, i.e. cos(2\u03b8k) = \u03bek/\u03bbk, sin(2\u03b8k) = \u2212\u0394/\u03bbk, where the transformed Hamiltonian takes the form (exercise) \u03bbk = (\u03942 + \u03be2 k)1/2, \u02c6H \u2212 \u03bc \u02c6N = k\u03c3 \u03bbk\u03b1 \u2020 k\u03c3\u03b1k\u03c3 + k (\u03bek \u2212 \u03bbk) + \u03942Ld g. (6.18) (6.19) This result shows that the elementary excitations, the Bogoliubov quasi-particles, cre\u2020 k\u03c3, have a minimum energy \u0394 known as the energy gap. The full dispersion ated by \u03b1 \u00b1\u03bbk is shown in the \ufb01gure above. Due to the energy gap separating \ufb01lled from empty quasiparticle states, elementary excitations are di\ufb03cult to excite at low temperatures, implying a rigidity of the ground state. To determine the ground state wavefunction one simply has to identify the vacuum state }, i.e the state that is annihilated by all the quasi-particle annihilation of the algebra {\u03b1k, \u03b1 operators \u03b1k\u03c3. This condition is met uniquely by the state \u2020 k |\u03a9s \u2261'k \u03b1k\u2191\u03b1\u2212k\u2193|\u03a9 \u223c cos \u03b8k \u2212 sin \u03b8kc \u2020 k\u2191c \u2020 \u2212k\u2193 |\u03a9,'k where |\u03a9 represents the vacuum state of the fermion operator algebra {ck, c }, and sin \u03b8k = (1 \u2212 \u03bek/\u03bbk)/2. Since the vacuum state of any algebra of canonically conjugate operators is unique, the state |\u03a9s must, up to normalization, be the vacuum state. From the representation given above, it is straightforward to verify that the normalization is unity (exercise). \u2020 k 22 It is instructive to compare with the previous analysis of Section 2.2 where the bosonic nature of the problem enforced diagonalization by a non-compact pseudo-unitary transformation. 23 If \u0394 \u2261 |\u0394|ei\u03c6 is not real, it can always be made so by the global gauge transformation ca \u2192 ei\u03c6/2ca, c\u2020 a \u2192 e\u2212i\u03c6/2c\u2020 a. Notice the similarity to the gauge freedom that led to Goldstone mode formation in the previous section! Indeed, we will see momentarily that the gauge structure of the superconductor has equally far-reaching consequences. Finally, we need to solve Eq. (6.17) self-consistently for the input parameter \u0394: 6.4 Superconductivity 273 \u0394 = g Ld g\u0394 2 k \u03c9D \u2212\u03c9D \u03a9s|c\u2212k\u2193ck\u2191|\u03a9s = \u2212 g Ld k sin \u03b8k cos \u03b8k = g 2Ld \u0394 (\u03942 + \u03be2 k)1/2 k \u03bd(\u03be)d\u03be (\u03942 + \u03be2)1/2 = g\u0394\u03bd sinh\u22121(\u03c9D/\u0394), (6.20) where we have assumed that the pairing interaction g uniformly extends over an energy scale \u03c9D (over which the density of states \u03bd is roughly constant). Rearranging this equation for \u0394, we obtain the important relation \u0394 = \u03c9D sinh(1/g\u03bd) g\u03bd1 2\u03c9Dexp. \u2212 1 g\u03bd (6.21) This is the second time that we have encountered the combination of energy scales on the right-hand side of the equation. Previously we identi\ufb01ed Tc = \u03c9Dexp[\u2212(g\u03bd)\u22121] as the transition temperature at which the Cooper instability takes place. Our current discussion indicates that Tc and the quasi-particle energy gap \u0394 at T = 0 coincide. In fact, that identi\ufb01cation might have been anticipated from our discussion above. At temperatures T < \u0394, thermal \ufb02uctuations are not capable of exciting quasi-particle states above the ground state. One thus expects that Tc \u223c \u0394 separates a low-temperature phase, characterized by the features of the anomalous pairing ground state, from a \u201cFermi-liquid-like\u201d high-temperature phase where free quasi-particle excitations prevail. In the mean-\ufb01eld approximation, the ground state |\u03a9s and its quasi-particle", " excitations formally diagonalize the BCS Hamiltonian. Before proceeding with the further development of the theory, let us pause to discuss a number of important properties of these states. Ground state k sin2 \u03b8 In the limit \u0394 \u2192 0, sin2 \u03b8k \u2192 \u03b8(\u03bc \u2212 k), and the ground state collapses to the \ufb01lled Fermi sea with chemical potential \u03bc. As \u0394 becomes nonzero, states in the vicinity of the Fermi surface rearrange themselves into a condensate of paired states. The latter involves the population of single-particle states with energy k > \u03bc. (This follows simply from the energy dependence of the weight function sin \u03b8k entering the de\ufb01nition of the ground state \u2013 see the \ufb01gure.) However, it is straightforward to show that, for any value g > 0, the total energy of the ground state, E|\u03a9s \u2261 \u03a9s| \u02c6H \u2212 \u03bc \u02c6N |\u03a9s = k(\u03bek \u2212 \u03bbk) + \u03942Ld/g, is lower than the energy E0 \u2261 2 \u03bek of the Fermi sphere when g = 0: cos2 \u03b8 k 0 |k|<pF EXERCISE To show that E|\u03a9s < E0, it is convenient to represent the ground state energy of the Fermi sea as E0 = lim\u0394\u21920 E|\u03a9s. Use this representation (and the solution of the mean\ufb01eld equation) to verify that the superconductor ground state energy lies below that of the uncorrelated Fermi sea. It is also instructive to ask for the minimum value E|\u03a9s may assume 274 Broken symmetry and collective phenomena upon variation of \u0394 for \ufb01xed g. Show that the solution of the variational equation \u2202\u0394E|\u03a9s = 0 leads back to the mean-\ufb01eld equation for \u0394 discussed above. Excitations It is very important to distinguish between quasi-particle states and \u201cexcitations.\u201d Quasiparticle states are the eigenstates of the BCS Hamiltonian. Their energy\u2013momentum relation is shown in the \ufb01gure on page 271. Notice that there is a positive\u2013 and a negative\u2013energy branch of quasi-particles. In the limit \u0394 \u2192 0, the quasi-particles evolve into ordinary \u2020 electrons. By contrast, the energy of excitations (as created by the operators \u03b1 k) is always positive. An excitation can be either the creation of a quasi-particle at positive energy or the elimination of a quasi-particle (the creation of a quasi-hole) at negative energy. (In the ground state, all negative-energy quasi-particle states are \ufb01lled.) As \u0394 \u2192 0, the excitation operators evolve into the operator algebra introduced in Eq. (2.21). Notice that the total number of excitations is equal to the number of quasi-particle states. However, their density of states (non-vanishing for positive energies only) is twice as large. This is because the dispersion of excitations is obtained by superimposing the positive branch of the quasi-particle spectrum on the sign-inverted negative branch. For a particle\u2013hole symmetric system, the quasi-particle spectrum is invariant under sign inversion, which implies that the two branches contribute equally to the density of excitations. After these general remarks we turn to the speci\ufb01c discussion of the excitations of the BCS superconductor. According to Eq. (6.19), the excitation spectrum is gapped, i.e. it takes a minimum energy \u0394 to create an excitation above the BCS ground state; the formerly (g = 0) continuous spectrum has acquired a gap. To better understand the pro\ufb01le of the spectrum, let us compute the density of excitations \u03c1(),24 in the vicinity of the Fermi surface, \u2248 \u03bc: \u03c1() = 1 Ld k\u03c3 \u03b4( \u2212 \u03bbk) = \u221e \u03b4 d\u03be \u2248 \u03bd s=\u00b11 0 1 Ld k\u03c3 \u03b4( \u2212 \u03bb(\u03be)) d\u03be \u03b4(\u03be \u2212 \u03bek)! \u03bd(\u03be) \u03be \u2212 s[2 \u2212 \u03942]1/2 = 2\u03bd\u0398( \u2212 \u0394) \u2202\u03be[\u03be2 + \u03942]1/2 (2 \u2212 \u03942)1/2. A schematic plot of the BCS quasi-particle density of states is shown in the \ufb01gure above. It is apparent that the spectral weight of the quasi-particles has been transferred from the Fermi surface to", " the interval [\u0394, \u221e]. The divergence at \u0394 signals that the majority of quasi-particle states populate the spectral region just above the gap. 24 To distinguish the density of states of the quasi-particles (\u03bd) from that of the excitations, we denote the latter by \u03c1. 6.4 Superconductivity 275 EXERCISE Integrate the result above to con\ufb01rm that that the excitation density of states is indeed twice as large as that of the quasi-particles. \u0394 d \u03c1() E\u0394\u2212\u2192 2E\u03bd. This demonstrates E The phenomenological analysis above explains important aspects of the physics of the BCS superconductor: the instability towards condensation, and the presence of a gap for quasiparticle excitations above the ground state. Indeed, it would be tempting to take the latter phenomenon as an explanation for the absence of electric resistivity below the transition temperature, i.e. if no low-lying quasi-particle excitations are available, should not external current be able to \ufb02ow through the system without energy dissipation? Yet that picture neglects the most important excitation of the system, i.e. the collective phase mode. Above we made the ad hoc decision to set the phase of the order parameter to zero. However, as with the super\ufb02uid, the phase represents a Goldstone mode and we must expect its presence to have important consequences. Indeed, it will turn out that the phase mode of the superconductor \u2013 in contrast to its counterpart in the \u201cneutral\u201d super\ufb02uid \u2013 is responsible for much of the electromagnetic phenomena displayed by the superconductor. INFO The discussion above may convey the impression that the amplitude of the superconducting order parameter is spatially uniform, with a value set by the mean \ufb01eld equation. It is important to realize, however, that there are superconductor species for which |\u0394| exhibits strong \ufb02uctuations in space. Since it is the amplitude of the order parameter that determines the size of the quasi-particle gap, \ufb02uctuations of |\u0394| lead to important phenomenological consequences, especially if there are regions for which |\u0394| \u2192 0. Without question the most important class of superconductors with spatially varying order parameters are the d-wave superconductors. Firstly, practically all high temperature superconductors belong to this family, i.e. d-wave superconductivity is a \ufb01eld of profound interest from both a fundamental and an applied point of view. Secondly, the physical properties of the d-wave superconductor di\ufb00er drastically from those of conventional superconductors. The reason is that it is \u201cgapless\u201d in the sense that its Brillouin zone contains four distinct points at which the order parameter vanishes. These \u201chot spots\u201d are exceptional in that quasi-particles of arbitrarily low energy coexist with the order parameter condensate. The list of many exotic phenomena induced by this coexistence \u2013 for a detailed discussion of which we refer to the literature \u2013 includes unconventional heat conduction and anomalous Josephson and Meissner e\ufb00ects. EXERCISE The order parameter of a lattice d-wave superconductor is given by \u0394k = \u03940(cos(kxa) \u2212 cos(kya)), where k = (kx, ky)T is a two-dimensional lattice momentum and a the lattice spacing.25 At two lines in momentum space, kx = \u00b1ky, the order parameter vanishes and low-energy quasi-particles persist. Assuming that the kinetic energy of the lattice problem is given by tk = \u2212t(cos(kxa) + cos(kya)), compute the quasi-particle energies of the Bogoliubov Hamiltonian \u02c6H = k((tk \u2212 \u03bc)\u03c33 + \u0394k\u03c31)\u03a8k. Show that, at four points in the Brillouin zone, k \u03a8\u2020 25 The origin of the designation \u201cd-wave\u201d is that it transforms like an l = 2 angular momentum state (sign change under a 90\u25e6 rotation) under the action of the rotation group. On the same basis, ordinary superconductors are sometimes called s-wave superconductors. Less frequently occurring are superconductors of p-wave or other angular momentum symmetry. 276 Broken symmetry and collective phenomena k = (\u00b11, \u00b11)\u03c0/2a, the quasi-particle energy vanishes. Linearize the Hamiltonian in momentum space around one of these \u201chot spots,\u201d say (+1, +1)\u03c0/2a, to show that it assumes the form \u02c6H++ = ta(kx + ky) \u03940a(kx \u2212 ky) \u03940a(k", "x \u2212 ky) \u2212ta(kx + ky). Apply a unitary transformation to bring \u02c6H++ to the form of a two-dimensional Dirac Hamiltonian \u02c6H++ \u2192 k1\u03c31 + k2\u03c32, where k1/2 = kx \u2213 ky, and we have rescaled k1 \u2192 (\u03940/t)k1 and set vF = 1. Show that the quasi-particle density of states of this Hamiltonian is given by \u03bd() \u223c ||. From a modern perspective, a comprehensive picture of the superconductor, encompassing the consequences of Goldstone mode formation, and the BCS structures discussed above, is most e\ufb03ciently developed within the \ufb01eld integral approach. It is to this framework that we now turn. Superconductivity from the \ufb01eld integral To investigate the BCS transition within the framework of the coherent state path integral, we start out from a coordinate representation of the BCS Hamiltonian (6.14), \u02c6HBCS = ddr c\u2020 \u03c3(r) (\u2212i\u2207 \u2212 eA)2 + e\u03c6 \u2212 \u03bc c\u03c3(r) \u2212 g ddr c \u2020 \u2191(r)c \u2020 \u2193(r)c\u2193(r)c\u2191(r), 1 2m where, once again, summation over repeated spin indices is implied. Anticipating the emergence of non-trivial electromagnetic phenomena, we have coupled the single-particle Hamiltonian to a vector potential A and a scalar potential \u03c6. The origin and physical consequences of these \ufb01elds are discussed somewhat later. Expressed in the form of the coherent state path integral, the corresponding quantum partition function takes the form Z = D( \u00af\u03c8, \u03c8) e\u2212S[ \u00af\u03c8,\u03c8],26 \u03b2 S[ \u00af\u03c8, \u03c8] = 0 d\u03c4 ddr \u00af\u03c8\u03c3 \u2202\u03c4 + ie\u03c6 + (\u2212i\u2207 \u2212 eA)2 \u2212 \u03bc \u03c8\u03c3 \u2212 g \u00af\u03c8\u2191 \u00af\u03c8\u2193\u03c8\u2193\u03c8\u2191, 1 2m (6.22) where \u03c8(r, \u03c4 ) denote Grassmann \ufb01elds. INFO In the \ufb01eld theoretical literature, the substitution \u2202\u03c4 \u2192 \u2202\u03c4 + ie\u03c6, \u2212i\u2207 \u2192 \u2212i\u2207 \u2212 eA, (6.23) is sometimes called a minimal coupling of an electromagnetic \ufb01eld. It is \u201cminimal\u201d in the sense that only orbital and potential coupling of the \ufb01eld are taken into account. (For example, the \ufb01eld\u2013spin interaction is neglected.) At the same time, the potential\u2013vector-potential coupling is complete enough to endow the theory with a local gauge invariance under U(1) transformations \u03c8 \u2192 ei\u03b8\u03c8, \u00af\u03c8 \u2192 e\u2212i\u03b8 \u00af\u03c8, \u03c6 \u2192 \u03c6 \u2212 e\u22121\u2202\u03c4 \u03b8, A \u2192 A + e\u22121\u2207\u03b8, (6.24) 26 Notice that the Eucilidean character of the imaginary time theory is responsible for the additional factor of \u201ci,\u201d i.e. as our \u03c6 transforms like a time-derivative under Wick rotation, it is given by (\u2212i) \u00d7 the conventional scalar potential. 6.4 Superconductivity 277 where \u03b8 = \u03b8(\u03c4, x) is an arbitrary space/time-dependent phase con\ufb01guration. The minimal coupling introduces the general quantum electrodynamical gauge principle into the theory. This should be compared with the discussion of the neutral super\ufb02uid where only invariance under global U(1) transformations was required. In the following, we will often use units wherein e = 1. As usual the quartic interaction of the \ufb01elds prevents the partition function from being evaluated explicitly. Moreover, anticipating the existence of a transition of the electron gas to a condensed phase in which electrons in the vicinity of the Fermi surface are paired, we can expect that a perturbative expansion in the coupling constant g will be inadequate. Motivated by the mean-\ufb01eld theory discussed above, we will instead introduce a bosonic \ufb01eld \u0394 to decouple the interaction, which will have the physical signi\ufb01cance of the order parameter. The decoupling is arranged using a Hubbard\u2013Stratonovich transformation in the Cooper channel (cf. the discussion on page 244) ) * exp g d\u03c4 ddr \u00af\u03c8\u2191 \u00af\u03c8\u2193\u03c8\u2193\u03c8\u2191 ) = D( \u00af\u0394, \u0394) exp \u2212", " d\u03c4 ddr 1 g |\u0394|2 \u2212 \u00af\u0394\u03c8\u2193\u03c8\u2191 + \u0394 \u00af\u03c8\u2191 \u00af\u03c8\u2193 *, where \u0394(r, \u03c4 ) represents a dynamically \ufb02uctuating complex \ufb01eld. Re\ufb02ecting the behavior of the bilinear \u03c8\u2193\u03c8\u2191, it obeys the periodic boundary condition, \u0394(0) = \u0394(\u03b2), i.e. \u0394 is a bosonic \ufb01eld variable. Were we to take \u0394 to be homogeneous in space and time, the quantum Hamiltonian corresponding to the action would coincide with that of the mean-\ufb01eld Hamiltonian considered in the previous section. Following that analysis (but not making the a priori assumption \u0394(r, \u03c4 ) = const.) we turn to theNambu spinor representation \u00af\u03a8 = \u00af\u03c8\u2191 \u03c8\u2193,, \u03c8\u2191 \u00af\u03c8\u2193 \u03a8 = comprising particle and hole degrees of freedom in a single object. Expressed in terms of the Nambu spinor, the partition function takes the form * ) Z = D( \u00af\u03c8, \u03c8) D( \u00af\u0394, \u0394) exp \u2212 d\u03c4 ddr |\u0394|2 \u2212 \u00af\u03a8 \u02c6G\u22121\u03a8, 1 g where \u239b \u239c \u239d $\u22121 # \u02c6G(p) 0 \u00af\u0394 # \u02c6G\u22121 = \u0394 $\u22121 \u02c6G(h) 0 \u239e \u239f \u23a0, (6.25) is known as the Gor\u2019kov Green function, and [ \u02c6G(p) [ \u02c6G(h) of the particle and hole respectively. 2m (\u2212i\u2207\u2212eA)2 +\u03bc, 2m (+i\u2207 \u2212 eA)2 \u2212 \u03bc represent the non-interacting Green functions 0 ]\u22121 = \u2212\u2202\u03c4 + ie\u03c6 + 1 0 ]\u22121 = \u2212\u2202\u03c4 \u2212ie\u03c6\u2212 1 INFO Computing the \u03a8-representation of the action for a general single-particle Hamiltonian 0 ]\u22121 = \u2212\u2202\u03c4 + \u02c6H T \u2212 \u03bc. (With \u2207T = \u2212\u2207, the \u02c6H, one \ufb01nds that [ \u02c6G(p) expression above is identi\ufb01ed as a special case of the more general form.) This representation is 0 ]\u22121 = \u2212\u2202\u03c4 \u2212 \u02c6H + \u03bc, [ \u02c6G(h) 278 Broken symmetry and collective phenomena actually very revealing: it tells us that the Green function of a hole is obtained from that of the electron by a sign change \u02c6H \u2192 \u2212 \u02c6H (the energy of a hole is the negative of the corresponding particle energy) followed by transposition \u2212 \u02c6H \u2192 \u2212 \u02c6H T, i.e. a quantum time reversal operation (a hole can be imagined as a particle propagating backwards in time). Noting that \u02c6pT = \u2212\u02c6p and \u02c6xT = \u02c6x, the pair of Green functions can equivalently be represented through [ \u02c6G(p) 0 ]\u22121 = \u2212\u2202\u03c4 \u2212 \u02c6H(\u02c6x, \u02c6p) + \u03bc, [ \u02c6G(h) 0 ]\u22121 = \u2212\u2202\u03c4 + \u02c6H(\u02c6x, \u2212\u02c6p) \u2212 \u03bc. (6.26) Bilinear in \u03a8, the Gaussian integration over the Grassmann \ufb01elds can now be performed straightforwardly, and yields the formal expression (cf. the analogous formula (6.4) for the normal electron system) Z = D( \u00af\u0394, \u0394) exp d\u03c4 ddr |\u0394|2 + ln det \u02c6G\u22121. (6.27) \u2212 1 g By introducing a Hubbard\u2013Stratonovich decoupling of the local interaction, we have succeeded in expressing the quantum partition function as a path integral over an auxiliary bosonic \ufb01eld \u0394. Further progress is possible only within some approximation. Mean-\ufb01eld theory # To identify a suitable \u201cplatform\u201d around which the action can be expanded perturbatively, we begin by identifying the mean-\ufb01eld con\ufb01gurations of \u0394. A variation of the action with respect to \u0394 generates the mean-\ufb01eld equation (for the di\ufb00erentiation of the \u201ctr ln\u201d with respect to \u0394 we refer to the analogous calculation in Eq. (6.5)), g\u22121 \u00af\u0394", "(\u03c4, x) \u2212 is a 2 \u00d7 2 matrix in Nambu space that takes the value tr of unity at position (i, j) and is zero otherwise. Assuming that con\ufb01gurations extremizing the action will be spatially and temporally homogeneous, i.e \u0394(\u03c4, x) \u2261 \u03940 = const., and, temporarily ignoring the dependence of the Green function on the \ufb01eld (\u03c6, A), the equation simpli\ufb01es to = 0, where Eph ij \u02c6G(\u03c4, x; \u03c4, x)Eph 12 $ \u00af\u03940 = tr \u23a3 1 g \u23a1 6 \u2212\u2202\u03c4 + \u22072 \u00af\u03940 2m + \u03bc 7\u22121 (\u03c4, x; \u03c4, x) \u2212 \u03bc 1 0 \u23a4 \u23a6 = T Ld p,n i\u03c9n \u2212 \u03bep \u00af\u03940 \u03940 i\u03c9n + \u03bep = T Ld p,n \u00af\u03940 p + |\u03940|2, \u03c92 n + \u03be2 21 \u03940 \u2212\u2202\u03c4 \u2212 \u22072 2m \u22121 where, in the second line, we have switched to a frequency\u2013momentum representation and, as usual, \u03bep = p2/2m \u2212 \u03bc. Rearranging the equation, we arrive at 1 1, n+\u03bb2 p 0)1/2 > 0 (cf. Eq. (6.18)). The Matsubara summation can be performed where \u03bbp = (\u03be2 by means of the summation techniques discussed on page 170 (see also the problem set of Chapter 4), after which one obtains, g = T Ld p + \u03942 p,n \u03c92 1 g = 1 Ld p 1 \u2212 2nF(\u03bbp) 2\u03bbp = \u03c9D \u2212\u03c9D d\u03be 1 Ld p \u03b4(\u03be \u2212 \u03bep)! \u03bd(\u03be) 1 \u2212 2nF(\u03bb(\u03be)) 2\u03bb(\u03be), 6.4 Superconductivity 279 where we have taken into account the fact that the range of the attractive interaction is limited by \u03c9D. Noting that the integrand is even in \u03be, and making use of the identity 1 \u2212 2nF() = tanh(/2T ), we arrive at the celebrated BCS gap equation 1 g\u03bd \u03c9D = d\u03be 0 tanh(\u03bb(\u03be)/2T ) \u03bb(\u03be), (6.28) where, as usual, we have assumed that the density of states \u03bd(\u03be) \u03bd is speci\ufb01ed by its value at the Fermi surface. For temperatures T \u03940, we may approximate tanh(\u03bb/2T ) 1 and we arrive back at the T = 0 gap equation (6.20) analyzed above. However, here we wish to be more ambitious and explore the fate of the gap as the temperature is increased. Intuitively, one would expect that, for large temperatures, thermal \ufb02uctuations will eventually wash out the gap. On the other hand, we know empirically that the onset of superconductivity has the character of a second order phase transition (see Chapter 8). Since the gap parameter \u03940 has the status of the order parameter of the transition \u2013 an identi\ufb01cation to be substantiated shortly \u2013 we must expect that the vanishing of \u03940(T ) occurs in a singular manner (in analogy to, e.g., the magnetization of a ferromagnet at the Curie temperature). Indeed, it turns out (see Problem 6.7) that the order parameter vanishes abruptly at the critical temperature of the BCS transition, Tc = const. \u00d7 \u03c9Dexp, (6.29) \u2212 1 g\u03bd where the numerical constant is O(1). Notice that this result is consistent with our perturbative analysis above. For temperatures slightly smaller than Tc (see Problem 6.7), \u03940 = const. \u00d7 Tc(Tc \u2212 T ), (6.30) scales as the square root of (Tc \u2212T )/Tc, i.e. the vanishing occurs with a diverging derivative, as is typical for second-order phase transitions. The interpolated temperature pro\ufb01le of the order parameter is shown in Fig. 6.6. Notice that (again, up to numerical factors) the critical temperature Tc coincides with the zero-temperature value of the gap \u03940(0). The square-root pro\ufb01le of the gap function has been accurately con\ufb01rmed by experiment (see Fig. 6.6). Having explored the large-scale pro\ufb01le of the gap function,", " we next turn our attention to the vicinity of the superconductor transition, i.e. to temperature regions \u03b4T = Tc \u2212 T T. Ginzburg\u2013Landau theory In the vicinity of the phase transition, the gap parameter \u0394 is small (in comparison with the temperature). This presents the opportunity to perturbatively expand the action (6.27) in \u0394. We will see that the expansion reveals much about the character of the superconducting transition and the nature of the collective excitations. Further, the expansion will make the connection to the neutral super\ufb02uid (as well as important di\ufb00erences) explicit. To keep the structure of the theory as transparent as possible, we will continue to ignore the coupling to the external \ufb01eld. Our task thus reduces to computing the expansion of 280 Broken symmetry and collective phenomena 0..4 Tin 33.5 Mc Tin 54 Mc BCS theory 0 0.4 T / Tc 0.8 Figure 6.6 Measurements of ultrasonic attenuation in a superconductor provide access to the ratio \u0394(T )/\u0394(0). The data here show the comparison of a tin alloy with the predicted square-root dependence of the mean-\ufb01eld theory. (Reprinted with permission from R. W. Morse and M. V. Bohm, Superconducting energy gap from ultrasonic attenuation measurements, Phys. Rev. 108 (1957), 1094\u20136. Copyright (1957) by the American Physical Society.) tr ln \u02c6G\u22121 in powers of \u0394. To facilitate the expansion, let us formally de\ufb01ne \u02c6G\u22121 and set \u02c6\u0394 \u2261 \u0394 0 \u2261 \u02c6G\u22121|\u0394=0, \u00af\u0394, so that # $ tr ln \u02c6G\u22121 = tr ln \u02c6G\u22121 0 (1 + \u02c6G0 \u02c6\u0394) = tr ln \u02c6G\u22121! const. 0 + tr ln [1 + \u02c6G0 \u02c6\u0394].! \u221e 2n tr( \u02c6G0 \u02c6\u0394)2n 1 n=0 \u2212 Here we have used the relation tr ln [ \u02c6A \u02c6B] = tr ln \u02c6A + tr ln \u02c6B.27 Further, note that only even contributions in \u02c6\u0394 survive. The constant contribution tr ln \u02c6G\u22121 recovers the free energy of the non-interacting electron gas and, for present purposes, provides an inessential contribution to the action. 0 To give this formal expansion some meaning, let us consider the second-order term in more detail. By substituting the explicit form of \u02c6G\u22121 0 tr ( \u02c6G0 \u02c6\u0394)2 = \u2212tr [ \u02c6G0]11\u0394[ \u02c6G0]22 \u00af\u0394 = \u2212 \u2212 1 2 it is straightforward to verify that T Ld p q [ \u02c6G0,p]11[ \u02c6G0,p\u2212q]22! p GpG\u2212p+q \u2212 T Ld \u0394(q) \u00af\u0394(q), where we have made use of the representation of the composite Green function \u02c6G0 in terms of the single-particle Green function G(p) \u2212p = p = Gp and the hole Green function G(h) p = \u2212G(p) 27 Notice that the relation tr ln [ \u02c6A \u02c6B] = tr ln \u02c6A + tr ln \u02c6B applies to non-commutative matrices. 6.4 Superconductivity 281 \u2212G\u2212p (cf. Eq. (6.26)). On combining with the \ufb01rst term in Eq. (6.27), we arrive at the quadratic action for the order parameter \ufb01eld, S(2)[\u0394, \u00af\u0394] = q \u0393\u22121 q |\u0394(q)|2, \u0393\u22121 q = 1 g \u2212 T Ld p GpG\u2212p+q. (6.31) This is our second encounter with the vertex function \u0393\u22121 q : in our perturbative analysis of the Cooper channel (cf. Eq. (6.15)) we had identi\ufb01ed the same expression. To appreciate the connection we should let the dust of our current technical operations settle and revise the general philosophy of the Hubbard\u2013Stratonovich scheme. The \ufb01eld \u0394 was introduced to decouple an attractive interaction in the Cooper channel. By analogy with the \ufb01eld \u03c6 used in the development of the", " RPA approximation to the direct channel, the action of the \ufb01eld \u0394 \u223c \u00af\u03c8\u2191 \u00af\u03c8\u2193 can be interpreted as the \u201cpropagator\u201d of the composite object \u00af\u03c8\u2191 \u00af\u03c8\u2193, i.e. a quadratic contraction \u223c \u00af\u0394\u0394 describes propagation in the Cooper channel \u223c \u00af\u03c8\u2191 \u00af\u03c8\u2193\u03c8\u2193\u03c8\u2191, as described by a four-point correlation function. This connection is made explicit by comparison of the quadratic action with the direct calculation of the Cooper four-point function above. However, in contrast to our discussion in Section 6.4 (where all we could do was to diagnose an instability as \u0393\u22121 \u2192 0), we are now in a position to comprehensively explore q=0 the consequences of the symmetry breaking. Indeed, \u0393\u22121 \u2192 0 corresponds to a sign change q=0 of the quadratic action of the constant order parameter mode \u0394(q = 0). In the vicinity of this point, the constant contribution to the action must scale as \u223c (T \u2212 Tc) from which one may conclude that the action assumes the form S(2)[\u0394, \u00af\u0394] = d\u03c4 ddr r(T ) 2 |\u0394|2 + O(\u2202\u0394, \u2202\u03c4 \u0394), where r(T ) \u223c T \u2212 Tc and O(\u2202\u0394, \u2202\u03c4 \u0394) denotes temporal and spatial gradients whose role will be discussed shortly. EXERCISE Use Eq. (6.16) and the expansion nF(, T ) \u2212 nF(, Tc) (T \u2212 Tc)\u2202T |T =Tc nF(, T ) = \u2212\u2202nF(, Tc)(T \u2212 Tc) T, to show that r(T ) = \u03bdt where t = T \u2212Tc Tc de\ufb01nes the reduced temperature. For temperatures below Tc, the quadratic action becomes unstable and \u2013 in direct analogy with our previous discussion of the super\ufb02uid condensate action \u2013 we have to turn to the fourth-order contribution, S(4), to ensure stability of the functional integral (see the \ufb01gure below where the upper/lower surface corresponds to temperatures above/below the transition). At orders n >2 of the expansion, spatial and temporal gradients can be safely 282 Broken symmetry and collective phenomena neglected (due to the smallness of \u0394 T, they will certainly be smaller than the gradient contributions to S(2)). For \u0394 = const., it is straightforward to verify that S(2n) = 1 2n tr ( \u02c6G0 \u02c6\u0394)2n = (\u2212)n 2n p = const. \u00d7 \u03bd|\u0394|2n \u03c9l 1 \u03c92n\u22121 l (GpG\u2212p)n |\u0394|2n \u03bd|\u0394|2n 2n 2n = const. \u00d7 \u03bdT, |\u0394| T \u03c9D \u03c9l \u2212\u03c9D d\u03be l + \u03be2)n (\u03c92 where \u201cconst.\u201d denotes only numerical conin the second stant factors. Once again, equality, we have expressed the Gor\u2019kov Green function through the respective particle and hole Green functions, and in the fourth equality we have noticed that, for \u03c9D T, the integral over the energy variable is dominated by the infrared divergence at small \u03be, i.e. d\u03be \u03be\u22122n. This esti- \u221e \u03c9l S Im \u0394 Re \u0394 d\u03be (\u03c92 l + \u03be2)\u2212n \u03c9D 0 mate tells us that the contributions of higher order to the expansion are (i) positive and (ii) small in the parameter |\u0394|/T 1. This being so, it is su\ufb03cient to retain only the fourth-order term (to counterbalance the unstable second-order term). We thus arrive at the e\ufb00ective action for the order parameter, S[\u0394, \u00af\u0394] = d\u03c4 ddr r(T ) 2 \u00af\u0394\u0394 + u(\u0394 \u00af\u0394)2 + O(\u2202\u0394, \u2202\u03c4 \u0394, |\u0394|6), (6.32) valid in the vicinity of the transition.28 In particular, one may note (see the \ufb01gure above) that the dependence of the action on (a constant) \u0394 mimics closely that of the condensate amplitude. INFO It is straightforward to include \ufb01nite spatial gradients in the derivation of the quadratic action S(2) (for details see Problem 6.7). The resulting action for static but", " spatially \ufb02uctuating con\ufb01gurations \u0394(r) takes the form SGL[\u0394, \u00af\u0394] = \u03b2 ddr ( r 2 |\u0394|2 + c 2 ) |\u2202\u0394|2 + u|\u0394|4, (6.33) where c \u223c \u03c10(vF/T )2. This result is known as the (classical) Ginzburg\u2013Landau action of the superconductor. It is termed \u201cclassical\u201d because (cf. our remarks on page 255) temporal \ufb02uctuations of \u0394 have been ignored. Notice that the form of the action might have been anticipated on symmetry grounds alone. Indeed, Eq. (6.33) was proposed by Ginzburg and Landau as an e\ufb00ective action for superconductivity some years before the advent of the microscopic theory.29 28 Here, the coupling constant u \u223c \u03c1T \u22123 shows only a weak temperature dependence in the vicinity of the transition. 29 V. L. Ginzburg and L. D. Landau, On the theory of superconductivity, Zh. Eksp. Teor. Fiz. 20 (1950), 1064\u201382. 6.4 Superconductivity 283 A generalization of the action to include temporal \ufb02uctuations leads to the time-dependent Ginzburg\u2013Landau theory, to be discussed below. Equation (6.32) makes the connection between the superconductor and super\ufb02uid explicit (cf. the condensate action (6.9)). Above Tc, r > 0 and the unique mean-\ufb01eld con\ufb01guration extremizing the action (6.32) is given by \u0394 = 0. However, below the critical temperature, r < 0 and a con\ufb01guration with non-vanishing Cooper pair amplitude \u03940 is energetically favorable: \u03b4S[\u0394, \u00af\u0394] \u03b4\u0394 = 0 \u21d2 \u00af\u03940 r 2 \u0394=\u03940 + 2u|\u03940|2 = 0 \u21d2 |\u03940| = \" \u2212r 4u \u223c Tc(Tc \u2212 T ), cf. the previous estimate Eq. (6.30). As with the super\ufb02uid, the mean-\ufb01eld equation determines only the modulus of the order parameter while the phase remains unspeci\ufb01ed. Below the transition, the symmetry of the action will be broken by the formation of a ground state with \ufb01xed global phase, e.g. \u03940 \u2208 R. This entails the formation of a phase-like Goldstone mode \u03b8, where con\ufb01gurations \u0394 = e2i\u03b8\u03940 explore deviations from the reference groundstate.30 Pursuing further the parallels with the super\ufb02uid, it would be tempting to conjecture that these phase \ufb02uctuations have a linear dispersion, i.e. that the system supports dissipationless supercurrents of charged particles: superconductivity. However, at this point, we have overstretched the analogies to our previous discussion. In fact, the argument above ignores the fact that the symmetry broken by the ground state of the super\ufb02uid was a global phase U(1). However, as explained on page 276, the microscopic action of the superconductor possesses a more structured local gauge U(1) symmetry. As we will discuss presently, this di\ufb00erence implies drastic phenomenological consequences. Action of the Goldstone mode The rami\ufb01cations of the local gauge symmetry can only be explored in conjunction with the electromagnetic \ufb01eld (\u03c6, A). We must, therefore, return to the ancestor action (6.27) where G now represents the full Gor\u2019kov Green function (6.25). For the present, there is no need to specify the origin of the electromagnetic \ufb01eld; it might represent an external experimental probe, or the background electromagnetic \ufb01eld controlled by the vacuum action, d\u03c4 ddr F\u03bc\u03bdF\u03bc\u03bd, whereF \u03bc\u03bd = \u2202\u03bcA\u03bd \u2212 \u2202\u03bdA\u03bc is the electromagnetic \ufb01eld i.e. SE.M. = 1 4 tensor.31 However, throughout, we will assume that the \ufb01eld is weak enough not to destroy the superconductivity, i.e. the mean modulus of the order parameter is still given by the value \u03940, as described by the analysis of the previous section. How, then, might an action describing the interplay of the phase degree of freedom and the electromagnetic \ufb01eld look? Below we will derive such an action explicitly by starting from the prototype Eq. (6.", "27). However, for the moment, let us stay on a less rigorous level 30 The motivation for transforming by 2\u03b8 is that, under a gauge transformation \u00af\u03c8 \u2192 ei\u03b8 \u00af\u03c8, the composite \ufb01eld \u0394 \u223c \u00af\u03c8 \u00af\u03c8 should acquire two phase factors. However, the introduction of that muliplicity factor is, of course, just a matter of convention; it can always be removed by a rescaling of the \ufb01eld \u03b8. 31 Notice that we are working within the framework of imaginary (Euclidean) time \ufb01eld theory, i.e. the de\ufb01nition of the \ufb01eld strength tensor does not involve the Minkowski metric (cf. the discussion on page 107). 284 Broken symmetry and collective phenomena and try to determine the structure of the action by symmetry reasoning. In doing so, we will be guided by a number of principles: The phase \u03b8 is a Goldstone mode, i.e. the action cannot contain terms that do not vanish in the limit \u03b8 \u2192 const. We will assume that gradients acting on the phase \u03b8 (but not necessarily the magnitude of the phase) and the electromagnetic potentials are small. That is, we will be content with determining the structure of the action to lowest order in these quantities. By symmetry, the action must not contain terms with an odd number of derivatives or mixed gradients of the type \u2202\u03c4 \u03b8\u2207\u03b8. Respecting the character of the microscopic model, the action must be rotationally invariant. The action must be invariant under the local gauge transformation Eq. (6.24). It may be con\ufb01rmed that the \ufb01rst three criteria would be satis\ufb01ed by the trial action % & S[\u03b8] = d\u03c4 ddr c1(\u2202\u03c4 \u03b8)2 + c2(\u2207\u03b8)2, where c1 and c2 are constants. However, such an action is clearly not invariant under a gauge shift of the phase, \u03b8(\u03c4, r) \u2192 \u03b8(\u03c4, r) +\u03d5(\u03c4, r). It can, however, be endowed with that quality by minimal substitution of the electromagnetic potential, i.e.32 S[\u03b8, A] = d\u03c4 ddr % c1(\u2202\u03c4 \u03b8 + \u03c6)2 + c2(\u2207\u03b8 \u2212 A)2 &. (6.34) To second order in gradients, this action uniquely describes the energy cost associated with phase \ufb02uctuations. When combined with the action SE.M. controlling \ufb02uctuations of the \ufb01eld (\u03c6, A), it should provide a general description of the low-energy electromagnetic properties of the superconductor. Notice, however, that the present line of argument does not \ufb01x the coupling constants c1,2. In particular, one cannot exclude the possibility that c1 or c2 vanishes (as would be the case, for example, in a non-superconducting system, cf. the problem set). To determine the values of c1,2, we need either to derive the action microscopically or to invoke further phenomenological input (see the Info block below). Either way one obtains c2 = ns/2m and c1 = \u03bd, where we have de\ufb01ned ns as the density of the Cooper pair condensate. (For a precise de\ufb01nition, see below.) In the following section we use the action (6.34) as a starting point to discuss the characteristic and remarkable electromagnetic phenomena displayed by superconductors. \u03b4S \u03b4A(\u03c4,r) INFO Beginning with coe\ufb03cient c2, let us brie\ufb02y discuss the phenomenological derivation of the coupling constants. The starting point is the observation that the functional derivative, = j(\u03c4, r), generates the expectation value of the current density operator. This relation follows quite generally from the fact that a vector potential couples to the action of a system of charged particles i = 1,..., N through the relation SA \u2261 d\u03c4 \u02d9ri \u00b7 A(ri) = d\u03c4 ddr \u03b4(r \u2212 ri) \u02d9ri \u00b7 A(\u03c4, r). i i 32 To keep the notation simple, we will henceforth set e = 1. 6.4 Superconductivity 285 i \u03b4(r \u2212 ri(\u03c4 )) \u02d9ri(\u03c4 ) is just the de\ufb01nition of the total particle current However, j = \u03b4S density. Indeed, it is straightforward to verify that, on the microscopic level (cf., e.g., Eq. (6.22)), \u03b4", "A(\u03c4,r) = 2 3 \u03b4S \u03b4A. = \u2212 e 2m \u00af\u03c8\u03c3(\u2212i\u2207 \u2212 A)\u03c8\u03c3 + [(i\u2207 \u2212 A) \u00af\u03c8\u03c3]\u03c8\u03c3 / \u2261 j, where j is the quantum current density operator. Staying for a moment on the microscopic side of the theory, let us assume that a certain fraction of the formerly uncorrelated electronic states participate in the condensate, i.e. one may write j = jn + js, where jn, the current carried by the normal states of the system, will not be of further concern to us, while js is the \u201csupercurrent\u201d carried by the condensate. Let us further assume that those states \u03c8s participating in the condensate carry a \u2018collective\u2019 phase \u03b8 with a non-vanishing average, i.e. \u03c8s = ei\u03b8 \u02dc\u03c8s, where the states \u02dc\u03c8s do not carry any structured phase information (the residual phase carried by the local amplitude \u02dc\u03c8s tends, on average, to zero). Then, concentrating on the phase information carried by the condensate and neglecting density \ufb02uctuations, 2 3 \u03b4S \u03b4A js \u2212 ns m \u2207\u03b8 \u2212 A, where ns \u2261 \u00af\u03c8s\u03c8s is the density of the condensate.. Now, let us evaluate the fundamental relation / \u03b4S \u03b4A = j on our trial action with its undeter- mined coupling constants: j!= 2 3 \u03b4S[ \u02dcA] \u03b4A = \u22122c2\u2207\u03b8 \u2212 A. Comparison with the phenomenological estimate for the expectation value of the (super)current operator above leads to the identi\ufb01cation c2 = ns/2m. Turning to the coupling constant c1, let us assume that the electron system has been subjected to a weak external potential perturbation \u03c6(\u03c4, r). Assuming that the potential \ufb02uctuates slowly enough to allow an adiabatic adjustment of the electron density (i.e. that it acts as a local modulation of the chemical potential), the particle density of the system would vary as \u03b4n(\u03c4, r) = \u03b4n(\u03bc + \u03c6(\u03c4, r)) \u2248 \u2202n \u2202\u03bc \u03c6(\u03c4, r) \u2248 \u03bd\u03c6(\u03c4, r), where we have approximately33 identi\ufb01ed \u2202\u03bcn with the single-particle density of states \u03bd. The ddr \u03c6(\u03c4, r)\u03b4n(\u03c4, r) = potential energy corresponding to the charge modulation is given by ddr \u03c62(\u03c4, r). Comparing this expression with our trial action \u2013 which contains the time\u2202\u03bcn integrated potential \u2013 we conclude that c1 = \u2202\u03bcn = \u03bd. Despite the integrity of the phenomenological arguments given above, some may feel ill at ease with all the liberal distribution of \u201cassumes.\u201d To complement this discussion, let us now recover the phase action using an explicit microscopic derivation. The construction detailed below represents a typical yet, until now, our most advanced \u201ccase study\u201d of a low-energy quantum \ufb01eld theory. Although formulated for the speci\ufb01c example of the BCS superconductor, many of its structural sub-units appear in other applications in basically the same form. This \u201cuniversality\u201d is our prime motivation for presenting the lengthy construction of the low-energy phase action of 33 For systems with strong inter-particle correlations, the thermodynamic density of states \u2202\u03bcn = 1 Ld \u2202\u03bcN may deviate signi\ufb01cantly from the single-particle density of states \u03bd = 1 Ld a \u03b4(a \u2212 \u03bc). 286 Broken symmetry and collective phenomena the superconductor in some detail. Our starting point is the Gor\u2019kov Green function appearing under the \u201ctr ln\u201d of the microscopic action (6.27), 2m (\u2212i\u2207 \u2212 A)2 + \u03bc \u03940e\u22122i\u03b8 \u03940e2i\u03b8 2m (\u2212i\u2207 \u2212 A)2 \u2212 \u03bc \u2212\u2202\u03c4 \u2212 i\u03c6 \u2212 1 \u2212\u2202\u03c4 + i\u03c6 + 1 \u02c6G\u22121 =, coupled to the full electromagnetic potential. To simplify the analysis, we have set the modulus of the order parameter to its constant mean-\ufb01eld value \u03940, i.e. concentrating on the Goldstone mode, we will neglect massive \ufb02uctuations \u0394 = \u03940 +\u03b4\u0394 around the extremum of the free energy. We next", " make use of the gauge freedom inherent in the theory to remove the phase dependence, and of the order parameter \ufb01eld. To do so, we introduce the unitary matrix \u02c6U \u2261 transform the Green function as ei\u03b8 \u2212i\u03b8 &'e \u02c6G\u22121 \u2192 \u02c6U \u02c6G\u22121 \u02c6U \u2020 = \u2212\u2202\u03c4 \u2212 i \u02dc\u03c6 \u2212 1 2m (\u2212i\u2207 \u2212 \u02dcA)2 + \u03bc \u03940 \u03940 \u2212\u2202\u03c4 + i \u02dc\u03c6 + 1 2m (\u2212i\u2207 \u2212 \u02dcA)2 \u2212 \u03bc, where the transformed electromagnetic potential is given by (exercise) \u02dc\u03c6 = \u03c6+\u2202\u03c4 \u03b8, \u02dcA = A\u2212\u2207\u03b8. (Reading the transformation in reverse, we conclude that \u2013 an important physical fact that should be remembered \u2013 the superconductor order parameter \ufb01eld is a gauge non-invariant quantity. Under gauge transformations it transforms as \u0394 \u2192 e2i\u03b8\u0394, as suggested by the de\ufb01nition of \u0394 \u223c \u00af\u03c8\u2191 \u00af\u03c8\u2193 as a pairing \ufb01eld. This fact implies that the order parameter itself cannot be an experimentally accessible observable.34) Owing to the unitary invariance of the trace, tr ln \u02c6G\u22121 = tr ln ( \u02c6U \u02c6G\u22121 \u02c6U \u2020), the gauge transformed and the original Green function, respectively, equivalently represent the theory. Save for the neglect of massive \ufb02uctuations \u03b4\u0394, our treatment of the theory thus far has been exact. However, to make further progress, we must resort to some approximations: assuming that both the electromagnetic potential and spatio-temporal \ufb02uctuations of the phase mode are small, let us expand the action in powers of ( \u02dc\u03c6, \u02dcA). In the literature, expansions of this type are known as gradient expansions, i.e. we are performing an expansion where the gradients \u2202\u03c4 \u03b8, \u2207\u03b8 and not the phase degree of freedom \u03b8 are assumed to be small. (Owing to its Goldstone mode character, \u03b8 can slide freely over the entire interval [0, 2\u03c0].) To facilitate the expansion, it will be useful to represent the 2 \u00d7 2 matrix structure of the Green function through a Pauli matrix expansion: \u02c6G\u22121 = \u2212\u03c30\u2202\u03c4 \u2212 \u03c33 i \u02dc\u03c6 + (\u2212i\u2207 \u2212 \u02dcA\u03c33)2 \u2212 \u03bc + \u03c31\u03940 = \u2212\u03c30\u2202\u03c4 \u2212 \u03c33 1 2m \u22072 \u2212 \u03bc \u2212 1 2m \u02c6G\u22121 0 + \u03c31\u03940 \u2212 i\u03c33 \u02dc\u03c6 + \u03c30[\u2207, \u02dcA]+ i 2m \u02c6X1 \u2212 \u03c33, \u02dcA2 1 2m \u02c6X2 where we have de\ufb01ned\u03c3 0 \u2261 1 as the unit matrix. Expressed in terms of these quantities, the expansion of the action to second order in the \ufb01eld \u02dcA takes the form S[ \u02dcA] =\u2212tr ln &'\u02c6G\u22121 0 \u2212 \u02c6X1 \u2212 \u02c6X2 &'&'= const. \u2212 tr ln 1 \u2212 \u02c6G0[ \u02c6X1 + \u02c6X2] = const. + tr \u02c6G0 \u02c6X1 S(1)[ \u02dcA] + tr \u02c6G0 \u02c6X2 + \u02c6G0 \u02c6X1G0 \u02c6X1 + \u00b7 \u00b7 \u00b7, (6.35) 1 2 S(2)[ \u02dcA] 34 This follows from the fundamental doctrine of electrodynamics that gauge transformations must not cause observable e\ufb00ects. 6.4 Superconductivity 287 where we have used the fact that \u02c6X1,2 are of \ufb01rst and second order in the \ufb01eld, respectively. (Structures of this type appear frequently in the construction of low-energy quantum \ufb01eld theories of many-body systems, i.e., after the introduction of some auxiliary \ufb01eld \u03c6 through a suitably devised Hubbard\u2013Stratonovich transformation, the microscopic Bose/Fermi degrees of freedom of the theory can be integrated out and one arrives at an action \u00b1tr ln( \u02c6G\u22121 0 + \u02c6X[\u03c6]), where \u02c6G is the non-interacting Green", " function of the problem and \u02c6X[\u03c6] is an operator depending on the new \ufb01eld. An expansion of the logarithm to \ufb01rst and second orders in \u02c6X then leads to structures similar to those given above.) Written more explicitly, the \ufb01rst-order action S(1) takes the form (exercise) S(1)[ \u02dcA] = T Ld p ( ) tr \u02c6G0,p \u02c6X1(p, p) = T Ld p tr \u02c6G0,p i\u03c33 \u02dc\u03c60 + \u03c30p \u00b7 \u02dcA0, i m where the subscripts 0 refer to the zero-momentum components of the \ufb01elds \u02dc\u03c6 and \u02dcA. Since the Green function \u02c6G0 is even in the momentum, the second contribution \u221d p vanishes by symmetry. Further (\u2202\u03c4 \u03b8)0 = 0 \u00b7 \u03b80 = 0, i.e. \u02dc\u03c60 = \u03c60 and S(1)[ \u02dcA] = iT \u03c60, where the Ld indices refer to particle\u2013hole space. To understand the meaning of this expression, notice that [ \u02c6G0,p]11 = \u00af\u03c8\u2191,p\u03c8\u2191,p0 gives the expectation value of the spin-up electron density operator on the background of a \ufb01xed-order parameter background. Similarly, \u2212[ \u02c6G0,p]22 = \u2212\u03c8\u2193,p \u00af\u03c8\u2193,p0 = + \u00af\u03c8\u2193,p\u03c8\u2193,p0 gives the spin-down density. Summation over frequencies and momenta recovers the full electron density: T Ld p([ \u02c6G0,p]11 \u2212 [ \u02c6G0,p]22) = N [ \u02c6G0,p]11 \u2212 [ \u02c6G0,p]22 Ld, or &'p S(1)[ \u02dcA] = iN \u03c60 = iN Ld d\u03c4 ddr \u03c6(\u03c4, r). Thus, the \ufb01rst contribution to our action simply describes the electrostatic coupling of the scalar potential to the total charge of the electron system. However, as with the Coulomb potential discussed earlier, the \u201ccorrect\u201d interpretation of this expression should rather suggest S(1) = 0. That is, the total electrostatic interaction of the potential with the electron system must be \u2013 by the overall charge neutrality of the system \u2013 compensated by an equally strong interaction with the positive counter charge of the ions (usually excluded for notational convenience). We thus turn to the discussion of the second-order contribution to the action S(2). The term containing \u02c6X2 is reminiscent in structure to the S(1) contribution discussed before. Thus, replacing \u02c6X1 by \u02c6X2, we may immediately infer that tr( \u02c6G0 \u02c6X2) = n 2m d\u03c4 ddr A2(\u03c4, r), (6.36) where we have de\ufb01ned asn \u2261 N/Ld the total particle density. This contribution, known as the 2m A2 to the electron diamagnetic term, derives from the familiar diamagnetic contribution 1 Hamiltonian. If it were only the diamagnetic contribution, an external \ufb01eld would lead to an increase of the energy. However, to obtain the complete picture, we need to include the magnetic \ufb01eld dependence of the operator \u02c6X1. Substituting for \u02c6X1, and noting that crossterms \u223c \u02dc\u03c6 p \u00b7 \u02dcA, being odd in momenta, vanish on integration, one obtains &'tr \u02c6G0 \u02c6X1 \u02c6G0 \u02c6X1 = 1 2 T 2Ld p,q tr \u2212 \u02c6G0,p\u03c33 \u02dc\u03c6q \u02c6G0,p\u03c33 \u02dc\u03c6\u2212q + 1 m2 \u02c6G0,p\u03c30p \u00b7 \u02dcAq \u02c6G0,p\u03c30p \u00b7 \u02dcA\u2212q. 288 Broken symmetry and collective phenomena Here, noting that we are already working at the second order of the expansion, the residual dependence of the Green functions \u02c6G0 on the small momentum variable q has been neglected.35 2m [p, \u02dcA]+ of the electron Hamiltonian, Alluding to its origin, i.e. the paramagnetic operator \u223c 1 the magnetic contribution to this expression is called the paramagnetic term. Paramagnetic contributions to the action describe a lowering", " of the energy in response to external magnetic \ufb01elds, i.e. the diamagnetic and the paramagnetic term act in competition. To proceed, it is convenient to change from an explicit matrix representation of the Gor\u2019kov Green function to an expansion in terms of Pauli matrices:36 \u02c6G0,p = [i\u03c30\u03c9n \u2212 \u03c33\u03bep + \u03c31\u03940]\u22121 = 1 p + \u03942 n + \u03be2 0 \u03c92 [\u2212i\u03c30\u03c9n \u2212 \u03c33\u03bep + \u03c31\u03940]. (6.37) On substituting into the equation above, noting that, for any rotationally invariant function F (p2) (exercise), p(p \u00b7 v) (p \u00b7 v)F (p2) = v\u00b7v d p p2F (p2), one obtains 1 2 tr = &'\u02c6G0 \u02c6X1 \u02c6G0 \u02c6X1 T Ld p,q 1 n + \u03bb2 p)2 (\u03c92 \u02dc\u03c6q \u02dc\u03c6\u2212q(\u2212\u03c92 n + \u03bb2 p \u2212 2\u03942 0) \u2212 p2 \u02dcAq \u00b7 \u02dcA\u2212q 3m2 (\u2212\u03c92 n + \u03bb2 p), where, as before, \u03bb2 0. We now substitute this result together with the diamagnetic contribution Eq. (6.36) back into the expansion (6.35), partially transform back to real space p + \u03942 p = \u03be2 q fqf\u2212q = d\u03c4 ddr f 2(\u03c4, r)2, and arrive at the action S[ \u02dcA] = d\u03c4 ddr ( T Ld p \u2212\u03c92 p \u2212 2\u03942 0 p)2 n + \u03bb2 n + \u03bb2 (\u03c92 c1 + n 2m \u2212 1 dm2 T Ld p c2 \u02dc\u03c62(\u03c4, r) p2(\u2212\u03c92 (\u03c92 n + \u03bb2 p) p)2 n + \u03bb2 ) \u02dcA2(\u03c4, r). This intermediate result identi\ufb01es the coupling constants c1,2. The last step of the derivation, i.e. the sum over the \u201cfast\u201d momenta p, is now a relatively straightforward exercise. Beginning with the frequency summations, one may note that the denominator has two isolated poles of second order at \u03c9n = \u00b1i\u03bbp. Applying the standard summation rules it is then straightforward to verify that (exercise) \u2212\u03c92 n + \u03bb2 (\u03c92 p \u2212 2\u03942 0 p)2 n + \u03bb2 = \u2212 1 2\u03bbp nF(\u2212\u03bbp) 2 \u03940 \u03bbp + n F(\u2212\u03bbp) \u03be2 p \u03bbp + (\u03bbp \u2194 \u2212\u03bbp) \u2248 \u2212 \u03942 0 2\u03bb3 p, \u2212\u03c92 (\u03c92 n + \u03bb2 p n + \u03bb2 p)2 = \u2212\u03b2[nF(\u03bbp)(1 \u2212 nF(\u03bbp)]. T T n n 35 i.e. we have set 36 In working with matrix pq( \u02c6Gp \u02dc\u03c6q \u02c6Gp+q \u02dc\u03c6\u2212q) \u2248 pq( \u02c6Gp \u02dc\u03c6q \u02c6Gp \u02dc\u03c6\u2212q). useful is it [v0\u03c30 + v \u00b7 \u03c3]\u22121 = ful identities include (i, j = 1, 2, 3; \u03bc, \u03bd = 1, 2, 3, 4) \u03c32 identity \u2212v2 [v0\u03c30 \u2212 v \u00b7 \u03c3], where v = (v0, v) is a four-component vector of coe\ufb03cients. Other usei = 1, i = j, [\u03c3i, \u03c3j ]+ = 0, \u03c3i\u03c3j = iijk\u03c3k, tr \u03c3\u03bc = 2\u03b4\u03bc,0. the matrix in mind operators keep v2 0 to 1 6.4 Superconductivity 289 3 T \u0394 2 1 5 0.2 0.15 0.1 0.05 0 10 0 \u03be \u0394 \u20135 ( \u201310 ) Figure 6.7 Plot of the function \u03b2nF( as a function of the dimensionless scales T /\u03940 and \u03be/\u03940. For T /\u03940 \u2192 0, the function vanishes (\u2192 perfect diamagnetic dn(1 \u2212 response). For T \u0394 n) = 1 (\u2192 cancellation of diamagnetic and paramagnetic response). At", " intermediate temperatures, 1, the function traces out a peak of width \u221d T and total weight \u03be2 + \u03942 0) \u03be2 + \u03942 0) 1 \u2212 nF( dn(1 \u2212 n) < 1, resulting in a partial survival of diamagnetism. Thus, one obtains the coupling constant of the potential contribution, c1 = \u2212 1 2Ld p \u03942 0 \u03bb3 p = \u2212 \u03bd 2 d\u03be \u03942 0 (\u03be2 + \u03942 0)3/2 = \u2212\u03bd, in accord with the previous estimate. With the magnetic contribution, the situation is more interesting. Converting the momentum sum to an energy integral, one obtains c2 = n 2m + \u03bd\u03bc dm d\u03be \u03b2[nF(\u03bb)(1 \u2212 nF(\u03bb)], where we have noted that the integrand is strongly peaked at the Fermi surface, i.e. that the factor p2 \u2248 2m\u03bc can be removed from under the integral. This expression illustrates the competition between the diamagnetic and paramagnetic contributions in the magnetic response of the system. At low temperatures, T \u03940, the positivity of p)1/2 \u2265 \u03940 implies that nF(\u03bbp) \u2248 0, i.e. approximate vanishing of the integral (see \u03bbp = (\u03942 0 + \u03be2 T \u03940\u2248 n 2m is weighted by the total density of the electron Fig. 6.7). Under these conditions, c2 gas, and the response of the system is governed by the diamagnetic term alone. Indeed, the diamagnetic response is known to be a hallmark of superconductivity; the superconductor tends to expel magnetic \ufb01elds, a phenomenon that culminates in the Meissner e\ufb00ect to be discussed shortly. 290 Broken symmetry and collective phenomena By contrast, for high temperatures, T \u03940, the integral extends over energy domains much larger than \u03940 and we can approximate \u2212 d\u03be \u03b2[nF(\u03bb)(1 \u2212 nF(\u03bb)] \u2248 \u2212 \u221e 0 \u221e d\u03be \u03b2[nF(\u03be)(1 \u2212 nF(\u03be)] = \u2212 d\u03be \u2202\u03benF(\u03be) = nF(0) \u2212 nF(\u221e) = 1, 0 to obtain37 c2 is typical for the response of normal conducting systems to external magnetic \ufb01elds. dm = 0. The near38 cancellation of dia- and paramagnetic contributions \u2212 \u03bd\u03bc 2m T \u0394\u2248 n At intermediate temperatures, the integral over the Fermi functions leads to a partial cancellation of the diamagnetic response. It is common to express that fact through the notation c2 = ns 2m, where the parameter ns \u2261 n + 2\u03bd\u03bc d d\u03be \u03b2[nF(\u03bb)(1 \u2212 nF(\u03bb)], is known as the super\ufb02uid density. Historically, the concept of a \u201csuper\ufb02uid density\u201d was introduced prior to the BCS theory, when a phenomenological model known as the two-\ufb02uid model presented the \u201cstate-of-the-art\u201d understanding of superconductivity. (Remember that the experimental discovery of superconductivity preceded its microscopic description by more than four decades!) The basic picture underlying this approach was that, below the transition, a fraction of the electron system condenses into a dissipationless super\ufb02uid of density ns, while the rest of the electrons remain in the state of a \u201cnormal\u201d Fermi liquid of density nn = n \u2212 ns. This simple model provided a phenomenological explanation of a large number of properties characteristic of superconductivity, prior to the development of the microscopic BCS theory. However, notwithstanding its success and its appealing simplicity, the two-\ufb02uid notion of a (T \u21920) \u2212\u2212\u2212\u2212\u2192 n at low temperatures cannot be maintained. Indeed, we have complete condensation ns seen that BCS superconductivity is a Fermi surface phenomenon, i.e. the bulk of the electrons are oblivious to the existence of an attraction mechanism at energies \u03bc \u00b1 \u03c9D and, therefore, will not enter a condensed state. Instead, our microscopic analysis produces a picture more subtle than the mere superposition of two \ufb02uids: as we saw above, the diamagnetic (paramagnetic) contribution to the response is provided by all quasi-particles (at the Fermi surface). In a normal metal, or, equivalently, a superconductor at T \u03940, quasi-particle excitations at the Fermi surface conspire to cancel the diamagnetic contribution of all other quasi-particles. However, at T \u03940 the existence of a quasi-particle energy", " gap at the Fermi surface blocks that compensation mechanism and a net diamagnetic signal remains. The far-reaching phenomenological consequences of the sustained diamagnetic contribution are discussed in the next section. 37 The last equality follows straightforwardly from the two de\ufb01nitions \u03bd n = 2 Ld p \u03b4(\u03bc \u2212 \u03bep) \u0398(\u03bc \u2212 \u03bep) = 2 (2\u03c0)d ddp \u03b4(\u03bc \u2212 \u03be) \u0398(\u03bc \u2212 \u03be). 38 Going beyond second lowest order of perturbation theory in A, a careful analysis of the coupling of a (small) magnetic \ufb01eld to the orbital degrees of freedom of the Fermi gas shows that the cancellation of diamagnetic and paramagnetic contributions is not perfect. The total response of the system is described by a weak diamagnetic contribution, \u03c7d, a phenomenon known as Landau diamagnetism. The diamagnetic orbital response is overcompensated by Pauli paramagnetism, i.e. the three times larger paramagnetic response of the Zeemancoupled electron spin, \u03c7p = \u22123\u03c7d. For large magnetic \ufb01elds, the situation changes totally, and more pronounced e\ufb00ects such as Shubnikov\u2013de Haas oscillations or even the quantum Hall e\ufb00ect are observed. 6.4 Superconductivity 291 Validity of the gradient expansion Before leaving this section let us discuss one last technical point. Above we have \u2013 without much of a justi\ufb01cation \u2013 expanded the phase action up to leading order in the gradients \u2202\u03c4 \u03b8 and \u2207\u03b8. Indeed, why is such a truncation permissible? This question arises whenever low-energy e\ufb00ective theories are derived from a microscopic parent theory by expanding in slow \ufb02uctuations, and it is worthwhile to address it in a general setting. Let us suppose we had performed some kind of Hubbard\u2013Stratonovich transformation to describe a system of interest in terms of an action S[\u03c6]. Let us further assume that the action is invariant under a shift of the \ufb01eld by a constant, \u03c6(x) \u2192 \u03c6(x) +\u03c6 0, i.e. that the action depends only on gradients \u2207\u03c6. (To keep the notation simple, we do not explicitly distinguish between spatial and temporal gradients.) An expansion of the action in the \ufb01eld gradients then leads to a series of the formal momentum-space structure S \u223c N (l0q)2\u03c6q\u03c6\u2212q + (l0q)4\u03c6q\u03c6\u2212q + \u00b7 \u00b7 \u00b7, q where N represents the large parameter of the theory,39 l0 is some microscopic reference scale of [length] needed to make the action dimensionless, and the ellipses stand for terms of higher order in q and/or \u03c6q. Now, using the fact that only \ufb01eld con\ufb01gurations with S \u223c 1 signi\ufb01cantly contribute to the \ufb01eld integral, we obtain the estimate \u03c6q \u223c 1\u221a l0q from the leading-order term N of the action. This means that terms of higher order in the \ufb01eld variable, N (l0q\u03c6q)n>2 \u223c N 1\u2212n/2, are small in inverse powers of the large parameter of the theory and can be neglected. Similarly, terms like N (l0q)n>2\u03c6q\u03c6\u2212q \u223c (l0q)n\u22122. As long as we are interested in large-scale \ufb02uctuations on scales q\u22121 l0, these terms, too, can be neglected. 1 Notice that our justi\ufb01cation for neglecting terms of higher order relies on two independent parameters; large N and the smallness of the scaling factor ql0. If N = 1 but still ql0 1, terms involving two gradients but large powers of the \ufb01eld \u223c q2\u03c6n>2 are no longer negligible. Conversely, if N 1 but one is interested in scales ql0 1, terms of second order in the \ufb01eld weighted by a large number of gradients \u223c qn>2\u03c62 q must be taken into account. An incorrect treatment of this point has been the source of numerous errors in the published literature! q Meissner e\ufb00ect and Anderson\u2013Higgs mechanism If you ask a person on the street to give a one-line de\ufb01nition of superconductivity, the answer will probably be that superconductors are materials showing no electrical resistance. However, to a physicist, that de\ufb01nition should not be altogether satisfactory", ". It highlights only one of many remarkable features of superconductors and does not have any predictive power. A better \u2013 if for most people incomprehensible \u2013 attempt at a de\ufb01nition would be to say that superconductivity arises when the quantum phases of a macroscopically large number of charged particles get locked into a collective degree of freedom. Indeed, that is exactly what the action (6.34) tells us: \ufb02uctuations of the phase of the condensate are penalized by a cost that scales with the (super\ufb02uid) density of the electron gas. This is to be contrasted to the situation in a normal metal where the action \u2013 the cancellation of the 39 In theories containing a large parameter N, that parameter mostly appears as a constant multiplying all, or at least several, operators of the action. For example, in the fermionic problems discussed above, where N was proportional to the density of states at the Fermi surface, the action contained a trace over all momentum states. The summation over these states then led to an overall factor N multiplying the action. 292 Broken symmetry and collective phenomena diamagnetic and the paramagnetic contribution! \u2013 usually does not contain the gradients of phase-like degrees of freedom. Indeed, the macroscopic phase rigidity of the BCS ground state wavefunction su\ufb03ces to explain a large number of non-trivial phenomena related to superconductivity. To see this, let us consider a simpli\ufb01ed version of the action (6.34). Speci\ufb01cally, let us assume (i) that the temperature is high enough to exclude quantum \ufb02uctuations of the phase, \u2202\u03c4 \u03b8 = 0, and (ii) that there are no electric \ufb01elds acting on our superconductor \u03c6 = 0, \u2202\u03c4 A = 0. INFO Why do we relate the presence of quantum \ufb02uctuations to temperature? Let us reiterate (cf. the remarks made on page 255) that a temporarily constant \ufb01eld variable acts like a classical degree of freedom. To understand why large temperatures inhibit temporal \ufb02uctuations superimposed on the classical sector, let us take the phase action of the superconductor as an example and write \u03b2 S[\u03b8] = \u03bd d\u03c4 0 ddr (\u2202\u03c4 \u03b8)2 + Scl[\u03b8], where the \ufb01rst term determines the temporal \ufb02uctuation behavior of the phase \ufb01eld, while Scl is the \u201cclassical\u201d contribution to the action, i.e. the contribution independent of time derivatives. Switching to a frequency representation, S[\u03b8] = \u03bd m,q \u03c92 m\u03b8m,q\u03b8\u2212m,\u2212q + Scl[\u03b8], from which one can infer that quantum \ufb02uctuations, i.e. modes \u03b8m=0, become inessential at large temperatures. More speci\ufb01cally, modes with non-vanishing Matsubara frequency can be m \u221d T 2 exceeds the characteristic neglected when the quantum \ufb02uctuation energy (density) \u03bd\u03c92 energy scales appearing in Scl[\u03b8]. To understand heuristically the temperature scaling of the \ufb02uctuation energy, remember that the Bose \ufb01eld \u03b8(\u03c4 ) obeys periodic boundary conditions \u03b8(0) = \u03b8(\u03b2). Inspection of Fig. 6.8 then shows that increasing the temperature, i.e. squeezing of the imaginary time interval [0, \u03b2], leads to a linear increase of the gradients \u2202\u03c4 \u03b8 \u221d T. Accordingly, the squared gradient appearing in the action increases quadratically with T. This mechanism con\ufb01rms the intuitive expectation that quantum \ufb02uctuations \u2013 i.e. fundamentally a low-energy phenomenon \u2013 should be damped out at increasing temperature. Under these conditions, the action simpli\ufb01es to S[A, \u03b8] = \u03b2 2 ddr # ns m (\u2207\u03b8 \u2212 A)2 + (\u2207 \u2227A) 2, $ ddr (\u2207 \u2227 where we have explicitly included the action 1 d\u03c4 4 A)2 = \u03b2 ddrB2 due to \ufb02uctuations of the magnetic \ufb01eld. As pointed out above, the action 2 is invariant under the gauge transformation A \u2192 A + \u2207\u03c6, \u03b8 \u2192 \u03b8 + \u03c6. One thus expects that integration over all realizations of \u03b8 \u2013 a feasible task since the action is quadratic \u2013 ddr F\u03bc\u03bdF \u03bc\u03bd \u03c6=0,A static = \u03b2 2 6.4 Superconductivity 293 \u03b8 incr. T \u03b2 1 \u03b2 2 \u03b2 Figure", " 6.8 Qualitative picture behind the quadratic \u223c T 2 energy increase of quantum \ufb02uctuations. As the temperature T is increased, the imaginary time interval [0, \u03b2 = T \u22121] gets squeezed. The same happens to the temporal pro\ufb01les of (quantum) \ufb02uctuating modes \u03b8(\u03c4 ). Consequently, gradients \u2202\u03c4 \u03b8 \u221d T increase linearly with temperature, and the energy density \u223c (\u2202\u03c4 \u03b8)2 \u221d T 2 grows quadratically. will produce a purely A-dependent, and gauge invariant, action S[A]. The integration over \u03b8 is most transparently formulated in momentum space where the action assumes the form S[A, \u03b8] = = \u03b2 2 \u03b2 2 q q ns m ns m (iq\u03b8q \u2212 Aq) \u00b7 (\u2212iq\u03b8\u2212q \u2212 A\u2212q) + (q \u2227 Aq) \u00b7 (q \u2227 A\u2212q) [\u03b8qq2\u03b8\u2212q \u2212 2i\u03b8qq \u00b7 A\u2212q + Aq \u00b7 A\u2212q] + (q \u2227 Aq) \u00b7 (q \u2227 A\u2212q). The integration over the \ufb01eld components \u03b8q is now straightforward and leads to an e\ufb00ective action for A : e\u2212S[A] \u2261 S[A] = \u03b2 2 q ns m D\u03b8 e\u2212S[A,\u03b8], where Aq \u00b7 A\u2212q \u2212 (q \u00b7 Aq)(q \u00b7 A\u2212q) q2 + (q \u2227 Aq) \u00b7 (q \u2227 A\u2212q). To bring this result into a more transparent form, let us split the vector potential into a longitudinal and a transverse component: Aq = Aq \u2212 q(q \u00b7 Aq) q2! A\u22a5 q + q(q \u00b7 Aq) q2!. A q (6.38) To motivate this decomposition, notice the following: The transverse component alone determines physical quantities, i.e. the magnetic \ufb01eld. (This follows from the relation Bq = iq \u2227 Aq and q \u2227 q = 0.) The transverse component is gauge invariant under transformations Aq \u2192 Aq + iq\u03c6q (but (q\u03c6q)\u22a5 = 0). In the language of longitudinal\u2013transverse components, the Coulomb gauge corresponds to a con\ufb01guration where A = 0. The terminology \u201clongitudinal component\u201d emphasizes the fact that F q is the projection of a vector \ufb01eld Fq onto the argument vector q. Correspondingly, the \u201ctransverse component\u201d is the orthogonal complement of the longitudinal component. 294 Broken symmetry and collective phenomena Applying some elementary rules of vector algebra, it is straightforward to verify that the e\ufb00ective action can be represented in the simple form S[A] = \u03b2 2 q + q2 ns m A\u22a5 q \u00b7 A\u22a5 \u2212q. (6.39) At this stage, it is useful to pause and review what has been achieved: (i) starting from a composite action containing the Goldstone mode \u03b8 and the gauge \ufb01eld A, we have arrived at an action for the gauge \ufb01eld alone. In a sense, the Goldstone mode has been absorbed into (the gauge degrees of freedom of) A. However, (ii) the coupling to the Goldstone mode has not left the gauge \ufb01eld una\ufb00ected. Indeed, S[A] has acquired a mass term proportional to the super\ufb02uid density, i.e., unlike the vacuum, the action of long-range \ufb01eld \ufb02uctuations Aq\u21920 no longer vanishes. That modi\ufb01cation has serious phenomenological consequences to be discussed shortly. (iii) The action is manifestly gauge invariant. INFO The analysis above shows that the spontaneous breaking of a global U(1) symmetry, and of a local gauge U(1) symmetry lead to very di\ufb00erent results. In the former (\u2192 neutral super\ufb02uids), the soft action S[\u03b8] of a phase-like Goldstone mode \u03b8 describes various long-range phenomena, such as supercurrent formation, etc. In the latter case (\u2192 superconductors or, more generally, charged super\ufb02uids), the system is described by a composite action S[A, \u03b8]. The ubiquitous gauge symmetry can then be employed to absorb the Goldstone mode into the D\u03b8\u2192 S[A]. The most important e\ufb00ect of the coupling A \u2194 \u03b8 is that, after gauge \ufb01eld S[A, \u03b8", "] integration over the latter, the former acquires a mass term. One may say that \u201cthe photon (vector potential) \ufb01eld has consumed the Goldstone mode to become massive.\u201d That principal mechanism was understood in 1964 by Higgs, wherefore it is called the Higgs mechanism or, crediting Anderson\u2019s pioneering discussion of gauge symmetry breaking in the context of superconductivity, the Anderson\u2013Higgs mechanism. Mass generation due to spontaneous gauge symmetry breaking is a very general phenomenon, i.e. not limited to the relatively simple context of the collective phase of the superconductor. The Higgs mechanism found perhaps its most signi\ufb01cant application in 1967 when Weinberg and Salam embedded it into their uni\ufb01ed theory of electromagnetic and weak interactions in particle physics. Although this is not a book about elementary particles, the role of the Higgs principle in the theory of electroweak interactions is of such fundametal importance to our understanding of the microscopic world that it is irresistible to brie\ufb02y discuss its implications.40 The standard model of high-energy physics describes the microscopic world in terms of a few generations of leptonic (electrons, e, electron neutrino, \u03bde, muon, \u03bc, etc.) and hadronic (the quarks, u, d, s, c, t, b) elementary particles that interact through the quanta of certain gauge \ufb01elds. In its original formulation, the model had one severe problem, that is, it did not know how to attribute mass to these particles. However, this stands in stark contrast to any kind of experimental observation. In particular, the quanta of the gauge \ufb01elds of the weak and strong interactions are known to be extremely heavy, with rest masses of O(102 GeV/c2). In view of the much lighter masses of typical composite hadrons \u2013 the proton weighs 938 MeV/c2 \u2013 the mass of the gauge quanta can certainly not be explained in terms of some \ufb01ctitious \ufb01ne structure mechanisms superimposed on the core of the standard model, i.e. a major modi\ufb01cation 40 For a pedagogical and much less super\ufb01cial discussion we refer, e.g., to L. H. Ryder, Quantum Field Theory (Cambridge University Press, 1996). 6.4 Superconductivity 295 was needed. It is now widely believed that the \u201ctrue\u201d principle behind mass generation lies in the (Anderson\u2013) Higgs mechanism, i.e. the spontaneous breakdown of a gauge symmetry. To sketch the principal idea of Weinberg and Salam, let us concentrate on a leptonic subsector of the theory, e.g. consider the two-component object \u03bde e, \u03a8 \u2261 comprising an electron neutrino and a (left-handed41) electron. Involving a charged particle, the Lagrangian controlling the dynamics of \u03a8 will surely possess a local U(1) gauge symmetry. However, on top of that, Weinberg and Salam proposed a much more far-reaching symmetry structure. Without going into detail we just mention that, building on principles proposed earlier (1954) by Yang and Mills, transformations \u03a8(x) \u2192 U (x)\u03a8(x), U (x) \u2208 SU(2), x \u2208 R4, locally mixing the two components \u03bde and e of the \u201cisospinor\u201d \u03a8 were introduced as a symmetry of the model, i.e. in analogy to the local U(1) gauge symmetry of quantum mechanics, it was postulated that the action S[\u03a8] possesses a local gauge symmetry under the group of SU(2) transformations. In combination with the standard U(1), the theory had thus been endowed with a composite U(1) \u00d7 SU(2) gauge structure. Physically, declaring a symmetry between the electron \u2013 interacting through electromagnetic forces \u2013 and the neutrino \u2013 weak interactions \u2013 was tantamount to a fusion of these types of interaction, i.e. the proposal of a theory of electroweak interactions. How can a theory de\ufb01ned through an action S[\u03a8] = dd+1x L(\u03a8, \u2202\u03bc\u03a8), containing the isospinor \u03a8 and its derivatives, be made invariant under non-abelian SU(2) gauge transformations? Referring for a more systematic discussion to Ryder,42 let us brie\ufb02y sketch the principal idea of non\u2013abelian gauge theory. We \ufb01rst notice that a fermion bilinear \u223c \u00af\u03a8\u2202\u03bc\u03a8 is generally not gauge invariant. Under a mapping \u03a8 \u2192 U \u03a8 it transforms to \u00af\u03a8(\u2202\u03bc + U \u22121\u2202\u03bcU )", "\u03a8. E.g. for U = ei\u03c6 \u2208 U(1) a \u201cstandard\u201d gauge transformation, the extra term U \u22121\u2202\u03bcU = i\u2202\u03bc\u03c6 \u2208 iR would be the ordinary \u201cderivative of a function,\u201d familiar from the gauge structure of quantum mechanics. More generally, for a non-abelian gauge transformation by an element U \u2208 G of a general group (e.g. G = SU(2)), the gauge term U \u22121\u2202\u03bcU \u2208 g is an element of the Lie algebra g of the group,43 i.e. the action picks up a matrix-valued extra contribution. To make the theory invariant, we have to introduce a gauge \ufb01eld, i.e. we generalize from \u2202\u03bc to a covariant derivative \u2202\u03bc + W\u03bc, whereW \u03bc \u2208 g. For example, for G = U(1), W\u03bc \u2261 A\u03bc \u2208 R is a (x)\u03c3a \u2208 su(2), the ordinary gauge \ufb01eld of quantum mechanics; for G = SU(2), W\u03bc = etc. Under a gauge transformation, the \ufb01eld W\u03bc transforms as W\u03bc \u2192 U W\u03bcU \u22121 \u2212 iU \u22121\u2202\u03bcU, i.e. in a way that makes the covariant bilinear, \u00af\u03a8(\u2202\u03bc + W\u03bc)\u03a8, invariant. ForG = U(1), A\u03bc \u2192 U A\u03bcU \u22121 \u2212 U \u2202\u03bcU \u22121 = A\u03bc \u2212 i\u2202\u03bc\u03c6 reduces to its familiar form. However, in the non-abelian case, a=1 \u03b1\u03bc 3 41 When viewed as a relativistic particle, the electron \ufb01eld has components of left and right chirality, but we shall not need to discuss that aspect any further. 42 Ryder, Quantum Field Theory. 43 The fact that U \u22121\u2202\u03bcU \u2208 g takes values in the Lie algebra of the gauge group can be proven by geometric considerations (for which we refer to textbooks of group theory). By way of example, consider an SU(2)a=1 \u03b1a(x)\u03c3a). It is straightforward to convince oneself that U \u22121\u2202\u03bcU is a linear valued \ufb01eld U (x) = exp(i combination of Pauli matrices with real-valued coe\ufb03cents, i.e. it takes values in the Lie algebra su(2). 3 296 Broken symmetry and collective phenomena the full structure on the right-hand side is needed to obtain invariance. The full action of the gauge theory then takes the form S[\u03a8, W ] = dd+1x L(\u03a8, (\u2202\u03bc \u2212 iW\u03bc)\u03a8) + S[W ], where the Lagrangian density contains the minimally coupled gauge \ufb01eld, and the action S[W ] describes the \ufb02uctuation behavior of W\u03bc.44 Within a fully quantum mechanical setting, both the \u201cmatter \ufb01eld\u201d \u03a8 and the gauge \ufb01eld W\u03bc are quantized. The \ufb01eld quanta of the non-abelian gauge \ufb01eld are described as vector bosons, where the attribute \u201cvector\u201d (somewhat misleadingly) refers to the higher-dimensional geometry of the \ufb01eld, and \u201cboson\u201d emphasizes the generally bosonic statistic of a quantum gauge \ufb01eld. We now have everything in place to turn back to the particular context of the electroweak interaction. Within the framework of the gauge theory, interactions between the particles e and \u03bde are mediated by the gauge \ufb01eld W\u03bc (as with a U(1) theory where interactions between the electrons can be described in terms of a \ufb02uctuating U(1) vector potential, cf. Section 6.2). Experimental analysis of typical weak interaction processes, such as the elastic collision, e+\u03bde \u2192 e +\u03bd e, indicates that the weak interaction forces are extraordinarily short-range, with a decay pro\ufb01le \u223c exp(\u221290 GeVr). However, according to the pure gauge theory, the propagator of W\u03bc should be long-range \u223c r\u22121. This is the most severe manifestation of the mass problem of the electroweak theory. In order to be consistent with experiment, a mechanism is needed that makes the gauge \ufb01eld (very) massive. Here is where the Higgs mechanism enters the stage. To solve the mass problem, Weinberg and Salam postulated the existence of a scalar (more precisely, a two-component \u201ciso-scalar\u201d", ") bosonic particle, the Higgs boson \u03c6. The action of the Higgs particle \u2013 again a postulate \u2013 is of generalized \u03c64 type, i.e. S[\u03c6, W\u03bc] = dd+1x 1 2 (\u2202\u03bc \u2212 W\u03bc)\u03c6\u2020(\u2202\u03bc \u2212 W\u03bc)\u03c6 \u2212 m2 2 \u03c6\u2020\u03c6 + g 2 (\u03c6\u2020\u03c6)2, where the minimal coupling to the gauge \ufb01eld provides the theory with local gauge invariance. The action of the Higgs has been deliberately designed so as to generate spontaneous symmetry breaking, i.e. the solution of the mean-\ufb01eld equations is given by |\u03c6| = ( m2 2g )1/2, with undetermined phase. In direct analogy to our discussion of the superconductor above, an integration over the phase degree of freedom (i.e. the Goldstone mode) then generates a mass term for the gauge \ufb01elds. In summary, Weinberg and Salam proposed an explanation of the short-rangeness of the weak interaction through the presence of an extra particle, i.e. a particle that does not belong to the standard hadronic or leptonic generations of the standard model. Since then, the hunt for the Higgs particle has been one of the big challenges for particle physics. In 1983 vector bosons of the predicted mass were for the \ufb01rst time observed in experiment45 (see Fig. 6.9), i.e. the existence of a massive gauge structure is now out of the question. However, the detection of the Higgs turned out to be a more di\ufb03cult task. In 2000 the N -experiment installed at the LEP (Large Electron Positron Collider) at CERN reported a \u201cshadowy\u201d Higgs signal in the e\u2212e+ scattering cross-section at the predicted energy interval. However, at that time, the designated live time of LEP had already expired and one month after 44 Typically, S[W ] will be given by generalization of the \ufb01eld strength tensor theory, i.e. the curvature tensor on G. dd+1x F\u03bc\u03bd F \u03bc\u03bd of the abelian 45 G. Arnison et al., Experimental observation of isolated large transverse energy electrons with associated missing \u221a energy at s = 540 GeV, Phys. Lett. B 122 (1983), 103\u201316. 6.4 Superconductivity 297 Figure 6.9 Computer-generated visualization of a scattering process recorded in the N -experiment at CERN. Hidden somewhere in the \u201cjet\u201d of particles generated as a result of the collision of the scattering particles should, hopefully, be the Higgs. the detection of the \ufb01rst suspicious signals the machine was indeed shut down. Unfortunately, after the closedown of LEP only a few accelerator projects o\ufb00er the perspective to participate in the hunt for the Higgs. Presently, with the \u201cSuperconducting Super Collider\u201d (SSC) disapproved by the American Congress, only the Stanford Linear Accelerator reaches the relevant energy scales. The situation may change when the Large Hadron Collider (LHC) at CERN commences its work. In the meantime the collected data recorded at LEP over the past years have been subjected to critical review. Frustratingly, it turned out that the data reported in 2000 did not pass the test of a careful re-examination, i.e. presently (2009) there is no direct evidence for a Higgs particle. In view of the fact that the Higgs generates the mass not only of the vector bosons but of all particles known to the standard model, much indeed hinges on the question of its existence. (Some people even call the Higgs the \u201cGod particle.\u201d) If it did not exist, our understanding of the microscopic world would be turned upside down. To conclude our discussion of BCS superconductivty, let us explore the phenomenological consequences of mass accumulation due to the Higgs mechanism. To this end, let us vary the action (6.39) with respect to A (keeping in mind the transversality condition q\u00b7A\u22a5 q = 0, we henceforth drop the superscript \u201c\u22a5\u201d), to obtain ( ns m + q2)Aq = 0, or \u2212 \u22072 A(r) = 0. ns m (6.40) 298 Broken symmetry and collective phenomena B \u03bb y I x Figure 6.10 On the Meissner e\ufb00ect: inside a superconductor (the shaded area), magnetic \ufb01elds decay exponentially. Microscopically, an external \ufb01eld existing outside a superconductor\u2013vacuum interface", " induces diamagnetic surface currents inside the superconductor. These currents generate a counter-\ufb01eld that diminishes the external \ufb01eld. Remembering that B = \u2207 \u2227 A, multiplication of this equation by \u2207\u2227 produces the \ufb01rst London equation \u2212 \u22072 B(r) = 0. ns m (6.41) For ns = 0, this equation does not have a non-vanishing constant solution, i.e. we conclude that A bulk superconductor cannot accommodate a magnetic \ufb01eld. This phenomenon is known as the Meissner e\ufb00ect. To understand what happens at the interface between the vacuum threaded by a constant magnetic \ufb01eld B0 and a superconductor, one can solve the London equation to obtain B(x) \u223c B0exp(\u2212x/\u03bb), where \" m ns, \u03bb = is known as the penetration depth and x is the direction perpendicular to the interface (see Fig. 6.10). The physical mechanism behind the Meissner phenomenon is as follows: above we saw that the magnetic response of a superconductor is fully diamagnetic. That is, in response to an external \ufb01eld, diamagnetic screening currents will be generated. The magnetic \ufb01eld generated by these currents counteracts the unwanted external \ufb01eld. To see this explicitly, we obtain the current density induced by the \ufb01eld by di\ufb00erentiating the \ufb01rst term of the action46 with respect to A: j(r) = \u03b4 \u03b4A(r) ddr ns 2m A2 = ns m A(r), (6.42) 46 Generally, the electrical current density induced by a \ufb01eld is obtained (cf. the remarks made on page 255 ) by di\ufb00erentiating the \ufb01eld/matter part of the action S[A] with respect to the vector potential, i.e. the purely \ufb01eld-dependent part of the action does not contribute to the current density. 6.4 Superconductivity 299 i.e. the current density is directly proportional to A. This is the second London equation. Since the vector potential and the magnetic \ufb01eld show the same decay pro\ufb01le (Eq. (6.40) and (6.41)), the current density also decays exponentially inside the superconductor. However, in doing so, it annihilates the external \ufb01eld. INFO To heuristically understand the incompatibility of magnetic \ufb01elds and superconductive pairing on a still more microscopic level, consider the real space representation of a Cooper pair state, r, r|k \u2191, \u2212k \u2193 \u223c e\u2212ikre+ik\u00b7r = const. The cancellation of the phases results from the fact that two electrons propagating with opposite momenta acquire opposite quantum phases. Thus, the pair state is a slowly \ufb02uctuating, and therefore stable, object. However, in the presence of a magnetic \ufb01eld, the phase factors have to be generalized to r, r|k \u2191, \u2212k \u2193 \u223c e\u2212i dr\u00b7(k\u2212eA)e\u2212i dr\u00b7(\u2212k\u2212eA) \u223c e2ieA\u00b7r, where we assumed that the vector potential varies only slowly across our observation region of O(|r|), i.e. the Cooper pair amplitude becomes an \u201cincoherent\u201d phase-dependent object. (Exercise: Employ the WKB approximation to convince yourself of the validity of this statement.) On the microscopic level, the lack of stability of the \ufb01eld-dependent Cooper amplitude is responsible for the aversion of the superconductor to magnetic \ufb01elds. It is interesting to explore how a strong magnetic \ufb01eld eventually makes its way into the superconductor. To understand the competition between superconductive ordering and magnetic \ufb01eld energy, we need to go back to the Ginzburg\u2013Landau action (6.33), i.e. to a description that involves both phase and amplitude of the order parameter (the latter detecting the presence or absence of a stable condensate). However, at the time when we derived that action, no attention had been paid to the electromagnetic properties of the system. Fortunately, after our general discussion of gauge invariance above, the minimal coupling of the system to the electromagnetic \ufb01eld is routine work. We simply have to remember that, under a gauge transformation, \u0394 \u2192 \u0394e2i\u03c6, i.e. \u2207\u0394 \u2207 \u00af\u0394 \u2192 (\u2207 + 2i\u2207\u03c6)\u0394(\u2207 \u2212 2i\u2207\u03c6) \u00af\u0394. The gauge", " invariant extension of Eq. (6.33) thus reads as SGL[\u0394, \u00af\u0394] = \u03b2 ddr ( r 2 |\u0394|2 + c 2 |(\u2207 \u2212 2iA)\u0394|2 + g|\u0394|4, ) where, as usual, A gauges as A \u2192 A + \u2207\u03c6. To monitor the fate of the order parameter as |A| \u221d |B| increases, consider the mean-\ufb01eld equation (exercise) r + c(\u2212i\u2207 \u2212 2A)2 + 4g|\u0394|2 \u0394 = 0. Here we assume that we are at temperatures below the zero-\ufb01eld superconductor transition, i.e. r < 0. Superconductive ordering exists when the equation has a non-vanishing solution \u0394. Now, the third contribution on the left-hand side is positive, so a solution can exist only if the \ufb01rst two terms add to a net negative contribution. This in turn requires the following condition on the eigenvalues of the minimally coupled operator, EV(\u2212i\u2207 \u2212 2A)2! < |r| c. Formally, (\u2212i\u2207\u22122A)2 is the kinetic energy operator of a particle of mass 1/2 and charge q = 2 in a uniform magnetic \ufb01eld. Its eigenvalues are the Landau levels, \u03c9c, n = 0, 1,..., familiar from elementary quantum mechanics. Here, \u03c9c is the cyclotron frequency, \u03c9c = qB/m = 4B. n + 1 2! 300 Broken symmetry and collective phenomena Thus, a \ufb01nite pairing amplitude can be obtained only if |r|/c is larger than the energy of the lowest Landau level, or B < Bc2 \u2261 |r| 2c. For magnetic \ufb01elds larger than this, the energy needed to expel the \ufb01eld is larger than the maximum gain of condensation energy S[\u0394] and superconductivity breaks down. There are superconductor materials \u2013 superconductors of type II \u2013 where the energy criterion opts for \ufb01eld penetration for \ufb01elds Bc2 > B > Bc1 lower than the critical \ufb01eld Bc2 speci\ufb01ed by the mean\ufb01eld criterion above. For these systems, the superconductor and the \ufb01eld \u201cmeet a compromise.\u201d That is, vortex tubes of quantized \ufb02ux penetrate the superconductor for a \ufb01eld strength smaller than Bc2 but larger than the critical \ufb01eld strength Bc1.47 These Abrikosov vortices usually arrange into a triangular vortex lattice. The \ufb01gure above (courtesy of U. Hartmann, University of Sarbr\u00a8ucken) shows an STM image of a vortex lattice in the type II superconductor NbSe2. The distance between vortex centers is about 50 nm. Each of the vortices in a \ufb02ux lattice contains magnetic \ufb02ux \u03a6 = n h where 1/h (or e/h in units where the electron charge is kept track of) is the magnetic \ufb02ux quantum. Inside the cores of the Abrikosov vortices, the superconducting order parameter is suppressed, but outside it still exists. For a discussion of the thermodynamics of superconductors not showing a mixed phase (superconductors of type I), we refer to the literature.48 To conclude this section let us discuss the most prominent superconducting phenomenon, absence of electrical resistivity. Assume we have chosen a gauge where an external electric \ufb01eld E is represented through E = \u2212i\u2202\u03c4 A (i.e. the static component of the potential vanishes). In this case, a time di\ufb00erentiation of the second London equation (6.42) gives \u2212i\u2202\u03c4 j = \u2212i ns m E. Continuing back to real times we conclude that m \u2202\u03c4 A = ns \u2202tj = ns m E, i.e. in the presence of an electric \ufb01eld the current increases linearly at a rate inversely proportional to the carrier mass and proportional to the carrier density. The unbound increase of current is indicative of ballistic \u2013 i.e. dissipationless \u2013 motion of the condensate particles inside the superconductor. Now, an unbound increasing current is clearly unphysical, i.e. what the relation above really tells us is that a superconductor cannot maintain nonvanishing \ufb01eld gradients. EXERCISE Assuming that each particle is subject", " to Newton\u2019s equation of motion m\u00a8r = E, obtain the current\u2013\ufb01eld relation above. How would the relation between \ufb01eld and current change if the equation of motion contained a friction term (modeling dissipation) m\u00a8r = \u2212 m \u03c4 \u02d9r + E? 47 A. A. Abrikosov, On the magnetic properties of superconductors of the second group, Soviet Physics JETP 5, 1174-83 (1957). 48 cf., e.g., L.D. Landau and E. M. Lifshitz, Course of Theoretical Physics, Vol. 9 - Statistical Physics 2, (Butterworth\u2013Heinemann, 1981). 6.5 Field theory of the disordered electron gas 301 6.5 Field theory of the disordered electron gas To close the chapter on broken symmetry and collective phenomena, we turn now to a \ufb01nal, detailed application of the \ufb01eld integral method involving the problem of electrons propagating in a disordered environment. As well as the importance attached to this general area in the recent literature, the quantum disorder problem presents an ideal arena in which to revise the diagrammatic and \ufb01eld theoretic methods developed in this and previous chapters. This being so, this section is structured in a manner that re\ufb02ects the di\ufb00erent techniques and much of our discussion is deliberately cast in the form of problem assignments. However, this section may be read equally as a complete and coherent text, depending on taste. Disorder in metals No semiconductor or metal of macroscopic49 extent is ever free of imperfections and impurities. Indeed, the e\ufb00ect of disorder on the phenomenology of metals or semiconductors could not be more varied: in some cases, disorder plays an essential role (for example, conventional light bulbs would not function without impurity scattering!), in others the e\ufb00ect is parasitic (imparting only a \u201cblurring\u201d of otherwise structured experimental data), or it conspires to give rise to completely unexpected types of electron dynamics (as is the case in the quantum Hall transition discussed in Chapter 9). With this in mind, a complete theory of electron transport must, by necessity, include diagnostic tools to understand whether disorder seriously a\ufb00ects the problem of interest. Moreover, such a theory should include some analytical machinery to deal with the cases where the answer is a\ufb03rmative. What kind of criteria should a successful theory of disordered conductors meet? A fair fraction of those problems where impurity scattering plays an essential role can be addressed in terms of in\ufb01nite-order perturbation theory. However, there are plenty of phenomena \u2013 Anderson localization, the quantum Hall e\ufb00ects, the combined theory of interactions and disorder, to mention just a few \u2013 where non-perturbative \ufb01eld theoretical methods are required. In this section, the foundation of a general approach to the disordered electron gas, extendable to both perturbative and non-perturbative schemes, is laid. However, before doing this, we must \ufb01rst clarify what is meant by a \u201ctheory\u201d of the disordered electron gas. Of course, while the problem may be considered formally as noninteracting, we will not be able to e\ufb00ect an exact diagonalization of the random Schr\u00a8odinger equation for an electron in a metal for a given realization of the impurity potential. (Indeed, were such a particular solution available, it would not convey much information.) Rather, one needs to develop a statistical approach wherein the system is described in terms of a few universal characteristics of the scattering landscape \u2013 the strength of the impurity 49 In ultraclean semiconducting devices, electrons may travel up to distances of several microns without experiencing impurity scattering. Even so, the \u201cchaotic\u201d scattering from the typically irregular boundaries of the system has an equally invasive e\ufb00ect on the charge carrier dynamics. 302 Broken symmetry and collective phenomena potential, the typical range of potential \ufb02uctuations, etc. In general, the analysis of generic properties involves averaging over microscopic realizations of the impurity potential. INFO In situations where the system is so large that di\ufb00erent regions behave as though they were statistically independent with respect to their microscopic impurity con\ufb01guration, properties become self\u2013averaging. In such cases, the con\ufb01guration average can be subsumed into a volume average of the individual system. As a rule of thumb, systems which behave in a selfaveraging manner must extend well beyond the phase coherence length \u03be \u2013 the length scale over which the quantum propagation remains phase coherent. (A more precise characterization of \u03be is given below.) Now, at low temperatures, \u03be(T ) grows rapidly", ", implying that even systems of near macroscopic extent (ca. O(1 \u03bcm) and more) can behave as though they were non-selfaveraging. Systems of this type are often termed mesoscopic,50 where \u201cmeso-\u201d alludes to the fact that such systems are macroscopic in extent yet microscopic in their re\ufb02ection of quantum mechanical character. Mesoscopic systems manifest a multitude of unusual quantum phenomena from localization to strong sample-to-sample \ufb02uctuations, some of which will be discussed below. The experimental and theoretical study of these phenomena is the central theme of mesoscopic physics. How might one set about modeling an impurity potential in statistical terms? One might, for example, propose that a single imperfection at position ri creates a potential Vimp(r \u2212 ri). Assuming that all impurities are, by and large, equivalent, the total perturbation experii Vimp(r \u2212 ri). Within enced by an electron at point r will then take the form V (r) \u2261 this framework (which prevails in the older literature) the disorder average amounts to integrating over the coordinates of the impurities, i.e. \u00b7 \u00b7 \u00b7 dis \u2261 L\u2212N d ddri (\u00b7 \u00b7 \u00b7). The disadvantage of this scheme is that its implementation is not so straightforward, especially in functional-based approaches. In practice, it is more convenient to think of the potential V as some function whose statistical properties are described by a probability measure P [V ]. Averages over the potential are then computed by performing the integral DV P [V ](\u00b7 \u00b7 \u00b7). In most 51 applications it is su\ufb03cient to implement a Gaussian \u00b7 \u00b7 \u00b7 dis =, where \u03b3 distribution (unnormalized), P [V ] = exp measures the strength of the potential and K describes its spatial correlation pro\ufb01le: ddr ddr V (r)K \u22121(r \u2212 r)V (r) \u2212 1 2\u03b32 N i=1 # $ V (r)V (r)dis = \u03b32K(r \u2212 r). (6.43) Very often one \ufb01nds that the \ufb01nite spatial correlation of V is inessential, in which case one may set K(r) = \u03b4(r), and P [V ]DV = exp ddr V 2(r) DV. (6.44) \u2212 1 2\u03b32 The freedom that one can exercise in modeling the scattering potentials re\ufb02ects the fact that, in a multiple scattering process, details of the potential are quickly erased. At any 50 Note that it has, however, become fashionable to label any system of a size of O(1 \u03bcm) \u201cmesoscopic,\u201d irrespective of whether or not its behavior is classical. 51 An important class of exceptions is presented by bosonic problems of general type where, for stability reasons, V (r)! > 0, while Gaussian distributed potentials categorically include negative \u201ctails.\u201d 6.5 Field theory of the disordered electron gas 303 rate, the short-range Gaussian distribution provides a convenient scheme both in purely diagrammatic perturbation theory and in functional approaches. As discussed above, our aim is to average the quantum expectation value of a certain observable O over the ensemble of disorder. Let us assume that the observable O can |J= 0 ln Z, be obtained by di\ufb00erentiation of the (functional) free energy, that is O = \u2212 \u03b4 \u03b4J where J represents some source \ufb01eld,52 i.e. Odis = \u2212 \u03b4 \u03b4J ln Z (6.45) dis. J= 0 This fundamental relation presents a technical challenge: should one \ufb01rst di\ufb00erentiate and only then average, one would need to compute integrals of the type \u2212 DV P [V ] 1 Z[V, J = 0] \u03b4 \u03b4J |J= 0 Z[V, J]. Due to the appearance of the function V in both the numerator and the denominator, integrals of this type are largely intractable. The problem is particularly acute in functional approaches where one intends to take the ensemble average at an early stage of the computation.53 To date, three di\ufb00erent approaches have been identi\ufb01ed in which the problem with the denominator can be circumvented: the supersymmetry approach,54 the Keldysh technique,55 and the replica trick.56 All of these approaches share the feature that they alter the de\ufb01nition of the functional partition function in such a way that (a) Z[J = 0] = 1 (i.e. the disorder dependence of the denominator disappears), while (b) Eq. (6.45) remains valid, and", " (c) the algebraic structure of Z[J] is left largely unchanged. Since the disorder appears linearly in the Hamiltonian, point (c) implies that we need to average functionals with actions linear in the potential V, an enterprise that turns out to be quite feasible. Let us exemplify this program on the (technically) simplest of the three approaches above, the replica trick. Consider the Rth power of the partition function, Z R. For integer R one may think of Z R as the partition function of R identical copies of the original system (see Fig. 6.11), hence the name \u201creplica\u201d trick. To appreciate the merit of this procedure, one may note the formal relations O = \u2212 \u03b4 \u03b4J ln Z[J] = \u2212 \u03b4 \u03b4J lim R\u21920 eR ln Z \u2212 1 1 R = \u2212 \u03b4 \u03b4J lim R\u21920 1 R Z R. The last equality tells us that the expectation value of observables can be obtained by performing computations with the Rth power of Z (instead of its logarithm). Crucially, expressed in the coherent state representation, the expression for the replicated partition 52 In the jargon of \ufb01eld theory, a source \ufb01eld represents a parameter (function) that, when linearly added to the exponent of a functional integral, can be used to generate the expectation value of observables by di\ufb00erentiation. For example, the parameter \u03bc is a source generating expectation values of the particle number: \u2212\u2202\u03bcF = \u02c6N. 53 In cases where the disorder can be treated perturbatively, one may \ufb01rst expand in powers of V and only then average; indeed, this is quite a viable strategy. 54 K. B. Efetov, Supersymmetry method in localisation theory, Sov. Phys. JETP 55 (1982), 514-21. 55 For a review see, e.g., A. Kamenev, Many body theory of non-equilibrium systems in Nanophysics: Coherence and Transport eds. H. Bouchiat et al. 177-246, (Elsevier, Amsterdam, 2005). 56 S. F. Edwards and P.W. Anderson, Theory of spin glasses, J. Phys. F 5 (1975), 965\u201374. 304 Broken symmetry and collective phenomena Z Z Z Z ZR... Figure 6.11 On the idea behind the replica trick; for an explanation, see the main text. dV )R = e dV, i.e. Z R will function involves an e\ufb00ective action which is still linear, (e contain the disorder linearly in the exponent and, therefore, will be comparatively easy to average. However, the replica-averaging procedure involves one unusual feature \u2013 at the end of the calculation, one must implement the analytic continuation R \u2192 0. More precisely, we \u03b4J R\u22121Z Rdis) for every integer R will have to compute a certain function (namely f (R) \u2261 \u03b4 and then analytically continue R \u2192 0. However, there is no guarantee that f (R) is analytic all the way down to R = 0.57 In other words, the method is poorly founded, which is why it is called the \u201creplica trick\u201d as opposed to, say, the \u201creplica theory.\u201d However \u2013 in view of its poorly justi\ufb01able theoretical standing, quite surprisingly \u2013 examples where the replica trick is known to fail are rare. To some extent, this success is explained by the fact that the method is exact as long as the disorder is treated perturbatively (a point to be clari\ufb01ed below). So long as a perturbatively accessible point in the parameter space is not too far away, the chances are that it will not fail. INFO The applicability of replica methods is not limited to the theory of disordered electron systems. Replicated \ufb01eld theories are of potential use whenever it comes to averaging the free energy functional of a disordered classical or quantum system. The method has been proven most fruitful (at times also controversial) in the theory of conventional and spin glasses.58 Replica \ufb01eld theory With this background, we return now to an analysis of the disordered electron system. The construction of the \ufb01eld integral begins with the representation of the replicated partition function as a coherent state \ufb01eld integral Z R[J] = D( \u00af\u03c8, \u03c8) exp \u2212 S[\u03c8a, \u00af\u03c8a, J], (6.46) R a=1 where \u03c8a, a = 1,..., R, denotes the Grassmann \ufb01eld representing partition function number a=1 D( \u00af\u03c8a", ", \u03c8a), and S[\u03c8a, \u00af\u03c8a, J] =S 0[\u03c8a, \u00af\u03c8a] +S int[\u03c8a, \u00af\u03c8a] + Ss[\u03c8a, \u00af\u03c8a, J]. a, D( \u00af\u03c8, \u03c8) \u2261 (Notice that the source J is the same for all replicas, i.e. it does not carry an index a.) Here r S0[ \u00af\u03c8a, \u03c8a] = n ddr \u00af\u03c8a n(r) \u2212i\u03c9n \u2212 \u2212 EF + V (r) \u03c8a n(r), (6.47) \u22072 2m 57 This uncertainty re\ufb02ects disorder-generated correlations between the replicated systems. Since Z Rdis = ZR dis, the function we wish to continue is not just a harmless power function. 58 For a review of the industry of replica-based theoretical approaches in these \ufb01elds we refer to the article by G. Parisi, Glasses, replicas and all that, in Les Houches \u2013 Ecole d\u2019Et\u00b4e de Physique Th\u00b4eorique, Vol. 77, ed. J.-L. Barrat et al. (Elsevier, 2004). 6.5 Field theory of the disordered electron gas 305 describes the non-interacting part of the disordered electron system (assumed, for simplicity, to be spinless), Sint speci\ufb01es the bare particle interaction, and Ss is the source-dependent part of the action (whose structure we need not specify for the moment). Notice that, before averaging, the action is \u201creplica diagonal,\u201d i.e. \ufb01elds \u03c8a and \u03c8b, a = b, do not interact with each other. Before proceeding with this expression, let us make one technical remark on the calculation of observables. Suppose we were interested in the expectation value of some operator which, in the non-replicated theory, assumes the form O = \u2212(\u03b4/\u03b4J) ln Z[J] = O( \u00af\u03c8, \u03c8)\u03c8/1\u03c8. Here, J represents some suitably devised source and, in the last equality, O( \u00af\u03c8, \u03c8) is the coherent state representation of the operator O. The denominator 1\u03c8 reminds us of the fact that a quantum thermal average involves an explicit normalization by the (functional) partition function. Within the replicated formalism, the same expectation value assumes the form O = \u2212 lim R\u21920 1 R \u03b4Z R[J] \u03b4J = lim R\u21920 1 R R a=1 O( \u00af\u03c8a, \u03c8a)\u03c8, (6.48) where the last equality follows from di\ufb00erentiating Eq. (6.46) with respect to the source J. Assuming that all observables are evaluated as in Eq. (6.48), we no longer need to keep an explicit reference to the source \ufb01eld J. Now, let us average the functional (6.46) over the distribution (6.44). A straightforward application of the Gaussian integral formula (3.19) leads us to the result \u23a1 \u23a4 Z R[J]dis = D( \u00af\u03c8, \u03c8) exp \u23a3\u2212 R a=1 Scl[\u03c8a, \u00af\u03c8a] \u2212 R a,b=1 Sdis[\u03c8a, \u03c8b, \u00af\u03c8a, \u00af\u03c8b] \u23a6, where Scl = S V =0 denotes the action of the non-disordered system, and Sdis[\u03c8a, \u03c8b, \u00af\u03c8a, \u00af\u03c8b] \u2261 \u2212 \u03b32 2 m(r) \u00af\u03c8b m(r)\u03c8a n(r)\u03c8b ddr \u00af\u03c8a mn n(r), (6.49) represents an e\ufb00ective quartic interaction generated by the disorder average. Notice the super\ufb01cial similarity between Sdis and an attractive short-range \u201cinteraction\u201d term. However, in contrast to a dynamically generated interaction, (a) Sdis does not involve frequency exchanging processes (the reason being, of course, that the scattering o\ufb00 static impurities is energy conserving), and (b) it describes interactions between particles of a di\ufb00erent replica index. To understand the physics behind the attractive inter-replica interaction, consider the potential landscape of a given impurity con\ufb01guration (see the \ufb01gure). Irrespective of their replica indices, all Feynman amplitudes will try to trace out", " those regions in con\ufb01guration space where the potential energy is low, i.e. there will be a tendency to propagate through the same regions in the potential landscape. (Recall that all replica \ufb01elds are confronted with the same potential pro\ufb01le.) On average, this looks as if the replica \ufb01elds were subject to an attractive interaction mechanism. 306 Broken symmetry and collective phenomena (b) In summary, one may account for the presence of quenched or static disorder by (a) representing replicating the formalism, observables as in Eq. (6.48), and (c) adding the replica non-diagonal contribution Eq. (6.49) to the action. This results in a theory wherein the disorder no longer appears explicitly. (Technically, the e\ufb00ective action has become translationally invariant.) The price to be paid is that the action now contains the non-linearity Eq. (6.49). This concludes our formulation of the quantum disorder problem as a functional integral. In the following section, we develop some intuition for the e\ufb00ects of disorder which draw on the formalism developed above. Basic notions of impurity scattering The most basic time scale characterizing the scattering of electrons from static impurities is the elastic scattering time \u03c4. (Here the term \u201celastic\u201d emphasizes that the scattering o\ufb00 static impurities may lead to the transfer of momentum but not of energy.) In the literature, the scattering time is often prematurely identi\ufb01ed as the typical time of \ufb02ight between neighboring impurities. However, this interpretation may be misleading. For example, a system may be (and often is) polluted by a dense accumulation of very weak scattering imperfections. In such cases, the scattering time may be parametrically larger than the time of \ufb02ight. Similarly, the scattering may be from the shallow Coulomb potential created by impurities spatially separated from the conductor (a situation generically realized in semiconductor heterostructures). In such cases, impurity positions are themselves not even well-de\ufb01ned. How, then, can the scattering time be de\ufb01ned unambiguously? And how does it relate to the microscopic characteristics of the impurity potential? To assimilate the general meaning of the scattering time, let us consider the quantum amplitude U (y, x; t) = y|exp(\u2212i \u02c6Ht)|x for a particle to propagate from a point x to a point y in a time t for a particular realization of the disorder potential. One may think of this amplitude as the sum of all Feynman paths connecting the points x and y. On its journey along each path, the particle may scatter (see Fig. 6.12), implying that the action of the path depends sensitively on the particular realization of impurities. For large separations, |x \u2212 y|, the scattering phase becomes a \u201cquasi-random\u201d function of the impurity con\ufb01guration. The same applies, of course, to the linear superposition of all paths, the net amplitude U. Let us now consider the impurity-averaged value of the transition amplitude U (x, y; t)dis. As we are averaging over a superposition of random phases, one may expect that the disorder average will be translationally invariant and, as a result of the random phase cancellation, rapidly decaying, U (x, y; t)dis \u223c exp(\u2212|x \u2212 y|/(2)). The decay constant of the averaged transition amplitude de\ufb01nes the elastic mean free path while the related time \u03c4 \u2261 /vF denotes the elastic scattering time. (We reiterate that only in systems of dilute strong scattering centers can this scale be identi\ufb01ed with the average spacing between 6.5 Field theory of the disordered electron gas 307 x y x y Figure 6.12 Showing the scattering of an electron o\ufb00 static imperfections. Inset: the corresponding Feynman diagram. impurities.) In the following, we develop a quantitative description of this \u201cdamping\u201d process. Technically, it is most convenient to explore the behavior of the transition amplitude in the imaginary-time formalism. More speci\ufb01cally, we shall consider the imaginary-time Green function59 G(x, y; \u03c4 ) \u2261 \u03c8(x, \u03c4 ) \u00af\u03c8(y, 0)\u03c8, where the averaging is over the Grassmann action (6.47). (To keep the notation simple, we do not explicitly keep track of the normalizing denominator 1\u03c8 in our notation.) As usual, it will be convenient to perform the intermediate steps of the computation in frequency\u2013momentum space. We thus represent the correlation function as G(x,", " y; \u03c4 ) = T Ld \u03c9n,p,p e\u2212i\u03c9n\u03c4 +ip\u00b7x\u2212ip\u00b7yGp,p;\u03c9n, (6.50) \u00af\u03c8np \u03c8. (Keep in where \u03c9n is a fermionic Matsubara frequency and G(p, p; \u03c9n) = \u03c8n,p mind that, prior to the impurity average, the system lacks translational invariance, i.e. the Green function will depend on two independent momentum arguments.) Following the general prescription developed in Section 6.5, the correlation function averaged over a Gaussian disorder distribution is then given by (cf. Eq. (6.48)) Gp,p,ndis = lim R\u21920 1 R R a=1 \u03c8a n,p \u00af\u03c8a n,p \u03c8\u03b4p,p, (6.51) where \u03c8a is the ath component of the R-fold replicated \ufb01eld and \u00b7 \u00b7 \u00b7 \u03c8 now stands for the functional average with an action including the interaction term Eq. (6.49). (Exercise: Consider why the averaged Green function is diagonal in momentum space.) To keep the discussion simple, we shall ignore dynamical interactions (i.e. Coulomb interaction, etc.) between the particles and assume that the non-disordered part of the action S[\u03c8a, \u00af\u03c8a] = n,p describes only free fermions. n,p In the remainder of this section, our objective is to explore the impact of the disorder generated interaction on the behavior of the Green function by means of diagrammatic n,p(\u2212i\u03c9n + p2 \u00af\u03c8a \u2212 EF)\u03c8a 2m 59 Representing the transition amplitude for the creation of a particle at point x followed by an annihilation process at y, this function generalizes the single-particle transition amplitude U to the presence of a continuum of particles; while all that has been said above about the disorder-generated attenuation remains valid. 308 Broken symmetry and collective phenomena a b a b (a) (b) (c) (d) Figure 6.13 (a) Impurity scattering vertex, (b) the \ufb01rst-order self-energy diagrams, (c) the secondorder self-energy diagrams, and (d) SCBA self-energy. With the last, the bold line represents the full Green function while the diagram states that the self-energy is computed neglecting all crossed lines (cf. the discussion in Section 5.3). perturbation theory. As mentioned above, since the disorder problem presents a useful arena to practice the methods developed in the text we will outline the program as a sequence of assignments while the detailed solution is given below. Following the general arguments of Section 5.3, the principal object of interest is the impurity-generated self-energy operator \u03a3. Let us prepare the analysis of this object by introducing a bit of diagrammatic notation. We depict the impurity scattering vertex de\ufb01ning the action (6.49) as in Fig. 6.13(a).60 As usual, setting p = (\u03c9n, p), the free-particle Green function G0,p \u2261 (i\u03c9n \u2212 \u03bep)\u22121 will be denoted by a \ufb01ne (directed) line. Using this notation, and following the rules of diagrammatic perturbation theory developed in Chapter 5: Q1: Consider the self-energy \u03a3(1) at \ufb01rst order in the impurity scattering (Fig. 6.13(b)). Show that the \u201cHartree-type\u201d diagram (right) does not contribute (in the replica limit!). Compute the real and imaginary parts of the \u201cFock\u201d (left) contribution to the self-energy. Show that, in dimensions d \u2265 2, Re \u03a3 actually diverges. Convince yourself (both formally and heuristically) that this divergence is an artefact of our modeling of the impurity potential by a \u03b4-correlated function (see Eq. (6.44) and the related discussion). Consider what could be the reason for the real part of the self-energy not playing a very important role. Q2: Turning to the second-order contribution \u03a3(2) (see Fig. 6.13(c)), convince yourself that, in dimensions d >1, the diagram with crossed impurity lines is parametrically smaller than the second contribution. (What is the small parameter of the expansion?) Q3: This motivates the computation of the self-energy in the self-consistent Born approximation (SCBA) (see Fig. 6.13(d)). Show that the SCBA equation can be approximately solved by the ansatz Im \u03a3(\u03c9n) = \u2212", "sgn(\u03c9n)/2\u03c4, where \u03c4 is a constant. Once again, identify the small parameter of the expansion. Q4: Put together, one thus obtains the impurity-averaged Green function Gpdis = (i\u03c9n \u2212 \u03bep + i sgn(n)/2\u03c4 )\u22121. Fourier transforming this result back to real time/space, justify the identi\ufb01cation of the self-energy with (one half of) the inverse scattering time. 60 In the diagrammatic literature, individual scattering events o\ufb00 the impurity potential V (r) are traditionally depicted by dashed lines (see Fig. 6.12, inset). Once we average a product of such vertices, V (r1)V (r2) \u00b7 \u00b7 \u00b7V (r2n), over the Gaussian disorder distribution (6.44), the coordinates {ri} become pairwise identi\ufb01ed in all possible combinations. This is usually indicated by connecting the corresponding vertex lines, i.e. a diagram with 2n dangling bonds becomes a sum of diagrams, each containing n interaction lines. In our pre-averaged replica \ufb01eld theory, we are using the corresponding interaction vertex from the outset. 6.5 Field theory of the disordered electron gas 309 Q5: Why is the replica method exact in perturbation theory? A1: Unlike the Fock diagram, where all replica indices are locked to the index of the incoming Green functions, the Hartree diagram contains one free replica summation. This summation yields an excess factor R that, in the limit R \u2192 0, vanishes. For the same reason, all diagrams with closed fermion loops (loops connected to the external \ufb01eld amplitudes only by impurity lines or not at all) do not contribute to the expansion. Technically, the excluded contributions represent vacuum diagrams,61 i.e. on the level of perturbation theory, the only62 e\ufb00ect of the replica limit is the elimination of all vacuum processes. We have thus shown that the replica theory exactly simulates the e\ufb00ect of the normalizing partition function present in the denominator of the unreplicated theory (cf. the discussion of the linked cluster theorem in Section 5.1). This proves \u2013 all on the level of perturbation theory \u2013 the equivalence of the representations. The representation of the disorder-generated interaction Eq. (6.49) in momentum space emphasizes the fact that the impurity scattering exchanges arbitrary momentum, but not frequency. A straightforward Wick contraction along the lines of our discussion in Section 5.3 then obtains the \ufb01rst-order contribution p G(0) p,n \u03b32 d \u03bd() i\u03c9n + EF \u2212 \u03b32P d \u03bd() EF \u2212 \u2212 i\u03c0\u03b32\u03bd sgn (\u03c9n), \u03a3(1) p = \u03b32 stands for the principal value integral. For d \u2265 2, the increase of the DoS \u03bd() as where P a function of renders the real part of the self-energy formally divergent. This divergence is an immediate consequence of the unbounded summation over p \u2013 which is an artefact of the model.63 In any case, the real part of the self-energy is not of prime interest to us: all that Re \u03a3pn = const. describes is a frequency- and momentum-independent shift of the energy. This shift can be absorbed into a rede\ufb01nition of the chemical potential and will not cause any observable e\ufb00ects. By contrast, the imaginary part Im \u03a3(1) p,n = \u2212\u03c0\u03b32\u03bd sgn (\u03c9n) describes the attenuation of the quasi-particle amplitude due to impurity scattering, a mechanism of great physical signi\ufb01cance. A2: The analysis of the second-order contribution \u03a3(2) parallels our discussion of the RPA in Chapter 5: the Green functions Gp are sharply peaked around the Fermi surface |p| = pF. (Since the Matsubara index n in p = (p, n) is conserved in impurity scattering, we will not always write it out explicitly.) Representing the diagram with non-crossing lines in 61 To understand this assertion, consider the non-replicated theory prior to the impurity average. Due to the absence of \u201creal\u201d interactions, any closed fermion loop appearing in the expansion must be a vacuum diagram. After taking the impurity average, the loop may become connected to the external amplitudes by an impurity line. However, it remains a vacuum loop and must cancel against the expansion of the normalization denominator, if we were crazy enough to formulate the numerator\u2013denominator expansion of the theory explicitly. 62 In a connected diagram, all", " replica indices are locked to the index a of the external \ufb01eld vertices. We thus obtain (symbolic notation) Gdis \u223c limR\u21920 \u00d7 const. = const. where the factor of R results from the summation over a and we have used the fact that the correlation function \u03c8a \u00af\u03c8a is independent of a (replica symmetry). \u03c8a \u00af\u03c8a \u221dlim R\u21920 63 In reality, the summation will be \ufb01nite because either (a) there is an underlying lattice structure (i.e. the psummation is limited to the Brillouin zone) or (b) the kernel K(r) describing the pro\ufb01le of the impurity potential varies on scales large in comparison with the Fermi wavelength. In this latter case, its Fourier transform has to be added to the de\ufb01nition of the scattering vertex above. The presence of this function then limits the p-summation to values |p \u2212 p| < \u03be\u22121 pF. 1 R R R a 310 Broken symmetry and collective phenomena p1,p2 p1,p2 G(0) [G(0) c. \u223c n.c. \u223c p2+p\u2212p1 p2 G(0) p1 G(0) p1 ]2G(0) momentum space, one obtains \u03a3(2) p2, restricting both momenta to the Fermi surface, i.e. |p1|, |p2| p F. By contrast, the contribution with crossed lines takes the form \u03a3(2). Since all three momentum arguments have to be tuned to values close to pF, only one summation runs freely over the Fermi surface. To estimate the relative weight of the two contributions, we need to know the width of the \u201cshell\u201d centered around the Fermi surface in which the Green functions assume sizeable values. Since |G| = [(EF \u2212 p2/2m)2 + (Im G\u22121)2]\u22121, the width of the Lorenzian pro\ufb01le is set by Im G\u22121. As long as we are working with the bare Green function, Im [G(0)]\u22121 = \u03c9n \u221d T is proportional to the temperature. However, a more physical approach is to anticipate that impurity scattering will broaden the width Im [G(0)]\u22121 \u223c \u03c4 \u22121 to a constant value (to be identi\ufb01ed momentarily as the inverse scattering time). Then, the relative weight of the two diagrams can be estimated as p2(d\u22121) /(pF(vF\u03c4 )\u22121)d\u22121 = (pF)d\u22121, where = vF\u03c4 is the elastic mean-free path, and the numerator and denominator estimate the volume in momentum space accessible to the p1,2 summations in the two diagrams. The important message to be taken away from this discussion is that, for weak disorder pF \u22121 (which we have assumed throughout), and in dimensions d > 1, scattering processes with crossed impurity lines are negligible. Under these conditions, we are entitled to evaluate the selfenergy (and, for that matter, all other observables) within the self-consistent Born, or non-crossing, approximation, i.e. an approximation that neglects processes with crossed \u201cinteraction\u201d lines. F A3: Drawing on the analogous discussion in Section 5.3, the SCBA for the self-energy is given by the diagram shown in Fig. 6.13(d). The corresponding analytical expression takes the form (cf. Eq. (5.40)) \u03a3p,n = \u03b32L\u2212d p 1 i\u03c9n + EF \u2212 p2/2m \u2212 \u03a3p,n. (6.52) Guided by our results obtained in the \ufb01rst order of perturbation theory, we may seek a solution of (the imaginary part of) this equation by the ansatz Im \u03a3p,n = \u2212 1 2\u03c4 sgn(\u03c9n). Substitution of this expression into the SCBA equation gives \u2212 1 2\u03c4 sgn(\u03c9n) \u03b32Im d \u03bd() \u2212\u03c0\u03b32\u03bd sgn (\u03c9n), i\u03c9n + EF \u2212 + i where we have assumed that EF\u03c4 1. We have thus arrived at the identi\ufb01cation of \u03c4 \u22121 = 2\u03c0\u03bd\u03b32 as the elastic scattering rate. (In the literature, the potential strength \u03b32 = 1/(2\u03c0\u03bd\u03c4 ) is usually de\ufb01ned through the scattering time, i.e. the parameter \u03b32 is expressed through \u03c4 from the outset.) Summarizing, we have obtained the", " important result 2\u03c4 sgn(\u03c9n) Gpdis = 1 i\u03c9n + EF \u2212 p2 2m + i 2\u03c4 sgn(\u03c9n). (6.53) A4: To compute the inverse Fourier transform of Eq. (6.50) one may trade the frequency summation for a complex contour integral. Taking account of the fact that the Green function has a cut along the entire real frequency axis (the non-analyticity of sgn(\u03c9n) \u2192 6.5 Field theory of the disordered electron gas 311 sgn(Im z) at Re z = 0), we chose the integration contour as shown in the \ufb01gure. This leads us to the representation G(x, y; \u03c4 )dis = p d 2\u03c0 e\u2212\u03c4 (1 \u2212 nF())eip\u00b7(x\u2212y) \u00d7 Im 1 EF + \u2212 p2 2m + i 2\u03c4. Here, we are using 1 \u2212 nF instead of nF as a pole function because e\u2212z\u03c4 (1 \u2212 nF(z)) is \ufb01nite for large |z|, wherease \u2212z\u03c4 nF(z) is not. It is now a straightforward matter to compute this expression by \ufb01rst (a) integrating out the angular degrees of freedom of p followed by (b) a contour integration over p (cf. the analogous calculation of the clean Green function in the exercise at the end of section 5.1). However, as we require only the asymptotic e\ufb00ect of the damping, we may follow a more e\ufb03cient route: the essential e\ufb00ect of the damping term is to shift the poles of the p-integration from the real axis into the complex plane. Equating the Green function denominator to zero and neglecting the parameter ( EF as long as the time parameter \u03c4 E\u22121 F is not extremely small), we \ufb01nd that the poles are located at p = pF \u00b1 i 2. The essential e\ufb00ect of this shift is that the exponentials exp(ip|x\u2212y|) have to be evaluated at the residues exp(ipF|x \u2212 y| \u2212 |x \u2212 y|/(2)). Consequently, the disorder averaged Green function is related to the Green function of the clean system Gcl by the exponential damping factor which we had surmised above, G(x, y; \u03c4 )dis = Gcl(x, y; \u03c4 )e\u2212|x\u2212y|/2. A5: We refer to the discussion in A1. This completes our discussion of the impurity-averaged single-particle Green function G(x, y; \u03c4 )dis. The latter has been shown to be short-range on the scale of the elastic mean free path. In the following section, we shall see that, by contrast, the impurity-averaged two-particle Green function acquires long-range correlations which encode the modes of density relaxation and provide a means to explore mechanisms of quantum interference which characterize the mesoscopic regime. Di\ufb00usion How do local \ufb02uctuations in the electron density \u03b4 \u02c6\u03c1(\u03c4, r) \u2261 \u02c6\u03c1(\u03c4, r) \u2212 \u02c6\u03c4, \u03c1(r) relax in a disordered environment?64 In a classical disordered system, this relaxation mechanism would be di\ufb00usive. Below, we shall see that the same characteristic behavior survives in the quantum picture. Formally, this is achieved by exploring the correlation function D(\u03c4, r) \u2261 64 Notice that the brackets \u00b7 \u00b7 \u00b7 represent the quantum average, and not yet the disorder average. 312 Broken symmetry and collective phenomena \u03b4 \u02c6\u03c1(\u03c4, r)\u03b4 \u02c6\u03c1(0, 0)dis, i.e. a function that describes how \u03b4 \u02c6\u03c1(\u03c4, r) changes in response to a density \ufb02uctuation at (r = 0, \u03c4 = 0). EXERCISE In a real time formulation, how is the correlation function D related to the quantum transmission probability density? INFO The di\ufb00usive relaxation process described by the function D extends over distances much larger than the decay length of the average single-particle Green function,. To identify the mechanism that underlies the stability of the four-point function, think of D(\u03c4, r) = \u00af\u03c8(\u03c4, r)\u03c8(0, 0) \u03c8(\u03c4, r) \u00af\u03c8(0, 0)dis as the product of the amplitudes for (a) the annihilation of a particle at space-time coordinate (0, 0) followed by particle creation at (\u03c4, r) and (b) the creation of a particle at (0, 0)", " followed by its annihilation at (\u03c4, r).65 The annihilation process initiating (a) may be interpreted alternatively as the creation of a hole, i.e. the composite process D describes the joint propagation of a particle and a hole amplitude through the medium (which explains why D is sometimes called the particle\u2013hole propagator). Now, against this background, let us temporarily switch to a real time description \u03c4 \u2192 it and draw on the intuition a\ufb00orded by the consideration of Feynman amplitudes. Speci\ufb01cally, let us consider the particle propagator as the sum of all Feynman paths \u03b1 connecting 0 and r (see Fig. 6.14). Similarly, the hole amplitude corresponds to a path-sum over all paths where each path \u03b2 is weighted by the negative of its classical action.66 We thus have the symbolic representation D \u223c d d \u03b1\u03b2 A\u03b1A\u2217 \u03b2exp i (S\u03b1() \u2212 S\u03b2()), where the notation emphasizes that the electron and hole have di\ufb00erent energies,. As with the single-particle propagator, the strong sensitivity of the actions S\u03b1,\u03b2 on the impurity potential implies that generic contributions (\u03b1, \u03b2) to the path double sum vanish upon impurity averaging. (S\u03b1()\u2212S\u03b1())), However, the \u201cdiagonal\u201d contribution (\u03b1, \u03b1) to the sum, Ddiag \u223c depends only weakly on the impurity potential, and will likely survive the averaging procedure. A glance at Fig. 6.14 (b) suggests that the diagonal contribution to D is but an elaborate quantum description of classical di\ufb00usion: two quantum amplitudes locally tied together, and propagating in a stochastic environment, should be capable of interpretation as a classically di\ufb00usive probability density. Indeed, this interpretation is corroborated by the fact that, upon approaching the classical limit \u2192 0, the dominance of the diagonal approximation becomes more pronounced. |A\u03b1|2exp( i \u03b1 However, notice that, beyond the straightforward diagonal con\ufb01guration, there are other contributions to the double path sum that are likely to be impervious to con\ufb01gurational averaging. For example, the two paths depicted in Fig. 6.14 (c) are globally di\ufb00erent but locally paired. The former attribute tells us that we are dealing with a fundamentally non-classical contribution to the correlation function, and the latter that the action di\ufb00erence S\u03b1 \u2212 S\u03b2 will be small. Indeed, 65 To understand the positioning of the quantum-thermal averaging brackets, consider the de\ufb01nition of D in terms of the operators \u03b4 \u02c6\u03c1, and take into account the fact that, of the two possible Wick contractions, one gets canceled due to the subtraction of \u02c6\u03c1. 66 Heuristically, the inversion of the sign of the action accounts for the fact that a hole is a \u201cmissing\u201d electron. More formally, it follows from the fact that the hole amplitude is obtained by complex conjugation of the particle amplitude. 6.5 Field theory of the disordered electron gas 313 (a) (b) (c) Figure 6.14 (a) A generic pair of Feynman paths contributing to the density\u2013density correlation function prior to averaging. (b) Particle and hole propagating along the same path in con\ufb01guration space \u2013 the \u201cdi\ufb00uson.\u201d (c) A non-classical contribution to the path sum. we shall see that processes like the one shown here lie at the root of the observed quantum phenomena that characterize the physics of disordered media. Below, we apply concepts very similar to those developed in Section 5.3 to understand the spatial long-range character of the four-point function. Further, we wish to elucidate the di\ufb00usive character of this correlation function. Q1: Before turning to explicit computations, let us derive two exact relations obeyed by the Fourier transform Dq,\u03c9m : Show that limq\u21920 Dq,0 = L\u2212d\u2202\u03bcN and D0,\u03c9m = 0, where N = \u02c6N \u03c8 denotes the number of particles in the system and, as usual, \u03bc \u2194 EF represents the chemical potential. Explain the origin of these two sum rules. Q2: Represent the four-point function in a momentum\u2013frequency form similar in structure to that of the correlation function C (4) introduced in Eq. (5.41). (Hint: It is convenient to ln Zdis|\u03bc(x)=\u03bc, where x \u2261 (\u03c4, r), and represent the correlation function as D(x) =\u03b4 2 \u03bc(x) =\u03bc (\u03c4, r) is the generalization of the chemical potential to a space-time", "-dependent source \ufb01eld.) \u03bc(x)\u03bc(0) Q3: To compute the two-particle correlation function, one may apply concepts similar to those introduced in Section 5.3. In doing so, we will bene\ufb01t from two major simpli\ufb01cations. Assuming that (pF)\u22121 1, we will work to leading order in this parameter (pF 1 plays a role similar to that of the large parameter N in Section 5.3). Secondly, we may make use of the fact that the momentum q Fourier conjugate to the argument |r| is much smaller than the Fermi momentum. Show that, under these conditions, the irreducible vertex, \u03930,q,p,p = (2\u03c0\u03bd\u03c4 )\u22121\u03b4\u03c9n,\u03c9, collapses to (the Fourier representation of) a single impurity vertex. (Since all \ufb01eld amplitudes that contribute in the replica limit carry the same replica index a, one may drop the replica structure from the notation.) Write down the Bethe\u2013Salpeter equation for the full vertex. n Q4: Solve the Bethe\u2013Salpeter equation to leading order in (pF)\u22121 (note |q| 1). (Hint: Notice that the two cases \u03c9n \u00b7 \u03c9n+m < 0 and \u03c9n \u00b7 \u03c9n+m > 0 behave in radically di\ufb00er- 314 Broken symmetry and collective phenomena p1 p2 \u2013p2 + q p1 \u2013p2 + q = \u2013p1 + q \u2013p1 + q p2 = + p \u2013p + q Figure 6.15 Maximally crossed contribution to the irreducible vertex. Second line: the corresponding Bethe\u2013Salpeter equation. In all diagrams, the upper (lower) Green functions carry frequency \u03c9n+m (\u03c9n). ent manners.) Use your result for the irreducible vertex to compute the density\u2013density correlation function. Q5: Referring to Fig. 6.14(c), the diagram suggests that a particle\u2013hole pair propagating along the same path in a disordered medium may split up, propagate along a closed loop in opposite directions, and later recombine. Provided that the action for traversal of a scattering path in the medium does not depend on the direction of propagation, one may expect the classical action di\ufb00erence of the composite process to be small.67 How would these processes \u2013 known as cooperon-modes, or just cooperons68 in the literature \u2013 manifest themselves in the present formalism? The cooperon describes a process wherein the two constituent partners propagate with near opposite momenta. Indeed, it turns out that, for p1 \u2212p2, the irreducible vertex contains a second contribution (besides the isolated impurity line) of more complex structure: a sum over diagrams with \u201cmaximally crossed\u201d impurity lines (see Fig. 6.15). This diagram, too, contains one Fermi sphere summation per impurity line. The most economic way to see this is to imagine the lower of the two fermion propagators twisted by 180\u25e6 (the second diagram in the \ufb01gure). Super\ufb01cially, it now resembles the previously explored vertex. The important di\ufb00erence, however, is that the arrows marking the fermion propagators now point in the same direction, i.e. as compared with the particle\u2013hole mode discussed above, the relative sense of traversal of the impurity sequence became reversed. Compute the cooperon contribution to the irreducible vertex. Discuss what would happen if the system was exposed to an external magnetic \ufb01eld. In spite of its structural similarity to the \u201cdi\ufb00uson mode\u201d identi\ufb01ed in Q4, why does the insertion of a single cooperon into the di\ufb00usive correlation function not give rise to a large correction? 67 Formally, the directional invariance of the action requires that the system is invariant under time-reversal. Time-reversal symmetry would be lost, for example, if the system were subject to an external magnetic \ufb01eld. In this case, the classical probability for propagation along a path depends on the sense of traversal. 68 The origin of the terminology \u201ccooperon\u201d is that these modes describe the dynamics of a two-particle process where the two constituents propagate in opposite direction (in analogy to the opposite momenta of the two electrons of a Cooper pair). 6.5 Field theory of the disordered electron gas 315 d\u03c4 ddr \u02c6\u03c1(r, \u03c4 )\u03c8 = A1: The functional expectation value of the particle number is given by N = ddr \u02c6\u03c1", "\u03c9m=0(r)\u03c8. Di\ufb00erentiating this expression with respect to \u03bc, d\u03c4 \u02c6\u03c1(r, \u03c4 ), and noting that the chemical potential couples to the action through \u03bc one obtains69 ddr \u2202\u03bcN = = ddr ddr [\u02c6\u03c10(r)\u02c6\u03c10(r)\u03c8 \u2212 \u02c6\u03c10(r)\u03c8\u02c6\u03c10(r)\u03c8] ddr ddr D\u03c9n=0(r \u2212 r) = Ld lim q\u21920 Dq,0. Particle number conservation demands that ddr D(r, \u03c4 ) = 0 or, equivalently, Dq=0,\u03c9m = 0. ddr \u03b4 \u02c6\u03c1(r, \u03c4 )\u03c8 = 0 at all times. Consequently, A2: It is straightforward to verify that the two-fold \u03bc-di\ufb00erentiation suggested above yields the correlation function D. Now, let us employ the replica formulation ln Zdis = Z R \u2212 1. Di\ufb00erentiating the right-hand side of this equation, one obtains limR\u21920 1 R D(x) = lim R\u21920 1 R \u00af\u03c8a(x)\u03c8a(x) \u00af\u03c8b(0)\u03c8b(0)\u03c8. To avoid the vanishing of this expression in the limit R \u2192 0, we need to connect operators \u00af\u03c8a and \u03c8b ( \u00af\u03c8b and \u03c8a) by fermion lines (thus enforcing a = b \u2013 otherwise the two-fold summation over a and b would produce an excessive factor R which would result in the vanishing of the expression in the replica limit). We thus obtain a structure similar to that discussed in Section 5.3 (cf. Fig. 6.16): two propagators connecting the points x and 0, where the role of the wavy interaction line of Section 5.3 is now played by the \u201cinteraction\u201d generated by the impurity correlator V V. As in Section 5.3, it will be convenient to formulate the diagrammatic analysis of the correlation function in momentum space. We thus substitute the Fourier decomposition \u03c8a(x) = (T /Ld)1/2 p into the de\ufb01nition of D, make use of the fact that \u00af\u03c8a \u00af\u03c8a \u221d \u03b4qq (momentum conservation in the averaged theory), and obtain p2+q \u03c8a \u03c8a p2 p1 \u03c8. The D(x) = T Ld impurity interaction is static, implying that the fermion frequency is conserved along each propagator: \u03c9n = \u03c9n, i.e. q eiq\u00b7xDq, whereD q = T Ld limR\u21920 R\u22121 p e\u2212ip\u00b7x\u03c8a \u00af\u03c8a p2+q\u03c8a p2 \u00af\u03c8a p1 p1+q p1+q \u03c8a p1p2 Dq = T Ld lim R\u21920 1 R \u03c9n p1p2 \u00af\u03c8a p1,\u03c9n \u03c8a p1+q,\u03c9n+m \u00af\u03c8a p2+q,\u03c9n+m \u03c8a p2,\u03c9n \u03c8. (6.54) A3: Only diagrams that contain one free summation over the Fermi surface per impurity line contribute to leading order in (pF)\u22121. (Formally, this is because each impurity line contributes a factor \u03b32 \u223c (\u03bd\u03c4 )\u22121 and the phase volume of a momentum summation over the Fermi surface is needed to compensate the density of states factor \u03bd in the denominator.) As in our discussion of the self-energy, this implies that diagrams with crossed impurity lines do not contribute to the irreducible vertex. (The only irreducible contribution without 69 From the expressions above it is, in fact, not quite clear why we set \ufb01rst \u03c9n = 0 and only then q = 0. That this is the correct order of limits can be seen by generalizing \u03bc \u2192 \u03bc(r) to a smoothly varying static \ufb01eld, evaluating the corresponding functional derivatives \u03b4/\u03b4\u03bc(r), and setting \u03bc = const. at the end of the calculation. 316 Broken symmetry and collective phenomena p1, n p1, n p2, n D = + p1 + q, n + m p1 + q, n + m p2 + q, n + m = + Figure 6.16 Diagrammatic expansion of the di\ufb00uson mode. crossings is the single impurity line.) Consequently, the diagrammatic expansion", " of the correlation function assumes a form shown in Fig. 6.16. The Bethe\u2013Salpeter equation for the impurity vertex (shaded in the \ufb01gure) reads \u0393p1,p2,q = 1 2\u03c0Ld\u03bd\u03c4 + 1 2\u03c0Ld\u03bd\u03c4 p Gp+q,n+mGp,n\u0393p,p2,q, (6.55) where the Gs denote the impurity-averaged single-particle Green functions evaluated in the SCBA and discussed in the previous section. A4: The correlation function D contains a summation over frequency. Assuming that \u03c9m > 0, let us organize the sum as D = D++ + D+\u2212 + D\u2212\u2212, where D++, D+\u2212, and D\u2212\u2212 denote, respectively, the contributions where {\u03c9n, \u03c9n+m} > 0, {\u03c9n+m > 0, \u03c9n < 0}, and {\u03c9n, \u03c9n+m} < 0. We begin by considering the most interesting term D+\u2212. Introducing the notation G\u00b1 \u2261 (EF \u2212 p2/2m \u00b1 i/2\u03c4 )\u22121, one may expand the integrand appearing in the p Bethe\u2013Salpeter equation to leading order in the small70 energies \u03c9n, \u03c9n+m and q \u00b7 p/m, and obtain Gp+qGp = G+ p G\u2212 p 1 \u2212 iG+ p \u03c9n+m \u2212 iG\u2212 p \u03c9n + (G+ p q \u00b7 p/m)2 + \u00b7 \u00b7 \u00b7, (6.56) where a term linear in q has been omitted as it will vanish upon integration over the angular coordinates of p. We next make the assumption (to be checked self-consistently) that, for small q, the vertex \u0393 = \u0393q does not depend on the \u201cfast\u201d momenta p, p. The integration over p \u2013 now decoupled from the vertex \u2013 can then be made with the help of two auxiliary identities:71 ddp f (|p|)p2 and ddp f (|p|)(v \u00b7 p)2 = v2 d L\u2212d % &n++1 % &n\u2212+1 G\u2212 p G+ p = 2\u03c0in\u2212\u2212n+ p (n+ + n\u2212)! n+! n\u2212! \u03bd\u03c4 n++n\u2212+1. 70 The inequality \u03c9n\u03c9n+m < 0 implies that |n| < |m|. Thus |\u03c9n|, |\u03c9m| \u03c4 \u22121, where it is assumed that the frequencies we are probing are much smaller than the scattering rate. 71 This can be checked by writing v \u00b7 p = vipi and using the symmetry of the integrand under the operations pi \u2192 \u2212pi and pi \u2192 pj. 6.5 Field theory of the disordered electron gas 317 The last identity is proven by converting the momentum sum into an integral, which is then performed by contour integration (exercise). Substituting the expansion (6.56), and employing the two auxiliary identities, one arrives at the relation 1 \u2212 \u03c4 \u03c9m \u2212 q2 \u03c4 2v2 d L\u2212d + 2\u03c0\u03bd\u03c4 1 2\u03c0\u03bd\u03c4 \u0393q = \u0393q F. Solving this equation for \u0393q, we obtain our \ufb01nal result for the vertex, \u0393q = 1 2\u03c0\u03bd\u03c4 2Ld 1 \u03c9m + Dq2, where D = vF/d de\ufb01nes the di\ufb00usion constant of a dirty metal. A few remarks about this result are in order. Firstly, one may note that the absence of (q, \u03c9m)-independent constants in the denominator results from a cancellation of two terms in the vertex equation. (We have met with a similar cancellation in our discussion of the vertex of the generalized \u03c64-theory in Chapter 5.) Thanks to this cancellation, \u0393(r, \u03c4 ) becomes a long-range object. Later, we will identify the absence of damping or mass-like terms in the denominator as a consequence of a fundamental symmetry in the problem. Secondly, one may note that \u0393 is a solution of the di\ufb00usion equation (\u2202\u03c4 \u2212 D\u22072)\u0393(\u03c4, r) = (2\u03c0Ld\u03c4 2)\u22121\u03b4d(r)\u03b4(\u03c4 ) (exercise), describing the manner in which a distribution initially centered at x = 0 spreads out in time. Alluding to this analogy, the vertex \u0393 has become known as the di\ufb00uson. However, more important than the terminology is the fact that we have succeeded in making the connection between particle\u2013", "hole propagation in a dirty medium and classical di\ufb00usive processes quantitative. To \ufb01nalize our calculation of the four-point function we need only add external Green functions to the vertex (see Fig. 6.16), add the \ufb01rst diagram (the empty bubble) shown in the \ufb01gure, and integrate over the momenta p1, p2. Expanding the corresponding Green functions to zeroth order in q, \u03c9n, \u03c9n+m (cf. Eq. (6.56)) and using the auxiliary identity above, one obtains D+\u2212 q = \u2212T \u2212\u03c9m<\u03c9n<0 2\u03c0\u03bd\u03c4 + (2\u03c0\u03bd\u03c4 )2 2\u03c0\u03bd\u03c4 2 1 \u03c9m + Dq2 \u2212 \u03bd\u03c9m \u03c9m + Dq2, where, in the \ufb01rst equality, the factors 2\u03c0\u03bd\u03c4 stem from the momentum integration over pairs of Green functions, and the second equality holds under the presumed conditions \u03c9m, Dq2 \u03c4 \u22121. (The overall minus sign comes from the fact that we are computing a closed fermion loop.) To complete the analysis, we need to compute the expressions D++/\u2212\u2212. The analysis of these objects is fairly simple inasmuch as the vertex actually collapses to a single impurity line. Since the two Green functions over which the sum in Eq. (6.55) is performed now have imaginary parts of the same sign, the contour integration over |p| gives zero. Similarly, the placement of the residual single-impurity line vertex into the second diagram of Fig. 6.16 leads to a vanishing contribution. Since the sum over |p1|, |p2| extends over Green functions with imaginary parts of the same sign, one would be tempted to say that, for the same reasons, the empty bubble diagram ought to vanish. However, this is not the case: the tricky 318 Broken symmetry and collective phenomena point is that, with regard to the \ufb01rst diagram, one must keep an eye on the frequency summation. Considering D++, all other diagrams correspond to frequency sums \u03c9n>0 F (\u03c9n), where F (\u03c9n) n1\u223c n\u2212k\u22654. However, the \ufb01rst diagram leads to an expression with only k = 2. This means that, after the frequency summation, one obtains a p-dependent expression that does not converge quickly enough to qualify for a safe contour integration procedure.72 To be speci\ufb01c: D++ q \u2212 T Ld p \u03c9n>0 1 (i\u03c9n \u2212 \u03bep + i 2\u03c4 )2 \u221e \u2212\u221e p = 1 2\u03c0iLd \u2212 1 2\u03c0iLd p 1 \u2212 \u03bep + i 2\u03c4, d nF() ( \u2212 \u03bep + i 2\u03c4 )2 where, as usual, \u03bep = p2/2m \u2212 EF and the frequency summation has been performed by integrating over the real axis and closing in the upper complex half plane, and in the last equality we have approximated nF() \u0398(\u2212). Similarly, an analogous computation obtains D\u2212\u2212 = (D++)\u2217, i.e. D++ + D\u2212\u2212 = i 2\u03c0Ld p 1 \u2212 \u03bep + i 2\u03c4 + c.c. = \u2212 1 \u03c0 d Im \u03bd() EF \u2212 + i/2\u03c4 \u03bd. Combining everything, we arrive at the \ufb01nal result, Dq = D++ q + D+\u2212 q + D\u2212\u2212 q = \u03bd 1 \u2212 |\u03c9m| |\u03c9m| + Dq2 = \u03bd Dq2 |\u03c9m| + Dq2. (Here we have used the fact that the result does not change under sign inversion of \u03c9m. Enthusiastic readers may wish to check this claim.) Summarizing, we have thus found that the density\u2013density correlation function assumes the form D(r, \u03c4 ) = T Ld q,\u03c9m eiq\u00b7r+i\u03c9m\u03c4 \u03bdDq2 |\u03c9m| + Dq2. This result makes the di\ufb00usive character of the density\u2013density correlation function manifest. Notice also that Dq obeys the two limiting conditions discussed in (A1). (Indeed, for a non-interacting fermion system, \u03bd = L\u2212d\u2202\u03bcN.) A5: Let us now consider the irreducible vertex in a sector in momentum space where the sum of the upper incoming momentum, p1, and lower incoming momentum, \u2212p1 + q, is small. (Momentum conservation then implies that the upper outgoing momentum, p2, and lower outgoing momentum,", " \u2212p2 + q, also sum to the small o\ufb00set q.) Further, let us assume that the Matsubara frequencies \u03c9n and \u03c9n+m carried by the upper and lower propagators are of opposite signs. Now, imagine the lower line is turned around in such a way that the diagram assumes the form of a ladder structure (still it remains \u201cirreducible\u201d because the notion of 72 Indeed, the |p|-summation only approximately extends over the entire real axis. We are thus well advised to do the frequency summation (which has true semi-in\ufb01nite support) prior to the momentum summation. 6.5 Field theory of the disordered electron gas 319 irreducibility has been de\ufb01ned for fermion lines of opposite orientation). The corresponding Bethe\u2013Salpeter equation takes the form \u0393C q = 1 2\u03c0\u03bd\u03c4 Ld + 1 2\u03c0\u03bd\u03c4 Ld p G\u2212p+q,n1+mG\u2212p,n1 \u0393C q, where \u0393C is the cooperon contribution to the irreducible vertex, and we have assumed that, for small external momenta, \u0393C is independent of the \u201cfast\u201d momenta. In the absence of an external magnetic \ufb01eld, G\u2212p = Gp is even, i.e. the Bethe\u2013Salpeter equation coincides exactly with Eq. (6.55) for the di\ufb00usion mode considered above. Consequently, we may infer that \u0393C q,\u03c9m = 1 2\u03c0\u03bd\u03c4 Ld 1 |\u03c9m| + Dq2. However, in the presence of a magnetic \ufb01eld, the inversion symmetry is lost. Let us, for a moment, assume that the vector potential A representing the magnetic \ufb01eld depends only weakly on its spatial coordinate. Neglecting commutators between the momenta and the vector potential, and de\ufb01ning Gp(A) = (i\u03c9n + EF \u2212 (p + A)2/2m + i/2\u03c4 sgn(\u03c9n))\u22121, (e = c = 1), one may then shift the fast momentum to obtain Gp+q(A)G\u2212p(A) = p\u2192p\u2212A Gp+q(A)Gp(\u2212A) \u2212\u2212\u2212\u2212\u2212\u2192 Gp+q(2A)Gp(0) = Gp+q\u22122A(0)Gp(0). In this way, one may see that the magnetic \ufb01eld can be absorbed into a rede\ufb01nition of the momentum di\ufb00erence q. In the presence of an external magnetic \ufb01eld, the cooperon then acquires the form \u0393C \u02c6q = 1 2\u03c0\u03bd\u03c4 Ld 1 |\u03c9m| + D(q + 2A)2. The notation here emphasizes that the simpli\ufb01ed argument above generalizes to \ufb01elds with arbitrary (yet smooth on the scale of the microscopic mean free path) space dependence. In that case, \u0393C is de\ufb01ned as the solution of the equation (|\u03c9m| + D(\u2212i\u2207r + 2A)2)\u0393(r, \u03c9m) = (2\u03c0\u03bd\u03c4 Ld)\u22121\u03b4d(r) (a formal solution of which is given by the right-hand side of the de\ufb01nition above). Crucially, the presence of the vector potential spoils the singularity of the cooperon mode in the limit q, \u03c9m \u2192 0. In other words, the magnetic \ufb01eld turns the cooperon into an exponentially decaying mode, as expected from our qualitative discussion above. To conclude this section, let us discuss the role of the cooperon mode within the larger framework of the \u201cquantum di\ufb00usive\u201d process. Neglecting the cooperon, we had identi\ufb01ed the density\u2013density correlation function \u2013 formally computed as a ladder sequence where each rung was given by the trivial single impurity line irreducible vertex \u2013 with a classical di\ufb00usive process. However, we have seen that coherent loop excursions traversed in opposite directions give rise to quantum corrections to this picture. Formally, these excursions were described by maximally crossed corrections to the irreducible vertex. We saw that, in the absence of external \ufb01elds, these corrections quantitatively resembled the di\ufb00usive form of the (leading-order) particle\u2013hole mode. (This could have been anticipated from Fig. 6.14(c). Although the impurity sequence is now", " traversed in an opposite sense, the cooperon excursion provides a di\ufb00usive contribution.) Does this formal similarity imply that the cooperon modes give a largeO (1) correction to the leading classical term? To understand why it 320 Broken symmetry and collective phenomena does not, we have to, once again, consider the phase space structure of the fast integration momenta. The di\ufb00usion mode is computed for an overall conserved small momentum di\ufb00erence between the upper and lower momenta. On the other hand, the cooperon mode requires the momentum sum of the upper incoming and lower outgoing momenta to be small. These two conditions have to be reconciled. It is straightforward to see that this leads to the loss of Fermi-sphere integration volume. In other words, the cooperon gives rise to a singular (in Dq2 + |\u03c9m|) correction to the di\ufb00usion mode. However, this correction is weighted by a small factor \u223c (pF)d\u22121. The reader may wonder why we left the cooperon mode calculation with the semiquantitative arguments above. The reason is that the actual embedding of the cooperon mode into the density\u2013density correlation function, i.e. the solution of the Bethe\u2013Salpeter equation for the full vertex in the presence of a cooperon-corrected irreducible vertex, is a tricky business. It requires a fair amount of diagrammatic skill to obtain the correct result (notably one that ful\ufb01lls the conservation laws discussed in A1). In the following section, we will see that this task is much more e\ufb03ciently tackled by functional methods, a feature not uncommon in the \ufb01eld-theoretic environment. Further, one may wonder what happens when the external momenta are tuned to zero. Eventually, the singularity of the cooperon mode will seriously compete with its small prefactor, so that qualitative corrections to classical di\ufb00usion must be expected. Indeed, it turns out that a proliferation of cooperon corrections leads to a slowing down of the di\ufb00usive propagation of electron densities. In a way, the cooperon describes the tendency of an electron\u2019s quantum amplitude to constructively self-interfere at the point of origin of the closed loop. Through an accumulation of such processes, the electron may eventually get stuck or localized. Qualitatively, this is the origin of the phenomenon of Anderson localization in metals. We shall return to this phenomenon from a more \ufb01eld-theoretical perspective below. Mean-\ufb01eld theory and spontaneous symmetry breaking Previously, we have seen that the Green function of the disordered electron gas contains a complex self-energy describing the damping of quasi-particle amplitudes by impurity scattering. Further, we have found that the more complex four-point function is a long-range object whose behavior is governed by soft modes of density relaxation: the di\ufb00uson and cooperon modes. The existence of the latter was intimately tied to the time-reversal invariance properties of the system. In concluding, we had anticipated that a complete theory of the weakly disordered conductor should be able to describe the embedding of di\ufb00uson and cooperon modes into structures of higher complexity. It also became quite clear that the construction of a diagrammatic approach to the problem would be a formidable task; rather one should aim to \ufb01nd a formulation wherein the di\ufb00uson and cooperon modes (in contrast to individual fermion Green functions) appear as fundamental degrees of freedom. Such considerations call for a \ufb01eld-theoretical construction. In this section, we utilize the machinery of the present chapter to construct the low-energy \ufb01eld theory of the disordered 6.5 Field theory of the disordered electron gas 321 p1 p2 p1 p2 p1 p1 + q p1 + q p2 + q \u2013p1 + q \u2013p2 + q p2 p2 + q Figure 6.17 The three di\ufb00erent low-momentum channels of the impurity vertex: exchange, Cooper, and direct. electron gas. In doing so, we establish contact with our previous results obtained from the diagrammatic perturbation theory, and we understand the phenomenon of soft mode formation in the disordered electron system from a wider perspective. As usual, our starting point is the replicated action of the non-interacting disordered electron gas derived in Section 6.5: \u2202\u03c4 \u2212 EF \u2212 \u22072 2m + V \u03c8 \u00b7\u00b7\u00b7 V\u2192 d\u03c4 ddr \u00af\u03c8 \u2202\u03c4 \u2212 EF \u2212 S[\u03c8, \u00af\u03c8] = d\u03c4 ddr \u00af\u03c8 + 1 4\u03c0\u03bd\u03c4 d\u03c4 d\u03c4 ddr ( \u00af\u03c8\u03c8)(\u03c4", " ) ( \u00af\u03c8\u03c8)(\u03c4 ), \u03c8 \u22072 2m (6.57) where the \ufb01elds \u03c8 = {\u03c8a} carry a replica index a = 1,..., R, and V denotes the disorder potential. The last term is generated by the ensemble averaging, where we have assumed that the potential is \u03b4-correlated and expressed its strength through the scattering time \u03c4. (For clarity, wherever possible, we will avoid spelling out the spatial argument r of the \ufb01elds.) Of course, our master strategy will be to decouple the disorder-generated interaction by a Hubbard\u2013Stratonovich transformation. However, before doing so, we need to decide which of the three low-momentum channels of the interaction vertex to keep (see Fig. 6.17). The \ufb01rst contribution (termed the \u201cexchange channel\u201d in our general discussion in Section 6.1) appeared as a principal building block in our diagrammatic discussion of the di\ufb00uson mode above; this contribution should surely be retained. Similarly, the second diagram (the Cooper channel) played an important role in the computation of the maximally crossed diagram and must be retained. The third diagram (the \u201cdirect channel\u201d) describes the scattering o\ufb00 impurities at low momentum transfer. There is no reason why such processes should play a distinguished role, so we will ignore this channel. Following the general discussion of Section 6.1, we should be prepared to introduce two di\ufb00erent Hubbard\u2013Stratonovich \ufb01elds, one for each impurity vertex. There is, however, an ingenious trick whereby the number of required Hubbard\u2013Stratonovich \ufb01elds can be reduced to one. The idea is to exploit the fact that the di\ufb00uson and the cooperon mode are linked to each other by the symmetry operation of time-reversal. (Recall that the cooperon mode described the traversal of an impurity path in a chronologically reversed order.) In the quantum mechanics of spinless particles (for the straightforward extension of the present discussion to particles with spin, we refer to Efetov73) the anti-unitary operation 73 K. B. Efetov, Supersymmetry and the theory of disordered metals, Adv. Phys. 32 (1983), 53\u2013127. 322 Broken symmetry and collective phenomena of time-reversal T maps operators \u02c6X onto their transposes \u02c6X \u2192 T \u02c6X \u2261 \u02c6X T.74 Notice that T \u02c6p = \u2212\u02c6p, i.e. T induces the momentum reversal distinguishing the di\ufb00uson and the cooperon vertex above. This latter observation suggests that one should consider a version of the theory \u201csymmetrized\u201d under time-reversal. Within the symmetrized description, the di\ufb00uson and the cooperon vertex should, in some sense, merge into a uni\ufb01ed object. The practical implementation of this idea goes as follows. S[ \u00af\u03c8, \u03c8] =\u2212 Let us return to the action prior to averaging and write d\u03c4 ddr \u00af\u03c8 \u02c6G\u22121\u03c8 = \u2212 1 2 # \u00af\u03c8 \u02c6G\u22121\u03c8 \u2212 \u03c8T \u02c6G\u22121 |\u2202\u03c4 \u2192\u2212\u2202\u03c4 d\u03c4 ddr d\u03c4 ddr # = \u2212 1 2 \u00af\u03c8 \u02c6G\u22121\u03c8 + ( \u00af\u03c8 \u02c6G\u22121\u03c8)T $ $ \u00af\u03c8T \u2261 \u2212 1 2 d\u03c4 ddr \u00af\u03a8 \u02c6G\u22121 \u02c6\u03a8. Here, \u02c6G\u22121 = \u2212\u2202\u03c4 \u2212 \u02c6H +EF and, in the second equality, we have used the fact that \u00af\u03c8 \u02c6G\u22121\u03c8 = ( \u00af\u03c8 \u02c6G\u22121\u03c8)T (simply because it is a number). In the third equality, we have expressed the transpose in terms of its constituents and used the fact that ( \u02c6G\u22121)T = \u02c6G\u22121 |\u2202\u03c4 \u2192\u2212\u2202\u03c4 (timereversal symmetry!).75 (The overall minus sign comes from the permutation of the Grassmann components of \u03c8.) Finally, we have de\ufb01ned \u03a8(\u03c4 ) \u2261 \u03c8(\u03c4 ) \u00af\u03c8T (\u2212\u03c4 ), \u00af\u03a8(\u03c4 ) \u2261 ( \u00af\u03c8(\u03c4 ), \u2212\u03c8T (\u2212\u03c4 )), to condense the action into a single bilinear form.76 Notice that the enlarged \ufb01elds \u00af\u03a8 and", " \u03a8 are no longer independent of each other; they are connected by the symmetry operation \u00af\u03a8(\u03c4 ) = \u2212\u03a8T (\u2212\u03c4 )(i\u03c3tr 2 ), (6.58) where \u03c3tr represents Pauli matrices acting in the newly introduced \u201ctime-reversal space.\u201d i Involving a transposition operation, Eq. (6.58) signals the fact that the \ufb01elds \u03a8 and \u00af\u03a8 resemble \u201creal\u201d rather than \u201ccomplex\u201d \ufb01elds (the quotes are used because we are dealing with Grassmann \ufb01elds). Being now symmetrized, an average of the \ufb01eld integral over the Gaussian distributed impurity potential gives S[\u03a8] = 1 2 d\u03c4 ddr \u00af\u03a8 \u2202\u03c4 \u2212 EF \u2212 \u22072 2m \u03a8 + 1 16\u03c0\u03bd\u03c4 d\u03c4 d\u03c4 ddr ( \u00af\u03a8\u03a8)(\u03c4 )( \u00af\u03a8\u03a8)(\u03c4 ). Cast in this form, we are now in a position to implement the usual programme to construct the low-energy \ufb01eld theory of the quantum disordered system: Q1: Introduce a Hubbard\u2013Stratonovich transformation to decouple the quartic interaction induced by the impurity average. Note that the corresponding Hubbard\u2013Stratonovich \ufb01eld Q 74 A more precise formulation is that there exists a representation (typically, the coordinate representation) wherein time-reversal amounts to transposition. 75 Note that, under time-reversal, the sign of the time-derivative is reversed: d\u03c4 ( \u00af\u03c8\u2202\u03c4 \u03c8)T = \u2212 d\u03c4 (\u2202\u03c4 \u03c8T ) \u00af\u03c8T = \u2212 d\u03c4 \u03c8T (\u2212\u2202\u03c4 ) \u00af\u03c8T. 76 Notice that the temporal minus sign in the \ufb01elds \u03c8T and \u00af\u03c8T is introduced to remove the unwanted sign multiplying \u2202\u03c4 in [ \u02c6G\u22121]T. Physically, it emphasizes the time-like character of the transformation. 6.5 Field theory of the disordered electron gas 323 must be matrix-valued \u2013 identify the set of indices on which the \ufb01elds depend. Finally, integrate out the fermions to determine the dependence of the e\ufb00ective action on the \ufb01eld Q. Q2: Starting with the e\ufb00ective action for Q, derive the corresponding mean-\ufb01eld equations. (Hint: You will \ufb01nd it convenient to switch to a frequency\u2013momentum representation before varying the action.) Q3: The mean-\ufb01eld equations can be solved by a simple matrix-diagonal ansatz, i.e. Q \u2261 \u039b. Motivate your ansatz and solve the equations. How does the result relate to the SCBA discussed in section 6.5? Q4: Assuming that the frequency contribution to the action \u223c \u00af\u03a8n\u03c9n\u03a8n is negligibly small, identify the global continuous symmetries of the problem, i.e. explore what happens if the fermion \ufb01elds are transformed as \u03a8 \u2192 T \u03a8, \u00af\u03a8 \u2192 \u00af\u03a8 \u00afT, where the matrices T and \u00afT are constant in space and must be chosen so as to respect the relation (6.58). Show that the diagonal solution of the mean-\ufb01eld equations derived in Q3 breaks the full symmetry of the problem. Identify the manifold of Goldstone modes of the problem. A1: To e\ufb00ect a Hubbard\u2013Stratonovich decoupling of the action, we must \ufb01rst isolate the Cooper and exchange contributions to the interaction vertex: Sdis 1 16\u03c0\u03bd\u03c4 d\u03c4 d\u03c4 ddr \u00af\u03a8(1)(\u03c4 )\u03a8(2)(\u03c4 ) \u00af\u03a8(1)(\u03c4 )\u03a8(2)(\u03c4 ) + \u00af\u03a8(1)(\u03c4 )\u03a8(2)(\u03c4 ) \u00af\u03a8(2)(\u03c4 )\u03a8(1)(\u03c4 ), where the convention is that two \ufb01elds carrying the same upper index form a slowly varying bilinear. Now, thanks to the symmetry Eq. (6.58), the two contributions to the action are identical. Indeed, \u00af\u03a8(1)\u03a8(2) = ( \u00af\u03a8(1)\u03a8(2))T = \u2212\u03a8(2)T \u00af\u03a8(1)T = \u03a8(2)T (i\u03c3tr 2 ) (i\u03c3tr 2 ) \u00af\u03a8(1)T = \u00af\u03a8(2)\u03a8(1),", " where in the last equality we have used the symmetry. Thus, we may focus on the decoud\u03c4 d\u03c4 ddr \u00af\u03a8(1)(\u03c4 )\u03a8(2)(\u03c4 ) \u00af\u03a8(2)(\u03c4 )\u03a8(1)(\u03c4 ). To pling of the single term Sdis = (8\u03c0\u03bd\u03c4 )\u22121 make further progress with this expression, let us introduce a composite index \u03b1 \u2261 (a, \u03c3) comprising the replica index a and an index\u03c3 = 1, 2 labeling components in time-reversal space. Using this notation, Sdis = 1 8\u03c0\u03bd\u03c4 = \u2212 1 8\u03c0\u03bd\u03c4 d\u03c4 d\u03c4 ddr \u00af\u03a8(1)\u03b1(\u03c4 )\u03a8(2)\u03b1(\u03c4 ) \u00af\u03a8(2)\u03b2(\u03c4 )\u03a8(1)\u03b2(\u03c4 ) d\u03c4 d\u03c4 ddr \u03a8(1)\u03b2(\u03c4 ) \u00af\u03a8(1)\u03b1(\u03c4 )\u03a8(2)\u03b1(\u03c4 ) \u00af\u03a8(2)\u03b2(\u03c4 ). To decouple the interaction, let us introduce an in\ufb01nite-dimensional, hermitian, matrix \ufb01eld Q = {Q\u03b1\u03b2(r; \u03c4, \u03c4 )}, slowly \ufb02uctuating in space and of the same structure as the dyadic product \u03a8 \u00af\u03a8 \u2261 {\u03a8\u03b1(\u03c4 ) \u00af\u03a8\u03b2(\u03c4 )}, i.e. Q = (i\u03c3tr 2 )\u22121. One may then multiply the ddr tr Q2] and perform a shift Q \u2192 Q + i\u03a8(1) \u00af\u03a8(1)/\u03c0\u03bd. DQ exp[\u2212(\u03c0\u03bd/8\u03c4 ) action by77 2 )QT (i\u03c3tr 77 Here, trQ2 \u2261 integral. d\u03c4 d\u03c4 Q\u03b1\u03b2 (\u03c4, \u03c4 )Q\u2217\u03b2\u03b1(\u03c4, \u03c4 ) and Hermiticity of Q is assumed to ensure the existence of the 324 Broken symmetry and collective phenomena This generates the identity e\u2212Sdis[ \u00af\u03a8,\u03a8] = DQ exp \u2212 \u03c0\u03bd 8\u03c4 ddr tr Q2 \u2212 i 4\u03c4 ddr \u00af\u03a8Q\u03a8, d\u03c4 d\u03c4 \u00af\u03a8\u03b1(\u03c4 )\u00d7 where we have made use of the fact that tr (Q\u03a8 \u00af\u03a8) = \u2212 \u00af\u03a8Q\u03a8 =\u2212 Q\u03b1\u03b2(\u03c4, \u03c4 )\u03a8\u03b2(\u03c4 ) and the superscripts (1) have been dropped. Combining terms, we arrive at the preliminary result78 Z = D\u03a8 DQ exp \u2212 \u03c0\u03bd 8\u03c4 ddr tr Q2 \u2212 1 2 = DQ exp ddr trQ2 + tr ln \u02c6G\u22121[Q]. \u2212 \u03c0\u03bd 8\u03c4 1 2 d\u03c4 ddr \u00af\u03a8 \u2212 \u02c6G\u22121[Q] \u2202\u03c4 \u2212 EF \u2212 \u22072 2m \u2212 i 2\u03c4! Q \u03a8 (6.59) A2: Fourier transforming the \ufb01eld Q(\u03c4, \u03c4 ) in its time arguments one obtains a matrix Q = {Qnn }, wheren, n index fermionic Matsubara frequencies. The frequency/momentum version of the action then assumes the form S[Q] = \u03c0\u03bdLd 8\u03c4 q tr Q(q)Q(\u2212q) \u2212 1 2 tr ln \u02c6G\u22121[Q], where \u02c6G\u22121[Q] \u2261 (i\u02c6\u03c9 \u2212 \u03bep + iQ/2\u03c4 )\u22121, \u02c6\u03c9 is the diagonal operator of Matsubara frequencies and all indices (including momentum indices) not written out explicitly are traced over. Varying the action with respect to the matrix Q, i.e. Q \u2192 Q + \u03b4Q, and requiring vanishing of contributions linear in \u03b4Q, one obtains the equation tr \u03c0\u03bdLdQ(q) \u2212 i \u02c6G[Q]p,p+q \u03b4Q(\u2212q) = 0. q p Holding for any matrix \u03b4Q, this equation is equivalent to the matrix mean-\ufb01eld equation \u03c0\u03bdQ(q) \u2212 i Ld \u02c6G[Q]p,p+q = 0. p A3: To seek solutions of the mean-\ufb01eld equation, let us apply the simplest ansatz, Q \u2261 \u039b, where the \u039b is diagonal in time-reversal and replica spaces, spatially uniform, and diagonal in Matsubara space. When this is substituted into the mean-\ufb01eld equation, the latter assumes the form \u03c0\u03bd\u039b = i Ld p 1 i\u03c9n \u2212 \u03bep", " + i 2\u03c4 \u039b, (6.60) reminiscent of the SCBA Eq. (6.52) for the average Green function (the identi\ufb01cation i\u039b/2\u03c4 = \u2212\u03a3 understood). Drawing on the earlier discussion, one may identify the solution \u039bn = \u22122i\u03c4 \u03a3n = sgn(\u03c9n) (where \u03a3n = \u2212i sgn(\u03c9n)/2\u03c4 ). 78 Notice the prefactor 1/2 generated by the Grassmann integration, i.e. \u00af\u03a8 \u02c6G\u22121\u03a8] \u221d det \u02c6G\u22121/2 = 2 tr ln \u02c6G\u22121] arises because the components of \u00af\u03a8 and \u03a8 are not independent. (To see this, assume that \u02c6G exp[ 1 is diagonalized and count the number of independent Grassmann integration variables.) D\u03a8 exp[\u2212 1 2 6.5 Field theory of the disordered electron gas 325 A4: In the mean-\ufb01eld approximation, \u02c6G becomes equivalent to the impurity-averaged Green function discussed in Section 6.5. But, clearly, this can not be the end of the story! We have seen in Section 6.5 that impurity scattering leads to di\ufb00usive phenomena that reach far beyond the damping of single-particle amplitudes. How are such structures re\ufb02ected in the \ufb01eld theory? To identify the di\ufb00usion modes, one must return to consider the symmetries of the theory (having in mind that symmetries tend to generate long-range low-energy excid \u00af\u03a8 \u02c6G\u22121[V ]\u03a8 and explore what happens tations). We thus go back to our initial action \u2212 as we transform \u03a8 \u2192 \u03a8 \u2261 T \u03a8, and \u00af\u03a8 \u2192 \u00af\u03a8 \u2261 \u00af\u03a8 \u00afT, whereT and \u00afT are matrices, constant in real space. Of course we require that the fundamental relation (6.58) be preserved under this transformation. This leads to the condition \u00afT = (i\u03c3tr 2 )\u22121. Moreover,T should be a symmetry of the problem, i.e. we require that \u00afT \u02c6G\u22121[V ]T \u02c6G\u22121[V ] \u00afT T = G\u22121[V ] or \u00afT = T \u22121. Here, we have used the fact that, owing to its constancy, T commutes with the (disordered) Hamiltonian. Nevertheless, \u00afT G\u22121[V ] \u02c6G\u22121[V ] \u00afT holds only approximately since, for a transformation Tnn of a general frequency structure, [T, \u02c6\u03c9] = 0. However, for transformations of physical interest (\ufb02uctuating slowly in time), this lack of commutativity is not 2 )\u22121, of much concern. Combining the two conditions above, we \ufb01nd that T = (i\u03c3tr which is the de\ufb01ning relation of the unitary symplectic group Sp(2 \u00b7 R \u00b7 (2M )). Here, we have assumed that [\u2212\u03c9max, \u03c9max] = [\u22122\u03c0M T, 2\u03c0M T ] is the range of 2M Matsubara frequencies over which the lack of commutativity of the symmetry group and the frequency operator can safely be neglected. 2 )T T (i\u03c3tr 2 )T T (i\u03c3tr With this background, we may note that the diagonal solution of the mean-\ufb01eld equations breaks the symmetry above. Indeed, the fact that T is a symmetry implies that, not only \u039b, but any con\ufb01guration Q \u2261 T \u039bT \u22121, solves the mean-\ufb01eld equation (6.60). The symmetry is broken, inasmuch as only a subgroup of transformations T0 \u2208 K \u2282 Sp(4RM ), T0\u039bT \u22121 0 = \u039b, leaves the diagonal saddle-point invariant. (It may be helpful to think of \u039b as some kind of spin, Sp(4RM ), the analog of the rotation group, and K the group of rotations around the spin axis.) The invariance group K is easily identi\ufb01ed. Projected \u221d \u00b11 is equal onto the positive/negative frequency range, the diagonal solution \u039b to the unit operator, i.e. transformations not mixing positive and negative frequencies will leave the saddle-point invariant. This means that the invariance subgroup is given by Sp(2RM ) \u00d7 Sp(2RM ) \u2282 Sp(4RM ) where the two factors are the invariance groups of the positive/negative frequency range.", " Following our general line of argument on page 257 we may conclude that the Goldstone mode manifold of the present problem is the coset space Sp(4RM )/(Sp(2RM ) \u00d7 Sp(2RM )). The elements of this space are conveniently parameterized as Q = T \u039bT \u22121, whereT = exp[W ] and the generators W anti-commute with \u039b, [W, \u039b]+ = 0.79 This condition is met by setting \u03c9n>/<0 W = \u2212B\u2020 B, (6.61) 79 Generators commuting with \u039b need not be taken into account inasmuch as they do not a\ufb00ect the diagonal saddle-point (or span the group K). 326 Broken symmetry and collective phenomena 2 ) eW (i\u03c3tr 2 )\u22121 = e\u2212W implies a second condition B = (i\u03c3tr where the block structure is in Matsubara frequency space (positive/negative frequencies) and B are 2M R-dimensional matrix \ufb01elds. The anti-Hermitian connection between the two blocks follows from the condition Q\u2020 = Q \u21d4 W \u2020 = \u2212W. Notice that 2 )\u22121. As one last (i\u03c3tr technical remark \u2013 of crucial importance for all that follows \u2013 let us notice that the matrices Q obey the nonlinear constraint Q2 = 1. (This follows from the fact that Q2 = (T \u039bT \u22121)2 n = sgn2(\u03c9n) = 1.) Field theories on manifolds with nonlinear constraints are and \u039b2 generally called nonlinear \u03c3-models.80 In the next section, we proceed to construct the low-energy action of the nonlinear \u03c3-model of the disordered electron gas. 2 )BT (i\u03c3tr Low-energy \ufb01eld theory In the previous section we constructed the backbone of a \ufb01eld-theoretical approach to the disordered electron gas, ran into a scenario of spontaneous symmetry breaking, and identi\ufb01ed the corresponding Goldstone mode manifold as Sp(4RM )/(Sp(2RM ) \u00d7 Sp(2RM )). In the following, we now proceed to construct the low-energy action of the Goldstone mode degrees of freedom. Our second objective is to understand the connections to the diagrammatic approach introduced above. Finally, we wish to develop an e\ufb03cient theoretical formalism, powerful enough to cope with the nonlinear proliferation of di\ufb00uson- and cooperon-type soft modes that govern the physics of disordered conductors at low temperatures. Our target is a low-energy action of the elements Q = T \u039bT \u22121 \u2208 Sp(4RM )/(Sp(2RM ) \u00d7 Sp(2RM )) of the Goldstone mode manifold. The action of this nonlinear \u03c3-model can be constructed by substituting the slowly \ufb02uctuating matrix con\ufb01guration Q = {Q\u03b1,\u03b1 nn (r)} into the action (6.59) and expanding in (a) gradients and (b) the small symmetry breaking induced by the lack of commutativity of the \u201crotation matrices\u201d T and the frequency operator \u02c6\u03c9 \u2013 a program which is instructive to follow.81 However, in the present section we shall pursue a more symmetry-oriented (but only slightly less rigorous) strategy of constructing the action. Q1: Using only symmetry arguments, identify the minimal (lowest number of derivatives) action S\ufb02[Q] measuring the cost of spatial \ufb02uctuations of the Q \ufb01eld. What is the structure of the action S\u03c9[Q] accounting for the non-commutativity of Q and the frequency operator? Q2: Our analysis so far has left the values of the two coupling constants of the theory undetermined. As we do not wish to perform an explicit gradient expansion, the only way to determine their values is to compare the predictions of the \ufb01eld-theoretical approach with those of another formalism. Indeed, we have seen in Section 6.5 that the long-range behavior of the density\u2013density correlation function is governed by speci\ufb01c types of soft modes: diffusons and cooperons. We also saw that, at low frequency scales, these modes were prone to 80 Other representatives of this family of \ufb01eld theories will be met in Chapters 8 and 9. 81 A yet more systematic approach would be to decompose the full \ufb01eld manifold Q = Qgs + Qm into a Goldstone mode con\ufb01guration Qgs = T \u039bT \u22121 and a massive contribution Qm and to integrate over the latter in the second order of perturbation theory. However, it can be shown that,", " in the present case, the contribution of the massive modes is inessential, i.e. we may work with the pure Goldstone theory from the outset. 6.5 Field theory of the disordered electron gas 327 \u201cinteract\u201d with each other. (A cooperon insertion into the di\ufb00uson mode can be interpreted as some kind of \u201cself-energy\u201d of the di\ufb00uson propagator.) Comparing this picture with the structure of the \ufb01eld theory, it is evident that the elementary di\ufb00usons/cooperons must be the Goldstone modes of the \u03c3-model. \u201cInteractions\u201d between di\ufb00usons and cooperons appear when we expand the theory to higher than quadratic order in the generators B. Expand the action (6.62) to second order in the generators B de\ufb01ned in Eq. (6.61). Comparing the structure of the quadratic action with that of the di\ufb00uson mode, \ufb01x the ratio of the coupling constantsc \ufb02 and c\u03c9. Q3: Having determined the ratio of the two coupling constants, one may now complete the analysis by \ufb01xing the absolute value of c\u03c9. We do so by introducing a source term S[\u03a8] \u2192 S[\u03a8] \u2212 1 2Ld \u00af\u03a8p,n\u03ban\u03a8p,n such that di\ufb00erentiation ddr \u00af\u03a8n\u03a8n = \u2212 1 Gp,n = \u2212\u03c0\u03bd sgn(\u03c9n), Z R = Im Im p,n lim R\u21920 1 R \u03b4 \u03b4\u03ban \u03ba=0 1 2Ld Ld Im p gives the density of states. Use the source \u02c6\u03ba = {\u03ban\u03b4nn } to determine the coupling constant c\u03c9. A1: As long as we ignore the symmetry breaking due to the frequency operator, the action is symmetric under transformations Q \u2192 \u02dcT Q \u02dcT \u22121, where \u02dcT \u2208 Sp(4RM ) is an element of ddr tr (\u2207Q)2 as the symmetry group. This requirement uniquely identi\ufb01es S\ufb02[Q] =c \ufb02 the dominant (lowest number of derivatives) contribution to the \ufb02uctuation action. The dominant operator describing the explicit breaking of the global symmetry will surely be of \ufb01rst order in \u02c6\u03c9: S\u03c9[Q] \u223c ddr tr(\u02c6\u03c9F [Q]), where F is some function. Since Q2 = 1, only ddr tr(\u02c6\u03c9Q). terms linear in Q contribute to the expansion of F and we may set S\u03c9[Q] = c\u03c9 Summarizing, one may deduce that the soft mode action acquires the form S[Q] = ddr tr % c\ufb02(\u2207Q)2 + c\u03c9 \u02c6\u03c9Q &. (6.62) As with other nonlinear \u03c3-models, the innocent appearance of the action above is somewhat deceptive: we need to remember that the \ufb01eld Q = eW \u039be\u2212W is a nonlinear object. An expansion of it in terms of the generators W will generate \u201cinteraction operators\u201d \u223c W n of arbitrary (even) order in n. The analysis of these interactions is a highly non-trivial task! A2: Substituting Q = eW \u039be\u2212W [W,\u039b]+=0 one obtains = S(2)[W ] = ddr tr e2W \u039b = (1 + 2W + 2W 2 + \u00b7 \u00b7 \u00b7) \u039b into the action, % \u22124c\ufb02(\u2207W )2 + 2c\u03c9 \u039b\u02c6\u03c9W 2 &. One may now express the matrices W in terms of the o\ufb00-diagonal blocks B and arrive at S(2)[B, B\u2020] = = % ddr tr 8c\ufb02\u2202iB\u2202iB\u2020 \u2212 2c\u03c9 \u02c6\u03c9(P +BB\u2020 \u2212 P \u2212B\u2020B) B\u03b1\u03b1 q,nn 8c\ufb02q2 \u2212 2c\u03c9|\u03c9n \u2212 \u03c9n | B \u2020\u03b1\u03b1 \u2212q,nn. & q,n>0,n<0 328 Broken symmetry and collective phenomena Here, P +/\u2212 is a projector onto the space of positive/negative Matsubara frequencies. In the second line, \u03b1 = (a, \u03c3) is the composite replica/time-reversal index introduced above. Importantly, the propagator of the B-\ufb01eld describes a mode of di", "\ufb00usive dynamics formed by the joint propagation of two states with (a) the Matsubara frequencies \u03c9n > 0 and\u03c9 n < 0 and (b) the momentum di\ufb00erence q. The analogy to our previous diagrammatic analysis of such processes is made complete by choosing 8c\ufb02 = cD, \u22122c\u03c9 = c, or S(2)[B, B\u2020] = c B\u03b1\u03b1 q,nn Dq2 + |\u03c9n \u2212 \u03c9n | B \u2020\u03b1\u03b1 \u2212q,nn, q,n>0,n<0 where c is an, as yet, undetermined global coupling constant. Before continuing, it is worthwhile taking a brief look at the time-reversal structure of the theory. Recalling that Q\u03c3\u03c3 \u223c \u03a8\u03c3 \u00af\u03a8\u03c3, one may conclude that Goldstone modes B with an index structure (\u03c3\u03c3) describe the joint propagation of a fermion propagating forwards in time (\u03c3 = 1) (or time-reversed [\u03c3 = 2]) with a hole propagator moving forwards in time (\u03c3 = 1) (or time reversed [\u03c3 = 2]). Put di\ufb00erently, modes B\u03c3\u03c3 o\ufb00-diagonal in time-reversal space (\u03c3 = \u03c3) represent cooperon modes, while modes B\u03c3\u03c3 re\ufb02ect the di\ufb00uson modes. In the absence of magnetic \ufb01elds, the actions of these two modes are identical (rotational invariance of the action in time-reversal space). However, had we included a magnetic \ufb01eld, this invariance would be lost. A3: Noting that the addition of the source to the action amounts to a shift \u02c6\u03c9 \u2192 \u02c6\u03c9 \u2212 i\u02c6\u03ba/Ld, one may conclude that the low-energy action changes to S\u03c9[Q] \u2192 c\u03c9 ddr tr [(\u02c6\u03c9 \u2212i\u02c6\u03ba/Ld)Q]. Di\ufb00erentiation of the Q-functional thus gives \u2212\u03c0\u03bd sgn(\u03c9n)!= lim R\u21920 1 R Im \u03b4 \u03b4\u03ban Z R = lim R\u21920 1 R Im i c\u03c9 Ld \u03ba=0 ddrtr Qnn = 2c\u03c9 sgn(\u03c9n) +\u00b7 \u00b7 \u00b7, where the ellipsis represents contributions arising from terms of higher order in the generators B. (In fact, one can show that all of these terms vanish in the replica limit.) Thus, c\u03c9 = \u2212\u03c0\u03bd/2 and one obtains the low-energy action of the nonlinear \u03c3-model, % & S[Q] = \u03c0\u03bd 2 ddr tr D 4 (\u2207Q)2 \u2212 \u02c6\u03c9Q. (6.63) In applications, one is often interested in situations involving an external electromagnetic \ufb01eld. How would such a \ufb01eld a\ufb00ect the structure of the action (6.63)? Although this question can be answered by microscopic calculation, a more time-e\ufb03cient strategy is to require that the action S[Q] be gauge invariant. Under a gauge transformation, \u03c8 \u2192 ei\u03c6\u03c8, the \ufb01elds \u03a8 enlarged by a time-reversal structure transform as (exercise) \u03a8 \u2192 ei\u03a6\u03a8, where \u03a6 = diag(\u03c6(\u03c4 ), \u2212\u03c6(\u2212\u03c4 )) (matrix structure in time-reversal space). This in turn implies that the \ufb01elds Q \u223c \u03a8 \u00af\u03a8 transform as Q \u2192 ei\u03a6Qe\u2212i\u03a6. Invariance under such transformations is ensured (exercise) if we generalize the action according to S[Q] = ddr tr ((\u2207 \u2212 i[A, ])Q)2 \u2212 (\u02c6\u03c9 + V )Q, (6.64) \u03c0\u03bd 2 D 4 where the vector potential A and scalar potential V transform in the standard manner A \u2192 A + \u2207\u03a6 and V \u2192 V + \u2202\u03c4 \u03a6. This way of coupling a gauge potential to an e\ufb00ective 6.6 Summary and outlook 329 action \u2013 to be discussed in more general terms in the next chapter \u2013 is known as minimal substitution. 6.6 Summary and outlook This concludes our preliminary discussion of functional mean-\ufb01eld methods. We have learned how the integration over microscopic quantum \ufb01elds can be traded for an integration over degrees of freedom adjusted to the low-energy characteristics of the system. We found that the functional dependence of the action on the new coordinates is usually highly nonlinear \u2013 the notorious \u201ctrace logarithms\u201d \u2013 and", " has to be dealt with by some kind of stationary phase analysis. While a \ufb01rst principles solution of the mean-\ufb01eld equations is often not possible, all applications discussed in this chapter shared the feature that solutions could be found by an \u201ceducated guess.\u201d Importantly, we realized that these solutions did not always display the full symmetry of the action. INFO However, it is important to realize that our discussion has excluded a number of problem classes with complex mean-\ufb01eld structure. For example, for certain lattice systems with discrete translational symmetry, the optimal mean-\ufb01eld solution shows staggering, i.e. changes sign under translation by one lattice spacing. Deceptively, these con\ufb01gurations usually exist in parallel to a solution that displays the full symmetry of the problem, but is of higher energy (and therefore unphysical), i.e. the \u201ctrue solution\u201d is camou\ufb02aged by the existence of an inferior but \u201cmore symmetric\u201d solution.82 Then, there are systems where the existence of meaningful stationary phase con\ufb01gurations is excluded by structural elements, such as frustration. Other systems, notably glasses (cf. the remarks on page 115), do possess an entire continuum of mean-\ufb01eld con\ufb01gurations, i.e. there is a macroscopic number of (metastable) extremal con\ufb01gurations that are very close in energy space but may be very far from each other in con\ufb01guration space. (For example, two such con\ufb01gurations may di\ufb00er by a restructuring of the atomic con\ufb01guration of a glass at remote places.) The dynamics of such systems is governed by the process of aging, i.e. transitions between di\ufb00erent metastable extremal con\ufb01gurations on very long timescales (witness the apparent rigidity of window glass!). It goes without saying that the \u201cmean-\ufb01eld con\ufb01gurations\u201d of glassy systems are irregular and not amenable to straightforward analytic calculation. Finally, even systems possessing a spatially homogeneous mean-\ufb01eld con\ufb01guration may host other solutions of physical signi\ufb01cance. Indeed, we have encountered a scenario of this type already in Chapter 3 above. Exploring quantum double-well tunneling, we observed that the imaginary time Euler\u2013Lagrange equations of a particle in a doubly degenerate harmonic well (\u2194 the \u201cmean-\ufb01eld equations\u201d) had a metastable constant solution (\u2194 the constant mean-\ufb01eld). However, in addition to that, we found instantons, i.e. temporally structured solutions of the mean-\ufb01eld equations. Super\ufb01cially, it seemed that, due to their non-vanishing action, instantons are irrelevant. However, for su\ufb03ciently long times, the energetic inferiority of the single instanton was over-compensated by statistical aspects, i.e. the fact that a continuum of instantons entered the partition function with approximately equal weight. In the language of statistical mechanics: 82 When dealing with problems whose lattice structure carries physical signi\ufb01cance, it is therefore a good idea to try a number of di\ufb00erent ans\u00a8atze, i.e. trial solutions that transform di\ufb00erently under translation by a unit cell. 330 Broken symmetry and collective phenomena the large entropy associated with instanton con\ufb01gurations outweighed the \ufb01nite energy (or, more precisely, action) to be paid for a single instanton. Later we will see that entropy/energy competitions of this type are realized in many other contexts of quantum and statistical \ufb01eld theory. For example, the phase actions of the BCS superconductor or the super\ufb02uid possess (approximate) extremal con\ufb01gurations wherein the phase winds around an integer multiple of 2\u03c0 as a \ufb01xed reference point in space is encircled (vortices). While the energy of an individual vortex solution may be high, the entropy corresponding to the choice of its center eventually, for su\ufb03ciently large temperatures, dominates and leads to a proliferation of vortices in the system (\u2194 the analog of the instanton gas). For a more quali\ufb01ed discussion of topologically non-trivial mean-\ufb01eld con\ufb01gurations we refer to Chapter 9. Finally, we have explored a number of di\ufb00erent applications showing the way low-energy e\ufb00ective actions describing \ufb02uctuations on", " top of a reference mean-\ufb01eld can be derived and evaluated. In cases where the mean-\ufb01eld breaks a continuous symmetry, these actions were partially soft, i.e. contained Goldstone modes. We saw that the presence of a Goldstone mode has a dominant e\ufb00ect on the observable physical properties of a system. As exempli\ufb01ed above, the theoretical machinery developed thus far enables us to tackle already highly non-trivial problems. However, to make the functional integral a universally applicable tool, two important gaps have yet to be \ufb01lled. Firstly, we have so far mostly been content with discussing general structures of condensed matter theory (i.e. much of the discussion has been limited to general manipulations on the level of the e\ufb00ective action.) However, in applications, one is typically interested in comparison with experimentally accessible data, i.e. we still need to provide links relating the \ufb01eld integral formalism to observables of experimental signi\ufb01cance. Secondly, our discussion of the \ufb02uctuation behavior of \ufb01eld theories has been quite limited. That is, we have typically expanded an action to second order in what we believed was the \u201cphysically relevant\u201d degree of freedom and computed the corresponding determinants. However, we also saw that some problems contain anharmonicities of profound signi\ufb01cance. For example, the quartic term in \u03c64-theory led to UV singularities whose nature remained largely obscure. We must, therefore, (i) develop criteria indicating in what circumstances a model can safely be limited to second order and, (ii) in cases where it cannot, learn how to deal with its anharmonic content. Beginning with the \ufb01rst, these two topics (which are not quite as unrelated as one might think) are discussed in Chapters 7 and 8. 6.7 Problems 6.7 Problems Peierls instability 331 Previously, in Problem 2.4 we have seen that the half-\ufb01lled one-dimensional lattice system is unstable towards the formation of a commensurate periodic lattice distortion. In fact, at zero temperature, a one-dimensional crystal comprising electrons and ions is always unstable against the formation of a lattice distortion which transforms the system to a dielectric. The aim of this problem is to study the nature of the lattice instability known as a Peierls distortion. Rudolph E. Peierls 1907\u201395 Born in Germany, Peierls did fundamental work during the early years of quantum mechanics. He also studied the physics of lattice vibrations (phonons) and is credited with the development of the concept of the \u201chole\u201d in condensed matter. In collaboration with Otto Frisch, Peierls was the \ufb01rst to realize that an atom bomb based on 235U would be feasible, and was engaged in the Manhattan Project. (Image c University of Cambridge, Cavendish Laboratory.) The starting point of the theory is the Euclidean action for a one-dimensional gas of electrons, Sel[\u03c8, \u00af\u03c8] = d\u03c4 dx \u00af\u03c8 \u2202\u03c4 \u2212 1 2m \u22022 x \u2212 \u03bc \u03c8. (Here, for simplicity, we take the electrons to be spinless.) Starting with a classical \ufb01eld theory of the lattice dynamics, the Euclidean action of the system of ions is given by Sph[u] = \u03c1 2 d\u03c4 dx 0 (\u2202\u03c4 u)2 + c2(\u2202xu)2 1, where u(\u03c4, x) denotes the scalar bosonic displacement \ufb01eld. Finally, to complete the theory, we suppose that the electron system is coupled to the distortion of the underlying lattice. In principle, one could try to derive a \u201crealistic\u201d model of the coupling. However, since we are interested in the general stability of the system, we will consider a phenomenological coupling which engages the \ufb01elds in the lowest order compatible with the symmetry of the interaction (for more details, see Problem 4.5). To this end, we introduce \u201cFr\u00a8ohlich\u2019s deformation potential approximation,\u201d wherein the coupling is taken to be proportional to the gradient of the deformation \u2202xu, Sel\u2212ph[\u03c8, \u00af\u03c8, u] = g d\u03c4 dx \u00af\u03c8\u03c8 \u2202xu. The full Euclidean action S = Sel + Sph + Sel\u2212ph represents an interacting theory. As such, an explicit and exact evaluation of the coherent state path integral is not feasible. Instead, we will develop an analysis in which the coupling will be treated perturbatively. In doing so, we will be able to explore the stability of the lattice system. (a) As a \ufffd", "\ufffd\ufffdrst step, integrate out the fermionic degrees of freedom \u03c8 and thereby obtain an e\ufb00ective action for the displacement \ufb01eld u. Assuming that the electron\u2013phonon 332 Broken symmetry and collective phenomena coupling constant g is small, expand the action up to second order in u. You will \ufb01nd that the coe\ufb03cient of u2 involves the density\u2013density response function \u03c7(q, \u03c9m). (b) In general, when seeking a saddle-point of the e\ufb00ective action, the \ufb01rst guess would be a homogeneous displacement \ufb01eld, u0(x, \u03c4 ) \u2261 u0. Here, u0 = 0 is such a solution. However, it is not necessarily the best solution. Show that the static solution u0(x, \u03c4 ) \u2261 u0 cos(2kFx+\u03d5) is energetically favorable (i.e. S[u = u0 cos(2kFx+\u03d5)]< S[u = 0]) below a certain critical temperature Tc. Thus, at low temperatures, the system is instable towards the formation of a static sinusoidal lattice distortion. Use the approximation \u03c7(2kF, 0) ln(\u03b2\u03c9D)/4\u03c0vF, where \u03c9D is the Debye frequency, to determine the transition temperature Tc. Answer: (a) Integrating out the electron degrees of freedom \u03c8, one obtains D[ \u00af\u03c8, \u03c8]e\u2212Sel\u2212Sel\u2212ph = det \u2202\u03c4 \u2212 \u22022 x 2m \u2212 \u03bc + g(\u2202xu). % & \u2202\u03c4 \u2212 (\u22022 x/2m) \u2212 \u03bc + g(\u2202xu). While the bare phonon action is diagonal Thus, the e\ufb00ective action for the displacement \ufb01eld u reads Se\ufb00 [u] = Sph[u] \u2212 in tr ln frequency\u2013momentum representation, the \u201ctr ln\u201d is not. Thus, the evaluation of the second term requires more care. De\ufb01ning the bare Green function of the electron system G\u22121 x/2m \u2212 \u03bc, an expansion up to second order in the coupling 0 = \u2202\u03c4 \u2212 \u22022 g gives tr ln[G\u22121 0 + g(\u2202xu)] tr ln G\u22121 \u03c9m,q q2\u03c7(q, \u03c9m)|u\u03c9m,q|2, where \u03c7(q, \u03c9m) = \u2212 T G0,p,\u03c9n G0,p+q,\u03c9n+\u03c9m and we note that the linear term of the L expansion vanishes by symmetry. Thus, to quadratic order in the displacement \ufb01eld, the e\ufb00ective action assumes the form \u2212 (g2/2) p\u03c9n 0 Se\ufb00 [u] q;\u03c9m \u03c1 2 (\u03c92 m + c2q2) \u2212 g2 2 q2\u03c7(q, \u03c9m) |u\u03c9m,q|2. (b) In the presence of the static periodic modulation, u\u03c9m,q = \u03c0u0[ei\u03d5\u03b4(q \u22122kF)+e \u2212i\u03d5\u03b4(q + F[\u03c1c2 \u2212 g2\u03c7(2kF, 0)]. Thus, not2kF)]\u03b4(\u03c9m) the action takes the value Se\ufb00 = 4\u03c02u2 ing that S[u = const.] = 0, the transition is realized when \u03c1c2 < 4\u03c02g2 \u03c7(2kF, 0). Substituting the expression for \u03c7, one obtains the relation for the critical temperature Tc: \u03c1c2 = g2 ln(\u03c9D/Tc) i.e. at temperatures T < Tc = \u03c9D e\u22124\u03c02vF\u03c1c2/g2, the system 4\u03c0vF becomes unstable against the formation of a periodic modulation of the lattice with a wavelength of \u03c0/kF. 0k2 Temperature pro\ufb01le of the BCS gap The following question concerns an exploration of the functional dependence of the BCS gap on temperature. In particular, we wish to understand its singular vanishing at the transition temperature (this being a hallmark of a second-order phase transition). 6.7 Problems 333 Consider the BCS gap equation (6.28), 1 g\u03bd = 0 \u03c9D/2T dx tanh(x2 + \u03ba2)1/2 (x2 + \u03ba2)1/2, where we have introduced a dimensionless integration", " variable and \u03ba \u2261 \u0394/2T. In spite of its innocent appearance, the temperature dependence of this equation is not straightforward to infer. Referring for a quantitative discussion to Abrikosov et al.,83 we here restrict ourselves to exploring the gap pro\ufb01le in the vicinity of the transition temperature. (a) To determine the value of Tc, we proceed somewhat indirectly assuming that this is determined through the condition \u0394(Tc) = 0. Use this criterion to obtain Eq. (6.29) for the critical temperature. (You may assume the hierarchy of energy scales \u03c9D Tc \u223c \u03940 \u0394(T Tc), where \u03940 = \u0394(T = 0).) (b) Now, let us derive the approximate pro\ufb01le Eq. (6.30) of the gap for temperatures T slightly smaller than Tc. To this end, add to and subtract from the right-hand side of. Then expand to leading order in the small the gap equation the integral parameters \u03b4T /Tc and \u0394/Tc, where\u03b4T = Tc \u2212 T > 0. dx tanh x x Answer: (a) For \u0394 = 0, \u03bb(\u03be) = (\u03be2 + \u03942)1/2 = |\u03be| and Eq. (6.28) assumes the form 1 g\u03bd = 0 \u03c9D/2Tc dx tanh x x, where we have introduced x \u2261 \u03be/2Tc as a dimensionless integration variable. The dominant contribution to the integral comes from the region x 1, where tanh x 1. As a result, one obtains 1/g\u03bd ln(\u03c9D/2Tc). Solving for Tc, we arrive at Eq. (6.29). (b) Adding and subtracting the integral given above, we have 1 g\u03bd \u03c9D/2T = 0 dx tanh(x2 + \u03ba2)1/2 (x2 + \u03ba2)1/2 \u2212 tanh x x 0 \u03c9D/2T dx tanh x x. + Arguing as in (a), the second integral can be estimated as ln(\u03c9D/2T ) \u2248 ln(\u03c9D/2Tc) + (\u03b4T /Tc) = (1/g\u03bd) + (\u03b4T /Tc), where we have expanded to linear order in \u03b4T. Thus, \u2212 \u03b4T Tc \u2248 0 \u03c9D/2T dx tanh(x2 + \u03ba2)1/2 (x2 + \u03ba2)1/2 \u2212 tanh x x. Now, the remaining integral can be split into a \u201clow-energy region\u201d 0 \u2264 x \u2264 1, and a \u201chigh-energy region\u201d 1 < x < \u03c9D/2T. Using the small-x expansion, tanh x x \u2212 x3/3, x>1\u2248 1, the second we \ufb01nd that the \ufb01rst region gives a contribution \u223c \u03ba2. With tanh x region contributes a term O(\u03ba2) which, however, is approximately independent of the 83 A. A. Abrikosov, L. P. Gorkov, and I. E. Dzyaloshinskii, Methods of Quantum Field Theory in Statistical Physics (Dover Publications, 1975). 334 Broken symmetry and collective phenomena large-energy cut-o\ufb00 \u03c9D/2T. Altogether, we obtain \u03b4T /Tc \u2248 const. \u00d7 \u03ba2 \u2248 const. \u00d7 (\u03942 c ) from which one obtains Eq. (6.30). 0/T 2 Fluctuation contribution to the Ginzburg\u2013Landau action of the superconductor In this short problem we derive the energy cost corresponding to large-scale spatial \ufb02uctuations of the order parameter of a BCS superconductor. The problem mainly serves technical training purposes. Consider the second-order contribution to the Ginzburg\u2013Landau action of the BCS superconductor Eq. (6.31). In Problem 4.5 we have seen that the frequency summation involved in the de\ufb01nition of the integral kernel evaluates to \u03c7c(\u03c9n, q) \u2261 \u2212 T Ld m,p Gp(i\u03c9m)G\u2212p+q(\u2212i\u03c9m + i\u03c9n) = 1 Ld p 1 \u2212 nF(\u03bep) \u2212 nF(\u03be\u2212p+q) i\u03c9n \u2212 \u03bep \u2212 \u03be\u2212p+q. Expand \u03c7c(0, q) to second order in q. (Hints: You may trade the momentum summation for an integral, and", " linearize the dispersion: \u03bep+q \u03bep +p\u00b7q/m (think why is this a permissible nF() = cT \u22122 (where the numerical simpli\ufb01cation). You may note the identity constant c = 7\u03b6(3)/2\u03c02 and \u03b6(x) = n=1 nx de\ufb01nes the \u03b6-function).) d\u22121\u22022 \u221e Answer: Using the fact that \u03bep = \u03be\u2212p, \u03c7c(0, q) = \u2212 ddp (2\u03c0)d 1 \u2212 nF(\u03bep+q/2) \u2212 nF(\u03bep\u2212q/2) \u03bep+q/2 + \u03bep\u2212q/2 \u2212 ddp (2\u03c0)d = \u03c7c(0, 0) \u2212 \u03bd\u03bcq2 12m 1 \u2212 2nF(\u03bep) \u2212 \u22022 2\u03bep \u03be nF(\u03bep) (q\u00b7p)2 4m2 d \u22121\u22022 nF() = \u03c7c(0, 0) \u2212 c 24 \u03bdv2 T 2 q2, F where c is a numerical constant and in the second equality we have used the fact that, for \u03be = O(T ), p2/2m \u03bc. Substituting this expansion into the quadratic action, we obtain the gradient term in Eq. (6.33). Coulomb blockade Technological advances have made it possible to manufacture small metallic or semiconducting devices of extension 1 \u03bcm and less. At low temperatures, the physics of these so-called quantum dots is predominantly in\ufb02uenced by charging e\ufb00ects. In this problem, we explore the impact of charging on the most basic characteristic of a quantum dot, the tunneling DoS. Consider a quantum dot weakly84 connected to an outside environment (see Fig. 6.18 for a semiconductor realization of such a setup). The addition of an electron to, or subtrac- 84 By \u201cweak\u201d we mean that the conductance of all external leads attached to the system is such that g < g0, where g0 = e2/h (25.8 k\u03a9)\u22121 is the quantum unit of conductance. 6.7 Problems 335 (a) GaAS AlxGa1\u2013xAs (b) 1 \u03bcm Figure 6.18 (a) Schematic picture of a con\ufb01ned two-dimensional electron gas (a quantum dot) formed at the interface between a GaAs and an AlGaAs layer. (b) Electron microscopic image of the \u201creal\u201d device. (Source: Courtesy of C. M. Marcus.) tion of one from, the device incurs an energy cost of order EC = e2/2C, where C is the capacitance of the system. This energy cost is o\ufb00set by an external gate potential. The discreteness of this charging energy leads to a plethora of observable physical phenomena. The most basic of these is a strong suppression of the DoS at the Fermi surface. For a non-interacting system, the single-particle DoS is de\ufb01ned as \u03c1() = tr(\u03b4( \u2212 \u02c6H)) = \u03c0 Im tr( \u02c6Gn)|i\u03c9n\u2192+i0, where \u02c6G(z) = (z \u2212 \u02c6H)\u22121 is the Green \u03c0 Im tr( \u02c6G( + i0)) = \u2212 1 \u2212 1 function. The tunneling DoS generalizes this de\ufb01nition to the interacting case: \u03bd() = \u03c0 Im tr( \u02c6Gn)|i\u03c9n\u2192+i0, where the Green function \u02c6Gn is de\ufb01ned85 as the Fourier transform \u2212 1 of the coherent state path integral, G\u03b1\u03b2(\u03c4 ) = Z \u22121 D(\u03c8, \u00af\u03c8) e\u2212S[ \u00af\u03c8,\u03c8] \u00af\u03c8\u03b2(\u03c4 )\u03c8\u03b1(0), (6.65) D(\u03c8, \u00af\u03c8) e\u2212S[ \u00af\u03c8,\u03c8], and the indices \u03b1, \u03b2 enumerate the eigenstates of the singlewhere Z = particle contribution to the Hamiltonian. Having an irregular structure, the eigenstates of the single-particle Hamiltonian are unknown. The simplest prototype Hamiltonian describing the joint e\ufb00ects of single particle dynamics and charging reads as \u02c6H = \u02c6H0+EC( \u02c6N \u2212N0)2, where \u02c6H0 = \u03b1a\u03b1 is the number operator and N0 represents the preferred", " number of particles (as set by the gate voltage). The action controlling the behavior of the Green function (6.65) is thus given by \u03b1a\u03b1, \u02c6N = \u03b1 \u03b1a\u2020 \u03b1 a\u2020 S[ \u00af\u03c8\u03b1, \u03c8\u03b1] = \u03b2 0 \u23a7 \u23a8 d\u03c4 \u23a9 \u03b1 \u00af\u03c8\u03b1(\u2202\u03c4 + \u03b1 \u2212 \u03bc)\u03c8\u03b1 + EC 72 \u00af\u03c8\u03b1\u03c8\u03b1 \u2212 N0 6 \u03b1 \u23ab \u23ac \u23ad. (6.66) (a) Introducing a bosonic \ufb01eld variable V (\u03c4 ), decouple the interaction by means of a Hubbard\u2013Stratonovich transformation. Bring the functional representation of the Green 85 The terminology of \u201ctunneling\u201d DoS is motivated by the fact that \u03bd() is an important building block in the calculation of tunneling currents. (Recall the \u201cGolden Rule\u201d: tunneling rates are obtained by multiplication of transition probabilities with state densities.) 336 Broken symmetry and collective phenomena DV e\u2212S[V ]Z [V ]G[V ] (b) Represent the \ufb01eld V as a sum over Matsubara components, V (\u03c4 ) = function into the form G\u03b1(\u03c4 ) = Z \u22121 \u03b1 (\u03c4 ), where G\u03b1 is the diagonal element of the Green function (the representation above implies that all o\ufb00-diagonal elements vanish), Z [V ] is the partition function, and G[V ] \u03b1 represents the Green function of the non-interacting system subject to an imaginary time-dependent potential iV (\u03c4 ). \u03c9m=0 e\u2212i\u03c9m\u03c4 Vm+ 2\u03c0kT + \u02dcV0, wherek \u2208 Z and \u02dcV0 \u2208 [0, 2\u03c0T ] (i.e. 2\u03c0kT + \u02dcV0 is but a complicated representation of the zeroth Matsubara mode). Show that all but the static component \u02dcV0 can be removed from the action S by a gauge transformation. Why can not \u02dcV0 be gauged, too? Explore the transformation behavior of the Green function and integrate over the non-zero mode components Vm=0. \u2212 \u03c0|x| 6 k=1 cos(kx) (c) Making use of the relation 2 + x2 4 + \u00b7 \u00b7 \u00b7, perform the Matsub\u03c9m=0. Show that the Green function can be expressed as G(\u03c4 ) = ara summation, F (\u03c4 ) \u02dcG(\u03c4 ), where the function F (\u03c4 ) = exp(\u2212EC(\u03c4 \u2212 \u03b2\u22121\u03c4 2)) is obtained by integration over the dynamical components of V, while \u02dcG is a non-interacting Green function averaged over the static component V0. k2 = \u03c02 (d) The remaining integration over the static component \u02dcV0 is achieved by the stationary phase method. Neglecting the weak dependence of the non-interacting Green function on \u02dcV0, derive and interpret the saddle-point equation. Approximate the functional by its value at the saddle-point (i.e. neglect quadratic \ufb02uctuations around the saddle-point value for \u02dcV0). As a result, obtain a representation G(\u03c4 ) = F (\u03c4 )G0(\u03c4 ) where G0 is a non-interacting Green function evaluated at a renormalized chemical potential. \u221e (e) Assuming EC T, compute an approximation of the Fourier transform of F (\u03c4 ). Use your result to obtain the zero-temperature DoS.86 (You may approximate Matsubara sums by integrals.) Answer: (a) Using the identity e\u2212EC d\u03c4 ( \u00af\u03c8\u03b1\u03c8\u03b1\u2212N0)2 \u03b1 = DV e \u2212 V 2 4EC \u2212iN0V +i \u00af\u03c8\u03b1V \u03c8\u03b1 \u03b1, the DV e\u2212S[V ]\u2212 d\u03c4 \u00af\u03c8\u03b1(\u2202\u03c4 +\u03b1\u2212\u03bc+iV )\u03c8\u03b1\u2212SJ [ \u00af\u03c8,\u03c8], \u03b1 quantum partition function takes the form Z = where S[V ] = d\u03c4 D( \u00af\u03c8, \u03c8) \u2212 iN0V V 2 4EC. Thus, G\u03b1(\u03c4 ) =Z \u22121 DV e\u2212S[V ] D( \u00af\u03c8, \u03c8) e\u2212 = Z \u22121 DV e\u2212S[V ]Z [V ]G[V ] \u03b1 (\u03c4 ), d\u03c4 \u03b1 \u00af\u03c8\u03b1(\u2202\u03c4 +\u03b1\u2212\u03bc+iV )\u03c8\u03b1 \u00af\u03c8\u03b1(\u03c4 )\u03c8\u03b1(0) where G[V ] is obtained from a non-interacting theory with action S[V", " ] \u2261 S|\u03bc\u2192\u03bc\u2212iV. 86 For a more elaborate analysis of \ufb01nite-temperature corrections, we refer to A. Kamenev and Y. Gefen, Zero bias anomaly in \ufb01nite-size systems, Phys. Rev. B 54 (1996), 5428\u201337. (b) The gauge transformation removing much of the time-dependent potential from the \u201ctr ln\u201d is de\ufb01ned by 6.7 Problems 337 \u00af\u03c8\u03b1(\u03c4 ) \u2212\u2192 \u00af\u03c8\u03b1(\u03c4 ) ei \u03c4 d\u03c4 (V (\u03c4 )\u2212 \u02dcV0); \u03c8\u03b1(\u03c4 ) \u2212\u2192 e\u2212i The zero-mode o\ufb00set \u02dcV0 has to be excluded from the transformation to preserve the time periodicity of the gauge factor (i.e. to make sure that the transformed \ufb01eld respects the time-antiperiodicity required of a Grassmann \ufb01eld). Substitution of the transformed \ufb01eld leads to (i) removal of the dynamic V -components from the action, S[V ] \u2192 S[ \u02dcV0], and (ii) the appearance of a gauge factor multiplying the pre-exponential terms. We thus obtain \u03c4 d\u03c4 (V (\u03c4 )\u2212 \u02dcV0)\u03c8\u03b1(\u03c4 ). G(\u03c4 ) = 1 Z DV D( \u00af\u03c8, \u03c8) e\u2212S[V ]e\u2212 d\u03c4 \u03b1 \u00af\u03c8\u03b1[\u2202\u03c4 \u2212\u03bc+i \u02dcV0+\u03b1]\u03c8\u03b1 \u00d7ei \u03c4 0 d\u03c4 (V (\u03c4 )\u2212 \u02dcV0) \u00af\u03c8\u03b1(\u03c4 )\u03c8\u03b1(0) \u03c4 0 d\u03c4 (V (\u03c4 )\u2212 \u02dcV0)Z [ \u02dcV0]G[ \u02dcV0] \u03b1 (\u03c4 ) DV e\u2212S[V ]ei d \u02dcV0 e \u2212 \u03b2 4EC 0 +i\u03b2N0 \u02dcV0 Z [ \u02dcV0]G[ \u02dcV0] \u02dcV 2 \u03b1 (\u03c4 ), 1 Z F (\u03c4 ) Z = = where we have omitted the 2\u03c0kT contribution to the zero mode (as it does not play much of a role in the context of this problem \u2013 for the physical meaning of the integers k see Chapter 9), the function'' \u2212 \u03b2 4EC VnV\u2212n+ Vn \u03c9n (exp(i\u03c9n\u03c4 )\u22121) = \u2212 2EC T \u03c92 n (1\u2212exp(\u2212i\u03c9n\u03c4 )), F (\u03c4 ) = dVn e n=0 e n=0 and in the second equality we have performed the Gaussian integral over Vn=0. (c) Using the formulae given above, the Matsubara summation gives \u22122ECT n=0 1 \u03c92 n (1 \u2212 exp(\u2212i\u03c9n\u03c4 )) = \u2212EC(|\u03c4 | \u2212\u03b2 \u22121\u03c4 2), i.e. F (\u03c4 ) = exp[\u2212EC(|\u03c4 | \u2212\u03b2 \u22121\u03c4 2)], and G\u03b1(\u03c4 ) = F (\u03c4 ) \u02dcG\u03b1(\u03c4 ), where \u02dcG\u03b1(\u03c4 ) = Z \u22121 d \u02dcV0 e \u2212 \u03b2 4EC 0 +i\u03b2N0 \u02dcV0 Z [ \u02dcV0]G[ \u02dcV0] \u02dcV 2 \u03b1 (\u03c4 ). (d) De\ufb01ning a \u02dcV0-dependent free energy by Z [ \u02dcV0] = exp(\u2212\u03b2F [ \u02dcV0]), noting that \u02dcV0 shifts the chemical potential, F [ \u02dcV0](\u03bc) =F (\u03bc \u2212 i \u02dcV0), and neglecting the \u02dcV0-dependence of G[ \u02dcV0] G, we obtain the saddle-point equation 0 = \u2202 \u2202 \u02dcV0 1 4EC \u02dcV 2 0 \u2212 iN0 \u02dcV0 \u2212 F (\u03bc \u2212 i \u02dcV0) = 1 2EC \u02dcV0 \u2212 iN0 + i \u02c6N \u03bc\u2212i \u02dcV0, F (\u03bc \u2212 i \u02dcV0) =\u2212 i\u2202\u03bcF (\u03bc \u2212 where in the second equality we have used the fact that. Substituting the solution of the saddle-point equation \u00afV0 = i \u02dcV0) =\u2212i \u02c6N \u03bc\u2212i \u02dcV0 2iEC(N0 \u2212 \u02c6N \u03bc\u2212i \u00afV0 ) amounts to replacing the chemical potential \u03bc by an e\ufb00ective chemical potential \u00af\u03bc = \u03bc + 2EC(N0 \u2212 \u02c6N \u00af\u03bc). As", " a preliminary result we thus obtain \u2202 \u2202 \u02dcV0 338 Broken symmetry and collective phenomena G\u03b1(\u03c4 ) = F (\u03c4 )G0\u03b1\u03b2, where the non-interacting Green function is evaluated at the renormalized chemical potential. (In passing we note that the condition for the applicability of the saddle-point approximation reads as (Exercise: Why?) 1/(2EC) \u2212 \u2202\u03bc \u02c6N \u00af\u03bc \u03b2.) (e) For EC T, the dominant contribution to the Fourier transform of F (\u03c4 ) comes from the boundary regions of the imaginary time interval, \u03c4 \u03b2 and \u03b2 \u2212 \u03c4 \u03b2. Linearizing the exponent of F in these regions we obtain \u03b2 \u221e Fm = 2EC C + \u03c92 m Using the fact that G0n\u03b1 = (i\u03c9n \u2212 \u03be\u03b1)\u22121, where\u03be \u03b1 = \u03b1 \u2212 \u00af\u03bc, we then obtain ei\u03c9m\u03c4 + e\u2212i\u03c9m\u03c4 d\u03c4 ei\u03c9m\u03c4 F (\u03c4 ) e\u2212EC \u03c4 = E2 d\u03c4 0 0. Gn\u03b1 = T m FmG0(n\u2212m)\u03b1 1 2\u03c0 d\u03c9 1 i\u03c9n \u2212 i\u03c9 \u2212 \u03be\u03b1 2EC C + \u03c92 = E2 1 i\u03c9n \u2212 EC sgn(\u03be\u03b1) \u2212 \u03be\u03b1. From this result we obtain the DoS \u03bd() = \u03b1 \u03b4( \u2212 EC sgn(\u03be\u03b1) \u2212 \u03be\u03b1) = d\u03c9 \u03bd0 (\u03c9)\u03b4( \u2212 EC sgn(\u03c9) \u2212 \u03c9) = \u03bd0( \u2212 EC sgn())\u0398(|| \u2212E C), where \u03bd0 is the DoS of the non-interacting systems. Due to the large charging energy, the single-particle DoS vanishes in a window of width 2EC centered around the Fermi energy: the Coulomb blockade. Particles of energy > EC larger than the charging threshold are free to enter the dot. However, in doing so they lose an amount EC of (charging) energy, which explains the energy shift in the factor \u03bd0. Action of a tunnel junction In the previous problem, we considered the physics of a perfectly isolated quantum dot. However, in practice (see, e.g., the dot depicted in the last problem) the system is usually connected to an external environment by some leads. It is the purpose of this problem to derive an e\ufb00ective action accounting for the joint e\ufb00ect of charging and the coupling to an environment. Consider a quantum dot connected to an external lead (it is straightforward to generalize to the presence of several leads). For simplicity, we model the latter as an ideal wave-guide, i.e. the eigenstates \u03c8a are plane waves whose detailed structure we need not specify. The composite system is described by an action S[\u03c8\u03b1, \u00af\u03c8\u03b1, \u03c8a, \u00af\u03c8a] = Sdot[\u03c8\u03b1, \u00af\u03c8\u03b1]+Slead[\u03c8a, \u00af\u03c8a]+ ST[\u03c8\u03b1, \u00af\u03c8\u03b1, \u03c8a, \u00af\u03c8a], where Sdot[\u03c8\u03b1, \u00af\u03c8\u03b1] is given by Eq. (6.66), Slead[ \u00af\u03c8a, \u03c8a] = \u03b2 a 0 d\u03c4 \u00af\u03c8a(\u2202\u03c4 + a \u2212 \u03bc)\u03c8a, and the coupling between dot and lead is described by ST[\u03c8\u03b1, \u00af\u03c8\u03b1, \u03c8a, \u00af\u03c8a] = a\u03b1 d\u03c4 \u00af\u03c8\u03b1T\u03b1a\u03c8a + h.c. Throughout we will assume that the coupling is su\ufb03ciently weak that contributions of O(T 4) to the e\ufb00ective action are negligibly small \u2013 the \u201ctunneling approximation.\u201d 6.7 Problems 339 (a) Proceeding as in the previous problem, decouple the charging interaction by a Hubbard\u2013 Stratonovich transformation. Integrate out the fermions and subject the problem to the same gauge transformation as used above to remove the dynamical contents of the Hubbard\u2013Stratonovich \ufb01eld V. You will observe that the gauge phase transforms the coupling matrices T. (b) Expand the action to leading (i.e. second) order in the coupling matrix elements T\u03b1a. (You may ignore the integration over the static component of the Hubbard\u2013Stratonovich \ufb01eld; as discussed above, it leads to merely a shift of the chemical potential.) Assuming that the single-particle DoS of dot and lead do not vary signi\ufb01cantly on the energy scales at which the \ufb01eld V \ufb02uctuates, determine the", " dependence of the tunneling term on the \u03c4 d\u03c4 (V (\u03c4 ) \u2212 \u02dcV0) and identify its coupling constant as the Golden gauge phase \u03c6(\u03c4 ) \u2261 Rule tunneling rate 4gT.87 (To obtain a \ufb01nite result, you will need to regularize the action by subtracting from the tunneling action Stun[\u03c6] the constant Stun[0].) Expressing the charging action Sc[V ] in terms of the gauge phase V = \u02d9\u03c6 + \u02dcV0 (and neglecting the constant o\ufb00set \u02dcV0), the complete dissipative tunneling action takes the form 6 S[\u03c6] = d\u03c4 7 \u2212 iN0 \u02d9\u03c6 \u2212 gT \u02d9\u03c62 4EC d\u03c4 d\u03c4 sin2((\u03c6(\u03c4 ) \u2212 \u03c6(\u03c4 ))/2) sin2(\u03c0T (\u03c4 \u2212 \u03c4 )). (6.67) INFO The action (6.67) was \ufb01rst derived by Ambegaokar, Eckern, and Sch\u00a8on.88 Crudely, its behavior mirrors that of the quantum dynamics of a particle on a ring with kinetic energy \u223c \u02d9\u03c62/EC and subject to a dissipative damping mechanism of strength \u223c gT. Physically, the latter describes the dissipation of the energy stored in dynamical voltage \ufb02uctuations V \u223c \u02d9\u03c6 into the microscopic degrees of freedom of the quasi-particle continuum. In the absence of dissipation, the action describes the ballistic motion of a quantum point particle on a ring. The ring topology re\ufb02ects the 2\u03c0-periodicity of the quantum phase, which in turn relates to the quantization of charge (recall that charge and phase are canonically conjugate). It is, thus, no surprise that the periodicity of the \u03c6-dynamics is the main source of charge quantization phenomena in the AES approach. For strong (gT > 1) dissipation, the particle begins to forget that it actually moves on a ring (i.e. full traversals of the ring get increasingly less likely). This damping manifests itself in a massive suppression of charge quantization phenomena. Indeed, for increasing coupling between lead and dot, the charge on the latter begins to \ufb02uctuate and is no longer e\ufb00ectively quantized. For a detailed account of the physical phenomena relating to the crossover from weak to strong charge quantization, we refer to one of several reviews.89 87 We denote the tunneling rate by gT because (see Problem 7.6.3) it is but (four times) the classical conductance of the tunneling barrier, measured in units of the conductance quantum e2/h. 88 U. Eckern, G. Sch\u00a8on, and V. Ambegaokar, Quantum dynamics of a superconducting tunnel junction, Phys. Rev. B 30 (1984), 6419\u201331. 89 See, e.g., I. L. Aleiner, P. W. Brouwer, and L. I. Glazman, Quantum e\ufb00ects in coulomb blockade, Phys. Rep. 358 (2002), 309-440. 340 Answer: Broken symmetry and collective phenomena (a) Decoupling the action and integrating over the fermionic degrees of freedom, one obtains the functional Z = exp(\u2212Se\ufb00 [V ]), where Se\ufb00 = Sc \u2212 tr ln = Sc \u2212 tr ln \u2202\u03c4 \u2212 \u03bc + iV + \u02c6d T \u2020 \u2202\u03c4 \u2212 \u03bc + iV0 + \u02c6d e\u2212i\u03c6T \u2020 T \u2202\u03c4 \u2212 \u03bc + \u02c6l T ei\u03c6 \u2202\u03c4 \u2212 \u03bc + \u02c6l (6.68) \u2212 iN0V ), \u02c6d = {\u03b1\u03b4\u03b1\u03b1 } and l = Here, as in the previous problem, Sc[V ] = {a\u03b4aa } contain the single-particle energies of dot and lead, respectively, the matrix d\u03c4 (V (\u03c4 ) \u2212 V0). In passing from the \ufb01rst structure is in dot/lead space, and \u03c6(\u03c4 ) = to the second equality we have subjected the argument of the \u201ctr ln\u201d to the unitary transformation (gauge transformation) described by the matrix diag(ei\u03c6, 1) (with the block structure in dot\u2013lead space). d\u03c4 ( V 2 4EC (b) Expanding the \u201ctr ln\u201d to second order in T, and regularizing by subtracting the constant Stun[0], we obtain Stun[\u03c6] = |T |2 \u03b1a n m", " G\u03b1,n(ei\u03c6)mGa,n+m(e\u2212i\u03c6)\u2212m \u2212 G\u03b1,nGa,n + Stun[0], (6.69) where G\u03b1/a,n = (i\u03c9n \u2212 \u02c6\u03b1/a + \u03bc)\u22121 are the Green functions of dot and lead, respectively, and the constant Stun[0] will be omitted throughout. Approximating the Green functions by G\u03b1/a,n = \u2212 d \u03c1d/l( + \u03bc) i\u03c9n + n + 2 \u03c92 \u2212\u03c0i\u03c1d,lsgn(\u03c9n), (6.70) \u03b1/a where \u03c1d,l \u2261 \u03c1d,l(\u03bc) is the density of states at energy \u03bc, we arrive at the result n,m gT 2 gT 2\u03c0T m Stun[\u03c6] = = (\u2212sgn(\u03c9n)sgn(\u03c9n+m) + 1)(ei\u03c6)m(e\u2212i\u03c6)\u2212m |\u03c9m|(ei\u03c6)m(e\u2212i\u03c6)\u2212m, (6.71) m(ei\u03c6)m(e\u2212i\u03c6)\u2212m = 1, gT = 2\u03c02\u03c1l\u03c1d|T |2 is proportional to the where we note that Golden Rule tunneling rate between dot and lead, and the appearance of a term \u223c |\u03c9m| is a clear signature of a dissipative damping mechanism. Using the fact that the Fourier transform of |\u03c9m| is given by (cf. Eq. (3.44)) \u03c0T sin\u22122(\u03c0T \u03c4 ), the tunneling action can be cast in a time-representation as Stun[\u03c6] =\u2212 gT 2 d\u03c4 d\u03c4 ei\u03c6(\u03c4 )\u2212i\u03c6(\u03c4 ) sin2(\u03c0T (\u03c4 \u2212 \u03c4 )) = gT d\u03c4 d\u03c4 sin2((\u03c6(\u03c4 ) \u2212 \u03c6(\u03c4 ))/2) sin2(\u03c0T (\u03c4 \u2212 \u03c4 )) + const. We \ufb01nally add the charging action to obtain the result Eq. (6.67). 6.7 Problems 341 Josephson junction Building on the results obtained in the previous problem, here we derive an e\ufb00ective action of a Josephson junction \u2013 a system comprising two superconductors separated by an insulating or normal conducting interface region. The problem includes a preliminary discussion of the physics of the Josephson junction, notably its current\u2013voltage characteristics. In Chapter 8, renormalization group methods will be applied to explore in detail the phenomenology of the system. Consider two superconducting quantum dots separated by a tunneling barrier. Generalizing the model discussed in the last two problems, we describe each dot by an action Si[ \u00af\u03c8i \u03b1, \u03c8i \u03b1, \u03c6i] = \u03b2 0 d\u03c4 \u03b1 \u00af\u03c8i \u03b1(\u2202\u03c4 + \u03bei\u03b1\u03c33 + ei\u03c6i(\u03c4 )\u03c33 \u0394\u03c31)\u03c8i \u03b1, i = 1, 2, 8 \u03b1\u2191, \u00af\u03c8i \u03b1 = (\u03c8i where \u03c8i \u03b1\u2193)T are Nambu spinors, \u03c3i Pauli matrices in particle\u2013hole space, and \u03c6i the phase of the order parameter on dot i. Noting that two dots form a capacitor, we assume the presence of a \u201ccapacitive interaction\u201d Sint = EC 4 d\u03c4 ( \u02c6N1 \u2212 \u02c6N2)2, where \u02c6Ni = system. Finally, the tunneling between the two dots is described by the action \u03b1 is the charge operator on doti, and 1 /2EC the capacitance of the \u00af\u03c8i \u03b1\u03c33\u03c8i \u03b1 ST[\u03c8\u03b1, \u00af\u03c8\u03b1] = d\u03c4 \u00af\u03c81 \u03b1(T\u03b1\u03b2\u03c33)\u03c82 \u03b2 + h.c., \u03b1\u03b2 where T\u03b1\u03b2 = \u03b1| \u02c6T |\u03b2 denotes the tunneling matrix elements between the single-particle states |\u03b1 and |\u03b2. Now, were it not for the presence of the superconducting order parameter, the low-energy physics of the system would again be described by the e\ufb00ective action (6.67). EXERCISE Convince yourself of the validity of this statement, i.e. check that the dot\u2013dot system can be treated along the same lines as the dot\u2013lead system considered above and trace the phase dependence of the various contributions to the action. (a) Turning to the superconducting case, show that, at an intermediate stage, the action is given by Se\ufb00 [V, \u03c6] = Sc[V", " ] \u2212 tr ln \u2202\u03c4 + ( \u02c6\u03be1 + i( \u02d9\u03c61 + V )/2)\u03c33 + \u0394\u03c31 T \u2020ei(\u03c61\u2212\u03c62)\u03c33/2 e\u2212i(\u03c61\u2212\u03c62)\u03c33/2T \u2202\u03c4 + ( \u02c6\u03be2 + i( \u02d9\u03c62 \u2212 V )/2)\u03c33 + \u0394\u03c31, where \u02c6\u03bei, i = 1, 2, comprise the single-particle energies of the system, and Sc[V ] = d\u03c4 V 2 is the charging action. At zeroth order in T, expand the action to (1/4EC) second order in the combinations \u02d9\u03c6i \u00b1 V. Vary your result with respect to \u02d9\u03c61,2 to obtain 342 Broken symmetry and collective phenomena the Josephson condition \u02d9\u03c61 = \u2212 \u02d9\u03c62 \u2261 \u2212 \u02d9\u03c6 and V = \u02d9\u03c6. Neglecting the e\ufb00ect of these massive quadratic \ufb02uctuations, we will rigidly impose this condition throughout. (Hint: You may assume that the characteristic frequencies \u03c9m carried by both the \ufb01elds \u03c6i and the voltage V are much smaller than those of the bare Green function. Use this assumption to keep your analysis on a schematic level, i.e. try to argue in general terms rather than performing the expansion in great detail.) (b) Expanding the action to second order in T (i.e. the leading order) and using the Joseph- son conditions, show that S = Sc + Stun, where Stun[\u03c6] = |T |2 tr \u03b1\u03b1\u03c9m\u03c9n G1,\u03b1\u03c9n (ei\u03c33\u03c6)mG2,\u03b1\u03c9n+\u03c9m (e\u2212i\u03c33\u03c6)\u2212m \u2212 (\u03c6 \u2194 0), Gi\u03b1\u03c9n = (i\u03c9n \u2212 \u03bei\u03b1\u03c33 \u2212 \u02c6\u0394\u03c31)\u22121 is the bare Gor\u2019kov Green function, and we again regularize the tunneling action by subtracting Stun[\u03c6 = 0]. Denoting the block diagonal/o\ufb00diagonal contributions to the Green function by Gi,d/o, respectively, the tunneling action splits into two contributions, (symbolically) tr(G1dei\u03c6\u03c33 G2de\u2212i\u03c6\u03c33 +G1oei\u03c6\u03c33 G2oe\u2212i\u03c6\u03c33 ). Show that, up to small corrections of O(\u03c9n/\u0394), the diagonal terms vanish and interpret this result. (Hint: Compare with the discussion of the previous problem.) (c) Turning to the particle/hole o\ufb00-diagonal sector, show that Stun[\u0394, \u03c6] = \u03b3 \u03b2 0 d\u03c4 cos(2\u03c6(\u03c4 )), \u03b3 = |T |2(\u03c0\u03c1)2\u0394. Combining everything, one obtains the action of the Josephson junction, S[\u03c6] = 1 4EC d\u03c4 \u02d9\u03c62 + \u03b3 d\u03c4 cos(2\u03c6(\u03c4 )) + \u0393Sdiss[\u03c6]. (6.72) While our analysis above suggests that the coe\ufb03cient of the dissipative term should be zero (on account of the absence of low-energy quasi-particle states which might act as a dissipative sink of energy), voltage \ufb02uctuations in \u201creal\u201d Josephson junctions do seem to be dissipatively damped, even at low \ufb02uctuation frequencies. Although there is no obvious explanation of this phenomenon, it is common to account for the empirically observed loss of energy by adding a dissipation term to the action. (d) Finally, explore the current\u2013voltage characteristics of the non-dissipative junction. To this end, perform a Hubbard\u2013Stratonovich transformation on the quadratic charging interaction. What is the physical meaning of the Hubbard\u2013Stratonovich auxiliary \ufb01eld? Interpret the action as the Hamiltonian action of a conjugate variable pair and compute the equations of motion. Show that the Josephson current \ufb02owing between the superconductors is given by I = \u22122\u03b3 sin(2\u03c6). (6.73) According to this equation, a \ufb01nite-order parameter phase di\ufb00erence causes the \ufb02ow of a static current, carried by Cooper pairs tunneling coherently across the barrier: the DC Josephson e\ufb00ect. Application of a \ufb01nite voltage di\ufffd", "\ufffd\ufffderence or, equivalently, the presence of a \ufb01nite charging energy, render the phase \u02d9\u03c6 = V dynamical. For a static 6.7 Problems 343 voltage di\ufb00erence, \u03c6 increases uniformly in time and the current across the barrier behaves as time-oscillatory: the AC Josephson e\ufb00ect. Finally, if the voltage di\ufb00erence becomes very large, V \u0394, the Fourier spectrum of \u03c6 contains frequencies |\u03c9m| > \u0394 (think about it). At these frequencies, phase variations have the capacity to create quasiparticle excitations which in turn may tunnel incoherently across the barrier (thereby paying a price in condensate energy but bene\ufb01ting from the voltage drop). The tunneling of independent quasi-particles is described by the dissipative term in the action (which, we recall, is negligible only at frequencies |\u03c9m| < \u0394). Answer: (a) The given result is proven by decoupling the capacitive interaction by a \ufb01eld V and integrating out the fermions. We next subject the \u201ctr ln\u201d to the gauge transformation (symbolic notation) 7 6 6 tr ln \u02c6G\u22121 1 T \u2020 T \u02c6G\u22121 2 \u2192 tr ln e\u2212i\u03c61\u03c33/2 e\u2212i\u03c62\u03c33/2 \u02c6G\u22121 1 T \u2020 T \u02c6G\u22121 2 7 ei\u03c61\u03c33/2 ei\u03c62\u03c33/2. This removes the phase dependence from the order parameters of the Green functions. At the same time, the tunneling matrices become phase-dependent and the diagonal terms of the Green functions acquire a contribution i \u02d9\u03c6i, as indicated in the formula above. An expansion of the T = 0 action to second order in \u03c6i and V gives an expression of the structure S[\u03c6i, V ] = i=1,2 d\u03c4 d\u03c4 ( \u02d9\u03c6i \u2212 (\u2212)iV )(\u03c4 )F (\u03c4 \u2212 \u03c4 )( \u02d9\u03c6i \u2212 (\u2212)iV )(\u03c4 ) = C d\u03c4 ( \u02d9\u03c6i \u2212 (\u2212)iV )2(\u03c4 ) +\u00b7 \u00b7 \u00b7, where the time dependence of the integral kernel F is determined by the bare Green function. Assuming that this \ufb02uctuates rapidly, we may expand as indicated in the second equality, where the ellipsis represents higher-order derivatives acting on the slow \ufb01elds \u03c6i, V and the constant C must be positive on stability grounds. A straightforward variation of the quadratic integral with respect to \u03c6i and V gives the Josephson conditions. (b) Decoupling the interaction operators by two Hubbard\u2013Stratonovich \ufb01elds Vi = \u02d9\u03c6i and integrating out the fermions, we obtain Se\ufb00 [\u03c6] = = Sc[\u03c6i] \u2212 tr ln Sc[\u03c6i] \u2212 tr ln i i \u2202\u03c4 \u2212 ( \u02c6\u03be1 \u2212 i \u02d9\u03c61)\u03c33 \u2212 \u0394\u03c31 T \u2020\u03c33 \u02c6G\u22121 1 ei\u0394\u03c6\u03c33 T \u2020 T e\u2212i\u0394\u03c6\u03c33 \u02c6G\u22121 2, T \u03c33 \u2202\u03c4 \u2212 ( \u02c6\u03be2 \u2212 i \u02d9\u03c62)\u03c33 \u2212 \u0394\u03c31 344 Broken symmetry and collective phenomena where \u02c6\u03bei denotes the matrix of energies of the two dots. An expansion to leading order in the o\ufb00-diagonal blocks obtains the preliminary action Stun[\u03c6]. Substituting the diagonal contribution to the Green function ( \u02c6Gi,d)n = (\u2212i\u03c9n \u2212 i + \u03942) into the tunneling action and comparing with Eq. (6.69), we \ufb01nd \u02c6\u03bei\u03c33)/(\u03c92 that the sum Eq. (6.70) becomes replaced by n + \u02c6\u03be2 (Gi,d)a,\u03c9n = d \u03c1i( + \u03bc) a \u2212i\u03c9n \u2212 \u03c33 n + 2 + \u03942 \u03c92 \u2212i\u03c0\u03c1i \u03c9n n + \u03942)1/2 (\u03c92. Comparing with Eq. (6.71), we obtain n \u2212 \u03c9n n + \u03942)1/2 (\u03c92 \u03c9n+m n+m + \u03942)1/2 (\u03c92 + 1 |\u03c9m|/\u03c0T, 0 + O(\u03c9m/\u0394", "), |\u03c9m| \u0394, |\u03c9m| \u0394, instead of a global factor |\u03c9m|/2\u03c0T. The physical interpretation of this result is that only high-frequency (\u03c9 > \u0394) \ufb02uctuations of the voltage \ufb01eld V = \u02d9\u03c6 have the capacity to overcome the superconductor gap and dissipate their energy by creating quasi-particle excitations. In contrast, low-frequency \ufb02uctuations do not su\ufb00er from dissipative damping. (c) Substituting the o\ufb00-diagonal term (Go,i)n(\u03c4 ) = \u2212\u0394\u03c31/(\u03c92 action and neglecting contributions of O(|\u03c9m|/\u0394), we \ufb01nd n + \u02c6\u03be2 i +\u03942) into the tunneling Stun[\u03c61] |T |2 \u03b1\u03b1 n,m |T |2(\u03c0\u03c1)2\u0394 2T = \u0394 \u03b1 + \u03942 n + \u03be2 \u03c92 m tr \u03c31(ei\u03c33\u03c6)m\u03c31(e\u2212i\u03c33\u03c6)\u2212m \u0394 \u03b1 + \u03942 tr n + \u03be2 \u03c92 \u03c31(ei\u03c33\u03c6)m\u03c31(e\u2212i\u03c33\u03c6)\u2212m = |T |2(\u03c0\u03c1)2\u0394 d\u03c4 cos(2\u03c6(\u03c4 )). (d) Think of the non-dissipative Josephson action as the action of a point particle with \u02d9\u03c62 and potential energy cos(2\u03c6). In this language, passage to the kinetic energy \u223c E\u22121 C Hubbard\u2013Stratonovich decoupled action S[\u03c6, N ] = EC d\u03c4 N 2 + \u03b3 d\u03c4 cos(2\u03c6(\u03c4 )) + i d\u03c4 N \u02d9\u03c6, amounts to a transition from the Lagrangian to the Hamiltonian picture. The notation emphasizes that the momentum conjugate to the phase variable is the number operator of the system. (More precisely, N measures the di\ufb00erence of the charge carried by the two superconductors; the total charge of the system is conserved.) Varying the action, we obtain the Hamilton equations \u02d9\u03c6 = i2ECN, \u02d9N = 2i\u03b3 sin(2\u03c6). Now, I = i\u2202\u03c4 N is the current \ufb02owing from one dot to the other, i.e. the second relation gives the Josephson current Eq. (6.73). The \ufb01rst relation states that, for a \ufb01nite charging energy, mismatches in the charge induce time variations in the phase. By virtue of the Josephson relation, such time variations are equivalent to a \ufb01nite voltage drop. 6.7 Problems 345 Field theory of the BCS to BEC crossover As we have seen in the main text, the formation of the superconducting and Bose\u2013Einstein condensates \u2013 the macroscopic occupation of a single quantum state \u2013 is characterized by two paradigms. In the BCS theory, the transition from a normal to a superconducting phase of electrons involves a pairing instability which takes place at a temperature Tc F where the Fermi energy F sets the degeneracy scale. In particular, the formation of Cooper pairs and their condensation occur simultaneously at the transition temperature. Similarly, the transition of a Bose gas to a Bose\u2013Einstein condensate (BEC) phase occurs at a temperature Tc comparable to the degeneracy temperature \u2013 in the bosonic system, the temperature at which the thermal length hv/kBT becomes comparable to the typical particle separation. However, the bosonic particles participating in the condensate are invariably molecules or composites involving an even number of elementary fermionic degrees of freedom, e.g. 4He, Rb atoms, etc. Usually, the corresponding dissociation temperature of the particles into their fermionic constituents Tdis is greatly in excess of the BEC transition temperature Tc \u2013 the condensate forms out of preformed bosons. The separation of the energy scales Tdis and Tc means that systems which undergo a BCS or BEC transition can be neatly classi\ufb01ed. However, in some systems (such as fermionic atom condensates), the transition temperature can be comparable to (or even tuned through) the molecular dissociation temperature. Such systems have the capacity to manifest a BEC to BCS crossover. The aim of this extended problem is to develop a \ufb01eld theory to describe the crossover. Although the study of the BEC to BCS crossover has a long history, the modern perspective on the subject can be traced back to the seminal work", "90 of Anthony Leggett. In this early work, it was shown that the smooth crossover from a BCS to a BEC state could be described within the framework of a single variational wavefunction. In the following, we will develop a \ufb01eld theory to explore the crossover in the framework of the original Hamiltonian considered by Leggett.91 Sir Anthony J. Leggett 1938\u2013 Co-recipient of the 2003 Nobel Prize in Physics (with Alexei A. Abrikosov and Vitaly L. Ginzburg) \u201cfor pioneering contributions to the theory of superconductors and super\ufb02uids.\u201d He has made important contributions to the theory of normal and super\ufb02uid helium liquids and other strongly coupled super\ufb02uids, macroscopic dissipative quantum systems, and the use of condensed systems to test the foundations of quantum mechanics. (Image c The Nobel Foundation.) 90 See, e.g., A. J. Leggett, Modern Trends in the Theory of Condensed Matter, A. Pekalski and R. Przystawa, eds. (Springer-Verlag, 1980) as well as the later work by P. Nozi`eres and S. Schmitt-Rink, J. Low Temp. Phys. 59 (1985), 195\u2013211. See also the related work by Keldysh and collaborators from the 1960s on the closely related problem of exciton condensation phenomena. 91 Our analysis will follow closely that described in M. Randeria, Crossover from BCS theory to Bose\u2013Einstein condensation, in Bose\u2013Einstein Condensation, A. Gri\ufb03n, D. W. Snoke, and S. Stringari eds. (Cambridge University Press, 1985). 346 Broken symmetry and collective phenomena Consider the Hamiltonian of a three-dimensional gas of spinful fermionic particles inter- acting through an attractive local (i.e. contact) pairwise interaction, \u02c6H \u2212 \u03bc \u02c6N = \u2020 k\u03c3ck\u03c3 \u2212 gL3 d3r c \u2020 \u2191c \u2020 \u2193c\u2193c\u2191, \u03bekc k\u03c3 where \u03bek = k \u2212 \u03bc, k = k2/2m, and the parameter g characterizes the strength of the interaction. (a) Starting with the coherent state path integral formulation of the quantum partition function, introduce a Hubbard\u2013Stratonovich \ufb01eld, decoupling the pair interaction and, integrating out the fermionic degrees of freedom, obtain an e\ufb00ective action involving the pairing \ufb01eld \u0394. (Select only the pairing channel in the decoupling.) (b) Varying the action with respect to \u0394, show that the transition temperature is determined by the saddle-point equation 1 g = k tanh(\u03bek/2Tc) 2\u03bek. An inspection of the momentum summation will con\ufb01rm the presence of a high-energy (UV) divergence. In the conventional BCS theory of the superconductor discussed in the main text, one may recall that this high-energy divergence was regularized by a physical cut-o\ufb00 derived from the pairing mechanism itself \u2013 the phonon exchange mechanism imposed a cut-o\ufb00 at the Debye frequency \u03c9D \u03bc, the highest phonon energy allowed by the crystal lattice, which restricted the range of the momentum sum to an energy shell around the Fermi level. In the present case, the UV divergence re\ufb02ects a pathology of the contact interaction. Had one chosen a more physical pair potential g(r \u2212 r) involving both a magnitude g and a range b, the momentum sum would have involved a soft cut-o\ufb00 at the momentum scale |k| \u223c1/b (exercise). Equivalently, to implement the physical cut-o\ufb00, one may replace the bare coupling constant g by the low-energy limit of the two-body T-matrix (in the absence of the surrounding medium). In three dimensions, this translates to the regularization92 m 4\u03c0a = \u2212 1 g 1 2k, + k where a denotes the s-wave scattering length. As a function of the (positive) bare interaction g, 1/a increases monotonically from \u2212\u221e for a very weak attraction to +\u221e for a very strong attraction. Beyond the threshold 1/a = 0, the two-body system develops a bound state with a size a and binding energy EB = 1/ma2. Therefore, one can identify the ratio of the typical particle separation and scattering length 1/kFa as the 92 To understand this result, one may note that, for an isolated two-body scattering involving a potential V, the scattering T-matrix obeys", " the Lippmann\u2013Schwinger equation \u02c6T = \u02c6V + \u02c6V \u02c6G0 \u02c6T, where \u02c6G0 denotes the bare Green function of the free particles. Applied to the potential V (r) =\u2212 gLd\u03b4(r), the diagonal k = 0 component of the T-matrix, which translates to the physical scattering length a, can be obtained from the self-consistent solution of the equation as T (0, 0) = 4\u03c0a/m = [\u2212(1/g) + k(1/2k)]\u22121. 6.7 Problems 347 dimensionless coupling constant where kF and F denote the Fermi wavevector and energy of the non-interacting system. With this interpretation, the saddle-point equation acquires the regularized form \u2212 m 4\u03c0a = k tanh(\u03bek/2Tc) 2\u03bek \u2212 1 2k. When combined with the equation for the particle density n(\u03bc, T ) = \u2212 1 L3 k [1 \u2212 tanh (\u03bek/2T )], valid in the mean-\ufb01eld approximation, this presents an equation \u2202F \u2202\u03bc for Tc and \u03bc. (c) In the weak coupling BCS limit, 1/kFa \u2192 \u2212\u221e, show that the chemical potential is \ufb01xed by the density such that \u03bc F and, making use of the identity \u221e 0 dz z1/2 tanh((z \u2212 1)/2t) 2(z \u2212 1) \u2212 1 2z = ln, 8\u03b3 \u03c0e2t where \u03b3 = eC and C denotes Euler\u2019s constant, show that the saddle-point equation translates to the condition Tc = 8\u03b3 \u03c0e2 Fexp \u03c0 2kF|a|. (d) In the strong coupling limit 1/kFa \u2192 +\u221e, the roles of the saddle-point equation and number density are reversed. The former now \ufb01xes the chemical potential \u03bc <0 while, super\ufb01cially, the latter determines Tc. In particular, making use of the identity, \u221e 0 dz z1/2 2, show that 1 2(z+1) = \u2212 \u03c0 \u2212 1 2z $ # \u03bc = \u2212 EB 2 = 1 2ma2, Tc EB EB F 2 ln 3/2. In fact, the apparent divergence of the transition temperature Tc in the strong coupling limit is a pathology of the mean-\ufb01eld analysis. The inferred value of Tc represents the dissociation energy of the pairs Tdis rather than the temperature scale at which coherence is established. To understand why, we have to turn to the analysis of \ufb02uctuations. EXERCISE Enthusiasts may enjoy exploring the saddle-point analysis of the condensed phase at zero temperature when the order parameter \u0394 acquires a non-zero expectation value. These results may be compared with the variational analysis based on the ground \u2212k\u2193)|\u03a9 described in Leggett\u2019s original work. state wavefunction |g.s. = Note that, in the low-density limit, the variational parameter vk describes the bound state wavefunction of a single pair. k(uk + vkc\u2020 k\u2191c\u2020 (e) Developing the action to quartic order in \u0394 show that the Ginzburg\u2013Landau expansion of the action takes the form S = F0 T + q \u03a0(q)|\u0394(q)|2 + ddr u 4 |\u0394(r)|4 + \u00b7 \u00b7 \u00b7, 348 Broken symmetry and collective phenomena where F0 denotes the free energy of the non-interacting Fermi gas, \u2212 m 4\u03c0a 1 \u2212 nf (\u03bek) \u2212 nF(\u03bek+q) i\u03c9m \u2212 \u03bek \u2212 \u03bek+q 1 2k \u03a0(q) = + k, denotes the pairing susceptibility, and u is a positive constant. In the weak coupling limit 1/kFa \u2192 \u2212\u221e, con\ufb01rm that the gradient expansion of the susceptibility leads to the Ginzburg\u2013Landau expansion discussed in Section 6.4. Conversely, in the strong coupling limit 1/kFa \u2192 +\u221e, show that \u03a0(q) \u03c0 2 \u221a \u03bd(F) 2FEB (\u2212i\u03c9m \u2212 2\u03bc + \u03c9B(q)), where \u03c9B = \u2212EB + q2/4m. Absorbing the prefactor into an overall rescaling of the \ufb01eld \u0394, the corresponding action can now be recognized as that of a weakly interacting Bose gas of composite particles with a mass 2 \u00d7 m and density n/2", " \u2013 cf. Section 6.3. The failure of the mean-\ufb01eld theory to infer the correct value of Tc now becomes clear. In the mean-\ufb01eld description, it is assumed that, at temperatures T > Tc, particles exist as unbound fermions. However, the action above shows that, for temperatures only slightly in excess of Tc, the particles already exist as bound pairs. The bulk transition takes place when these bound pairs condense. Once identi\ufb01ed as a weakly interacting Bose gas, one can immediately deduce that, in the strong coupling limit, Tc becomes independent of the scattering length and varies with density as n 2\u03b6(3/2) 2/3 \u03c0 m. Tc = INFO BEC\u2013BCS transition in fermionic alkali atomic gases and cold exciton liquids: As mentioned above, in the majority of condensed matter systems the dissociation energy Tdis is well separated from the transition temperature Tc, and the majority of condensates can be neatly classi\ufb01ed as being of BCS or BEC type. However, two systems which present the opportunity to explore crossover phenomena have been the subject of considerable interest in the experimental and theoretical literature. At low temperatures a \u201cquasi-equilibrium\u201d degenerate gas of electrons and holes forms a twocomponent plasma.93 At low densities, the constituent electrons and holes can bind to form neutral composite objects known as excitons. Comprising an electron and hole, these objects transform as bosons and, as such, have the capacity to undergo BEC. At high densities, the electrons and holes become unbound and exist as a two-component plasma. Yet, by exploiting the Coulomb interaction, at low enough temperatures the electrons and holes can condense into a collective BCS-like phase \u2013 the exciton insulator in which a quasi-particle energy gap develops at the Fermi surface. Leaving aside the capacity for other phases to develop (namely electron\u2013hole droplets or molecules), the particle density can be used as a parameter to mediate a crossover between a BEC and a BCS-like phase. (In fact, a careful study of the free energy at mean-\ufb01eld 93 Here the term \u201cquasi-equilibrium\u201d is used to acknowledge the fact that, in a conventional direct band-gap semiconductor, an electron\u2013hole plasma can lower its energy by recombination. If the recombination rate is slower than the equilibration time, an electron\u2013hole plasma may, in principle, acquire a quasi-equilibrium distribution. 6.7 Problems 349 shows the crossover to occur via a phase transition of high order. Whether the weak transition is smeared into a crossover by thermal \ufb02uctuations remains unclear.) To date, experimentalists have been unable to defeat the problems posed by fast radiative and Auger-assisted recombination processes to unambiguously realize such a condensate.94 A second example of a BEC to BCS crossover is presented by the atomic condensates. As mentioned in the text, the realization of BEC in atom condensates is now almost routine. Lately, there have been considerable e\ufb00orts targeted at realizing BCS-like condensates of fermionic atoms such as 6Li and 40K. Yet, it being charge neutral, the experimental identi\ufb01cation of a BCS phase presents considerable di\ufb03culties \u2013 a feature shared by the electron\u2013hole condensate. As a result experimentalists have used \u201cFeshbach resonance\u201d phenomena to tune the atomic pair interaction from weak to strong coupling, whence the atoms exist as tightly bound pairs. By monitoring the dynamics of BEC formation, attempts have been made to infer the properties of the ephemeral BCS-like phase.95 Answer: (a) Referring to Section 6.4, a decoupling of the pair interaction by a \ufb01eld \u0394 gives the partition function Z = D( \u00af\u0394, \u0394) exp \u2212 1 g d\u03c4 ddr |\u0394|2 + ln det \u02c6G\u22121, where \u02c6G\u22121 = \u2212\u2202\u03c4 \u2212 \u03bek \u00af\u0394 \u0394 \u2212\u2202\u03c4 + \u03bek, denotes the matrix Gor\u2019kov Green function. (b) Once again referring to Section 6.4, a variation of the action with respect to \u0394 gives the saddle-point equation \u00af\u0394(r, \u03c4 ) g # $ \u2212 tr \u02c6G(r, \u03c4 ; r, \u03c4 )Eph 12 = 0. In particular, for \u0394 \u2192 0, an integration over the Matsubara frequencies leads to the required saddle-point equation. (c) Firstly, in the weak coupling limit, we have that Tc \u03bc. Therefore", ", to leading order, one may note that \u03bc F. Then, in this approximation, applying the regularisation procedure outlined in the question, the saddle-point equation takes the form \u2212 m 4\u03c0a = \u221e 0 d \u03bd() tanh[( \u2212 F)/2Tc] 2( \u2212 F) \u2212 1 2, 94 For a general review of the \ufb01eld, we refer to Gri\ufb03n, Snoke, and Stringari, eds., Bose\u2013Einstein Condensation. 95 For a review of this general \ufb01eld, we refer to, e.g., M. Holland, S. J. J. M. F. Kokkelmans, M. L. Chiofalo, and R. Walser, Resonance super\ufb02uidity in a quantum degenerate Fermi gas, Phys. Rev. Lett. 87 (2001), 120406. 350 Broken symmetry and collective phenomena \u221a \u221a where \u03bd() = m3/2 2\u03c02 denotes the three-dimensional density of states. Making use of the given identity, and noting that m/4\u03c0a\u03bd(F) = \u03c0/2kFa, one obtains the required estimate for Tc. / (d) In the strong coupling limit, one expects a bound state to develop in advance of the transition. In the limit of low density F \u2192 0, the appearance of the bound state is signaled by the chemical potential reversing sign. For Tc |\u03bc|, the saddle-point equation becomes largely temperature-independent and acquires the form \u2212 m 4\u03c0a = d \u03bd() 1 2( + |\u03bc|) \u2212 1 2. Making use of the given identity, one obtains the required formula for the chemical potential. In the low-density limit, the latter asymptotes to (one half of) the bound state energy EB. Then, when \u03bc is substituted into the equation for the particle number, one obtains \u221e n 2 d \u03bd() exp 0 \u2212 + EB/2 T. F When equated with the expression for the number density, n = 0 d \u03bd(), a rearrangement of the equation obtains the required estimate for Tc. As mentioned in the question, here one must interpret Tc as the dissociation temperature, which may be \u2013 and, in the physical context, usually is \u2013 greatly in excess of the Bose-Einstein condensation temperature. (e) When properly regularized, the expansion of the action to second order follows directly the procedure outlined in Section 6.4. In the weak coupling limit, the chemical potential asymptotes to the Fermi energy of the non-interacting system, F. Here the gradient expansion of the pair susceptibility \u03a0(q) is strictly equivalent to that discussed in Section 6.4. By contrast, the strong coupling expansion requires further consideration. In this case, one may expect \u03bc < 0. For temperatures T |\u03bc|, noting that \u03bek+q/2 + \u03bek\u2212q/2 = 2(k \u2212 \u03bc) +q 2/4m one obtains \u03a0(q) k 1 i\u03c9m \u2212 2(k \u2212 \u03bc) \u2212 q2/4m + 1 2k \u2212 m 4\u03c0a. Then, making use of the identity proposed in part (d), one obtains \" \u03a0(q) = \u03c0 2 \u03bd(F)\u221a 2F \u22122\u03bc + q2 4m \u2212 i\u03c9m \u2212 m 4\u03c0a. Expanding the argument of the square root in deviations from the bound state energy EB, one obtains the required gradient expansion. Metallic magnetism Previously we have seen that the functional \ufb01eld integral provides a convenient framework in which a perturbation theory of the weakly interacting electron gas can be developed. By trading the \ufb01eld 6.7 Problems 351 operators of the electrons for the dressed photon \ufb01eld \u03c6, the diagrammatic series that comprises the RPA can be organized into a systematic expansion of the action in the charge, e. However, we have also seen that interactions can have a more striking e\ufb00ect on the electron liquid, initiating transitions to new electron phases. In the following, we will explore the transition of the electron gas to an itinerant magnetic phase \u2013 the Stoner transition.96 Historically, the Stoner transition has assumed a special place in the theoretical literature. Developments in statistical mechanics through the 1950s and 1960s highlighted the importance of \ufb02uctuation phenomena in the classi\ufb01cation and phenomenology of classical phase transitions (for more details, see Chapter 8). In the vicinity of a continuous classical phase transition, the collective properties of a thermodynamic system are characterized by a set of universal critical exponents. In the quantum mechanical system, a phase transition can be tuned by a change of", " an external parameter even at zero temperature \u2013 a \u201cquantum phase transition.\u201d In a seminal work, it was proposed by John Hertz97 that the region surrounding a quantum critical point was itself characterized by quantum critical phenomena. In this context, the problem of metallic magnetism presents a useful prototype \u2013 and the one used by Hertz to exemplify the phenomenology of quantum criticality. Lately, the class of heavy fermion materials has provided a rich experimental arena in which quantum critical phenomena have been observed and explored. In the following, we develop a low-energy theory of the interacting electron system and discuss the nature of the mean-\ufb01eld transition to the itinerant ferromagnetic phase. Later, following our discussion of the renormalization group methods in Chapter 8, we use the low-energy theory as a platform to discuss the general phenomenology of quantum criticality (see Problem 8.8.2). Our starting point is the lattice Hamiltonian for a non-interacting electron gas perturbed by a local \u201con-site\u201d Hubbard interaction, \u02c6H = \u02c6H0 + \u02c6HU where \u02c6H0 = p\u03c3 pc\u2020 p\u03c3cp\u03c3, \u02c6HU = U N i \u02c6ni\u2191 \u02c6ni\u2193. \u2020 i\u03c3ci\u03c3 denotes the number Here, the sum runs over the N lattice sites i and, as usual, \u02c6ni\u03c3 = c operator for spin \u03c3 on site i. The electron dispersion relation p (a function of the lattice geometry) as well as the dimensionality is, for the present, left unspeci\ufb01ed. As we have seen in Chapter 2, the phase diagram of the lattice Hubbard Hamiltonian is rich, exhibiting a range of correlated ground states depending sensitively on the density and strength of interaction. In the lattice system, commensurability e\ufb00ects can initiate charge or spin density wave instabilities while, at large U, the electron system can freeze into an insulating antiferromagnetic Mott\u2013Hubbard state. Conversely, in the following, we will show that, at low densities, the system may assume an itinerant (i.e. mobile) spin-polarized phase, the Stoner ferromagnet. The capacity of the interacting electron system to form a ferromagnetic phase re\ufb02ects the competition between the kinetic and interaction potential energies. Being forbidden 96 E. C. Stoner, Ferromagnetism, Rep. Prog. Phys. 11 (1947), 43-112. 97 J. A. Hertz, Quantum critical phenomena, Phys. Rev. B 14 (1976), 1165\u201384. 352 Broken symmetry and collective phenomena by the Pauli exclusion principle to occupy the same site, electrons of the same spin can escape the local Hubbard interaction. However, the same exclusion principle requires the system to occupy higher-lying single-particle states, raising the kinetic energy. When the total reduction in potential energy outweighs the increase in kinetic energy, a transition to a spin-polarized or ferromagnetic phase is induced. Once again, to facilitate the construction of a low-energy \ufb01eld theory of the magnetic transition, it is helpful to e\ufb00ect a Hubbard\u2013Stratonovich decoupling of the Hubbard interaction. For this purpose it is convenient to \ufb01rst separate the interaction into channels sensitive to the charge and spin densities, i.e. \u02c6HU = U 4 (\u02c6ni\u2191 + \u02c6ni\u2193)2 \u2212 U 4 i i (\u02c6ni\u2191 \u2212 \u02c6ni\u2193)2. Since we expect that \ufb02uctuations in the charge density channel will have little e\ufb00ect on the thermodynamic properties of the low-density system, we can therefore neglect their in\ufb02uence on the interaction, setting \u02c6HU \u2212U i = (\u02c6ni\u2191 \u2212 \u02c6ni\u2193)/2. i )2, where \u02c6Sz i( \u02c6Sz EXERCISE Here, for simplicity, we have isolated a component of the Hubbard interaction which couples to the spin degrees of freedom but violates the spin symmetry of the original interaction. How could the local interaction be recast in a manner which makes the spin symmetry explicit while isolating the coupling to the spin degrees of freedom? (a) Making use of the coherent state path integral, express the quantum partition function of the interacting system as a functional \ufb01eld integral. Decoupling the local quartic interaction by the introduction of a local scalar magnetization \ufb01eld mi(\u03c4 ), integrate out the fermionic degrees of freedom and show that the partition function takes the form Z = Z0 Dm exp \u03b2", " 0 \u2212 U 4 d\u03c4 i m2 i (\u03c4 ) + \u03c3 tr ln 1 \u2212 U 2 8 \u03c33 \u02c6m \u02c6G0, where Z0 and G0,p = (in \u2212 \u03bep)\u22121 denote, respectively, the quantum partition function and Green function of the non-interacting electron gas, the matrix \u02c6m = {mi(\u03c4 )\u03b4ij\u03b4(\u03c4 \u2212 \u03c4 )}, and\u03c3 i are Pauli spin matrices. Left in this form, the expression for the quantum partition function is formally exact, but seemingly unmanageable. Yet there is an expectation that the system will undergo a phase transition to a magnetic phase at some critical value of the interaction, U = Uc, signaled by the appearance of a non-zero expectation value of the magnetization \ufb01eld mi. This being so, a mean\ufb01eld description of the transition can be developed along two complementary lines. Firstly, when far below the transition, one expects that the quantum partition function is characterized by a well-developed saddle-point of the \ufb01eld integral. By minimizing the action with respect to mi, the corresponding saddle-point equation can be used to track the development of magnetic order (exercise). Alternatively, if the transition to the magnetic phase is of second order (i.e. the expectation value of the magnetization \ufb01eld grows continously from zero as the interaction is increased through Uc), a \ufb01eld theory of the system near the critical point can be developed as a perturbative expansion 6.7 Problems 353 of the action in powers of the magnetization \ufb01eld. It is this program to which we now turn. (b) Drawing on the RPA expansion of the weakly interacting electron gas discussed in the main text, expand the action to fourth order in the magnetization \ufb01eld. Subjecting the magnetic susceptibility to a gradient expansion, show that the action takes the general form (where dx \u2261 d\u03c4 ddx) S[m] = U 2\u03bd 4 q r + \u03be2q2 + |\u03c9n| v|q| |mq|2 + uT 4N dx m4(x) +\u00b7 \u00b7 \u00b7, where r = 1/(U \u03bd) \u2212 1, v = vFc, and c is some numerical constant.98 Identify the coe\ufb03cients of the expansion. (Hint: Recall the discussion of the Lindhard function on page 218; at fourth order in the expansion you will encounter the product of four fermion Green functions. Assuming that the momentum q carried by the magnetization \ufb01eld is much smaller than the typical momentum of the electronic single-particle states, you may approximate the product of Green functions by a constant whose value you need not specify.) Finally, rescaling the magnetization \ufb01eld and u, bring the action into the form S[m] = q 1 2 r + \u03be2q2 + |\u03c9n| v|q| |mq|2 + u 4 dx m4(x). (c) In the mean-\ufb01eld approximation, show that the system exhibits a (Stoner) transition to a spin-polarized phase when Uc\u03bd = 1. INFO Although the mean-\ufb01eld theory provides a good qualitative understanding of the nature of the transition, the Stoner criterion itself is unreliable. In the lattice system, the density of states is typically set by the bandwidth, i.e. \u03bd \u223c 1/t. Therefore, at the Stoner transition where U/t \u223c 1, the system enters the strongly correlated phase where the interaction cannot be considered as a small perturbation. In this regime, the electrons experience an e\ufb00ective interaction renormalized by the screening e\ufb00ect of the charge redistribution. Following Kanamori,99 an estimate for the e\ufb00ective interaction, Ue\ufb00 \u223c U/(1 + U/t), shows that, in the relevant regime, the Stoner criterion becomes replaced by the condition \u03bdt \u2265 1. Typically, for a smoothly varying density of states, \u03bd \u223c 1/t and the inequality is di\ufb03cult to satisfy. In practice, the Stoner transition to ferromagnetism tends to appear in materials where there is signi\ufb01cant enhancement of the density of states near the Fermi energy. 98 The validity of the gradient expansion of the action as an approximation relies on the benignancy of transverse \ufb02uctuations of the magnetization density. Recent results (D. Belitz, T. R. Kirkpatrick, and T. Vojta, Nonanalytic behavior of the spin susceptibility in clean Fermi systems, Phys. Rev.", " B 55 [1997], 9453\u201362) have cast some doubt on the integrity of this approximation. However, we should regard this pathology of the present scheme as an idiosyncrasy of the itinerant ferromagnetic system, while the present action provides a sound illustration of the guiding principles. 99 J. Kanamori, Electron correlation and ferromagnetism of transition metals, Prog. Theor. Phys. 30 (1963), 275\u201389. 354 Answer: Broken symmetry and collective phenomena (a) When cast as a \ufb01eld integral, the quantum partition function takes the form Z = D( \u00af\u03c8, \u03c8) exp \u2212 \u03b2 0 d\u03c4 p \u00af\u03c8p\u03c3(\u2202\u03c4 + \u03bep)\u03c8p\u03c3 \u2212 U 4 ( \u00af\u03c8i\u03c3\u03c3z \u03c3\u03c3 \u03c8i\u03c3 )2 i 8, where the sum over repeated spin indices is assumed. In this form, the interaction may be decoupled with the introduction of a commuting scalar \ufb01eld conjugate to the local magnetization density, 6 Z = Dm D( \u00af\u03c8, \u03c8)exp \u2212 d\u03c4 m2 i (\u03c4 ) + p \u00af\u03c8p\u03c3(\u2202\u03c4 + \u03bep)\u03c8p\u03c3 78 \u00af\u03c8i\u03c3\u03c3mi \u03c8i\u03c3. U 4 + U 2 i i Finally, integrating over the fermionic degrees of freedom, one obtains the required partition function. (b) The expansion of the action mirrors closely the RPA of the weakly interacting electron gas. Carried out to fourth order, it may be con\ufb01rmed straightforwardly that terms odd in powers of m vanish identically (a property compatible with the symmetry m \u2192 \u2212m), while the even terms lead to the expression Z = Z0 Dm e\u2212S[m], where S[m] = 1 2 q v2(q)|mq|2 + T 4N qi v4({qi}) 4' i=1 mqi \u03b4 i qi,0 + \u00b7 \u00b7 \u00b7 Focusing on the second-order contribution, one \ufb01nds that v2(q) = U \u03a0q represents the familiar fermion bubble or Lindhard function, 2 (1 \u2212 U \u03a0q) where \u03a0q = \u2212 T N k G0,kG0,k+q = \u2212 1 N k nF(k) \u2212 nF(k+q) i\u03c9n + \u03bek \u2212 \u03bek+q. As one would expect, the existence and nature of the magnetic instability depend sensitively on the susceptibility v2(q), which, in turn, depends on the detailed structure of the spectrum k. The static susceptibility is dominated by contributions to the momentum sum where \u03bek \u03bek+q. This involves the regions of reciprocal space where q is small or where, for some non-zero q = Q, \u03bek \u03bek+Q over a wide region of the Brillouin zone. The second condition re\ufb02ects a nesting symmetry where a translation by a constant wavevector leaves the spectrum invariant. It is in this situation, where commensurability e\ufb00ects become signi\ufb01cant, that magnetic or spin density waves develop. If, instead, the spectrum varies smoothly, so that nesting symmetry is absent, the susceptibility is maximized for q = 0. Leaving the spin density wave system for \u201cprivate investigation,\u201d we focus here on the channel q = 0. EXERCISE Explain why bipartite lattices lead naturally to nesting symmetry. 6.7 Problems 355 While, in general, the Lindhard function assumes a complicated structure re\ufb02ecting the detailed dispersion of the non-interacting system, for the free electron system k = k2/2m an exact expression can be derived explicitly (see Eq. (5.30)). In particular, |\u03c9n| at frequencies |\u03c9n|/|q|vF small, this expression has the expansion \u03a0q \u03a00,q \u2212 \u03bd v|q|, where v = cvF with a constant c depending on dimensionality. (In the three-dimensional system, c = \u03c0/2.) Similarly, for q small, one may expand the static susceptibility in gradients, i.e. \u03a00,q \u03bd[1 \u2212 \u03be2q2 + \u00b7 \u00b7 \u00b7] where\u03be \u223c 1/kF. Turning now to the quartic interaction, the general structure of v4 again involves a complex computation. However, since its e\ufb00ect near the critical point is merely to control the strength of the magnetization, in the following we may focus on the frequency- and momentum-independent contribution, setting v4({qi}) \u2192 v4(0) \u2261", " u = 2(U/2)4 \u03b2N p [G0(p)]4 \u223c U 4\u03bd, where \u03bd = \u22022 tive action assumes the form given above. =EF \u03bd(). Taken together with the quadratic interaction, the total e\ufb00ec- (c) In the leading approximation, an understanding of the model can be developed by ignoring the e\ufb00ect of \ufb02uctuations \u2013 spatial and temporal. In such a mean-\ufb01eld approximation, the \ufb01elds m are assumed independent of q (or x). In this case, the rescaled action takes the form S[m] \u03b2N = r 2 m2 + u 4 m4, where r = U (1 \u2212 U \u03bd)/2. Minimizing the action with respect to m, one \ufb01nds a Stoner transition to a spin-polarized ferromagnetic phase when r = 0, i.e. Uc\u03bd = 1. At values of U in excess of Uc, the (oversimpli\ufb01ed) mean-\ufb01eld analysis above predicts that the magnetization will grow in a continuous yet non-analytic manner, i.e. m = \u2212r/4u. 356 Broken symmetry and collective phenomena Functional bosonization The techniques introduced in this chapter are engaged to develop a functional-integral-oriented scheme of bosonizing the one-dimensional electron gas. Consider the interacting one-dimensional electron gas as described by the relativistic action Eq. (4.43) and the interaction contribution Eq. (4.44). Throughout, it will be convenient to formulate the last in a matrix representation, Sint[\u03c8\u2020, \u03c8] = 1 2 d\u03c4 dx \u02c6\u03c1+ \u02c6\u03c1\u2212 g4 g2 \u02c6\u03c1+ \u02c6\u03c1\u2212 \u2261 1 2 g2 g4 d\u03c4 dx \u02c6\u03c1+ \u02c6\u03c1\u2212 \u02c6g. \u02c6\u03c1+ \u02c6\u03c1\u2212 To probe the response of the system to external perturbations, we add to the action a source term Ssource[\u03c8, \u03c8\u2020, j, j\u2020] \u2261 s \u03c8s). By Grassmann di\ufb00erentiation with respect to the source \ufb01elds j and j\u2020, we may then generate any correlation function of interest. s js + j\u2020 s=\u00b1(\u03c8\u2020 d\u03c4 dx (a) Decouple the four-fermion interaction by introducing a two-component Hubbard\u2013 Stratonovich \ufb01eld \u03d5T = (\u03d5+, \u03d5\u2212). Show that the action can be written as S[ \u00af\u03c8, \u03c8, \u03d5] = 1 2 d2x \u03d5T \u02c6g\u22121\u03d5 \u2212 d2x \u00af\u03c8( \u2202 \u2212 i \u03d5)\u03c8, (6.74) \u2020 where \u03c8T = (\u03c8+, \u03c8\u2212), \u00af\u03c8 = (\u03c8 +), and we have switched to a covariant notation: x1 = \u03c4, x2 = x, d2x = dx1 dx2, and \u2202 = \u03c3\u03bc\u2202\u03bc and \u03d5 = \u03c3\u03bc\u03d5\u03bc. The space\u2013time components of the interaction \ufb01eld are de\ufb01ned by \u03d51 = 1 2 (\u03d5+ + \u03d5\u2212) and\u03d5 2 = 1 2i (\u03d5+ \u2212 \u03d5\u2212). \u2020 \u2212, \u03c8 The interaction \ufb01eld \u03d5 couples to the fermion action as a two-dimensional vector potential. As with any two-component vector, the coe\ufb03cients of \u03d5 can be decomposed into an irrotational and a divergenceless contribution (the Hodge decomposition): \u03d5\u03bc = \u2212(\u2202\u03bc\u03be + i\u03bc\u03bd\u2202\u03bd\u03b7). This is an interesting decomposition as it suggests that the vector potential can be removed from the action by a generalized gauge transformation. Indeed, the transformation \u03c8 \u2192 ei\u03be+i\u03b7\u03c33 \u03c8, \u00af\u03c8 \u2192 \u00af\u03c8e\u2212i\u03be\u2212i\u03b7\u03c33 naively removes the vector potential from the action (exercise: check this). A moment\u2019s thought identi\ufb01es the two transformations by \u03be and \u03b7 as the vectorial and axial gauge transformations discussed in Section 4.3 above. It turns out, however, that the transformation by \u03b7 is actually not permissible \u2013 a direct manifestation of the chiral anomaly. INFO There are di\ufb00erent ways to understand the origin of the chiral gauge symmetry violation. For instance, one may observe that the transformation leaves the action invariant,", " but not the measure d \u00af\u03c8 d\u03c8.100 This means that the functional integral as a whole lacks gauge invariance. Alternatively, one may integrate out the fermions and realize that the resulting \u201ctr ln\u201d exhibits problematic UV behavior. Applying a UV regularization scheme \u2013 of which, due to the importance of the Dirac operator in particle physics, there are many \u2013 one observes that the chiral gauge invariance gets lost.101 100 K. Fujikawa, Chiral anomaly and the Wess\u2013Zumino condition, Phys. Rev. D 31 (1985), 341\u201351. 101 For a detailed discussion of this point, see J. Zinn-Justin, Quantum Field Theory and Critical Phenomena (Oxford University Press, 1993). 6.7 Problems 357 Notice that both the lack of gauge invariance of the measure and the UV problems manifest themselves in an integral over (quantum) \ufb02uctuations, i.e. while the symmetry is preserved on the classical level, it gets lost in the quantum theory. This is the de\ufb01ning property of an anomaly. (b) To explore the consequences of the anomaly, integrate out the fermions and expand the resulting \u201ctr ln\u201d to second order in the \ufb01elds \u03d5\u00b1. Switching to a frequency\u2013momentum representation and approximating the Matsubara sum by an integral, one obtains an expression that is formally UV divergent. Regularize the integral by introducing a cut-o\ufb00 \u039b in momentum space. Show that the e\ufb00ective \u03d5-action reads as S[\u03d5] = 1 2 \u02dcq \u03d5T \u02dcq \u02c6g\u22121 + \u02c6\u03a0\u02dcq \u03d5\u2212\u02dcq, (6.75) where \u02dcq = (\u03c9, q) and \u02c6\u03a0\u02dcq \u2261 { \u03b4ss 2\u03c0 to represent the action as q \u2212is\u03c9+q }. Finally, introduce the \ufb01eld doublet \u0393T \u2261 (\u03be, \u03b7) S[\u0393] = 1 2 \u02dcq \u02dcq DT \u0393T \u02dcq \u02c6g\u22121 + \u02c6\u03a0\u02dcq D\u2212\u02dcq\u0393\u2212\u02dcq, (6.76) where the transformation matrix D\u02dcq \u2261 q \u2212 i\u03c9 \u2212q \u2212 i\u03c9 \u2212q \u2212 i\u03c9 q + i\u03c9 mediates between the \ufb01eld variables \u0393 and \u03d5 (exercise). We next turn our attention to the source terms. The integration over the original fermion variables generates a source contribution Ssource[\u03c8, \u03c8\u2020, j, j\u2020] D\u03c8\u2212\u2192 S[j, j\u2020, \u0393] = d2x d2x \u00afj(x) \u02c6G[\u0393](x, x)j(x) = d2x d2x (\u00afje\u2212i(\u03be+\u03b7\u03c33))(x) \u02c6G(x, x)(ei(\u03be+\u03b7\u03c33)j)(x), where x are space/time indices, the superscript [\u0393] indicates that the fermion Green function depends on the Hubbard\u2013Stratonovich interaction \ufb01elds, and in the last step we have applied the generalized gauge transformation above to transfer the (\u03be, \u03b7)-dependence to the source vectors j. The action above contains the free fermion Green function (a matrix in both space\u2013time and \u00b1-space) as an integration kernel. To proceed, notice that matrix elements of the fermion Green function can be obtained as correlation functions of a free bosonic theory. This connection was introduced in Problem 4.5 on the example of a speci\ufb01c free fermion correlation function. Generalizing the results of that problem, one may verify that (exercise) \u02c6Gss (x, x) = (2\u03c0a)\u22121e\u2212i(\u03d5+s\u03b8)(x)ei(\u03d5+s\u03b8)(x), where a is the lattice spacing and the action of the bosonic doublet \u039eT \u2261 (\u03c6, \u03b8) is given by (6.77) S0[\u039e] = 1 2 \u02dcq K\u02dcq\u039e\u2212\u02dcq, K\u02dcq \u2261 1 \u039eT \u03c0 (q2 \u2212 iq\u03c9\u03c31) (6.78) \u02dcq 358 Broken symmetry and collective phenomena i.e. a non-interacting variant of the Luttinger liquid action (cf. Eq. (4.49)). (c) Use the Fermi\u2013Bose correspondence to represent the generating function as a double \ufb01", "eld integral over \u0393 and \u039e. Next shift the integration variables \u039e to remove the \ufb01eld \u0393 from the source action and perform the quadratic integral over \u0393. Show that the \ufb01nal form of the action is given by (4.51). Summarizing, you have rediscovered the action of the interacting Luttinger liquid, and the boson representation of fermion correlation functions (the latter obtained by di\ufb00erentiation with respect to the source parameters j). While the present derivation is certainly less transparent than the one discussed in Section 4.3, it has the advantage of being more \u201cexplicit\u201d (inasmuch as we start from the standard \u201ctr ln,\u201d which is then subjected to manipulations standard in many-body \ufb01eld theory. On the other hand, the authors are not aware of applications where this aspect turned out to be of much practical relevance: usually, the standard bosonization approach is just \ufb01ne; and, where it is not, the formalism above would not be any better). Answer: (a) This is resolved by a straightforward exercise in Gaussian integration and reorganizing indices. (b) Integrating over fermions and momentarily forgetting about the impurity, we obtain the e\ufb00ective action S[\u03d5 d2x \u03d5T \u02c6g\u22121\u03d5 \u2212 tr ln( \u2202 \u2212 i \u03d5) d2x \u03d5T \u02c6g\u22121\u03d5 \u2212 1 2 tr( \u2202\u22121 \u03d5 \u2202\u22121 \u03d5) +O (\u03d54) \u02c6g\u22121 ss + \u03b4ss \u03a0s,\u02dcq \u03d5s,\u2212\u02dcq + O(\u03d54), \u03d5\u02dcq,s \u02dcq d2p ( + isp)\u22121( + \u03c9 + is(p + q))\u22121. Evidently, the where \u02dcq = (\u03c9, q) and \u03a0s,\u02dcq = structure of this integral poses a problem: while all poles of the integrand appear to be on one side of the real axis (so that analyticity arguments might suggest a vanishing of the integral), the double integral is manifestly divergent. We are, thus, confronted with a 0 \u00b7 \u221e con\ufb02ict and a regularization scheme is called for. (At [n >2]nd orders in the \u03d5 expansion, one \u201cfast\u201d momentum integration extends over n >2 energy denominators. These terms indeed vanish by analyticity, i.e. the second-order expansion of the logarithm is, in fact, exact.) To some extent, the choice of the regularization scheme is dictated by the physical context: in particle physics, relativistic covariance is sacred, and a rotationally invariant (Euclidean formalism!) regularization is required. However, in condensed matter physics, where the e\ufb00ective action is obtained by linearization of some band Hamiltonian, frequency and momentum play di\ufb00erent roles. While the integration domain of the former is in\ufb01nite, the latter is bounded to values |p| < \u039b, where \u039b is some 6.7 Problems 359 cuto\ufb00. To proceed, we \ufb01rst integrate over frequencies and then do the \ufb01nite integral over momentum: \u03a0s,\u02dcq = 1 \u03c9 + isq = \u2212 1 \u03c9 + isq \u039b dp 2\u03c0 \u039b \u2212\u039b \u2212\u039b i 2 d 2\u03c0 1 + isp \u2212 1 + \u03c9 + is(p + q) dp 2\u03c0 [sgn(sp) \u2212 sgn(s(p + q))] = 1 2\u03c0 q \u2212is\u03c9 + q. (Notice that the result is actually independent of the non-universal cuto\ufb00 \u039b.) Substituting this result into the action, we obtain Eq. (6.75). (c) Representing the fermion Green function as in Eq. (6.77), we obtain the local expression Z = D\u039e D\u0393 e\u2212S0[\u039e]\u2212S[\u0393]exp \u2212 d2x \u00afje\u2212i(\u03be+\u03c6+(\u03b7+\u03b8)\u03c33) + ei(\u03be+\u03c6+(\u03b7+\u03b8)\u03c33)j, where the non-universal factor 2\u03c0a has been absorbed in the de\ufb01nition of the source \ufb01elds. (To con\ufb01rm that the \u039e-integral faithfully reproduces the source action, one has to take into account the fact that exp(i(\u03c6 \u00b1 \u03b8))", " \u2194 \u03c8 is a Gaussian correlated variable.) The structure of the source term suggests a shift\u03c6 \u2192 \u03c6 \u2212 \u03be, \u03b8 \u2192 \u03b8 \u2212 \u03b7, or \u039e \u2192 \u039e \u2212 \u0393 for short. Denoting the now \u0393-independent source contribution by exp(\u2212Ssource[\u039e]), the D\u039e D\u0393 e\u2212S0[\u039e\u2212\u0393]\u2212S[\u0393]\u2212Ssource[\u039e]. We further partition function assumes the form Z = note that K\u02dcq = \u2212DT \u02dcq \u03a0\u02dcqD\u2212\u02dcq to obtain the integral over \u0393, S0[\u039e \u2212 \u0393] + S[\u0393] = 1 2 % \u02dcq \u02dcq K\u02dcq\u039e\u2212\u02dcq + \u0393T \u039eT \u02dcq (DT \u02dcq \u02c6g\u22121D\u2212\u02dcq)\u0393\u2212\u02dcq & \u02dcq K\u02dcq\u039e\u2212\u02dcq % \u039eT \u02dcq K\u02dcq \u2212 K\u02dcq(DT +\u039eT \u02dcq K\u02dcq\u0393\u2212\u02dcq + \u0393T & \u02dcq \u02c6g\u22121D\u2212\u02dcq)\u22121K\u02dcq \u039e\u2212\u02dcq D\u0393\u2212\u2192 1 2 = = 1 2 1 2\u03c0 \u02dcq \u039eT \u02dcq % K\u02dcq + 2q2(g4 \u2212 g2\u03c33) & \u039e\u2212\u02dcq \u02dcq (\u03d5, \u03b8)\u02dcq q2[1 + 2\u03c0(g4 \u2212 g2)] \u2212iq\u03c9 \u02dcq where we have used the fact thatK \u02dcqD\u22121 space/time, we obtain Eq. (4.51). \u2212iq\u03c9 q2[1 + 2\u03c0(g4 + g2)] \u03d5 \u03b8, \u2212\u02dcq \u2212\u02dcq = \u2212g(1 \u2212 i\u03c32). Transforming back to real 7 Response functions The chapter begins with a brief survey of concepts and techniques of experimental condensed matter physics. It will be shown how correlation functions provide a bridge between concrete experimental data and the theoretical formalism developed in previous chapters. Speci\ufb01cally we discuss \u2013 an example of outstanding practical importance \u2013 how the response of many-body systems to various types of electromagnetic perturbation can be described in terms of correlation functions and how these functions can be computed by \ufb01eld theoretical means. In the previous chapters we have introduced important elements of the theory of quantum many-body systems. Perhaps most importantly, we have learned how to map the basic microscopic representations of many-body systems onto e\ufb00ective low-energy models. However, to actually test the power of these theories, we need to understand how they can be related to experiment. This will be the principal subject of the present chapter. Modern condensed matter physics bene\ufb01ts from a plethora of sophisticated and highly re\ufb01ned techniques of experimental analysis including the following: electric and thermal transport; neutron, electron, Raman, and X-ray scattering; calorimetric measurements; induction experiments; and many more (for a short glossary of prominent experimental techniques, see Section 7.1.2 below). While a comprehensive discussion of modern experimental condensed matter would reach well beyond the scope of the present text, it is certainly profitable to attempt an identi\ufb01cation of some structures common to most experimental work in many-body physics. Indeed, we will need a discussion of this sort to construct meaningful links between the theoretical techniques developed above and experiment. 7.1 Crash course in modern experimental techniques 7.1.1 Basic concepts Crudely speaking, experimental condensed matter physics can be subdivided into three1 broad categories of analytical technique: experiments probing thermodynamic coe\ufb03cients; transport experiments; spectroscopy. 1 There are a few classes of experiment that do not \ufb01t comfortably into this three-fold scheme. These include scanning tunneling microscopy, a technique to be discussed in more detail below. 360 7.1 Crash course in modern experimental techniques 361 A summary of their utility, the basic experimental setup, the principal areas of application, and concrete realization of these families is given in the following section. (Readers who are totally unfamiliar with the basic notions of experimental many-body physics may \ufb01nd it useful to browse through that section before reading further.) The few occasional references to experimental data given in previous chapters were all to thermodynamic properties. The reason for this restriction was that the extraction of thermodynamic information from the theoretical formalism is relatively straightforward: one need only di\ufb00erentiate the partition function (alias the \ufb01eld integral) with respect to a few globally de\ufb01ned coe\ufb03cients (the temperature, homogeneous magnetic \ufb01eld, etc.). This simplicity has", " advantages but also limiting aspects: thermodynamic data are highly universal2 and, therefore, represent an important characteristic of a system. On the other hand, they contain information neither on spatial structures, nor on dynamical features. This means that thermodynamic data do not su\ufb03ce to fully understand the physics of a system. With the other two categories of experiment the situation is di\ufb00erent. Transport and spectroscopic measurements can be used to probe both static and dynamical features of a system; further, fully angle/frequency-resolved spectroscopic data contain detailed information on the spatio-temporal structure of the dominant excitations of a system or, in other words, on their dispersion relation. It is for these reasons that the focus in the present chapter will be on the last two of the experimental classes mentioned above. In spite of the wide diversity of present day analytical techniques, there are a few struc- tures common to all experimental probes of condensed matter: Firstly, the interaction of a many-body system with its environment is almost exclusively mediated by electromagnetic forces.3 Accordingly, most experiments subject the system under consideration to some external electromagnetic perturbation (a voltage drop, an in\ufb02ux of spin magnetic moments carried by a beam of neutrons, the local electric \ufb01eld formed at the tip of a scanning tunneling microscope, etc.). In a second step, the \u201cresponse\u201d of the system is then recorded by an appropriate detector or measuring device. Formally, the externally imposed perturbation is described by a (time-dependent) con- tribution to the Hamiltonian of the system, \u02c6HF = ddrF i (r, t) \u02c6X i(r). (7.1) Here, the coe\ufb03cients F i, sometimes referred to as generalized \u201cforces,\u201d represent a perturbation that couples to the system through some operators \u02c6X. For example, F i (r, t) = \u03c6(r, t) could represent a space- and time-dependent electric voltage coupling to the density of the electronic charge carriers in the system \u02c6X i = \u02c6\u03c1, etc. The use of the term \u201cperturbation\u201d is appropriate because the forces {F i } will, in general, be much weaker than the internal correlations of the system. The forces perturb the system out of its F i = 0 reference state. The measurable e\ufb00ect of this perturbation will be that certain operators \u02c6Xi, whose expectation values vanish 2 Remember that a few thermodynamic variables, i.e. numbers, su\ufb03ce to unambigously characterize the state of a homogeneous system in equilibrium. 3 An important exception involves heat conduction. 362 Response functions in the unperturbed state, build up non-vanishing expectation values Xi(r, t) = \u02c6Xi(r, t). (For example, in response to an external voltage F i = \u03c6, a current might begin to \ufb02ow, \u02c6Xi = \u02c6ji, etc.). The ultimate goal of any theory will be to understand and predict the functional dependence of the measured values of Xi on the forces F j. & In general, there is nothing one can say about that connection other than that Xi F j will be some functional of the forces. However, for a su\ufb03ciently weak force, the situation is simpler. In this case, one may expect that the functional relation between forces F j and the expectation values Xi is approximately linear, i.e. of the form % ddr dt \u03c7ij(r, t; r, t)F j(r, t) +O (F 2). (7.2) Xi(r, t) = 0 1 F j and {Xi} are externally adjustable/observable \u2013 either as an While the quantities experimental input/output, or as parameters in the theory \u2013 the integral kernel \u03c7 represents a purely intrinsic property of the system. It describes how the system \u201cresponds\u201d to the application of an external probe {F } (wherefore it is commonly referred to as a i response function) or generalized susceptibility. The functional pro\ufb01le of the response kernel is in turn determined by the dominant excitations of a system (notably its longrange excitations), i.e. our prime objects of interest. These considerations show that response functions play a principal role in promoting the dialog between experiment and theory. Experimentally, they will be measured by relating to the response {Xi}. Theory will attempt to predict the response behavior, the input ideally in a way that conforms with experimental observation. F j 0 1 7.1.2 Experimental methods To keep our discussion of the relation \u201cexperiment \u2194 theory\u201d less abstract, it is instructive to", " list a few prominent experimental techniques of condensed matter physics. Of course, the summary below can be no more than an introduction.4 Our intention is merely to illustrate the connection (perturbation response) through a few examples; indeed readers lacking a background in experimental many-body physics may welcome some motivation before plunging into the formalism of correlation functions and response developed below. Thermodynamic experiments Of the thermodynamic properties of a system that can be accessed in experiment, those most commonly investigated include the following: the speci\ufb01c heat, cv = \u2202U/\u2202T, i.e. the rate of change of the internal energy under a change of temperature; the magnetic susceptibility, \u03c7 = \u2202M/\u2202H, the change of magnetization in response to a (quasi-)static magnetic \ufb01eld; the (isothermal) compressibility, \u03ba = \u2212V \u22121\u2202V /\u2202p, the volume change in response to 4 For a short and pedagogical (if perhaps a bit outdated) introduction to a number of experimental approaches we refer to the classic text by N. W. Ashcroft and N. D. Mermin, Solid State Physics (Holt-Saunders International, 1983), while, for an up-to-date and detailed exposition of methods of spectroscopy, we refer to H. Kuzmany, Solid State Spectroscopy (Springer-Verlag, 1998). 7.1 Crash course in modern experimental techniques 363 the external pressure etc. Note that, strictly speaking, the magnetic susceptibility and the isothermal compressibility are both tensor quantities. The thermodynamic response functions are highly universal. (Remember that a few thermodynamic state variables su\ufb03ce to unambiguously characterize the state of a given system.) Speci\ufb01cally, for given values of chemical potential, magnetic \ufb01eld, pressure, etc., a calorimetric experiment will produce a one-dimensional function cv(T ). The lowtemperature pro\ufb01le of that function generally contains important hints as to the nature of the low-energy excitations of a system.5 However, the universality of thermodynamic data also implies a limitation: thermodynamic coe\ufb03cients contain information neither about the spatial \ufb02uctuations of a given system nor about its dynamics. Transport experiments I, IS IT, When subject to a gradient of a generalized \u201cvoltage\u201d U, a current \ufb02ows through a device (see \ufb01gure). Although, typically, the voltage is electrical, U = V, one can apply a temperature gradient U = \u0394T or even attach the sample to two \u201creservoirs\u201d of di\ufb00erent magnetizations U = \u0394M. One then records the current \ufb02ow induced by U. The corresponding current can be electrical, I carried by the charge of mobile carriers, the \u201cthermal current\u201d IT carried by their energy, or a \u201cspin current\u201d IS carried by their magnetic moments. Also notice that the current need not neccessarily be parallel to the voltage gradient. For example, in the presence of a perpendicular magnetic \ufb01eld, a voltage gradient will give rise to a transverse Hall current I\u22a5. The ratio of a current and a generalized voltage de\ufb01nes a conductance, g = I U.,T M I, IS IT, V, Conductance measurements represent the most common way to determine the transport behavior of a metal or the thermal conduction properties of insulators. A disadvantage is that conductance measurements are invasive, i.e. the system has to be attached to contacts. There are situations where the local injection process of charge carriers at the contact (rather than the bulk transport behavior in which one is interested) determines the value of the conductance. (For a further discussion of this point, we refer to Problem 7.6.1.) Spectroscopic experiments The general setup of a spectroscopic experiment is shown in Fig. 7.1. A beam of particles p \u2013 either massive (electrons, neutrons, muons, atoms, etc.), or the quanta of electromagnetic radiation \u2013 is generated at a source and then directed onto a sample. The kinematic information about the source beam is stored in the dispersion relation (k, \u03c9(k)).6 The particles 5 For example, the speci\ufb01c heat of the Fermi liquid, cv,Fermi liquid \u223c T, is linear in temperature, that of phonons, cv,phonon \u223c T 3, is cubic, while in a system without low-lying excitations, it vanishes exponentially (exercise: consider why). 6 For some sources, e.g. a laser, the preparation of a near-monochromatic source beam is (by now) standard.", " For others, such as neutrons, it requires enormous experimental skills (and a lot of money!). 364 Response functions D k\u2032,\u03c9 \u2032 S k, \u03c9 Figure 7.1 Basic setup of a spectroscopic experiment. A beam of electromagnetic radiation (or massive particles) of frequency\u2013momentum (\u03c9(k), k) is emitted by some source (S) and directed onto a target sample. The sample responds by emitting radiation according to some distribution P (\u03c9(k), k), which is, in turn, recorded by a detector (D). Notice that the emitted radiation can, but need not, contain the same type of particles as the source radiation. For example, light quanta may lead to the emission of electrons (photoemission spectroscopy). of the source beam then interact with constituents X of the sample to generate a secondary beam of scattered particles p. Symbolically, + p $ k, \u03c9(k) X $ K, \u03a9(K) \u2212\u2192 + p $ k, \u03c9(k) X $ K, \u03a9(K), where X represents the \ufb01nal state of the process inside the sample. Notice that the particles p leaving the sample need not be identical to those incident on the sample. (For example, in photoemission spectroscopy, X-ray quanta displace electrons from the core levels of atoms in a solid. Here p represent the light quanta, and p electrons.) Further, the dominant scattering process may be elastic (e.g. X-rays scattering o\ufb00 the static lattice structure) or inelastic (e.g. neutrons scattering o\ufb00 phononic excitations). In either case, the accessible information about the scattering process is stored in the frequency\u2013momentum distribution P (\u03c9(k), k) of the scattered particles, as monitored by a detector. From these data, one would like to restore the properties (i.e. the dispersion (\u03a9(K), K)) of the states inside the solid. This is where the detective work of spectroscopy begins. What we know is that the dispersions of the scattered particles and conof the (k, \u03c9(k)) stituents, and (K, \u03a9(k)), respecrelated are tively, sample Sir Chandrasekhara V. Raman 1888\u2013 1970 (left), Lord (John Rayleigh Strutt) William 1842\u20131919 (middle), Max von Laue 1879\u20131960 (right) Raman was awarded the Nobel Prize in Physics in 1930 \u201cfor his work on the scattering of light and for the discovery of the e\ufb00ect named after him.\u201d Lord Rayleigh was awarded the Nobel Prize in Physics in 1904 \u201cfor his investigations of the densities of the most important gases and for his discovery of argon in connection with these studies.\u201d Laue was awarded the Nobel Prize in Physics in 1914 \u201cfor his discovery of the di\ufb00raction of X-rays by crystals.\u201d (Image c The Nobel Foundation.) 7.1 Crash course in modern experimental techniques 365 through an energy\u2013momentum conservation law, i.e. k + K = k + K, \u03c9(k) + \u03a9(K) =\u03c9 (k) + \u03a9(K). According to this relation, a \u201cresonant\u201d peak in the recorded distribution P (\u03c9(k), k) signals the existence of an internal structure (for example, an excitation, or a lattice structure) of momentum \u0394K \u2261 K \u2212 K = k \u2212 k and frequency \u0394\u03a9 \u2261 \u03a9 \u2212 \u03a9 =\u03c9 \u2212 \u03c9. However, what sounds like a straightforward recipe in principle may be quite involved in practice: solid state components interact almost exclusively through electromagnetic forces. When charged particles are used as scattering probes, these interactions may actually turn out to be too strong. For example, a beam of electrons may interact strongly with the surface states of a solid (rather than probing its bulk), or the scattering amplitude may be the result of a complicated process of large order in the interaction parameters, and therefore di\ufb03cult to interpret. The obvious alternative \u2013 scattering of neutral particles \u2013 is met with its own problems (see below). Notwithstanding these di\ufb03culties, spectroscopy is one of the most important sources of experimental information in condensed matter physics. A number of prominent \u201csub-disciplines\u201d of solid state spectroscopy are summarized below: Raman spectroscopy: The inelastic scattering of visible light can be used to explore the dispersion of optical phonons (magnons, plasmons, or other electronic excitations). Such techniques require experimental skill to discriminate the \u201cRaman peak\u201d from the much larger \u201cRayleigh peak\u201d corresponding to the elastic scattering of light quanta.", " Infrared spectroscopy: The scattering of light in the infrared range can be used to explore the vibrational modes in polycrystalline solids and the band-gaps in semiconductors. X-ray crystallography: By measuring the angle-resolved intensity pro\ufb01le (the von Laue the crystalline that pattern), the elastic scattering of X-rays from the lattice ions of a solid can be used the structure of to infer substance. a (Notice typical wavelength of X-rays \u223c 10\u221210 m is of about the interatomic size of typical in solids.) Such spacings techniques of solid state spectroscopy have already acquired a long history dating back to 1913. Sir William H. Bragg and 1862\u20131942 (left) his Sir William Lawrence Bragg, 1890\u2013 1971 (right) Awarded the Nobel Prize in Physics in 1915 \u201cfor their the analysis of crystal structure by means of X-rays.\u201d (Image c The Nobel Foundation.) services son in X-ray/electron spectroscopy: A number of spectroscopic techniques are based on the fact that the ionization energies of atomic core levels lie in the X-ray range. In X-ray absorption spectroscopy the absorption of X-rays by a solid is measured as a function of the light frequency (see Fig. 7.2(a)). The absorption cross-section rises in a quasi- 366 Response functions (a) (b) (c) (d) Figure 7.2 The di\ufb00erent types of X-ray/electron spectroscopy. (a) X-ray absorption: the loss of X-ray radiation due to the ionization of core levels. (b) X-ray \ufb02uorescence: the recombination of valence electrons with previously X-ray-emptied core levels leads to the emission of radiation, which also lies in the X-ray range. The spectral analysis of this radiation contains information about the level structure of the system. (c) Photoemission spectroscopy (PES): detection of the frequencydependence of the electrons kicked out by X-ray core level ionization. (d) Auger spectroscopy: the energy emitted by a valence electron recombining with a core level is transmitted to a second valence electron which leaves the solid as a high-energy Auger electron. discontinuous manner whenever the energy of the X-rays becomes large enough to ionize an atomic core level of the atoms of the solid. Due to interatomic correlations, these energies di\ufb00er from the ionization energies of gaseous atoms, i.e. information about solid state binding energies is revealed. In X-ray \ufb02uorescence spectroscopy the radiation emitted by valence electrons recombining with core holes created by incident X-ray radiation is measured (Fig. 7.2(b)). This type of spectroscopy is frequently used to chemically analyze a sample, or to detect the presence of impurity atoms, i.e. di\ufb00erent elements have di\ufb00erent core/valence excitation energies. Peaks in the \ufb02uorescence spectrum at frequencies characteristic of individual atoms therefore identify the presence of these atoms in the target sample. In photoemission spectroscopy (PES) core electrons7 displaced by X-ray radiation are detected (Fig. 7.2(c)). The fully frequency/angle-resolved measurement of the photo-electron current, known as angle-resolved photoemission spectroscopy (ARPES), is one of the most important spectroscopic techniques in the experimental analysis of band structures. Auger spectroscopy is based on an interaction process of higher order (Fig. 7.2(d)). In this process, a core hole is created by irradiation by either X-rays or high-energy electrons. In a secondary process, part of the energy emitted by a recombining valence electron is transferred to another valence electron, which then leaves the atom and is detected. 7 To access valence electrons, soft X-ray radiation or hard UV radiation can be employed. 7.1 Crash course in modern experimental techniques 367 Neutron scattering: Thermal neutrons are scattered elastically or inelastically by a solid state target. Being a neutral particle, the neutron interacts only weakly with solid state constituents (i.e. magnetically, through its spin) and hence penetrates deeply into the sample. Owing to its particular energy dispersion, the neutron is tailor-made to the analysis of low-lying collective excitations (phonons, magnons, etc. \u2013 for example the data shown on page 78 were obtained by neutron spectroscopy). Just as with X-ray scattering, elastic neutron scattering can be employed to obtain crystallographic information. Unfortunately, the production of thermal neutrons requires a nuclear reactor, i.e. neutron scattering is an extremely expensive experimental enterprise", ". Magnetic resonance: A sample containing particles of non-vanishing moment is placed into a static (in practice, a slowly varying) magnetic \ufb01eld, strong enough to cause complete magnetic polarization. The sample is then exposed to an AC magnetic \ufb01eld perpendicular to the polarizing \ufb01eld. If the AC \ufb01eld frequency is in resonance with the Zeeman energy, magnetic transitions are resonantly induced. The observable e\ufb00ect is a strongly enhanced radiation loss of the AC \ufb01eld. In nuclear magnetic resonance (NMR) the nuclear spins of the sample are polarized. In solid state physics, NMR is applied to obtain information about the magnetic properties of the electronic states of the solid. Owing to the hyper\ufb01ne interaction of the electron spin and the nuclear spin, the e\ufb00ective magnetic \ufb01eld seen by the nucleus partially depends on the surrounding electron cloud. For example, in metals, the Pauli paramagnetic response of the electrons causes a characteristic shift of the spectral lines (as compared with the NMR spectra of nuclei in uncorrelated environments) known as the Knight shift. Analysis of this shift obtains information about the magnetic properties of the conduction electrons. Resonance spectroscopy of transitions between spin-polarized electron states, electron spin resonance (ESR), is frequently used in chemical analysis and molecular physics. spectroscopic techniques, e.g. M\u00a8ossbauer spectroscopy, For discussion of other positron\u2013electron annihilation spectroscopy (PES), muon scattering, electron energy loss spectroscopy (EELS), etc., we refer to the literature. Other experimental techniques There are a few experimental probes of condensed matter physics that do not \ufb01t comfortably into the three-fold \u201ctransport-thermodynamics-spectroscopy\u201d scheme discussed above. This applies, for example, to scanning tunneling microscopy (STM), a technique whose development by Binnig and Rohrer in the 1980s has triggered a revolution in the area of nanotechnology. 368 Response functions half Gerd Binnig 1947\u2013 and Heinrich Rohrer 1933\u2013 Awarded of one the 1986 Nobel Prize (together with E. Ruska, the the microscope) electron for \u201ctheir design of the scanning tunneling microscope.\u201d (Image Foundation.) The basic principle of STM is easily understood: a small tip, kept at a positive electrostatic potential, is brought in proximity to a surface. When the tip\u2013surface separation becomes comparable to atomic scales, electrons begin to tunnel from the substrate onto the tip. The resulting tunnel current is fed into a piezoelectric crystal that in turn levels the height of the tip. Through this mechanism, the surface\u2013tip separation can be kept constant, with an accuracy of fractions of typical atomic separations. A horizontal sweep of the tip then generates an accurate image of the surface pro\ufb01le. For example, Fig. 7.3 shows an STM image of a carbon nanotube. c The Nobel inventor of Figure 7.3 STM image of a carbon nanotube. (Figure courtesy of C. Dekker.) 7.2 Linear response theory In the previous section, we argued that condensed matter experiments typically probe the. Such linear (linear) response of a system to the application of weak perturbations response can be cast in terms of a generalized susceptibility \u03c7: Eq. (7.2). In the following we try to give the formal expression (7.2) a concrete meaning. Speci\ufb01cally,we relate the response function \u03c7 to microscopic elements of the theory familiar from previous chapters. However, before entering this discussion, let us list a few properties of \u03c7 that follow from common sense reasoning. F j 0 1 Causality: The generalized forces F j(t) cannot cause an e\ufb00ect prior to their action, i.e. \u03c7ij(r, r; t, t) = 0, t < t. Formally, we say that the response is retarded. If the system Hamiltonian does not explicitly depend on time (which will usually be the case), the response depends only on the di\ufb00erence of the time coordinates, \u03c7ij(r, r; t, t) = \u03c7ij(r, r; t \u2212 t). In this case it is convenient to Fourier transform the temporal convolution (7.2), i.e. to express the response in frequency space: Xi(r, \u03c9) = ddr \u03c7ij(r, r; \u03c9)F j(r, \u03c9) +O (F 2). (7.3) 7.2 Linear response theory 369 The important statement implicit in Eq. (7.3) is that, in the linear response regime, a (near) monochrom", "atic perturbation acting at a certain frequency \u03c9 will cause a response of the same frequency. For example, an AC voltage with frequency \u03c9 will drive an AC current of the same frequency, etc. We can read this statement in reverse to say that, if the system responds at frequencies = \u03c9, we have triggered a strong, nonlinear response. Indeed, it is straightforward to verify (exercise) that an expansion of the general functional R[F ] tonth order in F generates a response with frequency n\u03c9.8 According to Eq. (7.3), a peak in the response Xi(\u03c9) at a certain frequency \u03c9 indicates a local maximum of the response function, i.e. the presence of an intrinsic excitation with characteristic frequency \u03c9. For systems that are translationally invariant, the response function depends only on di\ufb00erences between spatial coordinates, \u03c7ij(r, r; t \u2212 t) = \u03c7ij(r \u2212 r; t \u2212 t). Spatial Fourier transformation then leads to the relation Xi(q, \u03c9) = \u03c7ij(q; \u03c9)F j(q, \u03c9) +O (F 2). (7.4) By analogy to what was said above about the frequency response, we conclude that a peak of the function Xi(q, \u03c9) signals the presence of an excitation with frequency \u03c9 and momentum q. We thus see that, at least in principle, linear response measurements are capable of exploring the full dispersion relation of the excitations of a system. This is as much as one can say on general grounds. In the following we will employ the \ufb01eld integral formalism to relate the response function to concrete microscopic properties of the system. EXERCISE Consider X-ray or neutron radiation probing a crystalline medium whose unit cells are spanned by vectors ai, i = 1,..., d. Show that the response function \u03c7 shows this periodicity through the condition \u03c7(k, k; \u03c9) \u221d \u03b4k\u2212k\u2212G, whereG belongs to the reciprocal lattice of the system. That is, the angle-resolved scattering pattern displays the full periodicity of the reciprocal lattice and, therefore, of the original lattice. This is, in a nutshell, the main principle behind spectroscopic crystallography. 7.2.1 Microscopic response theory We now set out to relate the response function to the microscopic elements of the theory. Previously we saw that, in quantum theory, the response signal X(t) should be interpreted as the expectation value of some (single-particle) operator9 \u02c6X = aXaa ca, wherec a aa c\u2020 8 While these \u201cside bands\u201d are usually negligible, they may become sizeable in, for example, laser-spectroscopic experiments. The \ufb01eld intensities reached by laser beams can be so large as to generate frequency-doubled or -tripled response signals. However, these phenomena, which belong to the realm of nonlinear optics, are beyond the scope of the present text. 9 For notational clarity, the indices i and r labeling a multi-component set of local response quantities Xi(r) will be dropped in this section (similarly with F j (r)). aa 370 Response functions may, as appropriate, represent bosonic or fermionic operators. Within the formalism of the \ufb01eld integral, the expectation value at imaginary times is thus given by X(\u03c4 ) = \u00af\u03c8a(\u03c4 )Xaa \u03c8a (\u03c4 ), (7.5) D( \u00af\u03c8, \u03c8) (\u00b7 \u00b7 \u00b7)exp{\u2212 S[F, \u00af\u03c8, \u03c8]} is the functional average where, as usual, \u00b7 \u00b7 \u00b7 = Z \u22121 over the action describing our system. The action of the system is given by S[F, \u00af\u03c8, \u03c8] = S0[ \u00af\u03c8, \u03c8] +\u03b4S [F, \u00af\u03c8, \u03c8], where S0 is the action of the unperturbed system and \u03b4S[F, \u00af\u03c8, \u03c8] = d\u03c4 \u02c6HF = d\u03c4 F (\u03c4 ) aa \u00af\u03c8a(\u03c4 )X aa \u03c8a (\u03c4 ), is the perturbation introduced by the action of the generalized force (cf. Eq. (7.1)). In practice, it is often convenient (for a better motivation, see the next section) to represent X(\u03c4 ) as a derivative of the free energy functional. To this end, let us formally couple our operator \u02c6X to a second \u201cgeneralized force\u201d and de\ufb01ne \u03b4S[F, \u00af\u03c8, \u03c8] \u2261 d\u03c4 F (\u03c4", " ) \u02c6X(\u03c4 ) = d\u03c4 F (\u03c4 ) \u00af\u03c8a(\u03c4 )Xaa \u03c8a (\u03c4 ), aa as a new element of our action. With S[F, F, \u00af\u03c8, \u03c8] =S 0[ \u00af\u03c8, \u03c8] + \u03b4S[F, \u00af\u03c8, \u03c8] + \u03b4S[F, \u00af\u03c8, \u03c8], we then have X(\u03c4 ) = \u2212 \u03b4 ln Z[F, F ], \u03b4F (\u03c4 ) F =0 where the notation Z[F, F ] = Z functionally depends on the two generalized forces. D( \u00af\u03c8, \u03c8) e\u2212S[F,F, \u00af\u03c8,\u03c8] indicates that the partition function Now, if it were not for the presence of the driving force F, the expectation value X would vanish. On the other hand, we also assume that F is weak in the sense that a linear approximation in F satisfactorily describes the measured value of X. Noting that the formal \ufb01rst-order expansion of a general functional G[F ] is given byG[ F ] G[0] + d\u03c4 \u03b4G[F ] \u03b4F (\u03c4 ) |F =0F (\u03c4 ), we can write X(\u03c4 ) \u2212 d\u03c4 \u03b42 \u03b4F (\u03c4 ) \u03b4F (\u03c4 ) F =F =0 ln Z[F, F ] F (\u03c4 ). Comparison with Eq. (7.2) then leads to the identi\ufb01cation \u03c7(\u03c4, \u03c4 ) = \u2212 \u03b42 \u03b4F (\u03c4 ) \u03b4F (\u03c4 ) ln Z[F, F ], F =F =0 of the response kernel. Carrying out the derivatives, we \ufb01nd \u03c7(\u03c4, \u03c4 ) = \u2212Z \u22121 + Z \u22121 \u03b42 \u03b4F (\u03c4 ) \u03b4F (\u03c4 ) \u03b4 \u03b4F (\u03c4 ) F =0 Z[F, F ] F =F =0 Z[0, F ] Z \u22121 \u03b4 \u03b4F (\u03c4 ) F =0 Z[F, 0]. 7.2 Linear response theory 371 Now, by construction, the second term in large parentheses is the functional expectation value \u02c6X(\u03c4 ) taken over the unperturbed action. We had assumed that this average vanishes, so that our preliminary, and still very formal, result for the response function is given by \u03c7(\u03c4, \u03c4 ) = \u2212Z \u22121 \u03b42 \u03b4F (\u03c4 ) \u03b4F (\u03c4 ) Z[F, F ]. F =F =0 (7.6) Performing the two derivatives, we obtain a more concrete representation of the response function in terms of a four-point correlation function: \u03c7(\u03c4, \u03c4 ) = \u2212 \u02c6X(\u03c4 ) \u02c6X (\u03c4 ) = \u2212 \u00af\u03c8a(\u03c4 ) \u02c6Xab\u03c8b(\u03c4 ) \u00af\u03c8a (\u03c4 ) \u02c6X ab \u03c8b (\u03c4 ). (7.7) ab ab EXERCISE Directly expand Eq. (7.5) to \ufb01rst order in the generalized force F to obtain this expression. As mentioned above, the usefulness of the derivative construction outlined above will become clear shortly. INFO Equation (7.7) indicates a connection between two seemingly very di\ufb00erent physical mechanisms. To disclose this relation, let us consider the case where the observed and the driving operator are equal: \u02c6X = \u02c6X. (Shortly, we will see that the important application to the electromagnetic response of a system falls into this category.) Using the vanishing of the equilibrium expectation values, \u02c6X(\u03c4 ) = 0, we can then rewrite (7.7) as \u03c7(\u03c4, \u03c4 ) = \u2212( \u02c6X(\u03c4 ) \u2212 \u02c6X(\u03c4 ))( \u02c6X(\u03c4 ) \u2212 \u02c6X(\u03c4 )). (7.8) This relation is called the \ufb02uctuation\u2013dissipation theorem. Indeed, the right-hand side of the relation clearly describes the quantum thermal \ufb02uctuation behavior of (the physical observable represented through) the operator \u02c6X. By contrast, the left-hand side is of dissipative nature. For example, for the case \u02c6X = j (see below), \u03c7 relates to the conductance of the system, i.e. the way in which the kinetic energy of a charge carrier is dissipated among the intrinsic excitations of the system. We might now proceed to evaluate this function by means of the machinery introduced in the previous chapters and that is, indeed, how the response function will be", " computed in practice. However, before doing this, we must face up to one more conceptual problem. What we are after is the real-time response X(t) to a real-time dynamical perturbation F (t). However, our functional integral formalism produces, naturally, an imaginary-time response \u03c7(\u03c4, \u03c4 ). In fact, we have frequently met with this problem before: while we are generally interested in real-time properties the formalism makes imaginary-time predictions. In previous chapters, we dealt with this problem by remembering that the imaginary-time setup could be obtained by analytical continuation t \u2192 \u2212i\u03c4 of the integration contour of a real-time functional integral. Reversing this \u201cWick rotation\u201d we argued that a real-time response X(t) can be extracted from an imaginary-time result X(\u03c4 ) by substitution \u03c4 \u2192 it. 372 Response functions In the majority of cases, this procedure indeed leads to correct results. However, sometimes one has to be more careful and that applies, in particular, to our present linear response calculus. Indeed, the simple substitution F (\u03c4 ) \u2192 F (t) = F (\u03c4 \u2192 it) is a crude shortcut of what mathematically should be a decent analytical continuation. Problems with this prescription arise when the answer F (\u03c4 ) generated by the functional integral contains singularities in the complex \u03c4 -plane.10 If a \ufb01ctitious contour interpolating between the limiting points \u03c4 and it inevitably crosses these singularities (see the \ufb01gure), the simple substitution prescription becomes a problem. The upshot of these considerations is that, before proceeding with the construction of the linear response formalism, we need to develop a better understanding of the mathematical structure of correlation functions. 7.3 Analytic structure of correlation functions It is the purpose of this section to clarify \u2013 at last \u2013 the connection between imaginary and real-time correlation functions. Throughout much of this section we will return to the traditional operator representation, i.e. expressions with circum\ufb02exes, \u02c6X, represent canonically quantized operators and \u00b7 \u00b7 \u00b7 = Z \u22121tr(\u00b7 \u00b7 \u00b7exp{\u2212 \u03b2[ \u02c6H \u2212 \u03bc \u02c6N ]}) represents the quantumthermal expectation value. Restricting ourselves to correlation functions of operators taken at two di\ufb00erent times,11 the general de\ufb01nition of the imaginary-time correlation function reads C \u03c4 X1X2 (\u03c41 \u2212 \u03c42) \u2261 \u2212T\u03c4 \u02c6X1(\u03c41) \u02c6X2(\u03c42) \u2261 \u2212 \u02c6X1(\u03c41) \u02c6X2(\u03c42), \u03b6 \u02c6X \u02c6X2(\u03c42) \u02c6X1(\u03c41), \u03c41 \u2265 \u03c42, \u03c42 > \u03c41, (7.9) where \u02c6Xi(c, c\u2020) are arbitrary operators represented in terms of either boson or fermion operators {c, c\u2020}. The time-dependence of these operators is de\ufb01ned through the imaginary-time Heisenberg representation, \u02c6X(\u03c4 ) \u2261 e\u03c4 ( \u02c6H\u2212\u03bc \u02c6N ) \u02c6Xe\u2212\u03c4 ( \u02c6H\u2212\u03bc \u02c6N ), (7.10) i.e. the imaginary-time analog of the real-time Heisenberg representation familiar from quantum mechanics.12 The role of the time-ordering operator T\u03c4, whose action is de\ufb01ned 10 More precisely, the quantity F (\u03c4 ) produced by evaluating the functional integral should be interpreted as a function de\ufb01ned on the imaginary axis of a complex-time domain. Analytical continuation then leads to a generalization F (z), where z may take values in some two-dimensional domain of the complex plane. What we imply when we substitute F (\u03c4 ) \u2192 F (\u03c4 \u2192 it) is that the analyticity domain includes the real axis, and that analytical continuation amounts to a simple re-substitution of the argument. Both assumptions may happen to be violated. 11 According to our discussion above, correlation functions of this type su\ufb03ce to explore the linear response of a system. Although the situation with nonlinear response signals is more complicated, much of what we are going to say below has general validity. 12 Within imaginary-time theory, it is customary to absorb the chemical potential into the dynamical evolution of an operator. 7.3 Analytic structure of correlation functions 373 through the second equality,13 is to chronologically order the two operators \u02c6X1(\u03c41) and \u02c6X2(\u03c42). Notice that for one-body operators \u02c6Xi = \u03b1[Xi]\u03b1\u03b2c\u03b2, the de\ufb01nition above describes \u2020", " the response functions discussed in the previous section. For \u02c6X1 = c\u03b1, \u02c6X2 = c \u03b2, Eq. (7.9) coincides with the one-body imaginary-time correlation functions (\u201cGreen functions\u201d) discussed earlier in this text, i.e. Eq. (7.9) and its real time descendants to be discussed next cover most of the correlation functions of relevance in many-body physics. \u03b1\u03b2 c\u2020 INFO Although we shall postpone the discussion of the connection to the \ufb01eld integral formalism for a while, it is instructive to compare the de\ufb01nition (7.9) with the \ufb01eld integral correlation function (7.7) de\ufb01ned above. Indeed, if the quantities \u02c6Xi(\u03c4i) in (7.9) were to be interpreted as the functional representation of second quantized operators, \u02c6X(a\u2020, a) \u2192 \u02c6X( \u00af\u03c8, \u03c8), the two correlation functions would coincide. The reason is that the time-ordering operation acting on the functional representation of an operator pair is redundant, T\u03c4 \u02c6X1( \u00af\u03c8(\u03c41), \u03c8(\u03c41)) \u02c6X2( \u00af\u03c8(\u03c42), \u03c8(\u03c42)) = \u02c6X1( \u00af\u03c8(\u03c41), \u03c8(\u03c41)) \u02c6X2( \u00af\u03c8(\u03c42), \u03c8(\u03c42)) (Exercise: Try to think why). In other words, the correlation function (7.9) reduces to (7.7) when the operators are represented within the \ufb01eld integral formalism. (The reason why T\u03c4 is not redundant for canonically quantized operators is that these have a non-vanishing commutator/anti-commutator at equal times.) In a manner that is di\ufb03cult to motivate in advance, we next introduce not one but three di\ufb00erent response functions of a real-time argument. Substituting in Eq. (7.9) real-time arguments for imaginary times, \u03c4 \u2192 it, we obtain the real-time response function C T X1X2 (t1 \u2212 t2) = \u2212iTt \u02c6X1(t1) \u02c6X2(t2), (7.11) where the factor of i has been introduced for later convenience, Tt chronologically orders real times, and \u02c6X(t) \u2261 eit( \u02c6H\u2212\u03bc \u02c6N ) \u02c6Xe\u2212it( \u02c6H\u2212\u03bc \u02c6N ) are real-time Heisenberg operators. While this expression appears to be the \u201cnatural\u201d generalization of (7.9), it is not our prime object of interest. Much more physical signi\ufb01cance is carried by the retarded response function C + X1X2 (t1 \u2212 t2) = \u2212i\u0398(t1 \u2212 t2)[ \u02c6X1(t1), \u02c6X2(t2)]\u03b6 \u02c6X, (7.12) i.e. an object that exists only for times t1 > t2 (hence the attribute \u201cretarded\u201d). The complementary time domain, t1 < t2, is described by the advanced response function C \u2212 X1X2 (t1 \u2212 t2) = +i\u0398(t2 \u2212 t1)[ \u02c6X1(t1), \u02c6X2(t2)]\u03b6 \u02c6X. (7.13) INFO What is the physical meaning of the real-time retarded response function? To address this question we need to reformulate the linear response arguments given above in the 13 The sign factor \u03b6 \u02c6X = \u00b11 depends on the statistics of \u02c6Xi, i.e. \u03b6 \u02c6X = 1 if \u02c6Xi are bosonic. Note that the operator \u02c6Xi is bosonic if {c, c\u2020} are Bose operators or if it is of even order in a fermion algebra. Conversely, \u03b6 \u02c6X = \u22121 if they are fermionic. 374 Response functions language of the canonical operator formalism. What we would like to compute is the expectation value X(t) = \u02c6X F (t), (7.14) building up in response to the presence of a weak perturbation F (t) \u02c6X in the Hamiltonian \u02c6H = \u02c6H0 + F (t) \u02c6X. In Eq. (7.14), the superscript \u201cF \u201d indicates that", " the time evolution of \u02c6X follows the full Hamiltonian (including the perturbation F (t) \u02c6X ). In contrast, the angular brackets represent a thermal average \u00b7 \u00b7 \u00b7 = Z \u22121 tr[(\u00b7 \u00b7 \u00b7 )exp{\u2212\u03b2 \u02c6H0}] that does not include the perturbation.14 The philosophy behind this convention is that, somewhere in the distant past, t \u2192 \u2212\u221e, the system was prepared in a thermal equilibrium distribution of the unperturbed Hamiltonian \u02c6H0. As time evolved, a perturbation \u221d F (t) was gradually switched on until it began to a\ufb00ect the expectation value of the dynamically evolved operator \u02c6X. (One need not be irritated by the somewhat arti\ufb01cial de\ufb01nition of the switching on procedure; all it tells us is that \u02c6X acts on a system in thermal equilibrium, an assumption that is not problematic if the perturbation is su\ufb03ciently weak.) To compute the expectation value, it is convenient to switch to a representation wherein the evolutionary changes due to the action of the perturbation are separated: X(t) = ( \u02c6U F )\u22121(t) \u02c6X(t) \u02c6U F (t), (7.15) (t) = \u02c6U \u22121 0 (t) \u02c6U (t), the evolution of \u02c6X follows the standard Heisenberg dynamics where \u02c6U F \u02c6X(t) = \u02c6U \u22121 0 (t) \u02c6X \u02c6U0(t), and \u02c6U ( \u02c6U0) generates the time evolution of the full (unperturbed) Hamiltonian. Using the de\ufb01ning equations of these evolution operators, it is straightforward to verify that \u02c6U F (t), i.e. the time evoluis controlled by the (Heisenberg representation of the) perturbation \u02c6X. According tion of \u02c6U F to conventional time-dependent quantum mechanical perturbation theory, the solution of this di\ufb00erential equation (with boundary condition \u02c6U F (t \u2192 \u2212\u221e) \u2192 1) is given by obeys the di\ufb00erential equation, dt \u02c6U F (t) =\u2212iF (t) \u02c6X (t) \u02c6U F t t \u02c6U F (t) = Ttexp \u2212i dt F (t) \u02c6X (t) 1 \u2212 i dt F (t) \u02c6X (t) +\u00b7 \u00b7 \u00b7 \u2212\u221e \u2212\u221e Substituting this result into Eq. (7.15) we obtain X(t) = \u2212i dt \u03b8(t \u2212 t)F (t) 4 [ \u02c6X(t), \u02c6X (t)] 5 = dt C + XX (t \u2212 t)F (t), i.e. the retarded response function turns out to generate the linear response of \u02c6X to the presence of the perturbation. In other words, the function C + is our prime object of interest, while all other correlation functions de\ufb01ned above play the (potentially important) role of supernumeraries. We next set out to explore the connection between the di\ufb00erent correlation functions de\ufb01ned above. In doing so, the principal question that should be at the back of our minds is \u201cHow do we obtain the retarded real time function C + provided we know the imaginary-time correlation function C \u03c4?\u201d The key to the answer of this question lies in a highly formal representation of the correlation functions C T,\u03c4,+,\u2212, known as the Lehmann representation. This representation is obtained by representing the correlation functions in terms of an exact eigenbasis {|\u03a8\u03b1} of the system: representing the trace entering the thermal expectation values as tr(\u00b7 \u00b7 \u00b7) = |\u03a8\u03b2\u03a8\u03b2| \u03a8\u03b1| \u00b7 \u00b7 \u00b7 |\u03a8\u03b1, and inserting a resolution of unity 1 = \u03b2 \u03b1 14 To simplify the notation, the chemical potential has been absorbed into the de\ufb01nition of \u02c6H. 7.3 Analytic structure of correlation functions 375 between the two operators appearing in the de\ufb01nition of the correlation function, it is straightforward to show that, e.g.15 C T (t) = \u2212iZ \u22121 \u03b1\u03b2 X1\u03b1\u03b2X2\u03b2\u03b1eit\u039e\u03b1\u03b2 \u0398(t)e\u2212\u03b2\u039e\u03b1 + \u03b6X \u0398(\u2212t)e\u2212\u03b2\u039e\u03b2. (7.", "16) Here, E\u03b1 is the eigenvalue corresponding to a state \u03a8\u03b1 and we have introduced the shorthand notations \u039e\u03b1 \u2261 E\u03b1 \u2212 \u03bcN\u03b1, \u039e\u03b1\u03b2 \u2261 \u039e\u03b1 \u2212 \u039e\u03b2, X\u03b1\u03b2 \u2261 \u03a8\u03b1| \u02c6X|\u03a8\u03b2. We next Fourier transform C T to \ufb01nd \u221e C T (\u03c9) = dt C T (t)ei\u03c9t\u2212\u03b7|t| = Z \u22121 \u2212\u221e \u03b1\u03b2 X1\u03b1\u03b2X2\u03b2\u03b1 e\u2212\u03b2\u039e\u03b1 \u03c9 + \u039e\u03b1\u03b2 + i\u03b7 \u2212 \u03b6 \u02c6X e\u2212\u03b2\u039e\u03b2 \u03c9 + \u039e\u03b1\u03b2 \u2212 i\u03b7, where the convergence-generating factor \u03b7 \u2013 which will play an important role throughout! \u2013 has been introduced to make the Fourier representation well-de\ufb01ned.16 Equation (7.16) is the Lehmann representation of the real time correlation function. What is the use of this representation? Clearly, Eq. (7.16) will be of little help for any practical purposes; the equation makes explicit reference to the exact eigenfunctions/states of the system. Should we have access to these objects, we would have a full solution of the problem anyway. Rather, the principal purpose of spectral resolutions such as (7.16) is to reveal exact connections between di\ufb00erent types of correlation functions and the analytical structure of these objects in general. To do so, we \ufb01rst need to compute the Lehmann representation of the other correlation functions. Proceeding as with the real time function above, it is straightforward to show that C T (\u03c9) C +(\u03c9) C \u2212(\u03c9) \u23ab \u23ac \u23ad = Z \u22121 \u03b1\u03b2 X1\u03b1\u03b2X2\u03b2\u03b1 \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 \u23a4 \u23a5 \u23a5 \u23a5 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 \u23ab \u23ac \u23ad i\u03b7. (7.17) \u2212 \u03b6 \u02c6X \u23ab \u23ac \u23ad i\u03b7 e\u2212\u03b2\u039e\u03b2 \u23a7 \u2212 \u23a8 + \u2212 \u23a9 \u03c9 + \u039e\u03b1\u03b2 e\u2212\u03b2\u039e\u03b1 \u23a7 \u23a8 + + \u2212 \u23a9 \u03c9 + \u039e\u03b1\u03b2 From this result, a number of important features of the correlation functions can be readily inferred. Anticipating the analytical structures alluded to above, we should think of C T,+,\u2212(z) as functions of a set of complex variables z. (The representations above apply to C T,+,\u2212(z = \u03c9) where\u03c9 is restricted to the real axis.) This extended interpretation allows us to view C T,+,\u2212 as complex functions with singularities in the immediate vicinity of the real axis. More speci\ufb01cally: The retarded correlation function C + has singularities for z = \u2212\u039e\u03b1\u03b2 \u2212 i\u03b7 slightly below the real axis. It is, however, analytic in the entire upper complex half plane Im(z) \u2265 0. 15 Wherever no confusion may arise, we omit the operator subscript CXX carried by the correlation functions. 16 Indeed, we can attach physical signi\ufb01cance to this factor. The switching on procedure outlined above can be implemented by attaching a small damping term exp(\u2212|t|\u03b7) to an otherwise purely oscillatory force. If we absorb dt F (t) (e\u2212t|\u03b7|ei\u03c9t)(\u00b7 \u00b7 \u00b7), this factor into the de\ufb01nition of all Fourier integrals, we arrive at the Fourier regularization mentioned above. dt (F (t)e\u2212t|\u03b7|)ei\u03c9t(\u00b7 \u00b7 \u00b7) \u2192 376 Response functions Conversely, the advanced correlation function C \u2212 has singularities above the real axis. It is analytic in the lower half plane Im(z) \u2264 0. Notice that C + and C \u2212 are connected through complex conjugation, C +(\u03c9) = % C \u2212(\u03c9) &\u2217. (7.18) The time-ordered correlation function has singularities on either side of the real axis (which makes it harder to analyze). The position of the singularities in the vicinity of the real axis contains important information about the fundamental excitations of the system (see Fig. 7.4). For example, consider the case where \u02c6X1 = c\u2020 a and \u02c6X2 = cb are one-particle creation and annihilation operators (no matter whether bosonic or fermionic).", " In this case, N\u03b1 \u2212 N\u03b2 \u2261 \u0394N = 1 (independent of the state indices \u03b1, \u03b2) and E\u03b1 \u2212 E\u03b2 is of the order of the single-particle energies of the system. (For a non-interacting system, E\u03b1 \u2212 E\u03b2 strictly coincides with the single-particle energies \u2013 exercise: why?) That is, the singularities of C T,+,\u2212 map out the single-particle spectrum of the system. This can be understood intuitively by remembering the meaning of the one-particle correlation function as the amplitude for creation of a state |a followed by the annihilation of a state |b at some later time. It is clear that the time Fourier transform of the amplitude, |a t\u2192 |b, becomes \u201clarge\u201d when the phase (\u223c \u03c9t) of the Fourier argument is in resonance with an eigenphase \u223c (E\u03b1 \u2212 E\u03b2)t supported by the system. (If you do not \ufb01nd this statement plausible, explore the simple example of a plane wave Hamiltonian.) Similarly, for a two-particle correlation function, \u02c6X1 \u223c c\u2020 acb, the energies E\u03b1 \u2212E\u03b2 describe the spectrum (the \u201cenergy cost\u201d) of two-particle excitations, etc. Notice that the single-particle spectrum can be continuous (in which case the functions C T,+,\u2212 have cuts parallel to the real axis), or discrete (isolated poles). Once one of the correlation functions is known, all others follow straightforwardly from a simple recipe: using the familiar Dirac identity, lim \u03b70 1 x \u00b1 i\u03b7 = \u2213i\u03c0\u03b4(x) +P 1 x, (7.19) singularities of C\u2013 singularities of C\u2013 Im z pole cut Re z Figure 7.4 Illustrating the singularities of advanced and retarded correlation functions in the complex plane. The points denote poles and the lines branch cuts. 7.3 Analytic structure of correlation functions 377 where P is the principal part, it is a straightforward matter to show that (exercise!) Re C T (\u03c9) = Re C +(\u03c9) = Re C \u2212(\u03c9), (7.20) and Im C T (\u03c9) = \u00b1Im C \u00b1(\u03c9) \u00d7 \u23a7 \u23aa\u23aa\u23a8 \u23aa\u23aa\u23a9 coth tanh \u03b2\u03c9 2 \u03b2\u03c9 2,, bosons, fermions, (7.21) i.e. the information stored in the three di\ufb00erent functions is essentially equivalent. After our discussion of the real-time correlation functions, the analysis of the imaginary-time function C \u03c4 is straightforward. The imaginary-time analog of Eq. (7.16) reads C \u03c4 (\u03c4 ) = \u2212Z \u22121 X1\u03b1\u03b2X2\u03b2\u03b1e\u039e\u03b1\u03b2 \u03c4 \u0398(\u03c4 )e\u2212\u03b2\u039e\u03b1 + \u03b6 \u02c6X \u0398(\u2212\u03c4 )e\u2212\u03b2\u039e\u03b2. (7.22) \u03b1\u03b2 Inspection of this representation for positive and negative times shows that C \u03c4 acquires the periodicity properties of the operators \u02c6X1,2: C \u03c4 (\u03c4 ) = \u03b6 \u02c6X C \u03c4 (\u03c4 + \u03b2), \u03c4 < 0. (7.23) Consequently C \u03c4 can be expanded in a Matsubara Fourier representation just like a con0 d\u03c4 C \u03c4 (\u03c4 )ei\u03c9n\u03c4, where, depending on the nature of the ventional operator, C \u03c4 (i\u03c9n) = operators \u02c6X1,2, \u03c9n may be a bosonic or a fermionic Matsubara frequency. Applying this transformation to the Lehmann representation (7.22), we obtain \u03b2 C \u03c4 (i\u03c9n) = Z \u22121 \u03b1\u03b2 % X1\u03b1\u03b2X2\u03b2\u03b1 i\u03c9n + \u039e\u03b1\u03b2 e\u2212\u03b2\u039e\u03b1 \u2212 \u03b6X e\u2212\u03b2\u039e\u03b2 &. (7.24) Our \ufb01nal task is to relate the four correlation functions de\ufb01ned through Eq. (7.17) and (7.24) to each other. To this end, we de\ufb01ne the \u201cmaster function\u201d C(z) = Z \u22121 \u03b1\u03b2 % X1\u03b1\u03b2X2\u03b2\u03b1 z + \u039e\u03b1\u03b2 e\u2212\u03b2\u039e\u03b1 \u2212 \u03b6X e\u2212\u03b2\u039e\u03b2 &, (7.25) depending on a complex argument z. When evaluated for z = \u03c9+, \u03c9\u2212, i\u03c9n, respectively, the function C(z) coincides withC +, C \u2212, C \u03c4. Further,C (z) is analytic everywhere except for the real axis. This knowledge su\ufffd", "\ufffd\ufffdces to construct the relation between di\ufb00erent correlation functions that was sought. Suppose then we had succeeded in computing C \u03c4 (i\u03c9n) = C(z = i\u03c9n) for all positive Matsubara frequencies.17 Further, let us assume that we had managed to \ufb01nd an analytic extension of C(z = i\u03c9n) \u2192 C(z) into the entire upper complex half plane Im z > 0. The evaluation of this extension on the in\ufb01nitesimally shifted real axis z = \u03c9 + i0 then coincides with the retarded Green function C +(\u03c9) (see \ufb01gure below). In other words, 17 Keep in mind that, in practical computations of this type, we will not proceed through the Lehmann represen- tation. 378 Response functions To \ufb01nd C +(\u03c9) we need to (i) compute C \u03c4 (i\u03c9n) for all positive Matsubara frequencies (e.g. by means of the thermal \ufb01eld integral) and then (ii) continue the result down to the real axis, i\u03c9n \u2192 \u03c9 + i0. The advanced Green function C \u2212 is obtained analogously, and by analytic extension the thermal correlation function C \u03c4 (i\u03c9n < 0), to frequencies with a negative o\ufb00set, \u03c9 \u2212 i0. \u03c9 n z These statements follow from a theorem of complex function theory stating that two analytic functions F1(z) and F2(z) coincide if F1(zn) = F2(zn) on a sequence {zi} with a limit point in the domain of analyticity. (In our case i\u03c9n \u2192 i\u221e is the limit point.) From inspection of (7.25) we already know that F1(i\u03c9n) \u2261 C +(\u03c9 \u2192 i\u03c9n) coincides with F2(i\u03c9n) = C \u03c4 (i\u03c9n). Thus any analytic extension of C \u03c4 must coincide with C + everywhere in the upper complex half plane, including the in\ufb01nitesimally shifted real axis. \u03c9 + i0 EXERCISE Writing z = \u03c9 \u00b1 i\u03b7, transform the spectral representation (7.25) back to the time d\u03c9 e\u2212i\u03c9tC(\u03c9 \u00b1 i\u03b7). Convince yourself that, for Im(z) positive (negative), domain: C(t) = 1 2\u03c0 the temporal correlation function C(t) contains a \u0398-function \u0398(t) (\u0398(\u2212t)). (Hint: Make use of Cauchy\u2019s theorem.) Importantly, the presence of this constraint does not hinge on \u03b7 being in\ufb01nitesimal. It even survives generalization to a frequency-dependent function \u03b7(\u03c9) > 0. (For the physical relevance of this statement, see below.) All that matters is that, for \u03b7 > 0, the function C(\u03c9 \u00b1 i\u03b7) is analytic in the upper (lower) complex half plane. This observation implies a very important connection between analyticity and causality: temporal correlation functions whose frequency representation is analytic in the upper (lower) complex half plane are causal (anticausal). (A time-dependent function is called \u201c(anti)causal\u201d if it vanishes for (positive) negative times.) How is the continuation process, required to \ufb01nd the retarded correlation function, carried out in practice? Basically, the answer follows from what was said above. If we know the correlation function C \u03c4 (i\u03c9n) for all positive Matsubara frequencies, and if that function remains analytic upon substitution C \u03c4 (i\u03c9n \u2192 z) of a general element of the full complex half plane, the answer is simple: we merely have to substitute i\u03c9n \u2192 \u03c9 + i0 into our result to obtain the retarded correlation function. Sometimes, however, we simply do not know C \u03c4 (i\u03c9n) for all positive frequencies. (For example, we may be working within an e\ufb00ective low-energy theory whose regime of validity is restricted to frequencies \u03c9n < \u03c9\u2217 smaller than some cut-o\ufb00 frequency.) In this case, we are in serious trouble. Everything then hinges on \ufb01nding a \u201cmeaningful\u201d model function that can be extended to in\ufb01nity and whose evaluation for small frequencies \u03c9n < \u03c9\u2217 coincides with our result; there are no generally applicable recipes for how to deal with such situations. INFO As a special case of great practical importance, let us brie\ufb02y explore the non-interacting single-particle Green function, i.e. the single-particle correlation function \u02c6X1 = ca, \u02c6X2 = c", "\u2020 a for a non-interacting system. (We assume that {|a} is an eigenbasis of the one-particle Hamiltonian.) As expected, these correlation functions assume a particularly simple form. In 7.3 Analytic structure of correlation functions 379 the non-interacting case, the eigenstates |\u03b1 = |a1, a2,... are symmetrical or antisymmetrical combinations of single-particle eigenstates |ai. Their energy is E\u03b1 = a1 + a2 + \u00b7 \u00b7 \u00b7, where a are the single-particle energies. Using the fact that E\u03b2 = E\u03b1 + a (exercise: why?) one then veri\ufb01es that the correlation function acquires the simple form C(z)| \u02c6X1=ca \u02c6X2=c \u2020 a \u2261 Ga(z) = 1 z \u2212 \u03bea, i.e. the partition function entering the de\ufb01nition of the general correlation function cancels against the thermally weighted summation over |\u03b1 (check!). Notice that the thermal version of this Green function, Ga(i\u03c9n) = (i\u03c9n \u2212 \u03bea)\u22121, appeared previously as the fundamental building block of perturbation theory. This is, of course, no coincidence: within the formalism of the \ufb01eld integral, the Green function appeared as the functional expectation value \u00af\u03c8a,n\u03c8a,n0 taken with respect to the Gaussian non-interacting action. But this object is just the functional representation of the operator correlation function considered above. Building on this representation, it is customary to introduce a Green function operator through the de\ufb01nition \u02c6G(z) \u2261 1 z + \u03bc \u2212 \u02c6H. By design, the eigenvalues of this operator \u2013 which are still functions of z \u2013 are given by the correlation function Ga(z) above. Numerous physical observables can be compactly represented in terms of the operator Green function. For example, using Eq. (7.19), it is straightforward to verify that the single-particle density of states of a non-interacting system is obtained as \u03c1() = \u2212 1 \u03c0 Im tr \u02c6G+(), (7.26) by taking the trace of the retarded Green function (operator). To illustrate the procedure of analytic continuation, let us consider a few elementary examples. 1. For the single-particle Green function ( \u02c6X1 = ca, \u02c6X2 = c\u2020 a) of an elementary excitation with energy a, Ga(\u03c9n) = (i\u03c9n \u2212 \u03bea)\u22121, the continuation amounts to a mere substitution, G+ a (\u03c9) = 1 \u03c9 + i0 \u2212 \u03bea. 2. We have seen that quasi-particle interactions lead to the appearance of a \u2013 generally complex \u2013 self-energy \u03a3(z): Ga(\u03c9n) \u2192 (i\u03c9n \u2212 \u03bea \u2212 \u03a3(i\u03c9n))\u22121, where we have simpli\ufb01ed the notation by suppressing the potential dependence of the self-energy on the Hilbert space index a. Extension down to the real axis leads to the relation G+ a (\u03c9) = 1 \u03c9+ \u2212 \u03bea \u2212 \u03a3(\u03c9+), (7.27) where \u03c9+ \u2261 \u03c9 + i0 and \u03a3(\u03c9+) is the analytic continuation of the function \u03a3(z) to the real axis. Although the speci\ufb01c structure of the self-energy depends on the problem under 380 Response functions consideration, a few statements can be made in general. Speci\ufb01cally, decomposing the self-energy into real and imaginary parts, we have Re \u03a3(\u03c9+) = + Re \u03a3(\u03c9\u2212), Im \u03a3(\u03c9+) = \u2212 Im \u03a3(\u03c9\u2212) < 0. (7.28) In words, the self-energy function has a cut on the real axis. Upon crossing the cut, its imaginary part changes sign. This important feature of the self-energy can be understood from di\ufb00erent perspectives. Formally, it follows from Eq. (7.18) relating the retarded and advanced Green functions through complex conjugation. More intuitively, the sign dependence of the imaginary part can be understood as follows. Suppose we start from a non-interacting imaginary-time formalism and gradually switch on interactions. The (Landau) principle of adiabatic continuity implies that nowhere in this process must the Green function \u2013 alias the propagator of the theory \u2013 become singular. This implies, in particular, that the combination i(\u03c9n \u2212 Im \u03a3i\u03c9n ) must not become zero, lest the dangerous real axis of the energy denominator be", " touched. The safeguard preventing the vanishing of the imaginary part of the energy denominator is that \u2212Im \u03a3 and \u03c9n have the opposite sign. Of course, this feature can be checked order by order in perturbation theory. Decomposing the self-energy into real and imaginary parts, \u03a3 = \u03a3 + i\u03a3, and transforming G+(\u03c9) back to the time domain we obtain G+(t) = e\u2212i\u03c9tG+(\u03c9) \u2248 e\u2212it(\u03bea+\u03a3)+t\u03a3 \u0398(t), d\u03c9 2\u03c0 where we have made the (over)simplifying assumption that the dependence of the selfenergy operator on \u03c9 is negligible: \u03a3(\u03c9) \u2248 \u03a3. EXERCISE Check the second equality above. If we interpret G+(t) as the amplitude for propagation in the state |a during a time interval t, and| G+|2 as the associated probability density, we observe that the probability to stay in state |a decays exponentially, |G+|2 \u221d e2t\u03a3 \u03c4 can be identi\ufb01ed as the inverse of the e\ufb00ective lifetime \u03c4 of state |a. The appearence of a \ufb01nite lifetime expresses the fact that, in the presence of interactions, single-particle states no longer represent stable objects, but rather tend to decay into the continuum of correlated manybody states. This picture will be substantiated in Section 7.3.1 below., i.e. 2\u03a3 \u2261 \u2212 1 3. Let us apply Eq. (7.26) to compute the BCS quasi-particle DoS of a superconductor. In Section 6.4 we saw that the thermal Gor\u2019kov Green function of a superconductor with spatially constant real order parameter is given by \u02c6G(i\u03c9n) = [i\u03c9n \u2212 ( \u02c6H \u2212 \u03bc)\u03c33 \u2212 \u0394\u03c31]\u22121. Switching to an eigenrepresentation and inverting the Pauli matrix structure, we obtain \u2212 1 \u03c0 tr \u02c6G(i\u03c9n) = 1 \u03c0 a tr i\u03c9n + \u03bea\u03c33 + \u0394\u03c31 \u03c92 n + \u03be2 a + \u03942 = 2i\u03c9n \u03c0 a 1 a + \u03942. n + \u03be2 \u03c92 Next, performing our standard change from a summation over eigenenergies to an integral, we arrive at \u2212 1 \u03c0 tr \u02c6G(i\u03c9n) 2i\u03c9n \u03c0 \u03bd d\u03be 1 n + \u03be2 + \u03942 = \u03c92 2i\u03c9n\u03bd n + \u03942 \u03c92, 7.3 Analytic structure of correlation functions 381 where, as usual, \u03bd denotes the normal density of states at the Fermi level. This is the quantity we need to continue to real frequencies. To this end we adopt the standard convention whereby the cut of the square root function is on the positive real axis, i.e. \u221a |r| for r positive real. Then, \u2212r + i0 =\u2212 \u2212r \u2212 i0 =i \u221a \u2212 1 \u03c0 tr \u02c6G(i\u03c9n \u2192 +) = \u221a 2+\u03bd \u2212+2 + \u03942 2\u03bd \u22122 \u2212 i0 sgn () + \u03942, where we anticipate that the in\ufb01nitesimal o\ufb00set of in the numerator is irrelevant (trace its fate!) and, making use of the fact that, for approaching the real axis, only the sign of the imaginary o\ufb00set matters: ( + i0)2 2 + 2i0 2 + 2i0sgn. Finally, taking the imaginary part of that expression, we arrive at the standard BCS form \u03bd() = Im 2\u03bd \u22122 \u2212 i0 sgn() + \u03942 = \u23a7 \u23a8 0, \u23a9 \u221a 2||\u03bd 2 \u2212 \u03942, || < \u0394, || > \u0394. 7.3.1 Sum rules and other exact identities In the next section we apply the analytical structures discussed above to construct a powerful theory of real time linear response. However, before doing so, let us stay for a moment with the formal Lehmann representation to disclose a number of exact identities, or \u201csum rules,\u201d obeyed by the correlation functions introduced in the previous section. Admittedly, this addition to our discussion above does not directly relate to the formalism of linear response, and readers wishing to proceed in a more streamlined manner are invited to skip this section at \ufb01rst reading. In fact, the formulae we are going to collect are not speci\ufb01c to any particular context and that is precisely their merit: identities based", " on the analytical structure of the Lehmann representation are exact and enjoy general applicability. They can be used (a) to obtain full knowledge of a correlation function from fragmented information \u2013 e.g. we saw in Eq. (7.20) and (7.21) how all three real time correlation functions can be deduced once any one of them is known \u2013 and, equally important, (b) to gauge the validity of approximate calculations. The violation of an exact identity within an approximate analysis is usually an indication of serious trouble, i.e. such deviations mostly lead to physically meaningless results. The spectral (density) function We begin by considering an object that carries profound physical signi\ufb01cance in its own right, especially in the area of strongly correlated fermion physics, i.e. (\u22122 times) the imaginary part of the retarded correlation function, A(\u03c9) \u2261 \u22122ImC +(\u03c9). (7.29) 382 Response functions Equation (7.29) de\ufb01nes the spectral function or spectral density function. Using Eq. (7.19) and the Lehmann representation (7.17) it is straightforward to verify that it has the spectral decomposition A(\u03c9) = 2\u03c0Z \u22121 X1\u03b1\u03b2X2\u03b2\u03b1 \u03b1\u03b2 % e\u2212\u03b2\u039e\u03b1 \u2212 \u03b6 \u02c6X e\u2212\u03b2\u039e\u03b2 & \u03b4(\u03c9 + \u039e\u03b1\u03b2). (7.30) INFO To understand the physical meaning of the spectral function let us consider the case where \u02c6X1,2 are single-particle creation/annihilation operators, \u02c6X1 = ca, \u02c6X2 = c\u2020 a. (We assume that the non\u2013interacting part \u02c6H0 of the Hamiltonian \u02c6H0 + \u02c6V is diagonalized in the basis {|a}.) For a non-interacting problem, \u02c6V = 0, it is then straightforward to show that Aa(\u03c9) = 2\u03c0\u03b4(\u03c9 \u2212 \u03bea), i.e. the spectral function is singularly peaked at the single-particle energy (measured from the chemical potential) of the state |a. Here, the subscript \u201ca\u201d indicates that we are dealing with a spectral function de\ufb01ned for a pair of one-body operators ca, c\u2020 a. Heuristically, the singular structure of Aa(\u03c9) can be understood by observing that, in the non-interacting case, the state c\u2020 a|\u03b1 obtained by adding a single particle |a to the many-particle state |\u03b1 is, again, an eigenstate of the system. In particular, it is orthogonal to all states by itself, whence the eigenstate summation over |\u03b2 contains only a single non-vanishing term. We say that the \u201cspectral weight\u201d carried by the (unit-normalized) state c\u2020 a|\u03b1 is concentrated on a single eigenstate of the system. What happens to this picture if interactions are restored? In this case, the addition c\u2020 a|\u03b1 of a single-particle state to a many-body eigenstate will, in general, no longer be an eigenstate of the system. In particular, there is no reason to believe that this state is orthogonal to all but one (N + 1) particle states |\u03b2. We have to expect that the spectral weight carried by the (still unit-normalized) state c\u2020 a|\u03b1 gets distributed over many, potentially a continuum, of states |\u03b2. It is instructive to explore the consequences of this phenomenon in the representation (7.27) where the e\ufb00ect of interactions has been lumped into a self-energy operator \u03a3. Taking the imaginary part of this expression, we \ufb01nd Aa(\u03c9) = \u22122 \u03a3(\u03c9) (\u03c9 \u2212 \u03bea \u2212 \u03a3(\u03c9))2 + (\u03a3(\u03c9))2, where \u03a3 and \u03a3 denote the real and imaginary part of the self-energy, respectively, and we have neglected the in\ufb01nitesimal imaginary o\ufb00set of \u03c9+ in comparison with the \ufb01nite imaginary contribution i\u03a3. The result above suggests that the net e\ufb00ect of interactions is an e\ufb00ective shift of the single-particle energy a \u2192 a + \u03a3(\u03c9) by the real part of the self-energy operator; interactions lead to a distortion of the single-particle energy spectrum, an e\ufb00ect that follows, for example, from straightforward perturbative reasoning. More importantly, the \u03b4-function obtained in the non-interacting case gets smeared into a Lorentzian (see Fig 7.", "5). In a sense, the spectral a |\u03b1 is distributed over a continuum of neighboring weight carried by the many-body states C + states, wherefore the spectral function loses its singular character. The width of the smearing interval is proportional to the imaginary part of the self-energy and, therefore, to the inverse of the lifetime \u03c4 discussed in the previous section. Notice that the smeared spectral function still obeys the normalization condition, d\u03c9 2\u03c0 Aa(\u03c9) = 1, as in the non-interacting case.18 This suggests an interpretation of A as a probability measure describing in what way the spectral weight carried by the state c\u2020 a|\u03b1 is spread out over the 18 Strictly speaking, we can integrate A only if the variance of \u03a3(\u03c9) over the interval [\u03bea + \u03a3 \u2212 \u03a3, \u03bea + \u03a3 + \u03a3] in which the Lorentzian is peaked is negligible (but see below). 7.3 Analytic structure of correlation functions 383 continuum of many-body states |\u03b2. To put that interpretation onto a \ufb01rm basis, not bound to the self-energy representation above, we consider the spectral decomposition (7.30). The positivity of all terms contributing to the right-hand side of the equation implies that Aa(\u03c9) > 0, a condition necessarily obeyed by any probability measure. To verify the general validity of the \u03c9 \u03b5 a \u03c9 V A Figure 7.5 Illustrating the meaning of the spectral function. For zero interaction, V = 0, the spectral weight is singularly concentrated on the single particle energies of the system. As interactions are switched on, the spectral weight is distributed among a continuum of states concentrated around a shifted center point. The width of the distribution increases with growing interaction strength. Dark solid curve: (schematic) shape of the spectral function if the self-energy carries a pronounced \u03c9 dependence. unit-normalization condition one may integrate Eq. (7.30) over \u03c9: ) ( Aa(\u03c9) =Z \u22121 ca\u03b1\u03b2c\u2020 a\u03b2\u03b1 e\u2212\u03b2\u039e\u03b1 \u2212 \u03b6ce\u2212\u03b2\u039e\u03b2 d\u03c9 2\u03c0 \u03b1\u03b2 \u239b = Z \u22121 \u239d cac\u2020 a \u03b1e\u2212\u03b2\u039e\u03b1 \u2212 \u03b6c \u03b1 c\u2020 aca \u03b2e\u2212\u03b2\u039e\u03b2 \u03b2 \u239e \u23a0 = Z \u22121 e\u2212\u03b2\u039e\u03b1 \u03b1| cac\u2020 \u03b1 \u03b1 aca a \u2212 \u03b6cc\u2020 \u2020 a]\u03b6c =1 [ca,c \u03b2 |\u03b1 = Z \u22121 \u03b1 1 = 1. e\u2212\u03b2\u039e\u03b1 Positivity and unit-normalization of Aa(\u03c9) indeed suggest that this function measures the distribution of spectral weight over the many-body continuum. To further substantiate this interpretation, consider the integral of the spectral function weighted by the Fermi- or Bose-distribution function: d\u03c9 2\u03c0 nF/B(\u03c9)Aa(\u03c9) =Z \u22121 = Z \u22121 = Z \u22121 \u03b1\u03b2 \u03b1\u03b2 \u03b2 ( ca\u03b1\u03b2c\u2020 a\u03b2\u03b1 e\u2212\u03b2\u039e\u03b1 \u2212 \u03b6ce\u2212\u03b2\u039e\u03b2 ) d\u03c9 \u03b4(\u03c9 + \u039e\u03b1\u03b2) 1 e\u03b2\u03c9 \u2212 \u03b6c ( ca\u03b1\u03b2c\u2020 e\u03b2\u039e\u03b2\u03b1 \u2212 \u03b6c a\u03b2\u03b1e\u2212\u03b2\u039e\u03b2 c\u2020 \u03b2 = \u02c6na, aca e\u2212\u03b2\u039e\u03b2 \u03b2 ) 1 e\u03b2\u039e\u03b2\u03b1 \u2212 \u03b6c which tells us that, if we weight the spectral density function of a single-particle state |a with the thermal distribution function, and integrate over all frequencies, we obtain the total occupation of that state. The relation d\u03c9 2\u03c0 nF/B(\u03c9)Aa(\u03c9) = \u02c6na, (7.31) 384 Response functions indeed states that Aa is a distribution function describing in what way the spectral weight of the state c\u2020 a|\u03b1 spreads over the continuum of exact eigenstates. A number of exact identities involving correlation functions are formulated in terms of the spectral function. We begin by showing that the spectral function carries the same information as the correlation function itself. (In view of the fact that A is obtained by removing the real part of C, this result might come as a surprise.) Indeed, starting from A(\u03c9) = i(C +(\u03c9) \u2212 C \u2212(\u03c9)), one may con\ufb01rm the relation C(z) = \u221e \u2212\u221e d\u03c9 2\u03c0 A(\u03c9) z \u2212 \u03c9. (7.32) Note that the \ufb01rst (second) term contributing to the right-hand side of the de\ufb01nition (a) is analytic in the upper (lower)", " half of the complex \u03c9-plane and (b) decays faster than \u03c9\u22121 for |\u03c9| \u2192 \u221e. For Im z > 0, the theorem of residues then implies that the C \u2212-term does not contribute to the integral. (Without enclosing singularities, the integration contour can be closed in the lower half plane.) As for the C +contribution, one may integrate over an in\ufb01nite semicircle \u03b3 closing in the upper half plane to obtain \u221e \u2212\u221e d\u03c9 2\u03c0 A(\u03c9) z \u2212 \u03c9 Im z>0= \u2212 1 2\u03c0i C +(\u03c9) z \u2212 \u03c9 d\u03c9 \u03b3 = C(z), (7.33) where the second identity relies on the analyticity of C + in the upper half plane. The case Im z < 0 is treated analogously. We thus \ufb01nd that knowledge of the imaginary part su\ufb03ces to reconstruct the full correlation function. (Notice, however, that the identity (7.32) is heavily \u201cnon-local\u201d; i.e. we need to know the spectral function for all \u03c9, including \u03c9 \u2192 \u00b1\u221e, to reconstruct the correlation function at a given value of z.) Considering the second equality in Eq. (7.33) and setting z = \u03c9+, one obtains C +(\u03c9) = \u2212 1 2\u03c0i d\u03c9 C +(\u03c9) \u03c9 \u2212 \u03c9 + i0. Representing the denominator under the integral in terms of the Dirac identity (7.19), and collecting terms, the identity assumes the form C +(\u03c9) = 1 \u03c0i d\u03c9C +(\u03c9)P 1 \u03c9 \u2212 \u03c9. It is customary to consider real and imaginary parts of this relation separately, whence one arrives at the celebrated Kramers\u2013Kronig or dispersion relations: Re C +(\u03c9) = 1 \u03c0 Im C +(\u03c9) = \u2212 1 \u03c0 d\u03c9 Im C +(\u03c9)P d\u03c9 Re C +(\u03c9)7.34). INFO To appreciate the physical content of the Kramers\u2013Kronig relations, let us anticipate our discussion below and note that the scattering amplitude of particles incident on a medium at 7.3 Analytic structure of correlation functions 385 energy \u03c9 is proportional to the retarded Green function C +(\u03c9). Taking as an example the scattering of electromagnetic radiation from a solid state sample, the Kramers\u2013Kronig relations then tell us that the real part of the scattering amplitude \u2013 the index of refraction \u2013 is proportional to the imaginary part \u2013 the index of absorption \u2013 integrated over all energies, i.e. the Kramers\u2013Kronig relations establish a connection between the two seemingly unrelated physical mechanisms of absorption and refraction. Other areas of application of dispersion relations include high energy physics, optics (both classical and quantum), and many more. In the next section we discuss some concrete applications of the Kramers\u2013Kronig relations in many-body physics. The dielectric function: a case study Equation (7.34) represents a \u201cmaster identity\u201d from which numerous other exact relations can be obtained. As an example of the derivation and usefulness of these identities, let us consider the frequency- and momentum-dependent dielectric function (q, \u03c9), i.e. the object describing the polarization properties of a medium in the presence of an electromagnetic \ufb01eld. In Section 5.2 we explored the dielectric function within the framework of the RPA approximation. However, as we are about to discuss exact relations, we should now be a bit more ambitious than that, i.e. we should base our discussion on a more generally valid representation of the dielectric function. Indeed, it is a straightforward exercise in linear response to show that (see Problem 7.6.2) (q, \u03c9) = 1 \u2212 V0(q) Ld d\u03c4 ei\u03c9m\u03c4 \u02c6n(q, \u03c4 )\u02c6n(\u2212q, 0)c \u22121 i\u03c9m\u2192\u03c9+, (7.35) where V0(q) = 4\u03c0e2/q2 is the bare Coulomb potential, \u02c6n\u02c6nc denotes the connected thermal average of two density operators \u02c6n(q, \u03c4 ) = c\u2020 q(\u03c4 )cq(\u03c4 ), and i\u03c9m \u2192 \u03c9+ indicates symbolically the analytical continuation to real frequencies. Heuristically, Eq. (7.35) can be understood by noting that 1/ = Ve\ufb00 /V0 measures the ratio between the e\ufb00ective potential felt by a test charge in a medium and the vacuum potential. The di\ufb00erence between these two quantities is due to the polar", "izability of the medium which, in turn, is a measure of its inclination to build up charge distortions \u03b4\u02c6n in response to the action of the potential operator \u223c dV0 \u02c6n. In linear response theory,19 \u03b4\u02c6n is given by the kernel d\u02c6n\u02c6nV0, i.e. the second term in Eq. (7.35). We thus observe that the (inverse of the) dielectric function is determined by the retarded correlation function C +(q, \u03c9) with \u02c6X1 = \u02c6X2 = \u02c6n. (For obvious reasons, this function is called the retarded density\u2013density response function. It appears as an important building block in many areas of many-body physics.) Building on the relation (q, \u03c9)\u22121 = 1 \u2212 V0(q) Ld C +(q, \u03c9), (7.36) 19 The linear response approximation is quite appropriate here because the standard de\ufb01nition of the dielectric function = limV0\u21920(V0/Veff ) implies an in\ufb01nitesimally weak external perturbation. 386 Response functions we next show how analyticity arguments and certain limit relations can be employed to derive strong test criteria for the dielectric functions. We \ufb01rst apply the Kramers\u2013Kronig relation to \u22121 \u2212 1 \u221d C + to obtain Re (q, \u03c9)\u22121 \u2212 1 = 1 \u03c0 d\u03c9 Im (q, \u03c9)\u22121 \u2212\u221e 1 \u03c9 \u2212 \u03c9. \u221e Using the fact that Im (q, \u03c9)\u22121 = \u2212Im (\u2212q, \u2212\u03c9)\u22121 = \u2212Im (q, \u2212\u03c9)\u22121, where the \ufb01rst identity holds for the Fourier transform of arbitrary real-valued functions and the second follows from real space symmetry, the integral can be brought to the compact form \u221e Re (q, \u03c9)\u22121 = 1 + 2 \u03c0 d\u03c9 Im (q, \u03c9)\u22121 0 \u03c9 \u03c92 \u2212 \u03c92. (7.37) We next consider this relation for the special case \u03c9 = 0 and |q| \u2192 0. The probing of this limit is motivated by the fact that the behavior of the static dielectric function (q, \u03c9 = 0) is signi\ufb01cantly more simple to analyze than that of the generic function (q, \u03c9). The reason is, of course, that a static external \ufb01eld does not prompt any dynamical response of the system. Indeed, one can show on general grounds (see the Info block below) that, in the static limit, (q, 0)\u22121 = (1 + 4\u03c0\u03bd|q|\u22122)\u22121 |q|\u21920\u2212\u2192 0. (7.38) Substitution of this result into Eq. (7.37) then leads to the identity E \u03bc \u221e lim q\u21920 0 d\u03c9 Im (q, \u03c9)\u22121 \u03c9 = \u2212 \u03c0 2. (7.39) Equation (7.39) is a typical example of a sum rule. For the derivation of a few more sum rules to be obeyed by the dielectric function see, e.g., the text by Mahan.20 \u2013Veff r INFO To understand the behavior of the dielectric function in the static limit, imagine a system of charged particles subject to a potential V0(r). Since V0 is static, it does not drive the system out of equilibrium. In particular, all fermionic quasi-particle states21 are \ufb01lled up to a uniform chemical potential \u03bc. This, however, necessitates a redistribution of charge. The reason is that the potential shifts the energies a of quasi-particle states |a according to the relation a \u2192 a \u2212 Ve\ufb00 (r), where Ve\ufb00 is the e\ufb00ective potential seen by the particles. (In order for this relation to make sense, the typical spatial extent of a quasi-particle must, of course, be smaller than the modulation range of V (r).) For low temperatures, states will be \ufb01lled up to an energy \u03bc\u2212Ve\ufb00 (r) (see the \ufb01gure). We can make this argument quantitative by introducing a distribution function ne\ufb00 (, r) \u2261 nF( \u2212 Ve\ufb00 (r)) locally controlling the occupation of quasi-particle states. In 20 G. Mahan, Many Particle Physics (Plenum Press, 1981). 21 We assume that we are dealing with a Fermi liquid, i.e. that we can think of the constituents of the", " system as fermionic quasi-particles, in the spirit of Landau\u2019s theory. 7.3 Analytic structure of correlation functions 387 a linear approximation, the induced charge density screening the external potential is then given by \u03c1ind(r) =\u2212 1 Ld a \u221e (ne\ufb00 (a, r) \u2212 nF(a)) = \u2212\u03bd d(ne\ufb00 (, r) \u2212 nF()) \u2212\u221e \u221e = \u2212\u03bd d(nF( \u2212 Ve\ufb00 (r)) \u2212 nF()) \u2212\u221e \u221e \u2248 \u03bd d \u2202nF()Ve\ufb00 (r) = \u2212\u03bdVe\ufb00 (r). \u2212\u221e To compute the di\ufb00erence between the external and the e\ufb00ective potential, respectively, we assume that the former had been generated through some charge density \u03c10: \u2212\u22022V0 = 4\u03c0\u03c10, or V0(q) = 4\u03c0q\u22122\u03c10(q). In contrast, the full potential Ve\ufb00 will be generated by a charge distribution \u03c1e\ufb00 comprising the external charge and the screening charge, Ve\ufb00 (q) = 4\u03c0q\u22122(\u03c10(q) +\u03c1 ind(q)) = V0(q) \u2212 4\u03c0\u03bdq\u22122Ve\ufb00 (q). We thus \ufb01nd that (q, 0) = V0(q) Ve\ufb00 (q) = 1 + 4\u03c0\u03bd|q|\u22122, as stated above. Indeed, we had obtained this result, generally known as Thomas\u2013Fermi screening, long before within the more microscopic framework of the RPA. In contrast, the merit of the present argument is that it is not based on any speci\ufb01c approximation scheme.22 While Eq. (7.39) was for the speci\ufb01c example of the dielectric function, the general construction recipe has much wider applicability: (a) a quantity of physical interest (here, the dielectric function) is represented in terms of a retarded response function which then (b) is substituted into a Kramers\u2013Kronig-type relation. This produces a frequency non-local connection between the response function at a given frequency to an integral (\u201csum\u201d) over all other frequencies. (c) This integral is evaluated for a reference frequency for which our reference quantity is generally known (here, \u03c9 \u2192 0). This produces an integral relation which should hold under very general conditions. As mentioned at the beginning of this section, sum rules play an important role as test criteria for physical approximation schemes. Experimental access to the spectral density function Earlier, we have seen that the spectral density function contains highly resolved microscopic information about a many-body system. But how do we access this information other than by theoretical model calculation? Interestingly, it turns out that the spectral function is not only central to theoretical analysis, but also directly related to a key experimental observable, the inelastic scattering cross-section. To appreciate this connection, consider again the prototypical setup of a scattering experiment shown in Fig. 7.1. The frequency- and angle-resolved scattering cross-section is a measure of the rate of transitions from the incoming state (, k) into an outgoing state (, k). To give the problem a quantum mechanical formulation, we \ufb01rst note that the full Hilbert space H of the system is the direct product of the Fock space, F, of the target system and the single-particle space H1 of the incoming particle species, H = F \u2297 H1. We assume that the interaction 22 However, the argument is a bit phenomenological and does rely on the Fermi liquid doctrine. 388 Response functions between the incoming particle and the constituent particles of the system is governed by an interaction Hamiltonian whose conventional (\ufb01rst quantized) real space representation reads \u02c6Hint = i V (\u02c6ri \u2212 \u02c6r). Here, \u02c6ri are the positions of the particles of the system and \u02c6r is the position of the incoming particle. For simplicity we assume that the interaction is point-like, i.e. V (\u02c6r\u2212\u02c6r) = C\u03b4(\u02c6r\u2212\u02c6r), where C is some constant. (The generalization to more general interaction potentials is straightforward.) The second quantized representation of the interaction operator then reads \u02c6Hint = C ddr \u03b4(\u02c6r \u2212 r)c\u2020(r)c(r), (7.40) where \u02c6r retains its signi\ufb01cance as a single-particle operator acting in the space H1. An alternative and, for all that follows, more convenient", " representation is given by \u02c6Hint = C ddr ddq (2\u03c0)d eiq(\u02c6r\u2212r)c\u2020(r)c(r) = C ddq (2\u03c0)d eiq\u00b7\u02c6r \u02c6\u03c1(q), (ddq/(2\u03c0)d)exp(iq \u00b7 r), and \u02c6\u03c1(q) \u2261 where we have made use of a plane wave representation of the \u03b4-function, \u03b4(r) = dr e\u2212iq\u00b7rc\u2020(r)c(r) describes density modulations in the target system of characteristic momentum q. Assuming, for simplicity, that the sample is kept at zero temperature (i.e. it is in its ground state), a scattering process of \ufb01rst order in the interaction Hamiltonian is described by the transition amplitude A(q) = \u03b2, k \u2212 q| \u02c6Hint|0, k, where |0 represents the ground state of the system, and |\u03b2 may be any exact eigenstate of the target system.23 Substitution of the representation of the interaction Hamiltonian above brings the transition amplitude to the form A(q) = \u03b2, k \u2212 q| \u02c6Hint|0, k \u221d \u03b2|\u02c6\u03c1q|0. (7.41) A \ufb01rst conclusion to be drawn from Eq. (7.41) is that the scattering amplitude probes density modulations in the bulk system. According to Fermi\u2019s Golden Rule, the transition rate associated with the scattering amplitude A(q) is given by P(q) = 2\u03c0 \u03b2 |\u03b2|\u02c6\u03c1(q)|0|2\u03b4(\u03c9 \u2212 \u039e\u03b20), (7.42) where \u039e\u03b20 = E\u03b2 \u2212 E0 > 0 is the excitation energy of |\u03b2 above the ground state and the \u03b4-function enforces energy conservation. The summation over \u03b2 re\ufb02ects the fact that only the beam of scattered particles is observed while the \ufb01nal state of the target remains unspeci\ufb01ed. 23 Using the nomenclature of Fig. 7.1, one might identify |0 with a ground state of zero collective momentum K = 0 and |\u03b2 = |K = q with a state that has absorbed the momentum of the scattered particle. The present discussion is more general in that it does not assume that the target eigenstates carry de\ufb01nite momentum. 7.4 Electromagnetic linear response 389 It is instructive to reformulate Eq. (7.42) in a number of di\ufb00erent ways. Representing the \u03b4-function as a time integral, 2\u03c0\u03b4(\u03c9) = dt exp(+i\u03c9t), one obtains P(q) = dt |\u03b2|\u02c6\u03c1(q)|0|2e+it(\u03c9\u2212\u039e\u03b2 ) \u03b2 0|ei( \u02c6H\u2212\u03bc \u02c6N )t \u02c6\u03c1(\u2212q)e\u2212i( \u02c6H\u2212\u03bc \u02c6N )t|\u03b2\u03b2|\u02c6\u03c1(q)|0 dt e+i\u03c9t0|ei( \u02c6H\u2212\u02c6\u03bcN )t \u02c6\u03c1(\u2212q)e\u2212i( \u02c6H\u2212\u03bc \u02c6N )t \u02c6\u03c1(q)|0 = dte+i\u03c9t0|\u02c6\u03c1(\u2212q, t)\u02c6\u03c1(q, 0)|0. \u03b2 dt e+i\u03c9t = = This clearly illustrates the connection between the observable scattering rate and the microscopic characteristics of the system, i.e. the rate P(q, \u03c9) is a measure of the dynamical propagation of density modulations of wavelength q at time scales \u223c \u03c9\u22121. To establish the connection to the previously developed apparatus of response functions, we introduce the abbreviation \u02c6\u03c1(q)\u03b1\u03b2 = \u03b1|\u02c6\u03c1(q)|\u03b2 and reformulate Eq. (7.42) as P(q) =\u22122 Im \u03b2 = \u22122 lim T \u21920 ImZ \u22121 \u03b1\u03b2 \u03c1(q)\u03b2\u03b1\u03c1(\u2212q)\u03b1\u03b2e\u2212\u03b2\u039e\u03b1 \u03c9+ + \u039e\u03b1\u03b2 \u03c1(q)\u03b20\u03c1(\u2212q)0\u03b2 \u03c9+ + \u039e0\u03b2 ImZ \u22121 \u03b1\u03b2 ImC +(\u03c9) = A(q, \u03c9), = \u22122 lim T \u21920 = \u22122 lim T \u21920 \u03c1(q)\u03b2\u03b1\u03c1(\u2212q)\u03b1\u03b2(e\u2212\u03b2\u039e\u03b1 \u2212 e\u2212\u03b2\u039e\u03b2 ) \u03c9+ + \u039e\u03b1\u03b2 where A(q", ", \u03c9) denotes the spectral density function evaluated for the density operators \u02c6X1 = \u02c6\u03c1(q), \u02c6X2 = \u02c6\u03c1(\u2212q). Here, we have made use of the fact that, for T \u2192 0, the Boltzmann weight exp(\u2212\u03b2\u039e\u03b1) projects onto the ground state. Similarly, for \u03c9 > 0, the contribution exp(\u2212\u03b2\u039e\u03b2) vanishes (exercise: why?). We thus note that information about the scattering cross-sections is also carried by a retarded real time response function. More speci\ufb01cally: The inelastic scattering cross-section for momentum transfer q and energy exchange \u03c9 is a direct probe of the spectral density function A(q, \u03c9). Although the derivation above did not exactly follow the linear response scheme, it had the same weak coupling perturbative \ufb02avor. (The golden rule is a \ufb01rst-order perturbative approximation!) While we derived our formula for the particular case of a short-range density coupling, it is clear that a more general beam\u2013target coupling mechanism would lead to an expression of the same architecture, i.e. a suitably de\ufb01ned retarded response function. 7.4 Electromagnetic linear response In the previous two sections we have assembled everything needed to compute the response of physical systems to moderately weak perturbations. We have learned how to linearize the response in the strength of the generalized force and to extract real time dynamical 390 Response functions information from imaginary time data. In this section we will illustrate the functioning of this formalism on undoubtedly its most important application, the response to a general electromagnetic \ufb01eld. The general setup of the problem is easily formulated: suppose a system of charged particles has been subjected to an electromagnetic signal represented through a scalar potential \u03c6(r, t) and/or a vector potential A(r, t). To simplify the notation, let us represent the perturbation through a (1 + d)-dimensional24 potential A\u03bc(x) = (\u03c6(x), A(x)), where x \u2261 (t, r) is a (1 + d)-dimensional space-time argument vector. The system will respond to this perturbation by a redistribution of charge, \u03c1(x) \u2261 \u02c6\u03c1(x), and/or the onset of current \ufb02ow j(x) \u2261 \u02c6j(x). Con\ufb01ning charge and vectorial current into a (1 + d)-dimensional generalized current vector j\u03bc = (\u03c1, j), our task is to identify the linear functional j = K[A] + O(A2) relating the current to its driving potential. Written more explicitly, j\u03bc(x) = t<t dx K\u03bc\u03bd(x, x)A\u03bd(x), (7.43) where the condition t < t indicates that the response is retarded. To compute the elements of the response tensor, we proceed according to the general recipe constructed in the preceding sections: we \ufb01rst employ the general formalism to derive the imaginary-time response K(r, r; i\u03c9n), and then continue to real frequencies, i\u03c9n \u2192 \u03c9 + i0. However, before formulating this programme in detail, let us derive two fundamental constraints to be ful\ufb01lled by the kernel K. Firstly, a gauge potential A\u03bc = \u2202\u03bcf (f is a function) cannot drive a physical current. Substituting this condition into Eq. (7.43), we \ufb01nd 0!= t<t dx K\u03bc\u03bd(x, x)\u2202\u03bdf (x) = \u2212 t<t dx (\u2202\u03bd x K\u03bc\u03bd(x, x))f (x). \u2190 \u2202\u03bd where the arrow indicates that Since f is arbitrary one may conclude that 0 = K\u03bc\u03bd the derivative acts from the right. Secondly, current conservation demands \u2202\u03bcj\u03bc = 0 or, transcribed to Eq. (7.43), 0!= x K\u03bc\u03bd(x, x)A\u03bd(x). Due to the arbitrariness of A\u03bc, t<t dx \u2202\u03bc this condition can only be generally valid if \u2192 \u2202\u03bc K\u03bc\u03bd = 0. Summarizing, Gauge invariance and particle number conservation demand the identity \u2192 \u2202\u03bc K\u03bc\u03bd = K\u03bc\u03bd \u2190 \u2202\u03bd= 0. We next turn to the derivation of the linear response kernel. Our starting point is an observation made already in Chapter 1 (and several times thereafter), namely the fact that 24 To not unnecessarily exclude important applications of the formalism to e\ufb00ectively two- or one-dimensional problems, the dimensionality of space is left unspeci\ufb01ed. 7.", "4 Electromagnetic linear response 391 the coupling of matter to the classical electromagnetic \ufb01eld is described by the relation25 j\u03bc = \u03b4Sc[A] \u03b4A\u03bc, (7.44) where the subscript \u201cc\u201d indicates that only the part of the action that couples matter to the \ufb01eld (but not the background action of the \ufb01eld) is di\ufb00erentiated. In this expression we are back to a Euclidean metric v\u03bcw\u03bc \u2261 v0w0 + v \u00b7 w (on account of the de\ufb01nition, j0 \u2261 i\u03c1, A0 \u2261 i\u03c6 of imaginary time current and vector potential, respectively). The additional i appearing in this convention mirrors the i of the temporal component of the imaginary-time vector (t, r) \u2192 (\u2212i\u03c4, r). Notice, however, that we are merely changing conventions, i.e. both the imaginary- and the real-time action contain the invariant contribution \u2212\u03c1\u03c6 + j \u00b7 A. To keep the notation simple, we will not indicate the di\ufb00erence between imaginary-time and real-time vectors explicitly; the appropriate convention is determined by the context in which we are operating. This appealingly simple structure (7.44) is all we need to write down a general expression for the linear response kernel. Comparison with Section 7.2.1 shows that F = F = A\u03bc, i.e. the generalized force, F, and the auxiliary \u201cforce\u201d F used to generate the expectation value of \u02c6j by di\ufb00erentiation are both given by A\u03bc. Equation (7.6) then tells us that K\u03bc\u03bd(x, x) = Z \u22121 \u03b42 \u03b4A\u03bc(x) \u03b4A\u03bd(x) Z[A] A=0, (7.45) which is one of the most important relations of microscopic response theory. Notice that, owing to the interchangeability of the two derivatives, K\u03bc\u03bd(x, x) =K \u03bd\u03bc(x, x). This symmetry in turn implies that \u2013 an observation repeatedly made before \u2013 gauge invariance, \u2190 \u2202\u03bd= 0, and particle number conservation, \u2192 \u2202\u03bc K\u03bc\u03bd = 0, represent di\ufb00erent sides of the K\u03bc\u03bd same coin. Equation (7.45) is a very general, but also very formal, result. To give it some meaning, it is instructive to evaluate the two derivatives for a few speci\ufb01c functionals Z[A], as described in the next section. Electromagnetic response of the microscopic theory As a particularly important example, let us consider the microscopic action of a Fermi or Bose system in the presence of an electromagnetic \ufb01eld. Ignoring the coupling of the \ufb01eld to the spin degrees of freedom, S[ \u00af\u03c8, \u03c8, A] = dx \u00af\u03c8\u03c3 \u2202\u03c4 + \u03c6 + (\u2212i\u2207 \u2212 A)2 \u2212 \u03bc + V0 \u03c8\u03c3 + Sint[ \u00af\u03c8, \u03c8], 1 2m 25 You may ask yourself why we use the relation (a) j\u03bc = \u03b4A\u03bc Sc[A] instead of a direct representation of linear coupling (b) Sc[A] = j\u03bcA\u03bc +... The reason is that in some prominent cases \u2013 the microscopic theory of charged particles discussed in Section 7.4 being the most prominent example \u2013 the current j\u03bc = j\u03bc[A] depends on A by itself. This means that A will enter the action non-linearly (viz. by a \u201cdiamagnetic term\u201d \u223c A2) and this in turn implies that (a) ceases to be a faithful representation of the more general (b). However, our theory of linear response is based on derivative operations such as (b) anyway and the above complication does not pose a signi\ufb01cant problem. 392 Response functions where V0 denotes a (\ufb01eld-independent) potential. With A0 = i\u03c6, Ai = (A)i, and 0 = i\u02c6\u03c1 = i \u00af\u03c8\u03c3\u03c8\u03c3, \u02c6jA i = (\u02c6j)i = \u02c6jA \u00af\u03c8\u03c3(\u2212i \u2194 \u2202i +Ai)\u03c8\u03c3, 1 2m (7.46) \u2194 \u2202i g \u2261 f \u2202ig \u2212 (\u2202if )g, we have \u02c6jA S[A] and the notation jA hints at where f the important fact that the current itself depends on the vector potential. Application of Eq. (7.45) then gives \u03bc = \u03b4 \u03b4A\u03bc K\u03bc", "\u03bd(x, x) =Z \u22121 \u03b42 \u03b4A\u03bc(x) \u03b4A\u03bd(x) \u03b4 \u03b4A\u03bd(x) A=0 A=0 D( \u00af\u03c8, \u03c8) e\u2212S[ \u00af\u03c8,\u03c8,A] D( \u00af\u03c8, \u03c8) \u02c6jA \u03bc (x) e\u2212S[ \u00af\u03c8,\u03c8,A] = \u2212Z \u22121 = \u2212\u03b4(x \u2212 x)\u03b4\u03bc\u03bd \u2202A\u03bc(x) \u03bc (x) + \u02c6jA \u02c6jA \u03bc (x)\u02c6jA \u03bd (x), A=0 where the angular brackets denote functional averaging. Setting A = 0, one obtains K\u03bc\u03bd(x, x) = \u2212 \u02c6\u03c1(x) m \u03b4(x \u2212 x)\u03b4\u03bc\u03bd(1 \u2212 \u03b4\u03bc0) + \u02c6jp \u03bc(x)\u02c6jp \u03bd (x). (7.47) The \ufb01rst term contributing to the response kernel is known as the diamagnetic term. Indeed, tracing back through the derivation, one discovers that this term originates from the diamagnetic contribution \u223c A2 to the Hamiltonian. Conversely, \u02c6jp \u2261 \u02c6j A=0 is called the paramagnetic contribution to the current operator. The functional expectation value \u223c \u02c6jp\u02c6jp de\ufb01nes the paramagnetic term. Notwithstanding its unappealing structure, Eq. (7.47) represents a strong result which encompasses virtually all aspects of electromagnetic linear response.26 For example, the diagonal vectorial components Kii, i = 1, 2,..., d, of the response tensor describe the longitudinal conductivity of the system (see Section 7.4.1). In cases where a magnetic \ufb01eld is present, the o\ufb00-diagonal components Ki=j measure the Hall conductivity. The temporal components K00 describe the density response of the system, a feature that is of importance, for example, in the analysis of scattering data, etc. Electromagnetic response of e\ufb00ective theories In cases where we are operating on a level beyond the microscopic description, i.e. within the framework of an e\ufb00ective low-energy theory, the structure of the response tensor may di\ufb00er from Eq. (7.47). To identify the \u201ce\ufb00ective\u201d response tensor, we need to carry out a canonical two-step programme: (i) identify the coupling of the electromagnetic potential to the relevant degrees of freedom (in cases where one has not kept track of the \ufb01eld dependence from the very beginning, it usually su\ufb03ces to minimally couple the \ufb01eld, i.e. to introduce the components A\u03bc so as to make the action gauge invariant); (ii) perform the two-fold derivative Eq. (7.45). 26 Except for spin-related phenomena. 7.4 Electromagnetic linear response 393 We illustrate the procedure on a familiar example, the phase action of the BCS superconductor, # S[\u03b8] = dx \u03bd(\u2202\u03c4 \u03b8)2 + $ (\u2207\u03b8)2. ns 2m (7.48) Under a gauge transformation induced by some function f (x), \u2202\u03bc\u03b8 \u2192 \u2202\u03bc\u03b8 + \u2202\u03bcf, i.e. the dx [\u03bd(\u2202\u03c4 \u03b8 \u2212 \u03c6)2 + gauge invariant extension of the action reads (cf. Eq. (6.34)) S[\u03b8, A] = 2m (\u2207\u03b8 \u2212 A)2]. Di\ufb00erentiating this result with respect to the components Ai, one obtains ns \u03b4ij \u2212 ns m Kij(x, x) = \u2212 ns m \u2202i\u03b8(x)\u2202j\u03b8(x) (7.49) $ #. EXERCISE Evaluate the correlator \u2202i\u03b8 \u2202j\u03b8 to show that the conductivity \u03c3ii(q = 0, \u03c9) diverges in the limit \u03c9 \u2192 0. The divergence of the conductivity can be traced back to the fact that, in a superconductor, there is no cancellation between diamagnetic and paramagnetic responses. Compute the response of the system to a static magnetic \ufb01eld. Show that, in the London gauge \u2207 \u00b7 A = 0, current and vector potential are proportional to each other, j \u221d A. Recall (cf. Section 6.4) that this result, in combination with the Maxwell equation \u2207 \u00d7H = 4\u03c0j, implies", " the Meissner e\ufb00ect. 7.4.1 Longitudinal conductivity of the disordered electron gas Equations (7.47) and (7.49) exemplify the structure of the response tensor after the master formula (7.45) has been applied. However, to obtain a concrete result for the conductivity, say, one must evaluate the two-(\u02c6\u03c1) or four-(\u02c6j\u02c6j) point correlation functions appearing in these expressions. Of course, the details of this last step of the program depend sensitively on the microscopic structure of the theory under consideration, i.e. no universally applicable computational recipe can be formulated. Notwithstanding these di\ufb00erences in details, a number of large-scale structural elements are recurrent in the mathematical analysis of the current\u2013current correlation functions of microscopic response theory. To illustrate these common structures, let us consider an example that is of considerable importance in its own right, namely the longitudinal AC conductivity of the electron gas. The AC conductivity is de\ufb01ned through the relation j(\u03c9) = \u03c3(\u03c9)E(\u03c9), (7.50) i.e. as the coe\ufb03cient relating the current density to a homogeneous external \ufb01eld oscillating with frequency \u03c9. In the absence of symmetry-breaking perturbations such as a strong magnetic \ufb01eld or system-intrinsic anisotropies, the current \ufb02ow will be in line with the \ufb01eld gradient. This justi\ufb01es the scalar, rather than a more complex tensorial ansatz for the conductivity in Eq. (7.50) (For a more complex situation, see Problem 7.6.4.) INFO Before entering the details of the linear response calculation, it is instructive to recapitulate how the conductivity can be obtained from common sense reasoning. To this end, let us look at the world from the persective of an individual conduction electron. In the presence of an electric \ufb01eld, the electron will be subject to two di\ufb00erent forces: the force of the \ufb01eld, \u2212eE, and a dissipative friction force \u2212 m \u03c4 \u02d9r inhibiting its free acceleration (deviating from our convention, 394 Response functions e = 1, we temporarily reintroduce the electron charge). The damping rate is essentially set by the scattering rate o\ufb00 static impurities, viz. 1/\u03c4. To a \ufb01rst approximation, the dynamics of the electron is thus described by the equation of motion m\u00a8r = \u2212eE(t) \u2212 m \u03c4 \u02d9r or, in Fourier representation, \u2212im\u03c9v(\u03c9) = eE(\u03c9) \u2212 m \u03c4 v(\u03c9). Solving for v one \ufb01nds that the current density j = \u2212nev (n is the particle density) is given by j(\u03c9) = \u03c3(\u03c9)E(\u03c9) with \u03c3(\u03c9) = ne2 m 1 \u2212 i\u03c9. 1 \u03c4 (7.51) In the limit \u03c9 \u03c4 \u22121, \u03c3(\u03c9) i ne2 m \u03c9\u22121, i.e. the electron current is determined by the ballistic motion of the electrons in the electric force \ufb01eld. Conversely, in the DC limit \u03c9 \u2192 0, one obtains the Drude formula \u03c3 = ne2\u03c4 m, i.e. the \ufb01eld drives a steady current density whose value is limited by the amount of momentum relaxation. Below we compute the conductivity under the simplifying assumption that Coulomb or other types of many-body interaction are negligible.27 By contrast, the discussion of the Info block above shows that any meaningful analysis of the conductivity must take account of the presence of static disorder; without disorder, the \ufb01eld would make the electrons freely accelerate, i.e. there would be no such thing as a steady current \ufb02ow in metals. Thus, what we should have in mind when we think about the conductivity is \u03c3V, the conductivity averaged over all realizations of a microscopic disorder potential.28 To obtain the conductivity, we will explore the current \ufb02ow in response to a vector potential A, where \u2202tA = \u2212E generates the electric \ufb01eld. Anticipating that, after averaging over impurities, the system will be e\ufb00ectively translationally invariant, the analysis is most economically carried out in Fourier space, i.e. we compute the expectation value j(q \u2192 0, \u03c9) in response to a potential A(q \u2192 0, \u03c9), where the limit q \u2192 0 indicates that we are interested in a spatially homogeneous external \ufb01e", "ld. Fourier transforming Eq. (7.47) in the di\ufb00erence of coordinates, x \u2212 x (cf. the remarks made in connection with Eq. (7.4)), one veri\ufb01es that the appropriate response kernel is given by K(q) = \u2212 1 m 1 Ld \u02c6\u03c1V + \u02c6jp i,q \u02c6jp i,\u2212q V, (7.52) where the vectorial index i is arbitrary and the double brackets indicate the two-fold average over the quantum thermal distribution and the disorder potential. Also notice that the connection \u03c9mA(q) = E(q) implies \u03c3(\u03c9) = \u2212 lim q\u21920 1 \u03c9m K(q). i\u03c9m\u2192\u03c9+i0 (7.53) 27 Nonetheless, the calculation will still be fairly technical. If you \ufb01nd the Drude conductivity of the electron gas too elementary an observable to justify these e\ufb00orts, please keep in mind that our prime motivation is methodological: technical operations very similar to those detailed here appear in practically every quantum response analysis. 28 A good question to ask is whether the conductivity of any particular realization of a disordered metal will di\ufb00er from the averaged conductivity, i.e. whether the conductivity is a \u201cself-averaging\u201d quantity. This leads one to the interesting problem of conductance statistics, a topic beyond the scope of the current text. 7.4 Electromagnetic linear response 395 To process this expression, we substitute the Fourier transform of Eq. (7.46) (exercise), \u02c6\u03c1 = p \u00af\u03c8\u03c3,p\u03c8\u03c3,p, \u02c6jp i,q = 1 2m p (2p + q)i \u00af\u03c8\u03c3,p\u03c8\u03c3,p+q, into (7.52) and apply Wick\u2019s theorem to compute the thermal expectation value. This leads to the expression K(q) = \u2212 2T Ld (2p + q)i(2p + q)i (2m)2 Gn+m,p+q,p+qGn,p,p V Gn,p,pV + \u03b4p,p m, n,p,p (7.54) where the factor of two accounts for the spin summation (for simplicity we assume that the Hamiltonian is spin-independent), Gn,p,p \u2261 \u00af\u03c8n,p\u03c8n,p is the thermal electron Green function for a particular realization of the disordered background, and the remaining bracket stands for the disorder average. (Notice that, before disorder averaging, the system lacks translational invariance, i.e. the single-particle Green function is not diagonal in momentum space.) A diagrammatic representation of the paramagnetic response kernel in terms of Green functions is shown in the upper part of the \ufb01gure below, where the wavy lines denote the current operator and the dashed lines symbolically represent the scattering o\ufb00 the static impurity potential (cf. our discussion of the impurity scattering and Green functions in Section 6.5.) To make further progress with this expression, we shall adopt an approximation that critical readers may \ufb01nd questionable: we will replace the average of the two-Green-function correlator by an impurity average of the individual Green functions, GG \u2192 GG.29 In Section 6.5 it has been shown that these averaged Green functions are given by Gn,p,p = \u03b4p,p Gp \u2261 \u03b4p,p i\u03c9n \u2212 \u03bep + i 2\u03c4 sgn \u03c9n, where \u03c4 de\ufb01nes the mean impurity scattering time. (Recall what has been said above about the meaning of the imaginary part of the Green function denominator as an e\ufb00ective inverse lifetime. In the present context, \u03c4 measures the time after which a particle with initial momentum p gets scattered into states of di\ufb00erent momentum.) Substituting this result into Eq. (7.54), we obtain K(q) = \u2212 2T Ld p 1 m Gp + 1 (2m)2 (2p + q)2 i Gp+qGp. (7.55) 29 Readers who have navigated through the disorder section in Chapter 6 are invited to critically assess the validity of this approximation. The result of this analysis is that, for hard impurity scattering potentials, i.e. for impurity potentials p|V |p = const. that scatter isotropically over the entire momentum shell, the approximation above becomes justi\ufb01able in the limit q \u2192 0. To understand why, construct any diagram involving impurity lines connecting the two Green functions. You will notice that, in this case, the", " two fast momenta carried by the current operators are no longer locked to each other. The subsequent angular integration over these momentum vectors then leads to a vanishing of the diagram. The physical mechanism behind this vanishing is that, in a disordered medium, the velocity vector of an electron changes rapidly, i.e. a velocity\u2013velocity (or current\u2013current) correlation function is highly susceptible to large-angle impurity scattering. 396 Response functions Looking at the structure of Eq. (7.54), one might be tempted to perform the frequency summation by one of the standard formulae. However, a moment\u2019s thought shows that this strategy will not work. The reason is that, in the present case, the extension of the Green function to the complex plane, Gp = 2\u03c4 sgn Im z Gp(z) =, has a cut along in turn, means that the product the real axis; this, Gp+q(z + i\u03c9m)Gp(z) has cuts at Im(z) = 0 and Im(z + i\u03c9m) = 0 (see the \ufb01gure). To circumvent this di\ufb03culty, we need to integrate over a contour that (a) encompasses all Matsubara frequencies and (b) avoids the cut lines. While there is no simple contour that does the job, a joint integration over the three contours \u03b31,2,3 shown in the \ufb01gure faithfully represents the Matsubara sum: z \u2212 \u03bep + i \u22121 \u03c9n GpGp+q = \u2212 \u03b2 2\u03c0i < \u03b31\u222a\u03b32\u222a\u03b33 dz nF(z)Gp(z)Gp+q(z + i\u03c9m). The three integrals contributing to the sum lead to very di\ufb00erent results: as shown in the exercise below, the two integrals over the contours \u03b31 and \u03b33 cancel against the diamagnetic contribution to the response tensor, i.e. the \ufb01rst term in Eq. (7.55). INFO To appreciate why the contribution from the contours \u03b31 and \u03b33 is unphysical, notice that, e.g. \u03b2 2\u03c0i, \u03b31 dz nF(z)Gp(z)Gp+q(z + i\u03c9m) = \u221e \u2212\u221e \u03b2 2\u03c0i dnF()Gp( + i0)Gp+q( + i\u03c9m) i\u03c9m\u2192\u03c9+i0 \u2212\u2192 \u221e \u2212\u221e \u03b2 2\u03c0i d nF()Gp( + i0)Gp+q( + \u03c9 + i0), i.e. the integral extends over a product of two retarded single-particle Green functions. Now, remember that at the end of the day we want to compute the conductance, i.e. a quantity that resembles a transition probability (namely the probability for electrons propagating under the in\ufb02uence of an applied electric \ufb01eld). In quantum mechanics, probabilities \u03c3 \u223c GG\u2217 appear as absolute squares of amplitudes, G. Indeed, we have seen before that the retarded Green function can be interpreted as a transition amplitude of quantum mechanical particles. We also saw that its complex conjugate, G\u2217, is an advanced Green function. Put di\ufb00erently, our admittedly very hand-waving argument indicates that the quantum dynamics of conduction should be described in terms of products of advanced and retarded Green functions. However, a product of two retarded (or two advanced) Green functions lacks an obvious physical interpretation. Indeed, these contributions cancel against the diamagnetic term. 7.4 Electromagnetic linear response 397 We thus focus on the integral over \u03b32: dz nF(z)Gp(z)Gp+q(z + i\u03c9m) \u03b2 2\u03c0i < \u03b32 i\u03c9m\u2192\u03c9+i0 \u2212\u2192 \u221e = \u03b2 2\u03c0i \u03b2 2\u03c0i = \u2212 \u03b2 2\u03c0i \u2212\u221e \u221e \u2212\u221e \u221e \u2212\u221e d nF() [\u2212Gp( \u2212 i0)Gp+q( + i\u03c9m) +G p( \u2212 i\u03c9m)Gp+q( + i0)] % d nF() \u2212Gp(\u2212)Gp+q(+ + \u03c9) +G p(\u2212 \u2212 \u03c9)Gp+q(+) & d [nF() \u2212 nF( + \u03c9)]Gp(\u2212)Gp+q(+ + \u03c9), where, in the \ufb01rst equality, we have used the symmetry nF( + i\u03c9m) = nF() of the distribution function under translation by bosonic", " frequencies. Substituting this result into Eq. (7.55) and (7.53), we arrive at the intermediate result \u03c3(\u03c9) = \u221e d \u2212\u221e 1 2\u03c0 nF() \u2212 nF( + \u03c9) \u03c9 2 Ldm2 p where G\u00b1 p () = \u2212 \u03bep \u00b1 i 2\u03c4 \u22121. i G\u2212 p2 p ()G+ p ( + \u03c9), (7.56) EXERCISE To complete our analysis, we need to understand better the cancellation of the diamagnetic term and, relatedly, the role played by the integration contours \u03b31,3. We \ufb01rst write K(q) = K(0) + [K(q) \u2212 K(0)] and observe that (i) the contribution of \u03b32 discussed above sits solely in the \ufb01nite q term, K(q) \u2212 K(0) (the contour \u03b32 collapses to zero at q = 0), while (ii) the diamagnetic term and the contribution of the curves \u03b31,3 are in K(0). (Finite q-corrections to the contour integrals along \u03b31,3 vanish due to the strong convergence of the integrands and the fact that all singularities lie on one side of the real axis: think about this point.) The cancellation of the diamagnetic term against the contribution of \u03b31,3 thus amounts to showing K(0) = 0. Introducing the notation Gn, \u2261 Gp|p=, show that K(0) = \u2212 2T m n d \u03bd() 1 + \u2202 Gn,, 2 d where \u03bd() = 1 p F (p)p by the isotropy of Ld space. Integrating by parts (why do the contributions of the integration boundaries vanish?), and using the fact that \u03bd() = const. \u00d7 (d\u22122)/2, show that K(0) = 0. p \u03b4( \u2212 p), and we used p F (p)p2 i = 2m d The analytical structure of this result actually reveals a number of important elements of the electron conduction process: an AC \ufb01eld of frequency \u03c9 creates electron\u2013hole pairs of excitation energy \u03c9. The phase volume accessible to these processes is measured by the -integral, weighted by the di\ufb00erence of two Fermi functions, i.e. Fermi statistics demands that the energy of the electron to be excited lies within a shell [\u03bc\u2212\u03c9, \u03bc] at the Fermi surface (see the \ufb01gure below). The dynamics of the excited electron is described by a retarded Green function of energy + \u03c9. 398 Response functions Conversely, the dynamics of the hole is described by the advanced Green function. Heuristically, this can be understood by noting that the advanced Green function describes the (\ufb01ctitious) propagation of an electron backwards in time. However, an electron propagating in a chronologically reversed direction can be interpreted as a hole moving forward in time. Thus, the advanced electronic Green function e\ufb00ectively represents a descriptor of hole dynamics. That the product of two Green functions is weighted by two operators pi/m re\ufb02ects the fact that current transport essentially depends on the velocity of the electrons. \u03c9 To perform the sum over p we note that, for a radially symmetric function F (\u03bep), ddp p2 i F (\u03bep) = 1 d ddp p2F (\u03bep) = 2m d d\u03be (\u03be + \u03bc)\u03bd(\u03be)F (\u03be). Identifying the function F with the product of our two Green functions, we obtain 2 Ld p i Gp(\u2212)Gp(+ + \u03c9) = p2 4m (2\u03c0)dd 4m\u03bc\u03bd (2\u03c0)dd \u221e \u2212\u03bc \u221e \u2212\u221e d\u03be (\u03be + \u03bc)\u03bd(\u03be)G\u2212 \u03be ()G+ \u03be ( + \u03c9) d\u03be G\u2212 \u03be ()G+ \u03be ( + \u03c9) = 1 \u2212 i\u03c9, 1 \u03c4 where, in the third equality, we have used the fact that the Green functions are strongly peaked on scales \u03c4 \u22121 \u03bc around \u03be = 0 implying that, to a very good approximation, the \u03be-integration can be extended to the entire real axis and the energy variation of the density of states \u03bd(\u03be) on the e\ufb00ective interval of integration is negligible. The last integral can be performed by elementary means or, more elegantly, by closing the \u03be-integration contour in the upper or lower half plane and using the analytic structure of the Green functions. d (nF() \u2212 Subst", "ituting the result into our formula for \u03c3(\u03c9), and using the fact that nF( + \u03c9))\u03c9\u22121 1, we arrive at the \ufb01nal result Eq. (7.51). At this point it is rewarding to pause for a moment to look back at a number of large-scale structures of the analysis: On the microscopic level, the two-point function \u02c6j\u02c6j of current operators is represented in terms of a four-point function \u223c \u00af\u03c8\u03c8 \u00af\u03c8\u03c8 of \ufb01eld operators. A moment\u2019s thought shows that four-point correlation functions generically appear as microscopic descriptors of the response functions of quantum single-particle operators (simply because, in second quantization, a single-particle operator maps onto a bilinear of \ufb01eld operators) and, therefore, of the vast majority of response functions. This observation implies a hierarchy of correlation functions. For example, in our non-interacting example, the retarded response correlation function was described in terms of a product of an advanced and a retarded microscopic single-particle correlation function.30 This almost trivial observation is of some importance because it entails 30 For interacting problems, the response functions no longer neatly split into a product of two single-particle Green functions. Instead, the four \ufb01eld operators describing the response functions can be connected by an arbitrarily complicated network of interaction vertices. However, ultimately, the propagators connecting this network are again the single-particle Green functions of the problem. 7.6 Problems 399 a separation of energy scales. The argument (\u03c9, q) of the response function is a measure of the frequency/momentum scale of the external perturbation. These parameters are to be distinguished from the argument (, p) of the single-particle propagators microscopically describing the response process. In many cases, the two sets of scales are parametrically di\ufb00erent. For example, in the problem discussed above, (\u03c9, q) is set by the resolution of an external electronic apparatus and, therefore, much slower than the frequency/momentum scale (, p) of electronic charge carriers at the Fermi surface. In these circumstances, the ratios \u03c9/ and |q|/|p| de\ufb01ne small expansion parameters which can be used in the approximate evaluation of the response kernel. (In other cases, e.g. spectroscopic analysis of microscopic single-particle excitations, it is essential that (\u03c9, q) and (, p) be of the same order.) 7.5 Summary and outlook This completes our preliminary survey of theory and application of correlation functions in many-body physics. We have seen that correlation functions, notably real-time retarded response functions, represent a principal interface between experiment and theory. The connection between these objects and experimental data \u2013 the latter represented through expectation values of certain operators \u2013 was established by a formalism known as linear response: assuming an experimentally imposed perturbation of a many-body system to be weak, its response can be analyzed within a controlled expansion scheme whose leadingorder term is identi\ufb01ed as a retarded correlation function. A number of general properties of the response function \u2013 e.g. causality \u2013 could be identi\ufb01ed from common sense reasoning. However, to develop a full understanding of this object, notably the connection between the imaginary time response function (the quantity produced by evaluation of the \ufb01eld integral) and its retarded real time counterpart (the quantity we are interested in), we had to go beyond that level. Indeed, we saw that the relation between various types of correlation function could be revealed by means of a powerful spectral representation that was simple, exact, and not dependent on any particular physical context (while not being particularly useful for practical computational purposes). The structural elements developed early in the chapter were \ufb01nally bundled into a powerful theory of electromagnetic linear response. We illustrated the application of this machinery on the example of the longitudinal conductivity of the electron gas and pointed out a number of computational elements common to most concrete microscopic linear response calculations. In the next chapter we return to the intrinsic development of the theory. We become acquainted with the renormalization group, a versatile analytical tool designed to analyze correlation functions (and other objects of physical interest) within the vast space left open between straightforward perturbation theory at one end and the few available exact evaluation schemes at the other. 400 Response functions 7.6 Problems 7.6.1 Orthogonality catastrophe This problem describes a scenario realized frequently in condensed matter in which a spatially localized quasi-particle is immersed in a large host system (the latter described by some N -body ground state wavefunction). The resulting wavefunction, \u201csingle-particle state \u2297 N -particle ground state,\u201d turns out to be largely orthog", "onal to the true ground state of the now (N + 1)-particle system. In practice, this means that it takes a very long time for the intruding particle to become accommodated to its new environment, a retardation mechanism known as the orthogonality catastrophe. Following Anderson\u2019s original line of argument we here explore the basic physical mechanism underlying the phenomenon. In the thermodynamic limit, Anderson31 has shown that the ground state |\u03a8 of a system in the presence of a local perturbation V (r) is orthogonal to the ground state |\u03a8 of the unperturbed system: |\u03a8|\u03a8|2 N\u2192\u221e\u2212\u2192 0. (Here, the local perturbation serves as a caricature of the interaction potential created by a new particle in the system.) This mechanism implies, for example, that transport coe\ufb03cients measured in setups where a sample is connected to external leads at local contacts often determine a \u201ccontact resistance\u201d (determined by the resistance of the sample to accommodate external charge carriers) rather than the bulk conduction properties of the system. (a) Consider a system of non-interacting fermions. Show that the overlap between the many-particle ground states of the unperturbed system and of the perturbed system is bounded from above: \u02dc\u03c7 \u2261 |\u03a8|\u03a8|2 < \u03c7 = exp \u23a1 \u23a3\u2212 1 2 \u23a4 n\u2264EF,m >EF |n|m|2 \u23a6 \u2261 exp[\u2212I], where |n and |m are single-particle states of the unperturbed and perturbed system, respectively, and EF is the Fermi energy. To obtain this inequality, recall that the noninteracting many-particle states are Slater determinants. What is the overlap between ground states? To simplify the resulting expression, use the fact that det A \u2264 1 ifA has normalized, but non-orthogonal rows. Here one may assume that, if the perturbation is small, |n|m|2 1 for n \u2264 EF. m >EF (b) We next compute the exponent I for the particularly simple case of a spherically symmetric system of radius R. Focusing on the sector of lowest angular momentum l = 0, the unperturbed states are given by \u03c6n(r) = Nn sin(knr)/knr, where kn = \u03c0n/R and the normalization Nn = kn/ 2\u03c0R. The asymptotic pro\ufb01le of the perturbed wavefunctions can be approximated as \u03c6 n(r) = Nn sin(knr + \u03b4m(1 \u2212 r/R))/knr, where \u03b4 is the \u221a 31 P. W. Anderson, Infrared catastrophe in Fermi gases with local scattering potentials, Phys. Rev. Lett. 18 (1967), 1049\u201351. 7.6 Problems 401 s-wave scattering phase shift. Use this ansatz to compute \ufb01rst the overlap matrix A and then the exponent I. Answer: (a) In the ground state, all single-particle levels with energies below the Fermi energy EF are \ufb01lled while all levels with higher energies are empty. Thus, the overlap of ground states is given by \u03c7 = det A, whereA nm = n|m and n, m < EF. De\ufb01ning a row|n|m|2, we conclude that normalized matrix \u02dcA by \u02dcAnm \u2261 Anm/Nn with Nn = \u03c7 = det \u02dcA Nn. We next take the logarithm of this relation to obtain Nn \u2264 + m n n \u239b \u239e ln \u03c7 \u2264 ln (Nn) = 1 2 \u239d ln |n|m|2 \u23a0. n<EF m <EF n<EF Using the closure relation m |mm| = 1 and expanding the logarithm in x = m >EF |n|m|2 1 we arrive at the approximation \u239b \u239b \u239e \u239d ln n<EF m <EF |n|m|2 \u23a0 = n<EF \u2212 \u239d1 \u2212 ln |n|m|2 \u239e \u23a0 m >EF |n|m|2 = \u22122I. n<EF,m >EF (b) The overlap matrix A is given by Anm = n|m = 4\u03c0NnNm R 0 sin dr r2 sin(knr) knr 1 kn \u2212 km + \u03b4m R kmr + \u03b4m(1 \u2212 r R ) kmr sin \u03b4m \u03c0(n \u2212 m) +\u03b4 m", "in turn implies that the overlap |\u03a8|\u03a8|2 vanishes as some negative power of N. sin2 \u03b4m (n\u2212m+ \u03b4m n<nF,m>nF \u03c0 )2 7.6.2 RPA dielectric function Much of the response of a system of charged fermions to an external electromagnetic perturbation is encoded in the dielectric function q. While the dielectric function cannot be computed rigorously, the most common approximation scheme \u2013 asymptotically exact in the limit of the in\ufb01nitely dense gas \u2013 is the RPA. It is the purpose of this problem to elaborate upon the various sporadic encounters with the RPA dielectric function we had in the text. Speci\ufb01cally, we will derive Eq. (7.35) and discuss the 402 Response functions \u03b5\u2013+ 1+ = \u2013 = 1+ Figure 7.6 Representation of the dielectric function in terms of the full density\u2013density correlation function (the dark-shaded bubble) and the interaction-irreducible density\u2013density correlation function (light-shaded bubble). generalization to a disordered electron gas. The problem is physically instructive and a good exercise in diagrammatic calculus. (a) Derive Eq. (7.35). To this end, consider two in\ufb01nitesimally weak (yet generally timedependent) test charges immersed in a system of charged fermions. Omitting the selfinteraction of the particles, determine the interaction correction to the free energy and use your result to compute the ratio between the vacuum and the actual interaction potential. Show that q = 1 \u2212 T V0(q) Ld \u22121 \u02c6nq \u02c6n\u2212qc, (7.57) where the subscript \u201cc\u201d indicates that only connected diagrams contribute. Analytic continuation to real frequencies obtains Eq. (7.35). (b) Equation (7.57) represents a rigorous yet quite formal result for the dielectric function; it involves the exact density\u2013density correlation function which is generally unknown. As a \ufb01rst step towards a more manageable expression, show that Eq. (7.35) is equivalent to q = 1 + T V0(q) Ld \u02c6nq \u02c6n\u2212qirr, (7.58) where \u02c6n\u02c6nirr is the interaction-irreducible density\u2013density response function, i.e. the sum of all diagrams that cannot be cut into two just by cutting one interaction line (i.e. \u02c6n\u02c6nirr is to the e\ufb00ective interaction what the self-energy is to the Green function). In order to show that Eq. (7.57) and (7.58) are equivalent, \ufb01nd a series expansion of C(q, i\u03c9m) in terms of V0 and \u03a0(q, i\u03c9m). The series is very similar to the RPA series. Note, however, that, in contrast to RPA, the bubble \u02c6n\u02c6nirr still contains interaction lines. A graphical representation might be helpful. Answer: 7.6 Problems 403 (a) If we represent the two test charges with two charge distributions \u03c11,2(q), the Coulomb interaction becomes Sint = T q[\u02c6nq +\u03c11,q +\u03c12,q]V0(q)[\u02c6n\u2212q +\u03c11,\u2212q +\u03c12,\u2212q]. An expan2Ld sion of the \ufb01eld integral (or, equivalently, the partition function) to lowest order in \u03c11,2 then produces the interaction contribution to the free energy, Fint[\u03c11, \u03c12] = \u2212 T 2 Ld q \u03c11,q V0(q) \u2212 T V 2 0 (q) Ld \u02c6n\u2212q \u02c6nqc \u03c12,\u2212q. The term in rectangular brackets determines the e\ufb00ective interaction potential Ve\ufb00 (q). Remembering that q = V0(q)/Ve\ufb00 (q), we obtain Eq. (7.57). (b) For a diagrammatic representation of the solution, see Fig. 7.6. 7.6.3 Electromagnetic response of a quantum dot In Problem 6.7 we considered a model of quantum dots coupled by tunnel contacts to external leads. There, we derived an e\ufb00ective action for this system but did not say much about its actual physical behavior. As a \ufb01rst step towards a better understanding of the electromagnetic properties of small quantum dots, we here derive a formula elucidating the classical response", " of the system. In Problem 8.8.2, we then apply renormalization group methods to study how quantum \ufb02uctuations a\ufb00ect the low-temperature physics of the system. (Before addressing this problem, refamiliarize yourself with the derivation of the dissipative tunneling action in Problem 6.7.) (a) Consider Eq. (6.67) for a quantum dot coupled to a lead. Assuming that a bias voltage iU (\u03c4 ) between lead and dot has been applied, show that the argument of the dissipative action 0 d\u03c4 U (\u03c4 ). Focusing on the regime of high temperatures, generalizes to \u03c6(\u03c4 ) \u2192 \u03c6(\u03c4 ) + expand the action to second order in \u03c6 (argue why, for high temperatures, anharmonic \ufb02uctuations of \u03c6 are small) and compute the 00-component of the linear response kernel, K00. Analytically continue back to real frequencies and show that the current \ufb02owing across the tunnel barrier in response to a voltage U (\u03c9) is given by \u03c4 I(\u03c9) = U (\u03c9) \u2212 2 EC i\u03c9. 2\u03c0 gT (7.59) (b) To understand the meaning of this result, notice that, classically, the dot is but a capacitor connected via a classical resistor to a voltage source. Apply Kirchho\ufb00\u2019s laws to obtain the current\u2013voltage characteristics of the classical system and compare with the result above. 404 Answer: Response functions (a) Consider the action of the system at an intermediate stage, immediately after the fermions have been integrated out (cf. Eq. (6.68)). In the presence of a bias voltage, the \u201ctr ln\u201d contains a term i(V + U ), where V is the Hubbard\u2013Stratonovich \ufb01eld decoupling the interaction. Removing the sum of these two \ufb01elds by the same gauge transformation as used in the derivation of the dissipative action, the tunneling matrix elements acquire the phase factor speci\ufb01ed above. At second order in the expansion in \u03c6, the action assumes the form S(2)[\u03c6, V ] = 1 4ECT m \u03c92 m |\u03c6m|2 + gT 4\u03c0T m |\u03c9m||\u03c6m + Um/i\u03c9m|2, where the Fourier identity |\u03c9m| \u2194 \u2212\u03c0T sin\u22122(\u03c0T \u03c4 ) has been used. Di\ufb00erentiating twice with respect to U, we obtain (K00)m = Z \u22121T \u03b42 \u03b4Um\u03b4U\u2212m 6 Z[U ] = \u2212 gT 2\u03c0|\u03c9m| 1 \u2212 gT |\u03c9m| 2\u03c0T 7 \u03c6m\u03c6\u2212m = \u2212 gT 2\u03c0|\u03c9m| 1 \u2212 gT |\u03c9m| 2\u03c0T 1 2EC T + gT |\u03c9m| \u03c92 m 2\u03c0T = \u2212 1 2\u03c0|\u03c9m| gT + 2EC. The 00-element of the linear response tensor describes changes in the particle number32 in response to an applied potential, \u03b4Nm = (K00)mUm. Noting that i\u2202\u03c4 \u03b4N = I is the current through the barrier, and substituting our result for K00, we obtain Im =. Analytical continuation from positive imaginary frequencies |\u03c9m| \u2192 Um \u2212i\u03c9 + 0 to real retarded frequencies gives Eq. (7.59). + 2EC |\u03c9m| \u22121 2\u03c0 gT (b) According to Kirchho\ufb00\u2019s laws, the sum of all voltage drops in the system must equal zero, 0 = U (\u03c9) +C \u22121\u03b4N (\u03c9) \u2212 RI(\u03c9), where C and R denote the capacitance and resistivity, respectively. Using the fact that \u03b4N = I/i\u03c9, and solving for I, we readily obtain I(\u03c9) = U (\u03c9) R\u22121/i\u03c9C, i.e. the current voltage characteristics of a classical RC-circuit. Identi\ufb01cations R = 2\u03c0/GT and C = 1/2EC bring us back to the formula derived microscopically above. 7.6.4 Hall conductivity In this problem we derive microscopic expressions for the current response of a (generally disordered) electron gas in the presence of a perpendicular magnetic \ufb01eld. Our results will be the starting point for the analysis of the quantum Hall e\ufb00ect in the next chapter. Consider a two-dimensional electron gas in the presence of a perpendicular magnetic \ufb01eld. 2m (\u02c6p\u2212 \u02c6A)", "2 + \u02c6V, where \u02c6V is a one-particle The system is described by the Hamiltonian \u02c6H = 1 32 Here, we notice that the factor of \u201cvolume\u201d discriminating between particle number and density is contained in our de\ufb01nition of the source variables. 7.6 Problems 405 potential (disorder and/or boundary potential) and \u02c6A = B 2 \u02c6r \u00d7 e3 is the vector potential of the external \ufb01eld. To compute the conductivity tensor, we start from the real-space33 expression of (cf. Eq. (7.47) and (7.53)), \u03c3ij = \u2212 1 \u03c9mL2 2 m d2x d2x \u03b4ij\u03b4(x \u2212 x)\u02c6\u03c10(x) + \u02c6jm,i(x)\u02c6j\u2212m,j(x), i\u03c9m\u2192+i0 where \u02c6ji is de\ufb01ned in Eq. (7.46).34 Using Wick\u2019s theorem to compute the expectation value of the current and density operators, we obtain \u03c3ij = 1 \u03c9m T L2 n d2x d2x \u22122 m \u03b4ij\u03b4(x \u2212 x)Gn(x, x) +(\u02c6vi \u02c6Gn+m)(x, x)(\u02c6vj \u02c6Gn)(x, x), i\u03c9m\u2192+i0 m (\u2212i\u2207 \u2212 \u02c6A) is the velocity operator and we have set the electron charge to where \u02c6v = 1 unity. Anticipating the presence of cuts on the real energy axis, the frequency summation is split into three parts, \u03c3 = \u03c3\u2212\u2212 + \u03c3+\u2212 + \u03c3++, where \u03c3\u00b1\u00b1 denotes the contribution for which the frequency indices n + m and n are positive or negative, respectively. (a) Beginning with the diagonal contribution to the conductivity tensor, \u03c3xx, show that the paramagnetic contributions to \u03c3\u2212\u2212 xx + \u03c3++ xx cancel against the diamagnetic term. (Hint: Use the fact that for n \u00b7 (n + m) > 0 the product of two Green functions is shortrange: Gn+mGnV = Gn+mV GnV, where, as usual, GV decays on the scale of the mean free path (i.e. unlike the combination Gn<0Gn+m>0, no long-range quantum interference processes contribute to the product of two Green functions of identical signature). Due to this short-rangeness, we may approximate the position operator entering the de\ufb01nition of the current operator by a c-number valued vector whose value is set by spatial arguments x x of G(x, x). Consequently, both \u02c6G and \u02c6j are diagonalized by the basis of momentum states.) (b) Owing to the vanishing of \u03c3++ xx and \u03c3\u2212\u2212 xx, the longitudinal contribution to the conducxx is obtained by summation over the sector of mixed frequency tance tensor \u03c3xx = \u03c3+\u2212 signature. Show that \u03c3xx 1 2\u03c0L2 d2x d2x (\u02c6vx \u02c6G+)(x, x)(\u02c6vx \u02c6G\u2212)(x, x), (7.60) where \u02c6G\u00b1 = (\u03bc \u00b1 i/2\u03c4 \u2212 \u02c6H)\u22121. 1 (c) Turning to the Hall coe\ufb03cients, we \ufb01rst note that (think why) \u03c3xy = \u2212\u03c3yx or \u03c3xy = m [ \u02c6H, \u02c6x], show that L2 \u2202B \u02c6N, where \u02c6N is the particle number operator. Convince yourself 2 ij\u03c3ij. Using the simpli\ufb01cations listed in (a) and the identity \u02c6v = i xy + \u03c3\u2212\u2212 \u03c3++ i yx 33 In view of the coordinate dependence of the vector potential, it is preferable to work in a real-space basis. 34 Do not confuse the vector potential of the external magnetic \ufb01eld with the weak vector potential employed to probe the electromagnetic response of the system; the current operators in the conductivity formula above contain the former but not the latter. 406 Response functions that the region of mixed frequency signature contributes as in the longitudinal case, i.e. that the total linear response Hall conductivity is given by \u03c3xy 1 4\u03c0L2 d2x d2x ij(\u02c6vi \u02c6G+)(x, x)(\u02c6vj \u02c6G\u2212)(x", ", x) + i L2 \u2202B \u02c6N. (7.61) (d) In Chapter 9, we apply the linear response formulae derived above to a \ufb01eld-theoretical analysis of the quantum Hall e\ufb00ect, the physics of a two-dimensional electron gas subject to a strong perpendicular \ufb01eld. Our starting point will be the replicated partition sum Z = D(\u03c8, \u00af\u03c8) exp(\u2212S[\u03c8, \u00af\u03c8]), where S[\u03c8, \u00af\u03c8] = d2x \u00af\u03c8(x) i\u03b4\u03c33 \u2212 \u03bc + (\u2212i\u2207 \u2212 A)2 + V (x) \u03c8(x), 1 2m and \u03c8 = {\u03c8ar} is a 2R-component \ufb01eld carrying a replica index r = 1,..., R and an index a = +, \u2212 distinguishing between \u201cretarded\u201d and \u201cadvanced\u201d \ufb01eld components. The Pauli matrix \u03c33 acts in the space of a-indices, thus determining the imaginary o\ufb00set of the inverse propagator in the action. To generate the linear response formulae derived above, we couple the action to a source: A \u2192 A + U \u22121\u2207U, where the matrix U = exp(i(x\u03bax\u03c31 + y\u03bay\u03c32) \u2297 P (1)) and P (1) = {\u03b4r1\u03b4r1} projects onto the \ufb01rst replica channel. Show that \u03c3xx = lim R\u21920 1 4\u03c0L2 \u22022 \u03bax\u03bax Z, \u03c3xy = lim R\u21920 \u03ba=0 1 4\u03c0iL2 \u22022 \u03bax\u03bay Z. \u03ba=0 (Hint: You will encounter terms \u223c \u00af\u03c8\u03c8 \u223c Re G+, which may be neglected.) Answer: (a) As with the previous analysis of the non-magnetic case, \u03c3++ xx = 1 \u03c9m T L2(2\u03c0)2 d2x d2p n>0 2 m G+ n (p, x) +v 2 x(p, x)(G+ n (p, x))2, i\u03c9m\u2192+i0 where we made use of the simplifying approximations above, v(p, x) = (p \u2212 A(x))/m n = (i\u03c9n \u2212 mv2/2 + \u03bc + i/2\u03c4 )\u22121. We (both p and x are vectors of c-numbers) and G+ may now shift p \u2192 p + A to remove the spatial dependence of the integrand, use the identity G2 = +mp\u22121 x \u2202px G, and integrate by parts to conclude that \u03c3++ xx = \u03c3\u2212\u2212 xx = 0. (b) It is an instructive exercise to obtain the result by contour integration. Introduce a contour that, avoiding the cuts on the real axis, runs around the frequency interval n(n + m) < 0 and expand the product of Green functions to leading order in i\u03c9m. Then take the DC limit. A simpli\ufb01ed, yet less rigorous, derivation goes as follows. For low frequencies, \u03c9m < 1/\u03c4, \u02c6Gn+m>0 \u02c6G+ and \u02c6Gn<0 \u02c6G\u2212 are approximately independent of the frequency index. Noting that \u03c9m/(2\u03c0T ) terms contribute to the frequency sum, we arrive at Eq. (7.60). 7.6 Problems 407 (c) Representing the frequency sums by a contour integral, xy + \u03c3\u2212\u2212 \u03c3++ yx = \u2212 1 4\u03c0i\u03c9mL2 (\u02c6vi \u02c6G)( + i\u03c9m, x, x)(\u02c6vj \u02c6G)( + i0, x, x) d2x d2x ij d nF() \u00d7 \u2212 (\u02c6vi \u02c6G) \u00d7( \u2212 i0, x, x)(\u02c6vj \u02c6G)( \u2212 i\u03c9m, x, x). i\u03c9m\u2192+i0 We next (i) use the identity \u02c6v = \u2212i[ \u02c6G, \u02c6x], (ii) expand the Green functions to leading order in i\u03c9m, and notice that the zeroth-order term in the expansion vanishes. This leads to the expansion xy + \u03c3\u2212\u2212 \u03c3++ yx = 1 4\u03c0L2 1 2\u03c0L2 = \u2212 i L2 d nF()ij tr \u02c6xi \ufffd", "\ufffdG( + i0)\u02c6vj \u02c6G( + i0) \u2212 c.c. d nF()\u2202 tr d nF()\u2202 tr \u02c6G( + i0)\u2202B \u02c6H \u2212 c.c. \u03b4( \u2212 \u02c6H)\u2202B \u02c6H = i L2 \u2202B d nF() tr \u03b4( \u2212 \u02c6H) = i L2 \u2202B \u02c6N, where in the second line we have used the approximate commutativity of the operators under the trace and the identity \u2202B \u02c6H = 1 2 ij \u02c6vi \u02c6xj. (d) To simplify the notation, we will discuss the longitudinal and Hall conductances separately. To generate the longitudinal conductance, we need to expand the generating functional to second order in \u03bax. Setting \u03bay = 0, we obtain U \u22121\u2207U = iP (1) \u2297 ex\u03bax and S[\u03c8, \u00af\u03c8] = S[\u03c8, \u00af\u03c8]|\u03ba=0 \u2212 i\u03bax d2x \u00af\u03c81\u03c31\u02c6vx\u03c81 \u2212 \u03ba2 x d2x \u00af\u03c81\u03c81. Twofold di\ufb00erentiation with respect to \u03bax then generates the expectation value B C 2 \u2212 d2x \u00af\u03c81\u03c31\u02c6vx\u03c81 + 2 d2x \u00af\u03c81\u03c81 lim R\u21920 1 L2 \u22022 \u03bax\u03bax Z = lim R\u21920 \u03ba=0 1 L2 \u2192 2 L2 d2x d2x (\u02c6vx \u02c6G+)(x, x)(\u02c6vx \u02c6G\u2212)(x, x) = 4\u03c0\u03c3xx, where in the second line we have noted that \u00af\u03c8P (1)\u03c8 \u223c G+ + G\u2212 generates the real part of the Green function which we decided to omit. Turning to the Hall conductivity, an expansion up to order \u03bax\u03bay gives U \u22121\u2207U = iP (1) (\u03bax\u03c31ex + \u03bay\u03c32ey + \u03bax\u03bay\u03c33(xey \u2212 yex)). Substituting this expression into the action, we obtain S[\u03c8, \u00af\u03c8] =S [\u03c8, \u00af\u03c8]|\u03ba=0 \u2212 i d2x \u00af\u03c81(\u03c31\u03bax\u02c6vx + \u03c32\u03bay \u02c6vy)\u03c81 \u2212i \u03bax\u03bay 2 d2x \u00af\u03c81\u03c33({\u02c6vy, \u02c6x} \u2212 {\u02c6vx, \u02c6y})\u03c81. 408 Response functions Then applying the derivative, we obtain @ lim R\u21920 1 L2 \u22022 \u03bax\u03bay 1 Z = lim L2 R\u21920 \u03ba=0 \u2212 d2x \u00af\u03c81\u03c31\u02c6vx\u03c81 d2x \u00af\u03c81\u03c32\u02c6vy\u03c81 A + i 2 \u2192 1 L2 d2x \u00af\u03c81\u03c33({\u02c6vy, \u02c6x} \u2212 {\u02c6vx, \u02c6y})\u03c81 d2x d2x ij(\u02c6vi \u02c6G+)(x, x)(\u02c6vj \u02c6G\u2212)(x, x) i \u22124 d2x ij \u02c6vi \u02c6xj Im G+(x, x) = 4\u03c0i\u03c3xy. 8 The renormalization group The method of the renormalization group (RG) provides theorists with powerful and e\ufb03cient tools to explore interacting theories, often in regimes where perturbation theory fails. Motivating our discussion with two introductory examples drawn from a classical and a quantum theory, we \ufb01rst become acquainted with the RG as a concept whereby nonlinear theories can be analyzed beyond the level of plain perturbation theory. With this background, we then proceed to discuss the idea and practice of RG methods in more rigorous and general terms, introducing the notion of scaling, dimensional analysis, and the connection to the general theory of phase transitions and critical phenomena. Finally, to conclude this chapter, we visit a number of concrete implementations of the RG program introduced and exempli\ufb01ed on a number of canonical applications. In Chapter 5, \u03c64-theory was introduced as an archetypal model of interacting continuum theories. Motivated by the existence of nonlinearities inherent in the model, a full perturbative scheme was developed, namely Wick contractions and their diagrammatic implementation. However, from a critical perspective,", " one may say that such perturbative approaches present only a limited understanding. Firstly, the validity of the \u03c64-action as a useful model theory was left unjusti\ufb01ed, i.e. the \u03c64-continuum description was obtained as a gradient expansion of, in that case, a d-dimensional Ising model. But what controls the validity of the low-order expansion? Indeed, the same question could be applied to any one of the many continuum approximations we have performed throughout the \ufb01rst chapters of the text. Secondly, having identi\ufb01ed a number of terms contributing to the perturbative expansion of the model (i.e. represented them in terms of momentum integrals over the non-interacting Green function), it rapidly became apparent that the expansion was seemingly uncontrolled: Those terms that were constructed explicitly (cf., e.g., Eq. (5.17) and the equation before) contained divergences at large momenta. We also saw that, in the transition region between the ferro- and the paramagnetic phase of the \u03c64-theory, these integrals were prone to the build\u2013up of infrared divergences. Worse still, we had to concede that we had no clue as to how to overcome these problems! Needless to say, such di\ufb03culties exempli\ufb01ed by the \u03c64-model are endemic in \ufb01eld theory. To better understand the origin and remedy of the problems identi\ufb01ed above, we must develop some new ideas. To this end, it is helpful to remember that the central driving force of all of our e\ufb00orts is the development of a better understanding of long-range characteristics. On the other hand, models such as \u03c64 exhibit \ufb02uctuations on all length-scales, while it was the short-scale \ufb02uctuations that were responsible for the majority of di\ufb03culties. In 409 410 The renormalization group principle, we already know how to deal with situations of this kind: our aim should be an e\ufb00ective theory of long-range \ufb02uctuations realized by integrating over all short-range \ufb02uctuations. Indeed, this program has been exempli\ufb01ed already on numerous examples where some slow \ufb01eld was coupled to rapidly \ufb02uctuating microscopic \ufb01elds, which were subsequently integrated out. L The problem with our current application is that the \u03c64-theory is unaware of any clear-cut separation into \u201cfast\u201d and \u201cslow\u201d degrees of freedom. Rather, all \ufb02uctuations, ranging from the shortest scales (of the order of some microscopic cuto\ufb00, a, limiting the applicability of the theory) to the longest scales (of the order of, say, the system size, L) are treated on the same footing. However, to nonetheless implement the scheme of \u201cintegrating over fast modes to generate an e\ufb00ective action of the slow degrees of freedom,\u201d one must declare arti\ufb01cially a certain length scale a(1) \u2261 ba > a, as the scale separating \u201cshort-wavelength \ufb02uctuations\u201d on scales [a, a(1)] from \u201clong-wavelength \ufb02uctuations\u201d on scales [a(1), L]. Having done so, one may proceed to integrate out the short-range \ufb02uctuations, thereby changing the action of the long-range degrees of freedom (see \ufb01gure). Since the short-range action is by no means simpler than the action of the long-range degrees of freedom, this step will likely involve some approximation. Indeed, the integration procedure may lead to a number of conceivable scenarios. For example, it may corrupt the algebraic structure of the long-wavelength action, leaving us with a theory fundamentally di\ufb00erent from the one with which we began. Alternatively, it may turn out that the e\ufb00ective action of the slow degrees of freedom is structurally similar to the original, in which case the entire e\ufb00ect of integrating over the fast \ufb02uctuations amounts to a changed set of coupling constants. a(2) a(1) a... If the latter, the procedure is well motivated: we have arrived at a theory identical to the original but for (a) a di\ufb00erent, or renormalized, set of coupling constants, and (b) an increased short-distance cut-o\ufb00 a \u2192 a(1) = ba. Evidently, one may then iterate this procedure; that is declare a new cut-o\ufb00 a(2) = ba(1) = b2a, integrate out \ufffd", "\ufffductuations on length scales [a(1), a(2)], etc. Along with the recursive integration of more \u201clayers of \ufb02uctuations,\u201d the coupling constants of the theory change, or \u201c\ufb02ow,\u201d until the cut-o\ufb00 a(n) \u223c L has become comparable to the length scales in which we are interested. Remarkably, we will see that the renormalized set of coupling constants in fact encodes much of the important information about the long-range behavior of the theory. Kenneth G. Wilson, 1936\u2013 Recipient of the 1982 Nobel Prize in Physics, awarded for \u201cdiscoveries he made in understanding how bulk matter undergoes phase transitions, i.e., sudden and profound structural changes resulting from variations in environmental conditions.\u201d Wilson\u2019s background ranges from elementary particle theory and condensed matter physics (critical phenomena and the Kondo problem) to quantum chemistry and computer science. (Image c The Nobel Foundation.) The general line of reasoning above summarizes much of the thinking behind the renormalization group. Of course, the approach would be quite useless had we to perform each The renormalization group 411 and every recursion step a(n\u22121) \u2192 a(n) explicitly. However, the utility of the program relies on the recursive reproduction of the model at each step: a single step already encodes all the information about the renormalization properties of the model. Cast in abstract terms, the conceptual idea may sound obscure. Therefore, before developing a general and formal framework for the renormalization group scheme, we shall acquire some intuition for the approach by focusing on two speci\ufb01c applications which encapsulate the basic concepts. INFO The formulation of renormalization group ideas has a long and varied history, re\ufb02ecting, in part, the versatility and generality of the method.1 Indeed, the advent of these ideas in the late 1960s and early 1970s marked the transition between two di\ufb00erent \u201cepochs.\u201d While hitherto researchers had focused primarily on the development and application of ever more-sophisticated perturbative techniques, the seventies remained under the spell of the renormalization program. In the second half of the 1960s, ideas to recursively generate \ufb02ows of coupling constants arose \u2013 apparently in independent developments \u2013 both in condensed matter and in particle physics. However, it took the insight of Kenneth Wilson to realize the full potential of the approach and to develop it into a widely applicable tool.2 It is probably fair to say that Wilson\u2019s original formulation of the approach, and its later extension by others, led to revolutionary progress in condensed matter physics, particle physics, and general statistical mechanics. Perhaps no less important, the RG concept turned out to be one of the major driving forces behind the partial uni\ufb01cation of these \ufb01elds. In fact, the (highly unfortunate) terminology of \u201crenormalization group\u201d discloses much of the historical origin of the approach. A widespread doctrine of late 1960s particle physics had been that, on a fundamental level, our world could be understood in terms of symmetries and their implementation through groups \u2013 the eight-fold way. In an attempt to absorb the newly developed RG approach into this general framework, it became dubbed the renormalization group. Of course, a linkage between group structures and renormalization methods would not have been drawn had it been utterly unjusti\ufb01ed. Indeed, one may argue that the sequence of RG transformations outlined above de\ufb01nes the structure of a semigroup.3 However, the connection between RG transformation and group algebraic structures is not only highly formal but also counterproductive. (It suggests a conceptual bond that simply is not there.) Besides, the group interpretation of the RG transformation is completely useless with regard to practical aspects. To introduce the conceptual foundation of the renormalization group, we \ufb01nd it helpful to draw initially on classical theories, later turning our attention to addressing the e\ufb00ects 1 A perspective on the development of the renormalization group can be found in the review article by M. E. Fisher, Renormalization group theory: its basis and formulation in statistical physics, Rev. Mod. Phys. 70 (1998), 653\u201381, or in the text by J. Cardy, Scaling and Renormalization in Statistical Physics (Cambridge University Press, 1996). 2 See K. Wilson, The renormalization group: critical phenomena and the Kondo problem, Rev. Mod. Phys. 47 (1975), 773\u2013840, still one of the best introductions to the approach! 3 To this end, one should interpret an individual RG transformation as an abstract mapping between two actions: S R\u2192 S, where S is the original action, S the action with increased short-distance cut-o\ufffd", "\ufffd, and R the mapping between them. One may then notice that these transformations can be iterated, or \u201cmultiplied,\u201d R \u25e6 R. A moment\u2019s thought shows that the composition law obeys the de\ufb01ning conditions of a semigroup: there is a unit transformation (nothing is integrated out) and the iteration is associative. (We are dealing with a semigroup because the RG transformation is irreversible, i.e. it does not have an inverse.) 412 The renormalization group of quantum \ufb02uctuations. Our \ufb01rst encounter with the renormalization group is, therefore, focused on the one-dimensional Ising model. 8.1 The one-dimensional Ising model To illustrate the implementation of the RG program on a simple example (indeed the simplest of examples), let us consider the one-dimensional Ising model de\ufb01ned through the classical microscopic Hamiltonian H = \u2212J N i=1 SiSi+1 \u2212 Hext N i=1 Si, where Si = \u00b11 denotes the (uniaxial) magnetization or spin of site i (periodic boundary conditions, SN +1 = S1, imposed), and Hext represents an external \ufb01eld. Throughout this section we will adhere to the discrete spin representation of the model, i.e. in contrast to Section 5.1, we will not map the model onto a continuum theory. However, this consideration does not impede our discussion of the RG program: One may simply regard \u2212\u03b2H[S] as a \u201cfunctional\u201d of the discrete \u201c\ufb01eld\u201d {Si}, and all steps outlined above can be carried out without substantial modi\ufb01cation. 8.1.1 Exact solution The feature of the one-dimensional Ising model which makes it of pedagogical interest in the present context is that it admits an exact solution, i.e. all of its macroscopically observable properties can be computed in closed form. Both the exact solution of the model and its RG formulation rely on its transfer matrix representation. One may notice that the Boltzmann weight of the system can be factorized according to the relation e\u2212\u03b2H = e N i=1(KSiSi+1+hSi) = N' i=1 T (Si, Si+1), where, for notational convenience, we have introduced the dimensionless parameters K \u2261 \u03b2J > 0 andh = \u03b2Hext, and the weight is de\ufb01ned through the relation T (S, S) = exp[KSS + h 2 (S + S)]. De\ufb01ning a two-component matrix T with elements T11 = T (1, 1), T12 = T (1, \u22121), T21 = T (\u22121, 1), and T22 = T (\u22121, \u22121), one may con\ufb01rm that the partition function of the system can be written as Z = {Si} e\u2212\u03b2H = N' {Si} i=1 T (Si, Si+1) = N' {ni} i=1 Tnini+1 = tr T N. Thus, we have managed to represent the partition function as a trace of the N th power of the two-dimensional \u201ctransfer\u201d matrix T.4 In this representation, the partition function 4 The terminology \u201ctransfer matrix\u201d originates in an interpretation of the Ising model as a \ufb01ctitious dynamical process in which a state Si is \u201ctransferred\u201d to a state Si+1, where the transition amplitude is given by T (Si, Si+1). 8.1 The one-dimensional Ising model 413 (a) (b) Figure 8.1 (a) On the absence of spontaneous symmetry breaking in the one-dimensional Ising model. No matter how low the temperature, the energy cost associated with the creation of a segment of \ufb02ipped spins is outweighed by the entropy gain. (b) In higher dimensions, entropic factors no longer have the capacity to overpower the extensive growth of energy associated with the formation of mismatched regions. may be presented in terms of the two eigenvalues, \u03bb\u00b1 = eK[cosh(h) \u00b1 (sinh2(h) + e\u22124K)1/2], of the transfer matrix T = eK+h e\u2212K e\u2212K eK\u2212h, + + \u03bbN as Z = tr T N = \u03bbN \u2212. Noting that \u03bb+ > \u03bb\u2212, one may see that, in the thermodynamic limit N \u2192 \u221e, the contribution of the latter may be neglected, so that Z N\u2192\u221e\u2212\u2192 \u03bbN +. Restoring the original microscopic parameters, in the thermodynamic limit, one obtains the free energy + ln", " Z = \u2212N J + T ln cosh(\u03b2Hext) + sinh2(\u03b2Hext) +e \u22124\u03b2J. (8.1) F \u2261 \u2212 1 \u03b2 With this result, one can compute the magnetization M by di\ufb00erentiation with respect to the magnetic \ufb01eld. As a result one obtains the magnetization per spin, m \u2261 M N + = sinh(\u03b2Hext) sinh2(\u03b2Hext) +e \u22124\u03b2J. (8.2) Two magnetization curves are shown in the \ufb01gure at two non-zero temperatures. Notice that, while the magnetization curves grow ever more steeply with decreasing temperature, the system does not magnetize at any non-zero temperature in the i.e., unlike absence of an external \ufb01eld, its higher-dimensional descendants, the one-dimensional model does not display spontaneous symmetry breaking. \u20131 \u20130.5 M 1 0.5 \u20130.5 \u20131 0.5 H 1 INFO There are many ways to understand the absence of spontaneous symmetry breaking in the one-dimensional Ising model (cf. our discussion of the connection with quantum mechanical tunneling in Section 3.3). Suppose that the system did exhibit a critical temperature Tc below which the system acquired long-range order with {Si} = 1 or {Si} = \u22121 (Fig. 8.1 (a)). Now, let us imagine that a segment of M consecutive spins were to \ufb02ip. In doing so, they would incur an energetic cost of O(2J), i.e. the energy associated with the unfavorable spin alignment at 414 The renormalization group the two ends of the domain of reversed spins. However, there are some N di\ufb00erent choices for placing the sector of M \ufb02ipped spins, i.e. the energy loss is counteracted by an entropic factor of \u223c T ln N (exercise). Thus, no matter how small T is, for large systems of large enough extent the free energy balance for domain creation is positive, implying that the system will be in a disordered phase at any non-zero temperature. EXERCISE By enumerating the number of spin con\ufb01gurations with the same energy, obtain a formal expression for the classical partition function of a one-dimensional Ising model of length N with periodic boundary conditions. (Hint: Consider the number of domain wall con\ufb01gurations.) Making use of Stirling\u2019s approximation ln n! n(ln n \u2212 1), determine the temperature dependence of the correlation length (domain size) at low temperatures. Con\ufb01rm that the system does not order at any \ufb01nite temperature. To what extent do these arguments carry over to Ising systems of higher dimensionality? Consider, for example, a two-dimensional variant of the model. Here the formation of a large connected region of M mismatched spins incurs an energy cost U \u223c M 1/2J. To understand why, one may note that the energy cost is proportional to the length of the one-dimensional boundary that encloses a (circular) domain of reversed spins. However, the entropic gain still scales as \u223c k ln N. (Allowing for less symmetric domains, the energy/entropy balance becomes more subtle. Yet the qualitative conclusion remains robust.) Thus, for the two-dimensional system, F \u223c JM 1/2 \u2212 T ln N. From this one can conclude that, over the range of temperatures for which this estimate is valid, no matter how small T, it is energetically unfavorable to \ufb02ip a thermodynamic M \u223c N x>0 number of spins. One may, therefore, conclude that the (d \u2265 2)dimensional Ising model does exhibit a phase transition into an ordered low-temperature phase. The phenomenon of spontaneous symmetry breaking occurs only in systems of su\ufb03ciently large dimensionality. The threshold dimension below which entropic mechanisms exclude spontaneous symmetry breaking is called the lower critical dimension dc. Our argument above indicates that the lower critical dimension of the Ising model and, more generally, The lower critical dimension of systems with discrete symmetries is dc = 1. Pursuing this theme, let us complete the discussion by addressing the question of the lower critical dimension for systems with a continuous symmetry (e.g. the U(1) symmetry of the super\ufb02uid phase, the O(3)-symmetry of the Heisenberg ferromagnet, etc.). To address this question, one may proceed in a manner analogous to that used in the Ising system, i.e. let us assume that a critical temperature exists below which the system is ordered. Using, for concreteness, the language of magnetic phenomena, we might say that the system with two-", "component spin degrees of freedom acquires a state of uniform magnetization S(r) = Se1, wheree 1 denotes the unit vector in the, say, 1-direction. Expanding the Hamiltonian \u03b2H[S] in \ufb02uctuations around this con\ufb01guration (and, for simplicity, taking a continuum limit), one obtains S[S] = cS2 2T ddr (\u2202\u03b8)2, 8.1 The one-dimensional Ising model 415 where \u03b8(r) denotes the polar angle measuring \ufb02uctuations around the ordered state and c is a coupling constant. One may now explore the thermal (or, if appropriate, quantum thermal) expectation value S1(r) of the 1-component of the spin variable anywhere in the system. Assuming that we are close to the ordered state, an expansion in \u03b8 gives S1(r) = S cos \u03b8 = S(1 \u2212 1 2 \u03b82 + \u00b7 \u00b7 \u00b7 ). We now proceed to check whether the close-to-ordered assumption was actually legitimate. For this, we have to verify that, with S1(r) = S \u2212 S \u03b82(r) + \u00b7 \u00b7 \u00b7, the contribution associated with 2 \ufb02uctuations is much smaller than the leading-order constant. Performing the Wick contraction with respect to the quadratic action, and switching to momentum space, one obtains S1(r) \u2248 S \" 1 \u2212 T 2cS2 q 1 q2 # \" \u2248 S 1 \u2212 T 2cS2 d L 2\u03c0 # ddq q2. d2q Crucially, in dimensions d \u2264 2, the integral is divergent. In the marginal case d = 2, q2 = \u03c0 ln(L/a), where we have used the fact that the momentum integral should be limited by a short(long-) wavelength cuto\ufb00 of the order of the inverse lattice spacing (the system size). In the thermodynamic limit, L \u2192 \u221e, the integral grows without bound, implying that the assumption of an ordered state was ill-founded no matter how small the temperature; the system is in a disordered state. Noting that nowhere did we rely on speci\ufb01cs of the spin system, we draw a conclusion known as the Mermin\u2013Wagner theorem: \u22121 a L\u22121 The lower critical dimension of systems with broken continuous symmetries is d = 2. Obviously, the divergence of the \ufb02uctuation integral is due to the fact that, for large wavelengths, and in low dimensions, the integration volume (alias the entropy) of \ufb02uctuations \u223c qd scales more slowly to zero than the energy cost \u223c q2. That is, as in the Ising model case, the Mermin\u2013 Wagner theorem5 can be understood as the statement of a competition of energy and entropy.6 8.1.2 Elements of scaling theory Returning to the one-dimensional Ising model, we may note that Eq. (8.2) represents a full solution of the problem. We have explicitly obtained the magnetization as a function of the magnetic \ufb01eld and the microscopic coupling constant of the model. Other thermodynamic characteristics, such as the magnetic susceptibility \u03c7 = \u2212\u22022 H F, can be generated by further di\ufb00erentiation with respect to H and/or T. (Here, and henceforth, we drop the subscript \u201cext\u201d from the de\ufb01nition of the external \ufb01eld, Hext, noting that it can always be discriminated from the Hamiltonian H by its context.) However, in the vast majority of physically interesting problems we will not be in possession of a closed analytical solution. This means that, before comparing the exact solution with the outcome of the RG program, we should reformulate the former in a universally applicable language, i.e. a code that can be used to 5 N. D. Mermin and H. Wagner, Absence of ferromagnetism or antiferromagnetism in one- or two-dimensional isotropic Heisenberg models, Phys. Rev. Lett. 17 (1966), 1133\u20136. 6 For completeness, we mention that the proof of the theorem (of which we gave a fairly abridged version here) in a subtle way relies on the fact that the symmetry in question is compact. For the (rare) class of systems with non-compact, e.g. hyperbolic, symmetries, the statement does not hold. 416 The renormalization group characterize the behavior of a model irrespective of the particular method by which this behavior has been analyzed. This objective leads us back to the familiar notion of correlation functions. All models of physical interest display non-trivial \ufb02uctuation behavior on large length", " scales. Indeed, in the vicinity of critical points marking the position of continuous phase transitions, one expects the onset of critical \ufb02uctuations, i.e. the accumulation of \ufb02uctuations on all length scales. (For a precise formulation of the terminology, see the section on phase transitions and critical phenomena below.) One of the central working hypotheses of the theory of \u201ccritical phenomena\u201d is that, in the vicinity of the critical points there is only one length scale of physical relevance, the correlation length \u03be. Within the context of the Ising model (cf. our previous discussion on page 199), \u03be is de\ufb01ned as the decay length of the correlation function, C(r1 \u2212 r2) \u2261 S(r1)S(r2) \u2212 S(r1)S(r2) \u223c exp \u2212, |r1 \u2212 r2| \u03be where we have switched to a continuum notation Si \u2192 S(r), to emphasize the large-distance character of \u03be. To relate this quantity to the thermodynamic characteristics, one may note that the magnetic susceptibility is given by \u03c7 = \u2212\u22022 H F H=0 = T \u22022 H ln Z H=0 = \u03b2 ddr ddr C(r\u2212r)! (S(r)S(r) \u2212 S(r)S(r)). (8.3) INFO In fact, this is yet another manifestation of the \ufb02uctuation\u2013dissipation theorem discussed in the previous chapter. A dissipative quantity (in this case, the susceptibility) is determined by the \ufb02uctuation behavior of the system. Comparison with the de\ufb01nition of the correlation function above identi\ufb01es \u03c7 \u223c \u03bed, i.e., in the one-dimensional system, the magnetic susceptibility is directly proportional to the length scale determining the decay of correlations in the system. So far, we have not yet made use of the speci\ufb01c results obtained for the Ising models above. However, to actually determine the correlation length for our present example, we need only di\ufb00erentiate Eq. (8.2) once again with respect to H, to obtain \u03be \u223c \u03c7 \u223c \u2202H | H=0 m \u223c e2\u03b2J, (8.4) i.e. the correlation length exponentially increases in the limit T \u2192 0 on a scale set by the microscopic \u201csti\ufb00ness constant\u201d, J. This result should not be too surprising: unlike its higher-dimensional counterparts, the one-dimensional Ising model does not display a \ufb01nitetemperature phase transition between a ferro- and a paramagnetic phase. It is only in the limit T \u2192 0 that long-range correlations develop. In the vicinity of a critical point, the diverging correlation length speci\ufb01es the singular thermodynamic properties of the system. More speci\ufb01cally, all observables X of dimensionality [length]DX should obey the scaling form X \u223c \u03beDX gX, 8.1 The one-dimensional Ising model 417 Figure 8.2 RG or block spin transformation of the Ising model illustrated for clusters of size b = 2. An r-fold iteration of the procedure reduces the degrees of freedom by a factor 2r. where gX is a dimensionless function. Let us explore this concept on our present example. The \u201creduced free energy,\u201d f (T ) \u2261 F T L, (8.5) has dimension L\u22121. Noting that N \u223c L, a straightforward low-temperature expansion of Eq. (8.1) indeed gives f (T ) \u2212 f (0) = \u03be\u22121 \u03be2h2 \u2261 \u03be\u22121g(\u03beh), (8.6) \u22121 \u2212 1 2 where we have subtracted the in\ufb01nite but inessential constant f (0) and assumed that 1 \u03be\u22121 h. (The scaling form above actually suggests that the magnetic \ufb01eld has dimension L\u22121, a prediction substantiated below.) We have thus found that the correlation length of the one-dimensional Ising system diverges according to Eq. (8.4) upon approaching zero temperature, and that the free energy obeys the scaling law (8.6). Of course, this is only a fraction of the full information stored in the exact solution. However, the reduced set of data has the striking advantage that it is of general relevance. Indeed, we saw above that the correlation length is directly related to measurable properties of the system such as the magnetic susceptibility. Similarly, the output of experiments on systems with long-range correlations is commonly encoded in the language of scaling relations. In other words, we have extracted that part of the information contained in the exact", " solution that carries universal relevance and can be compared with the output of other approaches. 8.1.3 Kadano\ufb00 \u2019s block spin RG With this background we are in a position to explore some of the conceptual foundations of the RG on a model application. According to the general scheme outlined at the beginning of the chapter, our aim is to devise an algorithm to recursively trace out parts of the short-scale \ufb02uctuations of the system and assess their in\ufb02uence on the remaining degrees of freedom. 418 The renormalization group Leo P. Kadano\ufb00 1934Theoretical physicist and applied mathematician who has contributed widely to research in the properties of matter, the development of urban areas, statistical models of physical systems, and the development of chaos in simple mechanical and \ufb02uid systems. He was instrumental in the development of the concepts of scale invariance and universality as they are applied to phase transitions. Following the program outlined by Kadano\ufb00\u2019s seminal work on the foundations of the RG, we will follow a strategy whereby this renormalization step may be e\ufb00ected by subdividing the spin chain into regular clusters of b neighboring spins (see Fig. 8.2). We may then proceed to sum over the 2b sub-con\ufb01gurations of each cluster, thereby generating an e\ufb00ective functional describing the inter-cluster energy balance. While it is clear that this energy functional exists, a far less obvious question to ask is whether it will again have the form of an e\ufb00ective Ising spin system. Remarkably, the answer is a\ufb03rmative: the Ising model is said to be \u201crenormalizable.\u201d The structural reproduction of the model implies that we can think of each cluster as some kind of meta-Ising spin, or block spin. More importantly, it guarantees that the renormalization step quali\ufb01es for iteration: in a second RG step, b block spins are grouped to form a new cluster (now comprising b2 of the microscopic spins) which are then traced out, etc. We next discuss how this algorithm is implemented in concrete terms. Within the transfer matrix approach, a cluster of b spins is represented through b transfer matrices T. Taking the partial trace over its degrees of freedom amounts to passing from these b matrices to the product T = T b. (By construction of the approach, the internal index summation involved in taking the product amounts to tracing out the degrees of freedom of the cluster.) The transition from the original partition function Z to the new partition function Z is de\ufb01ned through the relation ZN (K, h) = tr T N = tr (T b)N/b = tr (T )N/b = ZN/b(K, h), (8.7) where the notation makes the parametric dependence of the partition function on the size of the system, N, and on the coupling constants K, h explicit. Notice that the equation above makes the highly non-trivial assumption that the reduced trace, tr (T )N/b, can again be expressed as an Ising partition function or, equivalently, that the e\ufb00ective transfer matrix T has the same algebraic structure as the elementary matrices T. To con\ufb01rm the integrity of the transformation, one may explore the structure of the product matrices T for the simplest case of b = 2 block spins. Introducing the abbreviations u \u2261 e\u2212K, v \u2261 e\u2212h, we have T = eK+h e\u2212K e\u2212K eK\u2212h = u\u22121v\u22121 u u u\u22121v, while the product takes the form T \u2261 T 2 = u2 + u\u22122v\u22122 v + v\u22121 v + v\u22121 u2 + u\u22122v2!= C u\u22121v\u22121 u. u u\u22121v In the last equality we require that the new transfer matrix be of the same structure as the original. However, noting that this requirement will introduce three conditions (for the 8.1 The one-dimensional Ising model 419 v 0.1 0.2 u 0.9 v 1 0.8 0.6 0.4 0.2 (a) 0.2 0.4 0.6 0.8 1 u 0.8 (b) Figure 8.3 (a) The \ufb02ow of coupling constants of the one-dimensional Ising model generated by iteration of the RG transformation. The three lines shown are for starting values (u, v) = (0.01, 0.9999)(), (0.01, 0.999)(), and (0.01, 0.99)(). (b) Magni\ufb01cation of the zero-temperature \ufb01xed point region", ". three independent entries of the symmetric matrices T and T ), we are willing to tolerate the appearance of an overall multiplicative constant C.7 Having introduced this new parameter, we have enough freedom to solve the three equations, from which one \ufb01nds (exercise) u = \u221a v + v\u22121 (u4 + u\u22124 + v2 + v\u22122)1/4, v = \u221a \u221a u4 + v2 u4 + v\u22122, (8.8) \u221a v + v\u22121(u4 + u\u22124 + v2 + v\u22122)1/4. As a corollary, we remark that the possibility of and C = representing the new transfer matrix in the same algebraic structure as the old one implies that the transformed model again describes an Ising spin system (namely the spin system whose transfer matrix would be given by T ). However, the Hamiltonian \u03b2H of the new block spin system: is de\ufb01ned at a di\ufb00erent temperature, magnetic \ufb01eld and exchange constant (as described by the new values of the coupling constants (u, v)) and describes \ufb02uctuations on length scales that are twice as large as in the original system. In particular, the short-distance cuto\ufb00 has been doubled. To make further progress, one may focus on the two relevant parameters u and v and observe that the result of the block spin transformation can be represented as a discrete map u v = f1(u, v) f2(u, v), where the functions f1,2 are de\ufb01ned through Eq. (8.8). In Fig. 8.3 sequences of points generated by an iterative application of the map f are shown for di\ufb00erent values of \u201cinitial conditions\u201d (u0, v0). It is evident from the RG trajectories that the map f possesses two 7 Taking the product of the new transfer matrices, we see that this constant appears in the partition function as ln C which will be of no further Z \u223c CN/b, i.e. the free energy acquires an overall additive constant F \u223c \u2212 N T b signi\ufb01cance. 420 The renormalization group disjoint sets of \ufb01xed points, i.e. points (u\u2217, v\u2217) that remain invariant under the application of the map f : u\u2217 v\u2217 = f1(u\u2217, v\u2217) f2(u\u2217, v\u2217). Inspection of Eq. (8.8) shows that this is the case for (a) the point (u\u2217, v\u2217) = (0, 1), and (b) the line (u\u2217, v\u2217) = (1, v). The set of \ufb01xed points represents the most important structural characteristic of an RG analysis. They organize the space of \u201c\ufb02owing\u201d coupling constants into sectors of qualitatively di\ufb00erent behavior. In particular, one may note that, at a \ufb01xed point, all characteristics of the model, including its correlation length \u03be, remain invariant. On the other hand, we noticed above that an RG step is tantamount to doubling the fundamental length scale of the system. Consistency requires that either \u03be = 0 or \u03be = \u221e. In the present case, the line of \ufb01xed points is identi\ufb01ed with u = exp[\u2212\u03b2J] = 1, i.e. \u03b2 = 0. This is the limit of in\ufb01nitely large temperatures, at which we expect the model to be in a state of maximal thermal disorder, that is \u03be = 0. Besides the high-temperature \ufb01xed line, there is a zero-temperature \ufb01xed point (u, v) = (exp[\u2212\u03b2J], exp[\u2212\u03b2h]) = (0, 1) implying T \u2192 0 andh = 0. Upon approaching zero temperature, the system is expected to order and to build up long-range correlations, \u03be \u2192 \u221e. Notice, however, an important di\ufb00erence between the high- and the low-temperature set of \ufb01xed points: while the former is an attractive \ufb01xed point in the sense that the RG trajectories approach it asymptotically, the latter is a repulsive \ufb01xed point. No matter how low the temperature at which we start, the RG \ufb02ow will drive us into a regime of e\ufb00ectively higher temperature or lower ordering. (Of course, the physical temperature does not change under renormalization. All we are saying is that", " the block spin model behaves as an Ising model at a higher temperature than the original system.) To explore the low-temperature phase of the system quantitatively, we may linearize the RG map in the vicinity of the T = 0 \ufb01xed point. That is, condensing the \ufb01xed point coordinates into a two-component vector, x\u2217 \u2261 (u\u2217, v\u2217)T = (1, 0)T, and assuming that \u0394x parameterizes a small deviation from the \ufb01xed point, one may write x\u2217+\u0394x = f (x\u2217+\u0394x) \u2248 f (x\u2217) + \u2202xf \u00b7 \u0394x + O(\u0394x2). Now, drawing on the invariance of the \ufb01xed point, f (x\u2217) = x\u2217, we obtain the linearized map \u0394x = \u2202xf \u00b7 \u0394x + O(\u0394x2). To explore the linearized mapping in more detail, it is convenient to introduce yet another pair of variables, namely r \u2261 u4, s \u2261 v2, whereupon the RG transformation becomes rational, r = 2+s+s\u22121 r+s\u22121. Di\ufb00erentiating this map at (r, s) = (1, 0), it is straightforward to show that r+r\u22121+s+s\u22121, s = r+s \u0394r \u0394s 4 = \u0394r \u0394s. 2 Noting that a transformation with b = 4, say, is equivalent to a two-fold application of a b = 2 transformation, one may recast the relation above in the more general form \u0394r \u0394s = b2 b \u0394r \u0394s, (8.9) applicable to arbitrary b. 8.1 The one-dimensional Ising model 421 To make use of Eq. (8.9), let us once again consider the reduced free energy Eq. (8.5): f (\u0394r, \u0394s) \u2261 \u2212N \u22121 ln ZN (K(\u0394r), h(\u0394s)) \u2261 \u2212N \u22121 ln ZN (\u0394r, \u0394s) and reformulate Eq. (8.7) according to f (\u0394r, \u0394s) = \u2212 1 N ln ZN (\u0394r, \u0394s) = \u2212 1 N b This equation describes the \u201cscaling\u201d of the free energy density under block spin transformations (all in the linearizable low-temperature regime) or, equivalently, changes of the fundamental length scale at which we consider our model. The right-hand side of the equation describes how the model would look from a \u201cblurred\u201d perspective where all degrees of freedom on scales < b have been comprised in a single structural unit. ln ZN (\u0394r, \u0394s) = f (b2\u0394r, b\u0394s). 1 b Importantly, b is a free parameter without intrinsic signi\ufb01cance; it can be set to any desired value. For example, we may \ufb01nd it convenient to look at our model at scales where b2\u0394r = 1. With this choice one obtains f (\u0394r, \u0394s) = \u0394r1/2 f (1, \u0394s/\u0394r1/2) \u2261 \u0394r1/2 g(\u0394s/\u0394r1/2), where the dimensionless one-parameter function g is de\ufb01ned through the second equality. Finally, one can relate back to the physical parameters of the system: \u0394r = r \u2212 0 =r = u4 = e\u22124K, \u0394s = s \u2212 1 =v 2 \u2212 1 =e \u22122h \u2212 1 \u22122h, which brings us to the scaling relation f = e\u22122Kg(e2Kh). (8.10) This is the scaling form predicted by the RG analysis. Notice that a non-trivial statement is being made. The a priori dependence of the free energy on two independent parameters K, h has been reduced to a one-parameter function, multiplied by an overall prefactor. Indeed, we expect on general grounds that (see the discussion of the previous section) the reduced free energy should scale with the inverse of the correlation length \u03be, which in turn diverges upon approaching the zero-temperature \ufb01xed point. Comparing with Eq. (8.10), and noting that there are no reasons for the rescaled free energy g(x) =f (1, x) to diverge by itself, we conclude that the divergence of \u03be is driven by the prefactor, i.e. \u03be \u223c e2K, in agreement with the result of the exact analysis", ". With this identi\ufb01cation we obtain f = \u03be\u22121g(\u03beh), i.e. the magnetic \ufb01eld appears in conjunction with the correlation length and we have reproduced the exact asymptotic Eq. (8.6). Notice that there is no reason to be too irritated about the excessive appearance of vague proportionalities \u201c\u223c.\u201d As \ufb01xed points are approached, physical systems tend to build up all sorts of singular scales. The most fundamental of these is the correlation length, but the divergence of\u03be usually entails singular behavior of other physical quantities. These singularities characterize much of the observable behavior of a system both theoretically and, in fact, experimentally. In the immediate vicinity of a \ufb01xed or transition point, all but the strongest driving forces of singular scaling are of secondary importance. Indeed, this is why we could, for example, conclude from Eq. (8.10) that \u03be \u223c exp(2K) \u223c exp(2\u03b2J). We know on general grounds that f \u223c \u03be\u22121. On the other hand, (8.10) implies that the leading (i.e. exponential) driving force behind the divergence of that scaling factor must 422 The renormalization group reside in exp(2\u03b2J). This leads to the identi\ufb01cation \u03be \u223c exp(2\u03b2J), where all other factors of proportionality are of secondary importance. EXERCISE Apply the block spin RG procedure to the one-dimensional q-state Potts spin model N \u03b2H = \u2212K i=1 \u03b4si,si+1, wheres i = (1, 2,..., q). Identify all \ufb01xed points and note their stability. In Section 8.3.3 below we discuss these scaling arguments from a more rigorous and general perspective. However, before doing so, let us exemplify an RG analysis of a second case study. 8.2 Dissipative quantum tunneling Previously, we have explored the impact of the external environment on the quantum mechanical tunneling of a particle from a well. In particular, we have seen that, under a wide range of conditions, the in\ufb02uence of the environment can be captured by an ohmic |\u03c9n||\u03b8n|2, where the parameter contribution to the action, of the form Sdiss[\u03b8] = 1 \u03c0T g g\u22121 characterizes the bare viscosity. Indeed, we have seen that a similar dissipative structure appears in a number of di\ufb00erent settings, including the problem of impurity scattering in a quantum wire (see Problem 4.5). There, the dissipative term appeared as a result of integrating over the collective \ufb02uctuations of the bosonized \ufb01eld while pinning the \ufb01eld \u03b8 at the impurity. \u03c9n In the following, we exploit an RG program to uncover the e\ufb00ect of dissipation on the facility to tunnel. To be concrete, let us assume that the quantum mechanical particle inhabits a periodic potential, U (\u03b8) = c cos \u03b8. In this case, the quantum transition probability can be encapsulated by the quantum partition function Z = D\u03b8 exp(\u2212S[\u03b8]) where8 S[\u03b8] = 1 4\u03c0T g n |\u03c9n||\u03b8n|2 + c d\u03c4 cos(\u03b8(\u03c4 )). (8.11) To explore the physical behavior of the system described by the e\ufb00ective action (8.11), we might decide to focus on a physical observable that would then be calculated by taking appropriate functional averages with respect to the e\ufb00ective action S[\u03b8]. However, we shall instead proceed in a somewhat more indirect manner: following the general philosophy outlined at the beginning of the chapter, we begin by arbitrarily subdividing the set of all \ufb01elds {\u03b8} into short- and long-wavelength degrees of freedom. For example, assuming that the maximum frequency up to which the e\ufb00ective bosonic theory of the problem applies is given by \u039b, we might say that \ufb02uctuations on scales \u039b/b < |\u03c9n| < \u039b are \u201cfast\u201d (b >1) while those with |\u03c9n| < \u039b/b are \u201cslow.\u201d INFO Notice that this mechanism of subdivision of the degrees of freedom di\ufb00ers from that employed earlier. In the previous example, we carried out an RG transformation by integrating 8 Here, taking the viscosity 1/g to be large, we have neglected the in\ufb02uence of the kinetic contribution which serves only to regularize the UV behavior. d m 2 \u02d9\u03b82, 8", ".2 Dissipative quantum tunneling 423 out \ufb01eld \ufb02uctuations on short scales in real space, i.e. we performed a real space renormalization. Here, we chose to integrate over a high-lying sector in frequency space. Alluding to the status of frequency as the (d + 1)th component of a generalized momentum, this implementation of the RG transformation is called momentum shell renormalization. While the real space renormalization may seem more intuitive, it is usually less convenient to carry out analytically. (As regards numerical renormalization procedures, the situation is di\ufb00erent.) It goes without saying that the net outcome of any RG strategy must not depend on the speci\ufb01c choice of the integration procedure. Our \ufb01rst objective is to identify the e\ufb00ective action of the slow degrees of freedom after the fast \ufb02uctuations have been integrated out. To prepare this integration, it is useful to decompose a general \ufb01eld amplitude \u03b8(\u03c4 ) into a slow contribution \u03b8s(\u03c4 ) and its fast complementary \u03b8f (\u03c4 ) part, that is \u03b8(\u03c4 ) \u2261 \u03b8s(\u03c4 ) +\u03b8 f (\u03c4 ), where \u03b8s(\u03c4 ) \u2261 \u03c9n,s e\u2212i\u03c9n\u03c4 \u03b8n \u2261 d\u03c9 2\u03c0 s e\u2212i\u03c9\u03c4 \u03b8(\u03c9), \u03b8f (\u03c4 ) \u2261 \u03c9n,f e\u2212i\u03c9n\u03c4 \u03b8n \u2261 d\u03c9 2\u03c0 f e\u2212i\u03c9\u03c4 \u03b8(\u03c9). (8.12) \u039b/b\u2264|\u03c9n|<\u039b. Here we have introduced the abbreviations Anticipating that we shall be working at very low temperatures T \u039b, we note that one |\u03c9|<\u039b/b d\u03c9 and may represent the discrete frequency summations by integrals, f d\u03c9 \u2261 \u039b/b\u2264|\u03c9|<\u039b d\u03c9. (It is a good exercise to convince oneself of the legitimacy of this simpli\ufb01cation at every step of the construction below. Also notice that, with the continuum \ufb01elds, \u03b8(\u03c9) = \u03b8n/T.) |\u03c9n|<\u039b/b and s d\u03c9 \u2261 \u03c9n,s \u03c9n,f \u2261 \u2261 On substituting into the continuum representation of the action (8.11), S[\u03b8] = 1 4\u03c0g d\u03c9 2\u03c0 |\u03c9|<\u039b |\u03b8(\u03c9)|2|\u03c9| + c d\u03c4 cos(\u03b8(\u03c4 )), (8.13) one obtains S[\u03b8s, \u03b8f ] = Ss[\u03b8s] +S f [\u03b8f ] +S U [\u03b8s, \u03b8f ], where Ss,f [\u03b8s,f ] = d\u03c9 2\u03c0 s,f 1 4\u03c0g |\u03b8(\u03c9)|2|\u03c9|, SU [\u03b8s, \u03b8f ] = c d\u03c4 cos(\u03b8s(\u03c4 ) +\u03b8 f (\u03c4 )). To proceed with this expression, we will resort to an approximation that is di\ufb03cult to justify D\u03b8f e\u2212Sf [\u03b8f ](\u00b7 \u00b7 \u00b7), and in advance. De\ufb01ning e\u2212Seff [\u03b8s] \u2261 e\u2212Ss[\u03b8s]e\u2212SU [\u03b8s,\u03b8f ]f, where\u00b7 \u00b7 \u00b7 f \u2261 assuming that the coupling constant c is small, we approximate e\u2212Seff [\u03b8s] = e\u2212Ss[\u03b8s] 1 \u2212 SU [\u03b8s, \u03b8f ] +\u00b7 \u00b7 \u00b7 f \u2248 e\u2212Ss[\u03b8s]e\u2212SU [\u03b8s,\u03b8f ] f. (8.14) That is, by assuming that the coupling constant c is in some sense small, we expand in c \u2013 which appears to be similar to our previous perturbative approaches \u2013 only to reexponentiate it in the next step \u2013 a manipulation very di\ufb00erent from plain perturbation theory. Evidently, the validity of this step is bound to small values of the coupling constant. However, before attempting a more quali\ufb01ed justi\ufb01cation, let us tentatively accept the approximation above and explore its consequences. 424 The renormalization group Computing the average SU [\u03b8s, \u03b8f ] f = c = = c 2 c 2 D\u03b8f e\u2212Sf [\u03b8f ] d\u03c4 cos(\u03b8f (\u03c4 ) +\u03b8 s(\u03c4 )) d\u03c4 ei\u03b8s(\u03c4 ) D\u03b8f e", "\u2212 1 4\u03c0g f d\u03c9 2\u03c0 \u03b8(\u03c9)|\u03c9|\u03b8(\u2212\u03c9)ei f d\u03c9 2\u03c0 ei\u03c9\u03c4 \u03b8(\u03c9) + c.c. d\u03c4 ei\u03b8s(\u03c4 )e\u2212\u03c0g d\u03c9 2\u03c0 f |\u03c9|\u22121 + c.c. = c d\u03c4 cos(\u03b8s)e \u22122\u03c0g \u039b \u039b/b d\u03c9 2\u03c0\u03c9 = c d\u03c4 cos(\u03b8s)e\u2212g ln b = cb\u2212g d\u03c4 cos(\u03b8s), We arrive at the remarkable conclusion that the e\ufb00ective action for the slow \ufb01eld, Se\ufb00 [\u03b8s] = 1 4\u03c0g d\u03c9 2\u03c0 s |\u03b8(\u03c9)|2|\u03c9| + cb\u2212g d\u03c4 cos(\u03b8s), is structurally identical to the action we started out from; in other words, we have seen that the action is renormalizable. Nonetheless, Se\ufb00 di\ufb00ers from the initial action S above in two important respects: \ufb01rstly, the integration over the fast \ufb01elds induces a change of the coupling constant of the perturbation; secondly, the new action takes values on \ufb01eld con\ufb01gurations that \ufb02uctuate only on scales |\u03c9n| < \u039b/b. \u03c9 s \u03c9 \u03c9 \u039b \u039b / b The next step of the RG program requires the comparison of the model before and after the integration over the fast \ufb01elds. However, two model actions taking values on di\ufb00erent sets of \ufb01eld con\ufb01gurations cannot be sensibly compared. Rather, one must e\ufb00ect a rescaling of the fundamental unit of frequency/time, (\u03c4, \u03c9) \u2192 (\u03c4, \u03c9), such that the transformed \ufb01eld \u03b8s(\u03c9) \u2192 \u03b8(\u03c9) also \ufb02uctuates on all scales |\u03c9| < \u039b. We thus introduce a rescaled frequency variable according to \u03c9 \u2261 b\u03c9. This change of variables engages a number of secondary transformations. In order to keep the dimensionless combination \u03c9\u03c4 relating time to frequency invariant, one must set \u03c4 \u2261 b\u22121\u03c4. As to the transformation of the \ufb01eld variable \u03b8, there is some freedom (because \u03b8 is an integration variable that can be transformed arbitrarily). However, the Fourier representation Eq. (8.12) implies that, after the transformation of \u03b8(\u03c4 ) has been \ufb01xed, the transformation of \u03b8(\u03c9) follows from our previous rescaling of frequency/time. In order to keep the algebraic structure of the cosine operator invariant, we choose to de\ufb01ne \u03b8(\u03c4 ) \u2261 \u03b8s(\u03c4 ). Equation (8.12) then enforces the condition \u03b8(\u03c9) = b\u22121\u03b8(\u03c9). Substitution of the new variables \u03c9, \u03c4, \u03b8(\u03c4 ), and \u03b8(\u03c9) into the e\ufb00ective action then gives Se\ufb00 [\u03b8s] = S[\u03b8] \u2261 1 4\u03c0g d\u03c9 2\u03c0 |\u03c9|<\u039b |\u03b8(\u03c9)|2|\u03c9| + cb1\u2212g d\u03c4 cos(\u03b8(\u03c4 )). 8.2 Dissipative quantum tunneling 425 Notice that the \ufb01rst contribution to the action remained invariant, save for the fact that the integration now extends over the full interval |\u03c9| < \u039b. INFO In fact, one may note that, without any calculation, the transformation of the action under the rescaling of variables could have been anticipated from dimensional analysis: the de\ufb01nition \u03c9 = b\u03c9 implies that all contributions to the action of dimension [frequency]d change by a factor b\u2212d. Since\u03b8 (\u03c4 ) is a dimensionless phase, we have [\u03b8(\u03c4 )] = 1 while [\u03b8(\u03c9)] = [frequency]\u22121. Thus, the \ufb01rst term of the action has dimension 1 and remains invariant. The second operator carries the dimension [d\u03c4 ] = [frequency]\u22121 and, therefore, changes by a factor b. We are now in a position to compare the e\ufb00ective actions S[\u03b8] andS [\u03b8] before and after the integration over the fast modes. Obviously, the principal e\ufb00ect of the integration over fast modes is that the coupling constant of the periodic potential has changed according to the relation c \u2192 c(b) \u2261 cb1\u2212g. (8.15", ") (However, do not forget a point that is implicit in the notation: the new action describes \ufb02uctuations on slower frequency scales or larger temporal scales. That this di\ufb00erence is not manifest in the notation is due to the fact that we chose to measure the \u201cnew\u201d frequency continuum in rescaled variables \u03c9 = b\u03c9.) This \ufb01nding contains a preliminary answer to the question formulated above. To recapitulate, we asked how the presence of the dissipative term manifests itself in the facility to tunnel. We have found that an integration over short-scale \ufb02uctuations of the collective \ufb01eld \u03b8 alters the e\ufb00ective strength of the potential. (In the context of the quantum impurity problem in the wire, this change is a manifestation of the mutual in\ufb02uence of the Friedel scattering pattern and interactions in the electron gas. Indeed, it is apparent that in the non-interacting case, i.e. for g = 1 (cf. Eq. (4.52)), the coupling constant does not change. For repulsive (attractive) interactions, g < 1 (g > 1) the coupling increases (decreases) in accord with the qualitative picture formulated above.) However, at this stage, our result merely indicates how the coupling constant changes after one RG step. Notably, Eq. (8.15) depends in a non-universal manner on how we choose to dissect the frequency spectrum (the parameter b). However, ideally, we would like to know the value of the coupling constant after all degrees of freedom down to a certain infrared cuto\ufb00 scale \u03c9min have been integrated out. (For example, the role of \u03c9min might be taken by temperature, the oscillation frequency of an external perturbation, etc.) The general route towards obtaining this information is to iterate the RG step, i.e., setting c \u2261 c(0) and c \u2261 c(1), explore the sequence of coupling constants c(0) \u2192 c(1) \u2192 c(2) \u2192 \u00b7 \u00b7 \u00b7 In all that follows, it is convenient to think of this sequence as a kind of dynamical system. In each discrete \u201ctime step,\u201d the variable c changes according to c \u2192 c = cb1\u2212g = celn b(1\u2212g). Assuming that b = 1 + is very close to unity (which means that, in each step, only an asymptotically thin layer in frequency space is \u201cshaven o\ufb00 \u201d), we can encapsulate this information in the di\ufb00erential equation c\u2212c d ln b = c(1\u2212g). The evolution equation \u2248 dc d \u2248 dc 426 The renormalization group describing the \ufb02ow of the coupling constant under an in\ufb01nitesimal change of the control parameter, dc d ln b \u2261 \u03b2(c) = c(1 \u2212 g), (8.16) is known as a Gell-Mann\u2013Low equation. For historical reasons, the right-hand side of the equation is called the \u03b2-function. Equation (8.16) makes the interpretation of the RG-generated change of the coupling constant as a dynamical system manifest. Thinking of the parameter t \u2261 ln b as a \u201ctime coordinate,\u201d we can integrate the evolution equation to obtain9 c(t) =c (0)e(1\u2212g)t, where c(t = 0) = c(b = 1) has the status of the bare coupling constant of the theory. (Remember that, for b = 1, \u039be\ufb00 = \u039b/b = \u039b, which is the level of the unrenormalized theory.) Having inferred the RG \ufb02ow, one may wonder at what \u201ctime\u201d tmax should we stop the renormalization? What is the \u201creal\u201d value of the coupling constant? The answers to these questions are somewhat application-speci\ufb01c. For example, we might have coupled the system to an external perturbation of characteristic frequency \u03c9m. In that case, we might want to integrate out all degrees of freedom with frequency \u03c9n>m to then explore the e\ufb00ective low-energy theory at scales \u223c \u03c9m, i.e. we would set \u039be\ufb00 = \u03c9m, ort = ln(\u039b/\u03c9m). The e\ufb00ective theory would then look structurally identical to the microscopic model but with a renormalized coupling constant c = c(0)(\u039b/\u03c9m)1\u2212g \u221d \u03c9g\u22121 m. (Notice that both the bare constant c(0) and the cuto\ufb00 \u039b depend in a non-", "universal manner on microscopic elements of the model. The use of the proportionality sign indicates that we are not, in general, interested in these details but rather focus on the dependence of the coupling constant on the low-energy scale \u03c9m.) Alternatively, we might want to integrate out all degrees of freedom down to the lowest frequency allowed by Matsubara frequency quantization, \u039be\ufb00 = 2\u03c0T. In this case, c \u221d T g\u22121. For a system of \ufb01nite extent L, we might argue that the spectrum of the modes \u03b8(k) is quantized with \u03c9min = \u03c0v/L. In this case, the e\ufb00ective value of the coupling at the lowest frequencies would be c \u221d L1\u2212g. Summarizing, the RG \ufb02ow should be terminated at a low-energy scale determined by the speci\ufb01c problem under consideration. Our results so far on the behavior of the coupling constant are summarized in the bottom part of the \ufb01gure below. We have found that, for interaction parameters g > 1 (g < 1), the coupling constant c decreases (increases). The \u201cnon-interacting\u201d case g = 1 de\ufb01nes a \ufb01xed line where the coupling strength does not change. However, at this stage, we must remember that the analysis was based on a spurious expansion of the action to \ufb01rst order in the impurity operator. This implies that, in the repulsive case, even if the initial value of the coupling constant was small, it will soon \ufb02ow into a region where the perturbative RG analysis loses its meaning (indicated by a wavy line in the \ufb01gure). 9 Critical readers will notice that this result coincides with the change of the coupling constant obtained after the \ufb01rst RG step. This coincidence, however, is a consequence of the simple structure of the \u03b2-function in the present example. In more complicated cases, the change of the coupling constant after an in\ufb01nitesimal RG step will structurally di\ufb00er from the result obtained after following the RG \ufb02ow all the way down to the IR cuto\ufb00. The concept of a \u03b2-function and its integration are therefore indispensable elements of the theory. 8.2 Dissipative quantum tunneling 427 c \u221e repulsive attractive How do we know what happens after the coupling constant has disappeared into the forbidden zone? it turns out that, for the particular Remarkably, model in question, not only the regime of a weak potential but also the complementary case of an asymptotically large potential (the upper part of the \ufb01gure) is amenable to analytical investigation. As a result of a strong coupling perturbative expansion (see Problem 8.8.1) one may show that the coupling constant \ufb02ows to (away from) in\ufb01nite coupling for repulsive (attractive) interactions. It is thus tempting to conjecture a globally increasing/decreasing \ufb02ow continuously interpolating between the two asymptotic regimes. (Application of the machinery of conformal \ufb01eld theory to the problem has indeed shown this conjecture to be true.) 1/g 1 0 Therefore, when put together, we arrive at a remarkable conclusion. For g > 1, the strength of the potential \ufb02ows to zero, implying that the asymptotic dependence at zero temperature is free: the particle can tunnel freely between minima of the potential. However, for g < 1, the coupling constant \ufb02ows to in\ufb01nity, implying that the particle becomes localized, con\ufb01ned to a minimum of the potential. INFO Notwithstanding the limitation to small values of the coupling, we still have to discuss the consistency of the fast \ufb01eld integration, i.e. to show that the re-exponentiation of the potential operator after the averaging over fast \ufb02uctuations (see Eq. (8.14)) underlying the previous analysis is legitimate. According to Eq. (8.14), the functional average of the exponentiated action can e\ufb00ectively be replaced by the exponential of the averaged action, exp(\u2212Sc) \u2248 exp(\u2212Sc). To explore the integrity of this assumption, let us explore the relevance of a typical correction. For example, an expansion of the exponent to second order in the potential operator would lead to expressions of the type 2 c2 d\u03c4 d\u03c4 cos(\u03b8s(\u03c4 ) +\u03b8 f (\u03c4 )) cos(\u03b8s(\u03c4 ) +\u03b8 f (\u03c4 )) 3, c where, here, we used the connected average \u02c6A \u02c6Bc \u2261 \u02c6A \u02c6B \u2212 ", "\u02c6A \u02c6B since the square of the d cos(\u03b8))2 is already included in our previous scheme. At \ufb01rst sight, averaged action \u223c (c this expression looks worrisome. By averaging over the fast \ufb01eld \u03b8f we are bound to generate a composite operator that depends in a non-local manner on two time arguments \u03c4 and \u03c4. Re-exponentiation of these objects would lead to an action much more complicated than the original. However, before losing hope, let us have a closer inspection of the fast \ufb01eld average of the expression above. Representing the cos-functions as sums of exponentials, we are led to consider 428 The renormalization group expressions of the type. exp [\u00b1i(\u03b8s(\u03c4 ) +\u03b8 f (\u03c4 ))] exp 2 & \u00b1i(\u03b8s(\u03c4 ) +\u03b8 f (\u03c4 )) / 3 c 2'\u221d = exp f i exp \u22122\u03c0g d\u03c9 2\u03c0 f \" \u00b1ei\u03c9\u03c4 \u00b1 ei\u03c9\u03c4 \u03b8(\u03c9) \u2212 1 \u00b1 cos(\u03c9(\u03c4 \u2212 \u03c4 ))! # d\u03c9 2\u03c0 |\u03c9|\u22121 \u039b = b\u22122g exp \u22134\u03c0g d\u03c9 2\u03c0\u03c9 cos(\u03c9(\u03c4 \u2212 \u03c4 )) \u2212 1 \u2248 0. \u039b/b exp \u00b1i d\u03c9 2\u03c0 f \u2212 exp \u2212\u03c0g 3 2 ei\u03c9\u03c4 \u03b8(\u03c9) d\u03c9 2\u03c0 f |\u03c9|\u22121 2 Here, the \u00b1 signs in the \ufb01rst line indicate the four di\ufb00erent possible combinations of signs; in the second equality we have used our previous results on the integrals over the high-lying frequency shell; and in the crucial third equality, we have noticed that, typically, (\u03c4 \u2212 \u03c4 ) > b/\u039b, 1/\u039b such that the oscillatory term integrates to something close to zero and can be neglected in comparison with the constant. The estimate above is limited to time arguments |\u03c4 \u2212 \u03c4 |\u039b/b > 1 outside a narrow strip |\u03c4 \u2212 \u03c4 | \u223c b/\u039b. To show that the contribution of composite operators from these domains is also negligible, we employ an argument that is symptomatic for RG analyses: consider the connected average 0 c2 |\u03c4 \u2212\u03c4 |<b/\u039b. d\u03c4 d\u03c4 cos(\u03b8s(\u03c4 ) +\u03b8 f (\u03c4 )) cos(\u03b8s(\u03c4 ) +\u03b8 f (\u03c4 )) \u221d b d\u03c4 cos2(\u03b8s(\u03c4 )) + \u03b8f (\u03c4 ) \u221d b\u22124g+1 d\u03c4 cos2(\u03b8s(\u03c4 )) + const., / c 1 c where the notation highlights the fact that the integration area is proportional to b, and we have used the fact that, for those narrow time windows, the \ufb01eld integration will be oblivious to the di\ufb00erence between \u03b8(\u03c4 ) and\u03b8 (\u03c4 ). The second proportionality is obtained by averaging the integrand along the lines of our previous calculations. After the rescaling of time and frequency to restore the old cuto\ufb00, the operator gets multiplied by another factor of b, i.e. the overall scaling factor is given by b2(1\u22122g). This factor tells us that the composite operator generated by higher-order cumulative expansion of the action is of less operator relevance than the contributions we are keeping. For values of the interaction g \u2248 1 not too distant from unity, the relative value of the coupling constants of the anomalous operator (\u221d b2(1\u22122g)) and of the standard cos \u03b8 contribution (\u221d b1\u2212g) will scale to 0 as b is increased. A classi\ufb01cation of operators according to their relevance under the RG transformation indeed forms the general basis for the limitation of low-energy actions to few contributions, namely those contributions that promise to be of strongest scaling relevance as larger and larger \ufb01eld \ufb02uctuations are probed. We will discuss this point more systematically in the next section. Before leaving this section, let us make a few general observations about the renormalization procedure. We \ufb01rst notice that it would have been futile to attack the problem by the \u201cplain\u201d perturbation theory developed in Chapter 5. The reason is that the propagator of the (0 + 1)-dimensional e\ufb00ective theory, |\u03c9|\u22121, leads to logarithmic divergences when integrated over unbounded frequency intervals, i.e. the present theory is again plagued by the UV/IR divergences observed above in di\ufb00erent contexts. In Section 5.1 we had argued 8.3 Renormalization group: general theory 429 that,", " in principle, a way to overcome these problems was to introduce a UV and, if needed, an IR cuto\ufb00 into the theory. However, we soon dismissed this option because it seemed evident that it would lead to spurious non-universal cuto\ufb00 dependences of physical results. How do these observations relate to what we are doing presently? Obviously, the present10 version of the RG procedure also relies on the introduction of a cuto\ufb00 regularizing the logarithmic UV divergences mentioned above; apparently, the RG procedure shares a lot of structures with the perturbative approach. But, somehow, we managed to extract the information in which we were interested \u2013 the dependence of the potential strength on longrange system parameters \u2013 in a manner independent of the cuto\ufb00.11 The key to obtaining this information was to introduce not one, but an entire hierarchy of cuto\ufb00s and to integrate over each of these domains recursively. Now, a subtle and important point is that this procedure does not imply that the cuto\ufb00 or, more generally, short-scale \ufb02uctuations of the model have silently made their way out of the theory. After all, the UV divergences mentioned before are manifestations of a large \u201cphase volume\u201d of \ufb01eld \ufb02uctuations that are likely to somehow a\ufb00ect the behavior of the system. To understand the \u201cimplicit\u201d way through which these \ufb02uctuations enter our results, let us return to a remark made on page 425. There, we had noted that, upon scaling frequency/time, each operator changes according to its physical dimension. An operator carrying the dimension [time]d would acquire a scaling factor bd. The scaling dimension of an operator predicted by its \u201cphysical\u201d dimension is called the naive scaling dimension, the canonical scaling dimension, or, for obscure reasons, the engineering dimension. The designations indicate, however, that these dimensions are not the last word on the actual scaling behavior of an operator. Indeed, the net result of the RG analysis was that our operator of interest, d\u03c4 cos \u03b8, an object of engineering dimension 1, changes according to b1\u2212g. The correction to the naive scaling dimension (presently, g) is called the anomalous dimension of an operator. Its origin lies in the (cuto\ufb00-dependent) phase volume of \ufb02uctuations co-determining the change of an operator during each RG step. Put di\ufb00erently, \u22121, acts as a \u201cgray we can say that the cuto\ufb00 \u039b, by itself a quantity of dimension [time] eminence\u201d implicitly a\ufb00ecting the scaling behavior of an operator. The anomalous scaling dimensions of the theory e\ufb00ectively determine its long-range observable behavior and, therefore, represent quantities of prime interest. 8.3 Renormalization group: general theory Having discussed two extended examples, we are in a position to attempt a reasonably general outline of the RG strategy. Suppose we are given a \ufb01eld theory de\ufb01ned through the 10 Below we become acquainted with UV regularization procedures that are not based on introduction of a cuto\ufb00. 11 One may object that the solutions of the \u03b2-functions given above actually do contain the bare cuto\ufb00, through an initial condition; they also depend on the bare coupling strength and, possibly, other \u201cnon-universal\u201d parameters. However, that need not worry us: in most applications (both experimental and theoretical) one is interested not so much in the \u201cabsolute value\u201d of physical observables (as these usually depend on unknown material parameters anyway) but rather in the way these observables change as a relevant control parameter is varied. The important feature found above is that the rate at which the e\ufb00ective potential strength varies with temperature, say, is largely universal and cuto\ufb00-independent. 430 action The renormalization group N S[\u03c6] \u2261 gaOa[\u03c6], a=1 where \u03c6 is some (generally multi-component) \ufb01eld, ga are coupling constants and Oa[\u03c6] a certain set of operators. For concreteness, one may think of these operators as Oa = ddx (\u2207\u03c6)n\u03c6m, i.e. as space\u2013time local operators involving powers of the \ufb01eld and its derivatives \u2013 although more general structures are conceivable.12 By \u201crenormalization of the theory,\u201d we refer to a scheme to derive a set of Gell-Mann\u2013Low equations describing the change of the coupling constants {ga} as fast \ufb02uctuations of the theory are successively integrated out. 8.3.1 Gell", "-Mann\u2013Low equations There are a number of methodologically di\ufb00erent procedures whereby the set of \ufb02ow equations can be obtained from the microscopic theory. Here, we formulate this step in a language adjusted to applications in statistical \ufb01eld theory (as opposed to, say, particle physics). While there is considerable freedom in the actual implementation of the RG procedure, all methods share the feature that they proceed in a sequence of three more or less canonical steps. I: Subdivision of the \ufb01eld manifold In the \ufb01rst step, one may decompose the integration manifold {\u03c6} into a sector to be integrated out, {\u03c6f }, and a complementary set, {\u03c6s}. For example: We may proceed according to a generalized block spin scheme and integrate over all degrees of freedom located within a certain structural unit in the base manifold {x}. (This scheme is adjusted to lattice problems where {x} = {xi} is a discrete set of points. However, as pointed out above, even then it is di\ufb03cult to implement analytically.) We could decide to integrate over a certain sector in momentum space. When this sector is de\ufb01ned to be a shell \u039b/b \u2264 |p| < \u039b, one speaks of a momentum shell integration. Naturally, within this scheme, the theory will be explicitly cuto\ufb00-dependent at intermediate stages. Alternatively, we may decide to integrate over all high-lying degrees of freedom \u03bb\u22121 \u2264 |p|. In this case, we will of course encounter divergent integrals. An elegant way to handle these divergences is to apply dimensional regularization. Within this approach one formally generalizes from integer dimensions d to fractional values d \u00b1. One motivation for doing so (for another, see below) is that, miraculously, the formal extension of the characteristic integrals appearing during the RG step to non-integer dimensions are \ufb01nite. As long as one stays clear of the dangerous values d = integer one can then safely monitor the 12 In our previous example of the Luttinger liquid, there appeared an operator represented in space-time, this operator is highly non-local. (d\u03c9/2\u03c0)\u03b8(\u03c9)|\u03c9|\u03b8(\u2212\u03c9). When 8.3 Renormalization group: general theory 431 dependence of the integrals on the IR cuto\ufb00 \u03bb\u22121. For a good introduction to dimensional generalization we refer to the textbook by Ryder.13 For a discussion of alternative schemes, such as the introduction of short-distance real space cuto\ufb00s underlying the so-called operator product expansion, we refer to the literature (see, e.g., the excellent text by Cardy.1) II: RG step The second, and central, part of the program is to actually integrate over short\u2013range \ufb02uctuations. As exempli\ufb01ed above, this step usually involves approximations. In most cases, one will proceed by a so-called loop expansion, i.e. one organizes the integration over the fast \ufb01eld \u03c6f according to the number of independent momentum integrals \u2013 loops14 \u2013 that occur after the appropriate contractions. Of course, this strategy makes sense only if we can guarantee that the contribution of loops of higher orders is in some sense small, a precondition that is, alas, often di\ufb03cult to meet. At any rate, to engage loop numbers as an expansion parameter, we \ufb01rst need to understand the key role played by spacedimensionality in the present context. We return to this point in Section 8.4. Following the procedure, an expansion over the fast degrees of freedom gives an action S[\u03c6s] \u2261 O g a a[\u03c6s], a in which coupling constants of the remaining slow \ufb01elds are altered. Notice that the integration over fast \ufb01eld \ufb02uctuations may (and usually does) lead to the generation of \u201cnew\u201d operators, i.e. operators that have not been present in the bare action. In such cases one has to investigate whether the newly generated operators are \u201crelevant\u201d (see below) in their scaling behavior. If so, the appropriate way to proceed is to include these operators in the action from the very beginning (with an a priori undetermined coupling constant). One then veri\ufb01es whether the augmented action represents a complete system, i.e. one that does not lead to the generation of operators beyond those that are already present. If necessary, one has to repeat this step until a closed system is obtained. III: Rescaling One next rescales frequency/momentum so that the rescaled \ufb01eld amplitude \u03c6 \ufb02uctuates on", " the same scales as the original \ufb01eld \u03c6, i.e. one sets q \u2192 bq, \u03c9 \u2192 bz\u03c9. Here, the frequency renormalization exponent or dynamical exponent z may be unity, two, or sometimes a non-integer value, depending on the e\ufb00ective dispersion relating frequency and momentum. We \ufb01nally note that the \ufb01eld \u03c6, as an integration variable, may be rescaled arbitrarily. Using this freedom, we select a term in the action which we believe governs the behavior of the \u201cfree\u201d theory \u2013 in a theory with elastic coupling this might, 13 L. H. Ryder, Quantum Field Theory (Cambridge University Press, 1996). 14 For the de\ufb01nition of loops, see Section 5.1. 432 The renormalization group for example, be the leading-order gradient operator \u223c ddr(\u2207\u03c6)2 \u2013 and require that it be strictly invariant under the RG step. To this end we designate a dimension Ld\u03c6 for the \ufb01eld, chosen so as to compensate for the factor bx arising after the renormalization of the operator. The rescaling \u03c6 \u2192 bd\u03c6 \u03c6, is known as \ufb01eld renormalization. It renders the \u201cleading\u201d operator in the action scale invariant. As a result of all these manipulations, we obtain a renormalized action S[\u03c6] = a g a Oa[\u03c6], which is entirely described by the set of changed coupling constants, i.e. the e\ufb00ect of the RG step is fully encapsulated in the mapping g = \u02dcR(g), relating the old value of the vector of coupling constants, g = {ga}, to the renormalized one, g = {g }. By letting the control parameter, \u2261 ln b, of the RG step assume in\ufb01nitesimal a values, one can make the di\ufb00erence between bare and renormalized coupling constants arbitrarily small. It is then natural to express the di\ufb00erence g \u2212 g = \u02dcR(g) \u2212 g in the form of a generalized \u03b2-function or Gell-Mann\u2013Low equation dg d = R(g), (8.17) where the right-hand side is de\ufb01ned through the relation R(g) = lim\u21920 \u22121( \u02dcR(g) \u2212 g). INFO As mentioned at the beginning of the section, the formulation of the RG step above is actually not the only one possible. For instance, in high-energy physics, other renormalization schemes appear to be more natural. In this area of physics, there is actually no reason to believe in the existence of a well-de\ufb01ned \u201cbare\u201d action with \ufb01nite coupling constants. (Contrary to the situation in condensed matter physics, the bare action of quantum electrodynamics, say, is in principle inaccessible.) However, one may legitimately require that, after an integration over UV-divergent \ufb02uctuations, the \u201crenormalized\u201d coupling constants of the theory (which, in turn, determine observables such as the physical electron mass) are \ufb01nite. One may then postulate that the bare coupling constants of the theory are actually in\ufb01nite. The value of these in\ufb01nities is \ufb01ne-tuned so as to combine with the \ufb02uctuation-induced \u201cin\ufb01nities\u201d to realize \ufb01nite renormalized coupling constants. Alternatively, one may deliberately add extra operators, counter-terms, to the action which are designed so as to cancel divergences due to \ufb02uctuations. However, the net result of all these RG schemes (which are by and large equivalent) is a mapping describing the \ufb02ow of the coupling constants upon variation of a control parameter. 8.3 Renormalization group: general theory 433 Figure 8.4 The fractal Julia set is self-similar in the sense that any sub-region of it contains the full information of the original set. 8.3.2 Analysis of the Gell-Mann\u2013Low equation The Gell-Mann\u2013Low equation (8.17) represents the principal result of an RG analysis. Thinking of the control parameter as a kind of \u201c\ufb02ow parameter,\u201d one may identify this equation as a generalized dynamical system, namely the system describing the evolution of the e\ufb00ective coupling constants of a model upon changing length or time scales. As with any dynamical system, the prime structural characteristic of the set of equations (8.17) is the set of \ufb01xed points, i.e", ". the submanifold {g\u2217} of points in coupling constant space which are stationary under the \ufb02ow: R(g\u2217) = 0. Once the coupling constants are \ufb01ne-tuned to a \ufb01xed point, the system no longer changes under subsequent RG transformations. In particular it remains invariant under the change of space/time scale associated with the transformation. Alluding to the fact that they look the same no matter how large a magnifying glass is used, systems with this property are referred to as self-similar. (For example, fractals such as the Julia set shown in Fig. 8.4 are paradigmatic examples of self-similar systems; the magni\ufb01cation of any sub-region of the fractal looks identical to the full system.) Now, to each system, one can attribute at least one intrinsic length scale, namely the length \u03be determining the exponential decay of \ufb01eld correlations. However, the existence of a \ufb01nite, and pre-determined, intrinsic length scale clearly does not go together with invariance under scale transformations. We thus conclude that, at a \ufb01xed point, either \u03be = 0 (not so interesting), or \u03be = \u221e. However, a diverging correlation length \u03be \u2192 \u221e is a hallmark of 434 The renormalization group Figure 8.5 Showing the RG \ufb02ow in the vicinity of a \ufb01xed point with two irrelevant (\u03c61, \u03c63) and one relevant (\u03c62) scaling \ufb01elds. The manifold S de\ufb01ned through the vanishing of the relevant \ufb01eld, \u03c62 = 0, is called a critical surface. On this submanifold, the RG \ufb02ow is directed towards the \ufb01xed point f. Deviations o\ufb00 criticality make the system approach one of the stable \ufb01xed points f1 and f2. a second-order phase transition. We thus tentatively identify \ufb01xed points of the RG \ufb02ow as candidates for \u201ctransition points\u201d of the physical system. (For a more comprehensive review of phase transitions and the critical phenomena accompanying them, see the Info block starting on page 436 below.) This being so, it is natural to pay special attention to the behavior of the \ufb02ow in the immediate vicinity of the \ufb01xed-point manifolds. If the set of coupling constants, g, is only close enough to a \ufb01xed point, g\u2217, it will be su\ufb03cient to consider the linearized mapping R(g) \u2261 R((g \u2212 g\u2217) +g \u2217) W (g \u2212 g\u2217), Wab = \u2202Ra \u2202gb. g=g\u2217 To explore the properties of \ufb02ow, let us assume that we had managed to diagonalize the matrix W. Denoting the eigenvalues by \u03bb\u03b1, \u03b1 = 1,..., N, and theleft -eigenvectors15 by \u03c6\u03b1, we have \u03b1 W = \u03c6T \u03c6T \u03b1 \u03bb\u03b1. The advantage of proceeding via the unconventional set of left-eigenvectors is that it allows us to conveniently express the \ufb02ow of the physical coupling constants under renormalization. To this end, let v\u03b1 be the \u03b1th component of the vector g \u2212 g\u2217 when represented in the basis {\u03c6\u03b1}: v\u03b1 = \u03c6T \u03b1 (g \u2212 g\u2217). 15 Since there is no reason for W being symmetric, the left- and right-eigenvectors may be di\ufb00erent. 8.3 Renormalization group: general theory 435 These components display a particularly simple behavior under renormalization: dv\u03b1 d = \u03c6T \u03b1 d d (g \u2212 g\u2217) = \u03c6T \u03b1 W (g \u2212 g\u2217) = \u03bb\u03b1\u03c6T \u03b1 (g \u2212 g\u2217) = \u03bb\u03b1v\u03b1. Under renormalization, the coe\ufb03cients v\u03b1 change by a mere scaling factor \u03bb\u03b1, wherefore they are called scaling \ufb01elds \u2013 a somewhat unfortunate nomenclature. (The coe\ufb03cients v\u03b1 are actually not \ufb01elds but simply a set of -dependent coe\ufb03cients, the vector of coupling constants when expressed in the basis of eigenvectors \u03c6\u03b1.) These equations are trivially integrated to obtain v\u03b1() \u223c exp(\u03bb\u03b1). This result suggests a discrimination between at least three di\ufb00erent types of scaling \ufb01elds: For \u03bb\u03b1 > 0 the \ufb02ow is directed", " away from the critical point. The associated scaling \ufb01eld is said to be relevant (in the sense that it forcefully drives the system away from the critical region). In Fig. 8.5, v2 is a relevant scaling \ufb01eld. In the complementary case, \u03bb\u03b1 < 0, the \ufb02ow is attracted by the \ufb01xed point. Scaling \ufb01elds with this property (v1, v3) are said to beirrelevant. 16 Finally, scaling \ufb01elds which are invariant under the \ufb02ow, \u03bb\u03b1 = 0, are termed marginal.17 The distinction of relevant/irrelevant/marginal scaling \ufb01elds in turn implies a classi\ufb01cation of di\ufb00erent types of \ufb01xed points: Firstly, there are stable \ufb01xed points, i.e. \ufb01xed points whose scaling \ufb01elds are all irrelevant or, at worst, marginal. These points de\ufb01ne what we might call \u201cstable phases of matter\u201d: when you release a system somewhere in the parameter space surrounding any of these attractors, it will scale towards the \ufb01xed point and eventually sit there. Or, expressed in more physical terms, looking at the problem at larger and larger scales will make it more and more resemble the in\ufb01nitely correlated self-similar \ufb01xed-point con\ufb01guration. (Recall the example of the high-temperature \ufb01xed line of the one-dimensional Ising model encountered earlier.) By construction, the \ufb01xed point is impervious to moderate variations in the microscopic morphology of the system, i.e. it genuinely represents what one might call a \u201cstate of matter.\u201d Complementary to stable \ufb01xed points, there are unstable \ufb01xed points. Here, all scaling \ufb01elds are relevant (cf. the T = 0 \ufb01xed point of the 1-D Ising model). These \ufb01xed points represent the concept of a Platonic ideal: you can never get there and, even if you managed 16 The terminology \u201cirrelevant\u201d indicates that a scaling \ufb01eld of negative dimension usually does not play much of a physical role. There are, however, exceptions to this rule. For instance, it may happen that the free energy of the system depends in a singular manner on an irrelevant scaling variable \u2013 in which case the variable is called dangerously irrelevant. Dangerously irrelevant scaling variables not only strongly a\ufb00ect the outcome of the theory, but also invalidate the applicability of a number of established concepts of RG theory (such as the scaling laws to be discussed below). 17 A marginal scaling \ufb01eld corresponds to a direction in coupling constant space with vanishing partial derivative, \u2202\u03c6\u03b1 R|g\u2217=0 = 0. In this case, to obtain a re\ufb01ned picture, one sometimes considers the second-order derivative, \u22022 \u03b1. For x > 0 \u03c6\u03b1 (x <0) the \ufb01eld has the status of a marginally relevant (irrelevant) scaling \ufb01eld. It is relevant (irrelevant) on account of the non-vanishing direction of the \ufb02ow. However, it is also \u201cmarginal\u201d because the speed of the \ufb02ow decreases upon approaching the critical regime. R|g\u2217=0 \u2261 2x. In the vicinity of the \ufb01xed point, the scaling \ufb01eld then behaves as dv\u03b1 = xv2 436 The renormalization group to approach it closely, the harsh conditions of reality will make you \ufb02ow away from it. Although unstable \ufb01xed points do not correspond to realizable forms of matter, they are of importance inasmuch as they \u201corient\u201d the global RG \ufb02ow of the system. Finally, there is the generic class of \ufb01xed points with both relevant and irrelevant scaling \ufb01elds. These points are of particular interest inasmuch as they can be associated with phase transitions. To understand this point, we \ufb01rst note that the r eigenvectors \u03a6\u03b1 associated with irrelevant scaling \ufb01elds span the tangent space of an (r)-dimensional manifold known as the critical surface. (A schematic illustration for the case r = 2 is shown in Fig. 8.5.) This critical manifold forms the basin of attraction of the \ufb01xed point, i.e. whenever a set of physical coupling constants g is \ufb01ne-tuned so that g \u2208 S, the expansion in terms", " of scaling \ufb01elds contains only irrelevant contributions and the system will feel attracted to the \ufb01xed point as if it were a stable one. However, the smallest deviation from the critical surface introduces a relevant component driving the system exponentially away from the \ufb01xed point. A sketch of the resulting \ufb02ow is shown in Fig. 8.5 for the case of just one relevant scaling \ufb01eld. For example, in the case of the ferromagnetic phase transition \u2013 discussed in more detail in the next section \u2013 deviations from the critical temperature Tc are relevant. If we consider a system only slightly above or below Tc, it may initially (on intermediate length scales) appear to be critical. However, upon further increasing the scale, the relevant deviation will grow and drive the system away from criticality, either towards the stable high-temperature \ufb01xed point of the paramagnetic phase (T > Tc) or towards the ferromagnetic low-temperature phase (T < Tc). This picture actually suggests that systems with generic \ufb01xed points typically possess complementary stable \ufb01xed points, i.e. \ufb01xed points towards which the \ufb02ow is directed after it has left the critical region. We also notice that a scaling direction that is relevant at one \ufb01xed point (e.g. \u03a62 at the critical \ufb01xed point) may be irrelevant at others (\u03a62 at the high- and low-temperature \ufb01xed points). INFO The discussion above suggests that the concept of renormalization is intimately linked to the theory of phase transitions and critical phenomena, the traditional platform for the development of the subject in the literature. In view of the existing wealth of literature (and acknowledging the fact that we are approaching the \ufb01eld from a more operational perspective), we shall not endeavor to present another \u201cintroduction to the theory of critical phenomena.\u201d Rather, we will summarize in a concise, but hopefully self-contained, manner, those few tenets and principles that are necessary to place the concept of renormalization into a larger physical context. The most fundamental18 signature of a phase transition is its order parameter, M, i.e., a quantity whose value unambiguously identi\ufb01es the phase of the system. Examples from classical statistical mechanics include the magnetization for the ferromagnetic\u2013paramagnetic transition, the density for the liquid\u2013vapor transition, the order parameter amplitude for the BCS transition, etc. (However, to keep the terminology concrete, we shall mostly use the language of the ferromagnetic transition in the following.) 18 Notice that there are transitions whose order parameter is actually unknown. A famous example is the quantum Hall transition discussed in more detail in Section 9.3.4. 8.3 Renormalization group: general theory 437 m H Tc T Figure 8.6 Phase diagram of the ferromagnetic transition. Tuning a magnetic \ufb01eld at \ufb01xed temperature T < Tc through zero causes the magnetization to jump discontinuously: [0, Tc] is a line of \ufb01rst-order transitions. This line terminates in the unique second-order transition point of the system, (T = Tc, H = 0). Lowering the temperature at H = 0 causes the non-analytic, but continuous, development of a \ufb01nite magnetization at T < Tc. Transitions between di\ufb00erent phases of matter fall into two large categories.19 In \ufb01rst-order phase transitions the order parameter exhibits a discontinuous jump across the transition line while, in the complementary class of second-order transitions, the order parameter changes in a non-analytic but continuous manner. (The two cases are exempli\ufb01ed in Fig. 8.6 by the classical ferromagnet.) The phenomenology of second-order transitions is generally richer than that of \ufb01rst-order transitions. As a thermodynamic state variable, the order parameter is coupled to a conjugate \ufb01eld, H : M = \u2212\u2202H F, where F is the free energy. At a second-order transition, M changes nonanalytically, which means that the second-order derivative, a thermodynamic susceptibility, \u03c7 = \u2212\u22022 H F, develops a singularity. Now, you may recall from the discussion of the \ufb02uctuation dissipation theorem that the susceptibility is intimately linked to the \ufb01eld \ufb02uctuation behavior of the system. More precisely, \u03c7 is proportional to the integral over the correlation function C determining the \ufb02uctuation behavior of the \ufb01elds (cf. Eq. (8.3)). A divergence of the susceptibility implies the accumulation of in", "\ufb01nitely long-range \ufb01eld \ufb02uctuations. The divergence of the susceptibility goes hand in hand with non-analytic and/or singular behavior of all sorts of other physical quantities. In fact, an even stronger statement can be made. We have seen that, right at the transition/\ufb01xed point, the system is self-similar. This implies that the behavior of its various characteristics must be described by power laws. Referring for a more substantial discussion to Section 8.3.3 below, we here merely support this statement by a heuristic argument. Consider a function f (t), where f is representative of an observable of interest and t is a control parameter (a scaling \ufb01eld) determining the distance to the transition point. In the immediate vicinity of the transition point, f is expected to \u201cscale,\u201d i.e. under a change of the length scale x \u2192 x/b, t \u2192 tb\u2212Dt, the function f must, at most, change by a factor re\ufb02ecting its own scaling dimension. f (t) = bDf f (tb\u2212Dt ). (A more serious, structural change of the function would be in con\ufb02ict with asymptotic self-similarity.) Mathematically speaking, this equation amounts to homogeneity of the function f, equivalently expressed by f \u223c tDf /Dt. 19 Readers absolutely unfamiliar with the thermodynamics of phase transitions may wish to consult the corre- sponding section of a textbook on statistical mechanics. 438 The renormalization group The set of di\ufb00erent exponents characterizing the relevant power laws occurring in the vicinity of the transition are known as critical exponents. For at least four di\ufb00erent reasons, the set of critical exponents represents the most important structural \ufb01ngerprint of a transition: 1. They carry universal signi\ufb01cance, i.e. we do not have to invent a set of critical exponents for each transition anew. (For example, the divergence of the correlation length, \u03be \u223c |t|\u2212\u03bd, is characterized by a critical exponent commonly, and irrespective of the particular transition under consideration, denoted by \u03bd.) 2. The set of critical exponents carries the same information as the set of exponents of the scaling \ufb01elds, i.e. knowledge of the critical exponents is equivalent to the knowledge of the linear dynamical system characterizing the \ufb02ow in the transition region. (In fact, the set of critical exponents overdetermines the scaling \ufb01eld exponents, i.e. it contains redundancy. For example, of the six critical exponents characterizing the magnetic transition, only two are independent. The others are interrelated by20 scaling laws or exponent identities to be discussed below. 3. Critical exponents are fully universal; they are numbers depending, at most, on dimensionless characteristics such as the space-time dimensionality or number of components of the order parameter. 4. Perhaps most importantly, the critical exponents represent quantities that can be measured. In fact, their universality and structural importance make them quantities of prime experimental interest. In the following, let us brie\ufb02y enumerate the list of the most relevant exponents, \u03b1, \u03b2, \u03b3, \u03b4, \u03b7, \u03bd, and z.21 Although we shall again make use of the language of the magnetic transition, it is clear that (and, indeed, how) the de\ufb01nitions of most exponents generalize to other systems. \u03b1: In the vicinity of the critical temperature, the speci\ufb01c heat C = \u2212T \u22022 T F scales as C \u223c |t|\u2212\u03b1, where t = (T \u2212 Tc)/Tc measures the distance to the critical point. Note that, by virtue of this de\ufb01nition, a non-trivial statement has been made: although the phases above and below the transition are essentially di\ufb00erent, the scaling exponents controlling the behavior of C are identical. The same applies to most other exponents listed below. \u03b2: Approaching the transition temperature from below, the magnetization vanishes as M \u2261 \u2212\u2202H F \u223c (\u2212t)\u03b2. H0 \u03b3: The magnetic susceptibility behaves as \u03c7 \u2261 \u2202hM |h0 \u223c |t|\u2212\u03b3. \u03b4: At the critical temperature, t = 0, the \ufb01eld dependence of the magnetization is given by M \u223c |h|1/\u03b4. \u03bd: Upon approaching the transition point, the correlation length diverges as \u03be \u223c |t|\u2212\u03bd. \u03b7: This implies that the correlation function, * C(r) \u223c 1 |r|d\u2212", "2+\u03b7, exp[\u2212|r|/\u03be], |r| \u03be, |r| \u03be, crosses over from exponential to a power law scaling behaviour at the length scale \u03be. To motivate the power, one may notice that C \u223c \u03c6\u03c6 carries twice the dimension of the \ufb01eld 20 Unfortunately, the language used in the \ufb01eld of critical phenomena makes excessive use of the pre\ufb01x \u201cscaling.\u201d 21 Historically, the exponents are drawn from the \ufb01rst six letters of the Greek alphabet. The exceptional designation of the last exponent, z, betrays the fact that quantum dynamical \ufb02uctuations were considered only later. 8.3 Renormalization group: general theory 439 \u03c6. The engineering dimension of the latter follows from the requirement that the gradient ddr (\u2207\u03c6)2 be dimensionless: [\u03c6] =L (2\u2212d)/2, according to which C(r) has canonoperator \u223c ical dimension L2\u2212d. The exponent \u03b7, commonly referred to as theanomalous dimension of the correlation function, measures the mismatch between the observed and the canonical dimension. z: A quantum theory can, to a large extent, be viewed as a kind of classical theory in d + 1 dimensions. The theory is \u201cquantum critical\u201d if the e\ufb00ective classical theory contains a critical point. In the vicinity of that point large \ufb02uctuations are observed in both the d spatial directions and the temporal \u201cdirection.\u201d However, the di\ufb00erent physical origin of these dimensions manifests itself in the scaling being anisotropic. Denoting the correlation length in the temporal direction by \u03c4, we de\ufb01ne \u03c4 \u223c \u03bez, where deviations z = 1 in the dynamical exponent measure the degree of anisotropy. Now, a moment\u2019s thought shows that, of the six classical exponents, only a few can be truly independent. Previously we have noted that, modulo irrelevant perturbations, the \ufb02ow in the vicinity of a transition point is controlled by the relevant scaling \ufb01elds. Referring for a more quantitative discussion to Section 8.4 below, we anticipate that, for the magnetic transition, the magnetic \ufb01eld will certainly represent a relevant perturbation (a fact readily expressed by the positivity of the exponent \u03b4). Moreover, deviations from the critical temperature, t = 0, are also relevant.22 However, for the magnetic transition, that exhausts the list; in the asymptotic vicinity of the transition, the \ufb02ow is controlled by a two-dimensional dynamical system. This suggests that four constraining equations should reduce the set of six classical exponents to only two independent ones. Historically, these scaling laws were discovered one by one (at a time when the underlying connections to the system of \u201cscaling \ufb01elds\u201d had not been fully appreciated). For the sake of reference, these constraint equations (along with the names of the people who discovered them) are listed below. In Section 8.3.3 below, we exemplify how the scaling laws can be transparently derived from the intrinsic structure of the theory. Fisher Rushbrooke Widom Josephson \u03bd(2 \u2212 \u03b7) = \u03b3 \u03b1 + 2\u03b2 + \u03b3 = 2 \u03b2(\u03b4 \u2212 1) = \u03b3 2 \u2212 \u03b1 = \u03bdd For practical purposes, we need only compute/measure two exponents \u2013 no matter which \u2013 to fully specify the scaling structure of the theory. In the next section we discover that the dynamical system of scaling \ufb01elds encapsulates practically all information about \u201ccritical\u201d \ufb02uctuation phenomena accompanying a phase transition. However, for the moment, we shall restrict ourselves to the discussion of one more aspect of conceptual importance, namely universality. In fact, the majority of critical systems can be classi\ufb01ed into a relatively small number of universality classes. Crudely speaking, leaving apart more esoteric classes of phase transitions there are O(101) fundamentally di\ufb00erent types of \ufb02ow recurrently appearing in practical applications. This has to 22 If you \ufb01nd it di\ufb03cult to think of temperature as a \u201ccoupling constant,\u201d remember that, in our derivation of the \u03c64-model as the relevant theory of the magnetic transition, the coupling constant of the \u201cmass operator\u201d r ddr \u03c62 turned out to be proportional to the reduced temperature t = |T \u2212 Tc|/Tc. 440 The renormalization group be compared with the near in\ufb01nity of di\ufb00erent physical systems that display critical phenomena. Why, then, is it", " that the plethora of all these transitions can be grouped into a very limited set of di\ufb00erent universality classes? Remarkably, the origin of this universality can readily be understood from the concept of critical surfaces. systems theories Imagine, then, an experimentalist exploring a system that is known to exhibit a phase transition. Motivated by the critical phenomena that accompany phase transitions, the available control parameters Xi (temperature, pressure, magnetic \ufb01eld, etc.) will be varied until the system begins to exhibit large \ufb02uctuations. On a theoretical level, the variation of the control parameters determines the initial values of the coupling constants of the model (as they functionally depend on the Xis through their connection to the microscopic Hamiltonian). In Fig. 8.5 the curve in coupling constant space de\ufb01ned in this way is indicated by \u03b3. For microscopic parameters corresponding to a point above or below the critical manifold, the system asymptotically (i.e. when looked at at su\ufb03ciently large scales) falls into either the \u201chigh-\u201d or the \u201clow-temperature\u201d regime (as indicated by the curves branching out from \u03b3 in Fig. 8.5). However, eventually the trajectory through parameter space will intersect the critical surface. For this particular set of coupling constants, the system is critical. As we look at it on larger and larger length scales, it will be attracted by the \ufb01xed point at S, i.e. it will display the universal behavior characteristic of this particular point. This is the origin of universality: variation of the system parameters in a di\ufb00erent manner (or for that matter considering a second system with di\ufb00erent material constants) will generate a di\ufb00erent trajectory g\u03b1({Xi}) = \u03b3. However, as long as this trajectory intersects with S, it is guaranteed that the critical behavior will exhibit the same universal characteristics (controlled by the unique \ufb01xed point). universal predictions In fact a more far-reaching statement can be made. Given that there is an in\ufb01nity of systems exhibiting transition behavior (symbolically indicated by the row of boxes in the upper part of the \ufb01gure above) while there is only a very limited set of universality classes (the set of boxes on the left), many systems of very di\ufb00erent microscopic morphology must have the same universal behavior. More formally, di\ufb00erent microscopic systems must map onto the same critical low-energy theory. Examples of these coincidences include (to mention but a few entries of an endless list) the equivalence of the disordered Luttinger liquid to a Josephson junction (cf. Problem 6.7), the equivalence of models of planar magnets (see Section 8.6 below) to two-dimensional classical Coulomb plasmas, and the equivalence of the liquid\u2013gas transition to the ferromagnetic transition. (In all cases, \u201cequivalence\u201d means that the systems exhibit identical scaling behavior and, therefore, fall into the same universality class.) Further coincidences of this type will be encountered below. 8.3 Renormalization group: general theory 441 8.3.3 Scaling theory Previously, we have seen that the dynamical system of scaling \ufb01elds encodes a wealth of information on the large-scale structure and on the phases of a physical system. However, we have not yet established a connection between the concept of renormalization and concrete (i.e. experimentally accessible) data. This is the subject of the present section. Imagine, then, that we had represented some observable of experimental interest, X, in the language of the functional integral. According to the discussion of the previous chapter this means that we have managed to express X = C(pi, g\u03b1), p as the sum over an n-point correlation function C(pi, g\u03b1) = (\u00b7 \u00b7 \u00b7) \u03c6\u03c6 \u00b7 \u00b7 \u00b7\u03c6 \u03c6, where the (symbolic) notation indicates that C may depend both on the momentum scale at which it is evaluated (e.g. through the explicit momentum dependence of current operators, etc.) and on the coupling constants. The ellipsis (\u00b7 \u00b7 \u00b7) stand for optional algebraic elements entering the de\ufb01nition of the correlation function. We next build on our assumption of renormalizability of the theory, i.e. we make use of the fact that we can evaluate C before or after an RG step; the result must be the same. On the other hand, the RG transformation will, of course, not leave the individual constituents entering the de\ufb01nition of C invariant; it will change coupling constants, g\u03b1, the momenta pi, and the \ufb01eld amplitudes \u03c6 according to", " the prescriptions formulated in the previous section. Expressed in a single formula, C(pi, g\u03b1) = bnd\u03c6 C(pib, g\u03b1b\u03bb\u03b1 ), (8.18) where we have simpli\ufb01ed the notation by assuming that the coupling constants themselves scale (for, otherwise, the matrix elements of a linear transformation mediating between the coupling constants and the scaling \ufb01elds would appear). For notational convenience, let us also assume that the \ufb01xed point values of the coupling constants are speci\ufb01ed in such a way that g\u2217 = 0. The factor bnd\u03c6 accounts for the explicit rescaling of the n \ufb01elds entering the de\ufb01nition of C. Notice that Eq. (8.18) presents a remarkable statement. Although the three di\ufb00erent elements (\u03c6, pi, g\u03b1) contributing to the correlation function change under the transformation in seemingly unrelated manners, the net result of the concerted rescaling is nil. Indeed, Eq. (8.18) serves as a starting point for the derivation of various relations of immediate practical relevance. Let us return to a principle already employed in connection with the one-dimensional Ising model. For concreteness, imagine that we are working under conditions where there is just a Scaling functions 442 The renormalization group single relevant scaling \ufb01eld g1, while all g\u03b1>1 are irrelevant (or, for that matter, marginal). We can then write C(pi, g1, g\u03b1) = bnd\u03c6 C(pib, g1b\u03bb1, g\u03b1b\u03bb\u03b1 ) = g \u2212nd\u03c6/\u03bb1 1 g11\u2248 g \u22121/\u03bb1 1 C(pig, 1, 0) \u2261 g \u2212nd\u03c6/\u03bb1 1 C(pig \u22121/\u03bb1 1, 1, g\u03b1g \u2212\u03bb\u03b1/\u03bb1 1 ) \u2212nd\u03c6/\u03bb1 1 F (pig \u22121/\u03bb1 1 ). Here, we have used the freedom of arbitrarily choosing the parameter b to set g1b\u03bb1 = 1 while, in the third equality, we have assumed that we are su\ufb03ciently close to the transition that the dependence of C on irrelevant scaling \ufb01elds is inessential. The function F de\ufb01ned through the relation C(pi, g1) = g \u2212nd\u03c6/\u03bb1 1 F (pig \u22121/\u03bb1 1 ), (8.19) is an example of a scaling function. Alternatively (for example, if C represents a thermodynamic observable or a global transport coe\ufb03cient) we might be interested in the correlation function C(g1, g\u03b1) \u2261 C(pi = 0, g1, g\u03b1) at zero external momentum pi = 0. In this case, a typical question to ask would be the dependence of C on the most relevant and the second most relevant control parameter g2 (where we leave unspeci\ufb01ed whether g2 is relevant, marginal, or irrelevant). Following the same logic as above, one obtains C(g1, g2) = g \u2212nd\u03c6/\u03bb1 1 \u02dcF (g2g \u2212\u03bb2/\u03bb1 1 ), with some di\ufb00erent scaling function \u02dcF. INFO As an example particularly relevant to the comparison between analytical theory and numerics, we note the concept of \ufb01nite-size scaling. While analytical theories are most conveniently formulated in the thermodynamic limit, numerical simulations are carried out for systems of still very limited size. The need to compare theory and numerical simulations motivates the need to explicitly keep track of the system size under renormalization. Indeed, the system size L has dimension [length] and, therefore, gets rescaled as L \u2192 L/b. Setting L/b = 1, we obtain a scaling function with explicit system size dependence. G(g\u03b1, L) = Lnd\u03c6 Ffs(g\u03b1L\u03bb\u03b1 ), While the construction of any particular scaling function may be context-dependent, the principle behind the derivation is general: once the scaling behavior of a correlation function is known, the arbitrariness of the scaling parameter b can be used to reduce the number of independent variables by one. The reduced correlation function is called a scaling function. As with the response functions discussed in the previous chapter, scaling functions also represent a prime interface between theory and experiment. Experimentally, the measurement of an observable X in its dependence on a number of relevant system parameters, t and h say, results in a multi-parameter function X(t, h). In fact, a better way to think about this object is as a set of one-dimensional functions Xh(t) depending", " on a parameter h. (This is because, in experiment, one typically varies only a single control parameter, e.g. temperature at \ufb01xed magnetic \ufb01eld.) Scaling implies that all these functions collapse onto 8.3 Renormalization group: general theory 443 a generic one-dimensional23 pro\ufb01le, if only the data are plotted as a function of the relevant scaling parameter thx. This mechanism can be exploited in several di\ufb00erent ways. For example, if there is not yet a theory of the transition phenomenon in question, an experimentalist may empirically identify the relevant scaling parameters and pose the explanation of the observed scaling exponent x \u2013 by construction a fully universal number \u2013 as a challenge to theorists.24 Conversely, theorists may suggest a scaling exponent that can be put to the test by checking whether the experimental data collapse onto this exponent. Summarizing, one of the great virtues of the concept of scaling is that it condenses the information exchange between experiment and theory (and analytical theory and numerics for that matter) into a small set of universal numbers. INFO For the sake of completeness we mention that, especially in the \ufb01eld theoretical community, the information encapsulated in the scale-dependent correlation functions is often represented in a di\ufb00erent manner. Starting out from the relation C(pi, g\u03b1) = end\u03c6 C(pie, g\u03b1()), where we have set b = e, we can use the-independence of the left-hand side to write 0 = d end\u03c6 C(pie, g\u03b1()). (Notice that, here, we do not need to be in the asymptotic scaling regime, d i.e., for the sake of the present construction, the -dependence of the coupling constants need not be explicitly exponential.) We next carry out the -di\ufb00erentiation to obtain & & n de,\u03c6 +'\u03b7 2 + \u2202 + \u03b2\u03b1(g\u03b1)\u2202g\u03b1 C(pie, g\u03b1()) = 0. (8.20)'Here, de,\u03c6 is the engineering dimension of the \ufb01eld \u03c6 and \u03b7/2 =d \u03c6 \u2212de,\u03c6 its anomalous dimension (see the de\ufb01nition of \u03b7 in the Info block starting on page 436). Further, the partial derivative \u2202 acts on the explicit scale dependence of the momentum arguments (or any other explicitly scaledependent argument for that matter). Finally, \u03b2\u03b1(g\u03b1) is the \u03b2-function de\ufb01ned above. Equation (8.20) is known as a renormalization group equation. Both the RG equation and the scaling form that we used to derive it equivalently express the scaling behavior of the correlation function. Scaling functions and critical exponents Another important aspect of scaling theory is that it can be used to disclose relations between the seemingly independent25 critical exponents of the theory. For the sake of concreteness, let us consider the case of the ferromagnetic transition, i.e. a transition we have previously characterized in terms of six critical exponents \u03b1,..., \u03b7 (see page 438). However, the \ufb02ow in the vicinity of the magnetic \ufb01xed point is controlled by only two relevant scaling \ufb01elds, the (reduced) temperature t and the reduced magnetic \ufb01eld h \u2261 H/T. Neglecting 23 For an n-dimensional data set, the collapse is to an (n \u2212 1)-dimensional functional set. 24 Parenthetically, one may note that the empirical collapse of experimental data onto scaling functions requires a lot of skill. For example, if the data set consists of a number of functional \u201cpatches\u201d of only limited overlap, it is quite \u201ceasy\u201d to construct a scaling function of, in fact, almost any desired power law dependence. Data of this type tend to contain a lot of statistical uncertainty, which can easily lead to erroneous conclusions. 25 After all, the critical exponents describe the behavior of quite di\ufb00erent physical observables in the transition region. 444 The renormalization group irrelevant perturbations, we thus conclude that, under a renormalization group transformation, the reduced free energy f = F/T Ld will behave as26 f (t, h) = b\u2212df (tbyt, hbyh ). We next \ufb01x tbyt = 1 to reduce the number of independent variables to one: f (t, h) = td/yt \u02dcf (h/tyh/yt ). (8.21) Containing the complete thermodynamic information, Eq. (8.21) is all that we need to compute the critical exponents. Indeed, comparing with the de\ufb01nitions summarized on page 438, it is straightforward", " to show that \u03b1 = 2 \u2212 d yt, \u03b4 = yh d \u2212 yh, \u03b2 = \u03bd = d \u2212 yh yt, \u03b3 = 2yh \u2212 d yt, 1 yt, \u03b7 = 2 + d \u2212 2yh, \u23ab \u23aa\u23aa\u23ac \u23aa\u23aa\u23ad (8.22) from where follow the cross-relations summarized in the table on page 439 by direct comparison. These relations illustrate our previous assertion that, conceptually, the dimensions of the relevant scaling \ufb01elds have a more fundamental status than the critical exponents. EXERCISE Verify these statements. To obtain the \ufb01fth relation, the hyperscaling relation, notice that, under a change of scale, \u03be \u2192 b\u03be. On the other hand, we know that t \u223c \u03be\u22121/\u03bd. The sixth relation is obtained from Eq. (8.3) by a substitution of the de\ufb01nition of the spatial pro\ufb01le of the correlation function in terms of the critical exponent \u03b7 into the integral to obtain a relation between the critical exponents \u03b3 and \u03b7 (Fisher\u2019s scaling law). 8.4 RG analysis of the ferromagnetic transition In the previous section, we became acquainted with some fundamental elements of the structure of RG analyses, and their connection to the theory of critical phenomena. Being kept at a general and conceptual level, the discussion may have seemed somewhat abstract. Therefore, to elucidate the concepts introduced above, and to introduce some more elements of the RG, we turn now to a concrete application of the approach to the classical theory of the (uniaxial) ferromagnetic (or liquid\u2013gas) transition. In Section 5.1, the \u03c64-theory was identi\ufb01ed as an e\ufb00ective low-energy model of the ferromagnetic system. However, beyond the mean-\ufb01eld, we have not yet applied the model to explore the universal characteristics of the transition. In the following, we shall see that RG methods, and only RG methods, can be applied to successfully understand much of the intriguing behavior displayed by the (d >2)-dimensional Ising model in the vicinity of its phase transition. 26 Here we have made use of the fact that the reduced free energy does not carry an anomalous dimension. By de\ufb01nition, the free energy F = \u2212T ln Z does not change under renormalization (which after all, merely amounts to representing the number Z through functional integrals of di\ufb00erent space-time resolution). Thus, the scaling of the reduced free energy is entirely carried by the prefactor L\u2212d. 8.4 RG analysis of the ferromagnetic transition 445 8.4.1 Preliminary dimensional analysis The \ufb01rst question that we wish to address has a somewhat technical status: with what justi\ufb01cation was the Ising model represented in terms of the model action27 S[\u03c6] = ddr \u03c62 + (\u2207\u03c6)2 + \u03c64 \u2212 h\u03c6, (8.23) r 2 1 2 \u03bb 4! i.e. why was it possible to neglect both higher powers and gradients of the \ufb01eld \u03c6 that are surely present in the exact reformulation of the Ising problem in terms of \u03c6-variables? To rationalize the neglect of these terms, we proceed by dimensional analysis. Anticipating that the \u201creal\u201d dimensions carried by the operators in the action will be not too far from their engineering dimensions (see below), we begin by exploring the latter. We proceed along the lines of the general scheme outlined in the previous chapter and attribute a dimension (\u2207\u03c6)2 in the action. This entails the choice [\u03c6] = of unity to the leading gradient term L(2\u2212d)/2, from where it is straightforward to attribute engineering dimensions to all other operators: \u03c62 = L2, \u03c64 = L\u2212d+4, \u03c6n = Ld+(2\u2212d)n/2, (\u2207m\u03c6)2 = L2(1\u2212m). These relations convey much about the potential signi\ufb01cance of all structurally allowed operators: The engineering dimension of the non-gradient operator \u223c \u03c62 is positive in all dimensions, indicating general relevance. The \u03c64 operator is relevant (irrelevant) in dimensions d <4 ( d >4). This suggests that for d >4 a harmonic approximation (\u03bb = 0) of the model should be reasonable. It also gives us a preliminary clue as to how we might want to approach the \u03c64-model on a technical level: while for dimensions \u201cmuch\u201d smaller than", " d = 4 the interaction operator \u223c \u03c64 is strongly relevant, the dimension d = 4 itself is borderline. This suggests that we analyze the model at d = 4, or maybe \u201cclose\u201d28 to d = 4 where the \u03c64 operator is not yet that virulent, and then try to extrapolate to infer what happens at the \u201cphysical dimensions\u201d of d = 2 and 3. Operators \u03c6n>4 become relevant only in dimensions d <( \u22121/n + 1/2)\u22121 < 4. However, even below these threshold dimensions, operators of high powers in the \ufb01eld variable are \u03c64. This is the a posteriori much less relevant than the dominant non-harmonic operator justi\ufb01cation for the neglect of \u03c6n>4 operators in the derivation of the model. Similarly, operators with more than two gradients are generally irrelevant and can be neglected in all dimensions. In contrast, the operator \u03c6 coupling to the magnetic \ufb01eld carries dimension 1 + d/2 and is therefore always strongly relevant. 27 Generalizing our discussion from Section 5.1, we have incorporated a coupling to an external \ufb01eld. (Exercise: Recapitulate the construction of Section 5.1 to convince yourself that, to lowest order in an expansion in terms of \u03c6, coupling the system to a magnetic \ufb01eld leads to the fourth term of Eq. (8.23). In case you are too impatient to do this: justify the structure of the term on physical grounds.) 28 As we see shortly, the analysis of the problem is readily generalized to non-integer dimensions. 446 The renormalization group Dimensional analysis provides us with some valuable hints as to the importance of various operators appearing in the theory. It also indicates that, in the present context, dimension d = 4 might play a special role. Guided by this information, we now proceed to analyze the model in a sequence of steps of increasing sophistication. 8.4.2 Landau mean-\ufb01eld theory Given an action of the form (8.23), the \ufb01rst thing one might try is a mean-\ufb01eld analysis. That is, assuming that our coupling constants r and \u03bb are su\ufb03ciently large we might assume that the functional integral over \u03c6 is centered around solutions of the equation \u03b4S[\u03c6] \u03b4 \u00af\u03c6 = 0, or r \u00af\u03c6 + \u03bb 6 \u00af\u03c63 \u2212 h = 0, (8.24) where we have used the fact that the low-energy mean-\ufb01eld con\ufb01guration will be spatially constant. Just by inspecting the potential part of the \ufb01eld-free Lagrangian, r 4! \u03c64, it is clear that, depending on the sign of r, the mean-\ufb01eld equation possesses two fundamentally di\ufb00erent types of solution. For r > 0, the action has a global minimum at \u03c6 = 0, implying that \u00af\u03c6 = 0 is the unique mean-\ufb01eld (see Fig. 8.7). Noticing that the amplitude of \u03c6 represents a measure of the magnetization of the system (which is clear from the way the \u03c64-action was derived from the Ising model on page 196), we identify r > 0 as a phase of zero net magnetism, the paramagnetic phase. 2 \u03c62 + \u03bb In contrast, for r < 0, the action has two degenerate minima at non-zero values, \u00af\u03c6 = \u00b1\u03c60 \u2261 \u00b1(6|r|/\u03bb)1/2 (see Fig. 8.7). The system then has to make a choice as to whether it wants to sit in the ground state con\ufb01guration \u00af\u03c6 = \u03c60 or \u00af\u03c6 = \u2212\u03c60. This is the state of spontaneous symmetry breaking indicative of the low-temperature ferromagnetic phase. (Notice that, upon the switching on of a small magnetic \ufb01eld, the degeneracy between the two ground states is lifted and the system will populate a state of predetermined magnetization, \u00af\u03c6 = \u00b1\u03c60, depending on the sign of h.) The preliminary analysis above indicates that r has the status of a fundamental parameter tuning the system through the ferromagnetic transition. Indeed, the microscopic analysis in Section 5.1 had indicated that r \u223c T \u2212 Tc was a function of temperature that changed sign at some critical temperature Tc, the mean-\ufb01eld critical temperature of the transition. However, even if we did not know the microscopics, it would be clear that r(T ) is (i) some S S (a) r > 0 \u03c6 (b)", " r < 0 \u03c6 Figure 8.7 Action of the \u03c64-theory evaluated on a constant \ufb01eld con\ufb01guration above (a) and below (b) the critical point. 8.4 RG analysis of the ferromagnetic transition 447 Table 8.1 Critical exponents of the ferromagnetic transition obtained through di\ufb00erent methods. Experimental exponents represent cumulative data from various three-dimensional ferromagnetic materials. Exponent Experiment Mean-\ufb01eld Gaussian \u03b1 \u03b2 \u03b3 \u03b4 \u03bd \u03b7 0\u20130.14 0.32\u20130.39 1.3\u20131.4 4\u20135 0.6\u20130.7 0.05 0 1/2 1 3 \u2014 \u2014 1/2 1/4 1 5 1/2 0 1 1/6 1/3 7/6 4 7/12 0 5 0.109 0.327 1.238 4.786 0.631 0.037 Source: Data taken from K. Huang, Statistical Mechanics (Wiley, 1987). function of temperature which (ii) must have a zero at some temperature T = Tc (otherwise there would be no transition to begin with). Therefore, in the vicinity of T = Tc, we can set r \u223c T \u2212 Tc as our prime measure of the distance to the critical point. (This observation \u03c62 coupled is, in fact, in perfect agreement with our earlier observation that the operator to r is relevant \u2013 see the discussion in Section 8.3.2.) What can mean-\ufb01eld theory say about the prime descriptors of the transition, the critical exponents? Identifying the \ufb01eld amplitude \u03c6 (alias the magnetization) with the order parameter of the transition, and referring back to our list of exponents on page 438, the low-temperature pro\ufb01le is given by | \u00af\u03c6| = (12|r|/\u03bb)1/2 \u223c |t|1/2, implying that \u03b2 = 1/2. The exponent \u03b3 is obtained by di\ufb00erentiating Eq. (8.24) with respect to h. With \u03c7 \u223c \u2202h\u03c6, it is then straightforward to verify that, on approaching the critical point from either side of the transition, \u03c7 \u223c |t|\u22121, implying an exponent \u03b3 = 1. The action evaluated on the mean-\ufb01eld-con\ufb01guration takes the form S[ \u00af\u03c6] Ld = r 2 \u00af\u03c62 + \u03bb 4! \u00af\u03c64 \u223c \u03bb\u22121t2, 0, t < 0, t > 0. (8.25) With the mean-\ufb01eld free energy F = T S[ \u00af\u03c6] we \ufb01nd that the speci\ufb01c heat C = \u2212T 2\u22022 T F \u223c \u22022 t S behaves as a step function at the transition point, implying \u03b1 = 0. Right at the critical temperature, r = 0, the mean-\ufb01eld magnetization depends on h as \u00af\u03c6 \u223c h1/3, implying that \u03b4 = 3. Finally, the correlation length exponents \u03bd, \u03b7 cannot directly be computed from plain mean-\ufb01eld theory as they are tied to the spatial pro\ufb01le of \ufb02uctuating \ufb01eld con\ufb01gurations. For the sake of later comparison, the mean-\ufb01eld critical exponents are summarized in Table 8.1. At \ufb01rst sight the di\ufb00erences between the experimentally observed exponents (second column) and the mean-\ufb01eld exponents (third column) do not look too dramatic \u2013 apparently the primitive mean-\ufb01eld approach pursued here fares reasonably favorably \u2013 which, in view of the accumulation of pronounced \ufb02uctuations at the critical point, should come as something of a surprise. On the other hand we must keep in mind that the exponents describe singular power laws in the transition region. In view of that, the di\ufb00erence between 448 The renormalization group 1.3 and 1 does look quite signi\ufb01cant. At any rate, we should try to re\ufb01ne our theoretical understanding of the transition and search for the source of the discrepancy with experiment. 8.4.3 Gaussian model As a \ufb01rst improvement on the mean-\ufb01eld approximation, let us explore the e\ufb00ect of quadratic \ufb02uctuations around the constant \ufb01eld con\ufb01guration \u00af\u03c6. Approaching the transition point from above, we set \u00af\u03c6 = 0 and approximate the action through its quad", "ratic expansion29 S[\u03c6] \u2248 ddr r 2 \u03c62 + 1 2 (\u2207\u03c6)2 \u2212 h\u03c6. (8.26) In this form, one may e\ufb00ect the Gaussian integral over \ufb01eld \ufb02uctuations and evaluate the dependence of the free energy on the external parameters h and r. However, in anticipation of our analysis of the full problem below, we here pursue a slighly di\ufb00erent, renormalizationgroup-oriented approach, i.e., pretending that we did not know how to do the Gaussian integral, we subject the quadratic action to a momentum shell RG analysis. Proceeding along the lines of the canonical scheme, we split our \ufb01eld into fast and slow degrees of freedom \u03c6 = \u03c6s + \u03c6f resulting in the, now familiar, fragmentation of the action S[\u03c6s, \u03c6f ] =S s[\u03c6s] +S f [\u03c6f ] +S c[\u03c6s, \u03c6f ]. However, the crucial simpli\ufb01cation, characteristic of a Gaussian theory, is that the action Sc coupling fast and slow components vanishes (exercise), implying that the integration over the fast \ufb01eld merely leads to an inessential constant. The e\ufb00ect of the RG step on the action is then entirely contained in the rescaling of the slow action. According to our previous discussion, the scaling factors thus appearing are determined by the engineering dimensions of the operators appearing in the action, i.e. r \u2192 b2r and h \u2192 bd/2+1h. Using the fact that r \u223c t we can then readily write down the two relevant scaling dimensions of the problem, yt = 2 and yh = d/2 + 1. Comparison with Eq. (8.22) \ufb01nally leads to the list of exponents. Notice that the exponents now explicitly depend on the dimensionality of the system, a natural consequence of the fact that they describe the e\ufb00ect of spatial \ufb02uctuations. Table 8.1 contains the values of the exponents for a three-dimensional system. We cannot really say that the results are any better than those obtained by the mean-\ufb01eld analysis. Some exponents (e.g. \u03b4) agree better with the experimental data, while others (e.g. \u03b1) are decidedly worse. As a corollary to this section, we note that the Gaussian model possesses only one \ufb01xed point, namely r = h = 0, which in the context of \u03c64-theory is called the Gaussian \ufb01xed point. 29 The appearance of a linear term indicates that we are expanding not around the \u201ctrue\u201d mean-\ufb01eld, i.e. the exact solution of (8.24), but rather around the solution \u00af\u03c6 = 0 of the \ufb01eld-free system. However, in view of the fact that h has the status of an external perturbation, this choice of the reference con\ufb01guration is quite natural. 8.4 RG analysis of the ferromagnetic transition 449 8.4.4 Renormalization group analysis In the present analysis of the model, we have not really touched upon its principal source of complexity, namely the e\ufb00ect of the \u201cinteraction operator\u201d \u03c64 on the \ufb02uctuation behavior of the \ufb01eld. It seems likely that the neglect of this term is responsible for the comparatively poor predictive power both of the straightforward mean-\ufb01eld analysis, and of the Gaussian model. Indeed, the dimensional analysis of Section 8.4.1 indicated that the \u03c64 addition to the action becomes relevant below four dimensions. A more physical argument, to the same e\ufb00ect, is given in the Info block below. Although the solution of the general problem posed by the action (8.23) still appears to be hopelessly di\ufb03cult, there is one aspect we can turn to our advantage. While physical systems exist in integer dimensions d = 1,..., 4,..., there is actually no reason why we should not be allowed to evaluate our theory, i.e. the functional integral with action (8.23), in fractional dimensions. In the present context, this seemingly academic freedom turns out to be of concrete practical relevance. The point is that the nonlinear \u03c64 operator was found to be marginal at d = 4 and relevant below. One may thus expect that, in dimensions d = 4 \u2212, 1, the operator is relevant but not that relevant, i.e. one may expect that, for su\ufb03ciently small deviations", " o\ufb00 the threshold dimension, four, the theory knows of an expansion parameter, somehow related to, which will enable us to control the interaction operator. Of course, at the end of the day, we will have to \u201canalytically continue\u201d to dimensions of interest, = 1 or even = 2, but, for the present, we will see what we can learn from a d = 4 \u2212 representation of the theory. INFO Our previous analysis relied on the assumption that the \ufb01eld integration is tightly bound to the vicinity of the extrema of the action. But let us now ask under what conditions this assumption is actually justi\ufb01ed. We should develop some intuition as to the relative importance of the mean-\ufb01eld content of the theory and of the \ufb02uctuations around the mean-\ufb01eld. While there are several ways to proceed with this program, we will focus on the analysis of the magnetic susceptibility. (At this point, we should warn the reader that the arguments formulated below, while technically straightforward, are conceptually involved. The critical contemplation of the logical steps of the construction is time well invested.) Firstly, let us recall the de\ufb01nition of the susceptibility, \u03c7 = \u2212\u22022 H F \u223c ddr \u03c6(r)\u03c6(0)c \u223c G(k = 0), where we have used the fact that \u03c6(r)\u03c6(r) = G(r\u2212r) is the Green function of the model. Given this identi\ufb01cation, we note that a formal criterion of the transition \u2013 divergent susceptibility! \u2013 is synonymous with a singularity of the zero-momentum Green function. On the level of the Gaussian theory (see Eq. (8.26)) G(k) = (r + k2)\u22121, i.e. \u03c7 \u223c r\u22121. Anticipating troubling observations to come, we reiterate that the mean-\ufb01eld transition temperature is identi\ufb01ed by the condition r \u223c t = 0. Now, let us move on to explore corrections to the mean-\ufb01eld susceptibility on the level of a perturbative one-loop calculation. To this end, we recall that (if necessary, recapitulate the discussion of Section 5.1), due to the presence of the \u03c64 operator, the Green function acquires a self-energy which, at the one-loop level, is given by \u03a3 = \u2212 \u03bb 1 r+k2. 2 k 450 The renormalization group As a consequence, one can identify the susceptibility as \u03c7\u22121 \u223c (G(k = 0))\u2212 + k2. A \ufb01rst observation to be made is that non-Gaussian \ufb02uctuations (physically: interactions between harmonic \ufb02uctuations around the mean-\ufb01eld amplitude) lower the transition amplitude, i.e., setting r \u223c T \u2212 Tc, it now takes a smaller temperature T to reach the critical point; in accord with the intuitive expectation that \ufb02uctuations tend to \u201cdisorder\u201d the system. Frustratingly, one may also observe that the cuto\ufb00 \u039b is needed to prevent the \u201ccorrection\u201d, \u039b \u2212 \u03bb 2 k 1 r + k2 \u223c \u03bb ddk 1 r + k2, from diverging in dimensions d \u2265 2. To deal with this singularity, we have to realize that the e\ufb00ect of \ufb02uctuations is actually two-fold: the transition temperature gets shifted and the temperature dependence of the inverse susceptibility is apparently no longer simply linear (by virtue of the r-dependence of the integrand). The two e\ufb00ects can be disentangled by writing \u03c7\u22121 = \u02dcr + L 2\u03c0 \u03bb 2 d \u039b ddk 1 r + k2 \u2212 1 k2 \u2248 \u02dcr \u2212 \u03bb\u02dcr 2 L 2\u03c0 d \u039b ddk (\u02dcr + k2)k2, (8.27) where \u02dcr \u2261 r + d L 2\u03c0 \u03bb 2 \u039b ddk k2, represents the shifted transition temperature while the integral describing the deviation from the linear temperature dependence of the susceptibility is now UV-convergent in dimensions d <4. Notice that in the second equality of Eq. (8.27) we have replaced the parameter r in the integrand by the modi\ufb01ed parameter \u02dcr. To the accuracy of a one-loop calculation, this manipulation is permissible. Naively, it looks as if this sequence of manipulations has led to a catastrophe: the \ufb02uctuationrenormalized transition temperature appears to diverge as one sends the cuto\ufb00 to", " in\ufb01nity, clearly a nonsensical prediction! However, one may note that there was actually no justi\ufb01cation for identifying the physical transition temperature through the parameter r in the \ufb01rst place. This identi\ufb01cation was based on mean-\ufb01eld theory alone, i.e. an approach to the problem which neglected altogether the key e\ufb00ect of \ufb02uctuations. However, the bare parameter r appearing in the action carries as little \u201cuniversal\u201d meaning as the cuto\ufb00 \u039b, or any other microscopic system parameter for that matter! Once we have acknowledged this interpretation, we should then identify the transition temperature through the singularity of the macroscopically observable properties (e.g. divergence of the susceptibility leads to the vanishing of the modi\ufb01ed parameter \u02dcr at the one-loop level) while the microscopic parameters carry no signi\ufb01cance by themselves. EXERCISE This interpretation closely parallels the philosophy of renormalization in highenergy physics. There, the bare parameters of the action are fundamentally undetermined, while the inverse of the Green function at zero external momentum represents a physical observable, e.g. the mass of the electron. Since the loop corrections to this physical quantity appear to be in\ufb01nite (and the theory does not enjoy the luxury of the presence of a physically motivated cuto\ufb00), one postulates that the bare parameters of the action have been in\ufb01nite by themselves. These singularities are deliberately adjusted so as to cancel the divergence of the 8.4 RG analysis of the ferromagnetic transition 451 \ufb02uctuation corrections and to produce \ufb01nite \u201cphysical\u201d quantities. It is instructive to consult a textbook on renormalization in high-energy physics (such as Ryder30) to become acquainted with the functioning of this strategy, and with the enormous success it has had in the context of QED and other sub-branches of particle physics. We now turn to the second e\ufb00ect of the \ufb02uctuation correction, namely the deviation from the linear temperature dependence, as described by the integral contribution to Eq. (8.27). On dimensional grounds, the integral depends on the parameter \u02dcr as \u223c \u03bbLd \u02dcr(d\u22124)/2. The (mean-\ufb01eld + quadratic \ufb02uctuations) approach to the problem breaks down when this contribution becomes more important than the leading-order contribution to the susceptibility, i.e. for dimensions d \u2264 4. This observation is the essence of the so-called Ginzburg criterion. The criterion states that mean-\ufb01eld theory becomes inapplicable below the so-called upper critical dimension dc = 4. While we have derived this statement for the particular case of the \u03c64-model, it is clear that similar estimates can be performed for every nonlinear \ufb01eld theory, i.e. as with the lower critical dimension, the upper critical dimension also represents an important threshold separating the mean-\ufb01eld dominated d > dc from the \ufb02uctuation dominated d < dc behavior. Also notice that the analysis above conforms with our previous observation that the nonlinear \u03c64 operator is relevant in dimensions d < 4. (Convince yourself that the two lines of argument re\ufb02ect the same principle, namely the dependence of \ufb02uctuations on the accessible phase volume, as determined by the dimensionality of the system.) Before proceeding to the details of the RG program, let us try to predict a number of general elements of the \u03c64 phase diagram on dimensional grounds. We saw that in dimensions d >4 the \u03c64 operator is irrelevant and that the Gaussian model essentially dictates the behavior of the system. Speci\ufb01cally, for d >4, the Gaussian \ufb01xed point r = \u03bb = h = 0 is the only \ufb01xed point of the system. Below four dimensions, the \u03c64 operator becomes relevant and the emergence of a richer \ufb01xed point structure may be expected. However, for = 4 \u2212 d su\ufb03ciently small, we also expect that, whatever new \ufb01xed points appear, they should be close to the Gaussian point. This means that we can conduct our search for new \ufb01xed points within a double expansion in, and the small deviation of the coupling constants r, \u03bb, h around the Gaussian \ufb01xed point. (In fact, we will momentarily identify a third expansion parameter, namely the number of momentum loops appearing in fast-\ufb01eld integration.) Step I We next proceed to formulate the steps of the RG in detail. To keep things simple, the RG", " transformation will be carried out to lowest order in a triple expansion in, the coupling constants, and the number of momentum loops. The rationale behind the loop expansion can be best understood if we assume that the entire action31 is multiplied by a large parameter (which, in the case of a quantum theory, might be \u22121). The expansion in the number of loops is then equivalent to an expansion in the inverse of that parameter (for a quantum theory, an expansion away from the classical limit). 30 L. H. Ryder, Quantum Field Theory, (Cambridge University Press, 1996). 31 Before we rescaled the \ufb01elds so as to make the leading-order coe\ufb03cient equal to 1/2. 452 The renormalization group EXERCISE Verify this statement; to this end, notice that a diagram of nth order in perturbation theory in the \u03c64 vertex contains a prefactor an involving the large parameter. On the other hand, each of the I internal lines, or propagators, contained by the diagrams contributes a factor a\u22121, so that the overall power is an\u2212I. Next relate the number of internal lines to the number L of loops. Notice that each line corresponds to a momentum summation. However, the number of independent summations is constrained by the n \u03b4-functions carried by the vertices. Use this information to show that the overall power of the graph is a\u2212L+1, i.e. an expansion in L is equivalent to an expansion in the inverse of a. Let us now decompose the action in the standard manner, setting S[\u03c6s, \u03c6f ] = Sf [\u03c6f ] + Ss[\u03c6s] +S c[\u03c6s, \u03c6f ], where Sf [\u03c6f ] = Ss[\u03c6s] = Sc[\u03c6s, \u03c6f ] = \u03bb 4 r 2 r 2 \u03c62 f + \u03c62 s + 1 2 1 2 ddr ddr (\u2207\u03c6f )2, (\u2207\u03c6s)2 + \u03bb 4! \u03c64 s \u2212 h\u03c6s, ddr\u03c62 s \u03c62 f + \u00b7 \u00b7 \u00b7 Several approximations related to the loop order of the expansion are already imposed at this level. We have neglected terms of O(\u03c64 f ) because their contraction leads to two loop diagrams. The same applies to terms of O(\u03c6s\u03c63 s \u03c6f ) do not arise because the addition of a fast momentum and three slow momenta is incompatible with momentum conservation. f ) (exercise). Terms of O(\u03c63 Steps II and III To simplify the notation, let us rescale the momentum according to q \u2192 q/\u039b, implying that coordinates are measured in units of the inverse cuto\ufb00 r \u2192 r\u039b. With the coupling constants rescaled according to their engineering dimensions, r \u2192 r\u039b2, \u03bb \u2192 \u03bb\u039b4\u2212d, the action remains unchanged, while the fast and slow momenta are now integrated over the dimensionless intervals |qs| \u2208 [0, b\u22121] and| qf | \u2208 [b\u22121, 1], respectively. We next construct an e\ufb00ective action by integration over the fast \ufb01eld: e\u2212Seff [\u03c6s] = e\u2212Ss[\u03c6s] f. In performing the average over fast \ufb02uctuations, \u00b7 \u00b7 \u00b7 f, we shall (a) retain only contributions of one-loop order while (b) neglecting terms that lead to the appearance of \u03c6n>4 contributions 3 would lead to such a term.) To this in the action. (For example, the contraction level of approximation, one obtains e\u2212Sc[\u03c6s,\u03c6f ] s \u03c62 \u03c62 f : 9 s e\u2212Seff [\u03c6s] = e\u2212Ss[\u03c6s]exp \u2212 Sc[\u03c6s, \u03c6f ] f + 9 1 2 Sc[\u03c6s, \u03c6f ]2 :c f, where the superscript c denotes a connected average. (Exercise: It is instructive to check the consistency of this expansion for yourself.) The two diagrams corresponding to the contractions Sc[\u03c6s, \u03c6f ]f and Sc[\u03c6s, \u03c6f ]2c f are shown in parts (a) and (b) of the \ufb01gure below, respectively, where the external line segments indicate the passive \u03c6s amplitudes. 8.4 RG analysis of the ferromagnetic transition 453 According to the standard rules of perturbation theory, the \ufb01rst of the two diagrams, (a), evaluates to Sc[\u03c6s, \u03c6", "f ] f = \u03bb 4 ddq (2\u03c0)d 1 r + q2 f ddq (2\u03c0)d \u03c6s(q)\u03c6s(\u2212q). s We now consider the summation over fast momenta appearing in this expression. Using the fact that we are in the near vicinity of the critical point and anticipating that we are interested in no more than the expansion of the \u03b2-function for small values of the coupling, r+q2 = I1 \u2212 rI2, where we have we now expand the integrand to \ufb01rst order in r, introduced the shorthand notation, ddq (2\u03c0)d 1 f I\u03b1 \u2261 ddq (2\u03c0)d 1 q2\u03b1. f (8.28) These integrals are straightforwardly computed by switching to polar coordinates, I\u03b1 = \u03a9d 1 b\u22121 dq qd\u22122\u03b1\u22121 = \u03a9d d \u2212 2\u03b1 (1 \u2212 b2\u03b1\u2212d), where \u03a9d = (2\u03c0d/2/\u0393(d/2))/(2\u03c0)d denotes the volume of the d-dimensional unit sphere (measured in units of 2\u03c0). We thus \ufb01nd that, after the integration over fast modes, and the standard rescaling operation, q \u2192 bq, \u03c6 \u2192 b(d\u22122)/2\u03c6, the quadratic part of the action takes the form S(2)[\u03c6] = b2 2 r + \u03bb\u03a9d 2(d \u2212 2) (1 \u2212 b2\u2212d) \u2212 r\u03bb\u03a9d 2(d \u2212 4) (1 \u2212 b4\u2212d) ddr \u03c62. (8.29) (a) Turning to the second diagram (b) in the \ufb01gure, we notice that, owing to the presence of four external legs, its contribution will be proportional to \u03c64 s. Further, momentum conservation implies that the momenta carried by the internal lines of the diagram will depend on both the fast \u201cinternal\u201d momentum and the external momenta carried by the \ufb01elds \u03c6s. However, we can simplify the analysis by neglecting the dependence on the latter from the outset. The reason is that the integration over the internal momentum followed by Taylor expansion in the slow momenta would generate expressions of the structure F (q1, q2, q3)\u03c6(q1)\u03c6(q2)\u03c6(q3)\u03c6(\u2212q1 \u2212 q2 \u2212 q3), where q1,2,3 represent slow momenta and F is some polynomial. Taking account of the small momenta would thus generate derivatives acting on an operator of fourth order in \u03c6, a combination that we saw above is irrelevant. (b) Neglecting the external momenta, diagram (b) leads to the result 1 2 Sc[\u03c6s, \u03c6f ]2f \u03bb2 16 ddr \u03c64 s f ddq (2\u03c0)d 1 (r + q2)2 = \u03bb2I2 16 ddr \u03c64 s + O(\u03bb2r). Evaluating the integral and rescaling, we \ufb01nd that the quartic contribution to the renormalized action reads S(4)[\u03c6] = b4\u2212d \u03bb 4! \u2212 \u03bb2\u03a9d 16 1 \u2212 b4\u2212d d \u2212 4 ddr \u03c64. 454 The renormalization group Finally, there are no one-loop diagrams a\ufb00ecting the linear part of the action, i.e. S(1)[\u03c6] = hbd/2+1 ddr \u03c6, rescales according to its engineering dimension. Combining everything, we \ufb01nd that, to one-loop order, the coupling constants scale 2(d\u22124) (1 \u2212 b4\u2212d)), \u03bb \u2192 b4\u2212d(\u03bb \u2212 2(d\u22122) (1 \u2212 b2\u2212d) \u2212 r\u03bb\u03a9d 1\u2212b4\u2212d d\u22124 ), and h \u2192 hbd/2+1. We next set d = 4 \u2212 and evaluate the right-hand sides according to the relations r \u2192 b2(r + \u03bb\u03a9d 3 2 \u03bb2\u03a9d of these expressions to leading order in. With \u03a94\u2212 \u2248 \u03a94 = 1 8\u03c02, we thus obtain r \u2192 b2 r + \u03bb \u2192 (1 + ln b) \u03bb 32\u03c02 (1 \u2212 b\u22122) \u2212 r\u03bb \u03bb \u2212 3\u03bb2 16\u03c02 ln b, 16\u03c02 ln b h \u2192 hb3\u2212/2, which, setting b = e, lead to the Gell-Mann\u2013Low equations:", " dr d d\u03bb d dh d \u2212 r\u03bb 16\u03c02, \u03bb = 2r + 16\u03c02 = \u03bb \u2212 3\u03bb2 16\u03c02, 6 \u2212 h. 2 =, (8.30) These equations clearly illustrate the meaning of the -expansion. According to the second equation, a perturbation away from the Gaussian \ufb01xed point will initially grow at a rate set by the engineering dimension. While, on the level of the classical, zero-loop theory, \u03bb would grow inde\ufb01nitely, the one-loop contribution \u223c \u03bb2 stops the \ufb02ow at a value \u03bb \u223c. \u03b2 (\u03bb) 2, \u03bb\u2217 Equating the right-hand sides of Eq. (8.30) to zero (and temporarily ignoring the magnetic \ufb01eld), we indeed \ufb01nd that besides the Gaussian \ufb01xed point (r\u2217 1) = (0, 0) a nontrivial \ufb01xed point (r\u2217 2) = (\u2212 1 3 ) has appeared. Notice that, in accord with the schematic considerations made at the beginning of the section, the second \ufb01xed point is O() and coalesces with the Gaussian \ufb01xed point as is sent to zero. Plotting the \u03b2-function for the coupling constant \u03bb (see \ufb01gure), we further \ufb01nd that, for >0, \u03bb is relevant around the Gaussian \ufb01xed point but irrelevant at the non-trivial \ufb01xed point. 1, \u03bb\u2217 6, 16\u03c02 \u03b5 > 0 (\u03b5) \u03b5 < 0 \u03bb To understand the full \ufb02ow diagram of the system, one may linearize the \u03b2-function around both the Gaussian and the non-trivial \ufb01xed point. Denoting the linearized mappings by W1,2, we \ufb01nd 2 0 W1 = 1 16\u03c02 6 7, W2 = 2 \u2212 1 3 0 1+/6 16\u03c02 \u2212. 8.4 RG analysis of the ferromagnetic transition 455 non-trivial Gaussian unphysical r Figure 8.8 Phase diagram of the \u03c64-model as obtained from the -expansion. Figure 8.8 shows the \ufb02ow in the vicinity of the two \ufb01xed points, as described by the matrices W1,2 as well as the extrapolation to a global \ufb02ow chart. Notice that the critical surface of the system \u2013 the straight line interpolating between the two \ufb01xed points \u2013 is tilted with respect to the r \u223c temperature axis of the phase diagram. This implies that it is not the physical temperature alone that decides whether the system will eventually wind up in the paramagnetic (r 0) or ferromagnetic (r 0) sector of the phase diagram. Rather one has to relate temperature (\u223c r) to the strength of the nonlinearity (\u223c \u03bb) to decide on which side of the critical surface we are. For example, for strong enough \u03bb, even a system with r initially negative may eventually \ufb02ow towards the disordered phase. This type of behavior cannot be predicted from the mean-\ufb01eld analysis of the model (which would generally predict a ferromagnetic state for r < 0). Rather it represents a non-trivial e\ufb00ect of \ufb02uctuations. Finally notice that, while we can formally extend the \ufb02ow into the lower portion of the diagram, \u03bb <0, this region is actually unphysical. The reason is that, for \u03bb <0, the action is fundamentally unstable and, in the absence of a sixth-order contribution, does not describe a physical system. What are the critical exponents associated with the one-loop approximation? Of the two eigenvalues of W2, 2 \u2212 /3 and \u2212, only the former is relevant. As with the Gaussian \ufb01xed point, it is tied to the scaling of the coupling constant, r \u223c t, i.e. we have yt = 2 \u2212 /3 and, as before, yh = (d + 2)/2 = (6 \u2212 )/2. An expansion of the exponents summarized in Eq. (8.22) to \ufb01rst order in then yields the list + 12, \u03b7 = 0. If we are now reckless enough to extend the radius of the expansion to = 1, i.e. d = 3, we obtain the \ufb01fth column of Table 8.1. Apparently the agreement with the experimental results has improved \u2013 even in spite of the fact that we have driven the -expansion well beyond its range of applicability! (For =", " 1, terms of O(2) can, of course, no longer be neglected!) 456 The renormalization group How can one rationalize the success of the -expansion? Trusting in the principle that good theories tend to work well beyond their regime of applicability, we might simply speculate that nature seems to be sympathetic to the concept of renormalization and the loop expansion. Of course, a more quali\ufb01ed approach to the question is to explore what happens at higher order in the -expansion. Needless to say, the price to be paid for this ambition is that, at orders O(n>1), the analysis indeed becomes laborious. Nonetheless, the success of the \ufb01rst-order expansion prompted researchers to drive the -expansion up to \ufb01fth order! The results of this analysis are summarized in the last column of Table 8.1. In view of the fact that we are still extending a series beyond its radius of convergence,32 the level of agreement with the experimental data is striking. In fact, the exponents obtained by the -expansion even agree \u2013 to an accuracy better than one percent \u2013 with the exponents of the two-dimensional model,33 i.e. for a situation where the \u201csmall\u201d parameter has to be set to two. However, it is important to stress that the -expansion is not just a computational tool for the calculation of exponents. On a more conceptual level, its merit is that it enables one to explore the phase diagram of nonlinear theories in a more or less controlled manner. In fact, the -expansion not only is useful in the study of \ufb01eld theories close to the upper critical dimension (i.e. close to the mean-\ufb01eld threshold) but can equally well be applied to the analysis of systems in the vicinity of the lower critical dimension. In the following section, we consider a problem of this type, i.e. we will apply an -expansion around d = 2 to detect the onset of global thermal disorder in models with continuous symmetries. 8.5 RG analysis of the nonlinear \u03c3-model The scalar \ufb01eld theory encapsulates a wide class of systems encompassing a singlecomponent order parameter. However, throughout the text, we have encountered problems where the order parameter involves more than one component, e.g. the complex \ufb01eld associated with condensation phenomena, the matrix \ufb01eld associated with the quantum disordered metallic system, or the \ufb01eld theories involving spin. In such cases, one very often \ufb01nds that the low-energy content of the theory involves a projection which imparts a constraint to the \ufb01eld integral. In the context of condensation phenomena, we saw that, at low temperatures, one can neglect the massive amplitude \ufb02uctuations of the order parameter, while the collective \ufb02uctuations of the phase mode impacted signi\ufb01cantly on the low-energy properties of the system. In this case, the phase degree of freedom is constrained by its topology to lie on the unit circle. Similarly, if we neglect the \u201chigh-energy\u201d physics of local moment formation, classical and quantum spin theories are constrained by the normalization of the local spin. When subjected to an auxiliary constraint, theories that are otherwise free are known as nonlinear \u03c3-models. The aim of the present section is to apply methods of the RG to explore the critical properties of a general class of nonlinear 32 Indeed, it is believed that we are dealing with a series that is only asymptotically convergent. That is, beyond a certain order of the expansion, the agreement with the \u201ctrue\u201d exponents will presumably become worse. 33 The latter are known from the exact solution of the two-dimensional model, see L. Omsager, Crystal statistics I. A two-dimensional model with an order\u2013disorder transition. Phys. Rev. 65 (1944), 117\u201349. 8.5 RG analysis of the nonlinear \u03c3-model 457 \u03c3-models. Later, in the next section and in Chapter 9, we will consider the way topological considerations can impact on the nature of the system. Let us then consider the nonlinear \u03c3-model de\ufb01ned by the partition function Z = Dg e\u2212S[g], where S[g] = 1 \u03bb ddr tr % \u2207g\u2207g\u22121 &, (8.31) g \u2208 G takes values in some compact Lie group G and the integration d\u03bcg(x) extends over the Haar measure d\u03bcg of G. (Later in this section, we will specialize to the case G = O(3) relevant to the case of the three-component spin models, the Heisenberg magnet. Other", " realizations of G will be met in the next chapter.) Dg = x INFO The Haar measure d\u03bcg of a compact Lie group G is a (uniquely speci\ufb01ed) integration G d\u03bcg \u00b7 1 = 1, and (b) invariant under left and right measure that is (a) unit normalized, i.e. multiplication by a \ufb01xed group element: d\u03bcg f (g) = d\u03bcg f (gh\u22121) = d\u03bcg f (h\u22121g) G G G G d\u03bc(gh) f (g) = for any h \u2208 G. Upon translation of the integration variable g \u2192 hg or g \u2192 gh, these equations G d\u03bc(hg) f (g). Holding for any f, this implies assume the form that d\u03bc(hg) = d\u03bcg = d\u03bc(gh), which tells us that the Haar measure assigns equal volume density to any point on the group manifold. Owing to this homogeneity property, integrations over groups are almost always performed with respect to the Haar measure. Details on the explicit construction of this measure can be found in textbooks on Lie group theory.34 G d\u03bcg f (g) = As we have seen above, the presence of a continuous symmetry identi\ufb01es the lower critical dimension of the theory at dc = 2. At dimensions of dc and below, massless Goldstone \ufb02uctuations of the \ufb01eld destroy long-range order at any non-zero temperature. However, in the range 2 < d <4, one expects an ordered phase to persist at low temperatures, while the critical characteristics of the phase in the vicinity of the transition are \ufb02uctuation dominated. For reasons that will become clear, in the following we develop an -expansion around the lower critical dimension to detect the \ufb01xed point structure of the model. (Such an -expansion was proposed in a seminal work Alexander M. Polyakov 1945\u2013 He has made several important contributions to quantum \ufb01eld theory from non-abelian gauge theory to conformal \ufb01eld theory. His path integral formulation of string theory had profound and lasting impacts in the conceptual and mathematical understanding of the theory. He also played an important role in elucidating the conceptual framework behind renormalization independent of Kenneth Wilson\u2019s Nobel prize winning work. He formulated pioneering ideas in gauge/string duality long before the breakthrough of AdS/CFT using D-branes. 34 See, e.g., J. Fuchs and C. Schweigert, Symmetries, Lie Algebras and Representations, a Graduate Course for Physicists (Cambridge University Press, 1997). 458 The renormalization group by Polyakov.35). However, before turning to the actual formulation of the RG analysis, we need to do a bit of preparatory work. As usual, let us split the matrix \ufb01elds g into \u201cslow\u201d and \u201cfast\u201d components. This is achieved by de\ufb01ning g(r) = gs(r)gf (r), where gs(r) and gf (r) have momentum components in the ranges [0, \u039bb\u22121] and [\u039bb\u22121, \u039b], respectively. Substituting this decomposition into the action, one obtains S[gsgf ] = S[gs] +S [gf ] +S c[gs, gf ], where the coupling action Sc[gs, gf ] = ddr tr g\u22121 s \u2207gsgf \u2207g\u22121 f % & 2 \u03bb. (8.32) In the next step, one must integrate over the fast \ufb01eld gf. Since, of course, this integration cannot be performed exactly, we will have to resort to an approximation scheme based on the loop expansion. Yet, before doing so, we need to provide a little more background information on \ufb01eld integration over group-valued variables in general. 8.5.1 Field integrals over groups As with any other nonlinear integration space, to perform an integral over a group one \ufb01rst needs to introduce a set of suitable coordinates. Of course, with the choice of these variables one can exercise some freedom. However, in most applications, one will employ exponential coordinates, i.e. one may start out from g = eW, whereW \u2208 G lives in the Lie algebra of the group, and then expand W = i a \u03c0aT a in the Hermitian group generators. INFO Recall that the generators T a of a Lie group are M elements of the Lie algebra G speci\ufb01ed by the following two conditions. (i) The set {T a|a = 1", ",..., M } is a basis of G.36 (ii) The commutators translate to the relation [T a, T b] = \u2212if abcT c, where the so-called structure constants f abc encode the geometry of the group. For example, the structure constants of the three-dimensional (M = 3) groups SU(2) and O(3) are given by f abc = abc, a, b, c = 1, 2, 3, where abc is the antisymmetric tensor. For SU(2), the generators T a = 1 2 \u03c3a can be identi\ufb01ed with (one half of) the Pauli matrices, for O(3) with the angular momentum operators familiar from quantum mechanics. For later reference, we note that the generators of the group families U(N ), SU(N ) and O(N ) obey the completeness relations M a=1 M a=1 M a=1 ijT a T a kl = 1 2 \u03b4il\u03b4jk, U(N ), ijT a T a kl = 1 2 \u03b4il\u03b4jk \u2212 1 2N \u03b4ij\u03b4kl, SU(N ), ijT a T a kl = \u03b4il\u03b4jk \u2212 \u03b4ik\u03b4jl, O(N ). (8.33) (8.34) (8.35) 35 A. M. Polyakov, Interactions of Goldstone particles in two dimensions. Applications to ferromagnets and massive Yang\u2013Mills \ufb01elds, Phys. Lett. B 59 (1975), 79\u201381. 36 The Lie algebra of an M -dimensional Lie group is a vector space, so it has a basis. The \u201cvectors\u201d of this space are realized by (N \u00d7 N )-dimensional matrices Ta. 8.5 RG analysis of the nonlinear \u03c3-model 459 (Here, the upper summation limits are set by M = N 2 for G = U(N ), M = N 2 \u2212 1 for G = SU(N ), and M = N (N \u2212 1)/2 for G = O(N ) \u2013 exercise: why?) In general, the generators can be chosen so as to obey the normalization condition tr(T aT b) = c\u03b4ab, wherec is a normalization constant. Interpreting W |W \u2261tr(W W ) as a scalar product on the Lie algebra (exercise: convince yourself that | meets all the criteria required of a scalar product), the equation above tells us that the set {Ta} forms an orthonormal basis. In the cases U(N ), SU(N ) and O(N ), consistency with the orthonormalization relation above enforces the normalization c = 1/2, 1/2, and 1, i.e. tr(T aT b) = \u03b4ab 2, U(N ), SU(N ), tr(T aT b) = \u03b4ab, O(N ). \u23ab \u23aa\u23ac \u23aa\u23ad (8.36) Integration over all N real coe\ufb03cients \u03c0a will cover the full group G. Yet there remains one non-trivial problem to be solved: we need to understand how to express the Haar measure d\u03bcg \u2192 d\u03c0 J(\u03c0) in terms of the new integration variables. Here, the \u201c\ufb02at measure\u201d a d\u03c0a comprises the integration over all Lie algebra coe\ufb03cients while the function d\u03c0 = J(\u03c0) encapsulates both the geometry of the Haar measure and the Jacobian associated with the transformation g \u2192 a \u03c0aTa. Referring for a more detailed discussion of the function J to text books on group theory, we here merely note that its Taylor expansion in \u03c0 starts as J(\u03c0) = 1 + O(\u03c04) where theO (\u03c04) term will not play a role in the context of the one-loop analysis below. Thus, at the level of approximation discussed in this text, the integration measure of the fast-\ufb01eld components can be approximated by the \u201c\ufb02at\u201d measure d\u03c0. Using the RG application above as an example, let us now discuss how integrals over \ufb02uctuations in the group space are performed in practice. Substitution of the expansion gf = eW = 1 + W + W 2/2 +\u00b7 \u00b7 \u00b7 into S[gf ] andS c[gf, gs], gives a series of contributions n S(n) where S(n) is of nth order in the local \ufb01eld amplitudes W. As usual, we", " will S = organize our analysis around the fully solvable sector of the theory, i.e. the action quadratic in W. A straightforward substitution into S[gf ] gives S(2)[W ] =\u2212 (8.36) = 1 \u03bb 1 \u03bb ddr tr [\u2207W \u2207W ] = ddr \u2207\u03c0a\u2207\u03c0a = 1 2 p\u2208f 1 \u03bb ddr \u2207\u03c0a\u2207\u03c0b tr & % T aT b p\u03a0\u22121 \u03c0a p \u03c0a \u2212p, where we have used the normalization condition (8.36), the fast momentum sector and we have introduced \u03a0p \u2261 \u03bb fast \ufb01elds \u03c0a. p\u2208f stands for a summation over 2Ldp2 as the propagator of the Below, we engage integrals of the type tr(F (W (\u03c0))tr(G(W (\u03c0)) \u00b7 \u00b7 \u00b7 W, where F, G,... are functions of the generators W and \u00b7 \u00b7 \u00b7 \u03c0 \u2261 N d\u03c0 exp(\u2212S(2)[\u03c0])(\u00b7 \u00b7 \u00b7) denotes averaging over the quadratic action. These integrals may be evaluated with the help of Wick\u2019s theorem and the completeness relation (8.35). Wick\u2019s theorem states that we need to form all possible pairings of \u03c0-variables or, equivalently, W -matrices. Each pairing is of the form either tr(AW )tr(AW ) or tr(AW AW ), where the matrices A and A may contain W - 460 The renormalization group matrices themselves (however, when computing an individual pairing, these matrices are temporarily kept \ufb01xed). Speci\ufb01cally, for G = O(N ) the Gaussian integrals give37 p\u03c0a = \u2212\u03b4p,\u2212p \u03a0p tr(AT a) tr(AT a) p W tr(AT a)tr(AT a ) % (8.35) = \u2212\u03b4p,\u2212p \u03a0p tr(AWp)tr(AWp )W = \u2212\u03c0a tr(AA) \u2212 tr(AAT ) &, tr(AWpAWp )W = \u2212 >? p\u03c0a \u03c0a p = \u2212\u03b4p,\u2212p \u03a0p tr(AT aAT a) tr (AT aAT a W (8.35) = \u2212\u03b4p,\u2212p \u03a0p ) % tr(A)tr(A) \u2212 tr(AAT ) &. Integrals over W -matrices are computed by applying these contraction rules until all possible pairings have been exhausted. Probably the best way to become acquainted with this procedure is to work through an example.. EXERCISE Consider the expression X \u2261, where A, A \u2208 o(N ) belong to the Lie algebra of O(N ) (which implies, in particular, that they are traceless). Show that the pair contraction of the \ufb01rst W -matrix with the second (the third and the fourth) vanishes due to the tracelessness of A and A. Next perform the contractions (1\u22124)(2\u22123) and (1 \u2212 3)(2 \u2212 4) to obtain the result X = (N \u2212 2)\u03a0p\u03a0p+qtr tr(Aq,pWp+qW\u2212p)tr q,p Wp+q W\u2212p \u2212q,p+q \u2212 A!. Ap,q \u2212q,\u2212p A A!/ W 8.5.2 One-loop expansion With this background, let us now return to the actual RG program for the O(N )-model. As a \ufb01rst step, we need to identify all contractions that lead to no more than one fast momentum integration (one-loop order). Since the contraction of terms of O(W (x)n>4) will generate at least two fast integrations, it su\ufb03ces 38 to consider the action S[gs, W ] at order W 2. Substitution of gf = 1 + W + W 2/2 into the action (8.32) gives S(2) c [gs, W ] = 1 \u03bb ddr tr (\u03a6\u03bc[\u2207\u03bcW, W ]) \u2212 2iLd \u03bb q p\u2208f p\u03bc tr [\u03a6\u03bc,\u2212qWp+qW\u2212p], where we have introduced the abbreviation \u03a6\u03bc = g\u22121 s \u2202\u03bcgs and, in the last representation, neglected the small momentum in comparison with the fast, 2p + q 2p. To obtain all oneloop corrections, we have to expand the functional in", " powers of S(2) [gs, W ] and integrate c over W. However, since each power of S(2) comes with one derivative acting on a slow \ufb01eld, and terms of more than two such derivatives are irrelevant, it su\ufb03ces to consider terms of order O((S(2) )2). To one-loop order, the RG step thus e\ufb00ects the replacement 2 : 2 : 9 9 c c S[g] \u2192 S[gs] \u2212 ln 1 + S(2) c [gs, W ] W 1 2 S[gs] \u2212 1 2 S(2) c [gs, W ] W. 37 Although we are primarily interested in the case N = 3, it will be instructive to monitor the RG \ufb02ow for general values of N. 38 Convince yourself that contributions to the action of O(W 3) also do not contribute at the one-loop level. (A good way to do this is to invent some prototypical diagrammatic code that keeps track of the momentum \ufb02ow.) As in our previous analysis of the \u03c64-model, the expansion of S[gs, W ] to linear order in W vanishes. (Momentum conservation implies that it is not possible to integrate a single fast \ufb01eld against a single slow \ufb01eld.) 8.5 RG analysis of the nonlinear \u03c3-model 461 \u03a6 i k W j W (a) (b) i j k Figure 8.9 (a) Graphical visualization of the matrix vertex tr(\u03a6W W ). The line segments connecting the boxes represent the matrix indices. (b) One-loop diagram representing the contraction tr(\u03a6W W ) tr (\u03a6W W ) \u2192 N tr (\u03a6\u03a6). The factor of N stems from the free summation over the central index j. Written more explicitly, 9 S(2) c [gs, W ] 2 : W = q,q p,p\u2208f tr(Aq,pWp+qW\u2212p)tr(Aq,p Wp+q W\u2212p )W, where Ap,q = \u2212 2iLd Drawing on that analysis (see Fig. 8.9), one obtains \u03bb p\u03bc\u03a6\u03bc,\u2212q, i.e. an expression of the type considered in the exercise above. 9 S(2) c [gs, W ] 2 : W where the constant C = = \u2212 8(N \u2212 2)L2d (\u03a0p)2p\u03bcp\u03bd tr(\u03a6\u03bc,q \u03a6\u03bd,\u2212q) \u03bb2 \u22122(N \u2212 2) p\u2208f ddp (2\u03c0)d q q p\u03bcp\u03bd p4 Ld = \u2212C ddr tr [\u03a6\u03bc \u03a6\u03bc] = C ddr tr tr(\u03a6\u03bc,q\u03a6\u03bd,\u2212q) % \u2207gs\u2207g\u22121 s & = C\u03bbS[gs], 1 p2 = 2(N \u2212 2) d ddp (2\u03c0)d 2(N \u2212 2)\u03a9d(\u039bd\u22122 \u2212 (\u039b/b)d\u22122) (2\u03c0)dd(d \u2212 2) 2(N \u2212 2)\u03a9d d(2\u03c0)d \u039b \u039b/b dp pd\u22123 d=2+ (N \u2212 2) ln b 2\u03c0. Substituting this result back into the action, one obtains S[g] \u2192 1 \u2212 (N \u2212 2) ln b\u03bb 4\u03c0 S[gs] \u2192 1 \u2212 (N \u2212 2) ln b\u03bb 4\u03c0 bS[g], where in the second step we performed the rescaling of the momenta (i.e. we rescaled q \u2192 bq, in such a way that the formerly \u201cslow\u201d momenta again take values up to the cuto\ufb00 \u039b, and noted that \u2261 d \u2212 2. As expected, the RG reproduces the action up to a global 462 The renormalization group scaling factor. Absorbing this factor in a renormalized coupling constant, \u03bbr, we obtain \u03bb\u22121 r = 1 \u2212 (N \u22122) ln b\u03bb b\u03bb\u22121, or 2\u03c0 \u03bbr = 1 \u2212 (N \u2212 2) ln b\u03bb 4\u03c0 \u22121 b\u2212\u03bb ln b\u03bb (N \u2212 2)\u03bb 4\u03c0 \u2212 + \u03bb. Here, in the last equality, we have made use of the fact that both \u03bb and are small.39 Finally, di\ufb00erentiating this result one obtains the RG equation d ln \u03bb d ln b (N \u2212 2)\u03bb 4\u03c0 \u2212", " + O(\u03bb2, 2, \u03bb). (8.37) The \ufb01gure below shows the RG \ufb02ow for generic (N > 2) group dimensions and di\ufb00erent values of. As expected, for dimensions two and below, the coupling constant inexorably \ufb02ows towards a phase with large \u03bb where \ufb02uctuations are large (and our perturbative analysis loses its basis) \u2013 the \u201cdisordered\u201d type of behavior pervasive in the physics of low-dimensional systems with continuous symmetries. However, for dimensions d >2, the model exhibits a \ufb01xed point at \ufb01nite coupling strength \u03bbc = 2\u03c0/(N \u2212 2). Linearizing in the vicinity of the \ufb01xed point, one \ufb01nds a thermal exponent y\u03bb = from which, making use of the scaling relations, one can deduce a correlation length exponent \u03bd \u2248 1/ and a heat capacity exponent \u03b1 \u2248 2 \u2212 2/. EXERCISE To complete the analysis of the O(3) model, one may explore the magnetic exponent. ddr tr [g + g\u22121], show that, Introducing a magnetic \ufb01eld perturbation into the action, i.e. h under renormalization, h = byh h, where yh = 1 + N \u22123 2(N \u22122) + O(2). From this result, use the exponent identities to show that the correlation length exponent \u03b7 \u2248 /(N \u2212 2). \u03b2 (\u03bb) Notice that, for d = 2 and N = 2, the coupling constant does not renormalize. This is because, for O(2) \u2013 the abelian group of planar rotations \u2013 the action of the \u03c3-model simpli\ufb01es to d2r (\u2207\u03c6)2, where \u03c6(r) is the \ufb01eld of local rotation S = 1 \u03bb angles. This is a free action which is not renormalized by small \ufb02uctuations. (However, in the following section, we see that the situation is not quite as simple as it seems. The O(2) model admits large \ufb02uctuations \u2013 vortex con\ufb01gurations of the angular \ufb01eld \u2013 which do have a non-trivial impact on its behavior. However, being topological in nature, these excitations are beyond the scope of our present analysis.) d<2 d>2 \u03bb 39 The smallness of \u03bb is needed to justify the perturbative loop expansion. In the opposite regime \u03bb 1, the \ufb01elds g \ufb02uctuate wildly and perturbative expansions are not an option. 8.6 Berezinskii\u2013Kosterlitz\u2013Thouless transition 463 EXERCISE Repeat the analysis above for the SU(N )-model. First derive the intermediate identities, tr(AWp)tr(AWp )W = \u2212\u03b4p,\u2212p tr(AWpAWp )W = \u2212\u03b4p,\u2212p \u03a0p 2 \u03a0p 2 where \u03a0q = \u03bb/Ldp2, and tr(AA) \u2212 1 N tr(A)tr(A) tr(A)tr(A) \u2212 1 N tr(AA),, p\u03bcp\u03bc tr(\u03a6\u03bc,qWp+qW\u2212p)tr(\u03a6\u03bc,q Wp+q W\u2212p )W = N \u03a0p\u03a0p+q 4 p\u03bcp\u03bc tr(\u03a6\u03bc,q\u03a6\u03bc,\u2212q). Use these results to obtain the RG equation d ln \u03bb d ln b N \u03bb 8\u03c0 \u2212 + O(\u03bb2, 2, \u03bb). (8.38) Notice that the right-hand side of the SU(N = 2) equation coincides with the right-hand side of the O(N = 3) equation. This re\ufb02ects the (local) isomorphism of the groups O(3) and SU(2). 8.6 Berezinskii\u2013Kosterlitz\u2013Thouless transition The majority of theories discussed in earlier chapters were de\ufb01ned on essentially structureless target spaces \u2013 the real line, the complex plane, or, more generally, vector spaces of arbitrary dimensionality. Although we have encountered a number of models with more complex target manifolds (such as the superconductor, the Luttinger action with \ufb01elds de\ufb01ned on the unit circle, and the nonlinear \u03c3-model theories discussed in the previous section), the global geometric structure of the target space did not seem to play such an important role: the phase behavior of the O(N )-", "theory appears to di\ufb00er little from an unconstrained N -component vector \u03c64-theory. However, the large-scale geometry, or topology of the \ufb01eld manifold, may have a striking in\ufb02uence on the long-range behavior of a system, and it is one of the objectives of the present section to illustrate this phenomenon on a particular example. (A more comprehensive discussion of the role of geometry and topology in quantum \ufb01eld theory will be developed in the next chapter.) Consider then a two-dimensional square lattice with a phase-like variable exp(i\u03b8i) \u2208 S1 de\ufb01ned on each of its sites i. Demanding that the Hamiltonian or action of the system be periodic in all \u03b8i and minimal on homogeneous \ufb01eld con\ufb01gurations, the most elementary action we can formulate reads S[\u03b8] = \u2212J ij cos(\u03b8i \u2212 \u03b8j), (8.39) where ij denotes the usual sum over nearest neighbor sites of a lattice. Equation (8.39) is known as the action of the two-dimensional XY -model. (Its incarnation as a continuum \ufb01eld theory is the U(1) version of the nonlinear \u03c3-model discussed above.) To motivate the terminology, one may regard \u03b8i as an angle parameterizing the direction of a unimodular vector (a classical \u201cspin\u201d) in the two-dimensional XY -plane. Indeed, the model Eq. (8.39) 464 The renormalization group can be viewed as a two-dimensional descendant of the classical Heisenberg model, i.e. as a model where the Heisenberg spin is con\ufb01ned to a two-dimensional plane (a condition in fact realized for a number of magnetic materials). INFO Besides its natural occurrence in magnetism, the XY -model appears in many applications in condensed matter physics. By way of an example, we mention here the physics of granulated metals and the Josephson junction array. Imagine then a system of small metallic (or superconducting) islands (see \ufb01gure), which are connected by poorly conducting tunneling barriers. (Arrays of this structure are believed to mimic the mesoscopic morphology of structurally disordered metals. However, they can also be manufactured arti\ufb01cially, with islands as small as several nanometers.) On a microscopic level, the state of the system is essentially determined by the number Ni of electronic charges populating each island i. However, unless the inter-grain tunneling conductance, gT, is small (i.e. gT < e2/h, the fundamental quantum unit of conductance), temporal \ufb02uctuations of the charge degrees of freedom are vast. It is then more favorable to characterize the state of the system in terms of the variable canonically conjugate to the charge variable, i.e. the phase \u03b8i. (Remember that [\u03b8i, \u02c6Nj] = \u2212i\u03b4ij form a canonically conjugate pair.) The model action S[\u03b8] then, of course, depends on the microscopic structure of the system. However, for the case of a superconductor array, in the classical limit (where temporal \ufb02uctuations of the phase \ufb01eld \u03b8(\u03c4 ) are suppressed), and neglecting charging e\ufb00ects, the action collapses to that of the XY -action shown above. (Recapitulate the structure of the single-junction Josephson action discussed in Problem 6.7 to convince yourself that this is true.) Amongst the class of classical spin models, the two-dimensional XY -model assumes a unique status \u2013 a fact already hinted at in consideration of the RG in the previous section. To understand why, it is helpful to explore the high- and low-temperature expansions of the theory. In the high-temperature phase, J \u2192 0, one may expect that the classical partition function for the lattice system can be developed as a series expansion in J, Z = 2\u03c0'0 i d\u03b8i 2\u03c0 e\u2212S[\u03b8i] = 2\u03c0'' % d\u03b8i 2\u03c0 0 i ij 1 + J cos(\u03b8i \u2212 \u03b8j) +O (J 2). & 2\u03c0 Here each term in the product can be represented by a \u201cbond\u201d that connects neighboring sites i and j. To the lowest order in J, each bond on the lattice contributes a factor of either one or J cos(\u03b8i \u2212 \u03b8j). But, since 0 d\u03b81 cos(\u03b81 \u2212 \u03b82) = 0, any graph with a single bond emanating", " from a site vanishes. On the other hand, a site at which two bonds meet yields 0 (d\u03b82/2\u03c0) cos(\u03b81 \u2212 \u03b82) cos(\u03b82 \u2212 \u03b83) = (1/2) cos(\u03b81 \u2212 \u03b83). Therefore, in this way, a factor the classical partition function can be presented as a closed \u201cloop\u201d expansion. Similarly, the high-temperature expansion can be used to estimate the spin\u2013spin correlation function S0 \u00b7 Sr = cos(\u03b8r \u2212 \u03b80). To leading order, only those graphs that join sites 0 and r will survive, implying an exponential decay of the correlation function, 2\u03c0 S0 \u00b7 Sx \u223c |x| J 2 \u223c exp \u2212, |x| \u03be 8.6 Berezinskii\u2013Kosterlitz\u2013Thouless transition 465 with a correlation length \u03be\u22121 = ln(2/J ). Conversely, in the low-temperature phase, where \ufb02uctuations of neighboring phases \u03b8i are strongly penalized, the lattice model can be safely d2r J(\u2207\u03c6)2. Taking only harmonic \ufb02uctuations of replaced by the continuum theory 1 2 the \ufb01eld \u03c6(r), i.e. neglecting the topological constraint placed by the geometry, the spin\u2013 spin correlation function can be inferred from the rules of Gaussian integration, to wit S(0) \u00b7 S(r) = Reei(\u03b8(0)\u2212\u03b8(r)) = e\u2212(\u03b8(0)\u2212\u03b8(r))2/2. Since, in two dimensions, Gaussian \ufb02uctuations grow logarithmically, (\u03b8(0) \u2212 \u03b8(r))2/2 = ln(|r|/a)/2\u03c0J, where a denotes a short-distance cuto\ufb00 (e.g. the lattice spacing), one can infer a spin\u2013spin correlation function that decays algebraically, S(0) \u00b7 S(r) 1 a |r| 2\u03c0J, a phase of quasi-long-range order. The distinction between the nature of the asymptotic decays allows for the possibility of a \ufb01nite-temperature phase transition. However, these arguments are not speci\ufb01c to the XY -model: any continuous spin model will exhibit exponential decay of correlations at high temperature and, in two dimensions, a power law decay in a low-temperature Gaussian approximation. Now, formally, to establish that \u201cGaussian\u201d behavior persists at low temperatures, one must prove that it is not modi\ufb01ed by higher-order terms in the gradient expansion (including those induced by the nonlinear constraint, S2 i = 1). However, while the results of the (2 + )-expansion of the U(N ) theory show that the two-dimensional model \ufb02ows to a disordered phase for N > 2 (in accord with the Mermin\u2013Wagner theorem), for N = 2 the theory is non-committal. Indeed, in an extension of the perturbative RG to higher loop order, the \u03b2-function remains zero. So we are left with the question, is there indeed a phase transition that delineates between the high- and low-temperature asymptotics, and what is the mechanism by which the quasi-long-range order is destroyed at higher temperatures? Since the perturbative RG suggests that higher-order terms in the gradient expansion are not relevant, it is necessary to search for other relevant operators. 8.6.1 Vortices and the topological phase transition The gradient expansion describes the energy cost of small deformations around the ground state, and applies to con\ufb01gurations that can be continuously deformed to the uniformly ordered state. In separate and seminal works,40 Berezinskii, and Kosterlitz and Thouless, proposed that the disordering is facilitated by the \u201ccondensation\u201d of topological defects or vortices. Since the angle describing the orientation of a spin is de\ufb01ned up to an integer multiple of 2\u03c0, it is possible to construct spin con\ufb01gurations in which the traversal of a closed path will see the angle rotate by 2\u03c0n. The integer n de\ufb01nes a topological charge enclosed by the path. The discrete nature of the charge makes it impossible to \ufb01nd a continuous 40 V. L. Berezinskii, Violation of long range order in one-dimensional and two-dimensional systems with a continuous symmetry group. I. Classical systems, Sov. Phys. JETP 32 (1971), 493\u2013500; J. M. Kosterlitz and D. J.", " Thouless, Ordering, metastability, and phase transitions in two-dimensional systems, J. Phys. C 6 (1973), 1181\u2013203. 466 The renormalization group deformation that returns the state to the uniformly ordered con\ufb01guration in which the charge is zero. (More generally, topological defects arise in many models with a compact group describing the order parameter \u2013 e.g., a domain wall defect (e.g. an instanton) in a model with Z2 symmetry, or a \u201cskyrmion\u201d con\ufb01guration in an O(3) three-component spin Heisenberg ferromagnet; see the next chapter for a discussion of topological defects.) of the David J. Thouless, 1934\u2013 co-recipient with John Michael Kosterlitz 2000 Lars Onsager Prize \u201cfor the introduction of the theory of topological phase transitions, as well as their subsequent quantitative predictions by means of early and ingenious applications of the renormalization group, and advancing the understanding of electron localization and the behavior of spin glasses.\u201d The elementary defect, known in the present context as a vortex con\ufb01guration, has a unit \u201ccharge\u201d: in completing a circle centered on the defect, the orientation of the spin changes by \u00b12\u03c0 (see Fig. 8.10). If the radius r of the circle is su\ufb03ciently large, the variations in angle will be small and the underlying lattice structure of the spin model can be neglected. By symmetry \u2207\u03b8 has uniform magnitude and points along the azimuthal direction. The magnitude of the distortion \u2207\u03b8 \u00b7 dl = 2\u03c0n, i.e. can be obtained by integrating around a path that encloses the defect, r \u02c6ez \u00d7\u02c6er, where \u02c6er and \u02c6ez are unit vectors respectively in the plane and perpendicular \u2207\u03b8 = n to it. This continuum approximation fails close to the center (core) of the vortex where the lattice structure becomes important. = Although the defect-driven topological phase transition can be described within the framework of the RG (see below), the nature of the transition can be understood from a simpler and intuitive line of reasoning that one can \ufb01nd in the original paper by Kosterlitz and Thouless. The energy cost of a single vortex of charge n can be divided into contributions from the core region as well as from the relatively uniform distortions away from the center. The distinction between regions inside and outside the core is arbitrary (see below) and, for simplicity, we shall use a circle of radius a to distinguish the two, i.e. for a vortex of charge n, the associated action is given by Sn = Score n (a) + a J 2 d2r (\u2207\u03b8)2 = Score n (a) +\u03c0Jn 2 ln. L a The dominant contribution to the action arises from the region outside the core and diverges logarithmically with the system size L.41 The large energy cost associated with the defects inhibits their spontaneous formation at low temperatures and protects the integrity of the quasi-long-range ordered phase. To explore the range over which vortices are suppressed, one may explore the partition function for a con\ufb01guration with just a single vortex of unit 41 Notice that, if the spin degrees of freedom have three components or more, the energy cost of a defect is only \ufb01nite. Following the arguments below, one may re\ufb02ect on the implications of this fact for the inability of topological defects to e\ufb00ect a topological transition in systems with N > 2. 8.6 Berezinskii\u2013Kosterlitz\u2013Thouless transition 467 Figure 8.10 Spin con\ufb01gurations of the two-dimensional XY -model showing vortices of charge \u00b11. charge n = 1, Z1 \u2248 2 L a exp \u2212Score 1 (a) \u2212 \u03c0J ln. L a (8.40) Here the factor of (L/a)2 results from the con\ufb01gurational entropy associated with the range of possible vortex locations in an area of size L2. The entropy and energy of a vortex both grow as ln L, and the free energy is dominated by one or the other. At low temperatures (i.e. large J) the potential energy dominates over the entropy and Z1, a measure of the weight of con\ufb01gurations with a single vortex, vanishes. However, when J < Jc = 2/\u03c0, the entropic contribution overwhelms the potential and one may expect the spontaneous formation of vortices and a transition to the disordered phase. In fact one might expect that this estimate of Jc represents only a lower bound for the stability of", " the system towards the condensation of topological defects. This is because pairs (dipoles) of defects may appear at larger couplings. Consider a pair of charges \u00b11 separated by a distance d (see Fig. 8.10). Distortions far from the core |r| d can be obtained by superposing those of the individual vortices, i.e. \u2207\u03b8 = \u2207\u03b8(r + d/2) \u2212 \u2207\u03b8(r \u2212 d/2) \u2248 d \u00b7 \u2207(\u02c6er \u00d7\u02c6ez/|r|) \u223c d/|r|2. Integrating this distortion, one may infer that the energy cost of the con\ufb01guration is only \ufb01nite, and hence dipoles appear with the appropriate Boltzmann weight at any non-zero temperature. The low-temperature phase should therefore be visualized as a \u201cgas\u201d of tightly bound dipoles, their density and size increasing with temperature. The high-temperature phase constitutes a plasma of unbound vortices. To a large extent, the argument above conveys much of the phenomenology of the Berezinskii\u2013Kosterlitz\u2013Thouless (BKT) phase transition. However, one may justi\ufb01ably question the impact of harmonic \ufb02uctuations and dipole interactions in suppressing or even destroying the transition. To explore these e\ufb00ects, one has to turn to a more rigorous line of reasoning. To do so, we will once again draw on the RG. 468 The renormalization group 8.6.2 RG analysis of the BKT transition To prepare for the application of the RG, we must \ufb01rst reorganize the action to emphasize the di\ufb00erent channels of \ufb02uctuations. Fortunately, for the two-dimensional XY -model, the di\ufb00erent channels (harmonic and topological) can be neatly separated. To understand why, it is helpful to draw an analogy, noting that the distortion \ufb01eld, u \u2261 \u2207\u03b8, is similar to the velocity of a \ufb02uid. In the absence of vorticity, the \ufb02ow is potential, i.e. u = u0 = \u2207\u03c6, and \u2207 \u00d7 u0 = 0. The topological charge can be related to the vorticity \u2207 \u00d7 u by noting d2r \u02c6ez \u00b7 \u2207 \u00d7u, where the second integral is over the that, for any closed path, area enclosed by the path. Since the left-hand side is an integer multiple of 2\u03c0, one can set i ni\u03b42(r \u2212 ri), describing a collection of vortices of charge {ni} at locations \u2207 \u00d7 u = 2\u03c0\u02c6ez {ri}. The solution to this equation can be obtained by setting u = u0 \u2212 \u2207 \u00d7( \u02c6ez\u03c8), leading to dl \u00b7 u = = \u2207 \u00d7 u = \u02c6ez\u22072\u03c8 \u21d2 \u22072\u03c8 = 2\u03c0 ni\u03b42(r \u2212 ri). (8.41) i Thus, the \ufb01eld \u03c8 behaves as the potential due to a set of point charges {2\u03c0ni} while the i ni ln(|r \u2212 ri|), is simply a superposition of the potentials. Any twosolution, \u03c8(r) = dimensional distortion can therefore be written as u = u0 + u1 = \u2207\u03c6 \u2212 \u2207 \u00d7( \u02c6ez\u03c8), where u0 accommodates the harmonic \ufb01eld \ufb02uctuations while u1 includes the topological defects. The corresponding action, S[u] = 1 2 d2r Ju2, can be decomposed as S[u] = J 2 % d2r (\u2207\u03c6)2 \u2212 2\u2207\u03c6 \u00b7 \u2207 \u00d7 (\u02c6ez\u03c8) + (\u2207 \u00d7 \u02c6ez\u03c8)2 &. An integration by parts shows that the second term vanishes (\u2207 \u00b7 \u2207 \u00d7 u = 0), while the third term can be simpli\ufb01ed by noting that \u2207\u03c8 and \u2207 \u00d7( \u02c6ez\u03c8) are orthogonal vectors of equal length. Hence, making use of Eq. (8.41), the action associated with the topological defects takes the form St \u2261 J 2 d2r (\u2207 \u00d7 \u02c6ez\u03c8)2 = \u2212 J 2 d2r \u03c8\u22072\u03c8 = \u22122\u03c02J ninjC(ri \u2212 rj), ij where C(x) = ln |x|/2\u03c0 is the two-dimensional Coulomb", " potential. Once again, the unphysical divergence of the potential for i = j must be regularized by the core energy, whence St = i Score ni \u2212 4\u03c02J i<j ninjC(ri \u2212 rj). (8.42) Conveniently, the con\ufb01guration space of the XY -model can thus be partitioned into di\ufb00erent topological regions: one set arises from the degrees of freedom associated with the charges, {ni}, while the other re\ufb02ects the harmonic spin-wave excitations associated with the \ufb01eld \u03c6(r). The partition function of the lattice model is then approximately given by Z \u2248 Zs.w.Zt where Zs.w. = D\u03c6 exp \u2212 J 2 d2r (\u2207\u03c6)2, Zt = \u221e N =0 1 (N!)2 2N' i=1 6 7 d2ri e\u2212St, 8.6 Berezinskii\u2013Kosterlitz\u2013Thouless transition 469 and the ri are dimensionless integration variables ri = xi/a where xi is the position of the vortex centre. (Exercise: Consider why the combinatorial factors (N!)2 are needed to prevent overcounting of identical charge con\ufb01gurations.) To summarize, Zt describes a grand canonical ensemble of \u201ccharges\u201d with a two-dimensional Coulomb interaction. In fact, our derivation of the action relied on an integration by parts. The surface integral that was i ni. Therefore, one neglected in the process in fact grows with the system size as ln L must impose on the \ufb01eld con\ufb01gurations contributing to Zt the constraint of overall charge neutrality. In the following, to simplify our analysis further, we will limit considerations to only elementary excitations with ni = \u00b11.42 Then, if we de\ufb01ne the fugacity of the vortices as y0 \u2261 exp[\u2212Score \u00b11 ], the partition function Zt = \u221e N =0 y2N 0 (N!)2 6 2N' i=1 7 \u23a1 d2ri exp \u23a34\u03c02J i<j \u03c3i\u03c3jC(ri \u2212 rj) \u23a4 \u23a6, (8.43) reduces to that of a neutral Coulomb plasma of 2N unit charges ni = \u00b11. EXERCISE The low-temperature physics of many two-dimensional systems is e\ufb00ectively reduced to that of the XY -model or of the two-dimensional Coulomb plasma. For this reason, the equivalence XY -model \u21d4 Coulomb plasma demonstrated above represents an important result in its own right. In fact, there exists a third important system falling into this universality class: the two-dimensional sine\u2013Gordon model de\ufb01ned by the action S[\u03b8] = c 2 d2r \u2207\u03b8 \u2207\u03b8 + g d2r cos(\u03b8), (8.44) where \u03b8 is a real scalar \ufb01eld. To demonstrate its equivalence to the Coulomb gas model, expand the partition function in powers of g and show that Z = \u221e N =0 y2N 0 (N!)2 2N- i=1 0 1 ddri exp i (\u2212)i\u03b8(ri), 2N i=1 where y0 = g/2 and the angular brackets denote averaging over the free action. (To prove that positive and negative phases appear in equal numbers, consider the role of the integration over the zero mode \u03b8(r) = const.) Using the fact that \u03b8(r)\u03b8(r) = C(|r \u2212 r|)/c, and neglecting the in\ufb01nite self-interaction of the \ufb01elds at coinciding points, show that the partition function becomes identical to Eq. (8.43) when c = 1/8\u03c02J. Since Zs.w. and Zt involve independent degrees of freedom, they can be treated separately. Further, as the Gaussian partition function Zs.w. is everywhere analytic, any phase transition of the XY -model must originate from the Coulomb gas. Now, as mentioned above, in the low-temperature phase, charges appear only as tightly bound dipole pairs while, at high temperatures, the dipoles dissociate, forming a plasma. The two phases can be distinguished 42 A charge two vortex costs about as much energy as two unit-charge vortices. However, its entropy \u223c 2T ln(L/a) is much lower than the entropy \u223c 4T ln(L/a) of the two-vortex con\ufb01guration. This is why", " vortices of high winding number are negligible in the low-temperature limit. 470 The renormalization group by examining the interaction between two external test charges at some large separation X. In the absence of internal charges, the two particles interact by the bare Coulomb interaction C(X). The presence of internal charges provides a mechanism to screen the interaction, namely C(X)/\u03b5, where\u03b5 is an e\ufb00ective dielectric constant. With this interpretation, the transition with increasing temperature (or increasing y0) can be viewed as a transition from an insulating (dipolar) phase to a metallic (plasma) phase in which the external charges are completely screened and their e\ufb00ective interaction decays exponentially. To quantify this picture, one can compute the e\ufb00ective interaction between two external charges at r and r perturbatively in the fugacity y0. To lowest order, the e\ufb00ects of screening can be incorporated by including con\ufb01gurations with only two internal charges (positioned at s and s). In this case, the e\ufb00ective interaction can be written as e\u2212Seff.(r\u2212r) \u2261 e\u22124\u03c02JC(r\u2212r)t = from which one obtains (r \u2295 \u2212 &r ) +y 2 0 1 + y2 0 \u239e \u239b \u239d d2s d2s d2s d2s (s \u2295 \u2212 &s ) +O (y4 0) \u23a0 + O(y4 0), e\u2212Seff.(r\u2212r)+4\u03c02JC(r\u2212r) 1 + y2 0 = d2s d2y e\u22124\u03c02JC(s\u2212s)+4\u03c02J[C(r\u2212s)\u2212C(r\u2212s)\u2212C(r\u2212s)+C(r\u2212s)] + O(y4 0) d2s d2s e\u22124\u03c02JC(s\u2212s) + O(y4 0) e4\u03c02JD(r,r,s,s) \u2212 1 d2s d2s e\u22124\u03c02JC(s\u2212s) + O(y4 1 + y2 0 (8.45) 0). = 1 + y2 0 Here, D(r, r, s, s) = C(r \u2212 s) \u2212 C(r \u2212 s) \u2212 C(r \u2212 s) +C (r \u2212 s) denotes the interaction between the internal and external dipoles, while the direct interaction C(s \u2212 s) tends to keep the separation x = s \u2212 s small. De\ufb01ning the center of mass X = (s + s)/2, we can change variables to s = X\u2212x/2 ands = X+x/2, and expand the dipole\u2013dipole interaction in small x as D(r, r, s, s) \u2212x \u00b7 \u2207XC(r \u2212 X) +x \u00b7 \u2207XC(r \u2212 X) +O (x3). To the same order e4\u03c02JD(r,r,s,s) \u2212 1 \u22124\u03c02 Jx \u00b7 \u2207X(C(r \u2212 X) \u2212 C(r \u2212 X)) + 8\u03c04J 2[x \u00b7 \u2207X(C(r \u2212 X) \u2212 C(r \u2212 X))]2 + O(x3). Substituting this expression into Eq. (8.45), and changing integration variables, one \ufb01nds that the term linear in x integrates to zero while the angular average of (x \u00b7 \u2207XC)2 leads to x2(\u2207XC)2/2. Thus, to O(r4), one obtains e\u2212Seff (r\u2212r) e\u22124\u03c02JC(r\u2212r) \u00d7 1 + y2 0 (dx 2\u03c0 x)e\u22124\u03c02JC(x)8\u03c0J 2 x2 2 d2X (\u2207X(C(r \u2212 X) \u2212 C(r \u2212 X)))2. d2X [\u2207X(C(r\u2212X)\u2212C(r \u2212X))]2 = Using the identity \u22072C(r) = \u03b42(r), one may note that 2(C(r \u2212 r) \u2212 C(0)). Finally, absorbing the short-distance divergence into an appropriate 8.6 Berezinskii\u2013Kosterlitz\u2013Thouless transition 471 cuto\ufb00 with C(x) \u2192 ln(x/a)/2\u03c0, one arrives at the expression \u221e e\u2212Seff (r\u2212r) = e\u22124\u03c02JC(r\u2212r) 1 + 16\u03c05J 2y2", " 0C(r \u2212 r) 1 dx x3 e\u22122\u03c0J ln x + O(y4 0), where the variable x has been rescaled to absorb the short distance cuto\ufb00. Exponentiating the second term, one obtains the e\ufb00ective interaction Se\ufb00 (r \u2212 r) 4\u03c02Je\ufb00 C(r \u2212 r), where \u221e Je\ufb00 = J \u2212 4\u03c03J 2y2 0 1 dx x3\u22122\u03c0J + O(y4 0), (8.46) From this result, to leading order in y2 0, one can infer the dielectric constant of the medium as \u03b5 = J/Je\ufb00. However, while the integral remains convergent at large x, the perturbative correction is small. The breakdown of the perturbation theory for J < Jc = 2/\u03c0 occurs precisely at the point where the free energy of an isolated vortex changes sign. Indeed, this breakdown of the perturbation theory is reminiscent of that encountered in the \u03c64-theory for d < 4. This being so, one may propose a reorganization of the perturbation series into a RG calculation for the parameters J and y0. \u2192 b 1 + \u221e 1 The di\ufb03culty associated with the divergence at small J can be overcome by employing a renormalization procedure \ufb01rst used by Jos\u00b4e et al.43 By breaking the integral in Eq. (8.46) \u221e into two parts, i.e. b, the non-singular short-distance contribution to the integral can be evaluated and incorporated into J. This procedure can be carried out order by order in y0 even though the coe\ufb03cient of y2 0 is formally divergent. This leads to a new \u221e 0), where 3J \u22121 = J \u22121+4\u03c03y2 equation J \u22121 1 dx x3\u22122\u03c0J + b dx x3\u22122\u03c0J +O(y4 0). The variables in the remaining integral can be rescaled (x \u2192 x/b) to yield an equation O(y4 for J \u22121 e\ufb00 equivalent to Eq. (8.46) but with shifted and rescaled parameters J and y0: \u221e e\ufb00 = 3J \u22121+4\u03c03y2 0 0 b e\ufb00 = 3J \u22121 + 4\u03c033y2 J \u22121 0 dx x3\u22122\u03c0 J + O(y4 0), 1 where 3y0 = b2\u2212\u03c0J y. By choosing an in\ufb01nitesimal renormalization, i.e. b = e \u2248 1 +, one obtains the di\ufb00erential recursion relations dJ \u22121 d dy0 d = 4\u03c03y2 0 + O(y4 0), = (2 \u2212 \u03c0J)y0 + O(y3 0). (8.47) From these equations, one may note that the (inverse) coupling constant J \u22121 increases monotonically with while the recursion relation for y0 changes sign at J \u22121 c = \u03c0/2. At high temperatures, when J is small, y0 is relevant, while at lower temperatures it becomes irrelevant. The RG \ufb02ows, shown in Fig. 8.11(a), separate the parameter space into two regions. At low temperatures, and small y0, \ufb02ows terminate on a \ufb01xed line at y0 = 0 and Je\ufb00 \u2265 2/\u03c0. This is the insulating phase, in which only dipoles of \ufb01nite size occur (hence the vanishing of y0 under coarse-graining). The strength of the e\ufb00ective interaction is given by the point on the \ufb01xed line at which the \ufb02ow terminates. Flows that do not terminate on the 43 J. V. Jos\u00b4e, L. P. Kadano\ufb00, S. Kirkpatrick, and D. R. Nelson, Renormalization, vortices, and symmetry-breaking perturbations in the two-dimensional planar model, Phys. Rev. B 16 (1977), 1217\u201341. 472 The renormalization group Figure 8.11 Schematic RG \ufb02ow diagram for the XY -model. (a) Far from the critical point, and (b) close to the critical point. \ufb01xed line asymptote to larger values of J \u22121 and y0 where perturbation theory eventually breaks down. This is the signal of the high-temperature phase where vortices proliferate. The critical trajectory that separates the two regions of the phase diagram \ufb02ows to a \ufffd", "\ufffd\ufffdxed point at (J \u22121 c = \u03c0/2, y0 = 0). To explore the critical behavior at the transition, one must expand the recursion relations in the vicinity of this point. Setting t = J \u22121 \u2212 \u03c0/2, to lowest order, Eq. (8.47) simpli\ufb01es to dt d dy d = 4\u03c03y2 + O(ty2, y4), = 4 \u03c0 ty + O(t2y, y3). In contrast to those discussed in previous sections, the recursion relations here are inherently nonlinear in the vicinity of the \ufb01xed point (see Fig. 8.11(b)), suggesting that the critical behavior is likely to be novel. To uncover the structure, one may note that t2 \u2212 \u03c04y2 is a conserved quantity, i.e. the \ufb02ow proceeds along hyperbolae characterized by di\ufb00erent values of c \u2261 t2 \u2212 \u03c04y2. For c < 0, the focus of the hyperbola is along the y-axis, and the \ufb02ows proceed to (t, y) \u2192 \u221e. Conversely, hyperbolae with c >0 have foci along the t-axis, and have two branches in the half plane y \u2265 0: the branches for t < 0 \ufb02ow to the \ufb01xed line, while those in the t > 0 quadrant \ufb02ow to in\ufb01nity. The critical trajectory separating \ufb02ows to zero and in\ufb01nite y corresponds to c = 0, i.e. tc = \u2212\u03c02yc. Therefore, a small but \ufb01nite fugacity y0 reduces the critical temperature to J \u22121 c = \u03c0/2 \u2212 \u03c02y0 + O(y2 0). In terms of the original XY -model, the low-temperature phase is characterized by a line of \ufb01xed points with Je\ufb00 = lim\u2192\u221e J() \u2265 2/\u03c0. Here the phase correlations decay as a power law, i.e. cos(\u03b8(r) \u2212 \u03b8(0)) \u223c1/| r|\u03b7, with \u03b7 = 1/2\u03c0Je\ufb00 \u2264 1/4. Since the parameter c is negative in the low-temperature phase, and vanishes at the critical point, we can set c = \u2212b2(Tc \u2212 T ) close to the transition. In other words, the trajectory of initial points tracks a line (t(T ), y(T )). The resulting c = t2 \u2212 \u03c04y2 \u221d (Tc \u2212 T ) is a linear measure of the proximity of the phase transition. Under renormalization, such trajectories \ufb02ow to a \ufb01xed Tc \u2212 T. Thus, in the vicinity of the transition, the e\ufb00ective point at y = 0, and t = \u2212b interaction parameter, \u221a Je\ufb00 = 2 \u03c0 \u2212 4 \u03c02 lim \u2192\u221e t() = 2 \u03c0 + 4b \u03c02 Tc \u2212 T, exhibits a square root singularity. 8.6 Berezinskii\u2013Kosterlitz\u2013Thouless transition 473 ) = const. Tc = const. (using Ahns) Third sound: Hallock Mochel Rudnick Kosterlitz\u2013Thouless static theory 0 0.2 0.4 0.6 0.8 1.0 1.2 Tc (K) 1.4 1.6 1.8 2.0 2.2 Figure 8.12 Measurement of the super\ufb02uid density \u03c1s(Tc) as a function of Tc. The approximate linearity \u03c1s \u223c Tc demonstrates the universality of the sti\ufb00ness constant J at the transition. (Reprinted with permission from D. J. Bishop and J. D. Reppy, Study of the super\ufb02uid transition in twodimensional 4He \ufb01lms, Phys. Rev. Lett. 40 (1978), p. 1727\u201330. Copyright (1978) by the American Physical Society.) INFO The sti\ufb00ness, Je\ufb00, can be measured directly in experiments on super\ufb02uid \ufb01lms. As we have seen in Chapter 6, in the super\ufb02uid phase, the low temperature of neutral super\ufb02uids is described by a phase action with a sti\ufb00ness J = \u03b2\u03c1s/m2 determined by the super\ufb02uid density of the system. The density \u03c1s is measured by examining the changes in the inertia of a torsional oscillator; the", " super\ufb02uid fraction, \u03c1s experiences no friction and does not oscillate. Bishop and Reppy examined \u03c1s for a variety of super\ufb02uid \ufb01lms (of di\ufb00erent thickness, 4He concentration, etc.) wrapped around a torsional cylinder. By inferring the e\ufb00ective sti\ufb00ness J as a function of temperature, they found that, for all \ufb01lms, it undergoes a universal jump of 2/\u03c0 at the transition while the behavior of J for T < Tc was consistent with a square root singularity. They also observed (see Fig. 8.12) that the ratio \u03c1s(Tc)/Tc = Jcm2 = 2m2/\u03c0 is a universal quantity independent of the material parameters of the \ufb01lm. In the high-temperature phase, the correlations decay exponentially. In this phase the parameter c = t2 \u2212 \u03c04y2 = b2(T \u2212 Tc) is positive all along the hyperbolic trajectory. Here the recursion relation dt \u03c0 (t2 + b2(T \u2212 Tc)) can be integrated to give d = 4\u03c03y2 = 4 1 T \u2212 Tc t T \u2212 Tc arctan 4 \u03c0 \u221a \u221a b b. The integration must be terminated when t() \u223c y() \u223c 1 since, beyond there, the perturbative calculation is invalid. This occurs for a value \u2217 \u2248 \u03c0 2, where we have used the fact that arctan(1/b T \u2212 Tc) \u2248 \u03c0/2. The resulting correlation length is given by \u03c0 T \u2212Tc \u221a 4b \u221a \u03be \u2248 ael\u2217 \u2248 a exp, (8.48) \u221a \u03c02 T \u2212 Tc 8b 474 The renormalization group i.e. in contrast to the usual behavior described in previous sections, the divergence of the correlation length is not a power law, a consequence of the nonlinear nature of the recursion relations in the vicinity of the \ufb01xed point. Finally, vortices occur in bound pairs for distances smaller than \u03be while there can be an excess of vortices of one sign or the other at larger separations. The interactions between vortices at large distances can be obtained from the Debye\u2013H\u00a8uckel theory of polyelectrolytes. According to this theory, the free charges screen each other, leading to a screened Coulomb interaction exp(\u2212x/\u03be)C(x). On approaching the transition from the high-temperature side, the singular part of the free energy fsing \u221d \u03be\u22122 \u221d exp \u2212, \u221a \u03c02 T \u2212 Tc 4b (8.49) has only an essential singularity. All derivatives of this function are \ufb01nite at Tc. Thus the predicted heat capacity is quite smooth at the transition. Numerical results based on the RG equations indicate a smooth maximum in the heat capacity at a temperature higher than Tc corresponding to the point at which the majority of dipoles unbind. 8.7 Summary and outlook This concludes our preliminary introduction to the theory of the \u201crenormalization group.\u201d It should be emphasized that our approach to the subject was biased in that it introduced the RG in a manner tailored to applications in quantum statistical physics; a more particlephysics-oriented introduction (see Ryder13) would have focused on the principal issues of renormalizability, di\ufb00erent regularization schemes, renormalization through counter-terms and, of course, the many great triumphs renormalization group methods have had in QED and QCD. However, even with regard to statistical/condensed matter physics, our discussion has been far from complete (arguably the most serious omission being that of conformal symmetries \u2013 see the Info block below). We trust, however, that the discussion was substantial enough to convince the reader of the enormous power and versatility RG methods have in disclosing \u201cdeep\u201d physical information and to motivate further exploration of the subject. INFO While, for the most part, previous chapters were guided by symmetry principles, little has been said so far about the role of symmetries in connection with renormalization. At the same time it is quite obvious that, at a point of self-similarity, a physical system will have more symmetries than the ubiquitous global scale invariance: it will, at least, also be rotationally and translationally invariant: In fact, we expect these symmetries to be local symmetries, e.g. provided the system is described by a short-range e\ufb00ective theory, a local rescaling of coordinates r \u2192 b(r)r should not cause any e\ufb00", "ect; a point on which you might like to re\ufb02ect. More generally, we expect self-similar systems to be invariant under general conformal mappings r \u2192 r(r), i.e. mappings that leave angles between vectors locally invariant. (In a more mathematical language, a mapping is conformal if it leaves the metric tensor g\u03bc\u03bd (r) invariant, up \u03bc\u03bd (r) = \u03a9(r)g\u03bc\u03bd (r).) In general dimensions, the three classes of to an optional scale factor, i.e. g 8.8 Problems 475 transformations mentioned above, dilatations, rotations, and translations, essentially44 exhaust the list of conformal mappings. It turns out, however, that the information gain due to the presence of these symmetries is rather restricted. However, a radically di\ufb00erent situation arises in d = 2. Introducing complex coordinates, z = z0 + iz1, it can be shown45 that every holomorphic mapping z \u2192 w(z) is automatically conformal. This means that two-dimensional systems at a \ufb01xed point are invariant under a vast group of symmetry operations. These symmetry conditions can be employed \u2013 the subject of conformal \ufb01eld theory \u2013 to obtain very far-reaching information about the behavior of correlation functions. In fact, it is customary in two-dimensional conformal \ufb01eld theory to characterize the behavior of a theory \u201cindirectly,\u201d by the transformation characteristics of its correlation functions, rather than by specifying concrete microscopic actions. At any rate, a discussion of the foundations and applications of conformal \ufb01eld theory, and its rami\ufb01cations in various branches of mathematical physics, is beyond the scope of the present text. For a very accessible \ufb01rst introduction to the subject we refer to Cardy.1 Readers who want to learn more about the subject are referred to di Francesco et al.46 In Section 8.6 we saw the way large-scale topological defects render the physical behavior of a model, as seemingly innocuous as the two-dimensional XY -model, utterly non-trivial. Our observations made in connection with the XY -model exemplify a very general phenomenon: in physical systems admitting the formation of topologically stabilized excitations, a conspiracy of large-scale geometry-related structures and local \ufb02uctuations leads to intriguing physical e\ufb00ects. An introductory discussion of topological structures in quantum \ufb01eld theory, and of their rami\ufb01cations in concrete physical phenomena, is the subject of the next chapter. 8.8 Problems 8.8.1 Dissipative quantum tunneling: strong potential limit In Section 8.2, we considered a quantum particle subject to a periodic potential and a dissipative damping mechanism. A combination of perturbative methods and the renormalization group was applied to show that, for dissipation rates above a certain critical threshold, the external potential e\ufb00ectively grows. Eventually, however, the potential height will reach values at which a perturbative treatment is no longer valid. In this problem we explore the fate of the quantum particle in the high-potential limit. As a result, we obtain the upper portion of the \ufb02ow diagram discussed in Section 8.2. Let us consider the action (8.11) in the limit where the potential strength c is large in comparison with the high-frequency cuto\ufb00 \u039b of the theory. To explicitly implement the condition \u03c9n < \u039b, we add a \u201ckinetic energy term\u201d Skin[\u03b8] = l2m d\u03c4 \u02d9\u03b82 to the action, where 2 l is some \ufb01xed parameter of dimensionality [length]. The mass parameter is chosen such 44 The so-called special conformal transformation, r!\u2192 r/r2, also belongs to the family of conformal mappings. 45 Under a change of coordinates, z \u2192 w, the metric tensor changes according to g\u03bc\u03bd \u2192 \u2202w\u03bc \u2202w\u03bd \u2202z\u03c1 g\u03c3\u03c1. It is \u2202z\u03c3 straightforward to show (exercise) that the condition that the standard metric g\u03bc\u03bd = \u03b4\u03bc\u03bd change only by a global factor is equivalent to the Cauchy\u2013Riemann equations \u2202w1 \u2202z0 = \u2212 \u2202w1 \u2202z1. 46 P. di Francesco, P. Mathieu, and D. S\u00b4en\u00b4echal, Conformal Field Theory (Springer-Verlag, 1999). \u2202z1 and \u2202w0 \u2202z0 = \u2212 \u2202w0 476 The renormalization group that at the maximum value of the frequency, \u03c9n = \u039b, the kinetic", " energy term becomes of the same order as the dissipative term (i.e. for frequencies beyond \u039b, the system is no longer e\ufb00ectively described by Eq. (8.11)): ml2 \u223c 1/\u039bg. We are thus led to consider the imaginary time action S[\u03b8] = 1 4\u03c0T g \u03c9n |\u03c9n||\u03b8n|2 + d\u03c4 \u02d9\u03b82 + cos(\u03b8(\u03c4 )), ml2 2 describing a particle in a periodic potential and subject to a weak damping mechanism. Assuming the damping to be weak, we begin by exploring the solutions to the equations of motion of the undamped (g = \u221e) problem. Varying the action, we obtain the equations \u00af\u03b8 + sin(\u00af\u03b8) = 0, i.e. the Newton equation of the mathematical pendulum. of motion ml2d2 \u03c4 Although these equations can be solved in closed form, there is actually no need to do so. We need note only that extended periods where the classical particle rests at the maxima of the potential are interrupted by occasional events where it rolls through one of the potential minima (see the \ufb01gure overleaf). Throughout, it will be convenient to characterize the pro\ufb01le \u00af\u03b8 (see the bottom part of of a general trajectory \u00af\u03b8 in terms of its time-derivative h = d\u03c4 the \ufb01gure). For a function \u00af\u03b8 containing n+ instanton events and n\u2212 anti-instanton events, i=1 eif (\u03c4 \u2212 \u03c4i), where \u03c4i is the time of the event, ei = +1 (\u22121) for an instanton h(\u03c4 ) = (anti-instanton), n = n\u2212 + n+, andf (\u03c4 ) is a function that is peaked around zero, has a width \u223c c and integrates to 2\u03c0: \u221e \u2212\u221e d\u03c4 f (\u03c4 ) = h(\u221e) \u2212 h(\u2212\u221e) = 2\u03c0. n (a) Show that the action of a general instanton trajectory is given by S[\u00af\u03b8] = 1 4\u03c0T g n \u03c9m i,j=1 eiejei\u03c9m(\u03c4i\u2212\u03c4j ) |fm|2 |\u03c9m| + nSinst, where Sinst. is the action of a single instanton event. (b) Apply a Hubbard\u2013Stratonovich transformation to bring the partition function to the form Z = n {ei=\u00b11} e\u2212nSinst Dq e\u2212 g 4\u03c0T m |qm|2|\u03c9m| 1 n! n' i=1 d\u03c4i ei n i=1 eiq(\u03c4i), where we have neglected the contribution from the \ufb02uctuation determinant. (Hint: Use the fact that, in the frequency range of interest, fm fm=0 = 2\u03c0T.) (c) Sum over all con\ufb01gurations of the \u201ccharges\u201d ei to obtain 6 7 Z = Dq exp |qm|2|\u03c9m| + \u03b3 d\u03c4 cos(q(\u03c4 )), (8.50) \u2212 g 4\u03c0T \u03c9m where \u03b3 = 2e\u2212Sinst. Intriguingly, we have arrived at a functional integral with an action structurally equivalent to our starting point (8.11). The \u201conly\u201d di\ufb00erence is a change in g and c \u2192 \u03b3 = 2e\u2212Sinst. In words: very large values the coupling constants, g \u2192 gdual \u2261 1 of the potential c map onto a very small potential in the \u201cdual problem\u201d; dissipation strengths g > 1 map intog dual < 1 andvice versa. Use this information to convince yourself of the validity of the upper portion of the \ufb02ow diagram discussed in Section 8.2. 8.8 Problems 477 What we have achieved by the construction above is a duality transformation: a \ufb01eld theory subject to a strong potential (more generally, a \ufb01eld theory at strong interactions, or \u201cstrong coupling\u201d) has been mapped onto a weakly perturbed \ufb01eld theory (a theory at \u201cweak coupling\u201d). Duality transformations \u201cboost\u201d theories from one corner of their phase diagram into another and, therefore, establish links between sectors of the problem at hand that cannot be linked by perturbative means. This explains why the identi\ufb01cation of its dual partners represents an important step in the analysis of any \ufb01eld theory. The problem above is special in that it is self-dual, i.e. the theory obtained after the transformation is structurally equivalent to", " the starting theory. Answer: (a) The result is obtained by substitution of \u00af\u03b8m = \u2212 ifm \u03c9m (b) Substituting q(\u03c4 ) = m ei\u03c9m\u03c4 qm into the Hubbard\u2013Stratonovich transformed represen- i eiei\u03c9m\u03c4i into the action. tation and performing the Gaussian integral over qm, we arrive at n' Z = e\u2212nSinst 1 n! i=1 d\u03c4i e\u2212S[\u00af\u03b8], n {ei=\u00b11} i.e. the action of the previous line, integrated over all instanton coordinates and all \u201ccharge con\ufb01gurations\u201d {ei = \u00b11}. (c) Starting from the representation derived in (b) and reorganizing terms, we obtain Z = Dq e\u2212 g 4\u03c0T = Dq e\u2212 g 4\u03c0T |qm|2|\u03c9m| m |qm|2|\u03c9m| m n n 1 n! 1 n! e\u2212Sinst. d\u03c4 eieq(\u03c4 ) n e=\u00b11 n \u03b3 d\u03c4 cos(q(\u03c4 ). Resumming the exponential series, we obtain Eq. (8.50). 8.8.2 Quantum criticality So far, we have focused predominantly on scaling and critical phenomena associated with a thermally induced phase transition. However, as we have seen in previous chapters, the quantum system a\ufb00ords the possibility of a phase transition driven by some external parameter even in the zero-temperature system. For example, a transition from the superconducting to the normal phase of a metal can be driven by an external magnetic \ufb01eld. Similarly, the transition between a paramagnetic and a ferro- 478 The renormalization group or anti-ferromagnetic itinerant electron phase can be driven by changing the strength of the e\ufb00ective interaction (through, e.g., external pressure). In both cases, the application of the external parameter leads to a substantial rearrangement of the ground state \u2013 a quantum phase transition. If this is continuous, as with a thermally induced transition, one can expect the appearance of characteristic critical phenomena analogous to, but potentially di\ufb00erent from, the corresponding classical critical theory. The aim of the present problem is to explore the nature of the critical phenomena when the critical point is driven to zero temperature \u2013 a quantum critical point. Motivated by the coherent state formulation of the quantum partition function of an interacting system, it is tempting to associate directly the quantum statistical mechanics of a d-dimensional system with the classical statistical mechanics of a (d + 1)-dimensional system \u2013 the time coordinate would seem to just add one more dimension (albeit of \ufb01nite extent \u03b2). However, in most applications, time and space enter the action in an anisotropic manner. As a result, one must consider scalings along the time and space directions separately. This provides for a rich variety of new phenomena associated with the quantum critical point. Although a general theoretical phenomenology of the quantum critical system was formulated by Hertz47 soon after the development of the renormalization group method, it is only relatively recently that experimentalists have been able to access and explore in detail the quantum critical regime. Of these experimental developments, perhaps the most systematic has been in the exploration of metallic magnetism in the wide class of heavy fermion compounds. With this motivation in mind, let us then recall brie\ufb02y our discussion of the Stoner transition in Problem 6.7. INFO Without going into detail, we note that heavy fermions are compounds containing rareearth elements such as Ce or Yb, or actinide elements such as U (examples including UBe13, CeCu2Si2, and many more). Their inner shell conduction electrons often have e\ufb00ective masses several hundred times as large as that of bare electrons. As a consequence, the Fermi energy in heavy fermion materials is anomalously low (exercise). At low temperatures, many of these materials are magnetically ordered, others show strong paramagnetic behavior, and some display unconventional mechanisms of superconductivity. Starting with the quantum partition function of the lattice Hubbard system, a \ufb01eld theory of the Stoner transition was derived. Speci\ufb01cally, it was shown that, after an appropriate rescaling of space and time, the quantum partition function of the interacting electron system could be presented as a functional \ufb01eld integral over a scalar magnetization \ufb01eld m(x), weighted by the Euclidean time action48 S[m] = 1 2\u03b2V q \u03b4 + q2 + |\u03c9n| \u0393q |mq|2 + u 4 dx m4(x). In the quantum critical system, the parameter \u03b4 measures the", " distance from the critical point while the Landau damping factor \u0393q = \u0393|q| for the ferromagnetic transition and 47 J. A. Hertz, Quantum critical phenomena, Phys. Rev. B 14 (1976), 1165\u201384. 48 Here, to facilitate the RG, it is helpful to adopt the Fourier convention m(x) = T Ld dx m(x)e\u2212i(q\u00b7r\u2212\u03c9\u03c4 ), wheredx \u2261 ddr \u03b2 0 d\u03c4. q mqei(q\u00b7r\u2212\u03c9\u03c4 ), mq = 8.8 Problems 479 \u0393q = \u0393, a constant, in the antiferromagnetic case. Finally, to regularize the theory, one may suppose that the momentum integration is cut o\ufb00 at the lattice scale \u039b = 1/a. More generally, in the absence of an external magnetic \ufb01eld, one may consider a generalization from the scalar to a vectorial \ufb01eld m. However, keeping our discussion simple, we focus on a system with uniaxial symmetry (such as a metamagnetic system49) where the \ufb01eld is scalar (uniaxial). At the level of Landau mean-\ufb01eld theory, the magnetization \ufb01eld acquires a non-zero expectation value when \u03b4 < 0. In the classical system, our analysis of the \u03c64-theory (to which the present action collapses at high temperatures) has shown that, in dimensions d \u2265 du = 4, the critical system is controlled by the Gaussian \ufb01xed point while in dimensions 1 > d > du the critical properties are controlled by a new \ufb01xed point. Here, in the lowtemperature system, one can expect the critical properties to be in\ufb02uenced by quantum dynamical \ufb02uctuations. Drawing on the seminal work of Hertz and, later, by Millis,50 our aim here is to explore the rami\ufb01cations of scaling in the quantum critical system. (a) As a warm-up exercise, show that, up to a temperature-independent constant, the free energy of the Gaussian theory (i.e. \u03b4 > 0, u = 0) takes the form FGauss = Ld ddq (2\u03c0)d \u0393q 0 d\u03c9 \u03c0 \u03c9\u03b2 2 tan\u22121 \u03c9/\u0393q \u03b4 + q2. coth (Hint: Functional integration over the Gaussian \ufb01eld \ufb02uctuations leads to a formally divergent expression. To extract the \ufb01nite \u03b4-dependent contribution above, \ufb01rst di\ufb00erentiate, \u2202\u03b4F, to later regain the \u03b4-sensitive part of F by integration, d\u03b4 \u2202\u03b4F.) (b) Although, as we have seen above, the free energy can be obtained directly from the functional integral, to guide our analysis of the non-Gaussian theory it is helpful to develop the RG on the Gaussian model \ufb01rst. Using the standard coarse-graining procedure, m(q) = 0 < |q| < \u039b/b, ms(q), mf (q), \u039b/b \u2264 |q| < \u039b, show that the scaling exponent y\u03b4 = 2 and, from the temperature, identify the value of the dynamical exponent z for the ferromagnetic and antiferromagnetic models. Focusing on the Gaussian \ufb01xed point, show that the RG implies the scaling dimension yu = 4\u2212d\u2212z from which one obtains the upper critical dimension du = 4 \u2212 z. (c) To accommodate the e\ufb00ect of a general interaction U, one may make use of the pertur- bative RG expansion Z = Dms e\u2212S[ms]e\u2212U [ms,mf ]f, 49 As derived in Problem 6.7, the Stoner transition takes place from a paramagnetic to a ferromagnetic phase. However, experimentally, one often \ufb01nds that, while the paramagnetic phase is stable at zero external magnetic \ufb01eld, the magnetization can change abruptly at some non-zero \ufb01eld \u2013 a metamagnetic transition. The development of a Landau expansion of the action in the vicinity of the metamagnetic critical point leads to a theory with uniaxial symmetry. 50 A. J. Millis, E\ufb00ect of a non-zero temperature on quantum critical points in itinerant fermion systems, Phys. Rev. B 48 (1993), 7183\u201396. 480 The renormalization group Dmf e\u2212S(2)[mf ](\u00b7 \u00b7 \u00b7", " ) denotes averaging where U [ms, mf ] = 3u 2 over the quadratic action of the fast \ufb01eld. Keeping terms to \ufb01rst order in the quartic interaction,51 obtain the RG equations f and \u00b7 \u00b7 \u00b7 f = dx m2 s m2 dT (b) d ln b d\u03b4(b) d ln b du(b) d ln b = zT (b), = 2\u03b4(b) + 12u(b)f (2)[T (b)], = (4 \u2212 d \u2212 z)u(b), where f (2)[T (b)] = \u039bd\u03a9d\u0393\u039b 1 0 ds \u03c0 coth s\u0393\u039b\u03b2 2 s (\u03b4 + \u039b2)2 + s2 + z coth sz\u22122\u0393\u039b\u03b2 2 sd\u22123+z (\u03b4 + s2\u039b)2 + 1 denotes a dimensionless integral function of the temperature and \u03a9d is the surface of the d-dimensional unit ball divided by (2\u03c0)d. (Hint: Notice that, for a given value of the cuto\ufb00 \u039b, the integration domain in frequency\u2013momentum space is set by |\u03c9| < \u0393q, |q| \u2264 \u039b. This means that the \u201cshell\u201d of fast \ufb02uctuations obtained by a gradual lowering of the cuto\ufb00 is of the form shown in the \ufb01gure. Introduce the integration boundaries imposed on the fast-\ufb01eld integration after you have analytically continued to real frequencies.) \u03c9 \u0393\u039b \u0393\u039b / b /bz Already, from these di\ufb00erential recursion relations, one can recover much of the phenomenology of the quantum critical system. Integrating the last of the recursion relations, one \ufb01nds that for d + z \u2265 4, the interaction u(b) =ub 4\u2212(d+z) initially decreases under renormalization. Therefore, over at least some range of parameters, one may expect the theory to be controlled by the Gaussian \ufb01xed point T \u2217 = r\u2217 = 0. However, it is evident from the \ufb01rst of the scaling relations that the temperature increases under renormalization scaling as T (b) =T b z. Crucially, referring to the expression for f (2)[T (b)] (and, indeed, its higher-order descendants) \u2013 setting \u039b = \u0393\u039b = 1 for simplicity \u2013 one may see that the character of the renormalization changes when T (b) 1 (i.e. when b b0 = T \u22121/z). Here, the e\ufb00ect of the con\ufb01nement in the Euclidean time direction begins to impact. Whether this scaling regime is reached depends on the physical bare parameters. To stay within the scaling regime, we require \u03b4(b) < 1. Therefore, de\ufb01ning b1: \u03b4(b1) = 1, one may identify two regimes of behavior. Firstly, in a low-temperature regime where 1 b1 b0, the system is characterized by quantum critical behavior controlled by the T = 0 \u039b /b\u039b q 51 Comparing with the RG analysis of the standard \u03c64-model in Section 8.4.4, consider which \ufb02uctuation e\ufb00ects are kept at this level of approximation and which are not. 8.8 Problems 481 \ufb01xed point. Here, approximating f (2)[T (b)] f (2)[0], an integration of the scaling equation gives \u03b4(b) = b2 \u03b4 + 12uf (2)(0) z + d \u2212 2 \u2261 b2r, the renormalization of \u03b4 recording the e\ufb00ect of quantum dynamical and spatial \ufb02uctuations on the critical point. Referred through b1, this implies a \u201cquantum critical regime\u201d bounded by the inequality T rz/2. However, if this inequality is reversed (i.e. b1 > b0), the scaling behavior must be separated into regimes where b < b0 (T (b) < 1) and b > b0 (T (b) > 1). Noting that, for T (b) 1, f (2)[T (b)] \u221d T (b), the recursion relations may be simpli\ufb01ed (and, indeed, their nature elucidated) by de\ufb01ning v(b) = u(b)T (b), whence they take a more familiar form, d\u03b4(b) d ln b = 2\ufffd", "\ufffd(b) +Cv (b), dv(b) d ln b = (4 \u2212 d)v(b), with C relating the constant of proportionality. These are nothing but the di\ufb00erential recursion relations associated with the leading order of perturbation theory of the d-dimensional classical \u03c64-theory with the familiar upper critical dimension du = 4. Integrating these equations from b0 to b > b0, we obtain 2 4\u2212d 4\u2212d \u03b4(b) = \u03b4(b0) \u2212 Cv(b0) 2 \u2212 d b b0 + Cv(b0) 2 \u2212 d b b0, v(b) = v(b0). b b0 To determine the initial values (\u03b4, v)(b0), we need to integrate the low-temperature (b < b0, or T (b) < 1) scaling relations up to b0. Setting f (2)(T ezx) = f (2)(0) + [f (2)(T ezx) \u2212 f (2)(0)], the constant f (2)(0) recovers the result above while the term in rectangular brackets e\ufb00ects a temperature-dependent correction such that where B = 12 z \u03b4(b0) = T \u22122/z[r + BuT (d+z\u22122)/z], 0 dT T [2\u2212(d+2z)]/z[f (2)(T ) \u2212 f (2)(0)]. 1 v(b0) = uT (d+z\u22124)/z, We now have to distinguish between two very di\ufb00erent scenarios. (i) If the scaling is halted (\u03b4(b) = 1) while v(b) remains small, the perturbative expansion remains valid. However, (ii) if v(b) grows to order unity while \u03b4(b) remains small, the scaling behavior moves into a region potentially controlled by non-Gaussian \ufb02uctuations. From the equations above, the condition for Gaussian behavior, i.e. \u03b4(b) = 1 and v(b) 1, translates to the Ginzburg criterion uT (d+z\u22123)/z [r + (B + C)uT (d+z\u22122)/z]1/2 1. From the same result, one can infer a transition temperature Tc(r < 0, u) = \u2212 r (B + C)u z/(d+z\u22122). Notice that at the transition temperature we are always in regime (ii), i.e. a regime governed by strong non-Gaussian \ufb02uctuations. Gathered together, the results of this analysis are summarized in Fig. 8.13. 482 The renormalization group Tc = (\u03b4 \u2013 \u03b4 c)z / (d + z \u2013 2) T classical regime Gaussian ordered regime (\u03b4 \u2013 \u03b4 c)z / 2 disordered quantum regime \u03b4 \u03b4 c Figure 8.13 Phase diagram associated with a quantum critical point. The solid line denotes the locus of the transition temperature while the hatched area denotes the Ginzburg region where the scaling behavior is controlled by critical \ufb02uctuations. INFO Although this completes our formal discussion of the renormalization properties of the quantum critical theory, it is instructive to question how these results can be translated to experiment. From the scaling behavior, one can infer directly a number of physical manifestations. Of these, the most straightforward relates to the temperature dependence of the magnetic susceptibility. Focusing on the classical Gaussian regime, in the disordered phase, we have seen that the mass scales as \u03b4(b0) =T \u22122/z[r + BuT (d+z\u22122)/z]. The latter translates to a zero-\ufb01eld susceptibility \u03c7(T ) =\u03c7 0(r + BuT (d+z\u22122)/z). Once the bare temperature crosses into the disordered quantum regime, one expects the susceptibility to become temperature-independent. Now, in the clean system, magnetic \ufb02uctuations present the dominant mechanism of lowtemperature scattering. In this case, one can estimate their e\ufb00ect on the low-temperature resistivity. Referring to the literature, an estimate of the scattering rate based on the Born approximation gives 1 2\u03c4 = \u2212Im \u03a3R(kF, 0) g2 T vF 3 1 r + BuT (d+z\u22122)/z, where g denotes the coupling between the electrons and spin \ufb02uctuations. When the spin \ufb02uctuations dominate, one \ufb01nds a resistivity \u03c1(T ) \u223c T 3\u2212(d+z\u22122)/z while, in", " the opposite regime, the contribution from electron\u2013electron scattering dominates and one recovers the Fermi-liquid result \u03c1(T ) \u223c T 2. Answer: 8.8 Problems 483 (a) For the Gaussian theory, integration over the \ufb01elds m(q) and subsequent di\ufb00erentiation with respect to \u03b4 gives the free energy density \u2202\u03b4FGauss = Ld = \u2212Ld ddq (2\u03c0)d T \u03c9n 1 \u03b4 + q2 + |\u03c9n| \u0393q = Ld ddq (2\u03c0)d P \u0393q \u2212\u0393q d\u03c9 2\u03c0i tanh(\u03b2/2) \u03b4 + q2 \u2212 i \u03c9 \u0393q ddq (2\u03c0)d P \u0393q 0 d\u03c9 \u03c0 tanh(\u03b2/2) \u03c9/\u0393q (\u03b4 + q2)2 + (\u03c9/\u0393q)2. To arrive at the \ufb01rst integral, we have converted the sum over positive frequencies \u03c9n into a contour integral running parallel to the real axis, surrounding the pole at \u03c9m = 0 along an in\ufb01nitesimal semicircle in the upper complex half plane, and closing it at in\ufb01nity through the upper complex half plane as usual. Noting that the in\ufb01nitesimal integral around zero cancels against the contribution \u03c9m = 0, the integral reduces to the principal value integral given above. (The integration is restricted to |\u03c9| < \u0393q because outside this interval the low-frequency approximation of the polarization operator no longer applies.) Using the relation that \u2202\u03b4 tan\u22121 (\u03b4+q2)2+(\u03c9/\u0393q)2, we arrive at the required result. \u03c9/\u0393q \u03b4+q2 = \u2212 \u03c9/\u0393q (b) Since, at the level of the Gaussian theory, fast and slow \ufb02uctuations decouple, the e\ufb00ective action of the slow \ufb01elds is given by S[ms] = 1 2\u03b2 \u039b/b ddq (2\u03c0)d \u03c9n \u03b4 + q2 + |\u03c9n| \u0393q |ms(q)|2, i.e. the original action with lowered cuto\ufb00s. (The presence of a cuto\ufb00 in the frequency summation is implicit in the notation.) To restore the cuto\ufb00s to their original values, one must implement the rescaling q = qb, \u03c9 n = bz\u03c9n, where z denotes the dynamical exponent. Notice that the frequency scaling is tantamount to a rede\ufb01nition of temperature, T = bzT. The \ufb01nal step of the RG involves the renormalization of the magnetization \ufb01eld, m(q) = ms(q)/y. As a result, the action assumes the form S[m] = \u2212 1 2 \u039b ddq (2\u03c0)d T \u03c9 n \u03b4 + q2b\u22122 + b\u2212(d+z)y2|m(q)|2. |\u03c9 |bz n \u0393q/b Requiring that the gradient term remain invariant, one \ufb01nds that y = b(d+z+2)/2. Further, requiring that the damping term maintain the same form ensures that (1/\u0393q/b/bz)b\u2212(d+z)y2 = 1/\u0393q, \ufb01xing the dynamical exponent. In particular, for the ferromagnetic system (\u0393q = \u0393|q|), z = 3 while, for the antiferromagnetic system (\u0393q = \u0393), z = 2. Finally, applied to the coe\ufb03cient \u03b4, invariance of the action requires \u03b4 = y2b\u2212(d+z)\u03b4 = by\u03b4 \u03b4 with y\u03b4 = 2. As expected, the Gaussian \ufb01xed point lies at T \u2217 = \u03b4\u2217 = 0 and has the \ufb01xed point action S[m] = 1 2 \u039b ddq (2\u03c0)d 1 2\u03c0 \u0393\u039b d\u03c9 q2 + |m(q)|2, |\u03c9n| \u0393q 484 The renormalization group where \u0393\u039b sets the cuto\ufb00 in frequency direction. To identify the upper critical dimension, one must consider the e\ufb00ect of the quartic dx (\u2207m)2, interactions. Firstly, applied to the gradient component of the \ufb01xed action the relation m(x)", " = ms(x)/\u03b7 allows us to deduce the \ufb01eld renormalization \u03b7 = b(2\u2212d\u2212z)/2. Applied to the quartic interaction, the renormalization procedure leads to the relation u 4 dx m4(x) \u2192 u 4 bd+z\u03b74 dx m4(x), from which one obtains u = b4\u2212d\u2212zu. Therefore, in dimensions d + z < 4 the interaction coe\ufb03cient diminishes under renormalization and the properties of the system are governed by the mean-\ufb01eld exponents. Since the dynamical exponent z = 2 or 3, one \ufb01nds that the majority of physical systems fall above the upper critical dimension and are controlled by the Gaussian \ufb01xed point. (c) The \ufb01rst term in the u-expansion, (3u/2), gives rise to a renormalization of the coe\ufb03cient \u03b4. Evaluating this contribution, and expressing the frequency summation by a contour integral as in (a), we obtain dx m2 s m2 f m2 f = 1 \u03c0(2\u03c0)d \u2202\u039b dd q d\u03c9 coth \u03c9\u03b2 2 \u03c9/\u0393q (\u03b4 + q2)2 + (\u03c9/\u0393q)2, where the integral runs over the momentum shell \u2202\u039b (see the \ufb01gure on page 480). Taking each component of the integral in turn, one obtains m2 ddq (2\u03c0)d \u039bd\u22121\u03a9d \u0393\u039b d\u03c9 \u03c0 0 coth \u03c9\u03b2 2 \u03c9/\u0393\u039b (\u03b4 + \u039b2)2 + (\u03c9/\u0393\u039b)2 \u0393q \u2212 \u0393q bz coth \u0393q\u03b2 2 1 (\u03b4 + q2)2 + 1. Noting that \u039b(1 \u2212 e\u2212 ln b) \u039b lnb + \u00b7 \u00b7 \u00b7 and \u039b(1 \u2212 e\u2212z ln b) \u039bz ln b + \u00b7 \u00b7 \u00b7, the latter = f (2)[T (b)] ln b, where, applying the rescaling s = \u03c9/\u0393\u039b or can be expressed as m2 f s = |q|/\u039b as appropriate, one obtains the required expression for f (2)[T (b)]. Therefore, to \ufb01rst order in u, one obtains the scaling relation \u03b4 = \u03b4(1+2 ln b)+12uf (2)[T (b)] ln b. At \ufb01rst order in perturbation theory, the coupling constant u is not modi\ufb01ed by \ufb02uctuationinduced corrections, i.e. u changes according to its engineering dimension, u = u(1 + (4 \u2212 d \u2212 z) ln b). Similarly, T = bzT. Di\ufb00erentiating these expressions, we arrive at the required di\ufb00erential recursion relations. 8.8.3 RG analysis of the nonlinear \u03c3-model II Previously, we have argued that, below the lower critical dimension, Goldstone mode \ufb02uctuations become unbounded and induce a crossover of systems with spontaneously broken continuous symmetries into a \u201cdisordered\u201d state. Technically, the long-distance \ufb02uctuation behavior of Goldstone modes is described by nonlinear \u03c3-models with coset-space valued \ufb01elds. In this problem, we apply RG methods to show that, at and below the dimension two, these models indeed \ufb02ow towards a disordered phase. 8.8 Problems 485 Consider a situation where the symmetry of a model is broken from a \u201clarge\u201d compact group G down to some subgroup H. On physical grounds, we expect that (i) the action of the associated Goldstone modes g \u2208 G/H is space(-time) rotationally invariant and starts at O (two derivative operators). By symmetry (Exercise: Think why!) (ii) the action must be invariant under global transformations by elements of the full symmetry group, S[g] = S[\u02dcgg], where \u02dcg \u2208 G is constant. However, (iii) it must also be invariant under local transformations S[g] =S [gh], where h(x) \u2208 H is a \ufb01eld taking values in the \u201csmall\u201d group H. (This last condition simply states that g takes values in the (right) coset space G/H, i.e. is oblivious to (right) transformations g \u2192 gh.) How would one construct such an action in practice? An", " obvious recipe goes as follows. Take any reference ground state |n0. (Think of |n0 as an element of some vector space, e.g. a \u201cspin up\u201d state in a magnetic system.) De\ufb01ning the symmetry transformed states |n(x) = g(x)|n0, the unique G-invariant action is then given by S[n] = c ddx \u2207n|\u2207n. This action obviously meets all three criteria above. However, by making explicit reference to an arbitrarily chosen state (|n0) it partially obscures the symmetries of the model. Thus, before turning to the RG formulation, let us brie\ufb02y discuss an alternative, and more symmetry-oriented, formulation. Let \u039b \u2208 G be a group element that commutes with all h \u2208 H, i.e. h\u039bh\u22121 = \u039b.52 The set {Q = g\u039bg\u22121|g \u2208 G} is isomorphic to the coset space G/H (which is to say that two elements g and g = gh di\ufb00ering by an element of the small group are represented by the same Q). In the Q-language, the unique action meeting the criteria (i)\u2013(iii) above (exercise: check!) is given by S[Q] = 1 \u03bb ddr tr(\u2207Q \u2207Q). Below, we will explore the long-range behavior of the \ufb01eld integral DQ is the G invariant measure (DQ = D(gQg\u22121)) on the coset space. DQ exp(\u2212S[Q]), where INFO They being but di\ufb00erent representations of the same coset space G/H it is, of course, always possible to explicitly relate the n- to theQ -representation. Consider, for example, the case G = SU(2), and H = U(1). Representing the group G through Euler angles (see Section 3.3), g(\u03c6, \u03b8, \u03c8) = exp(i\u03c6 \u02c6S3) exp(i\u03b8 \u02c6S1) exp(i\u03c8 \u02c6S3), de\ufb01ne the (spin-coherent) states |n(\u03c6, \u03b8) = g(\u03c6, \u03b8)| \u2191, where| \u2191 is the state of maximal eigenvalue S3 = S, and we noted that the \u03c8 degree of freedom (spanning the small group) leaves |S3 invariant (up to a phase) and can be omitted. In Section 3.3 we used this representation to identify the phase space of spin as the coset space SU(2)/U(1), the 2 sphere. In the Q language, the same sphere is represented as Q = g\u03c33g\u22121, where \u02c6Si = \u03c3i/2 are identi\ufb01ed with the Pauli matrices. Again, the \u03c8 factor, commutative with \u03c33, drops out. The integration measure dQ = d\u03c6d\u03b8 sin \u03b8 reduces to the invariant integration measure on the sphere. 52 It being a subgroup of the (matrix-)group G, we can always \ufb01nd a representation wherein the group H is represented by block-diagonal matrices. In this representation, \u039b will be a diagonal matrix that contains identical elements on all sub-blocks de\ufb01ning H. 486 The renormalization group (a) By way of example, let us consider the case (G, H) = (SU(2), U(1)), relevant to, e.g., the classical Heisenberg magnet.53 De\ufb01ning g = gsgf where the fast \ufb01eld gf = exp(iW ), W = zE12 + \u00afzE21, z \u2208 C (explore the connection between this representation and the Euler angle representation above), and \u03a6\u03bc \u2261 g\u22121 s \u2202\u03bcgs, show that the freedom to transform gf by elements of the small group can be used to bring \u03a6s to an o\ufb00-diagonal form, [\u03a6s, \u03c33]+ = 0. Show that the expansion of the action to second order (\u2192 one-loop order) in the fast generators W is given by S[gs, W ] =S [gs] +S f [z, \u00afz] +S fs[gs, z, \u00afz], where Sf [z, \u00afz] = 8 \u03bb ddr \u2207z\u2207\u00afz and Sfs[gs, z, \u00afz] = \u2212 8 \u03bb ddr tr(\u03a6\u03bc(zE12 \u2212", " \u00afzE21)\u03a6\u03bc(zE12 \u2212 \u00afzE21) + \u03a6\u03bc\u03c33\u03a6\u03bc\u03c33z \u00afz). (b) Integrate over the \ufb01elds z, \u00afz to obtain the e\ufb00ective action S[gs] = 1 \u03bbr ddr tr(\u2207Qs\u2207Qs), where \u03bbr is the renormalized coupling constant. Di\ufb00erentiate this constant with respect to the scaling parameter b to obtain the RG equation of the \u03c3-model on the 2sphere, d ln \u03bb d ln b = \u03bb 8\u03c0 \u2212. (8.51) According to this equation, the coset \u03c3-model behaves very much like its group-valued cousin explored in Section 8.5: at dimensions d = 2 and below it \ufb02ows towards a disordered phase; at dimensions d >2 there is a phase transition separating a disordered regime from an ordered phase. Answer: ddr tr(\u2207[\u03a6]Qf \u2207[\u03a6]Qf ), where Qf = gf \u03c33g\u22121 (a) Substituting g = gsgf into the action and de\ufb01ning the shorthand notation \u03a6 \u2261 g\u22121 \u2207gs, we obtain S[gs, gf ] = 1 and the \u201ccovari\u03bb ant derivative\u201d \u2207[\u03a6]Q = \u2207Q + [\u03a6, Q]. We next transform the integration variables W \u2192 hW h\u22121, where h \u2208 H. (Notice that this really is a transformation of integration variables, i.e. the o\ufb00-diagonal form of the generators is preserved.) This transformation induces a change \u2207[\u03a6] \u2192 \u2207[\u03a6+h\u22121\u2207h] and we may choose h so as to remove the matrixdiagonal components of \u03a6. (Exercise: Explicitly construct a matrix h that does the job. Why can the matrix h be de\ufb01ned only locally and why does the absence of a globally de\ufb01ned h not matter in connection with a perturbative RG operation?) s f 53 Below the transition temperature, the symmetry of the Heisenberg magnet is broken from full rotational symlocally SU(2), and O(2) U(1), the metry, O(3), to rotations around the magnetization axis, O(2). Since O(3) example above covers the physics of the magnetic system. (We prefer to use the language of the unitary group, as it is somewhat easier to implement, technically.) A straightforward expansion to second order in W, Qf = \u03c33(1 + 2iW \u2212 2W 2 + \u00b7 \u00b7 \u00b7), 8.8 Problems 487 now leads to the expressions 4 Sf [W ] = \u03bb S[gs, W ] = \u2212 8 \u03bb ddr tr(\u2202\u03bcW \u2202\u03bcW ), ddr tr(\u03a6\u03bcW \u03c33\u03a6\u03bcW \u03c33 + \u03a6\u03bc\u03c33\u03a6\u03bc\u03c33W 2). (8.52) Substituting the z-representation of the generators W, we arrive at the action given above. (b) Every \ufb01eld \u03a6\u03bc contains a derivative acting on the slow \ufb01eld. Since we do not want to keep terms with more than two derivatives, the e\ufb00ective action is given byS e\ufb00 [gs] = S[gs] + S(2)[gs, z, \u00afz]. Switching to a momentum representation, Sf [z, \u00afz] = 8Ld p\u2208f zp \u00afzpp2, \u03bb and doing the Gaussian integrals, we obtain S(2)[gs, z, \u00afz] = \u2212 I1 4 ddr tr([\u03a6\u03bc, \u03c33]2) = \u2212 I1 4 ddr tr(\u2207Qs\u2207Qs), where the constant I1 has been de\ufb01ned in Eq. (8.28). Substituting this result back into the e\ufb00ective action, we obtain Se\ufb00 = 1 \u03bb \u2212 I1 4 ddr tr(\u2207Qs\u2207Qs) = ddr tr(\u2207Qs\u2207Qs), 1 \u03bbr where we have rescaled momenta and \u03bb\u22121 ing d = 2 + and expanding in, we obtain \u03bbr = \u03bb with respect to ln b, we obtain the RG equation (8.51). r = [\u03bb\u22121\u2212\u03a9d(1\u2212b2\u2212d)/4(d\u22122)]bd\u22122. Substitut. Di\ufffd", "\ufffd\ufffderentiating 1 \u2212 ln b + \u03bb 8\u03c0 ln b 8.8.4 Scaling theory of the Anderson metal insulator transition In Section 6.5 we derived the \ufb01eld theory of the disordered electron gas. Speci\ufb01cally, we showed that the low-energy physics of the system is governed by a phase of a spontaneously broken continuous symmetry. This being so we should expect d = 2 to be the \u201clower critical dimension\u201d of the disordered electron gas. At and below d = 2 Goldstone mode \ufb02uctuations are likely to drive a nominally conducting system into a phase governed by unbounded Goldstone mode \ufb02uctuations \u2013 an insulator in fact. In higher dimensions the e\ufb00ect of \ufb02uctuations is comparatively weaker and metallic phases have a chance to prevail. In this problem we apply RG methods to obtain a semi-quantitative understanding of the Anderson transition between these two phases. INFO Before turning to the technology, let us discuss the notion of Anderson localization in metals on a more qualitative level. Firstly, it is important to draw a distinction between qualitatively di\ufb00erent mechanisms of localization. Consider a d-dimensional electron system in the presence of static disorder. Evidently, the lowest energy states in a disordered system will be dictated by \u201coptimal \ufb02uctuations\u201d of the random impurity potential. These \u201ctail states\u201d can be interpreted as bound states localized in regions where the potential is unusually low and uniform. Instead, here we are interested in the localization properties of states with energies greatly in excess of the typical magnitude of the impurity potential, i.e. EF\u03c4 1. Here, the 488 The renormalization group mechanism of localization can derive only from the accumulation of interference corrections akin to the corrections discussed in Section 6.5 above. According to its energy E, Anderson proposed54 that a given state is either extended or localized depending on the degree of disorder. Following an argument due to Mott, it is also plausible that extended and localized states at the same energy do not coexist \u2013 in such circumstances, the admixture of the former due to arbitrarily small perturbations would lead to the delocalization of the latter. Moreover, it is natural that states at the band edge are more susceptible to the formation of bound or localized states. These considerations suggest a pro\ufb01le for the DoS in which the low-lying localized states are separated from higher-energy extended states by a mobility edge. The transition signaled by the mobility edge is known as the Anderson transition. Following the pioneering work of Anderson, it was almost two decades before a consistent theory of localization emerged. Its basis was a simple yet powerful scaling hypothesis that followed naturally from a conjecture made by Edwards and Thouless:55 the conductance g(L) of addimensional fragment of metal of size Ld (L \u03bbF) is determined by the sensitivity of the single-particle energy levels inside the sample to changes in the boundary conditions. Building on this assumption, Abrahams et al. proposed56 a one-parameter scaling hypothesis of localization. This hypothesis states that the conductance of a system of size bL is a function of g(L) only:g (bL) =f b(g(L)). The important point is that fb does not explicitly depend on the system size L. Independent of Thouless\u2019 argument, this restriction can be made plausible by dimensional analysis: since g is dimensionless, the only way for the parameter L to appear in f is through a dimensionless ratio L/a, where a is some compensating scale of dimensionality [length]. This, however, would imply explicit dependence of the conductance on a microscopic reference scale, which con\ufb02icts with the presumption of universal scaling behavior. Di\ufb00erentiation of the scaling form with respect to b gives the Gell\u2013Mann\u2013Low equation d ln g d ln L = \u03b2(g). The scaling function \u03b2(g) \u2261 f (g)/g is universal, i.e. independent of the microscopic properties of the sample, such as the bare microscopic conductance and /\u03bbF. While these predictions still have to \ufb01nd a truly rigorous mathematical proof, they de\ufb01nitely represent an important milestone in phenomenology. Moreover, the scaling picture \ufb01ts neatly into perturbative microscopic approaches to localization, such as the one discussed below. However, before turning to quantitative calculations, let us try to determine the pro\ufb01le of the \u03b2-function on qualitative grounds. For an ohmic conductor, g(L) \u223c Ld\u22122, implying that the scaling function takes a constant value \u03b2 = d \u2212 2. On the other hand, deep in the insulating regime, where states are exponentially localized, \u03b2(g) \u223c ln g (i.e. g", " \u223c e\u2212L/\u03be). A smooth interpolation between these limits (see Fig. 8.14) suggests that, below two dimensions, all states are localized, while, above, there is a critical conductance, g\u2217, marking a phase transition between a localized and a delocalized phase \u2013 the Anderson transition. At the time when this scaling behavior was proposed, it was not yet known that the low-energy theory of the disordered electron gas was a nonlinear \u03c3-model. However, with the bene\ufb01t of hindsight, one may interpret the phenomenology outlined above as a rami\ufb01cation of the notorious 54 P. W. Anderson, The absence of di\ufb00usion in certain random lattices, Phys. Rev. 109 (1958), 1492\u2013505. 55 J. T. Edwards and D. J. Thouless, Numerical studies of localization in disordered systems, J. Phys. C 5 (1972), 807\u201320. 56 E. Abrahams, P. W. Anderson, D. C. Licciardello, and T. V. Ramakrishnan, Scaling theory of localization: absence of quantum di\ufb00usion in two dimensions, Phys. Rev. Lett. 42 (1979), 673\u20136. 8.8 Problems \u03b2 489 ln (gc ln (g) Figure 8.14 Scaling function of the dimensionless conductance, \u03b2(g), in dimensions d = 1, 2, and 3. At large values of the conductance, the scaling function approaches the asymptote \u03b2(g) \u2192 d \u2212 2 corresponding to ohmic behavior. For very small values of conductance, the scaling function approaches \u03b2(g) \u2192 ln g, which is characteristic of insulating behavior. According to the weak localization expansion, a localization transition is predicted in dimensions greater than two. scaling behavior displayed by these models. Below, we show that the one-loop \u03b2-function of the replica \u03c3-model fully complies with the predictions of the one-parameter scaling theory. (a) Recapitulate the construction of the replica nonlinear \u03c3-model in Section 6.5. Convince yourself that, in order to explore the physical properties of a non-interacting disordered system, we actually do not need the full information contained in that model; the model introduced in Section 6.5 has been constructed to compute correlation functions \u02c6Gn \u02c6Gn for arbitrary Matsubara frequencies \u03c9n and \u03c9n. However, su\ufb03ciently farreaching information on the conduction behavior of a non-interacting system (cf. our discussion of linear response in Section 7.4) is contained in correlation functions \u223c \u02c6G+ \u02c6G\u2212 involving the product of a retarded and an advanced Green function \u02c6G\u00b1 = (EF \u00b1 (\u03c9/2 + i0) \u2212 \u02c6H)\u22121 at \ufb01xed energies EF \u00b1 \u03c9/2. Retracing the derivation of the \u03c3-model, convince yourself that correlation functions of this type can be obtained from the action Eq. (6.63), where the Q-matrices, Q = g\u039bg\u22121, are of dimension 4R (instead of 2RM as in the Matsubara case) and \u039b = {\u03c33\u03b4\u03b1\u03b1 } assumes the role of the matrix \u039b = {sgnnn \u03b4\u03b1\u03b1 }, i.e. the e\ufb00ectively in\ufb01nite-dimensional space of Matsubara frequencies has been reduced to a two-component space discriminating between retarded and advanced Green functions. (Here, \u03b1 = (a, \u03c3) are 2R-component indices comprising the replica index a and the two-component \u201ctime-reversal\u201d index \u03c3 introduced in Section 6.5.) (b) Focusing on the case of zero external frequency and introducing the abbreviation \u03bb\u22121 = \u03c0D\u03bd 8, we wish to explore the scaling behavior of the action S[Q] = 1 \u03bb ddr tr(\u2207Q)2. To do so, we follow the same logic as in the previous problem: we decompose Q = gsQf g\u22121 into fast and slow components and expand the former Qf = \u03c33(1+i2W \u22122W 2 + s 490 The renormalization group \u00b7 \u00b7 \u00b7). The matrices W anti-commute with \u03c33, and live in the Lie algebra sp(4R), i.e. they obey the condition W = \u2212\u03c3tr 2 where the Pauli matrices act in time-reversal space. These two conditions are encapsulated in the representation (block structure in advanced/retarded space) 2 W T \ufffd", "\ufffdtr W = B\u2020 B, B = \u2212\u03c3tr 2 BT \u03c3tr 2 \u2261 \u2212B\u03c4, (8.53) where the matrices B are of dimension 2R and we have used the abbreviated notation B\u03c4 \u2261 \u03c3tr 2. Substituting this representation into the action, we obtain S[gs, W ] = S[gs] +S f [W ] +S fs[gs, W ], where Sf and Sfs are given by Eq. (8.52). 2 BT \u03c3tr Perform the Gaussian integrals over the matrices W. First show that Bp\u03b1\u03b2B p\u03b2\u03b1 =. Use this result, and the symmetries of the B matrices, to verify the aux- \u2020 \u03bb\u03b4\u03b1\u03b1 \u03b4\u03b2\u03b2 \u03b4q,q 16Ldp2 iliary identities tr(BpAB \u2020 p A) = tr(BpAB\u2212p A) = \u03bb 16Ldp2 tr(A) tr(A), 16Ldp2 tr(AA\u03c4 ). \u03bb (8.54) (8.55) ddr tr(\u2207Q)2, where I1 has been Use these relations to show that S[gs, W ] = \u2212 I1 4 \u03bc = \u2212\u03a6\u03bc (exercise). Identify and neglect de\ufb01ned in Eq. (8.28). (Hint: Use the fact that \u03a6\u03c4 terms that vanish in the replica limit.) Adding this contribution to the core action S[gs] we obtain the same renormalized coupling constant as in the previous problem. Thus, the RG equation of the \u03c3-model of disordered systems (possessing spin rotation and time reversal invariance, to be precise) is again given by Eq. (8.51). In 2 + dimensions, the constant \u03bb \u221d (D\u03bd)\u22121 \u221d g\u22121 is proportional to the physical conductance and Eq. (8.51) describes the scaling of the latter. In accord with the phenomenological expectations summarized above, the scaling equation depends only on the conductance itself (one-parameter scaling) and predicts an (Anderson) transition point for d >2. For further discussion of the scaling behavior of the replica \u03c3-model and its rami\ufb01cations in the physics of disordered systems we refer to, e.g., Efetov.57 Answer: (a) Consider the Grassmann action \u00af\u03c8(EF + (\u03c9 + i0)\u03c33/2 \u2212 \u02c6H)\u03c8, where \u03c8 = {\u03c8\u00b1a} is a 2R-component Grassmann \ufb01eld and \u03c33 acts in the space of \u00b1-indices. By introducing suitably devised sources, this action can be employed to compute correlation functions involving an advanced and a retarded Green function. Recapitulating the derivation of the nonlinear \u03c3-model, we notice that nothing changes except that the in\ufb01nite vector of Matsubara frequencies \u02c6\u03c9 gets replaced by the two-dimensional matrix (\u03c9 + i0)\u03c33/2. Since the structure of the diagonal saddle-point, \u039b, is solely determined by the sign of the 57 K. B. Efetov, Supersymmetry in Disorder and Chaos (Cambridge University Press, 1996). 8.8 Problems 491 imaginary part of the frequency vector, \u039b = {sgnnn \u03b4aa } gets replaced by \u039b = {\u03c33\u03b4aa }. The Q matrices reduce to Q = g\u039bg\u22121, whereg \u2208 Sp(4R). \u03b1\u03b1 B\u03b1\u03b1 B q\u03b2\u03b1 = \u03bb \u2020 (b) Switching to an index representation, S[B, B\u2020] = 8Ld \u03b1\u03b1. From this rep\u03bb \u2020 resentation, one might naively conclude that Bq\u03b1\u03b2B 8Ldq2 \u03b4\u03b1\u03b1 \u03b4\u03b2\u03b2 \u03b4q,q with a right-hand side twice as large as the correct result. However, thanks to the relation 2, the matrix elements of B and B\u2020 are not independent complex variB = \u2212\u03c3tr ables. Using the symmetry relation (exercise), it is straightforward to show that all complex integration variables appear in pairs, i.e. the coupling constant e\ufb00ectively doubles. On the same basis, some matrix elements (Ba\u03c3,a \u00af\u03c3, where \u00af\u03c3 = (\u03c3 + 1 mod2) turn out to be real rather than complex. In either case, the Gaussian integration obtains a result twice as small as for independent complex variables. This explains the factor 1/16 in the given formula. 2 BT \u03c3tr The auxiliary identity (8.54) readily follows by representing the trace as a sum", " over indices and using the Gaussian formula for the matrix indices above. To prove Eq. (8.55), we \ufb01rst need to relate the BB contraction to a BB\u2020 contraction: tr(BpAB\u2212p A) = tr((\u03c3tr 2 ABpA\u03c3tr 2 )(\u03c3tr 2 A)\u03b1\u03b2Bp\u03b2\u03b3(A\u03c3tr 2 B\u2212p \u03c3tr 2 )\u03b3\u03b4(\u03c3tr 2 )) 2 B\u2212p \u03c3tr = (\u03c3tr = \u2212(\u03c3tr 2 A)\u03b1\u03b2Bp,\u03b2\u03b3(A\u03c3tr 2 )\u03b3\u03b4B \u2020 p\u03b1\u03b4 = \u2212 \u03bb 16Ldp2 (\u03c3tr 2 A)\u03b1\u03b2(\u03c3tr 2 AT )\u03b2\u03b1 = 2 )\u03b4\u03b1 = \u2212 \u03bb 16Ldp2 (\u03c3tr 16Ldp2 tr(AA\u03c4 ). \u03bb 2 A)\u03b1\u03b2(A\u03c3tr 2 )\u03b1\u03b2 Using these contraction rules to compute the second term in Eq. (8.52), tr(\u03a6\u03bc\u03c33\u03a6\u03bc\u03c33 (BB\u2020E11 +B\u2020BE22)) \u221d tr(1) \u221d R, we obtain an expression that vanishes in the replica s gs = \u2212\u03a6\u03bc limit. As for the \ufb01rst term, noting that \u03a6\u03c4 we obtain s = \u2202\u03bcg\u22121 g\u22121 s \u2202\u03bcgs = \u2202\u03bcg\u03c4 s g\u22121\u03c4 \u03bc = \u03c4 \u2212 8 \u03bb \u2192 \u2212 16 \u03bb ddr tr(\u03a6\u03bcW \u03c33\u03a6\u03bcW \u03c33) = \u2212 8 \u03bb ddr tr(\u03a6\u03bc(BE12 \u2212 B\u2020E21) \u00d7 \u03a6\u03bc(BE12 \u2212 B\u2020E21)) ddr tr(\u03a6\u03bcBE12\u03a6\u03bcBE12) = I1 = \u2212 I1 4 ddr tr(\u03a612 \u03bc (\u03a6\u03c4 \u03bc)21) ddr tr([\u03a6\u03bc, \u03c33][\u03a6\u03bc, \u03c33]). Here, the \u201c\u2192\u201d indicates that we are omitting terms that lead to traces of \u03a6ii = 0. We also note that, by virtue of the time reversal relation, contractions \u223c BB and \u223c B\u2020B\u2020 lead to identical results. Expressing \u03a6\u03bc in terms of the original variables Qs, we arrive at the formula above. 492 The renormalization group 8.8.5 Kondo e\ufb00ect: poor man\u2019s scaling In Problem 2.4 we saw that the low-energy properties of a metal, when it is coupled to a magnetic Anderson impurity, are encapsulated by an e\ufb00ective sd-Hamiltonian Eq. (2.51) involving itinerant carriers coupled to the spin of the magnetic impurity by an exchange coupling J. Applying a perturbative analysis, in Problem 5.5 we saw that the e\ufb00ect of the magnetic impurity \ufb02uctuations was to impart a contribution to the scattering rate which diverged at low temperatures \u2013 a phenomenon compatible with the observed increase in low-temperature resistivity. However, it being perturbative in character, little could be said about the e\ufb00ect of the magnetic impurity in the strong coupling regime. In the following, we will implement a perturbative RG scheme to monitor the \ufb02ow of the coupling constant and explore the low-temperature properties of the Kondo impurity system. Our starting point is the sd-Hamiltonian where, anticipating the capacity for anisotropy, we treat the bare components of the exchange coupling as independent parameters, \u02c6Hsd = k\u03c3 kc \u2020 k\u03c3ck\u03c3 + # Jz \u02c6Sz kk c \u2020 k\u2191ck\u2191 \u2212 c \u2020 k\u2193ck\u2193 + J+ \u02c6S+c \u2020 k\u2193ck\u2191 + J\u2212 \u02c6S\u2212c \u2020 k\u2191ck\u2193 $ where \u02c6S denotes the spin of the local moment formed by the d-electron system. In the following, we will implement an adaptation of the RG procedure which monitors the e\ufb00ect of high-energy (fast) degrees of freedom on the low-energy (slow). Our analysis is taken from the original paper by Anderson.58 (For a discussion of the generalization of the RG scheme and a much more lengthy discussion of the subject as a whole, we refer to the review by Hewson.59) To this end, let us divide the conduction band into high-lying electron/hole states that occupy an energy shell D/b < |k| < D, whereD denotes the bare bandwidth and b > 1, and the remaining states 0 < |k| < D/", "b. To eliminate the highlying excitations that fall within the band edge, we may implement a general procedure that mirrors the one exploited in the construction of the sd-Hamiltonian from the Anderson model (Problem 2.4). Speci\ufb01cally, let us write the total wavefunction of the Hamiltonian |\u03c8 as the sum of terms |\u03c80, |\u03c81, and| \u03c82 where |\u03c81 describes the component in which there are no conduction electrons/holes in the upper/lower band edge, |\u03c80 has at least one hole in the lower band edge, and |\u03c82 has at least one electron in the upper band edge. Following the discussion in Problem 2.4, eliminating |\u03c80 and |\u03c82 from the Schr\u00a8odinger equation 2 n=0 \u02c6Hmn|\u03c8n = E|\u03c8m, one obtains \u02c6H10 1 E \u2212 \u02c6H00 \u02c6H01 + \u02c6H11 + \u02c6H12 1 E \u2212 \u02c6H22 \u02c6H21 |\u03c81 = E|\u03c81. (a) Taking into account the process that scatters a conduction electron into the upper band edge, show that the term \u02c6H12 1 E \u2212 \u02c6H22 \u02c6H21|\u03c81 58 P. W. Anderson, A poor man\u2019s derivation of scaling laws for the Kondo problem, J. Phys. C 3 (1970), 2436\u201341. 59 A. C. Hewson, The Kondo Problem to Heavy Fermions (Cambridge University Press, 1993). 8.8 Problems 493 is associated with eight possible contributions, which can be grouped in pairs. (You may \ufb01nd it helpful to enumerate these possibilities diagrammatically.) In particular, show that there exists a contribution J+J\u2212 k skskf \u02c6S\u2212c \u2020 k s \u2191ckf \u2193 1 E \u2212 \u02c6H22 \u2020 \u02c6S+c kf \u2193cks\u2191|\u03c81, where the wavevectors kf (ks) index states which lie within (outside) the band edge. (b) Focusing on this contribution alone (for now), rearrange the operators to show that it takes the form J+J\u2212 k sks \u03bd0D(1 \u2212 1/b) \u02c6S\u2212 \u02c6S+c \u2020 k s \u2191cks\u2191 1 E \u2212 D + ks \u2212 \u02c6H0, k\u03c3 kc where \u03bd0 denotes the (assumed constant) density of states at the band edge and \u02c6H0 = \u2020 k\u03c3ck\u03c3 is the single-particle Hamiltonian of the band electrons. Now, measuring the energy with respect to the ground state of the non-interacting electron gas, \u02c6H0 may be approximated by zero. Finally, noting that, for spin S = 1/2, \u02c6S\u2212 \u02c6S+ = 1/2 \u2212 \u02c6Sz, the contribution takes the form J+J\u2212 k sks \u03bd0D(1 \u2212 1/b)(1/2 \u2212 \u02c6Sz)c \u2020 k s \u2191cks\u2191 1 E \u2212 D + ks. We leave it as an unanswered part of the exercise to con\ufb01rm that the parallel contribution from the process in which a hole is created in the lower band edge leads to the expression J+J\u2212 k sks \u03bd0D(1 \u2212 1/b)(1/2 + \u02c6Sz)ck s \u2191c \u2020 ks\u2191 1 E \u2212 D \u2212 ks. (Try to use arguments based on particle/hole symmetry rather than developing a \ufb01rst principles analysis.) Following a similar procedure, one may con\ufb01rm that the second class of spin conserving terms lead to the contributions J 2 z 4 k sks\u03c3 \u03bd0D(1\u22121/b)c \u2020 s\u03c3cks\u03c3 k 1 E \u2212 D + ks, J 2 z 4 k sks\u03c3 \u03bd0D(1\u22121/b)ck s\u03c3c \u2020 ks\u03c3 1 E \u2212 D \u2212 ks. This completes the analysis of contributions to the e\ufb00ective Hamiltonian for |\u03c81 which leave the electron and impurity spin unchanged. The four remaining contributions involve a spin \ufb02ip. Following a similar procedure to the one outlined above, one may identify two further contributions \u2212 JzJ+ 2 sks k JzJ+ 2 k sks \u03bd0D(1 \u2212 1/b) \u02c6S+c \u2020 k s \u2193cks\u2191 1 E \u2212 D + k", "s, \u03bd0D(1 \u2212 1/b) \u02c6S+ck s \u2191c \u2020 ks\u2193 1 E \u2212 D \u2212 ks, where we have used the identity \u02c6Sz \u02c6S+ = \u02c6S+/2. One may con\ufb01rm that the corresponding terms with the order reversed generate an equal contribution. 494 The renormalization group When combined with terms from a spin-reversed process, altogether one \ufb01nds that the components of the exchange constant become renormalized according to J\u00b1(b) =J \u00b1 \u2212 JzJ\u00b1\u03bd0D(1 \u2212 1/b) Jz(b) =J z \u2212 J+J\u2212\u03bd0D(1 \u2212 1/b,. Now, since we are interested in the renormalization of the low-energy properties, we may ignore E against D. Moreover, since the typical internal excitation energy k D, we may approximate these equations by the di\ufb00erential recursion relations dJ\u00b1 d = 2\u03bd0JzJ\u00b1, dJz d = 2\u03bd0J+J\u2212, where, as usual, b = e. (c) Starting with these di\ufb00erential recursion relations, show that the scaling trajectory obeys the relation J 2 z \u2212 J 2 \u00b1 = const. One may note that, on renormalization, the coupling constant Jz always increases. Therefore, in a ferromagnetic model Jz < 0 with bare coupling constants such that |Jz| < J\u00b1, J\u00b1 will scale to zero. Conversely, when the bare constants |Jz| > J\u00b1, both Jz and J\u00b1 \ufb02ow to strong coupling where the perturbation theory breaks down. For Jz = J\u00b1 \u2261 J > 0, an integration of the scaling equations dJ/d = 2\u03bd0J 2 gives 1 J \u2212 1 J() = 2\u03bd0 = 2\u03bd0 ln D() D. Therefore, we may set D exp \u2212 1 \u03bd0J = D() exp \u2212 1 \u03bd0J() \u2261 TK, where TK denotes the Kondo temperature. Answer: (a) Focusing on terms which scatter conduction electrons into the band edge, one may identify \u02c6H21 = # Jz \u02c6Sz ks,kf c \u2020 kf \u2191cks\u2191 \u2212 c \u2020 kf \u2193cks\u2193 + J+ \u02c6S+c \u2020 kf \u2193cks\u2191 + J\u2212 \u02c6S\u2212c \u2020 kf \u2191cks\u2193 $, where the wavevectors kf span the range of single-particle band energies k within the band edge while the components ks span the range of lower energies. Now, to evaluate the contribution of the term \u02c6H21(E \u2212 \u02c6H22)\u22121 \u02c6H21 to the e\ufb00ective Hamiltonian for |\u03c81 we must identify all processes that are closed. Since the action of the Hamiltonian must return us to the subspace in which the band edge states remain unoccupied, processes 8.8 Problems 495 involved in a virtual occupation of the band edge are constrained by the conservation of spin. From \u02c6H12 \u02c6H21, one may identify eight allowed processes, \u2020 (1) J\u2212 \u02c6S\u2212c k s (2) \u2212 Jz \u02c6Szc (3) Jz \u02c6Szc (4) Jz \u02c6Szc \u2020 k s \u2191ckf \u2191 \u00d7 J\u2212 \u02c6S\u2212c \u2191ckf \u2191 \u00d7 Jz \u02c6Szc \u2020 kf \u2193cks\u2191, \u2020 kf \u2193cks\u2191, \u2191ckf \u2193 \u00d7 J+ \u02c6S+c \u2193ckf \u2193 \u00d7 J+ \u02c6S+c \u2020 kf \u2191cks\u2193, \u2020 kf \u2191cks\u2191, \u2020 k s \u2020 k s and the four counterparts related by symmetry. Of the four kinds of process, terms (2) and (3) involve a spin-\ufb02ip process while terms (1) and (4) preserve the electron spin orientation. Each of these terms can be represented diagrammatically. When combined with the energy denominator, the particular contribution speci\ufb01ed in the question corresponds to the \ufb01rst process (1). (b) Since the band edge occupancy of the reference state |\u03c81 is zero, one can set kf c \u2020 k s \u2191ckf \u2193 1 E \u2212 \u02c6H22 c \u2020 kf \u2193cks\u2191|\u03c81 = \u03bd0D(1 \u2212 1/b)c \u2020 k s", " \u2191cks\u2191 1 E \u2212 D + ks, \u2212 \u02c6H0 where \u02c6H0 denotes the single-particle Hamiltonian of the band electrons, D D/b d\u03bd() \u03bd0D(1 \u2212 1/b), and we have set kf required formula. = D. As a result, one obtains the kf (c) Dividing the di\ufb00erential recursion relations, one obtains dJ\u00b1 dJz = Jz J\u00b1. Integrating this equation, one obtains the required scaling trajectory. 9 Topology x In this chapter we discuss low-energy theories with non-trivial forms of long-range order. We learn how to detect the presence of topologically non-trivial structures, and to understand their physical consequences. Topological terms (\u03b8-terms, Wess\u2013Zumino terms, and Chern\u2013Simons terms) are introduced as contributions to the action, a\ufb00ecting the behavior of low-energy \ufb01eld theories through the topology of the underlying \ufb01eld con\ufb01gurations. Applications discussed in this chapter include persistent currents, quantum spin chains, and the quantum Hall e\ufb00ects. In the preceding chapters we encountered a wide range of long-range orders, or, to put it more technically, di\ufb00erent types of mean\u2013\ufb01elds. Re\ufb02ecting the feature of (average) translational invariance, the large majority of these mean-\ufb01elds turned out to be spatially homogeneous. However, there have also been a number of exceptions: under certain conditions, mean-\ufb01eld con\ufb01gurations with long-range1 spatial textures are likely to form. One mechanism driving the formation of inhomogeneities is the perpetual competition of energy and entropy: being in con\ufb02ict with the (average) translational invariance of extended systems, a spatially non-uniform mean-\ufb01eld is energetically costly. On the other hand, this very \u201cdisadvantage\u201d implies a state of lowered degree of order, or higher entropy. (Remember, for example, instanton formation in the quantum double well: although energetically unfavorable, once it has been created it can occur at any \u201ctime,\u201d which brings about a huge entropic factor.) It then depends on the spatio-temporal extension of the system whether or not an entropic proliferation of mean-\ufb01eld textures is favorable. A second mechanism behind the formation of inhomogeneities can be the topological structure of the order parameter \ufb01eld; does the mean-\ufb01eld accommodate solutions that simply cannot be continuously deformed back into a uniform state? The XY -model discussed at the end of the preceding chapter conveniently illustrates this principle: a vortex cannot be eliminated by any smooth deformation of the \ufb01eld. One might argue that this irreducibility is associated with the behavior of the core region of the vortex, where local order breaks down and the mean-\ufb01eld theoretical description simply does not apply (i.e. the vortex cannot be removed by manipulations of the phase \ufb01eld alone). However, an alternative, and more generally valid, explanation of the phenomenon is that a vortex represents a \ufb01eld con\ufb01guration characterized by a non-vanishing winding number, i.e. an integer di\ufb00erent from 1 For example, sometimes a system may \ufb01nd it energetically favorable to develop a micro-texture optimally adjusted to the structure of the underlying Hamiltonian (a prominent example being charge density wave formation in one-dimensional systems). Yet even these structures exhibit a discrete translational invariance. 496 9.1 Example: particle on a ring 497 zero. But a non-vanishing integer cannot be continuously \u201cdeformed\u201d to zero. (The vortex example also illustrates that the two principles of topology and entropic proliferation tend to cooperate. Once the energy necessary to form a mean-\ufb01eld of non-vanishing winding number has been invested, the system can bene\ufb01t from the freedom to place the core region anywhere in the system.) The remarks above touch upon but two of many interesting aspects of systems with topologically non-trivial order parameters. In the next section we employ a trivial example (namely a free particle on a ring) to set the stage for the discussion of further phenomena hinging on topological concepts. Speci\ufb01cally, we introduce the concept of a topological term, i.e. an operator which a\ufb00ects the low-energy behavior of a theory solely on the basis of the topology of", " its \ufb01elds. Turning to the more systematic development of the theory, we then introduce homotopy as the key mathematical tool whereby \ufb01elds can be topologically classi\ufb01ed. This discussion provides the conceptual platform on which the rest of the chapter is based. It is followed by a discussion of di\ufb00erent classes of topological terms (\u03b8-terms, Wess\u2013Zumino terms, Chern\u2013Simons terms), along with a number of applications. It turns out that, whenever such terms are present in a theory, they tend to massively a\ufb00ect its long-range behavior. At the same time, topological terms are notoriously easy to overlook in \u201cstandard\u201d schemes of distilling low-energy theories from their microscopic origins. For this reason, some emphasis is placed on purely operational aspects; i.e. tricks that prevent one from missing the presence of a topological term! Before getting started, it is worthwhile emphasizing that we are about to plunge into a wide subject area that simply cannot be satisfactorily covered in a single chapter.2 Consequently, our discussion is example\u2013oriented and often regrettably super\ufb01cial (with regard to both physical depth and, especially, mathematical structures). In fact, the aim of the present text is to demystify the subject of topology in \ufb01eld theory, to arouse the interest of readers and to motivate them to proceed to more profound and substantial discussions in the literature! 9.1 Example: particle on a ring \u03a6 Consider the problem of a free quantum particle of charge e con\ufb01ned to one dimension and subject to periodic boundary conditions \u2013 a particle on a ring (see \ufb01gure). To make the problem somewhat more interesting, let us assume that the ring is threaded by a magnetic \ufb02ux \u03a6. Measuring the coordinate of the particle in terms of an angular variable \u03c6 \u2208 [0, 2\u03c0], the free Hamiltonian of the system thus takes the form ( = e = c = 1), \u02c6H = 1 2 (\u2212i\u2202\u03c6 \u2212 A)2, (9.1) 2 For an advanced text speci\ufb01cally targeted on topological considerations in physics, and the underlying mathe- matical structures, we refer to M. Nakahara, Geometry, Topology and Physics (IOP Publishing, 2003). 498 Topology where A = \u03a6/\u03a60 denotes the vector potential corresponding to the magnetic \ufb01eld (exercise), and \u03a60 = hc/e = 2\u03c0 represents the magnetic \ufb02ux quantum. (Here, for notational simplicity, we have set the radius of the ring and the particle mass to unity.) Periodicity implies that we are working on a Hilbert space of wavefunctions \u03c8 subject to the condition \u03c8(0) = \u03c8(2\u03c0). Of course, the problem de\ufb01ned by Eq. (9.1) is embarrassingly simple. One may readily verify that the eigenfunctions and spectrum of the Hamiltonian are given by \u03c8n(\u03c6) = 1\u221a 2\u03c0 exp(in\u03c6), n = 2 1 2 n \u2212 \u03a6 \u03a60, n \u2208 Z. (9.2) On the other hand, this very simplicity is somewhat deceptive;3 we shall see in a moment that many of the concepts of topological quantum \ufb01eld theory \ufb01nd a preliminary realization in the problem above. To explore these connections, let us reformulate the system in the language of the imag- inary time path integral (cf. Problem 3.5): Z = D\u03c6 e\u2212 d\u03c4 L(\u03c6, \u02d9\u03c6), (9.3) \u03c6(\u03b2)\u2212\u03c6(0)\u22082\u03c0Z where the boundary condition \u03c6(\u03b2)\u2212\u03c6(0) \u2208 2\u03c0Z expresses the fact that the phase is de\ufb01ned only up to integer multiples of 2\u03c0, and the Lagrangian is given by L(\u03c6, \u02d9\u03c6) = 1 2 EXERCISE Verify by Legendre transformation that the Hamiltonian corresponding to this Lagrangian is given by Eq. (9.1). Obtain the spectrum Eq. (9.2) from the path integral, i.e. represent the partition function in the form Z = n exp(\u2212\u03b2n). (Hint: You may \ufb01nd the Hubbard\u2013Stratonovich decoupling of the quadratic term useful.) \u02d9\u03c62 \u2212 iA \u02d9\u03c6. (9.4) Suppose that we were unaware of the exact solution of the problem. Our canonical approach to controlling the integral would be to subject the theory to", " a saddle-point analysis. The saddle-point (alias Euler\u2013Lagrange) equations of the action S[\u03c6] = 0 d\u03c4 L(\u03c6, \u02d9\u03c6), \u03b2 \u03b4S[\u03c6] \u03b4\u03c6(\u03c4 ) = 0 \u21d4 \u00a8\u03c6 = 0, have two interesting properties. (i) The vector potential does not enter the equations. On the other hand, we saw above that it does have a physical e\ufb00ect (the spectrum explicitly depends on A). We need to understand how these two seemingly contradictory observations can be reconciled with each other. (ii) There exists a whole family of solutions, \u03c6W (\u03c4 ) \u2261 W 2\u03c0\u03c4 /\u03b2. The action of these con\ufb01gurations, S[\u03c6W]|A= 0 = 1 2\u03b2 (2\u03c0W )2, varies discontinuously with W, i.e. by analogy with other cases where we found saddle-point 3 Note that the literature is full of erroneous statements on even this simple system. The most frequent of these is that the periodic boundary conditions of the problem force the \ufb02ux to be quantized in integer multiples of the \ufb02ux quantum \u03a6 = n\u03a60. This is, of course, incorrect. Even for non-integer \u03a6, the wavefunctions \u03c6n are perfectly periodic. The statement is probably triggered by \ufb02ux-quantization in superconducting systems (a phenomenon that relies on energetic, and not topological, considerations). 9.1 Example: particle on a ring 499 S1 \u03c4 \u03c6 0 \u03b2 S1 \u03c6 (\u03c4 Figure 9.1 Showing mappings \u03c6 : S1 \u2192 S1 of di\ufb00erent winding numbers. con\ufb01gurations de\ufb01ned by an integer index (cf. instanton solutions in the quantum double well), we may expect the di\ufb00erent solutions to be separated by huge energy barriers. However, the present problem is special insofar as, besides energy, there is a much more \u201cprofound\u201d principle separating con\ufb01gurations of di\ufb00erent W. To understand this point, let us note that, mathematically, the \ufb01eld \u03c6 is a mapping \u03c6 : S1 \u2192 S1, \u03c4 \u2192 \u03c6(\u03c4 ), from the unit circle S1 (imaginary time with periodic boundary conditions4) into another circle (\u03c6 is a phase!). Mappings of this type can be assigned to a winding number, W, i.e. the number of times \u03c6(\u03c4 ) winds around the unit circle as \u03c4 progresses from 0 to \u03b2: \u03c6(\u03b2) \u2212 \u03c6(0) = 2\u03c0W (see Fig. 9.1). Indeed, it is not possible to change W by a continuous deformation of \u03c6. Since continuity is a paradigm implicit in \ufb01eld integration, the integration over all functions \u03c6(\u03c4 ) can be organized as an integration over functions \u03c6(\u03c4 ) of di\ufb00erent winding numbers, or di\ufb00erent topological sectors: D\u03c6 e\u2212 d\u03c4 L(\u03c6, \u02d9\u03c6) = e2\u03c0iW A D\u03c6 e\u2212 1 2 d\u03c4 \u02d9\u03c62. (9.5) Z = W \u03c6(\u03b2)\u2212\u03c6(0)=2\u03c0W W \u03c6(\u03b2)\u2212\u03c6(0)=2\u03c0W Here, we have noted that the A-dependent term in the action, Stop[\u03c6] \u2261 iA \u03b2 0 d\u03c4 \u02d9\u03c6 = iA(\u03c6(\u03b2) \u2212 \u03c6(0)) = i2\u03c0W A, involves only the index of the topological sector of \u03c6. The representation (9.5) makes the topological aspects of the problem particularly transparent. Speci\ufb01cally, one may note that: 4 Strictly speaking, imaginary time should be identi\ufb01ed with a \u201ccircle\u201d of circumference \u03b2. However, for our present purposes, all that matters is that the periodic boundary conditions render the interval [0, \u03b2] isomorphic to a circle. 500 Topology The functional integral assumes the form of a sum of integrals over disjoint topological sectors. The contribution to the action, Stop, is our \ufb01rst example of a topological term. (More precisely, it belongs to the class of \u201c\u03b8-terms.\u201d) Since Stop is sensitive only to the topological sector of a \ufb01eld contribution, it cannot a\ufb00ect the equations of motion. This is because these probe how the action responds to an in\ufb01nitesimal variation of the \ufb01eld con\ufffd", "\ufffdguration, an operation that cannot change the winding number. (You may wish to ponder this point since it will be important.) However, the topological term does a\ufb00ect the outcome of the functional integration: it plays the role of a W -dependent \u201cphase,\u201d weighting the contribution of di\ufb00erent sectors. The fact that Stop knows only about the topological class of a \ufb01eld con\ufb01guration implies that it is impervious to any changes in the metric of the base manifold of the theory (in our case, imaginary time). For example, we might decide to measure time in di\ufb00erent units, i.e. \u03c4 \u2192 \u03b1\u03c4 = \u03c4, where \u03b1 is some scaling factor. This transformation leaves the topological term invariant. In particular, it remains form invariant under a change from imaginary to real time, \u03c4 \u2192 \u2212i\u03c4 = t. In both representations, Stop[\u03c6] = 2\u03c0iAW is purely imaginary. This, in fact, is a hallmark of topological terms; in both Euclidean and Minkowski space-time, their contribution to the action is always imaginary. While formulated for the (almost trivial) example of the particle on the ring, all of these features generalize to much more involved settings. However, to discuss these generalizations in a sensible manner, we need to provide somewhat more mathematical background. This will be the subject of the next section. INFO Owing to its simplicity, the system above frequently appears as an e\ufb00ective model in condensed matter physics. Examples we have encountered previously include the Josephson junction (where the presence of a condensate induces an additional cosine potential) and the physics of normal metal granules subject to strong charging (with the additional complication of dissipative damping of the \u03c6-\ufb02uctuations). Here we brie\ufb02y touch upon the physics of persistent currents as an example where the topological aspects of the problem play a particularly important role. Consider then a ring-shaped conductor subject to a magnetic \ufb02ux. According to a prediction by Byers and Yang5 the magnetic \ufb01eld induces an equilibrium current periodic in \u03a6 with period \u03a60. I(\u03a6) = \u2212 \u2202F (\u03a6) \u2202\u03a6, EXERCISE Remembering that a vector potential enters the free energy as \u223c dA \u00b7 j, derive the Byers and Yang formula. Show that, at zero temperature, the persistent current \ufb02owing in a perfectly clean one-dimensional metal of non-interacting fermions assumes the form of a \u03a60periodic sawtooth function, I(\u03a6) = 2\u03c0vF L [\u03a6/\u03a60], where [x] =x \u2212 n and n is the largest integer smaller than x. (Hint: For zero temperature, the free energy of a non-interacting system of particles is equal to the sum of all single-particle energies up to the Fermi energy. Notice that the current is 5 B. Byers and C. N. Yang, Theoretical considerations concerning quantized magnetic \ufb02ux in superconducting cylinders, Phys. Rev. Lett. 7 (1961), 46\u20139. 9.1 Example: particle on a ring 501 ).5 1.0 0.5 0.0 0.00 H (gauss) h/e h/2e 0.05 0.10 0.15 0.20 0.25 1/H (1/gauss) Figure 9.2 Power spectrum of the persistent current carried by a single gold loop (diameter O(1 \u03bcm), L/ \u223c 10). The data shows a peak at h/e and a small satellite modulation at h/2e. The magnitude of the current is some two orders of magnitude larger than that predicted by theory. (Reprinted with permission from V. Chandrasekhar, R. A. Wells, M. J. Brady, et al. Magnetic response of a single, isolated gold loop, Phys. Rev. Lett. 67 (1991), 3578\u201381. Copyright (1991) by the American Physical Society.) carried by the last occupied state, i.e. the currents \u2212\u2202\u03c6n carried by all levels beneath the Fermi energy cancel.) For a long time it has been believed that this statement was largely academic since surely even a moderate concentration of impurities6 would be su\ufb03cient to render the current undetectable. However, a simple consideration shows that this need not be the case: a gauge transformation \u03c8(\u03c6) \u2192 eiA\u03c6\u03c8(\u03c6) removes A from the Hamiltonian while changing the boundary conditions to \u03c8(0) = e2\u03c0iA\u03c8(2\u03c0). In the gauge-transformed", " picture, the presence of the magnetic \ufb01eld thus amounts to a twist in the boundary conditions of the wavefunctions, and the persistent current is a measure of the sensitivity of the spectrum to this twist. Now, there is no reason to believe that a wavefunction in a disordered system should be less sensitive to its boundary conditions than that of a clean system.7 Indeed, it is shown in Problem 9.7.1 that even rings of circumference L may carry a sizeable persistent current. In a series of beautiful experiments conducted in the early 1990s (see, e.g., Fig. 9.2), persistent currents in both \u201cballistic\u201d (L/ \u223c 1) and \u201cdirty\u201d (L/ \u223c 100) environments were indeed observed experimentally. Frustratingly, the measured current appears to be some two orders of magnitude larger than the theories of non-interacting particles would predict. This disturbing discrepancy led to the formulation of a plethora of theories of persistent currents in interacting/disordered systems. However, to date, the discrepancy between theory and experiment remains unresolved. 6 More precisely, it was believed that a sizeable current might only be observed if the circumference of the ring were smaller than the scattering mean free path. The arti\ufb01cial fabrication of rings of that quality requires semiconductor technology which was not available at the time. With regard to molecular rings, equilibrium currents had been predicted in the 1930s. However, the \ufb01elds needed to drive currents in molecules (O(105 T)) cannot be generated by the laboratory magnets currently available. 7 What does diminish this sensitivity is mechanisms destroying the coherence of wavefunctions (thermal noise, etc.) and strong localization, i.e. a wavefunction con\ufb01ned to a \ufb01nite portion of the ring will not sense changes in the boundary conditions. In fact, sensitivity with respect to changes in the boundary conditions has been used (largely by theorists) as a popular test for localization. 502 Topology 9.2 Homotopy 9.2.1 Generalities To appreciate the generalization of the structures above to more complicated contexts, we need to introduce a few more mathematical concepts. In Chapter 1 (see page 6), we de\ufb01ned \ufb01elds as mappings \u03c6 : M \u2192 T, z \u2192 \u03c6(z), from a \u201cbase manifold\u201d M \u2013 usually Rd or a subset thereof \u2013 into some \u201ctarget space\u201d T. By now we have gained enough experience to determine what are the \u201cmost frequent\u201d realizations of target spaces: indeed we have mostly been interested in the long-range behavior of Goldstone modes, i.e. long-range modes induced by a mechanism of symmetry breaking. In such cases (cf. page 257), T = G/H will be a coset space obtained by dividing the symmetry group of the problem, G, by some subgroup H \u2282 G stabilizing the mean-\ufb01eld around which the Goldstone modes \ufb02uctuate. Usually (but certainly not invariably) G will be one of the compact classical groups U(N ), O(N ), and Sp(N ), and H some subgroup thereof. INFO Why is the word \u201ctopology\u201d so often deployed in this chapter? In the example above, the most relevant characteristic of a \ufb01eld con\ufb01guration was its winding number, i.e. a quantity that does not change under any continuous deformation of a \ufb01eld, no matter how \u201clarge\u201d is this deformation. More generally, in this chapter we are concerned with features of \ufb01eld theories that essentially rely on the concept of continuity, but do not involve the notion of \u201cdistance\u201d or, more formally, of a metric. The most general mathematical structure for which the notion of continuity can be de\ufb01ned is a topological space:8 a mapping \u03c6 : X \u2192 Y between two topological spaces is called continuous if, for any open set U \u2282 Y, the set\u03c6 \u22121(U ) \u2282 X is open in X. This de\ufb01nition is not tied to the existence of a metric.9 This remark is not entirely academic: there are prominent physical spaces \u2013 the phase space of classical mechanics, for example \u2013 which do not possess a canonical metric but for which the notion of a continuous mapping certainly exists. Although our discussion below does not rely on the in-depth mathematics of di\ufb00erential topology, the frequent use of the attribute \u201ctopological\u201d emphasizes that what matters in this chapter is \u201ccontinuity minus metric.\u201d In fact, practical considerations also enable us to be a bit more speci\ufb01c as to the structure of the base manifold. Suppose our microscopic parent theory is de\ufffd", "\ufffd\ufffdned on some simply connected manifold M \u2282 Rd. As we shall be typically concerned with some kind of thermodynamic limit, M can be thought of as an \u201cin\ufb01nitely large\u201d object. On the level of the low-energy theory of the system, this requires that \u201csensible\u201d \ufb01eld con\ufb01gurations must 8 As a reminder, let X be a set and J = {Yi \u2282 X|i \u2208 I} a collection of its subsets. The pair (X, J ) is called a i\u2208J Yi \u2208 J, and, (c) for any \ufb01nite subset J \u2282 I, topological space if and only if, (a) {}, X \u2208 J, (b) forJ \u2282 I, i\u2208J Yi \u2208 J. The elements of J are called open subsets of X. 9 By contrast, in elementary mathematics courses, the continuity of a mapping \u03c6 is frequently de\ufb01ned via a metric, i.e. one relates the distance between two image points \u03c6(x) and\u03c6(y ) to that between the arguments x and y. approach a constant value on the boundary of M : i.e. \u03c6|\u2202M = const. (lest the action become in\ufb01nite). 9.2 Homotopy 503 As far as our e\ufb00ective \ufb01eld theory is concerned, we may thus identify all boundary points with a single point at \u201cin\ufb01nity.\u201d Geometrically, this implies that M can be compacti\ufb01ed to a large sphere (see the \ufb01gure). However, as the radius of this sphere carries no signi\ufb01cance (indeed, technically, it can be removed by a global rescaling of coordinates), we can, just the same, identify M Sd with the d-dimensional unit sphere. The practical upshot of all of these considerations is that, in practice, we shall mostly be interested in \ufb01elds \u03c6 : Sd \u2192 G/H, \u2192 \u03c6(z), z mapping a unit sphere into some coset space. We now turn to the discussion of topological aspects of such mappings. By analogy to our discussion of the previous section (where we considered mappings \u03c6 : S1 \u2192 S1 O(2)), let us consider two \ufb01elds \u03c61 and \u03c62 as topologically equivalent if they can be continuously deformed into each other. Technically, this condition amounts to the existence of a continuous mapping (a homotopy in the language of mathematics) \u03c6 : Sd \u00d7 [0, 1] \u2192 G/H, \u2192 \u02c6\u03c6(z, t), (z, t) such that \u02c6\u03c6(., 0) = \u03c61 and \u02c6\u03c62(., 1) = \u03c62. (Notice that \u02c6\u03c6 represents a mapping from (d+1)-dimensional space into G/H, a fact that will become important when we turn to the discussion of Wess\u2013Zumino terms below.) We denote the equivalence class of all \ufb01elds topologically equivalent to a given representative \u03c6 by [\u03c6]. (In the example discussed in the previous section, those equivalence classes would contain all \ufb01elds of a given winding number.) The set of all topological equivalence classes {[\u03c6]} of mappings \u03c6 : Sd \u2192 T is called the d-th homotopy group, \u03c0d(T ). INFO Some readers might wonder in what sense \u03c0d(T ) carries a group structure (rather than just being a set). To understand this point, it is convenient to deform our base manifold from a sphere to a d-dimensional unit cube I d = [0, 1]d. As far as topology is concerned, this is a permissible operation, if and only if and only if we identify the boundary \u2202I d of the cube with a single point on the sphere. For example, it is convenient to choose this point to be the representative of the \u201cin\ufb01nitely large\u201d boundary of our original base manifold. This choice \u2202Id = \u03c6\u2217 = const. be the constant \ufb01eld con\ufb01guration approached at the \u201cphysical requires that \u03c6| in\ufb01nity\u201d.10 10 The assumption of a unique asymptotic con\ufb01guration \u03c6\u2217 is less of a restriction than it may seem; any (constant) boundary \ufb01eld \u03c6 can be converted to \u03c6\u2217 by a", " global transformation acting on the \ufb01eld (an operation that does not leave the equivalence class). 504 Topology + = Figure 9.3 Concatenation of two two-dimensional \ufb01elds into a single one. The purpose of the cube construction is that it presents us with ways to glue two \ufb01elds \u03c61 and \u03c62 together to form a new \ufb01eld \u03c63 \u2261 \u03c61 \u2217 \u03c62 (see Fig. 9.3). For example, we might de\ufb01ne * \u03c63(x1, x2,..., xd) = x1 \u2208 [0, 1/2], \u03c61(2x1, x2,..., xd), \u03c62(2x1 \u2212 1, x2,..., xd), x1 \u2208 [1/2, 1]. One might justly object that, due to its explicit coordinate dependence, this is not an acceptable mathematical de\ufb01nition. However, as far as our discussion of homotopy is concerned, this objection is immaterial. Indeed, we may de\ufb01ne a group operation on \u03c0d(T ) by [\u03c61] \u2217 [\u03c62] \u2261 [\u03c63]. (9.6) This de\ufb01nition is canonical in the sense that any other coordinate convention in our concatenation operation above would not leave the equivalence class of \u03c63 (a point on which you might like to dwell). Similarly one may verify that \u201c\u2217\u201d obeys all the criteria for a group mapping. In short, the very possibility to \u201cconcatenate\u201d \ufb01elds induces a well-de\ufb01ned group operation on the set of topological equivalence classes. 9.2.2 Examples of homotopies The general group theoretical analysis of homotopies is a mathematical subject that reaches far beyond the scope of the present text. Here we restrict ourselves to the discussion of a few examples of practical interest. In simple cases, the homotopy group can be identi\ufb01ed by common sense reasoning. For example, in the previous section we saw that mappings S1 \u2192 S1 can be classi\ufb01ed in terms of winding numbers: \u03c01(S1) =Z. Similarly, it is clear that any mapping S1 \u2192 S2 \u2013 a closed curve on the 2-sphere \u2013 is continuously contractible to a point: \u03c01(S2) = \u2205. The same applies, by de\ufb01nition, to any curve in a simply connected space. Prominent examples of such spaces are the higher-dimensional spheres Sd>1 and SU(N ): \u03c01(Sd>1) =\u03c0 1(SU(N )) = \u2205. By contrast, the \ufb01rst homotopy groups of non-simply connected spaces are categorically non-trivial. For example, curves on the d-dimensional torus T d are classi\ufb01ed by (exercise: convince yourself of the veracity of this statement) \u03c01(T d. Turning to higher dimensions, it becomes more and more di\ufb03cult to identify homotopy classes simply by invoking one\u2019s imagination. One of the last intuitively accessible examples is \u03c02(S2) = Z: maps of the 2-sphere into itself can be classi\ufb01ed according to how often they Table 9.1 Homotopy groups of a number of frequently encountered mappings. A blank entry means that no general statements can be made. 9.3 \u03b8-0terms 505 S1 Z \u2205 \u2205 S2 \u2205 Z \u2205 Sd> \u2205 SU(2) SU(N ) \u2205 \u2205 \u2205 \u2205 \u2205 \u03c03(SU(2)) = Z a But \u03c03(S2) =Z. wrap around the sphere. This statement generalizes to \u03c0d(Sd) = Z, while \u03c0k(Sd>k) = \u2205. Interestingly, the situation can be non-trivial for mappings Sk \u2192 Sd<k. For example, Hopf has shown that \u03c03(S2) = Z. For a summary of these, and a few more results, see Table 9.1. Far from being complete, the list of examples in Table 9.1 fails to cover a number of homotopies of outstanding relevance. For example, in condensed matter physics (unlike particle physics where space and time are intertwined by relativistic covariance) we are typically confronted with a situation where time is separately compacti\ufb01ed to a circle \u2013", " imaginary time with periodic boundary conditions. On top of that, the \ufb01nite-action arguments outlined above motivate a compacti\ufb01cation of Rd \u2192 Sd of real space to a d-dimensional sphere. This implies that, in quantum statistical \ufb01eld theory, one often encounters the base manifold M S1 \u00d7 Sd (instead of Sd+1 as found in particle physics). Clearly the construction of homotopic groups \u03c0(S1 \u00d7 Sd, T ) corresponding to mappings S1 \u00d7 Sd \u2192 T is more complex than that involved in the de\ufb01nition of the group structures above. However, rather than dwelling on the near-endless \ufb01eld of homotopy theory, we now return to \ufb01eld theory and explore general implications of the homotopic classi\ufb01cation scheme.2 9.3 \u03b8-Terms Returning to \ufb01eld theory considerations, let us address the question of what can be learned from the concepts introduced in the previous section. Each \ufb01eld \u03c6 : M \u2192 T can be uniquely assigned to a certain homotopy class. Consequently, the functional integration de\ufb01ning a \ufb01eld theory can be organized as Z = D\u03c6W e\u2212S[\u03c6], W \u2208G D\u03c6W denotes integration over the homotopy class where G is the homotopy group and de\ufb01ned by a given element W \u2208 G \u2013 a \u201ctopological sector\u201d of the theory. It may happen that the action of our \ufb01eld theory, S[\u03c6] = S0[\u03c6] +S top[\u03c6], 506 Topology \u03c6 1 \u03c6 2 \u03c6 1*\u03c6 2 W1 W2 W1 + W2 Figure 9.4 The combination of two \ufb01elds with winding numbers W1 and W2 leads to a new \ufb01eld with winding number W1 + W2. contains a topological action component: i.e. a contribution Stop[\u03c6] \u2261 F (W ) that depends only on the topological class of the \ufb01eld \u03c6 (which, for integer-valued W, is sometimes called the topological charge of the con\ufb01guration). In such cases, one may pull the function F (W ) in front of the functional integral to write Z = e\u2212F (W ) D\u03c6W e\u2212S0[\u03c6]. (9.7) W \u2208G In the following, we would like to understand in what way the topological action depends on the index W. This question can also be addressed by the type of reasoning ubiquitous in this chapter. Suppose we are given two \ufb01elds \u03c61 and \u03c62 which are constant everywhere save for two well-localized regions of variation somewhere in space-time. Let us assume that the two regions where the \ufb01elds vary are \u201cin\ufb01nitely\u201d far away from each other (see Fig. 9.4). Now, suppose we had glued these \ufb01elds together (e.g. by the prescription formulated in the previous section) to form a new \ufb01eld \u03c61 \u2217 \u03c62. The in\ufb01nitely large separation of the two constituent \ufb01elds implies that they are completely \u201cuncorrelated,\u201d i.e. the action of the composite \ufb01eld S[\u03c61 \u2217 \u03c62] = S[\u03c61] +S [\u03c62] is obtained simply by adding the actions of the constituents. In particular, F (W1 + W2) = Stop[\u03c61 \u2217 \u03c62] =S top[\u03c61] +S top[\u03c62] = F (W1) + F (W2), where in the \ufb01rst equality we have used the fact that the composite \ufb01eld has topological index W1 + W2. The identity F (W1 + W2) = F (W1) + F (W2) tells us that the topological action is linear in the topological index. For example, consider the simple (and at the same time most frequently encountered) case where \u03c0(M, T ) Z, i.e. Wi \u2208 Z are just numbers. The linearity then uniquely determines the topological action, F [W ] = i\u03b8W, up to a constant (which we choose to be real lest the action become ill-de\ufb01ned at large values of the topological charge). These considerations tell us that the factor exp(\u2212F [W ]) = exp(\u2212i\u03b8W", " ) weighing the di\ufb00erent sectors assumes the form of a phase. Relatedly, the constant \u03b8 is usually referred to as a topological angle. (Since W is integer, \u03b8 is de\ufb01ned only mod 2\u03c0 \u2013 an angular 9.3 \u03b8-0terms 507 variable.) For historical reasons11 the topological action Stop is commonly referred to as a \u03b8term. Notwithstanding its simplicity, a drawback of this representation is that it explicitly relies on the decomposition of the \ufb01eld integral into a sum over distinct topological sectors. It would be much more desirable to work with a representation Stop[\u03c6] directly in terms of the \ufb01eld (rather than through its winding number). This would enable us to formulate the D\u03c6 exp(\u2212S0[\u03c6]\u2212Stop[\u03c6]), without the necessity to explicitly \ufb01eld integral more directly as sum over winding numbers. Indeed, it is almost always possible to represent the topological action as an integral over a topological Lagrangian density, Stop[\u03c6] = ddx Ltop[\u03c6, \u2202\u03bc\u03c6]. Unfortunately, no canonical recipe for the construction of this representation exists. However, this is not as serious a problem as one might suspect. For one thing, the Lagrangian densities of the \u201cusual suspects\u201d of topologically non-trivial \ufb01eld theories are known. Conversely, it is usually straightforward to deduce whether or not a given term in the action is a \u03b8-term in disguise. To illustrate these points, we next discuss the concept of a topological Lagrangian density on a few relevant examples. In fact, we have already encountered the simplest representative of a \u03b8-term in our analysis of the particle on a ring: Stop[\u03c6] \u2261 i\u03b8 \u03b2 0 d\u03c4 2\u03c0 \u02d9\u03c6 = iW \u03b8, where \u03b8 = 2\u03c0A was proportional to the magnetic \ufb02ux through the ring. We reiterate the key features of this term: it does not a\ufb00ect the equations of motion (a small distortion, or variation, of the \ufb01eld does not change its topological index, i.e. it leaves Stop invariant), and it is invariant under arbitrary coordinate reparameterizations \u03c4 \u2192 s = s(\u03c4 ). As a less simple example illustrating these features we now discuss a two-dimensional \ufb01eld theory. 9.3.1 A case study: \u03c02(S2) Consider a \ufb01eld theory in a two-dimensional compacti\ufb01ed space with the two-sphere as a target manifold. Important examples falling into this category include the theory of quantum spin chains (see Section 9.3.3), the two-dimensional classical Heisenberg model, and the \ufb01eld theory of the integer quantum Hall e\ufb00ect (see Section 9.3.4). Technically, the \ufb01elds of this theory are mappings n : S2 \u2192 S2, x \u2192 n(x), |n| = 1, 11 One of the \ufb01rst major applications where terms of this type played a dominant role was \u2019tHooft\u2019s analysis of SU(2) gauge \ufb01eld instantons in (3 + 1)-dimensional compacti\ufb01ed space-time. (G. \u2019tHooft, Magnetic monopoles in uni\ufb01ed gauge theories, Nucl. Phys. B79 (1974), 276\u201384.) There he systematically labeled the topological angle by \u03b8; hence the name \u03b8-term. 508 Topology and the relevant homotopy group is \u03c02(S2) Z. Let us suppose that the topological action of this \ufb01eld theory is given by Stop[n] = i\u03b8 4\u03c0 dx1 dx2 n \u00b7 (\u22021n \u00d7 \u22022n). (9.8) How does one verify an assertion of this type? A \ufb01rst condition for Stop to be of topological nature is an insensitivity to small variations of the \ufb01eld n. Suppose then we vary n by a small amount, i.e. n(x) \u2192 n(x) + a(x)Ran, where the functions a, a = 1, 2, 3, are in\ufb01nitesimal and Ra are the generators of rotations around the three coordinate axes. By integrating a few times by parts, it is straightforward to show that the variation of Stop assumes the form \u03b4Stop[n] = 3i\u03b8 4\u03c0 dx1 dx2 aRan \u00b7 (\u22021n \u00d7 \u22022n). However, Ran is perpendicular to n", " while (exercise) (\u22021n \u00d7 \u22022n) lies parallel, i.e. \u03b4Stop = 0. (As a corollary we note that Stop will not a\ufb00ect the equations of motion.) The invariance ofS top implies that we can evaluate its value on any convenient test \ufb01eld con\ufb01guration n0; each \ufb01eld that can be reached by continuous deformation of n0 will have the same topological action. Consider, for example, the family of \ufb01eld con\ufb01gurations n(W ) : R2 \u2192 S2, (x1, x2) \u2192 \u03c6 = W tan\u22121 x2 x1, \u03b8 = 2 tan\u22121 a2 1 + x2 x2 2, where we used polar coordinates to parameterize n(W )(\u03b8, \u03c6). For historical reasons,12 these \ufb01eld con\ufb01gurations are commonly referred to as skyrmions. (Notice that the su\ufb03x \u201c-on\u201d indicates that we are dealing with highly stable [particle-like] excitations.) A coarse visualization of the simplest skyrmion con\ufb01guration, n(1), is shown in the \ufb01gure: a texture of unit vectors varying on a scale set by the parameter a. Skyrmions of higher winding numbers, n(W >1), are di\ufb03cult to visualize. However, it is straightforward to verify (by substitution of the unit vector n(W )(\u03c6, \u03b8) into the integral (9.8)) that their topological charge is given by W. Equivalently, the topological action reads Stop[\u03c6(W )] = i\u03b8W. EXERCISE Verify these statements. Show that the topological charge is insensitive to coordinate changes on both the target and the base manifold. Try to invent other simple \ufb01eld con\ufb01gurations of non-vanishing topological charge. INFO A word on semantics: depending on the context, topologically non-trivial \ufb01eld con\ufb01gurations are described as solitons, instantons, skyrmions, etc. While there seem to be no 12 See T. H. R. Skyrme, A nonlinear \ufb01eld theory, Proc. Roy. Soc. London A 260 (1961), 127-38 where these excitations appeared in the context of an e\ufb00ective model of nuclear matter. 9.3 \u03b8-0terms 509 discernible systematics in this scheme, a rule of thumb is that topological excitations in dynamical quantum theories (i.e. theories where the base manifold represents space-time) are called \u201cinstantons.\u201d By contrast, \u201csolitons\u201d are topological solutions of classical equations of motions. However, this rule is also sometimes broken. The ubiquitous presence of the su\ufb03x \u201c-on\u201d re\ufb02ects a widespread tendency in physics to associate excitations that are protected from ordinary decay mechanisms (presently, by the presence of topological indices) with di\ufb00erent kinds of \u201cparticles\u201d. 9.3.2 Functional integration and topological textures: generalities How can our present understanding of topologically non-trival \ufb01eld con\ufb01gurations be extended to a working scheme of \ufb01eld integration? As we saw above, the di\ufb00erent topological sectors of the theory essentially lead their \u201cown lives.\u201d It is then an obvious idea to try to carry out the \u201ccanonical program\u201d of \ufb01eld integration (analysis of mean-\ufb01eld con\ufb01gurations integration over \ufb02uctuations) in each sector separately. Our \ufb01rst step would thus be to seek solutions of the equation \u03b4S[\u03c6] \u03b4\u03c6 = \u03b4S0[\u03c6] \u03b4\u03c6 \u03c6\u2208\u03c6W = 0. \u03c6\u2208\u03c6W (9.9) (The later equality expresses the fact that the topological action does not change under \ufb01eld variation.) In our prototypical example of Section 9.1, the solutions of these equations were readily identi\ufb01ed as \u03c6W (\u03c4 ) = 2\u03c0W \u03c4 /\u03b2. However, in general, \ufb01nding solutions of Eq. (9.9) for W = 0 is a task more complicated than the analysis of the W = 0 mean-\ufb01eld. (This is because \ufb01eld con\ufb01gurations with W = 0 generally", " exhibit some non-trivial spatial variations, i.e. we cannot rely on the standard homogeneity assumptions.) INFO However, there exists an elegant trick whereby the identi\ufb01cation of topologically non-trivial mean-\ufb01eld con\ufb01gurations can sometimes be drastically simpli\ufb01ed. Consider the expression 0 \u2264 1 2 d2x (\u2202\u03bcn + \u03bc\u03bd n \u00d7 \u2202\u03bd n) \u00b7 (\u2202\u03bcn + \u03bc\u03bd n \u00d7 \u2202\u03bd n) = d2x (\u2202\u03bcn \u00b7 \u2202\u03bcn + \u03bc\u03bd n \u00b7 (\u2202\u03bcn \u00d7 \u2202\u03bd n)), where the inequality simply expresses the fact that we are integrating a positive de\ufb01nite quantity. Now, we know that the second term in the latter integral yields just \u22128\u03c0 times the topological charge, i.e. W \u2264 1 8\u03c0 d2x \u2202\u03bcn \u00b7 \u2202\u03bcn = S0[\u03c6]. We thus conclude that W represents a lower bound for the action of \ufb01eld con\ufb01gurations of topological charge W. This limit is reached for extremal con\ufb01gurations \u2202\u03bcn + \u03bc\u03bd n \u00d7 \u2202\u03bd n = 0, (9.10) on which the integral above vanishes. Since any continuous variation of these \ufb01elds leads to a non-vanishing integral, they must be stationary \ufb01eld con\ufb01gurations. We have thus managed 510 Topology to reduce the identi\ufb01cation of stationary phase con\ufb01gurations to the solution of the \ufb01rst-order di\ufb00erential equation (9.10). In passing we note that Eq. (9.10) is most elegantly solved by introducing complex coordinates z \u2261 x1 + ix2 and representing n in terms of a stereographic projection: 1 +|w |2, n3 = where w \u2208 C. Straightforward substitution then shows that Eq. (9.10) assumes the form \u2202\u00afzw(z) = 0 (exercise). This means that any meromorphic function n1 + in2 = 2w 1 \u2212 |w|2 1 +|w |2, w = W- i=1 z \u2212 ai z \u2212 bi, solves the equation. To understand the identi\ufb01cation of the order of the product, W, with the topological charge, notice that the inverse z(w) will be a W -valued function, i.e. W indeed measures the number of times the sphere (w-space) is covered by the compacti\ufb01ed plane (zspace). (Exercise: Verify that, up to a di\ufb00erent choice of the coordinate axes, the skyrmion \u03c6(1) above is one of these extremal con\ufb01gurations.) As an attractive by-product, the scheme above automatically yields the action S0[ \u00af\u03c6] =W of the extremal con\ufb01gurations \u00af\u03c6. A slightly modi\ufb01ed variant of the same trick helps to \ufb01nd other topological mean-\ufb01eld con\ufb01gurations; e.g. the famous SU(2) instantons central to the analysis of \u03b8-vacua in QCD (for a pedagogical discussion of this subject see, e.g. Ryder.13). At this stage, the further course of action seems to be clear. We ought to compute the action of the mean-\ufb01eld solutions of a given topological index, and then integrate over \ufb02uctuations. In practice, however, things turn out to be not quite as straightforward. To understand what is going on, let us revisit a problem we discussed back in Chapter 3, namely the motion of a quantum particle in a steep periodic potential. The topological nature of that system follows from the fact that, (i) by virtue of the periodicity of the potential, V (x) =V (x + a), we can identify the con\ufb01guration space with a ring of periodicity a and, (ii) with exponentially large probability, at times t \u2192 \u00b1\u221e the quantum particle will rest in any one of the minima 13 L. H. Ryder, Quantum Field Theory (Cambridge University Press, 1996). W V Figure 9.5 Snapshot of two typical \ufb01eld con\ufb01gurations of the U(1) model. Dashed: minimal \ufb01eld con\ufb01guration with W = 4 instantons. Solid: \ufb01eld with 7 instantons and 7 \u2212 W = 3 anti-instantons. t 9.3 \u03b8-", "0terms 511 of the potential (see Fig. 9.5). This latter fact implies that the two points t = \u00b1\u221e forming the boundary of the base manifold can be identi\ufb01ed (in analogy to our discussion of higherdimensional compacti\ufb01cation above). Within the path integral approach to the problem, we are thus integrating over mappings S1 \u2192 S1; the spatial distance (in units of a) our particle traverses in the course of time translates to the winding number of the mappings. This number is a topological invariant, i.e. it does not change under any continuous deformation of the path. However, what may well happen is that, on its way from the minimum at x = 0 to another one at x = W a, the particle includes one or several detours (see Fig. 9.5). As is evident from the \ufb01gure, any of those non-direct paths can be continuously deformed to a straight tunneling path 0 \u2192 a \u2192 2a \u2192 \u00b7 \u00b7 \u00b7 \u2192 W a. To connect to the notions of topology, suppose that the individual tunneling events forming a path of total winding number W are widely separated in time. By analogy with our discussion in Section 9.2, we can then imagine the full path as the result of the superposition of V \u2265 W singleinstanton (W = 1) \ufb01eld con\ufb01gurations and V \u2212W anti-instantons (W = \u22121). The homotopic equivalence of a path with a non-vanishing number of anti-instantons to the direct path (no anti-instantons) amounts to the fact that an instanton and an anti-instanton annihilate, 1 + (\u22121) \u2192 0, when their temporal coordinates approach each other; only the di\ufb00erence between the numbers of instantons and anti-instantons is a topological invariant. For steep potentials, instantons and anti-instantons are widely separated in time, implying that correlations/annihilations are vanishingly improbable. For obvious reasons, such con\ufb01gurations are referred to as dilute instanton gases. However, for su\ufb03ciently shallow potentials, instantons begin to proliferate in number. The increase in the instanton \u201cdensity\u201d gives rise to correlation e\ufb00ects or \u201cinteractions\u201d between the instantons. (Technically, the interaction of two nearby tunneling/anti-tunneling events no longer is just the sum of the two partial actions, but contains correlation terms.) Of course, these instanton liquids are much more di\ufb03cult to describe than dilute instanton gases. Clearly, these phenomena are not limited to the one-dimensional example above. For general \u03c0(M, T ) Z, instantons exist as \u201cparticles\u201d (W > 0) and \u201canti-particles\u201d (W < 0). Only the di\ufb00erence in the number of particles and anti-particles is a topological quantum number. The general plan of an instanton analysis will, therefore, typically take the following form: Solve mean-\ufb01eld equations \u2013 \ufb01nd the instantons/anti-instantons. Analyze the dilute instanton phase. Identify correlations between instantons and explore when the diluteness assumption breaks down. If possible, try to understand the physics of the correlated instanton system. In Chapter 3, we exempli\ufb01ed this program on the U(1) example (albeit not emphasizing the underlying topology). A pedagogical discussion of the generalization to the S2-instanton gas (skyrmions/anti-skyrmions) can be found in the text by Polyakov.14 However, at this stage, we will not pursue further the complexities of the dense instanton systems. Rather 14 A. M. Polyakov, Gauge Fields and Strings (Harwood, 1987). 512 Topology we shall turn to the \u2013 long overdue \u2013 discussion of examples where S2-instanton formation plays an important physical role. 9.3.3 Spin chains In Section 2.2 we applied the Holstein\u2013Primako\ufb00 transformation to explore the dispersion relation (p) of excitations in ferro- and antiferromagnetic spin systems. Within this approach we found that, for antiferromagnetic systems, the dynamics of long-wavelength spin-wave excitations is characterized by a linear dispersion relation (p) \u223c vs|p|, (9.11) where vs is the spin wave velocity. However, let us recall that this result was based on a crude semi-classical expansion valid only to leading order in 1/S, whereS is the magnitude of the spins. Yet, what happens when", " a 1/S expansion is seemingly unjusti\ufb01ed? Do interaction processes between the elementary spin-waves signi\ufb01cantly renormalize the observable excitation spectrum? At any rate, we are dealing with a \u201cstrong coupling problem\u201d for which no obvious approximation scheme is in sight. In situations like this it is usually a good idea to identify an exactly solvable reference system where the physics is known. In the present context, the S = 1/2 antiferromagnetic spin chain plays this role. Indeed, we have seen (in Problem 2.415) that, at small wavevectors, the S = 1/2 antiferromagnet is equivalently described as a system of one-dimensional chiral fermions. We know that the excitations of the fermions (charge density waves) have a linear dispersion, i.e. they obey a one-dimensional variant of Eq. (9.11). This seems quite reassuring: both, S 1 and the exactly solvable point S = 1/2 are characterized by a linear dispersion. It is then, perhaps, not too bold to speculate that, in the analytically inaccessible intermediate regime S 1, interactions will also not corrupt the linear dispersion of antiferromagnetic spin chains. Surprisingly, though, this expectation does not conform with experimental observation. Neutron scattering experiments on one-dimensional spin 1/2 antiferromagnets have indeed shown that, in the vicinity of the N\u00b4eel ordering wavevector q = \u03c0/a, the dispersion is linear. However, spin S = 1 chains show altogether di\ufb00erent behavior! It turns out that these systems do not support low-energy magnetic excitations at all (see Fig. 9.6). More generally, the cumulative experimental \ufb01nding is that antiferromagnetic chains of half-integer spin do support a relativistic low-energy excitation, while chains of integer spins are gapped. 15 To be precise, in Problem 2.4 we applied a Jordan\u2013Wigner transformation to represent the S = 1/2 chain in terms of a half-\ufb01lled system of one-dimensional fermions. The fermion system contained an interaction term whose strength was determined by the anisotropy \u0394 of the magnetic correlations. In the XY -limit, \u0394 = 0 (vanishing coupling of the z-components), the fermion system becomes free. (That is, both the fermion system and its equivalent partner, the spin system, support long-range excitations.) For general \u0394, bosonization techniques can be applied to map the problem onto a two-dimensional sine\u2013Gordon model. This model falls into the universality class of the two-dimensional (!) classical XY -model (cf. discussion on page 469). Translated back to the context of the spin chain, the RG \ufb02ow behavior of the latter implies that, for values of the anisotropy all the way up to the Heisenberg limit \u0394 = 1, the system remains in a gapless phase (the interaction stays irrelevant). For further discussion of the spin 1/2 chain we refer to A. O. Gogolin, A. A. Nersesyan, and A. M. Tsvelik, Bosonization and Strongly Correlated Systems (Cambridge University Press, 1998). 9.3 \u03b8-0terms 513.0 0.8 0.6 0.4 0.2 0.0 0 0.00 ). u. l. r ( a Q 0.5 0.4 0.3 0.2 0.1 1.25 1.00 Qc (r.l.u.) 0.5 1.0 Qc (r.l.u.) 1.5 0.25 0.50 0.75 CsNiCl3 T = 6.2 K Ei = 30 meV 1.5 1.0 0.5 Qc (r.l.u.) 1.50 1.75 Figure 9.6 Neutron scattering data from the S = 1 compound CsNiCl3. The main panel shows the excitation energy for wavevectors close to the antiferromagnetic nesting vector Qc = \u03c0. In contrast to half integer spin systems, the spectrum is gapped. (Reprinted with permission from M. Kenzelmann, R. A. Cowley, W. J. L Buyers, et al., Properties of Haldane excitations and multiparticle states in the antiferromagnetic spin-1 chain compound CsNiCl3, Phys. Rev. B 66 (2002), 24407. Copyright (2002) by The American Physical Society.) As a rule, physical phenomena depending on the parity of an integer quantum number (presently, 2S being even or odd) tend to be of", " topological origin. To understand why topology appears in the present context recall that, classically, the con\ufb01guration space of a spin is a sphere (of radius S). The spin con\ufb01guration of a spin chain is thus described by some mapping from (1 + 1)-dimensional space-time into the sphere. Compacti\ufb01cation of space-time then leads to the mappings S2 \u2192 S2 discussed above. To substantiate this picture, let us start out from the quantum partition function of an isolated spin derived in Section 3.3: Z (1) = Dn eiS \u03b2 0 d\u03c4 LWZ(n,\u2202\u03c4 n), LWZ(n, \u2202\u03c4 n) = (1 \u2212 cos(\u03b8)) \u02d9\u03c6, where the integration extends over all paths n : \u03c4 \u2192 n(\u03c4 ) and (\u03c6, \u03b8) are two angles parameterizing the unit vector n. INFO In the light of the discussion above, the action of this integral looks rather suspicious: it is purely imaginary and remains invariant under reparameterizations of time \u03c4 \u2192 \u03c4 (\u03c4 ) \u2013 both hallmarks of topological terms. Indeed, LWZ[n, \u2202\u03c4 n] is the Lagrangian of a \u201cWess-Zumino action\u201d.16 For the topological character of these actions, and their connection to the \u03b8-terms discussed presently, see the next section. 16 Notice that, from the point of view of a purist, the notation LWZ(n, \u2202\u03c4 n) is problematic. As discussed earlier, the Wess\u2013Zumino action does not admit a globally coordinate invariant representation in terms of n and its derivatives. To formulate one, we have to deploy an explicit coordinate representation. For the underlying reason, see the next section. 514 Topology The generalization of the (0 + 1)-dimensional path integral to a (1 + 1)-dimensional \ufb01eld integral of the spin chain is a good exercise in guessing e\ufb00ective actions. As a warm-up exercise, we begin our discussion with the ferromagnetic spin chain. The interaction between the neighboring spins is mediated by an operator \u2212J \u02c6Si \u00b7 \u02c6Si+1 \u2212\u2192 \u2212JS2ni \u00b7 ni+1 \u2212\u2192 JS 2 (ni \u2212 ni+1)2, (9.12) where J is the positive exchange constant, i labels the sites on the chain, and the \ufb01rst arrow maps to the representation \u02c6S \u2192 Sn of the spin operators in the \ufb01eld integral language. In the second term we have noted that, up to irrelevant constants (n2 = 1), the interaction can be represented as a \u201cdiscrete derivative.\u201d Adding to this interaction term the Wess\u2013Zumino terms of the individual spins, we are led to the partition function Z = Dn exp(\u2212S[n]), with the e\ufb00ective action S[n] = d\u03c4 (ni \u2212 ni+1)2 + iSLWZ(ni, \u2202\u03c4 ni). (9.13) i JS2 2 Anticipating that, for a ferromagnetic system, the con\ufb01guration {ni} will typically be smooth, we may then take the continuum limit to arrive at the action Sferro[n] = a\u22121 d\u03c4 dx JS2a2 2 (\u2202n)2 + iSLWZ(n, \u2202\u03c4 n), (9.14) where a denotes the lattice spacing. The action (9.14) does not contain a \u03b8-term. EXERCISE Derive the equations of motion of this action. Show that the mean-\ufb01eld dispersion of the ferromagnetic chain, \u03c9 \u223c q2, is quadratic. (Hint: Recapitulate what we know about the variation of the Wess\u2013Zumino term from Section 3.3.) We now turn to the more interesting case of the antiferromagnetic spin chain. The exchange coupling is now negative, implying that the neighboring spins prefer antiparallel alignment. We thus start out from a con\ufb01gurational ansatz ni = (\u2212)in i, where n is the antiferromagnetic order parameter \ufb01eld. To derive the antiferromagnetic analog of the continuum action (9.14), we ought to substitute this ansatz into Eq. (9.13) and perform a gradient expansion. However, rather than going through the technical details of this expansion (see e.g., Jackson17), we will here \ufb01x the structure of the action by qualitative reasoning. For one thing, we know that the system supports a wave-like mode. The minimal action consistent with the global rotational", " invariance of the model (n \u2192 Rn, whereR \u2208 O(3)) and the presence of a wave-like mode is given by S0[n] = S 4 d\u03c4 dx 1 vs (\u2202xn)2 + vs(\u2202\u03c4 n)2, 17 J. D. Jackson, Classical Electrodynamics (Wiley, 1975). 9.3 \u03b8-0terms 515 where we have relabeled n \u2192 n for notational simplicity and vs = 2aJS is the spinwave velocity. Of course, the detailed structure of the coupling constants is beyond the scope of our present plausibility argument. However, what we can say is that the overall coupling constant must be proportional to a positive power of S. This is because, for large S, \u201cinteractions\u201d between the elementary spin waves ought to become weak. EXERCISE To show this, expand n around any preferential axis (e.g. by setting n1,2 \u2261 r1,2, n3 \u2261 (1 \u2212 r2 2 \u2208 [0, 1]). Expand the action in powers of ri and show that, for large S, the contribution of anharmonic terms (spin-wave \u201cinteractions\u201d) becomes small. 2)1/2, r2 1 \u2212 r2 1 + r2 Rescaling variables, \u03c4 \u2192 v S0[n] \u2192 1 \u03bb \u22121/2 s \u03c4 \u2261 x0, x \u2192 v1/2 s x \u2261 x1, d2x \u2202x\u03bc n \u00b7 \u2202x\u03bc n, \u03bb = 4/S, \u0393 (n) n (\u03c4) assumes the form of the action of the O(3) nonlinear \u03c3-model.18 However, as argued above, we expect the full action of the problem to contain not only a dynamical piece S0 but also a \u03b8-term. How can this be made plausible from the prototypical action (9.13)? We expect that, somehow, the \u03b8-term must appear as a descendant of the Wess\u2013Zumino action. In the ferromagnetic case, the Wess\u2013Zumino actions of the spins basically added to give a contribution that was not a \u03b8-term. However, in the present case, we are dealing with a staggered spin con\ufb01guration that leaves room for more 0 dLWZ(n, \u2202\u03c4 n) as interesting things to happen. Let us \ufb01rst recall the interpretation of iS the oriented area \u0393 [n] on the sphere swept out by the curve n(\u03c4 ) (see \ufb01gure). \u03c4 By an elementary geometrical consideration, \u0393 [\u2212n] = 4\u03c0 \u2212 \u0393 [n] =\u2212 \u0393 [n] mod 4\u03c0. This implies that the continuum version of the Wess\u2013Zumino action Stop[n] = iS i (\u2212)i \u03b2 0 d\u03c4 LWZ(ni, \u2202\u03c4 ni) = iS i (\u2212)i\u0393 [ni], evaluated on a staggered con\ufb01guration (\u2212)ini must contain a spatial derivative besides the temporal derivative inherent in LWZ (since for ni = const. the factor (\u2212)i would lead to a global cancellation). Indeed, the Wess\u2013Zumino actions of two neighboring con\ufb01gurations ni+1 and ni evaluate, respectively, to the areas bounded by the curves \u0393 [ni+1] and \u0393 [ni]. For |ni+1 \u2212 ni| small, the area di\ufb00erence can be approximated by \u0393 [ni+1] \u2212 \u0393 [ni] \u03b2 0 d\u03c4 ni \u00b7 ((ni+1 \u2212 ni) \u00d7 \u2202\u03c4 n). Summing over i and taking the continuum limit, we obtain (see \ufb01gure) Stop[n] = i\u03b8 d\u03c4 dx n \u00b7 (\u2202xn \u00d7 \u2202\u03c4 n), \u03b8 =, (9.15) S 2 18 Although the action above is commonly referred to as the action of the O(3)-model, a better terminology would be the (O(3)/O(2))-model. Indeed, the degrees of freedom of the theory span the 2-sphere S2 O(3)/O(2). Although the nonlinear \u03c3-model on the sphere has to be distinguished from the \u03c3-model on O(3) (cf. the discussion of Section 8.5 and Problem 8.8.3) we will follow the widespread convention to describe the model above as the O(3)-model. 516 Topology which we identify as our earlier representation of the \u03b8-term. (Notice that the integral d\u03c4 dx really represents an integral over the sphere. This is because we agreed that, for |\u03c4 | \u2192", " \u221e and/or |x| \u2192 \u221e, all trajectories approach a common reference point on S2.) Summarizing, we have obtained S0[n] + Stop[n] as the e\ufb00ective \u0394\u0393 ni +1 ni action of the antiferromagnet. In Problem 8.8.3, we have seen that at large length scales a two-dimensional system described by the action S0[n] \ufb02ows into a disordered phase. But how will the presence of the topological term modify this behavior? To get some idea of what might happen, let us reformulate the partition function as a sum over disjoint topological sectors, Z = DnW e2\u03c0iSW e\u2212S0[nW ], (9.16) W \u2208Z where nW denotes \ufb01eld con\ufb01gurations of winding number W. Equation (9.16) provides a preliminary explanation for the observation of the half-integer/integer spin staggering phenomena mentioned above. For integer spin, exp(2\u03c0iSW ) = 1 and the topological term is not operational. By contrast, for half-integer spin, exp(2\u03c0iSW ) = (\u2212)W, i.e. consecutive topological sectors are weighted by alternating signs. Notice that the topological term is susceptible to the parity of 2S and nothing else. To understand heuristically the consequences of this feature we emphasize that, at the mean-\ufb01eld level (this stabilized by large S), the partition function is governed by a mode with linear dispersion. Quantum \ufb02uctuations around this con\ufb01guration will alter its dispersion, potentially by the creation of an excitation gap. Now, in the integer case, these \ufb02uctuations additively (exp(\u2212S0) \u2208 R+) contribute to the partition function. For small S they may (and in fact do) totally mask the mean-\ufb01eld sector. By contrast, for half-integer spin, \ufb02uctuations contribute with alternating sign, thereby partially canceling each other; the mean-\ufb01eld sector has a better chance to survive. These observations form the basis of Haldane\u2019s conjecture,19 according to which spin chains of integer S will \ufb02ow into a disordered phase with no long-range excitations whilst, in chains of half integer spin, they remain in a gapless phase. Reassuringly, the predictions born out of these rather abstract constructions are in full agreement with neutron scattering measurements of the dispersion of various quasi-onedimensional magnets. But why should one, nonetheless, use the attribute \u201cconjecture\u201d (as opposed to, say, \u201ctheory\u201d)? The scenario above is based on a number of shaky suppositions. Most seriously, it is tacitly assumed that the \u03c3-model will seamlessly scale into the strong coupling region without changing its form. (That the model remains form invariant under renormalization is a perturbative prediction which can be trusted only for \u03bb 1.) Indeed, a subsequent analysis by A\ufb04eck and Haldane20 has shown that, en route to the strong coupling regime, much more drastic things happen. Speci\ufb01cally, the appropriate critical 19 F. D.M. Haldane, Nonlinear \ufb01eld theory of large-spin Heisenberg antiferromagnets: semiclassically quantized solitons of the one-dimensional easy-axis N\u00b4eel state, Phys. Rev. Lett. 50 (1983), 1153\u20136. 20 I. A\ufb04eck and F. D. M. Haldane, Critical theory of quantum spin chains, Phys. Rev. B 36 (1987), 5291\u2013300. 9.3 \u03b8-0terms 517 theory describing the behavior of half-integer chains at large length scales turns out to be not the O(3) nonlinear \u03c3-model but rather a \ufb01eld theory with SU(2)-valued degrees of freedom. We discuss this theory in Section 9.4.4 after the concept of Wess\u2013Zumino terms has been introduced. Introducing another group of phenomena where mechanisms of topology are crucial, we now turn to a discussion of the quantum Hall e\ufb00ects (QHE). The QHE belongs here because (a) it is surely a compulsory part of any modern treatise on condensed matter phenomena, and (b) it is, in many respects, a topological phenomenon. On the other hand, our discussion will lead us somewhat astray inasmuch as it involves, necessarily, a review of experimental observation, the elementary quantum mechanics of electrons in the presence of strong magnetic \ufb01elds, etc. Readers", " wishing to maintain a more streamlined discussion of topology in condensed matter \ufb01eld theory are therefore invited to skip the next section, and turn directly to the section on Wess\u2013Zumino terms below. 9.3.4 Integer quantum Hall e\ufb00ect In fact, it is quite misleading to talk about the quantum Hall e\ufb00ect: at the least, one should speak of the quantum Hall e\ufb00ect\u201cs\u201d \u2013 a spectrum of quite di\ufb00erent phenomena, almost unparalleled in diversity and conceptual depth. To get a preliminary impression of the phenomena, observe the raw experimental data displayed in Fig. 9.7. The \ufb01gure shows the Hall resistance \u03c1xy \u2261 RH and the longitudinal resistance \u03c1xx (the ragged curve in the bottom of the plot) of a two-dimensional electron gas as functions of a strong perpendicular magnetic \ufb01eld. Instead of a dull linearly increasing curve (the classical Hall resistance \u03c1xy) or an approximate constant (the Drude resistance \u03c1xx) one \ufb01nds a pro\ufb01le that could hardly be more structured. On close inspection of the data, one may notice a number of characteristic sub-structures: (1) Shubnikov\u2013de Haas oscillations at small magnetic \ufb01elds, followed by (2) the characteristic quantum Hall plateaus \u03c1xy = \u03bd\u22121h/e2 at rational \u201c\ufb01lling fractions\u201d (see below) \u03bd \u2208 Q to which the e\ufb00ect owes its name. These are accompanied (3) by a dramatic drop in the longitudinal resistance \u03c1xx. The functional form of the increase (4) from one plateau to the next is described by certain well-de\ufb01ned power laws as a function of temperature, indicative of a second-order phase transition \u2013 the zero-temperature quantum Hall transition. Barely visible, (5) a second generation of Shubnikov\u2013de Haas oscillations is observed at \u03bd = 1/2. Finally, there appears much structure in the pattern of rationals \u03bd for which plateaus are found. For example, for some low-lying rationals (such as \u03bd = 1/4) no plateau is formed. The set of rationals for which the e\ufb00ect occurs is known as the quantum Hall hierarchy. These are but a few of the most striking observations gathered under the label quantum Hall e\ufb00ect. Undoubtedly, to account even super\ufb01cially for all of these phenomena would present a task that is well beyond the scope of the present text. Rather, we will have to restrict ourselves to a brief (and, alas, painfully super\ufb01cial) discussion of a number of conceptual basics. In the present section we shall focus on the discussion of the conductance plateaus observed at integer \ufb01lling factors \u2013 the so-called integer quantum Hall e\ufb00ect 518 Topology 3 2.5 2 1..5 R 4/3 5/3 2/5 1/3 3/5 2/3 4/7 5/9 3/7 4/9 0 0 5/7 4/5 7/5 10 Magnetic field (T) 20 30 Figure 9.7 Hall conductance and longitudinal conductance of a two-dimensional electron gas as functions of a perpendicular magnetic \ufb01eld. For a discussion, see the main text. (Reprinted with permission from H. L. St\u00a8ormer, D. C. Tsui, and A. C Gossard, The fractional quantum Hall e\ufb00ect, Rev. Mod. Phys. 71 (1999), S298\u2013305. Copyright (1999) by the American Physical Society.) (IQHE). The generalization to plateaus at rational \ufb01lling fractions \u2013 which, curiously, hinges on altogether di\ufb00erent physical concepts \u2013 is discussed in Section 9.5.1 below. When speci\ufb01ed in units of the conductance quantum e2/h, the plateau conductance \u03c3xy = \u03c1\u22121 xy of the IQHE is an integer. By its very nature an integer cannot vary continuously (upon changing some physical control parameter). Indeed, more often than not, observables quantized in integer (or rational) units are linked to some topological origin. In the following, we will formulate two di\ufb00erent \u201cexplanations\u201d of the quantization phenomenon. The quotes indicate that, far from being rigorous, both lines of argument involve some degree of \u201cbootstrap character\u201d: assuming that the e\ufb00ect is of topological nature, we feel free to subject the \u201creal world\u201d arrangement of a QHE experiment to all kinds of abuses (deformation of the sample boundaries, etc.) to then discover", " that, yes, the Hall conductance emerges as a topological invariant. The backbone of a more rigorous approach to the problem is discussed in Section 9.3.7 below. 9.3.5 Background information on the IQHE Let us begin our discussion with a nutshell summary of the phenomenology of the IQHE. This is followed by a brief reminder of the phenomenology of Landau level quantization \u2013 formulated in a language that highlights the symmetries of the problem. 2DEG B Vx 9.3 \u03b8-0terms 519 Iy The QHE is observed in two-dimensional electron gases subject to a strong magnetic \ufb01eld. The prototypical setup of a QH experiment is shown in the \ufb01gure. A voltage drop Vx applied across the sample induces a current Ix in the x-direction plus, by virtue of the Lorentz force, a Hall current Iy in the y-direction. The relation between Ix and Iy determines the Hall conductance. More precisely, the key quantities of interest are the entries \u03c3xx and \u03c3xy determining the conductance tensor \u03c3xx \u2212\u03c3xy \u03c3 =, \u03c3xy \u03c3xx which is de\ufb01ned as usual by I = \u03c3V where I = (Ix, Iy)T, V = (Vx, Vy)T (note that, by symmetry, \u03c3xx = \u03c3yy). The inverse \u03c1 = \u03c3\u22121 de\ufb01nes the resistance tensor V = \u03c1I. For a system of linear extension L, conductance g and conductivity \u03c3 are connected by the relation g = \u03c3Ld\u22122. In two dimensions, the two quantities coincide. This implies that no device-geometry-dependent factors interfere when we pass from the basic quantity determined by the microscopic physics of the system (\u03c3) to experimental observables (g); otherwise, no completely universal Hall conductance could possibly be observed. In the context of the QHE, it is truly important to keep the tensorial structure of \u03c3 and \u03c1 in mind. For example, somewhat paradoxically, for \u03c3xy = 0, a vanishing of the longitudinal resistance \u03c1xx implies a vanishing (as opposed to a divergence) of the longitudinal conductance \u03c3xx. Within the context of the QHE, the natural unit for the strength of the applied magnetic \ufb01eld is the \u201c\ufb01lling fraction\u201d. This is de\ufb01ned as the ratio \u03bd \u2261 2\u03c0N l2 0 A, (9.17) where N denotes the number of electrons in the system, A is the sample area and l0 = \u03a60/2\u03c0|B| the magnetic length, i.e. the external magnetic \ufb02ux through the area l2 0 is equal to (2\u03c0 times) one \ufb02ux quantum.21 Experimentally, one \ufb01nds that (see the \ufb01gure overleaf, courtesy of D. Leadley), for \ufb01eld strengths close to an integer \ufb01lling fraction \u03bd \u2208 N, the Hall resistance is quantized, \u03c1xy = \u03bd\u22121h/e2 to an accuracy of O(10\u221210).22 At the same time, the longitudinal resistivity/conductivity drops by as much as 13 orders of magnitude. In passing we note that the rapid oscillations visible in the \ufb01gure at small \ufb01eld strengths represent the familiar Shubnikov\u2013de Haas oscillations. 21 One may recall that the \ufb02ux quantum is de\ufb01ned through the relation \u03a60 = h/e. In our standard units = e = c = 1, \u03a60 = 2\u03c0 and l0 = |B|\u22121/2. 22 Due to the striking precision of the experimental data, the unit of electrical resistance is nowadays maintained as h/e2 = 25 812.80 \u03a9 through quantum Hall measurements. 520 Topology 8 6 0 2 4 10 12 14 a 0.8 1.5 0.6 3.0 2.0 2.5 1.0 As topologi(presumed) cal phenomenon, we expect the quantization of the Hall conductance to be robust against sample imperfections and/or the presence of static disorder. What is more surprising is that the phenomenon, in fact, critically relies on the presence of disorder. To see this, let us for a moment assume that the opposite is true: we have a homogeneous electron gas accommodated by a translationally invariant device. We further assume the absence of external voltage gradients, E = 0, so that no current is \ufb02owing in the system, I = 0. Now suppose that we observe the system from a frame moving with velocity v in, say, the 1-direction", ". An experimentalist working in that frame would observe a current density j = \u2212v\u03c1e1, where \u03c1 is the density of the electron gas. Further, the Lorentz covariance of electrodynamics implies that one would measure a \ufb01nite electric \ufb01eld E = ve1 \u00d7 B = \u2212vBe2, where B = Be3 is the applied magnetic \ufb01eld. With j1 = \u03c312E2, we obtain \u03c312 = \u03c1B\u22121 for the Hall conductivity in the moving frame. Being independent of the boost velocity, v, this result holds in all moving frames, including the static frame, v \u2192 0. We conclude that, in any translationally invariant environment, the Hall conductivity is linearly related to the magnetic \ufb01eld. 10 Magnetic field (T) 0.4 0.2 0.0 0.5 0.0 14 12 4 6 0 2 8 Reciprocating the argument above, we see that the presence of (translational invariance breaking) disorder must be necessary for the observability of the QHE. In fact, we shall see in a moment that the e\ufb00ect is born out of a conspiracy of disorder induced localization and the phenomenon of Landau level quantization. However, before turning to the discussion of the combined e\ufb00ect of these two mechanisms, let us brie\ufb02y recapitulate the formation of Landau levels in a clean two-dimensional electron gas subject to a magnetic \ufb01eld. Let us temporarily consider a geometry where the electron gas assumes the form of a perfect disk.23 To explore the quantum mechanics of this problem, we represent the inplane vector potential in the so-called symmetric gauge, Ai = (B/2)ijxj, i = 1, 2 (where the coordinates are measured with respect to the center of the disk), whereupon the free electron Hamiltonian, \u02c6H = (\u02c6p \u2212 \u02c6A)2/(2m\u2217), assumes the form \u02c6H = 1 2m\u2217 \u2212i\u22021 \u2212 x2 2l2 0 2 + \u2212i\u22022 + 2 x1 2l2 0. (9.18) (To avoid confusion with the quantum number to appear shortly below, we designate the electron mass by m\u2217.) We wish to solve the Schr\u00a8odinger equation \u02c6H\u03c8n = n\u03c8n. This task is greatly simpli\ufb01ed by subjecting the eigenvalue problem to the similarity transformation 23 Remember our fundamental working hypothesis whereby the quantization phenomena forming the QHE will not depend on details of the geometry. 9.3 \u03b8-0terms # $ 521 \u02c6H \u2192 \u02c6H = S \u02c6HS\u22121, \u03c8n \u2192 \u03c8 the transformed problem \u02c6H \u03c8 n = S\u03c8n, where S = exp n = n\u03c8 1 4l2 0 n is governed by the e\ufb00ective Hamiltonian 1 + x2 x2 2. The reason is that 2 2 (x1 + ix2) + \u2212i\u22022 + (x1 + ix2), 1 2l2 0 \u02c6H = 1 2m\u2217 \u2212i\u22021 + i 1 2l2 0 i.e. an operator whose (vector) potential depends only on the linear combination x1 + ix2 rather than on two linearly independent coordinates x1 and x2. To bene\ufb01t from this simpli\ufb01cation, we may switch to complex coordinates, z = x1 + ix2, \u00afz = x1 \u2212 ix2, in which the Hamiltonian assumes the form (exercise) \u02c6H = 1 2m\u2217 \u22124\u2202z\u2202\u00afz + 2 z l2 0 \u2202z +. 1 l2 0 This Hamiltonian possesses a family of eigenstates \u03c8 n 1/2) = B Hamiltonian is diagonal on the states (n + m\u2217 (n + 1/2). Undoing the similarity transform, one can conclude that the original \u2261 zn with eigenvalues n = 1 m\u2217l2 0 \u03c8n = zne \u2212 1 4l2 0 z \u00afz, n = \u03c9c(n + 1/2), whose eigenvalues n are the celebrated Landau levels. The Landau levels di\ufb00er by integer multiples of the cyclotron frequency \u03c9c \u2261 B/m\u2217. On the other hand, we know that, for a F/2m\u2217. system of linear extension L, there are of O(kFL)2 states below the Fermi energy k2 This implies a typical level spacing", " \u223c 1/(m\u2217L)2 which is by a factor \u223c BL2 smaller than the spacing between Landau levels. Anticipating that the clean problem does not support energies other than n, one can conclude that the Landau levels must be hugely degenerate: each of them hosting BL2 = (L/l0)2 states. To reveal the origin of the massive degeneracy of the Landau levels, we have to identify a symmetry of the Hamiltonian (9.18) or, equivalently, a set of linearly independent operators commuting with \u02c6H. In the present context, these are the \u201cmagnetic translation operators\u201d k = \u2202z \u2212 \u00afz 4l2 0, \u00afk = \u2202\u00afz + z 4l2 0, \u00afk\u03c8n = 0. (In the absence of a magnetic \ufb01eld l0 \u2192 \u221e, these would be ordinary translation operators \u223c \u2212i\u2202i transformed to complex coordinates; hence the name magnetic \u201ctranslation\u201d operators.) It is straightforward (exercise) to verify the following properties: [ \u02c6H, k] = [ \u02c6H, \u00afk] = 0, [k, \u00afk] = 1 2l2 0. We next use these operators as generators for the creation of states degenerate with the reference states \u03c8n. To this end, let us de\ufb01ne T = exp k, U = exp i 4\u03c0l2 0 L \u00afk. 2\u03c0l2 0 L 522 Topology Using the commutation relations between k and \u00afk, one may verify that these operators obey the relation T U = U T exp 4\u03c02i, 2 l0 L sometimes referred to as the magnetic algebra. We now have all elements to construct the entire set of states: the fact that \u00afk annihilates \u03c8n implies that U \u03c8n = \u03c8n. Consider now the family of states \u03c8n,m \u2261 Nn,mT m\u03c8n, where Nn,m is a normalization factor. Since \u02c6H commutes with k (and therefore with T ), \u03c8n,m form eigenstates with eigenenergy n \u2013 they all populate the nth Landau level. However, we do not yet know whether \u03c8n,m actually form a set of linearly independent states. To prove their independence, one can employ Eq. (9.19) and U \u03c8n = \u03c8n to show that U \u03c8n,m = exp(\u2212i4\u03c02m(l0/L)2)\u03c8n,m. Therefore, as with Bloch states in a periodic potential, \u03c8n,m are eigenstates of a generalized translation operator. For m = 1,..., 2\u03c0(L/l0)2, the corresponding eigenvalues are di\ufb00erent, which proves that (9.19) {\u03c8n,m|0 \u2264 m <2\u03c0 (L/l0)2}, is a linearly independent set of eigenstates in the nth Landau level. The magnitude of this set 2\u03c0(L/l0)2 coincides with our estimate of the degeneracy of the Landau levels above, i.e. we have succeeded in constructing a complete eigenbasis of the magnetic Hamiltonian. 9.3.6 IQHE as a topological phenomenon Shortly after the experimental discovery of the IQHE,24 Laughlin presented an ingenious argument whereby the quantization of the Hall conductance could be explained under fairly general conditions.25 Slightly later it became clear26 that Laughlin\u2019s argument in fact implied a number of curious features of electrons subject to a magnetic \ufb01eld and static disorder. However, before turning to a more detailed discussion of Laughlin\u2019s ideas, and their subsequent re\ufb01nement by Halperin, let us \ufb01rst outline the basic skeleton of the argument. Given its complete universality, the quantization phenomenon must be insensitive to (continuous) deformations of the sample geometry. Using this freedom, Laughlin proposed to subject the basic quantum Hall geometry to the sequence of transformations indicated in the \ufb01gure. From a \u201cHall bar\u201d geometry we pass to an annular geometry of higher symmetry. 24 K. von Klitzing, G. Dorda, and M. Pepper, New method for high-accuracy determination of the \ufb01ne-structure constant based on quantized Hall resistance, Phys. Rev. Lett. 45 (1990), 494\u20137. 25 R. B. Laughlin, Quantized Hall conductivity in two dimensions, Phys. Rev. B 23 (1981), 5632\u20133. 26 See B. I. Halperin, Quantized Hall conductance, current-carrying", " edge states, and the existence of extended states in a two-dimensional disordered potential, Phys. Rev. B 25 (1982), 2185\u201390. 9.3 \u03b8-0terms 523 In the last step of the construction, the external bias voltage is replaced by the electromotive force generated by a weakly time-dependent \ufb02ux through the annulus. Laughlin then suggested monitoring the fate of the system upon an adiabatic (i.e. in\ufb01nitely slow) variation of the \ufb02ux threading the annulus. Before discussing the response of the system in more detail, let us try to motivate this idea. We know that, for speci\ufb01c values of the annular \ufb02ux, namely integer multiples of a \ufb02ux quantum \u03c6 = 2\u03c0n, the Hamiltonian of the system is gauge equivalent to the Hamiltonian in the absence of \ufb02ux. This is because an integer multiple of the \ufb02ux quantum can be removed by the gauge transformation \u03c8a \u2192 ein\u03b8\u03c8a acting on the wavefunctions of the system. (For non-integer \ufb02uxes, this transformation alters the boundary condition \u03c8(r, 2\u03c0) = \u03c8(r, 0) and, therefore, the Hilbert space of the problem.) \u03c6 = 0 \u03c6 = 2\u03c0 Now, let us see what happens as we gradually increase the \ufb02ux from \u03c6 = 0 to \u03c6 = 2\u03c0. The situation is visualized in the \ufb01gure where each line symbolically represents a basis of eigenstates of the Hamiltonian for a given value of the \ufb02ux. Assuming that a gauge transformation has been applied to move the \ufb02ux dependence of the problem to a change in the azimuthal boundary conditions (see our discussion in Section 9.1), the eigenstates for di\ufb00erent values of the \ufb02ux are truly distinct, i.e. as one increases the \ufb02ux from \u03c6 = 0, each state moves along a path \u201cperpendicular\u201d to the collective set of eigenstates, as shown schematically in the \ufb01gure. Eventually, for \u03c6 = 2\u03c0, we arrive back at the original \u03c6 = 0 basis. (This follows from the fact that \u02c6H|\u03c6=2\u03c0 can be mapped onto \u02c6H|\u03c6=0 by a gauge transformation that does not alter the boundary conditions.) That, however, does not necessarily imply that individual basis states map onto themselves upon the completion of the path \u03c6 = 0 \u2192 \u03c6 = 2\u03c0. I.e., while the set of eigenstates as a whole gets reproduced, permutations of individual states are consistent with the gauge invariance of the problem. EXERCISE If this statement does not make much sense to you consider, as an example, a clean one-dimensional ring subject to a magnetic \ufb02ux. Explore what happens to the eigenstates of \u02c6H = 1 2m (\u02c6p \u2212 A) as the \ufb02ux is increased from 0 to 2\u03c0. Show that the non-invariance of individual states is compatible with the invariance of the global spectrum. The non-invariance of individual states upon completion of a round trip back to a gaugeequivalent Hamiltonian is a phenomenon called spectral \ufb02ow. The spectral \ufb02ow of the eigenstates of our magnetic environment lies at the heart of Laughlin\u2019s argument. Specifically, we shall see that, upon the sending of a \ufb02ux quantum through the ring, n states radially centered at the inner perimeter are pushed above the Fermi energy (n is the number of Landau levels below the Fermi energy). At the same time, n states at the outer perimeter sink below the Fermi energy. To regain thermal equilibrium, the system responds by transferring n electrons from the inner to the outer perimeter. This process takes place during the time t0 it takes to adiabatically send the \ufb02ux quantum through the system, i.e. the transverse current I2 = n/t0 (remember, e = 1). The electromotive force driving the process is V1 = \u02d9\u03c6 = 2\u03c0/t0. The corresponding Hall conductance is therefore given by 524 Topology E EF l0 \u03c9 c \u03c4\u20131 Figure 9.8 Energy levels of a quantum Hall annulus as functions of the radial coordinate. For a discussion, see the main text. r \u03c312 = I2/V1 = n/2\u03c0. Expressed in physical units, this can be cast in the form \u03c312 = ne2/h \u2013 the quantum Hall e\ufb00ect! To substantiate", " this picture, let us consider a situation in which the Fermi level EF is placed somewhere between the \ufb01rst and the second Landau level. For simplicity we shall also make the (arti\ufb01cial but physically immaterial, cf. Halperin\u2019s paper26) assumption that the disorder is con\ufb01ned to the inner regions of the sample. We require the disorder strength, as set by the inverse elastic scattering time, \u03c4 \u22121, to be smaller than the separation between Landau levels, \u03c9c\u03c4 1. This condition is, in fact, necessary to prevent the Hall plateaus from becoming \u201cwashed out.\u201d As a result we obtain the level diagram shown in Fig. 9.8. The \ufb01gure schematically shows the energies of the single-particle states as functions of the radial coordinate of the annular region. In the bulk of the annulus, disorder leads to a broadening of the Landau levels to energy bands of width \u03c4 \u22121. At the outer/inner perimeter of the annulus, the con\ufb01ning potential pushes the levels energetically up. Crucially, this implies the presence of as many Fermi energy states \u2013 \u201cedge states\u201d \u2013 as there are occupied Landau levels (in our case, just one). These states, and not so much the bulk states buried deep below the Fermi energy, are likely to be the carriers of longitudinal currents in the system. As to the bulk states, let us presume that they are localized by disorder on a length scale \u03be much smaller than the circumference L of the system. (Intriguingly, we shall soon see that this assumption leads to a contradiction.) To explore the phenomenon of spectral \ufb02ow in this environment, we need to turn to a re\ufb01ned description where individual levels are resolved. This is the subject of the following. EXERCISE We wish to explore the radial structure of the states occupying the lowest Landau level. This task is most conveniently accomplished in a basis di\ufb00erent from the set {\u03c80,m} discussed above: revisit the construction on page 520 to show that all states \u03c60,m \u2261 N0,m \u00afzmexp[\u2212 1 z \u00afz] lie in the lowest Landau level. Switch to polar coordinates (r, \u03b8) 4l2 0 and verify that these states can be approximated as 9.3 \u03b8-0terms \u03c60,m(r, \u03b8) \u2248 Nn,me\u2212im\u03b8e \u2212 3 4l2 0 (r\u2212rm)2, 525 (9.20) where rm de\ufb01nes the area through which m \ufb02ux quanta pass; \u03c0r2 reference angle \u03b80, the azimuthal current carried by these states is given by mB = mh/e. Show that, for any I \u2261 1 m\u2217 \u221e 0 dr \u02c6j(r,\u03b80) \u00b7 e\u03b8 = 1 m\u2217 B m\u2217 \u221e dr |\u03c60,m|2 0 \u221e m r \u2212 Br 2 dr |\u03c60,m|2 (r \u2212 rm). (9.21) 0 The wavefunctions \u03c60,m(r, \u03b8) describe the system far away from its boundaries. They are symmetrically centered around rm and carry angular momentum L = m. From Eq. (9.21) and the symmetry around the center coordinate rm, we further conclude that the azimuthal current carried by these states vanishes, I = 0. However, in the vicinity of the boundaries, this picture becomes perturbed. Firstly, the con\ufb01ning potential will push the states \u03c60,m up in energy (see Fig. 9.8 where the circles represent the states centered at coordinates rm close to the inner boundary of the annulus). Secondly, the boundary potential will render the states radially asymmetric, implying that the integral (9.21) no longer vanishes: azimuthal currents \ufb02ow at the boundaries. The surface currents \ufb02owing at the inner/outer perimeter are opposite to each other. We next discuss what happens as some weakly time-dependent \ufb02ux \u02dc\u03c6 is sent through the \u02dc\u03c6/r. The vector potential generalized to annulus to generate an electromotive force E = dt the presence of the \ufb01eld E takes the form A = \u03c6(r)+ \u02dc\u03c6 2r e\u03b8, where\u03c6( r) = Br2. Referring for a more detailed discussion to Halperin\u2019s paper, we note that the \ufb02ux \u02dc\u03c6 adds to the background \ufb02ux \u03c6(r). As a consequence", ", the center coordinates rm(\u03c6 + \u02dc\u03c6) \u201ccontract.\u201d Once a full \ufb02ux quantum has been added to the system, \u02dc\u03c6 = 2\u03c0, the center coordinates have contracted by one unit, rm(\u03c6 + 2\u03c0) = rm\u22121(\u03c6), and the original set of levels (and therefore the spectrum of the system) is restored. However, the individual levels have changed \u2013 spectral \ufb02ow: at the inner edge of the system, one occupied level has been pushed above the Fermi energy, at the outer edge one empty level dived below the Fermi energy (see Fig. 9.8). To repair this energy imbalance, the system will want to transfer one electron from the inner edge to the outer edge. But how will it do that? Before the advent of the QHE, it had been common wisdom that the states of a two-dimensional electron gas in the presence of disorder (such as the states in the bulk of our annulus) are localized on a certain scale \u03be. For\u03be/L 1, these states would be completely oblivious to the presence of our driving \ufb02ux \u02dc\u03c6. This would imply that, as far as the action of \u02dc\u03c6 is concerned, the inner and the outer edge of the system are decoupled. How, then, would the system know that it ought to transfer one electron between the edges. (After all, this transfer has to be mediated through the bulk.) The only way out of this dilemma is to courageously postulate (as Halperin did) that, notwithstanding the presence of disorder, there must be at least one delocalized bulk state below the Fermi energy. Subsequently, it was indeed found that the localization length diverges upon approaching the center of the Landau band. The delocalized states in the center of the band establish the contact between the edges and may act as conduits of electronic charge. 526 Topology Laughlin\u2019s gauge argument hints at the topological nature of the QHE: independent of system-speci\ufb01c details, the addition of a \ufb02ux quantum through the annulus transfers an integer charge across the system. This suggests that the e\ufb00ect should be understandable in terms of some kind of topological index lurking behind the mapping of a parameter space (presently, the amount of \ufb02ux through the ring) into the Hilbert space of the problem. Subsequently, Avron and Seiler,27 and Thouless, Kohmoto, Nightingale, and den Nijs28 indeed succeeded in rigorously identifying the quantum Hall conductance as a topological invariant \u2013 the \ufb01rst Chern class of the U (1) principal bundle over the two-dimensional torus. However, a satisfactory discussion of these ideas, for which we would need to introduce much more background in di\ufb00erential topology, would lead us too far astray. Instead, we shall head back to our prime subject, the discussion of topological concepts in low-energy \ufb01eld theories of condensed matter systems. EXERCISE As preparation for the next section, refamiliarize yourself with the \ufb01eld theory approach to the disordered electron gas introduced in Section 6.5. 9.3.7 Field theory of the integer quantum Hall e\ufb00ect Laughlin\u2019s gauge argument, and its subsequent re\ufb01nement by others, helped to unravel many of the mysteries posed by the experimental discovery of the IQHE. Speci\ufb01cally it shed some light on the conspiracy of dissipationless edge currents, disorder induced localization of bulk states, and an exotic family of delocalized states in the formation of the e\ufb00ect. On the other hand, even with Laughlin\u2019s picture in store, we are still a long way from a more-than-schematic, quantitative understanding of the e\ufb00ect. A huge step towards a full microscopic theory of the IQHE \u2013 whose \ufb01nal structure still remains partly unknown \u2013 was taken by Pruisken when he adjusted the nonlinear \u03c3-model of disordered fermion systems so as to account for the presence of a strong magnetic \ufb01eld. In the next two subsections, we will reconstruct Pruisken\u2019s \ufb01eld theory, and employ it to gain more insight into the long-range behavior of the quantum Hall system. Pruisken\u2019s \ufb01eld theory: construction As we have seen in earlier chapters of this book, much of the information about a noninteracting disordered electron system is contained in correlation functions involving the product of a retarded and an advanced single-particle Green function, @ x1| 1 EF + \u03c9 + i0 \u2212 \u02c6H |x2x", "2 | 1 EF \u2212 \u03c9 \u2212 i0 \u2212 \u02c6H |x1. dis A Depending on the choice of coordinates and the energy argument \u03c9, these functions describe the conductance of the system, the statistics of its spectrum and many other characteristics. 27 J. E. Avron and R. Seiler, Quantization of the Hall conductance for general, multiparticle Schr\u00a8odinger Hamil- tonians, Phys. Rev. Lett. 54 (1985), 259\u201362. 28 D. J. Thouless, M. Kohmoto, M. P. Nightingale, and M. den Nijs, Quantized Hall conductance in a two- dimensional periodic potential, Phys. Rev. Lett. 49 (1982), 405\u20138. 9.3 \u03b8-0terms 527 We have also seen that the long-range behavior (scales L, much larger than the elastic scattering mean free path, ) of this correlation function can be extracted from a \ufb01eld theory whose action in d = 2 is of nonlinear \u03c3-model type,29 S[Q] = \u03c0\u03bd 4 # $ d2x D tr ( \u02c6\u2202\u03bcQ \u02c6\u2202\u03bcQ) \u2212 2\u03c9 tr (Q\u03c3ar 3 ). (9.22) For later convenience, let us recapitulate a few key features of the \ufb01eld theory de\ufb01ned by Eq. (9.22). The matrix \ufb01elds Qaa,ss (x) \u223c \u00af\u03c8as(x)\u03c8as (x) describe the behavior of a product of two fermion \ufb01eld amplitudes \u03c8as. The dynamics of the latter is controlled by the advanced/retarded single-particle Green function, \u00af\u03c8as(x)\u03c8as (x) \u223c (EF + s(\u03c9 + i0) \u2212 \u02c6H)\u22121(x, x)\u03b4ss \u03b4aa. Here, the two-component index s = +/\u2212, anda = 1,..., R refers to the replica index. The \ufb01elds Q take values in the coset space U(2R)/(U(R) \u00d7 U(R)).30 A concrete repis a Pauli matrix in 3 T \u22121, where T \u2208 U(2R) and \u03c3ar resentation is given by Q = T \u03c3ar two-dimensional advanced/retarded (ar) space. 3 For \u03c9 \u2192 0 (the limit we will concentrate on throughout) the action above is invariant under two distinct symmetry transformations: global transformations T \u2192 gT, where g \u2208 G \u2261 U(2R) is constant, and local transformations T \u2192 T h(x), where h(x) \u2208 H \u2261 U(R) \u00d7 U(R), i.e. the group of all matrices ful\ufb01lling the condition [h(x), \u03c3ar 3 ] = 0. The microscopic parent action from which Eq. (9.22) was derived was, in fact, rotationally invariant under the full group U(2R). That our \ufb01elds Q live in a smaller coset space signals the fact that, in a metallic system, this symmetry is spontaneously broken: the Qs are the Goldstone modes associated with the breakdown of the symmetry from U(2R) to U(R) \u00d7 U(R). Under a generalized31 gauge transformation, \u03c8 \u2192 ei\u03c6\u03c8, theQ s transform as Q \u2192 e\u2212i\u03c6Qei\u03c6. Gauge invariance then implies that the operators \u02c6\u2202\u03bc appearing in the e\ufb00ective action must be interpreted as covariant derivatives, \u02c6\u2202\u03bc = \u2202\u03bc \u2212 i[A\u03bc, ], where A\u03bc transforms as a non-abelian vector potential, A\u03bc \u2192 e\u2212i\u03c6A\u03bcei\u03c6 \u2212 ie\u2212i\u03c6\u2202\u03bcei\u03c6. (Notice, however, that the \u201cphysical vector potential\u201d generating the perpendicular magnetic \ufb01eld Aphys \u223c \u03b4ab\u03b4ss is diagonal in replica and ar spaces and, therefore, does not enter the covariant derivative.) In a series of famous papers, Pruisken32 extended the formalism above so as to account for the e\ufb00ect of a strong magnetic \ufb01eld. It turned out that a key player in the action of that 29 Here we are using the reduced variant of the model (see Problem 8.8.4) suitable to compute the product of a retarded and an advanced Green function. 30 Due to the presumed presence of a massive magnetic \ufb01eld, time reversal symmetry is broken and the introduction of", " a \u201ctime-reversal space\u201d unnecessary. 31 By \u201cgeneralized\u201d we mean that \u03c6 can be a matrix in ar space as well as in replica space. 32 For a review, see A. M. M. Pruisken, Field theory, scaling and the localization problem, in The Quantum Hall E\ufb00ect, ed. R. E. Prange and S. M. Girvin (Springer-Verlag, 1987). 528 Topology generalized \ufb01eld theory was a certain variant of a \u03b8-term. As with many other examples before, there are two ways to obtain this action: one may generalize the derivation of the \u03c3-model discussed in Section 6.5 for the presence of a strong magnetic \ufb01eld and derive the \u03b8-term from \ufb01rst principles. For the details of this not entirely straightforward program we refer to the original papers. Alternatively, one may guess the structure of the generalized action on the basis of symmetry arguments and \ufb01x the coupling constants by running some consistency checks. It is this second strategy that we shall pursue shortly. \u03b42 \u03b4A\u03bc \u03b4A\u03bd As discussed in Chapter 7, elements of the conductivity tensor \u03c3\u03bc\u03bd can be obtained by Z[A] of the partition or generating function with respect a two-fold di\ufb00erentiation to some generalized vector potential. Referring below for a more detailed discussion, all we shall aim for presently is to understand what kind of action is needed to produce a \ufb01nite Hall conductivity \u03c312 \u223c \u03b42 Z[A]. By symmetry, the mixed derivative computed on the \ufb01eld theory de\ufb01ned by Eq. (9.22) vanishes (a point that warrants some consideration). Rather, we have to look out for an operator comprising the two long derivatives \u02c6\u22021Q and \u02c6\u22022Q in a tr( \u02c6\u22021Q \u02c6\u22022Q). This single local expression. For example, we might contemplate a term like expression, however, is again not permitted by symmetry. The reason is that (unlike the rotationally invariant two-derivative operator in Eq. (9.22)) it is not form invariant under rotation of the coordinate axes (again something to think about). However, the next obvious choice, \u03b4A1 \u03b4A2 Stop[Q] = \u03b8\u03bc\u03bd d2x tr(Q \u02c6\u2202\u03bcQ \u02c6\u2202\u03bdQ), (9.23) does the job. For one thing, this term is rotationally invariant. Secondly, the de\ufb01nition of the long derivatives \u02c6\u2202\u03bc implies that it contains terms linear in the combination A1A2, from where we conclude that the (as yet undetermined) coupling constant \u03b8 must have something to do with the Hall conductivity. To be somewhat more speci\ufb01c, let us draw on Problem 7.6.4 where it has been shown that a source-vector potential suitable for the calculation of the conductance takes the form A\u03bc = U \u22121\u2202\u03bcU with U = exp(i(x1\u03ba1\u03c3ar 11 is a projector onto the \ufb01rst replica channel and \u03ba\u03bc are numbers). With this choice, 11) (here,E r 1 + x2\u03ba2\u03c3ar 2 ) \u2297 Er \u03c311 = lim R\u21920 1 4\u03c0L2 \u22022 \u03ba1\u03ba1 Z, \u03ba=0 \u03c312 = lim R\u21920 1 4\u03c0iL2 \u22022 \u03ba1\u03ba2 Z. \u03ba=0 We also know that the conventional Drude theory of a weakly disordered metal is obtained by setting Q = \u03c3ar 3. (Remember that \ufb02uctuations around the origin of the \ufb01eld space, \u03c3ar 3, describe mechanisms of localization, i.e. quantum e\ufb00ects beyond the Drude picture.) We next use this information to determine the coupling constants of the theory. To this end, we go to the Drude level (set Q = \u03c3ar 3 ), substitute the source A\u03bc into our long derivatives, and evaluate the action. This leads to (exercise) S[Q = \u03c3ar 3 ] = 2\u03c0L2D\u03bd(\u03ba2 1 + \u03ba2 2) + 16iL2\u03b8\u03ba1\u03ba2. Evaluation of the derivatives above on Z[A]| result \u03c30 exp(\u2212S[\u03c3ar 11 = 2\u03c0D\u03bd for the longitudinal Drude conductance \u03c30 Drude 3 ]) then readily leads to the 11 (which we knew anyway), 9.3", " \u03b8-0terms 529 and to the suspected identi\ufb01cation \u03b8 = \u2212\u03c30 Hall conductance. Summarizing, we obtain Pruisken\u2019s action of the IQHE, 12/8 of the coupling constant \u03b8 with the (Drude) S[Q] = 1 8 # $ d2x \u03c311 tr( \u02c6\u2202\u03bcQ \u02c6\u2202\u03bcQ) \u2212 \u03c312\u03bc\u03bdtr(Q \u02c6\u2202\u03bcQ \u02c6\u2202\u03bdQ). (9.24) What makes the identi\ufb01cation \u03b8 \u223c \u03c312 more interesting is that \u03b8 (and therefore the Hall conductance) actually plays the role of a topological angle; the second term in Eq. (9.24) is a topological term. That such a term might be present in our \ufb01eld theory follows from the fact that \u03c02(U(2R)/(U(R) \u00d7 U(R))) = Z. Unfortunately, we do not have the mathematical background to prove this result.33 We can, however, make it plausible; and we can demonstrate that Stop above is a representation of the corresponding \u03b8-term. To this end, let us temporarily focus on a single replica channel (a = 1, say) and consider the \ufb01eld con\ufb01guration \u02dcQ11(x) =U (x)\u03c3ar \u02dcQab(x) =\u03b4 ab\u03c3ar 3, a = 1, 3 U \u22121(x) \u2261 n(x) \u00b7 \u03c3ar, where U (x) \u2208 U(2). The second equality in the \ufb01rst line de\ufb01nes a unit vector n(x). It expresses the fact that the projection of the \ufb01eld space onto a single replica channel is isomorphic to S2.33 Evaluating Stop on this particular \ufb01eld con\ufb01guration, it is straightforward to verify that Stop[ \u02dcQ] =i d2x n \u00b7 (\u22021n \u00d7 \u22022n), an expression we identi\ufb01ed earlier (see Eq. (9.8)) as the topological term of a two-dimensional \ufb01eld theory on the sphere. We can now generalize from our particular \u02dcQ to \ufb01eld con\ufb01gurations Q = T (x) \u02dcQ(x)T \u22121(x) whereT (x) \u2208 U(2R). However, using arguments similar to those employed in Section 9.3.1, one may convince oneself that small variations T will not change the value of the action Stop. Indeed, Stop is the general \u03b8-term on the coset space U(2r)/(U(r) \u00d7 U(r)). \u03c312 2 Pruisken\u2019s \ufb01eld theory: long-range physics Now that we have \u201cderived\u201d the Pruisken action, the next question to ask is what to do with it. To begin with the bad news, the long-distance behavior of the model is still pretty much unknown and, in fact, a subject of ongoing research. (That this is the state of a\ufb00airs some 20 years after its derivation signals the fact that we are dealing with a very rich \ufb01eld theory.) However, by investing one\u2019s physical insight in the quantum Hall problem (and with a little bit of good will) quite a few things about the model can, nonetheless, be said. Suppose we were dealing with a system of annular geometry, similar to that discussed in the last section. Let us further assume that the Fermi energy lies in between the center of the nth and (n + 1)th Landau bands, so that there are no delocalized states at EF. As discussed in the previous section, the bulk of the sample is then pretty much impervious to external 33 Notice, however, that in the special case R = 1, U(2)/(U(1) \u00d7 U(1)) \u223c= SU(2)/U(1) \u223c= S2 is the 2\u2013sphere. What is non-trivial here is the generalization of \u03c02(U(2)/(U(1) \u00d7 U(1))) = \u03c02(S2) =Z to general R. 530 Topology perturbations, and the interesting Fermi-energy physics takes place at the boundaries (the inner and outer perimeters). Indeed, it is possible to rewrite the topological action as a pure boundary operator. A straightforward application of Stokes\u2019 theorem34 shows that Stop[T ] = \u03c312 2 ds \u00b7 tr(T \u03c3ar 3 \ufffd", "\ufffdT \u22121), \u2202M (9.25) where the integral runs over the boundary \u2202M of the annulus. Notice, however, that Eq. (9.25) is not represented in terms of the fundamental degrees of freedom of the theory 3 T \u22121). As (the Qs) but rather relies on a particular \u201ccoordinate\u201d representation (Q = T \u03c3ar discussed in Section 9.4, this is not accidental but rather re\ufb02ects a fundamental property of topological terms. For future reference, we also anticipate that the boundary descendant of the bulk \u03b8-term Eq. (9.23) is an example of a Wess\u2013Zumino term. 3 T \u22121 \u2192 T h\u03c3ar We next show that, under the conditions stated above, the boundary representation of the theory raises a consistency problem unless \u03c312 = integer. To appreciate the problem, recall that the bulk representation of the theory is invariant under local transformations T (x) \u2192 T (x)h(x), where h \u2208 H = U(R) \u00d7 U(R). This follows trivially from the fact that 3 h\u22121T \u22121 = Q are invariant. Now, under its degrees of freedom Q = T \u03c3ar our present working conditions \u2013 Fermi energy in a mobility gap \u2013 there are no Q-\ufb01eld excitations in the bulk of the system (the Qs describe mobile Fermi energy excitations), and the theory reduces to two decoupled boundary theories which must be separately invariant. In this respect, the only contribution that may potentially cause trouble is the topological boundary contribution Eq. (9.25) (since it is not a functional of the invariant degree of freedom, Q). Indeed, let us consider the speci\ufb01c choice h = exp(i\u03c3ar 3 \u03b8), where \u03b8 is the azimuthal coordinate of the system. Focusing on the e\ufb00ect of this transformation at the inner perimeter, say, we obtain Stop[T ] = \u03c312 2 = Stop[T ] + = Stop[T ] + 2\u03c0 0 R\u03c312 2 R\u03c312 2 3 \u2202\u03b8T \u22121) \u2192 \u03c312 2 d\u03b8 tr(T \u03c3ar 2\u03c0 2\u03c0 0 d\u03b8 tr(\u03c3ar 0 tr(\u03c3ar 3 ln h\u22121) 3 h\u2202\u03b8(h\u22121)) = Stop[T ] + 2\u03c0 0 = Stop[T ] \u2212 2\u03c0i\u03c312R. d\u03b8 tr(T h\u03c3ar 3 \u2202\u03b8(h\u22121T \u22121)) R\u03c312 2 2\u03c0 0 d\u03b8 \u2202\u03b8 tr(\u03c3ar 3 ln h\u22121) (9.26) The invariance of the exponentiated (!) boundary action requires that exp(2\u03c0iR\u03c312)!= 1,!= integer. Summarizing, we have seen that, for Fermi from which we conclude that \u03c312 energies in between two Landau band centers (that is, conditions where QH plateaus are experimentally observed), the intrinsic consistency of the theory requires quantization of the Hall conductance. A re\ufb01ned variant of such arguments (see Pruisken\u2019s aforementioned article) shows that the value of the conductance indeed coincides with the number of Landau levels below the Fermi energy. Notice also that the reasoning above is again \u201ctopological.\u201d This time, the key players are the winding numbers of the mappings S1 \u2192 U(1) \u00d7 U(1) from 34 For a review of the general form of Stokes\u2019 theorem, see Section 9.4.2 below. 9.3 \u03b8-0terms 531 the boundary manifold (topologically, a circle) into the projection of the local transformation group to a single replica channel (U(1) \u00d7 U(1)). EXERCISE As an exercise in group manipulations, extend the argument given above to more general transformations at the boundary. To this end, consider a general (but single-valued!) boundary transformation h(\u03b8) \u2208 H and show that we will not obtain information beyond the quantization criterion discussed above. Hint: Use the facts that U(r) = U(1) \u00d7 SU(r), and that \u2200R \u2208 SU(r) : 0 = ln det R = tr ln R. INFO The argument above focuses on the boundaries of the system. As a brief digression, let us show how the same information can be obtained from a bulk picture. In the bulk, all Fermi energy states are localized. Technically, this implies that the longitudinal conductance L\u03be\u2212\u2192 0, where the notation is meant to indicate that, in a ren", "ormalizarenormalizes to zero, \u03c311 tion group sense, the conductance scales to zero on length scales L \u03be much larger than the two-dimensional localization length. An alternative formulation of the same fact is to say that transformations \u03c8(x) \u2192 T (x)\u03c8(x), where T (x) \u2208 U(2R) \ufb02uctuates on scales \u03be, must leave the theory invariant.35 For \u03c311 \u2192 0, and frequency di\ufb00erences \u03c9 \u2192 0, the non-topological sector of the action vanishes and is therefore trivially oblivious to \ufb02uctuations of the Qs. However, this is not the case with the topological action. No matter how slowly it \ufb02uctuates in space, a topologically non-trivial con\ufb01guration Q(x) will have topological action S[Q] = 2\u03c0\u03c312n, where n is the corresponding winding number. The only way to make the theory generally impervious!= integer \u2013 the bulk variant of the quantization criterion. to Q-\ufb01eld \ufb02uctuations is to require \u03c312 Quantum Hall transition Having understood the basic quantization phenomenon, we might now ask what other features of the quantum Hall system can be extracted from the \ufb01eld theory approach. Given that the nonlinear \u03c3-model arguably represents the most powerful approach to disordered electron systems in general, we might, for example, be ambitious enough to seek more information about the nature of the delocalized states expected to reside at the centers of the Landau bands. As we saw above, these states do play a vital role in the formation of the QHE. However, as long as they are deeply buried below the Fermi energy, they in\ufb02uence the system rather indirectly. We should expect them to become much more vivid as the Fermi energy sweeps through the center of a Landau band. Indeed, as EF \u2192 n\u03c9c approaches a band center, the system begins to build up long-ranged correlations. This is because the physics at the Fermi energy is now controlled by ever more extended, or delocalized, states. Formally, EF \u2192 (n+1/2)\u03c9c goes along with a diverging correlation length \u03be, a phenomenon indicative of a second-order phase transition. Indeed, E\u2217 = (n + 1/2)\u03c9c marks the position of a very peculiar (and still not fully understood) quantum phase transition, the quantum Hall transition. 35 Indeed, we had argued earlier that metallic behavior goes along with a spontaneous breakdown of replica rotation symmetry. Our Qs are the Goldstone modes of this phenomenon. Conversely, localization is accompanied by a restoration of this symmetry; the Goldstone modes disappear and the theory will no longer respond to \ufb02uctuations of the Qs (or theT s for that matter). Further, notice that we are now concentrating on the \u201clarge\u201d transformation group of the model, U(2R), and not the \u201csmall\u201d invariance group U(R) \u00d7 U(R). 532 Topology delocalized localized E E Figure 9.9 On the behavior of the longitudinal conductance, Hall conductance, and density of states, \u03c1, at the quantum Hall transition. In Fig. 9.9, the behavior of the three most relevant players in the system, longitudinal conductance, Hall conductance, and density of states, is shown as functions of the Fermi energy. In the vicinity of the Landau band centers the system becomes critical. Right at a transition point, EF \u2261 E\u2217, (a) the longitudinal conductance \u03c311 = 1/2, (b) the Hall conductance \u03c312 = n + 1/2 is half integer, and (c) the correlation length \u03be characterizing the spatial pro\ufb01le of wavefunctions at the Fermi energy has diverged. The latter implies that, for any system of \ufb01nite size L, there is a whole range of energies around the critical value for which \u03be > L. Within this energy range, the wavefunctions are e\ufb00ectively delocalized. Naturally, the width of the band of delocalized states shrinks upon increase of the system size and, in the thermodynamic limit, approaches zero. More precisely, upon approaching the band center, the correlation length diverges as \u03be \u223c |E \u2212 E\u2217|\u2212\u03bd, where \u03bd de\ufb01nes the correlation length exponent and we have introduced \u0394E \u2261 |E \u2212 E\u2217| as a relevant scaling variable. For a given system size, the width of the energy band, \u0394E, of delocalized states is determined by the condition L \u223c \u03be(\u0394E), or \u0394E", " \u223c L\u22121/\u03bd. The number \u223c L\u22121/\u03bd+2, where \u03b4 \u223c L\u22122 is the twoof states within that window scales as N \u223c \u0394E \u03b4 dimensional level spacing. Unfortunately, there is still no reliable analytical prediction for \u03bd. (We shall see in a moment why this is so.) However, high-precision numerical analyses36 have shown that \u03be = 2.35 \u00b1 0.08. This implies that the number of states within the delocalized region diverges in the thermodynamic limit: even though the width of the delocalized energy window approaches zero, it hosts a continuum of extended states. We shall next explore the extent to which these features can be understood from the \ufb01eld theory approach. To this end, let us imagine the partition function formally expanded in 36 See, e.g., B. Huckestein, Scaling theory of the integer quantum Hall e\ufb00ect, Rev. Mod. Phys. 67 (1995), 357\u201396. 9.3 \u03b8-0terms 533 terms of the topological index of \ufb01eld con\ufb01gurations: Z = e2\u03c0iW \u03c312 ZW, W where ZW is the partition function reduced to the sector of \ufb01elds of winding number W, and Stop = 2\u03c0iW \u03c312 enters as a topological phase. In general, there is not much we can say about ZW, other than that it will be small: |Zn| \u03c3111 1. Indeed, an estimate similar to the one employed on page 509 yields (exercise) 0 \u2264 1 2 d2x tr [(\u2202\u03bcQ + i\u03bc\u03bdQ\u2202\u03bdQ)(\u2202\u03bcQ + i\u03bc\u03bbQ\u2202\u03bbQ)] = d2x tr [\u2202\u03bcQ \u2202\u03bcQ \u2212 i\u03bc\u03bdQ \u2202\u03bcQ \u2202\u03bdQ] = d2x tr(\u2202\u03bcQ \u2202\u03bcQ) \u2212 16\u03c0W. From here, we conclude that the non-topological contribution to the action obeys the inequality S0[Q] = \u03c311 8 d2x tr( \u02c6\u2202\u03bcQ \u02c6\u2202\u03bcQ) \u2265 2\u03c0W \u03c311, and that Zn \u223c exp(\u22122\u03c0W \u03c311) is weighted by a small \u201cenergetic\u201d factor. We have had ample opportunity to see that such factors can, in principle, be compensated for by large \u201centropic\u201d counterweights. However, Pruisken has shown that, in the present context, this does not happen. Similarly, the functional expectation values for longitudinal and Hall conductance can be organized in an instanton series: \u03c3ij = \u221e W =0 e2\u03c0iW \u03c30 12 C (W ) ij (\u03c30 11). ij The coe\ufb03cients C (W ) appearing in this series depend \u2013 in the exponentially sensitive way discussed above \u2013 on the longitudinal Drude conductance. Importantly, the righthand side of the series depends on the Drude values \u03c30 ij while the left-hand side sets the true renormalized conductance (i.e. the second-order derivative of the functional with respect to a generalized source-vector potential). There is not much more we can say about the structure of these series except for one important symmetry criterion: under a change of orientation of the coordinate system, \u03c312 changes sign while \u03c311 does not. Speci\ufb01cally, \u03c312(B) = \u2212\u03c321(B), \u03c311(B) = +\u03c322(B), which is an example of an Onsager relation. Consistency with these relations requires that the topological series be of the form \u03c311 = \u03c30 11 + \u03b4\u03c311 + \u221e W =1 cos(2\u03c0W \u03c30 12) a(W )(\u03c30 11), \u03c312 = \u03c30 12 + \u221e W =1 sin(2\u03c0W \u03c30 12) b(W )(\u03c30 11), (9.27) 534 Topology where a(W ) and b(W ) are expansion coe\ufb03cients. The notation emphasizes that, in the topologically trivial sector, W = 0, the longitudinal conductance may be subject to renormalization while \u03c312 remains unrenormalized.37 In view of these structures, let us now speculate a little on the renormalization characteristics of the model. Here, renormalization means, as usual, that we consider the outcome of the theory on ever increasing length scales. We have to keep in mind, however, that this program must be carried out for each topological sector separately: renormalization,", " i.e. the successive elimination of fast \ufb02uctuations, cannot change the topological index of a \ufb01eld con\ufb01guration. On the same footing a topological angle (presently, the coe\ufb03cient \u03c30 12 of the topological term) should not renormalize, at least not in a conventional sense. (For the cautious formulation, see below.) \u03c3 \u03c3 11 11 Consider, then, the \ufb02ow (\u03c312(\u03bb), \u03c311(\u03bb)) in the twodimensional parameter plane de\ufb01ned by longitudinal and Hall conductance as we increase a reference length scale \u03bb. From Eq. (9.27) it follows that there are two families of lines on which \u03c312 does not renor\u2208 N + 1/2. As \u2208 N and \u03c312 = \u03c30 malize: \u03c312 = \u03c30 12 12 to the \ufb01rst line, we do not expect any critical behavior in the vicinity of integer \u03c312. (Remember that the quantum Hall transition is observed at (\u03c312, \u03c311) = (N + 1/2, 1/2).) Rather we expect that, in the vicinity of integer Hall conductances, the system behaves pretty much like an ordinary two-dimensional electron gas. This implies (cf. our discussion above) that the Goldstone modes become gapped, \u03c311 scales to 0 and \u03c312 scales to an integer value: for (\u03c312, \u03c311) close to a line \u03c312 = n \u2208 N, the system will \ufb02ow towards n+1n+1 \u2208 N + 1/2. Again, the Hall the \ufb01xed point (n, 0). A special situation arises for \u03c312 = \u03c30 12 conductance does not renormalize, but now we are sitting on a critical surface (namely, E = E\u2217, corresponding to half-integer Hall conductance). At the transition point, the correlation length has diverged and the system has become metallic. Accordingly we should expect the longitudinal conductance to scale towards a \ufb01nite \ufb01xed point value \u03c3\u2217 11. Indeed, experimentally, one observes \u03c3\u2217 11 = O(1) at the transition points. The two-parameter \ufb02ow diagram discussed above was proposed in a seminal paper by Khmelnitskii.38 * 11 11 \u03c3 \u03c3 \u03c3 \u03c3 12 12 nn To theoretically understand the transition behavior, Pruisken and collaborators have derived renormalization group equations that take the lowest two topological sectors, W = 37 Renormalization cannot change the prefactor of the topological term, \u03c30 12. The contribution of the topologically trivial sector to the conductance is just this coe\ufb03cient. 38 D. E. Khmelnitskii, Quantization of Hall conductivity, JETP Lett. 38 (1983), 552\u20136. 0, 1, into account: 9.3 \u03b8-0terms \u03b211 \u2261 \u2202\u03c311 \u2202 ln L \u03b212 \u2261 \u2202\u03c312 \u2202 ln L = \u2212 1 2\u03c02\u03c311 \u2212 c\u03c311e\u22122\u03c0\u03c311 cos(2\u03c0\u03c312), = c\u03c311e\u22122\u03c0\u03c311 sin(2\u03c0\u03c312), 535 (9.28) where c > 0 is a numerical constant.39 These equations indeed have a family of \ufb01xed points (\u03c3\u2217 11), where \u03c3\u2217 = O(1) and \u03c3\u2217 \u2208 N + 1/2. In the vicinity of these points, \u03c312 (\u03c311) is a relevant (irrelevant) scaling variable. 12, \u03c3\u2217 12 Summarizing, Pruisken\u2019s approach appears to predict a parameter \ufb02ow as shown in the two-parameter phase diagram above. This diagram nicely conforms with experimental observations but, alas, there are some problems. For one thing, the interesting physics takes place in the vicinity of a \ufb01xed point value \u03c3\u2217 11 = O(1), well outside the regime of applicability of the \u03c3-model as such. (Remember that the derivation of this hinges on \u03c311 1.) One may trust in the principle that \u201cgood\u201d models (and Pruisken\u2019s model of the QHE certainly is good!) usually produce meaningful results even in parameter regimes where they no longer stand on safe ground. However, with the current problem, the chances are that this principle does not apply. Indeed, we have come across a very similar situation before: for large spin S 1, an antiferromagnetic spin chain is described by an O(", "3) nonlinear \u03c3-model. For topological angles \u03b8 = \u03c0 (corresponding to half-integer spin) the model is critical and \ufb02ows towards some strong coupling \ufb01xed point. The important observation now is that, in the vicinity of this point, the system is described no longer by an O(3) nonlinear \u03c3-model but by an altogether di\ufb00erent model: a \ufb01eld theory on the group manifold SU(2) with a topological term of Wess\u2013Zumino\u2013Witten type. Intriguingly, on its journey towards the strong coupling \ufb01xed point, the model manages to enlarge its \ufb01eld manifold from the coset locally O(3) space O(3)/O(2) S2 of the O(3) nonlinear \u03c3-model to the larger space SU(2) of the \ufb01xed point theory. Now, when projected onto a single replica channel, Pruisken\u2019s \u03c312 = 1/2 theory indeed reduces to an O(3) nonlinear \u03c3-model with topological angle \u03b8 = \u03c0. It is, therefore, quite conceivable that, at strong coupling, the general model also \ufb02ows towards a target model with a larger \ufb01eld manifold. Although nothing rigorous is known, a scenario to this e\ufb00ect has been outlined in the literature. At any rate, the applicability of Pruisken\u2019s RG equations in the vicinity of their \ufb01xed points remains questionable. INFO There is one other problem which should not be swept under the carpet: as discussed above, the absence of Goldstone modes in the localized phase requires the coupling constant of the topological term to be integer. This coupling constant is set by the Drude conductance \u03c30 12 and should not renormalize. In a way, we have to require that, for Fermi energies which lie between Landau levels, the Drude Hall conductance must already be integer, lest the theory run into a consistency problem. On the other hand, topological criteria do not require the physical conductance \u03c312 to be integer. This is exactly the opposite of our physical picture. We should 39 One might have expected the topological angle \u03c30 12, and not the physical conductance \u03c312, as an argument of the transcendental functions (sine and cosine). For a discussion of why the physical conductance appears there, we refer to Pruisken\u2019s article (Pruisken, in The Quantum Hall E\ufb00ect). 536 Topology expect that for any value of the Drude conductance (away from N + 1/2) the system will \ufb02ow towards an integer value of the physical conductance. (This is the \ufb02ow illustrated in the twoparameter \ufb02ow diagram above.) Essentially, the problem boils down to the fact that the coupling constant of the topological term does not renormalize. (If it did, it should \ufb02ow towards an integer value which could then justly be interpreted as the physical conductance.) To deal with this di\ufb03culty, Pruisken subjects the \ufb01elds to a transformation Q \u2192 U QU \u22121, whereU are the source \ufb01elds introduced above, and then discusses the structure of the theory after the \ufb01elds Q have been integrated out. This produces consistent results, but the procedure is highly implicit and, therefore, remains somewhat mysterious. Summarizing, Pruisken\u2019s \ufb01eld theory microscopically explains many of the intriguing aspects of the quantum Hall e\ufb00ect. Notably, it elucidates the interplay of disorder scattering and topological mechanisms (spectral \ufb02ow!) in the formation of the e\ufb00ect. Contrary to early expectations, though, it may well be that the actual quantum Hall transition lies outside the scope of the model. At any rate, a rigorous identi\ufb01cation of the universality class of this transition, not to mention a quantitative calculation of its critical exponents, remains an open problem. This concludes our discussion of \u03b8-terms in condensed matter \ufb01eld theory. In our brief survey, we were unable to discuss one of the most exciting applications of \u03b8-terms in \ufb01eld theory in general: \u2019tHooft\u2019s concept of \u03b8-vacua, and its relevance to understanding some of the most intriguing observed features of matter \u2013 CP- and T-violation. (However, the energetic reader is strongly encouraged to turn to a textbook on theoretical particle physics to learn more of this subject.) Rather, we proceed now to discuss another large and important family of topological \ufffd", "\ufffdeld theories. 9.4 Wess\u2013Zumino terms Almost every time that we met with a \u03b8-term in the previous chapter, a \ufb01eld theory with a Wess\u2013Zumino (WZ) term40 was just around the corner. Yet most condensed matter physicists appear to be only vaguely familiar with the ideas behind WZ \ufb01eld theory (in contrast to, say, the much more widely appreciated concept of \u03b8-terms). Perhaps the most important reason for this lack of appreciation is that the general meaning of a WZ term is di\ufb03cult to grasp in the \u201ctraditional\u201d languages familiar to condensed matter physicists. On the other hand, the relevant concepts become quite transparent once we venture to reformulate a few elements of \ufb01eld theory in the language of modern di\ufb00erential geometry. This reformulation will be the subject of the \ufb01rst half of this section. Not assuming any background knowledge, we begin with a crash course in di\ufb00erentiation on manifolds and exterior calculus. (Readers familiar with di\ufb00erential forms are invited to skip this introduction.) In Section 9.4.2 we 40 As to the terminology of Wess\u2013Zumino terms, there is no generally accepted convention. Historically, these topological terms \ufb01rst appeared in the work of Wess and Zumino. However, owing to Witten\u2019s seminal analysis of SU(N )-invariant chiral fermion systems (see below), they are often referred to as Wess\u2013Zumino\u2013Witten (WZW) terms. Yet another designation (especially popular in the Russian community) is Wess\u2013Zumino\u2013Novikov\u2013Witten (WZNW) terms. Except for the discussion of SU(N )-symmetric systems below, we stick here to the short variant of WZ terms. 9.4 Wess\u2013Zumino terms 537 \u03c6 x M U \u03c6 (x) T U\u2032 z z\u2032 z\u20131 z (\u03c6 (x)) V z\u2032 z\u2032(\u03c6 (x)) V\u2032 Figure 9.10 On the construction of coordinate representations of \ufb01eld manifolds. For a discussion, see the main text. then explain the general ideas behind WZ \ufb01eld theory before we turn to the discussion of a number of interesting applications. 9.4.1 A crash-course in di\ufb00erential geometry Figure 9.10 reiterates the mathematical backbone underlying most of our \ufb01eld theories: a \ufb01eld \u03c6 : M \u2192 T, x \u2192 \u03c6(x) is a mapping from a base-space M to a \ufb01eld space T. In practice, we mostly tend to nonchalantly identify \ufb01eld values \u03c6 \u2208 T and their coordinate representations z(\u03c6) \u2208 Rn (n is the dimension of T ). However, especially when it comes to the discussion of topological aspects, we must be very careful with such premature identi\ufb01cations. The point is that topologically non-trivial \ufb01eld spaces usually cannot be represented in terms of one globally de\ufb01ned system of coordinates. Take the sphere S2 as an example. You may choose the standard representation in terms of two angles (\u03b8, \u03c6), a stereographic projection onto a single complex variable z, or any other parameterization. Inevitably, there will be regions on S2 where the mapping \u201cS2 \u2192 coordinates\u201d becomes ill-de\ufb01ned. One may object that this ambiguity, manifesting itself only at a set of \u201cmeasure zero,\u201d41 cannot be of much practical signi\ufb01cance. Yet, with regard to topology this is not the case. Coordinate representations Given the importance of these singularities in the context of topological \ufb01eld theory, it is pertinent this time to discuss the construction of proper coordinate representations with mathematical rigor: the basis of each coordinate system is formed by a system of open 41 For example, in a system of polar coordinates, the problematic region is a line connecting the north and the south pole of the sphere. 538 Topology I subsets Ui \u2282 T chosen so that the union i Ui = T covers T.42 One next de\ufb01nes coordinate mappings zi : Ui \u2192 Vi \u2282 Rn from the patches Ui onto some open subsets Vi of Rn. The value zi(p) \u2208 Rn is a coordinate representation of the point p \u2208 T. To make this representation unique, we require the zis to have an inverse. The central point is that, on our manifold, there will be non", "-vanishing overlaps Ui \u2229 Uj = {}, i.e. points p \u2208 T that have more than one coordinate representation (see Fig. 9.10). Ambiguities between the di\ufb00erent systems are excluded by requiring that the maps zi \u25e6z\u22121 : zj(Ui \u2229 Uj) \u2192 zi(Ui \u2229 Uj) be di\ufb00eomorphisms (invertible and di\ufb00erentiable). Conceptually, the functions zi \u25e6 z\u22121 j mediate the change between di\ufb00erent coordinates. This being so, they must be as benign (di\ufb00erentiable) as possible. Notice, however, that it would be senseless to require the zis themselves to be di\ufb00erentiable, simply because in general T \u2282 Rn and the notion of di\ufb00erentiability need not even exist on T. j In the jargon of di\ufb00erential geometry, the coordinate maps zi are called charts of T while a fully covering collection {zi} is called an atlas. The existence of a proper atlas is, by de\ufb01nition, equivalent to the statement that T is a di\ufb00erentiable manifold. Example Consider T = S1, the unit circle. The one-dimensional manifold S1 has a natural embedding into R2 : S1 = {x \u2208 R2|x2 = 1}. We need a minimum of two charts to cover S1. For example, one may use z\u22121 : (0, 2\u03c0) \u2192 S1, z\u22121 : (0, 2\u03c0) \u2192 S1, z\u22121(\u03c6) = (cos(\u03c6), sin(\u03c6)), z\u22121(\u03c6) = (cos(\u03c6 + \u03c0), sin(\u03c6 + \u03c0)); z/z cover all of S1 except for the points (1, 0)/(\u22121, 0). Tangent space p vp Having discussed the coordinate representations of (\ufb01eld) manifolds, we are now in a position to lift elements integration, etc.) of standard calculus (di\ufb00erentiation, from Rn to manifolds. (Later, we identify topological terms as integrals over certain di\ufb00erentials on T.) We begin by introducing the tangent space TpT as a locally \ufb02at approximation to the manifold T at a point p \u2208 T (see the \ufb01gure). We then use this planar approximation to describe how functions de\ufb01ned on T vary in the neighborhood of p. To construct the tangent space, consider a curve \u03b3 : R \u2192 T with \u03b3(0) = p. It is tempting?\u2261 ds\u03b3(s)|s=0. However, (see the \ufb01gure) to de\ufb01ne a vector vp tangent to T at p by setting vp this \u201cde\ufb01nition\u201d is problematic because, in general, T \u2282 Rm, so that vp is not a decent vector. Nonetheless, the idea above is not far from the truth. To make it suitable, let us consider some function f \u2208 C(U ) de\ufb01ned on an open neighborhood U * p. (Here, C(X) TpT \u03b3 T 0 42 Of course, one can exercise much freedom in the choice of the Uis. For example, for the sphere, one might de\ufb01ne U1,2 to be any two overlapping \u201ccaps\u201d whose union covers T. Notice that overlaps are, in fact, unavoidable as we want to cover a compact set T by open subsets Ui. 9.4 Wess\u2013Zumino terms 539 denotes the space of smooth, real-valued functions de\ufb01ned on X.) We may then employ our curve \u03b3 to compute the directional derivative p (f ) \u2261 ds|s=0f (\u03b3(s)). v\u03b3 The notation on the right-hand side indicates that we have constructed a mapping that takes functions as arguments and produces a number.43 We next de\ufb01ne this mapping to be a tangent vector at point p. (Notice that the assignment \u201ccurve \u2192 tangent vector\u201d constructed in this way is not unique: two curves \u03b31 and \u03b32 tangent44 to each other at p will produce the same directional derivative, v\u03b31 p.) The set of all directional derivatives formed in this way de\ufb01nes the tangent (vector) space TpT at point p.45 p = v\u03b32 The de\ufb01nition of \u201cvectors\u201d given above may seem strange. (If you wait a", " while, its utility will become apparent!) However, given a coordinate function z we can meet the standard identi\ufb01cation \u201cvector \u2194 n-component object\u201d familiar from linear algebra. This is achieved by writing v\u03b3(f ) = ds|s=0f (z\u22121 \u25e6 z \u25e6 \u03b3(s)) \u2261 n i=1 \u2202if v\u03b3 i, where \u2202if \u2261 \u2202i(f \u25e6 z\u22121) is the ordinary partial derivative of the function f \u25e6 z\u22121 : V \u2282 Rn \u2192 R and v\u03b3 \u2208 R to be the ith component of v\u03b3 (in i the coordinate representation de\ufb01ned by z). Notice that the components of a tangent vector vi can alternatively be obtained as \u2261 ds|s=0zi(\u03b3(s)). We de\ufb01ne v\u03b3 i vi = v(zi), i.e. as the directional derivative of the ith component of the coordinate function. (Here, we have simpli\ufb01ed the notation by omitting the superscript reference to the curve \u03b3.) Relatedly, a coordinate system induces a natural basis of the tangent space, Tp. This, is de\ufb01ned by \u2200f \u2208 C(U ) : ei(f ) \u2261 \u2202if. Suppose that the reference point p is represented by two coordinate functions z and z. It is then straightforward to verify (exercise!) that the components of the tangent vector transform as v\u03b3 v i = \u2202zi \u2202z j j \u03b3 j. EXERCISE Compute the basis vectors e and e corresponding to the two charts forming the atlas of S1 discussed on page 538. Show that e = e = e\u03c6 where e\u03c6 is the azimuthal vector of a two-dimensional polar coordinate system, and the natural embedding of S1 into R2 is understood. 43 A purist might object that v\u03b3 p (f ) is just what we had earlier de\ufb01ned to be a functional, so that we should use square brackets [f ] for the argument. However, following standard convention in di\ufb00erential geometry, we shall here stick to (f ). 44 In di\ufb00erential geometry, one uses this criterion to de\ufb01ne the notion of tangency: two curves \u03b31,2 through p are tangent to each other (with the same tangent velocity) if, for all functions f, v\u03b31 p (f ) =v \u03b32 p (f ). 45 Mathematically, TpT is isomorphic to the space of all equivalence classes of curves through p, where two curves are called \u201cequivalent\u201d if they are tangent to each other. 540 Topology \u03c6 f\u02da \u03c6 f [\u03c6 \u2217(v)]\u03c6 (p) \u03c6 (p) T\u03c6 (p)T T p vp TpM M Figure 9.11 On the de\ufb01nition of the tangent mapping. For a discussion, see the main text. I The union T T \u2261 p\u2208T TpT of all local tangent spaces is called the tangent bundle of the manifold T.46 In fact, the tangent bundle is a di\ufb00erentiable manifold by itself. Its elements are given by (p, vp) \u2208 T T, where p \u2208 T and vp \u2208 TpT. (The dimension of T T is given by 2n, twice the dimension of T.) A mapping v : T \u2192 T T, p \u2192 (p, vp), smoothly assigning to each point of T a tangent vector is called a vector \ufb01eld on T. Now, suppose we are given two manifolds M and T. (Later on, the role of M will be played by the base space of \ufb01eld theory, a di\ufb00erential manifold by itself.) Further, suppose that there is a mapping (the \ufb01eld!) \u03c6 : M \u2192 T. The de\ufb01nition of tangent spaces above then implies the existence of an induced mapping, the so-called tangent mapping, \u03c6\u2217 : T M \u2192 T T, v \u2192 \u03c6\u2217(v). The image vector (\u03c6\u2217(v))p is de\ufb01ned by setting (\u03c6\u2217(v))p(f ) \u2261 v\u03c6\u22121(p)(f \u25e6 \u03c6), for any function de\ufb01ned in a neighborhood of p \u2208 T (see Fig. 9.11). Suppose we are given a system of coordinates w around p \u2208 M and z around \u03c6(p) \u2208 T. It is then straightforward to show that", " the coordinate representation of the vector (\u03c6\u2217(v))p is given by (\u03c6\u2217(v))i p = \u2202\u03c6i \u2202wj vj \u03c6\u22121(p), (9.29) where \u03c6i = zi(\u03c6) is a shorthand for the coordinate representation of \u03c6. The formula above explains why the mapping \u03c6\u2217 is sometimes referred to as the di\ufb00erential of the mapping \u03c6. It illustrates the general rule that everything taking place on the tangent spaces is a measure of local (or \u201cin\ufb01nitesimal\u201d) variations. 46 The authors hate the double-T notation T T, too. However, the pre\ufb01x T for tangent is a ubiquitous standard, and we want to keep emphasizing that the apparatus introduced here will be later applied to the target manifolds of \ufb01eld theory, T. 9.4 Wess\u2013Zumino terms 541 Di\ufb00erential forms A 1-form \u03c9p is a linear mapping \u03c9p : TpT \u2192 R (i.e. an element of the dual space of the vector space TpT ). Smoothly extending \u03c9p to a map \u03c9 globally de\ufb01ned on T T, we obtain a so-called di\ufb00erential 1-form (or, for brevity, just the 1-form). We will denote the space of 1-forms on T by \u039b1(T ). A number of important remarks on these de\ufb01nitions are in order: Most 1-forms that we shall encounter in practice are realized as di\ufb00erentials of func- tions: forf \u2208 C(T ) we de\ufb01ne the di\ufb00erential df by dfp(vp) \u2261 vp(f ). (Exercise: Convince yourself that, for manifolds T \u2282 Rn, this reduces to the standard de\ufb01nition of the di\ufb00erential with which you are familiar.) However, not every 1-form is a di\ufb00erential of a function. Consider, for example, the tangent basis of T = S1 constructed in the exercise on page 539. Let us de\ufb01ne a 1-form by setting \u03c9(e) = \u03c9(e) = 1, where the equality holds in the domain of overlap of the two charts. This 1-form cannot be represented as the di\ufb00erential of a single-valued function on S1 (exercise: why?). Finally, let us make a formal remark: for reasons that will become clear in a moment, functions f \u2208 C(T ) \u2261 \u039b0(T ) are sometimes referred to as 0-forms. Technically, a 1-form is a mapping \u03c9 : T T \u2192 \u039b0(T ), v \u2192 \u03c9(v), that maps vector \ufb01elds onto 0-forms. The value of the function \u03c9(v) at a pointp is given by \u03c9(v)(p) = \u03c9p(vp). Alternatively, we can say that the insertion of a vector \ufb01eld into a 1-form lowers the degree of the form from 1 down to 0. Given a system of local coordinates, z, each 1-form can be represented as \u03c9 = fi dzi, (9.30) where the coe\ufb03cient-functions are given by fi = \u03c9(ei) (a result that is instructive to check). A 1-form maps a single vector \ufb01eld onto a function. However, to describe the geometric structure of a manifold (distances, surface and volume elements, etc.) we need mappings that take more than one vector \ufb01eld as input. Mappings of this type are called tensors and de\ufb01ned as follows: a covariant tensor of rank r is a multi-linear mapping \u03c9 : T T \u00d7 \u00b7 \u00b7 \u00b7 \u00d7T T! r \u2192 \u039b0(T ), (v1,..., vr) \u2192 \u03c9(v1,..., vr). For example, some manifolds admit the de\ufb01nition of a metric. A metricg is a tensor of rank 2 that is positive (\u2200p \u2208 T : gp(vp, vp) > 0) and non-degenerate (\u2200wp : gp(vp, wp) = 0 \u21d2 vp = 0). A manifold with a metric is called a Riemannian manifold. We call gp(vp, vp) the length of the tangent vector vp. For later reference, we note that, in a system of local 542 Topology coordinates,", " g has the representation g = gijdzi \u2297 dzj. Here, gij = g(ei, ej) is themetric tensor of the manifold. EXERCISE Show that, in polar coordinates (r, \u03b8, \u03c6), the standard metric of R3, g = dx1 \u2297 dx1 + dx2 \u2297 dx2 + dx3 \u2297 dx3, assumes the form g = dr \u2297 dr + r2d\u03b8 \u2297 d\u03b8 + r2 sin2 \u03b8 d\u03c6 \u2297 d\u03c6. (9.31) However, by far most important in practical applications are tensors that are fully antisymmetric in their arguments: a p-form \u03c9 is a tensor that changes sign under odd permutations of its arguments: \u03c9(vS(1),..., vS(p)) = sgn(S)\u03c9(v1,..., vp), where S is an element of the permutation group. The space of all p-forms on T is denoted by \u039bp(T ). Given a p-form \u03c9 and a q-form \u03be we can produce a (p + q)-form \u03c9 \u2227 \u03be by the following rule: (\u03c9 \u2227 \u03be)(v1,..., vp+q) \u2261 1 p!q! S sgn(S) \u03c9(vS(1),..., vS(p)) \u03be(vS(p+1),..., vS(p+q)). The operation \u2227 : \u039bp(T ) \u00d7 \u039bq(T ) \u2192 \u039bp+q(T ) is called the exterior product of forms. (In passing we note that \u2227 de\ufb01nes the product of the Grassmann algebra \u039b(T ) \u2261 \u2295\u221e p=0\u039bp(T ) which we met already brie\ufb02y in Section 4.1.) EXERCISE Verify the following features: (a) \u03c9 \u2227 \u03c9 = 0 if \u03c9 \u2208 \u039bp and p is odd, (b) \u03c9 \u2227 \u03be = (\u2212)pq\u03be \u2227 \u03c9, and (c) \u03c9 \u2227 (\u03be \u2227 \u03b7) = (\u03c9 \u2227 \u03be) \u2227 \u03b7, permitting us to write just \u03c9 \u2227 \u03be \u2227 \u03b7 without brackets. Given a coordinate function z, each p-form has the unique coordinate representation \u03c9 = 1 p! \u03c9i1,\u00b7\u00b7\u00b7,ip dzi1 \u2227 \u00b7 \u00b7 \u00b7 \u2227dz ip, (9.32) where \u03c9i1,...,ip = \u03c9(ei1,..., eip ). To see this, notice that our de\ufb01nition of the exterior product above implies that (df1 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dfp)(v1,..., vp) = S sgn(S) df1(vS(1)) \u00b7 \u00b7 \u00b7df p(vS(p)). Also notice that the coe\ufb03cients \u03c9i1,...,ip are antisymmetric under odd exchange of the indices ij. This being so, Eq. (9.32) can alternatively be written as \u03c9 = \u03c9I1,...,Ip dzI1 \u2227 \u00b7 \u00b7 \u00b7 \u2227dz Ip, where, by convention, summations over capitalized indices are ordered, I1 < I2 < \u00b7 \u00b7 \u00b7 < Ip. Above we have seen that, computing the di\ufb00erential of a 0-form f \u2208 \u039b0(T ), we are led to a 1-form df \u2208 \u039b1(T ). This principle can be generalized to forms of arbitrary degree: let us de\ufb01ne the exterior derivative d by setting d\u03c9 \u2261 \u2202j\u03c9I1,...,Ip dzj \u2227 dzI1 \u2227 \u00b7 \u00b7 \u00b7 \u2227dz Ip, (9.33) where the coordinate representation Eq. (9.32) is understood. Obviously, d\u03c9 \u2208 \u039bp+1(T ), i.e. d can be interpreted as an operator on \u039b(T ) that raises the degree of forms by one. However, mathematically inclined people will object that it is unclear whether Eq. (9.33), an equation based on a speci\ufb01c coordinate representation, really represents a proper de\ufb01nition (i.e., is 9.4 Wess\u2013Zumino terms 543 it clear that the same rule, applied to a di\ufb00erent coordinate representation of \ufffd", "\ufffd, leads to the same form d\u03c9?). These worries are addressed in the following exercise. EXERCISE Show that (a) the de\ufb01nition Eq. (9.33) does not depend on the choice of coordinates, (b) d2 = 0, and (c) d(\u03c9 \u2227 \u03be) = d\u03c9 \u2227 \u03be + (\u2212)p\u03c9 \u2227 d\u03be, where p is the degree of \u03c9. In the following, two subspecies of di\ufb00erential forms will be of particular interest for us: we call a di\ufb00erential form \u03c9 \u2208 \u039bp(T ) closed if d\u03c9 = 0. Conversely, \u03c9 is called exact if \u03c9 = d\u03be, i.e. if \u03c9 is obtained as the exterior derivative of some (p \u2212 1)-form \u03be. Now, owing to the identity d2 = 0, every exact form is closed. Yet not every closed form is exact. EXERCISE Consider the 1-form \u03c9 on S1 discussed on page 541. Since \u03c9(e) = \u03c9(e) = 1, \u03c9 has the local (i.e. restricted to individual charts) representation \u03c9 = d\u03c6 or \u03c9 = d\u03c6. (The coordinates are those introduced on page 538.) Obviously, \u03c9 is closed. On the other hand, we have seen that there is no 0-form (function) f such that \u03c9 = df ; \u03c9 is not exact. The classi\ufb01cation of forms that are closed but not exact is a deep mathematical problem (the subject of cohomology theory). We shall return to this issue below when we discuss the geometry of topological terms. However, before doing so, we need to introduce one last concept of basic di\ufb00erential geometry. Previously, we have seen that a mapping \u03c6 : M \u2192 T between two manifolds induced a mapping \u03c6\u2217 from the vector \ufb01elds on M to those on T. In a very similar manner, \u03c6 gives rise to a mapping \u03c6\u2217 between forms on the two manifolds. This so-called pullback is de\ufb01ned by \u03c6\u2217 : \u039b(T ) \u2192 \u039b(M ), \u03c9 \u2192 \u03c6\u2217(\u03c9) \u2261 \u03c9 \u25e6 \u03c6\u2217. For example, for a 1-form \u03c9 \u2208 \u039b(T ) and v \u2208 T M, we have [\u03c6\u2217(\u03c9)](v) =\u03c9 (\u03c6\u2217(v)), etc. Notice that \u03c6\u2217 maps in a direction opposite to that of \u03c6, hence the name \u201cpullback.\u201d In a system of local coordinates z and w on M and T, respectively, the components of the 1-form \u03c6\u2217(\u03c9) are given by \u03c6\u2217(\u03c9)i = [\u03c6\u2217(\u03c9)](ei) = \u03c9(\u03c6\u2217(e i)) (9.29) = \u03c9 \u2202\u03c6k \u2202zi ek = \u2202\u03c6k \u2202zi \u03c9k, where {e i } is a basis of T. EXERCISE Check that the coordinate representation of the pullback of a p-form is given by \u03c6\u2217(\u03c9)I1,...,Ip = det \u2202\u03c6J1,..., \u2202\u03c6Jp \u2202zI1,..., \u2202zIp \u03c9J1,...,Jp. Also verify the useful formula \u03c6\u2217(\u03c9 \u2227 \u03be) = \u03c6\u2217\u03c9 \u2227 \u03c6\u2217\u03be, and the commutativity \u03c6\u2217 \u25e6 d = d \u25e6 \u03c6\u2217 of a pullback on an exterior derivative. (If you feel exhausted, just verify that \u03c6\u2217d\u03c9 = d(\u03c6\u2217\u03c9) holds when applied to forms \u03c9 \u2208 \u039b0,1(T ). The general proof proceeds along similar lines but is a bit more cumbersome.) Indeed, the pullback of forms is a very important operation since it enables us to de\ufb01ne... 544 Topology Integration on manifolds Consider a top-dimensional form \u03c9 \u2208 \u039bn(T ), i.e. a form whose rank is equal to the dimensionality of T. To begin with, let us assume that supp(\u03c9) \u2282 U (i.e. \u03c9p\u2208U = 0), where U \u2282 T is the domain of de\ufb01nition of a chart z. We then de\ufb01ne the integral of \u03c9 over U as U \u03c9 \u2261 V z\u22121\u03c9. (9", ".34) To make this de\ufb01nition somewhat less abstract, notice that \u03c9 can be represented as \u03c9 = f dz1 \u2227 \u00b7 \u00b7 \u00b7 \u2227dz n, where f is a function on T (exercise: why?). The de\ufb01nition above then assumes the form U f (x)dz1 \u2227 \u00b7 \u00b7 \u00b7 \u2227dz n = f (z)dz1 \u2227 \u00b7 \u00b7 \u00b7 \u2227dz n \u2261 f (z)dz1 \u00b7 \u00b7 \u00b7dz n. V V,Riemann Here, we have used the fact thatz \u22121(f (x)dz1 \u2227 \u00b7 \u00b7 \u00b7 \u2227dz n) = f (x(z))dz1 \u2227 \u00b7 \u00b7 \u00b7 \u2227dz n.47 In the (crucial) second equality, we declare the integral over the product of forms dz1 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dzn in V \u2282 Rn to be the ordinary Riemann integral. This identi\ufb01cation is meaningful because (dz1 \u2227 \u00b7 \u00b7 \u00b7 \u2227dz n)(v1,..., vn) measures the volume of the parallelepiped spanned by the vectors v1,..., vn (in the standard metric of Rn), i.e. it represents a sensible \u201cvolume element.\u201d48 To make the de\ufb01nition of the integral complete, we would have to discuss its extension to a global integral over the manifold. However, for practical reasons we shall not do so: for one thing, we are often enough dealing with manifolds T that admit a 1-chart atlas, in which case the de\ufb01nition above is su\ufb03cient. Only slightly worse is the situation where T can be covered by a single chart except for isolated singular points. (This is the case with, e.g., T = S1, S2 \u2013 exercise: why?) Since the integral is oblivious to \u201csets of measure 0,\u201d an integral over the domain of integration of such \u201cnearly complete\u201d charts is as good as an integral over all of T ; again, the de\ufb01nition above does the job. In the rare cases where one is dealing with an unpleasant manifold which does not belong to the two species above, one has to work harder and split up the support of the integrand by means of a so-called \u201cpartition of unity.\u201d However, in view of the relative rarity of such cases, and the fact that they are dealt with in every textbook on di\ufb00erential geometry, we limit attention to the \u201clocal\u201d de\ufb01nition above. 47 Do not be confused by the appearance of the same symbols dz1 \u2227 \u00b7 \u00b7 \u00b7 on both sides of the equations. On the left-hand side, dzi is referring to the di\ufb00erential of the ith coordinate function zi : T \u2192 R. On the right-hand side, zi : Rn \u2192 R simply projects a vector onto its ith component. 48 However, the discussion above sweeps one subtlety under the carpet: under a change of coordinates, z \u2192 w, \u03c9 = f (z)dz1\u2227 \u00b7 \u00b7 \u00b7 \u2227 dzn \u2192 f (z(w)) det (\u2202z/\u2202w) dw1 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dwn. Compatibility with the transition behavior of integration volume elements, dz1 \u00b7 \u00b7 \u00b7dz n \u2192 |det (\u2202z/\u2202W )| dw1 \u00b7 \u00b7 \u00b7dw n, requires det (\u2202z/\u2202W ) > 0. Coordinate systems with this property are said to have the same orientation. A manifold with an atlas of identically oriented charts is said to be orientable. (A prominent counterexample is presented by the M\u00a8obius strip.) The de\ufb01nition of the integral above implies that we have chosen a de\ufb01nite orientation. For future reference, we remark that, if \u03c6 : T1 \u2283 U1 \u2192 U2 \u2282 T2 is a di\ufb00eomorphism 9.4 Wess\u2013Zumino terms 545 between open subsets U1 and U2 of two manifolds T1 and T2, then \u03c9 = \u03c6\u2217\u03c9. U2 U1 (9.35) This is the generalization of the transformation law familiar from calculus to the integration on manifolds. (To understand this equation, notice that, if z : U2 \u2192 V is a chart of U2, then z \u25e6 \u03c6 : U1 \u2192 V is one of U1. The statement made by E", "q. (9.35) then follows from Eq. (9.34).) With the phalanx of de\ufb01nitions above, we are now \u2013 at last! \u2013 in a position to discuss the utility of di\ufb00erential forms in topological \ufb01eld theory. 9.4.2 From \u03b8- to Wess\u2013Zumino terms In Section 9.3, we have seen that, more often than not, the topological phase associated with non-vanishing \u201cwinding numbers\u201d could be given a representation in terms of an action Stop. However, no guidelines as to existence or non-existence of such representations have been given. We begin by discussing some principles behind the formulation of topological Lagrangian densities. This will bring us to a position where the connection to Wess\u2013Zumino terms can be established. The geometry of \u03b8-terms Let us \ufb01rst observe that most target manifolds T of topologically non-trivial \ufb01eld theories are Riemannian, i.e. come with a natural metric g. This may be the case because T \u2282 Rm is embedded into some Rm and inherits the natural metric of the latter (e.g. S2 \u2282 R3, etc.); or because T = G/H is a coset space of Lie groups49 (a situation characteristic of problems with spontaneous symmetry breaking); or, indeed, for some other reason. Most important is the fact that a metric g entails the existence of a canonical topdimensional form \u03c9 on T. Here, the attribute \u201ccanonical\u201d means the following: locally, each top-dimensional form can be represented as \u03c9 = f (z)dz1 \u2227 \u00b7 \u00b7 \u00b7 \u2227dz n, where zi, i = 1,..., n, are coordinate functions, n is the dimension of T, and f is some function. In general, the form of f depends on the choice of coordinates and cannot be globally speci\ufb01ed. However, on a Riemannian manifold, a canonical n-form with coordinate-invariant de\ufb01nition exists. To see this, let gijdzi \u2297 dzj be the metric. Further, let us de\ufb01ne \u221a gdz1 \u2227 \u00b7 \u00b7 \u00b7 \u2227dz n, \u03c9 = (9.36) where g = det{gij} is the determinant of the metric tensor. What makes \u03c9 special is that it has the same representation Eq. (9.36) in every coordinate system.50 49 Remember that a Lie group is a manifold with the additional structure of a group. As shown in textbooks of group theory, the group structure induces a metric. 50 To see this, recall that, under a change of coordinates z \u2192 W, dz1 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dzn \u2192 det A\u22121 dw1 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dwn, where A = (\u2202w/\u2202z). At the same time, gij \u2192 (AT gA)ij, i.e. g1/2 \u2192 det Ag1/2. The two determinants cancel each other, so that \u03c9 remains form invariant. 546 Topology The n-form \u03c9 is called the volume element of the manifold51 and VT \u2261 T \u03c9 is the volume of the (compact) Riemannian manifold T. Without loss of generality, we can set our unit of length so that VT = 1. EXERCISE For d = 3 and the metric Eq. (9.31), show that \u03c9 assumes the familiar form \u03c9 = 4\u03c0 r2 sin \u03b8 dr \u2227d\u03b8 \u2227d\u03c6 proportional to the three-dimensional volume element in polar coordinates. 1 We now claim that, for base and \ufb01eld manifolds of equal dimensionality dim M = dim T = d, Stop[\u03c6] = i\u03b8 M \u03c6\u2217\u03c9, (9.37) de\ufb01nes a coordinate invariant representation of the \u03b8-term. To get warmed up to this abstract representation, we \ufb01rst note that \u03c6\u2217\u03c9 is a top-dimensional form on M, i.e. we really have something to integrate and the notation makes sense. Now, let us consider a trivial \ufb01eld con\ufb01guration \u03c60(x \u2208 M ) = const. In this case, \u03c6\u2217 0\u03c9 = 0 (exercise: why?) and Stop[\u03c60] = 0, as one should expect. Next, let us assume that \u03c6 = \u03c61 is a di\ufb00eomorphic (1-1 and", " di\ufb00erentiable) covering of T. (The presumed existence of such a mapping amounts to the statement that we are dealing with a topologically non-trivial \ufb01eld theory.) In this case, the transformation law Eq. (9.35) holds and we get Stop[\u03c61] = i\u03b8 M \u03c6\u2217 1\u03c9 (9.35) = i\u03b8 T \u03c9 = 1 \u00d7 i\u03b8. Now consider a mapping \u03c6n that covers T W times (i.e. every point p \u2208 T is the image of W points x1,...,W \u2208 M ). One can show that Stop[\u03c6W ] =iW \u03b8, i.e. Stop indeed counts the winding number of \ufb01elds \u03c6. Rather than giving the proof of this statement for general W, let us consider a simple example. Let M = T = S2 and \u03c6W (\u03c6, \u03b8) = (W \u03c6, \u03b8).52 Then, \u03c9 = sin \u03b8 d\u03b8 \u2227 d\u03c6, and (check!) \u03c6\u2217 W \u03c9 = W sin \u03b8 d\u03b8 \u2227 d\u03c6. We thus obtainV S2 = 4\u03c0 and i\u03b8 M \u03c6\u2217 EXERCISE Let ni : S2 \u2192 R, i = 1, 2, 3, be the ith component of the unit vector de\ufb01ning a point p \u2208 S2. We consider the 2-form \u03c9 = n \u00b7 (dn \u2227 dn) \u2261 ijknidnj \u2227 dnk. Using the standard polar coordinate representation ni = ni(\u03b8, \u03c6), show that \u03c9 can alternatively be represented as \u03c9 = sin \u03b8 d\u03b8 \u2227 d\u03c6, which is the familiar volume element on the sphere (upon choosing an orientation so that z\u22121\u2217(d\u03b8 \u2227 d\u03c6) = d\u03b8 d\u03c6). Considering a \ufb01eldn : (x, y) \u2192 n(x, y) show that W \u03c9 = iW \u03b8, in agreement with the general rule. n\u2217\u03c9 = n \u00b7 (\u2202xn \u00d7 \u2202yn)dx \u2227 dy, M M (9.38) which we identify (again, for a de\ufb01nite orientation) with our earlier representation of the \u03b8-term of a \ufb01eld theory with S2-valued \ufb01elds. 51 To motivate this terminology, consider a basis in which the metric tensor gij = gi\u03b4ij is diagonal. The volume spanned by the (mutually orthogonal) tangent vectors e1,...,e n is then simply given by the product of their lengths, i.e. by g. This is precisely what we get when we evaluate \u03c9(e1,...,e n), i.e. the form \u03c9 measures the volume of the domain spanned by its arguments. gi = 52 Here we have adopted the usual abuse of notation; in principle we should write (z \u25e6 \u03c6W \u25e6 z\u22121)(\u03c6, \u03b8) = (W \u03c6, \u03b8), N i=1 \u221a \u221a where z and z are polar coordinate charts on M and T respectively. 9.4 Wess\u2013Zumino terms 547 \u03c6 \u03c6\u2032 M \u03c8 U T Figure 9.12 On the insensitivity of the integral representation Eq. (9.37) under \ufb01eld variations. For a discussion, see the main text. The discussion above shows that, when evaluated on certain reference con\ufb01gurations \u03c6W, the integral Eq. (9.37) yields the winding number W. To complete the identi\ufb01cation with our earlier representations of the \u03b8-term, we have to show that Eq. (9.37) does not change under continuous distortions of \u03c6W (i.e. that it responds to the topological sector, and nothing else). Readers not content with our assertion that this is the case may wish to navigate through the following argument (which, as a byproduct, nicely illustrates the power of geometric methods in topological \ufb01eld theory). INFO To show the invariance of Eq. (9.37) under continuous \ufb01eld deformations, let us consider two \ufb01eld con\ufb01gurations \u03c6 and \u03c6 which can be continuously deformed into each other. More speci\ufb01cally, we set \u03c6 = \u03c8 \u25e6 \u03c6, where \u03c8 : T \u2192 T is di\ufb00erent from unity only inside the domain U \ufffd", "\ufffd T of some chart, see Fig. 9.12. (This is no serious restriction as, by iterative deformations of this type, any \ufb01eld con\ufb01guration continuously deformable into \u03c6 can be reached.) We then have Stop[\u03c6] \u2212 Stop[\u03c6] =i\u03b8 M (\u03c6\u2217\u03c9 \u2212 (\u03c8 \u25e6 \u03c6)\u2217\u03c9) = i\u03b8 \u03c6\u2217(\u03c9 \u2212 \u03c8\u2217\u03c9). M Now \u03c9 \u2212 \u03c8\u2217\u03c9 \u2208 \u039bd(T ) is ad-form on T, di\ufb00erent from unity only locally (inside U ). This implies its exactness, i.e. the existence of a representation \u03c9 \u2212 \u03c8\u2217\u03c9 = d\u03ba, where \u03ba \u2208 \u039bd\u22121(T ).53 We thus \ufb01nd Stop[\u03c6] \u2212 Stop[\u03c6] =i\u03b8 \u03c6\u2217d\u03ba = i\u03b8 d(\u03c6\u2217\u03ba) =i\u03b8 \u03c6\u2217\u03ba = 0, M M \u2202M where, in the second equality, we have used the general commutativity of the exterior derivative and pullback and, in the third equality, Stokes\u2019 theorem (for a reminder, see below). In the fourth equality we have assumed that, on the boundary \u2202M of the base manifold (physically, in\ufb01nity), the \ufb01elds \u03c6|\u2202M approach a constant, so that \u03c6\u2217\u03ba = 0. The representation Eq. (9.37) sheds some light on the structure of \u03b8-terms on Riemannian54 \ufb01eld manifolds. However, this was, of course, not the only reason for maneuvering 53 The reason is that, locally, \u03c9 \u2212 \u03c8\u2217\u03c9 has a coordinate representation \u03c9 \u2212 \u03c8\u2217\u03c9 = f dz1 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dzn. With the ansatz \u03ba = gdz2 \u2227 \u00b7 \u00b7 \u00b7 \u2227 dzn, the equation d\u03ba = \u03c9 \u2212 \u03c8\u2217\u03c9 reduces to the di\ufb00erential equation \u22021g = f (if the special role ascribed to the coordinate direction 1 is irritating, notice that \u03ba is not uniquely de\ufb01ned; every \u03ba = \u03ba + \u03b7 with a closed (d\u03b7 = 0) form \u03b7 will do the job just as well), de\ufb01ned on some open interval of x1 (for \ufb01xed x2,..., xn). This is an ordinary di\ufb00erential equation which can be solved. 54 In fact, all we need to formulate the construction above is a canonical top-dimensional form, i.e. none of our arguments above actually relied on the fact that, in Riemannian geometry, this form happens to be the 548 Topology through the geometric constructions above. Our prime reason for introducing the invariant formulation is that it contains the key to understanding the connection to Wess\u2013Zumino terms. The geometry of Wess\u2013Zumino terms In this section, we discuss the general geometric principle common to all Wess\u2013Zumino (WZ) terms. We then explore how our previous sporadic encounters with WZ \ufb01eld theories \ufb01t into this scheme and discuss a few more applications. Note that \u03b8-terms appear in theories whose \ufb01eld manifold and base manifold are of equal dimensionality: dim M = dim T. In contrast, WZ terms are at home in \ufb01eld theories with dim M = dim T \u2212 1.55 How can a topological term be constructed that relies on this dimensional relation? As we shall see in a moment, the key to the answer lies in Stokes\u2019 theorem: d\u03c9 = \u03c9. C \u2202C (9.39) Here, C \u2282 M is a benign (smooth, orientable, etc.56) subset of a di\ufb00erentiable manifold M, \u2202C is its boundary, and d\u03c9 \u2208 \u039bn(M ) is top-dimensional on M (see \ufb01gure). INFO The proof of Eq. (9.39) is beyond the scope of our present discussion and we have to refer to textbooks on di\ufb00erential geometry. Nonetheless, a few remarks for readers not familiar with Stokes\u2019 theorem may be helpful. First note that \u2202C is a manifold by itself, with dim \u2202C = n\u22121. Thus, \u03c9 is top-dimensional on \u2202C and can be integrated. To gain some insight into the substance of Eq. (9.39) let", " us consider a few special cases. For example, let M be a three-dimensional manifold, n = 3. Consider the two-form \u03c9 = v1dx2 \u2227dx3 +v2dx3 \u2227dx1 +v3dx1 \u2227dx2 and interpret the coe\ufb03cients vi = vi(x) as the components of a three-dimensional vector \ufb01eld v = (v1, v2, v3). With d\u03c9 = div v dx1 \u2227 dx2 \u2227 dx3 \u2261 div v dV and the identi\ufb01cation \u03c9 = v \u00b7 dS, wheredS is the two-dimensional \u201csurface element,\u201d we obtain the standard representation of Gauss\u2019 law dV div v = dS \u00b7 v. M \u2202M Similarly, let M be some two-dimensional surface, n = 2, and \u03c9 = v1dx1 + v2dx2 + v3dx3. Identifying the surface element as above, we then have d\u03c9 = curl v \u00b7 dS and dS \u00b7 curl v = ds \u00b7 v, M \u2202M where v \u00b7 ds \u2261 \u03c9, ds is commonly known as a \u201cline element,\u201d and called Amp`ere\u2019s law. \u2202M is a line integral: this is volume element. For example, some \ufb01eld theories with dim(T ) = 2 live on manifolds with a so-called symplectic structure (which is nothing but a skew-symmetric, non-degenerate 2-form). This form may then take over the role of our \u03c9 above and, in perfect analogy to our discussion above, gives rise to a topological Lagrangian density. 55 More generally, WZ terms can be constructed if, in dimensions dim M + 1, the manifold T possesses a closed di\ufb00erential form. (In the special case dim M + 1 = dim T, this form will be the volume form of T.) 56 More precisely, C must be a so-called chain on M. However, for brevity, we shall keep the sloppy characterization above. 9.4 Wess\u2013Zumino terms 549 So suppose that we are dealing with a \ufb01eld theory de\ufb01ned on a d-dimensional base manifold M with (d + 1)-dimensional target manifold T. Without (too much) loss of generality, we may think of T = Sd+1 and M = Sd as unit spheres. INFO Although we have seen that spheres frequently appear as target spaces in \ufb01eld theory, the statement above may seem to be too restrictive \u2013 for example, what about the many \ufb01eld theories with group-valued target manifolds? The rationale behind emphasizing spherical target manifolds lies in an important statement of homotopy theory: the integration spaces we encounter in \ufb01eld theories with continuous symmetries are usually embedded in some su\ufb03ciently large U(N ). However, it can be shown that the topological structure of the group U(N \u2265 2) essentially reduces to that of its subgroup SU(2) \u2282 U(N ), i.e. mappings into U(N ) can be continuously deformed into mappings into SU(2). (If you \ufb01nd this di\ufb03cult to visualize, consider a two-dimensional plane with a circular hole as an example. Any curve in the plane can be continuously deformed to a curve on the circle surrounding the hole. In this sense, the circle preempts the topological content of the plane. Similar reduction mechanisms exist for higherdimensional manifolds.) However, SU(2) S3 is isomorphic to the 3-sphere. (Representing 3 j=1 xj\u03c3j, x\u03bc \u2208 R, the condition det(g) = 1 boils SU(2) as the group of matrices g = x0\u03c30 + i down to the relation \u03bc = 0.) This argument shows that, as far as topology is concerned, the spaces S1,2,3 basically exhaust the list of relevant target spaces. \u03bc=0 x2 3 As to the base manifold, the identi\ufb01cation M = Sd is motivated by the compacti\ufb01cation scheme discussed in Section 9.3 above. Notice that the target manifold T = Sd+1 is Riemannian and, therefore, comes with a canonical (d + 1)-form \u03c9. Given these prerequisites, our strategy will be to utilize, as much as possible, our previous understanding of the geometry of the \u03b8-term. To this end, let us interpret \u03c6(M ) as a ddimensional submanifold in T. Further, let \u0393+ \u2283 \u03c6(M ) be", " a subset of T on which the T volume form \u03c9 = d\u03ba+ can be represented as the exterior derivative of some d-form \u03ba+. EXAMPLE In fact, we may even assume that \u0393+ = Sd+1 \u2212 {p} covers all of our sphere except for one point p. For example, on the 2-sphere S2, the volume form \u03c9 = sin \u03b8 d\u03b8 \u2227 d\u03c6 is obtained from \u03ba+ = (1 \u2212 cos \u03b8)d\u03c6 as \u03c9 = d\u03ba+ everywhere except for the south pole. Similarly, with \u03ba\u2212 = \u2212(1 + cos \u03b8)d\u03c6, \u03c9 = d\u03ba\u2212 everywhere except for the north pole. Of course, the choice of the exclusion point is quite arbitrary. Notice, however, that there is no global representation \u03c9 = d\u03ba. If such a representation existed, Vol(T ) = \u2202T \u03ba = 0, because T is boundaryless, i.e. the volume of the sphere would vanish. T d\u03ba = T \u03c9 = \u2192 Sd+1 de\ufb01ned on the entire northern hemisphere, Sd+1 We now dimensionally extend the \ufb01eld \u03c6 : M = Sd \u2192 \u0393N \u2282 Sd+1 = T to a mapping \u02dc\u03c6 : Sd+1 N, ofS d+1. This is achieved N in a series of steps: \ufb01rstly, identify M = Sd as the equator of Sd+1 (see Fig. 9.13). We then introduce a (d + 1)-dimensional coordinate representation (s, x), s \u2208 [0, 1], of Sd+1 N. These coordinates are de\ufb01ned in such a way that (s = 1, x) parameterizes the equator Sd = M while lims\u21920(s, x) = pN is the north pole; otherwise, their choice is arbitrary. We \ufb01nally extend our \ufb01eld to a mapping \u02dc\u03c6 : Sd+1 \u2192 \u0393N, (s, x) \u2192 \u02dc\u03c6(s, x). Apart from the obvious consistency condition \u02dc\u03c6(s = 1, x) = \u03c6(x), the choice of this extension is, again, arbitrary. N 550 Topology Given this setup, let us de\ufb01ne SWZ[\u03c6] = iC \u02dc\u03c6\u2217\u03c9, Sd+1 N (9.40) as a trial candidate of a topological action (C = const.). Constructed in manifest analogy to the \u03b8-term Eq. (9.37), this expression is \u201ctopological\u201d in nature (it is scale and reparameterization invariant, etc.). Nonetheless, the de\ufb01nition does not look quite trustworthy. Notably, we have written \u201cSWZ[\u03c6]\u201d while the right-hand side of the de\ufb01nition involves the extension \u02dc\u03c6. To see that the integral is, in fact, independent of the particular choice of the extension, we use the fact that, on \u02dc\u03c6(Sd+1 N ) \u2282 \u0393N, the volume form can be written as \u03c9 = d\u03baN. This implies SWZ[\u03c6] = iC \u02dc\u03c6\u2217d\u03baN = iC d( \u02dc\u03c6\u2217\u03baN) = iC \u02dc\u03c6\u2217\u03baN = iC \u03c6\u2217\u03baN. M =Sd M Sd+1 N Sd+1 N Here we have made use of (i), in the second equality, the commutativity of the pullback and exterior derivative, (ii) in the third equality, Stokes\u2019 theorem, and, (iii) in the crucial fourth equality, that the integral over \u2202Sd+1 N = Sd = M depends only on the value of the boundary \ufb01eld \u02dc\u03c6(s = 1, x) = \u03c6(x). This proves the independence of the action SWZ on the extension scheme. At the same time, we have obtained the alternative representation SWZ[\u03c6] = iC M \u03c6\u2217\u03baN, (9.41) of the topological action. This form makes the independence of the action SWZ of the \ufb01eld extension manifest. For this, however, a price has had to be paid: Eq. (9.41) involves the d-form \u03baN which we saw is tied to a certain coordinate representation (owing to the absence of a global representation \u03c9 = d\u03ba) on the \ufb01eld manifold. In other words,", " the extensionindependent representation Eq. (9.41) necessarily involves the choice of a speci\ufb01c coordinate system of the \ufb01eld. INFO As an example, consider M = S1 and T = S2. Think of the latter as the space of unitvectors n. Our \ufb01eld is a mapping n : S1 \u2192 S2, t \u2192 n(t), where t \u2208 [0, 1] and periodic boundary (x,s) x ~\u03c6 \u03c6 \u0393+ ~ \u03c6 (x,s) \u03c6 (x) \u0393 \u2013 Figure 9.13 On the extension of a \ufb01eld theory de\ufb01ned on a d-dimensional base manifold M Sd to a (d + 1)-dimensional theory. 9.4 Wess\u2013Zumino terms 551 conditions n(0) = n(1) are understood. Extension to a \ufb01eld \u02dcn(s, t) then allows us to represent the action SWZ[n] as (see Eq. (9.38)) SWZ[n] = i 1 1 ds 0 0 C 4\u03c0 dt \u02dcn \u00b7 (\u2202s \u02dcn \u00d7 \u2202t \u02dcn). Now, except for the south pole, we can write the volume (or better to say area) form on S2 as \u03c9 = d[(1 \u2212 cos \u03b8)d\u03c6]. Then, using the fact that n\u2217[(1 \u2212 cos \u03b8)d\u03c6] = (1 \u2212 cos \u03b8(t))d\u03c6(t) = (1 \u2212 cos \u03b8(t))\u2202t\u03c6(t)dt, we obtain the alternative representation SWZ[n] =i 1 0 C 4\u03c0 dt (1 \u2212 cos \u03b8(t))\u2202t\u03c6(t). (9.42) Notice that this form explicitly uses the coordinate representation n \u2194 (\u03c6, \u03b8). Equations (9.40) and (9.41) are two di\ufb00erent representations of the Wess\u2013Zumino action. As discussed above, both have their advantages and disadvantages \u2013 the need to arti\ufb01cially extend the \ufb01eld vs. lack of representation invariance \u2013 which is why they are used interchangeably in the literature. Finally, we need to discuss one further important point; the value of the coupling constant C. As discussed above, the coupling constant of the \u03b8-term, the topological angle, was quite arbitrary. This is not the case with the WZ term. In fact, we shall see that the constant C is subject to quantization conditions. To understand why, recall that above we have chosen the northern hemisphere Sd+1 as the domain of integration of our extended \ufb01eld theory. Of course, the southern hemisphere Sd+1 would have provided just as good a choice. In this case, we would have de\ufb01ned N S S WZ[\u03c6] \u2261 \u2212iC \u02dc\u03c6\u2217\u03c9, Sd+1 S where \u02dc\u03c6 is a \ufb01eld extension to the southern hemisphere, and the extra minus sign takes care of the fact that the equator is the boundary of the oppositely oriented southern domain. Of course, our de\ufb01nition of the WZ action would be senseless if this ambiguity mattered. Therefore, SWZ[\u03c6] \u2212 S WZ[\u03c6] = iC \u02dc\u03c6\u2217\u03c9 + iC Sd+1 S Sd+1 N \u02dc\u03c6\u2217\u03c9 = iC \u02dc\u03c6\u2217\u03c9 = iCW, Sd+1 where W is integer. At \ufb01rst sight, the non-vanishing of this expression looks worrysome. However, since the action appears in the exponent no harm is done as long as the di\ufb00erence is equal to (i times) an integer multiple of 2\u03c0, i.e. the the coupling constant C obeys the quantization condition57 C = 2\u03c0k, k \u2208 Z. (9.43) The number k is called the level of the WZ theory. To summarize the main results of this section, we have found that: 57 Recall that the volume form \u03c9 underlying the construction has been normalized in such a way that For di\ufb00erent choices of the normalization, the value of C will change accordingly. Sd+1 \u03c9 = 1. 552 Topology The WZ action is closely allied to the \u03b8-term. It a\ufb00ords two di\ufb00erent representations where: Equation (9.40) involves a dimensional extension of the \ufb01eld, while Equation (9.41) relies", " on an explicit coordinate representation of the \ufb01eld. The coupling constant of the WZ term is quantized according to Eq. (9.43). Having discussed the structure and geometry of WZ theories, we now return to physics. Using a simple prototype system as an example, we begin by exploring how WZ terms enter low-energy theories of many-body quantum systems. 9.4.3 Example: magnetic moment coupled to fermions Consider a single energy level of a spinful fermion system. (One may think, for example, of a discrete level of an atom.) Let us assume that fermions inhabiting the level are coupled to a classical magnetic moment n. The coherent state action of this system is given by S[\u03c8, n] = \u03b2 0 d\u03c4 \u00af\u03c8(\u2202\u03c4 + \u03be + \u03b3n \u00b7 \u03c3)\u03c8, where \u03b3 is a coupling constant and, as usual, \u03be = \u2212 \u03bc. A complete speci\ufb01cation of the problem would have to include a term S[n] controlling the dynamics of the uncoupled magnetic moment. However, for the purposes of the present discussion, it is su\ufb03cient to consider the moment\u2013fermion coupling in isolation. INFO Actions of this type appear as building blocks of larger systems. For example, once generalized to a set of levels a, our model might describe a system of atomic shell electrons subject to Hund\u2019s rule coupling to a spin or orbital magnetic moment. Alternatively, the magnetic moment n might describe the Hubbard\u2013Stratonovich decoupling of some electron\u2013 electron interaction in a spinful channel. Integration over the fermion degrees of freedom brings us to the reduced action S[n] = \u2212tr ln(\u2202\u03c4 + \u03be + \u03b3n \u00b7 \u03c3). To proceed, let us write n = Re3, whereR \u2208 SO(3) is a rotation matrix, and then use the fact that a matrix U \u2208 SU(2) can be found58 such that n \u00b7 \u03c3 = (Re3) \u00b7 \u03c3 = U \u03c33U \u22121. (9.44) For example, it is straightforward to verify that, with the standard polar representation, n = (sin \u03b8 cos \u03c6, sin \u03b8 sin \u03c6, cos \u03b8)T, the choice U = e\u2212i \u03c6 2 \u03c33 e\u2212i \u03b8 2 \u03c31 e\u2212i \u03c8 2 \u03c33, (9.45) 58 This is, in fact, just a concrete realization of the correspondence between SU(2) and SO(3) familiar from group theory. 9.4 Wess\u2013Zumino terms 553 will su\ufb03ce. However, for the moment we shall not need an explicit representation of the rotation matrix. Rather, substituting the general expression Eq. (9.44) into the action, we obtain S[n] \u2192 S[U ] =\u2212tr ln( \u2202\u03c4 + \u03be + \u03b3U \u03c33U \u22121) = \u2212tr ln(U \u22121\u2202\u03c4 U + \u03be + \u03b3\u03c33) = \u2212tr ln(\u2202\u03c4 + \u03be + \u03b3\u03c33 + U \u22121 \u02d9U ), where, in the \ufb01rst equality, we have used the cyclic invariance of the trace and, in the last equality, we have de\ufb01ned \u2202\u03c4 U \u2261 \u02d9U. To proceed, we assume that the two energies |\u03be \u00b1\u03b3| T are well separated from the chemical potential, | \u00b1 \u03b3| T. We also assume that the frequency scales \u02dc\u03c9 on which the \ufb01elds n \u2194 U vary are so slow that they do not change the occupation of the fermionic levels: \u02dc\u03c9/\u03b3 1. In these circumstances, we may proceed by a straightforward expansion of the logarithm: S[U ] = tr( \u02c6G \u02c6U \u22121 \u02d9U ) + O(\u02dc\u03c9/\u03b3)2, where \u02c6G = (\u2212\u2202\u03c4 \u2212 \u03be \u2212 \u03b3\u03c33)\u22121 and we have anticipated (check!) that higher-order terms of the expansion will be small in the parameter \u02dc\u03c9/\u03b3. Switching to a frequency representation, S[U ] = n tr( \u02c6Gn( \u02c6U \u22121 \u02d9U )m=0) = \u2212 = \u2212 d\u03c4 tr nF(\u03be + \u03b3\u03c33) \u2212 1 2 d\u03c4 tr(nF(\u03be + \u03b3\u03c33)U \u22121\u2202\u03c4 U ) U \u22121\u2202\u03c4 U, where we have used the fact that the frequency sum of", " a fermionic Green function introduces the Fermi distribution function, nF. In the last line, we have shifted nF by 1/2, which is permissible because tr(U \u22121 \u02d9U ) is a boundary term: \u03b2 0 d\u03c4 tr(U \u22121 \u02d9U ) = \u03b2 0 d\u03c4 \u2202\u03c4 tr(ln U ) = tr ln(U )|\u03b2 0 = 0. (9.46) Now, if both levels are either occupied (nF(\u03be \u00b1 \u03b3) \u2212 1/2 \u2248 1/2) or unoccupied (nF(\u03be \u00b1 \u03b3) \u2212 % U \u22121\u2202\u03c4 U ) \u221d tr(U \u22121\u2202\u03c4 U ) is 1/2 \u2248 \u22121/2), the action vanishes59 since tr( again a boundary term. nF(\u03be + \u03b3\u03c33) \u2212 1 2 & Assuming, however, that \u03be + \u03b3 is empty while \u03be \u2212 \u03b3 is occupied we obtain S[U ]|\u03be+\u03b30\u03be\u2212\u03b3 \u2261 SWZ[U ] = \u2212 1 2 \u03b2 0 d\u03c4 tr(\u03c33U \u22121 \u02d9U ). (9.47) We now run a few tests on SWZ to identify it as a topological term of Wess\u2013Zumino type. For one thing, our base manifold M S1 is one-dimensional and SWZ involves a \ufb01rst-order derivative. This exempli\ufb01es the scale invariance characteristic of topological terms. The suspicion that SWZ is topological is corroborated by the fact that it is purely imaginary. (The time-derivative acts on a unitary matrix, i.e. something like a generalized \u201cphase\u201d \u2013 for more details, see below.) But what type of topological term are we dealing with? To get an answer to this question, we \ufb01rst need to determine the dimensionality of the target manifold T. Clearly, T = SU(2) in spite of the fact that we are temporarily using U \u2208 SU(2) for our 59 Up to corrections of O(exp(\u2212|\u03be \u00b1 \u03b3|/T )) which, in any case, are beyond the scope of the \ufb01rst-order expansion of the \u201ctr ln\u201d above. 554 Topology \ufb01elds: our original theory was de\ufb01ned for n \u2208 S2. Indeed, the representation Eq. (9.44) is invariant under the \u201cgauge transformation\u201d U \u2192 U exp(i\u03c8\u03c33). This means that the \u201ctrue\u201d \ufb01eld manifold is SU(2)/U(1) S2 as one should expect. Now, dim S2 = 2 = dim M + 1, so that SWZ is likely to be a Wess\u2013Zumino term. Indeed, we observe that, under the gauge transformation above, tr(\u03c33U \u22121 \u02d9U ) \u2192 tr(\u03c33U \u22121 \u02d9U ) + 2i \u02d9\u03c8, i.e. the topological density is not invariant. This indicates that we will not be able to \ufb01nd a coordinate-independent representation of SWZ in terms of n. As discussed in the previous chapter, the lack of parameterization invariance is a hallmark of WZ terms. (There is no need to worry, though, about the gauge invariance of the theory: the periodic boundary conditions imposed on U imply that \u03c8(\u03b2) =\u03c8 (0) + 2\u03c0n, where n is some winding number. Under the gauge transformation, the action thus changes by \u03b4S = \u22122\u03c0in, so that exp(\u2212\u03b4S) = 1 remains invariant.) To present SWZ in a less abstract form, we can substitute the angular representation Eq. (9.45) into Eq. (9.47). The straightforward evaluation of the derivative then gives SWZ[\u03c6, \u03b8] = \u2212 i 2 d\u03c4 (1 \u2212 cos \u03b8)\u2202\u03c4 \u03c6. (9.48) We identify this expression as the coordinate representation Eq. (9.42) of the WZ term arising in theories with base S1 and target S2. Recall that SWZ evaluates to \u2212i/2 times the area swept out by the closed curve (\u03c6, \u03b8) on the sphere. In fact, we have already met with the action Eq. (9.48) in Section 3.3 when we discussed the path integral of a single spin. That the core contribution to the action of that problem reappears here should not come as too much of a surprise: the classical magnetic moment \u221d", " n \u201censlaves\u201d the electron spin. In the limit of perfect alignment \u2013 realized under the conditions assumed above \u2013 we are left with the dynamics of a quantum spin whose dynamics are tied to that of n. The resulting action therefore coincides with that of a free spin described by the unit vector n. the In the Sir Michael Berry 1941\u2013 Theoretical physicist who has made groundbreaking contributions to the \ufb01eld of quantum nonlinear dynamics and optics. Berry introduced the concept of the Berry phase (or geometric phase as he himself prefers to call it) and explored its manifestations in various physical contexts. (Figure courtesy of Sir Michael Berry.) action literature, Eq. (9.48) is frequently referred to as the Berry phase action. To understand the rationale behind this terminology, let us brie\ufb02y recall (see the Info block below) a few facts about the quantum mechanical Berry phase. Consider a quantum particle subject to the Hamiltonian \u02c6H(x(t)), where the D-component vector x(t) \u2261 {xi(t)} parameterizes a weakly time-dependent contribution to \u02c6H. As shown by Berry, the dynamical phase acquired during the evolution of the particle can be expressed dt 0(t) +i\u03b3 (t)), where 0(t) is the energy of the instantaneous ground state as exp(\u2212i of \u02c6H(t). The \ufb01rst contribution to the exponent is the usual dynamical phase of quantum evolution. (Here, it is assumed that the time-dependence of \u02c6H is su\ufb03ciently weak for us to neglect transitions into excited levels; for a static Hamiltonian, this contribution reduces to 9.4 Wess\u2013Zumino terms 555 the familiar phase 0t.) The second contribution, \u03b3(t), is of geometric origin, inasmuch as it depends on the path traced out by the vector x in parameter space, but not on dynamical details. (In particular, it is independent of the velocity at which the parameter path is traversed.) The geometric phase assumes a particularly simple form if the parameter dependence of \u02c6H is periodic in time, x(0) = x(t). In this case, x(t) de\ufb01nes a closed curve and \u03b3 can be expressed as an integral over any surface in parameter space bounded by that curve. Speci\ufb01cally, for a spin S particle Zeeman-coupled to a weakly time-dependent magnetic \ufb01eld B(t) \u2261 n(t)B of constant magnitude, B, \u03b3 = S 2 d2x n \u00b7 \u22021n \u00d7 \u22022n, (9.49) is the area bounded by the curve n(t) on the 2-sphere (which, in this case, de\ufb01nes the parameter space). Comparison with Eq. (9.48) and the example discussed on page 538 indeed identi\ufb01es the WZ-term as the Berry phase of the spin problem.60 INFO A brief reminder of the Berry phase in quantum mechanics: Consider a particle governed by a Hamiltonian \u02c6H(x(t)). It is assumed that the time-dependence of the parameter vector x is adiabatic, which means that a particle initially prepared in the ground state |0(t = 0) of the Hamiltonian \u02c6H(x(0)) will remain in the instantaneous ground state |0(t) (i.e. the lowest eigenstate \u02c6H(x(t))|0(t) = (t)|0(t) of the operator \u02c6H(x(t))) throughout the entire observation time. (In practice, this means that the rate \u03c9 at which the parameters change in time must be much smaller than the excitation gap of the system.) We are interested in computing the dynamical phase corresponding to the time evolution of |0(t). To this end, let us represent the wavefunction of the particle as |\u03c8(t) = e\u2212i\u03c6(t)|0(t) and consider the time-dependent Schr\u00a8odinger equation \u02c6H(x(t))|\u03c8(t) = i\u2202t|\u03c8(t). Substitution of the representation above and multiplication by 0(t)| then leads to the equation \u2202t\u03c6 = (t) \u2212 i0(t)|\u2202t|0(t). Integrating over time and comparing with our discussion above, we 0 dt 0(t)|\u2202t |0(t) of the Berry phase (exercise: why is \u03b3 are led to the identi\ufb01cation \u03b3(t) =i real?). Now, the instantaneous ground state inherits its time dependence from the parameters", " x(t). We may thus write t \u03b3(t) = i t 0 dt 0(x(t))|\u2202xi(t)|0(x(t))\u2202txi(t) = i c dx 0(x)|\u2202x|0(x) = i 0|d0. c Here, the second integral has to be interpreted as a line integral in parameter space. It is taken along a curve c which starts at x(0), follows the evolution of the parameter vector, and ends at x(t). Importantly, the line integral depends only on the choice of \u03b3 but not on the velocity at which this curve is traversed (the dynamics of the process). In this sense, we are dealing with a phase of geometric origin. The third integral representation above emphasizes the geometric 60 In the discussion above, we considered a particle coupled to a weakly time-dependent magnetic moment \u2013 a canonical setup for the appearance of Berry phases. But why did the Berry phase/WZ-action also appear in our earlier discussion of a spin coupled to a \ufb01xed magnetic \ufb01eld? To answer this question, consider the world from the point of view of the spin. In the reference frame of the spin, the magnetic \ufb01eld is dynamical and a Berry phase term will be generated. Moreover, the area traced out by the \ufb01eld vector (i.e. the area determining the geometric phase) is equal to the area traced out by the spin (now we are back in the \ufb01xed-\ufb01eld reference frame) in its motion around the \ufb01eld axis. 556 Topology nature of the phase even more strongly: for any value of x, we have a state|0(x ). We may then construct the di\ufb00erential 1-form 0(x)|d0(x).61 The geometric phase is obtained by evaluating the integral of this form along the curve c. The advantage of the third representation above is that it suggests yet another formulation of the geometric phase, at least in cases where a closed path in parameter space is traversed. For a closed loop c, application of Stokes\u2019 theorem gives, \u03b3 = i 0|d0 = i d0| \u2227 |d0, c S (9.50) where S may be any surface in parameter space that is bounded by \u03b3. This last representation is aesthetic, and easy to memorize, but also a bit too compact to be of real computational use. To give it a more concrete meaning, we insert a spectral decomposition in terms of instantaneous eigenstates, \u03b3 = i S m=0 d0|m \u2227 m|d0. (Exercise: Why does the m = 0 term vanish, i.e. d0|0 \u2227 0|d0 = 0? Hint: Make use of the fact that d0|0 = d0|0 + 0|d0 = 0 and the skew-symmetry of the \u2227-product.) We now evaluate the equation 0 = m|d[( \u02c6H \u2212 0)|0] to obtain m|d0 = (0 \u2212 m)\u22121m|dH|0 or \u03b3 = i S m=0 0|dH|m \u2227 m|dH|0 (m \u2212 0)2. (9.51) This is about as far as we get in general terms. We have established the geometric nature of the Berry phase. However, inasmuch as it requires explicit knowledge of the spectrum of \u02c6H(x(t)), the actual calculation of the phase remains a di\ufb03cult problem. There are cases, however, where the calculation of the geometric phase reduces to a straightforward surface integral in parameter space. One such example is provided by a spin subject to a weakly time-dependent magnetic \ufb01eld. Consider the Hamiltonian \u02c6H = \u03bcn(x) \u00b7 \u03c3 \u2261 \u03bcU (x)\u03c33U \u22121(x), where U is the rotation matrix introduced in Eq. (9.44). The 2S + 1 instantaneous eigenstates of \u02c6H are given by U |S3, where\u03c3 3|S3 = S3|S3 and S3 = \u2212S,..., S is the azimuthal spin quantum number. To compute the Berry phase, we can consider the \ufb01rst of the two representations in Eq. (9.50). Noting that the ground state is given by |0 = U |S3 = \u2212S (we assume that the magnetic moment \u03b3 > 0) and parameterizing the rotation matrix as in Eq. (9.45), one veri\ufb01es that 0|d0 = \u2212S|U \u22121", "dU | \u2212S = iS(1 \u2212 cos \u03b8)d\u03c6. We thus obtain \u03b3 = S c d\u03c6 dt (1 \u2212 cos \u03b8) \u02d9\u03c6, which coincides (up to a Wick rotation t \u2192 \u2212i\u03c4, which we know (1 \u2212 cos \u03b8) = S does not a\ufb00ect topological terms) with the S = 1/2 WZ-action Eq. (9.48). : 9.4.4 Spin chains: beyond the semi\u2013classical limit In Section 9.3.3, we began to explore the physics of one-dimensional spin chains. Much of our analysis was limited to the case S 1, a semi-classical regime where quantum \ufb02uctuations 61 Should you \ufb01nd the representation too abstract, choose any basis {|\u03bb} and write 0(x)|d0(x) \u2261 0(x)|\u03bbd\u03bb|0(x), whered \u03bb|0(x) is the exterior derivative of the function \u03bb|0(x). 9.4 Wess\u2013Zumino terms 557 are weak. In this limit, the spin chain is described by an O(3) nonlinear \u03c3-model with a \u03b8term. This model, however, does not stay invariant under renormalization. It \ufb02ows towards a \u201cstrong coupling\u201d regime where \ufb02uctuations are large and the e\ufb00ective value of the spin (formally: the coupling constant of the gradient term in the model) becomes weak. Apart from the conjecture that, for half-integer/integer initial spin, the model \ufb02ows towards an ordered/disordered phase with gapless/gapped excitations, there was nothing we could say about the large-distance behavior of the model (or about the physics of chains consisting of small spins S = O(1)). In this section, we invest quite some e\ufb00ort in developing a fresh attack on the physics of the small S spin chain. This is motivated in part by the enormous amount of recent experimental activity on quasi-one-dimensional spin compounds. (\u201cReal\u201d spin chains are often realized as structural sub-units of transition metal compounds. In these systems, the spin \u2013 which is carried by Hund\u2019s rule coupled inner shell electrons of transition metal atoms \u2013 may reach as high as 7/2.) Another, and more theory-related, motivation for our study is that the low-energy physics of the spin chain is governed by a fascinating interplay of di\ufb00erent topologically non-trivial quantum \ufb01eld theories. Following A\ufb04eck and Haldane,62 the principal idea of our approach will be to exploit the equivalence of the antiferromagnetic spin chain to a one-dimensional Hubbard model at half \ufb01lling.63 The advantage gained by this digression is that the fermionic model is amenable to various analytical tools which (at least not in any known sense) do not apply to the spin-chain per se. We will \ufb01nd that the e\ufb00ective low-energy model describing the fermion system is the WZ \ufb01eld theory conjectured to be the \ufb01xed point theory of the \u03c3-model in Section 9.3.3. Fermion representation of the antiferromagnetic spin chain Consider a one-dimensional chain of equidistant (spacing a) lattice sites i. Each site hosts nc degenerate fermion states, or \u201corbitals.\u201d Consider a half-\ufb01lled situation, i.e. on average, each orbital is occupied by a single spin-(1/2) fermion. We now introduce some dynamics so that the low-energy physics of the system is equal to that of an antiferromagnetic spin chain with spin S = nc/2. (Notice that this construction is not quite as arti\ufb01cial as it may seem: the e\ufb00ective moments observed in \u201creal\u201d crystals are usually composite objects, composed of more than one elementary electron spin and stabilized by electron correlations.) To align the nc spins at each site, we introduce a strong (U T ) Hund\u2019s rule coupling \u02c6Hint = \u2212U = U 4 i i \u02c6Si \u00b7 \u02c6Si = \u2212 U 4 2 \u03c8\u03b1 ia \u2020\u03c8\u03b1 ib \u03c8\u03b1 ia \u2020\u03c3\u03b1\u03b2\u03c8\u03b2 ia \u2020 \u03c8\u03b1 ia \u03c3\u03b1\u03b2 \u03c8\u03b2 ia \u00b7 i \u2020 \u2212 \u03b4ab \u03c8\u03b2 ib \u03c8\u03b2 ia \u2212 \u03b4ba + \u03c8\u03b1 ia \u2020\u03c8\u03b1 i\u03b1 \u2212 nc 2 + const", ". (9.52) 62 I. A\ufb04eck and F. D. M. Haldane, Critical theory of quantum spin chains, Phys. Rev. B 36 (1987), 5291\u2013300. 63 Recall that, for strong interaction, the one-dimensional Hubbard model maps onto the so-called (t \u2212 J)-model which (at half \ufb01lling) reduces to the antiferromagnetic Heisenberg model (the spin chain). 558 Topology U = 0, J = 0 U > 0, J = 0 U > 0, J > 0 (a) (b) (c) Figure 9.14 (a) A chain of sites, each containing nc fermions on average. (b) A strong Hund\u2019s rule coupling maximizes the spin carried by each state to S = nc/2. (c) Upon the switching on of a \ufb01nite nearest neighbor hopping matrix element, the system becomes a spin S antiferromagnet. Here, a = 1,..., nc is the orbital index, \u03b1, \u03b2 are spin indices, and, to get from the \ufb01rst to the second line, we have used the identity \u03c3\u03b1\u03b2 \u00b7 \u03c3\u03b1\u03b2. Of special interest to our further discussion will be the last term in the second line. This Hubbard-type interaction tells us that each site favors a site occupancy of nc electrons (i.e half-\ufb01lled). Indeed, nc electrons are needed to manufacture a net spin of maximum weight S = nc/2 (see Fig. 9.14) and, thus, to optimize the Heisenberg interaction Eq. (9.52). \u03b4\u03b2\u03b1 \u2212 \u03b4\u03b1\u03b2\u03b4\u03b1\u03b2 = 2\u03b4\u03b1\u03b2 Let us now introduce a small amount of inter-site hopping: \u02c6H0 = \u2212 1 2 (ncJU )1/2 \u03c8\u03b1 ia \u2020\u03c8\u03b1 i+1a + h.c $, # i (9.53) where the constant J determines the hopping strength. In the limit J/U \u2192 0, the half-\ufb01lled system becomes equivalent to the spin S antiferromagnetic chain, \u02c6H = \u02c6H0 + \u02c6Hint \u02c6ni=nc\u2212\u2192 Haf = J i \u02c6Si \u00b7 \u02c6Si+1 + O(J/U ). (9.54) The easiest way to see this is to recall the situation in the standard Hubbard model (the nc = 1 variant of our present model) at half \ufb01lling (cf. Section 2.2). There, virtual deviations from half \ufb01lling led to an e\ufb00ective antiferromagnetic exchange coupling between the S = 1/2 spins carried by neighboring sites. The e\ufb00ective strength of this interaction was J \u223c t2/U, where t is the strength of the hopping term. Formally, the generalization of this mechanism to the case nc > 1 can be shown, for example, by subjecting the Hamiltonian \u02c6H to a canonical transformation eliminating the hopping term (all in complete analogy to the nc = 1 canonical transformation discussed in Section 2.2). We have, thus, established the equivalence between the strongly interacting J U fermion Hamiltonian and the spin chain. Now there comes a major conceptual jump \u2013 and admittedly one that is not backed up by quantitative reasoning: we postulate that the equivalence between the two systems pertains to the case J > xU, at least as far as the relevant long-range excitations are concerned. If this were not the case, there should be some kind of abrupt change (a phase transition) in the behavior of the system as the interaction 9.4 Wess\u2013Zumino terms 559 is increased. Although this scenario cannot be rigorously excluded, it seems unlikely to be taken too seriously. Non-abelian bosonization Let us, then, consider the low-energy physics of the weakly interacting (J > U ) fermion system. In fact, we shall begin by considering the totally non-interacting case.64 Switching from a lattice to a continuum description and linearizing around the two Fermi points (for details, see Section 2.2), we describe this prototypical system in terms of the action S[\u03c8\u2020, \u03c8] = s=\u00b11 dx d\u03c4 \u03c8\u2020r s (\u2212isvF\u2202x + \u2202\u03c4 ) \u03c8r s = d2x \u00af\u03c8r\u2202/\u03c8r, (9.55) where \u03c8s=\u00b11 are the left- and right-moving fermion \ufb01elds. In the latter equality we have set vF", " = 1 and switched to the Dirac notation (for details, see Section 4.3), and r = (a, \u03b1) = 1,..., 2nc \u2261 N is a composite index comprising spin and orbital components of the fermion \ufb01eld. In previous chapters, we have seen that the one-component (N = 1) variant of this model could be equivalently described in terms of a free bosonic action, S[ \u00af\u03c8, \u03c8] = d2x \u00af\u03c8\u2202/\u03c8 \u2194 S[\u03b8] = d2x \u2202\u03bc\u03b8 \u2202\u03bc\u03b8, (9.56) 1 2\u03c0 where the double arrow \u2194 indicates fermion operators O[ \u00af\u03c8, \u03c8] that all (currents, densities, etc.) can be expressed in terms of boson operators \u02dcO[\u03c6], and correlation functions O1O2 \u00b7 \u00b7 \u00b7 \u03c8 identically rewritten as \u02dcO1 \u02dcO2 \u00b7 \u00b7 \u00b7 \u03c6. can be Edward Witten, 1951\u2013 Mathematical physicist and string theorist. Awarded the 1990 Fields Medal for his ground breaking work in di\ufb00erential geometry. Witten contributed massively to the success of string theory. (Photo by Randall Hagadorn. Courtesy of the Institute of Advanced Study.) To what extent does this picture survive generalization to the manychannel case? The answer to this question was given in a seminal paper by Witten. Witten found that65 The free fermion action Eq. (9.55) can be equivalently described in terms of a two-dimensional nonlinear \u03c3-model with a Wess\u2013Zumino term. More precisely, he showed that S[ \u00af\u03c8, \u03c8] = d2x \u00af\u03c8r\u2202/\u03c8r \u2194 SWZW[g] = 1 8\u03c0 S2 d2x tr(\u2202\u03bcg\u2202\u03bcg\u22121) + \u0393 [g], (9.57) 64 As in our previous discussion of the single-channel case, it will turn out that the inclusion of interactions is straightforward once the e\ufb00ective bosonic degrees of freedom have been identi\ufb01ed. 65 E. Witten, Nonabelian bosonization in two dimensions, Commun. Math. Phys. 92 (1984), 455\u201372. 560 Topology where g \u2208 U(N ) and \u0393 [g] = \u2212 i 12\u03c0 B3 d3x ijk tr(g\u22121\u2202ig g\u22121\u2202jg g\u22121\u2202kg), (9.58) denotes the WZ action.66 On the right-hand side of Eq. (9.57), two-dimensional space-time has been compacti\ufb01ed to a 2-sphere S2. This sphere is then understood as the boundary of a three-dimensional unit ball, B3, which serves as the integration domain of the WZ functional. As in the one-component case, the double arrow in Eq. (9.57) implies equality of all correlation functions upon suitable identi\ufb01cation of operators (see the Info block below). Equation (9.58) is the multi-component or \u201cnon-abelian\u201d generalization of the prototypical bosonization identity (9.56). To understand the connection between the two equations, consider the restriction of U(N ) to its maximal abelian subgroup, i.e. the group of all diagonal matrices g = diag(ei2\u03b81 ). Evaluated on such con\ufb01gurations, the WZ term vanishes (exercise: show this) while the gradient term,,..., ei2\u03b8N S[\u03b8r] = N r=1 1 2\u03c0 d2x (\u2202\u03bc\u03b8r)2, (9.59) collapses to the sum of N free boson actions. This is the description we would have obtained had we applied Eq. (9.56) to each of the N fermion components individually (which, after all, is a perfectly legitimate thing to do!). But what, then, is the advantage of the generalized variant of bosonization Eq. (9.57)? Referring for a more detailed discussion to the Info block below, we here merely note that the action Eq. (9.57) possesses a huge symmetry group: transformations \u03c8s \u2192 gs\u03c8s, gs \u2208 U(N ), leave the action invariant, i.e. the symmetry group of the problem is given by U(N ) \u00d7 U(N ). Suppose we had bosonized each fermion in the standard abelian manner, thus arriving at the action Eq. (9", ".59). We might now ask how the symmetries of the problem \u2013 which of course must survive a change of representation \u2013 act in the \u03b8-language. Frustratingly, there is no answer to this question;67 the symmetries are no longer manifestly present and there is no direct way to bene\ufb01t from their existence. With the non-abelian generalization Eq. (9.57), this is not so. As we shall see in a moment, the symmetry group U(N )\u00d7U(N ) acts on the g-degrees of freedom by left\u2013right multiplication, g \u2192 g+gg\u22121 \u2212.68 In previous chapters we have emphasized time and again the tremendous importance of symmetries and the resulting conservation laws. Indeed it turns out that, in the present problem, the comparative complexity of the action Eq. (9.57) is far outweighed by the manifest presence of the symmetries. 66 Deviating from our earlier conventions, we will denote the WZ-action by \u0393[g] throughout (instead of the notation SWZ[g] used above). In doing so, we follow a standard literature convention. Also, the full action (including the gradient term) is commonly denoted by SWZW (where the last \u201cW\u201d credits Witten\u2019s contribution to two-dimensional WZ \ufb01eld theory), and the resemblance between the two symbols SWZ and SWZW may cause confusion. 67 One may object that a theory of free bosons is so simple that one need not care about the conservation laws introduced by symmetries. However, this argument is too shortsighted. It ignores the fact that most operators of interest are transcendental in the \u03b8s, i.e. the full theory is not quite as simple as Eq. (9.59) would suggest. 68 This provides another explanation of why the \u03b8-description is too narrow to accommodate the symmetry: a diagonal matrix g will not remain diagonal when acted upon by the transformation matrices. 9.4 Wess\u2013Zumino terms 561 INFO The proof of the bosonization identity (9.57) relies on an extension of the considerations summarized in Section 4.3 to the case of non-abelian symmetries. Although a detailed discussion of the construction (for which we refer to Witten\u2019s original, yet highly pedagogical, paper) would be beyond the scope of this text, let us brie\ufb02y summarize some of the principal ideas behind the method of non-abelian bosonization. Let us begin by recalling that, in the abelian case N = 1, we had two basic symmetries, \u03c8s \u2192 gs\u03c8s, whereg s = exp(i2\u03c6s), \u03c6s are constant phases and \u03c8s=\u00b11 the left- and right-moving fermion \ufb01elds. These symmetries express the independence of the left- and right-moving states or, equivalently, the chirality of the problem. A straightforward application of Noether\u2019s theorem (or, equivalently, a direct variation of the action) shows that the conservation laws induced by the chiral symmetry are \u2202z\u00afj = \u2202\u00afzj = 0, where j = \u03c8\u2020 +\u03c8+ and we have introduced complex coordinates z \u2261 1\u221a 2 (x0 + ix1), \u2202z = 1\u221a 2 In Section 4.3 we saw that, on the bosonic level, the chiral symmetries act as \u03b8 \u2192 \u03b8 + \u03c6+ \u2212 \u03c6\u2212. The conservation laws corresponding to these transformations read as \u2202z\u2202\u00afz\u03b8 = 0 (exercise: check this). At the same time we know that the transformation generated by \u03c6+ (\u03c6\u2212) generates the conservation law \u2202z\u00afj = 0 (\u2202\u00afzj = 0). Comparison with the two equations above leads to the identi\ufb01cation \u00afj = \u03c8\u2020 2\u03c0 has been included to obtain consistency with the de\ufb01nition of the vectorial current jv,0 = \u03c1 \u223c \u22021\u03b8 \u223c j + \u00afj.69 \u2202z\u03b8, where the factor of i/ \u2202\u00afz\u03b8 and j = \u03c8\u2020 \u2212\u03c8\u2212 \u2194 i\u221a 2\u03c0 \u2212\u03c8\u2212, \u00afj = \u03c8\u2020 +\u03c8+ \u2194 \u2212 i\u221a (\u22020 \u2212 i\u22021).69 \u221a 2\u03c0 We now reformulate these results in a manner amenable to generalization: introducing g \u2261 d2x \u2202\u03bcg \u2202\u03bcg\u22121, exp(i2\u03b8) \u2208 U(1), the bosonic action of", " Eq. (9.56) assumes the form S[g] = 1 8\u03c0 while the symmetries now act by conjugation: g \u2192 g+gg\u22121 \u2212. Here, ei2\u03c6\u00b1 \u2261 g\u00b1. Finally, the conserved currents can be de\ufb01ned as 1\u221a 8\u03c0 \u00afj = \u2212 1\u221a 8\u03c0 (\u2202\u00afzg)g\u22121. g\u22121\u2202zg, (9.60) j = At this stage, the ordering of the \u2013 abelian \u2013 factors g is of course arbitrary. However, this will change once we proceed to the multi-channel case. (Notice that the equations (9.60) are solved by g = g+(\u00afz)g\u2212(z), where the two independent factors g+ and g\u2212 describe the right/left-moving fermion states. This solution does not rely on the commutativity of the phases g and generalizes to the non-abelian case.) For N > 1, the phases gs generalize to two independent unitary matrices g\u00b1 \u2208 U(N ). The = 0, r, r = 1,..., N, where the cur- enlarged symmetry entails the conservation laws \u2202z\u2202\u00afzjrr + \u03c8r rents are given by jrr + and \u00afjrr = \u03c8r\u2020 = \u03c8r\u2020 \u2212 \u03c8r \u2212. EXERCISE To verify this last statement, consider the in\ufb01nitesimal unitary transformation Us = exp(i\u03b7W ) and expand the action Eq. (9.55) to \ufb01rst order in the Hermitian generator matrices W. What are the bosonic counterparts of these expressions? It is natural to generalize the phases g \u2208 U(1) of the abelian case to unitary matrices g \u2208 U(N ). As before, the symmetry group will act by left\u2013right-multiplication, i.e. g \u2192 g+gg\u22121 \u2212 and we expect the conserved currents to be given by Eq. (9.60).70 are related to the components of the vectorial currents discussed in Section 4.3 by jv0 \u223c j +\u00afj and jv1 \u223c 1 69 Throughout, it will be preferable to use this complex notation. Notice that the two conserved currents j and \u00afj i (j \u2212\u00afj). } and that \u2013 unlike in the abelian case \u2013 the ordering of the matrices g on the right-hand side of the de\ufb01nition is crucial. The factors in Eq. (9.60) are ordered in such a way that\u2202 \u00afzj = 0 and \u2202z\u00afj = 0 are compatible with each other. 70 Notice, however, that the currents are now matrices js = {jrr s 562 Topology We now need an action that (a) is invariant under the multiplicative action of the symmetry group and (b) for N = 1 reduces to \u2202\u03bcg\u2202\u03bcg\u22121. An obvious candidate would be S0[g] = 1 8\u03c0 d2x tr(\u2202\u03bcg\u2202\u03bcg\u22121). (9.61) Yet, for a number of reasons, Eq. (9.61) does not su\ufb03ce. Firstly, for this action, the conservation laws derived from the symmetry of the problem, read \u2202\u03bc(g\u22121\u2202\u03bcg) = 0, which is inconsistent with our result above, \u2202\u00afz(g\u22121\u2202zg) = 0. Secondly, we have seen in the previous chapter (cf. Section 8.5) that the nonlinear \u03c3-model Eq. (9.61) renormalizes at large length scales to smaller values of the coupling constant. At the same time, it is supposed to describe the fermionic action (9.55), which obviously does not renormalize. This tells us that S[g] alone does not su\ufb03ce to establish the boson\u2013fermion correspondence. But let us now inspect the second term in the action proposed by Witten, the two-dimensional WZ functional. The \ufb01rst thing we have to understand is why the second term of the action indeed represents a WZ functional in the sense of the discussion of Section 9.4.2.71 To construct a WZ term, we need a di\ufb00erential form \u03c9 on the target manifold that is closed, d\u03c9 = 0, but only locally exact (\u03c9 = d\u03ba only locally). In Problem 9.7.2 it is shown that, on a group-valued target manifold, these criteria are met by the form \u03c9 = tr(g\u22121dg \u2227 g", "\u22121dg \u2227 g\u22121dg). (Do not be confused by the ijklmno((g\u22121)ijdgjk \u2227(g\u22121)lmdgmn \u2227(g\u22121)nodgoi), where notation. What it really means is \u03c9 = dgjk is the di\ufb00erential form of the function gjk assigning to each element g \u2208 U(N ) its matrix B3 \u02dcg\u2217\u03c9 is component gjk.) The general theory developed in Section 9.4.2 then tells us that iC a WZ term where \u02dcg is a smooth extension of the \ufb01eld g : S2 \u2192 U(N ) to a \ufb01eld de\ufb01ned on the entire ball B3. Expressed in terms of some coordinate functions (x1, x2, x3) ofB 3, this expression becomes identical to Eq. (9.58). (As to the quantization of the coupling constant, see Problem 9.7.2.) In passing we note that the WZ functional is manifestly invariant under the action of the chiral symmetry group, as is required by the general structure of the theory. It is instructive to inspect the equations of motion obtained from the WZ term. In Problem 9.7.2 we show that, upon variation g \u2192 eW g (1 + W )g and expansion to \ufb01rst order in W, we obtain \u0393[(1 + W )g] \u2212 \u0393[g] = i 4\u03c0 d2x \u03bc\u03bd tr(W \u2202\u03bcg\u2202\u03bd g\u22121) +O (W 2). (9.62) A straightforward calculation shows that the variation of the gradient term S0 is given by S0[(1 + W )g] \u2212 S0[g] = \u2212 1 8\u03c0 d2x tr(W (g\u22022 \u03bcg\u22121 \u2212 (\u22022 \u03bcg)g\u22121)) + O(W 2). Combining these two results, we obtain the equations of motion \u22122i\u03bc\u03bd \u2202\u03bcg\u2202\u03bd g\u22121 + g\u22022 \u03bcg\u22121 \u2212 (\u22022 \u03bcg)g\u22121 = 0. A straightforward calculation shows that these equations are indeed equivalent to the relation \u2202z\u00afj \u221d \u2202z((\u2202\u00afzg)g\u22121) = 0. Summarizing, we have succeeded in \ufb01nding an action that (a) is manifestly chirally invariant and (b) produces the same conservation laws as the free fermion theory. Notice, however, that we have not \u201cproven\u201d the analogy Eq. (9.57); rather, our analysis was mostly based on drawing 71 Notice that, in Section 9.4.2, we focused on the case dim T = dim M + 1. Presently, however, dim M = 2 while dim T = dim U(N ) can become arbitrarily large. 9.4 Wess\u2013Zumino terms 563 illustrative analogies to the one-dimensional case. (For a more rigorous discussion, we refer to Gogolin et al.72) We conclude our preliminary survey of non-abelian bosonization by noting two crucial di\ufb00er- ences to the abelian case: In the abelian case, a free fermion action was mapped onto an equally free boson action. Being quadratic in the \ufb01elds, the two theories manifestly do not renormalize. In the non-abelian case, the situation is di\ufb00erent: although it is not obvious, the right\u2013hand side of Eq. (9.57) \u2013 a highly nonlinear functional of the group-valued \ufb01elds g \u2013 does not renormalize. For a proof of this feature we refer to Problem 9.7.3. Above, we have seen that bilinears \u223c \u03c8\u2020 \u00b1\u03c8\u00b1 composed of left- or right-moving fermions a\ufb00ord a representation in terms of the Bose \ufb01elds. Without proof, we mention that even bilinears involving fermions of di\ufb00erent chirality can be expressed in terms of bosonic \ufb01elds: \u03c8r\u2020 + \u03c8r \u2212 \u223c grr. (9.63) However, unlike in the abelian case (see Eq. (4.46)), no boson representation of individual fermion operators is known. This is quite unfortunate as it excludes the applicability of the formalism to several interesting \ufb01elds of investigation, notably the physics of fermions in the presence of disorder. In fact", ", the U(N ) action introduced in Eq. (9.57) de\ufb01nes two independent \ufb01eld theories at once: every matrix g \u2208 U(N ) can be decomposed as g = ei2\u03c6g into a matrix g \u2208 SU(N ) and a phase factor ei2N \u03c6 = det(g) \u2208 U(1). Substituting this decomposition into the action, we obtain S[g] = S[g] + N 2\u03c0 d2x (\u2202\u03bc\u03c6)2. S2 (9.64) Equation (9.64) tells us that the action decomposes into a WZW action for an SU(N )-valued \ufb01eld variable and an independent second action for the phase degree of freedom. Recalling that the invariance of the action under a homogeneous phase (gauge) transformation corresponds to the conservation of electric charge, we identify the phase action as that of the collective charge degrees of freedom (the charge density waves), while the SU(N )-action describes the spin degrees of freedom. Renormalization group \ufb02ow of the WZW model The action (9.57) describes a free fermion \ufb01xed point, if not in an obvious manner. To understand better the behavior of the WZW model under renormalization, let us generalize the free fermion action by introducing an arbitrary coupling constant \u03bb\u22121 in front of the gradient term. We thus consider the action S = 1 \u03bb d2x tr(\u2202\u03bcg \u2202\u03bcg\u22121) \u2212 i 12\u03c0 S2 d3x ijktr(g\u22121\u2202ig g\u22121\u2202jg g\u22121\u2202kg), (9.65) B3 where g \u2208 SU(N ) (and the complementary U(1) action is trivially free). Except for the presence of the WZ term, the model de\ufb01ned by this action is equivalent to the SU(N ) nonlinear \u03c3-model studied in Section 8.5. However, the latter is known not to have a \ufb01xed 72 A. O. Gogolin, A. A. Nersesyan, and A. M. Tsvelik, Bosonization and Strongly Correlated Systems (Cambridge University Press, 1998). 564 Topology point at \ufb01nite values of \u03bb. This tells us that the WZ term must have a crucial impact on the RG \ufb02ow. Suppose, then, we started renormalizing the model at small values of \u03bb. In this regime, the gradient term dominantly suppresses \ufb01eld \ufb02uctuations and we expect the WZ term to be of little signi\ufb01cance. Consequently, the RG \ufb02ow will initially resemble that of the standard SU(N ) model \u2013 towards larger values of \u03bb. Eventually, however, \u03bb and the coupling constant i/12\u03c0 of the WZ term will become of the same order. At this point, at last, the coupling constant will interfere with the \ufb02ow. As we know that \u03bb = 8\u03c0 de\ufb01nes a \ufb01xed point, we expect that it will simply truncate the \ufb02ow of the coupling constant. To con\ufb01rm this expectation, we need to go through the RG program, at least to one-loop order. Fortunately, however, the RG analysis of the WZW model (see Problem 9.7.3) almost exactly parallels that of the standard SU(N ) model. As a result, we can infer the scaling equation d\u03bb d ln b = N \u03bb 4\u03c0 1 \u2212 2 \u03bb 8\u03c0. (9.66) This result con\ufb01rms our qualitative expectation: the value \u03bb\u2217 = 8\u03c0 de\ufb01nes an (attractive) \ufb01xed point at which the upwards \ufb02ow of \u03bb comes to an end. Using the methods of conformal \ufb01eld theory, one can indeed show (see Witten\u2019s original paper65) that\u03bb \u2217 de\ufb01nes an exactly solvable reference point. (This is important additional information inasmuch as our one-loop analysis does not rigorously prove that \u03bb\u2217 is a \ufb01xed point.) WZW model of interacting fermions All we have accomplished so far is a highly complicated reformulation of the trivial free fermion problem. However, as we shall see in a moment, this exercise has been far from useless: as with the abelian case, it will turn out that the boson language is of unsurpassed e\ufb03ciency", " when it comes to the discussion of particle interactions. As usual, the most relevant particle interactions are mediated by certain four-fermion operators. Important constraints on the structure of these operators follow, once again, from the symmetries of the model. Above we have seen that the non-interacting model is invariant under transformations by U(2Nc) \u00d7 U(2Nc), where the \ufb01rst/second factor acts on spin and color indices of the left-/right-moving fermions. However, the Hubbard\u2013type interaction Eq. (9.52) reduces this symmetry. Inspection of the interaction operator shows that chiral symmetry gets lost (i.e. only transformations that act identically on the left-/right-moving sector are permitted, U(2Nc) \u00d7 U(2Nc) \u2192 U(2Nc)), and the remaining U(2Nc) \u2192 U(1) \u00d7 SU(2)\u00d7SU(Nc) gets reduced to symmetry transformations that act on the charge/spin/color sector separately. There are a number of continuum interaction operators that are compatible with the \u2261 sa \u03c8\u03b1 symmetries of the lattice system. Consider, for example, the bilinears jq s \u03c8\u2020\u03b1 \u2212 1 are the generators sa \u03c3i,\u03b1\u03b2\u03c8\u03b2 of U(Nc) transformations, and s = +/\u2212. These are the left-/right-moving components of the conserved currents jq,s,c = jq,s,c \u2212, generated by U(1), SU(2), and SU(Nc) sa, whereT j \u2208 u(Nc), j = 1, \u00b7 \u00b7 \u00b7, n2 c sa and jc s,j sa Taa \u03c8\u03b1 sa, js,i s + + jq,s,c \u2261 \u03c8\u2020\u03b1 \u2261 \u03c8\u2020\u03b1 9.4 Wess\u2013Zumino terms 565 transformations, respectively. As in the abelian case, interactions solely between left- or right-moving fermions are largely inessential. However, the operators \u03bbq = jq +jq \u2212, \u03bbs = tr(js +js \u2212), \u03bbc = tr(jc +jc \u2212), (9.67) are physically relevant and compatible with the symmetries of the model. Another relevant player is the \u201cumklapp operator\u201d \u03bbuk(\u03c8\u03b1\u2020 +a\u03c3\u03b1\u03b2 \u2212a ) + h.c.73 For the discussion of a few more allowed interaction operators, we refer to the original reference of A\ufb04eck and Haldane.62 \u2212a \u03c3\u03b1\u03b2 \u2020\u03b2 +a) (\u03c8\u03b1 2 \u03c8\u03b2 2 \u03c8 EXERCISE Show that the umklapp operator is invariant under the action of all three symmetry groups. Expressed in terms of the continuum \ufb01elds, the Hund\u2019s rules coupling Eq. (9.52) translates to a sum of the interaction operators listed above. The question we now have to answer is how these operators \u2013 which at su\ufb03ciently large strength will turn the free fermion model into the Heisenberg model in which we are interested \u2013 a\ufb00ect the long-distance dynamics of the model. Naturally, we shall address this question in the bosonized language developed in the previous section. As a warm-up to the case of arbitrary spin, we shall begin with the discussion of the colorless case, Nc = 1, i.e. the spin-1/2 chain. Writing g = ei2\u03c6g, whereg \u2208 SU(2) acts on the spin indices, and using Eq. (9.60) for g\u22121\u2202zg the bosonic representation of the currents, we obtain jq = \u2212 i\u221a for the spin current. Re\ufb02ecting its invariance under spin transformations, the bosonic representation of the umklapp operator reads \u223c \u03bbuk cos 4\u03c6. \u2202z\u03c6 and js = 1\u221a 8\u03c0 2\u03c0 EXERCISE To obtain the bosonic representation of the umklapp operator, substitute Eq. (9.63) into its de\ufb01nition and obtain \u223c \u03bbukei4\u03c6tr(g\u03c32gT \u03c32) + h.c. \u223c \u03bbuk cos(4\u03c6). Here the last equality is best proven by using the representation g = exp(iv \u00b7 \u03c3). The action generalized for the presence of these interaction operators takes the form S[\u03c6, g] = S[\u03c6] +S [g], where S[\u03c6] = 1 2\u03c0 (1 \u2212 \u03bbq) (\u2202\u03c6", ")2 + C\u03bbuk cos( \u221a 4\u03c6), d2x (9.68) S[g] = SWZW[g] +\u03bb s d2x tr(\u2202g\u2202g\u22121). Notably, the action still decouples into a spin and a charge sector (the separation of spin and charge characteristic of one-dimensional systems). Further, both S[\u03c6] andS [g] are old acquaintances: S[\u03c6] is the action of the two-dimensional sine\u2013Gordon model. In Section 8.6 we have seen that, for \u03bbq < 0 (repulsive interactions), it \ufb02ows towards a phase with a mass gap. This \ufb02ow is driven by the umklapp operator. What this tells us is that umklapp scattering leads to the presence of an excitation gap for charge density waves. This gap 73 Recall that umklapp scattering is the scattering of two fermion states of opposite spin from one point of the Fermi surface (\u00b1kF) to the other (\u2213kF). At half \ufb01lling, kF = \u03c0/2a, the momentum transferred in this process is of magnitude 2(\u03c0/2a \u2212 (\u2212\u03c0/2a)) = 2\u03c0/a = G, whereG = 2\u03c0/a is the reciprocal lattice vector. Since lattice momentum is conserved only up to multiples of G, umklapp scattering (at half \ufb01lling) is a permissible process. 566 Topology is nothing but the Mott\u2013Hubbard gap present in the spectrum of interacting fermions at half \ufb01lling. (Recall that the presence of umklapp processes is tied to the case of half \ufb01lling.) Turning to the action S[g], we note that the \u03bbs-perturbation merely renormalizes the coupling constant of the gradient term of SWZW, 1/8\u03c0 \u2192 (1/8\u03c0) + \u03bbs. However, in the previous section, we have seen that this change does not alter the long-range behavior of the system: irrespective of the value of the coupling constant, the model will \ufb02ow back towards the free fermion \ufb01xed point at \u03bb\u2217 = 1/8\u03c0. Summarizing, we have found that the Nc = 1 interacting fermion system builds up a Mott\u2013Hubbard gap for its charge excitations while the spin excitations are described by a critical WZW theory. Now, our entire analysis was based on the presumed equivalence (Nc = 1 interaction fermion system) \u21d4 (S = 1/2 Heisenberg chain). The existence of longrange excitations in the former then implies that the latter must be in an ordered phase (since a disordered phase would be de\ufb01ned by the absence of long-range excitations). This, however, is a result with which we are familiar.74 The truly interesting question is what happens for larger values of the spin (i.e. larger values of Nc). At Nc > 1, the structure of the theory gets signi\ufb01cantly more involved. Referring for a more detailed analysis to the original paper by A\ufb04eck and Haldane,62 we restrict ourselves here to a qualitative discussion of the most relevant aspects. Consider a matrix \ufb01eld g \u2208 SU(2) \u00d7 SU(Nc).75 Among the various interaction operators there is one that drives excitations of the color sector into a massive phase. This means that the soft excitations of the model can be parameterized as g = \u02c6g \u00d7 1Nc, where \u02c6g acts in the spin sector. Substituting these con\ufb01gurations into the free fermion reference action, we obtain SWZW[g] =N cSWZW[\u02c6g], where the prefactor Nc arises from the tracing out of the color sector. Remembering our discussion of the quantization of the WZ coupling constant in Section 9.4.2, we identify this model as a WZW model of level k = 2Nc. Among the plethora of conceivable perturbations of the critical theory, one operator family deserves special attention: tr(gn), where g \u2208 SU(2). (Here, we are writing g instead of \u02c6g for notational simplicity and n is some integer.) For two reasons, these are interesting operators: \ufb01rstly, they represent the most relevant perturbations of the theory (due to the absence of derivatives); secondly, they relate to an important discrete symmetry of the model. To understand why, let us return to the lattice version of the theory and consider the fermion representation of the spin operator Sj", " l. Decomposing the fermion into left- and right-moving components, \u03c8l = eikFl\u03c8+(xl) + e\u2212ikFl\u03c8\u2212(xl), s(xl)\u03c3j\u03c8s(xl) + [(\u2212)l\u03c8+(xl)\u2020\u03c3j\u03c8\u2212(xl) + h.c.] = where xl = la, we obtain Sj \u2212) + c(\u2212)l(g + g\u22121)]\u03c3j), where c and c are numerical constants. Here, we have tr([c(js used the relation eilkf x = ei(\u03c0/a)la = (\u2212)l as well as the bosonization identity Eq. (9.63). The second term under the trace is of particular interest. It tells us that translation by one site, l \u2192 l + 1, corresponds to a sign change of the \ufb01eld g. This observation can be read in di\ufb00erent ways. For example, contributions to the action that are not invariant l = \u03c8\u03b1\u2020 l \u03c3\u03b1\u03b2\u03c8\u03b2 + + js s \u03c8\u2020 l = 74 Recall that the S = 1/2 chain can be subjected to a Jordan\u2013Wigner transformation whereupon it becomes a model of spinless interacting fermions. The existence of gapless excitations (charge density waves) thus proves the existence of long-range order in the spin chain. 75 We will ignore the gapped charge sector throughout. 9.4 Wess\u2013Zumino terms 567 under g \u2192 \u2212g explicitly break translational invariance on the lattice. Similarly, a ground state that is not sign invariant cannot be translationally invariant, etc. After these preparatory remarks, let us return to the discussion of the operators tr(gn). As we are not interested in modeling situations where the Hamiltonian explicitly breaks translational invariance, only contributions with n even will be allowed. Physically, these operators correspond to products of n left-moving and n right-moving fermion states. In the case considered previously, Nc = 1, there is only one such contribution, namely the backscattering operator discussed above. However, for Nc > 1, terms tr(g2n\u2264Nc ) are d2x tr(g2).76 physically allowed. Let us, then, consider the action S[g] = NcSWZW + \u03bb Void of derivatives, the new contribution acts as a potential contribution to the action (although it is still invariant under the SU(2) symmetry transformation g \u2192 h\u22121gh). This being so, the physical behavior of the model crucially depends on the sign of the coupling constant \u03bb. For \u03bb < 0, the term \u03bb trg2 favors the mean-\ufb01eld con\ufb01gurations \u00afg = 1 or \u00afg = \u22121. Both ground states break the discrete sign inversion symmetry g \u2192 \u2212g and, therefore, they cannot be translationally invariant. Before identifying the physical meaning of these ground states, let us brie\ufb02y discuss the issue of \ufb02uctuations. Writing g = ei\u03c6j \u03c3j /2\u00afg and expanding to quadratic order in \u03c6, we obtain a mass term \u03bb d2x (\u03c6j)2. This means 2 that spin \ufb02uctuations around \u00afg are gapped. There is, indeed, one well known low-energy state of the spin chain that displays these features, the dimer phase. In this phase (see Fig. 9.15), spins at neighboring sites form spin singlets. Obviously, this state is not invariant under translation by one site (while translation by two sites is a symmetry). Further, there is no room for massless spin \ufb02uctuations, i.e. the system indeed shows an excitation gap. We next discuss the complementary case, \u03bb >0. In this case, con\ufb01gurations \u00afg with eigenvalues (i, \u2212i) are energetically favored. There is a continuous family of such states, namely \u00afg = exp(i\u03c0n \u00b7 \u03c3/2), where |n| = 1. Again, these states break translational invariance (g \u2192 \u2212g). However, they also break SU(2) symmetry g \u2192 hgh\u22121. This means that, unlike the dimerized case, the continuous spin rotation symmetry is spontaneously broken. Its breaking of both translational and spin rotational invariance identi\ufb01es \u00afg as the N\u00b4eel state of the chain. To describe the physics of the Goldstone modes corresponding to the symmetry broken state, we note that h exp(i\u03c0n \u00b7 \u03c3)h\u22121 = exp(", "i\u03c0n \u00b7 \u03c3), where the unit vector n = Rn and R \u2208 SO(3) is a rotation matrix canonically corresponding to the SU(2)-matrix h.77 Each n de\ufb01nes a new ground state, i.e. we have identi\ufb01ed the 2-sphere as the Goldstone mode manifold. EXERCISE Try to guess what the action of the Goldstone modes might be! To identify the Goldstone mode action, we substitute the soft \ufb01eld con\ufb01gurations exp(i\u03c0n(x) \u00b7 \u03c3/2) = in \u00b7 \u03c3 into the action. It is straightforward to verify that the gradient 76 The operator tr(g2) can be shown to be the most relevant of the family. 77 To any element h \u2208 SU(2), the prescription above assigns an element R \u2208 SO(3). The correspondence is 2\u20131 and not 1\u20131 because both h and \u2212h map onto the same R. 568 Topology (a) (b) Figure 9.15 Dimerized phase of the spin chain (a) vs. N\u00b4eel phase (b). term of the WZW action simply becomes S0[in \u00b7 \u03c3] = Nc d2x (\u2202n)2. As to the WZ 8\u03c0 functional, we show in Problem 9.7.2 that Nc\u0393[in\u00b7\u03c3] = Nc S2 n\u00b7\u22021n\u00d7\u22022n = \u03c0NcStop[n], 4 i.e. the \u03b8-term of the 2-sphere. Recalling that Nc = 2S, we conclude that, for large values of S (this is where the massive \ufb02uctuations around the Goldstone mode manifold can be neglected), the system is controlled by the action of the O(3) nonlinear \u03c3-model with topological angle 2\u03c0S, S[n] = 1 2\u03bb d2x (\u2202n)2 + 2\u03c0S Stop[n], (9.69) the same S[n] identi\ufb01ed earlier on semi-classical grounds as the low-energy action of the spin chain. This is now a good point to pause and consider what we have obtained. At \ufb01rst sight, it seems as if nothing much has been achieved: as a result of a long series of derivations we have arrived back at the semi-classical representation of the chain, the \u03c3-model. However, it turns out that it is the interplay between the WZW and the \u03c3-model that enables us to really understand the physics of the problem. To see this, let us consider the S = 1/2 chain in the presence of a next-nearest neighbor ferromagnetic coupling \u2013 admittedly an arti\ufb01cial model but, for the sake of the present argument, this does not matter. One can show that the long-range physics of this system is described by the O(3)-model with topological angle 2\u03c0S = \u03c0. On the other hand, we might have approached the problem via the WZW route discussed above. Within that context, the ferromagnetic perturbation turns out to be irrelevant (see A\ufb04eck and Haldane62), i.e. the long-range physics of the system is described by the critical free fermion WZW action. Comparing these two \ufb01ndings we can conclude that The O(3) nonlinear \u03c3-model with topological angle \u03c0 \u00d7 (odd integer) is equivalent to the SU(2) WZW action at critical coupling, (although no explicit \ufb01eld theoretical proof of this equivalence is known). This was the last missing piece of information needed to understand the long-range behavior of the spin chain. To summarize, we have found that (see Fig. 9.16) the antiferromagnetic spin S chain can be described in terms of a perturbed WZW action of level k = 2S. Depending on the sign of the most relevant perturbation \u223c \u03bb tr(g2) (which is set by the material parameters of the problem), this model can be either in a globally gapped phase \u2013 the dimer phase of the chain \u2013 or in a N\u00b4eel phase. The \ufb02uctuations superimposed on the N\u00b4eel phase are described by an O(3) model with topological angle 2\u03c0S. For integer spin, this model \ufb02ows towards a strong coupling phase, i.e. the spin chain is in a disordered state. However, for S half-integer, 9.5 Chern\u2013Simons terms 569 dimer phase spin S chain level 2S SU(2) WZ", "W model O(3) nonlinear \u03c3-model with topological angle 2\u03c0S half integer S level 1 SU(2) WZW model integer S massive phase ordered phase Figure 9.16 On the long-range physics of the antiferromagnetic spin chain. For a discussion, see the main text. it becomes equivalent to the level 1 WZW model at criticality, implying ordered behavior of the chain. Notice that none of the conclusions summarized above was established rigorously. More often than not we had to trust in the principle of adiabatic continuity, i.e. the belief that the physics of a system does not change qualitatively upon interpolating between the regime of weak interactions (where various approximation schemes work) to the regime of strong interactions (in which we are actually interested). Similarly, connections between di\ufb00erent theories were constructed on the basis of symmetry arguments and indirect reasoning (rather than by \u201chard-boiled\u201d calculations). In this way various pieces of evidence were pieced together to form a network that was intrinsically consistent and made physical sense. In recent years, this type of semi-quantitative research has become more and more prevalent in various areas of condensed matter physics. This development is driven by the increasing complexity of the questions and, relatedly, the absence of straightforward perturbative schemes. In the next section, we shall turn to another problem \ufb01eld where such \u201cdetective work\u201d has been successful, the fractional quantum Hall e\ufb00ect. 9.5 Chern\u2013Simons terms As our third and last example of topological \ufb01eld theories with relevance to condensed matter physics, we now turn to the discussion of Chern\u2013Simons (CS) theories. However, deviating from the strategy pursued in previous sections, this time we do not begin with a formal analysis of the underlying geometrical framework. Instead, we directly turn to a review of the master application of CS \ufb01eld theory in condensed matter physics, the fractional quantum Hall e\ufb00ect. It will then turn out that, once we have left the qualitative level and turned to the \ufb01eld integral description, we readily wind up in the basin of attraction of CS \ufb01eld theory. 570 Topology 9.5.1 Fractional quantum Hall e\ufb00ect (FQHE) In Section 9.3.4 we interpreted the QHE as a topological phenomenon. Depending on the chosen perspective, the integer number setting the Hall conductance can be interpreted as the number of occupied edge channels or, more formally, as the number of instanton excitations in Pruisken\u2019s \ufb01eld theory. Both \ufb01gures appear to be largely impervious to changes in the parameters of the system. In essence, the basic mechanism behind the formation of integer-valued Hall conductance had been understood shortly after the discovery of the e\ufb00ect. It thus came as a surprise when Tsui et al.78 discovered a sequence of plateaus at fractional values of the Hall conductance. More speci\ufb01cally, it turned out that: Fractional values \u03c3xy = \u03bd e2 m, of the Hall conductance are observed only in the purest samples. This indicates that, unlike with the integer e\ufb00ect, disorder does not stabilize the FQHE. h, \u03bd \u2261 n Not every rational n/m quali\ufb01es as a plateau value. The most prominent plateaus are observed for the \u201cprincipal sequence\u201d 1/m, where m is odd. More generally, plateaus have been observed for n m = p 2sp+1, s, p \u2208 N.79 Curiously, it has turned out that, at certain even-denominator fractions (formally, the limit p \u2192 \u221e in the hierarchy) the system largely behaves as if no magnetic \ufb01eld were present at all! For instance, in the vicinity of \u03bd = 1/2, pronounced Shubnikov\u2013de Haas oscillations (otherwise shown by Fermi liquids subject to a weak \ufb01eld) are observed. At \ufb01rst sight, the coexistence of phenomena of that degree of complexity with the (seemingly so robust) integer QHE may cause some consternation. But, then, let us recall that we are considering an isolated Landau level at fractional \ufb01lling! In other words, we are considering a hugely degenerate quantum state that is only partially populated. Within that environment, a macroscopic restructuring of the electron gas can be a\ufb00orded at little cost (which will be set by \u201cresidual\u201d mechanisms such as electron\u2013electron interactions and/or disorder scattering). On the same footing, it is clear that perturbative expansions", " around any given trial ground state will be pretty fragile. (Perturbations enjoy a huge phase volume while \u201cenergy denominators\u201d are small.) Keeping these things in mind, it is perhaps no longer surprising that dramatic things happen in the fractionally \ufb01lled Landau level. The considerations above suggest that elementary electrons \u2013 subject to the full strength of Coulomb interactions, at fully \u201cquenched\u201d kinetic energy \u2013 will hardly qualify as stable elementary excitations of the system. At the same time, the appearance of Fermi-liquid-like states at least for some \ufb01lling fractions suggests that the dominant players in the game are elementary fermionic particles. 78 D.C. Tsui, H.L. Stormer, and A.C. Gossard, Two-dimensional magnetotransport in the extreme quantum limit, Phys. Rev. Lett. 48 (1982), 1559\u201362. 79 In fact, the two-parameter \u201chierarchy\u201d de\ufb01ned by the right-hand side of this equation still is not general enough to account for all experimentally observed values. (The prominent exception is a fragile plateau at \u03bd = 5/2.) However, to explain these anomalous values of the Hall conductance, one has to keep track of the polarization of the electron spin, an extra level of complexity which we would here rather avoid. 9.5 Chern\u2013Simons terms 571 (a) (b) Figure 9.17 Illustrating the idea of the composite fermion approach. Imagine that an even number of \ufb02ux quanta constituting the magnetic \ufb01eld get tied to the electrons in the system. The new composite particles (\u201celectron + (2n) \ufb02ux quanta\u201d) continue to be fermions. They only see the remaining \u201cfree\u201d \ufb02ux quanta, i.e. a reduced external \ufb01eld (indicated by the lighter shading in part (b)). Remarkably, it is not at all di\ufb03cult80 to conceive of a fermionic quasi-particle picture wherein \u2013 at least at a mean-\ufb01eld level \u2013 the plethora of phenomena above admits a straightforward explanation: imagine the external magnetic \ufb01eld as a large number N\u03c6 of \ufb02ux tubes piercing the plane of the two\u2013dimensional electron gas (see Fig. 9.17). The ratio \u03bd \u2261 N/N\u03c6, where N denotes the number of electrons, de\ufb01nes the \ufb01lling fraction of the system. Since we are working under \u201cfractional\u201d conditions, 0 < \u03bd <1, the number of \ufb02ux quanta exceeds the number of charge carriers.81 Now assume that, by some mechanism, each electron matches up with an even number of \ufb02ux quanta to form a composite particle. (By way of example, consider \u03bd = 1/3, in which case there would be enough \ufb02ux quanta around to let every electron pair with two \ufb02ux tubes.) What can be said about the properties of these composite objects? Firstly, they would still be fermions. To understand why, recall that the statistics of particles can be probed by exchanging their position in space. However, our composite particles and the original electrons di\ufb00er only in the presence of an even number of integer \ufb02ux tubes. The \ufb02ux tubes give rise to additional phase factors, so that our new particles are fermionic: one speaks of composite fermions (CFs). The CFs see an e\ufb00ectively reduced external \ufb01eld. For example, for \u03bd = 1/3, each electron has absorbed two \ufb02ux quanta. Thus, the residual \ufb01eld seen by the CFs is three times lower than the original \ufb01eld. In other words, the number of remaining \ufb02ux quanta, N\u03c6 \u22122N = N, is equal to the number of CFs. Forgetting for a moment about the origin of the composite particles, we are considering a large number N of (composite) fermions subject to N \ufb02ux quanta, i.e. a situation where the integer QHE should arise. This picture suggests an interpretation of the FQHE as an IQHE of composite fermions.82 In the speci\ufb01c case of a half-\ufb01lled (\u03bd = 1/2) 80 However, our discussion should not deceive the reader about the fact that the composite fermion picture rep- resents the outgrowth of years of most intensive research! 81 If \u03bd > 1, \u03bd \u2212 [\u03bd], where [\u03bd] is the largest integer", " smaller than \u03bd, sets the \ufb01lling fraction of the highest occupied Landau level and our discussion applies to that level. 82 More generally, let us assume that every fermion binds 2s \ufb02ux quanta to it. Further suppose that N/(N\u03c6\u22122nN ) =p, i.e. that p Landau levels of the residual \ufb01eld are occupied. In this case, the CFs will also display the IQHE. This 572 Topology band, N\u03c6 \u2212 2N = 0, i.e. the CFs experience a mean-\ufb01eld of vanishing strength. This nicely conforms with the experimental observation of Fermi-liquid-like behavior (no QHE) close to half \ufb01lling. We have seen in Section 9.3.4 that, when a \ufb02ux quantum is adiabatically pushed through an annular quantum Hall geometry, an electron charge \ufb02ows from the inner to the outer perimeter of the sample. In a way, the phase vortex created by the addition of the gauge \ufb02ux leads to an expulsion of electronic charge. Similarly, the \ufb02ux tubes involved in the construction of the CF picture e\ufb00ectively carry a positive charge \u03bd. These \u201cscreen\u201d the charge carried by the bare electron, so that the CF is not exposed to the full strength of the unit-charge Coulomb interaction. In the history of the FQHE, the introduction of the CF picture by Jain83 was preceded by a number of other important developments. Shortly after the experimental discovery of the e\ufb00ect, Laughlin84 proposed a trial wavefunction which, in a close-to-optimal way,85 minimizes the Coulomb repulsion between the quasi-particles in the lowest Landau level. Most of the concepts central to the subsequent introduction of the CF picture \u2013 fractionally charged quasi-particles, incompressibility of the QH state, the importance of correlations, etc. \u2013 have e\ufb00ectively been motivated by this trial wavefunction. Later, the theory of the (many-body) CF system was formulated in terms of a Chern\u2013Simons type \ufb01eld theory.86 This e\ufb00ective \ufb01eld theory has become the basis of many subsequent analyses of FQHE phenomena. In this text, we shall turn the sequence of historical developments upside down: starting from a \ufb01eld theoretical description, we shall identify the CF degrees of freedom and then rediscover Laughlin\u2019s wavefunction. Once these structures are in place, the computation of the fractional Hall conductance will be little more than a straightforward exercise. 9.5.2 Chern\u2013Simons \ufb01eld theory: construction Consider the Hamiltonian of two-dimensional interacting electrons subject to a perpendicular magnetic \ufb01eld of strength B: \u02c6H = \u02c6H0 + \u02c6Hint where d2x a\u2020(x) 1 2m \u02c6H0 = Hint = 1 2 (\u2212i\u2202x + Aext)2 + V (x) a(x), d2x d2x (\u02c6\u03c1(x) \u2212 \u02c6\u03c10)V (x \u2212 x)(\u02c6\u03c1(x) \u2212 \u02c6\u03c10). happens for \ufb01lling fractions \u03bd = N/N\u03c6 = p/(2sp + 1). In essence, this simple picture explains the structure of the rationals where the FQHE is observed. 83 J. K. Jain, Composite-fermion approach for the fractional quantum Hall e\ufb00ect, Phys. Rev. Lett. 63 (1989), 199\u2013202. 84 R. B. Laughlin, Anomalous quantum Hall e\ufb00ect: an incompressible quantum \ufb02uid with fractionally charged excitations, Phys. Rev. Lett. 50 (1983), 1395\u20138. 85 For certain short-range correlated model interactions, the Laughlin wavefunction can even be shown to be an exact ground state, see F. D. M. Haldane, Fractional quantization of the Hall e\ufb00ect: a hierarchy of incompressible quantum \ufb02uid states, Phys. Rev. Lett. 51 (1983), 605\u20138. 86 A. L\u00b4opez and E. Fradkin, Fractional quantum Hall e\ufb00ect and Chern\u2013Simons gauge theories, Phys. Rev. B 44 (1991), 5246\u201362. 9.5 Chern\u2013Simons terms 573 2 (y, \u2212x)T is the vector potential of the external magnetic \ufb01eld (symHere, Aext = Bext metric gauge), V is a single-particle", " potential created by the presence of, for example, impurities, and \u02c6Hint describes the particle interaction on the background of a constant counter-density \u03c10. Singular gauge transformation After our discussion in the previous section, we do not expect bare electrons to be a useful reference for the construction of a low-energy \ufb01eld theory. Certainly, it will be more promising to start out from composite fermions as discussed above. As we have seen, a CF is an electron with 2s integer \ufb02ux tubes attached to it. Other fermions moving around the CF along a circular contour will acquire the winding phase 2s\u03c6 indicative of the presence of 2s \ufb02ux quanta. Within a \ufb01rst-quantized framework, these phase vortices can be attached87 to the position of each fermion by virtue of the \u201cgauge transformation\u201d \u23a1 \u23a4 \u03a8(x1,..., xN ) \u2192 \u03a8(x1,..., xN )exp \u23a3\u22122is i<j arg(xi \u2212 xj) \u23a6, (9.70) where \u03a8 is the many-body wavefunction, and arg(x) = tan\u22121(x2/x1) is the angle enclosed between x \u2208 R2 and the positive real axis. The transformation Eq. (9.70) becomes singular whenever two coordinates xi \u2192 xj approach each other, i.e. it does not represent an orthodox gauge transformation. In fact, the vector potential corresponding to the phase factor above, a = \u22122s\u2202x, creates a perpendicular magnetic \ufb01eld88 of strength (x1\u2212xi,1)e2\u2212(x2\u2212xi,2)e1 |x\u2212xi|2 i arg(x \u2212 xi) =\u2212 2s i b = ij\u2202xi aj = \u22124\u03c0s \u03b4(x \u2212 xi), (9.71) i.e. the \ufb01eld corresponding to 2sN \ufb02ux tubes centered at the coordinates of the fermions. Summarizing, the singular gauge transformation above converts the N fermions into a system of CFs (fermions with 2s \ufb02ux lines attached). i Derivation of the Chern\u2013Simons action Within the framework of the second quantization, the transformation Eq. (9.70) amounts to the replacement a\u2020(x) \u2192 a\u2020(x)exp \u22122is d2xarg (x \u2212 x)\u02c6\u03c1(x), where, as usual, \u03c1 = a\u2020a. EXERCISE Check that, with this de\ufb01nition, a quantum many-body wavefunction |\u03a8 \u2261 |0 transforms according to Eq. (9.70). (The index \u03bb refers to the states |\u03bb of a suitably N i=1 a\u2020 \u03bbi chosen single-particle basis [e.g. a basis of Landau states].) 87 In fact, the attachment of these phases is equivalent to placing \ufb02ux quanta at the winding center. 88 This would, of course, not be possible for a non-singular gauge transformation. 574 Topology Substituting the transformed operators into the Hamiltonian, one obtains \u02c6H0 \u2192 d2x a\u2020(x) 1 2m (\u2212i\u2202x + \u02c6A)2 + V (x) a(x), where \u02c6A = Aext + \u02c6a and \u02c6a(x) = \u22122s d2x (x1 \u2212 x 1)e2 \u2212 (x2 \u2212 x |x \u2212 x|2 2)e1 \u02c6\u03c1(x). (9.72) At this stage it is convenient to switch to a real-time \ufb01eld integral representation. We D( \u00af\u03c8, \u03c8) eiS[ \u00af\u03c8,\u03c8], where the actionS [ \u00af\u03c8, \u03c8] = thus introduce the partition function Z = N S0[ \u00af\u03c8, \u03c8] +S int[ \u00af\u03c8, \u03c8], with S0[ \u00af\u03c8, \u03c8] = Sint[ \u00af\u03c8, \u03c8] =\u2212 1 2 dt d2x \u00af\u03c8 i\u2202t + \u03bc \u2212 1 2m (\u2212i\u2202x + A[ \u00af\u03c8, \u03c8])2 \u2212 V (x) \u03c8, dt d2x d2x (\u03c1(x) \u2212 \u03c10)V (x \u2212 x)(\u03c1(x) \u2212 \u03c10), is obtained in the usual way by trading \ufb01eld operators for coherent state ampl", "itudes (specifically, \u03c1 = \u00af\u03c8\u03c8 and A[ \u00af\u03c8, \u03c8] \u2261 \u02c6A(a,a\u2020)\u2192(\u03c8, \u00af\u03c8)). Thanks to the presence of the vector potential A[ \u00af\u03c8, \u03c8], the kinetic energy operator has become a pretty unpleasant object, depending non-locally on up to six \ufb01eld amplitudes \u03c8, \u00af\u03c8. To avoid this complication89 let us shift the nonlinearities implied by the singular gauge transformation to some other place in the action. This can be achieved by promoting the vector potential to an integration variable whose value is set so as to generate the \ufb02ux pattern: i.e. multiply the partition function x,t \u03b4 (b(x, t) + 4\u03c0s\u03c1(x, t)), where b = ij\u2202ia\u22a5,j and the subscript \u201c\u22a5\u201d by 1 = N indicates that the integration extends only over transversal con\ufb01gurations of the vector potential (that is, con\ufb01gurations obeying \u2202iai = 0).90 As a result, we obtain the double functional integral, Da\u22a5'Z = N D( \u00af\u03c8, \u03c8) Da\u22a5 \u03b4 (b(x, t) + 4\u03c0s\u03c1(x, t)) exp \u2212S[ \u00af\u03c8, \u03c8, a\u22a5] = N x,t D( \u00af\u03c8, \u03c8) Da\u22a5 D\u03c6 exp iS[ \u00af\u03c8, \u03c8, a\u22a5] \u2212 i d2x dt \u03c6 \u2261 N D( \u00af\u03c8, \u03c8) Da\u22a5 D\u03c6 exp iSCF[\u03c8, \u00af\u03c8, a\u22a5, \u03c6] +i SCS[a\u22a5, \u03c6], b 4\u03c0s + \u03c1 \u03b8 2 where we have introduced the common shorthand notation \u03b8 \u2261 1/2\u03c0s, the action S[ \u00af\u03c8, \u03c8, a\u22a5] \u2261 S[ \u00af\u03c8, \u03c8] is obtained by replacing the \ufb01xed vector potential A[ \u00af\u03c8,\u03c8]\u2192Aext+a\u22a5 89 Technically, it is not advisable to disturb the structure of the most basic operator of the theory. 90 This latter condition is necessary because the \u03b4-distribution does not \ufb01x the longitudinal, or gauge, freedoms of the potential. In two dimensions, the decomposition of the vector potential into longitudinal (\u03d5) and transverse (\u03b8) components is achieved by setting ai = \u2202i\u03d5 + ij \u2202j \u03b8. 9.5 Chern\u2013Simons terms 575 a[ \u00af\u03c8, \u03c8] by the integration variable a\u22a5,91 SCF[\u03c8, \u00af\u03c8, a\u22a5] = d2x dt \u00af\u03c8 i\u2202t + \u03bc \u2212 \u03c6 + 1 2m SCS[a\u22a5] = \u2212 d2x dt \u03c6ij\u2202ia\u22a5,j, (\u2212i\u2202x + A)2 \u2212 V \u03c8 + Sint[ \u00af\u03c8, \u03c8], (9.73) (9.74) and A = Aext + a\u22a5. At this stage, we have ful\ufb01lled our intermediate goal; a \ufb01eld integral representation has been derived wherein particles are tightly bound to \ufb02uxes. There is, however, something unsatisfactory about the present representation of the theory: the way the variables a\u22a5 and \u03c6 enter the action (9.74) suggests an interpretation of the theory as one of fermions coupled to a (2 + 1)-dimensional electromagnetic gauge potential a\u22a5 \u2261 (\u03c6, a\u22a5). However, the action SCF + SCS falls short of the two standard criteria any decent theory of electromagnetism should obey: (i) in its present representation, it has nothing to say about the longitudinal degrees of freedom a of the vector potential and, relatedly, (ii) the contribution SCS is not gauge invariant (which means that, using a covariant representation x = {x\u03bc} \u2261 (x0 \u2261 t, x1, x2), under a transformation a\u03bc \u2192 a\u03bc + \u2202\u03bcf, \u03bc = 0, 1, 2, it changes value). We now claim that the action SCS possesses a natural gauge-invariant extension, namely the well-studied Chern\u2013Simons action92 SCS[a] = \u2212 d3x \u03bc\u03bd\u03c3a\u03bc\u2202\u03bda\u03c3. (9.75) Firstly, a straightforward integration by", " parts shows that SCS is gauge invariant. Secondly, one veri\ufb01es (exercise!) that, for a purely transverse con\ufb01guration a\u22a5 = (\u03c6, \u22022\u03b8, \u2212\u22021\u03b8), SCS[a\u22a5] reduces to the form given in Eq. (9.74). Put di\ufb00erently, the prototypical action (9.74) is but the gauge-invariant Chern\u2013Simons action evaluated in a particular gauge, namely the Coulomb or radiation gauge a = 0. The gauge-invariant extension of the theory is obtained by integration over all gauge sectors, Z = N D( \u00af\u03c8, \u03c8) Da exp iSCF[ \u00af\u03c8, \u03c8, a] +i SCS[a], \u03b8 4 (9.76) where SCF[\u03c8, \u00af\u03c8, a] = d3x \u00af\u03c8 i\u22020 + \u03bc \u2212 \u03c6 + 1 2m (\u2212i\u2202x + Aext \u2212 a)2 \u2212 V \u03c8 +Sint[ \u00af\u03c8, \u03c8], (9.77) 91 Notice that Eq. (9.72) is purely transversal, so that the replacement \u02c6a \u2192 a\u22a5 makes sense. 92 Recall that space-time vectorial components are de\ufb01ned as x\u03bc = (x0, x1, x2), while \u2202\u03bc = (\u2212\u22020, \u22021, \u22022). 576 Topology x1 x2 Figure 9.18 Di\ufb00erent ways of realizing constructive particle exchange along transmutation paths. is but the action (9.74) generalized to arbitrary gauge \ufb01eld con\ufb01gurations.93 In the following we will investigate what information can be obtained from the functional integral Eq. (9.76) about the physical behavior of the FQH system. Particle exchange in two dimensions Before embarking on this program, it may be of interest to discuss a few general aspects of Chern\u2013Simons \ufb01eld theory and particle transmutation statistics. In undergraduate quantum mechanics courses we learn to discriminate between particles with bosonic and fermionic statistics. On the formal level, the distinction between the two is met by considering the behavior of the wavefunction under an \u201cexchange of particles,\u201d namely \u03a8(..., x1,..., x2,...) = \u00b1\u03a8(..., x2,..., x1,...), (9.78) for bosons/fermions respectively. In fact, however, this de\ufb01nition should leave one with a certain feeling of uneasiness: what is actually meant by the phrase \u201cexchange of particles\u201d? Surely, the de\ufb01nition above does not imply a concrete physical prescription, i.e. strictly speaking it does not make sense. To appreciate the fact that we are not just discussing a formal subtlety, let us try to give the exchange of particles a more physical meaning. (The construction below follows closely an argument from chapter I.2 of Wilczek.94) Consider two quantum particles occupying positions x1, x2 in a two-dimensional system. Suppose we were interested in computing the amplitude for these particles to re-occupy the positions x1, x2 after some time t. Clearly, these amplitudes receive two distinct contributions, (i) x1 \u2192 x1, x2 \u2192 x2 and (ii) x1 \u2192 x2, x2 \u2192 x1 where x \u2192 y denotes the single-particle amplitude for propagation from x \u2192 y. Interpreting the second process as an operation of particle exchange, we are interested in identifying a \ufb01xed relative phase between (i) and (ii). Thinking about the total transition process in terms of a coherent double sum over single-particle paths, it is clear that contributions from (i) cannot be continuously deformed into those of type (ii). The path-sum falls into disconnected pieces, implying that there will be no variational (or classical) principles telling us about the relative phase. Yet, quantum mechanics itself provides us with 93 As usual with gauge theories, the integral Eq. (9.76) must be interpreted in a quali\ufb01ed sense: the very fact that the action is gauge invariant implies that the integration over all the di\ufb00erent gauge realizations yields the \u2013 in\ufb01nite \u2013 volume of the gauge sector. To give the functional integral some meaning, a gauge \ufb01xing contribution Sfix[a] has to be added to the action, i.e. a contribution that restricts the integration to a speci\ufffd", "\ufffdc reference gauge. Mostly, however, the presence of the gauge \ufb01xing action is not indicated explicitly. 94 F. Wilczek, Frictional Statistics and Anyons Superconductivity (World Scienti\ufb01c Publishing, 1990). 9.5 Chern\u2013Simons terms 577 an important clue: from the point of view of the \ufb01rst of the two particles, the net result of a (i)-process will be a rotation of particle 2 (around the position of particle 1) by an angle \u03c6 \u2208 2\u03c0Z. This is illustrated for \u03c6 = 0 (left) and \u03c6 = 2\u03c0 (right) in Fig. 9.18. Conversely, a (ii)-process corresponds to a rotation by \u03c6 \u2208 (2Z + 1)\u03c0. (For\u03c6 = \u03c0, see the center of the \ufb01gure.) Now, let us suppose that the topologically distinct processes di\ufb00er by some phase \u03ba(\u03c6). If we iterate transmutation processes, (x1, x2) t\u2192 (x2, x1) t \u2192 (x1, x2), the winding angles add while quantum mechanics requires that the topological phases multiply, \u03ba(\u03c6 + \u03c6) = \u03ba(\u03c6)\u03ba(\u03c6), implying that \u03ba(\u03c6) = exp(i\u03c8\u03c6), where \u03c8 \u2208 [0, 2] is some parameter.95 Speci\ufb01cally, a single exchange operation (x1, x2) \u2192 (x2, x1) corresponds to a phase ei\u03c0\u03c8. Let us compare this with the formal exchange de\ufb01nition above. According to Eq. (9.78), a twofold exchange, (x1, x2) \u2192 (x2, x1) \u2192 (x1, x2), leaves the wavefunction unaltered. In contrast, even the most elementary \u201cphysical\u201d exchange procedure corresponds to a winding angle \u03c6 = \u03c0 and, therefore, to a topological phase exp(2\u03c0i\u03c8). Only for the special choices \u03c8 = 0 (bosons) or \u03c8 = 1 (fermions) do we recover the result of the formal exchange. For a beautiful (yet non-path-integral-oriented) extension of the arguments above to a physically meaningful exchange prescription for general N -particle systems we refer to the seminal paper of Leinaas and Myrheim.96 Presently, all we need to appreciate is that a constructive exchange operation appears to leave more room for non-trivial (i.e. = 1, \u22121) transmutation statistics. Particles with \u03c8 = 0, 1 have been dubbed anyons, where the \u201cany\u201d stands for \u201cany exchange statistics.\u201d Skeptical readers may justly object that the overwhelming majority of particles observed in physics are of either bosonic or fermionic type. So where, then, does anyonic exchange statistics play a role? The short answer to this question is that our discussion above was critically tied to the two-dimensional system. In three or more dimensions, the winding angle \u03c6 is de\ufb01ned only mod 2\u03c0. For example, in the three-dimensional world, you might use the dimension perpendicular to the paper plane to contract the process shown in the right hand of Fig. 9.18, \u03c6 = 2\u03c0, to a \u03c6 = 0 type process. This forces \u03c8 to be an integer, i.e. \u03c8 = 0 (bosons) and \u03c8 = 1 (fermions) simply exhaust the list of possible options:97 in d = 2 anyons do not exist.98 Yet, in various two-dimensional applications, anyonic excitations do play an important role. For example, a theoretical approach to the FQHE competing with the CF approach discussed in the text is based on compositebosons.99 Although these are not anyons in the strict sense, here, too, the exchange statistics has been externally modi\ufb01ed (in this case to manufacture bosons from fermions). Indeed, for a while, attempts were made to link anyons to the subject 95 Since \u03c6 is a multiple of \u03c0, \u03c8 is de\ufb01ned only mod 2. 96 J. M. Leinaas and J. Myrheim, On the theory of identical particles, Il Nuovo Cimento B 37 (1977), 1\u201323. 97 In d = 1, particles cannot be exchanged anyway, which accounts for the interchangeability of bosonic and fermionic modelings. 98 Another argument can be developed to the same e\ufb00ect: according to the spin-statistics theorem, the exchange statistics of particles is intimately tied to the quantization of angular momentum", ". The latter, in turn, is a direct consequence of the SU(2) commutation relations [ \u02c6Ji, \u02c6Jj ] = iijkJk: the structure of the right-hand side \ufb01xes the dimension of the irreducible representations of SU(2) and, thereby, the particle statistics. Yet, when restricted to a two-dimensional world, the algebra of angular momentum becomes one-dimensional, i.e. abelian. The absence of angular momentum quantization then implies that the spin statistics theorem loses its meaning. 99 S. C. Zhang, T. H. Hansson, and S. Kivelson, E\ufb00ective \ufb01eld theory model for the fractional quantum Hall e\ufb00ect, Phys. Rev. Lett. 62 (1989), 82\u20135. 578 Topology of high-temperature superconductivity.100 Further, there is compelling evidence that the quasi-particle excitations in quasi-one-dimensional FQHE systems show fractional statistics, etc. Finally, \ufb01eld theories of anyons have been considered as prototypical model systems of matter \ufb01elds coupled to massive gauge \ufb01eld excitations. These applications are motivation enough to brie\ufb02y discuss the quantitative theoretical framework of anyon dynamics in twodimensional systems. Let us start out from the following functional setup: Z = Da D\u03c6 exp iS0[\u03c6] +i d3x j\u2191 \u03bca\u03bc \u2193 + SCS[a], i\u03b8 4 N D\u03c6 = where either D( \u00af\u03c8, \u03c8) may stand for a coherent state \ufb01eld integral, or D\u03c6 = i=1 D(xi, \u02d9xi) for a multiple path integral over the con\ufb01guration space trajectories {xi} i=1(1, \u02d9xi)\u03b4(x \u2212 xi(t)) is of N particles. In the latter case (see the exercise below), j(x) = the current density carried by the world lines of N particles. N EXERCISE Subject the \ufb01rst-quantized many-particle Hamiltonian \u02c6H = 2m + V (\u02c6xi)) to the singular gauge transformation (9.70). Then construct a path integral representation for the transition amplitude x1,...,x N | \u02c6U (t)|y1,...,y N. Show that the decoupling of the nonlinear gauge \ufb01eld contribution acquired by the Hamiltonian exactly parallels the \ufb01eld integral scheme discussed above. As before, the action contains the Chern\u2013Simons contribution Eq. (9.75) and a current vector potential coupling. The latter is given by the canonical expression i=1( \u02c6pi2 N dt i (\u2212\u03c6(xi) + \u02d9xi \u00b7 ai(xi)) \u2261 dt i a\u03bc(xi)j\u03bc(xi, \u02d9xi) \u2261 d3x a\u03bc(x)j\u03bc(x). In the \ufb01rst line, j( \u02d9x) \u2261 (1, \u02d9x) denotes the (2 + 1)-dimensional current carried by a point particle while, in the second line, j(x) \u2261 i(1, \u02d9xi)\u03b4(x \u2212 xi) represents the corresponding current density. (Upon space integration, d2x j\u03bc(x) = i j\u03bc( \u02d9xi).) We next employ the path integral variant to show that the statistical angle \u03b8 controls the transmutation statistics of the planar particle system. Let us consider a system of N = 2 particles initially (t = ti) prepared to occupy the coordinates (y1, y2). We want to analyze the transition amplitude y1, y2| \u02c6U (tf, ti)|y1, y2 to reoccupy the same state at a time tf. As in our qualitative construction above, the path integral describing this amplitude receives contributions from topologically distinct pairs of trajectories. (The world lines of two trajectories with winding angle \u03c6 = 0 (left) and \u03c6 = 2\u03c0 (right) are shown in the \ufb01gure.) To explore the impact of the statistical vector potential on the transition amplitude it is convenient to switch to an imaginary time formalism. Upon analytic continuation t \u2192 \u2212i\u03c4, a0 \u2192 ia0, the path integral for the two transition amplitudes assumes the form Z = D(x1, x2) exp \u2212S0[\u03c6] + d3x j\u03bca\u03bc \u2212 i\u03b8 4 SCS, 100 R. B. Laughlin, The relationship between high", "-temperature superconductivity and the fractional quantum Hall e\ufb00ect, Science 242 (1988), 525\u201333. 9.5 Chern\u2013Simons terms 579 \u2202\u03c4 where j\u03bc = i \u2202x\u03bc(\u03c4 ) \u03b4(x \u2212 x(\u03c4 )) is the (2 + 1)dimensional imaginary time current. The most important consequence of the transition t \u2192 \u2212i\u03c4 \u2208 \u2212i[0, \u03b2] is that the initial and \ufb01nal points of our two space-time trajectories now have to be identi\ufb01ed (the usual temporally periodic boundary conditions of the imaginary time formulation101). A glance at the \ufb01gure shows that this leads to a pair of two closed world line curves which, for a winding angle \u03c6 = 2\u03c0n, are n-fold intertwined.102 Focusing on the two most elementary variants \u03c6 = 0 and \u03c6 = 2\u03c0, we next evaluate the Chern\u2013Simons action on these trajectory pairs. t t x1 x1 x2 We begin by integrating out the statistical vector potential. For now, it will be convenient to perform this integration in the radiation gauge, \u2202\u03bca\u03bc = 0. (Remember that, to integrate over a gauge \ufb01eld, a speci\ufb01c gauge has to be chosen.) This gauge can be selected by adding a gauge \ufb01xing d3x (\u2202\u03bca\u03bc)2 to the action. In the limit \u03b1 \u2192 \u221e, con\ufb01gurations with \u2202\u03bca\u03bc no contribution \u03b1 longer contribute to the integration and the gauge is e\ufb00ectively \ufb01xed. Equivalently, one may limit the integration from the outset so that only the two components of a(q) perpendicular to q are integrated over. Either way, one \ufb01nds (exercise) a\u03bc(q)a\u03bd(q)a = 2 \u03b4q+q, \u03b8 where \u00b7 \u00b7 \u00b7 a denotes functional averaging over the (imaginary time) Chern\u2013Simons action. Using this result, we obtain \u03bc\u03bd\u03c3q\u03c3 q2 x2 Z = D(x1, x2) exp d3x d3x j\u03bc(x)j\u03bd(x) a\u03bc(x)a\u03bd(x)a = = 1 2 1 2 \u2212S0[\u03c6] + \u23a1 \u23a3\u2212S0[\u03c6] + qq D(x1, x2) exp D(x1, x2) exp \u2212S0[\u03c6] + j\u03bc(q)j\u03bd(q) a\u03bc(q)a\u03bd(q)a \u23a4 \u23a6 \u03bc\u03bd\u03c3 \u03b8 q j\u03bc(q) q\u03c3 q2 j\u03bd(\u2212q). Let us now try to understand the meaning of the last term in the action, the remnant of the Chern\u2013Simons integration. In fact, this term will turn out to be a topological invariant which tells us about the degree of knotting of the two integration paths. To see this, let us de\ufb01ne \u02dcj\u03bc \u2261 \u2212ij\u03bc = \u2202\u03c4 \u03bc(\u03c4 )\u03b4(x1 \u2212 xa xa 1(\u03c4 ))\u03b4(x2 \u2212 xa 2(\u03c4 )), where x = (\u03c4, x1, x2). Temporarily forgetting about the time-like origin of the \ufb01rst component, we may think of \u02dcj as the current vector \ufb01eld created by two loops in three-dimensional a=1,2 101 For the present, however, the parameter \u03b2 does not carry any physical signi\ufb01cance. 102 For the moment, we do not consider odd multiples \u03c6 = (2n + 1)\u03c0 (corresponding to a particle exchange) since for y1 = y2 these cases do not have a meaningful imaginary time extension. 580 Topology x0 1 x2 x1 Figure 9.19 Two intertwined current loops in (2+1)-dimensional Euclidean space\u2013time. space, each carrying a current I = 1 of unit103 strength (see Fig. 9.19). The key point now \u02dcj\u03c3 is the magnetic \ufb01eld created by is that, according to Amp`ere\u2019s law, b\u03bc = \u2212i\u03bc\u03bd\u03c3 a static104 current distribution \u02dcj. We may thus write the topological contribution to the action as q\u03bd q2 < aa d3x \u02dcja \u03bc(x)ba ds \u00b7 ba = \u2212 i \u03b8 (x) =\u2212 i \u03b8 aa Sa d\u03c4 aa dS \u00b7 curl ba dxa(\u03c4 ) d\u03c4 = \u2212 i \u03b8 \u00b7 ba (x) aa Sa dS", " \u02dcja Stop[x1, x2] =\u2212 i \u03b8 = \u2212 i \u03b8 = \u2212 2i \u03b8 \u03b3a aa I(\u03b3a, \u03b3a ). 1, xa 2 }, and Sa a surface spanned by \u03b3a. The Here, \u03b3a is a shorthand for the curve {\u03c4, xa crucial last line states that the topological action is proportional to the current I(\u03b3a, \u03b3a ) \ufb02owing through the area spanned by loop a due to the presence of loop a (see Fig. 9.19). Obviously, I(\u03b3a, \u03b3a) = 0. For a = a, I(\u03b3a, \u03b3a ) \u2208 Z is equal to the number of times \u03b3a pierces Sa or, in other words, the degree to which the two loops are intertwined. Speci\ufb01cally, amplitudes where one of the particles encircles the other n times acquire a phase \u03ba(2\u03c0n) = e2\u03c0in\u03c8, where \u03c8 = 1/(\u03c0\u03b8). We have thus found that the Chern\u2013Simons action does the book-keeping of the anyonic exchange phases discussed qualitatively above. In fact, it is relatively straightforward to extend our present two-particle analysis to N particles, or to the fully-\ufb02edged formalism of the coherent state \ufb01eld integral. In the speci\ufb01c case of the Chern\u2013Simons \ufb01eld theory of the FQHE, \u03b8 = (2\u03c0s)\u22121, i.e. fermions get transformed to (composite) fermions. In general, however, \u03b8 may be tuned so as to generate any form of exchange statistics. 103 To compute the strength of the current, one may integrate \u02dcj over a space-like surface x0 = \u03c4 = const. intersecting the loop (see Fig. 9.19). The integral gives I = dx1 dx2 \u02dcj0(x) = \u02dcj0(\u03c4, x1(\u03c4 ), x2(\u03c4 )) = 1. 104 Keep in mind that, in our present picture, the 0-direction of space no longer carries the signi\ufb01cance of time. 9.5 Chern\u2013Simons terms 581 9.5.3 Chern\u2013Simons \ufb01eld theory II: analysis Equation (9.76) de\ufb01nes an exact reformulation of the FQHE \ufb01eld integral in terms of a Chern\u2013Simons action. Trusting that the CS gauge \ufb01eld degree of freedom is su\ufb03ciently benign, we now proceed to our familiar program \u201cmean-\ufb01eld + \ufb02uctuations.\u201d We begin by subjecting the Coulomb interaction Sint to a Hubbard\u2013Stratonovich transformation, eiSint[ \u00af\u03c8,\u03c8] = D\u03c3 e i 2 d3x d3x \u03c3(x)[V \u22121(x\u2212x)\u03b4(x0\u2212x 0)]\u03c3(x)+i d3x ( \u02c6\u03c1(x)\u2212\u03c10)\u03c3(x), where V \u22121 denotes the inverse of the interaction kernel. Integration over the \u2013 now Gaussian \u2013 CF degrees of freedom \u03c8 then brings us to the partition function, Z = N Da D\u03c3 eiS[a,\u03c3], where S[a, \u03c3] = \u2212i tr ln i\u22020 + \u03bc \u2212 \u03c6 \u2212 \u03c3 + (\u2212i\u2207 + A)2 \u2212 V \u2212 \u03c10 d3x \u03c3(x) 1 2m + 1 2 d3x d3x \u03c3(x)V \u22121(x \u2212 x)\u03b4(x0 \u2212 x 0)\u03c3(x) + \u03b8 4 SCS[a]. (9.79) Starting from this representation, we now subject the theory to a mean-\ufb01eld analysis. Mean-\ufb01eld equations Let us seek for solutions of the equations \u03b4S[a, \u03c3] \u03b4a\u03bc(x) \u00af\u03c3,\u00afa = \u03b4S[a, \u03c3] \u03b4\u03c3(x) \u00af\u03c3,\u00afa = 0. Explicitly performing the di\ufb00erentiation with respect to a0 = \u03c6, one may see that the \ufb01rst of these equations translates to the \u2013 by now familiar \u2013 form \u03c1[\u00afa, \u00af\u03c3] = 1 4\u03c0s \u00afb, (9.80) 2m (\u2212i\u2207 + A)2 \u2212 V )\u22121(x, x) denotes the local density where \u03c1[a, \u03c3](x) =", "i( i\u22020 + \u03bc \u2212 \u03c6 \u2212 \u03c3 + 1 of CFs, and the notation emphasizes the functional dependence of \u03c1 on a and the Hubbard\u2013 Stratonovich potential \u03c3. The di\ufb00erentiation with respect to the space-like components a does not yield independent new information; all it gives us is two relations expressing the compatibility of Eq. (9.80) with the continuity equation. Finally, di\ufb00erentiation with respect to \u03c3 gives \u03c3(x) =\u2212 d2x V (x \u2212 x) (\u03c1[\u00afa, \u00af\u03c3](x) \u2212 \u03c10)| x 0=x0. (9.81) This equation also a\ufb00ords a transparent interpretation: on the mean-\ufb01eld level the potential V [\u03c1 \u2212 \u03c10] created by local density \ufb02uctuations compensates the interaction potential. In \u223c the absence of \ufb02uctuations of the external potential V (x), Eq. (9.80) and (9.81) possess the obvious homogeneous solution \u03c1[\u00afa, 0] = \u03c10 = const., \u03c3 = \u03c6 = 0, \u00afb = 4\u03c0s\u03c10 \u21d4 a = 2s\u03bdAext. The particle density is homogeneous, implying that no mean-\ufb01eld interaction \u03c3 is generated. The strength of the equally homogeneous CS mean-\ufb01eld \u00afb is set by the average density \u03c1 of 582 Topology composite fermions. As expected, the e\ufb00ective \ufb01eld seen by the CFs, B = Bext \u2212 b, turns out to be lower than that of the external magnetic \ufb01eld. Speci\ufb01cally, for a half-\ufb01lled band, \u03bd = 1/2, \u03c10 = Bext/4\u03c0. In this case, CFs with two \ufb02ux quanta attached, s = 1, experience a mean-\ufb01eld B of vanishing strength. Now let us investigate the stability of the mean-\ufb01eld with respect to \ufb02uctuations. In many respects the N CFs behave like ordinary fermions subject to a perpendicular \ufb01eld of strength B. In particular, they will undergo a Landau quantization where, typically, the highest Landau level will be only partially occupied. Foreseeably, these generic con\ufb01gurations will be highly susceptible to all sorts of \ufb02uctuation e\ufb00ects (i.e. poor candidates for mean-\ufb01eld schemes): the massive degeneracy of the Landau levels leads to \u201csmall-energy denominators,\u201d so that even a slight perturbation/\ufb02uctuation may cause dramatic e\ufb00ects. We thus expect our present mean-\ufb01eld scheme to work only in the vicinity of \ufb01lling fractions \u03bd where an integer number p of CF Landau levels are fully occupied. These \ufb01lling fractions are determined by the equation \u03bde\ufb00 = p or \u03a6e\ufb00 = 2\u03c0N/p, where \u03a6e\ufb00 = BL2 denotes the total e\ufb00ective \ufb02ux piercing the system. Using the relation B = Bext \u2212b where b = 4\u03c0sN L\u22122, it is straightforward to solve these equations to obtain \u03bd = 2\u03c0N BextL2 = p 2sp + 1, in agreement with the experimental observation. Summarizing, the mean-\ufb01eld analysis of the Chern\u2013Simons action con\ufb01rms the basic expectation formulated on page 571 that the fractional QHE can be interpreted as an integer QHE of composite fermions. INFO There is one more aspect of the theory that can be explored on the level of plain mean\ufb01eld theory, namely the charge of the composite fermions. To this end, let us consider the functional expectation value of the operator O(x, x) \u2261. \u00af\u03c8(x)\u03c8(x) /. The correlation function O describes the amplitude for creation of a CF at a space-time point x and its annihilation at x. Let us regard this amplitude as the coherent sum over all paths \u03b3 connecting x and x. Due to the presence of the external magnetic \ufb01eld, each such contribution : acquires an Aharonov\u2013Bohm phase \u03c6\u03b3 \u2261 \u2212qe\ufb00 ext, where the coe\ufb03cient qe\ufb00 de\ufb01nes the e\ufb00ective charge of the particle. (As we shall see, the presence of the statistical \ufb01eld makes qe", "\ufb00 di\ufb00erent from the bare electron value q = 1.) Performing the Gaussian integral over \u03c8, we represent O as \u03b3 dx\u03bc A\u03bc 2 O(x, x) = x|(i\u22020 + \u03bc \u2212 \u03c6 \u2212 \u03c3 + (\u2212i\u2207 + A)2 \u2212 V )\u22121|x 1 2m 3 a,\u03c3 1 = x|U (t, t)|xa,\u03c3 0 = x(t)=x x(t )=x MF\u2248 const. \u00d7 Dx exp iS0[x, \u03c3] \u2212 i dt (\u03c6 \u2212 A \u00b7 \u02d9x) a,\u03c3 Dx exp iS0[x, 0] \u2212 iqe\ufb00 dx \u00b7 Aext. x(t)=x x(t )=x 9.5 Chern\u2013Simons terms 583 In the \ufb01rst line, the angular brackets denote the functional averaging over the Chern\u2013Simons action (a) and the interaction kernel (\u03c3). Further, we have made use of the fact that the (retarded) real time Green function is equal to the time evolution operator U. The latter is represented as a path (as opposed to a \u201c\ufb01eld\u201d) integral. The action of the path integral contains a \ufb01eldindependent contribution S0, the coupling to the scalar \ufb01eld component \u03c6, and the canonical coupling to the vector potential A. Until now, all manipulations have been exact. In the crucial last line, we then evaluate the (a, \u03c3)-integration in a mean-\ufb01eld approximation. Assuming that \ufb02uctuations (lumped into the \u201cconst.\u201d in front of the integral) will be small, this amounts to a substitution (a, \u03c3) \u2192 (\u00afa, 0) in the action. Using the fact that aext = 2s\u03bdAext, we arrive at the last line, where the coupling constant qe\ufb00 = 1 \u2212 2s\u03bd = 1 1 + 2sp, (9.82) is identi\ufb01ed as the e\ufb00ective charge of the CF. The line of reasoning above tells us that CFs e\ufb00ectively carry fractional charge. The partial \u201cscreening\u201d of the bare electron charge, 0 < qe\ufb00 < 1, is explained by the tendency of the phase vortices to expel electronic charge. In the speci\ufb01c case, \u03bd = 1/2 qe\ufb00 = 0, the electron charge becomes completely screened by 2s = 2 vortices, a result that led some authors105 to interpret the \u03bd = 1/2 FQHE as a dynamic phenomenon of charge dipoles \u2013 each comprising one electron (q = 1) and two vortices (\u201cq = \u22122 \u00d7 1/2 =\u22121\u201d). Fluctuations We now proceed to explore the role of quadratic \ufb02uctuations around the homogeneous mean-\ufb01eld. In particular: What does the \ufb01eld theory have to say about the electromagnetic response of the system? What else (beyond the fractional transmutation statistics) does it tell us about the micro- scopic features of the composite fermions? To answer these questions, we do not have to go into much quantitative detail. Rather the ubiquitous condition of gauge invariance and the presence of an excitation gap (an integer number of CF Landau levels are fully occupied!) su\ufb03ce to \ufb01x the structure of the quadratic action. Shifting a \u2192 \u00afa + a and \u03c3 \u2192 \u00af\u03c3 + \u03c3, our goal is to expand the action to second order in the deviations (a, \u03c3). Let us begin by considering the interaction \ufb01eld \u03c3. A shift a0 = \u03c6 \u2192 \u03c6 + \u03c3 removes \u03c3 from the \u201ctr ln\u201d component of the action and makes it reappear in the Chern\u2013Simons action. More speci\ufb01cally, the Chern\u2013Simons acquires a linear contribution106 SCS[a] \u2192 SCS[a] \u2212 2 d3x \u03c3b so that the total \u03c3-expansion of the action now reads S[a, \u03c3] = S[a] \u2212 \u03b8 2 1 2 d3x \u03c3(x)b(x) + d3x d3x \u03c3(x)V \u22121(x \u2212 x)\u03b4(x0 \u2212 x 0)\u03c3(x). 105 A. Stern, B. I. Halperin, F. von Oppen, and S. H. Simon, Half-\ufb01lled Land", "au level as a Fermi liquid of dipolar quasiparticles, Phys. Rev. B 59 (1999), 12547\u201367. 106 Here b = ij \u2202iaj describes the \ufb02uctuations of the statistical magnetic \ufb01eld around its mean value \u00afb. 584 Topology After the straightforward Gaussian integration over \u03c3 we then obtain S[a, \u03c3] D\u03c3\u2212\u2192 S[a] + d3x d3x b(x)V (x \u2212 x)\u03b4(x0 \u2212 x 0)b(x). 1 2 The physical interpretation of the induced term is obvious: in the CS theory, \ufb02uctuations of the particle density are tied to \ufb02uctuations of the statistical magnetic \ufb01eld. Accordingly, spatial inhomogeneities of the statistical magnetic \ufb01eld get penalized by an interaction contribution, as described by the second term above. However, as far as the two basic questions above are concerned, we may temporarily forget about this contribution (see, however, the concluding remarks below) and focus attention on the action S[a]. Let us, then, consider the action S[a, A] \u2261 SCF[\u00afa + a + A] + \u03b8 4 SCS[\u00afa + a], where SCF[a] = \u2212i tr ln i\u22020 + \u03bc \u2212 a0 \u2212 1 2m (\u2212i\u2207 \u2212 a)2 \u2212 V, denotes the CF contribution to the action to which we have coupled a source potential A. As usual, a two-fold di\ufb00erentiation with respect to elements of A will later tell us about the relevant transport coe\ufb03cients of the system. Next, we develop the formal expansion S[a, A] \u2261 n S(n)[a, A], where S(n)[a, A] is of total order n in a and A. The zerothorder term S(0) describes the CF system on the mean-\ufb01eld level and will not be of further interest to us. The \ufb01rst-order term S(1) does not contain a as we are expanding around a stationary con\ufb01guration. Moreover, its A dependence S(1)[A] = iA\u03bc\u00afj\u03bc is inessential because the mean-\ufb01eld CF density \u00afj0 is structureless and a mean-\ufb01eld current \u00afj = 0 does not \ufb02ow. However, at the second-order level S(2), things start becoming more interesting. Formally, the second-order contribution can be represented as S(2)[a, A] = 1 2 d3x d3x (a + A)\u03bc(x)K\u03bc\u03bd(x, x)(a + A)\u03bd(x) + \u03b8 4 SCS[a], (9.83) \u03b4a\u03bc(x)\u03b4a\u03bd (x) where K\u03bc\u03bd(x, x) = \u03b42SCF[a]. By construction,K is but the linear response kernel discussed in Section 7.2. For the moment, all we need to recall about this object is that \u2190 it (a) is gauge invariant, \u2202\u03bcK \u03bc\u03bd = K \u03bc\u03bd \u2202 \u03bd= 0, (b) is generally short-range (K(q) can be expanded in powers of q), and (c) contains information about both the polarizability of the medium and its conductivity. a=\u00afa \u221e Building on property (b), one can expand the second-order action in derivatives: l=0 S(2,l)[a, A], where S(2,l)[a, A] is ofl th order in derivatives (\u22020, \u2202x) S(2)[a, A] = and, as usual, one may focus attention on the contribution with the least number of derivatives. As discussed before, no gauge-invariant zero derivative term S(2,0) can be constructed. However, with the contribution linear in the number of derivatives, S(2,1), the situation is more tricky. While in general there are no gauge invariant contributions of \ufb01rst order in q \u2194 \u2212i\u2207, terms nominally scaling as |q| for q \u2192 0 exist. (For example, in systems with a non-vanishing longitudinal conductivity, the action takes the form S(2,1)[A] \u223c A \u03bd(\u2212q), where D denotes the di\ufb00usion constant.) However, the system at hand is an insulator, so that no such term is present. \u03bc(q) q\u03bcq\u03bd \u2212q2 Dq2+i\u03c9 A Yet, in two-dimensional systems,", " there exists one more gauge-invariant \ufb01rst-order derivative term, namely the notorious Chern\u2013Simons term! Thus, we are led to the preliminary 9.5 Chern\u2013Simons terms 585 identi\ufb01cation S(2,1)[a, A] = c SCS[a + A] + \u03b8 4 SCS[a], where the \ufb01rst contribution is obtained by a \ufb01rst-order gradient expansion of the CF action while the second contribution has been present from the outset. Of course, the coupling constant c remains to be determined. This can be done by explicit \u2013 and quite laborious \u2013 calculation or by the following more elegant construction: remembering that SCS[a] = \u2212 q \u03bc\u03bd\u03c3a\u03bc(q)q\u03bda\u03c3(\u2212q), and comparing with Eq. (9.83), our \ufb01ndings so far translate to the relation d3x \u03bc\u03bd\u03c3a\u03bc\u2202\u03bda\u03c3 = \u2212i (9.84) K\u03bc\u03bd(q) = \u2212i2c \u03bc\u03c3\u03bdq\u03c3 + O(q2). INFO To fully describe the electromagnetism of the statistical gauge \ufb01eld, we should also include the second-order derivative contribution S(2,2) in our analysis: gauge-invariant contributions of second order are obtained as bilinears formed from elements of the \ufb01eld tensor F\u03bc\u03bd = \u2202\u03bca d3x (e \u00b7e +\u03c7b2), where e 0 is the electric \ufb01eld derived from the gauge \ufb01eld a, and and \u03c7 are the electric and magnetic permeabilities, respectively. Physically, the action S(2,2) describes the electric and magnetic susceptibility of the (mean-\ufb01eld) CF medium to the presence of gauge \ufb01eld \ufb02uctuations. However, as far as our present objectives are concerned, these e\ufb00ects turn out to be of lesser relevance. \u03bc. These terms are then combined into an action S(2,2) = 1 2 i \u2212 \u2202ia \u03bd \u2212\u2202\u03bd a i = \u22020a On the other hand, we know the conductivity of the system relates to the linear response 12 = \u2212i limq\u21920 \u03c9\u22121K12(\u03c9, q). For the present, we have to kernel through the relation \u03c30 12 = p interpret \u03c30 2\u03c0 as the quantized Hall conductivity carried by the CF system at the mean-\ufb01eld level. Comparison with Eq. (9.84) then leads to the identi\ufb01cation c = \u03c312/2, or S(2,1)[a, A] = \u03c30 12 2 SCS[a + A] + \u03b8 4 SCS[a]. Of course, \u03c30 12 does not coincide with the actual Hall conductance \u03c312 carried by the system \u2013 within our present level of approximation, the latter is obtained by two-fold di\ufb00erentiation with respect to A after the statistical gauge \ufb01eld has been integrated out. Now, there is a simple general formula telling us what happens to two quadratic actions (such as the Chern\u2013Simons actions) upon integration over one half of the \ufb01elds: c1S[a + b] +c 2S[a] Da\u2212\u2192 1 + c\u22121 c\u22121 2 \u22121 S[b]. (9.85) EXERCISE Of course, Eq. (9.85) can be proven by straightforward Gaussian integration over a. Da\u2212\u2192 S[b, \u00afa[b]], A more elegant procedure is based on the fact that, for any quadratic action, S[a, b] where \u00afa[b] is the solution of the mean-\ufb01eld equation \u03b4aS[a, b] = 0. To make use of this identity, let us formally write S[a] =a T Ka, whereK is a non-degenerate107 operator kernel. Find the solution of the mean-\ufb01eld equations corresponding to the left-hand side of Eq. (9.85) and show that, upon substitution back into the action, we obtain the right-hand side. If you are critical, you may object that our CS actions are not, in fact, non-degenerate: when evaluated on a=\u00afa 107 Otherwise the \ufb01eld integration would create a headache! 586 Topology a pure gauge con\ufb01guration a\u03bc = \u2202\u03bcf, they vanish. Nevertheless, you may convince yourself that Eq. (9.85) remains valid if the integration", " over a is performed in some \ufb01xed reference gauge. When applied to our action S(2,1) above, the auxiliary identity Eq. (9.85) tells us that S(2,1)[a, A] Da\u2212\u2192 Se\ufb00 [A] \u2261 \u03c312 2 \u03b8 )\u22121. The terminology \u201c\u03c312\u201d is justi\ufb01ed because, as seen above, the where \u03c312 \u2261 ( 1 \u03c30 12 constant in front of a CS source action is but the Hall conductance of the system. Recalling that \u03b8 = 1/2\u03c0s and \u03c30 SCS[A], 12 = p/2\u03c0, we \ufb01nd + 2 \u03c312 = 1 2\u03c0 p 1 + 2sp, i.e. we con\ufb01rm the expectation of fractional Hall quantization. Let us recall that this result is critically linked to the absence of a longitudinal conductance \u03c311. In our analysis above, alluding to the striking analogies that exist between CFs and ordinary electrons in a magnetic \ufb01eld, we simply postulated \u03c311 = 0 at CF \ufb01lling p. (Technically, this happened when we said that the CF \u201ctr ln\u201d does not support a longitudinal current\u2013voltage relation.) However, one may note that the actual problem of Anderson localization of CFs has not yet been attacked on a truly microscopic level. What makes the problem so di\ufb03cult is the massive inter-CF correlations induced by \ufb02uctuations of the statistical gauge \ufb01eld. In the absence of disorder, the full extent of this correlation mechanism does not yet become clear (at least not on the level of our simplistic \u201cmean-\ufb01eld plus quadratic \ufb02uctuations\u201d analysis). However, once external inhomogeneities are present, things instantly become more complicated. For example, a static impurity potential will be screened by an inhomogeneous CF distribution. This creates an accumulation of statistical \ufb02ux that in turn acts as a scattering center of CFs, etc. Thus, we readily wind up with a full-blown problem \u201cinteraction + disorder + strong magnetic \ufb01eld\u201d whose rigorous microscopic solution seems to be elusive. Nonetheless, all evidence suggests that eventually the CFs will be localized, so that the analogy (FQHE of fermions) \u2194 (IQHE of composite fermions) remains valid. INFO As mentioned above, the \ufb01eld theory approach to the FQHE was preceded by a number of other developments. Perhaps most importantly, shortly after the experimental discovery of the e\ufb00ect, Laughlin proposed a trial wavefunction \u2013 nowadays generally referred to as Laughlin\u2019s wavefunction108 \u2013 which did a fantastic job at explaining much of the phenomenology of the FQHE. With the bene\ufb01t of hindsight (!) it is not di\ufb03cult to motivate the structure of the Laughlin wavefunctions from a few simple considerations. Consider a clean FQHE system at a \ufb01lling fraction \u03bd = 1/(2s + 1) belonging to the principal series.109 Any ground state many-body eigenfunction should (a) be built by superposition of single-particle states belonging to the lowest 108 R. B. Laughlin, Anomalous quantum Hall e\ufb00ect: an incompressible quantum \ufb02uid with fractionally charged excitations, Phys. Rev. Lett. 50 (1983), 1395\u20138. 109 For an extension to more complex fractions, see F. D. M. Haldane, Fractional quantization of the Hall e\ufb00ect: a hierarchy of incompressible quantum \ufb02uid states, Phys. Rev. Lett. 51 (1983), 605\u20138, and B. I. Halperin, Statistics of quasiparticles and the hierarchy of fractional quantized Hall states, Phys. Rev. Lett. 52 (1984), 1583\u20136. 9.6 Summary and outlook 587 Landau level, (b) optimally account for the e\ufb00ects of Coulomb repulsion (recall that the strongest player in the Hamiltonian, the kinetic energy, is completely degenerate), and (c) obey Fermi statistics. As to (a), we have seen in Section 9.3.4 that any function of the form \u03c8(z) = f (z)exp(\u2212|z|2/(4l2 0)) automatically belongs to the lowest Landau level, provided that f (z) is analytic.110 So let us seek a manybody wavefunction of the form \u03a8 = F ({zi \u2212 zj})e \u2212 1 4l", "2 0 |zi|2 i, where F must depend only on di\ufb00erences of coordinates (the translational invariance of the clean system). The antisymmetry of the wavefunction (c) requires that F is skewsymmetric under any exchange zi \u2194 zj. Also, (b) F should vanish whenever two coordinates approach each other. Taken together, (a)\u2013(c) motivate the ansatz Robert B. Laughlin 1950\u2013 condensed matter physicist. Theoretical Laughlin was awarded one third of the 1998 Nobel Prsize for his groundbreaking contributions to the explanation of the FQHE. The complementary share went to the experimentalists Horst L. St\u00a8ormer (1949\u2013) and Dan C. Tsui (1939\u2013) for the experimental discovery of the e\ufb00ect. \u03a8(z1, \u00afz1,..., zN, \u00afzN ) = N (zi \u2212 zj)me |zi|2 i \u2212 1 4l2 0 i<j (9.86) with some a priori undetermined integer coe\ufb03cient m. Equation (9.86) de\ufb01nes Laughlin\u2019s wavefunction. Notice its high degree of universality (except for the integer m, the trial wavefunction does not contain a single adjustable parameter!) and the simplicity of its structure. Nonetheless, the ansatz Eq. (9.86) o\ufb00ers a straightforward explanation of many observable features of the FQHE. Speci\ufb01cally, it can be shown that: The integer m relates to the \ufb01lling factor as m = \u03bd\u22121 = 2s + 1. For certain types of short-range interaction, \u03c8 is an exact ground state of the Hamiltonian. Numerical analyses have shown that, even for the long-range Coulomb interaction, \u03c8 has a close to perfect overlap with the exact ground state. Single-particle excitations superimposed on \u03c8 are gapped and fractionally charged. In fact, one may readily rewrite \u03c8 in a way that suggests an interpretation in terms of composite fermions; simply factor out a power (zi \u2212 zj)m\u22121 to obtain - \u03c8 = i<j (zi \u2212 zj)2s\u03c8. m=1 Here, \u03c8m=1 is the wavefunction at an integer \ufb01lling factor while the prefactor adds 2s winding phases to each particle coordinate. In other words, the prefactor converts fermions to composite fermions, so that \u03c8 can be interpreted as a CF wavefunction at integer \ufb01lling. Indeed, this anticipates the \ufb01eld theoretical picture constructed above. 110 To be precise, in Section 9.3.4, we considered wavefunctions \u03c8(z) = f (\u00afz)exp(\u2212|z|2/(4l2 0)) with an antiholomorphic prefactor. One can switch from one form to the other by inversion of the external \ufb01eld B \u2192 \u2212B. For notational simplicity, we shall use the analytic form throughout. 588 Topology 9.6 Summary and outlook This concludes our introduction to topological quantum \ufb01eld theory.111 We have seen the way \u201clarge-scale\u201d geometric structures of quantum \ufb01elds may lead to intriguing physical phenomena \u2013 phenomena, in fact, whose understanding required the application of the entire spectrum of \ufb01eld integral techniques developed earlier in this text. Once again, this chapter could provide no more than a very preliminary impression of the diversity of topology-related quantum phenomena in condensed matter physics. Given the rapid development of the \ufb01eld, we encourage readers motivated to deepen their knowledge in topological condensed matter \ufb01eld theory to turn to the original literature. (For an excellent particle-physics-oriented introduction we refer to Ryder.13 A comprehensive discussion of topological textures in classical physics can be found in Efetov.112) 9.7 Problems 9.7.1 Persistent current of a disordered ring In the main text we have argued that, contrary to a long-standing belief, the presence of disorder does not con\ufb02ict with the formation of persistent currents in normal metal rings. Here we support this assertion by a microscopic calculation. Consider a quasi-one-dimensional ring of circumference L and transverse extension \u03bbF L\u22a5 <, where is the mean free path. We assume that the ring is pierced by a magnetic \ufb02ux \u03a6. Using the fact that L\u22a5/L 1, variations in transverse direction of the vector potential A may be neglected, i.e. A = (\u03a6/L)e, where e is the unit vector in the longitudinal direction and units e = =", " 1 are used. We are interested in the typical value Ityp \u2261 I 2 of the persistent current I = \u2212\u2202\u03a6F, where F is the \ufb02ux-dependent free energy.113 Assuming that the ring is metallic in the sense that its dimensionless conductance g 1, we will be content with computing the \ufb01rst contribution to Ityp in an expansion in powers of g\u22121. For notational simplicity, we assume the \ufb02ux \u03a6 to be measured in units of \u03a60 throughout. (a) Represent I(\u03a61)I(\u03a62) as a correlation function of two non-interacting Green functions. Without going into quantitative detail, convince yourself that the dominant contribution to the correlation function is given by the sum of diagrams shown in Fig. 9.20, where /m \u2261 \u2212\u02c6v and \u2202 is the derivative along the dots denote the operator the ring. Estimate the relative contributions of the four diagrams. (Hint: Recall the derivation of low-momentum di\ufb00usion modes in Section 6.5. Owing to the thinness of the ring, \ufb02uctuations of the di\ufb00uson/cooperon modes in the radial direction may safely be neglected.) \u2212i\u2202 \u2212 (\u03a6/L) 111 Strictly speaking, the terminology \u201ctopological quantum \ufb01eld theory\u201d is reserved for \ufb01eld theories whose behavior is solely determined by topological terms. However, we use it here in a more liberal sense to denote \ufb01eld theories where topological aspects play a signi\ufb01cant role. 112 K. B. Efetov, Supersymmetry in Disorder and Chaos (Cambridge University Press, 1996). 113 In the presence of static disorder, the sign of I sensitively depends on the impurity con\ufb01guration, i.e. I Ityp is much smaller than the typical value of the current for a \ufb01xed impurity con\ufb01guration. 9.7 Problems 589 c1 c2 d1 d2 Figure 9.20 Four diagrams contributing at leading order to the correlation function I(\u03a61)I(\u03a62) of the persistent current in a disordered ring. (b) To quantitatively compute the correlation function, we employ the formalism of the \ufb01eld integral. Using the fact that F = \u2212T limR\u21920 R (Z R \u2212 1), construct a generalization of the nonlinear \u03c3-model action S[Q] suitable to compute the current correlation function. (Hint: Introduce a two-component index t = 1, 2 discriminating between the two currents, and derive the form of the vector potential by minimal substitution, cf. Eq. (6.64), i.e. by recalling that the \ufb02ux dependence of the Hamiltonian may be formally removed by a gauge transformation.) 1 (c) As in Problem 8.8.4, introduce the parameterization Q = eiW \u039be\u2212iW = \u039b(1 \u2212 2iW \u2212 2W 2+\u00b7 \u00b7 \u00b7 ), where the generators W are given in Eq. (8.53) (block structure in Matsubara space). Notice that B = {Bnat\u03c3,nat\u03c3 } are matrices in Matsubara (n), replica (a), current (t), and time-reversal (\u03c3) spaces. Expand the action to second order in B. It is convenient to split B = Bd + Bc into contributions diagonal and o\ufb00-diagonal in time reversal space. Explore the symmetries of the constituent matrices Bd,c and perform the Gaussian integrals. Show that I(\u03a61)I(\u03a62) \u22022 \u03a61,\u03a62 T 2 q,\u03c9n>0,\u03c9n <0 \u0393q\u2212(\u03a61\u2212\u03a62)/L,nn + (\u03a61 \u2212 \u03a62 \u2192 \u03a61 + \u03a62), (9.87) where \u0393qnn \u2261 1 2\u03c0\u03bd\u03c4 % Dq2 + |\u03c9n \u2212 \u03c9n | &\u22121 is the di\ufb00usion mode. (d) The \ufb01nal step in the evaluation of the correlation function, the summation over q, \u03c9n, \u03c9n, is a good exercise in executing tricky integrals (here de\ufb01ned as integrals which have to be simpli\ufb01ed by physically motivated approximations). EXERCISE Di\ufb00erentiate with respect to the \ufb02ux before doing the frequency/momentum summations; identify four contributions corresponding to the diagrams above in Fig. 9.20. What prevents the", " expression above from vanishing is the discreteness of the momentum sum; were we able to approximate the sum by an integral, a shift q \u2192 q + \u03a6/L would remove the \ufb02ux dependence of the integrand. The validity of an integral approximation in turn depends on the value of |\u03c9|. For |\u03c9| > (2\u03c0)2D/L2 \u2261 Ec much larger than the magnitude of the lowest non-vanishing quantized q-mode (physically, the Thouless energy, i.e. the inverse time of di\ufb00usion around the ring), many modes contribute to the integral and a continuum approximation is valid. Thus, the dominant contribution to 590 Topology the correlation function comes from frequencies |\u03c9| < Ec, for which the discreteness of the sum really matters.114 Carry out the sum over momenta by the methods otherwise employed in performing frequency summations. Approximate the subsequent sum over Matsubara frequencies |\u03c9| < Ec by an integral. Show that, for small values of the \ufb02ux, Ityp \u223c Ec\u03a6(\u2212 ln(\u03a6))1/2. (Hint: Do not try to do the frequency integral rigorously; keeping the relevant information on board, simplify the integrand to obtain a manageable integral. As to the q-summation, use the freedom to do the \ufb02ux derivatives before or after the summation to obtain a contour integral that is as simple as possible.) The most important feature of this result is that it depends only algebraically on the disorder concentration, i.e. through EC \u223c D \u223c \u03c4. In contrast, the long-standing yet erroneous expectation that impurity scattering destroys the phase coherence required to maintain a stable persistent current inevitably leads to the prediction Ityp \u223c exp(\u2212L/). For experimentally \u201crealistic\u201d values L/ = O(102), the di\ufb00erence between the two results is dramatic. Irritatingly, the actually observed value of the current turns out to be roughly two orders of magnitude larger than our result above. Although the origin of this discrepancy is unknown, it is clear that it cannot be resolved within the framework of a non-interacting theory. Answer: L n tr( \u02c6Gn\u2202\u03a6 \u02c6H) = T (a) Using the fact that the free energy of a non-interacting Fermion system is given by n tr ln(i\u03c9n \u2212 \u02c6H), where \u02c6H = 1 2m (\u02c6p \u2212 A)2 + \u02c6V and \u02c6V represents the singleF = \u2212T particle potential of the problem (the sum of disorder and con\ufb01ning potentials), we n tr( \u02c6Gn\u02c6v). Expanding the Green function in obtain I(\u03a6) = \u2212T the impurity operator and constructing ladder diagrams at one-loop order, we identify the two cooperon (c1 and c2) and two di\ufb00uson (d1 and d2) diagrams in Fig. 9.20. Focusing on the cooperon sector, we note that the ladder diagram depends only on the sum q\u2212(\u03a61+\u03a62)/L of the fast momenta p\u2212(\u03a6i/L)e carried by the Green functions.115 More speci\ufb01cally, an individual cooperon ladder contributes a factor (see Section 6.5) \u223c \u0393q\u2212(\u03a61+\u03a62)/L,nn \u2261 1. Up to constants, the 2\u03c0\u03bd\u03c4 contribution of diagram c1 is thus given by \u0393q\u2212(\u03a61+\u03a62)/L,nn. Turning to diagram c2, we need to take into account the fact that the vector vertices \u223c v indicated by the dots are now integrated independently over the fast momenta (because they are separated by impurity lines; recall the momentum structure of a ladder diagram). In the limit of zero momentum di\ufb00erence, q \u2212 (\u03a61 + \u03a62)/L \u2192 0, the two Green function insertions carrying a dot vanish (think about it). This means that diagram c2 will be proportional D(q \u2212 (\u03a61 + \u03a62)/L)2 + |\u03c9n \u2212 \u03c9n | &\u22121 % 114 This statement conforms with the expectation that the minimal time t \u223c \u03c9\u22121 required to sense the \ufb02ux through a disordered ring is the di\ufb00usion time, i.e. the time for a quantum particle to traverse the ring at least once. 115 Due to the thinness of the ring, \ufb02uctuations of the cooperon in the transverse direction are negligible, i.e. the momentum q is a scalar quantity measuring \ufffd", "\ufffd\ufffductuations in the longitudinal direction. 9.7 Problems 591 to \u223c (q \u2212 (\u03a61 + \u03a62)/L)2\u03932 q\u2212(\u03a61+\u03a62)/L,nn, where the factor \u03932 accounts for the two ladders. The di\ufb00uson diagrams d1, d2 are obtained by replacing \u03a61 + \u03a62 \u2192 \u03a61 \u2212 \u03a62. (b) De\ufb01ning Z = D( \u00af\u03c8, \u03c8) exp(\u2212S[ \u00af\u03c8, \u03c8]) where S[ \u00af\u03c8, \u03c8] = ddr d\u03c4 \u00af\u03c8at(\u2202\u03c4 + \u02c6H(At) \u2212 \u03bc)\u03c8at, and At = (\u03a6t/L)e, we have the representation I(\u03a61)I(\u03a62) = limR\u21920 Z. All we need to do to deduce the structure of the low-energy action is to notice that the composite index (a, t) can be identi\ufb01ed with a \u201creplica index\u201d of doubled dimension 2R. Noting that the \ufb02ux dependence of the Hamiltonian may be formally (i.e. at the price of changing boundary conditions) removed by the transformation \u03c8 \u2192 ei\u03a6tr, where r is the coordinate along the ring, and recalling the discussion above Eq. (6.64), $ 4 ((\u2202 \u2212 i[A, ])Q)2 \u2212 \u02c6\u03c9Q we conclude that the action is given by S[Q] = \u03c0\u03bd 2 where the vector potential A = {\u03b4at,at (c) Using the symmetry relation W = \u2212\u03c3tr (\u03a6t/L)} \u2297 \u03c3tr 3. \u21d2 BT = \u2212\u03c3tr dr tr \u03a61,\u03a62 2, we verify that 2 B\u2020\u03c3tr 2 W \u03c3tr 2 # D, T 2 R2 \u22022 Bd = Etr 11bd \u2212 Etr 22b\u2217 d, Bc = Etr 12bc + Etr 21b\u2217 c, where bd,c = {bnat,nat the action and expanding to quadratic order, it is a straightforward matter to obtain } are complex matrices. Substituting this representation into d,c S(2)[B, B\u2020] = L T \u03c4 tr q,n>0,n<0 (bd)tt nn \u0393q\u2212(\u03a6t\u2212\u03a6t )/L,nn (b \u2020 d)tt nn +(bd \u2192 bc, \u03a6t \u2212 \u03a6t \u2192 \u03a6t + \u03a6t ). We \ufb01nally integrate over the \ufb01elds bd,c and arrive at'Z = q,n>0,n<0 (\u0393q\u2212(\u03a61\u2212\u03a62)/L,nn )R2 + (\u03a61 \u2212 \u03a62 \u2192 \u03a61 + \u03a62), where the factor of R2 counts the independent replica channels, constant factors C R2 R\u21920\u2192 1 have been omitted, and we also omitted those contributions to the integral that depend exclusively on \u03a61 or on \u03a62 (as they do not contribute to the two-fold derivative limR\u21920 R\u22122\u22022 [check!]). Di\ufb00erentiating with respect to the \ufb02ux and sending R \u2192 0, we obtain Eq. (9.87). \u03a61\u03a62 (d) Our strategy will be to do the momentum sum by contour integral methods. As it is not convenient to integrate over a function containing branch cuts (such as our logarithm), we do one of the \ufb02ux derivatives, \u2202\u03a61, say, \ufb01rst. Focussing on the di\ufb00uson contribution, and introducing the shorthand notation \u03a6 = \u03a61 \u2212 \u03a62, |\u03c9| = |\u03c9n \u2212 \u03c9 n |, we have \u2202\u03a62 L\u22121 q D(q \u2212 \u03a6/L) D(q \u2212 \u03a6/L)2 + |\u03c9| = \u2202\u03a62 = \u2212 1 2 < 1 4\u03c0i dz coth(zL/2) D(\u2212iz \u2212 \u03a6/L) D(\u2212iz \u2212 \u03a6/L)2 + |\u03c9| \u2202\u03a62 Im coth([(|\u03c9|/Ec)1/2 + i\u03a6]/2), 592 Topology where the integration contour is a circle at in\ufb01nity avoiding the two poles of the integrand.116 Notice that our result is 2\u03c0-period", "ic in the \ufb02ux \u03a6, as it should be. Turning to the frequency summation, we use the fact that |\u03c9|/Ec < 1 to approximate coth([(|\u03c9|/Ec)1/2 + i\u03a6]/2) 2[(|\u03c9|/Ec)1/2 + i(\u03a6 mod 2\u03c0)]\u22121. This leads to the expression \u22022 \u03a61\u03a62 T 2 q,n,n = \u2202\u03a62 (2\u03c0Ec)2\u03a6 \u0398(n)\u0398(\u2212n)\u0393q\u2212\u03a6/L,nn \u2212\u2202\u03a62 (2\u03c0)2 Im 1 0 dx x x + \u03a62 = (2\u03c0Ec)2\u2202\u03a62 \u03a6 \u2212 \u03a63 ln Ec 0 d\u03c9 \u03c9 (|\u03c9|/Ec)1/2 + i\u03a6. \u03a62 + 1 \u03a62 For small values of the \ufb02ux, the logarithm can be approximated as \u2212 ln(\u03a62) and \u2202\u03a62 \u03a63 ln(\u03a62) 3\u03a62 ln(\u03a62). Using this approximation and adding the cooperon contribution, we obtain I(\u03a61)I(\u03a62) \u2212(2\u03c0Ec)2 3(\u03a61 \u2212 \u03a62)2 ln((\u03a61 \u2212 \u03a62)2) \u2212 3(\u03a61 + \u03a62)2 ln((\u03a61 + \u03a62)2). Setting \u03a61 = \u03a62 we arrive at the required result. 9.7.2 Working with the SU (N ) Wess\u2013Zumino term In this problem, we develop some of the amazing mathematical features of the SU(N ) WZ term Eq. (9.58). The problem also illustrates the superiority of the coordinate-free calculus of di\ufb00erential forms in topological quantum \ufb01eld theory. (Expressed in the standard languages of calculus, all formulae below become intolerably messy \u2013 not to mention the fact that the underlying structures are much more di\ufb03cult to understand!) (a) WZ terms generally originate from a closed (but only locally exact) di\ufb00erential form on the target manifold \u03c9. Show that, on SU(N ), the form \u03c9 = tr(g\u22121dg \u2227 g\u22121dg \u2227 g\u22121dg) is closed. (Why is it only locally exact? This question will be addressed in part (c) below.) (b) Verify Eq. (9.61). Let us try to understand the normalization of the WZW action. To start with, let us recall a statement from group theory stated in the Info block on page 549: as far as topology is concerned, the subgroup SU(2) \u2282 SU(N ) preempts the structure of SU(N ), i.e. without loss of generality, we may restrict our discussion to the SU(2) WZ action. We next turn to the normalization of the WZW action. To start with, let us consider S2 as the boundary of the northern hemisphere S3+ of the 3-sphere. We de\ufb01ne the WZW term by integration overS 3+.117 To parameterize S3+, we introduce a third coordinate x3 (in addition to the two coordinates (x1, x2) parameterizing S2), such that x3 = 0 de\ufb01nes the north pole and x3 = \u03c0/2, the equatorial plane, S2. For any \ufb01eld g(x1, x2) \u2208 SU(2), the WZW action then becomes \u0393+[g] =iN dx3 \u0393[g, x3], where \u03c0/2 0 116 In spite of the weak convergence \u223c z\u22121 of the integrand at large values of z, it is permissible to do the integral in this manner: the second \ufb02ux derivative implicitly levels the decay up to \u223c z\u22122. 117 Topologically, this domain is equivalent to the three-dimensional unit ball B3 used in the text (just as the northern hemisphere of the 2-sphere is topologically equivalent to the unit disk B2). 9.7 Problems 593 dx1 dx2 \u03bc\u03bd\u03c3 tr(g\u22121\u2202\u03bcgg\u22121\u2202\u03bdgg\u22121\u2202\u03c3g), \u03bc, \u03bd, \u03c3 = 1, 2, 3, and g(x1, x2, x3) \u0393[g, x3] \u2261 is a smooth extension of g(x1, x2) fromS 2", " to S3+. But, of course, it would have been just as good to take the southern hemisphere S3\u2212 as our reference domain. To explore the consequences of this ambiguity, we extend the de\ufb01nition of the third coordinate in such a way that x3 = \u03c0 corresponds to the south pole of S3. In this case, the WZW \u03c0 action would assume the form \u0393\u2212[g] =\u2212iN \u03c0/2 dx3, \u0393 [g, x3], where \u0393[g, x3] is de\ufb01ned as above (albeit for an extension g(x1, x2, x3 < 0). For any reference \ufb01eld g(x1, x2), the ambiguity, \u0393+ \u2212 \u0393\u2212 = iN \u03c0 0 dx3 dx1 dx2 \u03bc\u03bd\u03c3tr(g\u22121\u2202\u03bcgg\u22121\u2202\u03bdgg\u22121\u2202\u03c3g)!= 2\u03c0n, must be an integer multiple of 2\u03c0.118 Consider, thus, the \ufb01eld g(x1, x2) \u2261 in(x1, x2) \u00b7 \u03c3, where n(x1, x2) is a unit vector. As its extension into the northern/southern hemisphere, we choose g(x1, x2, x3) = cos(x3) +i sin(x3)g(x1, x2) = exp(ix3n(x1, x2) \u00b7 \u03c3) \u2208 SU(2). This is an SU(2)-instanton, i.e. a mapping S3 \u2192 SU(2) that cannot be continuously deformed to the unit mapping. (c) Show that \u0393+[g] \u2212 \u0393\u2212[g] =iN 24\u03c02 \u00d7 (integer), which enforces N = 1/12\u03c0 as the normalization of the WZW action. As a corollary, we note that the form \u03c9 on which all of our discussion is based cannot be globally exact. (If it had been, Stokes\u2019 theorem would imply a vanishing of the integral over the boundary-less manifold S3.) (d) Show that, for the particular reference con\ufb01guration considered in part (c), g = in \u00b7 \u03c3 \u2208 SU(2), the WZW action reduces to the \u03b8-term for the unit-modular \ufb01eld n \u2208 S2 : \u0393[in \u00b7 \u03c3] = \u03c0Stop[n], where Stop[n] = 1 4\u03c0 S2 d2x n \u00b7 (\u22021n \u00d7 \u22022n). Answer: (a) Using the fact that g\u22121dg g\u22121 = \u2212dg\u22121 (why?) the form \u03c9 can be rewritten as \u03c9 = tr(dg\u22121 \u2227 dg \u2227 dg\u22121 g), i.e. d\u03c9 = \u2212tr(dg\u22121 \u2227 dg \u2227 dg\u22121 \u2227 dg). To show that this expression vanishes, one may make use of the fact that, for arbitrary matrix-valued forms, tr(\u03c91 \u2227 \u03c92) = (\u2212)deg(\u03c91)deg(\u03c92)tr(\u03c92 \u2227 \u03c91).119 Applied to d\u03c9, this yields d\u03c9 = + tr(dg \u2227 dg\u22121 \u2227 dg \u2227 dg\u22121) = tr(dg\u22121 \u2227 dg \u2227 dg\u22121 \u2227 dg) = \u2212d\u03c9, where in the last step we applied the formula g\u22121dg = \u2212dg\u22121 g to all derivatives. 118 Geometrically, this is the integral of the di\ufb00erential 3-form g\u2217\u03c9 (i.e. the pullback of the SU(2) form \u03c9 by the \ufb01eld g to a form onS 3) overS 3. 119 Proof: tr(\u03c9 \u2227 \u03b7) = ij \u03c9ij \u2227 \u03b7ji = (\u2212)deg(\u03c9)deg(\u03b7) ij \u03b7ji \u2227 \u03c9ij = (\u2212)deg(\u03c9)deg(\u03b7)tr(\u03b7 \u2227 \u03c9). 594 Topology (b) We want to explore what happens to the integral \u0393[g] =\u2212 i 12\u03c0 B3 g\u2217\u03c9 upon variation g \u2192 eW g (1 + W )g. Using the relation g\u22121dg \u2192 g\u22121dg + g\u22121dW g, we obtain \u0393[(1 + W )g] \u2212 \u0393[g] \u2212 i 4\u03c0 tr(g\u22121d", "W g \u2227 g\u22121dg \u2227 g\u22121dg) = = i 4\u03c0 i 4\u03c0 B3 tr(dW \u2227 dg \u2227 dg\u22121) = B3 S2 tr(W \u2227 dg \u2227 dg\u22121), i 4\u03c0 B3 % d tr(W \u2227 dg \u2227 dg\u22121) & where, in the last line, we have made use of Stokes\u2019 theorem. (c) By an elementary rearrangement of \ufb01elds, \u0393+ \u2212 \u0393\u2212 = 3iN \u03c0 0 dx3 dx1 dx2 ijtr(g\u22023g\u22121 \u2202ig \u2202jg\u22121), where the coordinates i, j = 1, 2. A straightforward calculation shows that g\u22023g\u22121 = \u2212in \u00b7 \u03c3. Using the auxiliary identities \u2202ig = i sin(x3)\u2202in \u00b7 \u03c3 and tr(n1 \u00b7 \u03c3 n2 \u00b7 \u03c3 n3 \u00b7 \u03c3) = 2in1 \u00b7 (n2 \u00d7 n3), we thus obtain, \u03c0 \u0393+ \u2212 \u0393\u2212 = 6iN dx3 sin2(x3) dx1 dx2 ijn \u00b7 (\u2202in \u00d7 \u2202jn) 0 = 6i\u03c0N dx1dx2n \u00b7 (\u22021n \u00d7 \u22022n) = 24i\u03c02 \u00d7 (integer), where the integer is the number of times the unit vector n wraps around the unit sphere. (d) By analogy with the calculation performed in (c), we obtain \u0393+[in \u00b7 \u03c3] = \u0393+ = 6i\u03c0N dx1 dx2 n \u00b7 (\u22021n \u00d7 \u22022n) = \u03c0Stop[n]. 9.7.3 Renormalization group analysis of the SU (N ) Wess\u2013Zumino model Before embarking on this problem, it is helpful to recapitulate the RG analysis of the SU(N ) nonlinear \u03c3-model discussed in Section 8.5. We want to study the RG \ufb02ow of the SU(N ) WZW model, as speci\ufb01ed by Eq. (9.65). As in our previous analyses of nonlinear \u03c3-models, we can split a general \ufb01eld con\ufb01guration into a slow and a fast part, and expand the latter as gf = 1 + W + W 2/2 + \u00b7 \u00b7 \u00b7, where g = gsgf W \u2208 su(N ), i.e. W lives in the Lie algebra su(N ) (the algebra of anti-Hermitian traceless matrices). To compute the one-loop RG equations, we need to expand the action to quadratic order in W and compute all contributions to the functional integral that contain (a) one fast momentum integration and (b) no more than two derivatives acting on a slow \ufb01eld. (a) Show that the expansion of the action S[gsgf ] to second order in the generators takes the form S[gf gs] = S[gs] +S [gs, W ] +S (2)[W ], where S(2)[W ] = \u2212 1 \u03bb S[gs, W ] = \u2212 1 \u03bb d2x tr(\u2202\u03bcW \u2202\u03bcW ), d2x tr S2 S2 s \u2202\u03bcgs \u2212 i\u03bb g\u22121 8\u03c0 (9.88) \u03bc\u03bdg\u22121 s \u2202\u03bdgs [\u2202\u03bcW, W ]. 9.7 Problems 595 (b) One-loop corrections to the action are obtained by expanding the functional to second order in S(2)[gs, W ] and integrating over W : S[g] \u2192 S[gs] \u2212 1 S[gs, W ]2W. Use the results of 2 Section 8.5 (in particular those derived in the exercises on page 463) to con\ufb01rm that (i) only the gradient term of the action is renormalized and (ii) the RG equation for its coupling constant is given by Eq. (9.66). Answer: (a) The \ufb01rst line of Eq. (9.88) and the \ufb01rst term in the second line are obtained by substitution of gsgf into the gradient term of the action (9.57) and expanding to second order in the generators W. The second term in S[gs, W ], a descendant of the WZ action, is best derived in the language of di\ufb00erential forms: substitution of (gsgf )\u22121d(gsgf ) = g\u22121 )gf into the pullback (", "gsgf )\u2217\u03c9 = tr(g\u22121dg \u2227 g\u22121dg \u2227 g\u22121dg)|g=gsgf f gives s dgs + (dgf )g\u22121 (g\u22121 f (gsgf )\u2217\u03c9 = g\u2217 s (\u03c9) + = g\u2217 s (\u03c9) + 3 2 3 2 % d tr(dg\u22121 s \u2227 dgs \u2227 [W, dW ]) + 3tr(dW \u2227 dW \u2227 g\u22121 s dgs) +O (W 3) tr([W, dW ] \u2227 g\u22121 s dgs) & + O(W 3). Application of Stokes\u2019 theorem thus leads to \u0393[gsgf ] =\u2212 B3 i 12\u03c0 = \u0393[gs] \u2212 i 8\u03c0 = \u0393[gs] \u2212 i 8\u03c0 (gsgf )\u2217\u03c9 = \u0393[gs] \u2212 i 8\u03c0 % d B3 tr([W, dW ] \u2227 g\u22121 s dgs) & + O(W 3) tr([W, dW ] \u2227 g\u22121 s dgs) +O (W 3) d2x\u03bc\u03bdtr([W, \u2202\u03bcW ] \u2227 g\u22121 s \u2202\u03bdgs) +O (W 3), S2 S2 where the O(W 2) term appears as the second contribution to the action S[gs, W ]. (b) De\ufb01ning \u03a6\u03bc \u2261 g\u22121 (\u2202\u03bc + i\u03bb s 2iLd \u03bb 8\u03c0 \u03bc\u03bd\u2202\u03bd)gs, the action S[gs, W ] assumes the form S[gs, W ] = pq p\u03bctr(\u03a6\u03bc\u03a6\u03bc). But for the di\ufb00erence in the de\ufb01nition of the \ufb01eld \u03a6, this is equal to the fast\u2013slow action of the standard SU(N ) model. Using the results derived in the exercises on page 463, we thus obtain \u2212 1 2 S[gs, W ]2W = \u2212 N ln b 8\u03c0 = \u2212 N ln b 8\u03c0 6 d2x tr(\u03a6\u03bc\u03a6\u03bc) 7 2 1 \u2212 \u03bb 8\u03c0 d2x tr(\u2202gs\u2202g\u22121 s ). This result con\ufb01rms that only the gradient term in the action is renormalized. By proceeding in direct analogy to the discussion of Section 8.5, it is a straightforward matter to derive the corresponding RG equation. The result is given by Eq. (9.66) 596 Topology V x x\u22a5 D \u2202D h(x) v Figure 9.21 On the formation of surface wave excitations in an FQHE droplet. For an explanation, see the main text. 9.7.4 Fractional quantum Hall e\ufb00ect: physics at the edge In Section 9.3.7 above, we have seen that the bulk physics of an integer quantum Hall system is intimately connected to that of its boundaries. These connections could be disclosed from both direct physical reasoning (the essence of the Laughlin\u2013Halperin semi-phenomenological approach) and an analysis of the behavior of the microscopic theory under gauge transformations. Following the seminal work of Wen,120 here we show that an equally close connection between bulk and boundaries exists for the FQHE. Again, we will be able to deduce the boundary physics by phenomenological, and by \ufb01eld-theory-oriented, reasoning. Consider a \ufb01nite FQHE system. For simplicity, we assume the system to be disc-shaped, although the details of the geometry will not be of relevance throughout. At the system boundaries, the two-dimensional electron gas is con\ufb01ned by a boundary potential V which we assume to be linear (see Fig. 9.21). In the \ufb01rst part of this problem, we want to describe the dynamics of edge excitations on phenomenological grounds. An important \ufb01rst observation is that the bulk system \u2013 presumed to sit at an FQHE plateau value \u2013 is in an \u201cincompressible state.\u201d This follows from the fact that, for an integer number of composite fermion Landau levels, the system does not support gapless excitations: \u2202\u03bc/\u2202N \u2192 \u221e. Now,121 \u2202\u03bc/\u2202N \u223c \u03ba\u22121 \u2261 \u2212V (\u2202P/\u2202V )N is proportional to the inverse compressibility, i.e. the lack of lowenergy excitations implies an incompressible state. We should, therefore, think of our system as a rigid \u201cliquid\u201d", " rather than as a gas. Given that the state is incompressible, the lowest-energy excitations of the system will be deformations of the boundary (similar to boundary distortions of a puddle of water). We may characterize these distortions by a surface density pro\ufb01le \u03c1(x), where x parameterizes the boundary.122 (a) To derive the boundary action on phenomenological grounds, proceed in two steps. First derive the energy of a boundary distortion \u03c1. Second, argue why a distortion pro\ufb01le \u03c1(x, t) propagates along the boundary at some constant velocity v. Derive an e\ufb00ective equation of motion for \u03c1 and use it to determine the canonical momentum associated 120 For a review, see X. G. Wen, Theory of the edge states in fractional quantum Hall e\ufb00ects, Int. J. Mod. Phys. B6 (1992), 1711\u201362. 121 G. Mahan, Many Particle Physics (Plenum Press, 1981). 122 More precisely, the geometric shape of the system is described by some height function h(x) (see Fig. 9.21). 2\u03c0 dx dx\u22a5, where 2\u03c0 h(x)dx, i.e. the Using the fact that the constant charge density of the system (cf. Eq. (9.17)) is given by \u03bdB x\u22a5 is the coordinate perpendicular to the system boundary, \u03c1(x)dx = \u03bdB dx\u22a5 dx = \u03bdB 2\u03c0 height pro\ufb01le is proportional to a density pro\ufb01le. h(x) 0 9.7 Problems 597 with the \u201cdisplacement \ufb01eld\u201d \u03c6 de\ufb01ned by \u2202x\u03c6 = 2\u03c0\u03c1. Show that the Hamiltonian action of the displacement \ufb01eld is given by S[\u03c6] = 1 4\u03c0\u03bd dx d\u03c4 v(\u2202x\u03c6)2 \u2212 i\u2202x\u03c6\u2202\u03c4 \u03c6, (9.89) the action of a \u201cchiral Luttinger liquid\u201d \u2013 the terminology follows from the fact that Eq. (9.89) describes the uni-directional propagation of a density \ufb01eld. (The connection to the Luttinger liquid is made explicit by decomposing the coordinate\u2013momentum pair describing one-dimensional charge density modes, (\u2202x\u03c6, \u03b8), into left- and right-moving components, \u03c6 = \u03c6L + \u03c6R, \u03b8 = \u03c6L \u2212 \u03c6R. Substituting this representation into the action (4.49), the latter decomposes into a left- and a right-moving chiral action.) (Hint: A boundary distortion costs energy because of the presence of a voltage gradient. The presence of a constant drift follows, e.g. from the \ufb01niteness of the Hall conductivity. Also remember Eq. (9.17) for the density of a quantum Hall system.) (b) We next derive the action from a complementary, bulk-oriented perspective. To this end, consider the CS action of the bulk system, Eq. (9.75). Throughout, it will be convenient (although not strictly necessary as one may translate all expressions back to the traditional representation) to use the language of di\ufb00erential forms. Interpreting a\u03bc as the coe\ufb03cients of a di\ufb00erential 1-form, a = a\u03bcdx\u03bc, Eq. (9.75) assumes the compact form SCS[a] = \u2212 a \u2227 da, D\u00d7S1 where the integral extends over the Cartesian product of the bulk of the system, D, and imaginary time S1. Show that SCS is gauge invariant up to a boundary term. To \u201ccure\u201d the gauge de\ufb01ciency of the action, we adopt a gauge-\ufb01xing condition a0|\u2202D = 0. Use this condition to integrate over the component a0 in the entire system. Show that this integration leads to the global constraint fij = 0, where fij are the realspace components of the \ufb01eld strength tensor f = da = \u2202\u03bca\u03bddx\u03bc \u2227 dx\u03bd. This condition implies that the real-space component of the vector potential a can be represented as a pure gauge, a = d\u03c6 \u2261 \u2202i\u03c6dxi, where we use the symbol d to denote the real space contribution to the exterior derivative. Plug this ansatz into the residual contribution to the action (after a0 has been integrated out) to reduce the \ufb01eld strength tensor to d\u03c4 \u2202x\u03c6\u2202", "\u03c4 \u03c6. Recalling that the CS action enters the boundary action S[\u03c6] = the theory as exp( i 4\u03c0\u03bd SCS),123 we conclude that the e\ufb00ective boundary action induced by the gauge non-invariance of SCS reads \u2202B dx Se\ufb00 [\u03c6] = \u2212 i 4\u03c0\u03bd dx d\u03c4 \u2202x\u03c6\u2202\u03c4 \u03c6. \u2202B 123 To be precise, the coupling constant 1/4\u03c0\u03bd appeared after (a) the system had been coupled to an external vector potential, (b) the coupling between the statistical vector potential and the matter degrees of freedom had been taken into account, and (c) the statistical vector potential had been integrated out (see Section 9.5.3). Here we assume that all these steps are implied, i.e. we should think of a as an external electromagnetic \ufb01eld and i 4\u03c0\u03bd SCS as the dominant (\ufb01rst-order derivative) contribution to its action. 598 Topology (c) One may recognize Se\ufb00 [\u03c6] as the second contribution to the action (9.89). The obvious next question to ask is why we did not obtain the full action of the chiral Luttinger liquid. Indeed, it takes only a moment\u2019s thought to realize that a \ufb01rst principles derivation of the full boundary action from the bulk theory is out of the question. The point is that the missing contribution \u223c v(\u2202x\u03c6)2 depends on the boundary velocity, which in turn is determined by the steepness of the boundary potential. However, the bulk action does not know of the structure of the boundary. We can, however, employ a trick to infuse the required information on the system boundaries into the bulk theory. (Although this manipulation is fairly ad hoc, it shows, at least, that the bulk and the boundary theory are not inconsistent.) Indeed, the gauge-\ufb01xing condition employed in the construction above involves a lot of arbitrariness. For example, instead of a0 = 0 the linear combination a0 + vax = 0 would have been just as good. To explore the consequences of the new gauge-\ufb01xing condition, let us introduce a new set of coordinates, \u02dcx = x\u2212vt, \u02dct = t, and \u02dcx\u22a5 = x\u22a5. Recalling the transformation behavior of di\ufb00erential forms (see page 537) under coordinate changes, compute the components of the vector potential a = \u02dca\u03bcd\u02dcx\u03bc. Show that the gauge \ufb01xing condition translates to \u02dca0 = 0. We now bene\ufb01t from the fact that we have expressed the CS term in a coordinateinvariant manner, i.e. it keeps the same form, no matter whether we express it in terms of the old or the new coordinates (a point on which to re\ufb02ect!). However, in the new coordinates, the gauge-\ufb01xing condition assumes the same form \u02dca0 as that considered in (b). This means that SCS can be reduced to a boundary contribution, which, however, is expressed in new coordinates. Finally, translate back to the old coordinates to obtain the full chiral Luttinger action. The message to be taken home from this construction is that a boundary-gauge-\ufb01xing condition can be used to establish the equivalence between the boundary Luttinger and the bulk CS description of the system. (d) Notwithstanding its innocuous appearance, the action (9.89) describes astonishingly rich physics. Referring for an in-depth discussion to the literature (see Wen\u2019s review article120), we here mention only a few of the characteristic features of the system described by Eq. (9.89). To establish contact with physically observable quantities, we \ufb01rst need to derive an expression for the electron operator c\u2020 (the chiral analog of the fermion relation (4.46)). Arguing as in Section 4.3 above, we start from the observation that the creation of an electron goes along with the creation of a unit charge. Building on the connection between the displacement \ufb01eld and the charge operator, show that the chiral bosonic representation of the fermion operator will contain a piece c\u2020 \u223c e\u2212i\u03bd\u22121\u03c6. To complete the manufacturing of a fermion operator, we need to ensure that c\u2020 obeys fermionic exchange statistics. Use the commutator relations of the \ufb01eld \u03c6 to show that ei\u03bd\u22121\u03c6(x)ei\u03bd\u22121\u03c6(x) = ei\u03bd\u22121\u03c6(x)ei\u03bd\u22121\u03c6(x)", " \u00d7 ei\u03c0\u03bd\u22121sgn(x\u2212x). This means that exp(i\u03bd\u22121\u03c6) describes a fermion provided that \u03bd = 1/m belongs to the principal sequence. 9.7 Problems 599 INFO But what will happen for a general \u03bd = n/m? It is, in fact, not clear how to repair the ansatz above so as to generatefermion statistics for general \ufb01lling fractions. Indeed, it turns out that the entire edge construction above is too narrow to describe the general case. We have seen in our discussion of the IQHE above that, in cases where p > 1 Landau levels are occupied, the edge hosts p edge channels. Similarly, an FQHE system with \ufb01lling fraction \u03bd = p/(2sp+1) (i.e. a system in which the composite fermions experience an e\ufb00ective \u03bd = 1/p IQHE) will have p chiral edge channels circulating at its boundaries. However, the construction of this edge channel hierarchy is beyond the scope of the present text and we refer to the literature. Notice that we may interprete i\u03bd\u22121\u03c6 = (ei\u03c6)\u03bd\u22121 as the \u03bd\u22121th power of the more elementary object ei\u03c6. According to the construction above, ei\u03c6 creates an entity of fractional charge e \u00d7 \u03bd. Also, the states created by ei\u03c6 obey fractional exchange statistics. Comparing with our discussion in the main text, we identify exp(i\u03c6) as the creator of the boundary variant of the fractionally charged Laughlin quasi-particles. Answer: (a) Owing to the presence of an (approximately linear) con\ufb01ning potential V = Ex\u22a5, a boundary distortion costs an energy (notation taken from footnote122) H = h(x) 0 dx (\u2202x\u03c6)2. dx h2(x) = E dx\u22a5\u03c3Ex\u22a5 = \u03bdEB 4\u03c0 dx 4\u03c0\u03bdB To understand why the density pro\ufb01le propagates along the boundary, notice that the (con\ufb01ning) electric \ufb01eld E perpendicular to the boundary will generate a Hall current density j \u223c \u03c312E = \u03c3B\u22121E tangential to the boundary. The total boundary current is obtained by integrating the current density from 0 to h(x), i.e. I \u223c \u03c1E/B = \u2202x\u03c6E/2\u03c0B. Integrating the continuity equation \u2202t\u03c1 = \u2202xj over x and substituting the identi\ufb01cation of the current above, we obtain the equation of motion \u2202t\u03c6 = v\u2202x\u03c6, where the velocity v \u223c E/B depends in a non-universal way on the boundary potential. This equation is solved by \u03c6 = \u03c6(x + vt), i.e. a uniformly propagating density distribution. Switching to momentum space, we may interpret \u02d9\u03c6k = \u03b4H/\u03b4\u03c0k = vik\u03c6k as an Hamiltonian equation of motion determining the momentum \u03c0 conjugate to \u03c6. Comparison124 with H = 2\u03c0\u03bd (\u2212ik)\u03c6\u2212k, i.e. the strange looking v 2\u03c0\u03bd prediction that the variable \u03c6k is canonically conjugate to its own derivative, k k2\u03c6k\u03c6\u2212k leads to the identi\ufb01cation \u03c0k = 1 [\u2202x\u03c6(x), \u03c6(x)] = 2\u03c0i\u03bd\u03b4(x \u2212 x). Remembering that the Hamiltonian action of a system is given by we arrive at the required result. (9.90) d\u03c4 (H \u2212 i\u03c0\u2202\u03c4 \u03c6), 124 Deviating from our standard conventions, we de\ufb01ne the Fourier transform by \u03c6k = L\u22121/2 dx eikx\u03c6(x). 600 Topology (b) Under a change of gauge, a \u2192 a + dg, whereg is a function, the CS action transforms as SCS[a + dg] =\u2212 (a + dg) \u2227 d(a + dg) = SCS[a] \u2212 D\u00d7S1 D\u00d7S1 dg \u2227 da = SCS[a] \u2212 dag, \u2202D\u00d7S1 where we noticed that dg \u2227 da = d(gda) and applied Stokes, theorem. To remove the boundary gauge ambiguity, we need to \ufb01x a gauge at \u2202D \u00d7S1, e.g. by setting a0|\u2202D = 0. Decomposing a = a0 + a and d = \u2202", "0d\u03c4 + \u2202idxi \u2261 d0 + d into temporal and real-space contributions, the action assumes the form SCS[a0, a] =\u2212 = \u2212 = \u2212 (a0 + a) \u2227 (d0 + d)(a0 + a) (a0 \u2227 da + a \u2227 d0a + a \u2227 da0 + a \u2227 da) (2a0 \u2227 da + a \u2227 d0a + a \u2227 da), D\u00d7S1 D\u00d7S1 D\u00d7S1 where we have used the skew-symmetry of the exterior product and in the crucial second equality integrated by parts.125 The boundary term corresponding to this integration vanishes due to the gauge-\ufb01xing condition a0|\u2202D = 0. Linearly coupled to the action, the temporal component a0 can be integrated out. As a result, we obtain a functional \u03b4-distribution globally enforcing the constraint da \u2261 f = 0, i.e. a vanishing of the spatial components of the \ufb01eld strength tensor. This in turn implies that a is a pure gauge, a = d\u03c6, where \u03c6 is some function. Substituting this representation into the action, we \u2202D\u00d7S1 (d\u03c6 \u2227 d0\u03c6) = obtain S[\u03c6] = \u2212 d\u03c4 \u2202x\u03c6 \u2202\u03c4 \u03c6, where in the last step we switched back to conventional notation. \u2202D dx (c) Using the fact that a = a\u03bcdx\u03bc = a\u03bc(\u2202x\u03bc/\u2202 \u02dcx\u03bd)d\u02dcx\u03bd \u2261 \u02dca\u03bdd\u02dcx\u03bd, and comparing coe\ufb03cients, we obtain the identi\ufb01cations \u02dcax = ax, \u02dcax\u22a5 = ax\u22a5, \u02dca0 = a0 + vax. \u201cForm invariance\u201d of the CS-action means that it does not matter whether we express it in terms of the old or the new coordinates: D\u00d7S1 d\u03c6 \u2227 d0d\u03c6 = \u2212 D\u00d7S1 d(d\u03c6 \u2227 d0\u03c6) = \u2212 SCS[a] = \u2212 (a\u03bcdx\u03bc) \u2227 dx\u03bd \u2227 (a\u03bddx\u03bd) = \u2212 (\u02dca\u03bcd\u02dcx\u03bc) \u2227 d\u02dcx\u03bd \u2227 (\u02dca\u03bdd\u02dcx\u03bd). \u2202 \u2202 \u02dcx\u03bd \u2202 \u2202x\u03bd Focusing on the second representation (wherein \u02dca0|\u2202D = 0), and repeating the analysis d\u02dct \u2202\u02dcx\u03c6 \u2202\u02dct\u03c6. of (b), we obtain the real-time representation of the action SCS = Finally, using the fact that \u2202x = \u2202\u02dcx and \u2202\u02dct = v\u2202x + \u2202t, we arrive at SCS[\u03c6] =. Switching back to imaginary time and attaching the \u2202D dx coupling constant, we obtain Eq. (9.89). v(\u2202x\u03c6)2 + \u2202x\u03c6 \u2202t\u03c6 \u2202D d\u02dcx dt % & 125 That is, we have applied Stokes\u2019 theorem: D\u00d7S1 a \u2227 da0 = \u2212 D\u00d7S1 [d(a \u2227 a0) \u2212 da \u2227 a0] =\u2212 \u2202D\u00d7S1 a \u2227 a0 + D\u00d7S1 da \u2227 a0 a0 | \u2202D =0 = da \u2227 a0. D\u00d7S1 9.7 Problems 601 (d) Since \u03c1 = 1 2\u03c0 \u2202x\u03c6, the presence of a unit charge localized at x0, \u03c1(x) =\u03c1( x \u2212 x0), amounts to the presence of a \u201ckink\u201d of height 2\u03c0 in \u03c6, i.e. \u03c6(x) = 2\u03c0\u03b8(x \u2212 x0). Now, according to Eq. (9.90), \u03c6 and \u2202x\u03c6/2\u03c0\u03bd form a canonically conjugate pair. This implies that Ud(x) \u2261 exp(\u2212id\u2202x\u03c6/2\u03c0\u03bd) acts like a unit translation operator on \u03c6: Ud(x)\u03c6(x) = \u03c6(x) +d\u03b4 (x \u2212 x). The unit-charge kink is generated by the action of dx U2\u03c0(x) = exp(\u2212i\u03c6/\u03bd) on the \ufb01eld \u03c6. Reformulating Eq. (9.90) as [\u03c6(x), \u03c6(x)] = i\u03c0\u03bd", " sgn(x \u2212 x), we \ufb01nd that \u221e x0 ei\u03bd\u22121\u03c6(x)ei\u03bd\u22121\u03c6(x) = ei\u03bd\u22121\u03c6(x) e\u2212i\u03bd\u22121\u03c6(x)ei\u03bd\u22121\u03c6(x)ei\u03bd\u22121\u03c6(x) = ei\u03bd\u22121\u03c6(x) ei\u03bd\u22121exp(\u2212i\u03bd\u22121[\u03c6(x), ])\u03c6(x) = ei\u03bd\u22121\u03c6(x)ei\u03bd\u22121\u03c6(x) \u00d7 ei\u03c0\u03bd\u22121 sgn(x\u2212x). 10 Nonequilibrium (classical) This chapter provides an introduction to nonequilibrium statistical (\ufb01eld) theory. In the following, we introduce a spectrum of concepts central to the description of many particle systems out of statistical equilibrium. We will see that key elements of the theory \u2013 Langevin theory and the formalism of the Fokker\u2013Planck equation \u2013 can be developed from the coherent framework of a path integral formalism. Applications discussed below include metastability, macroscopic quantum tunneling, driven di\ufb00usive lattices, and directed percolation. While the emphasis in this chapter is on classical nonequilibrium phenomena, the quantum theory of nonequilibrium systems will be discussed in the next chapter. The world is a place full of nonequilibrium phenomena: an avalanche sliding down a sandpile, a tra\ufb03c jam forming at rush hour, the dynamics of electrons inside a strongly voltage-biased electronic device, the turmoil of markets following an economic instability, the di\ufb00usion-limited reaction of chemical constituents and many others are examples of situations where a large number of correlated constituents evolves under \u201cout-of-equilibrium\u201d conditions. Statistical nonequilibrium physics is concerned with the identi\ufb01cation and understanding of universal structures that characterise these phenomena. Notwithstanding the existence of powerful principles of uni\ufb01cation, it is clear that a theory addressing the dazzling multitude of nonequilibrium phenomena must be multi-faceted. And indeed, nonequilibrium statistical physics is a strongly diversi\ufb01ed \ufb01eld comprising many independent sub-disciplines. But this is not the only remarkable feature of this branch of physics. Considering the development of physics in, say, the last two decades, one may observe that nonequilibrium physics is a \ufb01eld of disproportionate growth. Occasionally one may get the impression that many particle theory as a whole is drifting towards the nonequilibrium! INFO What does the attribute \u201cnonequilibrium\u201d actually de\ufb01ne? Perhaps the most straightforward de\ufb01nition identi\ufb01es it as the opposite of thermodynamic equilibrium, where the latter is de\ufb01ned by the following two conditions:1 An equilibrium system is characterized by a unique set of extensive and intensive variables which do not change in time. After isolation of the system from its environment, all the variables remain unchanged. The latter condition is necessary to distinguish equilibrium from stationary nonequilibrium states. For example, the particle distribution function of an electronic conductor subject to a 1 W. Ebeling and I. M. Sokolov, Statistical Thermodynamics and Stochastic Theory of Nonequilibrium Systems, (World Scienti\ufb01c,2005). 602 Nonequilibrium (classical) 603 strong time-independent voltage bias will be time-independent (\ufb01rst condition ful\ufb01lled), yet different from the equilibrium Gibbs distribution \u2013 a stationary nonequilibrium situation. However, after removal of the voltage carrying leads, or \u201cenvironment,\u201d it will relax back to the Gibbs form (second condition violated). The restrictiveness of the two conditions above vividly illustrates that thermodynamic equi- librium is a rare exception, a Platonic ideal scarcely realised in \u201creal life.\u201d What are the driving forces behind this development? For one thing, nonequilibrium statistical physics is a \ufb01eld of unusually high application potential. Many of the rapidly evolving interdisciplinary activities between physics and biology, or physics and the socio-economical sciences, expressly address nonequilibrium problems. But there is, of course, also intrinsic interest in the physics of nonequilibrium. For example, the response of an electronic device to shake-up by a sudden electromagnetic pulse evidently contains more structured information on its intrinsic dynamics than the (near equilibrium) linear response. The wealth of information encoded in the dynamics of strongly-driven many particle systems motivates the extension of existing (equilibrium) theories into the realm of nonequilibrium. Also, various many particle systems of interest are observed under nonequilibrium conditions (e.g. cold atom condensates), or intrinsically in a state of nonequilibrium (e.", "g. lasers). To understand the rich physical behavior of such systems, we need to develop a quantum theory of nonequilibrium phenomena. From what has been said above, it should be clear that we will not be able to adequately introduce the full body of nonequilibrium statistical physics in this textbook; we really have to restrict ourselves to an absolute minimum of essentials. In the following we focus on areas of nonequilibrium statistical physics that are of relevance speci\ufb01cally to condensed matter physics. Interesting developments relevant to, say, the physics of biological or socioeconomic systems are not covered. In line with the general philosophy of the book, the material is organized according to the conceptual structure of the theory (rather than by \ufb01elds of application). We start by introducing elements of probability theory and stochastic processes, Langevin theory, the theory of noise, and the theory of the Fokker Planck equation. It will then be seen that various connections between these elements can be understood from the unifying \ufb01eld theoretical framework of the Martin\u2013Siggia\u2013Rose\u2013Janssen\u2013de Dominicis (MSRJD) \ufb01eld integral. We apply the MSRJD formalism to introduce elements of the theory of dynamic phase transitions. In the next chapter we then identify the MSRJD \ufb01eld integral as the classical limit of a more general theory, the Keldysh \ufb01eld integral of quantum nonequilibrium statistical mechanics. INFO In this chapter, there is a lot of discussion about di\ufb00erent types of \u201cprobability.\u201d For the convenience of the reader, a few elements of probability theory are summarized below. (This material need not be studied in a coherent fashion, and can be used as a glossary whenever necessary.) Although we shall not enter a discussion of the foundations of probability theory, it may be worthwhile to note that the notion of \u201cprobability\u201d is di\ufb03cult to grasp. There exist di\ufb00erent and, in fact, controversial de\ufb01nitions of what is meant by \u201cthe probability\u201d of certain events. Here, we will sidestep these complications by assuming that the probabilities of certain basic events are a priori known (e.g. the probability of individual faces when throwing dice). The 604 Nonequilibrium (classical) task of probability theory then is to describe the probability of more complex, compound events (e.g. the probability of winning a certain game of dice). The probability P that a basic variable X take values xi, i = 1,... N (or continuous values x \u2208 [a, b]), is described by the probability distribution, discrete: pi \u2261 P (X = xi), pi \u2265 0, continuous: p(x)dx \u2261 P (x < X < x + dx), p(x) \u2265 0, N i=1 b a pi = 1, p(x) dx = 1. A variable whose de\ufb01ning properties are encapsulated in a probability distribution is called a random variable. To avoid discrete/continuous case distinctions, we may express the distribution of a discrete random variable in continuum form, p(x)dx \u2261 i \u03b4(x \u2212 xi)dx. Furthermore, we shall formulate our discussion for scalar random variables. The extension to multi-dimensional variables X \u2208 Rn should be obvious. Expectation values of functions f (X) depending on the random variable are de\ufb01ned as f (X) \u2261 dx p(x)f (x). Below we list a number of important examples of expectation values: The mean value X \u2261 dx p(x)x of the distribution and its moments Xn \u2261 X n \u2261 dx p(x)xn. Notice that the moments of distributions need not necessarily exist. For example, the Lorentzian distribution p(x) = 1 \u03c0 a a2+x2 does not have moments at all. The cumulants \u03bc1 \u2261 X1, \u03bc2 \u2261 X2 \u2212 X 2 1, \u03bc3 \u2261 X3 \u2212 3X1X2 + 2X 3 1, \u03bc4 \u2261..., (10.1) contain information about the degree of \u201ccorrelation\u201d in a probability distribution. They describe the way in which high moments di\ufb00er from products of moments of lesser degree. A systematic way to de\ufb01ne cumulants is by expansion of the cumulant generating function: (it)n lnexp(itX) = n! \u03bcn. Exercise: Think about connections between cumulants and the connectedness of diagrams resulting from the expansion of free energies (symbolic notation) F = \u2212T ln D\u03c6 exp(\u2212S[\u03c6]). \u221e i=1 The probability distribution may be formally represented as the expectation", " value of a \u03b4- distribution, p(x) = \u03b4(x \u2212 X). Far better than a useless tautology, this relation represents a vehicle to generate derived probability distributions: let F (X) be some function and de\ufb01ne Y \u2261 F (X). This is a new random variable which inherits its probability distribution p(y) from p(x). The connection between the two can be obtained as p(y) = \u03b4(y \u2212 Y ) = dx p(x)\u03b4(y \u2212 F (x)) = p(x(y)) \u2202x \u2202y, x=x(y) Nonequilibrium (classical) 605 where we assumed a unique functional relation y = F (x). This relation may also be obtained by direct transformation of the probability measure, p(x)dx = p(x(y))|\u2202x/\u2202y|dy \u2261 p(y)dy. In practice, however, the \u03b4-representation is often more convenient to use. The full set of information on a random variable is contained in the expectation value g(t) \u2261 exp(itX). (10.2) From this generating function, all moments can be obtained by power series expansion and all cumulants by expansion of ln g(t): g(t) = \u221e n=0 (it)n n! Xn, ln(g(t)) = \u221e n=0 (it)n n! \u03bcn. The full probability distribution can be obtained (exercise) from the generating function by Fourier transformation, p(x) = dt 2\u03c0 e\u2212itxg(t). For concreteness and later reference let us discuss a few distributions of outstanding impor- tance. The Gaussian distribution is given by the familiar expression p(x)dx = 1\u221a 2\u03c0\u03bc2 exp (x \u2212 \u03bc1)2 2\u03bc2 dx. (10.3) N n=1 xn that are obtained as the sum of Typically, it describes the probability of variables x = a large number of elementary random variables, xn, distributed according to some distribution \u02dcp with \ufb01rst and second cumulant \u02dc\u03bc1 and \u02dc\u03bc2, respectively. In this case \u2013 the statement of the central limit theorem2 \u2013 the probability of x is given by Eq. (10.3), with \ufb01rst and second cumulant \u03bc1 = N \u02dc\u03bc1 and \u03bc2 = N \u02dc\u03bc2, respectively. (All higher cumulants of the Gaussian distribution vanish. Exercise: verify this statement by showing that the generating function of the Gaussian distribution is again a Gaussian.) The ubiquity of additive random variables in nature explains the importance of Gaussians in natural sciences. The Poisson distribution p(m) describes the probability to observe m events in n 1 trials when the events are mutually uncorrelated3 and occur with low individual probability p. (As an example, consider the probability to count m cars passing a counter at a cross country road late at night in a time window of 100min when the passage probability per minute, p = 0.05.) 2 The proof of the central limit theorem amounts to an instructive miniature version of the large N -expansion methods introduced in chapter 5.3: p(x) =\u03b4 (x \u2212 n xn) = N n=1 n=1 dxn p(xn) dk eik(xn\u2212x) = dk exp N ln dx p(x)eikx \u2212 ixk = dk exp (N ln g(k) \u2212 ixk) = dk exp(N ik(\u03bc1 \u2212 x/N ) \u2212 k2\u03bc2/2 +O (k3) ) \u221a 1 2\u03c0N \u03bc2 exp (x \u2212 N \u03bc1)2 2N \u03bc2, where in the \ufb01rst line we used the Fourier representation of the \u03b4-function, g in the second line is the generating function of the distribution p, and in the crucial last equality we used the fact that anharmonic corrections to the quadratic k-exponent vanish in the large N limit. 3 Two random variables X and Y are said to be uncorrelated if XY = X Y. The correlation obviously vanishes if X and Y are independent random variables, i.e. if p(x, y) =p x(x)py(y). (The opposite conclusion, \u201clack of correlation \u2192 independence,\u201d is not valid in general.) 606 Nonequilibrium (classical) Assuming that the probability to observe the event in question at any individual trial is p 1 while n m, the Poisson distribution asymptotes to4 P (m) = (", "np)m m! exp(\u2212np). (10.4) The Poisson distribution has constant cumulants \u03bcl = np. (Exercise: compute the generating function to prove this result.) The Lorentzian distribution a.k.a. Breit\u2013Wigner distribution or Cauchy distribution is de\ufb01ned by p(x) = 1 \u03c0 a (x \u2212 x0)2 + a2. (10.5) In physics, the Lorentzian distribution describes the energy dependence of scattering resonances, (relatedly) the broadening of many particle spectral functions by interactions, the line shape distribution of damped electromagnetic modes, and many other phenomena governed by an interplay of driving/oscillation/damping. As mentioned above, its moments are unde\ufb01ned. Loosely speaking, the distribution is centered around x0, has widtha, and is exceptionally broad. In probability theory one is often interested in computing the composite probability p(x2, x1), from information about the probabilities p1(x1) andp 2(x2) of elementary random variables X1 and X2. In this context, a natural question to ask is \u201cwhat is the probability to obtain x2 provided x1 has been observed?\u201d By de\ufb01nition, the answer is given by the conditional probability, p(x2|x1). It is implicitly de\ufb01ned by p(x2, x1) = p(x2|x1)p1(x1). (10.6) Summing over all possible realizations of the random variable X2, we get the raw distribution of X2, dx1 p(x2, x1) = p(x2), or p(x2) = dx1 p(x2|x1)p1(x1). won\u2019t Considering the case where x1 by itself is a composite random variable, x2 \u2192 xn and x1 \u2192 (xn\u22121,..., x1), we obtain an important generalization of Eq. (10.6). Iteration of the de\ufb01nition then leads to p(xn,..., x1) = p(xn|xn\u22121,..., x1)p(xn\u22121|xn\u22122,..., x1)... p(x2|x1)p1(x1). (10.7) EXERCISE The characteristic function of a probability distribution P (x) is de\ufb01ned as g(t) = eitX = dx p(x)eitx. Compute the characteristic function for (a) the Gaussian distribution, p(x) = 1\u221a (b) the Lorentzian distribution, p(x) = 1 \u03c0 e\u2212 (x\u2212x0 )2, (x\u2212x0)2+a2, and (c) the Poisson distribution, p(x) = 2\u03c0\u03c3 2\u03c3 a 4 The probability to observe m successes is given by P (m) = nm and (1 \u2212 p)n\u2212m (1 \u2212 p)n = (1 \u2212 (np)/n)n exp(\u2212np). Combining these estimates, we obtain the Poisson distribution. (Exercise: show that the derivation becomes exact in the limit n \u2192 \u221e, p \u2192 0 at \ufb01xed product np. Show that for n \u2192 \u221e at \ufb01xed p the Gaussian distribution is obtained.) (1\u2212p)n\u2212mpm. Forn 1 and p 1, n m n m 10.1 Fundamental questions of (nonequilibrium) statistical mechanics 607 \u03bdx x! e\u2212\u03bd. Consider the distribution, p(x) = 1 that its characteristic function has the limiting behavior x(1\u2212x) 1\u221a \u03c0, forx \u2208 (0, 1) and zero otherwise. Show g(t) \u223c (1 \u2212 i)(i + eit) \u221a 2\u03c0t for t \u2192 \u221e. 10.1 Fundamental questions of (nonequilibrium) statistical mechanics From the discussion above, it should be obvious that thermodynamic equilibrium represents an exception, rather than a rule. The one reason why many particle theory has focused historically on the description of (near-)equilibrium states is that the latter are so much easier to describe. In thermal equilibrium we know that the state of an N -particle system is governed by the thermal density operator Z \u22121exp(\u2212\u03b2 \u02c6H). \u201cAll\u2019 that remains to be done is to compute physical observables from this distribution. However, in thermal nonequilibrium the distribution of the system in phase or Hilbert space is apriori unknown, a", " lack of knowledge that bears important consequences: Getting the relevant many particle distribution functions under control is an essential part of the theoretical challenge (and usually the \ufb01rst one to be addressed). Concepts which we tend to take for granted (the existence of a uniquely de\ufb01ned temperature, homogeneity of thermodynamic variables, etc.) need to be critically re-examined. There is no reason to hope that nonequilibrium statistical physics as a whole (whatever that is) will be nearly as universal as the equilibrium theory. Imagine, then, a situation where no apriori information about the state of a many particle system is available. How can we compute the distribution of its states (classical or quantum)? Indeed, even the formulation of the problem that needs to be solved is not entirely obvious. Ludwig Boltzmann 1844\u20131906 Austrian physicist famous for his pioneering contributions to statistical mechanics. Concepts such as Maxwell\u2013Boltzmann statistics, the Boltzmann distribution, or the logarithmic connection between entropy and probability remain foundations of this \ufb01eld. Boltzmann was one of the most important supporters of early atomic theory, at a time when the reality of atoms was still controversial. In principle, the state of a classical d-dimensional N -particle system can be represented by a point X \u2208 R2N d in 2N d dimensional phase space. The full information about its dynamics, and all derived physical properties is then contained in some high-dimensional Hamiltonian equation of motion. The above view is formally exact, but largely useless in practice. However, in the late nineteenth century, Boltzmann introduced an alternative picture which turned out to be much more powerful: Boltzmann suggested describing the system in terms of a \u201cswarm\u201d of N points in 2d-dimensional phase space, rather than by a single point in an incredibly high dimensional space. This idea paved the way to a statistical formulation of the problem and may be considered as the starting point of statistical mechanics as an independent branch 608 Nonequilibrium (classical) of physics. Indeed, we may trade much of the excessive \ufb01ne structure information encoded in the full coordinate data of the \u201cswarm\u201d by introducing a measure f (x, t)ddx, where x = (q, p) is a point in phase space and the dimensionless function f (x, t) is the (distribution of the) number of particles found at time t in the volume element d ddx = i=1 dqidpi (see the \ufb01gure), i.e. f is normalized according to p dp ddx f (x) = N, \u0393 dq q \u0393 is an integral over classical phase where space. The function f is the prime information carrier of the theory. Average values of physical observables X can be calculated according to the relation, X = ddx f (x, t)X(x), where X(x) is the phase space function representing the observable. For example, the energy carried by the system is obtained as E = H = ddx f (x, t)H(x), where H(x) is the Hamiltonian, etc. Since the state of the system is essentially described by its distribution function, we can reformulate the questions raised above as follows: How can we compute f for a given system if the Hamiltonian and an initial state f (., t) are speci\ufb01ed? However, before addressing this problem in full, we should be able to answer two less ambitious questions: what is the equilibrium structure of f, and how does the system approach that equilibrium con\ufb01guration if it is allowed to evolve unperturbed according to its own dynamics? To begin, let us try to understand the (time-independent) equilibrium form of the distribution function, f (x). Imagine a system prepared in some initial distribution. Think of the long-time density function f (x, t \u2192 \u221e) as a probabilistic object where averaging is over di\ufb00erent initial conditions. We assume that the system is weakly interacting, or gaseous, in the sense that the dynamics is essentially of single particle type. Collisions between particles, (x1, x2) \u2192 (x 2) will serve as a mechanism of relaxation towards an equilibrium state by exchange of energy and momentum between individual particles. Under these conditions, it is safe to assume that the distribution of particles will be independent in the sense that the joint probability to observe particles at x1 and x2, respectively, N 2p(x1, x2), factorizes into independent particle distributions N 2p(x1, x2) = (N p(x1)) \u00d7 (N p(x2)) \u2261 f (x1)f (x2). By virtue of this assumption, the function f acquires the status of a statistical distribution, fully", " describing the state of the system. 1, x EXERCISE Discuss: in what sense may this description become problematic once interactions become strong? 10.2 Langevin theory 609 1, x 1, x The fact that coordinate con\ufb01gurations (x1, x2) \u2194 (x 2) are coupled by an elementary scattering event implies conservation of probability, f (x1, x2) = f (x1)f (x2) = f (x 2) = f (x 2). Assuming that f (x) =f (H(x)) \u2261 f () is a function of energy, we conclude that probability conservation is compatible with energy conservation, 1 + 2 = 1 + 2, if ln f () = a + b is linear in energy. In this case, f (x1)f (x2) = exp(a(1 + 2) + 2b) = exp(a( 2) is indeed satis\ufb01ed. To \ufb01x the constants a and b we require normalization of the partition function and employ the equipartition theorem, i.e. we use the fact that in thermodynamic equilibrium, the expectation value of the energy of each of the 2N d degrees of freedom must equal T /2.5 Normalization requires (exercise) b = ln N \u2212 ln ddx exp(a) and from the equipartition theorem we obtain 2) + 2b) =f (x 1)f (x 1)f (x 1 + N dT!= dN \u0393 ddx eaH(x)H(x) \u0393 ddx eaH(x\u03b2, where, again, a quadratic Hamiltonian was assumed. We thus arrive at the conclusion: f (x) = N e\u2212\u03b2H(x) \u0393 dx e\u2212\u03b2H(x), (10.8) which is the famous Maxwell\u2013Boltzmann distribution. Exercise: Eq. (10.8) a direct descendant of the many particle Gibbs distribution? in what sense is The derivation above can tell us the form of the equilibrium distribution, but it is much too indirect to describe how equilibrium is actually approached. For that purpose we need to develop far more explicit ways to describe the dynamics of the distribution function. The construction of such theories is the subject of the next two sections. 10.2 Langevin theory Imagine a situation where, at time t = 0, some initial distribution f (x, 0) has been prepared. We wish to understand its dynamics. Speci\ufb01cally, we wish to understand how the Maxwell\u2013 Boltzmann distribution is approached if the system is kept in isolation. It will also be of interest to explore deviations from equilibrium if the system is externally perturbed. A purist might argue that phase space points x t\u2192 x(t) = exp(t{H,. })x develop according to the deterministic time evolution of Hamiltonian dynamics. This view will lead to s=0f (x(s), t) = {H, f (x, t)}. s=0f (x, t + s) = \u2212\u2202s f (x, t) = f (x(\u2212t), 0), or \u2202tf (x, t) = \u2202s Deterministic time evolution is thus described by the equation, (\u2202t \u2212 {H,. }) f (x, t) = 0. (10.9) Now, this equation clearly does not capture the irreversible approach to equilibrium. What it neglects is that we are actually concerned with a many particle system. The presence of other particles will a\ufb00ect the dynamics in two di\ufb00erent ways. 5 Here we assume that, close to thermal equilibrium, the dynamics of each particle will be approximately oscillatorlike, and energy comprises kinetic energy (d momentum degrees of freedom) and potential energy (d spatial degrees of freedom). 610 Nonequilibrium (classical) First, it will hinder the ballistic motion of individual particles and, in a coarse-grained perspective, will lead to friction. Second, repeated collisions with other particles will e\ufb00ectively act as a \ufb02uctuating force. As we shall see, these two mechanisms, dissipative friction and \ufb02uctuations are not independent. In thermal equilibrium they do, in fact, completely determine each other, a correlation we will identify as a manifestation of the \ufb02uctuation\u2013dissipation theorem. Paul Langevin 1872\u20131946 known who physicist French developed of concept Langevin dynamics. He is also known for his modern interpretation of para- and diamagnetism in terms of the electron spin. Langevin was a devoted antifascist and lost his academic position under the Vichy regime. (Shortly before his death he was rehabilitated.) the In a seminal work 6 the", " French physicist Paul Langevin proposed to describe the concerted action of dissipation and \ufb02uctuation in terms of a stochastic generalization of Newton\u2019s equation, the celebrated Langevin equation, M dtv + m\u03b3v \u2212 F = \u03be(t), (10.10) where v is the single particle velocity, F = \u2212\u2202qH a macroscopic force acting on the particles, \u03b3 a phenomenological friction coe\ufb03cient, and \u03be a randomly \ufb02uctuating force describing the e\ufb00ect of erratic pair collisions. Langevin proposed to model the force in terms of a short-range correlated Gaussian random variable with zero mean, \u03bei(t) = 0 and variance \u03bei(t)\u03bei (t) = A\u03b4ii \u03b4(t \u2212 t), (10.11) where A >0 is a constant. INFO The development of Langevin\u2019s theory was motivated by the phenomenon of Brownian motion. In 1827 the Scottish botanist, Robert Brown, observed7 unceasing erratic motion of pollen particles in aqueous immersion. First qualitative explanations of the phenomenon in terms of random particle collisions appeared in the late nineteenth century. Langevin suggested Eq. (10.10) as an e\ufb00ective equation of motion controlling the dynamics of \u201cmesoscopic\u201d particles, subject to inter-particle collisions. The erratic nature of solutions of the Langevin equation \u2013 qualitatively consistent with Brownian motion \u2013 is illustrated in Fig. 10.1. Before turning to the analysis of this equation, let us make a few general remarks: One of the important predictions of the theory is that, in thermal equilibrium, the friction coe\ufb03cient, \u03b3, and the \ufb02uctuation strength, A, are mutually dependent parameters: \ufb02uctuations and dissipative damping are twin e\ufb00ects \u2013 again, a manifestation of the \ufb02uctuation-dissipation theorem. 6 P. Langevin, Sur la th\u00b4eorie du mouvement Brownien, CR Hebd. Acad. Sci. 146, 530-3 (1908). 7 The phenomenon had, in fact, been observed earlier by the Dutch physician Jan Ingenhousz in 1785. 10.2 Langevin theory 611 Figure 10.1 A few two-dimensional solutions of Eq. (10.10) for di\ufb00erent \ufb02uctuation force con\ufb01gurations. The paths start at time t = 0 at x = y = 0. Qualitatively, they resemble the paths traced out by \ufb01ctitious \u201cBrownian particles\u201d. We may expect multiple particle collisions to cause e\ufb00ectively stochastic, or di\ufb00usive motion of individual particles. Indeed, we will identify the inverse A\u22121 as an e\ufb00ective measure of the di\ufb00usivity of the medium. A speci\ufb01c solution q(t) of the Langevin equation de\ufb01nes a realization of what is called a stochastic process. A stochastic process (cf. the more comprehensive discussion in Section 10.4 below) is described by a time-dependent random variable X(t). Here, \u201crandom\u201d means that no deterministic statements about the time dependence of X can be made. The best we can obtain is results on the probability distribution P (X(t)) of realizations of X. In the present context there are two alternatives: we may discuss the Langevin equation in terms of v[\u03be], i.e. speci\ufb01c realizations of the velocity, randomly dependent on the realization \u03be. Suitable averages over \u03be will then lead to a statistical description of velocity distributions, etc. This is the strategy originally proposed by Langevin. Alternatively, we may aim to describe the velocity distribution P [v] directly in terms of certain statistical propositions. This latter strategy was proposed by Einstein in his famous work on Brownian dynamics (see below). 10.2.1 Fluctuation\u2013Dissipation Theorem (FDT) To discuss the connection between dissipation and \ufb02uctuation in an unperturbed system, we consider the Langevin equation in the absence of driving forces, F = 0. Temporal Fourier \u03be(\u03c9). transformation v(\u03c9) = According to this relation, the \u201cinduced\u201d random variable v has zero mean, vi(t) = 0, dt exp(i\u03c9t)v(t), obtains the formal solution v(\u03c9) = 1 m(\u2212i\u03c9+\u03b3) 612 Nonequilibrium (classical) and second moment (exercise) v2 i (t) = \u221e \u2212\u221e A m2 d\u03c9 2\u03c0 1 \ufffd", "\ufffd2 + \u03b32 = A 2m2\u03b3. However, in thermal equilibrium, the equipartition theorem establishes a connection between /2!= T /2. Comthe average kinetic energy per degree of freedom and temperature, mv2 i parison with the result above then leads to the Einstein relation, A = 2m\u03b3T, (10.12) i.e. in thermal equilibrium, the variance of environmental \ufb02uctuating forces is proportional to the strength of dissipative (frictional) forces, and to temperature. INFO Let us brie\ufb02y reviewEinstein\u2019s derivation of Eq. (10.12). Einstein\u2019s approach is phenomenological but contains additional information inasmuch as it establishes a connection between the strength of \ufb02uctuation forces and the concept of di\ufb00usion. The basis of the argument is that, in a medium governed by frequent inter-particle collisions, the dynamics of individual particles will be di\ufb00usive. Similarly, externally applied forces will cause drift motion, rather than free ballistic acceleration. The \ufb01rst postulate implies that a density gradient, \u2202f, in the medium will lead to a di\ufb00usion current, jd = \u2212D\u2207f, (10.13) i.e. a current acting to restore a uniform density pro\ufb01le. Equation (10.13) is generally known as Fick\u2019s (\ufb01rst) law. Application of the continuity equation dtf = \u2212\u2207 \u00b7 jd then shows that \u2202tf = D\u0394f (Fick\u2019s second law) implying that the dynamics is indeed di\ufb00usive. The second postulate states that jext = \u03b3\u22121m\u22121f Fext, i.e. an external force generates a drift current proportional to the force, the density, and the inverse of the friction coe\ufb03cient. (This formula may be obtained by dimensional analysis, or by consideration of the stationary con\ufb01guration, \u2202tv = 0 obtained by averaging the Langevin equation (10.10).) In thermal equilibrium, di\ufb00usion current and external current must compensate each other, i.e. jd = \u2212jext, or D\u2207f = \u2212f \u2207V /m\u03b3, where we assumed that the force is generated by some potential, V. This equation for the density pro\ufb01le is solved by f \u223c exp(\u2212V /Dm\u03b3). However, compatibility with the Maxwell\u2013Boltzmann distribution (10.8) requires that f \u223c exp(\u2212V /T ), or D = T m\u03b3. (10.14) Equation (10.14) is the celebrated Einstein relation. Comparison with Eq. (10.12) \ufb01nally leads to the identi\ufb01cation, A = 2T 2 D, (10.15) i.e. the variance of the (microscopic) \ufb02uctuation forces in the medium is inversely proportional to its (macroscopic) di\ufb00usion constant. 10.2 Langevin theory 613 Figure 10.2 Three realizations of resonator modes subject to the interplay of \ufb02uctuations and dissipation. Left: RLC-electronic circuit; Middle: a resonator mode subject to radiation losses; Right: damped mechanical oscillator. 10.2.2 A brief compendium on noise Our discussion above conveys the important message, that dissipation or friction are intimately linked to the presence of \ufb02uctuating forces. However, unlike the everyday phenomenon of friction, the corresponding \ufb02uctuation forces are usually less noticeable. The reason is that friction acts in a directed way (cf. the action of a brake), while the response caused by \ufb02uctuations tends to average out. Yet, there are important exceptions to this rule. For example, high ampli\ufb01cation levels render the \ufb02uctuations, or noise caused by resistive elements in electronic circuits a quite noticeable (and usually unwelcome) side e\ufb00ect. Figure 10.2 shows three prominent setups where the interplay of \ufb02uctuations and dissipation is operational: an electronic RLC-circuit, an electromagnetic cavity mode experiencing radiative losses, and a damped oscillator. All of these systems are characterised by idealized non-dissipative dynamics that is harmonic. The presence of dissipation (resistor/radiative losses/mechanical damping) generates \ufb02uctuations. In the case of the macroscopic mechanical oscillators, these \ufb02uctuations will be largely inconsequential. However in the \ufb01rst two systems they cause noticeable e\ufb00ects. Johnson\u2013N", "yquist noise Let us explore the situation in the example of the RLC-circuit. In this case, we are concerned with the charge Q transmitted through the circuit (from which current and voltage drop across the resistive element are obtained as I = \u02d9Q and U = R \u02d9Q, respectively). In elementary courses, we learn that the dynamics of charge is governed by the equation, Ld2 t Q + Q C + RdtQ = Uext, (10.16) where Uext is the externally applied voltage. This is the equation of motion of a dissipatively damped oscillator. However, in view of our discussion above, we know that Eq. (10.16) neglects the e\ufb00ect of the \ufb02uctuations accompanying dissipation. A correct way to think about Eq. (10.16) is as an equation for the coordinate Q, averaged over realizations of those \ufb02uctuations. A realization speci\ufb01c generalization of Eq. (10.16) compatible with the 614 Nonequilibrium (classical) Figure 10.3 Left panel: Noise strength vs. resistance for wires made of di\ufb00erent materials. Right panel: Noise strength vs. temperature for the \u201cadvanced wire\u201d. The two data sets display nearperfect linearity, in agreement with the theory of Johnson\u2013Nyquist noise. Data taken from J. B. Johnson, Phys. Rev. 32, 97-109 (1928). \ufb02uctuation dissipation theorem is given by Ld2 t Q + Q C + RdtQ = Uext + Ujn(t), (10.17) where the correlator of the noise term is given by Ujn(t) = 0, Ujn(t)Ujn(t) = 2T R\u03b4(t \u2212 t). (10.18) The noise term in Eq. (10.17) can be interpreted as a time-dependent \ufb02uctuating voltage, additional to the external voltage. It is important to realize that this type of noise is an inevitable consequence of the presence of resistive elements in electronic circuits; it cannot be removed by \u201cimproving the quality\u201d of the device. The presence of a \ufb02uctuating voltage of strength |Ujn(\u03c9)|2 = 2T R was experimentally discovered by Johnson8 (cf. Fig. 10.3). Nyquist9 explained the phenomenon in terms of the thermal equipartition of energy of electromagnetic oscillator modes. Re\ufb02ecting this background, Ujn is alternatively denoted Johnson noise, Nyquist noise, Johnson\u2013Nyquist noise, or just thermal noise. INFO The strength of the noise correlator is \ufb01xed by slight modi\ufb01cation of the arguments used in Section 10.2.1 above: Think of a non-resistive LC-circuit as an electronic realization of the harmonic oscillator. The R = 0, U = 0 variant of Eq. (10.17) is obtained by variation of the \u201cLagrangian\u201d action S = i dt L 2 \u02d9Q2 \u2212 1 2C Q2, 8 J. B. Johnson, Thermal agitation of electricity in conductors, Phys. Rev. 32, 97-110 (1928). 9 H. Nyquist, Thermal agitation of electric charge in conductors, Phys. Rev. 32, 110-13 (1928). 10.2 Langevin theory 615 where L \u02d9Q2/2 and Q2/2C represent the kinetic (inductive) and potential (capacitive) energy, respectively. We expect an interplay of \ufb02uctuation and dissipation to establish an energy balance compatible with the equipartition theorem, \u02d9Q2 = L 2 1 2C Q2 = T 2, i.e. T /2 for both kinetic and potential energy. This energy balance should be established no matter how strong is the dissipation (R). In the absence of external biasing, Uext = 0, the temporal Fourier transform is given by Q(\u03c9) = Ujn(\u03c9) i\u03c9R \u2212 \u03c92L + C \u22121. Using Eq. (10.17) and (10.18), it is then straightforward to verify that \u02d9Q(t)2 = 2T R L 2 L 2 d\u03c9 2\u03c0 \u03c92 \u03c92R2 + (\u03c92L \u2212 C \u22121)2 = T 2. Similar reasoning shows that 1 2C Q2 = T 2. Although the voltage correlation Eq. (10.18) is consistent with the classical equipartition theorem, this very consistency generates an annoying problem: the second moment Vjn(t)2 \u221d \ufffd", "\ufffd(0) diverges, i.e. our analysis makes the unphysical prediction that a voltmeter measuring noise amplitudes will detect voltage spikes of arbitrary strength. Equivalently, we are making the unphysical prediction that the noise \ufb02uctuates with constant intensity at arbitrarily high frequencies. (Expressed more formally, the noise power S(Ujn, \u03c9) \u2261 lim T \u2192\u221e 1 T B T \u2212T dt Ujn(t)ei\u03c9t 2C = RT, (10.19) which is a measure of the \ufb02uctuation intensity at characteristic frequency \u03c9, remains constant.) This problem disappears if we replace the classical \ufb02uctuation energy of the harmonic 2 = T \u2192 \u03c9(e\u03c9/T \u2212 1)\u22121 by the energy of a quantum oscillator. (Exercise: oscillator, 2 \u00d7 T explore this point.) For frequencies much larger than temperature, \u03c9 T, the mode does not store energy and the problem with the diverging noise amplitude disappears.10 Shot noise Johnson noise is not the only \u201cinevitable\u201d source of noise. In electronic devices, the discreteness of charge quanta generates additional noise. A train of uncorrelated electrons running down a wire is an example of a Poisson process described in Section 10.4.2. The average number of electrons, n, passing through the wire in a time interval \u0394t (large in comparison to the mean passage time) de\ufb01nes the mean current, I = n \u0394t. 10 Notice that the expression \u03c9(e\u03c9/T \u2212 1)\u22121 excludes the vacuum or zero-point energy, \u03c9/2. If the vacuum energy is kept in the energy balance, the problem with the divergence in the integrated noise power reappears. One may reason that the exclusion of the vacuum energy is justi\ufb01ed because only physical transitions can participate in dissipative processes. For a more satisfactory picture, see below. 616 Nonequilibrium (classical) However, it is a characteristic feature of Poisson processes that the variance in the number of transmitted electrons var(n) = n, which means that the variance in the current, var(I) = 1 \u0394t I. Although the signal-to-noise ratio I2/var(I) = \u0394tI increases with increasing observation times and increasing mean current strength, the discreteness of charge remains a principal source of noise. In a continuum model of charge transport this shot noise11 can be modeled by addition of a suitable noise term to the right hand side of Eq. (10.16), in addition to the Johnson noise term. (In Section 11.6 of the next chapter we explore how this addition emerges in a microscopic construction.) Unlike Johnson noise, shot noise is independent of temperature and resistivity of the circuit. Both have in common that the noise power is largely independent of frequency. Other sources of noise A variety of physical processes produce noise. Empirically one \ufb01nds that these sources of noise can add to an overall noise signal that is (a) non-universal in that it depends on all sorts of system-speci\ufb01c details and (b) strongly frequency dependent (unlike the \u201cuniversal\u201d sources of Johnson and shot noise). It is customary to denote noise signals with a power spectrum S(\u03c9) \u223c \u03c9\u2212\u03b1, 0.5 < \u03b1 < 2, as 1/f -noise (\u201cf,\u201d because in engineering it is customary to use f instead of \u03c9 for frequency). For further discussion of 1/f -noise we refer to the literature.12 10.2.3 Fokker\u2013Planck equation I The network of connections revealed above and in Fig. 10.4 is of fundamental importance to statistical physics and deserves further investigation. To this end, we will adopt the second interpretation of the stochastic process v[\u03be](t) advocated above and work in terms of a probability distribution p(v, t). Analysis of the e\ufb00ective equations controlling the probability distribution leads us straight to the concept of the Fokker\u2013Planck equation, a dynamical equation whose importance to statistical physics can be compared to that of the Schr\u00a8odinger equation to quantum mechanics. EXERCISE Carefully think about the steps in the derivation below; it is exemplary of a general scheme to derive Fokker\u2013Planck evolution equations in the theory of stochastic processes. We wish to derive an equation controlling the evolution of the (conditional) probability p(v, t|v0, t0) to observe a particle velocity v at time t provided we started out with velocity 11 Shot noise was \ufb01rst observed in vacuum tubes, i.e. devices where electrons are \u201cshot\u201d from some cathode through empty space. 12 For instance, see P", ". Dutta and P. M. Horn, Low-frequency \ufb02uctuations in solids: 1/f-noise, Rev. Mod. Phys. 53, 497-516 (1981). 10.2 Langevin theory 617 Figure 10.4 Di\ufb00erent physical concepts revolving around the physics of the Langevin equation. v0 at time t0 < t. The evolution of the variable v is controlled by the Langevin equation (10.10), a \ufb01rst order di\ufb00erential equation in time. This means that the evolution v \u0394t\u2212\u2192 v from some initial con\ufb01guration v to v solely depends on the initial con\ufb01guration v (and, of course, on the speci\ufb01c realization of the random force, \u03be). However, it does not depend on the pre-history of the particle, i.e. the speci\ufb01c way in which it made it to the initial con\ufb01guration v. (Exercise: consider why this is a speci\ufb01c feature of a \ufb01rst order di\ufb00erential equation.) This feature, implies the \u201cconvolution\u201d property (again a statement to re\ufb02ect upon), p(v, t|v0, t0) = ddv p(v, t|v, t)p(v, t|v0, t0), (10.20) where t \u2208 [t0, t] is arbitrary and the second factor under the integral is the probability to \ufb01nd v at t for initial data (v0, t0). Weighing by the conditional probability to move on to v and summing over all intermediate con\ufb01gurations, v, we obtain the total probability (v0, t0) \u2192 (v, t). Equation (10.20) is the de\ufb01ning property of a Markovian process, i.e. a process which is fully determined by one-step probabilities. (For a more systematic discussion, see Section 10.4.2 below.) Evaluating Eq. (10.20) for the speci\ufb01c choice of time arguments t \u2192 t + \u03b4t and t \u2192 t, and suppressing the initial value argument (v0, t0) for notational simplicity, we obtain an equation for the incremental evolution of probability, p(v, t + \u03b4t) = ddu w(v \u2212 u, t; u, \u03b4t)p(v \u2212 u, t), where we have introduced the notation w(v, t; \u0394v, \u0394t) \u2261 p(v + \u0394v, t + \u0394t|v, t), for the probability of \u201ctransitions\u201d (v, t) \u2192 (v + \u0394v, t + \u0394t). We next make use of the fact that, for di\ufb00usive motion, the \u201cstep width\u201d u in a su\ufb03ciently small time window, \u03b4t, will be small in comparison to the length scales over which the probability distribution p changes considerably. It then makes sense to Taylor expand in the argument v \u2212 u around u, i.e. to expand in the underlined arguments in w(v \u2212 u, t; u, \u03b4t)p(v \u2212 u, t). It is important to appreciate that the legitimacy of this so called Kramers\u2013Moyal expansion hinges on the 618 Nonequilibrium (classical) kinematics of the process. In cases where one-step transitions are over large distances (as is the case in, say, ballistic collisions in dilute gases) other procedures have to be applied. One of these alternative approaches, Boltzmann kinematic theory, is discussed in Section (10.3) below. Performing the expansion, we obtain the in\ufb01nite order di\ufb00erential equation, p(v, t + \u03b4t) = (\u03b1(0)p)(v, t) \u2212 \u2202vi (\u03b1(1) i p)(v, t) + 1 2 \u22022 vi,vj (\u03b1(2) i,j p)(v, t) +..., (10.21) where \u03b1(0)(v, t) = \u03b1(1) i (v, t) = ddu w(v, t; u, \u03b4t), ddu w(v, t; u, \u03b4t) ui, \u03b1(2) i,j (v, t) = ddu w(v, t; u, \u03b4t) uiuj. To compute the coe\ufb03cients, w, we use the fact that for su\ufb03ciently", " small \u03b4t, the Langevin equation assumes the approximate form v(t + \u03b4t) = v \u2212 \u03b4t\u03b3v + \u03b4t m \u03be(t). The second and third term on the right\u2013hand side tell us how the random variable v(t + \u03b4t) relates to the random variable \u03be, provided v(t) =v was realized. This means that the transition probability is obtained as w(v, t; u, \u03b4t) = p(v + u, t + \u03b4t|v, t) = \u03b4(v + u \u2212 v(t + \u03b4t)) \u03be = \u03b4(u + \u03b4t\u03b3v \u2212 \u03b4t\u03be(t)/m) \u03be. D\u03be exp(\u2212 1 2A Now, the functional distribution function controlling the \ufb02uctuation force is given by, |\u03be(t)|2) (cf. Eq. (10.11)), p([\u03be]) = N where N ensures normalization and in the second representation we used \u03b4t as our fundamental time step discretization. The distribution factorizes, and the probability to \ufb01nd \u03be at time t is given by dt |\u03be(t)|2) \u2192 N D\u03be exp(\u2212 \u03b4t 2A t p(\u03be(t)) = N exp |\u03be(t)|2). \u2212 \u03b4t 2A Using this result, we get w(v, t; u, \u03b4t) = dd\u03be p(\u03be) \u03b4(u + \u03b4t\u03b3v \u2212 \u03b4t\u03be/m) = N exp \u2212 m2 2A\u03b4t |u + \u03b3\u03b4tv|2, where N \u2192 N m/\u03b4t is a renormalized constant ensuring normalization of the probability, ddu w(v, t; u, \u03b4t) = 1. With these results, it is straightforward to obtain the coe\ufb03cients \u03b1(n) as \u03b1(0)(v, t) = 1, \u03b1(1) i (v, t) = \u2212\u03b3\u03b4tvi, \u03b1(2) i,j (v, t) = \u03b4ij 2\u03b4tA m2 + (\u03b3\u03b4t)2vivj. 10.2 Langevin theory 619 Coe\ufb03cients \u03b1(n>2)(v, t) = O(\u03b4t2) are vanishingly small in the limit \u03b4t \u2192 0. Substitution of these results into Eq. (10.21), followed by \ufb01rst order expansion in \u03b4t leads to the Fokker\u2013 Planck equation,13 \u2202t \u2212 \u2202vi \u03b3vi \u2212 \u22022 vivi Dv p(v, t) = 0, (10.22) where Dv = A 2m2, is the di\ufb00usion constant of velocity. In equilibrium, Dv = T \u03b3/m (cf. Eq. (10.15)), which is related to the particle di\ufb00usion constant (10.14) through the relation Dv = \u03b32D. Eq. (10.22) is a partial second order di\ufb00erential equation for the probability distribution. It has to be solved with initial condition p(v, t = t0) = \u03b4(v \u2212 v0), i.e. Eq. (10.22) de\ufb01nes an initial value problem (much like the time-dependent Schr\u00a8odinger equation in quantum mechanics). The second and third contribution to the Fokker\u2013Planck operator describe the competition of drift and di\ufb00usion, respectively. The Fokker\u2013Planck equation, thus, represents the \u201ctheory\u201d behind the chain of connections shown in Fig. 10.4. Derived from the Langevin equation with its interplay of dissipation and \ufb02uctuation, it predicts di\ufb00usion and drift for the dominant transport mechanisms at large times. By way of example, the \ufb01gure above shows a set of trajectories with a constant drift contribution but subject to di\ufb00erent noise contributions leading to di\ufb00usion. In the following, we discuss the dynamics described by Eq. (10.22) from a number of di\ufb00erent perspectives. Speci\ufb01cally, we wish to understand the meaning of the di\ufb00usion and the drift term, the competition between the two, and at last, the approach to thermal equilibrium within the framework of Langevin theory. To understand the meaning of the drift term, consider the Langevin equation in the presence of a constant external driving force F. Browsing through", " the steps in the derivation of the Fokker\u2013Planck equation, one veri\ufb01es that the essential14 e\ufb00ect of this generalization is a shift \u03b1(1) \u2192 \u03b4t(\u2212\u03b3vi + m\u22121Fi) of the drift coe\ufb03cient. The Fokker\u2013Planck equation thus \u2202t \u2212 \u2202vi (\u03b3vi \u2212 m\u22121F ) \u2212 \u22022 p(v, t) = 0. Multiplying this equation by generalizes to vivi vi, and integrating over velocity we obtain, Dv i dtv + \u03b3v \u2212m \u22121F = 0, 13 Anticipating later generalizations to the case of v-dependent coe\ufb03cients, v, Dv, we place the latter to the right of the di\ufb00erential operators \u2202vi. 14 However, we also have to bear in mind that the presence of a constant external force drives the system out of equilibrium. (Switching o\ufb00 the force will change the state of the system, a criterion for a nonequilibrium situation.) This means that dissipation and \ufb02uctuation are no longer related to each other by the \ufb02uctuation\u2013 dissipation theorem. 620 Nonequilibrium (classical) a result that can equally be obtained by averaging the Langevin equation over noise. This equation describes the relaxation of the velocity expectation value to a stationary drift con\ufb01guration v \u2192 1 m\u03b3 F. Identifying that expectation value with the force-induced drift current, we obtain jd = F/m\u03b3, in agreement with the phenomenological reasoning above. The third term in Eq. (10.22) describes di\ufb00usion. The in\ufb02uence of the di\ufb00usion term on the dynamics is visualized in the \ufb01gure above for the case of particles in two dimensions subject to a uniform driving force F = const. \u00d7 (2ex + ey). Di\ufb00usion leads to a stochastic spreading around the (deterministic) drift trajectory. The corresponding distribution p(v) is predicted by the stationary long time limit of the Fokker\u2013Planck equation, (\u2202vi (\u03b3vi \u2212 m\u22121Fi) +D v\u22022 )p(v) = 0, or vivi \u03b3vi \u2212 m\u22121Fi + Dv\u2202vi p(v) = 0. This is solved by p(v) \u223c e\u2212 \u03b3, i.e. a \u201cdi\ufb00usion cloud\u201d centered around the drift trajectory. At F = 0, the di\ufb00usion constant Dv = T \u03b3/m assumes its equilibrium value and p(v) = e\u2212 m reduces to the Maxwell\u2013Boltzmann distribution. We thus draw the important conclusion that m\u03b3 F|2 |v\u2212 1 |v|2 2Dv 2T in the absence of external driving, a conspiracy of di\ufb00usion and drift sends the system into a state of thermal equilibrium. EXERCISE Our discussion above focused on the di\ufb00usive character of the velocity distribution, p(v, t). To describe the di\ufb00usive dynamics of the particle coordinates, q, let us consider a force-free (F = 0) Langevin equation in the overdamped limit, m\u03b3dtq = \u03be, i.e. a limit void of external forces where ballistic acceleration \u223c md2 t q is negligible. Show that the Fokker\u2013Planck equation for p(q, t) takes the form (\u2202t \u2212 D\u22022 q )p(q, t) = 0, (10.23) where D is given by Eq. (10.14). This con\ufb01rms the expectation that in thermal equilibrium the coordinate will perform drift-less di\ufb00usive motion whose level of agitation increases with temperature. EXERCISE Imagine that the single particle dynamics is governed by some Hamiltonian H(x) (as usual, x = (q, p)). Again, we model the in\ufb02uence of the environment by a combination of dissipation and \ufb02uctuation. This leads to the generalized phase space Langevin equation, \u2212 \u03b3p + \u03be. Show that the probability distribution p(x, t) is governed by dtq = \u2202H the phase space Fokker\u2013Planck equation, \u2202p, dtp = \u2212 \u2202H \u2202q dtp(x, t) = \u2212{H, p(x, t)} + \u2202 \u2202p \u03b3p + Dp \u2202 \u2202p p(x, t), (10.24) where Dp = m2Dv = T \u03b3m is the di", "\ufb00usion constant of momentum. The second operator on the right\u2013hand side describes the dissipation/\ufb02uctuation generalization of deterministic single particle dynamics (cf. Eq. (10.9)). 10.2 Langevin theory 621 q \u2212c(t) \u2212f (v)(t) Figure 10.5 Schematic showing the \u201cdaily routine\u201d of an amoeba. Left: energy intake at a rate q. Middle: dissipative energy losses due to basic metabolic activity, rate \u2212c(t), where (t) is the instantaneous energy stored in the organism. Right: motion at a velocity v leads to an energy loss proportional to both, (t) and a velocity dependent loss function, f (v). 10.2.4 Beyond equilibrium We have seen how the Langevin equation describes the approach to equilibrium: dissipative energy losses and the intake of energy through \ufb02uctuations balance each other in a way to form an equilibrium con\ufb01guration. However, the descriptive power of Langevin theory extends well-beyond the physics of equilibrium. E\ufb00ective Langevin equations with dissipation/\ufb02uctuation imbalances can be employed to describe rather complex out-of-equilibrium phenomena in physics, life sciences, and socio-economics. Active Brownian motion By way of example, we here discuss the construction of a minimalistic model to describe the phenomenon of collective biological motion (\u201cswarming\u201d). Of course, we do not aim to describe the actual behavior of a concrete species. Rather, our objective is to formulate an as-simple-as-possible model of collective intake of energy and its conversion into kinematic action. For the formulation of more powerful models and a survey of the recent literature, we refer to Ebeling and Sokolov.1 Our strategy is to endow the Brownian \u201cparticles\u201d (you may now think of them as amoebae or another not too complex animal) with a mechanism of self-propulsion. To this end, we imagine the particles endowed with an energy storage. The time-dependent energy level, (t), will (a) increase due to food intake at some rate, q, (b) decrease at rate \u2212c(t) due to metabolic activity (notice the assumption that energy loss is the larger the larger the energy stored in the organism), and (c) decrease due to kinematic activity (see Fig. 10.5). We assume that motion leads to a loss at rate \u2212(t)f (v), where f (v) \u2261 \u03bav2. The net change of the particle\u2019s energy is thus given by, dt = q \u2212 c \u2212 \u03bav2. We next assume that the kinematic energy loss is translated to a propelling force, F. The form of that force F = \u03bav follows from the condition that the work per time interval done by the force F\u00b7dr/dt = F\u00b7v must equal the kinematic energy loss, \u03bav2, and the assumption that F v. 622 Nonequilibrium (classical) The particle dynamics is then described by the generalized Langevin equation, mdtv + m\u03b3v = \u2212\u2207U + \u03bav + \u03be, where \u2207U represents an external potential and \u03b3 the background friction coe\ufb03cient. Further progress with this equation can be made if we assume that the population is in a balanced state in that the energy \u201cdepots\u201d are kept at equal \ufb01lling levels, on average: dt = 0, or = q c + \u03bav2. This condition can be used to replace in the Langevin equation and to obtain an e\ufb00ective Langevin equation governed by a mechanism of activated friction q\u03ba m(c + \u03bav2) mdtv + m\u03b3e\ufb00 (v)v = \u2212\u2207U + \u03be, \u03b3e\ufb00 (v) = \u03b3 \u2212 (10.25). An interesting situation arises if the background friction is small in that \u03b3 < \u03baq this condition, the e\ufb00ective friction vanishes for \ufb01nite velocities,15 mc. Under |v|2 = v2 0 \u2261 \u03baq \u2212 mc\u03b3 m\u03ba\u03b3. The condition |v|2 = v2 0 de\ufb01nes a sub-manifold in velocity space on which the dynamics becomes e\ufb00ectively Liouvillian, i.e. unhindered by friction. To understand how the particles arrange themselves around that manifold, we need to compute the velocity distribution function. Inspection of the derivation of the Fokker\u2013Planck equation shows that the generalization to activated friction amounts to a mere replacement \u03b3 \u2192 \u03b3e\ufb00", " (v) in Eq. (10.22). This means that the steady state probability distribution of the particles is determined by the equation, (Dv\u2202vi \u2212 \u03b3e\ufb00 (v)vi)p(v) = 0. Integration of this equation leads to p(v) = N exp \u2212 1 2Dv \u03b3|v|2 \u2212 q m ln(c + \u03ba|v|2), where N ensures normalization. In the absence of energy supply, q = 0, and the distribution reduces to the Maxwell\u2013Boltzmann form. However, in general, p(v) is far from equilibrium. The center of the distribution is determined by the condition \u2202|v|2 (exponent) = 0 which is solved by |v|2 = v2 0 is attractive in that it de\ufb01nes the center of the distribution function. The sharpness of the distribution is determined by the inverse of the di\ufb00usion constant, i.e. the more noisy the Langevin equation, the less rigid the \ufb01xation of the condition |v|2 = v2 0. 0. We conclude that the frictionless manifold |v|2 = v2 The model above \ufb01xes the modulus of the velocity of individual particles. However, to describe more complex patterns of collective biological motion, we need to (a) think about mechanisms controlling the direction of velocity, and (b) correlate the motion of many Swarms 15 In cases where the velocity v2 0 c\u03ba\u22121, the e\ufb00ective friction function \u03b3eff (v) c + cv2 may be e\ufb00ectively linearized in the squared velocity. Activated friction pro\ufb01les were employed by Rayleigh to describe the complex energy feedback of musical instruments. (J. W. S. Rayleigh, On the resultant of a large number of vibrations of the same pitch and of arbitrary phases, Phil. Mag. 10, 73-8 (1880).) 10.3 Boltzmann kinetic theory 623 particles to form collective entities. Crudely speaking, three prototypes of collective motion are frequently observed in nature: translational motion (a wandering herd of elephants), rotational motion (a school of \ufb01sh), and uncorrelated motion (bacteria, etc.). Theorists are trying to mimic such types of collective dynamics by extended e\ufb00ective friction models. Referring for a more comprehensive discussion to chapter 12 of Ebeling and Sokolov.1, we here restrict ourselves to a brief review of a few concepts relevant to this \ufb01eld. One aim is to characterize a swarm of N particles in terms of a center coordinate \u223c vi and relative motion. Often, the dynamics of the center is of limited concern. One may, then, decide to impose an external mean velocity V(t). The more interesting relative dynamics of the particles is governed by a combination of a one and a two-particle potential U (q1,..., qN ) = N i=1 U (1)(qi) + N i,j=1 U (2)(qi, qj). The interactions may be attractive (e.g. when one seeks to model motion of a school of \ufb01sh in open water), or repulsive (insects in a con\ufb01ned geometry). Collective and relative dynamics are described by the set of equations mdtvi + m\u03b3e\ufb00 (vi) +c (vi \u2212 V(t)) = \u2212\u2202qi U (q1,..., qN ) +\u03be (t), (10.26) where the term c(vi \u2212 V(t)), c > 0 drives the particle velocity towards the mean velocity of the swarm. This is about as much as can be said in general. Approximation schemes employed to describe speci\ufb01c types of motion include approximate decoupling of the equations into an equation for the center coordinate and N \u2212 1 equations for the relative coordinates, and mean \ufb01eld treatment of the pair interaction. For example, assuming that the mean \ufb01eld 2 (qi \u2212 Q)2, whereQ is the center of mass interaction potential is of the form, U (2) \u223c c coordinate, the equations for the relative coordinates decouple into the form mdtv i + m\u03b3e\ufb00 (v i) = \u2212cq i + \u03be(t), where the primes indicate relative coordinates, and we assumed that |v | V, so that the i friction force depends dominantly on the relative velocity. The four-dimensional dynamics predicted by these equations converges to a limit cycle on the frictionless manifold |v|2 = 0 and with radius speci\ufb01ed by c|q|2/2 =mv 2 v2", " 0/2 (equipartition of potential and kinetic energy). The dynamics of the many particle system then resembles the rotational motion of a swarm. For the discussion of other types of collective dynamics we refer to Ebeling and Sokolov.1. 10.3 Boltzmann kinetic theory Previously, we have explored the ways in which a sequence \u201cLangevin dynamics \u2192 Fokker\u2013 Planck equation \u2192 distribution functions\u201d leads to predictions on the macroscopic state of many particle systems, both in and out of equilibrium. A central step in the derivation of the Fokker\u2013Planck equation was the Kramers\u2013Moyal expansion, an expansion of the probability 624 Nonequilibrium (classical) of short time transitions x \u03b4t\u2192 x in the di\ufb00erence x \u2212 x. The Kramers\u2013Moyal expansion makes sense if, for su\ufb03ciently short \u03b4t, the hopping steps in phase space become small in comparison to the typical phase space coordinates themselves. More precisely, we assumed that the moments (x \u2212 x)n of the conditional probability p(x, t + \u03b4t|x, t) exist and scale with positive powers \u03b4tp(n). Only terms with p(n) \u2264 1 (in the speci\ufb01c case of Gaussian distributed noise, the \ufb01rst two) enter the Fokker\u2013Planck derivative operator. Put di\ufb00erently, the evolution of the probability distribution can be described in terms of a low order linear di\ufb00erential equation only if the nth moments of the transition probability are well behaved in that p(n) > 1 for all but the lowest ns. There are plenty of instances in many particle physics where this \u201clocality principle\u201d is violated. Below we derive an e\ufb00ective evolution equation that describes the dynamics of distribution functions under these circumstances. 10.3.1 Derivation of the Boltzmann equation By way of an example, consider a gaseous system of particles in a container. In the in\ufb01nitely dilute limit (where particle collisions can be neglected), the phase space coordinates of individual particles change according to Newtonian dynamics (cf. Eq. (10.9)). Speci\ufb01cally, the single particle energy is conserved, and (assuming the absence of smooth potential gradients) momentum is constrained to an energy shell |p|2/2m =. However, once interactions are taken into account, the rate at which the distribution f (x1, t) changes gets in\ufb02uenced by particle collisions: losses occur when particles at phase space point x2 scatter o\ufb00 particles of the reference momentum x1 into \ufb01nal states x 2 (cf. the \ufb01gure). We denote the corresponding two-particle transition rate by w(x 2ddx1ddx2, where the function w encapsulates kinematic constraints (energy and momentum conservation) of the collision. The dtf due to \u201cout\u201d processes then reads 2; x1, x2)ddx 1 and x 1ddx 1, x dtf1 out = d2 d1 d2 w(1, 2; 1, 2)f1f2, where we take into account the fact that the total transition rate depends linearly on the number of available collision partners in the two initial states, and introduce a (standard) notation wherein coordinate dependencies are indicated by numbers, ddx1 \u2192 d1, f (x1) \u2192 f1, etc. Gain occurs when particles get scattered into the reference state in collisions (1, 2) \u2192 (1, 2): in = dtf1 d2 d1 d2 w(1, 2; 1, 2)f1 f2. 10.3 Boltzmann kinetic theory The total occupation change due to collisions I[f ] \u2261 dtf1 what is commonly called the collision integral, \u2212 dtf1 in out is summed up in 625 I[f ] = d2 d1 d2 (w(1, 2; 1, 2)f1 f2 \u2212 w(1, 2; 1, 2)f1f2). (10.27) Notice that the collision integral is a nonlinear functional of the distribution functions, i.e. we are facing up to a nonlinear theory. The transition rates w are subject to certain symmetry relations which re\ufb02ect timereversal invariance and unitarity of the microscopic laws of physics. For example, microreversibility implies that a transition (1, 2) \u2192 (1, 2) must be as probable as the timereversed process (1T, 2T ) \u2192 (1T, 2T ), where 1T \u2261 (q1, p1)T = (q1, \u2212", "p1) is the time-reverse of a phase space point, i.e. w(1, 2; 1, 2) = w(1T, 2T ; 1T, 2T ). The consequences of unitarity are best exposed if we interpret the coe\ufb03cients w as classical limits of quantum transition probabilities. Identifying (1, 2) \u2194 |i as a classical approximation to a (coherent) state in two particle Hilbert space,16 and (1, 2) \u2194 |f as a \ufb01nal state, we have the identi\ufb01cation, w(1, 2; 1, 2) \u2194 |S\ufb01|2, where S\ufb01 is the scattering matrix. Unitarity means that |S\ufb01|2 = i |Sif |2 = 1, or i d1d2 w(1, 2; 1, 2) = d1d2 w(1, 2; 1, 2). Using this relation in the out-process, the collision integral can be transformed to I[f ] = d2 d1 d2 w(1, 2; 1, 2) (f1 f2 \u2212 f1f2). (10.28) Adding I[f ] \u2261 dtf1 out as the many particle contribution to the changes in the distribution function to the right\u2013hand side of Eq. (10.9), we obtain the celebrated Boltzmann transport equation,17 \u2212 dtf1 in (\u2202t \u2212 {H,. }) f (x, t) = I[f ]. (10.29) 10.3.2 Discussion of the Boltzmann equation What is the Boltzmann equation (10.29) good for? First, it has been, and still is, a very powerful tool in applied many particle physics. Before the advent of the more powerful 16 Here, we consider the particles as distinguishable, i.e. no (anti)symmetrization is implied. 17 In the Russian literature, the Boltzmann equation is usually called the \u201ckinetic equation.\u201d 626 Nonequilibrium (classical) techniques reviewed in previous chapters of this book, it used to be the principal tool to calculate physical observables in interacting many particle systems.18 Even today, Boltzmann equation-based approaches often represent the most economic and straightforward route to understanding the physics of interacting many particle systems. Second, the Boltzmann equation is of great conceptual value. Many of the principal questions raised in the beginning of the chapter have a relatively straightforward answer in terms of this equation. We \ufb01rst note (cf. the more schematic discussion before Eq. (10.8)) that the collision term vanishes in thermal equilibrium. In equilibrium, the distribution function is given by the Maxwell\u2013Boltzmann distribution Eq. (10.8). Energy conservation in elastic collisions, H(x1) +H (x2) = H(x 1) +H (x 2), means that f1f2 = f1 f2, i.e. vanishing of I[f ]. In equilibrium the losses and gains due to many particle collisions compensate each other and I[f ] does not change the distribution. (Since f (x) = f (H(x)), the single particle dynamics conserves f as well, {H, f } = 0, i.e. the Maxwell\u2013Boltzmann distribution is truly stationary under (10.29).) The Boltzmann H-Theorem However, it is not quite as easy to show that the collision term actually drives the distribution towards the thermal equilibrium con\ufb01guration. The de\ufb01ning property of the thermal equilibrium is that it maximizes entropy, S. In equilibrium statistical mechanics, it is usually taken for granted (the second law of thermodynamics) that many body relaxation processes increase entropy, dtS \u2265 0, before the maximum of the equilibrium con\ufb01guration is reached. However, a more general conceptual framework such as Boltzmann kinetic theory should be able to actually demonstrate the ways in which a microreversible theory leads to macroscopic irreversibility and entropy increase. INFO In view of the comparative complexity of this problem, the collision term is sometimes linearized in what is called the relaxation time approximation. One e\ufb00ects the following replacement of the complicated nonlinear collision integral, I[f ] \u2212\u2192 \u2212 1 \u03c4 (f \u2212 f0), (10.30) where f0 is the Maxwell\u2013Boltzmann distribution. The linearized approximation then describes a tendency to approach equilibrium at a rate set by \u03c4, a time scale interpreted as the mean collision time of interaction processes. The manner in which interactions increase entropy was demonstrated by Boltzmann in a famous construction known as the H-theorem. (It is known as the", " H-theorem rather than 18 For a detailed account of various applications and computational schemes revolving around the Boltzmann equation, see L.D. Landau and E. M. Lifshitz, Course of Theoretical Physics, Vol 10 - Physical Kinetics, (Butterworth-Heinemann, 1981). 10.3 Boltzmann kinetic theory 627 the S-theorem because Boltzmann called entropy H.) Consider the information-theoretic de\ufb01nition of entropy19 according to which S = \u2212ln(f /e), or S = \u2212 ddx f (x, t) ln(f (x, t)/e). The temporal change of S is then given by dtS = \u2212 ddx ln f \u2202tf = \u2212 ddx ln f [{H, f } + I[f ]]. Since the full phase space integral of any function is invariant under Hamiltonian \ufb02ow ddx ln(f ){H, f }, i.e. the change in (exercise: why?), we have 0 = entropy is entirely due to interactions. ddx {H, f ln(f /e)} = To explore how interactions do the job, consider the collision integral in its prototypical variant Eq. (10.27). For an arbitrary function \u03c6(x), and using the abbreviation d\u0393 \u2261 d1 d2 d1 d2, we have d1 \u03c6(1)I[f (1)] = 1 2 = = d\u0393 \u03c6(1) (w(1, 2; 1, 2)f1 f2 \u2212 w(1, 2; 1, 2)f1f2) d\u0393 (\u03c6(1) \u2212 \u03c6(1))w(1, 2; 1, 2)f1 f2 d\u0393 (\u03c6(1) + \u03c6(2) \u2212 \u03c6(1) \u2212 \u03c6(2))w(1, 2; 1, 2)f1 f2, (10.31) where in the \ufb01rst equality we relabeled coordinates (1, 2) \u2194 (1, 2) and the second equality is based on the symmetry 1 \u2194 2 under exchange of the collision partners. From this result, we can derive a few auxiliary identities. Setting \u03c6 = 1, we conclude that the collision term vanishes upon integration, 0 = d\u0393 I[f ] = d\u0393 w(1, 2; 1, 2)(f1 f2 \u2212 f1f2), 1.0 0.8 0.6 0.4 0.2 where the second representation Eq. (10.28) was used. Introducing the abbreviations 0.5 1.0 1.5 2.0 \u039b \u2261 f1 f2 f1f2, X \u2261 w(1, 2; 1, 2)f1f2, this can be written as 0 = d\u0393 X(\u039b \u2212 1). Setting \u03c6 = ln \u039b, Eq. (10.31) yields 0 = d\u0393 X\u039b ln(\u039b). Combination of these results \ufb01nally leads to dtS = 1 2 d\u0393 X(\u039b ln \u039b \u2212 \u039b + 1). 19 If you are not familiar with this de\ufb01nition, consult any advanced textbook on equilibrium statistical mechanics. Exercise: show that in thermal equilibrium, the above de\ufb01nition reduces to the standard de\ufb01nition of thermodynamics. 628 Nonequilibrium (classical) Now, the functions X and \u039b are manifestly positive. For positive \u039b, the combination \u039b ln \u039b\u2212 \u039b + 1 is positive as well (see the \ufb01gure or, alternatively, try to prove it). This demonstrates that elastic particle collisions indeed increase the entropy of the system. Mesoscopic evolution laws The arguments used in the derivation of the H-theorem are useful to derive \u201cmesoscopic\u2019 dynamical equations for observables of physical interest \u2013 densities, currents, and the like \u2013 i.e. equations that describe their behavior at length scales much larger than the collision mean free path l = \u03c4 v (where v is a typical particle velocity) yet smaller than macroscopic scales. These equations must re\ufb02ect the fundamental conservation laws of the system, the conservation of energy, particle number, momentum, etc. Now, interactions may change the energy or momentum of individual particles, but they will not change the total energy and momentum of a su\ufb03ciently large assembly of particles. To see this in quantitative terms, let \u03b4\u0393 \u2261 \u03b4V \u2295 R3 be a thin slice in phase space containing phase space points in the small volume element", " \u03b4V at coordinate q and of arbitrary momentum p \u2208 R3. The average particle, energy, and momentum density, respectively, are then given by \u23ab \u23ac \u23ad = \u03b4V \u22121 \u03b4\u0393 \u03c1(q, t) (q, t) \u03c0(q, t) ddx f (x, t) \u23a7 \u23a8 \u23a9 1 H(x) p. It is now straightforward to prove that interactions do not change these values. To this end, let us assume that the coe\ufb03cients w describe an elastic point interaction, i.e. an interaction local in space that respects energy and momentum conservation. In this case (exercise: think why), a restriction of Eq. (10.31) to the volume restricted phase space element \u03b4\u0393 holds, i.e. d1 \u03c6(1)I[f (1)] = \u03b4\u0393 1 2 d\u0393 (\u03c6(1) + \u03c6(2) \u2212 \u03c6(1) \u2212 \u03c6(1))w(1, 2; 1, 2)f1 f2. 1\u2208\u03b4\u0393 Now, particle, energy, and momentum conservation mean that, in the particular cases \u03c6 = 1,,\u03c0, the linear combinations \u03c6(1) + \u03c6(2) \u2212 \u03c6(1) \u2212 \u03c6(1) on the right\u2013hand side of the equation vanish. We thus conclude \u03b4\u0393 d1 I[f (1)] \u23a7 \u23a8 \u23a9 1 H(x) p = 0. This identity is the key to the formal derivation of conserving transport equations. For example, changes in the particle density are obtained as \u2202t\u03c1 \u03b4V = \u03b4\u0393 \u2202q\u03b1 H \u03b4\u0393 ddx \u2202tf = ddx {H, f } = \u03b4\u0393 ddx \u2202p\u03b1 f \u2212 \u03b4\u0393 ddx v\u03b1\u2202q\u03b1 f 0 \u2212 \u2202q\u03b1 j\u03b1, \u03b4\u0393 ddx (\u2202q\u03b1 H\u2202p\u03b1 f \u2212 \u2202p\u03b1 H\u2202q\u03b1 f ) 10.3 Boltzmann kinetic theory 629 where j\u03b1 \u2261 \u03b4V \u22121 obtain the particle density continuity equation, \u03b4\u0393 ddx v\u03b1f is the \u03b1-component of the particle current density. We thus \u2202t\u03c1 + \u2207 \u00b7 j = 0. If you \ufb01nd this result trivial, keep in mind that it has been derived for arbitrary distribution functions, and under speci\ufb01c (if reasonably general) assumptions on the nature of the microscopic interactions. Particle number conservation as well as the energy and momentum conservation laws discussed in the exercise below impose important consistency constraints on the validity of the theory. EXERCISE Show that, for a translationally invariant system, energy and momentum density obey the conservation laws \u2202t + \u2207 \u00b7 j = 0, \u03b4\u0393 ddx f Hv and \u03a0\u03b1\u03b2 \u2261 \u03b4V \u22121 where j \u2261 \u03b4V \u22121 and the momentum current tensor respectively. \u2202t\u03c0a + \u2202q\u03b2 \u03a0\u03b1\u03b2 = 0, \u03b4\u0393 ddx f mv\u03b1v\u03b2 are the energy current density (10.32) Beyond equilibrium: zero modes of the collision integral As with Langevin theory, the full potential of the formalism becomes visible once we go beyond the level of thermal equilibrium. We have seen that the collision integral favours a Maxwell\u2013Boltzmann distribution, f (x) =N exp(\u2212/T ). Importantly, however, it is indiscriminate as to the value of temperature, or the normalization factor N (which determines the particle number). In fact, if it were left to the collision integral alone, local equilibrium con\ufb01gurations described by spatially varying pro\ufb01les T \u2192 T (q), N \u2192 N (q) are perfectly balanced; all the collision integral requests is a logarithmically linear function of energy, \u223c exp(const. \u00d7 ). What prevents strong local variations of temperature and normalization is the left\u2013hand side of the Boltzmann equation, where Liouville \ufb02ow builds up correlations across the system. Once the system is perturbed out of equilibrium (by application of a strong electric \ufb01eld, say, or by connection to two thermal reservoirs kept at di\ufb00erent temperature), we may then expect a scenario wherein the collision integral aims to establish local thermal equilibrium, while the left hand side will try to compromise between the \u201ccost\u201d of spatial variations and the need to adjust to the conditions imposed by the external perturbation. The quantitative description of such competitions is the subject of Bolt", "zmann transport theory, a substantial \ufb01eld by itself (again, we refer to Landau and Lifshitz 18 for a very detailed exposure). In the following, we discuss a few basic elements of this theory and apply them to an example. To this end, let us imagine a system weakly perturbed out of equilibrium. We may then assume that its e\ufb00ective distribution function is of the form f = f 0 +..., where f 0 is a Maxwell\u2013Boltzmann distribution with uniform parameters and the ellipses represent a small deviation. The fact that f 0 depends on the phase space coordinates only through energy, (x) \u2261 H(x), suggests a representation f (x) = f 0() + \u2202f 0 \u2202 ()\u03c7(x) = f 0() \u2212 f 0() T \u03c7(x), 630 Nonequilibrium (classical) where \u03c7 describes deviations from equilibrium. Substituting this ansatz into the collision integral (10.28), and using the fact that f 0 2 = f 0 2, we obtain the linear integral operator I[\u03c7] = \u2212 f 0 1 T 2 (\u03c71 + \u03c72 \u2212 \u03c71 \u2212 \u03c72). d2 d1 d2 w(1, 2; 1, 2)f 0 (10.33) 1 f 0 1 f 0 This operator has a number of manifest \u201czero modes\u201d: for \u03c7(x) = c(q), \u03c7(x) = c(q) (p), \u03c7(x) = v(q) \u00b7 p, (10.34) with space-dependent c(q), v(q),20 we have I[\u03c7] = 0. These zero modes re\ufb02ect the invariance of the collision integral under local variations of the equilibrium parameters. For example, a local change in the particle density can be described by N \u2192 N 0 + \u03b4N (q). Linearity of the Maxwell\u2013Boltzmann distribution function in \u03b4N then gives f \u2192 f 0 + \u2202f 0 \u2202N \u03b4N (q) = f 0 \u2212 f 0 T T \u03b4N (q) N 0, i.e. with the identi\ufb01cation c(q) =T \u03b4N (q)/N 0, a variation as in the \ufb01rst equality of Eq. (10.34). EXERCISE Extend this argument to show that the second zero mode corresponds to local changes in temperature, and the third re\ufb02ects Gallilean invariance, i.e. invariance under Gallilean transformation to an inertial frame moving with velocity v(q). (Again, the locality of the collision integral entails that the boost velocity can be spatially varying.) Example: thermal conductivity of a gas of particles Even the linearized Boltzmann equation is a complicated integro-di\ufb00erential equation which is di\ufb03cult to solve in generality. Often, however, it is possible to resort to certain simplifying approximations \u2013 of which the relaxation time approximation is one of the more popular \u2013 whereupon the Boltzmann equation becomes a highly e\ufb03cient computational tool. In the following, we illustrate some of these ideas on the example of the thermal conductivity of a gas. Suppose we have imposed a temperature gradient T (q) on a gas of interacting particles. The gas will respond by the buildup of a heat current j (cf. Eq. (10.32)). For su\ufb03ciently weak temperature gradients, the current will depend linearly on \u2207T, and we may de\ufb01ne the thermal conductivity, \u03ba, through the relation, j = \u2212\u03ba\u2207T. (10.35) In the following, we outline, how this coe\ufb03cient may be obtained in a semi-phenomenological manner from the Boltzmann equation (for a more substantial discussion, cf. \u00a77 of Landau and Lifshitz.18 If the temperature gradient is small, we may assume a distribution of the form f (x) = f 0(x) + f 0(x) T (q) \u03c7(x), 20 Here we assume that the energy conserved in elastic collisions is a function of momentum only, at least on the microscopic length scales relevant to the interaction processes. 10.3 Boltzmann kinetic theory 631 where f 0(x) =N exp(\u2212(p)/T (q)) is a local Maxwell\u2013Boltzmann distribution with spatially varying temperature, and \u03c7 accounts for the distortion of the system out of the local equilibrium described by f 0. Since \u03c7 describes a mere readjustment of the particle concenddx f 0\u03c7 = 0. The heat current density carried by the", " system is given by tration, we have (cf. Eq. (10.32) j = 1 T \u03b4V ddx f 0\u03c7v, \u03b4\u0393 where we used the fact that the distribution f 0 does not support a current. We next substitute the above ansatz in the Boltzmann equation and linearize: noting that deviations out of (local) equilibrium are driven by \u2202qT, we anticipate that, to leading order, \u03c7 = O(\u2202qT ). Terms of higher order in \u2202qT will be neglected which means that, on the left hand side of the Boltzmann equation, the zeroth order approximation f f 0 can be used. On the right hand side, we meet with the linearized collision integral I[\u03c7]. Assuming stationarity of the distribution, we thus obtain \u2212{H, f 0} = v \u00b7 \u2202qf 0 = \u2202T f 0v \u00b7 \u2202qT = f0 T T v \u00b7 \u2202qT = I[\u03c7]. The linearity of I[\u03c7] in \u03c7 suggests an ansatz \u03c7 = (v \u00b7 \u2202qT )g(p), where g is a scalar function. While the solution of the linear integral equation for g remains a formidable task, things get a lot simpler if we resort to the phenomenological relaxation time approximation Eq. (10.30), I[\u03c7] = \u2212 f 0 T 1 \u03c4 (v \u00b7 \u2202qT )g(p), where \u03c4 is the characteristic time of relaxation processes in the medium. Substitution of this ansatz into the Boltzmann equation then readily leads to the solution g(p) = \u2212 (p)\u03c4 T. Substitution into the formula for the current then gives \u03c4 (v \u00b7 \u2202qT )v = \u2212 \u03c4 d j = \u2212\u03b4V \u22121 ddx f 0 T 2 \u03b4\u0393 ddp f 0 2 T v2\u2202qT, which leads to the result \u03c4 d \u03ba = ddp f 0 2 T v2 \u223c l \" T m, for the thermal conductivity. Here, l = \u03c4 \u00afv is the collision mean free path, and we have made use of the fact that the mean velocity of particles scales as \u00afv \u223c (T /m)1/2. (Typically, the mean free path depends only weakly on temperature.) Although our discussion above focused on a speci\ufb01c example, it re\ufb02ects a number of general principles in Boltzmann transport theory: In systems weakly perturbed out of equilibrium, it is convenient to use a representation f = f 0 + \u2202f 0\u03c7 for the distribution function, where f 0 is a (local) Maxwell\u2013Boltzmann distribution. 632 Nonequilibrium (classical) Typically, the left\u2013hand side of the Boltzmann equation contains the perturbing in\ufb02uences, while the right\u2013hand side (the collision term) aims to drive the system back to equilibrium. This suggests... an expansion of the left\u2013hand side to \ufb01rst order in the perturbations and to zeroth order in the distribution function. On the right\u2013hand side we linearize the collision integral, using the fact that I[f 0] = 0. The expansion of the left\u2013hand side then contains the perturbation in a form whose symmetries suggest a problem-adjusted ansatz for the correction, \u03c7. Substitution of this ansatz produces an e\ufb00ective equation for the strength of the correction which can be solved, e.g., in a relaxation time approximation. Programs of this type often provide an e\ufb03cient route to results otherwise obtained by microscopic linear response theory. However, problems far removed from equilibrium call for a more tailored approach and no general schemes can be formulated. 10.4 Stochastic processes In the previous sections we have introduced a few strategies to describe the approach to (as well as as departures from) thermal equilibrium. En route we have met with di\ufb00erent realizations of stochastic dynamics, or \u201cstochastic processes.\u201d Stochastic processes play a fundamental role in a whole spectrum of sciences, including the life sciences, engineering, socio-economic sciences, and many more. Equally important (for us as physicists) the languages in which stochastic processes are commonly described \u2013 the vocabulary includes \u201cprobability,\u201d \u201crate equations,\u201d \u201cMarkov approximations,\u201d etc. \u2013 are spoken by a large community of scientists, which is an ideal basis for interdisciplinary dialogue. 10.4.1 The notion of a stochastic process A process describes the temporal evolution of a certain state. It can be described in terms of a sequence {ai(ti)} where ti, i =", " 1,..., n are the discrete times at which the states are recorded and ai(ti) is a (generally multi-dimensional) state variable. A stochastic process21 is one where the evolution ai(ti) \u2192 ai+1(ti+1) involves elements of randomness. In the description of stochastic processes we then need to focus on the totality of all possible sequences {ai(ti)} where each realization gets assigned a certain probability p(an, tn;...; a1, t1). The probability function p carries the full information on the process, and this is the object we need to describe. Before moving on, it is worthwhile to distinguish between a few general types of stochastic processes. We speak of a stationary stochastic process if p does not change under simultaneous translation ti \u2192 ti + t0, i = 1,... n of all time arguments, i.e. if the theory is, on average, translationally invariant. Using the notion of conditional probabilities 21 For a state of the art introduction to theory and application of stochastic processes we refer to N.G. van Kampen, Stochastic Processes in Physics and Chemistry, (Elsevier,1992). 10.4 Stochastic processes 633 (cf. Eq. (10.7)), the joint probability function describing a stochastic process can be iteratively constructed as p(an, tn;...; a1, t1) = p(an, tn|an\u22121, tn\u22121;...; a1, t1) \u00d7p(an\u22121, tn\u22121|an\u22122, tn\u22122;...; a1, t1) \u00d7 \u00b7 \u00b7 \u00b7 \u00d7p( a2, t2|a1, t1) \u00d7 p(a1, t1). The process is called purely random if the probabilities p(aj, tj|aj\u22121, tj\u22121;...; a1, t1) = p(aj, tj) are independent of the history of events. In this case, n' p(an, tn;...; a1, t1) = p(aj, tj), j=1 is the product of n random numbers lacking intrinsic correlation. Evidently, purely random processes do not display very interesting dynamics. Next in the hierarchy of complexity are conditional probabilities for which p(an, tn|an\u22121, tn\u22121;...; a1, t1) = p(an, tn|an\u22121, tn\u22121). (10.36) Processes of this type are called Markov processes. Markov processes enjoy an enormous spectrum of applications in statistical sciences, which is why they deserve a separate discussion. 10.4.2 Markov processes In a Markov process, the passage to the state (an+1, tn+1) depends on the current state (an, tn), but not on the history that got us there, (an\u22121, tn\u22121;...). Put di\ufb00erently, a Markov process lacks memory. Markov processes owe their popularity to the fact that they optimally compromise between descriptive power and analytical tractability; if possible, one will try to (approximately) reduce a given process to an e\ufb00ective Markovian process. In doing so, it is important to keep in mind that22 the Markovian property holds only in an approximate sense. In modeling a process, the update times \u0394t = tn+1 \u2212 tn have to be chosen long enough to eradicate the short time memory of a process, yet short enough not to loose relevant aspects of the dynamics. (For example, in the process of Brownian motion, \u0394t must be chosen to be much23 larger than microscopic collision times.) Whether or not a process assumes a Markovian form generally depends on the coordinates used. For example, in a process where a velocity-like variable evolves in a Markovian process, 22 For a more substantial in-depth discussion, see van Kampen.21 23 To understand the meaning of \u201cmuch larger,\u201d suppose that \u0394t = O(collision time). If at a given instant, tn, the real space coordinate di\ufb00erence |xn \u2212 xn\u22121| happens to be atypically large, we know that the velocity at tn must have been high. It is then likely, that |xn+1 \u2212 xn| will also be large, i.e. the probability to reach xn+1 depends not just on the conditional probability p(xn+1, tn+1|xn, t), but also on the history of events before (xn, t", "). Averaging over a large number of collisions is necessary to average out this memory e\ufb00ect. 634 Nonequilibrium (classical) the corresponding coordinates \u2013 the integrals of velocity over time \u2013 may store long time memory and, therefore, be non-Markovian. (To see this, imagine a process of Brownian motion represented in coordinate space, q1 \u2192 q2 \u2192 \u00b7 \u00b7 \u00b7 \u2192 qn\u22122 \u2192 qn\u22121 \u2192 qn, where the coordinates are recorded at equal time steps. If the spacing |qn\u22121 \u2212 qn\u22122| was exceptionally large, the particle will have high velocity. Chances then are that the next step qn\u22121 \u2192 qn will also be large. In other words, the conditional probability p(qn|qn\u22121, qn\u22122,...) = p(qn|qn\u22121) and the process is not Markovian. Exercise: consider why velocity is a variable with less memory.) For a given process one will naturally try to identify coordinates that make it as \u201cMarkovian\u201d as possible. Chapman\u2013Kolmogorov relation and master equation The Markovian nature of a process can be expressed in a manner alternative to Eq. (10.36). Consider the probability to observe a at t given initial data (ai, ti) and some intermediate event (a, t), p(a, t; a, t; ai, ti). From this object, the \u201ctransition probability\u201d p(a, t; ai, ti) is obtained by integration over all realizations of the intermediate event, p(a, t; ai, ti) = da p(a, t; a, t; ai, ti) = da p(a, t|a, t; ai, ti) \u00d7 p(a, t; ai, ti), where we have used Eq. (10.7). For a Markovian process, p(a, t|a, t; ai, ti) = p(a, t|a, t), this reduces to the Chapman\u2013Kolmogorov relation p(a, t; ai, ti) = da p(a, t|a, t) \u00d7 p(a, t; ai, ti). (10.37) The usefulness of Eq. (10.37) becomes evident if we consider the case |t \u2212 t| |t \u2212 ti|. Eq. (10.37) then factorizes the description of the process into its three most relevant compounds, (a) our quantity of interest, i.e. the probability to get to a given we started at ai, (b) the probability to reach an intermediate stage a shortly before the \ufb01nal time, t, and (c) the short time transition probability a t\u2212t \u2212\u2192 a (see \ufb01gure). This iterative, or \u201ctransfer matrix type\u201d description of the process stands as the basis of most theories of Markovian dynamics. ai a, ti t, a t To push this iterative approach somewhat further, consider what happens if t \u2212 t \u2261 \u03b4t becomes (in\ufb01nitesimally) small. To a very good approximation, p(a, t|a, t \u2212 \u03b4t) \u03b4(a \u2212 a) + O(\u03b4t) will then be stationary. The term of O(\u03b4t) comprises losses out of a due to transition into some state a, and input due to transition from a. Denoting the probability of transitions a1 \u03b4t\u2192 a2 by W (a2|a1)\u03b4t, we may thus write p(a, t|a, t \u2212 \u03b4t) = 1 \u2212 \u03b4t da W (a|a) \u03b4(a \u2212 a) +\u03b4t W (a|a). 10.4 Stochastic processes 635 ai+1 ai ai\u22121 tn tn+1 Figure 10.6 The evolution of Markovian probability, as described by the master equation: probability gain of states ai (here considered as discrete entities, while the text uses continuum notation) due to in-processes is counteracted by loss due to out-processes. The sum over all feeds cancels, re\ufb02ecting probability conservation. Substituting this expression into Eq. (10.37), taking the limit \u03b4t \u2192 0 and suppressing the initial time argument for notational simplicity p(a, t|ai, ti) =p( a, t), we then obtain the so-called master equation24 (see Fig. 10.6). \u2202tp(a, t) = da [W (a", "|a)p(a, t) \u2212 W (a|a)p(a, t)]. (10.38) Notice that the master equation conserves the total probability, i.e. \u2202t da p(a, t) = 0. INFO In general, little can be said about the master equation. An important exception regards the master equation for physical systems that are closed and isolated in that no matter is exchanged with the outer world, and energy is conserved. The attribute \u201cphysical\u201d means that we consider systems whose microscopic dynamics is governed by some Hamiltonian. At large time scales, the solution of the master equation will relax to a time-independent equilibrium distribution function peq(a), typically a Maxwell\u2013Boltzmann distribution of the \u201cenergy\u201d correda [W (a|a)peq(a) \u2212 W (a|a)peq(a)]. This equation sponding to the state a. We thus have, 0 = should be interpreted as an e\ufb00ective sum rule obeyed by the transition matrix elements W (a|a). Using the invariance of microscopic Hamiltonian dynamics under time-reversal, it can in fact be shown24 that the above balance equation not only holds in an integral sense, but even for individual pairs (a, a), i.e. W (a|a)peq(a) = W (a|a)peq(a), or W (a|a) W (a|a) = peq(a) peq(a). (10.39) Equation (10.39) is known as the principle of detailed balance. 24 The term \u201cmaster equation\u201d was coined in A. Nordsieck, W. E. Lamb, and G. E. Uhlenbeck, On the theory of cosmic-ray showers I: the furry model and the \ufb02uctuation problem, Physica 7, 344-60 (1940), where a rate equation similar to Eq. (10.38) appeared as a fundamental equation of the theory. 636 Nonequilibrium (classical) The principle of detailed balance constrains the equilibrium behavior of systems containing di\ufb00erent compounds (as represented by the state variables a). For example, it states that, in a solution containing several chemical agents, the equilibrium reaction rates will be such that the reaction rates A \u2192 B between any one pair of agents A and B exactly compensate for the back reaction rates B \u2192 A. EXERCISE Kangaroo process: Consider the master equation, \u2202tp(a, t) = W (a|a)p(a, t) \u2212 W (a|a)p(a, t) da, with a factorizable transition matrix W (a|a) =u( a)v(a). What is kangaroo-like for the above transition matrix? Solve the master equation with the help of the Laplace transform p(a, z) = \u221e 0 dt eiztp(a, t). In a \ufb01rst step, show that the following relation holds p(a, z) = p(a, t = 0) + u(a)\u03c3(z) \u2212iz + v(a)/\u03c4, where 1/\u03c4 = for p(a, z) and solving the resulting algebraic equation. da u(a) and \u03c3(z) = da v(a)p(a, z). Determine \u03c3(z) by substituting the expression Kubo\u2013Anderson process: Taking v(a) \u2261 v, simplify the resulting equation for p(a, z) and compute the inverse Laplace transform to obtain the time evolution of the distribution p(a, t). Compare with the Ornstein\u2013Uhlenbeck process (Problem 10.9.2). Example: Gaussian process P (n, t) Let us illustrate the general concepts introduced above on two imporexamples. Consider a onetant dimensional random walk, i.e. a process wherein a particle performs random motion on a one-dimensional lattice (see \ufb01gure). Setting the lattice spacing to unity, the state variable a \u2261 n \u2208 Z takes integer values, and our goal is to compute the probability p(n, t|n0, t0). De\ufb01ning the probability density for an individual left-turn (right-turn) as q dt ((1 \u2212 q) dt), the master equation assumes the form 1 \u2212 q n n q t \u2202tp(n, t) = qp(n + 1, t) + (1 \u2212 q)p(n \u2212 1, t) \u2212 p(n, t). Anticipating smoothness of the probability distribution we may convert this equation into a di\ufb00erential equation. To this end we", " assume (without altering the problem too much) that the process takes place on a lattice of N points subject to periodic boundary conditions. De\ufb01ning the scaling variable x = n/N \u2208 [0, 1] and a rescaled probability distribution \u02dcp(x)dx = p(N x)dn = p(N x)(dn/dx)dx \u21d2 \u02dcp(x) =N p (N x), and Taylor expanding, 10.4 Stochastic processes 637 \u02dcp(x \u00b1 1/N ) = \u02dcp(x) \u00b1 N \u22121\u2202x \u02dcp(x) + (1/2)N \u22122\u22022 of a generalized di\ufb00usion equation (for notational simplicity, we write p instead of \u02dcp), x \u02dcp(x), the master equation assumes the form \u2202t + \u03b3\u2202x \u2212 D\u22022 x p(x, t) = 0, p(x, t0) = \u03b4(x \u2212 x0), with \u201cdrift coe\ufb03cient\u201d \u03b3 = (1 \u2212 2q)N \u22121, di\ufb00usion constant D = 1/2N 2, and initial point x0 = n0/N. This equation is solved by p(x, t) = dk 2\u03c0 e\u2212(i\u03b3k+Dk2)teik(x\u2212x0) = 1 2(\u03c0Dt)1/2 e\u2212 (x\u2212x0\u2212\u03b3t)2 4Dt, or, in terms of the original variables, p(n, t) = 1 (2\u03c0t)1/2 exp \u2212 (n \u2212 n0 \u2212 (1 \u2212 2q)t)2 2t. (Exercise: consider how you would solve the structurally similar Schr\u00a8odinger equation of a free particle.) General comments on Gaussian processes: Gaussian or \u201cnormal\u201d distributions typically arise when large numbers of statisticallyindependent events add to form a composite random variable. (In cases where the composite variable is obtained by multiplication of elementary random variables, logarithmically normal distributions result.) The Gaussian form of the probability distribution is a manifestation of the central limit theorem: we may think of n as a random variable obtained by summing t elementary random variables \u00b11 drawn from a bimodal distribution with probability p(1) = 1 \u2212 q and p(\u22121) = q. According to the central limit theorem, the result will be a Gaussian distributed variable centered around the mean t \u00d7 n0 + (1\u2212 2q)t and variance \u223c \u221a t. Random processes whose distributions are Gaussian are generally called Gaussian processes. The example illustrates that Gaussian processes are typically the results of the addition of a large number of elementary random variables. Example: Poisson process n Now consider a sequence of elementary events occurring in a manner (e.g. apples uncorrelated in time falling o\ufb00 a tree on an autumn day, signals of a Geiger counter exposed to weak radiation, etc.). The random interest is the number, variable of n, of events occurring in a certain time t (see \ufb01gure). Assuming that the probability of an event to occur in a short time window dt is \u03bddt, the master equation takes the form P (n, t) t \u2202tp(n, t) = \u03bdp(n \u2212 1, t) \u2212 \u03bdp(n, t). (10.40) 638 Nonequilibrium (classical) For de\ufb01niteness, we assume that this equation has to be solved with the boundary condition p(n, t = 0) =\u03b4 n,n0. Again, we may approximately solve this equation by assuming smoothness of the distribution and Taylor expansion. However, this procedure is a little bit too coarse to resolve the speci\ufb01c statistics relating to the discreteness of events. (Exercise: solve the equation by Taylor expansion and compare with the results below.) In the following, we employ concepts borrowed from quantum mechanics to construct a more accurate solution. Later on, we see that speci\ufb01c aspects of stochastic processes \u2013 the dynamics of very rare events, to mention one example \u2013 a\ufb00ord a general description in terms of \u201cquantum\u201d theory. (The quotes are used here because, of course, there is no true involved.) We start by rewriting the equation as \u2202tp(n, t) = \u03bd( \u02c6E\u22121 \u2212 1)p(n, t), where \u02c6Em is the translation operator in discrete n-space, i.e. \u02c6Emf (n) \u2261 f (n + m). Now, the translation operator a\ufb00", "ords a representation in terms of the \u201cmomentum\u201d operator \u02c6\u03c6 conjugate to the \u201cnumber operator\u201d n: With [ \u02c6\u03c6, \u02c6n] = \u2212i, we have \u02c6Em = eim \u02c6\u03c6. In the \u201cn-representation,\u201d the translation operator E\u22121 takes the form \u02c6E\u22121 = exp(\u2212\u2202n). Its eigenfunctions are given by fk(n) = (2\u03c0)\u22121/2exp(ikn) with eigenvalue \u03bbk = exp(\u2212ik). Finally, the initial con\ufb01guration of the probability distribution a\ufb00ords the spectral representation p(n, 0) = \u03b4n,n0 = 2\u03c0 e\u2212ik(n\u2212n0). By analogy with the solution of a quantum initial value problem, we may dk 2\u03c0 0 thus represent p(n, t) as p(n, t) = 2\u03c0 0 dk 2\u03c0 e\u2212\u03bdt(e\u2212ik\u22121)eik(n\u2212n0). Using the identity obtains the result (exercise) 2\u03c0 0 dk 2\u03c0 ei\u0394nk = \u03b4\u0394n,0, a straightforward Taylor expansion of the exponent p(n, t) = (\u03bdt)n\u2212n0 (n \u2212 n0)! e\u2212\u03bdt. (10.41) The process is described by a Poisson distribution (cf. Eq. (10.4) where \u03bdt \u2194 np represents the product of \u201cattempts\u201d (t \u2194 n) and \u201csuccess probability\u201d (\u03bd \u2194 p)), hence the name Poisson process. The \ufb01rst and second cumulant of the Poisson distribution are given by, respectively, \u03bc1 = \u03bdt, \u03bc2 = \u03bdt, i.e. centered around \u03bc1 = \u03bdt, the width of the distribution is given by \u223c (\u03bdt)1/2. The variable t is a bookkeeping index, which need not necessarily be physical time. For example, the statistics of energy levels on the energy axis of integrable quantum systems is usually described by a Poisson distribution, which re\ufb02ects the lack of inter-level correlations. EXERCISE One-step processes: The master equation of a one-step process is given by \u02d9pn = rn+1pn+1 + sn\u22121pn\u22121 \u2212 (rn + sn)pn. 10.4 Stochastic processes 639 To solve the master equation it is useful to make use of the generating function g(z, t) = \u221e n=\u2212\u221e znpn(t), where g(1, t) = 1 re\ufb02ects the conservation of probability. Show that for a given g, the probability follows from the inverse Mellin transform pn(t) =, 1 2\u03c0i dz z\u22121\u2212ng(z, t), (10.42) \u221a 4\u03c0t. where the integral is taken around a contour enclosing z = 0 (but no singularities of g.) (a) Symmetric random walk: Consider the special case rn = sn = 1 with the initial condition pn(0) = \u03b4n,0. From the master equation, derive a di\ufb00erential equation for g(z, t) and show that its solution is g(z, t) = exp((1/z + z \u2212 2)t). Solve the integral (10.42) in the saddle-point approximation and show that the resulting behavior for large times t \u2192 \u221e is di\ufb00usive, pn(t) e\u2212n2/4t/ (b) Furry process: Consider the case rn = 0 and sn = \u03b3n with the initial condition pn(0) = \u03b4n,1. Determine g(z, t). You can solve its partial di\ufb00erential equation with help of the method of characteristics. Using Eq. (10.42) show that the probabilities for n > 0 are given by pn(t) = e\u2212\u03b3t(1 \u2212 e\u2212\u03b3t)n\u22121. (c) Population growth: Consider the case rn = \u03b1n and sn = \u03b2n with the initial condition pn(0) = \u03b4n,m. Show that the probability that the population dies out at time t is given by m, where = e(\u03b1\u2212\u03b2)t. Discuss the short and long time limits. p0(t) =! \u03b1(1\u2212) \u03b2\u2212\u03b1 10.4.3 Fokker\u2013Planck equation II The master", " equation (10.38) is a linear integral equation for the probability density, p. It can be formally solved in terms of a spectral decomposition of the integral kernel {W (a|a)} (exercise). However, in cases where W (a|a) exhibits nontrivial dependence on the state variables a (which may be multi-dimensional!), an explicit eigenmode decomposition of p may be not attainable.25 However, in many cases of practical interest: The integral kernel W (a|a) decays rapidly as a function of the \u201cdistance\u201d |a \u2212 a|, i.e. for su\ufb03ciently short times \u03b4t, transitions between remote states are strongly suppressed. (Here, we assume that state space {a} comes with a meaningful notion of distance, which usually is the case.) We may anticipate smoothness of the distribution p(a) on the scales of variation of W. Under these conditions, a Kramers\u2013Moyal expansion can be applied to reduce the integral equation to the more benign form of a quasi-local equation (i.e. a di\ufb00erential equation). Repeating the sequence of steps carried out in Section 10.2.3, we change notation W (a|a) = Wb(a), b \u2261 a \u2212 a, where Wb(a) is the transition probability for a \u2192 a + b. In this notation, \u2202tp(a) = db [(Wbp)(a \u2212 b) \u2212 (W\u2212bp)(a)]. 25 Although a numerical diagonalization of W may still be the most e\ufb03cient route to obtain the (long time) pro\ufb01le of p. (10.43) (10.44) (10.45) 640 Nonequilibrium (classical) Using the relation, a (the index in Wb remains passive!) obtains the series db (Wb(a) \u2212 W\u2212b(a)) = 0, Taylor expansion of (Wbp)(a \u2212 b) in b around \u2202tp(a, t) = \u221e n=1 (\u2212)n n! \u2202n a (\u03b1np) (a, t), where the coe\ufb03cients \u03b1n are the moments of the transition probabilities, \u03b1n(a) = db Wb(a)bn. The Fokker\u2013Planck approximation \u2202tp(a, t) = \u2212\u2202a\u03b11(a) + with \u03b11 = negligible. As exempli\ufb01ed in Section 10.2.3 above: db Wb(a)b and \u03b12 = \u22022 a\u03b12(a) p(a, t), 1 2 db Wb(a)b2 assumes that all terms beyond n = 2 are The \ufb01rst derivative, or drift term \u223c \u2202a\u03b11p is \u201cdeterministic\u201d in that it describes the evolution of the averaged variable a. Indeed, \u2202ta = da a\u2202tp(a, t) = da a \u2212\u2202a\u03b11(a) + 1 2 \u22022 a\u03b12(a) p(a, t) = da \u03b11(a)p(a, t) = \u03b11, where, to prove this identity, we have applied an integration by parts. The second derivative, or di\ufb00usion term describes the di\ufb00usive spread of probability around the center of the distribution, a. In many cases, the Fokker\u2013Planck approximation does an excellent job at describing the probability distributions of stochastic processes. However, as with any other approximation scheme, it is not perfect. Some care must be exercised, e.g. if one is asking questions about the tails of probability distributions, \u201crare event\u201d regions where p has become small. In the next section, we discuss an example illustrating success and partial failure of the Fokker\u2013Planck approximation in a very simple setting. 10.4.4 Quality of the Fokker\u2013Planck approximation: an example The Fokker\u2013Planck approximation is routinely applied as a tool to explore analytically the dynamical evolution of probability distributions. It is therefore important to identify those situations where the approximation goes qualitatively wrong. Empirically, large deviations between the predictions of the Fokker\u2013Planck equation and results obtained by (numerical) iteration of the Master equation are frequently observed in the tails of the probability distribution. Often these tails carry information about \u201crare\u201d events far from the typical behavior of the system. Such rare events can be of profound importance. For example, in cases where the master equation describes the epidemic proliferation of a virus, one will likely be interested in the probability of exceptionally persistent 10.4 Stochastic processes 641 infections. Equally, in a master equation modeling a", " nuclear chain reaction, the eventuality of an uncontrolled chain reaction, no matter how unlikely, will be of de\ufb01nite interest, etc. In the following, we explore the shortcomings of the Fokker\u2013Planck approximation for the familiar example of the Poisson process. In Section 10.4.2 we introduced the master equation of the Poisson process and obtained Eq. (10.41) for its rigorous solution. Now, let us see what we would have got instead had we subjected the master equation to a Kramers\u2013Moyal expansion. Introducing a scaled variable, x = n/N, N 1, with associated continuum probability distribution \u02dcp(x, t)dx = p(n, t)dn \u21d2 \u02dcp(x, t) =p( xN, t)N, we may subject the equation to a second order Kramers\u2013Moyal expansion. For later reference, we note that this expansion amounts to second order expansion of the translation operator \u02c6E\u22121 \u2212 1 = e\u2212\u2202x/N \u2212 1 \u2212 1 N \u2202x \u2212 1 2N 2 \u22022 x. As a result, we obtain the Fokker\u2013Planck equation \u2202x \u2212 \u03bd \u02dcp(x, t) = 0, \u2202t + \u03bd N 2N 2 \u22022 x with p(x, 0) = \u03b4(x). This equation is solved by \u02dcp(x, t) =N ( 1 or, upon translating back to the original variable n, 2\u03c0\u03bdt )1/2exp(\u2212 1 2t\u03bd (N x \u2212 t\u03bd)2) p(n, t) = 1 2\u03c0\u03bdt 1/2 exp \u2212 1 2t\u03bd (n \u2212 t\u03bd)2. (10.46) This \u201cGaussian approximation\u201d displays the two principal characteristics of the Poisson distribution, mean value \u03bdt and width \u223c (\u03bdt)1/2. However, in the tails of the distribution, the approximation is poor. This is illustrated in (the logarithmic representation of) the \ufb01gure: At values of n about four times bigger than the mean value, in the far tails of the distribution, the Fokker\u2013Planck prediction is by many orders of magnitude o\ufb00 the true result. 0 -10 -20 -30 -40 10 20 30 n 40 The origins of the this error can be traced to the second order expansion of the operator \u02c6E1 = e\u2212\u2202x/N. Doing so, and replacing this operator by an e\ufb00ective long-range approximation wherein the \u201cmomentum operator\u201d \u2212i\u2202x is considered small, we forget about the integer nature of the variable n. (Exercise: why?) To obtain a better result in the far tail regions, we ought to keep the integrity of this operator intact. This makes the theory more complicated. However, pushing the quantum analogy further, and thinking of p(n, t) = p(n, t|0, 0) as a \u201ctransition amplitude\u201d we may turn the smallness of p to an advantage: using a path integral oriented language, and writing p \u223c exp(\u2212S) as the integral over an exponentiated \u201caction\u201d, we know that S 1 in the far tails. This suggests that semiclassical approximation schemes might be applicable. In the following, we formulate a semiclassical approach to the Poisson process, and demonstrate that it does an excellent job of describing the tail regions. 642 Nonequilibrium (classical) De\ufb01ning the \u201cHamiltonian operator\u201d, \u02c6H \u2261 \u2212\u03bd e\u2212i \u02c6p \u2212 1, the master equation assumes the form of a Schr\u00a8odinger equation for an imaginary time evolution operator, (\u2202t + \u02c6H)p(n, t) = 0, p(n, 0) = \u03b4n,0. Here, we are working in the system of original variables n, instead of the rescaled coordinate x. This equation has the formal solution p(n, t) = n|e\u2212t \u02c6H |0 = q(t)=n q(0)=0 D(q, p) e t 0 dt (ip \u02d9q\u2212H(p,q)), where n|n = \u03b4n,n is the scalar product in n-space, and in the second equality we switched to a path integral representation. To avoid the mixed appearance of real and imaginary terms in the exponent, we subject the integration over the momentum variable to a Wickrotation, p \u2192 \u2212ip. In the new variables, q(t)=n p(n, t) = D(q, p) e q", "(0)=0 t 0 dt (p \u02d9q\u2212H(q,p)), H(q, p) = \u2212\u03bd e\u2212p \u2212 1. We evaluate this path integral in the crudest form of a stationary phase approximation, t 0 dt (p \u02d9q \u2212 H(q, p)), q p(n, t) e\u2212S[q,p], S[q, p] = \u2212 where (q, p)(t) are solutions of the extremal equations with the con\ufb01guration space boundary conditions q(0) = 0, q(t) =n. Varying the action we obtain the equations of motion, dt q = \u2202pH = \u03bde\u2212p, dt p = \u2212\u2202qH = 0. The solution of these equations reads p(t) = \u2212 ln(n/\u03bdt), q(t) = n(t/t). n p The conserved energy is given by H = \u2212n/t + \u03bd. It is instructive to take a brief look at the \u201cphase portrait\u201d of these solutions shown in the \ufb01gure above. Conservation of momentum means that the \ufb02ow is parallel to the q-axis with velocity \u03bde\u2212p. Atp = 0, q(t) =\u03bdt, which re\ufb02ects the temporal evolution of the mean value n(t) of the Poisson process. The action corresponding to this mean trajectory vanishes, S(q, p = 0) = 0. Within the Hamiltonian approach it is possible to reach \ufb01nal con\ufb01gurations q(t) =n( t) very di\ufb00erent from t\u03bd. However, this comes at the price of a large action: the action of the solutions above is readily obtained as S[q, p] = n ln(n/\u03bdt) \u2212 n + \u03bdt. 10.5 Field theory I: zero dimensional theories 643 (Notice that this action assumes its minimum value S = 0 at n = \u03bdt.) The stationary phase approximation to the probability distribution thus reads p(n, t) e\u2212(n ln(n)\u2212n)+n ln(\u03bdt)\u2212\u03bdt = e\u2212(ln(n)\u2212n)(\u03bdt)ne\u2212\u03bdt (\u03bdt)n n! e\u2212\u03bdt, where the last approximate equality is based on an \u201cexponential\u201d approximation to the 2 ln(2\u03c0n)en ln n\u2212n \u223c en ln n\u2212n. The semiclassical approximation is Stirling formula, n! 1 shown in the \ufb01gure on p.641, dashed. In the center of the distribution, where S = 0 and semiclassical approximations are problematic, the Fokker\u2013Planck equation does a better job at approximating the Poisson distribution. However, in the tail regions, where S 1, semiclassics works well! The quality of the result may be improved by including quadratic \ufb02uctuations around the mean \ufb01eld trajectory, which then leads to an improved variant of Stirling\u2019s approximation. However, to \u201cexponential accuracy,\u201d the present approximation is su\ufb03cient. The take home message of this exercise is that variational methods can be a powerful tool in the study of \u201crare events\u201d in a stochastic process. Signi\ufb01cant deviations from the Fokker\u2013 Planck approach are to be expected if the Hamiltonian of the process is not harmonic in momenta (which means that the corresponding \u201cSchr\u00a8odinger equation\u201d di\ufb00ers from the second order Fokker\u2013Planck equation) due to, e.g., discreteness of the agents participating in the process. However, even if the Fokker\u2013Planck equation is valid, the semiclassical analysis can be a useful aid. The point is that the Fokker\u2013Planck equation \u2013 mathematically similar to an imaginary time Schr\u00a8odinger equation \u2013 usually de\ufb01es rigorous analytical solution. Much as WKB-type semiclassics is useful in quantum mechanics, the \u201csemiclassical\u201d approximation to a Fokker\u2013Planck equation can be an economic route to obtain information on large-action/rare-event statistics. 10.5 Field theory I: zero dimensional theories In the previous sections, we have introduced basic elements of classical nonequilibrium theory. We have seen that nonequilibrium systems can be described (see Fig. 10.7) on: a microscopic level, i.e. in terms of microscopic Hamiltonians and their interactions, a mesoscopic level, where the microscopic transition rates are lumped", " into either a stochastic di\ufb00erential equation of Langevin type, or a master equation, a continuum level. In our discussion so far, the continuum description was formulated in terms of the Fokker\u2013Planck equation, a second order partial di\ufb00erential equation in the state variables, and \ufb01rst order in time. Both, from a formal and a conceptual point of view, the description of nonequilibrium phenomena in terms of di\ufb00erential (Langevin, Fokker\u2013Planck) or integral (master) equations bears similarities to the Schr\u00a8odinger approach to quantum mechanics. It requires comparatively little theoretical machinery, a\ufb00ords a lean and reasonably intuitive description of many physical phenomena, but also su\ufb00ers from limitations. We had a \ufb01rst glimpse of the de\ufb01ciencies of the Fokker\u2013Planck approach when we explored the tail regions of the Poisson 644 Nonequilibrium (classical) microscopic Hamiltonian mesoscopic Langevin equation master equation continuum Fokker-Planck equation e l a c s Figure 10.7 Showing the three levels of description of nonequilibrium systems. process: To stay with the analogy above, the tails turned out to be described better in terms of a \u201cpath integral approach.\u201d Generally speaking, much of what has been said in the introductory chapters of this text by way of motivation of path integral and \ufb01eld theory oriented approaches to genuine quantum systems carries over to the present context. Speci\ufb01cally: In the description of rare events and large \ufb02uctuations, variational approaches and the analysis of extremal action con\ufb01gurations may be a powerful alternative to the solution of the Fokker\u2013Planck equation. So far, our discussion has been limited to statistical problems described in terms of only a few random variables \u2013 the analog of \u201czero dimensional\u201d quantum problems. However, many applications of interest are described in terms of many spatially resolved random variables \u2013 think of a \ufb01eld {n(r, t)} describing the local population size in a problem of evolutionary ecology. Much as in quantum \ufb01eld theory, \ufb01eld integration techniques are the method of choice to explore the ensuing phenomena of collective dynamics, criticality, etc. The Langevin equations explored above have been linear equations, on account of the oscillator dynamics of the underlying non-dissipative dynamics. Extremal principles derived from e\ufb00ective path integral descriptions are an e\ufb03cient method to explore the more interesting behavior shown by nonlinear stochastic di\ufb00erential equations. In this section we introduce a number of \ufb01eld theoretical tools that have been designed to study the behavior of classical nonequilibrium systems. For the sake of clarity, (a) this section is devoted exclusively to the discussion of formalism. Applications are addressed in the follow-up section. Furthermore, (b) we introduce the \ufb01eld theory approaches in the simple setting of zero-dimensional problems, i.e. as (0 + 1) dimensional path integrals. The extension to (d + 1)-dimensional \ufb01eld theories is discussed in Section 10.6. 10.5 Field theory I: zero dimensional theories 645 10.5.1 Martin\u2013Siggia\u2013Rose\u2013Janssen\u2013de Dominicis approach Let us begin our discussion by constructing a path integral reformulation of stochastic di\ufb00erential equations. A (\ufb01rst order) stochastic di\ufb00erential equation is an equation of the structure \u2202tn + f (n) = \u03be, (10.47) where n = n(t) is a time-dependent random variable, f = f (n(t)) a function, and \u03be = \u03be(t) a noise term. In some applications, \u03be = \u03be(n, t) depends on n. However, here we restrict ourselves to the case of linear noise, where \u03be does not depend on n.26 For simplicity, we assume that the noise is Gaussian with correlator, \u03be(t)\u03be = 0, \u03be(t)\u03be(t)\u03be = A\u03b4(t \u2212 t). (10.48) As a \ufb01rst step towards the construction of a path integral representation, we discretize time, n(t) \u2192 ni, i = 1,..., N, according to ni \u2212 ni\u22121 + \u0394t [f (ni\u22121) \u2212 \u03bei\u22121] = 0, (10.49) where \u0394t is the temporal discretization interval. Denoting a solution of the di\ufb00erential equation by n[\u03be], expectation values of observables O[", "n]\u03be may then be formally represented as O[n]\u03be = Dn O[n] \u03b4(n \u2212 n[\u03be])\u03be = Dn O[n] A \u03b4(X), \u03be @ \u03b4X \u03b4n i dni is the functional measure and \u03b4(n\u2212n[\u03be]) = i \u03b4(ni\u2212n[\u03be]i). In the second where Dn = equality, we have introduced a vector X = {Xi}, where Xi \u2261 ni \u2212ni\u22121 +\u0394t [f (ni\u22121) \u2212 \u03bei\u22121] }. The advantage of the discretizaand |\u03b4X/\u03b4n| is the (modulus) of the determinant of { \u2202Xi \u2202nj tion chosen above is that \u03b4X/\u03b4n is a triangular matrix with unit diagonal, i.e. the functional determinant equals unity. Thus, substituting the de\ufb01nition of X, O[n]\u03be = Dn O[n] \u03b4(\u2202tn + f (n) \u2212 \u03be)\u03be, (10.50) where \u2202tn = {ni \u2212 ni\u22121} is shorthand for the lattice time derivative. INFO The discretization leading to a unit functional determinant goes under the name Ito discretization. More general discretization schemes call for a treatment of the corresponding functional determinants. It is customary to represent these determinants in terms of Grassmann integrals. The resulting integrals jointly involve real integration variables (the functional representation of the \u03b4-function), and Grassmann variables. Interestingly, these integrations are coupled by a symmetry principle known as BRS-symmetry. For further discussion of BRS symmetries in the context of stochastic di\ufb00erential equations, we refer to J. Zinn-Justin, Quantum Field Theory and Critical Phenomena, (Oxford University Press, 1993). 26 For an application with nonlinear noise term, see Section 10.7.2 below. 646 Nonequilibrium (classical) Representing the \u03b4-function in terms of a Fourier integral, and switching back to a continuum notation, we arrive at the functional integral representation > O[n]\u03be = D(n, \u02dcn) O[n] ei dt \u02dcn(\u2202tn+f (n)\u2212\u03be)?, \u03be where the structure of the \u201caction\u201d suggests that the two \ufb01elds n and \u02dcn are \u201ccanonically conjugate\u201d to each other. We \ufb01nally average over the noise to arrive at the celebrated Martin\u2013Siggia\u2013Rose\u2013Janssen\u2013de Dominicis (MSRJD) functional integral O[n]\u03be = D(n, \u02dcn) O[n] exp dt i\u02dcn(\u2202tn + f (n)) \u2212 A 2 \u02dcn2. (10.51) Equivalently, we may use the path integral representation to compute the conditional probability p(n, t|n0, 0) (just set O[n] = \u03b4(n \u2212 n(t))) in terms of a path integral with \u201ccon\ufb01guration space boundary conditions\u201d n(t)=n p(n, t|n0, 0) = D(n, \u02dcn) exp dt i\u02dcn(\u2202tn + f (n)) \u2212 A 2 \u02dcn2. (10.52) n(0)=n0 Before proceeding, let us make a few remarks on the structure of the path integrals (10.51) and (10.52): Introducing a new set of variables (q, p) through (n, \u02dcn) \u2261 (q, \u2212ip), Eq. (10.51) assumes the form of an imaginary time phase space path integral, p(n, t|n0, 0) = q(t)=n q(0)=n0 D(q, p) e dt [p \u02d9q\u2212H(q,p)], H(q, p) = \u2212pf (q) \u2212 A 2 p2. (10.53) It is important to keep in mind that the integration over p extends over the imaginary axis. The action of the path integral vanishes on the line p = 0. That line generally hosts a solution of the classical equations of motion. Varying the action we indeed \ufb01nd that (q(t), p = 0), where \u02d9q = \u2212f (q) is a solution. Comparison with the Langevin equation shows that the dynamics describes the noise-averaged Langevin variable, q(t", ") \u2194 n(t), if \ufb02uctuations are weak: \u2202tn = \u2212f (n) \u2212f (n). Although the Hamiltonian does not assume the \u201cstandard form\u201d H = T (p) +V (q), it is quadratic in momenta. We may thus integrate over p to obtain p(n, t|n0, 0) = q(t)=n q(0)=n0 D(q) e\u2212 1 2A dt [\u2202tq+f (q)]2. (10.54) 10.5 Field theory I: zero dimensional theories 647 Multiplying out the square in the exponent, we arrive at a Lagrangian variant of the theory, the Onsager\u2013Machlup functional27 p(n, t|n0, 0) = q(t)=n q(0)=n0 D(q) e\u2212 dt L(q, \u02d9q), L(q, \u02d9q) = 1 2A \u02d9q2 + \u02d9q f (q) A + f (q)2 2A. (10.55) This is the Lagrangian of a one-dimensional point particle with \u201cmass\u201d m = 1/A which is coupled to a velocity-dependent one-dimensional \u201cvector potential\u201d of strength A(q) = f (q)/A, and a potential V (q) = f (q)2/2A. We may interpret the Onsager\u2013Machlup representation of the probability distribution as the formal analog of a Lagrangian path integral for a quantum time evolution operator. On general grounds, the \u201cSchr\u00a8odinger equation\u201d corresponding to this formulation reads (\u2202q + A(q))2 + V (q) \u03c6(q, t) = \u22022 q \u2212 \u2202qf (q) \u03c6(q, t) = 0, \u2202t \u2212 A 2 \u2202t \u2212 1 2m where \u03c6 is the analog of a wave function. Speci\ufb01cally, the \u201ctime evolution operator\u201d p(n, t|n0, 0) is a solution of the initial value problem \u2202t \u2212 A 2 \u22022 q \u2212 \u2202qf (q) p(q, t) = 0, p(n, 0) = \u03b4(n \u2212 n0), q=n which is nothing but the Fokker\u2013Planck description of the Langevin dynamics. We thus arrive at the important conclusion that the Fokker\u2013Planck equation is the \u201cSchr\u00a8odinger equation\u201d of the MSRJD path integral. The discussion above is overly-concise in that (a) we are not explicit regarding sign factors and imaginary is. For instance, the \u201cmomentum\u201d operator in our Schr\u00a8odinger equation reads \u2202q instead of the conventional \u2212i\u2202q, because the equation is derived from a theory with real exponent \u2212S[q], instead of iS[q]. More importantly, (b) we do not keep track of the order of derivatives (why \u2202qf (q), instead of f (q)\u2202q?). To monitor the correct ordering, one needs to keep track of the properly Ito-discretized sequential ordering of operators in the action (an instructive if somewhat technical exercise). INFO It may be instructive to observe the functioning of the concepts above on a miniature application. Consider a particle at rest in the bottom of a metastable potential con\ufb01guration (see inset of Fig. 10.8). Thermal activation may help the particle in surmounting the potential barrier to settle in an overall more favourable true energy minimum. We know that the corresponding thermal escape rates are controlled by Arrhenius factors. The detailed computation of such rates is of importance in, say, the chemistry of thermally activated reactions. For a given potential landscape, the functional integrals above can be a powerful aid in analyzing the corresponding 27 L. Onsager and S. Machlup, Fluctuations and irreversible processes, Phys. Rev. 91, 1505-12 (1953). 648 Nonequilibrium (classical) Figure 10.8 Inset: particle in a metastable potential well. Main \ufb01gure: phase space portrait of the corresponding MSRJD functional. For a discussion, see main text. escape kinematics. By way of example, here we discuss the emergence of Arrhenius activation factors from the functional integral formalism. Assuming overdamped dynamics, the equation of motion of a point particle in the well is given by \u03b3 \u02d9q + \u2202qV (q) = \u03be(t), where, by the \ufb02uctuation-diss", "ipation theorem, \u03be(t)\u03be(t) = 2\u03b3T \u03b4(t \u2212 t). Dividing by \u03b3, we obtain a stochastic di\ufb00erential equation of the type Eq. (10.47), with the identi\ufb01cation q \u2194 n, f = \u2202qV and noise strength A = 2T \u03b3\u22121. Let us assume that thermal escape will be an unlikely event in the sense that V /T 1, where \u0394V is the di\ufb00erence between the local maximum of the metastable well and its minimum. We thus expect a small escape probability p \u223c exp(\u2212V /T ). The largeness of the exponent \u2212 ln p \u223c V /T 1 then suggests an evaluation of the path integral (10.53) by semiclassical methods. In its crudest form, semiclassics means that we evaluate the Hamiltonian path integral at its stationary points and neglect all \ufb02uctuation e\ufb00ects. With H(q, p) = \u2212 p \u03b3 (\u2202qV (q) + T p), the Hamiltonian equations of motion of the action (10.53) read \u02d9q = \u2202pH(q, p) = \u2212 1 \u03b3 \u2202qV (q) \u2212 2T \u03b3 p, \u02d9p = \u2212\u2202qH(q, p) = 1 \u03b3 p\u22022 q V (q). For the purposes of the present discussion, we do not need to solve these equations; it is su\ufb03cient to inspect the phase portrait of the Hamiltonian. As we shall see, the topology of Hamiltonian \ufb02ow in phase space is essentially determined by the zero energy contours H(q, p) = 0. The structure of the Hamiltonian implies that one of these curves is speci\ufb01ed by (q, p = 0). On the line p = 0, the equations of motion reduce to \u03b3 \u02d9q = \u2212\u2202qV (q). This is drift motion down the potential gradient towards the local minimum. We 10.5 Field theory I: zero dimensional theories 649 note that the point of the local potential maximum q \u2261 q\u2217 de\ufb01nes an unstable point on that line, while q = 0 is stable. This observation entails two important consequences: (a) we expect that the equilibrium coordinates on the p = 0 line, 0 and q\u2217, must de\ufb01ne points of origin of motion qualitatively di\ufb00erent from downward drift q\u2217 \u2192 0. Speci\ufb01cally, we expect the option of noise-activated reverse motion 0 \u2192 q\u2217. If so, the equilibrium points must be the terminal points of another E = 0 trajectory. Indeed, the equation H(q, p) = 0 possesses a second solution, p = \u2212T \u22121\u2202qV. At the potential extrema \u2202qV = 0, or q = 0, q\u2217, this curve intersects the p = 0 line, which means that (q\u2217, 0) and (0, 0) are terminal points of phase space separatrices. (Recall that the principle of non-crossing phase space curves requires that the motion along the separatrices takes in\ufb01nite time. For example, you may convince yourself that in a strict mathematical sense, the downward drift trajectory q\u2217 \u2192 0 has in\ufb01nite passage time.) (b) The conservation of phase volume (Liouville\u2019s theorem) implies that the two phase space \ufb01xed points (0, 0) and (q\u2217, 0) are hyperbolic. This means that the point (0, 0) (stable in coordinate direction) is unstable in momentum direction, while (q\u2217, 0) (unstable in coordinate direction) is stable in momentum direction. Notice that the discussion above did not rest upon many speci\ufb01c characteristics of the escape problem. Indeed, the existence of E = 0 separatrices terminating in hyperbolic \ufb01xed points on the p = 0 line is a principle that carries over to many other \u201cnoise\u201d phase portraits. The separatrices largely determine the topology of phase \ufb02ow, as exempli\ufb01ed in Fig. 10.8. Speci\ufb01cally, the separatrix 0 \u2192 q\u2217 accounts for the possibility of noise-assisted motion against the potential gradient. Within our semiclassical framework, the probability for this happening p \u223c exp(\u2212S) is proportional to the action accumulated along this trajectory. The latter is calculated as S = \u2212 dt (p(q) \u02d9q \u2212 H(q, p)) = \u2217 q 0", " 1 T dq \u2202qV (q) = 1 T (V (q\u2217) \u2212 V (0)) = \u0394V T, where we have used H(q, p) = 0, and the de\ufb01ning equation of the separatrix. As expected, we obtain the Arrhenius weight as the dominant factor controlling the escape probability. Of course, much less heavy artillery is needed to arrive at just this result. However, as usual with functional approaches, the advantage of the above formalism is its extensibility to more complex problems. 10.5.2 Field integral representation of the master equation I In order to avoid potential confusion with the \u201cmomentum\u2019 variable below, the notation for probability are capitalized in this section, p \u2192 P. By generalization of the ideas exempli\ufb01ed in Section 10.4.4 on the Poisson process, we can derive a path integral representation for the master equations describing Markovian stochastic processes. Unsurprisingly, this path integral turns out to be a close ally of the MSRJD functional integral above. For simplicity, we consider a one step master equation, i.e. an equation where a discrete state variable n changes by no more than one unit \u00b11 in each time step.28 The 28 The straightforward generalization to \ufb01nite range equations leads to path integrals with more complicated actions. However, in many cases, the long time dynamics deriving from these integrals is qualitatively similar to the one of the single step process. 650 Nonequilibrium (classical) general form of a one step master equation reads # $ \u2202tP (n, t) = ( \u02c6E1 \u2212 1)f\u2212(n) + ( \u02c6E\u22121 \u2212 1)f+(n) P (n, t), (10.56) where E\u00b11f (n) \u2261 f (n \u00b1 1) acts by unit translation. As in our example application in Section 10.4.4, it is useful to think of the probability distribution as the analog of a quantum time evolution operator \u02c6P with \u201cmatrix elements\u201d P (n, t; n0, t0) \u2261 P (n, t|n0, 0) \u2261 P (n, t). Equation (10.56) then becomes an (imaginary time) evolution equation, \u2202t + \u02c6H(\u02c6q, \u02c6p) \u02c6P (t) = 0, \u02c6P (0) = 1, with the \u201dHamiltonian operator\u201d \u02c6H(\u02c6q, \u02c6p) = \u2212 e+i \u02c6p \u2212 1 f\u2212(\u02c6q) \u2212 e\u2212i \u02c6p \u2212 1 f+(\u02c6q). (10.57) (10.58) Here, it is understood that the eigenvalue spectrum of the \u201cposition operator,\u201d \u02c6q, will eventually be evaluated at discrete values n,29 and \u02c6p is canonically conjugate to \u02c6q. In the q representation, \u02c6p = \u2212i\u2202q. We proceed by representing the formal solution of Eq. (10.57) in terms of an imaginary time phase space path integral P (n, t; n0, t0) = n|e\u2212t \u02c6H |n0 = q(t)=n q(0)=n0 D(q, p) e dt (ip \u02d9q\u2212H (p,q)), where H (q, p) is the classical Hamiltonian corresponding to the operator Eq. (10.58). If it were not for the imaginary unit multiplying the p \u02d9q-term, the action of the path integral would be structurally identical to a Hamiltonian action. We may remove that discrepancy by introducing a new momentum variable p \u2192 \u2212ip, whereupon the path integral assumes the form P (n, t; n0, t0) = q(t)=n D(q, p) e dt (p \u02d9q\u2212H(p,q)), q(0)=n0 H(q, p) = \u2212 e+p \u2212 1 f\u2212(q) \u2212 e\u2212p \u2212 1 f+(q). (10.59) Here, H(p, q) \u2261 H (q, \u2212ip) is the Hamiltonian evaluated at the \u201cWick rotated\u201d momentum. At this stage it is instructive to make a few remarks on the structure of the path integral: It is important to remember that the integration over p extends over the imaginary axis. As with the path integral representation of Langevin dynamics Eq. (10.53), the action vanishes on the axis p = 0. Again, the", " line p = 0 accommodates a solution of the variational equations, (q(t), p = 0), where \u02d9q = f+(q) \u2212 f\u2212(q). Comparison with dtn = 29 One may worry that the initial condition of the operator P (q = n, t; n0, 0) = \u03b4(n \u2212 n0) con\ufb02icts with the discrete condition P (n, t|n0, 0) = \u03b4n,n0. Should this ever become an issue, one may switch to the framework of quantum mechanics of operators with discrete eigenvalue spectra. However, the continuous variables used here are better suited to the formulation of path integral methods below. 10.5 Field theory I: zero dimensional theories 651 f+(n) \u2212 f\u2212(n) \u2192 f+(n) \u2212 f\u2212(n) shows that this equation describes the dynamics of the \ufb02uctuationless theory, again in analogy with the Langevin theory. The Fokker\u2013Planck approximation to the master equation is obtained by quadratic expan- sion of the Hamilton operator, \u02c6H i\u02c6p(f+(\u02c6q) \u2212 f\u2212(\u02c6q)) + \u02c6p2 2 (f\u2212(\u02c6q) +f +(\u02c6q)). Substituting into Eq. (10.57), we obtain \u2202t + \u2202q(f+ \u2212 f\u2212)(q) \u2212 1 2 \u22022 q (f\u2212 + f+)(q) P (q, t) = 0. The functional integral Eq. (10.59) is strikingly similar to the MSRJD functional Eq. (10.53). The analogy becomes almost perfect if we replace the Hamiltonian of the former by the quadratic approximation above. Indeed, we may think of the discreteness of the state variables entering the master equation as a source of \u201cnoise.\u201d If we are only interested in structures \u0394q much larger than the \u201clattice spacing\u201d \u0394n = 1, the expansion of the Hamiltonian operator may be restricted to the \ufb01rst two non-vanishing terms. If we keep only the \ufb01rst term \u223c \u2202q, we obtain deterministic drift, \u02d9q = (f+ \u2212 f\u2212)(q) (as can be seen by integrating over p in Eq. (10.59)). The second term describes noise due to residual discreteness e\ufb00ects. This noise a\ufb00ects the time evolution of the probability distribution much like Langevin noise. 10.5.3 Doi\u2013Peliti operator technique Equation (10.59) is not the only \ufb01eld theory representation of the master equation. There exists another approach, the Doi\u2013Peliti operator technique30 which extensively draws upon the Fock space structures and coherent state \ufb01eld integration introduced in previous chapters. Although this technique can be introduced in rather general terms, it is tailored to situations where the weight functions f\u2212 in Eq. (10.56) are low order polynomials in n.31 By way of example, let us consider the case f+ = 0, f\u2212 = \u03bbn. The resulting master equation, \u2202tP (n, t) = \u03bb( \u02c6E1 \u2212 1)nP (n, t), describes extinction of a population whose individuals A \u03bb\u2192 0 perish at a constant rate \u03bb. To set the stage, let us identify an integer valued32 con\ufb01guration {n} with a bosonic Fock space state |n of occupation number, n. Next, let us introduce an algebra of Fock space operators through a|n = |n \u2212 1, \u00afa|n = (n + 1)|n + 1. (10.60) 30 M. Doi, Second quantization representation for classical many-particle system, J. Phys. A 9, 1465-77 (1976); M. Doi, Stochastic theory of di\ufb00usion-controlled reaction, J. Phys. A 9, 1479-95 (1976); L. Peliti, Path integral approach to birth-death processes on a lattice, J. Physique 46, 1469-83 (1985). 31 As with the formalism introduced in the previous section, the Doi\u2013Peliti operator technique a\ufb00ords straightforward generalization to transitions further than n \u00b1 1, as well as to multi-component/multi-dimensional situations. 32 The generalization to other discretization intervals as well as to multi-valued con\ufb01gurations should be obvious. 652 Nonequilibrium (classical) It is straightforward to verify that these operators obey canonical commutation", " relations, [a, \u00afa] = 1, i.e. they are related to the standard operator algebra33 used in second quantization by canonical transformation. The operators a and \u00afa can be used to represent the action of the linear operator on the right\u2013hand side of Eq. (10.56) in terms of the second quantized Hamiltonian (exercise), \u02c6H = \u2212\u03bb(\u00afa \u2212 \u00afaa). With this de\ufb01nition, it is straightforward to verify that P (n, t) \u2261 n|e\u2212t \u02c6H |n0, is a Fock space representation of the distribution function. (10.61) (10.62) We next apply the techniques introduced in Chapter 4 to represent the matrix element above in terms of a coherent state \ufb01eld integral. Proceeding in the usual manner, we decompose the time interval into N 1 steps, e\u2212t \u02c6H = (e\u2212\u03b4 \u02c6H )N, \u03b4 = t/N, note that the Hamiltonian Eq. (10.61) is properly normal ordered, and insert coherent state resolutions d( \u00af\u03c8, \u03c8) e\u2212 \u00af\u03c8\u03c8|\u03c8\u03c8|, where the coherent states |\u03c8 obey the relations of the identity, id. = a|\u03c8 = \u03c8|\u03c8, \u00afa|\u03c8 = \u2202\u03c8|\u03c8, \u03c8|\u00afa = \u03c8| \u00af\u03c8, \u03c8|a = \u2202 \u00af\u03c8 \u03c8|. This leads to the representation P (n, t) = D( \u00af\u03c8, \u03c8)e\u2212\u03b4 N \u22121 n=0 [\u03b4\u22121( \u00af\u03c8n\u2212 \u00af\u03c8n+1)\u03c8n+H( \u00af\u03c8n+1,\u03c8n)]+n ln \u03c8N +n0 ln \u00af\u03c80\u2212ln n!. Here, H( \u00af\u03c8, \u03c8) =\u2212\u03bb \u00af\u03c8(1 \u2212 \u03c8) is the coherent state representation of the (normal ordered) Hamiltonian and the last two terms in the action represent the boundary matrix elements n|\u03c8N = \u03c8n N = en ln \u03c8N, 1 0 = en0 ln \u00af\u03c80\u2212ln n0!. \u00af\u03c8n0 n! \u03c80|n0 = (10.63) Switching to a continuum notation, we arrive at the Doi\u2013Peliti functional integral P (n, t) = D( \u00af\u03c8, \u03c8)e\u2212 t 0 dt[\u2212\u2202t \u00af\u03c8\u03c8+H ( \u00af\u03c8,\u03c8)]+n ln \u03c8(t)+n0 ln \u00af\u03c8(0)\u2212ln n0!. (10.64) The coherent state \ufb01eld integral Eq. (10.64) de\ufb01nes a representation of the probability distribution which is an alternative to the phase space path integral Eq. (10.59). But how do these two representations relate to each other? The answer can be found by comparing the respective Hamiltonians. For f+ = 0 and f\u2212 = \u03bbn, the Hamiltonian Eq. (10.58) reduces to \u02c6H = \u2212\u03bb(e \u02c6p \u2212 1)\u02c6q, where we used the Wick rotated form of the momentum operator, \u02c6p \u2192 \u2212i\u02c6p. Comparison with Eq. (10.61) suggests the identi\ufb01cation \u00afa = e \u02c6p \u02c6q, a = e\u2212 \u02c6p, (10.65) 33 The latter is de\ufb01ned by a|n = n1/2|n \u2212 1, \u2020|n = (n + 1)1/2|n + 1. a 10.5 Field theory I: zero dimensional theories 653 Figure 10.9 Summary of di\ufb00erent \ufb01eld theory approaches to Langevin dynamics and stochastic processes. which, up to a constant, leads to an identi\ufb01cation of the two Hamiltonians. Eq. (10.65) is known as a Cole\u2013Hopf transformation. It is straightforward to verify, that the Cole\u2013 Hopf transformation is canonical, i.e. the relation [\u02c6p, \u02c6q] = 1 is compatible with [a, \u00afa] = 1. Substitution of the (coherent state version of the) transformation into the functional integral dt(\u2212(\u2202t \u00af\u03c8)\u03c8 + H( \u00af\u03c8, \u03c8)) Eq. (10.64) transforms the \u201cbulk contribution\u201d", " to the action, into that of Eq. (10.59). A careful treatment of the boundary term generates the e\ufb00ective boundary condition. q(t)=n q(0)=n0 EXERCISE Explicitly represent the coherent states above as exponentials of the operators a, \u00afa and verify the relations (10.63). Apply the Cole\u2013Hopf transformation to the time-discretized representation of the functional integral to prove the above boundary identity in the large nlimit. Carefully avoid usage of the hermitian adjoint operation \u2013 the operator \u00afa is di\ufb00erent from a\u2020! Summarizing, we have constructed two representations of P (n, t) which are related by a Cole\u2013Hopf transformation. Historically, the coherent state representation Eq. (10.64) was introduced prior to the path integral Eq. (10.59). (This is surprising inasmuch as Eq. (10.59) relies on a less formal construction.) It is probably not possible to decide categorically which of the two representations is the \u201cbetter.\u201d At least for certain types of stochastic processes, the Doi\u2013Peliti functional is controlled by a comparatively simple, if non-hermitian Hamiltonian operator. The price to be paid for this simplicity is that the integration variables \u03c8 and \u00af\u03c8 do not have a direct physical interpretation. The network of di\ufb00erent theories constructed in the previous sections is summarized in Fig. 10.9. So far, we have been focusing on the case of low-dimensional problems, where the state variable is a scalar or vectorial variable. In the next section, we upgrade the formalism to a genuine \ufb01eld theory describing multi-dimensional problems. 654 Nonequilibrium (classical) 10.6 Field theory II: higher dimensions Our aim in this section is to construct an e\ufb00ective \ufb01eld theory approach to higher dimensional systems showing dynamical or steady state nonequilibrium phenomena. To give the discussion some perspective, let us list a few categories of system we have in mind: Interacting lattice particle systems, i.e. systems of particles whose dynamics is governed by an interplay of single particle hopping and interactions. Important examples include exclusion processes, i.e. systems wherein a hardcore repulsion excludes multiple occupancy of lattice sites, and driven di\ufb00usive lattice gases, systems where a uniformly applied force stabilizes a steady state nonequilibrium con\ufb01guration. Reaction\u2013di\ufb00usion systems, i.e. lattice systems populated by particle species A, B,... which may undergo reactions. Setups of this type are common in the modeling of chemical reaction kinetics, the description of evolutionary or ecological dynamics, tra\ufb03c simulation and many more. Thinking, for example, about a \u201cfast\u201d chemical reaction A + B \u2192 C in solution, the factor limiting the total reaction throughput may be the speed at which the chemicals di\ufb00use through the reaction vessel. In this case, a combination of di\ufb00usion (conveniently modeled on a lattice) and local (on-site) reaction will be the minimal framework to describe the process. Important examples of particle reaction processes include birth \u2205 \u2192A, deathA \u2192 \u2205, transmutation A \u2192 B, death at contact, A + B \u2192 A, etc. Notice an important di\ufb00erence between the two categories above: while the particle number in interacting lattice systems is conserved, reaction\u2013di\ufb00usion systems contain non-conserved particle species. 10.6.1 Basic notions of dynamical critical phenomena What would be a good strategy to describe systems such as those listed above? On a fully microscopic level, the dynamics is governed by incredibly high dimensional equations of motion whose solution is out of the question. In practice we will, rather, turn to a description in terms of e\ufb00ective or coarse-grained variables. For example, in a system which may undergo a phase transition the order parameter \ufb01eld, m(x), of the transition would be an obvious choice of an e\ufb00ective variable; in a reaction di\ufb00usion system one might think of the coarse-grained concentration variables nA(x), nB(x),... of the particle species, etc. For de\ufb01niteness we will denote our e\ufb00ective variables by \u03c6a, a = 1,..., n throughout. As in our \u201czero-dimensional\u201d discussion in Section 10.2, the presence of microscopic \u201chidden degrees of freedom\u201d a\ufb00ects the dynamics of the e\ufb00ective variables in two di\ufb00erent ways: (i) a conspiracy of dissipative and macroscopic", " forces drives the e\ufb00ective variables towards local stationary con\ufb01gurations. However, (ii) we have observed that the presence of dissipative relaxation mechanisms is generally accompanied by microscopic stochastic \ufb02uctuations. Finally, (iii) for conserved e\ufb00ective variables, time-dependent changes, \u2202t\u03c6 must be consistent with a continuity relation \u2202t\u03c6 + \u2207 \u00b7 j = 0, where j is some current \ufb01eld. Before turning to the discussion of e\ufb00ective theories consistent with the principles (i)-(iii) above, it is worthwhile to formulate a set of good questions (observables) to ask. We are 10.6 Field theory II: higher dimensions 655 interested in out of equilibrium systems wherein particle correlations generate collective or even critical phenomena. As usual in many particle physics, we aim to identify \ufb01xed points separating phases with distinct universal properties. One may then proceed to an analysis of power law correlations emerging in the vicinity of critical points. In the next section, we set the stage for these types of discussion by introducing a few elements of the theory of dynamical critical phenomena. This material follows closely a seminal review article34 by Hohenberg and Halperin. Consider a system with some second order (continuous) phase transition. We measure the distance to the transition point in terms of some relevant scaling \ufb01eld \u03c4 (e.g. \u03c4 = (T \u2212Tc)/Tc, where T is temperature, and Tc the critical temperature.). We denote the order parameter of the transition by \u03c6. We assume that the order parameter is conjugate to a \ufb01eld h, in the sense that \u03c6a = \u2212\u2202hF [h], where F [h] is a suitably chosen functional (in thermal equilibrium, the free energy of the system). In the vicinity of a critical point, the system builds up long range correlation. For example, the instantaneous or static correlation function C(x) \u2261 \u03c6(x)\u03c6(0) is expected to show power law behavior C(x) = |x|\u2212(d\u22122+\u03b7)Y (|x|/\u03be), (10.66) where \u03be \u223c |\u03c4 |\u2212\u03bd is the correlation length and Y a dimensionless scaling function. The correlation length is a measure for the characteristic length scales of order parameter \ufb02uctuations. Turning to dynamics, we anticipate that the corresponding order parameter time scales will increase as t(\u03c4 ) \u223c \u03be(\u03c4 )z \u223c |\u03c4 |\u2212\u03bdz, (10.67) where the dynamical exponent z depends on how the order parameter relaxes in large correlation volumes. The growth of the relevant time scales is a phenomenon known as critical slowing down. Equation (10.67) can be converted to a statement on the dispersion of order parameter \ufb02uctuations. The characteristic frequencies corresponding to Eq. (10.67), \u03c9(\u03c4 ) \u223c t(\u03c4 )\u22121 \u223c \u03c4 \u03bdz, relate to the momenta |q| \u223c \u03be\u22121 \u223c |\u03c4 |\u03bd \u03c9 \u223c |q|zf (|q|\u03be), where f is a dimensionless function. Turning to the temporal \ufb02uctuation behavior of the order parameter, imagine a situation where \u03c6 has been prepared in some initial state and is subsequently allowed to relax. It is reasonable to describe the relaxation pro\ufb01le by the scaling ansatz \u03c6(t, \u03c4 ) = |\u03c4 |\u03b2g(t/t(\u03c4 )), where \u03b2 is the order parameter exponent and g a dimensionless scaling function with asymptotics g(x \u2192 \u221e) \u2192 const. (relaxation towards the static value (\u2212\u03c4 )\u03b2 below the transition point \u03c4 < 0, and zero above). For short times, t \u2192 0, the relaxation process cannot be 34 P. C. Hohenberg and B. I. Halperin, Theory of dynamic critical phenomena, Rev. Mod. Phys. 49, 435-79 (1977). 656 Nonequilibrium (classical) susceptible to the full extension of the correlation volume, and \u03c6(t, \u03c4 ) must be independent of \u03c4. This leads to the condition g(t/t(\u03c4 )) t\u21920\u223c |\u03c4 |\u2212\u03b2, or \u03c6(t) \u223c t\u2212\u03b1, \u03b1 = \u03b2/\u03bdz. We next de\ufb01ne two important correlation functions characterising the \ufb02uctuation/correlation behavior of the order parameter: The dynamic susceptibility, \u03c7, is de\ufb01ned through the linear response relation \u03c6(q) h\u21920= \u03c7(q)h(q), (10.68) where q = (q, \u03c9). Causality (i.", "e. the fact that \u03c6(t) can depend only on h(t < t)) means that \u03c7(\u03c9) is an analytic function in the upper complex \u03c9-plane. (Why was that?) The spatio-temporal \ufb02uctuation behavior of the order parameter is described by the dynamical correlation function, C(x) \u2261 \u03c6(x)\u03c6(0) \u2212 \u03c6(0)2, (10.69) where x = (x, t). As we review in Section 10.6.4 below, the classical \ufb02uctuation\u2013 dissipation theorem now states that, in thermal equilibrium, C(q, \u03c9) = 2T \u03c9 \u03c7(q, \u03c9). (10.70) Violations of this relation will be an important tool to diagnose nonequilibrium conditions. 10.6.2 Field theories of \ufb01nite dimensional Langevin systems Armed with the de\ufb01nitions above, we are now in a position to formulate an e\ufb00ective \ufb01eld theory approach to dynamical phenomena. We imagine that the energy stored in a particular order parameter con\ufb01guration is described by some energy functional H[\u03c6]. The question now is how to model the dynamics of the order parameter \ufb01eld. Following Hohenberg and Halperin (HH),34 we argue that: In a model for a system with non-conserved order parameter (model A, according to the conventions of HH), the order parameter will relax to the minima of the energy functional, i.e. the equations of motion will contain a contribution \u2202t\u03c6(x, t) = \u2212D \u03b4H[\u03c6] \u03b4\u03c6(x, t) +..., where D is a characteristic relaxation constant. In contrast, in a system with conserved order parameter (HH\u2019s model B), the continuity equation \u2202t\u03c6 + \u2207 \u00b7 j has to be obeyed. We assume that the current acts to evade regions of increased energy, j(x, t) = \u2212D\u2207 \u03b4H[\u03c6] \u03b4\u03c6(x,t) +..., where the ellipses represent \ufb02uctuating 10.6 Field theory II: higher dimensions 657 contributions to the current to be discussed in a moment. (Exercise: consider the meaning of this de\ufb01nition.)35 The continuity relation then assumes the form \u2202t\u03c6(x, t) = D\u0394 \u03b4H[\u03c6] \u03b4\u03c6(x, t) +.... The two evolution equations above can be subsumed as \u2202t\u03c6(x, t) = D\u0394n \u03b4H[\u03c6] \u03b4\u03c6(x,t) +..., where n = 0, 1 for models A and B, respectively. According to principle (ii) above, the relaxational character of the dynamics implies the presence of \ufb02uctuations. The full evolution equation, a generalized Langevin equation, thus reads as \u2202t\u03c6(x, t) = \u2212D(\u2212\u0394)n \u03b4H[\u03c6] \u03b4\u03c6(x, t) + \u03be(x, t), (10.71) where the noise is correlated as \u03be(x, t)\u03be(x, t) = 2\u03b4(t \u2212 t)K(x \u2212 x) \u2261 2K(x \u2212 x). (10.72) This correlator presumes that, on suitably chosen time scales, the noise can be modeled as Markovian. However, at this stage, not much can be said about the spatial correlation kernel K. We \ufb01nally note that correlation functions can be generated by addition of a \u201cconjugate \ufb01eld\u201d contribution to the Hamiltonian, H[\u03c6] \u2192 H[\u03c6] \u2212 ddx \u03c6(x, t)h(x, t). Functional averages O[\u03c6]\u03be may now be computed by straightforward generalization of the (0 + 1)-dimensional functional (10.50): O[\u03c6]\u03be = D\u03c6 O[\u03c6] B'\u03b4 x \u2202t\u03c6(x) +D (\u2212\u0394)n \u03b4H[\u03c6] \u03b4\u03c6(x) \u2212 \u03be(x) C. \u03be In practice, F [\u03c6](x, t) = F (\u03c6(x, t), \u2202x\u03c6(x, t),... ) will be a local function of the \ufb01eld and its spatial derivatives. Representing the \u03b4-constraint in terms of a Fourier integral", " over a \u201cmomentum \ufb01eld\u201d \u03c8 and integrating over the noise \ufb01eld we arrive at O[\u03c6]\u03be = D(\u03c6, \u03c8) O[\u03c6] ei dx \u03c8(x)(\u2202t\u03c6(x)+D(\u2212\u0394)n \u03b4H[\u03c6] \u03b4\u03c6(x) )\u2212 dxdx \u03c8(x)K(x\u2212x)\u03c8(x), 35 Typically, the dominant contribution (in the sense of dimensional relevancy) to the Hamiltonian will be a ddx \u03c62(x, t) +... In this case, the current reduces to quadratic form in the order parameter H[\u03c6] = r 2 j = \u2212Dr\u2207\u03c6 +..., which is a generalization of Fick\u2019s law. 658 Nonequilibrium (classical) dx \u2261 where generalization of the MSRJD\u2013functional ddxdt. A \u201cWick rotation\u201d \u03c8 \u2192 \u2212i\u03c8 \ufb01nally brings us to the \ufb01nite dimensional O[\u03c6]\u03be = D(\u03c6, \u03c8) O[\u03c6] e dx \u03c8(x)(\u2202t\u03c6(x)+D(\u2212\u0394)n \u03b4H[\u03c6] \u03b4\u03c6(x) )+ dxdx \u03c8(x)K(x\u2212x)\u03c8(x). (10.73) As in the zero\u2013dimensional cases above, we may integrate over the momentum \ufb01eld \u03c8 to obtain the generalized Onsager\u2013Machlup functional O[\u03c6]\u03be = D\u03c6 O[\u03c6] e \u2212 1 4 dx (\u2202t\u03c6(x)+D(\u2212\u0394)n \u03b4H[\u03c6] \u03b4\u03c6(x) )K\u22121(x\u2212x) \u2202t\u03c6(x)+D(\u2212\u0394)n \u03b4H[\u03c6] \u03b4\u03c6(x ), (10.74) where K \u22121 is the inverse of the noise kernel. INFO It is an instructive exercise to derive from Eq. (10.74) a functional variant of the Fokker\u2013Planck equation \u2013 \u201cfunctional\u201d because the distribution function we are after, Pt[\u03c6], is a functional of \u03c6, and the partial derivatives \u2202ni P (n, t) appearing in the standard Fokker\u2013Planck operator (cf. Eq. (10.22)) become functional derivatives. (To avoid the simultaneous appearance of rectangular and conventional brackets, we denote the time argument in pt[\u03c6] by an index.) Our starting point is the Onsager\u2013Machlup functional for the probability functional, 0 - x pt[\u03c6] \u2261 = \u03b4(\u03c6(x) \u2212 \u03c6(x, t)) D\u03c6 e\u2212 1 4 dxdx (\u2202t\u03c6(x)\u2212F [\u03c6](x))K \u22121(x\u2212x )(\u2202t\u03c6(x )\u2212F [\u03c6](x )), \u03c6(x)=\u03c6(x,t) where we have introduced the abbreviation F [\u03c6](x) \u2261 \u2212D(\u2212\u0394)n \u03b4H[\u03c6] argument of pt[\u03c6] while \u03c6(x, t) is the (d + 1)-dimensional integration variable. \u03b4\u03c6(x), and\u03c6 = {\u03c6(x)} is the As a \ufb01rst step towards the derivation of a Fokker\u2013Planck equation, we switch back to a time discretized representation, pt \u2192 pl, wherel = t/\u03b4 is an integer counting variable and \u03b4 a small time increment. The Ito-time discretized Onsager\u2013Machlup functional then reads as pl[\u03c6] = D\u03c6 e\u2212 \u03b4 4 l\u22121 k=0 O[\u03c6k+1,\u03c6k], (10.75) where O[\u03c6, \u03c6] is shorthand for O[\u03c6, \u03c6] \u2261 ddxddx (\u03c6 \u2212 \u03c6) \u2212 F [\u03c6] (x)K \u22121(x \u2212 x) 1 \u03b4 (\u03c6 \u2212 \u03c6) \u2212 F [\u03c6] (x). 1 \u03b4 \u03c6(x)=\u03c6l(x) We next have to realize that the integral over the \ufb01rst l \u22121 \ufb01elds\u03c6 k above de\ufb01nes the probability distribution pl\u22121: pl[\u03c6] = D", "\u03c6l\u22121 e\u2212 \u03b4 4 O[\u03c6,\u03c6l\u22121] pl\u22121[\u03c6l\u22121] = D\u03bc e\u2212 \u03b4 4 O[\u03c6,\u03c6\u2212\u03bc] pl\u22121[\u03c6 \u2212 \u03bc], where in the second equality we have renamed integration variables. Introducing the shorthand ddxddx f (x)K(x \u2212 x)g(x), we next represent the operator O[\u03c6, \u03c6 \u2212 \u03bc] as notation f |K|g \u2261 O[\u03c6, \u03c6 \u2212 \u03bc] = 4 \u03bc \u03b4 \u2212 F [\u03c6 \u2212 \u03bc] K \u03bc \u03b4 5 \u2212 F [\u03c6 \u2212 \u03bc]. 1 10.6 Field theory II: higher dimensions 659 4\u03b4 4 D\u03bc e\u2212 1 pl[\u03c6] = \u03bc|K|\u03bc (... ), an expansion to leading order in \u03b4 \u2013 the Using the abbreviation... \u2261 structure of the Gaussian weight entails that each power of \u03bc scales as \u03b41/2 \u2013 then obtains \u03bc|K| \u03b4F [\u03c6] \u03b4\u03c6 3 2 (\u03bc|K|F [\u03c6])2 \u2212 \u03b4 4 \u03bc \u03bc|K|F [\u03c6] \u2212 1 2 \u03bc \u03b42pl\u22121[\u03c6] \u03b4\u03c6\u03b4\u03c6 3 pl\u22121[\u03c6] \u2212 2 F [\u03c6]|K|F [\u03c6] F [\u03c6]|K|F [\u03c6]pl\u22121[\u03c6] = pl\u22121[\u03c6] \u2212 \u03b4 F [\u03c6 \u03b4pl\u22121[\u03c6] \u03b4\u03c6 \u03b4pl\u22121[\u03c6] \u03b4\u03c6 \u2212 \u03b4 \u03b4F [\u03c6] \u03b4\u03c6 \u03b42pl\u22121[\u03c6] \u03b4\u03c6\u03b4\u03c6 K. pl\u22121[\u03c6] \u2212 \u03b4 4 + \u03b4 F [\u03c6]|K|F [\u03c6]pl\u22121[\u03c6] + \u03b4tr 4 ; < Here, \u03b4p[\u03c6] \u03b4\u03c6\u03b4\u03c6 continuous time notation we obtain \u03b4p[\u03c6] \u03b4\u03c6(x)\u03b4\u03c6(x) \u2261. Dividing by \u03b4, taking the limit \u03b4 \u2192 0, and switching back to a \u2202tpt[\u03c6] + 2 F [\u03c6] 3 2 + 3 1 \u03b4F [\u03c6] \u03b4\u03c6 pt[\u03c6] \u2212 tr K \u03b42pt[\u03c6] \u03b4\u03c6\u03b4\u03c6 = 0, \u03b4pt[\u03c6] \u03b4\u03c6 or, in a coordinate resolved notation, \u2202tpt[\u03c6] \u2212 D ddx \u03b4 \u03b4\u03c6(x) (\u2212\u0394)n \u03b4H[\u03c6] \u03b4\u03c6(x) pt[\u03c6] \u2212 ddxddx K(x \u2212 x) \u03b42pt[\u03c6] \u03b4\u03c6(x)\u03b4\u03c6(x) = 0. (10.76) Equation (10.76) is a functional generalization of the Fokker\u2013Planck equation. Roughly speaking, (\u2212\u0394)n\u03b4H/\u03b4\u03c6 replaces the \u201cfriction coe\ufb03cient\u201d of the \ufb01nite dimensional equation, i.e. the instance that drives the distribution towards a relaxation equilibrium. The \u201cdi\ufb00usion term\u201d is controlled by the correlation matrix of the noise. So far, we have not speci\ufb01ed the noise kernel, K. However, we expect that in equilibrium it relates in a de\ufb01nite way to the drift term \u2013 the \ufb02uctuation dissipation theorem. By \u201cequilibrium\u201d we mean a time-independent distribution of generalized Maxwell\u2013Boltzmann type, peq[\u03c6] \u2261 N e\u2212 1 T H[\u03c6], (10.77) where N e\ufb00ects normalization. We now show that the condition that peq is a solution imposes a condition on the form of K. Substitution of Eq. (10.77) into (10.76) indeed obtains ddx \u03b4 \u03b4\u03c6(x) (\u2212\u0394)n \u03b4H[\u03c6] \u03b4\u03c6(x) ddxddxK(x \u2212 x) \u2212 D + 1 T + D T ddx(\u2212\u0394)n \u03b4H[\u03c6] \u03b4\u03c6(x) \u03b4H[\u03c6] \u03b4\u03c6(x) \u03b4 \u03b4\u03c6(x) \u03b4H", "[\u03c6] \u03b4\u03c6(x) \u2212 1 T \u03b4H[\u03c6] \u03b4\u03c6(x) \u03b4H[\u03c6] \u03b4\u03c6(x) = 0. This equation is solved provided that we declare that K(x \u2212 x) = DT (\u2212\u0394)n\u03b4(x \u2212 x). ddx((\u2202\u03c6)2 + Notice that for a Hamiltonian functional that is local in \u03c6 (think of H[\u03c6] = 1 2 r\u03c62 +... )) the functional derivatives appearing in the Fokker\u2013Planck equation are singular (for \u03b4\u03c6(x)\u03b4\u03c6(x) H[\u03c6] = \u2212(\u22022+r)\u03b4(0), etc.) and need regularization. However, for the purposes example, of our present discussion, a symbolic understanding of the derivatives is su\ufb03cient. \u03b42 To conclude, let us summarize several \u201ctake home messages\u201d implied by the construction above: (10.78) 660 Nonequilibrium (classical) The concept of a Fokker\u2013Planck equation a\ufb00ords a generalization to \ufb01nite dimensional prob- lems. However, such an equation involves an in\ufb01nite dimensional partial integro-di\ufb00erential equation which does not look very inviting. We do not have to solve the equation explicitly to verify that the underlying generalized Langevin equation relaxes to a Maxwell\u2013Boltzmann (or Gibbs) equilibrium con\ufb01guration, provided that the noise correlator is chosen appropriately. Finally, it is worth noting that the strategy pursued above, i.e. the construction of an integral recursion equation in discrete time space (Eq. (10.75)), followed by a Taylor expansion in the incremental variables, re\ufb02ects a general scheme for the derivation of transfer matrix type equations. 10.6.3 Field theory of \ufb01nite dimensional stochastic processes Problems in reaction kinetics or population dynamics are often encoded in terms of master equations de\ufb01ned on a \ufb01nite dimensional lattice, \u2202tpt[n] = n (W [n, n]pt[n] \u2212 W [n, n]pt[n]). (10.79) The notation Pt[n] indicates that the distribution depends on the \u201cfunction\u201d {ni} where i is a lattice point and ni a container variable comprising all relevant local information (the local concentrations Ai, Bi,... of chemicals, etc.). The transition rates, W [n, n] depend, likewise, on lattice functions. In many applications, transition rates W describe a competition of di\ufb00usive spreading (a random walk on a lattice) and particle reactions. Systems of this kind are termed reaction\u2013di\ufb00usion systems. Field theories for reaction\u2013di\ufb00usion systems can be obtained by straightforward generalization of the (0 + 1)-dimensional techniques introduced in Sections 10.5.1 and Eq. (10.61). By way of example, consider a pure extinction process, A \u03bb\u2192 \u2205 wherein a certain agent A perishes at a constant rate. De\ufb01ning ni \u2261 (# of As at lattice site i), we may then ask how the probability pt[n] evolves under the joint in\ufb02uence of annihilation and di\ufb00usive spreading. A \ufb01eld theory representation of this process can be obtained by generalization of the Doi\u2013Peliti operator algebra, ai|..., ni,... = |..., ni \u2212 1,..., \u00afai|..., ni,... = (ni + 1)|..., ni + 1,..., (10.80) where we represent the population {ni} in terms of a Fock space state. Again, the operator algebra obeys canonical commutation relations, [ai, \u00afaj] =\u03b4 ij. A straightforward extension of our discussion of Section 10.5.3 shows that the Hamiltonian generating the stochastic process above is given by \u02c6H = \u2212D ij \u00afai\u0394ijaj \u2212 \u03bb i \u00afai(1 \u2212 ai), (10.81) 10.6 Field theory II: higher dimensions 661 where \u0394ij is the lattice Laplacian and D a di\ufb00usion constant. The corresponding coherent state \ufb01eld integral action then reads S[ \u00af\u03c8, \u03c8] = ddxdt % \u00af\u03c8(\u2202t \u2212 D\u0394)\u03c8 \u2212 \u03bb \u00af\u03c8(1 \u2212 \u03c8) &, (10.82) where we have", " switched to a continuum notation. EXERCISE We may also derive a \ufb01eld theory by generalization of the techniques introduced in Section 10.5.2. Derive the corresponding action and show that it relates to Eq. (10.82) by a Cole\u2013Hopf transformation. 10.6.4 Fluctuation\u2013dissipation theorem (revisited) The \ufb01eld theories above represent our most general descriptions of non-equilibrium many particle systems. A question that frequently arises in applications is whether or not a given system is in thermal equilbrium. A straightforward answer to this question may be given as follows: compute the long time stationary probability distribution and check whether it is of Maxwell\u2013Boltzmann form. This procedure is generally valid but can be unnecessarily cumbersome in practice. (You have to compute the stationary distribution \ufb01rst, and this can be hard.) Often, it is more e\ufb03cient to check whether the \ufb02uctuation\u2013dissipation theorem \u2013 a hallmark of equilibrium systems \u2013 is ful\ufb01lled. Now, we have encountered the FDT in many di\ufb00erent forms and contexts. Given its practical importance, it is worthwhile to brie\ufb02y summarize some of the more important incarnations. The last of these, a formulation of the FDT for a linear variant of MSRJD \ufb01eld theory, plays an important role in the applications below. FDT I: Equilibrium linear response In Chapter 7 we discussed the apparatus of linear response for equilibrium systems. We considered the expectation value of certain observables \u02c6X(x)36 (with x = (x, t) as usual). Assuming that \u02c6X is conjugate to a generalized \u201cforce,\u201d F, and thatX (x) \u2261 \u02c6X(x) = 0 in an unbiased situation, we considered the linear relation X(\u03c9) = \u03c7(\u03c9)F (\u03c9) +O (F 2), (10.83) where \u03c7(\u03c9) de\ufb01nes a generalized susceptibility. Equation (10.83) establishes a connection between a macroscopic force (F ) and a macroscopic observable (X) in a \u201cbath\u201d containing a large number of microscopic degrees of freedom. In this sense, \u03c7 is a quantity probing \u201cdissipative\u201d physical processes. In Section 7.3 we showed that \u03c7(\u03c9) = C+(\u03c9) = \u2212i\u0398(t)[ \u02c6X(t), \u02c6X(0)], where C + was the retarded response function. We then established a connection between C + and another correlation function, the time ordered correlation function, C T (t) \u2261 \u2212iTt \u02c6X(t) \u02c6X(0), 36 Chapter 7 was formulated for quantum theories, but that aspect will not be of relevance throughout. 662 Nonequilibrium (classical) which describes \u201c\ufb02uctuations\u201d of the observable X. Speci\ufb01cally, we found that (cf. Eq. (7.21)) Im C T (\u03c9) = Im C +(\u03c9) coth(\u03b2\u03c9/2). (10.84) This is the quantum \ufb02uctuation dissipation theorem. We next wish to explore what happens to this relation in the classical limit. In the classical limit, \u02c6X \u2192 X(x) becomes a function in phase space (formally, the Wigner transform of the operator \u02c6X. For more information on the Wigner transform, see Section 11.5.1), the time dependence, \u02c6X(t) \u2192 X(x(t)) resides in the classical time evolution of the argument, and the time ordering of operators involved in the de\ufb01nition of C T becomes inessential. This means that the classical limit of the time ordered correlation function is given by where C T cl (t) = X(t)X(0), X(t)X(0) \u2261 ddx \u03c1(x)X((x(t))X(x). \u0393 It is not di\ufb03cult to verify that the Fourier transform Ccl(\u03c9) = Im Ccl(\u03c9) is purely imaginary. Likewise, it can be shown that the retarded response function approaches the form C + = C + cl \u2013 note the extra \u201c\u201d \u2013 with C + cl (t) = \u03c7cl(t) = \u0398(t){X(t), X(0)}, where the Poisson bracket replaces (\u2212(i/) times) the quantum commutator. (However, this expression is not too useful in applications. In practice, one will often try to obtain the linear response susceptibility by di\ufffd", "\ufffd\ufffderent means.) EXERCISE The expressions above, Poisson brackets replacing commutators, etc., contain the \u201cnatural\u201d classical counterparts of the quantum formulae. To actually verify their validity, employ the Wigner representation of the operator \u02c6X (cf. Section 11.5.1) and then take the classical limit by leading order expansion in. Finally, the Fourier argument \u03c9 in Eq. (10.84) has the dimension of a frequency. This means, T /\u03c9\u2212\u2192 2T that the bosonic distribution function must be understood as coth(\u03b2\u03c9/2) = \u03c9. Combining everything, we obtain the classical \ufb02uctuation dissipation theorem C(\u03c9) = 2T \u03c9 Im \u03c7(\u03c9), (10.85) where we removed the subscript \u201ccl.\u201d In the theory of steady state phenomena, it is often useful to consider a static variant of the FDT. To this end, let us de\ufb01ne the equal time correlation function as C \u2261 d\u03c9 2\u03c0 C(\u03c9) = X(0)X(0). 10.6 Field theory II: higher dimensions 663 The analytic properties of C(\u03c9) and\u03c7( \u03c9) (explore this point) then imply the static version of the FDT, C = T \u03c7(0). (10.86) FDT II: linear Langevin equations In Section 10.2.1 we discussed the FDT within the framework of linear Langevin theory. (Here \u201clinear\u201d means Langevin equations that are linear in the dynamical variable.) We saw that consistency with the equipartition law of equilibrium thermodynamics implies the presence of \ufb02uctuations in any dissipative medium. But how does this observation correlate with the linear response picture above? To understand this connection, let us look at the Langevin oscillator equation \u00a8q + \u03c92 0q + \u03b3 \u02d9q = \u03be, where \u03c90 is the oscillator frequency and mass has been set to unity. The FDT as discussed in Section 10.2.1 requires that \u03be(t)\u03be(t) = 2\u03b3T \u03b4(t \u2212 t). In Fourier space, this may be solved as q(\u03c9) = \u03c7(\u03c9)\u03be(\u03c9), \u03c7(\u03c9) = \u2212 1 \u03c92 \u2212 \u03c92 0 + i\u03b3\u03c9. (10.87) The notation indicates, that the kernel \u03c7 can be interpreted as the dynamical susceptibility describing the response of the coordinate q to the \u201cdriving force\u201d \u03be. The point to appreciate here is that the dynamical susceptibility of a system describes both its response to \ufb02uctuations (as in Langevin theory) and the response to external driving (as in linear response theory). The Fourier transform of the dynamical correlation function C(t) = q(t)q(0) reads as C(\u03c9) = 2\u03b3T 0)2 + (\u03c9\u03b3)2 = where we have used the relation \u03be(\u03c9)\u03be(\u03c9) = 2\u03b3T \u03b4(\u03c9 + \u03c9). Comparing Eq. (10.88) and Eq. (10.87) we see that the general \ufb02uctuation dissipation theorem Eq. (10.85) and the two approaches \u201cFDT from equipartition\u201d and \u201cFDT from linear response\u201d describe di\ufb00erent sides of the same coin. (\u03c92 \u2212 \u03c92 Im \u03c7(\u03c9), (10.88) 2T \u03c9 FDT III: MSRJD \ufb01eld theory In later sections, we explore the collective behavior of many particle systems by \ufb01eld theoretical methods. The FDT will turn out to be an important diagnostic tool to test whether a system is in equilibrium or not. Let us, then, explore how the FDT can be probed within the framework of \ufb01eld theory. To this end we consider the MSRJD functional Eq. (10.73) with noise correlator (10.78). While the FDT Eq. (10.85) is of general validity, we begin by considering linear (free) \ufb01eld theories. Assuming that the relaxation of \u03c6 is controlled by the quadratic Hamiltonian H[\u03c6] = ddx 1 2 (\u2202\u03c6)2 + r 2 \u03c62 \u2212 h\u03c6, 664 Nonequilibrium (classical) where the \ufb01eld h will be used to probe the linear response of the system, the corresponding MSRJD functional takes the form Z \u2261 D(\u03c6, \u03c8) e dx [\u03c8(\u2202t\u03c6+D(\u2212\u0394", ")n(\u2212\u0394\u03c6+r\u03c6\u2212h))+DT \u03c8(\u2212\u0394)n\u03c8]. Switching to momentum space and introducing a matrix notation, Z \u2261 D(\u03c6, \u03c8) exp \u2212S0[\u03c6, \u03c8] \u2212 D S0[\u03c6, \u03c8] = \u2212 1 2 dq (\u03c6, \u03c8)\u2212q dq \u03c8qq2nh\u2212q, q \u03c6 \u03c8, q (g\u2212)\u22121 2DT q2n 0 (g+)\u22121 \u22121 \u2261 dq \u2261 ddkd\u03c9 (2\u03c0)d+1, g\u00b1, and we are using an index notation where gq, instead of g(q) for improved notational clarity.37 Now, let us employ this functional to compute the dynamic susceptibility. Due to the evenness of S0 in the \ufb01elds, \u03c60 = 0, where... 0 \u2261 D(\u03c6, \u03c8) exp (\u2212S0[\u03c6, \u03c8]). To leading order in the driving \ufb01eld, we thus obtain \u2213i\u03c9 + Dk2n(k2 + r) q \u03c6q D dq \u03c6q q2n \u03c8\u2212q hq = Dg+ q q2nhq \u2261 \u03c7qhq. This leads to the identi\ufb01cation \u03c7(q) = Dq2ng+ tion function is given by q. On the other hand, the \ufb02uctuation correla- C(q) = dq\u03c6q\u03c6q 0 = 2DT q2ng+ q g\u2212 q = 2T \u03c9 Im\u03c7q, consistent with the FDT. This consideration shows that the free equilibrium \ufb01eld theory conforms with the FDT. However, it is much more challenging to show that this compatibility pertains to the interacting case. In practical applications, one will want to verify (for example by perturbative methods, or by renormalized perturbation theory) that (a) the equilibrium theory does obey the FDT. One may then proceed to (b) detect deviations from equilibrium by violations of the FDT. INFO In the theory of continuum systems, the correlation function C(q) \u2261 C(q, \u03c9) \u2261 S(q, \u03c9) is often referred to as the dynamic structure factor. This notation alludes to the fact that the function C(q, \u03c9) describes spatio-temporal \ufb02uctuations in the system at the characteristic length scale |q|\u22121 and time scale \u03c9\u22121. The FDT establishes a connection between the dynamic structure factor and the dynamic susceptibility. Likewise, the static version of the FDT Eq. (10.86) relates the static structure factor C(q) \u2261 S(q) to the static susceptibility. Finally, note that we may be confronted with a system that is kept at some predetermined noise level (possibly determined by the action of an underlying microscopic agent, or due to coupling to an external environment). One may then argue that an e\ufb00ective equilibrium emerges wherein the noise strength determines the e\ufb00ective temperature by way of the FDT condition. 37 If you \ufb01nd the sign-di\ufb00erence between the retarded and the advanced Green function unfamiliar, keep in mind that we are considering irreversible dynamics. Compute the time representation of g\u00b1 for the case n = 0, r = 0, to show that the sign properly distinguishes between retarded and advanced propagation. 10.7 Field theory III: applications 665 10.7 Field theory III: applications Having invested considerable e\ufb00ort into deriving various \ufb01eld theories of nonequilibrium systems, it is now time to apply the formalism. However, before doing so, let us identify a number of questions we may want to address: In Section 10.5.2 we assumed knowledge of a certain \u201cHamiltonian\u201d H[\u03c6] that drives the order parameter \ufb01eld \u03c6 to an equilibrium con\ufb01guration. But how can such a Hamiltonian be motivated and constructed for concrete physical systems? What can \ufb01eld theory say about the di\ufb00erent phases realized in and out of equilibrium? And what can be said about the transitions between these phases? The theories constructed above bear structural similarity to di\ufb00erent equilibrium theories discussed earlier in the text. To what extent can concepts developed earlier for equilibrium systems (the notion of upper and lower critical dimension, mean \ufb01eld theory, renormalized theory of \ufb02uctuations, etc.) be generalized to the present context? How do critical correlations re\ufb02ect in the dynamical behavior of a system? These are but a few questions that can be asked", ". Rather than attempting to \ufb01nd general answers, we shall address some of these questions on two speci\ufb01c examples. Even so, our discussion will be rather super\ufb01cial and can provide little more than a glimpse of the theory of nonequilibrium critical phenomena. (Some guides to further reading are given below.) 10.7.1 Driven di\ufb00usive lattice gases Imagine yourself confronted with the following exercise: engineer a genuine many particle system (a system containing particle dynamics and interaction) that is (a) simple, (b) couples to an e\ufb03cient mechanism driving an out-of-equilibrium situation, and (c) reduces to one of the \u201clarge\u201d universality classes of physics in the equilibrium limit. It is actually not so easy to come up with a good answer. One system that meets these criteria reasonably well is the driven di\ufb00usive lattice gas (see Fig. 10.10). As is indicated by the name, it involves a system of particles hopping on a lattice. The particles may hop from one site to the next at some rate t, and are subject to nearest neighbour repulsion as described by the Hamiltonian H[n] \u2261 J ninj. ij (10.89) Here ni,j \u2208 {0, 1} re\ufb02ecting that each site can be occupied by one particle at most. For simplicity we imagine periodic boundary conditions in all directions, i.e. the system lives on a d-dimensional torus. We next allow the particles to hop between nearest neighbour lattice sites (subject to the condition of no more than single occupancy). This will render the system dynamical and allow it to settle in a con\ufb01guration compromising between single particle dynamics and interactions. To see this more clearly, imagine the dynamics described in terms of a master equation. Interactions will in\ufb02uence transition rates in this equation, but not change the con\ufb01gurations by themselves. Above we have seen that the changes in the con\ufb01gurations 666 Nonequilibrium (classical) Figure 10.10 The driven di\ufb00usive lattice gas. A system of nearest neighbour interacting particles hopping on a lattice and subject to periodic boundary conditions. The system is driven by a constant force. (presently caused by hopping) a\ufb00ord an interpretation in terms of Langevin dynamics.38 This means, that we may think of the capacity to hop in terms of some e\ufb00ective thermal activation, or noise acting on the system. This noise will enable the gas to settle in an e\ufb00ective equilibrium con\ufb01guration favoured by the interaction, at an e\ufb00ective temperature set by the hopping rates. To drive the system out of equilibrium, we introduce a force acting uniformly along one of the directions of the torus. Moving against its action once around the torus (a closed loop) costs energy, i.e. the force is not conservative, and there is no corresponding potential. Rather, we may think about the force as an \u201celectromotive\u201d force due to piercing the torus by a time dependent magnetic \ufb01eld. The coupling to this \u201cwork reservoir\u201d gives rise to a nonequilibrium situation. Lack of a conservative potential invalidates descriptions of the system in terms of a Hamiltonian; this is the ultimate motivation to describe it in terms of master equations and e\ufb00ective Langevin equations. In this language, the e\ufb00ect of the force will be a bias in the hopping rates along one of the toroidal directions. An important preliminary conclusion to be drawn from this consideration is that the equilibrium physics of our system will fall into the Ising universality class. Indeed, we may introduce spin variables Si = 2ni \u2212 1 \u2208 {\u22121, 1}. Expressed in terms of the Sis the interaction Hamiltonian H[S] assumes the form of an Ising Hamiltonian, where the sign of J determines whether the coupling is ferromagnetic or antiferromagnetic. In equilibrium, 38 A quadratic approximation of the transition generators, the momenta conjugate to the occupation number variables, led to a quadratic contribution to the Hamiltonian action formally equivalent to the averaged noise term in the MSRJD functional. 10.7 Field theory III: applications 667 the system will be described by a Gibbs distribution, \u223c exp(\u2212H[S]/T ), at a temperature set by the hopping rates. Introduction of a \ufb01nite bias is expected to cause departures from the Ising/\u03c64-universality class. Summarizing, the driven di\ufb00usive lattice gas conforms with the three principles above: it is relatively simple, contains nonequilibrium driving, and relates to the big Ising universality class. Sadly, there", " do not seem to be any direct experimental realizations of this system, although it may be argued that aspects of its behavior re\ufb02ect in the physics of ionic conductors, microemulsions of charged particles, and DNA gel electrophoresis.39 Our theoretical approach to the system below is organized in a number of steps. After reviewing elements of its microscopic physics, we introduce a mesoscopic theory describing the approach to equilibrium in the unbiased case, as much as the principal e\ufb00ects of the driving mechanism. We expect the consequences of driving to be particularly strong if the system is tuned to its equilibrium transition point. Less evidently, however, it turns out that driving also causes remarkable phenomenological consequences above criticality (the \u201cdisordered\u201d phase). In our analysis below, we \ufb01rst address the latter regime, and then turn towards the more ambitious study of criticality. It should be emphasized that, to date, many aspects of driven di\ufb00usive systems remain poorly understood, and there is lack of direct experimental data. In our discussion of the system, we thus have to draw more liberally than usual from a mix of \ufb01eld theoretical evidence, simulation data, and common sense reasoning. The material assembled below is a fragmentary excerpt from the text by Schmittmann and Zia.39 Microscopic formulation Consider the master equation Eq. (10.79) for the evolution of the lattice gas population. In the absence of an external \ufb01eld, we have a Hamiltonian H[n] assigning to each con\ufb01guration n an energy. In equilibrium, the system will settle into a con\ufb01guration described by the distribution, peq[n] E=0= N exp(\u2212H[n]/T ). We do not have to specify the transition rates W [n, n] explicitly to say that compatibility with Eq. (10.90) implies the detailed balance relation (10.90) W [n, n] W [n, n] = peq[n] peq[n] = e\u2212\u03b2(H[n]\u2212H[n]). (10.91) This relation is satis\ufb01ed, e.g., by the choice W [n, n] = const. \u00d7 exp(\u2212\u03b2(H[n] \u2212 H[n])/2). More generally, any W [n, n] \u2261 f (\u03b2(H[n] \u2212 H[n])) with f (x)/f (\u2212x) = exp(\u2212x) will also do the job. In the presence of a driving \ufb01eld, we do not have a globally de\ufb01ned Hamiltonian any more. Locally, however, we may assign to a con\ufb01guration n an \u201cenergy\u201d that increases if we move against the \ufb01eld. Modeling the dynamics in terms of transitions n \u2192 n that involve only local changes in the con\ufb01guration, we may generalize according to W [n, n] \u2192 f (\u03b2(H[n] \u2212 H[n] +kaE )), 39 B. Schmittmann and R. K. P. Zia, Statistical Mechanics of Driven Di\ufb00usive Systems, (Academic Press,1995). 668 Nonequilibrium (classical) Figure 10.11 Three phases of the driven di\ufb00usive lattice gas. Left: disordered (unmagnetized) phase at high temperatures. Center: ordered (magnetized) islands at low temperatures and zero \ufb01eld. Right: ordered (magnetized) stripes at low temperatures in the presence of a driving \ufb01eld. where a is the lattice spacing and k the number of steps particles have to climb against the direction of the \ufb01eld E to get from con\ufb01guration n to n. Suppose we have modelled the system in terms of transition rules satisfying the above constraints. The corresponding master equation is the starting point for computer simulations. Given the lack of experimental data, and the relative unfamiliarity of the system, such simulations40 have played an important role in guiding analytical work: not surprisingly, it has been con\ufb01rmed that the \ufb01eld-free system falls into the Ising universality class. However, i ni \u2212 N contrary to the standard Ising model, the total magnetization M \u2261 (N is the total number of lattice sites) is conserved in our system. This means that, above a critical temperature Tc, the system will be in a \u201cdisordered\u201d phase, i.e. particles spread randomly across the lattice (see Fig. 10.11 left). Below Tc, we enter an ordered phase characterized by the presence of magnetized domains (particle", " clusters). The total area of spin up and spin down domains is determined by the concentration of particles in the system (Fig. 10.11 middle). A cursory inspection of numerical data may lead to the impression that above Tc, the application of a \ufb01eld has not much of an e\ufb00ect. We will see, however, that this is not the case. Finally, it has been found numerically that below Tc and for E = 0, the magnetized areas arrange in stripes parallel to the \ufb01eld direction (see Fig. 10.11 right). i Si \u2261 2 40 S. Katz, J. L. Lebowitz, and H. Spohn, Nonequilibrium steady states of stochastic lattice gas models of fast ionic conductors, J. Stat. Phys. 34, 497-537 (1984). 10.7 Field theory III: applications 669 Below we try to explain aspects of these observations analytically. In principle, master equations based on the transition rates W [n, n] may serve as a starting point towards the construction of an analytical theory, along the lines of our discussion in Section 10.6.3. However, the reduction of the resulting \u201cmicroscopic \ufb01eld theory\u201d to an e\ufb00ective long range theory will arguably be laborious. Instead, we will build our construction on a \u201cmesoscopic formulation\u201d wherein the dynamics of the site occupancy is described in coarse-grained terms. Mesoscopic formulation To provide coarse-grained description of the system, we describe its state in terms of its continuous magnetization \u03c6(x, t) \u2261 2n(x, t) \u2212 1, where... is symbolic for an average over many lattice sites. Let us consider the \ufb01eld free case \ufb01rst. We then know from our discussion in section 5.1, that the e\ufb00ective Hamiltonian governing the static pro\ufb01le of the magnetization variable in the absence of driving is the \u03c64-functional, H[\u03c6] = ddx (\u2202\u03c6)2 + r\u03c62 + \u03c64. (10.92) u 12 1 2 This Hamiltonian will determine the form of the Langevin equation. Second, we know that ddx \u03c6 = const, by virtue of particle number magnetization is a \u201cconserved quantity,\u201d conservation. Thus, our model belongs to class B in the classi\ufb01cation of Section 10.6.2, and we may write down an e\ufb00ective Langevin equation of motion, \u2202t\u03c6 = D\u0394 \u03b4H[\u03c6] \u03b4\u03c6 + \u2202 \u00b7 \u03be = D\u0394 r\u03c6 \u2212 \u22022\u03c6 + \u03c63 + \u2202 \u00b7 \u03be. u 6 (10.93) Here we have absorbed the derivatives appearing in the model B noise correlator Eq. (10.78) into the de\ufb01nition of a vectorial noise \ufb01eld \u03bei(x)\u03bei (x) = 2A\u03b4ii \u03b4(x \u2212 x), and the constant A determines the e\ufb00ective equilibrium temperature as A = DT. Let us now explore what happens once driving is switched on. As discussed above, the Hamiltonian cannot accommodate the non-conservative driving \ufb01eld. However, we may argue that the right\u2013hand side of the Langevin equation represents the divergence of a current. An external \ufb01eld will cause a current that (i) is proportional to the \ufb01eld E in direction and strength, and (ii) vanishes if neither particles nor holes are present. This motivates the ansatz, j = 4n(1 \u2212 n)E = (1 \u2212 \u03c62)E, (10.94) where a constant of proportionality has been absorbed in a rede\ufb01ned E. The addition of the divergence of this current to the right\u2013hand side of Eq. (10.93) will generate the bare version of a generalized Langevin equation. However, we must keep in mind that the strong anisotropy introduced by the \ufb01eld will, by way of renormalization, render the e\ufb00ective parameters of the Langevin anisotropic as well. This motivates the more general anisotropic representation \u2202t\u03c6 = D r\u22022 \u2212 \u03b1\u22024 + \u02c6r \u02c6\u22022 \u2212 \u02c6\u03b1 \u02c6\u22024 \u2212 \u03b2\u22022 \u02c6\u22022 \u03c6 + E\u2202\u03c62 + 1 3! u\u22022 + \u02c6u \u02c6\u22022 \u03c63 \u2212 \u2202", "\u03be \u2212 \u02c6\u2202 \u00b7 \u02c6\u03be, (10.95) 670 Nonequilibrium (classical) where quantities carrying a caret refer to the d \u2212 1 directions transverse to the \ufb01eld, and the various coe\ufb03cients account for the possibility of anisotropic renormalization. Accordingly, the e\ufb00ective noise correlator takes the form \u03be(x)\u03be(x) = 2A\u03b4(x \u2212 x), \u02c6\u03bei(x) \u02c6\u03bei (x) = 2 \u02c6A\u03b4ii \u03b4(x \u2212 x). (10.96) D(\u03c6, \u03c8) e\u2212S[\u03c6,\u03c8], The MSRJD functional of this equation (cf. Eq. (10.53)) is given by Z = where # S[\u03c6, \u03c8] = dx \u03c8 \u2202t\u03c6 \u2212 D + E\u2202\u03c62 + 1 3! r\u22022 \u2212 \u03b1\u22024 + \u02c6r \u02c6\u22022 \u2212 \u02c6\u03b1 \u02c6\u22024 \u2212 \u03b2\u22022 \u02c6\u22022 $ \u03c6 u\u22022 + \u02c6u \u02c6\u22022 \u03c63 + A(\u2202\u03c8)2 + \u02c6A( \u02c6\u2202\u03c8)2. (10.97) In the following sections we aim to extract some physical information from this seemingly formidable expression. Above criticality: consequences of FDT violation Let us consider the system at temperatures T > Tc far above its ordering transition. Drawing upon our experience with equilibrium physics, we expect the system to be in a disordered state, void of any long range (power law) correlations. Remarkably, however, the physics of the driven system turns out to be di\ufb00erent. We will see that the external \ufb01eld leads to long range correlations and critical phenomena even in the high temperature regime. Following the standard logic in the theory of continuous phase transitions we assume that, for T > Tc, the intrinsic nonlinear terms (such as \u223c \u03c8\u03c63) provide subleading contributions to the dynamics. Alluding to their low operator relevance, we will also neglect all terms containing more than two gradient operators. We thus consider the bare action S[\u03c6, \u03c8] = dx \u03c8 \u2202t\u03c6 \u2212 D % r\u0394\u03c6 + E\u2202\u03c62 & \u2212 A\u03c8\u0394\u03c8, where we have anticipated that the bare theory is isotropic, A = \u02c6A and r = \u02c6r and the driving term provides the only non-Gaussian contribution to the action. The importance of the driving term can be assessed by computing its engineering dimension. To this end, let us consider the e\ufb00ect of a rescaling q \u2192 q \u2261 qb\u22121. Requiring that all Gaussian terms in the action remain scale invariant, we obtain the relations q = b\u22121q, \u03c6 = b 2 \u03c6, d t \u2192 b\u22122t, \u03c8 = b 2 \u03c8. d Substituting the rescaled quantities into the action, we are led to the renormalization E = b\u2212 d 2 +1E. Thus, the upper critical dimension of our nonlinearity is dc = 2. Below d = 2, the driving term is relevant, at d = 2 it is marginal, and above irrelevant. However, it would be premature to conclude that in dimensions d > 2, driving has no physical e\ufb00ects! To appreciate this point, imagine that on the short distance scales |q| \u223cl where the renormalization process is started, our action is isotropic in that r(l) = \u02c6r(l) and 10.7 Field theory III: applications 671 A(l) = \u02c6A(l). In the course of renormalization, the e\ufb00ective strength of E will diminish (exponentially, if d > 2). However, the residual E(L) will also renormalize the coupling constants of the Gaussian model, and it will do so in an anisotropic manner. EXERCISE Identify the interaction processes that cause the renormalization of the coupling constants. You may want to formulate the RG equations? At large length scales, we thus have an e\ufb00ective Gaussian model, E = 0, which, however, is anisotropic. In other words, we have a family of Gaussian \ufb01xed points, each described by a free action S[\u03c6, \u03c8] = = 1 2 \u03c8 dx \u2202t\u03c6 \u2212 D r\u22022 + \u02c6r \ufffd", "\ufffd\u22022 \u03c6 + A(\u2202\u03c8)2 + \u02c6A( \u02c6\u2202\u03c8)2 dq (\u03c6, \u03c8)\u2212q (g+)\u22121 (g\u2212)\u22121 2(Ak2 + \u02c6A\u02c6k2) \u03c6 \u03c8, q q q = (\u2213i\u03c9 + D(rk2 + \u02c6r\u02c6k2))\u22121, and q = (k, \u03c9) withk = (k, \u02c6k)T. As we are going to where g\u00b1 show next, each of these actions exhibits \u201ccritical behavior\u201d in that it generates anisotropic power law correlations in the system. Indeed, consider the dynamic structure factor S(k, \u03c9) = dq \u03c6q\u03c6q = 2(Ak2 + \u02c6A\u02c6k2)g+ q g\u2212 q = 2 Ak2 + \u02c6A\u02c6k2 \u03c92 + D2(rk2 + \u02c6r\u02c6k2)2. (10.98) Integration over \u03c9 generates the static structure factor S(k) = d\u03c9 2\u03c0 S(k, \u03c9) = 1 D Ak2 + \u02c6A\u02c6k2 rk2 + \u02c6r\u02c6k2. (10.99) In equilibrium, the FDT enforces the fraction on the right hand side of this equation to equal unity. To appreciate this point, remember our discussion at the end of Section 10.6.2. We had seen that a (model B) system may relax to equilibrium if the noise correlator is weighted by a Laplace operator (cf. Eq. (10.78)) which compensates for the Laplace operator in Eq. (10.71). In the present context, both operators get renormalized to anisotropic second order di\ufb00erential operators. Speci\ufb01cally, the noiseor \u201c\ufb02uctuation\u201d Laplace operator gets renormalized according to \u0394f \u2192 k2 + \u02c6A \u02c6k2, while \u02c6k2. An equilibrium con\ufb01guration at some the \u201cdissipation\u201d operator becomes \u0394d \u2192 k2 + \u02c6r r e\ufb00ective temperature Te\ufb00 can be approached (cf. the construction above Eq. (10.78)), if and only if \u0394f \u221d \u0394d. While this does not necessarily imply isotropy in space, a minimal requirement reads k k A 0.6 0. \u2261 A\u02c6r \u02c6Ar = 1. 672 Nonequilibrium (classical) This condition entails S(k) = const.. Equivalently, violation of the FDT condition w = 1 implies singular (albeit \ufb01nite) behavior of the structure factor. The singularity can, e.g., be characterized by the ratio R \u2261 S(0, \u02c6k \u2192 0) S(k \u2192 0, 0). The anisotropy singularity of the structure factor implies power law correlations in coordinate space. To see this, let us consider the case d = 3 for concreteness and compute the Fourier transform S(x) \u2261 d3k (2\u03c0)3 eik\u00b7xS(k) \u221d d3k ei(xk+\u02c6x\u00b7\u02c6k) = \u2212(w\u22022 \u02dcx + \u22022 \u02c6x) d3k ei(\u02dcxk+\u02c6x\u00b7\u02c6k) 1 |k|2 k2 + \u02c6k2 A \u02c6A \u02c6r k2 + \u02c6k2 r \u02dcx + \u22022 \u02c6x) \u221d (w\u22022 1 (\u02dcx2 + \u02c6x2)1/2 d3k ei(\u02dcxk+\u02c6x\u00b7\u02c6k) wk2 + \u02c6k2 |k|2 \u221d = (w \u2212 1)\u22022 \u02dcx 1 (\u02dcx2 + \u02c6x2)1/2 = (w \u2212 1) 2\u02dcx2 \u2212 \u02c6x2 (\u02dcx2 + \u02c6x2)5/2, where \u02dcx = (\u02c6r/r)x is a rescaled coordinate and in the third line we discarded a contribution (\u22022 \u221d \u03b4(|x|), because our e\ufb00ective theory holds only for large |x| anyway. \u02dcx + \u22022 \u02c6x) We thus come to the surprising conclusion that 1 (\u02dcx2+\u02c6x2)1/2 the", " high temperature (\u201cdisordered\u201d) phase of the system supports power law correlations in coordinate space. These correlations re\ufb02ect the presence of a line of inequivalent Gaussian \ufb01xed point theories. INFO Can these power law correlations above criticality actually be observed in real systems? Anisotropies in the structure factor such as those shown in the \ufb01gure above have actually been measured in uniaxial ferromagnets with dipolar interactions41 However, these systems support long range correlations on the bare level, and it is not really clear whether they share a universality class with our short range Ising type benchmark model. However, singular behavior in correlation functions is de\ufb01nitely observable in Monte Carlo simulations40 of the system. It is also worth noting that the mechanisms whereby long range correlations were generated in our discussion above appear to be of rather general nature: to appreciate this point, consider the temporal decay of correlations in a di\ufb00usive environment C(0, t) \u223c t\u2212d/2. The dynamical exponent of di\ufb00usive dynamics is given by z = 2. Power counting arguments would then suggest that C(x, t) \u223c |x|\u2212d at \ufb01xed t. However, in equilibrium statistical mechanics, we are more accustomed to exponentially decaying correlations (power law correlations indicating criticality being an exception). Our discussion above signals, that the paradigm \u201cno long range correlations away from criticality\u201d may not have quite as generic a validity as is suggested by the theory of equilibrium. Rather, we saw that violations of the FDT may generate power law correlations far away from critical points. 41 J. Skalyo, B.C. Frazer, and G. Shirane, Ferroelectric-mode motion in KD2PO4, Phys. Rev. B 1, 278-86 (1970). 10.7 Field theory III: applications 673 EXERCISE Retrace the discussion above for an isotropic system with conserved di\ufb00usive dynamics. Convince yourself that the FDT enforces vanishing of power law correlations in the correlation function. Inasmuch as equilibrium (obedience to FDT) is an exception rather than the rule, we may expect that long range correlations in nature are of more frequent occurrence than what is implied by equilibrium statistical mechanics. The system at criticality At some critical Tc the system undergoes a transition into its ordered phase. Numerical analysis has shown that, in the presence of strong driving, this ordered phase contains stripes of uniform magnetization. The formation of stripes may be diagnosed by monitoring the behavior of the static structure factor upon changing the temperature. Speci\ufb01cally, we expect that: (i) the proximity to the phase transition point is characterized by some correlation length, \u03be. (ii) At length scales larger than \u03be, the structure factor will exhibit the \u223c |x|\u2212d scaling discussed in the previous section. (iii) At length scales smaller than \u03be we expect anisotropic scaling; correlations in the direction parallel to E will be qualitatively di\ufb00erent from those in the transverse direction. In the following, we tune the system right into the critical point and explore its correlation behavior. However, before doing so, we \ufb01rst have to identify the transition point. In equilibrium \u03c64-theory, the transition occurs when the (renormalized) coe\ufb03cient, of the quadratic term in the action \u223c \u03c62 vanishes. Presently, however, we have two such terms, with coe\ufb03cients r and \u02c6r, respectively. This raises the question whether the ordering transition occurs at (i) r = 0, \u02c6r = 0, or (ii) r = 0, \u02c6r = 0, or (iii) r = \u02c6r = 0. In the following, we go through a construction that self-consistently shows (ii) to be the correct variant.42 Let us assume that r = 0, \u02c6r = 0 characterizes the system at criticality. We begin by determining the engineering scaling dimensions of the various contributions to the action. Specifcially, we should like to identify the upper critical dimension, i.e. the dimension below which nonlinear terms become qualitatively important. To this end, let us use the freedom to independently scale q, \u02c6q, t, \u03c6, \u03c8 in a manner to leave the dominant quadratic contributions to the action invariant. De\ufb01ning the scale transformation \u02c6q \u2192 b \u02c6q, \u03c6 \u2192 bd\u03c6 \u03c6, q \u2192 b\u03c3q, t \u2192 b\u2212zt, \u03c8 \u2192 bd\u03c8 \u03c8, (10.100) 42 At \ufb01rst sight this claim may", " seem surprising. Since we expect ordering in the \ufb01eld direction, would not one expect the coe\ufb03cient corresponding to that direction to vanish? 674 Nonequilibrium (classical) let us consider the scaling behavior of di\ufb00erent contributions to the action Eq. (10.97), dx \u23a7 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 \u03c8\u2202t\u03c6 D \u03c8 \u02c6\u22024\u03c6 D \u03c8\u22022\u03c6 Dr \u03c8 \u02c6\u22022\u03c8 \u02c6A \u03c8\u22022\u03c8 A DE \u03c8\u2202\u03c62 Du \u03c8\u22022\u03c63 D\u02c6u \u03c8 \u02c6\u22022\u03c63 \u2192 b\u03c3+(d\u22121)+z dx \u23a7 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 b\u2212(d\u03c6+d\u03c8)\u2212z D \u03c8\u2202t\u03c6 \u03c8 \u02c6\u22024\u03c6 b\u2212(d\u03c6+d\u03c8)\u22124 D b\u2212(d\u03c6+d\u03c8)\u22122\u03c3 Dr \u03c8\u22022\u03c6 \u03c8 \u02c6\u22022\u03c8 \u02c6A b\u22122d\u03c8\u22122 b\u22122d\u03c8\u22122\u03c3 \u03c8\u22022\u03c8 A b\u2212(2d\u03c6+d\u03c8)\u2212\u03c3 DE \u03c8\u2202\u03c62 b\u2212(3d\u03c6+d\u03c8)\u22122\u03c3 Du \u03c8\u22022\u03c63 b\u2212(3d\u03c6+d\u03c8)\u22122 D\u02c6u \u03c8 \u02c6\u22022\u03c63. (10.101) We now choose the scaling dimensions d\u03c6, d\u03c8, z, and\u03c3 so as to make the \ufb01rst four contributions to the list above scale invariant. This generates the conditions z = 4, \u03c3 = 2, d\u03c8 = d + 3 2, d\u03c6 = d \u2212 1 2. (10.102) With these values, we can estimate the engineering dimensions of the remaining terms. Speci\ufb01cally, we \ufb01nd that (a) the noise term \u221d A is strongly irrelevant in all dimensions, (b) the interaction term \u221d u is less relevant (by two dimensions) than the interaction term \u221d \u02c6u, (c) the driving term \u221d E carries dimension (5 \u2212 d)/2, and (d) the interaction term \u221d \u02c6u carries dimension 3 \u2212 d. Speci\ufb01cally, point (c) implies that the model has upper critical dimension dc = 5. Notice that the high value of the upper critical dimension re\ufb02ects the strongly intrusive role of the driving \u2013 below d = 5, the physics of the model becomes \ufb02uctuation dominated, earlier than in the equilibrium model with its dc = 4. In addition, the driving term is more relevant than the interaction term \u221d \u02c6u, which becomes relevant only below d = 3. The estimate above suggests that we should drop the terms proportional to A, u, \u02c6u. We will thus model the critical system by the reduced e\ufb00ective action dx D \u03c8 D\u22121\u2202t \u2212 r\u22022 + \u02c6\u22024 \u03c6 + E\u2202\u03c62 + ( \u02c6\u2202\u03c8)2 dx (\u03c6, \u03c8) (g\u2212)\u22121 (g+)\u22121 \u22122 \u02c6\u22022 \u03c6 \u03c8 S[\u03c6, \u03c8] = = 1 2 + DE\u03c8\u2202\u03c62, (10.103) where g\u00b1 = (\u00b1\u2202t \u2212 Dr\u22022 + D \u02c6\u22024)\u22121, and we have absorbed the noise coe\ufb03cient \u02c6A in a rescaled \ufb01eld amplitude. In the following, we aim to understand the build\u2013up of anisotropies in the critical system at d = 5 \u2212 dimensions. To understand the rami\ufb01cations of scaling anisotropies, let", " us begin by computing the Fourier transform of the static structure factor \u2013 the spatial correlation function of the system \u2013 in a tree approximation. Comparing with Eq. (10.98) and (10.99), we obtain S(k) = d\u03c9 S(k, \u03c9) = d\u03c9 2\u03c0 2\u02c6k2 \u03c92 + D2(\u02c6r\u02c6k2 + \u02c6k4)2 = \u02c6k2 D(rk2 + \u02c6k4), 10.7 Field theory III: applications 675 and S(x) = \u223c ddk (2\u03c0)d ei(kx+\u02c6x\u00b7\u02c6k)S(k) = dd\u22121\u02c6k (2\u03c0)d\u22121 e\u2212\u02c6k2|x|r\u22121/2+i\u02c6x\u00b7\u02c6k \u223c |x|\u2212 d\u22121 ddk (2\u03c0)d ei(kx+\u02c6x\u00b7\u02c6k) 2 e \u2212 \u02c6x2 r1/2 4|x|. \u02c6k2 D(rk2 + \u02c6k4) This expression provides an a\u2013posteriori justi\ufb01cation of our choice of \u201cmass parameters\u201d r and \u02c6r at criticality. We have obtained a correlation function that decays as a power law in the longitudinal direction and exponentially in the transverse direction. This behavior conforms with the build\u2013up of critical \ufb02uctuations in the longitudinal direction, while the transverse direction remains in a disordered state. Perturbative RG How will critical \ufb02uctuations e\ufb00ect this picture? In the following, we perform a rudimentary RG analysis to show that below the upper critical dimension dc = 5, \ufb02uctuations conspire to strengthen the anisotropic character of the system. (For a full exposition of an RG analysis of this system, we refer to Schmittmann and Zia.39) As in Section 8.3, we aim to explore what happens as fast \ufb02uctuations successively get integrated out. To this end, we \ufb01rst introduce a graphical language representing the building blocks relevant to the perturbative RG (see the \ufb01gure.) \u03c6 \u03c8 \u03c6 \u03c6 \u2261 G11 \u03c6 \u03c8 \u2261 G12 \u2202\u03c8 \u03c62 EXERCISE Refamiliarize yourself with the plan of the perturbative RG discussed in Section 8.3. Decompose the \ufb01elds as \u03c6 = \u03c6s + \u03c6f into fast and slow sectors. To one loop order in the fast \ufb01elds, explore the di\ufb00erent diagrams contributing to the perturbative RG step. Speci\ufb01cally, convince yourself that no vertices other than those contained in the action Eq. (10.103) will be generated. Use power counting arguments to show that the one fast momentum integration becoming dangerous below d = 5 will act to renormalize the \u22022\u03c6\u03c8 vertex. The one diagram driving the renormalization of the theory is shown in the \ufb01gure to the right. To explore the e\ufb00ect of this contribution, we divide the \ufb01elds into fast and slow contributions. In view of the anisotropies inherent to the model, it is actually not so obvious how the space of coordinates (q, \u02c6q, \u03c9) is to be divided into a fast and a slow sector. On the other hand, the RG procedure should be insensitive to the detailed choice of an (in\ufb01nitesimally thin) layer of fast modes. (As an exercise, check it by inventing schemes di\ufb00erent from the one below.) We thus meet the technically convenient choice, fast: \u039b/b < |\u02c6q| < \u039b, (q, \u03c9) arbitrary, slow: |\u02c6q| < \u039b/b, (q, \u03c9) arbitrary. Denoting the 676 Nonequilibrium (classical) f dq, the diagram above renormalizes the coe\ufb03cient r integration over the fast modes by according to r \u2192 r + const. \u00d7 DE2 f = r + const. \u00d7 DE2 = r + const. \u00d7 E2 D f dd\u22121 \u02c6q f dd\u22121 \u02c6q dq G21(q)G22(q) = r + const. \u00d7 DE2 f dq g+ q \u02c6q2(g+g\u2212)q dqd\u03c9 \u02c6q2 (\u2212i\u03c9 + Drq2 + D \u02c6q4)2(i\u03c9 + Drq2 +", " D \u02c6q4) \u02c6q2 (rq2 + \u02c6q4)2 ln b + O(2), cE2 Dr1/2 r + dq where c = O(1) is a numerical constant. At the one-loop level, the vertex \u221d E does not renormalize; its renormalization stems from the the rescaling of momenta following the RG step, E \u2192 b(5\u2212d)/2E = eln b/2E. This latter rescaling does not change the r-term, as the \ufb01eld dimensions have been deliberately chosen so as to make this term (engineering) scale invariant. We thus obtain the two RG equations dE d ln b = 2 E, dr d ln b = cE2 Dr1/2. (10.104) Substituting the solution of the \ufb01rst equation, E = b/2E0 into the second, and solving the remaining ordinary di\ufb00erential equation, we obtain r \u223c b 2 3. This \ufb02uctuation induced scaling needs to be taken into account in an updated variant of the determination of scaling dimensions. Accounting for the scaling of r in Eq. (10.101), we \ufb01nd that the anisotropy exponent changes according to \u03c3 = 2 + 3. (10.105) This result adds to our overall observation that driving massively interferes with the physics of the system. By way of summary, we have seen that: In the disordered regime above criticality, external driving leads to the formation of new macroscopic structures which manifest themselves in new types of power law correlation. The strongly invasive character of driving is also re\ufb02ected in the increased value of the upper critical dimension, 4 \u2192 5 (as compared to the equilibrium system). At criticality, driving causes \ufb02uctuations which superimpose to strengthen the anisotropy of the system, Eq. (10.105). Qualitatively, the \ufb02uctuations generate patterns that appear to be strongly aligned in the \ufb01eld direction. For a deeper analysis of the system, and a discussion of its physics below criticality, we refer to a more comprehensive study of its scaling and renormalization behavior.39. 10.7.2 Directed percolation Amongst the di\ufb00erent universality classes of equilibrium statistical mechanics, the \u03c64- or Ising-class is the simplest. It owes its frequent occurrence in nature to the fact that its free energy function, a quartic polynomial of a real scalar variable, is the most basic one that leads to nontrivial critical behavior. In nonequilibrium statistical mechanics, the directed 10.7 Field theory III: applications 677 Figure 10.12 Cartoon of a directed percolation network with a given realization of open bonds, cf. ordinary, or isotropic percolation, where the network bonds are directionless. percolation class plays a similarly paramount role. In this section, we try to understand the importance of this universality class to nonequilibrium statistical mechanics, and explore a few of its critical properties. However, before doing so, it is useful to brie\ufb02y review directed percolation perse. (The material presented below is inspired by an excellent review by H. Hinrichsen.43) Directed Percolation: Phenomenology The term \u201cdirected percolation\u201d (DP) refers to a class of models that describe the spatially directed spreading of a substance through a medium \u2013 think, for example, of a liquid dispersing through porous rock under the in\ufb02uence of a gravitational force. Alternately, one may identify the driving direction with time, in which case directed percolation mimics, for example, the spreading of an epidemic, or related dynamical phenomena. The directed percolation model (see Fig. 10.12) is de\ufb01ned on a d-dimensional hypercubic lattice. A certain \u201cmain diagonal\u201d of the lattice is introduced as the identi\ufb01er of a sense of direction. The links of the lattice are then chosen to be open or closed with probability p and 1 \u2212 p, respectively. Like its cousin, the undirected, or isotropic percolation model, this system exhibits a phase transition at a critical value of the connectivity probability, p = pc, the directed percolation phase transition (see Fig. 10.13). The order parameter of this transition is the probability P\u221e that a randomly chosen site of the lattice is at the origin of an in\ufb01nitely large connected cluster. At p < pc, and using a language alluding to the liquid/porous rock application, the system is in a \u201cdry phase.\u201d", " At p > pc, it has become \u201cwet.\u201d Despite the simplicity of the model, the critical properties of the DP-transition are not fully understood. What can be said without any further calculation is that the critical value pc = pc(d) will depend on dimensionality. (In one dimension, any broken bond will truncate the cluster, i.e. pc(1) = 1; in in\ufb01nite dimensions, the unlimited number of options to go 43 H. Hinrichsen, Nonequilibrium critical phenomena and phase-transitions into absorbing states, Adv. Phys. 49, 815-958 (2000). 678 Nonequilibrium (classical) Figure 10.13 The two phases of a directed percolation network. Left, dry phase, wherein individual sites generate \ufb01nite size clusters; right, wet phase with in\ufb01nite size cluster generation. from any lattice site implies pc(d \u2192 \u221e) - 0.) Both, numerical analysis and \ufb01eld theory suggest that the corresponding critical exponents are complicated, and likely irrational. What gives the understanding of the DP transition increased relevance is that many reaction\u2013di\ufb00usion processes appear to fall into the DP universality class. To heuristically understand this connection, simply note that, upon interpreting the sense of direction as the \ufb02ow of time, the elementary compounds of a cluster a\ufb00ord a natural interpretation in terms of \u201creactions,\u201d or di\ufb00usion (cf. Fig. 10.14). The generation update on the DP network need not be modeled simply by drawing bond connectivities according to some \ufb01xed probabilities; more sophisticated update rules can be invented to describe a large set of dynamical setups. Algorithms implementing such update rules are called cellular automata. Many such automata exhibit a phase transition between two \u201cabsorbing states\u201d of the lattice, the empty state (above termed \u201cdry\u201d) and the totally occupied state (\u201cwet\u201d). One of the theoretical challenges in this \ufb01eld is to understand the universal properties of such transitions, and speci\ufb01cally their connection to the basic DP universality class. Below, we explore some of the universal properties of the DP transition by \ufb01eld theoretical methods. We also aim to understand the origins of the ubiquity of the DP class in reaction\u2013 di\ufb00usion kinetics. Elements of scaling theory Many physical properties of the DP transition can be conveniently described in terms of scaling laws. Introducing \u03c4 = p \u2212 pc as a scaling \ufb01eld measuring the distance from the transition point, the correlation behavior of the system is characterized by a correlation length \u03be \u223c \u03c4 \u2212\u03bd in its \u201cspatial directions,\u201d and a correlation \u201ctime\u201d t \u223c \u03bez \u223c \u03c4 \u2212\u03bdz in the orienting direction. The transition as such can be probed in terms of two candidate order 10.7 Field theory III: applications 679 Figure 10.14 Dynamical interpretation of a DP network. Each row of sites represents a generation of particles. The generation dynamical update then involves di\ufb00usion, birth and death, coagulation and o\ufb00spring creation. parameters. The \ufb01rst is the average number of active, or wet sites, \u03c1(t) \u2261 N \u22121 si(t), C B i where N is the total number of sites within one time layer, and si = 1 if site i is active, and zero otherwise. In the vicinity of the transition, \u03c1 is expected to exhibit power law scaling, \u03c1(t \u2192 \u221e) \u2261 \u03c1 \u223c \u03c4 \u03b4, \u03c4 < 0. More generally, the critical time dependence of order parameter \ufb02uctuations is described by the ansatz \u03c1(t) \u223c |\u03c4 |\u03b2g(t/t(\u03c4 )), with the short time asymptotics (cf. the discussion of Section 10.6.1), \u03c1(t) \u223c t\u2212\u03b1, \u03b1 = \u03b2/\u03bdz. Another possible order parameter is the in\ufb01nite cluster size probability P\u221e introduced above. We can generalize to a time-dependent quantity by declaring P (t) to be the average probability that a cluster started at t = 0 is still active at time t. Since the \u201cspatial\u201d size of the cluster grows inde\ufb01nitely with time, P\u221e = P (t \u2192 \u221e). The scaling behavior of P is described by P\u221e = \u03c4 \u03b2, \u03c4 < 0, and P (t) \u223c |\u03c4 |\u03b2 g(t/t(\u03c4 )), with short time limit P (t) \u223c t\u2212\u03b1, \u03b1 = \u03b2/\u03bdz. The four", " exponents \u03bd, z, \u03b2, \u03b2 describe the basic scaling behavior of the system. (For the elementary DP class, in\ufb01nite cluster size probability and overall wetness are equivalent characteristics, \u03b2 = \u03b2. For general phase transitions between absorbing states, however, these quantities can be di\ufb00erent.) An important derived quantity, similar in nature to the pair correlation functions of many particle theories, is the pair connectedness function C(x, t). This function is de\ufb01ned as 680 Nonequilibrium (classical) the probability that an active site at (0, 0) is connected to the site at (x, t) by an open path. The scaling hypothesis implies that C(x, t) = t\u03b8\u2212d/zF (|x|/t1/z, \u03c4 t1/\u03bdz). (10.106) The dimension \u03b8 can be \ufb01xed by running a few consistency checks: the probability that (x 0, t) is connected to (0, 0) is proportional to both the probability of \ufb01nding a cluster of temporal extension P (t), and the density of active sites, \u03c1(t). Here, the notation x 0 means that the \ufb01nal point x should lie inside the \u201cconical\u201d cluster spreading out from 0, i.e. C(0, t) t\u2192\u221e\u223c \u03c4 \u03b2+\u03b2. On the other hand, the expected stationarity of C(0, t \u2192 \u221e) requires that F (0, \u03c4 t1/\u03bdz)!\u223c t\u2212(\u03b8\u2212d/z), i.e. F (0, \u03c4 t1/\u03bdz) \u223c \u03c4 \u2212\u03bdz(\u03b8\u2212d/z). Comparing these two conditions, we obtain the identi\ufb01cation \u03c4 \u2212\u03bdz(\u03b8\u2212d/z)!\u223c \u03c4 \u03b2+\u03b2, or 10.107) Equations (10.106) and (10.107) describe much of the correlation behavior of the cluster. Notice, however, that the argument leading to Eq. (10.107) relied on the cluster spreading \u201cunderneath\u201d the initial point. However, in high dimensions, this condition may be violated, on account of the cluster spreading out in directions \u201corthogonal\u201d to |x \u2212 0|. Closer analysis shows that the validity of Eq. (10.107) is limited to dimensions below the upper critical dimension of the process. For the discussion of scaling functions higher in the hierarchy, and describing features such as the average extension of clusters, their volume, etc., we refer to Hinrichsen.43 In the next section, we discuss how \ufb01eld theoretical methods can be applied to describe the \ufb02uctuation behavior of the system and its critical exponents. Field theory Before analysing low energy \ufb01eld theories of the DP transition, we \ufb01rst have to construct one. On the face of it, it is not so obvious how to do this: we have not described DP in the language of master equations, nor by Langevin type evolution, so our standard recipes do not apply. Instead, we aim to construct a \u201cdynamical mean \ufb01eld equation\u201d of DP on phenomenological grounds. We then use this equation as input for the construction of a \ufb01eld theory, along the lines of Section 10.6.3. The mean \ufb01eld equation we are going to construct operates on the mean density of active sites, \u03c1(t). Adopting the dynamical interpretation of DP, \u03c1(t) increases at a rate that is proportional to the probability of open bonds, p, the density \u03c1(t) of active sites at time step n \u2212 1 that may act as germs of newly generated active sites, and the density of empty sites at time stepn that may be converted to active sites. The density decreases at a rate proportional to the density of active sites at the previous time step, and the probability of closed bonds, 1 \u2212 p. Combining all these considerations, we are led to the equation \u2202t\u03c1 = c1p\u03c1(1 \u2212 \u03c1) \u2212 c2(1 \u2212 p)\u03c1, where c1,2 are nonuniversal constants depending on the dimensionality of the system, and on the unit at which time is measured. We may always rescale density and time such that the equation assumes the form, \u2202t\u03c1 = (\u03bb \u2212 1)\u03c1 \u2212 \u03bb\u03c12. (10.108) 10.7 Field theory III: applications 681 The de\ufb01ning feature of this equation is that it possesses (a) a stationary con\ufb01guration \u03c1 = 0, and (b) a non-empty", " stationary state \u03c1 = \u03bb\u22121 \u03bb, provided \u03bb >1. The critical value \u03bbc = 1 at which the non-empty \u201cmean \ufb01eld\u201d begins to emerge marks the position of the continuous DP transition. Notice the simplicity of the DP mean \ufb01eld equation. It explains the ubiquity of the DP class: alluding to the principle of universality, one may suspect that processes described by a rate equation free of constant terms generically exhibit a DP transition, provided that the \ufb01rst two terms \u223c \u03c1 and \u223c \u03c12 admit a non-zero solution. Equation (10.108) \ufb01xes the mean \ufb01eld critical exponents of the system. Using the notation \u03c4 = \u03bb \u2212 \u03bbc = \u03bb \u2212 1 for the di\ufb00erence to the mean \ufb01eld critical point, near criticality \u03c1 \u03c4, i.e. \u03b2mf = 1. In the dry phase, \u03c4 < 1, the asymptotic temporal decay of the density is given by \u03c1 \u223c exp(\u2212|\u03c4 |t), which means that \u2212\u03bdz = 1. However, lacking any spatial structure, mean \ufb01eld theory cannot say anything about \u03bd (nor z) individually. The mean \ufb01eld equation (10.108) can be upgraded to a spatially resolved rate equation by upgrading \u03c1(t) \u2192 \u03c1(x, t) \u2261 \u03c1(x) to a local density pro\ufb01le and adding a di\ufb00usion term: \u2202t\u03c1 = D\u22022\u03c1 + \u03c4 \u03c1 \u2212 \u03bb\u03c12. (10.109) Dimensional analysis shows that \u03bdmf = 1/2. In combination with the results above (which survive generalization to a local equation \u2013 exercise) we then have the mean \ufb01eld, or tree level prediction, \u03b2mf = 1, \u03bdmf = 1/2, zmf = 2. (10.110) A \ufb01eld theory can now be constructed by interpreting Eq. (10.110) as a model A rate equation along the lines of our discussion of Section 10.6.2. To this end, we upgrade Eq. (10.109) to a noisy equation, where the noise is correlated as \u2202t\u03c1 = D\u22022\u03c1 + \u03c4 \u03c1 \u2212 \u03bb\u03c12 + \u03be, \u03be(x) = 0, \u03be(x)\u03be(x) = 2A\u03c1(x)\u03b4(x \u2212 x). (10.111) \u221a The crucial feature here is the scaling \u03be \u223c \u03c1. It re\ufb02ects the assumption that the system self-generates noise through its active sites. This type of noise must vanish in the empty state, \u03c1 = 0. Further, we are to interpret \u03be as a coarse-grained variable sampling \ufb02uctuations in a large number of active sites. According to the central limit theorem, these \ufb02uctuations \u03c1, which explains the above correlation law. Generally, noise that scales with some scale as power of the Langevin variable is termed multiplicative noise. \u221a We next subject Eq. (10.110) to our standard treatment of Langevin equations, thus genD(\u03c6, \u03c8) e\u2212S[\u03c6,\u03c8] erating the \ufb01eld theory representation of DP partition function, Z = where44 S[\u03c6, \u03c8] = dx % \u03c8(\u2202t \u2212 D\u22022 \u2212 \u03c4 )\u03c6 + \u03ba\u03c8\u03c6(\u03c6 \u2212 \u03c8) &, (10.112) 44 Curiously, the \ufb01eld theory Eq. (10.112) has a background in particle physics, where it is known as Reggeon \ufb01eld theory, see M. Moshe, Recent developments in Reggeon \ufb01eld theory, Phys. Rep. 37, 255-345 (1978) for a review. 682 Nonequilibrium (classical Figure 10.15 Three distinct phase portraits of directed percolation. Left: phase supporting a \ufb01nite concentration of active sites, center: phase transition, right: empty state. and \u03ba \u2261 (A\u03bb)1/2.(In deriving Eq. (10.112), we made use of the freedom to rescale \ufb01elds so as to make the coe\ufb03cients of the two nonlinear terms \u223c \u03c6\u03c82 and \u223c \u03c62\u03c8 equal.) We next explore (cf. Eqs. (10.100) and (10.101)) the behavior of the di\ufb00erent terms", " in the action under the rescaling transformation x \u2192 x/b, t \u2192 t/bz, \u03c6 \u2192 bd\u03c6 \u03c6, \u03c8 \u2192 bd\u03c8 \u03c8. (10.113) Referring to the action (10.97), we have dx \u23a7 \u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23a9 \u03c8\u2202t\u03c6 D \u03c8\u22022\u03c6 \u03c8\u03c6 \u03c4 \u03bb \u03c8\u03c62 A \u03c82\u03c6 \u2192 bd+z dx \u23a7 \u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23a9 b\u2212(d\u03c6+d\u03c8)\u2212z \u03c8\u2202t\u03c6 b\u2212(d\u03c6+d\u03c8)\u22122 D \u03c8\u22022\u03c6 b\u2212(d\u03c6+d\u03c8) b\u2212(2d\u03c6+d\u03c8) b\u2212(d\u03c6+2d\u03c8) \u03c8\u03c6 \u03c4 \u03bb \u03c8\u03c62 A \u03c82\u03c6. (10.114) Choosing z = 2 and d\u03c6 = d\u03c8 = d/2, the \ufb01rst two contributions to the action become (tree level) scale invariant. The operator \u223c \u03c4 \u03c8\u03c6 measuring the distance o\ufb00 criticality is strongly relevant with dimension 2. Finally, the nonlinear operators \u223c \u03bb\u03c8\u03c62 and \u223c A\u03c82\u03c6 carry dimension 4 \u2212 d. This identi\ufb01es the upper critical dimension of directed percolation as d = 4. Before turning to the discussion of the role of \ufb02uctuations below d = 4, let us brie\ufb02y recapitulate the mean \ufb01eld picture in the language of the e\ufb00ective \ufb01eld theory Eq. (10.112). As with the zero-dimensional problems discussed in Section 10.4, the phase structure of the system is essentially determined by the pattern of intersecting zero energy lines of the mean \ufb01eld Hamiltonian, H = \u03c8\u03c6(\u03c6 \u2212 \u03c8 \u2212 \u03c4 ). Depending on the sign of \u03c4, three distinct phases need to be distinguished (see Fig. 10.15). At \u03c4 > 0, the system is in its active phase: on the \ufb02uctuationless manifold \u03c8 = 0, the concentration variable \u03c6 is driven towards the stable con\ufb01guration \u03c6 = \u03c4. (Notice that only positive values of \u03c6 are physically meaningful.) At \u03c4 < 0, the empty state \u03c6 = 0 is stable, the inactive phase. The mean \ufb01eld phase transition between both phases happens at \u03c4 = 0. The phase space picture of directed 10.7 Field theory III: applications 683 percolation was introduced in Kamenev 45 as the basis for a far-reaching classi\ufb01cation of phase transitions in reaction\u2013di\ufb00usion systems. For further aspects of the utility of the Hamiltonian approach in this \ufb01eld, we refer to the original reference. Perturbative RG How will \ufb02uctuations alter this picture? Close to the upper critical dimension, in d = 4\u2212, renormalization group techniques can be applied to answer this question. Carrying out an RG analysis similar to that detailed in Section 10.7.1 (cf. Problem 10.9.4), one \ufb01nds that the two coupling constants \u03c4 and \u03ba \ufb02ow according to the equations C\u03ba2, 2 \u2212 3 4 d\u03c4 d ln b d\u03ba d ln b = \u2212DC\u03ba2 + \u03c4 = \u03ba 2 \u2212 3 2 C\u03ba2, (10.115) S4 (2\u03c0)4D2 and S4 is the area of the four dimensional unit-sphere. where the constant C = Within the above RG scheme, the di\ufb00usion constant, D, and the pre-factor of the frequency term \u03c9 are kept \ufb01xed. The dynamical exponent and the \u201c\ufb01eld renormalization\u201d \u03c6 \u2192 \u03c6b\u2212 4\u2212 2 +\u03c7 are given by, respectively, z = 2 \u2212 C\u03ba2 4, \u03c7 = C\u03ba2 4. 1/2 These equations possess a non-trivial \ufb01xed point at \u03c4 \u2217 = D which means that \ufb02uctuations shift the position of the DP transition to a non-vanishing value of \u03c4. At the \ufb01xed point, z = 2 \u2212 /12 and \u03c7 = /12. Linearization of the RG", " equations around the \ufb01xed point leads to 6 + O(2), \u03ba\u2217 = 3C = 2 \u2212 4 = \u2212\u02dc\u03ba, d\u02dc\u03c4 d ln b d\u02dc\u03ba d ln b \u02dc\u03c4 \u2212 \u02dc\u03baCD\u03ba\u2217 2 +, 4 where \u03c4 = \u03c4 \u2217 + \u02dc\u03c4 and \u03ba = \u03ba\u2217 + \u02dc\u03ba. The linearized equations imply that the coupling constant \u02dc\u03c4 is a relevant scaling \ufb01eld with \u02dc\u03c4 \u223c b2\u2212 4. This result can in turn be used to determine the 1-loop critical exponents: the scaling b\u22121 \u223c \u03be \u223c \u02dc\u03c4 \u2212\u03bd \u223c b\u2212\u03bd(2\u2212 4 ) leads 16 + O(2). The critical exponent \u03b2 follows from the scaling to the identi\ufb01cation \u03bd = 1 relation \u03c4 \u03b2!\u223c \u03c6 = b\u2212 4\u2212 4, we obtain the identi\ufb01cation \u03b2 = 1 \u2212 6 + O(2). Summarizing, the 1-loop RG analysis generates the list of exponents z = 2 \u2212 12 2 + 2 +\u03c7\u03c6(\u03c4 b2\u2212 \u03b2 = 1 \u2212 6 4 ). Setting b \u223c \u03c4 + O(2), + O(2), + O(2). 16 \u03bd = 1 2 \u2212 1 2\u2212 + How do these values compare with reality? Numerical simulations for (3 + 1)-dimensional clusters46 obtain (z, \u03bd, \u03b2) = (1.90(1), 0.581(5), 0.81(1)) which compares reasonably well with 45 A. Kamenev, Classi\ufb01cation of phase transitions in reaction\u2013di\ufb00usion models, Phys. Rev. E 74, 41101-17 (2006). 46 I. Jensen, Critical behavior of the three-dimensional contact process, Phys. Rev. E 45, R563-6 (1992). 684 Nonequilibrium (classical) the = 1 extrapolation of the exponents above, (z, \u03bd, \u03b2) = (1.92, 0.56, 0.83). A two-loop extension of the RG47 leads to excellent agreement. However, in lower dimensions, the situation is decidedly worse. For example, in the extreme case of (1+1)-dimensional percolation clusters, the RG exponents are o\ufb00 the results of simulations by ca. 40%. Even so, the \ufb01eld theory analysis sheds light on the physical mechanisms generating critical behavior, and this is information that cannot be obtained from direct simulations. 10.8 Summary and Outlook In this chapter, we have discussed foundations of nonequilibrium statistical mechanics, with emphasis put on later extensibility to quantum nonequilibrium theory. We have introduced the two major pathways to the description of nonequilibrium systems, Langevin dynamics and probabilistic formulations in terms of master equations. We saw that these two approaches a\ufb00ord a uni\ufb01ed description in terms of functional integrals, the MSRJD formalism. Interpreting the latter as a Hamiltonian functional integral, we looked at its Lagrangian partner \u2013 the Onsager\u2013Machlup functional \u2013 and recovered the Fokker\u2013Planck equation as its dynamical equation (similarly to the Schr\u00a8odinger equation being the dynamical equation of the quantum mechanical path integral). The arch stretching from the Langevin/master equations over the MSRJD\u2013/Onsager\u2013Machlup functional to the Fokker\u2013Planck equation provides a powerful framework wherein many nonequilibrium phenomena can be studied in a uni\ufb01ed setting. Speci\ufb01cally, the functional integral description suggested a variational approach (similar in nature to the WKB approach of quantum mechanics) to the description of large \ufb02uctuations: \ufb02uctuations in higher dimensional environments could be analyzed by concepts borrowed from the theory of critical phenomena, etc. In the next chapter, the focus is on another important aspect of the formalism, the interpretation of the MSRJD functional as the classical limit of a generalized quantum theory. 10.9 Problems 10.9.1 Wigner surmise The problem of eigenvalue correlations of Hermitian matrices drawn from random distributions is a subject with a long history dating back to studies of resonances in atomic nuclei. The following problem concerns the distribution of level spacings in a canonical 2 \u00d7 2 random Hamiltonian. The result, known as the Wigner surmise, captures much of the behavior of the general expression for random matrices of arbitrary rank. Consider a Hamiltonian H of the form of a 2 \u00d7 2 matrix parameterized as \u02c6H =. H1 \u0394 \u0394 H2 47 J. B. Bronzan and J. W. Dash, Higher order epsilon", " terms in the renormalization group approach to Reggeon \ufb01eld theory, Phys. Lett. B 51, 496498 (1974). 10.9 Problems 685 Suppose that the three real numbers H1, H2, and \u0394 are drawn from a Gaussian distribution P (H1, H2, \u0394) \u221d e\u2212 1. Derive the joint distribution P (E1, E2) of the two energy levels Ei, i = 1, 2, of the Hamiltonian \u02c6H. Using this, derive the Wigner surmise describing the distribution of the energy splitting s = |E1 \u2212 E2|, and show that it has the form 2 tr \u02c6H 2 P (s) = c1 s e\u2212c2s2, with some constants c1 and c2. Answer: The joint eigenvalue distribution is given by P (E+, E\u2212) \u223c dH1 dH2 d\u0394 \u03b4(E+ \u2212 \u03bb+)\u03b4(E\u2212 \u2212 \u03bb\u2212)e\u2212(E2 ++E2 \u2212)/2, 2 2 \u00b1 (\u03942 + ( H1\u2212H2 )2)1/2 denote the eigenvalues of the matrix H and we where \u03bb\u00b1 = H1+H2 assumed E+ > E\u2212 (in the opposite case, one has to interchange \u03bb+ \u2194 \u03bb\u2212). Integrating over (H1 + H2)/2 and setting \u03b4H = H1 \u2212 H2, one obtains P (E+, E\u2212) \u223c d\u03b4H d\u0394 \u03b4(E+ \u2212 E\u2212 \u2212 2(\u03942 + \u03b4H 2)1/2)e\u2212(E2 \u2212)/2. Finally, setting r = (\u03942 + \u03b4H 2)1/2, and integrating over \u2212)/2 \u223c the \u201cangular\u201d variable, one obtains P (E+, E\u2212) \u223c (E+ \u2212 E\u2212)e\u2212(E2 \u221e 0 dr r \u03b4(E+ \u2212 E\u2212 \u2212 2r)e\u2212(E2 ++E2 \u2212)/2. From this result, we obtain the distribution ++E2 ++E2 P (s) \u223c dE+ dE\u2212 P (E+, E\u2212)\u03b4(E+ \u2212 E\u2212 \u2212 s) \u223c se\u2212s2/8. E+>E\u2212 We therefore \ufb01nd that c2 = 1/8, and c1 = 1/4 is obtained from the normalisation condition. This result shows that the probability of \ufb01nding a degeneracy is vanishing i.e. levels repel. As a further exercise, show that if H is a complex Hermitian, i.e. both real and imaginary parts of \u0394 are drawn from a Gaussian distribution, the probability distribution vanishes as P (s) \u223c s2 for small s. 10.9.2 Ornstein-Uhlenbeck process The Ornstein-Uhlenbeck process48 is the \u201charmonic oscillator\u201d of nonequilibrium dynamics. It describes stochastic evolution in cases where the averaged dynamics of a Langevin particle is described by a linear \ufb01rst order di\ufb00erential equation. Realizations include the dynamics of a particle subject only to friction (but no \u201cexternal forces\u201d), or the dynamics of an overdamped particle in a harmonic external potential. Dissipative dynamics close to extremal potential points is often described by variants of this process. (a) Consider a Brownian particle in 1d whose velocity is governed by the stochastic di\ufb00er- ential equation \u02d9v + \u03b3v = f (t) m, (10.116) 48 G. E. Uhlenbeck and L. S. Ornstein, On the theory of Brownian motion, Phys. Rev. 36, 823-41 (1930). 686 Nonequilibrium (classical) where f is a Langevin force, and f (t)f (t) = A\u03b4(t \u2212 t). Determine the general solution of Eq. (10.116) for the velocity with the initial condition v(0) = v0, and calculate v(t) and v2(t). Using the long-time limit of the latter, identify the coe\ufb03cient A with the help of the equipartition theorem. (b) Consider the time-dependent velocity distribution p(v, t). It is governed by the Fokker\u2013 Planck equation \u2202 \u2202t + \u2202 \u2202v a1(v) \u2212 1 2 \u22022 \u2202v2 a2(v) p(v, t) = 0. (10.117) With the help of Eq.", " (10.117), derive equations for \u2202tv and \u2202tv2 and identify the coe\ufb03cients a1 and a2, e.g., by using the results of part (a). Show that the corresponding dve\u2212ikvp(v, t) is given Fokker\u2013Planck equation for the generating function g(k, t) = by \u2202 \u2202t + \u03b3k \u2202 \u2202k + \u03b3T m k2 g(k, t) = 0. (10.118) (c) The \ufb01rst order partial di\ufb00erential equation Eq. (10.118) can be solved for example with the method of characteristics. Show that a general solution is of the form g(k, t) = e\u2212 T 2m k2 \u03c6(ke\u2212\u03b3t). Determine the function \u03c6 from the initial condition p(v, t = 0) =\u03b4 (v \u2212 v0) and derive the distribution p(v, t). Discuss the short and long-time limits of your result. Answer: (a) Formally, integrating the equation, one obtains the solution, v(t) = e\u2212\u03b3t v0 + t e\u03b3t, dt f (t) m 0 where v0 \u2261 v(0). Then, averaging over the distribution for the Langevin force, since f = 0, we have v(t) = v0e\u2212\u03b3t. Similarly, making use of the expression for the correlator, we have v2(t) = e\u22122\u03b3t v2 0 + = e\u22122\u03b3t v2 0 + t dt dt f (t)f (t) e\u03b3(t+t) 0 A 2\u03b3m2 (e2\u03b3t \u2212 1) m2. In the long time limit, v2(t) = A v(t), one may further show that \u03b4v(t)\u03b4v(t) = T 2\u03b3m2 FDT= T m, i.e. A = 2\u03b3mT. Setting \u03b4v(t) = v(t) \u2212 m (e\u2212\u03b3|t\u2212t| \u2212 e\u2212\u03b3(t+t)). (b) Using the Fokker\u2013Planck equation, we have \u2202tv = \u2212 dvv\u2202v(a1P ) + dv 1 2 v\u22022 v (a2P ) = dva1P \u2212 1 2 dv \u2202v(a2P ) = a1. 10.9 Problems 687 From this result, we obtain a1(v) = \u2212\u03b3v. Similarly \u2202tv2 = \u2212 dvv2\u2202v(a1P ) + dv 1 2 v2\u22022 v (a2P ) = dv2va1P \u2212 dvv\u2202v(a2P ) = 2va1 + a2. 0e\u22122\u03b3t + A From part (a), we have \u2202tv2 = \u22122\u03b3v2 amount of algebra, we \ufb01nd that a2 = 2\u03b3T m. m2 e\u22122\u03b3t. As a result, after a small The derivation of Eq. (10.118) amounts to a straightforward substitution of the Fokker\u2013Planck equation into the Fourier integral de\ufb01ning g(k). (c) Using the method of characteristics, from the left-hand side of the equation in (b), we dt = \u03b3k and k(t) =ce \u03b3t. From the variation of g along the have \u2212\u03b3kdt + dk = 0, i.e. dk characteristic, dg g = \u2212 T m k dk, we obtain g(k, t) = const.e\u2212kBT k2/2m, i.e. g(k, t) = e\u2212kBT k2/2m\u03c6(ke\u2212\u03b3t). With the initial condition, g(k, 0) = e\u2212ikv0, we have \u03c6(k) =e \u2212ikv0 ekBT k2/2m. As a result, we obtain the solution g(k, t) = exp \u2212ikv0e\u2212\u03b3t \u2212 T 2m k2(1 \u2212 e\u22122\u03b3t). Taking the inverse transform, we thus obtain the probability distribution function, \u22121/2 p(v, t) = (1 \u2212 e\u22122\u03b3t) exp \u2212 (v \u2212 v0e\u2212\u03b3t)2 m (1 \u2212 e\u22122\u03b3t) 2 T. 2\u03c0T m + In the limit t \u2192 0, p(v, t) the limit t \u2192 \u221e, p(v, t) P eq", ".(v) = 1 2\u03c0Dt exp[\u2212 (v\u2212v0)2 4Dt m 2\u03c0T exp[\u2212 mv2 2T ]. ], with di\ufb00usion constant D = T m \u03b3. In 10.9.3 Ornstein-Uhlenbeck process revisited We here approach the Ornstein-Uhlenbeck process from the point of view of path integration. As with the harmonic oscillator in quantum mechanics, the path integral approach to the problem is cumbersome (as compared to a direct solution of the Fokker\u2013Planck equation.) However, the path integral of the Ornstein-Uhlenbeck process frequently appears as an approximate integral describing quadratic \ufb02uctuations around the stationary points of more complex integrals (much like the harmonic oscillator integral describes quadratic \ufb02uctuations in quantum mechanical path integrals). For this reason, it is well invested time to study the integral approach to the Ornstein-Uhlenbeck process. Once more, we write P instead of p to distinguish probability distributions from momenta p in phase space. 688 Nonequilibrium (classical) Consider the stochastic equation describing a particle in a harmonic potential with overdamped dynamics, \u02d9q + \u03c92 \u03b32. In the path integral representation, the probability distribution reads \u03b3 q = \u03be with \u03be(t)\u03be(t) = A\u03b4(t \u2212 t) whereA = 2T \u03c92 0 0 P (qf, t|qi, 0) = D[q, p]e\u2212S[q,p], q(t)=qf,q(0)=qi where S = \u2212 t 0 dt (p \u02d9q \u2212 H), and H = \u2212\u03b1pq \u2212 A 2 p2 with \u03b1 = \u03c92 \u03b3. 0 (a) Solve the Hamiltonian equations of motion and determine the corresponding classical action Scl. (b) In order to go beyond the leading semiclassical approximation P (qf, t|qi, 0) \u223c e\u2212S[q,p], we need to determine the \ufb02uctuation determinant. In Section 3.2 we had seen that the \ufb02uctuation determinant of a path integral assumes the form (cf. Eq. 3.28) ; 1\u221a 2\u03c0 \u2212 \u22022Scl \u2202qi\u2202qf, (10.119) where the replacement i \u2192 \u22121 of the pre-factor under the square root relates to the fact that we are now dealing with an imaginary time integral. However, the direct substitution of expression (10.119) for the \ufb02uctuation determinant would be premature: (10.119) applies to theories whose Hamiltonian is of \u201cconventional\u201d type, \u02c6H = \u02c6T + \u02c6U, where \u02c6T and \u02c6U are kinetic and potential energy, respectively. Interpreting q and p as operators with commutation relation [p, q] = 1, bring the Hamiltonian of the theory into this form. Show that this substitution changes the action to Scl \u2192 Scl \u2212 \u03b1 2 t. (If you feel uncomfortable about the interpretation of the function H in terms of operators, derive the Fokker\u2013Planck equation of the path integral, i.e. the analog of the Schr\u00a8odinger equation. Re-arrange operators so as to make the Fokker\u2013Planck operator manifestly hermitean. Then construct the path integral representation of that hermitean representation to arrive at the modi\ufb01ed action stated above. Think why the alternative representation corresponds to a \u201cmid-point\u201d discretization scheme, di\ufb00erent from the It\u02c6o representation used in the construction of the MSRJD functional.) (c) Evaluate the determinant Eq. (10.119) to obtain the result P (qf, t|qi, 0) = 2\u03b1 A 1/2 e \u2212 \u03b1 A (qi\u2212qf e\u03b1t )2 e2\u03b1t\u22121 (1 \u2212 e\u22122\u03b1t)1/2. Notice that the shift \u2212\u03b1t/2 we obtained as a result of the change in the discretization scheme cancels against a long-time divergent factor in the \ufb02uctuation determinant. Answer: (a) The classical equations of motion read \u02d9q = \u2202pH = \u2212\u03b1q \u2212 Ap, \u02d9p = \u2212\u2202qH = \u03b1p, 10.9 Problems 689 with the solution p(t) = p0e\u03b1t and q(t) = ce\u2212\u03b1t \u2212 Ap0 2\u03b1 = qi\u2212qf e\u03b1t one \ufb01nds \u2212 Ap0 some algebra, one", " obtains 1\u2212e2\u03b1t and c = qi\u2212qf e\u2212\u03b1t 2\u03b1 e\u03b1t. From the boundary conditions, 1\u2212e\u22122\u03b1t. Substituted into the classical action, after t Scl = \u2212 0 dt(p \u02d9q + \u03b1pq + p2) = A 2 \u03b1 A (qi \u2212 qf e\u03b1t)2 e2\u03b1t \u2212 1, with \u03b1/A = \u03b3/2T. (You may check for consistency that \u2202S/\u2202qi = pi.) (b) The form of the Hamiltonian used in the path integral is only correct for the It\u02c6o discretization. In order to apply the familiar techniques of quantum mechanics, we have to symmetrise the Hamiltonian by modifying the \u201cvector potential\u201d term, H = \u2212\u03b1pq \u2212 A 2 p2 \u2192 \u2212\u03b1 pq + qp 2 \u2212 \u03b1 2 \u2212 A 2 p2, where the symbol \u201c\u2192\u201d designates a correspondence that would apply if the functions p in the Hamiltonian were to be interpreted as an operator p = \u2202q. This leads to an additional factor in the exponential which is linear in t, Scl \u2192 Scl \u2212 \u03b1 2 t. More rigorously, we may observe that the full information on the path integral is stored in the Fokker\u2013 Planck equation \u2202t + \u03b1\u2202qq + A 2 \u22022 q P = 0. The \u201cFokker\u2013Planck\u201d operator can be rewritten as \u03b1 2 (\u2202qp + p\u2202q) + A 2 \u22022 q + \u03b1 2 = A 2 \u2202q + 2 \u03b1 A q \u2212 \u03b12 2A q2 + \u03b1 2, i.e. in a form resembling a harmonic oscillator Hamilton operator subject to a gauge potential \u03b1q/A. If we now apply the standard rules of path integral construction in quantum mechanics (a mid-point discretization to properly deal with the gauge potential understood) to the new Fokker\u2013Planck/Hamilton operator, we arrive at the imaginary time integral for our Hamiltonian function H (on the level of Hamiltonian functions, ordering is not an issue), shifted by the constant \u03b1t/2. The appearance of this constant can be attributed to a change in the discretization scheme. (c) A straightforward exercise in di\ufb00erentiation. 10.9.4 Directed percolation This problem is of more technical nature. Within the framework provided by the \ufb01eld theory Eq. (10.112), we explore the role of \ufb02uctuations in directed percolation slightly below the upper critical dimension, d = 4 \u2212. This leads us to a set of RG equations, whose physical signi\ufb01cance is discussed in the main text. 690 Nonequilibrium (classical) Phase transitions in the directed percolation universality class are described by the \ufb01eld theory Eq. (10.112). As with our previous examples, contractions of the \ufb01elds \u03c6, \u03c8 w.r.t. the free action of the theory, \u03ba = 0, generate a Green function, \u03c6q\u03c8q = gq\u03b4q,\u2212q, g\u22121 q = \u2212i\u03c9 + Dk2 \u2212 \u03c4, (10.120) where gq relates to the advanced and retarded Green functions used in earlier sections of \u2212q, andq \u2261 (\u03c9, k) is the four-momentum. the text as gq \u2261 g+ q = g\u2212 (a) Perturbation theory: In order to become familiar with the structure of perturbation theory that treats the interaction \u03ba as a small parameter, we consider the form of the perturbative corrections to the Green function and the vertex. Show that the Green function generalizes to g\u22121 \u2192 g\u22121 + \u03a3, where, to lowest order in \u03ba, the self-energy correction reads as \u03a3q = 2\u03ba2 ddk (2\u03c0)d 1 \u2212i\u03c9 + Dk2 + D(k \u2212 k)2 \u2212 2\u03c4. (10.121) Show that the lowest order correction to the coupling constant \u03ba \u2192 \u03ba + \u03b4\u03ba appears at third order and has the form, \u03b4\u03ba \u22122\u03ba3 ddk (2\u03c0)d 1 (Dk2 \u2212 \u03c4 )2. (10.122) Convince yourself that the diagrams contributing to the renormalization of \u03ba and \u03c4 correspond to the \u201ccon\ufb01guration space\u201d processes exempli\ufb01ed by the \ufb01gure above. (b) Renormalization group: Consider the theory regularized with a hard cuto\ufb00 \u039b for the (sp", "atial) momentum integral. We rescale (spatial) momentum by this cuto\ufb00 k \u2192 k/\u039b which means that the momentum integrals now extend over the support |k| \u22641 and all coupling constants are measured in units \u039bdx, whered x is the relevant engineering dimension. Integrate out perturbatively the spatial fast modes within the momentum shell (1, 1/b) with 0 < ln b 1. Show that in spatial dimension d = 4 \u2212, this modi\ufb01es the coupling constants in the following way ln b ln b,, + 1 x ln b, \u03ba \u2192 \u03ba (1 \u2212 2x ln b), \u03c4 \u2192 \u03c4 1 \u2212 where we introduced D \u03c4 x \u2261 S4 (2\u03c0)4 2, \u03ba D (10.123) and S4 is the area of the four-dimensional unit-sphere. Next, rescale coordinates and \ufb01elds as q \u2192 q/b, \u03c9 \u2192 \u03c9/bz, \u03c6 \u2192 \u03c6b\u2212 4\u2212 2 +\u03c7. Choose the dynamical exponent z and the 10.9 Problems 691 Figure 10.16 Skeleton structure of the two diagrams contributing to the renormalization of the propagator (left) and interaction vertex (right), respectively. \ufb01eld renormalization exponent \u03c7 so as to make the di\ufb00usion constant D and the coe\ufb03cient of frequency invariant under renormalization. Show that this condition generates the equations. The \u201cmass term\u201d \u03c4 and the interaction constant \u03ba \ufb02ow according to Eq. 10.115. The meaning of these equations is discussed in the main text. Answer: (a) In Fourier basis, the interaction contribution to the action S = S0 + Sint is given by Sint[\u03c6, \u03c8] = \u03ba dqdq \u03c8q\u03c6q (\u03c6\u2212q\u2212q \u2212 \u03c8\u2212q\u2212q ), where the measure dq = d\u03c9ddk/(2\u03c0)d+1. Expanding the action to second order in \u03ba, and applying the contraction rule Eq. (10.120), we obtain the self energy \u03a3q = \u22122\u03ba2 dq (gq\u2212q gq + gq+q gq ). d\u03c9 over gq+q qq vanishes because the integrand falls o\ufb00 as The frequency integration \u223c 1/\u03c92 and has no poles in the upper complex plane. Doing the contour integration over the remaining contribution, we readily obtain Eq. (10.121). The diagrammatic representation of this term is shown in Fig. 10.16, where the external \ufb01eld vertices (not contributing to the self energy) have been included for the sake of notational clarity. To obtain the renormalisation of the coupling constant of the interaction, \u03ba it is necessary to develop the third order of perturbation theory. The Wick contraction of these terms (up to the \u201cexternal \ufb01eld vertices\u201d entering the interaction operator, \u223c \u03c8\u03c62 and \u223c \u03c82\u03c6) obtains two one-loop diagrams. In one of these all Green functions have their poles on one side of the real axis, which implies vanishing upon frequency integration. The survivor diagram, whose graphical representation is shown in Fig. 10.16 has the analytic representation (\u22121)3 3! S3 int \u2192 \u2212 243 3! \u03ba3 dqdqdq \u03c6q \u03c8q (\u2212\u03c6q\u2212q\u2212q + \u03c8q\u2212q\u2212q ) dp g2 pg\u2212p, 692 Nonequilibrium (classical) where we neglected the \u201csmall\u201d momenta q, q, q in the arguments of the \u201cfast\u201d Green functions gp. The required renormalisation of \u03b4\u03ba thus reads \u03b4\u03ba = \u2212(2\u03ba)3 q g\u2212q. Doing the integral over frequency, we obtain Eq. (10.122). dq g2 (b) Consider the contribution \u03b4\u03a3 to Eq. (10.121) due to the integration over the fast momen1/b<|k|\u22641 ddk. An expansion of the self energy in \u201csmall\u201d correc- f ddk \u2261 tum layer tions to the fast momenta obtains \u03b4\u03a3q \u22122\u03ba2 + (i\u03c9 + 2\u03c4 )I\u22124 (2D)2 + k2 4D 2 d \u2212 1 I\u22124 +..., I\u22122 2D ddk (2\u03c0)d kn. Likewise, the fast momentum contribution to the where we introduced In \u2261 coupling constant correction Eq. (10.122) reads \u03b4\u03ba", " \u2212 2\u03ba3 f D2 I\u22124. In dimensions 4 \u2212, we have I\u22122 I\u22124 = S4 (2\u03c0)4 ln b. Substituting this result into the fast-\ufb02uctuation induced change of the Green function, g\u22121 \u2192 g\u22121 \u2212 \u03b4\u03a3, we obtain the required renormalization of coupling constants. The re-scaling of coordinates and \ufb01elds modi\ufb01es the renormalization of coupling con- stants according to \u03c9 \u2192 \u03c9 1 + \u03c4 \u2192 \u03c4 1 + ln b 2\u03c7 \u2212 x 2 2\u03c7 + z \u2212 Dx \u03c4 1 +, D \u2192 D 2\u03c7 + z \u2212 2 \u2212 x 4 ln b, \u2212 x ln b, \u03ba \u2192 \u03ba 1 + + 3\u03c7 + z \u2212 2 \u2212 2x ln b. 2 Invariance of the \ufb01rst two terms generates the required conditions on \u03c7 and z. We \ufb01nally substitute \u03c7 and z into the remaining two equations to obtain the RG equations (10.115). 11 Nonequilibrium (quantum) In the previous chapter we have seen that departures from equilibrium generate a plethora of new phenomena in statistical physics. While our discussion so far has been limited to classical phenomena, in this chapter we want to ask how quantum mechanics interferes with the conditions of nonequilibrium. Experience shows that, in some cases, nonequilibrium quantum systems can be understood in terms of reasonably straightforward quantization of their classical limits. In others, an interplay of quantum coherence and nonequilibrium driving leads to entirely new phenomena \u2013 such as lasing, a paradigm for a steady state nonequilibrium quantum system. At any rate, we have every reason to anticipate that quantum nonequilibrium physics will be as diverse and colourful as its classical counterpart. Even minor departures from equilibrium call for a theoretical description entirely di\ufb00erent from the \u2018Matsubara formalism\u2019 developed in earlier chapters of this book. The applicability of the latter is rigidly tied to the existence of a quantum grand canonical density operator, the hallmark of a many particle equilibrium system. But how then, can an functional theory of nonequilibrium quantum systems be developed? An obvious idea would be to once more go through the di\ufb00erent elements of classical nonequilibrium theory developed in the previous chapter, and subject every one of them to an individual quantization scheme. Luckily, there are more e\ufb03cient ways to achieve our goal: some decades ago, a many particle formalism suitable to describe nonequilibrium systems under the most general conditions1 was introduced by Keldysh.2 For a number of reasons, the Keldysh formalism will be the principal tool in our approach to quantum nonequilibrium physics: (i) it is highly \ufb02exible and can be applied to study the physics of practically any quantum system, in and out of equilibrium. (ii) Although the Keldysh technique was introduced in a pre-path integral era, it is tailor-made to functional integral formulations. (iii) The classical limit of that Keldysh \ufb01eld integral turns out to be the MSRJD functional integral extensively discussed in the previous chapter. In this sense, the Keldysh functional will be our \u2018theory of everything\u2019; from it the entire body of nonequilibrium theory, both quantum and classical can be derived by reduction. The discussion above motivates a strategy di\ufb00erent from that pursued in our introductory discussion of nonequilibrium physics above. In a \u201ctop-down\u201d approach, we begin by 1 By \u201cmost general,\u201d we mean that the Keldysh formalism enjoys applicability independent of the distribution (operator) describing the state of the system under consideration. 2 L. V. Keldysh, Diagram technique for nonequilibrium processes, Sov. Phys. JETP 20, 1018-26 (1965). 693 694 Nonequilibrium (quantum) a formal introduction of the Keldysh formalism. We alternatively interpret the theory as a generalization of the Matsubara formalism, and as a quantum extension of the MSRJD \ufb01eld integral. After these conceptual structures have been thoroughly introduced, we will turn to the discussion of applications. INFO In reference to the stochastic aspects of classical nonequilibrium processes, we began the last chapter with a short treatise on probability. Quantum mechanics, on the other hand, is a theory that is intrinsically probabilistic; the most we can hope for is to obtain results on the quantum probability to measure certain values of observables. The theory of quantum nonequilibrium processes, therefore, involves two \u201clayers\u201d of stochasticity, quantum probability and the extrinsic", " stochasticity of nonequilibrium processes. As it is important not to confuse these two, we begin by asking what sort of probability distributions might suitably generalize the distributions employed in the previous, classical chapter. To start, let us consider a classically deterministic system, prepared in a de\ufb01nite quantum state |\u03a8. The (quantum) expectation value to measure a certain observable is then given by ( \u02c6X) \u2261 \u03a8| \u02c6X|\u03a8 = n Xn|\u03a8n|2, whereX n is the nth eigenvalue of X (the generalization to a continuous spectrum is straightforward), \u03a8n \u2261 n|\u03a8, and|n is the nth eigenstate. We denote the quantum expectation value by ordinary brackets to distinguish it from the average over \ufb02uctuations below. The equation above suggests an interpretation of P \u02c6X (x) \u2261 n \u03b4(x \u2212 Xn)|\u03a8n|2 as a \u201cquantum probability\u201d distribution of the operator \u02c6X. This notation is suggestive in that it allows us to express the computation of quantum observables in formal analogy to the probabilistic formalism used in the previous chapter. For example, ( \u02c6X n) = dx P \u02c6X (x)xn, and P \u02c6X (x) = (\u03b4(x \u2212 \u02c6X)), etc. While the quantity P \u02c6X underpins the probabilistic nature of quantum mechanics, its usefulness rapidly reaches its limits.3 For our present purposes, an alternative representation of quantum expectation values will prove more useful: the above expression may equivalently be written as ( \u02c6X) = tr (\u02c6\u03c1|\u03a8 \u02c6X), where \u02c6\u03c1|\u03a8 \u2261 |\u03a8\u03a8| is a particular type of a quantum density operator, viz. a projector onto the one-dimensional space spanned by |\u03a8. The alternative representation can be readily generalized to the \u2013 more realistic \u2013 situation where the state of the system is not known with certainty. a pa|\u03a8a\u03a8a|, where{|\u03a8 a} is a complete orthogonal set, and the positive In general, \u02c6\u03c1 = a pa = 1, represent the probability to \ufb01nd the system in state |\u03a8a. In a coe\ufb03cients pa, many particle system, the inevitable presence of \ufb02uctuations \u2013 cf. the principles discussed in the previous chapter \u2013 may render the coe\ufb03cients pa e\ufb00ectively random. These \ufb02uctuations may be externally imposed, or caused by integration over microscopic degrees of freedom of the system. Summarily denoting the average over \ufb02uctuations by... f, we then have 4 5 \u02c6X \u2261 ( \u02c6X)f = tr (\u02c6\u03c1 \u02c6X) = tr (\u02c6\u03c1f \u02c6X) = f a paf \u03a8a| \u02c6X|\u03a8a. This representation illustrates the division of work between quantum and statistical \ufb02uctuations. In a canonical (basis-invariant) manner, all aspects of quantum mechanics \u2013 interference, wave coherence, etc. \u2013 are encapsulated in the mathematical properties of the density operator and 3 The function P \u02c6X suggests deceptively that the basic information on the state of a system is stored in the (positive de\ufb01nite) probability |\u03a8n|2. However, we know that the essence of quantum mechanics resides in wave function amplitudes. For example, it is not possible to unambiguously generalize the notion of P to that of a joint probability P \u02c6X, \u02c6Y (x, y) of two non-commutative operators. This means that positive de\ufb01nite probability distributions such as the one above will not su\ufb03ce to describe quantum processes. 11.1 Prelude: Quantum master equation 695 the trace operation. The statistics of \ufb02uctuations (both thermal, and externally imposed) is contained in the average over coe\ufb03cients,... f. We repeat that these latter \ufb02uctuations may e\ufb00ectively be generated by a quantum average (i.e. a trace operation) over microscopic degrees of freedom of the system. The discussion above suggests the following transcription of the probabilistic elements of classical nonequilibrium mechanics into quantum mechanics: Notice that if and only if \u02c6\u03c1 is an classical variable X values of X, x probability distribution p quantum hermitian operator \u02c6X eigenvalues of \u02c6X, x density operator \u02c6\u03c1 moments X n = dx p(x", ")xn moments \u02c6X n = tr(\u02c6\u03c1 \u02c6X n) operator function of \u02c6X (which will typically happen if \u02c6X \u2261 \u02c6H is energy), \u02c6X n = tr(\u02c6\u03c1 \u02c6X n) = dx \u03c1(x)xn, where \u03c1(x) are the eigenvalues of \u03c1( \u02c6X). In this particular case, we have a normalized ( dx \u03c1(x) = 1) and positive quantum probability distribution. In general, however, quantum statistical mechanics cannot be described in terms of positive de\ufb01nite probability functions. Rather, the relevant information is stored in density operators, and it is these objects we need to understand. 11.1 Prelude: Quantum master equation Before turning to the systematic construction of the Keldysh formalism, let us introduce a few elements of the quantum mechanics of nonequilibrium systems in more elementary terms. This discussion will get us acquainted with various \u201cquick and dirty\u201d routes to the description of nonequilibrium systems but will also motivate the construction of the more systematic theories below. (Readers wishing to proceed in a maximally streamlined manner may skip this discussion and directly turn to section 11.2.1.) We consider the standard setup of a \u201csystem\u201d coupled to a \u201cbath\u201d (cf. the \ufb01gure.) The bath may alternatively be interpreted as a large system of external degrees of freedom (e.g. the oscillator modes of an external electromagnetic \ufb01eld), or in terms of microscopic internal degrees of freedom a\ufb00ecting the more macroscopic (but still quantum) dynamics of the degrees of freedom we consider to be the \u201csystem.\u201d By straightforward adaptation of the principles discussed in the previous chapter, we interpret the degrees of freedom of the bath as unobservable. Integration over these degrees of freedom produces an e\ufb00ective, and foreseeably irreversible dynamics of the system. Our immediate goal is the derivation of the equations of motion governing this e\ufb00ective dynamics. 696 Nonequilibrium (quantum) 11.1.1 Derivation of the master equation Consider the density operator \u02c6\u03c1 describing the state of the total system in the product Hilbert space H \u2261 Hs \u2297 Hb of system (s) and bath (b). The reduced information in which we are interested is stored in the quantity \u02c6\u03c1s \u2261 trb(\u02c6\u03c1) : Hs \u2192 Hs, where trb denotes the trace (\u201cintegration\u201d) over the Hilbert space of the bath. The reduced density matrix \u02c6\u03c1s is a linear operator acting in the system Hilbert space Hs. The dynamics of the full system is controlled by the Hamiltonian operator \u02c6H \u2261 \u02c6Hs + \u02c6Hb + \u03b1 \u02c6Hi, where the coupling between system and bath, \u02c6Hi, is weighted by a dimensionless coupling constant \u03b1. We assume that, at some initial time t = 0 (a) the system and bath had been independent, and (b) that the bath was in a state of thermal equilibrium, \u02c6\u03c1(0) = \u02c6\u03c1s(0) \u2297 \u02c6\u03c1eq b. We now aim to formulate an equation of motion for the reduced density matrix. Re\ufb02ecting the applied relevance of this problem, plenty of di\ufb00erent strategies to achieve this goal have been formulated. Here, we derive the equation by a formally exact projector formalism,4 and then compare to other approaches. The dynamics of the density operator is governed by the equation of motion (\u2202t \u2212 \u02c6L)\u02c6\u03c1 = 0, where \u02c6L = \u02c6Ls + \u02c6Lb + \u03b1 \u02c6Li, and \u02c6Ls,b,i \u2261 \u2212i[ \u02c6Hs,b,i, ] are \u201cquantum Liouville operators.\u201d We next introduce the projector P \u2261 \u02c6\u03c1eq b trb(. ), where the trace operation acts on everything to the right, and unit normalization of \u02c6\u03c1eq is b assumed. (As an exercise, consider why this is a projector.) To simplify the derivation, we \u02c6Hi) = 0. assume that the thermal trace of the interaction operator over Hb vanishes,5 trb(\u02c6\u03c1eq b Our projector thus obeys the equations, P \u02c6Lb = \u02c6LbP = 0, [ \u02c6Ls, P] = 0, P \u02c6LiP = 0, (11.1) where the \ufb01rst follows from the cyclic invariance of the trace, the", " second should be obvious, and the third expresses the presumed vanishing of the interaction under the bath-trace. Introducing the complementary projector Q \u2261id \u2212 P, and the shorthand notation \u02c6\u03c1P \u2261 P \u02c6\u03c1, \u02c6\u03c1Q \u2261 Q\u02c6\u03c1, the quantum Liouville equation may now be split into two, \u2202t \u02c6\u03c1P = \u02c6Ls \u02c6\u03c1P + \u03b1P \u02c6Li \u02c6\u03c1Q, \u2202t \u02c6\u03c1Q = Q \u02c6L\u02c6\u03c1Q + \u03b1Q \u02c6Li \u02c6\u03c1P. (11.2) The second of these equations is solved by (exercise) \u02c6\u03c1Q(t) = \u03b1 t 0 dt etQ \u02c6LQ \u02c6Li \u02c6\u03c1P (t \u2212 t). 4 S. Nakajima, On quantum theory of transport phenomena: steady di\ufb00usion, Prog. Theory. Phys. 20, 948-59 (1958); R. Zwanzig, Ensemble method in the theory of irreversibility, J. Chem. Phys. 33, 1338-41 (1960). 5 Should this condition not be met by the interaction under consideration, the term trb( \u02c6\u03c1eq b interpreted as part of the system Hamiltonian operator. \u02c6Hi) = 0 can be 11.1 Prelude: Quantum master equation 697 Substitution of this result into the \ufb01rst equation obtains \u2202t \u02c6\u03c1P = \u02c6Ls \u02c6\u03c1P + \u03b12 t 0 dt P \u02c6LietQ \u02c6LQ \u02c6Li \u02c6\u03c1P (t \u2212 t), (11.3) or, employing the projector operation P, \u2202t \u02c6\u03c1s = \u02c6Ls \u02c6\u03c1s + \u03b12 > dt t 0 \u02c6LietQ \u02c6LQ \u02c6Li? b \u02c6\u03c1s(t \u2212 t), (11.4) where... b \u2261 trb((...)\u02c6\u03c1eq b ). Equation (11.4) is commonly denoted a generalized master equation. But in fact, the resemblance to a master equation is only formal. Rather, this equation (together with a complementary equation for \u02c6\u03c1Q) is an exact reformulation of the full quantum theory; the complete information on quantum unitary time evolution resides in the time non-local memory kernel on the right hand side. In contrast, the master equation generally describes timelocal (Markovian) and irreversible processes. To obtain a \u201ctrue\u201d master equation one may eliminate information in what e\ufb00ectively amounts to a Markovian approximation. Assuming that the time scales over which \u03c1s changes are large in comparison to the relaxation times t of the integral kernel, Eq. (11.4) reduces to a Markovian equation \u2202t \u02c6\u03c1s = ( \u02c6Ls + \u03b12 \u02c6X)\u02c6\u03c1s(t), where \u02c6X \u2261 \u221e 0 dt \u02c6LietQ \u02c6LQ \u02c6Lib. It is also customary to assume weak coupling, or small \u03b1, so that the \u03b1-dependence of the exponent can be neglected (a variant of a \u201cBorn approximation\u201d). The resulting simpli\ufb01cations are best discussed in the language of the prototypical equation (11.3): de\ufb01ning \u02c6L0 \u2261 \u02c6Ls + \u02c6Lb, the integral kernel in that equation reduces to P \u02c6LietQ \u02c6L0Q \u02c6Li \u02c6\u03c1P (t \u2212 t) = P \u02c6Liet \u02c6L0 \u02c6Li \u02c6\u03c1P (t \u2212 t). EXERCISE Use Eq. (11.1) above to verify this relation. Further, de\ufb01ning \u02c6H0 = \u02c6Hs + \u02c6Hb, and \u02c6O(t) \u2261 eit \u02c6H0 \u02c6Oe\u2212it \u02c6H0 for the interaction picture time representation of an operator \u02c6O, we have P \u02c6Liet \u02c6L0 \u02c6Li \u02c6\u03c1P (t \u2212 t) P \u02c6Li \u02c6Li(\u2212t) \u02c6\u03c1P (t \u2212 t). EXERCISE For an arbitrary operator \u02c6O, show that et \u02c6L0 \u02c6O = \u02c6O(t)et \u02c6L0. The equality above is approximate because", " we neglect the action of et \u02c6L0 on the (slow) operator \u02c6\u03c1P. Using this result, we \ufb01nd that Eq. (11.4) reduces to the (Markovian) quantum master equation \u2202t \u02c6\u03c1s = \u02c6Ls + \u03b12 > dt t 0 \u02c6Li(0) \u02c6Li(\u2212t)? \u02c6\u03c1s. b (11.5) 698 Nonequilibrium (quantum) 11.1.2 Example: oscillator coupled to a bath It is instructive to study the meaning of the master equation on a concrete example: consider a harmonic oscillator \u2013 the system \u2013 coupled to an assembly of other oscillators, the bath. We describe this setup in terms of the Hamiltonian \u02c6H = \u02c6Hs + \u02c6Hb + \u03b1 \u02c6Hi, where \u02c6Hs = a\u2020a +, \u02c6Hb = \u03c9k a \u2020 kak +, \u02c6Hi = cka \u2020 ka + h.c.. (11.6) 1 2 1 2 k k Using the fact that a(t) = e\u2212ita and ak(t) = e\u2212i\u03c9ktak, it is straightforward to verify that \u02c6\u03c1s = \u2212\u03b12 b \u221e |ck|2 dt 0 k \u02c6\u03c1saa\u2020 \u2212 a\u2020 \u02c6\u03c1sa aa\u2020 \u02c6\u03c1s \u2212 a\u2020 \u02c6\u03c1sa a\u2020a\u02c6\u03c1s \u2212 a\u02c6\u03c1sa\u2020 + \u02c6nk + 1 \u02c6\u03c1sa\u2020a \u2212 a\u02c6\u03c1sa\u2020 + \u02c6nk + 1 \u03b12 t dt 0 \u00d7? > \u02c6Li(0) \u02c6Li(\u2212t) e\u2212i(\u03c9k\u2212)t e+i(\u03c9k\u2212)t |ck|2\u03b4( \u2212 \u03c9k) \u02c6nk \u02c6nk = \u03b12\u03c0 k \u02c6nk \u00d7 \u03c0\u03b12|c|2\u03c1() 2a\u2020 \u02c6\u03c1sa \u2212 \u02c6\u03c1saa\u2020 \u2212 aa\u2020 \u02c6\u03c1s \u02c6n 2a\u2020 \u02c6\u03c1sa \u2212 \u02c6\u03c1saa\u2020 \u2212 aa\u2020 \u02c6\u03c1s + \u02c6nk + 1 2a\u02c6\u03c1sa\u2020 \u2212 \u02c6\u03c1sa\u2020a \u2212 a\u2020a\u02c6\u03c1s + \u02c6n + 1 2a\u02c6\u03c1sa\u2020 \u2212 \u02c6\u03c1sa\u2020a \u2212 a\u2020a\u02c6\u03c1s. \u2020 Here, \u02c6nk = a kak is the number operator of the bath, and we have neglected the subscript \u2018b\u2019 in... \u2261... b. In the fourth equality we introduced the spectral density of the bath, depend only on the \u03c1(\u03c9) \u2261 energy of the reference state. k \u03b4(\u03c9 \u2212 \u03c9k), and we assumed thatc k \u2261 c\u03c9k and \u02c6nk \u2261 \u02c6n\u03c9k EXERCISE In the second equality above we assumed that [\u02c6\u03c1s, a\u2020a] = 0. Verify that a relaxation of that assumption will lead to terms of the structure \u223c P is the principal value integral. Interpret these expressions in terms of an energy shift of the oscillator energy due to virtual transitions into the bath. \u02c6n\u03c9[a\u2020a, \u02c6\u03c1s], where P d\u03c9 |c\u03c9 |2\u03c1(\u03c9) \u03c9\u2212 Assuming once more commutativity of \u02c6\u03c1s with \u02c6n \u2261 a\u2020a, we have \u02c6Ls \u02c6\u03c1s = 0 and thence % \u2202t\u03c1s = \u03c0\u03b12|c|2\u03c1() + \u02c6n + 1 \u02c6n 2a\u2020 \u02c6\u03c1sa \u2212 \u02c6\u03c1saa\u2020 \u2212 aa\u2020 \u02c6\u03c1s & 2a\u02c6\u03c1sa\u2020 \u2212 \u02c6\u03c1sa\u2020a \u2212 a\u2020a\u02c6\u03c1s. (11.7) But what is the meaning of this equation? In fact, it will be instructive to look at Eq. (11.7) from a number of di\ufb00erent perspectives: the commutativity [\u02c6\u03c1s, \u02c6n] = 0 implies that \u02c6\u03c1s = \u02c6\u03c1s(\u02c6n). Using commutator relations such as a\u2020 \u02c6\u03c1s(\u02c6n) = \u02c6\u03c1s(\u02c6n \u2212 1)a\u2020, it is then straightforward to bring Eq. (11.7", ") into the form # \u2202t\u03c1s(\u02c6n) = 2\u03c0\u03b12|c|2\u03c1() \u02c6n (\u02c6n\u02c6\u03c1s(\u02c6n \u2212 1) \u2212 (\u02c6n + 1)\u02c6\u03c1(\u02c6n))! absorption $ + \u02c6n + 1 ((\u02c6n + 1)\u02c6\u03c1s(\u02c6n + 1)\u2212 \u02c6n\u02c6\u03c1s(\u02c6n))! emission. 11.1 Prelude: Quantum master equation 699 The right\u2013hand side of this equation now a\ufb00ords a straightforward interpretation in terms of absorption and emission of bath bosons (see \ufb01gure). The absorption term is proportional to the number of available bosons (\u02c6n) resonant with the oscillator frequency, the golden rule transition rate, \u223c |c|2\u03c1(), and to the number of absorbing bosons (plus one). Importantly, the absorption process contributes to the right\u2013hand side in a manner symptomatic of a master equation. The density operator (or better to say, the eigenvalue) \u02c6\u03c1s(n) increases due to an \u201cin\u201d process \u221d \u02c6\u03c1s(\u02c6n \u2212 1) and it diminishes due to an \u201cout\u201d process \u221d \u02c6\u03c1s(\u02c6n). There is also an emission process wherein bath bosons are resonantly created from oscillator bosons. Emission, too, enters the master equation as a sum of an in and an out process. A few more comments can be made on the structure of the master equation: The concerted appearance of in- and out- terms implies the conservation of the trace of the density operator (the quantum analogue of the conservation of probability): \u2202ttr(\u02c6\u03c1s) = 0 (exercise). Assume the bath is in equilibrium, \u02c6n = (exp(\u03b2) \u2212 1)\u22121. It is then straightforward to verify that the distribution \u02c6\u03c1s becomes stationary (the right\u2013hand side of the master equation vanishes) on the equilibrium con\ufb01guration \u02c6\u03c1s(\u02c6n) = Z \u22121e\u2212\u03b2\u02c6n, where Z = (1 \u2212 e\u2212\u03b2)\u22121 is the normalizing partition function. Interaction with the bath \u201cthermalizes\u201d the system at a temperature set by the bath temperature. Notice that equilibration relies on the interplay of emission and absorption processes. It is worth noting that the in-terms (out-terms) in the master equation derive from the terms \u223c a\u2020 \u02c6\u03c1sa, etc. (\u223c \u02c6\u03c1sa\u2020a, etc.) in the parent equation (11.7). To understand the heuristics behind this observation, recall that the interaction picture time evolution of the density operator is given by (symbolic notation throughout) \u02c6\u03c1s(t) = \u02c6U (t)\u02c6\u03c1s(0) \u02c6U \u2020(t), where \u02c6U is the time evolution operator. We may visualize the time evolution described by these operators by two lines in \u201ctime space,\u201d one directed forward and one backward, cf. Fig. 11.1. If we now probe the incremental change of \u02c6\u03c1s in time, i.e. compare \u02c6\u03c1s(t + \u0394t) with \u02c6\u03c1s(t), we may linearly expand either \u02c6U or \u02c6U \u2020 in \u03b1 \u02c6Hi. This produces a factor \u03b1 \u02c6Hi(t) acting either to the left or to the right of the time evolution operator, or an interaction Hamiltonian acting at the end of the forward or the backward time contour. However, to obtain a non-vanishing result, we need to expand to one more order. This gives a factor dt \u02c6Hi(t) which, again, may act to the left or the right of the density operator, or on \u223c \u03b1 the forward or backward contour. All in all, we obtain four contributions, as visualized in Fig. 11.1. Now, the contributions where \u02c6Hi acts on di\ufb00erent time contours (a) are formally described by a density operator sandwiched between creation and annihilation operators, and (b) physically change the state of the system (c) alluding to diagrammatic language, they resemble \u2018vertex corrections\u201d in a \u201ctwo-particle propagator. These are the in-terms. 700 Nonequilibrium (quantum) Figure 11.1 Evolution of the density operator (shaded box) along the double time contour. Left: out-processes (\u201cself-energy corrections\u201d),", " right: in-processes (\u201cvertex corrections\u201d). For the motivation for the \u201cleft\u201d orientation of the time arrow, see below. Conversely, the contributions where \u02c6Hi acts twice on the same contour, (a) are described by pairs of operators acting on one side of the density operator, and (b) do not change the state of the system (more precisely, it is \ufb01rst changed at t and then changed back at t.) (c) These out\u2013terms resemble self-energy corrections in a two-particle propagator. Indeed, we have seen on various occasions that the (imaginary part) of a self-energy correction represents a decay rate, i.e. the rate of an \u201cout-process.\u201d It is worth pointing out that our discussion of the quantum master equation so far has been rather super\ufb01cial (for a more substantial exposition, see Weiss,6 van Kampen,7 and Haake8). Its primary purpose was to motivate a number of concepts relevant to the physics of quantum nonequilibrium phenomena. We next turn to the construction of a \ufb01eld integral based theory which will enable us to describe nonequilibrium phenomena in a much broader setting. 11.2 Keldysh formalism: basics 11.2.1 The idea Before venturing into the construction of the Keldysh nonequilibrium theory, let us \ufb01rst try to summarize what kind of theory we are after: 1. Previously, we have argued that the object generalizing the probability distributions which pervade the previous chapter is the quantum density operator \u02c6\u03c1. Expectation values of observables are evaluated by taking the traces of hermitian operators against the dynamically evolved density operator. 6 U. Weiss, Quantum Dissipative Systems, (World Scienti\ufb01c Publishing, 1993). 7 N.G. van Kampen, Stochastic Processes in Physics and Chemistry, (Elsevier,1992). 8 F. Haake, Quantum Signatures of Chaos (Springer-Verlag, 2001). 11.2 Keldysh formalism: basics 701 2. That dynamics is described by the (adjoint action of the) quantum time evolution operator, \u02c6\u03c1 \u2192 U (t)\u02c6\u03c1 \u02c6U (t)\u2020. We may interpret \u02c6U ( \u02c6U \u2020) as the descriptor of quantum time evolution forward (backward) in time. The notion of \u201cchronological directedness\u201d becomes manifest in the interaction-picture representation of the evolution operator \u02c6U as a time ordered exponential \u02c6U = T exp(\u2212i 0 dt \u02c6Hint(t)). Conversely, we may generate \u02c6U \u2020 by time ordering along a reversed temporal contour, \u02c6U \u2020 = t dt \u02c6Hint(t)), where \u02dcT puts operators to the left that are chronologically \u02dcT exp(\u2212i earlier. 0 t 3. In many applications we may safely assume that, at some initial time (set as t = \u2212t0 for de\ufb01niteness) the system of interest was in a known state. For example, \u02c6\u03c1(\u2212t0) = exp(\u2212\u03b2 \u02c6H) may have been an equilibrium distribution, or \u02c6\u03c1(\u2212t0) = \u02c6\u03c1s \u2297 \u02c6\u03c1b in the case of a composite involving a system coupled to a bath. It is assumed that the interactions/correlations/nonequilibrium conditions rendering the situation nontrivial are slowly switched on at some later time. 4. From our experience with classical probability theory, we know that it is convenient to work with generating functions. For instance, in equilibrium statistical mechanics, one might wish to compute the Maxwell\u2013Boltzmann generating functional (symbolic notation) G(h) \u2261 Z \u22121 exp(\u2212\u03b2H) is the partition function, denotes an integration over all states of the system, H is the Hamiltonian and X an observable of interest. Di\ufb00erentiation with respect to the sources h generates moments of X and we have the normalization G(0) = 1. exp(\u2212\u03b2(H \u2212 hX)), where Z = How can the concept of generating functions be generalized to the quantum nonequilibrium case? Generalization to \u201cquantum\u201d is relatively obvious: replace functions by operators, and integrals by traces. But what about \u201cnonequilibrium\u201d? From what has been said above, it should be evident that we need to endow some known quantum distribution \u02c6\u03c1(0) with a sense of dynamics, in a manner that leads to a properly normalized, and generally time-dependent generating functional. In the mid\u20131960s, Keldysh formulated", " a theory2 that solved the problems alluded to above in one stroke.9 At \ufb01rst sight, Keldysh\u2019s conceptual ansatz may look strange, if not downright silly: suppose we are given an initial density operator \u02c6\u03c1 which described our system at some time t \u2261 \u2212t0 \u2192 \u2212\u221e in the distant past when the world was nice and easy \u2013 no interactions, stochastic \ufb02uctua- Leonid. V. Keldysh 1931\u2013 the LebeFormer Director of dev Physical Institute, Moscow. Keldysh has made seminal contributions to solid state theory from the physics of electron\u2013hole excitations in semiconductors to the development of techniques to explore quantum systems driven from equilibrium. Recipient of the Lenin Prize, Lomonosov Prize, and the Hewlett-Packard Prize. 9 For earlier work in the same spirit, see J. Schwinger, Brownian motion of a quantum oscillator, J. Math. Phys. 2, 407-32 (1961), L.P. Kadano\ufb00 and G. Baym, Quantum Statistical Mechanics (Benjamin, New York, 1962), R. P. Feynman and F. L. Vernon, The theory of a general quantum system interacting with a linear dissipative system, Ann. Phys. 24, 118-73 1963. 702 Nonequilibrium (quantum) s S/2 \u02c6H(t) \u2212 \u02c6H(t) t t0 \u02c6H \u02c6H0 \u2212 \u02c6H \u2212t0 Figure 11.2 Top: parameter dependence of the system Hamiltonian along an \u201cabstract\u201d Keldysh contour. Bottom: embedding of the contour in the complex time plane. tions, or similar complications. Our goal is to endow the density operator with a sense of dynamics. To this end, consider an abstract parameter interval, [0, S]. On this interval, we de\ufb01ne an operator valued function \u02c6H(s) according to the following rules: (i) \u02c6H(0) = \u2212 \u02c6H(S) = \u02c6H0, where \u02c6H0 is the \u201ctrivial\u201d Hamiltonian operator governing the system at time \u2212t0. (ii) As s increases, the non-trivial elements of the dynamics are gradually (adiabatically) switched on (cf. Fig. 11.2) until the Hamiltonian has reached its full form \u02c6H(s) = \u02c6H. (iii) Upon approaching the center of the parameter interval, we let the Hamiltonian diminish down to \u02c6H(S/2) = 0. (iv) Beyond s = S/2, the pro\ufb01le of \u02c6H is anti-symetrically continued, i.e. \u02c6H(s) = \u2212 \u02c6H(S \u2212 s) fors \u2208 [S/2, S]. Now consider the unitary operator \u02c6U \u2261 Tsexp(\u2212i 0 ds \u02c6H(s)), where Ts orders along the s-contour. This operator is but a complicated representation of unity, \u02c6U = 1. Heuristically, the triviality of \u02c6U follows from the fact that the dynamic \u201cphases\u201d \u223c ds \u02c6H(s) accumulated along the forward branch of the contour cancel against those from the backward branch. ds \u02c6H(s)) to More formally, one may consider the generalization \u02c6U (s2, s1) \u2261 Tsexp(\u2212i arbitrary segments on the contour. These operators satisfy the composition law (proof is analogous to the one for conventional time ordered operators) \u02c6U (s1, s2) \u02c6U (s2, s3) = \u02c6U (s1, s3), where s1 > s2 > s3. Thus, \u02c6U \u2261 \u02c6U (S, 0) = \u02c6U (S, S/2) \u02c6U (S/2, 0). But \u02c6U (S/2, 0) = \u02c6U (S, S/2)\u22121, by construction of the Hamiltonian. The equality \u02c6U = 1 trivially implies s2 s1 S Z \u2261 tr( \u02c6U \u02c6\u03c1) = 1, (11.8) on account of the normalization of \u02c6\u03c1. We therefore have an object that is obtained by dynamical evolution (1) of a known initial density operator (3), is time ordered (2), and manifestly normalized (4). What is not clear, however, is what the \ufffd", "\ufffdpartition function\u201d Z might be good for! 11.2 Keldysh formalism: basics 703 At the same time, the operator \u02c6U \u02c6\u03c1 is not too di\ufb00erent from the objects studied in the previous section. There, we had considered dynamically evolved density operators \u02c6U \u02c6\u03c1 \u02c6U \u2020, \u02c6U \u2261 \u02c6U (t0, \u2212t0). The trace over these operators, tr( \u02c6U \u02c6\u03c1 \u02c6U \u2020) = tr( \u02c6U \u2020 \u02c6U \u02c6\u03c1) =Z can be interpreted as a variant of the above partition function, where the curve [0, S] is parameterized in terms of a time argument running from \u2212t0 to t0 and back. Pictorially, (cf. Fig. 11.1), the trace operation can be visualized by a closure of the lines emanating from the initial density operator at a large time t0. To get a \ufb01rst hint at the usefulness of the closed contour construction, let us add a perturbation to \u02c6H(s) that breaks the symmetry between the contours: we generalize \u02c6H(s) \u2192 \u02c6H(s)+a\u03b4(s\u2212s0) \u02c6X, wheres 0 \u2208 [0, S/2] is in the \ufb01rst half of the contour, a is a number, and \u02c6X is some operator. Let \u02c6Uint be the time evolution operator in the \u201cinteraction representation\u201d with respect to this perturbation. This representation is de\ufb01ned in perfect analogy to the interaction representation of conventional time-dependent quantum mechanics, only that the \u201ctime integrals\u201d extend over the contour [0, S], instead of over an interval on the real time axis. Speci\ufb01cally, \u02c6U (a) = \u02c6U0 \u02c6Uint(a) = \u02c6Uint(a), where \u02c6U0 = 1 is the unperturbed 0 ds\u03b4(s \u2212 s0) \u02c6X(s) +O (a2) = 1 + ia \u02c6X(s0) + O(a2), evolution operator, and \u02c6Uint(a) = 1 + ia where \u02c6X(s0) = \u02c6U 0 (s0, 0) \u02c6XU0(s0, 0). Insertion of this result into the trace leads to the result S \u2020 \u2212i\u2202a tr( \u02c6U (a)\u02c6\u03c1) = tr( \u02c6U \u2020 0 (s0, 0) \u02c6X \u02c6U (s0, 0)\u02c6\u03c1) = tr( \u02c6X \u02c6\u03c1(s0)), a=0 where \u02c6\u03c1(s0) = \u02c6U (s0, 0)\u02c6\u03c1 \u02c6U \u2020(s0, 0) is the density operator evolved to s0 and we omitted the subscript 0. Thus, di\ufb00erentiation of the density operator with respect to the source generates expectation values of operators in the dynamically evolved state of the system. Put di\ufb00erently, Z(a) \u2261 tr(\u02c6\u03c1 \u02c6U (a)) is a normalized (Z(0) = 1) generating function for quantum expectation values of the operator \u02c6X in the dynamical state of the system \u2013 which need not be in equilibrium. It thus appears that the closed time formalism e\ufb03ciently addresses the points (1)-(4) listed above. This is the basic idea behind Keldysh\u2019s formalism. 11.2.2 Case study As a \ufb01rst step towards the construction of a functional integral implementation of the Keldysh partition function, we consider a miniature quantum system containing only a single bosonic state of energy \u03c9. In a rather natural way, the construction of the corresponding functional integral will introduce most concepts central to the Keldysh formalism, while the notation remains refreshingly simple. Consider, then, the partition function Z \u2261 tr T\u03b3e\u2212i \u03b3 dt \u02c6H(t) \u02c6\u03c10, where, setting Z0 = (1 \u2212 e\u2212\u03b2(\u03c9\u2212\u03bc))\u22121, \u02c6\u03c10 = Z \u22121 0 e\u2212\u03b2( \u02c6H\u2212\u03bc \u02c6N ), (11.9) (11.10) is the equilibrium density operator, \u02c6H = \u03c9a\u2020", "a the Hamiltonian operator and \u02c6N = a\u2020a, as usual. The operator T\u03b3 orders along the contour \u03b3 : (t = 0)\u2192 (t = T ) \u2192 (t = 0), and 704 Nonequilibrium (quantum) we denote the initial (\ufb01nal) time of the theory by t = 0 (t = T ) instead of the t = \u2212t0 (t = t0) used in the previous section. Finally, the time-dependent Hamiltonian operator reads \u02c6H(t) = \u02c6H sgn(T \u2212 t), where we ignore the stages of adiabatic switching on and o\ufb00 alluded to in the previous section. (The straightforward generalization to a smoothly varying \u02c6H(t) will render the notation slightly more complicated but is otherwise inconsequential.) EXERCISE Show that \u2020 \u00af\u03c8|eca a|\u03c8 = eec \u00af\u03c8\u03c8, (11.11) where \u00af\u03c8| and |\u03c8 are boson coherent states. Hint: di\ufb00erentiate by c to derive a \ufb01rst order di\ufb00erential equation and use the uniqueness of its solution. Let us now construct a coherent state \ufb01eld integral by the usual recipe \u2013 insertion of a large number 2N of coherent state resolutions of unity into a time-slice dissection of Z. This leads to expressions of the form, \u03c8\u2212 N \u03c8\u2212 | \u02c6U\u2212|\u03c8\u2212 |1|\u03c8+ N where = T /N and \u02c6U = e\u2212i \u02c6H. The functional integral representation then becomes | \u02c6U...| \u03c8+ 1 | \u02c6U\u2212...| \u03c8\u2212 | \u02c6U|\u03c8+ \u03c8\u2212 1 \u03c8+ N \u03c8+ 1 \u03c8+ N \u22121 N \u22121 N \u22121 N \u22121 1 |\u02c6\u03c10|\u03c8\u2212 N, Z = Z \u22121 0 D( \u00af\u03c8, \u03c8) e N j=2( \u00af\u03c8+ j [\u03c8+ j\u22121 \u2212\u03c8+ j \u2212i\u03c9\u03c8+ j\u22121]+ \u00af\u03c8\u2212 j [\u03c8\u2212 j\u22121 \u00d7 e\u2212 \u00af\u03c8+ 1 \u03c8+ 1 \u2212 \u00af\u03c8\u2212 1 \u03c8\u2212 1 +\u03ba \u00af\u03c8+ 1 \u03c8\u2212 N + \u00af\u03c8\u2212 1 \u03c8+ N, \u2212\u03c8\u2212 j +i\u03c9\u03c8\u2212 j\u22121]) (11.12) where \u03ba \u2261 exp(\u2212\u03b2(\u03c9 \u2212 \u03bc)). At this stage, we would normally take a continuum limit, N \u2192 \u221e at \ufb01xed T. In the present context, however, it is rewarding to stay for a while with the discrete representation, and interpret the exponent of the functional integral as a bilinear form i \u00af\u03c8G\u22121\u03c8, with the (2N ) \u00d7 (2N ) dimensional matrix kernel 1 \u2212a+ 1...... \u2212a\u22121 = i \u2212\u03ba 1 \u22121 1 \u2212a\u2212...... 1 \u2212a11.13) where a\u00b1 \u2261 1\u2213i\u03c9. Let us explore a few properties of this matrix. First, it is straightforward to verify that det(\u2212iG\u22121) = 1 \u2212 \u03ba(a+a\u2212)N. EXERCISE Verify this result. Hint: use the identity det = exp tr ln. Expand the logarithm in powers of the di\ufb00erence from the unit matrix, (\u2212iG\u22121 \u2212 1) (essentially the side-diagonal containing the coe\ufb03cients a\u00b1, plus the corner element \u03ba). Take the trace and re-sum the series into another logarithm. Exponentiation leads to the result. 11.2 Keldysh formalism: basics 705 Taking the limit N \u2192 \u221e, we obtain det(\u2212iG\u22121) = 1\u2212\u03ba(1+(\u03c9)2)N = 1\u2212\u03ba 1 + N (T \u03c9)2 N 2 N\u2192\u221e\u2212\u2192 1\u2212\u03ba = 1\u2212e\u2212\u03b2(\u2212\u03bc) = Z \u22121 0. Since Z = Z \u22121 Keldysh partition function. 0 det(\u2212iG\u22121)\u22121, this result proves the unit normalization of the functional In later applications we will need to compute expectation values \u00af\u03c8i\u03c8j... \u03c8k by Wick\u2019s theorem. For this purpose, we need to know the elementary contractions GCC ii = \u2212i\u03c8C i \u00af\u03c8C i, where C, C = \u00b1. To", " this end, let us introduce the block decomposition G\u22121 =. It is a straightforward (and M ++ M +\u2212 M \u2212+ M \u2212\u2212 worthwhile) exercise to show that % M CC &\u22121 ij = \u2212i\u0398(i \u2212 j)ai\u2212j \u00b1, where \u0398(n) = 1 if n \u2265 0 and zero otherwise. We may now use the general formulae for \u22121, where \u039e++ = the inversion of 2 \u00d7 2 block matrices to obtain G++ = (M ++ \u2212 \u039e++) ij = i\u03b4i1\u03b4jN aN \u22121 M +\u2212(M \u2212\u2212)\u22121M \u2212+. Substitution of \u039e++ \u2212 \u03ba into this formula followed by a straightforward series expansion in \u039e leads to the result ij = \u2212iai\u2212j G++ + \u0398(i \u2212 j) + \u03ba(a+a\u2212)N \u22121 1 \u2212 \u03ba(a+a\u2212)N \u22121. (11.14) INFO This formula a\ufb00ords an intuitive interpretation (cf. the \ufb01gure above). The matrix G++ is the amplitude for propagation between two discrete points j and i on the +-part of our closed time contour. To get from j to i we may either go directly, which is possible if i > j (the time ordering). In this case, we pick up i \u2212 j hopping amplitudes a+. This is the \ufb01rst term in the equation. Alternatively, we may go via round-trips through the \u2212-part of the contour. In this case, and no matter what the chronological ordering between i and j, we \ufb01rst go from j to N ((N \u2212 j) amplitudes a+), then proceed from 1 to N on the bottom part ((N \u2212 1) amplitudes a\u2212) go back to the upper part (a factor \u03ba), and \ufb01nally make it from 1 to i ((i \u2212 1) amplitudes a+). This interprets the \ufb01rst order in \u03ba contribution to the second term. Now, the going round can be repeated in\ufb01nitely. Each additional round trip gives a factor \u03baaN \u22121 +. Summation over all these processes generates the denominator of the equation. \u2212 aN \u22121 The other three propagator amplitudes can be computed in the same manner. Alternatively, one may just derive them by summing over all paths orbiting the closed contour. As a result 706 we obtain Nonequilibrium (quantum) ij = \u2212iai\u2212j G\u2212\u2212 \u2212 \u0398(i \u2212 j) + \u03ba(a\u2212a+)N \u22121 1 \u2212 \u03ba(a\u2212a+)N \u22121, G+\u2212 ij = \u2212i G\u2212+ ij = \u2212i + \u03baaN \u2212j \u2212 ai\u22121 1 \u2212 \u03ba(a\u2212a+)N \u22121, aN \u2212j + ai\u22121 1 \u2212 \u03ba(a+a\u2212)N \u22121. \u2212 Now, let us take the continuum limit. To this end, we assign to lattice points on the upper/lower contour a time variable (cf. Fig. 11.2) + : t = i, \u2212 : t = (N \u2212 i). Remembering that, in the large N limit, (a+a\u2212)N \u2192 1, and noting that \u2192 e\u2212i\u03c9t, \u2212 \u2192 ei\u03c9(T \u2212t), ai ai + we obtain the propagators in a continuous time representation tt = \u2212ie\u2212i\u03c9(t\u2212t)(\u0398(t \u2212 t) +n( \u03c9)), G++ tt = \u2212ie\u2212i\u03c9(t\u2212t)(\u0398(t \u2212 t) +n( \u03c9)), G\u2212\u2212 tt = \u2212ie\u2212i\u03c9(t\u2212t)n(\u03c9), G+\u2212 tt = \u2212ie\u2212i\u03c9(t\u2212t)(1 + n(\u03c9)), G\u2212+ (11.15) (11.16) where n(\u03c9) = \u03ba/(1 \u2212 \u03ba) = (e\u03b2(\u03c9\u2212\u03bc) \u2212 1) is the Bose distribution function. INFO In the literature, it is customary to denote the Green functions by the alternative notation G++ \u2261 GT, G\u2212\u2212 \u2261 G \u02dcT, G+\u2212 \u2261 G<, G\u2212+ \u2261 G>. As, however, this notation is not particularly easy to memorize, and the Green functions above will soon be replaced by di\ufb00erent functions, we here stick to the contour index notation. 11.2.3 Continuum \ufb01eld theory Having derived the continuum propagators, let us now cast the action in Eq.", " (11.12) in a continuum form. Using Eq. (11.15), we obtain Z = Z \u22121 0 D( \u00af\u03c8, \u03c8) ei T 0 dt \u00af\u03c8\u03c33(i\u2202t\u2212\u03c9)\u03c8, (11.17) where \u03c8 \u2261 (\u03c8+, \u03c8\u2212)T is a two component \ufb01eld in contour space and \u03c3i are Pauli matrices in that space. Equation (11.17) is our \ufb01rst prototype of a Keldysh functional integral. Notice that the continuum representation of the action ignores the boundary terms in Eq. (11.12) and speci\ufb01cally the information stored in the initial density operator. The proper way to interpret the sloppiness of the compact (and customary) notation of Eq. (11.17) is 11.2 Keldysh formalism: basics 707 as follows: the \ufb01eld theory is de\ufb01ned by its correlation functions, or propagators. The latter must be computed with proper boundary conditions, that is, with reference to the boundary data stored in the density operator \u03c10, and the gluing procedure at temporal \u201cin\ufb01nity\u201d t = T. Although we might encapsulate this information in a boundary contribution to the continuum action, the operationally safest way is by resorting to the discrete formulation above. Regardless, the result of such a calculation will be the set of Green functions (11.16). In other words, as long as we interpret these Green functions as the propagators of the action (11.17), the boundary conditions have been taken care of. (And of course, the propagators demonstrate that the upper and lower contour are not independent.) One more thing to notice is that the Green function G contains a certain degree of redundancy. Inspection of Eq. (11.16) shows, for example, that G+++G\u2212\u2212 = G+\u2212 +G\u2212+. De\ufb01ning the linear transformation U \u2261 1\u221a 2 1 1 1 \u22121, (11.18) we obtain the alternative representation of the block Green function Gtt \u2192G tt \u2261 U Gtt U \u2020 = = \u2212ie\u2212i\u03c9(t\u2212t) 1 2 1 1 1 \u22121 G++ G+\u2212 G\u2212+ G\u2212\u2212 1 + 2n(\u03c9) \u0398(t \u2212 t) \u2212\u0398(t \u2212 t) 0 \u2261 1 1 1 \u22121 tt GK G+ G\u2212 0, tt (11.19) where the three block Green functions are de\ufb01ned by10 tt \u2261 \u2212i\u0398(t \u2212 t)e\u2212i\u03c9(t\u2212t), G+ tt \u2261 +i\u0398(t \u2212 t)e\u2212i\u03c9(t\u2212t), G\u2212 tt \u2261 \u2212i(1 + 2n(\u03c9))e\u2212i\u03c9(t\u2212t). GK The denotation G\u00b1 indicates that we have met with the retarded and advanced Green function of quantum mechanics, i.e. (11.20) G+ G\u2212 t\u2212t = \u2212i\u0398(t \u2212 t)[a(t), a\u2020(t)], t\u2212t = +i\u0398(t \u2212 t)[a(t), a\u2020(t)], where a(t) = ei \u02c6Htae\u2212i \u02c6Ht = e\u2212i\u03c9ta are the Heisenberg evolved boson operators of our theory. The Keldysh Green function, GK, is a new acquaintance. Notice that GK, and onlyG K, does the book\u2013keeping of the initial distribution. We obtain a better understanding of the meaning of GK as we proceed. The correspondence GCC tt t \u2261 \u2212i\u03c8C \u2261 (U GU \u2020)CC = \u2212i\u03c8C tt t \u00af\u03c8C t. It is customary to denote the transformed \ufb01elds as t implies that GCC \u00af\u03c8C t ( \u00af\u03c8U \u2020)C \u2212i(U \u03c8)C = tt t \u03c8 = U \u03c8 = 1\u221a 2 \u03c8+ + \u03c8\u2212 \u03c8+ \u2212 \u03c8\u2212 \u2261 \u03c8c \u03c8q,, (11.21) 10 Or, expressed in terms of the block Green functions, G\u00b1 = G++ \u2212 G\u00b1\u2213 and GK = G++ + G\u2212\u2212. 708 Nonequilibrium (quantum) where \u03c8c and \u03c8q denote the classical and quantum \ufb01eld, respectively. To understand the origin of this notation, notice that the di\ufb00erence between classical and quantum dynamics is that the latter is governed by the product of two", " independent quantum amplitudes. It is di\ufb00erences between these amplitudes which makes quantum dynamics distinct from classical. This motivates calling (\u03c8+ \u2212\u03c8\u2212)/ 2 a classical \ufb01eld. 2 a quantum \ufb01eld and (\u03c8+ +\u03c8\u2212)/ \u221a \u221a With \u03c8 = U \u2020\u03c8 and \u00af\u03c8 = \u00af\u03c8U, and noting that U \u03c33U \u2020 = \u03c31, the theory assumes the form Z = Z \u22121 0 D( \u00af\u03c8, \u03c8) eiS[ \u00af\u03c8,\u03c8] where S[ \u00af\u03c8, \u03c8] \u2261 dt ( \u00af\u03c8c, \u00af\u03c8q) i\u2202t \u2212 \u03c9 i\u2202t \u2212 \u03c9 \u03c8c \u03c8q. (11.22) This representation has two undesirable and, in fact, interrelated aspects to it: (i) given that the action is purely imaginary, one may wonder whether the functional integral is properly convergent, and (ii) the action still does not contain explicit information on the initial distribution. These two de\ufb01ciencies can be repaired in one sweep. We \ufb01rst notice that the temporal exponents in Eq. (11.19) must be understood as in\ufb01nitesimally damped, exp(\u2212it\u03c9) \u2192 exp(\u2212it\u03c9 \u2212 \u03b4|t|), \u03b4 > 0. Fourier transformation of the Green functions then leads to \u221e dt eitG \u00b1,K t,0, (11.23) G\u00b1,K() = \u2212\u221e with G \u00b1 () = 1 \u00b1 \u2212 \u03c9, GK() = \u22122\u03c0iF ()\u03b4( \u2212 \u03c9) = F () % G+() \u2212 G\u2212() &. Here we have de\ufb01ned the function F () \u2261 1 + 2n() = coth. \u2212 \u03bc 2T (11.24) Notice that the temporal integration has been extended to in\ufb01nity, in spite of the fact that we are living on a contour [0, T ] of \ufb01nite duration. The rationale behind this extension is that the upper limit T must be chosen large enough that the \ufb01niteness of correlation times \u223c \u03b4\u22121 leads to \ufb01nite decay before the end of the contour is reached, T /\u03b4 1. INFO Equation (11.23) suggests two di\ufb00erent representations of the Keldysh Green function, both of which prove useful below. Noting that \u03b4( \u2212 \u03c9) \u2261 \u03c1() is the single particle density of states of our one-level system, we may write GK () \u2261 \u22122\u03c0iF ()\u03c1(). (11.25) Alternatively, we may notice that GK is an anti-hermitian operator. (Exercise: trace back the origin of GK to the basic propagators on the Keldysh contour to convince yourself that antihermiticity of GK is a general feature, not dependent on the simplistic nature of our model.) Equation (11.23) then suggests the ansatz \u02c6GK \u2261 \u02c6G+ \u02c6F \u2212 \u02c6F \u02c6G\u2212, (11.26) 11.2 Keldysh formalism: basics 709 where \u02c6G\u00b1 are the resolvent operators of the theory and \u02c6F is an hermitian operator. In the present context, insertion of the matrix elements G\u00b1 tt = G\u00b1 t\u2212t \u2194 G\u00b1() andF t\u2212t \u2194 F () leads back to Eq. (11.23). In general, we have the block structure \u02c6G = \u02c6GK \u02c6G\u2212 = \u02c6G+ 0 \u02c6G+ \u02c6F \u2212 \u02c6F \u02c6G\u2212 \u02c6G\u2212. \u02c6G+ 0 (11.27) Inserting these Fourier transforms, we obtain the energy representation of the action, S[\u03c8c, \u03c8q] = 6 d 2\u03c0 ( \u00af\u03c8c, \u00af\u03c8q) ( \u02c6G+)\u22121 # ( \u02c6G\u2212)\u22121 ( \u02c6G+)\u22121 \u02c6F \u2212 \u02c6F ( \u02c6G\u22121)\u22121 $ 7 \u03c8c \u03c8q, (11.28) where the matrix kernel de\ufb01nes the inverse of the Green function (11.27). In the speci\ufb01c context of our toy model where \u02c6F commutes with \u02c6G\u00b1, S[\u03c8c, \u03c8q] = d 2", "\u03c0 ( \u00af\u03c8c, \u00af\u03c8q) % + \u2212 \u03c9 \u2212 \u2212 \u03c9 2i\u03b4 coth \u2212\u03bc 2T & \u03c8c \u03c8q. (11.29) This formula makes the full signi\ufb01cance of the \u03b4-broadening manifest: The action now contains explicit reference to the distribution function, and we have arrived at a self\u2013contained representation. The functional integral is convergent. (Show that the imaginary part of the action is positive.) At \ufb01rst sight, keeping track of the in\ufb01nitesimal \u03b4 may seem to be a book\u2013keeping trick to get the distribution function into our description. Remember, however, that in physical systems, initially small convergence generating factors mostly get upgraded to \ufb01nite \u201cself energies.\u201d Similarly here. We will see that interactions will act to make the lower diagonal part of the inverse Green function \ufb01nite. The upper diagonal part, however, will categorically remain zero. Mindful of the discussion in the previous chapter, one may feel reminded of the structure of the MSRJD action. This coincidence is, of course, not accidental. In a sense discussed below, Eq. (11.29) de\ufb01nes an elementary quantum generalization of an MSRJD functional theory. 11.2.4 Generalization Before carrying on, it is useful to upgrade Eq. (11.29) to a functional \ufb01eld theory of generic interacting Bose systems. Nowhere in the derivation have we, in fact, made exclusive refer- 710 Nonequilibrium (quantum) ence to the simplistic nature of the model system. This means that our prototypical \ufb01eld theory Eq. (11.17) a\ufb00ords the generalization, Z = Z \u22121 D( \u00af\u03c8, \u03c8) eiS0[ \u00af\u03c8,\u03c8]+iSint[ \u00af\u03c8,\u03c8], where 0 S0[ \u00af\u03c8, \u03c8] \u2261 \u00af\u03c8CsC(i\u2202t \u2212 \u02c6H)\u03c8C, dtddr C dtddrddr Sint[ \u00af\u03c8, \u03c8] \u2261 \u2212 1 2 C sC ( \u00af\u03c8C\u03c8C)(r, t) V (r \u2212 r) ( \u00af\u03c8C\u03c8C)(r, t), (11.30) and sC is a sign factor assuming the values +1/\u22121 on the upper/lower contour, respectively, \u03c8 = {\u03c8C(r, t)}, \u02c6H is the single particle contribution and V (r \u2212 r) represents a two-particle interaction. We may now introduce classical and quantum \ufb01elds through the transformation (11.21). Substitution of the transformed \ufb01elds into the action leads to the representation S0[ \u00af\u03c8, \u03c8] \u2261 dtddr ( \u00af\u03c8c, \u00af\u03c8q) 6 0 ( \u02c6G+)\u22121 7 ( \u02c6G\u2212)\u22121 ( \u02c6G\u22121)K, \u03c8c \u03c8q (11.31) Sint[ \u00af\u03c8, \u03c8] \u2261 \u2212g dtddr \u00af\u03c8c \u00af\u03c8q(\u03c82 c + \u03c82 q ) + c.c., where, for notational simplicity, we assumed the interaction to be contact, V (r) = 2g\u03b4(r), \u02c6G\u00b1 = (i\u2202t \u00b1 i\u03b4 \u2212 \u02c6H)\u22121 are the retarded and the advanced Green functions, and ( \u02c6G\u22121)K \u2261 ( \u02c6G+)\u22121 \u02c6F \u2212 \u02c6F ( \u02c6G\u2212)\u22121, (11.32) is the Keldysh sector of the inverse of the Green function. Generalizing the discussion of the previous section, let us now interpret the meaning of the di\ufb00erent constituents of this action. Retarded and advanced Green function The in\ufb01nitesimal increments in Eq. (11.31) signify that the retarded and advanced Green functions obey the causality constraint G\u00b1(t, t) \u221d \u0398(\u00b1(t \u2212 t)).11 For a time-independent Hamiltonian, G\u00b1(t, t) = G\u00b1(t \u2212 t) = \u2213i\u0398(\u00b1(t \u2212 t))e\u2212i \u02c6H(t\u2212t), G\u00b1(\u03c9) = 1 \u03c9 \u00b1 i\u03b4 \u2212 \u02c6H. (11.33) If \u02c6H contains explicit time-dependence, we need to include a time ordering procedure", ", G\u00b1(t, t) = \u2213i\u0398(\u00b1(t \u2212 t))T\u00b1e\u2212i t t d\u02dct \u02c6H(\u02dct), where T\u00b1 time orders in chronological/anti-chronological direction. Finally, the Green functions obey the composition law G\u00b1(t, t) = G\u00b1(t, t)G\u00b1(t, t) and the limit behavior G+(t, t) +G \u2212(0, 0) = 0. 11 In the discrete representation used in the previous chapter, causality emerged as a consequence of (G+)\u22121 being a matrix containing one side diagonal below the main diagonal. This meant that G+ was a lower triangular matrix. Within the continuum formalism, we use imaginary increments to keep track of causality. 11.2 Keldysh formalism: basics 711 Keldysh Green function First notice a subtlety in the notation: ( \u02c6G\u22121)K is the Keldysh block of the inverse of the Green function, but not the inverse of the Keldysh Green function, ( \u02c6GK)\u22121. As discussed in the previous section, the anti-hermiticity of \u02c6GK motivates the ansatz (11.32), with an apriori undetermined hermitian matrix F. This object plays a crucial role in the theory; it is responsible for the book\u2013keeping on the state, or \u201cdistribution\u201d of our system. The task of determining \u02c6F gets drastically simpli\ufb01ed by the ansatz \u02c6F = {F (t, t)}, depending on time arguments, but independent of the Hilbert space indices of the problem (r in our current continuum \ufb01eld theory). INFO To motivate this ansatz for \u02c6F, we recall its origins in the non-interacting sector of the theory (cf. Eq. (11.25)) and interpret \u02c6F as a generalized (quasi-)particle distribution function. As long as interactions and external time dependencies do not play a role, it is natural to assume that \u02c6F () = \u02c6F (, \u02c6H) depends on the Hamiltonian of the system. The straightforward generalization of Eq. (11.25) to a higher dimensional Hilbert space \u02c6GK () =\u22122 \u03c0i \u02c6F ( \u02c6H, )\u03b4( \u2212 \u02c6H) then suggests the eigenfunction representation \u02c6GK (, \u03c9a) =\u22122 \u03c0i \u02c6F (, \u03c9a)\u03b4( \u2212 \u03c9a). The important point here is that the spectral \u03b4-function locks the eigenenergies \u03c9a to the energy argument. Since \u02c6F always appears in combination with the spectral function, dropping its Hilbert space index, \u02c6F (, \u03c9a) = \u02c6F () does not imply a loss of information. In a time-independent context, F (t, t) = F (t \u2212 t) and the Fourier transform F () signi\ufb01es the occupation of quasi-particles at energy. (For instance, F () = coth(T /2) = 1 + 2n() under equilibrium conditions, where n() is the Bose distribution function.) In a time dependent environment, it is usually a good idea to represent F by its Wigner transform, F (, t) \u2261 dt eit F (t + t/2, t \u2212 t/2). (11.34) The function F (., t) then represents the instantaneous particle distribution function at time t (as an exercise, think why!). We \ufb01nally note that retarded, advanced, and Keldysh Green function are obtained by computing the correlator \u2212i\u03c8\u03b1(r, t) \u00af\u03c8\u03b1 (r, t) = GK G+ G\u2212 0 (r, t; r, t). \u03b1\u03b1 (11.35) Interaction There is not much we can say about the interaction vertex at this stage in generality. Notice, however, that the interaction vanishes for \u03c8q = 0, which is a manifestation of the vanishing of the action of purely classical \ufb01elds.12 In later perturbative applications, it will 12 The particular two body interaction considered here also vanishes on a pure quantum con\ufb01guration. However, this vanishing does not re\ufb02ect a general structure. 712 Nonequilibrium (quantum) \u00af\u03c8c \u00af\u03c8q \u03c8c \u03c8q G+ = \u2212i\u03c8c \u00af\u03c8q G\u2212 = \u2212i\u03c8q \u00af\u03c8c GK = \u2212", "i\u03c8c \u00af\u03c8c Figure 11.3 Building blocks of bosonic Keldysh diagrammatic perturbation theory. Top: \ufb01eld vertices, middle: propagators, bottom: interaction vertices. be convenient to represent the interaction by a graphical code (see Fig. 11.3). Below we apply this language to explore the in\ufb02uence of interactions on the non-interacting sector of the theory. 11.2.5 Fluctuation dissipation theorem In the case of thermal equilibrium, the Keldysh Green function is fully determined by the retarded and advanced Green functions, \u02c6GK = \u02c6G+ \u02c6F \u2212 \u02c6F \u02c6G\u22121. In thermal equilibrium, the commutativity of \u02c6F = 1 + 2n = coth(/2T ) with \u02c6G\u00b1 implies \u02c6GK() = coth ( \u02c6G+() \u2212 \u02c6G\u2212()). (11.36) 2T The form of this dependence is reminiscent of the quantum \ufb02uctuation dissipation theorem reviewed in the previous chapter. To make the connection to the FDT explicit, let us proceed as in Section 10.6.4 and discuss the correlation functions \u02c6GK and \u02c6G\u00b1 in terms of their formal Lehmann representation. Beginning with the Keldysh Green function, we turn back to a representation in terms of contour \ufb01elds and have GK(t, 0) = \u2212i\u03c8c(t) \u00af\u03c8c(0) = 1 2 (\u03c8+(t) +\u03c8 \u2212(t))( \u00af\u03c8+(0) + \u00af\u03c8\u2212(0)), where we have omitted the Hilbert space indices of the \ufb01elds for notational clarity. Now, interpret a correlation function \u03c8C(t) \u00af\u03c8C (0) \u2194 T\u03b3aC(t)a \u2020 C (0), 11.2 Keldysh formalism: basics 713 as the coherent state functional integral representation of a contour ordered operator expectation value. In the latter expression, the angular brackets represent the equilibrium thermal expectation value... = tr((...)\u02c6\u03c10), and the combination (C, t) of the contour subscript and the time argument \ufb01xes the position of the operator on the closed time contour. We next employ the Heisenberg representation of the operators as aC(t) = e \u2212i \u2212t0 (C,t) dt \u02c6H(t) a e \u2212i (C,t) \u2212t0 dt \u02c6H(t), where the time integrals in the exponents are along the time contour, and we have dropped (C,t) dt \u02c6H(t), on account of the the contour time ordering operator T\u03b3e \u2212t0 presumed time-independence of \u02c6H. (Exercise: Consider why this convention conforms with the de\ufb01nition of the coherent state \ufb01eld integral, and the contour conventions of the Keldysh approach in general.) If (C, t) lies on a later point on the contour than (C, 0), this representation yields dt \u02c6H(t) = e (C,t) \u2212t0 \u2212i \u2212i @ aC(t)a \u2020 C (0) = \u2212t0 (C,t) dt \u02c6H(t) a e \u2212i e \u2212i (C,t) (C,0) dt \u02c6H(t) a\u2020 e \u2212i (C \u2212t0,0) dt \u02c6H(t) A. We may now insert complete sets of eigenfunctions of \u02c6H, {|\u03b1}, to obtain the Lehmann representation of these expectation values. For example, for C = +, C = \u2212, t <0, we obtain (cf. the \ufb01gure) \u2020 a \u2212(0)a+(t) = Z \u22121 0 \u03c1\u03b2|X\u03b1\u03b2|2ei\u039e\u03b1\u03b2 t, where X\u03b1\u03b2 \u2261 \u03b1|a|\u03b2, \u039e\u03b1 = E\u03b1 \u2212 \u03bcN\u03b1, \u039e\u03b1\u03b2 = \u039e\u03b1 \u2212 \u039e\u03b2, the thermal weighting factor \u03c1\u03b2 = e\u2212\u03b2\u039e\u03b2, and a summation over \u03b1, \u03b2 is implied. Treating all other combinations in the same manner, we \ufb01nd GK(t, 0) = \u2212iZ \u22121 0 |X\u03b1\u03b2|2ei\u039e\u03b1\u03b2 t(\u03c1\u03b1 + \u03c1\u03b2), which Fourier transforms to GK() =", " \u22122\u03c0iZ \u22121 0 |X\u03b1\u03b2|2\u03b4( + \u039e\u03b1\u03b2)(\u03c1\u03b1 + \u03c1\u03b2). Comparison with the expression above for the time ordered correlation function GT (t, 0) = \u2212iTta(t)a\u2020(0) leads us to the important conclusion, GK() = 2i. GT (). (11.37) The Keldysh Green function is equal to twice the imaginary part of the time ordered correlation function. In a similar manner, the Lehmann representations of the Green functions G\u00b1 can be analyzed to con\ufb01rm that the latter conform with the general de\ufb01nition of retarded and advanced response functions. These identi\ufb01cations show that Eq. (11.37) is but a manifestation of the quantum \ufb02uctuation dissipation theorem. 714 Nonequilibrium (quantum) 11.2.6 Classical limit I Readers who went through the previous chapter may have noticed the striking structural similarity between the Keldysh diagram language and the graphical code used in Section 10.5.1 to analyze the MSRJD functional integral. More generally, the blocktriangular form of the Green functions, containing retarded and advanced blocks, in conjunction with distribution functions, is reminiscent of the structure of the MSRJD theory. In this section, we explore the classical limit of the theory within the framework of a stationary phase analysis. A more comprehensive discussion of the connection to MSRJD theory is given in Section 11.3. Prior to taking semiclassical limits, we should let re-enter the theory. Remembering the S. In the limit origin of the action as a contour integrated quantum phase, we set iS \u2192 i \u2192 0, the functional will then be governed by the stationary con\ufb01gurations of its action \u2202S[\u03c8c, \u03c8q] \u2202\u03c8q = \u2202S[\u03c8c, \u03c8q] \u2202\u03c8c = 0. (11.38) These equations may have solutions for arbitrary (\u03c8c, \u03c8q) about which nothing much can be said in general terms. But there is always one that re\ufb02ects the classical physics of the problem. Recalling that an expansion of the action in \u03c8q starts at linear order (the purely classical action S[\u03c8c, 0] = 0), we identify this classical saddle point by the equation, classical saddle point: \u2202S[\u03c8c, \u03c8q] \u2202\u03c8q = 0. \u03c8q=0 (11.39) This saddle point is \u201cclassical\u201d in that it is de\ufb01ned on the contour \u03c8q = \u03c8+ \u2212 \u03c8\u2212 = 0, i.e. in a limit where the time evolution becomes classical. Speci\ufb01cally, for the interacting Bose action (11.30), the saddle point equation assumes the form i\u2202t \u2212 \u02c6H \u2212 V |\u03c8c|2 \u03c8c = 0. (11.40) This is the celebrated Gross\u2013Pitaevskii equation, a non-linear Schr\u00a8odinger equation describing the classical physics of the evolution of a bosonic order parameter amplitude \u03c8c in self\u2013interaction with its own density \u223c |\u03c8c|2. INFO We may think of the Gross\u2013Pitaevskii equation (GPE) as the time-dependent Ginzburg\u2013Landau equation of the transition into a super\ufb02uid state. Alternatively, it may be interpreted as a nonlinear Schr\u00a8odinger equation describing the wavefunction of a Bose\u2013Einstein condensate. Inhomogeneous solutions of this equation describe collective excitations of the condensate \ufb01eld. 11.2 Keldysh formalism: basics 715 By way of example, let us consider the case of a one-dimensional spatially homogeneous system, \u2212 \u03bc, where we have added a chemical potential term to \u02c6H. Passing to the \u201cenergy \u02c6H \u2192 \u2212 (\u2202x)2 representation\u201d of the equation, i\u2202t \u2192 E, 2m E + (\u2202x)2 2m + \u03bc \u2212 V |\u03c8c|2 \u03c8c = 0. The solution of lowest energy, E = 0, is the spatially homogeneous condensate amplitude (cf. the discussion in Section 6.3) |\u03c8c|2 = \u03bc V. However, the GPE also contains information on spatially inhomogeneous excitations of the condensate. For example, it is straightforward to verify that, for any x0, the function \u03c8c \u2261 \u03c80 tanh, (x \u2212 x0) \u03be (11.41) \u221a solves the GPE (", "cf. the \ufb01gure above.) Here, \u03c80 is a complex amplitude with |\u03c80|2 = \u03bc/g, and \u03be \u2261 / 4m\u03bc is the coherence length. The wave function (11.41) is called a dark soliton. The attribute \u201csoliton\u201d hints at the topological nature of the solution. The sign change at its origin x0 means that the phase of the soliton changes by \u03c0. This jump cannot be removed by continuous deformation, and in this sense the soliton is a topological defect. EXERCISE For a system with attractive interactions, V < 0, try to identify the bright soliton. One may then expand around the \u201cclassical\u201d saddle point to account for quantum and classical \ufb02uctuations. However, we \ufb01nd it more instructive to discuss these types of \ufb02uctuations in a slightly di\ufb00erent context below. This concludes our preliminary discussion of the generalities of the (bosonic) Keldysh formalism. Before carrying on, it may be worthwhile to recapitulate once more a few structures of the theory: We start at large negative times with a non-interacting system whose state is described by some initial distribution \u02c6F0. In this non-interacting limit, the Keldysh sector of the inverse Green function, ( \u02c6G\u22121)K is in\ufb01nitesimally small, and related to the retarded and advanced sector by the FDT. As interactions are switched on, the system will aim to identify its new distribution. The initial proportionality of ( \u02c6G\u22121)K to the initial imaginary damping terms i\u03b4 suggests that the Keldysh sector of the inverse Green function will no longer be in\ufb01nitesimally small; the emergence of \ufb01nite \u201cself energies\u201d, i\u03b4 \u2192 i Im \u03a3 will render it \ufb01nite. Put di\ufb00erently, the self energies of the theory, and in particular the Keldysh component of the self-energy, are expected to contain essential information on the state of the system. In the following, we explore the interaction-mediated formation of distribution functions on an instructive, and important example. 716 Nonequilibrium (quantum) 11.3 Particle coupled to an environment Let us go back to the general paradigm formulated in Section 11.1.1, a quantum system coupled to a bath. For concreteness, we will assume the system to be a single quantum particle, while the bath is an ensemble of harmonic oscillators. In the particular case where the dynamics of the particle is harmonic, this gets us back to the oscillator\u2013oscillator coupling studied in Section 11.1.2. Keldysh theory of a quantum particle Consider the quantum mechanics of a single point particle in one-dimensional space. In the distant past, we prepared the particle in some state (pure or mixed, it does not matter) and then let it evolve under the dynamics of its Hamiltonian operator \u02c6H = \u02c6p2 2m + V (\u02c6q). This situation can be described by an obvious modi\ufb01cation of our Keldysh functional theory. All we need to do is encapsulate the initial value information in a density operator \u02c6\u03c10, and replace the coherent state \ufb01eld integral by a Feynman path integral. As a result, we obtain the Keldysh partition function Zs = Z \u22121 0 = Z \u22121 0 = Z \u22121 0 D(q, p) ei C sC dt (pC \u2202tqC \u2212H(qC,pC )) C sC Dq ei dt ( m 2 (\u2202tqC )2\u2212V (qC )) Dq ei dt m\u2202tqq\u2202tqc\u2212V 1\u221a 2 (qc+qq) )+V 1\u221a 2 (qc\u2212qq). (11.42) To explore the classical limit of this expression, we re-introduce as iS \u2192 i\u22121S. Alluding to the discussion of the quantum-nature of con\ufb01guration di\ufb00erences q+ \u2212 q\u2212 \u223c qq above, we also rescale qc \u2192 qc. A \ufb01rst order expansion of the action in qc then identi\ufb01es the -independent (classical) sector of the action Sc[qc, qq] \u2261 dt qq \u2212m\u22022 t qc \u2212 \u221a 2 V 1\u221a 2 qc. Integrating over the quantum component, we obtain the constraint \u2212m\u22022 \u221a 0 or,", " upon rescaling coordinates qc/ 2 \u2261 q, Newton\u2019s equations of motion t qc\u2212 \u221a \u221a 2 \u2202qV (qc/ 2) = But of course, the formalism can do much more than recover deterministic classical motion. This is seen the moment we couple the system to its bath: m\u22022 t q = \u2212\u2202qV (q). (11.43) Coupling to an oscillator bath The bath as such is described by a partition function such as Eq. (11.31). Speci\ufb01cally, we assume the bath degrees of freedom to be non-interacting, V = 0, and oscillator-like, \u00af\u03c8 \u02c6H\u03c8 \u2261 \u00af\u03c8k \u03c9k \u03c8k, with a wave-like dispersion \u03c9k \u2261 c|k|. Thus, k Sb[ \u00af\u03c8, \u03c8] \u2261 \u00af\u03c8k,\u03c9 \u02c6G\u22121 k,\u03c9 \u03c8k,\u03c9, (11.44) d\u03c9 2\u03c0 k 11.3 Particle coupled to an environment 717 with \u03c8 = (\u03c8c, \u03c8q)T, and the oscillator Green function, \u02c6G\u22121 k,\u03c9 = 0 \u03c9 + i\u03b4 \u2212 \u03c9k \u03c9 \u2212 i\u03b4 \u2212 \u03c9k [2i\u03b4 coth(\u03c9/2T )]. The coupling between system and bath is modelled by the action Ssb[ \u00af\u03c8, \u03c8, q] \u2261 = dt d\u03c9 2\u03c0 k C k sCq(t) (\u03b3k\u03c8C,k(t) + c.c.) = dt \u03b3kqT \u03c9 \u03c31\u03c8k,\u2212\u03c9 + \u00af\u03b3k \u00af\u03c8T k,\u03c9\u03c31q\u2212\u03c9 k, q(t)T \u03c31 (\u03b3\u03c8k(t) + c.c.) (11.45) where \u03b3(k) is a set of complex coupling constants, and in the second line we switched to the Keldysh representation qT = (qc, qq). Integration over oscillator modes At this stage, the oscillators may be integrated out. Using the fact that \u2212i\u03c8 \u00af\u03c8T = \u02c6G (cf. Eq. (11.35)), we obtain the e\ufb00ective system action, Se\ufb00 = Ss + Sdiss, where the dissipative action is given by Sdiss[q] = \u2212 d\u03c9 2\u03c0 qT \u03c9 \u03c31 6 k 7 |\u03b3k|2 \u02c6Gk,\u03c9 \u03c31q\u2212\u03c9, and the Green function has the form \u02c6Gk,\u03c9 = [\u22122\u03c0i coth \u03c9 2T (\u03c9 \u2212 i\u03b4 \u2212 \u03c9k)\u22121 \u03b4(\u03c9 \u2212 \u03c9k)] (\u03c9 + i\u03b4 \u2212 \u03c9k)\u22121 0. To make progress with this expression, we assume that the coupling constant exhibits power law dependence on |k|, |\u03b3k| = \u03bb|k|\u03b1, where \u03bb >0 is a real coe\ufb03cient. Noting that the Green function enters in a frequency\u03c9 ( \u02c6G\u03c9 + \u02c6G\u2212\u03c9)q\u2212\u03c9, the summation over the k\u03c9 qT \u03c9 qT symmetrized form, \u03c9 modes then produces the expression (exercise) \u02c6G\u03c9q\u2212\u03c9 = 1 2 Sdiss[q] = d\u03c9 2\u03c0 qT \u03c9 \u02c6K\u03c9q\u03c9, where the dissipation kernel \u02c6K is given by 2\u03b1 \u02c6K\u03c9 = \u03c0\u03bd\u03bb2 \u03c9 c 0 i \u2212i 2i coth (11.46) (11.47). \u03c9 2T k \u03b4(\u03c9 \u2212 \u03c9k) and omitted a frequencyHere we have introduced the density of modes, \u03bd \u2261 independent contribution to K (which can be absorbed in the chemical potential). We may think of \u02c6K\u03c9 as a \u201cself-energy\u201d of the particle due to its coupling to the oscillator bath. The frequency dependence of K\u03c9 signals that Sdiss is a time-nonlocal contribution to the action. 718 Nonequilibrium (quantum) The temporal pro\ufb01le of the dissipation kernel depends on the coupling \u03b3. In the following, we concentrate on the case of Ohmic dissipation, where\u03b1 = 1/2 and \u02c6K\u03c9 = g \u2212i\u03c9 0 i\u03c9 2i\u03c9 coth \u03c9 2T, (11.48) with g \u2261 \u03c0\u03bd\u03bb2 c. INFO The power", " law \u03b1 = 1/2 is not quite as arbitrary as it may seem. Speci\ufb01cally, the choice |k| describes the frequent situation of dissipation due to coupling to an elastic \u03b3 = i\u03bb sgn(k) medium. To see this, recall the representation of the coherent state amplitudes in terms of oscillator coordinates and momenta, \u03c8k = 1\u221a 2 \u221a \u03c9kqk + pk. i\u221a \u03c9k Substitution into the coupling term generates the expression (\u03b3k\u03c8k + c.c.) \u223c sgn (k) |k| i \u221a \u03c9kqk + k \u223c k ikqk \u223c \u2202xq(x), x=0 k pk + c.c. i\u221a \u03c9k where, noting that functions q(x) and p(x) are real, we have made use of the identities \u00afqk = q\u2212k and \u00afpk = p\u2212k. The last term describes the local coupling of the system coordinate to the stress \u2202xq(x) acting on an elastic string. In many dissipative environments, macroscopic degrees of freedom are coupled to approximately harmonic microscopic modes in this way. Prominent examples include the coupling of the electric \ufb01eld applied to a resistor to harmonic electromagnetic modes in the medium. This coupling (cf. our discussion of Johnson noise in the previous chapter) is responsible for the denotation \u201cOhmic\u201d dissipation. We next explore the classical limit of the e\ufb00ective theory of the particle coupled to the bath. To this end, we let enter at three di\ufb00erent places in the functional, Langevin equation S, (i) : iS \u2192 i (ii) : coth(\u03c9/2T ) \u2192 coth(\u03c9/2T ) \u2192 2T \u03c9 (iii) : \u03c8q \u2192 \u03c8q., (11.49) In the semiclassical limit, and at \ufb01xed characteristic frequency \u03c9, the \u201ccoth\u201d can then be linearized as indicated in (ii). The extra power of appearing in the denominator of this expression compensates for the additional power in due to two factors qq in the quantum\u2013 quantum sector of the action. Summarizing, the e\ufb00ective classical action is given by Sc[qc, qq] = dt qq \u2212m\u22022 t qc \u2212 \u221a 2 V 1\u221a 2 qc \u2212 2g\u2202tqc + 4iT gq2 q. 11.3 Particle coupled to an environment 719 This has the characteristic structure of an MSRJD action. Comparison with the discussion of MSRJD functionals in the previous chapter shows that the (scaled) quantum \ufb01eld 2gqq plays a role identical to the MSRJD \u201cmomentum.\u201d Speci\ufb01cally, a \u201cLangevin\u201d equation13 can be derived by Hubbard\u2013Stratonovich decoupling of the qc term: Sc[qc, qq] \u2192 Sc[qc, qq, \u03be] = dt qq + i 8gT \u221a \u2212m\u22022 t qc \u2212 \u221a 2 V dt \u03be2. qc\u221a 2 \u2212 2g\u2202tqc + \u221a 2\u03be Introducing the rescaled variable q = qc/ equation 2 and integrating over qq, we obtain the Langevin where the noise correlator m\u22022 t q + V (q) + 2g\u2202tq = \u03be, \u03be(t)\u03be(t) = 4T g\u03b4(t \u2212 t), (11.50) (11.51) conforms with the FDT. To summarize, the coupling of the system to a bath renders the classical limit noisy. The e\ufb00ective equation of motion assumes the form of a Langevin equation, at a noise level that will drive the particle to an equilibrium distribution at the e\ufb00ective temperature of the bath. INFO Notice that the present theory solves a problem that plagued the purely classical theory discussed in the previous chapter. Discussing dissipative damping of voltage \ufb02uctuations in a resistor network, we had seen that Johnson noise is problematic inasmuch as the white noise correlator \u03be(t)\u03be(t) = 2RT \u03b4(t \u2212 t) will predict singularly strong noise levels \u03be(t)2. We had cited the expectation that these singularities will be cut o\ufb00 if we take into account the fact that the actual coupling is to a bath of quantum, rather than classical oscillators. This expectation is con\ufb01rmed by our present theory.", " Comparison with Eq. (11.48) and the subsequent steps of the derivation shows that the frequency representation of the actual noise correlator reads \u03be\u03c9\u03be\u03c9 = g \u03c0 \u03b4(\u03c9 + \u03c9)\u03c9 coth \u03c9 2T. In the classical limit \u2192 0, this reduces to the white noise limit 2T g transforms to Eq. (11.51). However, at large frequencies \u03c9 T, we obtain \u03c0 \u03b4(\u03c9 + \u03c9), which Fourier \u03b4(\u03c9 + \u03c9)|\u03c9|. \u03be\u03c9\u03be\u03c9 \u03c9T\u2212\u2192 g \u03c0 At \ufb01rst sight, it seems that the situation has become worse: the increase \u223c |\u03c9| implies ultraviolet behavior (short time correlations) even more malicious than in the white noise case. The resolution to this problem lies in a more careful interpretation of what is actually meant by the \u03c9 expression \u03bet\u03bet. The decomposition \u03c9 coth \u223c (1 + n(\u03c9)) + n(\u03c9) suggests that the voltage 2T \ufb02uctuations are due to absorption from (n) and emission into (n+1) the \ufb01eld of quantal resonator! 13 We put quotes around \u201cLangevin\u201d because the stochastic di\ufb00erential equations derived here contain second order time derivatives stemming from the acceleration terms in Newton\u2019s equations. (We are not working in the purely dissipative, overdamped limit.) The term \u201cLangevin equation\u201d is commonly reserved for \ufb01rst order equations, but this di\ufb00erence is of no relevance to the present discussion. 720 Nonequilibrium (quantum) modes of the oscillator system. Speci\ufb01cally, there is response to its vacuum \ufb02uctuations (the \u201c1\u201d in 1 + n). At low temperatures, T /\u03c9 \u2192 0, it is these \ufb02uctuations that cause our ultraviolet problems. On the other hand, it seems implausible that vacuum \ufb02uctuations will be probed in an actual measurement of noisy voltage \ufb02uctuations in a resistor. (The vacuum is a de\ufb01nite state, while the generation of \ufb02uctuations requires real transitions in the environment.) Closer inspection (cf. e.g., the pedagogical article G.B. Lesovik and R. Loosen, On the detection of \ufb01nite-frequency current \ufb02uctuations, JETP Lett. 65, 295-99 (1997)) shows that (i) the expression ( \u02c6Vt \u02c6Vt + \u02c6Vt \u02c6Vt corresponds to a symmetrized correlation function of the quantum \u03bet\u03bet \u2194 1 2 operators representing voltage \ufb02uctuations (exercise: to make this statement plausible, interpret \u03bet\u03bet \u223cG K as a Keldysh correlation function and use the connection between GK and time ordered correlation functions discussed in Section 11.2.4.) However, (ii) a concrete measurement will not correspond to this ordering. Speci\ufb01cally, it will not couple to the \ufb02uctuations of the vacuum. Once the vacuum contribution is removed, the ultraviolet problems disappear. For further discussion of this point, we refer to the literature. This concludes our discussion of the backbones of Keldysh \ufb01eld theory. We have introduced its fundamental generating functional, discussed the meaning of the propagators, and exposed the connections to the classical theories discussed in the previous chapter. However, to make the picture complete, we need to introduce the fermion variant of the Keldysh functional. This will be as necessary as it is uninspiring. As one may expect, the formal di\ufb00erences from the bosonic theory amount to a few sign factors here and there. For this reason, we will try to keep the discussion as concise as possible and restrict ourselves to listing the main di\ufb00erences from the bosonic theory. 11.4 Fermion Keldysh theory (a list of changes) 11.4.1 Single level As in the bosonic case, let us begin by considering a single level Hamiltonian \u02c6H = \u03c9a\u2020a, where a and a\u2020 are fermion operators. The de\ufb01nitions of the Keldysh partition function (11.9), and of the equilibrium density operator (11.10) remain unchanged, only that the normalizing factor is now given by Z0 = 1 + e\u2212\u03b2(\u03c9\u2212\u03bc). Again, we will want to represent the partition function in terms of (now fermionic) coherent states. It is straightforward to verify that the auxiliary identity Eq.", " (11.11) generalizes to the fermionic case, which means that the discrete functional Eq. (11.12) remains as it is but for a sign change\u03ba \u2192 \u2212\u03ba. As in the construction of equilibrium (Matsubara) functionals, this change represents the sign we pick up when we sweep one of the 2N coherent state amplitude through all others, as we do in the construction of the functional integral. Consequently, the inverse of the discrete fermionic Green function is given by Eq. (11.13), again up to a sign change \u03ba \u2192 \u2212\u03ba. Up to this sign factor, all remaining formulae of the discrete discussion of Section 11.2.2 generalize to the fermionic case. Speci\ufb01cally, det(\u2212iG\u22121) N\u2192\u221e\u2212\u2192 1 +\u03ba = 1 + e\u2212\u03b2(\u03c9\u2212\u03bc) = Z0 which proves normalization of the functional integral. Taking the limit 11.4 Fermion Keldysh theory (a list of changes) 721 N \u2192 \u221e, and switching to a continuum representation, we obtain the fermionic Green functions tt = \u2212ie\u2212i\u03c9(t\u2212t)(\u0398(t \u2212 t) \u2212 n(\u03c9)), G++ tt = \u2212ie\u2212i\u03c9(t\u2212t)(\u0398(t \u2212 t) \u2212 n(\u03c9)), G\u2212\u2212 tt = +ie\u2212i\u03c9(t\u2212t)n(\u03c9), G+\u2212 tt = \u2212ie\u2212i\u03c9(t\u2212t)(1 \u2212 n(\u03c9)), G\u2212+ n(\u03c9) = 1 1 + e\u03b2(\u03c9\u2212\u03bc), (11.52) where is the Fermion distribution. (The global plus sign appearing in G+\u2212 re\ufb02ects the sign factor picked up upon passage of the terminal point of the contour.) We may now pass to a continuum representation of the functional integral whose unregularized form (no account for in\ufb01nitesimal damping and boundary conditions) is given by the Grassmann version of Eq. (11.17). A glance at Eq. (11.52) shows that the redundancy in the theory which motivated the \u201cKeldysh rotation\u201d now assumes the form G++ + G\u2212\u2212 = G+\u2212 + G\u2212+. In the case of Grassmann variables, it is customary14 to account for this redundancy by way of a \ufb01eld transformation slightly di\ufb00erent from the bosonic Keldysh rotation above: using the independence of the Grassmann variables \u03c8 and \u00af\u03c8, we introduce new \ufb01elds by \u03c81 \u2261 1\u221a 2 \u00af\u03c81 \u2261 1\u221a 2 (\u03c8+ + \u03c8\u2212), ( \u00af\u03c8+ \u2212 \u00af\u03c8\u2212),, \u03c8 \u2261 Introducing the notation \u03c8 \u2261 the form of a non-unitary transformation \u03c8+ \u03c8\u2212 \u03c82 \u2261 1\u221a 2 \u00af\u03c82 \u2261 1\u221a 2 (\u03c8+ \u2212 \u03c8\u2212), ( \u00af\u03c8+ + \u00af\u03c8\u2212). (11.53) \u03c81 \u03c82, \u00af\u03c8 = ( \u00af\u03c8+, \u00af\u03c8\u2212), \u00af\u03c8 = ( \u00af\u03c81, \u00af\u03c82), this assumes \u03c8 \u2261 U \u03c8, \u00af\u03c8 \u2261 \u00af\u03c8\u03c33U T, where U has been de\ufb01ned in Eq. (11.18). (Due to the non-unitarity of the transformation, and the fact that Grassmann \ufb01elds are never \u201cclassical,\u201d we do not speak of classical or quantum components.) The form of the above transformation is motivated by the transformed contraction rule where \u02c6G = \u02c6G++ \u02c6G+\u2212 \u02c6G\u2212+ \u02c6G\u2212\u2212 \u2212i\u03c8 \u00af\u03c8 = U \u03c8 \u00af\u03c8\u03c33U T = U \u02c6GU T \u03c33 \u2261 \u02c6G,, and the transformed Green function assumes the form 6 \u02c6G+ \u02c6G = 7, \u02c6GK \u02c6G\u2212 (11.54) 14 A. I. Larkin and Yu. N. Ovchinnikov, Vortex motion in superconductors, in Nonequilibrium Superconductivity, eds. D. N. Langenberg and A. I. Larkin, (Elsevier, 1986). 722 Nonequilibrium (quantum) and the retarded, advanced, and Keldysh Green function are de\ufb01ned in analogy to Eq. (11.20), tt \u2261 \u2212i\u0398(t \u2212 t)e\u2212i\u03c9(t\u2212t", "), G+ tt \u2261 +i\u0398(t \u2212 t)e\u2212i\u03c9(t\u2212t), G\u2212 tt \u2261 \u2212i(1 \u2212 2n(\u03c9))e\u2212i\u03c9(t\u2212t). GK As we will mostly work with the transformed Green function \u02c6G, we omit the prime throughout, \u02c6G \u2192 \u02c6G. (11.55) 11.4.2 Generalization As with the bosonic theory, the form of the Green function Eq. (11.55) suggests the introduction of a fermionic distribution operator \u02c6F = {F ()}, with F () \u2261 1 \u2212 2n(). (11.56) Expressed as a Fourier transform, the Keldysh Green function then assumes the form \u02c6GK = \u02c6G+ \u02c6F \u2212 \u02c6F \u02c6G\u2212. (11.57) The generalization of the toy model above to a fully interacting theory is described by the Grassmann version of the functional Eq. (11.30). (The optional addition of spin-indices should be straightforward.) In contrast to the bosonic case, where mean \ufb01eld approaches to the native interacting theory lead to immediate results, most fermionic theories call for a Hubbard-Stratonovich decoupling of the interaction term. Prior to passing to the Keldysh rotated theory, we therefore introduce an auxiliary bosonic \ufb01eld to decouple the interaction as eiSint[ \u00af\u03c8,\u03c8] = D\u03c6 e\u2212 i 2 \u00d7 e+i dtddrddr C sC \u03c6C (r,t)V \u22121(r\u2212r)\u03c6C (r,t) dtddr C sC \u00af\u03c8C (r,t)\u03c6C (r,t)\u03c8C (r,t). We may now implement the Keldysh rotation to arrive at the e\ufb00ective action S[ \u00af\u03c8, \u03c8, \u03c6] = S[\u03c6] +S [ \u00af\u03c8, \u03c8] +S int[ \u00af\u03c8, \u03c8, \u03c6], where ddrdt \u03c6c(r, t)V \u22121(r \u2212 r)\u03c6q(r, t), S[\u03c6] = \u2212i S[ \u00af\u03c8, \u03c8] = ddrdt \u00af\u03c8 \u02c6G\u22121 0 \u03c8, 1\u221a 2 Sint[ \u00af\u03c8, \u03c8, \u03c6] = ddrdt \u00af\u03c8(\u03c6c\u03c30 + \u03c6q\u03c31)\u03c8. (11.58) Here 6 (G+ 0 )\u22121 \u02c6G\u22121 0 = 7, ( \u02c6G\u22121 ( \u02c6G\u2212 0 )K 0 )\u22121 (11.59) 11.5 Kinetic equation 723 is the inverse of the free Green function15 and ( \u02c6G\u00b1 that elements of the Green function are generated by the fermion variant of Eq. (11.35) 0 )\u22121 = (i\u2202t \u00b1 i0 \u2212 \u02c6H). We \ufb01nally note \u2212i\u03c8\u03b1(r, t) \u00af\u03c8\u03b1 (r, t) = G+ GK 0 G\u2212 (r, t; r, t). \u03b1\u03b1 (11.60) 11.5 Kinetic equation The action Eq. (11.58) may serve as the starting point of numerous applications. Here, we restrict ourselves to the discussion of one interaction phenomenon that is particularly relevant to the central theme of the chapter: the interaction mediated formation of e\ufb00ective particle distributions. As a byproduct, we also introduce elements of Keldysh perturbation theory, and compare to the \u201cequilibrium perturbation theory\u201d discussed in earlier chapters. The concepts and techniques introduced in this section are very important and are routinely applied in the analysis of out of equilibrium quantum systems. In carrying out the perturbative analysis, we are guided by two principles: \ufb01rst, we assume that the system (e.g. by way of high electron density) admits an e\ufb00ective RPA approach. Second, we think of the Hubbard\u2013Stratonovich \ufb01eld, \u03c6, as an independent, bosonic degree of freedom. That analogy becomes more substantial as we proceed and \u03c6 acquires its own dynamics. Speci\ufb01cally, the boson\u2013fermion analogy suggests not to integrate over the fermion degrees of freedom just yet (which would be an option as a result of the quadratic nature of the action Eq. (11.58)), but to study boson and", " fermion propagators separately. In Eq. (11.58), quasiparticles interact by the two-fermion oneboson interaction vertices contained in Sint. Our intermediate goal will be to obtain the emerging boson and fermion self energies \u02c6\u03a3f and \u02c6\u03a3b, respectively, in the RPA approximation, i.e. an approximation void of intersecting propagator lines. The topology of \u02c6\u03a3f is given by the \ufb01rst diagram shown in the \ufb01gure, with external legs removed, and all propagators representing full propagators, with self\u2013consistent account of the self energies. In a similar manner, \u02c6\u03a3b is obtained from the lower diagram. According to the general rules discussed above, the fermion self energies are obtained by contraction of interaction operators at \ufb01xed external \ufb01eld indices. Speci\ufb01cally, the fermion self-energy is de\ufb01ned as16 dxdx \u00af\u03c8(x)\u03a3f (x, x)\u03c8(x) = \u2212 i 2 dxdx \u00af\u03c8(x) > \u02c6\u03c6(x)\u03c8(x) \u00af\u03c8(x) \u02c6\u03c6(x)? \u03c8(x), where \u02c6\u03c6 \u2261 \u03c6c\u03c30 + \u03c6q\u03c31 and we have switched to the space-time abbreviation x \u2261 (r, t). The self\u2013consistency inherent to the RPA scheme requires that the contractions must be 15 The o\ufb00-diagonal block ( \u02c6G\u22121)K is given by ( \u02c6G\u22121)K = \u2212( \u02c6G+)\u22121 \u02c6GK ( \u02c6G\u2212)\u22121 = ( \u02c6G+)\u22121 \u02c6F \u2212 \u02c6F ( \u02c6G\u2212)\u22121 = 2i\u03b4 \u02c6F, where the last identity holds if \u02c6F commutes with ( \u02c6G\u00b1)\u22121. 16 Should you feel uncertain about this result, it is worthwhile investing some time to recapitulate the perturbative construction of self energies. 724 Nonequilibrium (quantum) interpreted to produce the full Green functions of the theory. Comparison of the two sides of the equation and usage of Eq. (11.60) leads to the identi\ufb01cation, > \u03a3f (x, x) = \u2212 i 2 = \u2212 i 2 \u02c6\u03c6(x)\u03c8(x) \u00af\u03c8(x) \u02c6\u03c6(x)? = \u2212 1 2 > \u02c6\u03c6(x) \u02c6G(x, x) \u02c6\u03c6(x)? \u02c6G(x, x)DK(x, x) + \u02c6G(x, x)\u03c31D+(x, x) +\u03c3 1 \u02c6G(x, x)D\u2212(x, x), where we have adopted the common convention to designate bosonic Green functions by D+,\u2212,K. Using this expression, the components of the self-energy operator \u02c6\u03a3f = 6 7, \u02c6\u03a3+ f 0 \u02c6\u03a3K f \u02c6\u03a3\u2212 f are identi\ufb01ed as f (x, x) = \u2212 i \u03a3\u00b1 2 f (x, x) = \u2212 i \u03a3K 2 = \u2212 i 2 G\u00b1(x, x)DK(x, x) +G K(x, x)D\u00b1(x, x) GK(x, x)DK(x, x) +G +(x, x)D+(x, x) +G \u2212(x, x)D\u2212(x, x) (11.61) GK(x, x)DK(x, x) + (G+(x, x) \u2212 G\u2212(x, x))(D+(x, x) \u2212 D\u2212(x, x)), where we have made use of the fact that causality implies G+(x, x)D\u2212(x, x) = 0, etc. Also notice that the causality of retarded and advanced Green functions, D\u00b1(x, x)G\u2213(x, x) = 0, causes the vanishing of the lower right\u2013hand block of the self-energy operator. EXERCISE Show that the bosonic self-energy \u02c6\u03a3b = 0 \u02c6\u03a3+ b, \u02c6\u03a3\u2212 b \u02c6\u03a3K b is given by b (x, x) = \u2212 i \u03a3\u00b1", " 2 b (x, x) = \u2212 i \u03a3K 2 = \u2212 i 2 & & & G\u2213(x, x)GK (x, x) +G \u00b1(x, x)GK (x, x) ', GK (x, x)GK (x, x) +G +(x, x)G\u2212(x, x) +G \u2212(x, x)G+(x, x) GK (x, x)GK (x, x) \u2212 (G+ \u2212 G\u2212)(x, x)(G+ \u2212 G\u2212)(x, x) '. (11.62)'In the present context, where the boson \ufb01eld actually represents \ufb02uctuations in a scalar potential, we may think of the self-energy as an e\ufb00ective polarization operator screening \ufb01eld \ufb02uctuations. 11.5.1 Quasiclassical theory We now have everything in place to address our main task, the computation of the fermion (or boson) distributions which develop in the interacting environment. The picture we should have in mind is that a system of fermions prepared in some initial distribution gets exposed to interactions. We wish to \ufb01nd out how the latter drive the system to a new state of equilibrium. The understanding of this mechanism is the basis for the description of out-ofequilibrium distributions forming, e.g. if the system is exposed to strong external \ufb01elds. 11.5 Kinetic equation Wigner transform 725 In a time-independent situation, all constituents of the theory depend only on time di\ufb00erences, \u0394t, and vary at time scales \u0394t \u223c \u22121 comparable to their characteristic energies. In cases where systematic time-dependent changes occur (e.g. in response to the formation of a nonequilibrium initial state) on time scales t \u0394t, it is convenient to pass to a Wigner transformed representation, \u02c6G(t1, t2) \u2192 G(, t), where t = (t1 + t2)/2 and is conjugate to \u0394t = t1 \u2212 t2. Technically, the Wigner representation of an arbitrary operator \u02c6g \u2261 {g(t, t)} is de\ufb01ned by the transform g(, t) \u2261 d\u0394t ei\u0394tg(t + \u0394t/2, t \u2212 \u0394t/2), g(t1, t2) = d 2\u03c0 e\u2212i(t1\u2212t2)g(, t). (11.63) In applications, we often need to know the Wigner representation of the product of two operators, \u02c6f \u02c6g. Using the de\ufb01nition above, it is straightforward to derive the Moyal product identity ( \u02c6f \u02c6g)(, t) = e\u2212 i 2 (\u2202t1 \u22022 \u2212\u22021 \u2202t2 ) f (1, t1)g(2, t2) 1=2= t1=t2=t = f (, t)g(, tf, g} +... (, t), (\u2202tf (, t)\u2202g(, t) \u2212 \u2202f (, t)\u2202tg(, t)) +... where in the third line we have introduced the Poisson bracket {f, g} = \u2202tf \u2202g \u2212 \u2202f \u2202tg. (11.64) (11.65) The scope of the Wigner transformation actually extends beyond the temporal variables: in general, the distribution functions may exhibit explicit space dependence \u02c6F \u2192 \u02c6F (r). (This will occur, say, if a system is connected to two reservoirs kept at di\ufb00erent temperatures, voltages, etc. In this case, gradients in the particle distributions may form.) We expect the characteristic scale of these variations to be much larger than the microscopic particle wavelengths. In this case, it is convenient to pass to the spatial Wigner transform, \u02c6G(r1, r2) \u2192 \u02c6G(r, p), where r = (r1 + r2)/2 and p is conjugate to \u0394r = r1 \u2212 r2, i.e. g(r, p) \u2261 d\u0394r e\u2212i\u0394r\u00b7pg(r + \u0394r/2, r \u2212 \u0394r/2), ddp (2\u03c0)d ei(r1\u2212r2)\u00b7pg(r, p), where the di\ufb00erent sign re\ufb02ects the conventions standard in spatial and temporal Fourier transforms, respectively. g(r1", ", r2) = (11.66) For later reference, a number of Wigner transform identities are summarized in table 11.1, where the spatial analogue of Eq. (11.65) is de\ufb01ned as {f, g} = \u2202rf \u2202pg \u2212 \u2202pf \u2202rg. 726 Nonequilibrium (quantum) Table 11.1 Wigner transform identities Temporal Spatial Operator \u02c6g(\u02c6t) \u02c6g(\u02c6) = \u02c6g(i\u2202t) [ \u02c6f, \u02c6g] tr(\u02c6g) Wigner representation g(t) g() \u2212i{f, g} +... ddt 2\u03c0 g(, t) Operator \u02c6g(\u02c6r) \u02c6g(\u02c6p) = \u02c6g(\u2212i\u2202r) [ \u02c6f, \u02c6g] tr(\u02c6g) Wigner representation g(r) g(p) i{f, g} +... ddrddp (2\u03c0)d g(r, p) In the following, it is convenient to bundle temporal and spatial transforms into a spatiotemporal Wigner transform, g(x, p) \u2261 d\u0394x ei\u0394xT \u03b7pg(x + \u0394x/2, x \u2212 \u0394x/2), g(x1, x2) = dp e\u2212i(x1\u2212x2)T gpg(x, p), (11.67) where x = (t, x)T = (x1 + x2)/2, p = (, p)T, the Minkovski metric \u03b7 = diag(1, \u22121d), and dx \u2261 dtddx, dp \u2261 dddp/(2\u03c0)d+1. Derivation of the kinetic equation We now set out to derive a di\ufb00erential equation describing the evolution of the distribution function. In view of the time scale separations alluded to above, it is convenient to formulate that equation in its Wigner representation. To this end, we consider the formal operator equation 6 \u02c6+ \u2212 \u02c6H \u2212 \u03a3+ f 0 \u2212\u03a3K f \u02c6\u2212 \u2212 \u02c6H \u2212 \u03a3\u2212 f 7 6 \u02c6G+ 0 \u02c6GK \u02c6G\u2212 7 =. 1 0 0 1 Writing \u02c6GK = \u02c6G+ \u02c6F \u2212 \u02c6F \u02c6G\u2212, this matrix equation is equivalent to the set (\u02c6 \u2212 \u02c6H \u2212 \u02c6\u03a3\u00b1) \u02c6G\u00b1 = 1 (\u02c6 \u2212 \u02c6H \u2212 \u02c6\u03a3+)( \u02c6G+ \u02c6F \u2212 \u02c6F \u02c6G\u2212) \u2212 \u03a3K \u02c6G\u2212 = 0. Multiplying the second equation by ( \u02c6G\u2212)\u22121 and using the \ufb01rst equation, we obtain \u02c6F (\u02c6 \u2212 \u02c6H \u2212 \u02c6\u03a3\u2212) \u2212 (\u02c6 \u2212 \u02c6H \u2212 \u02c6\u03a3+) \u02c6F \u2212 \u02c6\u03a3K = 0. At this stage, it is convenient to pass to the Wigner transform. Making use of the identities summarized in table 11.1, we obtain the kinetic equation (\u2202t + {H, }) F = i(\u03a3K f \u2212 (\u03a3+ f F \u2212 F \u03a3\u2212 f )) \u2261 Icol[F ], (11.68) where the equality is approximate in that higher order contributions to the Moyal product expansion have been neglected. In Eq. (11.68), F = F (, t, r, p) is a function of spatial and temporal variables. The left\u2013hand side of the kinetic equation describes the evolution of the distribution function F under the single particle dynamics. 11.5 Kinetic equation 727 The e\ufb00ect of particle interactions is encapsulated in the right\u2013hand side, which de\ufb01nes the collision term. To lowest order in the Moyal expansion, the operator products appearing in the collision term may be understood as products of Wigner functions, \u03a3+ f F \u2261 f (, t, r, p)F (, t, r, p), etc. As we will see in a moment, the term \u03a3K (\u03a3+F \u2212 F \u03a3\u2212) repre\u03a3+ sents an in-term (out-term), i.e. an interaction-mediated gain (loss) of particle occupancy. EXERCISE Compare the structure of the", " quantum kinetic equation to that of the classical Boltzmann equation discussed in Section 10.3. Try to anticipate the explicit structure of the collision term in our system of interacting electrons. Collision term So far, our discussion of the kinetic equation has been absolutely general; no reference to the speci\ufb01c mechanisms of self-energy generation has been made. To better understand the physics of the collision term, we now return to our example of interacting electrons. EXERCISE Show that the Wigner transform of composite operators with space\u2013time dependence g(x, x)f (x, x) is obtained as g(x, x)f (x, x) \u2192 dpg(x, p)f (x, p \u2212 p). (11.69) Also show that g(x, x)f (x, x) \u2192 dpg(x, p)f (x, p \u2212 p). (11.70) of n n+\u03c9 n+\u03c9 (1 \u2212 n+\u03c9) (1 \u2212 n ) (1 \u2212 n+\u03c9) The collision term governs the rates number occupation changes due to the collisions into and out of other states. This being so, we expect it to be determined by (a) the occupation numbers of the collision partners, and (b) the density of states participating in the interactions. We thus aim for a description in terms of occupation functions F or nf, andspectral functions A \u2261 \u22122 ImG +. Finally, to expose the fermion propagators contained in the self-energy of the boson propagator \u02c6D, we will aim to express the latter in terms of \u02c6\u03a3b. To this end, we note that the relation \u02c6D\u22121 = \u02c6V \u22121\u03c31 \u2212 \u02c6\u03a3b = implies that 0 \u02c6V \u22121\u2212 \u02c6\u03a3+ b \u02c6V \u22121\u2212 \u02c6\u03a3\u2212 b \u2212 \u02c6\u03a3K b \u02c6D\u00b1 = ( \u02c6V \u22121 \u2212 \u02c6\u03a3\u00b1 \u02c6D\u2212. \u02c6DK = \u02c6D+ \u02c6\u03a3K b b )\u22121 = \u02c6D+( \u02c6V \u22121 \u2212 \u03a3\u2213 b ) \u02c6D\u2212, 728 Nonequilibrium (quantum) We now have everything we need to compute the quasiclassical representation of the collision term in Eq. (11.68): Icol[F ](p) (11.61),(11.69) = # dq GK(p + q)DK(q) + (G+ \u2212 G\u2212)(p + q)(D+ \u2212 D\u2212)(q) 1 2 \u2212 (11.57) = \u2212 i 2 (G+ \u2212 G\u2212)(p + q)DK(q) +G K(p + q)(D+ \u2212 D\u2212)(q) dq D+(q)D\u2212(q)A(p + q) # \u00d7 (F (p + q) \u2212 F (p))\u03a3K b (q) + (\u03a3+ b \u2212 \u03a3\u2212 b )(q)(1 \u2212 F (p + q)F (p)) $ $ F (p) (11.62),(11.70) = # 1 4 dqdp D+(q)D\u2212(q)A(p + q)A(p + q)A(p) \u00d7 (F (p + q) \u2212 F (p))(1 \u2212 F (p + q)F (p)) \u2212 (F (p + q) \u2212 F (p))(1 \u2212 F (p + q)F (p) $. We now assume that F (p) = 1\u22122n depends only on energy, de\ufb01ne the transition probability T (,, \u03c9) \u2261 ddp (2\u03c0)d ddq (2\u03c0)d D+(q)D\u2212(q)A(p + q)A(p + q)A(p), and add and subtract a term nn+\u03c9n n+\u03c9 to arrive at the result Icol[F ](p) = d\u03c9d T (, ; \u03c9)(n(1 \u2212 n+\u03c9)n+\u03c9(1 \u2212 n ) \u2212 (1 \u2212 n)n+\u03c9(1 \u2212 n+\u03c9)n ). (11.71) Let us discuss the meaning of this expression: The structure of the collision term resembles that of the classical Boltzmann collision term discussed in Section 10.3. Quantum mechanics enters through the Pauli blocking factors 1 \u2212 n, etc. Much as the collision term of Boltzmann theory was annihilated by the Maxwell\u2013 Boltzmann distribution, the collision term Eq. (11.71) van", "ishes on the Fermi\u2013Dirac equation: interactions serve to install local equilibrium con\ufb01gurations, at a temperature T and chemical potential \u03bc which remain apriori unspeci\ufb01ed. Departures from equilibrium result from the competition between the left\u2013hand side and the right\u2013hand side of the kinetic equation. In many instances the system may end up in a compromise situation governed by a Fermi-Dirac distribution with spatially varying T (q) and\u03bc( q). Notice that the collision term vanishes in an approximation where the e\ufb00ects of screening in \u02c6D \u03c31 \u02c6V are ignored (technically, by ignoring the RPA self-energy of the boson line). Diagrammatically, the building blocks \u02c6D+\u2212 \u02c6D\u2212 = \u02c6D+( \u02c6\u03a3+\u2212 \u02c6\u03a3\u2212) \u02c6D\u2212 and \u02c6DK = \u02c6D+ \u02c6\u03a3K \u02c6D\u2212 are described by the boson insertion to the self-energy diagram shown in the bottom part of the \ufb01gure above. (Identify the two wavy lines with the self\u2013consistent RPA \u02c6DR and \u02c6DA, respectively, and the center bubble with the self-energy \u02c6\u03a3.) Heuristically, we may interpret this structure as the modulus square of the fundamental RPA decay amplitudes shown in the upper part of the \ufb01gure: the out\u2013term is due to the decay of a particle at energy into a particle hole excitation (energies and + \u03c9, respectively) and a particle at energy + \u03c9. This decay is mediated by the (now self\u2013consistently screened) interaction 11.6 A mesoscopic application 729 line at frequency \u03c9. In an analogous manner, the collapse of a particle\u2013hole excitation governs the \u201cin\u201d rate. These processes conserve energy and momentum, up to the smearing introduced by higher order decay processes. (Technically, the spectral functions A contain \u03b4-distributions \u03b4\u03a3( \u2212 (p)), where the subscript \u03a3 indicates their broadening by retarded self energies. As an exercise, consider how these self energies encapsulate the generation of higher order particle-hole decay processes, as described by the RPA approximation.) 11.6 A mesoscopic application In the previous sections we have set up the essential machinery to study quantum systems beyond thermal equilibrium. We have understood how interaction e\ufb00ects aim to drive a system back to a (local) equilibrium con\ufb01guration and how departures from that con\ufb01guration can, in principle, be described. However, we have not yet explored the actual physical e\ufb00ects a nonequilibrium environment has on the quantum phenomenology of a system. Of course, this latter question has many facets to it and cannot be addressed in general terms. Rather, we focus on one speci\ufb01c aspect, and we study it within the framework of a concrete application: quantum mechanics is largely about the coherent propagation of matter waves, and the resulting interference phenomena. We know that coherence is averse to any kind of noisy \ufb02uctuations in environmental degrees of freedom. At the same time, noise is an inevitable companion of any coupling to a dissipative medium, and it is likely to increase if that coupling generates a nonequilibrium state. In this section, we explore the formation of noise\u2013levels in a quantum system coupled to a nonequilibrium environment, and its destructive in\ufb02uence on quantum coherence. Questions of this type are attracting a lot of current attention: advances in experimentation make it possible to explore the physics of nanoscopic systems that are strongly in\ufb02uenced by quantum interference phenomena \u2013 from ultracold atom physics, quantum information devices, and spintronics, to single molecule electronics, and beyond. At the same time, the very smallness of these systems implies that they are easily driven out of equilibrium, or are in an out of equilibrium state per se. Studying the in\ufb02uence of nonequilibrium noise on quantum coherence has, therefore, emerged as a mainstream topic. 730 Nonequilibrium (quantum) 11.6.1 Out-of-equilibrium quantum dot In this section,17 we investigate the interplay between noise and coherence on a speci\ufb01c setup, the out-of-equilibrium metallic quantum dot. Consider a metallic or semiconducting island of mesoscopic size.18 The island is connected by tunnel electrodes to two leads, which are kept at a voltage di\ufb00erence V. This voltage gradient will be the primary instance by which the system is driven out of equilibrium. Finally, we assume capacitive coupling to a", " gate electrode (indicated by the horizontal plate in the \ufb01gure) which determines the electrostatically-preferred number of electrons on the dot. We describe this system by the Hamiltonian \u02c6H = x=L,R( \u02c6Hx + \u02c6Ht,x) + \u02c6Hd + \u02c6Hc, where \u02c6Hx = \u02c6Ht,x = \u02c6Hd = \u02c6Hc = ab a\u03bc \u03bc\u03bd EC 2 c\u2020 x,aHx,abcx,b, c\u2020 x,aTx,a\u03bcd\u03bc + h.c., d\u2020 \u03bcHd,\u03bc\u03bdd\u03bd, 72 d\u2020 \u03bcd\u03bc \u2212 N0. 6 \u03bc (11.72) Here cx=L/R,a create fermions in state |a of the left/right lead, respectively, d\u03bc creates fermions in state |\u03bc of the dot, and \u02c6Hx and \u02c6Hd are the corresponding single-particle Hamiltonians. The operators \u02c6Tx=L/R describe tunneling between the leads and the dot. Finally, EC is the electrostatic charging energy on the dot, and N0 de\ufb01nes its electrostaticallypreferred charge. (Notice that N0 need not be integer.) The voltage bias between the leads will be introduced momentarily through a chemical potential di\ufb00erence. EXERCISE Recapitulate the discussion of the equilibrium physics of this system in Problems 6.7 and 6.7. 17 This section is technical in nature. A few unanswered problems aimed at developing computational \ufb02uency in Keldysh \ufb01eld theory have been included in the text. (It is not necessary to tackle these problems at \ufb01rst reading.) 18 By \u201cmesoscopic,\u201d we mean that the temperature-dependent dephasing times over which quantum phase coherence is lost due to the interaction with environmental degrees of freedom exceed all time scales relevant to transport in the system. In practice, this means that we are studying systems of O(1\u03bcm) or less at temperatures of O(10(1\u22122)mK). System sizes/temperature ranges of this order are easily accessible to modern device technology. 11.6 A mesoscopic application 731 Generalizing the theoretical formulation introduced in Problems 6.7 and 6.7, we next aim to derive a Keldysh theory of the nonequilibrium system. Our starting point is the fermion x=L,R(Sx[ \u00af\u03c8x, \u03c8x] +S t,x[ \u00af\u03c8x, \u03c8x, \u00af\u03c8d, \u03c8d]) + Sd[ \u00af\u03c8d, \u03c8d, V ] +S c[V ], where action S[ \u00af\u03c8, \u03c8, V ] \u2261 Sx[ \u00af\u03c8x, \u03c8x] = dt \u00af\u03c8x 6 7 i\u2202t + i\u03b4 + EF \u2212 \u02c6Hx 0 6 7 2i\u03b4Fx i\u2202t \u2212 i\u03b4 + EF \u2212 \u02c6Hx \u03c8x, 7 \u02c6Tx 0 0 \u02c6Tx \u03c8d + \u00af\u03c8d 6 \u02c6T \u2020 x 0 0 \u02c6T \u2020 x \u03c8x, St,x[ \u00af\u03c8x, \u03c8x, \u00af\u03c8d, \u03c8d] = dt \u00af\u03c8x 6 dt \u00af\u03c8d Sd[ \u00af\u03c8d, \u03c8d, V ] = Sc[V ] = dt 1 2EC VcVq + N0Vq. i\u2202t + i\u03b4 + EF \u2212 \u02c6H0,d \u2212 Vc \u2212Vq/2 \u2212Vq/2 + 2i\u03b4Fd i\u2202t \u2212 i\u03b4 + EF \u2212 \u02c6H0,d \u2212 Vc Here, the distribution functions F L R = coth, \u00b1 V /2 2T 7 \u03c8d, (11.73) (11.74) contain the information on the biasing of the system and Vc,d are the classical and quantum components of a Hubbard-Stratonovich \ufb01eld decoupling the interaction. EXERCISE Adapt the Hubbard\u2013Stratonovich transformation introduced in Section 11.4.2 to the charging interaction in Eq. (11.72) to reproduce the action above. In Eq. (11.73), the Hubbard\u2013Stratonovich classical and quantum \ufb01elds Vc and Vd are de\ufb01ned as Vc \u2261 1 2 (V+ + V\u2212), Vq \u2261 V+ \u2212 V\u2212,", " in terms of the \ufb01elds V\u00b1 decoupling the interaction on the contours. This di\ufb00ers by a factor of 2 from earlier conventions and facilitates the interpretation of Vc as a classical degree of freedom. \u221a As with the equilibrium situation studied in Problem 6.7, it will be convenient to remove the \ufb02uctuating \u201cvoltage\u201d V by a gauge transformation on the dot: introducing the change of variables, \u03c8d \u2192 e\u2212i \u02c6\u03c6\u03c8d, \u00af\u03c8d \u2192 \u00af\u03c8dei \u02c6\u03c6, where \u02c6\u03c6 = \u03c6c + \u03c6q\u03c31/2 and the phases are de\ufb01ned by \u2202t\u03c6c,q = Vc,d, V disappears from the bulk action while the tunneling matrix becomes dynamical,19 \u02c6Tx \u2192 \u02c6Tx e\u2212i \u02c6\u03c6 \u2261 \u02dcTx. At this point, the fermions can be integrated out to generate the familiar \u201ctr ln.\u201d The rest of the derivation then proceeds as in problem 6.7: EXERCISE Adapt the derivation of Problem 6.7 to the present context: Expand the \u201ctr ln\u201d to second order in the tunneling matrix elements (recall what sort of assumptions may justify the second order truncation) to generate the tunneling action Stun[\u03c6] = i x tr( \u02c6Gx \u02dcTx \u02c6Gd \u02dcT \u2020 x ), 19 The gauge transformation also alters the distribution function Fd (exercise: how?). 732 Nonequilibrium (quantum) Evaluate this expression under the following simplifying assumptions: (i) the Hamiltonian operators \u02c6Hx and \u02c6Hd are diagonal in the bases |a and |\u03bc, respectively, (ii) in the range of relevant energies, the modulus |Tx,a\u03bc|2 \u2261 |Tx|2 does not signi\ufb01cantly depend on Hilbert space indices, and (iii) the spectral functions \u02c6Ax,d() = \u22122Im \u02c6G+ x,d() = 2\u03c0\u03bdx,d() are approximately energy independent. (Here \u03bdx,d are the single particle density of states of leads and dot, respectively.) Show that under these conditions, the tunneling action assumes the form Eq. (11.75). Hint: you may \ufb01nd it convenient to represent the Green functions as \u02c6G\u00b1() = d 2\u03c0 \u02c6A( ) \u00b1\u2212. The derivation sketched in the formulation of the exercise above then leads to the tunneling action Stun[\u03c6] = \u2212 gxtr(\u039bxe\u2212i \u02c6\u03c6\u039bdei \u02c6\u03c6), (11.75) i 4 x where \u039bx,d is a matrix with energy dependence, \u039bx,d \u2261 2i g+ gK x,d g\u2212 0, gK x,d \u2261 (g+ \u2212 g\u2212)Fx,d, (11.76) the unit spectral density auxiliary \u201cGreen function\u201d is de\ufb01ned by g\u00b1 \u2261 gx \u2261 4\u03c02\u03bdd\u03bdx|Tx|2, d 2\u03c0 1 \u00b1\u2212, and (11.77) de\ufb01nes the tunneling conductance between dot and lead x = L/R. Adding to this the phase representation of the charging action in Eq. (11.73), Sc[\u03c6] = dt \u2202t\u03c6c\u2202t\u03c6q + N0\u2202t\u03c6q, (11.78) 1 2EC we obtain a prototypical variant of the e\ufb00ective action in the phase representation, S[\u03c6] = Sc[\u03c6] +S tun[\u03c6]. In the case of strong interactions, phase \ufb02uctuations are largely controlled by the tunneling action. Speci\ufb01cally, in nearly insulated dots, gx 1, the phases \ufb02uctuate wildly, and no longer represent suitable degrees of freedom. (Exercise: interpret this phenomenon in terms of the voltages Vc,q = \u2202t\u03c6c,q; why do voltage \ufb02uctuations in the nearly decoupled dot become unimpeded?) As is usual in cases where a variable goes havoc, this limit is better described in terms of the canonically conjugate partner variable, which in the present context carries the signi\ufb01cance of a charge degree of freedom, n. (Indeed, the dot charge will \ufb02uctuate only moderately in the limit of small gx.) Technically, the charge variable is introduced by a", " Hubbard\u2013Stratonovich decoupling of the charging term. This leads to the action in the phase-charge representation, S[n, \u03c6] \u2261 dt (nc\u2202t\u03c6q + nq\u2202t\u03c6c \u2212 2Ec(nc + N0)nq) +S tun[\u03c6]. (11.79) One may think of the above transformation as passing from a Lagrangian variant of the theory \u2013 one variable, \u03c6, governed by an action with \u201cvelocity square\u201d term \u02d9\u03c62 \u2013 to a (n \u02d9\u03c6 \u2212 H(n, \u03c6)) de\ufb01ned in the phase space of variables (n, \u03c6). Notice Hamiltonian action that the classical component nc is conjugate to the quantum \u03c6q, and vice versa. 11.6 A mesoscopic application 733 But what about charge quantization? We should expect charge in the isolated dot gx \u2192 0 to be quantized to integer values, but our action above does not appear to know of this constraint. As in many other instances before, information on the quantization of n resides in the boundary conditions of its conjugate variable \u03c6. So far, we have ignored these boundary conditions. But as we are dealing with phases, we should be alert and expect windings by multiples of 2\u03c0 as \u03c6c and/or \u03c6q propagate along the Keldysh contour. EXERCISE To explore the topology of the phase \ufb01eld, it is best to go back to the framework of the closed Keldysh contour s \u2208 [0, S]. It is on this contour, where the Hubbard\u2013Stratonovich \ufb01eld \u03c6(s) is de\ufb01ned in the \ufb01rst place. Being a phase \ufb01eld, it is de\ufb01ned only up to multiples of 2\u03c0, i.e. we have the boundary conditions \u03c6(S) = \u03c6(0)+2\u03c0W, where the integer W has to be summed over (cf. our discussion of functional integrals over S1-valued \ufb01elds in Chapter 3). Implement the condition above by the ansatz, \u03c6(s) = \u02dc\u03c6(s) + 2\u03c0W 2T (s \u2212 T ), where \u02dc\u03c6 is periodic on the s-contour, and we assume the real time contour to be parameterized by t \u2208 [0, T ]. Show that this condition translates to an unconstrained \ufb01eld \u03c6c = \u02dc\u03c6c, and to \u03c6q(t) = \u02dc\u03c6q(t) + 2\u03c0W T (t \u2212 T ), (11.80) with Dirichlet conditions \u02dc\u03c6q(0) = \u02dc\u03c6q(T ) = 0. In the exercise above, it is shown that the information on phase windings resides in the quantum \ufb01eld, \u03c6q. Heuristically, quantization conditions for the \ufb01eld n may then be deduced as follows. Assuming that the winding number dependence of the tunneling action is negligible (the rationale being that in the limit T \u2192 \u221e the W -contribution to Eq. (11.80) is a nearly constant phase shift which is likely to cancel out in a term probing the local tunneling T dynamics), we observe that W enters the action through the contribution 2\u03c0W 1 0 dt nc. T 0 dt nc \u2208 Z: the temporal average of Summation over W then generates the condition 1 T classical charge on the dot is integer. But since the operators \u223c exp(i \u02c6\u03c6) in the action that change the charge do so in discrete steps,n c \u2192 nc \u00b1 1/2, and the numbers of positive (exp(i \u02c6\u03c6)) and negative (exp(\u2212i \u02c6\u03c6)) jumps are equal, this condition calls for an initial charge value nc(0) \u2208 Z. In the course of time, this initial value may change, but it does so in integer steps, i.e. the classical charge nc(t) is quantized. T EXERCISE If you \ufb01nd the argument above too vague, you may derive charge quantization in rigorous terms. Do not ignore the W -dependence of the tunneling term. Rather, expand exp(iS) schematically to nth order in Stun. The winding numbers then enter as a pure phase exp(2\u03c0iF W ), where F is an expression involving nc, and the discrete set of times at which the phases of the tunneling terms are evaluated. Summation over W generates", " the quantization condition F \u2208 Z. Analyse F to show that this condition is equivalent to the initial value quantization nc(0) \u2208 Z. In combination with the integer steps taken by the tunneling dynamics, this rigorously proves charge quantization. In e\ufb00ect, we may account for the windings of the phase \ufb01elds by limiting the integration path nc(t) to integer con\ufb01gurations, and thence forget about the winding numbers. 734 Nonequilibrium (quantum) 11.6.2 Dot distribution function Hidden in the action (11.79) sits the distribution function of the dot (cf. Eq. (11.75) and (11.76)). In principle, we should think about the distribution function as an observable of the theory, a quantity that determines the distribution of the particles on the dot after all degrees of freedom (including the interaction) have been integrated out. However, the results of that integration actually depend on the dot distribution function, i.e. the rigorous determination of the latter poses a recursive problem whose solution is likely out of reach. In practice, one will rather work with a trial distribution, which should of course be reasonably close to the actual distribution. In the present context, we will approximate Fd by the distribution function of the noninteracting dot. The rationale behind this ansatz is the assumption that the coupling to the leads will adjust an e\ufb00ective distribution on the dot, at time scales \u223c \u03bd|T |2, faster than the relaxation times due to interactions. The quality of such assumptions must of course be checked in retrospect. Trial dot distribution function But what will the distribution function on the dot, in the absence of interactions, look like? There are various ways to answer this question. Heuristically, one may argue that all single particle states on the dot will be occupied up to energy \u2212V /2, i.e. up to the threshold energy above which the band of the right lead begins to dry up (cf. the \ufb01gure.) At larger energies, states on the dot are hybridized with isoenergetic \ufb01lled states in the left lead, and empty states in the right lead. These coupling mechanisms are mutually independent, and occur at a strength gL,R, respectively. This suggests, that the distribution on the dot Fd = gLFL + gRFR gL + gR, (11.81) E n V E will be an additive superposition of the lead distributions. The corresponding distribution function nd = (\u2212Fd + 1)/2 assumes the form of a double step distribution (cf. the \ufb01gure.) EXERCISE Substantiate the argument above: To this end, consider the tunneling operators \u02c6Tx as a \u201cperturbation\u201d, and compute the corresponding self energies. Then consider the kinetic equation (11.68). For a stationary solution F, the left\u2013hand side of the equation vanishes. Show that the vanishing of the right\u2013hand side enforces the double step distribution. 11.6 A mesoscopic application 735 Readers who feel comfortable enough with these ways of \ufb01xing the distribution function may advance to the next section. For the bene\ufb01t of all others, we devote the rest of this section to showing how the (non-interacting) distribution function may be obtained directly from the e\ufb00ective action. To this end, we vary the tunneling action (11.75) in \u02c6\u03c6 and adjust \u039bd such that the non-interacting con\ufb01guration \u02c6\u03c6 = 0 is a stationary point. In the exercise below, we show the equivalence of this procedure to the kinetic equation approach above. We need to determine \u039bd such that the linear expansion of the action in \u02c6\u03c6 = \u03c6c + \u03c6q\u03c31 vanishes. It is straightforward to verify that variations in \u03c6c vanish regardless of the distribution function. As for the quantum component, we generalize \u03c6q\u03c31 \u2192 \u02c6\u03c6q \u2261 Re \u03c6q\u03c31 \u2212 Im \u03c6q\u03c32, which may be interpreted as a shift of the generator \u03c6q into the complex plane. It is then a straightforward exercise to verify that the variation of the action \u03b4 \u00af\u03c6S = 0 generates equation (11.81). But what might be the logic behind this variational approach? The answer is given in the following exercise: 0 EXERCISE The distribution function on the dot can be found by solution of an appropriate kinetic equation, ( \u02c6G\u22121 \u2212 \u03a3) \u02c6G = 1. In this exercise, we want to show that the component of this matrix equation relevant to the determination of Fd can be found by introduction of a suitable source. To this end, let", " us go back to the microscopic variant of the theory, Eq. (11.73), and generalize i\u03b4\u03c33 \u2192 i\u03b4 ei \u02c6\u03c6q \u03c33e\u2212i \u02c6\u03c6q in the dot action, Sd. Verify that the expansion of the action to \ufb01rst order in \u00af\u03c6q generates the Keldysh component of the kinetic equation. Under the assumption of constant spectral density (also underlying the derivation of the tunneling action), the vanishing of the linear variation becomes equivalent to the ful\ufb01lment of the kinetic equation. The stationary solution is given by Eq. (11.81). For notational simplicity, we assume contacts of equal transparency, throughout the rest of the section. gL = gR \u2261 g, Tunneling action We may use the result Eq. (11.81) to bring the tunneling action into a form more manageable than the abstract representation Eq. (11.75). To this end, let us introduce the notation c \u2261 ei\u03c6c cos(\u03c6q/2), s \u2261 ei\u03c6c sin(\u03c6q/2). We substitute ei \u02c6\u03c6 = c + is\u03c31 and e\u2212i \u02c6\u03c6 = \u00afc \u2212 i\u00afs\u03c31 into Eq. (11.75) and obtain Stun[\u03c6] = 2g = 2g dtdt (\u00afc, \u2212i\u00afs)t d\u03c9 2\u03c0 (\u00afc, \u2212i\u00afs)\u03c9 \u03a3\u2212 0 \u03a3+ \u03a3K \u03a3\u2212 0 \u03a3+ \u03a3K t,t \u03c9 c is t c is, \u03c9 (11.82) 736 Nonequilibrium (quantum) where (notice the similarity to the self-energy of a boson coupled to fermions Eq. (11.62)), \u03a3\u00b1 t\u2212t = \u03a3K t\u2212t = i 2 i 2 x x g\u00b1 t\u2212t gK x,t\u2212t + gK d,t\u2212t g\u2213 t\u2212t, x,t\u2212tgK gK d,t\u2212t \u2212 (g+ \u2212 g\u2212)t\u2212t (g+ \u2212 g\u2212)t\u2212t. Like so often, we focus on the damping contribution to the self energy, \u03a3+ \u2212 \u03a3\u2212. Using (g+ \u2212 g\u2212) = \u2212i, we obtain d 2\u03c0 (Fd, \u2212 Fx,\u2212\u03c9) = i (\u03a3+ \u2212 \u03a3\u2212)\u03c9 = (11.83) \u03c9 \u03c0 i 2, \u03a3K \u03c9 = i 2 x x d 2\u03c0 (1 \u2212 Fd,Fx,\u2212\u03c9) = i 4\u03c0 s=\u00b11 (\u03c9Fb(\u03c9) + (\u03c9 + sV )Fb(\u03c9 + sV )), where we used Eq. (11.74) and the relations, \u03c9)F ()) = \u03c9 thing, we obtain 2\u03c0 (1 \u2212 F ( \u2212 \u03c0 Fb(\u03c9), with F () = tanh(/2T ), and Fb(\u03c9) = coth(\u03c9/2T ). Combining every- 2\u03c0 (F ( + \u03c9) \u2212 F ()) = \u03c9 \u03c0, d d Stun[\u03c6] = g \u03c0 d\u03c9 2\u03c0 (\u00afc, \u2212i\u00afs)\u03c9 \u2212i\u03c9 0 i\u03c9 2iK(\u03c9) c is, \u03c9 (11.84) where K(\u03c9) \u2261 \u03c9 tunneling contribution Eq. (11.84) is the basis of all our further discussion. s=\u00b11(\u03c9 + sV )Fb(\u03c9 + sV ). The action Eq. (11.79) with the 2 Fb(\u03c9) + 1 4 11.6.3 Observables Before proceeding, we should choose an observable in terms of which the system\u2019s behavior will be described. While the system\u2019s total conductance (as opposed to the tunneling conductances of its individual dot-lead interfaces) might be the \ufb01rst option coming to one\u2019s mind, we here focus on the dot tunneling density of states, (TDoS). The latter is de\ufb01ned as20 \u03bd() \u2261 \u2212 1 \u03c0 Im tr ( \u02c6G+()) = \u2212 1 \u03c0 Im dt eittr ( \u02c6G+(t)). (11.85) What makes the TDoS an interesting observable to consider? The TDoS can be measured by tunneling spectroscopy. See Fig. 11.4 for an example of local probes of the TDoS in a mesoscopic superconductor\u2013normal\u2013superconductor (SNS) wire. (", "Barely discernible in the \ufb01gure is the straight normal wire, adjacent to a horseshoe shaped superconductor.) In this system, the TDoS is suppressed due to the superconductor proximity e\ufb00ect, the e\ufb00ect that a superconductor tends to induce pair correlations leading to a spectral gap in the adjacent normal metallic regions. The size of the gap as a function of voltage (energy) diminishes with the distance from the superconductor. 20 The notation assumes a steady state con\ufb01guration. In principle, \u03bd(, t) may be explicitly time dependent, via temporal changes in distribution functions. 11.6 A mesoscopic application 737 Figure 11.4 Tunneling density of states as a function of voltage (energy) in a mesoscopic SNS structure. The arrows mark the position where, in the normal wire, the TDoS was recorded. Figure taken from H. le Sueur, P. Joyez, H. Pothier, C. Urbina, and D. Esteve, Phase controlled superconducting proximity e\ufb00ect probed by tunneling spectroscopy, Phys. Rev. Lett. 100, 197002-6 (2008). At the same time, the TDoS is an observable of considerable conceptual interest. As is obvious from its de\ufb01nition, the function \u03bd() probes the amplitude of quasiparticle propagation in the system at time scales t \u223c \u22121. This propagation amplitude comes with a quantum dynamical phase. Out of thermal equilibrium, we expect this quantum phase to be susceptible to all kinds of nonequilibrium \ufb02uctuations. In this sense, the TDoS is an observable very close to the central theme of this chapter. More generally, the TDoS probes the general mechanism of the \u201corthogonality catastrophe\u201d: the quantum state \u201cbare quasiparticle plus unperturbed system\u201d forming right after the tunneling event will typically be very di\ufb00erent (\u201corthogonal\u201d) from the asymptotic stationary state in which the incoming particle has been accommodated into the system. The formation of that \ufb01nal state generally involves the re-adjustment of a large number of particles and may take much more time than, say, the tunneling process as such. This means that in strongly interacting systems, the TDoS is expected to show signi\ufb01cant structure at small energy scales \u2013 the zero bias anomaly. In fact, the Coulomb blockade discussed in Problem 6.7 can be viewed as an extreme manifestation of the zero bias anomaly: tunneling onto an isolated quantum dot creates a state that is energetically forbidden, or \u201csub-barrier,\u201d by an amount Ec. This means that the TDoS vanishes over an interval from = 0 to = EC. (Recall that energy is measured from the Fermi energy.) Equivalently, we may say that tunneling processes onto the dot are limited to short duration t \u223c E\u22121. Tunneling into a metallic system leads to more intricate behavior. The tunneling process generates an initial charge distribution which is energetically unfavourable (sub-barrier). In the course of time, that initial distribution will relax in what typically will be a di\ufb00usive spreading process. The action associated c 738 Nonequilibrium (quantum) to that process weights the quasiparticle propagator and hence the TDoS. For example in a two-dimensional setting, the long time relaxation action diverges,21 which implies singular behavior of the TDoS at low energies. The open quantum dot discussed in this section forms a compromise between a closed dot and a bulk metallic system. Below we try to understand the TDoS of this system, and the rami\ufb01cations of nonequilibrium \ufb02uctuations. However, before doing so, we \ufb01rst need to express \u03bd() in terms of our e\ufb00ective degrees of freedom. Fortunately, this is a straightforward task. It turns out to be best to work in the representation of the theory prior to the Keldysh rotation. Comparison with the formulae in Section 11.4.1 shows that \u03bd() = \u2212 1 2\u03c0i dt eit G\u2212+(t) \u2212 G+\u2212(t). To give the Green functions G\u00b1\u2213 some meaning, we just need to note two things: in the prest = ei(\u03c6\u2212(t)\u2212\u03c6+(t))G\u2212+ ence of interactions G\u2212+ \u00af\u03c8+ 0,tt, where G0 is the non-interacting Green function. For a single level, its form is shown in Eq. (11.52). For our many level dot this generalizes to t \u2192 \u2212iei(\u03c6", "\u2212(t)\u2212\u03c6+(t))\u03c8\u2212 \u00af\u03c8+ tt = \u2212i\u03c8\u2212 t t G\u2212+ 0,tt = \u2212i e\u2212i\u03bc(t\u2212t)(1 \u2212 nd(\u03bc)) = \u2212i d \u03bd()e\u2212i(t\u2212t)(1 \u2212 nd()) \u03bc \u22122\u03c0i\u03bdd(1 \u2212 nd)(t \u2212 t), where nd(t) is the Fourier transform of the function n() and we assumed constant noninteracting density of states \u03bdd as before.22 Combining this with the analogous construction for the Green function G\u00b1, we obtain \u03bd() = \u03bde() +\u03bd h(), where \u03bde() = \u03bdd Re dt eitei(\u03c6\u2212(t+\u00aft)\u2212\u03c6+(\u00aft))(1 \u2212 nd)(t), \u03bdh() = \u03bdd Re dt eitei(\u03c6+(t+\u00aft)\u2212\u03c6\u2212(\u00aft))nd(t). (11.86) Here \u03bde,h are to be interpreted as the contributions of electrons (holes) tunneling onto the dot, and \u00aft is an arbitrary reference time marking the beginning of the tunneling process. The hole contribution, \u03bdh, di\ufb00ers from\u03bd e in the replacement (1 \u2212 nd) \u2192 nd, and a sign change in the phase of the oscillatory exponential. 11.6.4 Open quantum dot In this section, we study the TDoS at large values of the tunneling conductance, g 1. In this limit, \ufb02uctuations of the phase \u03c6 are largely quenched. Conversely, the conjugate degree of freedom, the charge n, \ufb02uctuates strongly \u2013 the open quantum dot admits large variations of charges and currents. The discussion of the open system is rewarding in that 21 L.S. Levitov and A. V. Shytov, Semiclassical theory of the Coulomb anomaly, Pisma Zh. Eksp. Teor. Fiz. 66, 200-205 (1997) [JETP Lett. 66, 214-221 (1997)]. 22 Notice that (1 \u2212 nd)(t) =\u03b4 (t) \u2212 nd(t) is to be understood as the Fourier transform of 1 \u2212 nd(). 11.6 A mesoscopic application 739 it enables us to bridge between classical and quantum phenomena. On the one hand, the smallness of (quantum) \ufb02uctuations of the phase means that we stay close to the classical limit; in many respects, our system behaves like a classical RC-resistor unit, kept at voltage \u223c \u02d9\u03c6. On the other hand, Eq. (11.86) demonstrates that the same phase determines the quantum mechanics of the observable \u03bd. Indeed, we discover that the physics of the orthogonality catastrophe in the system can largely be understood in terms of classical voltage \ufb02uctuations. At large g, the tunneling action may be expanded to second order in \u03c6 around its sta- tionary con\ufb01guration \u03c6 = 0.23 As a result of this expansion we obtain S(2)[\u03c6] = 1 2EC dt \u2202t\u03c6c\u2202t\u03c6q + g 2\u03c0 d 2\u03c0 (\u03c6c, \u03c6q)\u03c9 \u2212i\u03c9 0 i\u03c9 iK(\u03c9) \u03c6c \u03c6q, \u03c9 (11.87) where the kernel K is de\ufb01ned in Eq. (11.84) and we used the phase representation of the charging action Eq. (11.78). We have also neglected the N0-contribution to the charging action, the formal reason being that for large g, topologically excited (\ufb01nite winding numbers dt \u2202t\u03c6q = W ) con\ufb01gurations of the phase \ufb01eld do not play a role. However, for W = 0, N0 0. (Exercise: what is the physical reason for the irrelevancy of the N0-term in the widely open quantum dot?) Except for the structure of the kernel K, this action is reminiscent of the standard MSRJD\u2013actions describing dissipative classical systems. To understand the rami\ufb01cations of this connection, we go one step backwards in the direction of classical physics and decouple the \u03c6q\u03c6q-term to obtain a Langevin-type system: S[\u03c6, \u03be] = dt \u03c6q \u2202t \u03c6c + \u03be + i \u03be\u2212\u03c9. (11.88) \u2212 1 2EC \u22022 t \u2212 2g 2\u03c0 d\u03c9 2\u03c0", " \u03be\u03c9 \u03c0 2gK(\u03c9) Classical resistor network Integration over \u03c6q in Eq. (11.88) freezes the classical \ufb01eld to the con\ufb01guration solving the equation where the noise \ufb01eld \u03be is correlated according to 1 2EC \u22022 t + 2g 2\u03c0 \u2202t \u03c6c = \u03be, \u03be(t)\u03be(t) = K(t \u2212 t), g \u03c0 d\u03c9 2\u03c0 e\u2212i\u03c9tK(\u03c9) is the Fourier transform of the kernel K. This is the Langevin and K(t) = equation for voltage \ufb02uctuations on the classical resistor network equivalent to our quantum dot. The network is depicted in the \ufb01gure where the central node represents the quantum dot, coupled by two resistors of resistance R = g\u22121 to a bias voltage source. The dot is also 23 Anharmonic \ufb02uctuations are suppressed in g\u22121, but weakly singular in the infrared time scales of the theory. This indicates scaling of the coupling constants (g, Ec) of the theory, which, however, we do not discuss in this text. 740 Nonequilibrium (quantum) coupled to a capacitor of capacitance C = 1/2EC. That capacitor is kept at a gate voltage VG = N0/C (which, however, we saw is irrelevant to the physics of the open dot). To make the connection to classical circuit dynamics more explicit, we change notation in the Langevin equation above according to24 g = 2\u03c0 R, EC = 1 2C, U = \u2202t\u03c6c, \u03be = C\u03b7. The Langevin equation describing voltage \ufb02uctuations o\ufb00 the static value V /2 then assumes the form \u2202tU + \u03b3U = \u03b7, (11.89) C K(t \u2212 t) and we have introduced the parameter \u03b3 \u2261 2 where \u03b7(t)\u03b7(t) = \u03b3 RC. The meaning of this equation is not di\ufb03cult to understand: in the absence of \ufb02uctuations, \u03b7 = 0, the voltage on the dot relaxes to its stationary value, V /2, at the RC-time scale characteristic for a capacitive shunt. (The relaxation time scale is actually 2/RC, because what matters is the resistance R/2 of a dot shunted by two parallel resistors.) In contrast to our previous studies of dissipative systems, the coupling to two distinct leads (\u201cbaths\u201d) implies that the voltage distribution on the dot cannot relax to an equilibrium con\ufb01guration. To explore this point, let us take a close look at the \ufb02uctuation correlator, K: In the absence of biasing, V = 0, and in the classical limit ()\u03c9 \u2192 0, we have K(\u03c9) =\u03c9 coth(\u03c9/2T ) \u03c9\u21920\u2212\u2192 2T, and \u03b7(t)\u03b7(t) = 2T \u03b3 C \u03b4(t \u2212 t). Comparison with our discussion of Langevin equations in the previous chapter then shows that the steady state distribution controlling the voltage on the dot takes the form P (V ) = N e\u2212 U 2 C 2T. This is the Maxwell\u2013Boltzmann distribution for a system with capacitive energy U 2C/2. Still assuming zero bias, V = 0, let us now explore the zero temperature limit. In this case, K(\u03c9) =\u03c9 coth(\u03c9/2T ) T \u21920\u2212\u2192 |\u03c9|. To understand the meaning of this limit, we may interpret the voltage \ufb02uctuations as due to the \ufb02uctuations of a large assembly of oscillators of arbitrary frequency \u03c9. (In \u201creality\u201d the \ufb02uctuations represent particle\u2013 hole excitations in the Fermi liquids constituting the system. However, for the sake of the argument, that concrete realization is not important.) In the classical limit T \u03c9, each spectral component stores \ufb02uctuation energy T (the equipartition theorem), which explains the above classical correlator. However, in quantum mechanics, the energy reads \u03c9(n(\u03c9)+1/2) = \u03c9 2 coth(\u03c9/2T ). This means that at small temperatures T, it gets replaced by |\u03c9|/2. Also notice that for \u03c9 > 0 the zero temperature energy is due to the factor 1/2 in the energy balance: the observable \ufb02uctuations are due to vacuum oscillations of the quantum baths. As discussed in the previous chapter, the equilibrium oscillator \ufb02uctuations leading to the statistics above are called Nyquist\u2013Johnson noise. 24 The factor of 2\u03c0 in", " the de\ufb01nition of R is explained as follows: g is the dimensionless conductance of the tunnel barriers. The physical conductance G \u2261 R\u22121 = ge2/h = (g/2\u03c0)/. In our units e2 = = 1 and this reduces to g = 2\u03c0/R. 11.6 A mesoscopic application 741 We next consider the consequences of \ufb01nite biasing, V > 0. For simplicity we keep the temperature at zero, T = 0. Inspection of Eq. (11.84) then reveals a crossover, K(\u03c9) = ) |\u03c9|, 1 2 |V |, |\u03c9| |V |, |\u03c9| |V |. The interpretation of this crossover is as follows: At large frequencies, |\u03c9| |V |, noise is dominated by the equilibrium \ufb02uctuations of high frequency oscillator modes. However, at lower frequencies, we start sensing the distortion out of equilibrium, due to the biasing of the dot. The latter causes a \ufb01nite mean current \ufb02ow through the system. This in turn generates noise, which will be predominantly shot noise. In the previous chapter we argued that if we de\ufb01ne the current passing through an elementary resistor in a time window [\u00aft, \u00aft + \u0394t] as I\u0394t = n/\u0394t in terms of the number of charges, n, passing through in a time \u0394t, the statistical variance of the current \u2013 incident charges assumed Poisson distributed \u2013 will be given by var(I\u0394t) = \u00afI/\u0394t, where \u00afI = I\u0394t is the statistical average of the current. We may write the instantaneous current through the resistor as I = \u00afI + \u03b4I, where\u03b4I (t)\u03b4I(t) = f \u03b4(t \u2212 t) describes short range correlated current \ufb02uctuations. Computing the variance, var(I\u0394t) = 1 (\u0394t)2 \u00aft+\u0394t \u00aft dtdt \u03b4I(t)\u03b4I(t) = f \u0394t, we \ufb01nd that f \u2261 \u00afI establishes compatibility to Poissonian shot noise. We now relate these \ufb01ndings to the Langevin equation describing our more complex two-resistor system. To this end we rewrite the latter as CdtU = \u2212 2 R U + C\u03b7, (11.90) i.e. an equation that relates the rate of changes in the charge of the dot (left\u2013hand side) to current \ufb02ow (right\u2013hand side). Speci\ufb01cally, C\u03b7 \u2261 \u03b4IL + \u03b4IR is to be interpreted as the sum of current \ufb02uctuations through the left and right tunnel resistor. Each of the two contributions Ix, x = L, R is expected to express Poissonian shot noise statistics, Ix(t)Ix(t) = \u00afI\u03b4(t \u2212 t) = V /(2R)\u03b4(t \u2212 t). Compatibility with our considerations above then requires \u03b7(t)\u03b7(t) = 1 C 2 (\u03b4IL(t)\u03b4IL(t) + \u03b4IR(t)\u03b4IR(t)) = in agreement with Eq. (11.89) and K(t \u2212 t) = (V /2)\u03b4(t \u2212 t): the noise in our system is obtained by superposition of the shot noise generated by the two resistive tunnel barriers. V RC 2 \u03b4(t \u2212 t), INFO The analysis of noise levels in compounds comprising several elementary Poissonian noise sources plays an important role in nonequilibrium mesoscopic physics. It is customary to quantify the noise in a composite system by a parameter known as the Fano factor. The Fano factor relates the DC-noise power in the system, S \u2261 2 dt \u03b4I(0)\u03b4I(t), (11.91) 742 Nonequilibrium (quantum) to the noise power, S0, of an elementary resistor unit with Poissonian statistics, F \u2261 S S0. (11.92) Here \u03b4I is the \ufb02uctuation contribution to the current in the system. To make these de\ufb01nitions more concrete, consider the elementary RC-unit shown in Fig. 10.2, \ufb01rst at the absence of external biasing V = 0, and at temperature T. The Langevin equation controlling current \ufb02ow in the system reads CdtU = \u2212R\u22121U + \u03b4I. The FDT requires that \u03b4I(t)\u03b4I(t) = 2 RT which means that, in equilibrium, Seq", " = 4T /R. Now consider the complementary case of zero temperature, T = 0, but \ufb01nite bias V across the resistor. In this case, the Poissonian statistics of charge transmission requires (cf. the discussion above) \u03b4I(t)\u03b4I(t) = \u00afI = V /R. This means that S0 = 2 \u00afI = 2V R. What is the Fano factor of the double barrier quantum dot discussed above? We de\ufb01ne I = U/R as the current whose statistics we wish to characterize. Comparison with Eq. (11.90) shows that \u03b4I = C\u03b7/2. From this identi\ufb01cation and Eq. (11.89) we \ufb01nd that, in equilibrium, the noise power S = 2T /R. This is the noise of a unit with total (series) resistance Rtot = 2R. Out of 4R \u03b4(t \u2212 t) implies S = V /2R = V /Rtot = \u00afI = S0/2, equilibrium, the correlator \u03b4I(t)\u03b4I(0) = V or F = 1/2. The system exhibits \ufb02uctuations lower than those of a single Poissonian resistor. The reason for this reduction is that the current \ufb02uctuations \u03b4I = (\u03b4IL + \u03b4IR)/2 are obtained by superposition of two statistically independent noise sources. Zero bias anomaly We next wish to explore the rami\ufb01cations of the resistor voltage \ufb02uctuations in the quantum mechanics of the TDoS. To this end, consider the electron contribution \u03bde to the density of dt U (t), we can rewrite states in Eq. (11.86). Using the fact that \u03c6c(t + \u00aft) \u2212 \u03c6c(\u00aft) = this expression as t+\u00aft \u00aft \u03bde() = \u03bdd Re dt eit = \u03bdd Re dt eit > > i 2 e i 2 e t+\u00aft \u00aft dt U (t)ei(\u03c6q(t+\u00aft)+\u03c6q(\u00aft)) t+\u00aft \u00aft dt U\u03b7(t)? \u03b7 (1 \u2212 nd)(t)? (1 \u2212 nd)(t) where U [\u03b7] is the solution to the di\ufb00erential equation (\u2202t + \u03b3) U\u03b7(t) = \u03b7 + 1 C (\u03b4(t \u2212 (\u00aft + t)) + \u03b4(t \u2212 t)). Once more, this equation is easy to interpret. What is new in comparison to the Langevin equation (11.89) describing the source-free action is the presence of the two \u03b4-functions on the right\u2013hand side. These \u03b4-functions re\ufb02ect a unit-jump in the classical charge upon the entry of a particle (amplitude) in the dot. The voltage on the dot is thus driven by the superposition of the noise \u03b7, and the entry peaks. In response it relaxes, as described by the left\u2013hand side of the Langevin equation. There is no reason to be puzzled by the appearance of two \u03b4-functions; depending on the sign of t, only one of them contributes to the pro\ufb01le of U\u03b7 in the time window [\u00aft, \u00aft + t] (or [\u00aft + t, \u00aft] for negative t). The reason is that the time evolution of U is retarded, i.e. the equation describes the forward evolution of U in causal 11.6 A mesoscopic application 743 response to driving sources. For example, for t > 0, only \u03b4(t \u2212 \u00aft) a\ufb00ects the evolution in the time window [\u00aft, \u00aft + t]. 2 t+\u00aft \u00aft We may interpret the phase e i dt U\u03b7(t) as the \u201cFeynman propagator\u201d of the tunneling amplitude. Our dot is largely structureless, i.e. the only contribution to the action of the particle is the \ufb02uctuating voltage U\u03b7. To understand its e\ufb00ect, we need to average the phase over noise \ufb02uctuations. This is most e\ufb03ciently done by substitution of the Fourier transform of U\u03b7, U\u03b7(\u03c9) = \u03b7(\u03c9) + 1 C \u2212i\u03c9 + \u03b3 ei\u03c9(\u00aft+t) + ei\u03c9t, and performing the Gaussian integral over \u03b7(\u03c9) using \u03b7(\u03c9)\u03b7(\u03c9) = 2\u03c0\u03b4(\u03c9", " + \u03c9) \u03b3 C K(\u03c9). As a result of a straightforward but not particularly illuminating calculation, we then obtain \u03bde() = \u03bdd Re dt eite\u2212 i 2\u03b3C sgn (t)(e\u2212\u03b3|t|\u22121)e\u2212S(t)(1 \u2212 nd)(t), where the noise action S(t) = \u03b3 C d\u03c9 2\u03c0 1 \u2212 cos(\u03c9t) \u03c92 (\u03c92 + \u03b32) K(\u03c9), (11.93) and the oscillatory exponential describes the decay of the entry voltage peak on the RCtime scale. The integral form of S(t) re\ufb02ects the \ufb01nite time interval over which the voltage \ufb02uctuations are monitored (the factor \u03c9\u22122(1 \u2212 cos(\u03c9t)), the RC-retarded nature of the voltage response to \ufb02uctuations (the factor ((\u03c92 + \u03b32)\u22121), and the statistics of the \ufb02uctuations themselves, K(\u03c9). The hole contribution, \u03bdh, di\ufb00ers from \u03bde by a sign change in the oscillatory phase, and in the replacement 1 \u2212 nd \u2192 nd. Adding the two contributions and using the fact that, in the open limit, the largeness of the dimensionless parameter \u03b3C = 2/R 1 permits a linearization in the exponential phase, we obtain \u03bd() = \u03bdd 6 1 \u2212 1 2\u03c0\u03b3C \u221e \u03c3=\u00b1 0 dt cos(( + \u03c3V /2)t) t 1 \u2212 e\u2212\u03b3t 7 e\u2212S(t), (11.94) where at zero temperature and equal barrier heights. \u03c3 ei\u03c3V /2t/(4\u03c0it) is the Fourier transform of the dot distribution function Eq. (11.81) In the unbiased limit, V = 0, the noise action can be estimated as S(t) V =0 ln 1 2\u03c0\u03b3C 1 + (\u03b3t)2. This result states that at large time scales, a conspiracy of RC-relaxation and noise \ufb02uctuations leads to a logarithmically diverging action for particle tunneling processes. The consequences become evident once we substitute S(t) into Eq. (11.94) for the tunneling 744 Nonequilibrium (quantum) DoS. Noting that the factors (1 \u2212 e\u2212\u03b3|t|) and cos(t) act as infrared and ultraviolet cuto\ufb00s, respectively, we obtain \u03bd \u03bdd 6 1 \u2212 1 \u03c0\u03b3C ||\u22121 \u03b3\u22121 dt t 7 6 e\u2212S(t) \u03bdd 1 \u2212 1 \u03c0\u03b3C ||\u22121 \u03b3\u22121 dt t 7 (\u03b3t) 1 \u03c0\u03b3C = \u03bdd(||/\u03b3) 1 \u03c0\u03b3C, or, using the de\ufb01nition of \u03b3 and R, V =0,1/RC \u03bd() \u03bdd \u03c0C|| g 1/g. (11.95) This is the celebrated zero bias anomaly. The accumulation of tunneling action at large time scales leads to a complete suppression of the tunneling DoS at small energies. Notice that the vanishing of the DoS at zero energy is a result of the weak accumulation of tunneling action, a phenomenon we may interpret as the long term maintenance of quantum coherence in the system. 1 0.9 2 3 1 0 How will this picture change as we turn on a \ufb01nite bias voltage to drive the system out of equilibrium? A glance at Eq. (11.94) shows that the voltage intervenes in two di\ufb00erent ways: it splits the centre energy of the zero bias anomaly into two, 0 \u2192 \u00b1V /2. This re\ufb02ects the shift in energy of the step distribution on the dot, relative to the Fermi distributions in the leads. Second, the voltage enhances the tunneling action. For example, at low energies, V, the action t\u22121 d\u03c9V /(\u03c92(\u03c92 +\u03b32)) \u221d RV |t|. This increases much more acquires a contribution \u221d (\u03b3/C) strongly, varying linearly in time, than the logarithmic increase we had in the equilibrium case. As a consequence, the coherent suppression of the DoS gets reduced and the zero bias anomaly peaks become less pronounced. In the \ufb01gure above, the resulting pro\ufb01le for the DoS is shown for a few values of the bias voltage. V 4 The discussion of this suppressive mechanism can be carried quite a bit further. For instance, the rounding of the zero bias anomaly peaks can be", " used as a means to de\ufb01ne a nonequilibrium dephasing rate, or one may study the role of the voltage biasing in the complementary regime of low tunneling conductances g 1 (where it turns out to be considerably less pronounced, cf. Problem 11.9.3). Here, however, we will not enter this more detailed discussion. Instead, and in view of the relatively technical character of this section, let us take a step back and summarize what we have found: The TDoS is a very \u201cquantum mechanical observable.\u201d Its low energy pro\ufb01le is determined by long time matter wave coherence, as described by the appropriate Feynman path integral. Nonetheless, the TDoS is susceptible to classical \ufb02uctuations in the environment. The latter a\ufb00ect the classical action of the tunneling particle and, thus, the phase of the quantum mechanical propagator. In equilibrium and at zero temperature, these \ufb02uctuations (then, zero point \ufb02uctuations) do not prevent the build\u2013up of long range coherence, a phenomenon that re\ufb02ects, e.g., in 11.7 Full counting statistics 745 the build\u2013up of the zero bias anomaly and a complete suppression of the TDoS. However, the installment of nonequilibrium conditions, even steady state nonequilibrium, goes along with increased noise levels, and this tends to suppress quantum coherence. Technically, the noise (Langevin noise \u03b7, in our above discussion) couples through a variable (voltage U\u03b7) in a manner that encapsulates dynamical aspects of the system, and may involve system speci\ufb01c retardation e\ufb00ects. In the case study above, we did not introduce this coupling mechanism by postulation of an external bath. Rather, both the collective variable of interest (voltage), and the noise sources emerged out of one microscopically de\ufb01ned system. Such noise generation mechanisms are typical for interacting many particle systems that are projected onto the dynamics of a few collective variables. 11.7 Full counting statistics Previously, we have seen that the statistics of currents may strongly a\ufb00ect the behavior of physical observables. It may also tell us about the internal structure of a system. For example, in the case study above, the noise level of currents (the Fano factor) signaled the presence of two tunneling barriers instead of just one. Statistical analysis of current \ufb02uctuations \u2013 both experimental and theoretical \u2013 has become a prominent tool of diagnostics to explore the microscopic nature of condensed matter systems through observable transport coe\ufb03cients. The cumulative keyword for these types of analysis is full counting statistics (FCS). In this section we introduce the notion of full counting statistics and apply it to elementary examples. A brief survey of more advanced applications is given in the end of the section.25 INFO The concept of full (photon) counting statistics was introduced in quantum optics (for a review, see e.g., L. Mandel and E. Wolf, Optical Coherence and Quantum Optics (Cambridge University Press, 1995).) Only much later have these ideas been transferred to the problem of transport in condensed matter (cf. L. S. Levitov and G. B. Lesovik, Charge distribution in quantum shot noise, Pis\u2019ma Zh. Eksp. Teor. Fiz. 58, 225-30 (1993) [JETP Lett. 58, 230-5 (1993)]). The rami\ufb01cations of this concept are included in the brief review at the end of the section. 11.7.1 Generalities Suppose we wish to characterize transport through a quantum system in terms of the charge transmitted in a time interval [\u00aft, \u00aft + \u0394t], \u02c6N \u2261 \u00aft+\u0394t \u00aft dt \u02c6I(t), (11.96) 25 For more comprehensive reviews, see W. Belzig, Full counting statistics in quantum contacts, Proceedings of the Summer School/Conference on Functional Nanostructures, Karlsruhe, 2003; L.S. Levitov The statistical theory of mesoscopic noise, Quantum Noise in Mesoscopic Physics: Proceedings of the NATO Advanced Research Workshop, Delft, the Netherlands, 2002, and M. Kindermann and Yu. V. Nazarov, Full counting statistics in electric circuits, in: Quantum Noise in Mesoscopic Physics, ed. by Yu. V. Nazarov, Kluwer (Dordrecht), p.403 (2002). 746 Nonequilibrium (quantum) where \u02c6I is the operator measuring current \ufb02ow through a section of the system. We may generate the expectation value of this operator by di\ufb00erentiation with respect to a suitably constructed source \ufb01eld. To", " this end, recall that the current density j(x), is obtained by di\ufb00erentiation of the action of the system with respect to its vector potential, A(x) (as usual x \u2261 (t, x) comprises space and time coordinates). Speci\ufb01cally, in the context of the Keldysh \ufb01eld theory j(x) =\u2212 i \u03b4 \u03b4A(x) Z [A \u2297 \u03c33/2], A=0 where the notation Z[A \u2297 \u03c33/2] indicates that the action is minimally coupled to a purely quantum vector potential (opposite signs on the two Keldysh contours). According to our general discussion of the Keldysh contour in Section 11.2.1, the coupling of A to \u03c33 makes it a suitable source variable.26 Building on this de\ufb01nition, we may introduce a source variable for currents by de\ufb01ning the vector potential, A(x, t) = \u03c7(t)e\u22a5 dd\u22121x \u03b4(x \u2212 x), (11.97) S where the surface integral con\ufb01nes the support of A to a planar section S of the system, e\u22a5 is normal to S (the generalization to curved sections is straightforward), and the de\ufb01nition of the counting \ufb01eld, \u03c7(t) \u2261 \u03c7\u0398(t \u2212 \u00aft)\u0398(\u00aft + \u0394t \u2212 t), implies a projection onto the counting time interval. Di\ufb00erentiation of Z[A \u2297 \u03c33/2] \u2261 Z(\u03c7) with respect to the source parameter \u03c7 then generates the relation \u2202 \u2202\u03c7 \u2212i \u03c7=0 Z(\u03c7) = \u00aft+\u0394t \u00aft S dS \u00b7 j(x, t) = \u02c6N. Further di\ufb00erentiation with respect to the source variable \u2013 and that is the prime advantage of the above construction \u2013 readily generates moments of the current: \u02c6N n = (\u2212i)n \u03b4n \u03b4\u03c7n Z(\u03c7). \u03c7=0 (11.98) This formula identi\ufb01es Z(\u03c7) as the moment generating function and its logarithm as the cumulant generating function. ln g(\u03c7) \u2261 ln Z(\u03c7), INFO Why would one want to know moments higher than the second of a current distribution? Suppose an experimentalist recorded the outcome of many runs of a charge transmission measurement. They might decide to communicate their results in a histogram \u201ccount number vs. events,\u201d i.e. data whose continuous idealization we interpret as a probability distribution. To a \ufb01rst approximation, the shape of this distribution is characterized by the \ufb01rst four cumulative moments, \u03bc1,...,4 (see Fig. 11.5): 26 To be more accurate, in Section 11.2.1 we introduce a source on the upper Keldysh contour. The sign inverted source on the lower contour generates the same observable. Coupling to \u03c33 adds the two contributions which are compensated for by the factor of 1/2 in the formula above. 11.7 Full counting statistics 747 Figure 11.5 Hypothetical current distribution interpolating through an (equally hypothetical) discrete data set and its characterization in terms of mean, width, skewness, and kurtosis. 1. moment, \u03bc1: theaverage, 2. moment, \u03bc2: thewidth, 3. moment, \u03bc3: theskewness of the distribution, 4. moment, \u03bc4: thekurtosis. High kurtosis implies a sharply peaked distribution with fat tails. Low kurtosis indicates a molli\ufb01ed distribution, with broader shoulders. The central limit theorem ensures that moments higher than the fourth are typically very small. 11.7.2 Realizations of current noise In this section we introduce two frequently encountered realizations of noisy current distributions in a schematic manner. We then explore how these prototypical forms relate to the \u201creal-life\u201d example of our resistively shunted quantum dot. Quantum point contact: consider a \u201cquantum point contact,\u201d i.e. an isolated scatterer embedded into a single-channel quantum conductor. This setup is not quite as academic as it may seem. By standard techniques of current device technology it is possible to manufacture few-channel quantum wires, and to introduce arti\ufb01cial imperfections or tunneling bridges (see the picture [courtesy of Nanocenter Basel] on the right for an example). Charge carriers which encounter the point contact get transmitted with probability T and re", "\ufb02ected with probability 1 \u2212 T. This means that the probability to transmit n charges in N events is given by the Bernoulli distribution p(n) = N n T n(1 \u2212 T )N \u2212n. 748 Nonequilibrium (quantum) The cumulant generating function of this distribution reads g(\u03c7) \u2261 ln ei\u03c7np(n) = N ln(1 + T (ei\u03c7 \u2212 1)). n A perfect single channel conductor has dimensionless conductance g = e2/h = 1/2\u03c0 (in our usual units e = = 1). If the system is biased, N = I\u0394t = V \u0394t/2\u03c0 charges will be incoming, i.e. the statistics of current in the quantum point contact (at zero temperature27) is described by ln g(\u03c7) = V \u0394t 2\u03c0 ln(1 + T (ei\u03c7 \u2212 1)). (11.99) Ohmic resistor: in an Ohmic resistor the situation is di\ufb00erent. Incidents of charge transmission are still uncorrelated, but we may not think of them as the result of a single scattering event. This suggests a description of charge transfer in terms of a Poisson distribution (cf. discussion on p.n 605), p(n) = e\u2212\u03bd \u03bdn n! \u2212\u2192 ln g(\u03c7) = \u03bd ei\u03c7 \u2212 1. (11.100) Bidirectional distribution: The current through a conductor connected to two terminals (for the generalization to multi\u2013terminal geometries, see25) is generally obtained by additive superposition of two counter\u2013propagating current \ufb02ows (see \ufb01gure). For simplicity, let us assume the two distributions pi(ni), i = 1, 2 of counter-propagating charges to be statistically independent. The distribution p(n) of the total number of transmitted charges, n = n1 \u2212 n2 can then be computed as follows: p(n) = n1n2 \u03b4n,n1\u2212n2 p1(n1)p2(n2) d\u03c7 2\u03c0 e\u2212i\u03c7(n\u2212n1+n2) e\u2212i\u03c7ng1(\u03c7)g2(\u2212\u03c7). = = n1n2 d\u03c7 2\u03c0 d\u03c71 2\u03c0 e\u2212i\u03c71n1 g(\u03c71) d\u03c72 2\u03c0 e\u2212i\u03c72n2 g2(\u03c72) This means that the generating function of the transmitted charge is the product g(\u03c7) \u2261 g1(\u03c7)g2(\u2212\u03c7) of the partial distributions. For example, in the case where the two constituent distributions are Poissonian, we \ufb01nd ln g(\u03c7) = \u03bd1 ei\u03c7 \u2212 1 + \u03bd2 e\u2212i\u03c7 \u2212 1, (11.101) where \u03bd1 and \u03bd2 determine the average rates. These rates will typically be set by the applied voltage bias, barrier transparencies, temperature, and the like. In the following, we return to the biased quantum dot studied in previous sections and explore the relevance of the general structures discussed above to this physical example. 27 For the generalization to \ufb01nite temperatures, cf. Refs.25. 11.7 Full counting statistics 749 11.7.3 Full counting statistics of the double barrier quantum dot Let us now return to consider the quantum dot system introduced in Section 11.6.1. Our goal is to better understand the statistics of current \ufb02ow. To this end, we will monitor the current through, say, the tunnel barriers connecting the dot to the left lead. The \ufb01rst thing we need is an appropriate counting \ufb01eld. Phenomenologically, we may argue that, in the presence of a vector potential Ae\u22a5 normal to the lead axis, the tunneling matrix elements TL,a\u03bc utilized in Section 11.6.1 will generalize to TL,a\u03bc \u2192 TL,a\u03bcexp(i dxA), where the integral runs over the transverse extension of the barrier. The de\ufb01nition Eq. (11.97) then means that the counting \ufb01eld couples to the tunneling matrix element on the Keldysh contour as TL,a\u03bc \u2192 TL,a\u03bcexp(i\u03c7(t)\u03c31/2). Notice that, as with the quantum component of the interaction \ufb01eld, the counting \ufb01eld couples to the tunneling matrix element. This observation is su\ufb03cient to determine the coupling of the counting \ufb01eld to the e\ufffd", "\ufffdective action without much further calculation. For the sake of a more symmetric notation, let us de\ufb01ne generalized matrix elements Tx,a\u03bcexp(i\u03c7x(t)\u03c31/2), x = L, R. (Later we will set \u03c7L \u2261 \u03c7, \u03c7R \u2261 0.) The tunneling action Eq. (11.75) then assumes the form Stun[\u03c7] = \u2212 i 4 \u039bx(\u03c7x)e\u2212i \u02c6\u03c6\u039bdei \u02c6\u03c6 \u039bx(\u03c7x) \u2261 ei \u03c7x 2 \u03c31 \u039bxe\u2212i \u03c7x 2 \u03c31, gxtr, x=L,R where we have generalized to barriers of di\ufb00erent tunnel conductance, gx. Following the logic of Section 11.6.2, our next step will be to determine the matrix \u039bd. As before, we do this by requiring stationarity under variations of \u03c6. (Exercise: think why this condition is equivalent to the condition of zero stationary current \ufb02ow onto/o\ufb00 the dot.) Variation of the action in the limit of vanishingly weak interaction, \u03c6 0, generates the condition # \u039bd, gx\u039bx(\u03c7x) $!= 0. x (11.102) Now, the transformations \u039b \u2192 exp(i\u03c7\u03c31/2)\u039b exp(\u2212i\u03c7\u03c31/2) render the matrices \u039bx(\u03c7x) non-triangular. This means that stationary con\ufb01gurations \u039bd will in general no longer be of the form Eq. (11.76). However, the transformation by the counting \ufb01eld does not alter the \u201cnormalization condition,\u201d \u039b2 = \u03c30 remains valid. It is straightforward to verify that the solution of Eq. (11.102) obeying this normalization is given by28 \u039bd = gL\u039bL(\u03c7L) +g R\u039bR(\u03c7R) R + gLgR[\u039bL(\u03c7L), \u039bR(\u03c7R)]+)1/2 (g2 L + g2. (Notice that the matrix in the denominator commutes with the numerator, i.e. the relative ordering of numerator and denominator is not an issue. Also notice that in the limit \u03c7 = 0, 28 In computing this result, we approximate the diagonal elements of the matrices \u039b in Eq. (11.76) by their imaginary parts, \u00b11. Diligent readers may wish to trace the fate of the real parts and check that they regularize the super\ufb01cially divergent constant C in Eq. (11.103) below. 750 Nonequilibrium (quantum) the solution reduces to that discussed in Section 11.6.2.) Substitution of this con\ufb01guration into the action leads to Stun[\u03c7] = \u2212 i 4 tr L + g2 g2 R + gLgR[\u039bL(\u03c7L), \u039bR(\u03c7R)]+ 1/2. ddt Assuming that the inverse of the counting time window (\u0394t)\u22121 is large in comparison to the energy scales relevant to the distribution functions, we will evaluate the trace over energy/time indices within the leading order Wigner approximation (cf. Section 11.5.1), 2\u03c0 (...), F (\u02c6) \u2192 F (), \u03c7(\u02c6t) \u2192 \u03c7(t), where F () and\u03c7( t) are ordinary functions tr(...) \u2192 of energy and time, respectively. As a result of a straightforward (if somewhat tedious) calculation, one \ufb01nds that the matrix [\u039bL(\u03c7L), \u039bR(\u03c7R)] is proportional to the unit matrix, and that the trace evaluates to Stun[\u03c7] = \u2212 i 2 (11.103) 1 dtd 2\u03c0 \u00d7 (gL + gR)2 + 4gLgR = \u2212 i\u0394t 2 d 2\u03c0 (gL + gR)2 + 4gLgR (ei(\u03c7L\u2212\u03c7R) \u2212 1)nL(1 \u2212 nR) + (ei(\u03c7R\u2212\u03c7L) \u2212 1)nR(1 \u2212 nL) 1 (ei\u03c7 \u2212 1)nL(1 \u2212 nR) + (e\u2212i\u03c7 \u2212 1)nR(1 \u2212 nL) 2 2 + C, where in the last equality we used \u03c7L(t) =\u03c7 (t) =\u03c7 \u0398", "(t \u2212 \u00aft)\u0398(\u00aft + \u0394t \u2212 t), and C is a constant. From this result we \ufb01nd the cumulant generating function ln g(\u03c7) ln exp(iS[\u03c7]) 1 (gL + gR)2 + 4gLgR (ei\u03c7 \u2212 1)nL(1 \u2212 nR) + (e\u2212i\u03c7 \u2212 1)nR(1 \u2212 nL) 2. ln g(\u03c7) = \u0394t 2 d 2\u03c0 (11.104) Now, let us try to make sense of this expression. Comparison with Eq. (11.101) shows that ln g contains the generating functions of two Poisson distributions as building blocks. To explore the meaning of this result, let us \ufb01rst consider the limit of zero temperature and voltage bias V, nL/R() = \u0398((+/\u2212) V 2 \u2212 ). The generating function then reduces to ln g(\u03c7) = \u0394tV 4\u03c0 (gL + gR)2 + 4gLgR (ei\u03c7 \u2212 1) 1 2 + C. According to Eq. (11.98), the \ufb01rst moment of the transmitted charge through the system is given by \u02c6N = \u2212i\u2202\u03c7 \u03c7=0 ln g(\u03c7) = \u0394tV 2\u03c0 gLgR gL + gR. Comparison with the de\ufb01nition of the conductance, G = \u02c6I/V = \u02c6N /V \u0394t gives G = gLgR 1 (gL+gR), which we recognize as the mean conductance of two tunnel conductances shunted 2\u03c0 in series (in units of the conductance quantum e2/h = 1/2\u03c0.) Turning to current statistics, let us consider the limit gL gR. We may then expand the square root to \ufb01rst order in the ratio gL/gR to obtain a bi-directional Poisson distribution Eq. (11.101), with (time\u2013 integrated) characteristic rates identi\ufb01ed as \u03bd1 = \u0394tgL d 2\u03c0 nL(1 \u2212 nR), \u03bd2 = \u0394tgL d 2\u03c0 nR(1 \u2212 nL). 11.7 Full counting statistics 751 These coe\ufb03cients may be interpreted as the integrated rate at which \ufb01lled states in the right lead scatter into empty states in the right lead, and vice versa. As one may expect, the statistics of the current is dominantly caused by the \u201cbottleneck\u201d in the system, i.e. the conductance of the weaker tunnel barrier, gL. At \ufb01nite temperature, \u03bd1 \u0394tgLV /2\u03c0, while \u03bd2 exp(\u2212V /T )\u0394tT /2\u03c0 shows that thermal activation is necessary to push charges against the voltage gradient. The second (cumulative) moment de\ufb01nes the noise power (cf. Eq. (11.91)), S0 = 2 \u0394t var ( \u02c6N ) = \u2212 2 \u0394t \u22022 \u03c7 \u03c7=0 ln g[\u03c7] = 2 \u0394t (\u03bd1 + \u03bd2) = 2 gL 2\u03c0 V coth(V /2T ). This shows how the noise interpolates between equilibrium noise, \u03b4I(t)\u03b4I(t) V \u21920\u223c gLT and the shot noise limit \u03b4I(t)\u03b4I(t) V T\u223c gL|V |. In the more general case of barriers of comparable transparency, gL gR, the current statistics is more complicated and described by the full expression Eq. (11.104). In principle, we might now include phase \ufb02uctuations to explore the interesting question as to how interaction e\ufb00ects will in\ufb02uence FCS. However, this topic29 lies beyond the scope of the present text. Instead, let us conclude by brie\ufb02y discussing a few more general aspects of full counting statistics. 11.7.4 General rami\ufb01cations of FCS Previously, we have introduced the concept of a counting \ufb01eld as a purely technical means to extract moments of observables from a Keldysh partition function. From a theoretical point of view, this has been an e\ufb03cient procedure, but it is also somewhat naive: when we speak about moments of quantum observables, we inevitably enter the di\ufb03cult territory of quantum measurement. First, we have to realize that, when talking about moments, what stands in the background is a repeated quantum measurement of", " observables. Second, in experiment, it is not the (fermion) current that is measured directly. Typically, it is \ufb02uctuations of secondary degrees of freedom, e.g. the bosonic modes of electromagnetic degrees of freedom coupling to the current that are detected. One thus needs to explore the connection between the original degrees of freedom (fermion current) and the detector degrees of freedom. These (deep) issues have been explored in the literature25 and we here restrict ourselves to a very super\ufb01cial discussion of some main ideas. An elegant (if not particularly realistic) way to model the coupling to a detector degree of freedom is by coupling the fermion degrees of freedom to a spin. The precession of the spin detector then encapsulates information on the transmitted fermion charge. Curiously, this procedure exactly amounts to what we have been doing in introducing our Keldysh counting \ufb01eld, and this connection is worked out in the following EXERCISE We here wish to explore in what sense our counting \ufb01eld, \u03c7, e\ufb00ectively models a spin detector variable (cf. the review by Levitov25 for an extended discussion). Consider a spin 29 cf. D. A. Bagrets and Y. V. Nazarov, Full counting statistics of charge transfer in Coulomb blockade system, Phys. Rev. B 67, 085316-32 (2003). 752 Nonequilibrium (quantum) 1/2 placed near our current carrying device and coupled to the latter through the Hamiltonian \u02c6Hc = ddrA \u00b7 j\u03c3z, where j denotes the fermion current density operator, A is a vector potential mediating the coupling to the spin, and we assume that only the z-component of the latter is coupled. Now, let us assume that the spin has been prepared in some initial state, so that the measurement of the operator \u03c3+ = (\u03c3x + i\u03c3y)/2 would lead to the expectation value \u03c3+(0)spin. The coupling to the fermion system makes the transverse spin components, probed by this expectation value precess. This precession is described by taking the expectation value at a later time, \u03c3+(t)spin. Show that the coupling to the fermion system above implies \u03c3+(t)spin =. ei \u02c6H(+A)te\u2212i \u02c6H(\u2212A)t / \u03c3+(0)spin, fermion where \u02c6H(A) is the fermion Hamiltonian operator minimally coupled to the vector potential, A. Express the fermion expectation value by a Keldysh partition function. The sign change in A relative to the contours then means that we have coupled the fermion system to a purely quantum vector potential, as we did in Section 11.7.1. Assuming that the spin-fermion coupling is local, i.e. that the vector potential is non-vanishing only in the immediate vicinity of a transverse section through the conductor, we arrive at a functional integral formally equivalent to the integral Z[\u03c7] employed above. The assumption of a spin detector is arti\ufb01cial in two respects. First, currents are not normally measured by spins. Second the coupling of our system of interest to a detector will in general imply a \u201cback action\u201d of the latter on the former, and this will a\ufb00ect the current statistics. These aspects are discussed in the review by Kindermann and Nazarov cited above.25 The authors argue that measurements on a (mesoscopic) device typically imply the embedding of the latter into a detector electromagnetic environment. With reasonable generality, the detector environment may be modelled as a linear electromagnetic i.e. a system whose relevant circuit (see \ufb01gure), degrees of freedom (currents, voltages, etc.) are controlled by some e\ufb00ective oscillator dynamics (cf. an LRC-circuit). The variables of the detector environment are linearly coupled to those of the mesoscopic device. (For example, the charge density on the latter may feel the potential \ufb02uctuations of the detector, etc.) One may then formulate an FCS functional for the composite system. Owing to their linearity, the degrees of freedom of the detector environment may be integrated out, and this leaves an e\ufb00ective functional in which the counting \ufb01eld and the primary variables of the mesoscopic device are coupled in a manner that will typically involve retardation e\ufb00ects. In this way, one has e\ufb00ectively described both a reasonably realistic readout procedure and, relatedly, the back action of the measurement device onto the microscopic system of interest. For further discussion of these issues we refer to the original literature. 11.8 Summary", " and outlook 753 11.8 Summary and outlook In this \ufb01nal chapter of the book, we have introduced the Keldysh \ufb01eld theory as a modern \u2013 in spite of its formulation in the mid\u201360s \u2013 tool to describe quantum nonequilibrium phenomena. It must be conceded that the Keldysh formalism comes with a relatively steep learning curve. However, once one gets used to the concept of two counter propagating time contours, one realizes that Keldysh \ufb01eld theory provides an enormously powerful and, indeed, intuitive framework to approach quantum nonequilibrium phenomena. Practically all other theoretical nonequilibrium tools (at least those familiar to the authors of this text) can be recovered as restrictions of the Keldysh approach. Speci\ufb01cally, we have discussed the derivation of quantum master and quantum kinetic equations, the connection to equilibrium \ufb01eld theory, the classical limit (which neatly connects to the MSRJD formalism central to the previous chapter), nonequilibrium variants of diagrammatic perturbation theory, and various other aspects. However, notwithstanding its indisputable power, one should not get carried away and declare the Keldysh formalism the new master tool, superior to the equilibrium concepts discussed in earlier chapters of the book. The \ufb02exibility of Keldysh theory may come as a burden when one is actually carrying more information about a problem than that which is called for. In situations where one is at equilibrium, or close to it, there is no need to explicitly keep track of the (then known) distribution function, and the Matsubara formulation may be the more e\ufb03cient option. However, at the time of writing of this book, quantum nonequilibrium problems are becoming more and more important, and this makes Keldysh theory an important concept in contemporary physics. At any rate, the best way for readers to proceed from here is to pick a nonequilibrium problem and conduct their own research! 11.9 Problems 11.9.1 Atom-\ufb01eld Hamiltonian The \u201catom-\ufb01eld-Hamiltonian\u201d is a simple model Hamiltonian of quantum optics. It reduces the interaction of atoms with electromagnetic \ufb01eld modes to a basic model of a two-level system (\u201cthe atom\u201d) coupled to an assembly of oscillator modes. The simplistic nature of this reduction notwithstanding, the atom-\ufb01eld Hamiltonian describes ample phenomenology, and it is a prominent model system of quantum optics. In this problem, we study the simplest variant of the system, the exactly solvable limit of the interaction with a single mode. The single mode limit describes fully coherent quantum phenomenology. We use it as a benchmark system to compare against the \u201cincoherent\u201d approximations underlying the quantum master equation of Section 11.1. In the follow up Problem 11.9.2 we then explore what happens upon generalization to multi-mode coupling. 754 Nonequilibrium (quantum) Consider an atom exposed to electromagnetic radiation. Assuming that the \ufb01eld modes predominantly couple two atomic states |a and |b (see \ufb01gure), and forgetting about the complications introduced by the polarization of the electromagnetic \ufb01eld, this setup may be described by the simple model Hamiltonian, \u02c6H = \u03c33 + \u03c9ka \u2020 kak + gk\u03c3+ak + \u00afgk\u03c3\u2212a, (11.105) \u2020 k 2 k k where is the energy di\ufb00erence between the excited state, |a, and the lower state, |b, the Pauli matrices \u03c3i act in the two-dimensional Hilbert space de\ufb01ned by these states, and \u03c3\u00b1 = (\u03c31 \u00b1 i\u03c32)/2, as usual. This atom-\ufb01eld Hamiltonian describes excitation processes |b \u2192 |a by \ufb01eld quantum absorption (at coupling constants gk), and the corresponding relaxation processes by quantum emission. We may simplify the problem further by assuming that only a single mode of the electromagnetic \ufb01eld satis\ufb01es the resonance condition \u03c9, that needs to be met to obtain signi\ufb01cant conversion e\ufb03ciency. This, then, leads to the single mode Hamiltonian \u02c6H = 2 \u03c33 + \u03c9a\u2020a + g \u03c3+a + \u03c3\u2212a\u2020, (11.106) where we have omitted the mode index \u201ck\u201d and gauged the coupling constant g so as to become real. (a) Considering the atom as our \u201csystem,\u201d and the mode as a \u201cbath,\u201d adapt the formalism introduced in Section 11.1 to derive an", " e\ufb00ective equation of motion for the reduced system density matrix, \u02c6\u03c1s. In doing so, try to be critical concerning the approximate elements of the construction, notably the Markovian approximation. (Hint: you will be met with a singular e\ufb00ective coupling constant \u223c \u03b4( \u2212 \u03c9). For the moment, do not worry about the regularization of this expression and treat it as a formal object.) Derive a closed expression for the diagonal elements of the reduced density operator \u03c1x \u2261 x|\u02c6\u03c1s|x, x = a, b and verify that the stationary limit \u03c1x,\u221e \u2261 \u03c1x(t \u2192 \u221e) obeys a detailed balance relation \u03c1a,\u221e \u03c1b,\u221e = n n + 1, (11.107) where n is the number of mode quanta and the expectation value is over the thermal distribution of the bath. Accordingly, the population imbalance between the levels approaches the limit \u0394\u03c1 \u2261 \u03c1a,\u221e \u2212 \u03c1b,\u221e = \u2212 1 2n + 1. (11.108) Discuss the meaning of this expression. Is it really appropriate for the situation at hand? If not, where do you think the approximation involved in its derivation failed? (b) Now, let us compare the predictions of the Markovian approximation to reality. To this end, solve the Schr\u00a8odinger equation for the Hamiltonian above. (Hint: use the fact that the Hamiltonian couples only very few of the states |x, n \u2261 |x \u2297 |n, x = a, b, where |n is an occupation eigenstate of the mode system. You will encounter a time dependent 11.9 Problems 755 1.0 0.5 - 0.5 5 10 15 Figure 11.6 Imbalance between the population of the two atomic states as a function of dimensionless time gt. Notice the aperiodic changes in the population of the states and the absence of relaxation towards a ground state con\ufb01guration \u0394\u03c1 = \u22121. n Schr\u00a8odinger equation that may be solved either by Fourier transform, or by educated x, n|x, n as the probability to \ufb01nd the system in state guessing.) Interpreting \u03c1x \u2261 |x, compute the population imbalance and compare to the predictions of the Markovian approach Eq. (11.108). If you have access to a computer, you may \ufb01nd it instructive to plot the result for the time dependent imbalance for di\ufb00erent values of temperature (presuming the mode distribution to be thermal), and frequency mismatch \u0394 = \u2212 \u03c9 as a function of the dimensionless time parameter gt. You will obtain patterns such as the one shown in Fig. 11.6. (c) Discuss in qualitative terms the origin of the discrepancies between the two approaches. Why is the Markovian approximation not appropriate under the present circumstances, and why does the exact solution not predict relaxation of an initially occupied state |a to the ground state, even at zero temperature. Answer: (a) In the interaction representation, a(t) = e\u2212i\u03c9ta, \u03c3\u00b1(t) = e\u00b1it\u03c3\u00b1, the interaction Hamil- tonian \u02c6Hi \u2261 g \u03c3+a + \u03c3\u2212a\u2020 reads \u02c6Hi(t) = g ei\u0394t\u03c3+a + e\u2212i\u0394t\u03c3\u2212a\u2020, where we de\ufb01ned the energy mismatch \u0394 \u2261 \u2212 \u03c9 between level splitting and mode frequency. De\ufb01ning \u02c6Li = \u2212i[ \u02c6Hi, ], it is then straightforward to verify that Eq. (11.4) assumes the form t dt e+i\u0394t (+n + 1[\u03c3+, \u03c3\u2212 \u02c6\u03c1s(t \u2212 t)] \u2212 n[\u03c3+, \u02c6\u03c1s(t \u2212 t)\u03c3\u2212]) \u2202t \u02c6\u03c1s = \u2212g2 \u2212 g2 t dt e\u2212i\u0394t (\u2212n + 1[\u03c3\u2212, \u02c6\u03c1s(t \u2212 t)\u03c3+] + n[\u03c3\u2212, \u03c3+ \u02c6\u03c1s(t \u2212 t)]), (11.109) where the expectation value is over the mode distribution. Assume that at some initial time t = 0, the density operator had diagonal form, \u02c6\u03c1s = \u03c1aP+ + \u03c1bP\u2212, 756 Nonequilibrium (quantum) where P\u00b1 are projectors onto the upper (a) or lower (b) state, and \u03c1a,b, \u03c1a + \u03c1b = 1 are the probabilities of occupation of these states. It is straightforward to check that the evolution equation preserves", " the diagonal form. Adopting a Markovian approximation (but is it appropriate?) wherein the time dependence of \u02c6\u03c1(s) under the integral above is considered negligible, the evolution equation for the coe\ufb03cients \u03c1a,b takes the form of a master equation \u2202t\u03c1a = \u0393 (n\u03c1b \u2212 n + 1\u03c1a), \u2202t\u03c1b = \u0393 (n + 1\u03c1a \u2212 n\u03c1b), (11.110) where the rate \u0393 = 2\u03c0g\u03b4(\u0394) is singular at resonance. This equation predicts an (irreversible) approach to a stationary limit satisfying the detailed balance relation Eq. (11.107). (b) It is evident that for any \ufb01xed n \u2265 0 the Hamiltonian acts in the two-dimensional space spanned by the states |a, n and |b, n + 1. Speci\ufb01cally, \u02c6Hi(t)|a, n = ge\u2212i\u0394t(n + 1)1/2|b, n + 1 and \u02c6Hi(t)|b, n + 1 = ge+i\u0394t(n + 1)1/2|a, n. Introducing wavefunctions by |\u03c8(t) = \u03c8a,n(t)|a, n + \u03c8b,n+1(t)|b, n + 1, the time dependent Schr\u00a8odinger equation i\u2202t\u03c8 = \u02c6Hi(t)\u03c8 assumes the form i\u2202t\u03c8a,n = ge+i\u0394t(n + 1)1/2 \u03c8b,n+1, i\u2202t\u03c8b,n+1 = ge\u2212i\u0394t(n + 1)1/2 \u03c8a,n. (11.111) These equations are solved by (devise your own solution strategy, there are several) 7 6 \u03c8a,n(t) = e+i \u0394 2 t \u03c8a,n(0) cos(\u03a9nt) \u2212 i 6 \u03c8b,n+1(t) = e\u2212i \u0394 2 t \u03c8b,n+1(0) cos(\u03a9nt) \u2212 i where we introduced the abbreviation 6 g(n + 1)1/2\u03c8b,n+1(0) + \u0394 2 \u03c8a,n(0) \u03a9n sin(\u03a9t) g(n + 1)1/2\u03c8a,n(0) \u2212 \u0394 2 \u03c8b,n+1(0) \u03a9n 7 sin(\u03a9t), \u03a9n \u2261 g2(n + 1) + 71/2 2. \u0394 2 \u221a Assuming for simplicity that the atom was initially in its excited state, we have \u03c8b,n+1 = \u03c1n, where \u03c1n is the nth eigenvalue of the mode density operator (e.g., 0 and\u03c8 a,n = \u03c1n = Z \u22121exp(\u2212\u03b2\u03c9(n + 1/2)) if the mode distribution is thermal). The exact result for the population imbalance then reads \u0394\u03c1 = n |\u03c8a,n|2 \u2212 |\u03c8b,n|2 = \u03c1n n 1 \u2212 2g2(n + 1) \u03a92 n sin2(\u03a9nt). This result is very di\ufb00erent from the one obtained within the Markovian approach, Eq. (11.108): no stationary limit is approached. A short period of decay of the initial value \u0394\u03c1(0) = 1 merges into a pattern of irregular \ufb02uctuations \u2013 a result of a superposition of contributions of incommensurate frequencies (see Fig. 11.6). In quantum optics, the phenomenon of transient near-recoveries of the initial value is known as 11.9 Problems 757 collapse and revival while the \ufb02uctuations a\ufb00ord an interpretation as Rabi oscillations. Clearly, the oscillatory pattern is a result of maintained quantum coherence and reversibility of the dynamics, we do not observe a systematic decay of the upper state into the lower. Notice in particular that even at zero temperature the atom does not relax by emission of \ufb01eld quanta: at T = 0 only the n = 0 term (zero \ufb01eld quanta) contributes to the sum above. This leads to oscillatory behavior of the density operator in which the initial state is recovered at regular intervals t \u223c g, but no relaxation. For further discussion of the \ufb02uctuation pattern we refer to Scully and Zubairy.30 Here, our main conclusion is that the prediction of irreversible dynamics derived in (a", ") is evidently incorrect. (c) The Markovian approximation fails because a single quantum oscillator mode does not behave as a bath. Indeed, the mode\u2013atom coupling is strongest at resonance, \u0394 = 0, when \u201csystem\u201d and \u201cbath\u201d \ufb02uctuate at comparable time scales. Technically this means that the (Markovian) assumption of constancy of the time evolution operator under the integral Eq. (11.109) is not justi\ufb01ed. The Markov approximation requires a short lived bath \u201cmemory,\u201d in the sense that bath correlation functions decay to zero at time scales much shorter than those at which the system density operator varies. This happens, e.g., if we couple to many modes instead of one (see the follow\u2013up problem.) In this case, the analog of Eq. (11.108) contains many uncorrelated contributions, and the dynamics becomes e\ufb00ectively irreversible.31 11.9.2 Atom-\ufb01eld Hamiltonian II: Weisskopf\u2013Wigner theory of spontaneous emission (Attack this problem after 11.9.1.) In the previous problem, we saw that the coupling of an atom to a single electromagnetic mode does not lead to radiative relaxation. The relaxation processes ubiquitous in the physics of atomic radiation must, then, be a consequence of the presence of many \ufb01eld modes. In this problem, we study how the large phase space of the multi\u2013mode system justi\ufb01es an approximation that renders the dynamics of the atom-\ufb01eld system e\ufb00ectively irreversible and does describe relaxation by radiation. The theory derived here plays an important role as a building block of, e.g., laser theory. (a) Consider the Hamiltonian Eq. (11.105) of a two-level atom coupled to a multi-mode \ufb01eld. Assuming zero temperature, derive the generalization of the Schr\u00a8odinger equation (11.111) for the initial con\ufb01guration |a \u2297 |0 to the multi-mode case (|0 is the zero temperature photon vacuum). By formal integration of the second equation, convert the system to a single integro-di\ufb00erential equation. Assuming that |gk|2 = g2(\u03c9k) depends only on the mode energy and that both the density of states of the bath modes and g2(\u03c9) change only negligibly over the inverse of the time scales at which the wave functions of 30 M.O. Scully and M.S. Zubairy, Quantum Optics, (Cambridge University Press, 1997). 31 Even then, it is not impossible that a rare accumulation of coherent contributions leads to an accidental population \u201crevival\u201d \u2013 after all, we are still dealing with unitary (reversible) dynamics. However, for su\ufb03ciently large numbers of contributing modes, such events are in\ufb01nitely improbable, and an e\ufb00ectively irreversible approximation is justi\ufb01ed. 758 Nonequilibrium (quantum) the problem vary \u2013 the Weisskopf\u2013Wigner approximation \u2013 derive an approximate solution of this equation. (b) For arbitrary temperature and initial population, attack the multi-mode problem by generalization of the projector formalism applied in Problem 11.9.1 (a) to the single mode case and compare to the results of the Weisskopf\u2013Wigner theory. Convince yourself that the approximation used there is equivalent to a Markovian approximation. Answer: (a) The initial state relevant to the description of an excited atom immersed into a zero temperature bath reads |a \u2297 |0, where |0 is the vacuum of the bath system. The \u2020 |0. Introducing a wave Hamiltonian couples this con\ufb01guration to the states |b \u2297 a k |0, the multi-mode generalization of the function by \u03c8 \u2261 \u03c80|a \u2297 |0 + Schr\u00a8odinger equation (11.111) reads k \u03c8k|b \u2297a \u2020 k i\u2202t\u03c80 = gkei\u0394kt\u03c8k, k i\u2202t\u03c8k = gke\u2212i\u0394kt\u03c80, where \u0394k = \u2212 \u03c9k. We integrate the second equation and substitute the result into the \ufb01rst equation to arrive at \u2202t\u03c80(t) = \u2212 t 0 g2 k k dt ei\u0394k(t\u2212t)\u03c80(t). Introducing the density of bath modes \u03c1(\u03c9) = k \u03b4(\u03c9 \u2212 \u03c9k), we obtain \u2202t\u03c80(t) = \u2212 d\u03c9 \u03c1(\u03c9", ")g(\u03c9) t 0 dt ei(\u2212\u03c9)(t\u2212t)\u03c80(t) \u22122\u03c0\u03c1g2\u03c80(t), where in the crucial second step we assumed approximate constancy of both \u03c1(\u03c9) \u03c1 d\u03c9 e\u2212i\u03c9(t\u2212t) = 2\u03c0\u03b4(t \u2212 t). The and g(\u03c9) g to evaluate the frequency integral as (irreversible) e\ufb00ective equation for \u03c80 is now trivially solved as32 \u03c80(t) = e\u2212\u03c0\u03c1g2t. This means that the population imbalance between the two atomic levels \u0394\u03c1 \u2261 |\u03c80|2 \u2212 k |\u03c8k|2 = \u22121 + 2e\u2212\u0393t, shows relaxation at the golden rule decay rate \u0393 \u2261 2\u03c0\u03c1g2. (11.112) 32 We count t 0 dt\u03b4(t \u2212 t) = 1/2 since the \u03b4-function lies at the boundaries of the integration domain. (b) Comparing to the discussion in the previous problem, we verify that the multi-mode 11.9 Problems 759 generalization of Eq. (11.109) reads d\u03c9 \u03c1(\u03c9)g2(\u03c9) t dt \u2202t \u02c6\u03c1s = \u2212 % \u00d7 e+i(\u2212\u03c9)t \u2212 e\u2212i(\u2212\u03c9)t (+n(\u03c9) + 1[\u03c3+, \u03c3\u2212 \u02c6\u03c1s(t \u2212 t)] \u2212 n(\u03c9)[\u03c3+, \u02c6\u03c1s(t \u2212 t)\u03c3\u2212]) (\u2212n(\u03c9) + 1[\u03c3\u2212, \u02c6\u03c1s(t \u2212 t)\u03c3+] + n(\u03c9)[\u03c3\u2212, \u03c3+ \u02c6\u03c1s(t \u2212 t)]) &, where n(\u03c9) is the boson distribution function. However, unlike in the previous problem, we now have justi\ufb01cation to assume near constancy of the density operator over the time scales at which the addition of uncorrelated bath contributions decays to zero. (Think why this is equivalent to the Weisskopf\u2013Wigner assumption of frequency independent \u03c1g2.) Doing the integral, we obtain the master equation (11.110), where n = n() is the mean number of bath quanta at the resonance energy and the decay rate is given by Eq. (11.112). Solution of this equation obtains the population imbalance \u0394\u03c1(t) = \u0394\u03c1(0) + 1 2n + 1 e\u2212\u0393(2n+1)t \u2212 1 2n + 1. In the particular case of initial occupancy of |a, \u0394\u03c1(0) = 1 and zero temperature, n() = 0, this reduces to the results obtained in (a). Summarizing our results, we have found that in the multi-mode case the dynamics of the system becomes e\ufb00ectively irreversible. Technically, \u201cmulti-mode\u201d means that we must be able to assume near constancy of the spectral density of bath modes over scales \u223c \u0393, i.e. the rate at which the density operator changes. Under this condition, the Markovian approximation (equivalent to the Weisskopf\u2013Wigner approximation) becomes justi\ufb01ed and the quantum master equation provides a convenient alternative to the direct solution of the Schr\u00a8odinger equation. 11.9.3 Keldysh theory of the Coulomb blockade (Recapitulate Section 11.6 before turning to this problem. If not stated otherwise, the notation of that section will be used throughout.) In this problem, we consider the out of equilibrium quantum dot discussed in Section 11.6 in a regime of near isolation from its environment (\u201cleads\u201d). Under these conditions, charge on the dot is quenched \u2013 the Coulomb blockade. We here discuss the dynamics of the formation of a Coulomb blockaded state, and its response to an external voltage bias. In Problems 6.7 and 6.7 we considered an equilibrium quantum dot in a state of perfect or near isolation from its environment. At low temperatures, the dot admits only the integer quantum of charges that minimizes its capacitive energy. This Coulomb blockade manifests itself in the partition function of the isolated quantum dot, Z = e\u2212 Ec T (n\u2212N0)2, (11.113) where the optimal number of charges is determined by the parameter N0 \u2208 R (which may be set by changing external gate voltages) and Ec is the charging energy. n 760 Nonequilibrium (quantum) At \ufb01rst sight, it may appear strange that the Keldysh-description of even this simplest signature of the Coulomb blockade is far from trivial! To understand why this is", " so, recall that the dynamics on the Keldysh contour builds on the thermal distribution of a non-interacting theory, while the partition sum above describes the thermal state of an interacting system. We thus need to describe the passage between a non-interacting and an interacting equilibrium state. Naively, one might hope that the interpolation between the two states occurs as interactions are gradually switched on in the dynamical evolution along the Keldysh contour. But this is not so. The switching on of interactions results in an increase in energy of the N -particle sector of Fock space, however, the distribution functions controlling the occupation of single particle states are left un-altered (think about this point.) We need to include a mechanism of relaxation-by-interaction to e\ufb00ect the formation of a thermal state similar to that described by Eq. (11.113). Once the mechanism of thermalization has been understood, we can describe structures way beyond the partition function Eq. (11.113). For example, we may explore what happens to the Coulomb blockade when the system is biased out of equilibrium by external voltage, or we may explore the dynamics of the approach to equilibrium after a sudden change in the state of the system, etc. But what, then, causes thermalization in the present context? It turns out that the most direct channel of relaxation is the coupling to the leads. Indeed, we have seen in Section 11.6.1 that the dot-lead coupling of the interacting systems reads (symbolic notation) \u00af\u03c8dT ei\u03c6\u03c8x, where\u03c8 d are the fermion \ufb01elds of the dot, \u03c8x=L,R, those of the leads, and \u02d9\u03c6 is the Hubbard\u2013Stratonovich \ufb01eld of the interaction. This means that a quasiparticle entering the dot creates an excitation in the \ufb01eld \u03c6 (physically: a voltage \ufb02uctuation). That \ufb02uctuation may in turn create a particle-hole excitation (via the vertex \u00af\u03c8xT \u2020e\u2212i\u03c6\u03c8d) in the dot/lead system. We conclude that the coupling vertices provide a mechanism for the excitation of particle-hole pairs, and hence of relaxation in energy. Below we explore the formation of the Coulomb blockade out of these processes, and the generalization to a nonequilibrium state. To keep the discussion simple, we assume vanishing gate voltage N0 = 0, and zero temperature T = 0 throughout. (Exercise: you may \ufb01nd it instructive to generalize to \ufb01nite gate voltage and temperatures.) (a) Our \ufb01rst step is purely technical: observing that at weak coupling to the external world, gT 1, \ufb02uctuations in the \ufb01eld \u03c6 are strong, we will trade the integration over \u03c6 for an integration over all charge states of the dot. (Recall that \u201ccharge\u201d is the degree of freedom conjugate to the phase \u03c6.) To this end, expand the action Eq. (11.79) in the tunneling action Eq. (11.82) and integrate over the phase degrees of freedom (hint: you may \ufb01nd it convenient to switch back to the contour variables \u03c6\u00b1 before doing the integration, and mind the quantization condition nc(0) \u2208 Z). to obtain the representation of the Keldysh partition function Z = \u221e m=0 \u2212 g 2 1 m! m {\u03c3} Dt e\u2212iEc dt (n2 +(t)\u2212n2 \u2212(t)) m' k=1 L\u03c32k\u22121,\u03c32k (t2k\u22121 \u2212 t2k), (11.114) 11.9 Problems 761 where Dt = {\u03c3} is a sum over all sign con\ufb01gurations \u03c3 \u2208 {\u22121, 1}2m, the integration measure 2m k=1 dtk and the charge pro\ufb01les n\u03c3(t) = n + \u03c3 2m k=1 (\u2212)k\u0398(t \u2212 tk)\u03b4\u03c3k,\u03c3. (11.115) Finally, the matrix elements of the kernel L are de\ufb01ned as L\u03c3\u03c3 = 1 4 \u03c3\u03c3\u03a3K + \u03c3\u03a3+ + \u03c3\u03a3\u2212, (11.116) where the self energies \u03a3K,\u00b1 have been introduced in Eq. (11.82). This representation expresses the partition function as a sum over quasiparticle inand out-tunneling events, connected by elements of the kernel L. We next need to make physical sense of this expansion.", " (b) The temporal entanglement of tunneling events makes a closed computation of the partition function impossible. However, for su\ufb03ciently small values of the tunneling conductance, tunneling processes containing intersecting or nested \u201cpropagator lines\u201d L(t, t) are negligibly small. Estimate the temporal range of the propagator L to derive a criterion for the applicability of the non-interacting blip approximation (NIBA) wherein tunneling events occur in a strictly sequential manner (cf. the \ufb01gure.) (c) The lack of entanglement of tunneling events in the NIBA makes the computation of the Keldysh partition function a lot easier. The basic picture now is that occasional charge tunneling events (so-called blips in the jargon of the \ufb01eld33) are interspersed into long periods of time wherein the charge contour pro\ufb01le stays in a diagonal state, n+ = n\u2212 = ncl \u2261 n (\u201csojourns.\u201d) During a blip, the quantum component n+ \u2212 n\u2212 = nq \u2261 \u03be \u2208 {\u22121, 0, 1} jumps to a value \u00b11, depending on the con\ufb01guration (\u03c32k\u22121, \u03c32k) of the tunneling event, and the sign of the time di\ufb00erence t2k\u22121 \u2212 t2k (see the \ufb01gure below). Building on this structure, and assuming zero biasing, V = 0, derive a master equation for the quantity P (n, t) \u2261 P (n, t|n0, 0), i.e. the probability that the system evolves into a charge state (n+, n\u2212) = (n, n) at time t, provided it started in (n0, n0) att = 0. To this end, interpret P (n, t) as the Keldysh \ufb01eld integral Eq. (11.114), subject to the constraint (cf. Eq. (11.115)) n\u03c3(t) = n and n\u03c3(0) = 0. Relate P (n, t) to P (n, t \u2212 \u03b4t), where \u03b4t E\u22121 is much bigger than the typical duration of blips, yet smaller than the average spacing between blips, \u0394t (gEc)\u22121. c 33 U. Weiss, Quantum Dissipative Systems, (World Scienti\ufb01c Publishing, 1993). 762 Nonequilibrium (quantum) Apply a continuum approximation (\u0394t)\u22121(P (n, t) \u2212 P (n, t\u2212\u0394t)) \u2202tP (n, t) to obtain the master equation # \u2202tP (n, t) = ( \u02c6E1 \u2212 1)Wn,n\u22121 $ + ( \u02c6E\u22121 \u2212 1)Wn,n+1 P (n, t), (11.117) where \u02c6E\u00b11f (n) \u2261 f (n \u00b1 1) are charge raising and lowering operators, the transition rates, Wn,n\u00b11 = g \u03c0 and Ec(n, n \u00b1 1)\u0398(Ec(n, n \u00b1 1)), Ec(n, n) \u2261 Ec(n2 \u2212 n2), (11.118) is the relative energy of di\ufb00erent charging states. (d) Discuss the approach to a Coulomb blockaded state, as described by the master equation (11.117). (e) Generalize to the case of \ufb01nite bias voltage V. Will the ground state occupancy change? Answer: (a) Expressed in terms of the contour representation \u03c6c = (\u03c6+ + \u03c6\u2212)/2, \u03c6q = \u03c6+ \u2212 \u03c6\u2212, the charging contribution to the action Sc = S g=0 reads as Sc[n, \u03c6] = dt n+\u2202t\u03c6+ \u2212 n\u2212\u2202t\u03c6\u2212 \u2212 Ec(n2 + \u2212 n2 \u2212), where we have introduced n\u00b1 = nc \u00b1 nq 2 The quantization condition on nc translates to n+(0) = n\u2212(0) \u2208 Z. The relative sign change between the \ufb01rst two terms tells us that the operator ei\u03c6+ raises the charge on the upper contour by one, n+ \u2192 n+ + 1 while ei\u03c6\u2212 lowers the charge on the lower contour by one, n\u2212 \u2192 n\u2212 \u2212 1. To make this action explicit, we \ufb01rst transform the tunneling action Eq. (11.84) to contour \ufb01eld", "s,. Stun[\u03c6] = g 2 dtdt (e\u2212i\u03c6+(t), e\u2212i\u03c6\u2212(t))L(t \u2212 t) 6 7, ei\u03c6+(t) ei\u03c6\u2212(t) where the matrix kernel L = {L\u03c3\u03c3 }, \u03c3, \u03c3 = \u00b11 is de\ufb01ned in Eq. (11.116). We may now expand exp(iStun) in powers of the coupling constant to obtain the series m eiStun[\u03c6] = \u221e m=0 1 m! ig 2 Dt ei 2m k=1(\u2212)k\u03c6\u03c3k (tk) m' k=1 L\u03c32k\u22121,\u03c32k (t2k\u22121 \u2212 t2k), {\u03c3} 11.9 Problems 763 Figure 11.7 A con\ufb01guration contributing to the expansion of the tunneling action: charge tunneling events at times..., t2k\u22121, t2k,... are weighted by matrix elements of the dissipation kernel. 2m where \u03c3 \u2208 {\u22121, 1}2m are sign con\ufb01gurations and Dt = k=1 dtk. The expansion weights in- and out-tunneling events at times t2k\u22121 and t2k, respectively, with elements of the kernel L (cf. Fig. 11.7.) An interpretation in terms of charge tunneling is readily established by integration over the phase \ufb01elds. Integrating the expansion above against the k=1(\u2212)k\u03b4(t \u2212 tk)\u03b4\u03c3k,\u03c3. Substicharging action Sc we obtain the constraint \u2202tn\u03c3 = \u03c3 tution of the solutions of this constraint Eq. (11.115) into the charging action we obtain the representation Eq. (11.114). 2m (b) Consider the kernel L(\u03c9) atV = 0 (exercise: con\ufb01rm that at \ufb01nite V the temporal range of L shrinks below its V = 0 value. This means that at V = 0 we obtain the most conservative estimates for the validity of the NIBA). The scaling L(\u03c9) \u223c \u03c9 (cf. Eq. (11.83)) implies L(t) \u223c t\u22122. This means that a charge tunneling event carries the statistical weight \u223c dt e\u00b1iEctt\u22122 \u223c EC. Its duration can be estimated as \u03b4t \u2261 dt e\u00b1iEctt\u22122t dt e\u00b1iEctt\u22122 \u223c E\u22121 c. The total statistical weight of a tunneling event occurring somewhere in a time window of duration t0 is then given by \u223c gECt0. This means that the average number of tunneling events in t0 is m m m m! (gECt0)m 1 m! (gECt0)m The temporal spacing between events follows as t0/m \u223c( gEc)\u22121, and this relates to the duration of the event as t0/\u03b4t \u223c g\u22121: at low tunneling g 1, the spacing between events exceeds their duration by far, and an approximation treating events in an uncorrelated sequential order becomes justi\ufb01ed. = gECt0. m (c) For E\u22121 c \u0394t (gEc)\u22121, the increment of P in the time window [t \u2212 \u0394t, t] will be determined by zero or one blip processes: P (n, t) P (n, \u2212\u0394t) + n+1 n=n\u22121 Cnn P (n, t \u2212 \u0394t), where the coe\ufb03cients Cnn are one-blip transition probabilities. An individual blip is characterized by its sign pro\ufb01le (\u03c3, \u03c3) (where the \ufb01gure on p762 shows that the connection between the signs and the post-blip increment in classical charge is given by 764 Nonequilibrium (quantum) (+, +), (\u2212, \u2212) :n \u2192 n, (+, \u2212) :n \u2192 n \u2212 1, (\u2212, +) : n \u2192 n + 1.), the center time of the blip t0 \u2208 [t \u2212 \u0394t, t] and its duration, s. Comparison with Eq. (11.114) shows that, e.g. Cn,n\u22121 = = ig\u0394t 2 ig\u0394t 2 ds eiEc(n\u22121,n)sL\u2212+(s) = ig\u0394t 2 L\u2212+(Ec(n \u2212 1, n)) (\u2212\u03a3K \u2212 \u03a3+ + \u03a3", "\u2212)(Ec(n \u2212 1, n)) = g\u0394t \u03c0 Ec(n \u2212 1, n)\u0398(Ec(n \u2212 1, n)), where the prefactor \u0394t results from the integration over the center time, and we introduced the relative charging energy Eq. (11.118). In an analogous manner, we obtain Cn,n+1 = Cnn = g\u0394t \u03c0 g\u0394t \u03c0 Ec(n + 1, n)\u0398(Ec(n + 1, n)), (\u2212Ec(n, n + 1)\u0398(Ec(n, n + 1))\u2212 Ec(n, n \u2212 1)\u0398(Ec(n, n \u2212 1))), where in computing Cnn it is important to keep in mind that \u03a3\u00b1(t) \u221d \u0398(\u00b1t) carry retarded and advanced causality. Substituting this result into the evolution equation above, dividing by \u0394t and taking the continuum limit, we obtain the master equation (11.117). (d) For our current parameter setting, N0 = 0, V = T = 0, the charge state n = 0 de\ufb01nes the energetic ground state. The master equation describes relaxation to this state at rates const. \u00d7 \u0393, where \u0393 \u2261 gEc/\u03c0 = 1/RC is inversely proportional to the RC-time of the system (the proportionality of these rates to g underpins the role of the leadcoupling as the source of relaxation) and the n-dependent constant prefactor increases with the distance o\ufb00 the ground state, n = 0. To show the relaxation in more explicit terms, let us assume that the relaxation has progressed to a level where only the charge states P\u22121,0,1 remain signi\ufb01cantly populated. The restriction of the master equation to this sub-system reads \u2202tP (0, t) = \u0393(P (1, t) +P (\u22121, t)), \u2202tP (\u00b11, t) = \u2212\u0393P (\u00b11, t), which is solved by P (\u00b11, t) = e\u2212\u0393tP (\u00b11, 0) and P (0, t) = 1 \u2212 P (1, 0) \u2212 P (\u22121, 0) + \u0393 0 t dt e\u2212\u0393t (P (1, 0) + P (\u22121, 0)). This solution describes the relaxation towards the state P (n, t) t\u2192\u221e\u2212\u2192 \u03b4n,0. (e) Finite voltages a\ufb00ect the theory through a rede\ufb01ned Keldysh self energy \u03a3K. Compar- ison with Eq. (11.83) shows that \u03a3K(V ) \u2212 \u03a3K(0) = i 2\u03c0 (V \u2212 |\u03c9|)\u0398(V \u2212 |\u03c9|). The transition rates thus change to Wn,n\u00b11 \u2192 Wn,n\u00b11 + g 4\u03c0 (V \u2212 |Ec(n \u00b1 1, n)|)\u0398(V \u00b1 |Ec(n \u2212 1, n)|). 11.9 Problems 765 This change is easy to interpret: for voltages |V | < |Ec(n \u00b1 1, n)| smaller than the charging energies, the transition rates of the unbiased problem remain unaltered. This means that the excess energy of external charge carriers \u223c V needs to exceed the charging energy to make charge transport through the dot possible; the Coulomb blockade is inert against moderate biasing.34 At large voltages, |V | |Ec(n \u00b1 1, n)|, the tunneling rates cross over to values \u223c gV /4\u03c0. For these rates, di\ufb00erent charging states become equally populated, at a rate \u223c gV set by the average current through the dot interfaces. 34 Closer inspection shows that this statement is not entirely correct: the NIBA neglects tunneling processes of higher order in a g-expansion. Among these, there are some (so-called elastic co-tunneling processes) for which the extra power in g gets rewarded by a suppression in Ec that is only algebraic (as compared to the complete \u223c exp(\u2212Ec/T ) T \u21920\u2212\u2192 0 suppression of the direct tunneling processes considered here). Although the description of elastic co-tunneling goes well beyond the scope of this text, we note that its contribution is proportional to the mean single particle levels spacing, \u0394, of the dot, i.e. the results above hold true in the double limit, T, \u0394 \u2192 0. Index 1/f -noise, 616 H-theorem, 626 \u03c64\u2013theory, 195 t \u2212 J Hamiltonian, 65 1-form, 540 activated friction,", "ano factor, 742 FDT (quantum), 712 Fermi energy, 53 Fermi liquid theory, 210 Fermi momentum, 53 Fermi sphere, 53 fermionization, 72 ferromagnetic coupling, 60 Feynman path integral, 95 Fick\u2019s law, 612, 657 \ufb01eld, 6 \ufb01eld renormalization, 432 \ufb01eld theory of directed percolation, 681 \ufb01lling fraction, 519 \ufb01nite size scaling, 442 \ufb01xed point, Gaussian, 448 \ufb01xed points, 420, 433 \ufb02uctuation\u2013dissipation theorem, 371, 611, 661 \ufb02uctuation\u2013dissipation theorem (classical), 656 Fock contribution, 213 Fock space, 43 Fokker\u2013Planck equation, 619, 659 four momentum, 211 fractional charge, 583 free propagator, 199 frequency renormalization, 431 frequency summation, 170 friction, 610 Friedel oscillations, 189 frustration, 80, 329 full counting statistics, 745 functional, 6 functional analysis, 11 functional average, 198 functional di\ufb00erentiation, 11 functional integrals, 100 gauge \ufb01xing, 576, 580 Gauss\u2019 law, 548 Gaussian process, 637 Gaussian distribution, 605 Gaussian functional integration, 104 Gaussian integration, 101 geckos, 29 Gell\u2013Mann\u2013Low equation, 426, 430 generalized Langevin equation, 657 generating function of a distribution, 605 geometric phase, 555 Gibbs distribution, 166 Ginzburg criterion, 451 Ginzburg\u2013Landau action, 282 Ginzburg\u2013Landau theory, 196 glasses, 115 golden rule, 388 Goldstone mode, 283 Goldstone modes, 259 Gor\u2019kov Green function, 277 Gor\u2019kov Hamiltonian, 271 gradient expansion, 286 graphene, 55 Grassmann algebra, 161, 542 Grassmann Gaussian integration, 164 Green function, 105, 198 Green function (non\u2013interacting), 378 Gross\u2013Pitaevskii equation, 714 Group integration, 458 768 Index Gutzwiller trace formula, 145 Haar measure, 136, 457 Haldane conjecture, 516 Hall conductivity, 404 Hall current, 363 Hamilton\u2019s extremal principle, 7 Hamiltonian density, 9 harmonic oscillator, 21 Hartree diagram, 212 heavy fermions, 478 Heisenberg ferromagnet, 258 Heisenberg Hamiltonian, 65 Heisenberg model, 76 Heisenberg representation, 372 Higgs boson, 296 Higgs mechanism, 294 Holstein\u2013Primako\ufb00 transformation, 78 homogeneity, 437 homotopy, 503 homotopy group, 503 Hubbard interaction, 60 Hubbard model, 61 Hubbard\u2013Stratonovich transformation, 197, 244 Hund\u2019s rule, 60, 553 hyperscaling, 444 independent random variable, 605 infrared spectroscopy, 365 instanton, 117 instanton gas, 118, 511 insulator, 54 interacting bose gas, 256 irreducible, 41 Ising model, 128, 196 isothermal compressibility, 362 itinerant magnet, 352 Ito discretization, 645 Jellium model, 52 Johnson noise, 614 Jordan\u2013Wigner transformation, 88 Josephson current, 343 Keldysh Green function, 707 Keldysh technique, 303 kinetic equation, 625, 726 kinetic theory, 623 Knight shift, 367 Kondo E\ufb00ect, 237 Kondo E\ufb00ect: Poor Man\u2019s Scaling, 492 Kondo problem, 91 Kondo temperature, 240, 494 Kramers\u2013Kronig relations, 384 Kramers\u2013Moyal expansion, 639 Kramers-Moyal expansion, 617 Kubo\u2013Anderson process, 636 kurtosis, 747 Lagrange\u2019s equation of motion, 7 Landau diamagnetism, 290 Landau levels, 520, 521 Landau mean\u2013\ufb01eld theory, 446 Landau\u2013Wilson model, 196 Laughlin wave function, 587 Lehmann representation, 374 level (of Wess\u2013Zumino action), 552 Lindhard function, 218, 354 linked cluster theorem, 206 local gauge invariance, 276 London equations, 298 longitudinal conductivity, 393 loop expansion, 431 loop order, 204 Lorentz gauge, 35 Lorentz invariance, 17 Lorentzian distribution, 606 lower critical dimension, 414 macroscopic quantum tunneling, 129 magnetic algebra, 522 magnetic length, 519 magnetic resonance, 367 magnetic susceptibility, 362 magnetic translation operator, 521 magnons, 79 manifolds, integration on, 544 many\u2013body path integral, 158 many-body wavefunction, 40", " Markov process, 633 Markov process, 617 Maslov index, 145 massless excitation, 24 massless modes, 259 master equation, 635 Matsubara frequencies, 168 Maxwell theory, 15 Maxwell\u2013Boltzmann distribution, 609 mean value of a distribution, 604 mean-\ufb01eld, 248 Meissner e\ufb00ect, 298 Mermin\u2013Wagner theorem, 415 mesoscopic physics, 302 mesoscopic systems, 302 metric tensor, 541 microreversibility, 625 minimal coupling, 276 minimal substitution, 329 Minkowski action, 107 mobility edge, 488 moments of a distribution, 604 monodromy matrix, 145 Mott\u2013Hubbard gap, 61, 566 Mott\u2013Hubbard transition, 62 Moyal product, 725 MSRJD\u2013functional, 658 MSRJD-functional, 646 multiplicative noise, 681 N\u00b4eel state, 80 naive scaling dimension, 429 Nambu spinor, 271, 277 nesting symmetry, 355 neutron scattering, 367 Noether current, 33, 179 Noether\u2019s theorem, 30 noise, 613 non\u2013abelian gauge theory, 295 non-abelian bosonization, 559 non-crossing approximation, 227 non-interacting blip approximation (NIBA), 761 nonequilibrium, 602 nonlinear \u03c3-model, 326 nonlinear optics, 369 normal ordering, 167 nuclear magnetic resonance, 367 Nyquist noise, 614 occupation number operator, 47, 48 Index 769 occupation number representation, 43 Ohmic dissipation, 132, 718 one-particle irreducible, 204 Onsager relation, 533 Onsager\u2013Machlup functional, 647, 658 order parameter, 436 orientation, 544 Ornstein\u2013Uhlenbeck process, 685 Ornstein-Uhlenbeck process, 687 oscillator bath, 130 p-form, 542 paramagnetic term, 288 particle indistinguishability, 40 particle number conservation, 73 partition function, 10 path integral (double well), 115 path integral (single well), 113 path integral for spin, 134 path integrals and statistical mechanics, 105 Pauli paramagnetism, 290 Peierls instability, 86 penetration depth, 298 periodic orbits, 145 persistent current, 500 phase transition, \ufb01rst order, 437 phase transition, second order, 437 phase transitions, 436 phonon modes, 24 phonons, 258 photoemission spectroscopy, 366 photon, 27 plasma frequency, 221 Plasma theory, 243 Poisson distribution, 605 polarization operator, 217 probability theory, 603 propagator, 98, 104, 198 Pruisken action, 529 pseudopotential, 51 pullback, 543 purely random stochastic process, 633 quantum billiard, 143 quantum chain, 19 quantum chaos, 142 quantum criticality, 351 quantum dots, 334 quantum electrodynamics, 24 quantum \ufb01elds, 19 quantum \ufb02uctuations, 255 quantum Hall e\ufb00ect, 290 quantum Hall transition, 531 quantum harmonic oscillator, 21 quantum optics, 754 quantum phase transition, 478 quantum vacuum, 28 quantum wire, 67 quasi long\u2013range order, 465 quasiclassical theory, 724 Rabi oscillations, 757 Raman spectroscopy, 365 random phase approximation, 217 random variable, 604 random walk, 636 reaction\u2013di\ufb00usion system, 660 reduced density matrix, 696 Reggeon \ufb01eld theory, 681 relaxation time approximation, 626 renormalization group equation, 443 renormalization, momentum space, 423 replica trick, 303 representation of a group, 41 repulsive \ufb01xed point, 420 resolvent operator, 198 response function, 362 retarded response, 368 retarded response function, 373 Riemannian manifold, 541 RPA, 250 saddle\u2013point approximation, 108 scaling \ufb01eld, irrelevant, 435 scaling \ufb01eld, marginal, 435 scaling \ufb01eld, marginally relevant, 435 scaling \ufb01eld, relevant, 435 scaling \ufb01elds, 435 scaling functions, 441 scaling laws, 438 scaling theory, 441 scaling theory of localization, 488 scanning tunneling microscopy (STM), 367 Schwinger boson representation, 88 Schwinger bosons, 88 screening, 250 sd\u2013Hamiltonian, 93 second quantization, 39 self energy, 250 self\u2013averaging, 302 self\u2013dual, 477 self\u2013similar, 433 self-energy operator, 224 semiclassics, 108 shot noise, 616, 741 Shubnikov\u2013de Haas oscillations, 290 sine\u2013Gordon equation, 152 sine\u2013Gordon model, 469 single particle density of states, 285 skyrmion, 508 Slater determinant, 41 Slater instability, 62 soft modes, 259 sojourn,", " 762 soliton, 151 sound waves, 8 source \ufb01eld, 303 speci\ufb01c heat, 362 spectral \ufb02ow, 523 spectral function, 382 spectroscopic experiments, 363 spiders, 29 spin coherent states, 137 spin density wave, 62 spin density wave (SDW), 75 spin operator, 48 spin quantization, 141 spin\u2013charge separation, 75, 90 spin\u2013waves, 77 spontaneous symmetry breaking, 257 stable \ufb01xed points, 435 staggering, 329 standard model, 294 stationary nonequilibrium, 602 stationary phase approximation, 108 stationary stochastic process, 632 statistics, 2 stochastic di\ufb00erential equation, 645 stochastic processes, 632 stochastic processes (one-step), 638 770 Index vertex operator, 200 vortex, 465, 466 vortices, 265, 300 Wannier states, 54 wave equation, 8 waveguide, 25 Weakly interacting electron gas, 243 Weisskopf\u2013Wigner theory, 758 Wess\u2013Zumino action, 551 Wick rotation, 107 Wick\u2019s theorem, 83, 103 Wigner transformation, 725 Wigner crystal, 209 Wigner surmise, 684 winding number, 153, 499 WZW model of level k, 567 X\u2013ray absorption spectroscopy, 365 X\u2013ray crystallography, 365 X\u2013ray electron spectroscopy, 365 X\u2013ray \ufb02uorescence spectroscopy, 366 zero bias anomaly, 737, 742 Stokes theorem, 548 Stone\u2013von Neumann theorem, 46 Stoner criterion, 353 Stoner transition, 351 structure constants, 458 SU(2), 135 Su-Shrie\ufb00er-Heeger model, 86 sum rule, 386 superconductivity, 265 superconductor energy gap, 272 supercurrent, 262 superexchange, 64 super\ufb02uid density, 290 super\ufb02uidity, 261 supersymmetry, 163, 303 susceptibility, 362 swarming, 621 switching on procedure, 374 symmetries, 2 T\u2013matrix approximation, 234, 237 tangent bundle, 539 tangent mapping, 540 tangent space, 539 target manifold, 6 thermal conductivity, 630, 631 thermal \ufb02uctuations, 256 thermal noise, 614 thermodynamic density of states, 285 thermodynamic experiments, 362 thermodynamic potential, 166 theta term, 507 Thomas\u2013Fermi screening, 220, 387 tight\u2013binding systems, 54 time dependent Ginzburg\u2013Landau theory, 283 time ordered correlation function, 372 time reversal, 314 time\u2013ordering operator, 372 Tomonaga\u2013Luttinger liquid, 76 topological action, 506 topological angle, 506 topological charge, 465, 506 topological excitation, 87 topological sector, 505 topological space, 502 topological term, 141 transfer matrix, 412 transport experiments, 363 tunneling, 129 tunneling density of states, 335 tunneling of quantum \ufb01elds, 125 two \ufb02uid model, 290 two particle propagator, 228 two\u2013parameter phase diagram, 535 type I superconductors, 300 type II superconductors, 300 umklapp scattering, 565 uncorrelated random variable, 605 universality, 2, 439 universality classes, 439 upper critical dimension, 451 vacuum graph, 206 vacuum state, 43 Van der Waals forces, 28 variational principles, 11 vector bosons, 296 vector current, 174 vector \ufb01eld, 540s when they try to solve a loop-the-loop problem using only an expression for centripetal force, perhaps with Newton\u2019s laws, but ignore the fact that energy is conserved too. In electrodynamics it more often comes from e.g. starting with the wave equation (correctly) but failing to re-insert individual Maxwell equations into the reasoning process, failing to use e.g. charge conservation, failing to recognize a physical constraint. After a long time and many tries (especially with Jackson problems, which are notorious for this) a student will often reach the perfect level of utter frustration and stop, scratch their head a bit, and decide to stop just doing math and try using a bit of physics, and half a page later the problem is solved. This is a valuable learning experience, but it is in some sense maximally painful. This short section is designed to help you at minimize that pain to at least some extent. In the following text some small e\ufb00ort will be made on occasion to di\ufb00erentiate the \u201cmathy\u201d parts of a demonstration or derivation from the \u201cphysicsy\u201d parts, so you can see where physics is being injected into a math result to obtain a new understanding, a new constraint or condition on an otherwise general solution, the next", " critical step on the true path to a desired solution to a problem. Students might well bene\ufb01t from marking up their texts or notes as they go along in the same way. What part of what you are writing down is \u201cjust math\u201d (and hence something you can reasonably expect your math skills to carry you through later if need be) and what part is physics and relies on your knowledge of physical laws, visualizable physical relationships, and intuition? Think about that as you proceed through this text. Part I Mathematical Physics 13 Chapter 2 Mathematical Prelude When I \ufb01rst started teaching classical electrodynamics, it rapidly became apparent to me that I was spending as much time teaching what amounted to remedial mathematics as I was teaching physics. After all, to even write Maxwell\u2019s equations down in either integral or di\ufb00erential form requires multivariate calculus \u2013 path integrals, surface integrals, gradients, divergences, curls. These equations are rapidly converted into inhomogeneous partial di\ufb00erential equations and their static and dynamic solutions are expanded in (multipolar) representations, requiring a knowledge of spherical harmonics and various hypergeometric solutions. The solutions are in many cases naturally expressed in terms of complex exponentials, and one requires a certain facility in doing e.g. contour integrals to be able to (for example) understand dispersion or establish representations between various forms of the Green\u2019s function. Green\u2019s functions themselves and Green\u2019s theorem emerge, which in turn requires a student to learn to integrate by parts in vector calculus. This culminates with the development of vector spherical harmonics, Hansen functions, and dyadic tensors in the integral equations that allow one to evaluate multipolar \ufb01elds directly. Then one hits theory of special relativity and does it all again, but now expressing everything in terms of tensors and the theory of continuous groups. It turns out that all the electrodynamics we worked so hard on is much, much easier to understand if it is expressed in terms of tensors of various rank1. 1Some parts are simpler still if expressed in terms of the geometric extension of the 15 We discover that it is essential to understand tensors and tensor operations and notation in order to follow the formulation of relativity theory and relativistic electrodynamics in a compact, workable form. This is in part because some of the di\ufb03culties we have encountered in describing the electric and magnetic \ufb01elds separately result from the fact that they are not, in fact, vector \ufb01elds! They are components of a second rank \ufb01eld strength tensor and hence mix when one changes relativistic frames. Tensors are indeed the natural language of \ufb01eld theories (and much else) in physics, one that is unfortunately not e\ufb00ectively taught where they are taught at all. The same is true of group theory. Relativity is best and most generally derived by looking for the group of all (coordinate) transformations that preserve a scalar form for certain physical quantities, that leave e.g. equations of motion such as the wave equation form invariant. There are strong connections between groups of transformations that conserve a property, the underlying symmetry of the system that requires that property to be conserved, and the labels and coordinatization of the physical description of the system. By e\ufb00ectively exploiting this symmetry, we can often tremendously simplify our mathematical description of a physical system even as we deduce physical laws associated with the symmetry. Unfortunately, it is the rare graduate student that already knows complex variables and is skilled at doing contour integrals, is very comfortable with multivariate/vector calculus, is familiar with the relevant partial di\ufb00erential equations and their basic solutions, has any idea what you\u2019re talking about when you introduce the notion of tensors and manifolds, has worked through the general theory of the generators of groups of continuous transformations that preserve scalar forms, or have even heard of either geometric algebra or Hansen multipoles. So rare as to be practically non-existent. I don\u2019t blame the students, of course. I didn\u2019t know it, either, when I was a student (if it can honestly be said that I know all of this now, for all that I try to teach it). Nevertheless \ufb01lling in all of the missing pieces, one student at a time, very de\ufb01nitely detracts from the \ufb02ow of teaching graded division algebra associated with complex numbers: \u201cgeometric algebra\u201d. This is the algebra of a class of objects that includes the reals, the complex numbers, and the quaternions \u2013 as well as generalized objects of what used to be called \u201cCli\ufb00ord algebra\u201d. I urge interested students to check out Lasenby\ufffd", "\ufffds lovely book on Geometric Algebra, especially the parts that describe the quaternionic formulation of Maxwell\u2019s equations. electrodynamics, while if one doesn\u2019t bother to \ufb01ll them in, one might as well not bother trying to teach the course at all. Over the years in between I\u2019ve tried many approaches to dealing with the missing math. The most successful one has been to insert little minilectures that focus on the math at appropriate points during the semester, which serve to both prepare the student and to give them a smattering of the basic facts that a good book on mathematical physics would give them, and to also require that the students purchase a decent book on mathematical physics even though the ones available tend to be encyclopediac and say far too much or omit whole crucial topics and thereby say far too little (or even both). I\u2019m now trying out a new, semi-integrated approach. This part of the book is devoted to a lightning fast, lecture note-level review of mathematical physics. Fast or not, it will endeavor to be quite complete, at least in terms of what is directly required for this course. However, this is very much a work in progress and I welcome feedback on the idea itself as well as mistakes of omission and commission as always. At the end of I list several readily available sources and references that I\u2019m using myself as I write it and that you might use independently both to learn this material more completely and to check that what I\u2019ve written is in fact correct and comprehensible. Chapter 3 Vector Calculus: Integration by Parts There is one essential theorem of vector calculus that is essential to the development of multipoles \u2013 computing the dipole moment. Jackson blithely integrates by parts (for a charge/current density with compact support) thusly: J d3x = (x\u2207 \u00b7 J )d3x \u2212 ZIR3 ZIR3 (3.1) Then, using the continuity equation and the fact that \u03c1 and J are preJ = sumed harmonic with time dependenct exp( i\u03c9t), we substitute \u2207 \u2212 \u00b7 \u2202\u03c1 \u2202t = i\u03c9\u03c1 to obtain: \u2212 ZIR3 where p is the dipole moment of the fourier component of the charge density distribution. ZIR3 \u2212 x\u03c1(x) d3x = i\u03c9p (3.2) J d3x = i\u03c9 \u2212 However, this leaves a nasty question: Just how does this integration by parts work? Where does the \ufb01rst equation come from? After all, we can\u2019t rely on always being able to look up a result like this, we have to be able to derive it and hence learn a method we can use when we have to do the same thing for a di\ufb00erent functional form. We might guess that deriving it will use the divergence theorem (or Green\u2019s theorem(s), if you like), but any naive attempt to make it do so will 19 lead to pain and su\ufb00ering. Let\u2019s see how it goes in this particularly nasty (and yet quite simple) case. Recall that the idea behind integration by parts is to form the derivative of a product, distribute the derivative, integrate, and rearrange: d(uv) = u dv + v du d(uv dv + b v du a Z b v du b u dv = (uv) a \u2212 Z | a a Z (3.3) where if the products u(a)v(a) = u(b)v(b) = 0 (as will often be the case and u and v have compact support) the process when a = \u201cthrows the derivative from one function over to the other\u201d:, b = \u2212\u221e \u221e a Z which is often extremely useful in evaluating integrals. a \u2212 Z b u dv = b v du (3.4) The exact same idea holds for vector calculus, except that the idea is to use the divergence theorem to form a surface integral instead of a boundary term. Recall that there are many forms of the divergence theorem, but they all map \u2207 to \u02c6n in the following integral form: \u2207...d3x ZV \u2192 IS/V \u02c6n...d2x (3.5) or in words, if we integrate any form involving the pure gradient operator applied to a (possibly tensor) functional form indicated by the ellipsis... in this equation, we can convert the result into an integral over the surface that bounds this volume, where the gradient operator is replaced by an outward directed normal but otherwise the functional form of the expression is preserved. So while the divergence theorem is: \u2207 \u00b7 ZV A d3x = A d2x \u02c6n \u00b7 IS/V there is a \u201cgradient theorem\ufffd", "\ufffd: \u2207f d3x = \u02c6nf d2x IS/V ZV and so on. (3.6) (3.7) To prove Jackson\u2019s expression we might therefore try to \ufb01nd a suitable product whose divergence contains J as one term. This isn\u2019t too easy, however. The problem is \ufb01nding the right tensor form. Let us look at the following divergence: \u2207 \u00b7 (xJ ) = \u2207x J + x\u2207 \u00b7 = Jx + x\u2207 J \u00b7 J \u00b7 (3.8) This looks promising; it is the x-component of a result we might use. However, if try to apply this to a matrix dyadic form in what looks like it might be the right way: \u2207 \u00b7 (xJ ) = (\u2207 x)J + x(\u2207 \u00b7 = 3J + x(\u2207 J ) \u00b7 J ) \u00b7 (3.9) we get the wrong answer. To assemble the right answer, we have to sum over the three separate statements: (xJ )) \u02c6x = (Jx + x\u2207 \u00b7 (yJ )) \u02c6y = + (Jy + y\u2207 (zJ )) \u02c6z = + (Jz + z\u2207 J ) \u02c6x (\u2207 + (\u2207 + (\u2207 \u00b7 \u00b7 \u00b7 or \u02c6xi (\u2207 \u00b7 (xiJ )) = J + x(\u2207 Xi J ) \u02c6y J ) \u02c6z J ) \u00b7 \u00b7 \u00b7 (3.10) (3.11) which is the sum of three divergences, not a divergence itself. If we integrate both sides over all space we get: \u02c6xi (\u2207 Xi ZIR3 \u02c6xi Xi ( \u02c6n ZS( ) \u221e (xiJ )) d3x = (xiJ )) d3x = \u00b7 \u00b7 Xi \u02c6xi0 = 0 = J d3x + J d3x + J d3x + J d3x + x(\u2207 x(\u2207 x(\u2207 x(\u2207 \u00b7 \u00b7 \u00b7 \u00b7 ZIR3 ZIR3 ZIR3 ZIR3 ZIR3 ZIR3 ZIR3 ZIR3 J )d3x (3.12) J )d3x (3.13) J )d3x (3.14) J )d3x (3.15) where we have used the fact that J (and \u03c1) have compact support and are zero everywhere on a surface at in\ufb01nity. We rearrange this and get: J d3x = x(\u2207 \u00b7 J )d3x \u2212 ZIR3 ZIR3 (3.16) which is just what we needed to prove the conclusion. This illustrates one of the most di\ufb03cult examples of using integration by parts in vector calculus. In general, seek out a tensor form that can be expressed as a pure vector derivative and that evaluates to two terms, one of which is the term you wish to integrate (but can\u2019t) and the other the term you want could integrate if you could only proceed as above. Apply the generalized divergence theorem, throw out the boundary term (or not \u2013 if one keeps it one derives e.g. Green\u2019s Theorem(s), which are nothing more than integration by parts in this manner) and rearrange, and you\u2019re o\ufb00 to the races. Note well that the tensor forms may not be trivial! Sometimes you do have to work a bit to \ufb01nd just the right combination to do the job. Chapter 4 Numbers (Note: This appendix is very much under development. Check back periodically.) It may seem silly to devote space to numbers as physicists by hypothesis love numbers, but the standard undergraduate training of physicists does not include a course in number theory per se, so most of what they are likely to know is gleaned from at most one course in complex numbers (math double majors and math minors excepted). This chapter makes no attempt to present an exhaustive review of number theory (however cool and worthy of a deeper treatment it might be) but instead con\ufb01nes itself to just a few points of direct relevance to electrodynamics. 4.1 Real Numbers Real numbers are of obvious importance in physics, and electrodynamics is no exception. Measured or counted quantities are almost invariably described in terms of real numbers or their embedded cousins, the integers. Their virtue in physics comes from from the fact that they form a (mathematical) \ufb01eld1 that is, they support the mathematical operations of addition, subtraction, multiplication and division, and it empirically turns out that physical laws turn out to be describable in terms of algebra", "ic forms based on (at least) real numbers. Real numbers form a group under or- 1Wikipedia: http://www.wikipedia.org/wiki/Field mathematics. ; 23 dinary multiplication and, because multiplication is associative and each element possesses a unique inverse, they form a division algebra2 A division algebra is one where any element other than zero can be divided into any other element to produce a unique element. This property of real numbers is extremely important \u2013 indeed it is the property that makes it possible to use algebra per se to solve for many physical quantities from relations expressed in terms of products and sums. The operational steps: b \u00b7 c\u2212 \u00b7 c\u2212 c = a 1 = a 1) = a 1 = a c) \u00b7 (c (\u2212 c\u2212 c\u2212 \u00b7 \u00b7 \u00b7 (4.1) are so pervasively implicit in our algebraic operations because they are all learned in terms of real numbers that we no longer even think about them until we run into them in other contexts, for example when a, b, c are matrices, with at least c being an invertible matrix. In any event real numbers are ideally suited for algebra because they form a \ufb01eld, in some sense the archetypical \ufb01eld, whereby physical law can be written down in terms of sums and products with measurable quantities and physical parameters represented by real numbers. Other \ufb01elds (or rings) are often de\ufb01ned in terms of either subsets of the real numbers or extensions of the real numbers, if only because when we write a symbol for a real number in an algebraic computation we know exactly what we can and cannot do with it. Real numbers are the basis of real \u201cspace\u201d and \u201ctime\u201d in physics \u2013 they are used to form an algebraic geometry wherein real numbers are spatiotemporal coordinates. This use is somewhat presumptive \u2013 spacetime cannot be 35 meters) \u2013 probed at distances shorter than the Planck length (1.616 and may be quantized and granular at that scale. Whatever this may or may not mean (close to nothing, lacking a complete quantum theory of gravity) it makes no meaningful di\ufb00erence as far as the applicability of e.g. calculus down to that approximate length scale, and so our classical assumption of smooth spacetime will be quite reasonable. 10\u2212 \u00d7 Are real numbers su\ufb03cient to describe physics, in particular classical 2Wikipedia: http://www.wikipedia.org/wiki/Division algebra.. electrodynamics? The answer may in some sense be yes (because classical measurable quantities are invariably real, as are components of e.g. complex numbers) but as we will see, it will be far easier to work over a di\ufb00erent \ufb01eld: complex numbers, where we will often view real numbers as just the real part of a more general complex number, the real line as just one line in a more general complex plane. As we will see, there is a close relationship between complex numbers and a two-dimensional Euclidean plane that permits us to view certain aspects of the dynamics of the real number valued measurable quantities of physics as the real projection of dynamics taking place on the complex plane. Oscillatory phenomena in general are often viewed in this way. 4.2 Complex Numbers The operation of taking the square root (or any other roots) of a real number has an interesting history which we will not review here. Two aspects of number theory that have grown directly out of exploring square roots are, however, irrational numbers (since the square root of most integers can be shown to be irrational) and imaginary numbers. The former will not interest us as we already work over at least the real numbers which include all rationals and irrationals, positive and negative. Imaginary numbers, however, are a true extension of the reals. Since the product of any two non-negative numbers is non-negative, and the product of any two negative numbers is similarly non-negative, we cannot \ufb01nd any real number that, when squared, is a negative number. This permits us to \u201cimagine\u201d a \ufb01eld of numbers where the square root of a nonzero negative number exists. Such a \ufb01eld cannot be identical to the reals already discussed above. It must contain the real numbers, though, in order to be closed under multiplication (as the square of an \u201cimaginary\u201d number is a negative real number, and the square of that real number is a positive real number). If we de\ufb01ne the unit imaginary number to be: i = +\u221a 1 \u2212 (4.2) such that i2 = 1 \u2212 \u00b1 (4.3) we can then form the rest of the \ufb01eld by scaling this imaginary unit through multiplication by a real number (to form the imaginary axis) and then generating the \ufffd", "\ufffd\ufffdeld of complex numbers by summing all possible combinations of real and imaginary numbers. Note that the imaginary axis alone does not form a \ufb01eld or even a multiplicative group as the product of any two imaginary numbers is always real, just as is the product of any two real numbers. However, the product of any real number and an imaginary number is always imaginary, and closure, identity, inverse and associativity can easily be demonstrated. The easiest way to visualize complex numbers is by orienting the real axis at right angles to the imaginary axis and summing real and imaginary \u201ccomponents\u201d to form all of the complex numbers. There is a one-to-one mapping between complex numbers and a Euclidean two dimensional plane as a consequence that is very useful to us as we seek to understand how this \u201cimaginary\u201d generalization works. We can write an arbitrary complex number as z = x+iy for real numbers x and y. As you can easily see, this number appears to be a point in a (complex) plane. Addition and subtraction of complex numbers are trivial \u2013 add or subtract the real and imaginary components separately (in a manner directly analogous to vector addition). Multiplication, however, is a bit odd. Given two complex numbers z1 and z2, we have: z = z1 \u00b7 so that the real and imaginary parts are z2 = x1x2 + i(x1y2 + y1x2) y1y2 \u2212 y1y2 z = x1x2 \u2212 \u211c z = x1y2 + y1x2 \u2111 (4.4) (4.5) (4.6) This is quite di\ufb00erent from any of the rules we might use to form the product of two vectors. It also permits us to form the so-called complex conjugate of any imaginary number, the number that one can multiply it by to obtain a purely real number that appears to be the square of the Euclidean length of the real and imaginary components z = x + iy z\u2217 = x iy 2 = z\u2217z = zz\u2217 = x2 + y2 | \u2212 z | (4.7) (4.8) (4.9) A quite profound insight into the importance of complex numbers can be gained by representing a complex number in terms of the plane polar coordinates of the underlying Euclidian coordinate frame. We can use the product of a number z and its complex conjugate z\u2217 to de\ufb01ne the amplitude z that is the polar distance of the complex number from the | complex origin. The usual polar angle \u03b8 can then be swept out from the positive real axis to identify the complex number on the circle of radius. | This representation can then be expressed in trigonometric forms as + iy = cos(\u03b8) + i | (cos(\u03b8) + i sin(\u03b8)) | ei\u03b8 | = = z | z | sin(\u03b8) z | | (4.10) (4.11) (4.12) where the \ufb01nal result can be observed any number of ways, for example by writing out the power series of eu = 1 + u + u2/2! +... for complex u = i\u03b8 and matching the real and imaginary subseries with those for the cosine and sine respectively. In this expression \u03b8 = tan\u2212 1 y x (4.13) determines the angle \u03b8 in terms of the original \u201ccartesian\u201d complex coordinates. Trigonometric functions are thus seen to be quite naturally expressible in terms of the exponentials of imaginary numbers. There is a price to pay for this, however. The representation is no longer single valued in \u03b8. In fact, it is quite clear that: z = ei\u03b8 | for any integer value of n. We usually avoid this problem initially by requiring \u03b8 \u03c0, \u03c0] (the \u201c\ufb01rst leaf\u201d) but as we shall see, this leads to problems when considering products and roots. (4.14) ( \u2212 z | 2n\u03c0 \u2208 \u00b1 It is quite easy to multiply two complex numbers in this representation: z1 = z2 = z = z1z2 = z1| | z2| | z1|| | ei\u03b81 ei\u03b82 z2| ei(\u03b81+\u03b82) (4.15) (4.16) (4.17) or the amplitude of the result is the product of the amplitudes and the phase of the result is the sum of the two phases. Since \u03b81 + \u03b82 may well be larger than \u03c0 even if the two angles individually are not, to stick to our resolution to keep the resultant", " phase in the range (\u03c0, \u03c0] we will have to form a suitable modulus to put it back in range. Division can easily be represented as multiplication by the inverse of a complex number: 1 z | and it is easy to see that complex numbers are a multiplicative group and division algebra and we can also see that its multiplication is commutative. (4.18) 1 = z\u2212 e\u2212 i\u03b8 | One last operation of some importance in this text is the formation of roots of a complex number. It is easy to see that the square root of a complex number can be written as: \u221az = z | | \u00b1 q ei\u03b8/2 = z | ei(\u03b8/2 \u00b1 | n\u03c0) q (4.19) for any integer n. We usually insist on \ufb01nding roots only within the \ufb01rst \u201cbranch cut\u201d, and return an answer only with a \ufb01nal phase in the range (pi, \u03c0]. There is a connection here between the branches, leaves, and topology \u2013 there is really only one actual point in the complex plane that corresponds to z; the rest of the ways to reach that point are associated with a winding number m that tells one how many times one must circle the origin (and in which direction) to reach it from the positive real axis. Thus there are two unique points on the complex plane (on the principle branch) that are square roots (plus multiple copies with di\ufb00erent winding numbers on other branches). In problems where the choice doesn\u2019t matter we often choose the \ufb01rst one reached traversing the circle in a counterclockwise direction (so that it has a positive amplitude). In physics choice often matters for a speci\ufb01c problem \u2013 we will often choose the root based on e.g. the direction we wish a solution to propagate as it evolves in time. Pursuing this general idea it is easy to see that z 1 n where n is an integer are the points 1 n ei(\u03b8/n \u00b1 2m\u03c0/n) z | | (4.20) where m = 0, 1, 2... as before. Now we will generally have n roots in the principle branch of z and will have to perform a cut to select the one desired while accepting that all of them can work equally well. 4.3 Contour Integration 4.4 Geometric Algebra Chapter 5 Partial Di\ufb00erential Equations (Note: This section is very much under development. Check back periodically.) 5.1 The Laplace Equation 5.2 The Helmholtz Equation 5.3 The Wave Equation 31 Chapter 6 Tensors 6.1 The Dyad and N -adic Forms There are two very di\ufb00erent ways to introduce the notion of a tensor. One is in terms of di\ufb00erential forms, especially the de\ufb01nition of the total di\ufb00erential. This form is ultimately the most useful (and we will dwell upon it below for this reason) but it is also algebraically and intuitively the most complicated. The other way is by contemplating the outer product of two vectors, otherwise known as a dyad. We will introduce the dyad in a two dimensional Euclidean space with Cartesian unit vectors, but it is a completely general idea and can be used in an arbitrary n-manifold within a locally Euclidean patch. Suppose one has a vector A = Ax \u02c6x + Ay \u02c6y and another vector B = Bx \u02c6x + By \u02c6y. If one simply multiplies these two vectors together as an outer product (ordinary multiplication with the distribution of the terms) one obtains the following result: AB = AxBx \u02c6x\u02c6x + AxBy \u02c6x\u02c6y + AyBx \u02c6y \u02c6x + AyBy \u02c6y \u02c6y (6.1) This product of vectors is called a dyadic, and each pair of unit vectors within is called a dyad. A dyad is an interesting object. Each term appears to be formed out of the ordinary multiplicative product of two numbers (which we can easily and fully compute and understand) followed by a pair of unit vectors that are juxtaposed. What, exactly does this juxtaposition of unit vectors mean? 33 We can visualize (sort of) what \u02c6x by itself is \u2013 it is a unit vector in the x direction that we can scale to turn into all possible vectors that are aligned with the x-axis (or into components of general vectors in the two dimensional space). It is not so simple to visualize what a dyad \u02c6x\u02c6x is in this way. The function of such a product becomes more apparent when we de\ufb01ne how it works. Suppose", " with take the inner product (or scalar product, or contraction) of our vector A with the elementary dyad \u02c6xhx. We can do this in either order (from either side): or (\u02c6x\u02c6x) = (A A \u00b7 \u00b7 \u02c6x)\u02c6x = Ax \u02c6x (\u02c6x\u02c6x) \u00b7 A = \u02c6x(\u02c6x \u00b7 A) = Ax \u02c6x (6.2) (6.3) We see that the inner product of a unit dyad \u02c6x\u02c6x with a vector serves to project out the vector that is the x-component of A (not the scalar magnitude of that vector Ax). The inner product of a dyad with a vector is a vector. What about the product of other dyads with A? (\u02c6x\u02c6y) A = \u02c6x(\u02c6y \u00b7 (\u02c6x\u02c6y) = (A A) = Ay \u02c6x \u00b7 \u02c6x)\u02c6y = Ax \u02c6y (6.4) (6.5) A \u00b7 \u00b7 which are not equal. In fact, these terms seem to create the new vector components that might result from the interchange of the x and y components of the vector A, as do (\u02c6y \u02c6x) A = Ax \u02c6y etc. \u00b7 Note well! Some of the dyads commute with respect to an inner product of the dyad with a vector, others (e.g. \u02c6x\u02c6y) do not! Our generalized dyadic multiplication produces what appear to be \u201cintrinsically\u201d non-commutative results when contracted with vectors on the left or the right respectively. This is in fact a break point \u2013 if we pursue this product in one direction we could easily motivate and introduce Geometric Algebra, in terms of which Maxwell\u2019s equations can be written in a compact and compelling form. However, even without doing this, we can arrive at that a compelling form (that is, in fact, quaternionic), so we will restrain ourselves and only learn enough about tensors to be able to pursue the usual tensor form without worrying about whether or how it can be decomposed in a division algebra. The thing to take out of the discussion so far is that in general the inner product of a dyad with a vector serves to project out the scalar amplitude of the vector on the left or the right and reconstruct a possibly new vector out of the remaining unit vector. Very shortly we are going to start writing relations that sum over basis vectors where the basis is not necessarily orthonormal (as this isn\u2019t really necessary or desireable when discussing curvilinear coordinate systems). To do this, I will introduce at this point the Einstein summation convention where writing a product with repeated indices implies summation over those indices: A = Ai \u02c6xi = Ai \u02c6xi Xi (6.6) You can see how the summation symbol is in some sense redundant unless for some reason we wish to focus on a single term in the sum. In tensor analysis this is almost never the case, so it is easier to just specify the exceptions. Note that we can form general dyadic forms directly from the unit dyads without the intermediate step of taking the outer product of particular vectors, producing terms like. We can also take another outer } product from the left or right with all of these forms producing tryads, (eight terms total). Furthermore we terms like can repeat all of the arguments above in higher dimensional spaces, e.g. \u02c6x\u02c6x, \u02c6x\u02c6y, \u02c6x\u02c6z,..., \u02c6z\u02c6z { \u02c6x\u02c6x\u02c6x, \u02c6x\u02c6y \u02c6x,...\u02c6y \u02c6x\u02c6y, \u02c6y \u02c6y \u02c6y { \u02c6x\u02c6x, \u02c6x\u02c6y, \u02c6y \u02c6x, \u02c6y \u02c6y {. } } There is a clear one-to-one correspondance of these monad unit vectors to speci\ufb01c column vectors, e.g.: 1 0 0 0 1 0 0 0 1 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u02c6x = \uf8eb \uf8ec \uf8ec \uf8ed \u02c6y = \uf8eb \uf8ec \uf8ec \uf8ed \u02c6z = \uf8eb \uf8ec \uf8ec \uf8ed (6.7) (6.8) (6.9)", " This correspondance continues through the various unit dyads, tryads \uf8f6 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u02c6x\u02c6x = \uf8eb \uf8ec \uf8ec \uf8ed \u02c6x\u02c6y = \uf8eb \uf8ec \uf8ec \uf8ed (6.10) (6.11) and so on. We will call all of these unit monads, dyads, tryads, and so on, as well as the quantities formed by multiplying them by ordinary numbers and summing them according to similar -adic type, tensors. As we can see, there are several ways of representing tensors that all lead to identical algebraic results, where one of the most compelling is the matrix representation illustrated above. Note well that the feature that di\ufb00erentiates tensors from \u201cordinary\u201d matrices is that the components correspond to particular -adic combinations of coordinate directions in some linear vector space; tensors will change, as a general rule, when the underlying coordinate description is changed. Let us de\ufb01ne some of the terms we will commonly use when working with tensors. The dimension of the matrix in a matrix representation of a tensor quantity we call its rank. We have special (and yet familiar) names for the \ufb01rst few tensor ranks: 0th rank tensor or scalar. This is an \u201cordinary number\u201d, which may at the very least be real or complex, and possibly could be numbers associated with geometric algebras of higher grade. It\u2019s characteristic de\ufb01ning feature is that is is invariant under transformations of the underlying coordinate system. All of the following are algebraic examples of scalar quantities: x, 1.182, \u03c0, Ax, A B... \u00b7 1st rank tensor or vector. This is a set of scalar numbers, each an amplitude corresponding to a particular unit vector or monad, and inherits its transformational properties from those of the underlying unit vec- tors. Examples: A = Ax \u02c6x + Ay \u02c6y, xi {, } \uf8f6, xi} { Ax Ay Az \u02c6z = \uf8eb \uf8ec \uf8ec \uf8ed where the i in e.g. xi does not correspond to a power but is rather a coordinate index corresponding to a contravariant (ordinary) vector where xi similarly corresponds to a covariant vector, and where covariance and contravariance will be de\ufb01ned below. \uf8f7 \uf8f7 \uf8f8 2nd rank tensor or D D matrix (where D is the dimension of the space, so the matrix has D2 components). Examples: Cxy \u02c6x\u02c6y, AB, \u21d4C, Aij, Aj \u00d7 i, Aij, Axx Axy Axz Ayx Ayy Ayz Azx Azy Azz \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u21d4A= \uf8eb \uf8ec \uf8ec \uf8ed where again in matrix context the indices may be raised or lowered to indicate covariance or contravariance in the particular row or column. 3rd and higher rank tensors are the D \u00d7 \u00d7 D D... matrices with a rank corresponding to the number of indices required to describe it. In physics we will have occassion to use tensors through the fourth rank occasionally, through the third rank fairly commonly, although most of the physical quantities of interest will be tensors of rank 0-2. For examples we will simply generalize that of the examples above, using \u21d4T as a generic tensor form or (more often) explicitly indicating its indicial form as in T111, T112,... or \u01ebijk. Using an indicial form with the Einstein summation convention is very powerful, as we shall see, and permits us to fairly simply represent forms that would otherwise involve a large number of nested summations over all coordinate indices. To understand precisely how to go about it, however, we have to \ufb01rst examine coordinate transformations. 6.2 Coordinate Transformations Suppose we have a coordinate frame K in D dimensions, where D will typically be 4 for relativistic spacetime (with the 0th coordinate equal to ct as usual) or 3 for just the spatial part. To simplify our notation, we will use roman characters such as i, j, k for the three-vector spatial-only part of a four-vector, and use greek characters such as \u00b5, \u03bd, \u03b3, \u03b4 for the entire fourvector (where recall, repeated indices imply summation over e.g. i = 1, 2, 3 or \u00b5 = 0, 1, 2, 3, hence the distinction as it can be", " used to de-facto restrict the summation range). Now suppose that we wish to transform to a new coordinate frame K \u2032. At this time we place very few restrictions on this transformation. The transformation might, therefore, translate, rotate, rescale or otherwise alter the original coordinate description. As we do this, our description of physical quantities expressed in the old coordinates must systematically change to a description in the new coordinates, since the actual physical situation being described is not altered by the change in coordinate frames. All that is altered is our point of view. Our \ufb01rst observation might be that it may not be possible to describe our physical quantities in the new frame if the transformation were completely general. For example, if the dimension of K \u2032 were di\ufb00erent (either larger or smaller than that of K) we might well be unable to represent some of the physics that involved the missing coordinate or have a certain degree of arbitrariness associated with a new coordinate added on. A second possible problem involves regions of the two coordinate frames that cannot be made to correspond \u2013 if there is a patch of the K frame that simply does not map into a corresponding patch of the K \u2032 frame we cannot expect to correctly describe any physics that depends on coordinates inside the patch in the new frame. These are not irrelevant mathematical issues to the physicist. A perpetual open question in physics is whether or not any parts of it involve additional variables. Those variables might just be \u201cparameters\u201d that can take on some range of values, or they might be supported only within spacetime scales that are too small to be directly observed (leaving us to infer what happens in these microscale \u201cpatches\u201d from observations made on the macroscale), they may be macroscopic domains over which frame transformations are singular (think \u201cblack holes\u201d) or they may be actual extra dimensions \u2013 hidden variables, if you like \u2013 in which interactions and structure can occur that is only visible to us in our four dimensional spacetime in projection. With no a priori reason to include or exclude any of these pos- sibilities, the wise scientist must be prepared to believe or disbelieve them all and to include them in the \u201cmix\u201d of possible explanations for otherwise di\ufb03cult to understand phenomena. However, our purposes here are more humble. We only want to be able to describe the relatively mundane coordinate transformations that do not involve singularities, unmatched patches, or additional or missing coordinate dimensions. We will therefore require that our coordinate transformations be one-to-one \u2013 each point in the spacetime frame K corresponds to one and only one point in the spacetime frame K \u2032 \u2013 and onto \u2013 no missing or extra patches in the K \u2032 frame. This su\ufb03ces to make the transformations invertible. There will be two very general classes of transformation that satisfy these requirements to consider. In one of them, the new coordinates can be reached by means of a parametric transformation of the original ones where the parameters can be continuously varied from a set of 0 values that describe \u201cno transformation\u201d. In the other, this is not the case. For the moment, let\u2019s stick to the \ufb01rst kind, and start our discussion by looking at our friends the coordinates themselves. By de\ufb01nition, the untransformed coordinates of an inertial reference frame are contravariant vectors. We symbolize contravariant components (not just 4-vectors \u2013 this discussion applies to tensor quantities on all manifolds on the patch of coordinates that is locally \ufb02at around a point) with superscript indices: xcontravariant = (x0, x1, x2, x3...) (6.12) where we are not going to discuss manifolds, curved spaces, tangent or cotangent bundles (much) although we will still use a few of these terms in a way that is hopefully clear in context. I encourage you to explore the references above to \ufb01nd discussions that extend into these areas. Note that I\u2019m using a non-bold x to stand for a four-vector, which is pretty awful, but which is also very common. Now let us de\ufb01ne a mapping between a point (event) x in the frame K and the same point x\u2032 described in the K \u2032 frame. x in K consists of a set of four scalar numbers, its frame coordinates, and we need to transform these four numbers into four new numbers in K \u2032. From the discussion above, we want this mapping to be a continuous function in both directions. That is: x0\u2032 = x0\u2032(x0, x1, x2...) (6.13) and x1\u2032 = x1\u2032(x0, x1, x2...) x2\u2032 = x2\u2032(x0, x1, x2...)... x", "0 = x0(x0\u2032, x1\u2032, x2\u2032...) x1 = x1(x0\u2032, x1\u2032, x2\u2032...) x2 = x2(x0\u2032, x1\u2032, x2\u2032...)... (6.14) (6.15) (6.16) (6.17) (6.18) (6.19) (6.20) have to both exist and be well behaved (continuously di\ufb00erentiable and so on). In the most general case, the coordinates have to be linearly independent and span the K or K \u2032 frames but are not necessarily orthonormal. We\u2019ll go ahead and work with orthonormal coordinate bases, however, which is \ufb01ne since non-orthnormal bases can always be othogonalized with Gram-Schmidt and normalized anyway. Given this formal transformation, we can write the following relation using the chain rule and de\ufb01nition of derivative: \u2202x0\u2032 \u2202x0 dx0 + \u2202x1\u2032 \u2202x0 dx0 + \u2202x2\u2032 \u2202x0 dx0 + \u2202x0\u2032 \u2202x1 dx1 + \u2202x1\u2032 \u2202x1 dx1 + \u2202x2\u2032 \u2202x1 dx1 + \u2202x0\u2032 \u2202x2 dx2 +... \u2202x1\u2032 \u2202x2 dx2 +... \u2202x2\u2032 \u2202x2 dx2 +... dx0\u2032 = dx1\u2032 = dx2\u2032 =... (6.21) (6.22) (6.23) where again, superscripts stand for indices and not powers in this context. We can write this in a tensor-matrix form: dx0\u2032 dx1\u2032 dx2\u2032... \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed = \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed \u2202x0\u2032 \u2202x0 \u2202x1\u2032 \u2202x0 \u2202x2\u2032 \u2202x0... \u2202x0\u2032 \u2202x1 \u2202x1\u2032 \u2202x1 \u2202x2\u2032 \u2202x1... \u2202x0\u2032 \u2202x1 \u2202x1\u2032 \u2202x1 \u2202x2\u2032 \u2202x1............... dx0 dx1 dx2... \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed (6.24) The determinant of the matrix above is called the Jacobean of the transformation and must not be zero (so the transformation is invertible. This matrix de\ufb01nes the di\ufb00erential transformation between the coordinates in the K and K \u2032 frame, given the invertible maps de\ufb01ned above. All \ufb01rst rank tensors that transform like the coordinates, that is to say according to this transformation matrix linking the two coordinate systems, are said to be contravariant vectors where obviously the coordinate vectors themselves are contravariant by this construction. We can signi\ufb01cantly compress this expression using Einsteinian summa- tion: dxi\u2032 = \u2202xi\u2032 \u2202xj dxj (6.25) in which compact notation we can write the de\ufb01nition of an arbitrary contravariant vector A as being one that transforms according to: Ai\u2032 = \u2202xi\u2032 \u2202xj Aj (6.26) There, that was easy! Chapter 7 Group Theory (Note: This appendix is very much under development. Check back periodically.) 7.1 De\ufb01nition of a Group 7.2 Groups of Transformation 43 Chapter 8 Math References \u2022 \u2022 \u2022 \u2022 www.grc.nasa.gov/WWW/K-12/Numbers/Math/documents/...... Tensors TM2002211716.pdf. This is a NASA white paper by Joseph C. Kolecki on the use of tensors in physics (including electrodynamics) and is quite lovely. It presents the modern view of tensors as entities linked both traditional bases and manifolds much as I hope to do here. Mathematical Physics by Donald H. Menzel, Dover Press, ISBN 0-48660056-4. This book was written", " in 1947 and hence presents both the \u201cold way\u201d and the \u201cnew way\u201d of understanding tensors. It is cheap (as are all Dover Press books) and actually is a really excellent desk reference for both undergraduate and graduate level classical physics in general! Section 27 in this book covers simple cartesian tensors, section 31 tensors de\ufb01ned in terms of transformations. Schaum\u2019s Outline series has a volume on vectors and tensors. Again an excellent desk reference, it has very complete sections on vector calculus (e.g. divergence theorem, stokes theorem), multidimensional integration (including de\ufb01nitions of the Jacobian and coordinate transformations between curvilinear systems) and tensors (the old way). http://www.mathpages.com/rr/s5-02/5-02.htm This presents tensors in terms of the manifold coordinate description and is actually quite lovely. It is also just a part of http://www.mathpages.com/, a rather huge collection of short articles on all sorts of really cool problems with absolutely no organization as far as I can tell. Fun to look over 45 and sometimes very useful. \u2022 \u2022 \u2022 \u2022 Wikipedia: http://www.wikipedia.org/wiki/Manifold Tensors tend to be described in terms of coordinates on a manifold. An n-dimensional manifold is basically a mathematical space which can be covered with locally Euclidean \u201cpatches\u201d of coordinates. The patches must overlap so that one can move about from patch to patch without ever losing the ability to describe position in local \u201cpatch coordinates\u201d that are Euclidean (in mathematese, this sort of neighborhood is said to be \u201chomeomorphic to an open Euclidean n-ball\u201d). The manifolds of interest to us in our discussion of tensors are di\ufb00erentiable manifolds, manifolds on which one can do calculus, as the transformational de\ufb01nition of tensors requires the ability to take derivatives on the underlying manifold. Wikipedia: http://www.wikipedia.org/wiki/Tensor This reference is (for Wikipedia) somewhat lacking. The better material is linked to this page, see e.g. Wikipedia: http://www.wikipedia.org/wiki/Covariant vector and Wikipedia: http://www.wikipedia.org/wiki/Contravariant vector and much more. http://www.mth.uct.ac.za/omei/gr/chap3/frame3.html This is a part of a \u201ccomplete online course in tensors and relativity\u201d by Peter Dunsby. It\u2019s actually pretty good, and is de\ufb01nitely modern in its approach. http://grus.berkeley.edu/ jrg/ay202/node183.html This is a section of an online astrophysics text or set of lecture notes. The tensor review is rather brief and not horribly complete, but it is adequate and is in the middle of other useful stu\ufb00. \u223c Anyway, you get the idea \u2013 there are plentiful resources in the form of books both paper and online, white papers, web pages, and wikipedia articles that you can use to really get to where you understand tensor algebra, tensor calculus (di\ufb00erential geometry), and group theory. As you do so you\u2019ll \ufb01nd that many of the things you\u2019ve learned in mathematics and physics classes in the past become simpli\ufb01ed notationally (even as their core content of course does not change). As footnoted above, this simpli\ufb01cation becomes even greater when some of the ideas are further extended into a general geometric division algebra, and I strongly urge interested readers to obtain and peruse Lasenby\u2019s book on Geometric Algebra. One day I may attempt to add a section on it here as well and try to properly unify the geometric algebraic concepts embedded in the particular tensor forms of relativistic electrodynamics. Part II Non-Relativistic Electrodynamics 49 Chapter 9 Plane Waves 9.1 The Free Space Wave Equation 9.1.1 Maxwell\u2019s Equations We begin with Maxwell\u2019s Equations (ME): \u2207 D \u2202B \u2202t = 0 (9.1) (9.2) (9.3) (9.4) in SI units, where D = \u01ebE and H = B/\u00b5. By this point, remembering these should be second nature, and you should really be able to freely go back and forth between these and their integral formulation, and derive/justify the Maxwell Displacement current in terms of charge conservation, etc. Note that there are two inhomogeneous (source-connected) equations and two homogeneous equations, and", " that the inhomogeneous forms are the ones that are medium-dependent. This is signi\ufb01cant for later, remember it. For the moment, let us express the inhomogeneous MEs in terms of just E = \u01ebD and B = H/\u00b5, explicitly showing the permittivity \u01eb and the permeability \u00b51: 1In SI units, now that Jackson 3rd \ufb01nally dropped the curs\u00b4ed evil of Gaussian units. 51 E = \u2207 \u2207 B \u00d7 \u2212 \u00b5\u01eb \u00b7 \u2202E \u2202t = \u00b5J \u03c1 \u01eb (9.5) (9.6) It is di\ufb03cult to convey to you how important these four equations are going to be to us over the course of the semester. Over the next few months, then, we will make Maxwell\u2019s Equations dance, we will make them sing, we will \u201cmutilate\u201d them (turn them into distinct coupled equations for transverse and longitudinal \ufb01eld components, for example) we will couple them, we will transform them into a manifestly covariant form, we will solve them microscopically for a point-like charge in general motion. We will (hopefully) learn them. For the next two chapters we will primarily be interested in the properties of the \ufb01eld in regions of space without charge (sources). Initially, we\u2019ll focus on a vacuum, where there is no dispersion at all; later we\u2019ll look a bit at dielectric media and dispersion. In a source-free region, \u03c1 = 0 and J = 0 and we obtain Maxwell\u2019s Equations in a Source Free Region of Space: \u2207 \u2207 \u2207 \u2207 \u00d7 \u00d7 B E + \u01eb\u00b5 \u2212 \u00b7 \u00b7 \u2202B \u2202t \u2202E \u22029.7) (9.8) (9.9) (9.10) where for the moment we ignore any possibility of dispersion (frequency dependence in \u01eb or \u00b5). 9.1.2 The Wave Equation After a little work (two curls together, using the identity: \u2207 \u00d7 (\u2207 \u00d7 a) = \u2207(\u2207 a) \u00b7 \u2212 \u2207 2a (9.11) Mostly, anyway. Except for relativity. and using Gauss\u2019 Laws) we can easily \ufb01nd that E and B satisfy the wave equation: (for u = E or u = B) where 2u \u2207 \u2212 1 v2 \u22022u \u2202t2 = 0 (9.12) (9.13) v = 1 \u221a\u00b5\u01eb. The wave equation separates2 for harmonic waves and we can actually write the following homogeneous PDE for just the spatial part of E or B: 2 + \u2207 2 + \u2207 \u03c92 v2! \u03c92 v2! E = B = \u2207 \u2207 2 + k2 E = 0 2 + k2 B = 0 (9.14) (9.15) where the time dependence is implicitly e\u2212 i\u03c9t and where v = \u03c9/k. This is called the homogeneous Helmholtz equation (HHE) and we\u2019ll spend a lot of time studying it and its inhomogeneous cousin. Note that it reduces in the k 0 limit to the familiar homogeneous Laplace equation, which is basically a special case of this PDE. \u2192 Observing that3: x x = ikneikn \u2207eikn \u00b7 \u00b7 (9.16) we can easily see that the wave equation has (among many, many others) a solution on IR3 that looks like: u(x, t) = u0ei(kn x \u00b7 \u2212 \u03c9t) where the wave number k = kn has the magnitude k = \u03c9 v = \u221a\u00b5\u01eb\u03c9 (9.17) (9.18) and determines the propagation direction of this plane wave. 2In case you\u2019ve forgotten: Try a solution such as u(x, t) = X(x)Y (y)Z(z)T (t), or (with a bit of inspiration) E(x)e\u2212i\u03c9t in the di\ufb00erential equation. Divide by u. You end up with a bunch of terms that can each be identi\ufb01ed as being constant as they depend on x, y, z, t separately. For a suitable choice of constants one obtains the following PDE for spatial part of harmonic waves. 3Yes, you should work this out termwise if you\u2019ve never done so before. Don\u2019t just take my word for anything. 9.1.3 Plane Waves Plane waves can propagate in any direction. Any superposition of these waves, for", " all possible \u03c9, k, is also a solution to the wave equation. However, recall that E and B are not independent, which restricts the solution in electrodynamics somewhat. To get a feel for the interdependence of E and B, let\u2019s pick k = that e.g.: E(x, t) = E+ei(kx \u2212 B(x, t) = B+ei(kx \u2212 \u03c9t) + E \u2212 \u03c9t) + B ei( \u2212 ei( \u2212 \u03c9t) kx \u2212 \u03c9t) kx \u2212 \u2212 k \u02c6x so \u00b1 (9.19) (9.20) which are plane waves travelling to the right or left along the x-axis for any complex E+, E. In one dimension, at least, if there is no dispersion we can construct a fourier series of these solutions for various k that converges to any well\u2013behaved function of a single variable., B+, B \u2212 \u2212 [Note in passing that: u(x, t) = f (x vt) + g(x + vt) (9.21) \u2212 for arbitrary smooth f (z) and g(z) is the most general solution of the 1dimensional wave equation. Any waveform that preserves its shape and travels along the x-axis at speed v is a solution to the one dimensional wave equation (as can be veri\ufb01ed directly, of course). How boring! These particular harmonic solutions have this form (verify this).] If there is dispersion (velocity a function of frequency) then the fourier superposition is no longer stable and the last equation no longer holds. Each fourier component is still an exponential, but their velocity is di\ufb00erent, and a wave packet spreads out it propagates. We\u2019ll look at this shortly to see how it works for a very simple (gaussian) wave packet but for now we\u2019ll move on. Note that E and B are connected by having to satisfy Maxwell\u2019s equations even if the wave is travelling in just one direction (say, in the direction of a unit vector n); we cannot choose the wave amplitudes separately. Suppose E(x, t) = Eei(kn x \u00b7 x B(x, t) = Bei(kn \u00b7 \u2212 \u2212 \u03c9t) \u03c9t) where E, B, and n are constant vectors (which may be complex, at least for the moment). Note that applying ( \u2207 2 + k2) to these solutions in the HHE leads us to: k2n \u00b7 n = \u00b5\u01eb\u03c92 = \u03c92 v2 (9.22) as the condition for a solution. Then a real b solution indicated above, with k = \u03c9 solution (but not the only one)! n = 1 leads to the plane wave v, which is the most familiar form of the \u00b7 This has mostly been \u201cmathematics\u201d, following more or less directly from the wave equation. The same reasoning might have been applied to sound waves, water waves, waves on a string, or \u201cwaves\u201d u(x, t) of nothing in particular. Now let\u2019s use some physics in the spirit suggested in the last section of the Syllabus and see what it tells us about the particular electromagnetic waves that follow from Maxwell\u2019s equations turned into the wave equation. These waves all satisfy each of Maxwell\u2019s equations separately. For example, from Gauss\u2019 Laws we see e.g. that: \u2207 x Eei(kn \u00b7 \u00b7 \u2207ei(kn x \u00b7 x nei(kn \u00b7 \u00b7 \u2212 \u2212 \u2212 \u2207 E ikE \u00b7 \u00b7 E = 0 \u03c9t) = 0 \u03c9t) = 0 \u03c9t) = 0 (9.23) or (dividing out nonzero terms and then repeating the reasoning for B): n \u00b7 E = 0 and n B = 0. \u00b7 (9.24) Which basically means for a real unit vector n that E and B are perpendicular to n, the direction of propagation! A plane electromagnetic wave is therefore a transverse wave. This seems like it is an important thing to know, and is not at all a mathematical conclusion of the wave equation per se. Repeating this sort of thing using one of the the curl eqns (say, Faraday\u2019s law) one gets: B = \u221a\u00b5\u01eb (n E) \u00d7 (9.25) (the i cancels, k/\u03c9 = 1/v = \u221a\u01eb\u00b5). This means that E and B have the same phase if n is real4 If n is real (and hence a unit vector), then we can introduce three real, mutually orthogonal unit vectors ( \u02c6\u01eb1, \u02c6", "\u01eb2, \u02c6n) and use them to express the \ufb01eld strengths: and E 1 = \u02c6\u01eb1E0, B1 = \u02c6\u01eb2\u221a\u00b5\u01ebE0 E 2 = \u02c6\u01eb2E\u20320, B2 = \u02c6\u01eb1\u221a\u00b5\u01ebE\u20320 \u2212 (9.26) (9.27) where E0 and E\u20320 are constants that may be complex. It is worth noting that E | | = v B | | (9.28) have the same dimensions and that the magnitude of the electric \ufb01eld is greater than that of the magnetic \ufb01eld to which it is coupled via Maxwell\u2019s Equations by a factor of the speed of light in the medium, as this will be used a lot in electrodynamics. These relations describe a wave propagating in the direction \u02c6n = \u02c6\u01eb1 \u00d7 \u02c6\u01eb2 = \u02c6\u01eb3. This follows from the (time-averaged) Poynting vector (for any particular component pair): S = = = = (E 1 2 1 2\u00b5 \u221a\u01eb\u00b5 2\u00b5 1 2 s H \u2217) \u00d7 (E B\u2217) \u00d7 (E \u00d7 vB\u2217) \u01eb \u00b5 | 2 \u02c6n E0 | (9.29) (9.30) (9.31) (9.32) Now, kinky as it may seem, there is no real5 reason that k = kn cannot be complex (while k remains real!) As an exercise, \ufb01gure out the complex vector of your choice such that n = 1. n \u00b7 (9.33) 4Whoops! You mean n doesn\u2019t have to be real? See below. Note also that we are assuming \u01eb and \u00b5 are real as well, and they don\u2019t have to be either. 5Heh, heh. Since I don\u2019t really expect you to do that (gotta save your strength for the real problems later) I\u2019ll work it out for you. Note that this is: n = nR + inI n2 R \u2212 nR \u00b7 n2 I = 1 nI = 0. (9.34) (9.35) (9.36) So, nR must be orthogonal to nI and the di\ufb00erence of their squares must be one. For example: nR = \u221a2\u02c6i nI = 1\u02c6j (9.37) works, as do in\ufb01nitely more More generally (recalling the properties of hyberbolics functions): n = \u02c6e1 cosh \u03b8 + i\u02c6e2 sinh \u03b8 (9.38) where the unit vectors are orthogonal should work for any \u03b8. Thus the most general E such that n E = 0 is \u00b7 E = (i\u02c6e1 sinh \u03b8 \u2212 \u02c6e2 cosh \u03b8)A + \u02c6e3B (9.39) where (sigh) A and B are again, arbitrary complex constants. Note that if n is complex, the exponential part of the \ufb01elds becomes: x ei(kn \u00b7 \u2212 \u03c9t) = e\u2212 knI xei(knR \u00b7 x \u00b7 \u2212 \u03c9t). (9.40) This inhomogeneous plave wave exponentially grows or decays in some direction while remaining a \u201cplane wave\u201d in the other (perpendicular) direction. Fortunately, nature provides us with few sources that produce this kind of behavior (Imaginary n? Just imagine!) in electrodynamics. So let\u2019s forget it for the moment, but remember that it is there for when you run into it in \ufb01eld theory, or mathematics, or catastrophe theory. Instead we\u2019ll concentrate on kiddy physics descriptions of polarization when n is a real unit vector, continuing the reasoning above. 9.1.4 Polarization of Plane Waves We\u2019ve really done all of the hard work already in setting things up above (and it wasn\u2019t too hard). Indeed, the E1 and E2 de\ufb01ned a few equations back are just two independent polarizations of a transverse plane wave. However, we need to explore the rest of the physics, and understand just what is going on in the whole electrodynamic \ufb01eld and not just the electric \ufb01eld component of same. Let\u2019s start by writing E in a fairly general way: Ei = \u02c6\u01ebiEiei(k x \u00b7 \u2212 \u03c9t), (", "9.41) Then we can return (as we will, over and over) to the curl equations to \ufb01nd: Bi = \u221a\u00b5\u01eb Ei k \u00d7 k (9.42) for i = 1, 2 and \u02c6\u01ebi a unit vector perpendicular to the direction of propagation n. Then generally, E(x, t) = (\u02c6\u01eb1E1 + \u02c6\u01eb2E2)ei(k x \u00b7 \u2212 \u03c9t) (9.43) 1 v \u02c6\u01eb1E2)ei(k x \u00b7 \u03c9t) B(x, t) = (\u02c6\u01eb2E1 \u2212 where E1 and E2 are (as usual) complex amplitudes since there is no reason (even in nature) to assume that the \ufb01elds polarized in di\ufb00erent directions have the same phase. (Note that a complex E corresponds to a simple phase shift in the exponential.) (9.44) \u2212 The polarization of the plane wave describes the relative direction, magnitude, and phase of the electric part of the wave. We have several well-known cases: 1. If E1 and E2 have the same phase (but di\ufb00erent magnitude) we have Linear Polarization of the E \ufb01eld with the polarization vector mak1 + E2 ing an angle \u03b8 = tan\u2212 2. Frequently we will choose coordinates in this case so that (say) E2 = 0. 1(E2/E1) with \u01eb1 and magnitude E = E2 q 2. If E1 and E2 have di\ufb00erent phases and di\ufb00erent magnitudes, we have It is fairly easy to show that the electric Elliptical Polarization. \ufb01eld strength traces out an ellipse in the 1, 2 plane. 3. A special case of elliptical polarization results when the amplitudes are out of phase by \u03c0/2 and the magnitudes are equal. In this case we have Circular Polarization. Since ei\u03c0/2 = i, in this case we have a wave of the form: E = E0 \u221a2 ( \u02c6\u01eb1 i\u02c6\u01eb2) = E0\u02c6\u01eb \u00b1. \u00b1 (9.45) where we have introduced complex unit helicity vectors such that: \u02c6\u01eb\u2217 \u00b1 \u00b7 \u02c6\u01eb\u2217 \u00b1 \u00b7 \u02c6\u01eb\u2217 \u00b1 \u00b7 \u02c6\u01eb = 0 \u2213 \u02c6\u01eb3 = 0 \u02c6\u01eb = 1 \u00b1 (9.46) (9.47) (9.48) As we can see from the above, elliptical polarization can have positive or negative helicity depending on whether the polarization vector swings around the direction of propagation counterclockwise or clockwise when looking into the oncoming wave. Another completely general way to represent a polarized wave is via the unit helicity vectors: E(x, t) = (E+\u02c6\u01eb+ + E x ) eik \u00b7 \u2212 \u03c9t \u02c6\u01eb \u2212 \u2212 (9.49) It is left as an exercise to prove this. Note that as always, E amplitudes! \u00b1 are complex I\u2019m leaving Stokes parameters out, but you should read about them on your own in case you ever need them (or at least need to know what they are). They are relevant to the issue of measuring mixed polarization states, but are no more general a description of polarization itself than either of those above. 9.2 Re\ufb02ection and Refraction at a Plane In- terface Suppose a plane wave is incident upon a plane surface that is an interface between two materials, one with \u00b5, \u01eb and the other with \u00b5\u2032, \u01eb\u2032. Incident Wave E = E0ei(k k B = \u221a\u00b5\u01eb x \u00b7 \u03c9t) \u2212 E \u00d7 k Refracted Wave E\u2032 = E\u2032 B\u2032 = q Re\ufb02ected Wave 0ei(k\u2032 x \u00b7 k\u2032 \u2212 \u03c9t) E\u2032 \u00b5\u2032\u01eb\u2032 \u00d7 k\u2032 E\u2032\u2032 = E\u2032\u2032 B\u2032\u2032 = \u221a\u00b5\u01eb 0ei(k k x \u00b7 \u03c9t) \u2212 E\u2032\u2032 \u00d7 k\u2032 (9.50) (9.51) (9.52) (9.53) (9.54) (9.55) where the re\ufb02ected wave and incident wave do not leave the \ufb01rst medium and hence retain speed v = 1/\u221a\u00b5\u01eb, \u00b5, \u01eb and k = \u03c9\u221a\u00b5\u01eb = \u03c9/v. The refracted wave changes to speed v\u2032 = 1/\u221a", "\u00b5\u2032\u01eb\u2032, \u00b5\u2032, k\u2032 = \u03c9\u221a\u00b5\u2032\u01eb\u2032 = \u03c9/v\u2032. [Note that the frequency \u03c9 of the wave is the same in both media! Ask yourself why this must be so as a kinematic constraint...] Our goal is to completely understand how to compute the re\ufb02ected and refracted wave from the incident wave. This is done by matching the wave across the boundary interface. There are two aspects of this matching \u2013 a static or kinematic matching of the waveform itself and a dynamic matching associated with the (changing) polarization in the medium. These two kinds of matching lead to two distinct and well-known results. 9.2.1 Kinematics and Snell\u2019s Law The phase factors of all three waves must be equal on the actual boundary itself, hence: (k \u00b7 x)z=0 = (k\u2032 \u00b7 x)z=0 = (k\u2032\u2032 x)z=0 \u00b7 (9.56) as a kinematic constraint for the wave to be consistent. That is, this has nothing to do with \u201cphysics\u201d per se, it is just a mathematical requirement for the wave description to work. Consequently it is generally covered even in kiddy-physics classes, where one can derive Snell\u2019s law just from pictures of incident waves and triangles and a knowledge of the wavelength shift associated with the speed shift with a \ufb01xed frequency wave. At z = 0, the three k\u2019s must lie in a plane and we obtain: k sin \u03b8incident = k\u2032 sin \u03b8refracted = k sin \u03b8re\ufb02ected n sin \u03b8incident = n\u2032 sin \u03b8refracted = n sin \u03b8re\ufb02ected (9.57) which is both Snell\u2019s Law and the Law of Re\ufb02ection, where we use k = \u03c9/v = n\u03c9/c to put it in terms of the index of refraction, de\ufb01ned by v = c/n. Note that we cancel \u03c9/c, using the fact that the frequency is the same in both media. 9.2.2 Dynamics and Re\ufb02ection/Refraction Now we do the dynamics, that is to say, the real physics. Real physics is associated with the equations of motion of the EM \ufb01eld, that is, with Maxwell\u2019s equations, which in turn become the wave equation, so dynamics is associated with the boundary value problem satis\ufb01ed by the (wave equation) PDEs. So what are those boundary conditions? Recall that the electric displacement perpendicular to the surface must be continuous, that the electric \ufb01eld parallel to the surface must be continuous, that the magnetic \ufb01eld parallel to the surface must be continuous and the magnetic induction perpendicular to the surface must be continuous. To put it another (more physical) way, the perpendicular components of the electric \ufb01eld will be discontinous at the surface due to the surface charge layer associated with the local polarization of the medium in response to the wave. This polarization is actually not instantaneous, and is a bulk response but here we will assume that the medium can react instantly as the wave arrives and that the wavelength includes many atoms so that the response is a collective one. These assumptions are valid for e.g. visible light incident on ordinary \u201ctransparent\u201d matter. Similarly, surface current loops cause magnetic induction components parallel to the surface to be discontinuously changed. Algebraically, this becomes (for E): \u01eb(E0 + E\u2032\u20320) (E0 + E\u2032\u20320) \u00b7 \u00d7 \u02c6n = \u01eb\u2032E\u20320 \u00b7 \u02c6n = E\u20320 \u00d7 \u02c6n \u02c6n (9.58) (9.59) where the latter cross product is just a fancy way of \ufb01nding E components. In most cases one wouldn\u2019t actually \u201cdo\u201d this decomposition algebraically, components one would just inspect the problem and write down the directly using a sensible coordinate system (such as one where \u02c6n = \u02c6z). and \u22a5 || \u22a5 Similarly for B: \u02c6n (9.60) (B0 + B\u2032\u20320) \u00b7 \u02c6n = B\u20320 \u00b7 \u02c6n = 1 \u00b5\u2032 1 \u00b5 \u02c6n \u00d7 (B0 + B\u2032\u20320) B\u20320 \u00d7 (where, recall, B = (k E)/(vk) etc.) Again, one usually would not use this cross product algebraically, but would simply formulate the problem in a convenient coordinate system and take advantage of the fact that: E0| v = \u221a\u00b5\u01eb", " (9.61) (9.62) B0| | E0| | = | \u00d7 Coordinate choice and Brewster\u2019s Law What, then, is a \u201cconvenient coordinate system\u201d? One where \u02c6n = \u02c6z is perpendicular to the surface is good for starters. The remaining two coordinates are selected to de\ufb01ne the plane of re\ufb02ection and refraction and its perpendicular. This is particularly useful because (as we shall see) the re\ufb02ected and refracted intensities depend on their polarization relative to the plane of scattering. Again, to motivate this before messing with the algebra, you hopefully are all familiar with the result taught at the kiddy-physics level known as Brewster\u2019s Law. The argument works like this: because the refracted ray consists of (basically) dipole re-radiation of the incident \ufb01eld at the surface and because dipoles do not radiate along the direction of the dipole moment, the polarization component with E in the scattering plane has a component in this direction. This leads to the insight that at certain angles the refracted ray will be completely polarized perpendicular to the scattering plane (Brewster\u2019s Law)! Our algebra needs to have this decomposition built in from the beginning or we\u2019ll have to work very hard indeed to obtain this as a result! Let us therefore treat rays polarized in or perpendicular to the plane of incidence/re\ufb02ection/refraction separately. E Perpendicular to Plane of Incidence \u00b7 \u00d7 \u02c6n | E | \u02c6n = 0 and The electric \ufb01eld in this case is perforce parallel to the surface and hence E = 1 (for incident, re\ufb02ected and refracted waves). Only two of the four equations above are thus useful. The E equation is trivial. The B equation requires us to determine the magnitude of the cross product of B of each wave with \u02c6n. Let\u2019s do one component as an example. Examining the triangle formed between B0 and \u02c6n for the incident waves (where \u03b8i is the angle of incidence and/or re\ufb02ection, and \u03b8r is the angle of refraction), we see that: B0 cos(\u03b8i) E0 cos(\u03b8i) 1 \u00b5| B0 \u00d7 \u02c6n | = = = 1 \u00b5 \u221a\u00b5\u01eb \u00b5 \u01eb \u00b5 s E0 cos(\u03b8i) (9.63) Repeating this for the other two waves and collecting the results, we obtain: E0 + E\u2032\u20320 = E\u20320 \u01eb\u2032 \u00b5\u2032 E\u2032\u20320 ) cos(\u03b8i) = s \u01eb \u00b5 s (E0 \u2212 E\u20320 cos(\u03b8r) (9.64) (9.65) This is two equations with two unknowns. Solving it is a bit tedious. We need: cos(\u03b8r) = = 1 1 q s \u2212 \u2212 sin2(\u03b8r) n2 2 sin2(\u03b8i) n\u2032 = q 2 n\u2032 \u2212 n2 sin2(\u03b8i) n\u2032 Then we (say) eliminate E\u20320 using the \ufb01rst equation: \u01eb \u00b5 s (E0 \u2212 E\u2032\u20320 ) cos(\u03b8i) = \u01eb\u2032 \u00b5\u2032 s (E0 + E\u2032\u20320 ) q 2 n\u2032 \u2212 n2 sin2(\u03b8i) n\u2032 Collect all the terms: (9.66) (9.67) (9.68) (9.69) \u01eb \u00b5 s E0 \uf8eb \uf8ed Solve for E\u2032\u20320 : cos(\u03b8i) 2 n\u2032 \u2212 \u01eb\u2032 \u00b5\u2032 q n2 sin2(\u03b8i) n\u2032 = \uf8f6 \u2212 s 2 n\u2032 \u2212 \u01eb\u2032 \u00b5\u2032 s q \uf8f8 n2 sin2(\u03b8i) + n\u2032 \u01eb \u00b5 s E\u2032\u20320 \uf8eb \uf8ed (9.70) cos(\u03b8i) \uf8f6 \uf8f8 E\u2032\u20320 = E0 q q \u01eb \u00b5 cos(\u03b8i) \u2212 \u01eb \u00b5 cos(\u03b8i) + \u221an\u20322 n2 sin2(\u03b8i) n\u2032 \u2212 \u221an\u20322 n2 sin2(\u03b8i) n\u2032 \u2212 \u01eb\u2032 \u00b5\u2032 \u01eb\u2032 \u00b5\u2032 q q (9.71) This expression can be simpli\ufb01ed after some tedious cancellations involv- ing n n\u2032 and either repeating the process or back-substituting to obtain : \u00b5\u01eb \u00b5\u2032\u01eb\u2032 = s", " (9.72) E\u2032\u20320 = E0 E\u20320 = E0 n cos(\u03b8i) \u00b5 \u00b5\u2032 \u2212 2 n\u2032 \u2212 q n cos(\u03b8i) + \u00b5 \u00b5\u2032 2 n\u2032 \u2212 q 2n cos(\u03b8i) n2 sin2(\u03b8i) (9.73) n2 sin2(\u03b8i) (9.74) n cos(\u03b8i) + \u00b5 \u00b5\u2032 2 n\u2032 \u2212 q n2 sin2(\u03b8i) E Parallel to Plane of Incidence Now the magnetic \ufb01eld is parallel to the surface so B = 1. This time three equations survive, but they cannot all be independent as we have only two unknowns (given Snell\u2019s law above for the re\ufb02ected/refracted waves). We might as well use the simplest possible forms, which are clearly \u02c6n = the ones where we\u2019ve already worked out the geometry, e.g. E0 \u00d7 E0 cos(\u03b8i) (as before for B0). The two simplest ones are clearly: \u02c6n = 0 and B | \u00d7 \u02c6n \u00b7 | E\u2032\u20320 ) cos(\u03b8i) = E\u20320 cos(\u03b8r) (E0 \u2212 \u01eb \u00b5 (E0 + E\u2032\u20320 ) = E\u20320 s \u01eb\u2032 \u00b5\u2032 s (9.75) (9.76) (from the second matching equations for both E and B above). It is left as a moderately tedious exercise to repeat the reasoning process for these two equations \u2013 eliminate either E\u20320 or E\u2032\u20320 and solve/simplify for the other, repeat or backsubstitute to obtain the originally eliminated one (or use your own favorite way of algebraically solving simultaneous equations) to obtain: E\u20320 = E0 E\u2032\u20320 = E0 2nn\u2032 cos(\u03b8i) \u00b5 \u00b5\u2032 n\u2032 \u00b5 \u00b5\u2032 n\u2032 \u00b5 \u00b5\u2032 n\u2032 2 cos(\u03b8i) + n 2 cos(\u03b8i) \u2212 2 cos(\u03b8i) + n 2 n\u2032 2 n\u2032 q n q 2 n\u2032 q n2 sin2(\u03b8i) n2 sin2(\u03b8i) n2 sin2(\u03b8i) \u2212 \u2212 \u2212 (9.77) (9.78) The last result that one should note before moving on is the important case of normal incidence (where cos \u03b8i = 1 and sin(\u03b8i) = 0). Now there should only be perpendicular solutions. Interestingly, either the parallel or perpendicular solutions above simplify with obvious cancellations and tedious eliminations to: E\u20320 = E0 E\u2032\u20320 = E0 2n n\u2032 + n n n\u2032 \u2212 n\u2032 + n (9.79) (9.80) Note well that the re\ufb02ected wave changes phase (is negative relative to the incident wave in the plane of scattering) if n > n\u2032. This of course makes sense \u2013 there are many intuitive reasons to expect a wave to invert its phase when re\ufb02ecting from a \u201cheavier\u201d medium starting with things one learns studying wave pulses on a string. If this doesn\u2019t make sense to you please ask for help. Intensity Without wanting to get all tedious about it, you should be able to compute the transmission coe\ufb03cient and re\ufb02ection coe\ufb03cient for all of these waves from these results. These are basically the fraction of the energy (per unit area per unit time) in the incident wave that is transmitted vs being re\ufb02ected by the surface. This is a simple idea, but it is a bit tricky to actually compute for a couple of reasons. One is that we only care about energy that makes it through the surface. The directed intensity of the wave (energy per unit area per unit time) is the Poynting vector S. We therefore have to compute the magnitude of the energy \ufb02ux through the surface S \u02c6n for each wave: \u00b7 Sn = S\u2032n = S\u2032\u2032| E0| 2 cos(\u03b8i) \u01eb\u2032 \u00b5\u2032 | \u01eb \u00b5| E\u20320| E\u2032\u20320 | 2 cos(\u03b8r) 2 cos(\u03b8i) (9.81) (9.82) (9.83) (9.84) This is easy only if the waves are incident one gets: \u22a5 to the surface, in which case T = = \u01eb\u2032\u00b5 \u01eb\u00b5\u2032 2 2 E\u20320| | E0| | I \u20320 I0 = s 4nn\u2032 (n\u2032 + n)2 R = = 2 2 E\u2032\u20320 | I \ufffd", "\ufffd\u20320 = | E0| I0 | n)2 (n\u2032 \u2212 (n\u2032 + n)2 In this case it is easy to verify that T + R = 1 as it should. (9.85) (9.86) (9.87) (9.88) Polarization Revisited: The Brewster Angle Note well the expression for the re\ufb02ected wave amplitude for in-plane polarization: E\u2032\u20320 = E0 \u00b5 \u00b5\u2032 n\u2032 \u00b5 \u00b5\u2032 n\u2032 2 cos(\u03b8i) n 2 n\u2032 \u2212 2 cos(\u03b8i) + n q 2 n\u2032 q n2 sin2(\u03b8i) n2 sin2(\u03b8i) \u2212 \u2212 (9.89) This amplitude will be zero for certain angles, namely those such that: \u00b5 \u00b5\u2032 n\u2032 2 cos(\u03b8i) = n 2 n\u2032 q n2 sin2(\u03b8i) \u2212 (9.90) Squaring both sides and restoring cosine term to its original form: 2 \u00b5 \u00b5\u2032! n\u2032 2 cos2(\u03b8i) = n2 cos2(\u03b8r) (9.91) We therefore expect the re\ufb02ected wave to vanish when \u00b5n\u2032 \u00b5\u2032n = cos(\u03b8r) cos(\u03b8i) (9.92) For optical frequencies \u00b5 is equivalent to: \u2248 \u00b5\u2032 (to simplify the algebra somewhat) and this From Snell\u2019s law this in turn is: n\u2032 cos(\u03b8i) = n cos(\u03b8r) n n\u2032 tan(\u03b8i) = n\u2032 n tan(\u03b8r) (9.93) (9.94) This trancendental equation can be solved by observation from its symmetry. It is true if and only if: The angle of incidence tan(\u03b8i) = n\u2032 n = cot(\u03b8r) \u03b8b = tan\u2212 1 n\u2032 n! (9.95) (9.96) is called Brewster\u2019s angle. At this re\ufb02ected and refracted wave travel at right angles with respect to one another according to Snell\u2019s law. This means that the dipoles in the second medium that are responsible for the re\ufb02ected wave are parallel to the direction of propagation and (as we shall see) oscillating dipoles to not radiate in the direction of their dipole moment! However, the result above was obtained without any appeal to the microscopic properties of the dielectric moments that actually coherently scatter the incident wave at the surface \u2013 it follows strictly as the result of solving a boundary value problem for electromagnetic plane waves. Students interested in optical \ufb01bers are encouraged to read further in Jackson, 7.4 and learn how the cancellation and reradiation of the waves to produce a re\ufb02ected wave at angles where total internal re\ufb02ection happens does not occur instantaneously at the refracting surface but in fact involves the penetration of the second medium some small distance by nonpropagating \ufb01elds. This in turn is related to polarization, dispersion, and skin depth, which we will now treat in some detail. 9.3 Dispersion Up to now, we have obtained all of our results with the assumption that the medium was free from dispersion. This just meant that we assumed that the index of refraction was constant as a function of frequency, so all wavelengths were similarly a\ufb00ected. Of course none of our results dependent particular strongly on this result, but in any event it is not correct. The permittivity (and to a lesser extent for transparent materials, the permeability) is a function of the frequency and thus the speed of light changes as we pass from one dielectric medium to another. Let us now \ufb01gure out what and how dispersion works. By the way, when I say that it \u201cisn\u2019t correct\u201d I\u2019m not asserting an opinion or mathematical conclusion. That\u2019s not how physics works. Rather it is always ultimately empirical: rainbows and prisms (not to mention opaque objects) remind us that most physical media are not free from dispersion. Understanding and modelling the dynamics of dispersion in a way that correctly explains these observed phenomena is a key step in the understanding of much modern physics, which involves the measurement and prediction of various susceptibilities (another form of the permittivity, basically, as you can see below) in both classical and quantum circumstances. A full under- standing of the particular dispersion of a physical medium is possible only in the context of quantum theory, but to understand the phenomenon itself we can fortunately rely on a rather simple model that exhibits all the essential features observed in macroscopic media. 9.3.1 Static Case Recall, (from sections 4.5", " and 4.6) that when the electric \ufb01eld penetrates a medium made of bound charges, it polarizes those charges. The charges themselves then produce a \ufb01eld that opposes, and hence by superposition reduces, the applied \ufb01eld. The key assumption in these sections was that the polarization of the medium was a linear function of the total \ufb01eld in the vicinity of the atoms. Linearity response was easily modelled by assuming a harmonic (linear) restoring force: F = m\u03c92 0x (9.97) \u2212 acting to pull a charge e from a neutral equilibrium in the presence of an electric \ufb01eld: m\u03c92 0x = eE (9.98) where E is the applied external \ufb01eld. The dipole moment of this (presumed) system is pmol = ex = e2 E. Real molecules, of course, have many bound m\u03c92 0 charges, each of which at equilibrium has an approximately linear restoring force with its own natural frequency, so a more general model of molecular polarizability is: \u03b3mol = 1 \u01eb0 e2 i mi\u03c92 i (9.99) Xi From the linear approximation you obtained an equation for the total polarization (dipole moment per unit volume) of the material: P = N \u03b3mol \u01eb0E + 1 3 P (9.100) (equation 4.68). This can be put in many forms. For example, using the de\ufb01nition of the (dimensionless) electric susceptibility: P = \u01eb0\u03c7eE (9.101). (9.102) we \ufb01nd that: \u03c7e = N \u03b3mol 1 N \u03b3mol 3 \u2212 The susceptibility is one of the most often measured or discussed quantities of physical media in many contexts of physics. However, as we\u2019ve just seen, in the context of waves we will most often have occasion to use polarizability in terms of the permittivity of the medium, \u01eb. In term of \u03c7e, this is: \u01eb = \u01eb0(1 + \u03c7e) (9.103) From a knowledge of \u01eb (in the regime of optical frequencies where \u00b5 \u2248 \u00b50 for many materials of interest) we can easily obtain, e. g. the index of refraction: n = or c v = \u221a\u00b5\u01eb \u221a\u00b50\u01eb0 \u2248 s \u01eb \u01eb0 \u2248 1 + 2N \u03b3mol 3 N \u03b3mol 1 3 n = v u u t 1 + \u03c7e q (9.104) (9.105) \u2212 if N and \u03b3mol are known or at least approximately computable using the (surprisingly accurate) expression above. So much for static polarizability of insulators \u2013 it is readily understandable in terms of real physics of pushes and pulls, and the semi-quantitative models one uses to understand it work quite well. However, real \ufb01elds aren\u2019t static, and real materials aren\u2019t all insulators. So we gotta 1. Modify the model to make it dynamic. 2. Evaluate the model (more or less as above, but we\u2019ll have to work harder. 3. Understand what\u2019s going on. Let\u2019s get started. 9.3.2 Dynamic Case The obvious generalization of the static model for the polarization is to assume a damped linear response to a harmonic (plane wave) driving electric \ufb01eld. That is, every molecule will be viewed as a collection of damped, driven (charged) harmonic oscillators. Magnetic and non\u2013linear e\ufb00ects will be neglected. This is valid for a variety of materials subjected to \u201cweak\u201d harmonic EM \ufb01elds6 which in practice (with optical frequencies) means nearly everything but laser light. The equation of motion7 for a single damped, driven harmonically bound charged electron is: = m \u00a8x + \u03b3 \u02d9x + \u03c92 h 0x i m\u03b3 \u02d9x is the velocity dependent dampwhere \u03b3 is the damping constant (so \u2212 ing force). If we assume that the electric \ufb01eld E and x are harmonic in time at frequency \u03c9 (or fourier transform the equation and \ufb01nd its solution for a single fourier component) and neglect the transients we get: eE(x, t) (9.106) \u2212 p = ex = \u2212 e2 m E\u03c9 \u03c92 \u2212 (\u03c92 0 \u2212 i\u03c9\u03b3) (9.107) for each electron8. Actually, we have N molecules/unit volume each with Z electrons where fi of them have frequencies and damp", "ing constants \u03c9i and \u03b3i, respectively then (since we will stick in the de\ufb01nitions P \u03c9 = \u01eb0\u03c7eE\u03c9 and (whew!) \u01eb = 1 + \u03c7e) \u01eb(\u03c9) = \u01eb0 1 + N e2 m fi \u03c92 (\u03c92 i \u2212 Xi \u2212 where the oscillator strengths satisfy the sum rule: i\u03c9\u03b3i)! (9.108) (9.109) fi = Z. Xi These equations (within suitable approximations) are valid for quantum theories, and indeed, since quantum oscillators have certain discrete frequencies, they seem to \u201cnaturally\u201d be quantum mechanical. 6Why? If you don\u2019t understand this, you need to go back to basics and think about expanding a potential well in a Taylor series about a particle\u2019s equilibrium position. The linear term vanishes because it is equilibrium, so the \ufb01rst surviving term is likely to be quadratic. Which is to say, proportional to x2 where x is the displacement from equilibrium, corresponding to a linear restoring force to lowest order. 7You do remember Newton\u2019s law, don\u2019t you? Sure hope so... 8I certainly hope you can derive this result, at least if your life depends on it. In quali\ufb01ers, while teaching kiddy physics, whenever. 9.3.3 Things to Note Before we go on, we should understand a few things: 1. \u01eb is now complex! The imaginary part is explicitly connected to the damping constant. 2. Consequently we can now see how the index of refraction n = c v = \u221a\u00b5\u01eb \u221a\u00b50\u01eb0, (9.110) can be also be complex. A complex index of refraction describes absorption (or ampli\ufb01cation!) and arises from the damping term in the electrons\u2019 EOM (or non\u2013linear, non\u2013equilibrium e\ufb00ects in lasers, which we will not consider here). This makes energy conservation kind of sense. Energy absorbed by the electrons and dissipated via the \u201cfrictional\u201d damping force is removed from the EM \ufb01eld as it propagates through the medium. This (complex dispersion of incident waves) is the basis for the \u201coptical\u201d description of scattering which is useful to nuclear physicists. 3. The term i\u03c9\u03b3 \u03c92 i \u2212 1 \u03c92 has a form that you will see again and again and again in your studies. It should be meditated upon, studied, dreamed about, mentally masticated and enfolded into your beings until you understand it. It is a complex equation with poles in the imaginary/real plane. It describes (very generally speaking) resonances. \u2212 It is useful to convert this into a form which has manifest real and imaginary parts, since we will have occasion to compute them in real problems one day. A bit of algebra gives us: 1 \u03c92 = i\u03c9\u03b3 \u2212 \u03c92 i \u2212 (\u03c92 (\u03c92 i \u2212 i \u2212 \u03c92) + i\u03c9\u03b3 \u03c92)2 + \u03c92\u03b32 4. If N is \u201csmall\u201d ( 1019 molecules/cc for a gas) \u03c7e is small (just like in the static case) and the medium is nearly transparent at most frequencies. \u223c 5. if N is \u201clarge\u201d ( 1023 molecules/cc for a liquid or solid) \u03c7e can be quite large in principle, and near a resonance can be quite large and complex! \u223c These points and more require a new language for their convenient de- scription. We will now pause a moment to develop one. 9.3.4 Anomalous Dispersion, and Resonant Absorp- tion Figure 9.1: Typical curves indicating the real and imaginary parts of \u01eb/\u01eb0 for an atom with three visible resonances. Note the regions of anomalous (descending) real dispersion in the immediate vicinity of the resonances, separated by large regions of normal (ascending) dispersion. The \u03b3i are typically small compared to the oscillator frequencies \u03c9i. (Just 1 for optical transitions 1 to \u03c9i \u223c to give you an idea, \u03b3i \u223c 1015 sec\u2212 109 sec\u2212 in atoms, with similar proportionalities for the other relevant transitions.) That means that at most frequencies, \u01eb(\u03c9) is nearly real Suppose we only have a few frequencies. Below the smallest \u03c9i, all the (real) terms in the sum are positive and Re \u01eb(\u03c9) > 1. As we increase \u03c9, one by one the terms in the sum become negative (in their real part) until beyond the highest frequency the entire sum", " and hence Re \u01eb(\u03c9) < 1. As we sweep past each \u201cpole\u201d (where the real part in the denominator of a single term is zero) that term increases rapidly in the real part, then dives through zero to become large and negative, then increases monotonically to zero. Meanwhile, its (usually small) imaginary part grows, reaching a peak just where the real part is zero (when \u01eb(\u03c9) is pure imaginary). In the vicinity of the pole, the contribution of this term can dominate the rest of the sum. We de\ufb01ne: Normal dispersion as strictly increasing Re \u01eb(\u03c9) with increasing \u03c9. This is the normal situation everywhere but near a pole. Anomalous dispersion as decreasing Re \u01eb(\u03c9) with increasing \u03c9. This is true only near a su\ufb03ciently strong pole (one that dominates the sum). At that point, the imaginary part of the index of refraction becomes (relatively) appreciable. Resonant Absorption occurs in the regions where Im \u01eb is large. We will parametrically describe this next. 9.3.5 Attenuation by a complex \u01eb Suppose we write (for a given frequency) Then k = \u03b2 + i \u03b1 2. E\u03c9(x) = eikx = ei\u03b2xe\u2212 \u03b1 2 x (9.111) (9.112) and the intensity of the (plane) wave falls o\ufb00 like e\u2212 damping of the plane wave in the medium. \u03b1x. \u03b1 measures the Let\u2019s think a bit about k9.113) where: n = c/v = \u221a\u00b5\u01eb \u221a\u00b50\u01eb0 (9.114) In most \u201ctransparent\u201d materials, \u00b5 Thus: \u2248 \u00b50 and this simpli\ufb01es to n = \u01eb/\u01eb0. k2 = \u03c92 c2 \u01eb \u01eb0 q (9.115) Nowever, now \u01eb has real and imaginary parts, so k may as well! In fact, using the expression for k in terms of \u03b2 and \u03b1 above, it is easy to see that: and Re k2 = \u03b22 \u03b12 4 \u2212 = \u03c92 c2 Re \u01eb \u01eb0 Im k2 = \u03b2\u03b1 = \u03c92 c2 Im \u01eb \u01eb0. (9.116) (9.117) As long as \u03b22 >> \u03b12 (again, true most of the time in trasparent materi- als) we can thus write: and \u03b1 \u03b2 \u2248 Im \u01eb(\u03c9) Re \u01eb(\u03c9) \u03b2 \u2248 (\u03c9/c) Re s \u01eb \u01eb0 (9.118) (9.119) This ratio can be interpreted as a quantity similar to Q, the fractional decrease in intensity per wavelength travelled through the medium (as opposed to the fractional decrease in intensity per period). To \ufb01nd \u03b1 in some useful form, we have to examine the details of \u01eb(\u03c9), which we will proceed to do next. When \u03c9 is in among the resonances, there is little we can do besides work out the details of the behavior, since the properties of the material can be dominated strongly by the local dynamics associated with the nearest, strongest resonance. However, there are two limits that are of particular interest to physicists where the \u201cresonant\u201d behavior can be either evaluated or washed away. They are the low frequency behavior which determines the conduction properties of a material far away from the electron resonances per se, and the high frequency behavior which is \u201cuniversal\u201d. 9.3.6 Low Frequency Behavior Near \u03c9 = 0 the qualitative behavior depends upon whether or not there is a \u201cresonance\u201d there. If there is, then \u01eb(\u03c9 0) can begin with a complex component that attenuates the propagation of EM energy in a (nearly static) applied electric \ufb01eld. This (as we shall see) accurately describes conduction and resistance. If there isn\u2019t, then \u01eb is nearly all real and the material is a dielectric insulator. \u2248 Suppose there are both \u201cfree\u201d electrons (counted by ff ) that are \u201cresonant\u201d at zero frequency, and \u201cbound\u201d electrons (counted by fb). Then if we start out with: \u01eb(\u03c9) = \u01eb0 1 + = \u01eb0 1 + + = \u01ebb + i\u01eb0 N e2 m N e2 m N e2 m (\u03c92 i \u2212 Xi (\u03c92 b \u2212 Xb fi \u03c92 fb \u03c92", " \u2212 \u2212 ff i\u03c9\u03b3i)! i\u03c9\u03b3b)! \u03c92 i\u03c9\u03b3f ) \u2212 ( Xf \u2212 N e2ff m\u03c9(\u03b30 \u2212 i\u03c9) where \u01ebb is now only the contribution from all the \u201cbound\u201d dipoles. We can understand this from \u2207 \u00d7 H = J + dD dt (9.120) (9.121) (Maxwell/Ampere\u2019s Law). Let\u2019s \ufb01rst of all think of this in terms of a plain old static current, sustained according to Ohm\u2019s Law: J = \u03c3E. (9.122) If we assume a harmonic time dependence and a \u201cnormal\u201d dielectric constant \u01ebb, we get: \u2207 \u00d7 H = (\u03c3 \u2212 i\u03c9 i\u03c9\u01ebb) E \u03c3 \u03c9 \u01ebb + i \u2212 = E. (9.123) On the other hand, we can instead set the static current to zero and consider all \u201ccurrents\u201d present to be the result of the polarization response D to the \ufb01eld E. In this case: \u2207 \u00d7 H = i\u03c9\u01ebE \u2212 = i\u03c9 \u2212 \u01ebb + i\u01eb0 N e2 m ff (\u03b30 \u2212 E i\u03c9)! (9.124) Equating the two latter terms in the brackets and simplifying, we obtain the following relation for the conductivity: \u03c3 = \u01eb0 nf e2 m 1 (\u03b30 \u2212. i\u03c9) (9.125) This is the Drude Model with nf = ff N the number of \u201cfree\u201d electrons per unit volume. It is primarily useful for the insight that it gives us concerning the \u201cconductivity\u201d being closely related to the zero-frequency complex part of the permittivity. Note that at \u03c9 = 0 it is purely real, as it should be, recovering the usual Ohm\u2019s Law. We conclude that the distinction between dielectrics and conductors is a matter of perspective away from the purely static case. Away from the static case, \u201cconductivity\u201d is simply a feature of resonant amplitudes. It is a matter of taste whether a description is better made in terms of dielectric constants and conductivity or complex dielectric. 9.3.7 High Frequency Limit; Plasma Frequency Way above the highest resonant frequency the dielectric constant takes on a simple form (factoring out \u03c9 >> \u03c9i and doing the sum to the lowest surviving order in \u03c9p/\u03c9. As before, we start out with: \u01eb(\u03c9) = \u01eb0 1 + N e2 m (\u03c92 i \u2212 Xi fi \u03c92 \u2212 i\u03c9\u03b3i)! 1 = \u01eb0 \uf8eb \uf8ed 1 \u01eb0 \u2248 \u01eb0 1 \u2248 N e2 \u03c92m \u2212 Xi N Ze2 \u03c92m! \u03c92 p \u03c92! \u2212 \u2212 fi (1 + i \u03b3i \u03c9 \u2212 \u03c92 \u03c92 ) \uf8f6 i \uf8f8 (9.126) where \u03c92 p = ne2 m. (9.127) This is called the plasma frequency, and it depends only on n = N Z, the total number of electrons per unit volume. The wave number in this limit is given by: ck = \u03c92 \u03c92 p \u2212 (9.128) q (or \u03c92 = \u03c92 p + c2k2). This is called a dispersion relation \u03c9(k). A large portion of contemporary and famous physics involves calculating dispersion relations (or equivalently susceptibilities, right?) from \ufb01rst principles. 10 10 Figure 9.2: The dispersion relation for a plasma. Features to note: Gap at k = 0, asymptoticallyx linear behavior. The dielectric constant is essentially real and slightly less than one. In certain physical situations (such as a plasma or the ionosphere) all the electrons are essentially \u201cfree\u201d (in a degenerate \u201cgas\u201d surrounding the positive charges) and resonant damping is neglible. In that case this relation can hold for frequencies well below \u03c9p (but well above the static limit, since plasmas are low frequency \u201cconductors\u201d). Waves incident on a plasma are re\ufb02ected and the \ufb01elds inside fall o\ufb00 exponentially away from the surface. Note that \u03b1p \u2248 2\u03c9p c (9.129) shows how electric \ufb02ux is expelled by the \u201cscreening\u201d electrons. The re\ufb02ectivity of metals is caused by essentially the same", " mechanism. At high frequencies, the dielectric constant of a metal has the form \u01eb(\u03c9) \u2248 \u01eb0(\u03c9) \u03c92 p \u03c92 \u2212 (9.130) where \u03c92 p = ne2/m\u2217 is the \u201cplasma frequency\u201d of the conduction electrons. m\u2217 is the \u201ce\ufb00ective mass\u201d of the electrons, introduced to describe the e\ufb00ects of binding phenomenologically. Metals re\ufb02ect according to this rule (with a very small \ufb01eld penetration length of \u201cskin depth\u201d) as long as the dielectric constant is negative; in the ultraviolet it becomes positive and metals can become transparent. Just one of many problems involved in making high ultraviolet, x\u2013ray and gamma ray lasers \u2014 it is so hard to make a mirror! 9.4 Penetration of Waves Into a Conductor \u2013 Skin Depth 9.4.1 Wave Attenuation in Two Limits Recall from above that: \u2207 \u00d7 H = i\u03c9\u01ebE = \u2212 i\u03c9 \u2212 \u01ebb + i \u03c3 \u03c9 E. Then: k2 = \u03c92 v2 = \u00b5\u01eb\u03c92 = \u00b5\u01ebb\u03c92 1 + i \u03c3 \u03c9\u01ebb (9.131) (9.132) Also k = \u03b2 + i \u03b1 2 so that k2 = \u03b22 \u03b12 4! \u2212 + i\u03b1\u03b2 = \u00b5\u01ebb\u03c92 1 + i \u03c3 \u03c9\u01ebb (9.133) Oops. To determine \u03b1 and \u03b2, we have to take the square root of a complex number. How does that work again? See the appendix on Complex Numbers... In many cases we can pick the right branch by selecting the one with the right (desired) behavior on physical grounds. If we restrict ourselves to the two simple cases where \u03c9 is large or \u03c3 is large, it is the one in the principle branch (upper half plane, above a branch cut along the real axis. From the last equation above, if we have a poor conductor (or if the frequency is much higher than the plasma frequency) and \u03b1 \u03b2, then: \u226a \u221a\u00b5\u01ebb\u03c9 \u00b5 \u01ebb \u03c3 \u2248 \u03b2 \u03b1 \u2248 s (9.134) (9.135) and the attenuation (recall that E = E0e\u2212 quency. \u03b1 E) is independent of fre2 ei\u03b2 \u02c6n \u00b7 The other limit that is relatively easy is a good conductor, \u03c3 that case the imaginary term dominates and we see that \u03b1 2 \u03b2 \u2248 \u03c9\u01ebb. In \u226b (9.136) or Thus \u03b2 \u03b1 \u00b5\u03c3\u03c9 2 2\u00b5\u03c3\u03c9 \u2248 r \u2248 q k = (1 + i) \u00b5\u03c3\u03c9 2 r x Recall that if we apply the \u2207 operator to Eeik(n \u00b7 \u2212 i\u03c9t we get: \u2207 \u00b7 ikE0 \u00b7 E0 \u00b7 E = 0 \u02c6n = 0 \u02c6n = 0 (9.137) (9.138) (9.139) (9.140) and \u2202B \u2202t \u2212 = \u2207 ~E \u00d7 i\u03c9\u00b5H 0 = i( \u02c6n E0)(1 + i) H 0 = = \u00d7 \u03c3\u03c9 \u00b5 \u00d7 \u00d7 \u03c3\u03c9 \u00b5 ( \u02c6n 1 \u03c9 s 1 \u03c9 s \u00b5\u03c3\u03c9 2 r 1 \u221a2 ( \u02c6n E0) (1 + i) E0)ei\u03c0/4 (9.141) so E0 and H 0 are not in phase (using the fact that i = ei\u03c0/2). In the case of superconductors, \u03c3 them is \u03c0/4. In this case H 0 \u226b magnetic. and the phase angle between E (show this!) and the energy is mostly \u2192 \u221e Finally, note well that the quantity = \u03b4 is an exponential damping length that describes how rapidly the wave attenuates as it moves into the conducting medium. \u03b4 is called the skin depth and we see that \u00b5\u03c3\u03c9 s (9.142) We will examine this quantity in some detail in the sections on waveguides and optical cavities, where it plays an important role. 9.5 Kramers-Kronig Relations We \ufb01nd KK relations by playing looped games with Fourier Transforms. We begin with the relation between the electric \ufb01eld and displacement at some particular frequency \u03c9: D(x, \u03c9) = \u01eb(\u03c9)E(x, \u03c9) (9.143) where we note the two (forward and backward) fourier transform relations: D(x, t) = D(x,", " \u03c9) = \u2212\u221e 1 \u221a2\u03c0 Z 1 \u221a2\u03c0 Z \u2212\u221e \u221e D(x, \u03c9)e\u2212 i\u03c9td\u03c9 \u221e D(x, t\u2032)ei\u03c9t\u2032 dt\u2032 (9.144) (9.145) and of course: E(x, t) = E(x, \u03c9) = 1 \u221a2\u03c0 Z 1 \u221a2\u03c0 Z \u2212\u221e \u2212\u221e \u221e E(x, \u03c9)e\u2212 i\u03c9td\u03c9 \u221e E(x, t\u2032)ei\u03c9t\u2032 dt\u2032 (9.146) (9.147) Therefore: D(x, t) = = 1 \u221a2\u03c0 Z 1 \u221a2\u03c0 Z \u2212\u221e \u221e \u221e \u01eb(\u03c9)E(x, \u03c9)e\u2212 i\u03c9td\u03c9 \u01eb(\u03c9)e\u2212 i\u03c9td\u03c9 \u221e E(x, t\u2032)ei\u03c9t\u2032 dt\u2032 1 \u221a2\u03c0 Z G(\u03c4 )E(x, t \u2212\u221e \u03c4 )d\u03c4 \u2212 (9.148) = \u01eb0 \u2212\u221e E(x, t) + \u221e Z \u2212\u221e where we have introduced the susceptibility kernel: G(\u03c4 ) = 1 2\u03c0 Z \u221e \u2212\u221e \u01eb(\u03c9) \u01eb0 \u2212 ( 1 ) e\u2212 i\u03c9\u03c4 d\u03c9 = 1 2\u03c0 Z \u2212\u221e \u221e \u03c7e(\u03c9)e\u2212 i\u03c9\u03c4 d\u03c9 (9.149) (noting that \u01eb(\u03c9) = \u01eb0(1 + \u03c7e(\u03c9))). This equation is nonlocal in time unless G(\u03c4 ) is a delta function, which in turn is true only if the dispersion is constant. To understand this, consider the susceptibility kernel for a simple one In this case, resonance model (more resonances are just superposition). recall that: \u03c7e = \u01eb \u01eb0 \u2212 1 = i\u03b30\u03c9 (9.150) so G(\u03c4 ) = \u03c92 p 2\u03c0 Z \u221e \u2212\u221e \u03c92 0 \u2212 e\u2212 i\u03c9\u03c4 d\u03c9 (9.151) i\u03b30\u03c9 \u2212 \u03c92 p \u03c92 \u2212 \u03c92 0 \u2212 1 \u03c92 This is an integral we can do using contour integration methods. We use the quadratic formula to \ufb01nd the roots of the denominator, then write the factored denominator in terms of the roots: or i\u03b3 \u03c91,2 = \u2212 \u00b1 q \u03b32 + 4\u03c92 0 \u2212 2 \u03c91,2 = \u2212 i\u03b3 2 \u00b1 \u03c90 1 s \u2212 \u03b32 4\u03c92 0 = \u2212 i\u03b3 2 \u00b1 \u03bd0 (9.152) (9.153) \u03c90 as long as \u03c90 \u226b \u03b3/2 (as is usually the case, remember \u03b2 and where \u03bd0 \u2248 \u03b1/2). Note that these poles are in the lower half plane (LHP) because of the sign of \u03b3 in the original harmonic oscillator \u2013 it was dissipative. This is important. Then G(\u03c4 ) = (2\u03c0i) \u03c92 p 2\u03c0 IC (\u03c9 1 \u03c91)(\u03c9 \u03c92) e\u2212 i\u03c9\u03c4 d\u03c9 (9.154) \u2212 \u2212 If we close the contour in the upper half plane (UHP), we have to restrict \u03c4 < 0 (why? because otherwise the integrand will not vanish on the contour at in\ufb01nity where \u03c9 has a positive imaginary part. Since it encloses no poles, G(\u03c4 < 0) vanishes, and we get no contribution from the future in the integral above for E. The result appears to be causal, but really we cheated \u2013 the \u201ccausality\u201d results from the damping term, which represents entropy and yeah, gives time an arrow here. But it doesn\u2019t really break the symmetry of time in this problem and if our model involved a dynamically pumped medium so that the wave experienced gain moving through it (an imaginary term that was positive) we would have had poles in the UHP and our expression for E would not be \u201ccausal\u201d. Really it is equally causal in both cases, because the fourier transforms involved sample all times anyway. If we close the integrand in the LHP, \u03c4 > 0 and if we do the rest of the (fairly straightforward) algebra we get: G(\u03c4 ) = \u03c92 pe\u2212 \u03b3\u03c4 2 sin(\u03bd0) \u03bd0 \u0398(\u03c4 ) (9.155) where the latter is a Heaviside function to enforce the \u03c4 > 0 constraint. Our last little exercise is to use", " complex variables and Cauchy\u2019s theorem again. We start by noting that D and E and G(\u03c4 ) are all real. Then we can integrate by parts and \ufb01nd things like: \u01eb(\u03c9) \u01eb0 \u2212 1 = i G(0) \u03c9 \u2212 G\u2032(0) \u03c92 +... (9.156) from which we can conclude that \u01eb( \u03c9) = \u01eb\u2217(\u03c9\u2217) and the like. Note the \u2212 even/odd imaginary/real oscillation in the series. \u01eb(\u03c9) is therefore analytic in the UHP and we can write: \u01eb(z) \u01eb0 \u2212 1 = 1 2\u03c0i IC \u01eb(\u03c9\u2032) 1 \u01eb0 \u2212 z \u03c9\u2032 \u2212 d\u03c9\u2032 (9.157) We let z = \u03c9 + i\u03b4 where \u03b4 0+ (or deform the integral a bit below the \u2192 singular point on the Re(\u03c9) axis). From the Plemlj Relation: 1 \u03c9 = P 1 \u03c9\u2032 \u2212 (see e.g. Wyld, Arfkin). If we substitute this into the integral above along the real axis only, do the delta-function part and subtract it out, cancel a factor of 1/2 that thus appears, we get: + i\u03c0\u03b4(\u03c9\u2032 \u03c9\u2032 \u2212 (9.158) \u03c9) \u2212 \u2212 i\u03b4 \u03c9 \u01eb(\u03c9) \u01eb0 = 1 + 1 i\u03c0 P \u221e Z \u2212\u221e \u01eb(\u03c9\u2032) 1 \u01eb0 \u2212 \u03c9 \u03c9\u2032 \u2212 d\u03c9\u2032 (9.159) Although this looks like a single integral, because of the i in the denominator it is really two. The real part of the integrand becomes the imaginary part of the result and vice versa. That is: Re Im \u01eb(\u03c9) \u01eb0! \u01eb(\u03c9) \u01eb0! = 1 + 1 \u03c0 P \u221e Z \u2212\u221e Re = 1 \u03c0 P \u2212 \u221e Z \u2212\u221e \u01eb(\u03c9\u2032) \u01eb0 \u03c9 Im \u03c9\u2032 \u2212 \u01eb(\u03c9\u2032) \u01eb0 \u03c9\u2032 \u2212 \u03c9 d\u03c9\u2032 1 \u2212 d\u03c9\u2032 (9.160) (9.161) These are the Kramers-Kronig Relations. They tell us that the dispersive and absorptive properties of the medium are not independent. If we know the entire absorptive spectrum we can compute the dispersive spectrum and vice versa. There is one more form of the KK relations given in Jackson, derived from the discovery above that the real part of \u01eb(\u03c9) is even in \u03c9 while the imaginary part is odd. See if you can derive this on your own for the fun of it all... 9.6 Plane Waves Assignment To start o\ufb00 the semester right, visit the Wikipedia and Mathworld websites and look up and understand: 1. Separation of variables 2. Spherical Harmonics 3. Bessel Functions 4. Spherical Bessel Functions 5. Green\u2019s Functions 6. Wave Equation 7. Plane Wave Just explore the kinds of things you can \ufb01nd there \u2013 I\u2019m discovering that these web references are rapidly becoming THE universal free textbook. It is actually amazing to watch it happen (and participate in it as time permits). Jackson, problems: 7.4, 7.6, 7.19, 7.21 Also, derive on your own all the principal results presented in these online lecture notes. It is easy to read and see me do it. It is not so easy to do it, even for me. Working through this, possibly several times until you really \u201cget it\u201d, will truly improve your understanding of how everything works. Chapter 10 Wave Guides 10.1 Boundary Conditions at a Conducting Surface: Skin Depth Let us consider for a moment what time dependent EM \ufb01elds look like at the surface of a \u201cperfect\u201d conductor. A perfect conductor can move as much charge instantly as is required to cancel all \ufb01elds inside. The skin depth \u03b4 = lim\u03c3 2/\u00b5\u01ebb\u03c3 = 0 as \u03b1 diverges \u2013 e\ufb00ectively all frequencies are \u201cstatic\u201d to a perfect conductor. This is how type I superconductors expel all \ufb01eld \ufb02ux. \u2192\u221e q If we examine the \ufb01elds in the vicinity of a boundary between a perfect conductor and a normal dielectric/diamagnetic material, we get: (D Dc) \u00b7 \u2212 \u02c6n = \u02c6n \u00b7 D = \u03a3 where Dc and Ec inside the", " conductor vanish. Similarly, (H \u02c6n \u00d7 \u2212 H c) = \u02c6n \u00d7 H = K (10.1) (10.2) (where in these expressions, \u03a3 is the surface charge density so we don\u2019t confuse it with the conductivity \u03c3, sigh, and similarly K is the surface current density). In addition to these two inhomogeneous equations that normal and parallel \ufb01elds at the surface to sources, we have the usual two homogeneous 87 equations: \u02c6n (B (E \u2212 \u2212 Bc) = 0 Ec) = 0 (10.3) (10.4) \u00b7 \u00d7 \u02c6n Note that these are pretty much precisely the boundary conditions for a static \ufb01eld and should come as no surprise. For perfect conductors, we expect the \ufb01elds inside to vanish, which in turn implies that E outside must be normal to the conducting surface and B outside must lie only parallel to the conducting surface, as usual. However, for materials that are not perfect conductors, the \ufb01elds don\u2019t vanish instantly \u201cat\u201d the mathematical surface. Instead they die o\ufb00 exponentially within a few multiples of the skin depth \u03b4. On scales large with respect to this, they will \u201clook\u201d like the static \ufb01eld conditions above, but of course within this cuto\ufb00 things are very di\ufb00erent. For one thing, Ohm\u2019s law tells us that we cannot have an actual \u201csurface layer of charge\u201d because for any \ufb01nite conductivity, the resistance scales like the cross-sectional area through which charge \ufb02ows. Consequently the real boundary condition on H precisely at the surface is: (H \u02c6n \u00d7 \u2212 H c) = 0 H = H c, || || (10.5) (10.6) || \u00d7 \u00d7 H) = ( \u02c6n \u02c6n. However, this creates a problem! If this \ufb01eld where H varies rapidly in some direction (and it does) it will generate an electric \ufb01eld according to Faraday\u2019s law! If the direction of greatest variation is \u201cinto the conductor\u201d (as the \ufb01eld is being screened by induced surface currents) then it will generate a small electric \ufb01eld parallel to the surface, one which is neglected (or rather, cannot occur) in the limit that the conductivity is in\ufb01nite. This electric \ufb01eld, in turn, generates a current, which causes the gradual cancellation of H as less and less the total bulk current is enclosed by a decending loop boundary. || If the conductivity is large but not in\ufb01nite, one way to \ufb01gure out what happens is to employ a series of successive approximations starting with the assumption of perfect conductivity and using it to generate a \ufb01rst order correction based on the actual conductivity and wavelength. The way it works is: 1. First, we assume that outside the conductor we have only E and from the statement of the boundary conditions assuming that the H \ufb01elds are instantly cancelled at the surface. \u22a5 || 2. Assume \u03b4 \u226a k\u2212 1 along the surface \u2013 the skin depth is much less than a wavelength and the \ufb01elds (whatever they may be) vanish across roughly this length scale, so we can neglect variation (derivatives) with respect to coordinates that lie along the surface compared to the coordinate perpendicular to the surface. 3. Use this approximation in Maxwell\u2019s Equations, along with the assumed boundary conditions for a perfect conductor, to derive relations between the \ufb01elds in the transition layer. 4. These relations determine the small corrections to the presumed bound- ary \ufb01elds both just outside and just inside the surface. The assumption of rapid variation only as one decends into the conductor is a key step, as we shall see. Thus (from 1): (H \u02c6n \u00d7 \u2212 H c) = 0 (10.7) (outside) = H or H because the result is boring if there are no \ufb01elds, right? (inside) = H || || || = 0, where the latter assumption is We both Ampere\u2019s law (assuming no displacement in the conductor to leading order) and Faraday\u2019s law to obtain relations for the harmonic \ufb01elds in terms of curls of each other: \u2207 \u2207 \u00d7 \u00d7 H c = \u03c3Ec = J Ec = \u2202Bc \u2202t \u2212 = i\u03c9\u00b5cH c become Ec = 1 \u03c3 \u2207 H c \u2207 Ec \u00d7", " \u00d7 1 \u00b5c\u03c9 H c = i \u2212 (10.8) (10.9) (10.10) (10.11) As we might expect, high frequencies create relatively large induced electric \ufb01elds as the magnetic \ufb01elds change, but high conductivity limits the size 6 of the supported electric \ufb01eld for any given magnetic \ufb01eld strength in a frequency independent way. Now we need to implement assumption 2 on the \u2207 operator. If we pick a coordinate \u03be to be perpendicular to the surface pointing into the conductor \u02c6n direction) and insist that only variations in this direction will be (in the signi\ufb01cant only on length scales of \u03b4: \u2212 \u2207 \u02c6n \u2248 \u2212 \u2202 \u2202\u03be then we get: Ec \u2248 \u2212 H c \u2248 i \u02c6n 1 \u03c3 1 \u00b5c\u03c9 \u2202H c \u2202\u03be! \u2202Ec \u2202\u03be! \u00d7 \u02c6n \u00d7 (10.12) (10.13) (Note well the deliberate use of approx to emphasize that there may well be components of the \ufb01elds in the normal direction or other couplings between the components in the surface, but those components do not vary particularly rapidly along the surface and so are not large contributors to the curl.) These two equations are very interesting. They show that while the magnitude of the \ufb01elds in the vicinity of the conducting surface may be large or small (depending on the charge and currents near the surface) the curls themselves are dominated by the particular components of Ec and H c that are in the plane perpendicular to \u02c6n (and each other) because the \ufb01eld strengths (whatever they are) are most rapidly varying across the surface. What this pair of equations ultimately does is show that if there is a magnetic \ufb01eld just inside the conductor parallel to its surface (and hence perpendicular to \u02c6n) H that rapidly varies as one descends, then there must be an electric \ufb01eld E that is its partner. Our zeroth approximation || boundary condition on H above shows that it is actually continuous across the mathematical surface of the boundary and does not have to be zero either just outside or just inside of it. However, in a good conductor the E \ufb01eld it produces is small. || || || This gives us a bit of an intuitive foundation for the manipulations of Maxwell\u2019s equations below. They should lead us to expressions for the cou- pled EM \ufb01elds parallel to the surface that self-consistently result from these two equations. We start by determining the component of H c (the total vector magnetic \ufb01eld just inside the conductor) in the direction perpendicular to the surface: H c = \u02c6n \u00b7 i \u00b5c\u03c9 \u02c6n \u00b7 ( \u02c6n \u00d7 \u2202Ec \u2202\u03be ) = 0 (10.14) \u02c6n \u2013 the magnetic \ufb01eld coupled This tells us that H c = H by Ec by Faraday\u2019s law lies in the plane of the conducting surface to lowest order. = ( \u02c6n H c) \u00d7 \u00d7 || Next we form a vector that lies perpendicular to both the normal and the magnetic \ufb01eld. We expect Ec to lie along this direction one way or the other. \u02c6n \u00d7 H c = \u02c6n 1 \u00b5c\u03c9 \u02c6n \u00d7 \u2202Ec \u2202\u03be! i \u00d7 \u2202 1 \u2202\u03be \u00b5c\u03c9 1 \u00b5c\u03c9 i \u2212 = i = \u2202Ec, || \u2202\u03be ( \u02c6n( \u02c6n Ec) \u00b7 \u2212 Ec) = \u02c6n( \u02c6n E) and Ec = Ec, \u22a5 (where Ec, ) and \ufb01nd that it does! The \u22a5 \u02c6n (+\u03be) direction picks fact that the electric \ufb01eld varies most rapidly in the out its component in the plane whatever it might be and relates it to the magnetic \ufb01eld direction also in the plane. + Ec, || \u2212 \u00b7 However, this does not show that the two conditions can lead to a selfsustaining solution in the absence of driving external currents (for example). To show that we have to substitute Ampere\u2019s law back into this: \u02c6n \u00d7 H c = \u02c6n H c = i i \u2212 1 \u00b5c\u03c9 1 \u00b5c\u03c9\u03c3 \u2202 \u2202\u03be \u2212 ( \u02c6n 1 \u03c3 \u22022Hc) \u2202\u03be2 \u00d7 ) ( \u02c6n \u00d7 \u2202Hc \u2202\u03be )! \u00d7 \u22022Hc) \u2202\u03be", "2 ( \u02c6n \u00d7 ) = i\u00b5c\u03c9\u03c3 \u02c6n \u2212 H c \u00d7 \u22022 \u2202\u03be2 ( \u02c6n \u00d7 Hc) + i\u00b5c\u03c9\u03c3 ( \u02c6n H c) = 0 \u00d7 or \u22022 \u2202\u03be2 ( \u02c6n \u00d7 H c) + 2i \u03b42 ( \u02c6n \u00d7 H c) = 0 (10.15) where we used the \ufb01rst result and substituted \u03b42 = 2/(\u00b5c\u03c9\u03c3). This is a well-known di\ufb00erential equation that can be written any of several ways. Let \u03ba2 = 2i \u03b42. It is equivalent to all of: ( \u22022 \u2202\u03be2 + \u03ba2)( \u02c6n \u00d7 H c) = 0 ( H c) \u02c6n = 0 \u22022 \u2202\u03be2 + \u03ba2)( \u02c6n \u00d7 \u00d7 \u22022 \u2202\u03be2 + \u03ba2)H \u22022 \u2202\u03be2 + \u03ba2)H c = 0 = 0 ( ( || Where: as noted above. The solution to this form is then: ( \u02c6n \u00d7 H c) \u00d7 \u02c6n = H || H c(\u03be) = H 0e\u00b1 \u03ba2\u03be \u221a \u2212 (10.16) (10.17) (10.18) (10.19) (10.20) (10.21) where H 0 is the magnetic \ufb01eld vector in the plane of the conductor at the surface and where this equation indicates how this value is attenuated as one decends into the conductor. As always, we have two linearly independent solutions. Either of them will work, and (given the already determined sign/branch associated with i\u03c9t) will ultimately have the physical interpretation the time dependence e\u2212 \u03be ( \u02c6n). of waves moving in the direction of +\u03be ( Let us pause for a moment to refresh our memory of taking the square root of complex numbers (use the subsection that treats this in the last chapter of these notes or visit Wikipedia of there is any problem understanding). \u02c6n) or in the direction of \u2212 \u2212 For this particular problem, \u221a 2i \u03b42 = (draw this out in pictures). We want the solution that propagates into the surface of the conductor, decending from the dielectric medium, which is the (10.22) 1 + i) \u03ba2 = ( \u2212 s\u2212 1 \u03b4 \u00b1 \u2212 positive branch: H c = H 0e\u221a \u2212 (consider ei\u03be/\u03b4 \u03c9t). \u2212 \u03ba2\u03be = H 0e = H 0e\u2212 1 \u03b4 ( \u03b4 ei \u03be \u2212 \u03be \u03b4 1+i)\u03be (10.23) Now we need to \ufb01nd an expression for Ec, which we do by backsubsti- tuting into Ampere\u2019s Law: \u02c6n \u2202H c \u2202\u03be! \u00d7 1 + i) ( \u02c6n ( \u2212 H 0) e 1 \u03b4 ( \u2212 1+i)\u03be \u00d7 Ec = = Ec = \u2212 \u2212 r 1 \u03c3 1 \u03b4\u03c3 \u00b5c\u03c9 2\u03c3 i)( \u02c6n (1 \u2212 \u00d7 H 0)e\u2212 \u03be \u03b4 ei \u03be \u03b4 (10.24) Note well the direction! Obviously \u02c6n Ec = 0, (in this approximation) so Ec must lie in the plane of the conductor surface, just like H! || \u00b7 As before (when we discussed \ufb01elds in a good conductor): \u2022 \u2022 \u2022 \u2022 \u2022 Ec, H c not in phase, but out of phase by \u03c0/4. Rapid decay as wave penetrates surface. Ec (\u03c3 \u201clarge\u201d, \u03b4 \u201csmall\u201d) so energy is primarily magnetic. H c \u226b \u02c6n \u2013 \ufb01elds are predominantly parallel to the sur\u02c6n face and mutually transverse, they propagate \u201cstraight into\u201d surface, attenuating rapidly as they go. H c \u22a5 Ec \u22a5 \u22a5 Recall: \u2212 at the surface. Since Ec lies approximately in the surface, this yields \u00d7 \u02c6n (E Ec) = 0 (10.25) E Ec \u2248 r \u2248 \u00b5c\u03c9 2\u03c3 (1 \u2212 i)( \u02c6n \u00d7 H 0)e\u2212 \u03be \u03b4 ei \u03be \u03b4 (10.26) just outside the surface \u2013 the \ufb01eld is approximately continuous! At this level of approximation, \u2207 E = i\u03c9B, E is parallel to the surface", ", and there is a small B to the surface of the same general order of magnitude as E. \u00d7 \u22a5 Since both E = 0 and H power \ufb02ow into the conductor! || 6 || 6 = 0 at the surface (\u03be = 0) there must be a dPin dA = 1 2 \u2212 Re (n (Ec \u00d7 \u00b7 H \u2217c)) = \u00b5c\u03c9\u03b4 4 | 2 H 0| (10.27) where we HOPE that it turns into heat. Let\u2019s see: J = \u03c3E = \u00b5c\u03c9\u03c3 2 (1 \u2212 i)( \u02c6n \u00d7 r H 0)e\u2212 \u03be(1 \u2212 i)/\u03b4 (10.28) so that the time averaged power loss is (from Ohm\u2019s Law): dP dV = 1 \u2206A dP d\u03be = 1 2 J \u00b7 E\u2217 = 1 2\u03c3 J \u00b7 J \u2217 \u2206P = \u2206A = \u2206A = \u2206A \u221e d\u03beJ J \u2217 0 1 2\u03c3 Z \u00b5c\u03c9 2 | \u00b5c\u03c9 4 | 2 2 H0| H0| \u00b7 \u221e 0 Z d\u03bee\u2212 2\u03be/\u03b4 (10.29) (10.30) which just happens to correspond to the \ufb02ux of the pointing vector through a surface \u2206A! Finally, we need to de\ufb01ne the \u201csurface current\u201d: K e\ufb00 = 0 Z \u221e J d\u03be = ( \u02c6n H) \u00d7 (10.31) where H is determined just outside(inside) of the surface of a \u201cperfect\u201d conductor in an idealized limit \u2013 note that we are just adding up the total current in the surface layer and that it all works out. Hopefully this exposition is complete enough (and correct enough) that any bobbles from lecture are smoothed out. You can see that although Jackson blithely pops all sorts of punch lines down in the text, the actual algebra of getting them, while straightforward, is not trivial! 10.2 Mutilated Maxwell\u2019s Equations (MMEs) We are now prepared to look at the propagation of waves in volumes of space bounded in some way by conducting surfaces. We\u2019ll generally assume that the conductors in question are \u201cperfect\u201d as far as boundary conditions on the dimensions of the volume in question are concerned. The place where this will lead to error is in the gradual attenuation of a propagating wave as it loses energy to the Joule heating of the surface of the bounding conductor, but this process will be slow relative to a wavelength and using the results of the previous section we can add this attenuation in by hand afterwards if necessary. Since we are going to have to solve boundary value problems for the wave equations for the coupled \ufb01eld components, we\u2019d better select a relatively simple geometry or we\u2019ll be here all semester. The two geometries we will examine are cylindrical waveguides where propagation is along the z axis of the cylinder and rectangular waveguides where the propagation is along the z axis of a waveguide with a rectangular cross-section in the x y plane of dimension a b. The transverse coordinates are therefore (\u03c1, \u03c6) or (x, y), respectively. \u00d7 \u2212 As usual, we will start by assuming that we\u2019re dealing with a harmonic i\u03c9t, write down Maxwell\u2019s equations in free wave with time dependence e\u2212 space (the cavity volume), turn them into wave equations for the \ufb01eld separately, note that the \ufb01elds are coupled by Maxwell\u2019s equations themselves, and impose boundary conditions. The only thing that is \u201cspecial\u201d about a cylinder is the form of the Laplacian and how we separate the laplacian to respect the boundary conditions. Let\u2019s skip ahead to the wave equation since by now everybody should be able to do this in their sleep: We look at propagation along z, making it \u201cplane-wave-like\u201d: 2 + \u00b5\u01eb\u03c92) ( \u2207 E or B { } = 0 (10.32) E(x, t) = E(\u03c1, \u03c6)e\u00b1 ikz i\u03c9t \u2212 B(x, t) = B(\u03c1, \u03c6)e\u00b1 ikz i\u03c9t \u2212 so that the wave equation becomes: 2 + (\u00b5\u01eb\u03c92 \u2212 k2) E or B { } = 0 \u2207 2 \u22a5 \u22022 \u2202z2 ). (Note that 2 = \u2207 Resolve \ufb01elds into components \u2207 \u2212 \u22a5 E = Ez \u02c6z + (\u02c6z and \u22a5 to z:", " E) \u02c6z \u00d7 || \u00d7 (10.33) (10.34) (10.35) = Ez + E \u22a5 B = Bz \u02c6z + (\u02c6z = Bz + B \u22a5 B) \u02c6z \u00d7 \u00d7 (10.36) (10.37) (10.38) (de\ufb01ning Ez and E in fairly obvious ways). Now we try to write Maxwell\u2019s equations in terms of these \ufb01eld components, assuming that the only z-dependence permitted is e\u00b1 etc. ikz. \u22a5 This isn\u2019t trivial to do \u2013 let\u2019s start with Faraday\u2019s law, for example: \u2207 \u00d7 E = \u2202B \u2202t \u2212 = i\u03c9B (10.39) If we project out the z component of both sides we get: \u02c6z \u00b7 ( \u2202Ez \u2202y \u2212 \u2202Ey \u2202z! \u02c6x + E) = i\u03c9Bz \u00b7 \u02c6z \u2202Ex \u2202z \u2212 (\u2207 \u00d7 \u2202Ez \u2202x! \u02c6y + \u02c6z ) = i\u03c9Bz = i\u03c9Bz \u2202Ex \u2202y! \u2202Ey \u2202x \u2212 \u2202Ey \u2202x \u2212 (\u2207 \u02c6z \u00b7 \u22a5 \u00d7 \u2202Ex \u2202y! E \u22a5 ) = i\u03c9Bz (10.40) as only the larly: \u22a5 components of the curl contribute to the z direction. Simi- \u02c6z \u00d7 ( \u2202Ez \u2202y \u2212 \u2202Ey \u2202z! \u02c6x + \u2202Ez \u2202y \u2212 \u2202Ey \u2202z! \u02c6y \u2212 \u2202E \u22a5 \u2202z (where \u02c6z B = \u02c6z B \u22a5 \u00d7 \u00d7, of course). \u00d7 \u02c6z \u2202Ex \u2202z \u2212 (\u2207 \u00d7 \u2202Ez \u2202x! \u02c6z \u2202Ey \u2202x \u2212 \u2202Ex \u2202z \u2212 \u2202Ex \u2202y! \u2202Ez \u2202x! E) = i\u03c9(\u02c6z B) \u00d7 \u02c6y + = i\u03c9(\u02c6z ) \u02c6x = i\u03c9(\u02c6z B \u22a5 \u00d7 B \u22a5 \u00d7 ) ) + i\u03c9(\u02c6z B \u22a5 \u00d7 ) = \u2207 Ez \u22a5 (10.41) Ouch! Looks like working through the curl termwise is a certain amount of pain! However, now that we\u2019ve done it once (and see how it goes) Ampere\u2019s law should be straightforward: \u02c6z and \u00b7 (\u2207 \u02c6z \u00b7 (\u2207 \u00d7 \u22a5 \u00d7 H) = B ) = \u22a5 i\u03c9Dz i\u03c9\u00b5\u01ebEz \u2212 \u2212 (\u2207 \u02c6z \u00d7 i\u03c9\u00b5\u01eb(\u02c6z H) = i\u03c9(\u02c6z \u00d7 E \u22a5 \u00d7 \u2212 ) = \u2207 Bz \u22a5 \u2202B \u22a5 \u2202z \u2212 D) \u00d7 Finally, we have Gauss\u2019s Law(s): \u2207 Ez \u2202z = 0 and identically, \u2207 E \u22a5 \u22a5 \u00b7 = \u2202Ez \u2202z \u2212 \u2207 B \u22a5 = \u22a5 \u00b7 \u2202Bz \u2202z \u2212 Let\u2019s collect all of these in just one place nowEz \u2202z \u2202Bz \u2202z i\u03c9\u00b5\u01ebEz \u2212 \u2212 \u2212 \u02c6z \u02c6z \u00b7 (\u2207 (\u2207 \u22a5 \u00d7 \u22a5 \u00d7 \u00b7 i\u03c9\u00b5\u01eb(\u02c6z + i\u03c9(\u02c6z B \u22a5 \u2202z \u2212 \u2202E \u22a5 \u2202z ) = i\u03c9Bz ) = \u2207 ) = \u2207 Bz Ez \u22a5 \u22a5 (10.42) (10.43) (10.44) (10.45) (10.46) (10.47) (10.48) Gee, only a few pages of algebra to obtain in a shortened way what Jackson just puts down in three short lines. Hopefully the point is clear \u2013 to \u201cget\u201d a lot of this you have to sooner or later work it all out, however long it may take you, or you\u2019ll end up memorizing (or trying to) all of Jackson\u2019s results. Something that most normal humans could never do in a lifetime of trying... Back to work, as there is still plenty to do. 10.", "3 TEM Waves Now we can start looking at waveforms in various cavities. Suppose we let Ez = Bz = 0. Then the wave in the cavity is a pure transverse electromagnetic (TEM) wave just like a plane wave, except that it has to satisfy the boundary conditions of a perfect conductor at the cavity boundary! Note from the equations above thatE \u22a5 = 0 = 0 from which we can immediately see that: and that E 2 \u2207 \u22a5 \u22a5 = 0 (10.49) \u2212 for some suitable potential that satis\ufb01es a propagating electrostatic wave. From the wave equation we see that: \u03c6 = 0. The solution looks like \u2207 \u22a5 \u22a5 2 E = \u2207\u03c6 (10.50) or \u00b5\u01eb\u03c92 = k2 (10.51) \u00b1 which is just like a plane wave (which can propagate in either direction, recall). k = \u03c9\u221a\u00b5\u01eb (10.52) Again referring to our list of mutilated Maxwell equations above, we see that: B ikE \u02c6z i\u03c9(\u02c6z \u03c9\u00b5\u01eb k \u221a\u00b5\u01eb(\u02c6z \u2212 \u2212 \u00b1 \u00d7 \u00d7 ) \u22a5 H H ) ) \u22a5 \u22a5 (10.53) or working the other way, that: B \u22a5 = \u221a\u00b5\u01eb(\u02c6z \u00b1 E ) \u22a5 \u00d7 (10.54) so we can easily \ufb01nd one from the other. TEM waves cannot be sustained in a cylinder because the surrounding (perfect, recall) conductor is equipotential. Therefore E. \u22a5 However, they are the dominant way energy is transmitted down a coaxial cable, where a potential di\ufb00erence is maintained between the central conductor and the coaxial sheathe. In this case the \ufb01elds are very simple, as the E is purely radial and the B \ufb01eld circles the conductor (so the energy goes which way?) with no z components. is zero as is B \u22a5 Finally, note that all frequencies are permitted for a TEM wave. It is not \u201cquantized\u201d by the appearance of eigenvalues due to a constraining boundary value problem. 10.4 TE and TM Waves Note well that we have written the mutilated Maxwell Equations so that the z components are all on the right hand side. If they are known functions, and if the only z dependence is the complex exponential (so we can do all the z-derivatives and just bring down a ik) then the transverse components E are determined! and B \u00b1 \u22a5 \u22a5 In fact (for propagation in the +z direction, e+ikz i\u03c9t): \u2212 ik(\u02c6z E \u22a5 \u00d7 ikE + i\u03c9(\u02c6z \u22a5 ) + i\u03c9\u02c6z \u00d7 \u00d7 \u00d7 B B E B \u22a5 \u22a5 \u22a5 \u22a5 (\u02c6z \u00d7 ik(\u02c6z (\u2207 \u02c6z \u00b7 \u22a5 \u00d7 ) = \u22a5 i\u03c9\u00b5\u01ebEz \u2212 ) = \u2207 ) = \u02c6z \u22a5 Ez \u2207 \u00d7 ) = i\u03c9B Ez \u22a5 + \u02c6z \u2207 Ez \u22a5 \u00d7 (10.55) (10.56) and ikB i\u03c9\u00b5\u01eb(\u02c6z \u22a5 \u2212 ikB E ) = \u2207 Bz \u22a5 Bz = i\u03c9\u00b5\u01eb(\u02c6z \u22a5 \u00d7 \u2207 i k2 \u03c9\u00b5\u01eb B \u22a5 \u2212 \u22a5 \u2212 k \u03c9\u00b5\u01eb \u2207 \u22a5 \u22a5 Bz = ik(\u02c6z \u00d7 E ) \u22a5 \u00d7 E ) \u22a5 i k2 \u03c9\u00b5\u01eb B \u22a5 \u2212 k \u03c9\u00b5\u01eb \u2207 \u22a5 Bz = i\u03c9B + \u02c6z \u22a5 \u2207 Ez \u22a5 \u00d7 or B E \u22a5 \u22a5 = = i \u00b5\u01eb\u03c92 \u2212 i \u00b5\u01eb\u03c92 \u2212 k2 (k\u2207 k2 (k\u2207 Bz + \u00b5\u01eb\u03c9(\u02c6z \u22a5 Ez \u2212 \u22a5 \u03c9(\u02c6z \u00d7 \u00d7 \u2207 \u2207 \u22a5 Ez)) Bz)) \u22a5 (10.57) (10.58) (10.59) (where we started with the second equation and eliminated \u02c6z the second equation just like the \ufb01rst). B \u22a5 \u00d7 to get Now comes the relatively tricky part. Recall the boundary conditions for a perfect conductor", ": \u02c6n \u00d7 \u02c6n (E \u2212 (B Ec) = \u02c6n \u00d7 Bc) = \u02c6n \u00b7 \u2212 \u00b7 \u02c6n \u00d7 \u02c6n \u00b7 They tell us basically that E (D) is strictly perpendicular to the surface and that B (H) is strictly parallel to the surface of the conductor at the surface of the conductor. This means that it is not necessary for Ez or Bz both to vanish everywhere inside the dielectric (although both can, of course, and result in a TEM wave or no wave at all). All that is strictly required by the boundary conditions is for Ez|S = 0 on the conducting surface S (it can only have a normal component so the z component must vanish). The condition on Bz is even weaker. It must lie parallel to the surface and be continuous across the surface (where H can discontinuously change because of K). That is: (10.60) \u2202Bz \u2202n |S = 0 (10.61) We therefore have two possibilities for non-zero Ez or Bz that can act as source term in the mutilated Maxwell Equations. 10.4.1 TM Waves Bz = 0 Ez|S = 0 The magnetic \ufb01eld is strictly transverse, but the electric \ufb01eld in the z direction only has to vanish at the boundary \u2013 elsewhere it can have a z component. (10.62) (10.63) Thus: E (\u00b5\u01eb\u03c92 (\u00b5\u01eb\u03c92 k2)E k2)E \u2212 \u2212 1 ik \u22a5 \u22a5 \u22a5 = i \u00b5\u01eb\u03c92 = ik\u2207 k2 (k\u2207 \u2212 Ez \u22a5 Ez = \u2207 \u22a5 Ez \u2212 \u22a5 \u03c9(\u02c6z \u00d7 \u2207 \u22a5 Bz)) which looks just perfect to substitute into: Bz + \u00b5\u01eb\u03c9(\u02c6z = i \u00b5\u01eb\u03c92 \u2212 = i\u00b5\u01eb\u03c9(\u02c6z \u00b5\u01eb\u03c9 k = \u00d7 (\u00b5\u01eb\u03c92 k2 (k\u2207 \u22a5 \u2207 Ez) \u22a5 k2)(\u02c6z \u2212 E ) \u22a5 \u00d7 B (\u00b5\u01eb\u03c92 (\u00b5\u01eb\u03c92 k2)B k2)B \u2212 \u2212 \u22a5 \u22a5 \u22a5 giving us: \u22a5 or (as the book would have it): B = \u00b5\u01eb\u03c9 k \u00b1 (\u02c6z E ) \u22a5 \u00d7 H \u22a5 = \u01eb\u03c9 k \u00b1 (\u02c6z E ) \u22a5 \u00d7 \u2207 \u22a5 Ez)) \u00d7 (10.64) (10.65) (10.66) (10.67) (where as usual the two signs indicate the direction of wave propagation). Of course, we still have to \ufb01nd at least one of the two \ufb01elds for this to do us any good. Or do we? Looking above we see: (\u00b5\u01eb\u03c92 k2)E \u2212 E \u22a5 \u22a5 = ik\u2207 \u22a5 \u00b1 (\u00b5\u01eb\u03c92 \u03c8 ik \u2212 = \u2207 \u03c8 \u22a5 k2) (10.68) Where \u03c8(x, y)eikz = Ez. This must satisfy the transverse wave function: k2) and the boundary conditions for a TM wave: + (\u00b5\u01eb\u03c92 \u2207 \u2212 \u22a5 2 \u03c8 = 0 (10.69) \u03c8 |S = 0 (10.70) TE Waves Ez = 0 (10.71) \u2202Bz \u2202n |S = 0 The electric \ufb01eld is strictly transverse, but the magnetic \ufb01eld in the zdirection can be nonzero. Doing exactly the same algebra on the same two equations as we used in the TM case, we get instead: (10.72) H \u22a5 = k \u00b5\u03c9 \u00b1 (\u02c6z E ) \u22a5 \u00d7 along with B \u22a5 = where \u03c8(x, y)eikz = Bz and \u00b1 (\u00b5\u01eb\u03c92 ik \u2212 \u2207 \u03c8 \u22a5 k2) (10.73) (10.74) k2) and the boundary conditions for a TE wave: + (\u00b5\u01eb\u03c92 \u2207 \u2212 \u22a5 2 \u03c8 = 0 (10.75) \u2202\u03c8 \u2202n |S = 0 (10.76) 10.4.2 Summary of TE/TM waves The transverse wave equation and boundary condition (dirichlet or neumann) are an eigenvalue problem. We can see two things right away. First of all: \u00b5", "\u01eb\u03c92 k2 \u2265 (10.77) or we no longer have a wave, we have an exponential function that cannot be made to satisfy the boundary conditions on the entire surface. Alternatively, v2 p = \u03c92 k2 \u2265 1 \u00b5\u01eb = v2 (10.78) which has the lovely property (as a phase velocity) of being faster than the speed of light in the medium! To proceed further in our understanding, we need to look at an actual example \u2013 we\u2019ll \ufb01nd that only certain kn = k0n for n = 1, 2, 3...ncuto\ufb00 will permit the boundary conditions to be solved, and we\u2019ll learn some important things about the propagating solutions at the same time. 10.5 Rectangular Waveguides Rectangular waveguides are important for two reasons. First of all, the Laplacian operator separates nicely in Cartesian coordinates, so that the boundary value problem that must be solved is both familiar and straightforward. Second, they are extremely common in actual application in physics laboratories for piping e.g. microwaves around as experimental probes. In Cartesian coordinates, the wave equation becomes: \u22022 \u2202x2 + \u22022 \u2202y2 + (\u00b5\u01eb\u03c92 k2)! \u2212 \u03c8 = 0 (10.79) This wave equation separates and solutions are products of sin, cos or exponential functions in each variable separately. To determine which combination to use it su\ufb03ces to look at the BC\u2019s being satis\ufb01ed. For TM waves, one solves for \u03c8 = Ez subject to Ez|S = 0, which is automatically true if: Ez(x, y) = \u03c8mn(x, y) = E0 sin m\u03c0x a sin n\u03c0y b (10.80) where a and b are the dimensions of the x and y sides of the boundary rectangle and where in principle m, n = 0, 1, 2.... However, the wavenumber of any given mode (given the frequency) is determined from: k2 = \u00b5\u01eb\u03c92 \u03c02 \u2212 m2 a2 + n2 b2! + (10.81) where k2 > 0 for a \u201cwave\u201d to exist to propagate at all. If either index m or n is zero, there is no wave, so the \ufb01rst mode that can propagate has a dispersion relation of: 11 = \u00b5\u01eb\u03c92 k2 \u03c02( 1 a2 + 1 b2 ) \u2212 (10.82) so that: \u03c9 \u03c0 \u221a\u00b5\u01eb s 1 a2 + Each combination of permitted m and n is associated with a cuto\ufb00 of this sort \u2013 waves with frequencies greater than or equal to the cuto\ufb00 can support propogation in all the modes with lower cuto\ufb00 frequencies. 1 b2 = \u03c9c,TM(11) (10.83) \u2265 If we repeat the argument above for TE waves (as is done in Jackson, which is why I did TM here so you could see them both) you will be led by nearly identical arguments to the conclusion that the lowest frequency mode cuto\ufb00 occurs for a > b, m = 1 and n = 0 to produce the Hz(x, y) = \u03c8(x, y) solution to the wave equation above. The cuto\ufb00 in this case is: \u03c9 \u2265 \u03c0 \u221a\u00b5\u01eb 1 a = \u03c9c,TE(10) < \u03c9c,TM(11) (10.84) There exists, therefore, a range of frequencies in between where only one TE mode is supported with dispersion: k2 = k2 10 = \u00b5\u01eb\u03c92 \u03c02 a2. \u2212 (10.85) Note well that this mode and cuto\ufb00 corresponds to exactly one-half a free-space wavelength across the long dimension of the waveguide. The wave solution for the right-propagating TE mode is: Hx = Ey = Hz = H0 cos eikz i\u03c9t \u2212 (10.86) \u03c0x a \u2202Hz \u2202x i\u00b5\u03c9a \u03c0 \u2212 = k2 ik \u00b5\u01eb\u03c92 \u00b5\u03c9 k k2 = \u03c02/a2 and E \u2212 Hx = H0 sin \u03c0x a i\u03c9t ika \u03c0 H0 sin eikz \u03c0x a = ik/\u03b32\u2207 \u2212 \u22a5 )) to get the last one. eikz \u2212 i\u03c9t (10.87) (10.88) \u03c8 to get the second of \u22a5 We used \u03b32 = \u00b5\u01eb\u03c92 = k these, and H \u2212 \u03c9\u00b5(\ufffd", "\ufffdz \u22a5 E \u22a5 \u00d7 There is a lot more one can study in Jackson associated with waveguides, but we must move on at this time to a brief look at resonant cavities (another important topic) and multipoles. 10.6 Resonant Cavities We will consider a resonant cavity to be a waveguide of length d with caps at both ends. As before, we must satisfy TE or TM boundary conditions on the cap surfaces, either Dirichlet in Ez or Neumann in Bz. In between, we expect to \ufb01nd harmonic standing waves instead of travelling waves. Elementary arguments for presumed standing wave z-dependence of: A sin kz + B cos kz (10.89) such that the solution has nodes or antinodes at both ends lead one to conclude that only: for p = 0, 1, 2... are supported by the cavity. For TM modes E must vanish on the caps because the nonzero Ez \ufb01eld must be the only E \ufb01eld component sustained, hence: \u22a5 k = p \u03c0 d Ez = \u03c8(x, y) cos p\u03c0z d (10.90) (10.91) For TE modes Hz must vanish as the only permitted \ufb01eld component is a non-zero H, hence: \u22a5 Hz = \u03c8(x, y) sin p\u03c0z d (10.92) Given these forms and the relations already derived for e.g. a rectangular cavity, one can easily \ufb01nd the formulae for the permitted transverse \ufb01elds, e.g.: E \u22a5 H \u22a5 for TM \ufb01elds and E H \u22a5 \u22a5 = = = = \u2212 p\u03c0 d(\u00b5\u01eb\u03c92 i\u01eb\u03c9 \u2212 \u00b5\u01eb\u03c92 \u2212 i\u00b5\u03c9 \u2212 \u00b5\u01eb\u03c92 p\u03c0 d(\u00b5\u01eb\u03c92 \u2212 \u2212 sin p\u03c0z d (\u02c6z p\u03c0z d k2) \u2212 k2 cos k2 sin cos k2) (\u02c6z p\u03c0z d p\u03c0z d \u2207 \u03c8 \u22a5 \u2207 \u03c8 \u22a5 \u2207 \u03c8) \u22a5 \u00d7 \u2207 \u03c8) \u22a5 \u00d7 for TE \ufb01elds, with \u03c8(x, y) determined as before for cavities. (10.93) (10.94) (10.95) (10.96) However, now k is doubly determined as a function of both p and d and as a function of m and n. The only frequencies that lead to acceptable solutions are ones where the two match, where the resonant k in the z direction corresponds to a permitted k(\u03c9) associated with a waveguide mode. I leave you to read about the de\ufb01nition of Q: Q = \u03c90 \u2206\u03c9 (10.97) or the fractional energy loss per cycle of the cavity oscillator in the limit where this quantity is small compared to the total energy. Note that \u2206\u03c9 is the full width at half maximum of the presumed resonant form (basically the same as was presumed in our discussions of dispersion, but for energy instead of \ufb01eld). I strongly advise that you go over this on your own \u2013 Q describes the damping of energy stored in a cavity mode due to e.g. the \ufb01nite conductivity of the walls or the partial transparency of the end caps to energy (as might exist in the case of a laser cavity). If you go into laser physics, you will very much need this. If not, you\u2019ll need to understand the general idea of Q to teach introductory physics and e.g. LRC circuits or damped driven harmonic oscillators, where it also occurs and should know it at least qualitatively for I added an optional problem for resonant cavities to the e.g. quali\ufb01ers. homework assignment in case you wanted something speci\ufb01c to work on while studying this. 10.7 Wave Guides Assignment Jackson 8.2,8.4(,8.6 optional) Chapter 11 Radiation Well, now we have learned a little about how to describe waves propagating through \u201cfree\u201d space \u2013 dielectric media, possibly bounded by a conducting surface. But how did they get there? Well, sit yourselves down and I\u2019ll tell you. They were radiated there by accelerating, time dependent charge\u2013 current distributions! And now we\u2019ll learn how... Note well! This treatment di\ufb00ers substantially from Jackson\u2019s, which actually kinda sucks. Ultimately it will be much simpler to understand and is consistently developed. However, it really is the same thing and one gets the same general expressions for the multipole", " \ufb01elds or potentials. 11.1 Maxwell\u2019s Equations, Yet Again Suppose we are given a system of classical charges that oscillate harmonically with time. Note that, as before, this can be viewed as the special case of the Fourier transform at a particular frequency of a general time dependent distribution; however, this is a very involved issue that we will examine in detail later in the semester. The form of the charge distribution we will study for the next few weeks is: \u03c1(x, t) = \u03c1(x)e\u2212 i\u03c9t (11.1) 107 J (x, t) = J (x)e\u2212 i\u03c9t. (11.2) The spatial distribution is essentially \u201carbitrary\u201d. Actually, we want it to have compact support which just means that it doesn\u2019t extend to in\ufb01nity in any direction. Later we will also want it to be small with respect to a wavelength. 11.1.1 Quickie Review of Chapter 6 Recall the following morphs of Maxwell\u2019s equations, this time with the sources and expressed in terms of potentials by means of the homogeneous equations. Gauss\u2019s Law for magnetism is: \u2207 \u00b7 B = 0 This is an identity if we de\ufb01ne B = \u2207 (\u2207 \u2207 \u00b7 \u00d7 Similarly, Faraday\u2019s Law is A: \u00d7 A) = 0 \u2207 \u00d7 E + E + \u2202\u2207 = 0 = 0 (E + ) = 0 \u2202B \u2202t A \u00d7 \u2202t \u2202A \u2202t \u2207 \u00d7 \u2207 \u00d7 and is satis\ufb01ed as an identity by a scalar potential such that: E + \u2202A \u2202t = E = \u2207\u03c6 \u2212 \u2207\u03c6 \u2212 \u2202A \u2202t \u2212 (11.3) (11.4) (11.5) (11.6) (11.7) (11.8) (11.9) Now we look at the inhomogeneous equations in terms of the potentials. Ampere\u2019s Law: \u2207 \u00d7 B = \u00b5(J + \u01eb \u2207 \u00d7 (\u2207 \u00d7 A) = \u00b5(J + \u01eb \u2202E \u2202t \u2202E \u2202t ) ) (11.10) (11.11) \u2207(\u2207 \u2207(\u2207 A) A) \u00b7 \u00b7 \u2212 \u2207 \u2212 \u2207 2A = \u00b5J + \u00b5\u01eb 2A = \u00b5J \u2202E \u2202t \u00b5\u01eb\u2207\u2202\u03c6 \u2212 \u2202t \u2212 \u00b5\u01eb \u22022A \u2202t2 2A \u2207 \u00b5\u01eb \u2212 \u22022A \u2202t2 = \u00b5J + \u2207(\u2207 \u2212 A + \u00b5\u01eb \u00b7 \u2202\u03c6 \u2202t ) Similarly Gauss\u2019s Law for the electric \ufb01eld becomes: \u2207 E = \u00b7 \u2202A \u2202t! A \u2212 \u2202\u2207 \u00b7 \u2202\u03c6 \u00b7 \u2212 2\u03c6 + \u2207 In the the Lorentz gauge, \u2207 \u00b7 A + \u00b5\u01eb \u2202\u03a6 \u2202t = 0 the potentials satisfy the following inhomogeneous wave equations: 2\u03a6 \u2207 \u2212 \u00b5\u01eb 2A \u2207 \u2212 \u00b5\u01eb \u22022\u03a6 \u2202t2 = \u22022A \u2202t2 = \u03c1 \u01eb \u2212 \u00b5J \u2212 (11.12) (11.13) (11.14) (11.15) (11.16) (11.17) (11.18) (11.19) (11.20) where \u03c1 and J are the charge density and current density distributions, respectively. For the time being we will stick with the Lorentz gauge, although the Coulomb gauge: \u2207 \u00b7 A = 0 (11.21) It is probably worth reminding is more convenient for certain problems. y\u2019all that the Lorentz gauge condition itself is really just one out of a whole family of choices. Recall that (or more properly, observe that in its role in these wave equations) \u00b5\u01eb = 1 v2 (11.22) where v is the speed of light in the medium. For the time being, let\u2019s just simplify life a bit and agree to work in a vacuum: \u00b50\u01eb0 = 1 c2 (11.23) so that: \u2207 2\u03a6 \u22022\u03a6 \u2202t2 = \u22022A \u2202t2 = If/when we look at wave sources embedded in a dielectric medium, we can always change back as the general formalism will not be any di\ufb00erent. 1 c2 1 c2 (11.25", ") (11.24) \u03c1 \u01eb0 \u00b50J 2A \u2207 \u2212 \u2212 \u2212 \u2212 11.2 Green\u2019s Functions for the Wave Equa- tion As by now you should fully understand from working with the Poisson equation, one very general way to solve inhomogeneous partial di\ufb00erential equations (PDEs) is to build a Green\u2019s function1 and write the solution as an integral equation. Let\u2019s very quickly review the general concept (for a further discussion don\u2019t forget WIYF,MWIYF). Suppose is a general (second order) linear partial di\ufb00erential operator on e.g. IR3 and one wishes to solve the inhomogeneous equation: D for f. f (x) = \u03c1(x) D (11.26) If one can \ufb01nd a solution G(x for a point source function2 : \u2212 x0) to the associated di\ufb00erential equation G(x, x0) = \u03b4(x x0) (11.27) D \u2212 1Note that this expression stands for: \u201cThe generalized point source potential/\ufb01eld developed by Green.\u201d A number of people criticize the various ways of referring to it \u2013 Green function (what color was that again? what shade of Green?), Greens function (a function made of lettuce and spinach and kale?), \u201ca\u201d Green\u2019s function (a singular representative of a plural class referenced as a singular object). All have problems. I tend to go with the latter of these as it seems least odd to me. 2 Note well that both the Green\u2019s \u201cfunction\u201d and the associated Dirac delta \u201cfunction\u201d are not functions \u2013 they are de\ufb01ned in terms of limits of a distribution in such a way that the interchange of limits and values of the integrals above make sense. This is necessary as both of the objects are singular in the limit and hence are meaningless without the limiting process. However, we\u2019ll get into real trouble if we have to write \u201cThe limit of the distribution de\ufb01ned by Green that is the solution of an inhomogeneous PDE with a then (subject to various conditions, such as the ability to interchange the di\ufb00erential operator and the integration) to solution to this problem is a Fredholm Integral Equation (a convolution of the Green\u2019s function with the source terms): f (x) = \u03c7(x) + G(x, x0)\u03c1(x0)d3x0 (11.28) ZIR3 where \u03c7(x) is an arbitrary solution to the associated homogeneous PDE: [\u03c7(x)] = 0 D This solution can easily be veri\ufb01ed: f (x) = \u03c7(x) + ZIR3 [\u03c7(x)] + G(x, x0)\u03c1(x0)d3x0 G(x, x0)\u03c1(x0)d3x0 D ZIR3 f (x) = D \u03c1(x0)d3x0 D f (x) = 0 + f (x) = 0 + f (x) = \u03c1(x) G(x, x0)\u03c1(x0)d3x0 ZIR3 D \u03b4(x \u2212 x0)\u03c1(x0)d3x0 ZIR3 D D D (11.29) (11.30) (11.31) (11.32) (11.33) (11.34) (11.35) It seems, therefore, that we should thoroughly understand the ways of building Green\u2019s functions in general for various important PDEs. I\u2019m uncertain of how much of this to do within these notes, however. This isn\u2019t really \u201cElectrodynamics\u201d, it is mathematical physics, one of the fundamental toolsets you need to do Electrodynamics, quantum mechanics, classical mechanics, and more. So check out Arfken, Wyld, WIYF, MWIYFand we\u2019ll content ourselves with a very quick review of the principle ones we need: 11.2.1 Poisson Equation The Green\u2019s function for the Poisson (inhomogeneous Laplace) equation: 2\u03c6 = \u2207 \u03c1 \u01eb0 \u2212 (11.36) source distribution that in the same limit approaches a unit source supported at a single point\u201d instead of just \u201cGreen\u2019s function\u201d. So we won\u2019t. is the solution to: 2G(x, x0) = \u03b4(x x0) \u2212 \u2207 (11.37) Thus G(x, x0) satis\ufffd", "\ufffdes the homogeneous Laplace PDE everywhere but at the single point x0. The solution to the Laplace equation that has the right degree of singularity is the \u201cpotential of a unit point charge\u201d: G(x, x0) = 1 \u2212 x | \u2212 4\u03c0 x0| located at x0. Hence: \u03c6(x) = \u03c70(x) + 1 4\u03c0\u01eb0 ZV \u03c1(x0) x0| x | \u2212 d3x0 which is just exactly correct. (11.38) (11.39) Note well that the inhomogeneous term \u03c70(x) solves the homogeneous It can be viewed as a Laplace equation and has various interpretations. \u201cboundary term\u201d (surface integral on S = \u2202V, the surface S bounding the volume V (Green\u2019s Theorem) or, as we shall see, as the potential of all the charges in the volume exterior to V, or as a gauge transformation of the potential. All are true, but the \u201cbest\u201d way to view it is as the potential of exterior charges as that is what it is in nature even when it is expressed, via integration by parts, as a surface integral, for a very sensible choice of asymptotic behavior of the potential. Note equally well that the Green\u2019s function itself has precisely the same gauge freedom, and can be written in its most general form as: 1 \u2212 x | 4\u03c0 x0| G(x, x0) = F (x, x0) + (11.40) \u2207 2F (x, x0) = \u2212 2 0F (x, x0) = 0 is any bilinear (symmetric in both where coordinates) solution to the Laplace equation! However, we will not proceed this way in this part of the course as it is in a sense unphysical to express the PDEs this way even though it does upon occasion facilitate the solution algebraically. \u2207 11.2.2 Green\u2019s Function for the Helmholtz Equation If we fourier transform the wave equation, or alternatively attempt to \ufb01nd i\u03c9t, we convert it into solutions with a speci\ufb01ed harmonic behavior in time e\u2212 the following spatial form: 2 + k2 \u03c6(x) = \u2207 \u03c1\u03c9 \u01eb0 \u2212 (11.41) (for example, from the wave equation above, where \u03c1(x, t) = \u03c1\u03c9(x)e\u2212 \u03c6(x, t) = \u03c6\u03c9(x)e\u2212 inhomogeneous Helmholtz equation (IHE). i\u03c9t, i\u03c9t, and k2c2 = \u03c92 by assumption). This is called the The Green\u2019s function therefore has to solve the PDE: 2 + k2 G(x, x0) = \u03b4(x \u2207 x0) \u2212 (11.42) \u2192 Once again, the Green\u2019s function satis\ufb01es the homogeneous Helmholtz equation (HHE). Furthermore, clearly the Poisson equation is the k 0 limit of the Helmholtz equation. It is straightforward to show that there are several functions that are good candidates for G. They are: x0| ) \u2212 x0| \u2212 x0 \u2212 x0| x0 x0| x cos(k | x 4\u03c0 | x e+ik | G+(x, x0) = \u2212 x 4\u03c0 | ik e\u2212 (x, x0) = \u2212 x 4\u03c0 | G0(x, x0) = \u2212 (11.43) (11.44) (11.45) \u2212 x | \u2212 G \u2212 \u2212 | | As before, one can add arbitrary bilinear solutions to the HHE, ( 2 + 0 + k2)F (x, x0) = 0 to any of these and the result is 2 In fact, these forms are related by this sort of k2)F (x, x0) = ( still a Green\u2019s function. transformation and superposition: \u2207 \u2207 G0(x, x0) = 1 2 (G+(x, x0) + G (x, x0)) \u2212 G+(x, x0) = F (x, x0) + G0(x, x0) = \u2212 i sin(k x 4\u03c0 | x | \u2212 x0| ) \u2212 x0| + G0(x, x0) or etc. In terms of any of these: \u03c6(x) = \u03c70(x) \u2212 = \u03c70(x) + 1 \u01eb0 ZV 1 4\u03c0\u01eb0 ZV \u03c1(x0)G(x, x0)d3x", "0 x0 | \u03c1(x0)eik x | \u2212 \u2212 x | x0| d3x0 (11.46) (11.47) (11.48) (11.49) (11.50) where ( \u2207 2 + k2)\u03c70(x) = 0 as usual. We name these three basic Green\u2019s functions according to their asymptotic time dependence far away from the volume V. In this region we expect to see a time dependence emerge from the integral of e.g. \u03c6(x, t) \u223c eikr \u2212 i\u03c9t (11.51) x where r =. This is an outgoing spherical wave. Consequently the Green\u2019s | | functions above are usually called the stationary wave, outgoing wave and incoming wave Green\u2019s functions. It is essential to note, however, that any solution to the IHE can be constructed from any of these Green\u2019s functions! This is because the form of the solutions always di\ufb00er by a homogeneous solution (as do the Green\u2019s functions) themselves. The main reason to use one or the other is to keep the form of the solution simple and intuitive! For example, if we are looking for a \u03c6(x, t) that is supposed to describe the radiation of an electromagnetic \ufb01eld from a source, we are likely to use an outgoing wave Green\u2019s function where if we are trying to describe the absorption of an electromagnetic \ufb01eld by a source, we are likely to use the incoming wave Green\u2019s function, while if we are looking for stationary (standing) waves in some sort of large spherical cavity coupled to a source near the middle then (you guessed it) the stationary wave Green\u2019s function is just perfect. [As a parenthetical aside, you will often see people get carried away in the literature and connect the outgoing wave Green\u2019s function for the IHE to the retarded Green\u2019s function for the Wave Equation (fairly done \u2013 they are related by a contour integral as we shall see momentarily) and argue for a causal interpretation of the related integral equation solutions. However, as you can clearly see above, not only is there no breaking of time symmetry, the resulting descriptions are all just di\ufb00erent ways of viewing the same solution! This isn\u2019t completely a surprise \u2013 the process of taking the Fourier transform symmetrically samples all of the past and all of the future when doing the time integral. As we will see when discussing radiation reaction and causality at the very end of the semester, if anything one gets into trouble when one assumes that it is always correct to use an outgoing wave or retarded Green\u2019s function, as the actual \ufb01eld at any point in space at any point in time is time reversal invariant in classical electrodynamics \u2013 absorption and emission are mirror processes and both are simultaneously occurring when a charged particle is being accelerated by an electromagnetic \ufb01eld.] 11.2.3 Green\u2019s Function for the Wave Equation This time we are interested in solving the inhomogeneous wave equation (IWE) 2 \u2207 \u2212 1 c2 \u22022 \u2202t2! \u03c6(x, t) = \u03c1(x, t) \u01eb0 \u2212 (11.52) (for example) directly, without doing the Fourier transform(s) we did to convert it into an IHE. Proceeding as before, we seek a Green\u2019s function that satis\ufb01es: 2 \u2207 \u2212 1 c2 \u22022 \u2202t2! G(x, t, x0, t0) = \u03b4(x x\u2032)\u03b4(t t\u2032). \u2212 \u2212 (11.53) The primary di\ufb00erences between this and the previous cases are a) the PDE is hyperbolic, not elliptical, if you have any clue as to what that means; b) it is now four dimensional \u2013 the \u201cpoint source\u201d is one that exists only at a single point in space for a single instant in time. Of course this mathematical description leaves us with a bit of an existential dilemna, as physicists. We generally have little trouble with the idea of gradually restricting the support of a distribution to a single point in space by a limiting process. We just squeeze it down, mentally. However, in a supposedly conservative Universe, it is hard for us to imagine one of those squeezed down distributions of charge just \u201cpopping into existence\u201d and then popping right out. We can\u2019t even do it via a limiting process, as it is a bit bothersome to create/destroy charge out of nothingness even gradually! We are left with the uncomfortable feeling that this particular de\ufb01nition is nonphysical in that it can describe", " no actual physical sources \u2013 it is by far the most \u201cmathematical\u201d or \u201cformal\u201d of the constructs we must use. It also leaves us with something to understand. One way we can proceed is to view the Green\u2019s functions for the IHE as being the Fourier transform of the desired Green\u2019s function here! That is, we can exploit the fact that: \u03b4(t \u2212 t0) = 1 2\u03c0 Z \u2212\u221e \u221e e\u2212 i\u03c9(t \u2212 t0)d\u03c9 (11.54) to create a Fourier transform of the PDE for the Green\u2019s function: 2 + k2 G(x, x0, \u03c9) = \u03b4(x x0)ei\u03c9t0 \u2212 (11.55) \u2207 (where I\u2019m indicating the explicit \u03c9 dependence for the moment). From the previous section we already know these solutions: G0(x, x0, \u03c9) = \u2212 x cos(k | x 4\u03c0 | x e+ik | G+(x, x0, \u03c9) = \u2212 x 4\u03c0 | ik e\u2212 (x, x0, \u03c9) = \u2212 x 4\u03c0 | x0| ) \u2212 x0| \u2212 x0 \u2212 ei\u03c9t0 x0| x0 x0| ei\u03c9t0 \u2212 x | \u2212 G \u2212 \u2212 | | ei\u03c9t0 (11.56) (11.57) (11.58) At this point in time 3 the only thing left to do is to Fourier transform back \u2013 to this point in time: G+(x, t, x0, t0) = | e\u2212 i\u03c9(t \u2212 t0)d\u03c9 (11.59) \u2212 x0 x0| \u221e \u2212 \u2212\u221e x | \u221e = \u2212\u221e 4\u03c0 1 2\u03c0 Z 1 2\u03c0 e+ik \u2212 x 4\u03c0 | \u2212 1 \u2212 x0| Z x | 1 \u2212 x 4\u03c0 | 1 2\u03c0 Z (t \u03b4 = \u2212 \u2212 x0| exp \u2212\u221e \u2212 \u2212 \u00d7 = ( \u221e \u2212 4\u03c0 t0) x | \u2212 x0| \u2212 i\u03c9 \u2212 x | x0 \u2212 c | e+i \u03c9 c | x x0 \u2212 |e\u2212 i\u03c9(t \u2212 t0)d\u03c9 (11.60) (t \" \u2212 t0) \u2212 x | \u2212 c x0| d\u03c9 (11.61) ) #! (11.62) (11.63) x0 \u2212 c | (x, t, x0, t0)) (11.64) \u2212 so that: G \u00b1 \u03b4 (x, t, x0, t0) = \u2212 (t G0(x, t, x0, t0) = 1 2 \u2212 4\u03c0 t0) x | x | \u2213 x0| \u2212 (G+(x, t, x0, t0) + G Note that when we set k = \u03c9/c, we basically asserted that the solution is being de\ufb01ned without dispersion! If there is dispersion, the Fourier transform will no longer neatly line up and yield a delta function, because the di\ufb00erent Fourier components will not travel at the same speed. In that case one 3Heh, heh, heh...:-) might still expect a peaked distribution, but not an in\ufb01nitely sharp peaked distribution. The \ufb01rst pair are generally rearranged (using the symmetry of the delta function) and presented as: G( \u00b1 )(x, t; x\u2032, t\u2032) = \u03b4(t\u2032 \u2212 t h vx | x | \u2212 c x\u2032 | i x\u2032 | \u2213 \u2212 (11.65) and are called the retarded (+) and advanced (-) Green\u2019s functions for the wave equation. The second form is a very interesting beast. It is obviously a Green\u2019s function by construction, but it is a symmetric combination of advanced and retarded. Its use \u201cmeans\u201d that a \ufb01eld at any given point in space-time (x, t) consists of two pieces \u2013 one half of it is due to all the sources in space in the past such that the \ufb01elds they emit are contracting precisely to the point x at the instant t and the other half is due to all of those same sources in space in the future such that the \ufb01elds currently emerging from the point x at t precisely arrive at them. According to this view, the \ufb01eld at all points in space-time is as much due to the charges in the future as it is those same charges in the past. Again it is worthwhile to note that any actual \ufb01eld con\ufffd", "\ufffdguration (solution to the wave equation) can be constructed from any of these Green\u2019s functions augmented by the addition of an arbitrary bilinear solution to the homogeneous wave equation (HWE) in primed and unprimed coordinates. We usually select the retarded Green\u2019s function as the \u201ccausal\u201d one to simplify the way we think of an evaluate solutions as \u201cinitial value problems\u201d, not because they are any more or less causal than the others. Cause may precede e\ufb00ect in human perception, but as far as the equations of classical electrodynamics are concerned the concept of \u201ccause\u201d is better expressed as one of interaction via a suitable propagator (Green\u2019s function) that may well be time-symmetric or advanced. A \ufb01nal note before moving on is that there are simply lovely papers (that we hope to have time to study) by Dirac and by Wheeler and Feynman that examine radiation reaction and the radiation \ufb01eld as constructed by advanced and retarded Green\u2019s functions in considerable detail. Dirac showed that the di\ufb00erence between the advanced and retarded Green\u2019s functions at the position of a charge was an important quantity, related to the change it made in the \ufb01eld presumably created by all the other charges in the Universe at that point in space and time. We have a lot to study here, in other words. Using (say) the usual retarded Green\u2019s function, we could as usual write an integral equation for the solution to the general IWE above for e.g. A(x, t): A(x, t) = \u03c7A(x, t) \u00b50 \u2212 ZV G+(x, t; x\u2032, t)J (x\u2032, t\u2032)d3x\u2032dt\u2032 (11.66) where \u03c7A solves the HWE. This (with \u03c7A = 0) is essentially equation (9.2), which is why I have reviewed this. Obviously we also have \u03c6(x, t) = \u03c7\u03c6(x, t) 1 \u01eb0 ZV \u2212 G+(x, t; x\u2032, t)\u03c1(x\u2032, t\u2032)d3x\u2032dt\u2032 (11.67) for \u03c6(x, t) (the minus signs are in the di\ufb00erential equations with the sources, note). You should formally verify that these solutions \u201cwork\u201d given the de\ufb01nition of the Green\u2019s function above and the ability to reverse the order of di\ufb00erentiation and integration (bringing the di\ufb00erential operators, applied from the left, in underneath the integral sign). Jackson proceeds from these equations by fourier transforming back into a k representation (eliminating time) and expanding the result to get to multipolar radiation at any given frequency. However, because of the way we proceeded above, we don\u2019t have to do this. We could just as easily start by working with the IHE instead of the IWE and use our HE Green\u2019s functions. Indeed, that\u2019s the plan, Stan... 11.3 Simple Radiating Systems Let us start by writing the integral equation for the vector potential A(x) where we presume that we\u2019ve already transformed the IWE into the IHE. We will choose to use the outgoing wave Green\u2019s function to make it clear that the \ufb01eld we are looking for is the one that the source is emitting, not one that it is absorbing. A(x) = +\u00b50 Z x eik | x 4\u03c0 | | x\u2032 x\u2032| \u2212 \u2212 J (x\u2032)d3x\u2032. (11.68) There is no inhomogeneous term if there are no boundaries with a priori known boundary conditions. Note that a more general solution would be one that allowed for absorption of incoming waves as well as the emission of outgoing waves, but that this would require knowing something about the sources outside the domain considered to be in\ufb01nite. We will talk about this later (scattering theory and the optical theorem). From A(x) we can easily \ufb01nd B or H: B = \u00b50B = \u2207 A \u00d7 (11.69) (by de\ufb01nition). Outside of the source, though (where the currents are all zero) Ampere\u2019s law tells us that: or or \u2207 H = i\u03c9D \u2212 \u00d7 \u2207 \u2207 \u00d7 \u00d7 B = B = i\u03c9\u00b50\u01eb0E \u03c9 c2 11.70) (11.71) (11.72) (11.73) k c E Doing the integral above can be quite di\ufb03cult in the general case.", " However, we\u2019ll \ufb01nd that for most reasonable, physical situations we will be able to employ certain approximations that will enable us to obtain a systematic hierarchy of descriptions that converge to the correct answer as accurately as you like, at the same time they increase our physical insight into the radiative processes. 11.3.1 The Zones Suppose the source lives inside a region of maximum size d \u03bb where \u03bb = 2\u03c0c/\u03c9. By that I mean that a sphere of radius d (about the origin) completely contains all charge\u2013current distributions. Then we can de\ufb01ne three zones of approximation: \u226a 1. The near (static) zone 2. The intermediate (induction) zone 3. The far (radiation) zone d << r << \u03bb d << r \u03bb \u223c d << \u03bb << r The \ufb01eld has very di\ufb00erent properties in these zones. We will brie\ufb02y discuss each of them. \u2022 \u2022 \u2022 (Atomic and Molecular) sources all live inside their own near zone at optical frequencies. If the atoms are in a liquid or solid, there is a near \ufb01eld interaction (implicitly alluded to in chapter 4 and 7) that may be important in determining optical dispersion and other observable phenomena. Only for microwave frequencies and less does the near zone become relevant on a macroscopic scale. For rf waves it becomes extremely relevant, as it may extend a hundred meters or more. The induction zone is an annoying region where most of the simple approximations fail. It is distinguished by not being either of the other two zones. The wave changes character completely inside this zone. Because condensed matter theory invariably has objects interacting within this zone it is important there, although it can only be crudely treated. Without doing an obnoxious amount of work, that is. The far zone is where we all live, most of the time, with respect to the major sources of EM radiation. Our detectors are many wavelengths away from the atoms. Our radios are many wavelengths away from the transmitters. And, generally, the sources are smaller, if not much smaller, than a wavelength4. In the far zone, the emitted EM \ufb01elds are characteristically transverse and fall o\ufb00 in amplitude as 1/r or faster, and often far enough away they look locally like plane waves! This is typical of radiation \ufb01elds from compact sources. We will spend most of our time considering solutions in the far zone. 11.3.2 The Near Zone Suppose that we are in the near zone. Then by de\ufb01nition 4We will learn to treat certain exceptions, believe me. k x | \u2212 | x\u2032 << 1 and eik x | \u2212 x\u2032 | 1 \u2248 This makes the integral equation into the \u201cstatic\u201d form already considis just the ered in chapter 5 (cf. equation (5.32)). We see that Green\u2019s function for the good old Poisson equation in this approximation and can be expanded in harmonic functions just like in the good old days: 1/4\u03c0 x | x\u2032 \u2212 \u2212 | G0(x, x\u2032) = XL 1 \u2212 2\u2113 + 1 \u2113 r\u2032 r\u2113+1 YL(\u02c6r)YL( \u02c6r\u2032)\u2217. (11.74) Note Well: I will use L (\u2113, m) freely and without warning in this course. The sum is over all \u2113, m. Hopefully, by now you know what they run over. If not, read the chapter in Wyld on spherical harmonics and review Jackson as well. This is important! \u2261 This means that (if you like) A(x) = lim 0 kr \u2192 XL 1 (2\u2113 + 1)r\u2113+1 YL(\u02c6r) Z J (x\u2032)r\u2032 \u2113YL( \u02c6r\u2032)\u2217d3r\u2032. (11.75) We will use expressions like this (derived from the multipolar expansion of the Green\u2019s function) frequently in what follows. For that reason I suggest that you study it carefully and be sure you understand it. Since (for \ufb01xed r outside the source) we see that this limit is reached (among other times) when lim k \u2192 0 \u2192 lim 0 kr \u2192 k 0 \u2192 (relative to the size of the source and point of measurement)! But then the IHE turns back into the Poisson equation (or inhomogeneous Laplace equation, ILE) as it should, come to think about it. The near \ufb01elds oscillate harmonically in time, but are spatially identical to the \ufb01elds produced by a \u201cstatic\u201d current with", " the given spatial distribution. That\u2019s why we also call the near zone the \u201cstatic zone\u201d. 11.3.3 The Far Zone Exactly the opposite is true in the far zone. Here kr >> 1 and the exponential oscillates rapidly. We can approximate the argument of the exponential as follows: x | x\u2032 | \u2212 = \u221ar2 + r\u2032 2 2 r = r 1 ( \u2212 2rn \u2212 n x\u2032 + \u00b7 x\u2032 \u00b7 2 r\u2032 r2 ) 1/2 = r n x\u2032 + (11.76) \u00b7 where we have assumed that r\u2032max < d << r and used a binomial expansion of the root sum. We neglect higher order terms. Note that this approximation is good independent of k and may be good even in the near zone. O \u2212 1 r Then (k)r lim vA(x) = \u00b50eikr 4\u03c0r Z In the far zone, the solution behaves like an outgoing spherical wave times an amplitude that depends on integral over the source that depends on angles in an intricate fashion. J (x\u2032)e\u2212 x\u2032 ik \u02c6n \u00b7 (11.77) d3x\u2032. \u2192\u221e At this point I could continue and extract lim (k)r \u2192\u221e vA(x) = \u00b50eikr 4\u03c0r ( \u2212 ik)n n! Z n X J (x\u2032)( \u02c6n \u00b7 x\u2032)nd3x\u2032 (11.78) (if the source is actually small enough to allow expansion of the exponential in a series5). This would give us a cheap introduction into multipoles. But it is so sloppy! Instead we are going to do it right. We will begin by reviewing the solutions to the homogeneous Helmholtz equation (which should really be discussed before we sweat solving the inhomogeneous equation, don\u2019t you think?) and will construct the multipolar expansion for the outgoing and incoming (and stationary) wave Green\u2019s function. Using this, it will be a trivial matter to write down a formally exact and convergent solution to the integral equation on all space that we can chop up and approximate as we please. This will provide a much more natural (and accurate) path to multipolar radiation. So let\u2019s start. 5Taylor? Power? Laurent? Who can remember... 11.4 The Homogeneous Helmholtz Equation Recall as you read this that WIYF and MWIYFin addition to the treatment of this available in Jackson, chapters 2, 3, 6, and 8 of Wyld, and doubtless Arfkin, Morse and Feshback, and probably six other sources if you look. Very important stu\ufb00, can\u2019t know it too well. Recall from above the Homogeneous Helmholtz Equation (HHE): We assume that6: 2 + k2)\u03c7(x) = 0. ( \u2207 \u03c7(x) = f\u2113(r)YL(\u03b8, \u03c6). XL (11.79) (11.80) We reduce the HHE with this assumption to the radial di\ufb00erential equation d2 dr2 + 2 r d dr \" + k2 \u2212 \u2113(\u2113 + 1) r2 If we substitute 1 r1/2 u\u2113(r) we transform this into an equation for u\u2113(r), f\u2113(r) = f\u2113(r) = 0. # (11.81) (11.82) d2 dr2 + 1 r d dr \" + k2 2)2 (\u2113 + 1 r2 # \u2212 u\u2113(r) = 0. (11.83) The is Bessel\u2019s di\ufb00erential equation. See Wyld, (2-6) or Jackson in various places (see key on back inside cover) for more detail. Or your own favorite Math Physics book. Two linearly independent solutions on IR3 minus the origin to this radial DE are: f\u2113(r) = j\u2113(kr) and f\u2113(r) = n\u2113(kr), (11.84) (11.85) 6This really isn\u2019t an assumption. We could equally well write 2 in spherical polar coordinates, separate variables, note that the angular ODEs have spherical harmonics as eigenstates (\u201cquantized\u201d by the requirement of single-valuedness on e.g. rotations of 2\u03c0 in \u03c6) and reconstruct the separated solution. But that\u2019s too much work and we already did it at least once in our lives, right? So we\u2019ll \u201cassume\u201d. \u2207 the spherical bessel function and spherical neumann functions respectively. They are both", " real, and hence are stationary in time (why?). The j\u2113(kr) are regular (\ufb01nite) at the origin while the n\u2113(kr) are irregular (in\ufb01nite) at the origin. This is in exact analogy with the situation for the homogeneous Laplace equation (which is a special case of this solution). The following is a MINIMAL table of their important properties. A better table can be found in Wyld between chps. 6 and 7 and in Morse and Feshbach (I can\u2019t remember which volume). 11.4.1 Properties of Spherical Bessel Functions Recursion Relation Let z\u2113(x) be either solution or a linear combination of the two. x is a complex scalar independent variable (in practice, x = kr). Then z\u2113+1(x) = 2\u2113 + 1 x z\u2113(x) 1(x). z\u2113 \u2212 \u2212 (11.86) This relation is stable for increasing \u2113 for z\u2113 = n\u2113. It is stable for decreasing \u2113 for z\u2113 = j\u2113. For that reason it is unstable in both directions for h\u00b1\u2113 (de\ufb01ned below). How would you make it? See Abramowitz and Stegun, Handbook of Mathmatical Functions for discussion of recursive algorithm and de\ufb01nition of power series expansions. The Lowest Few Functions sin(x) x sin(x) x2 \u2212 cos(x) x cos(x) x cos(x) \u2212 \u2212 x2 \u2212 sin(x) x j0(x) = j1(x) =... n0(x) = n1(x) =... (11.87) (11.88) (11.89) (11.90) Asymptotic Forms Small x: j\u2113(x) = lim 0 x \u2192 n\u2113(x) = lim 0 x \u2192 x\u2113 2\u2113\u2113! (2\u2113 + 1)! (2\u2113)! 1 x\u2113+1. 2\u2113\u2113! \u2212 (11.91) (11.92) Note that for small x (r << k) j\u2113(kr) is proportional to r\u2113 and n\u2113(kr) is proportional to 1/r\u2113+1, which are the regular and irregular solutions to the separated Laplace equation. This is the correct way to obtain the static limit. Large x: j\u2113(x) = n\u2113(x) = lim x \u2192\u221e lim x \u2192\u221e 1 x 1 x cos(x sin(x \u2212 \u2212 (\u2113 + 1) (\u2113 + 1) \u03c0 2 \u03c0 2 ) ). (11.93) (11.94) Note that both solutions are regular (go to zero smoothly) at in\ufb01nity and are the same (trig) function shifted by \u03c0/2 over x there. Note that they are not square integrable on IR3 (for your quantum course) but are still better than plane waves in that regard. Something to think about... Hankel Functions Examining the asymptotic forms, we see that two particular complex linear combinations of the stationary solution have the behavior, at in\ufb01nity, of an outgoing or incoming spherical wave when the time dependence is restored: h+ \u2113 (x) = j\u2113(x) + in\u2113(x) in\u2113(x) h\u2212\u2113 (x) = j\u2113(x) (= h1 (= h2 \u2113 (x)) \u2113 (x)) (11.95) (11.96) \u2212 the spherical hankel functions of the \ufb01rst (+) (outgoing) and second ( ) (incoming) kinds. Both of these solutions are singular at the origin like \u2212 1/x\u2113+1 (why?) and behave like i)\u2113+1 eix h+ \u2113 (x) = ( x \u2212 ix h\u2212\u2113 (x) = (i)\u2113+1 e\u2212 x lim x \u2192\u221e lim x \u2192\u221e (11.97) (11.98) at in\ufb01nity. Two particularly useful spherical hankel functions to know are the zeroth order ones: h+ 0 (x) = h\u22120 (x) = eix ix e\u2212 ix ix \u2212 (11.99) (11.100) Plane Wave Expansion Plane waves and free spherical waves both form an (on\u2013shell", ") complete orthnormal set on IR3 (with or without the origin). That means that one must be able to expand one in terms of the other. Plane waves can be expanded in terms of free spherical waves by: r = eikr cos(\u0398) eik \u00b7 = 4\u03c0i\u2113YL(\u02c6k)j\u2113(kr)YL(\u02c6r)\u2217. (11.101) XL \u0398). This is due to Lord Rayleigh and is sometimes called the Rayleigh expansion. Recall that \u0398 is the angle betwixt the ~r and the ~k and that cos(\u0398) = cos( \u2212 There is similarly an (integral) expression for j\u2113(kr) in terms of an integral over the eik r but we will not use it here. It follows from the completeness \u00b7 relation on page 214 in Wyld, the Rayleigh expansion, and the completeness relation on page 212. Derive it for homework (or \ufb01nd it somewhere and copy it, but you shouldn\u2019t have to). Check your result by \ufb01nding it somewhere. I think it might be somewhere in Wyld, but I know it is elsewhere. This will be handed in. 11.4.2 JL(r), NL(r), and H \u00b1L (r) For convenience, we de\ufb01ne the following: JL(r) = j\u2113(kr)YL(\u02c6r) NL(r) = n\u2113(kr)YL(\u02c6r) H \u00b1L (r) = h\u00b1\u2113 (kr)YL(\u02c6r) (11.102) (11.103) (11.104) These are the basic solutions to the HHE that are also eigenfunctions of L2 and Lz. Clearly there is an implicit label of k (or k2) for these solutions. A general solution (on a suitable domain) can be constructed out of a linear combination of any two of them. 11.4.3 General Solutions to the HHE On \u201cspherical\u201d domains (the interior and exterior of a sphere, or in a spherical shell) the completely general solution to the HHE can therefore be written in stationary form as: ALJL(r) + BLNL(r) (11.105) XL or (for scattering theory, mostly) in the outgoing wave form CLJL(r) + SLH + L (r). (11.106) XL Inside a sphere, BL and SL must be zero. Outside a sphere, or in a spherical annulus, all the coe\ufb03cients can be non\u2013zero unlike the situation for the Laplace equation (why?). [This should provoke deep thoughts about the fundamental signi\ufb01cance of the Laplace equation. Are there any \u201creally\u201d stationary sources in the dynamical, covariant, universe? Do we expect to have a contribution to the zero frequency charge/current density distribution in any region of space? What would this correspond to?] 11.4.4 Green\u2019s Functions and Free Spherical Waves We expect, for physical reasons7 that the wave emitted by a time dependent source should behave like an outgoing wave far from the source. Note that inside the bounding sphere of the source that need not be true. Earlier in this chapter, we used an \u201coutgoing wave Green\u2019s function\u201d to construct the solution to the IHE with this asymptotic behavior. Well, lo and behold: (x, x\u2032) = G \u00b1 ik 4\u03c0 \u2213 h\u00b10 (k x | \u2212 x\u2032 ) | (11.107) 7A cop\u2013out phrase if there ever was one. It translates as: because that\u2019s the way it turns out at the end. For stationary waves (useful in quantum theory) G0(x, x\u2032) = k 4\u03c0 n0(k x | \u2212 x\u2032 ). | (11.108) This extremely important relation forms the connection between free spherical waves (reviewed above) and the integral equation solutions we are interested in constructing. This connection follows from the addition theorems or multipolar expansions of the free spherical waves de\ufb01ned above. For the special case of L = (0, 0) these are: N0(r \u2212 r\u2032) = n0(k r\u2032) r | \u2212 1 \u221a4\u03c0 and H \u00b10 (r \u2212 r\u2032) = h\u00b10 (k r\u2032) r | \u2212 1 \u221a4\u03c0 = \u221a4\u03c0 NL(r>)JL(r<)\u2217 (11.109) XL = \u221a4\u03c0 H \u00b1L (r>)JL(r<)\u2217. (11.110) XL From this and the above,", " the expansion of the Green\u2019s functions in free spherical multipolar waves immediately follows: G0(r \u2212 r\u2032) = k NL(r>)JL(r<)\u2217 (11.111) XL and G (r \u00b1 \u2212 r\u2032) = ik \u2213 H \u00b1L (r>)JL(r<)\u2217. (11.112) XL Note Well: The complex conjugation operation under the sum is applied to the spherical harmonic (only), not the Hankel function(s). This is because the only function of the product YL(\u02c6r)YL(\u02c6r\u2032)\u2217 is to reconstruct the P\u2113(\u0398) via the addition theorem for spherical harmonics. Study this point in Wyld carefully on your own. These relations will allow us to expand the Helmholtz Green\u2019s functions exactly like we expanded the Green\u2019s function for the Laplace/Poisson equation. This, in turn, will allow us to precisely and beautifully reconstruct the multipolar expansion of the vector potential, and hence the EM \ufb01elds in the various zones exactly8. This ends our brief mathematical review of free spherical waves and we return to the description of Radiation. 8Well, in a uniformly convergent expansion, which is kind of exact, in the limit of an in\ufb01nite sum. In the mean time, it is a damn good approximation. Usually. 11.5 Electric Dipole Radiation Now that we have that under our belts we can address the multipolar expansion of the vector potential intelligently. To begin with, we will write the general solution for the vector potential in terms of the multipolar expansion for the outgoing wave Green\u2019s function de\ufb01ned above: A(r) = ik JL(r) XL + H + L (r) r 0 Z \u221e \u00b50J (r\u2032)HL(r\u2032)( )d3r\u2032 \u2217 r Z \u00b50J (r\u2032)JL(r\u2032)( )d3r\u2032 \u2217 (11.113) ) means that the YL(\u02c6r) is conjugated but the beswhere, by convention, ( \u2217 sel/neumann/hankel function is not. This is because the only point of the conjugation is to construct P\u2113(\u0398) from the m\u2013sum for each \u2113 via the addition theorem for spherical harmonics. We certainly don\u2019t want to change h+ into h\u2212, which changes the time dependent behavior of the solution9. Note that the integral over all space is broken up in such a way that the Green\u2019s function expansions above always converge. This solution is exact everywhere in space including inside the source itself! We can therefore simplify our notation by de\ufb01ning certain functions of the radial variable: A(r) = ik XL h CL(r)J(r) + SL(r)H + L (r) i. (11.114) In this equation, CL(r) = SL(r) = \u221e \u00b50J (r\u2032)HL(r\u2032)( )d3r\u2032 \u2217 \u00b50J (r\u2032)JL(r\u2032)( )d3r\u2032. \u2217 r Z r 0 Z (11.115) (11.116) Clearly SL(0) = 0 and for r > d, CL(r) = 0. At the origin the solution is completely regular and stationary. Outside the bounding sphere of the source distribution the solution behaves like a linear combination of outgoing spherical multipolar waves. From now on we will concentrate on the latter case, since it is the one relevant to the zones. 9This suggests that there are some interesting connections between the conjugation symmetry and time reversal symmetry. Too bad we won\u2019t have time to explore them. You may on your own, though. 11.5.1 Radiation outside the source Outside the bounding sphere of the source, A(r) = ik H + L (r) XL 0 Z \u221e \u00b50J (r\u2032)JL(r\u2032)( )d3r\u2032. \u2217 (11.117) At last we have made it to Jackson\u2019s equation 9.11, but look how elegant our approach was. Instead of a form that is only valid in the far zone, we can now see that this is a limiting form of a convergent solution that works in all zones, including inside the source itself! The integrals that go into the CL(r) and SL(r) may well be daunting to a person armed with pen and paper (depending on how nasty J (x\u2032) is) but they are very de\ufb01nitely computable with a computer! Now, we must use several interesting observations. First of all, JL(", "r) gets small rapidly inside d as \u2113 increases (beyond kd). This is the angular momentum cut\u2013o\ufb00 in disguise and you should remember it. This means that if J (r) is sensibly bounded, the integral on the right (which is cut o\ufb00 at r\u2032 = d) will get small for \u201clarge\u201d \u2113. In most cases of physical interest, kd << 1 by hypothesis and we need only keep the \ufb01rst few terms (!). In practically all of these cases, the lowest order term (\u2113 = 0) will yield an excellent approximation. This term produces the electric dipole radiation \ufb01eld. 11.5.2 Dipole Radiation Let us evaluate this term. It is (c. f. J9.13): r A(r) = \u00b50eikr 4\u03c0r Z (note: Y00(\u02c6r) = Y00(\u02c6r)\u2217 = 1/\u221a4\u03c0). If we integrate this term by parts (a surprisingly di\ufb03cult chore that will be an exercise) and use the continuity equation and the fact that the source is harmonic we get: J (r\u2032)d3r\u2032 (11.118) 0 where A(r) = i\u00b50\u03c9 4\u03c0 p eikr r \u2212 p = r\u2032\u03c1(r\u2032)d3r\u2032 Z (11.119) (11.120) is the electric dipole moment (see J4.8). Note that if we de\ufb01ne \u03c1(r) to be a \u201cprobability density\u201d for the electrons during a transition this expression is still valid. This is wonderfully simple. If only we could quit with the vector potential. Alas, no. We must reconstruct the electromagnetic \ufb01eld being radiated away from the source from the expressions previously given and B = \u2207 A \u00d7 E = \u2207 ic k \u00d7 B. After a tremendous amount of straightforward but nonetheless di\ufb03cult algebra that you will do and hand in next week (see problems) you will obtain: H = ck2 4\u03c0 p) (n \u00d7 eikr r 1 \u2212 1 ikr (11.121) and E = 1 4\u03c0\u01eb0 ( k2(n p) \u00d7 \u00d7 n eikr r + [3n(n p) \u00b7 \u2212 p] 1 r3 \u2212 ik r2! eikr ) (11.122) The magnetic \ufb01eld is always transverse to the radial vector. Electric dipole radiation is therefore also called transverse magnetic radiation. The electric \ufb01eld is transverse in the far zone, but in the near zone it will have a component (in the p direction) that is not generally perpendicular to n. Asymptotic properties in the Zones In the near zone we get: B = \u00b50H = i\u03c9\u00b50 4\u03c0 ( \u02c6n \u00d7 p) E = 1 4\u03c0\u01eb0 [3 \u02c6n( \u02c6n p) \u00b7 \u2212 p] 1 r2 1 r3 (11.123) (11.124) and can usually neglect the magnetic \ufb01eld relative to the electric \ufb01eld (it is smaller by a factor of kr << 1). The electric \ufb01eld is that of a \u201cstatic\u201d dipole (J4.13) oscillating harmonically. In the far zone we get: B = \u00b50H = E = \u2207 ic k \u00d7 ck2\u00b50 4\u03c0 ( \u02c6n B = c (B p) eikr r \u02c6n). \u00d7 \u00d7 (11.125) (11.126) This is transverse EM radiation. Expanded about any point, it looks just like a plane wave (which is how \u201cplane waves\u201d are born!). We are most interested, as you know, in the radiation zone and so we will focus on it for a moment. Energy radiated by the dipole Recall our old buddy the complex Poynting vector for harmonic \ufb01elds (J6.132): 1 2 S = Re E H \u2217. (11.127) { The factor of 1/2 comes from time averaging the \ufb01elds. This is the energy per unit area per unit time that passes a point in space. To \ufb01nd the time average power per solid angle, we must relate the normal area through which the energy \ufb02ux passes to the solid angle: \u00d7 } dAn = r2d\u2126 (11.128) and project out the appropriate piece of S, i. e. \u2014 n S. We get (with \u00b5 = 1) \u00b7 dP d\u2126 =", " 1 2 Re[r2n (E \u00b7 \u00d7 H\u2217)]. (11.129) where we must plug in E and H from the expressions above for the far \ufb01eld. After a bunch of algebra that I\u2019m sure you will enjoy doing, you will obtain: dP d\u2126 = c2 32\u03c02 s \u00b50 \u01eb0 k4 (n p) n 2. | \u00d7 \u00d7 | (11.130) The polarization of the radiation is determined by the vector inside the absolute value signs. By this one means that one can project out each component of p (and hence the radiation) before evaluating the square independently, if so desired. Note that the di\ufb00erent components of p need not have the same phase (elliptical polarization, etc.). If all the components of p (in some coordinate system) have the same phase, then p necessarily lies along a line and the typical angular distribution is that of (linearly polarized) dipole radiation: dP d\u2126 = c2 32\u03c02 s \u00b50 \u01eb0 k4 p 2 sin2 \u03b8 | | (11.131) where \u03b8 is measured between p and n. When you integrate over the entire solid angle (as part of your assignment) you obtain the total power radiated: P = c2k4 12\u03c0 s \u00b50 \u01eb0 | p 2 | (11.132) The most important feature of this is the k4 dependence which is, after all, why the sky is blue (as we shall see, never fear). Example: A centerfed, linear antenna In this antenna, d << \u03bb and I(z, t) = I0 1 2 z | | d! \u2212 e\u2212 i\u03c9t. (11.133) From the continuity equation (and a little subtle geometry), \u2207 \u00b7 J = dI dz = \u2202\u03c1\u2032(z)e\u2212 \u2202t \u2212 i\u03c9t = i\u03c9\u03c1\u2032(z) (11.134) and we \ufb01nd that the linear charge density (participating in the oscillation, with a presumed neutral background) is independent of z: \u03c1\u2032(z) = 2iI0 \u03c9d \u00b1 (11.135) \u2212 where the +/ sign indicates the upper/lower branch of the antenna and the \u2032 means that we are really treating \u03c1/(dxdy) (which cancels the related terms in the volume integral below). We can then evaluate the dipole moment of the entire antenna for this frequency: pz = d/2 d/2 Z \u2212 z\u03c1\u2032(z)dz = iI0d 2\u03c9. (11.136) The electric and magnetic \ufb01elds for r > d in the electric dipole approximation are now given by the previously derived expressions. The angular distribution of radiated power is dP d\u2126 = I 2 0 128\u03c02 s \u00b50 \u01eb0 (kd)2 sin2 \u03b8 and the total radiated power is P = 0 (kd)2 I 2 48\u03c0 s \u00b50 \u01eb0. (11.137) (11.138) Remarks. For \ufb01xed current the power radiated increases as the square of the frequency (at least when kd << 1, i. e. \u2013 long wavelengths relative to the size of the antenna). The total power radiated by the antenna appears as a \u201closs\u201d in \u201cOhm\u2019s Law\u201d for the antenna. Factoring out I 2 0 /2, the remainder must have the units of resistance and is called the radiation resistance of the antenna: Rrad = 2P I 2 0 = (kd)2 24\u03c0 s \u00b50 \u01eb0 \u2248 5(kd)2 ohms) (11.139) where we do the latter multiplication to convert the resulting units to ohms. Note that this resistance is there for harmonic currents even if the conductivity of the metal is perfect. Note further that by hypothesis this expression will only be valid for small values of Rrad. Good golly, this is wonderful. We hopefully really understand electric It would be truly sublime if all radiators dipole radiation at this point. were dipole radiators. Physics would be so easy. But (alas) sometimes the current distribution has no \u2113 = 0 moment and there is therefore no dipole term! In that case we must look at the next term or so in the multipolar expansions Lest you think that this is a wholly unlikely occurrance, please note that a humble loop carrying a current that varies harmonically is one such system. So let us proceed to: 11.6 Magnetic Dipole and Electric Quadrupole Radiation Fields The next term in the multipolar expansion is the \u2113 = 1 term: A(x)", " = ik\u00b50h+ 1 (kr) Y1,m(\u02c6r) 1 Xm= \u2212 1 0 Z When you (for homework, of course) \u221e J (x\u2032)j1(kr\u2032)Y1,m( \u02c6r\u2032)\u2217d3x\u2032 (11.140) 1. m\u2013sum the product of the Y\u2113,m\u2019s 2. use the small kr expansion for j1(kr\u2032) in the integral and combine it with the explicit form for the resulting P1(\u03b8) to form a dot product 3. cancel the 2\u2113 + 1\u2019s 4. explicitly write out the hankel function in exponential form you will get equation (J9.30, for \u2013 recall \u2013 distributions with compact support): A(x) = \u00b50 4\u03c0 eikr r 1 r \u2212 ik 0 Z \u221e J (x\u2032)(n x\u2032)d3x\u2032. \u00b7 (11.141) Of course, you can get it directly from J9.9 (to a lower approximation) as well, but that does not show you what to do if the small kr approximation is not valid (in step 2 above) and it neglects part of the outgoing wave! There are two important and independent pieces in this expression. One of the two pieces is symmetric in J and x\u2032 and the other is antisymmetric (get a minus sign when the coordinate system is inverted). Any vector quantity can be decomposed in this manner so this is a very general step: J (n \u00b7 x\u2032) = 1 2 [(n \u00b7 x\u2032)J + (n J )x\u2032] + \u00b7 1 2 (x\u2032 J ) n. \u00d7 \u00d7 (11.142) 11.6.1 Magnetic Dipole Radiation Let\u2019s look at the antisymmetric bit \ufb01rst, as it is somewhat simpler and we can leverage our existing results. The second term is the magnetization (density) due to the current J : (see J5.53, 5.54) so that ~ M = 1 2 (x \u00d7 J ) m = Z (x\u2032)d3x\u2032 ~ M (11.143) (11.144) where m is the magnetic dipole moment of the (fourier component of) the current. Considering only this antisymmetric term, we see that: AM1(x) = ik\u00b50 4\u03c0 (n \u00d7 m) eikr r 1 \u2212 1 ikr. (11.145) HMMMMMMM, (you had better say)! This looks \u201cjust like\u201d the expression for the magnetic \ufb01eld that resulted from the electric dipole vector potential. Sure enough, when you (for homework) crank out the algebra, you will show that B = \u00b50 4\u03c0 ( k2(n m) \u00d7 n eikr r \u00d7 + [3n(n m) \u00b7 \u2212 m] 1 r3 \u2212 ik r2! eikr ) (11.146) and E = 1 4\u03c0 s \u00b50 \u01eb0 \u2212 k2(n m) \u00d7 eikr r 1 \u2212 1 ikr. (11.147) Clearly, we don\u2019t need to discuss the behavior of the \ufb01elds in the zones since they are completely analogous. The electric \ufb01eld is always transverse, and the total \ufb01eld arises from a harmonic magnetic dipole. For this reason, this kind of radiation is called either magnetic dipole (M1) radiation or transverse electric radiation. For what it\u2019s worth, electric dipole radiation is also called (E1) radiation. However, this is only ONE part of the contribution from \u2113 = 1 terms in the Green\u2019s function expansion. What about the other (symmetric) piece? Oooo, ouch. 11.6.2 Electric Quadrupole Radiation Now let\u2019s to untangle the \ufb01rst (symmetric) piece. This will turn out to be a remarkably unpleasant job. In fact it is my nefarious and sadistic plan that it be so unpleasant that it properly motivates a change in approach to one that handles this nasty tensor stu\ufb00 \u201cnaturally\u201d. We have to evaluate the integral of the symmetric piece. We get: 1 2 Z [( \u02c6n \u00b7 x\u2032)J + ( \u02c6n \u00b7 J )x\u2032]d3x\u2032 = i\u03c9 2 Z \u2212 x\u2032( \u02c6n \u00b7 x\u2032)\u03c1(x\u2032)d3x\u2032 (11.148) The steps involved are: 1. integrate by parts (working to obtain divergences of J ). 2. changing \ufffd", "\ufffd \u00b7 (for a harmonic source). J into a \u03c1 times whatever from the continuity equation 3. rearranging and recombining. Don\u2019t forget the boundary condition at in\ufb01nity (J and \u03c1 have compact support)! You\u2019ll love doing this one... The vector potential is thus: 1 \u2212 \u2212 x\u2032( \u02c6n \u00b50ck2 8\u03c0 eikr r AE2(x) = x\u2032)\u03c1(x\u2032)d3x\u2032. 1 ikr Z Note that x\u2032 appears twice under the integral, and that its vector character similarly appears twice: once in x\u2032 itself and once in its projection on \u02c6n. The integral is the electric quadrupole moment of the oscillating charge density distribution and the resulting radiation \ufb01eld is called an electric quadrupole (radiation) \ufb01eld or an E2 radiation \ufb01eld (for short). (11.149) \u00b7 To get the \ufb01elds from this expression by taking its curl, and then the curl of its curl, is \u2013 ahem \u2013 most unpleasant. Jackson wimps out! Actually, taking the curls is no more di\ufb03cult than it was for the magnetic term, but untangling the integrals with the result is, because of the tensor forms that appear. Consequently we too will wimp out (in the comforting knowledge that we will shortly do this right and not wimp out to arbitrary order in a precise decomposition) and will restrict our attention to the far zone. There we need only consider the lowest order surviving term, which always comes from the curl of the exponential times the rest: B = ik( \u02c6n A) E = ik s ( \u02c6n A) \u02c6n. \u00d7 \u00d7 \u00d7 \u00b50 \u01eb0 If we keep only the lowest order terms of this we get B = ick2\u00b50 8\u03c0 \u2212 eikr r Z ( \u02c6n \u00d7 x\u2032)( \u02c6n \u00b7 x\u2032)\u03c1(x\u2032)d3x\u2032. (11.150) (11.151) (11.152) If we recall (from the beginning of Chapter 4) the discussion and de\ufb01nition of multipole moments, in particular the quadrupole moment tensor Q\u03b1\u03b2 = (3x\u2032\u03b1x\u2032\u03b2 \u2212 Z r\u2032 2\u03b4\u03b1\u03b2)\u03c1(x\u2032)d3x\u2032 (11.153) whose various components can be related to the \ufb01ve spherical harmonics with \u2113 = 2 (!) we can simplify matters. We can write the one messy integral in terms of another: where \u02c6n \u00d7 Z x\u2032( \u02c6n \u00b7 x\u2032)\u03c1(x\u2032)d3x\u2032 = 1 3 \u02c6n \u00d7 Q( \u02c6n) Q( \u02c6n) = Q\u03b1\u03b2n\u03b2 \u02c6x\u03b2. X\u03b2 (11.154) (11.155) Note that the \u201cvector\u201d Q( \u02c6n) (and hence the \ufb01elds) depends in both the magnitude and direction on the direction to the point of observation n as well as the properties of the source. With these de\ufb01nitions, B = ick3\u00b50 24\u03c0 eikr r \u2212 ( \u02c6n \u00d7 Q( \u02c6n)) (11.156) which looks (except for the peculiar form of Q) much like the E1 magnetic \ufb01eld. It is transverse. The electric \ufb01eld is obtained by appending n and is also transverse. Following exactly the same algebraic procedure as before, we \ufb01nd from \u00d7 E { and computing the \ufb02ux of the Poynting vector through a sphere of radius r as a function of angle that the angular power distribution is: (11.157) S = H \u2217 Re \u00d7 } 1 2 dP d\u2126 = c2 1152\u03c02 s \u00b50 \u01eb0 k6 ( \u02c6n | \u00d7 Q( \u02c6n)) \u02c6n 2 | \u00d7 (11.158) The angular distribution is too complicated to play with further unless you need to calculate it, in which case you will have to work it out. The total power can be calculated in a \u201cstraightforward\u201d way (to quote Jackson). First one changes the cross product to dot products using the second relation on the front cover and squares it. One then writes out the result in tensor components. One can then perform the angular integrals of the products of the components of the n (which is straightforward). Finally one term in the resulting expression goes away because Q\u03b1\u03b2 is traceless. The result is P = c", "2k6 1440\u03c0 s \u00b50 \u01eb0 2 Q\u03b1\u03b2 | X\u03b1,\u03b2 | (11.159) (note k6 frequency dependence). For the numerologists among you, note that there is almost certainly some sort of cosmic signi\ufb01cance in the 1440 in the denominator as this is the number of seconds in a day. Just kidding. For certain symmetric distributions of charge the general quadrupole moment tensor simpli\ufb01es still further. A typical case of this occurs when there is an additional, e. g. azimuthal symmetry such as an oscillating spheroidal In this case, the o\ufb00\u2013diagonal components of Q\u03b1\u03b2 distribution of charge. vanish and only two of the remaining three are independent. We can write Q33 = Q0, Q11 = Q22 = 1 2 \u2212 Q0 and the angular distribution of radiated power is dp d\u2126 = c2k6 512\u03c02 s \u00b50 \u01eb0 Q2 0 sin2 \u03b8 cos2 \u03b8 (11.160) (11.161) which is a four\u2013lobed radiation pattern characteristic of azimuthally symmetric sources. In this case it really is straightforward to integrate over the entire solid angle (or do the sum in the expression above) and show that: P = c2k6 960\u03c0 s \u00b50 \u01eb0 Q2 0. (11.162) At this point it should be clear that we are o\ufb00 on the wrong track. To quote Jackson: The labor involved in manipulating higher terms in (the multipolar expansion of A(x)) becomes increasingly prohibitive as the expansion is extended beyond the electric quadrupole terms. Some would say that we should have quit after the electric dipole or magnetic dipole. The problem has several roots. First, in the second and all succeeding terms in the expansion as written, the magnetic and electric terms are all mixed up and of di\ufb00erent tensorial character. This means that we have to project out the particular parts we want, which is not all that easy even in the simplest cases. Second, this approach is useful only when the wavelength is long relative to the source (kd << 1) which is not (always) physical for radio antennae. Third, what we have done is algebraically ine\ufb03cient; we keep having to do the same algebra over and over again and it gets no easier. Understanding the problem points out the way to solve it. We must start again at the level of the Green\u2019s function expansion, but this time we must construct a generalized tensorial multipolar expansion to use in the integral equation. After that, we must do \u201conce and for all\u201d the necessary curl and divergence algebra, and classify the resulting parts according to their formal transformation properties. Finally, we will reassemble the solution in the new vector multipoles and glory in its formal simplicity. Of course, the catch is that it is a lot of work at \ufb01rst. The payo\ufb00 is that it is general and systematically extendable to all orders. As we do this, I\u2019m leaving you to work out the various example problems in Jackson (e.g. section J9.4, 9.5) on your own. We\u2019ve already covered most of J9.6 but we have to do a bit more review of the angular part of the Laplace operator, which we largely skipped before. This will turn out to be key as we develop Multipolar Radiation Fields properly. 11.7 Radiation Assignment 1. Derive the integral expression for spherical bessel functions in terms of plane waves at the same wavenumber. 2. The addition theorems: N0(r \u2212 r\u2032) = n0(k r | \u2212 r\u2032) 1 \u221a4\u03c0 and = \u221a4\u03c0 NL(r>)JL(r<)\u2217 (11.163) XL H \u00b10 (r \u2212 r\u2032) = h\u00b10 (k r | \u2212 r\u2032) 1 \u221a4\u03c0 = \u221a4\u03c0 H \u00b1L (r>)JL(r<)\u2217. (11.164) XL are derived someplace, for both this special case and for the general case. Find at least one such place (for L = 0, 0), copy the derivation (with acknowledgement), and hand it in. If you work in a group, see how many places you can \ufb01nd it and compare. LEARN what you can from the process, that is, read the text accompanying the derivation(s) you \ufb01nd and try to understand it. Work it out. For extra credit, \ufb01nd in the literature the original paper that derives the general addition theorem. Hints: JMP, Danos and Max", "imon. Study it. 3. Derive the Green\u2019s function for the Helmholtz equation in free space (zero boundary conditions at in\ufb01nity). Do not use the addition theorem, since you do not (in principle) know its form yet and so do not know that it is a Neumann or Hankel function. Naturally, you can follow Wyld or Jackson or Arfken, but acknowledge your source and show the entire derivation. 4. Make a neat sheet with Everything You Never Wanted To Know About Spherical Bessel/Neumann/Hankel Functions but were Afraid Not To Ask on it. Don\u2019t hand it in, this will be your guide through life (for at least a few weeks). Do NOT simply photocopy my notes. Do it by hand. Pool your sheets with those of your friends \u2014 put together everything to make a \u201cbest\u201d sheet and then photocopy it. I use the term \u201csheet\u201d loosely. I expect it will \ufb01ll several (it did in my notes). 5. Using the addition theorem derived above (in the form of the Green\u2019s function) and the asymptotic relations on your worksheet, derive the static result for the vector potential A we previously obtained for the near \ufb01eld zone (my equation 66). Find the lowest order correction to this expression. This will, of course, involve \ufb01nding more out about spherical waves than I have so far told you! item Using the same addition theorem and the other asymptotic relations, derive an expression for the v.p. A in the far zone. Is there a correspondance of some sort with our previous result (Jackson 9.9)? 6. Show that A(x) = ikh+ 1 (kr) 1 Xm= \u2212 1 Y1,m(\u02c6r) J (x\u2032)j1(kr\u2032)Y1,m( \u02c6r\u2032)\u2217d3x\u2032 Z is equivalent to A(x) = eikr 4\u03c0r 1 r \u2212 ik J (x\u2032)(n Z x\u2032)d3x\u2032 \u00b7 for kd << 1. 7. Any vector quantity can be decomposed in a symmetric and an antisymmetric piece. Prove that, in the case of the \u2113 = 1 term derived above, the current term can be decomposed into J (n \u00b7 x\u2032) = 1 2 [(n \u00b7 x\u2032)J + (n J )x\u2032] + \u00b7 1 2 (x\u2032 J ) n \u00d7 \u00d7 8. Evaluate the antisymmetric piece. Show (from the magnetic dipole vector potential) that B = \u00b50 4\u03c0 ( k2(n m) \u00d7 n eikr r \u00d7 + [3n(n m) \u00b7 \u2212 m] 1 r3 \u2212 ik r2! eikr ) and E = 1 4\u03c0 s \u00b50 \u01eb0 \u2212 k2(n m) \u00d7 eikr r 1 \u2212 1 ikr. Remark upon the similarities and di\ufb00erences between this result and the electric dipole result. 9. Next start to evaluate the integral of the symmetric piece. Show that you get: 1 2 Z [(n \u00b7 x\u2032)J + (n \u00b7 J)x\u2032]d3x\u2032 = i\u03c9 2 Z \u2212 x\u2032(n \u00b7 x\u2032)\u03c1(x\u2032)d3x\u2032 The steps involved are: (a) integrate by parts (working to obtain divergences of J ). (b) changing \u2207 J into a \u03c1 times whatever from the continuity equa- tion (for a harmonic source). \u00b7 (c) rearranging and recombining. Don\u2019t forget the boundary condition at in\ufb01nity! 10. Homemade tables, part II. What you did for spherical bessel functions, do for spherical harmonics. In particular, derive the commutation rules for the raising and lowering operators from the cartesian commutation relations for L. From the commutation rules and LzY\u2113m = mY\u2113m derive on Y\u2113,m. the (normalized) action of L \u00b1 11. Jackson, problems 9.2, 9.3, 9.4 Chapter 12 Vector Multipoles As I noted just above, we\u2019re already half way through J9.6, which is mostly the review of spherical bessel, neumann, and hankel functions that we have just had. The remainder is a lightning review of scalar spherical harmonics. Since we\u2019re about to generalize that concept, we\u2019ll quickly go over the high parts. 12.1 Angular momentum and spherical har- monics The angular part of the Laplace operator 2 can be written: 1 r2 ( 1 sin \u03b8 \ufffd", "\ufffd \u2202\u03b8 sin \u03b8 \u2202 \u2202\u03b8! + \u22022 \u2202\u03c62 ) = L2 r2 \u2212 (12.1) \u2207 1 sin2 \u03b8 Eliminating r2 (to solve for the L2 di\ufb00erential equation) one needs to solve an eigenvalue problem: \u2212 L2\u03c8 = e\u03c8 (12.2) where e are the eigenvalues, subject to the condition that the solution be single valued on \u03c6 [0, 2\u03c0) and \u03b8 [0, \u03c0]. \u2208 \u2208 This equation easily separates in \u03b8, \u03c6. The \u03c6 equation is trivial \u2013 solutions periodic in \u03c6 are indexed with integer m. The \u03b8 equation one has to work 145 at a bit \u2013 there are constraints on the solutions that can be obtained for any given m \u2013 but there are many ways to solve it and at this point you should know that its solutions are associated Legendre polynomials P\u2113,m(x) where x = cos \u03b8. Thus the eigensolution becomes: L2Y\u2113m = \u2113(\u2113 + 1)Y\u2113m (12.3) where \u2113 = 0, 1, 2... and m = \u2113, mal(ized) on the solid angle 4\u03c0. \u2212 \u2113 + 1,..., \u2113 \u2212 \u2212 1, \u2113 and is typically orthonor- The angular part of the Laplacian is related to the angular momentum of a wave in quantum theory. In units where \u00afh = 1, the angular momentum operator is: L = 1 i (x \u00d7 \u2207) and L2 = L2 x + L2 y + L2 z (12.4) (12.5) Note that in all of these expressions L, L2, Lz, etc. are all operators. This means that they are applied to the functions on their right (by convention). When you see them appearing by themselves, remember that they only mean something when they are applied, so \u2207\u2019s out by themselves on the right are ok. The z component of L is: Lz = \u2202 \u2202\u03c6 i \u2212 and we see that in fact Y lm satis\ufb01es the two eigenvalue equations: and L2Y\u2113m = \u2113(\u2113 + 1)Y\u2113m LzY\u2113m = mY\u2113m (12.6) (12.7) (12.8) The Y lm\u2019s cannot be eigensolutions of more than one of the components of L at once. However, we can write the cartesian components of L so that they form an \ufb01rst rank tensor algebra of operators that transform the Y\u2113m, for a given \u2113, among themselves (they cannot change \u2113, only mix m). This is the hopefully familiar set of equations: L+ = Lx + iLy iLy = Lx \u2212 L \u2212 L0 = Lz (12.9) (12.10) (12.11) The Cartesian components of L do not commute. In fact, they form a nice antisymmetric set: which can be written in the shorthand notation [Li, Lj] = i\u01ebijkLk L = iL. L \u00d7 (12.12) (12.13) Consequently, the components expressed as a \ufb01rst rank tensor also do not commute among themselves: [L+, L \u2212 ] = 2Lz and [L \u00b1, Lz] = L \u00b1 \u2213 (12.14) (12.15) but all these ways of arranging the components of L commute with L2: [Li, L2] = 0 and therefore with the Laplacian itself: which can be written in terms of L2 as: 2, Li2 \u2202r2 (r ) \u2212 L2 r2 (12.16) (12.17) (12.18) As one can easily show either by considering the explict action of the actual di\ufb00erential forms on the actual eigensolutions Y\u2113m or more subtly by considering the action of Lz on L Y\u2113\u2113 (and showing that they behave like raising and lower operators for m and preserving normalization) one obtains: \u00b1 L+Y\u2113m = (\u2113 \u2212 m)(\u2113 + m + 1) Y\u2113,m+1 q L Y\u2113m = (\u2113 + m)(\u2113 \u2212 LzY\u2113m = mY\u2113m q m + 1) Y\u2113,m \u2212 1 \u2212 (12", ".19) (12.20) (12.21) Finally, note that L is always orthogonal to r where both are considered as operators and r acts from the left: L = 0. r \u00b7 (12.22) You will see many cases where identities such as this have to be written down in a particular order. Before we go on to do a more leisurely tour of vector spherical harmonics, we pause to motivate the construction. 12.2 Magnetic and Electric Multipoles Re- visited As we have now seen repeatedly from Chapter J6 on, in a source free region of space, harmonic electromagnetic \ufb01elds are divergenceless and have curls given by: \u2207 \u00d7 E = i\u03c9B = ikcB k c By massaging these a little bit (recall \u2207 and \u2207 be divergenceless solutions to the HHE: B = i \u2212 E. \u2207 \u00d7 \u00d7 \u00b7 2X \u00d7 X = 0 for X = E, B) we can easily show that both E and B must X) = \u2207(\u2207 \u2212 \u2207 (\u2207 X) \u00b7 (12.23) (12.24) 2 + k2)X = 0 ( \u2207 (12.25) If we know a solution to this equation for X = E we can obtain B from its curl from the equation above: \u2207 B = i \u03c9 and vice versa. However, this is annoying to treat directly, because of the vector charactor of E and B which complicate the description (as we have seen \u2013 transverse electric \ufb01elds are related to magnetic multipoles and vice versa). Let\u2019s eliminate it. (12.26) E \u00d7 \u2212 By considering the action of the Laplacian on the scalar product of r with a well\u2013behaved vector \ufb01eld X, 2(r \u00b7 \u2207 X) = r ( \u2207 \u00b7 2X) + 2\u2207 X \u00b7 (12.27) and using the divergenceless of E and B, we see that the scalars (r (r B) also satisfy the HHE: \u00b7 2 + k2)(r 2 + k2)(r ( \u2207 ( \u2207 E) = 0 B) = 0 \u00b7 \u00b7 E) and \u00b7 (12.28) (12.29) We already know how to write a general solution to either of these equations in terms of the spherical bessel, neumann, and hankel functions times spherical harmonics. Recall, that when we played around with multipole \ufb01elds, I kept emphasizing that electric n-pole \ufb01elds were transverse magnetic and vice versa? E) = 0 by de\ufb01nition, right? So now Well, transverse electric \ufb01elds have (r we de\ufb01ne a magnetic multipole \ufb01eld of order L by \u00b7 B(M ) L = r \u00b7 \u2113(\u2113 + 1) k g\u2113(kr)YL(\u02c6r) E(M ) L r \u00b7 = 0. (12.30) (12.31) Similarly, a electric multipole \ufb01eld of order L (which must be transverse magnetic) is any solution such that E(E) L = r \u00b7 \u2113(\u2113 + 1) k \u2212 f\u2113(kr)YL(\u02c6r) B(E) L = 0. r \u00b7 (12.32) (12.33) In these two de\ufb01nitions, g\u2113(kr) and f\u2113(kr) are arbitrary linear combinations of spherical bessel functions1, two at a time. Jackson uses the two hankel functions in (J9.113)k, but this is not necessary. Now, a little trickery. Using the curl equation for B we get: B(M ) L ) = k(r \u00b7 1 i r \u00b7 (\u2207 \u00d7 E(M ) L ) = 1 i (r \u2207) \u00b7 \u00d7 E(M ) L = L E(M ) L \u00b7 (12.34) E( so that L \ufb01elds. Ditto for L \u00b7 B(E) L \u00b7 LM ) is a scalar solution to the HHE for magnetic multipolar in the case of electric multipolar \ufb01elds. Thus, E(M ) L = \u2113(\u2113 + 1)g\u2113(kr)YL(\u02c6r) L \u00b7 (12.35) B(E) L. etc. for L \u00b7 1From now on, this term is generic unless clearly otherwise in context. Now we get really clever. Remember that r \u00b7 We have arranged things just so that if we write: L = 0. Also, L2 = L L", ". \u00b7 E(M ) L B(M ) L = g\u2113(kr)LYL(\u02c6r) E(12.36) (12.37) we exactly reconstruct the solutions above. Neato! This gives us a comf pletely general TE, MM EMF. A TM, EM EMF follows similarly with g and E B (and a minus sign in the second equation). \u2192 \u2194 This is good news and bad news. The good news is that this is a hell of a lot simpler than screwing around with symmetric and antisymmetric vector decompositions and integrations by parts ad nauseam. The radial part of the solutions is straightforward, and the angular part is written in a concise notation. The bad news is we have never seen that notation, good or bad, ever before. We have two choices. Either we can laboriously crank out the operator products and curls for each problem as we need to (which is really just as bad as what we have been doing) or we have to work out the algebra of these new objects once and for all so we can plug and chug out the most di\ufb03cult of answers with comparative ease. Guess which one we\u2019re about to do. 12.3 Vector Spherical Harmonics and Multi- poles Recall that L = ir \u2212 \u00d7 \u2207. (12.38) In systems with spin it is more This is an \u201corbital\u201d rotation operator. convenient in many cases to de\ufb01ne a \u201ctotal\u201d rotation operator that adds the orbital rotation operator to a \u201cspin\u201d rotation operator (de\ufb01ned below). Since total angular momentum (as opposed to orbital angular momentum) is a relativistically invariant quantity that appears \u201cnaturally\u201d in covariant kinematics, we are inspired to \ufb01nd a representation that is 1. A vector function of its coordinates. 2. Simultaneous eigenfunctions of J 2, L2, and Jz. 3. Possessed of certain desirable properties we will derive. Actually, \ufb01guring out something like this the \ufb01rst time is not quite so easy; it is full of false starts and exploring alternatives. After the fact, however, it is clear that this is the correct choice. It is also extremely useful in quantum theory. The total rotation operator is where J = L + S S = iI \u00d7 (12.39) (12.40) is the \u201cspin\u201d operator. Aside: The Spin Operator S in this expression is a tensor operator. It (like all operators) has no meaning by itself. It is, however, quite di\ufb00erent from the scalar operators you are used to. Among other things, when S operates on a vector A, it generates a new vector that points in a di\ufb00erent direction. Let us see this. In the de\ufb01nition of S, I is the identity tensor (unit diagonal matrix) and it is crossed into whatever sits on its right. To understand its action, let us evaluate its cartesian components acting on some vector A: A = i\u02c6x A \u00d7 SxA = iIx \u00d7 SyA = i\u02c6y A \u00d7 SxA = i\u02c6z A \u00d7 or (e. g.) SzA = i(Ax \u02c6y Ay \u02c6x). \u2212 (12.41) (12.42) (12.43) (12.44) Note that the action of a component of S on a vector A shifts the direction of A to a direction perpendicular to both S and the component. Only by considering the action of all the components can the total vector action of S on A in a given direction be evaluated. There are several important properties of S. The \ufb01rst is to note that it has the form of an angular momentum operator with a special action on vectors. If we form S2 and evaluate its action on A: S2A = = \u00d7 \u02c6x \u2212 n A \u2212{ (\u02c6x \u00d7 3A } \u2212 = 2A = s(s + 1)A A) + \u02c6y (\u02c6y \u00d7 \u00d7 A) + \u02c6z (\u02c6z \u00d7 \u00d7 A) o (12.45) for s = 1. S2 acting on any vector produces 2 times the same vector, telling us that a vector has \u201cspin angular momentum\u201d of 1. Note that this In connection is universal. electrodynamics (quantum or classical) the \u201cvector radiation \ufb01eld\u201d has spin one. In \ufb01eld theory a \u201cvector boson\u201d has spin 1. The spin operator thus formed is more general, because its action can be extended to higher rank tensors. (2nd rank tensor)", " gravitational \ufb01elds have spin 2. Scalar (0th rank tensor) \ufb01elds have spin 0. To treat more general cases, however, we have to work with tensor indices explicitly and you\u2019ll see enough of that in the section on relativity. Feel free to study this matter further. Louck and Biedenharn\u2019s book (Encycl. of Math Phys., see me for ref.) contains a much deeper discussion of this entire subject. It may seem that with such a peculiar structure, Sz can have no eigen- vectors. This is not the case. You should verify that (\u02c6x + i\u02c6y) 1 \u221a2 \u03c71 1 = \u03c70 \u2212 1 = \u02c6z 1 1 1 = \u221a2 \u03c7\u2212 (\u02c6x \u2212 i\u02c6y) are eigenvectors such that for ms = 1, 0, 1 and \u2212 Sz\u03c7ms 1 = ms\u03c7ms 1 S2\u03c7ms 1 = s(s + 1)\u03c7ms 1 (12.46) (12.47) (12.48) (12.49) (12.50) for s = 1. You should also verify the commutation relations for the components of S, that is, show that making it a \u201ctrue\u201d rotation/angular momentum operator. S = iS S \u00d7 (12.51) In addition, we will need to use the operators (etc.) and so that J 2 = JxJx + JyJy + JzJz, Jz = Lz + Sz L2 = LxLx + LyLy + LzLz J 2 = L2 + 2 + 2iL \u00d7 (12.52) (12.53) (12.54) (12.55) which can be proven as follows. Consider its action on A (as usual): J 2A = = = { { L2 + S2 + 2L A S } L2 + 2 + 2i[Lx(\u02c6x L2 + S2 + 2i(L \u00d7 \u00b7 \u00d7 where the meaning of the latter expression is hopefully now clear. { ) + Ly(\u02c6y A ) } ) + Lz(\u02c6z \u00d7 \u00d7 A )] } (12.56) Then we de\ufb01ne the vector spherical harmonics Ym j,\u2113 by: J 2Y m L2Y m JzY m j,\u2113 = j(j + 1)Y m j,\u2113 = \u2113(\u2113 + 1)Y m j,\u2113 = mY m j,\u2113. j,\u2113 j,\u2113 (12.57) (12.58) (12.59) Note that in order for the latter expression to be true, we might reasonably expect the vector spherical harmonics to be constructed out of sums of products of spherical harmonics and the eigenvectors of the operator Sz de\ufb01ned above. This is the vector analogue of constructing a spinor wavefunction in quantum theory. In addition, we normalize these orthogonal functions so that they are orthonormal as a dot product. This will allow us to use them to construct projections. Y m \u2217j,\u2113 (\u03b8, \u03c6) \u00b7 Z Y m\u2032 j\u2032,\u2113\u2032(\u03b8, \u03c6)d\u2126 = \u03b4jj\u2032\u03b4\u2113\u2113\u2032\u03b4mm\u2032 (12.60) We now need to derive the properties of these functions. We begin by applying J 2 to Y m j,\u2113 J 2Y m j,\u2113 = L2 + 2 + 2iL n \u00d7 o Y m j,\u2113 (12.61) so that we get 2iL \u00d7 Y m j,\u2113 = j(j + 1) { \u2212 \u2113(\u2113 + 1) Y m j,\u2113. 2 } \u2212 (12.62) Most of the later results will be based on this one, so understand it completely. If we take L \u00b7 L = iL we get: that L \u00d7 of both sides of (12.62), use a vector identity and recall [j(j + 1) \u2212 \u2113(\u2113 + 1)]L Y m j,\u2113 = 0. \u00b7 (12.63) Similarly, we form the vector product of L with both sides of (12.62): j(j + 1) { \u2212 \u2113(\u2113 + 1) L 2 } \u00d7 \u2212 Y m j,\u2113 = 2iL (L \u00d7 \u00d7 Y m j,\u2113). (12.64) To reduce this further,", " we must use the operator vector identity (which you should prove) (L L \u00d7 \u00d7 V) = L(L V) + iL \u00b7 V \u00d7 \u2212 L2V (12.65) and eliminate the L \u00d7 Y using (12.62). One gets: [j(j + 1) \u2212 \u2113(\u2113 + 1)] [j(j + 1) \u2113(\u2113 + 1) \u2212 \u2212 2] Y m j,\u2113 = 4\u2113(\u2113 + 1)Y m j,\u2113 \u2212 4L(L Y m j,\u2113. (12.66) \u00b7 If we eliminate the L Y (using the result above) we get the characteristic equation that is a constraint on the possible values of j and \u2113: \u00b7 where x3 \u2212 2x2 \u2212 4\u2113(\u2113 + 1)x = 0 x = j(j + 1) \u2113(\u2113 + 1) \u2212 (12.67) (12.68) by de\ufb01nition. The solutions to this factorizable cubic are: j = \u2113, \u2113 + 1, \u2113 1, \u2113 \u2212 \u2212 1, \u2113 \u2212 \u2212 \u2212 2, \u2113. \u2212 We only need to consider the solutions with positive j in this problem as the others are not independent in this case. Since \u2113 0 we only need consider the \ufb01rst three possibilities. \u2265 Solutions with j = \u2113 Then x = 0 and j(j + 1)Y m jj = L(L Y m jj) \u00b7 (12.69) from the third equation above. If we take the dot product of L with this relation, we get L2(L Y m jj) = j(j + 1)(L Y m jj) \u00b7 \u00b7 Y m jj \u221d and we thus see that L \u00b7 Yj,m and so: Y m jj = 1 j(j + 1) LYj,m (!) where we have normalized the result. q (12.70) (12.71) We have at last found something recognizable. This is precisely the combination of spherical harmonics and L we found in our brief excursion into multipoles! We see that we could have written the (e. g.) magnetic solution as E(M ) L B(M ) L = g\u2113(kr) \u2113(\u2113 + 1)Y m \u2113\u2113 = i \u03c9 \u2212 \u2207 q \u00d7 E(M ) L. (12.72) (12.73) With just a little more work (later) we will be able to obtain the curl part as a general result, which will really simplify life for us. It is a trivial exercise (left for the reader) to verify that JzY m jj = mY m jj. (12.74) One simply plugs in the explicit form of Jz and commutes the resultant Lz with L to cancel the \u201cspin\u201d part. Solutions with j = \u2113 If j = \u2113, we see from the equation after (12.62) that L Y = 0. To go further we have to go back to (12.62) and follow a di\ufb00erent line. If we multiply both sides by \u02c6r \u00b7 and \u02c6r \u00b7, \u00d7 \u2113(\u2113 + 1) 2] \u02c6r \u00b7 \u2212 Y m j\u2113 = 2i\u02c6r L \u00b7 \u00d7 Y m j\u2113 [j(j + 1) \u2212 and [j(j + 1) \u2113(\u2113 + 1) \u2212 2] \u02c6r \u00d7 \u2212 Y m j\u2113 = 2i\u02c6r (L \u00d7 \u00d7 Y m j\u2113) (12.75) (12.76) 6 6 We can reduce these with the vector identities (L \u02c6r \u00b7 \u00d7 A) = 2i\u02c6r A L \u00b7 \u2212 \u00b7 (\u02c6r \u00d7 A) (L \u02c6r \u00d7 \u00d7 A) = L(\u02c6r A) + i\u02c6r A. \u00d7 \u00b7 and You should get [j(j + 1) \u2212 \u2113(\u2113 + 1) + 2] \u02c6r Y m j\u2113 = \u00b7 2iL \u2212 (\u02c6r \u00b7 \u00d7 Y m j\u2113) and (12.77) (12.78) (12.79) [j(j + 1) \u2113(\u2113 + 1)] \u02c6r Y m j\u2113 = 2iL(\u02c6r Y m j\u2113). \u00d7 Finally, if we plug the", " second of these into the \ufb01rst and eliminate the cross product, we get the scalar equation: \u2212 \u00b7 (12.80) 1 4 [j(j + 1) \u2212 \u2113(\u2113 + 1)] [j(j + 1) \u2113(\u2113 + 1) + 2] (\u02c6r \u2212 Y m j\u2113) = L2(\u02c6r \u00b7 \u00b7 Y m j\u2113). (12.81) This implies that (\u02c6r Y m j\u2113) is a spherical harmonic: that is a constant Yk,m. What? This is not obvious to you? Well, just this once: \u00b7 \u00d7 j(j + 1) \u2212 2 \" \u2113(\u2113 + 1) j(j + 1) # \" \u2113(\u2113 + 1) \u2212 2 + 1 # = k(k + 1) (12.82) This has the solutions 1. k = j(j+1) 2. k = j(j+1) h h \u2113(\u2113+1) \u2113(\u2113+1) \u2212 2 \u2212 2 i i 1. \u2212 Since we already know that j = \u2113 1, we can investigate these two cases \u00b1 explicitly. The positive solutions (in both cases) are easily seen to be k = j. We can then construct the complete solutions, since Y m j,\u2113 = \u02c6r(\u02c6r Y m j,\u2113) \u00b7 \u02c6r \u00d7 \u2212 (\u02c6r \u00d7 Y m j,\u2113) (12.83) is an identity (related to the symmetric/antisymmetric decomposition and hence worth proving) and since we have already shown that Y m j,\u2113 = 2i [j(j + 1) \u02c6r \u00d7 \u2212 \u2113(\u2113 + 1)]\u2212 1 L(\u02c6r Y m j,\u2113) \u00b7 (12.84) with (\u02c6r \u00b7 Y m j,\u2113) a constant times Y\u2113,m. We get: Y m j,\u2113 = (constant) \u02c6r n 2i[j(j + 1) \u2212 \u2212 \u2113(\u2113 + 1)]\u2212 1(\u02c6r \u00d7 L) Y\u2113,m. (12.85) o An exercise will be to verify the normalization of the \ufb01nal solutions: 1 j(2j + 1) 1 Y m j,j 1 = \u2212 Y m j,j+1 = \u2212 \u2212 q q j \u02c6r + i\u02c6r [ \u2212 \u00d7 L]Y\u2113,m (12.86) (j + 1)(2j + 1) [(j + 1)\u02c6r + i\u02c6r \u00d7 L]Y\u2113,m. (12.87) You must also verify that they satisfy the equation for Jz. Finally, you are probably wondering why we have bothered to learn all = \u2113 cases in the \ufb01rst place. It is because of this about the j i\u2207 \u00d7 (Y m jjf (r)) = j + 1 2j + 1 s + (j + 1) f r + h s j 2j + 1 \u2212 h df dr j f r i + Y m j,j 1 \u2212 Y m j,j+1. (12.88) df dr i The action of the curl mixes the vector spherical harmonics. In fact, it acts to shift j by one in any permitted direction (see handout sheet). Therefore, in order to evaluate the entire EM \ufb01eld and express it compactly, one must use the notation of the vector spherical harmonics. You should prove this, and at leat one of the divergence equations for homework. You will need to get the components of the v.s.h. along and transverse to \u02c6r in order to do the vector algebra. This is not too bad, but (as we shall see) it is not the best we can do. By carefully de\ufb01ning a particular set of multipolar solutions, we can make our notation itself do almost all the work of doing the curls, etc. so that all we have to do at either end is translate a paticular problem into and out of the notation with the formal solution in hand. Next time we will do just that as we develop the Hansen Multipolar Solutions. 6 Chapter 13 The Hansen Multipoles We have already seen how if we let E or B be given by E or B = 1 \u2113(\u2113 + 1) q f\u2113(kr)LYL(\u02c6r) (13.1) then 1", ". Both the \ufb01elds given above and their partner \ufb01elds (given by the curl) have zero divergence. 2. The \ufb01elds given above are completely transverse, since \u02c6r erator). L = 0 (op- \u00b7 3. The partner \ufb01elds given by the curl are not purely transverse. 4. In order to be consistent, the \ufb01elds above are also the curls of the partner \ufb01elds. In fact, this follows from vector identities for divergenceless \ufb01elds. It is therefore sensible to de\ufb01ne, once and for all, a set of multipoles that embody these properties. In addition, anticipating a need to treat longitudinal \ufb01elds as well as transverse \ufb01elds, we will de\ufb01ne a third kind of multipoles with zero curl but non\u2013zero divergence. These will necessarily be \u201cconnected\u201d to sources (why?). We will call these \u201cpre-computed\u201d combinations of bessel functions, vector spherical harmonics, and their curls the Hansen Multipoles (following unpublished notes from L. C. Biedenharn as I have been unable to determine his original reference): 159 13.1 The Hansen Multipoles 13.1.1 The Basic Solutions ll. It is (by construction) purely transverse: \u02c6r The Hansen solutions to the vector HHE (that can expand the free space solutions for the vector potential or vector \ufb01elds) are as follows. M L is the (normalized) elementary solution consisting of a bessel function times LYL = Y m M L = 0. N L is the solution constructed by the taking the curl of M L. LL is the \u201clongitudinal\u201d solution constructed by taking the gradient of the scalar solution \u2013 it is left as an exercise to show that this still satis\ufb01es the HHE. The three of these pieces span the range of possible solutions and reconstruct an identity tensor that can be used to construct a vector harmonic Green\u2019s function expansion. \u00b7 This is summarized, with correction for factors of k introduced by the derivatives, here: 1 \u2113(\u2113 + 1) L (f\u2113(kr)YL(\u02c6r)) = M L = NL = \u2207 q i k i k ML \u00d7 \u2207(f\u2113(kr)YL(\u02c6r)) LL = \u2212 1 \u2113(\u2113 + 1) q f\u2113(kr)Y m ll (\u02c6r) (13.2) (13.3) (13.4) 13.1.2 Their Signi\ufb01cant Properties The virtue of the Hansen solutions is that they \u201cautomatically\u201d work to decompose \ufb01eld components into parts that are mutual curls (as required by Faraday/Ampere\u2019s laws for the \ufb01elds) or divergences (as required by Gauss\u2019s laws for the \ufb01elds): \u2207 \u00b7 \u2207 \u00b7 \u2207 M L = 0 N L = 0 LL = ikf\u2113(kr)YL(\u02c6r) \u00b7 (13.5) (13.6) (13.7) Hence M L and N L are divergenceless, while the divergence of LL is a scalar solution to the HHE! LL is related to the scalar \ufb01eld and the gauge invariance of the theory in an interesting way we will develop. Also: \u2207 \u2207 \u00d7 \u00d7 ikN L \u2212 M L = N L = ikM L LL = 0 \u2207 \u00d7 (13.8) (13.9) (13.10) which shows how M L and N L are now ideally suited to form the components of electric and magnetic multipole \ufb01elds mutually linked by Ampere\u2019s and Faraday\u2019s law. 13.1.3 Explicit Forms The beauty of the de\ufb01nitions above is that they permit us to do algebra that initially skips the following fully expanded forms in terms of the vector spherical harmonics. However ultimately one has to do computations, of course \u2013 there are no free lunches. The following results come from actually working out the gradients, divergences, and curls in the de\ufb01nitions: M L = f\u2113(kr)Y m \u2113\u2113 (13.11) N L = LL = \u2113 + 1 2\u2113 + 1 \u2113 2\u2113 + 1 f\u2113 \u2212 f\u2113 \u2212 s s 1(kr)Y m \u2113,\u2113", " \u2212 1 \u2212 s 1(kr)Y m \u2113,\u2113 \u2212 1 + s \u2113 2\u2113 + 1 \u2113 + 1 2\u2113 + 1 f\u2113+1(kr)Y m \u2113,\u2113+1 (13.12) f\u2113+1(kr)Y m \u2113,\u2113+1 (13.13) or (in di\ufb00erential form) M L = f\u2113(kr)Y m \u2113\u2113 N L = 1 kr d d(kr) (krf\u2113)(i\u02c6r Y m \u2113\u2113) \u00d7 \u2212 LL = \u2113(\u2113 + 1) q 1 kr (i\u02c6r \u00d7 f\u2113Y m \u2113\u2113) \u02c6r \u2212 (13.14) \u02c6r \u2113(\u2113 + 1)f\u2113YL (13.15) q d d(kr) f\u2113 YL (13.16) As we will see, these relations allow us to construct the completely general solution to the EM \ufb01eld equations in a way that is intuitive, reasonable, and mathematically and numerically tractible. In other words, we\u2019re (mostly) done with the grunt work and can begin to reap the rewards. What grunt work remains, you might ask? Well, there are a slew of identities and evaluations and relations developed from the de\ufb01nitions of the spherical harmonics themselves, the spherical bessel/neumann/hankel functions themselves, and the vector spherical harmonics and Hansen solutions that can be worked out and assembled in a table of sorts to simplify the actual process of doing algebra or computations using them. Such a table is presented at the end of this chapter, and proving relations on that table constitute most of the homework related to the chapter, since once this work is done doing actual computations for speci\ufb01c charge/current densities is reduced to quadratures (another way of saying \u201cexpressible as a bunch of de\ufb01nite integrals\u201d that can either be done analytically if they are relatively simple or numerically if not). Those rewards are most readily apparent when we construct the vector Green\u2019s function for the vector IHE. 13.2 Green\u2019s Functions for the Vector Helmholtz Equation The correct form for the Green\u2019s function for the vector Helmholtz equation is (where G \u00b1 (r, r\u2032) is a Green\u2019s function for the scalar IHE, that is: \u21d4G \u00b1 (r, r\u2032) =\u21d4I G (r, r\u2032) \u00b1 (r, r\u2032) = G \u00b1 ikR e\u00b1 4\u03c0R \u2212 (13.17) (13.18) r r\u2032 \u2212 for R =. The identity tensor transforms a vector on the right into | | the same vector, so this seems like a trivial de\ufb01nition. However, the point is that we can now expand the identity tensor times the scalar Green\u2019s function in vector spherical harmonics or Hansen functions directly! We get: \u21d4G \u00b1 (r, r\u2032) = = ik \u2213 ik \u2213 h\u00b1\u2113 (kr>)j\u2113(kr<)Y m j\u2113(\u02c6r)Y m \u2217j\u2113 (\u02c6r\u2032) Xj,\u2113,m M+ L (r>)M0 \u2217L (r<) + N+ L (r>)N0 \u2217L (r<) + XL L+ L (r>)L0 \u2217L (r<) (13.19) In all cases the \u201c*\u201ds are to be considered sliding, able to apply to the Y m only of either term under an integral. jl (\u02c6r) I do not intend to prove a key element of this assertion (that the products of the Y m jl (\u02c6r) involved reduce to Legendre polynomials in the angle between the arguments times the identity tensor) in class. Instead, I leave it as an exercise. To get you started, consider how similar completeness/addition theorems are proven for the spherical harmonics themselves from the given orthonormality relation. With these relations in hand, we end our mathematical digression into vector spherical harmonics and the Hansen solutions and return to the land of multipolar radiation. 13.3 Multipolar Radiation, revisited We will now, at long last, study the complete radiation \ufb01eld including the scalar, longitudinal, and transverse parts. Recall that we wish to solve", " the two equations (in the Lorentz gauge): 2 + k2 {\u2207 \u03a6(x) = } A(x) = } (r) \u03c1 \u01eb0 \u00b50J (x) \u2212 \u2212 2 + k2 {\u2207 with the Lorentz condition: \u2207 \u00b7 A + 1 c2 \u2202\u03a6 \u2202t = 0 (13.20) (13.21) (13.22) which is connected (as we shall see) to the continuity equation for charge and current. E and B are now (as usual) determined from the vector potential by the full relations, i. e. \u2013 we make no assumption that we are outside the region of sources: E = \u2212 B = \u2207 \u2207\u03a6 \u2212 A, \u00d7 \u2202A \u2202t (13.23) (13.24) Using the methods discussed before (writing the solution as an integral equation, breaking the integral up into the interior and exterior of the sphere of radius r, and using the correct order of the multipolar expansion of the Green\u2019s function in the interior and exterior regions) we can easily show that the general solution to the IHE\u2019s above is: \u03a6(r) = ik L (r)JL(r) + pint pext L (r)H + L (r) o where XL n pext L (r) = pint L (r) = r Z r 0 Z \u221e h+ \u2113 (kr\u2032)Y \u2217L (\u02c6r\u2032)\u03c1(r\u2032)d3r\u2032 j\u2113(kr\u2032)Y \u2217L (\u02c6r\u2032)\u03c1(r\u2032)d3r\u2032 (13.25) (13.26) (13.27) Outside the (bounding sphere of the) source, the exterior coe\ufb03cient is ) \u221e zero and the interior coe\ufb03cient is the scalar multipole moment pL = pint L ( of the charge source distribution, so that: \u03a6(r) = pLH + L (r) (13.28) XL This is an important relation and will play an signi\ufb01cant role in the implementation of the gauge condition below. ik \u01eb0 Similarly we can write the interior and exterior multipolar moments of the current in terms of integrals over the various Hansen functions to obtain a completely general expression for the vector potential A(r). To simplify matters, I am going to only write down the solution obtained outside the current density distribution, although the integration volume can easily be split into r< and r> pieces as above and an exact solution obtained on all space including inside the charge distribution. It is: A(r) = ik\u00b50 mLM + L (r) + nLN + L (r) + lLLL(r) o where XL n mL = J (r\u2032) nL = lL = Z Z Z J (r\u2032) J (r\u2032) M 0 L(r\u2032)\u2217d3r\u2032 N 0 L(r\u2032)\u2217d3r\u2032 L0 L(r\u2032)\u2217d3r\u2032 \u00b7 \u00b7 \u00b7 (13.29) (13.30) (13.31) (13.32) Note well that the action of the dot product within the dyadic form for the Green\u2019s function (expanded in Hansen solutions) reduces the dyadic tensor to a vector again. It turns out that these four sets of numbers: pL, mL, nL, lL are not independent. They are related by the requirement that the solutions satisfy the Lorentz gauge condition, which is a constraint on the admissible solutions. If we substitute these forms into the gauge condition itself and use the di\ufb00erential relations given above for the Hansen functions to simplify the results, we obtain: \u2207 \u00b7 A + \u2202\u03a6 1 c2 \u2202t pLH + L i\u03c9 c2\u01eb0 ikcpLH + L = 0 = 0 = 0 ik \u00b50lL\u2207 \u00b7 lL\u2207 XL ik XL n k2 \u2212 L+ L \u2212 L+ \u00b7 XL { L \u2212 lL \u2212 o H + L = 0 cpL} (13.33) where we used \u2207 L in the last step. If we multiply from the left by Y \u2217\u2113\u2032,m\u2032 and use the fact that the YL form a complete orthonormal set, we \ufb01nd the relation: L = ikH + L+ \u00b7 or lL \u2212 cpL = 0 lL = cpL (13.34) (13.35) This tells us that the e\ufb00ect of the scalar moments and the longitudinal moments are connected by the gauge", " condition. Instead of four relevant moments we have at most three. In fact, as we will see below, we have only two! Recall that the potentials are not unique \u2013 they can and do vary according to the gauge chosen. The \ufb01elds, however, must be unique or we\u2019d get di\ufb00erent experimental results in di\ufb00erent gauges. This would obviously be a problem! Let us therefore calculate the \ufb01elds. There are two ways to proceed. We can compute B directly from vA: B = \u2207 \u00d7 = ik\u00b50 A ml(\u2207 XL n M + L ) + nl(\u2207 N + L ) + ll(\u2207 \u00d7 \u00d7 L+ L ) o \u00d7 ikN + L ) + nl(ikM + L ) o = ik\u00b50 = k2\u00b50 ml( \u2212 mLN + XL n XL n L \u2212 nLM + L o (13.36) and use Ampere\u2019s Law, \u2207 B = \u00b50\u01eb0 \u2202E \u2202t = \u00d7 i\u03c9\u00b50\u01eb0E to \ufb01nd E: \u2212 E = \u2207 ic2 kc = ikc\u00b50 B \u00d7 mL(\u2207 \u00b50 XL n 1 \u00b50\u01eb0 \u00b50 \u01eb0 = ik s = = k2 s k2Z0 \u2212 \u2212 XL n XL n N + L ) \u00d7 \u2212 nL(\u2207 M + L ) o \u00d7 mL(ikM + L ) XL n \u2212 nL( ikN + L ) o \u2212 mLM + L + nLN + L mLM + L + nLN + L o. o (13.37) where Z0 = \u00b50 \u01eb0 is the usual impedance of free space, around 377 ohms. q Wow! Recall that the M waves are transverse, so the mL and nL are the magnetic (transverse electric) and electric (transverse magnetic) multipole moments respectively. The \ufb01eld outside of the source is a pure expansion in elementary transverse multipoles. (Later we will show that the (approximate) de\ufb01nitions we have used to date as \u201dmultipoles\u201d are the limiting forms of these exact de\ufb01nitions.) Note well that the actual \ufb01elds require only two of the basic hansen solutions \u2013 the two that are mutually transverse. Something happened to the longitudinal part and the dependence of the \ufb01eld on the scalar potential. To see just what, let us re-evaluate the electric \ufb01eld from: E = = = = \u2207\u03a6 \u2212 \u2207 \u2212 \u2212 ik \u01eb0 \u2202A \u2202t pLH + L! XL pL(\u2207H + L ) ik \u01eb0 \u2212 k2 \u01eb0 XL ( pL \u2212 XL ikc\u00b50\u01eb0 \u2212 1 c lL L+ L \u2212 k2Z0 + i\u03c9 ik\u00b50 mLM + L (r) + nLN + XL n lLL+ L ) \u2212 k2\u00b50c XL XL n mlM + L + nLN + L XL n o L (r) + lLLL(r) o! mlM + L + nLN + L o (13.38) (Note that we used \u03c9 = kc and \u2207H + gauge condition: L = ikL+ L.) From this we see that if the lL = cpL (13.39) is satis\ufb01ed, the scalar and longitudinal vector parts of the electric \ufb01eld cancel exactly! All that survives are the transverse parts: E = k2Z0 \u2212 mLM + L + nLN + L XL n o (13.40) as before. The Lorentz gauge condition is thus intimately connected to the vanishing of a scalar or longitudinal contribution to the E \ufb01eld! Also note that the magnitude of E is greater than that of B by c, the velocity of light. Now, we are interested (as usual) mostly in obtaining the \ufb01elds in the far zone, where this already simple expression attains a clean asymptotic form. Using the kr form of the hankel function, \u2192 \u221e h+ \u2113 (kr) = lim kr \u2192\u221e eikr (\u2113+1) i\u03c0 2 \u2212 kr (13.41) we obtain the limiting forms (for kr ): \u2192 \u221e M + L \u223c eikr (\u2113+1) i\u03c0 2 \u2212 kr Y m \u2113\ufffd", "\ufffd\ufffd = ( i)\u2113+1 eikr kr \u2212 Y m \u2113\u2113 (13.42) N + L \u223c eikr \u2113 i\u03c0 2 \u2212 kr \u2113 + 1 2\u2113 + 1 \uf8ee s Y m \u2113,\u2113 \u2212 1 + s \u2113 2\u2113 + 1 Y m \u2113,\u2113+1\uf8f9 (13.43) The bracket in the second equation can be simpli\ufb01ed, using the results \uf8f0 \uf8fb of the table I handed out previously. Note that \u2113 + 1 2\u2113 + 1 \uf8ee s Y m \u2113,\u2113 \u2212 1 + s \u2113 2\u2113 + 1 Y m \u2113,\u2113+1\uf8f9 = i(\u02c6r \u00d7 Y m \u2113\u2113) = i \u03c0 2 (\u02c6r e\u2212 \u2212 \u00d7 Y m \u2113\u2113) (13.44) \uf8f0 so that (still in the far zone limit) \uf8fb N + L \u223c \u2212 eikr (\u2113+1) i\u03c0 2 \u2212 kr (\u02c6r \u00d7 Y m \u2113\u2113) = i)\u2113+1 eikr ( kr \u2212 \u2212 (\u02c6r \u00d7 Y m \u2113\u2113). (13.45) Let us pause to admire this result before moseying on. This is just B = k2\u00b50 \u2212 eikr kr XL i)\u2113+1 ( \u2212 mL (\u02c6r { \u00d7 Y m \u2113\u2113) + nLY m \u2113\u2113} = k\u00b50 \u2212 i)\u2113+1 ( \u2212 { mL (\u02c6r Y m \u2113\u2113) + nLY m \u2113\u2113} \u00d7 (13.46) E = k2Z0 \u2212 i)\u2113+1 ( \u2212 mLY m { \u2113\u2113 \u2212 nL (\u02c6r Y m \u2113\u2113) } \u00d7 = kZ0 \u2212 i)\u2113+1 mLY m { \u2113\u2113 \u2212 nL (\u02c6r Y m \u2113\u2113) }. \u00d7 (13.47) eikr r eikr kr eikr r XL XL ( \u2212 XL If I have made a small error at this point, forgive me. Correct me, too. This is a purely transverse outgoing spherical wave whose vector character is \ufb01nally translucent, if not transparent. The power \ufb02ux in the outgoing wave is still not too easy to express, but it is a damn sight easier than it was before. At least we have the satisfaction of knowing that we can express it as a general result. Recalling (as usual) S = 1 2 Re(E H \u2217) \u00d7 (13.48) and that the power distribution is related to the \ufb02ux of the Poynting vector through a surface at distance r in a di\ufb00erential solid angle d\u2126: dP d\u2126 = 1 2 Re[r2 \u02c6n (E \u00b7 \u00d7 H \u2217)] we get S = k2 2r2 Z0Re \u00d7 i\u2113\u2032 \u2212 \" XL XL\u2032 \u02c6r m\u2217L\u2032 n \u00d7 \u2113 nL (\u02c6r mLY m { Y m\u2032 \u2217\u2113\u2032\u2113\u2032 \u2113\u2113 \u2212 + n\u2217L\u2032Y m\u2032 \u2217\u2113\u2032\u2113\u2032 oi (13.49) (13.50) Y m \u2113\u2113) } \u00d7 (Note: Units here need to be rechecked, but they appear to be consistent at \ufb01rst glance). This is an extremely complicated result, but it has to be, since it expresses the most general possible angular distribution of radiation (in the far zone). The power distribution follows trivially. We can, however, evaluate the total power radiated, which is a very useful number. This will be an exercise. You will need the results d2\u2126\u02c6r Y m \u2113\u2113 \u00d7 \u02c6r \u00d7 \u00b7 Z Y m\u2032 \u2217\u2113\u2032\u2113\u2032 = d2\u2126Y m \u2113\u2113 \u00b7 Y m\u2032 \u2217\u2113\u2032\u2113\u2032 Z = \u03b4\u2113\u2113\u2032\u03b4mm\u2032 (13.51) and d2\u2126\u02c6r Y m Y m\u2032 \u2217\u2113", "\u2032\u2113\u2032 = 0 \u2113\u2113 \u00d7 to evaluate typical terms. Using these relations, it is not too di\ufb03cult to show that \u00b7 Z (13.52) P = Z0 k2 2 2 + mL | 2 nL | | o (13.53) | XL n which is the sum of the power emitted from all the individual multipoles (there is no interference between multipoles!). Let us examine e.g. the electric multipolar moment nL to see how it 0 compares to the usual static results. Static results are obtained in the k (long wavelength) limit. In this limit e.g. j\u2113(kr) k\u2113r\u2113 and: \u2192 \u223c nL \u2248 ic s \u2113 + 1 \u2113 k\u2113 (2\u2113 + 1)! Z \u03c1(r)r\u2113Y\u2113,m(\u02c6r)d3r (13.54) The dipole term comes from \u2113 = 1. For a simple dipole: n1,m \u2248 \u03c1rY1,md3r 3 4\u03c0 e < r > e < r > ic k \u221a2 3 kc\u221a2 Z 3 s kc\u221a6 36\u03c0 ie \u221a6\u03c0\u03c9 i i \u2248 \u2248 \u2248 \u2212 < \u00a8r > (13.55) where we use < \u00a8r >= \u03c92 < r >. \u2212 In terms of this the average power radiated by a single electron dipole is: P = 1 2 e2 2 \u00a8r 6\u03c0\u01eb0c3! | | which compares well with the Larmor Formula: P = 2 3 e2 2 \u00a8r 4\u03c0\u01eb0c3! | | (13.56) (13.57) The latter is the formula for the instantaneous power radiated from a point charge as it is accelerated. Either \ufb02avor is the death knell of classical mechanics \u2013 it is very di\ufb03cult to build a model for a stable atom based on classical trajectories of an electron around a nucleus that does not involve acceleration of the electron in question. While it is not easy to see, the results above are essentially those obtained in Jackson (J9.155) except that (comparing e.g. J9.119, J9.122, and J91.165 to related results above) Jackson\u2019s aE,M (\u2113, m) moments di\ufb00er from the Hansen multipolar moments by factors of several powers of k. If one works hard enough, though, one can show that the results are identical, and even though Jackson\u2019s algebra is more than a bit Evil it is worthwhile to do this if only to validate the results above (where recall there has been a unit conversion and hence they do need validation). Another useful exercise is to recover our old friends, the dipole and quadrupole radiation terms of J9 from the exact de\ufb01nition of their respective moments. One must make the long wavelength approximation under the integral in the de\ufb01nition of the multipole moments, integrate by parts liberally, and use the continuity equation. This is quite di\ufb03cult, as it turns out, unless you have seen it before, so let us look at an example. Let us apply the methods we have developed above to obtain the radiation pattern of a dipole antenna, this time without assuming that it\u2019s length is small w.r.t. a wavelength. Jackson solves more or less the same problem in his section 9.12, so this will permit the direct comparison of the coe\ufb03cients and constants in the \ufb01nal expressions for total radiated power or the angular distribution of power. 13.4 A Linear Center-Fed Half-Wave Antenna Suppose we are given a center-fed dipole antenna with length \u03bb/2 (half-wave antenna). We will assume further that the antenna is aligned with the z axis and centered on the origin, with a current given by: I = I0 cos(\u03c9t) cos 2\u03c0z \u03bb (13.58) Note that in \u201creal life\u201d it is not easy to arrange for a given current because the current instantaneously depends on the \u201cresistance\u201d which is a function of the radiation \ufb01eld itself. The current itself thus comes out of the solution of an extremely complicated boundary value problem. For atomic or nuclear radiation, however, the \u201ccurrents\u201d are generally matrix elements associated with transitions and hence are known. In any event, the current density corresponding to this current is J = \u02c6zI0 cos 2\u03c0r \u03bb \u03b4(1 cos \u03b8 \u2212 | 2\u03c0r2 sin \u03b8 ) | J =", " 0 (13.59) (13.60) for r \u2264 \u03bb/4 and for r > \u03bb/4. When we use the Hansen multipoles, there is little incentive to convert this into a form where we integrate against the charge density in the antenna. Instead we can easily and directly calculate the multipole moments. The magnetic moment is mL = J \u00b7 0 Z I0 2\u03c0 Z = M 0 \u2217L d3r \u03bb/4 2\u03c0 0 Z cos(kr)j\u2113(kr) \u02c6z { \u00b7 Y m \u2217\u2113\u2113 (0, \u03c6) + \u02c6z Y m \u2217\u2113\u2113 (\u03c0, \u03c6) } \u00b7 d\u03c6 dr (13.61) (where we have done the integral over \u03b8). Now, Y m \u2113\u2113 = \u02c6z \u00b7 1 \u2113(\u2113 + 1) mYL q (Why? Consider (\u02c6z L)YL)...) and yet \u00b7 YL(0, \u03c6) = \u03b4m0 2\u2113 + 1 1/2 \" 4\u03c0 # YL(\u03c0, \u03c6) = ( 1)\u2113\u03b4m0 \u2212 2\u2113 + 1 \" 4\u03c0 # 1/2. (13.62) (13.63) (13.64) Consequently, we can conclude (m\u03b4m0 = 0) that mL = 0. (13.65) All magnetic multipole moments of this linear dipole vanish. Since the magnetic multipoles should be connected to the rotational part of the current density (which is zero for linear \ufb02ow) this should not surprise you. The electric moments are nL = J N 0 \u2217L d3r \u00b7 = Z I0 2\u03c0 Z 2\u03c0 \u03bb/4 0 Z cos(kr) \u2113 + 1 2\u2113 + 1 j\u2113 \u2212 1(kr) \u02c6z h s \uf8f1 \uf8f2 Y m \u2217\u2113,\u2113 \u2212 \u00b7 1(0, \u03c6) + \u02c6z \u00b7 0 \u2113 2\u2113 + 1 \u2212 s j\u2113+1(kr) \uf8f3 \u02c6z \u00b7 h Y m \u2217\u2113,\u2113+1(0, \u03c6) + \u02c6z Y m \u2217\u2113,\u2113 \u2212 \u00b7 1(\u03c0, \u03c6). \uf8fc \uf8fd i Y m \u2217\u2113,\u2113+1(\u03c0, \u03c6) d\u03c6 dr i (13.66) If we look up the de\ufb01nition of the v.s.h.\u2019s on the handout table, the z components are given by: \uf8fe \u02c6z Y m \u2217\u2113,\u2113 \u2212 \u00b7 1(0, \u03c6) = \u03b4m0 \u2113 4\u03c0 s \u02c6z Y m \u2217\u2113,\u2113 \u2212 \u00b7 1(\u03c0, \u03c6) = ( 1)\u2113 \u2212 \u2212 1\u03b4m0 \u2113 4\u03c0 s Y m \u2217\u2113,\u2113+1(0, \u03c6) = \u03b4m0 \u2212 s \u2113 + 1 4\u03c0 Y m \u2217\u2113,\u2113+1(\u03c0, \u03c6) = 1)\u2113 \u2212 ( \u2212 \u2212 1\u03b4m0 \u2113 + 1 4\u03c0 s \u02c6z \u00b7 \u02c6z \u00b7 so the electric multipole moments vanish for m = 0, and (13.67) (13.68) (13.69) (13.70) n\u2113,0 = I0\u03b4m0v u u t \u2113(\u2113 + 1) 4\u03c0(2\u2113 + 1) 1 + ( 1)\u2113+1 \u2212 \u03bb/4 cos(kr) j\u2113 h \u2212 0 Z 1(kr) + j\u2113+1(kr) dr. i (13.71) Examining this equation, we see that all the even \u2113 terms vanish! However, all the odd \u2113, m = 0 terms do not vanish, so we can\u2019t quit yet. We use the following relations: 1 + j\u2113+1 = j\u2113 \u2212 2\u2113 + 1 kr j\u2113 (the fundamental recursion relation), n0(kr) = cos(kr) kr \u2212 (true fact) and (13.72) (13.73) dz f\u2113(z)g\u2113\u2032(z) = Z z", "2 [\u2113\u2032(\u2113\u2032 + 1) \u2113(\u2113 + 1)] \u2212 f \u2032\u2113g\u2113\u2032 f\u2113g\u2032\u2113\u2032 \u2212 (13.74) 6 for any two spherical bessel type functions (a valuable thing to know that follows from integration by parts and the recursion relation). From these we get n\u2113,0 = \u03c0I0 2k \u03b4m0 2\u2113 + 1 4\u03c0\u2113(\u2113 + 1) s 1 + ( 1)\u2113+1 \u2212 j\u2113(\u03c0/2). (13.75) Naturally, there is a wee tad of algebra involved here that I have skipped. You shouldn\u2019t. Now, let\u2019s \ufb01gure out the power radiated from this source. Recall from above that: 2 + mL | 2 nL | | o P = = = k2 2 s k2 2 s \u03c0I 2 0 8 s | XL n \u00b50 \u01eb0 \u00b50 \u01eb0 X\u2113 odd | \u00b50 \u01eb0 X\u2113 odd 2 n\u2113,0| 2\u2113 + 1 \u2113(\u2113 + 1)! [j\u2113(\u03c0/2)]2 (13.76) Now this also equals (recall) 1 2I 2 0 Rrad, from which we can \ufb01nd the radia- tion resistance of the half wave antenna: Rrad = \u03c0 4 s \u00b50 \u01eb0 X\u2113 odd 2\u2113 + 1 \u2113(\u2113 + 1)! [j\u2113(\u03c0/2)]2. (13.77) We are blessed by this having manifest units of resistance, as we recognize our old friend Z0 = 377\u2126 (the impedance of free space) and a bunch of dimensionless numbers! In terms of this: q \u00b50 \u01eb0 \u2248 Rrad = Z0 \u03c0 4 X\u2113 odd 2\u2113 + 1 \u2113(\u2113 + 1)! [j\u2113(\u03c0/2)]2.! (13.78) We can obtain a good estimate of the magnitude by evaluating the \ufb01rst few terms. Noting that j1(\u03c0/2) = j3(\u03c0/2) = 2 2 2 \u03c0 2 \u03c0 60 \u03c02 \u2212 6 (13.79) (13.80) and doing some arithmetic, you should be able to show that Rrad = 73.1\u2126. Note that the ratio of the \ufb01rst (dipole) term to the third (octupole) term is 2 n3 n1 = = 2 6 2 60 \u03c02 \u2212 7 12 7 18 2 3 60 \u03c02 \u2212 0.00244 6 \u2248 That means that this is likely to be a good approximation (the answer is very nearly unchanged by the inclusion of the extra term). Even if the length of the antenna is on the order of \u03bb, the multipole expansion is an extremely accurate and rapidly converging approximation. That is, after all, why we use it so much in all kinds of localized source wave theory. However, if we plug in the \u201clong wavelength\u201d approximation we previ- ously obtained for a short dipole antenna (with d = \u03bb/2) we get: Rrad = (kd)2 24\u03c0 s \u00b50 \u01eb0 \u2248 48\u2126 (13.81) which is o\ufb00 by close to a factor of 50%. This is not such a good result. Using this formula with a long wavelength approximation for the dipole moment (only) of n1,0 \u2248 80\u2126, still o\ufb00 by 11%. I0 k s 2 3\u03c0 yields Rrad \u2248 (13.82) 13.5 Connection to Old (Approximate) Mul- tipole Moments To conclude our discussion of multipole \ufb01elds, let us relate the multipole moments de\ufb01ned and used above (which are exact) to the \u201cusual\u201d static, long wavelength moments we deduced in our earlier studies. Well, nL = J \u00b7 Z N0 \u2217L d3r (13.83) and N L = 1 \u2113(\u2113 + 1) \u2207 1 k (r \u00d7 \u00d7 \u2207)(f\u2113(kr)YL(\u02c6r)) q = 1 \u2113(\u2113 + 1r + 1!# (f\u2113(kr)YL(\u02c6r)) (13.84) (using the vector identity r + 1!# (", "13.85) to simplify). Then nL = k q 1 \u2212 \u2113(\u2113 + 1) k2 Z (r \u00b7 J )j\u2113(kr)Y \u2217L (\u02c6r)d3r + \u2207) (J \u00b7 \" Y \u2217L (\u02c6r) \u2202 \u2202r (rj\u2113(kr)) # d3r ) Z Now, (from the continuity equation) \u2207 \u00b7 J = i\u03c9\u03c1 so when we (sigh) integrate the second term by parts, (by using \u2207 \u00b7 (aB) = B \u2207a + a\u2207 B \u00b7 \u00b7 (13.86) (13.87) (13.88) so that (J \u2207) \u00b7 \" Y \u2217L (\u02c6r) \u2202 \u2202r (rj\u2113(kr)) # = \u2207 J Y \u2217L (\u02c6r) \u00b7\" \u2202 \u2202r (rj\u2113(kr)) Y \u2217L (\u02c6r) #\u2212 \u2202 \u2202r (rj\u2113(kr)) [\u2207 J ] \u00b7 (13.89) and the divergence theorem on the \ufb01rst term, \u2207 \u00b7 \" J Y \u2217L (\u02c6r) \u2202 \u2202r ZV (rj\u2113(kr)) # dV = Z\u2202V = 0 J Y \u2217L (\u02c6r) \u02c6n \u00b7 \" \u2202 \u2202r (rj\u2113(kr)) # dA (13.90) \u2192\u221e for sources with compact support to do the integration) we get nL = k q 1 \u2212 \u2113(\u2113 + 1) k2 Z (r \u00b7 J )j\u2113(kr)Y \u2217L (\u02c6r)d3r \u2212 (i\u03c9\u03c1(r)) Y \u2217L (\u02c6r) \" \u2202 \u2202r (rj\u2113(kr)) # d3r ) \u03c1(r) Y \u2217L (\u02c6r) \" \u2202 \u2202r (rj\u2113(kr)) # d3r Z ic = \u2113(\u2113 + 1) Z k q J )j\u2113(kr)Y \u2217L (\u02c6r)d3r (r \u00b7 (13.91) \u2212 \u2113(\u2113 + 1) Z q The electric multipole moment thus consists of two terms. The \ufb01rst term appears to arise from oscillations of the charge density itself, and might be expected to correspond to our usual de\ufb01nition. The second term is the contribution to the radiation from the radial oscillation of the current density. (Note that it is the axial or transverse current density oscillations that give rise to the magnetic multipoles.) Only if the wavelength is much larger than the source is the second term of lesser order (by a factor of ik c ). In that case we can write nL \u2248 ic \u2113(\u2113 + 1) Z \u03c1Y \u2217L \u2202 \u2202r (rj\u2113(kr))d3r. (13.92) Finally, using the long wavelength approximation on the bessel functions, q nL \u2248 ic (2\u2113 + 1)!! s \u2113 + 1 \u2113 k\u2113 \u03c1r\u2113Y \u2217L d3r Z ic (2\u2113 + 1)!! s \u2113 + 1 \u2113 k\u2113q\u2113,m \u2248 (13.93) (13.94) and the connection with the static electric multipole moments q\u2113,m is complete. In a similar manner one can establish the long wavelength connection between the mL and the magnetic moments of earlier chapters. Also note well that the relationship is not equality. The \u201capproximate\u201d multipoles need to be renormalized in order to \ufb01t together properly with the Hansen functions to reconstruct the EM \ufb01eld. 13.6 Angular Momentum Flux Let us consider the angular momentum radiated away with the electromagnetic \ufb01eld. The angular momentum \ufb02ux density is basically vr crossed into the momentum density S/c or: L = 1 2 Re r \u00d7 (E \u00d7 c ( H \u2217) ) (13.95) Into this expression we must substitute our expressions for E and H: k2Z0 E = \u2212 H = k2 XL n mLM + L + nLN + L XL n mLN + L \u2212 o nLM + L. o (13.96) (13.97) E = H = we get: H \ufffd", "\ufffd = E \u00d7 = If we try to use the asymptotic far \ufb01eld results: kZ0 \u2212 eikr r eikr r k \u2212 XL ( \u2212 XL i)\u2113+1 ( \u2212 mLY m { \u2113\u2113 \u2212 nL (\u02c6r Y m \u2113\u2113) } \u00d7 i)\u2113+1 mL (\u02c6r { \u00d7 Y m \u2113\u2113) + nLY m \u2113\u2113} (13.98) (13.99) k2Z0 r2 \u2113\u2032 i\u2113 \u2212 mLY m { Y m\u2032 \u2113\u2113(\u02c6r) nL (\u02c6r \u2212 + n\u2217L\u2032Y m\u2032 Y m \u2113\u2113(\u02c6r)) } \u00d7 \u00d7 \u2217\u2113\u2032\u2113\u2032 (\u02c6r) o Y m\u2032 \u02c6r \u00d7 \u00d7 \u2217\u2113\u2032\u2113\u2032 (\u02c6r) \u2113\u2113(\u02c6r) XL XL\u2032 \u02c6r m\u2217L\u2032 n k2Z0 r2 \u00d7 \u2113\u2032 i\u2113 \u2212 \u2217\u2113\u2032\u2113\u2032 (\u02c6r) mLm\u2217L\u2032Y m XL XL\u2032 n + mLn\u2217L\u2032Y m \u2113\u2113(\u02c6r) nLm\u2217L\u2032 (\u02c6r Y m\u2032 \u00d7 Y m \u2113\u2113(\u02c6r)) \u2212 \u00d7 nLn\u2217L\u2032 (\u02c6r Y m \u2113\u2113(\u02c6r)) Y m\u2032 \u2217\u2113\u2032\u2113\u2032 (\u02c6r) \u02c6r \u00d7 Y m\u2032 \u2217\u2113\u2032\u2113\u2032 (\u02c6r) o \u00d7 \u00d7. \u2217\u2113\u2032\u2113\u2032 (\u02c6r) \u2212 \u00d7 With some e\ufb00ort this can be shown to be a radial result \u2013 the Poynting vector points directly away from the source in the far \ufb01eld to leading order. Consequently, this leading order behavior contributes nothing to the angular momentum \ufb02ux. We must keep at least the leading correction term to the asymptotic result. (13.100) It is convenient to use a radial/tangential decomposition of the Hansen M L = 0). For the solutions. The M L are completely tangential (recall r N L we have: \u00b7 1 kr d dr N L(r) = (rf\u2113(kr)) (i\u02c6r Y m \u2113\u2113(\u02c6r)) \u2212 Using our full expressions for E and H \u2217: \u00d7 \u02c6r q \u2113(\u2113 + 1) kr fell(kr)YL(\u02c6r) (13.101) k2Z0 E = \u2212 H = k2 XL n mLM + L + nLN + L XL n mLN + L \u2212 o nLM + L o (13.102) (13.103) with this form substituted for N L and the usual form for M L we get: L = 1 2 Re r \u00d7 (E \u00d7 c ( H \u2217) ) r mLh+ \u2113 (kr)Y m \u2113\u2113(\u02c6r) = k4Z0 2c \u2212 Re \u00d7 d(rh+ \u2113 (kr)) dr XL XL\u2032 1 kr d(rh\u2212\u2113\u2032 (kr)) dr \" 1 kr +nL m\u2217L\u2032 \u00d7 \" (i\u02c6r \u00d7 Y m \u2113\u2113(\u02c6r)) \u2212 \u02c6r \u2113(\u2113 + 1) i\u02c6r ( \u2212 \u00d7 Y m\u2032 \u2217\u2113\u2032\u2113\u2032 (\u02c6r)) \u2212 \u2113(\u2113 + 1) q \u02c6r q h+ \u2113 (kr) kr h\u2212\u2113\u2032 (kr) kr YL(\u02c6r) # Y \u2217L\u2032(\u02c6r) # +n\u2217L\u2032h\u2212\u2113\u2032 (kr)Y m\u2032 \u2217\u2113\u2032\u2113\u2032 (\u02c6r) (13.104) All the purely radial terms in the outermost under the sum do not contribute to the angular momentum \ufb02ux density. The surviving terms are: L = k4Z0 2c \u2212 Re r mLm\u2217L\u2032h+ \u2113 (kr) q \u00d7 XL XL\u2032 d(rh+ \u2113 (kr)) 1 kr dr h+ \u2113", " (kr) kr h+ \u2113 (kr) kr q ((i\u02c6r Y m \u2113\u2113(\u02c6r) \u00d7 \u00d7 d(rh\u2212\u2113\u2032 (kr)) dr 1 kr q i\u02c6r \u00d7 \u2113(\u2113 + 1) q \u2113(\u2113 + 1)h\u2212\u2113\u2032 (kr)(\u02c6r Y m\u2032 \u2217\u2113\u2032\u2113\u2032 (\u02c6r) \u00d7 +nLm\u2217L\u2032 nLm\u2217L\u2032 \u2212 nLn\u2217L\u2032 \u2212 \u2113\u2032(\u2113\u2032 + 1) h\u2212\u2113\u2032 (kr) kr (Y m \u2113\u2113(\u02c6r) \u00d7 \u02c6r)Y \u2217L\u2032(\u02c6r) \u02c6r) \u2113\u2032(\u2113\u2032 + 1) h\u2212\u2113\u2032 (kr) kr Y \u2217L\u2032(\u02c6r) (\u02c6r \u00d7 Y m\u2032 \u2217\u2113\u2032\u2113\u2032 (\u02c6r)) (13.105) The lowest order term in the asymptotic form for the spherical bessel functions makes a contribution in the above expressions. After untangling the cross products and substituting the asymptotic forms, we get: L = k\u00b50 2r2 Re mLmL\u2032 \u2113\u2032(\u2113\u2032 + 1)i\u2113\u2032 \u2113Y \u2217L\u2032(\u02c6r)Y m \u2113\u2113(\u02c6r) \u2212 XL XL\u2032 nLm\u2217L\u2032 \u2212 +nLm\u2217L\u2032 +nLn\u2217L\u2032 \u2113(\u2113 + 1)i\u2113\u2032 \u2113(\u2113 + 1)i\u2113\u2032 \u2212 \u2212 q q \u2113(\u2113 + 1)i\u2113\u2032 \u2212 q \u2113Y \u2217L\u2032(\u02c6r) (\u02c6r \u2113YL(\u02c6r) \u02c6r \u2113YL(\u02c6r)Y m\u2032 \u00d7 \u2217\u2113\u2032\u2113\u2032 (\u02c6r) \u00d7 Y m \u2113\u2113(\u02c6r)) Y m\u2032 \u2217\u2113\u2032\u2113\u2032 (\u02c6r) (13.106) q The angular momentum about a given axis emitted per unit time is obtained by selecting a particular component of this and integrating its \ufb02ux through a distant spherical surface. For example, for the z-component we \ufb01nd (noting that r2 cancels as it should): dLz dt = k\u00b50 2 Re \u02c6z XL XL\u2032 Z... sin(\u03b8)d\u03b8d\u03c6 (13.107) \u00b7 where the brackets indicate the expression above. We look up the components of the vector harmonics to let us do the dot product and \ufb01nd: Y m \u2113\u2113 = \u02c6z \u00b7 m \u2113(\u2113 + 1) Y\u2113,m (13.108) (\u02c6r \u02c6z \u00b7 \u00d7 Y m \u2113\u2113 2\u2113 + 1 \u02c6z \u00b7 Y m \u2113,\u2113 \u2212 1 + s \u2113 2\u2113 + 1 \u02c6z \u00b7 Y m \u2113\u2113+1\uf8f9 (\u2113 + 1)(\u21132 \u2113(2\u2113 m2) \u2212 1)(2\u2113 + 1) \u2212 Y\u2113 \u2212 1,m \u2212 v u u t \uf8fb [(\u2113 + 1)2 m2]\u2113 \u2212 (2\u2113 + 1)(2\u2113 + 3)(\u2113 + 1) Y\u2113+1,m\uf8f9 \uf8fb (13.109) Doing the integral is now simple, using the orthonormality of the spher- ical harmonics. One obtains (after still more work, of course): Compare this to: dLz dt = k\u00b50 2 m XL 2 + mL| | 2 nL| | P = k2 2 Z0 2 + mL | 2 nL | | o | XL n term by term. For example: dLz(mL) dt = = k\u00b50m 2 ( 2 k2\u00b50c P (mL) = 2 mL| | ) m \u03c9 P (mL) (13.110) (13.111) (13.112) (where m in the fraction is the spherical harmonic m, not the", " multipole mL). In other words, for a pure multipole the rate of angular momentum about any given axis transferred is m/\u03c9 times the rate of energy transferred, where m is the angular momentum aligned with that axis. (Note that if we chose some other axis we could, with enough work, \ufb01nd an answer, but the algebra is only simple along the z-axis as the multipoles were originally de\ufb01ned with their m-index referred to this axis. Alternatively we could rotate frames to align with the new direction and do the entire computation over.) This is quite profound. If we insist, for example, that energy be transferred in units of \u00afh\u03c9, then angular momentum is also transferred in units of m\u00afh! 13.7 Concluding Remarks About Multipoles There are still many, many things we could study concerning multipoles and radiation. For example, we have not yet done a magnetic loop antenna, but doing one should now be straightforward (to obtain a magnetic dipole radiation \ufb01eld to leading order). Hmmm, sounds like a homework or exam problem to me... Still, I hope that this has left you with enough fundamentals that you: 1. Understand bessel functions; 2. Understand spherical harmonics; 3. Understand at least something about vector spherical harmonics; 4. Know what a \u201cmultipolar expansion\u201d is; 5. Know how to expand a variety of important Green\u2019s functions for vector and scalar Helmholtz equations (including the Poisson equation). 6. Know how to formulate an integral equation solution to these di\ufb00erential equations based on the Green\u2019s function, and at least formally solve it by partitioning the integral into domains of convergence. 7. Know how to describe the electromagnetic \ufb01eld at a variety of levels. These levels had better include the elementary description of the E1, E2, and M1 \u201cstatic\u201d levels as well as enough knowledge to be able to do it correctly for extended sources or sources where higher order moments are important, at least if your life or job or next paper depend on it. 8. Can pass prelims. If you feel de\ufb01cient in any of these areas, I recommend that you take the time to review and learn the material again, carefully. This has been the most important part of the course and is the one thing you should not fail to take out of here with you. I hope you have enjoyed it. 13.8 Table of Properties of Vector Harmon- ics 1. Basic De\ufb01nitions Y m \u2113\u2113 = Y m \u2113\u2113 \u2212 1 = Y m \u2113\u2113+1 = 1 \u2113(\u2113 + 1) LY\u2113,m 1 \u2113(2\u2113 + 1) 1 q \u2212 \u2212 q q \u2113\u02c6r + i\u02c6r [ \u2212 \u00d7 L] Y\u2113,m (\u2113 + 1)(2\u2113 + 1) [(\u2113 + 1)\u02c6r + i\u02c6r L] Y\u2113,m \u00d7 2. Eigenvalues (j, \u2113, m are integral): J 2Y m L2Y m JzY m j\u2113 = j(j + 1)Y m j\u2113 = \u2113(\u2113 + 1)Y m j\u2113 = mY m j\u2113 j\u2113 j\u2113 3. Projective Orthonormality: Y m j\u2113 \u00b7 Z Y m\u2032 \u2217j\u2032\u2113\u2032 d\u2126 = \u03b4jj\u2032\u03b4\u2113\u2113\u2032\u03b4mm\u2032 4. Complex Conjugation: Y m \u2217j\u2113 = ( 1)\u2113+1 \u2212 j( 1)mY \u2212 j\u2113 m \u2212 \u2212 5. Addition Theorem (LCB notes corrupt \u2013 this needs to be checked): Y m \u2217j\u2113 \u00b7 Y m\u2032 j\u2032\u2113\u2032 = n X 1)m+1 ( v \u2212 u u t 000 C jj\u2032n C \u2113\u2113\u2032n 0, \u2212 (2\u2113 + 1)(2\u2113\u2032 + 1)(2j\u2032 + 1)(2j + 1) 4\u03c0(2n + 1) \u00d7 m,m\u2032W (j\u2113j\u2032\u2113\u2032; n)Yn,(m\u2032 m) \u2212 6. For F any function of r only: \u2207 (Y m \u2113\u2113F ) = 0 \u00b7 (Y m \u2113\u2113 \u2212 \u2207 \u2207 \u00b7 \u00b7 1F ) = \u2113 2\u2113 + 1 \" (\u2113 s 1", ") F r \u2212 dF dr # \u2212 Y\u2113,m (Y m \u2113\u2113+1F ) = \u2113 + 1 2\u2113 + 1 \" s (\u2113 + 2) F r \u2212 dF dr # Y\u2113,m 7. Ditto: i\u2207 \u00d7 (Y m \u2113\u2113F ) = \u2113 + 1 2\u2113 + 1 \" s (\u2113 + 1) F r + dF dr # Y m \u2113\u2113 \u2212 1 + s \u2113 \u2113 2\u2113 + 1 \"\u2212 F r + dF dr # Y m \u2113\u2113+1 i\u2207 (Y m \u2113\u2113 \u2212 \u00d7 1F ) = \u2113 + 1 2\u2113 + 1 \" (\u2113 1) F r \u2212 dF dr # \u2212 Y m \u2113\u2113 \u2212s i\u2207 \u00d7 (Y m \u2113\u2113+1F ) = \u2113 2\u2113 + 1 \" s (\u2113 + 2) F r \u2212 dF dr # Y m \u2113\u2113 8. This puts the VSHs into vector form: Y m \u2113\u2113 = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed Y m \u2113\u2113 \u2212 \u2113\u2113+\u2113+m)(\u2113 m+1) \u2212 2\u2113(\u2113+1) Y\u2113,m \u2212 1 \u2212r m \u221a\u2113(\u2113+1) Y\u2113,m (\u2113 \u2212 m)(\u2113+m+1) 2\u2113(\u2113+1) Y\u2113,m+1 r r r r r (\u2113+m 1)(\u2113+m) 1) \u2212 2\u2113(2\u2113 \u2212 Y\u2113 \u2212 1,m \u2212 1 (\u2113 m)(\u2113+m) \u2212 \u2113(2\u2113 1) Y\u2113 \u2212 \u2212 1,m r (\u2113 1)(\u2113 \u2212 m \u2212 2\u2113(2\u2113 \u2212 1) \u2212 m) Y\u2113 \u2212 1,m+1 (\u2113 m+1)(\u2113 m+2) 2(\u2113+1)(2\u2113+3) Y\u2113+1,m \u2212 \u2212 \u2212 (\u2113 m+1)(\u2113+m+1) (\u2113+1)(2\u2113+3) Y\u2113+1,m \u2212 (\u2113+m+2)(\u2113+m+1) 2(\u2113+1)(2\u2113+3) Y\u2113+1,m+. Hansen Multipole Properties \u2207 \u00b7 \u2207 \u00b7 \u2207 M L = 0 N L = 0 LL = ikf\u2113(kr)YL(\u02c6r) \u00b7 \u2207 \u2207 \u00d7 \u00d7 ikN L \u2212 M L = N L = ikM L LL = 0 \u2207 \u00d7 10. Hansen Multipole Explicit Forms M L = f\u2113(kr)Y m \u2113\u2113 N L = LL = \u2113 + 1 2\u2113 + 1 \u2113 2\u2113 + 1 f\u2113 \u2212 f\u2113 \u2212 s s 1(kr)Y m \u2113,\u2113 \u2212 1 \u2212 s 1(kr)Y m \u2113,\u2113 \u2212 1 + s \u2113 2\u2113 + 1 \u2113 + 1 2\u2113 + 1 f\u2113+1(kr)Y m \u2113,\u2113+1 f\u2113+1(kr)Y m \u2113,\u2113+1 M L = f\u2113(kr)Y m \u2113\u2113 N L = 1 kr d d(kr) (krf\u2113)(i\u02c6r Y m \u2113\u2113) \u00d7 \u2212 LL = \u2113(\u2113 + 1) q 1 kr (i\u02c6r \u00d7 f\u2113Y m \u2113\u2113) \u02c6r \u2212 \u02c6r \u2113(\u2113 + 1)f\u2113YL q d d(kr) f\u2113 YL Chapter 14 Optical Scattering 14.1 Radiation Reaction of a Polarizable Medium Usually, when we consider optical scattering,", " we imagine that we have a monochromatic plane wave incident upon a polarizable medium embedded in (for the sake of argument) free space. The target we imagine is a \u201cparticle\u201d of some shape and hence is mathematically a (simply) connected domain with compact support. The picture we must describe is thus The incident wave (in the absence of the target) is thus a pure plane 185 wave: Einc = \u02c6\u01eb0E0eik \u02c6n0 H inc = \u02c6n0 \u00d7 r \u00b7 Einc/Z0. (14.1) (14.2) The incident wave induces a time dependent polarization density into the medium. If we imagine (not unreasonably) that the target is a particle or atom much smaller than a wavelength, then we can describe the \ufb01eld radiated from its induced dipole moment in the far zone and dipole approximation (see e.g. 4.122): Esc = 1 4\u03c0\u01eb0 H sc = \u02c6n \u00d7 p) \u00d7 ( \u02c6n k2 eikr r { Esc/Z0. k0 and \u02c6n = k \u02c6n \u02c6n \u2212 \u00d7 \u00d7 m/c } (14.3) (14.4) In these expressions, \u02c6n0 = k0 of the incident and scattered waves, respectively. k, while \u02c6\u01eb0, \u02c6\u01eb are the polarization We are interested in the relative power distribution in the scattered \ufb01eld (which should be proportional to the incident \ufb01eld in a way that can be made independent of its magnitude in a linear response/susceptibility approximation). The power radiated in direction \u02c6n with polarization \u02c6\u01eb is needed per unit intensity in the incident wave with \u02c6n0, \u02c6\u01eb0. This quantity is expressed as d\u03c3 d\u2126 ( \u02c6n, \u02c6\u01eb, \u02c6n0, \u02c6\u01eb0) = r2 1 2Z0 | 1 2Z0 | \u02c6\u01eb\u2217 \u00b7 \u02c6\u01eb\u22170 \u00b7 Esc| Einc| 2 2 (14.5) [One gets this by considering the power distribution: r2 \u02c6n n E \u00d7 (E \u00b7 \u00d7 H \u2217) o ( \u02c6n E) | \u00d7 dP d\u2126 = = 1 2 Re 1 2Z0 | 1 2Z0 | = 2 | as usual, where the latter relation steps hold for transverse EM \ufb01elds 7.1 and 7.2 only and where we\u2019ve projected out a single polarization from the incident and scattered waves so we can discuss polarization later.] (14.6) E This quantity has the units of area (r2) and is called the di\ufb00erential cross\u2013section: d\u03c3 d\u2126 = dP/d\u2126 dP0/dA \u221d dA d\u2126 \u223c A. (14.7) In quantum theory a scattering cross\u2013section one would substitute \u201cintensity\u201d (number of particles/second) for \u201cpower\u201d in this de\ufb01nition but it still holds. Since the units of angles, solid or not, are dimensionless, a cross\u2013 section always has the units of area. If one integrates the cross\u2013section around the 4\u03c0 solid angle, the resulting area is the \u201ce\ufb00ective\u201d cross\u2013sectional area of the scatterer, that is, the integrated are of its e\ufb00ective \u201cshadow\u201d. This is the basis of the optical theorem, which I will mention but we will not study (derive) for lack of time. The point in de\ufb01ning it is that it is generally a property of the scattering target that linearly determines the scattered power: dP d\u2126 = d\u03c3 d\u2126 \u00d7 I0 (14.8) where the last quantity is the intensity of the incident plane wave beam. The cross-section is independent (within reason) of the incident intensity and can be calculated or measured \u201conce and for all\u201d and then used to predict the power distribution for a given beam intensity. We need to use the apparatus of chapter 7 to handle the vector polarization correctly. That is, technically we need to use the Stokes parameters or something similar to help us project out of E a particular polarization component. Then (as can easily be shown by meditating on: Esc = \u02c6\u01eb\u2217 \u00b7 1 4\u03c0\u01eb0 k2 eikr r { \u02c6\u01eb\u2217 ( \u02c6n \u00b7 { p) \u02c6n \u00d7 \u2212 \u02c6n \u00d7 \u00d7 m", "/c }} (14.9) for a transverse \ufb01eld): d\u03c3 d\u2126 = r2 1 2Z0 | 1 2Z0 | \u02c6\u01eb\u2217 \u00b7 \u02c6\u01eb\u22170 \u00b7 2 2 = Esc| Einc| k4 (4\u03c0\u01eb0E0)2 | \u02c6\u01eb\u2217 \u00b7 p + ( \u02c6n \u02c6\u01eb\u2217) m/c 2. | \u00d7 \u00d7 (14.10) To get this result, we had to evaluate (using vector identities) ( \u02c6n \u02c6\u01eb\u2217 \u00b7 \u00d7 p) \u00d7 \u02c6n = \u02c6\u01eb\u2217 p \u00b7 and ( \u02c6n \u02c6\u01eb\u2217 \u00b7 \u00d7 m/c) = m \u2212 \u00b7 ( \u02c6n \u00d7 \u02c6\u01eb\u2217). From this we immediately see one important result: d\u03c3 d\u2126 \u221d k4 \u221d 1 \u03bb4. (14.11) (14.12) (14.13) This is called Rayleigh\u2019s Law; the scattering cross-section (and hence proportion of the power scattered from a given incident beam) by a polarizable medium is proportional to the inverse fourth power of the wavelength. Or, if you prefer, short wavelengths (still long with respect to the size of the scatterer and only if the dipole term in the scattering dominates) are scattered more strongly than long wavelengths. This is the original \u201cblue sky\u201d theory and probably the origin of the phrase! To go further in our understanding, and to gain some useful practice against the day you have to use this theory or teach it to someone who might use it, we must consider some speci\ufb01c cases. 14.2 Scattering from a Small Dielectric Sphere This is a relatively simple, and hence very standard problem. Now, we have no desire to \u201creinvent the sphere\u201d1 but it is important that you understand where our results come from. First of all, let us introduce dimensionless, scaled versions of the relative permeability and permittivity (a step that Jackson apparently performs in J10 but does not document or 1Hyuk, hyuk, hyuk... explain): \u01ebr = \u01eb(\u03c9)/\u01eb0 \u00b5r = \u00b5(\u03c9)/\u00b50 \u2248 where we assume that we are not at a resonance so that the spheres have normal dispersion and that these numbers are basically real. The latter is a good approximation for non-magnetic, non-conducting scatterers e.g. oxygen or nitrogen molecules. (14.14) (14.15) 1 If you refer back to J4.4, equation J4.56 and the surrounding text, you will see that the induced dipole moment in a dielectric sphere in terms of the relative permittivity is: p = 4\u03c0\u01eb0 \u01ebr \u2212 1 \u01ebr + 2 a3Einc (14.16) To recapitulate the derivation (useful since this is a common question on quali\ufb01ers and the like) we note that the sphere has azimuthal symmetry around the direction of E, so we can express the scalar potential inside and outside the sphere as \u03c6in = A\u2113r\u2113P\u2113(cos \u03b8) X\u2113 \u03c6out = B\u2113r\u2113 + C\u2113 X\u2113 1 r\u2113+1 (14.17) P\u2113(cos \u03b8). (14.18) We need to evaluate this. At in\ufb01nity we know that the \ufb01eld should be (to lowest order) undisturbed, so the potential must asymptotically go over to E0z = \u03c6out = lim r \u2192\u221e so we conclude that B1 = E0 and all other B\u2113>1 = 0. To proceed further, we must use the matching conditions of the tangential and normal \ufb01elds at the surface of the sphere: E0rP1(cos \u03b8) E0r cos \u03b8 = (14.19) \u2212 \u2212 \u2212 \u2212 1 a \u2212 \u2202\u03c6in \u2202\u03b8 (tangential component) and = 1 a \u2212 r=a \u01eb \u2212 \u2202\u03c6in \u2202r = \u01eb0 \u2212 r=a \u2202\u03c6out \u2202\u03b8 r=a \u2202\u03c6out \u2202r r=a (14.20) (14.21) (normal D onto E). Since this is the surface of a sphere (!) we can project out each spherical component if we wish and cause these equations", " to be satis\ufb01ed term by term. From the \ufb01rst (tangential) equation we just match \u03c6 itself: 1 a (A\u2113a\u2113) = 1 a B\u2113a\u2113 + C\u2113 1 a\u2113+1 or (using our knowledge of B\u2113) C1 a3 A1 = A\u2113 = E0 + \u2212 C\u2113 a2\u2113+1 From the second (normal) equation we get \u01ebrA1 = \u01ebrA\u2113 = 2 C1 E0 \u2212 a3 \u2212 (\u2113 + 1)C\u2113 a2\u2113+1 \u2212 \u2113 = 1 else \u2113 = 1 else. (14.22) (14.23) (14.24) (14.25) (14.26) The second equation of each pair are incompatible and have only the trivial A\u2113 = C\u2113 = 0 \u2113 = 1. (14.27) Only the \u2113 = 1 term survives. With a little work one can show that A1 = C1 = 3E0 2 + \u01ebr \u2212 \u01ebr \u2212 1 \u01ebr + 2 a3E0 so that \u03c6in = \u03c6out = E0r cos \u03b8 \u2212 3 \u01ebr + 2 E0r cos \u03b8 + \u2212 \u01ebr \u2212 1 \u01ebr + 2 E0 a3 r2 cos \u03b8. (14.28) (14.29) (14.30) (14.31) When we identify the second term of the external \ufb01eld with the dipole potential and compare with the expansion of the dipole potential \u03c6(r) = 1 4\u03c0\u01eb0 r p \u00b7 r3 (14.32) 6 we conclude that the induced dipole moment is: p = 4\u03c0\u01eb0 \u01ebr \u2212 1 \u01ebr + 2 a3E0\u02c6z. (14.33) as given above. There is no magnetic dipole moment, because \u00b5r = 1 and therefore the sphere behaves like a \u201cdipole antenna\u201d. Thus m = 0 and there is no magnetic scattering of radiation from this system. This one equation, therefore, (together with our original de\ufb01nitions of the \ufb01elds) is su\ufb03cient to determine the di\ufb00erential cross\u2013section: d\u03c3 d\u2126 = k4a6 \u01ebr \u2212 1 \u01ebr + 2 2 \u02c6\u01eb\u2217 | \u00b7 2 \u02c6\u01eb0| (14.34) where remember that \u01ebr(\u03c9) (for dispersion) and hopefully everybody notes the di\ufb00erence between dielectric \u01eb and polarization \u02c6\u01eb (sigh \u2013 we need more symbols). This equation can be used to \ufb01nd the explicit di\ufb00erential cross\u2013 sections given ( \u02c6n, \u02c6n0, \u02c6\u01eb, \u02c6\u01eb0), as desired. However, the light incident on the sphere will generally be unpolarized. Then the question naturally arises of whether the various independent polarizations of the incident light beam will be scattered identically. Or, to put it another way, what is the angular distribution function of radiation with a de\ufb01nite polarization? To answer this, we need to consider a suitable decomposition of the possible polarization directions. This decomposition is apparent from considering the following picture of the general geometry: Let \u02c6n, \u02c6n0 de\ufb01ne the plane of scattering. We have to \ufb01x \u02c6\u01eb(1) and \u02c6\u01eb(2) relative to this scattering plane and average over the polarizations in the 0 and \u02c6\u01eb(2) incident light, \u02c6\u01eb(1) choose the directions of polarization such that \u02c6\u01eb(2) = \u02c6\u01eb(2) 0 to the scattering plane and \u02c6\u01eb(1) = \u02c6\u01eb(1) 0 directions \u02c6n and \u02c6n0 respectively. The dot products are thus (also \ufb01xed relative to this plane). We can always is perpendicular are in it, and perpendicular to the 0 \u02c6\u01eb(1) \u02c6\u01eb(2) \u2217 \u2217 \u00b7 \u00b7 \u02c6\u01eb(1) 0 = \u02c6n \u02c6\u01eb(2) 0 = 1. \u00b7 \u02c6n0 = cos \u03b8 (14.35) (14.36) We need", " the average of the squares of these quantities. This is essentially (0, 2\u03c0). Alternatively, we can meditate 2. Thus (for the ) of scattering, averaging sin2 \u03c6 and cos2 \u03c6 over \u03c6 upon symmetry and conclude that the average is just 1 polarization in the plane ( ) and perpendicular to the plane ( k respectively) we have: \u22a5 \u2208 d\u03c3 k d\u2126 d\u03c3 \u22a5 d\u2126 = k4a6 2 cos2 \u03b8 2 (14.37) = k4a6 We see that light polarized perpendicular to the plane of scattering has no \u03b8 dependence, while light polarized in that plane is not scattered parallel to the direction of propagation at all (along \u03b8 = 0 or \u03c0). We will invert this statement in a moment so that it makes more sense. See the diagram below. (14.38) 2 1 2 \u01ebr \u2212 1 \u01ebr + 2 \u01ebr \u2212 1 \u01ebr + 2 Unfortunately, everything thus far is expressed with respect to the plane of scattering, which varies with the direction of the scattered light. If we de\ufb01ne the polarization \u03a0(\u03b8) of the scattered radiation to be \u03a0(\u03b8) = d\u03c3k d\u2126 d\u03c3\u22a5 d\u2126 \u2212 d\u2126 + \u03c3k d\u03c3\u22a5 d\u2126 = sin2 \u03b8 1 + cos2 \u03b8 (14.39) then we obtain a quantity that is in accord with our intuition. \u03a0(\u03b8) is maximum at \u03b8 = \u03c0/2. The radiation scattered through an angle of 90 degrees is completely polarized in a plane perpendicular to the plane of scattering. Finally, we can add the two pieces of the di\ufb00erential cross\u2013section to- gether: \u01eb 1 \u2212 \u01eb + 2 which is strongly and symmetrically peaked forward and backward. Finally, this is easy to integrate to obtain the total cross\u2013section: (1 + cos2 \u03b8) = k4a6 (14.40) d\u03c3 d\u2126 2 1 2 \u03c3 = 8\u03c0 3 k4a6 2 \u01ebr \u2212 1 \u01ebr + 2. (14.41) At last, we can put it all together. Molecules in the atmosphere behave, far from resonance, like itty\u2013bitty dielectric spheres to a remarkable approximation. Since blue light is scattered more strongly than red, light seen away from its direction of incidence (the sky and not the sun) is shifted in color from white to blue. When Mr. Sun is examined directly through a thick layer of atmosphere (at sunset) the blue is all scattered out and the remaining light looks red. Finally, light from directly overhead at sunup or sundown is polarized in a north\u2013south direction; at noon the light from the horizon is polarized parallel to the horizon (and hence is \ufb01ltered by vertical transmission axis polarized sunglasses. You should verify this at your next opportunity outdoors with a pair of polarized sunglasses, as this whole discussion is taught in elementary terms in second semester introductory physics courses. Don\u2019t say I never taught you anything2. The last remarks I would make concern the total cross\u2013section. Note that if we factor out a 4\u03c0a2 we get the \u201carea\u201d of the sphere times a pure (dimensionless) number (ka)4 associated with the relative size of the sphere radius and the wavelength and a second pure number involving only the dielectric properties of the medium: \u03c3 = (4\u03c0a2)(ka)4 2 2 3 \u01ebr \u2212 1 \u01ebr + 2. ) ( (14.42) This expression isn\u2019t any more useful than the one above, but it does make the role of the di\ufb00erent terms that contribute to the total scattering crosssection more clear. 2Even if it\u2019s true... 14.3 Scattering from a Small Conducting Sphere Perfect conductors are not just dielectrics where the electric \ufb01eld is completely zero inside. The electric \ufb01eld is exactly cancelled on the interior by the induced surface charge. As we have seen, this cancellation occurs close to the surface (within a few times the skin depth). However, the induced currents also tend to expel the time dependent magnetic \ufb01eld. We therefore have two modi\ufb01cation of our results from the previous section. The electric polarization will have a di\ufb00erent form, and there will be a contribution from the induced magnetic moment of the sphere as well. Recall (from J2.5) that the induced dipole moment on a conducting sphere is p = 4\u03c0\u01eb0a3Einc. (14.43) This is indeed the generalization of", " the result for p last time, as you should be able to derive in a few minutes of work. Either review that section or solve the boundary value problem where E is discontinuous at the surface and E = 0 on the surface to obtain: \u22a5 || \u03c6 = E0 \u2212 r \u2212 a3 r2! cos \u03b8 (14.44) from which we can easily extract this p. But, the magnetic \ufb01eld is also varying, and it induces an EMF that runs in loops around the magnetic \ufb01eld lines and opposes the change in magnetic \ufb02ux. Assuming that no \ufb01eld lines were trapped in the sphere initially, the induced currents act to cancel component of the magnetic \ufb01eld normal to the surface. The sphere thus behaves like a magnetically permeable sphere (see e.g. section J5.10 and J5.11, equations J5.106, J5.107, J5.115): M = m 4\u03c0a3/3 = 3 \u00b5 \u00b50 \u00b5 + 2\u00b50! \u2212 H inc with \u00b5r = \u00b5/\u00b50 = 0 so that: m = 2\u03c0a3H inc. \u2212 (14.45) (14.46) The derivation is again very similar to the derivation we performed last time, with suitably chosen boundary conditions on B and H. If we then repeat the reasoning and algebra for this case of the conducting sphere (substituting this p and m into the expression we derived for the di\ufb00erential cross\u2013section), we get: d\u03c3 d\u2126 = k4a6 \u02c6\u01eb0 \u2212 \u00b7 1 2 ( \u02c6n \u02c6\u01eb\u2217) ( \u02c6n0 \u00d7 \u00b7 \u00d7. (14.47) 2 \u02c6\u01eb0) \u02c6\u01eb\u2217 After much tedious but straightforward work, we can show (or rather you can show for homework) that: d\u03c3 k d\u2126 d\u03c3 \u22a5 d\u2126 = = 2 1 2 cos \u03b8 \u2212 1 2 cos \u03b8 1 \u2212 k4a6 2 k4a6 2 2 so that the total di\ufb00erential cross section is: d\u03c3 d\u2126 = k4a6 5 8 (1 + cos2 \u03b8) cos \u03b8) \u2212 and the polarization is: \u03a0(\u03b8) = 3 sin2 \u03b8 5(1 + cos2 \u03b8) 8 cos \u03b8 \u2212 (14.48) (14.49) (14.50) (14.51) Finally, integrating the di\ufb00erential cross section yields the total cross-section: \u03c3 = 10\u03c0k4a6 3 = (4\u03c0a2)(ka)4 2.5 3 \u223c \u03c3dielectric (14.52) Figure 14.1: Di\ufb00erential cross\u2013section and polarization of a small conducting sphere. for \u01ebr >> 1 curiously enough. What do these equations tell us? The cross\u2013section is strongly peaked backwards. Waves are re\ufb02ected backwards more than forwards (the sphere actually casts a \u201cshadow\u201d. The scattered radiation is polarized qualitatively alike the radiation scattered from the dielectric sphere, but with a somewhat di\ufb00erent angular distribution. It is completely polarized perpendicular to the scattering plane when scattered through an angle of 60\u25e6, not 90\u25e6. We see that dipole scattering will always have a characteristic k4 dependence. By know you should readily understand me when I say that this is the result of performing a multipolar expansion of the reaction \ufb01eld (essentially an expansion in powers of kd where d is the characteristic maximum extent of the system) and keeping the \ufb01rst (dipole) term. If one wishes to consider scattering from objects where kd 1 or greater, one simply has to consider higher order multipoles (and one must consider the proper multipoles instead of simple expansions in powers of kd). If kd >> 1 (which is the case for light scattering from macroscopic objects, \u223c radar scattering from airplanes and incoming nuclear missiles, etc) then a whole di\ufb00erent apparatus must be brought to bear. I could spend a semester (or a least a couple of weeks) just lecturing on the scattering of electromagnetic waves from spheres, let alone other shapes. However, no useful purpose would be so served, so I won\u2019t. If you ever need to \ufb01gure it out, you have the tools and can \ufb01nd and understand the necessary references. 14.4 Many Scatterers It is, however, worthwhile to spend a moment considering a collections of identical scatterers at \ufb01xed spatial positions. Each scatterer then acts identically, but is scattering an", " electromagnetic \ufb01eld with its own (spatially dependent) phase at a given moment of time. The scattered \ufb01elds then propagate freely, recombine, and form a total EM \ufb01eld that is measured by the detector. In order to evaluate the total di\ufb00erential cross\u2013section we must sum the \ufb01eld amplitudes times the appropriate phases, project out the desired polarization moments, and then square. A moment of quiet re\ufb02ection3 will convince you that in general: d\u03c3 d\u2126 = k4 (4\u03c0\u01eb0E0)2 where \u02c6\u01eb\u2217 Xj n pj + ( \u02c6n \u02c6\u01eb\u2217) \u00b7 \u00d7 \u00b7 mj/c xj eiq \u00b7 o (14.53) 2 q = k0 \u2212 accomodates the relative phase di\ufb00erence between the \ufb01eld emitted by the scatterers at di\ufb00erent locations. The geometry of this situation is pictured below. (14.54) k. In all directions but the forward direction, this depends on the distribution of scatterers and the nature of each scatterer. If we imagine all the scatterers to be alike (and assume that we are far from the collection) then this expression simpli\ufb01es: d\u03c3 d\u2126 = d\u03c30 d\u2126 F (q) (14.55) 3Sorry... Figure 14.2: Geometry of multiple scatterers. The relative phase of two sources depends on the projection of the di\ufb00erence in wave vectors onto the vector connecting the scatterers. where d\u03c30 is called a \u201cstructure factor\u201d: d\u2126 is the scattering cross-section of a single scatterer and the (q) F (q) = F = xj eiq \u00b7 (xj eiq \u00b7 Xj Xi,j 2 \u2212 xi). (14.56) (14.57) This last expression is 1 on the diagonal i = j. If the (e.g.) atoms are uniformly but randomly distributed, the sum of the o\ufb00-diagonal terms averages to zero and the total sum goes to N (the number of atoms). This is an incoherent superposition and the scattered intensitities add with negligible interference. If the atoms are instead on a regular lattice, then \u201cBragg\u201d scattering results. There will exist certain values of q that match the spacing between planes in such a way that whole rows of the matrix are 1. In those direction/wavelength combinations, the scattered intensity is of order N 2 and hence is much brighter. The scattered power distribution thus has bright spots in is corresponding to these directions, where constructive interference in the scattered waves occurs. Structure factor sums occur in many branches of physics. If you think about it for a moment, you can easily see that it is possible to do a structure factor sum using the Green\u2019s function expansions you have studied. In electrodynamics and quantum multiple scattering theory these sums appear frequently in association with spatially \ufb01xed structures (like crystal lattices or molecules). In \ufb01eld theory, lattice sums are sometimes used as a discretized approximation for the continuum, and \u201clattice gauge\u201d type \ufb01eld theories result. In these theories, our ability to do the structure factor sums is used to construct the Green\u2019s functions rather than the other way around. Either way, you should be familiar with the term and should think about the ways you might approach evaluating such a sum. We are now done with our discussion of scattering from objects per se. It is well worth your while to read J10.2 on your own. I have given you the semi\u2013quantitative argument for the blue sky; this section puts our simple treatment on \ufb01rmer ground. It also derives the perturbation theory of scattering (using the Born approximation), and discusses a number of interesting current research topics (such as critical opalescence). I will probably assign one problem out of this section to help you out. However, perturbative scattering is easier to understand, and more useful, in the context of (scalar) quantum theory and so I will skip this section, expecting that you will see enough of it there. You should also read J10.3. This presents one way to derive the Rayleigh expansion for a (scalar) plane wave in terms of free spherical waves (there are several). However, it goes further and addresses expansions of e.g. circularly polarized plane waves in terms of vector spherical harmonics! Lord knows why this is stuck o\ufb00 in this one section all by itself \u2013 I need to put the equivalent result for expansion in terms of Hansen solutions (which of course will", " be much more natural and will precompute most of the annoying parts of the algebra for us) in the sections on the Hansen functions and VSHs where it belongs, as it will actually be much simpler to understand there. J10.4 redoes scattering from a sphere \u201cright\u201d in terms of VSHs, and again, if we wished to pursue this we would need to redo this in terms of Hansen functions to keep it simple. The primary advantage of reading this chapter is that it de\ufb01nes the partial wave phase shifts of scattering from a sphere, quantities that are in use in precisely the same context in quantum scattering theory in e.g. nuclear physics. SO, if you plan to go into nuclear physics you are well advised to read this chapter as well and work through it. However, we cannot do this at this time because we had to go back and redo J7 and J8. Besides, we\u2019re doubtless a bit bored with multipoles and want to become excited again. We will therefore now move on to one of my favorite topics, relativity theory. Part III Relativistic Electrodynamics 201 Chapter 15 Special Relativity 15.1 Einstein\u2019s Postulates By this time I certainly hope that you are familiar with the two postulates, due to Einstein, that lead to the theory of special relativity. They are: 1. The laws of nature are invariant with respect to the uniform translation of the coordinate system in which they are measured. 2. The speed of light is independent of the motion of the source. Properly speaking, the second postulate is a consequence of the \ufb01rst, since if the speed of light depended on the motion of its source the laws of electrodynamics (which determine the speed of freely propagating electromagnetic waves) would depend on the inertial frame of the source, which contradicts the \ufb01rst postulate. For what it is worth, the \ufb01rst is not as obviously a consequence of the second: it seems entirely possible for some laws to depend on the velocity of the source and not contradict the second postulate, as long as they are not electrodynamical in nature. This has been the subject of considerable discussion, and I hesitate to state a religious view upon it. I will, however, point out that in the opinion of Dirac, at least \u2014 the discovery of the uniform 3\u25e6K blackbody background explicitly contradicted the \ufb01rst postulate but not the second. You might amuse yourself, some quiet evening, by considering experiments that would measure your absolute 203 velocity relative to the \u201crest\u201d frame of this radiation. The second postulate (which is all we need) thus seems to be the safer of the two upon which to base our reasoning. I strongly recommend that you read J11.1 \u2014 J11.2 on your own. They are \u201ctrue facts\u201d that will come in handy some day, and should astound and amaze you. Yes, Virginia, special relativity really really works. For our purposes, we will begin with a brief review of the basic Lorentz transformation and an introduction to four vectors. Because we will do it again (correctly) in a week or so we won\u2019t take long now. We will also study four\u2013velocity and four\u2013momentum. This will su\ufb03ce to give us the \u201c\ufb02avor\u201d of the theory and establish the geometricaly grounds for the matrix theory we will then derive. As an application of this, we will study Thomas precession brie\ufb02y and then go on to perform a detailed application of the theory to the dynamics of interacting charged particles and \ufb01elds. We will spend the rest of the semester on this subject, in one form or another. 15.2 The Elementary Lorentz Transformation To motivate the Lorentz transformation, recall the Galilean transformation between moving coordinate systems: vt x\u20321 = x1 \u2212 x\u20322 = x2 x\u20323 = x3 t\u2032 = t (15.1) (15.2) (15.3) (15.4) (where K is \ufb01xed and K \u2032 is moving in the 1\u2013direction at speed v). Then Fj = m\u00a8xj = m\u00a8x\u2032j = F \u2032j (15.5) or Newton\u2019s Laws are covariant with respect to the Gallilean transformation. But and so Thus if then \u2202 \u2202x1 = \u2202 \u2202x\u20321 + 1 v \u2202 \u2202t\u2032 + 1 v2 \u22022 \u2202t\u2032 2 + 2 v \u22022 \u2202x\u20321\u2202t\u2032 \u22022 2 \u2202x\u2032 1 \u22022 2 \u2202x\u2032 2 \u22022 2 \u2202x\u2032 3 \u22022", " \u2202t\u2032 2. = = \u22022 \u2202x2 1 \u22022 \u2202x2 2 \u22022 \u2202x2 3 \u22022 \u2202t2 = = 2 (\u2207 \u2212 1 c2 \u22022 \u2202t2 ) \u03c8 = 0 2 \u2032 \u2212 (\u2207 1 c2 \u22022 \u2202t\u2032 2 ) 1 \u03c8 = \u2212 v2 \u22022\u03c8 \u2202t\u2032 2 \u2212 2 v \u22022\u03c8 \u2202x\u20321\u2202t\u2032 = 0 (15.6) (15.7) (15.8) (15.9) (15.10) (15.11) (15.12) (!) so that the wave equation, and hence Maxwell\u2019s equations which lead directly to the wave equation in free space, are not covariant with respect to the Gallilean transformation! They already determine the 6 permitted velocity of a light wave, and do not allow that velocity to depend on anything but the properties of the medium through which the wave is transmitted. The simplest linear transformation of coordinates is that preserves the form of the wave equation is easy to determine. It is one that keeps the speed of the (light) wave equal in both the K and the K \u2032 frames. Geometrically, if a \ufb02ash of light is emitted from the (coincident) origins at time t = t\u2032 = 0, it will appear to expand like a sphere out from both coordinate origins, each in its own frame: and (ct)2 \u2212 (x2 + y2 + z2) = 0 \u2212 are simultaneous constraints on the equations. Most generally, (ct\u2032)2 (x\u2032 2 + y\u2032 2 + z\u2032 2) = 0 (ct)2 \u2212 (x2 + y2 + z2) = \u03bb2 (ct\u2032)2 h \u2212 (x\u2032 2 + y\u2032 2 + z\u2032 2) i (15.13) (15.14) (15.15) where, \u03bb(v) describes a possible change of scale between the frames. If we insist that the coordinate transformation be homogeneous and symmetric between the frames1, then \u03bb = 1. (15.16) Let us de\ufb01ne x0 = ct x1 = x x2 = y x3 = z (x4 = ict Minkowski metric) (15.17) (15.18) (15.19) (15.20) (15.21) Then we need a linear transformation of the coordinates that mixes x and (ct) in the direction of v in such a way that the length \u2206s2 = (x0)2 \u2212 (x2 1 + x2 2 + x2 3) (15.22) 1If we relax this requirement and allow for uniform expansions and/or contractions of the coordinate system, a more general group structure, the conformal group, results is conserved and that goes into the Gallilean transformation as v 0. If we continue to assume that v is in the 1 direction, this leads to the Lorentz transformation: \u2192 \u03b2x1) \u03b2x0) x\u20320 = \u03b3(x0 \u2212 x\u20321 = \u03b3(x1 \u2212 x\u20322 = x2 x\u20323 = x3 x1 = vt \u03b2 = \u2192 v c. 2 \u2206s2 = \u2206s\u2032 x2 0 \u2212 1 = \u03b32(x2 x2 0 \u2212 1) + \u03b32\u03b22(x2 x2 1 \u2212 x2 0) \u03b32(1 \u2212 \u03b22) = 1 (15.23) (15.24) (15.25) (15.26) (15.27) (15.28) (15.29) (15.30) where at x\u20321 = 0, Then leads to or so 1 \u03b3 = \u00b1 \u2212 (15.31) \u221a1 where we choose the + sign by convention. This makes \u03b3(0) = +1. Finally, \u03b22 as we all know and love. \u03b3(v) = 1 v2 c2 1 \u2212 q Now, let me remind you that when v << c, \u03b3(v) = 1 + 1 2 v2 c2 +... (15.32) (15.33) to lowest surviving order in v c. As we shall see, this is why \u201ckinetic energy\u201d in non\u2013relativistic systems (being de\ufb01ned as the total energy minus the potential energy and the rest mass energy) is the usual 1 2mv2. The inverse transformation (from K \u2032 to K) is also of some interest. x0 = \u03b3(x\u20320 + \u03b2x\u20321) x1 = \u03b3(x\u20321 +", " \u03b2x\u20320) x2 = x\u20322 x3 = x\u20323 (15.34) (15.35) (15.36) (15.37) which is perfectly symmetric, with v v. It appears that which frame is at \u201crest\u201d and which is moving is mathematically, at least, a matter of perspective. \u2192 \u2212 Finally, if we let ~\u03b2 = v c (15.38) (in an arbitrary direction) then we have but to use dot products to align the vector transformation equations with this direction: ~\u03b2 x\u20320 = \u03b3(x0 \u2212 \u03b3 x\u2032 = x + \u2212 \u03b22 \u00b7 1 x) (~\u03b2 \u00b7 x)~\u03b2 \u2212 \u03b3 ~\u03b2x0 (15.39) (15.40) I think that you should prove that this is correct as an exercise. Since the direction of ~\u03b2 is arbitrary, it should su\ufb03ce to show that this reduces the the form above for that direction and an arbitrary transverse direction. Solution: Note that = x k (~\u03b2 x)~\u03b2 \u00b7 \u03b22 (15.41) Lorentz transform it according to this rule and one gets (by inspection) = \u03b3(x x\u2032 k k \u2212 ~\u03b2x0) (15.42) as one should. The x\u20320 transform is obvious. Finally, the other two ( components do not get a contribution from \u03b3. That is, ) \u22a5 x\u2032 = (x + x k \u22a5 ) + \u03b3x x k \u2212 k \u2212 \u03b3 ~\u03b2x0 (15.43) (reconstructing the result above directly) QED. This is not the most general or convenient way to write the \ufb01nal transform. This is because \u03b3 and \u03b2 are both related functions; it should not be necessary to use two parameters that are not independent. Also, the limiting structure of the transformation is not at all apparent without considering the functional forms in detail. It is easy to see from the de\ufb01nition above that (\u03b32 \u2212 \u03b32\u03b22) = 1. (15.44) The range of \u03b2 is determined by the requirement that the transformation be non\u2013singular and its symmetry: 0 \u2264 \u03b2 < 1 so that 1 \u03b3 <. \u221e \u2264 (15.45) If we think about functions that \u201cnaturally\u201d parameterize these ranges, they are: where cosh2 \u03be \u2212 sinh2 \u03be = 1 \u03b2 = tanh \u03be = \u03b3 = cosh \u03be = \u03b3\u03b2 = sinh \u03be = e\u2212 e\u2212 1 2 1 2 \u03be e\u03be \u2212 \u03be + e\u03be (e\u2212 \u03be + e\u03be) (e\u2212 e\u03be) \u03be \u2212 [0, 1) \u2208 [1, ) \u221e [0, ). \u221e \u2208 \u2208 (15.46) (15.47) (15.48) (15.49) The parameter \u03be is called the boost parameter or rapidity. You will see this used frequently in the description of relativistic problems. You will also hear about \u201cboosting\u201d between frames, which essentially means performing a Lorentz transformation (a \u201cboost\u201d) to the new frame. This will become clearer later when we generalize these results further. To give you something to meditate upon, consider in your minds the formal similarity and x0 coordinates where between a \u201cboost\u201d and a \u201crotation\u201d between the x the rotation is through an imaginary angle i\u03be. Hmmmm. k To elucidate this remark a wee tad more, note that in this parameteri- zation, x x\u20320 = x0 cosh \u03be x\u2032 k = k x0 sinh \u03be + x. \u2212 = x \u2212 sinh \u03be cosh \u03be k x\u2032 \u22a5 \u22a5 (15.50) (15.51) (15.52) What is the 4 Does it look like a \u201chyperbolic rotation\u201d2 or what? 4 transformation matrix (in four dimensions) for this result? \u00d7 We have just determined that the (vector) coordinate system transforms a certain way. What, then, of vector \ufb01elds, or any other vector quantity? 2 \u201cHyperbolic\u201d because of the relative minus sign between x2 and ct2. More on this later. How do general vectors transform under a boost? This depends on the nature of the vectors. Many vectors, if not most, transform like the underlying coordinate description of the vectors. This includes the ones of greatest interest in physics. To make this obvious, we will have to generalize a vector quantity to four dimensions. 15.3 4-Vectors Note well that we are not YET introducing proper notation for", " co- and contravariant tensors as we don\u2019t know what that means. Actually the notation for ordinary coordinates should be x0, x1, x2, x3 and we will have to determine whether any given 4-vector quantity that is a function of the coordinates transforms like the coordinate or like a di\ufb00erential of the coordinate in order to determine if it is co- or contravariant. Similarly, we have not yet discussed how to form the various dyadic products of co- and contravariant vectors \u2013 some will form scalars, some vectors, some second rank tensors. In other words, the results below are all correct, but the notation sucks and this suckiness will make certain parts of doing the algebra more di\ufb03cult than it needs to be. I may rewrite this whole combined multichapter stretch, as I\u2019m not certain of the pegagogical value of presenting things incorrectly or in an elementary form and then correctly in an elegant form as Jackson does. In the meantime, please bear with the notation below allowing for the fact that much of it is just wrong. Coordinate 4\u2013vectors are (x0, x1, x2, x3). Arbitrary 4\u2013vectors are (A0, A1, A2, A3). If the \u201carbitrary\u201d vector transforms like the coordinates, then A) ~\u03b2 \u00b7 \u03b2A0) A\u20320 = \u03b3(A0 \u2212 = \u03b3(A A\u2032 k \u2212 k A\u2032 \u22a5 = A \u22a5 and \u2206A2 = A2 = A2 0 \u2212 0 \u2212 2 + A2 3) (A2 A 1 + A2 A \u00b7 (15.53) (15.54) (15.55) (15.56) Figure 15.1: The Light Cone: Past, now, future, and elsewhere. Events. The world line. is an invariant of the transformation. Note: whenever I boldface a vector quantity, I mean the 3D euclidean (cartesian) vector in ordinary space. In that case I will write the time (0) component explicitly. When I want to refer to a 4\u2013vector generically, I will not boldface it (e. g. \u2014 A vs A). Kids! Amaze your friends! Astound your neighbors! Show that A\u20320B\u20320 \u2212 A\u2032 B\u2032 = A0B0 \u2212 \u00b7 A \u00b7 B (15.57) is an invariant of the Lorentz transformation for arbitrary 4\u2013vectors A, B. This is (or will be) homework. Now, we have a few de\ufb01nitions of \u201cnew words\u201d to learn. Most of you probably already know them from one context or another, but we all need to agree at this point to talk a common language, so we will review the de\ufb01nitions carefully and avoid confusion. Electromagnetic signals (and anything else travelling at speed c) travel on the light cone. An event is a coordinate x = (x0, x). We are usually interested in causally connected events on a world line. This might be, for example, the trajectory of a massive particle (like one on the tip of your nose) with v < c. Causally connected world line trajectories must live inside the light cone of each event that lies upon them. Consider two events. If we de\ufb01ne the invariant interval 12 = c2(t1 \u2212 S2 t2)2 x1 \u2212 \u2212 | 2 x2| (15.58) then we have a timelike separation S2 \u21d2 12 > 0 t2)2 > c2(t1 \u2212 Both events are inside each other\u2019s light cone. These events can be \u201ccausally connected\u201d, because a light signal given o\ufb00 by one can reach the other from the \u201cinside\u201d. In this case, a suitable Lorentz transformation can make x\u20321 = x\u20322, but t\u20321 6 = t\u20322 always. x1 \u2212 | 2. x2| spacelike separation S2 2. x2| \u21d2 12 < 0 t2)2 < x1 \u2212 | c2(t1 \u2212 Both events are outside each other\u2019s light cone. These events are \u201ccausally disconnected\u201d, because a light signal given o\ufb00 by one can not reach the other. If nothing goes faster than light, then those particular events did not speak to one another. Note that this does not mean that earlier (and later) events on each world line to not connect. The events are disconnected, not the world lines themselves. In this case, a suitable Lorentz transformation can make t\u20321 = t\u20322, but x\u20321 6 = x", "\u20322 always. lightlike separation S2 \u21d2 12 = 0 t2)2 = c2(t1 \u2212 Both events are on each other\u2019s light cone. These events are \u201ccausally connected\u201d by electromagnetic radiation. The \ufb01eld produced by charges at one event are directly interacting with charges at the other event, and vice versa. x1 \u2212 | 2. x2| Note well that the event pairs considered above can be made spatially coincident, temporally coincident, or both, by suitably boosting the frame. Events with a timelike separation can be made spatially coincident. Events with a spacelike separation can be made to occur at the same time, or in either order. Events with a lightlike separation will always have a lightlike separation in all frames. We are about to run into a profound philosophical di\ufb03culty. Physics is dedicated to dynamics \u2013 typically solving initial value problems and hence predicting the dynamical evolution of systems in time. Unfortunately, we just eliminated time as an independent variable. By making it a part of our geometry, it is no longer available as an independent parameter that we can use to write traditional equations of motion. There are likely to other signi\ufb01cant consequences of this decision, as many of the quantities studied in physics are tensor forms de\ufb01ned with respect to spatial geometry. That is, when I compute \u201ccharge\u201d or \u201cmomentum\u201d or \u201celectric \ufb01eld\u201d or a \u201crotation matrix\u201d, I\u2019m computing 0th, 1st or 2nd rank tensors that inherit their directional character (or lack of it) from the underlying spatial coordinate system. Well, we\u2019ve just made that underlying coordinate system four dimensional and so quantities like \u201cmomentum\u201d and \u201celectric \ufb01eld\u201d will have to be reconsidered. We may need to \ufb01nd new \u201ctimelike\u201d coordinates to associate with some of these, and perhaps reclassify others as di\ufb00erent sorts of tensors. Finally, we need to recover a \u201ctime\u201d that can be used to write down some sort of equations of motion or we can\u2019t make a \u201cphysics\u201d. This will prove to be very di\ufb03cult. For one thing, we can no longer expect to be able to solve initial value problems, as time is now a symmetric coordinate. The trajectories of particles are determined by their relativistic interaction connections and di\ufb00erential \u201cequations of motion\u201d with boundary conditions on a closed four dimensional hypersurface at four\u2013in\ufb01nity! That means that it is impossible in principle to predict future trajectories from only a knowledge of those trajectories in the past. It is amazing how few people in physics are willing to internally acknowledge that fact. Accept it. It is true. You will be happier for it. Anyway, there are at least two ways around this (mathematical) di\ufb03culty. One is to introduce a \u201chypertime\u201d \u2013 yet another dimension containing a parameter that can serve us as time has served in the past3. This, however, introduces a \ufb01fth dimension which we need (currently) like a \ufb01fth wheel. Maybe God lives in hypertime, but there are in\ufb01nite di\ufb03culties associated with our trying to implement it in the complete absence of physical probes. Say hello to Plane Joe from Flatland. Leave it to masochistic theorists to play games with 10, 26, or even 4096 dimensional projective manifolds at least until you are ready to become one of them. The second way is to introduce the proper time. This is the time measured in the \u201crest frame\u201d of a particle as it moves along its world line. As such, it is still not an \u201cabsolute\u201d time like we are used to but it is the closest that we can come to it. Note well that proper time does not really solve our philosophical probIf it lems, because one must still ask how the \u201cparticle\u201d measures time. carries with it a little \u201cclock\u201d, that clock must have moving parts and some sort of associated period, and those parts have in turn their own proper time. If it is a point particle, its clock must either be in internal degrees of freedom \u2013 you begin to see why those theorists mentioned above work their 3Don\u2019t think too hard about this sentence or you\u2019ll start to go slightly nuts because it is self-referential and hence G\u00a8odelian. way up to higher dimensional spaces \u2013 or else the particle infers the passage of time from what it \u201csees\u201d of the rest of the Universe", " via its interaction connections and doesn\u2019t really have a proper time at all because it cannot have its own proper clock. It does, however, solve our immediate mathematical problem (that of \ufb01nding a suitable parameter in terms of which to describe the evolution of a system) so we\u2019ll go with it anyway. 15.4 Proper Time and Time Dilation Suppose we have a particle moving with a velocity v in a given coordinate system K. In a time dt (in that system) it moves dx = vdt. Then its invariant in\ufb01nitesimal interval is (ds)2 = (cdt)2 2 = c2dt2(1 dx | In the particular frame where the particle is at rest (dx\u2032 = 0) we de\ufb01ne the proper time to be (15.59) \u03b22). \u2212 | \u2212 so that d\u03c4 = dt\u2032 (ds)2 = c2(d\u03c4 )2. (15.60) (15.61) Thus the proper time is just the time experienced by the particle in its own rest frame. From the relations above, it is easy to see that d\u03c4 = dt 1 q \u03b22(t) = \u2212 dt \u03b3(t) (15.62) and to \ufb01nd the interval between two events on some world line it is necessary to integrate: t2 \u2212 t1 = = \u03c42 \u03c41 Z \u03c42 \u03c41 Z d\u03c4 1 \u03b22(\u03c4 ) \u2212 q \u03b3(\u03c4 )d\u03c4. (15.63) If \u03b2 is constant (so the frames are inertial) then we get the usual time dilation \u2206t = \u03b3\u2206\u03c4 (15.64) or \u2206\u03c4 = \u2206t \u03b3 (15.65) but this is not true if the particle is accelerating. Applying it without thought leads to the \u201ctwin paradox\u201d. However, the full integral relations will be valid even if the two particles are accelerating (so that \u03b2(\u03c4 )). You will need to evaluate these relations to solve the twin paradox for one of your homework problems. Finally, I want to note (without discussing it further at this time) that proper time dilation leads to a relativistic correction to the usual doppler shift. Or should I say that the non\u2013relativistic doppler shift is just a low velocity limit of the correct, time dilated result. Now that we have some notion of what an in\ufb01nitesimal time interval is, we could go ahead and try to de\ufb01nce 4\u2013dimensional generalizations of momentum and energy. First, however, we will learn how velocities Lorentz transform. 15.5 Addition of Velocities If we form the in\ufb01nitesimal version of the Lorentz transformation of coordinates: dx0 = \u03b3(dx\u20320 + \u03b2dx\u20321) dx1 = \u03b3(dx\u20321 + \u03b2dx\u20320) dx2 = dx\u20322 dx3 = dx\u20323 (15.66) (15.67) (15.68) (15.69) Point P is moving at velocity u\u2032 in frame K \u2032, which is in turn moving at velocity v = v\u02c61 with respect to the \u201crest\u201d frame K. We need to determine u (the velocity of P in K). We will express the problem, as usual, in coordito the direction of motion, exploiting the obvious azimuthal nates symmetry of the transformation about the \u02c61 direction. and \u22a5 k Note that ui = c dxi dx0 (15.70) Figure 15.2: P has u\u2032 = (u\u2032, \u03b8\u2032, \u03c6\u2032) in K \u2032 frame. K \u2032 is moving in the 1 direction at v = c\u03b2. \u03b3(v) changes frames. We want u(u, \u03b8, \u03c6). for i = 0... 3. Then u k = c \u03b3(dx\u20321 + \u03b2dx\u20320) \u03b3(dx\u20320 + \u03b2dx\u20321) dx\u2032 1 + \u03b2 dx\u2032 0 o 1 + \u03b2 dx\u2032 1 dx\u2032 v \u00b7 c2 o. = c n = Similarly, u \u22a5 (e.g. \u2014 u2) is given by u2 = cdx\u20322 \u03b3(dx\u20320 + \u03b2dx\u20321) = u\u20322 \u03b3(1 + u\u2032 v \u00b7 c2 or = u \u22a5 \u03b3 u \u22a5 1 + u\u2032 v \u00b7 c2 n o. We see, then, that the velocity changes in both the Note also that if and u\u2032 | | v | | << c, then v u\u2032 c2 << 1 \u00b7 (15.71) (15.72)", " (15.73) and the k \u22a5 directions. (15.74) and so that we recover the Gallilean result\u2032 k = u\u2032 \u22a5 + v. What about the other limit? If = c, then u\u2032 | | = c u | | (15.75) (15.76) (15.77) (15.78) as you should verify on your own. This is Einstein\u2019s second postulate! We have thus proven explicitly that the speed of light (and the speed of anything else travelling at the speed of light) is invariant under Lorentz coordinate transformations. This is their entire motivation. We observe that the three spatial components of \u201cvelocity\u201d do not seem to transform like a four vector. Both the components are mixed by a boost. We can, however, make the velocity into a four vector that does. We de\ufb01ne and the \u22a5 k U0 = U = = dx0 d\u03c4 = c\u03b3(u) dx d\u03c4 = u\u03b3(u) = dx0 dt dt d\u03c4 dx dt dt d\u03c4 (15.79) (15.80) where \u03b3(u) is evaluated using the magnitude of u. It is an exercise to show that this transforms like the coordinate 4\u2013vector x. Now we can \u201cguess\u201d that the 4\u2013momentum of a particle will be To prepare us for this, observe that U = (U0, U) = (\u03b3uc, \u03b3uu) are just the \u03b3u\u2013scaled \u201cvelocities\u201d of the particle: mU. \u223c (15.81) Figure 15.3: Note that \u03b3u \u2265 1 so that each component of the 4\u2013velocity is always \u201clarger\u201d than associated Cartesian components, even though (as usual) the length of the four velocity is invariant. What is its invariant length? 15.6 Relativistic Energy and Momentum We seek a relativistic generalization of momentum (a vector quantity) and energy. We know that in the low speed limit, v << c, p = mu E = E(0) + 1 2 mu2 (15.82) (15.83) where E(0) is a constant allowed by Newton\u2019s laws (since forces depend only on energy di\ufb00erences). The only possible form for this generalization of these equations consis- tent with our requirement that the laws of nature remain invariant are: p = (u)u M (15.84) E = (u), (15.85) E that is, the mass and the energy must become functions of the speed only, and leave the vector character of the velocity alone. A boost cannot change the direction of the momentum of a particle, and any (scalar) functional variation in its magnitude can be thrown into the \u201cmass\u201d term. This immediately yields the limiting forms: (0) = m (15.86) M \u2202 \u2202u2 (0) = E where we have assumed that there is no pathology in the functions at the origin. (15.87) m 2 There are several possible ways to evaluate the full forms of these functions. Jackson\u2019s (based on scattering theory) is tedious and conceals the structure of the result. Furthermore, after telling us that selecting clever initial directions with an eye to simplifying the algebra \u201clacks motivation\u201d he derives a result by selecting particular initial directions. The guy loves algebra, what can I say. Feel free to study his approach. It works. I, on the other hand, am too lazy to spend most of a period deriving a result that is \u201cobvious\u201d in the correct notation. I am therefore going to \u201cgive\u201d you the result and motivate it, and then verify it trivially be expressing it as a four\u2013vector. This works nearly as well and is not anywhere near as painful. We begin by considering elastic scattering theory. An elastic collision of two identical particles must conserve momentum and energy in all inertial frames. In the center of mass frame (which we will consider to be K \u2032) p\u2032ia + p\u2032ib = p\u2032f a + p\u2032f b E\u2032ia + E\u2032ib = E\u2032f a + E\u2032f b (15.88) (15.89) relate the intial and \ufb01nal momenta and energy of the two identical particles. Now, and u\u2032ia = v = u\u2032ib \u2212 uf a = v\u2032 = u\u2032f b (15.90) (15.91) by de\ufb01nition in the center of mass system. A moments quiet re\ufb02ection (egad, another pun!) should convince you that in terms of the general transformation: (", "v)v M (v)v = \u2212 M (v\u2032)v\u2032 M (v\u2032)v\u2032 \u2212 M (15.92) Figure 15.4: \u03b8\u2032 = 30\u25e6 and \u03b22 = 1 3. The dashed lines are the results of a Gallilean transformation from K \u2032 to K. Note that the scattering is more forward than expected because of the Lorentz contraction of the longitudinal distances seen by the particles. (v) + E E (v) = E (v\u2032) + (v\u2032). E (15.93) For what it is worth, if the collision is elastic and the particles are identical before and after the collision, v = v\u2032 and all the mass terms are the same. We will denote the scattering angle in K \u2032 as \u03b8\u2032. We thus begin with (v)v M (v)v = \u2212 M (v)v M (v)v \u2212 M (v) + (v) = (v) + (v) E E E E (15.94) (15.95) where v is the speed of the incoming and outgoing particles. Now, must be a scalar function of v, and in the limit v 0 must turn into \u2192 (v) M (v) = m. lim v \u2192 0 M (15.96) The only scalar function of v we have encountered so far with this behavior is \u03b3(v), so we should sensibly guess which has the exactly correct limiting behavior. (v) = \u03b3(v)m M (15.97) Thus p = \u03b3mu (15.98) is a reasonable guess to be the generalization of momentum we seek. It is easy to verify that this is a consistent choice, and that it indeed results in conservation of momentum in all inertial frames. To get the energy equation, we use the same approach. Recall that a binomial expansion of \u03b3 is given by \u03b3(v) = 1 lim 0 v \u2192 1/2 v2 \u2212 c2! \u2212 = 1 + 1 2 v2 c2 +... (15.99) We need to keep the \ufb01rst non\u2013constant term because we recall that physics is always \u201cindependent\u201d of absolute energy scale. Then it should be clear that lim v \u2192 0 E (v) = \u03b3(v) E (0) \u2248 E (0) + 1 2E (0) v2 c2 \u2248 E (0) + 1 2 mv2 (15.100) as it must in order to yield the low velocity limit of kinetic energy if and only if (0) = mc2. E (15.101) There are several questions to be answered at this point, some experimentally and some theoretically. We need to measure the rest masses and theoretically verify that only this transformation correctly preserves the energy momentum conservation laws in elastic collisions as required. Beyond that, there are still some uncertainties. For example, there could in principal be an additional constant energy added to the energy term that was not scaled by \u03b3 and the laws of physics would still be expressible, since they are not sensitive to absolute energy scale. We will take advantage of that freedom in several instances to add or subtract an in\ufb01nite theoretical constant in order to make the rest mass come out to the observed experimental mass m. This is called renormalization. To obtain the same result a di\ufb00erent way, we turn to the notation of 4\u2013vectors. We observe that the common factor of \u03b3 above in both E and p also occurs when one makes velocity into a four vector. This suggests that energy and momentum can similarly be made into four vectors that transform like the coordinates under a boost. If we try the combination p0 = mcU0 = p = mU E c we see that it works exactly. It results in an invariant p2 0 \u2212 p \u00b7 p = (mc)2. (15.102) (15.103) (15.104) It is easy to see the value of the invariant when v = 0; you should verify explicitly that it holds when v = 0 as well. Practically speaking, it su\ufb03ces to show that this length is invariant when one wishes to show that its components transform like the coordinates under the action of a boost (why is that?). The total energy can thus be expressed in terms of the three momentum as E = c2p2 + m2c4. (15.105) Finally, it is sometimes convenient to be able to get the velocity of the particle in terms of its energy and momentum q u = c2p E (15.106) which follows directly from the de\ufb01nitions. This completes our review of \u201celementary relativity theory\u201d. We shall now proceed to develop the theory in a new, geometric language which is suitable to", " our much more sophisticated needs. To do this, we will need to begin by generalizing the notion of a four dimensional vector space with a set of transformations that leave an appropriately de\ufb01ned \u201clength\u201d invariant. 6 Chapter 16 The Lorentz Group 16.1 The Geometry of Space\u2013Time Recall that a great deal of simpli\ufb01cation of the kinematics of classical non\u2013 relativistic mechanics occurs when one considers the group structure of transformations with respect to the underlying coordinates. Speci\ufb01cally, the group of inversions, translations and rotations of a given coordinate system leave the norm (length) of a given vector invariant. These transformations form the Euclidean group in three dimensions, E3. For those of you who led deprived childhoods, a group is a set of mathematical objects (a, b, c...) with a rule of composition, or group product, (a b) such that: G \u25e6 1. Every product of a pair of elements in the group is also in the group. ). This property is called closure. That is, if a, b then. The group must contain a special element called the identity I such that a \u25e6 I = a for all a. \u2208 G \u2208 G 3. Every element of the group must have an inverse, also in a \u2208 G then 1 a\u2212 \u2203 \u2208 G G such that a a\u2212 1 = I. \u25e6 If. G 4. The group product must be associative. That is, a c, a, b, c \u2200. \u2208 G (b \u25e6 \u25e6 c) = (a b) \u25e6 \u25e6 225 \u25e6 \u25e6 b = b If the group product commutes (a a) the group is said to be Abelian1 otherwise the group is said to be non\u2013Abelian, which is sensible enough. A Lie group is a continuous group2 such as the group of in\ufb01nitesimal transformations. It necessarily has an uncountable in\ufb01nity of elements. There are also discrete (but countably in\ufb01nite) groups, \ufb01nite groups, and everything in between. There are also \u201csemi\u2013groups\u201d (which do not, for example, contain an inverse). Finally, one can construct \u201cnon\u2013 associative\u201d structures like groups from non\u2013associative algebras like the octonions. Multiplication over the reals forms a continuous Abelian group. Rotations form a non\u2013Abelian Lie group. Multiplication over rational numbers forms a countably in\ufb01nite group. The set of rotations and inversions that leave a square invariant form a \ufb01nite (point) group. The \u201crenormalization group\u201d you will hear much about over the years is not a group but a semi\u2013group \u2014 it lacks an inverse. However, our purpose here is not, however, to study group theory per se. One could study group theory for four years straight and still only scratch the surface. It is somewhat surprising that, given the importance of group theory in physics, we don\u2019t o\ufb00er a single course in it, but then again, it\u2019s not that surprising... With that in mind, we can decide what we are looking for. We seek initially the set of transformations in four dimensions that will leave s2 = x2 (x x) (16.1) \u00b7 0 \u2212 invariant for a single event x with respect to a particular coordinate origin. These transformations form a group called the homogeneous Lorentz group. It consists of ordinary rotations in the spatial part, the Lorentz transformations we have just learned that mix space and time, and several discrete transformations such as space inversion(s) and time inversion. The set of transformations that leave the quantity s2(x, y) = (x0 \u2212 y0)2 \u2212 (x1 \u2212 y1)2 + (x2 \u2212 y2)2 + (x3 \u2212 n y3)2) o (16.2) invariant form the inhomogeneous Lorentz3 or Poincar\u00b4e group. It consists of the homogeneous group (including the \u201cimproper\u201d transformations 1Wikipedia: http://www.wikipedia.org/wiki/Abelian group. ; 2Wikipedia: http://www.wikipedia.org/wiki/Lie group., 3Wikipedia: http://www.wikipedia.org/wiki/Lorentz group., that include spatial re\ufb02ection and time reversal) and uniform translations of the origin. If anyone cares, the Lorentz group is the generalized orthogonal group O(1,3). The proper subgroup of the Lorentz group (the one that is simply", " connected spatially (no odd inversions) and contains the identity) is SO(1,3) the special orthogonal group. If time\u2019s direction is also preserved we add a +, SO+(1,3). This nomenclature is de\ufb01ned here for your convenience but of course the wikinote reference contains active links to a lot of this in detail. We will de\ufb01ne s(x, y) to be the norm of relativistic space\u2013time. This quantity may be considered to be the invariant \u201cdistance\u201d (squared) between two events, x and y, and of course is one of the fundamental objects associated with the construction of di\ufb00erentials. Since quantities that are unchanged by a geometric transformation are called scalars it is evident that s(x, y) is a 4\u2013scalar. Since the \ufb01rst postulate states that the laws of physics must be invariant under homogeneous (at least) Lorentz transformations, they must ultimately be based on Lorentz scalars. Indeed, the Lagrangian densities upon which \ufb01eld theories are based are generally constructed to be Lorentz scalars. This is a strong constraint on allowed theories. These scalars are, however, formed out of 4\u2013vectors (as we see above) or, more generally, the contraction of 4\u2013tensors. We must, therefore, determine the general transformation properties of a tensor of arbitrary rank to completely determine a theory. In the part of this book devoted to mathematical physics is an entire chapter that discusses tensors, in particular the de\ufb01nitions of covariant and contravariant tensors, how to contract (Einstein sum) pairs of tensors to form tensors of lower rank, and the role of the metric tensor in de\ufb01ning tensor coordinate frames and transformations thereupon. We will not repeat this review or introduction (depending on the student) and urge students to at this time spend an hour or so working through this chapter before continuing (even if you\u2019ve seen it before). 16.2 Tensors in 4 Dimensions Let us now consider the speci\ufb01c nature of tensors on four-dimensional spacetime. Tensors of rank k4 are categorized (for each coordinate index) by their transformation properties relative to a transformation of the underlying coordinate system x x\u2032 as de\ufb01ned above. This transformation is implicit in all the discussion below. \u2192 A scalar (tensor of rank zero) is unchanged by such a transformation. This is not a trivial statement! It is trivial for scalar numbers like \u03c0, no doubt, but in physics the interesting part of this requirement occurs when discussing the scalars that result algebraically from fully contracting products of tensors over all of their indices using the metric tensor. This will be made quite clear below. For a vector (tensor of rank one) we have two possibilities. Either it transforms like the coordinate itself and we have a contravariant vector (A0, A1, A2, A3) such that A\u03b1 = \u2202 \u00afx\u03b1 \u2202x\u03b2 A\u03b2 (16.3) (noting that all the indices are on top, along with the new primed coordinate). This makes the di\ufb00erential transformation relationship to the underlying ordinary (contravariant) coordinates explicit and is obviously an identity for those coordinates. Alternatively, we have a covariant vector (B0, B1, B2, B3) such that B\u03b1 = \u2202x\u03b2 \u2202 \u00afx\u03b1 B\u03b2 (16.4) (with the coordinate indices on top and the new primed coordinate on the bottom). Again, note that this is precisely what we expect \u2013 the transformation is in the opposite sense of that of the underlying coordinates. We need in both cases, of course, to \ufb01gure out the matrix of e.g. \u2202x\u03b2 \u2202x\u03b1 explicitly. 4The rank of a tensor is determined by the number of indices it has. Scalars are 0th rank, vectors are 1st rank, 2D matrices are 2nd rank, and our old friend \u01ebijk is a third rank fully antisymmetric tensor. In a moment we will see explicitly what exactly the di\ufb00erence is between these two types of \ufb01rst rank tensors. First, however, we should note that contravariant tensors of rank 2 transform like F \u03b1\u03b2 = \u2202 \u00afx\u03b1 \u2202x\u03b3 \u2202 \u00afx\u03b2 \u2202x\u03b4 F \u03b3\u03b4. Similarly, we have covariant tensors of rank 2 and mixed tensors of rank 2 G\u03b1\u03b2 = \u2202x\u03b3 \u2202 \u00afx\u03b1 \u2202x\u03b4 \u2202 \u00afx\u03b2 G\u03b3\ufffd", "\ufffd H \u03b1 \u03b2 = \u2202 \u00afx\u03b1 \u2202x\u03b3 \u2202x\u03b4 \u2202 \u00afx\u03b2 H \u03b3 \u03b4. (16.5) (16.6) (16.7) It is clearly a trivial exercise to determine the co/contra variant transformation properties of higher rank tensors. We can form higher rank tensors by means of an outer (dyadic) product, where we simply take two tensors of some rank and multiply them out componentwise, preserving products of any underlying basis vectors as they occur. For example we can construct a second rank tensor by: F \u03b1\u03b2 = A\u03b1B\u03b2 (16.8) where \u03b1 and \u03b2 run over the full range of index values. Note well that this de\ufb01nes a square matrix in this case of basis vector dyads as objects such as \u02c6x\u02c6x, \u02c6x\u02c6y,... occur. One important question is whether all e.g. second rank tensors can be written as products of \ufb01rst rank tensors. It is not the general case that this is possible, but in many of our uses of these ideas in physics it will In this case the generalized product forms a division algebra where be. we can factor e.g. second rank tensors into \ufb01rst rank tensors in various ways. Division algebras are discussed in the Mathematical Physics section as well, and interested students should return there to read about geometric algebras, the result of fully generalizing the notion of complex numbers to complex spaces of arbitrary dimension while preserving the factorizability of the algebraic objects. In addition to extending the rank of tensor objects by forming dyadic, triadic, or n-adic products of tensors, we can reduce the rank of tensors by means of a process called contraction. A contraction of two tensors is the result of setting two of the indices (typically they must be a covariant/contravariant pair) to be equal and performing the Einstein summation over the shared range. This reduces the rank of the expression by one relative to that of its constituents, hence the term \u201ccontraction\u201d. An expression can be contracted over several components at a time when doing algebra so second rank tensors can be contracted to form a 4-scalar, for example, or third rank tensors can be contracted to \ufb01rst. Our familiar notion of multiplying a vector by a matrix to produce a vector in proper tensor language is to form the outer product of the matrix (second rank tensor) and the vector (\ufb01rst rank tensor), set the rightmost indices to be equal and sum over that index to produce the resulting \ufb01rst rank tensor. Hence we de\ufb01ne our scalar product to be the contraction of a covari- ant and contravariant vector. A = B\u03b1A\u03b1 B \u00b7 (16.9) Note that I\u2019ve introduced a sort of \u201csloppy\u201d convention that a single quantity like B or A can be a four-vector in context. Clearly the expression on the right side is less ambiguous! Then: B\u2032 \u00b7 A\u2032 = \u2202 \u00afx\u03b1 \u2202x\u03b4 A\u03b4 = \u2202x\u03b3 \u2202 \u00afx\u03b1 B\u03b3 \u2202x\u03b3 \u2202x\u03b4 B\u03b3A\u03b4 = \u03b4\u03b3\u03b4B\u03b3A\u03b4 = B\u03b4A\u03b4 = B and the desired invariance property is proved. Hmmm, that was pretty easy! Maybe there is something to this notation thing after all! A \u00b7 (16.10) 16.3 The Metric Tensor The section above is still very generic and little of it depends on whether the tensors are three or four or ten dimensional. We now need to make them work for the speci\ufb01c geometry we are interested in, which is one where we will ultimately be seeking transformations that preserve the invariant interval: (ds)2 = (dx0)2 (dx1)2 (dx2)2 (dx3)2 \u2212 as this is the one that directly encodes an invariant speed of light. \u2212 \u2212 (16.11) From this point on, we must be careful not to confuse x x = x2 and x2 = y, etc. Contravariant indices should be clear from context, as should be powers. To simplify life, algebraically indices are always greek (4\u2013vector) or roman italic (3\u2013vector) while powers are still powers and hence are generally integers. \u00b7 Let us write this in terms of only contravariant pieces dx\u00b5. This requires that we introduce a relative minus sign when contracting out the components of the spatial part of the di\ufb00erential only. We can most easily encode this requirement into a special matrix (t", "ensor) called the metric tensor as: (ds)2 = g\u03b1\u03b2 dx\u03b1dx\u03b2 The tensor g obviously satis\ufb01es the following property: g\u03b1\u03b2 = g\u03b2\u03b1 (16.12) (16.13) (that is, it is symmetric) because the multiplication in the Einstein summation is ordinary multiplication and hence commutative. It is called the metric tensor because it de\ufb01nes the way length is measured. At this point if we were going to discuss general relativity we would have to learn what a manifold5 s. Technically, a manifold is a coordinate system that may be curved but which is locally \ufb02at. By locally \ufb02at I mean very speci\ufb01cally that one can cover the entire space with \u201cpatches\u201d in the neighborhood of points where the coordinate system is locally Euclidean (e.g. Cartesian). An example of a curved space manifold is the surface of a sphere (think the surface of the earth). When we look down at the 5Wikipedia: http://www.wikipedia.org/wiki/Manifold. i ground beneath our feet, it looks quite \ufb02at and we can draw triangles on it that appear to have interior angles that sum to \u03c0 and we can draw a map of (say) our county that more or less accurately encodes distances on the ground in terms of distances measured on the map. However, if we take too big a patch all of this breaks down. The angles in a triangle sum to strictly more than \u03c0 radians. Maps have to be distorted and chopped into pieces to correctly represent distances on the ground as distances on the \ufb02at 2-dimensional map. This is how a manifold works \u2013 we can work with it in the local neighborhood of any point as if it is \ufb02at, but if we go too far we have to work harder and correct for its curvature, where \u201ctoo far\u201d is obviously de\ufb01ned in terms of the scale of its curvature and some common sense. General relativity introduces the hypothesis that gravitational \ufb01elds bend space-time. However, this bending is very, very slight unless one is in a very strong gravitational \ufb01eld, and this bending preserves a local smoothness of space-time so that space-time, although it is no longer strictly Euclidean, is still a manifold and we can do all sorts of transformations in a very general way as long as we restrict the results to a locally \ufb02at patch. In our discussion of special relativity we will assume from the beginning that our space\u2013time is \ufb02at and not bent by strong gravitational \ufb01elds. In this case the metric tensor can be expressed in a very simple form. We will use the Lorentz metric (as opposed to the Minkowski metric that uses x4 = ict instead of x0). Using our de\ufb01nitions of the \u00b5 = 0, 1, 2, 3 coordinates, g in the di\ufb00erentials above is just: and we see that it is not just symmetric, it is diagonal. g00 = 1, g11 = g22 = g33 = 1 \u2212 (16.14) The contravariant and mixed metric tensors for \ufb02at space\u2013time are the \u2202x\u03b2 coordinate transformation matrices same (this follows by considering the \u2202x\u03b1 that de\ufb01ne co- and contra-variance): g\u03b1\u03b2 = g\u03b2 \u03b1 = g\u03b1\u03b2. (16.15) Finally, the contraction of any two metric tensors is the \u201cidentity\u201d tensor, g\u03b1\u03b3g\u03b3\u03b2 = \u03b4\u03b2 \u03b1 = \u03b4\u03b1\u03b2 = \u03b4\u03b1\u03b2. (16.16) Since we want (ds)2 to be (to contract to) a scalar, it is clear that: x\u03b1 = g\u03b1\u03b2x\u03b2 (16.17) x\u03b1 = g\u03b1\u03b2x\u03b2 (16.18) or the metric tensor can be used to raise or lower arbitrary indices, converting covariant indices to contravariant and vice\u2013versa: F \u00b5\u03b1\u03bd = g\u03b1\u03b2F \u00b5\u03bd \u03b2 (16.19) This is an important trick! Note well that in order to perform a contraction that reduces the rank of the expression by one, the indices being summed must occur as a co/contra pair (in either order). If both are covariant, or both are contravariant, one or the other must be raised or lowered by contracting it with the metric tensor before contracting the overall pair! We use this repeatedly in the algebra in sections below. Finally we are in a position to see how covariant and contravariant vectors di\ufb00er (in this metric). We have", " already seen that \u201cordinary\u201d vectors must linearly transform like contravariant vectors. Given a contravariant vector (A0, A1, A2, A3) we thus see that or A0 = A0, A1 = A1, A2 = \u2212 A2, A3 = \u2212 A3 \u2212 A\u03b1 = (A0, A), A\u03b1 = (A0, A). \u2212 (16.20) (16.21) Covariant vectors are just spatially inverted contravariant vectors. Note that this de\ufb01nition, together with our de\ufb01nition of the general scalar product, reconstructs the desired invariant: A = B\u03b1A\u03b1 = (B0A0 B \u00b7 B \u00b7 \u2212 A) (16.22) This tells us how ordinary quantities transform. However, we are also interested in how tensor di\ufb00erentials transform, since these are involved in the construction of a dynamical system. By considering the chain rule we see that \u2202 \u2202 \u00afx\u03b1 = \u2202x\u03b2 \u2202 \u00afx\u03b1 \u2202 \u2202x\u03b2 (16.23) or, di\ufb00erentiation by a contravariant coordinate transforms like a covariant vector operator. This is more or less the de\ufb01nition of covariant, in fact. Similarly, di\ufb00erentiation with respect to a covariant vector coordinate transforms like a contravariant vector operator. This also follows from the above by using the metric tensor, \u2202 \u2202x\u03b1 = g\u03b1\u03b2 \u2202 \u2202x\u03b2. (16.24) It is tedious to write out all of the pieces of partial derivatives w.r.t. various components, so we (as usual, being the lazy sorts that we are) introduce a \u201csimplifying\u201d notation. It does, too, after you get used to it. = ( \u2202\u03b1 = \u2202 \u2202x\u03b1 \u2202 \u2202x\u03b1 = ( Note that we have cleverly indicated the co/contra nature of the vector operators by the placement of the index on the bare partial. \u2202 \u2202x0, \u2212 \u2202 \u2202x0, +\u2207). (16.26) (16.25) \u2202\u03b1 = \u2207) We cannot resist writing down the 4\u2013divergence of a 4\u2013vector: \u2202\u03b1A\u03b1 = \u2202\u03b1A\u03b1 = \u2202A0 \u2202x0 + \u2207 \u00b7 A = 1 c \u2202A0 \u2202t + \u2207 A \u00b7 (16.27) which looks a lot like a continuity equation or a certain well\u2013known gauge condition. (Medidate on just what A\u00b5 would need to be for either of these equations to be realized as a four-scalar). Hmmmmmm, I say. Even more entertaining is the 4\u2013Laplacian, called the D\u2019Lambertian op- erator: 2 = \u2202\u03b1\u2202\u03b1 = \u22022 \u2202x02 \u2212 \u2207 = 1 c2 \u22022 \u2202t2 \u2212 \u2207 2 2 (16.28) (16.29) which just happens to be the (negative of the) wave operator! Hmmmmmmmm! By strange coincidence, certain objects of great importance in electrodynamics \u201cjust happen\u201d to be Lorentz scalars! Remember that I did say above that part of the point of introducing this lovely tensor notation was to make the various transformational symmetries of physical quantities manifest, and this appears to be true with a vengeance! That was the \u201ceasy\u201d part. It was all geometry. Now we have to do the messy part and derive the in\ufb01nitesimal transformations that leave scalars in this metric invariant. 16.4 Generators of the Lorentz Group Let x0 x1 x2 x3 \uf8f6 x = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed be a column vector. Note that we no longer indicate a vector by using a vector arrow and/or boldface \u2013 those are reserved for the spatial part of the four-vector only. Then a \u201cmatrix\u201d scalar product is formed in the usual way by (16.30) \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 (a, b) = \u02dcab (16.31) where \u02dca is the (row vector) transpose of a. The metrix tensor is just a matrix16.32) and g2 =\u21d4I. Finally, x0 x1 x2 x3 \u2212 \u2212 \u2212 x0 x1 x2 x3 \uf8f6 = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \ufffd", "\ufffd\ufffd gx = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed \uf8f6. (16.33) \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 In this compact notation we de\ufb01ne the scalar product in this metric to \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 be a \u00b7 b = (a, gb) = (ga, b) = \u02dcagb = a\u03b1g\u03b1\u03b2b\u03b2 = a\u03b1b\u03b1. (16.34) We seek the set (group, we hope) of linear transformations that leaves x invariant. Since this is the \u201cnorm\u201d (squared) of a four vec(x, gx) = x tor, these are \u201clength preserving\u201d transformations in this four dimensional metric. That is, we want all matrices A such that \u00b7 leaves the norm of x invariant, x\u2032 = Ax x\u2032 = \u02dcx\u2032gx\u2032 = \u02dcxgx = x x\u2032 \u00b7 x \u00b7 (16.35) (16.36) or or \u02dcx \u02dcAgAx = \u02dcxgx \u02dcAgA = g. Clearly this last condition is su\ufb03cient to ensure this property in A. Now, det = det )2 = det A | | g where the last equality is required. But det | 1 \u00b1 \u02dcAgA 1 \u2212 (det g | = = | | A | | is a constraint on the allowed matrices (transformations) A. There are thus two classes of transformations we can consider. The (16.40) det g | | = 0, so (16.37) (16.38) (16.39) proper Lorentz transformations with det A | | improper Lorentz transformations with det = +1; and A | | = 1. \u00b1 = \u2212 \u2212 1) and A = I (space and time inversion, with det Proper L. T.\u2019s contain the identity (and thus can form a group by themselves), but improper L. T.\u2019s can have either sign of the determinant. This is a signal that the metric we are using is \u201cinde\ufb01nite\u201d. Two examples of improper transformations that illustrate this point are spatial inversions (with det A | | In very general terms, the proper transformations are the continuously connected ones that form a Lie group, the improper ones include one or more inversions and are not equal to the product of any two proper transformations. The proper transformations are a subgroup of the full group \u2014 this is not true of the improper ones, which, among other things, lack the identity. With this in mind, let us review the properties of in\ufb01nitesimal linear transformations, preparatory to deducing the particular ones that form the homogeneous Lorentz group. = +1). A | | 16.4.1 In\ufb01nitesimal Transformations We seek (Lie) groups of continous linear transformations, x\u2032 = Tax (16.41) 6 or x\u2032 \u00b5 = f \u00b5(x; a) (16.42) for \u00b5 = 1, 2,... n. We require that the a = a1,..., ar are r real numbers r must be minimal (parameters) that characterize the transformation. (\u201cessential\u201d). Examples of transformations of importance in physics (that you should already be familiar with) include x\u2032 = Tdx = x + d (16.43) where d = (d1,..., dn). This is the (n parameter) translation group in n dimensions. Also, x\u2032 i = Rijxj (16.44) where R \u02dcR = I, det is the (three parameter) rotation group. > 0, i = 1, 2, 3 (16.45) R | | An in\ufb01nitesimal transformation in one of the parameters is de\ufb01ned by Ta(0)+\u01eb = I\u01eb + (\u01eb2). (16.46) O In this de\ufb01nition, a(0) are the (r\u2013parameter) values associated with the identity transformation I. These can be chosen to be zero by suitably choosing the parameter coordinates. The in\ufb01nitesimal parameters \u01ebu are taken to zero, so that \u01eb2 = \u01ebu\u01ebu (summed) is neglible. Thus where and Putting this all together, I\u01eb = I + \u01ebuQu Qu = f \u00b5u(x) \u2202 \u2202x\u00b5 f \u00b5u(x) = \u2202f \u00b5(x, a) \u2202au. a=a(0) x\u2032 =", " (Ta(0)+\u01eb)x = (I + \u01ebuQu) x = Ix + \u01ebuQux = x + \u01ebu \u2202f \u00b5(x, a) \u2202au \u2202x \u2202x\u00b5 a=a(0) (16.47) (16.48) (16.49) (16.50) (summed over \u00b5 = 0,..., 3 in four dimensional space\u2013time and u = 0,..., r). Thus (unsurprisingly) x\u2032 \u00b5 = x\u03bd\u03b4\u00b5 \u03bd + \u01ebu g\u00b5 \u03bd a=a(0) (16.51) \u2202f \u03bd \u2202au which has the form of the \ufb01rst two terms of a Taylor series. This is characteristic of in\ufb01nitesimal linear transformations. One can easily verify that I\u01ebI\u01eb\u2032 = I\u01eb\u2032I\u01eb (in\ufb01nitesimal transformations commute) and that 1 I \u2212 \u01eb = I \u01eb \u2212 (16.52) (16.53) (to order \u01eb2). They thus have an identity, an inverse, and can be shown to be associative. The continuous transformation group (mentioned above) follows immediately from making du (the displacement of coordinates) in\ufb01nitesimal and \ufb01nding \ufb01nite displacements by integration. The rotation group (matrices) are a little trickier. They are \u2202Rij \u2202ak where I\u01eb = I + gS \u02dcS = S, \u2212 Sij = \u01ebk (16.54) (16.55). a(0 The in\ufb01nitesimal S are antisymmetric and traceless (in 3D), so they have only three independent parameters (that are thus \u201cessential\u201d). We can write them generally as Sij = \u01ebijk d\u03c9k (16.56) where the d\u03c9k is the in\ufb01nitesimal parameter and where \u01ebijk is the antisymmetric unit tensor. Thus, if we see that dxi = x\u2032i \u2212 xi = Sijxj = \u01ebijk xj d\u03c9k (16.57) dx = x d\u03c9 \u00d7 (16.58) A moment of thought should convince you that d\u03c9 is the in\ufb01nitesimal (vector) rotation angle, with direction that points along the axis of rotation. To obtain the rotation group we must show that every rotation can be obtained by integrating Id\u03c9. This follows by writing an arbitrary rotation or product of rotations as a single rotation about a \ufb01xed axis. For d\u03c9 parallel to this axis \u2126, this is obviously true, as I show next. Since any rotation can be written this way, the rotations indeed form a group. The integration proceeds like: R\u2126 = lim \u2206\u03c9 (R\u2206\u03c9)\u2126/\u2206\u03c9 0 \u2192 where \u2206\u03c9 = \u2206\u03c9 | | and \u2126 =. We can parameterize this as \u2126 | | R\u2126 = lim m (I + \u2192\u221e 1 m \u2126S0)m = e\u2126S0 where (S0)ij = \u01ebijk \u2206\u03c9k \u2206\u03c9. (16.59) (16.60) (16.61) Believe it or not, this was one of the primary things we wanted to show in this aside. What it shows is that rotations about an arbitrary axis can be written as an exponential that can be thought of as the in\ufb01nite product of a series of in\ufb01nitesimal transformations where each transformation has various nice properties. With these known results from simpler days recalled to mind, we return to the homogeneous, proper Lorentz group. Here we seek the in\ufb01nitesimal linear transformations, etc. in four dimensions. Algebraically one proceeds almost identically to the case of rotation, but now in four dimensions and with the goal of preserving length in a di\ufb00erent metric. A general in\ufb01nitesimal transformation can be written compactly as: I\u01eb = I + gL (16.62) where (as before) \u02dcgL = gL (and hence gL is traceless), L is in\ufb01nitesimal, and where g is the usual metric tensor (that follows from all the annoying derivatives with respect to the parameters and coordinates). \u2212 Thus A = lim m \u2192\u221e I + m 1 m L = eL (16.63) de\ufb01nes the form of a general transformation matrix associated with a given", " \u201cdirection\u201d in the parameter space constructed from an in\ufb01nite product of in\ufb01nitesimal transformations, each of which is basically the leading term of a Taylor series of the underlying coordinate function transformation in terms of the parameters. This justi\ufb01es the \u201cansatz\u201d made by Jackson. The matrices L are called the generators of the linear transformation. Thus, whenever we write A = eL (16.64) where the L\u2019s are (to be) the generators of the Lorentz group transformations we should remember what it stands for. Let\u2019s \ufb01nd the distinct L. Each one 4 real, traceless matrix that is (as we shall see) antisymmetric in the is a 4 spatial part (since gL is antisymmetric from the above). \u00d7 To construct A (and \ufb01nd the distinct components of L) we make use of its properties. Its determinant is det A | | = det = eTr L = 1 \u00b1 (16.65) (eL) (This follows from doing a similarity transformation to put A in diagonal form. L is necessarily then diagonal. Similarity transformations do not alter the determinant, because det S\u2212 1M S = det 1 S\u2212 det det M | | S | | = det. M | | (16.66) If L is diagonal, then the last equation follows from the usual properties of the exponential and the de\ufb01nition of the exponential of a matrix.) If L is real then det 1 is excluded by this result. If L is traceless = A | | \u2212 (and only if, given that it is real), then det A | | = +1 (16.67) which is required to be true for proper Lorentz transformations (recall from last time). Making L a traceless 4x4 matrix therefore su\ufb03ces to ensure that we will \ufb01nd only proper Lorentz transformations. Think back to the requirement that: \u02dcAgA = g (16.68) in order to preserve the invariant interval where matrix. \u00d7 and L is a real, traceless, 4 If we multiply from the right by A\u2212 equivalent also to 16.69) 1 and the left by g, this equation is Since \u02dcA = e \u02dcL, A\u2212 1 = e\u2212 L, and I = g2: g \u02dcAg = A\u2212 1. or g \u02dcAg = eg2 \u02dcL = eg \u02dcLg = e\u2212 L g \u02dcLg = L. \u2212 (16.70) (16.71) (16.72) (This can also easily be proven by considering the \u201cpower series\u201d or product expansions of the exponentials of the associated matrices above, changing the sign/direction of the in\ufb01nitesimal series.) Finally, if we multiply both sides from the left by g and express the left hand side as a transpose, we get \u02dcgL = gL. \u2212 (16.73) From this we see that the matrix gL is traceless and antisymmetric as noted/expected from above. If we mentally factor out the g, we can without loss of generality write L as L01 L02 \u2212 L04 \u2212 L02 L03 L12 L13 L23 0 L23 0 L01 0 L12 L13 \u2212 This matrix form satis\ufb01es all the constraints we deduced above for the generators. Any L of this form will make an A that preserves the invariant interval (length) of a four vector. There are exactly six essential parameters as expected. Finally, if we use our intuition, we would expect that (16.74) \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8f6. the Lij for i, j = 1, 2, 3 form the rotation subgroup and describe physical rotations. So this is just great. Let us now separate out the individual couplings for our appreciation and easy manipulation. To do that we de\ufb01ne six fundamental matrices (called the generators of the group from which we can construct an arbitrary L and hence A. They are basically the individual matrices with unit or zero components that can be scaled by the six parameters L\u00b5\u03bd. The particular choices for the signs make certain relations work out nicely S1 = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed S2 = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed S3 = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed K1 = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed K2 = \uf8eb \uf8ec \ufffd", "\ufffd\ufffd \uf8ec \uf8ec \uf8ed K3 = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8. \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 (16.75) (16.76) (16.77) (16.78) (16.79) (16.80) The matrices Si generate rotations in the spatial part and the matrices Ki generate boosts. Note that the squares of these matrices are diagonal and either +1 or 1 in the submatrix involved = \uf8eb S2 \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed and 1 = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed etc. From this we can deduce that, \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 S3 K 3 Si i = i = Ki. \u2212 Note that these relations are very similar to the multiplication rules for unit pure complex or pure real numbers. The reason this is important is that if we form the dot product of a vector of these generators with a spatial vector (e\ufb00ectively decomposing a vector parameter in terms of these matrices) in the exponential expansion, the following relations can be used to reduce powers of the generators. and (\u02c6\u01eb S)3 = (\u02c6\u01eb \u00b7 S \u02c6\u01eb \u2212 \u00b7 K)3 = \u02c6\u01eb K \u00b7 \u00b7 In these expressions, \u02c6\u01eb an arbitrary unit vector, and these expressions e\ufb00ectively match up the generator axes (which were arbitrary) with the direction of the parameter vector for rotation or boost respectively. After the reduction (as we shall see below) the exponential is, in fact, a well-behaved and easily understood matrix! It is easy (and important!) to determine the commutation relations of these generators. They are: [Si, Sj] = \u01ebijkSk [Si, Kj] = \u01ebijkKk [Ki, Kj] = \u01ebijkSk. \u2212 (16.87) (16.88) (16.89) (16.81) (16.82) (16.83) (16.84) (16.85) (16.86) The \ufb01rst set are immediately recognizable. They tells us that \u201ctwo rotations performed in both orders di\ufb00er by a rotation\u201d. The second and third show that \u201ca boost and a rotation di\ufb00er by a boost\u201d and \u201ctwo boosts di\ufb00er by a rotation\u201d, respectively. In quotes because that is somewhat oversimpli\ufb01ed, but it gets some of the idea across. These are the generators for the groups SL(2, C) or O(1, 3). The latter is the group of relativity as we are currently studying it. \u00d7 A question that has been brought up in class is \u201cwhere is the factor i in the generators of rotation\u201d so that S S = iS as we might expect from considering spin and angular momentum in other contexts. It is there, but subtly hidden, in the fact that S2 Ii in the projective block of the rotation matrices only. Matrices appear to be a way to represent geometric algebras, as most readers of this text should already know from their study of the (quaternionic) Pauli spin matrices. We won\u2019t dwell on this here, but note well that the Pauli matrices I, \u03c31, \u03c32, \u03c33 are isomorphic to the unit quaternions 1, i, j, k via the mapping I k j, \u03c32\u03c33 \u2192 as the reader can easily verify6 Note well that: 1, \u03c31\u03c32 \u2192 i, \u03c33\u03c31 \u2192 i = \u2192 \u2212 \u03c33\u03c31 = 0 1 1 0! \u2212 (16.90) is both real and, not at all coincidentally, the structure of an S sub-block. With these de\ufb01nitions in hand, we can easily decompose L in terms of the S and the K matrices. We get: L = \u03c9 S \u03be K (16.91) \u00b7 where \u03c9 is a (\ufb01nite) rotation around", " an axis in direction \u02c6\u03c9 and where \u03be is a (\ufb01nite) boost in direction \u02c6\u03be. Thus the completely general form of A is \u2212 \u2212 \u00b7 A = e16.92) The (cartesian) components of \u03c9 and \u03be are now the six free parameters of the transformation. 6And should! That\u2019s right, you students, you know who I\u2019m talking to. So here\u2019s a question for you: Are I, \u03c33\u03c31 a real isomorphism to complex numbers? What would the various results of the introduction to complex numbers look like expressed in terms of these two matrices? What in particular does multiplying by a unimodular \u201ccomplex number\u201d such as cos(\u03b8)I + sin(\u03b8)\u03c33\u03c31 look like? Hmmm... veeeery interesting. Let us see that these are indeed the familiar boosts and rotations we are used to. After all, this exponential notation is not transparent. Suppose that \u03c9 = 0 and \u03be = \u03be \u02c6x. Then L = \u03beK1 and A = eL = I \u03beK1 + 1 2! \u2212 (\u03beK1)2 1 3! K1 sinh(\u03be) + K 2 K1(\u03be + \u2212 1 3! \u03be3 +...) + K 2 (\u03beK1)3 +... 1 2! 1 (I + 1 cosh(\u03beI \u2212 = (I \u2212 or (in matrix form) \u03be2 +...) (16.93) \uf8f6. (16.94) \u2212 cosh(\u03be) sinh(\u03be) 0 0 sinh(\u03be) 0 0 0 0 0 0 0 0 \u2212 cosh(\u03be) 0 0 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 A = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed which (ohmygosh!) is our old friend the Lorentz transformation, just like we derived it a la kiddy\u2013physics\u2013wise. As an exercise, show that the \u03c9 = \u03c9\u02c6x, \u03be = 0 result is a rotation around the x axis. Note that the step of \u201cadding and subtracting\u201d S2 1 is essential to reconstructing the series of the sine and cosine, just like the K1 was above for cosh and sinh. Now, a boost in an arbitrary direction is just A = e\u2212 \u03be K. \u00b7 We can certainly parameterize it by \u03be = \u02c6\u03b2 tanh\u2212 1 \u03b2 (16.95) (16.96) (since we know that \u03b2 = \u02c6\u03be tanh \u03be, inverting our former reasoning for [0, 1]. Then \u03b2 | | \u2208 A(\u03b2) = e\u2212 \u02c6\u03b2 \u00b7 K tanh\u22121 \u03b2. I can do no better than quote Jackson on the remainder: \u201cIt is left as an exercise to verify that... \u201d \u03b3\u03b21 \u03b3 \u2212 \u03b3\u03b21 1 + (\u03b3 1)\u03b22 1 \u2212 \u03b22 1)\u03b21\u03b22 \u03b22 (\u03b3 \u2212 \u2212 \u00b7 A(\u03b2) = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed \u2212 \u00b7 \u00b7 \u03b3\u03b22 \u2212 \u2212 \u2212 1)\u03b21\u03b22 \u03b22 (\u03b3 \u03b3\u03b23 \u00b7 \u00b7 \u00b7 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \u00b7 \u00b7 (16.97) (16.98) (etc.) which is just the explicit full matrix form of x0 \u2032 = \u03b3(x0 x\u2032 = x + \u2212 (\u03b3 x) \u00b7 1) (\u03b2 \u03b2 \u2212 \u03b22 x)\u03b2 \u00b7 \u2212 \u03b3\u03b2x0 (16.99) (16.100) from before. Now, we have enough information to construct the exact form of a simultaneous boost and rotation, but this presents a dual problem. When we go to factorize the results (like before) the components of independent boosts and rotations do not commute! If you like, A(\u03b2, 0)A(0, \u03c9) = A(0, \u03c9)A(\u03b2, 0) (16.101) and we cannot say anything trivial like A(\u03b2, \u03c9) = A(\u03b2, 0)A(0, \u03c9) (16.102) since it depends on the order they were performed in! Even worse, the product of two boosts is equal to a single boost and a rotation (if the boosts are not in the same direction)! The worst part, of course, is the algebra itself. A useful exercise for the algebraically", " inclined might be for someone to construct the general solution using, e.g. \u2013 mathematica. This suggests that for rotating relativistic systems (such as atoms or orbits around neutron stars) we may need a kinematic correction to account for the successive frame changes as the system rotates. The atom perceives itself as being \u201celliptically deformed\u201d. The conse- quences of this are observable. This is known as \u201cThomas precession\u201d. 16.5 Thomas Precession We must begin our discussion by noting that the magnetic moment of an electron is (according to the \u201cUhlenbeck-Goudsmit hypothesis\u201d) \u00b5 = ge 2mc s (16.103) 6 where s is the (half integer) spin of the electron in units of \u00afh and where g is the \u201cg\u2013factor\u201d introduced to accomodate two distinct results. The splitting of the observed spectra in an applied magnetic \ufb01eld B via the anomalous Zeeman interaction: UAZ = \u2212 ge 2mc B s \u00b7 (16.104) was correctly predicted only if g = 2. On the other hand (as we shall see), the simple classical argument that led to this result also led to a spin orbit interaction USO = g 2m2c2 (s 1 r dV dr L) \u00b7 (16.105) \u00d7 (where L = m(r v) is the orbital angular momentum of the electron) that was a factor of, curiously enough, 2 too large. That is, the \ufb01ne\u2013structure intervals observed in nature were only half the theoretically predicted values. If g = 1 was chosen instead, the splittings were correct but the Zeeman e\ufb00ect was then normal (instead of anomalous, as observed). I don\u2019t have the time to go into more detail on, for example, what the Zeeman e\ufb00ect (splitting of energy levels in an applied magnetic \ufb01eld) is. In any event, it is strictly a quantum e\ufb00ect, and you should study it soon in elementary quantum theory, if you haven\u2019t already. Thomas (who taught for years over at NC State) showed in 1927 that the discrepancy is due to a relativistic kinematic correction like that we previously considered. In a nutshell, the rest frame of the electron rotates as well as translates (boosts) and we must therefore take into account both kinematical e\ufb00ects. This results in an additional (Thomas) \u201cprecession\u201d of the frames. When Thomas precession is taken into account, not only are both the \ufb01ne structure and anomalous Zeeman e\ufb00ect in atoms accomodated, but a deeper understanding of the spin\u2013orbit interaction in nuclear physics (and rotating frames in general) also results. Let us begin by (na\u00b4\u0131vely) deriving the spin\u2013interaction energy. Suppose the electron moves with velocity v in external \ufb01elds E and B. Then the torque on the electron in its rest frame is just ds dt!rest frame = \u00b5 B\u2032 \u00d7 (16.106) where B\u2032 is the magnetic \ufb01eld in that frame. As we will show very soon, the magnetic \ufb01eld transforms like B\u2032 = B v c \u00d7 \u2212 E to order v2/c2. Then ds dt!rest frame = \u00b5 B \u2212 \u00d7 v c \u00d7 E. Associated with this torque there is an interaction energy 16.107) (16.108) (16.109) The electric force eE is very nearly the negative gradient of a spherically averaged potential energy V (r). For one electron atoms this is exact; it is a good approximation for all the others. Thus we will try using eE = r r dV dr \u2212 in the equation for the spin interaction energy: U \u2032 = ge 2mc s \u00b7 \u2212 B + g 2m2c2 (s \u00b7 L) 1 r dV dr (16.110) (16.111) (where L = m(r v) for the orbiting electron). This gives the anomalous Zeeman e\ufb00ect correctly (from the \ufb01rst term) but the spin orbit (\ufb01ne structure) splitting is a factor of two too large. Too bad! \u00d7 The error is, in a nutshell, that we have assumed the electron to be in a \u201crest\u201d frame (that is, a frame travelling in a straight line) when that frame is, in fact, rotating. There is an additional correction to vector quantities that arises from the rotation of the frame. This correction, in macroscopic systems, gives rise to things like coriolis force. Let us recall (from classical mechanics) that if a coordinate", " system rotates at some angular velocity \u03c9, the total rate of change of any vector quantity is given by dG dt!non = rot \u2212 dG dt!rest frame + \u03c9 G. \u00d7 (16.112) This is a geometric relation that says that a vector in a non\u2013rotating frame is related to the same vector expressed in a (rotating) \u201crest\u201d frame by adding its time rate of change in direction resulting from the rotation of the frame. A moment of quiet re\ufb02ection should convince you that this should have the magnitude d\u03b8 dt and should be perpendicular to \u03c9 and G. This just adds the rotation of the frame to the vector in the frame to get the vector in a non\u2013rotated frame. G Well, as I noted above, the expression we have given above for the time rate of change of the spin was correct for the \ufb01eld and moment expressed in the rest frame of the electron. In the lab (non\u2013rotating) frame, which is where we measure its energy, we therefore should have: ds dt!non rot \u2212 = s geB\u2032 2mc \u2212 \u03c9T! \u00d7 (16.113) where \u03c9T is the angular velocity of the precession of the frames. This adds a (s \u03c9T ) correction to the interaction energy: \u00b7 U = ge 2mc s \u00b7 \u2212 B + g 2m2c2 (s \u00b7 L) 1 r dV dr + (s \u03c9T ). \u00b7 (16.114) U is thus the laboratory potential energy of interaction. What, then, is the correct value of \u03c9T? To answer that we must consider carefully what de\ufb01nes the \u201crest\u201d frame of the accelerating electron. We will do so by chopping the motion of the If the electron is moving at velocity electron into in\ufb01nitesimal segments. v(t) = c\u03b2 at any instant of time t, then at t + \u03b4t the electron is moving at v(t) = c(\u03b2 + \u03b4\u03b2). To get from the lab frame (x) to the instantaneous rest frame of the electron (x\u2032) we must therefore boost: (at t) or x\u2032 = A(\u03b2)x x\u2032\u2032 = A(\u03b2 + \u03b4\u03b2)x (16.115) (16.116) (at t + \u03b4t). Note that for each of these transformations, there is no rotation, just the boost. The coordinate frame precession is going to be determined by the Lorentz transformation between these two (in\ufb01nitesimally separated) results: x\u2032\u2032 = AT x\u2032 (16.117) where (as I hope is obvious) AT = A(\u03b2 + \u03b4\u03b2)A\u2212 1(\u03b2) = A(\u03b2 + \u03b4\u03b2)A( \u03b2). \u2212 (16.118) To evaluate this (in the limit of vanishing \u03b4t) we will pick an initial \u03b2 along the 1 direction and add to it \u03b4\u03b2 in the 1\u20132 plane. Clearly this is general, for a suitable initial orientation of the coordinate system. Then \u03b3 \u03b3\u03b2 0 0 \u03b3\u03b2 ( \u2212 \u03b2) = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed and (keeping only \ufb01rst order terms in \u03b4\u03b2) \u2212 \u03b3 + \u03b33\u03b2\u03b4\u03b21 (\u03b3\u03b2 + \u03b33\u03b4\u03b21) \u03b3\u03b4\u03b22 0 \u2212 A(\u03b2 + \u03b4\u03b2\u03b3\u03b2 + \u03b33\u03b4\u03b21) \u03b3 + \u03b33\u03b2\u03b4\u03b21 1) \u03b2 \u03b4\u03b22 \u2212 0 (\u03b3 (16.119) \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \u03b3\u03b4\u03b22 0 \u2212 1) (\u03b3 \u03b2 \u03b4\u03b22 16.120) We multiply these matrices together to obtain: A(\u03b2 + \u03b4\u03b2) = 1 \u03b32\u03b4\u03b21 \u03b3\u03b4\u03b22 \u03b3 \u03b32\u03b4\u03b21 \u2212 \u2212 1 1) \u03b2 \u03b4\u03b22 \u2212 0 \u03b3\u03b4\u03b22 0 1) \u03b2 \u03b4\u03b22 0 \u2212 0 1 1 0 (\u03b3 (16.121Note that the action of A( we decompose this in terms of the S and K matrices, we get: \u03b2) is only in the upper left corner). Finally, if \u2212 AT = I \u03b3 1 \u2212 \u03b22! \u2212 (\u03b2 \u03b4\u03b2) S \u00b7 \u2212 \u00d7 (\u03b32\u03b4\u03b2 k + \u03b3\u03b4\u03b2 K ) \u22a5 \u00b7 (16.122) where \u03b4\u03b2 and \u03b4\u03b2 k to \u03b2, respectively. \u22a5 are the components of \u03b4\u03b2 parallel to and perpendicular To", " \ufb01rst order in \u03b4\u03b2, we see that the total transformation AT is equivalent to a boost and a rotation: AT = A(\u2206\u03b2)R(\u2206\u2126) (16.123) which can be performed in either order (because they are \u201cin\ufb01nitesimal\u201d and hence commute to \ufb01rst order. In this expression, and Obviously, and A(\u2206\u03b2) = I \u2206\u03b2 K \u00b7 \u2212 R(\u2206\u2126) = I \u2206\u2126 S. \u00b7 \u2212 \u2206\u03b2 = \u03b32\u03b4\u03b2 + \u03b3\u03b4\u03b2 \u22a5 k \u2206\u2126 = \u03b3 1 \u2212 \u03b22! (\u03b2 \u00d7 \u03b4\u03b2) = \u03b32 \u03b3 + 1 \u03b2 \u00d7 \u03b4\u03b2. (16.124) (16.125) (16.126) (16.127) Finally we see explicitly that at least for in\ufb01nitesimal transformations, a pure Lorentz boost A(\u03b2 + \u03b4\u03b2) is equivalent to a boost to an in\ufb01nitesimally di\ufb00ering frame A(\u03b2) followed by a simultaneous in\ufb01nitesimal boost and rotation. Now comes the tricky part. The equation of motion for the spin that we began with (in the \u201crest frame\u201d) can be expected to hold provided that the evolution of the rest frame is described by a series of in\ufb01nitesimal boosts alone (without rotations). In other words, we have to add the relativistic equivalent of counterrotating the frames (like we did above with the \u03c9T \u00d7 G term). These \u201crelativistically nonrotating coordinates\u201d are related to the instantaneous rest frame coordinates of the electron by the in\ufb01nitesimal boost { alone. In terms of the lab coordinates, x\u2032\u2032\u2032 = A(\u2206\u03b2) A(\u03b2)x = x\u2032 } x\u2032\u2032\u2032 = R( \u2206\u2126)A(\u03b2 + \u03b4\u03b2)x. \u2212 (16.128) (16.129) Thus the \u201crest\u201d system of coordinates of the electron are de\ufb01ned by x\u2032\u2032\u2032. \u2206\u2126 relative to the boosted laboratory axes x\u2032\u2032. If a They are rotated by physical vector G has a (proper) time rate of change of dG/d\u03c4 in the rest frame, the precession of the rest frame axes with respect to the laboratory makes the total time rate of change \u2212 dG dt!non = rot \u2212 dG dt!rest frame + \u03c9 G \u00d7 (16.130) as before with \u03c9T = lim 0 \u2192 (Recall that the connection to laboratory time is dG/dt = \u03b3\u2212 rest frame itself). \u00d7 c2 = \u03b4t \u2206\u2126 \u03b4t \u03b32 \u03b3 + 1 a v (16.131) 1dG/d\u03c4 in the The acceleration perpendicular to the instantaneous velocity appears in this expression because it is this quantity that produces the \u201crotation\u201d in the in\ufb01nitesimal transformation between frames that occured in the in\ufb01nitesimal time interval. Note that this is a purely kinematical e\ufb00ect, and has nothing to do with the laws of nature, just like the non-relativistic \u201ccoriolis force\u201d and \u201ccentrifugal force\u201d. If one wishes to relate the laws of nature as measured in some accelerating frame to those measured in a non-accelerating frame, then it is necessary to insert a \ufb01ctitious \u201cforce\u201d (or more properly interaction \u201cenergy\u201d) that is kinematic in origin. In this case, curiously enough, the laws of nature are known in the accelerating frame, and the \ufb01ctitious force appears in the lab frame, where it is not properly speaking \ufb01ctitious. However, it is still kinematic. That is, there is no actual energy associated with the \ufb01ctitious interaction (whatever that means); however, this interaction is necessary nonetheless if we wish to obtain the equation of motion from the energy equation alone without explicit consideration of the transformations of frames. To conclude, for electrons the acceleration is caused by the (screened) Coulomb force on the electron that keeps it bound. Thus 1 \u03c9T = \u2212 2c3 r \u00d7 m v 1 r dV dr 1 2m2c2 L = \u2212 1 r dV dr. (16.132) This has exactly the same form as the \u201crest frame\u201d spin orbit interaction with half the magnitude and the opposite sign. It beautifully cancels the extra factor of 2. The \ufb01", "nal result is: U \u2032 = ge 2mc s \u00b7 \u2212 B + (g 1) 2m2c2 (s \u2212 \u00b7 L) 1 r dV dr. (16.133) With g = 2, both the spin\u2013orbit interaction and the anomalous Zeeman e\ufb00ect are correctly predicted in accord with what is experimentally observed. Relativistic e\ufb00ects, which are generally thought of as being \u201csubtle\u201d, are not subtle at all when it comes to kinematics. The relativistic kinematic correction is as large as the other quantities naturally present independent of the particular orbit or speed of the electron. This e\ufb00ect is even more pronounced in atomic nuclei. There the electromagnetic forces are much weaker than the binding nuclear forces, and can be neglected to lowest order. However, even uncharged neutrons experience a spin\u2013orbit interaction \u00b7 L \u2212 1 r UrmN = dVN dr 1 2M 2c2 s that is now purely kinematic and has nothing whatsoever to do with the electromagnetic force! There will be a small electromagnetic correction to this for protons. This simple prediction is in reasonable agreement with what is observed in many nucleii for simple models for VN. Unfortunately, the system is actually so complicated that this simple minded, single particle description itself is not really valid. (16.134) This is just a drop in the proverbial bucket of accelerated systems. Clearly, accelerated, relativistic systems have a much more involved structure than that described by the Lorentz transformations alone. This becomes even more so when Einstein\u2019s revered equivalence principal is invoked, so that gravitational force and \u201creal\u201d acceleration are not (locally) distinguishable. But that is general relativity and far beyond the scope of this course. 16.6 Covariant Formulation of Electrodynam- ics We are now ready to get serious about electrodynamics. We have developed a beautiful, geometric system for describing the coordinates in terms of which electrodynamics must be formulated for the speed of light to be an invariant. We have developed a group of coordinate transformations that preserves that invariance. Now we must face up to the fact that our original equations of electrodynamics are not in a \u201ccovariant\u201d formulation that makes these constraints and transformation properties manifest. For example, we do not yet know how the electric and magnetic \ufb01elds themselves transform under a LT! Let us then reformulate our basic equations in 4\u2013tensor form. We will make the equations themselves 4\u2013scalars, 4\u2013vectors, or 4\u2013tensors of higher rank so that we can simply look at them and deduce their transformation properties. In addition, we will simplify the notation when possible. We begin at the beginning. All we really know about electromagnetic \ufb01elds is their (de\ufb01ned) action on a charged particle: dp dt = q E + v c \u00d7 B (16.135) (in 3\u2013vector notation). Well, we know that the 3\u2013vector momentum is just part of a 4\u2013vector momentum: p\u03b1 = (p0, p) = m(U 0, U) (16.136) (where p0 = E/c). Also, we don\u2019t know what \u201ct\u201d is (since that depends on the choice of frame) so we need to use \u201c\u03c4 \u201d instead in our de\ufb01nition. Thus we can write dp d\u03c4 = q c U 0E + U \u00d7. B (16.137) The left hand side tells us the rate of change of the (spatial) momentum, which is just part of a four vector. The time component should tell us how the energy changes with proper time: dp0 d\u03c4 = q c E. U \u00b7 (16.138) Now, if this energy\u2013force 4\u2013vector equation is to be covariant (so its transformed form is still a 4\u2013vector) then the right hand sides must form a 4\u2013vector too. Thus we must be able to express it (as a contraction of co and contra variant tensors) so that this property is \u201cmanifest\u201d. We know (experimentally) that charge is a Lorentz scalar; that is, charge is invariant under LT\u2019s. (U 0, U) forms a contravariant 4\u2013vector. From this we can deduce the 4\u2013tensor form for the electromagnetic \ufb01eld! E form the time component of a four vector, E Since the space parts U must be the time\u2013space part of a tensor of rank two. That is, \u00b7 U = F 0\u03b2U\u03b2. E \u00b7 (16.139) We could easily \ufb01", "nd B in a similar fashion and could eventually work out the electromagnetic \ufb01eld strength tensor. However, it is more constructive to keep on making four vectors, etc. out of the rest of the relations at hand. For example, we already have observed that the continuity equation is a covariant 4\u2013scalar: \u2202\u03c1 \u2202t + \u2207 \u00b7 J = 0. To make it\u2019s covariance manifest, we de\ufb01ne a 4\u2013current so that J \u03b1 = (c\u03c1, J) \u2202\u03b1J \u03b1 = 0 (16.140) (16.141) (16.142) is the continuity equation. Note that (as Jackson remarks) this only works because electric charge is a Lorentz invariant and so is a four\u2013dimensional volume element (since det A = +1). Next, consider the wave equations for the potentials in the Lorentz gauge (note well that Jackson for no obvious reason I can see still uses Gaussian units in this part of chapter 11, which is goiing to make this a pain to convert below \u2013 bear with me): 1 c2 \u22022\u03c6 \u2202t2 \u2212 \u2207 2\u03c6 = \u03c1 \u01eb0 = J 0 \u01eb0c = \u00b50 J 0 \u00b50\u01eb0c = \u00b50(cJ 0) so that: 1 c2 \u22022(\u03c6/c) \u2202t2 \u2212 \u2207 \u22022A \u2202t2 \u2212 \u2207 1 c2 2(\u03c6/c) = \u00b50J 0 2A = \u00b50J Therefore, if we form the 4\u2013vector potential A\u03b1 = ( \u03c6 c, A) then we can write the various 4\u2013relations: \u2202\u03b1A\u03b1 = 1 c \u2202A0 \u2202t + \u2207 A \u00b7 (16.143) (16.144) (16.145) (16.146) (16.147) (16.148) (which is the 4\u2013scalar Lorentz gauge condition) 2A\u03b1 = \u2202\u03b2\u2202\u03b2A\u03b1 = \u00b50J \u03b1 (16.149) (the 4\u2013vector inhomogeneous electromagnetic wave equation constructed from the 4\u2013scalar D\u2019Lambertian wave operator \u2013 the set of four wave equations for \u03c6 and the components of A above). Now we can construct the components of E and B from the covariant 4\u2013vector potential. For example, we know that: E = \u2207\u03c6 \u2212 \u2212 \u2202A \u2202t (16.150) where \u03c6 = cA0, so Ex = \u2202Ax \u2202(ct) \u2212 c \u2212 \u2202cA0 \u2202x = c(\u22020A1 \u2212 \u2212 \u22021A0) (16.151) and similarly, since B = \u2207 A: Bx = etc. \u00d7 \u2202Az \u2202y \u2212 \u2202Ay \u2202z = (\u22022A3 \u2212 \u2212 \u22023A2) (16.152) The components of the electric and magnetic \ufb01elds (all six of them) thus transform like the components of a second rank, antisymmetric, traceless \ufb01eld strength tensor7 : F \u03b1\u03b2 = \u2202\u03b1A\u03b2 \u2202\u03b2A\u03b1. \u2212 (16.153) In explicit component form, 0 Ex/c Ey/c Ez/c F \u03b1\u03b2 = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed \u2212 Ex/c 0 Bz By \u2212 Ey/c \u2212 Bz \u2212 0 Bx \u2212 Ez/c By Bx \u2212 0. \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 (16.154) 7Wikipedia: http://www.wikipedia.org/wiki/Electromagnetic tensor. Note that I\u2019m not completely comfortable with the signs for the covariant form of the potential in the Wikipedia article, although its main conclusions are sound enough. The tensor with two covariant indices (formed by two contractions with g) is obtained by replacing E with E. \u2212 0 Ex/c 0 Ey/c Bz Ez/c Ex/c Ey/c Ez/c Bz By \u2212 Bx 0 \u2212 By Bx 0 \u2212 F\u03b1\u03b2 = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed \u2212 \u2212 \u2212 (16.155). \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 Another important version of this tensor is the dual \ufb01eld strength \u03b1\u03b2. In terms of the totally antisymmetric tensor of the fourth rank tensor and the normal \ufb01eld strength tensor it is given by: F \u03b1\u03b2 =", " F 1 2 \u01eb\u03b1\u03b2\u03b3\u03b4F\u03b3\u03b4 = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed Bx 0 \u2212 Bx 0 Ez/c By \u2212 Bz Ey/c By \u2212 Ez/c 0 Ex/c \u2212 Bz \u2212 Ey/c \u2212 Ex/16.156) This is obtained from the basic contravariant \ufb01eld strength tensor by the E. Consideration of the section on magnetic substitutions E monopoles shows that this is indeed a particular duality transformation obtained in free space with the \u201crotation\u201d parameter equal to \u03c0/2 (in J6.151). B, B \u2192 \u2212 \u2192 Finally, we must write Maxwell\u2019s equations in covariant form. The inho- mogeneous equations are (recall) E = \u2207 1 c2 \u2207 B \u00d7 \u2212 \u00b7 \u2202E \u2202t = \u00b50J \u03c1 \u01eb0 (16.157) (16.158) The quantity on the right is proportional to the four current. The quantity on the left must therefore contract a 4\u2013derivative with the \ufb01eld strength tensor. You should verify that \u2202\u03b1F \u03b1\u03b2 = \u00b50J \u03b2 (16.159) exactly reconstructs the inhomogeneous equation for each component of J \u03b2. The homogeneous equations \u2207 \u2207 \u00d7 E + \u00b7 \u2202B \u2202t B = 0 = 0 (16.160) (16.161) also form a four vector (of zero\u2019s) and must hence be the contraction of a \ufb01eld strength tensor. But which one? Well, the second homogeneous equation requires that B B, so it must be the dual: E and both require that E \u2192 \u2212 \u2192 \u03b1\u03b2 = 0. \u2202\u03b1F (16.162) If we feel compelled to write everything in terms of the \ufb01eld strength tensor itself, this can be done. The result is the four equations \u2202\u03b1F \u03b2\u03b3 + \u2202\u03b2F \u03b3\u03b1 + \u2202\u03b3F \u03b1\u03b2 = 0 (16.163) where \u03b1, \u03b2, \u03b3 are any three of the four indices 0,1,2,3. However, this equation is a third rank tensor on the left, and its reduction by symmetry to a tensor of \ufb01rst rank is not manifest. It is ugly, if you will. Now that we have written Maxwell\u2019s equations (and the consequences of ME) in four dimensional form (remarking all the while that they are unusually beautiful and concise in this notation) we are done. Before we go on to deduce (from these results) how electric and magnetic \ufb01elds LT, however, we should complete the question with which we began the discussion, namely, how does Newton\u2019s law become covariant? The answer is (now that we know what the \ufb01eld strength tensor is) dp\u03b1 d\u03c4 = m dU \u03b1 d\u03c4 = q c F \u03b1\u03b2U\u03b2. (16.164) The time\u2013component equation is just the work\u2013energy theorem, and the space equations are Newton\u2019s law. As a postscript to our discussion (recalling that sometimes the \ufb01elds propagate in some medium and not free space) we note that in this case the homogeneous equation(s) remain unchanged, but the inhomgeneous equations are modi\ufb01ed (by using H and D instead of B and E). The inhomogeneous equation is then \u2202\u03b1G\u03b1\u03b2 = \u00b5J \u03b2 (16.165) where hopefully the de\ufb01nition of G\u03b1\u03b2 is obvious (that is, substitute v = 1/\u01eb\u00b5 for c throughout in appropriate places, or if you prefer recapitulate the entire derivation using H and D from the beginning). q Let us pause for a moment of religious silence and contemplate a great wonder of nature. This is the scientist\u2019s version of \u201cprayer in school\u201d. 16.7 The Transformation of Electromagnetic Fields Now that we have this in hand, we can easily see how to transform the electric and magnetic \ufb01elds when we boost a frame. Of course, that does not guarantee that the result will be simple. To convert F \u03b1\u03b2 from K to K \u2032, we must contract its indices with the transformation tensors, F \u2032 \u03b1\u03b2 = \u03b1 \u2202x\u2032 \u2202x\u03b3 \u03b2 \u2202x\u2032 \u2202x\u03b4 F \u03b3\u03b4. Note that since A is a linear transformation: A\u03b1 \u03b3 = \u03b1 \u2202x\u2032 \u2202x\u03b3 (16.166) (16.167) (where I have deliberately inserted a space to di\ufb00erent", "iate the \ufb01rst index from the second) we can write this in terms of the components of A as: F \u2032 \u03b1\u03b2 = A\u03b1 = A\u03b1 \u03b3F \u03b3\u03b4A \u03b2 \u03b3F \u03b3\u03b4 \u02dcA\u03b2 \u03b4 \u03b4 or (in a compressed notation): F \u2032 = AF \u02dcA (16.168) (16.169) This is just a speci\ufb01c case of the general rule that A can be used in general to transform any nth rank tensor by contracting it appropriately with each index. As we saw in our discussion of Thomas precession, we will have occasion to use this result for the particular case of a pure boost in an arbitrary direction that we can without loss of generality pick to be the 1 direction. Let\u2019s see how this goes. Recall that A for a pure boost in the one direction is the matrix formed with a lower right quadrant identity and an upper left \u03b3\u03b2 on the corners). Thus: so: quadrant 2 2 with \u03b3 on the diagonal and \u00d7 1F 10A 1 1 \u2212 1 + A0 \u03b32\u03b22 E1 c \u2212 = 01 = A0 0F 01A 1 F \u2032 \u03b32 E1 E\u20321 c \u2212 E\u20321 = (\u03b32 + \u03b32\u03b22)E1 E\u20321 = E1 c \u2212 (16.170) Note that we have extracted the ordinary cartesian components of E and B from F after transforming it. I leave the rest of them to work out yourself. You should be able to show that: E\u20321 = E1 E\u20322 = \u03b3(E2 \u2212 \u03b2B3) E\u20323 = \u03b3(E3 + \u03b2B2) B\u20321 = B1 B\u20322 = \u03b3(B2 + \u03b2E3) \u03b2E2) B\u20323 = \u03b3(B3 \u2212 (16.171) (16.172) (16.173) (16.174) (16.175) (16.176) The component of the \ufb01elds in the direction of the boost is unchanged, the perpendicular components of the \ufb01eld are mixed (almost as if they were space\u2013time pieces) by the boost. If you use instead the general form of A for a boost and express the components in terms of dot products, you should also show that the general transformation is given by: E\u2032 = \u03b3(E + \u03b2 B) \u00d7 B\u2032 = \u03b3(B \u03b2 \u2212 \u00d7 E) \u03b32 \u03b3 + 1 \u03b32 \u03b3 + 1 \u2212 \u2212 \u03b2(\u03b2 \u03b2(\u03b2 E) B). \u00b7 \u00b7 (16.177) (16.178) A purely electric or magnetic \ufb01eld in one frame will thus be a mixture of electric and magnetic \ufb01elds in another. We see that truly, there is little reason to distinguish them. We have to be a little careful, of course. If there is a monopolar (static) electric \ufb01eld in any frame, we cannot transform it completely into a magnetostatic \ufb01eld in another, for example. Why? Because the equations above will lead to some mixture for all \u03b2 < 1, and \u03b2 < 1 in nature as a constraint. I encourage you to review the example given in Jackson and meditate upon the remarks therein. We will not spend valuable class time on this, however. Instead we will end this, after all, purely mathematical/geometrical kinematical interlude (no Hamiltonians or Lagrangians = no physics) and do some physics. Let us deduce the covariant dynamics of relativistic particles in (assumed \ufb01xed) electromagnetic \ufb01elds. Chapter 17 Relativistic Dynamics 17.1 Covariant Field Theory We are interested in deducing the dynamics of point charged particles in e. \u2014 \ufb01xed) electromagnetic \ufb01elds. We already \u201cknow\u201d the \u201cgiven\u201d (i. answer, it is given by the covariant form of Newton\u2019s law, that is: dp\u03b1 d\u03c4 = m dU \u03b1 d\u03c4 = q c F \u03b1\u03b2U\u03b2. From this we can \ufb01nd the 4\u2013acceleration, dU \u03b1 d\u03c4 = q mc F \u03b1\u03b2U\u03b2 (17.1) (17.2) which we can integrate (in principle) to \ufb01nd the 4\u2013trajectory of the particle in question. However, this is not useful to us. Real physicists don\u2019t use Newton\u2019s law anymore. This is nothing against Newton, it is just that we need Hamilton\u2019s or Lagrange\u2019s formulation of dynamics in order to construct a quantum theory (or even an elegant", " classical theory). Our \ufb01rst chore, therefore, will be to generalize the arguments that lead to the Euler\u2013Lagrange or Hamilton equations of motion to four dimensions. 261 17.1.1 The Brute Force Way Recall that the Lagrangian path to the dynamics of a particle (which is most easily made covariant, since it uses (q(t), \u02d9q(t), t) as its variables) is based on the Action A = L(q(t), \u02d9q(t), t)dt. (17.3) t1 t0 Z By requiring that A be an extremum as a functional of the system trajectory, we obtain the Euler\u2013Lagrange equations d dt \u2202L \u2202 \u02d9qi \u2212 \u2202L \u2202qi! = 0. (17.4) These are equivalent to Newton\u2019s law for suitable de\ufb01nitions of L and the force. The simplest way to make this relativistic is to express it in terms of the proper time and then require that the action A be extremal in all frames. Then, A = \u03b3Ld\u03c4 (17.5) \u03c41 \u03c40 Z is the action, since dt = \u03b3d\u03c4. We now must remark 1. If the extremal condition is to be invariant with respect to LT\u2019s, then A must be invariant (and hence a 4\u2013scalar); 2. Therefore (since d\u03c4 is invariant) \u03b3L must be invariant; 3. Finally, since \u03b3 is just a number, L must be a 4\u2013scalar. This \ufb01nal conclusion greatly constrains the possible forms of L. Note well that it is not clear that this argument (from Jackson) is valid. \u03b3, while a number, is not invariant under boosts \u2013 indeed, it is parametrically related to the boost parameter! It is also perfectly clear that the \ufb01rst statement is false \u2013 while it is true that if A is a four-scalar with respect to boosts that its extremums must be preserved, it is equally true that this condition is not necessarily unique \u2013 all that is necessary is that a boost monotonically scale the action in such a way that the extremal property is preserved! A weaker (but still su\ufb03cient) argument might then be: If L is a 4-scalar, and \u03b3 is a monotonic function independent of the 4-coordinates, 4-velocities, and \u03c4, then the property of a given trajectory resulting in an extremum of the action is preserved. In my opinion this is clearer and still adequate for our purposes. L being a 4-scalar (0th rank tensor in 4-space w.r.t. the Lorentz transformation) is su\ufb03cient to produce an invariant extremum of the action A, even if the numerical values of A vary under a boost. To prove that it is also necessary very likely involves an exercise in the calculus of variations that distributes the derivatives over \u03b3L \u2013 similar exercises are already in the homework for this chapter in other contexts. Either way, we will now assert that the Lagrangian of a free particle must be a 4-scalar (and hence must be formed out of the full contraction of tensors of higher rank), and will remain alert in the work below for any sort of inconsistency that might be related to \u03b3. Obviously, we want it to reproduce the classical non\u2013relativistic theory in the appropriate limit, that is, a free particle should have constant energy and momentum, or, equivalently, 4\u2013velocity. The simplest (not \u201conly\u201d as Jackson states) Lorentz invariant function of the 4\u2013velocity is it\u2019s quadratic form: U \u03b1U\u03b1 = c2 (17.6) Of course, any polynomial functional of this quadratic is also a possible scalar, but they are not the thing to try \ufb01rst. Thus a reasonable guess for the Lagrangian is L = (constant)c2\u03b3\u2212 1 = mc2 \u2212 1 s \u2212 u2 c2. (17.7) If we now crunch through the Euler\u2013Lagrange equation we \ufb01nd that this choice for the constant leads to d dt (\u03b3mu) = 0 (17.8) which is indeed Newton\u2019s law for a free particle, but with the relativistic form of the three\u2013momentum. If one chooses a frame where the particle is initially at rest, a trajectory where it remains at rest will yield the least action (you should check this). This is because \u03b3\u2212 sign). 1 is maximal when \u03b2 = 0 (and the Lagrangian has", " a minus Now, suppose that the charged particle is in a electromagnet potential. If it were moving slowly in a scalar potential \u03a6 only, its potential energy would be V = q\u03a6. The non\u2013relativistic Lagrangian in this case should be V (where T is the free particle Lagrangian). The interaction part just T q\u03a6 in this non\u2013 of the relativistic Lagrangian must therefore reduce to relativistic limit. \u2212 \u2212 We must \ufb01nd a Lorentz invariant (scalar) form for \u03b3Lint that reduces to q\u03a6 for non\u2013relativistic velocities. Since \u03a6 is the time component of a four \u2212 vector potential, we can guess that the correct generalization must involve the four vector potential A\u03b1. Since it must be a scalar, it must involve the scalar product of A\u03b1 with some four vector(s). The only ones avaliable are x\u03b1 and U \u03b1. The correct \u03b3Lint depends only on the U \u03b1. If it depended on the coordinates as well, then the physics would not be translationally invariant and the results of our calculation might well depend on where we chose the origin. This does not seem reasonable. Once again, this does not uniquely determine it, it only determines the simplest (linear) form to within a sign and a constant: \u03b3Lint = \u2212 q c U\u03b1A\u03b1 Lint = q\u03a6 + \u2212 q c ~u \u00b7 ~A. (17.9) (17.10) or There could be additional terms involving polynomials of this quantity, the product A\u03b1A\u03b1 (which is indeed present in some theories) and other scalar reductions of \ufb01eld and charge/\ufb01eld tensor quantities. Linearity, in either the vector potential or the velocity, is an axiom and not logically necessary. The complete relativistic Lagrangian for a charged particle is thus L = mc2 u2 c2 + It should take you about one hour to show that this yields the correct relativistic Lorentz force law. The free particle part is obvious, the electric \ufb01eld is obvious. You will have to work a bit, using (17.11) q\u03a6. q c ~A \u2212 \u2212 \u2212 s ~u 1 \u00b7 d dt = \u2202 \u2202t + ~u \u2207 \u00b7! (17.12) ~u \u00d7 \u2212 (\u2207 ~A) out of the remainder. I suggest that you simply to squeeze work out the terms by expanding them as far as they go and reassembling the pieces, but some of you may know enough vector algebra to do it better ways. This will be on the next assignment, so feel free to start. \u00d7 The canonical momentum ~P conjugate to the position coordinates x is obtained (as usual) from This result, Pi = \u2202L \u2202ui = \u03b3mui + q c Ai. ~P = ~p + ~A q c (17.13) (17.14) (where ~p is the relativistic kinetic momentum of the particle) is extremely important to remember, as it is a necessary ingredient in the construction of either a quantum theory or an elegant classical theory. Placing the particle in a \ufb01eld alters its canonical \u201cmomentum\u201d. We make the Hamiltonian out of the Lagrangian and the canonical mo- mentum via \u00b7 The basic result here has too many variables. We must eliminate ~u in favor of ~A and ~P. Note that \u2212 H = ~P ~u L (17.15) (17.16) ~u = q ~A 2 c ~P ~P q c \u2212 \u2212 ~A r + m2c2 (something that is a wee chore to prove, of course, but it is straightforward algebra). With even more tedious algebra, you can show that the Hamiltonian is: H = (c ~P q \u2212 q ~A)2 + m2c4 + q\u03a6 = W. (17.17) From this result, Hamilton\u2019s equations of motion should reproduce the Lorentz force law. See that it does (although the relationship between the EL equations and Hamilton\u2019s equations makes the result obvious). Note that if we interpret the Hamiltonian (as usual) as the total energy W of the ~A) particle, this result is related to the free particle energy by ~p and the addition of the scalar potential energy q\u03a6. This is actually just a single change in the four\u2013vector momentum: ( ~P \u2192 \u2212 q c (W \u2212 q\u03a6)2 \u2212 (c ~P \u2212 q ~A)2 = m2c4 = p\u03b1p\u03b1 (17.18) (which has the usual form if p\u03b1 = E c, ~p", " = 1 c (W \u2212 e\u03a6), ~P q c ~A \u2212 (17.19) ). This also makes the invariance properties of the Hamiltonian manifest. It is really annoying to obtain the invariance properties of things after the fact. It is also annoying (although perhaps useful) to have the three vector coordinates ~x, ~u hanging around at this point. So let us rederive these results using only four\u2013vectors and suitable scalar reductions. 17.1.2 The Elegant Way We can write the free particle Lagrangian using only scalar reductions of suitable 4\u2013vectors: Lfree = \u2212 mc \u03b3 U\u03b1U \u03b1 q (which is still mc2\u03b3\u2212 \u2212 1). The action is thus A = mc \u2212 \u03c41 \u03c40 q Z U\u03b1U \u03b1d\u03c4. (17.20) (17.21) The variations on this action must be carried out subject to the constraint U\u03b1U \u03b1 = c2 (17.22) which severely limits the allowed solutions. We write this as d(U\u03b1U \u03b1) d\u03c4 U \u03b1 + U\u03b1 dU\u03b1 d\u03c4 dU\u03b1 d\u03c4 g\u03b1\u03b2g\u03b2\u03b1U \u03b1 + U\u03b1 U\u03b2 dU \u03b2 d\u03c4 + U\u03b1 2U\u03b1 U\u03b1 = 17.23) dU \u03b1 d\u03c4 dU \u03b1 d\u03c4 dU \u03b1 d\u03c4 dU \u03b1 d\u03c4 dU \u03b1 d\u03c4 Now, U\u03b1U \u03b1d\u03c4 = q dx\u03b1 d\u03c4 dx\u03b1 d\u03c4 s d\u03c4 = g\u03b1\u03b2dx\u03b1dx\u03b2 (17.24) q which is an in\ufb01nitesimal length in four\u2013space. The latter expression does not explicitly contain d\u03c4. We can thus parameterize the action in terms of a path\u2013parameter s that increases monotonically with \u03c4 but is otherwise arbitrary. Then A = mc \u2212 s1 s0 s Z g\u03b1\u03b2 dx\u03b1 ds dx\u03b2 ds ds. (17.25) We are clearly making progress. We have taken a perfectly good expression and made in unrecognizable. To make you a little happier, note that this has now got the form of A = \u02dcLds (17.26) Z where \u02dcL is a scalar \u201cLagrangian\u201d written in terms of an independent free parameter. This might be progress after all, since we have quashed the annoying \u03b3\u2212 1. If we now do the calculus of variations thing and get the Euler-Lagrange equations in four dimensions: d \u02dcL dx\u03b1 ds \u2202 \u2202\u03b1 \u02dcL = 0 \uf8f6 \u2212 (17.27) d ds \uf8eb \uf8ed \uf8f8 (for \u03b1 = 0, 4). Applying them to the Langrangian in this action, they turn out to be: mc \u2202 d ds o dx\u03b4 ds g\u03b4\u03b2 dx\u03b2 ds n dx\u03b1 ds \u2202 dx\u03b1 ds + dx\u03b1 ds dx\u03b2 ds dx\u03b2 ds dx\u03b1 ds dx\u03b2 ds dx\u03b2 ds mc 2 d ds \uf8f1 \uf8f2 mc q d \uf8f3 ds \uf8f1 \uf8f2 17.28) (17.29) = 0. (17.30) This still does not have the constraint above imposed on it. We impose the constraint by identifying ds with d\u03c4 in such a way that the constraint is simultaneously satis\ufb01ed: dx\u03b1 ds dx\u03b1 ds s ds = cd\u03c4 d d\u03c4 = c dx\u03b1 ds dx\u03b1 ds d ds q (17.31) (which requires both ds = d\u03c4 and U\u03b1U \u03b1 = c2). If you like, this constraint picks out of all possible path parameterizations the one that follows the proper time while keeping the four vector velocity scalar product Lorentz invariant. For free particles this is a lot of work, but it is paid back when we include an interaction. If we multiply the Euler-Lagrange equation (in terms of s) from the left by: c dx\u03b1 ds dx\u03b1 ds (17.32) and use the constraint to convert to \u03c4, the result (for the equation of motion) is: q m c dx\u03b2 ds dx\u03b2 ds d ds \uf8f1 \uf8f2 c dx\u03b2 ds dx\u03b2 ds d ds x\u03b1 q or q \uf8f3 d2x\u03b1 d\u03c4 2 = 0 which certainly looks it has the right form. m = 0 (17.33) \uf8fc \uf8fd \uf8fe (17.34) We can include an interaction. Just as before, \u03b3Lint must be a Lorentz scalar. When we make a parameterized version of the Lagrangian, the part under the integral", " must be a 4\u2013scalar. The covariant form of the result is (hopefully obviously) s1 A = \u2212 Z s0 \uf8f1 \uf8f2 mc g\u03b4\u03b2 d\u03b4 ds s dx\u03b2 ds + q c dx\u03b2 ds A\u03b2 The \u201cfour Lagrangian\u201d in this equation is \uf8f3 ds. (17.35) \uf8fc \uf8fd \uf8fe \u02dcL = mc \u2212 \uf8f1 \uf8f2 \uf8f3 g\u03b4\u03b2 d\u03b4 ds s dx\u03b2 ds + q c dx\u03b2 ds A\u03b2. \uf8fc \uf8fd \uf8fe (17.36) As before we construct the Euler-Lagrange equation. mc d ds \uf8f1 \uf8f2 dx\u03b1 ds dx\u03b2 ds dx\u03b2 ds + q c A\u03b1 q \uf8f3 \uf8fc \uf8fd \uf8fe q c x\u03b2 ds \u2212 \u2202\u03b1A\u03b2 = 0 Again we multiply through from the left by and convert to \u03c4 to get: c dx\u03b1 ds dx\u03b1 ds q m d2x\u03b1 d\u03c4 2 + q c dA\u03b1 d\u03c4 \u2212 q c dx\u03b2 d\u03c4 \u2202\u03b1A\u03b2 = 0 (17.37) (17.38) (17.39) (17.40) The derivative dA\u03b1/d\u03c4 is a bit jarring. However, if we express this total derivative in terms of partials we observe that: dA\u03b1 d\u03c4 = dx\u03b2 d\u03c4 \u2202 \u2202x\u03b2 A\u03b1 = dx\u03b2 d\u03c4 \u2202\u03b2A\u03b1 (17.41) Substituting, the equation of motion becomes: d(mU \u03b1) d\u03c4 = m d2x\u03b1 d\u03c4 2 = q c (\u2202\u03b1A\u03b2 \u2202\u03b2A\u03b1) \u2212 dx\u03b2 d\u03c4 = q c F \u03b1\u03b2U\u03b2. (17.42) which is, as expected, the Lorentz force law in covariant form! How lovely! To make a Hamiltonian in this notation, we must \ufb01rst make the canonical momentum: P \u03b1 = \u2202 \u02dcL x\u03b1 ds \u2212 \u2202 = mU \u03b1 + q c A\u03b1 (17.43) which is a covariant version of the complete set of interaction equations from the previous section (it does both energy and 3\u2013momentum). There are several ways to make a Hamiltonian (recall that in general there is what amounts to gauge freedom, minimally the ability to add an arbitrary constant which naturally does not a\ufb00ect the resulting di\ufb00erential equations). One is1: \u02dcH = U\u03b1P \u03b1 + \u02dcL 1Note that I\u2019ve rearranged this slightly to avoid having to do lots of stu\ufb00 with g (17.44) sandwiches below. Again, we must eliminate: in favor of P \u03b1, A\u03b1. Thus A\u03b1 (17.45) \u02dcL = mc \u2212 s 1 m2 P\u03b1 \u2212 q c A\u03b1 P \u03b1 q c \u2212 A\u03b1 \u2212 q mc P\u03b1 \u2212 q c A\u03b1 A\u03b1 (17.46) and \u02dcH = 1 m = 1 m P\u03b1 \u2212 \u2212 P\u03b1 \u2212 A\u03b1 q c q mc q c A\u03b1 \u2212 q c P \u03b1 P\u03b1 \u2212 P \u03b1 mc s 1 m2 P\u03b1 \u2212 q c A\u03b1 P \u03b1 q c \u2212 A\u03b1 A\u03b1 A\u03b1 (17.47) q A\u03b1 c c s P\u03b1 \u2212 \u2212 q c A\u03b1 P \u03b1 q c \u2212 A\u03b1 (17.48) \u2212 This Hamiltonian in four dimensions is no longer an energy since it is obviously a 4\u2013scalar and energy transforms like the time\u2013component of a four vector. However, it works. Hamilton\u2019s equations (in four dimensions) lead again directly to the relativistic Lorentz force law: dx\u03b1 d\u03c4 \u2202 \u02dcH \u2202x\u03b1 = = \u2202 \u02dcH \u2202P\u03b1 = \u2202\u03b1 \u02dcH = \u2212 dP \u03b1 d\u03c4 = \u2212 1 m q mc P \u03b1 \u2212 A\u03b1 q c q c A\u03b1 \u2202\u03b1A\u03b2 (17.50) P\u03b1 \u2212 (17.49) There is a bit of algebra involved in deriving this result. For example, one has to recognize that: p\u03b1 = mU \u03b1 = P \u03b1 q c A\u03b1 \u2212 (17.51) and p\u03b1p\u03b1 = m2c2 and apply this to eliminate unwanted terms judiciously, that is after di\ufb00erentiation. If you apply it too early (for example at the beginning) you observe the puzzling result that: \u02dcH = = 1 m 1 m = mc2 = 0 p\u03b1p\u03b1 c\u221ap\u03b1p\u03b1 \u2212 m2c", "2 c\u221am2c2 \u2212 mc2 \u2212 (17.52) which leads one to the very Zen conclusion that the cause of all things is Nothing (in four dimensions, yet)! We are left with a rather mysti\ufb01ed feeling that the algebraic hand is quicker than the eye. Somehow an equation whose four-scalar value is zero has a functional form, a structure, that leads to non\u2013zero, covariant equations of motion. Also (as already remarked) this Hamiltonian is not unique. Obviously one can add an arbitrary four-scalar constant to the equation and get no contribution from the derivatives (just as one can in nonrelativistic classical physics). There are other gauge freedoms \u2013 ultimately there several other ways of writing the Hamiltonian associated with the given Lagrangian; all of them yield a constant value that is not the energy when evaluated and yield the correct equations of motion when processed. Finally there exist what are called singular Lagrangians \u2013 Lagrangians for which the generalized coordinates do not always map into generalized conjugate variables! Dirac was (unsurprisingly) the \ufb01rst to formally identify this in the context of constrained systems (systems described by a Lagrangian and constraints with Lagrange multipliers for which the Hesse determinant vanishes); Bergmann (at Syracuse) also made major contributions to the formal development of the concept. However the roots of the problem date much further back to e.g. Noether\u2019s theorem. I have a couple of papers on this that I\u2019ve collected from the web, although the idea is also discussed in various monographs and textbooks on mechanics. It is worth pointing out that there was at one point considerable work being done here at Duke on the idea \u2013 N. Mukunda, Max Lohe, (both friends of mine) worked on the idea with Larry Biedenharn (my advisor); Biedenharn also published work with Louck on the subject, and of course Mukunda and Sudarshan\u2019s book on classical mechanics remains a \u201cclassic\u201d. Since Dirac\u2019s time the notion that the \u201cright\u201d uni\ufb01ed \ufb01eld theory will have certain interesting properties related to this has been batted around. This points out an ongoing problem in relativistic quantum theories. These theories are generally based on a Hamiltonian, but manifestly covariant Hamiltonians for a given system cannot in general be uniquely derived from \ufb01rst principles as the mapping between velocities and momenta is not always one-to-one. Thus even when a covariant Lagrangian density can be constructed, the associated Hamiltonian is not obvious or necessarily unique. This is just one (although it is one of the most fundamental) obstacles to overcome when developing a relativistic quantum \ufb01eld theory. 17.2 Motion of a Point Charge in a Static Magnetic Field Now that we have obtained the various covariant forms of the Lorentz force law, we can easily determine the trajectories of charged particles in various \ufb01xed \ufb01elds. In fact, we could have done this weeks ago (if not years) even without knowing the covariant forms. In a static magnetic \ufb01eld, the equations of motion are: dE dt d~p dt = 0 = q c ~v ~B \u00d7 (17.53) (17.54) for the energy and momentum, respectively (arranged like pieces of a four vector for clarity). Clearly the speed of the particle is constant since the force is perpendicular to the motion and does no work. \u03b3 is therefore also constant. Thus d~v dt = ~v ~\u03c9B \u00d7 where ~\u03c9B = q ~B \u03b3mc = qc ~B E (17.55) (17.56) is the gyration or precession (cyclotron) frequency. The motion described by this equation is a circular motion perpendicular to ~B coupled to a uniform motion parallel to ~B. This is too droll for words (and in fact you have probably already taught it to your kids in kiddy physics) but it does yield one important result. The magnitude of the momentum perpendicular to ~B is cp \u22a5 = qBa (17.57) where a is the radius of the circular helix. From this (in, for example, a bubble chamber, where the track can be photographed) and a knowledge (or guess) as the the charge, the transverse momentum can be measured. Measuring other things (like the rate of change of the curvature of the track) can yield the mass of the particle from a knowledge of its momentum. From these humble traces the entire picture we currently have of the sub\u2013atomic zoo has been built up. Sections 12.2-12.4 are too simple to waste time on. 12.5", "-12.6 are interesting but important only to plasma people. 12.7 is redundant of things we will do correctly later. Thus we skip to 12.8, leaving you to read any or all of the intermediate material on your own. We will skip 12.9. Finally, we will do 12.10\u201312.11 to complete chapter 12. 17.3 Building a Relativistic Field Theory We have not quite \ufb01nished the job of building a proper relativistic \ufb01eld theory of electromagnetism. That is because we would like to be able to obtain all of the equations of motion (that is, physics) describing the system from a covariant action principle. We have done that for the particles in the \ufb01elds, but what about the \ufb01elds themselves? In fact, since the particles produce (and hence modify) the \ufb01elds, we do not even have the correct solutions for the particles alone, yet. Let us see if we can develop a suitable Lagrangian for the \ufb01elds that leads, ideally, to Maxwell\u2019s equations. The Rules for building a \ufb01eld theory Lagrangian are of interest in and of themselves, since they are quite general. The rules are: 1. Take the position and velocity coordinates for continuous space time and replace them with \ufb01eld variables. 2. Label the \ufb01eld variables with discrete (coordinate direction) labels and with continuous (position) variables. 3. Replace the \u201cvelocity\u201d with the 4\u2013gradient. 4. Require the action to be stationary w.r.t. variations in the \ufb01eld vari- ables themselves and their gradients. That is, i \u2192 x\u03b1, k (17.58) qi \u2192 \u02d9qi \u2192 \u03c6k(x) \u2202\u03b1\u03c6k(x) L = Li(qi, \u02d9qi) Xi \u2202L \u2202 \u02d9qi! d dt = \u2192 Z L \u2202\u03b2 \u2202 \u2202L \u2202qi \u2192 L \u2202\u2202\u03b2\u03c6k = \u2202 L \u2202\u03c6k. (\u03c6k, \u2202\u03b1\u03c6k)d3x (17.59) (17.60) (17.61) (17.62) over time, making the When we make an action integral, we integrate total integral four dimensional. We therefore call the Lagrangian density in four dimensions. Note that the action will be covariant provided the Lagrangian density is a 4\u2013scalar. This is what I have meant whenever I have inadvertantly called the \u201cLagrangian\u201d a scalar. Good, clean, relativistic theories with or without particles are made out of scalar Lagrangian densities, not Lagrangians per se: L L A = d3xdt = Z Z L d4x. Z L (17.63) We now do the usual dance. We know that for the \ufb01elds must be a scalar. We also know that Maxwell\u2019s equations relate the \ufb01elds to the currents that produce them, and also link the electric and magnetic \ufb01elds. We thus need to build a theory out of F \u03b1\u03b2, A\u03b1, J \u03b1. Various ways we can do this include L and still messier pieces like F\u03b1\u03b2F \u03b1\u03b2 J\u03b1A\u03b1 F\u03b1\u03b2F \u03b1\u03b2 F\u03b1\u03b2J \u03b1A\u03b2. The \ufb01rst two terms are invariant under the transformations of the full Lorentz group. The third is not a scalar under inversion, but a pseudoscalar (odd under inversion). We reject it. The last is a mess. We reject it. We want a term quadratic in the 4-gradients of the \ufb01elds. This is the \ufb01rst term. We want a source term to couple the \ufb01elds and the particles. The second term does that. So, we try a Lagrangian density with just these two terms, with unknown constants Q and R that have to be determined so that they correctly reconstruct Maxwell\u2019s equations in whatever system of units we like: = L QF\u03b1\u03b2F \u03b1\u03b2 \u2212 RJ\u03b1A\u03b1. \u2212 (17.64) We need to take derivatives of write this Lagrangian in the form: L with respect to \u2202\u03b2A\u03b1, so it is useful to = L Qg\u03bb\u00b5g\u03bd\u03c3(\u2202\u00b5A\u03c3 \u2212 \u2202\u03c3A\u00b5)(\u2202\u03bbA\u03bd \u2202\u03bdA\u03bb) \u2212 \u2212 RJ\u03b1A\u03b1. \u2212 (17.65) When we form \u2202 equal to a", " pair of the indices above. We therefore get four terms: /\u2202(\u2202\u03b2A\u03b1) we get delta functions whenever \u03b1 and \u03b2 are L \u2202 L \u2202(\u2202\u03b2A\u03b1) = Qg\u03bb\u00b5g\u03bd\u03c3 \u2212 \u03b1F \u03bb\u03bd \u03b4\u00b5 \u03b2 \u03b4\u03c3 n \u2212 \u03b2 \u03b4\u00b5 \u03b4\u03c3 \u03b1F \u03bb\u03bd + \u03b4\u03bb \u03b2\u03b4\u03bd \u03b1F \u00b5\u03c3 \u03b2\u03b4\u03bb \u03b4\u03bd \u03b1F \u00b5\u03c3 \u2212 o(17.66) where the \ufb01rst two terms come from delta functions formed from the \ufb01rst term and the second two terms come from delta functions formed from the second term. g\u03b1\u03b2 is symmetric (in fact, diagonal). The F \u03b1\u03b2 is antisymmetric. When we do the sums against the \u03b4\u2013functions, the four terms make identical contributions: \u2202 L \u2202(\u2202\u03b2A\u03b1) = 4QF\u03b2\u03b1 = 4QF\u03b1\u03b2. \u2212 (17.67) The other part of the E\u2013L equation (which corresponds in position space to the \u201cpotential\u201d, or \u201cforce\u201d term) is \u2202 \u2202A\u03b1 = L RJ\u03b1. \u2212 (17.68) Therefore the equations of motion for the electromagnetic \ufb01eld can be written 4Q\u2202\u03b2F\u03b2\u03b1 = RJ\u03b1. (17.69) If one checks back in one\u2019s notes, one sees that this is indeed the covariant form of the inhomogeneous Maxwell\u2019s equations if Q = 1/4 and R = \u00b50: \u2202\u03b2F\u03b2\u03b1 = \u00b50J\u03b1 follows from the Lagrangian density: 1 4 \u2212 = L F\u03b1\u03b2F \u03b1\u03b2 \u2212 \u00b50J\u03b1A\u03b1. (17.70) (17.71) Therefore the Lagrangian we have constructed yields the inhomogeneous Maxwell equations, but not the homogeneous ones. That is okay, though, because we have constructed the F \u03b1\u03b2 in terms of the A\u03b1 in such a way that the homogeneous ones are satis\ufb01ed automatically! To observe that this miracle is true, we recall the covariant form of the homogeneous equations: Also, Thus \u03b1\u03b2 = 0. \u2202\u03b1F \u03b1\u03b2 = F 1 2 \u01eb\u03b1\u03b2\u03b3\u03b4F\u03b3\u03b4. \u03b1\u03b2 = \u2202\u03b1F 1 2 \u2202\u03b1\u01eb\u03b1\u03b2\u03b3\u03b4F\u03b3\u03b4 = \u2202\u03b1\u01eb\u03b1\u03b2\u03b3\u03b4\u2202\u03b3A\u03b4 = \u01eb\u03b1\u03b2\u03b3\u03b4\u2202\u03b1\u2202\u03b3A\u03b4 (17.72) (17.73) (17.74) is the \ufb01rst term. But \u2202\u03b1\u2202\u03b3 is symmetric, while \u01eb\u03b1\u03b2\u03b3\u03b4 is antisymmetric in the same two indices, so the contraction on the two indices vanishes (work it out term by term if you doubt it). \u03b1\u03b2 F Thus the homogeneous equations are satis\ufb01ed by our de\ufb01nition of quite independent of any dynamics. In four dimensions, all of the inhomogeneous source terms must appear in equations with the form of the inhomogeneous equation above, and only one of these equations can result from the action principle. The similarity transformation to the \ufb01elds we observe is thus the \u201cnatural\u201d form of the ME\u2019s, and in four dimensions the homogeneous equations are really not independent as far as the action principle is concerned. Note that this is fundamentally because our \ufb01eld strength tensor derives from the de\ufb01nition of the magnetic \ufb01eld as the curl of the vector \ufb01eld ~A (which is divergenceless) which is built into the de\ufb01nition. As a \ufb01nal quixotic note, observe that if we take the 4\u2013divergence of both sides of the inhomogeneous Maxwell equations: \u2202\u03b1\u2202\u03b2F\u03b2\u03b1 = \u00b50\u2202\u03b1J\u03b1 (17.75) the left hand side vanishes because again, a symmetric di\ufb00erential operator is contracted with a completely antisymmetric \ufb01eld strength tensor. Thus \u2202\u03b1J\u03b1 = 0, (17.76) which, by some strange coincidence, is the charge\u2013current conservation equation in four dimensions. Do you get the feeling that something very deep is going on? This is what I love about physics. Beautiful things are really beautiful! We will now blow o\ufb00 the \u201cproca\u201d Lagrangian, which would be appropriate if the photon had a mass. It doesn\u2019t, but if it did you would need to read this chapter", ". It might, of course, so you should probably read the chapter anyway, but it currently (bad pun) doesn\u2019t so I\u2019m going to make light of it (worse pun) and continue. If we had one more month, we would now study the covariant forms of the stress tensor. It is important, but it is also quite di\ufb03cult, and necessitates a broader discussion than we can now a\ufb00ord. To treat the subject properly, we would need to treat parts of chapter 17 simultaneously, and we would need to do a lot of algebra. This would mean that we would miss (in all probability) being able to learn the Li\u00b4enard\u2013Wiechart potential, which is far more important. We will therefore content ourselves with merely de\ufb01ning the stress tensor, remarking on some of its properties without proof, and moving on. You are responsible for working your way through this chapter, according to your needs, inclinations, and abilities, on your own. 17.4 The Symmetric Stress Tensor Imagine a big blob of jelly. Imagine poking it on a side. The whole thing wiggles and distorts, as the force of your poke acts on the entire blob of jelly. The mathematical mechanism that describes how your poke is distributed is calle the stress tensor of the material. It tells how energy and momentum are connected by the medium itself. The same concept can be generalized to a four dimensional medium, where the \u201cjelly\u201d is space time itself. Let us now study what an electromagnetic stress tensor is, and how it relates to electromagnetic \u201cpokes\u201d. Recall that pi = \u2202L \u2202 \u02d9qi (17.77) is the canonical momentum corresponding to the variable qi in an arbitrary Lagrangian. The Hamiltonian is given, in this case, by H = pi \u02d9qi \u2212 L Xi (17.78) If \u2202L/\u2202t = 0 then one can show that \u2202H/\u2202t = 0. For four as usual. dimensional \ufb01elds we should probably have a Lagrangian and Hamiltonian density whose 3\u2013integral are the usual Lagrangian and Hamiltonians. The Hamiltonian is the energy of a particle or system, so it should transform like the zeroth component of a four vector. Thus, since H = d3x Z H (17.79) and d4x = d0xd3x, then second rank tensor. If we de\ufb01ne the Hamiltonian density of a \ufb01eld, then Lagrangian density must transform like the time component of a in terms of the H H L = H Xk \u2202 L \u2202\u03c6k \u2202t \u2202 \u2202\u03c6k. \u2202t \u2212 L (17.80) Well, great! The \ufb01rst factor in the sum is the conjugate momentum by de\ufb01nition, and the second is the generalized \u201cvelocity\u201d. Since must transform like the time component of a second rank tensor (and the time derivative appears in this equation) it appears that the covariant generalization of the Hamiltonian density is something that puts a covariant derivative there, instead. We try H T \u03b1\u03b2 = Xk \u2202 L \u2202(\u2202\u03b1\u03c6k) \u2202\u03b2\u03c6k \u2212 g\u03b1\u03b2. L (17.81) This is called the canonical stress tensor, and is related to the stress tensor de\ufb01ned and studied in Chapter 6. This tensor has the covariant function of telling us how the energy and momentum carried by the electromagnetic \ufb01eld transform. What is this tensor? It is, in fact, highly non\u2013trivial. The best we can do is note that if we assume that only free \ufb01elds are present and that the free \ufb01elds are localized in some \ufb01nite region of space (neither assumption is particularly physical), then we can show that T 00d3x = Z 1 2 Z (\u01eb0E2 + 1 \u00b50 B2)d3x = E\ufb01eld (17.82) and T 0id3x = \u01eb0c (E Z Z B)id3x = \u00d7 1 c Z (E \u00d7 H)id3x = cP i \ufb01eld (17.83) which are the \u201cusual\u201d expressions for the energy and momentum of the free \ufb01eld. At least if I got the change to SI units right... What, you might ask, is this good for? Well, aside from this correspondance (which is full of holes, by the way), we can write the energy\u2013 momentum conservation law \ufffd", "\ufffd\u03b1T \u03b1\u03b2 = 0. (17.84) This is proven in Jackson, with a discussion of some of its shortcomings. One of these is that it is not symmetric. This creates di\ufb03culties when we consider the angular momentum carried by the \ufb01eld. Since the angular momentum density is important when we go to create photons (which must have quantized angular momenta), it is worthwhile to construct the symmetric stress tensor \u0398\u03b1\u03b2 = g\u03b1\u00b5F\u00b5\u03bbF \u03bb\u03b2 + 1 4 g\u03b1\u03b2F\u00b5\u03bbF \u00b5\u03bb (17.85) in terms of which we can correctly construct a covariant generalization of the energy momentum conservation law \u2202\u03b1\u0398\u03b1\u03b2 = 0 and the angular momentum tensor M \u03b1\u03b2\u03b3 = \u0398\u03b1\u03b2x\u03b3 \u0398\u03b1\u03b3x\u03b2 \u2212 (17.86) (17.87) which is therefore conserved. This form of the stress tensor can also be directly coupled to source terms, resulting in the covariant form of the work energy theorem for the combined system of particles and \ufb01elds. This is about all we will say about this at this time. I realize that it is unsatisfactory and apologize. If we had one more semester together, we could do it properly, but we don\u2019t. Therefore, it is on to 17.5 Covariant Green\u2019s Functions Just when you thought it was safe to go back into the classroom, along comes Jaws himself. Green\u2019s functions are your friends! The inhomogeneous Maxwell equations are now compactly written as \u2202\u03b1F \u03b1\u03b2 = \u00b50J \u03b2. From the de\ufb01nition of the \ufb01eld strength tensor, this is 2A\u03b2 \u2212 \u2202\u03b2(\u2202\u03b1A\u03b1) = \u00b50J \u03b2 (17.88) (17.89) If the potentials satisfy the Lorentz condition, \u2202\u03b1A\u03b1 = 0 and therefore 2A\u03b2 = \u00b50J \u03b2 (17.90) Do you get the feeling that there is something mystical about space\u2013time notations? Do you remember what a pain in the butt this was to derive the hard way? To solve this inhomogeneous di\ufb00erential equation, we construct simulta- neously a Green\u2019s function 2D(x, x\u2032) = \u03b4(4)(x x\u2032) \u2212 (17.91) and the associated integral equation over the source term: A\u03b1(x) = A\u03b1 I + \u00b50 d4x\u2032D(x x\u2032)J \u03b1(x\u2032) (17.92) Z (where the inhomogeneous term A\u03b1 I depends on the Green\u2019s function and is the \u201cboundary\u201d term or the free potential from inhomogeneous sources outside the region of integration). \u2212 Next week we will concentrate on the integral equation solutions themselves. Now let us see how to construct the appropriate (covariant) Green\u2019s function. As usual, the principle part of the Green\u2019s function can involve only the absolute distance between the points. Thus if y\u03b1 = x\u03b1 \u03b1 we seek solutions to \u2212 x\u2032 2D(y) = \u03b4(4)(y). (17.93) There are several ways we could go about solving this equation. They are all equivalent at some level or another. For example, we have already solved this equation for a single fourier component in Chapter 9. We could transform this result and obtain a four dimensional result. However, a more general procedure is to construct the solution from scratch. The four dimensional fourier transform of the desired Green\u2019s function is de\ufb01ned by D(y) = 1 (2\u03c0)4 Z d4k \u02dcD(k)e\u2212 ik y \u00b7 (17.94) where k y = k0y0 \u2212 \u00b7 k \u00b7 y. The four dimensional delta function is \u03b44(y) = 1 (2\u03c0)4 Z d4kd\u2212 ik y \u00b7 so (taking the 2 of D(y) under the integral and equating factors) We therefore know that the Green\u2019s function has the form \u02dcD(k) = \u2212 k 1 \u00b7. k (17.95) (17.96) 1 D(y) = \u2212 (2\u03c0)4 d4k Z e\u2212 k ik y \u00b7 k \u00b7. (17.97) The integrand in this expression is singular when k k = 0. Recall that the presence of singularities means that we have to decide how to treat them to get a well-de\ufb01ned result. There are several ways to do this, and each has a physical interpretation. If we integrate over the \u201ctime\u201d component k0 \ufb01", "rst, we get \u00b7 D(y) = 1 (2\u03c0)4 \u2212 Z d3keik y \u00b7 dk0 Z ik0y0 e\u2212 k2 0 \u2212 \u03ba2 (17.98) k | | = \u03ba. Now the singularities live in a single 1\u2013D integral that we can where easily evaluate via contour integration and the method of residues provided that we select a suitable contour. Let\u2019s do the integral carefully (in case your contour integration is bit rusty). Note that the poles of this integral are both real. This means that the integral is ambiguous \u2013 it can be assigned any of several possible values depending on how we choose to evaluation it. It is beyond the scope of these notes to evaluate the consequences of making and physically interpreting each of these choices. Instead we will choose to include both poles completely within a standard contour closed in the upper or lower half plane respectively, and then take limits such that the poles return to the real axis (Im z) \u0393 \u2212\u03ba\u2212 \u03b5i i\u03ba\u2212 \u03b5 k 0 (Re z) C Figure 17.1: Contours for evaluating the Green\u2019s function in 4\u2013dimensions. after the integral because this particular choice leads us very simply to the advanced and retarded forms of the Green\u2019s function that we already obtained when discussing the fourier transform of the incoming or outgoing spherical Green\u2019s functions for the Helmholtz equation. First we have to decide which way to close the contour. Examining the integrand, we note that if y0 = x0 0 > 0 the integrand vanishes on a x\u2032 \u2212 lower-half contour like C in the \ufb01gure above. We displace the poles down i\u01eb. Finally, let slightly so that they lie inside the contour \u0393: z = k0 + iki be a complex variable such that the real axis is k0. \u2192 \u00b1 \u00b1 \u2212 \u03ba \u03ba dz I\u0393 iy0z e\u2212 z2 \u2212 \u03ba2 = \u221e dk0 Z \u2212\u221e e\u2212 k2 0 \u2212 ik0y0 \u03ba2 + ZC dz iy0z e\u2212 z2 \u2212 \u03ba2 (17.99) As noted, the integral over C clearly vanishes for y0 > 0. Thus: \u221e dk0 Z \u2212\u221e ik0y0 e\u2212 k2 0 \u2212 \u03ba2 = dz I\u0393 e\u2212 z2 iy0z \u03ba2 \u2212 2\u03c0i)Res ( = lim \u2212 0 \u01eb \u2192 = = i\u03bay0 e\u2212 2\u03ba ( sin(\u03bay0) \u03ba 2\u03c0i \u2212 2\u03c0 \u2212 (z + izy0 e\u2212 i\u01eb))(z + (\u03ba + i\u01eb) (\u03ba \u2212 ei\u03bay0 \u2212 2\u03ba ) \u2212 (17.100) We can then write the Green\u2019s function as D(z) = = = = Z \u03b8(y0) (2\u03c0)3 \u03b8(y0) (2\u03c0)3 \u03b8(y0) (2\u03c0)2 \u03b8(y0) 2\u03c02R Z 0 Z 0 Z z sin(\u03baz0) d3keik \u00b7 \u03ba \u221e \u03ba2d\u03ba \u221e \u03ba2d\u03ba 1 1 Z \u2212 d(cos(\u03b8)) 2\u03c0 0 Z i\u03baR i\u03baR Z \u2212 d(i\u03baR cos(\u03b8)) d\u03c6ei\u03baR cos(\u03b8) sin(\u03bay0) \u03ba sin(\u03bay0) \u03ba ei\u03baR cos(\u03b8) i\u03baR 0 \u221e d\u03ba sin(\u03baR) sin(\u03bay0) (17.101) where R = x | x\u2032 | \u2212 is the spatial separation of the points x and x\u2032. Using a trig identity (or if you prefer expanding the sin\u2019s in terms of exponentials and multiplying out, then changing variables and exploiting we can the fact that only even terms survive) to extend the integral to write this as: \u2212\u221e D(z) = \u03b8(y0) 4\u03c0R 1 2\u03c0 Z \u221e d\u03ba ei(y0 \u2212 R)\u03ba \u2212\u221e \u2212 ei(y0+R)\u03ba. (17.102) These remaining integrals are just one dimensional Dirac delta functions. Evaluating, we get: Dr(x \u2212 x\u2032) = \u03b8(x0 x\u2032 0) \u2212 4\u03c0R \u03b4(x0 n 0 x\u2032 \u2212 \u2212 R) + \u03b4(x0 \u2212 x\u2032 0 + R) o (17.103) where we have now labelled it with \u201cr\u201d for \u201cretarded\u201d. The source event", " x\u2032 is always at an earlier time than the observation event x. This means that the domain of the support of the Heaviside function just happens to be disjoint from the support of the second delta function. We can therefore simplify this to: Dr(x \u2212 x\u2032) = \u03b8(x0 x\u2032 0) \u2212 4\u03c0R \u03b4(x0 0 x\u2032 R) \u2212 \u2212 (17.104) which is just what we got before from Fourier transforming the outgoing stationary wave Green\u2019s function, as it should be. If we had chosen the other contour, identical arguments would have led us to the advanced Green\u2019s function: Da(x \u2212 x\u2032) = \u03b8[ (x0 \u2212 \u2212 4\u03c0R x\u20320)] \u03b4(x0 \u2212 x\u20320 + R) (17.105) The other possible contours (enclosing only one or the other of the two singularities, using a contour that avoids the singularities on the real axis instead of displacing the singularities) would yield still other possible Green\u2019s functions. Just as an arbitrary normalized sum of outgoing and incoming Green\u2019s functions resulted in an acceptable Green\u2019s function before, an arbitrary sum of advanced and retarded Green\u2019s functions are acceptable here. However, the inhomogeneous term of the integral equation is a functional of the Green\u2019s function selected! For what it is worth, the Green\u2019s functions can be put in covariant form. One almost never uses them in that form, and it isn\u2019t pretty, so I won\u2019t bother writing it down. We can now easily write down formal solutions to the wave equation for arbitrary currents (not just harmonic ones): and A\u03b1(x) = A\u03b1 in(x) + \u00b50 d4x\u2032Dr(x Z A\u03b1(x) = A\u03b1 out(x) + \u00b50 d4x\u2032Da(x x\u2032)J \u03b1(x\u2032) (17.106) x\u2032)J \u03b1(x\u2032). (17.107) \u2212 \u2212 In these equations, the inhomogeneous terms are the radiation \ufb01eld incident upon (radiated from) the four-volume of space-time containing the fourcurrent that are not connected to the four-current in that four-volume by the retarded Green\u2019s function. Z It is a worthwhile exercise to meditate upon what might be a suitable form for the inhomogeneous terms if one considerst the integration fourvolume to be in\ufb01nite (with no inhomogeneous term at all) and then split the in\ufb01nite volume up into the interior and exterior of a \ufb01nite four-volume, as we did with incoming and outgoing waves before, especially when there are many charges and they are permitted to interact. Dirac noted that choosing a \u201cretarded\u201d Green\u2019s function, just as choosing an \u201coutgoing wave\u201d Green\u2019s function before, results in a somewhat misleading picture given that the actual physics is completely time-reversal symmetric (indeed, independent of using a mixed version of the Green\u2019s functions in either case). He therefore introduced the \u201cradiation \ufb01eld\u201d as the di\ufb00erence between the \u201coutgoing\u201d and the \u201cincoming\u201d inhomogenous terms given the contraint that the actual vector potential is the same regardless of the choice of Green\u2019s function used:: A\u03b1 radiation = A\u03b1 out \u2212 A\u03b1 in = 4\u03c0 c Z d4x\u2032D(x \u2212 x\u2032)J \u03b1(x\u2032) (17.108) where D(z) = Dr(z) Da(z). \u2212 (17.109) In some fundamental sense, only the radiation \ufb01elds are \u201cphysical\u201d \u2013 they are the change in the vector potential at an event produced symmetrically by any given four-current due to its past and its future motion. This is a critical aspect of the interpretation of radiation reaction as being produced by transfer of momentum both to a charge (event) from other charges in its past and from a charge to those same charges in its future. Chapter 18 Radiation from Point Charges To summarize from the last chapter, two useful Green\u2019s functions for the inhomogeneous wave equation: 2A\u03b1 = \u00b50J \u03b1 are Dr(x \u2212 x\u2032) = \u03b8(x0 x\u2032 0) \u2212 4\u03c0R \u03b4(x0 0 x\u2032 R) \u2212 \u2212 (the retarded Green\u2019s function) and Da(x \u2212 x\u2032) = \u03b8[ \u2212 (x0 \u2212 4\u03c0R x\u2032 0)] \u03b4(x0 \u2212 x\u2032 0 + R) (18.1) (18.2) (18.3", ") (the advanced Green\u2019s function). The integral equations associated with these Green\u2019s functions were: A\u03b1(x) = A\u03b1 in(x) + \u00b50 d4x\u2032Dr(x and A\u03b1(x) = A\u03b1 out(x) + \u00b50 Z Z d4x\u2032Da(x x\u2032)J \u03b1(x\u2032) (18.4) x\u2032)J \u03b1(x\u2032). (18.5) \u2212 \u2212 For the moment, let us ignore Dirac\u2019s observations and the radiation \ufb01eld and focus instead on only the \u201cnormal\u201d causally connected retarded potential produced by a single charged particle as it moves in the absence of external potentials. This potential is \u201ccausal\u201d in that the e\ufb00ect (the potential \ufb01eld) follows the cause (the motion of the charge) in time, where the advanced potential has the e\ufb00ect preceding the cause, so to speak. Let 287 me emphasize that this is not a particularly consistent assumption (again, we the theory is manifestly time symmetric so \u201cpast\u201d and \u201cfuture\u201d are pretty much arbitrary namings of two opposed directions), but it yields some very nice results, as well as some problems. In that case: A\u03b1(x) = \u00b50 d4x\u2032Dr(x Z x\u2032)J \u03b1(x\u2032) \u2212 where the four\u2013current of a point charge e is found from J(x\u2032, t) = c\u03c1(x\u2032, t) J (x\u2032, t) ) = ( ec \u03b4[x\u2032 ev\u03b4[x\u2032 ( r(t)] r(t)] ) \u2212 \u2212 in the lab/rest frame K or (in covariant form): where J \u03b1(x\u2032) = ec d\u03c4 U \u03b1(\u03c4 )\u03b4(4) ([x\u2032 Z U = \u03b3 c v! = dr d\u03c4 r(\u03c4 )]) \u2212 (18.6) (18.7) (18.8) (18.9) Note that the \u03b4 function in these expressions simply forces the particle to be found at the correct location at each (proper) time. The r(\u03c4 ) function is the trajectory of the particle. Its \u03c4 derivative is the four\u2013velocity. This yields (when the \u03b3\u2019s have all been accounted for) the rest frame expression. To do the integral, we need the \u201cmanifestly covariant\u201d form of the re- tarded Green\u2019s function. Note that: \u03b4[(x \u2212 \u2212 0)2 x\u2032 0 x\u2032 x \u2212 \u2212 | R)(x0 vx\u2032 2] | 0 + R)] x\u2032 x\u2032)2] = \u03b4[(x0 = \u03b4[(x0 1 2R = \u2212 [\u03b4(x0 \u2212 \u2212 R) + \u03b4(x0 \u2212 ). In terms of this, Dr is given by | \u2212 x\u20320 \u2212 (where R = x | \u2212 x\u2032 x\u20320 + R)] (18.10) 1 2\u03c0 \u2212 x\u2032) = Dr(x x\u20320)\u03b4[(x \u03b8(x0 \u2212 Again, the second delta\u2013function makes no contribution because of the opposing \u03b8\u2013function. Thus \u00b50c 2\u03c0 Z A\u03b1(x) = x\u20320)\u03b4([x (18.11) x\u2032]2) \u2212 x\u2032)2]. d4x\u2032\u03b8(x0 \u2212 e \u00d7 \u2212 d\u03c4 U \u03b1(\u03c4 )\u03b4(4) ([x\u2032 Z d\u03c4 U \u03b1(\u03c4 )\u03b8[x0 \u2212 \u2212 rx(\u03c4 )]\u03b4 = e\u00b50c 2\u03c0 Z r(\u03c4 )]) (18.12) [x n \u2212 r(\u03c4 )]2. (18.13) o The vector potential at a point gets a contribution only where\u2013when that point lies on the light cone in the future (picked out by the \u03b8 function) of the world line of the charge (picked out be the \u03b4 function). The contribution is proportional to eU \u03b1(\u03c4 ) at that (retarded) time. It dies o\ufb00 like 1/R, although that is obscured by the form of the \u03b4 function. To evaluate this (and discover the embedded R), we use the rule (from way back at the beginning of the book, p. 30 in J1.2) \u03b4(x xi) \u2212 (18.14) \u03b4[f (x)] = Xi df dx x=xi where the x = xi are the non\u2013degenerate zeros of f (x). f (x) is assumed to be \u201csmooth\u201d. Then if we let (18.15)", " (18.16) (18.17) (18.18) (which is zero when \u03c4 = \u03c4p in the past) then f (\u03c4 ) = [x r(\u03c4 )]2 \u2212 d d\u03c4 [x \u2212 r(\u03c4 )]2 = 2[x \u2212 \u2212 r(\u03c4 )]\u03b2U \u03b2(\u03c4 ) and therefore \u03b4([x \u2212 r(\u03c4 )]2) = \u03b4(\u03c4 \u2212 \u03c4p) r(\u03c4 )]\u03b2U \u03b2(\u03c4 ) | = \u03b4(\u03c4 \u03c4p) r(\u03c4 )]\u03b2U \u03b2(\u03c4 ) \u2212 2[x \u2212 2[x |\u2212 \u2212 From this we see that A\u03b1(x) = e\u00b50c 4\u03c0 U \u00b7 U \u03b1(\u03c4 ) [x \u2212 \u03c4 =\u03c4p r(\u03c4 )] where \u03c4p is the proper time in the past of x when the light cone of the charge contains the event x. This potential (and its other forms above) are called the Li\u00b4enard\u2013Wiechert potentials. In non\u2013covariant form, they are obtained from the identity (x U \u00b7 \u2212 r) = U0[x0 \u2212 = \u03b3cR(1 \u2212 \u00b7 r0(\u03c4p)] \u03b2 n) U \u00b7 \u2212 [x \u2212 r(\u03c4p)] (18.19) where n is a unit vector in the direction of x as usual. \u2212 r(\u03c4 ) and where \u03b2 = v(\u03c4 )/c Recall that A = (\u03c6/c, A). Thus: A0(x) = e\u00b50c 4\u03c0 \u03b3c \u03b3cR(1 \u03b2 \u00b7 \u2212 ret (18.20) \u02c6n) and \u03c6(x, t) = cA0 = e 4\u03c0\u01eb0 where all quantities (e.g. \u03b2, R) must be evaluated at the retarded time where the event x is on the light cone of a point on the particle trajectory. \u02c6n) (18.21) 1 \u03b2 R(1 \u2212 ret \u00b7 (18.22) ret Similarly A(x, t) = = e\u00b50c 4\u03c0 e 4\u03c0 s \u03b3cR(1 \u00b50 \u01eb0 R(1 \u03b3c\u03b2 \u03b2 \u2212 \u03b2 \u00b7 \u03b2 \u2212 \u00b7 ret \u02c6n) \u02c6n) where again things must be evaluated at retarded times on the particle trajectory. Note well that both of these manifestly have the correct nonrelativistic form in the limit << 1. \u03b2 | | We can get the \ufb01elds from the 4\u2013potential in any of these forms. However, the last few forms we have written are compact, beautiful, intuitive, and have virtually no handles with which to take vector derivatives. It is simpler to return to the integral form, where we can let \u2202\u03b1 act on the \u03b4 and \u03b8 functions. \u2202\u03b1A\u03b2 = e\u00b50c 2\u03c0 Z d\u03c4 U \u03b2(\u03c4 )\u03b8[x0 \u2212 r0(\u03c4 )]\u2202\u03b1\u03b4 [x r(\u03c4 )]2 \u2212 where \u2202\u03b1\u03b4[f ] = \u2202\u03b1f \u00b7 r(\u03c4 )]2. Then d df \u03b4[f ]! = \u2202\u03b1f d\u03c4 df \u00b7 d d\u03c4 \u00b7 \u03b4[f ]. Again, we let f = [x \u2212 \u2202\u03b1\u03b4[f ] = (x U \u00b7 \u2212 (x r)\u03b1 r) \u2212 d d\u03c4 \u2212 \u03b4[f ] (18.23) (18.24) (18.25) This is inserted into the expression above and integrated by parts: \u2202\u03b1A\u03b2 = = e\u00b50c 2\u03c0 Z \u2212 e\u00b50c 2\u03c0 Z d\u03c4 U \u03b2(\u03c4 )\u03b8[x0 \u2212 (x d U \u03b2(\u03c4 ) U d\u03c4 ( d\u03c4 r0(\u03c4 )] r)\u03b1 r) d d\u03c4 \u03b4[f ] \u2212 (x (x U \u00b7 r)\u03b1 \u2212 (x \u00b7 r) ) \u2212 \u2212 \u03b8[x0 \u2212 r0(\u03c4 )]\u03b4([x \u2212 r(\u03c4 )]2). (18.26) There is no contribution from the \u03b8 function because the derivative of a theta function is a delta function with the same arguments d d\u03c4 \u03b8(x0 \u2212 r0(\u03c4 )) = \u03b4[x0 \u2212 r0(\u03c4 )] (18.27) R2). This only gets a which constrains the other delta function to be \u03b4( contribution at R = 0 (on the world line of the charge), but we already feel uncomfortable about the \ufb01eld there, which we suspect is in\ufb01nite and meaningless, so we exclude this point from consideration. Anywhere else the result above is exact. \u2212 We can now", " do the integrals (which have the same form as the potential integrals above) and construct the \ufb01eld strength tensor: F \u03b1\u03b2 = e\u00b50c 4\u03c0 e (x U \u00b7 r) \u2212 (x d d\u03c4 ( \u2212 r)\u03b1U \u03b2 U \u2212 (x \u00b7 \u2212 \u2212 r) (x r)\u03b2U \u03b1 (18.28) ret This whole expression must be evaluated after the di\ufb00erentiation at the retarded proper time \u03c4p. ) This result is beautifully covariant, but not particularly transparent for all of that. Yet we will need to \ufb01nd explicit and useful forms for the \ufb01elds for later use, even if they are not as pretty. Jackson gives a \u201clittle\u201d list of ingredients (J14.12) to plug into this expression when taking the derivative to get the result, which is obviously quite a piece of algebra (which we will skip): E(x, t) = e\u00b50 4\u03c0c2 \" \u03b32(1 ( \u02c6n \u2212 \u2212 \u03b2 \u03b2) \u02c6n)3R2 #ret \u00b7 + \u02c6n e\u00b50 4\u03c0c3 \uf8ee \uf8f0 and B(x, t) = 1 c ( \u02c6n \u00d7 E) ( \u02c6n \u2212 \u03b2 \u00d7 (1 \u2212 \u00b7 \u02d9\u03b2 \u03b2) \u00d7 \u02c6n)3R \uf8f9 ret (18.29) \uf8fb (18.30) \u201cArrrgh, mateys! Shiver me timbers and avast!\u201d, you cry out in dismay. \u201cThis is easier? Nonsense!\u201d Actually, though, when you think about it (so think about it) the \ufb01rst term is clearly (in the low velocity, low acceleration limits) the usual static \ufb01eld: E \u2248 e 4\u03c0\u01eb0 \u02c6n R2 (18.31) Interestingly, it has a \u201cshort\u201d range and is isotropic. The second term is proportional to the acceleration of the charge; both 1 and hence are \u201clong E and B are transverse and the \ufb01elds drop o\ufb00 like R\u2212 range\u201d but highly directional. If you like, the \ufb01rst terms are the \u201cnear\u201d and \u201cintermediate\u201d \ufb01elds and the second is the complete \u201cfar\u201d \ufb01eld; only the far \ufb01eld is produced by the acceleration of a charge. Only this \ufb01eld contributes to a net radiation of energy and momentum away from the charge. With that (whew!) behind us we can proceed to discuss some important expressions. First of all, we need to obtain the power radiated by a moving charge. 18.1 Larmor\u2019s Formula If one is far (enough) away from the an accelerating charge in the right direction, the \ufb01eld is given by primarily by the second (acceleration) term. This is the \u201cusual\u201d transverse EM \ufb01eld. If the particle is moving slowly with respect to c (so \u03b2 << 1), then E = B = e 4\u03c0\u01eb0 1 c e 4\u03c0\u01eb0 1 c2 \u02c6n \u00d7 \u02c6n ( \u02c6n R \u02d9\u03b2 \u00d7 R \u00d7 ret \u02d9\u03b2) ret (18.32) (18.33) The energy \ufb02ux is given by the (instantaneous) Poynting vector: S = 1 \u00b50 (E B) \u00d7 = = = e2 16\u03c02\u01eb0 R2 e2 16\u03c02\u01eb0 R2 e2 16\u03c02\u01eb0 R2 1 \u00b50\u01eb0 1 c3 | 1 c3 | \u02c6n \u02c6n 1 c3 | \u02c6n ( \u02c6n \u02d9\u03b2) 2 \u02c6n | \u00d7 \u00d7 ( \u02c6n c2 \u02d9\u03b2) 2 \u02c6n | \u00d7 ( \u02c6n 2 \u02c6n \u02d9v) | \u00d7 \u00d7 \u00d7 (18.34) As always, the power cross-section (energy per unit solid angle) is dP d\u2126 = S \u02c6nR2 \u00b7 e2 16\u03c02\u01eb0 e2 16\u03c02\u01eb0 1 c3 | 1 c3 | = = \u02c6n ( \u02c6n 2 \u02d9v) | \u00d7 \u00d7 \u02d9v 2 sin2(\u0398) | (18.35) where \u0398 is the angle between \u02c6n and \u02d9v. Aha! we say. The characteristic sin2", " \u0398! Aha again! Inspecting the vector products, we see that the radiation is polarized in the plane of n, \u02d9v, perpendicular to n. Finally, the integral over angles yields 8\u03c0/3, so that P = e2 6\u03c0\u01eb0c3 | \u02d9v 2. | (18.36) This is the Larmor formula for the power radiated from a nonrelativistic accelerated point charge. This has a covariant generalization that is valid for any velocity of charge. First we factor out an m2 and convert this to momentum coordinates. Then we realize that the energy carried by this \ufb01eld (per unit time) is indeed related to the momentum by a factor of 1/c and convert the whole thing to 4\u2013vector form. Last, we convert t into (mv) dt e2 1 m2 6\u03c0\u01eb0c3 2 e2 d(mv) 6\u03c0\u01eb0m2c3 \u03b3d\u03c4 e2 \u03b22) 6\u03c0\u01eb0m2c3 (1 e2 6\u03c0\u01eb0m2c3 \u2212 dp\u03b1 e2 6\u03c0\u01eb0m2c3 d\u03c4! dp d\u03c4! dp\u03b1 d\u03c4 dp d\u03c4 1 c2 \u2212 2 2 \u2212 2 dE d\u03c4! (18.37) This can be written one more way, (substituting E = \u03b3mc2 and p = \u03b3mv and using some vector identities) due to Li\u00b4enard: P = e2 6\u03c0\u01eb0c3 \u03b36[( \u02d9\u03b2)2 (\u03b2 \u2212 \u00d7 \u02d9\u03b2)2] (18.38) We are all better people for knowing this. Why, you may ask, is this torture necessary? Because quite a few of you will spend unreasonable amounts of your lives calculating things like radiative losses in accelerators. After all, if we could build GeV accelerators in a little bitty ten foot ring it would be a whole lot cheaper than 6 billion bucks, plus in\ufb02ation. Unfortunately, nature says that if you try it the nasty thing will give o\ufb00 synchrotron radiation! Let us see that tanstaa\ufb021. The radiated power is proportional to the acceleration. The work is proportional to the tangential force times the velocity. Light particles accelerate the most for a given tangential force and have the highest velocity for a given energy; radiative losses are thus the most important for those particles at all energies. We will evaluate the radiative power loss for an electron in a linear accelerator. We begin with P = 2 e2 6\u03c0\u01eb0m2c3 dp dt! (18.39) \u2212 e is now really the charge on the electron. Since the accelerator is where linear, we can \ufb01nd the force directly from the rate at which work is done on the electron (otherwise we would have to include the force bending it in a curved path, which does no work). It is related to the \u201cgradient\u201d of the total energy, P = e2 6\u03c0\u01eb0m2c3 dE dx! 2. (18.40) For linear acceleration we don\u2019t care what the actual energy of the particle is; we only care how that energy changes with distance. We will turn this into a rate equation by using the chain rule: Prad = e2 6\u03c0\u01eb0m2c3 dE dx dE dt dt dx (18.41) Thus the ratio of power radiated to power supplied by the accelerator Pacc = dE/dt is: Prad Pacc = e2 6\u03c0\u01eb0m2c3 1 v dE dx \u2248 1 6\u03c0\u01eb0 e2/mc2 mc2 dE dx where the latter form is valid when the electron is travelling at v 1There Ain\u2019t No Such Thing As A Free Lunch. No kidding. (18.42) c. \u2248 This quantity will be less than one while the gain in energy in a distance 13 cm is of the order of mc2 =.5 MeV. That would e2/mc2 = 2.82 require a potential di\ufb00erence (or other force) on the order of 1014 MV/meter. 10\u2212 \u00d7 Maybe at the surface of a positron. Come to think of it, falling into a positron there comes a point where this is true and at that point the total mass energy of the pair is radiated away. But nowhere else. We can completely neglect radiative losses for linear acceleration simply because the forces required to produce the requisite changes in energy when the particle is moving at nearly the speed of light are ludicrously large. For a charged particle moving in a straight line, radiative losses are", " more important at low velocities. This is fortunate, or radios and the like with linear dipole antennas would not work! However, it is incovenient to build linear accelerators. That is because a linear accelerator long enough to achieve reasonable energies for electrons starts (these days) at around 100\u2013500 miles long. At that point, it is still not \u201cstraight\u201d because the earth isn\u2019t \ufb02at and we don\u2019t bother tunnelling out a secant. Also, it seems sensible to let a charged particle fall many times through the \u201csame\u201d potential, which is possible only if the accelerator is circular. Unfortunately, we get into real trouble when the accelerator is not straight. In a circular accelerator, there is a non\u2013zero force proportional to its velocity squared, even when little or no work is being done to accelerate the particle! In fact, the centripetal force on the particle is = \u03b3\u03c9 >> p | | 1 c dE d\u03c4 (18.43) dp d\u03c4 all of which increase as the speed of the particle increases. If we completely neglect the radiative loss due to tangential acceleration (which is completely negligible once relativistic velocities have been reached) we see that P = e2 6\u03c0\u01eb0m2c3 \u03b32\u03c92 2 = p | | e2c 6\u03c0\u01eb0r2 \u03b24\u03b34 (18.44) where we have used \u03c9 = (c\u03b2/r). The loss per revolution is obtained by multiplying by T (the period of a revolution). This yields \u2206E = 2\u03c0r c\u03b2 P = e2 3\u01eb0r \u03b23\u03b34 (18.45) which is still deadly if r is small and/or \u03b3 and \u03b2 are large. If one does some arithmetic (shudder), one can see that for high energy electrons (where \u03b2 1), this is \u2248 \u2206E(MeV) = 8.85 10\u2212 \u00d7 2 [E(GeV)]4 r(meters). (18.46) At around 1 GeV, one needs roughly 1/(10r) of that energy gain per cycle in order to turn (heh, heh) a net pro\ufb01t. That is not so bad, but the power of 4 says that at 10 GeV, one needs a gain per cycle of 1000/r GeV (!) in order to turn a pro\ufb01t. Now, it is true that the bigger the radius the longer the circumference (linearly) and the longer the circumference the more work one can do with a given \ufb01xed potential in a cycle. So in terms of force this relation is not as bad as it seems. But it is bad enough, because you still have to do the work, which costs you the same no matter how hard you have 100 meters or to push to do it. Clearly even at 10 GeV, an orbit of radius better is necessary. In electron\u2013positron storage rings, work must be done at this general rate just to keep the particles moving. \u223c Those of you who need to know can read section 14.3 on your own. The results are straightforward but algebraically tedious, and are of use only if you plan on studying accelerator design or neutron stars. Don\u2019t get me wrong. Nobel prizes have been won for accelerator design and may be again. Go for it. In Ditto for 14.4. This is highly readable and contains no algebra. a nutshell, a particle moving in a synchrotron emits its radiation in its instantaneous direction of motion (which is indeed perpendicular to the acceleration). Since it moves in a circle, a stationary observer in the plane of motion sees short bursts of radiation at the characteristic frequency c/r. The length (in time) of the pulses is L/c in time, and thus will contain (c/r)\u03b33 in a fourier decomposition of their \u201cwave frequencies up to c/L r/(2\u03b33) is the length of the pulse in space. For highly packet\u201d where L relativistic particles moving in big circles, the characteristic frequency can be many orders of magnitude smaller than the high frequency cut o\ufb00, as in AM radio frequencies to X\u2013rays or worse. Synchrotron radiation is a potential source of high frequency electromagnetic energy. \u223c \u2248 Of course, it isn\u2019t tunable or coherent (in fact, its highly incoherent since the spectrum is so wide!) and we\u2019d love to use the same kind of trick to make coherent, tunable, high frequency light. Some of you probably will use the same kind of trick before you leave, since free electron lasers produce energy from a similar principle (although with a totally di\ufb00erent spectrum!). Section 14.6 deals with the spectrum, and we will blow", " that o\ufb00, too. Su\ufb03ce it to say that it can be calculated, and you can learn how, if you need to. \u03c90\u03b33, and should take a peek at the You really should remember that \u03c9c \u2248 distribution curves. These curves let one detect synchrotron radiation from cosmological sources. These sources are generally charged particles falling into dark stars, radiation belts around planets, sunspots, or anyplace else that relativistic electrons are strongly accelerated in a circular, or helical, path. Finally, we will neglect 14.5 too, which analyzes radiation emitted by particles moving in wierd ways. Jackson is encyclopaediac, but we needn\u2019t be. We will come back into focus at section 14.7, Thomson Scattering of Radiation. This is scattering of radiation by charged particles and is closely related to Compton scattering. It is important, as it is a common phenomenon. 18.2 Thomson Scattering of Radiation Suppose that a plane wave of monochromatic electromagnetic radiation is incident on a free particle of charge e and mass m. The particle will experience a force from this \ufb01eld, and will accelerate. As it accelerates, it will emit radiation in di\ufb00erent directions, dispersing the incident beam. For a non\u2013relativistic particle accelerated by a force we can see that: dP d\u2126 = e2 16\u03c02\u01eb0 1 c3 | \u02c6\u01eb\u2217 \u02d9v 2 | \u00b7 (18.47) 2 sin2 \u0398 for a particular polarization perpendicular to (where | \u00b7 the plane of \u02c6n and \u02d9v). 2 = | \u02c6\u01eb\u2217 | \u02d9v | \u02d9v The (leading order) acceleration is due to the plane wave electric \ufb01eld with polarization \u02c6\u01eb0, wave vector k0, and Newton\u2019s Law: \u02d9v = e m E0\u02c6\u01eb0eik0 x \u00b7 \u2212 \u03c9t (18.48) If the charge moves much less than one wavelength during a cycle (true for all but the lightest particles and strongest \ufb01elds) then \u02d9v | |av = 1 2 Re ( \u02d9v \u02d9v\u2217) \u00b7 (18.49) Thus the average power \ufb02ux distribution is dP d\u2126!av = = 2 2 E0| c 32\u03c02\u01eb0 | e2 4\u03c0\u01eb0mc2 ) ( e2 mc2! 2 \u01eb0cE2 2 0 \u02c6\u01eb\u2217 | 2 \u02c6\u01eb0| \u00b7 \u02c6\u01eb\u2217 | 2 \u02c6\u01eb0| \u00b7 (18.50) This is clearly of the same general form as the scattering expressions we described and derived earlier. Since the result contains E2 0 it makes sense to divide out the incident intensity and thus obtain a di\ufb00erential cross section that works for all but the strongest \ufb01elds. We thus divide out the time-averaged \ufb02ux of the Poynting vector of the incident plane wave: hence I = \u01eb0cE2 0 2 d\u03c3 d\u2126 = 2 e2 4\u03c0\u01eb0mc2 ) ( \u02c6\u01eb\u2217 | 2 \u02c6\u01eb0| \u00b7 (18.51) (18.52) If we let the plane wave be incident along the z axis, let \u02c6n form an angle \u03b8 with that axis, and pick two polarization directions in and perpendicular to the ( \u02c6n, \u02c6z) plane (as before), and average over polarizations then this dot product yields: d\u03c3 d\u2126 = ( e2 4\u03c0\u01eb0mc2 ) 2 1 2 (1 + cos2 \u03b8). (18.53) as it did back in our earlier work on scattering, but now for a point particle. This is the Thomson formula for scattering of radiation by free charge. It works for X\u2013rays for electrons or \u03b3\u2013rays for protons. It does not work when the photon momentum and the recoil of the charged particle cannot be neglected. The integral of this, \u03c3T = 8\u03c0 3 ( 2 e2 4\u03c0\u01eb0mc2 ) (18.54) is called the Thomson cross\u2013section. It is 0.665 10\u2212 29 m2 for electrons. \u00d7 The quantity in parentheses has the units of length. If the total \u201cmass\u2013 energy\u201d of the electron were due to its charge being concentrated in a ball, then this would be the close order of the radius of that ball; it is called the classical electron radius. This number crops", " up quite frequently, so you should remember it. What it tells us is that even point particles have a \ufb01nite scattering cross-section that appears in this limit to be independent of the wavelength of the light scattered. However, this is not really true if you recall the approximations made \u2013 this expression will fail if the wavelength is on the same order as the classical radius, which is precisely where pair production becomes a signi\ufb01cant process quantum mechanically. In quantum mechanics, if the energy of the mc2 for the electron, signi\ufb01cant momentum is transincident photon \u00afh\u03c9 ferred to the electron by the collision and the energy of the scattered photon cannot be equal to the energy of the incident photon. Whatever a photon is... \u2248 We can actually \ufb01x that without too much di\ufb03culty, deriving the Compton scattering formula (which takes over from Thomson in this limit). This formula adds a wavelength/angle dependence to Thomson\u2019s general result and yields the Klien-Nishina formula, but this is beyond our scope in this course to derive or discuss in further detail. We are almost \ufb01nished with our study of electrodynamics. Our \ufb01nal object of study will be to to try to address the following observation: Accelerated charges radiate. Radiation accelerates charge. Energy must be conserved. These three things have not been consistently maintained in our treatments. We study one, then the other, and require the third to be true in only part of the dynamics. What is missing is radiation reaction. As charges accelerate, they radiate. This radiation carries energy away from the system. This, then means that a counterforce must be exerted on the charges when we try to accelerate them that damps charge oscillations. At last the folly of our ways is apparent. Our blind insistence that only retarded \ufb01elds are meaningful (so that we can imagine the \ufb01elds to be zero up to some time and then start moving a charge, which subsequently radiates) has left us with only one charge that can produce the \ufb01eld that produces the force that damps applied external forces \u2014 the charge itself that is radiating. No other charge produces a \ufb01eld that can act on this charge \u201cin time\u201d. We have invented the most sublime of violations of Newton\u2019s laws \u2013 an object that lift\u2019s itself up by its own bootstraps, an Aristotelian object that might even be able to come to rest on its own in the absence of external forces. Clearly we must investigate radiation reaction as a self\u2013force acting on an electron due to its own radiation \ufb01eld, and see if it is possible to salvage anything like a Newtonian description of even classical dynamics. We already know that Larmor radiation plus stable atoms spells trouble for Newton, but Newton still works classically, doesn\u2019t it? Let\u2019s take a look. Uh\u2013oh, you say. Wasn\u2019t the, well, wasn\u2019t everything singular on a point charge? Won\u2019t we get in\ufb01nities at every turn? How will we realize \ufb01nite results from in\ufb01nite \ufb01elds, potentials, self-energies, and so on? Yes! I cry with glee. That\u2019s the problem. Finally we will learn how to take a singular \ufb01eld, a singular charge, and in\ufb01nite energy, and make a physically realized (almost) radiation reaction force out of it. Chapter 19 Radiation Reaction 19.1 The Death of Classical Physics Thus far we have learned how to solve two kinds of problems. Either the \ufb01elds were assumed to be given, in which case the relativistic Lorentz force law yielded covariant equations of motion for a point charged massive particle interacting with these \ufb01elds or the trajectory of a charged, point particle was given and the \ufb01elds radiated by this particle were determined. This, however, was clearly not enough, or at least was not consistent. That is because (as a few simple mental problems will show) each of these processes is only half of an interaction \u2014 a complete, consistent \ufb01eld theory would include the self\u2013consistent interaction of a charged particle with the \ufb01eld in its vicinity, or better yet, the self-consistent interaction of all particles and \ufb01elds. We need to be able to calculate the total \ufb01eld (including the radiated \ufb01eld) at the position of any given point charge. Some of that \ufb01eld is due to the charge itself and some is due to the \ufb01eld produced by", " the other charges. But we do not know how to do this, really, since the one will a\ufb00ect the other, and there are clearly in\ufb01nities present. This sort of problem can also lead to Newtonian paradoxes, paradoxes that smack of the resurrection of Aristotelian dynamics. To see this, let us assume (non\u2013physically) that we have a Universe consisting of a single point charge orbiting around an uncharged gravitational mass (or some other force center that causes the charge to move in a bound orbit). In that case, the 301 point charge must (according to the laws of electrodynamics that we have thus far deduced) radiate energy and momentum into the electromagnetic \ufb01eld. As it accelerates, it must radiate. As it radiates, energy and momentum must be carried away from the point particle to \u201cin\ufb01nity\u201d. The particle must therefore decrease its total energy. If the particle is bound in an attractive, negative potential well, the only way that total energy can be conserved is if its total energy decreases. The particle must therefore spiral inwards the center, converting its potential energy into radiative energy in the \ufb01eld, until it reaches the potential minimum and comes to rest. There is only one di\ufb03culty with this picture. There is only one charged particle in the Universe, and it is interacting with only one attractive center. What acts to slow the particle down? This is a non\u2013question, of course \u2013 a thought experiment designed to help us understand where our equations of motion and classical picture are incomplete or inconsistent. The real universe has many charged particles, and they are all constantly interacting with all the other charged particles that lie within the \u201cevent horizon\u201d of an event relative to the time of the big bang, which is the set of the most distant events in space\u2013time in the past and in the future that can interact with the current event on the world line of each particle. It is the edge of the \u201cblack hole\u201d that surrounds us1. However, in our simplied Universe this question is very real. We have systematically rid ourselves of the \ufb01elds of all the other particles, so now we must \ufb01nd a \ufb01eld based on the particle itself that yields the necessary \u201cradiation reaction\u201d force to balance the energy\u2013momentum conservation equations. This approach will have many unsatisfactory aspects, but it works. First, when will radiation reaction become important? When the energy radiated by a particle is a reasonable fraction of the total relevant energy E0 of the particle under consideration. That is Erad \u223c 2 3c e2a2T 4\u03c0\u01eb0c2 (19.1) 1It is interesting to meditate upon the fact that your event horizon and my event horizon are not coincident, which leads in turn to an interesting problem with logical positivism. where a is the total (e.g. centripetal) acceleration and T is the period of the orbit associated with E0 or the time a uniform acceleration is applied. If Erad << E0 then we can neglect radiation reaction. As before, if a particle is uniformly (linearly) accelerated for a time \u03c4r, then we can neglect radiation reaction when E0 \u223c m(a\u03c4r)2 2 3c e2a2\u03c4r 4\u03c0\u01eb0c2 \u226b (19.2) Radiation reaction is thus only signi\ufb01cant when the opposite is true, when: \u03c4r \u223c 2 3c 2 3 e2 4\u03c0\u01eb0mc2 2 3 \u03c4e and a is large will radiation reaction be appreciable. For Only if \u03c4r \u223c 23 seconds. This was the situation we electrons this time is around 10\u2212 examined before for linear accelerators and electron\u2013positron anihillation. Only in the latter case is radiation reaction likely. re/c = (19.3) \u03c4e \u223c The second case to consider is where the acceleration is centripetal. Then the potential and kinetic energy are commensurate in magnitude (virial theorem) and where a if E0 \u223c m\u03c92 0d2 (19.4) 1/\u03c90. As before, we can neglect radiation reaction \u03c92 0d and \u03c4r \u223c \u223c m\u03c92 0d2 2 3c e2\u03c94 0d2 4\u03c0\u01eb0c2\u03c90 = \u03c90d2 2 3c e2 4\u03c0\u01eb0c2 \u226b (19.5) Radiation reaction is thus again signi\ufb01cant per cycle only if 1 \u03c90\u03c4r \u223c (ignoring factors of order one) where \u03c4r is given above \u2013 another way of saying the same thing. \u03c9\u2212 3) the time 0 \u03c4e associated with the motion,", " so only if this timescale corresponds to \u03c4r \u2248 will radiation reaction be signi\ufb01cant. is (within irrelevant factor of 2\u03c0 and 2 (19.6) 1 So far, our results are just a restatement of those we obtained discussing Larmor radiation except that we are going to be more interested in electrons in atomic scale periodic orbits rather than accelerators. Electrons in an atomic orbit would be constantly accelerating, so any small loss per cycle is summed over many cycles. A bit of very simple order-of-magnitude arithmetic will show you that radiative power loss need not be negligible as a rate compared to human timescales when \u03c9\u2212 is very small (e.g. order 0 15 seconds for e.g. optical frequency radiation). Charged particles of 10\u2212 (especially electrons) that move in a circle at a high enough (angular) speed do indeed radiate a signi\ufb01cant fraction of their energy per second when the loss is summed over many cycles. The loss per cycle may be small, but it adds up inexorably. 1 How do we evaluate this \u201cradiation reaction force\u201d that has no obvious physical source in the equations that remain? The easy way is: try to balance energy (and momentum etc) and add a radiation reaction force to account for the \u201cmissing energy\u201d. This was the approach taken by Abraham and Lorentz many moons ago. 19.2 Radiation Reaction and Energy Con- servation We know that F tot = m \u02d9v (19.7) is (nonrelativistic) Newton\u2019s 2nd Law for a charged particle being accelerated by a (for the moment, non\u2013electromagnetic) given external force. The work energy theorem dictates how fast the particle can gain kinetic energy if this is the only force acting. However, at the same time it is being acted on by the external force (and is accelerating), it is also radiating power away at the total rate: P (t) = e2 4\u03c0\u01eb0c2 \u02d9v2 mre \u02d9v2 c 2 3c 2 3 = m\u03c4r \u02d9v2 = (19.8) (the Larmor formula). These are the two pieces we\u2019ve thus far treated independently, neglecting the one to obtain the other. However, in order for Newton\u2019s law to correctly lead to the conservation of energy, the work done by the external force must equal the increase in kinetic energy plus the energy radiated into the \ufb01eld. Energy conservation for this system states that: Wext = \u2206Ee + \u2206Ef (19.9) or the total work done by the external force must equal the change in the total energy of the charged particle (electron) plus the energy that appears in the \ufb01eld. If we rearrange this to: Wext \u2212 \u2206Ef = \u2206Ee (19.10) and consider the electron only, we are forced to conclude that there must be another force acting on the electron, one where the total work done by the force decreases the change in energy of the electron and places the energy into the radiated \ufb01eld. We call that force F rad, the radiation reaction force. Thus (rewriting Newton\u2019s second law in terms of this force): F ext + F rad = m \u02d9v F rad = m \u02d9v F ext \u2212 (19.11) de\ufb01nes the radiation reaction force that must act on the particle in order for energy conservation to make sense. The reaction force has a number of necessary or desireable properties in order for us to not get into \u201ctrouble\u201d2. \u2022 \u2022 \u2022 We would like energy to be conserved (as indicated above), so that the energy that appears in the radiation \ufb01eld is balanced by the work done by the radiation reaction force (relative to the total work done by an external force that makes the charge accelerate). We would like this force to vanish when the external force vanishes, so that particles do not spontaneously accelerate away to in\ufb01nity without an external agent acting on them. We would like the radiated power to be proportional to e2, since the power and its space derivatives is proporotional to e2 and since the force magnitude should be dependent of the sign of the charge. 2Trouble such as particles capable of lifting themselves up by their own metaphorical bootstraps... \u2022 Finally, we want the force to involve the \u201ccharacteristic time\u201d \u03c4 (whereever it needs a parameter with the dimensions of time) since no other timescaled parameters are available. Let\u2019s start with the \ufb01rst of these. We want the energy radiated by some \u201cbound\u201d charge (one undergoing periodic", " motion in some orbit, say) to equal the work done by the radiation reaction force in the previous equation. Let\u2019s start by examining just the reaction force and the radiated power, then, and set the total work done by the one to equal the total energy radiated in the other, over a suitable time interval: t2 t1 Z F rad \u00b7 vdt = t2 \u2212 Z t1 P dt = t2 \u2212 Z t1 m\u03c4r \u02d9v \u02d9vdt \u00b7 (19.12) for the relation between the rates, where the minus sign indicates that the energy is removed from the system. We can integrate the right hand side by parts to obtain t2 t1 Z F rad \u00b7 vdt = t2 t1 Z m\u03c4r \u00a8v vdt \u00b7 \u2212 m\u03c4r( \u02d9v v) \u00b7 t2 t1 | (19.13) Finally, the motion is \u201cperiodic\u201d and we only want the result over a period; we can therefore pick the end points such that \u02d9v v = 0. Thus we get \u00b7 t2 t1 Z (F rad \u2212 m\u03c4r \u00a8v) \u00b7 vdt = 0. (19.14) One (su\ufb03cient but not necessary) way to ensure that this equation be satis\ufb01ed is to let F rad = m\u03c4r \u00a8v (19.15) This turns Newton\u2019s law (corrected for radiation reaction) into F ext = m \u02d9v = m( \u02d9v \u2212 F rad \u03c4r \u00a8v) \u2212 (19.16) This is called the Abraham\u2013Lorentz equation of motion and the radiation reaction force is called the Abraham\u2013Lorentz force. It can be made relativistic be converting to proper time as usual. Note that this is not necessarily the only way to satisfy the integral constraint above. Another way to satisfy it is to require that the di\ufb00erence be orthogonal to v. Even this is too speci\ufb01c, though. The only thing that is required is that the total integral be zero, and short of decomposing the velocity trajectory in an orthogonal system and perhaps using the calculus of variations, it is not possible to make positive statements about the necessary form of F rad. This \u201csu\ufb03cient\u201d solution is not without problems of its own, problems that seem unlikely to go away if we choose some other \u201csu\ufb03cient\u201d criterion. This is apparent from the observation that they all lead to an equation of motion that is third order in time. Now, it may not seem to you (yet) that that is a disaster, but it is. Suppose that the external force is zero at some instant of time t = 0. Then or \u03c4 \u00a8v \u02d9v \u2248 \u02d9v(t) = a0et/\u03c4 (19.17) (19.18) where a0 is the instantaneous acceleration of the particle at t = 0. \u00b7 Recalling that v \u02d9v = 0 at t1 and t2, we see that this can only be true if a0 = 0 (or we can relax this condition and pick up an additional boundary condition and work much harder to arrive at the same conclusion). Dirac had a simply lovely time with the third order equation. Before attacking it, though, let us obtain a solution that doesn\u2019t have the problems associated with it in a di\ufb00erent (more up-front) way. Let us note that the radiation reaction force in almost all cases will be very small compared to the external force. The external force, in addition, will generally be \u201cslowly varying\u201d, at least on a timescale compared 24 seconds. If we assume that F ext(t) is smooth (continuously to \u03c4r \u2248 F ext di\ufb00erentiable in time), slowly varying, and small enough that F rad \u226a we can use what amounts to perturbation theory to determine F rad and obtain a second order equation of motion. 10\u2212 Under these circumstances, we can assume that F ext \u2248 m \u02d9v, so that: F ext = m( \u02d9v \u2212 m \u02d9v \u2212 \u2248 \u03c4r \u00a8v) dF ext dt \u03c4r (19.19) or m \u02d9v = F ext + \u03c4r = F ext + \u03c4r dF ext dt \u2202 \u2202t ( + (v \u2207) \u00b7 ) F ext (19.20) This latter equation has no runaway solutions or acausal behavior as long as F ext is di\ufb00erentiable in space and time. We will defer the discussion of the covariant, structure free generalization of the Abraham\u2013Lorentz derivation until later.", " This is because it involves the use of the \ufb01eld stress tensor, as does Dirac\u2019s original paper \u2014 we will discuss them at the same time. What are these runaway solutions of the \ufb01rst (Abraham-Lorentz) equation of motion? Could they return to plague us when the force is not small and turns on quickly? Let\u2019s see... 19.3 Integrodi\ufb00erential Equations of Motion We seek solutions to the third order AL equation of motion that evolve into the \u201cnatural\u201d ones when the driving force is turned o\ufb00. In other words, radiation reaction must, by hypothesis, only damp the system and not drive it. Clearly even this requirement makes no sense when time reversal symmetry is considered. Once we fall into the trap of choosing retarded interaction only, we are sunk and anything we do to \ufb01x it will be a band\u2013aid. Let us introduce an \u201cintegrating factor\u201d into the equations of motion. If we assume (quite generally) that \u02d9v(t) = et/\u03c4ru(t) (19.21) where u(t) is to be determined, then the equations of motion simplify to m \u02d9u = 1 \u03c4r \u2212 e\u2212 t/\u03c4 F (t). (19.22) We can formally integrate this second equation, obtaining m \u02d9v(t) = C et/\u03c4r \u03c4r Z t e\u2212 t\u2032/\u03c4rF (t\u2032)dt\u2032 (19.23) The constant of integration is determined by our requirement that no runaway solutions exist! Note well that it is a constraint that lives in the future of the particle. In order to use this to \ufb01nd v(t), we must know the force F (t) for some time (of order \u03c4r) in the future! After this, the integrand is \u201ccut o\ufb00\u201d by the decaying exponential. This suggests that we can extend the integral to C = without di\ufb030, we recover Newton\u2019s law, as we should. To see \u221e culty. In the limit \u03c4r \u2192 this, let so that m \u02d9v(t) = s = 1 \u03c4r (t\u2032 t) \u2212 \u221e e\u2212 sF (t + \u03c4rs)ds. 0 Z The force is assumed to be slowly varying with respect to \u03c4 (or none of this makes sense, just as was the case above) so that a Taylor series expansion converges: F (t + \u03c4 s) = \u221e (\u03c4rs)2 n! a dnF (t) dtn Xn=0 which, upon substitution and integration over s, yields (19.24) (19.25) (19.26) (19.27) m \u02d9v = \u03c4 n r dnF dtn. \u221e Xn=0 \u2192 In the limit \u03c4 0 only the lowest order term survives. This is Newton\u2019s law without radiation reaction. The higher order terms are successive radiative corrections and matter only to the extent that the force varies in time. Note that this force obeys a \u201cLenz\u2019s Law\u201d sort of behavior; when the applied force is changed (say, increased) there is an additional \u201cforce\u201d in the direction of the change that acts on the particle. A particle moving in a circle has a force that changes direction but not magnitude. This change is (think about it) tangent to the motion and in the opposite direction. It acts to slow the charged particle down. Hmmmmmm. There are two extremely annoying aspects to this otherwise noble solution. First, as we have repeatedly noted, it requires a knowledge of F (t) in the future of the particle to obtain its acceleration now. Truthfully, this isn\u2019t really a problem \u2013 obviously this is absolutely equivalent to saying that F (t) can be expanded in a Taylor series (is an analytic function). Second, (and even worse) it responds to a force that is completely in its future with Figure 19.1: F (t), \u02d9v(t) and v(t) on a timescale of \u03c4r. Note that the particle \u201cpreaccelerates\u201d before \u201cthe force gets there\u201d, whatever that means. an acceleration now. It \u201cknows\u201d that a force is going to act on it before that force gets there. Mind you, not long before the force gets there. About 10\u2212 24 seconds before (for reasonable forces). Classically this is very bad, but quantum theory fuzzes physics over a much larger time scale. This is viewed by many physicists as an excuse for not working out a consistently causal classical theory. You can make up your own mind about that, but note well that even if the integro", "di\ufb00erential equation had involved past values of the force you should have been equally bothered \u2013 either one makes Newton\u2019s law nonlocal in time! Note well that we\u2019ve already seen (nonlocal) integrodi\ufb00erential equations in time in a somewhat similar context! Remember our derivation of of dispersion relations, in particular Kramers-Kronig? We had a kernel there that e\ufb00ectively sampled times in the future or past of a system\u2019s motion. This worked because we could integrate over frequencies with a constraint of analyticity \u2013 our \ufb01elds were presumed fourier decomposable. Fourier transforms are, of course, in\ufb01nitely continuously di\ufb00erentiable as long as we avoid sharp changes like (pure) heaviside function forces or \ufb01eld changes, and yes, they explicity provide a knowledge of the quantities in the future and past of their current values. I personally think that this is yet another aspect of the mistake made by requiring that our description of electrodynamics always proceed from the past into the future with a retarded interaction. As we have seen, this is silly \u2013 one could equally well use only advanced interactions or a mix of the two and the solutions obtained for a given boundary value problem will be identical, where the \u201cboundary\u201d is now a four-volume and hence requires future conditions to be speci\ufb01ed as well as the past conditions on a spatial three-surface bounding the four-volume. 19.4 Radiation Damping of an Oscillating Charge The most important application of the Abraham\u2013Lorentz force law is the radiation reaction of bound electrons in atoms as they radiate. This is the problem originally studied by Lorentz, in the context of a classical oscillator, and yes, we are returning to our discussion of dispersion but now with a physical model for why we expect there to be a damping term instead of a strictly phenomenological one. To simplify life, we consider a Lorentz \u201catom\u201d to be an electron on a spring with constant k = m\u03c92 0; a one\u2013dimensional classical oscillator with a resonant frequency \u03c90. If the oscillator is displaced from equilibrium, it radiates energy away and is simultaneously damped. This is a classical analogue of the emission of a photon by a quantum atom, which is accompanied by the atom entering a lower energy level. The equation of motion for the electron is (from the AL force law above, integrated as described for o\ufb00set times): \u00a8x(t) + \u03c92 0 0 Z \u221e e\u2212 sx(t + \u03c4 s)ds = 0 (19.28) where we have used Hooke\u2019s law. If we try the usual song and dance (assume that x(t) = x0e\u2212 \u03b1t we get the characteristic equation \u03b1t x0e\u2212 \u03b12 + \u03c92 0 0 Z \u221e e\u2212 (1+\u03b1\u03c4 )sds = 0. (19.29) In order for the integral to exist, it must damp at in\ufb01nity, so Re(1+\u03b1\u03c4 ) > 0. In that case, we get: \u03b12 + \u03c92 0 \u221e e\u2212 xdx = 0 \u2212 0 (1 + \u03b1\u03c4 ) Z \u03b12 + = 0 \u03c92 0 (1 + \u03b1\u03c4 ) \u03b12(1 + \u03b1\u03c4 ) + \u03c92 0 = 0 \u03c4 \u03b13 + \u03b12 + \u03c92 0 = 0 (\u03c4 \u03b1)3 + (\u03c4 \u03b1)2 + (\u03c90\u03c4 )2 = 0 0\u03c4 2 = 0 z3 + z2 + \u03c92 (19.30) where we\u2019ve de\ufb01ned z = \u03b1\u03c4. This is the same cubic that would arise directly from the original AL equation of motion but the restriction on the integral eliminates the \u201crun0\u03c4 2)/\u03c4 ) at the expense of introducing preacaway\u201d solutions (\u03b1 = celerated ones. There is no point in giving the physical roots in closed form here, but you should feel free to crank up e.g. Mathematica and take a look. (1 + \u03c92 \u2212 If \u03c90\u03c4 << 1 (which is the physically relevant range), then the \ufb01rst order result is whith and \u03b1 = \u0393 2 \u00b1 i(\u03c90 + \u2206\u03c9) \u0393 = \u03c92 0\u03c4 \u2206\u03c9 = 5 8 \u2212 0\u03c4 2. \u03c93 (19.31) (19.32) (19.33) The constant \u0393 is the decay constant and the \u2206\u03c9 is the level shift. Note that the radiative force both introduces damping and shifts the frequency, just like it does for", " a classical damped oscillator. If we evaluate the electric \ufb01eld radiated by such an oscillator (neglecting the transient signal at the beginning) we \ufb01nd that the energy radiated as a function of frequency is dI(\u03c9) d\u03c9 = I0 \u0393 2\u03c0 (\u03c9 1 \u2212 which is the characteristic spectrum of a broadened, shifted resonant line. \u2206\u03c9)2 + (\u0393/2)2 \u03c90 \u2212 (19.34) Figure 19.2: A typical broadened and shifted resonant line due to radiation reaction. This concludes our discussion of the consequences of radiation reaction. You will note that the derivations we have seen are not particularly satisfying or consistent. Now we will examine the \u201cbest\u201d of the derivations (Dirac\u2019s and Wheeler and Feynman\u2019s) and try to make some sense of it all. The following sections are alas still incomplete but will be added shortly. 19.5 Dirac\u2019s Derivation of Radiation Reac- tion 19.6 Wheeler and Feynman\u2019s Derivation of Radiation Reaction 19.7 My Own Field-Free Derivation of Radi- ation Reaction\u03b1), (4.45) where we have used the Maxwell equation \u2202\u03b1F \u03bd\u03b1 = 0. The latter form shows that the di\ufb00erence between the two tensors has vanishing divergence, \u2202\u03bd\u2202\u03b1(A\u00b5F \u03bd\u03b1) = 0, due to the antisymmetry of the \ufb01eld-strength tensor. We can always add the term (4.45) to the conserved tensor in Eq. (4.44) to form (\u2202\u00b5A\u03b1 \u2212 \u2202\u03b1A\u00b5) F \u03bd\u03b1 \u2212 \u00b5F\u03c1\u03c4 F \u03c1\u03c4 \u03b4\u03bd = F\u00b5\u03b1F \u03bd\u03b1 \u2212 \u00b5F\u03c1\u03c4 F \u03c1\u03c4 \u03b4\u03bd = T\u00b5 \u03bd, 1 \u00b50c 1 4 1 \u00b50c 1 4 1 c (4.46) without spoiling the conservation law. This tensor is now gauge-invariant and symmetric, and is the usual electromagnetic stress-energy tensor. If the kludge above is not to your taste, we note that there is a direct, but more complicated, way of extracting a symmetric, gauge-invariant stress-energy tensor from the action; see Part III General Relativity. 4.5 Symmetry of the stress-energy tensor (non-examinable) The stress-energy tensor is symmetric since g = N/c2 and \u03c3ij = \u03c3ji. We demonstrated above that this was true for electromagnetism but it is true generally. In this section we shall brie\ufb02y discuss why. Consider the energy and momentum conservation laws for an isolated system: \u2202 \u2202t \u2202gi \u2202t + \u2207 \u00b7 N = 0 + \u2202\u03c3ij \u2202xj = 0. 25 (4.47) (4.48) The usual 3D angular momentum of the system is L \u2261 x \u00d7 g d3x, (4.49) where the integral extends over all space. Taking the time derivative gives = ijk dLi dt = \u2212ijk xj d3x \u2202gk \u2202t \u2202\u03c3kl \u2202xl xj d3x \u2202 \u2202xl = \u2212ijk = \u2212ijk (xj\u03c3kl) d3x + ijk \u03c3kj d3x xj\u03c3kl dSl + ijk \u03c3kj d3x. (4.50) The \ufb01rst term on the right-hand side in the last line is the rate at which angular momentum is entering the (isolated) system from in\ufb01nity and and must vanish for a localised \ufb01eld con\ufb01guration. We then see that angular momentum of the isolated system is conserved provided that \u03c3ij is symmetric so that the last term on the right of Eq. (4.50) vanishes. Now consider a Lorentz transformation. The stress-energy tensor transforms as T \u00b5\u03bd = \u039b\u00b5 \u03b1\u039b\u03bd \u03b2T \u03b1\u03b2. (4.51) Considering the 1-2 component, we have \u03b1\u039b2 0\u039b2 \u03c3 \u03b2T \u03b1\u03b2 xy = \u039b1 = \u039b1 2T 02 + \u039b1 = \u2212\u03b3\u03b2T 02 + \u03b3T 12. 1\u039b2 2T 12 (standard boost) Repeating for the 2-1 component, we have yx = \u2212\u03b3\u03b2T 20 + \u03b3T 21. \u03c3 (4.52) (4.53) Symm", "etry of \u03c3ij in all inertial frames therefore requires T i0 = T 0i and so g = N/c2. A further consequence of the symmetry of the stress-energy tensor follows from considering the dynamics of the \u201ccentre of energy\u201d of the system, x d3x/ d3x. Since the total energy is conserved, the dynamics of the centre of energy follow from those 26 of x d3x; we have d dt xi d3x = xi d3x \u2202 \u2202t \u2202Nj \u2202xj \u2202 \u2202xj = \u2212 = \u2212 xi d3x (xiNj) d3x + Ni d3x = \u2212 xiNj dSj + Ni d3x. (4.54) If The \ufb01rst term on the right-hand side vanishes for a localised \ufb01eld distribution. N = gc2, then N d3x is constant by the conservation law for g [Eq. (4.48)], and we see from Eq. (4.54) that the centre of energy moves uniformly. In particular, in the zero momentum frame where g d3x = 0, the centre of energy remains at rest. The above argument shows that two theorems from mechanics for isolated systems \u2013 that the centre of energy moves uniformly and that angular momentum is conserved \u2013 are related in relativity. Indeed, one can introduce the object S\u00b5\u03b1\u03b2 \u2261 x\u03b1T \u00b5\u03b2 \u2212 x\u03b2T \u00b5\u03b1 = \u2212S\u00b5\u03b2\u03b1, (4.55) which transforms as a tensor under homogeneous Lorentz transformations. Then, for an isolated system, conservation of the stress-energy tensor implies that \u2202\u00b5S\u00b5\u03b1\u03b2 = \u03b4\u03b1 \u00b5 T \u00b5\u03b2 + x\u03b1 \u2202\u00b5T \u00b5\u03b2 =0 = T \u03b1\u03b2 \u2212 T \u03b2\u03b1 = 0. \u2212\u03b4\u03b2 \u00b5T \u00b5\u03b1 \u2212 x\u03b2 \u2202\u00b5T \u00b5\u03b1 =0 (4.56) Integrating this conservation law over space, the 0i component returns the centre-ofenergy theorem and the ij component returns 3D angular momentum conservation. 4.6 Stress-energy tensor of a plane electromagnetic wave As an example of computing the stress-energy tensor, we consider electromagnetic plane waves. Maxwell\u2019s equations admit wave-like solutions in free space: E = E0f (k \u00b7 x \u2212 \u03c9t) B = B0f (k \u00b7 x \u2212 \u03c9t) with \u03c9 = c|k|. (4.57) If the function f is sinusoidal, with period 2\u03c0, these are harmonic waves propagating at speed c with wavelength 2\u03c0/k (with k = |k|) and frequency \u03c9/(2\u03c0). At any time, the 27 phase of the wave, k \u00b7 x \u2212 \u03c9t, is constant on 3D planes perpendicular to the wavevector k. Since \u2207 \u00b7 E = 0 in free space, and \u2207 \u00b7 B = 0, we must have k \u00b7 E0 = 0 and k \u00b7 B0 = 0. Furthermore, from \u2207 \u00d7 E = \u2212\u2202B/\u2202t we have k \u00d7 E0 = \u03c9B0 \u21d2 \u02c6k \u00d7 E0 = cB0, (4.58) (4.59) where \u02c6k is a unit vector in the direction of k, and so c|B0| = |E0|. It follows that E0, B0 and k form a right-handed triad of vectors. We can write the \ufb01elds in covariant form as follows. Phase di\ufb00erences between two spacetime events must be Lorentz invariant so we can introduce the 4-wavector k\u00b5 = (\u03c9/c, k). (4.60) This is a null 4-vector, \u03b7\u00b5\u03bdk\u00b5k\u03bd = 0, since \u03c9 = ck. The phase can then be written covariantly as k \u00b7 x \u2212 \u03c9t = k\u00b5x\u00b5. Specialising to k along the x-direction of some inertial frame, and E0 along y (so that B0 is along z), the components of the \ufb01eld-strength tensor in that frame are F \u00b5\u03bd = B0f (k\u00b5x\u00b51 \u22124.61) where B0 = |B0|. Both invariants F \u00b5\u03bdF\u00b5\u03bd and F \u00b5\u03bdF\u00b5\u03bd vanish since E \u00b7 B = 0 and |E| = c|B|. The stress-energy tensor, Eq. (4.23), reduces to T \u00b5\u03bd = F \u00b5\u03b1F \u03bd It follows that \u03b1", "/\u00b50 since F \u00b5\u03bdF\u00b5\u03bd = 0. T \u00b5\u03bd = B2 0f 2(k\u00b5x\u00b5) \u00b50 = B2 0f 2(k\u00b5x\u00b5) \u00b50 \uf8eb \uf8ec \uf8ec \uf8ed \uf8eb \uf8ec \uf8ec \uf8ed 0 0 0 1 0 0 1 0 \u22121 \u2212 \uf8f6 \uf8f7 \uf8f7 \uf8f8 We can read o\ufb00 the energy density = B2 0f 2(k\u00b5x\u00b5)/\u00b50, 28 (4.62) (4.63) which receives equal contributions from the electric and magnetic \ufb01elds. The Poynting vector is along the x-direction, with Finally, the momentum \ufb02ux has only one non-zero component: Nx = cB2 0f 2(k\u00b5x\u00b5)/\u00b50. Note that \u03c3xx = B2 0f 2(k\u00b5x\u00b5)/\u00b50. \u03c3xx = Nx/c = cgx =. (4.64) (4.65) (4.66) These results are consistent with the photon picture in which photons with energy E = pc (where p is the magnitude of their momentum) propagate at speed c along the x-direction. The energy density is then c times the momentum density, and the momentum \ufb02ux is c times the momentum density3. Exercise: Consider re\ufb02ection of a plane electromagnetic wave, propagating along the xdirection, from a perfect conductor whose plane surface is at x = 0. The conductivity is so high that the \ufb01elds decay very rapidly inside the conductor and we can therefore approximate the \ufb01elds as being zero inside. The E \ufb01eld is tangent to the surface and, since the tangential component is continuous, the E \ufb01eld must be zero at the surface of the conductor. The B \ufb01eld is also tangential but is now discontinuous at the surface with the discontinuity driven by a surface current \ufb02owing in a thin layer on the surface. The total electric and magnetic \ufb01elds (sum of incident and re\ufb02ected waves) outside the conductor (x < 0) are therefore standing waves of the form Etot = B0c [f (kx \u2212 \u03c9t) \u2212 f (\u2212kx \u2212 \u03c9t)] \u02c6y Btot = B0 [f (kx \u2212 \u03c9t) + f (\u2212kx \u2212 \u03c9t)] \u02c6z. Show that the energy density is = B2 0 \u00b50 f 2(kx \u2212 \u03c9t) + f 2(\u2212kx \u2212 \u03c9t), (4.67) (4.68) which is the sum of the energy density in the incident and re\ufb02ected waves. For harmonic waves, the time-averaged energy density is the same at all x. Show also that the Poynting vector has a single non-zero component Nx = B2 0c \u00b50 f 2(kx \u2212 \u03c9t) \u2212 f 2(\u2212kx \u2212 \u03c9t), (4.69) which, again, is the sum of the incident and re\ufb02ected energy \ufb02uxes. The Poynting vector vanishes at the surface of the conductor at all times, so no energy enters the conductor. This 3The momentum carried through an area dS in the y-z plane in time dt is the total momentum of those photons within a volume cdtdS, i.e., \u03c3xxdSdt = gxcdtdS. 29 is still consistent with there being a surface current since the rate of energy dissipation by this current scales as the inverse of the conductivity and so vanishes in the limit of in\ufb01nite conductivity assumed here. At other locations, the time-averaged Poynting vector vanishes for harmonic waves. Finally, show that the Maxwell stress-tensor is diagonal with f 2(kx \u2212 \u03c9t) + f 2(\u2212kx \u2212 \u03c9t) \u03c3xx = \u03c3yy = B2 0 \u00b50 2B2 0 \u00b50 f (kx \u2212 \u03c9t)f (\u2212kx \u2212 \u03c9t), (4.70) and \u03c3zz = \u2212\u03c3yy. The \u03c3xx = is the sum of the incident and re\ufb02ected momentum \ufb02uxes. However, the \u03c3yy and \u03c3zz components arise purely from interference of the incident and re\ufb02ected waves. Since \u03c3yy and \u03c3zz only vary in the x-direction, the non", "-zero values are still consistent with the momentum density of the \ufb01eld and the conservation law (4.48). There is a force exerted on the conductor (\u201cradiation pressure\u201d) in the x-direction given per area by (4.71) \u03c3xx|x=0 = f 2(\u2212\u03c9t). 2B2 0 \u00b50 This follows from the integral form of the momentum conservation law, Eq. (4.14), noting the momentum density is zero inside the conductor. The incident momentum density at x = 0 is B2 0f 2(\u2212\u03c9t)/(\u00b50c), and so the radiation pressure is 2c times the incident momentum density. What is the physical origin of this force? It is the Lorentz force of the B \ufb01eld at the surface on the surface current distribution. To see that this is consistent, note that the surface current Js (de\ufb01ned such that the current through a given line element dl lying in the surface and perpendicular to Js is |Js|dl) needed to maintain the discontinuity in B is along the x-direction with \u00b50Js = By|x=0. The force per area on this current is JsBy|x=0/2 along the x-direction. The factor of 1/2 here accounts for the average value of the magnetic \ufb01eld across the depth of the current distribution. Using By|x=0 = 2B0(\u2212\u03c9t), we see that the Lorentz force per area is exactly \u03c3xx at the surface of the conductor, as required. 4.6.1 Radiation pressure of a photon \u201cgas\u201d We can obtain an isotropic distribution of radiation by forming a random superposition of incoherent plane waves propagating in all directions with equal intensity. (Incoherent waves have no lasting phase relation amongst themselves, e.g., the radiation emitted by two separated thermal sources.) Such radiation is the classical analogue of a photon Isotropy requires the (time-averaged) Poynting vector to vanish and that the gas. Maxwell stress tensor \u03c3ij = p\u03b4ij, where p is the pressure. The averaged stress-energy tensor is then diagonal with T \u00b5\u03bd = diag(, p, p, p). (4.72) 30 The electromagnetic stress energy tensor is necessarily trace-free3. (4.73) We see that for a photon gas the pressure is one-third of the energy density irrespective of the way that the energy is distributed over frequency (i.e., the radiation does not have to have a blackbody spectrum). 31 Part II Radiation of electromagnetic waves In this part we consider the production of electromagnetic radiation from time-dependent charge distributions. We \ufb01rst consider radiation from charge and current densities involving non-relativistic motions, as, for example, in radiation from a radio antenna. As we shall see, it is the acceleration of the charges that determines the radiation \ufb01eld. As an application of these ideas, we then brie\ufb02y consider scattering of electromagnetic radiation. To describe radiation from charges moving relativistically requires an alternative treatment, which we subsequently develop. This theory underlies extreme phenomena such as the synchrotron radiation emitted from charges circling relativistically in magnetic \ufb01elds. 5 Retarded potentials of a time-dependent charge distribution We begin by deriving the 4-potential generated causally by an arbitrary 4-current J \u00b5. To do this, we combine F\u00b5\u03bd = \u2202\u00b5A\u03bd \u2212 \u2202\u03bdA\u00b5 with the Maxwell equation to obtain where the operator \u2202\u00b5F \u00b5\u03bd = \u2212\u00b50J \u03bd, A\u00b5 \u2212 \u2202\u00b5(\u2202\u03bdA\u03bd) = \u2212\u00b50J \u00b5, \u2261 \u03b7\u00b5\u03bd\u2202\u00b5\u2202\u03bd = \u22072 \u2212 1 c2 \u22022 \u2202t2 (5.1) (5.2) (5.3) is the wave operator. It is very convenient to make a gauge choice (the Lorenz gauge) such that \u2202\u00b5A\u00b5 = 0. This is clearly a Lorentz-invariant statement, which has the bene\ufb01t of retaining covariance of the subsequent gauge-\ufb01xed equations under Lorentz transformations. It is always possible to make this gauge choice: starting in some general gauge we make a transformation A\u00b5 \u2192 A\u00b5 + \u2202\u00b5\u03c7 with \u03c7 chosen to satisfy \u03c7 = \u2212\u2202\u00b5A\u00b5. As we show below, we can always solve this equation to determine a suitable \u03c7. In the Lorenz gauge, Eq. (5.2) simpli\ufb01es", " to A\u00b5 = \u2212\u00b50J \u00b5 or \u22072 \u2212 1 c2 \u22022 \u2202t2 A\u00b5 = \u2212\u00b50J \u00b5, (5.4) which is a sourced wave equation. 32 We solve Eq. (5.4) by taking a Fourier transform in time, writing A\u00b5(t, x) = 1 \u221a 2\u03c0 \u221e \u2212\u221e \u02dcA\u00b5(\u03c9, x)ei\u03c9t d\u03c9, and similarly for J \u00b5. The sourced wave equation then becomes \u22072 + \u03c92 c2 \u02dcA\u00b5 = \u2212\u00b50 \u02dcJ \u00b5. (5.5) (5.6) This is a sourced Helmholtz equation with k2 = \u03c92/c2. As for Poisson\u2019s equation in electrostatics, we can solve this with a Green\u2019s function G(x; x) such that \u22072 + k2 G(x; x) = \u03b4(3)(x \u2212 x), so that \u02dcA\u00b5(\u03c9, x) = \u2212\u00b50 G(x; x) \u02dcJ \u00b5(\u03c9, x) d3x. (5.7) (5.8) To describe the \ufb01elds produced by localised sources, we require the Green\u2019s function to tend to zero as |x| \u2192 0. Homogeneity and isotropy require that G(x; x) is a function of |x \u2212 x| alone. Taking x = 0 and r = |x|, away from r = 0 we have 1 r2 \u2202 \u2202r r2 \u2202G \u2202r + k2G = 0. (5.9) The solution that tends to zero as r \u2192 \u221e is G = \u2212Ae\u2212ikr/r. (The reason for choosing e\u2212ikr rather than eikr is to do with causality and will be explained shortly.) We \ufb01x the normalisation A by integrating Eq. (5.7) over a sphere of radius centred on x, and taking the limit as \u2192 0. The integral of the k2G term goes like 3/ and so vanishes in the limit, leaving \u21d2 lim r\u21920 4\u03c0r2A so that A = 1/(4\u03c0) and lim r\u21920 e\u2212ikr 4\u03c0r2\u2202G/\u2202r = 1 r2 + ik e\u2212ikr r = 1, G(x; x) = \u2212 1 4\u03c0 e\u2212i\u03c9|x\u2212x|/c |x \u2212 x|. (5.10) (5.11) Inserting this Green\u2019s function into Eq. (5.8) and taking the inverse Fourier transform, gives A\u00b5(t, x) = \u00b50 4\u03c0 \u221e \u2212\u221e d\u03c9 \u221a 2\u03c0 d3x ei\u03c9(t\u2212|x\u2212x|/c) |x \u2212 x| \u02dcJ \u00b5(\u03c9, x). (5.12) 33 The integral over \u03c9 is the inverse Fourier transform of \u02dcJ \u00b5(\u03c9, x) and returns J \u00b5(tret, x) where the retarded time tret \u2261 t\u2212|x\u2212x|/c. This gives our \ufb01nal result for the 4-potential from an isolated time-dependent charge distribution: A\u00b5(t, x) = \u00b50 4\u03c0 J \u00b5(tret, x) |x \u2212 x| d3x. (5.13) Equation (5.13) has a simple physical interpretation. In components, it becomes \u03c6(t, x) = 1 4\u03c00 \u03c1(tret, x) |x \u2212 x| d3x, J(t, x) = \u00b50 4\u03c0 J(tret, x) |x \u2212 x| d3x. (5.14) If we were in a static situation, \u03c1(tret, x) = \u03c1(t, x), and similarly for the current, and these equations would become the usual potentials for electrostatics and magnetostatics (in the Coulomb gauge, \u2207 \u00b7 A = 0). The only di\ufb00erence for time-dependent charge distributions is that the sources have to be evaluated at the retarded time. Since the event at tret and x is on the past lightcone of the event at t and x, the \ufb01eld at any event x\u00b5 is generated by the superposition of the \ufb01elds due to currents at earlier times on the past lightcone of x\u00b5. This re\ufb02ects causality in that electromagnetic disturbances propagate at the speed of light and that only earlier", " events can in\ufb02uence the \ufb01eld at any spacetime point4. The potential A\u00b5 of Eq. (5.13) is called the retarded potential. We now see why we had to choose the e\u2212ikr/r form for the Green\u2019s function \u2013 had we chosen eikr/r we would have obtained the advanced potential for which the currents at t + |x \u2212 x|/c and x in\ufb02uence the \ufb01eld at t and x, i.e., acausal propagation. It is not obvious that Eq. (5.13) is covariant under Lorentz transformations, since we are integrating over space alone. To see that it is covariant, we rewrite it as a spacetime integral as follows: A\u00b5(x) = \u00b50 4\u03c0 J \u00b5(y) \u03b4(y0 \u2212 x0 + |x \u2212 y|) |x \u2212 y| d4y. (5.15) This looks more covariant since it involves the Lorentz-invariant measure d4y. However, we still have the issue of the delta-function. To see that this is Lorentz invariant, consider \u03b4 (\u03b7\u00b5\u03bd(x\u00b5 \u2212 y\u00b5)(x\u03bd \u2212 y\u03bd)) \u0398(x0 \u2212 y0), (5.16) where \u0398 is the Heaviside (unit-step) function, which equals unity if its argument is greater than zero and vanishes otherwise. The product of the two terms in Eq. (5.16) is therefore Lorentz invariant since the delta-function is non-zero only for null-separated x\u00b5 and y\u00b5, and the temporal ordering of such events is Lorentz invariant. We now show 4Recall that the temporal ordering of timelike and null-separated events is Lorentz invariant 34 that Eq. (5.16) is twice the term \u03b4(y0 \u2212 x0 + |x \u2212 y|)/|x \u2212 y| in question in Eq. (5.15). We do this by regarding the argument of the delta-function as a function of y0, noting that \u03b7\u00b5\u03bd(x\u00b5 \u2212 y\u00b5)(x\u03bd \u2212 y\u03bd) = |x \u2212 y|2 \u2212 (x0 \u2212 y0)2 = (|x \u2212 y| + x0 \u2212 y0)(|x \u2212 y| \u2212 x0 + y0), (5.17) and use the result in Eq. (3.55) for delta-functions with functions as their arguments. Summing over both roots of the argument, this gives \u03b4 (\u03b7\u00b5\u03bd(x\u00b5 \u2212 y\u00b5)(x\u03bd \u2212 y\u03bd)) = = \u03b4(y0 \u2212 x0 + |x \u2212 y|) 2|y0 \u2212 x0| \u03b4(y0 \u2212 x0 + |x \u2212 y|) 2|x \u2212 y| + + \u03b4(y0 \u2212 x0 \u2212 |x \u2212 y|) 2|y0 \u2212 x0| \u03b4(y0 \u2212 x0 \u2212 |x \u2212 y|) 2|x \u2212 y|. (5.18) The Heaviside function in Eq. (5.16) selects only the \ufb01rst term on the right, showing that \u03b4 (\u03b7\u00b5\u03bd(x\u00b5 \u2212 y\u00b5)(x\u03bd \u2212 y\u03bd)) \u0398(x0 \u2212 y0) = \u03b4(y0 \u2212 x0 + |x \u2212 y|) 2|x \u2212 y|, (5.19) as required. Equation (5.15) therefore involves a Lorentz-invariant integration of the 4-vector current over the past lightcone of the event x\u00b5, and so is properly covariant. Exercise: Noting that the gauge condition \u2202\u00b5A\u00b5 = 0 takes the form i\u03c9 \u02dcA0/c + \u2207 \u00b7 \u02dcA = 0 after Fourier transforming in time, verify that the right-hand side of Eq. (5.8) satis\ufb01es the gauge condition by virtue of charge conservation. 6 Dipole radiation As a \ufb01rst application of the retarded potential derived above, we consider radiation from some charge distribution in which all motions are non-relativistic (in some inertial frame). We shall be interested in the \ufb01elds at large distances compared to the spatial extent of the charge distribution. Speci\ufb01cally, consider the case where all charges are localised within some region of size a, and place the origin of spatial coordinates within this volume. To calculate the 4", "-potential at x, where r = |x| a, we Taylor expand |x \u2212 x| in Eq. (5.13) as |x \u2212 x \u2212 \u02c6x \u00b7 x + \u00b7 \u00b7 \u00b7, + \u00b7 \u00b7 \u00b7 35 (6.1) where \u02c6x is a unit vector in the direction of x. Note that the second term on the right is smaller than the \ufb01rst by O(a/r). Substituting into Eq. (5.13), we have A\u00b5(t, x) = \u00b50 4\u03c0r J \u00b5(tret, x) 1 + \u02c6x \u00b7 x r + \u00b7 \u00b7 \u00b7 d3x, (6.2) where tret = t\u2212r/c+\u02c6x\u00b7x/c+\u00b7 \u00b7 \u00b7. At r a, we can safely neglect the O(a/r) corrections in the round brackets in Eq. (6.2). However, the treatment of the dependence of tret on x depends on the rate at which the source varies. For an oscillating charge distribution, with angular frequency \u03c9, if a/c 1/\u03c9 then the variation in tret over the source is short compared to the period and can be handled perturbatively via a Taylor expansion of J \u00b5 in time. Note that the typical velocity of the charges is a\u03c9 and so we require this to be much less than c for an expansion to be valid. Equivalently, since the \ufb01elds radiated by a source oscillating at \u03c9 oscillate at the same frequency, the associated wavelength \u03bb must be large compared to a. For su\ufb03ciently slowly-varying sources, we can neglect the variation of tret across the source. The leading-order contribution to the magnetic vector potential is then A(t, x) = J(t \u2212 r/c, x) d3x. \u00b50 4\u03c0r We can express the right-hand side in terms of the time derivative of the electric dipole moment p of the charge distribution,5 where, recall from IB Electromagnetism, p(t) \u2261 \u03c1(t, x)x d3x. (6.3) (6.4) Charge conservation, \u2202\u03c1/\u2202t + \u2207 \u00b7 J = 0, implies that \u02d9p = = d dt \u2202\u03c1 \u2202t \u03c1(t, x)x d3x x d3x = \u2212 (\u2207 \u00b7 J)x d3x, so that \u02d9pi = \u2212 = = \u2202Jj \u2202xj \u2202xi \u2202xj Jj xi d3x d3x Ji d3x, (6.5) (6.6) 5We are using the same symbol for the electric dipole moment and the 3-momentum. This should not cause any confusion, with the meaning being clear from the context. 36 where the second equality follows from the generalised divergence theorem and noting that the surface term vanishes for a localised current distribution. We therefore have \u00b50 4\u03c0r \u02d9p(t \u2212 r/c). A(t, x) \u2248 (6.7) This is called the electric dipole approximation. For a source oscillating at frequency \u03c9, we have A \u223c e\u2212i\u03c9(t\u2212r/c)/r, i.e., outgoing spherical waves. We can calculate the magnetic \ufb01eld from B = \u2207 \u00d7 A. We have to be careful to remember that t \u2212 r/c depends on x, so that for some function f (t \u2212 r/c), \u2202f (t \u2212 r/c) \u2202xi = \u2212 = \u2212 1 c xi rc f (t \u2212 r/c) \u2202r \u2202xi f (t \u2212 r/c). It follows that B(t, x) = \u2212 = \u2212 \u00b50 4\u03c0r2 (\u2207r) \u00d7 \u02d9p(t \u2212 r/c) \u2212 \u00b50 \u00b50 \u02c6x \u00d7 \u02d9p(t \u2212 r/c) \u2212 4\u03c0r2 4\u03c0rc (\u2207r) \u00d7 \u00a8P(t \u2212 r/c) \u00b50 4\u03c0rc \u02c6x \u00d7 \u00a8p(t \u2212 r/c). (6.8) (6.9) Note that the \ufb01rst contribution to the \ufb01eld varies as 1/r2 while the second goes like 1/r. For a source oscillating at \u03c9, the relative size of the 1/r part compared to the 1/r2 part is O(r/\u03bb). We de\ufb01ne the far-\ufb01eld (sometimes known as the radiation", " zone or the wave zone) as the asymptotic region r \u03bb. In the far-\ufb01eld, the 1/r term dominates the magnetic \ufb01eld, and describes outgoing spherical waves of radiation: B(t, x) = \u2212 \u00b50 4\u03c0rc \u02c6x \u00d7 \u00a8p(t \u2212 r/c) (r \u03bb). (6.10) Note that the B \ufb01eld is perpendicular to \u02c6x and to \u00a8p, and that it is \u00a8p that determines the radiation \ufb01eld. For a set of charges {qi} located at {xi}, we have p = i qixi and so \u00a8p = qi\u00a8xi. i (6.11) This shows that it is acceleration of the charges that generates the dipole radiation \ufb01eld. We have to work a little harder for the electric potential since, with the same order of approximation as we used above for A, the electric potential \u03c6 involves \u03c1(t \u2212 r/c, x) d3x, i.e., the total charge Q at time t \u2212 r/c. Since Q is independent of time for an isolated charge distribution, the potential is also time-independent to this level of approximation (it is just the usual asymptotic Coulomb \ufb01eld). We therefore have 37 to work at higher order in a/\u03bb to \ufb01nd the part of \u03c6 that describes radiation6. It is straightforward to do so, but here we shall follow the simpler approach of using the Lorenz gauge condition to \ufb01nd \u03c6. If we use Eq. (6.7) in \u2207 \u00b7 A + c\u22122\u2202\u03c6/\u2202t = 0, we \ufb01nd \u2202\u03c6 \u2202t = \u00b50c2 4\u03c0r2 \u02c6x \u00b7 \u02d9p(t \u2212 r/c) + \u00b50 4\u03c0rc \u02c6x \u00b7 \u00a8p(t \u2212 r/c), (6.12) which integrates to give \u03c6(t, x) = Q 4\u03c00r + 1 4\u03c00r2 \u02c6x \u00b7 p(t \u2212 r/c) + 1 4\u03c00rc \u02c6x \u00b7 \u02d9p(t \u2212 r/c). (6.13) Here, we have included the Coulomb term from the constant charge Q, which appears as an integration constant. The second term is the usual dipole potential that would be generated by a static charge distribution, but here it depends on the electric dipole moment at the earlier time t \u2212 r/c. Typically, this term is smaller than the Coulomb term by O(a/r). It is the third term that describes the radiation component of the It varies as 1/r and is typically O(r/\u03bb) times the second term, and so potential. dominates in the far-\ufb01eld. We can form the electric \ufb01eld from E = \u2212\u2202A/\u2202t\u2212\u2207\u03c6. Equations (6.7) and (6.13) give terms varying as 1/r, 1/r2 and 1/r3. The part going as 1/r dominates in the far-\ufb01eld: E(t, x) = \u2212 [\u00a8p(t \u2212 r/c) \u2212 \u02c6x\u02c6x \u00b7 \u00a8p(t \u2212 r/c)] (r \u03bb) = \u02c6x \u00d7 [\u02c6x \u00d7 \u00a8p(t \u2212 r/c)]. (6.14) \u00b50 4\u03c0r \u00b50 4\u03c0r It originates from the time derivative of \u02d9p in A (Eq. 6.7) and the spatial derivative of \u02d9p(t \u2212 r/c) in \u03c6 (Eq. 6.13). This expression for the electric \ufb01eld in the far-\ufb01eld can also be obtained directly from the magnetic \ufb01eld with \u2207 \u00d7 B = c\u22122\u2202E/\u2202t. The electric \ufb01eld is perpendicular to \u02c6x and lies in the plane formed from \u02c6x and \u00a8p. We see that in the far-\ufb01eld, E = \u2212c\u02c6x \u00d7 B (r \u03bb), (6.15) so that |E| = c|B|, and E \u00b7 B = 0. This behaviour is just like for a plane wave with wavevector along \u02c6x. This makes sense since in the far-\ufb01eld, the radius r is so large compared to the wavelength \u03bb that on the scale of \u03bb the spherical wave", "fronts are e\ufb00ectively planar. 6The 3-current J \u223c \u03c1v is higher-order in v/c \u223c a/\u03bb than \u03c1, which is why we can apparently work to lower order in a/\u03bb when dealing with the vector potential. 38 6.1 Power radiated Energy is carried o\ufb00 to spatial in\ufb01nity from the source in the form of electromagnetic radiation. We can determine the rate at which energy is radiated by integrating the Poynting vector over a spherical surface at very large radius. Working in the far-\ufb01eld, we have E \u00d7 B B \u00d7 (\u02c6x \u00d7 B) |B|2\u02c6x 1 \u00b50 c \u00b50 c \u00b50 N = = = = \u00b50 (4\u03c0r)2c |\u02c6x \u00d7 \u00a8p|2\u02c6x. (6.16) This is radial, as expected. The magnitude of the Poynting vector varies as 1/r2, so the integral over a spherical surface of radius r is independent of r. If \u00a8p is along the z-axis, |\u02c6x \u00d7 \u00a8p| = sin \u03b8|\u00a8p|, where \u03b8 is the angle that \u02c6x makes with the z-axis. In this case, |\u00a8p|2 sin2 \u03b8 \u02c6x. (6.17) \u00b50 (4\u03c0r)2c The total radiated power then follows from N = N \u00b7 dS = = |\u00a8p|2 1 \u22121 sin2 \u03b8 dcos \u03b8 2\u03c0 0 d\u03c6 \u00b50 (4\u03c0)2c \u00b50 6\u03c0c |\u00a8p|2. (6.18) The power radiated per solid angle is |N|r2 and varies as sin2 \u03b8. The radiation is maximal in the plane perpendicular to \u00a8p and there is no radiation along the direction of \u00a8p. The radiation is linearly polarized, with the electric \ufb01eld in the plane containing \u02c6x and \u00a8p. The magnetic \ufb01eld is perpendicular to E. 6.2 Beyond the electric-dipole approximation For many applications involving non-relativistic motions, the electric-dipole approximation is su\ufb03cient. However, there are situations where this leading-order contribution vanishes, for example if p itself vanishes or is constant in time. In such situations it is necessary to expand the source J \u00b5(tret, x) to higher order in a/\u03bb. Retaining the next-order terms, we have A(t, x) \u2248 \u00b50 4\u03c0r \u02d9p(t \u2212 r/c) + \u00b50 4\u03c0rc 39 \u02d9J(t \u2212 r/c, x)\u02c6x \u00b7 x d3x. (6.19) The leading correction to the electric-dipole approximation involves the time derivative of the integral Jixj d3x. We can express this in terms of multipole moments of the charge and current distribution as follows. Since \u2202\u03c1/\u2202t + \u2207 \u00b7 J = 0, we have S (xjxk... xl) \u2207 \u00b7 J + \u2202\u03c1 \u2202t d3x = 0, (6.20) so that d dt \u03c1xjxk \u00b7 \u00b7 \u00b7 xl d3x = \u2212 = (xjxk... xl)\u2207 \u00b7 J d3x J \u00b7 \u2207 (xjxk... xl) d3x = [(Jjxk... xl) + \u00b7 \u00b7 \u00b7 + (xjxk... Jl)] d3x, (6.21) where we used the generalisation of the divergence theorem in going to the second line and dropped the surface term since it vanishes for a localised charge distribution. A special case of this result that is useful for our purposes here is (Jixj + Jjxi) d3x = d dt \u03c1xixj d3x. We can now write Jixj d3x = 1 2 Jixj d3x + d dt 1 2 \u03c1xixj d3x \u2212 Jjxi d3x, so that Ji(t, x)\u02c6x \u00b7 x d3x = 1 2 x \u00d7 J(t, x) d3x \u00d7 \u02c6x 1 6 d dt + i 3\u03c1(t, x)x ix j The \ufb01rst term on the right involves the magnetic dipole moment, m(t) \u2261 1 2 x \u00d7 J(t, x) d3x. (6.22) (6.23) \u02c6xj", ". (6.24) (6.25) The second term on the right is related to the time derivative of the trace-free electricquadrupole moment, Qij(t) \u2261 \u03c1(t, x)(3x ix j \u2212 \u03b4ij|x|2) d3x. (6.26) The trace term does not appear in Eq. (6.24), but it would only add a radial component (i.e., proportional to \u02c6x) to A and so would not contribute to the electric and magnetic \ufb01elds in the far-\ufb01eld (see the stucture of the second, dominant, term in Eq. 6.9). 40 Putting these results together, we can take A(t, x) \u2248 \u00b50 4\u03c0r \u02d9p(t \u2212 r/c) + \u00b50 4\u03c0rc \u02d9m(t \u2212 r/c) \u00d7 \u02c6x + 1 6 \u00b50 4\u03c0rc \u00a8Q(t \u2212 r/c)\u02c6x, (6.27) where Q\u02c6x denotes the contraction of the tensor Qij with \u02c6xj. The magnetic-dipole and electric-quadrupole terms are of the same order, but a/\u03bb down compared to the electric-dipole term. The magnetic \ufb01eld in the far-\ufb01eld follows from B = \u2207 \u00d7 A; keeping only the dominant terms that go as 1/r, we have B(t, x) \u2248 \u00b50 4\u03c0rc \u00a8p(t\u2212r/c)\u00d7 \u02c6x+ \u00b50 4\u03c0rc2 [ \u00a8m(t \u2212 r/c) \u00d7 \u02c6x]\u00d7 \u02c6x+ 1 6 \u00b50 4\u03c0rc2... Q(t \u2212 r/c)\u02c6x\u00d7 \u02c6x. (6.28) The electic \ufb01eld in the far-\ufb01eld is given by E = cB \u00d7 \u02c6x. The Poynting vector in the far-\ufb01eld is given by N = c \u00b50 |B|2\u02c6x, (6.29) as usual. If we integrate over all directions to determine the total radiated power, the cross-terms vanish between the various multipole contributions to the magnetic \ufb01eld leaving a sum of squares from the individual terms (exercise!): N \u00b7 dS = \u00b50 6\u03c0c |\u00a8p|2 + \u00b50 6\u03c0c3 | \u00a8m|2 + \u00b50 720\u03c0c3... Qij... Qij. (6.30) The power from the magnetic dipole and electric quadrupole are generally suppressed by (a/\u03bb)2 \u223c (v/c)2 compared to the electric dipole, unless the latter vanishes for some reason. 7 Scattering As an application of the material developed in the previous section on radiation of electromagnetic waves, we brie\ufb02y consider the issue of scattering. An incident electromagnetic wave will cause a free charge, such as an electron, to oscillate and the acceleration that accompanies the oscillation will cause the charge to radiate. In this manner, the incident radiation is scattered into other directions. A similar thing happens for radiation incident on neutral systems such as atoms or molecules. As we shall discuss further in Part III, there are now bound positive and negative charges that displace from their equilibrium positions in response to the applied \ufb01eld. These displacements generate an oscillating electric dipole moment that radiates, scattering the incident radiation. 41 7.1 Thomson scattering Thomson scattering describes the classical scattering of radiation from a free electron.7 We shall only consider incident \ufb01elds that are weak enough not to induce relativistic motions. In the frame in which the electron is at rest (on average), the equation of motion of the electron in the incident electromagnetic wave is simply me\u00a8x = \u2212eE, (7.1) where me is the mass of the electron, \u2212e is the charge, and E is the incident \ufb01eld evaluated at the location of the electron. For an incident wave at angular frequency \u03c9, we can evaluate the right-hand side of Eq. (7.1) at the average position of the electron provided that the wavelength of the incident wave is large compared to the amplitude of the oscillation induced upon the electron. In this case, the electron oscillates at frequency \u03c9 and with amplitude eE0/(me\u03c92), where E0 is the amplitude of the electric \ufb01eld in the incident wave. This oscillation amplitude will be small compared to the wavelength provided that eE0 mec\u03c9 1. (7.2) However, since the", " maximum speed of the electron is the product of the frequency and amplitude, i.e., eE0/(me\u03c9), Eq. (7.2) is equivalent to the requirement that the electron motion is non-relativistic and so will hold by assumption. The acceleration of the electron is therefore \u2212eE/me corresponding to an oscillating electric-dipole moment with It follows that the time-averaged radiation power is \u00a8p = e2E/me. dE dt = \u00b50 6\u03c0c |\u00a8p|2 = \u00b50 6\u03c0c e4E2 0 2m2 e. The time-averaged magnitude of the Poynting vector of the incident radiation is |N| = E2 0 2\u00b50c, (7.3) (7.4) (7.5) and the ratio of the radiated (scattered) power to this incident \ufb02ux de\ufb01nes the Thomson cross-section, \u03c3T = \u00b52 0e4 6\u03c0m2 e = e4 6\u03c02 0m2 ec4. (7.6) 7A quantum description is necessary when the energy of an incident photon is comparable to the rest mass energy of the electron. In this limit, the scattered radiation undergoes a frequency shift due to recoil of the electron. 42 This is the e\ufb00ective geometric cross-section of the electron to scattering. Note that it is independent of frequency. The numerical value of the Thomson cross-section is \u03c3T = 6.65 \u00d7 10\u221229 m2. The Thomson cross-section can be neatly written in the form where the classical electron radius re is de\ufb01ned by \u03c3T = 8\u03c0 3 r2 e, e2 4\u03c00re = mec2. (7.7) (7.8) (One can think of re as being proportional to the radius that a uniform, extended charge e would have if the rest-mass energy were all electrostatic.) 7.2 Rayleigh scattering The other type of scattering we shall consider here is scattering of long wavelength radiation (\u03bb physical size of scatterer) from charges bound within systems that have no net charge. This is known as Rayleigh scattering. We shall see in Part III that for bound charges, provided that the frequency of the incident wave is small compared to any resonant frequency (or transition frequency in a quantum description), the bound charges will collectively displace following the applied force from the incident wave. This induces an oscillating electric dipole moment of the form p = \u03b1E, where E is the incident electric \ufb01eld at the position of the (small) scatterer, and \u03b1 is the polarizability.. = dE dt The wavelength of the scattered radiation is the same as that of the incident radiation and is large compared to the size of the scatterer. We are therefore in the dipole limit, and the radiated power is given by Eq. (6.18). Since \u00a8p = \u2212\u03b1\u03c92E, the time-averaged radiated power is \u00b50 6\u03c0c Dividing by the time-averaged incident energy \ufb02ux gives the cross-section for Rayleigh scattering: \u03b12\u03c94E2 0 2 (7.9) \u03c3 = 1 \u03bb4. 0\u03b12\u03c94 \u00b52 6\u03c0 Note the 1/\u03bb4 wavelength dependence of the cross-section, so that short wavelengths (high frequencies) are scattered more than longer wavelengths. Rayleigh scattering underlies many natural phenomena. For example, the blue hue of the overhead sky is due to the (mostly) nitrogen and oxygen molecules in the atmosphere preferentially scattering the blue (short wavelength) component of sunlight. Furthermore, red sunsets follow from preferentially scattering blue light out of the line of sight towards the setting sun. (7.10) \u221d 43 8 Radiation from an arbitrarily moving point charge We now consider radiation from a single point charge moving arbitrarily, i.e., we relax the assumption of non-relativistic motion. Speci\ufb01cally, we consider a point charge q following a wordline y\u00b5(\u03c4 ) in spacetime, where \u03c4 is proper time. For such general motion, the dipole approximation may not be valid and its extension in terms of higher multipole moments will converge only very slowly. Here, we develop an alternative methodology to deal with this problem. 8.1 Lienard\u2013Weichert potentials and \ufb01elds Recall from Eq. (3.52) that the current 4-vector for a point particle can be written covariantly as J \u00b5(x) = qc \u03b4(4) (x \u2212 y(\u03c4 )) \u02d9y\u00b5(\u03c4 ) d\u03c4. (8.1) (Here, we are using the proper time", " \u03c4 to parameterise the path, rather than some general \u03bb.) Inserting this into Eq. (5.15) for the retarded potential, we \ufb01nd A\u00b5(x) = = \u00b50qc 4\u03c0 \u00b50qc 4\u03c0 d\u03c4 d4z \u03b4(4) (z \u2212 y(\u03c4 )) \u02d9y\u00b5(\u03c4 ) \u03b4(x0 \u2212 z0 \u2212 |x \u2212 z|) |x \u2212 z| \u02d9y\u00b5(\u03c4 ) \u03b4 (x0 \u2212 y0(\u03c4 ) \u2212 |x \u2212 y(\u03c4 )|) |x \u2212 y(\u03c4 )| d\u03c4, (8.2) where we performed the integration d4z using the de\ufb01ning property of the deltafunction. To make further progress, it is convenient to express the remaining 1D delta-function in covariant form using Eq. (5.19): \u03b4 (x0 \u2212 y0(\u03c4 ) \u2212 |x \u2212 y(\u03c4 )|) |x \u2212 y(\u03c4 )| = 2\u0398 R0(\u03c4 ) \u03b4 (\u03b7\u00b5\u03bdR\u00b5(\u03c4 )R\u03bd(\u03c4 )), (8.3) where we de\ufb01ned R\u00b5(\u03c4 ) = x\u00b5 \u2212 y\u00b5(\u03c4 ). The combination of the delta-function and Heaviside function selects the unique value of proper time when the charge\u2019s worldline intersects the past lightcone of x\u00b5. Denoting the proper time there by \u03c4\u2217 (which is a function of x\u00b5), and using we have d d\u03c4 [\u03b7\u00b5\u03bdR\u00b5(\u03c4 )R\u03bd(\u03c4 )] = \u22122\u03b7\u00b5\u03bdR\u00b5(\u03c4 ) \u02d9y\u03bd(\u03c4 ), \u03b4 (x0 \u2212 y0(\u03c4 ) \u2212 |x \u2212 y(\u03c4 )|) |x \u2212 y(\u03c4 )| = \u03b4(\u03c4 \u2212 \u03c4\u2217) |R\u00b5(\u03c4\u2217) \u02d9y\u00b5(\u03c4\u2217)|. 44 (8.4) (8.5) Finally, using this in Eq. (8.2), and integrating over \u03c4, we obtain the Leinard\u2013Wiechert potential in covariant form: A\u00b5(x) = \u00b50qc 4\u03c0 \u02d9y\u00b5(\u03c4\u2217) |R\u03bd(\u03c4\u2217) \u02d9y\u03bd(\u03c4\u2217)|. (8.6) Equation (8.6) is a beautifully compact expression for the 4-potential due to an arbitrarily moving charge. However, to gain intuition for this result, it is useful to unpack it into time and space components in some inertial frame. Then, \u02d9y\u00b5(\u03c4\u2217) = \u03b3(\u03c4\u2217) (c, v(\u03c4\u2217)), where v(\u03c4\u2217) is the 3-velocity at proper time \u03c4\u2217 and \u03b3(\u03c4\u2217) is the associated Lorentz factor. Furthermore, R\u00b5(\u03c4\u2217) = (c\u2206t(\u03c4\u2217), R(\u03c4\u2217)), where \u2206t(\u03c4\u2217) is the time di\ufb00erence between the current event and the time that the charge intersected the past lightcone, and R(\u03c4\u2217) \u2261 x \u2212 y(\u03c4\u2217). Note that c\u2206t(\u03c4\u2217) = |R(\u03c4\u2217)| = R(\u03c4\u2217). We now have |R\u03bd(\u03c4\u2217) \u02d9y\u03bd(\u03c4\u2217)| = c\u03b3(\u03c4\u2217)R(\u03c4\u2217) 1 \u2212 \u02c6R(\u03c4\u2217) \u00b7 v(\u03c4\u2217)/c, and the time and space components of Eq. (8.6) give \u03c6(t, x) = 4\u03c00R(\u03c4\u2217) A(t, x) = 4\u03c0R(\u03c4\u2217) q 1 \u2212 \u02c6R(\u03c4\u2217) \u00b7 v(\u03c4\u2217)/c \u00b50qv(\u03c4\u2217) 1 \u2212 \u02c6R(\u03c4\u2217) \u00b7 v(\u03c4\u2217)/c. (8.7) (8.8) (8.9) Due to the presence of the 1 \u2212 \u02c6R(\u03c4\u2217) \u00b7 v(\u03c4\u2217)/c factors in the denominators of Eqs. (8.8) and (8.9), the potentials (and hence \ufb01elds) are concentrated in the direction of motion, \u02c6R(\u03c4\u2217) \u221d v(\u03c4\u2217), for relativistic motion. This", " is a consequence of extreme aberration for a rapidly moving particle. If we adopt an inertial frame such that the particle is at rest in that frame at \u03c4\u2217, the Lienard\u2013Wiechert potentials simplify to A(t, x) = 0 and \u03c6(t, x) = q 4\u03c00Rrest(\u03c4\u2217) (rest-frame), (8.10) where Rrest(\u03c4\u2217) is the distance between the observation event and y\u00b5(\u03c4\u2217) in this restframe. These potentials are exactly what one would expect for a stationary point charge. If instead we adopt a frame in which the charge is not at rest at \u03c4\u2217, but is moving non-relativistically, the vector potential no longer vanishes but is approximately A(t, x) \u2248 \u00b50qv(\u03c4\u2217) 4\u03c0R(\u03c4\u2217) (non-relativistic). (8.11) This is consistent with the non-relativistic (dipole) approximation in Eq. (6.7) noting that dp/dt = qv for a point charge (where p is the electric dipole moment, not the 3-momentum!). 45 Example: The \ufb01elds of a charge in uniform motion. Consider a charge q moving with velocity v such that its position at time t is y(t) = vt. The potentials at (ct, x) are generated at the earlier time t\u2217, such that the spatial separation R = x \u2212 vt\u2217 and temporal separation \u2206t \u2261 t \u2212 t\u2217 are the component of a null vector: c\u2206t = |x \u2212 vt\u2217|. (8.12) For the uniform motion considered in this example, R (and so \u2206t) depends only on the displacement from the charge\u2019s current position, To see this, we note that the null condition becomes r \u2261 x \u2212 vt. c\u2206t = |r + v\u2206t|, which determines \u2206t in terms of r, and R = r + v\u2206t. We solve for \u2206t by squaring both sides of Eq. (8.14) to \ufb01nd c\u2206t/\u03b32 = r \u00b7 v/c + (r \u00b7 v)2/c2 + r2/\u03b32, (8.13) (8.14) (8.15) (8.16) where we have used \u2206t > 0 to select the appropriate root. The denominators in Eqs. (8.8) and (8.9) involve the combination R \u2212 R \u00b7 v/c. Noting that R = c\u2206t, we have so that R \u2212 R \u00b7 v/c = R \u2212 r \u00b7 v/c \u2212 v2\u2206t/c = c\u2206t/\u03b32 \u2212 r \u00b7 v/c, R \u2212 R \u00b7 v/c = (r \u00b7 v)2/c2 + r2/\u03b32. This allows us to write the potentials as \u03c6(t, x) = A(t, x) = \u03b3q 4\u03c00 r2 + \u03b32(r \u00b7 v)2/c2 \u00b50\u03b3qv 4\u03c0r2 + \u03b32(r \u00b7 v)2/c2. (8.17) (8.18) (8.19) (8.20) We can compute the \ufb01elds from B = \u2207\u00d7A and E = \u2212\u2202A/\u2202t\u2212\u2207\u03c6, recalling that r = x\u2212vt. For the electric \ufb01eld, this gives E = q\u03b3r 4\u03c00 [\u03b32(r \u00b7 v)2/c2 + r2]3/2, (8.21) and the magnetic \ufb01eld is B = v \u00d7 E/c2. We see that the electric \ufb01eld is radially out from the current position of the charge, and the magnetic \ufb01eld is azimuthal, and both are anisotropic. 46 For r along the direction of motion, the \ufb01eld is smaller by a factor of 1/\u03b32 than for a charge at rest at the current position, while perpendicular to this the \ufb01eld is enhanced by a factor of \u03b3 over that for a charge at rest. The \ufb01eld lines are thus concentrated close to the plane perpendicular to the motion for a rapidly moving particle. Note that this does not contradict our earlier \ufb01nding that the \ufb01elds are concentrated where \u02c6R is along v. If we ask how the \ufb01elds vary around a circle of", " radius r centred on the current position of the charge, we also have to take account that the \ufb01elds were \u201cgenerated\u201d at times t\u2217 that have a strong dependence on the direction \u02c6r. In particular, for \u02c6r along the direction of motion, c\u2206t becomes very large for a rapidly moving particle (going as 2r\u03b32 for v \u2248 c), since the particle almost keeps up with light signals emitted in the forward direction, and so the \ufb01elds were generated when the particle was very far away. This overcomes the intrinsic \ufb01eld concentration along \u02c6R = \u02c6v. Finally, we note that the \ufb01elds due to a uniformly-moving charge can also be computed from those for a static charge by Lorentz transformation (see IB Electromagnetism). To calculate the electric and magnetic \ufb01elds of an arbitrarily moving charge, we return to the covariant form of the Lienard\u2013Weichert potentials (Eq. 8.6) and form F\u00b5\u03bd = \u2202\u00b5A\u03bd \u2212 \u2202\u03bdA\u00b5. We have to remember that \u03c4\u2217 is a function of x\u00b5, so that, for example, \u2202\u00b5 \u02d9y\u03bd(\u03c4\u2217) = \u00a8y\u03bd(\u03c4\u2217)\u2202\u00b5\u03c4\u2217. We can evaluate \u2202\u00b5\u03c4\u2217 by di\ufb00erentiating the null condition \u03b7\u03b1\u03b2R\u03b1(\u03c4\u2217)R\u03b2(\u03c4\u2217) = 0 to \ufb01nd 0 = \u03b7\u03b1\u03b2\u2202\u00b5R\u03b1(\u03c4\u2217)R\u03b2(\u03c4\u2217) = \u03b7\u03b1\u03b2\u2202\u00b5 (x\u03b1 \u2212 y\u03b1(\u03c4\u2217)) R\u03b2(\u03c4\u2217) \u00b5 \u2212 \u02d9y\u03b1(\u03c4\u2217)\u2202\u00b5\u03c4\u2217 = \u03b7\u03b1\u03b2 \u03b4\u03b1 R\u03b2(\u03c4\u2217). Solving for \u2202\u00b5\u03c4\u2217 gives \u2202\u03c4\u2217 \u2202x\u00b5 = R\u00b5(\u03c4\u2217) R\u03bd(\u03c4\u2217) \u02d9y\u03bd(\u03c4\u2217). The other term we require in di\ufb00erentiating Eq. (8.6) is \u2202\u00b5 [R\u03bd(\u03c4\u2217) \u02d9y\u03bd(\u03c4\u2217)] = \u03b4\u03bd \u00b5 \u2212 \u02d9y\u03bd(\u03c4\u2217)\u2202\u00b5\u03c4\u2217 \u02d9y\u03bd(\u03c4\u2217) + R\u03bd(\u03c4\u2217)\u00a8y\u03bd(\u03c4\u2217)\u2202\u00b5\u03c4\u2217 = \u2202\u00b5\u03c4\u2217 [R\u03bd(\u03c4\u2217)\u00a8y\u03bd(\u03c4\u2217) \u2212 \u02d9y\u03bd(\u03c4\u2217) \u02d9y\u03bd(\u03c4\u2217)] + \u02d9y\u00b5(\u03c4\u2217) c2 + R\u03bd(\u03c4\u2217)\u00a8y\u03bd(\u03c4\u2217) + \u02d9y\u00b5(\u03c4\u2217). = \u2202\u00b5\u03c4\u2217 Putting these pieces together, we \ufb01nd (exercise!), on antisymmetrising, F\u00b5\u03bd(x) = \u2212 \u00b50qc 4\u03c0(cRrest) R\u00b5 \u00a8y\u03bd \u2212 R\u03bd \u00a8y\u00b5 (cRrest) + (R\u00b5 \u02d9y\u03bd \u2212 R\u03bd \u02d9y\u00b5) (cRrest)2 (c2 + R\u03b1 \u00a8y\u03b1), where all quantities on the right are evaluated at \u03c4\u2217. Here, we have used \u2212R\u00b5(\u03c4\u2217) \u02d9y\u00b5(\u03c4\u2217) = cRrest, 47 (8.22) (8.23) (8.24) (8.25) (8.26) to express F\u00b5\u03bd in terms of Rrest, the radial distance between x\u00b5 and y\u00b5(\u03c4\u2217) as measured in the frame in which the charge is at rest at \u03c4\u2217. Inspection of Eq. (8.26) reveals terms going like 1/R, that depend on the 4-acceleration, and a term going like 1/R2 that depends only on the 4-velocity of the charge. The 1/R terms describe radiation, while the 1/R2 term turns out to be the \ufb01eld that would be generated from a charge moving uniformly with the same velocity at \u03c4\u2217 (see below). Exercise: Extract the electric and magnetic \ufb01elds from the \ufb01eld-strength tensor in Eq. (8.26", ") in a general inertial frame, recalling that the 4-acceleration has components \u00a8y\u00b5 = \u03b32 \u03b32 c v \u00b7 a, \u03b32 c2 a \u00b7 vv + a. You should \ufb01nd (after a lengthy, but straightforward, calculation) that E(t, x) = q 4\u03c00(1 \u2212 v \u00b7 \u02c6R/c)3 \u02c6R \u2212 v/c \u03b32R2 + \u02c6R \u00d7 [( \u02c6R \u2212 v/c) \u00d7 a] Rc2, (8.27) (8.28) where all quantities on the right are evaluated at \u03c4\u2217. For the magnetic \ufb01eld, you should \ufb01nd B(t, x) = 1 c \u02c6R(\u03c4\u2217) \u00d7 E(t, x). (8.29) The exercise above derives the electric and magnetic \ufb01elds from F\u00b5\u03bd. Note the following points. \u2022 The magnetic \ufb01eld is always perpendicular to the electric \ufb01eld and to \u02c6R(\u03c4\u2217). \u2022 The radiation (1/R) part of the \ufb01elds is sourced by the 3-acceleration. The electric \ufb01eld has an angular dependence given by \u02c6R \u00d7 [( \u02c6R \u2212 v/c) \u00d7 a], and is therefore perpendicular to \u02c6R(\u03c4\u2217), like the magnetic \ufb01eld. The angular dependence is more complicated than for dipole radiation [Eq. (6.14)], due to the presence of the v/c correction to \u02c6R. However, in the non-relativistic limit (|v| c), the radiation \ufb01elds reduce to those derived in the dipole approximation in Sec. 6 (noting that d2p/dt2 = qa for a point charge). The magnitudes of the \ufb01elds satisfy the usual relation for radiation, |E| = c|B|, even for relativistic motion of the charge. \u2022 The 1/R2 contribution to the \ufb01elds are the same as for a charge moving uniformly with the same velocity at \u03c4\u2217. To see this, recall the following expression for the 48 electric \ufb01eld from a uniformly moving charge (given above in Eq. 8.21): E = q\u03b3r 4\u03c00 [\u03b32(r \u00b7 v)2/c2 + r2]3/2, (8.30) where r points from the current position of the charge to the observation point. We can express this back in terms of R(\u03c4\u2217) using Eq. (8.18) and r = R \u2212 v\u2206t = R( \u02c6R \u2212 v/c), (8.31) which recovers exactly the \ufb01rst (1/R2) term in Eq. (8.28). 8.2 Power radiated We now calculate the rate at which energy is radiated by an arbitrarily moving point charge. We shall \ufb01rst calculate this in the instantaneous rest frame of the charge and then use Lorentz transformations to deduce the rate in a general inertial frame. In the frame in which the particle is at rest at proper time \u03c4\u2217, the radiation part of electric \ufb01eld on the forward lightcone through y\u00b5(\u03c4\u2217) is, from Eq. (8.28), E(t, x) = q 4\u03c00Rrestc2 \u02c6R \u00d7 ( \u02c6R \u00d7 arest) (Instantaneous rest frame), (8.32) where arest is the 3-acceleration in the instantaneous rest frame. The magnetic \ufb01eld is B = \u02c6R \u00d7 E/c, so the Poynting vector is E \u00d7 B N = = = 1 \u00b50 1 c\u00b50 1 c\u00b50 E \u00d7 ( \u02c6R \u00d7 E) |E|2 \u02c6R, (8.33) which is radial. We can get the radiated power by integrating over a spherical surface of radius Rrest at time t (i.e., a section through the forward lightcone corresponding to a constant time surface in the instantaneous rest frame). Since the charge is at rest at \u03c4\u2217, the radiation emitted in proper time d\u03c4 around \u03c4\u2217 crosses this spherical surface in (coordinate) time8 dt = d\u03c4. It follows that the energy radiated in the instantaneous 8 This would not be true if the charge were moving at \u03c4\u2217. Generally, the time taken for the radiation emitted in d\u03c4\u2217 to cross a spherical surface of radius", " R(\u03c4\u2217) can be got from Eq. (8.23). Rearranging gives d\u03c4\u2217 = R\u00b5(\u03c4\u2217)dx\u00b5 R\u03bd(\u03c4\u2217) \u02d9y\u03bd(\u03c4\u2217). 49 rest frame in proper time d\u03c4 is dErest = d\u03c4 N \u00b7 dS = R2 restd\u03c4 c\u00b50 |E|2 d\u2126, (8.34) where d\u2126 is the element of solid angle (i.e., d\u2126 = d2 \u02c6R). Taking the rest-frame acceleration to be along the z-axis, we have | \u02c6R \u00d7 ( \u02c6R \u00d7 arest)| = sin \u03b8|arest|, so that on integrating over solid angle dErest d\u03c4 = \u00b50q2|arest|2 6\u03c0c. (8.35) (8.36) This simple result is exactly what we would have got by applying the dipole approximation, Eq. (6.18), in the instantaneous rest frame, with d2p/dt2 \u2192 qarest. This is not unexpected \u2013 the dipole approximation holds in situations where all velocities are non-relativistic and this is exactly true in the instantaneous rest frame. Indeed, the dipole approximation provides a useful short-cut to deriving Eq. (8.36). Now consider the 4-momentum radiated in time d\u03c4. Since equal power is emitted along \u02c6R and \u2212 \u02c6R in the instantaneous rest frame, the total 3-momentum of the radiation is zero. We therefore have dP \u00b5 = d\u03c4 1 c dErest d\u03c4, 0 (8.37) in the rest frame. Lorentz transforming to an inertial frame in which the charge has velocity v(\u03c4\u2217) at \u03c4\u2217, we have dE c = dP 0 = \u03b3(\u03c4\u2217)dP 0 = \u03b3(\u03c4\u2217)d\u03c4 c dErest d\u03c4. (8.38) The time elapsed between the events y\u00b5(\u03c4\u2217) and y\u00b5(\u03c4\u2217 + d\u03c4 ) in the general frame is dt = \u03b3(\u03c4\u2217)d\u03c4 by time dilation, hence dE dt = dErest d\u03c4. (8.39) Taking dx\u00b5 = (cdt, 0) gives d\u03c4\u2217 = \u03b3(\u03c4\u2217) dt 1 \u2212 \u02c6R(\u03c4\u2217) \u00b7 v(\u03c4\u2217)/c. This is just the usual Doppler e\ufb00ect. advertised. In the instantaneous rest frame, it reduces to d\u03c4\u2217 = dt, as 50 We see that the rate at which energy is radiated by the accelerated charge is Lorentz invariant and so, generally, the instantaneous power radiated is dE dt = \u00b50q2a\u00b5a\u00b5 6\u03c0c = \u00b50q2\u03b34 6\u03c0c a2 + \u03b32(a \u00b7 v/c)2, (8.40) where a\u00b5 is the acceleration 4-vector (so that a\u00b5a\u00b5 = |arest|2). This result is known as the relativistic Larmor formula. (Its non-relativistic limit was \ufb01rst worked out by Larmor in 1897.) Non-examinable aside: If the above argument is not to your taste, the same expression for the power radiated in a general frame can be obtained by direct calculation. The radiation part of the electric \ufb01eld is now given by E(t, x) = q 4\u03c00 \u02c6R \u00d7 [( \u02c6R \u2212 v/c) \u00d7 a] R(1 \u2212 v \u00b7 \u02c6R/c)3c2, (8.41) where all quantities on the right are evaluated at \u03c4\u2217. The Poynting vector is still given by Eq. (8.33), but when integrating this over the surface of a sphere of radius R(\u03c4\u2217) at time t, to determine the radiated power, we have to be careful to account for Doppler shifts that relate dt at the source to the increment in time for radiation to pass through the radius R(\u03c4\u2217). Following the arguments in the footnote on Page 49, and noting that the coordinate time between the events at \u03c4\u2217 and \u03c4\u2217 + d\u03c4 on the charge\u2019s worldline is \u03b3(\u03c4\u2217)d\u03c4, the power emitted per solid angle (i.e., energy per solid angle per coordinate time at source) is dE dtd\u2126 = 1 c\u00b50 1 \u2212 v \u00b7 \u02c6", "R c |E|2R2 = \u00b50q2 (4\u03c0)2c \u02c6R \u00d7 [( \u02c6R \u2212 v/c) \u00d7 a] 2 1 \u2212 v \u00b7 \u02c6R/c 5. (8.42) Note that the angular dependence of the radiated power is more complicated than the sin2 \u03b8 dependence in the instantaneous rest frame. In particular, for a particle moving relativistically, the power is signi\ufb01cantly concentrated in the direction of motion, \u02c6R \u221d v. After some vector algebra, Eq. (8.42) can be reduced to dE dtd\u2126 = \u00b50q2 (4\u03c0)2c \uf8eb \uf8ec \uf8ed2 ( \u02c6R \u00b7 a)(a \u00b7 v/c) 1 \u2212 v \u00b7 \u02c6R/c 4 + |a|2 3 \u2212 1 \u2212 v \u00b7 \u02c6R/c ( \u02c6R \u00b7 a)2 \u03b32 1 \u2212 v \u00b7 \u02c6R/c 5 \uf8f6 \uf8f7 \uf8f8. (8.43) Integrating over solid angle9 directly gives the \ufb01nal result in Eq. (8.40). 9A straightforward, but lengthy, way to do this is to align v with the z-axis and a in the x-z plane, and express the components of \u02c6R in spherical polar coordinates. 51 8.3 Synchrotron radiation We end our discussion of the radiation from arbitrarily moving charged particles with a speci\ufb01c example. Synchrotron radiation refers to the radiation emitted by a charged particle moving relativistically in circular motion (typically under the action of a magnetic \ufb01eld). Consider a particle of charge q and mass m moving in a circle with speed v perpendicular to a magnetic \ufb01eld of magnitude B. We saw in Sec. 3.1.1 that such a particle has angular velocity \u03c9 = qB/(\u03b3m) and so the acceleration has magnitude |a| = qBv \u03b3m, (8.44) and is perpendicular to v. The total instantaneous power radiated follows from Eq. (8.40), which here gives dE dt = \u00b50 6\u03c0c q2B\u03b3v m 2. (8.45) Note that this is proportional to the square of the 3-momentum of the particle. As discussed earlier, for a relativistic particle the emitted power is concentrated (or beamed ) into a narrow range of solid angles around the instantaneous velocity. The beaming is controlled by factors of 1 \u2212 v \u00b7 \u02c6R/c in the denominator of the emitted power (see e.g., Eq. 8.42). We can estimate the angular extent of the beamed radiation by asking for what angle \u03b8 between v and \u02c6R does (1 \u2212 v \u00b7 \u02c6R/c)\u22121 fall to one-half its peak value, i.e., 1 \u2212 \u03b2 cos \u03b8 = 2(1 \u2212 \u03b2), (8.46) where \u03b2 = v/c. Writing \u03b2 = 1 \u2212, and working to leading order in, we \ufb01nd \u03b82 = 2. We can write this in terms of the Lorentz factor \u03b3 noting that \u03b3\u22122 = 1 \u2212 \u03b22 = 2 + \u00b7 \u00b7 \u00b7, (8.47) so that \u03b8 \u2248 1/\u03b3. We see that for a relativistic particle with \u03b3 1, the emitted radiation is beamed into a cone in the direction of motion with an opening angle of the order of 1/\u03b3. Now consider observing the synchotron radiation at position x with the particle\u2019s orbit centred on the origin. For |x| large compared to the radius of the orbit, we are safely in the far-\ufb01eld and can also approximate \u02c6R with \u02c6x. The argument above tells us that only observers in the plane of the orbit will observe signi\ufb01cant radiation. Since the orbit is periodic with period \u03c9, so is the observed radiation. However, due to the beaming the radiation is received in short pulses (think of observing a rotating searchlight). We can estimate the duration of the pulses by noting that each pulse is emitted with duration \u2206t\u2217 at source, such that the particle moves through an angle equal to twice 52 the opening angle of the beamed radiation pattern, i.e., \u03c9\u2206t\u2217 = 2/\u03b3. In relating this to the duration of the observed pulse we have to taken account of the Doppler e\ufb00ect. Generally, an increment in retarded (coordinate) time corresponds to an increment in time at the observation point with (see the footnote", " on Page 49), so that for the duration of the pulse dt = (1 \u2212 \u02c6R \u00b7 v/c)dt\u2217 \u2206t = (1 \u2212 \u03b2)\u2206t\u2217 \u2248 1 \u03b33\u03c9. (8.48) (8.49) This means that the radiation is observed to have Fourier components up to roughly \u03b33\u03c9, and so a relativistic particle emits a broad frequency spectrum of radiation up to \u03b33 times the orbital frequency. 53 Part III Electromagnetism in media So far in this course we have considered isolated charges and currents in vacuum. However, for many practical applications we are interested in electric and magnetic \ufb01elds inside media, including solids, liquids and gases. Here, the story is rather more complicated. We have to consider the distribution of charges within the atoms/molecules that make up the medium, and the currents generated by the motion of these charges. Moreover, we need to consider how these distributions change in the presence of externallyapplied \ufb01elds. Generally, the electromagnetic \ufb01elds inside media are very complicated with signi\ufb01cant variation on the atomic scale. However, we are often only interested in the averaged properties of \ufb01elds in regions large compared to the atomic scale. For example, to consider the propagation of light of wavelength \u03bb \u223c O(500) nm through some medium, we can e\ufb00ectively average the medium on scales small compared to \u03bb but large compared to atomic scales, O(0.1) nm. This leads us to consider an e\ufb00ective or macroscopic version of the Maxwell equations that describe the \ufb01elds averaged over macroscopically-small but microscopically-large regions. In this \ufb01nal part of the course, we shall \ufb01rst brie\ufb02y introduce some of the basic electromagnetic properties of matter (Sec. 9). We then develop the macroscopic versions of Maxwell\u2019s equations in Sec. 10 by a spatial averaging procedure. The rest of the course applies these macroscopic equations to wave propagation in dielectrics (e.g., materials like glass; Secs 11 and 12) and conductors (Sec. 13). 9 Electromagnetic properties of matter We shall mostly be concerned with non-conducting media. There is still tremendous diversity in the electromagnetic properties of such materials, but we shall only be concerned with some of the simplest examples here. 9.1 Dielectric media An important class of insulating materials are called dielectrics. These have no mobile charges that can move freely in an applied \ufb01eld (unlike a conductor) but they nevertheless have a signi\ufb01cant e\ufb00ect on applied electric \ufb01elds. Generally, the individual atoms/moleclues that make up the material may have complicated time-dependent charge distributions. For example, water molecules have strong electric dipole mo- 54 ments and these vary on a short timescale due to thermal motion and internal vibration of the molecules. When averaged over macroscopic volumes, which contain a great many molecules, these dipole moments tend to cancel out in the absence of an applied \ufb01eld. However, if a \ufb01eld is applied, there will be coherent perturbations to the charge distribution of molecules within macroscopic volumes that will survive averaging. This phenomena is known as polarisation. For example, materials with intrinsic but randomly-oriented molecular dipole moments will tend to align their moments due to the torque of the applied \ufb01eld, while materials with no initial dipole moments will generally develop aligned moments as the applied \ufb01eld distorts the equilibrium charge distribution in each molecule. It is these coherent e\ufb00ects that determine the macroscopic behaviour of electromagnetic \ufb01elds in media. Toy model for atomic polarisability: Quantum theory is needed to describe the polarisation of atoms in applied \ufb01elds. However, the following simple classical toy model is useful to develop understanding and to give an order-of-magnitude estimate of the induced polarisation. Consider a point particle (the nucleus) with positive charge q, surrounded by an electron cloud of total charge \u2212q and radius a. Assume the electron cloud to have uniform charge density. At radius r from the centre of the cloud, the charge enclosed is \u2212q(r/a)3, and so the electric \ufb01eld at position x relative to the centre is Ecloud(x) = \u2212 q 4\u03c00a3 x. (9.1) If we apply a uniform external \ufb01eld E, the positive charge will displace from the centre of the cloud so that the total force on", " it is zero. (We ignore any distortion of the cloud.) If the displacement is d, we must have qd = 4\u03c00a3E \u21d2 p = 4\u03c00a3E. (9.2) Note that the induced dipole moment p is proportional to E. It will generally be the case that p is linear in E since this is the \ufb01rst term in a Taylor expansion of p(E) given that p(E = 0) = 0. Higher terms in the expansion are negligible if the applied \ufb01eld is much smaller than the typical \ufb01eld within the atom (around q/(4\u03c00a3) in this example) or, equivalently, the displacements are small compared to the atomic scale. If we de\ufb01ne the polarisability \u03b1 by p = \u03b1E, (9.3) we see that \u03b1 = 4\u03c00a3. We can estimate \u03b1 for hydrogen atoms by taking a \u223c 10\u221210 m (a typical atomic size from quantum theory) to \ufb01nd \u03b1/(4\u03c00) \u223c 10\u221230 m3; this is close to the measured value \u03b1/(4\u03c00) = 6.67 \u00d7 10\u221231 m3. The induced displacement d depends on the magnitude of the electric \ufb01eld. For a \ufb01eld of 105 V m\u22121 (i.e., 100 V over 1 mm), the displacement in hydrogen is d = 5 \u00d7 10\u221217 m, much smaller than the size of the atom. 55 9.2 Magnetic media Magnetic \ufb01elds are generated by charges in motion, i.e., currents. The leading-order magnetic \ufb01eld far away from a localised current distribution can be expressed in terms of the magnetic dipole moment m = 1 2 i qixi \u00d7 vi, (9.4) where the sum is over the charges, with the ith charge at position xi and having velocity vi. The continuum form of this is m = 1 2 (x \u00d7 J) d3x, (9.5) where the current density J(t, x) = i qivi(t)\u03b4(3)(x \u2212 xi(t)). The orbiting electrons in an atom/molecule may give rise to a net magnetic moment. There can also be a contribution from the intrinsic spin of the electrons \u2013 a purely quantum mechanical e\ufb00ect. The magnetic response of materials to applied \ufb01elds depends on whether the net magnetic moment of each constituent molecule vanishes or not in the absence of an applied magnetic \ufb01eld. The phenomenology is rather richer than the electrical response of dielectrics discussed above. \u2022 Diamagnetism. A material is diamagnetic if the net magnetic moments vanish when there is no applied \ufb01eld. Once a \ufb01eld is applied, the orbital motions are slightly distorted to generate magnetic moments that tend to be anti-parallel to the applied \ufb01eld. One can think of this in terms of Lenz\u2019s law: the induced current \ufb02ows in a direction to oppose the change in magnetic \ufb01eld. A simple (classical) toy model for diamagnetism is given below. Most metals are diamagnetic. \u2022 Paramagnetism. A material is paramagnetic if the constituent molecules have non-zero intrinsic magnetic moments. In this case, the magnetic moments will tend to be randomly oriented when there is no applied magnetic \ufb01eld, giving magnetic \ufb01elds inside the medium that vary rapidly in time and space. However, on averaging over a macroscopic volume, the \ufb01elds average down. Once an external magnetic \ufb01eld is applied, the intrinsic magnetic moments tend to align along the \ufb01eld due to the torque acting on them, giving a net e\ufb00ect that survives volume averaging. The response is strongly temperature dependent (as is the electric response for polar dielectric materials), being weaker at high temperatures due to the randomising e\ufb00ect of thermal motions. \u2022 Ferromagnetism. Ferromagnets are paramagnetic, but strong quantum-mechanical (exchange) interactions between the electrons in neighbouring constituents give radically di\ufb00erent behaviour. At temperatures below what is called the Curie 56 temperature, the interactions are strong enough to allow spontaneous magnetisation: the magnetic moments are aligned in macroscopically-large regions called domains, even in the absence of an applied \ufb01eld. Applying a \ufb01eld changes the domains, and the moments in di\ufb00erent domains tend to align along the applied \ufb01eld. When the external \ufb01eld is removed, some net alignment across domains remains producing a strong (intr", "insic) magnetic \ufb01eld. Ferromagnets are what we usually think of as \u201cmagnets\u201d. The most common example is iron. Toy model for atomic diamagnetism: Quantum theory is needed to describe properly the magnetic response of the electrons in an atom to an applied magnetic \ufb01eld10. However, the following simple classical calculation captures the basic physics at work. Consider a charge q of mass m orbiting about a nucleus at radius a. Let the orbit lie In the absence of an normal to the z-direction and the charge be moving anti-clockwise. applied magnetic \ufb01eld, the Coulomb force due to the nucleus, of magnitude FC, provides the centripetal force such that FC = mv2/a, (9.6) where v is the orbital speed. Now apply a magnetic \ufb01eld B along the positive z-axis. If we ignore any change in the shape and size of the orbit, the orbital speed will change to v + \u03b4v such that FC \u2212 q(v + \u03b4v)B = m(v + \u03b4v)2/a, (9.7) i.e., the change in the radial force due to the magnetic \ufb01eld equals the change in the centripetal force. Expanding to \ufb01rst order in \u03b4v and B gives As a result, the magnetic moment of the orbiting charge is changed by an amount \u03b4v = \u2212 qa 2m B. \u03b4m = \u2212 q2a2 4m B (9.8) (9.9) along the z-direction. Note that the change in magnetic \ufb01eld is anti-parallel to the applied \ufb01eld. If the particle were orbiting clockwise instead, the speed would increase by qaB/(2m), but the change in the magnetic moment would still be given by Eq. (9.9). Diamagnetism is a very weak e\ufb00ect. Taking q to be the charge of an electron, m the electron mass, a \u223c 10\u221210 m, and B = 1 T (a strong magnetic \ufb01eld in the laboratory), we \ufb01nd \u03b4m \u223c 10\u221228 A m2. This should be compared to the natural atomic unit of magnetic moment, the Bohr magneton: \u00b5B \u2261 e\u00afh 2me = 9.27 \u00d7 10\u221224 A m2. (9.10) 10Indeed, the Bohr\u2013van Leeuwen theorem forbids any non-zero thermal average of the magnetic moment for a classical system in thermal equilibrium. 57 (The Bohr magneton is approximately the magnetic moment due to the spin of an electron, and is also the z-component of the magnetic moment due to orbital motion of an electron with angular momentum Lz = \u00afh.) We see that, even for a strong \ufb01eld of 1 T, the induced magnetic moment is only O(10\u22125)\u00b5B. 10 Macroscopic Maxwell equations 10.1 Averaging the microscopic Maxwell equations In classical physics, on scales large compared to nuclear dimensions, O(10\u221215 m), we can treat atoms as collections of point particles. The microscopic Maxwell equations inside media are the usual ones, with the charges and currents due to a collection of (moving) point charges. We denote the rapidly-varying electric and magnetic \ufb01elds inside the medium by Emicro and Bmicro, so that \u2207 \u00b7 Emicro = \u2207 \u00d7 Emicro = \u2212 \u03c1micro 0 \u2202Bmicro \u2202t \u2207 \u00b7 Bmicro = 0 \u2207 \u00d7 Bmicro = \u00b50Jmicro + \u00b500 \u2202Emicro \u2202t, (10.1) (10.2) (10.3) (10.4) with \u03c1micro and Jmicro the microscopic charge and current densities, respectively. If we are only interested in lengthscales large compared to atomic dimensions, we can average the microscopic \ufb01elds to obtain smooth macroscopic \ufb01elds that we denote by E and B.11 The averaging is done by convolution with some smoothing function f (x), so that, for example, E(t, x) \u2261 Emicro(t, x) = f (x)Emicro(t, x \u2212 x) d3x. (10.5) We do not have to be precise about the choice of smoothing function, but it should be real, smoothly localised to some macroscopic region about x = 0, and integrate to unity over all space. It is usual to take f (x) to be spherically symmetric to preserve any directional characteristics of the medium. The averaging operation commutes with 11In the earlier parts of this course we were", " concerned with isolated charges in vacuum. In this situation, there is no distinction between the macroscopic and microscopic \ufb01elds well away from any charges. 58 di\ufb00erentiation in time and space, so that the microscopic Maxwell equations smooth to \u2207 \u00b7 E = \u2207 \u00d7 E = \u2212 \u03c1micro 0 \u2202B \u2202t \u2207 \u00b7 B = 0 \u2207 \u00d7 B = \u00b50Jmicro + \u00b500 \u2202E \u2202t, (10.6) (10.7) (10.8) (10.9) We split the microscopic charge and current densities into bound charges, which are bound to the atoms of the medium, and free charges, which are not bound. Examples of free charges include the mobile electrons in a conductor that carry the current in the presence of an applied electric \ufb01eld, and any additional charges that we have placed within or externally to the medium. We denote the macroscopic (smoothed) free charge and current densities by \u03c1free and Jfree, and similarly \u03c1bound and Jbound for the bound charges and currents. 10.1.1 Bound charge density If the material is composed of atoms or molecues, with the nth centred on xn(t), and each is made up of point charges {qi,n} at locations ri,n(t) relative to xn(t), we have \u03c1micro(t, x) = \u03c1free(t, x) + qi,n\u03b4(3)(x \u2212 xn(t) \u2212 ri,n(t)) = \u03c1free(t, x) + n i i n 59 qi,nf (x \u2212 xn(t) \u2212 ri,n(t)). (10.10) \u03c1bound(t,x) The relative locations ri,n are small in magnitude compared to the support of the smoothing function by assumption. We can therefore Taylor expand to obtain \u03c1bound(t, x) = qi,n [f (x \u2212 xn(t)) \u2212 ri,n(t) \u00b7 \u2207f (x \u2212 xn(t)) + \u00b7 \u00b7 \u00b7 ] n i = \u2212 qi,nri,n(t) \u00b7\u2207f (x \u2212 xn(t)) + \u00b7 \u00b7 \u00b7 n = \u2212\u2207 \u00b7 i pn(t) pn(t)f (x \u2212 xn(t)) + \u00b7 \u00b7 \u00b7 n = \u2212\u2207 \u00b7 n pn(t)\u03b4(3)(x \u2212 xn(t)) + \u00b7 \u00b7 \u00b7, (10.11) P(t,x) where we have assumed that the net charge i qi,n = 0. Here, we see that each atom/molecule may be replaced by its electric dipole moment pn(t), plus higher-order moments represented by the ellipsis in the above, for the purpose of calculating the smoothed bound charge density. We have introduced the macroscopic polarisation, P(t, x) \u2261 pn(t)\u03b4(3)(x \u2212 xn(t)), (10.12) n the divergence of which generates \u03c1bound. In principle, there are contributions to \u03c1bound from the higher multipoles but these are invariably negligible. For example, the contribution from electric quadrupole moments is smaller than that from electric dipoles by roughly a/lsmooth, where a is an atomic size and lsmooth is the smoothing length associated with f (x). As noted earlier, even if the material has non-zero molecular moments pn(t) in the absence of an applied \ufb01eld, these will generally be randomly oriented on macroscopic scales and will not contribute to P. (More carefully, the mean-squared contribution to P will go as 1/N where N 1 is the number of molecules within a smoothing volume.) What matters for P is the perturbations to the pn(t) due to the applied \ufb01eld. These will be coherent over scales large compared to the smoothing scale and so will not be averaged down by smoothing. Note that we have only discussed spatial smoothing, not smoothing in time. The latter is not required once we smooth in space since the resulting P(t, x) will only vary on the timescales of the applied \ufb01elds. Although intrinsic dipole moments pn(t) may vary over a range of timescales, these motions are incoherent on macroscopic scales and will average out in macroscopic volumes. 60 Bound surface charge We have seen that \u03c1bound = \u2212\u2207 \u00b7 P. Generally, this will be non-zero inside a material provided the polarisation is non-uniform. But note that, even", " if the polarisation is uniform inside the material, there will still be bound charges on the surface of the material. To see this, note that as we move from inside the material to the external vacuum, P will rapidly transition to zero over the smoothing scale lsmooth, producing a non-zero \u03c1bound localised near the surface. If we only care about length scales large compared to lsmooth, we can regard this as a surface charge density \u03c3bound, i.e., a charge per area of the surface, which we can determine from the divergence theorem: \u03c3bounddS = \u03c1bound d3x V = \u2212 P \u00b7 dS \u2202V = P \u00b7 dS. (10.13) (10.14) (10.15) Here, the integral is over a \u201cGaussian pillbox\u201d that straddles the transition region at the surface, and dS = \u02c6ndS is the outward-pointing area element. We see that the surface charge is \u03c3bound = P \u00b7 \u02c6n, (10.16) where P is the polarisation just inside the bulk of the material (but away from the transition region), and \u02c6n is the unit outward normal to the surface. 10.1.2 Bound current density The calculation of the bound current density proceeds similarly, but is rather more involved because of the vector nature of the current and the presence of molecular velocities vn(t) \u2261 \u02d9xn. Generally, we have Jmicro(t, x) = Jfree(t, x) + qi,n [vn(t) + vi,n(t)] \u03b4(3)(x \u2212 xn(t) \u2212 ri,n(t)) = Jfree(t, x) + n i i n qi,n [vn(t) + vi,n(t)] f (x \u2212 xn(t) \u2212 ri,n(t)). Jbound(t,x) (10.17) Here, vi,n \u2261 \u02d9ri,n and in the limit of non-relativistic velocities is the internal relative velocity of the ith charge in the nth molecule. We shall initially ignore the e\ufb00ects of the molecular velocities vn(t) since they are usually much smaller than the rapid internal 61 velocities (e.g., of electrons in the molecule) and tend not to be coherent over large smoothing volumes. In this approximation, we should also ignore the time dependence of f (x \u2212 xn(t)) since \u2202f (x \u2212 xn(t)) \u2202t = \u2212vn(t) \u00b7 \u2207f (x \u2212 xn(t)). (10.18) We then have Jbound(t, x) \u2248 qi,nvi,n(t) [f (x \u2212 xn(t)) \u2212 ri,n(t) \u00b7 \u2207f (x \u2212 xn(t)) + \u00b7 \u00b7 \u00b7 ] \u2248 \u2248 n i \u02d9pn(t)f (x \u2212 xn(t)) \u2212 n \u2202P(t, x) \u2202t \u2212 n i n i qi,nvi,n(t)ri,n(t) \u00b7 \u2207f (x \u2212 xn(t)) + \u00b7 \u00b7 \u00b7 qi,nvi,n(t)ri,n(t) \u00b7 \u2207f (x \u2212 xn(t)) + \u00b7 \u00b7 \u00b7. (10.19) The \ufb01rst term in the last line is the time derivative of the macroscopic polarization, and arises since changes in the molecular dipole moments are accompanied by a \ufb02ow of charge within the molecules. We can see this clearly from the continuity equation for the bound charge: given that \u03c1bound = \u2212\u2207 \u00b7 P + \u00b7 \u00b7 \u00b7, the divergence of Jbound must include the divergence of \u2202P/\u2202t. For the second term in Eq. (10.19), we use n i qi,nvi,n(t)ri,n(t) \u00b7 \u2207f (x \u2212 xn(t)) \u2248 \u2212 + 1 2 1 2 1 2 n i qi,nvi,n(t)ri,n(t) \u00b7 \u2207f (x \u2212 xn(t)) qi,nri,n(t)vi,n(t) \u00b7 \u2207f (x \u2212 xn(t)) n \u2202 \u2202t i n i qi,nri,n(t)ri,n(t) \u00b7 \u2207f (x \u2212 xn(t)) (10.20) where the approximation is to ignore the time dependence of f (x \u2212", " xn(t)). (This is similar to the rearrangement in Eq. 6.23.) The \ufb01rst two terms on the right combine to give \u2212\u2207 \u00d7 1 2 n i qi,nri,n(t) \u00d7 vi,n(t)f (x \u2212 xn(t)) = \u2212\u2207 \u00d7 mn(t)f (x \u2212 xn(t)) n = \u2212\u2207 \u00d7 n mn(t)\u03b4(3)(x \u2212 xn(t)), M(t,x) (10.21) 62 where mn is the molecular magnetic dipole moment, mn(t) \u2261 1 2 i qi,nri,n \u00d7 vi,n, (10.22) and M(t, x) is the macroscopic magnetisation. The \ufb01nal term on the right of Eq. (10.20) involves the molecular quadrupole moment tensor, Q n(t) = 3 i qi,nri,n(t) \u2297 ri,n(t). (10.23) We have 1 2 \u2202 \u2202t n i qi,nri,n(t)ri,n(t) \u00b7 \u2207f (x \u2212 xn(t)) = = 1 6 1 6 \u2202 \u2202t \u2202 \u2202t n Q n(t)\u2207f (x \u2212 xn(t)) \u2207 \u00b7 Q n(t)\u03b4(3)(x \u2212 xn(t)), n Q(t,x) (10.24) where Q(t, x) is the macroscopic quadrupole density. Note that \u2207 \u00b7 Q is shorthand for the vector with components \u2202Q ij/\u2202xj. Note further that the quadrupole moments here have not had the trace removed; we denote these with an additional prime to distinguish them from the trace-free quadrupole moments, e.g., Eq. (6.26). Putting these pieces together, we \ufb01nally \ufb01nd Jbound \u2248 \u2202P \u220210.25) We have already discussed the physical origin of the \ufb01rst term on the right, but what about the others? The third (i.e., quadrupole) term arises for the same reason as the \ufb01rst: if we include higher terms in the expansion of \u03c1bound in Eq. (10.11), the next term involves the divergence of \u2207 \u00b7 Q: \u03c1bound(t, x) = \u2212\u2207 \u00b7 P(t, x) \u2212 \u2207 \u00b7 Q(t, x) + \u00b7 \u00b7 \u00b7. 1 6 (10.26) The bound current must include the time derivative of the term in brackets to ensure charge conservation. However, we can also add divergence-free terms to Jbound since these would not a\ufb00ect the continuity equation. We see that the \ufb01rst such term is the curl of the magnetisation. One can think of the molecular dipole moments as loops of current; if these vary spatially, adjacent loops will not cancel perfectly and can give rise to a net bound current density in the bulk of the material. Often only the \ufb01rst two terms are retained in Eq. (10.25). As noted above, the quadrupole term is typically smaller by a/lsmooth than the electric polarization. The quadrupole and 63 magnetisation terms are each of the same order in a/lsmooth and the internal relative velocities, but retaining the magnetisation can be important in some situations, for example magnetostatics where the magnetic \ufb01eld induces magnetic dipole moments but no electric multipole moments. There is one loose end to tidy up \u2013 the neglect of the molecular velocities vn(t). It is left as an exercise to show that in this case there are additional contributions to the bound current that involve products of the vn(t) with the molecular electric dipole and quadrupole moments. As noted above, such terms are generally negligible. An exception is if the medium is in relativistic bulk motion in which case the vn(t) are coherent over a smoothing volume. This case can also be dealt with directly by Lorentz transformation. Bound surface current In the same way that the rapid variation in P at the surface of the material gives rise to a bound surface charge density, surface currents \ufb02ow at the surface due to the rapid variation in M. We can determine the surface current density12 Kbound by integrating Eq. (10.25) across an in\ufb01nitesimal plane surface whose normal is tangent to the surface of the material. The boundary of the in\ufb01nitesimal surface is", " a planar circuit that straddles the material surface. The long sides of this circuit are tangent to the surface of the material and have length \u2206l. One of the long sides lies just inside the material and is \u2206l, and the other (\u2212\u2206l) is just outside. The short sides have negligible length. Let the surface of the material have outward normal \u02c6n, and the normal to the circuit be \u02c6t. It follows that \u2206l = \u2206l\u02c6n \u00d7 \u02c6t. The integral of \u2207 \u00d7 M across the surface that spans the circuit gives a contribution to the surface current; by Stokes theorem, (\u2207 \u00d7 M) \u00b7 dS = M \u00b7 dl = M \u00b7 \u2206l, (10.27) where M is the magnetisation just inside the material (but away from the transition region). If the corresponding surface current is Kbound, we must have M \u00b7 \u2206l = Kbound \u00b7 (\u02c6t\u2206l) \u21d2 \u02c6t \u00b7 (M \u00d7 \u02c6n) = \u02c6t \u00b7 Kbound \u2200 \u02c6t \u22a5 \u02c6n. (10.28) It follows that Kbound = M \u00d7 \u02c6n. There is no further contribution to the surface current from the \u2202P/\u2202t term in Eq. (10.25) since the spanning surface has vanishing area. There is, however, in general a further contribution from the quadrupole term in Eq. (10.25). (10.29) 12Generally, a surface current density is represented by a vector K lying in the surface. The direction represents the direction the current \ufb02ows, and the magnitude gives the current per length through a line in the surface that is perpendicular to K. 64 As a simple application of Eq. (10.29), consider a uniformly magnetised rod, with M pointing along the rod. This is a simple model for a permanent bar magnet. The magnetisation is uniform inside the rod, so the only current is a surface current that \ufb02ows uniformly around the rod. The magnetic \ufb01eld is therefore equivalent to that of a \ufb01nite solenoid with a current per length |M|. 10.2 Electric displacement D Let us now substitute the free and bound charge density into the macroscopic Maxwell equation \u2207 \u00b7 E = \u03c1micro/0. Retaining only the contribution from the macroscopic polarisation to the bound charge, we have \u2207 \u00b7 E = 1 0 (\u03c1free \u2212 \u2207 \u00b7 P). If we de\ufb01ne the electric displacement, D(t, x) = 0E(t, x) + P(t, x), we obtain the simpler-looking equation \u2207 \u00b7 D = \u03c1free. (10.30) (10.31) (10.32) Of course, the complications of being inside a medium have not disappeared: they have been shifted to the relation between D and the macroscopic electric \ufb01eld E. As discussed earlier, it is often a good approximation to assume that the induced dipole moments are linear in the electric \ufb01eld. If the medium is also isotropic, then we must have13 P(t, x) = \u03c7e0E(t, x), (10.33) where the dimensionless constant \u03c7e is the electrical susceptibility of the medium. We can then write D = 0rE, (10.34) where r = 1 + \u03c7e is the relative permittivity of the medium. For dilute gases, r is close to unity, but in strongly polar substances it can be much larger (e.g., r \u2248 80 for water at room temperature). Inside a homogeneous medium, r is uniform and so Eq. (10.32) reduces to \u2207 \u00b7 E = \u03c1free 0r. (10.35) 13This is actually one simpli\ufb01cation too far. More generally, the polarization will be local in space, but non-local in time, i.e., P(t, x) will depend on E(t, x) at all times t \u2264 t. We shall discuss such dispersive media later in Sec. 12. 65 Combining this with the macroscopic Maxwell equation \u2207 \u00d7 E = \u2212\u2202B/\u2202t, we see that electrostatics inside an in\ufb01nite such medium amounts to reducing the \ufb01elds that would have been generated by the same free charges in vacuum by a factor r. The origin of this reduction is a screening of the free charges by the bound charges that arise from the divergence of the induced polarization. Example: a dielectric sphere with free charge Q at the centre. Consider a uniform", " and isotropic dielectric sphere of radius R and relative permittivity r. Let a free charge Q be placed at its centre. Inside the sphere, Eq. (10.32) implies that D = Q 4\u03c0r2 \u02c6x (r < R). (10.36) There are no free charges outside the sphere, so \u2207 \u00b7 D = 0 there. The boundary (or junction) condition on D follows from integrating \u2207 \u00b7 D = \u03c1free over a Gaussian pillbox at the surface. Since there is no free charge there, the perpendicular component of D is continous, so that Eq. (10.36) also holds for r > R. The electric \ufb01eld follows easily from D = 0rE. Inside the sphere, we have E = Q 4\u03c00rr2 \u02c6x (r < R). (10.37) This is indeed smaller than the value in vacuum by a factor of r. Since the polarization P = 0\u03c7eE is divergence-free for 0 < r < R, the only bound charges are at the centre and the surface of the sphere. The bound charge at the centre must combine with the free charge Q to give the electric \ufb01eld in Eq. (10.37), so that Qbound + Q = Q/r \u21d2 Qbound = Q 1 \u2212 r r. The surface charge at r = R follows from \u03c3bound = P \u00b7 \u02c6n, and so 4\u03c0R2\u03c3bound = Q r \u2212 1 r, (10.38) (10.39) where we have used \u03c7e = r \u2212 1. It follows that the total bound charge is zero, as it must be since the dielectric is electrically neutral. Finally, the electric \ufb01eld outside the sphere is simply E = Q 4\u03c00r2 \u02c6x (r > R). (10.40) This is the same as if the sphere were not there, a consequence of the spherical symmetry and the vanishing of the total bound charge. Note that the perpendicular component of E is discontinuous due to the presence of the surface charge. 66 10.3 Magnetising \ufb01eld H We now substitute Jmicro = Jfree + Jbound into the averaged Maxwell equation \u2207 \u00d7 B = \u00b50Jmicro + \u00b500 \u2202E \u2202t. (10.41) Retaining only the contributions to the bound current from electric and magnetic dipoles, we have \u2207 \u00d7 B = \u00b50Jfree + \u00b50 \u2207 \u00d7 M + \u2202P \u2202t + \u00b500 \u2202E \u2202t \u21d2 \u2207 \u00d7 (B \u2212 \u00b50M) = \u00b50Jfree + \u00b50 \u2202 \u2202t (0E + P). (10.42) If we introduce the magnetising \ufb01eld H with \u00b50H(t, x) \u2261 B(t, x) \u2212 \u00b50M(t, x), (10.43) and recalling the electric displacement D = 0E + P, we can write \u2207 \u00d7 H = Jfree + \u2202D \u2202t. (10.44) We see that the magnetising \ufb01eld plays a similar role to the electric displacement, in that the bound currents due to the magnetisation are absorbed into H. Note also how the contribution to the bound current from \u2202P/\u2202t got absorbed into \u2202D/\u2202t, leaving only the free current as a source. In linear, isotropic media the magnetisation can be written as \u00b50M(t, x) = \u03c7m 1 + \u03c7m B(t, x), (10.45) where \u03c7m is the dimensionless magnetic susceptibility. linearly related, It follows that B and H are also where the relative permeability is B(t, x) = \u00b50\u00b5rH(t, x), \u00b5r = 1 + \u03c7m. (10.46) (10.47) [The simplicity of this relation is reason for the rather odd-looking way that \u03c7m is introduced in Eq. (10.45).] The magnetic susceptibility is typically \u03c7m \u223c O(10\u22125). For diamagnets (M antiparallel to B), \u22121 < \u03c7m < 0 and so 0 < \u00b5r < 1; for paramagnets (M parallel to B), \u03c7m > 0 and so \u00b5r > 1. For ferromagnets, the relation between B and H is non-linear and depends on the history of preparation of the material. Finally, we gather together the macroscopic Maxwell equations for the averaged \ufb01elds E and 67 B and their associated \ufb01elds D and H inside continuous media: \ufffd", "\ufffd \u00b7 D = \u03c1free \u2202B \u2202 = Jfree + \u2202D \u2202t. (10.48) These can be solved given the free charges and currents once the constitutive relations relating D and H to E and B are speci\ufb01ed. These are dependent on the type of material; in the simplest linear, isotropic materials they are fully speci\ufb01ed by two constants, r and \u00b5r, with D = 0rE and B = \u00b50\u00b5rH. (10.49) Later, in Sec. 12, we shall brie\ufb02y consider more complicated relations where the relation between e.g., D and E is still linear and local in space, but non-local in time. 11 Electromagnetic waves in simple dielectric me- dia Maxwell\u2019s equations admit wavelike solutions in vacuum that propagate at the speed of light. In this section we extend this idea to the propagation of electromagnetic waves in simple linear, isotropic dielectric media. We shall also consider how waves re\ufb02ect o\ufb00 boundaries between two such materials. For the simple constitutive relations in Eq. (10.49), with r and \u00b5r constants, Maxwell\u2019s equations in the absence of any free charges or currents reduce to \u2207 \u00b7 E = 0 \u2202B \u2202 = r\u00b5r c2 \u2202E \u2202t. (11.1) Taking the curl of the second and fourth equations, and using that E and B are divergencefree, we recover wave equations: \u22072E = r\u00b5r c2 \u22022E \u2202t2 and \u22072B = r\u00b5r c2 \u22022B \u2202t2. The only e\ufb00ect of the medium is to change the propagation speed from c to v = \u221a c r\u00b5r. 68 (11.2) (11.3) The wave speed in the medium is often written in terms of the index of refraction n, with v = c/n, so that \u221a n = r\u00b5r. (11.4) As we have seen, in most materials \u00b5r \u2248 1, but r can be large making v rather less than c. We can look for plane-wave solutions of the wave equations with wavevector k and (angular) frequency \u03c9: E(t, x) = E0ei(k\u00b7x\u2212\u03c9t) and B(t, x) = B0ei(k\u00b7x\u2212\u03c9t), (11.5) where it is implicit that the \ufb01elds are the real part of the complex expressions on the righthand sides. Here, E0 and B0 are constant, but generally complex, polarization vectors. The \ufb01elds are divergence-free, which requires k \u00b7 E0 = 0 and k \u00b7 B0 = 0. To satisfy the wave equations (11.2), \u03c9 and k must be related by the dispersion relation \u03c92 = v2k2. Finally, to satisfy \u2207 \u00d7 E = \u2212\u2202B/\u2202t, the polarization vectors must be related by If all components of the wavevector are real14 we have B0 = \u02c6k \u00d7 E0/v. B0 = k \u00d7 E0/\u03c9. (11.6) (11.7) Finally, note that care is needed when forming products of \ufb01elds when the \ufb01elds are represented by the real part of complex quantities. For example, E \u00d7 B = = 1 2 E0ei(k\u00b7x\u2212\u03c9t) \u00d7 B0ei(k\u00b7x\u2212\u03c9t) E0 \u00d7 B0e2i(k\u00b7x\u2212\u03c9t) + E0 \u00d7 B\u2217 0, (11.8) where we used the identity (z1)(z2) = (z1z2 + z1z\u2217 2)/2 for any complex quantities z1 and z2. If we are only interested in time-averaged products of \ufb01elds (which is generally the case when discussing net \ufb02ows of energy and momentum), we have simply E\u00d7Bt = (E0\u00d7B\u2217 0)/2. Boundary conditions The wave solutions given above are valid in uniform media, but we shall also be interested in situations where the material properties are only piecewise constant. In this case, we have boundaries between regions with di\ufb00erent r and \u00b5r, and we need to know how the \ufb01elds are related on either side of a boundary in order to patch our wave solutions for the \ufb01elds together. We shall only consider the case where these are no free surface charges or currents. The junction conditions", " then follow from integrating Maxwell\u2019s equations (10.48), with \u03c1free and Jfree both zero, across the boundary in the usual way (see IB Electromagnetism). We start 14This need not be the case; we shall meet examples shortly where some components of k are imaginary. 69 with \u2207 \u00b7 D = 0 and \u2207 \u00b7 B = 0. Integrating these over a \u201cGaussian pillbox\u201d across the surface (whose unit normal is \u02c6n) gives \u02c6n \u00b7 (D1 \u2212 D2) = 0 \u02c6n \u00b7 (B1 \u2212 B2) = 0, (11.9) (11.10) where D1 and D2 are the electric displacements either side of the surface, and similarly for B1 and B2. In words, the normal component of the electric displacement and the normal component of the magnetic \ufb01eld are continuous across the boundary. Generally, this will mean that the normal component of the electric \ufb01eld is discontinuous. The discontinuity is sourced by bound surface charge due to the di\ufb00erent macroscopic polarizations P on either side of the boundary. For the parallel \ufb01eld components, we integrate around an in\ufb01nitesimal circuit with long sides parallel to and on either side of the boundary. This gives \u02c6n \u00d7 (E1 \u2212 E2) = 0 \u02c6n \u00d7 (H1 \u2212 H2) = 0. (11.11) (11.12) In words, the parallel components of the electric \ufb01eld and magnetising \ufb01eld are continuous across the boundary. Note that, generally, the parallel components of B will be discontinuous due to the bound surface current that \ufb02ows at the surface between two media with di\ufb00erent magnetisations M. 11.1 Re\ufb02ection and refraction We now consider the re\ufb02ection of waves from the plane boundary between two semi-in\ufb01nite, linear and isotropic dielectrics. We take the boundary surface to be x = 0 and consider a plane wave incident from x < 0. We shall set \u00b5r = 1 everywhere, but allow r to change discontinuously from r,I (x < 0) to r,T (x > 0). There is an associated change in refractive index from nI to nT. The incident wave will generally scatter of the boundary, giving rise to a re\ufb02ected wave and a transmitted wave. These must have the same frequency \u03c9 as the incident wave if we are to satisfy the boundary conditions for all times. We take the incident wavevector to lie in the x-y plane at an angle of incidence \u03b8I to the x-direction: kI = kI (cos \u03b8I \u02c6x + sin \u03b8I \u02c6y), (11.13) where kI = \u03c9/vI and the wave speed vI = c/nI = c/ wave is then \u221a r,I. The electric \ufb01eld of the incident We similarly write the re\ufb02ected wave as Einc(t, x) = EI ei(kI \u00b7x\u2212\u03c9t). Eref(t, x) = ERei(kR\u00b7x\u2212\u03c9t), 70 (11.14) (11.15) and the transmitted wave as Etrans(t, x) = ET ei(kT \u00b7x\u2212\u03c9t). (11.16) Note that \u03c92 = v2 T. The magnetic \ufb01elds follow from Eq. (11.7). Note, \ufb01nally, that the total electric \ufb01eld for x < 0 is given by the sum Einc + Eref, while the \ufb01eld for x > 0 is simply Etrans. R = v2 T k2 I k2 We need to apply the boundary conditions (11.9\u201311.12) at all points in the plane x = 0 and for all times. The only way we can satisfy the boundary conditions is if all three phase factors are equal for all points in the plane, i.e., kI \u00b7 x = kR \u00b7 x = kT \u00b7 x = 0 for x = 0. (11.17) It follows that the projections of all three wavevectors into the surface are equal, and so the re\ufb02ected and transmitted wavevectors also lie in the x-y plane. Writing we have It follows that kR = kI (\u2212 cos \u03b8R \u02c6x + sin \u03b8R \u02c6y) kT = kT (cos \u03b8T \u02c6x + sin \u03b8T \u02c6y), kI", " sin \u03b8I = kI sin \u03b8R = kT sin \u03b8T. \u03b8I = \u03b8R, (11.18) (11.19) (11.20) (11.21) so that the angle of incidence equals the angle of re\ufb02ection. If we further use \u03c9 = vI kI = vT kT, Eq. (11.20) implies that sin \u03b8I vI = sin \u03b8T vT or equivalently nI sin \u03b8I = nT sin \u03b8T. (11.22) This is the law of refraction, known as Snell\u2019s law. 11.2 Fresnel equations It is We now look at the amplitudes and phases of the re\ufb02ected and transmitted waves. simplest to consider separately the two possible linear polarizations of the incident wave: EI normal to the x-y plane and EI parallel to the x-y plane. The general case can be treated as a linear combination of these. 11.2.1 Normal polarization For normal polarization, the electric \ufb01eld of the incident wave is normal to the plane of incidence spanned by kI and the normal to the surface. In our set-up, this means we take EI = EI \u02c6z for some complex amplitude EI. The electric \ufb01elds of the re\ufb02ected and transmitted waves 71 are necessarily also along \u02c6z. Continuity of the perpendicular component of D is therefore identically satis\ufb01ed, while continuity of the parallel component of E requires15 EI + ER = ET. (11.23) The magnetic \ufb01elds follow from Eq. (11.7); we have BI = BR = BT = EI vI ER vI ET vT (sin \u03b8I \u02c6x \u2212 cos \u03b8I \u02c6y) (sin \u03b8I \u02c6x + cos \u03b8I \u02c6y) (sin \u03b8T \u02c6x \u2212 cos \u03b8T \u02c6y), (11.24) (11.25) (11.26) where we have used \u03b8R = \u03b8I. Continuity of the perpendicular component of B gives EI + ER vI sin \u03b8I = ET vT sin \u03b8T, (11.27) but with Snell\u2019s law this just reproduces EI + ER = ET. However, a new constraint does arise from continuity of the parallel component of H (which, here, means the parallel component of B is continuous since we are taking \u00b5r = 1): EI \u2212 ER vI cos \u03b8I = ET vT cos \u03b8T. (11.28) Combining Eqs (11.27) and (11.28), and expressing the velocities in terms of the refractive indices, we \ufb01nd ER EI ET EI = = nI cos \u03b8I \u2212 nT cos \u03b8T nI cos \u03b8I + nT cos \u03b8T 2nI cos \u03b8I nI cos \u03b8I + nT cos \u03b8T. Note that nT cos \u03b8T can be related to sin \u03b8I via Snell\u2019s law: nT cos \u03b8T = T \u2212 n2 n2 I sin2 \u03b8I. (11.29) (11.30) (11.31) The amplitude ratios (11.29) and (11.30) are plotted in Fig. 1 for nI = 1 and nT = 1.5, corresponding to a wave incident from air onto glass. 15If we retain components of the re\ufb02ected and transmitted electric \ufb01elds in the plane of incidence in the calculation, we have two further non-trivial boundary conditions from continuity of \u02c6n \u00b7 D and \u02c6n \u00d7 E. These can only be satis\ufb01ed if the electric \ufb01eld components of both waves in the plane of incidence vanish. 72 Figure 1: Ratio ER/EI (left) and ET /EI (right) as a function of the angle of incidence \u03b8I for light with normal polarization incident on the plane boundary between media with nI = 1 and nT = 1.5. 11.2.2 Parallel polarization In the case of parallel polarization, the incident electric \ufb01eld is parallel to the plane of incidence, so in our set-up lies in the x-y plane. The same will be true of the re\ufb02ected and transmitted electric \ufb01elds, so we can can write EI = EI (sin \u03b8I \u02c6x \u2212 cos \u03b8I \u02c6y) ER = ER (sin \ufffd", "\ufffdI \u02c6x + cos \u03b8I \u02c6y) ET = ET (sin \u03b8T \u02c6x \u2212 cos \u03b8T \u02c6y). The magnetic \ufb01eld of the three waves are all along the z-direction, with BI = \u2212 EI vI \u02c6z, BR = \u2212 ER vI \u02c6z, BT = \u2212 ET vT \u02c6z. (11.32) (11.33) (11.34) (11.35) Continuity of the perpendicular component of B is automatically satis\ufb01ed, while continuity of the parallel component of H gives (recall, \u00b5r = 1 here) EI + ER vI = ET vT. (11.36) For the electric \ufb01eld, continuity of the parallel component of E gives EI cos \u03b8I \u2212 ER cos \u03b8I = ET cos \u03b8T, (11.37) 73 Figure 2: Ratio ER/EI (left) and ET /EI (right) as a function of the angle of incidence \u03b8I for light with parallel polarization incident on the plane boundary between media with nI = 1 and nT = 1.5. Here, Brewster\u2019s angle, where ER = 0, is \u03b8B = 56.3\u25e6. while continuity of the perpendicular component of D requires (EI + ER) sin \u03b8I v2 I = ET sin \u03b8T v2 T. (11.38) This latter equation is equivalent to Eq. (11.36) on using Snell\u2019s law. Combining Eqs (11.36) and (11.37), we \ufb01nd the amplitude ratios ER EI ET EI = = nT cos \u03b8I \u2212 nI cos \u03b8T nI cos \u03b8T + nT cos \u03b8I 2nI cos \u03b8I nI cos \u03b8T + nT cos \u03b8I. (11.39) (11.40) The are plotted in Fig. 2 for nI = 1 and nT = 1.5. Note that for normal incidence (\u03b8I = 0), the amplitude ratios for both normal and parallel polarization reduce to ER EI ET EI = \u00b1 = nI \u2212 nT nI + nT 2nI nI + nT, (11.41) (11.42) where the plus sign in ER/EI is for normal polarization and the negative sign is for parallel polarization. 74 11.2.3 Brewster\u2019s angle It can be seen from Fig. 2 that for parallel polarization there is some value of \u03b8I for which there is no re\ufb02ected wave and everything is transmitted. This is called Brewster\u2019s angle, \u03b8B. To determine \u03b8B, we note from Eq. (11.39) that ER = 0 requires nT cos \u03b8I = nI cos \u03b8T. (11.43) Combining this with Snell\u2019s law, nI sin \u03b8I = nT sin \u03b8T, we see that we must have sin \u03b8I cos \u03b8I \u2212 sin \u03b8T cos \u03b8T = 0 \u21d2 sin 2\u03b8I = sin 2\u03b8T, (11.44) so that \u03b8I + \u03b8T = \u03c0/2. It follows that cos \u03b8T = sin \u03b8I and so, from Eq. (11.43), the Brewster angle is given by tan \u03b8B = nT /nI. (11.45) As an example, for the transmission from air to glass (nT \u2248 1.5), the Brewster angle is \u03b8B \u2248 56.3\u25e6. There is no such angle for which the re\ufb02ected wave vanishes in the case of normal polarization16. If unpolarized radiation (i.e., an incoherent mix of normal and parallel polarization) is incident at the Brewster angle, the re\ufb02ected radiation will be linearly polarized with its electric \ufb01eld normal to the plane of incidence. There is a simple physical way to understand the condition \u03b8I + \u03b8T = \u03c0/2 required for no re\ufb02ection. Consider an incident wave in vacuum. The re\ufb02ected wave is really the radiation produced by the oscillating electric dipole moments induced in the transmitting material by the incident wave. Each molecule produces electric dipole radiation as an outgoing spherical wave with its characteristic angular distribution. However, when we combine coherently the radiation from the very large number of dipoles, the spherical waves only constructively interfere along a direction at angle \u03b8I to the normal and so give a plane wave propagating in the vacuum region at", " angle \u03b8I. (The transmitted wave is the superposition of the radiation \ufb01eld and the \ufb01eld of the incident wave.) For electric dipole radiation, we have seen that there is no radiation along the axis of the dipole, which, for parallel polarization, is in the plane of incidence and normal to the direction of the transmitted wavevector kT. It follows that if \u03b8I + \u03b8T = \u03c0/2, the direction along which the dipole radiation constructively interferes is the same as the axis of the dipoles and so there is no re\ufb02ected wave. 16To see this, note from Eq. (11.29) that ER = 0 would require nI cos \u03b8I = nT cos \u03b8T. Combining with Snell\u2019s law gives tan \u03b8I = tan \u03b8T so that there can be no re\ufb02ection only if \u03b8I = \u03b8T. This is only possible if the refractive indices of the two materials are the same, in which case the lack of re\ufb02ection is trivial. 75 11.2.4 Total internal re\ufb02ection Consider the case where nI > nT, so \u03b8T > \u03b8I by Snell\u2019s law. For sin \u03b8I > nT /nI, we have sin \u03b8T > 1. How should we interpret this? To do so, note that kT = kT,x\u02c6x + kT,y \u02c6y \u03c92n2 T = = \u03c9 c I,y c2 \u2212 k2 n2 T \u2212 n2 I sin2 \u03b8I 1/2 \u02c6x + kI,y \u02c6y (Snell\u2019s law) 1/2 \u02c6x + \u03c9 c nI sin \u03b8I \u02c6y. (11.46) The \u02c6x component is imaginary for sin \u03b8I > nT /nI, i.e., \u03b8I > \u03b8C where the critical angle of incidence satis\ufb01es sin \u03b8C = nT /nI. In this case, let us write \u03c9 c T \u2212 n2 n2 I sin2 \u03b8I = \u00b1i\u03ba, (11.47) (11.48) where \u03ba is real and positive. Then ei(kT \u00b7x\u2212\u03c9t) = e\u2213\u03baxei(\u03c9nI sin \u03b8I y/c\u2212\u03c9t), (11.49) where we should choose \u2212\u03ba to ensure the solution remains regular as x \u2192 \u221e for a semi-in\ufb01nite transmitting medium. This solution describes wave propagation along the surface, but exponential decay into the medium for \u03b8I > \u03b8C. Such a wave is known as an evanescent wave. The situation with \u03b8I > \u03b8C is called total internal re\ufb02ection since all the energy in the incident wave is re\ufb02ected in this case. To see this, we return to Eqs (11.29) and (11.39) for the ratio ER/EI. These expressions still hold in the case \u03b8I > \u03b8C, but we should make the replacement \u03c9nT cos \u03b8T /c = kT,x = \u00b1i\u03ba. (11.50) It is straightforward then to show that |ER|2 = |EI|2 so that the time-averaged energy \ufb02ux of the incident wave equals that of the re\ufb02ected wave. One can also show that the evanescent wave inside the medium only transports energy parallel to the surface on average; there is no \ufb02ow into the medium. Total internal re\ufb02ection has many physical applications: it dictates the geometry in which we should cut diamonds to maximise their \u201csparkle\u201d, and is the reason that we can channel electromagnetic signals in a \ufb02exible way through optical \ufb01bres. 76 12 Dispersion In the discussion above, we assumed that r and \u00b5r were constants. However, in the presence of time-dependent \ufb01elds, the amplitude and phase of the induced polarisation P and magnetisation M will generally depend on the frequency. This gives rise to frequency-dependent r(\u03c9) and \u00b5r(\u03c9), a phenomenon known as dispersion. In this section, we shall see how this frequency dependence can arise with our simple toy model for atomic polarisability, and then explore some of the consequences of this frequency dependence on wave propagation. 12.1 Atomic polarisability revisited: a simple model for r(\u03c9) Earlier, we considered a simple model for atomic polarisability by thinking about how a rigid electron cloud displaced", " relative to the nucleus in the presence of a constant applied electric \ufb01eld. If the cloud has charge \u2212q, and is displaced by d relative to the nucleus, the restoring force on it is Frestore = \u2212 q2 4\u03c00a3 d = \u2212m\u03c92 0d. (12.1) In the second equality, we have written this restoring force in terms of the reduced mass of the nucleus-cloud system (very nearly the mass of the cloud since electrons are much lighter than neutrons and protons) and the natural oscillation frequency \u03c90. Now that we are considering the time-dependent response of the atom, we should include a phenomenological damping force Fdamping = \u2212\u03b3m \u02d9d, (12.2) where \u03b3 is some constant with dimensions of frequency. This term is included to allow the oscillating atoms to absorb energy, for example to mimic quantum excitation of the electrons in the atom. If we now apply a time-dependent electric \ufb01eld E(t) to the atom, the equation of motion for d is \u00a8d = \u2212 q m E(t) \u2212 \u03c92 0d \u2212 \u03b3 \u02d9d. (12.3) Here, we have assumed that the applied electric \ufb01eld varies slowly in space compared to the atomic scale, in which case the electric \ufb01eld E(t) can be evaluated at some \ufb01xed location within the atom. In this case, Eq. (12.3) becomes the equation of motion of a damped harmonic oscillator. We consider an applied \ufb01eld of the form E(t) = E0e\u2212i\u03c9t (again, it is left implicit that we take the real part), and look for solutions of the form d(t) = d0e\u2212i\u03c9t. This steady-state 77 solution is d0 = \u2212 qE0 m 1 0 \u2212 \u03c92 \u2212 i\u03b3\u03c9 \u03c92, so that the atomic polarisability p = \u03b1E, with p = \u2212qd, is \u03b1 = q2/m 0 \u2212 \u03c92 \u2212 i\u03b3\u03c9 \u03c92. (12.4) (12.5) Note that this is frequency dependent and, generally, complex. The latter means that the electric dipole moment need not oscillate in phase with the applied electric \ufb01eld. As \u03c9 \u2192 0, we recover the earlier result (9.2) for the polarizability in static \ufb01elds noting that m\u03c92 0 = q2/(4\u03c00a3). If the atoms have number density n, the macroscopic polarization is approximately P = np = n\u03b1E. Since P = 0(r \u2212 1)E, we have a frequency-dependent, complex relative permittivity r(\u03c9) = 1 + nq2 m0(\u03c92 0 \u2212 \u03c92 \u2212 i\u03b3\u03c9). (12.6) Although this classical toy model for atomic polarizability is clearly over-simpli\ufb01ed, it does capture the basic physics of dispersion. Normal dispersion and anomalous dispersion It is generally the case that \u03b3 \u03c90, so that the dimensionless quality factor of the oscillator Q \u2261 \u03c90/\u03b3 is large. In this case, the imaginary part of r(\u03c9) is small compared to the real part except for frequencies in the vicinity of \u03c90. For \u03c9 < \u03c90, the real part of r(\u03c9) is greater than unity, while for \u03c9 > \u03c90 it is less than one. At the resonant frequency \u03c90, r(\u03c9) is pure imaginary. This behaviour is illustrated in Fig. 3. Except in the vicinity of \u03c90, the real part of r(\u03c9) is an increasing function of frequency. Such behaviour is known as normal dispersion. Near the resonance, where the imaginary part of r(\u03c9) becomes appreciable, the real part decreases with frequency; this is referred to as anomalous dispersion. We shall see in the next section that for such frequencies there is signi\ufb01cant attenuation of waves as the electromagnetic energy in the wave is dissipated in the medium. This arises since the bound current \u2202P/\u2202t is almost in phase with the electric \ufb01eld near the resonance. 12.2 Electromagnetic waves in dispersive media We now explore some of the key di\ufb00erences between propagation of electromagnetic waves in dispersive and non-dispersive media. 78 Figure 3: Real part of r(\u03c9) \u2212 1 (left) and imaginary part (right) for \u03c90 = 1, nq2/(m0\u03c92 0) = 1 and quality factor Q = \u03c90/\u03b3 = 10.", " We look for plane-wave solutions of Maxwell\u2019s equations at (real) frequency \u03c9. It is generally a good approximation to take \u00b5r = 1, but we shall allow for a frequencydependent r(\u03c9). Speci\ufb01cally, we have E(t, x) = E0(\u03c9)ei(k\u00b7x\u2212\u03c9t) D(t, x) = 0r(\u03c9)E0(\u03c9)ei(k\u00b7x\u2212\u03c9t), and for the magnetic \ufb01eld B(t, x) = B0(\u03c9)ei(k\u00b7x\u2212\u03c9t) 1 \u00b50 H(t, x) = B0(\u03c9)ei(k\u00b7x\u2212\u03c9t). (12.7) (12.8) (12.9) (12.10) We assume there are no free charges or currents present, so two of Maxwell\u2019s equations are \u2207 \u00b7 (\u03c9)k \u00b7 E0(\u03c9) = 0 k \u00b7 B0(\u03c9) = 0. (12.11) (12.12) Provided that17 r(\u03c9) = 0, these equations imply that the electric and magnetic \ufb01elds are transverse to the wavevector k. The other Maxwell equations are \u2207 \u00d7 E = \u2212 \u2207 \u00d7 H = \u2202B \u2202t \u2202D \u2202t \u21d2 \u21d2 k \u00d7 E0(\u03c9) = \u03c9B0(\u03c9) k \u00d7 B0(\u03c9) = \u2212 r(\u03c9) c2 \u03c9E0(\u03c9). (12.13) (12.14) 17We shall meet an example where this condition does not hold in Sec. 13, where we consider wave propagation in conductors. 79 Taking the cross-product of the latter equation with k, and using k \u00b7 B0(\u03c9) = 0, we \ufb01nd the dispersion relation c2k2 = r(\u03c9)\u03c92. (12.15) As r(\u03c9) is generally complex, the wavevector will be also. We already met the idea of a wavevector with complex components in our discussion of total internal re\ufb02ection in Sec. 11.2.4, where we saw that it corresponds to a decay of the amplitude of the wave in space. (Note, however, that there k2 was still real.) If we consider a wave propagating in the z-direction, so that k = k\u02c6z for complex k, we have c(k1 + ik2) = (r,1 + ir,2)1/2\u03c9, (12.16) where k1 and k2 are the real and imaginary parts, respectively, of k, and similarly for r,1 and r,2. The resulting wave is given by the real part of E(t, x) = E0(\u03c9)e\u2212k2zei(k1z\u2212\u03c9t). (12.17) This is decaying along the positive z-direction if k2 > 0. Finally, we note a further consequence of an imaginary component of r(\u03c9): we see from \u03c9B0(\u03c9) = k \u00d7 E0(\u03c9) that there is a phase shift between the oscillations of the electric and magnetic \ufb01elds for complex k. 12.2.1 Phase and group velocity We have seen that for a wave of frequency \u03c9, a point of constant phase, such as a zero of the oscillations, has k1z \u2212 \u03c9t = const., where k1 is the real part of the wavevector. The speed of such a point of constant phase de\ufb01nes the phase velocity, vp = \u03c9/k1. (12.18) In a dispersive medium, vp will generally depend on frequency. This means that refraction will also be frequency dependent, allowing a prism to split white light into its constituent colours. A further consequence of a frequency-dependent vp comes from considering localised wave pulses or packets. These can be formed by superposing waves with a range of frequencies. Since the individual frequency components propagate at di\ufb00erent (phase) velocities, the shape of the pulse will typically change as it propagates (see below). An important property of a wavepacket is the speed at which the peak of the envelope, where the energy is concentrated, moves. A typical situation is to superpose waves with a range of frequencies centred around some central frequency. To simplify our discussion a little, we shall ignore the imaginary part of r(\u03c9) so that the wavevector is 80 real. If we invert the dispersion relation to view \u03c9 as a function of k, we can consider a wavepacket of the form E(t, z) = dk \u221a 2\u03c0 E0(k)ei", "(kz\u2212\u03c9(k)t), (12.19) where E0(k) is sharply peaked about some central wavenumber k0. As a concrete example, consider E0(k) \u221d exp[\u2212(k \u2212 k0)2/(2\u03c32)]. At t = 0, the wavepacket is simply the real part of the Fourier transform of E0(k), and so E(t = 0, z) \u221d cos(k0z)e\u2212z2\u03c32/2. (12.20) For the typical case of \u03c3 k0, this is a wide Gaussian modulation of a rapid oscillation at wavenumber k0. Now consider what happens at later times. For a sharply-peaked E0(k), we can expand the dispersion relation \u03c9(k) about k0, so the exponent kz \u2212 \u03c9(k)t = kz \u2212 \u03c9(k0)t \u2212 (k \u2212 k0) d\u03c9 dk k0 + \u00b7 \u00b7 \u00b7 = [k0z \u2212 \u03c9(k0)t] + (k \u2212 k0)[z \u2212 vg(k0)t] + \u00b7 \u00b7 \u00b7, (12.21) where we have introduced the group velocity vg = d\u03c9/dk. (12.22) If we retain just these \ufb01rst two terms, the \ufb01eld at time t is E(t, z) \u221d cos[k0(z \u2212 vp(k0)t)]e\u2212[z\u2212vg(k0)t]2\u03c32/2. (12.23) We see that the oscillation propagates at the phase velocity vp(k0), while the modulation propagates at the group velocity vg(k0). If the medium is non-dispersive, these two velocities are equal (and independent of k0). Moreover, in this case the expansion in Eq. (12.21) truncates at the order shown, and the wavepacket simply translates along at speed vp = vg without changing its shape. In contrast, for a dispersive medium vp = vg and the oscillation will propagate at a di\ufb00erent speed to the wavepacket modulation. Furthermore, inclusion of higher-order terms in the expansion in Eq. (12.21) will generally lead to a change in shape of the modulation as the wavepacket propagates (see Examples 3). In a dispersive medium, the refractive index is frequency dependent, n(\u03c9) = c/vp(\u03c9). (12.24) We can use this to relate the group and phase velocities (again ignoring any imaginary part of the wavenumber): 1 vg = dk d\u03c9 = d d\u03c9 n\u03c9 c = 1 vp + \u03c9 c dn d\u03c9. (12.25) 81 In regions of normal dispersion, dn/d\u03c9 > 0 and vg < vp; for anomalous dispersion, dn/d\u03c9 < 0 and vg > vp. In some situations, the phase velocity can exceed the speed of light in vacuum. An example of this occurs for the simple model of Eq. (12.6) for \u03c9 \u03c90. In this limit, r(\u03c9) \u2248 1 \u2212 nq2 m0\u03c92. Assuming that \u03c92 > nq2/(m0), the wavevector is approximately real with c2k2 = \u03c92 \u2212 nq2 m0. It follows that the phase velocity vp(\u03c9) = c 1 \u2212 nq2/(m0\u03c92), (12.26) (12.27) (12.28) which is greater than c. This should not cause alarm \u2013 a single-frequency wave extends throughout all of space and cannot be used to communicate any information. To do so requires sending a wave pulse, the peak of which will propagate at the group velocity. Here, vg(\u03c9)vp(\u03c9) = c2, so the group velocity is indeed less than c. 12.3 Causality and the Kramers\u2013Kronig relations A further consequence of the frequency dependence of r(\u03c9) is a non-locality in time of the connection between D(t, x) and E(t, x). To see this, consider writing a general D(t, x) in terms of its Fourier transform in time: D(t, x) = \u221e \u2212\u221e d\u03c9 \u221a 2\u03c0 \u02dcD(\u03c9, x)e\u2212i\u03c9t, (12.29) where \u02dcD(\u03c9, x) is the Fourier transform of the real \ufb01eld D(t, x) and so satis\ufb01es", " \u02dcD\u2217(\u03c9, x) = \u02dcD(\u2212\u03c9, x). We have seen that for oscillations at frequency \u03c9, the \ufb01eld D and E at each point in space are related by a factor 0r(\u03c9); it follows that D(t, x) = 0E(t, x) + 0 \u221e \u2212\u221e d\u03c9 \u221a 2\u03c0 [r(\u03c9) \u2212 1] \u02dcE(\u03c9, x)e\u2212i\u03c9t, (12.30) where \u02dcE(\u03c9, x) is the Fourier transform of E(t, x). Note that in extending to negative \u03c9, we must have \u2217 r(\u03c9) = r(\u2212\u03c9) from reality of D and E. In writing Eq. (12.30) in terms of r(\u03c9) \u2212 1, we have isolated the contribution to D from the polarization in 82 the material in the second term on the right-hand side. It is this term that can be non-local in time. If we express \u02dcE(\u03c9, x) as a Fourier integral, we \ufb01nd D(t, x) = 0E(t, x) + 0 = 0E(t, x) + 0 d\u03c9 \u221a 2\u03c0 [r(\u03c9) \u2212 1] dt \u221a 2\u03c0 dt G(t \u2212 t)E(t, x), \u221e \u2212\u221e \u221e \u2212\u221e E(t, x)e\u2212i\u03c9(t\u2212t) (12.31) where we have introduced the response function, G(t) = \u221e \u2212\u221e d\u03c9 2\u03c0 [r(\u03c9) \u2212 1]e\u2212i\u03c9t. (12.32) Note that this is simply the Fourier transform of r(\u03c9) \u2212 1. If r(\u03c9) is independent of frequency, G(t) \u221d \u03b4(t) and the connection between D and E is local in time. However, a frequency dependence of r(\u03c9) will introduce some non-locality. Example: response function for the simple model of atomic polarisability. For the model in Sec. 12.1, we have r(\u03c9) \u2212 1 = m0(\u03c92 0 \u2212 \u03c92 \u2212 i\u03b3\u03c9) nq2. (12.33) Considered as a function in the complex plane, this has simple poles at \u03c9\u00b1 where Note that these are both in the lower half plane. It follows that \u03c9\u00b1 = \u00b1(\u03c92 0 \u2212 \u03b32/4)1/2 \u2212 i\u03b3/2. G(t) = \u2212 nq2 m0 \u221e \u2212\u221e d\u03c9 2\u03c0 e\u2212i\u03c9t (\u03c9 \u2212 \u03c9+)(\u03c9 \u2212 \u03c9\u2212). (12.34) (12.35) We can evaluate this integral by contour integration. For t < 0, we can close the contour in the upper half-plane since e\u2212i\u03c9t then tends to zero exponentially fast with radius R along the semi-circle |\u03c9| = R. As there are no poles in the upper half-plane, we see that G(t) = 0 for t < 0. For t > 0, we must close the contour in the lower half-plane, in which case we pick up a contribution to the integral from the two poles. Summing the residues, we \ufb01nd G(t) = nq2 \u03c92 0 \u2212 \u03b32/4 m0 e\u2212\u03b3t/2 sin \u03c92 0 \u2212 \u03b32/4 t \u0398(t), (12.36) where \u0398(t) is the Heaviside function that ensures G(t) = 0 for t < 0. This is an oscillation at the natural (damped) frequency of the oscillator with a decay of characteristic time 1/\u03b3. The decay time is typically much longer than the oscillation period since \u03b3 \u03c90. This means that the non-locality of the response between D and E is con\ufb01ned to time di\ufb00erences of the order of 1/\u03b3. In a physical (i.e., quantum) model, \u03b3 is related to the frequency width of the spectral lines emitted by the atom in an excited state, and 1/\u03b3 \u223c 10\u22128 s. 83 In the above example, the response function vanishes for t < 0. This is essential for a causal response: the polarisation of the medium at time t cannot depend on the electric \ufb01eld at later times. If we consider evaluating Eq. (12.32) by contour integration (as we did in the example above), for t < 0 we can close the contour in the upper half-plane. If the response", " function is to vanish, so must the contour integral and hence the contour must not enclose any poles of the integrand. We see that, considered as a function in the complex plane, r(\u03c9) must be analytic in the upper half-plane for the response function to be causal. 12.3.1 Kramers\u2013Kronig relations The fact that r(\u03c9) is analytic in the upper half-plane can be used to relate the real and imaginary parts along the real axis. Consider the following principal-value integral along the real axis: I(\u03c9) = P \u221e \u2212\u221e [r(\u03c9) \u2212 1] \u03c9 \u2212 \u03c9 d\u03c9. (12.37) Taking the principal value18 is necessary since the integrand has a pole at \u03c9 = \u03c9. We can evaluate I(\u03c9) by contour integration by considering a closed contour that runs along the real axis from \u03c9 = \u2212\u221e, makes a diversion around the pole on the axis at \u03c9 = \u03c9 by following clockwise a semi-circle of radius \u03b4 (we shall ultimately consider the limit \u03b4 \u2192 0) centred on \u03c9 = \u03c9, and is completed by a semi-circle at in\ufb01nity in the upper half-plane. Since the integrand has no poles within this contour, the total integral is zero. Assuming that r(\u03c9) \u2212 1 vanishes as |\u03c9| \u2192 \u221e, there is no contribution from the large semi-circular path in the upper half-plane. It follows that I(\u03c9) + I1(\u03c9) = 0, where I1(\u03c9) is the contribution from the small semi-circular contour about \u03c9 = \u03c9. This can be evaluated by writing \u03c9 \u2212 \u03c9 = \u03b4ei\u03d5, so that I1(\u03c9) = lim \u03b4\u21920 0 [r (\u03c9 + \u03b4ei\u03d5) \u2212 1] \u03b4ei\u03d5 i\u03b4ei\u03d5d\u03d5 \u03c0 = \u2212i\u03c0[r(\u03c9) \u2212 1], which gives r(\u03c9) = 1 + 1 i\u03c0 P \u221e \u2212\u221e [r(\u03c9) \u2212 1] \u03c9 \u2212 \u03c9 d\u03c9. 18Recall that the principal value is the limit P \u221e \u2212\u221e f (\u03c9) d\u03c9 = lim \u03b4\u21920 \u03c9\u2212\u03b4 \u2212\u221e f (\u03c9) d\u03c9 + f (\u03c9) d\u03c9, \u221e \u03c9+\u03b4 when the integrand f (\u03c9) has a single pole at \u03c9 = \u03c9. 84 (12.38) (12.39) Finally, taking the real and imaginary parts gives r(\u03c9) = 1 + 1 \u03c0 P r(\u03c9) = \u2212 1 \u03c0 P \u221e d\u03c9 r(\u03c9) \u03c9 \u2212 \u03c9 [r(\u03c9) \u2212 1] \u03c9 \u2212 \u03c9 \u2212\u221e \u221e \u2212\u221e d\u03c9. (12.40) (12.41) These are the Kramers\u2013Kronig relations. They are of very general validity, following from causality of the response function alone. Note how they provide a link between the real and imaginary parts of r(\u03c9). This is of practical interest since, for example, if one can measure the imaginary part of r(\u03c9) by studying absorption as a function of frequency (recall that the imaginary part of r(\u03c9) is associated with absorption of radiation by the material since it describes oscillations of the macroscopic polarisation that are in phase quadrature with the applied electric \ufb01eld), one can infer the real part of r(\u03c9). The Kramers\u2013Kronig relations are sometimes written in an alternative form that involves only positive frequencies. Using the symmetry \u2217 r(\u03c9) = r(\u2212\u03c9) for real \u03c9, we deduce that along the real axis r(\u03c9) is an even function and r(\u03c9) is odd. With this property, Eqs (12.40) and (12.41) can be rewritten as r(\u03c9) = 1 + P r(\u03c9) = \u2212 P 2 \u03c0 2\u03c9 \u03c0 \u221e 0 \u221e 0 \u03c9r(\u03c9) \u03c92 \u2212 \u03c92 d\u03c9 [r(\u03c9) \u2212 1] \u03c92 \u2212 \u03c92 d\u03c9. (12.42) (12.43) 13 Electromagnetic waves in conductors So far, we have discussed the propagation of electromagnetic waves in insulating materials. We now consider the case of conductors where there can be free currents (and charges) due to the mobile electrons. We begin in Sec. 13.1 with a simple dynamical model \u2013 the Drude model \u2013 for the conductivity of a metal, before discussing in Sec. 13.2 the consequences of this", " model for wave propagation in conductors. 13.1 Drude model In a metal, the outermost electrons of the constituent atoms form a delocalised \u201csea\u201d of mobile electrons surrounding a lattice of ions (nuclei plus tightly-bound inner electrons). These electrons move in all directions and have a range of energies up to what is called the Fermi energy (typically a few eV), since the Pauli exclusion principle forbids them from all occupying the same quantum state. The Fermi energy is much 85 larger than the typical energy of thermal excitations, O(kBT ) \u223c 25 meV at temperature T = 300 K. A given electron has its velocity randomised by \u201ccollisions\u201d. A proper quantum treatment shows that such collisions occur due to anything that destroys the perfect periodicity of the ionic lattice; this includes thermal vibrations of the lattice, defects or impurities, and even the \ufb01nite extent of the sample. Here, we denote the mean time between collisions (often simply called the scattering time) by \u03c4. For reference, in a metal like copper \u03c4 \u223c 10\u221214 s. The mean distance that an electron moves between collisions is called the mean-free path. Taking the speed that corresponds to the Fermi energy, this gives typical velocities of 106 m s\u22121, and so mean-free paths of O(10\u22128) m. In the absence of an applied \ufb01eld, the average electron velocity at a point (i.e., averaged over all electrons present at that point) is zero and so is the free current. However, if we apply an electric \ufb01eld, the mobile electrons will be accelerated by the \ufb01eld and there will be a collective change in their velocity with time. This gives a non-zero average electron velocity and hence free current. The dynamics of the average electron velocity arises from the competition between acceleration from the applied \ufb01eld and collisions that tend to return the average velocity to zero. The Drude model 19 is a classical description of the non-relativistic average electron velocity v. In a uniform electric \ufb01eld E(t), it models the dynamics of v with me dv dt = \u2212eE \u2212 m v \u03c4, (13.1) where me is the electron mass and \u2212e is the charge. The \ufb01rst force on the right-hand side is the Lorentz force from the electric \ufb01eld (we can generally ignore the magnetic \ufb01eld). The second term on the right of Eq. (13.1) is a phenomenological drag force that describes the e\ufb00ect of collisions. Equation (13.1) assumes a uniform \ufb01eld, but is actually applicable also for spatiallyvarying \ufb01elds that vary slowly compared to the electron mean-free path. This is because at any point in space, the electrons there have typically travelled one mean-free path since their last collision (at which the direction of their velocity was randomised). The average electron velocity at a point is therefore determined by the integrated e\ufb00ect of the electric \ufb01eld at earlier times of order \u03c4, and throughout a region of space with size of the order of the mean-free path. Consider applying an oscillating electric \ufb01eld E(t) = [E(\u03c9)e\u2212i\u03c9t] with wavelength large compared to the electron mean-free path. We look for a solution of Eq. (13.1) for the average velocity of the form v(t) = [v(\u03c9)e\u2212i\u03c9t], which requires 1 \u03c4 E(\u03c9). (13.2) v(\u03c9) \u2212 i\u03c9 = \u2212 e me 19The model was proposed in 1900 by Drude. This was before the invention of quantum theory so he (wrongly!) envisaged collisions as arising from the mobile electrons Coulomb scattering o\ufb00 static charged ions. 86 The macroscopic free current at any point is Jfree(t, x) = \u2212enev(t, x), (13.3) where ne is the number density of the mobile electrons. It follows that the current also oscillates as J = [J(\u03c9)e\u2212i\u03c9t], with J(\u03c9) = nee2\u03c4 me 1 1 \u2212 i\u03c9\u03c4 E(\u03c9). This is just Ohm\u2019s law, but with a frequency-dependent optical conductivity where the DC conductivity is \u03c3(\u03c9) = \u03c3DC 1 \u2212 i\u03c9\u03c4, \u03c3DC = nee2\u03c4 me. (13.4) (13.5) (13.6) Only when the frequency is small compared to the scattering rate, i.e., \u03c9\u03c4 1, do we", " recover the DC conductivity. Note that the optical conductivity is generally complex, with \u03c3DC 1 + \u03c92\u03c4 2, \u03c3(\u03c9) = \u03c3(\u03c9) = and \u03c3DC\u03c9\u03c4 1 + \u03c92\u03c4 2. (13.7) Since J(\u03c9) = \u03c3(\u03c9)E(\u03c9), the real part of the optical conductivity generates current in phase with the applied \ufb01eld. This implies energy dissipation, with work being done on the mobile electrons by the oscillating electric \ufb01eld. However, this does not increase the energy of the electrons; rather the energy is transferred to vibrations of the ions through collisions. For \u03c9\u03c4 1, the optical conductivity is purely imaginary and independent of the scattering time \u03c4 (recall that \u03c3DC \u221d \u03c4 ). Here, the oscillations are so rapid that collisions are irrelevant and the mobile electrons respond to the electric \ufb01eld as if they were completely free electrons. 13.2 Propagation of waves in conductors We now consider the propagation of electromagnetic waves in conductors. We account for both the dielectric response of the lattice of ions, through a frequency-dependent permittivity r(\u03c9), and the macroscopic free current and charge density from the mobile electrons. We ignore any magnetic response (so take \u00b5r = 1). We look for plane-wave solutions of the familiar form E(t, x) = E0(\u03c9)ei(k\u00b7x\u2212\u03c9t) B(t, x) = B0(\u03c9)ei(k\u00b7x\u2212\u03c9t). 87 (13.8) (13.9) The electric displacement is D(t, x) = 0r(\u03c9)E0(\u03c9)ei(k\u00b7x\u2212\u03c9t), (13.10) and the free current Jfree(t, x) = \u03c3(\u03c9)E0(\u03c9)ei(k\u00b7x\u2212\u03c9t). The continuity equation for the free current, \u2202\u03c1free/\u2202t + \u2207 \u00b7 Jfree = 0, generally implies that there will be non-zero free charge density with (13.11) \u03c1free(t, x) = \u03c3(\u03c9) \u03c9 k \u00b7 E0(\u03c9)ei(k\u00b7x\u2212\u03c9t). Note that \u03c1free vanishes if the electric \ufb01eld is transverse, k \u00b7 E0(\u03c9) = 0. The Maxwell equation \u2207 \u00b7 D = \u03c1free gives 0r(\u03c9) + i \u03c3(\u03c9) \u03c9 k \u00b7 E0(\u03c9) = 0, and \u2207 \u00b7 B = 0 gives k \u00b7 B0(\u03c9) = 0. (13.12) (13.13) (13.14) We see that the magnetic \ufb01eld is always transverse and, generally, the electric \ufb01eld is also. The exception is if 0r(\u03c9) + i\u03c3(\u03c9)/\u03c9 = 0 for some frequency \u03c9; we shall discuss this case separately in Sec. 13.2.3. The remaining Maxwell equations are \u2207 \u00d7 H = Jfree + \u2202D/\u2202t, which implies that \u03c9 c2 \u03c3(\u03c9) 0\u03c9 r(\u03c9) + i E0(\u03c9), (13.15) k \u00d7 B0(\u03c9) = \u2212 and \u2207 \u00d7 E = \u2212\u2202B/\u2202t, which gives k \u00d7 E0(\u03c9) = \u03c9B0(\u03c9). (13.16) These equations have exactly the same form as those we met in Sec. 12.2, when discussing electromagnetic waves in dispersive media, provided that we replace r(\u03c9) with an e\ufb00ective permittivity: r(\u03c9) \u2192 e\ufb00 r (\u03c9) = r(\u03c9) + i \u03c3(\u03c9) 0\u03c9. (13.17) Here, the \ufb01rst term in e\ufb00 r (\u03c9) arises from the dielectric properties of the ions, while the second term arises from the free currents. We can use all of our previous results; in particular, the dispersion relation is c2k2 = e\ufb00 r (\u03c9)\u03c92. (13.18) We now explore the consequences of this dispersion relation taking the conductivity to be described by the Drude form (13.5). 88 13.2.1 Low-frequency behaviour If we write r(\u03c9) in terms of its real and imaginary parts, r(\u03c9) = r,1(\u03c9) + ir,2(\u03c9), we have e\ufb00 r (\u03c9) = r,1(\u03c9) + ir,2(\u03c9) + i 0\u03c9 \u03c3DC 1 \u2212 i\u03c9", "\u03c4. (13.19) For \u03c9\u03c4 1, the optical conductivity \u03c3(\u03c9) \u2248 \u03c3DC. Moreover, for the simple model of r(\u03c9) in Eq. (12.6), r,1 r,2 except near the resonance at \u03c90, and r,1 tends to a constant as \u03c9 \u2192 0. It follows that at su\ufb03ciently low frequencies, e\ufb00 r (\u03c9) \u2248 i \u03c3DC 0\u03c9. (13.20) At such low frequencies, it is the mobile electrons that totally dominate the electromagnetic properties of the conductor. Taking the wavevector k = k\u02c6z for complex k = k1 + ik2, we have, from Eq. (13.18), k1 + ik2 = \u00b1 \u00b50\u03c9\u03c3DC 2 (1 + i). (13.21) For propagation along the positive z-direction we should select the positive root, in which case where the skin-depth E(t, x) = E0(\u03c9)e\u2212z/\u03b4ei(z/\u03b4\u2212\u03c9t), \u03b4 \u2261 2 \u00b50\u03c9\u03c3DC. (13.22) (13.23) We see that we have a propagating wave, but with amplitude that decays along the propagation direction over a characteristic length given by the skin depth. Note that \u03b4 decreases with increasing frequency so that higher frequency electromagnetic \ufb01elds are more con\ufb01ned to the surfaces of conductors. The amplitude of the wave decays as energy is dissipated by the free current, which oscillates in phase with the electric \ufb01eld. We can \ufb01nd the magnetic \ufb01eld from k \u00d7 E0(\u03c9) = \u03c9B0(\u03c9), so that B0(\u03c9) = ei\u03c0/4 \u00b50\u03c3DC \u03c9 \u02c6z \u00d7 E0(\u03c9). (13.24) The magnetic \ufb01eld is \u03c0/4 out of phase with the electric \ufb01eld, and the amplitude of the magnetic \ufb01eld can be much larger than the electric \ufb01eld in good conductors at low frequency. 89 13.2.2 High-frequency behaviour We now consider high frequencies, \u03c9\u03c4 1. In this limit, collisions play no role and \u03c3(\u03c9) \u2248 i\u03c3DC/(\u03c9\u03c4 ), which is independent of \u03c4. We shall also assume that \u03c9 \u03c90 so that r(\u03c9) \u2248 1. It follows that \u03c3DC 0\u03c92\u03c4 \u03c92 p \u03c92, where we have introduced the plasma frequency e\ufb00 r \u2248 1 \u2212 = 1 \u2212 \u03c92 p \u2261 nee2 0me. (13.25) (13.26) In a typical metal, \u03c9p \u223c 1016 rad s\u22121. There is qualitatively di\ufb00erent behaviour for \u03c9 > \u03c9p or \u03c9 < \u03c9p, as follows. \u2022 \u03c9 > \u03c9p: in this case, e\ufb00 r (\u03c9) > 0 and the wavenumber k is real. We have simple wave propagation in the conductor with no dissipation (because Jfree and E are out of phase). This shows that metals are partially transparent above their plasma frequency. The waves are dispersive, with dispersion relation c2k2 = \u03c92 \u2212 \u03c92 p. The phase velocity is superluminal: vp(\u03c9) = c, 1 \u2212 \u03c92 p/\u03c92 (13.27) (13.28) but the group velocity is less than the speed of light since vg(\u03c9) = c2/vp(\u03c9). \u2022 \u03c9 < \u03c9p: in this limit20, e\ufb00 r (\u03c9) < 0 and the wavenumber k is purely imaginary. We therefore have an evanescent wave: wave propagation in a conductor is not possible below the plasma frequency (but above the collision rate). In this case, all of the energy of an electromagnetic wave that is incident on a conductor will be re\ufb02ected. Since the plasma frequency typically corresponds to ultraviolet radiation, optical light has frequency below the plasma frequency and is strongly re\ufb02ected from conductors. This is why metals are shiny. 20Given our assumptions that \u03c9\u03c4 1 and \u03c9 \u03c90, this requires \u03c9p max(\u03c4 \u22121, \u03c90). The condition on the collision rate is generally met, since \u03c4 \u22121 \u223c 1014 s\u22121, but for \u00afh\u03c90 of the order of atomic transition energies the condition on \u03c90 is not so obviously met. 90 13.2.3 Plasma oscillations We now return to Eq. (13.13), which", " we can rewrite as 0e\ufb00 r (\u03c9)k \u00b7 E0(\u03c9) = 0. (13.29) As noted earlier, if e\ufb00 r (\u03c9) vanishes for some \u03c9, there are solutions with k \u00b7 E0(\u03c9) = 0. Moreover, in this case, Eqs (13.14) and (13.15) give k \u00b7 B0(\u03c9) = 0 and k \u00d7 B0(\u03c9) = 0, which imply that the magnetic \ufb01eld vanishes. Equation (13.16) further tells us that k \u00d7 E0(\u03c9) = 0, so we have a new solution that is a longitudinal wave with k parallel to E. We can \ufb01nd oscillatory solutions of this form by considering the high-frequency limit of e\ufb00 r (\u03c9) in Eq. (13.25). This vanishes if \u03c9 = \u03c9p, giving solutions E(t, x) = E0(\u03c9p)ei(k\u00b7x\u2212\u03c9pt), (13.30) with k \u00d7 E0(\u03c9) = 0. Since these solutions are longitudinal waves, the charge density is non-zero and also oscillates at \u03c9p: \u03c1free(t, x) = i0k \u00b7 E0(\u03c9p)ei(k\u00b7x\u2212\u03c9pt), (13.31) where we have approximated r(\u03c9p) \u2248 1. Note that the wavevector k is arbitrary, so, more generally, we have E(t, x) = \u2212\u2207\u03c6(x)e\u2212i\u03c9pt, (13.32) for some potential \u03c6. The spatial pro\ufb01le of the free charge density is controlled by \u22072\u03c6 and is arbitrary. These oscillations are called plasma oscillations. 91 A Review of special relativity This appendix recaps the main ideas of special relativity and reviews the streamlined tensor notation that was introduced in IB Electromagnetism and is used throughout this course. Note that the metric signature adopted here is mostly positive, i.e., the Minkowki metric \u03b7\u00b5\u03bd = diag(\u22121, +1, +1, +1). Special relativity is based on the following two postulates: \u2022 The laws of physics are the same in all inertial frames21. \u2022 Light signals in vacuum propagate rectilinearly and with the same speed c in all inertial frames. The \ufb01rst postulate means that any equation of physics must take the same form when written in terms of spacetime coordinates in di\ufb00erent inertial frames. A.1 Lorentz transformations The consequence of these two postulates is that for two inertial frames, S and S, with spatial axes aligned and with S moving at speed v = \u03b2c along the positive x-axis of S and with origins coincident at t = t = 0, the spacetime coordinates of any event are related by the (standard) Lorentz transformation: ct = \u03b3(ct \u2212 \u03b2x) x = \u03b3(x \u2212 \u03b2ct) y = y z = z, (A.1) where \u03b3 \u2261 (1 \u2212 \u03b22)\u22121/2. The transformation for a general relative velocity v, with axes still aligned, can be obtained by \ufb01rst co-rotating the spatial axes of S and S so that the relative motion is along the x-axis, performing the above transformation, and then back-rotating. Such Lorentz transformations are called Lorentz boosts and they 21Inertial frames are frames of reference in which a freely-moving particle (i.e., one subject to no external forces) proceeds at constant velocity in accordance with Newton\u2019s \ufb01rst law. A frame of reference can be thought of as a set of test particles, all at rest with respect to each other, and each equipped with a synchronised clock. Laying the test particles out on a Cartesian grid and assigning spatial coordinates, spacetime coordinates can be assigned to any event in spacetime by the spatial coordinates of the test particle on whose wordline the event occurs, and by the time on the test particle\u2019s clock at the event. 92 contain no additional relative (spatial) rotation between the two frames. Note that the Lorentz transformation of Eq. (A.1) reduces to a Galilean transformation for \u03b2 1. By construction, Lorentz transformations preserve the spacetime interval between two events, i.e., (\u2206x)2 \u2212 (c\u2206t)2 = (\u2206x)2 \u2212 (c\u2206t)2. (A.2) If we write the spacetime coordinates as x\u00b5 = (ct, xi), with \u00b5 = 0 labelling the time component, and", " \u00b5 = i = 1, 2, 3 labelling the space components, we can write the invariant interval for in\ufb01nitesimal separations in the form ds2 = \u03b7\u00b5\u03bddx\u00b5dx\u03bd. (A.3) Here, \u03b7\u00b5\u03bd = diag(\u22121, 1, 1, 1) is the Minkowski metric and we have introduced the summation convention that repeated spacetime (Greek) indices, with one upper and one lower, are summed over. More generally, we de\ufb01ne homogeneous Lorentz transformations to be transformations from one system of spacetime coordinates x\u00b5 to another x\u00b5, with such that x\u00b5 = \u039b\u00b5 \u03bdx\u03bd, \u03b7\u00b5\u03bd\u039b\u00b5 \u03b1\u039b\u03bd \u03b2 = \u03b7\u03b1\u03b2. (A.4) (A.5) Equivalently, we can write this in matrix form as \u039bT \u03b7\u039b = \u03b7. Such transformations leave \u03b7\u00b5\u03bddx\u00b5dx\u03bd invariant since \u03b1\u039b\u03bd \u03b7\u00b5\u03bddx\u00b5dx\u03bd = \u03b7\u00b5\u03bd\u039b\u00b5 \u03b7\u03b1\u03b2 = \u03b7\u03b1\u03b2dx\u03b1dx\u03b2. \u03b2 dx\u03b1dx\u03b2 (A.6) Note that when we sum over repeated indices, the exact symbol we assign to the \u201cdummy\u201d index is irrelevant. For the standard transformation of Eq. (A.1), we have \u039b\u00b5 \u03bd = \uf8eb \uf8ec \uf8ec \uf8ed \u03b3 \u2212\u03b3\u03b2 0 0 \u2212\u03b3\u03b2 A.7) Homogeneous Lorentz transformations form a group. Closure in ensured since, for a composition of two transformations represented by the matrix product \u039b1\u039b2, we have (\u039b1\u039b2)T \u03b7(\u039b1\u039b2) = \u039bT = \u039bT = \u03b7. 2 \u039bT 2 \u03b7\u039b2 1 \u03b7\u039b1\u039b2 93 (A.8) Clearly, the set of homogeneous Lorentz transformations contains the identity, \u039b\u00b5 \u03bd = \u03bd, and every element is invertible22 since \u039bT \u03b7\u039b = \u03b7 implies that det(\u039b)2 = 1. We \u03b4\u00b5 shall be most interested in the subgroup of restricted homogeneous Lorentz transformations. These are transformations that are continuously connected to the identity, i.e., any element can be taken to the identity by smooth variation of its parameters. In particular, restricted Lorentz transformations have det(\u039b) = +1 and23 \u039b0 0 \u2265 1. Physically, they correspond to transformations between inertial frames that exclude space inversion or time reversal. Since \u03b7\u00b5\u03bd is symmetric, Eq. (A.5) puts 10 independent constraints on the 16 independent components of \u039b\u00b5 \u03bd. It follows that a general homogeneous Lorentz transformation is described by 16 \u2212 10 = 6 parameters. Three of these encode the relative velocity of the two frames (i.e., the boost part) and three any additional spatial rotation. The invariant interval \u2206s2 = \u03b7\u00b5\u03bd\u2206x\u00b5x\u03bd between two events allows us to classify all spacetime events in relation to some given event as being either spacelike separated (\u2206s2 > 0), timelike (\u2206s2 < 0) or null (lightlike, with \u2206s2 = 0). At any event, the set of null-separated events de\ufb01nes the light cone. Light rays emitted from some event and propagating in vacuum have worldlines that lie on the lightcone of the emission event. Massive particles follow worldlines for which the tangent vector is timelike (i.e., has negative norm) and at any event a massive particle will be moving within the lightcone in spacetime. For timelike- and null-separated events, the temporal ordering of the events (i.e., the sign of \u2206t) is Lorentz invariant so we can speak of the future and past light cone. This is not the case for spacelike-separated events. A.2 4-vectors In Newtonian physics, it is very convenient to write our equations in terms of 3D vectors (or tensors) to make manifest their transformation properties under spatial rotations. We now generalise this idea to spacetime 4-vectors and tensors. By expressing our equations in terms of such objects, we can be sure that we are writing down equations that are form invariant under Lorentz transformations and hence satisfy the postulates of special relativity. 22The inverse of a Lorentz transformation is also a Lorentz transformation since multiplying", " \u039bT \u03b7\u039b = \u03b7 on the left with (\u039bT )\u22121 and on the right with \u039b\u22121 implies that (\u039b\u22121)T \u03b7\u039b\u22121 = \u03b7. Physically, this makes sense since the inverse transforms back from x\u00b5 to x\u00b5. 23Note that the 00 element of Eq. (A.5) gives \u03b700 = \u03b700(\u039b0 0)2 + \u03b7ii(\u039bi 0)2 \u21d2 (\u039b0 0)2 = 1 + (\u039bi i 0)2 \u2265 1. i 94 We de\ufb01ne a 4-vector to be an object A\u00b5 that transforms like the displacement dx\u00b5 under Lorentz transformations, i.e., A\u00b5 \u2192 \u039b\u00b5 \u03bdA\u03bd for x\u00b5 \u2192 x\u00b5 = \u039b\u00b5 \u03bdx\u03bd. (A.9) It follows that \u03b7\u00b5\u03bdA\u00b5A\u03bd is invariant, de\ufb01ning a Lorentz-invariant norm. More generally, we can de\ufb01ne a Lorentz-invariant scalar product between two 4-vectors, A\u00b5 and B\u00b5 as \u03b7\u00b5\u03bdA\u00b5B\u03bd. We should properly call A\u00b5 a contravariant 4-vector to distinguish it from a covariant 4-vector. To motivate the introduction of the latter, consider the transformation of the di\ufb00erential operator \u2202/\u2202x\u00b5: the chain rule gives Since \u2202 \u2202x\u00b5 = \u2202x\u03bd \u2202x\u00b5 \u2202 \u2202x\u03bd. \u2202x\u03bd \u2202x\u00b5 \u2202x\u00b5 \u2202x\u03c1 \u039b\u00b5 \u03c1 = \u03b4\u03bd \u03c1, (A.10) (A.11) it follows that \u2202x\u03bd/\u2202x\u00b5 = (\u039b\u22121)\u03bd \u00b5, where the right-hand side is the matrix representing the inverse transformation. The Lorentz transformation law for partial derivatives is therefore \u2202 \u2202x\u00b5 = (\u039b\u22121)\u03bd \u00b5 \u2202 \u2202x\u03bd. (A.12) Note the distinction to the transformation law for a contravariant 4-vector. It is convenient to de\ufb01ne the quantities \u039b\u00b5 \u03bd \u2261 (\u039b\u22121)\u03bd \u00b5 ; (A.13) we then de\ufb01ne covariant 4-vectors, A\u00b5, to be objects that transform under Lorentz transformations as A \u00b5 = \u039b\u00b5 \u03bdA\u03bd. (A.14) Note the consistent index placement on both sides of all equations. We can express \u039b\u00b5 \u03bd in terms of \u039b\u00b5 \u03bd and the Minkowski metric directly by noting that \u039bT \u03b7\u039b = \u03b7 \u21d2 \u039b\u22121 = \u03b7\u22121\u039bT \u03b7, (A.15) or, in terms of components, \u039b\u00b5 \u03bd = (\u039b\u22121)\u03bd \u00b5 = \u03b7\u03bd\u03c1\u039b\u03c4 \u03c1\u03b7\u03c4 \u00b5 = \u03b7\u00b5\u03c4 \u03b7\u03bd\u03c1\u039b\u03c4 \u03c1, (A.16) where we have written the components of the inverse of the metric as \u03b7\u00b5\u03bd. Numerically, the components \u03b7\u00b5\u03bd and \u03b7\u00b5\u03bd are the same, and \u03b7\u00b5\u03c1\u03b7\u03c1\u03bd = \u03b4\u00b5 \u03bd. 95 (A.17) The contraction of a contravariant vector A\u00b5 and a covariant vector B\u00b5 is Lorentz invariant since A \u00b5B\u00b5 = \u039b\u00b5 \u03bdA\u03bd\u039b\u00b5 \u00b5\u039b\u00b5 \u03c4 B\u03c4 \u03c4 A\u03bdB\u03c4 = (\u039b\u22121)\u03bd = \u03b4\u03bd \u03c4 A\u03bdB\u03c4 = A\u03bdB\u03bd. (A.18) Every contravariant vector can be mapped with the Minkowski metric into an associated covariant vector and vice versa. For example, we can write A\u00b5 = \u03b7\u00b5\u03bdA\u03bd, A\u00b5 = \u03b7\u00b5\u03bdA\u03bd. (A.19) Explicitly, this means that A0 = \u2212A0 and Ai = Ai. We can easily check that \u03b7\u00b5\u03bdA\u03bd is a covariant vector by considering its transformation law: A \u00b5 = \u03b7\u00b5\u03bdA\u03bd = \u03b7\u00b5\u03bd\u039b\u03bd = \u03b7\u00b5\u03bd\u039b\u03bd \u03c1 \u039b\u00b5 \u03c1A\u03c1. = \u039b\u00b5 \u03c4 A\u03c4 \u03c4 \u03b7\u03c4 \u03c1 A\u03c1 (A.20) The scalar product between", " two contravariant vectors, A\u00b5 and B\u00b5, can then be written either as \u03b7\u00b5\u03bdA\u00b5B\u03bd, or as a contraction A\u00b5B\u00b5 between the contravariant A\u00b5 and the covariant B\u00b5 (or the other way around). For the covariant di\ufb00erential operator \u2202\u00b5 \u2261 \u2202/\u2202x\u00b5, we can form the Lorentz-invariant (scalar) operator \u2261 \u03b7\u00b5\u03bd\u2202\u00b5\u2202\u03bd = \u2212 \u22022 \u2202(ct)2 + \u22072. (A.21) This is the operator that appears in the wave equation for waves propagating at speed c. Alternative take on covariant and contravariant components of 4-vectors: In 3D Euclidean space we are used to thinking of vectors as geometric objects, the archetypal example being displacement vectors that connect two points in space. Let us generalise this idea to spacetime so that the 4-vector dx connects two neighbouring events24. By taking unit displacements along the coordinate axes of some inertial frame, we can introduce a basis of four vectors e\u00b5. Here, the subscript \u00b5 labels the element of the basis. In terms of this basis, we can write dx = dx\u00b5e\u00b5. (A.22) 24We shall temporarily denote 4-vectors by boldface in this brief interlude. 96 Under a Lorentz transformation (i.e., a passive reparameterisation of spacetime with a different set of inertial coordinates) the coordinate di\ufb00erentials dx\u00b5 transform but the 4-vector itself is unchanged. This means that the basis vectors must also change to compensate the change in the dx\u00b5. Indeed, since e\u00b5 = \u2202x/\u2202x\u00b5, we have e \u00b5 = \u2202x \u2202x\u00b5 = \u2202x\u03bd \u2202x\u00b5 \u2202x \u2202x\u03bd = \u2202x\u03bd \u2202x\u03bd e\u03bd = \u039b\u00b5 \u03bde\u03bd. (A.23) We introduce an inner product on spacetime; this is a bilinear symmetric function of two 4-vectors, A = A\u00b5e\u00b5 and B = B\u00b5e\u00b5, de\ufb01ned by It follows that we must have A \u00b7 B \u2261 \u03b7\u00b5\u03bdA\u00b5B\u03bd. e\u00b5 \u00b7 e\u03bd = \u03b7\u00b5\u03bd, and it is easy to verify that this is preserved under Lorentz transformations. The dual basis of 4-vectors, denoted e\u00b5, is de\ufb01ned by e\u00b5 \u00b7 e\u03bd = \u03b4\u00b5 \u03bd, (A.24) (A.25) (A.26) so that e\u00b5 = \u03b7\u00b5\u03bde\u03bd and e\u00b5 = \u03b7\u00b5\u03bde\u03bd. The dual basis transforms in the same way as the dx\u00b5 under Lorentz transformations, e\u00b5 = \u039b\u00b5 \u03bde\u03bd, to preserve the duality relation in Eq. (A.26). We can express a 4-vector A in terms of its components on either basis, i.e., A = A\u00b5e\u00b5 = A\u00b5e\u00b5. (A.27) As usual, we can extract the components by taking the inner product with the appropriate basis 4-vectors: A\u00b5 = A \u00b7 e\u00b5 and A\u00b5 = A \u00b7 e\u00b5. It follows that A\u00b5 = (A\u03bde\u03bd) \u00b7 e\u00b5 = \u03b7\u00b5\u03bdA\u03bd, and, similarly, A\u00b5 = \u03b7\u00b5\u03bdA\u03bd. Under Lorentz transformations\u039b\u00b5 \u03bde\u03bd) = \u039b\u00b5 \u03bdA\u03bd, (A.28) (A.29) so the A\u00b5 (the covariant components) transform in the same way as the coordinate basis vectors. This is the origin of the covariant and contravariant terminology for the components. We see that the covariant components A\u00b5 and the contravariant components A\u00b5 are an equivalent description of the same geometric object, the 4-vector A. Some 4-vectors are more naturally represented by their contravariant components, an obvious example being the dx\u00b5 of the spacetime displacement, while for others the contravariant components are more natural. An example of the latter are the \u2202/\u2202x\u00b5 of the spacetime gradient operator. 97 A.2.1 Examples of 4-vectors in relativistic kinematics Velocity 4-vector. Consider a point particle following a worldline x\u00b5(\u03c4 ), where \u03c4 is the particle", "\u2019s proper time (i.e., time as measured on a clock carried with the particle). The velocity 4-vector is de\ufb01ned by u\u00b5 = dx\u00b5/d\u03c4, (A.30) where we are dividing a 4-vector dx\u00b5 by the scalar d\u03c4. We can express the proper time in terms of the time in any inertial frame S by noting that c2d\u03c4 2 = \u2212\u03b7\u00b5\u03bddx\u00b5dx\u03bd = c2dt2 \u2212 dx2 = c2dt2(1 \u2212 v2/c2) = c2dt2/\u03b32, (A.31) where \u03b3 is the Lorentz factor associated with the particle\u2019s velocity v relative to S. It follows that d\u03c4 = dt/\u03b3, and so the components of the velocity 4-vector in S are u\u00b5 = \u03b3 d dt (ct, x) = (\u03b3c, \u03b3v). (A.32) Note the norm of u\u00b5: \u03b7\u00b5\u03bdu\u00b5u\u03bd = (\u03b7\u00b5\u03bddx\u00b5dx\u03bd)/(d\u03c4 )2 = \u2212c2. (A.33) Momentum 4-vector. For a massive particle of rest-mass m, the momentum 4-vector is obtained from the velocity 4-vector by multiplication by m: p\u00b5 \u2261 mu\u00b5 = (\u03b3mc, \u03b3mv). (A.34) The norm is \u03b7\u00b5\u03bdp\u00b5p\u03bd = \u2212m2c2. In some inertial frame, the components of p\u00b5 are E/c, the (total) energy measured in that frame, and p, the relativistic 3-momentum, i.e., p\u00b5 = (E/c, p) where E = \u03b3mc2 and p = \u03b3mv. (A.35) For an isolated system, the 4-momentum is conserved, i.e., the relativistic 3-momentum and energy are conserved. More generally, the 3-momentum of a particle is changed by a force F according to Newton\u2019s second law dp/dt = F. The interpretation of the time component of p\u00b5 as the energy is then consistent since di\ufb00erentiating \u03b7\u00b5\u03bdp\u00b5p\u03bd = \u2212m2c2 gives EdE/dt = c2p \u00b7 dp/dt = c2p \u00b7 F \u21d2 dE/dt = p \u00b7 F/(\u03b3m) = v \u00b7 F, 98 (A.36) where the quantity on the right of the \ufb01nal equality is the usual rate of working of the force. We can introduce a 4-vector force F \u00b5 as F \u00b5 = dp\u00b5/d\u03c4 ; then F \u00b5 = \u03b3, p, \u03b3F E c dE dt d dt \u03b3 c \u03b3 c = = v \u00b7 F, \u03b3F. (A.37) The 4-force is necessarily orthogonal to p\u00b5 (and so to u\u00b5) to preserve the norm of p\u00b5. Acceleration 4-vector. The 4-acceleration a\u00b5 is Constancy of the norm \u03b7\u00b5\u03bdu\u00b5u\u03bd = \u2212c2 implies that a\u00b5 is orthogonal to u\u00b5. The components of a\u00b5 are a\u00b5 = du\u00b5/d\u03c4. (A.38) a\u00b5 = \u03b3 (\u03b3c, \u03b3v) d dt = \u03b3 = \u03b32 c d\u03b3 dt \u03b32 c v + \u03b3a, d\u03b3 dt v \u00b7 a, \u03b32 c2 a \u00b7 vv + a, (A.39) where a = dv/dt is the usual 3-acceleration and we used d\u03b3/dt = \u03b33a \u00b7 v/c2, which follows from di\ufb00erentiating \u03b3\u22122 = 1 \u2212 v \u00b7 v/c2. In the instantaneous rest frame of the particle (i.e., the frame in which it is at rest at any instant), the components of the 4-acceleration are simply a\u00b5 = (0, arest). It follows that the norm of a\u00b5 is the square of the rest-frame acceleration: \u03b7\u00b5\u03bda\u00b5a\u03bd = a2 rest. Example: A particle moves along the x-axis of some inertial frame S with constant rest-frame acceleration \u03b1. At proper time \u03c4, suppose that the particle is at x(\u03c4 ) and t(\u03c4 ) in S and is moving with speed \u03b2c (along the x-axis). In its instantaneous rest frame, the components of", " the 4-acceleration are a\u00b5 = \u03b1(0, 1, 0, 0). Performing the inverse Lorentz transform back to S gives Since the velocity 4-vector is u\u00b5 = (\u03b3c, \u03b3c\u03b2, 0, 0), we have a\u00b5 = du\u00b5/d\u03c4 = (\u03b3\u03b2\u03b1, \u03b3\u03b1, 0, 0). d(\u03b3c) d\u03c4 = \u03b3\u03b2\u03b1 and c d(\u03b3\u03b2) d\u03c4 = \u03b3\u03b1. To solve these, we note that d\u03b3/d\u03c4 = \u03b33\u03b2d\u03b2/d\u03c4 ; combining with Eq. (A.41), we \ufb01nd d\u03b2 d\u03c4 = \u03b1 \u03b32c = \u03b1 c (1 \u2212 \u03b22). 99 (A.40) (A.41) (A.42) The solution with \u03b2 = 0 at \u03c4 = 0 is \u03b2(\u03c4 ) = tanh(\u03b1\u03c4 /c) and \u03b3(\u03c4 ) = cosh(\u03b1\u03c4 /c). (A.43) As expected, the speed asymptotes to c (\u03b2 = 1). We can solve for the wordline of the particle using dx d\u03c4 dx dt = \u03b3 = \u03b3c\u03b2 = c sinh(\u03b1\u03c4 /c). (A.44) The solution with x = c2\u03b1 at \u03c4 = 0 (for convenience) is x(\u03c4 ) = Similarly, dt/d\u03c4 = \u03b3 = cosh(\u03b1\u03c4 /c) gives t(\u03c4 ) = c2 \u03b1 c \u03b1 cosh(\u03b1\u03c4 /c). sinh(\u03b1\u03c4 /c), (A.45) (A.46) taking t = 0 at \u03c4 = 0. We see that the motion is hyperbolic in the x-t plane, and asymptotes to x = c|t| as |\u03c4 | \u2192 \u221e. The wordline is entirely within the \u201cRindler wedge\u201d, x > c|t|. Note that if we consider an observer at rest in S at the origin, sending out light signals to the accelerated particle, only signals emitted before t = 0 are ever received by the accelerated particle. A.3 4-tensors A tensor of type p q Such a tensor T \u00b5\u03bd... has p upper (contravariant) indices and q lower (covariant) indices. \u03c1\u03c3... transforms as T \u00b5\u03bd... \u03c1\u03c3... = \u039b\u00b5 \u03b1\u039b\u03bd \u03b2... \u039b\u03c1 \u03b3\u039b\u03c3 \u03b4... T \u03b1\u03b2... \u03b3\u03b4... (A.47) \u03bdx\u03bd. As with 4-vectors, one can raise and lower indices with \u03b7\u00b5\u03bd and under x\u00b5 = \u039b\u00b5 \u03b7\u00b5\u03bd, for example which takes a type-1 tensor to a type 2 1 T \u00b5\u03bd = \u03b7\u03bd\u03b1T \u00b5. 0 \u03b1 (A.48) The following rules apply to tensor manipulations. \u2022 Contractions may only be made between upper and lower indices. Contracting a p q tensor returns a p\u22121 tensor. For example T \u03b1 q\u22121 \u03b1 is a Lorentz scalar since T \u03c1 \u03b4 (A.49) T \u03b1 \u03b1 = \u039b\u03b1 \u03b4 \u03c1\u039b\u03b1 \u03b4\u03b4 \u03c1 = T \u03c1 \u03c1. 100 Note that the speci\ufb01c symbol assigned to a contracted (dummy) index is irrelevant. \u2022 Symmetrisation/antisymmetrisation may be performed only over upper or lower indices (but not mixed). Symmetrisation is denoted with round brackets and antisymmetrisation with square brackets: T (\u03b1\u03b2) \u2261 1 2 T \u03b1\u03b2 + T \u03b2\u03b1 and T [\u03b1\u03b2] \u2261 T \u03b1\u03b2 \u2212 T \u03b2\u03b1. 1 2 (A.50) \u2022 Contraction, symmetrisation and index raising/lowering commute with Lorentz transformations. For example, 2T (\u03b1\u03b2) = T \u03b1\u03b2 + T \u03b2\u03b1 = \u039b\u03b1 = \u039b\u03b1 = 2\u039b\u03b1 \u03c3\u039b\u03b2 \u03c1\u039b\u03b2 \u03c1\u039b\u03b2 \u03c1T \u03c3\u03c1 + \u039b\u03b2 \u03c3\u039b\u03b1 \u03c3 (T \u03c1\u03c3 + T \u03c3\u03c1) \u03c3T (\u03c1\u03c3). \u03c1T \u03c3\u03c1 (A.51) \u2022 Partial di\ufb00erentiation takes a p tensor to a p tensor. For example, under a q q+1 Lorentz transformation \u03b2 \u2202T \u03b1 \u2202x\u00b5 \u2192 \u03b2 \u2202T \u03b1 \u2202x\u00b5 = \u039b\u00b5 \u03c1\u039b\u03b1 \u03c3\ufffd", "\ufffd\u03b2 \u03c4 \u2202T \u03c3 \u03c4 \u2202x\u03c1, (A.52) and \u2202\u00b5T \u03b1 \u03b2 is a 1 tensor. 2 Finally, we return to the Minkowski metric. We now see that this is a 0 tensor that takes the same form in all Lorentz frames. This follows since (\u039b\u22121)T \u03b7\u039b\u22121 = \u03b7 implies 2 Similarly, \u03b7\u00b5\u03bd is a 2 tensor and 0 is of type 1. 1 \u03b7\u00b5\u03bd = \u039b\u00b5 \u03c1\u039b\u03bd \u03c3\u03b7\u03c1\u03c3. \u03b7\u00b5 \u03bd = \u03b7\u00b5\u03c1\u03b7\u03c1\u03bd = \u03b4\u00b5 \u03bd (A.53) (A.54) 1012h = 1 \u221a 2 hx d x. 13 A statistician will complain that I am confusing the average of a \ufb01nite sample (a million, in this case) with the \u201ctrue\u201d average (over the whole continuum). This can be an awkward problem for the experimentalist, especially when the sample size is small, but here I am only concerned with the true average, to which the sample average is presumably a good approximation. Thus the probability density (Equation 1.14) is \u03c1(x) = 1 \u221a 2 hx, (0 \u2264 x \u2264 h) (outside this range, of course, the probability density is zero). 1.3 Probability 13 We can check this result, using Equation 1.16: h 0 1 \u221a 2 hx d x = 1 \u221a 2 h 2x 1/2 h 0 = 1. The average distance (Equation 1.17) is h d x = 1 \u221a 2 which is somewhat less than h/2, as anticipated. 1 \u221a 2 x = hx x 0 x 3/2 2 3 h 0 h = h 3, Figure 1.7 shows the graph of \u03c1(x). Notice that a probability density can be in\ufb01nite, though probability itself (the integral of \u03c1) must of course be \ufb01nite (indeed, less than or equal to 1). \u03c1(x) 1 h 2 Figure 1.7: The probability density in Example 1.2: \u03c1(x) = 1 h x \u221a 2 hx. \u2217 Problem 1.1 For the distribution of ages in the example in Section 1.3.1: j 2 and j2. (a) Compute (b) Determine j for each j, and use Equation 1.11 to compute the standard deviation. (c) Use your results in (a) and (b) to check Equation 1.12. Problem 1.2 (a) Find the standard deviation of the distribution in Example 1.2. (b) What is the probability that a photograph, selected at random, would show a distance x more than one standard deviation away from the average? 14 CHAPTER 1 The Wave Function \u2217 Problem 1.3 Consider the gaussian distribution \u03c1(x) = Ae\u2212\u03bb(x\u2212a)2, where A, a, and \u03bb are positive real constants. (The necessary integrals are inside the back cover.) (a) Use Equation 1.16 to determine A. (b) Find x, x 2 (c) Sketch the graph of \u03c1(x)., and \u03c3. 1.4 NORMALIZATION We return now to the statistical interpretation of the wave function (Equation 1.3), which says that |(x, t)|2 is the probability density for \ufb01nding the particle at point x, at time t. It follows (Equation 1.16) that the integral of ||2 over all x must be 1 (the particle\u2019s got to be somewhere): +\u221e \u2212\u221e |(x, t)|2 d x = 1. (1.20) Without this, the statistical interpretation would be nonsense. However, this requirement should disturb you: After all, the wave function is supposed to be determined by the Schr\u00f6dinger equation\u2014we can\u2019t go imposing an extraneous condition on without checking that the two are consistent. Well, a glance at Equation 1.1 reveals that if (x, t) is a solution, so too is A(x, t), where A is any (complex) constant. What we must do, then, is pick this undetermined multiplicative factor so as to ensure that Equation 1.20 is satis\ufb01ed. This process is called normalizing the wave function. For some solutions to the Schr\u00f6dinger equation the integral is in\ufb01nite; in that case no multiplicative factor is going to make it 1. The same goes for the trivial solution = 0. Such non-normalizable solutions cannot", " represent particles, and must be rejected. Physically realizable states correspond to the square-integrable solutions to Schr\u00f6dinger\u2019s equation.14 But wait a minute! Suppose I have normalized the wave function at time t = 0. How do I know that it will stay normalized, as time goes on, and evolves? (You can\u2019t keep renormalizing the wave function, for then A becomes a function of t, and you no longer have a solution to the Schr\u00f6dinger equation.) Fortunately, the Schr\u00f6dinger equation has the remarkable property that it automatically preserves the normalization of the wave function\u2014 without this crucial feature the Schr\u00f6dinger equation would be incompatible with the statistical interpretation, and the whole theory would crumble. This is important, so we\u2019d better pause for a careful proof. To begin with, +\u221e \u2212\u221e d dt |(x, t)|2 d x = +\u221e \u2212\u221e \u2202 \u2202t |(x, t)|2 d x. (1.21) 14 Evidently (x, t) must go to zero faster than 1/ |x|, as |x| \u2192 \u221e. Incidentally, normalization only \ufb01xes the modulus of A; the phase remains undetermined. However, as we shall see, the latter carries no physical signi\ufb01cance anyway. \u221a (Note that the integral is a function only of t, so I use a total derivative (d/dt) on the left, but the integrand is a function of x as well as t, so it\u2019s a partial derivative (\u2202/\u2202t) on the right.) By the product rule, 1.4 Normalization 15 \u2202 \u2202t Now the Schr\u00f6dinger equation says that ||2 = \u2202 \u2202t \u2217 = \u2217 \u2202 \u2202t + \u2202\u2217 \u2202t. \u2202 \u2202t = i 2m \u2202 2 \u2202 x 2 \u2212 i V, and hence also (taking the complex conjugate of Equation 1.23) \u2202\u2217 \u2202t = \u2212 i 2m \u2202 2\u2217 \u2202 x 2 + i V \u2217, so \u2202 \u2202t ||2 = i 2m \u2217 \u2202 2 \u2202 x 2 \u2212 \u2202 2 2m \u2202 \u2202 x \u2212 \u2202\u2217 \u2202 x. (1.22) (1.23) (1.24) (1.25) The integral in Equation 1.21 can now be evaluated explicitly: +\u221e d dt |(x, t)|2 d x = i 2m But (x, t) must go to zero as x goes to (\u00b1) in\ufb01nity\u2014otherwise the wave function would not be normalizable.15 It follows that (1.26) \u2212\u221e \u2212. \u2202\u2217 \u2202 x \u2217 \u2202 \u2202 x +\u221e \u2212\u221e +\u221e d dt |(x, t)|2 d x = 0, \u2212\u221e (1.27) and hence that the integral is constant (independent of time); if is normalized at t = 0, it stays normalized for all future time. QED Problem 1.4 At time t = 0 a particle is represented by the wave function \u23a7 \u23aa\u23a8 (x, 0) = \u23aa\u23a9 A(x/a), 0 \u2264 x \u2264 a, A(b \u2212 x)/(b \u2212 a), a \u2264 x \u2264 b, otherwise, 0, where A, a, and b are (positive) constants. (a) Normalize (that is, \ufb01nd A, in terms of a and b). (b) Sketch (x, 0), as a function of x. (c) Where is the particle most likely to be found, at t = 0? (d) What is the probability of \ufb01nding the particle to the left of a? Check your result in the limiting cases b = a and b = 2a. (e) What is the expectation value of x? 15 A competent mathematician can supply you with pathological counterexamples, but they do not arise in physics; for us the wave function and all its derivatives go to zero at in\ufb01nity. 16 CHAPTER 1 The Wave Function \u2217 Problem 1.5 Consider the wave function (x, t) = Ae\u2212\u03bb|x| e\u2212i\u03c9t, where A, \u03bb, and \u03c9 are positive real constants. (We\u2019ll see in Chapter 2 for what potential (V ) this wave function satis\ufb01es the Schr\u00f6dinger equation.) (a) Normalize. (b) Determine the expectation values of x and x 2. (c) Find", " the standard deviation of x. Sketch the graph of ||2, as a function of x, and mark the points (x+\u03c3 ) and (x\u2212\u03c3 ), to illustrate the sense in which \u03c3 represents the \u201cspread\u201d in x. What is the probability that the particle would be found outside this range? 1.5 MOMENTUM For a particle in state, the expectation value of x is x = +\u221e \u2212\u221e x |(x, t)|2 d x. (1.28) What exactly does this mean? It emphatically does not mean that if you measure the position x ||2 d x is the average of the results you\u2019ll get. On of one particle over and over again, the contrary: The \ufb01rst measurement (whose outcome is indeterminate) will collapse the wave function to a spike at the value actually obtained, and the subsequent measurements (if they\u2019re performed quickly) will simply repeat that same result. Rather, x is the average of measurements performed on particles all in the state, which means that either you must \ufb01nd some way of returning the particle to its original state after each measurement, or else you have to prepare a whole ensemble of particles, each in the same state, and measure the positions of all of them: x is the average of these results. I like to picture a row of bottles on a shelf, each containing a particle in the state (relative to the center of the bottle). A graduate student with a ruler is assigned to each bottle, and at a signal they all measure the positions of their respective particles. We then construct a histogram of the results, which should match ||2, and compute the average, which should agree with x. (Of course, since we\u2019re only using a \ufb01nite sample, we can\u2019t expect perfect agreement, but the more bottles we use, the closer we ought to come.) In short, the expectation value is the average of measurements on an ensemble of identically-prepared systems, not the average of repeated measurements on one and the same system. Now, as time goes on, x will change (because of the time dependence of ), and we might be interested in knowing how fast it moves. Referring to Equations 1.25 and 1.28, we see that16 dx dt = x \u2202 \u2202t ||2 d x = i 2m \u2217 \u2202 \u2202 x x \u2202 \u2202 x \u2212 \u2202\u2217 \u2202 x d x. (1.29) 16 To keep things from getting too cluttered, I\u2019ll suppress the limits of integration (\u00b1\u221e). 1.5 Momentum 17 This expression can be simpli\ufb01ed using integration-by-parts:17 dx dt = \u2212 i 2m \u2217 \u2202 \u2202 x \u2212 \u2202\u2217 \u2202 x d x. (1.30) (I used the fact that \u2202 x/\u2202 x = 1, and threw away the boundary term, on the ground that goes to zero at (\u00b1) in\ufb01nity.) Performing another integration-by-parts, on the second term, we conclude: dx dt = \u2212 i m \u2217 \u2202 \u2202 x d x. (1.31) What are we to make of this result? Note that we\u2019re talking about the \u201cvelocity\u201d of the expectation value of x, which is not the same thing as the velocity of the particle. Nothing we have seen so far would enable us to calculate the velocity of a particle. It\u2019s not even clear what velocity means in quantum mechanics: If the particle doesn\u2019t have a determinate position (prior to measurement), neither does it have a well-de\ufb01ned velocity. All we could reasonably ask for is the probability of getting a particular value. We\u2019ll see in Chapter 3 how to construct the probability density for velocity, given ; for the moment it will suf\ufb01ce to postulate that the expectation value of the velocity is equal to the time derivative of the expectation value of position: v = dx dt. Equation 1.31 tells us, then, how to calculate v directly from. Actually, it is customary to work with momentum ( p = mv), rather than velocity: p = m dx dt = \u2212i \u2217 \u2202 \u2202 x d x. Let me write the expressions for x and p in a more suggestive way: x = p = \u2217 \u2217 [x] d x, \u2212i (\u2202/\u2202 x) d x. (1.32) (1.33) (1.34) (1.35) We say that the operator18 x \u201crepresents\u201d position, and the operator \u2212i (\u2202/\u2202 x) \u201crepresents\u201d momentum", "; to calculate expectation values we \u201csandwich\u201d the appropriate operator between \u2217 and, and integrate. 17 The product rule says that from which it follows that d d x ( f g) = f dg d x + d f d x g, b a f dg. Under the integral sign, then, you can peel a derivative off one factor in a product, and slap it onto the other one\u2014it\u2019ll cost you a minus sign, and you\u2019ll pick up a boundary term. 18 An \u201coperator\u201d is an instruction to do something to the function that follows; it takes in one function, and spits out some other function. The position operator tells you to multiply by x; the momentum operator tells you to differentiate with respect to x (and multiply the result by \u2212i). 18 CHAPTER 1 The Wave Function That\u2019s cute, but what about other quantities? The fact is, all classical dynamical variables can be expressed in terms of position and momentum. Kinetic energy, for example, is and angular momentum is T = 1 2 mv2 = p2 2m, L = r \u00d7 mv = r \u00d7 p (the latter, of course, does not occur for motion in one dimension). To calculate the expectation value of any such quantity, Q(x, p), we simply replace every p by \u2212i (\u2202/\u2202 x), insert the resulting operator between \u2217 and, and integrate: Q(x, p) = \u2217 Q(x, \u2212i \u2202/\u2202 x) d x. For example, the expectation value of the kinetic energy is \u2217 \u2202 2 \u2202 x 2 T = \u2212 2m 2 d x. (1.36) (1.37) Equation 1.36 is a recipe for computing the expectation value of any dynamical quantity, for a particle in state ; it subsumes Equations 1.34 and 1.35 as special cases. I have tried to make Equation 1.36 seem plausible, given Born\u2019s statistical interpretation, but in truth this represents such a radically new way of doing business (as compared with classical mechanics) that it\u2019s a good idea to get some practice using it before we come back (in Chapter 3) and put it on a \ufb01rmer theoretical foundation. In the mean time, if you prefer to think of it as an axiom, that\u2019s \ufb01ne with me. Problem 1.6 Why can\u2019t you do integration-by-parts directly on the middle expression in Equation 1.29\u2014pull the time derivative over onto x, note that \u2202 x/\u2202t = 0, and conclude that dx/dt = 0? \u2217 Problem 1.7 Calculate d p/dt. Answer: d p dt = \u2212. \u2202 V \u2202 x (1.38) This is an instance of Ehrenfest\u2019s theorem, which asserts that expectation values obey the classical laws.19 Problem 1.8 Suppose you add a constant V0 to the potential energy (by \u201cconstant\u201d I mean independent of x as well as t). In classical mechanics this doesn\u2019t change anything, but what about quantum mechanics? Show that the wave function picks up a time-dependent phase factor: exp (\u2212i V0t/). What effect does this have on the expectation value of a dynamical variable? 19 Some authors limit the term to the pair of equations p = m dx/dt and \u2212\u2202 V /\u2202 x = d p/dt. 1.6 The Uncertainty Principle 19 10 20 30 40 50 x (feet) Figure 1.8: A wave with a (fairly) well-de\ufb01ned wavelength, but an ill-de\ufb01ned position. 10 20 30 40 50 x (feet) Figure 1.9: A wave with a (fairly) well-de\ufb01ned position, but an ill-de\ufb01ned wavelength. 1.6 THE UNCERTAINTY PRINCIPLE Imagine that you\u2019re holding one end of a very long rope, and you generate a wave by shaking it up and down rhythmically (Figure 1.8). If someone asked you \u201cPrecisely where is that wave?\u201d you\u2019d probably think he was a little bit nutty: The wave isn\u2019t precisely anywhere\u2014it\u2019s spread out over 50 feet or so. On the other hand, if he asked you what its wavelength is, you could give him a reasonable answer: it looks like about 6 feet. By contrast, if you gave the rope a sudden jerk (Figure 1.9), you\u2019d get a relatively narrow bump traveling down the line. This time the \ufb01rst question (Where precisely is the wave?) is a sensible one, and the second (", "What is its wavelength?) seems nutty\u2014it isn\u2019t even vaguely periodic, so how can you assign a wavelength to it? Of course, you can draw intermediate cases, in which the wave is fairly well localized and the wavelength is fairly well de\ufb01ned, but there is an inescapable trade-off here: the more precise a wave\u2019s position is, the less precise is its wavelength, and vice versa.20 A theorem in Fourier analysis makes all this rigorous, but for the moment I am only concerned with the qualitative argument. This applies, of course, to any wave phenomenon, and hence in particular to the quantum mechanical wave function. But the wavelength of is related to the momentum of the particle by the de Broglie formula:21 p = h \u03bb = 2\u03c0 \u03bb. (1.39) Thus a spread in wavelength corresponds to a spread in momentum, and our general observation now says that the more precisely determined a particle\u2019s position is, the less precisely is its momentum. Quantitatively, \u03c3x \u03c3 p \u2265, 2 (1.40) 20 That\u2019s why a piccolo player must be right on pitch, whereas a double-bass player can afford to wear garden gloves. For the piccolo, a sixty-fourth note contains many full cycles, and the frequency (we\u2019re working in the time domain now, instead of space) is well de\ufb01ned, whereas for the bass, at a much lower register, the sixty-fourth note contains only a few cycles, and all you hear is a general sort of \u201coomph,\u201d with no very clear pitch. 21 I\u2019ll explain this in due course. Many authors take the de Broglie formula as an axiom, from which they then deduce the association of momentum with the operator \u2212i (\u2202/\u2202 x). Although this is a conceptually cleaner approach, it involves diverting mathematical complications that I would rather save for later. 20 CHAPTER 1 The Wave Function where \u03c3x is the standard deviation in x, and \u03c3 p is the standard deviation in p. This is Heisenberg\u2019s famous uncertainty principle. (We\u2019ll prove it in Chapter 3, but I wanted to mention it right away, so you can test it out on the examples in Chapter 2.) Please understand what the uncertainty principle means: Like position measurements, momentum measurements yield precise answers\u2014the \u201cspread\u201d here refers to the fact that measurements made on identically prepared systems do not yield identical results. You can, if you want, construct a state such that position measurements will be very close together (by making a localized \u201cspike\u201d), but you will pay a price: Momentum measurements on this state will be widely scattered. Or you can prepare a state with a de\ufb01nite momentum (by making a long sinusoidal wave), but in that case position measurements will be widely scattered. And, of course, if you\u2019re in a really bad mood you can create a state for which neither position nor momentum is well de\ufb01ned: Equation 1.40 is an inequality, and there\u2019s no limit on how big \u03c3x and \u03c3 p can be\u2014just make some long wiggly line with lots of bumps and potholes and no periodic structure. \u2217 Problem 1.9 A particle of mass m has the wave function mx 2/ +it (x, t) = Ae\u2212a, where A and a are positive real constants. (a) Find A. (b) For what potential energy function, V (x), is this a solution to the Schr\u00f6dinger equation? (c) Calculate the expectation values of x, x 2, p, and p2. (d) Find \u03c3x and \u03c3 p. Is their product consistent with the uncertainty principle? FURTHER PROBLEMS ON CHAPTER 1 Problem 1.10 Consider the \ufb01rst 25 digits in the decimal expansion of \u03c0 (3, 1, 4, 1, 5, 9,...). (a) If you selected one number at random, from this set, what are the probabilities of getting each of the 10 digits? (b) What is the most probable digit? What is the median digit? What is the average value? (c) Find the standard deviation for this distribution. Problem 1.11 [This problem generalizes Example 1.2.] Imagine a particle of mass m and energy E in a potential well V (x), sliding frictionlessly back and forth between the classical turning points (a and b in Figure 1.10). Classically, the probability of \ufb01nding the particle in the range d x (if, for example, you took a snapshot at a random time t) is equal to the fraction of the time T it takes to get from a", " to b that it spends in the interval d x: \u03c1(x) d x = dt T = (dt/d x) d x T = 1 v(x) T d x, (1.41) Further Problems on Chapter 1 21 V(x) E m a dx b x Figure 1.10: Classical particle in a potential well. where v(x) is the speed, and Thus T = T 0 dt = b a 1 v(x) d x. \u03c1(x) = 1 v(x)T. (1.42) (1.43) This is perhaps the closest classical analog22 to ||2. (a) Use conservation of energy to express v(x) in terms of E and V (x). (b) As an example, \ufb01nd \u03c1(x) for the simple harmonic oscillator, V (x) = kx 2/2. Plot \u03c1(x), and check that it is correctly normalized. (c) For the classical harmonic oscillator in part (b), \ufb01nd x,, and \u03c3x. x 2 \u2217\u2217 Problem 1.12 What if we were interested in the distribution of momenta ( p = mv), for \u221a to + the classical harmonic oscillator (Problem 1.11(b)). (a) Find the classical probability distribution \u03c1( p) (note that p ranges from \u2212 \u221a 2m E p2, and \u03c3 p. 2m E). (b) Calculate p, (c) What\u2019s the classical uncertainty product, \u03c3x \u03c3 p, for this system? Notice that this product can be as small as you like, classically, simply by sending E \u2192 0. But in quantum mechanics, as we shall see in Chapter 2, the energy of a simple harmonic k/m is the classical frequency. oscillator cannot be less than \u03c9/2, where \u03c9 = In that case what can you say about the product \u03c3x \u03c3 p? \u221a Problem 1.13 Check your results in Problem 1.11(b) with the following \u201cnumerical experiment.\u201d The position of the oscillator at time t is x(t) = A cos(\u03c9t). (1.44) 22 If you like, instead of photos of one system at random times, picture an ensemble of such systems, all with the same energy but with random starting positions, and photograph them all at the same time. The analysis is identical, but this interpretation is closer to the quantum notion of indeterminacy. 22 CHAPTER 1 The Wave Function You might as well take \u03c9 = 1 (that sets the scale for time) and A = 1 (that sets the scale for length). Make a plot of x at 10,000 random times, and compare it with \u03c1(x). Hint: In Mathematica, \ufb01rst de\ufb01ne x[t_] := Cos[t] then construct a table of positions: snapshots = Table[x[\u03c0 RandomReal[j]], {j, 10000}] and \ufb01nally, make a histogram of the data: Histogram[snapshots, 100, PDF, PlotRange \u2192 {0,2}] Meanwhile, make a plot of the density function, \u03c1(x), and, using Show, superimpose the two. Problem 1.14 Let Pab(t) be the probability of \ufb01nding the particle in the range (a < x < b), at time t. (a) Show that where d Pab dt = J (a, t) \u2212 J (b, t), J (x, t) \u2261 i 2m \u2202\u2217 \u2202 x \u2212 \u2217 \u2202 \u2202 x. What are the units of J (x, t)? Comment: J is called the probability current, because it tells you the rate at which probability is \u201c\ufb02owing\u201d past the point x. If Pab(t) is increasing, then more probability is \ufb02owing into the region at one end than \ufb02ows out at the other. (b) Find the probability current for the wave function in Problem 1.9. (This is not a very pithy example, I\u2019m afraid; we\u2019ll encounter more substantial ones in due course.) Problem 1.15 Show that \u221e \u2212\u221e d dt \u2217 1 2 d x = 0 for any two (normalizable) solutions to the Schr\u00f6dinger equation (with the same V (x)), 1 and 2. Problem 1.16 A particle is represented (at time t = 0) by the wave function a2 \u2212 x 2 (x, 0) = A 0,, \u2212a \u2264 x \u2264 +a, otherwise. (a) Determine the normalization", " constant A. (b) What is the expectation value of x? (c) What is the expectation value of p? (Note that you cannot get it from p = mdx/dt. Why not?) Further Problems on Chapter 1 23 (d) Find the expectation value of x 2. (e) Find the expectation value of p2. (f) Find the uncertainty in x (\u03c3x ). \u03c3 p (g) Find the uncertainty in p. (h) Check that your results are consistent with the uncertainty principle. \u2217\u2217 Problem 1.17 Suppose you wanted to describe an unstable particle, that spontaneously disintegrates with a \u201clifetime\u201d \u03c4. In that case the total probability of \ufb01nding the particle somewhere should not be constant, but should decrease at (say) an exponential rate: +\u221e P(t) \u2261 \u2212\u221e |(x, t)|2 d x = e\u2212t/\u03c4. A crude way of achieving this result is as follows. In Equation 1.24 we tacitly assumed that V (the potential energy) is real. That is certainly reasonable, but it leads to the \u201cconservation of probability\u201d enshrined in Equation 1.27. What if we assign to V an imaginary part: V = V0 \u2212 i, where V0 is the true potential energy and is a positive real constant? (a) Show that (in place of Equation 1.27) we now get d P dt = \u2212 2 P. (b) Solve for P(t), and \ufb01nd the lifetime of the particle in terms of. Problem 1.18 Very roughly speaking, quantum mechanics is relevant when the de Broglie wavelength of the particle in question (h/ p) is greater than the characteristic size of the system (d). In thermal equilibrium at (Kelvin) temperature T, the average kinetic energy of a particle is p2 2m (where k B is Boltzmann\u2019s constant), so the typical de Broglie wavelength is = 3 2 k B T \u03bb = \u221a h 3mk B T. (1.45) The purpose of this problem is to determine which systems will have to be treated quantum mechanically, and which can safely be described classically. (a) Solids. The lattice spacing in a typical solid is around d = 0.3 nm. Find the temperature below which the unbound23 electrons in a solid are quantum mechanical. Below what temperature are the nuclei in a solid quantum mechanical? (Use silicon as an example.) 23 In a solid the inner electrons are attached to a particular nucleus, and for them the relevant size would be the radius of the atom. But the outer-most electrons are not attached, and for them the relevant distance is the lattice spacing. This problem pertains to the outer electrons. 24 CHAPTER 1 The Wave Function Moral: The free electrons in a solid are always quantum mechanical; the nuclei are generally not quantum mechanical. The same goes for liquids (for which the interatomic spacing is roughly the same), with the exception of helium below 4 K. (b) Gases. For what temperatures are the atoms in an ideal gas at pressure P quantum mechanical? Hint: Use the ideal gas law (P V = N k B T ) to deduce the interatomic spacing. P 2/5. Obviously (for the gas to show quantum Answer: T < (1/k B) behavior) we want m to be as small as possible, and P as large as possible. Put in the numbers for helium at atmospheric pressure. Is hydrogen in outer space (where the interatomic spacing is about 1 cm and the temperature is 3 K) quantum mechanical? (Assume it\u2019s monatomic hydrogen, not H2.) h2/3m 3/5 2 TIME-INDEPENDENT SCHR\u00d6DINGER EQUATION 2.1 STATIONARY STATES In Chapter 1 we talked a lot about the wave function, and how you use it to calculate various quantities of interest. The time has come to stop procrastinating, and confront what is, logically, the prior question: How do you get (x, t) in the \ufb01rst place? We need to solve the Schr\u00f6dinger equation, i \u2202 \u2202t = \u2212 2 2m \u2202 2 \u2202 x 2 + V, (2.1) for a speci\ufb01ed potential1 V (x, t). In this chapter (and most of this book) I shall assume that V is independent of t. In that case the Schr\u00f6dinger equation can be solved by the method of separation of variables (the physicist\u2019s \ufb01rst line of attack on any partial differential equation): We look for solutions that are products, (x, t) = \u03c8(x) \u03d5(t), (2.2)", " where \u03c8 (lower-case) is a function of x alone, and \u03d5 is a function of t alone. On its face, this is an absurd restriction, and we cannot hope to obtain more than a tiny subset of all solutions in this way. But hang on, because the solutions we do get turn out to be of great interest. Moreover (as is typically the case with separation of variables) we will be able at the end to patch together the separable solutions in such a way as to construct the most general solution. For separable solutions we have \u2202 \u2202t = \u03c8 d\u03d5 dt, \u2202 2 \u2202 x 2 = d2\u03c8 d x 2 \u03d5 (ordinary derivatives, now), and the Schr\u00f6dinger equation reads i\u03c8 d\u03d5 dt = \u2212 2 2m d2\u03c8 d x 2 \u03d5 + V \u03c8\u03d5. 1 It is tiresome to keep saying \u201cpotential energy function,\u201d so most people just call V the \u201cpotential,\u201d even though this invites occasional confusion with electric potential, which is actually potential energy per unit charge. 26 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation Or, dividing through by \u03c8\u03d5: i 1 \u03d5 d\u03d5 dt = \u2212 2 2m 1 \u03c8 d2\u03c8 d x 2 + V. (2.3) Now, the left side is a function of t alone, and the right side is a function of x alone.2 The only way this can possibly be true is if both sides are in fact constant\u2014otherwise, by varying t, I could change the left side without touching the right side, and the two would no longer be equal. (That\u2019s a subtle but crucial argument, so if it\u2019s new to you, be sure to pause and think it through.) For reasons that will appear in a moment, we shall call the separation constant E. Then or and or i 1 \u03d5 d\u03d5 dt = E, d\u03d5 dt = \u2212 i E \u03d5, \u2212 2 2m 1 \u03c8 d2\u03c8 d x 2 + V = E, \u2212 2 2m d2\u03c8 d x 2 + V \u03c8 = E\u03c8. (2.4) (2.5) Separation of variables has turned a partial differential equation into two ordinary differential equations (Equations 2.4 and 2.5). The \ufb01rst of these is easy to solve (just multiply through by dt and integrate); the general solution is C exp (\u2212i Et/), but we might as well absorb the constant C into \u03c8 (since the quantity of interest is the product \u03c8\u03d5). Then3 \u03d5(t) = e\u2212i Et/. (2.6) The second (Equation 2.5) is called the time-independent Schr\u00f6dinger equation; we can go no further with it until the potential V (x) is speci\ufb01ed. The rest of this chapter will be devoted to solving the time-independent Schr\u00f6dinger equation, for a variety of simple potentials. But before I get to that you have every right to ask: What\u2019s so great about separable solutions? After all, most solutions to the (time dependent) Schr\u00f6dinger equation do not take the form \u03c8(x) \u03d5(t). I offer three answers\u2014two of them physical, and one mathematical: 2 Note that this would not be true if V were a function of t as well as x. 3 Using Euler\u2019s formula, ei\u03b8 = cos \u03b8 + i sin \u03b8, you could equivalently write \u03d5(t) = cos(Et/) + i sin(Et/); the real and imaginary parts oscillate sinusoidally. Mike Casper (of Carleton College) dubbed \u03d5 the \u201cwiggle factor\u201d\u2014it\u2019s the characteristic time dependence in quantum mechanics. 2.1 Stationary States 27 1. They are stationary states. Although the wave function itself, (x, t) = \u03c8(x)e\u2212i Et/, does (obviously) depend on t, the probability density, |(x, t)|2 = \u2217 = \u03c8 \u2217e+i Et/\u03c8e\u2212i Et/ = |\u03c8(x)|2, (2.7) (2.8) does not\u2014the time-dependence cancels out.4 The same thing happens in calculating the expectation value of any dynamical variable; Equation 1.36 reduces to Q(x, p) = \u03c8 \u2217 Q \u03c8 d x. (2.9) x, \u2212i d d x Every expectation value is constant in time; we might as well drop the factor \u03d5(t) altogether,", " and simply use \u03c8 in place of. (Indeed, it is common to refer to \u03c8 as \u201cthe wave function,\u201d but this is sloppy language that can be dangerous, and it is important to remember that the true wave function always carries that time-dependent wiggle factor.) In particular, x is constant, and hence (Equation 1.33) p = 0. Nothing ever happens in a stationary state. 2. They are states of de\ufb01nite total energy. In classical mechanics, the total energy (kinetic plus potential) is called the Hamiltonian: H (x, p) = p2 2m The corresponding Hamiltonian operator, obtained by the canonical substitution p \u2192 \u2212i(\u2202/\u2202 x), is therefore5 + V (x). (2.10) 2m Thus the time-independent Schr\u00f6dinger equation (Equation 2.5) can be written \u02c6H = \u2212 2 \u2202 2 \u2202 x 2 + V (x). \u02c6H \u03c8 = E\u03c8, (2.11) (2.12) and the expectation value of the total energy is H = \u03c8 \u2217 \u02c6H \u03c8 d x = E |\u03c8|2 d x = E ||2 d x = E. (2.13) (Notice that the normalization of entails the normalization of \u03c8.) Moreover, \u02c6H 2\u03c8 = \u02c6H \u02c6H \u03c8 = \u02c6H \u02c6H \u03c8 = E 2\u03c8, and hence H 2 = \u03c8 \u2217 \u02c6H 2\u03c8 d x = E 2 |\u03c8|2 d x = E 2. E\u03c8 = E So the variance of H is. (2.14) 4 For normalizable solutions, E must be real (see Problem 2.1(a)). 5 Whenever confusion might arise, I\u2019ll put a \u201chat\u201d (\u02c6) on the operator, to distinguish it from the dynamical variable it represents. 28 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation But remember, if \u03c3 = 0, then every member of the sample must share the same value (the distribution has zero spread). Conclusion: A separable solution has the property that every measurement of the total energy is certain to return the value E. (That\u2019s why I chose that letter for the separation constant.) 3. The general solution is a linear combination of separable solutions. As we\u2019re about to discover, the time-independent Schr\u00f6dinger equation (Equation 2.5) yields an in\ufb01nite collection of solutions (\u03c81(x), \u03c82(x), \u03c83(x),..., which we write as {\u03c8n(x)}), each with its associated separation constant (E1, E2, E3,... = {En}); thus there is a different wave function for each allowed energy: 1(x, t) = \u03c81(x)e\u2212i E1t/, 2(x, t) = \u03c82(x)e\u2212i E2t/,.... Now (as you can easily check for yourself) the (time-dependent) Schr\u00f6dinger equation (Equation 2.1) has the property that any linear combination6 of solutions is itself a solution. Once we have found the separable solutions, then, we can immediately construct a much more general solution, of the form (x, t) = \u221e n=1 cn\u03c8n(x)e\u2212i En t/. (2.15) It so happens that every solution to the (time-dependent) Schr\u00f6dinger equation can be written in this form\u2014it is simply a matter of \ufb01nding the right constants (c1, c2,...) so as to \ufb01t the initial conditions for the problem at hand. You\u2019ll see in the following sections how all this works out in practice, and in Chapter 3 we\u2019ll put it into more elegant language, but the main point is this: Once you\u2019ve solved the timeindependent Schr\u00f6dinger equation, you\u2019re essentially done; getting from there to the general solution of the time-dependent Schr\u00f6dinger equation is, in principle, simple and straightforward. A lot has happened in the past four pages, so let me recapitulate, from a somewhat different perspective. Here\u2019s the generic problem: You\u2019re given a (time-independent) potential V (x), and the starting wave function (x, 0); your job is to \ufb01nd the wave function, (x, t), for any subsequent time t. To do this you must solve the (time-dependent) Schr\u00f6dinger equation (", "Equation 2.1). The strategy is \ufb01rst to solve the time-independent Schr\u00f6dinger equation (Equation 2.5); this yields, in general, an in\ufb01nite set of solutions, {\u03c8n(x)}, each with its own associated energy, {En}. To \ufb01t (x, 0) you write down the general linear combination of these solutions: (x, 0) = \u221e n=1 cn \u03c8n(x); (2.16) 6 A linear combination of the functions f1(z), f2(z),... is an expression of the form f (z) = c1 f1(z) + c2 f2(z) + \u00b7 \u00b7 \u00b7, where c1, c2,... are (possibly complex) constants. 2.1 Stationary States 29 the miracle is that you can always match the speci\ufb01ed initial state7 by appropriate choice of the constants {cn}. To construct (x, t) you simply tack onto each term its characteristic time dependence (its \u201cwiggle factor\u201d), exp (\u2212i Ent/):8 (x, t) = \u221e n=1 cn\u03c8n(x)e\u2212i En t/ = \u221e n=1 cnn(x, t). (2.17) The separable solutions themselves, n(x, t) = \u03c8n(x)e\u2212i En t/, (2.18) are stationary states, in the sense that all probabilities and expectation values are independent of time, but this property is emphatically not shared by the general solution (Equation 2.17): the energies are different, for different stationary states, and the exponentials do not cancel, when you construct ||2. Example 2.1 Suppose a particle starts out in a linear combination of just two stationary states: (x, 0) = c1\u03c81(x) + c2\u03c82(x). (To keep things simple I\u2019ll assume that the constants cn and the states \u03c8n(x) are real.) What is the wave function (x, t) at subsequent times? Find the probability density, and describe its motion. Solution: The \ufb01rst part is easy: (x, t) = c1\u03c81(x)e\u2212i E1t/ + c2\u03c82(x) e\u2212i E2t/, |(x, t)|2 = where E1 and E2 are the energies associated with \u03c81 and \u03c82. It follows that c1\u03c81ei E1t/ + c2\u03c82ei E2t/ \u03c8 2 1 + 2c1c2\u03c81\u03c82 cos + c2 2 \u03c8 2 2 c1\u03c81e\u2212i E1t/ + c2\u03c82e\u2212i E2t/ (E2 \u2212 E1) t/. = c2 1 The probability density oscillates sinusoidally, at an angular frequency \u03c9 = (E2 \u2212 E1) /; this is certainly not a stationary state. But notice that it took a linear combination of stationary states (with different energies) to produce motion.9 7 In principle, any normalized function (x, 0) is fair game\u2014it need not even be continuous. How you might actually get a particle into that state is a different question, and one (curiously) we seldom have occasion to ask. 8 If this is your \ufb01rst encounter with the method of separation of variables, you may be disappointed that the solution takes the form of an in\ufb01nite series. Occasionally it is possible to sum the series, or to solve the time-dependent Schr\u00f6dinger equation without recourse to separation of variables\u2014see, for instance, Problems 2.49, 2.50, and 2.51. But such cases are extremely rare. 9 This is nicely illustrated in an applet by Paul Falstad, at www.falstad.com/qm1d/. 30 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation You may be wondering what the coef\ufb01cients {cn} represent physically. I\u2019ll tell you the answer, though the explanation will have to await Chapter 3: |cn|2 is the probability that a measurement of the energy would return the value En. (2.19) A competent measurement will always yield one of the \u201callowed\u201d values (hence the name), and |cn|2 is the probability of getting the particular value En.10 Of course, the sum of these probabilities should be 1: \u221e n=1 |cn|2 = 1, and the expectation value of the energy must be H = \u221e n=1 |cn|2 En. (", "2.20) (2.21) We\u2019ll soon see how this works out in some concrete examples. Notice, \ufb01nally, that because the constants {cn} are independent of time, so too is the probability of getting a particular energy, and, a fortiori, the expectation value of H. These are manifestations of energy conservation in quantum mechanics. \u2217 Problem 2.1 Prove the following three theorems: (a) For normalizable solutions, the separation constant E must be real. Hint: Write E (in Equation 2.7) as E0 + i (with E0 and real), and show that if Equation 1.20 is to hold for all t, must be zero. (b) The time-independent wave function \u03c8(x) can always be taken to be real (unlike (x, t), which is necessarily complex). This doesn\u2019t mean that every solution to the time-independent Schr\u00f6dinger equation is real; what it says is that if you\u2019ve got one that is not, it can always be expressed as a linear combination of solutions (with the same energy) that are. So you might as well stick to \u03c8s that are real. Hint: If \u03c8(x) satis\ufb01es Equation 2.5, for a given E, so too does its complex conjugate, and hence also the real linear combinations (\u03c8 + \u03c8 \u2217) and i (\u03c8 \u2212 \u03c8 \u2217). (c) If V (x) is an even function (that is, V (\u2212x) = V (x)) then \u03c8(x) can always be taken to be either even or odd. Hint: If \u03c8(x) satis\ufb01es Equation 2.5, for a given E, so too does \u03c8(\u2212x), and hence also the even and odd linear combinations \u03c8(x) \u00b1 \u03c8(\u2212x). 10 Some people will tell you that |cn |2 is \u201cthe probability that the particle is in the nth stationary state,\u201d but this is bad language: the particle is in the state, not n, and anyhow, in the laboratory you don\u2019t \u201c\ufb01nd the particle to be in a particular state,\u201d you measure some observable, and what you get is a number, not a wave function. V(x) 2.2 The In\ufb01nite Square Well 31 a x Figure 2.1: The in\ufb01nite square well potential (Equation 2.22). \u2217 Problem 2.2 Show that E must exceed the minimum value of V (x), for every normalizable solution to the time-independent Schr\u00f6dinger equation. What is the classical analog to this statement? Hint: Rewrite Equation 2.5 in the form d2\u03c8 d x 2 = 2m 2 [V (x) \u2212 E] \u03c8; if E < Vmin, then \u03c8 and its second derivative always have the same sign\u2014argue that such a function cannot be normalized. 2.2 THE INFINITE SQUARE WELL Suppose V (x) = 0, \u221e, 0 \u2264 x \u2264 a, otherwise (2.22) (Figure 2.1). A particle in this potential is completely free, except at the two ends (x = 0 and x = a), where an in\ufb01nite force prevents it from escaping. A classical model would be a cart on a frictionless horizontal air track, with perfectly elastic bumpers\u2014it just keeps bouncing back and forth forever. (This potential is arti\ufb01cial, of course, but I urge you to treat it with respect. Despite its simplicity\u2014or rather, precisely because of its simplicity\u2014it serves as a wonderfully accessible test case for all the fancy machinery that comes later. We\u2019ll refer back to it frequently.) Outside the well, \u03c8(x) = 0 (the probability of \ufb01nding the particle there is zero). Inside the well, where V = 0, the time-independent Schr\u00f6dinger equation (Equation 2.5) reads 2 \u2212 2m d2\u03c8 d x 2 = E\u03c8, or d2\u03c8 d x 2 = \u2212k2\u03c8, where k \u2261 \u221a 2m E. (2.23) (2.24) (By writing it in this way, I have tacitly assumed that E \u2265 0; we know from Problem 2.2 that E < 0 won\u2019t work.) Equation 2.24 is the classical simple harmonic oscillator equation; the general solution is 32 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation \u03c8(x) = A sin kx + B cos kx, (2.25) where A", " and B are arbitrary constants. Typically, these constants are \ufb01xed by the boundary conditions of the problem. What are the appropriate boundary conditions for \u03c8(x)? Ordinarily, both \u03c8 and d\u03c8/d x are continuous,11 but where the potential goes to in\ufb01nity only the \ufb01rst of these applies. (I\u2019ll justify these boundary conditions, and account for the exception when V = \u221e, in Section 2.5; for now I hope you will trust me.) Continuity of \u03c8(x) requires that \u03c8(0) = \u03c8(a) = 0, (2.26) so as to join onto the solution outside the well. What does this tell us about A and B? Well, \u03c8(0) = A sin 0 + B cos 0 = B, so B = 0, and hence \u03c8(x) = A sin kx. (2.27) Then \u03c8(a) = A sin ka, so either A = 0 (in which case we\u2019re left with the trivial\u2014nonnormalizable\u2014solution \u03c8(x) = 0), or else sin ka = 0, which means that ka = 0, \u00b1\u03c0, \u00b12\u03c0, \u00b13\u03c0,.... (2.28) But k = 0 is no good (again, that would imply \u03c8(x) = 0), and the negative solutions give nothing new, since sin(\u2212\u03b8 ) = \u2212 sin(\u03b8 ) and we can absorb the minus sign into A. So the distinct solutions are kn = n\u03c0 a, with n = 1, 2, 3,.... (2.29) Curiously, the boundary condition at x = a does not determine the constant A, but rather the constant k, and hence the possible values of E: En = 2k2 n 2m = n2\u03c0 22 2ma2. (2.30) In radical contrast to the classical case, a quantum particle in the in\ufb01nite square well cannot have just any old energy\u2014it has to be one of these special (\u201callowed\u201d) values.12 To \ufb01nd A, we normalize \u03c8:13 0 a |A|2 sin2(kx) d x = |A|2 a 2 = 1, so |A|2 = 2 a. \u221a This only determines the magnitude of A, but it is simplest to pick the positive real root: A = 2/a (the phase of A carries no physical signi\ufb01cance anyway). Inside the well, then, the solutions are \u03c8n(x) = 2 a sin x. n\u03c0 a (2.31) 11 That\u2019s right: \u03c8(x) is a continuous function of x, even though (x, t) need not be. 12 Notice that the quantization of energy emerges as a rather technical consequence of the boundary conditions on solutions to the time-independent Schr\u00f6dinger equation. 13 Actually, it\u2019s (x, t) that must be normalized, but in view of Equation 2.7 this entails the normalization of \u03c8(x). 2.2 The In\ufb01nite Square Well 33 \u03a8 1(x) \u03a8 2(x) \u03a8 3(x) a x a x a x Figure 2.2: The \ufb01rst three stationary states of the in\ufb01nite square well (Equation 2.31). As promised, the time-independent Schr\u00f6dinger equation has delivered an in\ufb01nite set of solutions (one for each positive integer n). The \ufb01rst few of these are plotted in Figure 2.2. They look just like the standing waves on a string of length a; \u03c81, which carries the lowest energy, is called the ground state, the others, whose energies increase in proportion to n2, are called excited states. As a collection, the functions \u03c8n(x) have some interesting and important properties: 1. They are alternately even and odd, with respect to the center of the well: \u03c81 is even, \u03c82 is odd, \u03c83 is even, and so on.14 2. As you go up in energy, each successive state has one more node (zero-crossing): \u03c81 has none (the end points don\u2019t count), \u03c82 has one, \u03c83 has two, and so on. 3. They are mutually orthogonal, in the sense that15 \u03c8m(x)\u2217 \u03c8n(x) d x = 0, (m = n). (2.32) Proof: \u03c8m(x)\u2217\u03c8n(x sin 0 a m\u03c0 a cos 0", " x sin x n\u03c0 a d x m \u2212 n a \u03c0 x \u2212 cos m \u2212 n) \u03c0 sin sin [(m \u2212 n) \u03c0 ] (m \u2212 n) 1 (m + n) \u03c0 sin \u2212 \u03c0 x m \u2212 n a \u2212 sin [(m + n) \u03c0 ] (m + n) = 0 Note that this argument does not work if m = n. (Can you spot the point at which it fails?) In that case normalization tells us that the integral is 1. In fact, we can combine orthogonality and normalization into a single statement: \u03c8m(x)\u2217 \u03c8n(x) d x = \u03b4mn, (2.33) 14 To make this symmetry more apparent, some authors center the well at the origin (running it now from \u2212a to +a). The even functions are then cosines, and the odd ones are sines. See Problem 2.36. 15 In this case the \u03c8s are real, so the complex conjugation (*) of \u03c8m is unnecessary, but for future purposes it\u2019s a good idea to get in the habit of putting it there. 34 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation where \u03b4mn (the so-called Kronecker delta) is de\ufb01ned by \u03b4mn = 0, m = n, 1, m = n. (2.34) We say that the \u03c8s are orthonormal. 4. They are complete, in the sense that any other function, f (x), can be expressed as a linear combination of them: f (x) = \u221e n=1 cn\u03c8n(x) = 2 a \u221e n=1 cn sin x. n\u03c0 a (2.35) 2/a sin(n\u03c0 x/a), but if I\u2019m not about to prove the completeness of the functions you\u2019ve studied advanced calculus you will recognize that Equation 2.35 is nothing but the Fourier series for f (x), and the fact that \u201cany\u201d function can be expanded in this way is sometimes called Dirichlet\u2019s theorem.16 \u221a The coef\ufb01cients cn can be evaluated\u2014for a given f (x)\u2014by a method I call Fourier\u2019s trick, which beautifully exploits the orthonormality of {\u03c8n}: Multiply both sides of Equation 2.35 by \u03c8m(x)\u2217, and integrate. \u03c8m(x)\u2217 f (x) d x = \u221e n=1 cn \u03c8m(x)\u2217 \u03c8n(x) d x = \u221e n=1 cn\u03b4mn = cm. (2.36) (Notice how the Kronecker delta kills every term in the sum except the one for which n = m.) Thus the nth coef\ufb01cient in the expansion of f (x) is17 cn = \u03c8n(x)\u2217 f (x) d x. (2.37) These four properties are extremely powerful, and they are not peculiar to the in\ufb01nite square well. The \ufb01rst is true whenever the potential itself is a symmetric function; the second is universal, regardless of the shape of the potential.18 Orthogonality is also quite general\u2014I\u2019ll show you the proof in Chapter 3. Completeness holds for all the potentials you are likely to encounter, but the proofs tend to be nasty and laborious; I\u2019m afraid most physicists simply assume completeness, and hope for the best. The stationary states (Equation 2.18) of the in\ufb01nite square well are n(x, t) = 2 a sin n\u03c0 a x e\u2212i(n2\u03c0 2/2ma2)t. (2.38) I claimed (Equation 2.17) that the most general solution to the (time-dependent) Schr\u00f6dinger equation is a linear combination of stationary states: 16 See, for example, Mary Boas, Mathematical Methods in the Physical Sciences, 3rd edn (New York: John Wiley, 2006), p. 356; f (x) can even have a \ufb01nite number of \ufb01nite discontinuities. 17 It doesn\u2019t matter whether you use m or n as the \u201cdummy index\u201d here (as long as you are consistent on the two sides of the equation, of course); whatever letter you use, it just stands for \u201cany positive integer.\u201d 18 Problem 2.45 explores this property. For further discussion, see John L. Powell and Bernd Crasemann, Quantum Mechanics (Addison-Wesley,", " Reading, MA, 1961), Section 5-7. 2.2 The In\ufb01nite Square Well 35 (x, t) = \u221e n=1 cn 2 a sin n\u03c0 a x e\u2212i(n2\u03c0 2/2ma2)t. (2.39) (If you doubt that this is a solution, by all means check it!) It remains only for me to demonstrate that I can \ufb01t any prescribed initial wave function, (x, 0) by appropriate choice of the coef\ufb01cients cn: (x, 0) = \u221e n=1 cn\u03c8n(x). The completeness of the \u03c8s (con\ufb01rmed in this case by Dirichlet\u2019s theorem) guarantees that I can always express (x, 0) in this way, and their orthonormality licenses the use of Fourier\u2019s trick to determine the actual coef\ufb01cients: a 0 2 a n\u03c0 a sin cn = x (x, 0) d x. (2.40) That does it: Given the initial wave function, (x, 0), we \ufb01rst compute the expansion coef\ufb01cients cn, using Equation 2.40, and then plug these into Equation 2.39 to obtain (x, t). Armed with the wave function, we are in a position to compute any dynamical quantities of interest, using the procedures in Chapter 1. And this same ritual applies to any potential\u2014the only things that change are the functional form of the \u03c8s and the equation for the allowed energies. Example 2.2 A particle in the in\ufb01nite square well has the initial wave function (x, 0) = Ax (a \u2212 x), (0 \u2264 x \u2264 a), for some constant A (see Figure 2.3). Outside the well, of course, = 0. Find (x, t). \u03a8(x, 0) Aa2 4 Figure 2.3: The starting wave function in Example 2.2. a x Solution: First we need to determine A, by normalizing (x, 0): 1 = so a 0 |(x, 0)|2 d x = |A|2 a x 2 (a \u2212 x)2 d x = |A|2 a5 30, 0 The nth coef\ufb01cient is (Equation 2.40) A =. 30 a5 36 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation d x x (a \u2212 x) d x a d x \u2212 x 2 sin x 0 \u2212 ax n\u03c0 n\u03c0 cos a (n\u03c0 x/a)2 \u2212 2 (n\u03c0/a)3 n\u03c0 a x cos x a 0 n\u03c0 a x a 0 (n\u03c0 )2 \u2212 2 (n\u03c0 )3 cos (n\u03c0 ) + a3 2 (n\u03c0 )3 cos (0) cn = a sin a n\u03c0 a x a x sin 0 a 2 a n\u03c0 30 a5 n\u03c0 a x n\u03c0 a sin 2 \u2212 x n\u03c0 a x sin a n\u03c0 \u2212 a3 n\u03c0 cos (n\u03c0 ) + a3 2 a \u221a a3 \u221a 0 15 15 a3 = 2 = 2 \u2212 2 \u221a = 2 15 a3 \u221a 15 = 4 = (n\u03c0 )3 [cos (0) \u2212 cos (n\u03c0 )] 0, n even, \u221a 15/ (n\u03c0 )3, n odd. 8 Thus (Equation 2.39): (x, t) = 2 \u03c0 30 a 3 n=1,3,5... 1 n3 sin n\u03c0 a x e\u2212in2\u03c0 2t/2ma2. Example 2.3 Check that Equation 2.20 is satis\ufb01ed, for the wave function in Example 2.2. If you measured the energy of a particle in this state, what is the most probable result? What is the expectation value of the energy? Solution: The starting wave function (Figure 2.3) closely resembles the ground state \u03c81 (Figure 2.2). This suggests that |c1|2 should dominate,19 and in fact \" 2! \u221a 8 15 \u03c0 3 |c1|2 = = 0.998555.... The rest of the coef\ufb01cients make up the difference:20 19 Loosely speaking, cn tells you the \u201camount of \u03c8n that is contained in.\u201d 20 You can look up the series and 1 16 + 1 36 + 1 56 + \u00b7 \u00b7 \u00b7 = 1 14 + 1 34 + 1 54 + \u00b7 \u00b7 \u00b7 = \u03c0 6 960 \u03c0 4 96 in math tables, under \u201cSums of Reciprocal Powers\u201d or \u201cRiemann", " Zeta Function.\u201d 2.2 The In\ufb01nite Square Well 37 \u221e n=1 |cn|2 =! \u221a 8 15 \" 2 \u03c0 3 \u221e n=1,3,5,... 1 n6 = 1. The most likely outcome of an energy measurement is E1 = \u03c0 22/2ma2\u2014more than 99.8% of all measurements will yield this value. The expectation value of the energy (Equation 2.21) is H = \u221e n=1,3,5,... \" 2! \u221a 8 15 n3\u03c0 3 n2\u03c0 22 2ma2 = 4802 \u03c0 4ma2 \u221e n=1,3,5,... 1 n4 = 52 ma2. As one would expect, it is very close to E1 (5 in place of \u03c0 2/2 \u2248 4.935)\u2014slightly larger, because of the admixture of excited states. Of course, it\u2019s no accident that Equation 2.20 came out right in Example 2.3. Indeed, this follows from the normalization of (the cns are independent of time, so I\u2019m going to do the proof for t = 0; if this bothers you, you can easily generalize the argument to arbitrary t). \u221e \"\u2217!! \u221e \" 1 = |(x, 0)|2 d x = cm\u03c8m(x) cn\u03c8n(x) d x = = \u221e \u221e m=1 \u221e n=1 \u221e n=1 m=1 m=1 n=1 c\u2217 mcn \u03c8m(x)\u2217 \u03c8n(x) d x c\u2217 mcn\u03b4mn = \u221e n=1 |cn|2. (Again, the Kronecker delta picks out the term m = n in the summation over m.) Similarly, the expectation value of the energy (Equation 2.21) can be checked explicitly: The time-independent Schr\u00f6dinger equation (Equation 2.12) says \u02c6H \u03c8n = En\u03c8n, (2.41) so H = \u2217 \u02c6H d x = cm\u03c8m \u2217 \u02c6H cn\u03c8n d x = c\u2217 mcn En \u03c8 \u2217 m \u03c8n d x = |cn|2 En. Problem 2.3 Show that there is no acceptable solution to the (time-independent) Schr\u00f6dinger equation for the in\ufb01nite square well with E = 0 or E < 0. (This is a special case of the general theorem in Problem 2.2, but this time do it by explicitly solving the Schr\u00f6dinger equation, and showing that you cannot satisfy the boundary conditions.) \u2217 Problem 2.4 Calculate x,, \u03c3x, and \u03c3 p, for the nth stationary state of the in\ufb01nite square well. Check that the uncertainty principle is satis\ufb01ed. Which state comes closest to the uncertainty limit?, p, x 2 p2 38 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation \u2217 Problem 2.5 A particle in the in\ufb01nite square well has as its initial wave function an even mixture of the \ufb01rst two stationary states: (x, 0) = A [\u03c81(x) + \u03c82(x)]. (a) Normalize (x, 0). (That is, \ufb01nd A. This is very easy, if you exploit the orthonormality of \u03c81 and \u03c82. Recall that, having normalized at t = 0, you can rest assured that it stays normalized\u2014if you doubt this, check it explicitly after doing part (b).) (b) Find (x, t) and |(x, t)|2. Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let \u03c9 \u2261 \u03c0 2/2ma2. (c) Compute x. Notice that it oscillates in time. What is the angular frequency of the oscillation? What is the amplitude of the oscillation? (If your amplitude is greater than a/2, go directly to jail.) (d) Compute p. (As Peter Lorre would say, \u201cDo it ze kveek vay, Johnny!\u201d) (e) If you measured the energy of this particle, what values might you get, and what is the probability of getting each of them? Find the expectation value of H. How does it compare with E1 and E2? Problem 2.6 Although the overall phase constant of the wave function is of no physical signi\ufb01cance (it cancels out whenever you", " calculate a measurable quantity), the relative phase of the coef\ufb01cients in Equation 2.17 does matter. For example, suppose we change the relative phase of \u03c81 and \u03c82 in Problem 2.5: # $ (x, 0) = A \u03c81(x) + ei\u03c6\u03c82(x), where \u03c6 is some constant. Find (x, t), |(x, t)|2, and x, and compare your results with what you got before. Study the special cases \u03c6 = \u03c0/2 and \u03c6 = \u03c0. (For a graphical exploration of this problem see the applet in footnote 9 of this chapter.) \u2217 Problem 2.7 A particle in the in\ufb01nite square well has the initial wave function (x, 0) = Ax, 0 \u2264 x \u2264 a/2, A (a \u2212 x), a/2 \u2264 x \u2264 a. (a) Sketch (x, 0), and determine the constant A. (b) Find (x, t). (c) What is the probability that a measurement of the energy would yield the value E1? (d) Find the expectation value of the energy, using Equation 2.21.21 21 Remember, there is no restriction in principle on the shape of the starting wave function, as long as it is normalizable. In particular, (x, 0) need not have a continuous derivative. However, if you try to calculate H using (x, 0)\u2217 \u02c6H (x, 0) d x in such a case, you may encounter technical dif\ufb01culties, because the second derivative of (x, 0) is ill de\ufb01ned. It works in Problem 2.9 because the discontinuities occur at the end points, where the wave function is zero anyway. In Problem 2.39 you\u2019ll see how to manage cases like Problem 2.7. 2.3 The Harmonic Oscillator 39 Problem 2.8 A particle of mass m in the in\ufb01nite square well (of width a) starts out in the state (x, 0) = A, 0, 0 \u2264 x \u2264 a/2, a/2 \u2264 x \u2264 a, for some constant A, so it is (at t = 0) equally likely to be found at any point in the left half of the well. What is the probability that a measurement of the energy (at some later time t) would yield the value \u03c0 22/2ma2? Problem 2.9 For the wave function in Example 2.2, \ufb01nd the expectation value of H, at time t = 0, the \u201cold fashioned\u201d way: H = (x, 0)\u2217 \u02c6H (x, 0) d x. Compare the result we got in Example 2.3. Note: Because H is independent of time, there is no loss of generality in using t = 0. 2.3 THE HARMONIC OSCILLATOR The paradigm for a classical harmonic oscillator is a mass m attached to a spring of force constant k. The motion is governed by Hooke\u2019s law, F = \u2212kx = m d2x dt 2 (ignoring friction), and the solution is x(t) = A sin(\u03c9t) + B cos(\u03c9t), where \u03c9 \u2261 k m is the (angular) frequency of oscillation. The potential energy is V (x) = 1 2 kx 2; (2.42) (2.43) its graph is a parabola. Of course, there\u2019s no such thing as a perfect harmonic oscillator\u2014if you stretch it too far the spring is going to break, and typically Hooke\u2019s law fails long before that point is reached. But practically any potential is approximately parabolic, in the neighborhood of a local minimum (Figure 2.4). Formally, if we expand V (x) in a Taylor series about the minimum: V (x) = V (x0) + V (x0) (x \u2212 x0) + 1 2 V (x0) (x \u2212 x0)2 + \u00b7 \u00b7 \u00b7, subtract V (x0) (you can add a constant to V (x) with impunity, since that doesn\u2019t change the force), recognize that V (x0) = 0 (since x0 is a minimum), and drop the higher-order terms (which are negligible as long as (x \u2212 x0) stays small), we get 40 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation V(x) x0 x Figure 2.4: Parabolic approximation (dashed curve) to an arbitrary potential, in the neighborhood of a local minimum. V (x) \u2248 1 2 V (x0) (", "x \u2212 x0)2, which describes simple harmonic oscillation (about the point x0), with an effective spring constant k = V (x0). That\u2019s why the simple harmonic oscillator is so important: Virtually any oscillatory motion is approximately simple harmonic, as long as the amplitude is small.22 The quantum problem is to solve the Schr\u00f6dinger equation for the potential V (x) = 1 2 (it is customary to eliminate the spring constant in favor of the classical frequency, using Equation 2.42). As we have seen, it suf\ufb01ces to solve the time-independent Schr\u00f6dinger equation: m\u03c92x 2 (2.44) \u2212 2 2m d2\u03c8 d x 2 + 1 2 m\u03c92x 2\u03c8 = E\u03c8. (2.45) In the literature you will \ufb01nd two entirely different approaches to this problem. The \ufb01rst is a straightforward \u201cbrute force\u201d solution to the differential equation, using the power series method; it has the virtue that the same strategy can be applied to many other potentials (in fact, we\u2019ll use it in Chapter 4 to treat the hydrogen atom). The second is a diabolically clever algebraic technique, using so-called ladder operators. I\u2019ll show you the algebraic method \ufb01rst, because it is quicker and simpler (and a lot more fun);23 if you want to skip the power series method for now, that\u2019s \ufb01ne, but you should certainly plan to study it at some stage. 2.3.1 Algebraic Method To begin with, let\u2019s rewrite Equation 2.45 in a more suggestive form: $ \u02c6p2 + (m\u03c9x)2 \u03c8 = E\u03c8, # 1 2m (2.46) 22 Note that V (x0) \u2265 0, since by assumption x0 is a minimum. Only in the rare case V (x0) = 0 is the oscillation not even approximately simple harmonic. 23 We\u2019ll encounter some of the same strategies in the theory of angular momentum (Chapter 4), and the technique generalizes to a broad class of potentials in supersymmetric quantum mechanics (Problem 3.47; see also Richard W. Robinett, Quantum Mechanics (Oxford University Press, New York, 1997), Section 14.4). 2.3 The Harmonic Oscillator 41 where \u02c6p \u2261 \u2212i d/d x is the momentum operator.24 The basic idea is to factor the Hamiltonian, # \u02c6H = 1 2m $ \u02c6p2 + (m\u03c9x)2. (2.47) If these were numbers, it would be easy: u2 + v2 = (iu + v) (\u2212iu + v). Here, however, it\u2019s not quite so simple, because \u02c6p and x are operators, and operators do not, in general, commute (x \u02c6p is not the same as \u02c6px, as we\u2019ll see in a moment\u2014though you might want to stop right now and think it through for yourself). Still, this does motivate us to examine the quantities \u02c6a\u00b1 \u2261 1\u221a 2m\u03c9 \u2213i \u02c6p + m\u03c9x (2.48) (the factor in front is just there to make the \ufb01nal results look nicer). Well, what is the product \u02c6a\u2212 \u02c6a+? \u02c6a\u2212 \u02c6a+ = 1 = 1 2m\u03c9 2m\u03c9 i \u02c6p + m\u03c9x # \u2212i \u02c6p + m\u03c9x \u02c6p2 + (m\u03c9x)2 \u2212 im\u03c9 x \u02c6p \u2212 \u02c6px $.. We call this the commutator of As anticipated, there\u2019s an extra term, involving x and \u02c6p; it is a measure of how badly they fail to commute. In general, the commutator of operators \u02c6A and \u02c6B (written with square brackets) is x \u02c6p \u2212 \u02c6px # $ \u02c6A, \u02c6B \u2261 \u02c6A \u02c6B \u2212 \u02c6B \u02c6A. In this notation, # \u02c6a\u2212 \u02c6a+ = 1 2m\u03c9 \u02c6p2 + (m\u03c9x)2 $ \u2212 i 2. x, \u02c6p We need to \ufb01gure out the commutator of x and \u02c6p. Warning: Operators are notoriously slippery to work with in the abstract, and you are bound to make mistakes unless you give them a \u201ctest function,\u201d f (x),", " to act on. At the end you can throw away the test function, and you\u2019ll be left with an equation involving the operators alone. In the present case we have: x, \u02c6p f (x) = x (\u2212i) = i f (x). d d x ( f ) \u2212 (\u2212i) d d x (x f ) = \u2212 Dropping the test function, which has served its purpose, x, \u02c6p = i. (2.49) (2.50) (2.51) (2.52) This lovely and ubiquitous formula is known as the canonical commutation relation.25 24 Put a hat on x, too, if you like, but since \u02c6x = x we usually leave it off. 25 In a deep sense all of the mysteries of quantum mechanics can be traced to the fact that position and momentum do not commute. Indeed, some authors take the canonical commutation relation as an axiom of the theory, and use it to derive \u02c6p = \u2212i d/d x. 42 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation With this, Equation 2.50 becomes or \u02c6a\u2212 \u02c6a+ = 1 \u03c9 \u02c6H + 1 2, \u02c6H = \u03c9 \u02c6a\u2212 \u02c6a+ \u2212 1 2. (2.53) (2.54) Evidently the Hamiltonian does not factor perfectly\u2014there\u2019s that extra \u22121/2 on the right. Notice that the ordering of \u02c6a+ and \u02c6a\u2212 is important here; the same argument, with \u02c6a+ on the left, yields In particular, \u02c6a+ \u02c6a\u2212 = 1 \u03c9 \u02c6H \u2212 1 2. \u02c6a\u2212, \u02c6a+ = 1. Meanwhile, the Hamiltonian can equally well be written \u02c6a+ \u02c6a\u2212 + 1 2 \u02c6H = \u03c9 (2.55) (2.56) (2.57). In terms of \u02c6a\u00b1, then, the Schr\u00f6dinger equation26 for the harmonic oscillator takes the form \u02c6a\u00b1 \u02c6a\u2213 \u00b1 1 2 \u03c8 = E\u03c8 \u03c9 (2.58) (in equations like this you read the upper signs all the way across, or else the lower signs). Now, here comes the crucial step: I claim that: \u02c6H \u03c8 = If \u03c8 satis\ufb01es the Schr\u00f6dinger equation with energy E (that is: E\u03c8), then \u02c6a+\u03c8 satis\ufb01es the Schr\u00f6dinger equation with energy (E + \u03c9): \u02c6H. = (E + \u03c9) \u02c6a+\u03c8 \u02c6a+\u03c8 Proof: \u02c6a+\u03c8 \u02c6H = \u03c9 \u02c6a+ \u02c6a\u2212 + 1 2 \u02c6a\u2212 \u02c6a+ + 1 2 = \u03c9 \u02c6a+ \u02c6a+\u03c8 = \u03c9 \u03c8 = \u02c6a+ \u03c9 \u02c6a+ \u02c6a+ \u02c6a\u2212 \u02c6a+ + 1 2 \u02c6a+ \u02c6a\u2212 + 1 + 1 2 \u02c6a+\u03c8 \u03c8 \u03c8 = \u02c6a+ \u02c6H + \u03c9 \u03c8 = \u02c6a+(E + \u03c9)\u03c8 = (E + \u03c9). QED (I used Equation 2.56 to replace \u02c6a\u2212 \u02c6a+ by in the second line. Notice that whereas the ordering of \u02c6a+ and \u02c6a\u2212 does matter, the ordering of \u02c6a\u00b1 and any constants\u2014 such as, \u03c9, and E\u2014does not; an operator commutes with any constant.) \u02c6a+ \u02c6a\u2212 + 1 \u02c6H = \u03c9 \u02c6a\u2212\u03c8 By the same token, \u02c6a\u2212\u03c8 is a solution with energy (E \u2212 \u03c9): \u02c6a+ \u02c6a\u2212 \u2212 1 \u02c6a\u2212\u03c8 2 \u02c6H \u2212 \u03c9 \u02c6a\u2212 \u02c6a+ \u2212 1 2 \u02c6a\u2212 \u02c6a+ \u2212 1 \u2212 1 2. \u02c6a\u2212\u03c8 = \u03c9 \u02c6a\u2212 = (E \u2212 \u03c9) = \u02c6a\u2212 = \u02c6a\u2212 \u03c9 \u03c8 \u03c8 \u03c8 = \u02c6a\u2212(E \u2212 \u03c9)\u03c8 26 I\u2019m getting tired of writing \u201ctime", "-independent Schr\u00f6dinger equation,\u201d so when it\u2019s clear from the context which one I mean, I\u2019ll just call it the \u201cSchr\u00f6dinger equation.\u201d 2.3 The Harmonic Oscillator 43 E E+3 h \u03c9 +\u03c8 \u00e2 3 E+2 h \u03c9 +\u03c8 \u00e2 2 E+ \u03c9 h \u00e2 +\u03c8 \u00e2 + E \u03c8 E\u2212 \u03c9 h \u00e2_\u03c8 E\u22122 \u03c9 h 2\u03c8 \u00e2_ \u00e2_ E0 \u03c8 0 Figure 2.5: The \u201cladder\u201d of states for the harmonic oscillator. Here, then, is a wonderful machine for generating new solutions, with higher and lower energies\u2014if we could just \ufb01nd one solution, to get started! We call \u02c6a\u00b1 ladder operators, because they allow us to climb up and down in energy; \u02c6a+ is the raising operator, and \u02c6a\u2212 the lowering operator. The \u201cladder\u201d of states is illustrated in Figure 2.5. But wait! What if I apply the lowering operator repeatedly? Eventually I\u2019m going to reach a state with energy less than zero, which (according to the general theorem in Problem 2.3) does not exist! At some point the machine must fail. How can that happen? We know that \u02c6a\u2212\u03c8 is a new solution to the Schr\u00f6dinger equation, but there is no guarantee that it will be normalizable\u2014it might be zero, or its square-integral might be in\ufb01nite. In practice it is the former: There occurs a \u201clowest rung\u201d (call it \u03c80) such that \u02c6a\u2212\u03c80 = 0. (2.59) We can use this to determine \u03c80(x): 1\u221a 2m\u03c9 d d x + m\u03c9x \u03c80 = 0, or d\u03c80 d x = \u2212 m\u03c9 x\u03c80. 44 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation This differential equation is easy to solve: d\u03c80 \u03c80 = \u2212 m\u03c9 x d x \u21d2 ln \u03c80 = \u2212 m\u03c9 2 x 2 + constant, \u03c80(x) = Ae\u2212 m\u03c9 2 x 2. so We might as well normalize it right away: 1 = |A|2 \u221a so A2 = m\u03c9/\u03c0, and hence \u221e \u2212\u221e e\u2212m\u03c9x 2/ d x = |A|2 \u03c0 m\u03c9, \u03c80(x) = 1/4 m\u03c9 \u03c0 e\u2212 m\u03c9 2 x 2. (2.60) To determine the energy of this state we plug it into the Schr\u00f6dinger equation (in the form of Equation 2.58), \u03c9 \u03c80 = E0\u03c80, and exploit the fact that \u02c6a\u2212\u03c80 = 0: \u02c6a+ \u02c6a\u2212 + 1/2 E0 = 1 2 With our foot now securely planted on the bottom rung (the ground state of the quantum oscillator), we simply apply the raising operator (repeatedly) to generate the excited states,27 increasing the energy by \u03c9 with each step: (2.61) \u03c9. \u03c8n(x) = An \u02c6a+ n \u03c80(x), with En = n + 1 2 \u03c9, (2.62) where An is the normalization constant. By applying the raising operator (repeatedly) to \u03c80, then, we can (in principle) construct all28 the stationary states of the harmonic oscillator. Meanwhile, without ever doing that explicitly, we have determined the allowed energies! Example 2.4 Find the \ufb01rst excited state of the harmonic oscillator. Solution: Using Equation 2.62, \u03c81(x) = A1 \u02c6a+\u03c80 = = A1 m\u03c9 \u03c0 A1\u221a 2m\u03c9 2m\u03c9 1/4 \u2212 d d x + m\u03c9x 1/4 m\u03c9 \u03c0 e\u2212 m\u03c9 2 x 2 xe\u2212 m\u03c9 2 x 2. (2.63) 27 In the case of the harmonic oscillator it is customary, for some reason, to depart from the usual practice, and number the states starting with n = 0, instead of n = 1. Of course, the lower limit on the sum in a formula such as Equation 2.17 should be altered accordingly. 28 Note that we obtain all the (normalizable) solutions by this procedure. For if there were some other solution, we could generate from it a second ladder, by repeated application of the raising and lowering operators", ". But the bottom rung of this new ladder would have to satisfy Equation 2.59, and since that leads inexorably to Equation 2.60, the bottom rungs would be the same, and hence the two ladders would in fact be identical. 2.3 The Harmonic Oscillator 45 We can normalize it \u201cby hand\u201d: |\u03c81|2 d x = |A1|2 2m\u03c9 m\u03c9 \u03c0 \u221e \u2212\u221e x 2e\u2212 m\u03c9 x 2 d x = |A1|2, so, as it happens, A1 = 1. I wouldn\u2019t want to calculate \u03c850 this way (applying the raising operator \ufb01fty times!), but never mind: In principle Equation 2.62 does the job\u2014except for the normalization. You can even get the normalization algebraically, but it takes some fancy footwork, so watch closely. We know that \u02c6a\u00b1\u03c8n is proportional to \u03c8n\u00b11, \u02c6a+\u03c8n = cn\u03c8n+1, \u02c6a\u2212\u03c8n = dn\u03c8n\u22121 (2.64) but what are the proportionality factors, cn and dn? First note that for \u201cany\u201d29 functions f (x) and g (x), \u221e \u2212\u221e \u221e \u2212\u221e \u02c6a\u00b1g f \u2217 d x = \u2217 \u02c6a\u2213 f g d x. (2.65) In the language of linear algebra, \u02c6a\u2213 is the hermitian conjugate (or adjoint) of \u02c6a\u00b1. Proof: \u221e \u2212\u221e \u02c6a\u00b1g f \u2217 d x = \u221e 1\u221a f \u2217 \u2213 d d x (d f /d x)\u2217 g d x (the boundary + m\u03c9x g d x, 2m\u03c9 \u2212\u221e f \u2217 (dg/d x) d x to \u2212 and integration by parts takes terms vanish, for the reason indicated in footnote 29), so \u00b1 d d x \u02c6a\u00b1g d x = \u221e \u221e f \u2217 \u2212\u221e 1\u221a \u221e \u2212\u221e 2m\u03c9 \u02c6a\u00b1 f \u2212\u221e \u2217 = g d x. QED \u2217 + m\u03c9x f g d x In particular, \u221e \u2212\u221e \u02c6a\u00b1\u03c8n \u2217 \u02c6a\u00b1\u03c8n d x = \u221e \u2212\u221e \u02c6a\u2213 \u02c6a\u00b1\u03c8n \u2217 \u03c8n d x. But (invoking Equations 2.58 and 2.62) \u02c6a+ \u02c6a\u2212\u03c8n = n\u03c8n, \u02c6a\u2212 \u02c6a+\u03c8n = (n + 1) \u03c8n, (2.66) so \u221e \u2212\u221e \u221e \u2212\u221e \u02c6a+\u03c8n \u02c6a\u2212\u03c8n \u2217 \u02c6a+\u03c8n \u2217 \u02c6a\u2212\u03c8n d x = |cn|2 |\u03c8n+1|2 d x = (n + 1) \u221e \u221e \u2212\u221e |\u03c8n|2 d x, d x = |dn|2 |\u03c8n\u22121|2 d x = n |\u03c8n|2 d x. \u2212\u221e 29 Of course, the integrals must exist, and this means that f (x) and g(x) must go to zero at \u00b1\u221e. 46 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation But since \u03c8n and \u03c8n\u00b11 are normalized, it follows that |cn|2 = n + 1 and |dn|2 = n, and hence30 \u02c6a+\u03c8n = \u221a n + 1 \u03c8n+1, \u02c6a\u2212\u03c8n = \u221a n \u03c8n\u22121. (2.67) Thus \u03c83 = 1\u221a 3 \u02c6a+\u03c81 = 1\u221a 2 \u03c81 = \u02c6a+\u03c80, \u03c82 = 1\u221a 2 3 \u03c80, \u03c84 = 1\u221a 4 \u02c6a+ 3 \u00b7 2 \u02c6a+\u03c82 = 1\u221a \u02c6a+ 2 \u03c80, \u02c6a+\u03c83 = 1\u221a 4 \u00b7 3 \u00b7 2 \u02c6a+ 4 \u03c80, and so on. Clearly \u03c8n = 1\u221a n! \u02c6a+ n \u03c80, which is to say that the normalization factor in", " Equation 2.62 is An = 1/ A1 = 1, con\ufb01rming our result in Example 2.4). \u221a (2.68) n! (in particular, As in the case of the in\ufb01nite square well, the stationary states of the harmonic oscillator are orthogonal: \u221e \u2212\u221e \u03c8 \u2217 m \u03c8n d x = \u03b4mn. (2.69) This can be proved using Equation 2.66, and Equation 2.65 twice\u2014\ufb01rst moving \u02c6a+ and then moving \u02c6a\u2212: \u221e \u221e \u02c6a+ \u02c6a\u2212 \u03c8 \u2217 m \u2212\u221e \u03c8n d x = n \u2212\u221e \u03c8 \u2217 m \u03c8n d x \u221e = \u02c6a\u2212\u03c8m \u2217 \u02c6a\u2212\u03c8n d x = \u221e \u2212\u221e \u02c6a+ \u02c6a\u2212\u03c8m \u2217 \u03c8n d x \u2212\u221e \u221e = m \u2212\u221e \u03c8 \u2217 m \u03c8n d x. \u03c8 \u2217 m Unless m = n, then, \u03c8n d x must be zero. Orthonormality means that we can again use Fourier\u2019s trick (Equation 2.37) to evaluate the coef\ufb01cients cn, when we expand (x, 0) as a linear combination of stationary states (Equation 2.16). As always, |cn|2 is the probability that a measurement of the energy would yield the value En. 30 Of course, we could multiply cn and dn by phase factors, amounting to a different de\ufb01nition of the \u03c8n ; but this choice keeps the wave functions real. Example 2.5 Find the expectation value of the potential energy in the nth stationary state of the harmonic oscillator. 2.3 The Harmonic Oscillator 47 Solution: \u221e 1 2 V = m\u03c92x 2 = 1 2 There\u2019s a beautiful device for evaluating integrals of this kind (involving powers of x or \u02c6p): Use the de\ufb01nition (Equation 2.48) to express x and \u02c6p in terms of the raising and lowering operators: \u03c8 \u2217 n x 2\u03c8n d x. m\u03c92 \u2212\u221e 2m\u03c9 x = \u02c6a+ + \u02c6a\u2212 ; \u02c6p = i m\u03c9 2 \u02c6a+ \u2212 \u02c6a\u2212. (2.70) In this example we are interested in x 2: x 2 = # \u02c6a+ 2m\u03c9 2 + \u02c6a+ \u02c6a\u2212 \u02c6a\u2212 \u02c6a+ + + 2 \u02c6a\u2212 $ So V = \u221e \u2212\u221e \u03c9 4 # \u02c6a+ \u03c8 \u2217 n 2 + \u02c6a+ \u02c6a\u2212 \u02c6a\u2212 \u02c6a+ + 2 \u02c6a\u2212 +. $ \u03c8n d x. But 2 \u03c8n is (apart from normalization) \u03c8n+2, which is orthogonal to \u03c8n, and the same 2 \u03c8n, which is proportional to \u03c8n\u22122. So those terms drop out, and we can use \u02c6a\u2212 \u02c6a+ goes for Equation 2.66 to evaluate the remaining two: V = \u03c9 4 (n + n + 1) = 1 2 \u03c9 n + 1 2. As it happens, the expectation value of the potential energy is exactly half the total (the other half, of course, is kinetic). This is a peculiarity of the harmonic oscillator, as we\u2019ll see later on (Problem 3.37). \u2217 Problem 2.10 (a) Construct \u03c82(x). (b) Sketch \u03c80, \u03c81, and \u03c82. (c) Check the orthogonality of \u03c80, \u03c81, and \u03c82, by explicit integration. Hint: If you exploit the even-ness and odd-ness of the functions, there is really only one integral left to do. \u2217 Problem 2.11 (a) Compute x, p,, and, for the states \u03c80 (Equation 2.60) and \u03c81 (Equation 2.63), by explicit integration. Comment: In this and other problems involving the m\u03c9/ x harmonic oscillator it simpli\ufb01es matters if you introduce the variable \u03be \u2261 and the constant \u03b1 \u2261 (m\u03c9/\u03c0 )1/4. x 2 p2 \u221a (b) Check the uncertainty principle for these", " states. (c) Compute T and V for these states. (No new integration allowed!) Is their sum what you would expect? 48 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation \u2217, and T, for the nth stationary state of the harProblem 2.12 Find x, p, monic oscillator, using the method of Example 2.5. Check that the uncertainty principle is satis\ufb01ed. x 2 p2, Problem 2.13 A particle in the harmonic oscillator potential starts out in the state (x, 0) = A [3\u03c80(x) + 4\u03c81(x)]. (a) Find A. (b) Construct (x, t) and |(x, t)|2. Don\u2019t get too excited if |(x, t)|2 oscillates at exactly the classical frequency; what would it have been had I speci\ufb01ed \u03c82(x), instead of \u03c81(x)?31 (c) Find x and p. Check that Ehrenfest\u2019s theorem (Equation 1.38) holds, for this wave function. (d) If you measured the energy of this particle, what values might you get, and with what probabilities? 2.3.2 Analytic Method We return now to the Schr\u00f6dinger equation for the harmonic oscillator, + 1 2 and solve it directly, by the power series method. Things look a little cleaner if we introduce the dimensionless variable m\u03c92x 2\u03c8 = E\u03c8, (2.71) 2m \u2212 2 d2\u03c8 d x 2 \u03be \u2261 m\u03c9 x; d2\u03c8 d\u03be 2 = \u03be 2 \u2212 K \u03c8, in terms of \u03be the Schr\u00f6dinger equation reads where K is the energy, in units of (1/2) \u03c9: K \u2261 2E \u03c9. (2.72) (2.73) (2.74) Our problem is to solve Equation 2.73, and in the process obtain the \u201callowed\u201d values of K (and hence of E). To begin with, note that at very large \u03be (which is to say, at very large x), \u03be 2 completely dominates over the constant K, so in this regime d2\u03c8 d\u03be 2 \u2248 \u03be 2\u03c8, (2.75) which has the approximate solution (check it!) \u03c8(\u03be ) \u2248 Ae\u2212\u03be 2/2 + Be+\u03be 2/2. (2.76) The B term is clearly not normalizable (it blows up as |x| \u2192 \u221e); the physically acceptable solutions, then, have the asymptotic form 31 However, x does oscillate at the classical frequency\u2014see Problem 3.40. 2.3 The Harmonic Oscillator 49 \u03c8(\u03be ) \u2192 ( )e\u2212\u03be 2/2, at large \u03be. This suggests that we \u201cpeel off\u201d the exponential part, \u03c8(\u03be ) = h(\u03be ) e\u2212\u03be 2/2, (2.77) (2.78) in hopes that what remains, h(\u03be ), has a simpler functional form than \u03c8(\u03be ) itself.32 Differentiating Equation 2.78, d\u03c8 d\u03be = dh d\u03be \u2212 \u03be h e\u2212\u03be 2/2, and d2\u03c8 d\u03be 2 = d2h d\u03be 2 \u2212 2\u03be dh d\u03be + (\u03be 2 \u2212 1)h e\u2212\u03be 2/2, so the Schr\u00f6dinger equation (Equation 2.73) becomes d2h d\u03be 2 \u2212 2\u03be dh d\u03be + (K \u2212 1) h = 0. I propose to look for solutions to Equation 2.79 in the form of power series in \u03be :33 h(\u03be ) = a0 + a1\u03be + a2\u03be 2 + \u00b7 \u00b7 \u00b7 = \u221e j=0 a j \u03be j. Differentiating the series term by term, dh d\u03be = a1 + 2a2\u03be + 3a3\u03be 2 + \u00b7 \u00b7 \u00b7 = \u221e j=0 ja j \u03be j\u22121, (2.79) (2.80) and d2h d\u03be 2 = 2a2 + 2 \u00b7 3a3\u03be + 3 \u00b7 4a4\u03be 2 + \u00b7 \u00b7 \u00b7 = \u221e j=0 ( j + 1) ( j + 2) a j+2\u03be j. Putting these into Equation 2.80, we \ufb01nd ( j + 1) ( j + 2) a j+2 \u2212 2 ja j + (K \u2212 1)", " a j \u03be j = 0. (2.81) \u221e j=0 It follows (from the uniqueness of power series expansions34) that the coef\ufb01cient of each power of \u03be must vanish, ( j + 1) ( j + 2) a j+2 \u2212 2 ja j + (K \u2212 1) a j = 0, 32 Note that although we invoked some approximations to motivate Equation 2.78, what follows is exact. The device of stripping off the asymptotic behavior is the standard \ufb01rst step in the power series method for solving differential equations\u2014see, for example, Boas (footnote 16), Chapter 12. 33 According to Taylor\u2019s theorem, any reasonably well-behaved function can be expressed as a power series, so Equation 2.80 ordinarily involves no loss of generality. For conditions on the applicability of the method, see Boas (footnote 16) or George B. Arfken and Hans-Jurgen Weber, Mathematical Methods for Physicists, 7th edn, Academic Press, Orlando (2013), Section 7.5. 34 See, for example, Arfken and Weber (footnote 33), Section 1.2. 50 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation and hence that a j+2 = () ( j + 2) a j. (2.82) This recursion formula is entirely equivalent to the Schr\u00f6dinger equation. Starting with a0, it generates all the even-numbered coef\ufb01cients: (5 \u2212 K ) (1 \u2212 K ) 12 2 a4 = a2 = a0, a2 = (5 \u2212 K ) (1 \u2212 K ) 24 a0,..., and starting with a1, it generates the odd coef\ufb01cients: a3 = (3 \u2212 K ) 6 a1, a5 = (7 \u2212 K ) 20 a3 = (7 \u2212 K ) (3 \u2212 K ) 120 a1,.... We write the complete solution as h(\u03be ) = heven(\u03be ) + hodd(\u03be ), (2.83) where heven(\u03be ) \u2261 a0 + a2\u03be 2 + a4\u03be 4 + \u00b7 \u00b7 \u00b7 is an even function of \u03be, built on a0, and hodd(\u03be ) \u2261 a1\u03be + a3\u03be 3 + a5\u03be 5 + \u00b7 \u00b7 \u00b7 is an odd function, built on a1. Thus Equation 2.82 determines h(\u03be ) in terms of two arbitrary constants (a0 and a1)\u2014which is just what we would expect, for a second-order differential equation. However, not all the solutions so obtained are normalizable. For at very large j, the recursion formula becomes (approximately) with the (approximate) solution a j+/2)!, for some constant C, and this yields (at large \u03be, where the higher powers dominate) 1 1 j! \u03be 2 \u03be 2 j \u2248 Ce\u03be 2. \u03be j \u2248 C ( j/2)! h(\u03be ) \u2248 C, then \u03c8 (remember \u03c8?\u2014that\u2019s what we\u2019re trying to calculate) goes Now, if h goes like exp \u03be 2/2 (Equation 2.78), which is precisely the asymptotic behavior we didn\u2019t want.35 like exp There is only one way to wiggle out of this: For normalizable solutions the power series must terminate. There must occur some \u201chighest\u201d j (call it n), such that the recursion formula spits out an+2 = 0 (this will truncate either the series heven or the series hodd; the other one must be zero from the start: a1 = 0 if n is even, and a0 = 0 if n is odd). For physically acceptable solutions, then, Equation 2.82 requires that K = 2n + 1, 35 It\u2019s no surprise that the ill-behaved solutions are still contained in Equation 2.82; this recursion relation is equivalent to the Schr\u00f6dinger equation, so it\u2019s got to include both the asymptotic forms we found in Equation 2.76. 2.3 The Harmonic Oscillator 51 for some positive integer n, which is to say (referring to Equation 2.74) that the energy must be En = n + 1 2 \u03c9, for n = 0, 1, 2,.... (2.84) Thus we recover, by a completely different method, the fundamental quantization condition we found algebraically in Equation 2.62. It seems at \ufb01rst rather surprising that the quantization of", " energy should emerge from a technical detail in the power series solution to the Schr\u00f6dinger equation, but let\u2019s look at it from a different perspective. Equation 2.71 has solutions, of course, for any value of E (in fact, it has two linearly independent solutions for every E). But almost all of these solutions blow up exponentially at large x, and hence are not normalizable. Imagine, for example, using an E that is slightly less than one of the allowed values (say, 0.49\u03c9), and plotting the solution: Figure 2.6(a). Now try an E slightly larger (say, 0.51\u03c9); the \u201ctail\u201d now blows up in the other direction (Figure 2.6(b)). As you tweak the parameter in tiny increments from 0.49 to 0.51, the graph \u201c\ufb02ips over\u201d at precisely the value 0.5\u2014only here does the solution escape the exponential asymptotic growth that renders it physically unacceptable.36 (a) \u22124 \u22122 (b) \u22124 \u22122 2 1.5 1 0.5 \u22120.5 \u22121 \u22121.5 \u22122 2 1.5 1 0.5 \u22120.5 \u22121 \u22121.5 \u22122 2 4 \u03be 2 4 \u03be Figure 2.6: Solutions to the Schr\u00f6dinger equation for (a) E = 0.49\u03c9, and (b) E = 0.51\u03c9. 36 It is possible to set this up on a computer, and discover the allowed energies \u201cexperimentally.\u201d You might call it the wag the dog method: When the tail wags, you know you\u2019ve just passed over an allowed value. Computer scientists call it the shooting method (Nicholas Giordano, Computational Physics, Prentice Hall, Upper Saddle River, NJ (1997), Section 10.2). See Problems 2.55\u20132.57. 52 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation Table 2.1: The \ufb01rst few Hermite polynomials, Hn(\u03be ). For the allowed values of K, the recursion formula reads a j+2 = \u22122 (n \u2212 j) ( j + 1) ( j + 2) a j. (2.85) If n = 0, there is only one term in the series (we must pick a1 = 0 to kill hodd, and j = 0 in Equation 2.85 yields a2 = 0): h0(\u03be ) = a0, and hence \u03c80(\u03be ) = a0e\u2212\u03be 2/2 (which, apart from the normalization, reproduces Equation 2.60). For n = 1 we take a0 = 0,37 and Equation 2.85 with j = 1 yields a3 = 0, so and hence h1(\u03be ) = a1\u03be, \u03c81(\u03be ) = a1\u03be e\u2212\u03be 2/2 (con\ufb01rming Equation 2.63). For n = 2, j = 0 yields a2 = \u22122a0, and j = 2 gives a4 = 0, so 1 \u2212 2\u03be 2 h2(\u03be ) = a0, and \u03c82(\u03be ) = a0 1 \u2212 2\u03be 2 e\u2212\u03be 2/2, and so on. (Compare Problem 2.10, where this last result was obtained by algebraic means). In general, hn(\u03be ) will be a polynomial of degree n in \u03be, involving even powers only, if n is an even integer, and odd powers only, if n is an odd integer. Apart from the overall factor (a0 or a1) they are the so-called Hermite polynomials, Hn(\u03be ).38 The \ufb01rst few of them are listed in Table 2.1. By tradition, the arbitrary multiplicative factor is chosen so that the coef\ufb01cient of the highest power of \u03be is 2n. With this convention, the normalized39 stationary states for the harmonic oscillator are \u03c8n(x) = 1/4 m\u03c9 \u03c0 1\u221a 2nn! Hn(\u03be ) e\u2212\u03be 2/2. (2.86) They are identical (of course) to the ones we obtained algebraically in Equation 2.68. 37 Note that there is a completely different set of coef\ufb01cients a j for each value of n. 38 The Hermite polynomials have been studied extensively in the mathematical literature, and there are many tools and tricks for working with them. A few of these are explored in Problem 2.16. 39 I shall not work out the normalization constant here; if", " you are interested in knowing how it is done, see for example Leonard Schiff, Quantum Mechanics, 3rd edn, McGraw-Hill, New York (1968), Section 13. 2.3 The Harmonic Oscillator 53 In Figure 2.7(a) I have plotted \u03c8n(x) for the \ufb01rst few ns. The quantum oscillator is strikingly different from its classical counterpart\u2014not only are the energies quantized, but the position distributions have some bizarre features. For instance, the probability of \ufb01nding the particle outside the classically allowed range (that is, with x greater than the classical amplitude for (a) \u03c8 n(x) \u23d0\u03c8 n(x)\u23d0 Potential x 0 0 x (b) \u23d0\u03c8 60(x)\u23d02 0.24 0.20 0.16 0.12 0.06 0.04 0.0 x Figure 2.7: (a) The \ufb01rst four stationary states of the harmonic oscillator. (b) Graph of |\u03c860|2, with the classical distribution (dashed curve) superimposed. 54 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation the energy in question) is not zero (see Problem 2.14), and in all odd states the probability of \ufb01nding the particle at the center is zero. Only at large n do we begin to see some resemblance to the classical case. In Figure 2.7(b) I have superimposed the classical position distribution (Problem 1.11) on the quantum one (for n = 60); if you smoothed out the bumps, the two would \ufb01t pretty well. Problem 2.14 In the ground state of the harmonic oscillator, what is the probability (correct to three signi\ufb01cant digits) of \ufb01nding the particle outside the classically allowed region? Hint: Classically, the energy of an oscillator is E = (1/2) ka2 = (1/2) m\u03c92a2, where a is the amplitude. So the \u201cclassically allowed region\u201d for an oscillator of energy E extends 2E/m\u03c92. Look in a math table under \u201cNormal Distribution\u201d or from \u2212 \u201cError Function\u201d for the numerical value of the integral, or evaluate it by computer. 2E/m\u03c92 to + % % Problem 2.15 Use the recursion formula (Equation 2.85) to work out H5(\u03be ) and H6(\u03be ). Invoke the convention that the coef\ufb01cient of the highest power of \u03be is 2n to \ufb01x the overall constant. \u2217\u2217 Problem 2.16 In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials. (a) The Rodrigues formula says that Hn(\u03be ) = (\u22121)n e\u03be 2 n d d\u03be e\u2212\u03be 2. (2.87) Use it to derive H3 and H4. (b) The following recursion relation gives you Hn+1 in terms of the two preceding Hermite polynomials: Hn+1(\u03be ) = 2\u03be Hn(\u03be ) \u2212 2n Hn\u22121(\u03be ). (2.88) Use it, together with your answer in (a), to obtain H5 and H6. (c) If you differentiate an nth-order polynomial, you get a polynomial of order (n \u2212 1). For the Hermite polynomials, in fact, dHn d\u03be = 2n Hn\u22121(\u03be ). (2.89) Check this, by differentiating H5 and H6. (d) Hn(\u03be ) is the nth z-derivative, at z = 0, of the generating function ; or, to put it another way, it is the coef\ufb01cient of zn/n! in the Taylor \u2212z2 + 2z\u03be exp series expansion for this function: e\u2212z2+2z\u03be = \u221e n=0 zn n! Hn(\u03be ). (2.90) Use this to obtain H1, H2, and H3. 2.4 THE FREE PARTICLE 2.4 The Free Particle 55 We turn next to what should have been the simplest case of all: the free particle (V (x) = 0 everywhere). Classically this would just be motion at constant velocity, but in quantum mechanics the problem is surprisingly subtle. The time-independent Schr\u00f6dinger equation reads or \u2212 2 2m d2\u03c8 d x 2 = E\u03c8,", " d2\u03c8 d x 2 = \u2212k2\u03c8, where k \u2261 \u221a 2m E. (2.91) (2.92) So far, it\u2019s the same as inside the in\ufb01nite square well (Equation 2.24), where the potential is also zero; this time, however, I prefer to write the general solution in exponential form (instead of sines and cosines), for reasons that will appear in due course: \u03c8(x) = Aeikx + Be\u2212ikx. (2.93) Unlike the in\ufb01nite square well, there are no boundary conditions to restrict the possible values of k (and hence of E); the free particle can carry any (positive) energy. Tacking on the standard time dependence, exp (\u2212i Et/), (x, t) = Ae ik x\u2212 k 2m t \u2212ik x+ k 2m t. + Be (2.94) Now, any function of x and t that depends on these variables in the special combination (x \u00b1 vt) (for some constant v) represents a wave of unchanging shape, traveling in the \u2213xdirection at speed v: A \ufb01xed point on the waveform (for example, a maximum or a minimum) corresponds to a \ufb01xed value of the argument, and hence to x and t such that x \u00b1 vt = constant, or x = \u2213vt + constant. Since every point on the waveform moves with the same velocity, its shape doesn\u2019t change as it propagates. Thus the \ufb01rst term in Equation 2.94 represents a wave traveling to the right, and the second represents a wave (of the same energy) going to the left. By the way, since they only differ by the sign in front of k, we might as well write k(x, t) = Ae i kx\u2212 k2 2m t, and let k run negative to cover the case of waves traveling to the left: \u221a k \u2261 \u00b1 2m E, with k > 0 \u21d2 traveling to the right, k < 0 \u21d2 traveling to the left. (2.95) (2.96) Evidently the \u201cstationary states\u201d of the free particle are propagating waves; their wavelength is \u03bb = 2\u03c0/ |k|, and, according to the de Broglie formula (Equation 1.39), they carry momentum p = k. The speed of these waves (the coef\ufb01cient of t over the coef\ufb01cient of x) is vquantum = |k| 2m = E 2m. (2.97) (2.98) 56 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation On the other hand, the classical speed of a free particle with energy E is given by E = (1/2) mv2 (pure kinetic, since V = 0), so vclassical = 2E m = 2vquantum. (2.99) Apparently the quantum mechanical wave function travels at half the speed of the particle it is supposed to represent! We\u2019ll return to this paradox in a moment\u2014there is an even more serious problem we need to confront \ufb01rst: This wave function is not normalizable: +\u221e \u2212\u221e \u2217 k k d x = |A|2 +\u221e \u2212\u221e d x = |A|2 (\u221e). (2.100) In the case of the free particle, then, the separable solutions do not represent physically realizable states. A free particle cannot exist in a stationary state; or, to put it another way, there is no such thing as a free particle with a de\ufb01nite energy. But that doesn\u2019t mean the separable solutions are of no use to us. For they play a mathematical role that is entirely independent of their physical interpretation: The general solution to the time-dependent Schr\u00f6dinger equation is still a linear combination of separable solutions (only this time it\u2019s an integral over the continuous variable k, instead of a sum over the discrete index n): (x, t) = 1\u221a 2\u03c0 +\u221e \u2212\u221e \u03c6(k)ei(kx\u2212 k2 2m t) dk. (2.101) \u221a 2\u03c0 is factored out for convenience; what plays the role of the coef\ufb01cient (The quantity 1/ \u03c6(k) dk.) Now this wave function can be 2\u03c0 cn in Equation 2.17 is the combination normalized (for appropriate \u03c6(k)). But it necessarily carries a range of ks, and hence a range of energies and speeds. We call it a wave packet.40 1/ \u221a In the generic quantum", " problem, we are given (x, 0), and we are asked to \ufb01nd (x, t). For a free particle the solution takes the form of Equation 2.101; the only question is how to determine \u03c6(k) so as to match the initial wave function: (x, 0) = 1\u221a 2\u03c0 +\u221e \u2212\u221e \u03c6(k)eikx dk. (2.102) This is a classic problem in Fourier analysis; the answer is provided by Plancherel\u2019s theorem (see Problem 2.19): f (x) = 1\u221a 2\u03c0 +\u221e \u2212\u221e F(k) eikx dk \u21d0\u21d2 F(k) = 1\u221a 2\u03c0 +\u221e \u2212\u221e f (x) e\u2212ikx d x. (2.103) F(k) is called the Fourier transform of f (x); f (x) is the inverse Fourier transform of F(k) (the only difference is the sign in the exponent).41 There is, of course, some restriction on the 40 Sinusoidal waves extend out to in\ufb01nity, and they are not normalizable. But superpositions of such waves lead to interference, which allows for localization and normalizability. 41 Some people de\ufb01ne the Fourier transform without the factor of 1/ \u221e \u2212\u221e F(k)eikx dk, spoiling the symmetry of the two formulas. f (x) = (1/2\u03c0 ) \u221a 2\u03c0. Then the inverse transform becomes allowable functions: The integrals have to exist.42 For our purposes this is guaranteed by the physical requirement that (x, 0) itself be normalized. So the solution to the generic quantum problem, for the free particle, is Equation 2.101, with 2.4 The Free Particle 57 \u03c6(k) = 1\u221a 2\u03c0 +\u221e \u2212\u221e (x, 0) e\u2212ikx d x. (2.104) Example 2.6 A free particle, which is initially localized in the range \u2212a < x < a, is released at time t = 0: (x, 0) = A, \u2212a < x < a, otherwise, 0, where A and a are positive real constants. Find (x, t). Solution: First we need to normalize (x, 0): \u221e a 1 = |(x, 0)|2 d x = |A|2 \u2212\u221e \u2212a Next we calculate \u03c6(k), using Equation 2.104: \u03c6(k) = 1\u221a 2\u03c0 = 1 \u221a k 1\u221a 2a \u2212a eika \u2212 e\u2212ika 2i \u03c0a a e\u2212ikx d x = 1 \u221a 2 = 1\u221a \u03c0a \u03c0a a. e\u2212ikx \u2212ik sin (ka) k \u2212a d x = 2a |A|2 \u21d2 A = 1\u221a 2a. Finally, we plug this back into Equation 2.101: \u221e (x, t) = 1 \u221a \u03c0 2a \u2212\u221e sin (ka) k i e kx\u2212 k2 2m t dk. (2.105) Unfortunately, this integral cannot be solved in terms of elementary functions, though it can of course be evaluated numerically (Figure 2.8). (There are, in fact, precious few cases in which the integral for (x, t) (Equation 2.101) can be carried out explicitly; see Problem 2.21 for a particularly beautiful example.) In Figure 2.9 I have plotted (x, 0) and \u03c6(k). Note that for small a, (x, 0) is narrow (in x), while \u03c6(k) is broad (in k), and vice versa for large a. But k is related to momentum, by Equation 2.97, so this is a manifestation of the uncertainty principle: the position can be well de\ufb01ned (small a), or the momentum (large a), but not both. 42 The necessary and suf\ufb01cient condition on f (x) is that \u221e \u2212\u221e |F(k)|2 dk is also \ufb01nite, and in fact the two integrals are equal. Some people call this Plancherel\u2019s theorem, leaving Equation 2.102 without a name.) See Arfken and Weber (footnote 33), Section 20.4. \u221e \u2212\u221e | f (x)|2 d x be \ufb01nite. (In that case 58 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation a\u23d0\u03a8(x, t)\u23d02 0.5 0.4 0.3", " 0.2 0.1 \u22126 \u22124 \u22122 0 2 4 6 x a Figure 2.8: Graph of |(x, t)|2 (Equation 2.105) at t = 0 (the rectangle) and at t = ma2/ (the curve). (a) \u03a8(x, 0) 1 2a (b) \u03c6 (k) a \u03c0/ \u2212a a x Figure 2.9: (a) Graph of (x, 0). (b) Graph of \u03c6(k). \u2212 \u03c0 a \u03c0 a k I return now to the paradox noted earlier: the fact that the separable solution k(x, t) travels at the \u201cwrong\u201d speed for the particle it ostensibly represents. Strictly speaking, the problem evaporated when we discovered that k is not a physically realizable state. Nevertheless, it is of interest to \ufb01gure out how information about the particle velocity is contained in the wave function (Equation 2.101). The essential idea is this: A wave packet is a superposition of sinusoidal functions whose amplitude is modulated by \u03c6 (Figure 2.10); it consists of \u201cripples\u201d contained within an \u201cenvelope.\u201d What corresponds to the particle velocity is not the speed of the individual ripples (the so-called phase velocity), but rather the speed of the envelope (the group velocity)\u2014which, depending on the nature of the waves, can be greater than, less than, or equal to, the velocity of the ripples that go to make it up. For waves on a string, the group velocity is the same as the phase velocity. For water waves it is one-half the phase velocity, as you may have noticed when you toss a rock into a pond (if you concentrate on a particular ripple, you will see it build up from the rear, move forward through the group, and fade away at the front, while the group as a whole propagates out at half that speed). What I need to show is that for the wave function of a free particle in quantum mechanics the group velocity is twice the phase velocity\u2014just right to match the classical particle speed. The problem, then, is to determine the group velocity of a wave packet with the generic form 2.4 The Free Particle 59 \u03c5 g \u03c5 p x Figure 2.10: A wave packet. The \u201cenvelope\u201d travels at the group velocity; the \u201cripples\u201d travel at the phase velocity. +\u221e \u2212\u221e k2/2m \u03c6(k) ei(kx\u2212\u03c9t) dk. (x, t) = 1\u221a 2\u03c0 In our case \u03c9 =, but what I have to say now applies to any kind of wave packet, regardless of its dispersion relation (the formula for \u03c9 as a function of k). Let us assume that \u03c6(k) is narrowly peaked about some particular value k0. (There is nothing illegal about a broad spread in k, but such wave packets change shape rapidly\u2014different components travel at different speeds, so the whole notion of a \u201cgroup,\u201d with a well-de\ufb01ned velocity, loses its meaning.) Since the integrand is negligible except in the vicinity of k0, we may as well Taylorexpand the function \u03c9(k) about that point, and keep only the leading terms: (2.106) \u03c9(k) \u2248 \u03c90 + \u03c9 0 (k \u2212 k0), where \u03c9 0 is the derivative of \u03c9 with respect to k, at the point k0. Changing variables from k to s \u2261 k \u2212 k0 (to center the integral at k0), we have +\u221e \u2212\u221e (x, t) \u2248 1\u221a 2\u03c0 = 1\u221a 2\u03c0 \u03c6(k0 + s) ei[(k0+s)x\u2212(\u03c90+\u03c9 0s)t] ds +\u221e ei(k0x\u2212\u03c90t) 0t) \u03c6(k0 + s) eis(x\u2212\u03c9 ds. (2.107) \u2212\u221e The term in front is a sinusoidal wave (the \u201cripples\u201d), traveling at speed \u03c90/k0. It is modulated by the integral (the \u201cenvelope\u201d), which is a function of x \u2212 \u03c9 0t, and therefore propagates at the speed \u03c9 0. Thus the phase velocity is while the group velocity is vphase = \u03c9 k, vgroup = d\u03c9 dk (2.108) (2.109) (both of them evaluated at k = k0). k2/2m In our case, \u03c9 =, so \u03c9/k = (k/2", "m), whereas d\u03c9/dk = (k/m), which is twice as great. This con\ufb01rms that the group velocity of the wave packet matches the classical particle velocity: vclassical = vgroup = 2vphase. (2.110) 60 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation \u2217 \u2217\u2217 Aeikx + Be\u2212ikx and [C cos kx + D sin kx] are equivalent ways Problem 2.17 Show that of writing the same function of x, and determine the constants C and D in terms of A and B, and vice versa. Comment: In quantum mechanics, when V = 0, the exponentials represent traveling waves, and are most convenient in discussing the free particle, whereas sines and cosines correspond to standing waves, which arise naturally in the case of the in\ufb01nite square well. Problem 2.18 Find the probability current, J (Problem 1.14) for the free particle wave function Equation 2.95. Which direction does the probability \ufb02ow? Problem 2.19 This problem is designed to guide you through a \u201cproof\u201d of Plancherel\u2019s theorem, by starting with the theory of ordinary Fourier series on a \ufb01nite interval, and allowing that interval to expand to in\ufb01nity. (a) Dirichlet\u2019s theorem says that \u201cany\u201d function f (x) on the interval [\u2212a, +a] can be expanded as a Fourier series: \u221e f (x) = n=0 # an sin n\u03c0 x a + bn cos $. n\u03c0 x a Show that this can be written equivalently as f (x) = \u221e n=\u2212\u221e cnein\u03c0 x/a. What is cn, in terms of an and bn? (b) Show (by appropriate modi\ufb01cation of Fourier\u2019s trick) that cn = 1 2a f (x) e\u2212in\u03c0 x/a d x. +a \u2212a (c) Eliminate n and cn in favor of the new variables k = (n\u03c0/a) and F(k) = \u221a 2/\u03c0 acn. Show that (a) and (b) now become \u221e f (x) = 1\u221a 2\u03c0 n=\u2212\u221e F(k) eikx k; F(k) = 1\u221a 2\u03c0 +a \u2212a f (x) e\u2212ikx d x, where k is the increment in k from one n to the next. (d) Take the limit a \u2192 \u221e to obtain Plancherel\u2019s theorem. Comment: In view of their quite different origins, it is surprising (and delightful) that the two formulas\u2014one for F(k) in terms of f (x), the other for f (x) in terms of F(k)\u2014have such a similar structure in the limit a \u2192 \u221e. Problem 2.20 A free particle has the initial wave function (x, 0) = Ae\u2212a|x|, 2.5 The Delta-Function Potential 61 where A and a are positive real constants. (a) Normalize (x, 0). (b) Find \u03c6(k). (c) Construct (x, t), in the form of an integral. (d) Discuss the limiting cases (a very large, and a very small). \u2217 Problem 2.21 The gaussian wave packet. A free particle has the initial wave function where A and a are (real and positive) constants. (x, 0) = Ae\u2212ax 2, (a) Normalize (x, 0). (b) Find (x, t). Hint: Integrals of the form ax 2+bx e\u2212 +\u221e \u2212\u221e d x can be handled by \u201ccompleting the square\u201d: Let y \u2261 that ax 2 + bx b2/4a. Answer: (x, t) = = y2 \u2212 2a \u03c0 1/4 1 \u03b3 \u221a a x + (b/2a), and note e\u2212ax 2/\u03b3 2, where \u03b3 \u2261 % 1 + (2iat/m). (2.111) (c) Find |(x, t)|2. Express your answer in terms of the quantity w \u2261 a/ 1 + (2at/m)2. Sketch ||2 (as a function of x) at t = 0, and again for some very large t. Qualitatively, what happens to ||2, as time goes on?, \u03c3x, and \u03c3 p. Partial answer: = a2, but it may take p2 some algebra to reduce it to this simple form. (d) Find x, p, x", " 2 p2, (e) Does the uncertainty principle hold? At what time t does the system come closest to the uncertainty limit? 2.5 THE DELTA-FUNCTION POTENTIAL 2.5.1 Bound States and Scattering States We have encountered two very different kinds of solutions to the time-independent Schr\u00f6dinger equation: For the in\ufb01nite square well and the harmonic oscillator they are normalizable, and labeled by a discrete index n; for the free particle they are non-normalizable, and labeled by a continuous variable k. The former represent physically realizable states in their own right, the latter do not; but in both cases the general solution to the time-dependent Schr\u00f6dinger equation is a linear combination of stationary states\u2014for the \ufb01rst type this combination takes the form of a sum (over n), whereas for the second it is an integral (over k). What is the physical signi\ufb01cance of this distinction? In classical mechanics a one-dimensional time-independent potential can give rise to two rather different kinds of motion. If V (x) rises higher than the particle\u2019s total energy (E) on either side (Figure 2.11(a)), then the particle is \u201cstuck\u201d in the potential well\u2014it rocks back 62 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation V(x) E Classical turning points (a) x V(x) E V(x) E Classical turning point x (b) x V(x) E Classical turning points (c) x Figure 2.11: (a) A bound state. (b) Scattering states. (c) A classical bound state, but a quantum scattering state. and forth between the turning points, but it cannot escape (unless, of course, you provide it with a source of extra energy, such as a motor, but we\u2019re not talking about that). We call this a bound state. If, on the other hand, E exceeds V (x) on one side (or both), then the particle comes in from \u201cin\ufb01nity,\u201d slows down or speeds up under the in\ufb02uence of the potential, and returns to in\ufb01nity (Figure 2.11(b)). (It can\u2019t get trapped in the potential unless there is some mechanism, such as friction, to dissipate energy, but again, we\u2019re not talking about that.) We call this a scattering state. Some potentials admit only bound states (for instance, the harmonic oscillator); some allow only scattering states (a potential hill with no dips in it, for example); some permit both kinds, depending on the energy of the particle. 2.5 The Delta-Function Potential 63 The two kinds of solutions to the Schr\u00f6dinger equation correspond precisely to bound and scattering states. The distinction is even cleaner in the quantum domain, because the phenomenon of tunneling (which we\u2019ll come to shortly) allows the particle to \u201cleak\u201d through any \ufb01nite potential barrier, so the only thing that matters is the potential at in\ufb01nity (Figure 2.11(c)): E < V (\u2212\u221e) and V (+\u221e) \u21d2 bound state, E > V (\u2212\u221e) or V (+\u221e) \u21d2 scattering state. (2.112) In real life most potentials go to zero at in\ufb01nity, in which case the criterion simpli\ufb01es even further: E < 0 \u21d2 bound state, E > 0 \u21d2 scattering state. (2.113) Because the in\ufb01nite square well and harmonic oscillator potentials go to in\ufb01nity as x \u2192 \u00b1\u221e, they admit bound states only; because the free particle potential is zero everywhere, it only allows scattering states.43 In this section (and the following one) we shall explore potentials that support both kinds of states. 2.5.2 The Delta-Function Well The Dirac delta function is an in\ufb01nitely high, in\ufb01nitesimally narrow spike at the origin, whose area is 1 (Figure 2.12): \u03b4(x) \u2261 0, \u221e, if x = 0 if x = 0, with +\u221e \u2212\u221e \u03b4(x) d x = 1. (2.114) Technically, it isn\u2019t a function at all, since it is not \ufb01nite at x = 0 (mathematicians call it a generalized function, or distribution).44 Nevertheless, it is an extremely useful construct in theoretical physics. (For example, in electrodynamics the charge density of a point charge is a \u03b4(x) Figure 2.12: The Dirac delta function (Equation", " 2.114). x 43 If you are irritatingly observant, you may have noticed that the general theorem requiring E > Vmin (Problem 2.2) does not really apply to scattering states, since they are not normalizable. If this bothers you, try solving the Schr\u00f6dinger equation with E \u2264 0, for the free particle, and note that even linear combinations of these solutions cannot be normalized. The positive energy solutions by themselves constitute a complete set. 44 The delta function can be thought of as the limit of a sequence of functions, such as rectangles (or triangles) of ever-increasing height and ever-decreasing width. 64 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation delta function.) Notice that \u03b4(x \u2212 a) would be a spike of area 1 at the point a. If you multiply \u03b4(x \u2212 a) by an ordinary function f (x), it\u2019s the same as multiplying by f (a), f (x) \u03b4(x \u2212 a) = f (a) \u03b4(x \u2212 a), (2.115) because the product is zero anyway except at the point a. In particular, +\u221e +\u221e f (x) \u03b4(x \u2212 a) d x = f (a) \u2212\u221e \u2212\u221e \u03b4(x \u2212 a) d x = f (a). (2.116) That\u2019s the most important property of the delta function: Under the integral sign it serves to \u201cpick out\u201d the value of f (x) at the point a. (Of course, the integral need not go from \u2212\u221e to +\u221e; all that matters is that the domain of integration include the point a, so a \u2212 to a + would do, for any > 0.) Let\u2019s consider a potential of the form V (x) = \u2212\u03b1\u03b4(x), (2.117) where \u03b1 is some positive constant.45 This is an arti\ufb01cial potential, to be sure (so was the in\ufb01nite square well), but it\u2019s delightfully simple to work with, and illuminates the basic theory with a minimum of analytical clutter. The Schr\u00f6dinger equation for the delta-function well reads it yields both bound states (E < 0) and scattering states (E > 0). \u2212 2 2m d2\u03c8 d x 2 \u2212 \u03b1\u03b4(x) \u03c8 = E\u03c8; We\u2019ll look \ufb01rst at the bound states. In the region x < 0, V (x) = 0, so d2\u03c8 d x 2 = \u2212 2m E 2 \u03c8 = \u03ba 2\u03c8, where \u221a \u22122m E. \u03ba \u2261 (2.118) (2.119) (2.120) (E is negative, by assumption, so \u03ba is real and positive.) The general solution to Equation 2.119 is \u03c8(x) = Ae\u2212\u03ba x + Be\u03ba x, but the \ufb01rst term blows up as x \u2192 \u2212\u221e, so we must choose A = 0: \u03c8(x) = Be\u03ba x, (x < 0). (2.121) (2.122) In the region x > 0, V (x) is again zero, and the general solution is of the form F exp (\u2212\u03ba x) + G exp (\u03ba x); this time it\u2019s the second term that blows up (as x \u2192 +\u221e), so \u03c8(x) = Fe\u2212\u03ba x, (x > 0). (2.123) It remains only to stitch these two functions together, using the appropriate boundary conditions at x = 0. I quoted earlier the standard boundary conditions for \u03c8: 1. \u03c8 2. d\u03c8/d x is always continuous; is continuous except at points where the potential is in\ufb01nite. (2.124) 45 The delta function itself carries units of 1 & length (see Equation 2.114), so \u03b1 has the dimensions energy\u00d7length. 2.5 The Delta-Function Potential 65 \u03c8(x) \u03ba \u03ba e\u2212\u03bax Figure 2.13: Bound state wave function for the delta-function potential (Equation 2.125). x In this case the \ufb01rst boundary condition tells us that F = B, so \u03c8(x) = Be\u03ba x, Be\u2212\u03ba x, (x \u2264 0), (x \u2265 0) ; (2.125) \u03c8(x) is plotted in Figure 2.13. The second boundary condition tells us nothing; this is (like the walls of the in\ufb01nite square well) the exceptional case where V is in\ufb01nite at the join, and it\u2019s", " clear from the graph that this function has a kink at x = 0. Moreover, up to this point the delta function has not come into the story at all. It turns out that the delta function determines the discontinuity in the derivative of \u03c8, at x = 0. I\u2019ll show you now how this works, and as a byproduct we\u2019ll see why d\u03c8/d x is ordinarily continuous. The idea is to integrate the Schr\u00f6dinger equation, from \u2212 to +, and then take the limit as \u2192 0: \u2212 2 2m + \u2212 d2\u03c8 x) \u03c8(x) d x = E + \u2212 \u03c8(x) d x (2.126) The \ufb01rst integral is nothing but d\u03c8/d x, evaluated at the two end points; the last integral is zero, in the limit \u2192 0, since it\u2019s the area of a sliver with vanishing width and \ufb01nite height. Thus d\u03c8 d x \u2261 lim \u21920 \u2202\u03c8 \u2202 x \u2212 + \u2202\u03c8 \u2202 x \u2212 + = 2m 2 lim \u21920 V (x) \u03c8(x) d x. (2.127) \u2212 Ordinarily, the limit on the right is again zero, and that\u2019s why d\u03c8/d x is ordinarily continuous. But when V (x) is in\ufb01nite at the boundary, this argument fails. In particular, if V (x) = \u2212\u03b1\u03b4(x), Equation 2.116 yields d\u03c8 d x = \u2212 2m\u03b1 2 \u03c8(0). For the case at hand (Equation 2.125), d\u03c8/d x = \u2212B\u03bae\u2212\u03ba x, for (x > 0), d\u03c8/d x = +B\u03bae+\u03ba x, for (x < 0), so d\u03c8/d x so d\u03c8/d x + \u2212 = \u2212B\u03ba, = +B\u03ba, and hence (d\u03c8/d x) = \u22122B\u03ba. And \u03c8(0) = B. So Equation 2.128 says and the allowed energy (Equation 2.120) is \u03ba = m\u03b1 2, E = \u2212 2\u03ba 2 2m = \u2212 m\u03b12 22. (2.128) (2.129) (2.130) 66 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation Finally, we normalize \u03c8: +\u221e |\u03c8(x)|2 d x = 2 |B|2 \u2212\u221e \u221e 0 e\u22122\u03ba x d x = |B|2 \u03ba = 1, so (choosing the positive real root): \u221a \u03ba = B = \u221a m\u03b1. (2.131) Evidently the delta function well, regardless of its \u201cstrength\u201d \u03b1, has exactly one bound state: \u03c8(x) = \u221a m\u03b1 e\u2212m\u03b1|x|/2 ; E = \u2212 m\u03b12 22. (2.132) What about scattering states, with E > 0? For x < 0 the Schr\u00f6dinger equation reads where d2\u03c8 d x 2 = \u2212 2m E 2 \u03c8 = \u2212k2\u03c8, \u221a 2m E k \u2261 is real and positive. The general solution is \u03c8(x) = Aeikx + Be\u2212ikx, (2.133) (2.134) and this time we cannot rule out either term, since neither of them blows up. Similarly, for x > 0, \u03c8(x) = Feikx + Ge\u2212ikx. The continuity of \u03c8(x) at x = 0 requires that F + G = A + B. (2.135) (2.136) The derivatives are d\u03c8/d x = ik d\u03c8/d x = ik Feikx \u2212 Ge\u2212ikx Aeikx \u2212 Be\u2212ikx, for (x > 0),, for (x < 0), so d\u03c8/d x so d\u03c8/d x + \u2212 = ik (F \u2212 G), = ik (A \u2212 B), and hence (d\u03c8/d x) = ik (F \u2212 G \u2212 A + B). Meanwhile, \u03c8(0) = (A + B), so the second boundary condition (Equation 2.128) says ik (F \u2212 G \u2212 A + B) = \u2212 2m\u03b1 2 (A + B), or, more compactly, F \u2212 G = A (1 + 2i\u03b2) \u2212 B (1 \u2212 2i\u03b2), where \u03b2 \u2261 m\u03b1 2k. (2.137) (2.138) Having imposed both boundary conditions,", " we are left with two equations (Equations 2.136 and 2.138) in four unknowns (A, B, F, and G)\u2014\ufb01ve, if you count k. Normalization won\u2019t help\u2014this isn\u2019t a normalizable state. Perhaps we\u2019d better pause, then, and examine the physical signi\ufb01cance of these various constants. Recall that exp (ikx) gives rise (when coupled with the wiggle factor exp (\u2212i Et/)) to a wave function propagating to the right, and exp (\u2212ikx) leads to a wave propagating to the left. It follows that A (in Equation 2.134) is the amplitude of a 2.5 The Delta-Function Potential 67 Aeikx Be\u2013ikx Feikx Ge\u2013ikx x Figure 2.14: Scattering from a delta function well. wave coming in from the left, B is the amplitude of a wave returning to the left; F (Equation 2.135) is the amplitude of a wave traveling off to the right, and G is the amplitude of a wave coming in from the right (see Figure 2.14). In a typical scattering experiment particles are \ufb01red in from one direction\u2014let\u2019s say, from the left. In that case the amplitude of the wave coming in from the right will be zero: G = 0 (for scattering from the left) ; (2.139) A is the amplitude of the incident wave, B is the amplitude of the re\ufb02ected wave, and F is the amplitude of the transmitted wave. Solving Equations 2.136 and 2.138 for B and F, we \ufb01nd B = i\u03b2 1 \u2212 i\u03b2 A, F = 1 1 \u2212 i\u03b2 A. (2.140) (If you want to study scattering from the right, set A = 0; then G is the incident amplitude, F is the re\ufb02ected amplitude, and B is the transmitted amplitude.) Now, the probability of \ufb01nding the particle at a speci\ufb01ed location is given by ||2, so the relative46 probability that an incident particle will be re\ufb02ected back is R \u2261 |B|2 |A|2 = \u03b22 1 + \u03b22. (2.141) R is called the re\ufb02ection coef\ufb01cient. (If you have a beam of particles, it tells you the fraction of the incoming number that will bounce back.) Meanwhile, the probability that a particle will continue right on through is the transmission coef\ufb01cient47 T \u2261 |F|2 |A|2 = 1 1 + \u03b22. Of course, the sum of these probabilities should be 1\u2014and it is: R + T = 1. (2.142) (2.143) 46 This is not a normalizable wave function, so the absolute probability of \ufb01nding the particle at a particular location is not well de\ufb01ned; nevertheless, the ratio of probabilities for the incident and re\ufb02ected waves is meaningful. More on this in the next paragraph. 47 Note that the particle\u2019s velocity is the same on both sides of the well. Problem 2.34 treats the general case. 68 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation Notice that R and T are functions of \u03b2, and hence (Equations 2.133 and 2.138) of E: R = 1 22 E/m\u03b12, 1 + T = 1 1 + (m\u03b12/22 E). (2.144) The higher the energy, the greater the probability of transmission (which makes sense). This is all very tidy, but there is a sticky matter of principle that we cannot altogether ignore: These scattering wave functions are not normalizable, so they don\u2019t actually represent possible particle states. We know the resolution to this problem: form normalizable linear combinations of the stationary states, just as we did for the free particle\u2014true physical particles are represented by the resulting wave packets. Though straightforward in principle, this is a messy business in practice, and at this point it is best to turn the problem over to a computer.48 Meanwhile, since it is impossible to create a normalizable free-particle wave function without involving a range of energies, R and T should be interpreted as the approximate re\ufb02ection and transmission probabilities for particles with energies in the vicinity of E. Incidentally, it might strike you as peculiar that we were able to analyze a quintessentially time-dependent problem (particle comes in, scatters off a potential, and \ufb02ies off to in\ufb01nity) using stationary states. After all, \u03c8 (in Equations 2.134 and 2.135) is simply a complex, time-independent, sinusoidal function, extending (", "with constant amplitude) to in\ufb01nity in both directions. And yet, by imposing appropriate boundary conditions on this function we were able to determine the probability that a particle (represented by a localized wave packet) would bounce off, or pass through, the potential. The mathematical miracle behind this is, I suppose, the fact that by taking linear combinations of states spread over all space, and with essentially trivial time dependence, we can construct wave functions that are concentrated about a (moving) point, with quite elaborate behavior in time (see Problem 2.42). As long as we\u2019ve got the relevant equations on the table, let\u2019s look brie\ufb02y at the case of a delta-function barrier (Figure 2.15). Formally, all we have to do is change the sign of \u03b1. This kills the bound state, of course (Problem 2.2). On the other hand, the re\ufb02ection and transmission coef\ufb01cients, which depend only on \u03b12, are unchanged. Strange to say, the particle is just as likely to pass through the barrier as to cross over the well! Classically, of course, a particle cannot make it over an in\ufb01nitely high barrier, regardless of its energy. In fact, classical scattering problems are pretty dull: If E > Vmax, then T = 1 and R = 0\u2014the particle V(x) = \u03b1\u03b4(x) Figure 2.15: The delta-function barrier. x 48 There exist some powerful programs for analyzing the scattering of a wave packet from a one-dimensional potential; see, for instance, \u201cQuantum Tunneling and Wave Packets,\u201d at PhET Interactive Simulations, University of Colorado Boulder, https://phet.colorado.edu. 2.5 The Delta-Function Potential 69 certainly makes it over; if E < Vmax then T = 0 and R = 1\u2014it rides up the hill until it runs out of steam, and then returns the same way it came. Quantum scattering problems are much richer: The particle has some nonzero probability of passing through the potential even if E < Vmax. We call this phenomenon tunneling; it is the mechanism that makes possible much of modern electronics\u2014not to mention spectacular advances in microscopy. Conversely, even if E > Vmax there is a possibility that the particle will bounce back\u2014though I wouldn\u2019t advise driving off a cliff in the hope that quantum mechanics will save you (see Problem 2.35). (a) Problem 2.22 Evaluate the following integrals: \u03b4(x + 2) d x. x 3 \u2212 3x 2 + 2x \u2212 1 +1 \u22123 \u221e [cos(3x) + 2] \u03b4(x \u2212 \u03c0 ) d x. 0 +1 \u22121 exp (|x| + 3)\u03b4(x \u2212 2) d x. (b) (c) Problem 2.23 Delta functions live under integral signs, and two expressions (D1(x) and D2(x)) involving delta functions are said to be equal if +\u221e +\u221e f (x) D1(x) d x = f (x) D2(x) d x, \u2212\u221e \u2212\u221e for every (ordinary) function f (x). (a) Show that \u03b4(cx) = 1 |c| \u03b4(x), where c is a real constant. (Be sure to check the case where c is negative.) (b) Let \u03b8 (x) be the step function: \u03b8 (x) \u2261 1, 0, x > 0, x < 0. (2.145) (2.146) (In the rare case where it actually matters, we de\ufb01ne \u03b8(0) to be 1/2.) Show that d\u03b8/d x = \u03b4(x). Problem 2.24 Check the uncertainty principle for the wave function in Equation 2.132. can be tricky, because the derivative of \u03c8 has a step discontinuity Hint: Calculating = at x = 0. You may want to use the result in Problem 2.23(b). Partial answer: (m\u03b1/)2. p2 p2 \u2217 \u2217 \u2217\u2217 Problem 2.25 Check that the bound state of the delta-function well (Equation 2.132) is orthogonal to the scattering states (Equations 2.134 and 2.135). \u2217 Problem 2.26 What is the Fourier transform of \u03b4(x)? Using Plancherel\u2019s theorem, show that 70 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation \u03b4(x) = 1 2\u03c0 +\u221e \u2212\u221e eikx dk. (2.147) Comment: This formula gives any respectable mathematician apoplexy. Although the", " integral is clearly in\ufb01nite when x = 0, it doesn\u2019t converge (to zero or anything else) when x = 0, since the integrand oscillates forever. There are ways to patch it up (for instance, you can integrate from \u2212L to +L, and interpret Equation 2.147 to mean the average value of the \ufb01nite integral, as L \u2192 \u221e). The source of the problem is that the delta function doesn\u2019t meet the requirement (square-integrability) for Plancherel\u2019s theorem (see footnote 42). In spite of this, Equation 2.147 can be extremely useful, if handled with care. \u2217\u2217 Problem 2.27 Consider the double delta-function potential V (x) = \u2212\u03b1 [\u03b4(x + a) + \u03b4(x \u2212 a)], where \u03b1 and a are positive constants. (a) Sketch this potential. (b) How many bound states does it possess? Find the allowed energies, for \u03b1 = 2/ma and for \u03b1 = 2/4ma, and sketch the wave functions. (c) What are the bound state energies in the limiting cases (i) a \u2192 0 and (ii) a \u2192 \u221e (holding \u03b1 \ufb01xed)? Explain why your answers are reasonable, by comparison with the single delta-function well. \u2217\u2217 Problem 2.28 Find the transmission coef\ufb01cient, for the potential in Problem 2.27. 2.6 THE FINITE SQUARE WELL As a last example, consider the \ufb01nite square well V (x) = \u2212V0, \u2212a \u2264 x \u2264 a, |x| > a, 0, (2.148) where V0 is a (positive) constant (Figure 2.16). Like the delta-function well, this potential admits both bound states (with E < 0) and scattering states (with E > 0). We\u2019ll look \ufb01rst at the bound states. In the region x < \u2212a the potential is zero, so the Schr\u00f6dinger equation reads where \u2212 2 2m d2\u03c8 d x 2 = E\u03c8, or d2\u03c8 d x 2 = \u03ba 2\u03c8, \u221a \u22122m E \u03ba \u2261 (2.149) is real and positive. The general solution is \u03c8(x) = A exp (\u2212\u03ba x) + B exp (\u03ba x), but the \ufb01rst term blows up (as x \u2192 \u2212\u221e), so the physically admissible solution is \u03c8(x) = Be\u03ba x, In the region \u2212a < x < a, V (x) = \u2212V0, and the Schr\u00f6dinger equation reads (x < \u2212a). (2.150) 2.6 The Finite Square Well 71 V(x) \u2013a a x \u2013V0 Figure 2.16: The \ufb01nite square well (Equation 2.148). \u2212 2 2m d2\u03c8 d x 2 \u2212 V0\u03c8 = E\u03c8, or d2\u03c8 d x 2 = \u2212l2\u03c8, where \u221a 2m (E + V0). l \u2261 (2.151) Although E is negative, for bound states, it must be greater than \u2212V0, by the old theorem E > Vmin (Problem 2.2); so l is also real and positive. The general solution is49 \u03c8(x) = C sin(lx) + D cos(lx), (\u2212a < x < a), (2.152) where C and D are arbitrary constants. Finally, in the region x > a the potential is again zero; the general solution is \u03c8(x) = F exp (\u2212\u03ba x) + G exp (\u03ba x), but the second term blows up (as x \u2192 \u221e), so we are left with \u03c8(x) = Fe\u2212\u03ba x, (x > a). (2.153) The next step is to impose boundary conditions: \u03c8 and d\u03c8/d x continuous at \u2212a and +a. But we can save a little time by noting that this potential is an even function, so we can assume with no loss of generality that the solutions are either even or odd (Problem 2.1(c)). The advantage of this is that we need only impose the boundary conditions on one side (say, at +a); the other side is then automatic, since \u03c8(\u2212x) = \u00b1\u03c8(x). I\u2019ll work out the even solutions; you get to do the odd ones in Problem 2.29. The cosine is even (and the sine is odd), so I\u2019m looking for solutions of the form \u23a7 \u23aa\ufffd", "\ufffd \u03c8(x) = \u23aa\u23a9 Fe\u2212\u03ba x, D cos(lx), \u03c8(\u2212x), (x > a), (0 < x < a), (x < 0). The continuity of \u03c8(x), at x = a, says Fe\u2212\u03baa = D cos(la), and the continuity of d\u03c8/d x says \u2212 \u03ba Fe\u2212\u03baa = \u2212l D sin(la). Dividing Equation 2.156 by Equation 2.155, we \ufb01nd that (2.154) (2.155) (2.156) 49 You can, if you like, write the general solution in exponential form (C eilx + De\u2212ilx ). This leads to the same \ufb01nal result, but since the potential is symmetric, we know the solutions will be either even or odd, and the sine/cosine notation allows us to exploit this right from the start. 72 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation tan z (z0/z)2\u22121 z0 \u03c0/2 \u03c0 3\u03c0/2 2\u03c0 5\u03c0/2 z Figure 2.17: Graphical solution to Equation 2.159, for z0 = 8 (even states). This is a formula for the allowed energies, since \u03ba and l are both functions of E. To solve for E, we \ufb01rst adopt some nicer notation: Let \u03ba = l tan(la). (2.157) z \u2261 la, and z0 \u2261 a \u03ba 2 + l2 % 2mV0. (2.158) = 2mV0/2, so \u03baa = z2 0 \u2212 z2, and According to Equations 2.149 and 2.151, Equation 2.157 reads tan z = (z0/z)2 \u2212 1. (2.159) This is a transcendental equation for z (and hence for E) as a function of z0 (which is a measure of the \u201csize\u201d of the well). It can be solved numerically, using a computer, or graphically, (z0/z)2 \u2212 1 on the same grid, and looking for points of intersection (see by plotting tan z and Figure 2.17). Two limiting cases are of special interest: 1. Wide, deep well. If z0 is very large (pushing the curve (z0/z)2 \u2212 1 upward on the graph, and sliding the zero crossing, z0, to the right) the intersections occur just slightly below zn = n\u03c0/2, with n odd; it follows (Equations 2.158 and 2.151) that En + V0 \u2248 n2\u03c0 22 2m(2a)2 (n = 1, 3, 5,...). (2.160) But E + V0 is the energy above the bottom of the well, and on the right side we have precisely the in\ufb01nite square well energies, for a well of width 2a (see Equation 2.30)\u2014 or rather, half of them, since this n is odd. (The other ones, of course, come from the odd wave functions, as you\u2019ll discover in Problem 2.29.) So the \ufb01nite square well goes over to the in\ufb01nite square well, as V0 \u2192 \u221e; however, for any \ufb01nite V0 there are only a \ufb01nite number of bound states. 2. Shallow, narrow well. As z0 decreases, there are fewer and fewer bound states, until \ufb01nally, for z0 < \u03c0/2, only one remains. It is interesting to note, however, that there is always one bound state, no matter how \u201cweak\u201d the well becomes. You\u2019re welcome to normalize \u03c8 (Equation 2.154), if you\u2019re interested (Problem 2.30), but I\u2019m going to move on now to the scattering states (E > 0). To the left, where V (x) = 0, we have 2.6 The Finite Square Well 73 \u03c8(x) = Aeikx + Be\u2212ikx, for (x < \u2212a), where (as usual) \u221a k \u2261 2m E. (2.161) (2.162) Inside the well, where V (x) = \u2212V0, \u03c8(x) = C sin(lx) + D cos(lx), for (\u2212a < x < a), (2.163) where, as before, \u221a 2m (E + V0). l \u2261 To the right, assuming there is no incoming wave", " in this region, we have \u03c8(x) = Feikx. (2.164) (2.165) Here A is the incident amplitude, B is the re\ufb02ected amplitude, and F is the transmitted amplitude.50 There are four boundary conditions: Continuity of \u03c8(x) at \u2212a says Ae\u2212ika + Beika = \u2212C sin(la) + D cos(la), (2.166) continuity of d\u03c8/d x at \u2212a gives # ik Ae\u2212ika \u2212 Beika $ = l [C cos(la) + D sin(la)] (2.167) continuity of \u03c8(x) at +a yields and continuity of d\u03c8/d x at +a requires C sin(la) + D cos(la) = Feika, l [C cos(la) \u2212 D sin(la)] = ik Feika. (2.168) (2.169) We can use two of these to eliminate C and D, and solve the remaining two for B and F (see Problem 2.32): B = i sin(2la) 2kl l2 \u2212 k2 F = e\u22122ika A (k2+l2) 2kl cos(2la) \u2212 i sin(2la) F,. (2.170) (2.171) The transmission coef\ufb01cient T = |F|2 / |A|2, expressed in terms of the original variables, is given by T \u22121 = 1 + V 2 0 4E (E + V0) sin2 2a % 2m(E + V0). (2.172) Notice that T = 1 (the well becomes \u201ctransparent\u201d) whenever the sine is zero, which is to say, when 50 We could look for even and odd functions, as we did in the case of bound states, but the scattering problem is inherently asymmetric, since the waves come in from one side only, and the exponential notation (representing traveling waves) is more natural in this context. 74 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation T 1 Figure 2.18: Transmission coef\ufb01cient as a function of energy (Equation 2.172). E % 2a 2m (En + V0) = n\u03c0, where n is any integer. The energies for perfect transmission, then, are given by En + V0 = n2\u03c0 22 2m (2a)2, (2.173) (2.174) which happen to be precisely the allowed energies for the in\ufb01nite square well. T is plotted in Figure 2.18, as a function of energy.51 \u2217 Problem 2.29 Analyze the odd bound state wave functions for the \ufb01nite square well. Derive the transcendental equation for the allowed energies, and solve it graphically. Examine the two limiting cases. Is there always an odd bound state? Problem 2.30 Normalize \u03c8(x) in Equation 2.154, to determine the constants D and F. Problem 2.31 The Dirac delta function can be thought of as the limiting case of a rectangle of area 1, as the height goes to in\ufb01nity and the width goes to zero. Show that the deltafunction well (Equation 2.117) is a \u201cweak\u201d potential (even though it is in\ufb01nitely deep), in the sense that z0 \u2192 0. Determine the bound state energy for the delta-function potential, by treating it as the limit of a \ufb01nite square well. Check that your answer is consistent with Equation 2.132. Also show that Equation 2.172 reduces to Equation 2.144 in the appropriate limit. Problem 2.32 Derive Equations 2.170 and 2.171. Hint: Use Equations 2.168 and 2.169 to solve for C and D in terms of F: C = sin(la) + i cos(la) eika F; D = cos(la) \u2212 i sin(la) eika F. k l k l Plug these back into Equations 2.166 and 2.167. Obtain the transmission coef\ufb01cient, and con\ufb01rm Equation 2.172. 51 This remarkable phenomenon was observed in the laboratory before the advent of quantum mechanics, in the form of the Ramsauer\u2013Townsend effect. For an illuminating discussion see Richard W. Robinett, Quantum Mechanics, Oxford University Press, 1997, Section 12.4.1. 2.6 The Finite Square Well 75 \u2217\u2217 \u2217 Problem 2.33 Determine the transmission coef\ufb01cient for a rectangular barrier (same as Equation 2.148, only with V (x) = +V", "0 > 0 in the region \u2212a < x < a). Treat separately the three cases E < V0, E = V0, and E > V0 (note that the wave function inside the barrier is different in the three cases). Partial answer: for E < V0,52 T \u22121 = 1 + V 2 0 4E (V0 \u2212 E) sinh2 Problem 2.34 Consider the \u201cstep\u201d potential:53 % 2a 2m(V0 \u2212 E). 0, V0, (a) Calculate the re\ufb02ection coef\ufb01cient, for the case E < V0, and comment on the x \u2264 0, x > 0. V (x) = answer. (b) Calculate the re\ufb02ection coef\ufb01cient for the case E > V0. (c) For a potential (such as this one) that does not go back to zero to the right of the barrier, the transmission coef\ufb01cient is not simply |F|2 / | A|2 (with A the incident amplitude and F the transmitted amplitude), because the transmitted wave travels at a different speed. Show that T = E \u2212 V0 E for E > V0. Hint: You can \ufb01gure it out using Equation 2.99, or\u2014more elegantly, but less informatively\u2014from the probability current (Problem 2.18). What is T, for E < V0? |F|2 |A|2 (2.175), (d) For E > V0, calculate the transmission coef\ufb01cient for the step potential, and check that T + R = 1. Problem 2.35 A particle of mass m and kinetic energy E > 0 approaches an abrupt potential drop V0 (Figure 2.19).54 V(x) \u2013V0 x Figure 2.19: Scattering from a \u201ccliff\u201d (Problem 2.35). 52 This is a good example of tunneling\u2014classically the particle would bounce back. 53 For interesting commentary see C. O. Dib and O. Orellana, Eur. J. Phys. 38, 045403 (2017). 54 For further discussion see P. L. Garrido, et al., Am. J. Phys. 79, 1218 (2011). 76 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation (a) What is the probability that it will \u201cre\ufb02ect\u201d back, if E = V0/3? Hint: This is just like Problem 2.34, except that the step now goes down, instead of up. (b) I drew the \ufb01gure so as to make you think of a car approaching a cliff, but obviously the probability of \u201cbouncing back\u201d from the edge of a cliff is far smaller than what you got in (a)\u2014unless you\u2019re Bugs Bunny. Explain why this potential does not correctly represent a cliff. Hint: In Figure 2.20 the potential energy of the car drops discontinuously to \u2212V0, as it passes x = 0; would this be true for a falling car? (c) When a free neutron enters a nucleus, it experiences a sudden drop in potential energy, from V = 0 outside to around \u221212 MeV (million electron volts) inside. Suppose a neutron, emitted with kinetic energy 4 MeV by a \ufb01ssion event, strikes such a nucleus. What is the probability it will be absorbed, thereby initiating another \ufb01ssion? Hint: You calculated the probability of re\ufb02ection in part (a); use T = 1 \u2212 R to get the probability of transmission through the surface. FURTHER PROBLEMS ON CHAPTER 2 Problem 2.36 Solve the time-independent Schr\u00f6dinger equation with appropriate boundary conditions for the \u201ccentered\u201d in\ufb01nite square well: V (x) = 0 (for \u2212a < x < +a), V (x) = \u221e (otherwise). Check that your allowed energies are consistent with mine (Equation 2.30), and con\ufb01rm that your \u03c8s can be obtained from mine (Equation 2.31) by the substitution x \u2192 (x + a) /2 (and appropriate renormalization). Sketch your \ufb01rst three solutions, and compare Figure 2.2. Note that the width of the well is now 2a. Problem 2.37 A particle in the in\ufb01nite square well (Equation 2.22) has the initial wave function (x, 0) = A sin3(\u03c0 x/a) (0 \u2264 x \u2264 a). Determine A, \ufb01nd (x, t), and calculate x, as a function of time. What is the", " expectation value of the energy? Hint: sinn \u03b8 and cosn \u03b8 can be reduced, by repeated application of the trigonometric sum formulas, to linear combinations of sin(m\u03b8 ) and cos(m\u03b8 ), with m = 0, 1, 2,..., n. Problem 2.38 (a) Show that the wave function of a particle in the in\ufb01nite square well returns to its original form after a quantum revival time T = 4ma2/\u03c0. That is: (x, T ) = (x, 0) for any state (not just a stationary state). (b) What is the classical revival time, for a particle of energy E bouncing back and forth between the walls? (c) For what energy are the two revival times equal?55 55 The fact that the classical and quantum revival times bear no obvious relation to one another (and the quantum one doesn\u2019t even depend on the energy) is a curious paradox; see D. F. Styer, Am. J. Phys. 69, 56 (2001). Further Problems on Chapter 2 77 Problem 2.39 In Problem 2.7(d) you got the expectation value of the energy by summing the series in Equation 2.21, but I warned you (in footnote 21) not to try it the \u201cold (x, 0)\u2217 \u02c6H (x, 0) d x, because the discontinuous \ufb01rst fashioned way,\u201d H = derivative of (x, 0) renders the second derivative problematic. Actually, you could have done it using integration by parts, but the Dirac delta function affords a much cleaner way to handle such anomalies. (a) Calculate the \ufb01rst derivative of (x, 0) (in Problem 2.7), and express the answer in terms of the step function, \u03b8 (x \u2212 a/2), de\ufb01ned in Equation 2.146. (b) Exploit the result of Problem 2.23(b) to write the second derivative of (x, 0) in terms of the delta function. (c) Evaluate the integral answer as before. (x, 0)\u2217 \u02c6H (x, 0) d x, and check that you get the same \u2217\u2217 Problem 2.40 A particle of mass m in the harmonic oscillator potential (Equation 2.44) starts out in the state (x, 0) = A 1 \u2212 2 2 m\u03c9 x e\u2212 m\u03c9 2 x 2, for some constant A. (a) Determine A and the coef\ufb01cients cn in the expansion of this state in terms of the stationary states of the harmonic oscillator. (b) In a measurement of the particle\u2019s energy, what results could you get, and what are their probabilities? What is the expectation value of the energy? (c) At a later time T the wave function is (x, T ) = B 1 + 2 2 m\u03c9 x e\u2212 m\u03c9 2 x 2, for some constant B. What is the smallest possible value of T? Problem 2.41 Find the allowed energies of the half harmonic oscillator V (x) = (1/2) m\u03c92x 2, \u221e, x > 0, x < 0. (This represents, for example, a spring that can be stretched, but not compressed.) Hint: This requires some careful thought, but very little actual calculation. \u2217\u2217\u2217 Problem 2.42 In Problem 2.21 you analyzed the stationary gaussian free particle wave packet. Now solve the same problem for the traveling gaussian wave packet, starting with the initial wave function (x, 0) = Ae\u2212ax 2 eilx, where l is a (real) constant. [Suggestion: In going from \u03c6(k) to (x, t), change variables to u \u2261 k \u2212 l before doing the integral.] Partial answer: 78 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation (x, t) = 1/4 1 \u03b3 2a \u03c0 e\u2212a(x\u2212lt/m)2/\u03b3 2 eil(x\u2212lt/2m) \u221a where \u03b3 \u2261 1 + 2iat/m, as before. Notice that (x, t) has the structure of a gaussian \u201cenvelope\u201d modulating a traveling sinusoidal wave. What is the speed of the envelope? What is the speed of the traveling wave? \u2217\u2217 Problem 2.43 Solve the time-independent Schr\u00f6dinger equation for a centered in\ufb01nite square well with a delta-function barrier in the middle: V (x) = \u03b1\u03b4(x), \u2212a < x < +a, |x| \u2265 a. \u221e, Treat", " the even and odd wave functions separately. Don\u2019t bother to normalize them. Find the allowed energies (graphically, if necessary). How do they compare with the corresponding energies in the absence of the delta function? Explain why the odd solutions are not affected by the delta function. Comment on the limiting cases \u03b1 \u2192 0 and \u03b1 \u2192 \u221e. Problem 2.44 If two (or more) distinct56 solutions to the (time-independent) Schr\u00f6dinger equation have the same energy E, these states are said to be degenerate. For example, the free particle states are doubly degenerate\u2014one solution representing motion to the right, and the other motion to the left. But we have never encountered normalizable degenerate solutions, and this is no accident. Prove the following theorem: In one dimension57(\u2212\u221e < x < \u221e) there are no degenerate bound states. [Hint: Suppose there are two solutions, \u03c81 and \u03c82, with the same energy E. Multiply the Schr\u00f6dinger equation for \u03c81 by \u03c82, and the Schr\u00f6dinger equation for \u03c82 by \u03c81, and subtract, to show that (\u03c82d\u03c81/d x \u2212 \u03c81d\u03c82/d x) is a constant. Use the fact that for normalizable solutions \u03c8 \u2192 0 at \u00b1\u221e to demonstrate that this constant is in fact zero. Conclude that \u03c82 is a multiple of \u03c81, and hence that the two solutions are not distinct.] Problem 2.45 In this problem you will show that the number of nodes of the stationary states of a one-dimensional potential always increases with energy.58 Consider two (real, normalized) solutions (\u03c8n and \u03c8m) to the time-independent Schr\u00f6dinger equation (for a given potential V (x)), with energies En > Em. (a) Show that d d x d\u03c8m d x \u03c8n \u2212 \u03c8m d\u03c8n d x = 2m 2 (En \u2212 Em) \u03c8m\u03c8n. 56 If two solutions differ only by a multiplicative constant (so that, once normalized, they differ only by a phase factor ei\u03c6 ), they represent the same physical state, and in this sense they are not distinct solutions. Technically, by \u201cdistinct\u201d I mean \u201clinearly independent.\u201d 57 In higher dimensions such degeneracy is very common, as we shall see in Chapters 4 and 6. Assume that the potential does not consist of isolated pieces separated by regions where V = \u221e\u2014two isolated in\ufb01nite square wells, for instance, would give rise to degenerate bound states, for which the particle is either in one well or in the other. 58 M. Moriconi, Am. J. Phys. 75, 284 (2007). Further Problems on Chapter 2 79 (b) Let x1 and x2 be two adjacent nodes of the function \u03c8m(x). Show that \u03c8 m (x2) \u03c8n(x2) \u2212 \u03c8 m (x1) \u03c8n(x1) = 2m 2 (En \u2212 Em) x2 x1 \u03c8m\u03c8n d x. \u2217\u2217 (c) If \u03c8n(x) has no nodes between x1 and x2, then it must have the same sign everywhere in the interval. Show that (b) then leads to a contradiction. Therefore, between every pair of nodes of \u03c8m(x), \u03c8n(x) must have at least one node, and in particular the number of nodes increases with energy. Problem 2.46 Imagine a bead of mass m that slides frictionlessly around a circular wire ring of circumference L. (This is just like a free particle, except that \u03c8(x + L) = \u03c8(x).) Find the stationary states (with appropriate normalization) and the corresponding allowed energies. Note that there are (with one exception) two independent solutions for each energy En\u2014corresponding to clockwise and counter-clockwise circulation; call them \u03c8 + (x). How do you account for this degeneracy, in n view of the theorem in Problem 2.44 (why does the theorem fail, in this case)? (x) and \u03c8 \u2212 n Problem 2.47 Attention: This is a strictly qualitative problem\u2014no calculations allowed! Consider the \u201cdouble square well\u201d potential (Figure 2.20). Suppose the depth V0 and the width a are \ufb01xed, and large enough so that several bound states occur. (a) Sketch the ground state wave function \u03c81 and the \ufb01rst excited state \u03c82, (i) for the case b = 0, (ii) for b \u2248 a, and (iii) for b", " a. (b) Qualitatively, how do the corresponding energies (E1 and E2) vary, as b goes from 0 to \u221e? Sketch E1(b) and E2(b) on the same graph. (c) The double well is a very primitive one-dimensional model for the potential experienced by an electron in a diatomic molecule (the two wells represent the attractive force of the nuclei). If the nuclei are free to move, they will adopt the con\ufb01guration of minimum energy. In view of your conclusions in (b), does the electron tend to draw the nuclei together, or push them apart? (Of course, there is also the internuclear repulsion to consider, but that\u2019s a separate problem.) V(x) a b a x Figure 2.20: The double square well (Problem 2.47). \u2013V0 80 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation Problem 2.48 Consider a particle of mass m in the potential \u23a7 \u23aa\u23a8 V (x) = \u23aa\u23a9 \u221e \u2212322/ma2 0 x < 0, 0 \u2264 x \u2264 a, x > a. (a) How many bound states are there? (b) In the highest-energy bound state, what is the probability that the particle would be found outside the well (x > a)? Answer: 0.542, so even though it is \u201cbound\u201d by the well, it is more likely to be found outside than inside! \u2217\u2217\u2217 Problem 2.49 (a) Show that (x, t) = 1/4 m\u03c9 \u03c0 exp! \u2212 m\u03c9 \u22122i\u03c9t + it m \" \u2212 2x0xe\u2212i\u03c9t satis\ufb01es the time-dependent Schr\u00f6dinger equation for the harmonic oscillator potential (Equation 2.44). Here x0 is any real constant with the dimensions of length.59 (b) Find |(x, t)|2, and describe the motion of the wave packet. (c) Compute x and p, and check that Ehrenfest\u2019s theorem (Equation 1.38) is satis\ufb01ed. \u2217\u2217 Problem 2.50 Consider the moving delta-function well: V (x, t) = \u2212\u03b1\u03b4(x \u2212 vt), where v is the (constant) velocity of the well. (a) Show that the time-dependent Schr\u00f6dinger equation admits the exact solution60 \u221a (x, t) = e\u2212m\u03b1|x\u2212vt|/2 e\u2212i (E+(1/2)mv2)t\u2212mvx /, m\u03b1 where E = \u2212m\u03b12/22 is the bound-state energy of the stationary delta function. Hint: Plug it in and check it! Use the result of Problem 2.23(b). (b) Find the expectation value of the Hamiltonian in this state, and comment on the result. \u2217\u2217 Problem 2.51 Free fall. Show that x + 1 2 (x, t) = 0 gt 2, t exp \u2212i mgt x + 1 6 gt 2 (2.176) satis\ufb01es the time-dependent Schr\u00f6dinger equation for a particle in a uniform gravitational \ufb01eld, V (x) = mgx, (2.177) 59 This rare example of an exact closed-form solution to the time-dependent Schr\u00f6dinger equation was discovered by Schr\u00f6dinger himself, in 1926. One way to obtain it is explored in Problem 6.30. For a discussion of this and related problems see W. van Dijk, et al., Am. J. Phys. 82, 955 (2014). 60 See Problem 6.35 for a derivation. Further Problems on Chapter 2 81 where 0(x, t) is the free gaussian wave packet (Equation 2.111). Find x as a function of time, and comment on the result.61 \u2217\u2217\u2217 Problem 2.52 Consider the potential V (x) = \u2212 2a2 m sech2(ax), where a is a positive constant, and \u201csech\u201d stands for the hyperbolic secant. (a) Graph this potential. (b) Check that this potential has the ground state and \ufb01nd its energy. Normalize \u03c80, and sketch its graph. \u03c80(x) = A sech(ax), (c) Show that the function \u03c8k(x) = A ik \u2212 a tanh(ax) ik + a eikx, \u221a (where k \u2261 energy E. Since tanh z \u2192 \u22121 as z \u2192 \u2212\u221e, 2m E/, as usual", ") solves the Schr\u00f6dinger equation for any (positive) \u03c8k(x) \u2248 Aeikx, for large negative x. This represents, then, a wave coming in from the left with no accompanying re\ufb02ected wave (i.e. no term exp (\u2212ikx)). What is the asymptotic form of \u03c8k(x) at large positive x? What are R and T, for this potential? Comment: This is a famous example of a re\ufb02ectionless potential\u2014every incident particle, regardless its energy, passes right through.62 Problem 2.53 The Scattering Matrix. The theory of scattering generalizes in a pretty obvious way to arbitrary localized potentials (Figure 2.21). To the left (Region I), V (x) = 0, so \u221a \u03c8(x) = Aeikx + Be\u2212ikx, where k \u2261 2m E. (2.178) Aeikx Be\u2013ikx V(x) Feikx Ge\u2013ikx x Region I Region II Region III Figure 2.21: Scattering from an arbitrary localized potential (V (x) = 0 except in Region II); Problem 2.53. 61 For illuminating discussion see M. Nauenberg, Am. J. Phys. 84, 879 (2016). 62 R. E. Crandall and B. R. Litt, Annals of Physics, 146, 458 (1983). 82 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation To the right (Region III), V (x) is again zero, so \u03c8(x) = Feikx + Ge\u2212ikx. In between (Region II), of course, I can\u2019t tell you what \u03c8 is until you specify the potential, but because the Schr\u00f6dinger equation is a linear, second-order differential equation, the general solution has got to be of the form (2.179) \u03c8(x) = C f (x) + Dg(x), where f (x) and g(x) are two linearly independent particular solutions.63 There will be four boundary conditions (two joining Regions I and II, and two joining Regions II and III). Two of these can be used to eliminate C and D, and the other two can be \u201csolved\u201d for B and F in terms of A and G: B = S11 A + S12G, F = S21 A + S22G. The four coef\ufb01cients Si j, which depend on k (and hence on E), constitute a 2 \u00d7 2 matrix S, called the scattering matrix (or S-matrix, for short). The S-matrix tells you the outgoing amplitudes (B and F) in terms of the incoming amplitudes ( A and G): B F = S11 S21 S12 S22 A G. (2.180) In the typical case of scattering from the left, G = 0, so the re\ufb02ection and transmission coef\ufb01cients are Rl = |B|2 |A|2 = |S11|2, Tl = G=0 |F|2 |A|2 = |S21|2. (2.181) G=0 For scattering from the right, A = 0, and Rr = |F|2 |G|2 = |S22|2, Tr = A=0 |B|2 |G|2 A=0 = |S12|2. (2.182) (a) Construct the S-matrix for scattering from a delta-function well (Equation 2.117). (b) Construct the S-matrix for the \ufb01nite square well (Equation 2.148). Hint: This requires no new work, if you carefully exploit the symmetry of the problem. \u2217\u2217\u2217 Problem 2.54 The transfer matrix.64 The S-matrix (Problem 2.53) tells you the outgoing amplitudes (B and F) in terms of the incoming amplitudes (A and G)\u2014Equation 2.180. For some purposes it is more convenient to work with the transfer matrix, M, which gives you the amplitudes to the right of the potential (F and G) in terms of those to the left (A and B): F G = M11 M12 M21 M22 A B. (2.183) 63 See any book on differential equations\u2014for example, John L. Van Iwaarden, Ordinary Differential Equations with Numerical Techniques, Harcourt Brace Jovanovich, San Diego, 1985, Chapter 3. 64 For applications of this method see, for instance, D. J. Grif\ufb01ths and C. A. Steinke, Am. J. Phys. 69, 137 (2001) or S. Das,", " Am. J. Phys. 83, 590 (2015). Further Problems on Chapter 2 83 M1 V = 0 V = 0 M2 x V = 0 Figure 2.22: A potential consisting of two isolated pieces (Problem 2.54). (a) Find the four elements of the M-matrix, in terms of the elements of the S-matrix, and vice versa. Express Rl, Tl, Rr, and Tr (Equations 2.181 and 2.182) in terms of elements of the M-matrix. (b) Suppose you have a potential consisting of two isolated pieces (Figure 2.22). Show that the M-matrix for the combination is the product of the two M-matrices for each section separately: M = M2M1. (2.184) (This obviously generalizes to any number of pieces, and accounts for the usefulness of the M-matrix.) (c) Construct the M-matrix for scattering from a single delta-function potential at point a: V (x) = \u2212\u03b1\u03b4(x \u2212 a). (d) By the method of part (b), \ufb01nd the M-matrix for scattering from the double delta- function V (x) = \u2212\u03b1 [\u03b4(x + a) + \u03b4(x \u2212 a)]. What is the transmission coef\ufb01cient for this potential? \u2217\u2217 Problem 2.55 Find the ground state energy of the harmonic oscillator, to \ufb01ve signi\ufb01cant digits, by the \u201cwag-the-dog\u201d method. That is, solve Equation 2.73 numerically, varying K until you get a wave function that goes to zero at large \u03be. In Mathematica, appropriate input code would be Plot[ Evaluate[ u[x] /. NDSolve[ {u[x] -(x2 - K)*u[x] == 0, u[0] == 1, u[0] == 0}, u[x], {x, 0, b} ] ], {x, a, b}, PlotRange -> {c, d} ] (Here (a, b) is the horizontal range of the graph, and (c, d) is the vertical range\u2014 start with a = 0, b = 10, c = \u221210, d = 10.) We know that the correct solution is K = 2n + 1, so you might start with a \u201cguess\u201d of K = 0.9. Notice what the \u201ctail\u201d of 84 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation the wave function does. Now try K = 1.1, and note that the tail \ufb02ips over. Somewhere in between those values lies the correct solution. Zero in on it by bracketing K tighter and tighter. As you do so, you may want to adjust a, b, c, and d, to zero in on the cross-over point. Problem 2.56 Find the \ufb01rst three excited state energies (to \ufb01ve signi\ufb01cant digits) for the harmonic oscillator, by wagging the dog (Problem 2.55). For the \ufb01rst (and third) excited state you will need to set u[0] == 0, u[0] == 1.) Problem 2.57 Find the \ufb01rst four allowed energies (to \ufb01ve signi\ufb01cant digits) for the in\ufb01nite square well, by wagging the dog. Hint: Refer to Problem 2.55, making appropriate changes to the differential equation. This time the condition you are looking for is u (1) = 0. Problem 2.58 In a monovalent metal, one electron per atom is free to roam throughout the object. What holds such a material together\u2014why doesn\u2019t it simply fall apart into a pile of individual atoms? Evidently the energy of the composite structure must be less than the energy of the isolated atoms. This problem offers a crude but illuminating explanation for the cohesiveness of metals. (a) Estimate the energy of N isolated atoms, by treating each one as an electron in the ground state of an in\ufb01nite square well of width a (Figure 2.23(a)). (b) When these atoms come together to form a metal, we get N electrons in a much larger in\ufb01nite square well of width N a (Figure 2.23(b)). Because of the Pauli exclusion principle (which we will discuss in Chapter 5) there can only be one electron (two, if you include spin, but let\u2019s ignore that) in each allowed state. What is the lowest energy for this system (Figure 2.23(b))? (a) (", "b) a N a Figure 2.23: (a) N electrons in individual wells of width a. (b) N electrons in a single well of width N a. Further Problems on Chapter 2 85 (c) The difference of these two energies is the cohesive energy of the metal\u2014the energy it would take to tear it apart into isolated atoms. Find the cohesive energy per atom, in the limit of large N. (d) A typical atomic separation in a metal is a few \u00c5ngstr\u00f6m (say, a \u2248 4 \u00c5). What is the numerical value of the cohesive energy per atom, in this model? (Measured values are in the range of 2\u20134 eV.) \u2217\u2217\u2217 Problem 2.59 The \u201cbouncing ball.\u201d65 Suppose V (x) = mgx, \u221e, x > 0, x \u2264 0. (2.185) (a) Solve the (time-independent) Schr\u00f6dinger equation for this potential. Hint: First convert it to dimensionless form: (2.186) \u2212 y(z) + z y(z) = y(z) \u221a 1/ \u221a a \u03c8(x) (the by letting z \u2261 ax and y(z) \u2261 a is just so y(z) is normalized with respect to z when \u03c8(x) is normalized with respect to x). What are the constants a and? Actually, we might as well set a \u2192 1\u2014this amounts to a convenient choice for the unit of length. Find the general solution to this equation (in Mathematica DSolve will do the job). The result is (of course) a linear combination of two (probably unfamiliar) functions. Plot each of them, for (\u221215 < z < 5). One of them clearly does not go to zero at large z (more precisely, it\u2019s not normalizable), so discard it. The allowed values of (and hence of E) are determined by the condition \u03c8(0) = 0. Find the ground state 1 numerically (in Mathematica FindRoot will do it), and also the 10th, 10. Obtain the corresponding normalization factors. Plot \u03c81(x) and \u03c810(x), for 0 \u2264 z < 16. Just as a check, con\ufb01rm that \u03c81(x) and \u03c810(x) are orthogonal. (b) Find (numerically) the uncertainties \u03c3x and \u03c3 p for these two states, and check that the uncertainty principle is obeyed. (c) The probability of \ufb01nding the ball in the neighborhood d x of height x is (of course) \u03c1Q(x) d x = |\u03c8(x)|2 d x. The nearest classical analog would be the fraction of time an elastically bouncing ball (with the same energy, E) spends in the neighborhood d x of height x (see Problem 1.11). Show that this is \u03c1C (x) d x = \u221a 2 mg E (E \u2212 mgx) d x, (2.187) or, in our units (with a = 1), \u221a 2 Plot \u03c1Q(x) and \u03c1C (x) for the state \u03c810(x), on the range 0 \u2264 x \u2264 12.5; superimpose the graphs (Show, in Mathematica), and comment on the result. ( \u2212 x) \u03c1C (x) = (2.188). 1 65 This problem was suggested by Nicholas Wheeler. 86 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation \u2217\u2217\u2217 Problem 2.60 The 1/x 2 potential. Suppose V (x) = \u2212\u03b1/x 2, \u221e, x > 0, x \u2264 0. (2.189) where \u03b1 is some positive constant with the appropriate dimensions. We\u2019d like to \ufb01nd the bound states\u2014solutions to the time-independent Schr\u00f6dinger equation d2\u03c8 d x 2 \u03c8 = E\u03c8 (2.190) \u03b1 x 2 2m 2 \u2212 \u2212 with negative energy (E < 0). (a) Let\u2019s \ufb01rst go for the ground state energy, E0. Prove, on dimensional grounds, that there is no possible formula for E0\u2014no way to construct (from the available constants m,, and \u03b1) a quantity with the units of energy. That\u2019s weird, but it gets worse.... (b) For convenience, rewrite Equation 2.190 as d2\u03c8 2\u03c8, where \u03b2 \u2261 2m\u03b1 2 and \u03ba \u2261 \u221a \u22122m E. (2.191) Show that if \u03c8(x) satis\ufb01es this equation with energy E", ", then so too does \u03c8(\u03bbx), with energy E = \u03bb2 E, for any positive number \u03bb. [This is a catastrophe: if there exists any solution at all, then there\u2019s a solution for every (negative) energy! Unlike the square well, the harmonic oscillator, and every other potential well we have encountered, there are no discrete allowed states\u2014and no ground state. A system with no ground state\u2014no lowest allowed energy\u2014would be wildly unstable, cascading down to lower and lower levels, giving off an unlimited amount of energy as it falls. It might solve our energy problem, but we\u2019d all be fried in the process.] Well, perhaps there simply are no solutions at all.... (c) (Use a computer for the remainder of this problem.) Show that \u221a \u03c8\u03ba (x) = A x Kig(\u03ba x), (2.192) \u221a satis\ufb01es Equation 2.191 (here Kig is the modi\ufb01ed Bessel function of order ig, and \u03b2 \u2212 1/4). Plot this function, for g = 4 (you might as well let \u03ba = 1 for g \u2261 the graph; this just sets the scale of length). Notice that it goes to 0 as x \u2192 0 and as x \u2192 \u221e. And it\u2019s normalizable: determine A.66 How about the old rule that the number of nodes counts the number of lower-energy states? This function has an in\ufb01nite number of nodes, regardless of the energy (i.e. of \u03ba). I guess that\u2019s consistent, since for any E there are always an in\ufb01nite number of states with even lower energy. (d) This potential confounds practically everything we have come to expect. The problem is that it blows up too violently as x \u2192 0. If you move the \u201cbrick wall\u201d over a hair, 66 \u03c8k (x) is normalizable as long as g is real\u2014which is to say, provided \u03b2 > 1/4. For more on this strange problem see A. M. Essin and D. J. Grif\ufb01ths, Am. J. Phys. 74, 109 (2006), and references therein. \u2217\u2217\u2217 Further Problems on Chapter 2 87 V (x) = \u2212\u03b1/x 2, \u221e, x > > 0, x \u2264, (2.193) it\u2019s suddenly perfectly normal. Plot the ground state wave function, for g = 4 and = 1 (you\u2019ll \ufb01rst need to determine the appropriate value of \u03ba), from x = 0 to x = 6. Notice that we have introduced a new parameter (), with the dimensions of length, so the argument in (a) is out the window. Show that the ground state energy takes the form E0 = \u2212 f (\u03b2), (2.194) \u03b1 2 for some function f of the dimensionless quantity \u03b2. Problem 2.61 One way to obtain the allowed energies of a potential well numerically is to turn the Schr\u00f6dinger equation into a matrix equation, by discretizing the variable x. (, with x j+1 \u2212 x j \u2261 x, and let Slice the relevant interval at evenly spaced points likewise V j \u2261 V (x j )). Then \u03c8 j+1 \u2212 \u03c8 j d\u03c8 d x \u2248 \u03c8 j+1 \u2212 \u03c8 j x, d2\u03c8 d x 2 \u2248 \u03c8 j \u2212 \u03c8 j\u22121 \u2212 (x)2 = \u03c8 j+1 \u2212 2\u03c8 j + \u03c8 j\u22121 (x)2. (2.195) (The approximation presumably improves as x decreases.) The discretized Schr\u00f6dinger equation reads + V j \u03c8 j = E\u03c8 j, (2.196) \u2212 or 2 2m \u03c8 j+1 \u2212 2\u03c8 j + \u03c8 j\u22121 (x)2 \u2212 \u03bb\u03c8 j+1 + 2\u03bb + V j \u03c8 j \u2212 \u03bb\u03c8 j\u22121 = E\u03c8 j, where \u03bb \u2261 2 2m (x)2. (2.197) In matrix form, H = E (2.198) where (letting v j \u2261 V j /\u03bb\u22121 \u22121 0 \u22121 0 0 H \u2261 \u03bb and \u22121 2 + v j \u22121 0 \u22121 2 + v j+1 0 0 \u22121 \u239e \u239f \u239f \u239f \u239f \u239f \u239f \u23a0... \u2261 \u239e \u239f \u239f \u239f \u239f \u239f \u239f \u23a0 \u239b \u239c \u239c \u239c \u239c \ufffd", "\ufffd\ufffd \u239c \u239d... \u03c8 j\u22121 \u03c8 j \u03c8 j+1... (2.199) (2.200) 88 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation (what goes in the upper left and lower right corners of H depends on the boundary conditions, as we shall see). Evidently the allowed energies are the eigenvalues of the matrix H (or would be, in the limit x \u2192 0).67 Apply this method to the in\ufb01nite square well. Chop the interval (0 \u2264 x \u2264 a) into N + 1 equal segments (so that x = a/(N + 1)), letting x0 \u2261 0 and x N +1 \u2261 a. The boundary conditions \ufb01x \u03c80 = \u03c8N +1 = 0, leaving \u239e \u239b \u2261 \u239c \u239d \u239f \u23a0. (2.201) \u03c81... \u03c8N (a) Construct the N \u00d7 N matrix H, for N = 1, N = 2, and N = 3. (Make sure you are correctly representing Equation 2.197 for the special cases j = 1 and j = N.) (b) Find the eigenvalues of H for these three cases \u201cby hand,\u201d and compare them with the exact allowed energies (Equation 2.30). (c) Using a computer (Mathematica\u2019s Eigenvalues package will do it) \ufb01nd the \ufb01ve lowest eigenvalues numerically for N = 10 and N = 100, and compare the exact energies. (d) Plot (by hand) the eigenvectors for N = 1, 2, and 3, and (by computer, Eigenvectors) the \ufb01rst three eigenvectors for N = 10 and N = 100. \u2217\u2217 Problem 2.62 Suppose the bottom of the in\ufb01nite square well is not \ufb02at (V (x) = 0), but rather V (x) = 500 V0 sin \u03c0 x a, where V0 \u2261 2 2ma2. Use the method of Problem 2.61 to \ufb01nd the three lowest allowed energies numerically, and plot the associated wave functions (use N = 100). Problem 2.63 The Boltzmann equation68 P(n) = 1 Z e\u2212\u03b2 En, Z \u2261 n e\u2212\u03b2 En, \u03b2 \u2261 1 k B T, (2.202) gives the probability of \ufb01nding a system in the state n (with energy En), at temperature T (k B is Boltzmann\u2019s constant). Note: The probability here refers to the random thermal distribution, and has nothing to do with quantum indeterminacy. Quantum mechanics will only enter this problem through quantization of the energies En. (a) Show that the thermal average of the system\u2019s energy can be written as \u00afE = n En P(n) = \u2212 \u2202 \u2202\u03b2 ln(Z ). (2.203) 67 For further discussion see Joel Franklin, Computational Methods for Physics (Cambridge University Press, Cambridge, UK, 2013), Section 10.4.2. 68 See, for instance, Daniel V. Schroeder, An Introduction to Thermal Physics, Pearson, Boston (2000), Section 6.1. Further Problems on Chapter 2 89 (b) For a quantum simple harmonic oscillator the index n is the familiar quantum number, and En = (n + 1/2) \u03c9. Show that in this case the partition function Z is Z = e\u2212\u03b2\u03c9/2 1 \u2212 e\u2212\u03b2\u03c9. (2.204) You will need to sum a geometric series. Incidentally, for a classical simple harmonic oscillator it can be shown that Zclassical = 2\u03c0/ (\u03c9\u03b2). (c) Use your results from parts (a) and (b) to show that for the quantum oscillator \u00afE = \u03c9 2 1 + e\u2212\u03b2\u03c9 1 \u2212 e\u2212\u03b2\u03c9. (2.205) For a classical oscillator the same reasoning would give \u00afEclassical = 1/\u03b2 = k B T. (d) A crystal consisting of N atoms can be thought of as a collection of 3N oscillators (each atom is attached by springs to its 6 nearest neighbors, along the x, y, and z directions, but those springs are shared by the atoms at the two ends). The heat capacity of the crystal (per atom) will therefore be Show that (in this model) C = 3k B C = 3 \u2202 \u00afE \u2202 T. 2 \u03b8E T e\u03b8E /T 2 e\u03b8E /T \u2212 1, (2.206) (2.207) where \u03b8E \u2261 \u03c9/k B is the so-called Einstein temperature. The same", " reasoning using the classical expression for \u00afE yields Cclassical = 3k B, independent of temperature. (e) Sketch the graph of C/k B versus T /\u03b8E. Your result should look something like the data for diamond in Figure 2.24, and nothing like the classical prediction. Cv Cp 2.0 1.5 1.0 0. 200 400 Temperature T (K) 600 800 1000 Figure 2.24: Speci\ufb01c heat of diamond (for Problem 2.63). From Semiconductors on NSM (http://www.ioffe.rssi.ru/SVA/NSM/Semicond/). 90 CHAPTER 2 Time-Independent Schr\u00f6dinger Equation Problem 2.64 Legendre\u2019s differential equation reads 1 \u2212 x 2 d f d x where is some (non-negative) real number. (a) Assume a power series solution, \u2212 2 x d2 f d x 2 + ( + 1) f = 0, (2.208) f (x) = \u221e n=0 an x n, and obtain a recursion relation for the constants an. (b) Argue that unless the series truncates (which can only happen if is an integer), the solution will diverge at x = 1. (c) When is an integer, the series for one of the two linearly independent solutions (either feven or fodd depending on whether is even or odd) will truncate, and those solutions are called Legendre polynomials P(x). Find P0(x), P1(x), P2(x), and P3(x) from the recursion relation. Leave your answer in terms of either a0 or a1.69 69 By convention Legendre polynomials are normalized such that P(1) = 1. Note that the nonvanishing coef\ufb01cients will take different values for different. 3 FORMALISM 3.1 HILBERT SPACE In the previous two chapters we have stumbled on a number of interesting properties of simple quantum systems. Some of these are \u201caccidental\u201d features of speci\ufb01c potentials (the even spacing of energy levels for the harmonic oscillator, for example), but others seem to be more general, and it would be nice to prove them once and for all (the uncertainty principle, for instance, and the orthogonality of stationary states). The purpose of this chapter is to recast the theory in more powerful form, with that in mind. There is not much here that is genuinely new; the idea, rather, is to make coherent sense of what we have already discovered in particular cases. Quantum theory is based on two constructs: wave functions and operators. The state of a system is represented by its wave function, observables are represented by operators. Mathematically, wave functions satisfy the de\ufb01ning conditions for abstract vectors, and operators act on them as linear transformations. So the natural language of quantum mechanics is linear algebra.1 But it is not, I suspect, a form of linear algebra with which you may be familiar. In an N dimensional space it is simplest to represent a vector, |\u03b1, by the N -tuple of its components, {an}, with respect to a speci\ufb01ed orthonormal basis: \u239b \u239e |3.1) \u239c \u239c \u239c \u239d a1 a2... aN the inner product, \u03b1|\u03b2, of two vectors (generalizing the dot product in three dimensions) is a complex number, \u03b1|\u03b2 = a\u2217 1 b1 + a\u2217 2 b2 + \u00b7 \u00b7 \u00b7 + a\u2217 N bN ; (3.2) linear transformations, T, are represented by matrices (with respect to the speci\ufb01ed basis), which act on vectors (to produce new vectors) by the ordinary rules of matrix multiplication: |\u03b2 = \u02c6T |\u03b1 \u2192 b = Ta = \u239c \u239c \u239c \u239d \u239b \u239e \u239b \u239e t11 t21... tN 1 t12 t22... tN 2 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 t1N t2N... tN N a1 a2... aN \u239f \u239f \u239f \u23a0 \u239c \u239c \u239c \u239d \u239f \u239f \u239f \u23a0. (3.3) But the \u201cvectors\u201d we encounter in quantum mechanics are (for the most part) functions, and they live in in\ufb01nite-dimensional spaces. For them the N -tuple/matrix notation is awkward, at best, and manipulations that are well behaved in the \ufb01nite-dimensional case can be", " 1 If you have never studied linear algebra, you should read the Appendix before continuing. 92 CHAPTER 3 Formalism problematic. (The underlying reason is that whereas the \ufb01nite sum in Equation 3.2 always exists, an in\ufb01nite sum\u2014or an integral\u2014may not converge, in which case the inner product does not exist, and any argument involving inner products is immediately suspect.) So even though most of the terminology and notation should be familiar, it pays to approach this subject with caution. The collection of all functions of x constitutes a vector space, but for our purposes it is much too large. To represent a possible physical state, the wave function must be normalized: ||2 d x = 1. The set of all square-integrable functions, on a speci\ufb01ed interval,2 f (x) such that b a | f (x)|2 d x < \u221e, (3.4) constitutes a (much smaller) vector space (see Problem 3.1(a)). Mathematicians call it L 2(a, b); physicists call it Hilbert space.3 In quantum mechanics, then: Wave functions live in Hilbert space. (3.5) We de\ufb01ne the inner product of two functions, f (x) and g(x), as follows: f |g \u2261 b a f (x)\u2217g(x) d x. (3.6) If f and g are both square-integrable (that is, if they are both in Hilbert space), their inner product is guaranteed to exist (the integral in Equation 3.6 converges to a \ufb01nite number).4 This follows from the integral Schwarz inequality:5 2 For us, the limits (a and b) will almost always be \u00b1\u221e, but we might as well keep things more general for the moment. 3 Technically, a Hilbert space is a complete inner product space, and the collection of square-integrable functions is only one example of a Hilbert space\u2014indeed, every \ufb01nite-dimensional vector space is trivially a Hilbert space. But since L2 is the arena of quantum mechanics, it\u2019s what physicists generally mean when they say \u201cHilbert space.\u201d By the way, the word complete here means that any Cauchy sequence of functions in Hilbert space converges to a function that is also in the space: it has no \u201choles\u201d in it, just as the set of all real numbers has no holes (by contrast, the space of all polynomials, for example, like the set of all rational numbers, certainly does have holes in it). The completeness of a space has nothing to do with the completeness (same word, unfortunately) of a set of functions, which is the property that any other function can be expressed as a linear combination of them. For an accessible introduction to Hilbert space see Daniel T. Gillespie, A Quantum Mechanics Primer (International Textbook Company, London, 1970), Sections 2.3 and 2.4. 4 In Chapter 2 we were obliged on occasion to work with functions that were not normalizable. Such functions lie outside Hilbert space, and we are going to have to handle them with special care. For the moment, I shall assume that all the functions we encounter are in Hilbert space. 5 For a proof, see Frigyes Riesz and Bela Sz.-Nagy, Functional Analysis (Dover, Mineola, NY, 1990), Section 21. In a \ufb01nite-dimensional vector space the Schwarz inequality, |\u03b1|\u03b2|2 \u2264 \u03b1|\u03b1 \u03b2|\u03b2, is easy to prove (see Problem A.5). But that proof assumes the existence of the inner products, which is precisely what we are trying to establish here. b a f (x)\u2217 g(xx)|2 d x b a |g(x)|2 d x. (3.7) 3.1 Hilbert Space 93 You can check for yourself that de\ufb01nition (Equation 3.6) satis\ufb01es all the conditions for an inner product (Problem 3.1(b)). Notice in particular that Moreover, the inner product of f (x) with itself, g | f = f | gx)|2 d x, (3.8) (3.9) is real and non-negative; it\u2019s zero only when f (x) = 0.6 A function is said to be normalized if its inner product with itself is 1; two functions are orthogonal if their inner product is 0; and a set of functions, { fn}, is orthonormal if they are normalized and mutually orthogonal: fm| fn = \u03b4mn. (3.10) Finally, a set of functions is complete if any other function (in Hilbert space) can be expressed", " as a linear combination of them: f (x) = \u221e n=1 cn fn(x). If the functions { fn(x)} are orthonormal, the coef\ufb01cients are given by Fourier\u2019s trick: cn = fn| f, (3.11) (3.12) as you can check for yourself. I anticipated this terminology, of course, back in Chapter 2. (The stationary states of the in\ufb01nite square well (Equation 2.31) constitute a complete orthonormal set on the interval (0, a); the stationary states for the harmonic oscillator (Equation 2.68 or 2.86) are a complete orthonormal set on the interval (\u2212\u221e, \u221e).) Problem 3.1 (a) Show that the set of all square-integrable functions is a vector space (refer to Section A.1 for the de\ufb01nition). Hint: The main point is to show that the sum of two square-integrable functions is itself square-integrable. Use Equation 3.7. Is the set of all normalized functions a vector space? (b) Show that the integral in Equation 3.6 satis\ufb01es the conditions for an inner product (Section A.2). 6 What about a function that is zero everywhere except at a few isolated points? The integral (Equation 3.9) would still vanish, even though the function itself does not. If this bothers you, you should have been a math major. In physics such pathological functions do not occur, but in any case, in Hilbert space two functions are considered equivalent if the integral of the absolute square of their difference vanishes. Technically, vectors in Hilbert space represent equivalence classes of functions. 94 CHAPTER 3 Formalism \u2217 Problem 3.2 (a) For what range of \u03bd is the function f (x) = x \u03bd in Hilbert space, on the interval (0, 1)? Assume \u03bd is real, but not necessarily positive. (b) For the speci\ufb01c case \u03bd = 1/2, is f (x) in this Hilbert space? What about x f (x)? How about (d/d x) f (x)? 3.2 OBSERVABLES 3.2.1 Hermitian Operators The expectation value of an observable Q(x, p) can be expressed very neatly in inner-product notation:7 Q = \u2217 \u02c6Q d x = | \u02c6Q. (3.13) Now, the outcome of a measurement has got to be real, and so, a fortiori, is the average of many measurements: Q = Q\u2217. But the complex conjugate of an inner product reverses the order (Equation 3.8), so \u02c6Q \u02c6Q =, (3.14) (3.15) and this must hold true for any wave function. Thus operators representing observables have the very special property that \u02c6Q f f \u02c6Q f f = for all f (x). We call such operators hermitian.8 Actually, most books require an ostensibly stronger condition: \u02c6Qg f = \u02c6Q f g for all f (x) and all g(x). (3.16) (3.17) But it turns out, in spite of appearances, that this is perfectly equivalent to my de\ufb01nition (Equation 3.16), as you will prove in Problem 3.3. So use whichever you like. The essential point is that a hermitian operator can be applied either to the \ufb01rst member of an inner product or to 7 Remember that \u02c6Q is the operator constructed from Q by the replacement p \u2192 \u2212i d/d x. These operators are linear, in the sense that a f (x) + bg(x) \u02c6Q = a \u02c6Q f (x) + b \u02c6Qg(x), for any functions f and g and any complex numbers a and b. They constitute linear transformations (Section A.3) on the space of all functions. However, they sometimes carry a function inside Hilbert space into a function outside it (see Problem 3.2(b)), and in that case the domain of the operator (the set of functions on which it acts) may have to be restricted (see Problem 3.48). 8 In a \ufb01nite-dimensional vector space hermitian operators are represented by hermitian matrices; a hermitian matrix is equal to its transpose conjugate: T = T\u2020 = \u02dcT \u2217. If this is unfamiiar to you please see the Appendix. the second, with the same result, and her", "mitian operators naturally arise in quantum mechanics because their expectation values are real: Observables are represented by hermitian operators. (3.18) 3.2 Observables 95 Well, let\u2019s check this. Is the momentum operator, for example, hermitian? \u02c6pg d x = \u2212i f \u2217g f \u2217 (\u2212ii d f d x \u2212\u221e \u221e \u2212\u221e dg d x \u2212\u221e \u02c6p f g. (3.19) I used integration by parts, of course, and threw away the boundary term for the usual reason: If f (x) and g(x) are square integrable, they must go to zero at \u00b1\u221e.9 Notice how the complex conjugation of i compensates for the minus sign picked up from integration by parts\u2014the operator d/d x (without the i) is not hermitian, and it does not represent a possible observable. The hermitian conjugate (or adjoint) of an operator \u02c6Q is the operator \u02c6Q\u2020 such that g for all f and g \u02c6Qg \u02c6Q\u2020 f. = f (3.20) A hermitian operator, then, is equal to its hermitian conjugate: \u02c6Q = \u02c6Q\u2020. \u2217 g Problem 3.3 Show that if \u02c6Q f are equivalent). Hint: First let h = f + g, and then let h = f + ig. h for all f and g (i.e. the two de\ufb01nitions of \u201chermitian\u201d \u2014Equations 3.16 and 3.17\u2014 for all h (in Hilbert space), then \u02c6Qh = h \u02c6Qh \u02c6Qg f = Problem 3.4 (a) Show that the sum of two hermitian operators is hermitian. (b) Suppose \u02c6Q is hermitian, and \u03b1 is a complex number. Under what condition (on \u03b1) is \u03b1 \u02c6Q hermitian? (c) When is the product of two hermitian operators hermitian? (d) Show that the position operator d2/d x 2 + V 2/2m \u2212 x are hermitian. \u02c6x and the Hamiltonian operator \u02c6H = \u2217 Problem 3.5 (a) Find the hermitian conjugates of x, i, and d/d x. (b) Show that \u2020 = \u02c6R\u2020 \u02c6Q\u2020 (note the reversed order), \u02c6Q \u02c6R (c \u02c6Q)\u2020 = c\u2217 \u02c6Q\u2020 for a complex number c. \u02c6Q + \u02c6R = \u02c6Q\u2020 + \u02c6R\u2020, and (c) Construct the hermitian conjugate of a+ (Equation 2.48). 9 As I mentioned in Chapter 1, there exist pathological functions that are square-integrable but do not go to zero at in\ufb01nity. However, such functions do not arise in physics, and if you are worried about it we will simply restrict the domain of our operators to exclude them. On \ufb01nite intervals, though, you really do have to be more careful with the boundary terms, and an operator that is hermitian on (\u2212\u221e, \u221e) may not be hermitian on (0, \u221e) or (\u2212\u03c0, \u03c0 ). (If you\u2019re wondering about the in\ufb01nite square well, it\u2019s safest to think of those wave functions as residing on the in\ufb01nite line\u2014they just happen to be zero outside (0, a).) See Problem 3.48. 96 CHAPTER 3 Formalism 3.2.2 Determinate States Ordinarily, when you measure an observable Q on an ensemble of identically prepared systems, all in the same state, you do not get the same result each time\u2014this is the indeterminacy of quantum mechanics. Question: Would it be possible to prepare a state such that every measurement of Q is certain to return the same value (call it q)? This would be, if you like, a determinate state, for the observable Q. (Actually, we already know one example: Stationary states are determinate states of the Hamiltonian; a measurement of the energy, on a particle in the stationary state n, is certain to yield the corresponding \u201callowed\u201d energy En.) Well, the standard deviation of Q, in a determinate state, would be zero, which is to say, 0 2 Q \u2212 Q 1 0 = \u02c6Q \u2212 q 1 2 = \u03c3 2 = \u02c6Q \u2212 q ", "\u02c6Q \u2212 q = 0. (3.21) (Of course, if every measurement gives q, their average is also q: Q = q. I used the fact that \u02c6Q (and hence also \u02c6Q \u2212 q) is a hermitian operator, to move one factor over to the \ufb01rst term in the inner product.) But the only vector whose inner product with itself vanishes is 0, so \u02c6Q = q. (3.22) This is the eigenvalue equation for the operator \u02c6Q; is an eigenfunction of \u02c6Q, and q is the corresponding eigenvalue: Determinate states of Q are eigenfunctions of \u02c6Q. (3.23) Measurement of Q on such a state is certain to yield the eigenvalue, q.10 Note that the eigenvalue is a number (not an operator or a function). You can multiply any eigenfunction by a constant, and it is still an eigenfunction, with the same eigenvalue. Zero does not count as an eigenfunction (we exclude it by de\ufb01nition\u2014otherwise every num\u02c6Q and all q). But ber would be an eigenvalue, since \u02c6Q 0 = q 0 = 0 for any linear operator there\u2019s nothing wrong with zero as an eigenvalue. The collection of all the eigenvalues of an operator is called its spectrum. Sometimes two (or more) linearly independent eigenfunctions share the same eigenvalue; in that case the spectrum is said to be degenerate. (You encountered this term already, for the case of energy eigenstates, if you worked Problems 2.44 or 2.46.) For example, determinate states of the total energy are eigenfunctions of the Hamiltonian: \u02c6H \u03c8 = E\u03c8, (3.24) which is precisely the time-independent Schr\u00f6dinger equation. In this context we use the letter E for the eigenvalue, and the lower case \u03c8 for the eigenfunction (tack on the wiggle factor exp to make it, if you like; it\u2019s still an eigenfunction of \u02c6H \u2212 i Et/. 10 I\u2019m talking about a competent measurement, of course\u2014it\u2019s always possible to make a mistake, and simply get the wrong answer, but that\u2019s not the fault of quantum mechanics. 3.3 Eigenfunctions of a Hermitian Operator 97 Example 3.1 Consider the operator \u02c6Q \u2261 i d d\u03c6, (3.25) where \u03c6 is the usual polar coordinate in two dimensions. (This operator might arise in a physical context if we were studying the bead-on-a-ring; see Problem 2.46.) Is \u02c6Q hermitian? Find its eigenfunctions and eigenvalues. Solution: Here we are working with functions f (\u03c6) on the \ufb01nite interval 0 \u2264 \u03c6 \u2264 2\u03c0, with the property that f (\u03c6 + 2\u03c0 ) = f (\u03c6), since \u03c6 and \u03c6 + 2\u03c0 describe the same physical point. Using integration by parts, g d\u03c6 = i f \u2217g \u02c6Q g g d\u03c6 = \u02c6Q f f \u2217 \u2212 = 2\u03c0 2\u03c0 i f i 2\u03c0 0 0 d f \u2217 d\u03c6 dg d\u03c6 0 (3.26), so \u02c6Q is hermitian (this time the boundary term disappears by virtue of Equation 3.26). The eigenvalue equation, has the general solution i d d\u03c6 f (\u03c6) = q f (\u03c6), f (\u03c6) = Ae\u2212iq\u03c6. Equation 3.26 restricts the possible values of the q: e\u2212iq2\u03c0 = 1 \u21d2 q = 0, \u00b11, \u00b12,.... The spectrum of this operator is the set of all integers, and it is nondegenerate. (3.27) (3.28) (3.29) \u02c6Q = d2/d\u03c62, where (as in Example 3.1) \u03c6 is the Problem 3.6 Consider the operator azimuthal angle in polar coordinates, and the functions are subject to Equation 3.26. Is \u02c6Q hermitian? Find its eigenfunctions and eigenvalues. What is the spectrum of \u02c6Q? Is the spectrum degenerate? 3.3 EIGENFUNCTIONS OF A HERMITIAN OPERATOR Our attention is thus directed to the eigenfunctions of hermitian operators (physically: determinate states of observables). These fall into two categories", ": If the spectrum is discrete (i.e. the eigenvalues are separated from one another) then the eigenfunctions lie in Hilbert space and they constitute physically realizable states. If the spectrum is continuous (i.e. the eigenvalues \ufb01ll out an entire range) then the eigenfunctions are not normalizable, and they do not represent possible wave functions (though linear combinations of them\u2014involving necessarily a spread in eigenvalues\u2014may be normalizable). Some operators have a discrete spectrum only (for example, the Hamiltonian for the harmonic oscillator), some have only a continuous spectrum (for example, the free particle Hamiltonian), and some have both a discrete part and a continuous part (for example, the Hamiltonian for a \ufb01nite square well). The discrete case is easier to 98 CHAPTER 3 Formalism handle, because the relevant inner products are guaranteed to exist\u2014in fact, it is very similar to the \ufb01nite-dimensional theory (the eigenvectors of a hermitian matrix). I\u2019ll treat the discrete case \ufb01rst, and then the continuous one. 3.3.1 Discrete Spectra Mathematically, the normalizable eigenfunctions of a hermitian operator have two important properties: Theorem 1 Their eigenvalues are real. Proof: Suppose \u02c6Q f = q f, (i.e. f (x) is an eigenfunction of \u02c6Q, with eigenvalue q), and11 0 1 \u02c6Q f 0 = \u02c6Q f 1 f \u02c6Q is hermitian. Then f q f | f = q\u2217 f | f (q is a number, so it comes outside the integral, and because the \ufb01rst function in the inner product is complex conjugated (Equation 3.6), so too is the q on the right). But f | f cannot be zero ( f (x) = 0 is not a legal eigenfunction), so q = q\u2217, and hence q is real. QED This is comforting: If you measure an observable on a particle in a determinate state, you will at least get a real number. Theorem 2 Eigenfunctions belonging to distinct eigenvalues are orthogonal. Proof: Suppose and \u02c6Q is hermitian. Then f \u02c6Qg = qg, \u02c6Q f = q f, \u02c6Qg = and g, so q f | g = q\u2217 f | g \u02c6Q f (again, the inner products exist because the eigenfunctions are in Hilbert space). But q is real (from Theorem 1), so if q = q it must be that f |g = 0. QED That\u2019s why the stationary states of the in\ufb01nite square well, for example, or the harmonic oscillator, are orthogonal\u2014they are eigenfunctions of the Hamiltonian with distinct eigenvalues. But this property is not peculiar to them, or even to the Hamiltonian\u2014the same holds for determinate states of any observable. Unfortunately, Theorem 2 tells us nothing about degenerate states. However, if two (or more) eigenfunctions share the same eigenvalue, any linear combination of them is itself an eigenfunction, with the same eigenvalue (Problem 3.7), and we can use the Gram\u2013Schmidt orthogonalization procedure (Problem A.4) to construct orthogonal eigenfunctions within each degenerate subspace. It is almost never necessary to do this explicitly (thank God!), but it can always be done in principle. So even in the presence of degeneracy the eigenfunctions q = q 11 It is here that we assume the eigenfunctions are in Hilbert space\u2014otherwise the inner product might not exist at all. 3.3 Eigenfunctions of a Hermitian Operator 99 can be chosen to be orthonormal, and we shall always assume that this has been done. That licenses the use of Fourier\u2019s trick, which depends on the orthonormality of the basis functions. In a \ufb01nite-dimensional vector space the eigenvectors of a hermitian matrix have a third fundamental property: They span the space (every vector can be expressed as a linear combination of them). Unfortunately, the proof does not generalize to in\ufb01nite-dimensional spaces. But the property itself is essential to the internal consistency of quantum mechanics, so (following Dirac12) we will take it as an axiom (or, more precisely, as a restriction on the class of hermitian operators that can represent observables): Axiom: The eigenfunctions of an observable operator are complete: Any function (in", " Hilbert space) can be expressed as a linear combination of them.13 Problem 3.7 (a) Suppose that f (x) and g(x) are two eigenfunctions of an operator \u02c6Q, with the same eigenvalue q. Show that any linear combination of f and g is itself an eigenfunction of \u02c6Q, with eigenvalue q. (b) Check that f (x) = exp (x) and g(x) = exp (\u2212x) are eigenfunctions of the operator d2/d x 2, with the same eigenvalue. Construct two linear combinations of f and g that are orthogonal eigenfunctions on the interval (\u22121, 1). Problem 3.8 (a) Check that the eigenvalues of the hermitian operator in Example 3.1 are real. Show that the eigenfunctions (for distinct eigenvalues) are orthogonal. (b) Do the same for the operator in Problem 3.6. 3.3.2 Continuous Spectra If the spectrum of a hermitian operator is continuous, the eigenfunctions are not normalizable, and the proofs of Theorems 1 and 2 fail, because the inner products may not exist. Nevertheless, there is a sense in which the three essential properties (reality, orthogonality, and completeness) still hold. I think it\u2019s best to approach this case through speci\ufb01c examples. Example 3.2 Find the eigenfunctions and eigenvalues of the momentum operator (on the interval \u2212\u221e < x < \u221e). Solution: Let f p(x) be the eigenfunction and p the eigenvalue: The general solution is \u2212 i d d x f p(x) = p f p(x). f p (x) = Aei px/. (3.30) 12 P. A. M. Dirac, The Principles of Quantum Mechanics, Oxford University Press, New York (1958). 13 In some speci\ufb01c cases completeness is provable (we know that the stationary states of the in\ufb01nite square well, for example, are complete, because of Dirichlet\u2019s theorem). It is a little awkward to call something an \u201caxiom\u201d that is provable in some cases, but I don\u2019t know a better way to do it. 100 CHAPTER 3 Formalism This is not square-integrable for any (complex) value of p\u2014the momentum operator has no eigenfunctions in Hilbert space. And yet, if we restrict ourselves to real eigenvalues, we do recover a kind of ersatz \u201corthonormality.\u201d Referring to Problems 2.23(a) and 2.26, \u221e \u2212\u221e f \u2217 p(x) f p(x) d x = |A|2 \u221e \u2212\u221e If we pick A = 1/ \u221a ei( p\u2212 p)x/ d x = |A|2 2\u03c0 \u03b4 p \u2212 p. (3.31) 2\u03c0, so that f p(x) = 1\u221a 2\u03c0 ei px/, 3.32) (3.33) then which is reminiscent of true orthonormality (Equation 3.10)\u2014the indices are now continuous variables, and the Kronecker delta has become a Dirac delta, but otherwise it looks just the same. I\u2019ll call Equation 3.33 Dirac orthonormality. Most important, the eigenfunctions (with real eigenvalues) are complete, with the sum (in Equation 3.11) replaced by an integral: Any (square-integrable) function f (x) can be written in the form \u221e f (x) = \u2212\u221e c( p) f p(x) dp = 1\u221a 2\u03c0 \u221e \u2212\u221e c( p) ei px/ dp. (3.34) The \u201ccoef\ufb01cients\u201d (now a function, c( p)) are obtained, as always, by Fourier\u2019s trick( p) dp = c( p) \u03b4 p \u2212 p dp = c. p (3.35) \u221e \u2212\u221e \u221e \u2212\u221e Alternatively, you can get them from Plancherel\u2019s theorem (Equation 2.103); indeed, the expansion (Equation 3.34) is nothing but a Fourier transform. The eigenfunctions of momentum (Equation 3.32) are sinusoidal, with wavelength \u03bb = 2\u03c0 p. (3.36) This is the old de Broglie formula (Equation 1.39), which I promised to justify at the appropriate", " time. It turns out to be a little more subtle than de Broglie imagined, because we now know that there is actually no such thing as a particle with determinate momentum. But we could make a normalizable wave packet with a narrow range of momenta, and it is to such an object that the de Broglie relation applies. What are we to make of Example 3.2? Although none of the eigenfunctions of \u02c6p lives in Hilbert space, a certain family of them (those with real eigenvalues) resides in the nearby \u201csuburbs,\u201d with a kind of quasi-normalizability. They do not represent possible physical states, but they are still very useful (as we have already seen, in our study of one-dimensional scattering).14 14 What about the eigenfunctions with nonreal eigenvalues? These are not merely non-normalizable\u2014they actually blow up at \u00b1\u221e. Functions in what I called the \u201csuburbs\u201d of Hilbert space (the entire metropolitan area is sometimes called a \u201crigged Hilbert space\u201d; see, for example, Leslie Ballentine\u2019s Quantum Mechanics: A Modern Development, 3.3 Eigenfunctions of a Hermitian Operator 101 Example 3.3 Find the eigenfunctions and eigenvalues of the position operator. Solution: Let gy(x) be the eigenfunction and y the eigenvalue: \u02c6x gy(x) = x gy(x) = y gy(x). (3.37) Here y is a \ufb01xed number (for any given eigenfunction), but x is a continuous variable. What function of x has the property that multiplying it by x is the same as multiplying it by the constant y? Obviously it\u2019s got to be zero, except at the one point x = y; in fact, it is nothing but the Dirac delta function: gy(x) = A\u03b4(x \u2212 y). This time the eigenvalue has to be real; the eigenfunctions are not square integrable, but again they admit Dirac orthonormality: \u221e \u221e x \u2212 y \u03b4(x \u2212 y) d x = |A|2 \u03b4 y \u2212 y. g\u2217 y(x) gy(x) d x = |A|2 \u03b4 \u2212\u221e \u2212\u221e (3.38) (3.39) (3.40) If we pick A = 1, so gy(x) = \u03b4(x \u2212 y), = \u03b4 y \u2212 y. then gy These eigenfunctions are also complete: gy \u221e f (x) = \u2212\u221e c(y) gy(x) dy = with \u221e \u2212\u221e c(y)\u03b4(x \u2212 y) dy, (3.41) (trivial, in this case, but you can get it from Fourier\u2019s trick if you insist). c(y) = f (y) (3.42) If the spectrum of a hermitian operator is continuous (so the eigenvalues are labeled by a continuous variable\u2014 p or y, in the examples; z, generically, in what follows), the eigenfunctions are not normalizable, they are not in Hilbert space and they do not represent possible physical states; nevertheless, the eigenfunctions with real eigenvalues are Dirac orthonormalizable and complete (with the sum now an integral). Luckily, this is all we really require. World Scienti\ufb01c, 1998) have the property that although they have no (\ufb01nite) inner product with themselves, they do admit inner products with all members of Hilbert space. This is not true for eigenfunctions of \u02c6p with nonreal eigenvalues. In particular, I showed that the momentum operator is hermitian for functions in Hilbert space, but the argument depended on dropping the boundary term (in Equation 3.19). That term is still zero if g is an eigenfunction of \u02c6p with a real eigenvalue (as long as f is in Hilbert space), but not if the eigenvalue has an imaginary part. In this sense any complex number is an eigenvalue of the operator \u02c6p, but only real numbers are eigenvalues of the hermitian operator \u02c6p\u2014the others lie outside the space over which \u02c6p is hermitian. 102 CHAPTER 3 Formalism Problem 3.9 (a) Cite a Hamiltonian from Chapter 2 (other than the harmonic oscillator) that has only a discrete spectrum. (b) Cite a Hamiltonian from Chapter 2 (other than the free particle) that has only a continuous spectrum. (c) Cite a Hamiltonian from Chapter 2 (other", " than the \ufb01nite square well) that has both a discrete and a continuous part to its spectrum. Problem 3.10 Is the ground state of the in\ufb01nite square well an eigenfunction of momentum? If so, what is its momentum? If not, why not? [For further discussion, see Problem 3.34.] 3.4 GENERALIZED STATISTICAL INTERPRETATION In Chapter 1 I showed you how to calculate the probability that a particle would be found in a particular location, and how to determine the expectation value of any observable quantity. In Chapter 2 you learned how to \ufb01nd the possible outcomes of an energy measurement, and their probabilities. I am now in a position to state the generalized statistical interpretation, which subsumes all of this, and enables you to \ufb01gure out the possible results of any measurement, and their probabilities. Together with the Schr\u00f6dinger equation (which tells you how the wave function evolves in time) it is the foundation of quantum mechanics. Generalized statistical interpretation: If you measure an observable Q(x, p) on a particle in the state (x, t), you are certain to get one of the eigenvalues of the hermitian operator \u02c6Q(x, \u2212i d/d x).15 If the spectrum of \u02c6Q is discrete, the probability of getting the particular eigenvalue qn associated with the (orthonormalized) eigenfunction fn(x) is |cn|2, where cn = fn|. (3.43) If the spectrum is continuous, with real eigenvalues q(z) and associated (Dirac-orthonormalized) eigenfunctions fz(x), the probability of getting a result in the range dz is |c(z)|2 dz where c(z) = fz|. (3.44) Upon measurement, the wave function \u201ccollapses\u201d to the corresponding eigenstate.16 15 You may have noticed that there is an ambiguity in this prescription, if Q(x, p) involves the product x p. Because \u02c6x and \u02c6p do not commute (Equation 2.52)\u2014whereas the classical variables x and p, of course, do\u2014it is not clear whether we should write \u02c6x \u02c6p or \u02c6p \u02c6x (or perhaps some linear combination of the two). Luckily, such observables are very rare, but when they do occur some other consideration must be invoked to resolve the ambiguity. 16 In the case of continuous spectra the collapse is to a narrow range about the measured value, depending on the precision of the measuring device. 3.4 Generalized Statistical Interpretation 103 The statistical interpretation is radically different from anything we encounter in classical physics. A somewhat different perspective helps to make it plausible: The eigenfunctions of an observable operator are complete, so the wave function can be written as a linear combination of them: (x, t) = cn(t) fn(x). (3.45) n (For simplicity, I\u2019ll assume that the spectrum is discrete; it\u2019s easy to generalize this discussion to the continuous case.) Because the eigenfunctions are orthonormal, the coef\ufb01cients are given by Fourier\u2019s trick:17 cn(t) = fn| = fn(x)\u2217 (x, t) d x. (3.46) Qualitatively, cn tells you \u201chow much fn is contained in,\u201d and given that a measurement has to return one of the eigenvalues of \u02c6Q, it seems reasonable that the probability of getting the particular eigenvalue qn would be determined by the \u201camount of fn\u201d in. But because probabilities are determined by the absolute square of the wave function, the precise measure is actually |cn|2. That\u2019s the essential message of the generalized statistical interpretation.18 Of course, the total probability (summed over all possible outcomes) has got to be one: n |cn|2 = 1, and sure enough, this follows from the normalization of the wave function: 1 = | = 2! cn fn = n n n c\u2217 ncn\u03b4nn = n \"! \"3 cn fn = n c\u2217 ncn = n |cn|2. n n fn fn c\u2217 ncn (3.47) (3.48) Similarly, the expectation value of Q should be the sum over all possible outcomes of the eigenvalue times the probability of getting that eigenvalue: Q = n qn |cn|2. Indeed, Q = \u02c6Q = 2! cn fn n! \" \u02c6Q n \" 3 cn fn, (3.49) (3.50", ") 17 Notice that the time dependence\u2014which is not at issue here\u2014is carried by the coef\ufb01cients; to make this explicit I write cn (t). In the special case of the Hamiltonian ( \u02c6Q = \u02c6H ), when the potential energy is time independent, the coef\ufb01cients are in fact constant, as we saw in Section 2.1. 18 Again, I am scrupulously avoiding the all-too-common claim \u201c|cn |2 is the probability that the particle is in the state fn.\u201d This is nonsense: The particle is in the state, period. Rather, |cn |2 is the probability that a measurement of Q would yield the value qn. It is true that such a measurement will collapse the state to the eigenfunction fn, so one might correctly say \u201c|cn |2 is the probability that a particle which is now in the state will be in the state fn subsequent to a measurement of Q\u201d... but that\u2019s a quite different assertion. 104 CHAPTER 3 Formalism but \u02c6Q fn = qn fn, so Q = n n c\u2217 ncnqn fn| fn = n n c\u2217 ncnqn\u03b4nn = n qn |cn|2. (3.51) So far, at least, everything looks consistent. Can we reproduce, in this language, the original statistical interpretation for position measurements? Sure\u2014it\u2019s overkill, but worth checking. A measurement of x on a particle in state must return one of the eigenvalues of the position operator. Well, in Example 3.3 we found that every (real) number y is an eigenvalue of x, and the corresponding (Dirac-orthonormalized) eigenfunction is gy(x) = \u03b4(x \u2212 y). Evidently c(y) = gy = \u221e \u2212\u221e \u03b4(x \u2212 y)(x, t) d x = (y, t), (3.52) so the probability of getting a result in the range dy is |(y, t)|2 dy, which is precisely the original statistical interpretation. What about momentum? In Example 3.2 we found the (Dirac-orthonormalized) eigenfunc- \u221a tions of the momentum operator, f p(x) = (1/ c( p) = f p = 1\u221a 2\u03c0 \u2212\u221e 2\u03c0 ) exp (i px/), so \u221e e\u2212i px/(x, t) d x. This is such an important quantity that we give it a special name and symbol: the momentum space wave function, ( p, t). It is essentially the Fourier transform of the (position space) wave function (x, t)\u2014which, by Plancherel\u2019s theorem, is its inverse Fourier transform: \u221e ( p, t) = 1\u221a 2\u03c0 (x, t) = 1\u221a 2\u03c0 \u2212\u221e \u221e \u2212\u221e e\u2212i px/(x, t) d x; ei px/( p, t) dp. (3.53) (3.54) (3.55) According to the generalized statistical interpretation, the probability that a measurement of momentum would yield a result in the range dp is |( p, t)|2 dp. (3.56) Example 3.4 A particle of mass m is bound in the delta function well V (x) = \u2212\u03b1\u03b4(x). What is the probability that a measurement of its momentum would yield a value greater than p0 = m\u03b1/? Solution: The (position space) wave function is (Equation 2.132) \u221a (x, t) = m\u03b1 e\u2212m\u03b1|x|/2 e\u2212i Et/ (where E = \u2212m\u03b12/22). The momentum space wave function is therefore ( p, t) = 1\u221a 2\u03c0 \u221a m\u03b1 e\u2212i Et/ \u221e \u2212\u221e e\u2212i px/e\u2212m\u03b1|x|/2 d x = 2 \u03c0 p3/2 0 e\u2212i Et/ p2 + p2 0 (I looked up the integral). The probability, then, is 2 \u03c0 p3 0 \u221e p0 1 p2 + p2 0 dp = 1 \u03c0 2 pp0 p2 + p2 0 + tan \u22121 3.5 The Uncertainty Principle 105 \u221e p0 p p0 = 1 4 \u2212 1 2\u03c0 = 0.0908 (again, I looked up the integral). Problem 3.11 Find the momentum-space wave function, ( p, t), for a particle in the ground state of the harmonic oscillator. What is the probability (to two signi\ufffd", "\ufffdcant digits) that a measurement of p on a particle in this state would yield a value outside the classical range (for the same energy)? Hint: Look in a math table under \u201cNormal Distribution\u201d or \u201cError Function\u201d for the numerical part\u2014or use Mathematica. Problem 3.12 Find ( p, t) for the free particle in terms of the function \u03c6(k) introduced in Equation 2.101. Show that for the free particle |( p, t)|2 is independent of time. Comment: the time independence of |( p, t)|2 for the free particle is a manifestation of momentum conservation in this system. \u2217 Problem 3.13 Show that x = \u2217 i \u2202 \u2202 p dp. (3.57) Hint: Notice that x exp(i px/) = \u2212i (\u2202/\u2202 p) exp(i px/), and use Equation 2.147. In momentum space, then, the position operator is i \u2202/\u2202 p. More generally, x, \u2212, t \u2202 p in position space; in momentum space. d x, dp, Q(x, p, t) = \u2217 \u02c6Q \u2217 \u02c6Q (3.58) In principle you can do all calculations in momentum space just as well (though not always as easily) as in position space. 3.5 THE UNCERTAINTY PRINCIPLE I stated the uncertainty principle (in the form \u03c3x \u03c3 p \u2265 /2), back in Section 1.6, and you have checked it several times, in the problems. But we have never actually proved it. In this section I will prove a more general version of the uncertainty principle, and explore some of its rami\ufb01cations. The argument is beautiful, but rather abstract, so watch closely. 3.5.1 Proof of the Generalized Uncertainty Principle For any observable A, we have (Equation 3.21): 0 \u02c6A \u2212 A \u02c6A \u2212 A 1 = f | f \u03c3 2 A = 106 CHAPTER 3 Formalism where f \u2261 \u02c6A \u2212 A. Likewise, for any other observable, B,. = g | g, where g \u2261 \u02c6B \u2212 B \u03c3 2 B Therefore (invoking the Schwarz inequality, Equation 3.7), | g \u2265 | f | g |2. Now, for any complex number z, |z|2 = [Re(z)]2 + [Im(z)]2 \u2265 [Im(z)]2 = 1 2i 2. z \u2212 z\u2217 Therefore, letting z = f |g, (3.59) (3.60) (3.61) 1 \u2265 1 2i 2. f |g \u2212 g| f \u03c3 2 A \u03c3 2 B But (exploiting the hermiticity of 0 in the \ufb01rst line) 0 \u02c6A \u2212 = \u02c6B \u2212 B \u02c6A \u2212 A \u02c6A \u2212 A 1 \u02c6A \u02c6B \u2212 \u02c6A B \u2212 \u02c6B A + A B \u02c6A \u02c6B \u02c6B \u02c6A \u2212 B \u2212 A 0 0 1 1 1 \u02c6A \u02c6B \u02c6A \u02c6B \u2212. \u02c6B \u2212 B + A B | (Remember, A and B are numbers, not operators, so you can write them in either order.) Similarly, g | f = \u2212 AB, \u02c6B \u02c6A so where f |g \u2212 g| f = \u02c6A \u02c6B \u02c6B \u02c6A \u2212 = \u02c6A, \u02c6B, # $ \u02c6A, \u02c6B \u2261 \u02c6A \u02c6B \u2212 \u02c6B \u02c6A is the commutator of the two operators (Equation 2.49). Conclusion: \u03c3 2 A \u03c3 2 B \u2265 0# $1 2. \u02c6A, \u02c6B 1 2i (3.62) This is the (generalized) uncertainty principle. (You might think the i makes it trivial\u2014isn\u2019t the right side negative? No, for the commutator of two hermitian operators carries its own factor of i, and the two cancel out;19 the quantity in parentheses is real, and its square is positive.) 19 More precisely, the commutator of two hermitian operators is itself anti-hermitian expectation value is imaginary (Problem 3.32). \u02c6Q\u2020 = \u2212 \u02c6Q, and its 3.5 The Uncertainty Principle 107 \u02c6A = x, and the second is momen- As", " an example, suppose the \ufb01rst observable is position \u02c6B = \u2212i d/d x tum So. We worked out their commutator back in Chapter 2 (Equation 2.52): \u02c6x, \u02c6p = i 2i i or, since standard deviations are by their nature positive, \u03c3x \u03c3 p \u2265. 2 (3.63) That\u2019s the original Heisenberg uncertainty principle, but we now see that it is just one application of a much more general theorem. There is, in fact, an \u201cuncertainty principle\u201d for every pair of observables whose operators do not commute\u2014we call them incompatible observables. Incompatible observables do not have shared eigenfunctions\u2014at least, they cannot have a complete set of common eigenfunctions (see Problem 3.16). By contrast, compatible (commuting) observables do admit complete sets of simultaneous eigenfunctions (that is: states that are determinate for both observables).20 For example, in the hydrogen atom (as we shall see in Chapter 4) the Hamiltonian, the magnitude of the angular momentum, and the z component of angular momentum are mutually compatible observables, and we will construct simultaneous eigenfunctions of all three, labeled by their respective eigenvalues. But there is no eigenfunction of position that is also an eigenfunction of momentum; these operators are incompatible. Note that the uncertainty principle is not an extra assumption in quantum theory, but rather a consequence of the statistical interpretation. You might wonder how it is enforced in the laboratory\u2014why can\u2019t you determine (say) both the position and the momentum of a particle? You can certainly measure the position of the particle, but the act of measurement collapses the wave function to a narrow spike, which necessarily carries a broad range of wavelengths (hence momenta) in its Fourier decomposition. If you now measure the momentum, the state will collapse to a long sinusoidal wave, with (now) a well-de\ufb01ned wavelength\u2014but the particle no longer has the position you got in the \ufb01rst measurement.21 The problem, then, is that the second measurement renders the outcome of the \ufb01rst measurement obsolete. Only if the wave function were simultaneously an eigenstate of both observables would it be possible to 20 This corresponds to the fact that noncommuting matrices cannot be simultaneously diagonalized (that is, they cannot both be brought to diagonal form by the same similarity transformation), whereas commuting hermitian matrices can be simultaneously diagonalized. See Section A.5. 21 Bohr and Heisenberg were at pains to track down the mechanism by which the measurement of x (for instance) destroys the previously existing value of p. The crux of the matter is that in order to determine the position of a particle you have to poke it with something\u2014shine light on it, say. But these photons impart to the particle a momentum you cannot control. You now know the position, but you no longer know the momentum. Bohr\u2019s famous debates with Einstein include many delightful examples, showing in detail how experimental constraints enforce the uncertainty principle. For an inspired account see Bohr\u2019s article in Albert Einstein: Philosopher-Scientist, edited by Paul A. Schilpp, Open Court Publishing Co., Peru, IL (1970). In recent years the Bohr/Heisenberg explanation has been called into question; for a nice discussion see G. Brum\ufb01el, Nature News https://doi.org/10.1038/nature.2012.11394. 108 CHAPTER 3 Formalism make the second measurement without disturbing the state of the particle (the second collapse wouldn\u2019t change anything, in that case). But this is only possible, in general, if the two observables are compatible. \u2217 Problem 3.14 (a) Prove the following commutator identities: \u02c6A, \u02c6C # \u02c6A + \u02c6B, \u02c6C # = # # $ $ $ # $ + $ \u02c6B, \u02c6C #, $ \u02c6A \u02c6B, \u02c6C = \u02c6A \u02c6B, \u02c6C + \u02c6A, \u02c6C \u02c6B. (b) Show that (c) Show more generally that x n, \u02c6p = inx n\u22121. f (x), \u02c6p = i d f d x, for any function f (x) that admits a Taylor series expansion. (d) Show that for the simple harmonic oscillator # $ \u02c6H, \u02c6a\u00b1 = \u00b1\u03c9 \u02c6a\u00b1. (3.64) (3.65) (3.66) (3.", "67) Hint: Use Equation 2.54. \u2217 Problem 3.15 Prove the famous \u201c(your name) uncertainty principle,\u201d relating the uncertainty in position (A = x) to the uncertainty in energy B = p2/2m + V : For stationary states this doesn\u2019t tell you much\u2014why not? \u03c3x \u03c3H \u2265 2m | p|. Problem 3.16 Show that two noncommuting operators cannot have a complete set of \u02c6P and \u02c6Q have a complete set of common common eigenfunctions. Hint: Show that if # eigenfunctions, then f = 0 for any function in Hilbert space. \u02c6P, \u02c6Q $ 3.5.2 The Minimum-Uncertainty Wave Packet We have twice encountered wave functions that hit the position-momentum uncertainty limit \u03c3x \u03c3 p = /2 : the ground state of the harmonic oscillator (Problem 2.11) and the Gaussian wave packet for the free particle (Problem 2.21). This raises an interesting question: What is the most general minimum-uncertainty wave packet? Looking back at the proof of the uncertainty principle, we note that there were two points at which inequalities came into the argument: Equation 3.59 and Equation 3.60. Suppose we require that each of these be an equality, and see what this tells us about. The Schwarz inequality becomes an equality when one function is a multiple of the other: g(x) = c f (x), for some complex number c (see Problem A.5). Meanwhile, in Equation 3.60 I threw away the real part of z; equality results if Re(z) = 0, which is to say, if Re f |g =Re(c f | f ) = 0. Now, f | f is certainly real, so this means the constant c must be pure imaginary\u2014let\u2019s call it ia. The necessary and suf\ufb01cient condition for minimum uncertainty, then, is 3.5 The Uncertainty Principle 109 g(x) = ia f (x), where a is real. For the position-momentum uncertainty principle this criterion becomes: \u2212i d d x \u2212 p = ia(x \u2212 x), (3.68) (3.69) which is a differential equation for as a function of x. Its general solution (see Problem 3.17) is (x) = Ae\u2212a(x\u2212x)2/2ei px/. (3.70) Evidently the minimum-uncertainty wave packet is a gaussian\u2014and, sure enough, the two examples we encountered earlier were gaussians.22 Problem 3.17 Solve Equation 3.69 for (x). Note that x and p are constants (independent of x). 3.5.3 The Energy-Time Uncertainty Principle The position-momentum uncertainty principle is often written in the form x p \u2265 ; 2 (3.71) x (the \u201cuncertainty\u201d in x) is loose notation (and sloppy language) for the standard deviation of the results of repeated measurements on identically prepared systems.23 Equation 3.71 is often paired with the energy-time uncertainty principle, t E \u2265. 2 (3.72) Indeed, in the context of special relativity the energy-time form might be thought of as a consequence of the position-momentum version, because x and t (or rather, ct) go together in the position-time four-vector, while p and E (or rather, E/c) go together in the energy-momentum four-vector. So in a relativistic theory Equation 3.72 would be a necessary concomitant to Equation 3.71. But we\u2019re not doing relativistic quantum mechanics. The Schr\u00f6dinger equation is explicitly nonrelativistic: It treats t and x on a very unequal footing (as a differential equation it is \ufb01rst-order in t, but second-order in x), and Equation 3.72 is emphatically not implied by Equation 3.71. My purpose now is to derive the energy-time uncertainty principle, and in 22 Note that it is only the dependence of on x that is at issue here\u2014the \u201cconstants\u201d A, a, x, and p may all be functions of time, and for that matter may evolve away from the minimal form. All I\u2019m asserting is that if, at some instant, the wave function is gaussian in x, then (at that instant) the uncertainty product is minimal. 23 Many casual applications of the uncertainty principle are actually based (often inadvertently) on a completely different\u2014and sometimes quite unjusti\ufb01ed\u2014measure of \u201cuncertainty", ".\u201d See J. Hilgevoord, Am. J. Phys. 70, 983 (2002). 110 CHAPTER 3 Formalism the course of that derivation to persuade you that it is really an altogether different beast, whose super\ufb01cial resemblance to the position-momentum uncertainty principle is actually quite misleading. After all, position, momentum, and energy are all dynamical variables\u2014measurable characteristics of the system, at any given time. But time itself is not a dynamical variable (not, at any rate, in a nonrelativistic theory): You don\u2019t go out and measure the \u201ctime\u201d of a particle, as you might its position or its energy. Time is the independent variable, of which the dynamical quantities are functions. In particular, the t in the energy-time uncertainty principle is not the standard deviation of a collection of time measurements; roughly speaking (I\u2019ll make this more precise in a moment) it is the time it takes the system to change substantially. As a measure of how fast the system is changing, let us compute the time derivative of the expectation value of some observable, Q (x, p, t): 0 1 Q d dt = d dt \u02c6Q = \u2202 \u2202t \u02c6Q 2 + \u2202 \u02c6Q \u2202t 3 + \u02c6Q \u2202 \u2202t. Now, the Schr\u00f6dinger equation says i \u2202 \u2202t = \u02c6H (where H = p2/2m + V is the Hamiltonian). So d dt Q = \u2212 1 i 0 \u02c6H 1 \u02c6Q + 1 i 0 1 \u02c6Q \u02c6H + 2 3. \u2202 \u02c6Q \u2202t But \u02c6H is hermitian, so \u02c6H | \u02c6Q = | \u02c6H \u02c6Q, and hence d dt Q = i 0# $1 + \u02c6H, \u02c6Q 2 3. \u2202 \u02c6Q \u2202t (3.73) This is an interesting and useful result in its own right (see Problems 3.18 and 3.37). It has no name, though it surely deserves one; I\u2019ll call it the generalized Ehrenfest theorem. In the typical case where the operator does not depend explicitly on time,24 it tells us that the rate of change of the expectation value is determined by the commutator of the operator with the Hamiltonian. In particular, if \u02c6Q commutes with \u02c6H, then Q is constant, and in this sense Q is a conserved quantity. Now, suppose we pick A = H and B = Q, in the generalized uncertainty principle (Equation 3.62), and assume that Q does not depend explicitly on t: $1 2 0# \u02c6H, \u02c6Q = 1 2i 2 = 1 2i dQ dt 2 2. dQ dt 2 \u03c3 2 H \u03c3 2 Q \u2265 Or, more simply, \u03c3H \u03c3Q \u2265 2. dQ dt (3.74) i 24 Operators that depend explicitly on t are quite rare, so almost always \u2202 \u02c6Q/\u2202t = 0. As an example of explicit time dependence, consider the potential energy of a harmonic oscillator whose spring constant is changing (perhaps the temperature is rising, so the spring becomes more \ufb02exible): Q = (1/2) m [\u03c9(t)]2 x 2. Let\u2019s de\ufb01ne E \u2261 \u03c3H, and Then t \u2261 \u03c3Q |dQ/dt|. E t \u2265, 2 3.5 The Uncertainty Principle 111 (3.75) (3.76) and that\u2019s the energy-time uncertainty principle. But notice what is meant by t, here: Since \u03c3Q = dQ dt t, t represents the amount of time it takes the expectation value of Q to change by one standard deviation.25 In particular, t depends entirely on what observable (Q) you care to look at\u2014the change might be rapid for one observable and slow for another. But if E is small, then the rate of change of all observables must be very gradual; or, to put it the other way around, if any observable changes rapidly, the \u201cuncertainty\u201d in the energy must be large. Example 3.5 In the extreme case of a stationary state, for which the energy is uniquely determined, all expectation values are constant in time (E = 0 \u21d2 t = \u221e)\u2014as in fact we noticed some time ago (see Equation 2.9). To make something happen you must take a linear combination of at least two stationary states\u2014say: (x,", " t) = a\u03c81(x)e\u2212i E1t/ + b\u03c82(x)e\u2212i E2t/. If a, b, \u03c81, and \u03c82 are real, |(x, t)|2 = a2 (\u03c81(x))2 + b2 (\u03c82(x))2 + 2ab\u03c81(x)\u03c82(x) cos. E2 \u2212 E1 t The period of oscillation is \u03c4 = 2\u03c0 / (E2 \u2212 E1). Roughly speaking, E = E2 \u2212 E1 and t = \u03c4 (for the exact calculation see Problem 3.20), so which is indeed \u2265 /2. E t = 2\u03c0, Example 3.6 Let t be the time it takes a free-particle wave packet to pass a particular point (Figure 3.1). Qualitatively (an exact version is explored in Problem 3.21), t = x/v = mx/ p. But E = p2/2m, so E = pp/m, and therefore, E t = pp m mx p = xp, which is \u2265 /2 by the position-momentum uncertainty principle. 25 This is sometimes called the \u201cMandelstam\u2013Tamm\u201d formulation of the energy-time uncertainty principle. For a review of alternative approaches see P. Busch, Found. Phys. 20, 1 (1990). 112 CHAPTER 3 Formalism \u0394x \u03c5 Figure 3.1: A free particle wave packet approaches the point A (Example 3.6). A x Example 3.7 The particle lasts about 10\u221223 s, before spontaneously disintegrating. If you make a histogram of all measurements of its mass, you get a kind of bell-shaped curve centered at 1232 MeV/c2, with a width of about 120 MeV/c2 (Figure 3.2). Why does the rest energy mc2 sometimes come out higher than 1232, and sometimes lower? Is this experimental error? No, for if we take t to be the lifetime of the particle (certainly one measure of \u201chow long it takes the system to change appreciably\u201d), E t = MeV \u221223 s 10 120 2 = 6 \u00d7 10 \u221222 MeV-s, whereas /2 = 3 \u00d7 10\u221222 MeV s. So the spread in m is about as small as the uncertainty principle allows\u2014a particle with so short a lifetime just doesn\u2019t have a very well-de\ufb01ned mass.26 1100 1200 1300 mass (MeV/c2) 1400 Figure 3.2: Measurements of the mass (Example 3.7). Notice the variety of speci\ufb01c meanings attaching to the term t in these examples: In Example 3.5 it\u2019s a period of oscillation; in Example 3.6 it\u2019s the time it takes a particle to pass a point; in Example 3.7 it\u2019s the lifetime of an unstable particle. In every case, however, t is the time it takes for the system to undergo \u201csubstantial\u201d change. It is often said that the uncertainty principle means energy is not strictly conserved in quantum mechanics\u2014that you\u2019re allowed to \u201cborrow\u201d energy E, as long as you \u201cpay it back\u201d 26 In truth, Example 3.7 is a bit of a fraud. You can\u2019t measure 10 \u221223 s on a stop-watch, and in practice the lifetime of such a short-lived particle is inferred from the width of the mass plot, using the uncertainty principle as input. However, the point is valid, even if the logic is backwards. Moreover, if you assume the is about the same size \u221223 sec is roughly the time it takes light to cross the particle, and it\u2019s hard to as a proton imagine that the lifetime could be much less than that. \u221215 m, then 10 \u223c 10 3.6 Vectors and Operators 113 in a time t \u2248 / (2E); the greater the violation, the briefer the period over which it can occur. Now, there are many legitimate readings of the energy-time uncertainty principle, but this is not one of them. Nowhere does quantum mechanics license violation of energy conservation, and certainly no such authorization entered into the derivation of Equation 3.76. But the uncertainty principle is extraordinarily robust: It can be misused without leading to seriously incorrect results, and as a consequence physicists are in the habit of applying it rather carelessly. \u2217 Problem 3.18 Apply Equation 3.73 to the following special cases: (a) Q = 1; (b) Q = H ; (c) Q = x; (d) Q = p. In each case, comment on the result, with particular reference to", " Equations 1.27, 1.33, 1.38, and conservation of energy (see remarks following Equation 2.21). Problem 3.19 Use Equation 3.73 (or Problem 3.18 (c) and (d)) to show that: (a) For any (normalized) wave packet representing a free particle (V (x) = 0), x moves at constant velocity (this is the quantum analog to Newton\u2019s \ufb01rst law). Note: You showed this for a gaussian wave packet in Problem 2.42, but it is completely general. (b) For any (normalized) wave packet representing a particle in the harmonic oscil-, x oscillates at the classical frequency. Note: lator potential You showed this for a particular gaussian wave packet in Problem 2.49, but it is completely general. 2 m\u03c92x 2 V (x) = 1 Problem 3.20 Test the energy-time uncertainty principle for the wave function in Problem 2.5 and the observable x, by calculating \u03c3H, \u03c3x, and dx/dt exactly. Problem 3.21 Test the energy-time uncertainty principle for the free particle wave packet in Problem 2.42 and the observable x, by calculating \u03c3H, \u03c3x, and dx/dt exactly. Problem 3.22 Show that the energy-time uncertainty principle reduces to the \u201cyour name\u201d uncertainty principle (Problem 3.15), when the observable in question is x. 3.6 VECTORS AND OPERATORS 3.6.1 Bases in Hilbert Space Imagine an ordinary vector A in two dimensions (Fig. 3.3(a)). How would you describe this vector to someone? You might tell them \u201cIt\u2019s about an inch long, and it points 20\u25e6 clockwise from straight up, with respect to the page.\u201d But that\u2019s pretty awkward. A better way would be 114 CHAPTER 3 Formalism y Ay A y\u2032 A \u2032 Ay x\u2032 x Ax (b) \u2032 Ax (c) A (a) Figure 3.3: (a) Vector A. (b) Components of A with respect to x y axes. (c) Components of A with respect to x y axes. to introduce cartesian axes, x and y, and specify the components of A: Ax = \u02c6\u0131 \u00b7 A, Ay = \u02c6j \u00b7 A (Fig. 3.3(b)). Of course, your sister might draw a different set of axes, x and y, and she would = \u02c6j \u00b7 A (Fig. 3.3(c))... but it\u2019s all the same report different components: A ( x vector\u2014we\u2019re simply expressing it with respect to two different bases. The vector itself lives \u201cout there in space,\u201d independent of anybody\u2019s (arbitrary) choice of coordinates. = \u02c6\u0131 \u00b7 A, A y \u2019 \u02c6\u0131, \u02c6j \u02c6\u0131, \u02c6j and \u2019 ( The same is true for the state of a system in quantum mechanics. It is represented by a vector, |S(t), that lives \u201cout there in Hilbert space,\u201d but we can express it with respect to any number of different bases. The wave function (x, t) is actually the x \u201ccomponent\u201d in the expansion of |S(t) in the basis of position eigenfunctions: (x, t) = x|S(t), (3.77) (the analog to \u02c6\u0131 \u00b7 A) with |x standing for the eigenfunction of \u02c6x with eigenvalue x.27 The momentum space wave function ( p, t) is the p component in the expansion of |S(t) in the basis of momentum eigenfunctions: ( p, t) = p|S(t) (3.78) (with | p standing for the eigenfunction of \u02c6p with eigenvalue p).28 Or we could expand |S(t) in the basis of energy eigenfunctions (supposing for simplicity that the spectrum is discrete): cn(t) = n|S(t) (3.79) \u02c6H \u2014Equation 3.46). But it\u2019s all the same (with |n standing for the nth eigenfunction of the functions and, and the collection of coef\ufb01cients {cn}, contain exactly state; the same information\u2014they are simply three different ways of identifying the same vector: 27 I hesitate to call it gx (Equation 3.39), because that is its form in the position basis, and the whole point here is to free ourselves from any particular basis. Indeed, when I \ufffd", "\ufffdrst de\ufb01ned Hilbert space as the set of square-integrable functions\u2014over x\u2014that was already too restrictive, committing us to a speci\ufb01c representation (the position basis). I want now to think of it as an abstract vector space, whose members can be expressed with respect to any basis you like. 28 In position space it would be f p (Equation 3.32). |S(t) \u2192 (y, t)\u03b4(x \u2212 y) dy = ( p, t) = cne\u2212i En t/\u03c8n(x). 1\u221a 2\u03c0 ei px/ dp (3.80) 3.6 Vectors and Operators 115 Operators (representing observables) are linear transformations on Hilbert space\u2014they \u201ctransform\u201d one vector into another: \u03b2 \u03b1. = \u02c6Q (3.81) Just as vectors are represented, with respect to an orthonormal basis {|en},29 by their components, |\u03b1 = en, an \u03b2 = en, bn en \u03b1, an = bn = en | \u03b2, (3.82) n n operators are represented (with respect to a particular basis) by their matrix elements30 en \u2261 Qmn. \u02c6Q em 0 1 In this notation Equation 3.81 says n bn |en = n \u02c6Q|en, an or, taking the inner product with |em, n bn em | en = 0 an em 1 \u02c6Q en, n and hence (since em|en = \u03b4mn) bm = n Qmnan. (3.83) (3.84) (3.85) (3.86) Thus the matrix elements of \u02c6Q tell you how the components transform.31 Later on we will encounter systems that admit only a \ufb01nite number N of linearly independent states. In that case |S(t) lives in an N -dimensional vector space; it can be represented as a column of (N ) components (with respect to a given basis), and operators take the form of ordinary (N \u00d7 N ) matrices. These are the simplest quantum systems\u2014none of the subtleties associated with in\ufb01nite-dimensional vector spaces arise. Easiest of all is the two-state system, which we explore in the following example. 29 I\u2019ll assume the basis is discrete; otherwise n becomes a continuous index and the sums are replaced by integrals. 30 This terminology is inspired, obviously, by the \ufb01nite-dimensional case, but the \u201cmatrix\u201d will now typically have an in\ufb01nite (maybe even uncountable) number of elements. 31 In matrix notation Equation 3.86 becomes b = Qa (with the vectors expressed as columns), by the ordinary rules of matrix multiplication\u2014see Equation A.42. 116 CHAPTER 3 Formalism Example 3.8 Imagine a system in which there are just two linearly independent states:32 |1 = and |2 = 1 0 0 1. The most general state is a normalized linear combination: |S = a|1 + b|2 =, with |a|2 + |b|2 = 1. a b The Hamiltonian can be expressed as a (hermitian) matrix (Equation 3.83); suppose it has the speci\ufb01c form h g where g and h are real constants. If the system starts out (at t = 0) in state |1, what is its state at time t? H = g h, Solution: The (time-dependent) Schr\u00f6dinger equation33 says i d dt As always, we begin by solving the time-independent Schr\u00f6dinger equation: |S(t) = \u02c6H |S(t). (3.87) \u02c6H |s = E|s; (3.88) \u02c6H. The characteristic equation that is, we look for the eigenvectors and eigenvalues of determines the eigenvalues: h \u2212 E g g h \u2212 E det = (h \u2212 E)2 \u2212 g2 = 0 \u21d2 h \u2212 E = \u2213g \u21d2 E\u00b1 = h \u00b1 g. Evidently the allowed energies are (h + g) and (h \u2212 g). To determine the eigenvectors, we write h \u00b1 g) \u21d2 h\u03b1 + g\u03b2 = (h \u00b1 g)\u03b1 \u21d2 \u03b2 = \u00b1\u03b1, so the normalized eigenvectors are |s\u00b1 = 1\u221a 2. 1 \u00b11 Next we expand the initial state as a linear combination of eigenvectors of the Hamiltonian: = 1\u221a 2 (|s+ + |s\u2212). |S(0) = 1 0 32 Technically, the", " \u201cequals\u201d signs here mean \u201cis represented by,\u201d but I don\u2019t think any confusion will arise if we adopt the customary informal notation. 33 We began, back in Chapter 1, with the Schr\u00f6dinger equation for the wave function in position space; here we generalize it to the state vector in Hilbert space. 3.6 Vectors and Operators 117 Finally, we tack on the standard time-dependence (the wiggle factor) exp (\u2212i Ent/): |S(t) = 1\u221a 2 e\u2212i ht/ # $ e\u2212i(h+g)t/|s+ + e\u2212i(h\u2212g)t/|s\u2212 1 1 e\u2212igt/ + eigt/ 1 \u22121 e\u2212i ht/ e\u2212igt/ + eigt/ e\u2212igt/ \u2212 eigt/ = e\u2212i ht/ = 1 2 = 1 2 cos(gt/) \u2212i sin(gt/). If you doubt this result, by all means check it: Does it satisfy the time-dependent Schr\u00f6dinger equation (Equation 3.87)? Does it match the initial state when t = 0?34 Just as vectors look different when expressed in different bases, so too do operators (or, in the discrete case, the matrices that represent them). We have already encountered a particularly nice example: \u02c6x (the position operator) \u2192 \u02c6p (the momentum operator) \u2192 (in position space), (in momentum space); x i\u2202/\u2202 p \u2212i\u2202/\u2202 x p (in position space), (in momentum space). (\u201cPosition space\u201d is nothing but the position basis; \u201cmomentum space\u201d is the momentum basis.) If someone asked you, \u201cWhat is the operator, \u02c6x, representing position, in quantum mechanics?\u201d you would probably answer \u201cJust x itself.\u201d But an equally correct reply would be \u201ci \u2202/\u2202 p,\u201d and the best response would be \u201cWith respect to what basis?\u201d I have often said \u201cthe state of a system is represented by its wave function, (x, t),\u201d and this is true, in the same sense that an ordinary vector in three dimensions is \u201crepresented by\u201d the triplet of its components; but really, I should always add \u201cin the position basis.\u201d After all, the state of the system is a vector in Hilbert space, |S(t); it makes no reference to any particular basis. Its connection to (x, t) is given by Equation 3.77: (x, t) = x|S(t). Having said that, for the most part we do in fact work in position space, and no serious harm comes from referring to the wave function as \u201cthe state of the system.\u201d 3.6.2 Dirac Notation Dirac proposed to chop the bracket notation for the inner product, \u03b1|\u03b2, into two pieces, which he called bra, \u03b1|, and ket, |\u03b2 (I don\u2019t know what happened to the c). The latter is a vector, but what exactly is the former? It\u2019s a linear function of vectors, in the sense that when it hits a vector (to its right) it yields a (complex) number\u2014the inner product. (When an operator hits a vector, it delivers another vector; when a bra hits a vector, it delivers a number.) In a function space, the bra can be thought of as an instruction to integrate: f | = f \u2217 [\u00b7 \u00b7 \u00b7 ] d x, 34 This is a crude model for (among other things) neutrino oscillations. In that context |1 represents (say) the electron neutrino, and |2 the muon neutrino; if the Hamiltonian has a nonvanishing off-diagonal term (g) then in the course of time the electron neutrino will turn into a muon neutrino (and back again). 118 CHAPTER 3 Formalism with the ellipsis [\u00b7 \u00b7 \u00b7 ] waiting to be \ufb01lled by whatever function the bra encounters in the ket to its right. In a \ufb01nite-dimensional vector space, with the kets expressed as columns (of components with respect to some basis), |\u03b1 \u2192 \u239e \u239f \u239f \u239f \u23a0, \u239b \u239c \u239c \u239c \u239d a1 a2... an (3.89) (3.90) (3.94) (3.95) (3.96) the bras are rows: \u03b2| \u2192 1 b\u2217 b\u2217 2,... b\ufffd", "\ufffd n and \u03b2|\u03b1 = b\u2217 another vector space\u2014the so-called dual space. 2a2+\u00b7 \u00b7 \u00b7+b\u2217 1a1+b\u2217 nan is the matrix product. The collection of all bras constitutes The license to treat bras as separate entities in their own right allows for some powerful and pretty notation. For example, if |\u03b1 is a normalized vector, the operator \u02c6P \u2261 |\u03b1\u03b1| (3.91) picks out the portion of any other vector that \u201clies along\u201d |\u03b1: \u02c6P |\u03b2 = (\u03b1| \u03b2) |\u03b1; we call it the projection operator onto the one-dimensional subspace spanned by |\u03b1. If {|en} is a discrete orthonormal basis, then em|en = \u03b4mn, n |enen| = 1 (3.92) (3.93) (the identity operator). For if we let this operator act on any vector |\u03b1, we recover the expansion of |\u03b1 in the {|en} basis: (en |\u03b1)| en = |\u03b1. Similarly, if {|ez} is a Dirac orthonormalized continuous basis, ez|ez = \u03b4 z \u2212 z, n then |ezez| dz = 1. Equations 3.93 and 3.96 are the tidiest ways to express completeness. Technically, the guts of a ket or a bra (the ellipsis in | \u00b7 \u00b7 \u00b7 or \u00b7 \u00b7 \u00b7 |) is a name\u2014the name of the vector in question: \u201c\u03b1,\u201d or \u201cn,\u201d or for that matter \u201cAlice,\u201d or \u201cBob.\u201d It is endowed with no intrinsic mathematical attributes. Of course, it may be helpful to choose an evocative name\u2014 for instance, if you\u2019re working in the space L 2 of square-integrable functions, it is natural to name each vector after the function it represents: | f. Then, for example, we can write the de\ufb01nition of a hermitian operator as we did in Equation 3.17: 0 1 \u02c6Q f f 0 = \u02c6Q f 1 f. 3.6 Vectors and Operators 119 Strictly speaking, in Dirac notation this is a nonsense expression: operators act on vectors, not on names. The left side should properly be written as 0 1 f here is a name, and, \u02c6Q f really means \u201cthe bra dual to \u02c6Q | f,\u201d but what \u02c6Q f f but what are we to make of the right side? is its name? I suppose we could say 0 the name of the vector \u02c6Q | f, but that\u2019s a mouthful. However, since we have chosen to name each vector after the function it represents, and since we do know how \u02c6Q acts on the function (as opposed to the name) f, this in fact becomes35 \u02c6Q f |, and we are OK after all.36 An operator takes one vector in Hilbert space and delivers another: \u02c6Q|\u03b1 = |\u03b2. The sum of two operators is de\ufb01ned in the obvious way, \u02c6Q + \u02c6R |\u03b1 = \u02c6Q |\u03b1 + \u02c6R |\u03b1, and the product of two operators is \u02c6Q \u02c6R |\u03b1 = \u02c6Q \u02c6R |\u03b1 (3.97) (3.98) (3.99) (\ufb01rst apply \u02c6R to |\u03b1, and then apply \u02c6Q to what you got\u2014being careful, of course, to respect their ordering). Occasionally we shall encounter functions of operators. They are typically de\ufb01ned by the power series expansion: e \u02c6Q \u2261 1 + \u02c6Q + 1 2 \u02c6Q2 + 1 3! \u02c6Q3 + \u00b7 \u00b7 \u00b7 1 1 \u2212 \u02c6Q 1 + \u02c6Q ln \u2261 1 + \u02c6Q + \u02c6Q2 + \u02c6Q3 + \u02c6Q4 + \u00b7 \u00b7 \u00b7 \u2261 \u02c6Q \u2212 1 2 \u02c6Q2 + 1 3 \u02c6Q3 \u2212 1 4 \u02c6Q4 + \u00b7 \u00b7 \u00b7 (3.100) (3.101) (3.102) and so on. On the right-hand side we have only sums and products, and we know how to handle them. Problem 3.23 Show that projection operators are idempotent: eigenvalues of \u02c6P, and characterize its eigenvectors. \u02c6P 2 = \u02c6P. Determine the 35 Note that \u02c6Q", " f | = f | \u02c6Q\u2020, by virtue of Equation 3.20. 36 Like his delta function, Dirac\u2019s notation is beautiful, powerful, and obedient. You can abuse it (everyone does), and it won\u2019t bite. But once in a while you should pause to ask yourself what the symbols really mean. 120 CHAPTER 3 Formalism Problem 3.24 Show that if an operator orthonormal basis satisfy Qmn = Q\u2217 transpose conjugate. \u02c6Q is hermitian, then its matrix elements in any nm. That is, the corresponding matrix is equal to its Problem 3.25 The Hamiltonian for a certain two-level system is \u02c6H = (|1 1| \u2212 |2 2| + |1 2| + |2 1|), where |1, | 2 is an orthonormal basis and is a number with the dimensions of energy. Find its eigenvalues and eigenvectors (as linear combinations of |1 and |2). What is the matrix H representing \u02c6H with respect to this basis? \u2217 Problem 3.26 Consider a three-dimensional vector space spanned by an orthonormal basis |1, |2, |3. Kets |\u03b1 and |\u03b2 are given by |\u03b1 = i| 1 \u2212 2 |2 \u2212 i| 3, |\u03b2 = i| 1 + 2|3. (a) Construct \u03b1| and \u03b2| (in terms of the dual basis 1|, 2|, 3|). (b) Find \u03b1|\u03b2 and \u03b2|\u03b1, and con\ufb01rm that \u03b2|\u03b1 = \u03b1|\u03b2\u2217. (c) Find all nine matrix elements of the operator \u02c6A \u2261 |\u03b1\u03b2|, in this basis, and construct the matrix A. Is it hermitian? Problem 3.27 Let \u02c6Q be an operator with a complete set of orthonormal eigenvectors: \u02c6Q |en = qn| en (n = 1, 2, 3,...). (a) Show that \u02c6Q can be written in terms of its spectral decomposition: \u02c6Q = qn |enen|. (3.103) n Hint: An operator is characterized by its action on all possible vectors, so what you must show is that 4 \u02c6Q|\u03b1 = qn|en en| |\u03b1, n for any vector |\u03b1. (b) Another way to de\ufb01ne a function of \u02c6Q is via the spectral decomposition: f ( \u02c6Q) = f (qn) |enen|. (3.104) Show that this is equivalent to Equation 3.100 in the case of e \u02c6Q. n 3.6 Vectors and Operators 121 Problem 3.28 Let \u02c6D = d/d x (the derivative operator). Find (a) (b) sin \u02c6D x 5. 1 1 \u2212 \u02c6D/2 cos(x). \u2217\u2217 # $ Problem 3.29 Consider operators \u02c6A, \u02c6B \u02c6C = instance, \u02c6x and \u02c6p). (a) Show that but do commute with their commutator: \u02c6A and \u02c6B that do not commute with each other \u02c6A, \u02c6C = 0 (for \u02c6B, \u02c6C = # $ $ # # $ \u02c6An, \u02c6B = n \u02c6An\u22121 \u02c6C. Hint: You can prove this by induction on n, using Equation 3.65. (b) Show that $ # e\u03bb \u02c6A, \u02c6B = \u03bb e\u03bb \u02c6A \u02c6C, where \u03bb is any complex number. Hint: Express e\u03bb \u02c6A as a power series. (c) Derive the Baker\u2013Campbell\u2013Hausdorff formula:37 \u02c6Be\u2212 \u02c6C/2. \u02c6A+ \u02c6B = e \u02c6A e e Hint: De\ufb01ne the functions \u02c6f (\u03bb) = e \u03bb \u02c6A+ \u02c6B, \u02c6g(\u03bb) = e\u03bb \u02c6A e\u03bb \u02c6B e\u2212\u03bb2 \u02c6C/2. Note that these functions are equal at \u03bb = 0, and show that they satisfy the same differential equation: d \u02c6f /d\u03bb = \u02c6g. Therefore, the functions are themselves equal for all \u03bb.38 \u02c6f and d \u02c6g/d\u03bb = \u02c6A + \ufffd", "\ufffdB \u02c6A + \u02c6B 3.6.3 Changing Bases in Dirac Notation The advantage of Dirac notation is that it frees us from working in any particular basis, and makes transforming between bases seamless. Recall that the identity operator can be written as a projection onto a complete set of states (Equations 3.93 and 3.96); of particular interest are the position eigenstates |x, the momentum eigenstates | p, and the energy eigenstates (we will assume those are discrete) |n: 37 This is a special case of a more general formula that applies when \u02c6A and \u02c6B do not commute with \u02c6C. See, for example, Eugen Merzbacher, Quantum Mechanics, 3rd edn, Wiley, New York (1998), page 40. 38 The product rule holds for differentiating operators as long as you respect their order: # $ \u02c6A(\u03bb) \u02c6B(\u03bb) d d\u03bb = \u02c6A (\u03bb) \u02c6B(\u03bb) + \u02c6A(\u03bb) \u02c6B (\u03bb). (3.105) 122 CHAPTER 3 Formalism 1 = d x |x x|, 1 = dp | p p|, 1 = |n n|. Acting on the state vector |S(t) with each of these resolutions of the identity gives |S(t) = d x |x x | S(t) \u2261 (x, t) |x d x, |S(t) = dp | p p | S(t) \u2261 ( p, t) | p dp, |S(t) = n |n n | S(t) \u2261 cn(t) |n. (3.106) (3.107) Here we recognize the position-space, momentum-space, and \u201cenergy-space\u201d wave functions (Equations 3.77\u20133.79) as the components of the vector |S(t) in the respective bases. Example 3.9 Derive the transformation from the position-space wave function to the momentum-space wave function. (We already know the answer, of course, but I want to show you how this works out in Dirac notation.) Solution: We want to \ufb01nd ( p, t) = p|S(t) given (x, t) = x|S(t). We can relate the two by inserting a resolution of the identity: ( p, t) = p | S(t) d x |x x| S(t(t) d x p | x (x, t) d x. (3.108) Now, x| p is the momentum eigenstate (with eigenvalue p) in the position basis\u2014what we called f p(x), in Equation 3.32. So \u2217 = \u2217 = 1\u221a = e\u2212i p x/. p x x p f p(x) 2 \u03c0 Plugging this into Equation 3.108 gives ( p, t) = which is precisely Equation 3.54. 1\u221a 2 \u03c0 e\u2212i p x/ (x, t) d x, Just as the wave function takes different forms in different bases, so do operators. The 3.6 Vectors and Operators 123 position operator is given by in the position basis, or \u02c6x \u2192 x \u02c6x \u2192 i \u2202 \u2202 p in the momentum basis. However, Dirac notation allows us to do away with the arrows and stick to equalities. Operators act on kets (for instance, \u02c6x |S(t)); the outcome of this operation can be expressed in any basis by taking the inner product with an appropriate basis vector. That is, = action of position operator in x basis = x (x, t), (3.109) \u02c6x S(t) x or \u02c6x S(t) p = action of position operator in p basis = i \u2202 \u2202 p. (3.110) In this notation it is straightforward to transform operators between bases, as the following example illustrates. Example 3.10 Obtain the position operator in the momentum basis (Equation 3.110) by inserting a resolution of the identity on the left-hand side. Solution: \u02c6x S(t) p = = x S(t) x d x \u02c6x p p |x| x x | S(t) d x, \u02c6x |x = x |x where I\u2019ve used the fact that |x is an eigenstate of \u02c6x out of the inner product (it\u2019s just a number) and \u02c6x S(t) p = x p | x (x", ", t) d x ; x can then be pulled (x, t) d x = x e\u2212i p x\u2212i p x/ \u221a 2 \u03c0 (x, t) d x. Finally we recognize the integral as ( p, t) (Equation 3.54). Problem 3.30 Derive the transformation from the position-space wave function to the \u201cenergy-space\u201d wave function (cn(t)) using the technique of Example 3.9. Assume that the energy spectrum is discrete, and the potential is time-independent. 124 CHAPTER 3 Formalism FURTHER PROBLEMS ON CHAPTER 3 \u2217 Problem 3.31 Legendre polynomials. Use the Gram\u2013Schmidt procedure (Problem A.4) to orthonormalize the functions 1, x, x 2, and x 3, on the interval \u22121 \u2264 x \u2264 1. You may recognize the results\u2014they are (apart from normalization)39 Legendre polynomials (Problem 2.64 and Table 4.1). Problem 3.32 An anti-hermitian (or skew-hermitian) operator is equal to minus its hermitian conjugate: \u02c6Q\u2020 = \u2212 \u02c6Q. (3.111) (a) Show that the expectation value of an anti-hermitian operator is imaginary. (b) Show that the eigenvalues of an anti-hermitian operator are imaginary. (c) Show that the eigenvectors of an anti-hermitian operator belonging to distinct eigenvalues are orthogonal. (d) Show that the commutator of two hermitian operators is anti-hermitian. How about the commutator of two anti-hermitian operators? (e) Show that any operator \u02c6Q can be written as a sum of a hermitian operator \u02c6A and an anti-hermitian operator \u02c6B, and give expressions for \u02c6A and \u02c6B in terms of \u02c6Q and its adjoint \u02c6Q\u2020. Problem 3.33 Sequential measurements. An operator \u02c6A, representing observable A, has two (normalized) eigenstates \u03c81 and \u03c82, with eigenvalues a1 and a2, respectively. Operator \u02c6B, representing observable B, has two (normalized) eigenstates \u03c61 and \u03c62, with eigenvalues b1 and b2. The eigenstates are related by \u03c81 = (3\u03c61 + 4\u03c62) /5, \u03c82 = (4\u03c61 \u2212 3\u03c62) /5. (a) Observable A is measured, and the value a1 is obtained. What is the state of the system (immediately) after this measurement? (b) If B is now measured, what are the possible results, and what are their probabili- ties? (c) Right after the measurement of B, A is measured again. What is the probability of getting a1? (Note that the answer would be quite different if I had told you the outcome of the B measurement.) \u2217\u2217\u2217 Problem 3.34 (a) Find the momentum-space wave function n( p, t) for the nth stationary state of the in\ufb01nite square well. (b) Find the probability density |n( p, t)|2. Graph this function, for n = 1, n = 2, n = 5, and n = 10. What are the most probable values of p, for large n? Is this what you would have expected?40 Compare your answer to Problem 3.10. (c) Use n( p, t) to calculate the expectation value of p2, in the nth state. Compare your answer to Problem 2.4. 39 Legendre didn\u2019t know what the best convention would be; he picked the overall factor so that all his functions would go to 1 at x = 1, and we\u2019re stuck with his unfortunate choice. 40 See F. L. Markley, Am. J. Phys. 40, 1545 (1972). Further Problems on Chapter 3 125 Problem 3.35 Consider the wave function (x, 0) = 1\u221a 2n\u03bb ei2\u03c0 x/\u03bb, \u2212n\u03bb < x < n\u03bb, 0, otherwise, where n is some positive integer. This function is purely sinusoidal (with wavelength \u03bb) on the interval \u2212n\u03bb < x < n\u03bb, but it still carries a range of momenta, because the oscillations do not continue out to in\ufb01nity. Find the momentum space wave function ( p, 0). Sketch the graphs of |(x, 0)|2 and |( p, 0", ")|2, and determine their widths, wx and w p (the distance between zeros on either side of the main peak). Note what happens to each width as n \u2192 \u221e. Using wx and w p as estimates of x and p, check that the uncertainty principle is satis\ufb01ed. Warning: If you try calculating \u03c3 p, you\u2019re in for a rude surprise. Can you diagnose the problem? Problem 3.36 Suppose (x, 0) = A x 2 + a2, (\u2212\u221e < x < \u221e) for constants A and a. (a) Determine A, by normalizing (x, 0). (b) Find x,, and \u03c3x (at time t = 0). (c) Find the momentum space wave function ( p, 0), and check that x 2 it is normalized. (d) Use ( p, 0) to calculate p,, and \u03c3 p (at time t = 0). (e) Check the Heisenberg uncertainty principle for this state. p2 \u2217 Problem 3.37 Virial theorem. Use Equation 3.73 to show that d dt \u2202 V \u2202 x x p = 2T \u2212 x, (3.112) where T is the kinetic energy (H = T + V ). In a stationary state the left side is zero (why?) so 2T = x. (3.113) d V d x This is called the virial theorem. Use it to prove that T = V for stationary states of the harmonic oscillator, and check that this is consistent with the results you got in Problems 2.11 and 2.12. Problem 3.38 In an interesting version of the energy-time uncertainty principle41 t = \u03c4/\u03c0, where \u03c4 is the time it takes (x, t) to evolve into a state orthogonal to (x, 0). Test this out, using a wave function that is a linear combination of two some (arbitrary) potential: (x, 0) = (orthonormal) 1/ 2 stationary states of [\u03c81(x) + \u03c82(x)]. \u221a 41 See L. Vaidman, Am. J. Phys. 60, 182 (1992) for a proof. 126 CHAPTER 3 Formalism \u2217\u2217 Problem 3.39 Find the matrix elements n|x|n and n| p|n in the (orthonormal) basis of stationary states for the harmonic oscillator (Equation 2.68). You already calcuin Problem 2.12; use the same technique for lated the \u201cdiagonal\u201d elements the general case. Construct the corresponding (in\ufb01nite) matrices, X and P. Show that (1/2m) P2 + (m\u03c92/2)X2 = H is diagonal, in this basis. Are its diagonal elements what you would expect? Partial answer: n = n n x n = 2m\u03c9 \u221a \u221a n\u03b4n,n\u22121 + n\u03b4n,n\u22121. (3.114) \u2217\u2217 Problem 3.40 The most general wave function of a particle in the simple harmonic oscillator potential is (x, t) = n cn \u03c8n(x) e\u2212i En t/. Show that the expectation value of position is x = C cos(\u03c9t \u2212 \u03c6), where the real constants C and \u03c6 are given by \u221e! \" Ce\u2212i\u03c6 = 2 m\u03c9 n=0 \u221a n + 1 c\u2217 n+1 cn. Thus the expectation value of position for a particle in the harmonic oscillator oscillates at the classical frequency \u03c9 (as you would expect from Ehrenfest\u2019s theorem; see problem 3.19(b)). Hint: Use Equation 3.114. As an example, \ufb01nd C and \u03c6 for the wave function in Problem 2.40. Problem 3.41 A harmonic oscillator is in a state such that a measurement of the energy would yield either (1/2) \u03c9 or (3/2) \u03c9, with equal probability. What is the largest possible value of p in such a state? If it assumes this maximal value at time t = 0, what is (x, t)? \u2217\u2217\u2217 Problem 3.42 Coherent states of the harmonic oscillator. Among the stationary states of the harmonic oscillator (Equation 2.68) only n = 0 hits the uncertainty limit ; in general, \u03c3x \u03c3 p = (2n + 1) /2, as you found in Problem 2.12. \u03c3x \u03c3 p = /2 But certain linear combinations (known as coherent states) also minimize the uncertainty product. They are (as it turns out) eigenfunctions of the lowering operator", ":42 (the eigenvalue \u03b1 can be any complex number). a\u2212 |\u03b1 = \u03b1| \u03b1 42 There are no normalizable eigenfunctions of the raising operator. Further Problems on Chapter 3 127, p, (a) Calculate x, in the state |\u03b1. Hint: Use the technique in Example 2.5, and remember that a+ is the hermitian conjugate of a\u2212. Do not assume \u03b1 is real. x 2 p2 (b) Find \u03c3x and \u03c3 p; show that \u03c3x \u03c3 p = /2. (c) Like any other wave function, a coherent state can be expanded in terms of energy eigenstates: |\u03b1 = \u221e n=0 cn |n. Show that the expansion coef\ufb01cients are cn = \u03b1n \u221a n! (d) Determine c0 by normalizing |\u03b1. Answer: exp (e) Now put in the time dependence: c0. \u2212 |\u03b1|2 /2. |n \u2192 e\u2212i En t/ |n, and show that |\u03b1 (t) remains an eigenstate of a\u2212, but the eigenvalue evolves in time: \u03b1(t) = e\u2212i\u03c9t \u03b1. So a coherent state stays coherent, and continues to minimize the uncertainty product. (f) Based on your answers to (a), (b), and (e), \ufb01nd x and \u03c3x as functions of time. It helps if you write the complex number \u03b1 as \u03b1 = C m \u03c9 2 ei \u03c6 for real numbers C and \u03c6. Comment: In a sense, coherent states behave quasiclassically. (g) Is the ground state (|n = 0) itself a coherent state? If so, what is the eigenvalue? Problem 3.43 Extended uncertainty principle.43 The generalized uncertainty principle (Equation 3.62) states that \u03c3 2 A \u03c3 2 B \u2265 1 4 C2, # $ \u02c6A, \u02c6B where \u02c6C \u2261 \u2212i. (a) Show that it can be strengthened to read C2 + D2 3.115) where \u02c6D \u2261 \u02c6A \u02c6B + \u02c6B \u02c6A \u2212 2 A B. Hint: Keep the Re(z) term in Equation 3.60. 43 For interesting commentary and references, see R. R. Puri, Phys. Rev. A 49, 2178 (1994). 128 CHAPTER 3 Formalism (b) Check Equation 3.115 for the case B = A (the standard uncertainty principle is trivial, in this case, since \u02c6C = 0; unfortunately, the extended uncertainty principle doesn\u2019t help much either). Problem 3.44 The Hamiltonian for a certain three-level system is represented by the matrix \u239b = where a, b, and c are real numbers. (a) If the system starts out in the state |S(0) = what is |S(t)? (b) If the system starts out in the state |S(0) = what is |S(t)? \u239b \u239d \u239e Problem 3.45 Find the position operator in the basis of simple harmonic oscillator energy states. That is, express n \u02c6x S(t) in terms of cn(t) = n|S(t). Hint: Use Equation 3.114. Problem 3.46 The Hamiltonian for a certain three-level system is represented by the matrix \u239b Two other observables, A and B, are represented by the matrices \u239e \u23a0, \u239e \u239b \u239d where \u03c9, \u03bb, and \u03bc are positive real numbers. (a) Find the eigenvalues and (normalized) eigenvectors of H, A, and B. (b) Suppose the system starts out in the generic state \u239e \u239b |S(0) = \u239d \u23a0, c1 c2 c3 with |c1|2 + |c2|2 + |c3|2 = 1. Find the expectation values (at t = 0) of H, A, and B. Further Problems on Chapter 3 129 (c) What is |S(t)? If you measured the energy of this state (at time t), what values might you get, and what is the probability of each? Answer the same questions for observables A and for B. \u2217\u2217 Problem 3.47 Supersymmetry. Consider the two operators \u02c6p 2m \u02c6A\u2020 = \u2212i + W (x) \u02c6A = i and \u221a \u221a \u02c6p 2m + W (x), (3.116)", " for some function W (x). These may be multiplied in either order to construct two Hamiltonians: \u02c6H1 \u2261 \u02c6A\u2020 \u02c6A = \u02c6p2 2m + V1(x) and \u02c6H2 \u2261 \u02c6A \u02c6A\u2020 = \u02c6p2 2m + V2(x); (3.117) V1 and V2 are called supersymmetric partner potentials. The energies and eigenstates of \u02c6H1 and \u02c6H2 are related in interesting ways.44 (a) Find the potentials V1(x) and V2(x), in terms of the superpotential, W (x). (b) Show that if \u03c8 (1) n, then \u02c6A\u03c8 (1) (1) is an n n (x) is an eigenstate of \u02c6H1 with eigenstate of the same eigenvalue. The two Hamiltonians therefore have essentially identical spectra. \u02c6H2 with the same eigenvalue. Similarly, show that if \u03c8 (2) n, then \u02c6A\u2020\u03c8 (2) (2) \u02c6H2 with eigenvalue E is an eigenstate of \u02c6H1 with eigenvalue E is an eigenstate of n n (c) One ordinarily chooses W (x) such that the ground state of \u02c6H1 satis\ufb01es \u02c6A\u03c8 (1) 0 (x) = 0, (3.118) (1) 0 and hence E ground state wave function, \u03c8 (1) 0 \u02c6H2 actually has one less eigenstate than \u02c6H1, and is missing the eigenvalue E = 0. Use this to \ufb01nd the superpotential W (x), in terms of the 0 means that (1) 0.) (x). (The fact that \u02c6A annihilates \u03c8 (1) (d) Consider the Dirac delta function well, V1(x) = m\u03b12 22 \u2212 \u03b1\u03b4 (x), (3.119) (the constant term, m\u03b12/22, is included so that E state (Equation 2.132) (1) 0 = 0). It has a single bound \u03c8 (1) 0 (x) = \u221a m\u03b1 exp # \u2212 m\u03b1 2 $ |x|. (3.120) Use the results of parts (a) and (c), and Problem 2.23(b), to determine the superpotential W (x) and the partner potential V2(x). This partner potential is one that you will likely recognize, and while it has no bound states, the supersymmetry between these two systems explains the fact that their re\ufb02ection and transmission coef\ufb01cients are identical (see the last paragraph of Section 2.5.2). 44 Fred Cooper, Avinash Khare, and Uday Sukhatme, Supersymmetry in Quantum Mechanics, World Scienti\ufb01c, Singapore, 2001. 130 CHAPTER 3 Formalism \u2217\u2217 Problem 3.48 An operator is de\ufb01ned not just by its action (what it does to the vector it is applied to) but its domain (the set of vectors on which it acts). In a \ufb01nite-dimensional vector space the domain is the entire space, and we don\u2019t need to worry about it. But for most operators in Hilbert space the domain is restricted. In particular, only functions such that \u02c6Q f (x) remains in Hilbert space are allowed in the domain of \u02c6Q. (As you found in Problem 3.2, the derivative operator can knock a function out of L 2.) A hermitian operator is one whose action is the same as that of its adjoint45 (Problem 3.5). But what is required to represent observables is actually something more: the domains of \u02c6Q and \u02c6Q\u2020 must also be identical. Such operators are called self-adjoint.46 (a) Consider the momentum operator, \u02c6p = \u2212id/d x, on the \ufb01nite interval 0 \u2264 x \u2264 a. With the in\ufb01nite square well in mind, we might de\ufb01ne its domain as the set of f (0) = f (a) = 0 (it goes without saying that functions f (x) such that f (x) f and \u02c6p f (x) are in L 2 (0, a)). Show that \u02c6p is hermitian:, with \u02c6p\u2020 = \u02c6p. But is it self-adjoint? Hint: as", " long as f (0) = f (a) = 0, there is no restriction on g(0) or g(a)\u2014the domain of \u02c6p\u2020 is much larger than the domain of \u02c6p.47 \u02c6p f \u02c6p\u2020g = g (b) Suppose we extend the domain of \u02c6p to include all functions of the form f (a) = \u03bb f (0), for some \ufb01xed complex number \u03bb. What condition must we then impose on the domain of \u02c6p\u2020 in order that \u02c6p be hermitian? What value(s) of \u03bb will render \u02c6p self-adjoint? Comment: Technically, then, there is no momentum operator on the \ufb01nite interval\u2014or rather, there are in\ufb01nitely many, and no way to decide which of them is \u201ccorrect.\u201d (In Problem 3.34 we avoided the issue by working on the in\ufb01nite interval.) (c) What about the semi-in\ufb01nite interval, 0 \u2264 x < \u221e? Is there a self-adjoint momentum operator in this case?48 \u2217\u2217 Problem 3.49 (a) Write down the time-dependent \u201cSchr\u00f6dinger equation\u201d in momentum space, for a free particle, and solve it. Answer: exp \u2212i p2t/2m ( p, 0). (b) Find ( p, 0) for the traveling gaussian wave packet (Problem 2.42), and construct ( p, t) for this case. Also construct |( p, t)|2, and note that it is independent of time. by evaluating the appropriate integrals involving, and (c) Calculate p and p2 compare your answers to Problem 2.42. (d) Show that H = p2 /2m + H 0 (where the subscript 0 denotes the stationary gaussian), and comment on this result. 45 Mathematicians call them \u201csymmetric\u201d operators. 46 Because the distinction rarely intrudes, physicists tend to use the word \u201chermitian\u201d indiscriminately; technically, we should always say \u201cself-adjoint,\u201d meaning \u02c6Q = \u02c6Q\u2020 both in action and in domain. 47 The domain of \u02c6Q is something we stipulate; that determines the domain of \u02c6Q\u2020. 48 J. von Neumann introduced machinery for generating self-adjoint extensions of hermitian operators\u2014or in some cases proving that they cannot exist. For an accessible introduction see G. Bonneau, J. Faraut, and B. Valent, Am. J. Phys. 69, 322 (2001); for an interesting application see M. T. Ahari, G. Ortiz, and B. Seradjeh, Am. J. Phys. 84, 858 (2016). 4 QUANTUM MECHANICS IN THREE DIMENSIONS 4.1 THE SCHR\u00d6DINGER EQUATION The generalization to three dimensions is straightforward. Schr\u00f6dinger\u2019s equation says i \u2202 \u2202t = \u02c6H ; the Hamiltonian operator \u02c6H is obtained from the classical energy 1 2 mv2 + V = 1 2m p2 x + p2 y + p2 z + V by the standard prescription (applied now to y and z, as well as x): \u2202 \u2202 x py \u2192 \u2212i px \u2192 \u2212i \u2202 \u2202 y pz \u2192 \u2212i,, \u2202 \u2202z, or for short. Thus where p \u2192 \u2212i\u2207, i \u2202 \u2202t = \u2212 2 2m \u22072 + V, \u2207 \u2202 y2 + \u2202 2 \u2202z2, (4.1) (4.2) (4.3) (4.4) (4.5) is the Laplacian, in cartesian coordinates. The potential energy V and the wave function are now functions of r = (x, y, z) and t. The probability of \ufb01nding the particle in the in\ufb01nitesimal volume d3r = d x d y dz is |(r, t)|2 d3r, and the normalization condition reads ||2 d3r = 1, (4.6) with the integral taken over all space. If V is independent of time, there will be a complete set of stationary states, n(r, t) = \u03c8n(r) e\u2212i En t/, (4.7) 132 CHAPTER 4 Quantum Mechanics in Three Dimensions where the spatial wave function \u03c8n satis\ufb01es the time-independent Schr\u00f6dinger equation: \u2212 2 2m \u22072", "\u03c8 + V \u03c8 = E\u03c8. (4.8) The general solution to the (time-dependent) Schr\u00f6dinger equation is (4.9) with the constants cn determined by the initial wave function, (r, 0), in the usual way. (If the potential admits continuum states, then the sum in Equation 4.9 becomes an integral.) (r, t) = cn\u03c8n(r) e\u2212i En t/, \u2217 Problem 4.1 operators r and p: (a) Work out all of the canonical commutation relations for components of the x, px = i\u03b4i j,, and so on. Answer: py, pz ri, r j x, py pi, r j ri, p j x, y pi, p j (4.10) = 0,, = \u2212, =, where the indices stand for x, y, or z, and rx = x, ry = y, and rz = z. (b) Con\ufb01rm the three-dimensional version of Ehrenfest\u2019s theorem, d dt r = 1 m p, and d dt p = \u2212\u2207V. (4.11) (Each of these, of course, stands for three equations\u2014one for each component.) Hint: First check that the \u201cgeneralized\u201d Ehrenfest theorem, Equation 3.73, is valid in three dimensions. (c) Formulate Heisenberg\u2019s uncertainty principle in three dimensions. Answer: \u03c3x \u03c3 px \u03c3y\u03c3 py but there is no restriction on, say, \u03c3x \u03c3 py. \u2265 /2, \u2265 /2, \u2265 /2, \u03c3z\u03c3 pz (4.12) \u2217 Problem 4.2 Use separation of variables in cartesian coordinates to solve the in\ufb01nite cubical well (or \u201cparticle in a box\u201d): V(x, y, z) = 0, \u221e, x, y, z all between 0 and a; otherwise. (a) Find the stationary states, and the corresponding energies. (b) Call the distinct energies E1, E2, E3,..., in order of increasing energy. Find E1, E2, E3, E4, E5, and E6. Determine their degeneracies (that is, the number of different states that share the same energy). Comment: In one dimension degenerate bound states do not occur (see Problem 2.44), but in three dimensions they are very common. (c) What is the degeneracy of E14, and why is this case interesting? 4.1.1 Spherical Coordinates Most of the applications we will encounter involve central potentials, for which V is a function only of the distance from the origin, V(r) \u2192 V(r ). In that case it is natural to adopt 4.1 The Schr\u00f6dinger Equation 133 Figure 4.1: Spherical coordinates: radius r, polar angle \u03b8, and azimuthal angle \u03c6. spherical coordinates, (r, \u03b8, \u03c6) (Figure 4.1). In spherical coordinates the Laplacian takes the form1 \u22072 = 1 r 2 \u2202 \u2202r r 2 \u2202 \u2202r + 1 r 2 sin \u03b8 \u2202 \u2202\u03b8 sin \u03b8 \u2202 \u2202\u03b8 + 1 r 2 sin2 \u03b8. \u2202 2 \u2202\u03c62 (4.13) In spherical coordinates, then, the time-independent Schr\u00f6dinger equation reads \u2212 2 2m 1 r 2 \u2202 \u2202r r 2 \u2202\u03c8 \u2202r + 1 r 2 sin \u03b8 \u2202 \u2202\u03b8 sin \u03b8 \u2202\u03c8 \u2202\u03b8 + 1 r 2 sin2 \u03b8 \u2202 2\u03c8 \u2202\u03c62 + V \u03c8 = E\u03c8. (4.14) We begin by looking for solutions that are separable into products (a function of r times a function of \u03b8 and \u03c6): \u03c8(r, \u03b8, \u03c6) = R(r ) Y(\u03b8, \u03c6). (4.15) Putting this into Equation 4.14, we have \u2212 2 2m r 2 d R dr Y r 2 d dr + R r 2 sin \u03b8 \u2202 \u2202\u03b8 sin \u03b8 \u2202Y \u2202\u03b8 + R r 2 sin2 \u03b8 \u2202 2Y \u2202\u03c62 + V RY = E RY. Dividing by Y R and multiplying by \u22122mr 2/2: r 2 d R dr 1 R d dr \u2212 2mr 2 2 [V (r ) \u2212 E] + 1 Y 1 sin \u03b8 \ufffd", "\ufffd \u2202\u03b8 sin \u03b8 \u2202Y \u2202\u03b8 + 1 sin2 \u03b8 \u2202 2Y \u2202\u03c62 = 0. 1 In principle, this can be obtained by change of variables from the cartesian expression 4.5. However, there are much more ef\ufb01cient ways of getting it; see, for instance, M. Boas, Mathematical Methods in the Physical Sciences 3rd edn, Wiley, New York (2006), Chapter 10, Section 9. 134 CHAPTER 4 Quantum Mechanics in Three Dimensions The term in the \ufb01rst curly bracket depends only on r, whereas the remainder depends only on \u03b8 and \u03c6; accordingly, each must be a constant. For reasons that will appear in due course,2 I will write this \u201cseparation constant\u201d in the form ( + 1): r 2 d R dr [V(r ) \u2212 E] = ( + 1) ; \u2212 2mr 2 2 1 R (4.16) d dr 1 Y 1 sin \u03b8 \u2202 \u2202\u03b8 sin \u03b8 \u2202Y \u2202\u03b8 + 1 sin2 \u03b8 \u2202 2Y \u2202\u03c62 = \u2212 ( + 1). (4.17) Problem 4.3 (a) Suppose \u03c8(r, \u03b8, \u03c6) = Ae\u2212r/a, for some constants A and a. Find E and V(r ), assuming V(r ) \u2192 0 as r \u2192 \u221e. (b) Do the same for \u03c8(r, \u03b8, \u03c6) = Ae\u2212r 2/a2, assuming V(0) = 0. 4.1.2 The Angular Equation Equation 4.17 determines the dependence of \u03c8 on \u03b8 and \u03c6; multiplying by Y sin2 \u03b8, it becomes: sin \u03b8 \u2202 \u2202\u03b8 sin \u03b8 \u2202Y \u2202\u03b8 + \u2202 2Y \u2202\u03c62 = \u2212 ( + 1) sin2 \u03b8 Y. (4.18) You might recognize this equation\u2014it occurs in the solution to Laplace\u2019s equation in classical electrodynamics. As always, we solve it by separation of variables: Y(\u03b8, \u03c6) = (\u03b8 ) (\u03c6). (4.19) Plugging this in, and dividing by, sin \u03b8 d d\u03b8 sin \u03b8 d d\u03b8 1 + ( + 1) sin2 \u03b8 + 1 d2 d\u03c62 = 0. The \ufb01rst term is a function only of \u03b8, and the second is a function only of \u03c6, so each must be a constant. This time3 I\u2019ll call the separation constant m2: sin \u03b8 d d\u03b8 sin \u03b8 d d\u03b8 + ( + 1) sin2 \u03b8 = m2; (4.20) 1 The \u03c6 equation is easy: 1 d2 d\u03c62 = \u2212m2. d2 d\u03c62 = \u2212m2 \u21d2 (\u03c6) = eim\u03c6. (4.21) (4.22) 2 Note that there is no loss of generality here\u2014at this stage could be any complex number. Later on we\u2019ll discover that must in fact be an integer, and it is in anticipation of that result that I express the separation constant in a way that looks peculiar now. 3 Again, there is no loss of generality here, since at this stage m could be any complex number; in a moment, though, we will discover that m must in fact be an integer. Beware: The letter m is now doing double duty, as mass and as a separation constant. There is no graceful way to avoid this confusion, since both uses are standard. Some authors now switch to M or \u03bc for mass, but I hate to change notation in mid-stream, and I don\u2019t think confusion will arise, a long as you are aware of the problem. 4.1 The Schr\u00f6dinger Equation 135 Actually, there are two solutions: exp(im\u03c6) and exp(\u2212im\u03c6), but we\u2019ll cover the latter by allowing m to run negative. There could also be a constant factor in front, but we might as well absorb that into. Incidentally, in electrodynamics we would write the azimuthal function () in terms of sines and cosines, instead of exponentials, because electric \ufb01elds are real. But there is no such constraint on the wave function, and exponentials are a lot easier to work with. Now, when \u03c6 advances by 2\u03c0, we return to the same point in space (see Figure 4.1), so it is natural to require that4 (\u03c6 + 2\u03c0 ) = (\u03c6). (", "4.23) In other words, exp[im(\u03c6 + 2\u03c0 )] = exp(im\u03c6), or exp(2\u03c0im) = 1. From this it follows that m must be an integer: m = 0, \u00b11, \u00b12,.... The \u03b8 equation, sin \u03b8 d d\u03b8 sin \u03b8 d d\u03b8 # ( + 1) sin2 \u03b8 \u2212 m2 $ + = 0, may not be so familiar. The solution is (\u03b8 ) = A P m (cos \u03b8 ), where P m is the associated Legendre function, de\ufb01ned by5 (x) \u2261 (\u22121)m P m 1 \u2212 x 2 m/2 m P(x), d d x and P(x) is the th Legendre polynomial, de\ufb01ned by the Rodrigues formula: P(x) \u2261 1 24.24) (4.25) (4.26) (4.27) (4.28) For example, P0(x) = 1, P2(x) = 1 4 \u00b7 2 d d x P1(x, d d x x 2 \u2212 1 3x 2 \u2212 1, 2 = 1 2 and so on. The \ufb01rst few Legendre polynomials are listed in Table 4.1. As the name suggests, P(x) is a polynomial (of degree ) in x, and is even or odd according to the parity of. But (x) is not, in general, a polynomial6\u2014if m is odd it carries a factor of P m 1 \u2212 x 2: \u221a 4 This is more slippery than it looks. After all, the probability density (||2) is single valued regardless of m. In Section 4.3 we\u2019ll obtain the condition on m by an entirely different\u2014and more compelling\u2014argument. 5 Some books (including earlier editions of this one) do not include the factor (\u22121)m in the de\ufb01nition of Pm. Equation 4.27 assumes that m \u2265 0; for negative values we de\ufb01ne ( \u2212 m)! ( + m)! \u2212m A few books (including earlier versions of this one) de\ufb01ne P convention used by Mathematica. (x) = (\u22121)m \u2212m P Pm (x). = Pm. I am adopting now the more standard 6 Nevertheless, some authors call them (confusingly) \u201cassociated Legendre polynomials.\u201d 136 CHAPTER 4 Quantum Mechanics in Three Dimensions Table 4.1: The \ufb01rst few Legendre polynomials, P (x): (a) functional form, (b) graph. = 1 = x P0 P1 = P2 = P3 = P4 = P5 1 2 1 2 1 8 1 8 (3x2 \u2212 1) (5x3 \u2212 3x) (35x4 \u2212 30x2 + 3) (63x5 \u2212 70x3 + 15x) (a) P0 1 0.5 P1 P2 \u22121 1 x P3 \u22121 (b) P 0 2 (x) = 1 2 3x 2 \u2212 1, P 1 2 (x) = \u2212 1 \u2212 x 2 1/2 d d x 1 2 3x 2 \u2212 1 % = \u22123x 1 \u2212 x 2, P 2 2 (x 3x cos \u03b8 ) etc. (On the other hand, what we need is P m is always a polynomial in cos \u03b8, multiplied\u2014if m is odd\u2014by sin \u03b8. Some associated Legendre functions of cos \u03b8 are listed in Table 4.2.) 1 \u2212 cos2 \u03b8 = sin \u03b8, so P m (cos \u03b8 ), and \u221a Notice that must be a non-negative integer, for the Rodrigues formula to make any sense; = 0. For any given, then, there are (2 + 1) moreover, if m >, then Equation 4.27 says P m possible values of m: = 0, 1, 2,... ; m = \u2212, \u2212 + 1,..., \u22121, 0, 1,..., \u2212 1,. (4.29) But wait! Equation 4.25 is a second-order differential equation: It should have two linearly independent solutions, for any old values of and m. Where are all the other solutions? Answer: They exist, of course, as mathematical solutions to the equation, but they are physically unacceptable, because they blow up at \u03b8 = 0 and/or \u03b8 = \u03c0 (see Problem 4.5). Now, the volume element in spherical coordinates7 is d3r = r 2 sin \u03b8 dr d\u03b8 d\u03c6 = r", " 2 dr d, where d \u2261 sin \u03b8 d\u03b8 d\u03c6, (4.30) so the normalization condition (Equation 4.6) becomes |\u03c8|2 r 2 sin \u03b8 dr d\u03b8 d\u03c6 = |R|2 r 2 dr |Y |2 d = 1. It is convenient to normalize R and Y separately: \u03c0 \u221e |R|2 r 2 dr = 1 and 2\u03c0 0 0 0 7 See, for instance, Boas (footnote 1), Chapter 5, Section 4. |Y |2 sin \u03b8 d\u03b8 d\u03c6 = 1. (4.31) 4.1 The Schr\u00f6dinger Equation 137 (cos \u03b8 ): (a) functional form, (b) Table 4.2: Some associated Legendre functions, P m (in these plots r tells you the magnitude of the function graphs of r = in the direction \u03b8 ; each \ufb01gure should be rotated about the z axis). (cos 3 cos2 \u03b8 \u2013 1) 1 2 P 1 1 = \u2013sin \u03b8 P 3 3 = \u201315 sin \u03b8 (1 \u2013 cos2 \u03b8 ) P 0 1 = cos \u03b8 P 2 3 = 15 sin2 \u03b8 cos \u03b8 P 2 2 = 3 sin2 \u03b8 P 1 2 = \u2013 3 sin \u03b8 cos \u03b8 = \u2013 sin \u03b8 (5 cos2 \u03b8 \u2013 1) 3 2 = (5 cos3 \u03b8 \u2013 3 cos a) z z z P 1 1 (\u03b8) P 1 2 (\u03b8) P 2 2 (\u03b8) z z z P 0 0 (\u03b8) P 0 1 (\u03b8) P 0 2 (\u03b8) (b) Table 4.3: The \ufb01rst few spherical harmonics, Y m (\u03b8, \u03c6). The normalized angular wave functions8 are called spherical harmonics: / Y m (\u03b8, \u03c6) = (2 + 1) 4\u03c0 ( \u2212 m)! ( + m)! eim\u03c6 P m (cos \u03b8 ), (4.32) As we shall prove later on, they are automatically orthogonal: \u03c0 2\u03c0 0 0 Y m (\u03b8, \u03c6) # \u2217 $ Y m (\u03b8, \u03c6) sin \u03b8 d\u03b8 d\u03c6 = \u03b4\u03b4mm, (4.33) In Table 4.3 I have listed the \ufb01rst few spherical harmonics. \u2217 Problem 4.4 Use Equations 4.27, 4.28, and 4.32, to construct Y 0 are normalized and orthogonal. 0 and Y 1 2. Check that they 8 The normalization factor is derived in Problem 4.63. 138 CHAPTER 4 Quantum Mechanics in Three Dimensions Problem 4.5 Show that (\u03b8 ) = A ln tan(\u03b8/2) satis\ufb01es the \u03b8 equation (Equation 4.25), for = m = 0. This is the unacceptable \u201csecond solution\u201d\u2014what\u2019s wrong with it? \u2217 \u2217\u2217 Problem 4.6 Using Equation 4.32 and footnote 5, show that \u2212m = (\u22121)m Y Y m \u2217. Problem 4.7 Using Equation 4.32, \ufb01nd Y Table 4.2, but you\u2019ll have to work out P satisfy the angular equation (Equation 4.18), for the appropriate values of and m. 3 from from Equations 4.27 and 4.28.) Check that they (\u03b8, \u03c6). (You can take P 2 (\u03b8, \u03c6) and Y 2 3 Problem 4.8 Starting from the Rodrigues formula, derive the orthonormality condition for Legendre polynomials: 1 \u22121 P(x)P(x) d x = 2 2 + 1 \u03b4. (4.34) Hint: Use integration by parts. 4.1.3 The Radial Equation Notice that the angular part of the wave function, Y (\u03b8, \u03c6), is the same for all spherically symmetric potentials; the actual shape of the potential, V(r ), affects only the radial part of the wave function, R(r ), which is determined by Equation 4.16: r 2 d R dr d dr \u2212 2mr 2 2 This simpli\ufb01es if we change variables: Let [V(r ) \u2212 E] R = ( + 1) R. (4.35) so that R = u/r, d R/dr = hence u(r ) \u2261 r R(r ), r (du/dr ) \u2212 u /r 2, (d/dr ) r 2 (d R/dr ) = r d2u/dr 2, and", " (4.36) \u2212 2 2m d2u dr 2 + V + 2 2m ( + 1) r 2 u = Eu. (4.37) This is called the radial equation;9 it is identical in form to the one-dimensional Schr\u00f6dinger equation (Equation 2.5), except that the effective potential, Veff = V + 2 2m ( + 1) r 2, (4.38) 9 Those ms are masses, of course\u2014the separation constant m does not appear in the radial equation. 4.1 The Schr\u00f6dinger Equation 139 contains an extra piece, the so-called centrifugal term,. It tends to throw the particle outward (away from the origin), just like the centrifugal (pseudo-)force in classical mechanics. Meanwhile, the normalization condition (Equation 4.31) becomes ( + 1) /r 2 2/2m \u221e 0 |u|2 dr = 1. That\u2019s as far as we can go until a speci\ufb01c potential V(r ) is provided. Example 4.1 Consider the in\ufb01nite spherical well, V(r ) = 0, \u221e, r \u2264 a; r > a. Find the wave functions and the allowed energies. (4.39) (4.40) Solution: Outside the well the wave function is zero; inside the well, the radial equation says d2u dr 2 = ( + 1) r 2 \u2212 k2 u, where \u221a k \u2261 2m E. (4.41) (4.42) Our problem is to solve Equation 4.41, subject to the boundary condition u(a) = 0. The case = 0 is easy: d2u dr 2 = \u2212k2u \u21d2 u(r ) = A sin(kr ) + B cos(kr ). But remember, the actual radial wave function is R(r ) = u(r )/r, and [cos(kr )] /r blows up as r \u2192 0. So10 B = 0. The boundary condition then requires sin(ka) = 0, and hence ka = N \u03c0, for some integer N. The allowed energies are E N 0 = N 2\u03c0 22 2ma2, (N = 1, 2, 3,...) (4.43) (same as for the one-dimensional in\ufb01nite square well, Equation 2.30). Normalizing u(r ) yields A = 2/a: \u221a 2 a Notice that the radial wave function has N \u2212 1 nodes (or, if you prefer, N \u201clobes\u201d). u N 0 = sin. N \u03c0r a (4.44) The general solution to Equation 4.41 (for an arbitrary integer ) is not so familiar: u(r ) = Ar j(kr ) + Br n(kr ), (4.45) 10 Actually, all we require is that the wave function be normalizable, not that it be \ufb01nite: R(r ) \u223c 1/r at the origin is normalizable (because of the r 2 in Equation 4.31). For a compelling general argument that u(0) = 0, see Ramamurti Shankar, Principles of Quantum Mechanics, 2nd edn (Plenum, New York, 1994), p. 342. For further discussion see F. A. B. Coutinho and M. Amaku, Eur. J. Phys. 30, 1015 (2009). 140 CHAPTER 4 Quantum Mechanics in Three Dimensions where j(x) is the spherical Bessel function of order, and n(x) is the spherical Neumann function of order. They are de\ufb01ned as follows: j(x) \u2261 (\u2212x) 1 x d d x sin x x ; n(x) \u2261 \u2212 (\u2212x) 1 x d d x cos x x. (4.46) For example, j0(x) = sin x x j1(x) = (\u2212x) 1 x d d x ; n0(x) = \u2212 cos x x ; sin x x = sin x x 2 \u2212 cos x x ; j2(x) = (\u2212x)2 1 x 2 sin x x d d x = 3 sin x \u2212 3x cos x \u2212 x 2 sin cos x \u2212 sin x x 3 ; and so on. The \ufb01rst few spherical Bessel and Neumann functions are listed in Table 4.4. For small x (where sin x = x \u2212 x 3/3! + x 5/5! \u2212 \u00b7 \u00b7 \u00b7 and cos x = 1 \u2212 x 2/2 + x 4/4! \u2212 \u00b7 \u00b7 \u00b7 ), j0(x) \u2248 1; n0(x) \u2248 1 x ; j1(x) \u2248 x 3 ;", " j2(x) \u2248 x 2 15 ; etc. Notice that Bessel functions are \ufb01nite at the origin, but Neumann functions blow up at the origin. Accordingly, B = 0, and hence R(r ) = A j(kr ). (4.47) There remains the boundary condition, R (a) = 0. Evidently k must be chosen such that j(ka) = 0; (4.48) that is, (ka) is a zero of the th-order spherical Bessel function. Now, the Bessel functions are oscillatory (see Figure 4.2); each one has an in\ufb01nite number of zeros. But (unfortunately for us) they are not located at nice sensible points (such as multiples of \u03c0 ); they have to be computed numerically.11 At any rate, the boundary condition requires that k = 1 a where \u03b2N is the N th zero of the th spherical Bessel function. The allowed energies, then, are given by \u03b2N, (4.49) E N = 2 2ma2 \u03b22 N. (4.50) It is customary to introduce the principal quantum number, n, which simply orders the allowed energies, starting with 1 for the ground state (see Figure 4.3). The wave functions are 11 Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, Dover, New York (1965), Chapter 10, provides an extensive listing. Table 4.4 The \ufb01rst few spherical Bessel and Neumann functions, asymptotic forms for small x. j(x) and n(x); 4.1 The Schr\u00f6dinger Equation 141 = \u20130 = \u20131 = \u20132 = \u20133 2 4 6 8 10 12 14 x 0.6 0.5 0.4 0.3 0.2 0.1 0 \u22120.1 \u22120.2 \u22120.3 Figure 4.2: Graphs of the \ufb01rst four spherical Bessel functions. \u03c8nm(r, \u03b8, \u03c6) = An j \u03b2N r a Y m (\u03b8, \u03c6), (4.51) with the constant An to be determined by normalization. As before, the wave function has N \u2212 1 radial nodes.12 12 We shall use this notation (N \u2212 1 as a count of the number of radial nodes, n for the order of the energy) with all central potentials. Both n and N are by their nature integers (1, 2, 3,... ); n is determined by N and (conversely, N is determined by n and ), but the actual relation can (as here) be complicated. In the special case of the Coulomb potential, as we shall see, there is a delightfully simple formula relating the two. 142 CHAPTER 4 Quantum Mechanics in Three Dimensions Figure 4.3: Energy levels of the in\ufb01nite spherical well (Equation 4.50). States with the same value of N are connected by dashed lines. Notice that the energy levels are (2 + 1)-fold degenerate, since there are (2 + 1) different values of m for each value of (see Equation 4.29). This is the degeneracy to be expected for a spherically symmetric potential, since m does not appear in the radial equation (which determines the energy). But in some cases (most famously the hydrogen atom) there is extra degeneracy, due to coincidences in the energy levels not attributable to spherical symmetry alone. The deeper reason for such \u201caccidental\u201d degeneracy is intriguing, as we shall see in Chapter 6. Problem 4.9 (a) From the de\ufb01nition (Equation 4.46), construct n1(x) and n2(x). (b) Expand the sines and cosines to obtain approximate formulas for n1(x) and n2(x), valid when x 1. Con\ufb01rm that they blow up at the origin. Problem 4.10 (a) Check that Ar j1(kr ) satis\ufb01es the radial equation with V(r ) = 0 and = 1. (b) Determine graphically the allowed energies for the in\ufb01nite spherical well, (N + 1/2)2. Hint: First j1(x) = 0 \u21d2 x = tan x. Plot x and tan x on the same graph, and when = 1. Show that for large N, E N 1 \u2248 show that locate the points of intersection. 2\u03c0 2/2ma2 4.2 The Hydrogen Atom 143 \u2217\u2217 Problem 4.11 A particle of mass m is placed in a \ufb01nite spherical well: V(r ) = \u2212V0, 0, r \u2264 a; r", " > a. Find the ground state, by solving the radial equation with = 0. Show that there is no bound state if V0a2 < \u03c0 22/8m. 4.2 THE HYDROGEN ATOM The hydrogen atom consists of a heavy, essentially motionless proton (we may as well put it at the origin), of charge e, together with a much lighter electron (mass me, charge \u2212e) that orbits around it, bound by the mutual attraction of opposite charges (see Figure 4.4). From Coulomb\u2019s law, the potential energy of the electron13 (in SI units) is V(r ) = \u2212 e2 4\u03c0 0, 1 r and the radial equation (Equation 4.37) says \u2212 e2 4\u03c0 0 d2u dr 2 2me 2 \u2212 + + 1 r 2 2me ( + 1) r 2 u = Eu. (4.52) (4.53) (The effective potential\u2014the term in square brackets\u2014is shown in Figure 4.5.) Our problem is to solve this equation for u(r ), and determine the allowed energies. The hydrogen atom is such an important case that I\u2019m not going to hand you the solutions this time\u2014we\u2019ll work them out in detail, by the method we used in the analytical solution to the harmonic oscillator. (If any step in this process is unclear, you may want to refer back to Section 2.3.2 for a more complete explanation.) Incidentally, the Coulomb potential (Equation 4.52) admits continuum states (with E > 0), describing electron-proton scattering, as well as discrete bound states, representing the hydrogen atom, but we shall con\ufb01ne our attention to the latter.14 \u2212e (electron) r +e (proton) Figure 4.4: The hydrogen atom. 13 This is what goes into the Schr\u00f6dinger equation\u2014not the electric potential (e/4\u03c0 0r ). 14 Note, however, that the bound states by themselves are not complete. 144 CHAPTER 4 Quantum Mechanics in Three Dimensions Veff r Figure 4.5: The effective potential for hydrogen (Equation 4.53), if > 0. 4.2.1 The Radial Wave Function Our \ufb01rst task is to tidy up the notation. Let \u03ba \u2261 \u221a \u22122me E. (For bound states, E is negative, so \u03ba is real.) Dividing Equation 4.53 by E, we have 1 \u2212 mee2 2\u03c0 02\u03ba ( + 1) (\u03bar )2 1 (\u03bar ) d2u dr 2 1 \u03ba 2 u. = + This suggests that we introduce so that \u03c1 \u2261 \u03bar, and \u03c10 \u2261 mee2 2\u03c0 02\u03ba, 1 \u2212 = d2u d\u03c12 \u03c10 \u03c1 + ( + 1) \u03c12 u. (4.54) (4.55) (4.56) Next we examine the asymptotic form of the solutions. As \u03c1 \u2192 \u221e, the constant term in the brackets dominates, so (approximately) The general solution is d2u d\u03c12 = u. u(\u03c1) = Ae\u2212\u03c1 + Be\u03c1, but e\u03c1 blows up (as \u03c1 \u2192 \u221e), so B = 0. Evidently, u(\u03c1) \u223c Ae\u2212\u03c1, (4.57) (4.58) for large \u03c1. On the other hand, as \u03c1 \u2192 0 the centrifugal term dominates;15 approximately, then: The general solution (check it!) is d2u d\u03c12 = ( + 1) \u03c12 u. u(\u03c1) = C\u03c1+1 + D\u03c1\u2212, 15 This argument does not apply when = 0 (although the conclusion, Equation 4.59, is in fact valid for that case too). But never mind: All I am trying to do is provide some motivation for a change of variables (Equation 4.60). 4.2 The Hydrogen Atom 145 but \u03c1\u2212 blows up (as \u03c1 \u2192 0), so D = 0. Thus u(\u03c1) \u223c C\u03c1+1, (4.59) for small \u03c1. The next step is to peel off the asymptotic behavior, introducing the new function v(\u03c1): u(\u03c1) = \u03c1+1e\u2212\u03c1v(\u03c1), (4.60) in the hope that v(\u03c1) will turn out to be simpler than u(\u03c1). The \ufb01rst indications are not auspicious: du d\u03c1 = \u03c1e\u2212\u03c1 ( + 1 \u2212 \u03c1)v + \u03c1 dv d\u03c1, and = \u03c1e\u2212\u03c1 \u22122 \u2212 2 + \u03c1 + ( + 1) \u03c1 d2u d\u03c12 v + 2 (", " + 1 \u2212 \u03c1). dv d\u03c1 + \u03c1 d2v d\u03c12 In terms of v(\u03c1), then, the radial equation (Equation 4.56) reads \u03c1 d2v d\u03c12 + 2 ( + 1 \u2212 \u03c1) dv d\u03c1 + [\u03c10 \u2212 2 ( + 1)] v = 0. (4.61) Finally, we assume the solution, v(\u03c1), can be expressed as a power series in \u03c1: v(\u03c1) = \u221e j=0 c j \u03c1 j. (4.62) Our problem is to determine the coef\ufb01cients (c0, c1, c2,...). Differentiating term by term: dv d\u03c1 = \u221e j=0 jc j \u03c1 j\u22121 = \u221e j=0 ( j + 1) c j+1\u03c1 j. (In the second summation I have renamed the \u201cdummy index\u201d: j \u2192 j + 1. If this troubles you, write out the \ufb01rst few terms explicitly, and check it. You may object that the sum should now begin at j = \u22121, but the factor ( j + 1) kills that term anyway, so we might as well start at zero.) Differentiating again, d2v d\u03c12 = \u221e j=0 j ( j + 1) c j+1\u03c1 j\u22121. Inserting these into Equation 4.61, \u221e j=0 j ( j + 1) c j+1\u03c1 j + 2 ( + 1) \u221e j=0 j ( j + 1) c j+1\u03c1 j \u22122 \u221e j=0 jc j \u03c1 j + [\u03c10 \u2212 2 ( + 1)] \u221e j=0 c j \u03c1 j = 0. Equating the coef\ufb01cients of like powers yields j ( j + 1) c j+1 + 2 ( + 1) ( j + 1) c j+1 \u2212 2 jc j + [\u03c10 \u2212 2 ( + 1)] c j = 0, 146 CHAPTER 4 Quantum Mechanics in Three Dimensions or: c j+1 = 2 ( j + + 1) \u2212 \u03c10 ( j + 1) ( j + 2 + 2) c j. (4.63) This recursion formula determines the coef\ufb01cients, and hence the function v(\u03c1): We start with c0 (this becomes an overall constant, to be \ufb01xed eventually by normalization), and Equation 4.63 gives us c1; putting this back in, we obtain c2, and so on.16 Now let\u2019s see what the coef\ufb01cients look like for large j (this corresponds to large \u03c1, where the higher powers dominate). In this regime the recursion formula says17 c j+1 \u2248 2 j j( j + 1, so c j \u2248 2 j j! c0. (4.64) Suppose for a moment that this were the exact result. Then v(\u03c1) = c0 \u221e j=0 2 j j! \u03c1 j = c0e2\u03c1, and hence u(\u03c1) = c0\u03c1l+1e\u03c1, which blows up at large \u03c1. The positive exponential is precisely the asymptotic behavior we didn\u2019t want, in Equation 4.57. (It\u2019s no accident that it reappears here; after all, it does represent the asymptotic form of some solutions to the radial equation\u2014they just don\u2019t happen to be the ones we\u2019re interested in, because they aren\u2019t normalizable.) (4.65) There is only one escape from this dilemma: The series must terminate. There must occur some integer N such that cN \u22121 = 0 but cN = 0 (4.66) (beyond this all coef\ufb01cients vanish automatically).18 In that case Equation 4.63 says De\ufb01ning 2 (N + ) \u2212 \u03c10 = 0. n \u2261 N +, (4.67) 16 You might wonder why I didn\u2019t use the series method directly on u(\u03c1)\u2014why factor out the asymptotic behavior before applying this procedure? Well, the reason for peeling off \u03c1+1 is largely aesthetic: Without this, the sequence would begin with a long string of zeros (the \ufb01rst nonzero coef\ufb01cient being c+1); by factoring out \u03c1+1 we obtain a series that starts out with \u03c10. The e\u2212\u03c1 factor is more critical\u2014if you don\u2019t pull that out, you get a three-term recursion formula, involving c j+2, c j+1", " and c j (try it!), and that is enormously more dif\ufb01cult to work with. 17 Why not drop the 1 in j + 1? After all, I\u2019m ignoring 2 ( + 1) \u2212 \u03c10 in the numerator, and 2 + 2 in the denominator. In this approximation it would be \ufb01ne to drop the 1 as well, but keeping it makes the argument a little cleaner. Try doing it without the 1, and you\u2019ll see what I mean. 18 This makes v(\u03c1) a polynomial of order (N \u2212 1), with (therefore) N \u2212 1 roots, and hence the radial wave function has N \u2212 1 nodes. we have But \u03c10 determines E (Equations 4.54 and 4.55): \u03c10 = 2n. E = \u2212 2\u03ba 2 2m = \u2212 mee4 8\u03c0 22 0 2\u03c12 0, so the allowed energies are 4.2 The Hydrogen Atom 147 (4.68) (4.69) En = \u2212 2 e2 4\u03c0 0 me 22 1 n2 = E1 n2, n = 1, 2, 3,.... (4.70) This is the famous Bohr formula\u2014by any measure the most important result in all of quantum mechanics. Bohr obtained it in 1913 by a serendipitous mixture of inapplicable classical physics and premature quantum theory (the Schr\u00f6dinger equation did not come until 1926). Combining Equations 4.55 and 4.68, we \ufb01nd that where \u03ba = mee2 4\u03c0 02 1 n = 1 an, a \u2261 4\u03c0 02 mee2 = 0.529 \u00d7 10 \u221210 m is the so-called Bohr radius.19 It follows (again, from Equation 4.55) that \u03c1 = r an. The spatial wave functions are labeled by three quantum numbers (n,, and m):20 \u03c8nm(r, \u03b8, \u03c6) = Rn(r ) Y m (\u03b8, \u03c6), where (referring back to Equations 4.36 and 4.60) Rn(r ) = 1 r \u03c1+1e\u2212\u03c1v(\u03c1), (4.71) (4.72) (4.73) (4.74) (4.75) and v(\u03c1) is a polynomial of degree n \u2212 \u2212 1 in \u03c1, whose coef\ufb01cients are determined (up to an overall normalization factor) by the recursion formula c j+) ( j + 1) ( j + 2 + 2) c j. (4.76) 19 It is customary to write the Bohr radius with a subscript: a0. But this is cumbersome and unnecessary, so I prefer to leave the subscript off. 20 Again, n is the principal quantum number; it tells you the energy of the electron (Equation 4.70). For unfortunate historical reasons is called the azimuthal quantum number and m the magnetic quantum number; as we\u2019ll see in Section 4.3, they are related to the angular momentum of the electron. 148 CHAPTER 4 Quantum Mechanics in Three Dimensions The ground state (that is, the state of lowest energy) is the case n = 1; putting in the accepted values for the physical constants, we get:21 E1 = \u2212 2 e2 4\u03c0 0 me 22 = \u221213.6 eV. (4.77) In other words, the binding energy of hydrogen (the amount of energy you would have to impart to the electron in its ground state in order to ionize the atom) is 13.6 eV. Equation 4.67 forces = 0, whence also m = 0 (see Equation 4.29), so \u03c8100(r, \u03b8, \u03c6) = R10(r )Y 0 0 (\u03b8, \u03c6). (4.78) The recursion formula truncates after the \ufb01rst term (Equation 4.76 with j = 0 yields c1 = 0), so v(\u03c1) is a constant (c0), and Normalizing it, in accordance with Equation 4.31: R10(r ) = c0 a e\u2212r/a. \u221e 0 |R10|2r 2 dr = |c0|2 a2 \u221a \u221e 0 e\u22122r/a r 2 dr = |c0|2 a 4 = 1, \u221a so c0 = 2/ a. Meanwhile, Y 0 0 = 1/ 4\u03c0, and hence the ground state of hydrogen is If n = 2 the energy is \u03c8100(r, \u03b8, \u03c6) = 1\u221a \u03c0a3 e\u2212r/a. E2 = \u221213.6 eV 4 = \u2212", "3.40 eV; (4.79) (4.80) (4.81) this is the \ufb01rst excited state\u2014or rather, states, since we can have either = 0 (in which case m = 0) or = 1 (with m = \u22121, 0, or +1); evidently four different states share this same energy. If = 0, the recursion relation (Equation 4.76) gives c1 = \u2212c0 (using j = 0), and c2 = 0 (using j = 1), so v(\u03c1) = c0 (1 \u2212 \u03c1), and therefore 1 \u2212 r 2a R20(r ) = c0 2a c j e\u2212r/2a. (4.82) are completely different for different quantum (Notice that the expansion coef\ufb01cients numbers n and.) If = 1 the recursion formula terminates the series after a single term; v(\u03c1) is a constant, and we \ufb01nd (In each case the constant c0 is to be determined by normalization\u2014see Problem 4.13.) R21(r ) = c0 4a2 r e\u2212r/2a. (4.83) 21 An electron volt is the energy acquired by an electron when accelerated through an electric potential of 1 volt: 1 eV = 1.60 \u00d7 10 \u221219 J. 4.2 The Hydrogen Atom 149 For arbitrary n, the possible values of (consistent with Equation 4.67) are = 0, 1, 2,..., n \u2212 1, (4.84) and for each there are (2 + 1) possible values of m (Equation 4.29), so the total degeneracy of the energy level En is d(n) = n\u22121 =0 (2 + 1) = n2. (4.85) In Figure 4.6 I plot the energy levels for hydrogen. Notice that different values of carry the same energy (for a given n)\u2014contrast the in\ufb01nite spherical well, Figure 4.3. (With Equation 4.67, dropped out of sight, in the derivation of the allowed energies, though it does still affect the wave functions.) This is what gives rise to the \u201cextra\u201d degeneracy of the Coulomb potential, as compared to what you would expect from spherical symmetry alone (n2 = 1, 4, 9, 16,..., as opposed to (2 + 1) = 1, 3, 5, 7,...). The polynomial v(\u03c1) (de\ufb01ned by the recursion formula, Equation 4.76) is a function well known to applied mathematicians; apart from normalization, it can be written as where (\u03c1) = L 2+1 n\u2212\u22121 (2\u03c1), L p q (x) \u2261 (\u22121) p p d d x L p+q (x) (4.86) (4.87) Scattering States (E > 0) Bound States (E < 0 Figure 4.6: Energy levels for hydrogen (Equation 4.70); n = 1 is the ground state, with E1 = \u221213.6 eV; an in\ufb01nite number of states are squeezed in between n = 5 and n = \u221e; E\u221e = 0 separates the bound states from the scattering states. Compare Figure 4.3, and note the extra (\u201caccidental\u201d) degeneracy of the hydrogen energies. 150 CHAPTER 4 Quantum Mechanics in Three Dimensions Table 4.5: The \ufb01rst few Laguerre polynomials. L0(x) = 1 L1(x) = \u2212x + 1 L2(x) = 1 2 x2 \u2212 2x + 1 L3(x) = \u2212 1 6 x3 + 3 2 x2 \u2212 3x + 1 L4(x) = 1 24 x4 \u2212 2 3 x3 + 3x2 \u2212 4x + 1 L5(x) = \u2212 1 120 x5 + 5 24 x4 \u2212 5 3 x3 + 5x2 \u2212 5x + 1 L6(x) = 1 720 x6 \u2212 1 20 x5 + 5 8 x4 \u2212 10 3 x3 + 15 2 x2 \u2212 6x + 1 Table 4.6: Some associated Laguerre polynomials. L0 0(x) = 1 L2 0(x) = 1 L0 1(x) = \u2212x + 1 L2 1(x) = \u2212x + 3 L0 2(x) = 1 2 x2 \u2212 2x + 1 L2 2(x) = 1 2 x2 \u2212 4x + 6 L1 0(x) = 1 L3 0(x) = 1 L1", " 1(x) = \u2212x + 2 L3 1(x) = \u2212x + 4 L1 2(x) = 1 2 x2 \u2212 3x + 3 L3 2(x) = 1 2 x2 \u2212 5x + 10 is an associated Laguerre polynomial, and Lq (x) \u2261 ex q! d d x q e\u2212x x q (4.88) is the qth Laguerre polynomial.22 The \ufb01rst few Laguerre polynomials are listed in Table 4.5; some associated Laguerre polynomials are given in Table 4.6. The \ufb01rst few radial wave functions are listed in Table 4.7, and plotted in Figure 4.7.) The normalized hydrogen wave functions are23 22 As usual, there are rival normalization conventions in the literature. Older physics books (including earlier editions of this one) leave off the factor (1/q!). But I think it is best to adopt the Mathematica standard (which sets Lq (0) = 1). As the names suggest, Lq (x) and L p q (x) are polynomials (of degree q) in x. Incidentally, the associated Laguerre polynomials can also be written in the form q (x) = x \u2212 pex L p q! q d d x \u2212x x p+q e. 23 If you want to see how the normalization factor is calculated, study (for example), Leonard I. Schiff, Quantum Mechanics, 2nd edn, McGraw-Hill, New York, 1968, page 93. In books using the older normalization convention for the Laguerre polynomials (see footnote 22) the factor (n + )! under the square root will be cubed. 4.2 The Hydrogen Atom 151 Table 4.7: The \ufb01rst few radial wave functions for hydrogen, Rn(r ). R10 = 2a\u22123/2 exp(\u2212r/a) R20 = 1 \u221a 2 a\u22123/2 1 \u2212 1 2 r a exp(\u2212r/2a) R21 = 1 \u221a 2 6 a\u22123/2 r a exp(\u2212r/2a) R30 = 2 \u221a 3 3 a\u22123/2 1 \u2212 2 3 r a R31 = 8 \u221a 27 6 a\u22123/ 27 r a exp(\u2212r/3a) 2 exp(\u2212r/3a) R32 = 4 \u221a 81 30 a\u22123/2 2 r a exp(\u2212r/3a) a\u22123/2 1 \u2212 R40 = 192 R41 = 5 \u221a 16 15 a\u22123/2 1 \u2212 1 4 r a R42 = 1 \u221a 64 5 a\u22123/2 1 \u2212 1 12 r a + 1 80 exp(\u2212r/4a) 3 exp(\u2212r/4a) exp(\u2212r/4a) R43 = 1 \u221a 768 35 a\u22123/2 3 r a exp(\u2212r/4a) / \u03c8nm = 3 (n \u2212 \u2212 1)! 2n (n + )! 2 na e\u2212r/na # 2r na L 2+1 n\u2212\u22121 $ (2r/na) Y m (\u03b8, \u03c6). (4.89) They are not pretty, but don\u2019t complain\u2014this is one of the very few realistic systems that can be solved at all, in exact closed form. The wave functions are mutually orthogonal: \u03c8 \u2217 nm \u03c8nm r 2 dr d = \u03b4nn\u03b4\u03b4mm. (4.90) This follows from the orthogonality of the spherical harmonics (Equation 4.33) and (for n = n) from the fact that they are eigenfunctions of \u02c6H with distinct eigenvalues. Visualizing the hydrogen wave functions is not easy. Chemists like to draw density plots, in which the brightness of the cloud is proportional to |\u03c8|2 (Figure 4.8). More quantitative (but perhaps harder to decipher) are surfaces of constant probability density (Figure 4.9). The quantum numbers n,, and m can be identi\ufb01ed from the nodes of the wave function. The number of radial nodes is, as always, given by N \u2212 1 (for hydrogen this is n \u2212 \u2212 1). For each radial node the wave function vanishes on a sphere, as can be seen in Figure 4.8. The quantum 152 CHAPTER 4 Quantum Mechanics in Three Dimensions ) r ( n R 2 / 3 a 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 10 20 30 31 10 21 21 32", " 32 0 2 4 30 20 6 8 10 12 14 30 32 16 31 r/a 18 Figure 4.7: Graphs of the \ufb01rst few hydrogen radial wave functions, Rn(r ). (2,0,0) (3,1,0) (4,0,0) (4,1,0) (4,2,0) (4,3,0) Figure 4.8: Density plots for the \ufb01rst few hydrogen wave functions, labeled by (n,, m). Printed by permission using \u201cAtom in a Box\u201d by Dauger Research. You can make your own plots by going to: http://dauger.com. |\u03c8 |2 200 |\u03c8 |2 300 |\u03c8 |2 210 |\u03c8 |2 310 |\u03c8 |2 211 |\u03c8 |2 320 4.2 The Hydrogen Atom 153 |\u03c8 |2 322 |\u03c8 |2 311 |\u03c8 |2 321 Figure 4.9: Shaded regions indicate signi\ufb01cant electron density (|\u03c8|2 > 0.25/nm3) for the \ufb01rst few hydrogen wave functions. The region 0 < \u03c6 < \u03c0/2 has been cut away; |\u03c8|2 has azimuthal symmetry in all cases. number m counts the number of nodes of the real (or imaginary) part of the wave function in the \u03c6 direction. These nodes are planes containing the z axis on which the real or imaginary part of \u03c8 vanishes.24 Finally, \u2212 m gives the number of nodes in the \u03b8 direction. These are cones about the z axis on which \u03c8 vanishes (note that a cone with opening angle \u03c0/2 is the x \u2212 y plane itself). \u2217 Problem 4.12 Work out the radial wave functions R30, R31, and R32, using the recursion formula (Equation 4.76). Don\u2019t bother to normalize them. 24 These planes aren\u2019t visible in Figure 4.8 or 4.9, since these \ufb01gures show the absolute value of \u03c8, and the real and imaginary parts of the wave function vanish on different sets of planes. However, since both sets contain the z axis, the wave function itself must vanish on the z axis for m = 0 (see Figure 4.9). 154 CHAPTER 4 Quantum Mechanics in Three Dimensions \u2217 \u2217 Problem 4.13 (a) Normalize R20 (Equation 4.82), and construct the function \u03c8200. (b) Normalize R21 (Equation 4.83), and construct \u03c8211, \u03c8210, and \u03c821\u22121. Problem 4.14 (a) Using Equation 4.88, work out the \ufb01rst four Laguerre polynomials. (b) Using Equations 4.86, 4.87, and 4.88, \ufb01nd v(\u03c1), for the case n = 5, = 2. (c) Find v(\u03c1) again (for the case n = 5, = 2), but this time get it from the recursion formula (Equation 4.76). \u2217 Problem 4.15 (a) Find r and (b) Find x and r 2 x 2 answers in terms of the Bohr radius. for an electron in the ground state of hydrogen. Express your for an electron in the ground state of hydrogen. Hint: This requires no new integration\u2014note that r 2 = x 2 + y2 + z2, and exploit the symmetry of the ground state. x 2 in the state n = 2, = 1, m = 1. Hint: this state is not symmetrical in (c) Find x, y, z. Use x = r sin \u03b8 cos \u03c6. Problem 4.16 What is the most probable value of r, in the ground state of hydrogen? (The answer is not zero!) Hint: First you must \ufb01gure out the probability that the electron would be found between r and r + dr. 0 1 Problem 4.17 Calculate, in the ground state of hydrogen. Hint: This takes two pages and six integrals, or four lines and no integrals, depending on how you set it up. To do it the quick way, start by noting that [z, [H, z]] = 2z H z \u2212 H z2 \u2212 z2 H.25 z \u02c6H z Problem 4.18 A hydrogen atom starts out in the following linear combination of the stationary states n = 2, = 1, m = 1 and n = 2, = 1, m = \u22121: (r, 0) = 1\u221a 2 (\u03c82 1 1 + \u03c82 1 \u22121). (a)", " Construct (r, t). Simplify it as much as you can. (b) Find the expectation value of the potential energy, V. (Does it depend on t?) Give both the formula and the actual number, in electron volts. 25 The idea is to reorder the operators in such a way that \u02c6H appears either to the left or to the right, because we know (of course) what \u02c6H \u03c8100 is. 4.2.2 The Spectrum of Hydrogen 4.2 The Hydrogen Atom 155 In principle, if you put a hydrogen atom into some stationary state nm, it should stay there forever. However, if you tickle it slightly (by collision with another atom, say, or by shining light on it), the atom may undergo a transition to some other stationary state\u2014either by absorbing energy, and moving up to a higher-energy state, or by giving off energy (typically in the form of electromagnetic radiation), and moving down.26 In practice such perturbations are always present; transitions (or, as they are sometimes called, quantum jumps) are constantly occurring, and the result is that a container of hydrogen gives off light (photons), whose energy corresponds to the difference in energy between the initial and \ufb01nal states:! \" E\u03b3 = Ei \u2212 E f = \u221213.6 eV 1 n2 i \u2212 1 n2 f. (4.91) Now, according to the Planck formula,27 the energy of a photon is proportional to its frequency: Meanwhile, the wavelength is given by \u03bb = c/\u03bd, so! E\u03b3 = h\u03bd. 1 \u03bb = R 1 n2 f \u2212 1 n2 i \", where R \u2261 me 4\u03c0 c3 2 e2 4\u03c0 0 = 1.097 \u00d7 107 m \u22121 (4.92) (4.93) (4.94) is known as the Rydberg constant. Equation 4.93 is the Rydberg formula for the spectrum of hydrogen; it was discovered empirically in the nineteenth century, and the greatest triumph of Bohr\u2019s theory was its ability to account for this result\u2014and to calculate R in terms of the lie in the ultraviolet; fundamental constants of nature. Transitions to the ground state they are known to spectroscopists as the Lyman series. Transitions to the \ufb01rst excited state n f = 2 fall in the visible region; they constitute the Balmer series. Transitions to n f = 3 (the Paschen series) are in the infrared; and so on (see Figure 4.10). (At room temperature, most hydrogen atoms are in the ground state; to obtain the emission spectrum you must \ufb01rst populate the various excited states; typically this is done by passing an electric spark through the gas.) n f = 1 \u2217 Problem 4.19 A hydrogenic atom consists of a single electron orbiting a nucleus with Z protons. (Z = 1 would be hydrogen itself, Z = 2 is ionized helium, Z = 3 is doubly ionized lithium, and so on.) Determine the Bohr energies En(Z ), the binding energy E1(Z ), the Bohr radius a(Z ), and the Rydberg constant R(Z ) for a hydrogenic atom. (Express 26 By its nature, this involves a time-dependent potential, and the details will have to await Chapter 11; for our present purposes the actual mechanism involved is immaterial. 27 The photon is a quantum of electromagnetic radiation; it\u2019s a relativistic object if there ever was one, and therefore outside the scope of nonrelativistic quantum mechanics. It will be useful in a few places to speak of photons, and to invoke the Planck formula for their energy, but please bear in mind that this is external to the theory we are developing. 156 CHAPTER 4 Quantum Mechanics in Three Dimensions ) 1.0 \u22122.0 \u22123.0 \u22124.0 \u22125.0 \u22126.0 \u22127.0 \u22128.0 \u22129.0 \u221210.0 \u221211.0 \u221212.0 \u221213.0 \u221214.0 Lyman series Paschen series Balmer series n\u221e 6 5 4 3 2 1 Figure 4.10: Energy levels and transitions in the spectrum of hydrogen. your answers as appropriate multiples of the hydrogen values.) Where in the electromagnetic spectrum would the Lyman series fall, for Z = 2 and Z = 3? Hint: There\u2019s nothing much to calculate here\u2014in the potential (Equation 4.52) e2 \u2192 Z e2, so all you have to do is make the same substitution in all the \ufb01nal results. Problem 4.20 Consider the earth\u2013sun system as a gravitational analog to the hydrogen atom. (a) What is the potential energy function (replacing Equation 4", ".52)? (Let m E be the mass of the earth, and M the mass of the sun.) (b) What is the \u201cBohr radius,\u201d ag, for this system? Work out the actual number. (c) Write down the gravitational \u201cBohr formula,\u201d and, by equating En to the classical ro/ag. From this, energy of a planet in a circular orbit of radius ro, show that n = estimate the quantum number n of the earth. % (d) Suppose the earth made a transition to the next lower level (n \u2212 1). How much energy (in Joules) would be released? What would the wavelength of the emitted photon (or, more likely, graviton) be? (Express your answer in light years\u2014is the remarkable answer28 a coincidence?) 28 Thanks to John Meyer for pointing this out. 4.3 Angular Momentum 157 4.3 ANGULAR MOMENTUM As we have seen, the stationary states of the hydrogen atom are labeled by three quantum numbers: n,, and m. The principal quantum number (n) determines the energy of the state (Equation 4.70); and m are related to the orbital angular momentum. In the classical theory of central forces, energy and angular momentum are the fundamental conserved quantities, and it is not surprising that angular momentum plays an important) role in the quantum theory. Classically, the angular momentum of a particle (with respect to the origin) is given by the formula which is to say,29 L = r \u00d7 p, (4.95) L x = ypz \u2212 zpy, L y = zpx \u2212 x pz, L z = x py \u2212 ypx. (4.96) The corresponding quantum operators30 are obtained by the standard prescription px \u2192 \u2212i\u2202/\u2202 x, py \u2192 \u2212i\u2202/\u2202 y, pz \u2192 \u2212i\u2202/\u2202z. In this section we\u2019ll obtain the eigenvalues of the the angular momentum operators by a purely algebraic technique reminiscent of the one we used in Chapter 2 to get the allowed energies of the harmonic oscillator; it is all based on the clever exploitation of commutation relations. After that we will turn to the more dif\ufb01cult problem of determining the eigenfunctions. 4.3.1 Eigenvalues The operators L x and L y do not commute; in fact L x, L y = = ypz \u2212 zpy, zpx \u2212 x pz \u2212 ypz, x pz ypz, zpx \u2212 zpy, zpx + zpy, x pz. (4.97) From the canonical commutation relations (Equation 4.10) we know that the only operators here that fail to commute are x with px, y with py, and z with pz. So the two middle terms drop out, leaving L x, L y = ypx pz, z + x py z, pz = i x py \u2212 ypx = iL z. (4.98) L y, L z, but there is no need to calculate Of course, we could have started out with these separately\u2014we can get them immediately by cyclic permutation of the indices (x \u2192 y, y \u2192 z, z \u2192 x): L z, L x or L x, L y = iL z; L y, L z = iL x ; L z, L x = iL y. (4.99) These are the fundamental commutation relations for angular momentum; everything follows from them. 29 Because angular momentum involves the product of position and momentum, you might worry that the ambiguity addressed in Chapter 3 (footnote 15, page 102) would arise. Fortunately, only different components of r and p are multiplied, and they commute (Equation 4.10). 30 To reduce clutter (and avoid confusion with the unit vectors \u02c6\u0131, \u02c6j, \u02c6k, \u02c6r, \u02c6\u03b8, \u02c6\u03c6) I\u2019m going to take the hats off operators for the rest of the chapter. 158 CHAPTER 4 Quantum Mechanics in Three Dimensions Notice that L x, L y, and L z are incompatible observables. According to the generalized uncertainty principle (Equation 3.62), 2i 2 iL z = 2 4 L z2, or \u03c3L x \u03c3L y \u2265 2 |L z|. (4.100) It would therefore be futile to look for states that are simultaneously eigenfunctions of L x and L y. On the other hand, the square of the total angular momentum4.101) does commute with (\u2212iL z) + (\u2212iL z) L y + L z = 0. L z, L x iL y + + iL y L z, L", " x L z L z (I used Equation 3.65 to reduce the commutators; of course, any operator commutes with itself.) It follows that L 2 also commutes with L y and L z: L 2, L x L 2, L y L 2, L z (4.102) = 0, = 0, = 0, or, more compactly, L 2, L = 0. (4.103) So L 2 is compatible with each component of L, and we can hope to \ufb01nd simultaneous eigenstates of L 2 and (say) L z: L 2 f = \u03bb f and L z f = \u03bcf. (4.104) We\u2019ll use a ladder operator technique, very similar to the one we applied to the harmonic oscillator back in Section 2.3.1. Let L \u00b1 \u2261 L x \u00b1 i L y. (4.105) Its commutator with L z is L z, L \u00b1 = L z, L x \u00b1 i L z, L y = iL y \u00b1 i (\u2212iL, so Also (from Equation 4.102) L z, L \u00b1 = \u00b1L \u00b1. L 2, L \u00b1 = 0. (4.106) (4.107) I claim that if f is an eigenfunction of L 2 and L z, so also is L \u00b1 f : Equation 4.107 says L 2 f = L \u00b1 (\u03bb f ) = \u03bb (L \u00b1 f ), L 2 (L \u00b1 f ) = L \u00b1 (4.108) so L \u00b1 f is an eigenfunction of L 2, with the same eigenvalue \u03bb, and Equation 4.106 says L z (L \u00b1 f ) = (L z L \u00b1 \u2212 L \u00b1 L zL \u00b1 f + L \u00b1 (\u03bcf ) = (\u03bc \u00b1 ) (L \u00b1 f ), (4.109) 4.3 Angular Momentum 159 h ft \u03bc + 3 h L3 + f L+ \u03bc + 2 h L2 + f \u03bc + h L L_ f L2_ f L3_ f L_ h fb Figure 4.11: The \u201cladder\u201d of angular momentum states. is an eigenfunction of L z with the new eigenvalue \u03bc \u00b1. We call L + the raising so L \u00b1 f operator, because it increases the eigenvalue of L z by, and L \u2212 the lowering operator, because it lowers the eigenvalue by. For a given value of \u03bb, then, we obtain a \u201cladder\u201d of states, with each \u201crung\u201d separated from its neighbors by one unit of in the eigenvalue of L z (see Figure 4.11). To ascend the ladder we apply the raising operator, and to descend, the lowering operator. But this process cannot go on forever: Eventually we\u2019re going to reach a state for which the z-component exceeds the total, and that cannot be.31 There must exist a \u201ctop rung\u201d, ft, such that32 L + ft = 0. (4.110) Let be the eigenvalue of L z at the top rung (the appropriateness of the letter \u201c\u201d will appear in a moment): L z ft = ft ; L 2 ft = \u03bb ft. (4.111) + L2 z, but L2 x = f |L2 x f = L x f |L x f \u2265 0 (and likewise for L y ), so 32 Actually, all we can conclude is that L+ ft is not normalizable\u2014its norm could be in\ufb01nite, instead of zero. Problem + L2 31 Formally, L2 = L2 x y + \u03bc2 \u2265 \u03bc2. \u03bb = L2 x + L2 y 4.21 explores this alternative. 160 CHAPTER 4 Quantum Mechanics in Three Dimensions NowiL z), or, putting it the other way around. (4.112) It follows that L 2 ft = ft = 0 + 22 + 2 ft = 2 ( + 1) ft, and hence \u03bb = 2 ( + 1). This tells us the eigenvalue of L 2 in terms of the maximum eigenvalue of L z. Meanwhile, there is also (for the same reason) a bottom rung, fb, such that L \u2212 fb = 0. Let \u00af be the eigenvalue of L z at this bottom rung: L z fb = \u00af fb; L 2 fb = \u03bb fb. (4.113) (4.114) (4.115) Using Equation 4.112, we have L 2 fb = fb = 0 + 2 \u00af2 \u2212 2 \u00af fb = 2 \u00af( \u00af \u2212 1) fb, and therefore 4.116), so", " either \u00af = + 1 Comparing Equations 4.113 and 4.116, we see that ( + 1) = \u00af (which is absurd\u2014the bottom rung would be higher than the top rung!) or else \u00af = \u2212. (4.117) So the eigenvalues of L z are m, where m (the appropriateness of this letter will also be clear in a moment) goes from \u2212 to +, in N integer steps. In particular, it follows that = \u2212 + N, and hence = N /2, so must be an integer or a half-integer. The eigenfunctions are characterized by the numbers and m4.118) where = 0, 1/2, 1, 3/2,... ; m = \u2212, \u2212 + 1,..., \u2212 1,. (4.119) For a given value of, there are 2+1 different values of m (i.e. 2+1 \u201crungs\u201d on the \u201cladder\u201d). Some people like to illustrate this with the diagram in Figure 4.12 (drawn for the case = 2). The arrows are supposed to represent possible angular momenta (in units of )\u2014they 6 = 2.45), and their z components are the all have the same length allowed values of m (\u22122, \u22121, 0, 1, 2). Notice that the magnitude of the vectors (the radius of ( + 1) >, except the sphere) is greater than the maximum z component! (In general, for the \u201ctrivial\u201d case = 0.) Evidently you can\u2019t get the angular momentum to point perfectly ( + 1) (in this case \u221a \u221a \u221a 4.3 Angular Momentum 161 Lz L 2 1 0 \u22121 \u22122 Ly Lx Figure 4.12: Angular momentum states (for = 2). along the z direction. At \ufb01rst, this sounds absurd. \u201cWhy can\u2019t I just pick my axes so that z points along the direction of the angular momentum vector?\u201d Well, to do that you would have to know all three components simultaneously, and the uncertainty principle (Equation 4.100) says that\u2019s impossible. \u201cWell, all right, but surely once in a while, by good fortune, I will just happen to aim my z axis along the direction of L.\u201d No, no! You have missed the point. It\u2019s not merely that you don\u2019t know all three components of L; there just aren\u2019t three components\u2014a particle simply cannot have a determinate angular momentum vector, any more than it can simultaneously have a determinate position and momentum. If L z has a well-de\ufb01ned value, then L x and L y do not. It is misleading even to draw the vectors in Figure 4.12\u2014at best they should be smeared out around the latitude lines, to indicate that L x and L y are indeterminate. I hope you\u2019re impressed: By purely algebraic means, starting with the fundamental commutation relations for angular momentum (Equation 4.99), we have determined the eigenvalues of L 2 and L z\u2014without ever seeing the eigenfunctions themselves! We turn now to the problem of constructing the eigenfunctions, but I should warn you that this is a much messier business. Just so you know where we\u2019re headed, I\u2019ll let you in on the punch line: f m \u2014the eigenfunctions of L 2 and L z are nothing but the old spherical harmonics, which we came upon by a quite different route in Section 4.1.2 (that\u2019s why I chose the same letters and m, of course). And I can now explain why the spherical harmonics are orthogonal: They are eigenfunctions of L 2 and L z) belonging to distinct eigenvalues (Theorem 2, Section 3.3.1). hermitian operators = Y m \u2217 Problem 4.21 The raising and lowering operators change the value of m by one unit: f m+1 f m\u22121 L \u2212 f m L + f m = = Bm Am, (4.120) and Bm where Am are constants. Question: What are they, if the eigenfunctions are to be normalized? Hint: First show that L \u2213 is the hermitian conjugate of L \u00b1 (since L x and L y are observables, you may assume they are hermitian... but prove it if you like); then use Equation 4.112. Answer: 162 CHAPTER 4 Quantum Mechanics in Three Dimensions % % Am = Bm = ( + 1) \u2212 m (m + 1) = ( + 1) \u2212 m (m \u2212 1) = % %", " ( \u2212 m) ( + m + 1), ( + m) ( \u2212 m + 1). (4.121) Note what happens at the top and bottom of the ladder (i.e. when you apply L + to f L \u2212 to f ). or \u2212 \u2217 Problem 4.22 (a) Starting with the canonical commutation relations for position and momentum (Equation 4.10), work out the following commutators: = \u2212ix, = iy, = i py, = \u2212i px, L z, x L z, px = 0 = 0. L z, z L z, pz L z, y L z, py = iL y directly from Equation 4.96. L z, p2 (where, of course, r 2 = x 2 + y2 + z2 (4.122) (b) Use these results to obtain (c) Find the commutators L z, L x and and p2 = p2 x + p2 y (d) Show that the Hamiltonian H = + p2 L z, r 2 z ). + V commutes with all three components of L, provided that V depends only on r. (Thus H, L 2, and L z are mutually compatible observables.) p2/2m \u2217\u2217 Problem 4.23 (a) Prove that for a particle in a potential V(r) the rate of change of the expectation value of the orbital angular momentum L is equal to the expectation value of the torque: where d dt L = N, N = r \u00d7 (\u2212\u2207V ). (This is the rotational analog to Ehrenfest\u2019s theorem.) (b) Show that dL/dt = 0 for any spherically symmetric potential. (This is one form of the quantum statement of conservation of angular momentum.) 4.3.2 Eigenfunctions First of all we need to rewrite L x, L y, and L z in spherical coordinates. Now, L = \u2212i (r \u00d7 \u2207), and the gradient, in spherical coordinates, is:33 \u2207 = \u02c6r \u2202 \u2202r + \u02c6\u03b8 1 r \u2202 \u2202\u03b8 + \u02c6\u03c6 1 r sin \u03b8 \u2202 \u2202\u03c6 ; (4.123) 33 George Arfken and Hans-Jurgen Weber, Mathematical Methods for Physicists, 7th edn, Academic Press, Orlando (2013), Section 3.10. 4.3 Angular Momentum 163 \u2202 \u2202\u03b8 \u02c6r \u00d7 \u02c6\u03c6 + 1 sin \u03b8 \u2202 \u2202\u03c6. meanwhile, r = r \u02c6r, so \u02c6r \u00d7 \u02c6r But L = \u2212i \u02c6r \u00d7 \u02c6\u03b8 = 0, \u02c6r \u00d7 \u02c6r r \u02c6r \u00d7 \u02c6\u03b8 + \u2202 \u2202r \u02c6r \u00d7 \u02c6\u03c6 = \u02c6\u03c6, and = \u2212 \u02c6\u03b8 (see Figure 4.1), and hence L = \u2212i \u02c6\u03c6 \u2202 \u2202\u03b8 \u2212 \u02c6\u03b8 1 sin \u03b8 \u2202 \u2202\u03c6. The unit vectors \u02c6\u03b8 and \u02c6\u03c6 can be resolved into their cartesian components: \u02c6\u03b8 = (cos \u03b8 cos \u03c6) \u02c6\u0131 + (cos \u03b8 sin \u03c6) \u02c6j \u2212 (sin \u03b8 ) \u02c6k; \u02c6\u03c6 = \u2212 (sin \u03c6) \u02c6\u0131 + (cos \u03c6) \u02c6j. Thus So and \u2212 sin \u03c6 \u02c6\u0131 + cos \u03c6 \u02c6j \u2202 \u2202\u03b8 L = \u2212i \u2212 cos \u03b8 cos \u03c6 \u02c6\u0131 + cos \u03b8 sin \u03c6 \u02c6j \u2212 sin \u03b8 \u02c6k 1 sin \u03b8 \u2202 \u2202\u03c6. L x = \u2212i \u2212 sin \u03c6 L y = \u2212i + cos \u03c6 \u2202 \u2202\u03b8 \u2202 \u2202\u03b8 \u2212 cos \u03c6 cot \u03b8 \u2212 sin \u03c6 cot \u03b8,, \u2202 \u2202\u03c6 \u2202 \u2202\u03c6 L z = \u2212i \u2202 \u2202\u03c6. We shall also need the raising and lowering operatorsi (\u2212 sin \u03c6 \u00b1 i cos \u03c6) \u2202 \u2202\u03b8 \u2212 (cos \u03c6 \u00b1 i sin \u03c6) cot \u03b8 But cos \u03c6 \u00b1 i sin \u03c6 = e\u00b1i\u03c6, so L \u00b1 = \u00b1e\u00b1i\u03c6 \u2202 \u2202\u03b8 \u00b1 i cot \u03b8. \u2202 \u2202\u03c6 In particular (Problem 4.24(a)):", " L + L \u2212 = \u22122 and hence (Problem 4.24(b)): \u2202 2 \u2202\u03b8 2 + cot \u03b8 \u2202 \u2202\u03b8 + cot2 \u03b8 \u2202 2 \u2202\u03c62 + i \u2202 \u2202\u03c6, L 2 = \u22122 1 sin \u03b8 \u2202 \u2202\u03b8 sin \u03b8 \u2202 \u2202\u03b8 + 1 sin2 \u03b8 \u2202 2 \u2202\u03c62. (4.124) (4.125) (4.126) (4.127) (4.128) (4.129) (4.130) (4.131) (4.132). \u2202 \u2202\u03c6 We are now in a position to determine f m (\u03b8, \u03c6). It\u2019s an eigenfunction of L 2, with eigenvalue 2 ( + 1): 164 CHAPTER 4 Quantum Mechanics in Three Dimensions L 2 f m = \u22122 1 sin \u03b8 \u2202 \u2202\u03b8 sin \u03b8 \u2202 \u2202\u03b8 + 1 sin2 \u03b8 \u2202 2 \u2202\u03c62 = 2 ( + 1) f m f m. But this is precisely the \u201cangular equation\u201d (Equation 4.18). And it\u2019s also an eigenfunction of L z, with the eigenvalue m: L z f m = \u2212i \u2202 \u2202\u03c6 = m f m f m, but this is equivalent to the azimuthal equation (Equation 4.21). We have already solved this system of equations! The result (appropriately normalized) is the spherical harmonic, (\u03b8, \u03c6). Conclusion: Spherical harmonics are the eigenfunctions of L 2 and L z. When we Y m solved the Schr\u00f6dinger equation by separation of variables, in Section 4.1, we were inadvertently constructing simultaneous eigenfunctions of the three commuting operators H, L 2, and L z: H \u03c8 = E\u03c8, L 2\u03c8 = 2 ( + 1) \u03c8, L z\u03c8 = m\u03c8. (4.133) Incidentally, we can use Equation 4.132 to rewrite the Schr\u00f6dinger equation (Equation 4.14) more compactly: \u22122 1 2mr 2 \u2202 \u2202r r 2 \u2202 \u2202r + L 2 \u03c8 + V \u03c8 = E\u03c8. There is a curious \ufb01nal twist to this story: the algebraic theory of angular momentum permits (and hence also m) to take on half -integer values (Equation 4.119), whereas separation of variables yielded eigenfunctions only for integer values (Equation 4.29).34 You might suppose that the half-integer solutions are spurious, but it turns out that they are of profound importance, as we shall see in the following sections. Problem 4.24 (a) Derive Equation 4.131 from Equation 4.130. Hint: Use a test function; otherwise you\u2019re likely to drop some terms. (b) Derive Equation 4.132 from Equations 4.129 and 4.131. Hint: Use Equation 4.112. Problem 4.25 (a) What is L +Y (b) Use the result of (a), together with Equation 4.130 and the fact that L zY (\u03b8, \u03c6), up to a normalization constant.? (No calculation allowed!) to determine Y = Y, (c) Determine the normalization constant by direct integration. Compare your \ufb01nal answer to what you got in Problem 4.7. \u2217 \u2217 Problem 4.26 In Problem 4.4 you showed that % (\u03b8, \u03c6) = \u2212 Y 1 2 15/8\u03c0 sin \u03b8 cos \u03b8 ei\u03c6. Apply the raising operator to \ufb01nd Y 2 2 (\u03b8, \u03c6). Use Equation 4.121 to get the normalization. 34 For an interesting discussion, see I. R. Gatland, Am. J. Phys. 74, 191 (2006). 4.4 Spin 165 \u2217\u2217 Problem 4.27 Two particles (masses m1 and m2) are attached to the ends of a massless rigid rod of length a. The system is free to rotate in three dimensions about the (\ufb01xed) center of mass. (a) Show that the allowed energies of this rigid rotor are En = 2 2I n (n + 1), (n = 0, 1, 2,...), where I = m1m2 (m1 + m2) a2 is the moment of inertia of the system. Hint: First express the (classical) energy in terms of the angular momentum. (b) What are the normalized", " eigenfunctions for this system? (Let \u03b8 and \u03c6 de\ufb01ne the orientation of the rotor axis.) What is the degeneracy of the nth energy level? (c) What spectrum would you expect for this system? (Give a formula for the frequencies of the spectral lines.) Answer: \u03bd j = j/2\u03c0 I, j = 1, 2, 3,.... (d) Figure 4.13 shows a portion of the rotational spectrum of carbon monoxide (CO). What is the frequency separation (\u03bd) between adjacent lines? Look up the masses of 12C and 16O, and from m1, m2, and \u03bd determine the distance between the atoms. 10 30 50 \u03c5 (cm\u20131) 70 Figure 4.13: Rotation spectrum of CO. Note that the frequencies are in spectroscopist\u2019s units: inverse centimeters. To convert to Hertz, multiply by c = 3.00 \u00d7 1010 cm/s. Reproduced by permission from John M. Brown and Allan Carrington, Rotational Spectroscopy of Diatomic Molecules, Cambridge University Press, 2003, which in turn was adapted from E. V. Loewenstein, Journal of the Optical Society of America, 50, 1163 (1960). 4.4 SPIN In classical mechanics, a rigid object admits two kinds of angular momentum: orbital (L = r\u00d7p), associated with motion of the center of mass, and spin (S = I \u03c9), associated with motion about the center of mass. For example, the earth has orbital angular momentum attributable to its annual revolution around the sun, and spin angular momentum coming from its daily rotation about the north\u2013south axis. In the classical context this distinction is largely a matter of convenience, for when you come right down to it, S is nothing but the sum total of the 166 CHAPTER 4 Quantum Mechanics in Three Dimensions \u201corbital\u201d angular momenta of all the rocks and dirt clods that go to make up the earth, as they circle around the axis. But a similar thing happens in quantum mechanics, and here the distinction is absolutely fundamental. In addition to orbital angular momentum, associated (in the case of hydrogen) with the motion of the electron around the nucleus (and described by the spherical harmonics), the electron also carries another form of angular momentum, which has nothing to do with motion in space (and which is not, therefore, described by any function of the position variables r, \u03b8, \u03c6) but which is somewhat analogous to classical spin (and for which, therefore, we use the same word). It doesn\u2019t pay to press this analogy too far: The electron (as far as we know) is a structureless point, and its spin angular momentum cannot be decomposed into orbital angular momenta of constituent parts (see Problem 4.28).35 Suf\ufb01ce it to say that elementary particles carry intrinsic angular momentum (S) in addition to their \u201cextrinsic\u201d angular momentum (L). The algebraic theory of spin is a carbon copy of the theory of orbital angular momentum, beginning with the fundamental commutation relations:36 Sx, Sy = iSz, Sy, Sz = iSx, Sz, Sx = iSy. It follows (as before) that the eigenvectors of S2 and Sz satisfy37 S2|s m = 2s (s + 1) |s m Sz|s m = m|s m ; ; and % S\u00b1 |s m = s (s + 1) \u2212 m (m \u00b1 1) |s (m \u00b1 1), (4.134) (4.135) (4.136) where S\u00b1 \u2261 Sx \u00b1 i Sy. But this time the eigenvectors are not spherical harmonics (they\u2019re not functions of \u03b8 and \u03c6 at all), and there is no reason to exclude the half-integer values of s and m: s = 0, 1 2, 1s, \u2212s + 1,..., s \u2212 1, s. (4.137) It so happens that every elementary particle has a speci\ufb01c and immutable value of s, which we call the spin of that particular species: \u03c0 mesons have spin 0; electrons have spin 1/2; photons have spin 1; baryons have spin 3/2; gravitons have spin 2; and so on. By contrast, the orbital angular momentum quantum number l (for an electron in a hydrogen atom, say) can take on any (integer) value you please, and will change from one to another when the system is perturbed. But s is \ufb01xed, for any given particle, and this makes the theory of spin comparatively simple.38 35 For a contrary interpretation, see Hans C. Ohanian, \u201cWhat is Spin?\ufffd", "\ufffd, Am. J. Phys. 54, 500 (1986). 36 We shall take these as postulates for the theory of spin; the analogous formulas for orbital angular momentum (Equation 4.99) were derived from the known form of the operators (Equation 4.96). Actually, they both follow from rotational invariance in three dimensions, as we shall see in Chapter 6. Indeed, these fundamental commutation relations apply to all forms of angular momentum, whether spin, orbital, or the combined angular momentum of a composite system, which could be partly spin and partly orbital. 37 Because the eigenstates of spin are not functions, I will switch now to Dirac notation. By the way, I\u2019m running out of letters, so I\u2019ll use m for the eigenvalue of Sz, just as I did for L z (some authors write ml and ms at this stage, just to be absolutely clear). 38 Indeed, in a mathematical sense, spin 1/2 is the simplest possible nontrivial quantum system, for it admits just two basis states (recall Example 3.8). In place of an in\ufb01nite-dimensional Hilbert space, with all its subtleties and complications, we \ufb01nd ourselves working in an ordinary two-dimensional vector space; instead of unfamiliar differential equations and fancy functions, we are confronted with 2 \u00d7 2 matrices and two-component vectors. For this reason, some authors begin quantum mechanics with the study of spin. (An outstanding example is John 4.4 Spin 167 Problem 4.28 If the electron were a classical solid sphere, with radius rc = e2 4\u03c0 0mc2, (4.138) (the so-called classical electron radius, obtained by assuming the electron\u2019s mass is attributable to energy stored in its electric \ufb01eld, via the Einstein formula E = mc2), and its angular momentum is (1/2), then how fast (in m/s) would a point on the \u201cequator\u201d be moving? Does this model make sense? (Actually, the radius of the electron is known experimentally to be much less than rc, but this only makes matters worse.)39 4.4.1 Spin 1/2 By far the most important case is s = 1/2, for this is the spin of the particles that make up ordinary matter (protons, neutrons, and electrons), as well as all quarks and all leptons. Moreover, once you understand spin 1/2, it is a simple matter to work out the formalism for, which we call spin up (informally, \u2191), any higher spin. There are just two eigenstates: 1 1 2 1 2 and particle can be represented by a two-element column matrix (or spinor):, spin down (\u2193). Using these as basis vectors, the general state40 of a spin-1/2 \u2212 1 2 1 1 2 \u03c7 = with representing spin up, and a b = a\u03c7+ + b\u03c7\u2212, \u03c7+ = \u03c7\u2212 = 1 0 0 1 (4.139) (4.140) (4.141) for spin down. With respect to this basis the spin operators become 2 \u00d7 2 matrices,41 which we can work out by noting their effect on \u03c7+ and \u03c7\u2212. Equation 4.135 says S. Townsend, A Modern Approach to Quantum Mechanics, 2nd edn, University Books, Sausalito, CA, 2012.) But the price of mathematical simplicity is conceptual abstraction, and I prefer not to do it that way. 39 If it comforts you to picture the electron as a tiny spinning sphere, go ahead; I do, and I don\u2019t think it hurts, as long as you don\u2019t take it literally. 40 I\u2019m only talking about the spin state, for the moment. If the particle is moving around, we will also need to deal with its position state (), but for the moment let\u2019s put that aside. 41 I hate to be fussy about notation, but perhaps I should reiterate that a ket (such as |s m) is a vector in Hilbert space (in this case a (2s + 1)-dimensional vector space), whereas a spinor \u03c7 is a set of components of a vector, 1 1, in the case of spin 1, displayed as a column. Physicists with respect to a particular basis 2 2 1 = \u03c7+, but technically this confuses a vector (which lives \u201cout there\u201d in sometimes write, for instance, 2 Hilbert space) with its components (a string of numbers). Similarly, Sz (for example) is an operator that acts on kets; it is represented (with respect to the chosen basis) by a matrix Sz (sans serif), which multiplies spinors\u2014but again, Sz = Sz, though perfectly", " intelligible, is sloppy language and 1 168 CHAPTER 4 Quantum Mechanics in Three Dimensions S2\u03c7+ = 3 4 2\u03c7+ and S2\u03c7\u2212 = 3 4 2\u03c7\u2212. (4.142) If we write S2 as a matrix with (as yet) undetermined elements, S2 = c e, d f then the \ufb01rst equation says so c = (3/4)2 and e = 0. The second equation says or, or = d f, 0 2 3 4 so d = 0 and f = (3/4) 2. Conclusion: S2 = 3 4 2 1 0 0 1. Similarly, from which it follows that Sz\u03c7+ = 2 \u03c7+, Sz\u03c7\u2212 = \u2212 2 \u03c7\u2212, Sz = 1 0 0 \u22121. 2 Meanwhile, Equation 4.136 says S+\u03c7\u2212 = \u03c7+, S\u2212\u03c7+ = \u03c7\u2212, S+\u03c7+ = S\u2212\u03c7\u2212 = 0, so S+ = 0 1 0 0, S\u2212 = 0 1 0 0. (4.143) (4.144) (4.145) (4.146) Now S\u00b1 = Sx \u00b1 i Sy, so Sx = (1/2) (S+ + S\u2212) and Sy = (1/2i) (S+ \u2212 S\u2212), and hence 1 0 0 \u2212i i 0 0 1 Sx = Sy =,. 2 2 (4.147) Since Sx, Sy, and Sz all carry a factor of /2, it is tidier to write S = (/2)\u03c3, where 0 1 1 0 \u03c3x \u2261, \u03c3y \u2261 0 \u2212i i 0, \u03c3z \u2261 1 0 0 \u22121. (4.148) These are the famous Pauli spin matrices. Notice that Sx, Sy, Sz, and S2 are all hermitian matrices (as they should be, since they represent observables). On the other hand, S+ and S\u2212 are not hermitian\u2014evidently they are not observable. The eigenspinors of Sz are (or course): \u03c7+ =, eigenvalue + ; \u03c7\u2212 = 1 0 4.4 Spin 169, eigenvalue \u2212 0 1. 2 (4.149) 2 2 If you measure Sz on a particle in the general state \u03c7 (Equation 4.139), you could get +/2, with probability |a|2, or \u2212/2, with probability |b|2. Since these are the only possibilities, |a|2 + |b|2 = 1 (4.150) (i.e. the spinor must be normalized: \u03c7 \u2020\u03c7 = 1).42 But what if, instead, you chose to measure Sx? What are the possible results, and what are their respective probabilities? According to the generalized statistical interpretation, we need to know the eigenvalues and eigenspinors of Sx. The characteristic equation is \u2212\u03bb /2 /2 \u2212\u03bb = 0 \u21d2 \u03bb2 = 2 2 \u21d2 \u03bb = \u00b1. 2 Not surprisingly (but it gratifying to see how it works out), the possible values for Sx are the same as those for Sz. The eigenspinors are obtained in the usual way so \u03b2 = \u00b1\u03b1. Evidently the (normalized) eigenspinors of Sx are \u03c7 (x) + =, eigenvalue + ; \u03c7 (x) \u2212 =, eigenvalue \u2212 \u221a 2 \u221a 2 1/ \u22121/ \u221a \u221a 2 2 1/ 1/ 2. (4.151) As the eigenvectors of a hermitian matrix, they span the space; the generic spinor \u03c7 (Equation 4.139) can be expressed as a linear combination of themx) + + \u03c7 (x) \u2212. (4.152) \u03c7 = If you measure Sx, the probability of getting +/2 is (1/2) |a + b|2, and the probability of getting \u2212/2 is (1/2) |a \u2212 b|2. (Check for yourself that these probabilities add up to 1.) Example 4.2 Suppose a spin-1/2 particle is in the state \u03c7 = 1\u221a 6 1 + i 2. What are the probabilities of getting +/2 and \u2212/2, if you measure Sz and Sx? Solution: Here a = (1 + i) / \u221a 2 = 1/3, and the probability of getting \u2212/2 is 6 is 6, so for Sz the probability of getting +/2 2 = 2/3. For 2/ (1 + i) / 6 and b = 2/ \u221a \u221a \u221a 6 42", " People often say that |a|2 is the \u201cprobability that the particle is in the spin-up state,\u201d but this is bad language; what they mean is that if you measured Sz, |a|2 is the probability you\u2019d get /2. See footnote 18, page 103. 170 CHAPTER 4 Quantum Mechanics in Three Dimensions Sx the probability of getting +/2 is (1/2) (\u22121 + i) / getting \u2212/2 is (1/2) is \u221a \u221a 6 (3 + i) / 2 = 5/6, and the probability of 2 = 1/6. Incidentally, the expectation value of Sx which we could also have obtained more directly: (1 \u2212 i) / Sx = \u03c7 \u2020Sx \u03c7 = \u221a 6 6 2/ \u221a 0 /2 /2 0 (1 + i) / \u221a 2/ 6 \u221a 6 =. 3 I\u2019d like now to walk you through an imaginary measurement scenario involving spin 1/2, because it serves to illustrate in very concrete terms some of the abstract ideas we discussed back in Chapter 1. Let\u2019s say we start out with a particle in the state \u03c7+. If someone asks, \u201cWhat is the z-component of that particle\u2019s spin angular momentum?\u201d, we can answer unambiguously: +/2. For a measurement of Sz is certain to return that value. But if our interrogator asks instead, \u201cWhat is the x-component of that particle\u2019s spin angular momentum?\u201d we are obliged to equivocate: If you measure Sx, the chances are \ufb01fty-\ufb01fty of getting either /2 or \u2212/2. If the questioner is a classical physicist, or a \u201crealist\u201d (in the sense of Section 1.2), he will regard this as an inadequate\u2014not to say impertinent\u2014response: \u201cAre you telling me that you don\u2019t know the true state of that particle?\u201d On the contrary; I know precisely what the state of the particle is: \u03c7+. \u201cWell, then, how come you can\u2019t tell me what the xcomponent of its spin is?\u201d Because it simply does not have a particular x-component of spin. Indeed, it cannot, for if both Sx and Sz were well-de\ufb01ned, the uncertainty principle would be violated. At this point our challenger grabs the test-tube and measures the x-component of the particle\u2019s spin; let\u2019s say he gets the value +/2. \u201cAha!\u201d (he shouts in triumph), \u201cYou lied! This particle has a perfectly well-de\ufb01ned value of Sx : /2.\u201d Well, sure\u2014it does now, but that doesn\u2019t prove it had that value, prior to your measurement. \u201cYou have obviously been reduced to splitting hairs. And anyway, what happened to your uncertainty principle? I now know both Sx and Sz.\u201d I\u2019m sorry, but you do not: In the course of your measurement, you altered the particle\u2019s state; it is now in the state \u03c7 (x) +, and whereas you know the value of Sx, you no longer know the value of Sz. \u201cBut I was extremely careful not to disturb the particle when I measured Sx.\u201d Very well, if you don\u2019t believe me, check it out: Measure Sz, and see what you get. (Of course, he may get +/2, which will be embarrassing to my case\u2014but if we repeat this whole scenario over and over, half the time he will get \u2212/2.) To the layman, the philosopher, or the classical physicist, a statement of the form \u201cthis particle doesn\u2019t have a well-de\ufb01ned position\u201d (or momentum, or x-component of spin angular momentum, or whatever) sounds vague, incompetent, or (worst of all) profound. It is none of these. But its precise meaning is, I think, almost impossible to convey to anyone who has not studied quantum mechanics in some depth. If you \ufb01nd your own comprehension slipping, from time to time (if you don\u2019t, you probably haven\u2019t understood the problem), come back to the spin-1/2 system: It is the simplest and cleanest context for thinking through the conceptual paradoxes of quantum mechanics. Problem 4.29 (a) Check that the spin matrices (Equations 4.145 and 4.147) obey the fundamental commutation relations for angular momentum, Equation 4.134. 4.4 Spin 171 (b) Show that the Pauli spin matrices (Equation 4", ".148) satisfy the product rule \u03c3 j \u03c3k = \u03b4 jk + i jkl \u03c3l, (4.153) where the indices stand for x, y, or z, and jkl is the Levi-Civita symbol: +1 if jkl = 123, 231, or 312; \u22121 if jkl = 132, 213, or 321; 0 otherwise. l \u2217 Problem 4.30 An electron is in the spin state \u03c7 = A. 3i 4 (a) Determine the normalization constant A. (b) Find the expectation values of Sx, Sy, and Sz. (c) Find the \u201cuncertainties\u201d \u03c3Sx, \u03c3Sy, and \u03c3Sz. Note: These sigmas are standard deviations, not Pauli matrices! (d) Con\ufb01rm that your results are consistent with all three uncertainty principles (Equation 4.100 and its cyclic permutations\u2014only with S in place of L, of course). \u2217 \u2217 \u2217\u2217 Problem 4.31 For the most general normalized spinor \u03c7 (Equation 4.139), compute Sx, S2 Sy x. Check that, Sz,, and S2 y S2 z S2 z S2 y S2 x S2., + + = Problem 4.32 (a) Find the eigenvalues and eigenspinors of Sy. (b) If you measured Sy on a particle in the general state \u03c7 (Equation 4.139), what values might you get, and what is the probability of each? Check that the probabilities add up to 1. Note: a and b need not be real! (c) If you measured S2 y, what values might you get, and with what probabilities? Problem 4.33 Construct momentum along an arbitrary direction \u02c6r. Use spherical coordinates, for which representing the component of spin angular the matrix Sr \u02c6r = sin \u03b8 cos \u03c6 \u02c6\u0131 + sin \u03b8 sin \u03c6 \u02c6j + cos \u03b8 \u02c6k. Find the eigenvalues and (normalized) eigenspinors of Sr. Answer: e\u2212i\u03c6 sin(\u03b8/2) cos(\u03b8/2) ei\u03c6 sin(\u03b8/2) \u2212 cos(\u03b8/2) \u03c7 (r ) + = ; \u03c7 (r ) \u2212 = (4.154) (4.155). Note: You\u2019re always free to multiply by an arbitrary phase factor\u2014say, ei\u03c6\u2014so your answer may not look exactly the same as mine. 172 CHAPTER 4 Quantum Mechanics in Three Dimensions Problem 4.34 Construct the spin matrices for a particle of spin 1. Hint: How many eigenstates of Sz are there? Determine the action of Sz, S+, and S\u2212 on each of these states. Follow the procedure used in the text for spin 1/2. Sx, Sy, and Sz 4.4.2 Electron in a Magnetic Field A spinning charged particle constitutes a magnetic dipole. Its magnetic dipole moment, \u03bc, is proportional to its spin angular momentum, S: \u03bc = \u03b3 S; (4.156) the proportionality constant, \u03b3, is called the gyromagnetic ratio.43 When a magnetic dipole is placed in a magnetic \ufb01eld B, it experiences a torque, \u03bc \u00d7 B, which tends to line it up parallel to the \ufb01eld (just like a compass needle). The energy associated with this torque is44 H = \u2212\u03bc \u00b7 B, (4.157) so the Hamiltonian matrix for a spinning charged particle, at rest45 in a magnetic \ufb01eld B, is H = \u2212\u03b3 B \u00b7 S, (4.158) where S is the appropriate spin matrix (Equations 4.145 and 4.147, in the case of spin 1/2). Example 4.3 Larmor precession: Imagine a particle of spin 1/2 at rest in a uniform magnetic \ufb01eld, which points in the z-direction: B = B0 \u02c6k. The Hamiltonian (Equation 4.158) is H = \u2212\u03b3 B0Sz = \u2212 \u03b3 B0 2 0 1 0 \u22121. The eigenstates of H are the same as those of Sz: \u03c7+, with energy E+ = \u2212 (\u03b3 B0) /2, \u03c7\u2212, with energy E\u2212 = + (\u03b3 B0) /2. (4.159) (4.160) (4.161) 43 See, for example, David J. Grif\ufb01ths, Introduction to Electrodynamics, 4th edn (Pearson, Boston, 2013), Problem 5.58. Class", "ically, the gyromagnetic ratio of an object whose charge and mass are identically distributed is q/2m, where q is the charge and m is the mass. For reasons that are fully explained only in relativistic quantum theory, the gyromagnetic ratio of the electron is (almost) exactly twice the classical value: \u03b3 = \u2212e/m. 44 Grif\ufb01ths (footnote 43), Problem 6.21. 45 If the particle is allowed to move, there will also be kinetic energy to consider; moreover, it will be subject to the Lorentz force (qv \u00d7 B), which is not derivable from a potential energy function, and hence does not \ufb01t the Schr\u00f6dinger equation as we have formulated it so far. I\u2019ll show you later on how to handle this (Problem 4.42), but for the moment let\u2019s just assume that the particle is free to rotate, but otherwise stationary. 4.4 Spin 173 The energy is lowest when the dipole moment is parallel to the \ufb01eld\u2014just as it would be classically. Since the Hamiltonian is time independent, the general solution to the time-dependent Schr\u00f6dinger equation, i \u2202\u03c7 \u2202t = H\u03c7, (4.162) can be expressed in terms of the stationary states: \u03c7 (t) = a\u03c7+e\u2212i E+t/ + b\u03c7\u2212e\u2212i E\u2212t/ = aei\u03b3 B0t/2 be\u2212i\u03b3 B0t/2. The constants a and b are determined by the initial conditions: \u03c7 (0) = a b, (of course, |a|2 + |b|2 = 1). With no essential loss of generality46 I\u2019ll write a = cos(\u03b1/2) and b = sin(\u03b1/2), where \u03b1 is a \ufb01xed angle whose physical signi\ufb01cance will appear in a moment. Thus \u03c7 (t) = cos(\u03b1/2) ei\u03b3 B0t/2 sin(\u03b1/2) e\u2212i\u03b3 B0t/2. (4.163) To get a feel for what is happening here, let\u2019s calculate the expectation value of S, as a function of time: cos (\u03b1/2)e\u2212i\u03b3 B0t/2 = Sx = \u03c7 (t)\u2020 Sx \u03c7 (t) 0 1 sin (\u03b1/2)ei\u03b3 B0t/2 2 1 0 cos (\u03b1/2)ei\u03b3 B0t/2 sin (\u03b1/2)e\u2212i\u03b3 B0t/2 Similarly, and = 2 sin \u03b1 cos (\u03b3 B0t). Sy = \u03c7 (t)\u2020Sy\u03c7 (t) = \u2212 2 sin \u03b1 sin(\u03b3 B0t), Sz = \u03c7 (t)\u2020Sz\u03c7 (t) = 2 cos \u03b1. (4.164) (4.165) (4.166) Thus S is tilted at a constant angle \u03b1 to the z axis, and precesses about the \ufb01eld at the Larmor frequency \u03c9 = \u03b3 B0, (4.167) 46 This does assume that a and b are real; you can work out the general case if you like, but all it does is add a constant to t. 174 CHAPTER 4 Quantum Mechanics in Three Dimensions \u03c9 z \u03b1 \u2329S\u2329 y x Figure 4.14: Precession of S in a uniform magnetic \ufb01eld. just as it would classically47 (see Figure 4.14). No surprise here\u2014Ehrenfest\u2019s theorem (in the form derived in Problem 4.23) guarantees that S evolves according to the classical laws. But it\u2019s nice to see how this works out in a speci\ufb01c context. Example 4.4 The Stern\u2013Gerlach experiment: In an inhomogeneous magnetic \ufb01eld, there is not only a torque, but also a force, on a magnetic dipole:48 F = \u2207 (\u03bc \u00b7 B). (4.168) This force can be used to separate out particles with a particular spin orientation. Imagine a beam of heavy neutral atoms,49 traveling in the y direction, which passes through a region of static but inhomogeneous magnetic \ufb01eld (Figure 4.15)\u2014say B(x, y, z) = \u2212\u03b1x \u02c6\u0131 + (B0 + \u03b1z) \u02c6k, (4.169) 47 See, for instance, Richard P. Feynman and Robert B. Leighton, The Feynman Lectures on Physics (Addison-Wesley, Reading, 1964), Volume II, Section 34-3.", " Of course, in the classical case it is the angular momentum vector itself, not just its expectation value, that precesses around the magnetic \ufb01eld. 48 Grif\ufb01ths (footnote 43), Section 6.1.2. Note that F is the negative gradient of the energy (Equation 4.157). 49 We make them neutral so as to avoid the large-scale de\ufb02ection that would otherwise result from the Lorentz force, and heavy so we can construct localized wave packets and treat the motion in terms of classical particle trajectories. In practice, the Stern\u2013Gerlach experiment doesn\u2019t work, for example, with a beam of free electrons. Stern and Gerlach themselves used silver atoms; for the story of their discovery see B. Friedrich and D. Herschbach, Physics Today 56, 53 (2003). z x y 4.4 Spin 175 Spin up Figure 4.15: The Stern\u2013Gerlach apparatus. Magnet Spin down where B0 is a strong uniform \ufb01eld and the constant \u03b1 describes a small deviation from homogeneity. (Actually, what we\u2019d prefer is just the z component of this \ufb01eld, but unfortunately that\u2019s impossible\u2014it would violate the electromagnetic law \u2207 \u00b7 B = 0; like it or not, the x component comes along for the ride.) The force on these atoms is50 \u2212Sx \u02c6\u0131 + Sz F = \u03b3 \u03b1 \u02c6k. But because of the Larmor precession about B0, Sx oscillates rapidly, and averages to zero; the net force is in the z direction: Fz = \u03b3 \u03b1Sz, (4.170) and the beam is de\ufb02ected up or down, in proportion to the z component of the spin angular momentum. Classically we\u2019d expect a smear (because Sz would not be quantized), but in fact the beam splits into 2s +1 separate streams, beautifully demonstrating the quantization of angular momentum. (If you use silver atoms, all the inner electrons are paired, in such a way that their angular momenta cancel. The net spin is simply that of the outermost\u2014 unpaired\u2014electron, so in this case s = 1/2, and the beam splits in two.) The Stern\u2013Gerlach experiment has played an important role in the philosophy of quantum mechanics, where it serves both as the prototype for the preparation of a quantum state and as an illuminating model for a certain kind of quantum measurement. We tend casually to assume that the initial state of a system is known (the Schr\u00f6dinger equation tells us how it subsequently evolves)\u2014but it is natural to wonder how you get a system into a particular state in the \ufb01rst place. Well, if you want to prepare a beam of atoms in a given spin con\ufb01guration, you pass an unpolarized beam through a Stern\u2013Gerlach magnet, and select the outgoing stream you are interested in (closing off the others with suitable baf\ufb02es and shutters). Conversely, if you want to measure the z component of an atom\u2019s spin, you send it through a Stern\u2013Gerlach apparatus, and record which bin it lands in. I do not claim that this is always the most practical way to do the job, but it is conceptually very clean, and hence a useful context in which to explore the problems of state preparation and measurement. Problem 4.35 In Example 4.3: (a) If you measured the component of spin angular momentum along the x direction, at time t, what is the probability that you would get +/2? (b) Same question, but for the y component. (c) Same, for the z component. 50 For a quantum mechanical justi\ufb01cation of this equation see Problem 4.73. 176 CHAPTER 4 Quantum Mechanics in Three Dimensions \u2217\u2217 Problem 4.36 An electron is at rest in an oscillating magnetic \ufb01eld B = B0 cos(\u03c9t) \u02c6k, where B0 and \u03c9 are constants. (a) Construct the Hamiltonian matrix for this system. (b) The electron starts out (at t = 0) in the spin-up state with respect to the x axis (that is: \u03c7 (0) = \u03c7 (x) + ). Determine \u03c7 (t) at any subsequent time. Beware: This is a time-dependent Hamiltonian, so you cannot get \u03c7 (t) in the usual way from stationary states. Fortunately, in this case you can solve the time-dependent Schr\u00f6dinger equation (Equation 4.162) directly. (c) Find the probability of getting \u2212/2, if you measure Sx. Answer: sin2 \u03b3 B0 2\u03c9 sin(\u03c9t). (", "d) What is the minimum \ufb01eld (B0) required to force a complete \ufb02ip in Sx? 4.4.3 Addition of Angular Momenta Suppose now that we have two particles, with spins s1 and s2. Say, the \ufb01rst is in the state |s1 m1 and the second in the state |s2 m2. We denote the composite state by |s1 s2 m1 m2: S(1)2 |s1 s2 m1 m2 = s1 (s1 + 1) 2 |s1 s2 m1 m2, S(2)2 |s1 s2 m1 m2 = s2 (s2 + 1) 2 |s1 s2 m1 m2, S(1) |s1 s2 m1 m2 = m1 |s1 s2 m1 m2, z S(2) |s1 s2 m1 m2 = m2 |s1 s2 m1 m2. z Question: What is the total angular momentum, S = S(1) + S(2), (4.171) (4.172) of this system? That is to say: what is the net spin, s, of the combination, and what is the z component, m? The z component is easy: Sz |s1 s2 m1 m2 = S(1) z |s1 s2 m1 m2 + S(2) z = (m1 + m2) |s1 s2 m1 m2 = m |s1 s2 m1 m2, |s1 s2 m1 m2, so m = m1 + m2; (4.173) (4.174) it\u2019s just the sum. But s is much more subtle, so let\u2019s begin with the simplest nontrivial example. Example 4.5 Consider the case of two spin-1/2 particles\u2014say, the electron and the proton in the ground state of hydrogen. Each can have spin up or spin down, so there are four possibilities in all:51 51 More precisely, the composite system is in a linear combination of the four states listed. For spin 1/2 I \ufb01nd the arrows more evocative than the four-index kets, but you can always revert to the formal notation if you\u2019re worried about it. 4.4 Spin 177 m = 1, m = 0, m = 0, |\u2191\u2191 = |\u2191\u2193 = |\u2193\u2191 = 1 2 \u22121 2 This doesn\u2019t look right: m is supposed to advance in integer steps, from \u2212s to +s, so it appears that s = 1\u2014but there is an \u201cextra\u201d state with m = 0., m = \u22121. |\u2193\u2193 = \u22121 2 1 2 One way to untangle this problem is to apply the lowering operator, S\u2212 = S (1) \u2212 + S (2) \u2212 to the state |\u2191\u2191, using Equation 4.146: S\u2212 |\u2191\u2191 = (1) \u2212 |\u2191 S |\u2191 + |\u2191 (2) \u2212 |\u2191 S = ( |\u2193 ) |\u2191 + |\u2191 ( |\u2193 ) = (|\u2193\u2191 + |\u2191\u2193 ). Evidently the three states with s = 1 are (in the notation |s m): \u23a7 \u23aa\u23a8 \u23aa\u23a9 |\u2191\u2191 |1 1 = |1 0 = 1\u221a 2 |\u2193\u2193 |1 \u22121 = (|\u2191\u2193 + |\u2193\u2191 ) \u23ab \u23aa\u23ac \u23aa\u23ad s = 1 (triplet). (4.175) (As a check, try applying the lowering operator to |1 0; what should you get? See Problem 4.37(a).) This is called the triplet combination, for the obvious reason. Meanwhile, the orthogonal state with m = 0 carries s = 0: |0 0 = 1\u221a 2 (|\u2191\u2193 \u2212 |\u2193\u2191 ) s = 0 (singlet). (4.176) (If you apply the raising or lowering operator to this state, you\u2019ll get zero. See Problem 4.37(b).) I claim, then, that the combination of two spin-1/2 particles can carry a total spin of 1 or 0, depending on whether they occupy the triplet or the singlet con\ufb01guration. To con\ufb01rm this, I need to prove that the triplet states are eigenvectors of S2 with eigenvalue 22, and the singlet is an eigenvector of S2 with eigenvalue 0. Now, S", "(1) + S(2) S(1) + S(2) + 2S(1) \u00b7 S(2). (4.177) S2 = S(1) S(2) 2 2 = + \u00b7 Using Equations 4.145 and 4.147, we have S(1) \u00b7 S(2) |\u2191\u2193 = = = 2 2 4 (2 |\u2193\u2191 \u2212 |\u2191\u2193 ). S(1) x |\u2191 S(2) x |\u2193 + S(1) y |\u2191 S(2) y |\u2193 + S(1) z |\u2191 S(2) z |\u2193 |\u2193 |\u2191 + |\u2193 |\u2191 + |\u2191 2 i 2 \u2212i 2 2 \u2212 2 |\u2193 178 CHAPTER 4 Quantum Mechanics in Three Dimensions Similarly, It follows that S(1) \u00b7 S(2) (|\u2193\u2191 ) = 2 4 (2 |\u2191\u2193 \u2212 |\u2193\u2191 ). S(1) \u00b7 S(2) |1 0 = 2 4 1\u221a 2 and (2 |\u2193\u2191 \u2212 |\u2191\u2193 + 2 |\u2191\u2193 \u2212 |\u2193\u2191 ) = 2 4 |1 0, (4.178) S(1) \u00b7 S(2) |0 0 = 2 4 1\u221a 2 (2 |\u2193\u2191 \u2212 |\u2191\u2193 \u2212 2 |\u2191\u2193 + |\u2193\u2191 ) = \u2212 32 4 |0 0. (4.179) Returning to Equation 4.177 (and using Equation 4.142), we conclude that S2 |1 0 = 32 4 + 32 4 + 2 2 4 |1 0 = 22 |1 0, (4.180) so |1 0 is indeed an eigenstate of S2 with eigenvalue 22; and S2 |0 0 = + 32 4 so |0 0 is an eigenstate of S2 with eigenvalue 0. (I will leave it for you to con\ufb01rm that |1 1 and |1 \u22121 are eigenstates of S2, with the appropriate eigenvalue\u2014see Problem 4.37(c).) |0 0 = 0, 32 4 32 4 (4.181) \u2212 2 What we have just done (combining spin 1/2 with spin 1/2 to get spin 1 and spin 0) is the simplest example of a larger problem: If you combine spin s1 with spin s2, what total spins s can you get?52 The answer53 is that you get every spin from (s1 + s2) down to (s1 \u2212 s2)\u2014or (s2 \u2212 s1), if s2 > s1\u2014in integer steps: s = (s1 + s2), (s1 + s2 \u2212 1), (s1 + s2 \u2212 2),..., |s1 \u2212 s2|. (4.182) (Roughly speaking, the highest total spin occurs when the individual spins are aligned parallel to one another, and the lowest occurs when they are antiparallel.) For example, if you package together a particle of spin 3/2 with a particle of spin 2, you could get a total spin of 7/2, 5/2, 3/2, or 1/2, depending on the con\ufb01guration. Another example: If a hydrogen atom is in the state \u03c8nm, the net angular momentum of the electron (spin plus orbital) is + 1/2 or \u2212 1/2; if you now throw in spin of the proton, the atom\u2019s total angular momentum quantum number is + 1,, or \u2212 1 (and can be achieved in two distinct ways, depending on whether the electron alone is in the + 1/2 con\ufb01guration or the \u2212 1/2 con\ufb01guration). The combined state |s m with total spin s and z-component m will be some linear combination of the composite states |s1 s2 m1 m2: C s1s2s |s m = m1+m2=m m1m2m |s1 s2 m1 m2 (4.183) (because the z-components add, the only composite states that contribute are those for which m1 + m2 = m). Equations 4.175 and 4.176 are special cases of this general form, 52 I say spins, for simplicity, but either one (or both) could just as well be orbital angular momentum (for which, however, we would use the letter ). 53 For a proof you must look in a more advanced text; see, for instance, Claude Cohen-Tannoudji, Bernard Diu, and Franck Lalo\u00eb, Quantum Mechanics, Wiley, New York (1977), Vol. 2, Chapter", " X. Table 4.8: Clebsch\u2013Gordan coef\ufb01cients. (A square root sign is understood for every entry; the minus sign, if present, goes outside the radical.) 4.4 Spin 179 1 1/2 \u00d7 1/2 +1 +1/2 +1/2 1 \u22121/2 +1/2 +1/2 \u22121/2 1 0 0 0 1/2 1/2 1/2 \u22121/2 \u22121/2 \u22121/2 1 \u22121 1 3/2 1 \u00d7 1/2 +3/2 +1 +1/2 1 \u22121/2 +1 +1/2 0 3 2 \u00d7 1 +3 +2 +1 1 +2 0 +1 +1 2 1 \u00d7 1 +2 +1 +1 1 +1 0 0 +1 2 +1 1 +1 1/2 1/2 1/2 \u22121/2 +1 \u22121 0 0 \u22121 +1 3/2 1/2 +1/2 +1/2 1/3 2/3 2/3 \u22121/3 \u22121/2 +1/2 0 \u22121 3 +2 2 +2 2/3 1/3 2/3 \u22121/3 \u22121 +2 +1 0 +1 0 1/2 3/2 \u22121/2 \u22121/2 1/3 2/3 1/3 \u22122/3 \u22121 \u22121/2 3/2 \u22123/2 1 3 +1 1/15 8/15 6/15 2 +1 1/3 1/6 \u22121/2 1 +1 3/5 \u22123/10 1/10 \u22121 0 +1 +1 0 \u22121 2 0 1 0 0 0 1/6 2/3 1/6 1/2 0 \u22121/2 1/3 \u22121/3 1/3 0 \u22121 0 \u22121 2 \u22121 1 \u22121 1/2 1/2 1/2 \u22121/2 \u22121 \u22121 2 \u22122 1 2 \u00d7 1/2 +2 1/2 5/2 +5/2 1 +2 \u22121/2 +1 +1/2 3/2 5/2 3/2 +3/2 4/5 1/5 4/5 \u22121/5 +1 \u22121/2 0 +1/2 3/2 5/2 +1/2 +1/2 3/5 2/5 3/5 \u22122/5 \u22121/2 +1/2 0 \u22121 3/2 \u00d7 1/2 +3/2 +1/2 2 +2 1 +3/2 \u22121/2 +1/2 +1/2 2 +1 1 +1 3/4 1/4 3/4 \u22121/4 +1/2 \u22121/2 \u22121/2 +1/2 3/2 5/2 \u22121/2 \u22121/2 2/5 3/5 2/5 \u22123/5 \u22121 \u22121/2 \u22122 +1/2 3/2 5/2 \u22123/2 \u22123/2 1/5 4/5 1/5 \u22124/5 \u22122 \u22121/2 5/2 \u22125/2 1 2 0 1 0 1/2 1/2 1/2 \u22121/2 \u22121/2 \u22121/2 \u22123/2 +1/2 2 \u22121 1 \u22121 1/4 3/4 1/4 \u22123/4 \u22121/2 \u22123/2 2 \u22122 1 5/2 \u22121/2 3/10 3/5 1/10 3/2 \u22121/2 1/2 \u22121/2 8/15 \u22121/15 \u22122/5 1/6 \u22121/3 1/2 \u22121/2 \u22121 \u22123/2 0 3/2 5/2 \u22123/2 \u22123/2 2/5 3/5 2/5 \u22123/5 \u22123/2 \u22121 5/2 \u22125/2 1 5/2 3/2 \u00d7 1 +5/2 +3/2 +1 1 +3/2 +1/2 0 +1 3 0 1/5 3/5 1/5 2 0 1/2 0 \u22121/2 1 0 3/10 \u22122/5 3/10 \u22121 0 +1 0 \u22121 \u22122 3/2 5/2 +3/2 +3/2 3/5 2/5 3/5 \u22122/5 \u22121 +3/2 +1/2 0 \u22121/2 +1 5/2 +1/2 1/10 3/5 3/10 3/2 +1/2 2/5 1/15 \u22128/15 1/2 +1/2 1/2 \u22121/3 1/6 \u22121 0 +1 +", "1/2 \u22121/2 \u22123/2 3 \u22121 2 \u22121 1 \u22121 6/15 8/15 1/15 1/2 \u22121/6 \u22121/3 1/10 \u22123/10 3/5 \u22121 \u22121 \u22122 0 3 \u22122 2 \u22122 1/3 2/3 1/3 \u22122/3 \u22122 \u22121 3 \u22123 1 with s1 = s2 = 1/2. The constants C s1s2s m1m2m are called Clebsch\u2013Gordan coef\ufb01cients. A few of the simplest cases are listed in Table 4.8.54 For example, the shaded column of the 2 \u00d7 1 table tells us that |3 0 = 1\u221a 5 |2 1 |1 \u22121 + 3 5 |2 0 |1 0 + 1\u221a 5 |2 \u22121 |1 1. If two particles (of spin 2 and spin 1) are at rest in a box, and the total spin is 3, and its z (1) could return the value (with probability 1/5), or component is 0, then a measurement of S z 0 (with probability 3/5), or \u2212 (with probability 1/5). Notice that the probabilities add up to 1 (the sum of the squares of any column on the Clebsch\u2013Gordan table is 1). These tables also work the other way around: |s1 s2 m1 m2 = C s1s2s m1m2m |s m, (m = m1 + m2). (4.184) s For example, the shaded row in the 3/2 \u00d7 1 table tells us that 15 \u2212 If you put particles of spin 3/2 and spin 1 in the box, and you know that the \ufb01rst has m1 = 1/2 and the second has m2 = 0 (so m is necessarily 1/2), and you measured the total spin, s, you could get 5/2 (with probability 3/5), or 3/2 (with probability 1/15), or 1/2 (with probability 1/3). Again, the sum of the probabilities is 1 (the sum of the squares of each row on the Clebsch\u2013 Gordan table is 1). If you think this is starting to sound like mystical numerology, I don\u2019t blame you. We will not be using the Clebsch\u2013Gordan tables much in the rest of the book, but I wanted you to know 54 The general formula is derived in Arno Bohm, Quantum Mechanics: Foundations and Applications, 2nd edn, Springer, 1986, p. 172. 180 CHAPTER 4 Quantum Mechanics in Three Dimensions where they \ufb01t into the scheme of things, in case you encounter them later on. In a mathematical sense this is all applied group theory\u2014what we are talking about is the decomposition of the direct product of two irreducible representations of the rotation group into a direct sum of irreducible representations (you can quote that, to impress your friends). \u2217 Problem 4.37 \u221a (a) Apply S\u2212 to |1 0 (Equation 4.175), and con\ufb01rm that you get (b) Apply S\u00b1 to |0 0 (Equation 4.176), and con\ufb01rm that you get zero. (c) Show that |1 1 and |1 \u22121 (Equation 4.175) are eigenstates of S2, with the 2 |1 \u22121. appropriate eigenvalue. Problem 4.38 Quarks carry spin 1/2. Three quarks bind together to make a baryon (such as the proton or neutron); two quarks (or more precisely a quark and an antiquark) bind together to make a meson (such as the pion or the kaon). Assume the quarks are in the ground state (so the orbital angular momentum is zero). (a) What spins are possible for baryons? (b) What spins are possible for mesons? \u2217 Problem 4.39 Verify Equations 4.175 and 4.176 using the Clebsch\u2013Gordan table. Problem 4.40 (a) A particle of spin 1 and a particle of spin 2 are at rest in a con\ufb01guration such that the total spin is 3, and its z component is. If you measured the z-component of the angular momentum of the spin-2 particle, what values might you get, and what is the probability of each one? Comment: Using Clebsch\u2013Gordan tables is like driving a stick-shift\u2014scary and frustrating when you start out, but easy once you get the hang of it. (b) An electron with spin down is in the state \u03c8510 of the hydrogen atom. If you could measure", " the total angular momentum squared of the electron alone (not including the proton spin), what values might you get, and what is the probability of each? (1) Problem 4.41 Determine the commutator of S2 with S z Generalize your result to show that # $ S2, S(1) = 2i S(1) \u00d7 S(2) (where S \u2261 S(1) + S(2)).. (4.185) (1) does not commute with S2, we cannot hope to \ufb01nd states that are Comment: Because S z simultaneous eigenvectors of both. In order to form eigenstates of S2 we need linear com(1) binations of eigenstates of S. This is precisely what the Clebsch\u2013Gordan coef\ufb01cients z (in Equation 4.183) do for us. On the other hand, it follows by obvious inference from Equation 4.185 that the sum S(1) + S(2) does commute with S2, which only con\ufb01rms what we already knew (see Equation 4.103).] 4.5 Electromagnetic Interactions 181 4.5 ELECTROMAGNETIC INTERACTIONS 4.5.1 Minimal Coupling In classical electrodynamics55 the force on a particle of charge q moving with velocity v through electric and magnetic \ufb01elds E and B is given by the Lorentz force law: F = q (E + v \u00d7 B). (4.186) This force cannot be expressed as the gradient of a scalar potential energy function, and therefore the Schr\u00f6dinger equation in its original form (Equation 1.1) cannot accommodate it. But in the more sophisticated form i \u2202 \u2202t = \u02c6H (4.187) there is no problem. The classical Hamiltonian for a particle of charge q and momentum p, in the presence of electromagnetic \ufb01elds is56 H = 1 2m (p \u2212 qA)2 + q\u03d5, (4.188) where A is the vector potential and \u03d5 is the scalar potential: E = \u2212\u2207\u03d5 \u2212 \u2202A/\u2202t, B = \u2207 \u00d7 A. Making the standard substitution p \u2192 \u2212i\u2207, we obtain the Hamiltonian operator57 \u02c6H = 1 2m and the Schr\u00f6dinger equation becomes (\u2212i\u2207 \u2212 qA)2 + q\u03d5, (4.189) (4.190) i \u2202 \u2202t = 1 2m (\u2212i\u2207 \u2212 qA)2 + q\u03d5. (4.191) This is the quantum implementation of the Lorentz force law; it is sometimes called the minimal coupling rule.58 \u2217\u2217\u2217 Problem 4.42 (a) Using Equation 4.190 and the generalized Ehrenfest theorem (3.73), show that = 1 m (p \u2212 qA). d r dt (4.192) Hint: This stands for three equations\u2014one for each component. Work it out for, say, the x component, and then generalize your result. 55 Readers who have not studied electrodynamics may want to skip Section 4.5. 56 See, for example, Herbert Goldstein, Charles P. Poole, and John Safko, Classical Mechanics, 3rd edn, Prentice Hall, Upper Saddle River, NJ, 2002, page 342. 57 In the case of electrostatics we can choose A = 0, and q\u03d5 is the potential energy V. 58 Note that the potentials are given, just like the potential energy V in the regular Schr\u00f6dinger equation. In quantum electrodynamics (QED) the \ufb01elds themselves are quantized, but that\u2019s an entirely different theory. 182 CHAPTER 4 Quantum Mechanics in Three Dimensions (b) As always (see Equation 1.32) we identify dr /dt with v. Show that59 m dv dt = qE + q 2m (p \u00d7 B \u2212 B \u00d7 p) \u2212 q2 m (A \u00d7 B). (4.193) (c) In particular, if the \ufb01elds E and B are uniform over the volume of the wave packet, show that m dv dt = q (E + v \u00d7 B), (4.194) so the expectation value of v moves according to the Lorentz force law, as we would expect from Ehrenfest\u2019s theorem. \u2217\u2217\u2217 Problem 4.43 Suppose A = B0 2 x \u02c6j \u2212 y\u02c6\u0131, and \u03d5 = K z2, where B0 and K are constants. (a) Find the \ufb01elds E and", " B. (b) Find the allowed energies, for a particle of mass m and charge q, in these \ufb01elds. Answer: E (n1, n2) = n1 + 1 2 \u03c91 + \u221a n2 + 1 2 \u03c92, (n1, n2 = 0, 1, 2,...), (4.195) where \u03c91 \u2261 q B0/m and \u03c92 \u2261 2q K /m. Comment: In two dimensions (x and y, with K = 0) this is the quantum analog to cyclotron motion; \u03c91 is the classical \u03c91, are called n1 + 1 cyclotron frequency, and \u03c92 is zero. The allowed energies, 2 Landau Levels.60 4.5.2 The Aharonov\u2013Bohm Effect In classical electrodynamics the potentials A and \u03d5 are not uniquely determined; the physical quantities are the \ufb01elds, E and B.61 Speci\ufb01cally, the potentials \u03d5 \u2261 \u03d5 \u2212 \u2202 \u2202t, A \u2261 A + \u2207 (4.196) (where is an arbitrary real function of position and time) yield the same \ufb01elds as \u03d5 and A. (Check that for yourself, using Equation 4.189.) Equation 4.196 is called a gauge transformation, and the theory is said to be gauge invariant. In quantum mechanics the potentials play a more direct role (it is they, not the \ufb01elds, that appear in the Equation 4.191), and it is of interest to ask whether the theory remains gauge invariant. It is easy to show (Problem 4.44) that 59 Note that p does not commute with B, so (p \u00d7 B) = \u2212 (B \u00d7 p), but A does commute with B, so (A \u00d7 B) = \u2212 (B \u00d7 A). 60 For further discussion see Leslie E. Ballentine, Quantum Mechanics: A Modern Development, World Scienti\ufb01c, Singapore (1998), Section 11.3. 61 See, for example, Grif\ufb01ths (footnote 43), Section 10.1.2. 4.5 Electromagnetic Interactions 183 \u2261 eiq/ (4.197) satis\ufb01es Equation 4.191 with the gauge-transformed potentials \u03d5 and A (Equation 4.196). Since differs from only by a phase factor, it represents the same physical state,62 and in this sense the theory is gauge invariant. For a long time it was taken for granted that there could be no electromagnetic in\ufb02uences in regions where E and B are zero\u2014any more than there can be in the classical theory. But in 1959 Aharonov and Bohm63 showed that the vector potential can affect the quantum behavior of a charged particle, even when the particle is con\ufb01ned to a region where the \ufb01eld itself is zero. Example 4.6 Imagine a particle constrained to move in a circle of radius b (a bead on a wire ring, if you like). Along the axis runs a solenoid of radius a < b, carrying a steady electric current I (see Figure 4.16). If the solenoid is extremely long, the magnetic \ufb01eld inside it is uniform, and the \ufb01eld outside is zero. But the vector potential outside the solenoid is not zero; in fact (adopting the convenient gauge condition \u2207 \u00b7 A = 0),64 A = 2\u03c0r \u02c6\u03c6, (r > a), (4.198) where = \u03c0a2 B is the magnetic \ufb02ux through the solenoid. Meanwhile, the solenoid itself is uncharged, so the scalar potential \u03d5 is zero. In this case the Hamiltonian (Equation 4.190) becomes # \u22122\u22072 + q2 A2 + 2iqA \u00b7 \u2207 $ \u02c6H = 1 2m (4.199) (Problem 4.45(a)). But the wave function depends only on the azimuthal angle \u03c6 (\u03b8 = \u03c0/2 and r = b), so \u2207 \u2192 (d/d\u03c6), and the Schr\u00f6dinger equation reads \u02c6\u03c6/b 2 1 2m \u2212 2 b2 d2 d\u03c62 + q 2\u03c0 b + i q \u03c0 b2 d d\u03c6 \u03c8(\u03c6) = E\u03c8(\u03c6). (4.200) This is a linear differential equation with constant coef\ufb01cients: d2\u03c8 d\u03c62 \u2212 2i\u03b2 d\u03c8 d\u03c6 + \u03c8 = 0, where \u03b2 \u2261", " q 2\u03c0 and \u2261 2mb2 E 2 \u2212 \u03b22. (4.201) (4.202) 62 That is to say, r, dr/dt, etc. are unchanged. Because depends on position, p (with p represented by the operator \u2212i\u2207) does change, but as you found in Equation 4.192, p does not represent the mechanical momentum (mv) in this context (in Lagrangian mechanics p = mv + qA is the so-called canonical momentum). 63 Y. Aharonov and D. Bohm, Phys. Rev. 115, 485 (1959). For a signi\ufb01cant precursor, see W. Ehrenberg and R. E. Siday, Proc. Phys. Soc. London B62, 8 (1949). 64 See, for instance, Grif\ufb01ths (footnote 43), Equation 5.71. 184 CHAPTER 4 Quantum Mechanics in Three Dimensions I b q B a Figure 4.16: Charged bead on a circular ring through which a long solenoid passes. Solutions are of the form with \u03c8 = Aei\u03bb\u03c6, \u03bb = \u03b2 \u00b1 \u03b22 + = \u03b2 \u00b1 b \u221a 2m E. Continuity of \u03c8(\u03c6), at \u03c6 = 2\u03c0, requires that \u03bb be an integer: \u221a \u03b2 \u00b1 b 2m E = n, (4.203) (4.204) (4.205) and it follows that En = 2 2mb2 2 n \u2212 q 2\u03c0, (n = 0, \u00b11, \u00b12,...). (4.206) The solenoid lifts the two-fold degeneracy of the bead-on-a-ring (Problem 2.46): positive n, representing a particle traveling in the same direction as the current in the solenoid, has a somewhat lower energy (assuming q is positive) than negative n, describing a particle traveling in the opposite direction. More important, the allowed energies clearly depend on the \ufb01eld inside the solenoid, even though the \ufb01eld at the location of the particle is zero!65 More generally, suppose a particle is moving through a region where B is zero (so \u2207 \u00d7 A = 0), but A itself is not. (I\u2019ll assume that A is static, although the method can be generalized to time-dependent potentials.) The Schr\u00f6dinger equation, 65 It is a peculiar property of superconducting rings that the enclosed \ufb02ux is quantized: = (2\u03c0 /q) n, where n is an integer. In that case the effect is undetectable, since En = is just another integer. (Incidentally, the charge q here turns out to be twice the charge of an electron; the superconducting electrons are locked together in pairs.) However, \ufb02ux quantization is enforced by the superconductor (which induces circulating currents to make up the difference), not by the solenoid or the electromagnetic \ufb01eld, and it does not occur in the (nonsuperconducting) example considered here. 2/2mb2 n + n n + n 2, and 4.5 Electromagnetic Interactions 185 can be simpli\ufb01ed by writing where 1 2m (\u2212i\u2207 \u2212 qA)2 = i \u2202 \u2202t, = eig, g(r) \u2261 q r O r A \u00b7 dr, (4.207) (4.208) (4.209) and O is some (arbitrarily chosen) reference point. (Note that this de\ufb01nition makes sense only when \u2207 \u00d7 A = 0 throughout the region in question66\u2014otherwise the line integral would depend on the path taken from O to r, and hence would not de\ufb01ne a function of r.) In terms of, the gradient of is \u2207 = eig (i\u2207g) + eig \u2207 ; but \u2207g = (q/) A, so and it follows that (\u2212i\u2207 \u2212 qA) = \u2212ieig\u2207, (\u2212i\u2207 \u2212 qA)2 = \u22122eig\u22072 (4.210) (4.211) (Problem 4.45(b)). Putting this into Equation 4.207, and cancelling the common factor of eig, we are left with \u2212 2 2m \u22072 = i \u2202 \u2202t. (4.212) Evidently satis\ufb01es the Schr\u00f6dinger equation without A. If we can solve Equation 4.212, correcting for the presence of a (curl-free) vector potential will be trivial: just tack on the phase factor eig. Aharonov and Bohm proposed an experiment in which a beam of electrons is split", " in two, and they pass either side of a long solenoid before recombining (Figure 4.17). The beams are kept well away from the solenoid itself, so they encounter only regions where B = 0. But A, which is given by Equation 4.198, is not zero, and the two beams arrive with different phases:67 g = q A \u00b7 dr = q 2\u03c0 r \u02c6\u03c6 d\u03c6 \u02c6\u03c6 \u00b7 1 r = \u00b1 q 2. (4.213) 66 The region in question must also be simply connected (no holes). This might seem like a technicality, but in the present example we need to excise the solenoid itself, and that leaves a hole in the space. To get around this we treat each side of the solenoid as a separate simply-connected region. If that bothers you, you\u2019re not alone; it seems to have bothered Aharanov and Bohm as well, since\u2014in addition to this argument\u2014they provided an alternative solution to con\ufb01rm their result (Y. Aharonov and D. Bohm, Phys. Rev. 115, 485 (1959)). The Aharonov\u2013Bohm effect can also be cast as an example of Berry\u2019s phase (see Chapter 11), where this issue does not arise (M. Berry, Proc. Roy. Soc. Lond. A 392, 45 (1984)). 67 Use cylindrical coordinates centered on the axis of the solenoid; put O on the incoming beam, and let \u03c6 run 0 < \u03c0 on one side and 0 > \u2212\u03c0 on the other, with r > a always. 186 CHAPTER 4 Quantum Mechanics in Three Dimensions B I A Beam split Beam recombined Figure 4.17: The Aharonov\u2013Bohm effect: The electron beam splits, with half passing either side of a long solenoid. Solenoid The plus sign applies to the electrons traveling in the same direction as A\u2014which is to say, in the same direction as the current in the solenoid. The beams arrive out of phase by an amount proportional to the magnetic \ufb02ux their paths encircle: phase difference = q. (4.214) This phase shift leads to measurable interference, which has been con\ufb01rmed experimentally by Chambers and others.68 What are we to make of the Aharonov\u2013Bohm effect? It seems our classical preconceptions are simply mistaken: There can be electromagnetic effects in regions where the \ufb01elds are zero. Note, however, that this does not make A itself measurable\u2014only the enclosed \ufb02ux comes into the \ufb01nal answer, and the theory remains gauge invariant.69 \u2217\u2217 Problem 4.44 Show that (Equation 4.197) satis\ufb01es the Schr\u00f6dinger equation (Equation 4.191 with the potentials \u03d5 and A (Equation 4.196). Problem 4.45 (a) Derive Equation 4.199 from Equation 4.190. (b) Derive Equation 4.211, starting with Equation 4.210. 68 R. G. Chambers, Phys. Rev. Lett. 5, 3 (1960). 69 Aharonov and Bohm themselves concluded that the vector potential has a physical signi\ufb01cance in quantum mechanics that it lacks in classical theory, and most physicists today would agree. For the early history of the Aharonov\u2013Bohm effect see H. Ehrlickson, Am. J. Phys. 38, 162 (1970). Further Problems on Chapter 4 187 FURTHER PROBLEMS ON CHAPTER 4 \u2217 Problem 4.46 Consider the three-dimensional harmonic oscillator, for which the potential is V(r ) = 1 2 (a) Show that separation of variables in cartesian coordinates turns this into three onedimensional oscillators, and exploit your knowledge of the latter to determine the allowed energies. Answer: m\u03c92r 2. (4.215) n + 3 2 En = \u03c9. (4.216) (b) Determine the degeneracy d (n) of En. \u2217\u2217\u2217 Problem 4.47 Because the three-dimensional harmonic oscillator potential (see Equation 4.215) is spherically symmetrical, the Schr\u00f6dinger equation can also be handled by separation of variables in spherical coordinates. Use the power series method (as in Sections 2.3.2 and 4.2.1) to solve the radial equation. Find the recursion formula for the coef\ufb01cients, and determine the allowed energies. (Check that your answer is consistent with Equation 4.216.) How is N related to n in this case? Draw the diagram analogous to Figures 4.3 and 4.6, and determine the degeneracy of nth", " energy level.70 \u2217\u2217 Problem 4.48 (a) Prove the three-dimensional virial theorem: 2 T = r \u00b7 \u2207V (4.217) (for stationary states). Hint: refer to Problem 3.37. (b) Apply the virial theorem to the case of hydrogen, and show that T = \u2212En; V = 2En. (4.218) (c) Apply the virial theorem to the three-dimensional harmonic oscillator (Problem 4.46), and show that in this case T = V = En/2. (4.219) \u2217\u2217\u2217 Problem 4.49 Warning: Attempt this problem only if you are familiar with vector calculus. De\ufb01ne the (three-dimensional) probability current by generalization of Problem 1.14: J \u2261 i 2m \u2207\u2217 \u2212 \u2217 \u2207 (a) Show that J satis\ufb01es the continuity equation \u2207 \u00b7 J = \u2212 \u2202 \u2202t ||2,. (4.220) (4.221) 70 For some damn reason energy levels are traditionally counted starting with n = 0, for the harmonic oscillator. That con\ufb02icts with good sense and with our explicit convention (footnote 12), but please stick with it for this problem. 188 CHAPTER 4 Quantum Mechanics in Three Dimensions which expresses local conservation of probability. It follows (from the divergence theorem) that 8 J \u00b7 da = \u2212 d dt where V is a (\ufb01xed) volume and S is its boundary surface. In words: The \ufb02ow of probability out through the surface is equal to the decrease in probability of \ufb01nding the particle in the volume. ||2 d3r, (4.222) V S (b) Find J for hydrogen in the state n = 2, l = 1, m = 1. Answer: 64\u03c0 ma5 r e\u2212r/a sin \u03b8 \u02c6\u03c6. (c) If we interpret mJ as the \ufb02ow of mass, the angular momentum is L = m (r \u00d7 J) d3r. Use this to calculate L z for the state \u03c82 1 1, and comment on the result.71 \u2217\u2217\u2217 Problem 4.50 The (time-independent) momentum space wave function in three dimen- sions is de\ufb01ned by the natural generalization of Equation 3.54: \u03c6(p) \u2261 1 (2\u03c0 )3/2 e\u2212i(p\u00b7r)/\u03c8 (r) d3r. (4.223) (a) Find the momentum space wave function for the ground state of hydrogen (Equation 4.80). Hint: Use spherical coordinates, setting the polar axis along the direction of p. Do the \u03b8 integral \ufb01rst. Answer: 3/2 \u03c6(p) = 1 \u03c0 2a 1 1 + (ap/)2. 2 (4.224) (b) Check that \u03c6(p) is normalized. (c) Use \u03c6(p) to calculate (d) What is the expectation value of the kinetic energy in this state? Express your answer as a multiple of E1, and check that it is consistent with the virial theorem (Equation 4.218)., in the ground state of hydrogen. p2 \u2217\u2217\u2217 Problem 4.51 In Section 2.6 we noted that the \ufb01nite square well (in one dimension) has at least one bound state, no matter how shallow or narrow it may be. In Problem 4.11 you showed that the \ufb01nite spherical well (three dimensions) has no bound state, if the potential is suf\ufb01ciently weak. Question: What about the \ufb01nite circular well (two 71 Schr\u00f6dinger (Annalen der Physik 81, 109 (1926), Section 7) interpreted eJ as the electric current density (this was before Born published his statistical interpretation of the wave function), and noted that it is time-independent (in a stationary state): \u201cwe may in a certain sense speak of a return to electrostatic and magnetostatic atomic models. In this way the lack of radiation in [a stationary] state would, indeed, \ufb01nd a startlingly simple explanation.\u201d (I thank Kirk McDonald for calling this reference to my attention.) Further Problems on Chapter 4 189 dimensions)? Show that (like the one-dimensional case) there is always at least one bound state. Hint: Look up any information you need about Bessel functions, and use a computer to draw the graphs. Problem 4.52 (a) Construct the spatial wave function (\u03c8) for hydrogen in the state n = 3, = 2, m = 1. Express your answer as a function", " of r, \u03b8, \u03c6, and a (the Bohr radius) only\u2014no other variables (\u03c1, z, etc.) or functions (Y, v, etc.), or constants (A, c0, etc.), or derivatives, allowed (\u03c0 is okay, and e, and 2, etc.). (b) Check that this wave function is properly normalized, by carrying out the appropriate integrals over r, \u03b8, and \u03c6. (c) Find the expectation value of r s in this state. For what range of s (positive and negative) is the result \ufb01nite? Problem 4.53 (a) Construct the wave function for hydrogen in the state n = 4, = 3, m = 3. Express your answer as a function of the spherical coordinates r, \u03b8, and \u03c6. (b) Find the expectation value of r in this state. (As always, look up any nontrivial integrals.) (c) If you could somehow measure the observable L 2 x y on an atom in this state, what value (or values) could you get, and what is the probability of each? + L 2 Problem 4.54 What is the probability that an electron in the ground state of hydrogen will be found inside the nucleus? (a) First calculate the exact answer, assuming the wave function (Equation 4.80) is correct all the way down to r = 0. Let b be the radius of the nucleus. (b) Expand your result as a power series in the small number \u2261 2b/a, and show that the lowest-order term is the cubic: P \u2248 (4/3) (b/a)3. This should be a suitable approximation, provided that b a (which it is). (c) Alternatively, we might assume that \u03c8 (r ) is essentially constant over the (tiny) volume of the nucleus, so that P \u2248 (4/3) \u03c0 b3|\u03c8(0)|2. Check that you get the same answer this way. (d) Use b \u2248 10\u221215 m and a \u2248 0.5 \u00d7 10\u221210 m to get a numerical estimate for P. Roughly speaking, this represents the \u201cfraction of its time that the electron spends inside the nucleus.\u201d Problem 4.55 (a) Use the recursion formula (Equation 4.76) to con\ufb01rm that when = n \u2212 1 the radial wave function takes the form Rn(n\u22121) = Nnr n\u22121e\u2212r/na, and determine the normalization constant Nn by direct integration. (b) Calculate r and for states of the form \u03c8n(n\u22121)m. r 2 190 CHAPTER 4 Quantum Mechanics in Three Dimensions (c) Show that the \u201cuncertainty\u201d in r (\u03c3r ) is r / 2n + 1 for such states. Note that the fractional spread in r decreases, with increasing n (in this sense the system \u201cbegins to look classical,\u201d with identi\ufb01able circular \u201corbits,\u201d for large n). Sketch the radial wave functions for several values of n, to illustrate this point. \u221a Problem 4.56 Coincident spectral lines.72 According to the Rydberg formula (Equation 4.93) the wavelength of a line in the hydrogen spectrum is determined by the principal quantum numbers of the initial and \ufb01nal states. Find two distinct pairs {ni, n f } that yield the same \u03bb. For example, {6851, 6409} and {15283, 11687} will do it, but you\u2019re not allowed to use those! Problem 4.57 Consider the observables A = x 2 and B = L z. (a) Construct the uncertainty principle for \u03c3A\u03c3B. (b) Evaluate \u03c3B in the hydrogen state \u03c8nm. (c) What can you conclude about x y in this state? Problem 4.58 An electron is in the spin state 1 \u2212 2i 2 (a) Determine the constant A by normalizing \u03c7. (b) If you measured Sz on this electron, what values could you get, and what is the \u03c7 = A. probability of each? What is the expectation value of Sz? (c) If you measured Sx on this electron, what values could you get, and what is the probability of each? What is the expectation value of Sx? (d) If you measured Sy on this electron, what values could you get, and what is the probability of each? What is the expectation value of Sy? \u2217\u2217\u2217 Problem 4.59 Suppose two spin-1/2 particles are known to be in the singlet con\ufb01guration (1) a be the component of the spin angular", " momentum of particle (2) b be the component (Equation 4.176). Let S number 1 in the direction de\ufb01ned by the vector a. Similarly, let S of 2\u2019s angular momentum in the direction b. Show that 0 (2) S(1) a S b 1 = \u2212 2 4 cos \u03b8, (4.225) where \u03b8 is the angle between a and b. \u2217\u2217\u2217 Problem 4.60 (a) Work out the Clebsch\u2013Gordan coef\ufb01cients for the case s1 = 1/2, s2 = anything. Hint: You\u2019re looking for the coef\ufb01cients A and B in 1 2 s2 |s m = A 1 2 s2 1 2 m + 1 2 1, 72 Nicholas Wheeler, \u201cCoincident Spectral Lines\u201d (unpublished Reed College report, 2001). Further Problems on Chapter 4 191 such that |s m is an eigenstate of S2. Use the method of Equations 4.177 through (for instance) does to |s2 m2, refer back to 4.180. If you can\u2019t \ufb01gure out what S Equation 4.136 and the line before Equation 4.147. Answer: (2) x / / A = s2 \u00b1 m + 1/2 2s2 + 1 ; B = \u00b1 s2 \u2213 m + 1/2 2s2 + 1, where the signs are determined by s = s2 \u00b1 1/2. (b) Check this general result against three or four entries in Table 4.8. Problem 4.61 Find the matrix representing Sx for a particle of spin 3/2 (using as your basis the eigenstates of Sz). Solve the characteristic equation to determine the eigenvalues of Sx. \u2217\u2217\u2217 Problem 4.62 Work out the spin matrices for arbitrary spin s, generalizing spin 1/2 (Equations 4.145 and 4.147), spin 1 (Problem 4.34), and spin 3/2 (Problem 4.61). Answer: Sz = Sx = 2 Sy = 2 0 s \u2212 1 0...... \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u2212s...... 0 \u239b \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239d \u239b \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239d bs 0 bs 0 bs\u22121 0 0 0...... 0 0 0 0 0 \u2212ibs ibs 0 0... 0 0 0 ibs\u22121 0... 0 0 0 bs\u22121 0 bs\u22122... 0 0 0 0 bs\u22122 0... ibs\u22121 0 ibs\u22122... 0 0 0 0 \u2212ibs\u22122 0... 0 0 0 0 0 0... 0 b\u2212s+... b\u2212s+1 0 0 0 0 0... 0 ib\u2212s+... \u2212ib\u2212s+1 0 where % (s + j) (s + 1 \u2212 j ). b j \u2261 \u2217\u2217\u2217 Problem 4.63 Work out the normalization factor for the spherical harmonics, as follows. From Section 4.1.2 we know that = K m Y m eim\u03c6 P m (cos \u03b8 ) ; 192 CHAPTER 4 Quantum Mechanics in Three Dimensions the problem is to determine the factor K m (which I quoted, but did not derive, in Equation 4.32). Use Equations 4.120, 4.121, and 4.130 to obtain a recursion relation giving K m+1 up to an overall constant, C () \u2261 K 0. Finally, use the result of Problem 4.25 to \ufb01x the constant. You may \ufb01nd the following formula for the derivative of an associated Legendre function useful:. Solve it by induction on m to get K m in terms of +1 \u2212 mx P m. (4.226) Problem 4.64 The electron in a hydrogen atom occupies the combined spin and position state % R21 1/3 Y 0 1 \u03c7+ + % 2/3 Y 1 1 \u03c7\u2212. (a) If you measured the orbital angular momentum squared (L 2), what values might you get, and what is the probability of each? (b) Same for the z component of orbital angular momentum (L z). (c) Same for the spin angular momentum squared (S2). (d) Same for the z component of spin angular momentum (Sz). Let J \u2261 L + S be the total angular momentum. (e) If you measured J 2, what values might you get, and what is the probability of each? (", "f) Same for Jz. (g) If you measured the position of the particle, what is the probability density for \ufb01nding it at r, \u03b8, \u03c6? (h) If you measured both the z component of the spin and the distance from the origin (note that these are compatible observables), what is the probability per unit r for \ufb01nding the particle with spin up and at radius r? \u2217\u2217 Problem 4.65 If you combine three spin-1/2 particles, you can get a total spin of 3/2 or 1/2 (and the latter can be achieved in two distinct ways). Construct the quadruplet and the two doublets, using the notation of Equations 4.175 and 4.176: \u23a7 \u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23a9 1 2 \u22123 2 =?? =?? =?? =?? \u23ab \u23aa\u23aa\u23aa\u23aa\u23ac \u23aa\u23aa\u23aa\u23aa\u23ad 4 =?? =?? 4 =?? =?? 1 2 1 \u22121 2 1 1 2 2 \u2212quadrupletdoublet 1) (doublet 2) = |\u2191\u2191\u2191 ; apply the lowering operator to get the Hint: The \ufb01rst one is easy: other states in the quadruplet. For the doublets you might start with the \ufb01rst two in the singlet state, and tack on the third: 3 2 Further Problems on Chapter 4 193 (|\u2191\u2193 \u2212 |\u2193\u2191) |\u2191. 1 2 1 2 = 1\u221a 2 1 1 2. Note: the Take it from there two doublets are not uniquely determined\u2014any linear combination of them would still carry spin 1/2. The point is to construct two independent doublets. 2 is orthogonal to 1 2 1 2 1 and to 1 2 3 2 make sure 1 2 Problem 4.66 Deduce the condition for minimum uncertainty in Sx and Sy (that is, equality \u2265 (/2) |Sz|), for a particle of spin 1/2 in the generic in the expression \u03c3Sx state (Equation 4.139). Answer: With no loss of generality we can pick a to be real; then the condition for minimum uncertainty is that b is either pure real or else pure imaginary. \u03c3Sy \u2217\u2217 Problem 4.67 Magnetic frustration. Consider three spin-1/2 particles arranged on the corners of a triangle and interacting via the Hamiltonian H = J (S1 \u00b7 S2 + S2 \u00b7 S3 + S3 \u00b7 S1), (4.227) where J is a positive constant. This interaction favors opposite alignment of neighboring spins (antiferromagnetism, if they are magnetic dipoles), but the triangular arrangement means that this condition cannot be satis\ufb01ed simultaneously for all three pairs (Figure 4.18). This is known as geometrical \u201cfrustration.\u201d (a) Show that the Hamiltonian can be written in terms of the square of the total spin, S2, where S = i Si. (b) Determine the ground state energy, and its degeneracy. (c) Now consider four spin-1/2 particles arranged on the corners of a square, and interacting with their nearest neighbors: H = J (S1 \u00b7 S2 + S2 \u00b7 S3 + S3 \u00b7 S4 + S4 \u00b7 S1). (4.228) In this case there is a unique ground state. Show that the Hamiltonian in this case can be written $ S2 \u2212 (S1 + S3)2 \u2212 (S2 + S4)2. (4.229) # H = 1 2 J What is the ground state energy? 3 1 2 4 1 3 2 Figure 4.18: The \ufb01gure shows three spins arranged around a triangle, where there is no way for each spin to be anti-aligned with all of its neighbors. In contrast, there is no such frustration with four spins arranged around a square. 194 CHAPTER 4 Quantum Mechanics in Three Dimensions \u2217\u2217 \u2217\u2217 Problem 4.68 Imagine a hydrogen atom at the center of an in\ufb01nite spherical well of radius b. We will take b to be much greater than the Bohr radius (a), so the low-n states are not much affected by the distant \u201cwall\u201d at r = b. But since u (b) = 0 we can use the method of Problem 2.61 to solve the radial equation (4.53) numerically. (a) Show that v j (in Problem 2.61) takes the form v j = \u2212 2\u03b2 where \u03b2 \u2261 +. ( + 1", ") j 2 j b (N + 1) a (b) We want r a (so as to sample a reasonable number of points within the potential) and a b (so the wall doesn\u2019t distort the atom too much). Thus 1 \u03b2\u22121 N. Let\u2019s use \u03b2 = 1/50 and N = 1000. Find the three lowest eigenvalues of H, for = 0, = 1, and = 2, and plot the corresponding eigenfunctions. Compare the known (Bohr) energies (Equation 4.70). Note: Unless the wave function drops to zero well before r = b, the energies of this system cannot be expected to match those of free hydrogen, but they are of interest in their own right as allowed energies of \u201ccompressed\u201d hydrogen.73 Problem 4.69 Find a few of the Bohr energies for hydrogen by \u201cwagging the dog\u201d (Problem 2.55), starting with Equation 4.53\u2014or, better yet, Equation 4.56; in fact, why not use Equation 4.68 to set \u03c10 = 2n, and tweak n? We know that the correct solutions occur when n is a positive integer, so you might start with n = 0.9, 1.9, 2.9, etc., and increase it in small increments\u2014the tail should wag when you pass 1, 2, 3,.... Find the lowest three ns, to four signi\ufb01cant digits, \ufb01rst for = 0, and then for = 1 and = 2. Warning: Mathematica doesn\u2019t like to divide by zero, so you might change \u03c1 to (\u03c1 + 0.000001) in the denominator. Note: u (0) = 0 in all cases, but u (0) = 0 only for \u2265 1 (Equation 4.59). So for = 0 you can use u (0) = 0, u (0) = 1. For > 0 you might be tempted to use u (0) = 0 and u (0) = 0, but Mathematica is lazy, and will go for the trivial solution u(\u03c1) \u2261 0; better, therefore, to use (say) u (1) = 1 and u (0) = 0. Problem 4.70 Sequential Spin Measurements. (a) At time t = 0 a large ensemble of spin-1/2 particles is prepared, all of them in the spin-up state (with respect to the z axis).74 They are not subject to any forces or torques. At time t1 > 0 each spin is measured\u2014some along the z direction and others along the x direction (but we aren\u2019t told the results). At time t2 > t1 their spin is measured again, this time along the x direction, and those with spin up (along x) are saved as a subensemble (those with spin down are discarded). Question: Of those remaining (the subensemble), what fraction had spin up (along z or x, depending on which was measured) in the \ufb01rst measurement? 73 For a variety of reasons this system has been much studied in the literature. See, for example, J. M. Ferreyra and C. R. Proetto, Am. J. Phys. 81, 860 (2013). 74 N. D. Mermin, Physics Today, October 2011, page 8. Further Problems on Chapter 4 195 (b) Part (a) was easy\u2014trivial, really, once you see it. Here\u2019s a more pithy generalization: At time t = 0 an ensemble of spin-1/2 particles is prepared, all in the spin-up state along direction a. At time t1 > 0 their spins are measured along direction b (but we are not told the results), and at time t2 > t1 their spins are measured along direction c. Those with spin up (along c) are saved as a subensemble. Of the particles in this subensemble, what fraction had spin up (along b) in the \ufb01rst measurement? Hint: Use Equation 4.155 to show that the probability of getting spin up (along b) in the \ufb01rst measurement is P+ = cos2(\u03b8ab/2), and (by extension) the probability of getting spin up in both measurements is P++ = cos2(\u03b8ab/2) cos2(\u03b8bc/2). Find the other three probabilities (P+\u2212, P\u2212+, and P\u2212\u2212). Beware: If the outcome of the \ufb01rst measurement was spin down, the relevant angle 1 + tan2(\u03b8ab/2) tan2(\u03b8bc/2) is now the supplement of \u03b8bc", ". Answer: \u22121. Problem 4.71 In molecular and solid-state applications, one often uses a basis of orbitals aligned with the cartesian axes rather than the basis \u03c8nm used throughout this chapter. For example, the orbitals \u03c82 px (r, \u03b8, \u03c6) = \u03c82 py (r, \u03b8, \u03c6) = \u03c82 pz (r, \u03b8, \u03c6) = 1\u221a 32 \u03c0 a3 1\u221a 32 \u03c0 a3 1\u221a 32 \u03c0 a3 x a y a z a e\u2212r/2a e\u2212r/2a e\u2212r/2a are a basis for the hydrogen states with n = 2 and = 1. (a) Show that each of these orbitals can be written as a linear combination of the orbitals \u03c8nm with n = 2, = 1, and m = \u22121, 0, 1. (b) Show that the states \u03c82 pi are eigenstates of the corresponding component of angular momentum: \u02c6Li. What is the eigenvalue in each case. (c) Make contour plots (as in Figure 4.9) for the three orbitals. In Mathematica use ContourPlot3D. Problem 4.72 Consider a particle with charge q, mass m, and spin s, in a uniform magnetic \ufb01eld B0. The vector potential can be chosen as A = \u2212 1 2 r \u00d7 B0. (a) Verify that this vector potential produces a uniform magnetic \ufb01eld B0. (b) Show that the Hamiltonian can be written H = p2 2 m + q \u03d5 \u2212 B0 \u00b7(\u03b3o L + \u03b3 S) + q2 8 m # r 2 B2 0 $ \u2212(r \u00b7 B0)2, (4.230) where \u03b3o = q/2m is the gyromagnetic ratio for orbital motion. Note: The term linear in B0 makes it energetically favorable for the magnetic moments 196 CHAPTER 4 Quantum Mechanics in Three Dimensions (orbital and spin) to align with the magnetic \ufb01eld; this is the origin of paramagnetism in materials. The term quadratic in B0 leads to the opposite effect: diamagnetism.75 Problem 4.73 Example 4.4, couched in terms of forces, was a quasi-classical explanation for the Stern\u2013Gerlach effect. Starting from the Hamiltonian for a neutral, spin-1/2 particle traveling through the magnetic \ufb01eld given by Equation 4.169, H = p2 2 m \u2212 \u03b3 B \u00b7 S, use the generalized Ehrenfest theorem (Equation 3.73) to show that m d2 dt 2 z = \u03b3 \u03b1 Sz. Comment: Equation 4.170 is therefore a correct quantum-mechanical statement, with the understanding that the quantities refer to expectation values. Problem 4.74 Neither Example 4.4 nor Problem 4.73 actually solved the Schr\u00f6dinger equation for the Stern\u2013Gerlach experiment. In this problem we will see how to set up that calculation. The Hamiltonian for a neutral, spin-1/2 particle traveling through a Stern\u2013Gerlach device is H = p2 2 m \u2212 \u03b3 B \u00b7 S where B is given by Equation 4.169. The most general wave function for a spin-1/2 particle\u2014including both spatial and spin degrees of freedom\u2014is76 (r, t) = +(r, t) \u03c7+ + \u2212(r, t) \u03c7\u2212. (a) Put (r, t) into the Schr\u00f6dinger equation H = i \u2202 \u2202t to obtain a pair of coupled equations for \u00b1. Partial answer: \u2212 2 2 m \u22072 + \u2212 2 \u03b3 (B0 + \u03b1 zt +. (b) We know from Example 4.3 that the spin will precess in a uniform \ufb01eld B0 \u02c6k. We can factor this behavior out of our solution\u2014with no loss of generality\u2014by writing \u00b1(r, t) = e\u00b1i \u03b3 B0 t/2 \u02dc(r, t). Find the coupled equations for \u02dc\u00b1. Partial answer: 2 \u2212 2 m \u22072 \u02dc+ \u2212 \u03b3 \u03b1 z \u02dc+ + \u03b3 \u03b1 x e\u2212i \u03b3 B0 t \u02dc\u2212 = i 2 2 \u2202 \u2202t \u02dc+. 75 That\u2019s not obvious but we\u2019ll prove it in Chapter 7. 76 In this notation, |+(r)|2 d3r gives the probability of \ufb01nding the particle in the vicinity of r with spin up, and similarly measuring its spin along the z axis to be up", ", and similarly for |\u2212(r)|2 d3r with spin down. Further Problems on Chapter 4 197 (c) If one ignores the oscillatory term in the solution to (b)\u2014on the grounds that it averages to zero (see discussion in Example 4.4)\u2014one obtains uncoupled equations of the form \u2212 2 2 m \u22072 \u02dc\u00b1 + V\u00b1 \u02dc\u00b1 = i \u2202 \u2202t \u02dc\u00b1. Based upon the motion you would expect for a particle in the \u201cpotential\u201d V\u00b1, explain the Stern\u2013Gerlach experiment. Problem 4.75 Consider the system of Example 4.6, now with a time-dependent \ufb02ux (t) through the solenoid. Show that with (t) = 1\u221a 2 \u03c0 ei n \u03c6 e\u2212i f (t) f (t) = 1 t 0 2 2 m b2! n \u2212 t q 2 \u03c0 \" 2 dt is a solution to the time-dependent Schr\u00f6dinger equation. Problem 4.76 The shift in the energy levels in Example 4.6 can be understood from classical electrodynamics. Consider the case where initially no current \ufb02ows in the solenoid. Now imagine slowly increasing the current. (a) Calculate (from classical electrodynamics) the emf produced by the changing \ufb02ux and show that the rate at which work is done on the charge con\ufb01ned to the ring can be written \u03c9 2 \u03c0 where \u03c9 is the angular velocity of the particle. dW d = \u2212q, (b) Calculate the z component of the mechanical angular momentum,77 Lmechanical = r \u00d7 m v = r \u00d7(p \u2212 q A), for a particle in the state \u03c8n in Example 4.6. Note that the mechanical angular momentum is not quantized in integer multiples of!78 (4.231) (c) Show that your result from part (a) is precisely equal to the rate at which the stationary state energies change as the \ufb02ux is increased: d En/d. 77 See footnote 62 for a discussion of the difference between the canonical and mechanical momentum. 78 However, the electromagnetic \ufb01elds also carry angular momentum, and the total (mechanical plus electromagnetic) is quantized in integer multiples of. For a discussion see M. Peshkin, Physics Reports 80, 375 (1981) or Chapter 1 of Frank Wilczek, Fractional Statistics and Anyon Superconductivity, World Scienti\ufb01c, New Jersey (1990). 5 IDENTICAL PARTICLES 5.1 TWO-PARTICLE SYSTEMS For a single particle, (r, t) is a function of the spatial coordinates, r, and the time, t (I\u2019ll ignore spin, for the moment). The state of a two-particle system is a function of the coordinates of particle one (r1), the coordinates of particle two (r2), and the time: Its time evolution is determined by the Schr\u00f6dinger equation: (r1, r2, t). i \u2202 \u2202t = \u02c6H, (5.1) (5.2) where H is the Hamiltonian for the whole works: 2 2 \u02c6H = \u2212 \u22072 1 \u2212 \u22072 2 + V (r1, r2, t) 2m2 (the subscript on \u2207 indicates differentiation with respect to the coordinates of particle 1 or particle 2, as the case may be). The statistical interpretation carries over in the obvious way: 2m1 (5.3) |(r1, r2, t)|2 d3r1 d3r2 is the probability of \ufb01nding particle 1 in the volume d3r1 and particle 2 in the volume d3r2; as always, must be normalized: (5.4) |(r1, r2, t)|2 d3r1 d3r2 = 1. (5.5) For time-independent potentials, we obtain a complete set of solutions by separation of variables: where the spatial wave function (\u03c8) satis\ufb01es the time-independent Schr\u00f6dinger equation: (r1, r2, t) = \u03c8(r1, r2)e\u2212i Et/, (5.6) \u2212 2 2m1 \u03c8 \u2212 \u22072 1 2 2m2 \u22072 2 \u03c8 + V \u03c8 = E\u03c8, (5.7) and E is the total energy of the system. In general, solving Equation 5.7 is dif\ufb01cult, but two special cases can be reduced to one-particle problems: 1. Noninteracting particles. Suppose the particles do not interact with one", " another, but each is subject to some external force. For example, they might be attached to two different springs. In that case the total potential energy is the sum of the two: V (r1, r2) = V1(r1) + V2(r2), (5.8) 5.1 Two-Particle Systems 199 and Equation 5.7 can be solved by separation of variables: \u03c8(r1, r2) = \u03c8a(r1)\u03c8b(r2). Plugging Equation 5.9 into Equation 5.7, dividing by \u03c8(r1, r2), and collecting the terms in r1 alone and in r2 alone, we \ufb01nd that \u03c8a(r1) and \u03c8b(r2) each satisfy the one-particle Schr\u00f6dinger equation: (5.9) \u2212 \u2212 2 2m1 2 2m2 \u22072 1 \u22072 2 \u03c8a(r1) + V1(r1)\u03c8a(r1) = Ea\u03c8a(r1), \u03c8b(r2) + V2(r2)\u03c8b(r2) = Eb\u03c8b(r2), (5.10) and E = Ea + Eb. In this case the two-particle wave function is a simple product of one-particle wave functions, (r1, r2, t) = \u03c8a(r1)\u03c8b(r2)e\u2212i(Ea +Eb)t/ \u03c8a(r1)e\u2212i Ea t/ = \u03c8b(r2)e\u2212i Ebt/ = a(r1, t)b(r2, t), (5.11) and it makes sense to say that particle 1 is in state a, and particle 2 is in state b. But any linear combination of such solutions will still satisfy the (time-dependent) Schr\u00f6dinger equation\u2014for instance (r1, r2, t) = 3 5 a(r1, t) b(r2, t) + 4 5 c(r1, t) d (r2, t). (5.12) In this case the state of particle 1 depends on the state of particle 2, and vice versa. If you measured the energy of particle 1, you might get Ea (with probability 9/25), in which case the energy of particle 2 is de\ufb01nitely Eb, or you might get Ec (probability 16/25), in which case the energy of particle 2 is Ed. We say that the two particles are entangled (Schr\u00f6dinger\u2019s lovely term). An entangled state is one that cannot be written as a product of single-particle states.1 2. Central potentials. Suppose the particles interact only with one another, via a potential that depends on their separation: V (r1, r2) \u2192 V (|r1 \u2212 r2|). (5.13) The hydrogen atom would be an example, if you include the motion of the proton. In this case the two-body problem reduces to an equivalent one-body problem, just as it does in classical mechanics (see Problem 5.1). In general, though, the two particles will be subject both to external forces and to mutual interactions, and this makes the analysis more complicated. For example, think of the two electrons in a helium atom: each feels the Coulomb attraction of the nucleus (charge 2e), and at the same time they repel one another: V (r1, r2) = 1 4\u03c0 0 \u2212 2e2 |r1| \u2212 2e2 |r2| + e2 |r1 \u2212 r2|. (5.14) We\u2019ll take up this problem in later sections. 1 The classic example of an entangled state is two spin-1/2 particles in the singlet con\ufb01guration (Equation 4.176). 200 CHAPTER 5 Identical Particles \u2217\u2217 Problem 5.1 A typical interaction potential depends only on the vector displacement r \u2261 r1 \u2212r2 between the two particles: V (r1, r2) \u2192 V (r). In that case the Schr\u00f6dinger equation separates, if we change variables from r1, r2 to r and R \u2261 (m1r1 + m2r2) / (m1 + m2) (the center of mass). (a) Show that r1 = R + (\u03bc/m1) r, r2 = R \u2212 (\u03bc/m2) r, and \u22071 = (\u03bc/m2) \u2207R + \u2207r, \u22072 = (\u03bc/m1) \u2207R \u2212 \ufffd", "\ufffdr, where \u03bc \u2261 m1m2 m1 + m2 (5.15) is the reduced mass of the system. (b) Show that the (time-independent) Schr\u00f6dinger equation (5.7) becomes \u2212 2 2 (m1 + m2) \u22072 R \u03c8 \u2212 2 2\u03bc \u22072 r \u03c8 + V (r)\u03c8 = E\u03c8. (c) Separate the variables, letting \u03c8(R, r) = \u03c8R(R)\u03c8r (r). Note that \u03c8R satis\ufb01es the one-particle Schr\u00f6dinger equation, with the total mass (m1 + m2) in place of m, potential zero, and energy E R, while \u03c8r satis\ufb01es the one-particle Schr\u00f6dinger equation with the reduced mass in place of m, potential V (r), and energy Er. The total energy is the sum: E = E R + Er. What this tells us is that the center of mass moves like a free particle, and the relative motion (that is, the motion of particle 2 with respect to particle 1) is the same as if we had a single particle with the reduced mass, subject to the potential V. Exactly the same decomposition occurs in classical mechanics;2 it reduces the two-body problem to an equivalent one-body problem. Problem 5.2 In view of Problem 5.1, we can correct for the motion of the nucleus in hydrogen by simply replacing the electron mass with the reduced mass. (a) Find (to two signi\ufb01cant digits) the percent error in the binding energy of hydrogen (Equation 4.77) introduced by our use of m instead of \u03bc. (b) Find the separation in wavelength between the red Balmer lines (n = 3 \u2192 n = 2) for hydrogen and deuterium (whose nucleus contains a neutron as well as the proton). (c) Find the binding energy of positronium (in which the proton is replaced by a positron\u2014positrons have the same mass as electrons, but opposite charge). (d) Suppose you wanted to con\ufb01rm the existence of muonic hydrogen, in which the electron is replaced by a muon (same charge, but 206.77 times heavier). Where (i.e. at what wavelength) would you look for the \u201cLyman-\u03b1\u201d line (n = 2 \u2192 n = 1)? 2 See, for example, Jerry B. Marion and Stephen T. Thornton, Classical Dynamics of Particles and Systems, 4th edn, Saunders, Fort Worth, TX (1995), Section 8.2. 5.1 Two-Particle Systems 201 Problem 5.3 Chlorine has two naturally occurring isotopes, Cl35 and Cl37. Show that the vibrational spectrum of HCl should consist of closely spaced doublets, with a splitting given by \u03bd = 7.51 \u00d7 10\u22124\u03bd, where \u03bd is the frequency of the emitted photon. Hint: Think k/\u03bc, where \u03bc is the reduced mass (Equation of it as a harmonic oscillator, with \u03c9 = 5.15) and k is presumably the same for both isotopes. \u221a 5.1.1 Bosons and Fermions Suppose we have two noninteracting particles, number 1 in the (one-particle) state \u03c8a(r), and number 2 in the state \u03c8b(r). In that case \u03c8(r1, r2) is the product (Equation 5.9): \u03c8(r1, r2) = \u03c8a(r1)\u03c8b(r2). (5.16) Of course, this assumes that we can tell the particles apart\u2014otherwise it wouldn\u2019t make any sense to claim that number 1 is in state \u03c8a and number 2 is in state \u03c8b; all we could say is that one of them is in the state \u03c8a and the other is in state \u03c8b, but we wouldn\u2019t know which is which. If we were talking classical mechanics this would be a silly objection: You can always tell the particles apart, in principle\u2014just paint one of them red and the other one blue, or stamp identi\ufb01cation numbers on them, or hire private detectives to follow them around. But in quantum mechanics the situation is fundamentally different: You can\u2019t paint an electron red, or pin a label on it, and a detective\u2019s observations will inevitably and unpredictably alter its state, raising the possibility that the two particles might have secretly switched places. The fact is, all electrons are utterly identical, in a way that no two classical objects can ever be. It\u2019s not just that we don\u2019t know which electron is which;", " God doesn\u2019t know which is which, because there is really no such thing as \u201cthis\u201d electron, or \u201cthat\u201d electron; all we can legitimately speak about is \u201can\u201d electron. Quantum mechanics neatly accommodates the existence of particles that are indistinguishable in principle: We simply construct a wave function that is noncommittal as to which particle is in which state. There are actually two ways to do it: \u03c8\u00b1(r1, r2) = A [\u03c8a(r1)\u03c8b(r2) \u00b1 \u03c8b(r1)\u03c8a(r2)] ; (5.17) the theory admits two kinds of identical particles: bosons (the plus sign), and fermions (the minus sign). Boson states are symmetric under interchange, \u03c8+(r2, r1) = \u03c8+(r1, r2); fermion states are antisymmetric under interchange, \u03c8\u2212(r2, r1) = \u2212\u03c8\u2212(r1, r2). It so happens that all particles with integer spin are bosons, and all particles with half integer spin are fermions. (5.18) This connection between spin and statistics (bosons and fermions have quite different statistical properties) can be proved in relativistic quantum mechanics; in the nonrelativistic theory it is simply taken as an axiom.3 3 It seems strange that relativity should have anything to do with it, and there has been a lot of discussion as to whether it might be possible to prove the spin-statistics connection in other ways. See, for example, Robert C. Hilborn, Am. J. Phys. 63, 298 (1995); Ian Duck and E. C. G. Sudarshan, Pauli and the Spin-Statistics Theorem, World Scienti\ufb01c, Singapore (1997). For a comprehensive bibliography on spin and statistics see C. Curceanu, J. D. Gillaspy, and R. C. Hilborn, Am. J. Phys. 80, 561 (2010). 202 CHAPTER 5 Identical Particles It follows, in particular, that two identical fermions (for example, two electrons) cannot occupy the same state. For if \u03c8a = \u03c8b, then \u03c8\u2212(r1, r2) = A [\u03c8a(r1)\u03c8a(r2) \u2212 \u03c8a(r1)\u03c8a(r2)] = 0, and we are left with no wave function at all.4 This is the famous Pauli exclusion principle. It is not (as you may have been led to believe) a weird ad hoc assumption applying only to electrons, but rather a consequence of the rules for constructing two-particle wave functions, applying to all identical fermions. Example 5.1 Suppose we have two noninteracting (they pass right through one another... never mind how you would set this up in practice!) particles, both of mass m, in the in\ufb01nite square well (Section 2.2). The one-particle states are n\u03c0 a (where K \u2261 \u03c0 22/2ma2). If the particles are distinguishable, with number 1 in state n1 and number 2 in state n2, the composite wave function is a simple product: En = n2 K \u03c8n(x) = 2 a sin x, \u03c8n1n2 (x1, x2) = \u03c8n1 (x1)\u03c8n2 (x2), En1n2 = n2 1 + n2 2 K. For example, the ground state is \u03c811 = 2 a \u03c0 x1 a \u03c0 x2 a sin, E11 = 2K ; sin the \ufb01rst excited state is doubly degenerate: \u03c812 = 2 a \u03c0 x1 a sin \u03c821 = 2 a sin 2\u03c0 x1 a sin sin 2\u03c0 x2 a \u03c0 x2 a,, E12 = 5K, E21 = 5K ; and so on. If the two particles are identical bosons, the ground state is unchanged, but the \ufb01rst excited state is nondegenerate: \u221a 2 a \u03c0 x1 a \u03c0 x2 a 2\u03c0 x2 a 2\u03c0 x1 a + sin sin sin sin (still with energy 5K ). And if the particles are identical fermions, there is no state with energy 2K ; the ground state is \u03c0 x1 a \u03c0 x2 a 2\u03c0 x2 a 2\u03c0 x1 a \u2212 sin 2 a sin sin sin \u221a, and its energy is 5K. 4 I\u2019m still leaving out the spin, don\u2019t forget\u2014if this bothers you (after all", ", a spinless fermion is an oxymoron), assume they\u2019re in the same spin state. I\u2019ll show you how spin affects the story in Section 5.1.3 \u2217 Problem 5.4 (a) If \u03c8a and \u03c8b are orthogonal, and both are normalized, what is the constant A in Equation 5.17? (b) If \u03c8a = \u03c8b (and it is normalized), what is A? (This case, of course, occurs only for 5.1 Two-Particle Systems 203 bosons.) Problem 5.5 (a) Write down the Hamiltonian for two noninteracting identical particles in the in\ufb01nite square well. Verify that the fermion ground state given in Example 5.1 is an eigenfunction of \u02c6H, with the appropriate eigenvalue. (b) Find the next two excited states (beyond the ones given in the example)\u2014wave functions, energies, and degeneracies\u2014for each of the three cases (distinguishable, identical bosons, identical fermions). 5.1.2 Exchange Forces To give you some sense of what the symmetrization requirement (Equation 5.17) actually does, I\u2019m going to work out a simple one-dimensional example. Suppose one particle is in state \u03c8a(x), and the other is in state \u03c8b(x), and these two states are orthogonal and normalized. If the two particles are distinguishable, and number 1 is the one in state \u03c8a, then the combined wave function is \u03c8(x1, x2) = \u03c8a(x1)\u03c8b(x2); (5.19) if they are identical bosons, the composite wave function is (see Problem 5.4 for the normalization) \u03c8+(x1, x2) = 1\u221a 2 and if they are identical fermions, it is \u03c8\u2212(x1, x2) = 1\u221a 2 [\u03c8a(x1)\u03c8b(x2) + \u03c8b(x1)\u03c8a(x2)] ; (5.20) [\u03c8a(x1)\u03c8b(x2) \u2212 \u03c8b(x1)\u03c8a(x2)]. (5.21) Let\u2019s calculate the expectation value of the square of the separation distance between the two particles, (x1 \u2212 x2) +. x1x2 (5.22) Case 1: Distinguishable particles. For the wave function in Equation 5.19, 2 d x1 2 d x2 = \u03c8b \u03c8a x 2 x2 x1 = x 2 1 x 2 1 a (the expectation value of x 2 in the one-particle state \u03c8a), \u03c8a x1 2 d x1 \u03c8b x2 2 d x2 = x 2 b, x 2 2 and x1x2 = \u03c8a x1 2 d x1 x1 \u03c8b x2 2 d x2 = x a x2 x b. x 2 2 = 204 CHAPTER 5 Identical Particles In this case, then x1 \u2212 x2 \u2212 2 x a x b. a (Incidentally, the answer would\u2014of course\u2014be the same if particle 1 had been in state \u03c8b, and particle 2 in state \u03c8a.) b d Case 2: Identical particles. For the wave functions in Equations 5.20 and 5.21, (5.23 # \u03c8a x 2 1 \u03c8b x 2 1 x 2 1 \u03c8a x1 2 d x1 x1 2 d x1 x1 \u2217\u03c8b \u2217\u03c8a x1 \u03c8b x1 \u03c8b \u03c8a x2 2 d x2 x2 2 d x2 d x1 x1 d x1 $ \u00b1 0 \u00b1 0 = 1 2 \u03c8b x2 x2 \u03c8a \u2217\u03c8a \u2217\u03c8b x2 d x2 x2 d x2. a x 2 + x 2 b Similarly, (Naturally, x 2 2 = x 2 1 x1x2, since you can\u2019t tell them apart.) But 2 d x1 = 1 2 \u03c8a x1 \u03c8b x1 x1\u03c8a x1 x1\u03c8b x1 x1 2 d x1 x1 \u2217\u03c8b \u2217\u03c8a x1 x1 + \u00b1 \u00b1 \u03c8b x2 \u03c8a x2 x2 2 d x2 x2 2 d x2 x2\u03c8b d x1 d x1 x2\u03c8a \u2217\u03c8a \u2217\u03c8b x2 d x2 x2 d x2 x", "2 x2 = 1 2 (xaxb + xbxa \u00b1 xabxba \u00b1 xbaxab) = xaxb \u00b1 |xab|2, x ab \u2261 x\u03c8a(x)\u2217\u03c8b(x) d x. 0 x1 \u2212 x2 \u2212 2xaxb \u2213 2|xab|2. (5.24) (5.25) where Thus Comparing Equations 5.23 and 5.25, we see that the difference resides in the \ufb01nal term ab 2; (5.26) identical bosons (the upper signs) tend to be somewhat closer together, and identical fermions (the lower signs) somewhat farther apart, than distinguishable particles in the same two states. Notice that x ab vanishes unless the two wave functions actually overlap: if \u03c8a(x) is zero wherever \u03c8b(x) is nonzero, the integral in Equation 5.24 is zero. So if \u03c8a represents an electron 5.1 Two-Particle Systems 205 in an atom in Chicago, and \u03c8b represents an electron in an atom in Seattle, it\u2019s not going to make any difference whether you antisymmetrize the wave function or not. As a practical matter, therefore, it\u2019s okay to pretend that electrons with non-overlapping wave functions are distinguishable. (Indeed, this is the only thing that allows chemists to proceed at all, for in principle every electron in the universe is linked to every other one, via the antisymmetrization of their wave functions, and if this really mattered, you wouldn\u2019t be able to talk about any one unless you were prepared to deal with them all!) The interesting case is when the overlap integral (Equation 5.24) is not zero. The system behaves as though there were a \u201cforce of attraction\u201d between identical bosons, pulling them closer together, and a \u201cforce of repulsion\u201d between identical fermions, pushing them apart (remember that we are for the moment ignoring spin). We call it an exchange force, although it\u2019s not really a force at all5\u2014no physical agency is pushing on the particles; rather, it is a purely geometrical consequence of the symmetrization requirement. It is also a strictly quantum mechanical phenomenon, with no classical counterpart. \u2217 Problem 5.6 Imagine two noninteracting particles, each of mass m, in the in\ufb01nite square well. If one is in the state \u03c8n (Equation 2.28), and the other in state \u03c8l (l = n), calculate (x1 \u2212 x2)2, assuming (a) they are distinguishable particles, (b) they are identical bosons, and (c) they are identical fermions. \u2217\u2217 Problem 5.7 Two noninteracting particles (of equal mass) share the same harmonic oscillator potential, one in the ground state and one in the \ufb01rst excited state. (a) Construct the wave function, \u03c8(x1, x2), assuming (i) they are distinguishable, (ii) they are identical bosons, (iii) they are identical fermions. Plot |\u03c8(x1, x2)|2 in each case (use, for instance, Mathematica\u2019s Plot3D). (b) Use Equations 5.23 and 5.25 to determine (c) Express each \u03c8(x1, x2) in terms of the relative and center-of-mass coordinates r \u2261 x1 \u2212 x2 and R \u2261 (x1 + x2) /2, and integrate over R to get the probability of \ufb01nding the particles a distance |r | apart: for each case. (x1 \u2212 x2)2 P(|r |) = 2 |\u03c8(R, r )|2 d R (the 2 accounts for the fact that r could be positive or negative). Graph P (r ) for the three cases. (d) De\ufb01ne the density operator by n(x) = 2 i=1 \u03b4(x \u2212 xi ) ; n(x) d x is the expected number of particles in the interval d x. Compute n(x) for each of the three cases and plot your results. (The result may surprise you.) 5 For an incisive critique of this terminology see W. J. Mullin and G. Blaylock, Am. J. Phys. 71, 1223 (2003). 206 CHAPTER 5 Identical Particles Problem 5.8 Suppose you had three particles, one in state \u03c8a(x), one in state \u03c8b(x), and one in state \u03c8c(x). Assuming \u03c8a, \u03c8b", ", and \u03c8c are orthonormal, construct the three-particle states (analogous to Equations 5.19, 5.20, and 5.21) representing (a) distinguishable particles, (b) identical bosons, and (c) identical fermions. Keep in mind that (b) must be completely symmetric, under interchange of any pair of particles, and (c) must be completely anti-symmetric, in the same sense. Comment: There\u2019s a cute trick for constructing completely antisymmetric wave functions: Form the Slater determinant, whose \ufb01rst row is \u03c8a(x1), \u03c8b(x1), \u03c8c(x1), etc., whose second row is \u03c8a(x2), \u03c8b(x2), \u03c8c(x2), etc., and so on (this device works for any number of particles).6 5.1.3 Spin It is time to bring spin into the story. The complete state of an electron (say) includes not only its position wave function, but also a spinor, describing the orientation of its spin:7 When we put together the two-particle state,8 \u03c8(r)\u03c7. \u03c8(r1, r2)\u03c7 (1, 2) (5.27) (5.28) it is the whole works, not just the spatial part, that has to be antisymmetric with respect to exchange: \u03c8(r1, r2)\u03c7 (1, 2) = \u2212\u03c8(r2, r1)\u03c7 (2, 1). (5.29) Now, a glance back at the composite spin states (Equations 4.175 and 4.176) reveals that the singlet combination is antisymmetric (and hence would have to be joined with a symmetric spatial function), whereas the three triplet states are all symmetric (and would require an antisymmetric spatial function). Thus the Pauli principle actually allows two electrons in a given position state, as long as their spins are in the singlet con\ufb01guration (but they could not be in the same position state and in the same spin state\u2014say, both spin up). \u2217\u2217 Problem 5.9 In Example 5.1 and Problem 5.5(b) we ignored spin (or, if you prefer, we assumed the particles are in the same spin state). (a) Do it now for particles of spin 1/2. Construct the four lowest-energy con\ufb01gurations, and specify their energies and degeneracies. Suggestion: Use the notation |s m, where \u03c8n1n2 is de\ufb01ned in Example 5.1 and |s m in Section 4.4.3.9 \u03c8n1n2 6 To construct a completely symmetric con\ufb01guration, use the permanent (same as determinant, but without the minus signs). 7 In the absence of coupling between spin and position, we are free to assume that the state is separable in its spin and spatial coordinates. This just says that the probability of getting spin up is independent of the location of the particle. In the presence of coupling, the general state would take the form of a linear combination: \u03c8+(r)\u03c7+ + \u03c8\u2212(r)\u03c7\u2212, as in Problem 4.64. 8 I\u2019ll let \u03c7 (1, 2) stand for the combined spin state; in Dirac notation it is some linear combination of the states |s1 s2 m1 m2. I assume that the state is again a simple product of a position state and a spin state; as you\u2019ll see in Problem 5.10, this is not always true when three or more electrons are involved\u2014even in the absence of coupling. 9 Of course, spin requires three dimensions, whereas we ordinarily think of the in\ufb01nite square well as existing in one dimension. But it could represent a particle in three dimensions that is con\ufb01ned to a one-dimensional wire. (b) Do the same for spin 1. Hint: First work out the spin-1 analogs to the spin-1/2 singlet and triplet con\ufb01gurations, using the Clebsch\u2013Gordan coef\ufb01cients; note which of them are symmetric and which antisymmetric.10 5.1 Two-Particle Systems 207 5.1.4 Generalized Symmetrization Principle I have assumed, for the sake of simplicity, that the particles are noninteracting, the spin and position are decoupled (with the combined state a product of position and spin factors), and the potential is time-independent. But the fundamental symmetrization/antisymmetrization", " requirement for identical bosons/fermions is much more general. Let us de\ufb01ne the exchange operator, \u02c6P, which interchanges the two particles:11 (5.31) (5.32) \u02c6P |(1, 2) =|(2, 1). (5.30) Clearly, \u02c6P 2 = 1, and it follows (prove it for yourself) that the eigenvalues of \u02c6P are \u00b11. Now, if the two particles are identical, the Hamiltonian must treat them the same: m1 = m2 and V (r1, r2, t) = V (r2, r1, t). It follows that \u02c6P and \u02c6H are compatible observables, # $ \u02c6P, \u02c6H = 0, and hence (Equation 3.73) 0 1 \u02c6P d dt = 0. = \u2014then it will stay that way forever. The symmetrization axiom says that for identical If the system starts out in an eigenstate of \u02c6P\u2014symmetric \u22121 particles the state is not merely allowed, but required to satisfy, or antisymmetric \u02c6P \u02c6P = 1 |(1, 2) = \u00b1|(2, 1), (5.33) with the plus sign for bosons, and the minus sign for fermions.12 If you have n identical particles, of course, the state must be symmetric or antisymmetric under the interchange of any two: |(1, 2,..., i,..., j,..., n) = \u00b1|(1, 2,..., j,..., i,..., n), (5.34) This is the general statement, of which Equation 5.17 is a special case. 10 This problem was suggested by Greg Elliott. 11 \u02c6P switches the particles (1 \u2194 2); this means exchanging their positions, their spins, and any other properties they might possess. If you like, it switches the labels, 1 and 2. I claimed (in Chapter 1) that all our operators would involve multiplication or differentiation; that was a lie. The exchange operator is an exception\u2014and for that matter so is the projection operator (Section 3.6.2). 12 It is sometimes alleged that the symmetrization requirement (Equation 5.33) is forced by the fact that \u02c6P and \u02c6H commute. This is false: It is perfectly possible to imagine a system of two distinguishable particles (say, an electron and a positron) for which the Hamiltonian is symmetric, and yet there is no requirement that the state be symmetric (or antisymmetric). But identical particles have to occupy symmetric or antisymmetric states, and this is a new fundamental law\u2014on a par, logically, with Schr\u00f6dinger\u2019s equation and the statistical interpretation. Of course, there didn\u2019t have to be any such things as identical particles; it could have been that every single particle in the universe was distinguishable from every other one. Quantum mechanics allows for the possibility of identical particles, and nature (being lazy) seized the opportunity. (But don\u2019t complain\u2014this makes matters enormously simpler!) 208 CHAPTER 5 Identical Particles \u2217\u2217 Problem 5.10 For two spin-1/2 particles you can construct symmetric and antisymmetric states (the triplet and singlet combinations, respectively). For three spin-1/2 particles you can construct symmetric combinations (the quadruplet, in Problem 4.65), but no completely anti-symmetric con\ufb01guration is possible. (a) Prove it. Hint: The \u201cbulldozer\u201d method is to write down the most general linear combination: \u03c7 (1, 2, 3) = a |\u2191\u2191\u2191 + b| \u2191\u2191\u2193 + c |\u2191\u2193\u2191 + d| \u2191\u2193\u2193 + e |\u2193\u2191\u2191 + f | \u2193\u2191\u2193 + g |\u2193\u2193\u2191 + h| \u2193\u2193\u2193. What does antisymmetry under 1 \u2194 2 tell you about the coef\ufb01cients? (Note that the eight terms are mutually orthogonal.) Now invoke antisymmetry under 2 \u2194 3. (b) Suppose you put three identical noninteracting spin-1/2 particles in the in\ufb01nite square well. What is the ground state for this system, what is its energy, and what is its degeneracy? Note: You can\u2019t put all three in the position state \u03c81 (why not?); you\u2019ll need two", " in \u03c81 and the other in \u03c82. But the symmetric con\ufb01guration [\u03c81(x1)\u03c81(x2)\u03c82(x3) + \u03c81(x1)\u03c82(x2)\u03c81(x3) + \u03c82(x1)\u03c81(x2)\u03c81(x3)] is no good (because there\u2019s no antisymmetric spin combination to go with it), and you can\u2019t make a completely antisymmetric combination of those three terms.... In this case you simply cannot construct an antisymmetric product of a spatial state and a spin state. But you can do it with an appropriate linear combination of such products. Hint: Form the Slater determinant (Problem 5.8) whose top row is \u03c81(x1) |\u2191 1, \u03c81(x1) |\u2193 1, \u03c82(x1) |\u2191 1. (c) Show that your answer to part (b), properly normalized, can be written in the form (1, 2, 3) = 1\u221a 3 [(1, 2) \u03c6(3) \u2212 (1, 3) \u03c6(2) + (2, 3) \u03c6(1)] where (i, j) is the wave function of two particles in the n = 1 state and the singlet spin con\ufb01guration, (i, j) = \u03c81(xi ) \u03c81 x j \u2191i \u2193 j \u2212 \u221a \u2193i \u2191 j 2, (5.35) and \u03c6(i) is the wave function of the ith particle in the n = 2 spin up state: \u03c6(i) = \u03c82(xi ) |\u2191i. Noting that (i, j) is antisymmetric in i \u2194 j, check that (1, 2, 3) is antisymmetric in all three exchanges 1 \u2194 2, 2 \u2194 3, and 3 \u2194 1. \u2217\u2217 Problem 5.11 In Section 5.1 we found that for noninteracting particles the wave function can be expressed as a product of single-particle states (Equation 5.9)\u2014or, for identical particles, as a symmetrized/antisymmetrized linear combination of such states (Equations 5.20 and 5.21). For interacting particles this is no longer the case. A famous example is the Laughlin wave function,13 which is an approximation to the ground state of N electrons 13 \u201cRobert B. Laughlin\u2014Nobel Lecture: Fractional Quantization.\u201d Nobelprize.org. Nobel Media AB 2014. http://www.nobelprize.org/nobel_prizes/physics/laureates/1998/laughlin-lecture.html. 5.2 Atoms 209 con\ufb01ned to two dimensions in a perpendicular magnetic \ufb01eld of strength B (the setting for the fractional quantum Hall effect). The Laughlin wave function is \u23a1 \u23a4 \u03c8(z1, z2 where q is a positive odd integer and z j \u2261 N; \u23a3 z j \u2212 zk q \u23a6 exp j<zk|2 (Spin is not at issue here; in the ground state all the electrons have spin down with respect to the direction of B, and that is a trivially symmetric con\ufb01guration.) (a) Show that \u03c8 has the proper antisymmetry for fermions. (b) For q = 1, \u03c8 describes noninteracting particles (by which I mean that it can be written as a single Slater determinant\u2014see Problem 5.8). This is true for any N, but check it explicitly for N = 3. What single particle states are occupied in this case? (c) For values of q greater than 1, \u03c8 cannot be written as a single Slater determinant, and describes interacting particles (in practice, Coulomb repulsion of the electrons). It can, however, be written as a sum of Slater determinants. Show that, for q = 3 and N = 2, \u03c8 can be written as a sum of two Slater determinants. Comment: In the noninteracting case (b) we can describe the wave function as \u201cthree particles occupying the three single-particle states \u03c8a, \u03c8b and \u03c8c,\u201d but in the interacting case (c), no corresponding statement can be made; in that case, the different Slater determinants that make up \u03c8 correspond to occupation of different sets of single-particle states. 5.2 ATOMS A neutral atom, of atomic number Z, consists of a heavy nucleus,", " with electric charge Z e, surrounded by Z electrons (mass m and charge \u2212e). The Hamiltonian for this system is14 \u02c6H = Z j=1 \u2212 2 2m \u22072 j \u2212 1 4\u03c0 0 Z e2 r j + 1 2 1 4\u03c0 0 Z j=k e2 r j \u2212 rk. (5.36) The term in curly brackets represents the kinetic plus potential energy of the jth electron, in the electric \ufb01eld of the nucleus; the second sum (which runs over all values of j and k except j = k) is the potential energy associated with the mutual repulsion of the electrons (the factor 14 I\u2019m assuming the nucleus is stationary. The trick of accounting for nuclear motion by using the reduced mass (Problem 5.1) works only for the two-body problem; fortunately, the nucleus is so much heavier than the electrons that the correction is extremely small even in the case of hydrogen (see Problem 5.2(a)), and it is smaller still for other atoms. There are more interesting effects, due to magnetic interactions associated with electron spin, relativistic corrections, and the \ufb01nite size of the nucleus. We\u2019ll look into these in later chapters, but all of them are minute corrections to the \u201cpurely coulombic\u201d atom described by Equation 5.36. 210 CHAPTER 5 Identical Particles of 1/2 in front corrects for the fact that the summation counts each pair twice). The problem is to solve Schr\u00f6dinger\u2019s equation, \u02c6H \u03c8 = E\u03c8, (5.37) for the wave function \u03c8(r1, r2,..., rZ ).15 Unfortunately, the Schr\u00f6dinger equation with Hamiltonian in Equation 5.36 cannot be solved exactly (at any rate, it hasn\u2019t been), except for the very simplest case, Z = 1 (hydrogen). In practice, one must resort to elaborate approximation methods. Some of these we shall explore in Part II; for now I plan only to sketch some qualitative features of the solutions, obtained by neglecting the electron repulsion term altogether. In Section 5.2.1 we\u2019ll study the ground state and excited states of helium, and in Section 5.2.2 we\u2019ll examine the ground states of higher atoms. Problem 5.12 (a) Suppose you could \ufb01nd a solution \u03c8(r1, r2,..., rZ ) to the Schr\u00f6dinger equation (Equation 5.37), for the Hamiltonian in Equation 5.36. Describe how you would construct from it a completely symmetric function, and a completely antisymmetric function, which also satisfy the Schr\u00f6dinger equation, with the same energy. What happens to the completely antisymmetric function if \u03c8(r1, r2,..., rZ ) is symmetric in (say) its \ufb01rst two arguments (r1 \u2194 r2)? (b) By the same logic, show that a completely antisymmetric spin state for Z electrons is impossible, if Z > 2 (this generalizes Problem 5.10(a)). 5.2.1 Helium After hydrogen, the simplest atom is helium (Z = 2). The Hamiltonian, \u02c6H = \u2212 2 2m \u22072 1 \u2212 1 4\u03c0 0 2e2 r1 + \u2212 2 2m \u22072 2 \u2212 1 4\u03c0 0 2e2 r2 + 1 4\u03c0 0 e2 |r1 \u2212 r2|, (5.38) consists of two hydrogenic Hamiltonians (with nuclear charge 2e), one for electron 1 and one for electron 2, together with a \ufb01nal term describing the repulsion of the two electrons. It is this last term that causes all the trouble. If we simply ignore it, the Schr\u00f6dinger equation separates, and the solutions can be written as products of hydrogen wave functions: \u03c8(r1, r2) = \u03c8nm(r1)\u03c8nm(r2), (5.39) only with half the Bohr radius (Equation 4.72), and four times the Bohr energies (Equation 4.70)\u2014if you don\u2019t see why, refer back to Problem 4.19. The total energy would be E = 4 (En + En), where En = \u221213.6/n2 eV. In particular, the ground state would be \u03c80(r1, r2) = \u03c8100(r1)\u03c8100(r2) = 8 \u03c0a3 e\u22122(r1+r2)/a, (5.40) (5.41) 15 Because the Hamiltonian (5", ".36) makes no reference to spin, the product \u03c8(r1, r2,..., rZ )\u03c7 (s1, s2,..., sZ ) still satis\ufb01es the Schr\u00f6dinger equation. However, for Z > 2 such product states cannot in general meet the (anti-)symmetrization requirement, and it is necessary to construct linear combinations, with permuted indices (see Problem 5.16). But that comes at the end of the story; for the moment we are only concerned with the spatial wave function. 5.2 Atoms 211 (Equation 4.80), and its energy would be E0 = 8 (\u221213.6 eV) = \u2212109 eV. (5.42) Because \u03c80 is a symmetric function, the spin state has to be antisymmetric, so the ground state of helium should be a singlet con\ufb01guration, with the spins \u201coppositely aligned.\u201d The actual ground state of helium is indeed a singlet, but the experimentally determined energy is \u221278.975 eV, so the agreement is not very good. But this is hardly surprising: We ignored electron\u2013electron repulsion, which is certainly not a small contribution. It is clearly positive (see Equation 5.38), which is comforting\u2014evidently it brings the total energy up from \u2212109 to \u221279 eV (see Problem 5.15). The excited states of helium consist of one electron in the hydrogenic ground state, and the other in an excited state: \u03c8nm\u03c8100. (5.43) (If you try to put both electrons in excited states, one immediately drops to the ground state, releasing enough energy to knock the other one into the continuum (E > 0), leaving you with a helium ion (He+) and a free electron. This is an interesting system in its own right\u2014 see Problem 5.13\u2014but it is not our present concern.) We can construct both symmetric and antisymmetric combinations, in the usual way (Equation 5.17); the former go with the antisymmetric spin con\ufb01guration (the singlet)\u2014they are called parahelium\u2014while the latter require a symmetric spin con\ufb01guration (the triplet)\u2014they are known as orthohelium. The ground state is necessarily parahelium; the excited states come in both forms. Because the symmetric spatial state brings the electrons closer together (as we discovered in Section 5.1.2), we expect a higher interaction energy in parahelium, and indeed, it is experimentally con\ufb01rmed that the parahelium states have somewhat higher energy than their orthohelium counterparts (see Figure 5.1). Problem 5.13 (a) Suppose you put both electrons in a helium atom into the n = 2 state; what would the energy of the emitted electron be? (Assume no photons are emitted in the process.) (b) Describe (quantitatively) the spectrum of the helium ion, He+. That is, state the \u201cRydberg-like\u201d formula for the emitted wavelengths. Problem 5.14 Discuss (qualitatively) the energy level scheme for helium if (a) electrons were identical bosons, and (b) if electrons were distinguishable particles (but with the same mass and charge). Pretend these \u201celectrons\u201d still have spin 1/2, so the spin con\ufb01gurations are the singlet and the triplet. \u2217\u2217 Problem 5.15 (a) Calculate (1/ |r1 \u2212 r2|) for the state \u03c80 (Equation 5.41). Hint: Do the d3r2 integral \ufb01rst, using spherical coordinates, and setting the polar axis along r1, so that |r1 \u2212 r2| = r 2 1 + r 2 2 \u2212 2r1r2 cos \u03b82. The \u03b82 integral is easy, but be careful to take the positive root. You\u2019ll have to break the r2 integral into two pieces, one ranging from 0 to r1, the other from r1 to \u221e. Answer: 5/4a. 212 CHAPTER 5 Identical Particles Parahelium Orthohelium 1S 1P 1D 1F 3S 3P 3D 3F 0 4P 4D 4F 3P 3D 2P 4S 3S 2S 4S \u20131 4P 4D 4F 3P 3D 3S \u20132 \u2013 2S \u20134 \u20135 2P (1S at \u201324.5 eV) Figure 5.1: Energy level diagram for helium (the notation is explained in Section 5.2.2). Note that parahelium energies are uniformly higher than their orth", "ohelium counterparts. The numerical values on the ): 4 \u00d7 (\u221213.6) eV = \u221254.4 eV; to vertical scale are relative to the ground state of ionized helium (He get the total energy of the state, subtract 54.4 eV. + (b) Use your result in (a) to estimate the electron interaction energy in the ground state of helium. Express your answer in electron volts, and add it to E0 (Equation 5.42) to get a corrected estimate of the ground state energy. Compare the experimental value. (Of course, we\u2019re still working with an approximate wave function, so don\u2019t expect perfect agreement.) Problem 5.16 The ground state of lithium. Ignoring electron\u2013electron repulsion, construct the ground state of lithium (Z = 3). Start with a spatial wave function, analogous to Equation 5.41, but remember that only two electrons can occupy the hydrogenic ground state; the third goes to \u03c8200.16 What is the energy of this state? Now tack on the spin, and antisymmetrize (if you get stuck, refer back to Problem 5.10). What\u2019s the degeneracy of the ground state? 16 Actually, = 1 would do just as well, but electron\u2013electron repulsion favors = 0, as we shall see. 5.2.2 The Periodic Table 5.2 Atoms 213 The ground state electron con\ufb01gurations for heavier atoms can be pieced together in much the same way. To \ufb01rst approximation (ignoring their mutual repulsion altogether) the individual electrons occupy one-particle hydrogenic states (n,, m), called orbitals, in the Coulomb potential of a nucleus with charge Z e. If electrons were bosons (or distinguishable particles) they would all shake down to the ground state (1, 0, 0), and chemistry would be very dull indeed. But electrons are in fact identical fermions, subject to the Pauli exclusion principle, so only two can occupy any given orbital (one with spin up, and one with spin down\u2014or, more precisely, in the singlet con\ufb01guration). There are n2 hydrogenic wave functions (all with the same energy En) for a given value of n, so the n = 1 shell has room for two electrons, the n = 2 shell holds eight, n = 3 takes 18, and in general the nth shell can accommodate 2n2 electrons. Qualitatively, the horizonal rows on the Periodic Table correspond to \ufb01lling out each shell (if this were the whole story, they would have lengths 2, 8, 18, 32, 50, etc., instead of 2, 8, 8, 18, 18, etc.; we\u2019ll see in a moment how the electron\u2013electron repulsion throws the counting off). With helium, the n = 1 shell is \ufb01lled, so the next atom, lithium (Z = 3), has to put one electron into the n = 2 shell. Now, for n = 2 we can have = 0 or = 1; which of these will the third electron choose? In the absence of electron\u2013electron interactions, they have the same energy (the Bohr energies depend on n, remember, but not on ). But the effect of electron repulsion is to favor the lowest value of, for the following reason. Angular momentum tends to throw the electron outward, and the farther out it gets, the more effectively the inner electrons screen the nucleus (roughly speaking, the innermost electron \u201csees\u201d the full nuclear charge Z e, but the outermost electron sees an effective charge hardly greater than e). Within a given shell, therefore, the state with lowest energy (which is to say, the most tightly bound electron) is = 0, and the energy increases with increasing. Thus the third electron in lithium occupies the orbital (2,0,0).17 The next atom (beryllium, with Z = 4) also \ufb01ts into this state (only with \u201copposite spin\u201d), but boron (Z = 5) has to make use of = 1. Continuing in this way, we reach neon (Z = 10), at which point the n = 2 shell is \ufb01lled, and we advance to the next row of the periodic table and begin to populate the n = 3 shell. First there are two atoms (sodium and magnesium) with = 0, and then there are six with = 1 (aluminum through argon). Following argon there \u201cshould\u201d be 10 atoms with n = 3 and = 2; however, by this time the screening effect is so strong that it overlaps the next shell; potassium (Z = 19) and calcium (Z = 20) choose n = 4, = 0, in preference to n = 3", ", = 2. After that we drop back to pick up the n = 3, = 2 stragglers (scandium through zinc), followed by n = 4, = 1 (gallium through krypton), at which point we again make a premature jump to the next row (n = 5), and wait until later to slip in the = 2 and = 3 orbitals from the n = 4 shell. For details of this intricate counterpoint I refer you to any book on atomic physics.18 I would be delinquent if I failed to mention the archaic nomenclature for atomic states, because all chemists and most physicists use it (and the people who make up the Graduate Record Exam love this sort of thing). For reasons known best to nineteenth-century spectroscopists, = 0 is called s (for \u201csharp\u201d), = 1 is p (for \u201cprincipal\u201d), = 2 is d (\u201cdiffuse\u201d), and = 3 is f (\u201cfundamental\u201d); after that I guess they ran out of imagination, because it now 17 This standard argument has been called into question by W. Stacey and F. Marsiglio, EPL, 100, 43002 (2012). 18 See, for example, Ugo Fano and L. Fano, Basic Physics of Atoms and Molecules, Wiley, New York (1959), Chapter 18, or the classic by Gerhard Herzberg, Atomic Spectra and Atomic Structure, Dover, New York (1944). 214 CHAPTER 5 Identical Particles continues alphabetically (g, h, i, skip j, just to be utterly perverse, k, l, etc.).19 The state of a particular electron is represented by the pair n, with n (the number) giving the shell, and (the letter) specifying the orbital angular momentum; the magnetic quantum number m is not listed, but an exponent is used to indicate the number of electrons that occupy the state in question. Thus the con\ufb01guration (1s)2 (2s)2 (2 p)2 (5.44) tells us that there are two electrons in the orbital (1,0,0), two in the orbital (2,0,0), and two in some combination of the orbitals (2,1,1), (2,1,0), and (2,1,\u22121). This happens to be the ground state of carbon. In that example there are two electrons with orbital angular momentum quantum number 1, so the total orbital angular momentum quantum number, L (capital L\u2014not to be confused with the L denoting n = 2\u2014instead of, to indicate that this pertains to the total, not to any one particle) could be 2, 1, or 0. Meanwhile, the two (1s) electrons are locked together in the singlet state, with total spin zero, and so are the two (2s) electrons, but the two (2 p) electrons could be in the singlet con\ufb01guration or the triplet con\ufb01guration. So the total spin quantum number S (capital, again, because it\u2019s the total) could be 1 or 0. Evidently the grand total (orbital plus spin), J, could be 3, 2, 1, or 0 (Equation 4.182). There exist rituals, known as Hund\u2019s Rules (see Problem 5.18) for \ufb01guring out what these totals will be, for a particular atom. The result is recorded as the following hieroglyphic: 2S+1 L J, (5.45) (where S and J are the numbers, and L the letter\u2014capitalized, because we\u2019re talking about the totals). The ground state of carbon happens to be 3 P0: the total spin is 1 (hence the 3), the total orbital angular momentum is 1 (hence the P), and the grand total angular momentum is zero (hence the 0). In Table 5.1 the individual con\ufb01gurations and the total angular momenta (in the notation of Equation 5.45) are listed, for the \ufb01rst four rows of the Periodic Table.20 \u2217 Problem 5.17 (a) Figure out the electron con\ufb01gurations (in the notation of Equation 5.44) for the \ufb01rst two rows of the Periodic Table (up to neon), and check your results against Table 5.1. Table 5.1 Ground-state electron con\ufb01gurations for the \ufb01rst four rows of the Periodic Table. Element Z 1 H 2 He 3 Li 4 Be 5 B 6 C Configuration (1s) (1s)2 (He)(2s) (He)(2s)2 (He)(2s)2(2p)", " (He)(2s)2(2p)2 2S1/2 1S0 2S1/2 1S0 2P1/2 3P0 19 The shells themselves are assigned equally arbitrary nicknames, starting (don\u2019t ask me why) with K : The K shell is n = 1, the L shell is n = 2, M is n = 3, and so on (at least they\u2019re in alphabetical order). 20 After krypton\u2014element 36\u2014the situation gets more complicated (\ufb01ne structure starts to play a signi\ufb01cant role in the ordering of the states) so it is not for want of space that the table terminates there. 5.2 Atoms 215 7 N 8 O F 9 10 Ne 11 Na 12 Mg 13 Al Si 14 P 15 16 S 17 Cl 18 Ar 19 K 20 Ca Sc 21 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr (He)(2s)2(2p)3 (He)(2s)2(2p)4 (He)(2s)2(2p)5 (He)(2s)2(2p)6 (Ne)(3s) (Ne)(3s)2 (Ne)(3s)2(3p) (Ne)(3s)2(3p)2 (Ne)(3s)2(3p)3 (Ne)(3s)2(3p)4 (Ne)(3s)2(3p)5 (Ne)(3s)2(3p)6 (Ar)(4s) (Ar)(4s)2 (Ar)(4s)2(3d ) (Ar)(4s)2(3d )2 (Ar)(4s)2(3d )3 (Ar)(4s)(3d )5 (Ar)(4s)2(3d )5 (Ar)(4s)2(3d )6 (Ar)(4s)2(3d )7 (Ar)(4s)2(3d )8 (Ar)(4s)(3d )10 (Ar)(4s)2(3d )10 (Ar)(4s)2(3d)10(4p) (Ar)(4s)2(3d )10(4p)2 (Ar)(4s)2(3d )10(4p)3 (Ar)(4s)2(3d )10(4p)4 (Ar)(4s)2(3d )10(4p)5 (Ar)(4s)2(3d )10(4p)6 4S3/2 3P2 2P3/2 1S0 2S1/2 1S0 2P1/2 3P0 4S3/2 3P2 2P3/2 1S0 2S1/2 1S0 2D3/2 3F2 4F3/2 7S3 6S5/2 5D4 4F9/2 3F4 2S1/2 1S0 2P1/2 3P0 4S3/2 3P2 2P3/2 1S0 (b) Figure out the corresponding total angular momenta, in the notation of Equation 5.45, for the \ufb01rst four elements. List all the possibilities for boron, carbon, and nitrogen. \u2217\u2217 Problem 5.18 (a) Hund\u2019s \ufb01rst rule says that, consistent with the Pauli principle, the state with the highest total spin (S) will have the lowest energy. What would this predict in the case of the excited states of helium? (b) Hund\u2019s second rule says that, for a given spin, the state with the highest total orbital angular momentum (L), consistent with overall antisymmetrization, will have the lowest energy. Why doesn\u2019t carbon have L = 2? Hint: Note that the \u201ctop of the ladder\u201d (ML = L) is symmetric. (c) Hund\u2019s third rule says that if a subshell (n, ) is no more than half \ufb01lled, then the lowest energy level has J = |L \u2212 S|; if it is more than half \ufb01lled, then J = L + S has the lowest energy. Use this to resolve the boron ambiguity in Problem 5.17(b). (d) Use Hund\u2019s rules, together with the fact that a symmetric spin state must go with an antisymmetric position state (and vice versa) to resolve the carbon and nitrogen ambiguities in Problem 5.17(b", "). Hint: Always go to the \u201ctop of the ladder\u201d to \ufb01gure out the symmetry of a state. 216 CHAPTER 5 Identical Particles Problem 5.19 The ground state of dysprosium (element 66, in the 6th row of the Periodic Table) is listed as 5 I8. What are the total spin, total orbital, and grand total angular momentum quantum numbers? Suggest a likely electron con\ufb01guration for dysprosium. 5.3 SOLIDS In the solid state, a few of the loosely-bound outermost valence electrons in each atom become detached, and roam around throughout the material, no longer subject only to the Coulomb \ufb01eld of a speci\ufb01c \u201cparent\u201d nucleus, but rather to the combined potential of the entire crystal lattice. In this section we will examine two extremely primitive models: \ufb01rst, the \u201celectron gas\u201d theory of Sommerfeld, which ignores all forces (except the con\ufb01ning boundaries), treating the wandering electrons as free particles in a box (the three-dimensional analog to an in\ufb01nite square well); and second, Bloch\u2019s theory, which introduces a periodic potential representing the electrical attraction of the regularly spaced, positively charged, nuclei (but still ignores electron\u2013electron repulsion). These models are no more than the \ufb01rst halting steps toward a quantum theory of solids, but already they reveal the critical role of the Pauli exclusion principle in accounting for \u201csolidity,\u201d and provide illuminating insight into the remarkable electrical properties of conductors, semi-conductors, and insulators. 5.3.1 The Free Electron Gas Suppose the object in question is a rectangular solid, with dimensions lx, ly, lz, and imagine that an electron inside experiences no forces at all, except at the impenetrable walls: V (x, y, z) = 0, \u221e, 0 < x < lx, 0 < y < ly, 0 < z < lz; otherwise. (5.46) The Schr\u00f6dinger equation, \u2212 2 2m \u22072\u03c8 = E\u03c8, separates, in Cartesian coordinates: \u03c8(x, y, z) = X (x)Y (y)Z (z), with \u2212 2 2m d2 X d x 2 = Ex X, \u2212 2 2m d2Y dy2 = E yY, \u2212 2 2m d2 Z dz2 = Ez Z, and E = Ex + E y + Ez. Letting \u221a kx \u2261 2m Ex, ky \u2261 % 2m E y, kz \u2261 \u221a 2m Ez, we obtain the general solutions X (x) = Ax sin(kx x) + Bx cos(kx x), Y (y) = Ay sin ky y + By cos ky y, Z (z) = Az sin(kz z) + Bz cos(kz z). The boundary conditions require that X (0) = Y (0) = Z (0) = 0, so Bx = By = Bz = 0, and X (lx ) = Y = Z (lz) = 0, so ly kxlx = nx \u03c0, kyly = n y\u03c0, kzlz = nz\u03c0, (5.47) where each n is a positive integer: nx = 1, 2, 3,..., n y = 1, 2, 3,..., nz = 1, 2, 3,.... (5.48) 5.3 Solids 217 (5.49) (5.50) The (normalized) wave functions are \u03c8nx n y nz = and the allowed energies are / sin nx \u03c0 lx x sin y sin, z nz\u03c0 lz n y\u03c0 ly 8 lxlylz Enx n y nz =! 2\u03c0 2 2m + n2 x l2 x \" + n2 n2 y z l2 l2 y z kx, ky, kz, = 2k2 2m,. 3\u03c0/ly where k is the magnitude of the wave vector, k \u2261, If you imagine a three-dimensional space, with axes kx, ky, kz, and planes drawn in at,..., and at kz = (\u03c0/lz), \u03c0/ly kx = (\u03c0/lx ), (2\u03c0/lx ), (3\u03c0/lx ),..., at ky = (2\u03c0/lz), (3\u03c0/lz),..., each intersection point represents a distinct (one-particle)", " stationary state (Figure 5.2). Each block in this grid, and hence also each state, occupies a volume 2\u03c0/ly \u03c0 3 lxlylz = \u03c0 3 V kz (5.51) ky kx Figure 5.2: Free electron gas. Each intersection on the grid represents a stationary state. The shaded volume is one \u201cblock,\u201d and there is one state (potentially two electrons) for every block. 218 CHAPTER 5 Identical Particles of \u201ck-space,\u201d where V \u2261 lxlylz is the volume of the object itself. Suppose our sample contains N atoms, and each atom contributes d free electrons. (In practice, N will be enormous\u2014on the order of Avogadro\u2019s number, for an object of macroscopic size\u2014whereas d is a small number\u20141, 2, or 3, typically.) If electrons were bosons (or distinguishable particles), they would all settle down to the ground state, \u03c8111.21 But electrons are in fact identical fermions, subject to the Pauli exclusion principle, so only two of them can occupy any given state. They will \ufb01ll up one octant of a sphere in k-space,22 whose radius, kF, is determined by the fact that each pair of electrons requires a volume \u03c0 3/V (Equation 5.51): Thus where 1 8 4 3 \u03c0 k3 F = N d 2. \u03c0 3 V 3\u03c1\u03c0 2 1/3, kF = \u03c1 \u2261 N d V (5.52) (5.53) is the free electron density (the number of free electrons per unit volume). The boundary separating occupied and unoccupied states, in k-space, is called the Fermi surface (hence the subscript F). The corresponding energy is the Fermi energy, E F ; for a free electron gas, E F = 2 2m 3\u03c1\u03c0 2 2/3. (5.54) The total energy of the electron gas can be calculated as follows: A shell of thickness dk (Figure 5.3) contains a volume kz dk k ky kx Figure 5.3: One octant of a spherical shell in k-space. 21 I\u2019m assuming there is no appreciable thermal excitation, or other disturbance, to lift the solid out of its collective ground state. If you like, I\u2019m talking about a \u201ccold\u201d solid, though (as you will see in Problem 5.21(c)), typical solids are still \u201ccold,\u201d in this sense, far above room temperature. 22 Because N is such a huge number, we need not worry about the distinction between the actual jagged edge of the grid and the smooth spherical surface that approximates it. 4\u03c0 k2 dk, 1 8 5.3 Solids 219 so the number of electron states in the shell is 2 (1/2) \u03c0 k2 dk \u03c0 3/V = V \u03c0 2 k2 dk. Each of these states carries an energy 2k2/2m (Equation 5.50), so the energy of the electrons in the shell is d E = 2k2 2m V \u03c0 2 k2 dk, (5.55) and hence the total energy of all the \ufb01lled states is kF Etot = k4 dk = 2V 2\u03c0 2m 2k5 F V 10\u03c0 2m This quantum mechanical energy plays a role rather analogous to the internal thermal energy (U ) of an ordinary gas. In particular, it exerts a pressure on the walls, for if the box expands by an amount d V, the total energy decreases: V \u22122/3. (5.56) = 0 2 5/3 3\u03c0 2 N d 10\u03c0 2m d Etot = \u2212 2 3 2 3\u03c0 2 N d 10\u03c0 2m 5/3 V \u22125/3 d V = \u2212 2 3 Etot d V V, and this shows up as work done on the outside (dW = P d V ) by the quantum pressure P. Evidently P = 2 3 Etot V = 2 3 2k5 F 10\u03c0 2m = 3\u03c0 2 2/3 2 5m \u03c15/3. (5.57) Here, then, is a partial answer to the question of why a cold solid object doesn\u2019t simply collapse: There is a stabilizing internal pressure, having nothing to do with electron\u2013electron repulsion (which we have ignored) or thermal motion (which we have excluded), but is strictly quantum mechanical, and derives ultimately from the antisymmetrization requirement for the wave functions of identical fermions. It is sometimes called degeneracy pressure, though \u201cexclusion pressure\u201d might be a", " better term.23 Problem 5.20 Find the average energy per free electron (Etot/N d), as a fraction of the Fermi energy. Answer: (3/5) E F. Problem 5.21 The density of copper is 8.96 g/cm3, and its atomic weight is 63.5 g/mole. (a) Calculate the Fermi energy for copper (Equation 5.54). Assume d = 1, and give your answer in electron volts. (b) What is the corresponding electron velocity? Hint: Set E F = (1/2) mv2. Is it safe to assume the electrons in copper are nonrelativistic? (c) At what temperature would the characteristic thermal energy (k B T, where k B is the Boltzmann constant and T is the Kelvin temperature) equal the Fermi energy, for copper? Comment: This is called the Fermi temperature, TF. As long as the actual temperature is substantially below the Fermi temperature, the material can be 23 We derived Equations 5.52, 5.54, 5.56, and 5.57 for the special case of an in\ufb01nite rectangular well, but they hold for containers of any shape, as long as the number of particles is extremely large. 220 CHAPTER 5 Identical Particles regarded as \u201ccold,\u201d with most of the electrons in the lowest accessible state. Since the melting point of copper is 1356 K, solid copper is always cold. (d) Calculate the degeneracy pressure (Equation 5.57) of copper, in the electron gas model. Problem 5.22 Helium-3 is fermion with spin 1/2 (unlike the more common isotope helium4 which is a boson). At low temperatures (T TF ), helium-3 can be treated as a Fermi gas (Section 5.3.1). Given a density of 82 kg/m3, calculate TF (Problem 5.21(c)) for helium-3. Problem 5.23 The bulk modulus of a substance is the ratio of a small decrease in pressure to the resulting fractional increase in volume: B = \u2212V d P d V. Show that B = (5/3) P, in the free electron gas model, and use your result in Problem 5.21(d) to estimate the bulk modulus of copper. Comment: The observed value is 13.4 \u00d7 1010 N/m2, but don\u2019t expect perfect agreement\u2014after all, we\u2019re neglecting all electron\u2013 nucleus and electron\u2013electron forces! Actually, it is rather surprising that this calculation comes as close as it does. 5.3.2 Band Structure We\u2019re now going to improve on the free electron model, by including the forces exerted on the electrons by the regularly spaced, positively charged, essentially stationary nuclei. The qualitative behavior of solids is dictated to a remarkable degree by the mere fact that this potential is periodic\u2014its actual shape is relevant only to the \ufb01ner details. To show you how it goes, I\u2019m going to develop the simplest possible model: a one-dimensional Dirac comb, consisting of evenly spaced delta-function spikes (Figure 5.4).24 But \ufb01rst I need to introduce a powerful theorem that vastly simpli\ufb01es the analysis of periodic potentials. A periodic potential is one that repeats itself after some \ufb01xed distance a: V (x + a) = V (x). (5.58) Bloch\u2019s theorem tells us that for such a potential the solutions to the Schr\u00f6dinger equation, \u2212 2 2m d2\u03c8 d x 2 + V (x)\u03c8 = E\u03c8, can be taken to satisfy the condition \u03c8(x + a) = eiqa\u03c8(x), (5.59) (5.60) 24 It would be more natural to let the delta functions go down, so as to represent the attractive force of the nuclei. But then there would be negative energy solutions as well as positive energy solutions, and that makes the calculations more cumbersome (see Problem 5.26). Since all we\u2019re trying to do here is explore the consequences of periodicity, it is simpler to adopt this less plausible shape; if it comforts you, think of the nuclei as residing at \u00b1a/2, \u00b13a/2, \u00b15a/2,.... 5.3 Solids 221 for some constant q (by \u201cconstant\u201d I mean that it is independent of x; it may well depend on E).25 In a moment we will discover that q is in fact real, so although \u03c8(x) itself is not periodic, |\u03c8(x)|2 is: |\u03c8(", "x + a)|2 = |\u03c8(x)|2, (5.61) as one would certainly expect.26 Of course, no real solid goes on forever, and the edges are going to spoil the periodicity of V (x), and render Bloch\u2019s theorem inapplicable. However, for any macroscopic crystal, containing something on the order of Avogadro\u2019s number of atoms, it is hardly imaginable that edge effects can signi\ufb01cantly in\ufb02uence the behavior of electrons deep inside. This suggests the following device to salvage Bloch\u2019s theorem: We wrap the x axis around in a circle, and connect it onto its tail, after a large number N \u2248 1023 of periods; formally, we impose the boundary condition \u03c8(x + N a) = \u03c8(x). (5.62) It follows (from Equation 5.60) that so ei N qa = 1, or N qa = 2\u03c0 n, and hence ei N qa\u03c8(x) = \u03c8(x), q = 2\u03c0 n N a, (n = 0, \u00b11, \u00b12,...). (5.63) In particular, q is necessarily real. The virtue of Bloch\u2019s theorem is that we need only solve the Schr\u00f6dinger equation within a single cell (say, on the interval 0 \u2264 x < a); recursive application of Equation 5.60 generates the solution everywhere else. Now, suppose the potential consists of a long string of delta-function spikes (the Dirac comb): V (x) = \u03b1 N \u22121 j=0 \u03b4 (x \u2212 ja). (5.64) (In Figure 5.4 you must imagine that the x axis has been \u201cwrapped around\u201d, so the N th spike actually appears at x = \u2212a.) No one would pretend that this is a realistic model, but remember, it is only the effect of periodicity that concerns us here; the classic Kronig\u2013Penney model27 V(x) \u22122a \u2212a 0 a 2a 3a 4a 5a 6a x Figure 5.4: The Dirac comb, Equation 5.64. 25 The proof of Bloch\u2019s theorem will come in Chapter 6 (see Section 6.2.2). 26 Indeed, you might be tempted to reverse the argument, starting with Equation 5.61, as a way of proving Bloch\u2019s theorem. It doesn\u2019t work, for Equation 5.61 alone would allow the phase factor in Equation 5.60 to be a function of x. 27 R. de L. Kronig and W. G. Penney, Proc. R. Soc. Lond., ser. A, 130, 499 (1930). 222 CHAPTER 5 Identical Particles used a repeating rectangular pattern, and many authors still prefer that one.28 In the region 0 < x < a the potential is zero, so \u2212 2 2m d2\u03c8 d x 2 = E\u03c8, d2\u03c8 d x 2 = \u2212k2\u03c8, \u221a k \u2261 2m E, or where as usual. The general solution is (5.65) \u03c8(x) = A sin(kx) + B cos(kx), (0 < x < a). (5.66) According to Bloch\u2019s theorem, the wave function in the cell immediately to the left of the origin is \u03c8(x) = e\u2212iqa {A sin[k (x + a)] + B cos[k (x + a)]}, (\u2212a < x < 0). (5.67) At x = 0, \u03c8 must be continuous, so B = e\u2212iqa [A sin(ka) + B cos(ka)] ; (5.68) its derivative suffers a discontinuity proportional to the strength of the delta function (Equation 2.128, with the sign of \u03b1 switched, since these are spikes instead of wells): k A \u2212 e\u2212iqak [A cos(ka) \u2212 B sin(ka)] = 2m\u03b1 2 B. Solving Equation 5.68 for A sin(ka) yields A sin(ka) = # eiqa \u2212 cos(ka) $ B. Substituting this into Equation 5.69, and cancelling k B, we \ufb01nd # $ # eiqa \u2212 cos(ka) which simpli\ufb01es to 1 \u2212 e\u2212iqa cos(ka) $ + e\u2212iqa sin2(ka) = 2m\u03b1 2k sin(ka), cos(qa) = cos(ka) + m\u03b1 2k sin(ka). (5.69) (5.70) (5.", "71) This is the fundamental result, from which all else follows.29 Equation 5.71 determines the possible values of k, and hence the allowed energies. To simplify the notation, let z \u2261 ka, and \u03b2 \u2261 m\u03b1a 2, so the right side of Equation 5.71 can be written as f (z) \u2261 cos(z) + \u03b2 sin(z) z. (5.72) (5.73) 28 See, for instance, David Park, Introduction to the Quantum Theory, 3nd edn, McGraw-Hill, New York (1992). 29 For the Kronig\u2013Penney potential (footnote 27, page 221), the formula is more complicated, but it shares the qualitative features we are about to explore. 5.3 Solids 223 ) z ( f 1 0 \u20131 z 0 \u03c0 2\u03c0 3\u03c0 4\u03c0 Figure 5.5: Graph of f (z) (Equation 5.73) for \u03b2 = 10, showing allowed bands separated by forbidden gaps. The constant \u03b2 is a dimensionless measure of the \u201cstrength\u201d of the delta function. In Figure 5.5 I have plotted f (z), for the case \u03b2 = 10. The important thing to notice is that f (z) strays outside the range (\u22121, +1), and in such regions there is no hope of solving Equation 5.71, since |cos (qa)|, of course, cannot be greater than 1. These gaps represent forbidden energies; they are separated by bands of allowed energies. Within a given band, virtually any energy is allowed, since according to Equation 5.63, qa = 2\u03c0 n/N, where N is a huge number, and n can be any integer. You might imagine drawing N horizontal lines on Figure 5.5, at values of cos (2\u03c0 n/N ) ranging from +1 (n = 0) down to \u22121 (n = N /2), and back almost to +1 (n = N \u2212 1)\u2014at this point the Bloch factor eiqa recycles, so no new solutions are generated by further increasing n. The intersection of each of these lines with f (z) yields an allowed energy. Evidently there are N states in each band, so closely spaced that for most purposes we can regard them as forming a continuum (Figure 5.6). So far, we\u2019ve only put one electron in our potential. In practice there will be N d of them, where d is again the number of \u201cfree\u201d electrons per atom. Because of the Pauli exclusion principle, only two electrons can occupy a given spatial state, so if d = 1, they will half \ufb01ll the \ufb01rst band, if d = 2 they will completely \ufb01ll the \ufb01rst band, if d = 3 they half \ufb01ll the second band, and so on. (In three dimensions, and with more realistic potentials, the band structure may be more complicated, but the existence of allowed bands, separated by forbidden gaps, persists\u2014band structure is the signature of a periodic potential.30) Now, if the topmost band is only partly \ufb01lled, it takes very little energy to excite an electron to the next allowed level, and such a material will be a conductor (a metal). On the other hand, if the top band is completely \ufb01lled, it takes a relatively large energy to excite an electron, since it has to jump across the forbidden zone. Such materials are typically insulators, though if the gap is rather narrow, and the temperature suf\ufb01ciently high, then random thermal energy can knock an electron over the hump, and the material is a semiconductor (silicon and germanium are examples).31 In the free electron model all solids should be metals, since there are no 30 Regardless of dimension, if d is an odd integer you are guaranteed to have partially-\ufb01lled bands and you would expect metallic behavior. If d is an even integer, it depends on the speci\ufb01c band structure whether there will be partially-\ufb01lled bands or not. Interestingly, some materials, called Mott insulators, are nonconductors even though d is odd. In that case it is the interactions between electrons that leads to the insulating behavior, not the presence of gaps in the single-particle energy spectrum. 31 Semiconductors typically have band gaps of 4 eV or less, small enough that thermal excitation at room temperature (kB T \u2248 0.025 eV) produces perceptible conductivity. The conductivity of a semiconductor can be controlled by 224 CHAPTER 5 Identical Particles E band gap band gap band Figure 5.6: The allowed energies for a periodic potential form essentially continuous bands. large gaps in the spectrum of allowed energies. It takes the band theory", " to account for the extraordinary range of electrical conductivities exhibited by the solids in nature. Problem 5.24 (a) Using Equations 5.66 and 5.70, show that the wave function for a particle in the periodic delta function potential can be written in the form >? \u03c8(x) = C sin(kx) + e\u2212iqa sin[k (a \u2212 x)], (0 \u2264 x \u2264 a). (Don\u2019t bother to determine the normalization constant C.) (b) At the top of a band, where z = j\u03c0, (a) yields \u03c8(x) = 0/0 (indeterminate). Find the correct wave function for this case. Note what happens to \u03c8 at each delta function. Problem 5.25 Find the energy at the bottom of the \ufb01rst allowed band, for the case \u03b2 = 10, correct to three signi\ufb01cant digits. For the sake of argument, assume \u03b1/a = 1 eV. doping: including a few atoms of larger or smaller d; this puts some \u201cextra\u201d electrons into the next higher band, or creates some holes in the previously \ufb01lled one, allowing in either case for weak electric currents to \ufb02ow. Further Problems on Chapter 5 225 \u2217\u2217 Problem 5.26 Suppose we use delta function wells, instead of spikes (i.e. switch the sign of \u03b1 in Equation 5.64). Analyze this case, constructing the analog to Figure 5.5. This requires no new calculation, for the positive energy solutions (except that \u03b2 is now negative; use \u03b2 = \u22121.5 for the graph), but you do need to work out the negative energy solutions (let \u22122m E/ and z \u2261 \u2212\u03baa, for E < 0); your graph will now extend to negative z). \u03ba \u2261 How many states are there in the \ufb01rst allowed band? \u221a Problem 5.27 Show that most of the energies determined by Equation 5.71 are doubly degenerate. What are the exceptional cases? Hint: Try it for N = 1, 2, 3, 4,..., to see how it goes. What are the possible values of cos (qa) in each case? Problem 5.28 Make a plot of E vs. q for the band structure in Section 5.3.2. Use \u03b1 = 1 (in units where m = = a = 1). Hint: In Mathematica, ContourPlot will graph E(q) as de\ufb01ned implicitly by Equation 5.71. On other platforms the plot can be obtained as follows: \u2022 Choose a large number (say 30,000) of equally-spaced values for the energy in the range E = 0 and E = 30. \u2022 For each value of E, compute the right-hand side of Equation 5.71. If the result is between \u22121 and 1, solve for q from Equation 5.71 and record the pair of values {q,E} and {-q,E} (there are two solutions for each energy). You will then have a list of pairs {{q1,E1},{q2,E2},...} which you can plot. FURTHER PROBLEMS ON CHAPTER 5 Problem 5.29 Suppose you have three particles, and three distinct one-particle states (\u03c8a(x), \u03c8b(x), and \u03c8c(x)) are available. How many different three-particle states can be constructed (a) if they are distinguishable particles, (b) if they are identical bosons, (c) if they are identical fermions? (The particles need not be in different states\u2014\u03c8a(x1)\u03c8a(x2)\u03c8a(x3) would be one possibility, if the particles are distinguishable.) Problem 5.30 Calculate the Fermi energy for electrons in a two-dimensional in\ufb01nite square well. Let \u03c3 be the number of free electrons per unit area. Problem 5.31 Repeat the analysis of Problem 2.58 to estimate the cohesive energy for a three-dimensional metal, including the effects of spin. 226 CHAPTER 5 Identical Particles Problem 5.32 Consider a free electron gas (Section 5.3.1) with unequal numbers of spin-up and spin-down particles (N+ and N\u2212 respectively). Such a gas would have a net magnetization (magnetic dipole moment per unit volume) M = \u2212 (N+ \u2212 N\u2212) V \u03bcB \u02c6k = M \u02c6k, (5.74) where \u03bcB = e /2 me is the Bohr magneton. (The minus sign is there, of course, because the charge of the electron is negative.) (", "a) Assuming that the electrons occupy the lowest energy levels consistent with the number of particles in each spin orientation, \ufb01nd Etot. Check that your answer reduces to Equation 5.56 when N+ = N\u2212. (b) Show that for M/\u03bcB \u03c1 \u2261 (N+ + N\u2212) /V (which is to say, |N+ \u2212 N\u2212| (N+ + N\u2212)), the energy density is 2 Etot = 1 V 3 \u03c0 2 \u03c1 10 \u03c0 2 m 5/3 1 + 5 9 M \u03c1 \u03bcB 2. The energy is a minimum for M = 0, so the ground state will have zero magnetization. However, if the gas is placed in a magnetic \ufb01eld (or in the presence of interactions between the particles) it may be energetically favorable for the gas to magnetize. This is explored in Problems 5.33 and 5.34. Problem 5.33 Pauli paramagnetism. If the free electron gas (Section 5.3.1) is placed in a uniform magnetic \ufb01eld B = B \u02c6k, the energies of the spin-up and spin-down states will be different:32! \" E \u00b1 nx n y nz = 2 \u03c0 2 2 m n2 x l2 x + n2 y l2 y + n2 z l2 z \u00b1 \u03bcB B. There will be more spin-down states occupied than spin-up states (since they are lower in energy), and consequently the system will acquire a magnetization (see Problem 5.32). (a) In the approximation that M/\u03bcB \u03c1, \ufb01nd the magnetization that minimizes the total energy. Hint: Use the result of Problem 5.32(b). (b) The magnetic susceptibility is33 \u03c7 = \u03bc0 d M d B. Calculate the magnetic susceptibility for aluminum compare the experimental value34 of 22 \u00d7 10\u22126. \u03c1 = 18.1 \u00d7 1022 cm\u22123 and 32 Here we are considering only the coupling of the spin to the magnetic \ufb01eld, and ignoring any coupling of the orbital motion. 33 Strictly speaking, the susceptibility is d M/d H, but the difference is negligible when, as here, \u03c7 1. 34 For some metals, such as copper, the agreement is not so good\u2014even the sign is wrong: copper is diamagnetic (\u03c7 < 0). The explanation for this discrepancy lies in what has been left out of our model. In addition to the Further Problems on Chapter 5 227 Problem 5.34 The Stoner criterion. The free-electron gas model (Section 5.3.1) ignores the Coulomb repulsion between electrons. Because of the exchange force (Section 5.1.2), Coulomb repulsion has a stronger effect on two electrons with antiparallel spins (which behave in a way like distinguishable particles) than two electrons with parallel spins (whose position wave function must be antisymmetric). As a crude way to take account of Coulomb repulsion, pretend that every pair of electrons with opposite spin carries extra energy U, while electrons with the same spin do not interact at all; this adds E = U N+ N\u2212 to the total energy of the electron gas. As you will show, above a critical value of U, it becomes energetically favorable for the gas to spontaneously magnetize (N+ = N\u2212); the material becomes ferromagnetic. (a) Rewrite E in terms of the density \u03c1 and the magnetization M (Equation 5.74). (b) Assuming that M/\u03bcB \u03c1, for what minimum value of U is a non-zero magnetization energetically favored? Hint: Use the result of Problem 5.32(b). \u2217\u2217\u2217 Problem 5.35 Certain cold stars (called white dwarfs) are stabilized against gravitational collapse by the degeneracy pressure of their electrons (Equation 5.57). Assuming constant density, the radius R of such an object can be calculated as follows: (a) Write the total electron energy (Equation 5.56) in terms of the radius, the number of nucleons (protons and neutrons) N, the number of electrons per nucleon d, and the mass of the electron m. Beware: In this problem we are recycling the letters N and d for a slightly different purpose than in the text. (b) Look up, or calculate, the gravitational energy of a uniformly dense sphere. Express your answer in terms of G (the constant of universal gravitation), R, N, and M (the mass of a nucleon). Note that the gravitational energy is negative. (c) Find the radius for which the total energy, (a) plus (b), is a minimum. Answer: R = 9\u03c0 4 2/3 2d", "5/3 Gm M 2 N 1/3. (Note that the radius decreases as the total mass increases!) Put in the actual numbers, for everything except N, using d = 1/2 (actually, d decreases a bit as the atomic number increases, but this is close enough for our purposes). Answer: R = 7.6 \u00d7 1025 N \u22121/3 m. (d) Determine the radius, in kilometers, of a white dwarf with the mass of the sun. (e) Determine the Fermi energy, in electron volts, for the white dwarf in (d), and compare it with the rest energy of an electron. Note that this system is getting dangerously relativistic (see Problem 5.36). paramagnetic coupling of the spin magnetic moment to an applied \ufb01eld there is a coupling of the orbital magnetic moment to an applied \ufb01eld and this has both paramagnetic and diamagnetic contributions (see Problem 4.72). In addition, the free electron gas model ignores the tightly-bound core electrons and these also couple to the magnetic \ufb01eld. In the case of copper, it is the diamagnetic coupling of the core electrons that dominates. 228 CHAPTER 5 Identical Particles \u2217\u2217\u2217 Problem 5.36 We can extend the theory of a free electron gas (Section 5.3.1) to the relativistic domain by replacing the classical kinetic energy, E = p2/2m, with the p2c2 + m2c4 \u2212 mc2. Momentum is related to the wave relativistic formula, E = vector in the usual way: p = k. In particular, in the extreme relativistic limit, E \u2248 pc = ck. (a) Replace 2k2/2m in Equation 5.55 by the ultra-relativistic expression, ck, and % calculate Etot in this regime. (b) Repeat parts (a) and (b) of Problem 5.35 for the ultra-relativistic electron gas. Notice that in this case there is no stable minimum, regardless of R; if the total energy is positive, degeneracy forces exceed gravitational forces, and the star will expand, whereas if the total is negative, gravitational forces win out, and the star will collapse. Find the critical number of nucleons, Nc, such that gravitational collapse occurs for N > Nc. This is called the Chandrasekhar limit. Answer: 2.04 \u00d7 1057. What is the corresponding stellar mass (give your answer as a multiple of the sun\u2019s mass). Stars heavier than this will not form white dwarfs, but collapse further, becoming (if conditions are right) neutron stars. (c) At extremely high density, inverse beta decay, e\u2212+ p+ \u2192 n+\u03bd, converts virtually all of the protons and electrons into neutrons (liberating neutrinos, which carry off energy, in the process). Eventually neutron degeneracy pressure stabilizes the collapse, just as electron degeneracy does for the white dwarf (see Problem 5.35). Calculate the radius of a neutron star with the mass of the sun. Also calculate the (neutron) Fermi energy, and compare it to the rest energy of a neutron. Is it reasonable to treat a neutron star nonrelativistically? Problem 5.37 An important quantity in many calculations is the density of states G(E): G(E) d E \u2261 number of states with energy between E and E + d E. For a one-dimensional band structure, G(E) d E = 2, dq 2 \u03c0/N a where dq/(2 \u03c0/N a) counts the number of states in the range dq (see Equation 5.63), and the factor of 2 accounts for the fact that states with q and \u2212q have the same energy. Therefore 1 N a G(E) = 1 \u03c0 1 |d E/dq|. (a) Show that for \u03b1 = 0 (a free particle) the density of states is given by 1 N a Gfree(E) = 1 \u03c0 m 2E. (b) Find the density of states for \u03b1 = 0 by differentiating Equation 5.71 with respect to q to determine d E/dq. Note: Your answer should be written as a function of E only (well, and \u03b1, m,, a, and N ) and must not contain q (use k as a shorthand for \u221a 2m E/, if you like). Further Problems on Chapter 5 229 (c) Make a single plot showing G(E) /N a for both \u03b1 = 0 and \u03b1 = 1 (in units where m = = a = 1). Comment: The divergences at the band edges are examples of van Hove singularities.35 \u2217\u2217\u2217 Problem 5.", "38 The harmonic chain consists of N equal masses arranged along a line and connected to their neighbors by identical springs: \u02c6H = \u2212 N 2 2 m j=1 \u2202 \u2202 x 2 j + N j=1 1 2 m \u03c92 x j+1 \u2212 x j 2, where x j is the displacement of the jth mass from its equilibrium position. This system (and its extension to two or three dimensions\u2014the harmonic crystal) can be used to model the vibrations of a solid. For simplicity we will use periodic boundary conditions: x N +1 = x1, and introduce the ladder operators36 / N e\u00b1 i 2 \u03c0 j k/N \u02c6ak\u00b1 \u2261 1\u221a N m \u03c9k 2 x j \u2213 2 m \u03c9k \u2202 \u2202 x j (5.75) j=1 where k = 1,..., N \u2212 1 and the frequencies are given by \u03c9k = 2 \u03c9 sin \u03c0 k N. (a) Prove that, for integers k and k between 1 and N \u2212 1, 1 N 1 N N j=1 N j=1 ei 2 \u03c0 j(k\u2212k)/N = \u03b4k,k ei 2 \u03c0 j(k+k)/N = \u03b4k,N \u2212k. Hint: Sum the geometric series. (b) Derive the commutation relations for the ladder operators: \u02c6ak\u2212, \u02c6ak+ = \u03b4k,k and \u02c6ak\u2212, \u02c6ak\u2212 = \u02c6ak+, \u02c6ak+ = 0. (5.76) 35 These one-dimensional Van Hove singularities have been observed in the spectroscopy of carbon nanotubes; see J. W. G. Wild\u00f6er et al., Nature, 391, 59 (1998). 36 If you are familiar with the classical problem of coupled oscillators, these ladder operators are straightforward to construct. Start with the normal mode coordinates you would use to decouple the classical problem, namely qk = 1\u221a N N j=1 \u2212i 2 \u03c0 j k/N x j. e The frequencies \u03c9k are the classical normal mode frequencies, and you simply create a pair of ladder operators for each normal mode, by analogy with the single-particle case (Equation 2.48). 230 CHAPTER 5 Identical Particles (c) Using Equation 5.75, show that x j = R + 1\u221a N N \u22121 k=1 / \u02c6ak\u2212 + \u02c6aN \u2212k+ ei 2 \u03c0 j k/N 2 m \u03c9k N \u22121 \u2202 \u2202 x j where \u221a N m \u03c9k 2 \u02c6ak\u2212 \u2212 \u02c6aN \u2212k+ ei 2 \u03c0 j k/N k=1 j x j /N is the center of mass coordinate. (d) Finally, show that \u02c6H = \u2212 2 2 (N m) \u2202 2 \u2202 R2 + \u02c6ak+ \u02c6ak\u2212 + 1 2. \u03c9k N \u22121 k=1 Comment: Written in this form above, the Hamiltonian describes N \u2212 1 independent oscillators with frequencies \u03c9k (as well as a center of mass that moves as a free particle of mass N m). We can immediately write down the allowed energies: E = \u2212 2 K 2 2 (N m) N \u22121 + \u03c9k nk + 1 2 k=1 where K is the momentum of the center of mass and nk = 0, 1,... is the energy level of the kth mode of vibration. It is conventional to call nk the number of phonons in the kth mode. Phonons are the quanta of sound (atomic vibrations), just as photons are the quanta of light. The ladder operators ak+ and ak\u2212 are called phonon creation and annihilation operators since they increase or decrease the number of phonons in the kth mode. Problem 5.39 In Section 5.3.1 we put the electrons in a box with impenetrable walls. The same results can be obtained using periodic boundary conditions. We still imagine the electrons to be con\ufb01ned to a box with sides of length lx, ly, and lz but instead of requiring the wave function to vanish on each wall, we require it to take the same value on opposite walls: \u03c8(x, y, z) = \u03c8(x + lx, y, z) = \u03c8 x, y + ly, z = \u03c8(x, y, z + lz). In this case we can represent the wave functions as traveling waves, ei(kx x+ky y+kz z), ei k\u00b7r = 1%", " 1% \u03c8 = lx ly lz lx ly lz rather than as standing waves (Equation 5.49). Periodic boundary conditions\u2014while certainly not physical\u2014are often easier to work with (to describe something like electrical current a basis of traveling waves is more natural than a basis of standing waves) and if you are computing bulk properties of a material it shouldn\u2019t matter which you use. Further Problems on Chapter 5 231 (a) Show that with periodic boundary conditions the wave vector satis\ufb01es kx lx = 2 nx \u03c0, ky ly = 2 n y \u03c0, kz lz = 2 nz \u03c0, where each n is an integer (not necessarily positive). What is the k-space volume occupied by each block on the grid (corresponding to Equation 5.51)? (b) Compute kF, E F, and Etot for the free electron gas with periodic boundary conditions. What compensates for the larger volume occupied by each k-space block (part (a)) to make these all come out the same as in Section 5.3.1? 6 SYMMETRIES & CONSERVATION LAWS 6.1 INTRODUCTION Conservation laws (energy, momentum, and angular momentum) are familiar from your \ufb01rst course in classical mechanics. These same conservation laws hold in quantum mechanics; in both contexts they are the result of symmetries. In this chapter we will explain what a symmetry is and what it means for something to be conserved in quantum mechanics\u2014and show how the two are related. Along the way we\u2019ll investigate two related properties of quantum systems\u2014energy level degeneracy and the selection rules that distinguish allowed from \u201cforbidden\u201d transitions. What is a symmetry? It is some transformation that leaves the system unchanged. As an example consider rotating a square piece of paper, as shown in Figure 6.1. If you rotate it by 30\u25e6 about an axis through its center it will be in a different orientation than the one it started in, but if you rotate it by 90\u25e6 it will resume its original orientation; you wouldn\u2019t even know it had been rotated unless (say) you wrote numbers on the corners (in which case they would be permuted). A square therefore has a discrete rotational symmetry: a rotation by n\u03c0/2 for any integer n leaves it unchanged.1 If you repeated this experiment with a circular piece of paper, a rotation by any angle would leave it unchanged; the circle has continuous rotational symmetry. We will see that both discrete and continuous symmetries are important in quantum mechanics. Now imagine that the shapes in Figure 6.1 refer not to pieces of paper, but to the boundaries of a two-dimensional in\ufb01nite square well. In that case the potential energy would have the same rotational symmetries as the piece of paper and (because the kinetic energy is unchanged by a rotation) the Hamiltonian would also be invariant. In quantum mechanics, when we say that a system has a symmetry, this is what we mean: that the Hamiltonian is unchanged by some transformation, such as a rotation or a translation. 6.1.1 Transformations in Space In this section, we introduce the quantum mechanical operators that implement translations, inversions, and rotations. We de\ufb01ne each of these operators by how it acts on an arbitrary function. The translation operator takes a function and shifts it a distance a. The operator that accomplishes this is de\ufb01ned by the relation 1 A square of course has other symmetries as well, namely mirror symmetries about axes along a diagonal or bisecting two sides. The set of all transformations that leave the square unchanged is called D4, the \u201cdihedral group\u201d of degree 4. 6.1 Introduction 233 \u03c0 2 \u03b1 Figure 6.1: A square has a discrete rotational symmetry; it is unchanged when rotated by \u03c0/2 or multiples thereof. A circle has continuous rotational symmetry; it is unchanged when rotated by any angle \u03b1. \u03c8 (x) \u03c8 \u2032(x) x\u2013a x Figure 6.2: A wave function \u03c8(x) and the translated wave function \u03c8 (x) = \u02c6T (a) \u03c8(x). Note that the value of \u03c8 at x is equal to the value of \u03c8 at x \u2212 a. \u02c6T (a) \u03c8(x) = \u03c8 (x) = \u03c8(x \u2212 a). (6.1) The sign can be confusing at \ufb01rst; this equation says that the translated function \u03c8 at x is equal to the untranslated function \u03c8 at x \u2212 a (Figure 6.2)\u2014the function itself has been shifted to the right by an amount a. The operator", " that re\ufb02ects a function about the origin, the parity operator in one dimension, is de\ufb01ned by \u02c6 \u03c8(x) = \u03c8 (x) = \u03c8(\u2212x). The effect of parity is shown graphically in Figure 6.3. In three dimensions parity changes the sign of all three coordinates: \u02c6 \u03c8(x, y, z) = \u03c8(\u2212x, \u2212y, \u2212z).2 Finally, the operator that rotates a function about the z axis through an angle \u03d5 is most naturally expressed in polar coordinates as \u02c6Rz(\u03d5) \u03c8(r, \u03b8, \u03c6) = \u03c8 (r, \u03b8, \u03c6) = \u03c8(r, \u03b8, \u03c6 \u2212 \u03d5). (6.2) 2 The parity operation in three dimensions can be realized as a mirror re\ufb02ection followed by a rotation (see Probrotation. We lem 6.1). In two dimensions, the transformation \u03c8 (x, y) = \u03c8(\u2212x, \u2212y) is no different from a 180 \u25e6 will use the term parity exclusively for spatial inversion, \u02c6 \u03c8(r) = \u03c8(\u2212r), in one or three dimensions. 234 CHAPTER 6 Symmetries & Conservation Laws \u03c8 \u2032 (x) \u03c8 (x) \u2013x x Figure 6.3: A function \u03c8(x) and the function \u03c8 (x) = \u02c6 \u03c8(x) = \u03c8(\u2212x) after a spatial inversion. The value of \u03c8 at x is equal to the value of \u03c8 at \u2212x. \u03c8 (r,\u03c6) \u03c8 \u2032(r,\u03c6) \u03d5 Figure 6.4: A function \u03c8(r, \u03c6) and the rotated function \u03c8 (r, \u03c6) = \u03c8(r, \u03c6 \u2212 \u03d5) after a counter-clockwise rotation about the vertical axis by an angle \u03d5. When we take up the study of rotations in Section 6.5, we will introduce expressions for rotations about arbitrary axes. The action of the rotation operator on a function \u03c8 is illustrated in Figure 6.4. \u2217 Problem 6.1 Consider the parity operator in three dimensions. (a) Show that \u02c6 \u03c8(r) = \u03c8 (r) = \u03c8(\u2212r) is equivalent to a mirror re\ufb02ection followed by a rotation. (b) Show that, for \u03c8 expressed in polar coordinates, the action of the parity operator is \u02c6 \u03c8(r, \u03b8, \u03c6) = \u03c8(r, \u03c0 \u2212 \u03b8, \u03c6 + \u03c0 ). (c) Show that for the hydrogenic orbitals, \u02c6 \u03c8nm(r, \u03b8, \u03c6) = (\u22121) \u03c8nm(r, \u03b8, \u03c6). That is, \u03c8nm is an eigenstate of the parity operator, with eigenvalue (\u22121). Note: This result actually applies to the stationary states of any central potential V (r) = V (r ). For a central potential, the eigenstates may be written in the sepa (\u03b8, \u03c6) where only the radial function Rn\u2014which plays no rable form Rn (r ) Y m role in determining the parity of the state\u2014depends on the speci\ufb01c functional form of V (r ). 6.2 The Translation Operator 235 6.2 THE TRANSLATION OPERATOR Equation 6.1 de\ufb01nes the translation operator. We can express \u02c6T (a) in terms of the momentum operator, to which it is intimately related. To that end, we replace \u03c8(x \u2212 a) by its Taylor series3 \u02c6T (a) \u03c8(x) = \u03c8(x \u2212 a) = = \u221e n=0 1 n! \u2212i a (\u2212a)n dn d x n \u03c8(x) \u221e 1 n! n=0 n \u02c6p \u03c8(x). The right-hand side of this equation is the exponential function,4 so \u02c6T (a) = exp \u2212 i a \u02c6p. We say that momentum is the \u201cgenerator\u201d of translations.5 Note that \u02c6T (a) is a unitary operator:6 \u02c6T (a)\u22121 = \u02c6T (\u2212a) = \u02c6T (a)\u2020. (6.3) (6.4) The \ufb01rst equality is obvious physically (the inverse operation of shifting something to the right is shifting it by an equal amount to", " the left), and the second equality then follows from taking the adjoint of Equation 6.3 (see Problem 6.2). \u2217 Problem 6.2 Show that, for a Hermitian operator \u02c6Q, the operator \u02c6U = exp Hint: First you need to prove that the adjoint is given by \u02c6U \u2020 = exp that \u02c6U \u2020 \u02c6U = 1. Problem 3.5 may help. # \u2212i \u02c6Q # i \u02c6Q $ $ is unitary. ; then prove 6.2.1 How Operators Transform So far I have shown how to translate a function; this has an obvious graphical interpretation via Figure 6.2. We can also consider what it means to translate an operator. The transformed operator \u02c6Q is de\ufb01ned to be the operator that gives the same expectation value in the untranslated state \u03c8 as does the operator \u02c6Q in the translated state. \u02c6Q \u02c6Q 3 I\u2019m assuming that our function has a Taylor series expansion, but the \ufb01nal result applies more generally. See Problem 6.31 for the details. 4 See Section 3.6.2 for the de\ufb01nition of the exponential of an operator. 5 The term comes from the study of Lie groups (the group of translations is an example). If you\u2019re interested, an introduction to Lie groups (written for physicists) can be found in George B. Arfken, Hans J. Weber, and Frank E. Harris, Mathematical Methods for Physicists, 7th edn, Academic Press, New York (2013), Section 17.7. 6 Unitary operators are discussed in Problem A.30. A unitary operator is one whose adjoint is also its inverse: \u02c6U \u02c6U \u2020 = \u02c6U \u2020 \u02c6U = 1. 236 CHAPTER 6 Symmetries & Conservation Laws There are two ways to calculate the effect of a translation on an expectation value. One could actually shift the wave function over some distance (this is called an active transformation) or one could leave the wave function where it was and shift the origin of our coordinate system by the same amount in the opposite direction (a passive transformation). The operator \u02c6Q is the operator in this shifted coordinate system. Using Equation 6.1, 0 \u03c8 \u02c6T \u2020 \u02c6Q \u02c6T 1 \u03c8 0 = \u03c8 \u02c6Q 1 \u03c8. (6.5) Here I am using the fact that the adjoint of an operator is de\ufb01ned such that, if \u02c6T | f \u2261 |T f, then T f | = f | \u02c6T \u2020 (see Problem 3.5). Because Equation 6.5 is to hold for all \u03c8, it follows that \u02c6Q = \u02c6T \u2020 \u02c6Q \u02c6T. (6.6) The transformed operator for the case \u02c6Q = \u02c6x is worked out in Example 6.1. Figure 6.5 illustrates the equivalence of the two ways of carrying out the transformation. Example 6.1 Find the operator \u02c6x obtained by applying a translation through a distance a to the operator \u02c6x. That is, what is the action of \u02c6x, as de\ufb01ned by Equation 6.6, on an arbitrary f(x)? Solution: Using the de\ufb01nition of \u02c6x (Equation 6.6) and a test function f(x) we have \u02c6x f(x) = \u02c6T \u2020(a) \u02c6x \u02c6T (a) f(x), and since \u02c6T \u2020(a) = \u02c6T (\u2212a) (Equation 6.4), \u02c6x f(x) = \u02c6T (\u2212a) \u02c6x \u02c6T (a) f(x). From Equation 6.1 and from Equation 6.1 again, \u02c6T (\u2212a) = (x + a) f(x), so \u02c6x f(x) = \u02c6T (\u2212a) x f (x \u2212 a), x f (x \u2212 a) \u02c6x f(x) = (x + a) f(x). Finally we may read off the operator \u02c6x = \u02c6x + a. As expected, Equation 6.7 corresponds to shifting the origin of our coordinates to the left by a so that positions in these transformed coordinates are greater by a than in the untransformed coordinates. (6.7) In Problem 6.3 you will apply a translation to", " the momentum operator to show that \u02c6p = \u02c6T \u2020 \u02c6p \u02c6T = \u02c6p: the momentum operator is unchanged by this transformation. Physically, this is because the particle\u2019s momentum is independent of where you place the origin of your coordinates, depending only on differences in position: p = m d x/dt. Once you know how the position and momentum operators behave under a translation, you know how any operator does, since 6.2 The Translation Operator 237 (a) (b) (c) b b a a+b a a+b b x x x\u2032 Figure 6.5: Active vs. passive transformations: (a) depicts the original function, (b) illustrates an active transformation in which the function is shifted to the right by an amount a, and (c) illustrates a passive transformation where the axes are shifted to the left by an amount a. A point on the wave a distance b from the origin before the transformation is a distance a + b from the origin after the transformation in either (b) or (c); this is the equivalence of the two pictures. \u02c6Q \u02c6x, \u02c6p = \u02c6T \u2020 \u02c6Q \u02c6x, \u02c6p \u02c6T = \u02c6Q \u02c6x, \u02c6p = \u02c6Q \u02c6x + a, \u02c6p. (6.8) Problem 6.4 will walk you through the proof. Problem 6.3 Show that the operator \u02c6p obtained by applying a translation to the operator \u02c6p is \u02c6p = \u02c6T \u2020 \u02c6p \u02c6T = \u02c6p. 238 CHAPTER 6 Symmetries & Conservation Laws Problem 6.4 Prove Equation 6.8. You may assume that Q series \u02c6Q \u02c6x, \u02c6p = \u221e \u221e amn \u02c6x m \u02c6pn \u02c6x, \u02c6p can be written in a power for some constants amn. m=0 n=0 6.2.2 Translational Symmetry So far we have seen how a function behaves under a translation and how an operator behaves under a translation. I am now in a position to make precise the notion of a symmetry that I mentioned in the introduction. A system is translationally invariant (equivalent to saying it has translational symmetry) if the Hamiltonian is unchanged by the transformation: \u02c6H = \u02c6T \u2020 \u02c6H \u02c6T = \u02c6H. Because \u02c6T is unitary (Equation 6.4) we can multiply both sides of this equation by \u02c6T to get \u02c6H \u02c6T = \u02c6T \u02c6H. Therefore, a system has translational symmetry if the Hamiltonian commutes with the translation operator: # $ \u02c6H, \u02c6T = 0. (6.9) For a particle of mass m moving in a one-dimensional potential, the Hamiltonian is According to Equation 6.8, the transformed Hamiltonian is \u02c6H = \u02c6p2 2 m + V (x). \u02c6H = \u02c6p2 2 m + V (x + a) so translational symmetry implies that V (x + a) = V (x). (6.10) Now, there are two very different physical settings where Equation 6.10 might arise. The \ufb01rst is a constant potential, where Equation 6.10 holds for every value of a; such a system is said to have continuous translational symmetry. The second is a periodic potential, such as an electron might encounter in a crystal, where Equation 6.10 holds only for a discrete set of as; such a system is said to have discrete translational symmetry. The two cases are illustrated in Figure 6.6. DISCRETE TRANSLATIONAL SYMMETRY AND BLOCH\u2019S THEOREM What are the implications of translational symmetry? For a system with a discrete translational symmetry, the most important consequence is Bloch\u2019s theorem; the theorem speci\ufb01es the form taken by the stationary states. We used this theorem in Section 5.3.2; I will now prove it. 6.2 The Translation Operator 239 V (x) V (x) a x x Figure 6.6: Potentials for a system with continuous (top) and discrete (bottom) translational symmetry. In the former case the potential is the same when shifted right or left by any amount; in the latter case the potential is the same when shifted right or left by an integer multiple of a. In Section A.5", " it is shown that if two operators commute, then they have a complete set of simultaneous eigenstates. This means that if the Hamiltonian is translationally invariant (which is to say, if it commutes with the translation operator), then the eigenstates \u03c8(x) of the Hamiltonian can be chosen to be simultaneously eigenstates of \u02c6T : \u02c6H \u03c8(x) = E \u03c8(x), \u02c6T (a) \u03c8(x) = \u03bb \u03c8(x), where \u03bb is the eigenvalue associated with \u02c6T. Since \u02c6T is unitary, its eigenvalues have magnitude 1 (see Problem A.30), which means that \u03bb can be written as \u03bb = ei \u03c6 for some real number \u03c6. By convention we write \u03c6 = \u2212qa where q is called the crystal momentum. Therefore, the stationary states of a particle of mass m moving in a periodic potential have the property \u03c8(x \u2212 a) = e\u2212i qa \u03c8(x). (6.11) 240 CHAPTER 6 Symmetries & Conservation Laws There is a more illuminating way to write Equation 6.11:7 \u03c8(x) = ei q x u(x) (6.12) where u(x) is a periodic function of x: u(x + a) = u(x) and ei q x is a traveling wave (recall that a traveling wave by itself describes a free particle\u2014Section 2.4) with wavelength 2 \u03c0 /q. Equation 6.12 is Bloch\u2019s theorem and it says that the stationary states of a particle in a periodic potential are periodic functions multiplying traveling waves. Note that just because the Hamiltonian is translationally invariant, that doesn\u2019t mean the stationary states themselves are translationally invariant, it simply means that they can be chosen to be eigenstates of the translation operator. Bloch\u2019s theorem is truly remarkable. It tells us that the stationary states of a particle in a periodic potential (such as an electron in a crystal) are, apart from a periodic modulation, traveling waves. As such, they have a nonzero velocity.8 This means that an electron could travel through a perfect crystal without scattering! That has dramatic implications for electronic conduction in solids. CONTINUOUS TRANSLATIONAL SYMMETRY AND MOMENTUM CONSERVATION If a system has continuous translation symmetry then the Hamiltonian commutes with \u02c6T (a) for any choice of a. In this case it is useful to consider an in\ufb01nitesimal translation \u02c6T (\u03b4) = e\u2212i \u03b4 \u02c6p/ \u2248 1 \u2212 i \u03b4 \u02c6p, where \u03b4 is an in\ufb01nitesimal length.9 If the Hamiltonian has continuous translational symmetry, then it must be unchanged under any translation, including an in\ufb01nitesimal one; equivalently it commutes with the translation operator, and hence # $ \u02c6H, \u02c6T (\u03b4) = \u02c6H, 1 \u2212 i \u03b4 \u02c6p = 0 \u21d2 # $ \u02c6H, \u02c6p = 0. So if the Hamiltonian has continuous translational symmetry, it must commute with the momentum operator. And if the Hamiltonian commutes with momentum, then according to the \u201cgeneralized Ehrenfest\u2019s theorem\u201d (Equation 3.73) $1 0# d dt p = i \u02c6H, \u02c6p = 0. (6.13) 7 It is clear that Equation 6.12 satis\ufb01es Equation 6.11. In Problem 6.5 you\u2019ll prove that they are in fact equivalent statements. 8 For a delightful proof using perturbation theory, see Neil Ashcroft and N. David Mermin, Solid State Physics, Cengage, Belmont, 1976 (p. 765), after you have completed Problem 6.6 and studied Chapter 7. 9 For the case of continuous symmetries, it is often much easier to work with the in\ufb01nitesimal form of the transformation; any \ufb01nite transformation can then be built up as a product of in\ufb01nitesimal transformations. In particular, the \ufb01nite translation by a is a sequence of N in\ufb01nitesimal translations with \u03b4 = a/N in the limit that N \u2192 \u221e: lim N \u2192\u221e 1 \u2212 i N a N 1 \u02c6p \u2212 i a \u02c6p. = exp For a proof see R. Shankar, Basic Training in Mathematics: A Fitness Program for Science Students, Plenum Press, New York", ", 1995 (p.11). 6.2 The Translation Operator 241 This is a statement of momentum conservation and we have now shown that continuous translational symmetry implies that momentum is conserved. This is our \ufb01rst example of a powerful general principle: symmetries imply conservation laws.10 Of course, if we\u2019re talking about a single particle of mass m moving in a potential V (x), the only potential that has continuous translational symmetry is the constant potential, which is equivalent to the free particle. And it is pretty obvious that momentum is conserved in that case. But the analysis here readily extends to a system of interacting particles (see Problem 6.7). The fact that momentum is conserved in that case as well (so long as the Hamiltonian is translationally invariant) is a highly nontrivial result. In any event, the point to remember is that conservation of momentum is a consequence of translational symmetry. Problem 6.5 Show that Equation 6.12 follows from Equation 6.11. Hint: First write \u03c8(x) = ei q x u(x), which is certainly true for some u(x), and then show that u(x) is necessarily a periodic function of x. \u2217\u2217\u2217 Problem 6.6 Consider a particle of mass m moving in a potential V (x) with period a. We know from Bloch\u2019s theorem that the wave function can be written in the form of Equation 6.12. Note: It is conventional to label the states with quantum numbers n and q as \u03c8nq (x) = e\u2212i q x unq (x) where Enq is the nth energy for a given value of q. (a) Show that u satis\ufb01es the equation \u2212 2 2 m d2unq d x 2 \u2212 i2 q m dunq d x + V (x) unq = Enq \u2212 2 q2 2 m unq. (6.14) (b) Use the technique from Problem 2.61 to solve the differential equation for unq. You need to use a two-sided difference for the \ufb01rst derivative so that you have a Hermitian matrix to diagonalize: d\u03c8. For the potential in the interval d x 0 to a let \u2248 \u03c8 j+1\u2212\u03c8 j\u22121 2x V (x) = \u2212V0 0 a/4 < x < 3 a/4 else with V0 = 20 2/2 m a2. (You will need to modify the technique slightly to account for the fact that the function unq is periodic.) Find the lowest two energies for the following values of the crystal momentum: q a = \u2212\u03c0, \u2212\u03c0/2, 0, \u03c0/2, \u03c0. Note that q and q + 2 \u03c0/a describe the same wave function (Equation 6.12), so there is no reason to consider values of q a outside of the interval from \u2212\u03c0 to \u03c0. In solid state physics, the values of q inside this range constitute the \ufb01rst Brillouin zone. (c) Make a plot of the energies E1q and E2q for values of q between \u2212\u03c0/a and \u03c0/a. If you\u2019ve automated the code that you used in part (b), you should be able to show a large number of q values in this range. If not, simply plot the values that you computed in (b). 10 In the case of a discrete translational symmetry, momentum is not conserved, but there is a conserved quantity closely related to the discrete translational symmetry, which is the crystal momentum. For a discussion of crystal momentum see Steven H. Simon, The Oxford Solid State Basics, Oxford, 2013, p.84. 242 CHAPTER 6 Symmetries & Conservation Laws Problem 6.7 Consider two particles of mass m1 and m2 (in one dimension) that interact via a potential that depends only on the distance between the particles V (|x1 \u2212 x2|), so that the Hamiltonian is \u2202 2 \u2202 x 2 1 Acting on a two-particle wave function the translation operator would be + V (|x1 \u2212 x2|). \u2202 2 \u2202 x 2 2 \u02c6H = \u2212 2 m2 2 m1 2 2 \u2212 \u02c6T (a) \u03c8(x1, x2) = \u03c8(x1 \u2212 a, x2 \u2212 a). (a) Show that the translation operator can be written \u02c6P, \u02c6T (a) = e\u2212 i a where \u02c6P = \u02c6p1 + \u02c6p2 is the total momentum. (b) Show that the total momentum is conserved for this system. 6.3 CONSERVATION LAWS In", " classical mechanics the meaning of a conservation law is straightforward: the quantity in question is the same before and after some event. Drop a rock, and potential energy is converted into kinetic energy, but the total is the same just before it hits the ground as when it was released; collide two billiard balls and momentum is transferred from one to the other, but the total remains unchanged. But in quantum mechanics a system does not in general have a de\ufb01nite energy (or momentum) before the process begins (or afterward). What does it mean, in that case, to say that the observable Q is (or is not) conserved? Here are two possibilities: \u2022 First de\ufb01nition: The expectation value Q is independent of time. \u2022 Second de\ufb01nition: The probability of getting any particular value is independent of time. Under what conditions does each of these conservation laws hold? Let us stipulate that the observable in question does not depend explicitly on time: \u2202 Q/\u2202t = 0. In that case the generalized Ehrenfest theorem (Equation 3.73) tells us that the \u02c6Q commutes with the expectation value of Q is independent of time if The operator Hamiltonian. It so happens that the same criterion guaranatees conservation by the second de\ufb01nition. I will now prove this result. Recall that the probability of getting the result qn in a measurement of Q at time t is (Equation 3.43) where fn is the corresponding eigenvector: evolution of the wave function is (Equation 2.17) P(qn) = | fn | (t)|2, (6.15) \u02c6Q | fn = qn | fn.11 We know that the time 11 If the spectrum of (i) n for i = 1, 2,... qn | f \u02c6Q is degenerate there are distinct eigenvectors with the same eigenvalue qn : \u02c6Q | f (i) n =, then we need to sum over those states: P(qn ) = 0 (i) n f (t) 1 2. i Except for the sum over i the proof proceeds unchanged. |(t) = m e\u2212i Em t/ cm |\u03c8m, 6.4 Parity 243 2. 2 where the |\u03c8n are the eigenstates of \u02c6H, and therefore P(qn) = e\u2212i Em t/ cm fn|\u03c8m m Now the key point: since \u02c6Q and \u02c6H commute we can \ufb01nd a complete set of simultaneous eigenstates for them (see Section A.5); without loss of generality then | fn = |\u03c8n. Using the orthonormality of the |\u03c8n, P(qn) = e\u2212i Em t/ cm \u03c8n | \u03c8m = |cn|2, m which is clearly independent of time. 6.4 PARITY 6.4.1 Parity in One Dimension A spatial inversion is implemented by the parity operator \u02c6 ; in one dimension, \u02c6 \u03c8(x) = \u03c8 (x) = \u03c8(\u2212x). Evidently, the parity operator is its own inverse: \u02c6 \u22121 = \u02c6 ; in Problem 6.8 you will show that it is Hermitian: \u02c6 \u2020 = \u02c6. Putting this together, the parity operator is unitary as well: \u02c6 \u22121 = \u02c6 = \u02c6 \u2020. Operators transform under a spatial inversion as \u02c6Q = \u02c6 \u2020 \u02c6Q \u02c6. (6.16) (6.17) I won\u2019t repeat the argument leading up to Equation 6.17, since it is identical to the one by which we arrived at Equation 6.6 in the case of translations. The position and momentum operators are \u201codd under parity\u201d (Problem 6.10): \u02c6x = \u02c6 \u2020 \u02c6x \u02c6 = \u2212 \u02c6x, \u02c6p = \u02c6 \u2020 \u02c6p \u02c6 = \u2212 \u02c6p, (6.18) (6.19) and this tells us how any operator transforms (see Problem 6.4): \u02c6Q \u02c6x, \u02c6p = \u02c6 \u2020 \u02c6Q \u02c6x, \u02c6p \u02c6 = \u02c6Q \u2212 \u02c6x, \u2212 \u02c6p. A system has inversion symmetry if the Hamiltonian is unchanged by a parity transforma- tion: \u02c6H = \u02c6 \u2020 \u02c6H \u02c6 = \u02c6", "H, or, using the unitarity of the parity operator, # $ = 0. (6.20) \u02c6H, \u02c6 244 CHAPTER 6 Symmetries & Conservation Laws If our Hamiltonian describes a particle of mass m in a one-dimensional potential V (x), then inversion symmetry simply means that the potential is an even function of position: V (x) = V (\u2212x). The implications of inversion symmetry are two: First, we can \ufb01nd a complete set of simultaneous eigenstates of \u02c6 and \u02c6H. Let such an eigenstate be written \u03c8n; it satis\ufb01es \u02c6 \u03c8n(x) = \u03c8n(\u2212x) = \u00b1 \u03c8n(x), since the eigenvalues of the parity operator are restricted to \u00b11 (Problem 6.8). So the stationary states of a potential that is an even function of position are themselves even or odd functions (or can be chosen as such, in the case of degeneracy).12 This property is familiar from the simple harmonic oscillator, the in\ufb01nite square well (if the origin is placed at the center of the well), and the Dirac delta function potential, and you proved it in general in Problem 2.1. Second, according to Ehrenfest\u2019s theorem, if the Hamiltonian has an inversion symmetry then d dt = i 0# $1 \u02c6H, \u02c6 = 0 so parity is conserved for a particle moving in a symmetric potential. And not just the expectation value, but the probability of any particular outcome in a measurement, in accord with the theorem of Section 6.3. Parity conservation means, for example, that if the wave function of a particle in a harmonic oscillator potential is even at t = 0 then it will be even at any later time t; see Figure 6.7. \u2217 Problem 6.8 (a) Show that the parity operator \u02c6 is Hermitian. (b) Show that the eigenvalues of the parity operator are \u00b11. 6.4.2 Parity in Three Dimensions The spatial inversion generated by the parity operator in three dimensions is The operators \u02c6r and \u02c6p transform as \u02c6 \u03c8(r) = \u03c8 (r) = \u03c8(\u2212r). \u02c6r = \u02c6 \u2020 \u02c6r \u02c6 = \u2212\u02c6r, \u02c6p = \u02c6 \u2020 \u02c6p \u02c6 = \u2212 \u02c6p. Any other operator transforms as \u02c6r, \u02c6p \u02c6Q = \u02c6 \u2020 \u02c6Q \u02c6r, \u02c6p \u02c6 = \u02c6Q \u2212\u02c6r, \u2212 \u02c6p. (6.21) (6.22) (6.23) 12 For bound (normalizable) states in one dimension, there is no degeneracy and every bound state of a symmetric potential is automatically an eigenstate of parity. (However, see Problem 2.46.) For scattering states, degeneracy does occur. 6.4 Parity 245 0(\u03be ) = A \u03be 2 e\u2212\u03be 2/2 Figure 6.7: This \ufb01lmstrip shows the time evolution of a particular wave function for a particle in the harmonic oscillator potential. The solid and dashed curves are the real and imaginary parts of the wave function respectively, and time increases from top to bottom. Since parity is conserved, a wave function which is initially an even function of position (as this one is) remains an even function at all later times. Example 6.2 Find the parity-transformed angular momentum operator \u02c6L = \u02c6 \u2020 \u02c6L \u02c6, in terms of \u02c6L. Solution: Since L = r \u00d7 p, Equation 6.23 tells us that \u02c6L = \u02c6 \u2020 \u02c6L \u02c6 = \u02c6r \u00d7 \u02c6p = (6.24) We have a special name for vectors like \u02c6L, that are even under parity. We call them pseudovectors, since they don\u2019t change sign under parity the way \u201ctrue\u201d vectors, such as \u02c6r or \u02c6p, = \u02c6r \u00d7 \u02c6p = \u02c6L. \u2212\u02c6r \u00d7 \u2212 \u02c6p 246 CHAPTER 6 Symmetries & Conservation Laws do. Similarly, scalars that are odd under parity are called pseudoscalars, since they do not behave under parity the way that \u201ctrue\u201d scalars (such as \u02c6r \u00b7 \u02c6r which is even under parity) do", ". See Problem 6.9. Note: The labels scalar and vector describe how the operators behave under rotations; we will de\ufb01ne these terms carefully in the next section. \u201cTrue\u201d vectors and pseudovectors behave the same way under a rotation\u2014they are both vectors. In three dimensions, the Hamiltonian for a particle of mass m moving in a potential V (r) will have inversion symmetry if V (\u2212r) = V (r). Importantly, any central potential satis\ufb01es this condition. As in the one-dimensional case, parity is conserved for such systems, and the eigenstates of the Hamiltonian may be chosen to be simultaneously eigenstates of parity. In Problem 6.1 you proved that the eigenstates of a particle in a central potential, written \u03c8nm(r, \u03b8, \u03c6) = Rn(r ) Y m (\u03b8, \u03c6), are eigenstates of parity:13 \u02c6 \u03c8nm(r, \u03b8, \u03c6) = (\u22121) \u03c8nm(r, \u03b8, \u03c6). \u2217 Problem 6.9 (a) Under parity, a \u201ctrue\u201d scalar operator does not change: \u02c6 \u2020 \u02c6f \u02c6 = \u02c6f, whereas a pseudoscalar changes sign. Show therefore that? > \u02c6, \u02c6f scalar, whereas \u02c6A, \u02c6B operators \u02c6A and \u02c6B is de\ufb01ned as (b) Similarly, a \u201ctrue\u201d vector changes sign >? \u2261 \u02c6A \u02c6B + \u02c6B \u02c6A. = 0 for a pseudoscalar. Note: the anti-commutator of two # $ \u02c6, \u02c6f = 0 for a \u201ctrue\u201d \u02c6 \u2020 \u02c6V \u02c6 = \u2212 \u02c6V, whereas a pseudovector is unchanged. Show therefore that vector and = 0 for a pseudovector. $ # \u02c6, \u02c6V? > \u02c6, \u02c6V = 0 for a \u201ctrue\u201d 6.4.3 Parity Selection Rules Selection rules tell you when a matrix element is zero based on the symmetry of the situation. Recall that a matrix element is any object of the form b| \u02c6Q|a; an expectation value is a special case of a matrix element with a = b = \u03c8. One operator whose selection rules are physically important is the electric dipole moment operator \u02c6pe = q \u02c6r. The selection rules for this operator\u2014the operator itself is nothing more than the charge of the particle times its position\u2014determine which atomic transitions are allowed and which are forbidden (see Chapter 11). It is odd under parity since the position vector \u02c6r is odd: 13 Note that Equation 6.24 could equivalently be written as and therefore also \u02c6L2 # \u02c6, \u02c6L $ = 0. The fact that parity commutes with every is the reason you can \ufb01nd simultaneous eigenstates component of the angular momentum of \u02c6L2, \u02c6L z, and \u02c6. \u02c6 \u2020 \u02c6pe \u02c6 = \u2212 \u02c6pe. 6.4 Parity 247 (6.25) Now consider the matrix elements of the electric dipole operator between two states \u03c8nm and \u03c8nm (we label the corresponding kets |nm and |nm). Using Equation 6.25 we have nm \u02c6pe nm = \u2212 0 nm 0 = \u2212 nm = (\u22121)++1 1 \u02c6 \u02c6 \u2020 \u02c6pe nm (\u22121) \u02c6pe (\u22121) \u02c6pe nm 1 nm nm. From this we see immediately that \u02c6pe nm nm = 0 if + is even. (6.26) This is called Laporte\u2019s rule; it says that matrix elements of the dipole moment operator vanish between states with the same parity. The reasoning by which we obtained Equation 6.26 can be generalized to derive selection rules for any operator, as long as you know how that operator transforms under parity. In particular, Laporte\u2019s rule applies to any operator that is odd under parity. The selection rule for an operator that is even under parity, such as \u02c6L, is derived in Problem 6.11. Problem 6.10 Show that the position and momentum operators are odd under parity. That is, prove Equations 6.18, 6.19, and, by extension, 6.21 and 6.22. \u2217 \u02c6L between two de\ufb01nite-parity states", ": Problem 6.11 Consider the matrix elements of nm| \u02c6L|nm. Under what conditions is this matrix element guaranteed to vanish? Note that the same selection rule would apply to any pseudovector operator, or any \u201ctrue\u201d scalar operator. Problem 6.12 Spin angular momentum, \u02c6S, is even under parity, just like orbital angular momentum \u02c6L: \u02c6 \u2020 \u02c6S \u02c6 = \u02c6S or = 0. (6.27) $ # \u02c6, \u02c6S Acting on a spinor written in the standard basis (Equation 4.139), the parity operator becomes a 2 \u00d7 2 matrix. Show that, due to Equation 6.27, this matrix must be a constant times the identity matrix. As such, the parity of a spinor isn\u2019t very interesting since both spin states are parity eigenstates with the same eigenvalue. We can arbitrarily choose that parity to be +1, so the parity operator has no effect on the spin portion of the wave function.14 14 However, it turns out that antiparticles of spin 1/2 have opposite parity. Thus the electron is conventionally assigned parity +1, but the positron then has parity \u22121. 248 CHAPTER 6 Symmetries & Conservation Laws \u2217 Problem 6.13 Consider an electron in a hydrogen atom. (a) Show that if the electron is in the ground state, then necessarily pe = 0. No calculation allowed. (b) Show that if the electron is in an n = 2 state, then pe need not vanish. Give an example of a wave function for the energy level n = 2 that has a non-vanishing pe and compute pe for this state. 6.5 ROTATIONAL SYMMETRY 6.5.1 Rotations About the z Axis The operator that rotates a function about the z axis by an angle \u03d5 (Equation 6.2) \u02c6Rz(\u03d5) \u03c8(r, \u03b8, \u03c6) = \u03c8 (r, \u03b8, \u03c6) = \u03c8(r, \u03b8, \u03c6 \u2212 \u03d5) is closely related to the z component of angular momentum (Equation 4.129). By the same reasoning that led to Equation 6.3, (6.28) \u02c6Rz(\u03d5) = exp \u02c6L z, (6.29) \u2212 i \u03d5 and we say that \u02c6L z is the generator of rotations about the z axis (compare Equation 6.3). How do the operators \u02c6r and \u02c6p transform under rotations? To answer this question we use the in\ufb01nitesimal form of the operator: \u02c6Rz(\u03b4) \u2248 1 \u2212 i \u03b4 \u02c6L z. Then the operator \u02c6x transforms as \u02c6x = \u02c6R\u2020 \u02c6x \u02c6R \u2248 # \u2248 \u02c6x + i \u03b4 \u02c6L z 1 \u2212 i \u03b4 \u02c6x \u02c6L z 1 + i \u03b4 $ \u02c6L z, \u02c6x \u2248 \u02c6x \u2212 \u03b4 \u02c6y (I used Equation 4.122 for the commutator). Similar calculations show that \u02c6y = \u02c6y + \u03b4 \u02c6x and \u02c6z = \u02c6z. We can combine these results into a matrix equation \u239b \u239b \u239e \u239b \u239e \u239e \u239d \u23a0 = \u239d \u23a0 \u239d \u23a0. 1 \u2212\u03b4 \u03b4 1 0 0 0 0 1 \u02c6x \u02c6y \u02c6z \u02c6x \u02c6y \u02c6z That doesn\u2019t look quite right for a rotation. Shouldn\u2019t it be \u02c6x cos \u03d5 \u2212 sin \u03d5 0 \u02c6y sin \u03d5 0 \u02c6z 1 0 cos \u03d5 0 \u23a0 = \u239e \u239b \u239d \u239d \u239b \u239e \u239b \u23a0 \u239d \u239e \u23a0? \u02c6x \u02c6y \u02c6z (6.30) (6.31) Yes, but don\u2019t forget, we are assuming \u03d5 \u2192 \u03b4 is in\ufb01nitesimal, so (dropping terms of order \u03b42 and higher) cos \u03d5 \u2192 1 and sin \u03d5 \u2192 \u03b4.15 15 To go the other way, from in\ufb01nitesimal to \ufb01nite, see Problem 6.14. 6.5 Rotational Symmetry 249 \u2217", "\u2217 Problem 6.14 In this problem you will establish the correspondence between Equations 6.30 and 6.31. (a) Diagonalize the matrix16 to obtain the matrix M = 1 \u03d5/N \u2212\u03d5/N 1 \u22121, \u22121 is the unitary matrix whose columns are the (normalized) eigenvectors M = S M S where S of M. (b) Use the binomial expansion to show that limN \u2192\u221e entries e\u2212i\u03d5 and ei\u03d5 on the diagonal. M N is a diagonal matrix with (c) Transform back to the original basis to show that M MN = S \u22121 N S lim N \u2192\u221e lim N \u2192\u221e agrees with the matrix in Equation 6.31. 6.5.2 Rotations in Three Dimensions Equation 6.29 can be generalized in the obvious way to a rotation about an axis along the unit vector n: \u02c6Rn(\u03d5) = exp \u2212 i \u03d5 n \u00b7 \u02c6L (6.32) Just as linear momentum is the generator of translations, angular momentum is the generator of rotations. Any operator (with three components) that transforms the same way as the position operator under rotations is called a vector operator. By \u201ctransforms the same way\u201d we mean that \u02c6V = D \u02c6V where D is the same matrix as appears in \u02c6r = D \u02c6r. In particular for a rotation about the z axis, we would have (Equation 6.31) \u239e \u239f \u23a0 = \u239b \u239d \u239b \u239c \u239d \u02c6V x \u02c6V y \u02c6V z cos \u03d5 \u2212 sin \u03d5 0 sin \u03d5 0 1 0 cos \u03d5 0 \u239e \u23a0 \u239b \u239c \u239d \u239e \u239f \u23a0. \u02c6Vx \u02c6Vy \u02c6Vz This transformation rule follows from the commutation relations17 $ # \u02c6Li, \u02c6V j = i i jk \u02c6Vk, (6.33) 16 See Section A.5. 17 The Levi-Civita symbol i jk is de\ufb01ned in Problem 4.29. 250 CHAPTER 6 Symmetries & Conservation Laws Table 6.1: Operators are classi\ufb01ed as vectors or scalars based on their commutation relations with \u02c6L, which encode how they transform under a rotation, and as pseudoor \u201ctrue\u201d quantities based on their commutation relations with \u02c6, which encode how they transform under a spatial inversion. The curly brackets in the \ufb01rst column denote the anti-commutator, de\ufb01ned in Problem 6.9. To include the spin \u02c6S in this table, one simply replaces \u02c6Li everywhere it appears in the third column with \u02c6Ji = \u02c6Li + \u02c6Si (Problems 6.12 and 6.32, respectively, discuss the effect of parity and rotations on spinors). \u02c6S, like \u02c6L, is then a pseudovector and \u02c6p \u00b7 \u02c6S is a pseudoscalar. true vector \u02c6V pseudovector \u02c6V true scalar \u02c6f pseudoscalar \u02c6f? $ parity > \u02c6, \u02c6Vi # \u02c6, \u02c6Vi # $ \u02c6, \u02c6f? > \u02c6, \u02c6f = jk = i i jk \u02c6Vk \u02c6Vk # $ rotations $ # \u02c6Li, \u02c6V j \u02c6Li, \u02c6V j $ \u02c6Li, \u02c6f $ \u02c6Li, \u02c6f # # = 0 = 0 examples \u02c6r, \u02c6p \u02c6L \u02c6r \u00b7 \u02c6r (see Problem 6.16), and we may take Equation 6.33 as the de\ufb01nition of a vector operator. So far we have encountered three such operators, \u02c6r, \u02c6p and \u02c6L: $ # \u02c6Li, \u02c6r j = i i jk \u02c6rk, $ # \u02c6Li, \u02c6p j = i i jk \u02c6pk, $ # \u02c6Li, \u02c6L j = i i jk \u02c6L k (see Equations 4.99 and 4.122). A scalar operator is a single quantity that is unchanged by rotations; this is equivalent to", " saying that the operator commutes with \u02c6L: # $ \u02c6Li, \u02c6f = 0. (6.34) We can now classify operators as either scalars or vectors, based on their commutation relations with \u02c6L (how they transform under a rotation), and as \u201ctrue\u201d or pseudo-quantities, based on their commutators with \u02c6 (how they transform under parity). These results are summarized in Table 6.1.18 CONTINUOUS ROTATIONAL SYMMETRY For a particle of mass m moving in a potential V (r), the Hamiltonian \u02c6p2 2 m is rotationally invariant if V (r) = V (r ) (the central potentials studied in Section 4.1.1). In this case the Hamiltonian commutes with a rotation by any angle about an arbitrary axis + V (r) \u02c6H = 18 Of course, not every operator will \ufb01t into one of these categories. Scalar and vector operators are simply the \ufb01rst two instances in a hierarchy of tensor operators. Next come second-rank tensors (the inertia tensor from classical mechanics or the quadrupole tensor from electrodynamics are examples), third-rank tensors, and so forth. 6.5 Rotational Symmetry 251 # \u02c6H, \u02c6Rn(\u03d5) $ = 0. In particular, Equation 6.35 must hold for an in\ufb01nitesimal rotation Rn(\u03b4) \u2248 1 \u2212 i \u03b4 n \u00b7 \u02c6L, which means that the Hamiltonian commutes with the three components of L: # $ \u02c6H, \u02c6L = 0. What, then, are the consequences of rotational invariance? From Equation 6.36 and Ehrenfest\u2019s theorem 0# d dt L = i $1 \u02c6H, \u02c6L = 0 (6.35) (6.36) (6.37) for a central potential. Thus, angular momentum conservation is a consequence of rotational invariance. And beyond the statement 6.37, angular momentum conservation means that the probability distributions (for each component of the angular momentum) are independent of time as well\u2014see Section 6.3. Since the Hamiltonian for a central potential commutes with all three components of angular momentum, it also commutes with \u02c6L 2. The operators \u02c6H, \u02c6L z, and \u02c6L 2 form a complete set of compatible observables for the bound states of a central potential. Compatible means that they commute pairwise $ \u02c6H, \u02c6L 2 # # # \u02c6H, \u02c6L z \u02c6L z, \u02c6L 2 $ $ = 0, = 0, = 0, (6.38) so that the eigenstates of \u02c6H can be chosen to be simultaneous eigenstates of \u02c6L 2 and \u02c6L z. \u02c6H \u03c8nm = En \u03c8nm, \u02c6L z \u03c8nm = m \u03c8nm, \u02c6L 2 \u03c8nm = ( + 1) 2 \u03c8nm. Saying they are complete means that the quantum numbers n,, and m uniquely specify a bound state of the Hamiltonian. This is familiar from our solution to the hydrogen atom, the in\ufb01nite spherical well, and the three-dimensional harmonic oscillator, but it is true for any central potential.19 \u2217 Problem 6.15 Show how Equation 6.34 guarantees that a scalar is unchanged by a rotation: \u02c6f = \u02c6R\u2020 \u02c6f \u02c6R = \u02c6f. 19 This follows from the fact that the radial Schr\u00f6dinger equation (Equation 4.35) has at most a single normalizable solution so that, once you have speci\ufb01ed and m, the energy uniquely speci\ufb01es the state. The principal quantum number n indexes those energy values that lead to normalizable solutions. 252 CHAPTER 6 Symmetries & Conservation Laws \u2217 Problem 6.16 Working from Equation 6.33, \ufb01nd how the vector operator \u02c6V transforms for an in\ufb01nitesimal rotation by an angle \u03b4 about the y axis. That is, \ufb01nd the matrix D in \u02c6V = D \u02c6V. Problem 6.17 Consider the action of an in\ufb01nitesimal rotation about the n axis of an angular momentum eigenstate \u03c8nm. Show that \u02c6Rn(\u03b4) \u03c8nm = Dm m \u03c8n m, m and \ufb01nd the complex numbers D", "m m (they will depend on \u03b4, n, and as well as m and m). This result makes sense: a rotation doesn\u2019t change the magnitude of the angular momentum (speci\ufb01ed by ) but does change its projection along the z axis (speci\ufb01ed by m). 6.6 DEGENERACY Symmetry is the source of most20 degeneracy in quantum mechanics. We have seen that a symmetry implies the existence of an operator \u02c6Q that commutes with the Hamiltonian # $ \u02c6H, \u02c6Q = 0. (6.39) So why does symmetry lead to degeneracy in the energy spectrum? The basic idea is this: if = \u02c6Q |\u03c8n is a stationary state with the same energy. we have a stationary state |\u03c8n, then |\u03c8 n The proof is straightforward: = \u02c6H \u02c6Q |\u03c8n = \u02c6Q \u02c6H |\u03c8n = \u02c6Q En |\u03c8n = En \u02c6Q |\u03c8n = En \u03c8 n \u02c6H \u03c8 n. For example, if you have an eigenstate of a spherically-symmetric Hamiltonian and you rotate that state about some axis, you must get back another state of the same energy. You might think that symmetry would always lead to degeneracy, and that continuous symmetries would lead to in\ufb01nite degeneracy, but that is not the case. The reason is that the two states |\u03c8n and |\u03c8 might be the same.21 As an example, consider the Hamiltonian for the n harmonic oscillator in one dimension; it commutes with parity. All of its stationary states are either even or odd, so when you act on one with the parity operator you get back the same state 20 When we can\u2019t identify the symmetry responsible for a particular degeneracy, we call it an accidental degeneracy. In most such cases, the degeneracy turns out to be no accident at all, but instead due to symmetry that is more dif\ufb01cult to identify than, say, rotational invariance. The canonical example is the larger symmetry group of the hydrogen atom (Problem 6.34). 21 This is highly non-classical. In classical mechanics, if you take a Keplerian orbit there will always be some axis about which you can rotate it to get a different Keplerian orbit (of the same energy) and in fact there will be an in\ufb01nite number of such orbits with different orientations. In quantum mechanics, if you rotate the ground state of hydrogen you get back exactly the same state regardless of which axis you choose, and if you rotate one of the states with n = 2 and = 1, you get back a linear combination of the three orthogonal states with these quantum numbers. 6.6 Degeneracy 253 you started with (perhaps multiplied by \u22121, but that, physically, is the same state). There is therefore no degeneracy associated with inversion symmetry in this case. In fact, if there is only a single symmetry operator \u02c6Q (or if there are multiple symmetry operators that all commute), you do not get degeneracy in the spectrum. The reason is the same theorem we\u2019ve now quoted many times: since \u02c6Q and \u02c6H commute, we can \ufb01nd simultaneous eigenstates |\u03c8n of \u02c6Q and \u02c6H and these states transform into themselves under the symmetry operation: \u02c6Q |\u03c8n = qn |\u03c8n. call them \u02c6Q and But what if there are two operators that commute with the Hamiltonian, but do not commute with each other? In this case, degeneracy in the energy spectrum is \u02c6 inevitable. Why? First, consider a state |\u03c8 that is an eigenstate of both \u02c6H and \u02c6Q with eigenvalues En and qm respectively. Since \u02c6H and \u02c6 commute we know that the state |g = \u02c6 |\u03c8 is also an \u02c6H with eigenvalue En. Since \u02c6Q and \u02c6 do not commute we know (Section A.5) eigenstate of \u02c6Q, \u02c6 that there cannot exist a complete set of simultaneous eigenstates of all three operators and \u02c6H. Therefore, there must be some |\u03c8 such that \u02c6 |\u03c8 is distinct from |\u03c8 speci\ufb01cally, it is not an eigenstate of \u02c6 meaning that the energy level En is at least doubly degenerate. The presence of multiple non-commuting symmetry operators guarantees degeneracy of the energy spectrum. This is precisely the situation we have encountered in the", " case of central potentials. Here the Hamiltonian commutes with rotations about any axis ( or equivalently with the generators \u02c6L x, \u02c6L y, and \u02c6L z but those rotations don\u2019t commute with each other. So we know that there will be degeneracy in the spectrum of a particle in a central potential. The following example shows exactly how much degeneracy is explained by rotational invariance. Example 6.3 Consider an eigenstate of a central potential \u03c8nm with energy En. Use the fact that the Hamiltonian for a central potential commutes with any component of \u02c6L, and therefore also with \u02c6L + and \u02c6L \u2212, to show that \u03c8nm\u00b11 are necessarily also eigenstates with the same energy as \u03c8nm.22 Solution: Since the Hamiltonian commutes with \u02c6L \u00b1 we have \u02c6H \u02c6L \u00b1 \u2212 \u02c6L \u00b1 \u02c6H \u03c8nm = 0, so or \u02c6H \u02c6L \u00b1 \u03c8nm = \u02c6L \u00b1 \u02c6H \u03c8nm = En \u02c6L \u00b1 \u03c8nm \u02c6H \u03c8nm\u00b11 = En \u03c8nm\u00b11 \u221a (I canceled the constant ( + 1) \u2212 m (m \u00b1 1) from both sides in the last expression). This argument could obviously be repeated to show that \u03c8nm\u00b12 has the same energy as 22 Of course, we already know the energies are equal since the radial equation, Equation 4.35, does not depend on m. This example demonstrates that rotational invariance is behind the degeneracy. 254 CHAPTER 6 Symmetries & Conservation Laws \u03c8nm\u00b11, and so on until you\u2019ve exhausted the ladder of states. Therefore, rotational invariance explains why states which differ only in the quantum number m have the same energy, and since there are 2 + 1 different values of m, 2 + 1 is the \u201cnormal\u201d degeneracy for energies in a central potential. Of course, the degeneracy of hydrogen (neglecting spin) is n2 (= 1, 4, 9,...) (Equation 4.85) which is greater than 2 + 1 (= 1, 3, 5,...).23 Evidently hydrogen has more degeneracy than is explained by rotational invariance alone. The source of the extra degeneracy is an additional symmetry that is unique to the 1/r potential; this is explored in Problem 6.34.24 In this section we have focused on continuous rotational symmetry, but discrete rotational symmetry, as experienced (for instance) by an electron in a crystal, can also be of interest. Problem 6.33 explores one such system. Problem 6.18 Consider the free particle in one dimension: \u02c6H = \u02c6p2/2 m. This Hamiltonian has both translational symmetry and inversion symmetry. (a) Show that translations and inversion don\u2019t commute. (b) Because of the translational symmetry we know that the eigenstates of \u02c6H can be chosen to be simultaneous eigenstates of momentum, namely f p(x) (Equation 3.32). Show that the parity operator turns f p(x) into f\u2212 p(x); these two states must therefore have the same energy. (c) Alternatively, because of the inversion symmetry we know that the eigenstates of \u02c6H can be chosen to be simultaneous eigenstates of parity, namely 1\u221a \u03c0 cos p x and 1\u221a \u03c0 sin p x. Show that the translation operator mixes these two states together; they therefore must be degenerate. Note: Both parity and translational invariance are required to explain the degeneracy in the free-particle spectrum. Without parity, there is no reason for f p(x) and f\u2212 p(x) to have the same energy (I mean no reason based on symmetries discussed thus far... obviously you can plug them in to the time-independent Schr\u00f6dinger equation and show it\u2019s true). Problem 6.19 For any vector operator \u02c6V one can de\ufb01ne raising and lowering operators as \u02c6V\u00b1 = \u02c6Vx \u00b1 i \u02c6Vy. (a) Using Equation 6.33, show that $ $ # # \u02c6L z, \u02c6V\u00b1 \u02c6L 2, \u02c6V\u00b1 = \u00b1 \u02c6V+. = 2 2 \u02c6V\u00b1 \u00b1 2 \u02c6V\u00b1 \u02c6L z \u2213 2 \u02c6Vz \u02c6L \u00b1. 23 I don\u2019t mean that they necessarily occur in this order. Look back at the in\ufb01nite", " spherical well (Figure 4.3): starting with the ground state the degeneracies are = 1, 3, 5, 1, 7, 3, 9, 5,.... These are precisely the degrees of degeneracy we expect for rotational invariance (2 + 1 for integer ) but the symmetry considerations don\u2019t tell us where in the spectrum each degeneracy will occur. 24 For the three-dimensional harmonic oscillator the degeneracy is n (n + 1) /2 = 1, 3, 6, 10,... (Problem 4.46) which again is greater than 2 + 1. For a discussion of the additional symmetry in the oscillator problem see D. M. Fradkin, Am. J. Phys. 33, 207 (1965). 6.7 Rotational Selection Rules 255 (b) Show that, if \u03c8 is an eigenstate of \u02c6L 2 and \u02c6L z with eigenvalues ( + 1) 2 and respectively, then either \u02c6V+\u03c8 is zero or \u02c6V+\u03c8 is also an eigenstate of \u02c6L 2 and \u02c6L z with eigenvalues( + 1) ( + 2) 2 and( + 1) respectively. This means that, acting on a state with maximal m =, the operator \u02c6V+ either \u201craises\u201d both the and m values by 1 or destroys the state. 6.7 ROTATIONAL SELECTION RULES The most general statement of the rotational selection rules is the Wigner\u2013Eckart Theorem; as a practical matter, it is arguably the most important theorem in all of quantum mechanics. Rather than prove the theorem in full generality I will work out the selection rules for the two classes of operators one encounters most often: scalar operators (in Section 6.7.1) and vector operators (in Section 6.7.2). In deriving these selection rules we consider only how the operators behave under a rotation; therefore, the results of this section apply equally well to \u201ctrue\u201d scalars and pseudoscalars, and those of the next section apply equally well to \u201ctrue\u201d vectors and pseudeovectors. These selection rules can be combined with the parity selection rules of Section 6.4.3 to obtain a larger set of selection rules for the operator. 6.7.1 Selection Rules for Scalar Operators \u02c6f with the three components of angular The commutation relations for a scalar operator momentum (Equation 6.34) can be rewritten in terms of the raising and lowering operators as # # $ \u02c6L z, \u02c6f $ \u02c6L \u00b1, \u02c6f # $ \u02c6L 2, \u02c6f = 0 = 0 = 0. (6.40) (6.41) (6.42) \u02c6f by sandwiching these commutators between two states of We derive selection rules for de\ufb01nite angular momentum, which we will write as |n m and |n m. These might be hydrogenic orbitals, but they need not be (in fact they need not even be eigenstates of any Hamiltonian but I\u2019ll leave the quantum number n there so they look familiar); we require only that |n m is an eigenstate of \u02c6L 2 and \u02c6L z with quantum numbers and m respectively.25 Sandwiching Equation 6.40 between two such states gives 0 # 1 n m \u02c6L z, \u02c6f $ n m = 0 or and therefore 0 n m \u02c6L z 1 n m \u02c6f \u2212 0 n m \u02c6f \u02c6L \u02c6f n m 1 = 0 (6.43) using the hermiticity of \u02c6L z vanish unless m \u2212 m \u2261 m = 0. Repeating this procedure with Equation 6.42 we get. Equation 6.43 says that the matrix elements of a scalar operator 25 Importantly, they satisfy Equations 4.118 and 4.120. 256 CHAPTER 6 Symmetries & Conservation Laws \u02c6L 2, \u02c6f \u02c6f \u02c6L 2 \u02c6f 1 1 1 = 0 = 0 = 0. (6.44) \u2212 ( + 1) n m 1 n m 0 \u02c6L 2 \u02c6f n m + 1 This tells us that the matrix elements of a scalar operator vanish unless \u2212 \u2261 = 0.26 These, then, are the selection rules for a scalar operator: = 0 and m = 0. However, we can get even more information about the matrix elements from the remaining commutators: (I\u2019ll just do the + case and leave the \u2212 case for Problem 6.20) 1 0 n m", " \u02c6f m \u2212 1 \u02c6L + \u02c6f n m 1 n m 1 \u2212 Am 0 n Bm # 0 n m 0 n m \u02c6f n m $ n m n m \u02c6L +, \u02c6f \u02c6f \u02c6L + n (m + 1, (6.45) where (from Problem 4.21) % % Am = Bm = ( + 1) \u2212 m (m + 1), and (I also used the fact that \u02c6L \u00b1 is the Hermitian conjugate of \u02c6L \u2213: \u03c8| \u02c6L \u00b1 = L \u2213 \u03c8|.)27 Both terms in Equation 6.45 are zero unless m = m + 1 and =, as we proved in Equations 6.43 = Am and 6.44. When these conditions are satis\ufb01ed, the two coef\ufb01cients are equal and Equation 6.45 reduces to \u02c6f n (m + 1) n (m + 1) n m Bm+1 ( + 1) \u2212 m (m \u2212 1). n m \u02c6f (6.46) = 0 1 0 1. Evidently the matrix elements of a scalar operator are independent of m. The results of this section can be summarized as follows: \u02c6f n m n m n = \u03b4 \u03b4mm 6.47) The funny-looking matrix element on the right, with two bars, is called a reduced matrix element and is just shorthand for \u201ca constant that depends on n,, and n, but not m.\u201d Example 6.4 (a) Find r 2 for all four of the degenerate n = 2 states of a hydrogen atom. Solution: From Equation 6.47 we have, for the states with = 1, the following equality1 r 2 2 1 \u2212. \u2261 26 The other root of the quadratic integers this isn\u2019t possible. 27 Since \u02c6L x and \u02c6L y are Hermitian, + 1 \u2212 ( + 1) = 0 is = \u2212( + 1); since and are non-negative \u02c6L\u2020 \u00b1 = \u02c6L x \u00b1 i \u02c6L y \u2020 = \u02c6L\u2020 x \u00b1(\u2212i) \u02c6L\u2020 y = \u02c6L x \u2213 i \u02c6L y = \u02c6L\u2213. 6.7 Rotational Selection Rules 257 To calculate the reduced matrix element we simply pick any one of these expectation values\u03c8210(r )|2 d3r \u221e 0 r 4 |R21(r )|2 dr Y 0 1 2 (\u03b8, \u03c6) d. The spherical harmonics are normalized (Equation 4.31), so the angular integral is 1, and the radial functions Rn are listed in Table 4.7, giving @ r 2 @ @ 2 1 a2 e\u2212r/a dr = 30 a2. 1 24 a3 @ @ @ That determines three of the expectation values. The \ufb01nal expectation value is \u03c8200(r )|2 d3r r 4 |R20(r )|2 dr 1 \u2212 1 2 r 4 1 2 a3 Y 0 0 2 d (\u03b8, \u03c6) 2 r a e\u2212r/a dr = = = \u221e 0 \u221e 0 = 42 a2. Summarizing: 0 200 r 2 1 200 = 42 a2, 211 210 r 2 r 2 211 r 2 210 21 \u2212 1 21 \u2212 1 \u23ab \u23ac \u23ad = 30 a2. (6.48) (b) Find the expectation value of r 2 for an electron in the superposition state |\u03c8 = 1\u221a 2 (|200 \u2212 i |211). Solution: We can expand the expectation value as 0 \u03c8 r 2 1 \u03c8 (200| + i 211|) r 2 (|200 \u2212 i |211) 0 1 1 0 0 + i 211 r 2 200 \u2212 i 200 r 2 1 211 = 1 2 = 1 2 + r 2 200 200 r 2 211 211 0 1. From Equation 6.47 we see that two of these matrix elements vanish, and @ @ @ 20 20 0 + 21 1 @ @ @ r 2 @ @ @ 21 = 36 a2. (6.49) 258 CHAPTER 6 Symmetries & Conservation Laws Problem 6.20 Show that the commutator $ \u02c6L +, \u02c6f tion 6.46, as does the commutator # # $ \u02c6L \u2212, \u02c6f = 0. = 0 leads to the same rule, Equa- \u2217 Problem 6.21 For an electron in the hydrogen state \u03c8 = 1\u221a 2", " (\u03c8211 + \u03c821\u22121), \ufb01nd r after \ufb01rst expressing it in terms of a single reduced matrix element. 6.7.2 Selection Rules for Vector Operators We now move on to the selection rules for a vector operator \u02c6V. This is signi\ufb01cantly more work than the scalar case, but the result is central to understanding atomic transitions (Chapter 11). We begin by de\ufb01ning, by analogy with the angular momentum raising and lowering operators, the operators28 \u02c6V\u00b1 \u2261 \u02c6Vx \u00b1 i \u02c6Vy. Written in terms of these operators, Equation 6.33 becomes # $ # \u02c6L z, \u02c6Vz \u02c6L z, \u02c6V\u00b1 \u02c6L \u00b1, \u02c6V\u00b1 # \u02c6L \u00b1, \u02c6Vz \u02c6L \u00b1, \u02c6V\u2213 # # $ $ $ $ = 0 = \u00b1 \u02c6V\u00b1 = 0 = \u2213 \u02c6V\u00b1 = \u00b1 2 \u02c6Vz (6.50) (6.51) (6.52) (6.53) (6.54) as you will show in Problem 6.22(a).29 Just as for the scalar operator in Section 6.7.1, we sandwich each of these commutators between two states of de\ufb01nite angular momentum to derive (a) conditions under which the matrix elements are guaranteed to vanish and (b) relations between matrix elements with differing values of m or different components of \u02c6V. 0 From Equation 6.51, n m n m \u02c6L z \u02c6V\u00b1 1 0 \u2212 n m \u02c6V\u00b1 \u02c6L z 1 n m 0 n m \u02c6V\u00b1 n m 1, = \u00b1 28 The operators \u02c6V\u00b1 are, up to constants, components of what are known as spherical tensor operators of rank 1, written \u02c6T (k) q where k is the rank and q the component of the operator: (1) \u02c6T \u00b11 = \u2213 1\u221a 2 \u02c6V\u00b1 (1) \u02c6T 0 = \u02c6Vz. Similarly, the scalar operator f treated in Section 6.7.1 is a rank-0 spherical tensor operator: (0) \u02c6T 0 29 Equations 6.51\u20136.54 each stand for two equations: read the upper signs all the way across, or the lower signs. = \u02c6f. and since our states are eigenstates of \u02c6L z, this simpli\ufb01es to m \u2212(m \u00b1 1) 0 n m \u02c6V\u00b1 n m 6.7 Rotational Selection Rules 259 1 = 0. (6.55) Equation 6.55 says that either m = m \u00b1 1, or else the matrix element of \u02c6V\u00b1 must vanish. Equation 6.50 works out similarly (see Problem 6.22) and this \ufb01rst set of commutators gives us the selection rules for m: 0 0 n m 0 \u02c6V+ \u02c6Vz n m \u02c6V unless m = m + 1 unless m = m unless m = m \u2212 1. (6.56) (6.57) (6.58) 0 Note that, if desired, these expressions can be turned back into selection rules for the x- and y-components of our operator, since \u02c6Vx \u02c6Vy \u02c6V\u2212 \u02c6V\u2212 \u02c6V+ \u02c6V$ 1$ #0 # The remaining commutators, Equations 6.52\u20136.54, yield a selection rule on and relations among the nonzero matrix elements. As shown in Problem 6.24, the results may be summarized as30 0 0 nm nm 0 nm \u02c6V+ \u02c6V\u2212 \u02c6Vz 1 1 1 nm nm nm = \u2212 \u221a = \u221a 2 C 1 m1m 2 C 1 m\u22121m m0m (6.59) (6.60) (6.61) The constants C j1 j2 J m1 m2 M in these expressions are precisely the Clebsch\u2013Gordan coef\ufb01cients that appeared in the addition of angular momenta (Section 4.4.3). The Clebsch\u2013Gordan coef\ufb01cient C j1 j2 J m1 m2 M vanishes unless M = m1 + m2 (since the z-components of angular momentum add) and unless J = j1 + j2, j1 + j2 \u2212 1,...,", " | j1 \u2212 j2| (Equation 4.182). In particular, the matrix elements of any component of a vector operator, nm| \u02c6Vi |nm, are nonzero only if = 0, \u00b11, and m = 0, \u00b11. (6.62) 30 A warning about notation: In the selection rules for the scalar operator r, @ @ n nm @ @ r r = \u03b4 \u03b4 m n n mm, and for a component (say z) of the vector operator r, n m z nm = C 1 m0m n @ @ r @ @ n, the two reduced matrix elements are not the same. One is the reduced matrix element for r and one is the reduced matrix element for r, and these are different operators that share the same name. You could tack on a subscript n to distinguish between the two if that helps keep them straight and v 260 CHAPTER 6 Symmetries & Conservation Laws Example 6.5 Find all of the matrix elements of \u02c6r between the states with n = 2, = 1 and n = 3, = 2: 3 2 m ri 2 1 m where m = \u22121, 0, 1, m = \u22122, \u22121, 0, 1, 2, and ri = x, y, z. Solution: With the vector operator \u02c6V = \u02c6r, our components are Vz = z, V+ = x + i y, and V\u2212 = x \u2212 i y. We start by calculating one of the matrix elements, 320 | z | 210 = = \u03c8320(r) r cos \u03b8 \u03c82 1 0(r) d3r R32(r )\u2217 r R21(r ) r 2 dr (\u03b8, \u03c6)\u2217 Y 0 2 cos \u03b8 Y 0 1 (\u03b8, \u03c6) d \u221a 3 a. = 212 33 57 From Equation 6.61 we can then determine the reduced matrix element 212 33 57 Therefore 320 | z | 210 = C 1 1 2 0 0 0 32 V 21 \u221a 3 a = 2 3 32 V 21. 32 V 21 = 212 34 \u221a 2 57 a. (6.63) We can now \ufb01nd all of the remaining matrix elements from Equations 6.59\u20136.60 with the help of the Clebsch\u2013Gordan table. The relevant coef\ufb01cients are shown in Figure 6.8. The nonzero matrix elements are 0 0 0 32 \u2212 2 0 1 1 320 320 0 0 0 322 211 210 \u02c6V+ \u02c6V+ 1 21 \u2212 1 1 211 210 321 \u02c6V+ \u02c6V\u2212 \u02c6V\u2212 1 21 \u2212 1 1 211 32 \u2212 1 \u02c6V\u2212 \u02c6Vz \u02c6Vz 1 21 \u2212 1 320 \u02c6Vz 210 321 1 0 1 0 32 \u2212 112 112 2 C 112 011 2 C 112 \u2212110 32 V 21 = \u2212 \u221a 2 32 V 21 32 V 21 = \u2212 32 V 21 32 V 21 = \u2212 1\u221a 3 32 V 21 \u221a \u221a \u221a = = = 2 C 1 12 1\u221210 32 V 21 = 1\u221a 3 32 V 21 2 C 1 1 2 0\u22121\u22121 32 V 21 = 32 V 21 \u221a 2 32 V 21 = C 112 101 = C 112 000 32 V 21 = 2 C 1 1 2 \u22121\u22121\u22122 32 V 21 = 1\u221a 2 2 3 32 V 21 = 1\u221a 2 32 V 21 = = C 11 2 \u221210\u22121 32 V 21 32 V 21 32 V 21, 6.7 Rotational Selection Rules 261 with the reduced matrix element given by Equation 6.63. The other thirty-six matrix elements vanish due to the selection rules (Equations 6.56\u20136.58 and 6.62). We have determined all forty-\ufb01ve matrix elements and have only needed to evaluate a single integral. I\u2019ve left the matrix elements in terms of V+ and V\u2212 but it\u2019s straightforward to write them in terms of x and y using the expressions on page 259. 1\u00d71 +1 +1 +1 0 2 +2 1 0 +1 2 +1 1/2 1/2 1 +1 1/2 \u20131/2 +1 \u20131 0 0 \u20131 +1 2 0 1/6 2/3 1/6 1 0 0 0 1/2 1/3 0 \u20131/3 \u20131/2 1/3 0 \u20131 \u20131 0 2 \u20131 1/2 1/2 1 \u20131 1/2 2 \u20131/2 \u20132 \u20131 \u20131 1 Figure 6.8: The Clebsch\u2013Gordan coef\ufb01cients for 1 \u2297 1. It is no coincidence that the", " Clebsch\u2013Gordan coef\ufb01cients appear in Equations 6.59\u20136.61. States have angular momentum, but operators also carry angular momentum. A scalar operator (Equation 6.34) has = 0\u2014it is unchanged by a rotation\u2014just as a state of angular momentum 0 is unchanged. A vector operator (Equation 6.33) has = 1; its three components transform into each other under a rotation in the same way the triplet of states with angular momentum = 1 transform into each other.31 When we act on a state with an operator, we add together the angular momentum of the state and the operator to obtain the angular momentum of the resultant state; this addition of angular momenta is the source of the Clebsch\u2013Gordan coef\ufb01cients in Equations 6.59\u20136.61.32 \u2217 Problem 6.22 (a) Show that the commutation relations, Equations 6.50\u20136.54, follow from the definition of a vector operator, Equation 6.33. If you did Problem 6.19 you already derived one of these. (b) Derive Equation 6.57. 31 In the case of the position operator \u02c6r, this correspondence is particularly evident when we rewrite the operator with the help of Table 4.3: x \u00b1 i y = r sin \u03b8 e 8 \u03c0 3 \u00b11 Y 1 (\u03b8, \u03c6) \u00b1i \u03c6 = \u2213r 32 Since C 0 m0m = \u03b4 z = r cos \u03b8 = r 4 \u03c0 3 mm \u03b4, one could rewrite the selection rules for a scalar operator (Equation 6.47) as 0 m0m nm @ @ @n (\u03b8, \u03c6). \u02c6f @ @ @ 262 CHAPTER 6 Symmetries & Conservation Laws (6.64) (6.65) Problem 6.23 The Clebsch\u2013Gordan coef\ufb01cients are de\ufb01ned by Equation 4.183. Adding together two states with angular momentum j1 and j2 produces a state with total angular momentum J according to |J M = C j1 j2 J m1 m2 M | j1 j2 m1 m2. m1,m2 (a) From Equation 6.64, show that the Clebsch\u2013Gordan coef\ufb01cients satisfy C j1 j2 J m1 m2 M = j1 j2 m1 m2 | J M. (b) Apply \u02c6J\u00b1 = \u02c6J (1) \u00b1 + \u02c6J (2) \u00b1 to Equation 6.64 to derive the recursion relations for Clebsch\u2013Gordan coef\ufb01cients: m1 m2 M+1 J C j1 j2 J AM J C j1 j2 J B M m1 m2 M\u22121 m1\u22121 m2 M = Bm1 j1 = Am1 j1 C j1 j2 J C j1 j2 J m+1 m2 M + Bm2 j2 + Am2 j2 C j1 j2 J C j1 j2 J m1 m2+1 M m1 m2\u22121 M (6.66) \u2217\u2217\u2217 Problem 6.24 (a) Sandwich each of the six commutation relations in Equations 6.52\u20136.54 between \u02c6V. As an nm| and |nm to obtain relations between matrix elements of example, Equation 6.52 with the upper signs gives n m \u2212 1 V+ n m Bm = Am n m V+ n m + 1. (b) Using the results in Problem 6.23, show that the six expressions you wrote down in part (a) are satis\ufb01ed by Equations 6.59\u20136.61. \u2217 Problem 6.25 Express the expectation value of the dipole moment pe for an electron in the hydrogen state \u03c8 = 1\u221a 2 (\u03c8211 + \u03c8210) in terms of a single reduced matrix element, and evaluate the expectation value. Note: this is the expectation value of a vector so you need to compute all three components. Don\u2019t forget Laporte\u2019s rule! 6.8 TRANSLATIONS IN TIME In this section we study time-translation invariance. Consider a solution (x, t) to the timedependent Schr\u00f6dinger equation \u02c6H (x, t) = i \u2202 \u2202t (x, t). We can de\ufb01ne the operator that propagates the wave function forward in time, \u02c6U (t) by \u02c6U (t) (x, 0) = (x, t) ; (6.67) \u02c6U (t", ") can be expressed in terms of the Hamiltonian, and doing so is straightforward if the Hamiltonian is not itself a function of time. In that case, expanding the right-hand side of Equation 6.67 in a Taylor series gives33 6.8 Translations in Time 263 \u02c6U (t) (x, 0) = (x, t) = \u221e (x, t) t=0 t n \u2202 n \u2202t n 1 n! n \u02c6H t (x, 0). n=0 \u2212 i = \u221e n=0 1 n! (6.68) (6.69) Therefore, in the case of a time-independent Hamiltonian, the time-evolution operator is34 \u02c6U (t) = exp \u2212 i t \u02c6H. (6.71) We say that the Hamiltonian is the generator of translations in time. Note that \u02c6U (t) is a unitary operator (see Problem 6.2). The time-evolution operator offers a compact way to state the procedure for solving the time-dependent Schr\u00f6dinger equation. To see the correspondence, write out the wave function at time t = 0 as a superposition of stationary states \u02c6H \u03c8n = En \u03c8n : (x, 0) = n cn \u03c8n(x). Then (x, t) = \u02c6U (t) (x, 0) = cn n \u02c6U (t) \u03c8n(x) = cn e\u2212i \u02c6H t/ \u03c8n(x) = cn e\u2212i En t/ \u03c8n(x). n n In this sense Equation 6.71 is shorthand for the process of expanding the initial wave function in terms of stationary states and then tacking on the \u201cwiggle factors\u201d to obtain the wave function at a later time (Section 2.1). 33 Why is this analysis limited to the case where \u02c6H is independent of time? Whether or not \u02c6H depends on time, \u02c6H is time dependent then the second derivative of is Schr\u00f6dinger\u2019s equation says i \u02d9 = \u02c6H. However, if given by \u22022 \u2202t 2 = \u2202 \u2202t 1 i \u02c6H = 1 i \u2202 \u02c6H \u2202t \u2212 1 2 \u02c6H 2 and higher derivatives will be even more complicated. Therefore, Equation 6.69 only follows from Equation 6.68 when \u02c6H has no time dependence. See also Problem 11.23. 34 This derivation assumes that the actual solution to Schrodinger\u2019s equation, (x, t), can be expanded as a Taylor series in t, and nothing guarantees that. B. R. Holstein and A. R. Swift, A. J. Phys. 40, 829 (1989) give an innocentseeming example where such an expansion does not exist. Nonetheless, Equation 6.71 still holds in such cases as long as we de\ufb01ne the exponential function through its spectral decomposition (Equation 3.103): \u02c6U (t) \u2261 n \u2212i En t/ |\u03c8n \u03c8n |. e (6.70) See also M. Amaku et al., Am. J. Phys. 85, 692 (2017). 264 CHAPTER 6 Symmetries & Conservation Laws 6.8.1 The Heisenberg Picture Just as for the other transformations studied in this chapter, we can examine the effect of applying time translation to operators, as well as to wave functions. The transformed operators are called Heisenberg-picture operators and we follow the convention of giving them a subscript H rather than a prime: \u02c6Q H (t) = \u02c6U \u2020(t) \u02c6Q \u02c6U (t). (6.72) Example 6.6 A particle of mass m moves in one dimension in a potential V (x): \u02c6H = \u02c6p2 2 m + V (x). Find the position operator in the Heisenberg picture for an in\ufb01nitesimal time translation \u03b4. Solution: From Equation 6.71, Applying Equation 6.72, we have \u02c6U (\u03b4) \u2248 1 \u2212 i \u03b4 \u02c6H. \u02c6x H (\u03b4) \u2248 1 + i \u2248 \u02c6x \u2212 i \u03b4 \u03b4 # \u02c6H \u2020 \u02c6x, \u02c6H 1 \u2212 i \u02c6x $ \u2248 \u02c6x \u2212 i \u02c6H i \u02c6p m \u03b4 \u03b4 so \u02c6x H (\u03b4)", " \u2248 \u02c6x H (0) + 1 m \u02c6pH (0) \u03b4 (making use of the fact that the Heisenberg-picture operators at time 0 are just the untransformed operators). This looks exactly like classical mechanics: x(\u03b4) \u2248 x(0) + v(0) \u03b4. The Heisenberg picture illuminates the connection between classical and quantum mechanics: the quantum operators obey the classical equations of motion (see Problem 6.29). Example 6.7 A particle of mass m moves in one dimension in a harmonic-oscillator potential: \u02c6p2 2 m m \u03c92 x 2. + 1 2 \u02c6H = Find the position operator in the Heisenberg picture at time t. Solution: Consider the action of \u02c6x H on a stationary state \u03c8n. (Introducing \u03c8n allows us to replace the operator e\u2212i \u02c6H t/ with the number e\u2212i En t/, since e\u2212i \u02c6H t/ \u03c8n = e\u2212i En t/ \u03c8n.) Writing \u02c6x in terms of raising and lowering operators we have (using Equations 2.62, 2.67, and 2.70) \u02c6x H (t) \u03c8n(x) = \u02c6U \u2020(t) \u02c6x \u02c6U (t) \u03c8n(x) = ei \u02c6H t/ 2 m \u03c9 \u02c6a+ + \u02c6a\u2212 e\u2212i \u02c6H t/ \u03c8n(x) 6.8 Translations in Time 265 \u03c8n(x \u221a + = = = = \u02c6a+ + \u02c6a\u2212 #\u221a e\u2212i En t/ ei \u02c6H t/ e\u2212i En t/ ei \u02c6H t/ #\u221a e\u2212i En t/ n + 1 ei En+1 t/ \u03c8n+1(x) $ n + 1 \u03c8n+1(x) + $ \u221a n \u03c8n\u22121(x) n ei En\u22121 t/ \u03c8n\u22121(x) 2 m \u03c9 #\u221a n + 1 ei \u03c9 t \u03c8n+1(x) + \u221a n e\u2212i \u03c9 t \u03c8n\u22121(x) $. (6.73) Thus35 \u02c6x H (t) = 2 m \u03c9 # ei \u03c9 t \u02c6a+ + e\u2212i \u03c9 t \u02c6a\u2212 $. Or, using Equation 2.48 to express \u02c6a\u00b1 in terms of \u02c6x and \u02c6p, \u02c6x H (t) = \u02c6x H (0) cos(\u03c9 t) + 1 m \u03c9 \u02c6pH (0) sin(\u03c9 t). (6.74) As in Example 6.6 we see that the Heisenberg-picture operator satis\ufb01es the classical equation of motion for a mass on a spring. In this book we have been working in the Schr\u00f6dinger picture, so-named by Dirac because it was the picture that Schr\u00f6dinger himself had in mind. In the Schr\u00f6dinger picture, the wave function evolves in time according to the Schr\u00f6dinger equation \u02c6H (x, t) = i \u2202 \u2202t (x, t). The operators \u02c6x = x and \u02c6p = \u2212i \u2202x have no time dependence of their own, and the time dependence of expectation values (or, more generally, matrix elements) comes from the time dependence of the wave function:36 0 1 0 \u02c6Q = (t) \u02c6Q 1 (t). In the Heisenberg picture, the wave function is constant in time, H (x) = (x, 0), and the operators evolve in time according to Equation 6.72. In the Heisenberg picture, the time dependence of expectation values (or matrix elements) is carried by the operators. 0 0 1 1 \u02c6Q = H \u02c6Q H (t) H Of course, the two pictures are entirely equivalent since 0 (t) \u02c6Q 1 (t) 0 = (0) \u02c6U \u2020 \u02c6Q \u02c6U 1 (0) \u02c6Q H (t) H 1. = H. 0 35 Since Equation 6.73 holds for any stationary state \u03c8n and since the \u03c8n constitute a complete set of states, the operators must in fact be identical. 36 I am assuming that \u02c6Q, like \u02c6x or ", "\u02c6p, has no explicit time dependence. 266 CHAPTER 6 Symmetries & Conservation Laws A nice analogy for the two pictures runs as follows. On an ordinary clock, the hands move in a clockwise direction while the numbers stay \ufb01xed. But one could equally well design a clock where the hands are stationary and the numbers move in the counter-clockwise direction. The correspondence between these two clocks is roughly the correspondence between the Schr\u00f6dinger and Heisenberg pictures, the hands representing the wave function and the numbers representing the operator. Other pictures could be introduced as well, in which both the hands of the clock and the numbers on the dial move at intermediate rates such that the clock still tells the correct time.37 Problem 6.26 Work out correspondence with the classical equation of motion. \u02c6pH (t) for the system in Example 6.7 and comment on the Problem 6.27 Consider a free particle of mass m. Show that the position and momentum operators in the Heisenberg picture are given by \u02c6x H (t) = \u02c6x H (0) + 1 m \u02c6pH (t) = \u02c6pH (0). \u02c6pH (0) t \u2217 \u2217\u2217 Comment on the relationship between these equations and the classical equations of $ motion. Hint: you will \ufb01rst need to evaluate the commutator ; this will allow you \u02c6x, \u02c6U to evaluate the commutator \u02c6x, \u02c6H n $ # #. 6.8.2 Time-Translation Invariance If the Hamiltonian is time-dependent one can still write the formal solution to the Schr\u00f6dinger equation in terms of the time-translation operator, \u02c6U : (x, t) = \u02c6U (t, t0) (x, t0), (6.75) but \u02c6U no longer takes the simple form 6.71.38 (See Problem 11.23 for the general case.) For an in\ufb01nitesimal time interval \u03b4 (see Problem 6.28) \u02c6U (t0 + \u03b4, t0) \u2248 1 \u2212 i \u02c6H (t0) \u03b4. (6.76) Time-translation invariance means that the time evolution is independent of which time interval we are considering. In other words \u02c6U (t1 + \u03b4, t1) = \u02c6U (t2 + \u03b4, t2) (6.77) for any choice of t1 and t2. This ensures that if the system starts in state |\u03b1 at time t1 and evolves for a time \u03b4 then it will end up in the same state |\u03b2 as if the system started in the same state |\u03b1 at time t2 and evolved for the same amount of time \u03b4; i.e. the experiment proceeds the same on Thursday as it did on Tuesday, assuming identical conditions. Plugging Equation 6.76 into Equation 6.77 we see that the requirement for this to be true is \u02c6H (t1) = \u02c6H (t2), and since 37 Of these other possible pictures the most important is the interaction picture (or Dirac picture) which is often employed in time-dependent perturbation theory. 38 And is a function of both the initial time t0 and the \ufb01nal time t, not simply the amount of time for which the wave function has evolved. this must hold true for all t1 and t2, it must be that the Hamiltonian is in fact time-independent after all (for time-translation invariance to hold): 6.8 Translations in Time 267 \u2202 \u02c6H \u2202t = 0. In that case the generalized Ehrenfest theorem says $1 0# 0 1 d dt \u02c6H = i \u02c6H, \u02c6H 3 2 + \u2202 \u02c6H \u2202t = 0. Therefore, energy conservation is a consequence of time-translation invariance. We have now recovered all the classical conservation laws: conservation of momentum, energy, and angular momentum, and seen that they are each related to a continuous symmetry of the Hamiltonian (spatial translation, time translation, and rotation, respectively). And in quantum mechanics, discrete symmetries (such as parity) can also lead to conservation laws. Problem 6.28 Show that Equations 6.75 and 6.76 are the solution to the Schr\u00f6dinger equation for an in\ufb01nitesimal time \u03b4. Hint: expand (x, t) in a Taylor series. \u2217 Problem 6.29 Differentiate Equation 6.72 to obtain the Heisenberg equations of motion i d dt \u02c6Q H (t)", " = # $ \u02c6Q H (t), \u02c6H (6.78) (for \u02c6Q and \u02c6H independent of time).39 Plug in \u02c6Q = \u02c6x and \u02c6Q = \u02c6p to obtain the differential equations for \u02c6x H and \u02c6pH in the Heisenberg picture for a single particle of mass m moving in a potential V (x). \u2217\u2217\u2217 Problem 6.30 Consider a time-independent Hamiltonian for a particle moving in one dimension that has stationary states \u03c8n(x) with energies En. (a) Show that the solution to the time-dependent Schr\u00f6dinger equation can be written (x, t) = \u02c6U (t) (x, 0) = K x, x, t x, 0 d x, where K x, x, t, known as the propagator, is K x\u2212i En t/ \u03c8n(x). (6.79) n K 2 is the probability for a quantum mechanical particle to travel x, x, t Here from position x to position x in time t. (b) Find K for a particle of mass m in a simple harmonic oscillator potential of frequency \u03c9. You will need the identity 39 For time-dependent \u02c6Q and \u02c6H the generalization is # i d dt \u02c6Q H (t) = $ \u02c6Q H (t), \u02c6HH (t) + \u02c6U \u2020 \u2202 \u02c6Q \u2202t \u02c6U. 268 CHAPTER 6 Symmetries & Conservation Laws 1\u221a 1 \u2212 z2 exp \u2212 \u03be 2 + \u03b72 \u2212 2 \u03be \u03b7 z 1 \u2212 z2 = e\u2212\u03be 2 e\u2212\u03b72 \u221e n=0 zn 2n n! Hn(\u03be ) Hn(\u03b7). (c) Find (x, t) if the particle from part (a) is initially in the state40 (x, 0) = e\u2212a(x\u2212x0)2. 1/4 2 a \u03c0 Compare your answer with Problem 2.49. Note: Problem 2.49 is a special case with a = m\u03c9/2. (d) Find K for a free particle of mass m. In this case the stationary states are continuous, not discrete, and one must make the replacement n in Equation 6.79. \u221e \u2192 dp \u2212\u221e (e) Find (x, t) for a free particle that starts out in the state (x, 0) = 1/4 2 a \u03c0 e\u2212a x 2. Compare your answer with Problem 2.21. FURTHER PROBLEMS ON CHAPTER 6 Problem 6.31 In deriving Equation 6.3 we assumed that our function had a Taylor series. The result holds more generally if we de\ufb01ne the exponential of an operator by its spectral decomposition, \u02c6T (a) = e\u2212i a p/ | p p| dp, (6.80) rather than its power series. Here I\u2019ve given the operator in Dirac notation; acting on a position-space function (see the discussion on page 123) this means \u02c6T (a) \u03c8(x) = e\u2212i a p/ f p(x) ( p) dp, (6.81) \u221e \u2212\u221e where ( p) is the momentum space wave function corresponding to \u03c8(x) and f p(x) is de\ufb01ned in Equation 3.32. Show that the operator \u02c6T (a), as given by Equation 6.81, applied to the function \u221a \u03c8(x) = \u03bb e\u2212\u03bb |x| (whose \ufb01rst derivative is unde\ufb01ned at x = 0) gives the correct result. 40 The integrals in (c)\u2013(e) can all be done with the following identity: \u221e \u2212\u221e \u2212ax 2+bx d x = e which was derived in Problem 2.21. eb2/4a, \u03c0 a Further Problems on Chapter 6 269 \u2217\u2217 Problem 6.32 Rotations on spin states are given by an expression identical to Equation 6.32, with the spin angular momentum replacing the orbital angular momentum: # \u2212i Rn(\u03d5) = exp n \u00b7 S $. \u03d5 In this problem we will consider rotations of a spin-1/2 state. (a) Show that (a \u00b7 \u03c3 ) (a \u00d7 b) \u00b7 \u03c3, where the \u03c3i are the Pauli spin matrices and a and b are ordinary vectors. Use the result of Problem", " 4.29. (b) Use your result from part (a) to show that # \u2212i \u03d5 exp $ n \u00b7 S = cos \u03d5 2 \u2212 i sin \u03d5 2 n \u00b7 \u03c3. Recall that S = (/2) \u03c3. (c) Show that your result from part (b) becomes, in the standard basis of spin up and spin down along the z axis, the matrix \u03d5 2 Rn = cos 1 0 0 1 \u2212 i sin \u03d5 2 cos \u03b8 sin \u03b8 ei \u03c6 sin \u03b8 e\u2212i \u03c6 \u2212 cos \u03b8 where \u03b8 and \u03c6 are the polar coordinates of the unit vector n that describes the axis of rotation. (d) Verify that the matrix Rn in part (c) is unitary. (e) Compute explicitly the matrix S = R\u2020 Sx R where R is a rotation by an angle \u03d5 about the z axis and verify that it returns the expected result. Hint: rewrite your x in terms of Sx and Sy. result for S (f) Construct the matrix for a \u03c0 rotation about the x axis and verify that it turns an up x spin into a down spin. (g) Find the matrix describing a 2 \u03c0 rotation about the z axis. Why is this answer surprising?41 Problem 6.33 Consider a particle of mass m in a two-dimensional in\ufb01nite square well with sides of length L. With the origin placed at the center of the well, the stationary states can be written as \u03c8nx n y (x, y) = 2 L sin nx \u03c0 L x \u2212 L 2 sin n y \u03c0 L y \u2212 L 2, with energies Enx n2 x + n2 y, for positive integers nx and n y. 41 For a discussion of how this sign change is actually measured, see S. A. Werner et al., Phys. Rev. Lett. 35, 1053 (1975). 270 CHAPTER 6 Symmetries & Conservation Laws (a) The two states \u03c8ab and \u03c8ba for a = b are clearly degenerate. Show that a rotation by 90\u25e6 counterclockwise about the center of the square carries one into the other, \u02c6R \u03c8ab \u221d \u03c8ba, and determine the constant of proportionality. Hint: write \u03c8ab in polar coordinates. (b) Suppose that instead of \u03c8ab and \u03c8ba we choose the basis \u03c8+ and \u03c8\u2212 for our two degenerate states: \u03c8\u00b1 = \u03c8ab \u00b1 \u03c8ba\u221a 2. Show that if a and b are both even or both odd, then \u03c8+ and \u03c8\u2212 are eigenstates of the rotation operator. (c) Make a contour plot of the state \u03c8\u2212 for a = 5 and b = 7 and verify (visually) that it is an eigenstate of every symmetry operation of the square (rotation by an integer multiple of \u03c0/2, re\ufb02ection across a diagonal, or re\ufb02ection along a line bisecting two sides). The fact that \u03c8+ and \u03c8\u2212 are not connected to each other by any symmetry of the square means that there must be additional symmetry explaining the degeneracy of these two states.42 \u2217\u2217\u2217 Problem 6.34 The Coulomb potential has more symmetry than simply rotational invariance. This additional symmetry is manifest in an additional conserved quantity, the Laplace\u2013Runge\u2013Lenz vector \u02c6M = \u02c6p \u00d7 \u02c6L \u2212 \u02c6L \u00d7 \u02c6p 2 m + V (r ) r, where V (r) is the potential energy, V (r ) = \u2212e2/4 \u03c0 0 r.43 The complete set of commutators for the conserved quantities in the hydrogen atom is # $ (i) (ii) (iii) (iv) (v) # # # # = 0 = 0 \u02c6H, \u02c6Mi $ \u02c6H, \u02c6Li \u02c6Li, \u02c6L j \u02c6Li, \u02c6M j $ $ $ \u02c6Mi, \u02c6M j = \u02c6L k \u02c6Mk = i i jk = i i jk i jk L k i \u02c6H. 2 m The physical content of these equations is that (i) M is a conserved quantity, (ii) L is a conserved quantity, (iii) L is a vector, and (iv) M is a vector ((v) has no obvious interpretation). There are two additional relations between the quantities \u02c6L, \u02c6M, and \u02c6H. They are 42 See F. Leyvraz, et al., Am. J. Phys. 65", ", 1087 (1997) for a discussion of this \u201caccidental\u201d degeneracy. 43 The full symmetry of the Coulomb Hamiltonian is not just the obvious three-dimensional rotation group (known to mathematicians as SO(3)), but the four-dimensional rotation group (SO(4)), which has six generators (L and M). (If the four axes are w, x, y, and z, the generators correspond to rotations in each of the six orthogonal planes, wx, wy, wz (that\u2019s M) and yz, zx, x y (that\u2019s L). Further Problems on Chapter 6 271 \u02c6M 2 = 2 e2 4 \u03c0 0 + 2 m \u02c6H \u02c6L 2 + 2 \u02c6M \u00b7 \u02c6L = 0. (vi) (vii) (a) From the result of Problem 6.19, and the fact that \u02c6M is a conserved quantity, we \u02c6M+ \u03c8n = cn \u03c8n(+1)(+1) for some constants cn. Apply (vii) to the know that state \u03c8n to show that \u02c6Mz \u03c8n = \u2212 1\u221a 2 1\u221a + 1 cn \u03c8n(+1). (b) Use (vi) to show that \u02c6M\u2212 \u02c6M+ \u03c8n = e2 \u03c8n \u2212 \u02c6M 2 z \u03c8n. (c) From your results to parts (a) and (b), obtain the constants cn. You should \ufb01nd that cn is nonzero unless = n \u2212 1. Hint: Consider |M+\u03c8nm|2d3r and use the fact that M\u00b1 are Hermitian conjugates. Figure 6.9 shows how the degenerate states of hydrogen are related by the generators \u02c6L and \u02c6M. \u00d8 \u02c6M+ \u02c6M+ 300 \u02c6M+ 311 \u02c6L\u2212 310 \u02c6L\u2212 31\u22121 322 321 320 \u02c6L\u2212 \u02c6L\u2212 \u02c6L\u2212 32\u22121 \u02c6L\u2212 32\u22122 Figure 6.9: The degenerate n = 3 states of the hydrogen atom, and the symmetry operations that connect them. 272 CHAPTER 6 Symmetries & Conservation Laws \u2217\u2217 Problem 6.35 A Galilean transformation performs a boost from a reference frame S to a reference frame S moving with velocity \u2212v with respect to S (the origins of the two frames coincide at t = 0). The unitary operator that carries out a Galilean transformation at time t is \u02c6(v, t) = exp \u2212 i v t \u02c6p \u2212 m \u02c6x. (a) Find \u02c6x = \u02c6\u2020 \u02c6x \u02c6 and \u02c6p = \u02c6\u2020 \u02c6p \u02c6 for an in\ufb01nitesimal transformation with velocity \u03b4. What is the physical meaning of your result? (b) Show that \u02c6(v, t) = exp i m v2 \u02c6T (vt) m v2 t. = \u02c6T (vt) exp where \u02c6T is the spatial translation operator (Equation 6.3). You will need to use the Baker\u2013Campbell\u2013Hausdorff formula (Problem 3.29). (c) Show that if is a solution to the time-dependent Schr\u00f6dinger equation with Hamiltonian \u02c6p2 2 m then the boosted wave function = \u02c6(v, t) is a solution to the time-dependent Schr\u00f6dinger equation with the potential V (x) in motion: + V (x) \u02c6H = d/dt \u02c6A = e \u02c6A e Note: d \u02c6A/dt \u02c6p2 2 m \u02c6H = only if + V (x \u2212 v t). $ # \u02c6A, d \u02c6A/dt = 0. (d) Show that the result of Problem 2.50(a) is an example of this result. Problem 6.36 A ball thrown through the air leaves your hand at position r0 with a velocity of v0 and arrives a time t later at position r1 traveling with a velocity v1 (Figure 6.10). Suppose we could instantaneously reverse the ball\u2019s velocity when it reaches r1. Neglecting air resistance, it would retrace the path that took it from r0 to r1 and arrive back at r0 after another time t had passed, traveling with a velocity \u2212v0. This is an example of time-reversal invariance\u2014", "reverse the motion of a particle at any point along its trajectory and it will retrace its path with an equal and opposite velocity at all positions. Why is this called time reversal? After all, it was the velocity that was reversed, not time. Well, if we showed you a movie of the ball traveling from r1 to r0, there would be no way to tell if you were watching a movie of the ball after the reversal playing forward, or a movie of the ball before the reversal playing backward. In a time-reversal invariant system, playing the movie backwards represents another possible motion. Further Problems on Chapter 6 273 y y \u03c5 0 \u03c5 1 \u2212\u03c5 1 x x \u2212\u03c5 0 Figure 6.10: A ball thrown through the air (ignore air resistance) is an example of a system with time-reversal symmetry. If we \ufb02ip the velocity of the particle at any point along its trajectory, it will retrace its path. A familiar example of a system that does not exhibit time-reversal symmetry is a charged particle moving in an external magnetic \ufb01eld.44 In that case, when you reverse the velocity of the particle, the Lorentz force will also change sign and the particle will not retrace its path; this is illustrated in Figure 6.11. y \u03c5 0 \u03c5 1 x y \u2212\u03c5 1 x Figure 6.11: An external magnetic \ufb01eld breaks time-reversal symmetry. Shown is the trajectory of a particle of charge +q traveling in a uniform magnetic \ufb01eld pointing into the page. If we \ufb02ip the particle\u2019s velocity from v1 to \u2212v1 at the point shown, the particle does not retrace its path, but instead moves onto a new circular orbit. 44 By external magnetic \ufb01eld, I mean that we we only reverse the velocity of our charge q, and not the velocities of the charges producing the magnetic \ufb01eld. If we reversed those velocities as well, the magnetic \ufb01eld would also switch directions, the Lorentz force on the charge q would be unchanged by the reversal, and the system would in fact be time-reversal invariant. 274 CHAPTER 6 Symmetries & Conservation Laws The time-reversal operator \u02c6 is the operator that reverses the momentum of the particle (p \u2192 \u2212p), leaving its position unchanged. A better name would really be the \u201creversal of the direction of motion\u201d operator.45 For a spinless particle, the timereversal operator \u02c6 simply complex conjugates the position-space wave function46 \u02c6 (x, t) = \u2217(x, t). (6.82) (a) Show that the operators \u02c6x and \u02c6p transform under time reversal as \u02c6x = \u02c6\u22121 \u02c6x \u02c6 = \u02c6x \u02c6p = \u02c6\u22121 \u02c6p \u02c6 = \u2212 \u02c6p. Hint: Do this by calculating the action of \u02c6x and \u02c6p on an arbitrary test function f(x). (b) We can write down a mathematical statement of time-reversal invariance from our discussion above. We take a system, evolve it for a time t, reverse its momentum, and evolve it for time t again. If the system is time-reversal invariant it will be back where it started, albeit with its momentum reversed (Figure 6.10). As an operator statement this says \u02c6U (t) \u02c6 \u02c6U (t) = \u02c6. If this is to hold for any time interval, it must hold in particular for an in\ufb01nitesimal time interval \u03b4. Show that time-reversal invariance requires # $ \u02c6, \u02c6H = 0. (6.83) (c) Show that, for a time-reversal invariant Hamiltonian, if \u03c8n(x) is a stationary state with energy En, then \u03c8 \u2217 (x) is also a stationary state with the same energy En. If n the energy is nondegenerate, this means that the stationary state can be chosen as real. (d) What do you get by time-reversing a momentum eigenfunction f p(x) (Equation 3.32)? How about a hydrogen wave function \u03c8nm(r, \u03b8, \u03c6)? Comment on each state\u2019s relation to the untransformed state and verify that the transformed and untransformed states share the same energy, as guaranteed by (c). Problem 6.37 As an angular momentum, a particle\u2019s spin must \ufb02ip", " under time reversal (Problem 6.36). The action of time-reversal on a spinor (Section 4.4.1) is in fact 45 See Eugene P. Wigner, Group Theory and its Applications to Quantum Mechanics and Atomic Spectra (Academic Press, New York, 1959), p. 325. 46 Time reversal is an anti-unitary operator. An anti-unitary operator satis\ufb01es f | g = f | g\u2217 \u02c6 (a |\u03b1 + b |\u03b2) = a \u2217 \u02c6 |\u03b1 + b \u2217 \u02c6 |\u03b2 whereas a unitary operator satis\ufb01es the same two equations without the complex conjugates. I won\u2019t de\ufb01ne the adjoint of an anti-unitary operator; instead I use \u02c6\u22121 for an anti-unitary operator where we might have used \u02c6U \u2020 or \u02c6U \u22121 interchangeably for a unitary operator. Further Problems on Chapter 6 275 \u02c6 a b = \u2212b\u2217 a\u2217 (6.84) so that, in addition to the complex conjugation, the up and down components are interchanged.47 (a) Show that \u02c62 = \u22121 for a spin-1/2 particle. (b) Consider an eigenstate |\u03c8n of a time-reversal invariant Hamiltonian (Equa = \u02c6 |\u03c8n is also an eigenstate tion 6.83) with energy En. We know that |\u03c8 n and |\u03c8n are of the same state (meaning |\u03c8 = c |\u03c8n for some complex constant c) or they are n distinct states. Show that the \ufb01rst case leads to a contradiction in the case of a spin-1/2 particle, meaning the energy level must be (at least) two-fold degenerate in that case. \u02c6H with the same energy En. There two possibilities: either |\u03c8 n Comment: What you have proved is a special case of Kramer\u2019s degeneracy: for an odd number of spin-1/2 particles (or any half-integer spin for that matter), every energy level (of a time-reversal-invariant Hamiltonian) is at least two-fold degenerate. This is because\u2014as you just showed\u2014for half-integer spin a state and its time-reversed state are necessarily distinct.48 47 For arbitrary spin, \u02c6 = e \u2212i \u03c0 \u02c6Sy / \u02c6K (6.85) where the \ufb01rst term is a rotation by \u03c0 about the y axis and \u02c6K is the operator that performs the complex conjugation. 48 What about in the case of a spin-0 particle\u2014does time-reversal symmetry tell us anything interesting? Actually it does. For one thing, the stationary states can be chosen as real; you proved this back in Problem 2.2 but we now see that it is a consequence of time-reversal symmetry. Another example is the degeneracy of the energy levels in a periodical potential (Section 5.3.2 and Problem 6.6) for states with crystal momentum q and \u2212q. This can be ascribed to inversion symmetry if the potential is symmetric, but the degeneracy persists even when inversion symmetry is absent (try it out!); that is a result of time-reversal symmetry. Part II Applications \u25c8 7 TIME-INDEPENDENT PERTURBATION THEORY 7.1 NONDEGENERATE PERTURBATION THEORY 7.1.1 General Formulation Suppose we have solved the (time-independent) Schr\u00f6dinger equation for some potential (say, the one-dimensional in\ufb01nite square well): H 0\u03c8 0 n = E 0 n \u03c8 0 n, obtaining a complete set of orthonormal eigenfunctions = \u03b4nm, (7.1) (7.2) and the corresponding eigenvalues E 0 n. Now we perturb the potential slightly (say, by putting a little bump in the bottom of the well\u2014Figure 7.1). We\u2019d like to \ufb01nd the new eigenfunctions and eigenvalues: H \u03c8n = En\u03c8n, (7.3) but unless we are very lucky, we\u2019re not going to be able to solve the Schr\u00f6dinger equation exactly, for this more complicated potential. Perturbation theory is a systematic procedure for obtaining approximate solutions to the perturbed problem, by building on the known exact solutions to the unperturbed case. To begin with we write the new Hamiltonian as the sum of two terms: H = H 0 + \u03bb", "H, (7.4) where H is the perturbation (the superscript 0 always identi\ufb01es the unperturbed quantity). For the moment we\u2019ll take \u03bb to be a small number; later we\u2019ll crank it up to 1, and H will be the true Hamiltonian. Next we write \u03c8n and En as power series in \u03bb: + \u00b7 \u00b7 \u00b7 ; \u03c8n = \u03c8 0 n + \u03bb\u03c8 1 n + \u03bb2\u03c8 2 n (7.5) En = E 0 n n is the \ufb01rst-order correction to the nth eigenvalue, and \u03c8 1 + \u03bb2 E 2 n + \u03bbE 1 n + \u00b7 \u00b7 \u00b7. Here E 1 to the nth eigenfunction; E 2 Equations 7.5 and 7.6 into Equation 7.3, we have: n and \u03c8 2 n is the \ufb01rst-order correction n are the second-order corrections, and so on. Plugging (7.6) # H 0 + \u03bbH \u03c8 0 n + \u03bb\u03c8 1 n + \u03bb2\u03c8 + \u03bbE 1 n + \u03bb2 + \u03bb\u03c8 1 n + \u03bb2\u03c8 2 n + \u00b7 \u00b7 \u00b7, $ 280 CHAPTER 7 Time-Independent Perturbation Theory V (x) a x Figure 7.1: In\ufb01nite square well with small perturbation. or (collecting like powers of \u03bb): H 0\u03c8 0 n + \u03bb H 0\u03c8 1 n + H \u03c8 0 n + \u03bb2 H 0\u03c8 \u03bb0 To lowest order1, \ufb01rst order \u03bb1 + + \u03bb2 + \u00b7 \u00b7 \u00b7. this yields H 0\u03c8 0 n = E 0 n \u03c8 0 n, which is nothing new (Equation 7.1). To To second order \u03bb2, H 0\u03c8 0\u03c8 7.7) (7.8) and so on. (I\u2019m done with \u03bb, now\u2014it was just a device to keep track of the different orders\u2014so crank it up to 1.) 7.1.2 First-Order Theory Taking the inner product of Equation 7.7 with \u03c8 0 integrating), n (that is, multiplying by 0 \u03c8 0 n 1 H 0\u03c8 and But H 0 is hermitian, so 0 \u03c8 0 n H 0\u03c8 1 n 1 0 = H 0\u03c8 As always (footnote 34, page 49) the uniqueness of power series expansions guarantees that the coef\ufb01cients of like powers are equal. 7.1 Nondegenerate Perturbation Theory 281 and this cancels the \ufb01rst term on the right. Moreover, \u03c8 0 n |\u03c8 0 n = 1, so2 7.9) This is the fundamental result of \ufb01rst-order perturbation theory; as a practical matter, it may well be the most frequently used equation in quantum mechanics. It says that the \ufb01rst-order correction to the energy is the expectation value of the perturbation, in the unperturbed state. Example 7.1 The unperturbed wave functions for the in\ufb01nite square well are (Equation 2.31) 2 a Suppose we perturb the system by simply raising the \u201c\ufb02oor\u201d of the well a constant amount V0 (Figure 7.2). Find the \ufb01rst-order correction to the energies. (x) = \u03c8 0 n sin x. n\u03c0 a V (x) V0 a x Figure 7.2: Constant perturbation over the whole well. Solution: In this case H = V0, and the \ufb01rst-order correction to the energy of the nth state is V0 \u03c8 0 n = V0 \u03c8 0 n = V0. \u03c8 0 n + V0; they are simply lifted by the amount The corrected energy levels, then, are En \u2248 E 0 n V0. Of course! The only surprising thing is that in this case the \ufb01rst-order theory yields the exact answer. Evidently for a constant perturbation all the higher corrections vanish.3 2 In this context it doesn\u2019t matter whether we write \u03c8 0 n |H \u03c8 0 (with the extra vertical bar), because n we are using the wave function itself to label the state. But the latter notation is preferable, because it frees us from this convention. For instance, if we used |n to denote the nth state of the harmonic oscillator (Equation 2.86), H |n makes sense, but |H n is unintelligible", " (operators act on vectors/functions, not on numbers). |H |\u03c8 0 n or \u03c8 0 n 3 Incidentally, nothing here depends on the speci\ufb01c nature of the in\ufb01nite square well\u2014the same holds for any potential, when the perturbation is a constant. 282 CHAPTER 7 Time-Independent Perturbation Theory On the other hand, if the perturbation extends only half-way across the well (Figure 7.3), then E 1 n = 2V0 a a/2 0 sin2 n\u03c0 a x d x = V0 2. In this case every energy level is lifted by V0/2. That\u2019s not the exact result, presumably, but it does seem reasonable, as a \ufb01rst-order approximation. V (x) V0 a 2 a x Figure 7.3: Constant perturbation over half the well. Equation 7.9 is the \ufb01rst-order correction to the energy; to \ufb01nd the \ufb01rst-order correction to the wave function we rewrite Equation 7.77.10) The right side is a known function, so this amounts to an inhomogeneous differential equation for \u03c8 1 n (like any other function) can be expressed as a linear combination of them: n. Now, the unperturbed wave functions constitute a complete set, so \u03c8 1 \u03c8 1 n = m=n c(n) m \u03c8 0 m. (7.11) (There is no need to include m = n in the sum, for if \u03c8 1 (\u03c8 1 n ), for any constant \u03b1, and we can use this freedom to subtract off the \u03c8 0 n satis\ufb01es Equation 7.10, so too does n term.4) If + \u03b1\u03c8 0 n we could determine the coef\ufb01cients c (n) m, we\u2019d be done. 4 Alternatively, a glance at Equation 7.5 reveals that any \u03c8 0 combined with the \ufb01rst term. In fact, the choice c Equation 7.5\u2014is normalized (to \ufb01rst order in \u03bb): \u03c8n |\u03c8n = \u03c8 0 n but the orthonormality of the unperturbed states means that the \ufb01rst term is 1 and \u03c8 1 n as \u03c8 1 (n) n = 0 ensures that \u03c8n \u2014with 1 as the coef\ufb01cient of \u03c8 0 |\u03c8 0 n n might as well be pulled out and n in + \u03bb2 (\u00b7 \u00b7 \u00b7 ) + \u00b7 \u00b7 \u00b7, |\u03c8 1 = 0, as long n + \u03c8 0 |\u03c8 1 n n = \u03c8 0 |\u03c8 0 n n n component in \u03c8 1 + \u03bb \u03c8 1 n |\u03c8 0 n n has no \u03c8 0 n component. Well, putting Equation 7.11 into Equation 7.10, and using the fact that \u03c8 0 m satis\ufb01es the 7.1 Nondegenerate Perturbation Theory 283 unperturbed Schr\u00f6dinger equation (Equation 7.1), we have E 0 m \u2212 E 0 n c(n=n Taking the inner product with \u03c8 0 l, c(n=. If l = n, the left side is zero, and we recover Equation 7.9; if l = n, we get 0 E 0 l \u2212 E 0 n (n, or so c(n) m = \u03c8 1 n = m=7.12) (7.13). Notice that the denominator is safe (since there is no coef\ufb01cient with m = n) as long as the unperturbed energy spectrum is nondegenerate. But if two different unperturbed states share the same energy, we\u2019re in serious trouble (we divided by zero to get Equation 7.12); in that case we need degenerate perturbation theory, which I\u2019ll come to in Section 7.2. That completes \ufb01rst-order perturbation theory: The \ufb01rst-order correction to the energy, E 1 n, n, is given by is given by Equation 7.9, and the \ufb01rst-order correction to the wave function, \u03c8 1 Equation 7.13. \u2217 Problem 7.1 Suppose we put a delta-function bump in the center of the in\ufb01nite square well: H = \u03b1\u03b4(x \u2212 a/2), where \u03b1 is a constant. (a) Find the \ufb01rst-order correction to the allowed energies. Explain why the energies are not perturbed for even n. (", "b) Find the \ufb01rst three nonzero terms in the expansion (Equation 7.13) of the correction to the ground state, \u03c8 1 1. \u2217 Problem 7.2 For the harmonic oscillator V (x) = (1/2)kx 2, the allowed energies are En = (n + 1/2)\u03c9, (n = 0, 1, 2,...), \u221a where \u03c9 = slightly: k \u2192 (1 + ) k. (Perhaps we cool the spring, so it becomes less \ufb02exible.) k/m is the classical frequency. Now suppose the spring constant increases 284 CHAPTER 7 Time-Independent Perturbation Theory (a) Find the exact new energies (trivial, in this case). Expand your formula as a power series in, up to second order. (b) Now calculate the \ufb01rst-order perturbation in the energy, using Equation 7.9. What is H here? Compare your result with part (a). Hint: It is not necessary\u2014in fact, it is not permitted\u2014to calculate a single integral in doing this problem. Problem 7.3 Two identical spin-zero bosons are placed in an in\ufb01nite square well (Equation 2.22). They interact weakly with one another, via the potential V (x1, x2) = \u2212aV0\u03b4(x1 \u2212 x2) (where V0 is a constant with the dimensions of energy, and a is the width of the well). (a) First, ignoring the interaction between the particles, \ufb01nd the ground state and the \ufb01rst excited state\u2014both the wave functions and the associated energies. (b) Use \ufb01rst-order perturbation theory to estimate the effect of the particle\u2013particle interaction on the energies of the ground state and the \ufb01rst excited state. 7.1.3 Second-Order Energies Proceeding as before, we take the inner product of the second-order equation (Equation 7.8) with 0\u03c8. Again, we exploit the hermiticity of H 0: 0 0 1 H 0\u03c8 2 n \u03c8 0 n = H 0\u03c8, so the \ufb01rst term on the left cancels the \ufb01rst term on the right. Meanwhile, \u03c8 0 n we are left with a formula for E 2 n: |\u03c8 0 n = 1, and 7.14) But \u03c8 0 n \u03c8 1 n = m=n \u03c8 0 n \u03c8 0 m c(n) m = 0 (because the sum excludes m = n, and all the others are orthogonal), so (n=n m=n or, \ufb01nally=n This is the fundamental result of second-order perturbation theory. \u03c8 0 m, (7.15) We could go on to calculate the second-order correction to the wave function, the third-order correction to the energy, and so on, but in practice Equation 7.15 is ordinarily as far as it is useful to pursue this method.5 \u03c8 2 n 7.1 Nondegenerate Perturbation Theory 285 \u2217\u2217 Problem 7.4 Apply perturbation theory to the most general two-level system. The unperturbed Hamiltonian is H0 = E 0 a 0 0 E 0 b and the perturbation is H = \u03bb Vaa Vab Vba Vbb with Vba = V \u2217 that will later be set to 1. ab, Vaa and Vbb real, so that H is hermitian. As in Section 7.1.1, \u03bb is a constant (a) Find the exact energies for this two-level system. (b) Expand your result from (a) to second order in \u03bb (and then set \u03bb to 1). Verify that the terms in the series agree with the results from perturbation theory in Sections 7.1.2 and 7.1.3. Assume that Eb > Ea. (c) Setting Vaa = Vbb = 0, show that the series in (b) only converges if. Vab \u2212 E 0 a E 0 b < 1 2 Comment: In general, perturbation theory is only valid if the matrix elements of the perturbation are small compared to the energy level spacings. Otherwise, the \ufb01rst few terms (which are all we ever calculate) will give a poor approximation to the quantity of interest and, as shown here, the series may fail to converge at all, in which case the \ufb01rst few terms tell us nothing. \u2217 Problem 7.5 (a) Find the second-order correction to the energies for the potential in Problem 7.", "1. Comment: You can sum the series explicitly, obtaining \u22122m(\u03b1/\u03c0 n)2 for odd n. E 2 n 5 In the short-hand notation Vmn \u2261 \u03c8 0 m E 1 n = Vnn, E 2 n = |H |\u03c8 0 n |Vnm |2 nm m=n, mn \u2261 E 0 m \u2212 E 0 n, the \ufb01rst three corrections to the nth energy are, E 3 n = l,m=n Vnl Vlm Vmn nl nm \u2212 Vnn m=n |Vnm |2 2 nm. The third-order correction is given in Landau and Lifschitz, Quantum Mechanics: Non-Relativistic Theory, 3rd edn, Pergamon, Oxford (1977), page 136; the fourth and \ufb01fth orders (together with a powerful general technique for obtaining the higher orders) are developed by Nicholas Wheeler, Higher-Order Spectral Perturbation (unpublished Reed College report, 2000). Illuminating alternative formulations of time-independent perturbation theory include the Dalgarno\u2013Lewis method and the closely related \u201clogarithmic\u201d perturbation theory (see, for example, T. Imbo and U. Sukhatme, Am. J. Phys. 52, 140 (1984), for LPT, and H. Mavromatis, Am. J. Phys. 59, 738 (1991), for Delgarno\u2013Lewis). 286 CHAPTER 7 Time-Independent Perturbation Theory (b) Calculate the second-order correction to the ground state energy for the potential in Problem 7.2. Check that your result is consistent with the exact solution. E 2 0 \u2217\u2217 Problem 7.6 Consider a charged particle in the one-dimensional harmonic oscillator potential. Suppose we turn on a weak electric \ufb01eld (E), so that the potential energy is shifted by an amount H = \u2212q E x. (a) Show that there is no \ufb01rst-order change in the energy levels, and calculate the second-order correction. Hint: See Problem 3.39. (b) The Schr\u00f6dinger equation can be solved directly in this case, by a change of vari-. Find the exact energies, and show that they are ables: x \u2261 x \u2212 consistent with the perturbation theory approximation. q E/m\u03c92 \u2217\u2217 Problem 7.7 Consider a particle in the potential shown in Figure 7.3. (a) Find the \ufb01rst-order correction to the ground-state wave function. The \ufb01rst three nonzero terms in the sum will suf\ufb01ce. (b) Using the method of Problem 2.61 \ufb01nd (numerically) the ground-state wave function and energy. Use V0 = 4 2/m a2 and N = 100. Compare the energy obtained numerically to the result from \ufb01rst-order perturbation theory (see Example 7.1). (c) Make a single plot showing (i) the unperturbed ground-state wave function, (ii) the numerical ground-state wave function, and (ii) the \ufb01rst-order approximation to the ground-state wave function. Note: Make sure you\u2019ve properly normalized your numerical result, 1 = |\u03c8(x)|2 d x \u2248 N i=1 |\u03c8i |2 x. 7.2 DEGENERATE PERTURBATION THEORY a and \u03c8 0 ) If the unperturbed states are degenerate\u2014that is, if two (or more) distinct states (\u03c8 0 b (b) (Equation 7.12) and E 2 share the same energy\u2014then ordinary perturbation theory fails: c a a H = 0\u2014a loop\u03c8 0 (Equation 7.15) blow up (unless, perhaps, the numerator vanishes, a hole that will be important to us later on). In the degenerate case, therefore, there is no reason to trust even the \ufb01rst-order correction to the energy (Equation 7.9), and we must look for some other way to handle the problem. Note this is not a minor problem; almost all applications of perturbation theory involve degeneracy. \u03c8 0 b 7.2.1 Two-Fold Degeneracy Suppose that H 0\u03c8 0 a = E 0\u03c8 0 a, H 0\u03c8 0 b = E 0\u03c8, (7.16) 7.2 Degenerate Perturbation Theory 287 E E0 Figure 7.4: \u201cLifting\u201d of a degeneracy by a perturbation. 1 \u03bb (7.17) (7.18) with", " \u03c8 0 a and \u03c8 0 b both normalized. Note that any linear combination of these states, is still an eigenstate of H 0, with the same eigenvalue E 0: \u03c8 0 = \u03b1\u03c8 0 a + \u03b2\u03c8 0 b, H 0\u03c8 0 = E 0\u03c8 0. H will \u201cbreak\u201d (or \u201clift\u201d) the degeneracy: As we increase \u03bb Typically, the perturbation (from 0 to 1), the common unperturbed energy E 0 splits into two (Figure 7.4). Going the other direction, when we turn off the perturbation, the \u201cupper\u201d state reduces down to one linear combination of \u03c8 0 b, and the \u201clower\u201d state reduces to some (orthogonal) linear combination, but we don\u2019t know a priori what these \u201cgood\u201d linear combinations will be. For this reason we can\u2019t even calculate the \ufb01rst-order energy (Equation 7.9)\u2014we don\u2019t know what unperturbed states to use. a and \u03c8 0 The \u201cgood\u201d states are de\ufb01ned as the limit of the true eigenstates as the perturbation is switched off (\u03bb \u2192 0) but that isn\u2019t how you \ufb01nd them in realistic situations (if you knew the exact eigenstates you wouldn\u2019t need perturbation theory). Before I show you the practical techniques for calculating them, we\u2019ll look at an example where we can take the \u03bb \u2192 0 limit of the exact eigenstates. Example 7.2 Consider a particle of mass m in a two-dimensional oscillator potential H 0 = p2 2 m + 1 2 x 2 + y2 m \u03c92 to which is added a perturbation H = m \u03c92 x y. The unperturbed \ufb01rst-excited state (with E 0 = 2 \u03c9) is two-fold degenerate, and one basis for those two degenerate states is \u03c8 0 a = \u03c80(x) \u03c81(y) = 2 \u03c0 m \u03c9 y e\u2212 m \u03c9 2 x 2+y2 288 CHAPTER 7 Time-Independent Perturbation Theory \u03c8 0 b = \u03c81(x) \u03c80(y) = 2 \u03c0 m \u03c9 x e\u2212 m \u03c9 2 x 2+y2, (7.19) where \u03c80 and \u03c81 refer to the one-dimensional harmonic oscillator states (Equation 2.86). To \ufb01nd the \u201cgood\u201d linear combinations, solve for the exact eigenstates of H = H 0 + H and take their limit as \u2192 0. Hint: The problem can be solved by rotating coordinates 7.20) Solution: In terms of the rotated coordinates, the Hamiltonian is m (1 + ) \u03c92 \u2202 y2 + 1 2 + m (1 \u2212 ) \u03c92 y2. This amounts to two independent one-dimensional oscillators. The exact solutions are \u03c8mn = \u03c8 + m m are one-dimensional oscillator states with frequencies \u03c9\u00b1 = \u03c8 \u2212 n x y, where \u03c8 \u00b1 respectively. The \ufb01rst few exact energies, m + 1 2 Emn = \u03c9+ + n + 1 2 \u03c9\u2212, \u221a 1 \u00b1 \u03c9 (7.21) are shown in Figure 7.5. E h \u03c9 3 2 1 0.0 0.1 0.2 0.3 0.4 0.5 Figure 7.5: Exact energy levels as a function of for Example 7.2. The two states which grow out of the degenerate \ufb01rst-excited states as is increased have m = 0, n = 1 (lower state) and m = 1, n = 0 (upper state). If we track these states back to = 0 (in that limit \u03c9+ = \u03c9\u2212 = \u03c9) we get lim \u21920 \u03c801(x) = lim \u2192 = \u03c80 m \u03c9 x \u2212 y \u221a 2 x 2+y2 e\u2212 m \u03c9 2 x \u2212 y \u221a 2 \u03c81 x + y \u221a 2 \u2212\u03c8 0 + \u03c8 0 a b\u221a 2 =. 7.2 Degenerate Perturbation Theory 289 \u03c810(x) = lim \u21920 \u03c8 0 a + \u03c8 0 b\u221a 2. Therefore the \u201cgood\u201d states for this problem are \u00b1 \u2261 17.22) (7.23) In this example we were able to \ufb01nd the exact eigenstates of H and then turn off the perturbation to see what states they evolve from. But how do we \ufffd", "\ufffd\ufffdnd the \u201cgood\u201d states when we can\u2019t solve the system exactly? For the moment let\u2019s just write the \u201cgood\u201d unperturbed states in generic form (Equation 7.17), keeping \u03b1 and \u03b2 adjustable. We want to solve the Schr\u00f6dinger equation, H \u03c8 = E\u03c8, (7.24) with H = H 0 + \u03bbH and E = E 0 + \u03bbE 1 + \u03bb2 + \u03bb\u03c8 1 + \u03bb2\u03c8 2 + \u00b7 \u00b7 \u00b7. (7.25) Plugging these into Equation 7.24, and collecting like powers of \u03bb (as before) we \ufb01nd H 0\u03c8 0 + \u03bb H \u03c8 0 + H 0\u03c8 1 + \u00b7 \u00b7 \u00b7 = E 0\u03c8 0 + \u03bb E 1\u03c8 0 + E 0\u03c8 1 + \u00b7 \u00b7 \u00b7. But H 0\u03c8 0 = E 0\u03c8 0 (Equation 7.18), so the \ufb01rst terms cancel; at order \u03bb1 we have H 0\u03c8 1 + H \u03c8 0 = E 0\u03c8 1 + E 1\u03c8 0. (7.26) Taking the inner product with \u03c8 0 a : 0 0 1 H 0\u03c8 Because H 0 is hermitian, the \ufb01rst term on the left cancels the \ufb01rst term on the right. Putting in Equation 7.17, and exploiting the orthonormality condition (Equation 7.16), we obtain, or, more compactly, where \u03b1Waa + \u03b2Wab = \u03b1 E 1, 0 Wi i, j = a, b). (7.27) (7.28) Similarly, the inner product with \u03c8 0 b yields \u03b1Wba + \u03b2Wbb = \u03b2 E 1. (7.29) Notice that the W s are (in principle) known\u2014they are just the \u201cmatrix elements\u201d of H, with a and \u03c8 0 respect to the unperturbed wave functions \u03c8 0 b. Written in matrix form, Equations 7.27 and 7.29 are = E 1. (7.30) \u03b1 \u03b2 Waa Wab Wba Wbb BC W A D \u03b1 \u03b2 290 CHAPTER 7 Time-Independent Perturbation Theory The eigenvalues of the matrix W give the \ufb01rst-order corrections to the energy and the corresponding eigenvectors tell us the coef\ufb01cients \u03b1 and \u03b2 that determine the \u201cgood\u201d states.6 The Appendix (Section A.5) shows how to obtain the eigenvalues of a matrix; I\u2019ll reproduce those steps here to \ufb01nd a general solution for E 1. First, move all the terms in Equation 7.30 to the left-hand side. Waa \u2212 E 1 Wba Wab Wbb \u2212 E 1 \u03b1 \u03b2 = 0 (7.31) E 1 This equation only has non-trivial solutions if the matrix on the left is non-invertible\u2014that is to say, if its determinant vanishes: Waa \u2212 E 1 Wba Wab Wbb \u2212 E 1 = (Waa \u2212 E1) (Wbb \u2212 E1) \u2212 |Wab|2 = 0, (7.32) where we used the fact that Wba = W \u2217 ab. Solving the quadratic, \u00b1 = 1 E 1 2 Waa + Wbb \u00b1 (Waa \u2212 Wbb)2 + 4 |Wab|2. (7.33) This is the fundamental result of degenerate perturbation theory; the two roots correspond to the two perturbed energies. Example 7.3 Returning to Example 7.2, show that diagonalizing the matrix W gives the same \u201cgood\u201d states we found by solving the problem exactly. Solution: We need to calculate the matrix elements of W. First, Waa = (x, y) H \u03c8 0 \u03c8 0 a a (x, y) d x d y = m \u03c92 |\u03c80(x)|2 x d x |\u03c80(y)|2 y dy = 0 (the integrands are both odd functions). Similarly, Wbb = 0, and we need only compute Wab = (x, y) H \u03c8 0 \u03c8 0 a b (x, y) d x d y = m \u03c92 \u03c80(x) x \u03c81(x) d x \u03c81(y) y \u03c80(y) dy. These two integrals are equal, and recalling (Equation 2.70) x = 2 m \u03c9 (a+ + a\u2212) 6 This assumes that the eigenvalues of W are distinct so that", " the degeneracy lifts at \ufb01rst order. If not, any choice of \u03b1 and \u03b2 satis\ufb01es Equation 7.30; you still don\u2019t know what the good states are. The \ufb01rst-order energies are correctly given by Equation 7.33 when this happens, and in many cases that\u2019s all you require. But if you need to know the \u201cgood\u201d states\u2014for example to calculate higher-order corrections\u2014you will have to use second-order degenerate perturbation theory (see Problems 7.39, 7.40, and 7.41) or employ the theorem of Section 7.2.2. 7.2 Degenerate Perturbation Theory 291 2 (a+ + a\u2212) \u03c81(x) d x 2 = \u03c9. 2 we have Wab = m \u03c92 \u03c80(x) 2 m \u03c9 = \u03c9 2 Therefore, the matrix W is \u03c80(x) \u03c80(x. The (normalized) eigenvectors of this matrix are 1\u221a 2 1\u221a 2 and 1 1. \u22121 1 These eigenvectors tell us which linear combination of \u03c8 0 a and \u03c8 0 b are the good states: \u00b1 = 1, just as in Equation 7.23. The eigenvalues of the matrix W, give the \ufb01rst-order corrections to the energy (compare 7.33). E 1 = \u00b1 \u03c9 2, (7.34) If it happens that Wab = 0 in Equation 7.30 then the two eigenvectors are 1 0 and 0 1 and the energies, 0 E 1 + = Waa = \u2212 = Wbb = 7.35) are precisely what we would have obtained using nondegenerate perturbation theory (Equation 7.9). We have simply been lucky: The states \u03c8 0 b were already the \u201cgood\u201d linear combinations. Obviously, it would be greatly to our advantage if we could somehow guess the \u201cgood\u201d states right from the start\u2014then we could go ahead and use nondegenerate perturbation theory. As it turns out, we can very often do this by exploiting the theorem in the following section. a and \u03c8 0 7.2.2 \u201cGood\u201d States Theorem: Let A be a hermitian operator that commutes with H 0 and H. If \u03c8 0 a and b (the degenerate eigenfunctions of H 0) are also eigenfunctions of A, with distinct \u03c8 0 eigenvalues, A\u03c8 0 a = \u03bc\u03c8 0 a, A\u03c8 0 b = \u03bd\u03c8 0 b, and \u03bc = \u03bd, then \u03c8 0 a and \u03c8 0 b are the \u201cgood\u201d states to use in perturbation theory. 292 CHAPTER 7 Time-Independent Perturbation Theory Proof: Since H (\u03bb) = H 0+\u03bb H and A commute, there exist simultaneous eigenstates \u03c8\u03b3 (\u03bb) where H (\u03bb) \u03c8\u03b3 (\u03bb) = E(\u03bb) \u03c8\u03b3 (\u03bb) and A \u03c8\u03b3 (\u03bb) = \u03b3 \u03c8\u03b3 (\u03bb). (7.36) 0 The fact that A is hermitian means 1 A\u03c8\u03b3 (\u03bb) 1 \u03c8\u03b3 (\u03bb) 1 \u03c8\u03b3 (\u03bb) (\u03b3 \u2212 \u03bc = A\u03c8 0 a 0 = \u03bc\u2217 \u03c8 0 a 1 \u03c8\u03b3 (\u03bb) \u03c8\u03b3 (\u03bb) = 0, 1 (7.37) (7.38) (making use of the fact that \u03bc is real). This holds true for any value of \u03bb and taking the limit as \u03bb \u2192 0 we have 0 \u03c8 0 a 1 \u03c8\u03b3 (0) = 0 unless \u03b3 = \u03bc, and similarly 0 \u03c8 0 b 1 \u03c8\u03b3 (0) = 0 unless \u03b3 = \u03bd. a and \u03c8 0 Now the good states are linear combinations of \u03c8 0 From above it follows that either \u03b3 = \u03bc, in which case \u03b2 = good state is simply \u03c8 0 a, or \u03b3 = \u03bd and the good state is \u03c8 0 \u03c8 0 b b. QED b : \u03c8\u03b3 (0) = \u03b1 \u03c8 0 a \u03c8\u03b3 (0) + \u03b2 \u03c8 0 b. = 0 and the Once we identify the \u201cgood\u201d states, either by solving Equation 7.30 or by applying this theorem, we can use these \u201cgood\u201d states as our unperturbed states and apply ordinary nondegenerate perturbation theory.7 In most cases, the operator A will be suggested by symmetry; as you saw in Chapter 6, symmetries", " are associated with operators that commute with H \u2014 precisely what are required to identify the good states. Example 7.4 Find an operator A that satis\ufb01es the requirements of the preceding theorem to construct the \u201cgood\u201d states in Examples 7.2 and 7.3. Solution: The perturbation H has less symmetry than H 0. H 0 had continuous rotational symmetry, but H = H 0 + H is only invariant under rotations by integer multiples of \u03c0. For A, take the operator R(\u03c0 ) that rotates a function counterclockwise by an angle \u03c0. Acting on our stats \u03c8a and \u03c8b we have R(\u03c0 ) \u03c8 0 a R(\u03c0 ) \u03c8 0 b (\u2212x, \u2212y) = \u2212\u03c8 0 a (\u2212x, \u2212y) = \u2212\u03c8 0 b (x, y) = \u03c8 0 a (x, y) = \u03c8 0 b (x, y), (x, y). That\u2019s no good; we need an operator with distinct eigenvalues. How about the operator that interchanges x and y? This is a re\ufb02ection about a 45\u25e6 diagonal of the well. Call this operator D. D commutes with both H 0 and H, since they are unchanged when you switch x and y. Now, 7 Note that the theorem is more general than Equation 7.30. In order to identify the good states from Equation 7.30, the energies E 1 \u00b1 need to be different. In some cases they are the same and the energies of the degenerate states split at second, third, or higher order in perturbation theory. But the theorem allows you to identify the good states in every case. 7.2 Degenerate Perturbation Theory 293 D \u03c8 0 a D \u03c8 0 b (x, y) = \u03c8 0 a (x, y) = \u03c8 0 b (y, x) = \u03c8 0 b (y, x) = \u03c8 0 a (x, y), (x, y). So our degenerate eigenstates are not eigenstates of D. But we can construct linear combinations that are: \u03c8 0 \u00b1 \u2261 \u00b1\u03c8 0 a + \u03c8 0 b. (7.39) Then D \u00b1\u03c8 0 a + \u03c8 0 b. These are \u201cgood\u201d states, since they are eigenstates of an operator D with distinct eigenvalues (\u00b11), and D commutes with both H 0 and H. Moral: If you\u2019re faced with degenerate states, look around for some hermitian operator A that commutes with H 0 and H ; pick as your unperturbed states ones that are simultaneously eigenfunctions of H 0 and A (with distinct eigenvalues). Then use ordinary \ufb01rst-order perturbation theory. If you can\u2019t \ufb01nd such an operator, you\u2019ll have to resort to Equation 7.33, but in practice this is seldom necessary. Problem 7.8 Let the two \u201cgood\u201d unperturbed states be, \u03c8 0 + \u03b2\u00b1\u03c8 0 b \u00b1 = \u03b1\u00b1\u03c8 0 a where \u03b1\u00b1 and \u03b2\u00b1 are determined (up to normalization) by Equation 7.27 (or Equation 7.29). Show explicitly that \u03c8 0 (a) \u03c8 0 \u00b1 are orthogonal \u03c8 0 \u03c8 0 = 0; (bc) \u00b1 \u00b1 \u00b1 given by Equation 7.33. \u2212 = 0 \u00b1, with E 1 H H +|\u03c8 0 ; Problem 7.9 Consider a particle of mass m that is free to move in a one-dimensional region of length L that closes on itself (for instance, a bead that slides frictionlessly on a circular wire of circumference L, as in Problem 2.46). (a) Show that the stationary states can be written in the form \u03c8n(x) = 1\u221a L e2\u03c0inx/L, (\u2212L/2 < x < L/2), where n = 0, \u00b11, \u00b12,..., and the allowed energies are En = 2 m 2. n\u03c0 L Notice that\u2014with the exception of the ground state (n = 0)\u2014these are all doubly degenerate. (b) Now suppose we introduce the perturbation H = \u2212V0e\u2212x 2/a2, 294 CHAPTER 7 Time-Independent Perturbation Theory where a L. (This puts a little \u201cdimple\u201d in the potential at x = 0, as though we bent the wire slightly to make a \u201ctrap\u201d.) Find the \ufb01rst-order correction to En, using Equation 7.33. Hint: To evaluate the integrals, exploit", " the fact that a L to extend the limits from \u00b1L/2 to \u00b1\u221e; after all, H is essentially zero outside \u2212a < x < a. (c) What are the \u201cgood\u201d linear combinations of \u03c8n and \u03c8\u2212n, for this problem? (Hint: use Eq. 7.27.) Show that with these states you get the \ufb01rst-order correction using Equation 7.9. (d) Find a hermitian operator A that \ufb01ts the requirements of the theorem, and show that the simultaneous eigenstates of H 0 and A are precisely the ones you used in (c). 7.2.3 Higher-Order Degeneracy In the previous section I assumed the degeneracy was two-fold, but it is easy to see how the method generalizes. In the case of n-fold degeneracy, we look for the eigenvalues of the n \u00d7 n matrix H (7.40) 0 1. Wi j = \u03c8 0 j \u03c8 0 i For three-fold degeneracy (with degenerate states \u03c8 0 b, and \u03c8 0 c to the energies E 1 are the eigenvalues of W, determined by solving \u03b1 \u03b2 \u03b3 Waa Wab Wac Wba Wbb Wbc Wca Wcb Wcc \u23a0 = E 1 a, \u03c8 0 \u03b1 \u03b2 \u03b3 \u239b \u239d \u239e \u239b \u239d \u239e \u239d \u23a0 \u239b ) the \ufb01rst-order corrections \u239e \u23a0, (7.41) and the \u201cgood\u201d states are the corresponding eigenvectors:7.42) Once again, if you can think of an operator A that commutes with H 0 and H, and use the simultaneous eigenfunctions of A and H 0, then the W matrix will automatically be diagonal, and you won\u2019t have to fuss with calculating the off-diagonal elements of W or solving the characteristic equation.9 (If you\u2019re nervous about my generalization from two-fold degeneracy to n-fold degeneracy, work Problem 7.13.) Problem 7.10 Show that the \ufb01rst-order energy corrections computed in Example 7.3 (Equation 7.34) agree with an expansion of the exact solution (Equation 7.21) to \ufb01rst order in. Problem 7.11 Suppose we perturb the in\ufb01nite cubical well (Problem 4.2) by putting a delta function \u201cbump\u201d at the point (a/4, a/2, 3a/4): H = a3V0\u03b4(x \u2212 a/4)\u03b4(y \u2212 a/2)\u03b4(z \u2212 3a/4). Find the \ufb01rst-order corrections to the energy of the ground state and the (triply degenerate) \ufb01rst excited states. 8 If the eigenvalues are degenerate, see footnote 6. 9 Degenerate perturbation theory amounts to diagonalization of the degenerate part of the Hamiltonian; see Problems 7.34 and 7.35. 7.3 The Fine Structure of Hydrogen 295 \u2217 Problem 7.12 Consider a quantum system with just three linearly independent states. Suppose the Hamiltonian, in matrix form, is H = V0 \u239b \u239d ( where V0 is a constant, and is some small number ( 1). (a) Write down the eigenvectors and eigenvalues of the unperturbed Hamiltonian ( = 0). (b) Solve for the exact eigenvalues of H. Expand each of them as a power series in, up to second order. (c) Use \ufb01rst- and second-order non-degenerate perturbation theory to \ufb01nd the approximate eigenvalue for the state that grows out of the nondegenerate eigenvector of H0. Compare the exact result, from (b). (d) Use degenerate perturbation theory to \ufb01nd the \ufb01rst-order correction to the two initially degenerate eigenvalues. Compare the exact results. Problem 7.13 In the text I asserted that the \ufb01rst-order corrections to an n-fold degenerate energy are the eigenvalues of the W matrix, and I justi\ufb01ed this claim as the \u201cnatural\u201d generalization of the case n = 2. Prove it, by reproducing the steps in Section 7.2.1, starting with \u03c8 0 = n j=1 \u03b1 j \u03c8 0 j (generalizing Equation 7.17), and ending by showing that the analog to Equation 7.27 can be interpreted as the", " eigenvalue equation for the matrix W. 7.3 THE FINE STRUCTURE OF HYDROGEN In our study of the hydrogen atom (Section 4.2) we took the Hamiltonian\u2014called the Bohr Hamiltonian\u2014to be HBohr = \u2212 2 2m \u22072 \u2212 e2 4\u03c0 0 1 r (7.43) (electron kinetic energy plus Coulombic potential energy). But this is not quite the whole story. We have already learned how to correct for the motion of the nucleus: Just replace m by the reduced mass (Problem 5.1). More signi\ufb01cant is the so-called \ufb01ne structure, which is actually due to two distinct mechanisms: a relativistic correction, and spin-orbit coupling. Compared to the Bohr energies (Equation 4.70), \ufb01ne structure is a tiny perturbation\u2014smaller by a factor of \u03b12, where \u03b1 \u2261 e2 4\u03c0 0c \u2248 1 137.036 (7.44) is the famous \ufb01ne structure constant. Smaller still (by another factor of \u03b1) is the Lamb shift, associated with the quantization of the electric \ufb01eld, and smaller by yet another order of magnitude is the hyper\ufb01ne structure, which is due to the interaction between the magnetic 296 CHAPTER 7 Time-Independent Perturbation Theory Table 7.1: Hierarchy of corrections to the Bohr energies of hydrogen. Bohr energies: Fine structure: Lamb shift: Hyperfine splitting: of order of order of order of order \u03b12mc2 \u03b14mc2 \u03b15mc2 (m /mp)\u03b1 4mc2 dipole moments of the electron and the proton. This hierarchy is summarized in Table 7.1. In the present section we will analyze the \ufb01ne structure of hydrogen, as an application of time-independent perturbation theory. Problem 7.14 (a) Express the Bohr energies in terms of the \ufb01ne structure constant and the rest energy mc2 of the electron. (b) Calculate the \ufb01ne structure constant from \ufb01rst principles (i.e., without recourse to the empirical values of 0, e,, and c). Comment: The \ufb01ne structure constant is undoubtedly the most fundamental pure (dimensionless) number in all of physics. It relates the basic constants of electromagnetism (the charge of the electron), relativity (the speed of light), and quantum mechanics (Planck\u2019s constant). If you can solve part (b), you have the most certain Nobel Prize in history waiting for you. But I wouldn\u2019t recommend spending a lot of time on it right now; many smart people have tried, and all (so far) have failed. 7.3.1 The Relativistic Correction The \ufb01rst term in the Hamiltonian is supposed to represent kinetic energy: and the canonical substitution p \u2192 \u2212i\u2207 yields the operator T = 1 2 mv2 = p2 2m, T = \u2212 2 2m \u22072. (7.45) (7.46) But Equation 7.45 is the classical expression for kinetic energy; the relativistic formula is mc2 T = \u2212 mc2. 1 \u2212 (v/c)2 (7.47) The \ufb01rst term is the total relativistic energy (not counting potential energy, which we aren\u2019t concerned with at the moment), and the second term is the rest energy\u2014the difference is the energy attributable to motion. We need to express T in terms of the (relativistic) momentum, p = mv, 1 \u2212 (v/c)2 (7.48) instead of velocity. Notice that m2v2c2 + m2c4 1 \u2212 (v/c)2 p2c2 + m2c4 = so 1 \u2212 (v/c)2 7.3 The Fine Structure of Hydrogen 297 = m2c4 1 \u2212 (v/c)2 = T + mc2 2, T = p2c2 + m2c4 \u2212 mc2. (7.49) This relativistic equation for kinetic energy reduces (of course) to the classical result (Equation 7.45), in the nonrelativistic limit p mc; expanding in powers of the small number ( p/mc), we have T = mc2 1 + 2 p mc \u2212 1 = mc2 1 + 1 2 2 p mc 4 \u2212 1 8 p mc \u00b7 \u00b7 \u00b7 \u2212 1 = p2 2m \u2212 p4 8m3c2 + \u00b7 \u00b7 \u00b7. The lowest-order10 relativistic correction to the Hamiltonian is therefore H r = \u2212 p4 8m3c2. (7.50) (7.51) In \ufb01r", "st-order perturbation theory, the correction to En is given by the expectation value of H in the unperturbed state (Equation 7.9): 8m3c2 \u03c8 1 p4\u03c8 = \u2212 1 8m3c2 0 1 p2\u03c8. p2\u03c8 Now, the Schr\u00f6dinger equation (for the unperturbed states) says p2\u03c8 = 2m(E \u2212 V )\u03c8, (7.52) (7.53) and hence11 E 1 r = \u2212 1 2mc2 0 1 (E \u2212 V )2 # = \u2212 1 2mc2 0 1$ E 2 \u2212 2E V + V 2. (7.54) So far this is entirely general; but we\u2019re interested in hydrogen, for which V (r ) = \u2212 (1/4\u03c0 0) e2/r : E 1 r = \u2212 1 2mc2 E 2 n + 2En + 1 r e2 4\u03c0 0 e2 4\u03c0 0 2 1 r 2, (7.55) where En is the Bohr energy of the state in question. To complete the job, we need the expectation values of 1/r and 1/r 2, in the (unperturbed) state \u03c8nm (Equation 4.89). The \ufb01rst is easy (see Problem 7.15): 1 r = 1 n2a, (7.56) 10 The kinetic energy of the electron in hydrogen is on the order of 10 eV, which is minuscule compared to its rest energy (511,000 eV), so the hydrogen atom is basically nonrelativistic, and we can afford to keep only the lowestorder correction. In Equation 7.50, p is the relativistic momentum (Equation 7.48), not the classical momentum mv. It is the former that we now associate with the quantum operator \u2212i\u2207, in Equation 7.51. 11 An earlier edition of this book claimed that p4 is not hermitian for states with = 0 (calling into question the maneuver leading to Equation 7.54). That was incorrect\u2014 p4 is hermitian, for all (see Problem 7.18). 298 CHAPTER 7 Time-Independent Perturbation Theory where a is the Bohr radius (Equation 4.72). The second is not so simple to derive (see Problem 7.42), but the answer is12 1 r 2 = 1 ( + 1/2) n3a2. (7.57) It follows that E 1 r = \u2212 1 2mc2 E 2 n + 2En e2 4\u03c0 0 1 n2a + 2 e2 4\u03c0 0 1 ( + 1/2) n3a2, or, eliminating a (using Equation 4.72) and expressing everything in terms of En (using Equation 4.70): E 1 r = \u2212 (En)2 2mc2 4n + 1/2 \u2212 3. (7.58) Evidently the relativistic correction is smaller than En, by a factor of about En/mc2 = 2 \u00d7 10\u22125. You might have noticed that I used non-degenerate perturbation theory in this calculation (Equation 7.52), in spite of the fact that the hydrogen atom is highly degenerate. But the perturbation is spherically symmetric, so it commutes with L 2 and L z. Moreover, the eigenfunctions of these operators (taken together) have distinct eigenvalues for the n2 states with a given En. Luckily, then, the wave functions \u03c8nm are the \u201cgood\u201d states for this problem (or, as we say, n,, and m are the good quantum numbers), so as it happens the use of nondegenerate perturbation theory was legitimate (see the \u201cMoral\u201d to Section 7.2.1). From Equation 7.58 we see that some of the degeneracy of the nth energy level has lifted. The (2 + 1)-fold degeneracy in m remains; as we saw in Example 6.3 it is due to rotational symmetry, a symmetry that remains intact with this perturbation. On the other hand, the \u201caccidental\u201d degeneracy in has disappeared; since its source is an additional symmetry unique to the 1/r potential (see Problem 6.34), we expect that degeneracy to be broken by practically any perturbation. \u2217 Problem 7.15 Use the virial theorem (Problem 4.48) to prove Equation 7.56. Problem 7.16 In Problem 4.52 you calculated the expectation value of r s in the state \u03c8321. Check your answer for the special cases s = 0 (trivial), s = \u22121 (Equation 7.56), s", " = \u22122 (Equation 7.57), and s = \u22123 (Equation 7.66). Comment on the case s = \u22127. \u2217\u2217 Problem 7.17 Find the (lowest-order) relativistic correction to the energy levels of the onedimensional harmonic oscillator. Hint: Use the technique of Problem 2.12. 12 The general formula for the expectation value of any power of r is given in Hans A. Bethe and Edwin E. Salpeter, Quantum Mechanics of One- and Two-Electron Atoms, Plenum, New York (1977), p. 17. 7.3 The Fine Structure of Hydrogen 299 \u2217\u2217\u2217 Problem 7.18 Show that p2 is hermitian, for hydrogen states with = 0. Hint: For such states \u03c8 is independent of \u03b8 and \u03c6, so p2 = \u2212 2 r 2 d dr r 2 d dr (Equation 4.13). Using integration by parts, show that 0 1 p2g f = \u22124\u03c0 2 r 2 f dg dr \u2212 r 2g d f dr \u221e 0 0 + p2 f 1 g. Check that the boundary term vanishes for \u03c8n00, which goes like \u03c8n00 \u223c near the origin. 1 \u221a \u03c0 (na)3/2 exp (\u2212r/na) The case of p4 is more subtle. The Laplacian of 1/r picks up a delta function (see, for example, D. J. Grif\ufb01ths, Introduction to Electrodynamics, 4th edn, Eq. 1.102). Show that $ # e\u2212kr = \u22074 \u2212 4k3 r + k4 e\u2212kr + 8\u03c0 k \u03b43 (r), and con\ufb01rm that p4 is hermitian.13 7.3.2 Spin-Orbit Coupling Imagine the electron in orbit around the nucleus; from the electron\u2019s point of view, the proton is circling around it (Figure 7.6). This orbiting positive charge sets up a magnetic \ufb01eld B, in the electron frame, which exerts a torque on the spinning electron, tending to align its magnetic moment (\u03bc) along the direction of the \ufb01eld. The Hamiltonian (Equation 4.157) is H = \u2212\u03bc \u00b7 B. (7.59) To begin with, we need to \ufb01gure out the magnetic \ufb01eld of the proton (B) and the dipole moment of the electron (\u03bc). B, L \u2212e r +e Figure 7.6: Hydrogen atom, from the electron\u2019s perspective. 13 Thanks to Edward Ross and Li Yi-ding for \ufb01xing this problem. 300 CHAPTER 7 Time-Independent Perturbation Theory \u00b5, S r q, m Figure 7.7: A ring of charge, rotating about its axis. The Magnetic Field of the Proton. If we picture the proton (from the electron\u2019s perspective) as a continuous current loop (Figure 7.6), its magnetic \ufb01eld can be calculated from the Biot\u2013Savart law: B = \u03bc0 I 2r, with an effective current I = e/T, where e is the charge of the proton and T is the period of the orbit. On the other hand, the orbital angular momentum of the electron (in the rest frame of the nucleus) is L = r mv = 2\u03c0 mr 2/T. Moreover, B and L point in the same direction (up, in Figure 7.6), so (I used c = 1/ \u221a e mc2r 3 0\u03bc0 to eliminate \u03bc0 in favor of 0.) B = 1 4\u03c0 0 L. (7.60) The Magnetic Dipole Moment of the Electron. The magnetic dipole moment of a spinning charge is related to its (spin) angular momentum; the proportionality factor is the gyromagnetic ratio (which we already encountered in Section 4.4.2). Let\u2019s derive it, this time, using classical electrodynamics. Consider \ufb01rst a charge q smeared out around a ring of radius r, which rotates about the axis with period T (Figure 7.7). The magnetic dipole moment of the ring is de\ufb01ned as the current (q/T ) times the area \u03c0r 2 : \u03bc = q\u03c0r 2 T. mr 2 times the If the mass of the ring is m, its angular momentum is the moment of inertia angular velocity (2\u03c0/T ):. T S = 2\u03c0 mr 2 The gyromagnetic ratio for this con\ufb01guration is evidently \u03bc/S = q/2m. Notice that it is", " independent of r (and T ). If I had some more complicated object, such as a sphere (all I require is that it be a \ufb01gure of revolution, rotating about its axis), I could calculate \u03bc and S by chopping it into little rings, and adding up their contributions. As long as the mass and the charge are distributed in the same manner (so that the charge-to-mass ratio is uniform), the gyromagnetic ratio will be the same for each ring, and hence also for the object as a whole. Moreover, the directions of \u03bc and S are the same (or opposite, if the charge is negative), so.3 The Fine Structure of Hydrogen 301 j = 7/2 j = 5/2 j = 3/2 j = 1/2 j = 5/2 j = 3/2 j = 1/2 j = 3/2 j = 1/2 j = 1/2 l = 0 (S) l = 1 (P) l = 2 (D) l = 3 (F) Figure 7.8: Energy levels of hydrogen, including \ufb01ne structure (not to scale). \u03bc = S. q 2m (7.61) That was a purely classical calculation, however; as it turns out the electron\u2019s magnetic moment is twice the classical value: \u03bc e = \u2212 e m S. (7.62) The \u201cextra\u201d factor of 2 was explained by Dirac, in his relativistic theory of the electron.14 Putting all this together, we have H = e2 4\u03c0 0 1 m2c2r 3 S \u00b7 L. But there is a serious fraud in this calculation: I did the analysis in the rest frame of the electron, but that\u2019s not an inertial system\u2014it accelerates, as the electron orbits around the nucleus. You 14 We have already noted that it can be dangerous to picture the electron as a spinning sphere (see Problem 4.28), and it is not too surprising that the naive classical model gets the gyromagnetic ratio wrong. The deviation from the classical expectation is known as the g-factor: \u03bc = g (q/2m) S. Thus the g-factor of the electron, in Dirac\u2019s theory, is exactly 2. But quantum electrodynamics reveals tiny corrections to this: ge is actually 2 + (\u03b1/\u03c0 ) + \u00b7 \u00b7 \u00b7 = 2.002.... The calculation and measurement (which agree to exquisite precision) of the so-called anomalous magnetic moment of the electron were among the greatest achievements of twentieth-century physics. 302 CHAPTER 7 Time-Independent Perturbation Theory can get away with this if you make an appropriate kinematic correction, known as the Thomas precession.15 In this context it throws in a factor of 1/2:16 H so = e2 8\u03c0 0 1 m2c2r 3 S \u00b7 L. (7.63) This is the spin-orbit interaction; apart from two corrections (the modi\ufb01ed gyromagnetic ratio for the electron and the Thomas precession factor\u2014which, coincidentally, exactly cancel one another) it is just what you would expect on the basis of a naive classical model. Physically, it is due to the torque exerted on the magnetic dipole moment of the spinning electron, by the magnetic \ufb01eld of the proton, in the electron\u2019s instantaneous rest frame. Now the quantum mechanics. In the presence of spin-orbit coupling, the Hamiltonian no longer commutes with L and S, so the spin and orbital angular momenta are not separately conserved (see Problem 7.19). However, H so does commute with L 2, S2 and the total angular momentum J \u2261 L + S, (7.64) and hence these quantities are conserved (Equation 3.73). To put it another way, the eigenstates of L z and Sz are not \u201cgood\u201d states to use in perturbation theory, but the eigenstates of L 2, S2, J 2, and Jz are. Now J 2 = (L + S) \u00b7 (L + S) = L 2 + S2 + 2L \u00b7 S, so \u2212 S2, (7.65) and therefore the eigenvalues of L \u00b7 S are j ( j + 1) \u2212 ( + 1) \u2212 s (s + 1). 2 2 In this case, of course, s = 1/2. Meanwhile, the expectation value of 1/r 3 (see Problem 7.43)17 is 1 r 3 = 1 ( + 1/2) ( + 1) n3a3, (7.66) 15 One way of thinking of it is that the electron is continually stepping from one inertial system to another; Thomas precession amounts to the cumulative effect of all these Lorentz transformations. We could", " avoid the whole problem, of course, by staying in the lab frame, in which the nucleus is at rest. In that case the \ufb01eld of the proton is purely electric, and you might wonder why it exerts any torque on the electron. Well, the fact is that a moving magnetic dipole acquires an electric dipole moment, and in the lab frame the spin-orbit coupling is due to the interaction of the electric \ufb01eld of the nucleus with the electric dipole moment of the electron. Because this analysis requires more sophisticated electrodynamics, it seems best to adopt the electron\u2019s perspective, where the physical mechanism is more transparent. 16 More precisely, Thomas precession subtracts 1 from the gyromagnetic ratio (see R. R. Haar and L. J. Curtis, Am. J. Phys., 55, 1044 (1987)). 17 In Problem 7.43 the expectation values are calculated using the hydrogen wave functions \u03c8(nm)\u2014that is, eigenstates of L z \u2014whereas we now want eigenstates of Jz \u2014which are linear combinations of m = m j + 1 2 and m = m j \u2212 1 is independent of m, it doesn\u2019t matter. r s 2. But since 7.3 The Fine Structure of Hydrogen 303 and we conclude that E 1 so = H so = e2 8\u03c0 0 1 m2c2 2/2 j ( j + 1) \u2212 ( + 1) \u2212 3/4 ( + 1/2) ( + 1) n3a3, or, expressing it all in terms of En:18 E 1 so = (En)2 mc2 n j ( j + 1) \u2212 ( + 1) \u2212 3/4 ( + 1/2) ( + 1) 4. (7.67) It is remarkable, considering the totally different physical mechanisms involved, that the. Adding E 2 relativistic correction and the spin-orbit coupling are of the same order n them together, we get the complete \ufb01ne-structure formula (see Problem 7.20): /mc2 (En)2 2mc2 3 \u2212 4n j + 1/2. E 1 fs = (7.68) Combining this with the Bohr formula, we obtain the grand result for the energy levels of hydrogen, including \ufb01ne structure: En j = \u2212 13.6 eV n2 1 + \u03b12 n2 n j + 1/2 \u2212 3 4. (7.69) Fine structure breaks the degeneracy in (that is, for a given n, the different allowed values of do not all carry the same energy), but it still preserves degeneracy in j (see Figure 7.8). The z-component eigenvalues for orbital and spin angular momentum (ml and ms) are no longer \u201cgood\u201d quantum numbers\u2014the stationary states are linear combinations of states with different values of these quantities; the \u201cgood\u201d quantum numbers are n,, s, j, and m j.19 L \u00b7 S, L 2 Problem 7.19 Evaluate the following commutators: (a) [L \u00b7 S, L], (b) [L \u00b7 S, S], (c) [L \u00b7 S, J], (d). Hint: L and S satisfy the fundamental commutation relations for angular momentum (Equations 4.99 and 4.134), but they commute with each other. L \u00b7 S, J 2 L \u00b7 S, S2, (e), (f) 18 The case = 0 looks problematic, since we are ostensibly dividing by zero. On the other hand, the numerator is also zero, since in this case j = s; so Equation 7.67 is indeterminate. On physical grounds there shouldn\u2019t be any spin-orbit coupling when = 0. In any event, the problem disappears when the spin-orbit coupling is added to the relativistic correction, and their sum (Equation 7.68) is correct for all. If you\u2019re feeling uneasy about this whole calculation, I don\u2019t blame you; take comfort in the fact that the exact solution can be obtained by using the (relativistic) Dirac equation in place of the (nonrelativistic) Schr\u00f6dinger equation, and it con\ufb01rms the results we obtain here by less rigorous means (see Problem 7.22.) 19 To write | jm j (for given and s) as a linear combination of | s m ms we would use the appropriate Clebsch\u2013 Gordan coef\ufb01cients (Equation 4.183). 304 CHAPTER 7 Time-Independent Perturbation Theory \u2217 \u2217\u2217 Problem 7.20 Derive the \ufb01ne structure formula (Equation 7.68) from the relativistic correction (Equation 7.", "58) and the spin-orbit coupling (Equation 7.67). Hint: Note that j = \u00b1 1/2 (except for = 0, where only the plus sign occurs); treat the plus sign and the minus sign separately, and you\u2019ll \ufb01nd that you get the same \ufb01nal answer either way. Problem 7.21 The most prominent feature of the hydrogen spectrum in the visible region is the red Balmer line, coming from the transition n = 3 to n = 2. First of all, determine the wavelength and frequency of this line according to the Bohr theory. Fine structure splits this line into several closely-spaced lines; the question is: How many, and what is their spacing? Hint: First determine how many sublevels the n = 2 level splits into, and \ufb01nd fs for each of these, in eV. Then do the same for n = 3. Draw an energy level diagram E 1 showing all possible transitions from n = 3 to n = 2. The energy released (in the form of a photon) is (E3 \u2212 E2) + E, the \ufb01rst part being common to all of them, and E (due to \ufb01ne structure) varying from one transition to the next. Find E (in eV) for each transition. Finally, convert to photon frequency, and determine the spacing between adjacent spectral lines (in Hz)\u2014not the frequency interval between each line and the unperturbed line (which is, of course, unobservable), but the frequency interval between each line and the next one. Your \ufb01nal answer should take the form: \u201cThe red Balmer line splits into (???) lines. In order of increasing frequency, they come from the transitions (1) j = (???) to j = (???), (2) j = (???) to j = (???),.... The frequency spacing between line (1) and line (2) is (???) Hz, the spacing between line (2) and line (3) is (???) Hz,....\u201d Problem 7.22 The exact \ufb01ne-structure formula for hydrogen (obtained from the Dirac equation without recourse to perturbation theory) is20 \u239b \u23a1 \u23a4 \u22121/2 \u23a7 \u23aa\u23a8 En j = mc2 \u23aa\u23a9 \u23a2 \u23a3/2) + ( j + 1/2)2 \u2212 \u03b12 \u239e 2 \u23a0 \u23a5 \u23a6 \u23ab \u23aa\u23ac. \u2212 1 \u23aa\u23ad Expand to order \u03b14 (noting that \u03b1 1), and show that you recover Equation 7.69. 7.4 THE ZEEMAN EFFECT When an atom is placed in a uniform external magnetic \ufb01eld Bext, the energy levels are shifted. This phenomenon is known as the Zeeman effect. For a single electron, the perturbation is21 H Z = \u2212 \u03bc l + \u03bc s \u00b7 Bext, (7.70) 20 Bethe and Salpeter (footnote 12, page 298), page 238. 21 This is correct to \ufb01rst order in B. We are ignoring a term of order B2 in the Hamiltonian (the exact result was calculated in Problem 4.72). In addition, the orbital magnetic moment (Equation 7.72) is proportional to the mechanical angular momentum, not the canonical angular momentum (see Problem 7.49). These neglected terms give corrections of order B2, comparable to the second-order corrections from H Z. Since we\u2019re working to \ufb01rst order, they are safe to ignore in this context. 7.4 The Zeeman Effect 305 where \u03bc s = \u2212 e m S (Equation 7.62) is the magnetic dipole moment associated with electron spin, and \u03bc l = \u2212 e 2m L (Equation 7.61) is the dipole moment associated with orbital motion.22 Thus H Z = e 2m (L + 2S) \u00b7 Bext. (7.71) (7.72) (7.73) The nature of the Zeeman splitting depends critically on the strength of the external \ufb01eld in comparison with the internal \ufb01eld (Equation 7.60) that gives rise to spin-orbit coupling. If Bext Bint, then \ufb01ne structure dominates, and H Z can be treated as a small perturbation, whereas if Bext Bint, then the Zeeman effect dominates, and \ufb01ne structure becomes the perturbation. In the intermediate zone, where the two \ufb01elds are comparable, we need the full machinery of degenerate perturbation theory, and it is necessary", " to diagonalize the relevant portion of the Hamiltonian \u201cby hand.\u201d In the following sections we shall explore each of these regimes brie\ufb02y, for the case of hydrogen. Problem 7.23 Use Equation 7.60 to estimate the internal \ufb01eld in hydrogen, and characterize quantitatively a \u201cstrong\u201d and \u201cweak\u201d Zeeman \ufb01eld. 7.4.1 Weak-Field Zeeman Effect If Bext Bint, \ufb01ne structure dominates; we treat HBohr + H f s as the \u201cunperturbed\u201d Hamiltonian and H Z as the perturbation. Our \u201cunperturbed\u201d eigenstates are then those appropriate to \ufb01ne structure: |njm j and the \u201cunperturbed\u201d energies are En j (Equation 7.69). Even though \ufb01ne structure has lifted some of the degeneracy in the Bohr model, these states are still degenerate, since the energy does not depend on m j or. Luckily the states |njm j are the \u201cgood\u201d states for treating the perturbation H Z (meaning we don\u2019t have to write down the W matrix for Z \u2014it\u2019s already diagonal) since H H Z commutes with Jz (so long as we align Bext with the z axis) and L 2, and each of the degenerate states is uniquely labeled by the two quantum numbers m j and. In \ufb01rst-order perturbation theory, the Zeeman correction to the energy is n j m j \u02c6k \u00b7 L + 2S, n j m j Bext = 7.74) where, as mentioned above, we align Bext with the z axis to eliminate the off-diagonal elements of W. Now L + 2S = J + S. Unfortunately, we do not immediately know the expectation value of S. But we can \ufb01gure it out, as follows: The total angular momentum J = L + S is constant (Figure 7.9); L and S precess rapidly about this \ufb01xed vector. In particular, the (time) average value of S is just its projection along J: Save = (S \u00b7 J) J 2 J. (7.75) 22 The gyromagnetic ratio for orbital motion is just the classical value (q/2m)\u2014it is only for spin that there is an \u201cextra\u201d factor of 2. 306 CHAPTER 7 Time-Independent Perturbation Theory S J L Figure 7.9: In the presence of spin-orbit coupling, L and S are not separately conserved; they precess about the \ufb01xed total angular momentum, J. But L = J \u2212 S, so L 2 = J 2 + S2 \u2212 2J \u00b7 S, and hence S \u00b7 J = 1 2 J 2 + S2 \u2212 ) + s (s + 1) \u2212 ( + 1), (7.76) from which it follows that23 1 + S \u00b7 J J 2 L + 2S = = J 1 + j ( j + 1) \u2212 ( + 1) + s (s + 1) 2 j ( j + 1) J. (7.78) The term in square brackets is known as the Land\u00e9 g-factor, gJ.24 The energy corrections are then where E 1 Z = \u03bcB gJ Bextm j, \u03bcB \u2261 e 2m = 5.788 \u00d7 10 \u22125 eV/T (7.79) (7.80) is the so-called Bohr magneton. Recall (Example 6.3) that degeneracy in the quantum number m is a consequence of rotational invariance.25 The perturbation H Z picks out a speci\ufb01c direction in space (the direction of B) which breaks the rotational symmetry and lifts the degeneracy in m. 23 While Equation 7.78 was derived by replacing S by its average value, the result is not an approximation; L + 2S and J are both vector operators and the states are angular-momentum eigenstates. Therefore, the matrix elements can be evaluated by use of the Wigner\u2013Eckart theorem (Equations 6.59\u20136.61). It follows (Problem 7.25) that the matrix elements are proportional: 0 njm j L + 2 S njm 1 j = gJ 0 njm j J njm 1 j (7.77) and the constant of proportionality gJ is the ratio of reduced matrix elements. All that remains is to evaluate gJ : see Claude Cohen-Tannoudji, Bernard Diu, and Franck Lalo\u00eb, Quantum Mechanics, Wiley, New York (1977), Vol. 2, Chapter X. 24 In", " the case of a single electron, where j = \u00b1 1 25 That example speci\ufb01cally treated orbital angular momentum, but the same argument holds for the total angular 2, gJ = (2 j + 1) / (2 + 1). momentum. 7.4 The Zeeman Effect 307 E \u221213.6 (1 + \u03b1 2/4) eV \u03bc BBext = 1/2 mj = \u22121/2 mj Figure 7.10: Weak-\ufb01eld Zeeman splitting of the ground state of hydrogen; the upper line has slope \u22121. slope 1, the lower line m j = \u22121/2 m j = 1/2 has The total energy is the sum of the \ufb01ne-structure part (Equation 7.69) and the Zeeman contribution (Equation 7.79). For example, the ground state (n = 1, = 0, j = 1/2, and therefore gJ = 2) splits into two levels: \u2212 13.6 eV 1 + \u03b12/4 \u00b1 \u03bcB Bext, (7.81) with the plus sign for m j = 1/2, and minus for m j = \u22121/2. These energies are plotted (as functions of Bext) in Figure 7.10. \u2217 Problem 7.24 Consider the (eight) n = 2 states, |2 j m j. Find the energy of each state, under weak-\ufb01eld Zeeman splitting, and construct a diagram like Figure 7.10 to show how the energies evolve as Bext increases. Label each line clearly, and indicate its slope. Problem 7.25 Use the Wigner\u2013Eckart theorem (Equations 6.59\u20136.61) to prove that the matrix elements of any two vector operators, V and W, are proportional in a basis of angular-momentum eigenstates7.82) = \u03b1. Comment: With replaced by j (the theorem holds regardless of whether the states are eigenstates of orbital, spin, or total angular momentum), V = L + 2 S and W = J, this proves Equation 7.77. 7.4.2 Strong-Field Zeeman Effect If Bext Bint, the Zeeman effect dominates26 and we take the \u201cunperturbed\u201d Hamiltonian to be HBohr + H Z and the perturbation to be H fs. The Zeeman Hamiltonian is 26 In this regime the Zeeman effect is also known as the Paschen\u2013Back effect. 308 CHAPTER 7 Time-Independent Perturbation Theory H Z = e 2m Bext (L z + 2Sz), and it is straightforward to compute the \u201cunperturbed\u201d energies: Enml ms = \u2212 13.6 eV n2 + \u03bcB Bext (m + 2ms). (7.83) The states we are using here: |nmms are degenerate, since the energy does not depend on, and there is an additional degeneracy due to the fact that, for example, m = 3 and ms = \u22121/2 or m = 1 and ms = 1/2 have the same energy. Again we are lucky; |nmms are the \u201cgood\u201d states for treating the perturbation. The \ufb01ne structure Hamiltonian H fs commutes with both L 2 and with Jz (these two operators serve as A in the theorem of Section 7.2.2); the \ufb01rst operator resolves the degeneracy in and the second resolves the degeneracy from coincidences in m + 2 ms = m j + ms. In \ufb01rst-order perturbation theory the \ufb01ne structure correction to these levels is E 1 fs = n m ms H r + H so n m ms. (7.84) The relativistic contribution is the same as before (Equation 7.58); for the spin-orbit term (Equation 7.63) we need S \u00b7 L = Sx L x + (7.85) = L x = L x = 0 for eigenstates of Sz and L z). Putting all this + Sz L z = 2ml ms L y Sy (note that Sx = Sy together (Problem 7.26), we conclude that E 1 fs = 13.6 eV n3 \u03b12 \u2212 ( + 1) \u2212 mms ( + 1/2) ( + 1). (7.86) 3 4n (The term in square brackets is indeterminate for = 0; its correct value in this case is 1\u2014see Problem 7.28.) The total energy is the sum of the Zeeman part (Equation 7.83) and the \ufb01ne structure contribution (Equation 7.86). Problem 7.26 Starting with Equation 7.84, and using Equations", " 7.58, 7.63, 7.66, and 7.85, derive Equation 7.86. \u2217\u2217 Problem 7.27 Consider the (eight) n = 2 states, |2 m ms. Find the energy of each state, under strong-\ufb01eld Zeeman splitting. Express each answer as the sum of three terms: the Bohr energy, the \ufb01ne-structure (proportional to \u03b12), and the Zeeman contribution (proportional to \u03bcB Bext). If you ignore \ufb01ne structure altogether, how many distinct levels are there, and what are their degeneracies? Problem 7.28 If = 0, then j = s, m j = ms, and the \u201cgood\u201d states are the same (|n ms) for weak and strong \ufb01elds. Determine E 1 Z (from Equation 7.74) and the \ufb01ne structure energies (Equation 7.69), and write down the general result for the = 0 Zeeman effect\u2014 regardless of the strength of the \ufb01eld. Show that the strong-\ufb01eld formula (Equation 7.86) reproduces this result, provided that we interpret the indeterminate term in square brackets as 1. 7.4.3 Intermediate-Field Zeeman Effect In the intermediate regime neither H equal footing, as perturbations to the Bohr Hamiltonian (Equation 7.43): Z nor H fs dominates, and we must treat the two on an 7.4 The Zeeman Effect 309 + H fs H = H Z I\u2019ll con\ufb01ne my attention here to the case n = 2 (you get to do n = 3 in Problem 7.30). It\u2019s not obvious what the \u201cgood\u201d states are, so we\u2019ll have to resort to the full machinery of degenerate perturbation theory. I\u2019ll choose basis states characterized by, j, and m j.27 Using the Clebsch\u2013 Gordan coef\ufb01cients (Problem 4.60 or Table 4.8) to express | j m j as linear combinations of | s ml ms,28 we have: (7.87). = 0 : = 1 : \u03c81 \u2261 \u03c82 \u2261 1 2 1 2 1 2 \u22121 2 \u2212/3 \u221a \u03c83 \u2261 \u03c84 \u2261 \u03c85 \u2261 3 2 3 2 3 2 3 2 \u2212 = \u2212 \u03c86 \u2261 \u03c87 \u2261 1/3 1 1 2 1 1 1 2 2 In this basis the nonzero matrix elements of H 1/3 \u221a \u03c88 \u2261 = \u2212 3 2 2/3 \u22121 2 \u22121 2 \u221a = 2 0 1 2 \u22121 1 2 \u22121 1 2, \u2212/3 2/3 2/3 1/1 2 \u22121 2 \u22121 2 \u22121 2 tion 7.68; H 7.29): \u239b fs are all on the diagonal, and given by EquaZ has four off-diagonal elements, and the complete matrix \u2212W is (see Problem \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239d 5\u03b3 \u2212 \u03b2 0 0 0 0 0 0 0 0 5\u03b3 + \u2212 2\u03b2 0 0 0 0 0 0 0 0 \u03b3 + 2\u03b2 5\u03b3 \u2212 5\u03b3 + 1 3 \u03b2 27 You can use, m, ms states if you prefer\u2014this makes the matrix elements of H fs more dif\ufb01cult; the W matrix will be more complicated, but its eigenvalues (which are independent of basis) are the same either way. Z easier, but those of H 28 Don\u2019t confuse the notation | s m ms in the Clebsch\u2013Gordan tables with |n j m j (in Section 7.4.1) or |n m ms (in Section 7.4.2); here n is always 2, and s (of course) is always 1/2. 310 CHAPTER 7 Time-Independent Perturbation Theory Table 7.2: Energy levels for the n = 2 states of hydrogen, with \ufb01ne structure and Zeeman splitting. = E2 = E2 = E2 = E2 = E2 = E2 = E2 = E2 1 2 3 4 5 6 7 8 \u03b2 \u2212 5\u03b3 + \u03b2 \u2212 5\u03b3 \u2212 \u03b2 \u2212 \u03b3 + 2\u03b2 \u2212 \u03b3 \u2212 2\u03b2 \u2212 3\u03b3 + \u03b2/2 + \u221a4\u03b3 2 + (2/3)\u03b3\u03b2 + \u03b22/4 \u2212 3\u03b3 + \u03b2/2 \u2212 \u221a4\u03b3 2 + (2/3)\u03b3\u03b2 + \u03b22/4 \u2212 3\u03b3 \u2212 \u03b2/2 + \ufffd", "\ufffd4\u03b3 2 \u2212 (2/3)\u03b3\u03b2 + \u03b22/4 \u2212 3\u03b3 \u2212 \u03b2/2 \u2212 \u221a4\u03b3 2 \u2212 (2/3)\u03b3\u03b2 + \u03b22/4 where \u03b3 \u2261 (\u03b1/8)2 13.6 eV and \u03b2 \u2261 \u03bcB Bext. The \ufb01rst four eigenvalues are already displayed along the diagonal; it remains only to \ufb01nd the eigenvalues of the two 2 \u00d7 2 blocks. The characteristic equation for the \ufb01rst of these is \u03bb2 \u2212 \u03bb (6\u03b3 \u2212 \u03b2) + 5\u03b3 2 \u2212 11 3 \u03b3\u03b2 = 0, and the quadratic formula gives the eigenvalues: \u03bb\u00b1 = 3\u03b3 \u2212 (\u03b2/2) \u00b1 4\u03b3 2 + (2/3) \u03b3\u03b2 + \u03b22/4. (7.88) The eigenvalues of the second block are the same, but with the sign of \u03b2 reversed. The eight energies are listed in Table 7.2, and plotted against Bext in Figure 7.11. In the zero-\ufb01eld limit (\u03b2 = 0) they reduce to the \ufb01ne structure values; for weak \ufb01elds (\u03b2 \u03b3 ) they reproduce what you got in Problem 7.24; for strong \ufb01elds (\u03b2 \u03b3 ) we recover the results of Problem 7.27 (note the convergence to \ufb01ve distinct energy levels, at very high \ufb01elds, as predicted in Problem 7.27). Problem 7.29 Work out the matrix elements of H given in the text, for n = 2. Z and H fs, and construct the W matrix \u2217\u2217\u2217 Problem 7.30 Analyze the Zeeman effect for the n = 3 states of hydrogen, in the weak, strong, and intermediate \ufb01eld regimes. Construct a table of energies (analogous to Table 7.2), plot them as functions of the external \ufb01eld (as in Figure 7.11), and check that the intermediate-\ufb01eld results reduce properly in the two limiting cases. Hint: The Wigner\u2013 Eckart theorem comes in handy here. In Chapter 6 we wrote the theorem in terms of the orbital angular momentum but it also holds for states of total angular momentum j. In particular, 0 j m j V z jm for any vector operator V (and L + 2 S is a vector operator). E Weak Intermediate Strong 7.5 Hyper\ufb01ne Splitting in Hydrogen 311 Figure 7.11: Zeeman splitting of the n = 2 states of hydrogen, in the weak, intermediate, and strong \ufb01eld regimes. \u03bc BBext 7.5 HYPERFINE SPLITTING IN HYDROGEN The proton itself constitutes a magnetic dipole, though its dipole moment is much smaller than the electron\u2019s because of the mass in the denominator (Equation 7.62): \u03bc p = g pe 2m p S p, \u03bc e = \u2212 e me Se. (7.89) (The proton is a composite structure, made up of three quarks, and its gyromagnetic ratio is not as simple as the electron\u2019s\u2014hence the explicit g-factor (g p), whose measured value is 5.59 as opposed to 2.00 for the electron.) According to classical electrodynamics, a dipole \u03bc sets up a magnetic \ufb01eld29 B = \u03bc0 4\u03c0r 3 3 \u03bc \u00b7 \u02c6r \u02c6r \u2212 \u03bc + 2\u03bc0 3 \u03bc\u03b43(r). (7.90) So the Hamiltonian of the electron, in the magnetic \ufb01eld due to the proton\u2019s magnetic dipole moment, is (Equation 7.59) 3 \u2212 S p \u00b7 Se S p \u00b7 \u02c6r S p \u00b7 Se\u03b43(r). (7.91) H hf = \u03bc0g pe2 8\u03c0 m pme + \u03bc0g pe2 3m pme Se \u00b7 \u02c6r r 3 29 If you are unfamiliar with the delta function term in Equation 7.90, you can derive it by treating the dipole as a spinning charged spherical shell, in the limit as the radius goes to zero and the charge goes to in\ufb01nity (with \u03bc held constant). See D. J. Grif\ufb01ths, Am. J. Phys., 50, 698 (1982). 312 CHAPTER 7 Time-Independent Perturbation Theory According to perturbation theory, the \ufb01rst-order correction to the energy (Equation 7.9) is the expectation value of the perturbing Hamiltonian: 2 \u2212 S p \u00b7 Se 3 3 S p \u00b7 \ufffd", "\ufffdr Se \u00b7 \u02c6r r 3 E 1 hf = \u03bc0g pe2 8\u03c0 m pme \u03bc0g pe2 3m pme + S p \u00b7 Se |\u03c8(0)|2. (7.92) In the ground state (or any other state for which = 0) the wave function is spherically symmetric, and the \ufb01rst expectation value vanishes (see Problem 7.31). Meanwhile, from Equation 4.80 we \ufb01nd that |\u03c8100(0)|2 = 1/, so \u03c0a3 \u03bc0g pe2 3\u03c0 m pmea3 in the ground state. This is called spin-spin coupling, because it involves the dot product of two spins (contrast spin-orbit coupling, which involves S \u00b7 L). S p \u00b7 Se (7.93) E 1 hf =, In the presence of spin-spin coupling, the individual spin angular momenta are no longer conserved; the \u201cgood\u201d states are eigenvectors of the total spin, S \u2261 Se + S p. As before, we square this out to get S p \u00b7 Se = 1 2 S2 \u2212 S2 e \u2212 S2 p. (7.94) (7.95) = (3/4) 2. In the triplet state But the electron and proton both have spin 1/2, so S2 e (spins \u201cparallel\u201d) the total spin is 1, and hence S2 = 22; in the singlet state the total spin is 0, and S2 = 0. Thus = S2 p E 1 hf = 4g p4 3m pm2 ec2a4 +1/4, \u22123/4, (triplet) ; (singlet). (7.96) Spin-spin coupling breaks the spin degeneracy of the ground state, lifting the triplet con\ufb01guration and depressing the singlet (see Figure 7.12). The energy gap is (7.97) (7.98) E = 4g p4 = 5.88 \u00d7 10 \u22126eV. ec2a4 The frequency of the photon emitted in a transition from the triplet to the singlet state is 3m pm2 Unperturbed = 1420 MHz, \u03bd = E h Triplet \u0394E Singlet Figure 7.12: Hyper\ufb01ne splitting in the ground state of hydrogen. and the corresponding wavelength is c/\u03bd = 21 cm, which falls in the microwave region. This famous 21-centimeter line is among the most pervasive forms of radiation in the universe. Further Problems on Chapter 7 313 Problem 7.31 Let a and b be two constant vectors. Show that sin \u03b8 d\u03b8 d\u03c6 = 4\u03c0 3 b \u00b7 \u02c6r a \u00b7 \u02c6r (a \u00b7 b) (7.99) (the integration is over the usual range < 2\u03c0 ). Use this result to demonstrate that 2 3 S p \u00b7 \u02c6r Se \u00b7 \u02c6r r 3 3 \u2212 S p \u00b7 Se = 0, for states with = 0. Hint: \u02c6r = sin \u03b8 cos \u03c6 \u02c6\u0131 + sin \u03b8 sin \u03c6 \u02c6j + cos \u03b8 \u02c6k. Do the angular integrals \ufb01rst. Problem 7.32 By appropriate modi\ufb01cation of the hydrogen formula, determine the hyper\ufb01ne splitting in the ground state of (a) muonic hydrogen (in which a muon\u2014same charge and g-factor as the electron, but 207 times the mass\u2014substitutes for the electron), (b) positronium (in which a positron\u2014same mass and g-factor as the electron, but opposite charge\u2014substitutes for the proton), and (c) muonium (in which an anti-muon\u2014same mass and g-factor as a muon, but opposite charge\u2014substitutes for the proton). Hint: Don\u2019t forget to use the reduced mass (Problem 5.1) in calculating the \u201cBohr radius\u201d of these exotic \u201catoms,\u201d but use the actual masses in the gyromagnetic ratios. Incidentally, the answer you get for positronium (4.82 \u00d7 10\u22124 eV) is quite far from the experimental value (8.41 \u00d7 10\u22124 eV); the large discrepancy is due to pair annihilation, which contributes an extra (3/4) E, and does not occur (of course) in ordinary hydrogen, muonic hydrogen, or muonium.30 e+ + e\u2212 \u2192 \u03b3 + \u03b3 FURTHER PROBLEMS ON CHAPTER 7 Problem 7.33 Estimate the correction to the ground state energy of hydrogen due", " to the \ufb01nite size of the nucleus. Treat the proton as a uniformly charged spherical shell of radius b, so the potential energy of an electron inside the shell is constant: \u2212e2/ (4\u03c0 0b); this isn\u2019t very realistic, but it is the simplest model, and it will give us the right order of magnitude. Expand your result in powers of the small parameter (b/a), where a is the Bohr radius, and keep only the leading term, so your \ufb01nal answer takes the form E E = A (b/a)n. Your business is to determine the constant A and the power n. Finally, put in b \u2248 10\u221215 m (roughly the radius of the proton) and work out the actual number. How does it compare with \ufb01ne structure and hyper\ufb01ne structure? 30 For details see Grif\ufb01ths, footnote 29, page 311. 314 CHAPTER 7 Time-Independent Perturbation Theory Problem 7.34 In this problem you will develop an alternative approach to degenerate perturbation theory. Consider an unperturbed Hamiltonian H 0 with two degenerate states b (energy E 0), and a perturbation H. De\ufb01ne the operator that projects31 a and \u03c8 0 \u03c8 0 onto the degenerate subspace: \u03c8 0 b \u03c8 0 a PD = + (7.100 The Hamiltonian can be written H = H 0 + H = \u02dcH 0 + \u02dcH (7.101) where \u02dcH 0 = H 0 + PD H PD, (7.102) \u02dcH 0 as the \u201cunperturbed\u201d Hamiltonian and \u02dcH as the perturbation; \u02dcH 0 is nondegenerate, so we can use ordinary nondegenerate \u02dcH = H \u2212 PD H PD. The idea is to treat as you\u2019ll soon discover, perturbation theory. (a) First we need to \ufb01nd the eigenstates of \u02dcH 0. i. Show that any eigenstate \u03c8 0 eigenstate of \u02dcH 0 with the same eigenvalue. n (other than \u03c8 0 a or \u03c8 0 b ) of H 0 is also an ii. Show that the \u201cgood\u201d states with \u03b1 and \u03b2 determined by solving Equation 7.30) are eigenstates of \u02dcH 0 with energies E 0 + E 1 \u00b1. (b) Assuming that E 1 + and E 1 \u2212 are distinct, you now have a nondegenerate unperturbed Hamiltonian \u02dcH 0 and you can do nondegenerate perturbation theory using the perturbation \u02dcH. Find an expression for the energy to second order for the states \u03c8 0 \u00b1 in (ii). Comment: One advantage of this approach is that it also handles the case where the unperturbed energies are not exactly equal, but very close:32 E 0 b. In a this case one must still use degenerate perturbation theory; an important example of this occurs in the nearly-free electron approximation for calculating band structure.33 \u2248 E 0 Problem 7.35 Here is an application of the technique developed in Problem 7.34. Consider the Hamiltonian H0 = \u239b \u239d \u2217 b\u2217 a 0 c\u2217 b c 0 (a) Find the projection operator PD (it\u2019s a 3 \u00d7 3 matrix) that projects onto the subspace spanned by \u239b \u239d = 1 \u03c8 0 a 1 0 0 \u239e \u23a0 and \u239b \u239d = Then construct the matrices \u02dcH 0 and \u02dcH. 31 See page 118 for a discussion of projection operators. 32 See Problem 7.4 for a discussion of what close means in this context. 33 See, for example, Steven H. Simon, The Oxford Solid State Basics (Oxford University Press, 2013), Section 15.1. 0 (b) Solve for the eigenstates of \u02dcH and verify... Further Problems on Chapter 7 315 i. that its spectrum is nondegenerate, ii. that the nondegenerate eigenstate of H0 \u239b is also an eigenstate of \u02dcH with the same eigenvalue. (c) What are the \u201cgood\u201d states, and what are their energies, to \ufb01rst order in the perturbation? Problem 7.36 Consider the isotropic three-dimensional harmonic oscillator (Problem 4.46). Discuss the effect (in \ufb01rst order) of the perturbation H = \u03bbx 2 yz (for some constant \u03bb) on (a) the ground state; (b) the (triply degenerate) \ufb01r", "st excited state. Hint: Use the answers to Problems 2.12 and 3.39. \u2217\u2217\u2217 Problem 7.37 Van der Waals interaction. Consider two atoms a distance R apart. Because they are electrically neutral you might suppose there would be no force between them, but if they are polarizable there is in fact a weak attraction. To model this system, picture each atom as an electron (mass m, charge \u2212e) attached by a spring (spring constant k) to the nucleus (charge +e), as in Figure 7.13. We\u2019ll assume the nuclei are heavy, and essentially motionless. The Hamiltonian for the unperturbed system is H 0 = 1 2m p2 1 + 1 2 kx 2 1 + 1 2m p2 2 + 1 2 kx 2 2. The Coulomb interaction between the atoms is \u2212 e2 \u2212 e2 H = 1 4\u03c0 0 e2 R R \u2212 x1 R + x2 + e2 R \u2212 x1 + x2. (7.103) (7.104) +e \u2212e +e \u2212e x1 R x2 Figure 7.13: Two nearby polarizable atoms (Problem 7.37). (a) Explain Equation 7.104. Assuming that |x1| and |x2| are both much less than R, show that H \u223c= \u2212 e2x1x2 2\u03c0 0 R3. (7.105) (b) Show that the total Hamiltonian (H 0 plus Equation 7.105) separates into two harmonic oscillator Hamiltonians: 316 CHAPTER 7 Time-Independent Perturbation Theory H = 1 2m + + 1 p2 2 k \u2212 e2 2\u03c0 0 R3 x 2 + + 1 2m \u2212 + 1 p2 2 k + e2 2\u03c0 0 R3 under the change variables x 2 \u2212, (7.106) (7.107) x\u00b1 \u2261 1\u221a 2 p\u00b1 = 1\u221a 2 (c) The ground state energy for this Hamiltonian is evidently (x1 \u00b1 x2), which entails ( p1 \u00b1 p2). E = 1 2 (\u03c9+ + \u03c9\u2212), where \u03c9\u00b1 = / e2/2\u03c0 0 R3 k \u2213 m. (7.108) Without the Coulomb interaction it would have been E0 = \u03c90, where \u03c90 = \u221a, show that k/m. Assuming that k e2/2\u03c0 0 R3 V \u2261 E \u2212 E0 \u223c= \u2212 8m2\u03c93 0 2 e2 2\u03c0 0. 1 R6 (7.109) Conclusion: There is an attractive potential between the atoms, proportional to the inverse sixth power of their separation. This is the van der Waals interaction between two neutral atoms. (d) Now do the same calculation using second-order perturbation theory. Hint: The (x2), where \u03c8n(x) is a one-particle (x1) \u03c8n2 unperturbed states are of the form \u03c8n1 oscillator wave function with mass m and spring constant k; V is the secondorder correction to the ground state energy, for the perturbation in Equation 7.105 (notice that the \ufb01rst-order correction is zero).34 \u2217\u2217 Problem 7.38 Suppose the Hamiltonian H, for a particular quantum system, is a function of some parameter \u03bb; let En(\u03bb) and \u03c8n(\u03bb) be the eigenvalues and eigenfunctions of H (\u03bb). The Feynman\u2013Hellmann theorem35 states that \u2202 En \u2202\u03bb = \u03c8n \u2202 H \u2202\u03bb \u03c8n (7.110) (assuming either that En is nondegenerate, or\u2014if degenerate\u2014that the \u03c8ns are the \u201cgood\u201d linear combinations of the degenerate eigenfunctions). (a) Prove the Feynman\u2013Hellmann theorem. Hint: Use Equation 7.9. (b) Apply it to the one-dimensional harmonic oscillator, (i) using \u03bb = \u03c9 (this yields a formula for the expectation value of V ), (ii) using \u03bb = (this yields T ), and (iii) using \u03bb = m (this yields a relation between T and V ). Compare your answers to Problem 2.12, and the virial theorem predictions (Problem 3.37). 34 There is an interesting fraud in this well-known problem. If you expand H to order 1/R5, the extra term has a nonzero expectation value in the ground state of H 0, so there is a nonzero \ufb01rst-order perturbation, and the dominant contribution goes like 1/R5, not 1/R6. The model gets", " the power \u201cright\u201d in three dimensions (where the expectation value is zero), but not in one. See A. C. Ipsen and K. Splittorff, Am. J. Phys. 83, 150 (2015). 35 Feynman obtained Equation 7.110 while working on his undergraduate thesis at MIT (R. P. Feynman, Phys. Rev. 56, 340, 1939); Hellmann\u2019s work was published four years earlier in an obscure Russian journal. Further Problems on Chapter 7 317 Problem 7.39 Consider a three-level system with the unperturbed Hamiltonian \u239b H0 = \u239d 7.111) (a > c) and the perturbation \u239b \u239d \u23a0. V V 0 H = (7.112) 0 0 0 0 V \u2217 V \u2217 Since the (2 \u00d7 2) matrix W is diagonal (and in fact identically 0) in the basis of states (1, 0, 0) and (0, 1, 0), you might assume they are the good states, but they\u2019re not. To see this: (a) Obtain the exact eigenvalues for the perturbed Hamiltonian H = H0 + H (b) Expand your results from part (a) as a power series in |V | up to second order. (c) What do you obtain by applying nondegenerate perturbation theory to \ufb01nd the energies of all three states (up to second order)? This would work if the assumption about the good states above were correct.. Moral: If any of the eigenvalues of W are equal, the states that diagonalize W are not unique, and diagonalizing W does not determine the \u201cgood\u201d states. When this happens (and it\u2019s not uncommon), you need to use second-order degenerate perturbation theory (see Problem 7.40). Problem 7.40 If it happens that the square root in Equation 7.33 vanishes, then E 1 + = E 1 \u2212; the degeneracy is not lifted at \ufb01rst order. In this case, diagonalizing the W matrix puts no restriction on \u03b1 and \u03b2 and you still don\u2019t know what the \u201cgood\u201d states are. If you need to determine the \u201cgood\u201d states\u2014for example to calculate higher-order corrections\u2014you need to use second-order degenerate perturbation theory. (a) Show that, for the two-fold degeneracy studied in Section 7.2.1, the \ufb01rst-order correction to the wave function in degenerate perturbation theory is \u03c8 1 = m=a,b \u03b1 Vma + \u03b2 Vmb b) Consider the terms of order \u03bb2 (corresponding to Equation 7.8 in the nondegenerate case) to show that \u03b1 and \u03b2 are determined by \ufb01nding the eigenvectors of the matrix W2 (the superscript denotes second order, not W squared) where # $ W2 = i j m=a, and that the eigenvalues of this matrix correspond to the second-order energies E 2. (c) Show that second-order degenerate perturbation theory, developed in (b), gives the correct energies to second order for the three-state Hamiltonian in Problem 7.39. \u2217\u2217 Problem 7.41 A free particle of mass m is con\ufb01ned to a ring of circumference L such that \u03c8(x + L) = \u03c8(x). The unperturbed Hamiltonian is 318 CHAPTER 7 Time-Independent Perturbation Theory H 0 = \u2212 2 2 m d2 d x 2, to which we add a perturbation 2 \u03c0 x L (a) Show that the unperturbed states may be written H = V0 cos. (x) = 1\u221a L for n = 0, \u00b11, \u00b12 and that, apart from n = 0, all of these states are two-fold degenerate. ei 2 \u03c0 n x/L \u03c8 0 n (b) Find a general expression for the matrix elements of the perturbation: 0 H mn = c) Consider the degenerate pair of states with n = \u00b11. Construct the matrix W and calculate the \ufb01rst-order energy corrections, E 1. Note that the degeneracy does not lift at \ufb01rst order. Therefore, diagonalizing W does not tell us what the \u201cgood\u201d states are. (d) Construct the matrix W2 (Problem 7.40) for the states n = \u00b11, and show that the degeneracy lifts at second order. What are the good linear combinations of the states with n = \u00b11? (e) What are the energies, accurate to second order, for these", " states?36 \u2217\u2217 Problem 7.42 The Feynman\u2013Hellmann theorem (Problem 7.38) can be used to determine the expectation values of 1/r and 1/r 2 for hydrogen.37 The effective Hamiltonian for the radial wave functions is (Equation 4.53) d2 dr 2 and the eigenvalues (expressed in terms of )38 are (Equation 4.70) \u2212 e2 4\u03c0 0 ( + 1) r 2 H = \u2212 2m 2m 1 r 2 2 +, 32\u03c0 22 0 (a) Use \u03bb = e in the Feynman\u2013Hellmann theorem to obtain 1/r. Check your result En = \u2212 me4 2 (N + )2. against Equation 7.56. 1/r 2 (b) Use \u03bb = to obtain. Check your answer with Equation 7.57. 36 See D. Kiang, Am. J. Phys. 46 (11), 1978 and L.-K. Chen, Am. J. Phys. 72 (7), 2004 for further discussion of this problem. It is shown that each degenerate energy level, E 0 \u00b1n, splits at order 2n in perturbation theory. The exact solution to the problem can also be obtained as the time-independent Schr\u00f6dinger equation for H = H 0 + H 1 reduces to the Mathieu equation. 37 C. S\u00e1nchez del Rio, Am. J. Phys., 50, 556 (1982); H. S. Valk, Am. J. Phys., 54, 921 (1986). 38 In part (b) we treat as a continuous variable; n becomes a function of, according to Equation 4.67, because N, which must be an integer, is \ufb01xed. To avoid confusion, I have eliminated n, to reveal the dependence on explicitly. \u2217\u2217\u2217 \u2217\u2217\u2217 Problem 7.43 Prove Kramers\u2019 relation:39 + s 4 \u2212 (2s + 1) a s + 1 n2 r s\u22121 r s Further Problems on Chapter 7 319 # (2 + 1)2 \u2212 s2 $ r s\u22122 a2 = 0, (7.113) which relates the expectation values of r to three different powers (s, s \u2212 1, and s \u2212 2), for an electron in the state \u03c8nm of hydrogen. Hint: Rewrite the radial equation (Equation 4.53) in the form ( + 1) r 2 + 1 n2a2 r s\u22121 ur su and use it to express, and tion by parts to reduce the second derivative. Show that and dr in terms of r s, u = 2/ (s + 1) \u2212 2 ar ur s+1u dr = \u2212 ur su u, ur su dr. Take it from there. r s\u22122. Then use integra- r s\u22121, dr = \u2212 (s/2) Problem 7.44 (a) Plug s = 0, s = 1, s = 2, and s = 3 into Kramers\u2019 relation (Equation 7.113). Note that you could continue, and r 2 to obtain formulas for inde\ufb01nitely, to \ufb01nd any positive power., r, r \u22121 r 3 (b) In the other direction, however, you hit a snag. Put in s = \u22121, and show that all r \u22123 you get is a relation between r \u22122 and. (c) But if you can get by some other means, you can apply the Kramers relation to obtain the rest of the negative powers. Use Equation 7.57 (which, and check your answer against is derived in Problem 7.42) to determine Equation 7.66. r \u22123 r \u22122 Problem 7.45 When an atom is placed in a uniform external electric \ufb01eld Eext, the energy levels are shifted\u2014a phenomenon known as the Stark effect (it is the electrical analog to the Zeeman effect). In this problem we analyze the Stark effect for the n = 1 and n = 2 states of hydrogen. Let the \ufb01eld point in the z direction, so the potential energy of the electron is H S = eEextz = eEextr cos \u03b8. Treat this as a perturbation on the Bohr Hamiltonian (Equation 7.43). (Spin is irrelevant to this problem, so ignore it, and neglect the \ufb01ne structure.) (a) Show that the ground state energy is not affected by this perturbation, in \ufb01rst order. (b) The \ufb01rst excited state is four-fold degenerate: \u03c8200, \u03c8211, \u03c8210, \u03c821\u22121. Using degenerate perturbation theory", ", determine the \ufb01rst-order corrections to the energy. Into how many levels does E2 split? (c) What are the \u201cgood\u201d wave functions for part (b)? Find the expectation value of the, in each of these \u201cgood\u201d states. Notice that the electric dipole moment results are independent of the applied \ufb01eld\u2014evidently hydrogen in its \ufb01rst excited state can carry a permanent electric dipole moment. = \u2212er pe 39 This is also known as the (second) Pasternack relation. See H. Beker, Am. J. Phys. 65, 1118 (1997). For a proof based on the Feynman\u2013Hellmann theorem (Problem 7.38) see S. Balasubramanian, Am. J. Phys. 68, 959 (2000). 320 CHAPTER 7 Time-Independent Perturbation Theory Hint: There are lots of integrals in this problem, but almost all of them are zero. So study each one carefully, before you do any calculations: If the \u03c6 integral vanishes, there\u2019s not much point in doing the r and \u03b8 integrals! You can avoid those integrals altogether if you use the selection rules of Sections 6.4.3 and 6.7.2. Partial answer: W13 = W31 = \u22123ea Eext; all other elements are zero. \u2217\u2217\u2217 Problem 7.46 Consider the Stark effect (Problem 7.45) for the n = 3 states of hydrogen. There are initially nine degenerate states, \u03c83m (neglecting spin, as before), and we turn on an electric \ufb01eld in the z direction. (a) Construct the 9 \u00d7 9 matrix representing the perturbing Hamiltonian. Partial 3a, 3 1 \u00b11 |z| 3 2 \u00b11 = 6a, 3 1 0 |z| 3 2 0 = \u22123 \u221a \u221a answer: 3 0 0 |z| 3 1 0 = \u22123 \u2212 (9/2) a. (b) Find the eigenvalues, and their degeneracies. Problem 7.47 Calculate the wavelength, in centimeters, of the photon emitted under a hyper\ufb01ne transition in the ground state (n = 1) of deuterium. Deuterium is \u201cheavy\u201d hydrogen, with an extra neutron in the nucleus; the proton and neutron bind together to form a deuteron, with spin 1 and magnetic moment the deuteron g-factor is 1.71. \u03bc d = gd e 2md Sd ; \u2217\u2217\u2217 Problem 7.48 In a crystal, the electric \ufb01eld of neighboring ions perturbs the energy levels of an atom. As a crude model, imagine that a hydrogen atom is surrounded by three pairs of point charges, as shown in Figure 7.14. (Spin is irrelevant to this problem, so ignore it.) z d3 q3 d3 q3 d2 d1 q2 x q1 d1 q1 d2 q2 y Figure 7.14: Hydrogen atom surrounded by six point charges (crude model for a crystal lattice); Problem 7.48. Further Problems on Chapter 7 321 (a) Assuming that r d1, r d2, and r d3, show that H = V0 + 3 \u03b21x 2 + \u03b22 y2 + \u03b23z2 \u2212 (\u03b21 + \u03b22 + \u03b23) r 2, where, \u03b21d2 1 + \u03b22d2 2 and V0 = 2 \u03b2i \u2261 \u2212 e 4\u03c0 0 qi d3 i (b) Find the lowest-order correction to the ground state energy. (c) Calculate the \ufb01rst-order corrections to the energy of the \ufb01rst excited states (n = 2). Into how many levels does this four-fold degenerate system split, (i) in the case of cubic symmetry, \u03b21 = \u03b22 = \u03b23; (ii) in the case of tetragonal symmetry, \u03b21 = \u03b22 = \u03b23; (iii) in the general case of orthorhombic symmetry (all three different)? Note: you might recognize the \u201cgood\u201d states from Problem 4.71. + \u03b23d2 3. Problem 7.49 A hydrogen atom is placed in a uniform magnetic \ufb01eld B0 = B0 \u02c6z (the Hamiltonian can be written as in Equation 4.230). Use the Feynman\u2013Hellman theorem (Problem 7.38) to show that \u2202 En \u2202 B0 where the electron\u2019s magnetic dipole moment40 (orbital plus spin) is = \u2212 \u03c8n |\u03bcz| \u03c8n (7.114) \u03bc = \u03b3o L", "mechanical + \u03b3 S. The mechanical angular momentum is de\ufb01ned in Equation 4.231. Note: From Equation 7.114 it follows that the magnetic susceptibility of N atoms in a volume V and at 0 K (when they\u2019re all in the ground state) is41 \u03c7 = \u03bc0 \u2202 M \u2202 B0 = \u2212 N V \u03bc0 \u2202 2 E0 \u2202 B2 0, (7.115) where E0 is the ground-state energy. Although we derived Equation 7.114 for a hydrogen atom, the expression applies to multi-electron atoms as well\u2014even when electron\u2013electron interactions are included. Problem 7.50 For an atom in a uniform magnetic \ufb01eld B0 = B0 \u02c6z, Equation 4.230 gives H = Hatom \u2212 B0 (\u03b3o L z + \u03b3 Sz) + e2 8 m B2 0 x 2 i + y2 i, Z i=1 where L z and Sz refer to the total orbital and spin angular momentum of all the electrons. 40 For most purposes we can take this to be the magnetic moment of the atom as well. The proton\u2019s larger mass means that its contribution to the dipole moment is orders of magnitude smaller than the electron\u2019s contribution. 41 See Problem 5.33 for the de\ufb01nition of magnetic susceptibility. This formula does not apply when the ground state is degenerate (see Neil W. Ashcroft and N. David Mermin, Solid State Physics (Belmont: Cengage, 1976), p. 655); atoms with non-degenerate ground states have J = 0 (see Table 5.1). 322 CHAPTER 7 Time-Independent Perturbation Theory (a) Treating the terms involving B0 as a perturbation, compute the shift of the ground state energy of a helium atom to second order in B0. Assume that the helium ground state is given by \u03c80 = \u03c8100(r1) \u03c8100(r2) |\u2191 \u2193 \u2212 |\u2193\u2191 \u221a 2 where \u03c8100 refers to the hydrogenic ground state (with Z = 2). (b) Use the results of Problem 7.49 to calculate the magnetic susceptibility of helium. Given a density of 0.166 kg/m3, obtain a numerical value for the susceptibility. Note: The experimental result is \u22121.0 \u00d7 10\u22129 (the negative sign means that helium is a diamagnet). The results can be brought closer by taking account of screening, which increases the orbital radius (see Section 8.2). \u2217\u2217\u2217 Problem 7.51 Sometimes it is possible to solve Equation 7.10 directly, without having to n in terms of the unperturbed wave functions (Equation 7.11). Here are two expand \u03c8 1 particularly nice examples. (a) Stark effect in the ground state of hydrogen. (i) Find the \ufb01rst-order correction to the ground state of hydrogen in the presence of a uniform external electric \ufb01eld Eext (see Problem 7.45). Hint: Try a solution of the form A + Br + Cr 2 e\u2212r/a cos \u03b8 ; your problem is to \ufb01nd the constants A, B, and C that solve Equation 7.10. (ii) Use Equation 7.14 to determine the second-order correction to the ground state energy (the \ufb01rst-order correction is zero, as you found in Problem 2 7.45(a)). Answer: \u2212m 3a2eEext/2. (b) If the proton had an electric dipole moment p, the potential energy of the electron in hydrogen would be perturbed in the amount H = \u2212 ep cos \u03b8 4\u03c0 0r 2. (i) Solve Equation 7.10 for the \ufb01rst-order correction to the ground state wave function. (ii) Show that the total electric dipole moment of the atom is (surprisingly) zero, to this order. (iii) Use Equation 7.14 to determine the second-order correction to the ground state energy. What is the \ufb01rst-order correction? Problem 7.52 Consider a spinless particle of charge q and mass m constrained to move in the x y plane under the in\ufb02uence of the two-dimensional harmonic oscillator potential V (x, y) = 1 2 m\u03c92 x 2 + y2. (a) Construct the ground state wave function, \u03c80(x, y), and write down its energy. Do the same for the (degenerate) \ufb01rst excited states. (b) Now imagine that we turn on a weak magnetic \ufb01eld of magnitude B0 pointing in the z-", "direction, so that (to \ufb01rst order in B0) the Hamiltonian acquires an extra term Further Problems on Chapter 7 323 H = \u2212\u03bc \u00b7 B = \u2212 q 2m (L \u00b7 B) = \u2212 q B0 2m x py \u2212 y px. Treating this as a perturbation, \ufb01nd the \ufb01rst-order corrections to the energies of the ground state and \ufb01rst excited states. \u2217\u2217\u2217 Problem 7.53 Imagine an in\ufb01nite square well (Equation 2.22) into which we introduce a delta-function perturbation, H (x) = \u03bb \u03b4(x \u2212 x0), where \u03bb is a positive constant, and 0 < x0 < a (to simplify matters, let x0 = pa, where 0 < p < 1).42 (a) Find the \ufb01rst-order correction to the nth allowed energy (Equation 2.30), assuming \u03bb is small. (What does \u201csmall\u201d mean, in this context?) (b) Find the second-order correction to the allowed energies. (Leave your answer as a sum.) (c) Now solve the Schr\u00f6dinger equation exactly, treating separately the regions 0 \u2264 x < x0 and x0 < x < a, and imposing the boundary conditions at x0. Derive the transcendental equation for the energies: un sin(un) + sin( pun) sin Here \u2261 2ma\u03bb/2, un \u2261 kna, and kn \u2261 reproduces your result from part (a), in the appropriate limit. (1 \u2212 p) un \u221a = 0 (E > 0). (7.116) 2m En/. Check that Equation 7.116 \u221a (d) Everything so far holds just as well if \u03bb is negative, but in that case there may be an additional solution with negative energy. Derive the transcendental equation for a negative-energy state: v sinh(v) + sinh( pv) sinh = 0 (E < 0), (7.117) (1 \u2212 p) v \u22122m E/. Specialize to the symmetrical case p = 1/2, where v \u2261 \u03baa and \u03ba \u2261 and show that you recover the energy of the delta-function well (Equation 2.132), in the appropriate regime. (e) There is in fact exactly one negative-energy solution, provided that || > 1/. First, prove this (graphically), for the case p = 1/2. (Below that critical value there is no negative-energy solution.) Next, by computer, plot the solution v, as a function of p, for = \u22124.1, \u22125, and \u221210. Verify that the solution only exists within the predicted range of p. p (1 \u2212 p) (f) For p = 1/2, plot the ground state wave function, \u03c81 (x), for = 0, \u22122, \u22123, \u22123.5, \u22124.5, \u22125, and \u221210, to show how the sinusoidal shape (Figure 2.2) evolves into the exponential shape (Figure 2.13), as the delta function well \u201cdeepens.\u201d43 \u2217\u2217\u2217 Problem 7.54 Suppose you want to calculate the expectation value of some observable, in the nth energy eigenstate of a system that is perturbed by H : 0 1 \u02c6 \u03c8n. = \u03c8n 42 We adopt the notation of Y. N. Joglekar, Am. J. Phys. 77, 734 (2009), from which this problem is drawn. 43 For the corresponding analysis of the delta function barrier (positive \u03bb) see Problem 11.34. 324 CHAPTER 7 Time-Independent Perturbation Theory Replacing \u03c8n by its perturbation expansion, Equation 7.5,44 1$ # The \ufb01rst-order correction to is therefore #0 1 = 2Re \u03c8 0 n 1$ \u02c6 \u03c8 1 n, or, using Equation 7.13, 1 = 2 Re 0 \u03c8 0 n m=n \u02c6 \u03c8 0 m E 0 n H 10 \u03c8 0 m \u2212 E 0 m + \u03bb2 (\u00b7 \u00b7 \u00b7 ) + \u00b7 \u00b7 \u00b7. 1 \u03c8 0 n (7.118) (assuming the unperturbed energies are nondegenerate, or that we are using the \u201cgood\u201d basis states). (a) Suppose = H (the perturbation itself). What does Equation 7.118 tell us in this case? Explain (carefully) why this is consistent with Equation 7.15. (b) Consider a particle of charge q (maybe an electron in a hydrogen", " atom, or a pith ball connected to a spring), that is placed in a weak electric \ufb01eld Eext pointing in the x direction, so that H = \u2212q Eextx. The \ufb01eld will induce an electric dipole moment, pe = q x, in the \u201catom.\u201d The expectation value of pe is proportional to the applied \ufb01eld, and the proportionality factor is called the polarizability, \u03b1. Show that 2q2 (7.119) 2. m=n Find the polarizability of the ground state of a one-dimensional harmonic oscillator. Compare the classical answer. (c) Now imagine a particle of mass m in a one-dimensional harmonic oscillator with a small anharmonic perturbation45 H = \u2212 1 6 \u03ba x 3. (7.120) Find x (to \ufb01rst order), in the nth energy eigenstate. Answer: 2m2\u03c93 Comment: As the temperature increases, higher-energy states are populated, and the particles move farther (on average) from their equilibrium positions; that\u2019s why most solids expand with rising temperature. \u03ba/ n + 1 2. Problem 7.55 Crandall\u2019s Puzzle.46 Stationary states of the one-dimensional Schr\u00f6dinger equation ordinarily respect three \u201crules of thumb\u201d: (1) the energies are nondegenerate, 44 In general, Equation 7.5 does not deliver a normalized wave function, but the choice c guarantees normalization to \ufb01rst order in \u03bb, which is all we require here (see footnote 4, page 282). (n) n = 0 in Equation 7.11 45 This is just a generic tweak to the simple harmonic oscillator potential, 1 2 kx 2; \u03ba is some constant, and the factor of \u22121/6 is for convenience. 46 Richard Crandall introduced me to this problem. Further Problems on Chapter 7 325 (2) the ground state has no nodes, the \ufb01rst excited state has one node, the second has two, and so on, and (3) if the potential is an even function of x, the ground state is even, the \ufb01rst excited state is odd, the second is even, and so on. We have already seen that the \u201cbead-on-a-ring\u201d (Problem 2.46) violates the \ufb01rst of these; now suppose we introduce a \u201cnick\u201d in at the origin: H = \u2212\u03b1\u03b4(x). (If you don\u2019t like the delta function, make it a gaussian, as in Problem 7.9.) This lifts the degeneracy, but what is the sequence of even and odd wave functions, and what is the sequence of node numbers? Hint: You don\u2019t really need to do any calculations, here, and you\u2019re welcome to assume that \u03b1 is small, but by all means solve the Schr\u00f6dinger equation exactly if you prefer. \u2217\u2217\u2217 Problem 7.56 In this problem we treat the electron\u2013electron repulsion term in the helium Hamiltonian (Equation 5.38) as a perturbation, H = 1 4\u03c0 0 e2 |r1 \u2212 r2|. (This will not be very accurate, because the perturbation is not small, in comparison to the Coulomb attraction of the nucleus... but it\u2019s a start.) (a) Find the \ufb01rst-order correction to the ground state, \u03c80(r1, r2) = \u03c8100(r1)\u03c8100(r2). (You have already done this calculation, if you worked Problem 5.15\u2014only we didn\u2019t call it perturbation theory back then.) (b) Now treat the \ufb01rst excited state, in which one electron is in the hydrogenic ground state, \u03c8100, and the other is in the state \u03c8200. Actually, there are two such states, depending on whether the electron spins occupy the singlet con\ufb01guration (parahelium) or the triplet (orthohelium):47 \u03c8\u00b1(r1, r2) = 1\u221a 2 [\u03c8100(r1)\u03c8200(r2) \u00b1 \u03c8200(r1)\u03c8100(r2)]. Show that where K \u00b1 J ), \u03c8100(r1)\u03c8200(r2)H \u03c8100(r1)\u03c8200(r2) d3r1 d3r2, J \u2261 2 \u03c8100(r1)\u03c8200(r2)H \u03c8200(r1)\u03c8100(r2) d", "3r1 d3r2. Evaluate these two integrals, put in the actual numbers, and compare your results with Figure 5.2 (the measured energies are \u221259.2 eV and \u221258.4 eV).48 47 It seems strange, at \ufb01rst glance, that spin has anything to do with it, since the perturbation itself doesn\u2019t involve spin (and I\u2019m not even bothering to include the spin state explicitly). The point, of course, is that an antisymmetric spin state forces a symmetric (position) wave function, and vice versa, and this does affect the result. 48 If you want to pursue this problem further, see R. C. Mass\u00e9 and T. G. Walker, Am. J. Phys. 83, 730 (2015). 326 CHAPTER 7 Time-Independent Perturbation Theory Problem 7.57 The Hamiltonian for the Bloch functions (Equation 6.12) can be analyzed with perturbation theory by de\ufb01ning H 0 and H such that H 0 un0 = En0 un0, unq = Enq unq. H 0 + H In this problem, don\u2019t assume anything about the form of V (x). (a) Determine the operators H 0 and H (express them in terms of \u02c6p). (b) Find Enq to second order in q. That is, \ufb01nd expressions for An, Bn, and Cn (in terms of the En0 and matrix elements of \u02c6p in the \u201cunperturbed\u201d states un0) Enq \u2248 An + Bn q + Cn q2. (c) Show that the constants Bn are all zero. Hint: See Problem 2.1(b) to get started. Remember that un0(x) is periodic. Comment: It is conventional to write Cn = 2/2 m\u2217 of particles in the nth band since then, as you\u2019ve just shown, n where m\u2217 n is the effective mass 2 q2 2 m\u2217 n just like the free particle (Equation 2.92) with k \u2192 q. Enq \u2248 constant + 8 THE VARIATIONAL PRINCIPLE 8.1 THEORY Suppose you want to calculate the ground state energy, Egs, for a system described by the Hamiltonian H, but you are unable to solve the (time-independent) Schr\u00f6dinger equation. The variational principle will get you an upper bound for Egs, which is sometimes all you need, and often, if you\u2019re clever about it, very close to the exact value. Here\u2019s how it works: Pick any normalized function \u03c8 whatsoever; I claim that Egs \u2264 \u03c8|H |\u03c8 \u2261 H. (8.1) That is, the expectation value of H, in the (presumably incorrect) state \u03c8 is certain to overestimate the ground state energy. Of course, if \u03c8 just happens to be one of the excited states, then obviously H exceeds Egs; the point is that the same holds for any \u03c8 whatsoever. Proof: Since the (unknown) eigenfunctions of H form a complete set, we can express \u03c8 as a linear combination of them:1 \u03c8 = cn\u03c8n, with H \u03c8n = En\u03c8n. Since \u03c8 is normalized, 1 = \u03c8|\u03c8 = n cm\u03c8m 3 cn\u03c8n = n m n c\u2217 mcn \u03c8m|\u03c8n = n |cn|2, 2 m (assuming the eigenfunctions themselves have been orthonormalized: \u03c8m|\u03c8n = \u03b4mn). Meanwhile, 2 H = cm\u03c8m cn\u03c8n = 3 H c\u2217 m Encn \u03c8m|\u03c8n = En|cn|2. m m But the ground state energy is, by de\ufb01nition, the smallest eigenvalue, Egs \u2264 En, and hence n n n H \u2265 Egs |cn|2 = Egs, which is what we were trying to prove. n 1 If the Hamiltonian admits scattering states, as well as bound states, then we\u2019ll need an integral as well as a sum, but the argument is unchanged. 328 CHAPTER 8 The Variational Principle This is hardly surprising. After all, \u03c8 might be the actual wave function (at, say, t = 0). If you measured the particle\u2019s energy you\u2019d be certain to get one of the eigenvalues of H, the smallest of which is Egs, so the average of multiple measurements (H ) cannot be lower than Egs. Example 8.1 Suppose we want to \ufb01nd the ground state energy for", " the one-dimensional harmonic oscillator: 2 d2 d x 2 Of course, we already know the exact answer in this case (Equation 2.62): Egs = (1/2) \u03c9; but this makes it a good test of the method. We might pick as our \u201ctrial\u201d wave function the gaussian, + 1 2 m\u03c92x 2. H = \u2212 2m \u03c8(x) = Ae\u2212bx 2, where b is a constant, and A is determined by normalization: 1 = |A|2 \u221e \u2212\u221e e\u22122bx 2 d x = |A|2 \u03c0 2b \u21d2 A = 1/4. 2b \u03c0 Now where, in this case 2m |A|2 \u221e \u2212\u221e e\u2212bx 2 d2 d x 2 e\u2212bx 2 d x = 2b 2m, and so V = 1 2 m\u03c92|A|2 \u221e \u2212\u221e e\u22122bx 2 x 2 d x = m\u03c92 8b, H = 2b 2m + m\u03c92 8b. (8.2) (8.3) (8.4) (8.5) (8.6) According to Equation 8.1, this exceeds Egs for any b; to get the tightest bound, let\u2019s minimize H : d db H = 2 2m \u2212 m\u03c92 8b2 = 0 \u21d2 b = m\u03c9 2. Putting this back into H, we \ufb01nd H min = 1 2 \u03c9. (8.7) In this case we hit the ground state energy right on the nose\u2014because (obviously) I \u201cjust happened\u201d to pick a trial function with precisely the form of the actual ground state (Equation 2.60). But the gaussian is very easy to work with, so it\u2019s a popular trial function, even when it bears little resemblance to the true ground state. Example 8.2 Suppose we\u2019re looking for the ground state energy of the delta function potential: 8.1 Theory 329 2 d2 d x 2 Again, we already know the exact answer (Equation 2.132): Egs = \u2212m\u03b12/22. As before, we\u2019ll use a gaussian trial function (Equation 8.2). We\u2019ve already determined the normalization, and calculated T ; all we need is \u2212 \u03b1\u03b4(x). H = \u2212 2m V = \u2212\u03b1|A|2 \u221e \u2212\u221e e\u22122bx 2 \u03b4(x) d x = \u2212\u03b1 2b \u03c0. Evidently H = 2b 2m \u2212 \u03b1 2b \u03c0, and we know that this exceeds Egs for all b. Minimizing it, d db H = 2 2m \u2212 \u221a \u03b1 2\u03c0 b = 0 \u21d2 b = 2m2\u03b12 \u03c0 4. So which is indeed somewhat higher than Egs, since \u03c0 > 2. H min = \u2212 m\u03b12 \u03c0 2, (8.8) (8.9) I said you can use any (normalized) trial function \u03c8 whatsoever, and this is true in a sense. However, for discontinuous functions it takes some fancy footwork to assign a sensible meaning to the second derivative (which you need, in order to calculate T ). Continuous functions with kinks in them are fair game, however, as long as you are careful; the next example shows how to handle them.2 Example 8.3 Find an upper bound on the ground state energy of the one-dimensional in\ufb01nite square well (Equation 2.22), using the \u201ctriangular\u201d trial wave function (Figure 8.1):3 \u23a7 \u23aa\u23a8 \u03c8(x) = \u23aa\u23a9 0 \u2264 x \u2264 a/2, Ax, A (a \u2212 x), a/2 \u2264 x \u2264 a, 0, otherwise, where A is determined by normalization: 1 = |A|2 a/2 0 x 2 d x + a a/2 (a \u2212 x)2 d x = |A|2 a3 12 \u21d2 A = 2 a. 3 a (8.10) (8.11) 2 For a collection of interesting examples see W. N. Mei, Int. J. Math. Educ. Sci. Tech. 30, 513 (1999). 3 There is no point in trying a function (such as the gaussian) that extends outside the well, because you\u2019ll get V = \u221e, and Equation 8.1 tells you nothing. 330 CHAPTER 8 The Variational Principle \u03c8(x) a/2 a x Figure 8.1: Triangular trial wave function for the in\ufb01nite", " square well (Equation 8.10). d\u03c8/dx A \u2212A a/2 a x Figure 8.2: Derivative of the wave function in Figure 8.1. In this case \u23a7 \u23aa\u23a8 \u23aa\u23a9 d\u03c8 d x = A, 0 \u2264 x \u2264 a/2, \u2212A, a/2 \u2264 x \u2264 a, 0, otherwise, as indicated in Figure 8.2. Now, the derivative of a step function is a delta function (see Problem 2.23(b)): d2\u03c8 d x 2 and hence = A\u03b4(x) \u2212 2A\u03b4(x \u2212 a/2) + A\u03b4(x \u2212 a), (8.12) H = \u2212 = \u2212 2 A 2m 2 A 2m \u03b4(x) \u2212 2\u03b4(x \u2212 a/2) + \u03b4(x \u2212 a) \u03c8(x) d x \u03c8(0) \u2212 2\u03c8(a/2) + \u03c8(a) = 2 A2a 2m = 122 2ma2. (8.13) The exact ground state energy is Egs = \u03c0 22/2ma2 (Equation 2.30), so the theorem works 12 > \u03c0 2. Alternatively, you can exploit the hermiticity of \u02c6p: 1 0 H = 1 2m 2 = \u02c6p\u03c8 = 1 2m \u02c6p2 a/2 = 1 2m \u02c6p\u03c8 a (A)2 d x + (\u2212A)2 d x = 2m 0 a/2 a 0 \u2212i d\u03c8 d x 2 2m \u2217 \u2212i d\u03c8 d x A2a = 122 2ma2 8.1 Theory 331 d x. (8.14) \u2217 \u2217\u2217 The variational principle is extraordinarily powerful, and embarrassingly easy to use. What a physical chemist does, to \ufb01nd the ground state energy of some complicated molecule, is write down a trial wave function with a large number of adjustable parameters, calculate H, and tweak the parameters to get the lowest possible value. Even if \u03c8 has little resemblance to the true wave function, you often get miraculously accurate values for Egs. Naturally, if you have some way of guessing a realistic \u03c8, so much the better. The only trouble with the method is that you never know for sure how close you are to the target\u2014all you can be certain of is that you\u2019ve got an upper bound.4 Moreover, as it stands the technique applies only to the ground state (see, however, Problem 8.4).5 Problem 8.1 Use a gaussian trial function (Equation 8.2) to obtain the lowest upper bound you can on the ground state energy of (a) the linear potential: V (x) = \u03b1|x|; (b) the quartic potential: V (x) = \u03b1x 4. Problem 8.2 Find the best bound on Egs for the one-dimensional harmonic oscillator using a trial wave function of the form where A is determined by normalization and b is an adjustable parameter. \u03c8(x) = A x 2 + b2, Problem 8.3 Find the best bound on Egs for the delta function potential V (x) = \u2212\u03b1\u03b4(x), using a triangular trial function (Equation 8.10, only centered at the origin). This time a is an adjustable parameter. Problem 8.4 (a) Prove the following corollary to the variational principle: If = 0, then H \u2265 Efe, where Efe is the energy of the \ufb01rst excited state. Comment: If we can \ufb01nd a trial function that is orthogonal to the exact ground state, we can get \u03c8|\u03c8gs 4 In practice this isn\u2019t much of a limitation, and there are sometimes ways of estimating the accuracy. The binding energy of helium has been calculated to many signi\ufb01cant digits in this way (see for example G. W. Drake et al., Phys. Rev. A 65, 054501 (2002), or Vladimir I. Korobov, Phys. Rev. A 66, 024501 (2002). 5 For a systematic extension of the variational principle to the calculation of excited state energies see, for example, Linus Pauling and E. Bright Wilson, Introduction to Quantum Mechanics, With Applications to Chemistry, McGrawHill, New York (1935, paperback edition 1985), Section 26. 332 CHAPTER 8 The Variational Principle an upper bound on the \ufb01rst excited state. In general, it\u2019s dif\ufb01cult to be sure that \u03c8 is orthogonal to \u03c8gs, since (presumably) we don", "\u2019t know the latter. However, if the potential V (x) is an even function of x, then the ground state is likewise even, and hence any odd trial function will automatically meet the condition for the corollary.6 (b) Find the best bound on the \ufb01rst excited state of the one-dimensional harmonic oscillator using the trial function \u03c8(x) = Axe\u2212bx 2. Problem 8.5 Using a trial function of your own devising, obtain an upper bound on the ground state energy for the \u201cbouncing ball\u201d potential (Equation 2.185), and compare it with the exact answer (Problem 2.59): Egs = 2.33811 1/3. 2 mg22 & Problem 8.6 (a) Use the variational principle to prove that \ufb01rst-order non-degenerate perturbation theory always overestimates (or at any rate never underestimates) the ground state energy. (b) In view of (a), you would expect that the second-order correction to the ground state is always negative. Con\ufb01rm that this is indeed the case, by examining Equation 7.15. 8.2 THE GROUND STATE OF HELIUM The helium atom (Figure 8.3) consists of two electrons in orbit around a nucleus containing two protons (also some neutrons, which are irrelevant to our purpose). The Hamiltonian for this system (ignoring \ufb01ne structure and smaller corrections) is: |r1 \u2212r2 | \u2212e r2 \u2212e r1 +2e Figure 8.3: The helium atom. 6 You can extend this trick to other symmetries. Suppose there is a Hermitian operator A such that [A, H ] = 0. The ground state (assuming it is nondegenerate) must be an eigenstate of A; call the eigenvalue \u03bb: A \u03c8gs = \u03bb \u03c8gs. If you choose a variational function \u03c8 that is an eigenstate of A with a different eigenvalue: A \u03c8 = \u03bd \u03c8 with \u03bb = \u03bd, you can be certain that \u03c8 and \u03c8gs are orthogonal (see Section 3.3). For an application see Problem 8.20. 8.2 The Ground State of Helium 333 H = \u2212 2 2m \u22072 1 + \u22072 2 \u2212 e2 4\u03c0 0 2 r1 + 2 r2 \u2212 1 |r1 \u2212 r2|. (8.15) Our problem is to calculate the ground state energy, Egs. Physically, this represents the amount of energy it would take to strip off both electrons. (Given Egs it is easy to \ufb01gure out the \u201cionization energy\u201d required to remove a single electron\u2014see Problem 8.7.) The ground state energy of helium has been measured to great precision in the laboratory: Egs = \u221278.975 eV (experimental). (8.16) This is the number we would like to reproduce theoretically. It is curious that such a simple and important problem has no known exact solution.7 The trouble comes from the electron\u2013electron repulsion, Vee = e2 4\u03c0 0 1 |r1 \u2212 r2|. (8.17) If we ignore this term altogether, H splits into two independent hydrogen Hamiltonians (only with a nuclear charge of 2e, instead of e); the exact solution is just the product of hydrogenic wave functions: \u03c80(r1, r2) \u2261 \u03c8100(r1)\u03c8100(r2) = 8 \u03c0a3 e\u22122(r1+r2)/a, (8.18) and the energy is 8E1 = \u2212109 eV (Equation 5.42).8 This is a long way from \u221279 eV, but it\u2019s a start. To get a better approximation for Egs we\u2019ll apply the variational principle, using \u03c80 as the trial wave function. This is a particularly convenient choice because it\u2019s an eigenfunction of most of the Hamiltonian: Thus where9 H \u03c80 = (8E1 + Vee) \u03c80. H = 8E1 + Vee, Vee = e2 4\u03c0 0 8 \u03c0a3 2 e\u22124(r1+r2)/a |r1 \u2212 r2| d3r1 d3r2. (8.19) (8.20) (8.21) I\u2019ll do the r2 integral \ufb01rst; for this purpose r1 is \ufb01xed, and we may as well orient the r2 coordinate system so that the polar axis lies along r1 (see Figure 8.4). By the law of cosines", ", |r1 \u2212 r2| = r 2 1 + r 2 2 \u2212 2r1r2 cos \u03b82, (8.22) 7 There do exist exactly soluble three-body problems with many of the qualitative features of helium, but using non-Coulombic potentials (see Problem 8.24). 8 Here a is the ordinary Bohr radius and En = \u221213.6/n2 eV is the nth Bohr energy; recall that for a nucleus with atomic number Z, En \u2192 Z 2 En and a \u2192 a/Z (Problem 4.19). The spin con\ufb01guration associated with Equation 8.18 will be antisymmetric (the singlet). 9 You can, if you like, interpret Equation 8.21 as \ufb01rst-order perturbation theory, with H = Vee (Problem 7.56(a)). However, I regard this as a misuse of the method, since the perturbation is comparable in size to the unperturbed potential. I prefer, therefore, to think of it as a variational calculation, in which we are looking for a rigorous upper bound on Egs. 334 CHAPTER 8 The Variational Principle z2 r1 |r1 \u2212r2 | \u03b8 2 r2 \u03c6 2 y2 x2 Figure 8.4: Choice of coordinates for the r2-integral (Equation 8.21). and hence I2 \u2261 e\u22124r2/a |r1 \u2212 r2| d3r2 = e\u22124r2/a r 2 1 + r 2 2 \u2212 2r1r2 cos \u03b82 r 2 2 sin \u03b82 dr2 d\u03b82 d\u03c62. (8.23) The \u03c62 integral is trivial (2\u03c0 ); the \u03b82 integral is \u03c0 0 sin \u03b82 \u2212 2r1r2 cos \u03b82 + r 2 2 r 2 1 d\u03b82 = r 2 1 + r 2 2 \u2212 2r1r2 cos \u03b82 r1r2 \u03c0 0 = 1 r1r2 = 1 r1r2 r 2 1 + r 2 2 + 2r1r2 \u2212 r 2 1 [(r1 + r2) \u2212 |r1 \u2212 r2|] = + r 2 2 \u2212 2r1r2 2/r1, 2/r2, r2 < r1, r2 > r1. Thus I2 = 4\u03c0 = \u03c0a3 8r1 1 r1 1 \u2212 0 r1 e\u22124r2/ar 2 2 dr2 + 1 + 2r1 a e\u22124r2/ar2 dr2 \u221e r1 It follows that Vee is equal to e2 4\u03c0 0 8 \u03c0a3 1 + 2r1 a 1 \u2212 e\u22124r1/a. e\u22124r1/a e\u22124r1/ar1 sin \u03b81 dr1d\u03b81d\u03c61. The angular integrals are easy (4\u03c0 ), and the r1 integral becomes \u221e r e\u22124r/a \u2212 0 r + 2r 2 a e\u22128r/a dr = 5a2 128. (8.24) (8.25) 8.2 The Ground State of Helium 335 Finally, then, and therefore Vee = 5 4a e2 4\u03c0 0 = \u2212 5 2 E1 = 34 eV, H = \u2212109 eV + 34 eV = \u221275 eV. (8.26) (8.27) Not bad (remember, the experimental value is \u221279 eV). But we can do better. We need to think up a more realistic trial function than \u03c80 (which treats the two electrons as though they did not interact at all). Rather than completely ignoring the in\ufb02uence of the other electron, let us say that, on the average, each electron represents a cloud of negative charge which partially shields the nucleus, so that the other electron actually sees an effective nuclear charge (Z ) that is somewhat less than 2. This suggests that we use a trial function of the form \u03c81(r1, r2) \u2261 Z 3 \u03c0a3 We\u2019ll treat Z as a variational parameter, picking the value that minimizes H. (Please note that in the variational method we never touch the Hamiltonian itself \u2014the Hamiltonian for helium is, and remains, Equation 8.15. But it\u2019s \ufb01ne to think about approximating the Hamiltonian as a way of motivating the choice of the trial wave function.) e\u2212Z (r1+r2)/a. (8.28) This wave function is an eigenstate of the \u201cunperturbed\u201d Hamiltonian (neglecting electron repulsion), only with", " Z, instead of 2, in the Coulomb terms. With this in mind, we rewrite H (Equation 8.15) as follows: H = \u2212 \u22072 1 2 2m + e2 4\u03c0 0 + \u22072 2 (Z \u2212 2) r1 \u2212 e2 Z 4\u03c0 0 r1 (Z \u2212 2) r2 + + Z r2. + 1 |r1 \u2212 r2| The expectation value of H is evidently H = 2Z 2 E1 + 2 (Z \u2212 2) 1 r e2 4\u03c0 0 + Vee. (8.29) (8.30) Here 1/r is the expectation value of 1/r in the (one-particle) hydrogenic ground state \u03c8100 (with nuclear charge Z ); according to Equation 7.56, 1 r = Z a. (8.31) The expectation value of Vee is the same as before (Equation 8.26), except that instead of Z = 2 we now want arbitrary Z \u2014so we multiply a by 2/Z : Vee = 5Z 8a e2 4\u03c0 0 = \u2212 5Z 4 E1. (8.32) Putting all this together, we \ufb01nd # 2Z 2 \u2212 4Z (Z \u2212 2) \u2212 (5/4) Z H = $ # $ E1 = \u22122Z 2 + (27/4) Z E1. (8.33) 336 CHAPTER 8 The Variational Principle According to the variational principle, this quantity exceeds Egs for any value of Z. The lowest upper bound occurs when H is minimized: H = \u22124Z + (27/4) E1 = 0, d d Z from which it follows that Z = 27 16 = 1.69. (8.34) This seems reasonable; it tells us that the other electron partially screens the nucleus, reducing its effective charge from 2 down to about 1.69. Putting in this value for Z, we \ufb01nd 6 H = 1 2 3 2 E1 = \u221277.5 eV. (8.35) The ground state of helium has been calculated with great precision in this way, using increasingly complicated trial wave functions, with more and more adjustable parameters.10 But we\u2019re within 2% of the correct answer, and, frankly, at this point my own interest in the problem begins to wane.11 Problem 8.7 Using Egs = \u221279.0 eV for the ground state energy of helium, calculate the ionization energy (the energy required to remove just one electron). Hint: First calculate the ground state energy of the helium ion, He+, with a single electron orbiting the nucleus; then subtract the two energies. \u2217 Problem 8.8 Apply the techniques of this Section to the H\u2212 and Li+ ions (each has two electrons, like helium, but nuclear charges Z = 1 and Z = 3, respectively). Find the effective (partially shielded) nuclear charge, and determine the best upper bound on Egs, for each case. Comment: In the case of H\u2212 you should \ufb01nd that H > \u221213.6 eV, which would appear to indicate that there is no bound state at all, since it would be energetically favorable for one electron to \ufb02y off, leaving behind a neutral hydrogen atom. This is not entirely surprising, since the electrons are less strongly attracted to the nucleus than they are in helium, and the electron repulsion tends to break the atom apart. However, it turns out to be incorrect. With a more sophisticated trial wave function (see Problem 8.25) it can be shown that Egs < \u221213.6 eV, and hence that a bound state does exist. It\u2019s only barely bound, however, and there are no excited bound states,12 so H\u2212 has no discrete spectrum (all transitions are to and from the continuum). As a result, it is dif\ufb01cult to study in the laboratory, although it exists in great abundance on the surface of the sun.13 10 The classic studies are E. A. Hylleraas, Z. Phys. 65, 209 (1930); C. L. Pekeris, Phys. Rev. 115, 1216 (1959). For more recent work, see footnote 4. 11 The \ufb01rst excited state of helium can be calculated in much the same way, using a trial wave function orthogonal to the ground state. See Phillip J. E. Peebles, Quantum Mechanics, Princeton U.P., Princeton, NJ (1992), Section 40. 12 Robert N. Hill, J. Math. Phys. 18, 2316 (1977). 13 For further discussion see Hans A. Bethe and Edwin E. Salpeter, Quantum Mechanics of One- and Two-Electron Atoms, Plenum, New York (1977), Section 34.", " 8.3 THE HYDROGEN MOLECULE ION 8.3 The Hydrogen Molecule Ion 337 + 2, Another classic application of the variational principle is to the hydrogen molecule ion, H consisting of a single electron in the Coulomb \ufb01eld of two protons (Figure 8.5). I shall assume for the moment that the protons are \ufb01xed in position, a speci\ufb01ed distance R apart, although one of the most interesting byproducts of the calculation is going to be the actual value of R. The Hamiltonian is H = \u2212 2 2m \u22072 \u2212 e2 4\u03c0 0 1 r + 1 r, (8.36) where r and r are the distances to the electron from the respective protons. As always, our strategy will be to guess a reasonable trial wave function, and invoke the variational principle to get a bound on the ground state energy. (Actually, our main interest is in \ufb01nding out whether this system bonds at all\u2014that is, whether its energy is less than that of a neutral hydrogen atom plus a free proton. If our trial wave function indicates that there is a bound state, a better trial function can only make the bonding even stronger.) To construct the trial wave function, imagine that the ion is formed by taking a hydrogen atom in its ground state (Equation 4.80), \u03c80(r) = 1\u221a \u03c0a3 e\u2212r/a, (8.37) bringing the second proton in from \u201cin\ufb01nity,\u201d and nailing it down a distance R away. If R is substantially greater than the Bohr radius, the electron\u2019s wave function probably isn\u2019t changed very much. But we would like to treat the two protons on an equal footing, so that the electron has the same probability of being associated with either one. This suggests that we consider a trial function of the form \u03c8 = A \u03c80(r ) + \u03c80. (8.38) r (Quantum chemists call this the LCAO technique, because we are expressing the molecular wave function as a linear combination of atomic orbitals.) Our \ufb01rst task is to normalize the trial function: 1 = |\u03c8|2 d3r = |A|2 + \u03c80 r 2 d3r + 2 \u03c80(r )2 d3r d3r. r \u03c80(r )\u03c80 (8.39) \u2212e r\u2032 r +e R +e Figure 8.5: The hydrogen molecule ion, H + 2. 338 CHAPTER 8 The Variational Principle The \ufb01rst two integrals are 1 (since \u03c80 itself is normalized); the third is more tricky. Let \u03c80(r ) \u03c80 r I \u2261 = 1 \u03c0a3 e\u2212(r +r )/a d3r. (8.40) (8.41) (8.42) Picking coordinates so that proton 1 is at the origin and proton 2 is on the z axis at the point R (Figure 8.6), we have % r = r 2 + R2 \u2212 2r R cos \u03b8, and therefore I = 1 \u03c0a3 \u221a e\u2212r/ae\u2212 r 2+R2\u22122r R cos \u03b8/ar 2 sin \u03b8 dr d\u03b8 d\u03c6. The \u03c6 integral is trivial (2\u03c0 ). To do the \u03b8 integral, let y \u2261 d = 2y dy = 2r R sin \u03b8 d\u03b8. Then y2 \u221a r 2 + R2 \u2212 2r R cos \u03b8, so that \u03c0 0 e\u2212 \u221a r 2+R2\u22122r R cos \u03b8/a sin \u03b8 d\u03b8 = 1 r R r +R |r \u2212R| e\u2212y/a y dy # e\u2212(r +R)/a (r + R + a) \u2212 e\u2212|r \u2212R|/a (|r \u2212 R| + a) $. = \u2212 a r R The r integral is now straightforward: \u2212e\u2212R/a \u221e I = 2 a2 R \u221e 0 (r + R + a)e\u22122r/ar dr + e\u2212R/a R 0 (R \u2212 r + a) r dr + e R/a (r \u2212 R + a) e\u22122r/ar dr. R Evaluating the integrals, we \ufb01nd (after some algebraic simpli\ufb01cation), I = e\u2212R/8.43) z 2 R 1 r\u2032 = r 2 + R2 \u22122r R cos \u03b8 r \u03b8 y x Figure 8.6: Coordinates for the calculation of I (Equation 8", ".40). 8.3 The Hydrogen Molecule Ion 339 I is called an overlap integral; it measures the amount by which \u03c80(r ) overlaps \u03c80 (notice that it goes to 1 as R \u2192 0, and to 0 as R \u2192 \u221e). In terms of I, the normalization factor (Equation 8.39) is r |A|2 = 1 2 (1 + I ). (8.44) Next we must calculate the expectation value of H in the trial state \u03c8. Noting that 2 1 r (where E1 = \u221213.6 eV is the ground state energy of atomic hydrogen)\u2014and the same with r in place of r \u2014we have \u03c80(r ) = E1\u03c80(r ) \u22072 \u2212 e2 4\u03c0 0 2m \u2212 H \u03c8 = A \u2212 2 2m = E1\u03c8 \u2212 A \u22072 \u2212 e2 4\u03c0 0 e2 4\u03c0 0 1 r 1 + 1 r r \u03c80 (r ) + 1 r It follows that H = E1 \u2212 2|A|2 e2 4\u03c0 0 \u03c80(r ) \u03c80(r ) 1 r \u03c80(r ) + \u03c80 r r \u03c80 + \u03c80(r ). \u03c80 r. 1 r I\u2019ll let you calculate the two remaining quantities, the so-called direct integral, D \u2261 a \u03c80(r ) \u03c80(r ), 1 r and the exchange integral, X \u2261 a \u03c80(r ) \u03c80 r. 1 r The results (see Problem 8.9) are \u22122R/a, and 1 + R a X = e\u2212R/a. Putting all this together, and recalling (Equations 4.70 and 4.72) that E1 = \u2212 (1/2a), we conclude: H = 1 + 2 E1. (D + X ) (1 + I ) (8.45) (8.46) (8.47) (8.48) (8.49) e2/4\u03c0 0 (8.50) According to the variational principle, the ground state energy is less than H. Of course, this is only the electron\u2019s energy\u2014there is also potential energy associated with the proton\u2013proton repulsion: Vpp = e2 4\u03c0 0 1 R = \u2212 2a R E1. (8.51) 340 CHAPTER 8 The Variational Principle F(x) 0 \u22120.5 \u22121 \u22121.2 Equilibrium 1 2 3 4 5 x 6 Figure 8.7: Plot of the function F (x), Equation 8.52, showing existence of a bound state. Thus the total energy of the system, in units of \u2212E1, and expressed as a function of x \u2261 R/a, is less than 4 F (x) = \u22121 + 2 x 1 \u2212 (2/3)x 2 1 + x + (1/3)x 2 1 + e\u2212x + (1 + x) e\u22122x e\u2212x. (8.52) \u2217 \u2217\u2217 This function is plotted in Figure 8.7. Evidently bonding does occur, for there exists a region in which the graph goes below \u22121, indicating that the energy is less than that of a neutral atom plus a free proton (\u221213.6 eV). It\u2019s a covalent bond, with the electron shared equally by the two protons. The equilibrium separation of the protons is about 2.4 Bohr radii, or 1.3 \u00c5 (the experimental value is 1.06 \u00c5). The calculated binding energy is 1.8 eV, whereas the experimental value is 2.8 eV (the variational principle, as always, over estimates the ground state energy\u2014and hence under estimates the strength of the bond\u2014but never mind: The essential point was to see whether binding occurs at all; a better variational function can only make the potential well even deeper. Problem 8.9 Evaluate D and X (Equations 8.46 and 8.47). Check your answers against Equations 8.48 and 8.49. Problem 8.10 Suppose we used a minus sign in our trial wave function (Equation 8.38): r \u03c8 = A \u03c80(r ) \u2212 \u03c80 Without doing any new integrals, \ufb01nd F(x) (the analog to Equation 8.52) for this case, and construct the graph. Show that there is no evidence of bonding.14 (Since the variational principle only gives an upper bound, this doesn\u2019t prove that bonding cannot occur for such a state, but it certainly doesn\u2019t look promising.) (8.53). 14 The wave", " function with the plus sign (Equation 8.38) is called the bonding orbital. Bonding is associated with a buildup of electron probability in between the two nuclei. The odd linear combination (Equation 8.53) has a node at the center, so it\u2019s not surprising that this con\ufb01guration doesn\u2019t lead to bonding; it is called the anti-bonding orbital. 8.4 The Hydrogen Molecule 341 \u2217\u2217\u2217 Problem 8.11 The second derivative of F(x), at the equilibrium point, can be used to estimate the natural frequency of vibration (\u03c9) of the two protons in the hydrogen molecule ion (see Section 2.3). If the ground state energy (\u03c9/2) of this oscillator exceeds the binding energy of the system, it will \ufb02y apart. Show that in fact the oscillator energy is small enough that this will not happen, and estimate how many bound vibrational levels there are. Note: You\u2019re not going to be able to obtain the position of the minimum\u2014still less the second derivative at that point\u2014analytically. Do it numerically, on a computer. 8.4 THE HYDROGEN MOLECULE Now consider the hydrogen molecule itself, adding a second electron to the hydrogen molecule ion we studied in Section 8.3. Taking the two protons to be at rest, the Hamiltonian is 2 \u22072 1 + e2 4 \u03c0 0 + \u22072 2 \u02c6H = \u2212 1 2 m r12 1 are the distances of electron 1 from each proton and r2 and r where r1 and r 2 are the distances of electron 2 from each proton; as shown in Figure 8.8. The six potential energy terms describe the repulsion between the two electrons, the repulsion between the two protons, and the attraction of each electron to each proton r1 \u2212 1 r2 + 1 R (8.54) For the variational wave function, associate one electron with each proton, and symmetrize: \u03c8+(r1, r2) = A+ \u03c80(r1) \u03c80 r 2 + \u03c80 r 1 \u03c80(r2). (8.55) We\u2019ll calculate the normalization A+ in a moment. Since this spatial wave function is symmetric under interchange, the electrons must occupy the antisymmetric (singlet) spin state. Of course, we could also choose the trial wave function r 2 \u03c80(r1) \u03c80 \u03c8\u2212(r1, r2) = A\u2212 \u03c80(r2) \u2212 \u03c80 r 1 (8.56) in which case the electrons would be in a symmetric (triplet) spin state. These two variational wave functions constitute the Heitler\u2013London approximation.15 It is not obvious which of \u2212e r1 r12 r2 \u2212 Figure 8.8: Diagram of H2 showing the distances on which the potential energy depends. 15 W. Heitler and F. London, Z. Phys. 44, 455 (1928). For an English translation see Hinne Hettema, Quantum Chemistry: Classic Scienti\ufb01c Papers, World Scienti\ufb01c, New Jersey, PA, 2000. 342 CHAPTER 8 The Variational Principle Equations 8.55 or 8.56 would be energetically favored, so let\u2019s calculate the energy of each one, and \ufb01nd out.16 First we need to normalize the wave functions. Note that |\u03c8\u00b1(r1, r2)|2 = A2 \u00b1 # \u03c80(r1)2 \u03c80 r 2 r 2 \u00b1 2 \u03c80(r1) \u03c80 2 + \u03c80 r 1 \u03c80 r 1 2 \u03c80(r2)2 $ \u03c80(r2). (8.58) Normalization requires 1 = = A2 \u00b1 |\u03c8\u00b1(r1, r2)|2 d3r1 d3r2 r 2 \u03c80(r1)2 d3r1 \u03c80 \u00b1 2 \u03c80(r1) \u03c80 r 1 d3r1 2 d3r2 + r 2 \u03c80 r 1 \u03c80 2 d3r1 \u03c80(r2)2 d3r2 \u03c80(r2) d3r2. (8.59) The individual orbitals are normalized and the overlap integral was given the symbol I and calculated in Equation 8.43. Thus A8.60) To calculate the expectation value of the energy, we will start with the kinetic energy of particle 1. Since \u03c80 is the ground state of the hydrogen Hamiltonian, the same trick that brought us to Equation", " 8.45 gives \u2212 2 2 m \u22072 1 \u03c8\u00b1 = A\u00b1 \u2212 2 2 m \u22072 1 \u03c80(r1) \u00b1 \u2212 r 2 \u03c80 \u22072 1 \u03c80 r 1 2 2 m \u03c80(r2) = A\u00b1 \u00b1 E1 + E1 + e2 4 \u03c0 0 r1 e2 4 \u03c0 0 r 1 \u03c80(r1) \u03c80 r 1 \u03c80 \u03c80(r2) r 2 = E1 \u03c8\u00b1 + e2 4 \u03c0 0 a A\u00b1 a r1 \u03c80(r1) \u03c80 r 2 \u00b1 a r 1 r 1 \u03c80 \u03c80(r2). Taking the inner product with \u03c8\u00b1 then gives \u2212 2 2 m \u22072 1 = E1 + \u03c80 \u00b1 A2 \u00b1 e2 4 \u03c0 0 a a r \u03c80(r1) 1 r1 \u03c80(r1) \u03c80(r2) \u03c80(r1) r 2 a r1 \u03c80 \u03c80 r 2 \u03c80 r 2 16 Another natural variational wave function consists of placing both electrons in the bonding orbital studied in Section 8.3: \u03c8(r1, r2) = \"!! r \u03c80(r1) + \u03c80 1 2 (1 + I ) \u221a r \u03c80(r2) + \u03c80 2 2 (1 + I ) \u221a \", (8.57) also paired with a singlet spin state. If you expand this function you\u2019ll see that half the terms\u2014such as \u03c80(r1) \u03c80(r2)\u2014involve attaching two electrons to the same proton, which is energetically costly because of the electron\u2013electron repulsion in Equation 8.54. The Heitler\u2013London approximation, Equation 8.55, amounts to dropping the offending terms from Equation 8.57. \u00b1 \u03c80(r1) + \u03c80 r 1 \u03c80 \u03c80 a r 1 a r 1 8.4 The Hydrogen Molecule 343 \u03c80 r 2 \u03c80(r2) r 1 \u03c80(r2) | \u03c80(r2). r 1 (8.61) These inner products were calculated in Section 8.3 and the kinetic energy of particle 1 is \u2212 2 2 m \u22072 1 = E1 + e2 8.62) The kinetic energy of particle 2 is of course the same, so the total kinetic energy is simply twice Equation 8.62. The calculation of the electron\u2013proton potential energy is similar; you will show in Problem 8.13 that \u2212 e2 4 \u03c0 0 r1 = \u2212 1 2 e2 8.63) and the total electron\u2013proton potential energy is four times this amount. The electron\u2013electron potential energy is given by Vee = = e2 4 \u03c0 0 a e2 4 \u03c0 0 a A2 \u00b1 |\u03c8\u00b1(r1, r2)|2 a r12 \u03c80(r1)2 a r12 d3r1 d3r2 r 2 \u03c80 2 d3r1 d3r2 + \u03c80 r 1 2 a r12 \u00b1 2 \u03c80(r1) \u03c80 \u03c80(r2)2 d3r1 d3r2 r 1 \u03c80(r2) \u03c80 a r12 d3r1 d3r2. (8.64) r 2 The \ufb01rst two integrals in Equation 8.64 are equal, as you can see by interchanging the labels 1 and 2. We will give the two remaining integrals the names D2 = X2 = |\u03c80(r1)|2 a r12 r 1 \u03c80(r1) \u03c80 \u03c80 a r12 r 2 2 d3r1 d3r2 r 2 \u03c80(r2) \u03c80 so that Vee = e2 4 \u03c0 0 a D2 \u00b1 X2 1 \u00b1 I 2. d3r1 d3r2 (8.65) (8.66) (8.67) The evaluation of these integrals is discussed in Problem 8.14. Note that the integral D2 is just the electrostatic potential energy of two charge distributions \u03c11 = |\u03c80(r1)|2 and \u03c12 = |\u03c80(r2)|2. The exchange term X2 has no such classical counterpart. When we add all of the contributions to the energy\u2014the kinetic energy, the electron\u2013proton potential energy, the electron\u2013electron potential energy, and the proton\u2013proton potential energy (", "which is a constant, e2/4 \u03c0 0 R)\u2014we get 344 CHAPTER 8 The Variational Principle H \u00b1 = 2 E1 1 \u2212 a R + 2 D \u2212 D2 \u00b1(2 I X \u2212 X2) 1 \u00b1 I 2. (8.68) A plot of H + and H \u2212 is shown in Figure 8.9. Recall that the state \u03c8+ requires placing the two electrons in the singlet spin con\ufb01guration, whereas \u03c8\u2212 means putting them in a triplet spin con\ufb01guration. According to the \ufb01gure, bonding only occurs if the two electrons are in a singlet con\ufb01guration\u2014something that is con\ufb01rmed experimentally. Again, it\u2019s a covalent bond. Locating the minimum on the plot, our calculation predicts a bond length of 1.64 Bohr radii (the experimental value is 1.40 Bohr radii), and suggests a binding energy of 3.15 eV (whereas the experimental value is 4.75 eV). The trends here follow those of the Hydrogen molecule ion: the calculation overestimates the bond length and underestimates the binding energy, but the agreement is surprisingly good for a variational calculation with no adjustable parameters. The difference between the singlet and triplet energies is called the exchange splitting J. In the Heitler\u2013London approximation it is J = H + \u2212 H \u2212 = 4 E1 (D2 \u2212 2 D) I 2 \u2212(X2 \u2212 8.69) which is roughly \u221210 eV (negative because the singlet is lower in energy) at the equilibrium separation. This means a strong preference for having the electron spins anti-aligned. But in this treatment of H2 we\u2019ve left out completely the (magnetic) spin\u2013spin interaction between the electrons\u2014remember that the spin\u2013spin interaction between the proton and the electron is what leads to hyper\ufb01ne splitting (Section 7.5). Were we right to ignore it here? Absolutely: applying Equation 7.92 to two electrons a distance R apart, the energy of the spin\u2013spin interaction is something like 10\u22124 eV in this system, \ufb01ve orders of magnitude smaller than the exchange splitting. \u00e1 \u00e1H E1 \u20130.5 \u20131.0 \u20131.5 \u20132.0 \u20132.5 1 2 33 4 5 R/a 6 Figure 8.9: The total energy of the singlet (solid curve) and triplet (dashed curve) states for H2, as a function of the separation R between the protons. The singlet state has a minimum at around 1.6 Bohr radii, representing a stable bond. The triplet state is unstable and will dissociate, as the energy is minimized for R \u2192 \u221e. 8.4 The Hydrogen Molecule 345 This calculation shows us that different spin con\ufb01gurations can have very different energies, even when the interaction between the spins is negligible. And that helps us understand ferromagnetism (where the spins in a material align) and anti-ferromagnetism (where the spins alternate). As we\u2019ve just seen, the spin\u2013spin interaction is way too weak to account for this\u2014 but the exchange splitting isn\u2019t. Counterintuitively, it\u2019s not a magnetic interaction that accounts for ferromagnetism, but an electrostatic one! H2 is a sort of inchoate anti-ferromagnet where the Hamiltonian, which is independent of the spin, selects a certain spatial ground state and the spin state comes along for the ride, to satisfy the Fermi statistics. Problem 8.12 Show that the antisymmetric state (Equation 8.56) can be expressed in terms of the molecular orbitals of Section 8.3\u2014speci\ufb01cally, by placing one electron in the bonding orbital (Equation 8.38) and one in the anti-bonding orbital (Equation 8.53). Problem 8.13 Verify Equation 8.63 for the electron\u2013proton potential energy. \u2217\u2217\u2217 Problem 8.14 The two-body integrals D2 and X2 are de\ufb01ned in Equations 8.65 and 8.66. To evaluate D2 we write \u03c80 r 2 2 (r2) d3r2 D2 = = \u22122 e R2+r 2 2 \u22122 R r2 cos \u03b82/a \u03c0 a3 (r2) r 2 2 dr2 sin \u03b82 d\u03b82 d\u03c62 where \u03b82 is the angle between R and r2 (Figure 8.8), and (r2) \u2261 |\u03c80(r1)|2 a |r1 \u2212 r2| d", "3r1. (a) Consider \ufb01rst the integral over r1. Align the z axis with r2 (which is a constant vector for the purposes of this \ufb01rst integral) so that (r2) = 1 \u03c0 a3 ae\u22122 r1/ r1 r2 cos \u03b81 r 2 1 dr1 sin \u03b81 d\u03b81 d\u03c61. Do the angular integration \ufb01rst and show that 1 + a r2 (r2) = a r2 (b) Plug your result from part (a) back into the relation for D2, and show that e\u22122 r2/a. \u2212 2 D2 = a R \u2212 e\u22122 R/ + 11 8 + a R. (8.70) Again, do the angular integration \ufb01rst. Comment: The integral X2 can also be evaluated in closed form, but the procedure is rather involved.17 We will simply quote the result, 17 The calculation was done by Y. Sugiura, Z. Phys. 44, 455 (1927). 346 CHAPTER 8 The Variational Principle X2 = e\u22122 R/a 5 \u2212 23 20 \u03b3 + log 15 \" 2 \u02dcI I \u2212 4 R a Ei \u2212 2 R a Ei \u02dcI I \u2212 2 \u23a4 \u23a6, (8.71) where \u03b3 = 0.5772... is Euler\u2019s constant, Ei(x) is the exponential integral e\u2212t t and \u02dcI is obtained from I by switching the sign of R: Ei(x) = \u2212 \u2212x dt, \u221e \u02dcI = e R/8.72) \u2217 Problem 8.15 Make a plot of the kinetic energy for both the singlet and triplet states of H2, as a function of R/a. Do the same for the electron-proton potential energy and for the electron\u2013electron potential energy. You should \ufb01nd that the triplet state has lower potential energy than the singlet state for all values of R. However, the singlet state\u2019s kinetic energy is so much smaller that its total energy comes out lower. Comment: In situations where there is not a large kinetic energy cost to aligning the spins, such as two electrons in a partially \ufb01lled orbital in an atom, the triplet state can come out lower in energy. This is the physics behind Hund\u2019s \ufb01rst rule. FURTHER PROBLEMS ON CHAPTER 8 Problem 8.16 (a) Use the function \u03c8(x) = Ax (a \u2212 x) (for 0 < x < a, otherwise 0) to get an upper bound on the ground state of the in\ufb01nite square well. (b) Generalize to a function of the form \u03c8(x) = A [x (a \u2212 x)] p, for some real number p. What is the optimal value of p, and what is the best bound on the ground state energy? Compare the exact value. Answer: 5 + 2 2/2ma2. 6 \u221a Problem 8.17 (a) Use a trial wave function of the form \u03c8 (x) = A cos (\u03c0 x/a), \u2212a/2 < x < a/2, 0, otherwise, to obtain a bound on the ground state energy of the one-dimensional harmonic oscillator. What is the \u201cbest\u201d value of a? Compare H min with the exact energy. Note: This trial function has a \u201ckink\u201d in it (a discontinuous derivative) at \u00b1a/2; do you need to take account of this, as I did in Example 8.3? (b) Use \u03c8(x) = B sin (\u03c0 x/a) on the interval (\u2212a, a) to obtain a bound on the \ufb01rst excited state. Compare the exact answer. Further Problems on Chapter 8 347 \u2217\u2217 Problem 8.18 (a) Generalize Problem 8.2, using the trial wave function18 \u03c8(x) = A x 2 + b2, n for arbitrary n. Partial answer: The best value of b is given by b2 = m\u03c9 n (4n \u2212 1) (4n \u2212 3) 2 (2n + 1) 1/2. (b) Find the least upper bound on the \ufb01rst excited state of the harmonic oscillator using a trial function of the form \u03c8(x) = Bx x 2 + b2. n Partial answer: The best value of b is given by b2 = m\u03c9 n (4n \u2212 5) (4n \u2212 3) 2 (2n + 1) 1/2. (c) Notice that the bounds", " approach the exact energies as n \u2192 \u221e. Why is that? Hint: Plot the trial wave functions for n = 2, n = 3, and n = 4, and compare them with the true wave functions (Equations 2.60 and 2.63). To do it analytically, start with the identity ez = lim n\u2192\u221e 1 + z n n. Problem 8.19 Find the lowest bound on the ground state of hydrogen you can get using a gaussian trial wave function \u03c8(r) = Ae\u2212br 2, where A is determined by normalization and b is an adjustable parameter. Answer: \u221211.5 eV. Problem 8.20 Find an upper bound on the energy of the \ufb01rst excited state of the hydrogen atom. A trial function with = 1 will automatically be orthogonal to the ground state (see footnote 6); for the radial part of \u03c8 you can use the same function as in Problem 8.19. \u2217\u2217 Problem 8.21 If the photon had a nonzero mass be replaced by the Yukawa potential, m\u03b3 = 0, the Coulomb potential would V(r) = \u2212 e2 4\u03c0 0 e\u2212\u03bcr r, (8.73) where \u03bc = m\u03b3 c/. With a trial wave function of your own devising, estimate the binding energy of a \u201chydrogen\u201d atom with this potential. Assume \u03bca 1, and give your answer correct to order (\u03bca)2. 18 W. N. Mei, Int. J. Educ. Sci. Tech. 27, 285 (1996). 348 CHAPTER 8 The Variational Principle Problem 8.22 Suppose you\u2019re given a two-level quantum system whose (timeindependent) Hamiltonian H 0 admits just two eigenstates, \u03c8a (with energy Ea), and \u03c8b (with energy Eb). They are orthogonal, normalized, and nondegenerate (assume Ea is the smaller of the two energies). Now we turn on a perturbation H, with the following matrix elements: H \u03c8a \u03c8b \u03c8b \u03c8a H H H \u03c8b \u03c8b \u03c8a (8.74) = h, = 0; \u03c8a = = where h is some speci\ufb01ed constant. (a) Find the exact eigenvalues of the perturbed Hamiltonian. (b) Estimate the energies of the perturbed system using second-order perturbation theory. (c) Estimate the ground state energy of the perturbed system using the variational principle, with a trial function of the form \u03c8 = (cos \u03c6) \u03c8a + (sin \u03c6) \u03c8b, where \u03c6 is an adjustable parameter. Note: Writing the linear combination in this way is just a neat way to guarantee that \u03c8 is normalized. (8.75) (d) Compare your answers to (a), (b), and (c). Why is the variational principle so accurate, in this case? Problem 8.23 As an explicit example of the method developed in Problem 8.22, consider \u02c6k, for which the Hamiltonian is an electron at rest in a uniform magnetic \ufb01eld B = Bz (Equation 4.158): H 0 = eBz m Sz. (8.76) The eigenspinors, \u03c7a and \u03c7b, and the corresponding energies, Ea and Eb, are given in Equation 4.161. Now we turn on a perturbation, in the form of a uniform \ufb01eld in the x direction: H = eBx m (a) Find the matrix elements of H, and con\ufb01rm that they have the structure of (8.77) Sx. Equation 8.74. What is h? (b) Using your result in Problem 8.22(b), \ufb01nd the new ground state energy, in second- order perturbation theory. (c) Using your result in Problem 8.22(c), \ufb01nd the variational principle bound on the ground state energy. \u2217\u2217\u2217 Problem 8.24 Although the Schr\u00f6dinger equation for helium itself cannot be solved exactly, there exist \u201chelium-like\u201d systems that do admit exact solutions. A simple example19 is \u201crubber-band helium,\u201d in which the Coulomb forces are replaced by Hooke\u2019s law forces: 19 For a more sophisticated model, see R. Crandall, R. Whitnell, and R. Bettega, Am. J. Phys 52, 438 (1984). Further Problems on Chapter 8 349 H = \u2212 2 2m \u22072 1 + \u22072 2 + 1 2 m\u03c92 m\u03c92|r1 \u2212 r2", "|2. (8.78) (a) Show that the change of variables from r1, r2, to u \u2261 1\u221a v \u2261 1\u221a 2 2 (r1 + r2), (r1 \u2212 r2), (8.79) turns the Hamiltonian into two independent three-dimensional harmonic oscillators: H = \u2212 2 2m \u22072 u + 1 2 m\u03c92u2 + \u2212 2 2m v + 1 \u22072 2 (1 \u2212 \u03bb) m\u03c92v2. (8.80) (b) What is the exact ground state energy for this system? (c) If we didn\u2019t know the exact solution, we might be inclined to apply the method of Section 8.2 to the Hamiltonian in its original form (Equation 8.78). Do so (but don\u2019t bother with shielding). How does your result compare with the exact answer? Answer: H = 3\u03c9 (1 \u2212 \u03bb/4). \u2217\u2217\u2217 Problem 8.25 In Problem 8.8 we found that the trial wave function with shielding (Equation 8.28), which worked well for helium, is inadequate to con\ufb01rm the existence of a bound state for the negative hydrogen ion. Chandrasekhar20 used a trial wave function of the form \u03c8(r1, r2) \u2261 A [\u03c81(r1)\u03c82(r2) + \u03c82(r1)\u03c81(r2)], (8.81) where / Z 3 1 \u03c0a3 \u03c81(r ) \u2261 e\u2212Z1r/a, and \u03c82(r ) \u2261 / Z 3 2 \u03c0a3 e\u2212Z2r/a. (8.82) In effect, he allowed two different shielding factors, suggesting that one electron is relatively close to the nucleus, and the other is farther out. (Because electrons are identical particles, the spatial wave function must be symmetrized with respect to interchange. The spin state\u2014which is irrelevant to the calculation\u2014is evidently antisymmetric.) Show that by astute choice of the adjustable parameters Z1 and Z2 you can get H less than \u221213.6 eV. Answer: \u2212x 8 + 2x 7 + 1 x 6 + y6 2 where x \u2261 Z1 + Z2 and y \u2261 2 Z1 Z2. Chandrasekhar used Z1 = 1.039 (since this is larger than 1, the motivating interpretation as an effective nuclear charge cannot be sustained, but never mind\u2014it\u2019s still an acceptable trial wave function) and Z2 = 0.283. x 3 y4 + 11 8 x 5 y2 \u2212 1 8 x 6 y2 \u2212 1 2 H = E1 x y6 \u2212 1 2 y8 \u221a, Problem 8.26 The fundamental problem in harnessing nuclear fusion is getting the two particles (say, two deuterons) close enough together for the attractive (but short-range) nuclear force to overcome the Coulomb repulsion. The \u201cbulldozer\u201d method is to heat 20 S. Chandrasekhar, Astrophys. J. 100, 176 (1944). 350 CHAPTER 8 The Variational Principle the particles up to fantastic temperatures, and allow the random collisions to bring them together. A more exotic proposal is muon catalysis, in which we construct a \u201chydrogen molecule ion,\u201d only with deuterons in place of protons, and a muon in place of the electron. Predict the equilibrium separation distance between the deuterons in such a structure, and explain why muons are superior to electrons for this purpose.21 \u2217\u2217\u2217 Problem 8.27 Quantum dots. Consider a particle constrained to move in two dimensions in the cross-shaped region shown in Figure 8.10. The \u201carms\u201d of the cross continue out to in\ufb01nity. The potential is zero within the cross, and in\ufb01nite in the shaded areas outside. Surprisingly, this con\ufb01guration admits a positive-energy bound state.22 y a \u2212a \u2212a a x Figure 8.10: The cross-shaped region for Problem 8.27. (a) Show that the lowest energy that can propagate off to in\ufb01nity is Ethreshold = \u03c0 22 8ma2 ; any solution with energy less than that has to be a bound state. Hint: Go way out one arm (say, x a), and solve the Schr\u00f6dinger equation by separation of variables; if the wave function propagates out to in\ufb01nity, the dependence on x must take the form exp (ikx x) with kx > 0. 21 The classic paper on muon-catalyzed fusion is J. D.", " Jackson, Phys. Rev. 106, 330 (1957); for a more recent popular review, see J. Rafelski and S. Jones, Scienti\ufb01c American, November 1987, page 84. 22 This model is taken from R. L. Schult et al., Phys. Rev. B 39, 5476 (1989). For further discussion see J. T. Londergan and D. P. Murdock, Am. J. Phys. 80, 1085 (2012). In the presence of quantum tunneling a classically bound state can become unbound; this is the reverse: A classically unbound state is quantum mechanically bound. Further Problems on Chapter 8 351 (b) Now use the variational principle to show that the ground state has energy less than Ethreshold. Use the following trial wave function (suggested by Jim McTavish): \u03c8(x, y) = A e\u2212\u03b1, \u23a7 \u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23a9 cos(\u03c0 x/2a) + cos(\u03c0 y/2a) cos(\u03c0 x/2a)e\u2212\u03b1|y|/a, cos(\u03c0 y/2a)e\u2212\u03b1|x|/a, 0, |x| \u2264 a and |y| \u2264 a |x| \u2264 a and |y| > a |x| > a and |y| \u2264 a elsewhere. Normalize it to determine A, and calculate the expectation value of H. Answer:! \" H = 2 ma2 \u03c0 2 8 \u2212 1 \u2212 (\u03b1/4) 8/\u03c0 2 + (1/2\u03b1) 1 +. Now minimize with respect to \u03b1, and show that the result is less than Ethreshold. Hint: Take full advantage of the symmetry of the problem\u2014you only need to integrate over 1/8 of the open region, since the other seven integrals will be the same. Note however that whereas the trial wave function is continuous, its derivatives are not\u2014there are \u201croof-lines\u201d at the joins, and you will need to exploit the technique of Example 8.3.23 Problem 8.28 In Yukawa\u2019s original theory (1934), which remains a useful approximation in nuclear physics, the \u201cstrong\u201d force between protons and neutrons is mediated by the exchange of \u03c0 -mesons. The potential energy is V(r ) = \u2212r0V0 e\u2212r/r0 r, (8.83) where r is the distance between the nucleons, and the range r0 is related to the mass of the meson: r0 = /m\u03c0 c. Question: Does this theory account for the existence of the deuteron (a bound state of the proton and the neutron)? The Schr\u00f6dinger equation for the proton/neutron system is (see Problem 5.1): \u2212 2 2\u03bc \u22072\u03c8(r) + V (r )\u03c8(r) = E\u03c8(r), (8.84) where \u03bc is the reduced mass (the proton and neutron have almost identical masses, so call them both m), and r is the position of the neutron (say) relative to the proton: r = rn \u2212 r p. Your task is to show that there exists a solution with negative energy (a bound state), using a variational trial wave function of the form \u03c8\u03b2 (r) = Ae\u2212\u03b2r/r0. (8.85) 23 W.-N. Mei gets a somewhat better bound (and avoids the roof-lines) using \u2212\u03b1 x 2+y2 /a2 \u03c8(x, y) = Ae \u23a7 \u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23a9 but the integrals have to be done numerically., 1 \u2212 x 2 y2/a4 1 \u2212 x 2/a2, 1 \u2212 y2/a2, 352 CHAPTER 8 The Variational Principle (a) Determine A, by normalizing \u03c8\u03b2 (r). (b) Find the expectation value of the Hamiltonian \u03c8\u03b2. Answer: H = \u2212 2 2\u03bc \u22072 + V in the state E(\u03b2) = 2 2\u03bcr 2 0 \u03b22 1 \u2212 4\u03b3\u03b2 (1 + 2\u03b2)2, where \u03b3 \u2261 2\u03bcr 2 0 2 V0. (8.86) (c) Optimize your trial wave function, by setting d E(\u03b2)/d\u03b2 = 0. This tells you \u03b2 as a function of \u03b3 (and hence\u2014everything else being constant\u2014of V0), but let\u2019s use it instead to eliminate \u03b3 in favor of \u03b2: Emin = 2 2", "\u03bcr 2 0 \u03b22 (1 \u2212 2\u03b2) (3 + 2\u03b2). (8.87) (d) Setting 2/2\u03bcr 2 0 = 1, plot Emin as a function of \u03b2, for 0 \u2264 \u03b2 \u2264 1. What does this tell you about the binding of the deuteron? What is the minimum value of V0 for which you can be con\ufb01dent there is a bound state (look up the necessary masses)? The experimental value is 52 MeV. Problem 8.29 Existence of Bound States. A potential \u201cwell\u201d (in one dimension) is a function V(x) that is never positive (V(x) \u2264 0 for all x), and goes to zero at in\ufb01nity (V(x) \u2192 0 as x \u2192 \u00b1\u221e).24 (a) Prove the following Theorem: If a potential well V1(x) supports at least one bound state, then any deeper/wider well (V2(x) \u2264 V1(x) for all x) will also support at least one bound state. Hint: Use the ground state of V1, \u03c81(x), as a variational test function. (b) Prove the following Corollary: Every potential well in one dimension has a bound state.25 Hint: Use a \ufb01nite square well (Section 2.6) for V1. (c) Does the Theorem generalize to two and three dimensions? How about the Corollary? Hint: You might want to review Problems 4.11 and 4.51. Problem 8.30 Performing a variational calculation requires \ufb01nding the minimum of the energy, as a function of the variational parameters. This is, in general, a very hard problem. However, if we choose the form of our trial wave function judiciously, we can develop an ef\ufb01cient algorithm. In particular, suppose we use a linear combination of functions \u03c6n(x): \u03c8(x) = N n=1 cn \u03c6n(x), (8.88) \u2217\u2217 24 To exclude trivial cases, we also assume it has nonzero area d x = 0). Notice that for the purposes of this problem neither the in\ufb01nite square well nor the harmonic oscillator is a \u201cpotential well,\u201d though both of them, of course, have bound states. 25 K. R. Brownstein, Am. J. Phys. 68, 160 (2000) proves that any one-dimensional potential satisfying \u221e \u2212\u221e V (x) d x \u2264 0 admits a bound state (as long as V (x) is not identically zero)\u2014even if it runs positive in some places. V (x Further Problems on Chapter 8 353 where the cn are the variational parameters. If the \u03c6n are an orthonormal set (\u03c6m|\u03c6n = \u03b4mn), but \u03c8(x) is not necessarily normalized, then H is \u03c8 | H | \u03c8 \u03c8 | \u03c8 (8.89) \u03b5 = = mn c\u2217 n m Hmn cn |cn|2 where Hmn = \u03c6m|H |\u03c6n. Taking the derivative with respect to c\u2217 result equal to 0) gives26 j (and setting the n H jn cn = \u03b5 c j, recognizable as the jth row in an eigenvalue problem: \u239b \u239c \u239c \u239c \u239d H11 H12 H21 H22...... HN 1 HN 2 \u00b7 \u00b7 \u00b7 H1N \u00b7 \u00b7 \u00b7 H2N...... \u00b7 \u00b7 \u00b7 HN N \u239e \u239f \u239f \u239f \u23a0 = \u03b5 \u239e \u239b \u239f \u239f \u239f \u23a0 \u239c \u239c \u239c \u239d c1 c2... cN (8.90) (8.91) \u239e \u239f \u239f \u239f \u23a0. \u239b \u239c \u239c \u239c \u239d c1 c2... cN The smallest eigenvalue of this matrix H gives a bound on the ground state energy and the corresponding eigenvector determines the best variational wave function of the form 8.88. (a) Verify Equation 8.90. (b) Now take the derivative of Equation 8.89 with respect to c j and show that you get a result redundant with Equation 8.90. (c) Consider a particle in an in\ufb01nite square well of width a, with a sloping \ufb02oor: \u23a7 \u23a8 V (x) = \u23a9 \u221e", " V0 x/a \u221e x < 0, 0 \u2264 x \u2264 a, x > a. Using a linear combination of the \ufb01rst ten stationary states of the in\ufb01nite square well as the basis functions, sin n \u03c0 x a, 2 a \u03c6n = determine a bound for the ground state energy in the case V0 = 100 2/m a2. Make a plot of the optimized variational wave function. [Note: The exact result is 39.9819 2/m a2.] 26 Each c j, being complex, stands for two independent parameters (its real and imaginary parts). One could take derivatives with respect to the real and imaginary parts, \u2202 \u2202 Re c j E = 0 and \u2202 \u2202 Im c j E = 0, but it is also legitimate (and simpler) to treat c j and c\u2217 j as the independent parameters: \u2202 \u2202 c j E = 0 and \u2202 \u2202 c\u2217 j E = 0. You get the same result either way. 9 THE WKB APPROXIMATION The WKB (Wentzel, Kramers, Brillouin)1 method is a technique for obtaining approximate solutions to the time-independent Schr\u00f6dinger equation in one dimension (the same basic idea can be applied to many other differential equations, and to the radial part of the Schr\u00f6dinger equation in three dimensions). It is particularly useful in calculating bound state energies and tunneling rates through potential barriers. The essential idea is as follows: Imagine a particle of energy E moving through a region where the potential V(x) is constant. If E > V, the wave function is of the form \u03c8(x) = Ae\u00b1ikx, with k \u2261 2m (E \u2212 V )/. % The plus sign indicates that the particle is traveling to the right, and the minus sign means it is going to the left (the general solution, of course, is a linear combination of the two). The wave function is oscillatory, with \ufb01xed wavelength (\u03bb = 2\u03c0/k) and unchanging amplitude (A). Now suppose that V(x) is not constant, but varies rather slowly in comparison to \u03bb, so that over a region containing many full wavelengths the potential is essentially constant. Then it is reasonable to suppose that \u03c8 remains practically sinusoidal, except that the wavelength and the amplitude change slowly with x. This is the inspiration behind the WKB approximation. In effect, it identi\ufb01es two different levels of x-dependence: rapid oscillations, modulated by gradual variation in amplitude and wavelength. By the same token, if E < V (and V is constant), then \u03c8 is exponential: \u03c8(x) = Ae\u00b1\u03ba x, with \u03ba \u2261 2m (V \u2212 E)/. % And if V (x) is not constant, but varies slowly in comparison with 1/\u03ba, the solution remains practically exponential, except that A and \u03ba are now slowly-varying functions of x. Now, there is one place where this whole program is bound to fail, and that is in the immediate vicinity of a classical turning point, where E \u2248 V. For here \u03bb (or 1/\u03ba) goes to in\ufb01nity, and V (x) can hardly be said to vary \u201cslowly\u201d in comparison. As we shall see, a proper handling of the turning points is the most dif\ufb01cult aspect of the WKB approximation, though the \ufb01nal results are simple to state and easy to implement. 9.1 THE \u201cCLASSICAL\u201d REGION The Schr\u00f6dinger equation, \u2212 2 2m d2\u03c8 d x 2 + V (x)\u03c8 = E\u03c8, 1 In Holland it\u2019s KWB, in France it\u2019s BWK, and in England it\u2019s JWKB (for Jeffreys). 9.1 The \u201cClassical\u201d Region 355 V(x) E Turning points Classical region x Figure 9.1: Classically, the particle is con\ufb01ned to the region where E \u2265 V (x). can be rewritten in the following way: where d2\u03c8 d x 2 = \u2212 p2 2 \u03c8, % p(x) \u2261 2m [E \u2212 V (x)] (9.1) (9.2) is the classical formula for the (magnitude of the) momentum of a particle with total energy E and potential energy V (x). For the moment, I\u2019ll assume that E > V (x), so that p(x) is real; we call this the \u201cclassical\u201d region, for obvious reasons\u2014classically the particle is con\ufb01ned to this range of x (see Figure 9.1). In general", ", \u03c8 is some complex function; we can express it in terms of its amplitude, A(x), and its phase, \u03c6(x)\u2014both of which are real: Using a prime to denote the derivative with respect to x, \u03c8(x) = A(x)ei\u03c6(x). = A + i A\u03c6 ei\u03c6, d\u03c8 d x and # = d2\u03c8 d x 2 A + 2i A\u03c6 + i A\u03c6 \u2212 A $ 2 \u03c6 ei\u03c6. Putting this into Equation 9.1: A + 2i A\u03c6 + i A\u03c6 \u2212 A \u03c6 2 = \u2212 p2 2 A. (9.3) (9.4) (9.5) This is equivalent to two real equations, one for the real part and one for the imaginary part: A \u2212 A \u03c6 2 = \u2212 p2 2 A, or A = A \u03c6 2 \u2212 p2 2, and 2A\u03c6 + A\u03c6 = 0, or A2\u03c6 = 0. (9.6) (9.7) 356 CHAPTER 9 The WKB Approximation Equations 9.6 and 9.7 are entirely equivalent to the original Schr\u00f6dinger equation. The second one is easily solved: A2\u03c6 = C 2, or A = C \u221a |\u03c6|, (9.8) where C is a (real) constant. The \ufb01rst one (Equation 9.6) cannot be solved in general\u2014so here comes the approximation: We assume that the amplitude A varies slowly, so the A term is 2 and p2/2.) In negligible. (More precisely, we assume that A/A is much less than both that case we can drop the left side of Equation 9.6, and we are left with 2 = p2 2 = \u00b1 p \u03c6 \u03c6 or,, d\u03c6 d x and therefore \u03c6(x) = \u00b1 1 p(x) d x. (9.9) (I\u2019ll write this as an inde\ufb01nite integral, for now\u2014any constant of integration can be absorbed into C, which thereby becomes complex. I\u2019ll also absorb a factor of.) Then \u221a Notice that \u03c8(x) \u2248 C \u221a p(x) e\u00b1 i p(x) d x. |\u03c8(x)|2 \u2248 |C|2 p(x), (9.10) (9.11) which says that the probability of \ufb01nding the particle at point x is inversely proportional to its (classical) momentum (and hence its velocity) at that point. This is exactly what you would expect\u2014the particle doesn\u2019t spend long in the places where it is moving rapidly, so the probability of getting caught there is small. In fact, the WKB approximation is sometimes derived by starting with this \u201csemi-classical\u201d observation, instead of by dropping the A term in the differential equation. The latter approach is cleaner mathematically, but the former offers a more illuminating physical rationale. The general (approximate) solution, of course, will be a linear combination the two solutions in Equation 9.10, one with each sign. Example 9.1 Potential well with two vertical walls. Suppose we have an in\ufb01nite square well with a bumpy bottom (Figure 9.2): V (x) = some speci\ufb01ed function, \u221e, (0 < x < a), (otherwise). Inside the well (assuming E > V (x) throughout) we have \u03c8(x) \u2248 1\u221a # C+ei\u03c6(x) + C\u2212e\u2212i\u03c6(x) p(x) $, or, more conveniently, \u03c8(x) \u2248 1\u221a p(x) [C1 sin \u03c6(x) + C2 cos \u03c6(x)], (9.12) (9.13) (9.14) V(x) 9.1 The \u201cClassical\u201d Region 357 Figure 9.2: In\ufb01nite square well with a bumpy bottom. a x where (exploiting the freedom noted earlier to impose a convenient lower limit on the integral)2 \u03c6(x9.15) 0 Now, \u03c8(x) must go to zero at x = 0, and therefore (since \u03c6(0) = 0) C2 = 0. Also, \u03c8(x) goes to zero at x = a, so Conclusion: \u03c6(a) = n\u03c0 (n = 1, 2, 3,...). a 0 p(x) d x = n\u03c0. (9.16) (9", ".17) This quantization condition determines the (approximate) allowed energies. For instance, if the well has a \ufb02at bottom (V (x) = 0), then p(x) = \u221a 2m E (a constant), and Equation 9.17 says pa = n\u03c0, or En = n2\u03c0 22 2ma2, which is the old formula for the energy levels of the in\ufb01nite square well (Equation 2.30). In this case the WKB approximation yields the exact answer (the amplitude of the true wave function is constant, so dropping A cost us nothing). \u2217 Problem 9.1 Use the WKB approximation to \ufb01nd the allowed energies (En) of an in\ufb01nite square well with a \u201cshelf,\u201d of height V0, extending half-way across (Figure 7.3): \u23a7 \u23aa\u23a8 V (x) = \u23aa\u23a9 V0, 0, \u221e, (0 < x < a/2), (a/2 < x < a), (otherwise). 2 We might as well take the positive sign, since both are covered by Equation 9.13. 358 CHAPTER 9 The WKB Approximation \u2261 (n\u03c0 )2 /2ma2 (the nth allowed energy Express your answer in terms of V0 and E 0 n > V0, but do not assume that for the in\ufb01nite square well with no shelf). Assume that E 0 1 En V0. Compare your result with what we got in Section 7.1.2, using \ufb01rst-order perturbation theory. Note that they are in agreement if either V0 is very small (the perturbation theory regime) or n is very large (the WKB\u2014semi-classical\u2014regime). \u2217\u2217 Problem 9.2 An alternative derivation of the WKB formula (Equation 9.10) is based on an expansion in powers of. Motivated by the free-particle wave function, \u03c8 = A exp (\u00b1i px/), we write \u03c8(x) = ei f (x)/, where f (x) is some complex function. (Note that there is no loss of generality here\u2014any nonzero function can be written in this way.) (a) Put this into Schr\u00f6dinger\u2019s equation (in the form of Equation 9.1), and show that f 2 + p2 = 0. i f \u2212 (b) Write f (x) as a power series in : f (x) = f0(x) + f1(x) + 2 f2(x) + \u00b7 \u00b7 \u00b7, and, collecting like powers of, show that 2 = p2 etc. 1 (c) Solve for f0(x) and f1(x), and show that\u2014to \ufb01rst order in \u2014you recover Equation 2 9.10. Note: The logarithm of a negative number is de\ufb01ned by ln(\u2212z) = ln(z) + in\u03c0, where n is an odd integer. If this formula is new to you, try exponentiating both sides, and you\u2019ll see where it comes from. 9.2 TUNNELING So far, I have assumed that E > V, so p(x) is real. But we can easily write down the corresponding result in the non-classical region (E < V )\u2014it\u2019s the same as before (Equation 9.10), only now p(x) is imaginary:3 \u03c8(x) \u2248 \u221a e\u00b1 1 C | p(x)| | p(x)| d x. (9.18) Consider, for example, the problem of scattering from a rectangular barrier with a bumpy top (Figure 9.3). To the left of the barrier (x < 0), \u03c8(x) = Aeikx + Be\u2212ikx, (9.19) 3 In this case the wave function is real, and the analogs to Equations 9.6 and 9.7 do not follow necessarily from Equation 9.5, although they are still suf\ufb01cient. If this bothers you, study the alternative derivation in Problem 9.2. 9.2 Tunneling 359 V(x) A B E 0 F a x Figure 9.3: Scattering from a rectangular barrier with a bumpy top. A F 0 a x Figure 9.4: Qualitative structure of the wave function, for scattering from a high, broad barrier. where A is the incident amplitude, B is the re\ufb02ected amplitude, and k \u2261 Section 2.5). To the right of the barrier (x >", " a), \u221a 2m E/ (see where F is the transmitted amplitude. The transmission probability is \u03c8(x) = Feikx, T = |F|2 |A|2. In the tunneling region (0 \u2264 x \u2264 a), the WKB approximation gives \u03c8(x) \u2248 \u221a C | p(x)| 1 e x 0 | p(x )| d x + \u221a D | p(x)| x 0 e\u2212 1 | p(x )| d x. (9.20) (9.21) (9.22) If the barrier is very high and/or very wide (which is to say, if the probability of tunneling is small), then the coef\ufb01cient of the exponentially increasing term (C) must be small (in fact, it would be zero if the barrier were in\ufb01nitely broad), and the wave function looks something like4 Figure 9.4. The relative amplitudes of the incident and transmitted waves are determined essentially by the total decrease of the exponential over the nonclassical region: \u223c e\u2212 1 a 0 | p(x )| d x, |F| |A| so T \u2248 e\u22122\u03b3, with \u03b3 \u2261 1 a 0 | p(x)| d x. (9.23) 4 This heuristic argument can be made more rigorous\u2014see Problem 9.11. 360 CHAPTER 9 The WKB Approximation Example 9.2 Gamow\u2019s theory of alpha decay.5 In 1928, George Gamow (and, independently, Condon and Gurney) used Equation 9.23 to provide the \ufb01rst successful explanation of alpha decay (the spontaneous emission of an alpha particle\u2014two protons and two neutrons\u2014by certain radioactive nuclei).6 Since the alpha particle carries a positive charge (2e), it will be electrically repelled by the leftover nucleus (charge Z e), as soon as it gets far enough away to escape the nuclear binding force. But \ufb01rst it has to negotiate a potential barrier that was already known (in the case of uranium) to be more than twice the energy of the emitted alpha particle. Gamow approximated the potential energy by a \ufb01nite square well (representing the attractive nuclear force), extending out to r1 (the radius of the nucleus), joined to a repulsive Coulombic tail (Figure 9.5), and identi\ufb01ed the escape mechanism as quantum tunneling (this was, by the way, the \ufb01rst time that quantum mechanics had been applied to nuclear physics). V(r) E \u2212V0 Coulomb repulsion r1 r2 r Nuclear binding Figure 9.5: Gamow\u2019s model for the potential energy of an alpha particle in a radioactive nucleus. If E is the energy of the emitted alpha particle, the outer turning point (r2) is determined by 2Z e2 r2 The exponent \u03b3 (Equation 9.23) is evidently7 / 1 4\u03c0 0 = E. \u221a \u03b3 = 1 r2 2m r1 1 4\u03c0 0 2Z e2 r The integral can be done by substitution r2 r2 r 2m E, and the result is r1 \u2212 1 dr. \u2212 E dr = letr \u2261 r2 sin2 u % \u03b3 = \u221a 2m E r2 \u03c0 2 \u2212 sin \u22121 \u2212 r1 (r2 \u2212 r1) r1 r2. (9.25) (9.24) 5 For a more complete discussion, and alternative formulations, see B. R. Holstein, Am. J. Phys. 64, 1061 (1996). 6 For an interesting brief history see E. Merzbacher, \u201cThe Early History of Quantum Tunneling,\u201d Physics Today, August 2002, p. 44. 7 In this case the potential does not drop to zero on the left side of the barrier (moreover, this is really a three- dimensional problem), but the essential idea, contained in Equation 9.23, is all we really need. Typically, r1 r2, and we can simplify this result using the small angle approximation (sin \u2248 ): \u221a \u03b3 \u223c= 2m E $ \u221a r2 \u2212 2 r1r2 # \u03c0 2 = K1 Z \u221a E % Zr1, \u2212 K2 where and K1 \u2261 K2 \u2261 \u03c0 \u221a 2m e2 4\u03c0 0 1/2 \u221a 4 m e2 4\u03c0 0 = 1.980 MeV1/2, = 1.485 fm \u22121/2. 9.2 Tunneling 361 (9.26) (9.27) (9.28) (One fermi (fm) is 10\u221215 m, which is about the size of a typical nucleus.)", " If we imagine the alpha particle rattling around inside the nucleus, with an average velocity v, the time between \u201ccollisions\u201d with the \u201cwall\u201d is about 2r1/v, and hence the frequency of collisions is v/2r1. The probability of escape at each collision is e\u22122\u03b3, so the probability of emission, per unit time, is (v/2r1) e\u22122\u03b3, and hence the lifetime of the parent nucleus is about e2\u03b3. \u03c4 = 2r1 v Unfortunately, we don\u2019t know v\u2014but it hardly matters, for the exponential factor varies over a fantastic range (twenty-\ufb01ve orders of magnitude), as we go from one radioactive nucleus to another; relative to this the variation in v is pretty insigni\ufb01cant. In particular, if you plot the logarithm of the experimentally measured lifetime against 1/ E, the result is a beautiful straight line (Figure 9.6),8 just as you would expect from Equations 9.26 and 9.29. (9.29 12 8 4 0 \u22124 \u22128 4 U238 Th232 236 230 234 5 E [MeV] 232 228 230 228 226 6 7 Figure 9.6: Graph of the logarithm of the half-life energy of the emitted alpha particle), for isotopes of uranium and thorium. \u03c41/2 = \u03c4 ln 2 \u221a versus 1/ E (where E is the 8 This \ufb01gure is reprinted by permission from David Park, Introduction to the Quantum Theory, 3rd edn, Dover Publications, New York (2005); it was adapted from I. Perlman and J. O. Rasmussen, \u201cAlpha Radioactivity,\u201d Encyclopedia of Physics, Vol. 42, Springer (1957). 362 CHAPTER 9 The WKB Approximation \u2217 \u2217\u2217 Problem 9.3 Use Equation 9.23 to calculate the approximate transmission probability for a particle of energy E that encounters a \ufb01nite square barrier of height V0 > E and width 2a. Compare your answer with the exact result (Problem 2.33), to which it should reduce in the WKB regime T 1. Problem 9.4 Calculate the lifetimes of U238 and Po212, using Equations 9.26 and 9.29. Hint: The density of nuclear matter is relatively constant (i.e. the same for all nuclei), so (r1)3 is proportional to A (the number of neutrons plus protons). Empirically, \u223c= (1.07 fm) A1/3. r1 (9.30) The energy of the emitted alpha particle can be deduced by using Einstein\u2019s formula E = mc2 : E = m pc2 \u2212 md c2 \u2212 m\u03b1c2, (9.31) where m p is the mass of the parent nucleus, md is the mass of the daughter nucleus, and m\u03b1 is the mass of the alpha particle (which is to say, the He4 nucleus). To \ufb01gure out what the daughter nucleus is, note that the alpha particle carries off two protons and two neutrons, so Z decreases by 2 and A by 4. Look up the relevant nuclear masses. To estimate v, use E = (1/2) m\u03b1v2; this ignores the (negative) potential energy inside the nucleus, and surely underestimates v, but it\u2019s about the best we can do at this stage. Incidentally, the experimental lifetimes are 6 \u00d7 109 yrs and 0.5 \u00b5s, respectively. Problem 9.5 Zener Tunneling. In a semiconductor, an electric \ufb01eld (if it\u2019s large enough) can produce transitions between energy bands\u2014a phenomenon known as Zener tunneling. A uniform electric \ufb01eld E = \u2212E0 \u02c6\u0131, for which H = \u2212e E0 x, makes the energy bands position dependent, as shown in Figure 9.7. It is then possible for an electron to tunnel from the valence (lower) band to the conduction (upper) band; this phenomenon is the basis for the Zener diode. Treating the gap as a potential barrier through which the electron may tunnel, \ufb01nd the tunneling probability in terms of Eg and E0 (as well as m,, e). 9.3 THE CONNECTION FORMULAS In the discussion so far I have assumed that the \u201cwalls\u201d of the potential well (or the barrier) are vertical, so that the \u201cexterior\u201d solution is simple, and the boundary conditions trivial. As it turns out, our main results (Equations 9.17 and 9.23) are reasonably accurate even when the edges are not so abrupt (indeed", ", in Gamow\u2019s theory they were applied to just such a case). Nevertheless, it is of some interest to study more closely what happens to the wave function at a turning point (E = V ), where the \u201cclassical\u201d region joins the \u201cnonclassical\u201d region, and 9.3 The Connection Formulas 363 E g (a) x (b) x Figure 9.7: (a) The energy bands in the absence of an electric \ufb01eld. (b) In the presence of an electric \ufb01eld an electron can tunnel between the energy bands. Turning point Linearized potential V(x) E Patching region Classical region 0 Nonclassical region x Figure 9.8: Enlarged view of the right-hand turning point. the WKB approximation itself breaks down. In this section I\u2019ll treat the bound state problem (Figure 9.1); you get to do the scattering problem for yourself (Problem 9.11).9 For simplicity, let\u2019s shift the axes over so that the right hand turning point occurs at x = 0 (Figure 9.8). In the WKB approximation, we have \u23a7 \u23a8 # \u03c8(x) \u2248 \u23a9 i 1\u221a Be p(x) | p(x)| De\u2212 1 1\u221a 0 x p(x ) d x + Ce\u2212 i | p(x )| d x, x 0 0 x p(x ) d x $, x < 0, x > 0. (9.32) (Assuming V (x) remains greater than E for all x > 0, we can exclude the positive exponent in this region, because it blows up as x \u2192 \u221e.) Our task is to join the two solutions at the boundary. But there is a serious dif\ufb01culty here: In the WKB approximation, \u03c8 goes to in\ufb01nity at the turning point (where p(x) \u2192 0). The true wave function, of course, has no such wild behavior\u2014as anticipated, the WKB method simply fails in the vicinity of a turning point. And yet, it is precisely the boundary conditions at the turning points that determine the 9 Warning: The following argument is quite technical, and you may wish to skip it on a \ufb01rst reading. 364 CHAPTER 9 The WKB Approximation allowed energies. What we need to do, then, is splice the two WKB solutions together, using a \u201cpatching\u201d wave function that straddles the turning point. \u03c8 p Since we only need the patching wave function in the neighborhood of the origin, we\u2019ll approximate the potential by a straight line: V (x) \u2248 E + V (0)x, (9.33) and solve the Schr\u00f6dinger equation for this linearized V : \u2212 2 2m d2\u03c8 p d x 2 + (E + V (0)x)\u03c8 p = E\u03c8 p, or where d2\u03c8 p d x 2 = \u03b13x\u03c8 p, \u03b1 \u2261 1/3. 2m 2 V (0) The \u03b1s can be absorbed into the independent variable by de\ufb01ning so that z \u2261 \u03b1x, d2\u03c8 p dz2 = z\u03c8 p. (9.34) (9.35) (9.36) (9.37) This is Airy\u2019s equation, and the solutions are called Airy functions.10 Since the Airy equation is a second-order differential equation, there are two linearly independent Airy functions, Ai(z) and Bi(z). They are related to Bessel functions of order 1/3; some of their properties are listed in Table 9.1 and they are plotted in Figure 9.9. Evidently the patching wave function is a linear combination of Ai(z) and Bi(z): \u03c8 p(x) = aAi(\u03b1x) + bBi(\u03b1x), (9.38) for appropriate constants a and b. Now, \u03c8 p is the (approximate) wave function in the neighborhood of the origin; our job is to match it to the WKB solutions in the overlap regions on either side (see Figure 9.10). These overlap zones are close enough to the turning point that the linearized potential is reasonably accurate (so that \u03c8 p is a good approximation to the true wave function), and yet far enough away from the turning point that the WKB approximation is reliable.11 In the overlap regions Equation 9.33 holds, and therefore (in the notation of Equation 9.35) p(x) \u2248 % 2m(E \u2212 E \u2212 V (0)x) = \u03b13/2 \u221a \u2212x. (9.39) 10 Classically", ", a linear potential means a constant force, and hence a constant acceleration\u2014the simplest nontrivial motion possible, and the starting point for elementary mechanics. It is ironic that the same potential in quantum mechanics yields stationary states that are unfamiliar transcendental functions, and plays only a peripheral role in the theory. Still, wave packets can be reasonably simple\u2014see Problem 2.51 and especially footnote 61, page 81. 11 This is a delicate double constraint, and it is possible to concoct potentials so pathological that no such overlap region exists. However, in practical applications this seldom occurs. See Problem 9.9. 9.3 The Connection Formulas 365 Table 9.1: Some properties of the Airy functions. Differential Equation: d 2y dz2 = zy. Solutions: Linear combinations of Airy functions, Ai(z) and Bi(z). Integral Representation: Ai(z) = 1 \u03c0 Bi(z) = 1 \u03c0 Asymptotic Forms: \u221e s3 3 cos ( + sz) ds, \u221e[e\u2212 s3 3 0 0 + sz + sin( + sz)]ds. s3 3 Ai(z) ~ 1 2\u221a\u03c0z1/4 2 e\u2212 z3/2 3 Bi(z) ~ 1 \u221a\u03c0z1/4 2 e z3/2 3 z >> 0; Ai(z) ~ Bi(z) ~ 1 \u221a\u03c0(\u2212z)1/4 1 \u221a\u03c0(\u2212z)1/4 2 3 \u03c0 4 sin[ (\u2212z)3/2 + ] cos[ (\u2212z)3/2 + ] \u03c0 4 2 3 z << 0. Bi(z) 2.0 1.8 1.6 1.4 1.2 1.0 0.8 \u22128 \u22126 \u22123 \u22127 \u22125 \u22124 \u22122 \u22121 3 z 0.6 0.4 0.2 1 2 \u22120.2 \u22120.4 \u22120.6 \u22120.8 \u22121.0 0.5 0.4 0.3 0.2 0.1 0 \u22120.1 \u22120.2 \u22120.3 \u22120.4 \u22120.5 ) z ( i A \u22128 \u22126 \u22124 \u22122 0 2 4 z Figure 9.9: Graph of the Airy functions. In particular, in overlap region 2, x p d x \u2248 \u03b13/\u03b1x)3/2, and therefore the WKB wave function (Equation 9.32) can be written as \u03c8(x) \u2248 \u221a D \u03b13/4x 1/4 e\u2212 2 3 (\u03b1x)3/2. (9.40) 366 CHAPTER 9 The WKB Approximation \u03c8 WKB \u03c8 WKB \u03c8 p Overlap 1 Turning point Overlap 2 x Figure 9.10: Patching region and the two overlap zones. Patching region Meanwhile, using the large-z asymptotic forms12 of the Airy functions (from Table 9.1), the patching wave function (Equation 9.38) in overlap region 2 becomes \u03c8 p(x) \u2248 \u221a 2 a \u03c0 (\u03b1x)1/4 e\u2212 2 3 (\u03b1x)3/2 + \u221a b \u03c0 (\u03b1x)1/4 2 3 (\u03b1x)3/2. e Comparing the two solutions, we see that a = 4\u03c0 \u03b1 D, and b = 0. (9.41) (9.42) Now we go back and repeat the procedure for overlap region 1. Once again, p(x) is given by Equation 9.39, but this time x is negative, so x 0 x p d x \u2248 2 3 (\u2212\u03b1x)3/2 and the WKB wave function (Equation 9.32) is \u03c8(x) \u2248 \u221a # 1 \u03b13/4(\u2212x)1/4 Bei 2 3 (\u2212\u03b1x)3/2 + Ce\u2212i 2 3 (\u2212\u03b1x)3/2 $. (9.43) (9.44) Meanwhile, using the asymptotic form of the Airy function for large negative z (Table 9.1), the patching function (Equation 9.38, with b = 0) reads \u03c8 p(x) \u2248 \u221a a \u03c0 (\u2212\u03b1x)1/4 a \u03c0(\u2212\u03b1x)1/4 sin 1 2i = \u221a (\u2212\u03b1x)3/2 + 2 3 # ei\u03c0/4ei 2 3 \u03c0 4 (\u2212\u03b1x)3/2 \u2212 e\u2212i\u03c0/4e\u2212i 2 3 (\u2212\u03b1x)3/2 $. (9.45) 12 At \ufb01rst glance it seems absurd to use a large-z", " approximation in this region, which after all is supposed to be reasonably close to the turning point at z = 0 (so that the linear approximation to the potential is valid). But notice that the argument here is \u03b1x, and if you study the matter carefully (see Problem 9.9) you will \ufb01nd that there is (typically) a region in which \u03b1x is large, but at the same time it is reasonable to approximate V (x) by a straight line. Indeed, the asymptotic forms of the Airy functions are precisely the WKB solutions to Airy\u2019s equation, and since we are already using \u03c8WKB in the overlap region (Figure 9.10) it is not really a new approximation to do the same for \u03c8 p. 9.3 The Connection Formulas 367 Comparing the WKB and patching wave functions in overlap region 1, we \ufb01nd a \u221a ei\u03c0/4 = B \u221a \u03b1 \u03c0 2i and \u2212a \u221a \u03c0 2i e\u2212i\u03c0/4 = C \u221a \u03b1, or, putting in Equation 9.42 for a: B = \u2212iei\u03c0/4 D, and C = ie\u2212i\u03c0/4 D. (9.46) These are the so-called connection formulas, joining the WKB solutions at either side of the turning point. We\u2019re done with the patching wave function now\u2014its only purpose was to bridge the gap. Expressing everything in terms of the one normalization constant D, and shifting the turning point back from the origin to an arbitrary point x2, the WKB wave function (Equation 9.32) becomes \u03c8(x) \u2248 \u23a7 \u23aa\u23a8 \u23aa\u23a9 # 2D\u221a p(x) sin D\u221a | p(x)| exp x p 1 # \u2212 1 x2 x p x x2 $ < x2; x > x2. (9.47) Example 9.3 Potential well with one vertical wall. Imagine a potential well that has one vertical side (at x = 0) and one sloping side (Figure 9.11). In this case \u03c8(0) = 0, so Equation 9.47 says 0 x2 p(x) d x + \u03c0 4 = n\u03c0, (n = 1, 2, 3,...), x2 0 p(x9.48) 1 or V(x) E x2 x Figure 9.11: Potential well with one vertical wall. 368 CHAPTER 9 The WKB Approximation For instance, consider the \u201chalf-harmonic oscillator\u201d, V (x) = 2 m\u03c92x 2, 1 \u221e, (x > 0), (otherwise). (9.49) In this case where p(x) = 2m E \u2212 (1/2)m\u03c92x 2 = m\u03c9 x 2 2 \u2212 x 2, is the turning point. So 0 x2 p(x) d x = m\u03c9 x2 = 1 \u03c9 2E m x2 m\u03c9x 2 2 = \u03c0 E 2\u03c9, and the quantization condition (Equation 9.48) yields, 7 2 2n \u2212 1 2 En = \u03c9 = 3 2,... \u03c9., 11 2 (9.50) In this particular case the WKB approximation actually delivers the exact allowed energies (which are precisely the odd energies of the full harmonic oscillator\u2014see Problem 2.41). Example 9.4 Potential well with no vertical walls. Equation 9.47 connects the WKB wave functions at a turning point where the potential slopes upward (Figure 9.12(a)); the same reasoning, applied to a downward-sloping turning point (Figure 9.12(b)), yields (Problem 9.10) \u23a7 \u23aa\u23a8 D \u221a | p(x)| exp # 2D \u221a p(x) sin # \u2212 1 p x1 x 1 p x x x1 (x) \u2248 \u23aa\u23a9,, x < x1; x > x1. (9.51) V(x) E x E E x2 (a) x1 (b) x1 x2 (c) Figure 9.12: Upward-sloping and downward-sloping turning points. In particular, if we\u2019re talking about a potential well (Figure 9.12(c)), the wave function in the \u201cinterior\u201d region (x1 < x < x2) can be written either as \u03b82(x) \u2261 1 sin \u03b82(x), where \u03c8(x) \u2248 2D p(x) d x + x \u221a \u03c0 x2 4 p, x (Equation 9.47), or", " as \u03c8(x) \u2248 \u22122D \u221a p(x) sin \u03b81(x), where \u03b81(x) \u2261 \u2212 1 x x1 x p d x \u2212 \u03c0 4, 9.3 The Connection Formulas 369 (Equation 9.51). Evidently the arguments of the sine functions must be equal, modulo \u03c0 :13 \u03b82 = \u03b81 + n\u03c0, from which it follows that p(x) d x = \u03c0, with n = 1, 2, 3,.... (9.52) x2 x1 n \u2212 1 2 This quantization condition determines the allowed energies for the \u201ctypical\u201d case of a potential well with two sloping sides. Notice that it differs from the formulas for two vertical walls (Equation 9.17) or one vertical wall (Equation 9.48) only in the number that is subtracted from n (0, 1/4, or 1/2). Since the WKB approximation works best in the semiclassical (large n) regime, the distinction is more in appearance than in substance. In any event, the result is extraordinarily powerful, for it enables us to calculate (approximate) allowed energies without ever solving the Schr\u00f6dinger equation, by simply evaluating one integral. The wave function itself has dropped out of sight. \u2217\u2217 Problem 9.6 The \u201cbouncing ball\u201d revisited. Consider the quantum mechanical analog to the classical problem of a ball (mass m) bouncing elastically on the \ufb02oor.14 (a) What is the potential energy, as a function of height x above the \ufb02oor? (For negative x, the potential is in\ufb01nite\u2014the ball can\u2019t get there at all.) (b) Solve the Schr\u00f6dinger equation for this potential, expressing your answer in terms of the appropriate Airy function (note that Bi(z) blows up for large z, and must therefore be rejected). Don\u2019t bother to normalize \u03c8(x). (c) Using g = 9.80 m/s2 and m = 0.100 kg, \ufb01nd the \ufb01rst four allowed energies, in joules, correct to three signi\ufb01cant digits. Hint: See Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions, Dover, New York (1970), page 478; the notation is de\ufb01ned on page 450. (d) What is the ground state energy, in eV, of an electron in this gravitational \ufb01eld? How high off the ground is this electron, on the average? Hint: Use the virial theorem to determine x. \u2217 Problem 9.7 Analyze the bouncing ball (Problem 9.6) using the WKB approximation. (a) Find the allowed energies, En, in terms of m, g, and. (b) Now put in the particular values given in Problem 9.6(c), and compare the WKB approximation to the \ufb01rst four energies with the \u201cexact\u201d results. (c) About how large would the quantum number n have to be to give the ball an average height of, say, 1 meter above the ground? 13 Not 2\u03c0 \u2014an overall minus sign can be absorbed into the normalization factors D and D 14 For more on the quantum bouncing ball see Problem 2.59, J. Gea-Banacloche, Am. J. Phys. 67, 776 (1999), and N. Wheeler, \u201cClassical/quantum dynamics in a uniform gravitational \ufb01eld\u201d, unpublished Reed College report (2002). This may sound like an awfully arti\ufb01cial problem, but the experiment has actually been done, using neutrons (V. V. Nesvizhevsky et al., Nature 415, 297 (2002)).. 370 CHAPTER 9 The WKB Approximation \u2217 Problem 9.8 Use the WKB approximation to \ufb01nd the allowed energies of the harmonic oscillator. \u2217\u2217 \u2217\u2217\u2217 Problem 9.9 Consider a particle of mass m in the nth stationary state of the harmonic oscillator (angular frequency \u03c9). (a) Find the turning point, x2. (b) How far (d) could you go above the turning point before the error in the linearized potential (Equation 9.33, but with the turning point at x2) reaches 1%? That is, if V (x2 + d) \u2212 Vlin(x2 + d) V (x2) = 0.01, what is d? (c) The asymptotic form of Ai(z) is accurate to 1% as long as z", " \u2265 5. For the d in part (b), determine the smallest n such that \u03b1d \u2265 5. (For any n larger than this there exists an overlap region in which the linearized potential is good to 1% and the large-z form of the Airy function is good to 1%.) Problem 9.10 Derive the connection formulas at a downward-sloping turning point, and con\ufb01rm Equation 9.51. Problem 9.11 Use appropriate connection formulas to analyze the problem of scattering from a barrier with sloping walls (Figure 9.13). Hint: Begin by writing the WKB wave function in the form \u03c8(x) \u2248 \u23a7 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 # 1\u221a p(x) 1\u221a | p(x)| 1\u221a p(x) Ae\u2212 i 1 Ce x1 x x x1 p(x ) d x + Be | p(x )| d x + De i x1 x p(x ) d x, $ \u2212 1 x x1 | p(x )| d x x x2 i Fe p(x ) d x, x < x1;, x1 < x < x2; (9.53) x > x2. Do not assume C = 0. Calculate the tunneling probability, T = |F|2/|A|2, and show that your result reduces to Equation 9.23 in the case of a broad, high barrier. V(x) E x1 x2 x Figure 9.13: Barrier with sloping walls. Further Problems on Chapter 9 371 \u2217\u2217 Problem 9.12 For the \u201chalf-harmonic oscillator\u201d (Example 9.3), make a plot comparing the normalized WKB wave function for n = 3 to the exact solution. You\u2019ll have to experiment to determine how wide to make the patching region. Note: You can do the integrals of p(x) by hand, but feel free to do them numerically. You\u2019ll need to do the integral of |\u03c8WKB|2 numerically to normalize the wave function. FURTHER PROBLEMS ON CHAPTER 9 \u2217\u2217 Problem 9.13 Use the WKB approximation to \ufb01nd the allowed energies of the general power-law potential: V (x) = \u03b1|x|\u03bd, where \u03bd is a positive number. Check your result for the case \u03bd = 2. Answer:15 \u23a1 En = \u03b1 \u23a3(n \u2212 1/2) \u03c0 2m\u03b1 \u03bd + 3 1 2 \u03bd + 1 1 \u23a4 2\u03bd \u03bd+2 \u23a6. (9.54) \u2217\u2217 Problem 9.14 Use the WKB approximation to \ufb01nd the bound state energy for the potential # in Problem 2.52. Compare the exact answer: \u2212 (9/8) \u2212 $ \u221a 1/ 2 2a2/m. Problem 9.15 For spherically symmetrical potentials we can apply the WKB approximation to the radial part (Equation 4.37). In the case l = 0 it is reasonable16 to use Equation 9.48 in the form r0 p(r ) dr = (n \u2212 1/4) \u03c0, (9.55) where r0 is the turning point (in effect, we treat r = 0 as an in\ufb01nite wall). Exploit this formula to estimate the allowed energies of a particle in the logarithmic potential 0 V (r ) = V0 ln (r/a) (for constants V0 and a). Treat only the case l = 0. Show that the spacing between the levels is independent of mass. Partial answer: En+1 \u2212 En = V0 ln. n + 3/4 n \u2212 1/4 15 As always, the WKB result is most accurate in the semi-classical (large n) regime. In particular, Equation 9.54 is not very good for the ground state (n = 1). See W. N. Mei, Am. J. Phys. 66, 541 (1998). 16 Application of the WKB approximation to the radial equation raises some delicate and subtle problems, which I will not go into here. The classic paper on the subject is R. Langer, Phys. Rev. 51, 669 (1937). 372 CHAPTER 9 The WKB Approximation \u2217\u2217 Problem 9.16 Use the WKB approximation in the form of Equation 9.52, r2 p(r ) dr = n \u2212 1/2 \u03c0 (9.56) r1 to estimate the", " bound state energies for hydrogen. Don\u2019t forget the centrifugal term in the effective potential (Equation 4.38). The following integral may help: Answer: a % (x \u2212 a)(b \u2212 x En \u2248 \u221213.6 eV n \u2212 (1/2) + \u221a ( + 1). 2 (9.57) (9.58) I put a prime on n, because there is no reason to suppose it corresponds to the n in the Bohr formula. Rather, it orders the states for a given, counting the number of nodes in the radial wave function.17 In the notation of Chapter 4, n = N = n \u2212, (Equation 4.67). Put this in, expand the square root and compare your result to the Bohr formula + \u00b7 \u00b7 \u00b7 \u221a \u2217\u2217\u2217 Problem 9.17 Consider the case of a symmetrical double well, such as the one pictured in Figure 9.14. We are interested in bound states with E < V (0). V(x) E Figure 9.14: Symmetric double well; Problem 9.17. x1 x2 x (a) Write down the WKB wave functions in regions (i) x > x2, (ii) x1 < x < x2, and (iii) 0 < x < x1. Impose the appropriate connection formulas at x1 and x2 (this has already been done, in Equation 9.47, for x2; you will have to work out x1 for yourself), to show that \u23a7 \u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23a9 D\u221a | p(x)| exp # 2D\u221a p(x) sin # 1 D\u221a | p(x)| # \u2212 1 p x x2 d x x $ $, x2 (x )|d x + sin \u03b8 e\u2212 1 x1 x 2 cos \u03b8 e 1 \u03c8(x) \u2248 where \u03b8 \u2261 1 x2 x1 p(x) d x. x1 x | p(x )|d x $, (i) (ii) (iii) (9.59) 17 I thank Ian Gatland and Owen Vajk for pointing this out. Further Problems on Chapter 9 373 (b) Because V (x) is symmetric, we need only consider even (+) and odd (\u2212) wave functions. In the former case \u03c8 (0) = 0, and in the latter case \u03c8(0) = 0. Show that this leads to the following quantization condition: where tan \u03b8 = \u00b12e\u03c6, \u03c6 \u2261 1 x1 \u2212x1 p x d x. (9.60) (9.61) Equation 9.60 determines the (approximate) allowed energies (note that E comes into x1 and x2, so \u03b8 and \u03c6 are both functions of E). (c) We are particularly interested in a high and/or broad central barrier, in which case \u03c6 is large, and e\u03c6 is huge. Equation 9.60 then tells us that \u03b8 must be very close to a half-integer multiple of \u03c0. With this in mind, write \u03b8 = (n + 1/2) \u03c0 +, where || 1, and show that the quantization condition becomes \u2212\u03c6. (d) Suppose each well is a parabola:18 V (x) = 2 m\u03c92 (x + a)2, 1 2 m\u03c92 (x \u2212 a)2, 1 x < 0, x > 0. Sketch this potential, \ufb01nd \u03b8 (Equation 9.59), and show that 2\u03c0 e\u2212\u03c6. (9.62) (9.63) (9.64) Comment: If the central barrier were impenetrable (\u03c6 \u2192 \u221e), we would simply have two detached harmonic oscillators, and the energies, En = (n + 1/2) \u03c9, would be doubly degenerate, since the particle could be in the left well or in the right one. When the barrier becomes \ufb01nite (putting the two wells into \u201ccommunihave slightly lower energy, cation\u201d), the degeneracy is lifted. The even states and the odd ones have slightly higher energy. \u03c8 + n \u03c8 \u2212 n (e) Suppose the particle starts out in the right well\u2014or, more precisely, in a state of the form (x, 0) = 1 which, assuming the phases are picked in the \u201cnatural\u201d way, will be concentrated in the right well. Show that it oscillates back and forth between the wells, with a period \u03c4 = 2\u03c0 2 \u03c9 e\u03c6. (9.65", ") 18 Even if V (x) is not strictly parabolic in each well, this calculation of \u03b8, and hence the result (Equation 9.64) will V (x0)/m, where x0 is the position of be approximately correct, in the sense discussed in Section 2.3, with \u03c9 \u2261 the minimum. % 374 CHAPTER 9 The WKB Approximation (f) Calculate \u03c6, for the speci\ufb01c potential in part (d), and show that for V (0) E, \u03c6 \u223c m\u03c9a2/. Problem 9.18 Tunneling in the Stark Effect. When you turn on an external electric \ufb01eld, the electron in an atom can, in principle, tunnel out, ionizing the atom. Question: Is this likely to happen in a typical Stark effect experiment? We can estimate the probability using a crude one-dimensional model, as follows. Imagine a particle in a very deep \ufb01nite square well (Section 2.6). (a) What is the energy of the ground state, measured up from the bottom of the well? Assume V0 2/ma2. Hint: This is just the ground state energy of the in\ufb01nite square well (of width 2a). (b) Now introduce a perturbation H = \u2212\u03b1x (for an electron in an electric \ufb01eld E = \u2212Eext\u02c6\u0131 we would have \u03b1 = eEext). Assume it is relatively weak. Sketch the total potential, and note that the particle can now tunnel out, in the direction of positive x. \u03b1a 2/ma2 (c) Calculate the tunneling factor \u03b3 (Equation 9.23), and estimate the time it would /3\u03b1, \u03c4 = take for the particle to escape (Equation 9.29). Answer: \u03b3 = 8ma2/\u03c0 8mV 3 0 e2\u03b3. (d) Put in some reasonable numbers: V0 = 20 eV (typical binding energy for an outer electron), a = 10\u221210 m (typical atomic radius), Eext = 7 \u00d7 106 V/m (strong laboratory \ufb01eld), e and m the charge and mass of the electron. Calculate \u03c4, and compare it to the age of the universe. Problem 9.19 About how long would it take for a (full) can of beer at room temperature to topple over spontaneously, as a result of quantum tunneling? Hint: Treat it as a uniform cylinder of mass m, radius R, and height h. As the can tips, let x be the height of the center above its equilibrium position (h/2). The potential energy is mgx, and it topples when x reaches the critical value x0 = R2 + (h/2)2 \u2212 h/2. Calculate the tunneling probability (Equation 9.23), for E = 0. Use Equation 9.29, with the thermal energy ((1/2) mv2 = (1/2) k B T ) to estimate the velocity. Put in reasonable numbers, and give your \ufb01nal answer in years.19 V(x) E \u2013a a x Figure 9.15: Re\ufb02ection from a barrier (Problem 9.20). 19 R. E. Crandall, Scienti\ufb01c American, February, 1997, p. 74. Further Problems on Chapter 9 375 Problem 9.20 Equation 9.23 tells us the (approximate) transmission probability for tunneling through a barrier, when E < Vmax\u2014a classically forbidden process. In this problem we explore the complementary phenomenon: re\ufb02ection from a barrier when E > Vmax (again, a classically forbidden process). We\u2019ll assume that V (x) is an even analytic function, that goes to zero as x \u2192 \u00b1\u221e (Figure 9.15). Question: What is the analog to Equation 9.23? (a) Try the obvious approach: assume the potential vanishes for |x| \u2265 a, and use the WKB approximation (Equation 9.13) in the scattering region: \u23a7 \u23aa\u23aa\u23a8 \u03c8(x) \u23aa\u23aa\u23a9 = Aeikx + Be\u2212ikx, C+ei\u03c6(x) + C\u2212e\u2212i\u03c6(x) \u2248 1\u221a p(x) = Ceikx, x < a,, \u2212a < x < a, (9.66) x > a. Impose the usual boundary conditions at \u00b1a, and solve for the re\ufb02ection probability, R = |B|2/|A|2. Unfortunately, the result (R = 0) is uninformative. It\u2019s true that", " the R is exponentially small (just as the transmission coef\ufb01cient is, for E < Vmax), but we\u2019ve thrown the baby out with the bath water\u2014this approximation is simply too drastic. The correct formula is R = e\u22122\u03bb, where \u03bb \u2261 2 y0 0 p(i y) dy (9.67) and y0 is de\ufb01ned by p(i y0) = 0. Notice that \u03bb (like \u03b3 in Equation 9.23) goes like 1/; it is in fact the leading term in an expansion in powers of : \u03bb = c1/ + c2 + c3 + c42 + \u00b7 \u00b7 \u00b7. In the classical limit ( \u2192 0), \u03bb and \u03b3 go to in\ufb01nity, so R and T go to zero, as expected. It is not easy to derive Equation 9.67,20 but let\u2019s look at some examples. (b) Suppose V (x) = V0 sech2(x/a), for some positive constants V0 and a. Plot V (x),. Plot plot p(i y) for 0 \u2264 y \u2264 y0, and show that \u03bb = (\u03c0a/) R as a function of E, for \ufb01xed V0.. Plot V (x), and express \u03bb in terms of an elliptic 1 + (x/a)2 integral. Plot R as a function of E. (c) Suppose V (x) = V0/ 2m E \u2212 2mV0 \u221a \u221a 20 L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, Pergamon Press, Oxford (1958), pages 190\u2013191. R. L. Jaffe, Am. J. Phys. 78, 620 (2010) shows that re\ufb02ection (for E > Vmax) can be regarded as tunneling in momentum space, and obtains Equation 9.67 by a clever analog to the argument yielding Equation 9.23. 10 SCATTERING 10.1 INTRODUCTION 10.1.1 Classical Scattering Theory Imagine a particle incident on some scattering center (say, a marble bouncing off a bowling ball, or a proton \ufb01red at a heavy nucleus). It comes in with energy E and impact parameter b, and it emerges at some scattering angle \u03b8 \u2014see Figure 10.1. (I\u2019ll assume for simplicity that the target is symmetrical about the z axis, so the trajectory remains in one plane, and that the target is very heavy, so its recoil is negligible.) The essential problem of classical scattering theory is this: Given the impact parameter, calculate the scattering angle. Ordinarily, of course, the smaller the impact parameter, the greater the scattering angle. b \u03b8 Scattering center z Figure 10.1: The classical scattering problem, showing the impact parameter b and the scattering angle \u03b8. Example 10.1 Hard-sphere scattering. Suppose the target is a billiard ball, of radius R, and the incident particle is a BB, which bounces off elastically (Figure 10.2). In terms of the angle \u03b1, the impact parameter is b = R sin \u03b1, and the scattering angle is \u03b8 = \u03c0 \u2212 2\u03b1, so b = R sin = R cos \u03c0 2 \u2212 \u03b8 2 \u03b8 2 Evidently 2 cos\u22121(b/R), 0, \u03b8 = (b \u2264 R), (b \u2265 R).. (10.1) (10.2) 10.1 Introduction 377 b \u03b1 \u03b1 \u03b8 R \u03b1 Figure 10.2: Elastic hard-sphere scattering. d\u03c3 \u03c6 b z d\u03a9 \u03b8 d\u03b8 Figure 10.3: Particles incident in the area d\u03c3 scatter into the solid angle d. More generally, particles incident within an in\ufb01nitesimal patch of cross-sectional area d\u03c3 will scatter into a corresponding in\ufb01nitesimal solid angle d (Figure 10.3). The larger d\u03c3 is, the bigger d will be; the proportionality factor, D(\u03b8 ) \u2261 d\u03c3/d, is called the differential (scattering) cross-section:1 d\u03c3 = D(\u03b8 ) d. (10.3) 1 This is terrible language: D isn\u2019t a differential, and it isn\u2019t a cross-section. To my ear, the words \u201cdifferential crosssection\u201d would attach more naturally to d\u03c3. But I\u2019m afraid we\u2019re stuck with this terminology. I should also warn you that the notation D(\u03b8 ) is nonstandard\u2014most people just call it d\u03c3/d (which makes Equation 10.3 look like a tautology). I think", " it will be less confusing if we give the differential cross-section its own symbol. 378 CHAPTER 10 Scattering In terms of the impact parameter and the azimuthal angle \u03c6, d\u03c3 = b db d\u03c6 and d = sin \u03b8 d\u03b8 d\u03c6, so D(\u03b8 ) = b sin \u03b8 (Since \u03b8 is typically a decreasing function of b, the derivative is actually negative\u2014hence the absolute value sign.) db d\u03b8 (10.4). Example 10.2 Hard-sphere scattering (continued). In the case of hard-sphere scattering (Example 10.1) db d\u03b8 = \u2212 1 2 R sin, \u03b8 2 so D(\u03b8 ) = R cos(\u03b8/2) sin \u03b8 R sin(\u03b8/2) 2 = R2 4. This example is unusual, in that the differential cross-section is independent of \u03b8. The total cross-section is the integral of D(\u03b8 ), over all solid angles: \u03c3 \u2261 D(\u03b8 ) d; (10.5) (10.6) (10.7) roughly speaking, it is the total area of incident beam that is scattered by the target. For example, in the case of hard-sphere scattering, \u03c3 = R2/4 d = \u03c0 R2, (10.8) which is just what we would expect: It\u2019s the cross-sectional area of the sphere; BB\u2019s incident within this area will hit the target, and those farther out will miss it completely. But the virtue of the formalism developed here is that it applies just as well to \u201csoft\u201d targets (such as the Coulomb \ufb01eld of a nucleus) that are not simply \u201chit-or-miss\u201d. Finally, suppose we have a beam of incident particles, with uniform intensity (or luminosity, as particle physicists call it) L \u2261 number of incident particles per unit area, per unit time. (10.9) The number of particles entering area d\u03c3 (and hence scattering into solid angle d), per unit time, is d N = L d\u03c3 = L D(\u03b8 ) d, so D(10.10) This is sometimes taken as the de\ufb01nition of the differential cross-section, because it makes reference only to quantities easily measured in the laboratory: If the detector subtends a solid angle d, we simply count the number recorded per unit time (the event rate, d N ), divide by d, and normalize to the luminosity of the incident beam. 10.1 Introduction 379 \u2217\u2217\u2217 Problem 10.1 Rutherford scattering. An incident particle of charge q1 and kinetic energy E scatters off a heavy stationary particle of charge q2. (a) Derive the formula relating the impact parameter to the scattering angle.2 Answer: b = (q1q2/8\u03c0 0 E) cot(\u03b8/2). (b) Determine the differential scattering cross-section. Answer: D(\u03b8 ) = q1q2 16\u03c0 0 E sin2 (\u03b8/2) 2. (10.11) (c) Show that the total cross-section for Rutherford scattering is in\ufb01nite. 10.1.2 Quantum Scattering Theory In the quantum theory of scattering, we imagine an incident plane wave, \u03c8(z) = Aeikz, traveling in the z direction, which encounters a scattering potential, producing an outgoing spherical wave (Figure 10.4).3 That is, we look for solutions to the Schr\u00f6dinger equation of the generic form \u03c8(r, \u03b8 ) \u2248 A eikz + f (\u03b8 ) eikr r, for large r. (10.12) (The spherical wave carries a factor of 1/r, because this portion of |\u03c8|2 must go like 1/r 2 to conserve probability.) The wave number k is related to the energy of the incident particles in the usual way: eikr \u03b8 z eikz Figure 10.4: Scattering of waves; an incoming plane wave generates an outgoing spherical wave. 2 This isn\u2019t easy, and you might want to refer to a book on classical mechanics, such as Jerry B. Marion and Stephen T. Thornton, Classical Dynamics of Particles and Systems, 4th edn, Saunders, Fort Worth, TX (1995), Section 9.10. 3 For the moment, there\u2019s not much quantum mechanics in this; what we\u2019re really talking about is the scattering of waves, as opposed to particles, and you could even think of Figure 10.4 as a picture of water waves encountering a rock, or (better, since we\u2019re interested in three-dimensional scattering) sound waves bouncing off a basketball. 380 CH", "APTER 10 Scattering d\u03c3 \u03c5 \u03c5 dt Figure 10.5: The volume d V of incident beam that passes through area d\u03c3 in time dt. \u221a k \u2261 2m E. (10.13) (As before, I assume the target is azimuthally symmetrical; in the more general case f would depend on \u03c6 as well as \u03b8.) The whole problem is to determine the scattering amplitude f (\u03b8 ); it tells you the probability of scattering in a given direction \u03b8, and hence is related to the differential cross-section. Indeed, the probability that the incident particle, traveling at speed v, passes through the in\ufb01nitesimal area d\u03c3, in time dt, is (see Figure 10.5) d P = |\u03c8incident|2 d V = |A|2 (v dt) d\u03c3. But this is equal to the probability that the particle scatters into the corresponding solid angle d: |A|2 | f |2 r 2 from which it follows that d\u03c3 = | f |2 d, and hence d P = |\u03c8scattered|2 d V = (v dt) r 2 d, D(\u03b8 ) = d\u03c3 d = | f (\u03b8 )|2. (10.14) Evidently the differential cross-section (which is the quantity of interest to the experimentalist) is equal to the absolute square of the scattering amplitude (which is obtained by solving the Schr\u00f6dinger equation). In the following sections we will study two techniques for calculating the scattering amplitude: partial wave analysis and the Born approximation. Problem 10.2 Construct the analogs to Equation 10.12 for one-dimensional and twodimensional scattering. 10.2 PARTIAL WAVE ANALYSIS 10.2.1 Formalism As we found in Chapter 4, the Schr\u00f6dinger equation for a spherically symmetrical potential V (r ) admits the separable solutions \u03c8(r, \u03b8, \u03c6) = R(r )Y m (\u03b8, \u03c6), (10.15) where Y m equation (Equation 4.37): is a spherical harmonic (Equation 4.32), and u(r ) = r R(r ) satis\ufb01es the radial \u2212 2 2m d2u dr 2 + V (r ) + 2 2m ( + 1) r 2 u = Eu. (10.16) 10.2 Partial Wave Analysis 381 Intermediate region V \u2245 0 Radiation zone (kr >> 1) V \u2260 0 Scattering region Figure 10.6: Scattering from a localized potential: the scattering region (dark), the intermediate region, where V = 0 (shaded), and the radiation zone (where kr 1). At very large r the potential goes to zero, and the centrifugal contribution is negligible, so The general solution is d2u dr 2 \u2248 \u2212k2u. u(r ) = Ceikr + De\u2212ikr ; the \ufb01rst term represents an outgoing spherical wave, and the second an incoming one\u2014for the scattered wave we want D = 0. At very large r, then, R(r ) \u223c eikr r, as we already deduced (on physical grounds) in the previous section (Equation 10.12). That\u2019s for very large r (more precisely, for kr 1; in optics it would be called the radiation zone). As in one-dimensional scattering theory, we assume that the potential is \u201clocalized,\u201d in the sense that exterior to some \ufb01nite scattering region it is essentially zero (Figure 10.6). In the intermediate region (where V can be ignored but the centrifugal term cannot),4 the radial equation becomes d2u dr 2 \u2212 ( + 1) r 2 u = \u2212k2u, (10.17) and the general solution (Equation 4.45) is a linear combination of spherical Bessel functions: u(r ) = Ar j(kr ) + Br n(kr ). (10.18) However, neither j (which is somewhat like a sine function) nor n (which is a sort of generalized cosine function) represents an outgoing (or an incoming) wave. What we need are the linear combinations analogous to eikr and e\u2212ikr ; these are known as spherical Hankel functions: (1) (x) \u2261 j(x) + in(x); h (2) (x) \u2261 j(x) \u2212 in(x). h (10.19) 4 What follows does not apply to the Coulomb potential, since 1/r goes to zero more slowly than 1/r 2, as r \u2192 \u221e, and the centrifugal term does not dominate in this region. In this sense the Coulomb potential is not localized", ", and partial wave analysis is inapplicable. 382 CHAPTER 10 Scattering Table 10.1: Spherical Hankel functions, h (1) (x) and h (2) (x). (1) (kr ) (the HanThe \ufb01rst few spherical Hankel functions are listed in Table 10.1. At large r, h (2) kel function of the \ufb01rst kind) goes like eikr /r, whereas h (kr ) (the Hankel function of the second kind) goes like e\u2212ikr /r ; for outgoing waves, then, we need spherical Hankel functions of the \ufb01rst kind: R(r ) \u223c h (1) (kr ). (10.20) The exact wave function, in the exterior region (where V (r ) = 0), is \u23a7 \u23a8 \u23a9eikz + \u03c8(r, \u03b8, \u03c6) = A (1) (kr ) Y m (\u03b8, \u03c6) C,m h \u23ab \u23ac \u23ad,m. (10.21) The \ufb01rst term is the incident plane wave, and the sum (with expansion coef\ufb01cients C,m) is the scattered wave. But since we are assuming the potential is spherically symmetric, the wave \u223c eim\u03c6). function cannot depend on \u03c6.5 So only terms with m = 0 survive (remember, Y m Now (from Equations 4.27 and 4.32) Y 0 (\u03b8, \u03c6) = 2 + 1 4\u03c0 P(cos \u03b8 ), (10.22) where P is the th Legendre polynomial. It is customary to rede\ufb01ne the expansion coef\ufb01cients C,0 \u2261 i +1k : \u221a 4\u03c0 (2 + 1) a \u03c8(r, \u03b8 ) = A eikz + k i +1 (2 + 1) a h (1) (kr ) P(cos \u03b8 ). (10.23) 4 \u221e =0 You\u2019ll see in a moment why this peculiar notation is convenient; a is called the th partial wave amplitude. For very large r, the Hankel function goes like (\u2212i)+1 eikr /kr (Table 10.1), so \u03c8(r, \u03b8 ) \u2248 A eikz + f (\u03b8 ), (10.24) eikr r 5 There\u2019s nothing wrong with \u03b8 dependence, of course, because the incoming plane wave de\ufb01nes a z direction, breaking the spherical symmetry. But the azimuthal symmetry remains; the incident plane wave has no \u03c6 dependence, and there is nothing in the scattering process that could introduce any \u03c6 dependence in the outgoing wave. where f (\u03b8 ) = \u221e =0 (2 + 1) a P(cos \u03b8 ). (10.25) 10.2 Partial Wave Analysis 383 This con\ufb01rms more rigorously the general structure postulated in Equation 10.12, and tells us how to compute the scattering amplitude, f (\u03b8 ), in terms of the partial wave amplitudes (a). The differential cross-section is D(\u03b8 ) = | f (\u03b8 )|2 = (2 + 1) 2 + 1 a\u2217 a P(cos \u03b8 ) P(cos \u03b8 ), (10.26) and the total cross-section is \u03c3 = 4\u03c0 \u221e =0 (2 + 1) |a|2. (10.27) (I used the orthogonality of the Legendre polynomials, Equation 4.34, to do the angular integration.) 10.2.2 Strategy All that remains is to determine the partial wave amplitudes, a, for the potential in question. This is accomplished by solving the Schr\u00f6dinger equation in the interior region (where V (r ) is not zero), and matching it to the exterior solution (Equation 10.23), using the appropriate boundary conditions. The only problem is that as it stands my notation is hybrid: I used spherical coordinates for the scattered wave, but cartesian coordinates for the incident wave. We need to rewrite the wave function in a more consistent notation. Of course, eikz satis\ufb01es the Schr\u00f6dinger equation with V = 0. On the other hand, I just argued that the general solution to the Schr\u00f6dinger equation with V = 0 can be written in the form A,m j(kr ) + B,m n(kr ) Y m (\u03b8, \u03c6).,m In particular, then, it must be possible to express eikz in this way. But eikz is \ufb01nite at", " the origin, so no Neumann functions are allowed in the sum (n(kr ) blows up at r = 0), and since z = r cos \u03b8 has no \u03c6 dependence, only m = 0 terms occur. The resulting expansion of a plane wave in terms of spherical waves is known as Rayleigh\u2019s formula:6 eikz = \u221e =0 i (2 + 1) j(kr )P(cos \u03b8 ). (10.28) Using this, the wave function in the exterior region (Equation 10.23) can be expressed entirely in terms of r and \u03b8 : \u03c8(r, \u03b8 ) = A # i (2 + 1) \u221e =0 j(kr ) + ik a h $ (1) (kr ) P(cos \u03b8 ). (10.29) 6 For a guide to the proof, see George Arfken and Hans-Jurgen Weber, Mathematical Methods for Physicists, 7th edn, Academic Press, Orlando (2013), Exercises 15.2.24 and 15.2.25. 384 CHAPTER 10 Scattering Example 10.3 Quantum hard-sphere scattering. Suppose The boundary condition, then, is V (r ) = \u221e, 0, (r \u2264 a), (r > a). \u03c8(a, \u03b8 ) = 0, (10.30) (10.31) so # i (2 + 1) \u221e =0 j(ka) + ik a h (1) (ka) $ P(cos \u03b8 ) = 0 (10.32) for all \u03b8, from which it follows (Problem 10.3) that j(ka) (1) (ka) a = i k h. In particular, the total cross-section (Equation 10.27) is \u03c3 = 4\u03c0 k2 \u221e =0 (2 + 1) j(ka) (1) (ka) h (10.33) (10.34) 2. That\u2019s the exact answer, but it\u2019s not terribly illuminating, so let\u2019s consider the limiting case of low-energy scattering: ka 1. (Since k = 2\u03c0/\u03bb, this amounts to saying that the wavelength is much greater than the radius of the sphere.) Referring to Table 4.4, we note that n(z) is much larger than j(z), for small z, so j(z) (1) (z) h = j(z) j(z) + in(z) \u2248 \u2212i j(z) n(z) \u2248 \u2212i 2!z/ (2 + 1)! \u2212 (2)!z\u2212\u22121/2! = i 2 + 1 2 2! (2)! z2+1, (10.35) and hence \u03c3 \u2248 4\u03c0 k2 \u221e =0 1 2 + 1 4 2! (2)! (ka)4+2. But we\u2019re assuming ka 1, so the higher powers are negligible\u2014in the low-energy approximation the scattering is dominated by the = 0 term. (This means that the differential cross-section is independent of \u03b8, just as it was in the classical case.) Evidently \u03c3 \u2248 4\u03c0a2, (10.36) for low energy hard-sphere scattering. Surprisingly, the scattering cross-section is four times the geometrical cross-section\u2014in fact, \u03c3 is the total surface area of the sphere. This \u201clarger effective size\u201d is characteristic of long-wavelength scattering (it would be true in optics, as well); in a sense, these waves \u201cfeel\u201d their way around the whole sphere, whereas classical particles only see the head-on cross-section (Equation 10.8). 10.3 Phase Shifts 385 Problem 10.3 Prove Equation 10.33, starting with Equation 10.32. Hint: Exploit the orthogonality of the Legendre polynomials to show that the coef\ufb01cients with different values of must separately vanish. \u2217\u2217 Problem 10.4 Consider the case of low-energy scattering from a spherical delta-function shell: V (r ) = \u03b1\u03b4(r \u2212 a), f (\u03b8 ), the differential where \u03b1 and a are constants. Calculate the scattering amplitude, cross-section, D (\u03b8 ), and the total cross-section, \u03c3. Assume ka 1, so that only the = 0 term contributes signi\ufb01cantly. (To simplify matters, throw out all = 0 terms right from the start.) The main problem, of course, is to determine C0. Express your answer in terms of the dimensionless quantity \u03b2 \u2261 2ma\u03b1/2. Answer: \u03c3 = 4\u03c0a2", "\u03b22/ (1 + \u03b2)2. 10.3 PHASE SHIFTS Consider \ufb01rst the problem of one-dimensional scattering from a localized potential V (x) on the half-line x < 0 (Figure 10.7). I\u2019ll put a \u201cbrick wall\u201d at x = 0, so a wave incident from the left, is entirely re\ufb02ected \u03c8i (x) = Aeikx (x < \u2212a) \u03c8r (x) = Be\u2212ikx (x < \u2212a). (10.37) (10.38) Whatever happens in the interaction region (\u2212a < x < 0), the amplitude of the re\ufb02ected wave has got to be the same as that of the incident wave (|B| = |A|), by conservation of probability. But it need not have the same phase. If there were no potential at all (just the wall at x = 0), then B = \u2212A, since the total wave function (incident plus re\ufb02ected) must vanish at the origin: eikx \u2212 e\u2212ikx V \u03c8(x) = A Be\u2013ikx Aeikx \u2013a (V (x) = 0). (10.39) x Figure 10.7: One-dimensional scattering from a localized potential bounded on the right by an in\ufb01nite wall. 386 CHAPTER 10 Scattering If the potential is not zero, the wave function (for x < \u2212a) takes the form \u03c8(x) = A eikx \u2212 ei(2\u03b4\u2212kx) (V (x) = 0). (10.40) The whole theory of scattering reduces to the problem of calculating the phase shift7 \u03b4 (as a function of k, and hence of the energy E = 2k2/2m), for a speci\ufb01ed potential. We do this, of course, by solving the Schr\u00f6dinger equation in the scattering region (\u2212a < x < 0), and imposing appropriate boundary conditions (see Problem 10.5). The advantage of working with the phase shift (as opposed to the complex number B) is that it exploits the physics to simplify the mathematics (trading a complex quantity\u2014two real numbers\u2014for a single real quantity). carries no angular momentum in the z direction (Rayleigh\u2019s formula contains no terms with m = 0), but it includes all values of the total angular momentum ( = 0, 1, 2,...). Because angular momentum is conserved (by a spherically symmetric potential), each partial wave (labelled by a particular ) scatters independently, with (again) no change in amplitude8\u2014only in phase. If there is no potential at all, then \u03c8 = Aeikz, and the th partial wave is (Equation 10.28) Now let\u2019s return to the three-dimensional case. The incident plane wave Aeikz \u03c8 () = Ai (2 + 1) j(kr ) P(cos \u03b8 ) (V (r ) = 0). (10.41) But (from Equation 10.19 and Table 10.1) # j(x) = 1 2 (1) (x) + h (2) (x) h $ \u2248 1 2x # (\u2212i)+1 ei x + i +1e\u2212i x $ (x 1). (10.42) So for large r \u03c8 () \u2248 A (2 + 1) 2ikr # eikr \u2212 (\u22121) e\u2212ikr $ P(cos \u03b8 ) (V (r ) = 0). (10.43) The second term inside the square brackets represents an incoming spherical wave; it comes from the incident plane wave, and is unchanged when we now introduce a potential. The \ufb01rst term is the outgoing wave; it picks up a phase shift (due to the scattering potential): \u03c8 () \u2248 A # (2 + 1) 2ikr ei(kr +2\u03b4) \u2212 (\u22121) e\u2212ikr $ P(cos \u03b8 ) (V (r ) = 0). (10.44) Think of it as a converging spherical wave (the e\u2212ikr term, due exclusively to the h component in eikz), which is phase shifted an amount \u03b4 on the way in, and again \u03b4 on the way out (1) (hence the 2), emerging as an outgoing spherical wave (the eikr term, due to the h part of eikz plus the scattered wave). (2) In Section 10.2.1 the whole theory was expressed in terms of the partial wave amplitudes a; now we have formulated it in terms of the phase shifts \u03b4. There must be a connection between the two. Indeed,", " comparing the asymptotic (large r ) form of Equation 10.23 \u03c8 () \u2248 A (2 + 1) 2ikr # eikr \u2212 (\u22121) e\u2212ikr $ + (2 + 1) r a eikr P(cos \u03b8 ) (10.45) 7 The 2 in front of \u03b4 is conventional. We think of the incident wave as being phase shifted once on the way in, and again on the way out; \u03b4 is the \u201cone way\u201d phase shift, and the total is 2\u03b4. 8 One reason this subject can be so confusing is that practically everything is called an \u201camplitude\u201d: f (\u03b8 ) is the \u201cscattering amplitude\u201d, a is the \u201cpartial wave amplitude\u201d, but the \ufb01rst is a function of \u03b8, and both are complex numbers. I\u2019m now talking about \u201camplitude\u201d in the original sense: the (real, of course) height of a sinusoidal wave. 10.3 Phase Shifts 387 with the generic expression in terms of \u03b4 (Equation 10.44), we \ufb01nd9 a = 1 2ik e2i\u03b4 \u2212 1 = 1 k ei\u03b4 sin(\u03b4). (10.46) It follows in particular (Equation 10.25) that and (Equation 10.27) f (\u03b8 ) = 1 k \u221e =0 (2 + 1) ei\u03b4 sin(\u03b4) P(cos \u03b8 ) (10.47) \u03c3 = 4\u03c0 k2 \u221e =0 (2 + 1) sin2(\u03b4). (10.48) Again, the advantage of working with phase shifts (as opposed to partial wave amplitudes) is that they are easier to interpret physically, and simpler mathematically\u2014the phase shift formalism exploits conservation of angular momentum to reduce a complex quantity a (two real numbers) to a single real one \u03b4. Problem 10.5 A particle of mass m and energy E is incident from the left on the potential \u23a7 \u23aa\u23a8 0, \u2212V0, \u221e, V (x) = \u23aa\u23a9 (x < \u2212a), (\u2212a \u2264 x \u2264 0), (x > 0). \u221a (a) If the incoming wave is Aeikx (where k = 2m E/), \ufb01nd the re\ufb02ected wave. Answer: Ae\u22122ika k \u2212 ik cot k + ik cot ka ka e\u2212ikx, where k = % 2m (E + V0)/. (b) Con\ufb01rm that the re\ufb02ected wave has the same amplitude as the incident wave. (c) Find the phase shift \u03b4 (Equation 10.40) for a very deep well (E V0). Answer: \u03b4 = \u2212ka. Problem 10.6 What are the partial wave phase shifts (\u03b4) for hard-sphere scattering (Example 10.3)? Problem 10.7 Find the S-wave ( = 0) partial wave phase shift \u03b40(k) for scattering from a delta-function shell (Problem 10.4). Assume that the radial wave function u(r ) goes to 0 as r \u2192 0. Answer: cot(ka) + \u2212 cot \u22121 ka \u03b2 sin2(ka), where \u03b2 \u2261 2m\u03b1a 2. 9 Although I used the asymptotic form of the wave function to draw the connection between a and \u03b4, there is nothing approximate about the result (Equation 10.46). Both of them are constants (independent of r ), and \u03b4 means the phase shift in the asymptotic region (where the Hankel functions have settled down to e\u00b1ikr /kr ). 388 CHAPTER 10 Scattering 10.4 THE BORN APPROXIMATION 10.4.1 Integral Form of the Schr\u00f6dinger Equation The time-independent Schr\u00f6dinger equation, can be written more succinctly as \u2212 2 2m \u22072\u03c8 + V \u03c8 = E\u03c8, \u22072 + k2 \u03c8 = Q, where \u221a k \u2261 2m E and Q \u2261 2m 2 V \u03c8. (10.49) (10.50) (10.51) This has the super\ufb01cial appearance of the Helmholtz equation; note, however, that the \u201cinhomogeneous\u201d term (Q) itself depends on \u03c8. Suppose we could \ufb01nd a function G (r) that solves the Helmholtz equation with a delta function \u201csource\u201d: \u22072 + k2 G(r) = \u03b43(r). (10.52", ") (10.53) Then we could express \u03c8 as an integral: \u03c8(r) = G(r \u2013 r0)Q(r0) d3r0. For it is easy to show that this satis\ufb01es Schr\u00f6dinger\u2019s equation, in the form of Equation 10.50: \u22072 + k2 # $ \u03c8(r) = \u22072 + k2 G(r \u2013 r0) Q(r0) d3r0 = \u03b43(r \u2013 r0)Q(r0) d3r0 = Q(r). G (r) is called the Green\u2019s function for the Helmholtz equation. (In general, the Green\u2019s function for a linear differential equation represents the \u201cresponse\u201d to a delta-function source.) Our \ufb01rst task10 is to solve Equation 10.52 for G(r). This is most easily accomplished by taking the Fourier transform, which turns the differential equation into an algebraic equation. Let G(r) = 1 (2\u03c0 )3/2 eis\u00b7rg(s) d3s. (10.54) Then But \u22072 + k2 G(r) = 1 (2\u03c0 )3/2 # \u22072 + k2 $ eis\u00b7r g(s) d3s. and (see Equation 2.147) \u22072eis\u00b7r = \u2212s2eis\u00b7r, \u03b43(r) = 1 (2\u03c0 )3 eis\u00b7r d3s, (10.55) (10.56) 10 Warning: You are approaching two pages of heavy analysis, including contour integration; if you wish, skip straight to the answer, Equation 10.65. 10.4 The Born Approximation 389 s r \u03b8 \u03c6 Figure 10.8: Convenient coordinates for the integral in Equation 10.58. so Equation 10.52 says 1 (2\u03c0 )3/2 It follows11 that \u2212s2 + k2 eis\u00b7rg (s) d3s = 1 (2\u03c0 )3 eis\u00b7r d3s. g(s) = 1 k2 \u2212 s2. (2\u03c0 )3/2 Putting this back into Equation 10.54, we \ufb01nd: G(r) = 1 (2\u03c0 )3 eis\u00b7r 1 k2 \u2212 s2 d3s. (10.57) (10.58) Now, r is \ufb01xed, as far as the s integration is concerned, so we may as well choose spherical coordinates (s, \u03b8, \u03c6) with the polar axis along r (Figure 10.8). Then s \u00b7 r = sr cos \u03b8, the \u03c6 integral is trivial (2\u03c0 ), and the \u03b8 integral is \u03c0 0 eisr cos \u03b8 sin \u03b8 d\u03b8 = \u2212 eisr cos \u03b8 isr \u03c0 0 = 2 sin(sr ) sr. Thus G(r) = 1 (2\u03c0 )2 2 r \u221e 0 s sin(sr ) k2 \u2212 s2 ds = 1 4\u03c0 2r \u221e \u2212\u221e s sin(sr ) k2 \u2212 s2 ds. (10.59) (10.60) The remaining integral is not so simple. It pays to revert to exponential notation, and factor the denominator: G(r) = i 8\u03c0 2r \u221e \u2212\u221e seisr (s \u2212 k) (s + k) ds \u2212 \u221e \u2212\u221e se\u2212isr (s \u2212 k) (s + k) ds = i 8\u03c0 2r (I1 \u2212 I2). These two integrals can be evaluated using Cauchy\u2019s integral formula: 8 f (z) (z \u2212 z0) dz = 2\u03c0i f (z0), (10.61) (10.62) 11 This is clearly suf\ufb01cient, but it is also necessary, as you can easily show by combining the two terms into a single integral, and using Plancherel\u2019s theorem, Equation 2.103. 390 CHAPTER 10 Scattering Im(s) Figure 10.9: Skirting the poles in the contour integral (Equation 10.61). s = \u2212k s = +k Re(s) Figure 10.10: Closing the contour in Equations 10.63 and 10.64. (a) (b) if z0 lies within the contour (otherwise the integral is zero). In the present case the integration is along the real axis, and it passes right over the pole singularities at \u00b1k. We have to decide how to skirt the poles\u2014I\u2019ll go", " over the one at \u2212k and under the one at +k (Figure 10.9). (You\u2019re welcome to choose some other convention if you like\u2014even winding seven times around each pole\u2014you\u2019ll get a different Green\u2019s function, but, as I\u2019ll show you in a minute, they\u2019re all equally acceptable.)12 For each integral in Equation 10.61 I must \u201cclose the contour\u201d in such a way that the semicircle at in\ufb01nity contributes nothing. In the case of I1, the factor eisr goes to zero when s has a large positive imaginary part; for this one I close above (Figure 10.10(a)). The contour encloses only the singularity at s = +k, so 8 I1 = seisr s + k 1 s \u2212 k ds = 2\u03c0i seisr s + k = i\u03c0 eikr. (10.63) s=k In the case of I2, the factor e\u2212isr goes to zero when s has a large negative imaginary part, so we close below (Figure 10.10(b)); this time the contour encloses the singularity at s = \u2212k (and it goes around in the clockwise direction, so we pick up a minus sign): 8 se\u2212isr s \u2212 k 1 s + k ds = \u22122\u03c0i se\u2212isr s \u2212 k s=\u2212k = \u2212i\u03c0 eikr. (10.64) I2 = \u2212 Conclusion: G(r) = i 8\u03c0 2r # i\u03c0 eikr \u2212 $ \u2212i\u03c0 eikr = \u2212 eikr 4\u03c0r. (10.65) 12 If you are unfamiliar with this technique you have every right to be suspicious. In truth, the integral in Equation 10.60 is simply ill-de\ufb01ned\u2014it does not converge, and it\u2019s something of a miracle that we can make sense of it at all. The root of the problem is that G(r) doesn\u2019t really have a legitimate Fourier transform; we\u2019re exceeding the speed limit, here, and just hoping we won\u2019t get caught. 10.4 The Born Approximation 391 This, \ufb01nally, is the Green\u2019s function for the Helmholtz equation\u2014the solution to Equation 10.52. (If you got lost in all that analysis, you might want to check the result by direct differentiation\u2014see Problem 10.8.) Or rather, it is a Green\u2019s function for the Helmholtz equation, for we can add to G(r) any function G0(r) that satis\ufb01es the homogeneous Helmholtz equation: \u22072 + k2 G0(r) = 0; (10.66) clearly, the result (G + G0) still satis\ufb01es Equation 10.52. This ambiguity corresponds precisely to the ambiguity in how to skirt the poles\u2014a different choice amounts to picking a different function G0(r). Returning to Equation 10.53, the general solution to the Schr\u00f6dinger equation takes the form \u03c8(r) = \u03c80(r) \u2212 m 2\u03c0 2 eik|r\u2212r0| |r \u2212 r0| V (r0)\u03c8(r0) d3r0, (10.67) where \u03c80 satis\ufb01es the free-particle Schr\u00f6dinger equation, \u22072 + k2 \u03c80 = 0. (10.68) Equation 10.67 is the integral form of the Schr\u00f6dinger equation; it is entirely equivalent to the more familiar differential form. At \ufb01rst glance it looks like an explicit solution to the Schr\u00f6dinger equation (for any potential)\u2014which is too good to be true. Don\u2019t be deceived: There\u2019s a \u03c8 under the integral sign on the right hand side, so you can\u2019t do the integral unless you already know the solution! Nevertheless, the integral form can be very powerful, and it is particularly well suited to scattering problems, as we\u2019ll see in the following section. Problem 10.8 Check that Equation 10.65 satis\ufb01es Equation 10.52, by direct substitution. Hint: \u22072 (1/r ) = \u22124\u03c0 \u03b43(r).13 \u2217\u2217 Problem 10.9 Show that the ground state of hydrogen (Equation 4.80) satis\ufb01es the integral form of the Schr\u00f6dinger equation, for the appropriate V and E (note that E is negative, so k = i\u03ba, where \u03ba \u2261 \u22122m E/). \u221a", " 10.4.2 The First Born Approximation Suppose V (r0) is localized about r0 = 0\u2014that is, the potential drops to zero outside some \ufb01nite region (as is typical for a scattering problem), and we want to calculate \u03c8(r) at points far away from the scattering center. Then |r| |r0| for all points that contribute to the integral in Equation 10.67, so |r \u2013 r0|2 = r 2 + r 2 0 \u2212 2r \u00b7 r0 \u2248 r 2 1 \u2212 2 and hence r \u00b7 r0 r 2, (10.69) 13 See, for example, D. Grif\ufb01ths, Introduction to Electrodynamics, 4th edn (Cambridge University Press, Cambridge, UK, 2017), Section 1.5.3. 392 CHAPTER 10 Scattering Let then and therefore |r \u2013 r0| \u2248 r \u2212 \u02c6r \u00b7 r0. k \u2261 k \u02c6r ; eik|r\u2212r0| \u2248 eikr e\u2212ik\u00b7r0, eik|r - r0| |r \u2212 r0| \u2248 eikr r e\u2212ik\u00b7r0. (10.70) (10.71) (10.72) (10.73) (In the denominator we can afford to make the more radical approximation |r \u2013 r0| \u2248 r ; in the exponent we need to keep the next term. If this puzzles you, try including the next term in the expansion of the denominator. What we are doing is expanding in powers of the small quantity (r0/r ), and dropping all but the lowest order.) In the case of scattering, we want \u03c80(r) = Aeikz, (10.74) representing an incident plane wave. For large r, then, \u03c8(r) \u2248 Aeikz \u2212 m 2\u03c0 2 eikr r e\u2212ik\u00b7r0 V (r0)\u03c8(r0) d3r0. (10.75) This is in the standard form (Equation 10.12), and we can read off the scattering amplitude: f (\u03b8, \u03c6) = \u2212 m 2\u03c0 2 A e\u2212ik\u00b7r0 V (r0)\u03c8(r0) d3r0. (10.76) This is exact.14 Now we invoke the Born approximation: Suppose the incoming plane wave is not substantially altered by the potential; then it makes sense to use \u03c8(r0) \u2248 \u03c80(r0) = Aeikz0 = Aeik\u00b7r0, where k \u2261 k \u02c6z, (10.77) (10.78) inside the integral. (This would be the exact wave function, if V were zero; it is essentially a weak potential approximation.15) In the Born approximation, then, f (\u03b8, \u03c6) \u2248 \u2212 m 2\u03c0 2 ei(k\u2212k)\u00b7r0 V (r0) d3r0. (10.79) (In case you have lost track of the de\ufb01nitions of k and k, they both have magnitude k, but the former points in the direction of the incident beam, while the latter points toward the detector\u2014 see Figure 10.11; is the momentum transfer in the process.) k \u2212 k 14 Remember, f (\u03b8, \u03c6) is by de\ufb01nition the coef\ufb01cient of Aeikr /r at large r. 15 Typically, partial wave analysis is useful when the incident particle has low energy, for then only the \ufb01rst few terms in the series contribute signi\ufb01cantly; the Born approximation is more useful at high energy, when the de\ufb02ection is relatively small. 10.4 The Born Approximation 393 \u03ba = k\u2032\u2212k k = kr\u02c6 \u03b8 k\u2032 = kz\u02c6 Figure 10.11: Two wave vectors in the Born approximation: k scattered direction. points in the incident direction, k in the In particular, for low energy (long wavelength) scattering, the exponential factor is essentially constant over the scattering region, and the Born approximation simpli\ufb01es to f (\u03b8, \u03c6) \u2248 \u2212 m 2\u03c0 2 V(r) d3r, (low energy). (10.80) (I dropped the subscript on r, since there is no likelihood of confusion at this point.) Example 10.4 Low-energy soft-sphere scattering.16 Suppose V (r) = V0, 0, (r \u2264 a), (r > a). In this case the low-energy scattering amplitude is f (\u03b8, \u03c6) \u2248 \u2212 m 2\u03c0 2", " V0 \u03c0a3, 4 3 (independent of \u03b8 and \u03c6), the differential cross-section is and the total cross-section is d\u03c3 d = | f |2 \u2248 2mV0a3 32 2, \u03c3 \u2248 4\u03c0 2mV0a3 32 2. (10.81) (10.82) (10.83) (10.84) For a spherically symmetrical potential, V (r) = V (r )\u2014but not necessarily at low energy\u2014the Born approximation again reduces to a simpler form. De\ufb01ne \u03ba \u2261 k \u2212 k, (10.85) 16 You can\u2019t apply the Born approximation to hard-sphere scattering (V0 = \u221e)\u2014the integral blows up. The point is that we assumed the potential is weak, and doesn\u2019t change the wave function much in the scattering region. But a hard sphere changes it radically\u2014from Aeikz to zero. 394 CHAPTER 10 Scattering and let the polar axis for the r0 integral lie along \u03ba, so that k \u2212 k \u00b7 r0 = \u03bar0 cos \u03b80. Then f (\u03b8 ) \u2248 \u2212 m 2\u03c0 2 ei\u03bar0 cos \u03b80 V (r0)r 2 0 sin \u03b80 dr0 d\u03b80 d\u03c60. (10.86) (10.87) The \u03c60 integral is trivial (2\u03c0 ), and the \u03b80 integral is one we have encountered before (see Equation 10.59). Dropping the subscript on r, we are left with f (\u03b8 ) \u2248 \u2212 2m 2\u03ba \u221e 0 r V (r ) sin(\u03bar ) dr, (spherical symmetry). (10.88) The angular dependence of f is carried by \u03ba; in Figure 10.11 we see that \u03ba = 2k sin(\u03b8/2). (10.89) Example 10.5 Yukawa scattering. The Yukawa potential (which is a crude model for the binding force in an atomic nucleus) has the form V (r ) = \u03b2 e\u2212\u03bcr r, where \u03b2 and \u03bc are constants. The Born approximation gives f (\u03b8 ) \u2248 \u2212 2m\u03b2 2\u03ba e\u2212\u03bcr sin(\u03bar ) dr = \u2212 2m\u03b2 \u03bc2 + \u03ba 2. 2 \u221e 0 (10.90) (10.91) (You get to work out the integral for yourself, in Problem 10.11.) Example 10.6 Rutherford scattering. If we put in \u03b2 = q1q2/4\u03c0 0, \u03bc = 0, the Yukawa potential reduces to the Coulomb potential, describing the electrical interaction of two point charges. Evidently the scattering amplitude is f (\u03b8 ) \u2248 \u2212 2mq1q2 4\u03c0 02\u03ba 2, or (using Equations 10.89 and 10.51): f (\u03b8 ) \u2248 \u2212 q1q2 16\u03c0 0 E sin2(\u03b8/2). The differential cross-section is the square of this: d\u03c3 d = q1q2 16\u03c0 0 E sin2(\u03b8/2) 2, (10.92) (10.93) (10.94) which is precisely the Rutherford formula (Equation 10.11). It happens that for the Coulomb potential classical mechanics, the Born approximation, and quantum \ufb01eld theory all yield the same result. As they say in the computer business, the Rutherford formula is amazingly \u201crobust.\u201d 10.4 The Born Approximation 395 \u2217 \u2217\u2217 \u2217 Problem 10.10 Find the scattering amplitude, in the Born approximation, for soft-sphere scattering at arbitrary energy. Show that your formula reduces to Equation 10.82 in the low-energy limit. Problem 10.11 Evaluate the integral in Equation 10.91, to con\ufb01rm the expression on the right. Problem 10.12 Calculate the total cross-section for scattering from a Yukawa potential, in the Born approximation. Express your answer as a function of E. Problem 10.13 For the potential in Problem 10.4, (a) calculate f (\u03b8 ), D(\u03b8 ), and \u03c3, in the low-energy Born approximation; (b) calculate f (\u03b8 ) for arbitrary energies, in the Born approximation; (c) show that your results are consistent with the answer to Problem 10.4, in the appropriate regime. 10.4.3 The Born Series The Born approximation is similar in spirit to the impulse approximation in classical scattering theory. In the impulse approximation we begin by pretending that the particle keeps going in a straight line (Figure 10.12), and compute the transverse impulse that would be delivered to it in that case: I = F\ufffd", "\ufffd\ufffd dt. (10.95) If the de\ufb02ection is relatively small, this should be a good approximation to the transverse momentum imparted to the particle, and hence the scattering angle is \u03b8 \u2248 tan \u22121(I / p), (10.96) where p is the incident momentum. This is, if you like, the \u201c\ufb01rst-order\u201d impulse approximation (the zeroth-order is what we started with: no de\ufb02ection at all). Likewise, in the zeroth-order Born approximation the incident plane wave passes by with no modi\ufb01cation, and what we b Actual trajectory F\u22a5 \u03b8 Scattering center Figure 10.12: The impulse approximation assumes that the particle continues unde\ufb02ected, and calculates the transverse momentum delivered. 396 CHAPTER 10 Scattering explored in the previous section is really the \ufb01rst-order correction to this. But the same idea can be iterated to generate a series of higher-order corrections, which presumably converge to the exact answer. The integral form of the Schr\u00f6dinger equation reads \u03c8(r) = \u03c80(r) + g(r \u2013 r0)V (r0)\u03c8(r0) d3r0, (10.97) where \u03c80 is the incident wave, g(r) \u2261 \u2212 m 2\u03c0 2 eikr r (10.98) is the Green\u2019s function (into which I have now incorporated the factor 2m/2, for convenience), and V is the scattering potential. Schematically, \u03c8 = \u03c80 + gV \u03c8. (10.99) Suppose we take this expression for \u03c8, and plug it in under the integral sign: \u03c8 = \u03c80 + gV \u03c80 + gV gV \u03c8. (10.100) Iterating this procedure, we obtain a formal series for \u03c8: \u03c8 = \u03c80 + gV \u03c80 + gV gV \u03c80 + gV gV gV \u03c80 + \u00b7 \u00b7 \u00b7. (10.101) In each integrand only the incident wave function (\u03c80) appears, together with more and more powers of gV. The \ufb01rst Born approximation truncates the series after the second term, but it is pretty clear how one generates the higher-order corrections. The Born series can be represented diagrammatically as shown in Figure 10.13. In zeroth order \u03c8 is untouched by the potential; in \ufb01rst order it is \u201ckicked\u201d once, and then \u201cpropagates\u201d out in some new direction; in second order it is kicked, propagates to a new location, is kicked again, and then propagates out; and so on. In this context the Green\u2019s function is sometimes called the propagator\u2014it tells you how the disturbance propagates between one interaction and the next. The Born series was the inspiration for Feynman\u2019s formulation of relativistic quantum mechanics, which is expressed entirely in terms of vertex factors (V ) and propagators (g), connected together in Feynman diagrams. Problem 10.14 Calculate \u03b8 (as a function of the impact parameter) for Rutherford scattering, in the impulse approximation. Show that your result is consistent with the exact expression (Problem 10.1(a)), in the appropriate limit +... Figure 10.13: Diagrammatic interpretation of the Born series (Equation 10.101). \u2217\u2217\u2217 \u2217\u2217\u2217 \u2217\u2217 Further Problems on Chapter 10 397 Problem 10.15 Find the scattering amplitude for low-energy soft-sphere scattering in the second Born approximation. Answer: # \u2212 2mV0a3/32 1 \u2212 4mV0a2/52 $. FURTHER PROBLEMS ON CHAPTER 10 Problem 10.16 Find the Green\u2019s function for the one-dimensional Schr\u00f6dinger equation, and use it to construct the integral form (analogous to Equation 10.66). Answer: \u03c8(x) = \u03c80(x) \u2212 im 2k \u221e \u2212\u221e eik|x\u2212x0| V (x0)\u03c8(x0) d x0. (10.102) Problem 10.17 Use your result in Problem 10.16 to develop the Born approximation for one-dimensional scattering (on the interval \u2212\u221e < x < \u221e, with no \u201cbrick wall\u201d at the origin). That is, choose \u03c80(x) = Aeikx, and assume \u03c8(x0) \u2248 \u03c80(x0) to evaluate the integral. Show that the re\ufb02ection coef\ufffd", "\ufffd\ufffdcient takes the form: \u221e R \u2248 2 m 2k e2ikx V (x) d x \u2212\u221e. (10.103) 2 Problem 10.18 Use the one-dimensional Born approximation (Problem 10.17) to compute the transmission coef\ufb01cient (T = 1 \u2212 R) for scattering from a delta function (Equation 2.117) and from a \ufb01nite square well (Equation 2.148). Compare your results with the exact answers (Equations 2.144 and 2.172). Problem 10.19 Prove the optical theorem, which relates the total cross-section to the imaginary part of the forward scattering amplitude: \u03c3 = 4\u03c0 k Im f (0). (10.104) Hint: Use Equations 10.47 and 10.48. Problem 10.20 Use the Born approximation to determine the total cross-section for scattering from a gaussian potential V(r) = V0e\u2212\u03bcr 2/a2. Express your answer in terms of the constants V0, a, and m (the mass of the incident particle), and k \u2261 2m E/, where E is the incident energy. \u221a Problem 10.21 Neutron diffraction. Consider a beam of neutrons scattering from a crystal (Figure 10.14). The interaction between neutrons and the nuclei in the crystal is short ranged, and can be approximated as 398 CHAPTER 10 Scattering V (r) = 2 \u03c0 2 b m i \u03b43(r \u2212 ri ), where the ri are the locations of the nuclei and the strength of the potential is expressed in terms of the nuclear scattering length b. k \u03b8 k\u2032 z Figure 10.14: Neutron scattering from a crystal. (a) In the \ufb01rst Born approximation, show that d\u03c3 d = b2 2 e\u2212i q\u00b7ri i where q \u2261 k \u2212 k. (b) Now consider the case where the nuclei are arranged on a cubic lattice with spacing a. Take the positions to be ri = l a \u02c6\u0131 + m a \u02c6j + n a \u02c6k where l, m, and n all range from 0 to N \u22121, so there are a total of N 3 nuclei.17 Show that d\u03c3 d = b2 sin2(N qx a/2) sin2(qx a/2) sin2 sin2 N qy a/2 qy a/2 sin2(N qz a/2) sin2(qz a/2). (c) Plot sin2(N qx a/2) sin2(qx a/2) as a function of qx a for several values of N (N = 1, 5, 10) to show that the function describes a series of peaks that become progressively sharper as N increases. 1 N (d) In light of (c), in the limit of large N the differential scattering cross section is negligibly small except at one of these peaks: q = Gmn = 2 \u03c0 a l \u02c6\u0131 + m \u02c6j + n \u02c6k 17 It makes no difference that this crystal isn\u2019t \u201ccentered\u201d at the origin: shifting the crystal by R amounts to adding R to each of the ri, and that doesn\u2019t affect d\u03c3/d. After all, we\u2019re assuming an incident plane wave, which extends to \u00b1\u221e in the x and y directions. Further Problems on Chapter 10 399 for integer l, m, and n. The vectors Glmn are called reciprocal lattice vectors. Find the scattering angles (\u03b8 ) at which peaks occur. If the neutron\u2019s wavelength is equal to the crystal spacing a, what are the three smallest (nonzero) angles? Comment: Neutron diffraction is one method used, to determine crystal structures (electrons and x-rays can also be used and the same expression for the locations of the peaks holds). In this problem we looked at a cubic arrangement of atoms, but a different arrangement (hexagonal for example) would produce peaks at a different set of angles. Thus from the scattering data one can infer the underlying crystal structure. \u2217\u2217\u2217 Problem 10.22 Two-dimensional scattering theory. By analogy with Section 10.2, develop partial wave analysis for two dimensions. (a) In polar coordinates (r, \u03b8 ) the Laplacian is \u2202 2 \u2202r 2 \u2202 2 \u2202 x 2 \u2202 2 \u2202 y2 \u22072 = + = + 1 r \u2202 \u2202r + 1 r 2 \u2202 2 \u2202\u03b8 2. (10.105) Find the separable solutions to the (time-independent) Schr\u00f6dinger equation, for a potential with az", "imuthal symmetry (V (r, \u03b8 ) \u2192 V (r )). Answer: where j is an integer, and u \u2261 \u2212 2 2m d2u dr 2 + V (r ) + 2 2m j 2 \u2212 1/4 r 2 \u03c8(r, \u03b8 ) = R(r ) ei j\u03b8, \u221a r R satis\ufb01es the radial equation (10.106) u = Eu. (10.107) (b) By solving the radial equation for very large r (where both V (r ) and the centrifugal term go to zero), show that an outgoing radial wave has the asymptotic form R(r ) \u223c eikr \u221a r, (10.108) \u221a where k \u2261 2m E/. Check that an incident wave of the form Aeikx satis\ufb01es the Schr\u00f6dinger equation, for V (r ) = 0 (this is trivial, if you use cartesian coordinates). Write down the two-dimensional analog to Equation 10.12, and compare your result to Problem 10.2. Answer: \u03c8(r, \u03b8 ) \u2248 A eikx + f (\u03b8 ), for large r. (10.109) eikr \u221a r (c) Construct the analog to Equation 10.21 (the wave function in the region where V (r ) = 0 but the centrifugal term cannot be ignored). Answer: \u23a7 \u23ab \u23a8 \u23ac \u23a9eikx + \u03c8(r, \u03b8 ) = A (kr )ei j\u03b8 c j H (1) j \u221e \u23ad j=\u2212\u221e, (10.110) (1) where H j order j.18 is the Hankel function (not the spherical Hankel function!) of 18 See Mary Boas, Mathematical Methods in the Physical Sciences, 3rd edn (Wiley, New York, 2006), Section 12.17. 400 CHAPTER 10 Scattering (d) For large z, Use this to show that H (1) j (z) \u223c % 2/\u03c0 e\u2212i\u03c0/4 (\u2212i) j ei z \u221a z. (10.111) % f (\u03b8 ) = 2/\u03c0 k e\u2212i\u03c0/4 \u221e j=\u2212\u221e (\u2212i) j c j ei j\u03b8. (10.112) (e) Adapt the argument of Section 10.1.2 to this two-dimensional geometry. Instead of the area d\u03c3, we have a length, db, and in place of the solid angle d we have the increment of scattering angle |d\u03b8 |; the role of the differential cross-section is played by D(\u03b8 ) \u2261, db d\u03b8 and the effective \u201cwidth\u201d of the target (analogous to the total cross-section) is 2\u03c0 B \u2261 D(\u03b8 ) d\u03b8. Show that 0 D(\u03b8 ) = | f(\u03b8 )|2, and =\u2212\u221e (10.113) (10.114) (10.115) (f) Consider the case of scattering from a hard disk (or, in three dimensions, an in\ufb01nite cylinder19) of radius a: V (r ) = \u221e, 0, (r \u2264 a), (r > a). (10.116) By imposing appropriate boundary conditions at r = a, determine B. You\u2019ll need the analog to Rayleigh\u2019s formula: eikx = \u221e j=\u2212\u221e (i) j J j (kr ) ei j\u03b8 (10.117) (where J j is the Bessel function of order J ). Plot B as a function of ka, for 0 < ka < 2. Problem 10.23 Scattering of identical particles. The results for scattering of a particle from a \ufb01xed target also apply to the scattering of two particles in the center of mass frame. With \u03c8(R, r) = \u03c8R(R) \u03c8r (r), \u03c8r (r) satis\ufb01es \u2212 2 \u03bc \u22072\u03c8r + V (r ) \u03c8r = Er \u03c8r (10.118) 19 S. McAlinden and J. Shertzer, Am. J. Phys. 84, 764 (2016). Further Problems on Chapter 10 401 (see Problem 5.1) where V (r ) is the interaction between the particles (assumed here to depend only on their separation distance). This is the one-particle Schr\u00f6dinger equation (with the reduced mass \u03bc in place of m). Figure 10.15: Scattering of identical particles. (a) Show that if the two partic", "les are identical (spinless) bosons, then \u03c8r (r) must be an even function of r (Figure 10.15). (b) By symmetrizing Equation 10.12 (why is this allowed?), show that the scattering amplitude in this case is f B(\u03b8 ) = f (\u03b8 ) + f (\u03c0 \u2212 \u03b8 ) where f (\u03b8 ) is the scattering amplitude of a single particle of mass \u03bc from a \ufb01xed target V (r ). (c) Show that the partial wave amplitudes of f B vanish for all odd powers of. (d) How are the results of (a)\u2013(c) different if the particles are identical fermions (in a triplet spin state). (e) Show that the scattering amplitude for identical fermions vanishes at \u03c0/2. (f) Plot the logarithm of the differential scattering cross section for fermions and for bosons in Rutherford scattering (Equation 10.93).20 20 Equation 10.93 was derived by taking the limit of Yukawa scattering (Example 10.5) and the result for f (\u03b8 ) is missing a phase factor (see Albert Messiah, Quantum Mechanics, Dover, New York, NY (1999), Section XI.7). That factor drops out of the cross-section for scattering from a \ufb01xed potential\u2014giving the correct answer in Example 10.6\u2014but would show up in the cross-section for scattering of identical particles. 11 QUANTUM DYNAMICS So far, practically everything we have done belongs to the subject that might properly be called quantum statics, in which the potential energy function is independent of time: V (r, t) = V (r). In that case the (time-dependent) Schr\u00f6dinger equation, i \u2202 \u2202t = \u02c6H, can be solved by separation of variables: (r, t) = \u03c8(r)e\u2212i Et/, where \u03c8(r) satis\ufb01es the time-independent Schr\u00f6dinger equation, \u02c6H \u03c8 = E\u03c8. (11.1) (11.2) (11.3) (11.4), |cn|2 Because the time dependence of separable solutions is carried by the exponential factor e\u2212i Et/, which cancels out when we construct the physically relevant quantity ||2, all probabilities and expectation values (for such states) are constant in time. By forming linear combinations of these stationary states we obtain wave functions with more interesting time dependence, (r, t) = cn\u03c8n(r)e\u2212i En t/, but even then the possible values of the energy (En), and their respective probabilities are constant. If we want to allow for transitions (quantum jumps, as they are sometimes called) between one energy level and another, we must introduce a time-dependent potential (quantum dynamics). There are precious few exactly solvable problems in quantum dynamics. However, if the time-dependent part of the Hamiltonian is small (compared to the time-independent part), it can be treated as a perturbation. The main purpose of this chapter is to develop time-dependent perturbation theory, and study its most important application: the emission or absorption of radiation by an atom. Problem 11.1 Why isn\u2019t it trivial to solve the time-dependent Schr\u00f6dinger equation (11.1), in its dependence on t? After all, it\u2019s a \ufb01rst-order differential equation. (a) How would you solve the equation (for f(t)), if k were a constant? d f dt = k f (b) What if k is itself a function of t? (Here k(t) and f (t) might also depend on other variables, such as r\u2014it doesn\u2019t matter.) (c) Why not do the same thing for the Schr\u00f6dinger equation (with a time-dependent 11.1 Two-Level Systems 403 Hamiltonian)? To see that this doesn\u2019t work, consider the simple case \u02c6H (t) = \u02c6H1, \u02c6H2, (0 < t < \u03c4 ), (t > \u03c4 ), where \u02c6H1 and \u02c6H2 are themselves time-independent. If the solution in part (b) held for the Schr\u00f6dinger equation, the wave function at time t > \u03c4 would be # (t) = e \u2212i but of course we could also write $ \u02c6H1\u03c4 + \u02c6H2(t\u2212\u03c4 ) /(0), (t) = e\u2212i \u02c6H2(t\u2212\u03c4 )/(\u03c4 ) = e\u2212i \u02c6H2(t\u2212\u03c4 )/e\u2212", "i \u02c6H1\u03c4/(0). Why are these generally not the same? [This is a subtle matter; if you want to pursue it further, see Problem 11.23.] 11.1 TWO-LEVEL SYSTEMS To begin with, let us suppose that there are just two states of the (unperturbed) system, \u03c8a and \u03c8b. They are eigenstates of the unperturbed Hamiltonian, \u02c6H 0: \u02c6H 0\u03c8a = Ea\u03c8a, and \u02c6H 0\u03c8b = Eb\u03c8b, and they are orthonormal: \u03c8 j Any state can be expressed as a linear combination of them; in particular, (i, j = a, b). = \u03b4i j, \u03c8i (11.5) (11.6) (0) = ca\u03c8a + cb\u03c8b. (11.7) The states \u03c8a and \u03c8b might be position-space wave functions, or spinors, or something more exotic\u2014it doesn\u2019t matter. It is the time dependence that concerns us here, so when I write (t), I simply mean the state of the system at time t. In the absence of any perturbation, each component evolves with its characteristic wiggle factor: (t) = ca\u03c8ae\u2212i Ea t/ + cb\u03c8be\u2212i Ebt/. (11.8) Informally, we say that |ca|2 is the \u201cprobability that the particle is in state \u03c8a\u201d\u2014by which we really mean the probability that a measurement of the energy would yield the value Ea. Normalization of requires, of course, that |ca|2 + |cb|2 = 1. (11.9) 11.1.1 The Perturbed System \u02c6H (t). Since \u03c8a and \u03c8b constitute a Now suppose we turn on a time-dependent perturbation, complete set, the wave function (t) can still be expressed as a linear combination of them. The only difference is that ca and cb are now functions of t: 404 CHAPTER 11 Quantum Dynamics (t) = ca(t)\u03c8ae\u2212i Ea t/ + cb(t)\u03c8be\u2212i Ebt/. (11.10) (I could absorb the exponential factors into ca(t) and cb(t), and some people prefer to do it this way, but I think it is nicer to keep visible the part of the time dependence that would be present even without the perturbation.) The whole problem is to determine ca and cb, as functions of time. If, for example, the particle started out in the state \u03c8a(ca(0) = 1, cb(0) = 0), and at some later time t1 we \ufb01nd that ca(t1) = 0, cb(t1) = 1, we shall report that the system underwent a transition from \u03c8a to \u03c8b. We solve for ca(t) and cb(t) by demanding that (t) satisfy the time-dependent Schr\u00f6dinger equation, \u02c6H = i \u2202 \u2202t, where \u02c6H = \u02c6H 0 + \u02c6H (t). (11.11) From Equations 11.10 and 11.11, we \ufb01nd: \u02c6H 0\u03c8a e\u2212i Ea t/ + cb ca \u02c6H 0\u03c8b # e\u2212i Ebt/ + ca \u02c6H \u03c8a e\u2212i Ea t/ e\u2212i Ebt/ = i + cb \u02c6H \u03c8b \u2212 i Ea + ca\u03c8a \u02d9ca\u03c8ae\u2212i Ea t/ + \u02d9cb\u03c8be\u2212i Ebt/ \u2212 i Eb e\u2212i Ebt/ $. e\u2212i Ea t/ + cb\u03c8b In view of Equation 11.5, the \ufb01rst two terms on the left cancel the last two terms on the right, and hence \u02c6H \u03c8a \u02d9ca\u03c8ae\u2212i Ea t/ + \u02d9cb\u03c8be\u2212i Ebt/ e\u2212i Ea t/ + cb e\u2212i Ebt/ = i. (11.12) \u02c6H \u03c8b ca To isolate \u02d9ca, we use the standard trick: Take the inner product with \u03c8a, and exploit the orthogonality of \u03c8a and \u03c8b (Equation 11.6): 0 \u02c6H \u03c8a e\u2212i Ea t/ + cb \u03c8", "a ca 1 0 \u03c8a 1 \u02c6H \u03c8b e\u2212i Ebt/ = i \u02d9cae\u2212i Ea t/. For short, we de\ufb01ne 0 H i j \u2261 1 ; (11.13) \u02c6H \u03c8 j \u2217 \u03c8i note that the hermiticity of \u02c6H entails H ji we conclude that: = H i j. Multiplying through by \u2212 (i/) ei Ea t/, \u02d9ca = \u2212 i # ca H aa + cb H abe\u2212i(Eb\u2212Ea )t/ $. Similarly, the inner product with \u03c8b picks out \u02d9cb: \u02c6H \u03c8a e\u2212i Ea t/ + cb \u02c6H \u03c8b \u03c8b ca 0 0 1 1 \u03c8b e\u2212i Ebt/ = i \u02d9cbe\u2212i Ebt/, and hence \u02d9cb = \u2212 i # cb H bb + ca H baei(Eb\u2212Ea )t/ $. (11.14) (11.15) Equations 11.14 and 11.15 determine ca(t) and cb(t); taken together, they are completely equivalent to the (time-dependent) Schr\u00f6dinger equation, for a two-level system. Typically, the diagonal matrix elements of \u02c6H vanish (see Problem 11.5 for the general case): H aa = H bb (11.16) = 0. 11.1 Two-Level Systems 405 \u2217 \u2217 If so, the equations simplify: \u02d9ca = \u2212 i abe\u2212i\u03c90t cb, H \u02d9cb = \u2212 i baei\u03c90t ca, H where \u03c90 \u2261 Eb \u2212 Ea. (I\u2019ll assume that Eb \u2265 Ea, so \u03c90 \u2265 0.) (11.17) (11.18) Problem 11.2 A hydrogen atom is placed in a (time-dependent) electric \ufb01eld E = E(t) \u02c6k. i j of the perturbation \u02c6H = eE z between the ground Calculate all four matrix elements H state (n = 1) and the (quadruply degenerate) \ufb01rst excited states (n = 2). Also show that H = 0 for all \ufb01ve states. Note: There is only one integral to be done here, if you exploit ii oddness with respect to z; only one of the n = 2 states is \u201caccessible\u201d from the ground state by a perturbation of this form, and therefore the system functions as a two-state con\ufb01guration\u2014assuming transitions to higher excited states can be ignored. Problem 11.3 Solve Equation 11.17 for the case of a time-independent perturbation, assuming that ca(0) = 1 and cb(0) = 0. Check that |ca(t)|2+|cb(t)|2 = 1. Comment: Ostensibly, this system oscillates between \u201cpure \u03c8a\u201d and \u201csome \u03c8b.\u201d Doesn\u2019t this contradict my general assertion that no transitions occur for time-independent perturbations? No, but the reason is rather subtle: In this case \u03c8a and \u03c8b are not, and never were, eigenstates of the Hamiltonian\u2014a measurement of the energy never yields Ea or Eb. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end, \u03c8a and \u03c8b are eigenstates of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time T \u2014this doesn\u2019t affect the calculations, but it allows for a more sensible interpretation of the result. \u2217\u2217 Problem 11.4 Suppose the perturbation takes the form of a delta function (in time): \u02c6H = \u02c6U \u03b4(t); assume that Uaa = Ubb = 0, and let Uab = U \u2217 \u2261 \u03b1. If ca (\u2212\u221e) = 1 and cb (\u2212\u221e) = 0, ba \ufb01nd ca(t) and cb(t), and check that |ca(t)|2 + |cb(t)|2 = 1. What is the net probability (Pa\u2192b for t \u2192 \u221e) that a transition occurs? Hint: You", " might want to treat the delta function as the limit of a sequence of rectangles. Answer: Pa\u2192b = sin2 (|\u03b1| /). 11.1.2 Time-Dependent Perturbation Theory So far, everything is exact: We have made no assumption about the size of the perturbation. But if \u02c6H is \u201csmall,\u201d we can solve Equation 11.17 by a process of successive approximations, as follows. Suppose the particle starts out in the lower state: 406 CHAPTER 11 Quantum Dynamics If there were no perturbation at all, they would stay this way forever: ca(0) = 1, cb(0) = 0. Zeroth Order: c(0) a (t) = 1, (0) c b (t) = 0. (11.19) (11.20) (I\u2019ll use a superscript in parentheses to indicate the order of the approximation.) To calculate the \ufb01rst-order approximation, we insert the zeroth-order values on the right side of Equation 11.17: First Order: (1) dc a dt = 0 \u21d2 c(1) a = \u2212 i baei\u03c90t \u21d2 c H (1) b (1) dc b dt (t) = 1; = \u2212 i 0 t H ba t ei\u03c90t dt. (11.21) Now we insert these expressions on the right side of Equation 11.17 to obtain the second-order approximation: Second Order: (2) dc a dt (t(2) a abe\u2212i\u03c90t H t 0 H ab t \u2212 i 0 e\u2212i\u03c90t t t ei\u03c90t dt \u21d2 t t H ba ei\u03c90t dt dt, (11.22) H ba 0 (1) (2) while cb is unchanged (c (t) = c b b the second-order correction would be the integral part alone.) (t)). (Notice that c (2) a (t) includes the zeroth-order term; In principle, we could continue this ritual inde\ufb01nitely, always inserting the nth-order approximation into the right side of Equation 11.17, and solving for the (n + 1)th order. The zeroth order contains no factors of \u02c6H, the \ufb01rst-order correction contains one factor of \u02c6H, the secondorder correction has two factors of \u02c6H, and so on.1 The error in the \ufb01rst-order approximation is = 1 (the exact coef\ufb01cients must, of course, obey evident in the fact that 2 is equal to 1 to \ufb01rst order in \u02c6H, which is all (t) Equation 11.9). However, we can expect from a \ufb01rst-order approximation. And the same goes for the higher orders. (1) + c b 2 (1) c b (1) a (t) (1) a (t) c c 2 2 (t) + Equation 11.21 can be written in the form (1) c b (t) e\u2212i Eb t/ = \u2212 i t e\u2212i Eb(t\u2212t )/ H ba t e\u2212i Ea t / dt (11.23) 0 (where I\u2019ve restored the exponential we factored out in Equation 11.10). This suggests a nice pictorial interpretation: reading from right to left, the system remains in state a from time 0 to time t (picking up the \u201cwiggle factor\u201d e\u2212i Ea t /), makes a transition from state a to state b at time t, and then remains in state b until time t (picking up the \u201cwiggle factor\u201d e\u2212i Eb(t\u2212t )/). This process is represented in Figure 11.1. (Don\u2019t take the picture too literally: there is no 1 Notice that ca is modi\ufb01ed in every even order, and cb in every odd order; this would not be true if the perturbation included diagonal terms, or if the system started out in a linear combination of the two states. 11.1 Two-Level Systems 407 b a Figure 11.1: Pictorial representation of Equation 11.23. time a a + Figure 11.2: Pictorial representation of Equation 11.24. b a sharp transition between these states; in fact, you integrate over all the times t at which this transition can occur.) This interpretation of the perturbation series is especially illuminating at higher orders and for multi-level systems, where the expressions become complicated. Consider Equation 11.22, which", " can be written c(2) a (t) e\u2212i Ea t/ = e\u2212i Ea t/ + 2 t t \u2212 i \u00d7 H ab t 0 0 e\u2212i Eb(t \u2212t )/ H e\u2212i Ea(t\u2212t )/ t ba e\u2212i Ea t / dt dt. (11.24) The two terms here describe a process where the system remains in state a for the entire time, and a second process where the system transitions from a to b at time t and then back to a at time t. Graphically, this is shown in Figure 11.2. With the insight provided by these pictures, it is easy to write down the general result for a multi-level system:2 c(2) n (t) e\u2212i En t/ = \u03b4ni e\u2212i Ei t/ + 2 t 0 \u2212i t + \u2212 i e\u2212i En(t\u2212t )/ Hni t e\u2212i Ei t / dt t e\u2212i En(t\u2212t ) t e\u2212i Em(t \u2212t )/ Hnm 0 m \u00d7 Hm i 0 e\u2212i Ei t / dt dt. For n = i, this is represented by the diagram in Figure 11.3. The \ufb01rst-order term describes a direct transition from i to n, and the second-order term describes a process where the transition occurs via an intermediate (or \u201cvirtual\u201d) state m. (11.25) t = H bb 2 \u2217\u2217 Problem 11.5 Suppose you don\u2019t assume H aa = 0. (a) Find ca(t) and cb(t) in \ufb01rst-order perturbation theory, for the case ca(0) = 1, (t) c (b) There is a nicer way to handle this problem. Let = 1, to \ufb01rst order in \u02c6H. cb(0) = 0. Show that (1) c b (1) a (t) 2 + da \u2261 e i t 0 H aa(t )dt ca, db \u2261 e i 0 H t bb (t )dt cb. (11.26) 2 Perturbation theory for multi-level systems is treated in Problem 11.24. 408 CHAPTER 11 Quantum Dynamics n m n i i + Figure 11.3: Pictorial representation of Equation 11.25 for n = i. Show that \u02d9da = \u2212 i ei\u03c6 H abe\u2212i\u03c90t db; \u02d9db = \u2212 i e\u2212i\u03c6 H baei\u03c90t da, where \u03c6(t) \u2261 1 0 t t H aa t \u2212 H bb dt. (11.27) (11.28) So the equations for da and db are identical in structure to Equation 11.17 (with an extra factor ei\u03c6 tacked onto \u02c6H ). (c) Use the method in part (b) to obtain ca(t) and cb(t) in \ufb01rst-order perturbation theory, and compare your answer to (a). Comment on any discrepancies. Problem 11.6 Solve Equation 11.17 to second order in perturbation theory, for the general case ca(0) = a, cb(0) = b. Problem 11.7 Calculate ca(t) and cb(t), to second order, for the perturbation in Problem 11.3. Compare your answer with the exact result. Problem 11.8 Consider a perturbation to a two-level system with matrix elements H ab = H ba = \u221a \u03b1 \u03c0 \u03c4 e\u2212(t/\u03c4 )2, H aa = H bb = 0. where \u03c4 and \u03b1 are positive constants with the appropriate units. \u2217 \u2217\u2217 \u2217 (a) According to \ufb01rst-order perturbation theory, if the system starts off in the state ca = 1, cb = 0 at t = \u2212\u221e, what is the probability that it will be found in the state b at t = \u221e? (b) In the limit that \u03c4 \u2192 0, H ab from part (a) and compare the result of Problem 11.4. = \u03b1 \u03b4(t). Compute the \u03c4 \u2192 0 limit of your expression (c) Now consider the opposite extreme: \u03c90 \u03c4 1. What is the limit of your expression from part (a)? Comment: This is an example of the adiabatic theorem (Section 11.5.2). 11.1.3 Sinusoidal Perturbations Suppose the perturbation has sinusoidal time dependence: \u02c6H (r, t", ") = V (r) cos(\u03c9t), (11.29) 11.1 Two-Level Systems 409 so that where H ab = Vab cos(\u03c9t), Vab \u2261 \u03c8a |V | \u03c8b. (11.30) (11.31) (As before, I\u2019ll assume the diagonal matrix elements vanish, since this is almost always the case in practice.) To \ufb01rst order (from now on we\u2019ll work exclusively in \ufb01rst order, and I\u2019ll dispense with the superscripts) we have (Equation 11.21): t t \u03c9t ei\u03c90t # ei(\u03c90+\u03c9)t + ei(\u03c90\u2212\u03c9)t $ cos Vba 0 ei(\u03c90+\u03c9)t \u2212 1 \u03c90 + \u03c9 cb(t) \u2248 \u2212 i = \u2212 Vba 2 dt = \u2212 i Vba 2 + ei(\u03c90\u2212\u03c9)t \u2212 1 \u03c90 \u2212 \u03c9 That\u2019s the answer, but it\u2019s a little cumbersome to work with. Things simplify substantially if we restrict our attention to driving frequencies (\u03c9) that are very close to the transition frequency (\u03c90), so that the second term in the square brackets dominates; speci\ufb01cally, we assume: (11.32) dt 0. \u03c90 + \u03c9 |\u03c90 \u2212 \u03c9|. (11.33) This is not much of a limitation, since perturbations at other frequencies have a negligible probability of causing a transition anyway. Dropping the \ufb01rst term, we have $ # ei(\u03c90\u2212\u03c9)t/2 \u2212 e\u2212i(\u03c90\u2212\u03c9)t/2 cb(t) \u2248 \u2212 Vba 2 Vba = \u2212i ei(\u03c90\u2212\u03c9)t/2 \u03c90 \u2212 \u03c9 (\u03c90 \u2212 \u03c9) t/2 sin \u03c90 \u2212 \u03c9 ei(\u03c90\u2212\u03c9)t/2. (11.34) The transition probability\u2014the probability that a particle which started out in the state \u03c8a will be found, at time t, in the state \u03c8b\u2014is Pa\u2192b(t) = |cb(t)|2 \u2248 sin2 |Vab|2 2 (\u03c90 \u2212 \u03c9) t/2 (\u03c90 \u2212 \u03c9)2. (11.35) The most remarkable feature of this result is that, as a function of time, the transition probability oscillates sinusoidally (Figure 11.4). After rising to a maximum of |Vab|2 /2 (\u03c90 \u2212 \u03c9)2\u2014necessarily much less than 1, else the assumption that the perturbation is \u201csmall\u201d would be invalid\u2014it drops back down to zero! At times tn = 2n\u03c0/ |\u03c90 \u2212 \u03c9|, where n = 1, 2, 3,..., the particle is certain to be back in the lower state. If you want to maximize your chances of provoking a transition, you should not keep the perturbation on for a long period; you do better to turn it off after a time \u03c0/ |\u03c90 \u2212 \u03c9|, and hope to \u201ccatch\u201d the system in the upper state. In Problem 11.9 it is shown that this \u201c\ufb02opping\u201d is not an artifact of perturbation theory\u2014it occurs also in the exact solution, though the \ufb02opping frequency is modi\ufb01ed somewhat. As I noted earlier, the probability of a transition is greatest when the driving frequency is close to the \u201cnatural\u201d frequency, \u03c90.3 This is illustrated in Figure 11.5, where Pa\u2192b is plotted as a function of \u03c9. The peak has a height of (|Vab| t/2)2 and a width 4\u03c0/t; evidently it 3 For very small t, Pa\u2192b(t) is independent of \u03c9; it takes a couple of cycles for the system to \u201crealize\u201d that the perturbation is periodic. 410 CHAPTER 11 Quantum Dynamics P(t) 2\u03c0 \u2212 \u03c9| |\u03c9 0 4\u03c0 \u2212 \u03c9| 0 |\u03c9 6\u03c0 \u2212 \u03c9| 0 |\u03c9 2 |Vab \u03c9 ( 0 | \u2212 \u03c9 ) h t Figure 11.4: Transition probability as a function of time, for a sinusoidal perturbation (Equation 11.35). P(\u03c9) \u03c9 0 (\u03c9 0 \u2212 2\u03c0/t) (\u03c9 0 + 2\u03c0/t) \u03c9 Figure 11.5: Transition probability as a function of driving frequency (Equation 11.35). gets higher and", " narrower as time goes on. (Ostensibly, the maximum increases without limit. However, the perturbation assumption breaks down before it gets close to 1, so we can believe the result only for relatively small t. In Problem 11.9 you will see that the exact result never exceeds 1.) \u2217\u2217 Problem 11.9 The \ufb01rst term in Equation 11.32 comes from the ei\u03c9t /2 part of cos(\u03c9t), and the second from e\u2212i\u03c9t /2. Thus dropping the \ufb01rst term is formally equivalent to writing \u02c6H = (V /2) e\u2212i\u03c9t, which is to say, H ba = Vba 2 e\u2212i\u03c9t, H ab = Vab 2 ei\u03c9t. (11.36) (The latter is required to make the Hamiltonian matrix hermitian\u2014or, if you prefer, to pick out the dominant term in the formula analogous to Equation 11.32 for ca(t).) Rabi noticed that if you make this so-called rotating wave approximation at the beginning of the calculation, Equation 11.17 can be solved exactly, with no need for perturbation theory, and no assumption about the strength of the \ufb01eld. (a) Solve Equation 11.17 in the rotating wave approximation (Equation 11.36), for the usual initial conditions: ca(0) = 1, cb(0) = 0. Express your results (ca(t) and cb(t)) in terms of the Rabi \ufb02opping frequency, \u03c9r \u2261 1 2 (\u03c9 \u2212 \u03c90)2 + (|Vab| /)2. (11.37) (b) Determine the transition probability, Pa\u2192b(t), and show that it never exceeds 1. Con\ufb01rm that |ca(t)|2 + |cb(t)|2 = 1. 11.2 Emission and Absorption of Radiation 411 (c) Check that Pa\u2192b(t) reduces to the perturbation theory result (Equation 11.35) when the perturbation is \u201csmall,\u201d and state precisely what small means in this context, as a constraint on V. (d) At what time does the system \ufb01rst return to its initial state? 11.2 EMISSION AND ABSORPTION OF RADIATION 11.2.1 Electromagnetic Waves An electromagnetic wave (I\u2019ll refer to it as \u201clight\u201d, though it could be infrared, ultraviolet, microwave, x-ray, etc.; these differ only in their frequencies) consists of transverse (and mutually perpendicular) oscillating electric and magnetic \ufb01elds (Figure 11.6). An atom, in the presence of a passing light wave, responds primarily to the electric component. If the wavelength is long (compared to the size of the atom), we can ignore the spatial variation in the \ufb01eld;4 the atom, then, is exposed to a sinusoidally oscillating electric \ufb01eld E = E0 cos(\u03c9t) \u02c6k (11.38) (for the moment I\u2019ll assume the light is monochromatic, and polarized along the z direction). The perturbing Hamiltonian is5 H = \u2212q E0z cos(\u03c9t), where q is the charge of the electron.6 Evidently7 H ba = \u2212\u2118 E0 cos(\u03c9t), where \u2118 \u2261 q\u03c8b | z | \u03c8a. (11.39) (11.40) z Electric field x Magnetic field Figure 11.6: An electromagnetic wave. y Direction of propagation 4 For visible light \u03bb \u223c 5000 \u00c5, while the diameter of an atom is around 1 \u00c5, so this approximation is reasonable; but it would not be for x-rays. Problem 11.31 explores the effect of spatial variation of the \ufb01eld. 5 The energy of a charge q in a static \ufb01eld E is \u2212q E \u00b7 dr. You may well object to the use of an electrostatic formula for a manifestly time-dependent \ufb01eld. I am implicitly assuming that the period of oscillation is long compared to the time it takes the charge to move around (within the atom). 6 As usual, we assume the nucleus is heavy and stationary; it is the wave function of the electron that concerns us. 7 The letter \u2118 is supposed to remind you of electric dipole moment (for which, in electrodynamics, the letter p is customarily used\u2014in this context it is rendered as a squiggly \u2118 to avoid confusion with momentum). Actually, \u2118 is the off-diagonal matrix element of the z component of", " the dipole moment operator, qr. Because of its association with electric dipole moments, radiation governed by Equation 11.40 is called electric dipole radiation; it is overwhelmingly the dominant kind, at least in the visible region. See Problem 11.31 for generalizations and terminology. 412 CHAPTER 11 Quantum Dynamics Typically, \u03c8 is an even or odd function of z; in either case z |\u03c8|2 is odd, and integrates to zero (this is Laporte\u2019s rule, Section 6.4.3; for some examples see Problem 11.2). This licenses \u02c6H vanish. Thus the interaction of our usual assumption that the diagonal matrix elements of light with matter is governed by precisely the kind of oscillatory perturbation we studied in Section 11.1.3, with Vba = \u2212\u2118 E0. (11.41) 11.2.2 Absorption, Stimulated Emission, and Spontaneous Emission If an atom starts out in the \u201clower\u201d state \u03c8a, and you shine a polarized monochromatic beam of light on it, the probability of a transition to the \u201cupper\u201d state \u03c8b is given by Equation 11.35, which (in view of Equation 11.41) takes the form Pa\u2192b(t) = 2 sin2 |\u2118| E0 (\u03c90 \u2212 \u03c9) t/2 (\u03c90 \u2212 \u03c9)2. (11.42) In this process, the atom absorbs energy Eb \u2212 Ea = \u03c90 from the electromagnetic \ufb01eld, so it\u2019s called absorption. (Informally, we say that the atom has \u201cabsorbed a photon\u201d (Figure 11.7(a).) Technically, the word \u201cphoton\u201d belongs to quantum electrodynamics\u2014the quantum theory of the electromagnetic \ufb01eld\u2014whereas we are treating the \ufb01eld itself classically. But this language is convenient, as long as you don\u2019t read too much into it.) I could, of course, go back and run the whole derivation for a system that starts off in the upper state (ca(0) = 0, cb(0) = 1). Do it for yourself, if you like; it comes out exactly the same\u2014except that this time we\u2019re calculating Pb\u2192a = |ca(t)|2, the probability of a transition down to the lower level: Pb\u2192a(t) = 2 sin2 |\u2118| E0 (\u03c90 \u2212 \u03c9) t/2 (\u03c90 \u2212 \u03c9)2. (11.43) (It has to come out this way\u2014all we\u2019re doing is switching a \u2194 b, which substitutes \u2212\u03c90 for \u03c90. When we get to Equation 11.32 we now keep the \ufb01rst term, with \u2212\u03c90 + \u03c9 in the denominator, and the rest is the same as before.) But when you stop to think of it, this is an absolutely astonishing result: If the particle is in the upper state, and you shine light on it, it can make a transition to the lower state, and in fact the probability of such a transition is exactly the same as for a transition upward from the lower state. This process, which was \ufb01rst predicted by Einstein, is called stimulated emission. In the case of stimulated emission the electromagnetic \ufb01eld gains energy \u03c90 from the atom; we say that one photon went in and two photons came out\u2014the original one that caused the transition plus another one from the transition itself (Figure 11.7(b)). This raises the possibility of ampli\ufb01cation, for if I had a bottle of atoms, all in the upper state, and triggered it with a single incident photon, a chain reaction would occur, with the \ufb01rst photon producing two, these two producing four, and so on. We\u2019d have an enormous number of photons coming out, all with the same frequency and at virtually the same instant. This is the principle behind the laser (light b a b a b a (a) Absorption (b) Stimulated emission (c) Spontaneous emission Figure 11.7: Three ways in which light interacts with atoms: (a) absorption, (b) stimulated emission, (c) spontaneous emission. 11.2 Emission and Absorption of Radiation 413 ampli\ufb01cation by stimulated emission of radiation). Note that it is essential (for laser action) to get a majority of the atoms into the upper state (a so-called population inversion), because absorption (which costs one photon) competes with stimulated emission (which creates one); if you started with an even mixture of the two states, you", "\u2019d get no ampli\ufb01cation at all. There is a third mechanism (in addition to absorption and stimulated emission) by which radiation interacts with matter; it is called spontaneous emission. Here an atom in the excited state makes a transition downward, with the release of a photon, but without any applied electromagnetic \ufb01eld to initiate the process (Figure 11.7(c)). This is the mechanism that accounts for the normal decay of an atomic excited state. At \ufb01rst sight it is far from clear why spontaneous emission should occur at all. If the atom is in a stationary state (albeit an excited one), and there is no external perturbation, it should just sit there forever. And so it would, if it were really free of all external perturbations. However, in quantum electrodynamics the \ufb01elds are nonzero even in the ground state\u2014just as the harmonic oscillator (for example) has nonzero energy (to wit: \u03c9/2) in its ground state. You can turn out all the lights, and cool the room down to absolute zero, but there is still some electromagnetic radiation present, and it is this \u201czero point\u201d radiation that serves to catalyze spontaneous emission. When you come right down to it, there is really no such thing as truly spontaneous emission; it\u2019s all stimulated emission. The only distinction to be made is whether the \ufb01eld that does the stimulating is one that you put there, or one that God put there. In this sense it is exactly the reverse of the classical radiative process, in which it\u2019s all spontaneous, and there is no such thing as stimulated emission. Quantum electrodynamics is beyond the scope of this book,8 but there is a lovely argument, due to Einstein,9 which interrelates the three processes (absorption, stimulated emission, and spontaneous emission). Einstein did not identify the mechanism responsible for spontaneous emission (perturbation by the ground-state electromagnetic \ufb01eld), but his results nevertheless enable us to calculate the spontaneous emission rate, and from that the natural lifetime of an excited atomic state.10 Before we turn to that, however, we need to consider the response of an atom to non-monochromatic, unpolarized, incoherent electromagnetic waves coming in from all directions\u2014such as it would encounter, for instance, if it were immersed in thermal radiation. 11.2.3 Incoherent Perturbations The energy density in an electromagnetic wave is11 u = 0 2, E 2 0 (11.44) 8 For an accessible treatment see Rodney Loudon, The Quantum Theory of Light, 2nd edn (Clarendon Press, Oxford, 1983). 9 Einstein\u2019s paper was published in 1917, well before the Schr\u00f6dinger equation. Quantum electrodynamics comes into the argument via the Planck blackbody formula, which dates from 1900. 10 For an alternative derivation using \u201cseat-of-the-pants\u201d quantum electrodynamics, see Problem 11.11. 11 David J. Grif\ufb01ths, Introduction to Electrodynamics, 4th edn, (Cambridge University Press, Cambridge, UK, 2017), Section 9.2.3. In general, the energy per unit volume in electromagnetic \ufb01elds is u = (0/2) E 2 + (1/2\u03bc0) B2. For electromagnetic waves, the electric and magnetic contributions are equal, so and the average over a full cycle is (0/2 cos2(\u03c9t), 0, since the average of cos2 (or sin2) is 1/2. 414 CHAPTER 11 Quantum Dynamics where E0 is (as before) the amplitude of the electric \ufb01eld. So the transition probability (Equation 11.43) is (not surprisingly) proportional to the energy density of the \ufb01elds: Pb\u2192a(t) = 2u 02 |\u2118|2 sin2 (\u03c90 \u2212 \u03c9) t/2 (\u03c90 \u2212 \u03c9)2. (11.45) But this is for a monochromatic wave, at a single frequency \u03c9. In many applications the system is exposed to electromagnetic waves at a whole range of frequencies; in that case u \u2192 \u03c1(\u03c9)d\u03c9, where \u03c1(\u03c9)d\u03c9 is the energy density in the frequency range d\u03c9, and the net transition probability takes the form of an integral:12 4 \u221e Pb\u2192a(t) = 2 02 |\u2118|2 \u03c1(\u03c9) 0 sin2 (\u03c90 \u2212 \u03c9) t/2 (\u03c90 \u2212 \u03c9)2 d\u03c9. (11.46) The term in curly brackets is sharply peaked about \u03c90 (Figure 11.5), whereas \ufffd", "\ufffd(\u03c9) is ordinarily quite broad, so we may as well replace \u03c1(\u03c9) by \u03c1(\u03c90), and take it outside the integral: Pb\u2192a(t) \u2248 2 |\u2118|2 02 \u03c1 (\u03c90) \u221e sin2 (\u03c90 \u2212 \u03c9) t/2 (\u03c90 \u2212 \u03c9)2 d\u03c9. (11.47) 0 Changing variables to x \u2261 (\u03c90 \u2212 \u03c9) t/2, extending the limits of integration to x = \u00b1\u221e (since the integrand is essentially zero out there anyway), and looking up the de\ufb01nite integral we \ufb01nd \u221e \u2212\u221e sin2 x x 2 d x = \u03c0, Pb\u2192a(t) \u2248 \u03c0 |\u2118|2 02 \u03c1(\u03c90)t. (11.48) (11.49) This time the transition probability is proportional to t. The bizarre \u201c\ufb02opping\u201d phenomenon characteristic of a monochromatic perturbation gets \u201cwashed out\u201d when we hit the system with an incoherent spread of frequencies. In particular, the transition rate (R \u2261 d P/dt) is now a constant: Rb\u2192a = |\u2118|2 \u03c1(\u03c90). (11.50) \u03c0 02 Up to now, we have assumed that the perturbing wave is coming in along the y direction (Figure 11.6), and polarized in the z direction. But we are interested in the case of an atom bathed in radiation coming from all directions, and with all possible polarizations; the energy in the \ufb01elds (\u03c1(\u03c9)) is shared equally among these different modes. What we need, in place of |\u2118|2, is the average of |\u2118 \u00b7 \u02c6n|2, where \u2118 \u2261 q\u03c8b | r | \u03c8a (11.51) (generalizing Equation 11.40), and the average is over all polarizations and all incident directions. 12 Equation 11.46 assumes that the perturbations at different frequencies are independent, so that the total transition probability is a sum of the individual probabilities. If the different components are coherent (phase-correlated), then we should add amplitudes (cb(t)), not probabilities (|cb(t)|2), and there will be cross-terms. For the applications we will consider the perturbations are always incoherent. 11.2 Emission and Absorption of Radiation 415 z n\u02c6 \u03b8 k\u02c6 \u03c6 y x Figure 11.8: Axes for the averaging of \u2118 \u00b7 \u02c6n 2. \u02c6k The averaging can be carried out as follows: Choose spherical coordinates such that the is along z, and the vector \u2118 direction of propagation de\ufb01nes the spherical angles \u03b8 and \u03c6 (Figure 11.8).13 (Actually, \u2118 is \ufb01xed, here, and we\u2019re averaging over all \u02c6k and \u02c6n consistent with \u02c6k \u22a5 \u02c6n\u2014which is to say, over all \u03b8 and \u03c6. But we might as well integrate over all directions of \u2118, keeping \u02c6k and \u02c6n \ufb01xed\u2014it amounts to the same thing.) Then is along x, the polarization \u02c6n \u2118 \u00b7 \u02c6n = \u2118 cos \u03b8, (11.52) and \u2118 \u00b7 \u02c6n 2 ave = 1 4\u03c0 |\u2118|2 4 \u03c0 = |\u2118|2 cos2 \u03b8 sin \u03b8 d\u03b8 d\u03c6 \u2212 cos3 \u03b8 3 (2\u03c0 ) = 1 3 \u03c0 0 |\u2118|2. (11.53) Conclusion: The transition rate for stimulated emission from state b to state a, under the in\ufb02uence of incoherent, unpolarized light incident from all directions, is Rb\u2192a = \u03c0 302 |\u2118|2 \u03c1(\u03c90), (11.54) where \u2118 is the matrix element of the electric dipole moment between the two states (Equation 11.51), and \u03c1(\u03c90) is the energy density in the \ufb01elds, per unit frequency, evaluated at \u03c90 = (Eb \u2212 Ea) /. 13 I\u2019ll treat \u2118 as though it were real, even though in general it will be complex. Since \u2118 \u00b7 \u02c6n 2 = Re(\u2118) \u00b7 \u02c6n + iIm(\u2118) \u00b7 \u02c6n 2 = Re(\u2118", ") \u00b7 \u02c6n 2 + Im(\u2118) \u00b7 \u02c6n 2 we can do the whole calculation for the real and imaginary parts separately, and simply add the results. In Equation 11.54 the absolute value signs denote both the vector magnitude and the complex amplitude: |\u2118|2 = |\u2118x |2 + \u2118y 2 + |\u2118z |2. 416 CHAPTER 11 Quantum Dynamics 11.3 SPONTANEOUS EMISSION 11.3.1 Einstein\u2019s A and B Coef\ufb01cients Picture a container of atoms, Na of them in the lower state (\u03c8a), and Nb of them in the upper state (\u03c8b). Let A be the spontaneous emission rate,14 so that the number of particles leaving the upper state by this process, per unit time, is Nb A.15 The transition rate for stimulated emission, as we have seen (Equation 11.54), is proportional to the energy density of the electromagnetic \ufb01eld: Bba\u03c1(\u03c90), where Bba = \u03c0 |\u2118|2 /302; the number of particles leaving the upper state by this mechanism, per unit time, is Nb Bba\u03c1(\u03c90). The absorption rate is likewise proportional to \u03c1(\u03c90)\u2014call it Bab\u03c1(\u03c90); the number of particles per unit time joining the upper level is therefore Na Bab\u03c1(\u03c90). All told, then, d Nb dt = \u2212Nb A \u2212 Nb Bba\u03c1(\u03c90) + Na Bab\u03c1(\u03c90). (11.55) Suppose these atoms are in thermal equilibrium with the ambient \ufb01eld, so that the number of particles in each level is constant. In that case d Nb/dt = 0, and it follows that \u03c1(\u03c90) = A (Na/Nb) Bab \u2212 Bba. (11.56) On the other hand, we know from statistical mechanics16 that the number of particles with energy E, in thermal equilibrium at temperature T, is proportional to the Boltzmann factor, exp (\u2212E/k B T ), so and hence Na Nb = e\u2212Ea /k B T e\u2212Eb/k B T = e\u03c90/k B T, \u03c1(\u03c90) = A e\u03c90/k B T Bab \u2212 Bba. But Planck\u2019s blackbody formula17 tells us the energy density of thermal radiation: \u03c1(\u03c9) = \u03c0 2c3 \u03c93 e\u03c9/k B T \u2212 1 ; comparing the two expressions, we conclude that and Bab = Bba A = \u03c93 0 \u03c0 2c3 Bba. (11.57) (11.58) (11.59) (11.60) (11.61) 14 Normally I\u2019d use R for a transition rate, but out of deference to der Alte everyone follows Einstein\u2019s notation in this context. 15 Assume that Na and Nb are very large, so we can treat them as continuous functions of time and ignore statistical \ufb02uctuations. 16 See, for example, Daniel Schroeder, An Introduction to Thermal Physics (Pearson, Upper Saddle River, NJ, 2000), Section 6.1. 17 Schroeder, footnote 16, Section 7.4. 11.3 Spontaneous Emission 417 Equation 11.60 con\ufb01rms what we already knew: the transition rate for stimulated emission is the same as for absorption. But it was an astonishing result in 1917\u2014indeed, Einstein was forced to \u201cinvent\u201d stimulated emission in order to reproduce Planck\u2019s formula. Our present attention, however, focuses on Equation 11.61, for this tells us the spontaneous emission rate (A)\u2014which is what we are looking for\u2014in terms of the stimulated emission rate (Bba\u03c1(\u03c90))\u2014which we already know. From Equation 11.54 we read off \u03c0 302 and it follows that the spontaneous emission rate is Bba = |\u2118|2, A = |\u2118|2 \u03c93 0 3\u03c0 0c3. (11.62) (11.63) Problem 11.10 As a mechanism for downward transitions, spontaneous emission competes with thermally stimulated emission (stimulated emission for which blackbody radiation is the source). Show that at room temperature (T = 300 K) thermal stimulation dominates for frequencies well below 5 \u00d7 1012 Hz, whereas spontaneous emission dominates for frequencies well above 5 \u00d7 1012 Hz. Which mechanism dominates for visible light? Problem 11.11 You could derive the spontaneous emission rate (Equation 11.63) without the detour through Einstein\u2019s A and B coef\ufb01cients if you knew the ground state energy density of the", " electromagnetic \ufb01eld, \u03c10(\u03c9), for then it would simply be a case of stimulated emission (Equation 11.54). To do this honestly would require quantum electrodynamics, but if you are prepared to believe that the ground state consists of one photon in each classical mode, then the derivation is fairly simple: (a) To obtain the classical modes, consider an empty cubical box, of side l, with one corner at the origin. Electromagnetic \ufb01elds (in vacuum) satisfy the classical wave equation18 1 c2 \u2202 2 \u2202t 2 \u2212 \u22072 f (x, y, z, t) = 0, where f stands for any component of E or of B. Show that separation of variables, and the imposition of the boundary condition f = 0 on all six surfaces yields the standing wave patterns fnx,n y,nz = A cos (\u03c9t) sin nx \u03c0 l x sin n y\u03c0 l y sin, z nz\u03c0 l with \u03c9 = \u03c0 c l n2 x + n2 y + n2 z. There are two modes for each triplet of positive integers corresponding to the two polarization states. nx, n y, nz = 1, 2, 3,..., 18 Grif\ufb01ths, footnote 11, Section 9.2.1. 418 CHAPTER 11 Quantum Dynamics (b) The energy of a photon is E = h\u03bd = \u03c9 (Equation 4.92), so the energy in the mode nx, n y, nz is \u03c0 c l What, then, is the total energy per unit volume in the frequency range d\u03c9, if each mode gets one photon? Express your answer in the form Enx,n y,nz + n2 y + n2 z = 2 n2 x. 1 l3 d E = \u03c10(\u03c9) d\u03c9 and read off \u03c10(\u03c9). Hint: refer to Figure 5.3. (c) Use your result, together with Equation 11.54, to obtain the spontaneous emission rate. Compare Equation 11.63. 11.3.2 The Lifetime of an Excited State Equation 11.63 is our fundamental result; it gives the transition rate for spontaneous emission. Suppose, now, that you have somehow pumped a large number of atoms into the excited state. As a result of spontaneous emission, this number will decrease as time goes on; speci\ufb01cally, in a time interval dt you will lose a fraction A dt of them: d Nb = \u2212ANb dt, (11.64) (assuming there is no mechanism to replenish the supply).19 Solving for Nb(t), we \ufb01nd: Nb(t) = Nb(0)e\u2212At ; (11.65) evidently the number remaining in the excited state decreases exponentially, with a time constant \u03c4 = 1 A. (11.66) We call this the lifetime of the state\u2014technically, it is the time it takes for Nb(t) to reach 1/e \u2248 0.368 of its initial value. I have assumed all along that there are only two states for the system, but this was just for notational simplicity\u2014the spontaneous emission formula (Equation 11.63) gives the transition rate for \u03c8b \u2192 \u03c8a regardless of what other states may be accessible (see Problem 11.24). Typically, an excited atom has many different decay modes (that is: \u03c8b can decay to a large number of different lower-energy states, \u03c8a1, \u03c8a2, \u03c8a3,... ). In that case the transition rates add, and the net lifetime is \u03c4 = 1 A1 + A2 + A3 + \u00b7 \u00b7 \u00b7. (11.67) 19 This situation is not to be confused with the case of thermal equilibrium, which we considered in the previous section. We assume here that the atoms have been lifted out of equilibrium, and are in the process of cascading back down to their equilibrium levels. 11.3 Spontaneous Emission 419 Example 11.1 Suppose a charge q is attached to a spring and constrained to oscillate along the x axis. Say it starts out in the state |n (Equation 2.68), and decays by spontaneous emission to state |n. From Equation 11.51 we have \u2118 = q n x n \u02c6\u0131. You calculated the matrix elements of x back in Problem 3.39: n x n = \u221a n \u03b4n,n\u22121 + \u221a n \u03b4n,n\u22121, 2m\u03c9 where \u03c9 is the natural frequency of the oscillator (I no longer need this letter for the frequency of the stimulating radiation). But we\u2019re talking about emission,", " so n must be lower than n; for our purposes, then, n 2m\u03c9 Evidently transitions occur only to states one step lower on the \u201cladder\u201d, and the frequency of the photon emitted is \u03b4n,n\u22121 \u02c6\u0131. \u2118 = q (11.68) \u03c90 = En \u2212 En = (n + 1/2) \u03c9 \u2212 n + 1/2 \u03c9 n \u2212 n = \u03c9 = \u03c9. (11.69) Not surprisingly, the system radiates at the classical oscillator frequency. The transition rate (Equation 11.63) is and the lifetime of the nth stationary state is A = nq2\u03c92 6\u03c0 0mc3, \u03c4n = 6\u03c0 0mc3 nq2\u03c92. (11.70) (11.71) Meanwhile, each radiated photon carries an energy \u03c9, so the power radiated is A\u03c9: P = q2\u03c92 6\u03c0 0mc3 (n\u03c9), or, since the energy of an oscillator in the nth state is E = (n + 1/2) \u03c9, P = q2\u03c92 6\u03c0 0mc3 This is the average power radiated by a quantum oscillator with (initial) energy E. E \u2212 1 2 \u03c9. (11.72) For comparison, let\u2019s determine the average power radiated by a classical oscillator with the same energy. According to classical electrodynamics, the power radiated by an accelerating charge q is given by the Larmor formula:20 P = q2a2 6\u03c0 0c3. (11.73) 20 See, for example, Grif\ufb01ths, footnote 11, Section 11.2.1. 420 CHAPTER 11 Quantum Dynamics For a harmonic oscillator with amplitude x0, x(t) = x0 cos(\u03c9t), and the acceleration is a = \u2212x0\u03c92 cos(\u03c9t). Averaging over a full cycle, then, \u03c94 q2x 2 0 12\u03c0 0c3 But the energy of the oscillator is E = (1/2) m\u03c92x 2 P =. 0, so x 2 0 = 2E/m\u03c92, and hence P = q2\u03c92 6\u03c0 0mc3 This is the average power radiated by a classical oscillator with energy E. In the classical limit ( \u2192 0) the classical and quantum formulas agree;21 however, the quantum formula (Equation 11.72) protects the ground state: If E = (1/2) \u03c9 the oscillator does not radiate. (11.74) E. Problem 11.12 The half-life of an excited state is the time it would take for half the atoms in a large sample to make a transition. Find the relation between t1/2 and \u03c4 (the \u201clifetime\u201d of the state). t1/2 \u2217 Problem 11.13 Calculate the lifetime (in seconds) for each of the four n = 2 states of hydrogen. Hint: You\u2019ll need to evaluate matrix elements of the form \u03c8100 |x| \u03c8200, \u03c8100 |y| \u03c8211, and so on. Remember that x = r sin \u03b8 cos \u03c6, y = r sin \u03b8 sin \u03c6, and z = r cos \u03b8. Most of these integrals are zero, so inspect them closely before you start calculating. Answer: 1.60 \u00d7 10\u22129 seconds for all except \u03c8200, which is in\ufb01nite. 11.3.3 Selection Rules The calculation of spontaneous emission rates has been reduced to a matter of evaluating matrix elements of the form \u03c8b | r | \u03c8a. As you will have discovered if you worked Problem 11.13, (if you didn\u2019t, go back right now and do so!) these quantities are very often zero, and it would be helpful to know in advance when this is going to happen, so we don\u2019t waste a lot of time evaluating unnecessary integrals. Suppose we are interested in systems like hydrogen, for which the Hamiltonian is spherically symmetrical. In that case we can specify the states with the usual quantum numbers n,, and m, and the matrix elements are nm r nm. Now, r is a vector operator, and we can invoke the results of Chapter 6 to obtain the selection rules22 \u2261 \u2212 = \u00b11, m \u2261 m \u2212 m = 0 or \u00b1 1. (11.75) 21 This is an example of Bohr\u2019s Correspondence Principle. In fact, if we express P in terms of the energy above the ground state, the two formulas are identical. 22 See Equation 6.62 (Equation 6.26 eliminates = 0), or derive them from scratch using Problems 11.", "14 and 11.15. 11.3 Spontaneous Emission 421 Figure 11.9: Allowed decays for the \ufb01rst four Bohr levels in hydrogen. These conditions follow from symmetry alone. If they are not met, then the matrix element is zero, and the transition is said to be forbidden. Moreover, it follows from Equations 6.56\u20136.58 that \u23a7 \u23aa\u23a8 \u23aa\u23a9 if m = m, if m = m \u00b1 1, then then nm x nm nm x nm nm and = = \u00b1i nm z y nm nm = 0. nm nm y = 0, (11.76) So it is never necessary to compute the matrix elements of both x and y; you can always get one from the other. Evidently not all transitions to lower-energy states can proceed by electric dipole radiation; most are forbidden by the selection rules. The scheme of allowed transitions for the \ufb01rst four Bohr levels in hydrogen is shown in Figure 11.9. Notice that the 2S state (\u03c8200) is \u201cstuck\u201d: it cannot decay, because there is no lower-energy state with = 1. It is called a metastable state, and its lifetime is indeed much longer than that of, for example, the 2P states (\u03c8211, \u03c8210, and \u03c821\u22121). Metastable states do eventually decay, by collisions, or by \u201cforbidden\u201d transitions (Problem 11.31), or by multiphoton emission. Problem 11.14 From the commutators of L z with x, y, and z (Equation 4.122): L z, x = iy, L z, y = \u2212ix, L z, z = 0, (11.77) obtain the selection rule for m and Equation 11.76. Hint: Sandwich each commutator between nm| and |nm. \u2217\u2217 Problem 11.15 Obtain the selection rule for as follows: (a) Derive the commutation relation # # $$ L 2, L 2, r = 22 rL 2 + L 2r. (11.78) Hint: First show that # $ L 2, z = 2i x L y \u2212 y L x \u2212 iz. Use this, and (in the \ufb01nal step) the fact that r \u00b7 L = r \u00b7 (r \u00d7 p) = 0, to demonstrate that # # $$ L 2, L 2, z = 22 z L 2 + L 2z. The generalization from z to r is trivial. 422 CHAPTER 11 Quantum Dynamics (b) Sandwich this commutator between nm| and |nm, and work out the implica- tions. \u2217\u2217 Problem 11.16 An electron in the n = 3, = 0, m = 0 state of hydrogen decays by a sequence of (electric dipole) transitions to the ground state. (a) What decay routes are open to it? Specify them in the following way: |300 \u2192 |nm \u2192 \u2192 \u00b7 \u00b7 \u00b7 \u2192 |100. nm (b) If you had a bottle full of atoms in this state, what fraction of them would decay via each route? (c) What is the lifetime of this state? Hint: Once it\u2019s made the \ufb01rst transition, it\u2019s no longer in the state |300, so only the \ufb01rst step in each sequence is relevant in computing the lifetime. 11.4 FERMI\u2019S GOLDEN RULE In the previous sections we considered transitions between two discrete energy states, such as two bound states of an atom. We saw that such a transition was most likely when the \ufb01nal energy satis\ufb01ed the resonance condition: E f = Ei + \u03c9, where \u03c9 is the frequency associated with the perturbation. I now want to look at the case where E f falls in a continuum of states (Figure 11.10). To stick close to the example of Section 11.2, if the radiation is energetic enough it can ionize the atom\u2014the photoelectric effect\u2014exciting the electron from a bound state into the continuum of scattering states. We can\u2019t talk about a transition to a precise state in that continuum (any more than we can talk about someone being precisely 16 years old), but we can compute the probability that the system makes a transition to a state with an energy in some \ufb01nite range E about E f. That is given by the integral of Equation 11.35 over all the \ufb01nal states: 4 E f +E/2 E f \u2212E/2 |Vin|2 2 sin2 (\u03c90 \u2212 \u03c9) t/2 (\u03c90 \u2212 \u03c9)2 \u03c1(En)", " d En, (11.79a) (b) Figure 11.10: A transition (a) between two discrete states and (b) between a discrete state and a continuum of states. 11.4 Fermi\u2019s Golden Rule 423 where \u03c90 = (En \u2212 Ei ) /. The quantity \u03c1(E) d E is the number of states with energy between E and E + d E; \u03c1(E) is called the density of states, and I\u2019ll show you how it\u2019s calculated in Example 11.2. At short times, Equation 11.79 leads to a transition probability proportional to t 2, just as for a transition between discrete states. On the other hand, at long times the quantity in curly brackets in Equation 11.79 is sharply peaked: as a function of En its maximum occurs at E f = Ei + \u03c9 and the central peak has a width of 4\u03c0 /t. For suf\ufb01ciently large t, we can therefore approximate Equation 11.79 as23 2 Vi f 2 \u03c1 E f P = \u221e sin2 \u2212\u221e (\u03c90 \u2212 \u03c9) t/2 (\u03c90 \u2212 \u03c9)2 d En. The remaining integral was already evaluated in Section 11.2.3: P = 2 \u03c0 Vi f 2 2 \u03c1 E f t. (11.80) The oscillatory behavior of P has again been \u201cwashed out,\u201d giving a constant transition rate:24 R = 2 \u03c0 Vi f 2 2 \u03c1 E f. (11.81) Equation 11.81 is known as Fermi\u2019s Golden Rule.25 Apart from the factor of 2\u03c0/, it says that the transition rate is the square of the matrix element (this encapsulates all the relevant information about the dynamics of the process) times the density of states (how many \ufb01nal states are accessible, given the energy supplied by the perturbation\u2014the more roads are open, the faster the traf\ufb01c will \ufb02ow). It makes sense. 23 This is the same set of approximations we made in Equations 11.46\u201311.48. 24 In deriving Equation 11.35, our perturbation was \u02c6H = V cos(\u03c9 t) \u2192 V 2 \u2212i \u03c9 t e since we dropped the other (off-resonance) exponential. That is the source of the two inside the absolute value in Equation 11.81. Fermi\u2019s Golden rule can also be applied to a constant perturbation, \u02c6H = \u02c6V, if we set \u03c9 = 0 drop the 2: R = 2 \u03c0 Vi f 2 \u03c1 E f. 25 It is actually due to Dirac, but Fermi is the one who gave it the memorable name. See T. Visser, Am. J. Phys. 77, 487 (2009) for the history. Fermi\u2019s Golden Rule doesn\u2019t just apply to transitions to a continuum of states. For instance, Equation 11.54 can be considered an example. In that case, we integrated over a continuous range of perturbation frequencies\u2014not a continuum of \ufb01nal states\u2014but the end result is the same. 424 CHAPTER 11 Quantum Dynamics Example 11.2 Use Fermi\u2019s Golden Rule to obtain the differential scattering cross-section for a particle of mass m and incident wave vector k scattering from a potential V (r) (Figure 11.11). k detector k\u2032 V = 0 \u2044 Figure 11.11: A particle with incident wave vector k is scattered into a state with wave vector k. Solution: We take our initial and \ufb01nal states to be plane waves: \u03c8i = 1\u221a l3 ei k\u00b7r and \u03c8 f = 1\u221a l3 ei k\u00b7r. (11.82) Here I\u2019ve used a technique called box normalization; I place the whole setup inside a box of length l on a side. This makes the free-particle states normalizable and countable. Formally, we want the limit l \u2192 \u221e; in practice l will drop out of our \ufb01nal expression. Using periodic boundary conditions,26 the allowed values of k are k = 2 \u03c0 l nx \u02c6\u0131 + n y \u02c6j + nz (11.83) \u02c6k for integers nx, n y, and nz. Our perturbation is the scattering potential, \u02c6H = V (r), and the relevant matrix element is V f i = \u03c8 \u2217 f (r ) V (r) \u03c8i (r) d3r = 1 l3 ei(k\u2212", "k)\u00b7r V (r) d3r. (11.84) We need to determine the density of states. In a scattering experiment we measure the number of particles scatterred into a solid angle d. We want to count the number of states with energies between E and E + d E, with wave vectors k lying inside d. In k space these states occupy a section of a spherical shell of radius k and thickness dk that subtends a solid angle d; it has a volume k2 dk d and contains a number of states27 \u03c1(E) d E = k2 dk d (2 \u03c0/l)3 3 = l 2 \u03c0 k2 dk d E d E d. 26 Periodic boundary conditions are discussed in Problem 5.39. In the present context we use periodic boundary conditions\u2014as opposed to impenetrable walls\u2014because they admit traveling-wave solutions. 27 Each state in k-space \u201coccupies\u201d a volume of(2 \u03c0/l)3, as shown in Problem 5.39. 11.4 Fermi\u2019s Golden Rule 425 Since E = 2 k2/2 m this gives \u03c1(E) = \u221a 3 l 2 \u03c0 2 m3 E 3 d. (11.85) From Fermi\u2019s Golden Rule, the rate at which particles are scattered into the solid angle d is28 Ri\u2192d = 2 \u03c0 1 l6 ei(k\u2212k)\u00b7r V (r) d3r 2 3 l 2 \u03c0 2 m3 E f 3 d. This is closely related to the differential scattering cross section: d\u03c3 d = Ri\u2192d Ji d (11.86) where Ji is the \ufb02ux (or probability current) of incident particles. For an incident wave of the form \u03c8i = A ei k\u00b7r, the probability current is (Equation 4.220). and Ji = |A|2 v = 1 l3 k m d\u03c3 d = \u2212 m 2 \u03c0 2 ei(k\u2212k)\u00b7r V (r) d3r 2. (11.87) (11.88) This is exactly what we got from the \ufb01rst Born approximation (Equation 10.79). \u2217\u2217\u2217 Problem 11.17 In the photoelectric effect, light can ionize an atom if its energy ( \u03c9) exceeds the binding energy of the electron. Consider the photoelectric effect for the ground state of hydrogen, where the electron is kicked out with momentum k. The initial state of the electron is \u03c80(r ) (Equation 4.80) and its \ufb01nal state is29 as in Example 11.2. \u03c8 f = 1\u221a l3 ei k\u00b7r, (a) For light polarized along the z direction, use Fermi\u2019s Golden Rule to compute the rate at which electrons are ejected into the solid angle d in the dipole approximation.30 Answer : Ri\u2192d = 256 0 E 2 0 a3 (k a)3 1 +(k a)2 6 cos2 \u03b8 d. 28 See footnote 24. 29 This is an approximation; we really should be using a scattering state of hydrogen. For an extended discussion of the photoelectric effect, including comparison to experiment and the validity of this approximation, see W. Heitler, The Quantum Theory of Radiation, 3rd edn, Oxford University Press, London (1954), Section 21. 30 The result here is too large by a factor of four; correcting this requires a more careful derivation of the matrix element for radiative transitions (see Problem 11.30). Only the overall factor is affected though; the more interesting features (the dependence on k and \u03b8 ) are correct. 426 CHAPTER 11 Quantum Dynamics Hint:To evaluate the matrix element, use the following trick. Write z ei k\u00b7r = \u2212i d dkz ei k\u00b7r, pull d/dkz outside the integral, and what remains is straightforward to compute. (b) The photoelectric cross section is de\ufb01ned as \u03c3 (k) = Ri\u2192all \u03c9 0 E 2 0 c 1 2 where the quantity in the numerator is the rate at which energy is absorbed ( \u03c9 = 2k2 \u2212 E1 per photoelectron) and the quantity in the denominator is the intensity 2m of the incident light. Integrate your result from (a) over all angles to obtain Ri\u2192all, and compute the photoelectric cross section. (c) Obtain a numerical value for the photoelectric cross section for ultraviolet light of wavelength 220 \u00c5 (n.b. this is the wavelength of the incident light, not the scattered electron). Express your answer in mega-barns Mb = 10\u221222 m2. 11.5 THE AD", "IABATIC APPROXIMATION 11.5.1 Adiabatic Processes Imagine a perfect pendulum, with no friction or air resistance, oscillating back and forth in a vertical plane. If you grab the support and shake it in a jerky manner the bob will swing around chaotically. But if you very gently move the support (Figure 11.12), the pendulum will continue to swing in a nice smooth way, in the same plane (or one parallel to it), with the same amplitude. This gradual change of the external conditions de\ufb01nes an adiabatic process. Notice that there are two characteristic times involved: Ti, the \u201cinternal\u201d time, representing the motion of the system itself (in this case the period of the pendulum\u2019s oscillations), and Te, the \u201cexternal\u201d time, over which the parameters of the system change appreciably (if the pendulum were mounted on a rotating platform, for example, Te would be the period of the platform\u2019s motion). An adiabatic process is one for which Te Ti (the pendulum executes many oscillations before the platform has moved appreciably).31 What if I took this pendulum up to the North Pole, and set it swinging\u2014say, in the direction of Portland (Figure 11.13). For the moment, pretend the earth is not rotating. Very gently (that is, adiabatically), I carry it down the longitude line passing through Portland, to the equator. At this point it is swinging north-south. Now I carry it (still swinging north\u2013south) part way around the equator. And \ufb01nally, I take it back up to the North Pole, along the new longitude line. The pendulum will no longer be swinging in the same plane as it was when I set out\u2014 indeed, the new plane makes an angle with the old one, where is the angle between the southbound and the northbound longitude lines. More generally, if you transport the pendulum around a closed loop on the surface of the earth, the angular deviation (between the initial plane of the swing and the \ufb01nal plane) is equal to the solid angle subtended by the path with respect to the center of the sphere, as you can prove for yourself if you are interested. Incidentally, the Foucault pendulum is an example of precisely this sort of adiabatic transport around a closed loop on a sphere\u2014only this time instead of me carrying the pendulum 31 For an interesting discussion of classical adiabatic processes, see Frank S. Crawford, Am. J. Phys. 58, 337 (1990). 11.5 The Adiabatic Approximation 427 Figure 11.12: Adiabatic motion: If the case is transported very gradually, the pendulum inside keeps swinging with the same amplitude, in a plane parallel to the original one. Pendulum \u0398 Portland Equator Figure 11.13: Itinerary for adiabatic transport of a pendulum on the surface of the earth. around, I let the rotation of the earth do the job. The solid angle subtended by a latitude line \u03b80 (Figure 11.14) is = sin \u03b8 d\u03b8 d\u03c6 = 2\u03c0 (\u2212 cos \u03b8 ) \u03b80 0 = 2\u03c0 (1 \u2212 cos \u03b80). (11.89) Relative to the earth (which has meanwhile turned through an angle of 2\u03c0 ), the daily precession of the Foucault pendulum is 2\u03c0 cos \u03b80\u2014a result that is ordinarily obtained by appeal to Coriolis forces in the rotating reference frame,32 but is seen in this context to admit a purely geometrical interpretation. 32 See, for example, Jerry B. Marion and Stephen T. Thornton, Classical Dynamics of Particles and Systems, 4th edn, Saunders, Fort Worth, TX (1995), Example 10.5. Geographers measure latitude (\u03bb) up from the equator, rather than down from the pole, so cos \u03b80 = sin \u03bb. 428 CHAPTER 11 Quantum Dynamics z \u03b8 0 Figure 11.14: Path of a Foucault pendulum, in the course of one day. \u221a L/g; if the length is now gradually changing, the period will be 2\u03c0 The basic strategy for analyzing an adiabatic process is \ufb01rst to solve the problem with the external parameters held constant, and only at the end of the calculation allow them to vary (slowly) with time. For example, the classical period of a pendulum of (\ufb01xed) length L is L(t)/g. A more 2\u03c0 subtle subtle example occurred in our discussion of the hydrogen molecule ion (Section 8.3). We began by assuming that the nuclei were at rest, a \ufb01xed distance R", " apart, and we solved for the motion of the electron. Once we had found the ground state energy of the system as a function of R, we located the equilibrium separation and from the curvature of the graph we obtained the frequency of vibration of the nuclei (Problem 8.11). In molecular physics this technique (beginning with nuclei at rest, calculating electronic wave functions, and using these to obtain information about the positions and\u2014relatively sluggish\u2014motion of the nuclei) is known as the Born\u2013Oppenheimer approximation. \u221a 11.5.2 The Adiabatic Theorem In quantum mechanics, the essential content of the adiabatic approximation can be cast in the form of a theorem. Suppose the Hamiltonian changes gradually from some initial form \u02c6H (0) to some \ufb01nal form \u02c6H (T ). The adiabatic theorem33 states that if the particle was initially in the nth eigenstate of \u02c6H (0), it will be carried (under the Schr\u00f6dinger equation) into the nth eigenstate of \u02c6H (T ). (I assume that the spectrum is discrete and nondegenerate throughout the transition, so there is no ambiguity about the ordering of the states; these conditions can be relaxed, given a suitable procedure for \u201ctracking\u201d the eigenfunctions, but I\u2019m not going to pursue that here.) Example 11.3 Suppose we prepare a particle in the ground state of the in\ufb01nite square well (Figure 11.15(a)): \u03c8 i (x) = 2 a \u03c0 a sin. x (11.90) 33 The adiabatic theorem, which is usually attributed to Ehrenfest, is simple to state, and it sounds plausible, but it is not easy to prove. The argument will be found in earlier editions of this book, Section 10.1.2. 11.5 The Adiabatic Approximation 429 If we now gradually move the right wall out to 2a, the adiabatic theorem says that the particle will end up in the ground state of the expanded well (Figure 11.15(b)): 1 a (apart from a phase factor, which we\u2019ll discuss in a moment). Notice that we\u2019re not talking about a small change in the Hamiltonian (as in perturbation theory)\u2014this one is huge. All we require is that it happen slowly. \u03c8 f (x) = (11.91) sin x, \u03c0 2a \u03c8 i(x) \u03c8 f(x) \u03c8 i(x) 2a x a (a) a (b) 2a x a (c) 2a x Figure 11.15: (a) Particle starts out in the ground state of the in\ufb01nite square well. (b) If the wall moves slowly, the particle remains in the ground state. (c) If the wall moves rapidly, the particle is left (momentarily) in its initial state. Energy is not conserved here\u2014of course not: Whoever is moving the wall is extracting energy from the system, just like the piston on a slowly expanding cylinder of gas. By contrast, if the well expands suddenly, the resulting state is still \u03c8 i (x) (Figure 11.15c), which is a complicated linear combination of eigenstates of the new Hamiltonian (Problem 11.18). In this case energy is conserved (at least, its expectation value is); as in the free expansion of a gas (into a vacuum) when the barrier is suddenly removed; no work is done. According to the adiabatic theorem, a system that starts out in the nth eigenstate of the initial Hamiltonian ( \u02c6H (0)) will evolve as the nth eigenstate of the instantaneous Hamiltonian ( \u02c6H (t)), as the Hamiltonian gradually changes. However, this doesn\u2019t tell us what happens to the phase of the wave function. For a constant Hamiltonian it would pick up the standard \u201cwiggle factor\u201d e\u2212i En t/, but the eigenvalue En may now itself be a function of time, so the wiggle factor naturally generalizes to ei\u03b8n (t), where \u03b8n(t) \u2261 \u2212 1 t t dt. En (11.92) 0 This is called the dynamic phase. But it may not be the end of the story; for all we know there may be an additional phase factor, \u03b3n(t), the so-called geometric phase. In the adiabatic limit, then, the wave function at time t takes the form34 n(t) = ei\u03b8n (t)ei\u03b3n (t)\u03c8n(t), (11.93) 34 I\ufffd", "\ufffdm suppressing the dependence on other variables; only the time dependence is at issue here. 430 CHAPTER 11 Quantum Dynamics where \u03c8n(t) is the nth eigenstate of the instantaneous Hamiltonian, \u02c6H (t)\u03c8n(t) = En(t)\u03c8n(t). (11.94) Equation 11.93 is the formal statement of the adiabatic theorem. Of course, the phase of \u03c8n(t) is itself arbitrary (it\u2019s still an eigenfunction, with the same eigenvalue, whatever phase you choose), so the geometric phase itself carries no physical signi\ufb01cance. But what if we carry the system around a closed cycle (like the pendulum we hauled down to the equator, around, and back to the north pole), so that the Hamiltonian at the end is identical to the Hamiltonian at the beginning? Then the net phase change is a measurable quantity. The dynamic phase depends on the elapsed time, but the geometric phase, around an adiabatic closed cycle, depends only on the path taken.35 It is called Berry\u2019s phase:36 \u03b3B \u2261 \u03b3 (T ) \u2212 \u03b3 (0). (11.95) Example 11.4 Imagine an electron (charge \u2212e, mass m) at rest at the origin, in the presence of a magnetic \ufb01eld whose magnitude (B0) is constant, but whose direction sweeps out a cone, of opening angle \u03b1, at constant angular velocity \u03c9 (Figure 11.16): z \u03b1 \u03c9 B \u2013e y x Figure 11.16: The magnetic \ufb01eld sweeps around in a cone, at angular velocity \u03c9 (Equation 11.96). B(t) = B0 # sin \u03b1 cos(\u03c9t)\u02c6\u0131 + sin \u03b1 sin(\u03c9t) \u02c6j + cos \u03b1 \u02c6k $. (11.96) The Hamiltonian (Equation 4.158) is B \u00b7 S = eB0 \u02c6H (t) = e 2m m \u03c91 cos \u03b1 ei\u03c9t sin \u03b1 2 = sin \u03b1 cos(\u03c9t)\u03c3x + sin \u03b1 sin(\u03c9t)\u03c3y + cos \u03b1\u03c3z e\u2212i\u03c9t sin \u03b1 \u2212 cos \u03b1, (11.97) 35 As Michael Berry puts it, the dynamic phase answers the question \u201cHow long did your trip take?\u201d and the geometric phase, \u201cWhere have you been?\u201d 36 For more on this subject see Alfred Shapere and Frank Wilczek, eds., Geometric Phases in Physics, World Scienti\ufb01c, Singapore (1989); Andrei Bernevig and Taylor Hughes, Topological Insulators and Topological Superconductors, Princeton University Press, Princeton, NJ (2013), Chapter 2. where \u03c91 \u2261 eB0 m. The normalized eigenspinors of \u02c6H (t) are \u03c7+(t) = cos(\u03b1/2) ei\u03c9t sin(\u03b1/2), and \u03c7\u2212(t) = e\u2212i\u03c9t sin(\u03b1/2) \u2212 cos(\u03b1/2) ; 11.5 The Adiabatic Approximation 431 (11.98) (11.99) (11.100) they represent spin up and spin down, respectively, along the instantaneous direction of B(t) (see Problem 4.33). The corresponding eigenvalues are E\u00b1 = \u00b1 \u03c91 2. Suppose the electron starts out with spin up, along B(0): \u03c7 (0) =. cos(\u03b1/2) sin(\u03b1/2) (11.101) (11.102) The exact solution to the time-dependent Schr\u00f6dinger equation is (Problem 11.20): \u239b # cos(\u03bbt/2) \u2212 i # cos(\u03bbt/2) \u2212 i $ \u239e sin(\u03bbt/2) $ sin(\u03bbt/2) cos(\u03b1/2)e\u2212i\u03c9t/2 sin(\u03b1/2)e+i\u03c9t/2 \u23a0, (11.103) \u03c7 (t) = \u239d where (\u03c91\u2212\u03c9) \u03bb (\u03c91+\u03c9) \u03bb \u03bb \u2261 \u03c92 + \u03c92 1 \u2212 2\u03c9\u03c91 cos \u03b1, (11.104) or, expressing it as a linear combination of \u03c7+ and \u03c7\u2212: \u03c7 (t) = cos + i (\u03c91 \u2212 \u03c9 cos \u03b1) \u03bb sin \u2212 i \u03bbt 2 \u03c9 \u03bb sin \u03b1 sin \u03bbt 2 \u03bbt 2 e\u2212i\u03c9t/2\u03c7+(t) e+i\u03c9t/2\u03c7", "\u2212(t). (11.105) Evidently the (exact) probability of a transition to spin down (along the current direction of B) is |\u03c7 (t) | \u03c7\u2212(t)|2 = \u03c9 \u03bb sin \u03b1 sin \u03bbt 2 2. (11.106) The adiabatic theorem says that this transition probability should vanish in the limit Te Ti, where Te is the characteristic time for changes in the Hamiltonian (in this case, 1/\u03c9) and Ti is the characteristic time for changes in the wave function (in this case, /(E+ \u2212 E\u2212) = 1/\u03c91). Thus the adiabatic approximation means \u03c9 \u03c91: the \ufb01eld rotates slowly, in comparison with the phase of the (unperturbed) wave functions. In the adiabatic regime \u03bb \u2248 \u03c91 (Equation 11.104), and therefore |\u03c7 (t) | \u03c7\u2212(t)|2 \u2248 sin \u03b1 sin \u2192 0, (11.107) \u03c9 \u03c91 2 \u03bbt 2 432 CHAPTER 11 Quantum Dynamics as advertised. The magnetic \ufb01eld leads the electron around by its nose, with the spin always pointing in the direction of B. By contrast, if \u03c9 \u03c91 then \u03bb \u2248 \u03c9, and the system bounces back and forth between spin up and spin down (Figure 11.17). \u23a2\u2329\u03c7(t)\u23a2\u03c7_(t)\u232a \u23a22 1 (\u03c9 sin \u03b1)2 \u03bb 2\u03c0/\u03bb 4\u03c0/\u03bb 6\u03c0/\u03bb 8\u03c0/\u03bb t Figure 11.17: Plot of the transition probability, Equation 11.106, in the non-adiabatic regime (\u03c9 \u03c91). \u2217 Problem 11.18 A particle of mass m is in the ground state of the in\ufb01nite square well (Equation 2.22). Suddenly the well expands to twice its original size\u2014the right wall moving from a to 2a\u2014leaving the wave function (momentarily) undisturbed. The energy of the particle is now measured. (a) What is the most probable result? What is the probability of getting that result? (b) What is the next most probable result, and what is its probability? Suppose your measurement returned this value; what would you conclude about conservation of energy? (c) What is the expectation value of the energy? Hint: If you \ufb01nd yourself confronted with an in\ufb01nite series, try another method. Problem 11.19 A particle is in the ground state of the harmonic oscillator with classical frequency \u03c9, when suddenly the spring constant quadruples, so \u03c9 = 2\u03c9, without initially changing the wave function (of course, will now evolve differently, because the Hamiltonian has changed). What is the probability that a measurement of the energy would still return the value \u03c9/2? What is the probability of getting \u03c9? Answer: 0.943. \u2217\u2217 Problem 11.20 Check that Equation 11.103 satis\ufb01es the time-dependent Schr\u00f6dinger equation for the Hamiltonian in Equation 11.97. Also con\ufb01rm Equation 11.105, and show that the sum of the squares of the coef\ufb01cients is 1, as required for normalization. Further Problems on Chapter 11 433 \u2217 Problem 11.21 Find Berry\u2019s phase for one cycle of the process in Example 11.4. Hint: Use Equation 11.105 to determine the total phase change, and subtract off the dynamical part. You\u2019ll need to expand \u03bb (Equation 11.104) to \ufb01rst order in \u03c9/\u03c91. Problem 11.22 The delta function well (Equation 2.117) supports a single bound state (Equation 2.132). Calculate the geometric phase change when \u03b1 gradually increases from \u03b11 to \u03b12. If the increase occurs at a constant rate (d\u03b1/dt = c), what is the dynamic phase change for this process?37 FURTHER PROBLEMS ON CHAPTER 11 \u2217\u2217\u2217 Problem 11.23 In Problem 11.1 you showed that the solution to d f dt = k(t) f (t) (where k(t) is a function of t) is f (t) = eK (t) f (0), where K (t) \u2261 0 t t k dt. This suggests that the solution to the Schr\u00f6dinger equation (11.1) might be (t) = e \u02c6G(t)(0), where \u02c6G(t) \u2261 \u2212 i t t \u02c6H dt. 0 (11", ".108) It doesn\u2019t work, because \u02c6H (t) is an operator, not a function, and \u02c6H (t1) does not (in general) commute with \u02c6H (t2). (a) Try calculating i \u2202/\u2202t, using Equation 11.108. Note: as always, the exponenti- ated operator is to be interpreted as a power series: Show that if # \u02c6G, \u02c6H \u02c6G \u2261 1 + \u02c6G + 1 2 \u02c6G \u02c6G + 1 3! \u02c6G \u02c6G \u02c6G + \u00b7 \u00b7 \u00b7. e $ = 0, then satis\ufb01es the Schr\u00f6dinger equation. (b) Check that the correct solution in the general case (t) = 1 + t 0 \u2212 i \u02c6H(t1) dt1 + \u2212 i 2 t 0 $ = 0 # \u02c6G, \u02c6H is \u02c6H(t1) \u02c6H(t2) dt2 dt1 t1 0 \u2212 i + 3 t 0 \u02c6H(t1) t1 0 \u02c6H(t2) t2 0 (11.109) \u02c6H (t3) dt3 dt2 dt1 + \u00b7 \u00b7 \u00b7 (0). 37 If \u03c8n (t) is real, the geometric phase vanishes. You might try to beat the rap by tacking an unnecessary (but perfectly legal) phase factor onto the eigenfunctions: \u03c8 (t) \u2261 ei\u03c6n \u03c8n (t), where \u03c6n is an arbitrary (real) function. Try it. n You\u2019ll get a nonzero geometric phase, all right, but note what happens when you put it back into Equation 11.93. And for a closed loop it gives zero. 434 CHAPTER 11 Quantum Dynamics UGLY! Notice that the operators in each term are \u201ctime-ordered,\u201d in the sense \u02c6H appears at the far left, followed by the next latest, and so that the latest on (t \u2265 t1 \u2265 t2 \u2265 t3 \u00b7 \u00b7 \u00b7 ). Dyson introduced the time-ordered product of two operators: # T \u02c6H (ti ) \u02c6H $ t j \u2261 \u02c6H (ti ) \u02c6H \u02c6H t j, t j \u02c6H (ti ), ti \u2265 t j t j \u2265 ti (11.110) or, more generally, # $ \u02c6H (t1) \u02c6H (t2) \u00b7 \u00b7 \u00b7 \u02c6H (tn) T \u2261 \u02c6H t ji t j2 \u02c6H \u00b7 \u00b7 \u00b7 \u02c6H t jn, (11.111) where t j1 (c) Show that \u2265 t j2 \u2265 \u00b7 \u00b7 \u00b7 \u2265 t jn. $ # T \u02c6G \u02c6G = \u2212 2 2 t 0 \u02c6H (t1) t1 0 \u02c6H (t2) dt2 dt1, and generalize to higher powers of \u02c6G. In place of \u02c6Gn, in equation 11.108, we really want T $ : \u02c6Gn # (t) = T # e\u2212 i $ t 0 \u02c6H(t ) dt (0). (11.112) This is Dyson\u2019s formula; it\u2019s a compact way of writing Equation 11.109, the formal solution to Schr\u00f6dinger\u2019s equation. Dyson\u2019s formula plays a fundamental role in quantum \ufb01eld theory.38 \u2217\u2217 Problem 11.24 In this problem we develop time-dependent perturbation theory for a multi- level system, starting with the generalization of Equations 11.5 and 11.6: \u02c6H0\u03c8n = En\u03c8n, \u03c8n|\u03c8m = \u03b4nm. (11.113) At time t = 0 we turn on a perturbation H (t), so that the total Hamiltonian is \u02c6H = \u02c6H0 + \u02c6H (t). (11.114) 38 The interaction picture is intermediate between the Heisenberg and Schr\u00f6dinger pictures (see Section 6.8.1). In the interaction picture, the wave function satis\ufb01es the \u201cSchr\u00f6dinger equation\u201d where the interaction- and Schr\u00f6dinger-picture operators are related by i d dt |I (t) = \u02c6H I (t) |I (t), and the wave functions", " satisfy (t) = ei \u02c6H0t/ \u02c6H (t)e \u2212i \u02c6H0t/ \u02c6H I |I (t) = ei \u02c6H0t/ |(t). If you apply the Dyson series to the Schr\u00f6dinger equation in the interaction picture, you end up with precisely the perturbation series derived in Section 11.1.2. For more details see Ramamurti Shankar, Principles of Quantum Mechanics, 2nd edn, Springer, New York (1994), Section 18.3. Further Problems on Chapter 11 435 (a) Generalize Equation 11.10 to read and show that where (t) = \u02d9cm = \u2212 i cn(t)\u03c8ne\u2212i En t/, n n cn H mnei(Em \u2212En )t/, 0 \u03c8m 1 \u02c6H \u03c8n. H mn \u2261 (11.115) (11.116) (11.117) (b) If the system starts out in the state \u03c8N, show that (in \ufb01rst-order perturbation theory) cN (t) \u2248 1 \u2212 i t dt. H N N t 0 (11.118) and cm(t) \u2248 \u2212 i t 0 t ei(Em \u2212E N )t / dt, H m N (m = N ). (11.119) H 2 sin2 (c) For example, suppose \u02c6H is constant (except that it was turned on at t = 0, and switched off again at some later time T ). Find the probability of transition from state N to state M (M = N ), as a function of T. Answer: (E N \u2212 E M ) T /2 (E N \u2212 E M )2 (d) Now suppose \u02c6H is a sinusoidal function of time: \u02c6H = V cos(\u03c9t). Making the usual assumptions, show that transitions occur only to states with energy E M = E N \u00b1 \u03c9, and the transition probability is PN \u2192M = |VM N |2 sin2 (E N \u2212 E M \u00b1 \u03c9) T /2 (E N \u2212 E M \u00b1 \u03c9)2 (11.121) (11.120) M N 4.. (e) Suppose a multi-level system is immersed in incoherent electromagnetic radiation. Using Section 11.2.3 as a guide, show that the transition rate for stimulated emission is given by the same formula (Equation 11.54) as for a two-level system. Problem 11.25 For the examples in Problem 11.24 (c) and (d), calculate cm(t), to \ufb01rst order. Check the normalization condition: |cm(t)|2 = 1, (11.122) m and comment on any discrepancy. Suppose you wanted to calculate the probability of remaining in the original state \u03c8N ; would you do better to use |cN (t)|2, or 1 \u2212 m=N |cm(t)|2? 436 CHAPTER 11 Quantum Dynamics Problem 11.26 A particle starts out (at time t = 0) in the N th state of the in\ufb01nite square well. Now the \u201c\ufb02oor\u201d of the well rises temporarily (maybe water leaks in, and then drains out again), so that the potential inside is uniform but time dependent: V0(t), with V0(0) = V0(T ) = 0. (a) Solve for the exact cm(t), using Equation 11.116, and show that the wave function changes phase, but no transitions occur. Find the phase change, \u03c6(T ), in terms of the function V0(t). (b) Analyze the same problem in \ufb01rst-order perturbation theory, and compare your answers. Comment: The same result holds whenever the perturbation simply adds a constant (constant in x, that is, not in t) to the potential; it has nothing to do with the in\ufb01nite square well, as such. Compare Problem 1.8. \u2217 \u2217\u2217\u2217 Problem 11.27 A particle of mass m is initially in the ground state of the (one-dimensional) in\ufb01nite square well. At time t = 0 a \u201cbrick\u201d is dropped into the well, so that the potential becomes \u23a7 \u23aa\u23a8 V (x) = \u23aa\u23a9 V0, 0, \u221e, 0 \u2264 x \u2264 a/2, a/2 < x \u2264 a, otherwise, where V0 E1. After a time T, the brick is removed, and the energy of", " the particle is measured. Find the probability (in \ufb01rst-order perturbation theory) that the result is now E2. Problem 11.28 We have encountered stimulated emission, (stimulated) absorption, and spontaneous emission. How come there is no such thing as spontaneous absorption? Problem 11.29 Magnetic resonance. A spin-1/2 particle with gyromagnetic ratio \u03b3, \u02c6k, precesses at the Larmor frequency \u03c90 = at rest in a static magnetic \ufb01eld B0 \u03b3 B0 (Example 4.3). Now we turn on a small transverse radiofrequency (rf) \ufb01eld, Brf, so that the total \ufb01eld is cos(\u03c9t) \u02c6\u0131 \u2212 sin(\u03c9t) \u02c6j B = Brf cos(\u03c9t) \u02c6\u0131 \u2212 Brf sin(\u03c9t) \u02c6j + B0 \u02c6k. (11.123) a(t) b(t) \u02d9a = i 2 (a) Construct the 2 \u00d7 2 Hamiltonian matrix (Equation 4.158) for this system. (b) If \u03c7 (t) = is the spin state at time t, show that ei\u03c9t b + \u03c90a ; \u02d9b = i 2 e\u2212i\u03c9t a \u2212 \u03c90b, (11.124) where \u2261 \u03b3 Brf is related to the strength of the rf \ufb01eld. (c) Check that the general solution for a(t) and b(t), in terms of their initial values a0 and b0, is Further Problems on Chapter 11 437 a(t) = a0 cos b(t) = b0 cos \u03c9t/2 \u03c9t/2 where + i + i \u03c9t/2 \u03c9 [a0 (\u03c90 \u2212 \u03c9) + b0] sin \u03c9t/2 \u03c9 [b0 (\u03c9 \u2212 \u03c90) + a0] sin ei\u03c9t/2 e\u2212i\u03c9t/2 \u03c9 \u2261 (\u03c9 \u2212 \u03c90)2 + 2. (11.125) (d) If the particle starts out with spin up (i.e. a0 = 1, b0 = 0), \ufb01nd the probability of a transition to spin down, as a function of time. Answer: P(t) = \u2019 2/ ( \u03c9t/2. sin2 (e) Sketch the resonance curve, (\u03c9 \u2212 \u03c90)2 + 2 P(\u03c9) = 2 (\u03c9 \u2212 \u03c90)2 + 2, (11.126) as a function of the driving frequency \u03c9 (for \ufb01xed \u03c90 and ). Note that the maximum occurs at \u03c9 = \u03c90. Find the \u201cfull width at half maximum,\u201d \u03c9. (f) Since \u03c90 = \u03b3 B0, we can use the experimentally observed resonance to determine the magnetic dipole moment of the particle. In a nuclear magnetic resonance (nmr) experiment the g-factor of the proton is to be measured, using a static \ufb01eld of 10,000 gauss and an rf \ufb01eld of amplitude 0.01 gauss. What will the resonant frequency be? (See Section 7.5 for the magnetic moment of the proton.) Find the width of the resonance curve. (Give your answers in Hz.) \u2217\u2217 Problem 11.30 In this problem we will recover the results Section 11.2.1 directly from the Hamiltonian for a charged particle in an electromagnetic \ufb01eld (Equation 4.188). An electromagnetic wave can be described by the potentials A = E0 \u03d5 = 0 \u03c9 sin(k \u00b7 r \u2212 \u03c9 t), where in order to satisfy Maxwell\u2019s equations, (E0 \u00b7 k = 0) and of course travel at the speed of light (\u03c9 = c |k|). (a) Find the electric and magnetic \ufb01elds for this plane wave. (b) The Hamiltonian may be written as H 0 + H where H 0 is the Hamiltonian in the absence of the electromagnetic wave and H is the perturbation. Show that the perturbation is given by the wave must be transverse \u02c6H (t) = e 2 i m \u03c9 ei k\u00b7r E0 \u00b7 \u02c6p e\u2212\u2212i k\u00b7r E0 \u00b7 \u02c6p ei \u03c9 t, (11.127) plus a term proportional to E 2 to absorption and the second to emission. 0 that we will ignore. Note: the \ufb01rst term corresponds (c) In the dipole approximation we set", " ei k\u00b7r \u2248 1. With the electromagnetic wave polarized along the z direction, show that the matrix element for absorption is then 438 CHAPTER 11 Quantum Dynamics Vba = \u2212 \u03c90 \u03c9 \u2118 E0. Compare Equation 11.41. They\u2019re not exactly the same; would the difference effect our calculations in Section 11.2.3 or 11.3? Why or why not? Hint: To turn the matrix element of p into a matrix element of r, you need to prove the following identity: i m = \u02c6p. \u02c6H 0, \u02c6r # $ \u2217\u2217\u2217 Problem 11.31 In Equation 11.38 I assumed that the atom is so small (in comparison to the wavelength of the light) that spatial variations in the \ufb01eld can be ignored. The true electric \ufb01eld would be E (r, t) = E0 cos(k \u00b7 r \u2212 \u03c9t). If the atom is centered at the origin, then k \u00b7 r 1 over the relevant volume (|k| = 2\u03c0/\u03bb, so k \u00b7 r \u223c r/\u03bb 1), and that\u2019s why we could afford to drop this term. Suppose we keep the \ufb01rst-order correction: (11.128) E (r, t) = E0 [cos(\u03c9t) + (k \u00b7 r) sin(\u03c9t)]. (11.129) The \ufb01rst term gives rise to the allowed (electric dipole) transitions we considered in the text; the second leads to so-called forbidden (magnetic dipole and electric quadrupole) transitions (higher powers of k \u00b7 r lead to even more \u201cforbidden\u201d transitions, associated with higher multipole moments).39 (a) Obtain the spontaneous emission rate for forbidden transitions (don\u2019t bother to average over polarization and propagation directions, though this should really be done to complete the calculation). Answer: \u02c6n \u00b7 r b (11.130) \u02c6k \u00b7 r 0 a 1 2. Rb\u2192a = q2\u03c95 \u03c0 0c5 (b) Show that for a one-dimensional oscillator the forbidden transitions go from level n to level n \u2212 2, and the transition rate (suitably averaged over \u02c6n and \u02c6k) is R = q2\u03c93n (n \u2212 1) 15\u03c0 0m2c5. (11.131) (Note: Here \u03c9 is the frequency of the photon, not the oscillator.) Find the ratio of the \u201cforbidden\u201d rate to the \u201callowed\u201d rate, and comment on the terminology. (c) Show that the 2S \u2192 1S transition in hydrogen is not possible even by a \u201cforbidden\u201d transition. (As it turns out, this is true for all the higher multipoles as well; the dominant decay is in fact by two-photon emission, and the lifetime is about a tenth of a second.40 ) 39 For a systematic treatment (including the role of the magnetic \ufb01eld) see David Park, Introduction to the Quantum Theory, 3rd edn (McGraw-Hill, New York, 1992), Chapter 11. 40 See Masataka Mizushima, Quantum Mechanics of Atomic Spectra and Atomic Structure, Benjamin, New York (1970), Section 5.6. Further Problems on Chapter 11 439 \u2217\u2217\u2217 Problem 11.32 Show that the spontaneous emission rate (Equation 11.63) for a transition from n, to n, in hydrogen is e2\u03c93 I 2 3\u03c0 0c3 \u00d7 \u23a7 \u23aa\u23a8 \u23aa\u23a9 +1 2+1, = + 1, 2+1 = \u2212 1, (11.132) where \u221e I \u2261 0 r 3 Rn(r )Rn(r ) dr. (11.133) (The atom starts out with a speci\ufb01c value of m, and it goes to any of the states m consistent with the selection rules: m = m + 1, m, or m \u2212 1. Notice that the answer is independent of m.) Hint: First calculate all the nonzero matrix elements of x, y, and for the case = + 1. From these, determine the z between |nm and quantity nm n, + 1, m r nm 2 + r nm 2 + r nm 2. n, + 1, m + 1 n, + 1, m \u2212 1 Then do the same for = \u22121. You may \ufb01nd useful the following recursion formulas (which hold for m \u2265 0):41 (2 + 1) x P m (x) = ( + m) P m \u22121 (x) = P m+1 1 \u2212", " x 2 P m +1 (x) + ( \u2212 m + 1) P m +1 (x) \u2212 P m+1 \u22121 (2 + 1) % (x), (x), (11.134) (11.135) and the orthogonality relation Equation 4.33. Problem 11.33 The spontaneous emission rate for the 21-cm hyper\ufb01ne line in hydrogen (Section 7.5) can be obtained from Equation 11.63, except that this is a magnetic dipole transition, not an electric one:42 where \u03bc e = \u2212 e me Se, \u03bc p = 5.59 e 2m p S p are the magnetic moments of the electron and proton (Equation 7.89), and |0, |1 are the singlet and triplet con\ufb01gurations (Equations 4.175 and 4.176). Because m p me, the proton contribution is negligible, so \u03c93 0e2 3\u03c0 0c5m2 e Work out |1| Se |0|2 (use whichever triplet state you like). Put in the actual numbers, to determine the transition rate and the lifetime of the triplet state. Answer: 1.1 \u00d7 107 years. |1| Se |0|2. A = 41 George B. Arfken and Hans J. Weber, Mathematical Methods for Physicists, 7th edn, Academic Press, San Diego (2013), p. 744. 42 Electric and magnetic dipole moments have different units\u2014hence the factor of 1/c (which you can check by dimensional analysis). 440 CHAPTER 11 Quantum Dynamics \u2217\u2217\u2217 Problem 11.34 A particle starts out in the ground state of the in\ufb01nite square well (on the interval 0 \u2264 x \u2264 a). Now a wall is slowly erected, slightly off-center:43 V (x) = f (t)\u03b4, x \u2212 a 2 \u2212 where f (t) rises gradually from 0 to \u221e. According to the adiabatic theorem, the particle will remain in the ground state of the evolving Hamiltonian. (a) Find (and sketch) the ground state at t \u2192 \u221e. Hint: This should be the ground state of the in\ufb01nite square well with an impenetrable barrier at a/2 +. Note that the particle is con\ufb01ned to the (slightly) larger left \u201chalf\u201d of the well. (b) Find the (transcendental) equation for the ground state energy at time t. Answer: where z \u2261 ka, T \u2261 ma f(t)/2, \u03b4 \u2261 2/a, and k \u2261 2m E/. z sin z = T [cos z \u2212 cos(z\u03b4)], \u221a (c) Setting \u03b4 = 0, solve graphically for z, and show that the smallest z goes from \u03c0 to 2\u03c0 as T goes from 0 to \u221e. Explain this result. (d) Now set \u03b4 = 0.01 and solve numerically for z, using T = 0, 1, 5, 20, 100, and 1000. (e) Find the probability Pr as a function of z and \u03b4. Answer: Pr = 1/ 1 \u00b1 \u03b4 \u2212 (1/z) sin(z(1 \u00b1 \u03b4)) ically for the T \u2019s and \u03b4 in part (d). Comment on your results. that the particle is in the right \u201chalf\u201d of the well,, where I\u00b1 \u2261. Evaluate this expression numer- z (1 \u2213 \u03b4) /2 1 + (I+/I\u2212) sin2 (e) Plot the ground state wave function for those same values of T and \u03b4. Note how it gets squeezed into the left half of the well, as the barrier grows.44 \u2217\u2217\u2217 Problem 11.35 The case of an in\ufb01nite square well whose right wall expands at a constant velocity (v) can be solved exactly.45 A complete set of solutions is n(x, t) \u2261 2 w sin n\u03c0 w x ei mvx 2\u22122E i nat /2w, (11.136) where w(t) \u2261 a + vt is the width of the well and E i \u2261 n2\u03c0 22/2ma2 is the n nth allowed energy of the original well (width a). The general solution is a linear combination of the \u2019s: \u221e (x, t) = cnn(x, t), (11.137) n=1 whose coef\ufb01cients cn are independent of t. (a) Check that Equation 11.136 satis\ufb01es the time-dependent Schr\u00f6dinger equation, with the appropriate boundary conditions.", " 43 Julio Gea-Banacloche, Am. J. Phys. 70, 307 (2002) uses a rectangular barrier; the delta-function version was suggested by M. Lakner and J. Peternelj, Am. J. Phys. 71, 519 (2003). 44 Gea-Banacloche (footnote 43) discusses the evolution of the wave function without using the adiabatic theorem. 45 S. W. Doescher and M. H. Rice, Am. J. Phys. 37, 1246 (1969). Further Problems on Chapter 11 441 (b) Suppose a particle starts out (t = 0) in the ground state of the initial well: (x, 0) = \u03c0 a sin 2 a. x Show that the expansion coef\ufb01cients can be written in the form \u03c0 cn = 2 \u03c0 0 e\u2212i\u03b1z2 sin(nz) sin(z) dz, (11.138) where \u03b1 \u2261 mva/2\u03c0 2 is a dimensionless measure of the speed with which the well expands. (Unfortunately, this integral cannot be evaluated in terms of elementary functions.) (c) Suppose we allow the well to expand to twice its original width, so the \u201cexternal\u201d time is given by w(Te) = 2a. The \u201cinternal\u201d time is the period of the timedependent exponential factor in the (initial) ground state. Determine Te and Ti, and \u2248 1 show that the adiabatic regime corresponds to \u03b1 1, so that exp over the domain of integration. Use this to determine the expansion coef\ufb01cients, cn. Construct (x, t), and con\ufb01rm that it is consistent with the adiabatic theorem. \u2212i\u03b1z2 (d) Show that the phase factor in (x, t) can be written in the form t \u03b8 (t) = \u2212 1 t dt, (11.139) E1 0 where En(t) \u2261 n2\u03c0 22/2mw2 is the nth instantaneous eigenvalue, at time t. Comment on this result. What is the geometric phase? If the well now contracts back to its original size, what is Berry\u2019s phase for the cycle? \u2217\u2217\u2217 Problem 11.36 The driven harmonic oscillator. Suppose the one-dimensional harmonic oscillator (mass m, frequency \u03c9) is subjected to a driving force of the form F(t) = m\u03c92 f (t), where f (t) is some speci\ufb01ed function. (I have factored out m\u03c92 for notational convenience; f (t) has the dimensions of length.) The Hamiltonian is H (t) = \u2212 2 2m \u2202 2 \u2202 x 2 + 1 2 m\u03c92x 2 \u2212 m\u03c92x f (t). (11.140) Assume that the force was \ufb01rst turned on at time t = 0: f (t) = 0 for t \u2264 0. This system can be solved exactly, both in classical mechanics and in quantum mechanics.46 (a) Determine the classical position of the oscillator, assuming it started from rest at the origin (xc(0) = \u02d9xc(0) = 0). Answer: t xc(t) = \u03c9 f 0 \u03c9 sin t \u2212 t dt. (11.141) t (b) Show that the solution to the (time-dependent) Schr\u00f6dinger equation for this oscillator, assuming it started out in the nth state of the undriven oscillator ((x, 0) = \u03c8n(x) where \u03c8n(x) is given by Equation 2.62), can be written as 46 See Y. Nogami, Am. J. Phys. 59, 64 (1991), and references therein. 442 CHAPTER 11 Quantum Dynamics (x, t) = \u03c8n(x \u2212 xc)e # i \u2212 n+ 1 2 \u03c9t+m \u02d9xc(x\u2212 xc 2 )+ m\u03c92 2 0 f (t )xc(t )dt t $. (11.142) (c) Show that the eigenfunctions and eigenvalues of H (t) are \u03c9 \u2212 1 2 \u03c8n(x, t) = \u03c8n(x \u2212 f ); n + 1 2 En(t) = m\u03c92 f 2. (11.143) (d) Show that in the adiabatic approximation the classical position (Equation 11.141) reduces to xc(t) \u2248 f (t). State the precise criterion for adiabaticity, in this con- \u03c9 text, as a constraint on the time derivative of f. Hint: Write sin as", " (1/\u03c9) and use integration by parts. d/dt t \u2212 t t \u2212 t \u03c9 cos (e) Con\ufb01rm the adiabatic theorem for this example, by using the results in (c) and (d) to show that (x, t) \u2248 \u03c8n(x, t)ei\u03b8n (t)ei\u03b3n (t). (11.144) Check that the dynamic phase has the correct form (Equation 11.92). Is the geometric phase what you would expect? Problem 11.37 Quantum Zeno Paradox.47 Suppose a system starts out in an excited state \u03c8b, which has a natural lifetime \u03c4 for transition to the ground state \u03c8a. Ordinarily, for times substantially less than \u03c4, the probability of a transition is proportional to t (Equation 11.49): Pb\u2192a = t \u03c4. (11.145) If we make a measurement after a time t, then, the probability that the system is still in the upper state is Pb(t) = 1 \u2212 t \u03c4. (11.146) Suppose we do \ufb01nd it to be in the upper state. In that case the wave function collapses back to \u03c8b, and the process starts all over again. If we make a second measurement, at 2t, the probability that the system is still in the upper state is 2 1 \u2212 t \u03c4 \u2248 1 \u2212 2t \u03c4, (11.147) which is the same as it would have been had we never made the \ufb01rst measurement at t (as one would naively expect). However, for extremely short times, the probability of a transition is not proportional to t, but rather to t 2 (Equation 11.46):48 Pb\u2192a = \u03b1t 2. (11.148) 47 This phenomenon doesn\u2019t have much to do with Zeno, but it is reminiscent of the old adage, \u201ca watched pot never boils,\u201d so it is sometimes called the watched pot effect. 48 In the argument leading to linear time dependence, we assumed that the function sin2 (t/2) / 2 in Equation 11.46 was a sharp spike. However, the width of the \u201cspike\u201d is of order \u03c9 = 4\u03c0/t, and for extremely short t this \u03c1 (\u03c9) d\u03c9. assumption fails, and the integral becomes t 2/4 Further Problems on Chapter 11 443 (a) In this case what is the probability that the system is still in the upper state after the two measurements? What would it have been (after the same elapsed time) if we had never made the \ufb01rst measurement? (b) Suppose we examine the system at n regular (extremely short) intervals, from t = 0 out to t = T (that is, we make measurements at T /n, 2T /n, 3T /n,..., T ). What is the probability that the system is still in the upper state at time T? What is its limit as n \u2192 \u221e? Moral of the story: Because of the collapse of the wave function at every measurement, a continuously observed system never decays at all!49 \u2217\u2217\u2217 Problem 11.38 The numerical solution to the time-independent Schr\u00f6dinger equation in Problem 2.61 can be extended to solve the time-dependent Schr\u00f6dinger equation. When we discretize the variable x, we obtain the matrix equation H = i d dt. The solution to this equation can be written (t + t) = U(t) (t). (11.149) (11.150) If H is time independent, the exact expression for the time-evolution operator is50 U(t) = e\u2212i H t/ (11.151) and for t small enough, the time-evolution operator can be approximated as U(t) \u2248 1 \u2212 i H t. (11.152) While Equation 11.152 is the most obvious way to approximate U, a numerical scheme based on it is unstable, and it is preferable to use Cayley\u2019s form for the approximation:51 U(t) \u2248 2 i t 2 i t Combining Equations 11.153 and 11.150 we have 1 \u2212 1 2 (t + t11.153) H (t). t i (11.154) 49 This argument was introduced by B. Misra and E. C. G. Sudarshan, J. Math. Phys. 18, 756 (1977). The essential result has been con\ufb01rmed in the laboratory: W. M. Itano, D. J. Heinzen, J. J. Bollinger, and D. J. Wineland, Phys. Rev. A 41, 2295 (1990). Unfortunately, the", " experiment is not as compelling a test of the collapse of the wave function as its designers hoped, for the observed effect can perhaps be accounted for in other ways\u2014see L. E. Ballentine, Found. Phys. 20, 1329 (1990); T. Petrosky, S. Tasaki, and I. Prigogine, Phys. Lett. A 151, 109 (1990). 50 If you choose t small enough, you can actually use this exact form. Routines such as Mathematica\u2019s MatrixExp can be used to \ufb01nd (numerically) the exponential of a matrix. 51 See A. Goldberg et al., Am. J. Phys. 35, 177 (1967) for further discussion of these approximations. 444 CHAPTER 11 Quantum Dynamics This has the form of a matrix equation M x = b which can be solved for the unknown x = (t + t). Because the matrix M = 1 + 1 H is tri-diagonal,52 ef\ufb01cient algorithms exist for doing so.53 (a) Show that the approximation in Equation 11.153 is accurate to second order. That is, show that Equations 11.151 and 11.153, expanded as power series in t, agree up through terms of order (t)2. Verify that the matrix in Equation 11.153 is unitary. 2 i t As an example, consider a particle of mass m moving in one dimension in a simple harmonic oscillator potential. For the numerical part set m = 1, \u03c9 = 1, and = 1 (this just de\ufb01nes the units of mass, time, and length). (b) Construct the Hamiltonian matrix H for N + 1 = 100 spatial grid points. Set the spatial boundaries where the dimensionless length is \u03be = \u00b110 (far enough out that we can assume that the wave function vanishes there for low-energy states). By computer, \ufb01nd the lowest two eigenvalues of H, and compare the exact values. Plot the corresponding eigenfunctions. Are they normalized? If not, normalize them before doing part (c). (c) Take (0) = (\u03c80 + \u03c81) / 2 (from part (b)) and use Equation 11.154 to evolve the wave function from time t = 0 to t = 4 \u03c0/\u03c9. Create a movie (Animate, in Mathematica) showing Re((t)), Im((t)), and |(t)|, together with the exact result. Hint: You need to decide what to use for t. In terms of the number of time steps Nt, Nt t = 4 \u03c0/\u03c9. In order for the approximation of the exponential to hold, we need to have E t/ 1. The energy of our state is of order \u03c9, and therefore Nt 4 \u03c0. So you will need at least (say) 100 time steps. \u221a \u2217\u2217\u2217 Problem 11.39 We can use the technique of Problem 11.38 to investigate time evolution when the Hamiltonian does depend on time, as long as we choose t small enough. Evaluating H at the midpoint of each time step we simply replace Equation 11.154 with54 1 + 1 2 (t + t) = 1 \u2212 1 2 (t Consider the driven harmonic oscillator of Problem 11.36 with f (t) = A sin( t), (11.155) (11.156) where A is a constant with the units of length and is the driving frequency. In the following we will set m = \u03c9 = = A = 1 and look at the effect of varying. Use the same parameters for the spatial discretization as in Problem 11.38, but set 52 A tri-diagonal matrix has nonzero entries only along the diagonal and one space to the right or left of the diagonal. 53 Use your computing enviroment\u2019s built-in linear equation solver; in Mathematica that would be x = Linear- Solve[M, b]. To learn how it actually works, see A. Goldberg et al., footnote 51. 54 C. Lubich, in Quantum Simulations of Complex Many-Body Systems: From Theory to Algorithms, edited by J. Grotendorst, D. Marx, and A. Muramatsu (John von Neumann Institute for Computing, J\u00fclich, 2002), Vol. 10, p. 459. Available for download from the Neumann Institute for Computing (NIC) website. Further Problems on Chapter 11 445 Nt = 1000. For a particle that starts off in the ground state at t = 0, create a movie showing the numerical and exact solutions as well as the instantaneous ground state from t = 0 to t = 2 \u03c0/ for (a) = \u03c9/5. In line with the adiab", "atic theorem, you should see that the numerical solution is close (up to a phase) to the instantaneous ground state. (b) = 5 \u03c9. In line with what you\u2019ve learned about sudden perturbations, you should see that the numerical solution is barely affected by the driving force. (c) = 6 \u03c9/5. 12 AFTERWORD Now that you have a sound understanding of what quantum mechanics says, I would like to return to the question of what it means\u2014continuing the story begun in Section 1.2. The source of the problem is the indeterminacy associated with the statistical interpretation of the wave function. For (or, more generally, the quantum state\u2014it could be a spinor, for example) does not uniquely determine the outcome of a measurement; all it tells us is the statistical distribution of possible results. This raises a profound question: Did the physical system \u201cactually have\u201d the attribute in question prior to the measurement (the so-called realist viewpoint), or did the act of measurement itself \u201ccreate\u201d the property, limited only by the statistical constraint imposed by the wave function (the orthodox position)\u2014or can we duck the issue entirely, on the grounds that it is \u201cmetaphysical\u201d (the agnostic response)? According to the realist, quantum mechanics is an incomplete theory, for even if you know everything quantum mechanics has to tell you about the system (to wit: its wave function), still you cannot determine all of its features. Evidently there is some other information, unknown to quantum mechanics, which (together with ) is required for a complete description of physical reality. The orthodox position raises even more disturbing problems, for if the act of measurement forces the system to \u201ctake a stand,\u201d helping to create an attribute that was not there previously,1 then there is something very peculiar about the measurement process. Moreover, in order to account for the fact that an immediately repeated measurement yields the same result, we are forced to assume that the act of measurement collapses the wave function, in a manner that is dif\ufb01cult, at best, to reconcile with the normal evolution prescribed by the Schr\u00f6dinger equation. In light of this, it is no wonder that generations of physicists retreated to the agnostic position, and advised their students not to waste time worrying about the conceptual foundations of the theory. 1 This may be strange, but it is not mystical, as some popularizers would like to suggest. The so-called wave\u2013particle duality, which Niels Bohr elevated to the status of a cosmic principle (complementarity), makes electrons sound like unpredictable adolescents, who sometimes behave like adults, and sometimes, for no particular reason, like children. I prefer to avoid such language. When I say that a particle does not have a particular attribute before its measurement, I have in mind, for example, an electron in the spin state \u03c7 = ; a measurement of the xcomponent of its angular momentum could return the value /2, or (with equal probability) the value \u2212/2, but until the measurement is made it simply does not have a well-de\ufb01ned value of Sx. 1 0 12.1 THE EPR PARADOX 12.1 The EPR Paradox 447 In 1935, Einstein, Podolsky, and Rosen2 published the famous EPR paradox, which was designed to prove (on purely theoretical grounds) that the realist position is the only tenable one. I\u2019ll describe a simpli\ufb01ed version of the EPR paradox, due to David Bohm (call it EPRB). Consider the decay of the neutral pi meson into an electron and a positron: \u03c0 0 \u2192 e\u2212 + e+. Assuming the pion was at rest, the electron and positron \ufb02y off in opposite directions (Figure 12.1). Now, the pion has spin zero, so conservation of angular momentum requires that the electron and positron occupy the singlet spin con\ufb01guration: 1\u221a 2 (|\u2191\u2193 \u2212 |\u2193\u2191). (12.1) If the electron is found to have spin up, the positron must have spin down, and vice versa. Quantum mechanics can\u2019t tell you which combination you\u2019ll get, in any particular pion decay, but it does say that the measurements will be correlated, and you\u2019ll get each combination half the time (on average). Now suppose we let the electron and positron \ufb02y far off\u201410 meters, in a practical experiment, or, in principle, 10 light years\u2014and then you measure the spin of the electron. Say you get spin up. Immediately you know that someone 20 meters (or 20 light years) away will get spin down, if that person examines the positron. To the realist, there\u2019s nothing surprising about this\u2014the electron really", " had spin up (and the positron spin down) from the moment they were created... it\u2019s just that quantum mechanics didn\u2019t know about it. But the \u201corthodox\u201d view holds that neither particle had either spin up or spin down until the act of measurement intervened: Your measurement of the electron collapsed the wave function, and instantaneously \u201cproduced\u201d the spin of the positron 20 meters (or 20 light years) away. Einstein, Podolsky, and Rosen considered such \u201cspooky action-at-adistance\u201d (Einstein\u2019s delightful term) preposterous. They concluded that the orthodox position is untenable; the electron and positron must have had well-de\ufb01ned spins all along, whether quantum mechanics knows it or not. The fundamental assumption on which the EPR argument rests is that no in\ufb02uence can propagate faster than the speed of light. We call this the principle of locality. You might be tempted to propose that the collapse of the wave function is not instantaneous, but \u201ctravels\u201d at some \ufb01nite velocity. However, this would lead to violations of angular momentum conservation, for if we measured the spin of the positron before the news of the collapse had reached it, there would be a \ufb01fty\u2013\ufb01fty probability of \ufb01nding both particles with spin up. Whatever you might think of such a theory in the abstract, the experiments are unequivocal: No such violation occurs\u2014the (anti-)correlation of the spins is e\u2212 \u03c00 e+ Figure 12.1: Bohm\u2019s version of the EPR experiment: A \u03c0 0 at rest decays into an electron\u2013positron pair. 2 A. Einstein, B. Podolsky, and N. Rosen, Phys. Rev. 47, 777 (1935). 448 CHAPTER 12 Afterword perfect. Evidently the collapse of the wave function\u2014whatever its ontological status\u2014is instantaneous.3 Problem 12.1 Entangled states. The singlet spin con\ufb01guration (Equation 12.1) is the classic example of an entangled state\u2014a two-particle state that cannot be expressed as the product of two one-particle states, and for which, therefore, one cannot really speak of \u201cthe state\u201d of either particle separately.4 You might wonder whether this is somehow an artifact of bad notation\u2014maybe some linear combination of the one-particle states would disentangle the system. Prove the following theorem: Consider a two-level system, |\u03c6a and |\u03c6b, with example, |\u03c6a might represent spin up and |\u03c6b spin down.) The two-particle state = \u03b4i j. (For \u03c6i |\u03c6 j \u03b1|\u03c6a(1)|\u03c6b(2) + \u03b2|\u03c6b(1)|\u03c6a(2) (with \u03b1 = 0 and \u03b2 = 0) cannot be expressed as a product for any one-particle states |\u03c8r and |\u03c8s. |\u03c8r (1)|\u03c8s(2), Hint: Write |\u03c8r and |\u03c8s as linear combinations of |\u03c6a and |\u03c6b. Problem 12.2 Einstein\u2019s Boxes. In an interesting precursor to the EPR paradox, Einstein proposed the following gedanken experiment:5 Imagine a particle con\ufb01ned to a box (make it a one-dimensional in\ufb01nite square well, if you like). It\u2019s in the ground state, when an impenetrable partition is introduced, dividing the box into separate halves, B1 and B2, in such a way that the particle is equally likely to be found in either one.6 Now the two boxes are moved very far apart, and a measurement is made on B1 to see if the particle is in that box. Suppose the answer is yes. Immediately we know that the particle will not be found in the (distant) box B2. (a) What would Einstein say about this? (b) How does the Copenhagen interpretation account for it? What is the wave function in B2, right after the measurement on B1? 3 Bohr wrote a famous rebuttal to the EPR paradox (Phys. Rev. 48, 696 (1935)). I doubt many people read it, and certainly very few understood it (Bohr himself later admitted that he had trouble making sense of his own argument), but it was a relief that the great man had solved the problem, and everybody else could go back to business. It was not until the mid-1960s that most physicists began to worry seriously about the EPR paradox. 4 Although the term \u201centanglement\u201d is usually applied to", " systems of two (or more) particles, the same basic notion can be extended to single particle states (Problem 12.2 is an example). For an interesting discussion see D. V. Schroeder, Am. J. Phys. 85, 812 (2017). 5 See T. Norsen, Am. J. Phys. 73, 164 (2005). 6 The partition is inserted rapidly; if it is done adiabatically the particle may be forced into the (however slightly) larger of the two, as you found in Problem 11.34. 12.2 BELL\u2019S THEOREM 12.2 Bell\u2019s Theorem 449 Einstein, Podolsky, and Rosen did not doubt that quantum mechanics is correct, as far as it goes; they only claimed that it is an incomplete description of physical reality: The wave function is not the whole story\u2014some other quantity, \u03bb, is needed, in addition to, to characterize the state of a system fully. We call \u03bb the \u201chidden variable\u201d because, at this stage, we have no idea how to calculate or measure it.7 Over the years, a number of hidden variable theories have been proposed, to supplement quantum mechanics;8 they tend to be cumbersome and implausible, but never mind\u2014until 1964 the program seemed eminently worth pursuing. But in that year J. S. Bell proved that any local hidden variable theory is incompatible with quantum mechanics.9 Bell suggested a generalization of the EPRB experiment: Instead of orienting the electron and positron detectors along the same direction, he allowed them to be rotated independently. The \ufb01rst measures the component of the electron spin in the direction of a unit vector a, and the second measures the spin of the positron along the direction b (Figure 12.2). For simplicity, let\u2019s record the spins in units of /2; then each detector registers the value +1 (for spin up) or \u22121 (spin down), along the direction in question. A table of results, for many \u03c0 0 decays, might look like this: electron +1 +1 \u22121 +1 \u22121... positron \u22121 +1 +1 \u22121 \u22121... product \u22121 +1 \u22121 \u22121 +1... Bell proposed to calculate the average value of the product of the spins, for a given set of detector orientations. Call this average P (a, b). If the detectors are parallel (b = a), we recover the original EPRB con\ufb01guration; in this case one is spin up and the other spin down, so the product is always \u22121, and hence so too is the average: e\u2212 a \u03c00 b e+ Figure 12.2: Bell\u2019s version of the EPRB experiment: detectors independently oriented in directions a and b. 7 The hidden variable could be a single number, or it could be a whole collection of numbers; perhaps \u03bb is to be calculated in some future theory, or maybe it is for some reason of principle incalculable. It hardly matters. All I am asserting is that there must be something\u2014if only a list of the outcomes of every possible experiment\u2014associated with the system prior to a measurement. 8 D. Bohm, Phys. Rev. 85, 166, 180 (1952). 9 Bell\u2019s original paper (Physics 1, 195 (1964), reprinted as Chapter 2 in John S. Bell, Speakable and Unspeakable in Quantum Mechanics, Cambridge University Press, UK (1987)) is a gem: brief, accessible, and beautifully written. 450 CHAPTER 12 Afterword By the same token, if they are anti-parallel (b = \u2212a), then every product is +1, so P(a, a) = \u22121. For arbitrary orientations, quantum mechanics predicts P(a, \u2212a) = +1. P(a, b) = \u2212a \u00b7 b (12.2) (12.3) (12.4) (see Problem 4.59). What Bell discovered is that this result is incompatible with any local hidden variable theory. The argument is stunningly simple. Suppose that the \u201ccomplete\u201d state of the electron\u2013 positron system is characterized by the hidden variable(s) \u03bb (\u03bb varies, in some way that we neither understand nor control, from one pion decay to the next). Suppose further that the outcome of the electron measurement is independent of the orientation (b) of the positron detector\u2014which may, after all, be chosen by the experimenter at the positron end just before the electron measurement is made, and hence far too late for any subluminal message to get back to the electron detector. (This is the locality assumption.) Then there exists some function A(a, \u03bb) which determines the result of an electron measurement, and some other function B(b, \ufffd", "\ufffd) for the positron measurement. These functions can only take on the values \u00b11:10 A(a, \u03bb) = \u00b11; B(b, \u03bb) = \u00b11. When the detectors are aligned, the results are perfectly (anti)-correlated: A(a, \u03bb) = \u2212B(a, \u03bb), regardless of the value of \u03bb. Now, the average of the product of the measurements is (12.5) (12.6) P(a, b) = \u03c1(\u03bb)A(a, \u03bb)B(b, \u03bb) d\u03bb, (12.7) where \u03c1(\u03bb) is the probability density for the hidden variable. (Like any probability density, it is \u03c1(\u03bb) d\u03bb = 1, but beyond this we real, nonnegative, and satis\ufb01es the normalization condition make no assumptions about \u03c1(\u03bb); different hidden variable theories would presumably deliver quite different expressions for \u03c1.) In view of Equation 12.6, we can eliminate B: P(a, b) = \u2212 \u03c1(\u03bb)A(a, \u03bb)A(b, \u03bb) d\u03bb. (12.8) If c is any other unit vector, P(a, b) \u2212 P(a, c) = \u2212 \u03c1(\u03bb) A(a, \u03bb)A(b, \u03bb) \u2212 A(a, \u03bb)A(c, \u03bb) d\u03bb. (12.9) 10 This already concedes far more than a classical determinist would be prepared to allow, for it abandons any notion that the particles could have well-de\ufb01ned angular momentum vectors with simultaneously determinate components. The point of Bell\u2019s argument is to demonstrate that quantum mechanics is incompatible with any local deterministic theory\u2014even one that bends over backwards to be accommodating. Of course, if you reject Equation 12.5, then the theory is manifestly incompatible with quantum mechanics. Or, since [A(b, \u03bb)]2 = 1: P(a, b) \u2212 P(a, c) = \u2212 \u03c1(\u03bb) 1 \u2212 A(b, \u03bb)A(c, \u03bb) A(a, \u03bb)A(b, \u03bb) d\u03bb. (12.10) 12.2 Bell\u2019s Theorem 451 it follows from Equation 12.5 that But A(b, \u03bb)A(c, \u03bb)] \u2265 0, so |P(a, b) \u2212 P(a, c)| \u2264 or, more simply: |A(a, \u03bb)A(b, \u03bb)| = 1; moreover \u03c1 (\u03bb) [1 \u2212 \u03c1(\u03bb) 1 \u2212 A(b, \u03bb)A(c, \u03bb) d\u03bb, |P(a, b) \u2212 P(a, c)| \u2264 1 + P(b, c). (12.11) (12.12) This is the famous Bell inequality. It holds for any local hidden variable theory (subject only to the minimal requirements of Equations 12.5 and 12.6), for we have made no assumptions whatever as to the nature or number of the hidden variable(s), or their distribution (\u03c1). But it is easy to show that the quantum mechanical prediction (Equation 12.4) is incompatible with Bell\u2019s inequality. For example, suppose the three vectors lie in a plane, and c makes a 45\u25e6 angle with a and b (Figure 12.3); in this case quantum mechanics says P(a, b) = 0, P(a, c) = P(b, c) = \u22120.707, which is patently inconsistent with Bell\u2019s inequality: 0.707 \u2264 1 \u2212 0.707 = 0.293. With Bell\u2019s modi\ufb01cation, then, the EPR paradox proves something far more radical than its authors imagined: If they are right, then not only is quantum mechanics incomplete, it is downright wrong. On the other hand, if quantum mechanics is right, then no hidden variable theory is going to rescue us from the nonlocality Einstein considered so preposterous. Moreover, we are provided with a very simple experiment to settle the issue once and for all.11 Many experiments to test Bell\u2019s inequality were performed in the 1960s and 1970s, culminating in the work of Aspect, Grangier, and Roger.12 The details do not concern us here (they b c 45\u00b0 45\u00b0 a Figure 12.3: An orientation of the detectors that demonstrates quantum violations of Bell\u2019s inequality. 11 It is an embarrassing historical fact that Bell\u2019s theorem, which is now universally recognized as one of the most profound discoveries of the twentieth century, was barely", " noticed at the time, with the exception of an inspired fringe element. For a fascinating account, see David Kaiser, How the Hippies Saved Physics, W. W. Norton, New York, 2011. 12 A. Aspect, P. Grangier, and G. Roger, Phys. Rev. Lett. 49, 91 (1982). There were logically possible (if implausible) loopholes in the Aspect experiment, which were gradually closed over the ensuing years; see J. Handsteiner et al., Phys. Rev. Lett. 118, 060401 (2017). It is now possible to test Bell\u2019s inequality in the undergraduate laboratory: D. Dehlinger and M. W. Mitchell, Am. J. Phys. 70, 903 (2002). 452 CHAPTER 12 Afterword actually used two-photon atomic transitions, not pion decays). To exclude the remote possibility that the positron detector might somehow \u201csense\u201d the orientation of the electron detector, both orientations were set quasi-randomly after the photons were already in \ufb02ight. The results were in excellent agreement with the predictions of quantum mechanics, and inconsitent with Bell\u2019s inequality by a wide margin.13 Ironically, the experimental con\ufb01rmation of quantum mechanics came as something of a shock to the scienti\ufb01c community. But not because it spelled the demise of \u201crealism\u201d\u2014most physicists had long since adjusted to this (and for those who could not, there remained the possibility of nonlocal hidden variable theories, to which Bell\u2019s theorem does not apply).14 The real shock was the demonstration that nature itself is fundamentally nonlocal. Nonlocality, in the form of the instantaneous collapse of the wave function (and for that matter also in the symmetrization requirement for identical particles) had always been a feature of the orthodox interpretation, but before Aspect\u2019s experiment it was possible to hope that quantum nonlocality was somehow a nonphysical artifact of the formalism, with no detectable consequences. That hope can no longer be sustained, and we are obliged to reexamine our objection to instantaneous action-at-a-distance. Why are physicists so squeamish about superluminal in\ufb02uences? After all, there are many things that travel faster than light. If a bug \ufb02ies across the beam of a movie projector, the speed of its shadow is proportional to the distance to the screen; in principle, that distance can be as large as you like, and hence the shadow can move at arbitrarily high velocity (Figure 12.4). However, the shadow does not carry any energy, nor can it transmit any information from one point on the screen to another. A person at point X cannot cause anything to happen at point Y by manipulating the passing shadow. On the other hand, a causal in\ufb02uence that propagated faster than light would carry unacceptable implications. For according to special relativity there exist inertial frames in which such a signal propagates backward in time\u2014the effect preceding the cause\u2014and this leads to inescapable logical anomalies. (You could, for example, arrange to kill your infant grandfather. Think about it... not a good idea.) The question is, are the superluminal in\ufb02uences predicted by quantum mechanics and detected by Aspect causal, in this sense, or are they somehow ethereal enough (like the bug\u2019s shadow) to escape the philosophical objection? 13 Bell\u2019s theorem involves averages and it is conceivable that an apparatus such as Aspect\u2019s contains some secret bias which selects out a nonrepresentative sample, thus distorting the average. In 1989, an improved version of Bell\u2019s theorem was proposed, in which the contrast between the quantum prediction and that of any local hidden variable theory is even more dramatic. See D. Greenberger, M. Horne, A. Shimony, and A. Zeilinger, Am. J. Phys. 58, 1131 (1990) and N. D. Mermin, Am. J. Phys. 58, 731, (1990). An experiment of this kind suitable for an undergraduate laboratory has been carried out by Mark Beck and his students: Am. J. Phys. 74, 180 (2006). 14 It is a curious twist of fate that the EPR paradox, which assumed locality in order to prove realism, led \ufb01nally to the demise of locality and left the issue of realism undecided\u2014the outcome (as Bell put it) Einstein would have liked least. Most physicists today consider that if they can\u2019t have local realism, there\u2019s not much point in realism at all, and for this reason nonlocal hidden variable theories occupy a rather peripheral niche. Still, some authors\u2014notably Bell himself, in Speakable and Unspeakable in Quantum Mechanics (footnote 9 in this", " chapter)\u2014argue that such theories offer the best hope of bridging the conceptual gap between the measured system and the measuring apparatus, and for supplying an intelligible mechanism for the collapse of the wave function. \u03c5 Projector Bug 12.2 Bell\u2019s Theorem 453 Y \u03c5\u00b4 X Screen Figure 12.4: The shadow of the bug moves across the screen at a velocity v screen is far enough away. greater than c, provided the Well, let\u2019s consider Bell\u2019s experiment. Does the measurement of the electron in\ufb02uence the outcome of the positron measurement? Assuredly it does\u2014otherwise we cannot account for the correlations in the data. But does the measurement of the electron cause a particular outcome for the positron? Not in any ordinary sense of the word. There is no way the person manning the electron detector could use his measurement to send a signal to the person at the positron detector, since he does not control the outcome of his own measurement (he cannot make a given electron come out spin up, any more than the person at X can affect the passing shadow of the bug). It is true that he can decide whether to make a measurement at all, but the positron monitor, having immediate access only to data at his end of the line, cannot tell whether the electron has been measured or not. The lists of data compiled at the two ends, considered separately, are completely random. It is only later, when we compare the two lists, that we discover the remarkable correlations. In another reference frame the positron measurements occur before the electron measurements, and yet this leads to no logical paradox\u2014the observed correlation is entirely symmetrical in its treatment, and it is a matter of indifference whether we say the observation of the electron in\ufb02uenced the measurement of the positron, or the other way around. This is a wonderfully delicate kind of in\ufb02uence, whose only manifestation is a subtle correlation between two lists of otherwise random data. We are led, then, to distinguish two types of in\ufb02uence: the \u201ccausal\u201d variety, which produce actual changes in some physical property of the receiver, detectable by measurements on that subsystem alone, and an \u201cethereal\u201d kind, which do not transmit energy or information, and for which the only evidence is a correlation in the data taken on the two separate subsystems\u2014 a correlation which by its nature cannot be detected by examining either list alone. Causal in\ufb02uences cannot propagate faster than light, but there is no compelling reason why ethereal ones should not. The in\ufb02uences associated with the collapse of the wave function are of the latter type, and the fact that they \u201ctravel\u201d faster than light may be surprising, but it is not, after all, catastrophic.15 15 An enormous amount has been written about Bell\u2019s theorem. My favorite is an inspired essay by David Mermin in Physics Today (April 1985, page 38). An extensive bibliography will be found in L. E. Ballentine, Am. J. Phys. 55, 785 (1987). 454 CHAPTER 12 Afterword \u2217\u2217 Problem 12.3 One example16 of a (local) deterministic (\u201chidden variable\u201d) theory is... classical mechanics! Suppose we carried out the Bell experiment with classical objects (baseballs, say) in place of the electron and proton. They are launched (by a kind of double pitching machine) in opposite directions, with equal and opposite spins (angular momenta), Sa and Sb = \u2212Sa. Now, these are classical objects\u2014their angular momenta can point in any direction, and this direction is set (let\u2019s say randomly) at the moment of launch. Detectors placed 10 meters or so on either side of the launch point measure the spin vectors of their respective baseballs. However, in order to match the conditions for Bell\u2019s theorem, they only record the sign of the component of S along the directions a and b: A \u2261 sign (a \u00b7 Sa), B \u2261 sign (b \u00b7 Sb). Thus each detector records either +1 or \u22121, in any given trial. In this example the \u201chidden variable\u201d is the actual orientation of Sa, speci\ufb01ed (say) by the polar and azimuthal angles \u03b8 and \u03c6: \u03bb = (\u03b8, \u03c6): z S \u03b8 \u03c6 a \u03b7 b x Figure 12.5: Axes for Problem 12.3. y (a) Choosing axes as in the \ufb01gure, with a and b in the x\u2013y plane and a along the x axis, verify that A(a, \u03bb) B(b, \u03bb) = \u2212sign [cos(\u03c6) cos(\u03c6 \u2212 \u03b7)],", " where \u03b7 is the angle between a and b (take it to run from \u2212\u03c0 to +\u03c0 ). (b) Assuming the baseballs are launched in such a way that Sa is equally likely to point in any direction, compute P(a, b). Answer: (2|\u03b7|/\u03c0 ) \u2212 1. (c) Sketch the graph of P(a, b), from \u03b7 = \u2212\u03c0 to +\u03c0, and (on the same graph) the quantum formula (Equation 12.4, with \u03b8 \u2192 \u03b7). For what values of \u03b7 does this hidden variable theory agree with the quantum-mechanical result? (d) Verify that your result satis\ufb01es Bell\u2019s inequality, Equation 12.12. Hint: The vectors a, b, and c de\ufb01ne three points on the surface of a unit sphere; the inequality can be expressed in terms of the distances between those points. 16 This problem is based on George Greenstein and Arthur G. Zajonc, The Quantum Challenge, 2nd edn., Jones and Bartlett, Sudbury, MA (2006), Section 5.3. 12.3 MIXED STATES AND THE DENSITY MATRIX 12.3.1 Pure States 12.3 Mixed States and the Density Matrix 455 In this book we have dealt with particles in pure states, |\u2014a harmonic oscillator in its nth stationary state, for instance, or in a speci\ufb01c linear combination of stationary states, or a free particle in a gaussian wave packet. The expectation value of some observable A is then 0 \u02c6A 1 A = ; (12.13) it\u2019s the average of measurements on an ensemble of identically-prepared systems, all of them in the same state |. We developed the whole theory in terms of | (a vector in Hilbert space, or, in the position basis, the wave function). But there are other ways to formulate the theory, and a particularly useful one starts by de\ufb01ning the density operator,17 With respect to an orthonormal basis element of the matrix A representing the operator \u02c6A is an operator is represented by a matrix; the i j \u02c6\u03c1 \u2261 ||. \u2019 ( |e j 0 \u02c6A e j 1. Ai j = ei (12.14) (12.15) (12.16) (12.17) (12.18) (12.19) (12.20) In particular, the i j element of the density matrix \u03c1 is \u02c6\u03c1 = \u03c1i j = ei e j ei e j. The density matrix (for pure states) has several interesting properties: \u03c12 = \u03c1, \u03c1\u2020 = \u03c1, (it is idempotent), (it is hermitian), (its trace is 1). Tr (\u03c1) = \u03c1ii = 1, The expectation value of an observable A is i A = Tr(\u03c1A). We could do everything using the density matrix, instead of the wave function, to represent the state of a particle. Example 12.1 In the standard basis 1 0, 0 1, e2 = e1 = (12.21) representing spin up and spin down along the z direction (Equation 4.149), construct the density matrix for an electron with spin up along the x direction. Solution: In this case | = \u221a \u221a 1/ 1/ 2 2 (12.22) 17 It\u2019s actually the \u201cprojection operator\u201d onto the state |\u2014see Equation 3.91. 456 CHAPTER 12 Afterword (Equation 4.151). So \u03c111 = 1 0 1 0 0 1 0 1 \u03c112 = \u03c121 = \u03c122 = \u221a \u221a \u221a \u221a \u221a \u221a \u221a \u221a 1/ 1/ 1/ 1/ 1/ 1/ 1/ 1/ 1/ \u221a 1/ \u221a 1/ \u221a 1/ \u221a \u221a 2 1/ \u221a 2 1/ \u221a 2 1/ \u221a 2 1\u221a 2 = 1\u221a 2 = 1\u221a 2 = 1\u221a 2 1\u221a 2 1\u221a 2 1\u221a 2 1,,,, and hence \u03c1 = 1/2 1/2. 1/2 1/2 Or, more ef\ufb01ciently, \u03c1 = || = \u221a \u221a 1/ 1/ 2 2 \u221a 1/ \u221a 2 1/ 2 = 1/2 1/2. 1/2 1/2 Note that \u03c1 is hermitian, its trace is 1, and 1/2 1/2 1/2 1/2 1/2 1/2 \u03c12 = 1/2 1/2 = 1/2 1", "/2 1/2 1/2 = \u03c1. (12.23) (12.24) (12.25) Problem 12.4 (a) Prove properties 12.17, 12.18, 12.19, and 12.20. (b) Show that the time evolution of the density operator is governed by the equation # $. \u02c6H, \u02c6\u03c1 = i d \u02c6\u03c1 dt (12.26) (This is the Schr\u00f6dinger equation, expressed in terms of \u02c6\u03c1.) Problem 12.5 Repeat Example 12.1 for an electron with spin down along the y direction. 12.3.2 Mixed States In practice it is often the case that we simply don\u2019t know the state of the particle. Suppose, for example, we are interested in an electron emerging from the Stanford Linear Accelerator. It might have spin up (along some chosen direction), or it might have spin down, or it might be in some linear combination of the two\u2014we just don\u2019t know.18 We say that the particle is in a mixed state.19 18 I\u2019m not talking about any fancy quantum phenomenon (Heisenberg uncertainty or Born indeterminacy, which would apply even if we knew the precise state); I\u2019m talking here about good old-fashioned ignorance. 19 Do not confuse a linear combination of two pure states, which itself is still a pure state (the sum of two vectors in Hilbert space is another vector in Hilbert space) with a mixed state, which is not represented by any (single) vector in the Hilbert space. 12.3 Mixed States and the Density Matrix 457 How should we describe such a particle? I could simply list the probability, pk, that it\u2019s in each possible state |k. The expectation value of an observable would now be the average of measurements taken over an ensemble of systems that are not identically prepared (they are not all in the same state); rather, a fraction pk of them is in each (pure) state |k: 1 0 \u02c6A k. k A = pk k There\u2019s a slick way to package this information, by generalizing the density operator: \u02c6\u03c1 \u2261 pk|kk|. k Again, it becomes a matrix when referred to a particular basis: \u03c1i j = pkei | k k e j. k The density matrix encodes all the information available to us about the system. Like any probabilities, 0 \u2264 pk \u2264 1 and k pk = 1. (12.27) (12.28) (12.29) (12.30) The density matrix for mixed states retains most of the properties we identi\ufb01ed for pure states: \u03c1\u2020 = \u03c1, Tr (\u03c1) = 1, A = Tr(\u03c1A). $ # i d \u02c6\u03c1, (if dt \u02c6H, \u02c6\u03c1 = dpk dt but \u03c1 is idempotent only if it represents a pure state: \u03c12 = \u03c1 = 0 for all k) (12.31) (12.32) (12.33) (12.34) (12.35) (indeed, this is a quick way to test whether the state is pure). Example 12.2 Construct the density matrix for an electron that is either in the spin-up state or the spindown state (along z), with equal probability. \u03c1 = Solution: In this case p1 = p2 = 1/2, so pk|kk| = 1 2 0 0 0 1/2 Note that \u03c1 is hermitian, and its trace is 1, but k 1/2 0 0 0 1 0 = + \u03c12 = this is not a pure state/2 0 = 0 1/2. (12.36) 1/2 0 0 1/2 1/2 0 1/4 0 = 0 1/4 = \u03c1; (12.37) 0 1/2 458 CHAPTER 12 Afterword Problem 12.6 (a) Prove properties 12.31, 12.32, 12.33, and 12.34. (b) Show that Tr (c) Show that \u03c12 = \u03c1 if and only if \u03c1 represents a pure state. \u2264 1, and equal to 1 only if \u03c1 represents a pure state. \u03c12 Problem 12.7 (a) Construct the density matrix for an electron that is either in the state spin up along x (with probability 1/3) or in the state spin down along y (with probability 2/3). (b) Find Sy for the electron in (a). Problem 12.8 (a) Show that the most general density matrix for a spin-1/2 particle can be written in terms of three real numbers (a1, a2", ", a3): \u03c1 = 1 2 (a1 \u2212 ia2) (1 \u2212 a3) where \u03c31, \u03c32, \u03c33 are the three Pauli matrices. Hint: It has to be hermitian, and its trace must be 1. (1 + a3) (a1 + ia2) (1 + a \u00b7 \u03c3 ), = 1 2 (12.38) (b) In the literature, a is known as the Bloch vector. Show that \u03c1 represents a pure state if and only if |a| = 1, and for a mixed state |a| < 1. Hint: Use Problem 12.6(c). Thus every density matrix for a spin-1/2 particle corresponds to a point in the Bloch sphere, of radius 1. Points on the surface are pure states, and points inside are mixed states. (c) What is the probability that a measurement of Sz would return the value +/2, if the tip of the Bloch vector is at (i) the north pole (a = (0, 0, 1)), (ii) the center of the sphere (a = (0, 0, 0)), (iii) the south pole (a = (0, 0, \u22121))? (d) Find the spinor \u03c7 representing the (pure) state of the system, if the Bloch vector lies on the equator, at azimuthal angle \u03c6. 12.3.3 Subsystems There is another context in which one might invoke the density matrix formalism: an entangled state, such as the singlet spin con\ufb01guration of an electron/positron pair, 1\u221a 2 (|\u2191\u2193 \u2212 |\u2193\u2191). (12.39) Suppose we are interested only in the positron: what is it\u2019s state? I cannot say... a measurement could return spin up (\ufb01fty\u2013\ufb01fty probability) or spin down. This has nothing to do with ignorance; I know the state of the system precisely. But the subsystem (the positron) by itself does not occupy a pure state. If I insist on talking about the positron alone, the best I can do is to tell you its density matrix: 1/2 0 0 1/2, (12.40) \u03c1 = representing the 50/50 mixture. 12.4 The No-Clone Theorem 459 Of course, this is the same as the density matrix representing a positron in a speci\ufb01c (but unknown) spin state (Example 12.2). I\u2019ll call it a subsystem density matrix, to distinguish it from an ignorance density matrix. The EPRB paradox illustrates the difference. Before the electron spin was measured, the positron (alone) was represented by the \u201csubsystem\u201d density matrix (Equation 12.40); when the electron is measured the positron is knocked into a de\ufb01nite state... but we (at the distant positron detector) don\u2019t know which. The positron is now represented by the \u201cignorance\u201d density matrix (Equation 12.36). But the two density matrices are identical! Our description of the state of the positron has not been altered by the measurement of the electron\u2014all that has changed is our reason for using the density matrix formalism. 12.4 THE NO-CLONE THEOREM Quantum measurements are typically destructive, in the sense that they alter the state of the system measured. This is how the uncertainty principle is enforced in the laboratory. You might wonder why we don\u2019t just make a bunch of identical copies (clones) of the original state, and measure them, leaving the system itself unscathed. It can\u2019t be done. Indeed, if you could build a cloning device (a \u201cquantum Xerox machine\u201d), quantum mechanics would be out the window. For example, it would then be possible to send superluminal messages using the EPRB apparatus.20 Say the message to be transmitted, from the operator of the electron detector (conventionally \u201cAlice\u201d) to the operator of the positron detector (\u201cBob\u201d), is either \u201cyes\u201d (\u201cdrop the bomb\u201d) or \u201cno.\u201d If the message is to be \u201cyes,\u201d Alice measures Sz (of the electron). Never mind what result she gets\u2014all that matters is that she makes the measurement, for this means that the positron is now in the pure state \u2191 or \u2193 (never mind which). If she wants to say \u201cno,\u201d she measures Sx, and that means the positron is now in the de\ufb01nite state \u2190 or \u2192 (never mind which). In", " any case, Bob makes a million clones of the positron, and measures Sz on half of them, and Sx on the other half. If the \ufb01rst group are all in the same state (all \u2191 or all \u2193), then Alice must have measured Sz, and the message is \u201cyes\u201d (the Sx group should be a 50/50 mixture). If all the Sx measurements yield the same answer (all \u2190 or all \u2192), then Alice must have measured Sx, and the message is \u201cno\u201d (in that case the Sz measurements should be a 50/50 mixture). It doesn\u2019t work, because you can\u2019t make a quantum Xerox machine, as Wootters, Zurek, and Dieks proved in 1982.21 Schematically, we want the machine to take as input a particle in state |\u03c8 (the one to be copied), plus a second particle in state |X (the \u201cblank sheet of paper\u201d), and spit out two particles in the state |\u03c8 (original plus copy): |\u03c8|X \u2192 |\u03c8|\u03c8. Suppose we have made a device that successfully clones the state |\u03c81: |\u03c81|X \u2192 |\u03c81|\u03c81, (12.41) (12.42) 20 Starting around 1975, members of the so-called \u201cFundamental Fysiks Group\u201d proposed a series of increasingly ingenious schemes for faster-than-light communication\u2014inspiring in turn a succession of increasingly sophisticated rebuttals, culminating in the no-clone theorem, which \ufb01nally put a stop to the whole misguided enterprise. For a fascinating account, see Chapter 11 of Kaiser\u2019s How the Hippies Saved Physics (footnote 11, page 451). 21 W. K. Wootters and W. H. Zurek, Nature 299, 802 (1982); D. Dieks, Phys. Lett. A 92, 271 (1982). 460 CHAPTER 12 Afterword and also works for state |\u03c82: |\u03c82|X \u2192 |\u03c82|\u03c82 (12.43) (|\u03c81 and |\u03c82 might be spin up and spin down, for example, if the particle is an electron). So far, so good. But what happens when we feed in a linear combination |\u03c8 = \u03b1|\u03c81 + \u03b2|\u03c82? Evidently we get22 |\u03c8|X \u2192 \u03b1|\u03c81|\u03c81 + \u03b2|\u03c82|\u03c82, (12.44) which is not at all what we wanted\u2014what we wanted was |\u03c8|X \u2192 |\u03c8|\u03c8 = [\u03b1|\u03c81 + \u03b2|\u03c82][\u03b1|\u03c81 + \u03b2|\u03c82] = \u03b12|\u03c81|\u03c81 + \u03b22|\u03c82|\u03c82 + \u03b1\u03b2[|\u03c81|\u03c82 + |\u03c82|\u03c81]. (12.45) You can make a machine to clone spin-up electrons and spin-down electrons, but it will fail for any nontrivial linear combinations (such as eigenstates of Sx ). It\u2019s as though you bought a Xerox machine that copies vertical lines perfectly, and also horizontal lines, but completely distorts diagonals. The no-clone theorem turned out to have an importance well beyond \u201cmerely\u201d protecting quantum mechanics from superluminal communication (and hence an inescapable con\ufb02ict with special relativity).23 In particular, it opened up the \ufb01eld of quantum cryptography, which exploits the theorem to detect eavesdropping.24 This time Alice and Bob want to agree on a key for decoding messages, without the cumbersome necessity of actually meeting face-toface. Alice is to send the key (a string of numbers) to Bob via a stream of carefully prepared photons.25 But they are worried that their nemesis, Eve, might try to intercept this communication, and thereby crack the code, without their knowledge. Alice prepares a string of photons in four different states: linearly polarized (horizontal |H and vertical |V ), and circularly polarized (left |L and right |R), which she sends to Bob. Eve hopes to capture and clone the photons en route, sending the originals along to Bob, who will be none the wiser. (Later on, she knows, Alice and Bob will compare notes on a sample of the photons, to make sure there has been no tampering\u2014that\u2019s why she has to clone them perfectly, to go undetected.) But the no-clone theorem guarantees that Eve\u2019s Xerox machine will fail;26 Alice and Bob will catch the eavesdropping when they compare the samples. (They will", " then, presumably, discard that key.) 22 This assumes that the device acts linearly on the state |\u03c8, as it must, since the time-dependent Schr\u00f6dinger equation (which presumably governs the process) is linear. 23 The no-clone theorem is one of the foundations for quantum information theory, \u201cteleportation,\u201d and quantum computation. For a brief history and a comprehensive bibliography, see F. W. Strauch, Am. J. Phys. 84, 495 (2016). 24 For a brief summary, see W. K. Wootters and W. H. Zurek, Physics Today, February 2009, page 76. 25 Electrons would do, but traditionally the story is told using photons. By the way, there is no entanglement involved, and they\u2019re not in a hurry\u2014this has nothing to do with EPR or superluminal signals. 26 If Alice and Bob were foolish enough to use just two orthogonal photon states (say, |H and |V ), then Eve might get lucky, and use a quantum Xerox machine that does faithfully clone those two states. But as long as they include nontrivial linear combinations (such as |R and |L), the cloning is certain to fail, and the eavesdropping will be detected. 12.5 SCHR\u00d6DINGER\u2019S CAT 12.5 Schr\u00f6dinger\u2019s Cat 461 The measurement process plays a mischievous role in quantum mechanics: It is here that indeterminacy, nonlocality, the collapse of the wave function, and all the attendant conceptual dif\ufb01culties arise. Absent measurement, the wave function evolves in a leisurely and deterministic way, according to the Schr\u00f6dinger equation, and quantum mechanics looks like a rather ordinary \ufb01eld theory (much simpler than classical electrodynamics, for example, since there is only one \ufb01eld (), instead of two (E and B), and it\u2019s a scalar). It is the bizarre role of the measurement process that gives quantum mechanics its extraordinary richness and subtlety. But what, exactly, is a measurement? What makes it so different from other physical processes?27 And how can we tell when a measurement has occurred? Schr\u00f6dinger posed the essential question most starkly, in his famous cat paradox:28 A cat is placed in a steel chamber, together with the following hellish contraption.... In a Geiger counter there is a tiny amount of radioactive substance, so tiny that maybe within an hour one of the atoms decays, but equally probably none of them decays. If one decays then the counter triggers and via a relay activates a little hammer which breaks a container of cyanide. If one has left this entire system for an hour, then one would say the cat is living if no atom has decayed. The \ufb01rst decay would have poisoned it. The wave function of the entire system would express this by containing equal parts of the living and dead cat. At the end of the hour, then, the wave function of the cat has the schematic form \u03c8 = 1\u221a 2 (\u03c8alive + \u03c8dead). (12.46) The cat is neither alive nor dead, but rather a linear combination of the two, until a measurement occurs\u2014until, say, you peek in the window to check. At that moment your observation forces the cat to \u201ctake a stand\u201d: dead or alive. And if you \ufb01nd him to be dead, then it\u2019s really you who killed him, by looking in the window. Schr\u00f6dinger regarded this as patent nonsense, and I think most people would agree with him. There is something absurd about the very idea of a macroscopic object being in a linear combination of two palpably different states. An electron can be in a linear combination of spin up and spin down, but a cat simply cannot be in a linear combination of alive and dead. But how are we to reconcile this with quantum mechanics? 27 There is a school of thought that rejects this distinction, holding that the system and the measuring apparatus should be described by one great big wave function which itself evolves according to the Schr\u00f6dinger equation. In such theories there is no collapse of the wave function, but one must typically abandon any hope of describing individual events\u2014quantum mechanics (in this view) applies only to ensembles of identically prepared systems. See, for example, Philip Pearle, Am. J. Phys. 35, 742 (1967), or Leslie E. Ballentine, Quantum Mechanics: A Modern Development, 2nd edn, World Scienti\ufb01c, Singapore (1998). 28 E. Schr\u00f6dinger, Naturwiss. 48, 52 (1935); translation by Josef M. Jauch, Foundations of Quantum Mechanics,", " Addison-Wesley, Reading, MA (1968), page 185. 462 CHAPTER 12 Afterword The Schr\u00f6dinger cat paradox forces us to confront the question \u201cWhat constitutes a \u2018measurement,\u2019 in quantum mechanics\u201d? Does the \u201cmeasurement\u201d really occur when we peek in the keyhole? Or did it happen much earlier, when the atom did (or did not) decay? Or was it when the Geiger counter registered (or did not) the decay, or when the hammer did (or did not) hit the vial of cyanide? Historically, there have been many answers to this question. Wigner held that measurement requires the intervention of human consciousness; Bohr thought it meant the interaction between a microscopic system (subject to the laws of quantum mechanics) and a macroscopic measuring apparatus (described by classical laws); Heisenberg maintained that a measurement occurs when a permanent record is left; others have pointed to the irreversible nature of a measurement. The embarrassing fact is that none of these characterizations is entirely satisfactory. Most physicists would say that the measurement occurred (and the cat became either alive or dead) well before we looked in the window, but there is no real consensus as to when or why. And this still leaves the deeper question of why a macroscopic system cannot occupy a linear combination of two clearly distinct states\u2014a baseball, say, in a linear combination of Seattle and Toronto. Suppose you could get a baseball into such a state, what would happen to it? In some ultimate sense the macroscopic system must itself be described by the laws of quantum mechanics. But wave functions, in the \ufb01rst instance, represent individual elementary particles; the wave function of a macroscopic object would be a monstrously complicated composite structure, built out of the wave functions of its 1023 constituent particles. And it is subject to constant bombardment from the environment29\u2014subject, that is, to continuous \u201cmeasurement\u201d and the attendant collapse. In this process, presumably, \u201cclassical\u201d states are statistically favored, and in practice the linear combination devolves almost instantaneously into one of the ordinary con\ufb01gurations we encounter in everyday life. This phenomenon is called decoherence, and although it is still not entirely understood it appears to be the fundamental mechanism by which quantum mechanics reduces to classical mechanics in the macroscopic realm.30 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 In this book I have tried to tell a consistent and coherent story: The wave function () represents the state of a particle (or system); particles do not in general possess speci\ufb01c dynamical properties (position, momentum, energy, angular momentum, etc.) until an act of measurement intervenes; the probability of getting a particular value in any given experiment is determined by the statistical interpretation of ; upon measurement the wave function collapses, so that an immediately repeated measurement is certain to yield the same result. There are other possible interpretations\u2014nonlocal hidden variable theories, the \u201cmany worlds\u201d picture, \u201cconsistent histories,\u201d ensemble models, and others\u2014but I believe this one is conceptually the simplest, 29 This is true even if you put it in an almost complete vacuum, cool it down practically to absolute zero, and somehow shield out the cosmic background radiation. It is possible to imagine a single electron avoiding all contact for a signi\ufb01cant time, but not a macroscopic object. 30 See, for example, M. Schlosshauer, Decoherence and the Quantum-to-Classical Transition, Springer, (2007), or W. H. Zurek, Physics Today, October, 2014, page 44. 12.5 Schr\u00f6dinger\u2019s Cat 463 and certainly it is the one shared by most physicists today.31 It has stood the test of time, and emerged unscathed from every experimental challenge. But I cannot believe this is the end of the story; at the very least, we have much to learn about the nature of measurement and the mechanism of collapse. And it is entirely possible that future generations will look back, from the vantage point of a more sophisticated theory, and wonder how we could have been so gullible. 31 See Daniel Styer et al., Am. J. Phys. 70, 288 (2002). APPENDIX LINEAR ALGEBRA Linear algebra abstracts and generalizes the arithmetic of ordinary vectors, as we encounter them in \ufb01rst-year physics. The generalization is in two directions: (1) we allow the scalars to be complex numbers, and (2) we do not restrict ourselves to three dimensions. A.1 VECTORS A vector space consists of a set of vectors (|\u03b1, |\u03b2, |\u03b3,...), together with a set of scalars (a, b, c,...),1 which is closed2 under two operations:", " vector addition and scalar multiplication. \u2022 Vector Addition The \u201csum\u201d of any two vectors is another vector: Vector addition is commutative: |\u03b1 + |\u03b2 = |\u03b3. |\u03b1 + |\u03b2 = |\u03b2 + |\u03b1, and associative: |\u03b1 + (|\u03b2 + |\u03b3 ) = (|\u03b1 + |\u03b2) + |\u03b3. There exists a zero (or null) vector,3 |0, with the property that |\u03b1 + |0 = |\u03b1, (A.1) (A.2) (A.3) (A.4) for every vector |\u03b1. And for every vector |\u03b1 there is an associated inverse vector (| \u2212 \u03b1),4 such that |\u03b1 + |\u2212\u03b1 = |0. (A.5) 1 For our purposes, the scalars will be ordinary complex numbers. Mathematicians can tell you about vector spaces over more exotic \ufb01elds, but such objects play no role in quantum mechanics. Note that \u03b1, \u03b2, \u03b3... are not (ordinarily) numbers; they are names (labels)\u2014\u201cCharlie,\u201d for instance, or \u201cF43A-9GL,\u201d or whatever you care to use to identify the vector in question. 2 That is to say, these operations are always well-de\ufb01ned, and will never carry you outside the vector space. 3 It is customary, where no confusion can arise, to write the null vector without the adorning bracket: |0 \u2192 0. 4 This is funny notation, since \u03b1 is not a number. I\u2019m simply adopting the name \u201c\u2212Charlie\u201d for the inverse of the vector whose name is \u201cCharlie.\u201d More natural terminology will suggest itself in a moment. A.1 Vectors 465 \u2022 Scalar Multiplication The \u201cproduct\u201d of any scalar with any vector is another vector: a|\u03b1 = |\u03b3. Scalar multiplication is distributive with respect to vector addition: a (|\u03b1 + |\u03b2) = a|\u03b1 + a|\u03b2, and with respect to scalar addition: (a + b) |\u03b1 = a|\u03b1 + b|\u03b1. It is also associative with respect to the ordinary multiplication of scalars: a (b|\u03b1) = (ab) |\u03b1. Multiplication by the scalars 0 and 1 has the effect you would expect: (A.6) (A.7) (A.8) (A.9) 0|\u03b1 = |0 ; 1|\u03b1 = |\u03b1. (A.10) Evidently |\u2212\u03b1 = (\u22121) |\u03b1 (which we write more simply as \u2212|\u03b1. There\u2019s a lot less here than meets the eye\u2014all I have done is to write down in abstract language the familiar rules for manipulating vectors. The virtue of such abstraction is that we will be able to apply our knowledge and intuition about the behavior of ordinary vectors to other systems that happen to share the same formal properties. A linear combination of the vectors |\u03b1, |\u03b2, |\u03b3,..., is an expression of the form a|\u03b1 + b|\u03b2 + c|\u03b3 + \u00b7 \u00b7 \u00b7. (A.11) A vector |\u03bb is said to be linearly independent of the set |\u03b1, |\u03b2, |\u03b3,..., if it cannot be written as a linear combination of them. (For example, in three dimensions the unit vector \u02c6k is linearly independent of \u02c6\u0131 and \u02c6j, but any vector in the x y plane is linearly dependent on \u02c6\u0131 and \u02c6j.) By extension, a set of vectors is \u201clinearly independent\u201d if each one is linearly independent of all the rest. A collection of vectors is said to span the space if every vector can be written as a linear combination of the members of this set.5 A set of linearly independent vectors that spans the space is called a basis. The number of vectors in any basis is called the dimension of the space. For the moment we shall assume that the dimension (n) is \ufb01nite. With respect to a prescribed basis |e1, |e2,..., |en, any given vector |\u03b1 = a1|e1 + a2|e2 + \u00b7 \u00b7 \u00b7 + an|en, is uniquely represented by the (ordered) n-tuple of its components: |\u03b1 \u2194 (a1, a2,..., an). (A.12) (A.13) (A.14) 5 A set of vectors that spans the space is also called complete, though I personally reserve that word for the in\ufb01nite- dimensional case, where subtle questions of convergence may arise. 466 Appendix Linear", " Algebra It is often easier to work with the components than with the abstract vectors themselves. To add vectors, you add their corresponding components: |\u03b1 + |\u03b2 \u2194 (a1 + b1, a2 + b2,..., an + bn); to multiply by a scalar you multiply each component: c|\u03b1 \u2194 (ca1, ca2,..., can) ; the null vector is represented by a string of zeroes: |0 \u2194 (0, 0,..., 0) ; and the components of the inverse vector have their signs reversed: |\u2212\u03b1 \u2194 (\u2212a1, \u2212a2,..., \u2212an). (A.15) (A.16) (A.17) (A.18) The only disadvantage of working with components is that you have to commit yourself to a particular basis, and the same manipulations will look very different to someone using a different basis. Problem A.1 Consider the ordinary vectors in three dimensions complex components. ax \u02c6\u0131 + ay \u02c6j + az \u02c6k, with (a) Does the subset of all vectors with az = 0 constitute a vector space? If so, what is its dimension; if not, why not? (b) What about the subset of all vectors whose z component is 1? Hint: Would the sum of two such vectors be in the subset? How about the null vector? (c) What about the subset of vectors whose components are all equal? \u2217 Problem A.2 Consider the collection of all polynomials (with complex coef\ufb01cients) of degree < N in x. (a) Does this set constitute a vector space (with the polynomials as \u201cvectors\u201d)? If so, suggest a convenient basis, and give the dimension of the space. If not, which of the de\ufb01ning properties does it lack? (b) What if we require that the polynomials be even functions? (c) What if we require that the leading coef\ufb01cient (i.e. the number multiplying x N \u22121) be 1? (d) What if we require that the polynomials have the value 0 at x = 1? (e) What if we require that the polynomials have the value 1 at x = 0? Problem A.3 Prove that the components of a vector with respect to a given basis are unique. A.2 INNER PRODUCTS In three dimensions we encounter two kinds of vector products: the dot product and the cross product. The latter does not generalize in any natural way to n-dimensional vector spaces, but the former does\u2014in this context it is usually called the inner product. The inner product A.2 Inner Products 467 of vectors |\u03b1 and |\u03b2 is a complex number (which we write as \u03b1|\u03b2), with the following properties: \u03b2 | \u03b1 = \u03b1 | \u03b2\u2217, \u03b1 | \u03b1 \u2265 0, and \u03b1 | \u03b1 = 0 \u21d4 |\u03b1 = |0, \u03b1| (b |\u03b2 + c |A.19) (A.20) (A.21) Apart from the generalization to complex numbers, these axioms simply codify the familiar behavior of dot products. A vector space with an inner product is called an inner product space. Because the inner product of any vector with itself is a non-negative number (Equation A.20), its square root is real\u2014we call this the norm of the vector: % \u03b1 \u2261 \u03b1|\u03b1; (A.22) it generalizes the notion of \u201clength.\u201d A unit vector (one whose norm is 1) is said to be normalized (the word should really be \u201cnormal,\u201d but I guess that sounds too judgmental). Two vectors whose inner product is zero are called orthogonal (generalizing the notion of \u201cperpendicular\u201d). A collection of mutually orthogonal normalized vectors, \u03b1i \u03b1 j = \u03b4i j, (A.23) is called an orthonormal set. It is always possible (see Problem A.4), and almost always convenient, to choose an orthonormal basis; in that case the inner product of two vectors can be written very neatly in terms of their components: \u03b1|\u03b2 = a\u2217 1 b1 + a\u2217 2 b2 + \u00b7 \u00b7 \u00b7 + a\u2217 n bn, the norm (squared) becomes \u03b1 | \u03b1 = |a1|2 + |a2|2 + \u00b7 \u00b7 \u00b7 + |an|2, and the components themselves are ai = ei |\u03b1. (A.24) (A.25) (A.26) (These results generalize the familiar formulas a\u00b7b = ax bx +ayby +azbz, |a|", "2 = a2 +a2 z, x and ax = \u02c6\u0131 \u00b7 a, ay = \u02c6j \u00b7 a, az = \u02c6k \u00b7 a, for the three-dimensional orthonormal basis \u02c6\u0131, \u02c6j, \u02c6k.) From now on we shall always work in orthonormal bases, unless it is explicitly indicated otherwise. Another geometrical quantity one might wish to generalize is the angle between two vectors. In ordinary vector analysis cos \u03b8 = (a \u00b7 b) / |a| |b|, but because the inner product is in general a complex number, the analogous formula (in an arbitrary inner product space) does not de\ufb01ne a (real) angle \u03b8. Nevertheless, it is still true that the absolute value of this quantity is a number no greater than 1, +a2 y |\u03b1 | \u03b2|A.27) This important result is known as the Schwarz inequality; the proof is given in Problem A.5. So you can, if you like, de\ufb01ne the angle between |\u03b1 and |\u03b2 by the formula / cos A.28) 468 Appendix Linear Algebra \u2217 Problem A.4 Suppose you start out with a basis (|e1, |e2,..., |en) that is not orthonormal. The Gram\u2013Schmidt procedure is a systematic ritual for generating from it an orthonormal basis (|e 1,..., |e n (i) Normalize the \ufb01rst basis vector (divide by its norm): ). It goes like this:, |e 2 e 1 = |e1 e1.. e 1 e 1 This vector is orthogonal to |e 1 (ii) Find the projection of the second vector along the \ufb01rst, and subtract it off: e2 |e2 \u2212 ; normalize it to get |e 2 and |e (iii) Subtract from |e3 its projections along |e : 1 2 e e e3 e3 e |e3 \u2212 \u2212 2 2 1 ; normalize it to get |e. And so on. 3 Use the Gram\u2013Schmidt procedure to orthonormalize the 3-space basis |e1 = (1 + i) and |e 2 This is orthogonal to |e 1 \u02c6\u0131 + (1) \u02c6j + (i) \u02c6k, |e2 = (i) \u02c6\u0131 + (3) \u02c6j + (1) \u02c6k, |e3 = (0) \u02c6\u0131 + (28) \u02c6j + (0) \u02c6k. e 1.. Problem A.5 Prove the Schwarz inequality (Equation A.27). Hint: Let |\u03b3 = |\u03b2 \u2212 (\u03b1|\u03b2 / \u03b1|\u03b1) |\u03b1, and use \u03b3 |\u03b3 \u2265 0. Problem A.6 Find the angle (in the sense of Equation A.28) between the vectors |\u03b1 = (1 + i) \u02c6\u0131 + (1) \u02c6j + (i) \u02c6k and |\u03b2 = (4 \u2212 i) \u02c6\u0131 + (0) \u02c6j + (2 \u2212 2i) \u02c6k. Problem A.7 Prove the triangle inequality: (|\u03b1 + |\u03b2) \u2264 \u03b1 + \u03b2. A.3 MATRICES Suppose you take every vector (in 3-space) and multiply it by 17, or you rotate every vector by 39\u25e6 about the z axis, or you re\ufb02ect every vector in the x y plane\u2014these are all examples of linear transformations. A linear transformation6 takes each vector in a vector space and \u201ctransforms\u201d it into some other vector (|\u03b1 \u2192 |\u03b1 = \u02c6T |\u03b1), subject to the condition that the operation be linear: \u02c6T \u02c6T (a|\u03b1 + b|\u03b2) = a \u02c6T |\u03b1 + b \u02c6T |\u03b2, (A.29) for any vectors |\u03b1, |\u03b2 and any scalars a, b. If you know what a particular linear transformation does to a set of basis vectors, you can easily \ufb01gure out what it does to any vector. For suppose that 6 In this chapter I\u2019ll use a hat (\u02c6) to denote linear transformations; this is not inconsistent with my convention in the text (putting hats on operators), for (as we shall see) quantum operators are linear transformations. A.3 Matrices 469 \u02c6T |e1 = T11|e1 + T21|e2 + \u00b7 \u00b7 \u00b7 + Tn1|en, \u02c6T |e2 = T12", "|e1 + T22|e2 + \u00b7 \u00b7 \u00b7 + Tn2|en, \u00b7 \u00b7 \u00b7 \u02c6T |en = T1n|e1 + T2n|e2 + \u00b7 \u00b7 \u00b7 + Tnn|en, or, more compactly, e j \u02c6T = n i=1 Ti j |ei, ( j = 1, 2,..., n). (A.30) If |\u03b1 is an arbitrary vector, |\u03b1 = a1|e1 + a2|e2 + \u00b7 \u00b7 \u00b7 + an|en = n j=1 a j |e j, (A.31) then \u02c6T |\u03b1 = n j=1 \u02c6T |e j = a j n n j=1 i=1 a j Ti j |ei = \u239b n n \u239d i=1 j=1 \u239e Ti j a j \u23a0 |ei. (A.32) Evidently \u02c6T takes a vector with components a1, a2,..., an into a vector with components7 n a i = Ti j a j. (A.33) j=1 Thus the n2 elements Ti j uniquely characterize the linear transformation \u02c6T (with respect to a given basis), just as the n components ai uniquely characterize the vector |\u03b1 (with respect to that basis): \u02c6T \u2194 (T11, T12,..., Tnn). If the basis is orthonormal, it follows from Equation A.30 that 0 \u02c6T 1 e j. Ti j = ei It is convenient to display these complex numbers in the form of a matrix:8 \u239b \u239c \u239c \u239c \u239d T = T11 T12 T21 T22...... Tn1 Tn2... T1n... T2n...... Tnn \u239e \u239f \u239f \u239f \u23a0. (A.34) (A.35) (A.36) The study of linear transformations reduces, then, to the theory of matrices. The sum of two is de\ufb01ned in the natural way: linear transformations \u02c6S + \u02c6T \u02c6S + \u02c6T |\u03b1 = \u02c6S|\u03b1 + \u02c6T |\u03b1; (A.37) 7 Notice the reversal of indices between Equations A.30 and A.33. This is not a typographical error. Another way of putting it (switching i \u2194 j in Equation A.30) is that if the components transform with Ti j, the basis vectors transform with T ji. 8 I\u2019ll use boldface capital letters, sans serif, to denote square matrices. 470 Appendix Linear Algebra this matches the usual rule for adding matrices (you add the corresponding elements): The product of two linear transformations succession\u2014\ufb01rst \u02c6T, then \u02c6S: U = S + T \u21d4 Ui j = Si j + Ti j. (A.38) \u02c6S \u02c6T is the net effect of performing them in \u03b1 = \u02c6T |\u03b1; \u03b1 \u03b1 = \u02c6S = \u02c6S \u02c6T |\u03b1 = \u02c6S \u02c6T |\u03b1. (A.39) What matrix U represents the combined transformation \u02c6U = \u02c6S \u02c6T? It\u2019s not hard to work it out: \" \u239e \u239b! a i = Evidently n n n Si j a j = Si j T jkak = n n \u239d Si j T jk \u23a0 ak = n j=1 j=1 k=1 k=1 j=1 k=1 U = ST \u21d4 Uik = n j=1 Si j T jk Uikak. (A.40) \u2014this is the standard rule for matrix multiplication: to \ufb01nd the ikth element of the product, you look at the ith row of S, and the kth column of T, multiply corresponding entries, and add. The same prescription allows you to multiply rectangular matrices, as long as the number of columns in the \ufb01rst matches the number of rows in the second. In particular, if we write the n-tuple of components of |\u03b1 as an n \u00d7 1 column matrix (or \u201ccolumn vector\u201d):9 a \u2261 \u239e \u239f \u239f \u239f \u23a0, \u239b \u239c \u239c \u239c \u239d a1 a2... an (A.41) (A.42) the transformation rule (Equation A", ".33) can be expressed as a matrix product: Now some matrix terminology: a = Ta. \u2022 The transpose of a matrix (which we shall write with a tilde: \u02dcT) is the same set of elements, but with rows and columns interchanged. In particular, the transpose of a column matrix is a row matrix: \u02dca = a1 a2... an. (A.43) For a square matrix taking the transpose amounts to re\ufb02ecting in the main diagonal (upper left to lower right): \u239e \u239b \u02dcT = \u239c \u239c \u239c \u239d T11 T21 T12 T22...... T1n T2n... Tn1... Tn2...... Tnn \u239f \u239f \u239f \u23a0. (A.44) 9 I\u2019ll use a boldface lower-case letters, sans serif, for row and column matrices. A (square) matrix is symmetric if it is equal to its transpose; it is antisymmetric if this operation reverses the sign: A.3 Matrices 471 (A.45) \u2022 The (complex) conjugate of a matrix (which we denote, as usual, with an asterisk, T\u2217 ), symmetric : \u02dcT = T; antisymmetric : \u02dcT = \u2212T. consists of the complex conjugate of every element: \u239e \u239b T\u2217 = \u239c \u239c \u239c \u239d 12 22 11 T \u2217 T \u2217 21 T \u2217 T \u2217...... n1 T \u2217 T \u2217 n2... T \u2217 1n... T \u2217 2n...... T \u2217 nn \u239f \u239f \u239f \u23a0 ; a\u2217 = \u239e \u239f \u239f \u239f \u23a0. (A.46) \u239b \u239c \u239c \u239c \u239d a\u2217 1 a\u2217 2... a\u2217 n A matrix is real if all its elements are real, and imaginary if they are all imaginary: real : T\u2217 = T; imaginary : T\u2217 = \u2212T. (A.47) \u2022 The hermitian conjugate (or adjoint) of a matrix (indicated by a dagger, T\u2020) is the transpose conjugate: \u2217 T\u2020 \u2261 \u02dcT = \u239b \u239c \u239c \u239c \u239d 22 21 11 T \u2217 T \u2217 12 T \u2217 T \u2217...... 1n T \u2217 T \u2217 2n \u239e \u239f \u239f \u239f \u23a0... T \u2217 n1... T \u2217 n2...... T \u2217 nn ; a\u2020 \u2261 \u02dca \u2217 = a\u2217 1 a\u2217 2... a\u2217 n. (A.48) A square matrix is hermitian (or self-adjoint) if it is equal to its hermitian conjugate; if hermitian conjugation introduces a minus sign, the matrix is skew hermitian (or anti-hermitian): hermitian : T\u2020 = T; skew hermitian : T\u2020 = \u2212T. (A.49) In this notation the inner product of two vectors (with respect to an orthonormal basis\u2014 Equation A.24), can be written very neatly as a matrix product: \u03b1|\u03b2 = a\u2020b. (A.50) Notice that each of the three operations de\ufb01ned in this paragraph, if applied twice, returns you to the original matrix. Matrix multiplication is not, in general, commutative (ST = TS); the difference between the two orderings is called the commutator:10 S, T \u2261 ST \u2212 TS. The transpose of a product is the product of the transposes in reverse order: GST = \u02dcT\u02dcS, (see Problem A.11), and the same goes for hermitian conjugates: ST \u2020 = T\u2020S\u2020. (A.51) (A.52) (A.53) 10 The commutator only makes sense for square matrices, of course; for rectangular matrices the two orderings wouldn\u2019t even be the same size. 472 Appendix Linear Algebra The identity matrix (representing a linear transformation that carries every vector into itself) consists of ones on the main diagonal, and zeroes everywhere else: \u239b 1 \u239c 0 \u239c \u239c... \u239d...... 0... 0 0... 1 I \u2261 In other", " words, Ii j = \u03b4i j. The inverse of a (square) matrix (written T\u22121) is de\ufb01ned in the obvious way:11 T\u22121T = T T\u22121 = I. A matrix has an inverse if and only if its determinant12 is nonzero; in fact, T\u22121 = 1 det T \u02dcC, (A.54) (A.55) (A.56) (A.57) where C is the matrix of cofactors (the cofactor of element Ti j is (\u22121)i+ j times the determinant of the submatrix obtained from T by erasing the ith row and the jth column). A matrix that has no inverse is said to be singular. The inverse of a product (assuming it exists) is the product of the inverses in reverse order: \u22121 = T\u22121S ST \u22121. (A.58) A matrix is unitary if its inverse is equal to its hermitian conjugate:13 unitary : U\u2020 = U\u22121. (A.59) Assuming the basis is orthonormal, the columns of a unitary matrix constitute an orthonormal set, and so too do its rows (see Problem A.12). Linear transformations represented by unitary matrices preserve inner products, since (Equation A.50) \u03b1 \u03b2 = a\u2020b = (Ua)\u2020 (Ub) = a\u2020U\u2020Ub = a\u2020b = \u03b1|\u03b2. (A.60) \u2217 Problem A.8 Given the following two matrices1 1 2 0 2i \u22122i, (d) \u02dcA, (e) A\u2217 A,i 0 1 2 3, (f) A\u2020, (g) det (B), and (h) B\u22121. Check compute: (a) A + B, (b) A B, (c) that B B\u22121 = 1. Does A have an inverse? 11 Note that the left inverse is equal to the right inverse, for if AT = I and TB = I, then (multiplying the second on the left by A and invoking the \ufb01rst) we get B = A. 12 I assume you know how to evaluate determinants. If not, see Mary L. Boas, Mathematical Methods in the Physical Sciences, 3rd edn (John Wiley, New York, 2006), Section 3.3. 13 In a real vector space (that is, one in which the scalars are real) the hermitian conjugate is the same as the transpose, \u22121. For example, rotations in ordinary 3-space are represented by and a unitary matrix is orthogonal: \u02dcO = O orthogonal matrices. A.4 Changing Bases 473 \u2217 \u2217 Problem A.9 Using the square matrices in Problem A.8, and the column matrices \u239b \u239e \u239b a = \u239d \u23a0, b = \u239d i 2i 2 \u239e \u23a0, 2 (1 \u2212 i) 0 \ufb01nd: (a) Aa, (b) a\u2020b, (c) \u02dcaBb, (d) a b\u2020. Problem A.10 By explicit construction of the matrices in question, show that any matrix T can be written (a) as the sum of a symmetric matrix S and an antisymmetric matrix A; (b) as the sum of a real matrix R and an imaginary matrix M; (c) as the sum of a hermitian matrix H and a skew-hermitian matrix K. Problem A.11 Prove Equations A.52, A.53, and A.58. Show that the product of two unitary matrices is unitary. Under what conditions is the product of two hermitian matrices hermitian? Is the sum of two unitary matrices necessarily unitary? Is the sum of two hermitian matrices always hermitian? Problem A.12 Show that the rows and columns of a unitary matrix constitute orthonormal sets. Problem A.13 Noting that det(\u02dcT) = det(T), show that the determinant of a hermitian matrix is real, the determinant of a unitary matrix has modulus 1 (hence the name), and the determinant of an orthogonal matrix (footnote 13) is either +1 or \u22121. A.4 CHANGING BASES The components of a vector depend, of course, on your (arbitrary) choice of basis, and so do the elements of the matrix representing a linear transformation. We might inquire how these numbers change when we switch to a different basis. The old basis vectors, |ei are\u2014like all vectors\u2014linear", " combinations of the new ones, | fi : |e1 = S11| f1 + S21| f2 + \u00b7 \u00b7 \u00b7 + Sn1| fn, |e2 = S12| f1 + S22| f2 + \u00b7 \u00b7 \u00b7 + Sn2| fn, \u00b7 \u00b7 \u00b7 |en = S1n| f1 + S2n| f2 + \u00b7 \u00b7 \u00b7 + Snn| fn, (for some set of complex numbers Si j ), or, more compactly, 474 Appendix Linear Algebra |e j = n i=1 Si j | fi, ( j = 1, 2,..., n). (A.61) This is itself a linear transformation (compare Equation A.30),14 and we know immediately how the components transform: a f i = n j=1 Si j ae j, (A.62) (where the superscript indicates the basis). In matrix form a f = Sae. (A.63) What about the matrix representing a linear transformation \u02c6T \u2014how is it modi\ufb01ed by a change of basis? Well, in the old basis we had (Equation A.42) ae = Teae, and Equation A.63\u2014multiplying both sides by S a f = Sa e = S Evidently \u22121\u2014entails15 ae = S = STeS \u22121a f. Teae \u22121a f, so T f = S TeS \u22121. (A.64) \u22121 for some (nonIn general, two matrices (T1 and T2) are said to be similar if T2 = ST1S singular) matrix S. What we have just found is that matrices representing the same linear transformation, with respect to different bases, are similar. Incidentally, if the \ufb01rst basis is orthonormal, the second will also be orthonormal if and only if the matrix S is unitary (see Problem A.16). Since we always work in orthonormal bases, we are interested mainly in unitary similarity transformations. While the elements of the matrix representing a given linear transformation may look very different in the new basis, two numbers associated with the matrix are unchanged: the determinant and the trace. For the determinant of a product is the product of the determinants, and hence T f det S TeS \u22121 S Te det det = det S \u22121 = det And the trace, which is the sum of the diagonal elements, Tr(T) \u2261 m i=1 Tii, has the property (see Problem A.17) that Tr(T1T2) = Tr(T2T1), = detTe. (A.65) (A.66) (A.67) 14 Notice, however, the radically different perspective: In this case we\u2019re talking about one and the same vector, referred to two completely different bases, whereas before we were thinking of a completely different vector, referred to the same basis. \u22121 certainly exists\u2014if S were singular, the | fi s would not span the space, so they wouldn\u2019t constitute 15 Note that S a basis. (for any two matrices T1 and T2), so T f Tr = Tr A.5 Eigenvectors and Eigenvalues 475 S TeS \u22121 TeS \u22121S Te. = Tr = Tr (A.68) Problem A.14 Using the standard basis \u02c6\u0131, \u02c6j, \u02c6k for vectors in three dimensions: (a) Construct the matrix representing a rotation through angle \u03b8 (counterclockwise, looking down the axis toward the origin) about the z axis. (b) Construct the matrix representing a rotation by 120\u25e6 (counterclockwise, looking down the axis) about an axis through the point (1,1,1). (c) Construct the matrix representing re\ufb02ection through the x y plane. (d) Check that all these matrices are orthogonal, and calculate their determinants. \u02c6\u0131, \u02c6j, \u02c6k, construct the matrix Tx representing a rotation Problem A.15 In the usual basis through angle \u03b8 about the x axis, and the matrix Ty representing a rotation through angle \u03b8 about the y axis. Suppose now we change bases, to \u02c6\u0131 = \u02c6j, \u02c6j = \u2212\u02c6\u0131, \u02c6k = \u02c6k. Construct \u22121 are what the matrix S that effects this change of basis, and check that STx S you would expect. \u22121 and STyS Problem A.16 Show that similarity preserves matrix multiplication (that is, if AeBe = Ce, then A f B f = C f ). Similarity does not, in general, preserve symmetry, reality, or hermiticity;", " show, however, that if S is unitary, and He is hermitian, then H f is hermitian. Show that S carries an orthonormal basis into another orthonormal basis if and only if it is unitary. \u2217 Problem A.17 Prove that Tr(T1T2) = Tr(T2T1). It follows immediately that Tr(T1T2T3) = Tr(T2T3T1), but is it the case that Tr(T1T2T3) = Tr(T2T1T3), in general? Prove it, or disprove it. Hint: The best disproof is always a counterexample\u2014the simpler the better! A.5 EIGENVECTORS AND EIGENVALUES Consider the linear transformation in 3-space consisting of a rotation, about some speci\ufb01ed axis, by an angle \u03b8. Most vectors (with tails at the origin) will change in a rather complicated way (they ride around on a cone about the axis), but vectors that happen to lie along the axis have very simple behavior: They don\u2019t change at all ( \u02c6T |\u03b1 = |\u03b1). If \u03b8 is 180\u25e6, then vectors which lie in the the \u201cequatorial\u201d plane reverse signs ( \u02c6T |\u03b1 = \u2212 |\u03b1). In a complex vector space16 every linear transformation has \u201cspecial\u201d vectors like these, which are transformed into scalar multiples of themselves: 16 This is not always true in a real vector space (where the scalars are restricted to real values). See Problem A.18. 476 Appendix Linear Algebra \u02c6T |\u03b1 = \u03bb|\u03b1; (A.69) they are called eigenvectors of the transformation, and the (complex) number \u03bb is their eigenvalue. (The null vector doesn\u2019t count, even though in a trivial sense it obeys Equation A.69 for any \u02c6T and any \u03bb; technically, an eigenvector is any nonzero vector satisfying Equation A.69.) Notice that any (nonzero) multiple of an eigenvector is still an eigenvector, with the same eigenvalue. With respect to a particular basis, the eigenvector equation assumes the matrix form or Ta = \u03bba, (a = 0), (T \u2212 \u03bbI) a = 0. (A.70) (A.71) (Here 0 is the (column) matrix whose elements are all zero.) Now, if the matrix (T \u2212 \u03bbI) had an inverse, we could multiply both sides of Equation A.71 by (T \u2212 \u03bbI)\u22121, and conclude that a = 0. But by assumption a is not zero, so the matrix (T \u2212 \u03bbI) must in fact be singular, which means that its determinant is zero: det(T \u2212 \u03bbI) = = 0. (A.72) (T11 \u2212 \u03bb) T21... Tn1 T12 (T22 \u2212 \u03bb)... Tn2......... T1n T2n... (Tnn \u2212 \u03bb) Expansion of the determinant yields an algebraic equation for \u03bb: Cn\u03bbn + Cn\u22121\u03bbn\u22121 + \u00b7 \u00b7 \u00b7 + C1\u03bb + C0 = 0, (A.73) where the coef\ufb01cients Ci depend on the elements of T (see Problem A.20). This is called the characteristic equation for the matrix; its solutions determine the eigenvalues. Notice that it\u2019s an nth-order equation, so (by the fundamental theorem of algebra) it has n (complex) roots.17 However, some of these may be multiple roots, so all we can say for certain is that an n \u00d7 n matrix has at least one and at most n distinct eigenvalues. The collection of all the eigenvalues of a matrix is called its spectrum; if two (or more) linearly independent eigenvectors share the same eigenvalue, the spectrum is said to be degenerate. To construct the eigenvectors it is generally easiest simply to plug each \u03bb back into Equation A.70 and solve \u201cby hand\u201d for the components of a. I\u2019ll show you how it goes by working out an example. Example A.1 Find the eigenvalues and eigenvectors of the following matrix: \u239b M = \u239d \u239e 2 \u22122i 1 0 \u22122 i 2i 0 \u22121 \u23a0. (A.74) 17 It is here that the case of real vector spaces becomes more awkward, because the characteristic equation need not have any (real) solutions at all.", " See Problem A.18. Solution: The characteristic equation is (2 \u2212 \u03bb) \u22122i 1 0 (i \u2212 \u03bb) 0 \u22122 2i (\u22121 \u2212 \u03bb) A.5 Eigenvectors and Eigenvalues 477 = \u2212\u03bb3 + (1 + i) \u03bb2 \u2212 i\u03bb = 0, (A.75) and its roots are 0, 1, and i. Call the components of the \ufb01rst eigenvector (a1, a2, a3); then \u239e \u239e \u239b \u239e \u239e \u239b \u239b \u239d 2 \u22122i 1 0 \u22122 i 2i 0 \u22121 a1 a2 a3 \u23a0 \u239d \u23a0 = a1 a2 a3 which yields three equations: 2a1 \u2212 2a3 = 0, \u22122ia1 + ia2 + 2ia3 = 0, a1 \u2212 a3 = 0. The \ufb01rst determines a3 (in terms of a1): a3 = a1; the second determines a2: a2 = 0; and the third is redundant. We may as well pick a1 = 1 (since any multiple of an eigenvector is still an eigenvector): a(1) = \u23a0, for \u03bb1 = 0. (A.76) \u239e \u239b 1 \u239d 0 1 For the second eigenvector (recycling the same notation for the components) we have \u239b \u239d 2 \u22122i 1 0 \u22122 i 2i 0 \u22121 a1 a2 a3 \u239e \u239b \u239e \u239b \u239e \u239b \u239e \u23a0 \u239d \u23a0 = 1 \u239d \u23a0 = \u239d \u23a0, a1 a2 a3 a1 a2 a3 which leads to the equations with the solution a3 = (1/2) a1, a2 = \u239b a1; this time I\u2019ll pick a1 = 2, so a(2) = \u239d \u23a0, for \u03bb2 = 1. (A.77) 2a1 \u2212 2a3 = a1, \u22122ia1 + ia2 + 2ia3 = a2, a1 \u2212 a3 = a3, (1 \u2212 i) /2 \u239e 2 1 \u2212 i 1 Finally, for the third eigenvector, \u239b \u239d 2 \u22122i 1 0 \u22122 i 2i 0 \u22121 \u239e \u23a0 = i \u239e \u23a0 \u239b \u239d a1 a2 a3 \u239b \u239d a1 a2 a3 \u239e \u23a0 = \u239b \u239d ia1 ia2 ia3 \u239e \u23a0, which gives the equations 2a1 \u2212 2a3 = ia1, \u22122ia1 + ia2 + 2ia3 = ia2, a1 \u2212 a3 = ia3, 478 Appendix Linear Algebra whose solution is a3 = a1 = 0, with a2 undetermined. Choosing a2 = 1, we conclude \u239e a(3) = \u23a0, for \u03bb3 = i. (A.78) \u239b 0 \u239d 1 0 If the eigenvectors span the space (as they do in the preceding example), we are free to use them as a basis: \u02c6T | f1 = \u03bb1| f1, \u02c6T | f2 = \u03bb2| f2, \u00b7 \u00b7 \u00b7 \u02c6T | fn = \u03bbn| fn. In this basis the matrix representing \u02c6T takes on a very simple form, with the eigenvalues strung out along the main diagonal, and all other elements zero: \u239b \u239c \u239c \u239c \u239d T = \u239e \u239f \u239f \u239f \u23a0, \u03bb1 0... 0 0 \u03bb2... 0...... 0 0...... \u03bbn and the (normalized) eigenvectors are \u239e \u239f \u239f \u239f \u239f \u239f \u23a0, \u239b 1 \u239c 0 \u239c \u239c 0 \u239c \u239c... \u239d \u239c \u239c... \u239d \u239f \u239f... \u23a0 1. (A.79) (A.80) A matrix that can be brought to diagonal form (Equation A.79) by a change of basis is said to be diagonalizable (evidently a matrix is diagonal", "izable if and only if its eigenvectors span the space). The similarity matrix that effects the diagonalization can be constructed by using the eigenvectors (in the old basis) as the columns of S \u22121: \u22121 S = i j a( j) (A.81). i \u239e 0 \u23a0, 1 0 Example A.2 In Example A.1, \u239b \u22121 = \u239d S 2 1 0 (1 \u2212 i) 1 1 so (using Equation A.57) \u239b \u239d \u22121 1 (i \u2212 1) \u239e \u23a0 ; 2 \u22121 0 0 1 (1 \u2212 i) S = A.5 Eigenvectors and Eigenvalues 479 you can check for yourself that \u239b 1 \u239d 0 0 Sa(1) = \u239e \u23a0, Sa(2 Sa(3 and \u239b SMS \u2212 There\u2019s an obvious advantage in bringing a matrix to diagonal form: it\u2019s much easier to work with. Unfortunately, not every matrix can be diagonalized\u2014the eigenvectors have to span the space. If the characteristic equation has n distinct roots, then the matrix is certainly diagonalizable, but it may be diagonalizble even if there are multiple roots. (For an example of a matrix that cannot be diagonalized, see Problem A.19.) It would be handy to know in advance (before working out all the eigenvectors) whether a given matrix is diagonalizable. A useful suf\ufb01cient (though not necessary) condition is the following: A matrix is said to be normal if it commutes with its hermitian conjugate: $ normal : = 0. (A.82) # N\u2020, N Every normal matrix is diagonalizable (its eigenvectors span the space). In particular, every hermitian matrix is diagonalizable, and so is every unitary matrix. Suppose we have two diagonalizable matrices; in quantum applications the question often arises: Can they be simultaneously diagonalized (by the same similarity matrix S)? That is to say, does there exist a basis all of whose members are eigenvectors of both matrices? In this basis, both matrices would be diagonal. The answer is yes if and only if the two matrices commute, as we shall now prove. (By the way, if two matrices commute with respect to one basis, they commute with respect to any basis\u2014see Problem A.23.) We \ufb01rst show that if a basis of simultaneous eigenvectors exists then the matrices commute. Actually, it\u2019s trivial in the (simultaneously) diagonal form... \u03bdn TV = = 0 \u03bd2... 0 \u239c \u239c \u239c \u239d \u239b \u239c \u239c \u239c \u239d \u03bb1 0... 0 \u03bb1\u03bd1 0... 0 0 \u03bb2... 0 \u239f \u239f \u239f \u23a0 \u239c \u239c \u239c \u239d \u03bd1 0... 0 \u00b7 \u00b7 \u00b7 0 \u00b7 \u00b7 \u00b7 0...... \u00b7 \u00b7 \u00b7 \u03bbn \u00b7 \u00b7 \u00b7 0 0 \u00b7 \u00b7 \u00b7 \u03bb2\u03bd2 0......... \u00b7 \u00b7 \u00b7 \u03bbn\u03bdn 0 \u239e \u239f \u239f \u239f \u23a0 = VT. (A.83) The converse is trickier. We start with the special case where the spectrum of T is nondegenerate. Let the basis of eigenvectors of T be labeled a(i) We assume T, V T a(i) = \u03bbi a(i). = 0 and we want to prove that a(i) is also an eigenvector of V. T, V a(i) = 0, or T V a(i) \u2212 V T a(i) = 0 (A.84) (A.85) 480 Appendix Linear Algebra and from Equation A.84 T V a(i) = \u03bbi V a(i). (A.86) Equation A.86 says that the vector b(i) \u2261 V a(i) is an eigenvector of T with eigenvalue \u03bbi. But by assumption, the spectrum of T is nondegenerate and that means that b(i) must be (up to a constant) a(i) itself. If we call the constant \u03bdi, b(i) = V a(i) = \u03bdi a(i), (A.87) so a(i) is an eigenvector of V. All that remains is to relax the assumption of nondegeneracy. Assume now that T has at least one degenerate eigenvalue such that both", " a(1) and a(2) are eigenvectors of T with the same eigenvalue \u03bb0: T a(1) = \u03bb0 a(1) T a(2) = \u03bb0 a(2). We again assume that the matrices T and V commute, so T, V T, V a(1) = 0 a(2) = 0, which leads to the conclusion (as in the nondegenerate case) that both b(1) \u2261 V a(1) and b(2) \u2261 V a(2) are eigenvectors of T with eigenvalue \u03bb0. But this time we can\u2019t say that b(1) is a constant times a(1) since any linear combination of a(1) and a(2) is an eigenvector of T with eigenvalue \u03bb0. All we know is that b(1) = V a(1) = c11 a(1) + c21 a(2) b(2) = V a(2) = c12 a(1) + c22 a(2) for some constants ci j. So a(1) and a(2) are not eigenvectors of V (unless the constants c12 and c21 just happen to vanish). But suppose we choose a different basis of eigenvectors \u02dca, (1) = d11 a(1) + d21 a(2) \u02dca (2) = d12 a(1) + d22 a(2), \u02dca (1) (2) for some constants di j, such that \u02dca and \u02dca are eigenvectors of V: (A.88) (i) = \u03bdi (A.89) The \u02dcas are still eigenvectors of T, with the same eigenvalue \u03bb0, since any linear combinations of a(1) and a(2) are. But can we construct linear combinations (A.88) that are eigenvectors of V \u2014 how do we get the appropriate coef\ufb01cients di j? Answer: We solve the eigenvalue problem18 (i). \u02dca V \u02dca d1i d2i = \u03bdi d1i d2i A c11 c21 c12 c22 BC C D. (A.90) 18 You might worry that the matrix C is not diagonalizable, but you need not. The matrix C is a 2\u00d72 block of the transformation \u02c6V written in the basis a(i); it is diagonalizable by virtue of the fact that V itself is diagonalizable. A.5 Eigenvectors and Eigenvalues 481 I\u2019ll let you show (Problem A.24) that the eigenvectors \u02dca constructed in this way satisfy Equation A.88, completing the proof.19 What we have seen is that, when the spectrum contains degeneracy, the eigenvectors of one matrix aren\u2019t automatically eigenvectors of a second commuting matrix, but we can always choose a linear combination of them to form a simultaneous basis of eigenvectors. (i) \u2217 Problem A.18 The 2 \u00d7 2 matrix representing a rotation of the x y plane is T = cos \u03b8 \u2212 sin \u03b8 cos \u03b8 sin \u03b8. (A.91) Show that (except for certain special angles\u2014what are they?) this matrix has no real eigenvalues. (This re\ufb02ects the geometrical fact that no vector in the plane is carried into itself under such a rotation; contrast rotations in three dimensions.) This matrix does, however, have complex eigenvalues and eigenvectors. Find them. Construct a matrix S that diagonal- izes T. Perform the similarity transformation explicitly, and show that it reduces T to diagonal form. STS \u22121 Problem A.19 Find the eigenvalues and eigenvectors of the following matrix: M = Can this matrix be diagonalized? 1 0 1 1. Problem A.20 Show that the \ufb01rst, second, and last coef\ufb01cients in the characteristic equation (Equation A.73) are: Cn = (\u22121)n, Cn\u22121 = (\u22121)n\u22121 Tr(T), and C0 = det(T). (A.92) For a 3 \u00d7 3 matrix with elements Ti j, what is C1? Problem A.21 It\u2019s obvious that the trace of a diagonal matrix is the sum of its eigenvalues, and its determinant is their product (just look at Equation A.79). It follows (from Equations A.65 and A.68) that the same holds for any diagonalizable matrix. Prove that in", " fact det(T) = \u03bb1\u03bb2 \u00b7 \u00b7 \u00b7 \u03bbn, Tr(T) = \u03bb1 + \u03bb2 + \u00b7 \u00b7 \u00b7 + \u03bbn, for any matrix. (The \u03bb\u2019s are the n solutions to the characteristic equation\u2014in the case of multiple roots, there may be fewer linearly-independent eigenvectors than there are solutions, but we still count each \u03bb as many times as it occurs.) Hint: Write the characteristic equation in the form (A.93) (\u03bb1 \u2212 \u03bb) (\u03bb2 \u2212 \u03bb) \u00b7 \u00b7 \u00b7 (\u03bbn \u2212 \u03bb) = 0, and use the result of Problem A.20. 19 I\u2019ve only proved it for a two-fold degeneracy, but the argument extends in the obvious way to a higher-order degeneracy; you simply need to diagonalize a bigger matrix C. 482 Appendix Linear Algebra Problem A.22 Consider the matrix 1 1 i 1. M = (a) Is it normal? (b) Is it diagonalizable? Problem A.23 Show that if two matrices commute in one basis, then they commute in any basis. That is: Te 1, Te. (A.94) Hint: Use Equation A.64. Problem A.24 Show that the \u02dca computed from Equations A.88 and A.90 are eigenvectors of V. \u2217 Problem A.25 Consider the matrices \u239b \u239d A = 1 4 4 \u22122 \u22121 2 \u22122 \u22122 \u22121 \u22122 1 (a) Verify that they are diagonalizable and that they commute. (b) Find the eigenvalues and eigenvectors of A and verify that its spectrum is nonde- generate. (c) Show that the eigenvectors of A are eigenvectors of B as well. Problem A.26 Consider the matrices \u239b \u239d A = 2 2 \u22121 \u22121 2 2 \u22121 \u22121 2 \u22121 2 5 \u22121 2 (a) Verify that they are diagonalizable and that they commute. (b) Find the eigenvalues and eigenvectors of A and verify that its spectrum is degener- ate. (c) Are the eigenvectors that you found in part (b) also eigenvectors of B? If not, \ufb01nd the vectors that are simultaneous eigenvectors of both matrices. A.6 HERMITIAN TRANSFORMATIONS In Equation A.48 I de\ufb01ned the hermitian conjugate (or \u201cadjoint\u201d) of a matrix as its transposeconjugate: T\u2020 = \u02dcT. Now I want to give you a more fundamental de\ufb01nition for the hermitian conjugate of a linear transformation: It is that transformation \u02c6T \u2020 which, when applied to the \u2217 A.6 Hermitian Transformations 483 \ufb01rst member of an inner product, gives the same result as if \u02c6T itself had been applied to the second vector: 0 \u02c6T \u2020\u03b1 1 \u03b2 0 = \u03b1 1 \u02c6T \u03b2, (A.95) (for all vectors |\u03b1 and |\u03b2).20 I have to warn you that although everybody uses it, this is lousy notation. For \u03b1 and \u03b2 are not vectors (the vectors are |\u03b1 and |\u03b2), they are names. In particular, they are endowed with no mathematical properties at all, and the expression \u201c \u02c6T \u03b2\u201d is literally nonsense: Linear transformations act on vectors, not labels. But it\u2019s pretty clear what the notation means: \u02c6T \u03b2 is the name of the vector \u02c6T |\u03b2, and \u02c6T \u2020\u03b1|\u03b2 is the inner product of the vector \u02c6T \u2020|\u03b1 with the vector |\u03b2. Notice in particular that whereas for any scalar c. \u03b1 | c\u03b2 = c \u03b1 | \u03b2, c\u03b1 | \u03b2 = c\u2217 \u03b1 | \u03b2, (A.96) (A.97) If you\u2019re working in an orthonormal basis (as we always do), the hermitian conjugate of a linear transformation is represented by the hermitian conjugate of the corresponding matrix; for (using Equations A.50 and A.53), 0 \u03b1 1 \u02c6T \u03b2 = a\u2020T b = T\u2020a \u2020b = 0 \u02c6T \u2020\u03b1 1 \u03b2. (A.98) So the terminology is consistent, and we can speak interchangeably in the language of transformations or of matrices. In quantum mechanics, a fundamental role is played by hermitian transformations. The eigenvectors and eigenvalues of a hermitian transformation have three", " crucial \u02c6T \u2020 = \u02c6T properties: 1. The eigenvalues of a hermitian transformation are real. Proof: Let \u03bb be an eigenvalue of \u02c6T : \u02c6T |\u03b1 = \u03bb|\u03b1, with |\u03b1 = |0. Then 1 0 \u02c6T \u03b1 \u03b1 = \u03b1 | \u03bb\u03b1 = \u03bb \u03b1 | \u03b1. Meanwhile, if \u02c6T is hermitian, then 0 1 0 \u02c6T \u03b1 \u03b1 = \u02c6T \u03b1 1 \u03b1 = \u03bb\u03b1 | \u03b1 = \u03bb\u2217 \u03b1 | \u03b1. But \u03b1|\u03b1 = 0 (Equation A.20), so \u03bb = \u03bb\u2217, and hence \u03bb is real. QED 2. The eigenvectors of a hermitian transformation belonging to distinct eigenvalues are orthogonal. Proof: Suppose \u02c6T |\u03b1 = \u03bb|\u03b1 and \u02c6T |\u03b2 = \u03bc|\u03b2, with \u03bb = \u03bc. Then 1 0 \u02c6T \u03b2 \u03b1 = \u03b1 | \u03bc\u03b2 = \u03bc \u03b1 | \u03b2, 20 If you\u2019re wondering whether such a transformation necessarily exists, that\u2019s a good question, and the answer is \u201cyes.\u201d See, for instance, Paul R. Halmos, Finite Dimensional Vector Spaces, 2nd edn, van Nostrand, Princeton (1958), Section 44. 484 Appendix Linear Algebra and if \u02c6T is hermitian, 0 1 \u02c6T \u03b2 \u03b1 = 0 \u02c6T \u03b1 1 \u03b2 = \u03bb\u03b1 | \u03b2 = \u03bb\u2217 \u03b1 | \u03b2. But \u03bb = \u03bb\u2217 (from (1)), and \u03bb = \u03bc, by assumption, so \u03b1|\u03b2 = 0. QED 3. The eigenvectors of a hermitian transformation span the space. As we have seen, this is equivalent to the statement that any hermitian matrix can be diagonalized. This rather technical fact is, in a sense, the mathematical support on which much of quantum mechanics leans. It turns out to be a thinner reed than one might have hoped, because the proof does not carry over to in\ufb01nite-dimensional vector spaces. Problem A.27 A hermitian linear transformation must satisfy \u03b1| \u02c6T \u03b2 = \u02c6T \u03b1|\u03b2 for all vectors |\u03b1 and |\u03b2. Prove that it is (surprisingly) suf\ufb01cient that \u03b3 | \u02c6T \u03b3 = \u02c6T \u03b3 |\u03b3 for all vectors |\u03b3. Hint: First let |\u03b3 = |\u03b1 + |\u03b2, and then let |\u03b3 = |\u03b1 + i|\u03b2. \u2217 Problem A.28 Let a) Verify that T is hermitian. (b) Find its eigenvalues (note that they are real). (c) Find and normalize the eigenvectors (note that they are orthogonal). (d) Construct the unitary diagonalizing matrix S, and check explicitly that it diagonal- izes T. (e) Check that det(T) and Tr(T) are the same for T as they are for its diagonalized form. \u2217\u2217 Problem A.29 Consider the following hermitian matrix: \u239e \u239b T = \u239d i 2 \u2212i 2 1 \u2212i \u23a0. 1 i 2 (a) Calculate det (T) and Tr(T). (b) Find the eigenvalues of T. Check that their sum and product are consistent with (a), in the sense of Equation A.93. Write down the diagonalized version of T. (c) Find the eigenvectors of T. Within the degenerate sector, construct two linearly independent eigenvectors (it is this step that is always possible for a hermitian matrix, but not for an arbitrary matrix\u2014contrast Problem A.19). Orthogonalize them, and check that both are orthogonal to the third. Normalize all three eigenvectors. (d) Construct the unitary matrix S that diagonalizes T, and show explicitly that the similarity transformation using S reduces T to the appropriate diagonal form. Problem A.30 A unitary transformation is one for which \u02c6U \u2020 \u02c6U = 1. (a) Show that unitary transformations preserve inner products, in the sense that \u02c6U \u03b1| \u02c6U \u03b2 = \u03b1 | \u03b2, for all vectors |\u03b1, |\u03b2. A.6 Hermitian Transformations 485 (b) Show that the eigenvalues of a unitary transformation have modulus 1. (c) Show that the eigenvectors of a unitary transformation belonging to distinct eigenvalues are orthogonal. \u2217\u2217", " geometry and field. Astrophysics of black holes. The laws of black-hole dynamics. Orbits in \"equatorial plane\" of Kerr-Newman black hole. electromagnetic 883 887 911 Horizons are generated by nonterminating null geodesics. Roger Penrose. 936 Stephen W. Hawking. 926 938 35.1.1 Transverse-traceless part of a wave. 948 36\u00b71.1 36.2. 36.3. 37.1./ 37.2. 37.3. 37.4. Gravitational waves from pulsating neutron stars. 984 Analysis of burst of radiation from impulse event. 987 Radiation from several binary star systems. 990 1007 force for gravitational-wave accelera Derivation of equations of motion of detec tor. Lines of tions. Use of cross-section for wave-dominated detec tor. Vibrating, shape. resonant detector arbitrary 1020 1011 1031 of 38\u00b71.1 38.2. 38.3. 39.1./ 39.2. 39.3. 39.4. 39.5. 40'1'1 40.2. 40.3. 40.4. Technology of the 1970's confronted with rela 1048 tivistic phenomena. Baron Lorand von Eotvos. Robert Henry Dicke. 1053 1051 of 1070 the ten The theories of Dicke-Brans-Jordan and Ni. Heuristic description of post-Newtonian parameters. Post-Newtonian expansion of cients. Summary of the PPN formalism. PPN parameters used in the literature: a transla tor's guide. 1072 the metric coeffi 1092 1093 1077 results on deflection of 1104 Experimental radio waves. Experimental results on radar time-delay. Experimental sion. 1112 Catalog of experiments. perihelion results 1129 on light and 1109 preces 41.1. Spinor representation of simple tensors. 1154 42.1. 42.2. 42.3. The hinges where \"angle of rattle\" is concen 1169 trated in two, three, and four dimensions. Flow diagrams for Regge calculus. Synthesis of higher-dimensional skeleton geome tries out of lower-dimensional ones. 1176 11 71 43.1. Geometrodynamics compared with particle dy namics. 1181 44.1. 44.2. 44.3. 44.4. 44.5. 1197 Collapse of universe compared and contrasted with collapse of atom. Three levels of gravitational collapse. Relation of spin ~ to geometrodynamics. Bucket-of-dust concept of pregeometry. Pregeometry 1211 tions. 1204 1205 proposi calculus 1201 the as of FIGURES 1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 1.7. 1.8. 1.9. 1.10. 1.11. 1.12. 2.1. 2.2. 2.3. 2.4. 2.5. 2.6. 2.7. 2.8. 2.9. 4.1. 4.2. 4.3. 4.4. 4.5. 4.6. 4.7. 5.1. 5.2. 5.3. 7 4 6 11 Spacetime compared with the surface of an apple. World-line crossings mark events. Two systems of coordinates for same events. Mere coordinate singularities. Singularities in the coordinates on a 2-sphere. 12 The Roll-Krotkov-Dicke experiment. Testing for a local Path of totality of an ancient eclipse. Good clock versus bad clock. \"Acceleration of the separation\" of nearby geo desics. Separation of geodesics in a 3-manifold. Satellite period and Earth density. inertial frame. 39 25 27 14 18 31 31 49 From bilocal vector to tangent vector. Different curves, same tangent vector. 50 Velocity 4-vector resolved into components. A l-form pierced by a vector. Gradient as l-form. Addition of l-forms. Vectors and their corresponding l-forms. Lorentz basis. The centrifuge and the photon. 56 57 55 63 60 52 58 100 104 Faraday 2-form. Faraday form creates a l-form out of 4-veloc ity. Spacelike slices through Faraday. Faraday and its dual, Maxwell. Maxwell 2-form for charge at rest. Mechanism of radiation. l-form. Simple types of 111 123 107 109 106 6.3. 6.4. 7.1. 8.1. 8.2. 8.3. 8.4. 9.1. 9.2. 9.3. 10.", " in a Future null supernova explosion. M inkowski spacetime depicted in coordinates that are finite at infinity. 919 Schwarzschild spacetime. Reissner-Nordstr.0m spacetime depicted in coor dinates that are finite at infinity. Spacetime diagrams for selected causal relation ships. Black holes in an asymptotically flat spacetime. The horizon produced by spherical collapse of a star. Spacetime diagram used to prove the second law of black-hole dynamics. 924 925 922 932 921 35.1.1 Plane electromagnetic waves. 35.2. 35.3. Plane gravitational waves. Exact plane-wave solution. 953 959 952 36.1'1 Why gravitational radiation is ordinarily weak. 36.2. 36.3. 37'1'1 37.2. 37.3. 37.4. 37.5. 38.1. 38.2. 38.3. 976 Spectrum given off Schwarzschild black hole. Slow-motion source. 997 983 in head-on plunge into a vibrating bar detector. Reference frame for 1005 1013 Types of detectors. Separation between geodesics responds to a gravitational wave. 1014 Vibrator tion. Hammer blow of a gravitational wave on a noisy detector. responding to linearly polarized radia 1037 1022 The Pound-Rebka-Snider measurement of gravi tational redshift on the Earth. Brault's determination of the redshift of the 0 1 line of sodium from the sun. The Turner-Hill search for a dependence of proper clock rate on velocity relative to distant mat ter. 1059 1057 1065 40.1.1 Bending of trajectory near the sun. 40.2. Coordinates used in calculating the deflection of 1100 light. 1101 XXVI GRAVITATION 40.3'1 Coordinates for calculating the relativistic time 40.4. 41.1. 41.2. 41.3. 41.4. 41.5. 41.6. 41.7. 41.8. 1106 delay. Laser measurement of Earth-moon separation. 1130 1137 1136 1138 two rotations seen in terms of Com bination of rotations of 90\u00b0 about axes that diverge by 90\u00b0. Rotation depicted as two reflections. Composition of reflections. Law of composition of rotations epitomized in a spherical triangle. \"Orientation-entanglement relation\" between a cube and its surroundings. A 720\u00b0 rotation is equivalent 1149 Spinor as flagpole plus flag. Direction in space represented on the complex plane. to no rotation. 1157 1148 1139 1161 42.1. 42.2. 43.1. 43.2. 43.3. 44.1. 44.2. 44.3. 44.4. 44.5. A 2-geometry approximated by a polyhedron. 1168 Cycle of building blocks associated with a single hinge. 1170 the simplicial Superspace in 1182 Space, spacetime, and superspace. Electron motion affected by field fluctuations. 1190 approximation. 1183 1200 Wormhole picture of electric charge. Gravitation as the metric elasticity of space. 1207 What pregeometry is not. Black-box model 1213 A mind full of geometrodynamics. reprocessing of universe. 1210 1219 for ACKNOWLEDGMENTS Deep appreciation goes to all who made this book possible. A colleague gives us a special lecture so that we may adapt it into one of the chapters of this book. Another investigator clears up for us the tangled history of the production of matter out of the vacuum by strong tidal gravitational forces. A distant colleague telephones in references on the absence of any change in physical constants with time. One student provides a problem on the energy density of a null electromagnetic field. Another supplies curves for effective potential as a function of distance. A librarian writes abroad to get us an article in an obscure publication. A secretary who cares types the third revision of a chapter. Editor and illustrator imaginatively solve a puzzling problem of presentation. Repeat in imagination such instances of warm helpfulness and happy good colleagueship times beyond count. Then one has some impression of the immense debt we owe to over a hundred-fifty colleagues. Each face is etched in our mind, and to each our gratitude is heartfelt. Warm thanks we give also to the California Institute of Technology, the Dublin Institute for Advanced Studies, the Institute for Advanced Study at Princeton, Kyoto University, the University of Maryland, Princeton University, and the University of Texas at Austin for hospitality during the writing of this book. We are grateful to the Academy of Sciences of the U.S.S.R., to Moscow University, and to our Soviet colleagues for their hospitality and the opportunity to become better acquainted in June-July 1971 with Soviet work in gravitation physics. For assistance in", " the research that went into this book we thank the National Science Foundation for grants (GP27304 and 28027 to Caltech; GPl7673 and GP8560 to Maryland; and GP3974 and GP7669 to Princeton); the U.S. Air Force Office of Scientific Research (grant AF49-638-1545 to Princeton); the U.S. National Aeronautics and Space Agency (grant NGR 05-002-256 to Caltech, NSG 210-002-010 to Maryland); the Alfred P. Sloan Foundation for a fellowship awarded to one of us (K.S.T.); and the John Simon Guggenheim Memorial Founda tion and All Souls College, Oxford, England, for fellowships awarded to another of us (C.W.M.). GRAVITATION PART I SPACETIME PHYSICS Wherein the reader is led, once quickly (\u00a7 1.1), then again more slowly, down the highways and a few byways of Einstein's geometrodynamics without benefit of a good mathematkal CQmpass. GEOMETRODYNAMICS IN BRIEF CHAPTER 1 \u00a71.1. THE PARABLE OF THE APPLE and seeing the fall of an apple, as his niece told me, One day in the year 1666 Newton had gone to the country, let himself be led into a deep meditation on the cause which thus draws every object along a line whose extension would pass almost through the center of the Earth. VOLTAIRE (1738) Once upon a time a student lay in a garden under an apple tree reflecting on the difference between Einstein's and Newton's views about gravity. He was startled by the fall of an apple neiirby. As he looked at the apple, he noticed ants beginning to run along its surface (Figure l.l). His curiosity aroused, he thought to investigate the principles of navigation followed by an ant. With his magnifying glass, he noted one track carefully, and, taking his knife, made a cut in the apple skin one mm above the track and another cut one mm below it. He peeled off the resulting little highway of skin and laid it out on the face of his book. The track ran as straight as a laser beam along this highway. No more economical path could the ant have found to cover the ten cm from start to end of that strip of skin. Any zigs and zags or even any smooth bend in the path on its way along the apple peel from starting point to end point would have increased its length. \"What a beautiful geodesic.\" the student commented. His eye fell on two ants starting off from a common point P in slightly different directions. Their routes happened to carry them through the region of the dimple at the top of the apple. one on each side of it. Each ant conscientiously pursued 4 1. GEOMETRODYNAMICS IN BRIEF --- Figure 1.1. The Riemannian geometry of the spacetime of general relativity is here symbolized by the two-dimen sional geometry of the surface of an apple. The geodesic tracks followed by the ants on the apple's surface symbolize the world line followed through spacetime by a free particle. In any sufficiently localized region of spacetime, the geometry can be idealized as flat, as symbolized on the apple's two-dimensional surface' by the straight-line course of the tracks viewed in the magnifying glass (\"local Lorentz character\" of geometry of spacetime). In a region of greater extension, the curvature of the manifold (four-dimen sional spacetime in the case of the real physical world; curved two-dimensional geometry in the case of the apple) makes itself felt. Two tracks (J and (fl, originally diverging from a commOn point <1', later approach, cross, and go off in very different directions. In Newtonian theory this effect is ascribed to gravitation acting at a distance from a center of attraction, symbolized here by the stem of the apple. According to Einstein a particle gets its moving orders locally, from the geometry of spacetime right where it is. Its instructions are simple: to follow the straightest possible track (geodesic). Physics is as simple as it could be locally. Only because spacetime is curved in the large do the tracks cross. Geome trodynamics, in brief, is a double story of the effect of geometry on matter (causing originally divergent geodesics to cross) and the effect of matter on geometry (bending of spacetime initiated by concentration of mass, symbolized by effect of stem on nearby surface of apple). his geodesic. Each went as straight on his strip of appleskin as he possibly could. Yet because of the curvature of the dimple itself, the two tracks not only crossed but emerged in very different directions. \"What happier illustration of", " Einstein's geometric theory of gravity could one possibly ask?\" murmured the student. \"The ants move as if they were attracted by the apple stem. One might have believed in a Newtonian force at a distance. Yet from nowhere does an ant get his moving orders except from the local geometry along his track. This is surely Einstein's concept that all physics takes place by 'local action.' What a difference from Newton's 'action at a distance' view of physics! Now I understand better what this book means.\" And so saying, he opened his book and read, \"Don't try to describe motion relative to faraway objects. Physics is simple only when analyzed locally. And locally Einstein's local view of physics contrasted with Newton's \"action at a distance\" Physics is simple only when analyzed locally \u00a71.2. SPACETIME WITH AND WITHOUT COORDINATES 5 the world line that a satellite follows [in spacetime, around the Earth) is already as straight as any world line can be. Forget all this talk about 'deflection' and 'force of gravitation.' I'm inside a spaceship. Or I'm floating outside and near it. Do I feel any 'force of gravitation'? Not at all. Does the spaceship 'feel' such a force? No. Then why talk about it? Recognize that the spaceship and I traverse a region of spacetime free of all force. Acknowledge that the motion through that region is already ideally straight.\" The dinner bell was ringing, but still the student sat, musing to himself. \"Let me see if I can summarize Einstein's geometric theory of gravity in three ideas: (I) locally, geodesics appear straight: (2) over more extended regions of space and time, geodesics originally receding from each other begin to approach at a rate governed by the curvature of spacetime, and this effect of geometry on matter is what we mean today by that old word 'gravitation': (3) matter in turn warps geometry. The dimple arises in the apple because the stem is there. I think I see how to put the whole story even more briefly: Space acts on matter, telling it how to move. In turn, telling it how to curve. In other words, matter here,\" matter reacts back on space, he said, rising and picking up the apple by its stem, \"curves space here. To produce a curvature in space here is to force a curvature in space there,\" he went on, as he watched a lingering ant busily following its geodesic a finger's breadth away from is Einstein's the apple's stem. \"Thus matter here influences matter there. That explana tion for 'gravi ta tion.''' Then the dinner bell was quiet. and he was gone, with book, magnifying glass-and apple. \u00a71.2. SPACETIME WITH AND WITHOUT COORDINATES Now it came to me:... the independence of the gravitational acceleration from the nature of the falling substance, may be expressed as follows: In a gravitational field (of small spatial extension) things behave as they do in a space free of gravitation.... This happened in 1908. Why were another seven years required for the construction of the general theory of relativity? The main reason lies in the fact that it is not so easy to free oneself from the idea that coordinates must have an immediate metrical meaning. ALBERT EINSTEIN [in Schilpp (1949), pp. 65-67.] Space tells matter how to move Matter tells space how to curve Nothing is more distressing on first contact with the idea of \"curved spacetime\" than the fear that every simple means of measurement has lost its power in this unfamiliar context. One thinks of oneself as confronted with the task of measuring the shape of a gigantic and fantastically sculptured iceberg as one stands \\vith a meter stick in a tossing rowboat on the surface of a hea\\'ing ocean. Were it the rowboat itself whose shape were to be measured. the procedure would be simple enough. One would draw it up on shore. turn it upside down. and drive tacks in lightly at strategic points here and there on the surface. The measurement of distances from tack to Problem: how to measure in curved spacetime 6 1. GEOMETRODYNAMICS IN BRIEF Figure 1.2. The crossing of straws in a barn full of hay is a symbol for the world lines that fill up spacetime. By their crossings and bends. these world lines mark events with a uniqueness beyond all need of coordinate systems or coordinates. Typical events symbolized in the diagram, from left to right (black dots). are: absorption of a photon; reemission of a photon; collision between a particle and a particle: collision between a photon and a particle; another collision between a photon and a particle; explosion of a firecracker; and collision of a particle from outside with one of the fragments", " senior occupant, so one can and often does attach arbitrary names to specific events in spacetime. as in Box 1.1. Coordinates. however, are convenient. How else from the great thick catalog of events, randomly listed, can one easily discover that along a certain world line one will first encounter event Trinity. then Baker, then Mike, then Argus-but not the same events in some permuted order? To order events, introduce coordinates! (See Figure 1.3.) Coordinates are four indexed numbers per event in spacetime; on a sheet of paper, only two. Trinity acquires coordinates In christening events with coordinates, one demands smoothness but foregoes every thought of mensuration. The four numbers for an eventare nothing but an elaborate kind of telephone number. Compare their \"telephone\" numbers to discover whether two events are neighbors. But do not expect to learn how many meters separate them from the difference in their telephone numbers! Nothing prevents a subscriber from being served by competing telephone systems, nor an event from being catalogued by alternative coordinate systems (Figure 1.3). Box 1.1 illustrates the relationships between one coordinate system and another, as well as the notation used to denote coordinates and their transforma tions. Choose two events, known to be neighbors by the nearness of their coordinate values in a smooth coordinate system. Draw a little arrow from one~,,:ent to the other. Such an arrow is called a vector. (It is a well-defined concept in fiilt spacetime, or in curved spacetime in the limit of vanishingly small length; for finite lengths in curved spacetime, it must be refined and made precise, under the new name \"tangent vector,\" on which see Chapter 9.) This vector, like events, can be given a name. But whether named \"John\" or \"Charles\" or \"Kip,\" it is a unique, well defined geometrical object. The name is a convenience, but the vector exists even without it. Just as a quadruple of coordinates is a particularly useful name for the event \"Trinity\" (it can be used to identify what other events are nearby), so a quadruple of \"components\" l. Box 1.1 MATHEMATICAL NOTATION FOR EVENTS, COORDINATES, AND VECTORS Events are denoted by capital script, one-letter Latin names such as Sometimes subscripts are used: Coordinates of an event ~1' are denoted by or by or more abstractly by where it is understood that Greek indices can take on any value 0, 1, 2. or 3. Time coordinate (when one of the four is picked to play this role) '!J', ;2, d, ~.1i. '!F0' ~:i\", ~1'6' tU'). x(:1'). y(:1'), :('!F), XO(~11), x'(::I'). x 2(:1'). x 3et'), x!L(~P) or x\"(:1'), Space coordinates are and are sometimes denoted by It is to be understood that Latin indices take on values I, 2. or 3. x'(~:I'), X2(~1'), X3(~1') xi(:\u00b7P) or Xk(~1') or.... Shorthand notation: One soon tires of writing explicitly the functional depen dence of the coordinates, x il (:1'); so one adopts the shorthand notation for the coordinates of the event ~'P, and for the space coordinates. One even begins to think of x il as representing the event :1' itself, but must remind oneself that the values of xo, x', x 2, x 3 depend not only on the choice of ~P but also on the arbitrary choice of coordinates! Other coordinates for the same event :1' may be denoted EXA~IPLE: In Figure 1.3 (XO, Xl) = (77.2.22.6) and (xo. Xl) = (18.5.51.4) refer to the same event. The bars, priines, and hats distinguish one coordinate system from another: by putting them on the indices rather than on the x\u00b7s. we simplify later notation. Transformation from one coordinate system to another is achieved by the four functions which are denoted more succinctly Separation vector* (little arrow) reaching from one event i! to neighboring event ~.p can be denoted abstractly by It can also be characterized by the coordinate-value ditferencest between :'1' and i! (called \"components\" of the vector) Transformation of components of a vector from one coordinate system to another is achieved by partial derivatives of transformation equations since ~\" = x\"U'p) - X\"(~') = (2x\"i'x ll", ". Its neighborhood is a one-dimensional ball, but is not yet known to be one-dimensional (\"line segment\"). The boundaries of this object are two points. This circumstance tells that the intervening manifold is one-dimensional; therefore the previous mani fold was two-dimensional; and so on. The dimensionality of the original manifold is equal to the number of points employed in the construction. For spacetime, the dimensionality is 4. This kind of mathematical reasoning about dimensionality makes good sense at the everyday scale of distances, at atomic distances (10-8 cm), at nuclear dimensions (10- 13 cm), and even at lengths smaller by several powers of ten, if one judges by the concord between prediction and observation in quantum electrodynamics at high r = 1.8m --r-----1r----t---t----+------- r = 2.2m 2.4m 2.6m 2.8m 3.0m r::: 111 t = _ m Figure 1.4. Howa mere coordinare singulariry arises. Above: A coordinare sysrem becomes singular when rhe \"cells in rhe egg crare\" are squashed ro zero volume. Below: An example showing such a singulariry in rhe Schwarzschild coordinares r, t offen used ro describe rhe geomerry around a black hole (Chaprer 31). For simpliciry rhe angular coordinares 0, \u00a2 have been suppressed. The singulanry shows itself in two ways. Firsr, all rhe poinrs along rhe doffed line. while quire disrincr one from anorher, are designared by rhe same pair of(r. t) values: namely. r = 2m. t = 00. The coordinares provide no way ro disringuish rhese points. Second. rhe \"cells in rhe egg crare.\" of which one is shown grey in rhe diagram. collapse ro zero conrenr ar rhe doffed line. In summary, rhere is norhing srrange abour rhe geomerry ar rhe doffed line: all rhe singulariry lies in rhe coordinare sysrem (\"poor sysrem ofrelephone numbers\"). No confusion should be permiffed ro arise from rhe accidenral circumsrance rhar rhe t coordinare affains an infinire value on rhe doffed line. No such infiniry would occur if I were replaced by rhe new coordinare t. defined by (t/2m) = ran(i/2m). When t = x. rhe new coordinare t is t = ':Tn!. The r, t coordinares srill provide no way ro disringuish rhe poinrs along rhe dorred line. They sriIl give \"cells in rhe egg crare\" collapsed ro zero conrenr along rhe dOffed line. Figure 1.5. Singularities in familiar coordinates on the two-sphere can be eliminated by covering the sphere with two overlapping coordinate patches. A. Spherical polar coordinate~, singular at the North and South Poles, and discontinuous at the international date line. B. Projection of the Euclidean coordinates of the Euclidean two-plane, tangent at the North Pole, onto the sphere via a line running to the South Pole; coordinate singularity at the South Pole. C. Coverage of two-sphere by two overlapping coordinate patches. One, constructed as in B, covers without singularity the northern hemisphere and also the southern tropics down to the Tropic of Capricorn. The other (grey) also covers without singularity all of the tropics and the southern hemisphere besides. Breakdown in smoothness of spacetime at Planck length energies (corresponding de Broglie wavelength 10-16 em). Moreover, classical general relativity thinks of the spacetime manifold as a deterministic structure, completely well-defined down to arbitrarily small distances. Not so quantum general relativity or \"quantum geometrodynamics.\" It predicts violent fluctuations in the geometry at distances on the order of the Planck length, L * = (fiGj C3 )1/2 = [(1.054 X 10-27 g cm2 jsec)(6.670 X 10-8 cm3jg sec2)j1/2 X X (2.998 X 1010 cmjsect 3/ 2 (1.1) = 1.616 X 10-33 em. No one has found any way to escape this prediction. As nearly as one can estimate, these fluctuations give space at small distances a \"multiply connected\" or \"foamlike\" character. This lack of smoothness may well deprive even the concept of dimension ality itself of any meaning at the Planck scale of distances. The", " further exploration of this issue takes one to the frontiers of Einstein's theory (Chapter 44). Ifspacetime at small distances is far from the mathematical model of a continuous manifold, is there not also at larger distances a wide gap between the mathematical \u00a71.3. WEIGHTLESSNESS 13 idealization and the physical reality? The infinitely dense collection of light rays and of world lines of infinitesimal test particles that are to define all the points of the manifold: they surely are beyond practical realization. Nobody has ever found a particle that moves on timelike world lines (finite rest mass) lighter than an electron. A collection of electrons, even if endowed with zero density of charge (e+ and e world lines present in equal numbers) will have a density of mass. This density will curve the very manifold under study. Investigation in infinite detail means unlimited density, and unlimited disturbance of the geometry. However, to demand investigatability in infinite detail in the sense just described is as out of place in general relativity as it would be in electrodynamics or gas dynamics. Electrodynamics speaks of the strength of the electric and magnetic field at each point in space and at each moment of time. To measure those fields, it is willing to contemplate infinitesimal test particles scattered everywhere as densely as one pleases. However, the test particles do not have to be there at all to give the field reality. The field has everywhere a clear-cut value and goes about its deterministic dynamic evolution willy-nilly and continuously, test particles or no infinitesimal test particles. Similarly with the geometry of space. infinitesimal In conclusion, when one deals with spacetime in the context of classical physics, one accepts (I) the notion of \"infinitesimal test particle\" and (2) the idealization that the totality of identifiable events forms a four-dimensional continuous manifold. Only at the end of this book will a look be taken at some of the limitations placed by the quantum principle on one's way of speaking about and analyzing spacetime. \u00a71.3. WEIGHTLESSNESS \"Gravity is a great mystery. Drop a stone. See it fall. Hear it hit. No one understands why.\" What a misleading statement! Mystery about fall? What else should the stone do except fall? To fall is normal. The abnormality is an object standing in the way of the stone. If one wishes to pursue a \"mystery,\" do not follow the track of the falling stone. Look instead at the impact, and ask what was the force that pushed the stone away from its natural \"world line,\" (Le., its natural track through space time). That could lead to an interesting issue of solid-state physics, but that is not the topic of concern here. Fall is synonymous with weightlessness: absence of any force to drive the object away from its normal track through space time. Travel aboard a freely falling elevator to experience weightlessness. Or travel aboard a spaceship also falling straight toward the Earth. Or, more happily, travel aboard a spaceship in that state of steady fall toward the Earth that marks a circular orbit. In each case one is following a natural track through spacetime. is. Free fall The traveler has one chemical composition, the spaceship another; yet they travel together, the traveler weightless in his moving home. Objects of such different nuclear constitution as aluminum and gold fall with accelerations that agree to better than one part in lOll, according to Roll, Krotkov, and Dicke (1964), one of the most important null experiments in all physics (see Figure 1.6). Individual molecules fall in step, too, with macroscopic objects [Estermann, Simpson, and Stern (1938\u00bb): and so do individual neutrons [Dabbs, Harvey, Paya, and Horstmann (1965\u00bb). individual (coJllillued all page 16) Difficulty in defining geometry even at classical distances? No; one must accept geometry at classical distances as meaningful Free fall of motion is the natural state All objects fall with the same acceleration 14 1. GEOMETRODYNAMICS IN BRIEF Figure 1.6. Principle of the Roll-Krotkov-Dicke experiment. which showed that the grm'itational accelerations of gold and aluminum are equal to I part in 1011 or better (Princeton, 1964). In the upper lefthand corner, equal masses of gold and aluminum hang from a supporting bar. This bar in turn is supported at its midpoint. If both objects fall toward the sun with the same acceleration of g = 0.59 cm/sec2 \u2022 the bar does not turn. If the Au mass receives a higher acceleration. g + 8g. then the gold end of the bar starts to turn toward the sun in the Earth-fixed frame. Twoelve hours laler the sun is on the other side. pulling the other way. The ahernating torque", " lends itself to recognition against a background of noise because of its precise 24-hour period. Unhappily. any substamial mass nearby, such as an experimenter, located at M. will produce a torque that swamps the effect sought. Therefore the actual arrangement was as shown in the body of the figure. One gold weight and two aluminum weights were supported at the three corners of a horizontal equilateral triangle. 6 cm on a side (three-fold axis of symmetry. giving zero response to all the simplest nonuniformities in the gravitational field). Also. the observers performed all operations remotely to eliminate their own gravitational effects\". To detect a rotation of the torsion balance as small as -10-\" rad without disturbing the balance. Roll. Krotkov, and Dicke reflected a very weak light beam from the optically flat back face of the quartz triangle. The image of the source slit fell on a wire of about the same size as the slit image. The light transmitted past the wire fell on a photomultiplier. A separate oscillator circuit drove the wire back and forth across the image at 3,000 hertz. When the image was centered perfectly, only even harmonics of the oscillation frequency appeared the fundamental in the light intensity. However, when the image was displaced slightly to one side. frequency appeared in the light intensity. The electrical output of the photomultiplier then contained a 3,OOO-hertz component. The magnitude and sign of this component were determined automatically. Equally automatically a proportional D.C. voltage was applied to the electrodes shown in the diagram. It restored the torsion balance to its zero position. The D.C. voltage required to restore the balance to its zero position was recorded as a measure of the torque acting on the pendulum. This torque was Fourier-analyzed over a period of many days. The magnitude of the Fourier component of 24-hour period indicated a ratio 8g/g = (0.96 \u00b1 1.04) X 10-11 \u2022 Aluminum and gold thus fall with the same acceleration, despite their important differences summarized in the table. Ratios Number of neutrons Number of prolons Mass of kinetic energy of K-electron Rest mass of eleclron Eleetroslatic mass-energy of nucleus Mass of atom AI 1.08 Au 1.5 0.005 0.16 0.001 0.004 The theoretical implications of this experiment will be discussed in greater detail in Chapters 16 and 38. Braginsky and Panov (1971) at Moscow University performed an experiment identical in principle to that of Dicke-Roll-Krotkov, but with a modified experimental set-up. Comparing the accelerations of platinum and aluminum rather than of gold and aluminum, they say that 8g/g ~ I X 10-12\u2022 \"Other perturbations had to be, and were, guarded against. (I) A bit of iron on the torsion balance as big as 10-3 cm on a side would have contribu ted, in the Earth's magnetic field, a torque a hundred times greater than the measured torque. (2) The unequal pressure of radiation on the two sides of a mass would have produced an unacceptably large perturbation if the temperature difference between these two sides had exceeded 10-4 OK. (3) Gas evolution from one side ofa mass would have propelled it like a rocket. If the rate of evolution were as great as 10-8 g/day, the calculated force would have been - 10- 7 g cm/sec2, enough to affect the measurements. (4) The rotation was measured with respect to the pier that supported the equipment. As a guarantee that this pier did not itself rotate, it was anchored to bed rock. (5) Electrostatic forces were eliminated; otherwise they would have perturbed the balance. 15 /'--- I \\ \\... ~ 0.59 em! sec~ 0.59 cm!sec2 '\\ } 0.59 cm/secl Au Earth 0 \u2022 M AI Fused quartz fiber coaled with a thin aluminum film to eliminate e1ecrrostat\"rc fields Light emer!!es fro;n fixed slit Sun al o.\\.H. 16 1. GEOMETRODYNAMICS IN BRIEF electrons [Witteborn and Fairbank (1967\u00bb) and individual mu mesons [Beall (1970\u00bb). What is more, not one of these objects has to see out into space to know how to move. Contemplate the interior ofa spaceship, and a key, penny, nut, and pea by accident or design set free inside. Shielded from all view of the world outside by the walls of the vessel, each object stays at rest relative to the vessel. Or it moves through the room in a straight line with uniform velocity. That is the lesson which experience shouts out. Forego talk of acceleration! That, paradox", "ically, is the lesson of the circumstance that \"all objects fall with the same acceleration.\" Whose fault were those accelera tions, after all? They came from allowing a groundbased observer into the act. The Box 1.2 MATERIALS OF THE MOST DIVERSE COMPOSITION FALL WITH THE SAME ACCELERATION (\"STANDARD WORLD LINE\") Aristotle: \"the downward movement of a mass of gold or lead, or of any other body endowed with weight, is quicker in proportion to its size.\" Pre-Galilean literature: metal and wood weights fall at the same rate. Galileo: (I) \"the variation of speed in air between balls of gold, lead, copper, porphyry, and other heavy materials is so slight that in a fall of 100 cubits [about 46 meters] a ball of gold would surely not outstrip one of copper by as much as four fingers. Having observed this, I came to the con clusion that in a medium totally void of resistance all bodies would fall with the same speed.\" (2) later experiments of greater precision \"diluting gravity\" and finding same time of descent for different objects along an inclined plane. Newton: inclined plane replaced by arc of pendu lum bob; \"time of fall\" for bodies of different composition determined by comparing time of oscillation of pendulum bobs of the two materials. Ultimate limit of precision in such experiments limited by problem of determining effective length (acceleration) = (2'7T/pe each of riod)2(length). pendulum: Lorand von Eotvos, Budapest, 1889 and 1922: compared on the rotating earth the vertical defined by a plumb bob of one material with the vertical defined by a plumb bob of other material. The two hanging masses, by the two unbroken threads them, were drawn along identical that support world lines through spacetime (middle of the labo ratory of Eotvos!). If cut free, would they also follow identical tracks through spacetime (\"normal world line of test mass\")? If so, the acceleration that draws the actual world line from the normal free-fall world line will have a standard value, a. The experiment of Eotvos did not try to test agree ment on the magnitude of a between the two masses. Doing so would have required (1) cutting the threads and (2) following the fall of the two masses. Eotvos renounced this approach in favor of a static observation that he could make with greater precision, comparing the direction of a for the two masses. The direction of the supporting thread, so his argument ran, reveals the direction in which the mass is being dragged away from its normal world line of \"free fall\" or \"weightless ness.\" This acceleration is the vectorial resultant of (I) an acceleration of magnitude g, directed outward against so-called gravity, and (2) an ac celeration directed toward the axis of rotation of the earth, of magnitude w2 R sin 8 (w, angular ve- \u00a7 1.3. WEIGHTLESSNESS 17 push of the ground under his feet was driving him away from a natural world line. Through that flaw in his arrangements, he became responsible for all those accelera tions. Put him in space and strap rockets to his legs. No difference!* Again the responsibility for what he sees is his. Once more he notes that \"all objects fall with '''No difference\" spelled out amounts to Einstein's (1911) principle of the local equivalence between a \"gravitational field\" and an acceleration: \"We arrive at a very satisfactory Interpretation of this law of experience, if we assume that the systems K and K' are physically exactly equivalent, that is, if we assume that we may just as well regard the system K as being in a space free from gravitational fields, if we then regard K as uniformly accelerated. This assumption of exact physical equivalence makes it impossible for us to speak of the absolute acceleration of the system of reference, just as the usual theory of relativity forbids us to talk of the absolute velocity of a system; and it makes the equal falling of all bodies in a gravitational field seem a matter of course.\" locity; R, radius of earth; 8, polar angle measured from North Pole to location of experiment). This centripetal acceleration has a vertical component _w 2 R sin2 8 too small to come into discussion. The important component is w2 R sin 8 cos 8, di rected northward and parallel to the surface of the earth. It deflects the thread by the angle horizontal acceleration vertical acceleration ing 0.59 cm!sec2 produced by the sun's attraction. Reported Ig(Au) - g(Al)l!g less than 1 X 10-11. See Figure 1.6. Braginsky and Panov, Moscow, 1971: like Roll, Krotkov, and Dicke", ", remains always similar and immovable.\" But how does one give meaning to Newton's absolute space, find its cornerstones, mark out its straight lines? In the real world of gravitation, no particle ever follows one of Newton's straight lines. His ideal geometry is beyond observation. \"A comet going past the sun is deviated from an ideal straight line.\" No. There is no pavement on which to mark out that line. The \"ideal straight line\" is a myth. It never happened, and it never will. \"It required a severe struggle [for Newton) to arrive at the concept of independent and absolute space, indispensible for the development of theory.... Newton's decision was, in the contemporary state of science, the only possible one, and particularly the only fruitful one. But the subsequent development of the problems, proceeding in a roundabout way which no one could then possibly foresee, has shown that the resistance of Leibniz and Huygens, intuitively well-founded but supported by inadequate argu ments, was actually justified.... It has required no less strenuous exertions subsequently to overcome this concept [of absolute space)\" [A. EINSTEIN (1954\u00bb). What is direct and simple and meaningful, according to Einstein, is the geometry in every local inertial reference frame. There every particle moves in a straight line with uniform velocity. Define the local inertial frame so that this simplicity occurs for the first few particles (Figure 1.7). In the frame thus defined, every other free particle is observed also to move in a straight line with uniform velocity. Collision and disintegration processes follow the laws of conservation of momentum and energy of special relativity. That all these miracles come about, as attested by tens of thousands of observations in elementary particle physics, is witness to the inner workings of the machinery of the world. The message is easy to summarize: (l) physics is always and everywhere locally Lorentzian; i.e., locally the laws of special relativity are valid; (2) this simplicity shows most clearly in a local Lorentz frame of reference (\"inertial frame of reference\"; Figure 1.7); and (3) to test for a local Lorentz frame, test for weightlessness! \u00a71.4. LOCAL LORENTZ GEOMETRY, WITH AND WITHOUT COORDINATES On the surface of an apple within the space of a thumbprint, the geometry is Euclidean (Figure 1.1; the view in the magnifying glass). In spacetime, within a limited region, the geometry is Lorentzian. On the apple the distances between point and point accord with the theorems of Euclid. In spacetime the intervals (\"proper distance,\" \"proper time\") between event and event satisfy the corresponding theo rems of Lorentz-Minkowski geometry (Box 1.3). These theorems lend themselves (continued 011 page 23) Newton's absolute space is unobservable, nonexistent But Einstein's loca,l inertial frames exist, are simple In local inertial frames, physics is Lorentzian Local Lorentz geometry is the spacetime analog of local Euclidean geometry. 20 1. GEOMETRODYNAMICS IN BRIEF Box 1.3 LOCAL LORENTZ GEOMETRY AND LOCAL EUCLIDEAN GEOMETRY: WITH AND WITHOUT COORDINATES 1. Local Euclidean Geometry What does it mean to say that the geometry of a tiny thumbprint on the apple is Euclidean? A. Coordinatejree language (Euclid): Given a line de. Extend it by an equal distance e:z. Let ~t3 be a point not on d:;:\": but equidistant from d and :=. Then B. (Theorem of Pythagoras; also other theo rems of Euclidean geometry.) Language of coordinates (Descartes): ~13 From any point d to any other point there is a distance s given in suitable (Eucli dean) coordinates by stl,,l = [x 1Ut3) - x 1(d)j2 + [x 2(:t3) - x 2(d)j2. If one succeeds in finding any coordinate system where this is true for all points d and ~13 in the thumbprint, then one is guar anteed that (i) this coordinate system is locally Euclidean, and (ii) the geometry of the apple's surface is locally Euclidean. II. Local Lorentz Geometry What does it mean to say that the geometry of a sufficiently limited region of spacetime in the real physical world is Lorentzian? A. Coordinatejree language (Robb 1936): Let d:z be the world line of a free particle. Let ~B be an event not on this world line. Let a", " light ray from ~3 strike d:z at the event f2. Let a light ray take off from such an earlier event <3' along d:!? that it reaches ('11. Then the proper distance Stl'yJ (spacelike separation) or proper time Ttl!,) (timelike separation) is given by tI---~ {/ t' \u2022 \",' x2 = 12 /\" J 1/ J II{/ N II -~ (\"fj -~..::t II -~ V') -~ {/ \u00a71.4. GEOMETRY IS LOCALLY LORENTZIAN 21 :E!2 1 + x 1 - X '3' (/ o x i.e ~ V V I~ V') -~ \\0 II -~ rII -~ 00 -~ 0\\ II -~ Proof of above criterion for local Lorentz geometry, using coordinate methods in the local Lorentz frame where particle remains at rest: 1(nl = t2 - X2 = (t - X)(t + x) = 1 (/'}'1(/Z. B. Language ofcoordinates (Lorentz, Poincare, Minkowski, Einstein): From any event d to any other nearby event ~i3, there is a proper distance Sd<tJ or proper time 1(/,,! given in suitable (local Lorentz) coordinates by S(/'fl2 = -1(/i = - [XO(gj) - XO(d)]2 + [x 1(gJ) - x 1(d)j2 + [x 2(gJ) - x 2(d)j2 + [X3(~:B) - x 3(d)j2. If one succeeds in finding any coordinate system where this is locally true for all neighboring events d and ~'B, then one is guaranteed that (i) this coordinate system is locally Lorentzian, and (ii) the geometry of spacetime is locally Lorentzian. III. Statements of Fact The geometry of an apple's surface is locally Eu clidean everywhere. The geometry of spacetime is locally Lorentzian everywhere. Box 1.3 (continued) IV. Local Geometry in the Language of Modern Mathematics - local Euclidean coordinates (on apple) or local Lorentz coordinates (in spacetime). A. The metric for any manifold: At each point on the apple, at each event of spacetime, indeed, at each point of any \"Riemannian manifold.\" there exists a geo metrical object called the metric tensor g. lt is a machine with two input slots for the insertion of two vectors: slot 1 slot 2 + + ). g( Ifone inserts the same vector u into both slots, one gets out the square of the length of u: g(u, u) = u 2 If one inserts two different vectors, u and v (it matters not in which order!), one gets out a number called the \"scalar product of u on v\" and denoted u\u00b7 v: \u2022 g(u, v) = g(v, u) = u\u00b7 v = v\u00b7 u. The metric is a linear machine: g(2u + 3w, v) = 2g(u, v) + 3g(w, v), g(u, av + bw) = ag(u, v) + bg(u, w). Consequently, in a given (arbitrary) coordi nate system, its operation on two vectors can be written in terms of their components as a bilinear expression: g(u, v) = ga(3uav(3 (implied summation on 0:, 13) = gnu1v1 + g12u1v2 + g21 u2v1 +.... The quantities ga(3 = g(3a (0: and 13 running from 0 to 3 in spacetime, from 1 to 2 on the apple) are called the \"components ofgin the given coordinate system.\" B. Components ofthe metric in local Lorentz and local Euclidea~ frames: To connect the metric with our previous de scriptions of the local geometry, introduce 7,,'.' v Let ( be the separation vector reaching from d to ~jj. lts components in the local Eucli dean (Lorentz) coordinates are ~a = xa(~13) - xa(d) (cf. Box 1.1). Then the squared length of Ud,';' which is the same as the squared distance from d tO~1j, must be (cf. LB. and 11.B. above) (. ( = g\u00ab(, () = ga(3~a~(3 = sdi = (~1)2 +", " (e)2 on apple = _ (~O)2 + (~1f + (~2)2 + (e)2 in spacetime. Consequently, the components of the met ric are i.e., gfr(3 = 0a(3 gu = g22 = 1, g12 = g21 = 0; on apple, in local Euclidean coordinates; goo = -I, gOk = 0, gjk = 0jk in spacetime, in local Lorentz coordinates. These special components of the metric in local Lorentz coordinates are written here and hereafter as g,,{3 or 1/a(3' by analogy with the Kronecker delta 0a(3' In matrix notation: f3---+ o 1 3 2 o -,-------------------------~ \u00a71.5. TIME 23 to empirical test in the appropriate, very special coordinate systems: Euclidean coordinates in Euclidean geometry; the natural generalization of Euclidean coordi nates (local Lorentz coordinates; local inertial frame) in the local Lorentz geometry of physics. However, the theorems rise above all coordinate systems in their content. They refer to intervals or distances. Those distances no more call on coordinates for their definition in our day than they did in the time of Euclid. Points in the great pile of hay that is spacetime; and distances between these points: that is geometry! State them in the coordinate-free language or in the language of coordi nates: they are the same (Box 1.3). \u00a7 1.5. TIME Time is defined so that motion looks simple. Time is awake when all things sleep. Time stands straight when all things fall. Time shuts in all and will not be shut. Is, was, and shall be are Time's children. a Reasoning, be witness, be stable. the Mahabarata (ca. A.D. 400) VYASA, Relative to a local Lorentz frame, a free particle \"moves in a straight line with uniform velocity.\" What \"straight\" means is clear enough in the model inertial reference frame illustrated in Figure 1.7. But where does the \"uniform velocity\" come in? Or where does \"velocity\" show itself? There is not even one clock in the drawing! A more fully developed model of a Lorentz reference frame will have not only holes, as in Fig. 1.7, but also clock-activated shutters over each hole. The projectile can reach its target only if it (l) travels through the correct region in space and (2) gets through that hole in the correct interval of time (\"window in time\"). How then is time defined? Time is defined so that motion looks simple! No standard of time is more widely used than the day, the time from one high noon to the next. Take that as standard, however, and one will find every good clock or watch clashing with it, for a simple reason. The Earth spins on its axis and also revolves in orbit about the sun. The motion of the sun across the sky arises from neither effect alone, but from the two in combination, different in magnitude though they are. The fast angular velocity of the Earth on its axis (roughly 366.25 complete turns per year) is wonderfully uniform. Not so the apparent angular velocity of the sun about the center of the Earth (one turn per year). It is greater than average by 2 per cent when the Earth in its orbit (eccentricity 0.017) has come 1 per cent closer than average to the sun (Kepler's law) and lower by 2 per cent when the Earth is 1 per cent further than average from the sun. In the first case, the momentary rate of rotation of the sun across the sky, expressed in turns per year, is approximately 366.25 - (l + 0.02); The time coordinate of a local Lorentz frame is so defined that motion looks simple 24 in the other, 1. GEOMETRODYNAMICS IN BRIEF 366.25 - (1 - 0.02). Taking the \"mean solar day\" to contain 24 X 3.600 = 86,400 standard seconds, one sees that, when the Earth is 1 per cent closer to (or further from) the sun than average, then the number of standard seconds from one high noon to the next is greater (or less) than normal by 0.02 (drop in turns per year) 365.25 (turns per year on average) 86400, sec _ 47. sec. This is the bookkeeping on time from noon to noon. No standard of time that varies so much from one month to another is acceptable. If adopted, it would make the speed of light vary from month to month! This lack of uniformity, once recognized (and it was already recognized by the ancients), forces one to abandon the solar day as", " the standard of time; that day does not make motion look simple. Turn to a new standard that eliminates the motion of the Earth around the sun and concentrates on the spin of the Earth about its axis: the sidereal day, the time between one arrival of a star at the zenith and the next arrival of that star at the zenith. Good! Or good, so long as one's precision of measurement does not allow one to see changes in the intrinsic angular velocity of the Earth. What clock was so bold as first to challenge the spin of the Earth for accuracy? The machinery of the heavens. Halley (1693) and later others, including Kant (1754), suspected something was amiss from apparent discrepancies between the paths of totality in eclipses of the sun, as predicted by Newtonian gravitation theory using the standard of time then current, and the location of the sites where ancient Greeks and Romans actually recorded an eclipse on the day in question. The moon casts a moving shadow in space. On the day of a solar eclipse, that shadow paints onto the disk of the spinning Earth a black brush stroke, often thousands of kilometers in length, but of width generally much less than a hundred kilometers. He who spins the globe upon the table and wants to make the shadow fall rightly on it must calculate back meticu lously to determine two key items: (1) where the moon is relative to Earth and sun at each moment on the ancient day in question; and (2) how much angle the Earth has turned through from then until now. Take the eclipse of Jan. 14, A.D. 484, as an example (Figure 1.8), and assume the same angular velocity for the Earth in the intervening fifteen centuries as the Earth had in 1900 (astronomical reference point). One comes out wrong. The Earth has to be set back by 30\u00b0 (or the moon moved from its computed position, or some combination of the two effects) to make the Athens observer fall under the black brush. To catch up those 30\u00b0 (or less, if part of the effect is due to a slow change in the angular momentum of the moon), the Earth had to tum faster in the past than it does today. Assigning most of the discrepancy to terrestrial spin-down (rate of spin-down compatible with modern atomic-clock evidence), and assuming a uniform rate of slowing from then to now 25 I I I I o 200 400 km. I A: 'Observed' track WE 20 0 E.. B: Computed track if length of day constant Figure 1.8. Calculated path of totality for the eclipse of January 14, A.D. 484 (left; calculation based on no spin-down of Earth relative to its 1900 angular velocity) contrasted with the same path as set ahead enough to put the center of totality (at sunrise) at Athens [displacement very close to 30\u00b0; actual figure of deceleration adopted in calculations, 32.75 arc sec/(century)2). This is \"undoubtedly the most reliable of all ancient European eclipses,\" according to Dr. F. R. Stephenson, of the Department of Geophysics and Planetary Physics of the University of Newcastle upon Tyne, who most kindly prepared this diagram especially for this book. He has also sent a passage from the original Greek biography of Proclus of Athens (died at Athens A.D. 485) by Marinus of Naples, reading, \"Nor were there portents wanting in the year which preceded his death; for example, such a great eclipse of the Sun that night seemed to fall by day. For a profound darkness arose so that stars even appeared in the sky. This happened in the eastern sky when the Sun dwelt in Capricorn\" [from Westermann and Boissonade (1878\u00bb). Does this 30\u00b0 for this eclipse, IOgether with corresponding amounts for other eclipses, represent the \"right\" correction? \"Right\" is no easy word. From one total eclipse of the sun in the Mediterranean area to another is normally many years. The various provinces of the Greek and Roman worlds were far from having a uniform level of peace and settled life, and even farther from having a uniform standard of what it is to observe an eclipse and put it down for posterity. If the scores of records of the past are unhappily fragmentary, even more unhappy has been the willingness of a few uncritical \"investigators\" in recent times to rush in and identify this and that historical event with this and that calculated eclipse. Fortunately, by now a great literature is available on the secular deceleration of the Earth's rotation, in the highest tradition of critical scholarship, both astronomical and historical. In addition to the books of o. Neugebauer (1959) and Munk and MacDonald (1960). the paper of Curott (1966), and items cited by these workers, the following are key items. (For direction to them, we thank", " Professor Otto Neuge bauer-no relation to the other Neugebauer cited below!) For the ancient records, and for calculations of the tracks of ancient eclipses, F. K. Ginzel (1882, 1883, 1884): for an atlas of calculated eclipse tracks, Oppolzer (1887) and Ginzel (1899); and for a critical analysis of the e\\idence. P. V. Neugebauer (1927, 1929, and 1930). This particular eclipse was chosen rather than any other because of the great reliability of the historical record of it. 26 1. GEOMETRODYNAMICS IN BRIEF (angular velocity correction proportional to first power of elapsed time: angle cor rection itself proportional to square of elapsed time), one estimates from a correction of 30\u00b0 or 2 hours 1,500 years ago the following corrections for intermediate times: 30\u00b0 /10 2, or 1.2 min 150 years ago, 30\u00b0 /10'\\ or 0.8 sec 15 years ago. Thus one sees the downfall of the Earth as a standard of time and its replacement by the orbital motions of the heavenly bodies as a better standard: a standard that does more to \"make motion look simple.\" Astronomical time is itself in turn today being supplanted by atomic time as a standard of reference (see Box lA, \"Time Today\"). Look at a bad clock for a good view of how time is defined. Let t be time on a \"good\" clock (time coordinate of a local inertial frame); it makes the tracks of free particles through the local region of spacetime look straight. Let T(t) be the reading of the \"bad\" clock; it makes the world lines of free particles through the local region of spacetime look curved (Figure 1.9). The old value of the acceleration, translated into the new (\"bad\") time, becomes Good clocks make spacetime trajectories of free particles look straight 0= d 2 x =.!J....(dT dX) = d dt2 dt dT dt 2 T dx + (dT)2 d dt2 dT dt dT2 2 x. To explain the apparent accelerations of the particles, the user of the new time introduces a force that one knows to be fictitious: (l.2) It is clear from this example of a \"bad\" time that Newton thought of a \"good\" time when he set up the principle that \"Time flows uniformly\" (d 2T/dt2 = 0). Time is defined to make motion look simple! The principle of uniformity, taken by itself, leaves free the scale of the time variable. The quantity T = at + b satisfies the requirement as well as t itself. The history of timekeeping discloses many choices of the unit and origin of time. Each one required some human action to give it sanction, from the fiat of a Pharaoh to the communique of a committee. In this book the amount of time it takes light to travel one centimeter is decreed to be the unit of time. Spacelike intervals and timelike intervals are measured in terms of one and the same geometric unit: the centimeter. Any other decision would complicate in analysis what is simple in nature. No other choice would live up to Minkowski's words, \"Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.\" Our choice of unit for measuring time: the geometrodynamic centimeter. \u00a71.5. TIME t( I 27 tT I -x~ -x~ Figure 1.9. Good clock (left) vs. bad clock (right) as seen in the maps they give of the same free particles moving through the same region of spacetime. The world lines as depicted at the right give the impression that a force is at work. The good definition of time eliminates such fictitious forces. The dashed lines connect corresponding instants on the two time scales. One can measure time more accurately today than distance. Is that an argument this the geometrodynamic standard centimeter against taking the elementary unit to be the centimeter? No, provided that definition of the centimeter is accepted: is the fraction 1/(9.460546 X 1017 ) (1.3) of the interval between the two \"effective equinoxes\" that bound the tropical year 1900.0. The tropical year 1900.0 has already been recognized internationally as the fiducial interval by reason of its definiteness and the precision with which it is known. Standards committees have defined the ephemeris second so that 31,556,925.974 sec make up that standard interval. Were the speed of light known with perfect precision, the standards committees could have given in the same breath the number of centimeters in the standard interval. But it isn't; it is", " known to only six decimals. Moreover, the international centimeter is defined in terms of the orange-red wave length of Kr86 to only nine decimals (16,507.6373 wavelengths). Yet the standard second is given to 11 decimals. We match the standard second by arbitrarily defining the geometrodynamic standard centimeter so that 9.4605460000 X 1017 such centimeters are contained in the standard tropical year 1900.0. The speed of light then becomes exactly 9.4605460000 X 1017 31,556,925.974 geometrodynamlc cm/sec. \u2022 (1.4) This is compatible with the speed oflight, as known in 1967, in units of \"international cm/sec\": 29,979,300,000 \u00b1 30,000 international cm/sec. ~----------... Box 1.4 TIME TODAY Prior to 1956 the second was defined as the frac tion 1/86,400 of the mean solar day. From 1956 to 1967 the\u00b7 \"second\" meant ephemeris second, 1/(31,556,925.9747) the OOhOOmOOs December 31, 1899. defined of as the fraction year the tropical (Systeme International) Since 1967 the standard second has been the SI second, defined as 9,192,631,770 periods of the unperturbed micro wave transition between the two hyperfine levels of the ground state of CS133. Like the foregoing evolution of the unit for the the evolution of a time coordinate time interval, has been marked by several stages. Universal time, UTO, is based on the count of days as they actually occurred historically; in other words, on the actual spin of the earth on its axis; historically, on mean solar time (solar position as corrected by the \"equation of time\"; i.e., the faster travel of the earth when near the sun than when far from the sun) as determined at Greenwich Observatory. UTI, the \"navigator's time scale,\" is the same time as corrected for the wobble of the earth on its axis (..1t - 0.05 sec). UT2 is UTI as corrected for the periodic fluc tuations of unknown origin with periods of one half year and Qne year (..1t - 0.05 sec; measured to 3 ms in one day). Ephemeris Time, ET (as defined by the theory of gravitation and by astronomical observations and calculations), is essentially determined by the orbital motion of the earth around the sun. \"Measurement uncertainties limit the realization of accurate ephemeris time to about 0.05 sec for a nine-year average.\" Coordinated Universal Time (UTC) is broadcast on stations such as V,rWV. It was adopted interna tionally in February 1971 to become effective Jan uary 1,1972. The clock rate is controlled by atomic clocks to be as uniform as possible for one year (atomic time is measured to -0.1 microsec in 1 min, with diffusion rates of 0.1 microsec per day for ensembles of clocks), but is changed by the infrequent addition or deletion of a second-called a \"leap second\"-so that UTC never differs more than 0.7 sec from the navigator's time scale, UTI. Time suspended for a second ~ June 30. All Time will stand still throughout the world for one ~ccond at mid night. radio time signal~ will insert a.. leap!iecond.. to bring Greenwich Mean Time into line with the earth's los~ of three thousandths of a second a day. The signal from the Royal Green wich Ohservatory to Broadcasting House at midnight GMT (I am BST July I) wH1 be six short pips marking the seconds 55 to 60 inclu sive, followed by a lengthened sig nal at the following second to mark tho new minute. THE TIMES Wednesday June 21 1972 The foregoing account is abstracted from J. A. Barnes (1971). The following is extracted from a table (not official at time of receipt), kindly sup plied by the Time and Frequency Division of the U.S. National Bureau of Standards in Boulder, Colorado. Timekeeping capabilities of some familiar clocks are as follows: Tuning fork wrist watch (1960), 1 min/mo. Quartz crystal clock (1921-1930), 1 fLsec/day, 1 sec/yr. Quartz crystal wrist watch (1971), 0.2 sec/2 mos., 1 sec/yr. Cesium beam (atomic resonance, C S133), (1952 1955), 0.1 fLsec/day, 0.5 fLsec/mo. Rubidium gas cell (Rb8i resonance), (1957), 0.1 fLsec/ day, 1-5 fLsec/mo. Hydrogen maser (1960), 0.Q1 fLsec/2 hr, 0.", "1 fLsec/day. Methane stabilized laser (1969), 0.Q1 fLsec/IOO sec. \u00a71.6. CURVATURE 29 Recent measurements [Evenson et at. (1972)] change the details of the foregoing 1967 argument, but not the principles. \u00a71.6. CURVATURE Gravitation seems to have disappeared. Everywhere the geometry of spacetime is locally Lorentzian. And in Lorentz geometry, particles move in a straight line with constant velocity. Where is any gravitational deflection to be seen in that? For answer, turn back to the apple (Figure 1.1). Inspect again the geodesic tracks of the ants on the surface of the apple. Note the reconvergence of two nearby geodesics that originally diverged from a common point. What is the analog in the real world of physics? What analogous concept fits Einstein's injunction that physics is only simple when analyzed locally? Don't look at the distance from the spaceship to the Earth. Look at the distance from the spaceship to a nearby spaceship! Or, to avoid any possible concern about attraction between the two ships, look at two nearby test particles in orbit about the Earth. To avoid distraction by the nonlocal element (the Earth) in the situation, conduct the study in the interior of a spaceship, also in orbit about the Earth. But this region has already been counted as a local inertial frame! What gravitational physics is to be seen there? None. Relative to the spaceship and therefore relative to each other, the two test particles move in a straight line with uniform velocity, to the precision of measurement that is contemplated (see Box 1.5, \"Test for Flatness\"). Now the key point begins to appear: precision of measurement. Increase it until one begins to discern the gradual acceleration of the test particles away from each other, if they lie along a common radius through the center of the Earth; or toward each other, if their separation lies perpendicular to that line. In Newtonian language, the source of these accelerations is the tide-pro ducing action of the Earth. To the observer in the spaceship, however, no Earth is to be seen. And following Einstein, he knows it is important to analyze motion locally. He represents the separation of the new test particle from the fiducial test particle by the vector ~k(k = 1,2,3; components measured in a local Lorentz frame). For the acceleration of this separation, one knows from Newtonian physics what he will find: if the Cartesian z-axis is in the radial direction, then (1.5) Proof: In Newtonian physics the acceleration of a single particle toward the center of the Earth in conventional units of time is Gmconv/r2, where G is the Newtonian constant of gravitation, 6.670 X 10-8 cm3/g sec2 and mconv is the mass of the Earth in conventional units of grams. In geometric units of time (cm of light-travel time). Gravitation is manifest in relative acceleration of neighboring test particles 30 1. GEOMETRODYNAMICS IN BRIEF the acceleration is Gl11eonv / c2r 2 \u2022 When the two particles are separated by a distance ~ perpendicular to r, the one downward acceleration vector is out of line with the other by the angle ~/r. Consequently one particle accelerates toward the other by the stated amount. When the separation is parallel to r, the relative acceleration is given by evaluating the Newtonian acceleration at r and at r + ~, and taking the difference (~times d/dr) Q.E.D. In conclusion, the \"local tide-producing acceleration\" of Newtonian gravitation theory provides the local description of gravitation that Einstein bids one to seek. What has this tide-producing acceleration to do with curvature? (See Box 1.6.) the apple or, better, at a sphere of radius a (Figure 1.10). The Look again at separation of nearby geodesics satisfies the \"equation of geodesic deviation,\" d2~/ds2 + R~ = O. (1.6) Here R = l/a2 is the so-called Gaussian curvature of the surface. For the surface of the apple, the same equation applies, with the one difference that the curvature R varies from place to place. Relative acceleration is caused by curvature Box 1.5 TEST FOR FLATNESS 1. Specify the extension in space L (cm or m) and extension in time T (cm or m of light travel time) of the region under study. (d/dr)(m/r 2 ) = -2m/r3 = -0.888 cm/(6.37 X 108 cm)3 = -3.44 X 10-2; cm-2 2. Specify the precision o~ with which one can measure the separation of test particles in this region.", " 3. Follow the motion of test particles moving along initially parallel world lines through this region of spacetime. 4. When the world lines remain parallel to the precision o~ for all directions of travel, then one says that \"in a region so limited and to a precision so specified, spacetime is flat.\" EXAMPLE: Region just above the surface of the earth, 100 m X 100 m X 100 m (space extension), followed for 109 m of light-travel 3 sec). Mass of Earth, meonv = 5.98 X 102; g, m = (0.742 X 10-28 cm/g) X (5.98 X 102; g) = 0.444 cm [see eq. (1.12)]. Tide-producing accelera tion R Z OzO (relative acceleration in z-direction of two test particles initially at rest and separated from each other by 1 cm of vertical elevation) is time (Teonv (\"cm of relative displacement per cm of light travel time per cm of light-travel time per cm of vertical separation\"). Two test particles with a ver tical separation ~z = 104 cm acquire in the time t = 1011 cm (difference between time and proper time negligible for such slowly moving test parti cles) a relative displacement o~z = -~Rzozot2~Z = 1.72 X 10-2; cm-2(lOll cm)2 104 cm = 1.72 mm. (Change in relative separation less for other direc tions of motion). When the minimum uncertainty o~ attainable in a measurement over a 100 m spacing is \"worse\" than this figure (exceeds 1.72 mm), then to this level of precision the region of spacetime under consideration can be treated as is flat. When the uncertainty in measurement \"better\" (less) than 1.72 mm, then one must limit attention to a smaller region of space or a shorter interval of time or both, to find a region of space time that can be regarded as flat to that precision. \u00a7 1.6. CURVATURE 31 Figure 1.10. Curvature as manifested in the \"acceleration of the separation\" of two nearby geodesics. Two geodesics, originally parallel, and separated by the distance (\"geodesic deviation\") ~o, are no longer parallel when followed a distance s. The separation is ~ := ~o cos <P := ~o cos (sla), where a is the radius of the sphere. The separation follows the equation of simple harmonic motion, rJ2~/ds2 + (lla2) ~ := 0 (\"equation of geodesic devia tion\"). The direction of the separation vector, (, is fixed fully by its orthogon ality to the fiducial geodesic. Hence, no reference to the direction of ( is needed or used in the equation of geodesic deviation: only the magni tude ~ of ( appears there, and only the magnitude, not direction, of the relative acceleration appears. / \"'- \"- / \" / I--<p~\\ \\ o ---'---s In a space of more than two dimensions, an equation of the same general form applies, with several differences. In two dimensions the direction of acceleration of one geodesic relative to a nearby, fiducial geodesic is fixed uniquely by the demand that their separation vector, f, be perpendicular to the fiducial geodesic (see Figure 1.10). Not so in three dimensions or higher. There f can remain perpendicular to the fiducial geodesic but rotate about it (Figure 1.11). Thus, to specify the relative acceleration uniquely, one must give not only its magnitude, but also its direction. The relative acceleration in three dimensions and higher, then, is a vector. Call it \"D2f /ds2,\" and call its four components \"D2~a:/ds2.\" Why the capital D? Why not \"d2~a:/ ds2\"? Because our coordinate system is completely arbitrary (cf. \u00a7 1.2). The twisting and turning of the coordinate lines can induce changes from point to point in the components ~a: of f, even if the vector f is not changing at all. Consequently, the accelerations of the components d2~a:/ ds2 are generally not equal to the compo nents D2~a:/ ds2 of the acceleration! How, then, in curved spacetime can one determine the components D2~a:/ds2 of the relative acceleration? By a more complicated version of the equation of geodesic deviation (1.6). Differential geometry (Part III of this book) provides us with a geometrical object called the Riemann curvature tensor, \"Riemann.\" Riemann is (continued on page 34) Curvature is characterized by Riemann tensor", " Figure 1.11. The separation vector ( between two geodesics in a curved three dimensional manifold. Here ( can not only change its length from point to point. but also rotate at a varying rate about the fiducial geodesic. Consequently. the relative acceleration of the geodesics must be characterized by a direction as well as a magnitude: it must be a vector, D~(/ds2. ~------- Box 1.6 CURVATURE OF WHAT? Nothing seems more attractive at first glance than the idea that gravitation is a manifestation of the curvature of space (A)-, and nothing more ridicu lous at a second glance (B). How can the tracks of a ball and of a bullet be curved so differently if that curvature arises from the geometry of space? No wonder that great Riemann did not give the world a geometric theory of gravity. Yes, at the age of 28 (June 10, 1854) he gave the world the mathematical machinery to define and calcu late curvature (metric and Riemannian geometry). Yes, he spent his dying days at 40 working to find a unified account of electricity and gravitation. But if there was one reason more than any other why he failed to make the decisive connection between that he gravitation and curvature, thought of space and the curvature of space, not it was this, of spacetime and the curvature of spacetime. To make that forward step took the forty years to special relativity (1905: time on the same footing as space) and then another ten years (1915: gen the eral relativity). Depicted in spacetime (C), tracks of ball and bullet appear to have compara ble curvature. In fact, however, neither track has any curvature at all. They both look curved in (C) only because one has forgotten that the spacetime they reside in is itself curved-curved precisely enough to make these tracks the straightest lines in existence (\"geodesics\"). If it is at first satisfying to see curvature, and curvature of spacetime at that, coming to the fore in so direct a way, then a little more reflection produces a renewed sense of concern. Curvature with respect to what? Not with respect to the labo- Photograph of stars when sun (eclipsed bv moon) lies \" a~ indicated...\" Photograph of stars when sun swims elsewhere t I I I I l' I I I I A. Bending of light by the sun depicted as a conse quence of the curvature of space near the sun. Ray of light pursues geodesic, but geometry in which it travels is curved (actual travel takes place in spacetime rather than space; correct deflection is twice that given by above elementary picture). Deflection inversely propor tional to angular separation between star and center of sun. See Box 40.1 for actual deflections observed at time of an eclipse. [32] ratory. The earth-bound laboratory has no simple status whatsoever in a proper discussion. First, it is no Lorentz frame. Second, even to mention the earth makes one think of an action-at-a-distance version of gravity (distance from center of earth to ball or bullet). In contrast, it was the whole point of Einstein that physics looks simple only when analyzed locally. To look at local physics, however, means to compare one geodesic of one test particle with geodesics of other test particles traveling (1) nearby with (2) nearly the same di rections and (3) nearly the same speeds. Then one can \"look at the separations between these nearby test particles and from the second time-rate of change of these separations and the 'equation of geodesic deviation' (equation 1.8) read out the curvature of spacetime.\" z Ball 1 - - - - - -10 m -----~..I....., 5 mlsec \", \"- \\ \\ 500 mlsec _______ \\ -l \\ 5 X 10-4 T 5m m B. Tracks of ball and bullet through space as seen in laboratory have very different curvatures. r light-tra vel time- meters 0 C. Tracks of ball and bullet through spacetime, as re corded in laboratory, have comparable curvatures. Track compared to arc of circle: (radius) = (horizontal distance)2/8 (rise). [33] 34 1. GEOMETRODYNAMICS IN BRIEF the higher-dimensional analog of the Gaussian curvature R of our apple's surface. Riemann is the mathematical embodiment of the bends and warps in spacetime. And Riemann is the agent by which those bends and warps (curvature of spacetime) produce the relative acceleration of geodesics. Riemann, like the metric tensor 9 of Box 1.3, can be thought of as a", " family of machines, one machine residing at each event in spacetime. Each machine has three slots for the insertion of three vectors: slot slot 2 slot 3 + + + Riemann ( ). Choose a fiducial geodesic (free-particle world line) passing through an event 2, and denote its unit tangent vector (particle 4-velocity) there by u = dx/dr; components, ua = dx\"/dr. (1.7) Choose another, neighboring geodesic, and denote by ( its perpendicular separation from the fiducial geodesic. Then insert u into the first slot of Riemann at 2, ( into the second slot, and u into the third. Riemann will grind for awhile; then out will pop a new vector, Riemann (u, (, u). Riemann tensor, through equation of geodesic deviation, produces relative accelerations The equation of geodesic deviation states that this new vector is the negative of the relative acceleration of the two geodesics: D2( /dr2 + Riemann (u, (, u) = O. (1.8) The Riemann tensor, like the metric tensor (Box 1.3), and like all other tensors, is a linear machine. The vector it puts out is a linear function of each vector inserted into a slot: Riemann (2u, aw + bv, 3,) = 2 X a X 3 Riemann (u, w,') + 2 X b X 3 Riemann (u, v,,). (1.9) Consequently, in any coordinate system the components of the vector put out can be written as a \"trilinear function\" of the components of the vectors put in: r = Riemann (u, v, w) r\" = R\"/3Y3 u(3 vy w3. (1.10) (Here there is an implied summation on the indices {3, y, 8; cf. Box 1.1.) The 4 X 4 X 4 X 4 = 256 numbers R\"(3Y3 are called the \"components of the Riemann tensor in the given coordinate system.\" In terms of components, the equation of geodesic deviation states (1.8') \u00a71.6. CURVATURE 35 In Einstein's geometric theory of gravity, this equation of geodesic deviation summarizes the entire effect of geometry on matter. It does for gravitation physics what the Lorentz force equation, Equation of geodesic deviation is analog of Lorentz force law (Lll) does for electromagnetism. See Box 1.7. The units of measurement of the curvature are cm-2 just as well in spacetime as on the surface of the apple. Nothing does so much to make these units stand out clearly as to express mass in \"geometrized units\": Geometrized units m(cm) = (G/c2)mconv(g) = (0.742 X 10-28 cm/g)mconv(g). (1.12) Box 1.7 EQUATION OF MOTION UNDER THE INFLUENCE OF A GRAVITATIONAL FIELD AND AN ELECTROMAGNETIC FIELD, COMPARED AND CONTRASTED Electromagnetism [Lorentz force, equation (1.11)] Gravitation [Equation of geodesic deviation (1.8,)] Acceleration is defined for one particle? Yes Acceleration defined how? Actual world line compared to world line of uncharged \"fiducial\" test particle passing through same point with same 4-velocity. No Already an uncharged test particle, which can't accelerate relative to itself! Acceleration measured relative to a nearby test particle as fiduciary standard. Acceleration depends on all four components of the 4-veloci ty of the particle? Universal acceleration for all test particles in same locations with same 4-velocity? Yes No: is proportional to elm Yes Yes Driving field Electromagnetic field Riemann curvature tensor Ostensible number of distinct components of driving field Actual number when allowance is made for symmetries of tensor Names for more familiar of these components 4 X 4 = 16 4' = 256 6 3 electric 3 magnetic 20 6 components of local Newtonian tide-producing acceleration 36 1. GEOMETRODYNAMICS IN BRIEF This conversion from grams to centimeters by means of the ratio G/c 2 = 0.742 X 10-28 cm/g is completely analogous to converting from seconds to centimeters by means of the ratio 9.4605460000 X 1017 cm c =..:.....:...:....:.:-:~-::-:-::-=-:=--'='::-:-:..-..;:.:", "..c.;.. 31,556,925.974 sec (see end of \u00a7 1.5). The sun, which in conventional units has meonv = 1.989 X 1033 g, has in geometrized units a mass m = 1.477 km. Box 1.8 gives further discussion. Using geometrized units, and using the Newtonian theory of gravity, one can readily evaluate nine of the most interesting components of the Riemann curvature tensor near the Earth or the sun. The method is the gravitational analog of deter mining the electric field strength by measuring the acceleration of a slowly moving test particle. Consider the separation between the geodesics of two nearby and slowly moving (v ~c) particles at a distance r from the Earth or sun. In the standard, nearly inertial coordinates of celestial mechanics, all components of the 4-velocity of the Components of Riemann tensor evaluated from relative accelerations of slowly moving particles Box 1.8 GEOMETRIZED UNITS Throughout this book, we use \"geometrized units,\" in which the speed of light c, Newton's gravita tional constarit G, and Boltzman's constant k are all equal to unity. The following alternative ways to express the number l.0 are of great value: 1.0 = c = 2.997930... X 1010 cm/sec 1.0 = G/c 2 = 0.7425 X 10- 28 cm/g; 1.0 = G/c4 = 0.826 X 10-49 cm/erg; l.0 = Gk/c4 = 1.140 X 10-65 cm/K; 1.0 = C2/GI/2 = 3.48 X 1024 cm/gauss- I \u2022 One can multiply a factor of unity, expressed in anyone of these ways, into any term in any equa tion without affecting the validity of the equation. Thereby one can convert one's units of measure from grams to centimeters to seconds to ergs to.... For example: Mass of sun = Me:> = 1.989 X 1033 g ) = (1.989 X 1033 g) X (G/c 2 = 1.477 X 105 cm = (1.989 X 1033 g) X (c 2 = 1.788 X 1054 ergs. ) The standard unit, in terms of which everything is measured in this book, is centimeters. However, occasionally conventional units are used; in such cases a subscript \"conv\" is sometimes, but not always, appended to the quantity measured: M0conv = 1.989 X 1033 g. \u00a71.7. EFFECT OF MATTER ON GEOMETRY 37 fiducial test particle can be neglected except dxoldT = 1. The space components of the equation of geodesic deviation read Comparing with the conclusions of Newtonian theory, equations (1.5), we arrive at the following information about the curvature of spacetime near a center of mass: (1.13) 0\",0 RZ\u2022\u2022\u2022 RYozo RZ\u2022\u2022\u2022 = mlr3 RIIOYO RZ\u2022\u2022\u2022 = 0 RZo\" OyO RZ\u00b7\u00b7o RYozo RZo\" = 0 0\",0 Oz zO 110 0 mlr3 0 0 0 -2mlr3 (1.14) (units cm- 2). Here and henceforth the caret or \"hat\" is used to indicate the compo nents of a vector or tensor in a local Lorentz frame of reference (\"physical compo nents,\" as distinguished from components in a general coordinate system). Einstein's theory will determine the values of the other components of curvature (e.g., RZzzz = -mlr 3); but these nine terms are the ones of principal relevance for many applications of gravitation theory. They are analogous to the components of the electric field in the Lorentz equation of motion. Many of the terms not evaluated are analogous to magnetic field components-ordinarily weak unless the source is in rapid motion. This ends the survey of the effect of geometry on matter (\"effect of curvature of apple in causing geodesics to cross\"-especially great near the dimple at the top, just as the curvature of spacetime is especially large near a center of gravitational attraction). Now for the effect of matter on geometry (\"effect of stem of apple in causing dimple\")! \u00a71.7. EFFECT OF MAnER ON GEOMETRY The weight of any heavy body of known weight at a particular distance from the center of the world varies according to the variation of its distance therefrom; so that as often as it is removed from the center, it becomes heavier, and when brought the relation of gravity to gravity is as the relation of distance to distance from the center. is lighter. On this account, near", " to it, AL KHAZINi (Merv, A.D. 1115), Book of the Balance of Wisdom Figure 1.12 shows a sphere of the same density, p == 5.52 g/cm3, as the average density of the Earth. A hole is bored through this sphere. Two test particles, A and E, execute simple harmonic motion in this hole, with an 84-minute period. Therefore their geodesic separation (, however it may be oriented, undergoes a simple periodic motion with the same 84-minute period: j = x or y or z. (I.I 5) 38 1. GEOMETRODYNAMICS IN BRIEF Box 1.9 GAll LEO GAll LEI Pisa, February 15, 1564-Arcetri, Florence, January 8, 1642 UfflZI Gallery, Florence \"In questions of science the authority of a thousand is not worth the humble reasoning of a single individual. \" GAll LEO GALILEI (1632) \"The spaces described by a body falling from rest with a uniformly accelerated motion are to each other as the squares of the time intervals employed in traversing these distances. \" GAll LEO GAll LEI (1638) \"Everything that has been said before and imagined bV other people [about the tides] is in my opinion completely invalid. But among the great men who have philosophised about this marvellous effect of nature the one who surprised me the most is Kepler. More than other people he was a person of independent genius, sharp, and had in his hands the motion of the earth. He later pricked up his ears and became interested in the action of the moon on the water, and in other occult phenomena, and similar childishness. \" GALILEO GAll LEI (1632) body of the Moon.. \"It is a most beautiful and delightful sight to behold [with the new telescope] the the Moon certainly does not possess a smooth and polished surface, but one rough and uneven... full of vast protuberances, deep chasms and sinuosities... stars in myriads, which have never been seen before and which surpass the old, previously known, stars in number more than ten times. I have discovered four planets, neither known nor observed by anyone of the astronomers before my time... got rid of disputes about the Galaxy or Milky Way, and made its nature clear to the very senses, not to say to the understanding... the galaxy is nothing else than a mass of luminous stars planted together in clusters... the number of small ones is quite beyond determination-the stars which have been called by everyone of the astronomers up to this day nebulous are groups of small stars set thick together in a wonderful way. \" GALILEO GAll LEI IN SIDEREUS NUNCIUS (1610) \"So the principles which are set forth in this treatise will, when taken up by thoughtful minds, lead to many another more remarkable result; and it is to be believed that it will be so on account of the nobility of the subiect, which is superior to any other in nature. \" GAll LEO GAll LEI (1638) \u00a71.7. EFFECT OF MATTER ON GEOMETRY 39 tI I A 5.52 g/cm3 Figure 1.12. Test particles A and B move up and down a hole bored through the Earth, idealized as of uniform density. At radius r, a parti cle feels Newtonian acceleration \u00b7,-r d d \u00b7,r- tFr c2 dlconv 2 G (mass inside radius r) r2 = - c2 _ (G )(4\\7 - - r2c2 3 pconvr 3) Consequently, each particle oscillates in simple harmonic mo tion with precisely the same angular frequency as a satellite, grazing the model Earth, traverses its circular orbit:.,( 2 w- cm- ) == -., p cm--). 4\\7 ( 3 \" 4\\7G 3 w-eonv(sec -) == -3- Peony g, em' ). (/ _\" Comparing this actual motion with the equation of geodesic deviation (1.13) for slowly moving particles in a nearly inertial frame, we can read off some of the curvature components for the interior of this model Earth. The Riemann tensor inside the Earth 0\",0 RX.., RYoxo RZo;'fl R\"'oiJo RYoiJo RX... RY... R\"fl.ifl I RZ... = (4'iTpj3) 0 0 OyO OzO OzO 0 I 0 0 0 I (1.16) This example illustrates how the curvature of spacetime is connected to the distribu tionof matter. Let a gravitational wave", " from a supernova pass through the Earth. Idealize the Earth's matter as so nearly incompressible that its density remains practically un changed, The wave is characterized by ripples in the curvature of spacetime. propa gating with the speed of light. The ripples will show up in the components RiokO of the Riemann tensor. and in the relative acceleration of our two test particles. The left side of equation (1.16) will ripple: but the right side will not. Equation ( 1.16) will break down. No longer will the Riemann curvature be generated directly and solely by the Earth's matter. Nevertheless. Einstein tells us. a part of equation (1.16) is undisturbed by the Effect of gravitational wave on Riemann tensor 40 waves: its trace 1. GEOMETRODYNAMICS IN BRIEF (1.1 7) Einstein tensor introduced Even in the vacuum outside the Earth this is valid; there both sides vanish rcf. (1.14)]. More generally, a certain piece of the Riemann tensor, called the Einstein tensor and denoted Einstein or G, is always generated directly by the local distribution of matter. Einstein is the geometric object that generalizes Roo, the lefthand side Box 1.10 ISAAC NEWTON Woolsthorpe, Lincolnshire, England, December 25, 1642 Kensington, London, March 20, 1726 \"The description of right lines and circles, upon which geometry is founded, belongs to mechanics. Geometry does not teach us to draw these lines, but requires them to be drawn. \" [FROM P. 1 OF NEWTON'S PREFACE TO THE FIRST (1687) EDITION OF THE PRINCIPIA) \"Absolute space, in its own nature, without relation to anything external, remains always similar and immovable true, and mathematical time, \"Absolute, flows of itself, and from its own nature, equably without relation to anything external. \" [FROM THE SCHOLIUM IN THE PRINCIPIA) \"I have not been able to discover the cause of those properties of gravity from phenomena, and I frame no hypotheses; for whatever is not reduced from the phenomena is to be called an hypothesis; and hypotheses... have no place in experimental philosophy.... And to us it is enough that gravity does really exist, and act according to the laws which we have explained, and abundantly serves to account for all the motions of the celestial bodies, and of our sea. \" [FROM THE GENERAL SCHOLIUM ADDED AT THE END OF THE THIRD BOOK OF THE PRINCIPIA IN THE SECOND EDITION OF 1713; ESPECIALLY FAMOUS FOR THE PHRASE OFTEN QUOTED FROM NEWTON'S ORIGINAL LATIN. \"HYPOTHESES NON FINGO.\") \"And the same year [1665 or 1666] I began to think of gravity extending to the orb of the Moon, and having found out.... All this was in the two plague years of 1665 and 1666, for in those days I was in the prime of my age for invention, and minded Mathematicks and Philosophy more than at any time since.\" [FROM MEMORANDUM IN NEWTON'S HANDWRITING ABOUT HIS DISCOVERIES ON FLUXIONS, THE BINOMIAL THEOREM, OPTICS, DYNAMICS. AND GRAVITY, BELIEVED TO HAVE BEEN WRITTEN ABOUT 1714. AND FOUND BY ADAMS ABOUT 1887 IN THE \"PORTSMOUTH COLLECTION\" OF NEWTON PAPERS) \u00a7 1. 7, EFFECT OF MATTER ON GEOMETRY 41 of equation (1.17). Like Roo, Einstein is a sort of average of Riemann over all directions. Generating Einstein and generalizing the righthand side of (1.16) is a geometric object called the stress-energy tensor of the matter. It is denoted T. No coordinates are need to define Einstein, and none to define T; like the Riemann tensor, Riemann, and the metric tensor, g, they exist in the complete absence of coordinates. Moreover, in natu're they are always equal, aside from a factor of 8'77: Einstein = G = 8'77T. (1.18) St ress-energy tensor introduced \"For hypotheses ought,,, to explain the properties of things and not attempt to predetermine them except in so far as they can be an aid to experiments,\" [FROM LETIER OF NEWTON TO I, M, PARDIES. 1672. AS QUOTED IN THE CAJORI NOTES AT THE END OF NEWTON (1687). P. 673) \"That one body may", " act upon another at a distance through a vacuum, without the mediation of any thing else, by and through which their action and force may be conveyed from one to another, is to me so great an absurdity, that I believe no man, who has in philosophical matters a competent faculty of thinking, can ever fall into it. \" [PASSAGE OFTEN QUOTED BY MICHAEL FARADAY FROM LETTERS OF NEWTON TO RICHARD BENTLY, 1692-1693, AS QUOTED IN THE NOTES OF THE CAJORI EDITION OF NEWTON (1687), P. 643) \"The attractions of gravity, magnetism, and electricity, reach to very sensible distances, and so have been observed.., ; and there may be others which reach to so small distances as hitherto escape observation;,,, some force, which in immediate contract is exceeding strong, at small distances performs the chemical operations above-mentioned, and reaches not far from the particles with any sensible effect. \" [FROM QUERY 31 AT THE END OF NEWTON'S OPTICKS (1730)) \"What is there in places almost empty of matter, and whence is it that the sun and planets gravitate towards one another, without dense matter between them? Whence is it that nature doth nothing in vain; and whence arises all that order and beauty which we see in the world? To what end are comets, and whence is it that planets move all one and the same way in orbs concentrick, while comets move all manner of ways in orbs very excentrick; and what hinders the fixed stars from falling upon one another?\" [FROM QUERY 28) \"He is not eternity or infinity, but eternal and infinite; He is not duration or space, but He endures and is present. He endures forever, and is everywhere present; and by existing always and everywhere, He constitutes duration and space,,,, And thus much concerning God; to discourse of whom from the appearances of things, does certainly belong to natural philosophy, \" [FROM THE GENERAL SCHOLIUM AT THE END OF THE PRINCIPIA (1687)) 42 1. GEOMETRODYNAM ICS IN BRIEF Einstein field equation: how matter generates curvature This Einstein field equation, rewritten in terms of components in an arbitrary coordi nate system. reads Gaf! = 8r.-Tap \u00b7 (1.19) The Einstein field equation is elegant and rich. No equation of physics can be written more simply. And none contains such a treasure of applications and conse quences. The field equation shows how the stress-energy of matter generates an average curvature (Einstein = G) in its neighborhood. Simultaneously, the field equation is a propagation equation for the remaining, anisotropic part of the curvature: it governs the external spacetime curvature of a static source (Earth); it governs the generation of gravitational waves (ripples in curvature of spacetime) by stress-energy in motion; and it governs the propagation of those waves through the universe. The field equation even contains within itself the equations of motion (\"Force = Consequences of Einstein field equation Box 1.11 ALBERT EINSTEIN Ulm, Germany. March 14. 1879- i Princeton. New Jersey. April 18. 1955 : LIbrary of E. T. Hochschule. Zunch AcademIa des SCIences. Pans ArchIves of Cahfornla Instttute of Technology SEAL: Courtesy of the LewIs and Rosa Strauss FoundatIon and Pnnceton UnIversIty Press \u00a71.7. EFFECT OF MATTER ON GEOMETRY 43 mass X acceleration\") for the matter whose stress-energy generates the curvature. Those were some consequences of G = 8'7TT. Now for some applications. The field equation governs the motion of the planets in the solar system; it governs the deflection of light by the sun; it governs the collapse of a star to form a black hole; it determines uniquely the external spacetime geometry of a black hole (\"a black hole has no hair\"); it governs the evolution of spacetime singularities at the end point of collapse; it governs the expansion and recontraction of the universe. And more; much more. In order to understand how the simple equation G = 8'7TT can be so all powerful, it is desirable to backtrack, and spend a few chapters rebuilding the entire picture of spacetime, of its curvature, and of its laws, this time with greater care, detail, and mathematics. Thus ends this survey of the effect of geometry on matter, and the reaction of matter back on geometry, rounding out the parable of the apple. Applications of Einstein field equation \"What really interests me is whether God had any choice in the creation of the world\" EINSTEIN TO", " AN ASSISTANT. AS QUOTED BY G. HOLTON (1971). P. 20 \"But the years of anxious searching in the dark, with their intense longing, their alternations of confidence and exhaustion, and the final emergence into the light-only those who have experienced it can understand that\" EINSTEIN. AS QUOTED BY M. KLEIN (1971). P. 1315 \"Of all the communities available to us there is not one I would want to devote myself to, except for the society of the true searchers, which has very few living members at any time... \" EINSTEIN LETTER TO BORN. QUOTED BY BORN (1971). P. 82 \"I am studying your great works and-when I get stuck anywhere-now have the pleasure of seeing your friendly young face before me smiling and explaining\" EINSTEIN. LETTER OF MAY 2. 1920. AFTER MEETING NIELS BOHR \"As far as the laws of mathematics refer to reality, as they are certain, they are not certain; and as far they do not refer to reality. \" 'The most incomprehensible thing about the world is that it is comprehensible. \" EINSTEIN. IN SCHILPP (1949). P, 112 EINSTEIN (1921). P. 28 EXERCISES 44 1. GEOMETRODYNAMICS IN BRIEF Exercise 1.1. CURVATURE OF A CYLINDER Show that the Gaussian curvature R of the surface of a cylinder is zero by showing that geodesics on that surface (unroll!) suffer no geodesic deviation. Give an independent argu ment for the same conclusion by employing the formula R = l/PIPZ' where PI and Pz are the principal radii of curvature at the point in question with respect to the enveloping Euclidean three-dimensional space. Exercise 1.2. SPRING TIDE VS. NEAP TIDE Evaluate (1) in conventional units and (2) in geometrized units the magnitude of the Newton ian tide-producing acceleration RmOno(m, n = 1,2,3) generated at the moon (m eODV = 7.35 X 1025 g, r = 3.84 X 10 10 cm) and (2) the sun (m eODV = 1.989 X 1033 g, r = 1.496 X 10 13 cm). By what factor do you expect spring tides to exceed neap tides? the Earth by (1) Exercise 1.3. KEPLER ENCAPSULATED A small satellite has a circular frequency w(cm- 1 ) in an orbit of radius r about a central object of mass m(cm). From the known value of w, show that it is possible to determine neither r nor m individually, but only the effective \"Kepler density\" of the object as averaged over a sphere of the same radius as the orbit. Give the formula for wZ in terms of this Kepler density. It is a reminder of the continuity of history that Kepler and Galileo (Box 1.9) wrote back and forth, and that the year that witnessed the death of Galileo saw the birth of Newton (Box 1.10). After Newton the first dramatically new synthesis of the laws of gravitation came from Einstein (Box 1.11). And what the dead had no speech for, when living, the communication They can tell you, being dead; Of the dead is tongued with fire beyond the language of the living. T. S. ELIOT. in LITTLE GIDDING (1942) I measured the skies Now the shadows I measure Skybound was the mind Earthbound the body rests JOHANN ES KEPLER. d. November 15, 1630. He wrote his epitaph in Latin; it is translated by Coleman (1967). p. 109. Ubi materia, ibi geometria. JOHANNES KEPLER PART II PHYSICS IN FLAT SPACETIME Wherein the reader meets an old friend, Special Relativity, outfitted in new, mod attire, and becomes more intimately acquainted with her charms. CHAPTER 2 FOUNDATIONS OF SPECIAL RELATIVITY In geometric and physical applications, it always turns out that a quantity is characterized not only by its tensor order, but also by symmetry. HERMAN WEYL (1925) Undoubtedly the most striking development of geometry during the last 2,000 years is the continual expansion of the concept \"geometric object.\" This concept began by comprising only the few curves and surfaces of Greek synthetic geometry; it was stretched, during the Renaissance, to cover the whole domain of those objects defined by analytic geometry; more recently, it has been extended to cover the boundless universe treated by point-set theory. KARL MENGER. IN SCHILPP (1949", "he next few sections will introduce several geometric objects, and show the roles they playas representatives of physical quantities in flat spacetime. Every physical quantity can be described by a geometric object laws of physics can be All expressed geometrically 49 \u00a72.3. VECTORS A.~ Two events 9('\\ = 0.7) C/,\\=l curved space) D/ Vector Vd~ Vector Vd~ conceived as arrow Cl~ or ~ - Cl (not valid in,\\ = 0 conceived as Parametrized line 9(,\\) = Cl +,\\(~ - Cl) (valid in c!ed space) Figure 2.1. From vector as connector of two points to vector as derivative (\"tangent vector\"; a local rather than a bilocal concept). \u00a72.3. VECTORS Begin with the simplest idea of a vector (Figure 2.1 B): an arrow extending from Ways of defining vector: one spacetime event Cl (\"tail\") to another event?iJ (\"tip\"). Write this vector as As arrow v(/qj =?iJ - Cl (or Cl~B). For many purposes (including later generalization to curved spacetime) other com pletely equivalent ways to think of this vector are more convenient. Represent the arrow by the parametrized straight line 9(;\\) = Cl + ;\\(?iJ - Cl), with ;\\ = 0 the tail As parametrized straightHine of the arrow, and;\\ = I its tip. Form the derivative of this simple linear expression for 9(;\\): (d/d;\\)[Cl + ;\\(~B - Cl)] =?iJ - Cl = 9(1) - 9(0) (tip) - (tail) V{/!ff' This result allows one to replace the idea of a vector as a 2-point object (\"bilocal\") by the concept of a vector as a I-point object (\"tangent vector\"; local): (2,1) As derivative of point along curve Example: if 9(T) is the straight world line of a free particle, parametrized by its proper time, then the displacement that occurs in a proper time interval of one second gives an arrow u = 9( I) - 9(0). This arrow is easily drawn on a spacetime diagram. It accurately shows the 4-velocity of the particle. However, the derivative formula u = d9/dT for computing the same displacement (I) is more suggestive of the velocity concept and (2) lends itself to the case of accelerated motion. Thus, given a world line 9(T) that is not straight, as in Figure 2.2, one must first form d9/dT, and only thereafter draw the straight line 9(0) + ;\\(d':P /dT)O of the arrow u = d9/dT to display the 4-velocity u. 50 2. FOUNDATIONS OF SPECIAL RELATIVITY 3 7 \\,=6------- -2 A=I Figure 2.2. Same tangent vector derived from two very different curves. That parame trized straight line is also drawn which best fits the two curves at ';'i'o' The tangent vector reaches from 0 to I on this straight line. Components of a vector components of the 4-velocity in a specific Lorentz reference frame: The reader may be unfamiliar with this viewpoint. More familiar may be the Uo =.!!!.... = I dT ~'. dx i u'=-=-=== dT ~' vi (2.2) where Vi = dx i jdt = components of \"ordinary velocity,\" 1'2 = (V.1')2 + (V Y)2 + (V Z )2. Even the components (2.2) of 4-velocity may seem slightly unfamiliar if the reader is accustomed to having the fourth component of a vector be multiplied by a factor i = v=T If so, he must adjust himself to new notation. (See \"Farewell to 'ict,'\" Box 2.1.) More fundamental than the components of a vector is the vector itself. It is a geometric object with a meaning independent of all coordinates. Thus a particle has a world line 9(T), and a 4-velocity u = d9/dT, that have nothing to do with any coordinates. Coordinates enter the picture when analysis on a computer is required (rejects vectors; accepts numbers). For this purpose one adopts a Lorentz frame with orthonormal basis vectors (Figure 2.3) eo, e l, e2' and ea' Relative to the origin e of this frame, the world line has a coordinate description Expressed relative to the same Lorentz frame, the 4-velocity of the particle is Basis vectors \u00a72", ".4. METRIC TENSOR 51 Box 2.1 FAREWELL TO \"iet\" One sometime participant in special relativity will to the sword: \"x 4 = ict.\" This have to be put imaginary coordinate was invented to make the geometry of spacetime look formally as little different as possible from the geometry of Eu clidean space; to make a Lorentz transformation look on paper like a rotation; and to spare one the distinction that one otherwise is forced to make between quantities with upper indices (such as the components piJ. of the energy-momentum vector) and quantities with lower indices (such as the components PiJ. of the energy-momentum I-form). However, it is no kindness to be spared this latter distinction. Without it, one cannot know whether a vector (\u00a72.3) is meant or the very different geo metric object that is a I-form (\u00a72.5). Moreover, there is a significant difference between an angle on which everything depends periodically (a rota tion) and a parameter the increase of which gives rise to ever-growing momentum differences (the \"velocity parameter\" of a Lorentz transformation; Box 2.4). If the imaginary time-coordinate hides from view the character of the geometric object being dealt with and the nature of the parameter in a transformation, it also does something even it hides the completely different more serious: metric structure (\u00a72.4) of + + + geometry and - + + + geometry. In Euclidean geometry, when the distance between two points is zero, the two points must be the same point. In Lorentz-Min kowski geometry, when the interval between two events is zero, one event may be on Earth and the other on a supernova in the galaxy M3l, but their separation must be a null ray (piece of a light cone). The backward-pointing light cone at a given event contains all the events by which that event can be influenced. The forward-pointing light cone contains all events that it can influence. The multi tude of double light cones taking off from all the events of spacetime forms an interlocking causal structure. This structure makes the machinery of the physical world function as it does (further comments on this structure in Wheeler and Feyn man 1945 and 1949 and in Zeeman 1964). If in a region where spacetime is flat, one can hide this structure from view by writing with x 4 = ict, no one has discovered a way to make an imaginary coordinate work in the general curved spacetime manifold. If \"x4 = ict\" cannot be used there, it will not be used here. In this chapter and hereafter, as throughout the literature of general relativity, a real time coordinate is used, XO = t = ctconv (superscript 0 rather than 4 to avoid any possibility of confusion with the imagi nary time coordinate). The components w\" of any other vector w in this frame are similarly defined as the coefficients in such an expansion, Expansion of vector in terms of basis Notice: the subscript a on e\" tells which vector, not which component! w = w\"e\". (2.4) \u00a72.4. THE METRIC TENSOR The metric tensor, one recalls from part IV of Box 1.3, is a machine for calculating the squared length of a single vector, or the scalar product of two different vectors. 52 2. FOUNDATIONS OF SPECIAL RELATIVITY 3 2 y \u2022 3 2 x 3 Figure 2.3. The 4-velocity of a particle in flat spacetime. The 4-velocity u is the unit vector (arrow) tangent to the particle's world line-one tangent vector for each event on the world line. In a specific Lorentz coordinate system, there are basis vectors of unit length, which point along the four coordinate axes: eo, ej,eZ' e 3. The 4-velocity, like any vector, can be expressed as a sum of components along the basis vectors: u = uOeo + u1ej + uZez + u3e 3 = uaea. More precisely, the metric tensor 9 is a machine with two slots for inserting vectors slot I slot 2 + + ). g( Upon insertion, the machine spews out a real number: g(u, v) = \"scalar product of u and v,\" also denoted u\u00b7 v. g(u, u) = \"squared length of u,\" also denoted u 2\u2022 (2.5) (2.6) Moreover, this number is independent of the order in which the vectors are inserted (\"symmetry of metric tensor\"), g(u, v) = g(v, u); and it is linear in the vectors inserted g(au + bv, w) =g(w, au + bv)", " =ag(u, w) + bg(v, w). (2.7) (2.8) Because the metric \"machine\" is linear, one can calculate its output, for any input, Metric defined as machine for computing scalar products of vectors \u00a72.5. DIFFERENTIAL FORMS 53 as follows, if one kriows only what it does to the basis vectors e a of a Lorentz frame. (l) Define the symbols (\"metric coefficients\") 'r/ap by Metric coefficients (2.9) (2) Calculate their numerical values from the known squared length of the separation vector ( =..:1xae a between two events: (L1s)2 = - (,jxO)2 + (..:1x 1)2 + (..:1x2f + (..:1x3f = g(.axaea,,jxPep ) =..:1xa,jxPg (ea, ep) =..:1xa,jxP'r/aP for every choice of,jxa -1 o o o 000 100 010 001 in any Lorentz frame. (2.10) (3) Calculate the scalar product of any two vectors u and v from u\u00b7 v = g(u, v) = g(uaea, vPep) = uavPg(ea, e p); u\u00b7 v = uavP'r/aP = -uovo + U1V1 + U2V2 + U3V3. (2.11) Scalar products computed from components of vectors That one can classify directions and vectors in spacetime into \"timelike\" (negative squared length), \"spacelike\" (positive squared length), and \"null\" or \"lightlike\" (zero squared length) is made possible by the negative sign on the metric coefficient 'r/oo' Box 2.2 shows applications of the above ideas and notation to two elementary problems in special relativity theory. \u00a72.5. DIFFERENTIAL FORMS Vectors and the metric tensor are geometric objects that are already familiar from Chapter I and from elementary courses in special relativity. Not so familiar, yet equally important, is a third geometric object: the \"differential form\" or \"] form.\" Consider the 4-momentum p of a particle, an electron, for example. To spell out one concept of momentum, start with the 4-velocity, u = d?i'/dT, of this electron (\"spacetime displacement per unit of proper time along a straightline approximation of the world line\"). This is a vector of unit length. Multiply by the mass m of the particle to obtain the momentum vector p =mu. But physics gives also quite another idea of momentum. It associates a de Broglie wave with each particle. Moreover, this wave has the most direct possible physical significance. Diffract this wave from a crystal lattice. From the pattern of diffraction, one can determine not merely the length of the de Broglie waves, but also the pattern in space made by surfaces of equal, integral phase ep = 7, ep = 8, ep = 9,.... This The 1-form illustrated by de Broglie waves 54 2. FOUNDATIONS OF SPECIAL RELATIVITY Box 2.2 WORKED EXERCISES USING THE METRIC Exercise: Show that the squared length of a test particle's 4-velocity u is -1. Soilition: In any Lorentz frame. using the components (2.2), one calculates as follows u 2 = g(u, u) = U\"l1/31/o./3 = _(11\u00b0)2 + (u 1)2 + (u 2f + (11 3)2 I = ----ry +---) =-1. 1,2 I - 1'- I - I'~ Exercise: Show that the rest mass of a particle is related to its energy and momen tum by the famous equation or, equivalently (geometrized units!), First Solution: The 4-momentum is defined by p = mu, where u is the 4-velocity and m is the rest mass. Consequently, its squared length is Second Solution: In the frame of the observer, where E and p are measured, the 4-momentum splits into time and space parts as pO = E, hence, its squared length is But in the particle's rest frame, p splits as hence, its squared length is p2 = _m 2\u2022 But the squared length is a geometric object defined independently of any coordinate system; so it must be the same by whatever means one calculates it: \u00a72.5. DIFFERENTIAL FORMS 55 Figure 2.4. The vector separation v = '", "1' '1'0 between two neighboring events '1'0 and '1'; a I-form 0\"; and the piercing of 0\" by v to give the number (0\", v) = (number of surfaces pierced) = 4.4 (4.4 \"bongs of bell\"). When 0\" is made of surfaces of constant phase, </> = 17, </> = 18, </> = 19,... of the de Broglie wave for an electron, then (0\", v) is the phase difference between the events '1'0 and '1'. Note that 0\" is not fully specified by its surfaces; an orientation is also necessary. Which direction from surface to surface is \"positive\"; i.e., in which direction does </> increase? pattern of surfaces, given a name \"ii.,\" provides the simplest illustration one can easily find for a I-form. The pattern of surfaces in spacetime made by such a I-form: what is it good for? Take two nearby points in spacetime, 9 and 9 0, Run an arrow v = 9 - 9 0 from 9 0 to 9. It will pierce a certain number of the de Broglie wave's surfaces of integral phase, with a bong of an imaginary bell at each piercing. The number of surfaces pierced (number of \"bongs of bell\") is denoted Vector pierces 1-form (ii., v); I-form pierced ~ [vector that pierces in this example it equals the phase difference between tail (90) and tip (9) of v, See Figure 2.4. Normally neither 9 0 nor 9 will lie at a point of integral phase. Therefore one can and will imagine, as uniformly interpolated between the surfaces of integral phase, an infinitude of surfaces with all the intermediate phase values. With their aid, the precise value of (ii., v) = ep(9) - ep(90 ) can be determined. To make the mathematics simple, regard ii. not as the global pattern of de Broglie wave surfaces, but as a local pattern near a specific point in spacetime. Just as the vector u = d9/dT represents the local behavior of a particle's world line (linear approximation to curved line in general), so the I-form ii. represents the local form The 1-form viewed as family of flat equally spaced surfaces 56 2. FOUNDATIONS OF SPECIAL RELATIVITY Ii = dc>, with three extra -----\"......,.---< surfaces interleaved to show its structure more clearly Figure 2.5. This is a dual-purpose figure. (a) It illustrates the de Broglie wave I-form Ii at an event '1'0 (family of equally spaced, flat surfaces, or \"hyperplanes\" approximating the surfaces of constant phase). (b) It illustrates the gradient d<i> of the function </> (concept defined in \u00a72.6), which is the same oriented family of flat surfaces Ii = d<f>. At different events, Ii = d<f> is different-different orientation of surfaces and different spacing. The change in </> between the tail and tip of the very short vector v is equal to the number of surfaces of d</> pierced by v, (d</>, v); it equals -0.5 in this figure. The 1-form viewed as linear function of vectors of the de Broglie wave's surfaces (linear approximation; surfaces flat and equally spaced; see Figure 2.5). Regard the I-form ii. as a machine into which vectors are inserted, and from which numbers emerge. Insertion of v produces as output (ii., v). Since the surfaces of ii. are flat and equally spaced, the output is a linear function of the input: (ii., au + bv) = a(ii., u) + b(ii., v). (2.l2a) This, in fact, is the mathematical definition of a I-form: a Iform is a linear, real valued function of vectors; i.e., a linear machine that takes in a vector and puts out a number. Given the machine ii., it is straightforward to draw the corresponding surfaces in spacetime. Pick a point?Yo at which the machine is to reside. The surface of ii. that passes through?Yo contains points?Y for which (ii.,?Y -?Yo) = 0 (no bongs of bell). The other surfaces contain points with (ii.,?Y -?Yo) = -+ 1, -+2, -+3,.... \u00a72.5. DIFFERENTIAL FORMS a Positive sense -----. 57 u Positive sense +I lu I I I I + t I I Positive i sense I Figure 2.6. The addition of two I\u00b7forms, a and P, to produce the I-form u. Required is a pictorial", " construction that starts from the surfaces of a and P, e.g., (a, 'Y 'Yo) =... -1,0, 1,2,..., and constructs those of u = a + p. Such a construction, based on linearity (2.I2b) of the addition process, is as follows. (I) Pick several vectors u, v,... that lie parallel to the surfaces of P (no piercing!), but pierce precisely 3 surfaces of a; each of these must then pierce precisely 3 surfaces of u: (u, u) = (a + P, u) = (a, u) = 3. (2) Pick several other vectors w,... that lie parallel to the surfaces of a but pierce precisely 3 surfaces of P; these will also pierce precisely 3 surfaces of u. (3) Construct that unique family of equally spaced surfaces in which u, v,..., w,... all have their tails on one surface and their tips on the third succeeding surface. Sometimes I-forms are denoted by boldface, sans-serif Latin letters with tildes over them, e.g., k; but more often by boldface Greek letters, e.g., a, P, CT. The output of a I-form CT, when a vector u is inserted, is called \"the value of CT on u\" or \"the contraction of CT with u.\" Also, I-forms, like any other kind of function, can be added. The I-form aa + bP Addition of 1-forms is that machine (family of surfaces) which puts out the following number when a vector u is put in: (aa + bP,u) = a(a,u) + b(P,u). (2.12b) Figure 2.6 depicts this addition in terms of surfaces. One can verify that the set of all I-forms at a given event is a \"vector space\" in the abstract, algebraic sense of the term. Return to a particle and its de Broglie wave. Just as the arrow p = md?Y/dT represents the best linear approximation to the particle's actual world line near?Yo' so the flat surfaces of the I-form k provide the best linear approximation to the curved surfaces of the particle's de Broglie wave, and k itself is the linear function that best approximates the de Broglie phase ep near?Yo: epW) = epWo) + (k,?Y - 9 0 ) + terms of higher order in (9 - 9 0 ), (2.13) 58 4 2, FOUNDATIONS OF SPECIAL RELATIVITY B, j A, A /'7 ~fJ' 3 ~ \\P\",;'h\u00b7, \\ sense \\ 2 /FIJI sense ~,0fB Positive CIJt of A.c+-7?jib Positi\\'e ~ \"sense~' -~ 2 3 4 2 3 4 GBB~E'E x -------Positive sense of i p\",;/}i+1 -- sense of 0 Figure 2.7. Several vectors, A, B, C, D, E, and corresponding l-forms,4, S, C, 0, E. The process of drawing D corresponding to a given vector U is quite simple. (I) Orient the surfaces of D orthogonal to the vector to U must pierce no surfaces of D is perpendicular U. (0 = U\u00b7 V = (D, V\u00bb and must therefore lie in a surface of D.) (2) Space the surfaces of D so that the number of surfaces pierced by some arbitrary vector Y (e.g., Y = U) is equal to y. U. (Why? Because any vector V that Note that in the figure the surfaces of S are, indeed, orthogonal to B; those of Care, indeed, orthogonal to C, etc. If they do not look so, that is because the reader is attributing Euclidean geometry, not Lorentz geometry, to the spacetime diagram. He should recall, for example, that because C is a null vector, it is orthogonal to itself (C' C = 0), so it must itself lie in a surface of the I-form C. Confused readers may review spacetime diagrams in a more elementary text, e.g., Taylor and Wheeler (1966). Actually, the de Broglie I-form ii: and the momentum vector p contain precisely the same information, both physically (via quantum theory) and mathematically. To see their relationship. relabel the surfaces of ii: by n X phase, thereby obtaining the \"momentum Iform\"p. Pierce this I-form with any vector v, and find the result (exercise 2.1) that p'v = (p, v", "). (2.14) the projection of v on the 4-momentum vector p equals the number of In words:'surfaces it pierces in the 4-momentum I-form p. Examples: Vectors v lying in a surface of p (no piercing) are perpendicular to p (no projection); p itself pierces p2 = _ m 2 surfaces of p. Corresponding to any vector p there exists a unique I-form (linear function of vectors) p defined by equation (2.14). And corresponding to any I-form p, there exists a unique vector p defined by its projections on all other vectors, by equation (2.14), Figure 2.7 shows several vectors and their corresponding I-forms. Physical correspondence between 1-forms and vectors Mathematical correspondence between 1-forms and vectors \u00a72.6. GRADIENTS AND DIRECTIONAL DERIVATIVES 59 A single physical quantity can be described equally well by a vector p or by the corresponding I-form p. Sometimes the vector description is the simplest and most natural; sometimes the I-form description is nicer. Example: Consider a I-form representing the march of Lorentz coordinate time toward the future-surfaces X o =...,7,8, 9,.... The corresponding vector points toward the past [see Figure 2.7 or equation (2.14)]; its description of the forward march of time is not so nice! One often omits the tilde from the I-form p corresponding to a vector p, and uses the same symbol p for both. Such practice is justified by the unique correspond ence (both mathematical and physical) between p and p. Exercise 2.1. Show that equation (2.14) is in accord with the quantum-mechanical properties ofa de Broglie wave, EXERCISE 1J! = e1q, = exp [i(k: x - wt)]. Gradient of a function as a 1-form \u00a72.6. GRADIENTS AND DIRECTIONAL DERIVATIVES There is no simpIer I-form than the gradient, \"df,\" of a functionf Gradient a I-form? How so? Hasn't one always known the gradient as a vector? Yes, indeed, but only because one was not familiar with the more appropriate I-form concept. The more familiar gradient is the vector corresponding, via equation (2.14), to the I-form gradient. The hyperplanes-representing dJ at a point rjJo are just the level surfaces ofJitself, except for flattening and adjustment to equal spacing (Figure 2.5; identify Jhere with ep there). More precisely, they are the level surfaces of the linear function that approximates J in an infinitesimal neighborhood of fjlo. Why the name \"gradient\"? Because dJ describes the first order changes in J in the neighborhood of '!l0: J('!l) = JWo) + (df, fjl - 9 0>+ (nonlinear terms). (2.15) [Compare the fundamental idea of \"derivative\" of something as \"best linear ap proximation to that something at a point\"-an idea that works even for functions whose values and arguments are infinite dimensional vectors! See, e.g., Dieudonne (1960).] Take any vector v; construct the curve ~)(A) defined by '!leA) - '!Po ~ AV; and differentiate the function J along this curve: The \"differential operator,\" (2.16a) (2.16b) Directional derivative operator defined 60 2. FOUNDATIONS OF SPECIAL RELATIVITY which does this differentiating, is called the \"directional derivative operator along the vector v.\" The directional derivative ovf and the gradient df are intimately related, as one sees by applying Ov to equation (2.15) and evaluating the result at the point rjJ0: ov/= (df,dCJ'/dA) = (df,v). (2.17) This result, expressed in words, is: df is a linear machine for computing the rate of change off along any desired vector v. Insert v into df, the output (\"number of surfaces pierced; number of bongs of bell\") is ovf-which, for sufficiently small v, is simply the difference in f between tip and tail of v. \u00a72. 7. COORDINATE REPRESENTATION OF GEOMETRIC OBJECTS Basis 1-forms In flat spacetime, special attention focuses on Lorentz frames. The coordinates XO(CJ'), x 1(9), x 2(CJ'), x 3(CJ') of a Lorentz frame are functions; so their gradients can be calculated. Each of the resulting \"basis I-forms,\" has as its", " hyperplanes the coordinate surfaces x a = const; see Figure 2.8. Conse quently the basis vector e a pierces precisely one surface of the basis I-form w a, (2.18) I ~ ~ 2 \u2022 Positive : sense I Positive sense Figure 2.8. The basis vectors and I-forms of a particular Lorentz coordinate frame. The basis I-forms are so laid out that \u00a72.7. COORDINATE REPRESENTATION OF GEOMETRIC OBJECTS 61 while the other three basis vectors lie parallel to the surfaces of wa and thus pierce none: (2.19) (One says that the set of basis I-forms {wa} and the set of basis vectors {e,8} are the \"duals\" of each other if they have this property.) Just as arbitrary vectors cim be expanded in terms of the basis ea, v = vaea, so arbitrary I-forms can be expanded in terms of w,8: (2.20) Expansion of 1-form in terms of basis The expansion coefficients (J,8 are called \"the components of CT on the basis w,8.\" These definitions produce an elegant computational formalism, thus: Calculate how many surfaces of CT are pierced by the basis vector e a ; equations (2.19) and (2.20) give the answer: Calculation and manipulation of vector and 1-form components i.e., Similarly, calculate (w a, v) for any vector v = e,8v,8; the result is (2.21a) (2.21 b) Multiply equation (2.21 a) by va and sum, or multiply (2.21b) by (Ja and sum; the result in either case is (2.22) This provides a way, using components, to calculate the coordinate-independent value of (CT, v). Each Lorentz frame gives a coordinate-dependent representation of any geometric object or relation: v is represented by its components va; CT, by its components (Ja; a point '3', by its coordinates x a; the relation (CT, v) = 17.3 by (Java = 17.3. To find the coordinate representation of the directional derivative operator 0v' rewrite equation (2.16b) using elementary calculus \u00b0v =C;D~o =~~t~OalOng~(h)-~O= h~C:a); va; see equation (2.3) the result is In particular, [eal,B = (w,B,ea) = 8,Ba) is the directional derivative along a basis vector e a 0a = oeo = %xa. This should also be obvious from Figure 2.8. (2.23) Directional derivative in terms of coordinates (components (2.24) 62 2. FOUNDATIONS OF SPECIAL RELATIVITY Components of gradient The components of the gradient I-form df which are denoted fa are calculated -easily using the above formulas: (2.25a) fa = (df, eo.) [standard way to calculate components; equation (2.2Ia)] =Caf = of/ox a [by relation (2.17) between directional derivative and gradient] [by equation (2.24 )]. Thus, in agreement with the elementary calculus idea of gradient, the components of df are just the partial derivatives along the coordinate axes: i.e., df = CiJf/oxQ) dx Q. (2.25b) fa = if/cxa; (Recall: w a = dx a.) The formula df = (of/ox a rigorous version of the \"differential\" of elementary calculus; see Box 2.3. Q ) dx a suggests, correctly, that df is Other important coordinate representations for geometric relations are explored in the following exercises. EXERCISES Derive the following computationally useful formulas: Exercise 2.2. LOWERING INDEX TO GET THE 1-FORM CORRESPONDING TO A VECTOR The components Uo. of the I-form Ii that corresponds to a vector u can be obtained by \"lowering an index\" with the metric coefficients 1jo.fJ: Exercise 2.3. RAISING INDEX TO RECOVER THE VECTOR One can return to the components of u by raising indices. the matrix lI1jo.fJlI is defined as the inverse of lI1jo.fJlI, and happens to equal lI1jo.fJlI: 1jo.fJ = 1jafJ for all a, (3. Exercise 2.4. VARIED ROUTES TO THE SCALAR PRODUCT The scalar product of u with v can be calculated in any of the following ways: (2.26a) (2.", "26b) (2.27) (2.28) \u00a72.8. CENTRIFUGE AND THE PHOTON 63 Box 2.3 DIFFERENTIALS The \"exterior derivative\" or \"gradient\" df of a function f is a more rigorous version of the ele mentary concept of \"differential.\" In elementary textbooks, one is presented with the differential df as representing \"an infinitesimal change in the functionf(9)\" associated with some infinitesimal displacement of the point '3'; but one will recall that the displacement of '3' is left arbi infinitesimal. Thus df represents a trary, albeit change in f in some unspecified direction. But this is precisely what the exterior derivative df represents. Choose a particular, infinitesimally long displacement v of the point '3'. Let the dis- placement vector v pierce df to give the number (df, v) = avi That number is the change off in going from the tail of v to its tip. Thus df, before it has been pierced to give a number, represents the change off in an unspecified direction. The act of piercing df with v is the act of making explicit the direction in which the change is to be measured. The only failing of the textbook presen tation, then, was its suggestion that dfwas a scalar or a number; the explicit recognition of the need for specifying a direction v to reduce df to a num ber (df, v) shows that in fact df is a I-form, the gradient ofI \u00a72.8. THE CENTRIFUGE AND THE PHOTON Vectors, metric, I-forms, functions, gradients, directional derivatives: all these geo metric objects and more are used in flat spacetime to represent physical quantities; and all the laws of physics must be expressible in terms of such geometric objects. As an example, consider a high-precision redshift experiment that uses the Moss bauer effect (Figure 2.9). The emitter and the absorber of photons are attached to Geometric objects ~n action: example of centrifuge and photon Absorber at time of emission Absorber at time of absorption Emitter at time of emission w Figure 2.9. The centrifuge and the photon. the rim of a centrifuge at points separated by an angle a, as measured in the inertial laboratory. The emitter and absorber are at radius r as measured in the laboratory, and the centrifuge rotates with angular velocity w. PROBLEM: What is the redshift measured, in terms of w. r, and a? 64 2. FOUNDATIONS OF SPECIAL RELATIVITY SOLUTION: Let ue be the 4-velocity of the emitter at the event of emission of a given photon; let U a be the 4-velocity of the absorber at the event of absorption; and letp be the 4-momentum of the photon. All three quantities are vectors defined without reference to coordinates. Equally coordinate-free are the photoIL energies E e and Ea measured by emitter and absorber. No coordinates are needed to describe the fact that a specific emitter emitting a specific photon attribute~ to it the energy E e; and no coordinates are required in the geometric formula (2.29) for Ee. [That this formula works can be readily verified by recalling that, in the emitter's frame, ue o = I and u/ = 0; so in accordance with the identification \"(time component of 4-momentum) = (en ergy.\"] Analogous to equation (2.29) is the purely geometric formula for the absorbed energy. The ratio of absorbed wavelength to emitted wavelength is the inverse of the energy ratio (since E = hv = he/A): This ratio is most readily calculated in the inertial laboratory frame (2.30) (Here and throughout we use boldface Latin letters for three-dimensional vectors in a given Lorentz frame; and we use the usual notation and formalism of three dimensional, Euclidean vector analysis to manipulate them.) Because the magnitude of the ordinary velocity of the rim of the centrifuge, v = wr, is unchanging in time, 0 are equal, and the magnitudes-but not the directions-of ue and ua 0 and Ua ue are equal: From the geometry of Figure 2.9, one sees that ue makes the same angle with p as does Ua \u2022 Consequently, p. ue = p. ua'and Aabsorbed/Aemitted = 1. There is no redshift! Notice that this solution made no reference whatsoever to Lorentz transforma tions-they have not even been discussed yet in this book! The power of the geomet ric, coordinate-free viewpoint is evident! _\u00a7_2._8_._C_E_N_", "T_R_IF_U_G_E_A_N_D_T_H_E_P_H_O_T_O_N 6_5 ~ One must have a variety of coordinate-free contacts between theory and experiment in order EXERCISES to use the geometric viewpoint. One such contact is the equation E = -p' u for the energy of a photon with 4-momentum p, as measured by an observer with 4-velocity u. Verify the following other points of contact. Exercise 2.5. ENERGY AND VELOCITY FROM 4-MOMENTUM A particle of rest mass m and 4-momentum p is examined by an observer with 4-velocity u. Show that just as (a) the energy he measures is so (b) the rest mass he attributes to the particle is E=-p'u; (c) the momentum he measures has magnitude!PI = [(p' U)2 + (p' p)]1/2; (d) the ordinary velocity v he measures has magnitude (2.31) (2.32) (2.33) (2.34) where!PI and E are as given above; and (e) the 4-vector v, whose components in the observer's Lorentz frame are VO = 0, Vi = (dx i / dt)ror particle = ordinary velocity, is given by p + (p' u)u v = -=----\"--\"---. -p'U (2.35) TEMPERATURE GRADIENT Exercise 2.6. To each event f2 inside the sun one attributes a temperature T(f2), the temperature measured by a thermometer at rest in the hot gas there. Then T(f2) is a function; no coordinates are required for its definition and discussion. A cosmic ray from outer space flies through the sun with 4-velocity u. Show that, as measured by the cosmic ray's clock, the time derivative of temperature in its vicinity is In a local Lorentz frame inside the sun, this equation can be written Why is this result reasonable? (2.36) (2.37) Lorentz transformations: of coordinates 66 2. FOUNDATIONS OF SPECIAL RELATIVITY \u00a72.9. LORENTZ TRANSFORMATIONS To simplify computations, one often works with the components of vectors and I-forms, rather than with coordinate-free language. Such component manipulations sometimes involve transformations from one Lorentz frame to another. The reader is already familiar with such Lorentz transformations; but the short review in Box 2.4 will refresh his memory and acquaint him with the notation used in this book. The key entities in the Lorentz transformation are the matrices IIAQ',s1l and IIA,sa,ll; the first transforms coordinates from an unprimed frame to a primed frame, while the second goes from primed to unprimed X a' - Aa' x,s,e-' - (2.38) Since they go in opposite directions, each of the two matrices must be the inverse of the other: Aa' A,s,s y' -. ~a' u y\" A,s Aa' -~,s y - a' U y. (2.39) From the coordinate-independent nature of 4-velocity, u = (dxajdT)ea, one readily derives the expressions Of basis vectors (2.40) for the basis vectors of one frame in terms of those of the other; and from other geoinetric equations, such as v = eava = e,s'v.B', (CT, v) = 0ava = 0,s,v,s', CT = 0awa = 0,B'w,s', one derives transformation laws Of basis 1-forms Of components w a' - Aa' w,s,s -, (2.41 ) (2.42) (2.43) One need never memorize the index positions in these transformation laws. One need only line the indices up so that (1) free indices on each side of the equation are in the same position; and (2) summed indices appear once up and once down. Then all will be correct! (Note: the indices on A always run \"northwest to southeast.\") Box 2.4 LORENTZ TRANSFORMATIONS Rotation of Frame of Reference by Angle 6 in x-y Plane Slope s = tan ();. () SID = 1 S ( + s 2)1/2 : cos () = 1 1 ( + s 2)1/2 y t = [ x = X COS () - y sin () ~:T+JCOS6 1,0;;;;;;'-'.... x [== t x == x cos () + Y sin () y = - x", " sin () + Y cos () z==z All signs follow from sign of this term. Positive by inspection of point '3'. Combination of Two Such Rotations or Boost of Frame of Reference by Velocity Parameter a in z-t Plane Velocity {3 = tanh a; cosh a = (l _ {32)1/2 == \"y\" I ~-_-Io.._z tan () == velocity {3 = tanh a x.x t = [cosh a + zsinh a x=x y=y z = fsinh a + zcosh a [== t cosh a - z sinh a x==x J==Y z == - t sinh a + z cosh a '----I\"\"\"' All signs follow from sign of this term. Positive because object at rest at z == 0 in rocket frame moves in direction of increasing z in lab frame. Matrix notation: x IL =AILvxv, XV == AV xILIL IIAIL.II == cosh a 0 0 sinh a sinh a cosh a:,IIAvILII == cosh shih a 0 0 -sinh a 0 0 cosh a Box 2.4 (continued) Energy-momentum 4-vector E = {cosh a + pi sinh a pZ =pz pll =pll pZ = E sinh a + pi cosh a Charge density-current 4-vector p = p cosh a +l F =j: j1I =jll f = Psinh a +ii cosh a sinh a Aberration, incoming photon: /32)1/2 sin 8 sin () = --=-..J:. = -'--~'----_ /3 cos 8 (1 - p E I cos 8 - /3 pz. cos () = -- =-----=-/3 cos 8 I tan (()12) = ea tan (812) E /32) 1/2 sin () sin if = - P.1.. = (1 E - pi E cos () + /3 cos () =-- =----I + /3 cos () - I + /3 cos () tan (fiI2) = e-a tan (()12) Combination of Two Boosts in Same Direction or General Combinations of Boosts and Rotations Spinor formalism of Chapter 41 Poincare Transformation Condition on the Lorentz part of this transformation: dS'2 = l1a'p' dxa' dx P' = ds 2 = 11 1L,AILa'A'P' dxa'dx p' or ATl1 A = 11 (matrix equation, with Tindicating \"transposed,\" or rows and columns interchanged). Effect of transformation on other quantities: ulL = AlLa,ua' plL = AlLa,pa' FIL' = AILa'A'p,Fa'P' ea' = elLAlLa, w a' = Aa' WIL IL U = ea'u a' = elLu lL = u ua' = uILAlLa,; Pa' = PILAlLa,; Fa,p' = FIL,AILa'A'P'; (4-velocity) (4-momentum) (electromagnetic field) (basis vectors); (basis I-forms); (the 4-velocity vector). \u00a72.10. COLLISIONS 69 Exercise 2.7. BOOST IN AN ARBITRARY DIRECTION An especially useful Lorentz transformation has the matrix components EXERCISE 1 - o-y=~, Ao'AO'j = Aro = -f3yni, Ark = Ak'i = (y I)nin k + {jik, AlLv, = (same as AV'IL but with 13 replaced by -13), where 13, nl, n 2, and n 3 are parameters, and n 2 =(n l )2 + (n 2 )2 + (n 3)2 = 1. Show (a) that this does satisfy the condition AT1jA = 1j required of a Lorentz transformation (see Box 2.4); (b) that the primed frame moves with ordinary velocity f3n as seen in the unprimed frame; (c) that the unprimed frame moves with ordinary velocity - f3n (i.e., VI' = - f3n l, v2' = - f3n 2, v3'= -f3n3 ) as seen in the primed frame; and (d) that for motion in the z direction, the transformation matrices reduce to the familiar form (2.44) y o o -f3y o 0 I 0 o 1 o 0 -f3y", " f3y 0 0 0 0 f3y 0 0 y (2.45) \u00a72.10. COLLISIONS Whatever the physical entity, whether it is an individual mass in motion, or a torrent of fluid, or a field of force, or t~e geometry of space itself, it is described in classical general relativity as a geometric object of its own characteristic kind. Each such object is built directly or by abstraction from identifiable points, and needs no coordinates for its representation. It has been seen how this coordinate-free description translates into, and how it can be translated out of, the language of coordinates and compo nents, and how components in a local Lorentz frame transform under a Lorentz transformation. Turn now to two elementary applications of this mathematical machinery to a mass in motion. One has to do with short-range forces (collisions, this section); the other, with the long-range electromagnetic force (Lorentz force law, next chapter). In a collision, all the change in momentum is concentrated in a time that is short compared to the time of observation. Moreover, the target is typically so small, and quantum mechanics so dominating, that a probabilistic description is the right one. A quantity do = (~) d!2 d!2 8 (2.46) gives the cross section (cm2) for scattering into the element of solid angle d!2 at the deflection angle 8; a more complicated expression gives the probability that the Scattering of particles Conservation of energy-momentum in a collision 70 2. FOUNDATIONS OF SPECIAL RELATIVITY original particle will enter the aperture d!2 at a given polar angle () and azimuth ep and with energy E to E + dE, while simultaneously products of reaction also emerge into specified energy intervals and into specified angular apertures. It would be out of place here to enter into the calculation of such cross sections, though it is a fascinating branch of atomic physics. It IS enough to note that the cross section is an area oriented perpendicular to the line of travel of the incident particle. Therefore it is unaffected by any boost of the observer in that direction, provided of course that energies and angles of emergence of the particles are transformed in accordance with the magnitude of that boost (\"same events seen in an altered reference system\"). Over and above any such detailed account of the encounter as follows from the local dynamic analysis, there stands the law of conservation of energy-momentum: original particles, J final particles, K (2.47) Out of this relation, one wins without further analysis such simple results as the following. (1) A photon traveling as a plane wave through empty space cannot split (not true for a focused photon!). (2) When a high-energy electron strikes an electron at rest in an elastic encounter, and the two happen to come off sharing the energy equally, then the angle between their directions of travel is less than the Newtonian value of 90 0, and the deficit gives a sim pIe measure of the energy of the primary. (3) When an electron makes a head-on elastic encounter with a proton, the formula for the fraction of kinetic energy transferred has three rather different limiting forms, according to whether the energy of the electron is nonrelativistic, relativistic, or extreme-relativistic. (4) The threshold for the production of an (e+, e-) pair by a photon in the field of force of a massive nucleus is 2me. (5) The threshold for the production of an (e+, e-) pair by a photon in an encounter with an electron at rest \u2022 is 4me (or 4me f when account is taken of the binding of the e+e-e- system in a very light \"molecule\"). All these results (topics for independent projects!) and more can be read out of the law of conservation of energy-momentum. For more on this topic, see Blaton (1950), Hagedorn (1964), and Chapter 4 and the last part of Chapter 5 of Sard (1970). CHAPTER 3 THE ELECTROMAGNETIC FIELD The rotating armatures of every generator and every motor in this age of electricity are steadily proclaiming the truth of the relativity theory to al/ who have ears to hear. LEIGH PAGE (1941) \u00a73.1. THE LORENTZ FORCE AND THE ELECTROMAGNETIC FIELD TENSOR At the opposite extreme from an impulsive change of momentum in a collision (the last topic of Chapter 2) is the gradual change in the momentum of a charged particle under the action of electric and magnetic forces (the topic treated here). Let electric and magnetic fields act on a system of charged particles. The accelera tions of the particles reveal the electric and magnetic field strengths. In other words, the Lorentz force law, plus measurements on the components of acceleration of test particles, can be viewed as defining", " the components of the electric and magnetic fields. Once the field components are known from the accelerations of a few test particles, they can be used to predict the accelerations of other test particles (Box 3.1). Thus the Lorentz force law does double service (1) as definer of fields and (2) as predicter of motions. that Here and elsewhere in science, as stressed not least by Henri Poincare, that view is out of date which used to say, \"Define your terms before you proceed. \" All the laws and theories of physics, including the Lorent= force law, have this deep and subtle they both define the concepts they use (here B and E) and make character, statements about these concepts. Contrariwise, the absence of some body of theory, law, and principle deprives one of the means properly to define or even to use concepts. Any forward step in human knowledge is truly creative in this sense: that theory, concept, law, and method of measurement-forever inseparable-are born into the world in union. Lorentz force as definer of fields and predicter of motions 72 3. THE ELECTROMAGNETIC FIELD Box 3.1 LORENTZ FORCE LAW AS BOTH DEFINER OF FIELDS AND PREDICTER OF MOTIONS How one goes about determining the components of the field from measurements of accelerations is not different in principle for electromagnetism and for gravitation. Compare the equations in the two cases: d 2xa -d2 = - Faf3uf3 in a Lorentz frame, (1) e m T and D2~a - -? - = -Ra\"Y8Uf3f!u8 in any coordinate system. dT-,... (2) To make explicit the simpler procedure for elec tromagnetism will indicate in broad outline how one similarly determines all the components of Raf3y8 for gravity. Begin by asking how many test particles one needs to determine the three compo nents of B and the three components of E in the neighborhood under study. For one particle, three components of acceleration are measurable; for a second particle, three more. Enough? No! The information from the one duplicates in part the information from the other. The proof? Whatever the state of motion of the first test particle, pick one's Lorentz frame to be moving the same way. Having zero velocity in this frame, the particle has a zero response to any magnetic field. The electric field alone acts on the particle. The three compo nents of its acceleration give directly the three components Ez' Ell' Ez of the electric field. The second test particle cannot be at rest if it is to do more than duplicate the information provided by the first test particle. Orient the x-axis of the frame of reference parallel to the direction of motion of this second particle, which will then respond to and measure the components B II and Bz of the magnetic field. Not so Bz! The acceleration in the x-direction merely remeasures the already once measured Ez. To evaluate Bz' a third test particle is required, but it then gives duplicate information about the other field components. The alternative? Use all N particles simultaneously and on the same democratic footing, both in the evaluation of the six Faf3 and in the testing of the Lorentz force, by applying the method of least squares. Thus, write the discrepancy between predicted and observed acceleration of the Kth particle in the form U K _..!!- F Uf3,K = Sa K a' a m af3 (3) Take the squared magnitude of this discrepancy and sum over all the particles s = 2: 1jaf3 SaaKSaK f3' k (4) In this expression, everything is regarded as known except the six Faf3. Minimize with respect to these six unknowns. In this way, arrive at six equations for the components of Band E. These equations once solved, one goes back to (3) to test the Lor entz force law. The 6 X 6 determinant of the coefficients in the equation for the Faf3 automatically vanishes when there are only two test particles. The same line of reasoning permits one to determine the minimum number of test particles required to determine all the components of the Riemann curvature tensor. \u00a73.1. LORENTZ FORCE AND ELECTROMAGNETIC FIELD TENSOR 73. The Lorentz force law, written in familiar three-dimensional notation,with E = electric field, B = magnetic field, v = ordinary velocity of particle, p = momentum of particle, e = charge of particle, reads The three-dimensional version of the Lorentz force law (dpjdt) = e(E + v x B). (3.1) Useful though this version of the equation may be, it is far", " from the geometric spirit of Einstein. A fully geometric equation will involve the test particle's energy-mo mentum 4-vector, p, not just the spatial part p as measured in a specific Lorentz frame; and it will ask for the rate of change of momentum not as measured by a specific Lorentz observer (djdt), but as measured by the only clock present apriori in the problem: the test particle's own clock (djdT). Thus, the lefthand side of a fully geometric equation will read dpjdT =. The righthand side, the Lorentz 4-force, must also be a frame-independent object. It will be linear in the particle's 4-velocity u, since the frame-dependent expression dp dT e _~ (E + vxB) = e(uOE + uxB) v I - v2 (3.2a) is linear in the components of u. Consequently, there must be a linear machine named Faraday, or F, or \"electromagnetic field tensor,\" with a slot into which one inserts the 4-velocity of a test particle. The output of this machine, multiplied by the particle's charge, must be the electromagnetic 4-force that it feels: Electromagnetic field tensor defined Geometrical version of Lorentz force law\u00b7 dpjdT = eF(u). (3.3) By comparing this geometric equation with the original Lorentz force law (equa tion 3.2a), and with the companion energy-change law = I dE vT=V2 dt I -==__ eE\u00b7 v = eE\u00b7 u, vT=V2 (3.2b) one can read off the components of F in a specific Lorentz frame. The components of dpjdT are dpajdT, and the components of eF(u) can be written (definition of Pf3!) eFaf3u f3. Consequently (3.4) must reduce to equations (3.2a,b). Indeed it does if one makes the identification /3=0 (3=1 (3=2 (3=3 IIFa II =f3 0:=0 0:=1 0:=2 0:=3 0 E r Ell E z Er 0 -Bz B II Ell Br 0 -Br E z -BII Br 0 (3.5) Components of electromagnetic field tensor 74 3. THE ELECTROMAGNETIC FIELD More often seen in the literature are the \"covariant components,\" obtained by lowering an index with the metric components: F\"f3 = l1 ay F 'Yf3; 0 11F\"f3 11 = Er Ell \u00a3z -Ex 0 -Bz BII -Ell Bz 0 -Bx -Ezil -B ~x II (3.6) (3.7) This matrix equation demonstrates the unity of the electric and magnetic fields. Neither one by itself, E or B, is a frame-independent, geometric entity. But merged together into a single entity, F = Faraday, they acquire a meaning and significance that transcends coordinates and reference frames. EXERCISE Exercise 3.1. Derive equations (3.5) and (3.7) for the components of Faraday by comparing (3.4) with (3.2a,b), and by using definition (3.6). Examples of tensors \u00a73.2. TENSORS IN ALL GENERALITY A digression is in order. Now on the scene are several different tensors: the metric tensor 9 (\u00a72.4), the Riemann curvature tensor Riemann (\u00a7 1.6), the electromagnetic field tensor Faraday (\u00a73.l). Each has been defined as a linear machine with input slots for vectors, and with an output that is either a real number, e.g., g(u, v), or a vector, e.g., Riemann (u, v, w) and Faraday (u). Should one make a distinction between tensors whose outputs are scalars, and tensors whose outputs are vectors? No! A tensor whose output is a vector can be reinterpreted trivially as one whose output is a scalar. Take, for example, Fara day = F. Add a new slot for the insertion of an arbitrary I-form u, and gears and wheels that guarantee the output F(u, u) = (u, F(u) = real number. (3.8) Then permit the user to choose whether he inserts only a vector, and gets out the vector F(..., u) = F(u), or whether he inserts a 'form and a vector, and gets out the number F(u, u).", " The same machine will do both jobs. Moreover, in terms of components in a given Lorentz frame, both jobs are achieved very simply: F(..., u) is a vector with components F\"-f3uf3; F(u, u) is the number (u, F(..., u) = o,,Faf3u f3. (3.9) \u00a73.2. TENSORS 75 By analogy, one defines the most general tensor H and its rank (~ as 'follows: H is a linear machine with n input slots for n I-forms, and m input slots for m vectors; given the requested input, it puts out a real number denoted Definition of tensor as linear machine that converts vectors and 1-forms into numbers H(CT,..4,..., p, u, v,..., w).., \\ n I-forms m vectors (3.10) For most tensors the output changes when two input vectors are interchanged, Riemann(CT, U, v, w) ::j:. Riemann(CT, v, U, w), (3.11 ) or when two input I-forms are interchanged. Choose a specific tensor S, of rank cD for explicitness. Into the slots of S, insert the basis vectors and I-forms of a specific Lorentz coordinate frame. The output is a \"component of S in that frame\": This defines components. Knowing the components in a specific frame, one can easily calculate the output produced from' any input forms and vectors: S(CT, p, v) = S(IJaWa, Pf3wf3, uYey) = IJaPf3uYS(wa, w f3, e y) = saf3 yIJaPf3UY. (3.13) Tensor\"s machine action expressed in terms of components (3.12) Components of a tensor And knowing the components of S in one Lorentz frame (unprimed), plus the Lorentz transformation matrices IIAa'f311 and IIAf3 a'II which link that frame with another (primed), one can calculate the components in the new (primed) frame. As shown in exercise 3.2, one need only apply a matrix to each index of S, lining up the matrix indices in the logical manner SIl'v' \"h' - saf3 All' Av' AY a Y f3 \"h\" (3.14) Lorentz transformation of components A slight change of the internal gears and wheels inside the tensor enables one of its I-form slots to accept a vector. All that is necessary is a mechanism to convert an input vector n into its corresponding I-form ii and then to put that I-form into the old machinery. Thus, denoting the modified tensor by the same symbol S as was used for the original tensor, one demands S(CT, n, v)= S(CT, ii, v); or, in component notation sa IJ n f3 uY = S af3 a n uy. Y a f3 f3y a (3.15) Modifying a tensor to accept either a vector or a 1-form into each slot (3.15') This is achieved if one raises and lowers the indices of S using the components of the metric: Sap. Y -.\"p.f3Sa \" f3y' (3.16) Raising and lowering indices (See exercise 3.3 below.) By using the same symbol S for the original tensor and 76 3. THE ELECTROMAGNETIC FIELD the modified tensor, one allows each slot to accept either a I-form or a vector, so one loses sight of whether S is a CD tensor, or a CD tensor, or a @ tensor, or a m tensor; one only distinguishes its total rank, 3. Terminology: an \"upstairs index\" is called \"contravariant\"; a \"downstairs\" index is called \"covariant.\" Thus in sa.{jy' \"a\" is a contravariant index, while \"{3\" and \"y\" are covariant indices. Because tensors are nothing but functions, they can be added (if they have the same rank!) and multiplied by numbers in the usual way: the output of the rank-3 tensor as + bO, when vectors u, v, ware put in, is (as + bO)(u, v, w) as(u, v, w) + bO(u, v, w). (3.17) Several other important operations on tensors are explored in the following exercises. They and the results of the exercises will be used freely in the material that follows. EXERCISES Exercise 3.2", ". TRANSFORMATION LAW FOR COMPONENTS OF A TENSOR From the transformation laws for components of vectors and I-forms, derive the transforma tion law (3.14). Exercise 3.3. RAISING AND LOWERING INDICES Derive equations (3.16) from equation (3.15') plus the law n a = Tfa/3n/3 for getting the components of the I-form Ii from the components of its corresponding vector n. Exercise 3.4. TENSOR PRODUCT Given any two vectors u and v, one defines the second-rank tensor u I8l v (\"tensor product of u with v\") to be a machine, with two input slots, whose output is the number (u I8l v)(O\", A) = (0\", u)(A, v) when I-forms 0\" and A are inserted. Show that the components of T = u I8l v are the products of the components of u and v: (3.18) (3.19) Extend the definition to several vectors and forms, (u I8l v I8l P I8l w)(O\", A, n, 0 = (0\", u)(A, v)(P, n)(C w), (3.20) and show that the product rule for components still holds: s = u I8l v I8l P I8l w has components sPo\\r = uPou\"f3).wr. (3.21 ) Exercise 3.5. BASIS TENSORS Show that a tensor M with components Ma/3 -/ in a given Lorentz frame can be reconstructed from its components and from the basis I-forms and vectors of that frame as follows: (3.22) (For a special case of this, see Box 3.2.) Box 3.2 THE METRIC IN DIFFERENT LANGUAGES A. Geometric Language 9 is a linear, symmetric machine with two slots for insertion of vectors. When vectors u and v are inserted, the output of 9 is their scalar product: g(u, v) = u' v. B. Component Language 'T/p.\" are the metric components. They are used to calculate the scalar product of two vectors from components in a specific Lorentz frame: C. Coordinate-Based Geometric Language The metric 9 can be written, in terms of basis I-forms of a specific Lorentz frame, as [see equations (2.18) and (3.22)]. D. Connection to the Elementary Concept of Line Element Box 2.3 demonstrated the correspondence between the gradient df of a function, and the elementary concept df of a differential change of f in some unspecified direction. There is a similar correspondence between the metric, written as 'T/p.\" dxP. \u00ae dx\", and the elementary concept of \"line element,\" written as ds 2 = 'T/p.\" dxp. dx\". This elementary line element, as expounded in many special relativity texts, repre sents the squared length of the displacement \"dxp.\" in an unspecified direction. The metric'T/p.\" dxP. \u00ae dx\" does the same. Pick a specificinfinitesimal displacement vector (', and insert it into the slots of 'T//l\" dx/l \u00ae dx\". The output will be e = 'T/p.\"~/l~\", the squared length of the displacement. Before (' is inserted, 'T/p.\" dxP. \u00ae dx\" has the potential to tell the squared length of any vector; the insertion of (' converts potenti ality into actuality: the numerical value of (2. Because the metric 'T/p.\" dx/l \u00ae dx\" and the line element ds 2 = 'T/p.\" dxp. dx\" perform this same function of representing the squared length of an unspecified infinitesimal displacement, there is no conceptual distinction between them. One sometimes uses the symbols ds 2 to denote the metric; one sometimes gets pressed and writes it as ds2 = 'T/p.\" dX/l dx\", omitting the \"\u00ae\"; and one sometimes even gets so pressed as to use nonbold characters, so that no notational distinction remains at all between metric and elementary line element: 9 = ds 2 = ds 2 = 'T//l\" dxP. dx\". 78 3. THE ELECTROMAGNETIC FIELD Exercise 3.6. Faraday MACHINERY AT WORK An observer with 4-velocity u picks out three directions in spacetime that are orthogonal and purely spatial (no time part) as seen in his frame. Let 8j, 8 2,83 be unit vectors in those directions and let them be oriented in a righthanded way (8j \u2022 8 2 X 8 3 = + I in three-di", " mensional language). Why do the following relations hold? 8J' U = 0, What vectors are to be inserted in the two slots of the electromagnetic field tensor Faraday if one wants to get out the electric field along 8J as measured by this observer? What vectors must be inserted to get the magnetic field he measures along 8j? \u00a73.3. THREE-PLUS-ONE VIEW VERSUS GEOMETRIC VIEW The power of the geometric view of physics Example of electromagnetism Great computational and conceptual power resides in Einstein's geometric view of physics. Ideas that seem complex when viewed in the everyday \"space-plus-time\" or \"3 + I\" manner become elegant and simple when viewed as relations between geometric objects in four-dimensional spacetime. Derivations that are difficult in 3 + I language simplify in geometric language. The electromagnetic field is a good example. In geometric language, it is described by a second-rank, antisymmetric tensor (\"2-form\") F, which requires no coordinates for its definition. This tensor produces a 4-force on any charged particle given by dpjdT = eF(u). It is all so simple! By contrast, consider the \"3 + I\" viewpoint. In a given Lorentz frame, there is an electric field E and a magnetic field B. They push on a particle in accordance with dpjdt = e(E + v x B). Transformation law for electric and magnetic fields But the values of p, E, l', and B all change when one passes from the given Lorentz frame to a new one. For example, the electric and magnetic fields viewed from a rocket ship (\"barred\" frame) are related to those viewed in the laboratory (\"un barred\" frame) by (3.23) (Here \"II\" means component along direction of rocket's motion; \"1.\" means perpen dicular component; and {3; = dxirocketjdt is the rocket's ordinary velocity.) The analogous transformation laws for the particle's momentum p and ordinary velocity \u00a73.4. MAXWELL'S EQUATIONS 79 v, and for the coordinate time t, all conspire-as if by magic, it seems, froni the 3 + 1 viewpoint-to maintain the validity of the Lorentz force law in all frames. Not only is the geometric view far simpler than the 3 + 1 view, it can even derive the 3 + 1 equations with greater ease than can the 3 + 1 view itself. Consider, for example, the transformation law (3.23) for the electric and magnetic fields. The geometric view derives it as follows: (l) Orient the axes of the two frames so their relative motion is in the z-direction. (2) Perform a simple Lorentz transformation (equation 2.45) on the components of the electromagnetic field tensor: = (1 - Ell = Ex = F3fj = A\"3A.8oF\".8 = y2F30 + f32 y2Fo3 f32)y2 F30 = F30 = Ex = Eil' Ex = FlO = A\"rA.8oF\".8 = yFlO + f3yF13 = y(Ex etc. (3.24) f3B y ), By contrast, the 3 + I view shows much more work. A standard approach is based on the Lorentz force law and energy-change law (3.2a,b), written in the slightly modified form - dZ) m-=e E -+O -+B - -B -, Y dT (- dt x dT - dJ x dT d'X dT d 2'X dT2 (3.25)... (three additional equations).... It proceeds as follows (details omitted because of their great length!): (1) Substitute for the d2'XjdT 2, etc., the expression for these quantities in terms of the d 2xjdT2, \u2022\u2022 \u2022 (Lorentz transformation). (2) Substitute for the d 2xjdT2, \u2022\u2022\u2022 the expression for these accelerations in terms of the laboratory E and B (Lorentz force law). (3) In these expressions, wherever the components dxjdT of the 4-velocity in the laboratory frame appear, substitute expressions in terms of the 4-velocities in the rocket frame (inverse Lorentz transformation). (4) In (3.25) as thus transformed, demand equality of left and right sides for all values of the d'Xj dT, etc. (validity for all test particles). (5) In this way arrive at the expressions (3.23) for the E and jj in terms of the E and B. The contrast in difficulty is obvious! \u00a73", ".4. MAXWELL'S EQUATIONS Turn now from the action of the field on a charge, and ask about the action of a charge on the field, or, more generally, ask about the dynamics of the electromagnetic 80 3. THE ELECTROMAGNETIC FIELD field, charge or no charge. Begin with the simplest of Maxwell's equations in a specific Lorentz frame, the one that says there are no free magnetic poles: Magnetodynamics derived from magnetostatics V' B oB\" 0 d\u00b7 lvB= - -+ -+ -=. oy oB\" ax oB\" oz (3.26) This statement has to be true in all Lorentz frames. It is therefore true in the rocket frame: (3.27) For an infinitesimal Lorentz transformation in the x-direction (nonrelativistic ve locity 13), one has (see Box 2.4 and equations 3.23) B\" = BII + f3E\", li\" = B\", a a a ax = ax + f3 ai, a a oj! - oy' a =oz a oz (3.28) (3.29) Substitute into the condition of zero divergence in the rocket frame. Recover the original condition of zero divergence in the laboratory frame, plus the following information (requirement for the vanishing of the coefficient of the additional arbitrary small velocity 13): (3.30) Had the velocity of transformation been directed in the y- or z-directions, a similar equation would have been obtained for aBu/at or oBz/ot. In the language of three dimensional vectors, these three equations reduce to the one equation oB - + V X E at oB - + curl E = O. at (3.31 ) How beautiful that (l) the principle of covariance (laws of physics are the same in every Lorentz reference system, which is equivalent to the geometric view of physics) plus (2) the principle that magnetic tubes of force never end, gives (3) Maxwell's dynamic law for the time-rate of change of the magnetic field! This suggests that the magnetostatic law V \u2022 B = 0 and the magnetodynamic law oB/ot + V X E = 0 must be wrapped up together in a single frame-independent, geometric law. In terms of components of the field tensor F, that geometric law must read (3.32) since this reduces to V \u2022 B = 0 when one takes a = 1,13= 2, y = 3; and it reduces to oB/ot + V X E = 0 when one sets any index, e.g., a, equal to zero (see exercise 3.7 below). In frame-independent geometric language, this law is written (see \u00a73.5, exercise 3.14, and Chapter 4 for notation) Magnetodynamics and magnetostatics unified in one geometric law \u00a73.5. WORKING WITH TENSORS dF = 0, or, equivalently, V\u00b7 *F = 0; 81 (3.33) and it says, \"Take the electromagnetic 2-form F (a geometric object defined even in absence of coordinates); from it construct a new geometric object dF (called the \"exterior derivative of F\"); dF must vanish. The details of this coordinate-free process will be spelled out in exercise 3.15 and in \u00a74.5 (track 2). Two of Maxwell's equations remain: the electrostatic equation and the electrodynamic equation V\u00b7E= 4'7Tp, aEjat - V X B = -4'7TJ. (3.34) (3.35) like the magnetostatic and magnetodynarnic equations, are actually two They, different parts of a single geometric law. Written in terms of field components, that law says Electrodynamics and electrostatics unified in one geometric law where the 'components of the \"4-current\" J are Fa./3./3 - 4'7TJa., JO = p = charge density, (J\\ P, J3) = components of current density. (3.36) (3.37) Written in coordinate-free, geometric language, this electrodynamic law says - d*F = 4'7T *J or, equivalently, V. F = 4'7TJ. (3.38) (For full discussion, see exercise 3.15 and \u00a74.5, which is on Track 2.) Exercise 3.7. MAXWELL'S EQUATIONS Show, by explicit examination of components, that the geometric laws EXERCISE do reduce to Maxwell's equations (3.26), (3.31), (3.34), (3.35), as claimed above. \u00a73.5 WORKING WITH TENSORS Another mathematical digression is needed. Given an arbitrary", " tensor field, S, of arbitrary rank (choose rank = 3 for concreteness), one can construct new tensor fields by a variety of operations. Ways to produce new tensors from old: One operation is the gradient V. (The symbol d is reserved for gradients of scalars, df, and for \"exterior derivatives of differential forms;\" a Track-2 in which case Vf Gradient 82 3. THE ELECTROMAGNETIC FIELD concept, on which see \u00a74.5.) Like 5, V5 is a machine. It has four slots, whereas 5 has three. It describes how 5 changes from point to point. Specifically, if one desires to know how the number 5(u, v, w) for fixed u, v, w changes under a displacement (', one inserts u, v, w, (' into the four slots of V 5: V5(u, v, w, n o(5(u, v, w) with u, v, w fixed ~ + [value of 5(u, v, w) at tip of (] [value of 5(u, v, w) at tail of n - (3.39) Contraction Divergence In component notation in a Lorentz frame, this says - S - etl3y.8 uetv13 Wl\"t8... Thus, the Lorentz-frame components of V5 are nothing but the partial derivatives of the components of 5. Notice that the gradient raises the rank of a tensor by 1 (from 3 to 4 for 5). Contraction is another process that produces a new tensor from an old one. It seals off (\"contracts\") two of the old tensor's slots, thereby reducing the rank by two. Specifically, if R is a fourth-rank tensor and M is obtained by contracting the first and third slots of R, then the output of M is given by (definition!) M(u, v) = L R(eet' u, wet, v). 3 et=O (3.40) Here eet and wet are the basis vectors and I-forms ofa specific but arbitrary Lorentz coordinate frame. It makes no difference which frame is chosen; the result will always be the same (exercise 3.8 below). In terms of components in any Lorentz frame, equation (3.40) says (exercise 3.8) so that (3.41 ) Thus, in terms of components, contraction amounts to putting one index up and the other down, and then summing on them. Divergence is a third process for creating new tensors from old. It is accomplished by taking the gradient, then contracting the gradient's slot with one of the original slots: (divergence of 5 on first slot) V. 5 is a machine such that V' 5(u, v) = V5(w et, U, v, eet) = Set13'1.etu13v'l; (3.42) i.e. V' 5 has components SetP'I.et' \u00a73.5. WORKING WITH TENSORS 83 Transpose is a fourth, rather trivial process for creating' new tensors. It merely Transpose interchanges two slots: N obtained by transposing second and third slots of S = N(u, v, w) = S(u, w, v). (3.43) Symmetrization and antisymmetrization are fifth and sixth processes for producing new tensors from old. A tensor is completely symmetric if its output is unaffected by an interchange of two input vectors or I-forms: Symmetrization and antisymmetrization S(u, v, w) = S(v, u, w) = S(v, w, u) = (3.44a) It is completely antisymmetric if it reverses sign on each interchange of input S(u, v, w) = -S(v, u, w) = +S(v, w, u) =.... (3.44b) Any tensor can be symmetrized or antisymmetrized by constructing an appropriate linear combination of it and its transposes; see exercise 3.12. Wedge product is a seventh process for producing new tensors from old. It is merely Wedge product an antisymmetrized tensor product: given two vectors u and v, their wedge product, the \"bivector\" u /\\ v, is defined by Bivector u /\\ v u \u00ae v - v \u00ae u; (3.45a) similarly, the \"2jorm\" a /\\ p constructed from two I-forms is 2-form a /\\ P a \u00ae p - P \u00ae a. (3.45b) From three vectors u,", " v, w one constructs the \"trivector\" Trivector u /\\ v /\\ w (u /\\ v) /\\ w _ u /\\ (v /\\ w) =u \u00ae v \u00ae w + terms that guarantee complete antisymmetry = 3.45c) -v\u00aeu\u00aew-u\u00aew\u00aev-w\u00aev\u00aeu. From I-forms a, p, y one similarly constructs the \"3-forms\" a /\\ p /\\ y. The wedge product gives a simple way to test for coplanarity (linear dependence) of vectors: if u and v are collinear, so u = av, then u /\\ v = av /\\ v = 0 (by antisymmetry of\" /\\ \"). If w is coplanar with u and v so w = au + bv (\"collapsed box\"), then w /\\ u /\\ v = au /\\ u /\\ v + bv /\\ u /\\ v = O. The symbol\" /\\\" is called a \"hat\" or \"wedge\" or \"exterior product sign.\" Its proper ties are investigated in Chapter 4. Taking the dual is an eighth process for constructing new tensors. It plays a fundamental role in Track 2 of this book, but since it is not needed for Track 1, its definition and properties are treated only in the exercises (3.14 and 3.15). Dual 84 3. THE ELECTROMAGNETIC FIELD Because the frame-independent geometric notation is somewhat ambiguous (which slots are being contracted? on which slot is the divergence taken? which slots are being transposed?), one often uses component notation to express coordinate-inde pendent, geometric relations between geometric objects. For example, J/3Y = sa./3y.a. means \"J is a tensor obtained by taking the divergence on the first slot of the tensor S\". Also, vY = (F FiJ.P).Y = (F FiJ.P) - iJ.P iJ.P TJ/3Y./3 Index gymnastics means \"v is a vector obtained by (1) constructing the tensor product F \u00ae F of F with itself, (2) contracting F \u00ae F on its first and third slots, and also on its second and fourth, (3) taking the gradient of the resultant scalar function, (4) converting that gradient, which is a I-form, into the corresponding vector.\" \"Index gymnastics,\" the technique of extracting the content from geometric equations by working in component notation and rearranging indices as required, must be mastered if one wishes to do difficult calculations in relativity, special or general. Box 3.3 expounds some of the short cuts in index gymnastics, and exercises 3.8-3.18 offer practice. EXERCISES Exercise 3.8. CONTRACTION IS FRAME-INDEPENDENT Show that contraction, as defined in equation (3.40), does not depend on which Lorentz frame eO. and wo. are taken from. Also show that equation (3.40) implies Exercise 3.9. DIFFERENTIATION (a) Justify the formula by considering the special case J.L = 0, v = I. (b) Explain why Exercise 3.10. MORE DIFFERENTIATION (a) Justify the formula, by writing out the summation u/Lu/L - l1/LpU/LUP explicitly. (b) Let aindicate a variation or small change, and justify the formula \u00a73.5. WORKING WITH TENSORS 85 Box 3.3 TECHNIQUES OF INDEX GYMNASTICS Equation Name and Discussion Computing components. Computing other components. Reconstructing the rank-@ version of S. Reconstructing the rank-a) version of S. [Recall: one does not usually distinguish between the various versions; see equa tion (3.15) and associated discussion.] Raising an index. Lowering an index. Contraction of S to form a new tensor M. Tensor product of S with M to form a new tensor T. Squared length of vector A produced by forming tensor product A \u00ae A and then contracting, which is the same as forming the corresponding I-formA and thenpiercing:A2 = (A, A) = AaAa. The matrix formed from the metric's \"covariant components,\" is the inverse of that formed from its \"contravariant II1Ja/lIl, components,\" lI1Ja/l11. Equivalently, raising one index of the metric 1Ja/l produces the Kronecker delta. Gradient of N to form a new tensor S. Divergence of N to form a new tensor R. Taking gradients and raising or lowering indices", " are operations that commute. Contravariant index on a gradient is obtained by raising covari ant index. Gradient of a tensor product; says V(R \u00ae M) = Transpose (VR \u00ae M) + R \u00ae VM. Antisymmetrizing a tensor F to produce a new tensor G. Symmetrizing a tensor F to produce a new tensor H. Forming the rank-3 tensor that is dual to a vector (exercise 3.14). Forming the antisymmetric rank-2 tensor that is dual to a given antisymmetric rank-2 tensor (exercise 3.14). Forming the I-form that is dual to an antisymmetric rank-3 tensor (exercise 3.14). 86 3. THE ELECTROMAGNETIC FIELD Exercise 3.11. SYMMETRIES Let AIL\" be an antisymmetric tensor so that AIL' = -A\"IL; and let SIL\" be a symmetric tensor so that SIL' = S\u00b7\u00b7IL. (a) Justify the equations AIL\u2022.SIL' = 0 in two ways: first, by writing out the sum explicitly (all sixteen terms) and showing how\"the terms in the sum cancel in pairs; second, by giving an argument to justify each equals sign in the following string: (b) Establish the following two identities for any arbitrary tensor VIL \u2022. : Exercise 3.12. SYMMETRIZATION AND ANTISYMMETRIZATION To \"symmetrize\" a tensor, one averages it with all of its transposes. The components of the new, symmetrized tensor are distinguished by round brackets: One \"antisymmetrizes\" a tensor (square brackets) similarly: (3.46) (3.47) (a) Show that such symmetrized and antisymmetrized tensors are, indeed, symmetric and antisymmetric under interchange of the vectors inserted into their slots: V(a.By)uav.B w Y = + V(a.By)vau.BwY = \"', V[a.BYluav.BwY = - V[a.BYlvau.BwY =.... (b) Show that a second-rank tensor can be reconstructed from its symmetric and antisym metric parts, (3.48) but that a third-rank tensor cannot; V(a.BY) and V[a.BYl contain together \"less information\" than Va.BY' \"Young diagrams\" (see, e.g., Messiah [1961], appendix D) describe other symme tries, more subtle than these two, which contain the missing information. (c) Show that the electromagnetic field tensor satisfies F(aP) = 0, (3.49a) (d) Show that Maxwell's \"magnetic\" equations can be rewritten in the form F[a.B,Yl = O. (3.49b) \u00a73.5. WORKING WITH TENSORS Exercise 3.13. LEVI-CIVITA TENSOR 87 The \"Levi-Civita tensor\" & in spacetime is a fourth-rank, completely antisymmetric tensor: &(n, u, v, w) changes sign when any two of the vectors are interchanged. (3.50a) Choose an arbitrary but specific Lorentz frame, with 8 0 pointing toward the future and with 8 1, 8 2, 83 a righthanded set of spatial basis vectors. The covariant components of & in this frame are [Note: In an n-dimensional space, & is the analogous completely antisymmetric rank-n tensor. Its components are (3.50b) when computed on a \"positively oriented,\" orthonormal basis 8 1,...,8nol (a) Use the antisymmetry to show that f\"j3Y8 = 0 unless a, {3, y, S are all different, _(+ I for even permutations of 0, I, 2, 3, and I for odd permutations. f.\".0\"\"'.\".2\"\"3 - _ (b) Show that f'7TO'1Tl'iT2'iTJ = - f'1TO'1Tl '1T2'1T3' (3.50c) (3.50d) (3.50e) (3.50f) (c) By means of a Lorentz transformation show that eCi1Jyr, and f;;1Jyr, have these same values in any other Lorentz frame with eo pointing toward the future and with 8y", ", ez, 8 3 a righthanded set. Hint: show that from ATl1 A = 11, show that detIA;;.1 = \u00b1 I; and verify that the determinant is + I for trans formations between frames with 8 0 and 8ij future-pointing, and with 8 1, 8 2, 8 3 and 8y, 8 2, 8} right handed. (d) What are the components of & in a Lorentz frame with past-pointing 8ij? with lefthanded 8y, 8 2\u2022 8\"3? (e) From the Levi-Civita tensor, one can construct several \"permutation tensors.\" In index (3.50g) S\"j3y It'~ =-f\"j3YPflt'~p; S\"j3 =.!. S\"j3~ Itl'~ It' - 2 = - ~\"j3~Pf 2. 1t'~P' notation: Show that: + I if a{3y is an even permutation of J.LvA. S\"j3yIt.~ = - I if a{3y is an odd permutation of J.LvA, [ o otherwise; (3.50h) (3.50i) (3.50j) (3.50k) 88 3. THE ELECTROMAGNETIC FIELD saj3\"p = sa\"Sj3., - Sa pSj3\" + I if af3 is an even permutation of j.LV, = - I if af3 is an odd permutation of j.LV, [ o otherwise; Sa _ { + I if a = j.L, 0 otherwise. \" - - - (3.501) (3.50m) Exercise 3.14. DUALS From any vector J, any second-rank antisymmetric tensor F(Faj3 = F[aj3])' and any third-rank antisymmetric tensor B(Baj3y = B[aj3yj), one can construct new tensors defined by (3.5 I) One calls *J the \"dual\" of J, *F the dual of F, and *B the dual of B. [A previous and entirely distinct use of the word \"dual\" (\u00a72.7) called a set of basis one-forms {w a } dual to a set of basis vectors {ea} if (w a, ej3) = saw Fortunately there are no grounds for confusion between the two types of duality. One relates sets of vectors to sets of I-forms. The other relates antisymmetric tensors of rank p to antisymmetric tensors of rank 4 - p.] (a) Show that **J = J, **F = -F, **B = B. (3.52) so (aside from sign) one can recover any completely antisymmetric tensor H from its dual *H by taking the dual once again, **H. This shows that Hand *H contain precisely the same information. (b) Make explicit this fact of same-information-content by writing out the components *Aaj3y in terms of Aa, also *Faj3 in terms of Faj3, also *Ba in terms of Baj3y. Exercise 3.15. GEOMETRIC VERSIONS OF MAXWELL EQUATIONS Show that, if F is the electromagnetic field tensor, then V' *F = 0 is a geometric frame-in dependent version of the Maxwell equations Similarly show that V' F = 4r.J (divergence on second slot of F) is a geometric version of Faj3,j3 = 4r.Ja. Exercise 3.16. CHARGE CONSERVATION From Maxwell's equations Faj3,j3 = 4r.Ja, derive the \"equation of charge conservation\" Ja = O.,a (3.53) Show that this equation does, indeed, correspond to conservation of charge. It will be studied further in Chapter 5. Exercise 3.17. VECTOR POTENTIAL The vector potential A of electromagnetic theory generates the electromagnetic field tensor via the geometric equation F = - (antisymmetric part of VA), i.e., (3.54) (3.54') \u00a73.5. WORKING WITH TENSORS 89 (a) Show that the electric and magnetic fields in a specific Lorentz frame are given by B= V XA, E = -CiA/Cit - VAo. (b) Show that F will satisfy Maxwell's equations if and only if A satisfies (3.55) (3.56) (c) Show that \"gauge transformations\" ANEW = A OLD + dc[>, c", "[>'= arbitrary function, (3.57) leave F unaffected. (d) Show that one can adjust the gauge so that V'A =0 (\"Lorentz gauge\"), DA = -4r.J. Here 0 is the wave operator (\"d'Alembertian\"): (3.58a) (3.58b) (3.59) Exercise 3.18. DIVERGENCE OF ELECTROMAGNETIC STRESS-ENERGY TENSOR From an electromagnetic field tensor F, one constructs a second-rank, symmetric tensor T (\"stress-energy tensor,\" to be studied in Chapter 5) as follows: Tit\" = -L(Fp.aF\" -ll1lt\"F Fa f3 ) a 4r. af3 4. (3.60) As an exercise in index gymnastics: (a) Show that V' T has components (b) Manipulate this expression into the form T \" It.\" - 4r. - -L [- F Fa\" p.a.\" _ 1 Faf3 (F 2 F)] : af3.1t + lta.f3 + f31t. a'F note that the first term of (3.62) arises directly from the second term of (3.6 I). (c) Use Maxwell's equations to conclude that (3.61) (3.62) (3.63) y CHAPTER 4 ELECTROMAGNETISM AND DIFFERENTIAL FORMS The ether trembled at his agitations In a manner so familiar that I only need to say, In accordance with Clerk Maxwell's six equations It tickled peoples' optics far away. You can feel the way it's done, You may trace them as they run- dy by dy less df3 by dz is equal KdX/dt... While the curl of (X, Y, Z) is the minus d/dt of the vector (a, b, c): From The Revolution of the Corpuscle, written by A. A. Robb (to the tune of The Interfering Parrott) for a dinner of the research students of the Cavendish Laboratory in the days of the old mathematics. This chapter is all Track 2. It is needed as preparation for \u00a7\u00a7 14.5 and 14.6 (computation of curvature using differential forms) and for Chapter 15 (Bianchi identities and boundary of a boundary), but is not needed for the rest of the book. \"'\" \u00a74.1. EXTERIOR CALCULUS Stacks of surfaces, individually or intersecting to make \"honeycombs,\" \"egg crates,\" and other such structures (\"differential forms\"), give unique insight into the geometry of electromagnetism and gravitation. However, such insight comes at some cost in time. Therefore, most readers should skip this chapter and later material that depends on it during a first reading of this book. Analytically speaking, differential forms are completely antisymmetric tensors; pictorially speaking, they are intersecting stacks of surfaces. The mathematical formalism for manipulating differential forms with ease, called \"exterior calculus,\" is summarized concisely in Box 4.1; its basic features are illustrated in the rest of this chapter by rewriting electromagnetic theory in its language. An effective way to tackle this chapter might be to (1) scan Box 4.1 to get the flavor of the formalism; (2) read the rest of the chapter in detail; (3) restudy Box 4.1 carefully; (4) get practice in manipulating the formalism by working the exercises.* (continued on page 99) \u2022 Exterior calculus is. treated in greater detail than here by: E. Cartan (1945); de Rham (1955); Nickerson, Spencer, and Steenrod (1959); Hauser (1970); Israel (1979); especially Flanders (1963, relatively easy, with many applications); Spivak (1965, sophomore or junior level, but fully in tune with modem mathematics); H. Cartan (1970); and Choquet-Bruhat (l968a). \u00a74.1. EXTERIOR CALCULUS 91 Box 4.1 DIFFERENTIAL FORMS AND EXTERIOR CALCULUS IN BRIEF The fundamental definitions and formulas of exterior calculus are summarized here for ready reference. Each item consists of a general statement (at left of page) plus a leading application (at right of page). This formalism is applicable not only to spacetime, but also to more general geometrical systems (see heading of each section). No attempt is made here to demonstrate the internal consistency of the formalism, nor to derive it from any set of definitions and axioms. For a systematic treatment that does so, see, e.g., Spivak (1965), or Misner and Wheeler (1957", " 1 /\\ a 2 /\\ \u2022.\u2022 /\\ ap is /\\ a 2 (q1 /\\ q2 /\\... /\\ qP, a 1 /\\... /\\ ap )' d. The Jacobian determinant of a set of p functions ) with respect to p of their arguments fk(xl,...,xn is ( dp /\\ dj2 /\\... /\\ djP, a'!J /\\ a'!J /\\... /\\ a'!J) ax1 ax2 axp _ d II( afk )11 = a(jl,j2,..,fP) 2 a(x, x,..., x P). ax'. et 1 \u2022 5. Simple forms. a. A simple p-form is one that can be written as a wedge product of pI-forms: q = a /\\ p /\\ '\" P, /\\ y., p factors. b. A simple p-form a /\\ p /\\ '\" /\\ Y is repre sented by the intersecting families of surfaces of a, p,...,y (egg-crate structure) plus a sense of circulation (orientation). Applications: a. In four dimensions (e.g., spacetime) all O-forms, 1 forms, 3-forms, and 4-forms are simple. A 2-form F is generally a sum of two simple forms, e.g., F = - e dt /\\ dx + h dy /\\ dz; it is simple if and only if F /\\ F = O. b. A set of I-forms a, p,..., y is linearly dependent (one a linear combination of the others) if and only if a/\\p/\\\u00b7\u00b7\u00b7/\\y=O (egg crate collapsed). B. Exterior Derivative (applicable to any \"differentiable manifold,\" with or without metric) 1. d produces a (p + 1)-form dq from a p-form q. 2. Effect of d is defined by induction using the 94 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Box 4.1 (continued) (Chapter 2) definition of df, and f a function (0 form), plus d(a /\\ f!) = da /\\ p + (-I)Pa /\\ dp, P q d 2 = dd = O. Two applications d(a /\\ dP) = da /\\ dp. For the p-form cp, with cp = <l>li,...i.1 dxit /\\... /\\ dxi., one has (alternative and equivalent definition of dcp) C. Integration (applicable to any \"differentiable manifold,\" with or without metric) I. Pictorial interpretation. Text and pictures of Chapter 4 interpret fa (inte gral of specified I-form a along specified curve from specified starting point to specified end point) as \"number of a-surfaces pierced on that route\"; similarly, they interpret fcp (integral of specified 2-form cp over specified bit of surface on which there is an assigned sense of circulation or \"orien tation\") as \"number of cells of the honeycomb-like structure cp cut through by that surface\"; similarly for the egg-crate-like structures that represent 3 forms; etc. 2. Computational rules for integration. To evaluate fa, the integral of a p-form a = (Xlit...i.l(xl,..., x n ) dxit /\\... /\\ dxi., over a p-dimensional surface, proceed in two steps. a. Substitute a parameterization of the surface, X k(Al,..., AP) into a, and collect terms in the form a = alAi) dA1 /\\ '\" /\\ dAP (this is a viewed as a p-form in the p-dimen sional surface); \u00a74.1. EXTERIOR CALCULUS 95 b. Integrate f a =f a(AJ) clA l clA2... clAP using elementary definition of integration. Example: See equations (4.12) to (4.14). 3. The differential geometry of integration. Calculate fa for a p-form a as follows. a. Choose the p-dimensional surface S over which to integrate. b. Represent S by a parametrization giving the generic point of the surface as a function of the parameters, &'(;\\1, ;\\2,... ;\\P). This fixes the ori entation. The same function with ;\\1 ~ ;\\2, &,(;\\2,;\\ 1,...,;\\P), describes a different (i.e., op positely", " oriented) surface, - S. c. The infinitesimal parallelepiped is tangent to the surface. The number of cells of a it slices is /a o'!P /\\... /\\ o&') \\. 0;\\ P '0;\\ 1 A~ 1 411~ A~ P \u2022\u2022 \u2022 \"-11\\ \u2022 This number changes sign if two of the vectors o&'10;\\ k are interchanged, as for an oppositely oriented surface. d. The above provides an interpretation motivat ing the definition f a=ff... f (a.2J!.../\\ o&' /\\... /\\ o&'), 0;\\1 0;\\2 o;\\P - d;\\1 d;\\2... clAp. This definition is identified with the computa tional rule of the preceding section (C.2) in exercise 4.9. An application Integrate a gradient df along a curve, '!P(;\\) from 91(0) to '!P(1): f df =f <df, d:l'I clA) d;\\ =f (dfld;\\) d;\\ 1 I (I o =f[~P(1)] - f['!P(O)]. e. Three different uses for symbol \"d\": First, light face d in explicit derivative expressions such as 96 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Box 4.1 (continued) dlda, or dflda, or d9 Ida; neither numerator nor denominator alone has any meaning, but only the full string of symbols. Second, lightface d inside an integral sign; e.g., Ifda. This is an instruction to perform integration, and has no meaning whatsoever without an integral sign; \"I... d...\" lives as an indivisible unit. Third, sans-serif d; e.g., d alone, or df, or da. This is an exterior derivative, which converts a p-form into a (p + I)-form. Sometimes lightface d is used for the same purpose. Hence, d alone, or df, or dx, is always an exterior derivative unless coupled to an I sign (second use), or coupled to a I sign (first use). 4. The generalized Stokes theorem (see Box 4.6). a. Let a'Y be the closed p-dimensional boundary of a (p + I)-dimensional surface 'Y. Let q be a p-form defined throughout 'Y. Then [integral ofp-form q over boundary a'Y equals integral of (p + I)-form dq over interior 'Y]. b. For the sign to come out right, orientations of 'Y and a'Y must agree in this sense: choose coordinates yO, yl,...,yP on a portion of 'Y, with yO specialized so yO ::; 0 in 'Y, and yO = 0 at the boundary a'Y; then the orientation a'!J /\\ a'!J /\\... /\\ a'!J ayp ayo ayl for 'Y demands the orientation for a'Y. c. Note: For a nonorientable surface, such as a Mobius strip, where a consistent and continuous choice of orientation is impossible, more intri cate mathematics is required to give a definition of \"a\" for which the Stokes theorem holds. Applications: Includes as special cases all integral theo rems for surfaces of arbitrary dimension in spaces of arbitrary dimension, with or without metric, generaliz- \u00a74.1. EXTERIOR CALCULUS 97 ing all versions of theorems of Stokes and Gauss. Exam ples: a. 'V a curve, o'V its endpoints, C1 =f a O-form (func tion): f df= f (df/d\"A)dA =f f=f(l) -f(O). 1 'V 0 3'V b. 'Va 2-surface in 3-space, o'V its closed-curve bound ary, val-form; translated into Euclidean vector notation, the two integrals are f dv =f (V X v)\u00b7 dS; f v = f v\u00b7 dl. 'V 'V cl'V 3'V c. Other applications in \u00a7\u00a75.8, 20.2, 20.3, 20.5, and exercises 4.10, 4.11, 5.2, and below. D. Algebra II (applicable to any vector space with metric) 1. Norm of a pjorm. lIall 2 = 0:. - ['too.'p[. o:it...i p\u2022 Two applications: Norm of a I-form equals its squared length", ", lIall 2 = a \u2022 a. Norm of electromagnetic 2-form or Faraday: IIFII2 = B2 _ \u00a32. 2. Dual of a pjorm. a. In an n-dimensional space, the dual of a p-form a is the (n - p)-form *a, with components ( *0:) kt...k n _ p - o:litoo.ipl E il...i p kt...kn-p\u00b7 b. Properties of duals: **a = (-I)P-1a in spacetime; a /\\ *a = lIall 2e in general. c. Note: the definition of e (exercise 3.13) entails choosing an orientation of the space, i.e., decid ing which orthonormal bases (l) are \"right handed\" and thus (2) have e(e1,...,en) = + 1. Applications a. For f a O-form, *f = fe, and ffd(volume) = f*f b. Dual of charge-current I-form J is charge-current 3-form *J. The total charge Q in a 3-dimensional hypersurface region S is Q(S) = f *J. s 98 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Box 4.1 (continued) Conservation of charge is stated locally by d*J = O. Stokes' Theorem goes from this differential conserva tion law to the Integral conservation law, o =f d*J::=f aT 'V *J. This law is of most interest when a'Tf\" = 52 - 51 con sists of the future 52 and past 51 boundaries of a in which case it states Q(5 2) = spacetime region, Q(5 1 ); see exercise 5.2. c. Dual of electromagnetic field tensor F = Faraday is *F = Maxwell. From the d*F = 417 *J Maxwell find 417Q = 417fs *J = fs d*F = fas *F. equation, 3. Simple forms revisited. a. The dual of a simple form is simple. b. Egg crate of *(1 is perpendicular to egg crate of (1 = a /\\ p /\\... /\\ P in this sense: (1) pick any vector V lying in intersection of surfaces of (1 \u00aba, V) = (P, V) =... = (P, V) = 0); (2) pick any vector W lying in intersection of surfaces of *(1; (3) then V and Ware necessarily perpendicu lar: V\u00b7 W = O. Example: (1 = 3 dt is a simple I-form in spacetime. a. *(1 = -3 dx /\\ dy /\\ dz is a simple 3-form. b. General vector in surfaces of (1 is V = VZez + Vllell + VZez\u00b7 c. General vector in intersection of surfaces of *(1 is d. W\u00b7 V=O. \u00a74.2. ELECTROMAGNETIC 2-FORM AND LORENTZ FORCE 99 \u00a74.2. ELECTROMAGNETIC 2-FORM AND LORENTZ FORCE The electromagnetic field tensor, Faraday = F, is an antisymmetric second-rank tensor (i.e., 2-form). Instead of expanding it in terms of the tensor products of basis I-forms, the exterior calculus prefers to expand in terms of antisymmetrized tensor products (\"exterior products,\" exercise 4.1): F = i Fa /3 dx a /\\ dx/3, dx a /\\ dx/3 =dx a \u00ae dx/3 - dx/3 \u00ae dx a \u2022 (4.1) (4.2) Any 2-form (antisymmetric, second-rank tensor) can be so expanded. The symbol \" /\\\" is variously called a \"wedge,\" a \"hat,\" or an \"exterior product sign\"; and dx a /\\ dx/3 are the \"basis 2-forms\" of a given Lorentz frame (see \u00a73.5, exercise 3.12, and Box 4.1). There is no simpler way to illustrate this 2-form representation of the electromag netic field than to consider a magnetic field in the x-direction: (4.3) FyZ = -F\"y = Bx' F = Bx dy /\\ dz. The I-form dy = grad y is the set of surfaces (actually hypersurfaces) y = 18 (all t, x, z),y = 19 (all t, x, z),y= 20 (all t, x, z), etc.;", " and surfaces uniformly interpolated between them. Similarly for the I-form dz. The intersection between these two sets of surfaces produces a honeycomb-like structure. That structure becomes a \"2-form\" when it is supplemented by instructions (see arrows in Figure 4.1) that give a \"sense of circulation\" to each tube of the honeycomb (order of factors in the \"wedge product\" of equation 4.2; dy /\\ dz = -dz /\\ dy). The 2-form F in the example differs from this \"basis 2-form\" dy /\\ dz only in this respect, that where dy /\\ dz had one tube, the field 2-form has B x tubes. When one considers a tubular structure that twists and turns on its way through spacetime. one must have more components to describe it. The 2-form for the general electromagnetic field can be written as F=~~/\\~+~~/\\~+~~/\\~+~~/\\~ + By dz /\\ dx + Bz dx /\\ dv (4.4) Electromagnetic 2-form expressed in terms of exterior products A 2-form as a honeycomb of tubes with a sense of circulation (6 components. 6 basis 2-forms). A I-form is a machine to produce a number out of a vector (bongs of a bell as the vector pierces successive surfaces). A 2-form is a machine to produce a number out of an oriented surface (surface with a sense of circulation indicated on it: Figure 4.1, lower right). The meaning is as clear here as it is in elementary magnetism: A 2-form as a machine to produce a number out of an oriented surface 100 z z x dy x dz y r--------.L-~~y z - t~ul\\v/ 1 ~ u x dy 1\\ dz v y x Figure 4.1. Construction of the 2-form for the electromagnetic field F = Bz dy 1\\ dz out of the I-forms dy and dz by \"wedge multiplication\" (formation of honeycomb-like structure with sense of circulation indicated by arrows). A 2-form is a \"machine to construct a number out of an oriented surface\" (illustrated by sample surface enclosed by arrows at lower right; number of tubes intersected by this surface is f (this surface) F= 18; Faraday's concept of \"magnetic flux\"). This idea of 2-form machinery can be connected to the \"tensor as-machine\" idea of Chapter 3 as follows. The shape of the oriented surface over which one integrates F does not matter, for small surfaces. All that affects JF is the area of the surface, and its orientation. Choose two vectors, u and v, that lie in the surface. They form two legs of a parallelogram, whose orientation (u followed by v) and area are embodied in the exterior product u 1\\ v. Adjust the lengths of u and v so their parallelogram, u 1\\ v, has the same area as the surface of integration. Then f. surface F =f u 1\\ v I F = F(u, v). machinery idei1 of. this chapter]-----J t'- - -'t-machinery idea ~ of Chapter 3 Exercise: derive this result, for an infinitesimal surface u 1\\ v and for general F, using the formalism of Box 4.1. \u00b7\"'\\ \u00a74.2, ELECTROMAGNETIC 2\u00b7FORM AND LORENTZ FORCE 101 the number of Faraday tubes cu't by that surface. The electromagnetic 2-form F or Faraday described by such a \"tubular structure\" (suitably abstracted; Box 4.2) has a reality and a location in space that is independent of all coordinate systems and all artificial distinctions between \"electric\" and \"magnetic\" fields. Moreover, those tubes provide the most direct geometric representation that anyone has ever been able to give for the machinery by which the electromagnetic field acts on a charged particle. Take a particle of charge e and 4-velocity Let this particle go through a region where the electromagnetic field is described by the 2-form F = Bz dy /\\ dz (4.6) (4.5) of Figure 4.1. Then the force exerted on the particle (regarded as a I-form) is the contraction of this 2-form with the 4-velocity (and the charge); Lorentz force as contraction of electromagnetic 2-form with particle's 4-velocity p = dp/dT = eF(u) = e(F, u), (4.7) as one sees by direct evaluation, letting the two factors in the 2-form act in turn on the tangent vector u: p := eBz(dy /\\", " dz, u) := eBz{dy(dz,u) - dZ(dy,u)} := eBx{dy(dz,uZez ) - dZ(dy,uVev)} or Comparing coefficients of the separate basis I-forms on the two sides of this equa tion, one sees reproduced all the detail of the Lorentz force exerted by the magnetic field Bz : (4.8) dpv. dz Pv := dT := eBx dT'(4.9) By simple extension of this line of reasoning to the general electromagnetic field, one concludes that the time-rate ofchange ofmomentum (ljorm) is equal to the charge multiplied by the contraction of the Faraday with the 4.velocity. Figure 4.2 illustrates pictorially how the 2-form, F, serves as a machine to produce the I-form, p, out of the tangent vector, eu. (continued on page 105) 102 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Box 4.2 ABSTRACTING A 2-FORM FROM THE CONCEPT OF \"HONEYCOMB LIKE STRUCTURE,\" IN 3-SPACE AND IN SPACETIME Open up a cardboard carton containing a dozen bottles, and observe the honeycomb structure of intersecting north-south and east-west cardboard separators between the bottles. That honeycomb structure of \"tubes\" (\"channels for bottles\") is a fairly apt illustration of a 2-form in the context of everyday 3-space. It yields a number (number of tubes cut) for each choice of smooth element of 2-surface slicing through the three-dimensional structure. However, the intersecting cardboard separators are rather too specific. All that a true 2-form can ever give is the number of tubes sliced through, not the \"shape\" of the tubes. Slew the carton around on the floor by 45 0 \u2022 Then half the separators run NW-SE and the other half run NE-SW, but through a given bit of 2-surface fixed in 3-space the count of tubes is unchanged. There fore, one should be careful to make the concept of tubes in the mind's eye abstract enough that one envisages direction of tubes (vertical in the example) and density of tubes, but not any specific location or orientation for the tube walls. Thus all the following representations give one and the same 2-form, CT: CT = B dx /\\ dy; CT = B(2 dx) /\\ (~ dY) (NS cardboards spaced twice as close as before: EW cardboards spaced twice as wide as before); CT = Bd(X0) /\\ d(X;I) (cardboards rotated through 45 0 ) ; a dx + f3 dy /\\ Y dx + 0 dy CT-B----= f3y)1/2 f3y)1/2 (ao - (ao - - (both orientation and spacing of \"cardboards\" changing from point to point, with all four functions, 0:, /1, y, and 8, depending on position). What has physical reality, and constitutes the real geometric object, is not anyone of the I-forms just encountered individually, but only the 2-form CT itself. This circumstance helps to explain why in the physical literature one sometimes refers to \"tubes of force\" and sometimes to \"lines of force.\" The two terms for the same structure have this in common, that each yields a number when sliced by a bit of surface. The line-of-force picture has the advantage of not imposing on the mind any specific structure of \"sheets of cardboard\"; that is, any specific decomposition of the 2-form into the product of I-forms. However, that very feature is also a disadvantage, for in a calculation one often finds it useful to have a well-defined representa tion of the 2-form as the wedge product of I-forms. Moreover, though it the tube picture, abstract must be if it is to be truthful, also has this advan tage, that the orientation of the elementary tubes (sense of circulation as indicated by arrows in Figures 4.1 and 4.5, for example) lends itself to ready visualization. Let the \"walls\" of the tubes therefore remain in all pictures drawn in this book as a reminder that 2-forms can be built out of I-forms; but let it be understood here and here after how manyfold are the options for the indi vidual I-forms! Turn now from three dimensions to four, and find that the concept of \"honeycomb-like struc ture\" must be made still more abstract. In three dimensions the arbitrariness of the decomposition of the 2-form into I-forms showed in the slant and packing of the \"card", "boards,\" but had no effect on the verticality,of the \"channels for the bottles\" (\"direction of Faraday lines of force or tubes of \u00a74.2. ELECTROMAGNETIC 2\u00b7FORM AND LORENTZ FORCE 103 --------------------------------------..~ force\"); not so in four dimensions, or at least not in the generic case in four dimensions. In special cases, the story is almost as simple in four dimensions as in three. An example of a special case is once again the 2-form C1 = B dx /\\ dy, with all the options for decomposition into I-forms that have already been mentioned, but with every option giving the same \"direction\" for the tubes. If the word \"direction\" now rises in status from \"tube walls unpierced by motion in the direction of increasing z\" to \"tube walls un pierced either by motion in the direction of in creasing z, or by motion in the direction of in creasing t, or by any linear combination of such motions,\" that is a natural enough consequence of the same adding the new dimension. Moreover, simplicity prevails for an electromagnetic plane wave. For example, let the wave be advancing in the z-direction, and let the electric polarization point in the x-direction; then for a monochromatic wave, one has and all components distinct from these equal zero. Faraday is F = FOl dt /\\ dx + F31 dz /\\ dx = Eo cos w(z - t) d(z - t) /\\ dx, which is again representable as a single wedge product of two I-forms. Not so in general! The general 2-form in four dimensions consists of six distinct wedge products, F = FOl dt /\\ dx + F02 dt /\\ dy + '\" + F23 dy /\\ dz. It is too much to hope that this expression will reduce in the generic case to a single wedge prod uct of two I-forms (\"simple\"2-form). It is not even true that it will. It is only remarkable that it can be reduced from six exterior products to two (de tails in exercise 4.1); thus, Each product n i /\\ (i individually can be visual ized as a honeycomb-like structure like those de picted in Figures 4.1, 4.2, 4.4, and 4.5. Each such structure individually can be pictured as built out of intersecting sheets (I-forms), but with such de tails as the tilt and packing of these I-forms ab stracted away. Each such structure individually gives a number when sliced by an element of surface. What counts for the 2-form F, however, is neither the number of tubes of n 1 /\\ (l cut by the surface, nor the number of tubes of n2 /\\ (2 cut by the surface, but only the sum of the two. This sum is what is referred to in the text as the \"number of tubes of F\" cut by the surface. The contribution of either wedge product individua~y the de is not well-defined, for a simple reason: composition of a six-wedge-product object into two wedge products, miraculous though it seems, is actually far from unique (details in exercise 4.2). In keeping with the need to have two products of I-forms to represent the general 2-form note that the vanishing of dF (\"no magnetic charges\") does not automatically imply that d(n 1 /\\ (1) or d(n2 /\\ (2) separately vanish. Note also that any spacelike slice through the general 2-form F (re duction from four dimensions to three) can always be represented in terms of a honeycomb-like structure (\"simple\" 2-form in three dimensions; Faraday's picture of magnetic tubes of force). Despite the abstraction that has gone on in see ing in all generality what a 2-form is, there is no bar to continuing to use the term \"honeycomb-like structure\" in a broadened sense to describe this object; and that is the practice here and hereafter. y F - 3 + n - 206'8 2 - n - Contract eu with - - - - - - - - - -....-~ ~ 6J~:4 t]1j p+2 p+1 P t]P+12 Figure 4.2. The Faraday or 2-form F of the electromagnetic field is a machine to produce a I-form (the time-rate of change of momentum p of a charged particle) out of a tangent vector (product of charge e of the particle and its 4-velocity u). In spacetime the general 2-form is the \"superposition\" (see Box 4.2) of two structures like that illustrated at the top of this diagram, the tubes of the first being tilted", " and packed as indicated, the tubes of the second being tilted in another direction and having a different pack ing density. \u00a74.3. ELECTROMAGNETISM IN TERMS OF FORMS 105 \u00a74.3. FORMS ILLUMINATE ELECTROMAGNETISM, AND ELECTROMAGNETISM ILLUMINATES FORMS All electromagnetism allows itself to be summarized in the language of 2-forrns, honeycomb-like \"structures\" (again in the abstract sense of \"structure\" of Box 4.2) of tubes filling all spacetime, as well when spacetime is curved as when it is flat. In brief, there are two such structures, one Faraday = F, the other Maxwell = *F, each dual (\"perpendicular,\" the only place where metric need enter the discussion) to the other, each satisfying an elementary equation: (\"no tubes of Faraday ever end\") and dF = 0 d*F = 417 *J (4.10) (4.11 ) (\"the number of tubes of Maxwell that end in an elementary volume is equal to the amount of electric charge in that volume\"). To see in more detail how this machinery shows up in action, look in turn at: (1) the definition of a 2-forrn; (2) the appearance of a given electromagnetic field as Faraday and as Maxwell; (3) the Maxwell structure for a point-charge at rest; (4) the same for a point-charge in motion; (5) the nature of the field of a charge that moves uniformly except during a brief instant of acceleration; (6) the Faraday structure for the field of an oscillating dipole; (7) the concept of exterior derivative; (8) Maxwell's equations in the language offorms; and (9) the solution of Maxwell's equations in flat spacetime, using a I-form A from which the Lienard-Wiechert 2-forrn F can be calculated via F = dA. A 2-forrn, as illustrated in Figure 4.1, is a machine to construct a number (\"net number of tubes cut\") out of any \"oriented 2-surface\" (2-surface with \"sense of circulation\" marked on it): Preview of key points in electromagnetism A 2-form as machine for number of tubes cut nUmber) of tubes = f ( cut surface F (4.12) For example, let the 2-forrn be the one illustrated in Figure 4.1 Number of tubes cut calculated in one example and let the surface of integration be the portion of the surface of the 2-sphere t = constant, bounded between () = 70\u00b0 and () = 110\u00b0 and x 2 + y2 + Z2 = a2 between cp = 0\u00b0 and cp = 90\u00b0 (\"Atlantic region of the tropics\"). Write, y = a sin () sin cp, z = a cos (), dy = a (cos () sin cp cJ() + sin () cos cp dcp), dz = -a sin () cJ(), dy /\\ dz = a2 sin2() cos cp cJ() /\\ dcp. (4.13) 106. 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS B Figure 4.3. Spacelike slices through Faraday, the electromagnetic 2-form, a geometric object, a honeycomb of tubes that pervades all spacetime (\"honeycomb\" in the abstract sense spelled out more precisely in Box 42). The surfaces in the drawing do not look like a 2-form (honeycomb), because the second family of surfaces making up the honeycomb extends in the spatial direction that is suppressed from the drawing. Diagram A shows one spacelike slice through the 2-form (time increases upwards in the diagram). In diagram B, a projection of the 2-form on this spacelike hypersurface gives the Faraday tubes of magnetic force. in this three-dimensional geometry (if the suppressed dimension were restored, the tubes would be tubes, not channels between lines). Diagram C shows another spacelike slice (hypersurface of simultaneity for an observer in a different Lorentz frame). Diagram D shows the very different pattern of magnetic tubes in this reference system. The demand that magnetic tubes of force shall not end (V. B = 0), repeated over and over for every spacelike slice through Faraday, gives everywhere the result OB/Of = - V X E. Thus (magnetostatics) + (covariance) -+ (magnetodynamics). Similarly-see Chapters 17 and 21 (geometrostatics) + (covariance) -+ (geometrodynamics). The structure d() /\\ d() looks like a \"collapsed egg-crate\" (Figure 1.4, upper right) and has", " zero content, a fact formally evident from the vanishing of a /\\ p = - P /\\ a when a and p are identical. The result of the integration, assuming constant Bz' is f surface F = a2Bzi 110\u00b7 sin2 () d()f 70\u00b7 O\u00b7 90\u00b7 COS cp dcp (4.14) It is not so easy to visualize a pure electric field by means of its 2-form F (Figure 4.4, left) as it is to visualize a pure magnetic field by means of its 2-form F (Figures 4.1,4.2,4.3). Is there not some way to treat the two fields on more nearly the same footing? Yes, construct the 2-form *F (Figure 4.4, right) that is dual (\"perpendicular\"; Box 4.3; exercise 3.14) to F \u00a74.3. ELECTROMAGNETISM IN TERMS OF FORMS 107 OF y f7 r - __ y,z x Figure 4.4. The Faraday structure F = 2\" F~. dx~ 1\\ dx' = 2\" F OI dt 1\\ dx + 2\" FlO dx 1\\ dt = Ez dx 1\\ dt I I I associated with an electric field in the x-direction, and the dual (\"perpendicular\") Maxwell honeycomb like 2-form I *F = 2\" * F~. dx~ 1\\ dx' = *F23 dx 2 1\\ dx3 = FOI dx2 1\\ dx3 = FlO dx 2 1\\ dx3 = Ez dy 1\\ dz. Represent in geometric form the field of a point-charge of strength e at rest at the origin. Operate in flat space with spherical polar coordinates: The electric field in the r-direction being Er = ejr Z, it follows that the 2-form F or Faraday is (4.15) F = IF dxlJ. /\\ dx' = -Erdt /\\ dr = - ez dt /\\ dr. 2 IJ.V r Its dual, according to the prescription in exercise 3.14, is Maxwell: Maxwell = *F = e sin B dB /\\ dfP, (4.16) Pattern of tubes in dual structure Mexwel/ for point-charge at rest (4.17) as illustrated in Figure 4.5. Take a tour in the positive sense around a region of the surface of the sphere illustrated in Figure 4.5. The number of tubes of *F encompassed in the route will be precisely nUmber) (SOlid) of tubes = e angle. ( The whole number of tubes of *F emergent Over the entire sphere will be 47Te, in conformity with Faraday's picture of tubes of force. 108 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Box 4.3 DUALITY OF 2-FORMS IN SPACETIME Given a general 2:.form (containing six exterior or wedge products) F = Exdx /\\ dt + E\"dy /\\ dt +... + Bzdx /\\ dy, one gets to its dual (\"perpendicular\") by the prescription *F = -Bxdx /\\ dt -... + E\"dz /\\ dx + Ezdx /\\ dy. Duality Rotations Note that the dual of the dual is the negative of the original 2-form; thus **F = -Exdx /\\ dt -... -Bzdx /\\ dy = -F. In this sense * has the same property as the imaginary number i: ** = ii = -1. Thus one can write e*'\" = cos a + *sin a. This operation, applied to F, carries attention from the generic 2-form in its simplest representation (see exercise 4.1) F = Exdx /\\ dt + Bxdy /\\ dz to another \"duality rotated electromagnetic field\" e*\"'F = (Ex cos a - Bx sin a) dx /\\ dt + (Bx cos a + Ex sin a) dy /\\ dz. If the original field satisfied Maxwell's empty-space field equations, so does the new field. With suitable choice of the \"complexion\" a, one can annul one of the two wedge products at any chosen point in spacetime and have for the other Field of a point-charge in motion How can one determine the structure of tubes associated with a charged particle moving at a uniform velocity? First express *F in rectangular coordinates moving with the particle (barred coordinates in this comoving \"rocket\" frame of reference; unbarred coordinates will be used later for a laboratory frame of reference). The relevant steps can be listed: (a) *F = e sin B dB /\\ dip = -e(d cos B", ") /\\ dip; \u00a74.3. ELECTROMAGNETISM IN TERMS OF FORMS 109 Figure 4.5. The field of 2-fonns Maxwell = of = e sin (J dO 1\\ d<P that describes the electromagnetic field of a charge e at rest at the origin. This picture is actually the intersection of of with a 3-surface of constant time t; i.e., the time direction is sup pressed from the picture. (b) (c) fP = arctan ~ ; x _ xdy- Jdx dfP= 22'x +Y COS() =-; z r -d(cosB) = -!!T +~ (xdx+ Jdy+ TdZ); r r (d) combine to find *F = (e/r 3)(xdy /\\ dT + Jeff /\\ dx + Tdx /\\ dy) (4.18) (electromagnetic field of point charge in a comoving Cartesian system; spherically symmetric). Now transform to laboratory coordinates: velocity parameter a velocity /3 = tanh a 1 ---;::====:::;: = cosh a, VI - /32 /3 VI - /32 = sinh a t = t cosh a - x sinh a, ~ = - t si~ a + x cosh a, [ z = z; Y =y r = [(x cosh a t sinh a)2 + y 2 + z2jl/2; (a) (b) (c) *F = (e/r 3)[(x cosh a - t sinh a) dy /\\ dz + Y dz /\\ (cosh a dx - sinh a dt) + z(cosh a dx - sinh a dt) /\\ ~vl; (4.19) 110 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS (d) compare with the general dual 2-form, ~=~~A~+~~A~+~~A~ + Bz dt A dx + Bv dt A dy + Bz dt Adz; and get the desired individual field components (e) Er = (e/r 3)(x cosh a E v = (e/r 3)y cosh cr, Ez = (e/r 3)z cosh a, ( t sinh a), Br = 0, By = -(e/r 3)z sinh cr, Bz = (e/r 3)y sinh 0:. One can verify that the invariants B2 _ E2 = 1 F px/3 2 a/3'1 E\u00b7 B = -F *Fa/3 4 a/3. (420) (4.21 ) (4.22) have the same value in the laboratory frame as in the rocket frame, as required. Note that the honeycomb structure of the differential form is not changed when one goes from the rocket frame to the laboratory frame. What changes is only the mathematical formula that describes it. \u00a74.4. RADIATION FIELDS The Maxwell structure of tubes associated with a charge in uniform motion is more remarkable than it may seem at first sight, and not only because of the Lorentz contraction of the tubes in the direction of motion. The tubes arbitrarily far away move on in military step with the charge on which they center, despite the fact that there is no time for information \"emitted\" from the charge \"right now\" to get to the faraway tube \"right now.\" The structure of the faraway tubes \"right now\" must therefore derive from the charge at an earlier moment on its uniform-motion, straight-line trajectory. This circumstance shows up nowhere more clearly than in what happens to the field in consequence of a sudden change, in a short time,,'h, from one uniform velocity to another uniform velocity (Figure 4.6). The tubes have the standard patterns for the two states of motion, one pattern within a sphere of radius r, the other outside that sphere, where r is equal to the lapse of time (\"cm of light-travel time\") since the acceleration took place. The necessity for the two patterns to fit together in the intervening zone, of thickness..dr =..dr, forces the field there to be multiplied up by a \"stretching factor,\" proportional to r. This factor is responsible for the well-known fact that radiative forces fall off inversely only as the first power of the distance (Figure 4.6). When the charge continuously changes its state of motion, the structure of the electromagnetic field, though based on the same simple principles as those illustrated in Figure 4.6, nevertheless looks more complex. The following is the Faraday 2-form How an acceleration causes radiation \u00a74.4. RADIATION FIELDS 1 1 1 ~----------------", "--~:~~ -----..--- +/3 -/3 Figure 4.6. Mechanism of radiation. J. J. Thomson's way to understand why the strength of an electromagnetic wave falls only as the inverse first power of distance r and why the amplitude of the wave varies (for low velocities) as sin (J (maximum in the plane perpendicular to the line of acceleration). The charge was moving to the left at uniform velocity. Far away from it, the lines of force continue to move as if this uniform velocity were going to continue forever (Coulomb field of point.charge in slow motion). However, closer up the field is that of a point-change moving to the right with uniform velocity (1/r 2 dependence of strength upon distance). The change from the one field pattern to another is confined to a shell of thickness.17 located at a distance r from the point of acceleration (amplification of field by \"stretching factor\" r sin (J J/3/JT; see text). We thank C. Teitelboim for the construction of this diagram. for the field of an electric dipole of magnitude PI oscillating up and down parallel to the z-axis: Field of an oscillating dipole F = Ex dx /\\ dl +... + Bx d)' /\\ dz +... = real part of {PIeiwr-iwt 1. -:-1 -? dr /\\ dt + SIn,.. () ( 1 -3 r [2 cos () cJ() /\\ d 2 w r iw ) r- iw? r- ( r t - ), ; ', gives E r gives E 9 + sin () (-~w _~) dr /\\ r cJ()]} r- r gives B\u00a2 (4.23) 112 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS and the dual 2-form Maxwell = *F is *F = -Bx dx /\\ dt -... + Ex dy /\\ d= +... = real part of {Pleiwr-iwt [sin 0 ( - ~w - ~) dt /\\ rsin 0 de> r r- gives B\u00a2 + 2 cos 0 C~ - ~~)r dO /\\ r sin 0 dep, I gives E r + sin 0(-1_ iw - ~)r sin 0 d<j> /\\ dr]). r3 r2 r (4.24) gives Eo \u00a74.5. MAXWELL'S EQUATIONS The general 2-form F is written as a superposition of wedge products with a factor ~, F = 1- F dx/l /\\ dx' 2 JlP'(4.25) because the typical term appears twice, once as Fz\" dx /\\ dy and the second time as F\"x dy /\\ dx, with F\"x = - Fz\" and dy /\\ dx = - dx /\\ dy. If differentiation (\"taking the gradient\"; the operator d) produced out of a scalar a I-form, it is also true that differentiation (again the operator d, but now generally known under Cartan's name of \"exterior differentiation\") produces a 2-form 'out of the general I-form; and applied to a 2-form produces a 3-form; and applied to the form of the highest order that spacetime will a 3-form produces a 4-form, accommodate. Write the general.fform as (4.26) Taking exterior derivative where the coefficient ep\"I\"Z\"'''f' like the wedge product that follows it, is antisym metric under interchange of any two indices. Then the exterior derivative of t/J is Take the exterior derivative of Faraday according to this rule and find that it vanishes, not only for the special case of the dipole oscillator, but also for a general electromagnetic field. Thus, in the coordinates appropriate for a local Lorentz frame, one has (4.27) \u00a74.5. MAXWELL'S EQUATIONS dF = d(Ezdx /\\ dt +... + Bzdy /\\ dz +... ) = __z dt + __z dx + __z dy + __z dz aE ) az aE ay +... (5 more such sets of 4 terms each).... aE ax aE at ( /\\ dx /\\ dt 113 (4.28) Note that such a term as dy /\\ dy /\\ dz is automatically zero (\"collapse of egg-crate cell when stamped on\"). Collect the terms that do not vanish and find dF = __z + __\" + __z dx /\\ dy /\\ dz aB ax ( aB ay aB ) az + ( aBz at + - aEz ay -", " - aE,,) az dt /\\ dy /\\ dz + ( aB\" at + - aEz az - - aEz ) ax dt /\\ dz /\\ dx aE + __z + __\" ax aB at ( z dt /\\ dx /\\ dy. aE ) ay (4.29) Each term in this expression is familiar from Maxwell's equations and div B = V\u00b7 B = 0 curlE = V xE=-B. Each vanishes, and with their vanishing Faraday itself is seen to have zero exterior derivative: dF=O. (4.30) In other words, \"Faraday is a closed 2-form\"; \"the tubes of F nowhere come to an end.\" Faraday structure: tubes nowhere end A similar calculation gives for the exterior derivative of the dual2-form Maxwell the result d*F = d(-Bzdx /\\ dt - '\" + Ezdy /\\ dz +... ) aE) = __x + __\" + __z dx /\\ dy /\\ dz az aE ay aE ax ( - - aBz ay + - aB,,) az dt /\\ dy /\\ dz + ( aEz at +.., = 417(p dx /\\ dy /\\ dz - Jx dt /\\ dy /\\ dz - J\" dt /\\ dz /\\ dx - Jz dt /\\ dx /\\ dy) = 4'ii *J; d*F = 417 *J. Maxwell structure: density of tube endings given by charge-current 3-form (4.31) 114 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS In empty space this exterior derivative, too, vanishes; there Maxwell is a closed 2-form; the tubes of *F, like the tubes of F, nowhere come to an end. In a region where charge is present, the situation changes. Tubes of Maxwell take their origin in such a region. The density of endings is described by the 3-form *J = charge, a \"collection of eggcrate cells\" collected along bundles of world lines. The two equations and dF= 0 d*F=4'iT*J summarize the entire content of Maxwell's equations in geometric language. The forms F = Faraday, and *F = Maxwell, can be described in any coordinates one pleases-or in a language (honeycomb and egg-crate structures) free of any reference whatsoever to coordinates. Remarkably, neither equation makes any reference whatsoever to metric. As Hermann Weyl was one of the most emphatic in stressing (see also Chapters 8 and 9), the concepts of form and exterior derivative are metric free. Metric made an appearance only in one place, in the concept of duality (\"perpendicularity\") that carried attention from F to the dual structure *F \u00a74.6. EXTERIOR DERIVATIVE AND CLOSED FORMS The words \"honeycomb\" and \"egg crate\" may have given some feeling for the geometry that goes with electrodynamics. Now to spell out these concepts more clearly and illustrate in geometric terms, with electrodynamics as subject matter, what it means to speak of \"exterior differentiation.\" Marching around a boundary, yes; but how and why and with what consequences? It is helpful to return to functions and I-forms, and see them and the 2-forms Faraday and Maxwell and the 3-form charge as part of an ordered progression (see Box 4.4). Two-forms are seen in this box to be of two kinds: (I) a special 2-form, known as a \"closed\" 2-form, which has the property that as many tubes enter a closed 2-surface as emerge from it (exterior derivative of2-form zero; no 3-form derivable from it other than the trivial zero 3-form!); and (2) a general 2-form, which sends across a closed 2-surface a non-zero net number of tubes, and therefore permits one to define a nontrivial3-form (\"exterior derivative of the 2-form\"), which has precisely as many egg-crate cells in any closed 2-surface as the net number of tubes of the 2-form emerging from that same closed 2-surface (generalization of Faraday's concept of tubes of force to the world of spacetime, curved as well as flat). (continued on page 120) Duality: the only place in electromagnetism where metric must enter Closed 2-form contrasted with general 2-form \u00a74.6. EXTERIOR DERIVATIVE AND CLOSED FORMS 115 Box 4.4 THE PROGRESSION OF FORMS AND EXTERIOR DERIVATIVES O-Form or Scalar, f", " An example in the context of 3-space and Newto nian physics is temperature, T(x,y, z), and in the context of spacetime, a scalar potential, ep(t, x,y, z). From Scalar to 1-Form Take the gradient or \"exterior derivative\" of a scalar fto obtain a special I-form, y = df Com ments: (a) Any additive constant included in f is erased in the process of differentiation; the quan tity n in the diagram at the left is unknown and irrelevant. (b) The I-form y is special in the sense that surfaces in one region \"mesh\" with surfaces in a neighboring region (\"closed I-form\"). (c) Line integral f~ df is independent of path for any class of paths equivalent to one another under continu ous deformation. (d) The I-form is a machine to produce a number (\"bongs of bell\" as each succes sive integral surface is crossed) out of a displace ment (approximation to concept of a tangent vector). \\ \\ General 1-Form f3 = f3 a dx a This is a pattern of surfaces, as illustrated in the diagram at the right; i.e., a machine to produce a number (\"bongs of bell\"; <p, u\u00bb out of a vector. A I-form has a reality and position in space inde pendent of all choice of coordinates. Surfaces do not ordinarily mesh. Integral fp around indicated closed loop does not give zero (\"\"more bongs than antibongs\"). From 1-Form to 2-Form ( = df3 = - - dx'\" /\\ dx a af3a ax'\" ( is a pattern of honeycomb-like cells, with a di rection of circulation marked on each, so stationed 116 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Box 4.4 (continued) that the number of cells encompassed in the dotted closed path is identical to the net contribution (excess of bongs over antibongs) for the same path in the diagram of P above. The \"exterior deriva tive\" is defined so this shall be so; the generalized Stokes theorem codifies it. The word\u00b7 \"exterior\" comes from the fact that the path goes around the periphery of the region under analysis. Thus the 2-form is a machine to get a number (number of tubes, \u00ab, u /\\ v\u00bb ou t of a bit of surface (u /\\ v) that has a sense of circulation indicated upon it. The 2-form thus defined is special in this sense: a rubber sheet \"supported around its edges\" by the dotted curve or any other closed curve is crossed by the same number of tubes when; (a) it bulges up in the middle; (b) it is pushed down in the middle; (c) it experiences any other continu ous deformation. The Faraday or 2-form F of electromagnetism, always expressible as F = dA (A = 4-potential, a I-form), also has always this special property (\"conservation of tubes\"). O-Form to 1-Form to 2-Form? No! Go from scalar f to I-form y = df The next step to a 2-form a is vacuous. The net contribution of the line integral fy around the dotted closed path is automatically zero. To reproduce that zero result requires a zero 2-form. Thus a = dy = ddf has to be the zero 2-form. This result is a special in stance of the general result dd = O. Again, this is a honeycomb-like structure, and again a machine to get a number (number of tubes, <U, u /\\ v\u00bb out of a surface (u /\\ v) that has a sense of circulation indicated On it. It is general in the sense that the honeycomb structures in one region do not ordinarily mesh with those \u00a74.6. EXTERIOR DERIVATIVE AND CLOSED FORMS 117 in a neighboring region. In consequence, a closed 2-surface, such as the box-like surface indicated by dotted lines at the right, is ordinarily crossed by a non-zero net number of tubes. The net num ber of tubes emerging from such a closed surface is, however, exactly zero when the 2-forrn is the exterior derivative of a I-form. /f --/~ II \u2022 / /... t'\\: --I \\ rC-- - - \u2022.. J.\u2022\u2022\u2022.\u2022\u2022...; i----/ I / I I / I,/ 1/ 1,/. L_.. 1/ -----J :'. From 2-Form to 3-Form JJ = dq = oO'la,81 dxY 1\\ dx a 1\\ dx,8", ", where dxY 1\\ dx a 1\\ dx,8 =3! dx[Y \u00ae dx a \u00ae dx,8l oxY This egg-crate type of structure is a machine to get a number (number of cells (p, U 1\\ v 1\\ w\u00bb from a volume (volume U 1\\ v 1\\ w within which one counts the cells). A more complete diagram would provide each cell and the volume of inte gration itself with an indicator of orientation (analogous to the arrow of circulation shown for cells of the 2-form). The contribution of a given cell to the count of cells is + I or -I, according as the orientation indicators have same sense or opposite sense. The number of egg-crate cells of p = dq in any given volume (such as the volume indicated by the dotted lines) is tailored to give precisely the same number as the net number of tubes of the 2-form q (diagram above) that emerge from that volume (generalized Stokes theorem). For electromagnetism, the exterior derivative of Faraday or 2-form F gives a null 3-form, but the exterior derivative of Maxwell or 2-form *F gives 4'1T times the 3-form *J of charge: *J = p dx 1\\ dy 1\\ dz - Jz dt 1\\ dy 1\\ dz - Jv dt 1\\ dz 1\\ dx - Jz dt 1\\ dx 1\\ dy. y 118 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Box 4.4 (continued) From 1-Form to 2-Form to 3-Form? No! Starting with a I-form (electromagnetic 4-potential), construct its exterior deriva tive. the 2-form F = dA (Faraday). The tubes in this honeycomb-like structure never end. So the number of tube endings in any elementary volume, and with it the 3-form dF = ddA, is automatically zero. This is another example of the general result that dd =0. From 2-Form to 3-Form to 4-Form? No! Starting with 2-form *F (Maxwell), construct its exterior derivative, the 3-form 4'17 *J. The cells in this egg-crate type of structure extend in a fourth dimension (\"hypertube\"). The number of these hypertubes that end in any elementary 4-vol ume, and with it the 4-form d(4'17 *J) = dd*F, is automatically zero, still another example of the general result that dd = O. This result says that aJ ) d * dt 1\\ dx 1\\ dy 1\\ dz = 0 az aJ ax aJ ay ap at ( (\"law of conservation of charge\"). Note: This implies dt 1\\ dx 1\\ dy 1\\ dz = c. From 3-Form to 4-Form T = dJJ = oV 1a OX~,8YI dx~ /\\ dx a /\\ dx,8 /\\ dxY This four-dimensional \"super-egg-crate\" type structure is a machine to get a number (number of cells, (7, n 1\\ u 1\\ v 1\\ w\u00bb from a 4-volume n 1\\ u 1\\ v 1\\ w. \u00a74.6. EXTERIOR DERIVATIVE AND CLOSED FORMS 119 From 4-Form to 5-Form? No! Spacetime, being four-dimensional, cannot accommodate five-dimensional egg-crate structures. At least two of the dxll's in dx\" 1\\ dx/3 1\\ dx Y 1\\ dx 8 1\\ dx' must be the same; so, by antisymmetry of\" 1\\,\" this \"basis 5-form\" must vanish. Results of Exterior Differentiation, Summarized O-form I-form 2-form 3-form 4-form 5-form? f df ddf=O No! A F= dA dF = ddA =0 *F 4'17 *J = d*F d(4'17 *J) = dd*F = 0 \"T = dtI P dT=O dp =0 New Forms from Old by Taking Dual (see exercise 3.14) Dual of scalar f is 4-form: *f = f dxo 1\\ dx 1 1\\ dx 2 1\\ dx3 = fe. Dual of I-form J is 3-form: *J = JO dx 1 1\\ dx 2 1\\ dx3 - j l dx 2 1\\ dx3 1\\ dxo + j2 dx3 1\\ dxo 1\\ dx 1 - J3 dxo 1\\ dx 1 1\\ dx2. Dual of 2-form F is 2-form:", " *F = FI\"/3l f \"/3lll v l F\"/3 = l)\"Al)/3 8FA8 \u2022 dx ll 1\\ dx V where'Dual of 3-form K is I-form: *K = K012 dx3 - K123 dxo + K230 dx 1 - K301 dx2, where K,,/3Y = l)\"~/3vl)YAKIlVA' Dual of 4-form L is a scalar: L = L 0123 dxo 1\\ dx 1 1\\ dx2 1\\ dx3 ; *L = L0123 = -L0123' Note I: This concept of duality between one form and another is to be distinguished from the concept of duality between the vector basis e\" and the Ijorm basis w\" of a given frame. The two types of duality have nothing whatsoever to do with each other! 120 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Box 4.4 (continued) Note 2: In spacetime, the operation of taking the dual, applied twice, leads back to the original form for forms of odd order, and to the negative thereof for forms of even order. In Euclidean 3-space the operation reproduces the original form, regardless of its order. Duality Plus Exterior Differentiation Start with scalar cf;. Its gradient dcf; is a I-form. Take its dual, to get the 3-forrn *dcf;. Take its exterior derivative, to get the 4-form d *dcf;. Take its dual, to get the scalar Dcf; = -*d *dcf;. Verify by index manipulations that D as defined here is the wave operator; i.e., in any Lorentz frame, Dcf; = cf;,<>'<> = -(a 2cf;/at2) + V 2cf;. Start with 110rm A. Get 2-form F = dA. Take its dual *F = *dA, also a 2-form. Take its exterior derivative, obtaining the 3-form d*F (has value 4'17 *J in electro magnetism). Take its dual, obtaining the I-form *d*F = *d*dA = 4'17J (\"Wave equation for electromagnetic 4-potential\"). Reduce in index notation to F,v = A,v - A,v = 4'17J. \",JI. JI.,\" JI.\" JI. [More in Flanders (1963) or Misner and Wheeler (1957); see also exercise 3.17.] \u00a74.7. DISTANT ACTION FROM LOCAL LAW Differential forms are a powerful tool in electromagnetic theory, but full power requires mastery of other tools as well. Action-at-a-distance techniques (\"Green's functions,\" \"propagators\") are of special importance. Moreover, the passage from Maxwell field equations to electromagnetic action at a distance provides a preview of how Einstein's local equations will reproduce (approximately) Newton's l/r2 law. In flat spacetime and in a Lorentz coordinate system, express the coordinates of particle A as a function of its proper time a, thus: d 2all da 2 _ \"Il( ) - a a. (4.32) Dirac found it helpful to express the distribution of charge and current for a particle of charge e following such a motion as a superposition of charges that momentarily \u00a74.7. DISTANT ACTION FROM LOCAL LAW 121 flash into existence and then flash out of existence. Any such flash has a localization in space and time that can be written as the product of four Dirac delta functions [see, for example, Schwartz (1950-1951), Lighthill (1958)]: World line of charge regarded as succession of flash-on, flash-off charges 84(X Il - all) = 8[xO - dl(a)] 8[x1 - a1(a)] 8[x2 - a2(a)] 8[x3 - a3(a)]. (4.33) Here any single Dirac function 8(x) (\"symbolic function\"; \"distribution\"; \"limit of a Gauss error function\" as width is made indefinitely narrow and peak indefinitely high, with integrated value always unity) both (1) vanishes for x f:. 0, and (2) has the integral J:: 8(x) dx = 1. Described in these terms, the density-current vector for the particle has the value (\"superposition of flashes\") Jil = ef 84[x\" - a\"(a)]all(a) da. The density-current (4.34) drives the electromagnetic field, F. Write F = dA to satisfy automatically half of Maxwell's equations (dF = dd", "A = 0): (4.34) In flat space, the remainder of Maxwell's equations (d*F = 4'1T *J) become or (4.35) (4.36) Make use of the freedom that exists in the choice of 4-potentials A\" to demand aA\"-=0 ax\" (4.37) (Lorentz gauge condition; see exercise 3.17). Thus get The density-current being the superposition of \"flashes,\" the effect (A) of this density-current can be expressed as the superposition of the effects E of elementary flashes; thus AIl(X) =f E[x - a(a)]all(a) da, (4.39) The solution of the wave equation where the \"elementary effect\" E (\"kernel\"; \"Green's function\") satisfies the equation (4.38) The electromagnetic wave equation One solution is the \"half-advanced-plus-half-retarded potential,\" DE(x) = -4'1T 84(x). (4.40) (4.41) 122 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS It vanishes everywhere except on the backward and forward light cones, where it has equal strength. Normally more useful is the retarded solution, R(x) = {~E(X) if XO > 0, if XO < 0, (4.42) which is obtained by doubling (4.41) in the region of the forward light cone and nullifying it in the region of the backward light cone. All electrodynamics (Coulomb forces, Ampere's law, electromagnetic induction, radiation) follows from the simple expression (4.39) for the vector potential [see, e.g., Wheeler and Feynman (1945) and (1949), also Rohrlich (1965)]. EXERCISES Exercise 4.1. GENERIC LOCAL ELECTROMAGNETIC FIELD EXPRESSED IN SIMPLEST FORM In the laboratory Lorentz frame, the electric field is E, the magnetic field B. Special cases are: (I) pure electric field (B = 0); (2) pure magnetic field (E = 0); and (3) \"radiation field\" or \"null field\" (E and B equal in magnitude and perpendicular in direction). All cases other than (I), (2), and (3) are \"generic.\" In the generic case, calculate the Poynting density of flow of energy E X B/4\" and the density of energy (E2 + B2)/S\". Define the direction of a unit vector n and the magnitude of a velocity parameter a by the ratio of energy flow to energy density: 2E X B n tanh 2a = E2 + B2. View the same electromagnetic field in a rocket frame moving in the direction of n with the velocity parameter a (not 2a; factor 2 comes in because energy flow and energy density are components, not of a vector, but of a tensor). By employing the formulas for a Lorentz transformation (equation 3.23), or otherwise, show that the energy flux vanishes in the rocket frame, with the consequence that E and B are parallel. No one can prevent the z-axis from being put in the direction common to E and B. Show that with this choice of direction, Faraday becomes (only two wedge products needed to represent the generic local field; \"canonical representa tion\"; valid in one frame, valid in any frame). Exercise 4.2. FREEDOM OF CHOICE OF 1-FORMS IN CANONICAL REPRESENTATION OF GENERIC LOCAL FIELD Deal with a region so small that the variation of the field from place to place can be neglected. Write Faraday in canonical representation in the form where PA (A = I or II) and qA are scalar functions of position in spacetime. Define a \"canonical transformation\" to new scalar functions of position PA and qA by way of the \"equation of transformation\" \u00a74.7. DISTANT ACTION FROM LOCAL LAW 123 v=df (\"curl-free\") v= hdf (\"rotation-free\") (has rotation) Figure 4.7. Some simple types of I-forms compared and contrasted. where the \"generating function\" S of the transformation is an arbitrary function of the qA and the qA: (a) Derive expressions for the two PA'S and the two p;r's in terms of S by equating coefficients of dqI, dqIl, dqT. dqlT individually on the two sides of the equation of trans formation. (b) Use these expressions for the PA's and pis to show that F = dPA /\\ dqA and F = dpA. /\\", " dqA, ostensibly different, are actually expressions for one and the same 2-form in terms of alternative sets of I-forms. Exercise 4.3. A CLOSED OR CURL-FREE 1-FORM IS A GRADIENT Given a I-form u such that du = 0, show that u can be expressed in the form u = df, wherefis some scalar. The I-form u is said to be \"curl-free,\" a narrower category of I-form than the \"rotation-free\" I-form of the next exercise (expressible as u = h df), and it in turn is narrower (see Figure 4.7) than the category of \"I-forms with rotation\" (not expressible in the form u = h df). When the I-form u is expressed in terms of basis I-forms dx a, multiplied by corresponding components Ga, show that \"curl-free\" implies G[a,13l = O. Exercise 4.4. CANONICAL EXPRESSION FOR A ROTATION-FREE 1-FORM In three dimensions a rigid body turning with angular velocity w about the z-axis has components of velocity vy= wx, and vr = -wy. The quantity curl v = V X v has z-com ponent equal to 2w, and all other components equal to zero. Thus the scalar product of v and curl I' vanishes: The same concept generalizes to four dimensions, and lends itself to expression in coordinate-free language, as the requirement that a certain 3-form must vanish: dv /\\ v = O. 124 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Any I-form v satisfying this condition is said to be \"rotation-free.\" Show that a I-form is rotation-free if and only if it can be written in the form v = h df, where hand f are scalar functions of position (the \"Frobenius theorem\"). Exercise 4.5. FORMS ENDOWED WITH POLAR SINGULARITIES List the principal results on how such forms are representable, such as and the conditions under which each applies [for the meaning and answer to this exercise, see Lascoux (1968)]. Exercise 4.6. THE FIELD OF THE OSCILLATING DIPOLE Verify that the expressions given for the electromagnetic field of an oscillating dipole in equations (4.23) and (4.24) satisfy dF = 0 everywhere and d*F = 0 everywhere except at the origin. Exercise 4.7. THE 2-FORM MACHINERY TRANSLATED INTO TENSOR MACHINERY This exercise is stated at the end of the legend caption of Figure 4.1. Exercise 4.8. PANCAKING THE COULOMB FIELD Figure 4.5 shows a spacelike slice, t = const, through the Maxwell of a point-charge at rest. By the following pictorial steps, verify that the electric-field lines get compressed into the transverse direction when viewed from a moving Lorentz frame: (I) Draw a picture of an equatorial slice (8 =?T/2; t, r, cf> variable) through Maxwell = *F. (2) Draw various space like slices, corresponding to constant time in various Lorentz frames, through the result ant geometric structure. (3) Interpret the intersection of Maxwell = *F with each Lorentz slice in the manner of Figure 4.3. Exercise 4.9. COMPUTATION OF SURFACE INTEGRALS In Box 4.1 the definition fa = f\u00b7\u00b7\u00b7f (a, ;0 A... A ;~) d'A1... d'AP is given for the integral of a p-form a over a p-surface P('A 1, \u2022\u2022\u2022,'A P) in n-dimensional space. From this show that the following computational rule (also given in Box 4.1) works: (I) substitute the equation for the surface, into a and collect terms in the form (2) integrate fa = f\u00b7\u00b7.f a('A1,...,'AP)d'A1... d'A P using the elementary definition of integration. \u00a74.7. DISTANT ACTION FROM LOCAL LAW 125 Exercise 4.10. WHITAKER'S CALUMOID, OR, THE LIFE OF A LOOP Take a closed loop, bounding a 2-dimensional surface S. It entraps a certain flux of Faraday tPF :::: IsF (\"magnetic tubes\") and a certain flux of Maxwell tPM :::: Is*F (\"electric tubes\"). (a) Show that the fluxes tPF and tPM depend only on the choice of loop, and not on the choice of the surface S bounded by the loop, if and only if d", "F :::: d*F :::: 0 (no magnetic charge; no electric charge). Hint: use generalized Stokes theorem, Boxes 4.1 and 4.6. (b) Move the loop in space and time so that it continues to entrap the same two fluxes. Move it forward a little more here, a little less there, so that it continues to do so. In this way trace out a 2-dimensional surface (\"calumoid\"; see E. T. Whitaker 1904) '3' :::: '3'(a, b); the elementary bivector in this surface, E :::: 0'3'loa /\\ 0'3'lob xIJ. satisfies (F, E) :::: 0 and (*F, E) :::: O. :::: xIJ.(a, b). Show that (c) Show that these differential equations for xIJ.(a, b) can possess a solution, with given :::: xIJ.(a,O) for the initial location of the loop, if dF :::: 0 and d*F :::: 0 initial condition xIJ. (no magnetic charge, no electric charge). (d) Consider a static, uniform electric field F:::: -Ex dt /\\ dx. Solve the equations, (F, E) :::: 0 and (*F, E) :::: 0 to find the equation '3'(a, b) for the most general calumoid. [Answer: y :::: y(a), z:::: z(a), x:::: x(b), t:::: t(b).] Exhibit two special cases: (i) a calumoid that lies entirely in a hypersurface of constant time [loop moves at infinite velocity; analogous to super-light velocity of point of crossing for two blades of a pair of scissors]; (ii) a calumoid whose loop remains forever at rest in the t, x, y, z Lorentz frame. Exercise 4.11. DIFFERENTIAL FORMS AND HAMILTONIAN MECHANICS Consider a dynamic system endowed with two degrees of freedom. For the definition of this system as a Hamiltonian system (special case: here the Hamiltonian is independent of time), one needs (I) a definition of canonical variables (see Box 4.5) and (2) a knowledge of the Hamiltonian H as a function of the coordinates ql, q2 and the canonically conjugate momenta Pl' P2' To derive the laws of mechanics, consider the five-dimensional space of Pl, P2' ql, q2, and t, and a curve in this space leading from starting values of the five coordinates (subscript A) to final values (subscript B), and the value B 1:::: f Pl dql + P2 d q 2 - H(p, q) dt :::: f W B A A of the integral 1 taken along this path. The difference of the integral for two \"neigh boring\" paths enclosing a two-dimensional region S, according to the theorem of Stokes (Boxes 4.1 and 4.6), is 81:::: \u00a2.w :::: f dw. s The principle of least action (principle of \"extremal history\") states that the representative point of the system must travel along a route in the five-dimensional manifold (route with tangent vector d'3' Idt) such that the variation vanishes for this path: i.e., dW(...,d'3'Idt) :::: 0 (2-form dw with a single vector argument supplied, and other slot left unfilled, gives the I-form in 5-space that must vanish). This fixes only the direction of d!i'ldt; its magnitude can be normalized by requiring (dt, ti':Pjdt) :::: 1. (a) Evaluate dw from the expression w :::: Pi dqi - H dt. (b) Set ti'.i'/dt:::: qJ(o'3'loqi) + N0'3'loPi) + l(o':Plot), and expand dw(...,d'3'ldt):::: 0 in terms of the basis {dpi' dqk, dt}. 126 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Box 4.5 METRIC STRUCTURE AND HAMILTONIAN OR \"SYMPLECTIC STRUCTURE\" COMPARED AND CONTRASTED Metric structure Symplectic structure I. Physical application 2. Canonical structure Geometry of spacetime (... '...) = '\"ds 2\" = -dt \u00ae dt e = dpl 1\\ dql + dP2 1\\ dq2 Hamiltonian mechanics", " +dx\u00aedx+dv\u00aedl' + dz \u00ae d:. \u2022 3. Nature of \"metric\" Symmetric Antisymmetric 4. Name for given coordinate system and any other set of four coordinates in which metric has same form 5. Field equation for this metric Lorentz coordinate system System of \"canonically\" (or \"dynamically\") conjugate coordinates R~,a{J = 0 (zero Riemann curvature; flat spacetime) de = 0 ('\"closed 2-form\"; condition automatically satisfied by expression above). 6. The four-dimensional manifold Spacetime Phase space 7. Coordinate-free description of the structure of this manifold Riemann = 0 8. Canonical coordinates distinguished from other coordinates (allowable but less simple) Make metric take above form Make metric take above form (item 2) (item 2) (c) Show that this five-dimensional equation can be written in the 4-dimensional phase space of {qi,pd as e(..., d'3' j dt) = dH, where e is the 2-form defined in Box 4.5. (d) Show that the components of'e(..., d'3' jdt) = dH in the {qi,Pk} coordinate system are the familiar Hamilton equations. Note that this conclusion depends only on the form assumed for e, so that one also obtains the standard Hamilton equations in any other phase-space coordinates {iii, jid (\"canonical variables\") for which Exercise 4.12. SYMMETRY OPERATIONS AS TENSORS We define the meaning of square and round brackets enclosing a set of indices as follows: \u00a74.7. DISTANT ACTION FROM LOCAL LAW 127 Box 4.6 BIRTH OF STOKES' THEOREM Central to the mathematical formulation of electromagnetism are the theorems of Gauss (taken up in Chapter 5) and Stokes. Both today appear together as one unity when expressed in the language of forms. In earlier times the unity was not evident. Everitt (1970) recalls the history of Stokes' theorem: \"The Smith's Prize paper set by [G. C.] Stokes [Lucasian Professor of Mathematics] and taken by Maxwell in [February] 1854... 5. Given the centre and two points of an ellipse, and the length of the major axis, find its direction by a geometrical construction. 6. Integrate the differential equation (a2 _ x 2) dy 2 + 2xydydx + (a2 _ y2) dx2 = O. Has it a singular solution? 7. In a double system of curves of double curvature, a tangent is always drawn at the variable point P; shew that, as P moves away from an arbitrary fixed point Q, it must begin to move along a generating line of an elliptic cone having Q for vertex in order that consecutive tangents may ultimately intersect, but that the conditions of the problem may be impossible. 8. If X, Y, Z be functions of the rectangular co-ordinates x, y, z, dS an element of any limited surface, I, m, n the cosines of the inclinations of the normal at dS to the axes, dsan element of the bounding line, shew that f.J{ /(dZ _ dY) + m(dX _ dZ) + n(dY _ dX)}dS dx dx dy dy dz dz dZ) = X ds + Y ds + Z ds f( dx dy ds, the differential coefficients of X, Y, Z being partial, and the single integral being taken all round the perimeter of the surface marks the first appearance in print of the formula connecting line and surface integrals now known as Stokes' theorem. This was of great importance to Maxwell's development of electromagnetic theory. The earliest explicit proof of the theorem appears to be that given in a letter from Thomson to Stokes dated July 2, 1850.\" [Quoted in Campbell and Garnett (1882), pp. 186-187.] 128 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Here the sum is taken over all permutations 'IT of the numbers I, 2,..., p, and (- I)\"\" is + I or - I depending on whether the permutation is even or odd. The quantity V may have other indices, not shown here, besides the set of p indices aI' a 2,...,ap'but only this set of indices is affected by the operations described here. The numbers \"I' 'lT 2,. \u2022., 'lTp are the numbers 1,2,...,p rearranged according to the permutation \". (Casesp = 2,3 were treated in exercise 3.12", " they too contribute to the river. How can the flow of the river be quantified? By means of a linear machine: the stress-energy tensor T. Choose a small, three-dimensional parallelepiped in spacetime with vectors A, B, C for edges (Figure 5.1). Ask how much 4-momentum crosses that volume in r \\.. \u00a75.1. TRACK-1 OVERVIEW 131 Box 5.1 CHAPTER 5 SUMMARIZED A. STRESS-ENERGY TENSOR AS A MACHINE At each event in spacetime, there exists a stress-energy tensor. It is a machine that contains a knowledge of the energy density, momentum density, and stress as measured by any and all observers at that event. Included are energy, momentum, and stress associated with all forms of matter and all nongravitational fields. The stress-energy tensor is a linear, symmetric machine with two slots for the insertion of two vectors: T(...,...). Its output, for given input, can be summarized as follows. (1) Insert the 4-ve1ocity u of an observer into one of the slots; leave the other slot empty. The output is T(u,...) = T(...,u) = - density of 4-momentum, \"dpjdV,\" i.e., 4-momentum per unit of three-dimensional volume, as measured in observer's Lorentz frame at event where T is chosen i.e., T\"/3u/3 = Tp\"u/3 = -(dp<>jdV) for observer with 4-velocity U<>. (2) Insert 4-velocity of observer into one slot; insert an arbitrary unit vector n into the other slot. The output is. Tun=Tnu=- (, ) (, ) component, \"n' dpjdV\", of ) 4-momentum density along the n directIOn, as measured In..,.. observer's Lorentz frame ( i.e., T<>/3u<>n/3 = T<>/3n<>u/3 = -n/Ldp/LjdV (3) Insert 4-ve1ocity of observer into both slots. The output is the density of mass energy that he measures in his Lorentz frame: T(u, u) = (mass-en~rgy per u~it volume.as measured). In frame with 4-ve1oclty u (4) Pick an observer and choose two space1ike basis vectors, ej and e k, of his Lorentz frame. Insert e j and ek into the slots of T The output is the i,k component of the stress as measured by that observer: 1jk = T(ei, e k) = Tki = T(ek, e j ) ) E to side x k + E, i-component of force acting from side x k across a unit surface area with perpendicular direction e k - ( k-COmponent of force acting from side x j E to side xi + E, = across a unit surface area with perpendicular direction e i ( - ). 132 5. STRESS-ENERGY TENSOR AND CONSERVATION LAWS Box 5.1 (continued) B. STRESS-ENERGY TENSOR FOR A PERFECT FLUID One type of matter studied extensively later in this book is a \"perfect fluid.\" A perfect fluid is a fluid or gas that (1) moves through spacetime with a 4-velocity u which may vary from event to event, and (2) exhibits a density of mass-energy p and an isotropic pressure p in the rest frame of each fluid element. Shear stresses, anisotropic pressures, and viscosity must be absent, or the fluid is not perfect. The stress-energy tensor for a perfect fluid at a given event can be constructed from the metric tensor, g, the 4-ve1ocity, u, and the rest-frame density and pressure, p and p: T = (p + p)u \u00ae u + pg, In the fluid's rest frame, the components of this stress-energy tensor have the expected form (insert into a slot of T, as 4-velocity of observer, just the fluid's 4-velocity): T\"/3u/3 = [(p + p)u\"u/3 + p8\"/3]u/3 = -(p + p)u\" + pu\" = -pu\"; I.e., also TO/3u/3 = -p = -(mass-energy density) = -dpo/dV, Tj/3u/3 = 0 = -(momentum density", ") = -dpiJdV; T;k = T(ej' ek) = p8jk = stress-tensor components. C. CONSERVATION OF ENERGY-MOMENTUM In electrodynamics the conservation of charge can be expressed by the differential equation a(charge density)/at + V' (current density) = 0; i.e., JO,o + V' J = 0; energy-momentum can be expressed by the fundamental geometric law i.e., V' J = O. Similarly, conservation of i.e. J\".\" = 0; V\u00b7T=O. (Because T is symmetric, it does not matter on which slot the divergence is taken.) This law plays an important role in gravitation theory. \u00a75.2. THREE-VOLUMES AND STRESS-ENERGY TENSOR 133 ~y x Figure 5.1. The \"river\" of 4-momentum flowing through spacetime, and three different 3-volumes across which it flows. (One dimension is suppressed from the picture; so the 3-volumes look like 2-volumes.) The first 3-volume is the interior of a cubical soap box momentarily at rest in the depicted Lorentz frame. Its edges are Lez'Ley, Le.: and its volume I-form, with \"positive\" sense toward future (\"standard orientation\"), is E = L3 dl = - Vu( V = L3 = volume as measured in rest frame; u = - dl = 4-velocity of box). The second 3-volume is the \"world sheet\" swept out in time LiT by the top of a second cubical box. The box top's edges are Lez and Le.: and its volume I-form, with \"positive\" sense away from the box's interior, in direction of increasing y, is E = U LiT ely = (f LiTO\" \u00abf = U = area of box top; 0\" = dy = unit I-form containing world tUbe). The third 3-volume is an arbitrary one, with edges A, B, C and volume I-form I~ = (\"\"lJyAaBIJCY. its positive sense (i.e., from its \"negative side\" toward its \"positive side\"). To calculate the answer: (1) Construct the \"volume I-form\" Mathematical representation of 3-volumes (5.1) the parallelepiped lies in one of the I-form surfaces, and the positive sense across the parallelepiped is defined to be the positive sense of the I-form E. (2) Insert this volume I-form into the second slot of the stress-energy tensor T. The result is T(..., E) = p = emp~ slot momentum crossing from negative side toward positive side. ) ( (5.2) (3) To get the projection of the 4-momentum along a vector w or I-form a, insert the volume I-form E into the second slot and w or a into the first: T(w,E) = w'p, T(a, E) = (a,p). (5.3) This defines the stress-energy tensor. Momentum crossing a 3-volume calculated, using stress-energy tensor Interior of a soap box: 134 5. STRESS-ENERGY TENSOR AND CONSERVATION LAWS The key features of 3-volumes and the stress-energy tensor are encapsulated by the above three-step procedure. But encapsulation is not sufficient; deep under standing is also required. To gain it, one must study special cases, both of 3-volumes and of the operation of the stress-energy machinery. A Special Case A soap box moves through spacetime. A man at an event tj'0 on the box's world line peers inside it, and examines all the soap, air, and electromagnetic fields it \u20acontains. He adds up all their 4-momenta to get a grand total Pboxat90' How much is this grand total? One can calculate it by noting that the 4-momentum inside the box at ~Po is precisely the 4-momentum crossing the box from past toward future there (Figure 5.1). Hence, the 4-momentum the man measures is Pboxat90 = T(..., E), (5.4) Its volume l-form where E is the box's volume I-form at?lo' But for such a soap box, E has a magnitude equal to the box's volume Vas measured by a man in its momentary rest frame, and the box itself lies in one of the hyperplanes of E; equivalently, Its 4-momentum content where u is the soap box's", " 4-velocity at?lo (minus sign because u, regarded as a I-form, has positive sense toward the past, Uo < 0); see Box 5.2. Hence, the total 4-momentum inside the box is E = -Vu, (5.5) Pboxat90 = T(..., - Vu) = - VT(..., u), or, in component notation, (5.6) (5.6') The energy in the box, as measured in its rest frame, the 4-momentum on the box's 4-velocity: is minus the projection of E = -U'Pbox at?o = + VTc<f!ucruf! = VT(u, u); so Its energy density energy density as1 measured in box's =.~ = T(u, u). rest frame ( (5.7) Another Special Case A man riding with the same soap box opens its top and pours out some soap. In a very small interval of time LIT, how much total 4-momentum flows out of the box? Box 5.2 THREE-DIMENSIONAL VOLUMES A. General Parallelepiped 1. Edges of parallelepiped are three vectors A, B, C. One must order the edges; e.g., \"A is followed by B is followed by C.\" B (One dimension, that orthogonal to the parallelepiped, is suppressed here.) 2. Volume trivector is defined to be A 1\\ B 1\\ C. It enters into the sophisticated theory of volumes (Chapter 4), but is not used much in the elementary theory. 3. Volume I-form is defined by ~/l = f.p.a!3yAaB!3CY. (A, B, C must appear here in standard order as chosen in step 1.) Note that the vector \"corresponding\" to X and the volume trivector are related by X = - *(A 1\\ B 1\\ C). 4. Orientation of the volume is defined to agree with the orientation of its I-form X. More specifically: the edges A, B, C lie in a hyper plane of X\u00abX, A) = (X, B) = (X, C) = 0; no \"bongs of bell\"). Thus, the volume itself is one of X's hyperplanes! The positive sense moving away from the volume is defined to be the positive sense of X. Note: reversing the order of A, B, C reverses the positive sense! (One dimension, that along which C extends, is suppressed here.) 5. The \"standard orientation\" for a spacelike 3-volume has the positive sense of the I-form X toward the future, corresponding toA, B, C forming a righthanded triad of vectors. B. 3-Volumes of Arbitrary Shape Can be analyzed by being broken up into union of parallelepipeds. C. Interior of a Soap Box (Example) 1. Analysis in soap box's rest frame. Pick an event on the box's world line. The box's three edges there are three specific vectors A, B, C. In the box's rest frame they are purely spatial: AD = BO = Co = O. Hence, the volume I-form has components ~ i = 0 and Box 5.2 (continued) Al A2 A31 ~ 0 = (OiikAiBiCk = det BI B2 B3 CI C2 C3\\ =A \u2022 (B X C), in the standard notation of 3-dimensional vector analysis; = + V (V = volume of box) if (A, B, C) are righthand ordered (positive sense of E toward future; standard orientation); = - V (V = volume of box) if(A, B, C) are lefthand ordered (positive sense of E toward past). 2. This result reexpressed in geometric language: Let u be the box's 4-velocity and V be its volume, as measured in its rest frame. Then either E = - Vu, in which case the \"positive side\" of the box's 3-surface is the future side, and its edges are ordered in a righthanded manner-the standard orien tation; or else E = + Vu, in which case the \"positive side\" is the past side, and the box's edges are ordered in a lefthanded manner. D. 3-Volume Swept Out in Time LiT by Two-Dimensional Top of a Soap Box (Example) 1. Analysis in box's rest frame: Pick an event on box's world line. There the two edges of the box top are vectors A and B. In the box's rest frame, orient the space axes so that A", " and B lie in the y,z-plane. During the lapse of a proper time.1T, the box top sweeps out a 3-volume whose third edge is u.1T (u = 4-velocity of box). In the box's rest-frame, with ordering \"A followed by B followed by u.1T,\" the volume I-form has components, and ~ 1 = (likoAiB k.1TUO= -(01i0i Bk.1T = -tl.1T (tl = area of box top) if (ez'A, B) are righthand ordered = +tl.1T (tl = area of box top) if (ez'A, B) are lefthand ordered. (Note: No standard orientation can be defined in this case, because E can be carried continuously into - E by purely spatial rotations.) 2. This result reexpressed in geometric language: Let tl be the area of the box top as measured in its rest frame; and let u be a unit I-form, one of whose surfaces contains the box top and its 4-velocity (i.e., contains the box top's \"world sheet\"). Orient the positive sense of u with the (arbitrarily chosen) positive sense of the box-top 3-volume. Then E = tl.1TU. \u00a75.3. COMPONENTS OF STRESS-ENERGY TENSOR 137 To answer this question, consider the three-dimensional volume swept out during LiT by the box's opened two-dimensional top (\"world sheet of top\"). The 4-momen tum asked for is the 4-momentum that crosses this world sheet in the positive sense (see Figure 5.1); hence, it is The top of a soap box: Pflowsout =.T(...,E), (5.8) where E is the world sheet's volume I-form. Let tl be the area of the box top, and. Its volume 1-form u be the outward-oriented unit I-form, whose surfaces contain the world sheet (i.e., contain the box top and its momentary 4-velocity vector). Then E = tl LiTU (see Box 5.2); so the 4-momentum that flows out during LiT is Pflowsout = tl LiT T(..., u). (5.9) (5.10) Its 4-momentum that flows across \u00a75.3. COMPONENTS OF STRESS-ENERGY TENSOR Like all other tensors, the stress-energy tensor is a machine whose definition and significance transcend coordinate systems and reference frames. But anyone ob server, locked as he is into some one Lorentz frame, pays more attention to the components of T than to T itself. To each component he ascribes a specific physical significance. Of greatest interest, perhaps, is the \"time-time\" component. It is the total density of mass-energy as measured in the observer's Lorentz frame: interpretation of Physical stress-energy tensor's components: Too = - To0 ~ TOo = T(eo, eo) = density of mass-energy (5.11) roo: energy density (cr. equation 5.7, with the observer's 4-velocity u replaced by the basis vector eo = u). The \"spacetime\" components TiO can be interpreted by considering the interior of a soap box at rest in the observer's frame. If its volume is V, then its volume I-form is E = - Vu = + V dt; and the JL-component of 4-momentum inside it is piJ. = (dxiJ.,p) = T(dxiJ., E) = VT(dxiJ., dt) = Vpo. Thus, the 4-momentum per unit volume is or, equivalently: TOo = density of mass-energy (units: g/cm3, or erg/cm3, or cm- 2); TiO = density of j-component of momentum (units: g (em/sec) cm-3, or cm- 2). (5.12a) (5.13a) (5.13b) rio: momentum density The components TiJ. k can be interpreted using a two-dimensional surface of area tl, at rest in the observer's frame with positive normal pointing in the k-direction. y 138 5. STRESS-ENERGY TENSOR AND CONSERVATION LAWS During the lapse of time Lit, this 2-surface sweeps out a 3-volume with volume I-form E = {/ Lit dx k (see Box 5.2). The JI.-component of 4-momentum that crosses the", " 2-surface in time Lit is the flux of 4-momentum (4-momentum crossing a unit surface oriented Thus, perpendicular to e k, in unit time) is TOk: energy flux Ti k : stress or, equivalently: '['Ok = k-component of energy flux (units: erg/cm2 sec, or cm- 2); Tik = i, k component of \"stress\" _ k-component of flux ofi-component of momentum =i-component of force produced by fields and matter at x k E: acting on fields and matter at x k + E: across a unit surface, the perpendicular to which is e k (units: dynes/cm2, or cm-2). - (5.12b) (5.13c) (5.13d) (Recall that \"momentum transfer per second\" is the same as \"force.\") The stress-energy tensor is necessarily symmetric, Ta/3 = T/3 a ; but the proof of this will be delayed until several illustrations have been examined. \u00a75.4. STRESS-ENERGY TENSOR FOR A SWARM OF PARTICLES Consider a swarm of particles. Choose some event'!! inside the swarm. Divide the particles near'!! into categories, A = 1, 2,..., in such a way that all particles in the same categof'j have the same properties: rest mass; 4-velocity; 4-momentum. Number-flux vector for swarm of particles defined Let N(A) be the number of category-A particles per unit volume, as measured in the particles' own rest frame. Then the \"number-flux vector\" S(A), defined by (5.14) has components with simple physical meanings. In a frame where category-A parti cles have ordinary velocity vw, the meanings are: \u00a75.5. STRESS-ENERGY TENSOR FOR A PERFECT FLUID S~) = N(A)u~) = N(A) [l - V(A)2t 1 / 2 = number density; ~~ Number density in particles' rest frame Lorentz contraction factor for volume S(A) = N(A)u(A) = S~)v(A) = flux of particles. 139 (5.l5a) (5.l5b) Consequently, the 4-momentum density has components Tl1) =pLi)S~) = m(A)uLi)N(A)U~) =m(A)N(A)uLi)u~); and the flux of JI.-component of momentum across a surface with perpendicular direction e j is T(1) = pLi)Sk) = m(A)uLi)N(A)ulA) = m(A)N(A)uLi)ulA)\u00b7 These equations are precisely the JI., 0 and JI.,j components of the geometric, frame independent equation (5.16) Stress-energy tensor for swarm of particles The total number-flux vector and stress-energy tensor for all particles in the swarm near?l are obtained by summing over all categories: T = ~ m(A)N(A)u(A) \u00ae u(A) = ~ P(A) \u00ae S(A). A A (5.17) (5.18) \u00a75.5. STRESS-ENERGY TENSOR FOR A PERFECT FLUID There is no simpler example of a fluid than a gas of noninteracting particles (\"ideal gas\") in which the velocities of the particles are distributed isotropically. In the Lorentz frame where isotropy obtains, symmetry argues equality of the diagonal space-space components of the stress-energy tensor, Ideal gas defined and vanishing of all the Off-diagonal components. Moreover, (5.19) represents a product: the number of particles per unit volume, multiplied by velocity in the x-direction (giving flux in the x-direction) and by momentum in the x-direction, (5.19) 140 5. STRESS-ENERGY TENSOR AND CONSERVATION LAWS giving the standard kinetic-theory expression for the pressure, p. Therefore, the stress-energy tensor takes the form 5.20) in this special Lorentz frame-the \"rest frame\" of the gas. Here the quantity p has nothing directly to do with the rest-masses of the constituent particles. It measures the density of rest-pius-kinetic energy of these particles. Rewrite (5.20) in terms of the 4-velocity ua = (1, 0,0,0) of the fluid in the gas's rest frame, and find p 0 0", " 0 0 0 0 0 TaP = 0p 0 0 0 p = puaup + p(YJaP + uaup ), or, in frame-independent, geometric language T = pg + (p + p)u \u00ae u. (5.21 ) Expression (5.21) has general application. It is exact for the \"ideal gas\" just consid ered. It is also exact for any fluid that is \"perfect\" in the sense that it is free of such transport processes as heat conduction and viscosity, and therefore (in the rest frame) free of shear stress (diagonal stress tensor; diagonal components identical, because if they were not identical, a rotation of the frame of reference would reveal fluid, density p of presence of shear stress). However, for a general perfect mass-energy as measured in the fluid's rest frame includes not only rest mass plus kinetic energy of particles, but also energy of compression, energy of nuclear binding, and all other sources of mass-energy [total density of mass-energy as it might be determined by an idealized experiment, such as that depicted in Figure 1.12, with the sample mass at the center of the sphere, and the test particle executing oscillations of small amplitude about that location, with w2 = (4'IT/3)p]. \u00a75.6. ELECTROMAGNETIC STRESS-ENERGY Faraday, with his picture of tensions along lines of force and pressures at right angles to them (Figure 5.2), won insight into new features of electromagnetism. In addition to the tension E2/8'IT (or B2/8'IT) along lines of force, and an equal pressure at right angles, one has the Poynting flux (E X B)/4'IT and the Maxwell expression for the Stress-energy tensor for ideal gas or perfect fluid Perfect fluid defined \u00a75.7. SYMMETRY OF THE STRESS-ENERGY TENSOR 141 Figure 5.2. Faraday stresses at work. When the electromagnet is connected to an alternating current, the aluminum ring flies into the air. energy density, (\u00a32 + B2)/87T. All these quantities find their places in the Maxwell stress-energy tensor, defined by (5.22) Stress-energy tensor for electromagnetic field Exercise 5.1. EXERCISE Show that expression (5.22), evaluated in a Lorentz coordinate frame, gives roo = (\u00a32.-t- B2)/87T, Tik =J... [_(\u00a3i\u00a3k + BiBk) +.!. (\u00a32 + B2) 8ik]. TOi = TiO= (\u00a3 X B)i/47T, 47T 2 (5.23) Show that the stress tensor does describe a tension (\u00a32 + B2)/87T along the field lines and a pressure (\u00a32 + B2)/87T perpendicular to the field lines, as stated in the text. \u00a75.7. SYMMETRY OF THE STRESS-ENERGY TENSOR All the stress-energy tensors explored above were symmetric. That they could not have been otherwise one sees as follows. Calculate in a specific Lorentz frame. Consider first the momentum density (components TiO) and the energy flux (components TOi). They must be equal because energy = mass (\"E = Mc 2 = M\"): Proof that stress-energy tensor is symmetric TOi = (energy flux) = (energy density) X (mean velocity of energy flow)i = (mass density) X (mean velocity of mass flow)i = (momentum density) = Tio. (5.24) Only the stress tensor Ti k remains. For it, one uses the same standard argument as in Newtonian theory. Consider a very small cube, of side L, mass-energy TOOL3, 142 5. STRESS-ENERGY TENSOR AND CONSERVATION LAWS and moment of inertia _7\"loL5. With the space coordinates centered at the cube, the expression for the z-component of torque exerted on the cube by its surround ings is (L/2) + '----\"\" (pz[2)'-.. -'(y~~o;:~~n::t)(a~:e~o) (y~~of~~~n::tVa~:e~o ) - x face A- x face +x face +x face (-FYL2) (L/2) (FY[2) (-L/2), I~ ~ (X~~of~~~n::t)fa~e;e~o) (X~~of:~n::tVa~:e~o ) +y face ~y face - y face A-y face Since", " the torque decreases only as L3 with decreasing L, while the moment of inertia decreases as L5, the torque will set an arbitrarily small cube into arbitrarily great angular acceleration-which is absurd. To avoid this, the stresses distribute them selves so the torque vanishes: Put differently, if the stresses were not so distributed, the resultant infinite angular accelerations would instantaneously redistribute them back to equilibrium. This condition of torque balance, repeated for all other pairs of directions, is equivalent to symmetry of the stresses: (5.25) \u00a75.8. CONSERVATION OF 4-MOMENTUM: INTEGRAL FORMULATION Energy-momentum conservation has been a cornerstone of physics for more than a century. Nowhere does its essence shine forth so clearly as in Einstein's geometric formulation of it (Figure 5.3,a). There one examines a four-dimensional region of spacetime 'V bounded by a closed, three-dimensional surface a'V. As particles and fields flow into 'V and later out, they carry 4-momentum. Inside 'V the particles collide, break up, radiate; radiation propagates, jiggles particles, produces pairs. But at each stage in this complex maze of physical processes, total energy-momentum is conserved. The energy-momentum lost by particles goes into fields; the energy momentum lost by fields goes into particles. So finally, when the \"river\" of 4-mo mentum exits from 'V, it carries out precisely the same energy-momentum as it carried in. Restate this equality by asking for the total flux of 4-momentum outward across a'V. Count inflowing 4-momentum negatively. Then \"inflow equals outflow\" means \"total outflow vanishes\": Integral conservation law for 4-momentum: f T\u00b7 d 3E = 0 C'1 a'Y \\ -d \\0 s... ;;:;0 ~ / /J/f/ /III '~~~~, /, (a) '\\ _-r r- /.......-----\"7'iI\u00a3----,----''-/ \"ll-l- ---::\"i\"\"\"\"'---+---4---X \\ t \\ 1 \\ a-r = a('l-2 - -1'1) = 5 - 5. I (c) ~ I I /, \\ I ----l-----t---- X I I t a-Y = 5 - 5 +~\"1 x I I '\",,..., \u2022I I -1' 52 51....., \\ I... / X a'r = 52 - 51 (b) I / / / I \"\"\"--__-\"T\"-'... I I I \\ \\ - - - - - \"'V ___+-+ ------~ X I\u2022 (d) Figure 5.3. (a) A four-dimensional region of spacetime '1' bounded by a closed three-dimensional surface a-r. The positive sense of a'r is defined to be everywhere outward (away from -1'). Conservation of energy momentum demands that every bit of 4-momentum which flows into -1' through c-r must somewh~re flow back out; none can g~t lost inside; the total flux of 4-momentum across c-l - in the positive (outward) sense must be zero: the interior contains no \u00b7-sinks.\" Equivalently. (e) 1 P\"'d:lJ:u=O. a,.. Figures (b). (c). (d). and (e) depict examples to which the text applies this law of conservation of 4-momentum. All symbols -1' (or S) in these figures mean spacetime volumes (or spacelike 3-volumes) with standard orientations. The dotted arrows indicate the positive sense of the closed surface cor used in the text's discussion of 4-momentum conservation. How cor is construct~d from the surfaces 5 and :\"1 is indicated by formulas below the figures. For example. in case (b). cT = 52 - 51 m~ans that c-T' is mad~ by joining together S~ with its stand:trd orientation and 51 with reversed orientation. 144 5. STRESS-ENERGY TENSOR AND CONSERVATION LAWS Total flux of 4-momentum outward across a closed three-dimensional surface must vanish. (5.26) To calculate the total outward flux in the most elementary of fashions, approximate the closed 3-surface a'Y by a large number of flat 3-volumes (\"boiler plates\") with positive direction oriented outward (away from 'V). Then Ptotalout = boiler plates A (5.27) where Ew IS the volume I-form of boiler plate A. Equivalently, notation III component", " p,utotalout = ~ T,ua2:Wa' A (5.27') To be slightly more sophisticated about the calculation, take the limit as the number of boiler plates goes to infinity and their sizes go to zero. The result is an integral (Box 5.3, at the end of this section), p,utotalout = f T,ua d 32:a = O. aT (5.28) Think of this (like all component equations) as a convenient way to express a coordinate-independent statement: Ptotalout =f T' d a'V 3 E = O. (5.29) To be more sophisticated yet (not recommended on first reading of this book) and to simplify the computations in practical cases, interpret the integrands as exterior differential forms (Box 5.4, at the end of this section). But however one calculates it, and however one interprets the integrands, the statement of the result is simple: the total flux of 4-momentum outward across a closed 3-surface must vanish. Several special cases of this \"integral conservation law,\" shown in Figure 5.3, are instructive. There shown, in addition to the general case (a), are: Case (b) The closed 3-surface a'Y is made \\lP of two slices taken at constant time t of a specific Lorentz frame, plus timelike surfaces at \"infinity\" that join the two slices together. The surfaces at infinity do not contribute to ~a\" Tf.W- d 32:a if the stress-energy tensor dies out rapidly enough there. The boundary a'Y of the standard-oriented 4-volume 'Y, by definition, has its positive sense away from 'Y. This demands nonstandard Special cases of integral conservation law: \u00a75.8. CONSERVATION LAWS: INTEGRAL FORMULATION 145 orientation of 3 1 (positive sense toward past), as is indicated by writing 0'V = 32 - 31; and it produces a sign flip in the evaluation of the hypersurface integral 0= f p:p. d 3Ip. = - f POdxdydz + f P:odxdydz. Clo/ SI S2 Because p.0 is the density of 4-momentum, this equation says total 4-momentum in ) all of space at time t 1 ( =f p:o dxdydz SI f total 4-momentum in) = all of space at time t 2 = S ( 2 (5.30) P.0 dxdy dz. Total 4\u00b7momentum conserved in time Case (c) Here one wants to compare hypersurface integrals over 3 and S, which are slices of constant time, t = const and l = const in two different Lorentz frames. To form a closed surface, one adds time-like hypersurfaces at infinity and assumes they do not contribute to the integral. The orientations fit together smoothly and give a closed surface a'V = S - 3 + (surfaces at infinity) only if one takes 'V = 'V2 'VI-i.e., only if one uses the nonstandard 4-volume orientation in 'V l' (See part A.I of Box 5.3 for \"standard\" versus \"non-standard\" orientation.) The integral conservation law then gives o =LT' d 3E - iT' d 3E, S S or, equivalently, L T\u00b7 d3E = (totaI4-momentum p on S) S =f T' d 3E = (totaI4-momentum p on 3). S Total 4-momentum the same in all Lorentz frames (5.31 ) This says that observers in different Lorentz frames measure the same total 4-mo mentum p. It does not mean that they measure the same components (pa t:. p'Y.); rather, it means they measure the same geometric vector a vector whose components are connected by the usual Lorentz transformation law pa = Aaiip lJ \u2022 (5.32) y Total 4-momentum independent of hypersurface where measured 146 Case (d) 5. STRESS-ENERGY TENSOR AND CONSERVATION LAWS Here the contribution to the integral comes entirely from two arbitrary spacelike hypersurfaces, SA and SB' cutting all the way across spacetime. As in cases (a) and (b), the integral form of the conservation law says (5.33) i.e., the total 4-momentum on a spacelike slice through spacetime is independent of the specific slice chosen-so long as the energy-momentum flux across the \"hypersur face at infinity\" connecting SA and SB is zero. Case (e) This", " case concerns a box whose walls oscillate and accelerate as time passes. The three-dimensional boundary a'Y is made up of (1) the interior S of the box, at an initial moment of time t = constant in the box's initial Lorentz frame, taken with nonstandard orientation; (2) the interior S of the box, at t = constant in its final Lorentz frame, with standard orientation; (3) the 3-volume '!j swept out by the box's two-dimensional faces between the initial and final states, with positive sense oriented outward. The integral conservation law f (J'jf\" T' d 3E = 0 says Change with time of 4-momentum in a box equals flux of 4-momentum across its faces ( total 4-mOEJentum) _ (total 4-momentum) in box at S in box at S = (total 4-momentum that enters ~ox through). (5.34) its faces between states Sand S \u00a75.9. CONSERVATION OF 4-MOMENTUM: DIFFERENTIAL FORMULATION Complementary to any \"integral conservation law in flat spacetime\" is a \"differential conservation law\" with identical information content. To pass back and forth between them, one can use Gauss's theorem. Gauss's theorem in four dimensions, applied to the law of 4-momentum conserva tion, converts the surface integral of Tf.W- into a volume integral of p.a,a: 0= f p a d 3I a = f Tf.W-,adtdxdydz. (J'jf\" '\"- (5.35) (See Box 5.3 for elementary discussion; Box 5.4 for sophisticated discussion.) If the integral of Tf.W-,a is to vanish, as demanded, for any and every 4-volume 'Y, then Tf.W-,a must itself vanish everywhere in spacetime: Tf.W-,a = 0; i.e., V' T = 0 everywhere. (5.36) (continued on page 152) Differential conservation law for 4-momentum: V' T = 0 \u00a75.9. CONSERVATION LAWS: DIFFERENTIAL FORMULATION 147 Box 5.3 VOLUME INTEGRALS. SURFACE INTEGRALS. AND GAUSS'S THEOREM IN COMPONENT NOTATION A. Volume Integrals in Spacetime 1. By analogy with three-dimensional space, piped\" with vector edges A, S, C, 0 is the volume of a \"hyperparallele 4-volume [J = \u00a3O:/3yaAO:B /3 CYD a = det = *(A /\\ S /\\ C /\\ D). Here, as for 3-volumes, orientation matters; interchange of any two edges reverses the sign of [J. The standard orientation for any 4-volume is the one which makec; [J positive; thus, 8 0 /\\ 8 1 /\\ 8 2 /\\ 8 3 has standard orientation if 8 0 points toward the future and 8 1, 8 2, 8 3 are a righthanded triad. 2. The \"volume element\" whose edges in a specific, standard-oriented Lorentz frame are AO: = (Lit, 0, 0, 0), Bo: = (0, Llx, 0, 0), CO: = (0,0, Lly, 0), Do: = (0,0,0, Liz) has a 4-volume, according to the above definition, given by Ll4[J = \u00a30123 Lit Llx Lly Liz = Lit Llx Lly Liz. 3. Thus, the volume integral of a tensor S over a four-dimensional region 'V of spacetime, defined as Lim number Of) (elementarv) vol~e.s it elementary volumes In', ->00 ( Satcenterofil (volume of (/), can be calculated in a Lorentz frame by 148 5. STRESS-ENERGY TENSOR AND CONSERVATION LAWS Box 5.3 (continued) B. Integrals over 3-Surfaces in Spacetime 1. Introduce arbitrary coordinates a, b, c on the three-dimensional surface. The elementary volume bounded by coordinate surfaces ao < a < ao + Lla, Co < c < Co + Llc has edges bo < b < bo + Lib, A\" = ax\" Lla BP = ax P Lib CY = ax Y Llc' ac'ab'aa'so its volume I-form is 2. The integral of a tensor S over the 3-surface S thus has components An equivalent formula involving a Jacobian is often used (see exercise 5.5", "... _ I atmosphere 10-3 g/cm3 P 106 dynes/cm2 = 109 cm2 _ 10-12 c2 = 10-12 <{ 1. 10-3 g/cm3 sec2 The stress-energy tensor for such a fluid has components TOO = (p + p)UOUO_ P ::::: p, TOi = TiO= (p + p)UOui ::::: pvi, Tik = (p + p)Uiu k + P 8ik ::::: pviv k + p 8;k; and the equation of motion V. T = 0 has components TOO,o + TO;,; = ap/at + V' (pv) = 0 (\"equation of continuity\"); and Tio,o + Tik,k = a(pvi)/at + a(pvivk)/ax k + ap/axi = 0, or, equivalently (by combining with the equation of continuity), (5037a) (5037b) (5037c) (5038a) av - + (v' V)v = - - Vp at I p (\"Euler's equation\"). (5038b) Box 5.5 derives and discusses these results from the Newtonian viewpoint. As a second application of V. T = 0, consider a composite system: a block of rubber with electrically charged beads imbedded in it, interacting with an electro magnetic field. The block of rubber vibrates, and its accelerating beads radiate electromagnetic waves; at the same time, incoming electromagnetic waves push on the beads, altering the pattern of vibration of the block of rubber. The interactions shove 4-momentum back and forth between beaded block and electromagnetic field. Newtonian fluid characterized by Ivii ~ 1. P ~ P Stress-energy tensor and equation of motion for a Newtonian fluid Application of..,. T = 0 to an electrically charged. vibrating rubber block \u00a75.10. APPLICATIONS OF \u2022\u2022 T = 0 153 Box 5.5 NEWTONIAN HYDRODYNAMICS REVIEWED Consider a classical, nonrelativistic, perfect fluid. Apply Newton's law F = rna to that is, to a small fixed mass of fluid followed in its progress a \"fluid particle\"; through space:.!!...- (momentum per unit mass) = (force per unit mass) dt or = (force per unit volume) (density) = - (gradient of pressure) (density) dv -= --Vp. dt 1 p (1) Translate from time-rate of change following the fluid to time-rate of change as measured at a fixed location, finding rate of Change). h. tIme r 11 fl'd. 10 owmg UI WIt ( or or (rate of Change). h. = Wit fi xe tIme at + ocatlOn d l'l've OCIty (Of flUId). '... f h rate 0 c ange (WIth pOSItIOn) (2) (Latin indices run from 1 to 3; summation convention; upper and lower indices used indifferently for space dimensions in flat space!) This is Euler'sfundamenral equation for the hydrodynamics of a perfect fluid. Two further equations are needed to complete the description of a perfect fluid. One states the absence of heat transfer by requiring that the specific entropy (entropy per unit mass) be constant for each fluid \"particle\": ds _ 0 dt -, or as- + (v'V)s = o. at The final equation expresses the conservation of mass: ap - + V' (pv) = 0'ct or (3) (4) 154 5. STRESS-ENERGY TENSOR AND CONSERVATION LAWS Box 5.5 (continued) it is analagous in every way to the equation that expresses conservation of charge in electrodynamics and that bears the same name, \"equation of continuity.\" The Newtonian stress-energy tensor, like its relativistic counterpart, is linked to conservation of momentum and mass. Therefore examine the time-rate of change of the density of fluid momentum, PVi'contained in a unit volume; thus, (5) Momentum flows into the little volume element on the left (\"force equals time-rate of change of momentum\") and out on the right; similarly at the other faces. Therefore the right hand side of (5) must represent the divergence of this momentum flux: Consequently, we take for the momentum flux itself a(pVi)/ot = - Tik,k' Tik = Tik = PViVk \"convection\" + 8ikP '--..--' \"push\" For the momentum density, the Newtonian value is 'J'Oi = T", "io = pvi \u2022 (6) (7) (8) With this notation, the equation for the time-rate of change of momentum becomes and with TOo = p, the equation of continuity reads aTiIJ./axIJ. = 0; (9) (10) In conclusion, these Newtonian considerations give a reasonable approximation to the relativistic stress-energy tensor: p : pvi (11) \u00a75.10. APPLICATIONS OF..,. T = 0 155 The 4~momentum of neither block nor field is conserved; neither V' 1(block) nor V' 1(emfieldl vanishes. But total 4-momentum must be conserved, so V' (1(block) + 1(emfieldl) must vanish. (5.39) For a general electromagnetic field interacting with any source, V' 1(emfieldl has the form Tr:mfieldl,v = -FILO'.JO'.. (5.40) (This was derived in exercise 3.18 by combining TILv,v = 0 with expression 5.22 for the electromagnetic stress-energy tensor, and with Maxwell's equations.) For our beaded block, J is the 4-current associated with the vibrating, charged beads, and F is the electromagnetic field tensor. The time component of equation (5.40) reads 11:m fieldl v = - FOkJk = - E \u2022 J __ (rate at which electric field E does WOrk) on a unit volume of charged beads -. (5.41 ) For comparison, 11b\u00b0lock),O is the rate at which the block's energy density changes 11~lOCk),i is the contribution of the block's energy flux to this rate of with time, change of energy density, and consequently their difference 11WOCk),v has the meaning n v (block), v rate at which mass-energy of block per) = unit volum: increases due t.o actions other than mternal mechamcal forces between one part of block and another (. (5.42) Hence, the conservation law says that the mass-energy of the block increases at precisely the same rate as the electric field does work on the beads. A similar result holds for momentum: Tt:mfieldl,vek = -FkvJvek = -(JOE + J X B) __ (LOrentz force per unit VOlUme) - acting on beads'rate at which momentum per unit VOlUme) TtwoCk),vek = of block increases due to actions ( other than its OWn stresses (5.43) ; (5.44) so the conservation law says that the rate of change of the momentum of the block equals the force of the electromagnetic field on its beads. 156 5. STRESS-ENERGY TENSOR AND CONSERVATION LAWS \u00a75.11. ANGULAR MOMENTUM Angular momentum defined and its integral conservation law derived The symmetry, TIL' = T\"IL, of the stress-energy tensor enables one to define a conserved angular momentum lO'./3, analogous to the linear momentum pO'.. The angular momentum is defined relative to a specific but arbitrary origin-an event tl with coordinates, in a particular Lorentz frame, The angular momentum about d is defined using the tensor xO'.(tl) = aO'.. (5.45) (5.46) (Note that xO'. aO'. is the vector separation of the \"field point\" xO'. from the \"origin\" tl; TO'.Y is here evaluated at the \"field point\".) Because of the symmetry of T, jO'./3Y has vanishing divergence: = T/3O'. _ TO'./3 = O. Consequently, its integral over any closed 3-surface vanishes f jO'./3Y d3I y = 0 a\u00b7v (5.47) (5.48) (\"integral form of the law of conservation of angular momentum\"). The integral over a spacelike surface of constant time t is lO'./3 =f jO'./30 dx dy dz =f [(xO'. - aO'.)T/30 - (x/3 - a/3)TO'.O] dx dy dz. (5.49) Recalling that T/30 is momentum density, one sees that (5.49) has the same form as the equation \"i = r X p\" of Newtonian theory. Hence the name \"total angular momentum\" for lO'./3. Various aspects of this conserved angular momentum, including the tie to its Newtonian cousin, are explored in Box 5.6. EXERCISES Exercise 5.2. CHARGE CONSERVATION Exercise 3.", "roid of the system, at his Lorentz time XO = t and in his own Lorentz frame, by XW) = (IIPO)f xo= t xiTOo d3x in Lorentz frame where u = 0'1'loxo. (5.56) This centroid depends on (i) the particular system being studied, (ii) the 4-velocity u of the observer, and (iii) the time t at which the system is observed. (a) Show that the centroid moves with a uniform velocity corresponding to the 4-velocity U = P/M. (5.57) (5.57') Note that this \"4-velocity of centroid\" is independent of the 4-velocity u used in defining the centroid. (b) The centroid associated with the rest frame of the system (i.e., the centroid defined with u = U) is called the center of mass; see Box 5.6. Let (u be a vector reaChing from any event on the center-of-mass world line to any event on the world line of the centroid associated with 4-velocity u; thus the components of (u in any coordinate system are Show that (u satisfies the equation ~~ = X~ - Xu' (5.58) (5.59) [Hint: perform the calculation in a Lorentz frame where u = 2'1'/oxo.) (c) Show that, as seen in the rest-frame of the system at any given moment of time. the above equation reduces to the three-dimensional Euclidean equation ~u = -(v X S)/M, (5.59') where v = u/uo is the ordinary velocity of the frame associated with the centroid. (d) Assume that the energy density measured by any observer anywhere in spacetime is 162 5. STRESS-ENERGY TENSOR AND CONSERVATION LAWS non-negative (u' T' u :2: 0 for all timelike u). In the rest frame of the system, construct the smallest possible cylinder that is parallel to 8 and that contains the entire system (Tap = 0 everywhere outside the cylinder). Show that the radius '0 of this cylinder is limited by Thus, a system with given intrinsic angular momentum 8 and given mass M has a minimum possible size 'Omin = 181/M as measured in its rest frame. '0:2: 18 1/M. (5.60) ---------~ CHAPTER 6 ACCELERATED OBSERVERS The objective world simply is; it does not happen. Only to the gaze of my consciousness, crawling upward along the life line [world line] of my body, does a section of this world come to life as a fleeting image in space which continuously changes in time. HERMAN WEYL (1949. p. 116) \u00a76.1. ACCELERATED OBSERVERS CAN BE ANALYZED USING SPECIAL RELATIVITY It helps in analyzing gravitation to consider a situation where gravity is mocked up by acceleration. Focus attention on a region so far from any attracting matter, and so free of disturbance, that (to some proposed degree of precision) spacetime there can be considered to be flat and to have Lorentz geometry. Let the observer acquire the feeling that he is subject to gravity, either because ofjet rockets strapped to his legs or because he is in a rocket-driven spaceship. How does physics look to him? Dare one answer this question? At this early stage in the book, is one not too ignorant of gravitation physics to predict what physical effects will be measured by an observer who thinks he is in a gravitational field, although he is really in an accelerated spaceship? Quite the contrary; special relativity was developed precisely to predict the physics of accelerated objects-e.g., the radiation from an accelerated charge. Even the fantastic accelerations anuclear __ v 2/r __ 1031 cm/sec2 -- 1028 \"earth gravities\" suffered by a neutron bound in a nucleus. and the even greater accelerations met in high-energy particle-scattering events, are routinely and accurately treated within Accelerated motion and accelerated observers can be analyzed using special relativity Box 6.1 GENERAL RELATIVITY IS BUILT ON SPECIAL RELATIVITY A tourist in a powered interplanetary rocket feels \"gravity.\" Can a physicist by local effects convince him that this \"gravity\" is bogus? Never, says Ein stein's principle of the local equivalence of gravity and accelerations. But then the physicist will make no errors if he deludes himself into treating true gravity as a local illusion caused by acceleration. Under this delusion, he barges ahead and solves gravitational problems by using special relativity: if he is clever enough to divide every problem into a network of local questions, each solvable under such", " a delusion, then he can work out all influ- ences of any gravitational field. Only three basic principles are invoked: special-relativity physics, the equivalence principle, and the local nature of physics. They are simple and clear. To apply them, however, imposes a double task: (I) take space time apart into locally flat pieces (where the prin ciples are valid), and (2) put these pieces together again into a comprehensible picture. To undertake this dissection and reconstitution, to see curved dynamic spacetime inescapably take form, and to see the consequences for physics: that is general relativity. the framework of special relativity. The theoretician who confidently applies special relativity to antiproton annihilations and strange-particle resonances is not about to be frightened off by the mere illusions of a rocket passenger who gullibly believed the travel brochures advertising \"earth gravity all the way.\" When spacetime is flat, move however one will, special relativity can handle the job. (It can handle bigger jobs too; see Box 6.1.) The essential features of how special relativity handles the job are summarized in Box 6.2 for the benefit of the Track-l reader, who can skip the rest of the chapter, and also for the benefit of the Track-2 reader, who will find it useful background for the rest of the chapter. Box 6.2 ACCELERATED OBSERVERS IN BRIEF An accelerated observer can carry clocks and measuring rods with him, and can use them to set up a reference frame (coordinate system) in his neighborhood. His clocks, if carefully chosen so their structures are affected negligibly by acceler ation (e.g., atomic clocks), will tick at the same rate as unaccelerated clocks moving momentarily along with him: time interval ticked off L1 = by observer's clocks as he T - moves a vector displacement ) ( ( along his world line _ _ - [ g\u00ab(, ()] 1/2. And his rods, if chosen to be sufficiently rigid, will measure the same lengths as \u00a76.1. ACCELERATED OBSERVERS VIA SPECIAL RELATIVITY 165 momentarily comoving, unaccelerated rods do. (For further discussion, see \u00a7 16.4, and Boxes 16.2 to 16.4.) Let the observer's coordinate system be a Cartesian latticework of rods and clocks, with the origin of the lattice always on his world line. He must keep his latticework small: I=(spatial dim.ensions) \u00ab( the acceleration measure~ )-1 --l. of lattlCe by accelerometers he carnes g At distances I away from his world line, strange things of dimensionless magnitude gl happen to his lattice-e.g., the acceleration measured by accelerometers differs from g by a fractional amount -gl (exercise 6.7); also, clocks initially synchronized with the clock on his world line get out of step (tick at different rates) by a fractional amount -gl (exercise 6.6). (Note that an acceleration of one \"earth gravity\" corre sponds to g-1 _ 10- 3 sec2/cm _ 10 18 cm - 1 light-year, so the restriction 1\u00ab l/g is normally not severe.) To deduce the results of experiments and observations performed by an accelerated observer, one can analyze them in coordinate-independent, geometric terms, and then project the results onto the basis vectors of his accelerated frame. Alternatively, one can analyze the experiments and observations in a Lorentz frame, and then transform to the accelerated frame. As deduced in this manner, the results of experiments performed locally (at 1\u00ab l/g) by an accelerated observer differ from the results of the same experiments performed in a Lorentz frame in only three ways: (1) There are complicated fractional differences of order gl \u00ab 1 mentioned above, that can be made negligible by making the accelerated frame small enough. (2) There are Coriolis forces of precisely the same type as are encountered in Newtonian theory (exercise 6.8). These the observer can get rid of by carefully preventing his latticework from rotating-e.g., by tying it to gyroscopes that he accelerates with himself by means of forces applied to their centers of mass (no torque!). Such a nonrotating latticework has \"Fermi-Walker trans ported\" basis vectors (\u00a76.5), where u = 4-velocity, and a = duldT = 4-acceleration. (3) There are inertial forces of precisely the same type as are encountered in Newtonian theory (exercise 6.8). These are due to the observer's acceleration, and he cannot get rid of them except by stopping his accelerating. (1) 166 6. ACCELERATED OBSERVERS \u00a76.2.", " HYPERBOLIC MOTION Study a rocket passenger who feels \"gravity\" because he is being accelerated in flat spacetime. Begin by describing his motion relative to an inertial reference frame. His 4-velocity satisfies the condition u 2 = - I. To say that it is fixed in magnitude is to say that the 4-acceleration, is orthogonal to the 4-velocity: a = du/dT, 0= (d/dT)(-1/2) = (d/dT)(~ u.u) = a\u00b7u. (6.1) (6.2) This equation implies that a O = 0 in the rest frame of the passenger (that Lorentz frame, where, at the instant in question, u = eo); in this frame the space components of aIL reduce to the ordinary definition of acceleration, a i = d 2x i /dt 2\u2022 From the components aIL = (0; a i ) in the rest frame, then, one sees that the magnitude of the acceleration in the rest frame can be computed as the simple invariant Consider, for simplicity, an observer who feels always a constant acceleration g. Take the acceleration to be in the Xl direction of some inertial frame, and take x 2 = x 3 = O. The equations for the motion of the observer in that inertial frame become The rest of this chapter is Track 2. It depends on no preceding Track-2 material. It is needed as preparation for (1) the mathematical (2) analysis of gyroscopes in curved spacetime (exercise 19.2, \u00a740. 7), and the mathematical theory of the proper reference frame of an accelerated observer (\u00a713.6). It will be helpful in many applications of gravitation theory (Chapters 1 B-40). \u00b0 dt -=u, dT Write out the three algebraic equations UILU ulLa IL = -1, lL = -uoao + ula l = 0, and Solve for the acceleration, finding (6.3) (6.4) These linear differential equations can be solved immediately. The solution, with a suitable choice of the origin, reads ( = g-l sinh p, (6.5) Uniformly accelerated observer moves on hyperbola in spacetime diagram (2 = g-2. The world line is a hyperbola in a spacetime diagram Note that x 2 (\"hyperbolic motion\"; Figure 6.1). Several interesting aspects of this motion are \u00a76.2. HYPERBOLIC MOTION 167 \"- \"- \"- \"- \"- \"- \"- \"- \"- / / /U II / / / / / / \"\" / / - - I I I - - - - l IE - - / / / / / // ////// IV I - r t -....---x Figure 6.1. Hyperbolic motion. World line of an object that (or an ob server who) experiences always a fixed acceleration g with respect to an inertial frame that is instantaneously comoving (different inertial frames at different instants!). The 4-acceler alion a is everywhere orthogonal (Lorentz geometry!) to the 4-velocity u. treated in the exercises. Let the magnitude of the constant acceleration g be the acceleration of gravity, g = 980 cm/sec2 experienced on earth: g ~ (l03 cm/sec2)/ (3 X 1010 cm/sec)2 = (3 X 107 sec\u00b7 3 X 1010 cm/sec)-l = (llight-year)-l. Thus the observer will attain relativistic velocities after maintaining this acceleration for something like one year of his own proper time. He can outrun a photon if he has a head start on it of one light-year or more. Exercise 6.1. A TRIP TO THE GALACTIC NUCLEUS EXERCISES Compute the proper time required for the occupants of a rocket ship to travel the -30,000 light-years from the Earth to the center of the Galaxy. Assume that they maintain an acceleration of one \"earth gravity\" (103 cm/sec2 ) for half the trip, and then decelerate at one earth gravity for the remaining half. Exercise 6.2. ROCKET PAYLOAD What fraction of the initial mass of the rocket can be payload for the journey considered in exercise 6.l? Assume an ideal rocket that converts rest mass into radiation and ejects all the radiation out the back of the rocket with 100 per cent efficiency and perfect collimation. Exercise 6.3. TWIN PARADOX (a) Show that. of all timelike world lines connecting two events {/ and ~IJ, the one with the longest lapse of proper time is the unaccelerated one. (Hint: perform the calculation in the inertial frame of the unaccelerated world line.) (", "b) One twin chooses to move from {/ to ~lJ along the unaccelerated world.line. Show that the other twin, by an appropriate choice of accelerations, can get from (/ to ~IJ in arbitrarily small proper time. (c) If the second twin prefers to ride in comfort. with the acceleration he feels never exceeding one earth gravity, g, what is the shortest proper time-lapse he can achieve between {/ and ~iJ? Express the answer in terms of g and the proper time-lapse.1T measured by the unaccelerated twin. (d) Evaluate the answer numerically for several interesting trips. 168 6. ACCELERATED OBSERVERS Exercise 6.4. RADAR SPEED INDICATOR A radar set measures velocity by emitting a signal at a standard frequency and comparing it with the frequency of the signal reflected back by another object. This redshift measurement is then converted, using. the standard special-relativistic formula, into the corresponding velocity, and the radar reads out this velocity. How useful is this radar set as a velocity-mea suring instrument for a uniformly_ accelerated observer? (a) Consider this problem first for the special case where the object and the radar set are at rest with respect to each other at the instant the radar pulse is reflected. Compute the redshift I + Z = we/wo that the radar set measures in this case, and the resulting (incorrect) velocity it infers. Simplify by making use of the symmetries of the situation. (b) Now consider the situation where the object has a non-zero velocity in the momentary rest frame of the observer at the instant it reflects the radar pulse. Compute the ratio of the actual relative velocity to the velocity read out by the radar set. Exercise 6.5. RADAR DISTANCE INDICATOR Use radar as a distance-measuring device. The radar set measures its proper time T between the instant at which it emits a pulse and the later instant when it receives the reflected pulse. It then performs the simple computation L o = T/2 and supplies as output the \"distance\" L o. How accurate is the output reading of the radar set for measuring the actual distance L to the object, when used by a uniformly accelerated observer? (L is defined as the distance in the momentary rest frame of the observer at the instant the pulse is reflected, which is at the observer's proper time halfway between emitting and receiving the pulse.) Give a correct formula relating L o=T/2 to the actual distance L. Show that the reading L o becomes infinite as L approaches g-l, where g is the observer's acceleration, as measured by an\u00b7 accelerometer he carries. Difficulties in constructing \"the coordinate system of an accelerated observer\": Breakdown in communication between observer and events at distance I > (acceleration)-l \u00a76.3. CONSTRAINTS ON SIZE OF AN ACCELERATED FRAME It is very easy to put together the words \"the coordinate system of an accelerated observer,\" but it is much harder to find a concept these words might refer to. The most useful first remark one can make about these words is that, if taken seriously, they are self-contradictory. The definite article \"the\" in this phrase suggests that one is thinking of some unique coordinate system naturally associated with some specified accelerated observer, such as one whose world line is given in equation (6.5). If the coordinate system is indeed natural, one would expect that the coordi na tes of any event could be determined by a sufficiently ingenious observer by sending and receiving light signals. But from Figure 6.1 it is clear that the events composing one quarter of all spacetime (Zone III) can neither send light signals to, nor receive light signals from, the specified observer. Another half of spacetime suffers lesser disabilities in this respect: Zone II cannot send to the observer, Zone IV cannot receive from him. It is hard to see how the observer could define in any natural way a coordinate system covering events with which he has no causal relationship, which he cannot see, and from which he cannot be seen! Difficulties also occur when one considers an observer who begins at rest in one frame, is accelerated for a time, and maintains thereafter a constant velocity, at rest in some other inertial coordinate system. Do his motions define in any natural way \u00a76.4. UNIFORMLY ACCELERATED OBSERVER 169 Figure 6.2. World line of an observer who has undergone a brief period of acceleration. In each phase of motion at constant velocity, an inertial coordinate system can be set up. However, there is no way to reconcile these discordant coordinates in the region of overlap (beginning at distance g-l to the left of the region of acceleration). a coordinate system? Then this coordinate system (1) should be the inert", "ial frame x/L in which he was at rest for times X o less than O. and (2) should be the other inertial frame x/L' for times x o' > T in which he was at rest in that other frame. Evidently some further thinking would be required to decide how to define the coordinates in the regions not determined by these two conditions (Figure 6.2). More serious, however, is the fact that these two conditions are inconsistent for a region of spacetime that satisfies simultaneously X o < 0 and x o' > T. In both examples of accelerated motion (Figures 6.1 and 6.2), the serious difficulties about defining a coordinate system begin only at a finite distance g-1 from the world line of the accelerated observer. The problem evidently has no solution for distances from the world line greater than g-l. It does possess a natural solution in the immediate vicinity of the observer. This solution goes under the name of \"Fermi-Walker transported orthonormal tetrad.\" The essential idea lends itself to simple illustration for hyperbolic motion, as follows. \u00a76.4. THE TETRAD CARRIED BY A UNIFORMLY ACCELERATED OBSERVER An infinitesimal version of a coordinate system is supplied by a \"tetrad,\" or \"moving frame\" (Cartan's \"repere mobile\"), or set of basis vectors eo\" e l \" e 2\" e 3, (subscript tells which vector, not which component of one vector!) Let the time axis be the time axis of a comoving inertial frame in which the observer is momentarily at rest. Thus the zeroth basis vector is identical with his 4-velocity: eo' = u. The space axes e 2 and e 3 are not affected by Lorentz transformations in the I-direction. Therefore take e 2, and e3' to be the unit basis vectors of the all-encompassing Lorentz frame relative to which the hyperbolic motion of the observer has already been described in equations (6.5): e 2, = e 2 ; e3' = e 3 \u2022 The remaining basis vector, e 1\" orthogonal to the other three, is parallel to the acceleration vector, e 1, = g-Ia [see equation (6.4)]. There is a more satisfactory way to characterize this moving frame: the time axis eo' is the observer's 4-velocity, so he is always at rest in this frame; and the Natural coordinates inconsistent at distance I > (acceleration)-l Orthonormal tetrad of basis vectors carried by uniformly accelerated observer y Orthonormal tetrad of arbitrarily accelerated observer: should be \"nonrotating\" \"Nonrotating\" means rotation only in timelike plane of 4-velocity and 4-acceleration Mathematics of rotation in 3-space 170 6. ACCELERATED OBSERVERS other three vectors e l, are chosen in such a way as to be (l) orthogonal and (2) nonrotating. These basis vectors are: (eoY' = (cosh gr; sinh gr, 0, 0); (el,)1' = (sinh g7; cosh gr, 0, 0); (e2,)1' = (0; 0, 1, 0); (e3.)1l = (0; 0, 0, 1). (6.6) There is a simple prescription to obtain these four basis vectors. Take the four basis vectors eo, e 1, e 2, e 3 of the original global Lorentz reference frame, and apply to them a simple boost in the I-direction, of such a magnitude that eO' comes into coincidence with the 4-velocity of the observer. The fact that these vectors are all orthogonal to each other and of unit magnitude is formally stated by the equation (6.7) \u00a76.5. THE TETRAD FERMI-WALKER TRANSPORTED BY AN OBSERVER WITH ARBITRARY ACCELERATION Turn now from an observer, or an object, in hyperbolic motion to one whose acceleration, always finite, varies arbitrarily with time. Here also we impose three criteria on the moving, infinitesimal reference frame, or tetrad: (1) the basis vectors el\" of the tetrad must remain orthonormal [equation (6.7)]~ (2) the basis vectors must form a rest frame for the observer at each instant (eo' = u); and (3) the tetrad should be \"nonrotating.\" This last criterion requires discussion. The basis vectors of the tetrad at any proper time 7 must be related to the basis vectors eo, e 1, e 2, e 3 of some given inertial frame by a Lorentz transformation", " el',(7) = Ap1\"(7)ep. Therefore the basis vectors at two successive instants must also be related to each other by a Lorentz transformation. But a Lorentz transformation can be thought of as a \"rotation\" in spacetime. The 4-velocity u, always of unit magnitude, changes in direction. The very concept of acceleration therefore implies \"rotation\" of velocity 4-vector. How then is the re quirement of \"no rotation\" to be interpreted? Demand that the tetrad el',(7) change from instant to instant by precisely that amount implied by the rate of change of u = eO', and by no additional arbitrary rotation. In other words, (1) accept the inevitable pseudorotation in the timelike plane defined by the velocity 4-vector and the acceleration, but (2) rule out any ordinary rotation of the three space vectors. Nonrelativistic physics describes the rotation of a vector (components Vi) by an instantaneous angular velocity vector (components Wi)' This angular velocity appears in the formula for the rate of change of v, (6.8) For the extension to four-dimensional spacetime, it is helpful to think of the rotation \u00a76.5. FERMI-WALKER TRANSPORT 171 as occurring in the plane perpendicular to the angular velocity vector \"'. Thus rewrite (6.8) as where (6.9) (6.10) has non-zero components only in the plane of the rotation. In other words, to speak of \"a rotation in the (1, 2)-plane\" is more useful than to speak of a rotation about the 3-axis. The concept of \"plane of rotation\" carries over to four dimensions. There a rotation in the (1,2)-plane will leave constant not only the va but also the Vo component of the velocity. The four-dimensional definition of a rotation is Mathematics of rotation in spacetime with (6.11 ) To test the appropriateness of this definition of a generalized rotation or infinitesimal Lorentz transformation, verify that it leaves invariant the length of the 4-vector: (6.12) The last expression vanishes because gllv is antisymmetric, whereas vllV V is symme tric. Note also that the antisymmetric tensor gllv (\"rotation matrix\"; \"infinitesimal Lorentz transformation\") has 4 X 3/2 = 6 independent components. This number agrees with the number of components in a finite Lorentz transformation (three parameters for rotations, plus three parameters for the components of a boost). The \"infinitesimal Lorentz transformation\" here must (1) generate the appropriate Lor entz transformation in the timelike plane spanned by the 4-velocity and the 4-accel eration, and (2) exclude a rotation in any other plane, in particular, in any spacelike plane. The unique answer to these requirements is i.e., a = a /\\ u. (6.13) Apply this rotation to a spacelike vector w orthogonal to u and a, (u\u00b7 w = 0 and a\u00b7 w = 0). Immediately compute gllv wv = O. Thus verify the absence of any space rotation. Now check the over-all normalization of gllv in equation (6.13). Apply the infinitesimal Lorentz transformation to the velocity 4-vector u of the observer. Thus insert Vil = u ll in (6.11). It then reads This result is an identity, since u\u00b7 u = -1 and u\u00b7 a = O. A vector v that undergoes the indicated infinitesimal Lorentz transformation, is said to experience \"Fermi-Walker transport\" along the world line of the observer. (6.14) Fermi-Walker law of transport for oononrotating OO tetrad of basis vectors carried by an accelerated observer 172 6. ACCELERATED OBSERVERS. / ----~-+---t_ /. / Figure 6.3. Construction of spacelike hyperplanes (dashed) orthogonal to the world line (heavy line) of an accelerated particle at selected moments along that world line. Note crossing of hyperplanes at distance g-I(T) from the world line. (time-dependent acceleration!) The natural moving frame associated with an accelerated observer consists of four orthonormal vectors, each of which is Fermi- Walker transported along the world line and one of which is 80' = u (the 4-velocity of the observer). Fermi-Walker transport of the space basis vectors 8j' can be achieved in practice by attaching them to gyroscopes (see Box 6.2 and exercise 6.9). \u00a76.6. THE LOCAL COORDINATE SYSTEM OF AN ACC", "ELERATED OBSERVER Tetrad used to construct \"local coordinate system of accelerated observer\" Extend this moving frame or \"infinitesimal coordinate system\" to a \"local coordinate system\" covering a finite domain. Such local coordinates can escape none of the problems encountered in \"hyperbolic motion\" (Figure 6.1) and \"briefly accelerated motion\" (Figure 6.2). Therefore the local coordinate system has to be restricted to a region within a distance g-1 of the observer, where these problems do not arise. Figure 6.3 illustrates the construction of the local coordinates ~p.'. At any given proper time T the observer sits at a specific event :1'(T) along his world line. Let the displace ment vector, from the origin of the original inertial frame to his position :1'(T), be Z(T). At :1'(T) the observer has three spacelike basis vectors 8 1'(T), 82'(T), 83'(T). The point :1'(T) plus those basis vectors define a spacelike hyperplane. The typical point of this hyperplane can be represented in the form x = ~1'81'(T) + e 82'(T) + ~3'83'(T) + Z(T) (6.15) = (separation vector from origin of original inertial frame). Here the three numbers ~k' play the role of Euclidean coordinates in the hyperplane. This hyperplane advances as proper time unrolls. Eventually the hyperplane cuts through the event :1'0 to which it is desired to assign coordinates. Assign to this event as coordinates the numbers ~' = T, ~k' given by (6.15). Call these four numbers \u00a76,6. LOCAL COORDINATES FOR ACCELERATED OBSERVER 173 \"coordinates relative to the accelerated observer.\" In detail, the prescription for the determination of these four coordinates consists of the four equations (6.16) in which the xll- are considered as known, and the coordinates T, ~k' are considered unknowns. At a certain distance from the accelerated world line, successive spacelike hyper planes, instead of advancing with increasing T, will be retrogressing. At this distance, and at greater distances, the concept of \"coordinates relative to the accelerated observer\" becomes ambiguous and has to be abandoned. To evaluate this distance, note that any sufficiently short section of the world line can be approximated by a hyperbola (\"hyperbolic motion with acceleration g\"), where the time-dependent acceleration g(T) is given by the equation g2 = all-all-' Apply the above general prescription to hyperbolic motion, arriving at the equa tions X O = (g-l + ~l')sinh(g~'), Xl = (g-l + ~l')cosh(g~'), x2 = ~2', x3 = ~3'. (6.17) The surfaces of constant ~' are the hyperplanes with XO / Xl = tanh g~' sketched in Figure 6.4. Substitute expressions (6.17) into the Minkowski formula for the line element to find (6.18) Local coordinate system for uniformly accelerated observer ---jll':::-+=:--t----+:::::--I-7--*:--~()' = 0 Figure 6.4. Local coordinate system associated with an ob server in hyperbolic motion (heavy black world line). The local coordinate system fails for e' less than _g-I. y 174 6. ACCELERATED OBSERVERS The coefficients of d~Il' de' in this expansion are not the standard Lorentz metric components. The reason is clear. The ~Il' do not form an inertial coordinate system. However, at the coefficients reduce to the standard form: Therefore these \"local coordinates\" approximate a Lorentz coordinate system in the immediate neighborhood of the observer. the position of the observer, ~l' = 0, EXERCISES Exercise 6.6. CLOCK RATES VERSUS COORDINATE TIME IN ACCELERATED COORDINATES Let a clock be attached to each grid point, (~l', ~~', ~3') = constant, of the local coordinate system of an accelerated observer. Assume for simplicity that the observer is in hyperbolic motion. Use equation (6.18) to show that proper time as measured by a lattice clock differs from coordinate time at its lattice point: (Of course, very near the observer, at F < g-l, the discrepancy is negligible.) Exercise 6.7. ACCELERATION OF LATTICE POINTS IN ACCELERATED COORDINATES Let an accelerometer be attached to each grid point of the local coordinates of", " an observer in hyperbolic motion. Calculate the magnitude of the acceleration measured by the acceler ometer at (~l', e', ~3) Exercise 6.8. OBSERVER WITH ROTATING TETRAD An observer moving along an arbitrarily accelerated world line chooses not to Fermi-Walker transport his orthonormal tetrad. Instead, he allows it to rotate. The antisymmetric rotation tensor \u00b0 that enters into his transport law splits into a Fermi-Walker part plus a spatial rotation part: illlV =,allu V - aVull, +,UaWp(aPIl\", ilrFW) ilrSB) w = a vector orthogonal to 4-velocity u. (6.19) (6.20) (a) The observer chooses his time basis vector to be eo' = u. Show that this choice is permitted by his transport law (6.19), (6.20). (b) Show that ilrSR) produces a rotation in the plane perpendicular to u and w-i.e., that O(SB)\"U = 0, O(SR)\"W = O. (6.21 ) (c) Suppose the accelerated observer Fermi-Walker transports a second orthonormal tetrad e a\". Show that the space vectors of his first tetrad rotate relative to those of his second tetrad with angular velocity vector equal to w. Hint: At a moment when the tetrads coincide, show that (in three-dimensional notation, referring to the 3-space orthogonal to the observer's world line): d(e,_ - e,.,)/dT = '\" X e,.. (6.22) \u00a76.6. LOCAL COORDINATES FOR ACCELERATED OBSERVER 175 (d) The observer uses the same prescription [equation (6.16)] to set up local coordinates based on his rotating tetrad as for his Fermi-Walker tetrad. Pick an event 2 on the observer's world line, set T =\u00b0there, and choose the original inertial frame of prescription (6.16) so (1) it comoves with the accelerated observer at 2, (2) its origin is at 2, and (3) its axes coincide with the accelerated axes at 2. Show that these conditions translate into Zfl(O) = 0, ea,(O) = ea' (6.23) (e) Show that near 2, equations (6.16) for the rotating, accelerated coordinates reduce to: XO = ~o' + ak,~k'~O' + o([~a'j3); xi = e + -.L ai~O'2 + (iklWk~I'~O' + O(W'j3). 2 (6.24) (f) A freely moving particle passes through the event 2 with ordinary velocity ~. as measured in the inertial frame. By transforming its straight world line xi = uixo to the accelerated, rotating coordinates, show that its coordinate velocity and acceleration there are: (d2~i' /d~O'2)at~ = -ai - 2\u20acikIwkul + 2viak uk. --..inertial ~ coriots \"'- - -.. - - \" -. - relLivisitc (6.25) acceleration acceleration correction to inertial acceleration Exercise 6.9. THOMAS PRECESSION Consider a spinning body (gyroscope, electron,...) that accelerates because forces act at its center of mass. Such forces produce no torque; so they leave the body's intrinsic angular momentum vector S unchanged, except for the unique rotation in the u 1\\ a plane required to keep S orthogonal to the 4-velocity u. Mathematically speaking, the body Fermi- Walker transports its angular momentum (no rotation in planes other than u 1\\ a): dS/dT = (u 1\\ a)\u00b7 S. (6.26 ) This transport law applies to a spinning electron that moves in a circular orbit of radius r around an atomic nucleus. As seen in the laboratory frame, the electron moves in the x, y-plane with constant angular velocity, w. At time t = 0, the electron is at x = r, y = 0; and its spin (as treated classically) has components so = 0, s\" =_I_ n Vi' SY = 0, sz = -.Ln. 2 Calculate the subsequent behavior of the spin as a function oflaboratory time. SIl(t). Answer: S\" = ~fJ (cos wt cos wyt + y sin wt sin wyt); SY = ~ n (sin wt cos wyt - y cos w", "t sin wyt); sz = 1. n; 2 u = wr; so = - ~ fJ uy sin wyt; y = (I - U2 )-1/2. (6.27) y 176 6. ACCELERATED OBSERVERS Rewrite the time-dependent spatial part of this as S\" + iSY =....!!..- [e-;(Y-llw/ + i(l - y)sin(wyt)e iw /]. v'2 The first term rotates steadily in a retrograde direction with angular velocity WTbomas = (y - I)w (6.28) (6.29) It is called the Thomas precession. The second term rotates in a righthanded manner for part of an orbit (0 < wyt < 17) and in a lefthanded manner for the rest (17 < wyt < 217). I < I). It Averaged in time, it does nothing. Moreover, in an atom it is very small (y must be present, superimposed on the Thomas precession, in order to keep s\u00b7 u = S\u00b7 u - SOuo = 0, and (6.30) (6.31 ) It comes into play with righthanded rotation when S\u00b7 u is negative; it goes out of play when S\u00b7 u = 0; and it returns with lefthanded rotation when S\u00b7 u turns positive. The Thomas precession can be understood, alternatively, as a spatial rotation that results from the combination of successive boosts in slightly different directions. [See, e.g., exercise 103 of Taylor and Wheeler (1966).] For an alternative derivation of the Thomas precession (6.29) from \"spinor formalism,\" see \u00a741.4. --------------------------------~ ~ CHAPTER 7 INCOMPATIBILITY OF GRAVITY AND SPECIAL RELATIVITY \u00a77.1. ATTEMPTS TO INCORPORATE GRAVITY INTO SPECIAL RELATIVITY The discussion of special relativity so far has consistently assumed an absence of gravitational fields. Why must gravity be ignored in special relativity? This chapter describes the difficulties that gravitational fields cause in the foundations of special relativity. After meeting these difficulties, one can appreciate fully the curved-space time methods that Einstein introduced to overcome them. Start, then, with what one already knows about gravity, Newton's formulation of its laws: d 2x i /dt 2 = -af/J/ax i V2f/J = 4'17Gp., (7.1 ) (7.2) \\. This chapter is entirely Track 2. It depends on no preceding Track-2 material. It is not needed as preparation for arw later chapter, but will be helpful in Chapter 18 (weak gravitational fields). and in Chapters 38 and 39 (experimental tests and other theories of gravity). These equations cannot be incorporated as they stand into special relativity. The equation of motion (7.1) for a particle is in three-dimensional rather than four-di mensional form; it requires modification into a four-dimensional vector equation for d 2xlJ./dT2 \u2022 Likewise, the field equation (7.2) is not Lorentz-invariant, since the appearance of a three-dimensional Laplacian operator instead of a four-dimensional d'Alembertian operator means that the potential f/J responds instantaneously to changes in the density p at arbitrarily large distances away. In brief, Newtonian gravitational fields propagate with infinite velocity. One's first reaction to these problems might be to think that they are relatively straigh tforward to correct. Exercises at the end of this section study some relatively straightforward generalizations of these equations, in which the gravitational poten tial f/J is taken to be first a scalar, then a vector, and finally a symmetric tensor field. Each of these theories has significant shortcomings. and all fail to agree with obser vations. The best of them is the tensor theory (exercise 7.3, Box 7.1). which, however, Newton's gravitational must be modified into four-dimensional, geometric form laws All straightforward modifications are unsatisfactory y Best modification (tensor theory in flat spacetime) is internally inconsistent; when repaired, it becomes general relativity. EXERCISES 178 7. INCOMPATIBILITY OF GRAVITY AND SPECIAL RELATIVITY is internally inconsistent and admits no exact solutions. This difficulty has been attacked in recent times by Gupta (1954, 1957, 1962), Kraichnan (1955), Thirring (1961), Feynman (1963), Weinberg (1965), Deser (1970). They show how the flat space tensor theory may be modified within the spirit of present-day relativistic field theory to overcome these inconsistencies. By this field-theory route (part 5 of", " Box 17.2), they arrive uniquely at standard 1915 general relativity. Only at this end point does one finally recognize, from the mathematical form of the equations, that what ostensibly started out as a flat-space theory of gravity is really Einstein's theory, with gravitation being a manifestation of the curvature of spacetime. This book follows Einstein's line of reasoning because it keeps the physics to the fore. EXERCISES ON FLAT\u00b7SPACETIME THEORIES OF GRAVITY The following three exercises provide a solid challenge. Happily, all three require similar techniques, and a solution to the most difficult one (exercise 7.3) is presented in Box 7.1. Therefore, it is reasonable to proceed as follows. (a) Work either exercise 7.1 (scalar theory of gravity) or 7.2 (vector theory of gravity), skimming exercise 7.3 and Box 7.1 (tensor theory of gravity) for outline and method, not for detail, whenever difficulties arise. (b) Become familiar with the results of the other exercise (7.2 or 7.1) by discussing it with someone who has worked it in detail. (c) Read in detail the solution to exercise 7.3 as presented in Box 7.1, and compare with the computed results for the other two theories. (d) Develop computa tional power by checking some detailed computations from Box 7.1. Exercise 7.1. SCALAR GRAVITATIONAL FIELD. A. Consider the variational principle 81 = 0, where f/J f 1 = -m e tfJ dz'\" dz/3 )112 dt.. -11\"'/3 dt.. ( dt.., (7.3) Here m = (rest mass) and z\"'(t..) = (parametrized world line) for a test particle in the scalar gravitational field f/J. By varying the particle's world line, derive differential equations governing the particle's motion. Write them using the particle's proper time as the path parameter, so that u'\" = dz\"'/dr satisfies ua u/311\"'/3 = -I. B. Obtain the field equation for f/J(x) implied by the variational principle 81 = 0, where 1 = JE d4x and (7.4) Show that the second term here gives the same integral as that studied in part A (equa tion 7.3). Discussion: The field equations obtained describe how a Single particle of mass m generates the scalar field. If many particles are present, one includes in 13 a term - Jme tfJ 84[x - Z(T)] dT for each particle. C. Solve the field equation of part B, assuming a single source particle at rest. Also assume that e tfJ = I is an adequate approximation in the neighborhood of the particle. Then check this assumption from your solution; i.e., what value does it assign to e tfJ at the surface of the earth? (Units with c = I are used throughout; one may also set G = I, if one wishes.) \u00a77.1. GRAVITY BURSTS OUT OF SPECIAL RELATIVITY 179 D. Now treat the static, spherically symmetric field rp from part C as the field of the sun acting as a given external field in the variational principle of part A, and study the motion of a planet determined by this variational principle. Constants of motion are available from the spherical symmetry and time-independence of the integrand. Use spherical coordinates and assume motion in a plane. Derive a formula for the perihelion precession of a planet. E. Pass to the limit of a zero rest-mass particle in the equations of motion of part A. Do this by using a parameter A different from proper time, so chosen that kP. = dxP./ dA is the energy-momentum vector, and by taking the limit m -+ \u00b0with kO = ym = E remain ing finite (so UO = Y -+ 00). Use these equations to show that the quantities qp. = kP.etfJ are constants of motion, and from this deduce that there is no bending of light by the sun in this scalar theory. Exercise 7.2. VECTOR GRAVITATIONAL FIELD, rp p. A. Verify that the variational principle 8I =\u00b0gives Maxwell's equations by varying Ap.' and the Lorentz force law by varying ZP.(T), when (7.5) rp p..' (7.6) Here Fp.. is an abbreviation for A.,p. - Ap.\u2022\u2022. Hint: to vary Ap.(x), rewrite the last term as a spacetime integral by introducing a delta function 84[x - Z", "(T)] as in exercise 7.1, parts A and B. B. Define, by analogy to the above, a vector gravitational field rpp. with GIJ.\" =rp '.p. - using a variational principle with 1 = + - - G Gp.\u00b7 d 4x + -mf--=.1!:. dT + m rpP.-d dT. f f I 167iG p.. dzP. dz dT dT dzP. T I 2 (Note: ifmany particles are present, one must augment 1 by terms ~mJ(dzP./dT)(dzp./dT) dT + mJrpp.(dzP./dT) dT for each particle.) Find the \"Coulomb\" law in this theory, and verify that the coefficients of the terms in the variational principle have been chosen reasonably. C. Compute the perihelion precession in this theory. D. Compute the bending of light in this theory (i.e., scattering of a highly relativistic particle UO = Y -+ 00), as it passes by the sun, because of the sun's rp p. field. E. Obtain a formula for the total field energy corresponding to the Lagrangian implicit in part B. Use the standard method of Hamiltonian mechanics, with is the Lagrangian density and L =Jf d3x is the Lagrangian. The corresponding Hamil f tonian density (= energy density) is Show that vector gravitational waves carry negative energy. Exercise 7.3. SYMMETRIC TENSOR GRAVITATIONAL FIELD, hp.,. = h.p. Here the action principle is, as for the vector field, 81 = 0, with 1 = l tie1d + 1particle + linteraction' 1particle is the same as for the vector field: Iparticle = '2 m I f dziJ. dz dT 7: dT. (7.7) 180 7. INCOMPATIBILITY OF GRAVITY AND SPECIAL RELATIVITY However, ltield and linteraction are different: with h- - h p.. = p.. - I 2\"11p.. hU'u'. -Ifl TP.\u00b7 d4 Ip.. 2 interaction - I X. Gote that J?ne h her~, s not an h (7.8a) (7.8b) (7.8c) (7.9) Here TP.\u00b7 is the stress-energy tensor for all nongravitational fields and matter present. For a system of point particles (used throughout this exercise), (7.10) A. Obtain the equations of motion of a particle by varying ZP.(T) linteraction) = O. Express your result in terms of the \"gravitational force field\" in 8(lparticle + (7.11) derived from the tensor gravitational potentials hp.. = h.w B. Obtain the field equations from 8(lfield + linteraction) = 0; express them in terms of Discuss gauge invariance, and the condition hp.<JI\" = O. C. Find the tensor gravitational potentials hp.: due to the sun (treated as a point mass). D. Compute the perihelion precession. E. Compute the bending of light. F. Consider a gravitational wave (7.12) What conditions are imposed by the field equations? By the gauge condition hP.\u00b7 = AP.\u00b7 exp( ik\"x\"). hP.\".\" = O? Show that, by further gauge transformations hp.. --+ hp.. + ~p.., + ~\"p. that preserve the hp.\" \" = 0 restrictions, further conditions (7.13) (7.14) (7.15) (7.16) can be imposed, where u\" is a fixed, timelike vector. It is sufficient to consider the case, obtained by a suitable choice of reference frame, where u\" = (1; 0,0,0) and k\" = (w; 0,0, w). G. From the Hamiltonian density (7.17) for the field, show that the energy density of the waves considered in part F is positive. H. Compute TP.\u00b7 \u2022\u2022 for the stress-energy tensor of particles TP.\u00b7 that appears in the action integral I. Does P\u00b7.\u2022 vanish (e.g., for the earth in orbit around the sun)? Why? Show that the coupled equations for fields and particles obtained from 8I = 0 have no solutions. (continued on page 187) Box 7.1 AN ATTEMPT TO DESCRIBE GRAVITY BY A SYMMETRIC TENSOR FI", "ELD IN FLAT SPACETIME [Solution to exercise 7.3] Attempts to describe gravity within the framework of special relativity would natu rally begin by considering the gravitational field to be a scalar (exercise 7.1) as it is in Newtonian theory, or a vector (exercise 7.2) by analogy to electromagnetism. Only after these are found to be deficient (e.g., no bending of light in either theory; negative-energy waves in the vector theory) would one face the computational complexities of a symmetric tensor gravitational potential, hllv = hv!l' which has more indices. The foundations of the most satisfactory of all tensor theories of gravity in flat spacetime are laid out at the beginning of exercise 7.3. The choice of the Lagrangian made there (equations 7.8) is dictated by the demand that hllv be a \"Lorentz covari ant, massless, spin-two field.\" The meaning of this demand, and the techniques of special relativity required to translate it into a set offield equations, are customarily found in books on elementary particle physics or quantum field theory; see, e.g., Wentzel (1949), Feynman (1963), or Gasiorowicz (1966). Fierz and Pauli (1939) were the first to write down this Lagrangian and investigate the resulting theory. The conclusions of the theory are spelled out here in the form of a solution to exercise 7.3. A. Equation of Motion for a Test Particle (exercise 7.3A) Carry out the integration in equation (7.9), using the particle stress-energy tensor of equation (7.10), to find 1 =1 p+i - particle + interaction - 2 m I -1 I( l1 w + IlV Z Z h )'Il'vd T, (1) where Then compute M p + i. and find that the coefficient of the arbitrary variation in path 8z Il vanishes if and only if Rewrite this equation of motion in the form r 11 1l,. + Il\" Z +!-\"x{3- Z ( h ) \u00b7\u00b7v +0I'(3 - 0 - \u2022 (2) where r lla{3 is defined in equation (7.11). B1. Field Equations (exercise 7.3B) Use l fie1d and linteraction in the forms given in equations (7.8) and (7.9); but for the quickest and least messy derivation. do not use the standard Euler-Lagrange equations. Instead. compute directly the first-order change 8t~r produced by a small 182 7. INCOMPATIBILITY OF GRAVITY AND SPECIAL RELATIVITY Box 7.1 (continued) variation 8ha{3 of the field. For the second term of t'(, it is clear (by relabeling dummy indices as needed) that varying each factor gives the same result, so the two terms from the product rule combine: 8(h,ah/l{3{3) = 2h/l{3{38h,1.1'. JL {X, JLa, A similar result holds for the first term of t'(, in view of the identity all); /ll' = G/lpb/l V which holds for the \"bar\" operation of equations (7.8); each side here is just a/lpb/l V - ~a/l/lbP\". Consequently,, Next use this expression in Mfjeld; and, by an integration by parts, remove the derivatives from 8hw'giving Mfie1d = (32.\".GrIf [h p{3,a,a 8hP{3 - 2h/l{3,{3'1.I' 8h/lal d 4x. To find the coefficient of 8hw in this expression, write (from equation 7.8c) (3) and then rearrange and relabel dummy (summation) indices to obtain Mfie1d = (32.\".G)-1f [h/l{3,a,a 8h v{3 - iii/l{3,{3,a 8h/lal d 4x. By combining this with Minteraetion = ~T/lP 8h/l pd 4x, and by using the symmetry 8hjJ.P = 8h pIl' obtain The definition made in equation (7.12) allows this to be rewritten as B2 \u00b7 Gauge Invariance (exercise 7.3B, continued) The symmetries, of H/lll'P{3 imply an identity (4) (4') analogous to F/lP,p", "/l 0 in electromagnetism. Thus T/lP,p = 0 is required of the sources, just as is J/l'/l = 0 in electromagnetism (exercise 3.16). These identities make the field equations (4') too weak to fix h w \u00a77.1. GRAVITY BURSTS OUT OF SPECIAL RELATIVITY 183 completely. In particular, by direct substitution in equations (4), one verifies that to any solution one can add a gauge field h (gauge) = I:. ~\" h (gauge) ~\" - ~~,,, I:. ~~,,, + I:. ~\"'~' + I:. ~\"'~ - 1:.01 1J~,,'i),a' (5) without changing PP. Let ~/.L vanish outside some finite spacetime volume, but be otherwise arbitrary. Then h/.Lp and h/.Lp = h/.Lp + h/.L}gauge) both satisfy the source equation (4) for the same source T/.LP and the same boundary conditions at infinity. We therefore expect them to be physically equivalent. Bya specialization of the gauge analogous to the \"Lorentz\" specializationAOI.a = 0 of electromagnetism (equation 3.58a; exercise 3.17), one imposes the condition h/.La = o.,a This reduces the field equations (4) to the simple d'Alembertian form (6) (7) (see exercise 18.2). Here and henceforth we set G = 1 (\"geometrized units\"). C. Field of a Point Mass (exercise 7.3C) For a static source, the wave equation (7) reduces to a Laplace equation V 2h/.Lp = -16'17Tw'The stress-energy tensor for a static point mass (equation 7.10) is TJo = M8 3(x) and pk = O. Put this into the Laplace equation, solve for h/.Lp, and use equation (7.8c) to obtain h/.Lp' The result is: hOO = 2M/r; (8) (see equation l8.15a). D. Perihelion Precession (exercise 7.30) Direct substitution of the potential (8) into the equations of motion (2) is tedious and not very instructive. Variational principles are popular in mechanics because they simplify such calculations. Return to the basic variational principle 8lp + i = 0 (equation 1), and insert the potential (8) for the sun. Convert to spherical coordinates so oriented that the orb'it lies in the equatorial (8 = '17/2) plane: IV+i =f L dT; L = ~ m[ -(1 - 2Mr- l )i:!. + (1 + 2Mr- I)(;2 + r2<1>2)]. (9) (10) 184 7. INCOMPATIBILITY OF GRAVITY AND SPECIAL RELATIVITY Box 7.1 (continued) From the absence of explicit t-, \u00a2-, and,,-dependence in L, infer three constants of motion: the canonical momenta Pt = -my = aL/ai (this defines y) and (this defines 0:); and the Hamiltonian H = xlL(a L/axJ1.) - L, which can be set equal to -~m by appropriate normalization of the path parameter T. From these constants of the motion, derive an orbit equation as follows: (I) calculate H = -!m in terms of r, f, 4>, and i; (2) eliminate i and ~ in favor of the constants y and 0:; (3) as in Newtonian orbit problems, define u = M/r, and write du. - = \"7\" = - -. = - - (I + 2u)r; d\u00a2 Mf r2ep M 0: it ep (4) in H, eliminate f in favor of du/d\u00a2 via the above equation, and eliminate r in favor of u; (5) solve for du/dep. The result is dU)2 d\u00a2 ( + u2 = (y2 - I + 2u)0: 2 M2 [I + 2u] I - 2u. (II) Neglecting cubic and higher powers of u = GM/c 2r -- (I - y2) in this equation, derive the perihelion shift. (For details of method, see exercise 40.4, with the y and 0: of this box renamed E and r, and with the y and f3 of that exercise set equal to I and 0.) The resulting shift per", " orbit is L1ep = 8'17M/ro + O([M/roF). This is 1the prediction of general relativity; and it disagrees with the observations on Mercury (see Box 40.3). (12) E. Bending of Light (exercise 7.3E) The deflection angle for light passing the sun is, on dimensional grounds, a small quantity, L1ep -- Md R0 -- 10-6 ; from the outset, one makes approximations based on this smallness. A diagram of the photon trajectory, set in the x, z-plane, shows that, for initial motion parallel to the z-axis, the deflection angle can be expressed in terms of the final momentum as L1ep = P./pz- Compute the finalp., by an integral along the trajectory, p., =f +\"\" (dp.,/dz) dz, -\"\" \u00a77.1. GRAVITY BURSTS OUT OF SPECIAL RELATIVITY 185 x Pinitial pz treating pz as essentially constant. This computation requires generalization of the equation of motion (2) to the case of zero rest mass. To handle the limit m ---+- 0, introduce a new parameter A. = T/m; then p,. = m(dz\"/dT) = dz\"/dA.. Also define P,. = (11,.\" + h,.\")p\", since this quantity appears simply in equation (2) and agrees with P,. in the limit r ---+- 00, where one will need to evaluate it. Then equation (2) reads, for any m, including m = 0, dP,. _ ~ dA. a {3 - 2 ha{3,,.p P. On the righthand side here, since ha{3,,. is small, a crude approximation to p,. suffices; p 1 = p2 = 0, pO = p3 = dz/dA. = w = constant. Thus, dP1 _ 1 dA. -\"2 (hoo + 2h03 + h33),1w 2 and For the sun, hoo := h33 = 2M/r, and h03 = 0 (equation 8), so For light grazing the sun, 1 = R0, this gives L1<1> = 4M0 / R0 radians = 1\".75, which is also the prediction of general relativity, and is consistent with the observations (see Box 40.1). F. Gravitational Waves (exercise 7.3F) The field equations (4) and gauge properties (5) of the present flat-spacetime theory the treatment of are identical gravitational waves using linearized theory, which is presented in \u00a7\u00a7 18.2,35.3, and 35.4, applies here. to those of Einstein's \"linearized theory.\" Thus, G. Positive Energy of the Waves (exercise 7.3G) Computing a general formula for :7( from equation (7.17) is tedious, but it is sufficient to consider only the special case of a plane wave (equation 7.13)-or better still, Box 7.1 (continued) a plane wave with only h12 = h21 =fez t). Any gravitational wave can be con structed as a superposition of such plane waves. First compute the Langrangian for this case. According to equation (7.8), it reads Now the full content of the formula (7.17) defining J( is precisely the following: start from the Lagrangian; keep all terms that are quadratic in time derivatives; omit all terms that are linear in time derivatives; and reverse the sign of terms that contain no time derivatives. The result is (14) which is positive. H. Self-Inconsistency of the Theory (exercise 7.3H) From equation (7.10), find But i)4(X - z) depends only on the difference x ll - Zll, so - alaz\" can replace alax\" when acting on the i)-function. Noting that rewrite Til\",\" as P\",\" = -mf ill(dldT) i)4[X - Z(T)] dT = +mf ill i)4[X - Z(T)] dT. (The last step is obtained by an integration by parts.) Thus P\" \" =\u00b0holds if and only If ill = 0. But ill =\u00b0means that the gravitational fields ha~e no effect on the motion of the particle. But this contradicts the equation of motion (2), which follows from the theory's variational principle. Thus, this tensor theory of gravity is incon sistent. [Stated briefly, equation (4) requires P\" \" = 0, while equation (2) excludes it.], The fact that, in", " this theory, gravitating bodies cannot be affected by gravity, also holds for bodies made of arbitrary stress-energy (e.g., rubber balls or the Earth). Since all bodies gravitate, since the field equations imply P\",\" = 0, and since this \"equation of motion for stress-energy\" implies conservation of a body's totaI4-mo mentum pll = fTllo d3x, no body can be accelerated by gravity. The Earth cannot be attracted by the sun; it must fly off into interstellar space! Straightforward steps to repair this inconsistency in the theory lead inexorably to general relativity (see Box 17.2 part 5). Having adopted general relativity as the theory of gravity, one can then use the present flat-spacetime theory as an approximation to it (\"Linearized general relativity\"; treated in Chapters 18, 19, and 35; see especially discussion at end of\u00a718.3). \u00a77.2. RED SHIFT FROM ENERGY CONSERVATION 187 \u00a77.2. GRAVITATIONAL RED SHIFT DERIVED FROM ENERGY CONSERVATION Einstein argued against the existence of any ideal, straight-line reference system such as is assumed in Newtonian theory. He emphasized that nothing in a natural state of motion, not even a photon, could ever give evidence for the existence or location of such ideal straight lines. That a photon must be affected by a gravitational field Einstein (1911) showed Gravitational redshift derived from the law of conservation of energy, applied in the context of Newtonian gravitation theory. Let a particle of rest mass m start from rest in a gravitational field g at point d and fall freely for a distance h to point qJ. It gains kinetic energy mgh. Its total energy, including rest mass, becomes from energy considerations m + mgh. (7.18) Now let the particle undergo an annihilation at qJ, converting its total rest mass plus kinetic energy into a photon of the same total energy. Let this photon travel upward in the gravitational field to d. If it does not interact with gravity, it will have its original energy on arrival at d. At this point it could be converted by a suitable apparatus into another particle of rest mass m (which could then repeat the whole process) plus an eXCess energy mgh that costs nothing to produce. To avoid this contradiction of the principal of conservation of energy, which can also be stated in purely classical terms, Einstein saw that the photon must suffer a red shift. The energy of the photon must decrease just as that of a particle does when it climbs out of the gravitational field. The photon energy at the top and the bottom of its path through the gravitational field must therefore be related by (7.19) The drop in energy because of work done against gravitation implies a drop in frequency and an increase in wavelength (red shift; traditionally stated in terms of a red shift parameter, Z = &/A); thus, 1 + Z =~ = hVbottom = Ebottom = 1 + gh. hvtop Abottom Etop (7.20) The redshift predicted by this formula has been verified to 1 percent by Pound and Snider (1964, 1965), refining an experiment by Pound and Rebka (1960). \u00a77.3. GRAVITATIONAL REDSHIFT IMPLIES SPACETIME IS CURVED the An argument by Schild (1960, 1962, 1967) yields an important conclusion: existence of the gravitational redshift shows that a consistent theory of gravity cannot be constructed within the framework of special relativity. Assume gravity is described by an (unspecified) field in flat spacetime... 188 7. INCOMPATIBILITY OF GRAVITY AND SPECIAL RELATIVITY A -r-.. -r-.. B Figure 7.1. Successive pulses of light rising from height \"._ to height \"2 = '\" + h against the gravitational field of the earth. The paths y\" and Y2 must be exactly congruent, whether sloped at 45\u00b0 (left) or having variable slope (right). Whereas Einstein's argument (last section) was formulated in Newtonian theory, Schild's is formulated in special relativity. It analyzes gravitational redshift experi ments in the field of the Earth, using a global Lorentz frame tied to the Earth's center. It makes no demand that free particles initially at rest remain at rest in this global Lorentz frame (except far from the Earth, where gravity is negligible). On the contrary, it demands that free particles be accelerated relative to the Lorentz frame by the Earth's gravitational field. It is indifferent to the mathematical nature of that field (scalar, vector, tensor,...), but it does insist that the gravitational accelerations agree with experiment. And, of course, it de!Jlands that proper lengths and times be governed by the metric of special relativity", ". Schild's argument proceeds as follows. Consider one observer at rest on the Earth's surface at height Zl' and a second above the Earth's surface at height Z2 = Zl + h (Figure 7.1). The observers may verify that they are at rest relative to each other and relative to the Earth's Lorentz frame by, for instance, radar ranging to free particles that are at rest in the Earth's frame far outside its gravitational field. The bottom experimenter then emits an electromagnetic signal of a fixed standard frequency W b which is received by the observer on top. For definiteness, let the signal be a pulse exactly N cycles long. Then the interval of time* 8Tbot required for the emission of the pulse is given by 2'T1N = W b 8Tbot\u2022 The observer at the top is then to receive these same N cycles of the electromagnetic wave pulse and measure the time interval* 8Ttop required. By the definition of \"frequency,\" it satisfies 2'T1N = W t 8Ttop'The redshift effect, established by experiment (for us) or by energy conserva tion (for Einstein), shows W t < wb; consequently the time intervals are different, 8Ttop > 8Tbot\u2022 Transfer this information to the special-relativity spacetime diagram of the experiment (Figure 7.1). The waves are light rays; so one might show them as traveling along 45 0 null lines in the spacetime diagram (Figure 7.1,A). In this *Proper time equals Lorentz coordinate time for both observers, since they are at rest in the Earth's Lorentz frame. \u00a77.4. RED SHIFT AS EVIDENCE FOR EQUIVALENCE PRINCIPLE 189 simplified but slightly inadequate form of the argument, one reaches a contradiction by noticing that here one has drawn a parallelogram in Minkowski spacetime in which two of the sides are unequal, Ttop > Tbat> whereas a parallelogram in flat Minkowski spacetime cannot have opposite sides unequal. One concludes that special relativity cannot be valid over any sufficiently extended region. Globally, spacetime, as probed by the tracks of light rays and test particles, departs from flatness (\"curva ture\"; Parts III and IV of this book), despite the fine fit that Lorentz-Minkowski flatness gives to physics locally. Figure 7.1,B, repairs an oversimplification in this argument by recognizing that the propagation of light will be influenced by the gravitational field. Therefore photons might not follow straight lines in the diagram. Consequently, the world lines 11 and 12 of successive pulses are curves. However, the gravitational field is static and the experimenters do not move. Therefore nothing in the experimental setup changes with time. Whatever the path 11' the path 12 must be a congruent path of exactly the same shape, merely translated in time. On the basis of this congruence and the fact that the observers are moving on parallel world lines, one would again conclude, if flat Minkowski geometry were valid, that Tbot = Ttop' thus contradicting the observed redshift experiment. The experimenters do not need to understand the propagation oflight in a gravitational field. They need only use their radar apparatus to verify the fact that they are at rest relative to each other and relative to the source of the gravitational field. They know that, whatever influence the gravitational field has on their radar apparatus, it will not be a time-dependent influence. Moreover, they do not have to know how to compute their separation in order to verify that the separation remains constant. They only need to verify that the round-trip time required for radar pulses to go out to each other and back is the same every time they measure it. Schild's redshift argument does not reveal what kind of curvature must exist, or whether the curvature exists in the neighborhood of the observational equipment or some distance away from it. It does say, however, quite unambigously, that the flat spacetime of special relativity is inadequate to describe the situation, and it should therefore motivate one to undertake the mathematical analysis of curvature in Part III. This assumption is incompatible with gravitational redshift Conclusion: spacetime is curved \u00a77.4. GRAVITATIONAL REDSHIFT AS EVIDENCE FOR THE PRINCIPLE OF EQUIVALENCE Einstein (1908, 1911) elevated the idea of the universality of gravitational interactions to the status of a fundamental principle of eqUivalence, that all effects of a uniform gravitational field are identical to the effects ofa uniform acceleration ofthe coordinate system. This principle generalized a result of Newtonian gravitation theory, in which a uniform acceleration of the coordinate system in equation (7.1) gives rises to a Principle of equivalence: a uniform gravitational field is indistinguishable from a uniform acceleration of a reference frame 190 7. INCOMPATIBILITY OF GRAV", "ITY AND SPECIAL RELATIVITY supplementary uniform gravitational field. However, the Newtonian theory only gives this result for particle mechanics. Einstein's principle of equivalence asserts that a similar correspondence will hold for all the laws of physics, including Max well's equations for the electromagnetic field. The rules ofthe game-the \"scientific method\"-require that experimental support be sought for any new theory or principle, and Einstein could treat the gravitational redshift as the equivalent of experimental confirmation of his principle of equival ence. There are two steps in such a confirmation: first, the theory or principle must predict an effect (the next paragraph describes how the equivalence principle implies the redshift); second, the predicted effect must be observed. With the Pound-Rebka Snider experiments, one is in much better shape today than Einstein was for this second step. Einstein himself had to rely on the experiments supporting the general concept of energy conservation, plus the necessity of a redshift to preserve energy conservation, as a substitute for direct experimental confirmation. The existence of the gravitational redshift can be deduced from the equivalence principle by considering two experimenters in a rocket ship that maintains a constant acceleration g. Let the distance between the two observers be h in the direction of the acceleration. Suppose for definiteness that the rocket ship was at rest in some inertial coordinate system when the bottom observer sent off a photon. It will require time t = h for the photon to reach the upper observer. In that time the top observer acquires a velocity v = gt = gh. He will therefore detect the photon and observe a Doppler redshift z = v = gh. The results here are therefore identical to equation (7.20). The principle of equivalence ofcourse requires that, if this redshift is observed in an experiment performed under conditions of uniform acceleration in the absence of gravitational fields, then the same redshift must be observed by an experiment performed under conditions where there is a uniform gravitational field, but no acceleration. Consequently, by the principle of equivalence, one can derive equation (7.20) as applied to the gravitational situation. \u00a77.5. LOCAL FLATNESS, GLOBAL CURVATURE The equivalence principle helps one to discern the nature of the spacetime curvature, whose existence was inferred from Schild's argument. Physics is the same in an accelerated frame as it is in a laboratory tied to the Earth's surface. Thus, an Earth-bound lab can be regarded as accelerating upward, with acceleration g, relative to the Lorentz frames in its neighborhood.* Equivalently, relative to the lab and the Earth's surface, all Lorentz frames must accelerate downward. But the downward (radial) direction is different at different latitudes and longitudes. Hence, local Lorentz frames, initially at rest with respect to each other but on opposite sides of the Earth, subsequently fall toward the center and go flying through each other. Clearly they cannot be meshed to form the single global Lorentz frame, tied to the *This upward acceleration of the laboratory, plus equation (6.18) for the line element in an accelerated coordinate system, explains the nonequality of the bottom and top edges of the parallelograms in Figure 7.1.'Gravitational redshift derived from equivalence principle Equivalence principle implies nonmeshing of local Lorentz frames near Earth (spacetime curvature!) \u00a77.5. LOCAL FLATNESS, GLOBAL CURVATURE 191 Earth, that was assumed in Schild's argument. This nonmeshing of local Lorentz frames, like the nonmeshing of local Cartesian coordinates on a curved 2-surface, is a clear manifestation of spacetime curvature. Geographers have similar problems when mapping the surface of the earth. Over small areas, a township or a county, it is easy to use a standard rectangular coordinate system. However, when two fairly large regions are mapped, each with one coordi nate axis pointing north, then one finds that the edges of the maps overlap each other best if placed at a slight angle (spacetime analog: relative velocity of two local Lorentz frames meeting at center of Earth). It is much easier to start from a picture of a spherical globe, and then talk about how small flat maps might be used as good approximations to parts of it, than to start with a huge collection of small maps and try to piece them together to build up a picture of the globe. The exposition of the geometry of spacetime in this book will therefore take the first approach. Now that one recognizes that the problem is to fit together local, flat spacetime descriptions of physics into an over-all view of the universe, one should be happy to jump, in the next chapter, into a broadscale study of geometry. From this more advantageous viewpoint, one can then face the problem of discussing the relationship between the local inertial coordinate systems appropriate to two nearby regions that have slightly different", " gravitational fields. There are actually two distinguishable ways in which geometry enters the theory of general relativity. One is the geometry of lengths and angles in four-dimensional spacetime, which is inherited from the metric structure ds 2 of special relativity. Schild's argument already shows (without a direct appeal to the equivalence princi ple) that the special-relativistic ideas of length and angle must be modified. The modified ideas of metric structure lead to Riemannian geometry, which will be treated in Chapters 8 and 13. However, geometry also enters general relativity because of the equivalence principle. An equivalence principle can already be stated within Newtonian gravitational theory, in which no concepts of a spacetime metric enter, but only the Euclidean metric structure of three-dimensional space. The equivalence-principle view of Newtonian theory again insists that the local standard of reference be the freely falling particles. This requirement leads to the study of a spacetime geometry in which the curved world lines of freely falling particles are defined to be locally straight. They play the role in a curved spacetime geometry that straight lines play in flat spacetime. This \"affine geometry\" will be studied in Chapters 10-12. It leads to a quantitative formulation of the ideas of \"covariant derivative\" and \"curvature\" and even \"curvature of Newtonian spacetime\"! Nonmeshing of local Lorentz frames motivates study of geometry Two types of geometry relevant to spacetime: Riemannian geometry (lengthS and angles) Affine geometry (\"straight lines\" and curvature) PART III THE MATHEMATICS OF CURVED SPACETIME Wherein the reader is exposed to the charms of a new temptress Modern Differential Geometry-and makes a decision: to embrace her for eight full chapters; or, to escape after one. having drunk his fill, CHAPTER 8 DIFFERENTIAL GEOMETRY: AN OVERVIEW I am coming more and more to the conviction that the necessity of our geometry cannot be demonstrated, at least neither by, nor the human intellect.... geometry should be ranked, not with arithmetic, which is purely aprioristic, but with mechanics. (1817) for, We must confess in all humility that while number is a product of our mind alone, space has a reality beyond the mind whose rules we cannot completely prescribe. (1830) CARL FRIEDRICH GAUSS \u00a78.1. AN OVERVIEW OF PART III Gravitation is a manifestation of spacetime curvature, and that curvature shows up in the deviation of one geodesic from a nearby geodesic (\"relative acceleration of test particles\"). The central issue of this part of the book is clear: How can one quantify the \"separation,\" and the \"rate of change\" of \"separation, \"of two \"geodesics\" in \"curved\" spacetime? A clear, precise answer requires new concepts. \"Separation\" between geodesics will mean \"vector.\" But the concept of vector as employed in flat Lorentz spacetime (a bilocal object: point for head and point for tail) must be sharpened up into the local concept of tangent vector, when one passes to curved spacetime. Chapter 9 does the sharpening. It also reveals how the passage to curved spacetime affects I-forms and' tensors. It takes one tool (vectors in curved geometry, Chapter 9) to define \"separation\" clearly as a vector; it takes another tool (parallel transport in curved spacetime, Chapter 10) to compare separation vectors at neighboring points and to define the \"rate of change of separation.\" No transport, no comparison; no comparison, no meaning to the term \"rate of change\"! The notion of parallel transport founds itself Concepts to be developed in Part III: Tangent vector Geodesic Covariant derivative Geodesic deviation Spacetime curvature This cha pter: a Track-1 overview of differential geometry 196 B. DIFFERENTIAL GEOMETRY: AN OVERVIEW on the idea of \"geodesic,\" the world line of a freely falling particle. The special mathematical properties of a geodesic are explored in Chapter 10. That chapter uses geodesics to define parallel transport, uses parallel transport to define covariant derivative, and-completing the circle~uses covariant derivative to describe geo desics. Chapter 11 faces up to the central issue: geodesic deviation (\"rate of change of separation vector between two geodesics\"), and its use in defining the curvature of spacetime. But to define curvature is not enough. The man who would understand gravity deeply must also see curvature at work, producing relative accelerations of particles in_ Newtonian spacetime (Chapter 12); he must learn how, in Einstein spacetime, distances (metric) determine completely the curvature and the law of parallel trans port (Chapter 13);", " he must be the master of powerful tools for computing curvature (Chapter 14); and he must grasp the geometric significance of the algebraic and differential symmetries of curvature (Chapter 15). Unfortunately, such deep understanding requires time-far more time than one can afford in a ten-week or fifteen-week course, far more than a lone reader may wish to spend on first passage through the book. For the man who must rush on rapidly, this chapter contains a \"Track-I\" overview of the essential mathematical tools (\u00a7\u00a78.4-8.7). From it one can gain an adequate, but not deep, understanding ofspacetime curvature, of tidal gravitational forces, and of the mathematics ofcurved spacetime. This overview is also intended for the Track-2 reader; it will give him a taste of what is to come. The ambitious reader may also wish to consult other introductions to differential geometry (see Box 8.1). Box 8.1 BOOKS ON DIFFERENTIAL GEOMETRY current There are several mathematics texts that may be consulted for a more detailed and extensive dis cussion of modern differential geometry along the line taken here. Bishop and Goldberg (1968) is the no. 1 reference. Hicks (1965) could be chosen as a text, with standard graduate-level O'Neill (1966) at the undergraduate level intro ducing many of the same topics without presuming that the reader finds easy and obvious the current style in which pure mathematicians think and write. Auslander and MacKenzie (1963) at a somewhat more advanced level also allow for the reader to whom differential equations are more familiar than homomorphisms. Willmore (1959) is easy to read but presents no challenge, and leads to little progress in adapting to the style of current mathematics. Trautman (1965) and Misner (1964a, 1969a) are introductions somewhat similar to ours, except for deemphasis of pictures; like ours, they are aimed at the student of relativity. Flanders (1963) is easy and useful as an introduction to exterior differential forms; it also gives examples of their application to a wide variety of topics in physics and engineering. \u00a78.2. TRACK 1 VERSUS TRACK 2 197 \u00a78.2. TRACK 1 VERSUS TRACK 2: DIFFERENCE IN OUTLOOK AND POWER Nothing is more wonderful about the relation between Einstein's theory of gravity and Newton's theory than this, as discovered by Elie Cartan (1923,1924): that both that in both this theories lend themselves to des0iption in terms of curvature; curvature is governed by the density of mass-energy; and that this curvature allows itself to be defined and measured without any use of or reference to any concept of metric. The difference between the two theories shows itself up in this: Einstein's theory in the end (or in the beginning, depending upon how one presents it!) does define an interval between every event and every nearby event; Newton's theory not only does not, but even says that any attempt to talk of spacetime intervals violates Newton's laws. This being the case, Track 2 will forego for a time (Chapters 9-12) any use of a spacetime metric (\"Einstein interval\"). It will extract everything possible from a metric-free description of spacetime curvature (all of Newton's theory; important parts of Einstein's theory). Geodesic deviation is a measurer and definer of curvature, but the onlooker is forbidden to reduce a vector description of separation to a numerical measure of distance (no metric at this stage of the analysis): what an impossible situation! Nevertheless, that is exactly the situation with which Chapters 9-12 will concern themselves: how to do geometry without a metric. Speaking physically, one will overlook at this stage the fact that the geometry of the physical world is always and everywhere locally Lorentz, and endowed with a light cone, but one will exploit to the fullest the Galileo-Einstein principle of equivalence: in any given locality one can find a frame of reference in which every neutral test particle, whatever its velocity, is free of acceleration. The tracks of these neutral test particles define the geodesics of the geometry. These geodesics provide tools with which one can do much: define parallel transport (Chapter 10), define covariant derivative (Chapter 10), quantify geodesic deviation (Chapter 11), define spacetime curvature (Chapter 11), and explore Newtonian gravity (Chapter 12). Only after this full exploitation of metric-free geodesics will Track 2 admit the Einstein metric back into the scene (Chapters 13-15). But to forego use of the metric is a luxury which Track 1 can ill afford; too little time would be left for relativistic stars, cosmology, black holes, gravitational waves, experimental tests, and the dynamics of geometry. Therefore, the Track-l overview in this chapter keeps the Einstein metric throughout", ". But in doing so, it pays a heavy price: (1) no possibility ofseeing curvature at work in Newtonian spacetime (Chapter 12); (2) no possibility of comparing and contrasting the geometric structures of Newtonian spacetime (Chapter 12) and Einstein spacetime (Chapter 13), and hence no possibility of grasping fully the Newtonian-based motivation for the Einstein field equations (Chapter 17); (3) no possibility of understanding fulry the mathematical interrelationships of \"geodesic,\" \"parallel transport,\" \"covariant derivative,\" \"cur vature,\" and \"metric\" (Chapters 9,10,11,13); (4) no possibility of introducing the mathematical subjects \"differential topology\" (geometry without metric or covariant Preview of Track-2 differential geometry What the Track-l miss reader will 198 8. DIFFERENTIAL GEOMETRY: AN OVERVIEW derivative, Chapter 9) and \"affine geometry\" (geometry with covariant derivative but no metric, Chapters 10 and 11), subjects which find major application in modern analytical mechanics [see, e.g., Arnold and Avez (1968); also exercise 4.11 of this book],in Lie group theory with its deep implications for elementary particle physics [see, e.g., Hermann (1966); also exercises 9.12, 9.13, 10.16, and 11.12 of this book], in the theory and solution of partial differential equations [see, e.g., Sternberg (1969)], and, of course, in gravitation theory. \u00a78.3. THREE ASPECTS OF GEOMETRY: PICTORIAL, ABSTRACT, COMPONENT Gain the power in \u00a78,4 and Chapter 9 to discuss tangent vectors, I-forms, tensors in curved spacetime; gain the power in \u00a78.5 and Chapter 10 to parallel-transport vectors, to differentiate them, to discuss geodesics; use this power in \u00a78.7 and Chapter 11 to discuss geodesic deviation, to define curvature;.... But full power this will be only if it can be exercised in three ways: in pictures, in abstract notation, and in component notation (Box 8.3). Elie Cartan (Box 8.2) gave new insight into both Geometry from three viewpoints: pictorial, abstract, component Box 8.2 ELIE CARTAN, 1869-1951 Elie Cartan is a most remarkable figure in recent mathematical history. One learns from his obitu ary [Chern and Chevalley (1952)] that he was born a blacksmith's son in southern France, and proved the value of government scholarship aid by rising through the system to a professorship at the Sor bonne in 1912 when he was 43. At the age of32 he invented the exterior derivative [Cartan (1901)], which he used then mostly in differential equations and the theory of Lie groups, where he had al ready made significant contributions. He was about fifty when he began applying it to geometry, and sixty before Riemannian geometry specifically was the object of this research, including his text [Cartan (1928)], which is still reprinted and worth studying. Although universally recognized, his work did not find responsive readers until he neared retirement around 1940, when the \"Bour baki\" generation of French mathematicians began to provide a conceptual framework for (among other things) Cartan's insights and methods. This made Cartan communicable and teachable as his own writings never were, so that by the time of his death at 82 in 1951 his influence was obviously dominating the revolutions then in full swing in all the fields (Lie groups, differential equations, and differential geometry) in which he had prim arily worked. The modern, abstract, coordinate-free approach to geometry, which is used extensively in this book, is due largely to Elie Cartan. He also discovered the geometric approach to Newtonian gravity that is developed and exploited in Chapter 12. ~.....\u00b7 - :.. \u00a78...._3_._T_H_R_E_E_A_S_P_E_CT_S_O_F_G_E_O_M_E_T_R_y 1_9_9 Box 8.3 THREE LEVELS OF DIFFERENTIAL GEOMETRY (1) Purely pictorial treatment of geometry: tangent vector is conceived in terms of the separation of two points in the limit in which the points are indefinitely close; vectors are added and subtracted locally as in flat space; vectors at distinct points are compared by par allel transport from one point to another; transport is accomplished by a \"Schild's ladder construction\" of geodesics (Box 10.2); this parallel diagrams, yes; algebra, no; it is tied conceptually as closely as possible to the world", " of test particles and measure ments. (2) Abstract differential geometry: treats a tangent vector as existing in its Own right, without necessity to give its break down into components, A = AOeo + Alel + A2ez + A3e3, just as one is accustomed nowadays in elec tromagnetism to treat the electric vector E, without having to write out its components; uses a similar approach to differentiation (compare gradienfoperator V of elementary vector analysis, as distinguished from coor dinate-dependent pieces of such an opera tor, such as a/ax, a/ay, etc.); is the quickest, simplest mathematical scheme one knows to derive general results in differ ential geometry. (3) Differential geometry as expressed in the lan guage of components: is indispensible in programming large parts of general relativity for a computer; is convenient or necessary or both when one is dealing even at the level of elementary algebra with the most simple applications of relativity, from the expansion of the Fried mann universe to the curvature around a static center of attraction. Newtonian gravity (Chapter 12) and the central geometric simplicity of Einstein's field equations (Chapter 15), because he had full command of all three methods of doing differential geometry. Today, nO One has full power to communicate with others about the subject who cannot express himself in all three languages. Hence the interplay between the three forms of expression in what follows. It is not new to go back and forth between the three languages, as witnesses the textbook treatment of the velocity and acceleration of a planet in Kepler motion around the sun. The velocity is written (The hats '\".\" On ei- and eJ, signify that these are unit vectors.) The acceleration is (8.1) Planetary orbit as example of three viewpoints a =- = - e'+ --e' + V T dv\u00a2 dt \u00a2 dV T dt T dl' dt. de' _ dt T + v\u00a2=.\u00a3... de' dt (8.2) 200 8. DIFFERENTIAL GEOMETRY: AN OVERVIEW r 1=17 Sun ~ -..1 _ Figure 8.1. A Keplerian orbit in the sun's gravitational field, as treated using the standard Euclidean-space version of Newtonian gravity. The basis vectors themselves change from point to point along the orbit [equations (8.3)J. This figure illustrates the pictorial aspect of differential geometry. Later (exercise 8.5) it will illustrate the concepts of \"covariant derivative\" and \"connection coefficients.\" The unit vectors are turning (Figure 8.1) with the angular velocity w = dcp/dt; so de;. - = we- =-edt dq; dt <I> <1>' de:::;;:- = -we;. = - -;It e;.. dcp Thus the components of the acceleration have the values a;' = dv;' _ v\u00a2 dcp = d. dt dt 2 r _ r(dCP)2 dt2 dt and (8.3) (8.4a) (8.4b) Here is the acceleration in the language of components; a was the acceleration in abstract language; and Figure 8.1 shows the acceleration as an arrow. Each of these three languages will receive its natural generalization in the coming sections and chapters from two-dimensional flat space (with curvilinear coordinates) to four-dimensional curved spacetime, and from spacetime to more general manifolds (see \u00a79.7 on manifolds). Turn now to the Track-l overview of differential geometry. \u00a78.4. TENSOR ALGEBRA IN CURVED SPACETIME 201 \u00a78.4. TENSOR ALGEBRA IN CURVED SPACETIME To see spacetime curvature at work, examine tidal gravitational forces (geodesic deviation); and to probe these forces, make measurements in a finite-sized laboratory. Squeeze the laboratory to infinitesimal size; all effects ofspacetime curvature become infinitesimal~ the physicist cannot tell whether he is in flat spacetime or curved spacetime. Neither can the mathematician, in the limit as his domain of attention squeezes down to a single event, ':fo' At the event ':fo (in the infinitesimal laboratory) both physicist and mathematician can talk of vectors, of I-forms, of tensors~ and no amount of spacetime curvature can force the discussion to change from its flat-space form. A particle at ':f 0 has a 4-momentum p, with squared length Tensor algebra: (1) occurs in infinitesimal neighborhood of an event (2) is same in curved spacetime as in flat p2 =", " P' P = g(p,p) = _m 2. The squared length, as always, is calculated by inserting p into both slots of a linear machine, the metric g at ':fo. The particle also has a 4-acceleration a at ':fo; and, if the particle is charged and freely moving, then a is produced by the electromag netic field tensor F: ma = eF(..., u). In no way can curvature affect such local, coordinate-free, geometric relations. And in no way can it prevent one from introducing a local Lorentz frame at ':fo' and from performing standard, flat-space index manipulations in it: P 2 _ pap p 1J.- ap - _ pap a' But local Lorentz frames are not enough for the man who would calculate in curved spacetime. Non-Lorentz frames (nonorthonormal basis vectors {e a }) often simplify calculations. Fortunately, no effort at all is required to master the rules of \"index mechanics\" in an arbitrary basis at a fixed event ':fo' The rules are identical to those in flat spacetime, except that (1) the covariant Lorentz components 1Jap of the metric are replaced by (2) the contravariant components 1Jap are replaced by gaP, where gap ea' e p g(ea, ep); Le., g gPy - 8 y. - a' ap (8.5) (8.6) (8.6') (3) the Lorentz transformation matrix IIAa'pll and its inverse IIAP a' II are replaced by an arbitrary but nonsingular transformation matrix 1IL\"\"pll and its inverse IILP\"\"II: (3) rules for component manipulation change slightly when.using nonorthonormal basis Components of metric Transformation of basis (8.7) (8.8) 202 8. DIFFERENTIAL GEOMETRY: AN OVERVIEW (4) in the special case of \"coordinate bases,\" e a = a9/ax a, e{J' = a9/ax fJ ', (8.9) Components of Levi-Civita - tensor ana (5) the Levi-Civita tensor &, like the metric tensor, has components that depend--- on how nonorthonormal the basis vectors are (see exercise 8.3): if eo points toward the future and e l, e 2, e 3 are righthanded, then f afJys = (- g)1/2[0:,8y8], f afJYS = g-lfafJys = - (- g)-1/2[a,8y8], where [0:,8y8] is the completely antisymmetric symbol + 1 if 0:,8y8 Is' an even permutation of 0123, [0:,8y8]_ -1 if 0:,8y8 is an odd permutation of 0123, 1 o if 0:,8y8 are not all different, and where g is the determinant of the matrix IgafJ \" (8.10a) (8.10b) (8.11) Read Box 8.4 for full discussion and proofs; work exercise 8.1 below for fuller understanding and mastery. Several dangers are glossed over in this discussion. In flat spacetime one often does not bother to say where a vector, I-form, or tensor is located. One freely moves geometric objects from event to event without even thinking. Of course, the unwritten rule of transport is: hold all lengths and directions fix~d while moving; i.e., hold i.e., \"parallel-transport\" the object. But in all Lorentz-frame components fixed; Box 8.4 TENSOR ALGEBRA Al A FIXED EVENT IN AN ARBITRARY BASIS A. Bases Tangent-vector basis: Pick eo' e l, e 2, e 3 at 9 0 arbitrarily-but insist they be linearly independent. \"Dual basis\" for I-forms: The basis {ea } determines a I-form basis {wa } (its \"dual basis\") by [see equation (2.19)]. Geometric interpretation (Figure 9.2): e 2, e 3, eo lie parallel to surfaces of w l ; e l pierces precisely one surface of w l. Function interpretation: ( wa, e fJ>= 801.fJ determines the value of wa on any vector u = ufJefJ (number of \"bongs of bell\" as u pierces wa): \u00a78.4. TENSOR ALGEBRA IN CURVED SPACETIME 203 (wa, u) = (wa, ufJefJ ) = U", "anholonomic\") some cap y\u00a5-o (see exercise 9.9). \u00a78.4. TENSOR ALGEBRA IN CURVED SPACETIME 205 curved spacetime there is no global Lorentz coordinate system in which to hold components fixed; and objects initially parallel, after \"parallel transport\" along different curves cease to be parallel (\"geodesic deviation\"; Earth's meridians, parallel at equator, cross at north and south poles). Thus, in curved spacetime one must not blithely move a geometric object from point to point, without carefully specifying how it is to be moved and by what route. Each local geometric object has its own official place of residence (event ':fo); it can interact with other 05jects residing there (tensor algebra); but it cannot interact with any object at another event 2, until it has been carefully transported from ':fo to 2. This line of reasoning, pursued further, leads one to speak of the \"tangent space\" at each event, in which that event's vectors (arrows) and I-forms (families ofsurfaces) lie, and in which its tensors (linear machines) operate. One even draws heuristic pictures of the tangent space, as in Figure 9.1 (p. 231). Another danger in curved spacetime is the temptation to regard vectors as arrows linking two events (\"point for head and point for tail\")-Le., to regard the tangent space of Figure 9.1 as lying in spacetime itself. This practice can be useful for heuristic purposes, but it is incompatible with complete mathematical precision. (How is the tangent space to be molded into a warped surface?) Four definitions of a vector were given in Figure 2.1 (page 49): three definitions relying on \"point for head and point for tail\"; one, \"d':f/ d\"A.\", purely local. Only the local definition is wholly viable in curved spacetime, and even it can be improved upon, in the eyes of mathematicians, as follows. There is a one-to-one correspondence (complete \"isomorphism\") between vectors u and directional derivative operators au' The concept of vector is a bit fuzzy, but \"directional derivative\" is perfectly well-defined. To get rid of all fuzziness, exploit the isomorphism to the full: define the tangent vector u to be equal to the corre sponding directional derivative (8.12) (This practice, unfamiliar as it may be to a physicist at first, has mathematical power; so this book will use it frequently. For a fuller discussion, see \u00a79.2.) Vectors and tensors must not be moved blithely from point to point Tangent space defined Definitions of vector in curved spacetime: (1) as d'!i'/dA (2) as directional derivative Exercise 8.1. PRACTICE WITH TENSOR ALGEBRA Let t, x, y, z be Lorentz coordinates in flat spacetime, and let EXERCISES o= cos- 1(z/r), ep = tan- 1(y/x) be the corresponding spherical coordinates. Then eo = o'.i'lot, e T = o'.i'for, e 9 = o'.i'/00, is a coordinate basis, and o'!i' e6 = at' o'.i' e;.=--;;;:-, o'.i' I e -= - - r sin 0 oep if> is a noncoordinate basis. 206 8. DIFFERENTIAL GEOMETRY: AN OVERVIEW (a) Draw a picture of e9' e\u00a2, e9' and e \u00a2 at several different points on a sphere of constant t, r. [Answer for e9' eo should resemble Figure 9.1.] (b) What are the I-form bases {w\"} and {we.} dual to these tangent-vector bases? [Answer: WO = dt. w T = dr. we = dO. w o = d9; wi! = dt. wi = dr, we = rdO, w\u00a2 = r sin 0 d9.] (c) What is the transformation matrix linking the original Lorentz frame to the spherical_ coordinate frame {e,,}? [Answer: nonzero components are LZ r = sin 0 cos 9, U T = sin 0 sin cp, aLZ = -=- = cosO. ar r 9 = rcosOcoS9, L Z LV 9 = r cos 0 sin 9, Z _ oz -. 0 L 9 - an - - r SIn, L Z \u00a2 = - r sin 0 sin cp, U\u00a2 = rsinOcos9.] (d) Use this transformation matrix to calculate the metric components g\"p in the spherical coordinate basis", ". and invert the resulting matrix to get g\"p. [Answer: goo = -I, gOO = -I, gTT = I, gTT = I, g99 = r2 g99 = r-2,, g\u00a2\u00a2 = r2 sin2 0, g\u00a2\u00a2 = r- 2 sin- 2 0, all other g\"p = O. all other g\"p = 0.] 1)\"13; i.e. that (e) Show that the noncoordinate basis {ea} is orthonormal everywhere; i.e., that gaP = 9 = -wo\u00ae W o+ wi \u00ae w T+ w 9 \u00ae w 9 + w\u00a2 \u00ae w\u00a2. (f) Write the gradient of a function f in terms of the spherical coordinate and noncoordi nate bases. [Answer: df = af dt + of dr + af dO + of dcp ao acp at alf\u00b7 at ar alf\u00b7 or I alf\u00b7 I ~. =-w O + -w T + __ w 9 + -.--w\u00a2.] r 00 r SIn 0 acp (g) What are the components of the Levi-Civita tensor in the spherical coordinate and noncoordinate bases? [Answer for coordinate basis: \u20acOT9\u00a2 = -\u20acr09\u00a2 = +\u20acT90\u00a2 = '\" = r 2 sin 0, \u20acOr9\u00a2 = _\u20acT09\u00a2 = +\u20acT9\u00a2O =... = _r- 2 sin- t 0.]. Exercise 8.2. COMMUTATORS Take the mathematician's viewpoint that tangent vectors and directional derivatives are the same thing, u =au. Let u and v be two vector fields, and define their commutator in the manner familiar from quantum mechanics (a) Derive the following expression for [u, v], valid in any coordinate basis, (8.13a) (8.13b) Thus, despite the fact that it looks like a second-order differential operator, [u, v] is actually of first order-i.e., it is a tangent vector. (b) For any basis {e,,}, one defines the \"commutation coefficients\" cpy'\" and cpY\" by (8.14) Show that cpr\" = cpY\" = 0 for any coordinate basis. \u00a78.5. PARALLEL TRANSPORT AND COVARIANT DERIVATIVE 207 (c) Calculate cpr\" for the spherical noncoordinate basis of exercise 8.1. [Answer: All vanish except cri/i = -ce/l = -I/r, cr/I> = -c;il> = -I/r, ceil> = -cif,/I> = -cotO/r.] Exercise 8.3. COMPONENTS OF LEVI-CIVITA TENSOR IN NONORTHONORMAL FRAME (a) Show that expressions (8.10) are the components of I: in an arbitrary basis, with 8 0 pointing toward the future and 8 1, 8 2,83 right-handed. [Hints: (I) Review the discussion of I: in Lorentz frames, given in exercise 3.13. (2) Calculate ea{3ya and e a {3ya by transforming from a local Lorentz frame {8it}, e.g., e a{3ya - - Lit LP L~ LP eO',\u00b7 a P.PAp\u00b7 y a {3 (3) Show that these expressions reduce to (4) Show, from the transformation law for the metric components, that (detIUit\")2 detllga {311 = -I. (5) Combine these results to obtain expressions (8.10).] (b) Show that the components of the permutation tensors [defined by equations (3.50h) (3.50j)] have the same values [equations (3.50k)-(3.50m)] in arbitrary frames as in Lorentz frames. Additional exercises on tensor algebra: exercises 9.3 and 9.4 (page 234). \u00a78.5. PARALLEL TRANSPORT, COVARIANT DERIVATIVE, CONNECTION COEFFICIENTS, GEODESICS The vehicle that carries One from classical mechanics to quantum mechanics is the correspondence principle. Similarly, the vehicle between flat spacetime and curved spacetime is the equivalence principle: \"The laws of physics are the same in any local Lorentz frame of curved spacetime as in a global Lorentz frame of flat space time.\" But to apply the equiv~lence principle, one must first have a mathematical representation of a local Lorentz frame. The obvious choice is this: A local Lorentz frame at a given event", " Cfo is the closest thing there is to a global LOrentz frame at that event; i.e., it is a coordinate system in which Equivalence principle as vehicle between flat spacetime and curved Local Lorentz frame: mathematical representation (8.l5a) and in which gp.p holds as tightly as possible to 1Jp.p in the neighborhood of <}'o: More tightly than this it cannot hold in general [gp.p,a,8(&'o) cannot be set to zero]; spacetime curvature forces it to change. [Combine \u00a7 11.5 with equations (8.24) and (8.44).] (8.l5b) Parallel transport defined Covariant derivative defined 208 8. DIFFERENTIAL GEOMETRY: AN OVERVIEW An observer in a local Lorentz frame in curved spacetime can compare vectors and tensors at neighboring events, just as he would in flat spacetime. But to make the comparison, he must parallel-transport them to a common event. For him the act of parallel transport is simple: he keeps all Lorentz-frame components fixed, just as if he were in flat spacetime. But for a man without a local Lorentz frame perhaps with no coordinate system or basis vectors at all-parallel transport is less trivial. He must either ask his Lorentz-based friend the result, or he must use a more sophisticated technique. One technique he can use-a \"Schild's ladder\" construction that requires no coordinates or basis vectors whatsoever-is described in \u00a71O.2 and Box 10.2. But the Track-l reader need not master Schild's ladder. He can always ask a local Lorentz observer what the result of any given parallel transport iS,or he can use general formulas worked out below. Comparison by parallel transport is the foundation on which rests the gradient of a tensor field, VT. No mention of parallel transport was made in \u00a73.5, where the gradient was first defined, but parallel transport occurred implicitly: one defined VTin such a way that its components were T\"'p,y = aT\"'plaxY [for Ta CD tensor]; i.e., one asked VT to measure how much the Lorentz-frame components of Tchange from point to point. But \"no change in Lorentz components\" would have meant \"parallel transport,\" so one was implicitly asking for the change in T relative to what T would have been after pure parallel transport. To codify in abstract notation this concept pf differentiation, proceed as follows. First define the \"covariant derivative\" V u T of T along a curve 9(;\\'), whose tangent vector is u = d9Id;\\.: (V T) u at\u00a7'(OI - - Lim {T[9(e)]parallel-transportedto\u00a7'(01 - T[9(O)]}.....0 e. (8.16) Gradient defined (See Figure 8.2 for the special case where T is a vector field v.) Then define VT to be the linear machine, that gives V u T when u is inserted into its last slot: (8.17) The result is the same animal (\"gradient\") as was defined in \u00a73.5 (for proof see exercise 8.8). But this alternative definition makes clear the relationship to parallel transport, including the fact that Vu T = 0 --- T is parallel-transported along u = d9 I d;\\,. (8.18) In a local Lorentz frame, the components of VT are directional derivatives of the components of T: TP\"',r Not so in a general basis. If {ep(9)} is a basis that varies arbitrarily but smoothly from point to point, and {w\"'(9)} is its dual basis, then VT = V(TP\",ep \u00ae w\"') will contain contributions from Vep and VW\"', as well as from VTP\", dTP\", = TP\"\"ywY. To quantify the contributions from Vep and Vw\"', i.e., to quantify the twisting, turning, expansion, and contraction of the basis vectors and I-forms, one defines \"connection coefficients\": Connection coefficients defined \u00a78.5. PARALLEL TRANSPORT AND COVARIANT DERIVATIVE 209 '\\=3 Figure 8.2. Definition of the covariant derivative \"\"uv\" of a vector field v along a curve <j'(,\\), with tangent vector u == d<j'ldA: (I) choose a point <j'(0) on the curve, at which to evaluate \"uv. (2) Choose a nearby point <j'(f) on the curve. (3) Parallel-transport V[<j'(f)] along the curve back to <j", " deep into gravitation theory; and its rules of manipulation \u00a78.5. PARALLEL TRANSPORT AND COVARIANT DERIVATIVE 211 are simple enough to be learned easily. By contrast, the abstract notation (VT, V u T, etc.) is poorly suited to complex calculations; but it possesses great conceptual power. This contrast shows clearly in the way the two notations handle the concept of geodesic. A geodesic of spacetime is a curve that is straight and uniformly parame trized, as measured in each local Lorentz frame along its way. If the geodesic is timelike, then it is a possible world line for a freely falling particle, and its uniformly ticking parameter \"A. (called \u00abaffine parameter\") is a multiple of the particle's proper time;\"A. = aT + b. (Principle of equivalence: test particles move on straight lines in local Lorentz frames, and each particle's clock ticks at a uniform rate as measured by any Lorentz observer.) This definition of geodesic is readily translated into abstract, coordinate-free language: a geodesic is a curve 9(\"A.) that parallel-transports its tangent vector u = d9ld\"A. along itself- (8.25) (See Figure 10.1.) What could be simpler conceptually? But to compute the geodesic, given an initial event 9 0 and initial tangent vector u(9o) there, one must use the component formalism. Introduce a coordinate system x\"'(9), in which u'\" = dx'\"I d\"A., and write the component version of equation (8.25) as Geodesic and affine parameter defined _ D(dx\"'ld\"A.) _ d(dx\"'ld\"A.) d\"A. - d\"A. o - ('\" dXIJ.) dxY d\"A. + r IJ.Y d\"A. [see equation (8.23), with one less index on Tl; i.e., (8.26) Geodesic equatio.n This geodesic equation can be solved (in principle) for the coordinates of the geodesic, x\"'(\"A.), when initial data [x'\" and dx'\"Id\"A. at \"A. = \"A.ol have been specified. The geodesics of the Earth's surface (great circles) are a foil against which one can visualize connection coefficients; see Figure 8.3. The material of this section is presented more deeply and from a different view point in Chapters 10 and 13. The Track-2 reader who plans to study those chapters is advised to ignore the following exercises. The Track-l reader who intends to skip Chapters 9-15 will gain necessary experience with the component formalism by working exercises 8.4-8.7. Less important to him, but valuable nonetheless, are exercises 8.8-8.15, which develop the formalism of covariant derivatives and con nection coefficients in a systematic manner. The most important results of these exercises will be summarized in Box 8.6 (pages 223 and 224). Exercise 8.4. PRACTICE IN WRITING COMPONENTS OF GRADIENT Rewrite the following quantities in terms of ordinary derivatives <ly =0ej= 'Vy/) and \"rT\" correction terms: (a) T;y where T is a function. (b) Ta;y where Tis a vector. (c) Ta;y where T is a I-form. (d) Tapa';y. [Answer: EXERCISES (al T;y:::: T.r (d) Tapa':,:::: Tapa'.,! + rafJyTfJpa' - (b) Ta;,:::: Ta,y + rafJ,pJ. rfJpyTafJa' - (c) Ta;y:::: Ta,'! - rfJayT(J' rfJayTa p/ + r'fJ,Tapl.] 212 p v -~-----,.-- Figure 8.3. The why of connection coefficients, schematically portrayed. The aviator pursuing his great circle route from Peking to vancouver finds himself early going north, but later going south, although he is navigating the straightest route iliat is at all open to him (geodesic). The apparent change in direction indicates a turning. not in his route, but in the system of coordinates with respect to which his route is described. The vector v of his velocity (a vector defined not on spacetime but rather on the Earth's two-dimensional surface), carried forward by parallel transport from an earlier moment to a later moment, finds itself in agreement with the velocity that he is then pursuing; or. in the abstract language of coordinate-free differential geometry. the covariant derivative\"v", "V vanishes along the route (\"equation of a geodesic\"). Though v is in this sense constant, the individual pieces of which the navigator considers this vector to be built, v = u8e B+ V\u00a2e\u00a2. are not constant. In the language of components, the quantities uB and u'\" are changing along the route at a rate that annuls the covariant derivative of v; thus or \"vv = a = a4'e\u00a2 + aBeB= 0, In this sense the connection coefficients r imn serve as \"turning coefficients\" to tell how fast to \"turn\" the components of a vector in order to keep that vector constant (against the turning influence of the base vectors). vectors along the plane's route: Alternatively, the navigator can use an \"automatic pilot system\" which parallel-transports its own base \"VeB' = \"ve\"\" = 0; solid vectors at (! become dotted vectors at!ii. Then the components of v must be kept fixed to achieve a great-circle route. dV\u00a2' dUB' - -= - -=0 ; dt dt and the turning coefficients are used to describe the turning of the lines of latitude and longitude relative to this parallel-transported basis: \"VeB = emrmenUn, \"\"s\", = emr m\",,,vn. The same turning coefficients enter into both viewpoints. The only difference is in how these coefficients are used. \u00a78.5. PARALLEL TRANSPORT AND COVARIANT DERIVATIVE 213 Exercise 8.5. A SHEET OF PAPER IN POLAR COORDINATES The two-dimensional metric for a flat sheet of paper in polar coordinates (r, B) is ds 2 = dr2 + r2dep2-or, in modern notation, 9 = dr \u00ae dr + r2 dep \u00ae dep. (a) Calculate the connection coefficients using equations (8.24). [Answer: r T\u00a2\u00a2 = -r; r\u00a2T\u00a2 = r\u00a2\u00a2T = l/r; all others vanish.] (b) Write down the geodesic equation in (r, ep).coordinates. (c) Solve this geodesic equation for r(A) and ep(A), and show that the solution is a uniformly r(depldA)2 = 0; d2epldA 2 + (2/r)(drldA)(depldA) = 0.] parametrized straight line (x =r cos ep = aA + b for some a and b; y =r sin ep =jA + k for the noncoordinate basis e, - e T= a'!'lor, e \u00a2 = r-Ie\u00a2 r- 1 o<J'loep, w T= dr, w\u00a2 = r dep is orthonormal, and that (w a, ep) = {jap. Then calculate the connection coefficients of this basis from a knowledge [part (a)] of the connection of the coordinate basis. [Answer: some j and k). (d) Verifx. that [Answer: d2rldA 2 r\u00a2\u00a2, = (w\u00a2, V,e\u00a2) = (rdep, V.(r-1e\u00a2) = r(dep, (VTr- 1)e\u00a2 + r-I(VTe,;) = r(dep, -r-2e\u00a2) + (dep, VTe,) = -r- I + r\u00a2\u00a2T = -r- I + r- I = 0; similarly, r\u00a2T\u00a2 = + l/r, r T\u00a2\u00a2 = -l/r; all others vanish.] (e) Consider the Keplerian orbit of Figure 8.1 and \u00a78.3 as a nongeodesic curve in the sun's two-dimensional, Euclidean, equatorial plane. In place of the old notation dvldt, de,ldt, etc., use the new notation Vvv, Vve T, etc. Then v = dr:P Idt is the tangent to the orbit, and a = Vv v is the acceleration. Derive equations (8.4) for aT and a\u00a2 using component manipu lations and connection coefficients in the orthonormal basis. Exercise 8.6. SPHERICAL COORDINATES IN FLAT SPACETIME The spherical noncoordinate basis tea} of Exercise 8.1 was orthonormal, gap = 1)a.8' but had nonvanishing commutation coefficients [part (c) of Exercise 8.2]. (a) Calculate the connection coefficients for this basis, using equations (8.24). [Answer: r il \u2022\u2022 - rL - _r T,. - _r T\u2022. \u00a2\u00a2 - T\u00a2 -.9 -", " II,., T9 r\u00a2~\u00b7 u\u00a2 - - r il \u2022\u2022 - cot BI,., \u00a2\u00a2 - all others vanish.] (b) Write down expressions for V;,e p in terms of e y, and verify the correctness of these expressions by drawing sketches of the basis vectors on a sphere of constant t and r. [Answer: V\",eil = (cotBlr)e<i>' All others vanish.] (C) Calculate the divergence of a vector, V\u00b7 A = Aa:a, in this basis. [Answer: This answer should be familiar from flat-space vector analysis.] Exercise 8.7. SYMMETRIES OF CONNECTION COEFFICIENTS From equation (8.24b), the symmetry of the metric, and the antisymmetry (c{3fP. = -c,{3p') 214 8. DIFFERENTIAL GEOMETRY: AN OVERVIEW of the commutation coefficients. show that: ro1f3y] = 0 (last two indices are symmetric) in a coordinate basis: r(aJj)y = 0 (first two indices are antisymmetric) in a globally orthonormal basis. gail = 1)0/3\u00b7 SYSTEMATIC DERIVATION OF RESULTS IN \u00a78.5 Exercise 8.8. NEW DEFINITION OF VT COMPARED WITH OLD DEFINITION The new definition of VT is given by equations (8.16) and (8.17). Use the fact that parallel transport keeps local-Lorentz components fixed to derive, from (8.16), the Lorentz-frame equation VuT = T/3o,yuYe/3 0 W O \u2022 From this and equation (8.17), infer that the Lorentzframe components of V Tare T/3 0, y-which accords with the old definition of V T. Exercise 8.9. CHAIN RULE FOR VuT (a) Use calculations in' a local Lorentz frame to show that \"Vu\" obeys the standard chain rule for derivatives: Here A and B are arbitrary vectors, I-forms, or tensors; andfis an arbitrary function. [Hint: assume for concreteness that A is a (I) tensor and B is a vector. Then this equation reads, in Lorentz-frame component notation, (8.27') (b) Rewrite equation (8.27) in component notation in an arbitrary basis. [Answer: same as (8.27'), except \",\" is replaced everywhere by\";\". But note that/;aua = faua, because the function f \"has no components to correct\".]. Exercise 8.10. COVARIANT DERIVATIVE COMMUTES WITH CONTRACTION (a) Let S be a mtensor, Using components in a local Lorentz frame show that Vu (contraction on slots I and 2 of S) = (contraction on slots I and 2 ofVuS). (8.28) [Hint: in a local Lorentz frame this equation makes the trivial statement (2: SOo(3) u Y = 2: (soo/3,yuY).] a,Y a Exercise 8.11. ALGEBRAIC PROPERTIES OF V Use calculations in a local Lorentz frame to show that for all tangent vectors u, v and numbers G, b; also that for any two tensor fields Sand M of the same rank; also that (8.29) (8.30) \u00a78.5. PARALLEL TRANSPORT AND COVARIANT DERIVATIVE VuW - Vwu = [u, w], - - -..'215 (8.31 ) t,ommutato, of U,od w;] 4 discussed in exercise 8.2 for any two vector fields u and w. Exercise 8.12. CONNECTION COEFFICIENTS FOR 1-FORM BASIS Show that the same connection coefficients r a py that describe the changes in {e p } from point to point [definition (8.19a)] also describe the changes in twa}, except for a change in sign [equation (8.19b)]. {Answer: (I) (wa,ep) = sap is a constant function (0 or I, de pending on whether ex = {3). (2) Thus, Vy(w a, e p) = 0e (w a, e p) = o. (3) But (w a, e p) is the contraction of w a \u00ae e p' so equation (8.28) implie; 0 = Vy(contraction of w a \u00ae e p) = contraction of [Vy(w a \u00ae e", " p)]. (4) Apply the chain rule (8.27) to conclude 0 = contraction of [(Vyw a) \u00ae e p + w a \u00ae (Vyep)] = (Vyw a, e p) + (w a, Vyep). (5) Finally, use defini tion (8.l9a) to arrive at the desired result, (8.19b).} Exercise 8,13. \"rT\" CORRECTION TERMS FOR TPa;y Derive equation (8.21) for TPa;y in an arbitrary basis by first calculating the components of VuT for arbitrary u, and by then using equation (8.17) to infer the components of VT, [Answer: (I) Use the chain rule (8.27) to get VuT = Vu(TPaep \u00ae w a) = (VuTPa)e p \u00ae w a + TPa(Vuep) \u00ae w a + T 13ae p \u00ae (Vuw a). (2) Write u in terms of its components, u = uYe y; use linearity of Vu in u from equation (8.29), to get Vu = UyVy; and use this in Vu T: VuT = u1 { TPa,yep \u00ae w a + TPa(Vye p) \u00ae w a + TPaep \u00ae (Vyw a )}. (3) Use equations (8.19a,b), rewritten as (8.32) to put Vu T in the form VuT = uY{TPa,yep \u00ae w a + rfJpyTPaefJ \u00ae w a - rafJyTPaep \u00ae w fJ }. (4) Rename dummy indices so that the basis tensor e p \u00ae w a can be factored out: (5) By comparison with Vu T = V T(...,..., u) = (TI3 a;yuY)ep \u00ae w a, read off the value of TPa;r] Exercise 8.14. METRIC IS COVARIANTLY CONSTANT Show on physical grounds (using properties of local Lorentz frames) that Vg = 0 (8.33) 216 8. DIFFERENTIAL GEOMETRY: AN OVERVIEW or, equivalently, that Vug = 0 for any vector u. Then deduce as a mathematical consequence the obviously desirable product rule [Answer: (I) As discussed following equation (8.18), the components ofVg in a local Lorentz frame are gp.I',a' Just use 9 for T in that discussion. But these components all vanish by equation (8.15b). Therefore equation (8.33) holds in this frame, and-as a tensor equation-in all frames. (2) The product rule is also a tensor equation, true immediately via components in a local Lorentz frame. (3) Prove the product rule also the hard way, to see where equation (8.33) enters. Use the chain rule of exercise 8.9 to write Vu(g 0 A \u00ae B) = (Vug) 0 A \u00ae B + 9 \u00ae (V~) \u00ae B + 9 \u00ae A \u00ae (VuB). Use tquation (8.33) to drop one term, then contract, forming A \u2022 B = contraction (g \u00ae A \u00ae B) and the other inner products. Exercise 8.10 is used to justify commuting the contraction with Vu on the lefthand side.] Exercise 8.15. CONNECTION COEFFICIENTS IN TERMS OF METRIC Use the fact that the metric is covariantly constant [equation (8.33)] to derive equation (8.24b) for the connection coefficients. Treat equation (8.24c) as a definition of Tp. py in terms of Ta py\u2022 [Answer: (I) Calculate the components of Vg in an arbitrary frame: gap;y = 0 = gap,y - TP.aygp.P - TP.pygp.ix = gap,y - T pay - T apy; thereby conclude that gap,y = 2T(a{3)y' (Round brackets denote symmetric part.) (2) Construct the metric terms in the claimed answer for T p.py: I'2 (gp.p,y + gp.y,p - gpy,p.) = T(p.{3)y + T(P.YlP - T(PYlp. I= '2 [Tp. py + Tpp.y + Tp.yp + Typ. p - Tpyp. - Typp.] = Tp. py + (-Tp.[pYI + Tp[p.yl + Ty[p.PI)\u00b7 (", "3) Infer from equation (8.31), with u and w chosen as two basis vectors (u =e\"\" w = e v) that i.e., Tp [p.vl I. - __ cPo 2 p.v, - (8.34) (4) This, combined with step (2) yields the desired formula for Tp. py.] \u00a78.6. LOCAL LORENTZ FRAMES: MATHEMATICAL DISCUSSION 217 \u00a78.6: LOCAL LORENTZ FRAMES: MATHEMATICAL DISCUSSION An observer falling freely in curved spacetime makes measurements in his local Lorentz frame. What he discovers has been discussed extensively in Parts I and II of this book. Try now to derive his basic discoveries from the formalism of the last section. Pick an event '3'0 on the observer's world line. His local Lorentz frame there is Local Lorentz frame: a coordinate system x a('3') in which (Lorentz metric at '3'0)' and in which (8.35a) (8.35b) (metric as Lorentz as possible near '3'0). [See equation (8.15).] In addition, by virtue of equations (8.24), (no \"correction terms\" in covariant derivatives). Of course, the observer must be at rest in his local Lorentz frame; i.e., his world line must be Xo varying. (8.37) (8.36) Query: Equations (8.35) to (8.37) guarantee that the observer is at rest in a local Lorentz frame. Do they imply that he is freely falling? (They should!) Answer: Calculate the observer's 4-acceleration 8 = dujdr (notation of chapter 6) = Vuu (notation of this chapter). His 4-velocity, calculated from equation (8.37) is Origin falls freely along a geodesic so his 4-acceleration is u = (dxajdr)ea = (dxOjdr)eo = eo; because u and eo both1t have unit length r [ 8 = Vuu = Voeo = raooea = 0 at '3'0. (8.38) (8.39) Thus, he is indeed freely falling (8 = 0); and he moves along a geodesic (Vuu = 0). Query: Do freely falling particles move along straight lines (d 2x ajdr 2 = 0) in the observer's local Lorentz frame at '3'0? (They should!) Answer: A freely falling particle experiences zero 4-acceleration Freely falling particles move on straight lines I = V ti par c e 8 U Uparticle par c e ti I = 0; i.e., it parallel-transports its 4-velocity; i.e., it moves along a geodesic of spacetime 218 8. DIFFERENTIAL GEOMETRY; AN OVERVIEW with affine parameter equal to its proper time. The geodesic equation for its world line, in local Lorentz coordinates, says d 2x a dxIJ. dx p - -= _ra - - - dr\u00b7 dr 2 IJ.P dr = 0 at '!fa. Basis vectors at origin are Fermi-Walker transported The particle's world line is, indeed, straight at '3'0' Query: Does the freely falling observer Fermi-Walker-transport his spatial basis vectors e;; i.e., can he attach them to gyroscopes that he carries? (He should be able to!) Answer: Fermi-Walker transport (Box 6.2) would say dej dr t old notation t new notation But u = eo, eo' e j = 0, and a = 0 for the observer; so Fermi-Walker transport in this case reduces to parallel transport along eo: thus \"'oe; = O. This is, indeed, how e; is transported through 0/0' because \"'oe; = rajOe a = 0 at '3'0' \u00a78.7. GEODESIC DEVIATION AND THE RIEMANN CURVATURE TENSOR \"Gravitation is a manifestation of spacetime curvature, and that curvature shows up in the deviation of one geodesic from a nearby geodesic (relative acceleration of test particles).\" To make this statement precise, first quantify the \"deviation\" or \"relative acceleration\" of neighboring geodesics. Focus attention on a family of geodesics '3'(>-.., n); see Figure 804. The smoothly varying parameter n (\"selector parameter\") distinguishes one geodesic from the next. For fixed n, '", "3'(>-.., n) is a geodesic with affine parameter>-.. and with tangent vector u = 0'3'jo>-..; thus \"'uu = 0 (geodesic equation). The vector n = 0'3'jon (8040) (8041 ) measures the separation between points with the same value of>-.. on neighboring geodesics. An observer falling freely along the \"fiducial geodesic\" n = 0 watches a test particle fall along the \"test geodesic\" n = I. The velocity of the test particle relative \u00a78.7. GEODESIC DEVIATION AND RIEMANN CURVATURE TENSOR 219 / I \"\"\").. 1\\r I / / / / / n:::: O Figure 8.4. A family of geodesics ~(A, n). The selector parameter n tells \"which\" geodesic; the affine parameter A tells \"where\" on a given geodesic. The separation vector n =a~ jan at a point!1'(A, 0) along the fiducial geodesic, n = 0, reaches (approximately) to the point ~(A, I) with the same value of A on the test geodesic, n = I. to him he quantifies by \"\"un. This relative velocity, like the separation vector n, is an arbitrary \"initial condition,\" Not arbitrary, however, is the \"relative accelera tion,\" \"\"u \"\"un of the test particle relative to the observer (see Boxes 11.2 and 11.3). It would be zero in flat spacetime. In curved spacetime, it is given by \"\"u \"\"un + Riemann (..., u, n, u) = 0, (8.42) Riemann curvature tensor defined by relative acceleration of geodesics or, in component notation, (8.43) This equation serves as a definition of the \"Riemann curvature tensor;\" and it can also be used to derive the following expressions for the components of Riemann Components of Riemann in a coordinate basis: (8.44) Effects of curvature 220 8. DIFFERENTIAL GEOMETRY: AN OVERVIEW (For proof, read Box 1104, Box 11.5. and exercise 11.3, in that order.) For a glimpse of the man who first analyzed the curvature of spaces with three and more dimen sions, see Box 8.5. Spacetime curvature causes not only geodesic devi~tion, but als? route dependence in parallel transport (parallel transport around a closed curve changes a vector or to commute [equation tensor-Box 11.7); it causes covariant derivatives to fail (8044)]; and it prevents the existence of a global Lorentz coordinate system (\u00a711.5). At first sight one might think Riemann has 4 X 4 X 4 X 4 = 256 independent components. But closer examination (\u00a7 13.5) reveals a variety of symmetries Symmetries of Riemann (8045) Box 8.5 GEORG FRIEDRICH BERNHARD RIEMANN September 17, 1826, Breselenz, Hanover-July 20, 1866, Selasca, Lake Maggiore With his famous doctoral thesis of 1851, \"Founda tions for a general theory of functions of a single complex variable,\" Riemann founded one branch of modem mathematics (the theory of Riemann surfaces); and with his famous lecture of three years later founded another (Riemannian geom.etry). These and other writings will be found in his collected works, edited by H. Weber (1953). \"The properties which distinguish space from other conceivable triply-extended magnitudes are only to be deduced from experience.... At every point the three-directional measure of curvature can have an arbitrary value if only the effective curvature of every measurable region of space does not differ noticeably from zero.\" [G. F. B. Riemann, \"On the hypotheses that lie at the foun dations of geometry,\" Habilitationsvorlesung of June 10, 1854, on entry into the philosophical faculty of the University of Gottingen.] Dying of tuberculosis twelve years later, occu- \u00a78.7. GEODESIC DEVIATION AND RIEMANN CURVATURE TENSOR 221 (antisymmetry on first two indices; antisymmetry on last two; symmetry under exchange of first pair with last pair; vanishing of completely antisymmetric parts). These reduce Riemann (in four dimensions) from 256 to 20 independent compo nents. Besides these algebraic symmetries, Riemann possesses differential symmetries called \"Bianchi identities,\" which have deep geometric significance (", "Chapter 15). From Riemann one can form several other curvature tensors by contraction. The easiest to form are the \"Ricci curvature tensor,\" (8.46) Bianchi identities \\ pied with an attempt at a unified explanation of gravity and electromagnetism, Riemann commun icated to Betti his system of characterization of multiply-connected topologies (which opened the door to the view of electric charge as \"lines of force trapped in the topology of space\"), making use of numbers that today are named after Betti but that are identified with a symbol, Rrl' that honors Rie mann. \"A more detailed scrutiny of a surface might disclose that what we had considered an elemen-\u00b7 tary piece in reality has tiny handles attached to it which change the connectivity character of the piece, and that a microscope of ever greater mag nification would reveal ever new topological com plications of this type, ad infinitum. The Riemann point of view allows, also for real space, topologi cal conditions entirely different from those realized by Euclidean space. I believe that only on the basis of the freer and more general conception of geom etry which had been brought out by the develop ment of mathematics during the last century, and with an open mind for the imaginative possibilities which it has revealed, can a philosophically fruitful attack upon the space problem be undertaken.\" H. Weyl (1949, p. 91). \"But... physicists were still far removed from such a way of thinking; space was still, for them, a rigid, homogeneous something, susceptible of no change or conditions. Only the genius of Riemann, solitary and uncomprehended, had already won its way by the middle of the last century to a new conception of space, in which space was deprived of its rigidity, and in which its power to take part in physical events was recognized as possible.\" A. Einstein (1934, p. 68). Riemann formulated the first known model for superspace (for which see Chapter 43), a super space built, however, not of the totality of all 3 geometries with positive definite Riemannian met ric (the dynamic arena of Einstein's general relativity), but\u00b7of all conformally equivalent closed Riemannian 2-geometries of the same topology, a type of superspace known today as Teichmiiller space, for more on Riemann's contributions to which and the subsequent development of which, see the chapters by L. Bers and 1. A. Wheeler in Gilbert and Newton (1970). Ricci curvature tensor Scalar curvature Einstein curvature tensor 222 8. DIFFERENTIAL GEOMETRY: AN OVERVIEW R = Ra jJ..J' - jJ..D:JI - - r a r a + r a ~in a coordinate frame] jJ..J',o: jJ..o:,J' - po- and the \"scalar curvature,\" r /3 jJ..J' - r a f3v r/3 jJ..Q' (8.47) (8.48) But of much greater geometric significance is the \"Einstein curvature tensor\" Gil =1 lla /3YR p(Jl.' - 2 E /3Y 2 Evap(J - - RIl _loll R 2 v\u00b7 v (8.49) Of all second-rank curvature tensors one can form by contracting Riemann, only Einstein = G retains part of the Bianchi identities (8.46): it satisfies Contracted Bianchi identities (8.50) For the beautiful geometric meaning of these \"contracted Bianchi identities\" (\"the boundary of a boundary is zero\"), see Chapter 15. Box 8.6 summarizes the above equations describing curvature, as well as the fundamental equations for covariant derivatives. EXERCISE [The following exercises from Track 2 are appropriate for the Track-l reader who wishes to solidfy his understanding of curvature: 11.6, 11.9, 11.10, 13.7-11, and 14.3.] Exercise 8.16. SOME USEFUL FORMULAS IN COORDINATE FRAMES In any coordinate frame, define g to be the determinant of the matrix gaB [equation 8.11]. Derive the following relations, valid in any coordinate frame. (a) Contraction of connection coefficients: [Hint: Use the results of exercise 5.5.] (b) Components of Ricci tensor: (c) Divergence of a vector Aa or antisymmetric tensor Fa/3: (8.51a) (8.51b) (8.51 c) (d) Integral of a scalar field'!' over the proper volume of a 4-dimensional region '\"V: 1'!' d(pro", " \"Einstein Curvature Tensor R = (contraction of Ricci) G = Ricci -1 gR \"Symmetries of Curvature Tensors Bianchi Identities \"Contracted Bianchi Identities Geodesic Deviation Parallel Transport around closed curve (\u00a7IIA) Vu v~ + Riemann (..., u. n, u) = 0 SA + Riemann (... A, u, v) = 0 if u, v are edges of curve \"If metric is absent, these starred formulas cannot be formulated. All other formulas are valid in absence of metric. ------------...~ CHAPTER 9 DIFFERENTIAL TOPOLOGY frame. In analytic geometry. many relations which are independent of any frame must be expressed with respect to some particular It is therefore preferable to devise new methods methods which lead directly to intrinsic properties without any mention of coordinates. The development of the topology of general spaces and of the objects which occur in them. as well as the development of the geometry of general metric spaces. are steps in this direction. KARL MENGER, in Schilpp (1949), p. 467. \u00a79.1. GEOMETRIC OBJECTS IN METRIC-FREE, GEODESIC-FREE SPACETIME Curved spacetime without metric or geodesics or parallel transport, i.e., \"differential topology,\" is the subject of this easy chapter. It is easy because all the necessary geometric objects (event, curve, vector, I-form, tensor) are already familiar from flat spacetime. Yet it is also necessary, because one's viewpoint must be refined when one abandons the Lorentz metric of flat spacetime. Events The primitive concept of an event '3' (Figure 1.2) needs no refinement. The essential property here is identifiability, which is not dependent on the Lorentz metric struc ture of spacetime. This chapter is entirely Track 2. It depends on no preceding Track-2 material. It is needed as preparation for (1) Chapters 10-13 (differential geometry; Newtonian gravity). and (2) Box 30.1 (mixmaster cosmology). It will be helpful in (1) Chapter 14 (calculation of curvature) and in (2) Chapter 15 (Bianchi identities). y 226 9. DIFFERENTIAL TOPOLOGY Metric is abandoned Curves Geometric concepts must be refined Again no refinement. A \"curve\" 9(A.) is also too primitive to care whether spacetime has a metric-except that, with metric gone, there is no concept of \"proper length\" along the curve. This is in accord with Newton's theory of gravity, where one talks of the lengths of curves in \"space,\" but never in \"spacetime.\" Vectors Here refinement is needed. In special relativity one could dress primitive (\"identifi able\") events in enough algebraic plumage to talk of vectors as differences '3' - 2 between \"algebraic\" events. Now the plumage is gone, and the old bilocal (\"point for head and point for tail\") version of a vector must be replaced by a purely local version (\u00a79.2). Also vectors cannot be moved around; each vector must be attached to a specific event (\u00a7\u00a79.2 and 9.3). 1-Forms Almost no refinement needed, except that, with metric gone, there is no way to tell which I-form corresponds to a given vector (no way to raise and lower indices), and each I-form must be attached to a specific event (\u00a79.4). Tensors Again almost no refinement, except that each slot of a tensor is specific: if it accepts vectors, then it cannot accommodate I-forms, and conversely (no raising and lower ing of indices); also, each tensor must be attached to a specific event (\u00a79.5). \u00a79.2. \"VECTOR\" AND \"DIRECTIONAL DERIVATIVE\" REFINED INTO TANGENT VECTOR Flat spacetime can accommodate several equivalent definitions of a vector (\u00a72.3): a vector is an arrow reaching from an event '3'0 to an event 2 0; it is the parameterized '3'0) extending from '3'0 at A. = 0 to 2 0 at A. = I; straight line, '3'(A.) = '3'0 + A.(20 it is the rate of change of the point '3'(A.) with increasing A., d'3'/ dA.. With Lorentz metric gone, the \"arrow\" definition and the \"parametrized-straight line\" definition must break down. By what route is the arrow or line to be laid out between '3'0 and 2 01 There is no concept of straightness; all routes are equally", " straight or bent. Such fuzziness forces one to focus on the \"rate-of-change-of-point-along-curve\" Old definitions of vector break down when metric is abandoned \u00a79.2. TANGENT VECTOR AND DIRECTIONAL DERIVATIVE 227 Box 9.1 TANGEl'lT VECTORS AND TANGENT SPACE ~~\"n ~(P) ~(T/) A tangent vector d'3'I d\"A. is defined to be \"the limit, when N ----+ 00, of N times the displacement of '3' as \"A. ranges from 0 to liN.\" One cannot think of this final displacement d'3'Id\"A. as lying in space time; fuzziness forbids (no concept of straightness). Instead, one visualizes d'3'I d\"A. as lying in a \"tan gent plane\" or \"tangent space,\" which makes con tact with spacetime only at '3'(0), the event where d'3'I d\"A. is evaluated. All other tangent vectors at '3'(O)-e.g., d'3'ldp, d'3'ldri, d'3'ld~-lie in this same tangent space. To make precise these concepts of tangent space and tangent vector,one may regard spacetime as embedded in a flat space of more than four di- mensions. One can then perform the limiting process that leads to d'3'I d\"A., using straight arrows in the flat embedding space. The result is a higher dimensional analog of the figure shown above. that But such a treatment is dangerous. It suggests, the tangent vector d'3'Id\"A. and the falsely, tangent space at '3'0 depend on how the embedding is done, or depend for their existence on the em bedding process. They do not. And to make clear that they do not is one motivation for defining the directional derivative operator \"dldA.\" to be the tangent vector, rather than using Cartan's more pictorial concept \"d'3'I d\"A.\". definition, d'3'I d\"A.. It, under the new name \"tangent vector,\" is explored briefly in Box 9.1, and in greater depth in the following paragraphs. Even \"d'3'I d\"A.\" is a fuzzy definition of tangent vector, most mathematicians would argue. More acceptable, they suggest, is this definition: the tangent vector u to a curve '3'(\"A.) is the directional derivative operator along that curve u = au = (dld\"A.)a!ongcurve. (9.1) Tangent vector equals directional derivative operator? Preposterous! A vector started out as a happy, irresponsible trip from '3'0 to 20- It ended up loaded with the social responsibility to tell how something else changes at '3'0. At what point did the vector get saddled with this unexpected load? And did it really change its character all that much, as it seems to have done? For an answer, go back and try Best new definition: \"tangent vector equals directional derivative operator\" u = d/d\"A 228 9. DIFFERENTIAL TOPOLOGY to redo the \"rate-of-change-of-point\" definition, d'3'I d\"A., in the form of a limiting process: O. Choose a curve g>(\"A.) whose tangent vector dg>Id\"A. at \"A. = 0 is desired. 1. Take the displacement of g> as \"A. ranges from 0 to I; that is not d'j'I d\"A.. 2. Take twice the displacement of g> as \"A. ranges from 0 to!; that is not dg>I d\"A.. N. Take N times the displacement of '3' as \"A. ranges from 0 to 1IN; that is not d'3'Id\"A.. 00. Take the limit of such displacements as N ----+ 00; that is dg>Id\"A.. This definition has the virtue that d'3'I d\"A. describes the properties of the curve '3'(\"A.), not over the huge range from \"A. = 0 to \"A. = 1, where the curve might be doing wild things, but only in an infinitesimal neighborhood of the point g>0 = g>(0). The deficiency in this definition is that no meaning is assigned to steps I, 2,..., N,..., so there is nothing, yet, to take the limit of. To make each \"displacement of '3''' a definite mathematical object in a space where \"limit\" has a meaning, one can imagine the", " original manifold to be a low-dimensional surface in some much higher-dimensional flat space. Then '3'(11N) '3'(0) is just a straight arrow connect ing two points, i.e. a segment of a straight line, which, in general, will not lie in the surface itself-see Box 9.1. The resulting mental picture of a tangent vector makes its essential properties beautifully clear, but at the cost of some artifacts. The picture relies on a specific but arbitrary way of embedding the manifold of interest (metric free spacetime) in an extraneous flat space. In using this picture, one must ignore everything that depends on the peculiarities of the embedding. One must think like the chemist, who uses tinkertoy molecular models to visualize many essential prop erties of a molecule clearly, but easily ignores artifacts of the model (colors of the atoms, diameters of the pegs, its tendency to collapse) that do not mimic quantum mechanical reality. Elie Cartan's approach to differential geometry, including the d'3'I d\"A. idea of a tangent vector, suggests that he always thought of manifolds as embedded in flat spaces this way, and relied on insights that he did not always formalize to separate the essential geometry of these pictures from their embedding-dependent details. Acceptance of his methods of calculation came late. Mathematicians, who mistrusted their own ability to distinguish fact from artifact, exacted this price for acceptance: stop talking about the movement of the point itself, and start dealing only with concrete measurable changes that take place within the manifold, changes in any or all scalar functions f as the point moves. The limiting process then reads: O. Choose a curve '3'(\"A.) whose tangent vector at \"A. = 0 is desired. 1. Compute the number f['3'(1)] - f['3'(O)], which measures the change inf as the point '3'(\"A.) moves from '3'0 = '3'(0) to 2 0 = '3'(1). 2. Compute 2{f['3'(!)] - f['3'(O)]}, which is twice the change inf as the point goes from '3'(0) to '3'(!). N. Compute N{f['3'(1/N)]- f['3'(O)]}, which is N times the change infas the point goes from '3'(0) to '3'(1 IN). Alternative definition. u = d':f'/d\"A. requires embedding in flat space of higher dimensio nality Refinement of d'J'/ d\"A into d/d\"A \u00a79.2. TANGENT VECTOR AND DIRECTIONAL DERIVATIVE 229 00. Same in the limit as N ----+ 00: (change in f) = dfldA. O. The vector is not itself the change inf It is instead the operation dldA, which, when applied to f, gives the change dfldA. Thus tangent vector = dldA [cf. definition (9.1)]. The operation dldA clearly involves nothing but the last steps N ----+ 00 in this limiting process, and only those aspects of these steps that are independent of f But this means it involves the infinitesimal displacements of the point '3' and nothing more. One who wishes both to stay in touch with the present and to not abandon Cartan's deep geometric insight (Box 9.1) can seek to keep alive a distinction between: (A) the tangent vector itself in the sense of Cartan, the displacement d'3'IdA of a point; and (B) the \"tangent vector operator,\" or \"directional derivative operator,\" telling what happens to a function in this displacement: (tangent vector operator) =dldA. However, pres~nt practice drops (or, if one will, \"slurs\") the word \"operator\" in (B), and uses the phrase \"tangent vector\" itself for the operator, as will be the practice here from now on. The ideas (A) and (B) should also slur or coalesce in one's mind, so that when one visualizes an embedding diagram with arrows drawn tangent to the surface, one always realizes that the arrow characterizes an infinitesimal motion of a point d'3'IdA that takes place purely within the surface, and when one thinks of a derivative operator dldA, one always visualizes this same infinitesimal motion of a point in the manifold, a motion that must occur in constructing any derivative dldA as both \"a df('3')ldA. In this sense, one should regard a vector d'3'IdA displacement that carries", " attention from one point to another\" and \"a purely geo metric object built on points and nothing but points.\" The hard-nosed physicist may still be inclined to say \"Tangent vector equals directional derivative operator? Preposterous!\" Perhaps he will be put at ease by another argument. He is asked to pick an event '3'0' At that event he chooses any set of four noncoplanar vectors (vectors defined in whatever way seems reasonable to him); he names them eo, e 1, e2, e3 ; and he uses them as a basis on which to expand all other vectors at '3'0: (9.2) oSo along his He is asked to construct the four directional derivative operators 0a four basis vectors. As in flat spacetime, so also here; the same expansion coefficients that appear in u = uaea also appear in the expansion for the directional derivative: Isomorphism between directional derivatives and vectors (9.3) 230 9. DIFFERENTIAL TOPOLOGY Hence, every relation between specific vectors at ~i'\u00b0 induces an identical relation between their differential operators: u = aw + bv ~ uQ = awQ + buQ ~ Ou = a Ow + b ov' (9.4) There is a complete \"isomorphism\" between the vectors and the corresponding directional derivatives. So how can the hard-nosed physicist deny the hard-nosed mathematician the right to identify completely each tangent vector with its direc tional derivative? No harm is done; no answer to any computation can be affected. This isomorphism extends to the concept \"tangent space.\" Because linear relations (such as Ou = a Ow + b ov) among directional derivatives evaluated at one and the same point Po are meaningful and obey the usual addition and multiplication rules, these derivative operators form an abstract (but finite-dimensional) vector space called the tangent space at '3'0' In an embedding picture (Box 9.1) one uses these derivatives (as operators in the flat embedding space) to construct tangent vectors u = ou'!!' v = o.;f, in the form of straight arrows. Thereby one identifies the abstract tangent space with the geometrically visualized tangent space. \u00a79.3. BASES, COMPONENTS, AND TRANSFORMATION LAWS FOR VECTORS Tangent space defined Coordi nate-induced basis defined An especially useful basis in the tangent space at an event '3'0 is induced by any coordinate system [four functions, x\u00b0('3'), x l ('3'), x 2('3'), x 3('3')]: 80 _ ~ = curve with constant (Xl, x 2, x3) directional derivative along the) ( ) ox 1 2 x,x,x 3 ( and wIth parameter {\\ = X '0 \u2022 at &'0' (9.5) \u00b0 8 2 = - -2'ox (See Figure 9.1.) Changes of basis: transformation matrices defined A transformation from one basis to another in the tangent space at '3'0' like any change of basis in any vector space, is produced by a nonsingular matrix, (9.6) and, as always (including the Lorentz frames of flat spacetime), the components of a vector must transform by the inverse matrix (9.7) (9.8) \u00a79.4. '-FORMS 231 Figure 9.1. The basis vectors induced, by a coordinate sys tem, into the tangent space at each event. Here a truncated, two-dimensional spacetime is shown (two other dimensions suppressed), with coordi nates X(P) and \"'(P), and with corresponding basis vectors a/aX and a/a\",. This \"inverse\" transformation law guarantees compatibility between the expansions u = ea,ua' and u = e f3 uf3 : u = ea,ua' = (eyLYa,)(La'f3uf3) = e/>Yf3uf3 = e f3 uf3. In the special case of transformations between coordinate-induced bases, the transformation matrix has a simple form: a -;--;- = -;--;- --13 (by usual rules of calculus), uxa ax 13 a ux a ax so Lf3 a' - (a 1310 a') X X at event Po where tangent space lies' (9.9) (Note: this generalizes the Lorentz-transformation law x f3 = Af3a,xa', which has the differential form Af3a' = ax f3 laxa'; also, it provides a good way to remember the signs in the A matrices.) \u00a79.4. 1-FORMS When the Lorentz metric is removed from spacetime, one must sharpen up the concept of a", " I-form u by insisting that it, like any tangent vector u, be attached to a specific event '3'0 in spacetime. The family of surfaces representing u resides in the tangent space at '3'0' not in spacetime itself. The piercing of surfaces of u by an arrow u to produce the number (u, u) (\"bongs of bell\") occurs entirely in the tangent space. 232 9. DIFFERENTIAL TOPOLOGY ~---------Positive sense Figure 9.2. The basis vectors ea and dual basis I-forms wf3 in the tangent space of an event '17 0, The condition (w f3,ea>= of3 a dictates that the vectors e2 and e3 lie parallel to the surfaces of wI, and that eI extend from one surface of WI to the next (precisely 1.00 surfaces pierced). Notice that this picture could fit perfectly well into a book on X-rays and crystallography. There the vectors e1' e 2, e3 would be the edges of a unit cell of the crystal; and the surfaces of wI, w 2, w 3 would be the surfaces of unit cells. Also, for an X-ray diffraction experiment, with wavelength of radiation and orientation of crystal appropriately adjusted, the successive surfaces of WI would produce Bragg reflection. For other choices of wavelength and orientation, the surfaces of w 2 or w 3 would produce Bragg reflection. Dual basis of 1-forms defined Given any set of basis vectors {eo, e1, e2, e3 } at an event '3'0' one constructs the \"dual basis\" of I-forms {WO, wI, w 2, w 3} by choosing the surfaces of w/3 such that that (9.10) See Figure 9.2. A marvelously simple formalism for calculating and manipulating components of tangent vectors and I-forms then results: Component-manipulation formulas 0\" == (J/3w/3 u a == (w a, u) (definition of components of u), (definition of components of 0\"), (way to calculate components of u), (way to calculate components of 0\"), (way to calculate (0\", u) using components), (transformation law for I-form basis, corresponding to equation 9.6), (transformation law for I-form components). (9.1Ia) (9.11 b) (9.11c) (9.1Id) (9.11 e) (9.11 f) (9.1Ig) (Exercise 9.1 below justifies these equations.) \u00a79.5. TENSORS 233 In the absence of a metric, there is no way to pick a specific I-form ii at an event CJ'0 and say that it corresponds to a specific tangent vector u at CJ'0' The correspond ence set up in flat spacetime, Correspondence between vectors and 1-forms rubbed out (ii, v) = u' v for all v, was rubbed out when\"\u00b7\" was rubbed out. Restated in component language: the raising of an index, u a = 1/af3 uf3, is i!Jlpossible because the 1/af3 do not exist; similarly, lowering of an index, uf3 = 1/f3aua, is impossible. The I-form gradient dfwas introduced in \u00a72.6 with absolutely no reference to Gradient of a function metric. Consequently, it and its mathematical formalism are the same here, without metric, as there with metric, except that, like all other I-forms, df now resides in the tangent space rather than in spacetime itself. For example, there is no change in the fundamental equation relating the projection of the gradient to the directional derivative: (df, u) = ouf = u[f]. r old notation fOIl t Ldirectional derivativeJ t rnew notation;] '-lrecall u = ou. Similarly, there are no changes in the component equations, (expansion of df in arbitrary basis), (way to calculate components of dj), if {ea } is a coordinate basis, (9.12) (9.13a) (9.13b) except that they work in arbitrary bases, not just in Lorentz bases. And, as in Lorentz frames, so also in general: the one-form basis {dx a } and the tangent-vector basis {oloxa}, which are induced into a tangent space by the same coordinate system, are the duals of each other, (9.14) (See exercise 9.2 for proofs.) Also, most aspects of Cartan's \"Exterior Calculus\" (parts A, B, C of Box 4.1) are left unaffected by the removal of metric.", " \u00a79.5. TENSORS A tensor S, in the absence of Lorentz metric, differs from the tensors of flat, Lorentz spacetime in two ways. (l) S must reside at a specific event <jJ0' just as any vector or I-form must. (2) Each slot of S is specific; it will accept either vectors or I-forms, but not both, because it has no way to convert a I-form ii into a \"corresponding Specificity of tensor slots 234 9. DIFFERENTIAL TOPOLOGY vector\" u as it sends u through its linear machinery. Thus, if 5 is a Wtensor 5(..., '..,...), insert I-form hereJ r linsert vector here (9.15 ) insert vector here then it cannoT be converted alternatively to a (i) tensor, or a @ tensor, or a (~) tensor by the procedure of\u00a73.2. In component language, the indices of 5 cannot be raised and lowered. Except for these two restrictions (attachment to a specific event; specificity of slots), a tensor 5 is the same linear machine as ever. And the algebra of component manipulations is the same: saf3y = 5(w a, e 5 = Sa 5(0', U, v) = saf3y(Jauf3vY., e y) f3y e a @ w f3 @ wy, f3 (5, w a, ef3 must all reside at same event) (9.16) (9.17) (9.18) EXERCISES Exercise 9.1. COMPONENT MANIPULATIONS Derive equations (9.11c) through (9.llg) from (9.10), (9.1 la, b), (9.6), (9.7), and (9.8). Exercise 9.2. COMPONENTS OF GRADIENT, AND DUALITY OF. COORDINATE BASES In an arbitrary basis, define!a by the expansion (9.13a). Then combine equations (9.lld) and (9.12) to obtain the meth~d (9.13b) of computing!,a' Finally, combine equations (9.12) and (9.13b) to show that the bases {dx a } and {%x/3} are the duals of each other. Exercise 9.3. PRACTICE MANIPULATING TANGENT VECTORS Let 'f0 be the point with coordinates (x = 0, y = I, z = 0) in a three-dimensional space; and define three curves through '3'0 by '3'(11.) = (A, I, A), '3'm = (sin~, cos~, no '3'(p) = (sinh p, cosh p, p + p3). # (a) Compute (d/dA)f, (d/dnf, and (d/dp)f for the function f = x 2 y2 + Z2 at the point '3'o' (b) Calculate the components of the tangent vectors d/dA, d/d~, and d/dp at '3'0' using the basis {a/ox, %y, %z}. Exercise 9.4. MORE PRACTICE WITH TANGENT VECTORS In a three-dimensional space with coordinates (x,y, z), introduce the vector field v =y2 a/ox - x a/oz, and the functions f = xy, g = Z3. Compute (a) v[f] (b) v[g] (c) v[fg] (d) fv[g] - gv[f] (e) v[j2 + g2] (j) v{v[f]} \u00a79.6. COMMUTATORS AND PICTORIAL TECHNIQUES 235 Exercise 9.5. PICTURE OF BASIS 1-FORMS INDUCED BY COORDINATES In the tangent space of Figure 9.1, draw the basis I-forms do/ and dX induced by the 1f;, X-coordinate system. Exercise 9.6. PRACTICE WITH DUAL BASES In a three-dimensional space with spherical coordinates r, 0, cp, one often likes to use, instead of the basis a/or, 0/00, %cp, the basis a ei = or' a I e----rsinO ocp' \u00a2> - (a) What is the I-form basis {wi, wi!, w\u00a2} dual sphere r= I, draw pictures of the bases {%r,%O,%cp}.", " and {wi, wi!, w\u00a2}. to this tangent-vector basis? (b) On the {ei,ei!,e\u00a2}, {dr,dO,dC>}, \u00a79.6. COMMUTATORS AND PICTORIAL TECHNIQUES A vector U o given only at one point '3'0 suffices to compute the derivative uo[/l ouo!, which is simply a number associated with the point <j'o' In contrast, a vector field u provides a vector u('3')-which is a differential operator ou(,r)-at each point '3' in some region of spacetime. This vector field opera tes on a function I to produce ouf A second vector field v can not just a number, but another function u[/l perfectly well operate on this new function, to produce yet another function Does this function agree with the result of applying v first and then u? Equivalently, does the \"commutator\" [u, vJ[/l u{v[/J} - v{u[fJ} (9.19) Commutator defined vanish? The simplest special case is when,u and v are basis vectors of a coordinate system,, v = %x p. Then the commutator does vanish, because partial deriva u = %xa tives always commute: But in general the commutator is nonzero, as one sees from a coordinate-based calculation: y 236 9. DIFFERENTIAL TOPOLOGY Commutator of two vector fields is a vector field Notice however, that the commutator [u, v], like u and v themselves, is a vector field, i.e., a linear differential operator at each event: [u, v] = (u[v,B]- v[u,BJ)- = (uav,B,a o ox,B 0 vau,B ) -.,a ox,B (9.20) l Such results should be familiar from quantum theory's formalism for angular mo mentum operators (exercise 9.8). The three levels of geometry-pictorial, abstract, and component-yield three different insights into the commutator. (1) The abstract expression [u, v] suggests the close connection to quantum theory, and brings to mind the many tools devel oped there for handling operators. But recall that the operators of q~antum theory need not be first-order differential operators. The kinetic energy is second order and the potential is zeroth order in the familiar Schrodinger equation. Only first-order operators are vectors. (2) The component expression uav,B,a - vau,B,a' valid in any coordinate basis, brings the commutator into the reaches of the powerful tools of index mechanics. (3) The pictorial representation of [u, v] (Box 9.2) reveals its fundamental role as a \"closer of curves\"-a role that will be important in Chapter II's analysis of curvature. Commutators find application in the distinction between a coordinate-induced basis, {ea} = {o joxa}, and a noncoordinate basis. Because partial derivatives always commute, lea' e,B] = [ojoxa, ojox,B] = 0 in any coordinate basis. (9.21 ) Commutator as a \"closer of curves\" Box 9.2 THE COMMUTATOR AS A CLOSER OF QUADRILATERALS A. Pictorial Representation in Flat Spacetime 1. For ease of visualization, consider flat spacetime, so the two vector fields uW) and v{g\u00bb can be laid out in spacetime itself. 2. Choose an event ~1~0 where the commutator [u, v] is to be calculated. 3. Give the names g>1' g>2' g>3' g>4 to the events pictured in the diagram. 4. Then the vector g>4 - g>3' which measures how much the four-legged curve fails to close, can be expressed in a coordinate basis as ':Jo turns out to be [u, vI,.) \u00a79.6. COMMUTATORS AND PICTORIAL TECHNIQUES 237 '3'4 - '3'3 = [u('3'O) + v('3'l)] = [v('3'1) - v('3'O)] = (V f3 uae,,)<? = [U, v]<? + errors.,a fJ 0 [U('3'2) + v('3'O)] [U('3'2) - u('3'O)] (UP aVae,,)<? + errors fJ 0, o 4terms such as vf3,l' p ullu Pe", " ] f3 5. Notice that if u and v are halved everywhere, then [u, v] is cut down by a factor of 4, while the error terms in the above go down by a factor of 8. Thus, [u, v] represents accurately the gap in the four-legged curve (\"quadrilateral\") in the limit where u and v are sufficiently short; i.e., [u, v] \"closes the quadrila teral\" whose edges are the vectors fields u and v.. B. Pictorial Representation in Absence of Metric, or in Curved Spacetime with a Metric A:: - 0.5 A === _ 0.5 A = 0 1. The same picture must work, but now one dares not (at least initially) layout the vector fields in spacetime itself. Instead one lays out two families of curves: the curves for which u('3') is the tangent vector; and the curves for which v('1') is the tangent vector. 2. The gap \"'1'4 - '3'3\" in the four-legged curve can be characterized by the difference 1('3'4) - 1('3'3) in the values of an arbitrary function at '1'4 and '3'3' That difference is, in a coordinate basis, 238 9. DIFFERENTIAL TOPOLOGY Box 9.2 (continued) f(~1\\) - f(?P 3) = Jf(~'P4) - f(~1~ 1)] (.r.\"V a + ; f.af3 V\"Vf3t. 1 + [j(~1\\) - fer 0)] (.r.\"U\" + ; f.af3 U\"Uf3t \u2022 0 -Jf(~f2) - f(~1~O)] [j(?P3) - f ( 2)] (.r.\"V\"+ ;f.\"f3 V\"Vf3), (.r.\"U\"+ ;f.\"f3U\"Uf3) \"0 9 2 = [(f.\"Va),f3Uf3 (f.aUa),f3Vf3]~,O + \"cubic errors\" = [(Uf3 V\",f3 - Vf3U\",f3)Of/OX\"]90 + \"cubic errors\" = {[U, V][f]},r + \"cubic errors.\" o Here \"cubic errors\" are cut down by a factor of 8, while [U, v]f is cut down by one of 4, whenever u and v are cut in half. 3. The result says that [u, v] is a tangent vector at tj'0 that describes the separation between the points '3'3 and!1~4' Its description gets arbitrarily accurate when u and v get arbitrarily short. Thus, [u, v] closes the quadrilateral whose edges are the projections of u and v into spacetime. - C. Philosophy of Pictures 1. Pictures are no substitute for computation. Rather, they are useful for (a) suggesting geometric relationships that were previously unsuspected and that one verifies subsequently by computation; (b) interpreting newly learned geometric results.. 2. This usual noncomputational role of pictures permits one to be sloppy in drawing them. No essential new insight was gained in part B over part A, when one carefully moved the tangent vectors into their respective tangent spaces, and permitted only curves to lie in spacetime. Moreover, the original picture (part A) was clearer because of its greater simplicity. 3. This motivates one to draw \"sloppy\" pictures, with tangent vectors lying in spacetime itself-so long as one keeps those tangent vectors short and occa sionally checks the scaling of errors when the lengths of the vectors are halved. Vanishing commutator: a test for coordinate bases Commutation coefficients defined Physicists' notation vs. mathematicians' notation \u00a79.6. COMMUTATORS AND PICTORIAL TECHNIQUES 239 Conversely, if one is given a field of basis vectors (\"frame field\") {eaW)}, but one does not know whether a coordinate system {xaW)} exists in which {ea} = {a /ox a}, one can find out by a simple test: calculate all (4 X 3)/2 = 6 commutators rea' e.e]; if they all vanish, then there exists such a coordinate system. If not, there doesn't. Stated more briefly, {eaW)} is a coordinate-induced basis if and only If [ea, e.e] = 0 lor all e a and e p' (See exercise 9.9 for proof; see \u00a7 11.5 for an important application.) Coordinate-induced bases are", " sometimes called \"holonomic.\" In an \"anholonomic basis\" (noncoordinate basis), one defines the commutation coefficients cp. p a by (9.22) They enter into the component formula for the commutator of arbitrary vector fields u and v: (9.23) (see exercise 9.10). [Warning! In notation for functions and fields, mathematicians and physicists often use the same symbols to mean contradictory things. The physicist may write 1 when considering the length of some critical component in an instrument he is designing, then switch to leT) when he begins to analyze its response to temperature changes. Thus 1 is a number, whereas leT) is a function. The mathematician, in contrast, will write I for a function that he may be considering as an element in some infinite-dimensional function space. Once the function is supplied with an argument, he then contemplates I(x), which is merely a number: the value of I at the point x. Caught between these antithetical rituals of the physics and mathematics sects, the authors have adopted a clear policy: vacillation. Usually physics-sect statements, like \"On a curve g'(A)...,\" are used; and the reader can translate them himself into mathematically precise language: \"Consider a curve 8 on which a typical point is g' = 8(A); on this curve....\" But on occasion the reader will encounter a pedan tic-sounding paragraph written in mathematics-sect jargon (Example: Box 23.3). Such paragraphs deal with concepts and relationships so complex that standard physics usage would lead to extreme confusion. They also should prevent the reader from becoming so conditioned to physics usage that he is allergic to the mathematical literature, where great advantages of clarity and economy of thought are achieved by consistent reliance on wholly unambiguous notation.] Exercise 9.7. PRACTICE WITH COMMUTATORS Compute the commutator [e~. e cd of the vector fields EXERCISES Express your result as a linear combination of e 8 and e \u00a2. 17 e \u00b7= - - r sin e 179' I \u00a2 240 9. DIFFERENTIAL TOPOLOGY Exercise 9.8. ANGULAR MOMENTUM OPERATORS In Cartesian coordinates qf three-dimensional Euclidean space, one defines three \"angular mome11lum operators\" (vector fields) L j by Draw a picture of these three vector fields. Calculate their commutators both pictorially and analytically. Exercise 9.9. COMMUTATORS AND COORDINATE-INDUCED BASES Let u and v be vector fields in spacetime. Show that in some neighborhood of any given point there exists a coordinate system for which u = %.xl, if and only if u and v are linearly independent and commute: [u,v] = O. First make this result plausible from the second figure in Box 9.2; then prove it mathe matically. Note: this result can be generalized to four arbitrary vector fields eo, e 1, e 2, e 3 \u2022 There exists a coordinate system in which e\" = a/ox\" if and only if eo, e 1, e2' e 3 are linearly independent and [e I\" e p] = 0 for all pairs e 1\" e I\" Exercise 9.10. COMPONENTS OF COMMUTATOR IN NON-COORDINATE BASIS Derive equation (9.23). Exercise 9.11. LIE DERIVATIVE The \"Lie derivative\" of a vector field vW) along a vector field uW) is defined by ~uv= [u, v]. (9.24) Draw a space-filling family of curves (a \"congruence'') on a sheet of paper. Draw an arbitrary vector vat an arbitrary point 'J' 0 on the sheet. Transport that vector along the curve through '!Po by means of the \"Lie transport law\" ~uv = 0, where u = d/dt is the tangent to the curve. Draw the resulting vector v at various points 'J'(t) along the curve. Exercise 9.12. A CHIP OFF THE OLD BLOCK (a) Prove the Jacobi identity [u, [v, w]] + [v, [w, u]] + [w, [u, v]] = 0 (9.25) by picking out all terms of the form au avow, showing that they add to zero, and arguing from symmetry that all other terms, e.g., Ow au Ov terms, must similarly cancel. (b) State this identity in index form. (c) Draw a picture corresponding to this identity (see Box 9.2). \u00a79.7. MANIFOLDS AND DIFFERENTIAL TOPOLOGY Spacetime is not the only arena in which the ideas of this chapter can be applied. Points", ", curves, vectors, I-forms, and tensors exist in any \"differentiable manifold.\" \u00a79.7. MANIFOLDS AND DIFFERENTIAL TOPOLOGY 241 Their use to study differentiable manifolds constitutes a branch ofmathematics called \"differential topology\"-hence the title of this chapter. The mathematician usually begins his development of differential topology by introducing some very primitive concepts, such as sets and topologies of sets, by building a fairly elaborate framework out of them, and by then using that framework to define the concept of a differentiable manifold. But most physicists are satisfied with a more fuzzy, intuitive definition of manifold: roughly speaking, an n-dimen sional differentiable manifold is a set of \"points\" tied together continuously and differentiably, so that the points in any sufficiently small region can be put into a one-to-one correspondence with an open set of points of R\". [Rn is the number space of n dimensions, i.e., the space of ordered n-tuples (xl, x 2, \u2022\u2022\u2022,xn ).] That corre spondence furnishes a coordinate system for the neighborhood..A few examples will convey the concept better than this definition. Elementary examples (Euclidean 3-spaces, the surface of a sphere) bring to mind too many geometric ideas from richer levels of geometry; so one is forced to contemplate something more complicated. Let R3 be a three-dimensional number space with the usual advanced-calculus ideas of continuity and differentiability. Points \u20ac of R3 are triples, \u20ac = (~1' ~2' ~3)' of real numbers. Let a ray '3' in R3 be any half-line from the origin consisting of all \u20ac of the form \u20ac = A7J for some fixed 7J i' 0 and for all positive real numbers A > O. (See Figure 9.3.) A good example of a differentiable manifold then is the set S2 of all distinct rays. If1 is a real-valued function with a specific value 1('3') for any ray '3' [so one writes f S2 ----+ R: '3' ----+ 1('3')], it should be intuitively (or even demonstrably) clear that we can define what we mean by saying 1 is continuous or differentiable. In this sense S2 itself is continuous and differen tiable. Thus S2 is a manifold, and the rays '3' are the points of S2. There are many other manifolds that differential topology finds indistinguishable from S2. The simplest is the two-dimensional spherical surface (2-sphere), which is the standard representation of S2; it is the set of points \u20ac of R3 satisfying (~1)2 + (~2)2 + (~3)2 = 1. Clearly a different point '3' of S2 (one ray in R3) intersects each point of this standard 2-sphere surface, and the correspondence is continuous and differentiable in either direction (ray to point; point to ray). The same is true for <:tny ellipsoidal surface in R3 enclosing the origin, and for any other surface enclosing the origin that has Differentiable manifold \"defined\" Examples of differentiable manifolds The manifold 8 2 Figure 9.3. Three different representations of the differentiable manifold S2. The first is the set of all rays emanating from the origin; the second is the sphere they intersect; the third is an odd shaped, closed surface that each ray intersects precisely once. The manifold P The manifold 80(3) (rotation group) Affine geometry and Riemannian geometry defined EXERCISES 242 9. DIFFERENTIAL TOPOLOGY a different ray through each point of itself. They each embody the same global continuity and differentiability concepts. and represent the same abstract differ entiable manifold S2, the 2-sphere. They, and the bundle of rays we started with, all have the same geometric properties at this rudimentary level of geometry. A two-dimensional manifold that has a different geometric structure at this level (a different \"differentiable structure\") is the torus T2, the surface of a donut. There is no way to imbed this surface smoothly in R3 so that a distinct ray tjJ E S2 intersects each of its points; there is no invertible and differentiable correspondence between T2 and S2. Another example of a manifold is the rotation group SO(3), whose points ':P are all the 3 X 3 orthogonal matrices of unit determinant, so,:P = IIP;jll with '!JT'!J = I and det ~1~ = I. This is a three-dimensional space (one often uses the three Euler angle parameters in computations), where differential ideas (e.g., angular velocity", ") are employed; hence, it is a manifold. So is the Lorentz group. The differentiability of a manifold (i.e., the possibility of defining differentiable functions on it) permits one to introduce coordinate systems locally, if not globally, and also curves, tangent spaces, tangent vectors, I-forms, and tensors, just as is done for spacetime. But the mere fact that a manifold is differentiable does not mean that such concepts as geodesics, parallel transport, curvature, metric, or length exist in it. These are additional layers of structure possessed by some manifolds, but not by all. Roughly speaking, every manifold has smoothness properties and topology, but without additional structure it is shapeless and sizeless. That branch of mathematics which adds geodesics, parallel transport, and curva ture (shape) to a manifold is called affine geometry; that branch which adds a metric is called Riemannian geometry. They will be studied in the next few chapters. EXERCISES ON THE ROTATION GROUP As the exposition of differential geometry becomes more and more sophisticated in the the exercises will return time and again to the rotation group as an following chapters, example of a manifold. Then, in Box 30.1, the results developed in these exercises will be used to analyze the \"Mixmaster universe,\" which is a particularly important cosmological solution to Einstein's field equation. Before working these exercises, the reader may wish to review the EUler-angie parametri zation for rotation matrices, as treated, e.g., on pp. 107-109 of Goldstein (1959). Exercise 9.13. ROTATION GROUP: GENERATORS Let :l{t be three 3 X 3 matrices whose components are (Kt)mn = ltmn' (a) Display the matrices.'l{l' (.'h\"\"1) 2, (.'l{ 1)3, and (.X1)4. (b) Sum the series (9.26) Show that (';lz(O) is a rotation matrix and that it produces a rotation through an angle o about the x-axis. \u00a79.7. MANIFOLDS AND DIFFERENTIAL TOPOLOGY 243 (c) Show similarly that qilz(cf\u00bb = exp(X'3cf\u00bb and qilu(X) = exp(X'2X) are roiation matrices, the z- and y-axes, they produce rotations through angles cf> and X about and that respectively. (d) Explain why 'J' = qilz(l[;)qilz(O)~'ilz(cf\u00bb defines the Euler-angle coordinates, l[;, 0, cf> for the generic element 'J' E SO(3) of the rotation group. (e) Let e be the curve 'J' = qilz(t) through the identity matrix, e(O) =.'! E SO(3). Show that its tangent, (de /dt)(O) =6'(0) does not vanish by computing (:(0)/12' where /12 is the function /12('3') = P12' whose value is the 12 matrix element of 'J'. (f) Define a vector field 8 3 on SO(3) by letting 8 3('3') be the tangent (at t = 0) to the curve e(t) =!'ilz(t)'J' through 'J'. Show that 8 3('3') is nowhere zero. Note: 8 3('3') is called the \"generator of rotations about the z-axis,\" because it points from 'J' toward neigh boring rotations,!'ilz<t)'J', which differ from 'J' by a rotation about the z-axis. (g) Show that 8 3 = (0/0l[;)8</>' (h) Derive the following formulas, valid for t ~ 1: qilz(t)qilz(l[;}r'!lz(O)!'ilz(cf\u00bb =!'ilz(l[; qilu(t)'1lz<l[;)!'ilz(O)qilz<cf\u00bb = qilz(l[; + t cos l[; cot O)!'ilZ<0 + t sin l[;)!'ilz<cf> - tsinl[;cotO)!'ilz(O + tcosl[;)!'ilz(ep + tsinl[;/sinO); t cos l[;/sin 0). (i) Define 8 1('3') and 8 2('3') to be the tangent vectors (at t =", " 0) to the curves e(t) = qilz(t)'J' and e(t) =!'ilu(t)'J', respectively. Show that o 8 1 = cosl[;- - SIOl[; cot 00. ( 00 ol[; I 0) - - - - -, sin 0 ocf> 8 2 = SIOl[;-O + cosl[; cotO- - -'-O:l. 0) (0 I ol[; sm ucf> 0 o. 8 1 and 8 2 are the \"generators of rotations about the x- and y-axes.\" Exercise 9.14. ROTATION G\"OUP: STRUCTURE CONSTANTS Use the three vector fields constructed in the last exercise, ( 0 0 ol[; o 8 = cosl[;- - SIOl[; cot 00 8 2 = sinl[;~o + cosl[;(cotO~ - ~O+), I 0) - -. - -, smO ocf>. 1 ol[; SIO ucf> o (9.27) as basis vectors for the manifold of the rotation group. The above equations express this \"basis of generators\" in terms of the Euler-angle basis. Show that the commutation coefficients for this basis are (9.28) independently of location ':'J in the rotation group. These coefficients are also called the structure constants of the rotation group. / CHAPTER 10 AFFINE GEOMETRY: GEODESICS, PARALLEL TRANSPORT, AND COVARIANT DERIVATIVE Galilei's Principle of Inertia is sufficient in itself to prove conclusively that the world is affine in character. HERMANN WEYL This chapter is entirely Track 2. Chapter 9 is necessary preparation for it. It will be needed as preparation for (11 Chapters 11-1 3 (differential geometry; Newtonian gravitYl, (21 the second half of Chapter 14 (calculation of curvaturel, and (31 the details, but not the message, of Chapter 15 (Bianchi identitiesl. Freely falling particles and their clocks \u00a710.1. GEODESICS AND THE EQUIVALENCE PRINCIPLE Free fall is the \"natural state of motion,\" so natural, in fact, that the path through spacetime of a freely falling, neutral test body is independent of its structure and composition (the \"weak equivalence principle\" of Einstein, Eotvos, Dicke; see Box 1.2 and \u00a738.3). Picture spacetime as filled with free-fall trajectories. Pick an event. Pick a velocity there. They determine a unique trajectory. Be more precise. Ask for the maximum amount of information tied up in each trajectory. Is it merely the sequence of points along which the test body falls? No; there is more. Each test body can carry a clock with itself (same kind of clock-\"good\" clock in sense of Figure 1.9-regardless of structure or composition of test body). The clock ticks as the body moves, labeling each event on its trajectory with a number: the time A the body was there. Result: the free-fall trajectory is not just a sequence of points; it is a parametrized sequence, a \"curve\"!7'(A). But is the parametrization unique? Not entirely. Quite arbitrary are (1) the choice of time origin,!7'(O); and (2) the units (centimeters, seconds, furlongs,...) in which clock time A is measured. Hence, A is unique only up to linear transformations Anew = aA01d + b; (10.1) \u00a710.2. PICTORIAL APPROACH 245 A==5 A==4 A=3 A=2 A = 1 A=O. Figure 10.1. A geodesic viewed as a rule for \"straight-on parallel transport.\" Pick an event Po' and a tangent vector u == d/dA there. Construct the unique geodesic P(A) that (I) passes through Po: P(O) = Po; and (2) has u as its tangent vector there: (dP /dAh.=o = u. This geodesic can be viewed as a rule for picking up u from P(O) and laying it down again at its tip, P(I), in as straight a manner as possible, and for then picking it up and laying it down as straight as possible again at P(2), U.=l = (dP/dA h.=l; U.=2 = (dP /dAh.:2; etc. This sequence of \"straight as possible,\" \"taiI-on-tip", "\" transports gives meaning to the idea that (dP/dAh.=17 and u = (dP /dAh.=o are \"the same vector\" at different points along the geodesic; or, equivalently. that one has been obtained from the other by \"straight-on parallel transport.\" b (\"new origin of clock time\") is a number independent of location on this specific free-fall trajectory, and a (\"ratio of new units to old\") is also. In the curved spacetime of Einstein (and in that of Cartan-Newton, Chapter 12), these parametrized free-fall trajectories are the straightest of all possible curves. Consequently, one gives these trajectories the same name, \"geodesics,\" that mathe maticians use for the straight lines of a,curved manifold; and like the mathematicians, one uses the name \"afftrze parameter\" for the parameter A along a free-fall geodesic. Equation (10.1) then says \"the affine parameter of a geodesic is unique up to linear transformations.\" The affine parameter (\"clock time\") along a geodesic has nothing to do, it priori, with any pletric. It exists even in the absence of metric (e.g., in Cartan-Newtonian spacetime). It gives one a method for comparing the separation between events on a geodesic (~ and (J are \"twice as far apart\" as ~ and!?2 if [A r;; - Ad1= 2[A qz - A;:?D. But the affine parameter measures relative separations only along its own geodesic. It has no means of reaching off the geodesic. The above features of geodesics, and others, are summarized in Figure 10.1 and Box 10.1. \u00a710.2. PARALLEL TRANSPORT AND COVARIANT DERIVATIVE: PICTORIAL APPROACH Two test bodies, initially falling through spacetime on parallel, neighboring geodesics, get pushed toward each other or apart by tidal gravitational forces (spacetime curvature). To quantify this statement, one must quantify the concepts of \"parallel\" and \"rate of acceleration away from each other.\" Begin with parallelism. Geodesic defined as a free-fall trajectory Affine parameter defined as clock time along free-fall trajectory Box 10.1 GEODESICS Geodesic in brief Geodesic: gravitation physics in context of 246 10. GEODESICS, PARALLEL TRANSPORT. AND COVARIANT DERIVATIVE Give point, give tangent vector; get unique, affine-parametrized curve (\"geodesic\"). World line of a neutral test particle (\"Einstein's geometric theory of gravity\"; also \"Cartan's translation into geometric terms of Newton's theory of gravity\"): (I) \"given point\": some event on this world line; (2) \"given vector\": vector (\"displacement per unit increase of parameter\") tangent to world line at instant defined by that event; (3) \"unique curve\": every neutral test particle with a specified initial position and a specified initial velocity follows the same world line, regardless of its composition and regardless of its mass (small; test mass!; \"weak equivalence principle of Einstein-Eotvos-Dicke\"); (4) \"affine parameter\": in Cartan-Newton theory, Newton's \"universal time\" (which is measured by \"good\" clocks); in the real physical world, \"proper time\" (as measured by a \"good\" clock) along a timelike geodesic; (5) \"parametrized curve\": (a) affine parameter unique up to a transfor mation of the form A~ ciA + b, where a and b are constants (no arbitrariness along a given geodesic other than zero of parameter and unit of parameter); or equivalently (b) given any three events (1, &3, e on the geodesic, one can find by well-determined physical con struction (\"clocking\") a unique fourth event 6j) on the geodesic such that (Aq) - Ae) is equal to (\\11 - Aa); or equivalently (c) [differential version] given a tangent vector with components (dx\"/dA)a at point (1, one can find by physical construction (again \"clocking\") \"the same tangent vector\" at point e with uniquely determined components (dx\" /dA)e (vector \"equal\"; components ordinarily not equal because of twisting and turning of arbitrary base vectors be tween (1 and e). Comparison of vectors at different events by parallel transport Consider two neighboring events (1 and &3 connected by a curve", " g'(A). A vector va lies in the tangent space at (1, and a vector V,il lies in the tangent space at &3. How can one say whether va and V!il are parallel, and how can one compare their lengths? The equivalence principle gives an answer: an observer travels (using rocket power as necessary) through spacetime along the world line g'(A). He carries the vector va with himself as he moves, and he uses flat-space Newtonian or Minkows kian standards to keep it always unchanging (flat-space physics is valid locally \u00a7 10.2, PICTORIAL APPROACH 247 according to the equivalence principle!). On reaching event ~ the observer compares his \"parallel-transported vector\" v(/ with the vector v ll. If they are identical, then the original vector vd was (by definition) parallel to v~ll' and they had the same length. (No metric means no way to quantify length; nevertheless, parallel transport gives a way to compare length!) The equivalence principle entered this discussion in a perhaps unfamiliar way, applied to an observer who may be accelerated, rather than to one who is freely falling. But one cannot evade a basic principle by merely confronting it with an intricate application. (Ingenious perpetual-motion machines are as impossible as simpleminded ones!) The equivalence principle states that no local measurement that is insensitive to gravitational tidal forces can detect any difference whatsoever between flat and curved spacetime. The spaceship navigator has an inertial guidance system (accelerometers, gyroscopes, computers) capable of preserving an inertial reference frame in flat spacetime; and in flat spacetime it can compute the attitude and velocity of any object in the spaceship relative to a given inertial frame. The purchaser may specify whether he wants a guidance computer programmed with the laws of zero-gravity Newtonian mechanics, or with those of special-relativity physics. Use this same guidance system-including the same computer program-in curved spacetime. A vector is being parallel transported if the guidance system's computer says it is not changing. Will the result of transport in this way be independent of the curve used to link (J and ~i3? Clearly yes, in gravity-free spacetime, since this is a principal performance criterion that the purchaser of an inertial guidance system can demand of the manufacturer. But in a curved spacetime, the answer is \"NO!\" If vd agrees with V\"l after parallel transport along one curve, it need not agree with v~ after parallel transport along another. Spacetime curvature produces discrepancies. But one is not ready to study and quantify those discrepancies (Chapter 11), until one has developed the mathematical formalism of parallel transport, which, in turn, cannot be done until one has made precise the \"flat-space standards for keeping the vector Vii always unchanging\" as it is transported along a curve. The flat-space standards are made precise in Box 10.2. They lead to (1) a \"Schild's ladder\" construction for performing parallel transport; (2) the concept \"covariant derivative,\" Vuv, of a vector field v along a curve with tangent u; (3) the \"equation of motion\" Vuu = 0 for a geodesic, which states that \"a geodesic parallel transports its own tangent vector along itself;\" and (4) a link between the tangent spaces at adjacent events (Figure 10.2). \u00a710.3. PARALLEL TRANSPORT AND COVARIANT DERIVATIVE: ABSTRACT APPROACH From the \"Schild's ladder\" construction of Box 10.2, one learns the following properties of spacetime's covariant derivative: (continued on page 252) Parallel transport defined using inertial guidance systems and equivalence principle Result of parallel transport depends on route Schild's ladder for performing transport; its parallel consequences 248 10. GEODESICS. PARALLEL TRANSPORT. AND COVARIANT DERIVATIVE Box 10.2 FROM GEODESICS TO PARALLEL TRANSPORT TO COVARIANT DIFFERENTIATION TO GEODESICS TO... \"Parallel transport\" as defined by geodesics A. Transport any sufficiently short stretch of a curve 11~:r (Le., any tangent vector) parallel to itself along curve 11('i3 to point!t, as follows: {/ /'~tJ / / / / / / / / / / :~ll\\/ 9l / ':r {/...--. I I. Take some point i)/l along 11!'iJ close to 11. Take geodesic Xi)/l through X and i)/l. Take any affine parametrization A of Xi)/l and define a unique point ~ by the condition A~ =!(A~t +", " A + S = C; (2) A, S, Ceach paral lel transported with himself by freely falling then A + S = C (3) (inertial) always. Any other result would violate the equivalence principle! observer; \u00a7 10.2. PICTORIAL APPROACH 251 1. Consequence of this (as seen by following through definition of covariant derivative, and by noting that any vector u can be regarded as the tangent vector to a freely falling world line): for any vector u and vector fields v and w. 2. Consequence of this, combined with sym metry of covariant derivative, and with additivity of the \"closer of quadrilaterals\" [u, v]: (See exercise 10.1.) This can be inferred, alternatively, from the equivalence princi ple: in a local inertial frame, as in special relativity or Newtonian theory, the change in v along u + n should equal the sum of the changes along u and along n. 3. Consequence of above: choose n to be a multiple of u; thereby conclude F. The \"Schild's ladder\" construction process for parallel transport (beginning of this box), ap plied to the tangent vector of a geodesic (exer cise 10.6) guarantees: a geodesic parallel trans ports vector along itself. Translated into covariant-derivative language: its own tangent u = d/d\"A is a tangent) vector to a curve, and = a geodesic Vuu =0 (the curve is) (. Thus closes the circle: geodesic to parallel transport to covariant derivative to geodesic. Geodesics as defined by parallel transport or covariant differentiation Covariant derivative: ba~c properties 252 10. GEODESICS, PARALLEL TRANSPORT, AND COVARIANT DERIVATIVE Symmetry: Chain rule: Vuv - Vvu = [u, v] for any vector fields u and v; Vu(jv) =/Vuv + v au/for any functionJ, vector field v, and vector u; Additivity: Vu(v + w) = Vuv + Vuw for any vector fields v and w, and vector u; V au + bn V = a V u v + b Vn v for any vector field v, vectors or vector fields u and n, and numbers or functions a and b. (lO.2a) (lO.2b) (lO.2c) (lO.2d) Figure 10.2. The link between the tangent spaces at neighboring points, made possible by a parallel-transport law. Choose basis vectors e 1 and e 2 at the event d. Parallel transport them to a neighboring event!iJ. (Schild's ladder for transport of e 1 is shown in the Ilgure.) Then any other vector v that is parallel transported from d to!i' will have the same components at the two events (parallel transport cannot break the legs of a triangle; see Box 10.2): nm e numerically as at d] v = v le 1 + v2e 2 at d = v = v le 1 + v2e 2 at!Xl. ~el transported from d to!Xl] Thus, parallel transport provides a unique and complete link between the tangent space at d and the tangent space at!Xl. It identifies a unique vector at!Xl with each vector at d in a way that preserves all algebraic relations. Similarly (see \u00a71O.3), it identifies a unique I-form at!Xl with each I-form at d, and a unique tensor at!Xl with each tensor at d, preserving all algebraic relations such as (0', v) = 19.9 and $(0', v, w) = 37 I. Actually, all this is true only in the limit when d and!Xl are arbitrarily close to each other. When d and!Xl are close but not arbitrarily close, the result of parallel transport is slightly different for different paths; so the link between the tangent spaces is slightly nonunique. But the differences decrease by a factor of 4 each time the affine-parameter distance between d and!Xl is cut in half; see Chapter I I. \u00a7 10.3. ABSTRACT APPROACH 253 Any \"rule\" V, for producing new vector fields from old, that satisfies these four conditions, is called by differential geometers a \"symmetric covariant derivative.\" Such a rule is not inherent in the more primitive concepts (Chapter 9) of curves, vectors, tensors, etc. In the arena of a spacetime laboratory, there are as many ways of defining a covariant derivative rule V as there are of rearranging", " sources of the gravitational field. Different free-fall trajectories (geodesics) result from different distributions of masses. Given the geodesics of spacetime, or of any other manifold, one can construct a unique corresponding covariant derivative by the Schild's ladder procedure of Box 10.2. Given any covariant derivative, one can discuss parallel transport via the equation dvld>\" Vuv = 0 -<==>- the vector field v is parallel transported along the vector u = did>..; (10.3) Equation for parallel transport and one can test whether any curve is a geodesic via Vuu = 0 -<==>- the curve g'(>\") with tangent vector u = did>\" parallel transports its own tangent vector u (10.4) -<==>- g'(>\") is a geodesic. Thus a knowledge of all geodesics is completely equivalent to a knowledge of the covariant derivative. The covariant derivative V generalizes to curved spacetime the flat-space gradient V. Like its flat-space cousin, it can be viewed as a machine for producing a number (q, Vuv) out of a I-form q, a vector u, and a vector field v. This machine viewpoint is explored in Box 10.3. Note there an important fact: despite its machine nature, V is not a tensor; it is a nontensorial geometric object. In curved as in flat spacetime, V can be applied not only to vector fields, but also to functions, I-form fields, and tensor fields. Its action on functions is defined in the obvious manner: Knowledge of all geodesics is equivalent to knowledge of covariant derivative Cova riant derivative generalizes flat-space gradient Action of covariant derivative on functions. tensors l-forms. and VI df; (10.5) Its action on I-form fields and tensor fields is defined by the curved-space generali zation of equation (3.39): VS is a linear machine for calculating the change in output of S, from point to point, when \"constant\" (i.e., parallel transported) vectors are inserted into its slots. Example: the gradient of a (?) tensor, i.e., of a I-form field q. Pick an event g'0; pick two vectors u and v in the tangent space at g'0; construct from v a \"constant\" vector field v(g') by parallel transport along the direction of u, Vuv = O. Then Vq is a (g) tensor, and Vuq is a mtensor defined at '!J~o by (10.6) where u = did>... This defines V q and Vuq, because it states their output for any (continued on page 257) 254 10. GEODESICS. PARALLEL TRANSPORT. AND COVARIANT DERIVATIVE Box 10.3 COVARIANT DERIVATIVE VIEWED AS A MACHINE; CONNECTION COEFFICIENTS AS ITS COMPONENTS A. The Machine View 1. The covariant derivative operator V, like most other geometric objects, can be regarded as a machine with slots. There is one such machine at each event ':l'o in spacetime. In brief, the machine interpretation of V at ':'J~o says V( q...... v(~1~), u) _ (q, Vuv)....... :Th' d I I t--- ~ Ir sot: IOsert iF' I'-1'---'S--d---Jtled I Irst sot: Insert a I-form 0\" that resides in the tan_ent space at ~'1'0 I _ t: econ sot; IOsert a vector field v(~:P) defined on a neighborhood of '!Po I a vector u that resides in the tangent space at (oj' 0 I I - - - -., new vector: 1A \"the covariant derivative of the vector field v wi th {espect to u.\" [Note: this slot notation for V serves no useful purpose except to emphasize the \"machine\"-nature of V. This box is the only place it will be used.] 2. Geometrically, the output of the machine, (q, Vuv), is obtained as follows: (a) Calculate the rate of change of v, Vuv, along the vector u; when u and v are infinitesimally small, the calculation can be represented pictorially: vat tail of u \\~ _ vattipofu vat tip of u, after parallel transport back to tail ( Vuv. a vector that resides in the tangent space at '.'f0 (b) Count how many surfaces of the I-", "form q are pierced by the vector Vuv (piercing occurs in tangent space at ':'J'o) \u00a7 10.3. ABSTRACT APPROACH 255 This number is the output ~f the machine V, when q, v(!7') and u are inserted into its slots. 3. Another, equivalent, statement of covariant derivative as a machine. Leave first slot empty (no mention of any I-form q); get a new vector field from original vector field v: V(, v(!7'), u) Vuv ~ empty = \"covariant derivative of vector field v along vector u.\" 4. A third machine operation. Leave first and third slots empty (no mention of any I-form q; no mention of any vector u along which to differentiate); get a m tensor field from original vector field v; V(, v(!7'),.:...:-;.- ) Vv.:...:-;.- empty empty = \"covariant derivative\" or \"gradient\" of vector field v. This tensor field, Vv, is the curved-space generalization of the flat-space Vv studied in \u00a73.5. It has two slots (the two left empty in its definition). Its output for given input is empty Vv(q, u) = (q, Vuv). 5. Summary of the quantities defined above: (a) V is a covariant derivative operator; to get a number from it, insert q, v(~), and u; the result is (q, Vuv). (b) Vv is the gradient of v; to get a number from it, insert q and u; the result is (q, Vuv) [same as in (a)]. (c) Vuv is the covariant derivative ofv along u; to get a number from it, insert q; the result is (q, Vuv) [same as in (a) and (b)]. B. How V Differs from a Tensor The machine V differs from a tensor in two ways. (1) The middle slot of V will not accept a vector; it demands a vector field-the vector field that is to be differ entiated. (2) V is not a linear machine (whereas a tensor must be linear!): 256 10. GEODESICS. PARALLEL TRANSPORT. AN D COVARIANT DERIVATIVE Box 10.3 (continued) V (a(j.j(':f)v(tJ'), bu) _ (aq, Vbufv) =abf<q, Vuv) +,ab(q, v) Vuf fthis would be absent if V were a\" t [J L linear machine. C, The \"Connection Coefficients\" as Components of V Given a tensor S of rank (D, a basis of tangent vectors {e\"J at the event tJ'0 where S resides, and the dual basis of I-forms {w\"}, one defines the components of S by One defines the components of V similarly, except that for V one needs not only a basis {e,,} at the event '3'0' but also a basis {e\"W)} at each event '3' in its neighbor hood: r\"f3Y components of V = V(w\", ef3('3'), e y) (w\", V eyef3) ~ (\"a-component of change in basis vector ef3' when) in evaluating ef3 one moves from tail to tip of e y. - These components of V are called the \"connection coefficients\" of the basis {e,,}. They are the \"coordinate representation\" of the covariant derivative operator V. The covariant derivative operator V and the connection coefficients fa II~ provide different mathematical representations of the same geometric animal? Preposterous! The one animal runs from place to place and barks, or at least bites (takes difference, for example, between vector fields at one place and at a nearby place). The other animal, endowed with forty faces (see exercise 10.9) sits quietly at one spot. It would be difficult for two animals to look more different. Yet they do the same jobs in any world compatible with the equivalence principle: (1) they summarize the properties of all geodesics that go through the point in question; and, so doing, (2) they provide a physical means (parallel transport) to compare the values of vector fields and tensor fields at two neighboring events. \u00a7 10.3. ABSTRACT APPROACH 257 given input vectors v and u. If v(&') is not constrained to be \"constant\" along u = dld\"A., then (did>...) (C1, v) has contributions from both the change in v and the change in C1:", " (l0.7) (see exercise 103). Similarly, if S is a mtensor field, then its gradient V S is a (}) tensor field defined as follows. Pick an event &'0; pick three vectors u, v, W, and a I-form C1 in the tangent space at &'0; tum v, W, and C1 into \"constant\" vector fields and a \"constant\" I-form field near &'0 by means of parallel transport (Vuv = Vuw = V uC1 = 0 at &'0); then define VS(C1, v, w, u) (VuS)(C1, v, w) = 0u[S(C1, v, w)]. V u[S(C1, v, w)] (l0.8) Exercise 10.1. ADDITIVITY OF COVARIANT DIFFERENTIATION Show that the commutator (\"closer of quadrilaterals\") is additive: EXERCISES [u, v + w) = [u, v) + [u, w); [u + n, v) = [u, v) + [n, v). Use this result, the additivity condition Vu(v + w) = Vuv + Vuw, and symmetry of the covariant derivative, Vuv - Vvu = [u, v), to prove that Exercise 10.2. CHAIN RULE FOR COVARIANT DIFFERENTIATION Use pictures. and the \"take-the-difference-and-take-the-limit\" definition of Vuv (Box 10.2) to show that (10.9) Exercise 10.3. ANOTHER CHAIN RULE Derive equation (10.7), using the \"take-the-difference-and-take-the-limit\" definitions of derivatives. Hint: Before taking the differences, parallel transport CT[~j'(.\\.\u00bb) and v[':i'(.\\.\u00bb) back from :7'(.\\.) to :7'(0). Exercise 10.4. STILL ANOTHER CHAIN RULE Show that, as in flat spacetime, so also in curved spacetime, (10.10) Write down the more familiar component version of this equation in flat spacetime. Solution to first part of exercise: Choose I-forms CT and p at the event ':i'0 in question, and extend them along the vector u = did.\\. by parallel transport, VuP = VuCT = O. Then 258 10. GEODESICS, PARALLEL TRANSPORT, AND COVARIANT DERIVATIVE [Vu(v Q9 w\u00bb)(p. a) = it.. [(v Q9 w)(p, al) d= d>\" [(p, v>(a. w\u00bb d(p,v> = d>\" (a, w> + (p, v> = (p, Vuv>(a, w> + (p, v>(a, Vuw> d(a,w> d>\" (def of Vu on a tensor) (def of tensor product \"Q9\") (chain rule for derivatives) = [(Vuv) Q9 w)(p, a) + [v Q9 (Vuw\u00bb)(p, a) (by equation 10.7 with p, a const) (def of tensor product\" Q9\"). Exercise 10.5. ONE MORE CHAIN RULE Show, using techniques similar to those in exercise lOA, that Exercise 10.6. GEODESIC EQUATION Use the \"Schild's ladder\" construction process for parallel transport (beginning of Box 10.2) to show that a geodesic parallel transports its own tangent vector along itself (end of Box 10.2). \u00a710.4. PARALLEL TRANSPORT AND COVARIANT DERIVATIVE: COMPONENT APPROACH The pictorial approach motivates the mathematics; the abstract approach makes the pictorial ideas precise; but usually one must use the component approach in order to actually do complex calculations. To work with components, one needs a set of basis vectors {ea } and the dual set of basis I-forms {w a }. In flat spacetime a single such basis suffices; all events can use the same Lorentz basis. Not so in curved spacetime! There each event has its own tangent space, and each tangent space requires a basis of its own. As one travels from event to event, comparing their bases via parallel transport, one sees the bases twist and turn. They must do so. In no other way can they accommodate themselves to the curvature of spacetime. Bases at points '3'0 and '3'1' which are the same when", " compared by parallel transport along one curve, must differ when compared along another curve (see \"Curvature\"; Chapter 11). To quantify the twisting and turning of a \"field\" of basis vectors {ea('3')} and forms {w a('3')}, use the covariant derivative. Examine the changes in vector fields along a basis vector e/3' abbreviating (def of V/3); (10.12) and especially examine the rate of change of some basis vector: V /3ea \u2022 This rate of change is itself a vector, so it can be expanded in terms of the basis: \u00a7 10.4. COMPONENT APPROACH 259 V/3ea = eJJ.rJJ.a/3 note reversal of order of it and {3! (def of rJJ.a/3); \"---\"' '-\"-' (10.13) Connection coefficients defined and the resultant \"connection coefficients\" r JJ.a/3 can be calculated by projection on the basis I-forms: (10.14) (See exercise 10.7; also Box IOJ.) Because the basis I-forms are \"locked into\" the basis vectors \u00abW V, e a) = oVa)' these same connection coefficients r V at3 tell how the I-form basis changes from point to point: V/3W V = -Fva/3w a,, e a) = -Fva/3' v (V/3w (10.15) (10.16) (See exercise 10.8.) The connection coefficients do even more. They allow one to calculate the compo nents of the gradient of an arbitrary tensor S. In a Lorentz frame of flat spacetime, the components of V S are obtained by letting the basis vectors e a = 0'3' jox a = ojox a act on the components of S. Thus for a mtensor field S one finds that VS has components sa/3y,8 = O~8 [sa/3Y]' Not so in curved spacetime, or even in a non-Lorentz basis in flat spacetime. There twist, expand, and contract, so even if S were constant the basis vectors tum, (VS = 0), its components on the twisting basis vectors would vary. The connection coefficients, properly applied, will compensate for this twisting and turning. As one learns in exercise 10.10, the components of VS, called sa/3Y;8 so that (10.17) can be calculated from those of S by the usual flat-space method, plus a correction applied to each index (i.e., to each basis vector): Components of the gradient of a tensor field [\"+\" when correcting \"up\" indicf Gnterchange and sum J Ardifferentiating index] Lon index being corrected sa/3Y;8 = sa/3y,8 + SJJ./3yFaJJ.8 - sa JJ.yFJJ./38 [\" -\" when correcting \"down\" indices~~ ~ sa/3JJ.FJJ.y8' (10.18) pnterchange and sum Lon index being corrected differentiatingl J index Here (10.19) Components of the covariant derivative of a tensor field 260 10. GEODESICS, PARALLEL TRANSPORT, AND COVARIANT DERIVATIVE Equation (10.18) looks complicated; but it is really very simple, once the pattern has been grasped. Just as one uses special notation, sa/3y;8' for the components of VS, so one introduces special notation, Dsa/3y!d\"A., for components of the covariant derivative VuS along u = dld\"A.: (10.20) Since for any f this reduces to Chain rule for gradient The power of the component approach shows up clearly when one discusses chain rules for covariant derivatives. The multitude of abstract-approach chain rules (equations lO.2b, 10.7, 10.10, 10.11) all boil down into a single rule for components: The gradient operation \";\" obeys the standard partial-differentiation chain rule of ordinary calculus. Example: (fva);/L =!/Lva + fva;/L Y =f because f has no indices to correct],/L (contract this with u/L to get chain rule 1O.2b). Another example: (O'ava);/L = O'a;/Lva + O'ava;/L ~ = (O'ava)./L because O'ava has no free indices to correct] (contract this with u/L to get chain rule 10.7). Another example: (1O.22a) (1O.22b) (1O.22c) (contract this with u/L to get chain rule 10.11", "). Another example: see Exercise (10.12) below. EXERCISES Exercise 10.7. COMPUTATION OF CONNECTION COEFFICIENTS Derive equation (10.14) for F/LafJ from equation (10.13). Exercise 10.8. CONNECTION FOR 1-FORM BASIS Derive equations (10.15) and (10.16), which relate V13 w Pto FPafJ, from equation (10.14). Hint: use equation (10.7). \u00a710.4. COMPONENT APPROACH 261 Exercise 10.9. SYMMETRY OF CONNECTION COEFFICIENTS Show that the symmetry of spacetime's covariant derivative (equation 1O.2a) is equivalent to the following symmetry condition on the connection coefficients: (antisymmetric part of Fl'a(3) =t(Fl'a/3 - FI'/3a) =F/L - [a(3) - -l(WI' [ - 2 l>' ~~ = - \"2 ca/3 I I'. (10.23) I,ommutoto, of b,,;, vomo\"J As a special case, F I'a/3 is symmetric in a and f3 when a coordinate basis (ea = %xa) is used. Show that in a coordinate basis this symmetry reduces the number of independent connection coefficients at each event from 4 X 4 X 4 = 64 to 4 X 10 = 40. Exercise 10.10. COMPONENTS OF GRADIENT Derive equation (10.18) for the components of the gradient, sa/3Y;~' Hint: Expand S in terms of the given basis, and then evaluate the righthand side of for an arbitrary vector u. Use the chain rules (1O.2b) and (10.11). By comparing the result with read off the components sa/3Y;~' VuS = sa/3y;~u~ea Q9 w/3 Q9 w Y\u2022 Exercise 10.11. DIVERGENCE Let T be a @ tensor field, and define the divergence on its second slot by the same process as in flat spacetime: V. T = contraction of VT; i.e., (V. T)a = rafl ;/3' ( 10.24) Write the components Ta/3;/3 in terms of Ta/3,/3 plus correction terms for each of the two indices of T. [Answer: Exercise 10.12. VERIFICATION OF CHAIN RULE Let sa/3 y be components of a mtensor field, and M/ be components of a mtensor field. By contracting these tensor fields, one obtains a vector field sa/3 yM/3 Y' The chain rule for the divergence of this vector field reads Verify the validity of this chain rule by expressing both sides of the equation in terms of directional derivatives (, e) plus connection-coefficient corrections. Hint: the left side becomes ( S\"/3 M Y) Y /3 ;a - (sa/3 M Y) + Fa (SI'/3 M Y), Y /3 Y /3 I'a.Ct,. t \"\"sa/3 y,a M flY + S\"/3y M/3Y,a 1 L{~y chain rule for directional derivativ':.J y 262 10. GEODESICS. PARALLEL TRANSPORT. AND COVARIANT DERIVATIVE The right side has many more correction terms (three on S\"1l y;,,: two on M{3 Y;,,), but they must cancel against each other, leaving only one. Exercise 10.13. TRANSFORMATION LAW FOR CONNECTION COEFFICIENTS Let {e,,} and {el\"} be two different fields of basis vectors related by the transformation law Show that the corresponding connection coefficients are related by F\"'{3'y' =,L\"'pL I'{3,L\"y, FPI\"; + L\"'I'LI'{3',y' standard transformation law for components of a tensor ( 10.25) (10.26) Exercise 10.14. POLAR COORDINATES IN FLAT 2-DIMENSIONAL SPACE On a sheet of paper draw an (r, ep) polar coordinate system. At neighboring points, draw the basis vectors e T = a/or and e\", r- 1 a/aep. (a) Use this picture, and Euclid's version of parallel transport, to justify the relations VTe T = 0, VTe\", = 0, (b) From these relations write down the connection coefficients. (c) Let A = ATeT + A\"'e\", be a vector field. Show", " that its divergence, V' A = A'\\., = ACe,a + riri<irAi<, can be calculated using the formula (which should be familiar to most readers). Geodesic equation: abstract version \u00a710.5. GEODESIC EQUATION Geodesics-the parametrized paths of freely falling particles-were the starting point of this chapter. From them parallel transport was constructed (Schild's ladder; Box 10.2); and parallel transport in turn produced the covariant derivative and its connection coefficients. Given the covariant derivative, one recovered the geodesics: they were the curves whose tangent vectors, U = d'3' jdA, satisfy Vuu = 0 (u is parallel transported along itself). Let a coordinate system {x\"('3')} be given. Let it induce basis vectors e\" = ajax\" into the tangent space at each event. Let the connection coefficients r\"/3Y for this \"coordinate basis\" be given. Then the component version of the \"geodesic equation\" Vuu = 0 becomes a differential equation for the geodesic X\"(A): (1) u = -!!.... = dx\"_a_ dA ax\" =- dA components of u are u\" = dA ; dx\" \u00a7 10.5. GEODESIC EQUATION 263 (2) then components of Vuu = 0 are 0= ua;l3u13 = (u a,13 + raYl3uY)ul3 a I dx a ) dx 13 = a;e\\ dA. dA. + r a dxY dx 13 YI3 dA. dA.' which reduces to the differential equation (10.27) Component version This component version of the geodesic equation gives an analytic method (\"translation\" of Schild's ladder) for constructing the parallel transport law from a knowledge of the geodesics. Pick an event '3'0 and set up a coordinate system in its neighborhood. Watch many clock-carrying particles pass through (or arbitrarily close to) '3'0' For each particle read off the values of d 2x a / dA. at '3'0' Insert all the data for many particles into equation (10.27), and solve for the connec tion coefficients. Do not be disturbed that only the symmetric part of r aYI3 is obtained thereby; the antisymmetric part, r a[YI31' vanishes identically in any coordinate frame! (See exercise 10.9.) Knowing r a YI3' use them to parallel transport any desired vector along any desired curve through '3'0: / dA.2 and dx a How to construct parallel transport law from knowledge of geodesics dv a ra dA. + Yl3 v Y dx 13 - 0 dA. -. (10.28) Exercise 10.15. COMPONENTS OF PARALLEL-TRANSPORT LAW Show that equation (10.28) is the component version of the law for parallel transporting a vector v along the curve P(.\\.) with tangent vector u = dPjd.\\.. EXERCISES Exercise 10.16. GEODESICS IN POLAR COORDINATES In rectangular coordinates on a flat sheet of paper, Euclid's straight lines (geodesics) satisfy d 2xjd.\\.2 = d 2yjd.\\.2 = O. Transform this geodesic equation into polar coordinates (x = r cos cp, y = r sin cp); and read off the resulting connection coefficients by comparison with equation (10.27). These are the connection coefficients for the coordinate basis (ajar, ajacp). From them calculate the connection coefficients for the basis a e; = ar' I 2 e\u00b7 - - r a\u00a2' \u00a2 - The answer should agree with the answer to part (b) of Exercise 10.14. Hint: Use such relations as y 264 10. GEODESICS, PARALLEL TRANSPORT, AND COVARIANT DERIVATIVE Exercise 10.17. ROTATION GROUP: GEODESICS AND CONNECTION COEFFICIENTS [Continuation of exercises 9.13 and 9.14.] In discussing the rotation group, one must make a clear distinction between the Euclidean space (coordin_ates x,y, z; basis vectors a/ax, a/ay, a/az) in which the rotation matrices act. and the group manifold 50(3) (coordinates l/;, 0, ep; coordinate basis a/al/;, 0/00, 2/a\u00a2; basis of \"generators\" e 1, e 2, e 3 ), whose points Pare rotation matrices. (a) Pick a vector n = n", "\" a/ax + n Y a/ay + n Z a/az in Euclidean space. Show that (10.29) is a rotation matrix that rotates the axes of Euclidean space by an angle about the direction n. (::rj are matrices defined in exercise 9.13.) (b) In the group manifold 50(3), pick a point (rotation matrix) P, and pick a tangent vector u = u\"e\" at P. Let u be a vector in Euclidean space with the same components as u has in 50(3): u = u1 a/ax + u2 a/ay + u3 a/oz. (10.30) Show that u is the tangent vector (at t = 0) to the curve (10.31) The curve e(t) through the arbitrary point P with arbitrary tangent vector u = (de /dt)t =0 is a very special curve: every point on it differs from P by a -rotation tJlu(t) about one and the same direction u. No other curve in 50(3) with \"starting conditions\" {P, u} has such beautiful simplicity. Hence it is natural to decree that each such e(t) is a geodesic of the group manifold 50(3). This decree adds new geometric structure to 50(3); it converts 50(3) from a differentiable manifold into something more special: an affine manifold. One has no guarantee that an arbitrarily chosen family of curves in an arbitrary manifold can be decreed to be geodesics. Most families of curves simply do not possess the right geometric properties to function as geodesics. Most will lead to covariant derivatives that violate one or more of the fundamental conditions (10.2). To learn whether a given choice of geodesics is possible, one can try to derive connection coefficients r\"/3Y (for some given basis) corresponding to the chosen geodesics. If the derivation is successful, the choice of geodesics was a possible one. If the derivation produces inconsistencies, the chosen family of curves have the wrong geometric properties to function as geodesics. (c) For the basis of generators {e,,} derive connection coefficients corresponding to the chosen geodesics, 2(t) = <:ilu(t)P, of 50(3). Hint: show that the components u\" = (w''\", u) of the tangent u = de /dt to a given geodesic are independent of position e(t) along the geodesic. Then use the geodesic equation Vuu = 0, expanded in the basis {e,,}, to calculate the symmetric part of the connection r\"(/3Y)' Finally use equation (10.23) to calculate r\"r/3Y)' [Answer: (10.32) where \u00a3\"/3Y is the completely antisymmetric symbol with \u00a3123 = + I. This answer is inde pendent of location P in 50(3)!] ----------------------------.....;...\"\"...~ CHAPTER 11 GEODESIC DEVIATION AND SPACETIME CURVATURE \u00a711.1. CURVATURE, AT LAST! Spacetime curvature manifests itself as gravitation, by means of the deviation of one geodesic from a nearby geodesic (relative acceleration of test particles). Let the geodesics of spacetime be known. Then the covariant derivative V and its connection coefficients r a /3y are also known. How, from this information, does one define, calculate, and understand geodesic deviation and spacetime curvature? The answer unfolds in this chapter, and is summarized in Box 11.1. To disclose the answer one must (l) define the \"relative acceleration vector\" Vu Vun, which meas ures the deviation of one geodesic from another (\u00a7 11.2); (2) derive an expression in terms of V or r a /3Y for the \"Riemann curvature tensor,\" which produces the geodesic deviation (\u00a7 11.3); (3) see Riemann curvature at work, producing changes in vectors that are parallel transported around closed circuits (\u00a7 11.4); (4) see Riemann curvature test whether spacetime is flat (\u00a7 11.5); and (5) construct a special coordinate system, \"Riemann normal coordinates,\" which is tied in a special way to the Riemann curvature tensor (\u00a7 11.6). This chapter is entirely Track 2. Chapters, 9 and 10 are necessary preparation for it. It will be needed as preparation for (1) Chapters 12 and 13 (Newtonian gravity; Riemannian geometry). (2) the second half of Chapter 14 (calculation of curvature). and (3) the details, but not the message,", " of Chapter 15 (Bianchi identities). Overview of chapter \u00a711.2. THE RELATIVE ACCELERATION OF NEIGHBORING GEODESICS Focus attention on a family of geodesics (Figure 11.1). Let one geodesic be distin guished from another by the value of a \"selector parameter\" n. The family includes not only geodesics n = 0, 1, 2,... but also geodesics for all intervening values of Geometry of a family of geodesics: Selector parameter 266 11. GEODESIC DEVIATION AND SPACETIME CURVATURE Box 11.1 GEODESIC DEVIATION AND RIEMANN CURVATURE IN BRIEF \"Geodesic separation\" n is displacement (tangent vector) from point on fiducial geodesic to point on nearby geodesic characterized by same value of affine parameter.\\. Geodesic separation changes with respect to.\\ (i.e., changes along the tangent vector u = djd.\\) at a rate given by the equation ofgeodesic deviation V u Vun + Riemann (..., u, n, u) = \u00b0 (1) (second-order equation; see \u00a7\u00a7 1.6 and 1.7; Figures 1.10, 1.11, 1.12). In terms of components of the Riemann tensor the driving force (\"tidal graviational force\") is Riemann (..., u, n, u) = eaRa/3Y8u/3nYu8. The components of the Riemann curvature ten sor in a coordinate frame are given in terms of the connection coefficients by the formula (2) a R /3y8 - ar a/3Y _ ar a/38 - - - - ax8 axY + r a/Ly r /L/38 - r a/L8 r /L/3Y' (3) This curvature tensor not only quantifies the concept of \"tidal gravitational force,\" but also enters into Einstein's law, by which \"matter tells spacetime how to curve.\" That law, to be studied in later chapters, takes the following operational computational form in a given coordinate system: (a) Write down trial formula for dynamic evol ution of metric coefficients gJLP with time. (b) Calculate the connection coefficients from r a /LV - - ga/3r\u00b7 /3/L'\" = ~(ag/3v + ag/3/L _ ag/Lv) 2 axIL ax V ax/3 (4) (5) r /3/LV (derived in Chapter 13). (c) Calculate Riemann curvature tensor from equation (3). (d) Calculate Einstein curvature tensor from (6) (geometric significance in Chapter 15). into Einstein's equations (Chapter (e) Insert 17): G/LV = \u00b0 G/LV = 87rT/Lv (empty space), (when mass-energy is present).. (f) Test whether the trial formula for the dy namic evolution of the geometry was cor rect, and, if not, change it so it is. Affine parameter n. The typical point '3' on the typical geodesic will be a continuous, doubly differen tiable function of the selector parameter n and the affine parameter.\\; thus '3' = '3'(.\\, n). (11.1 ) Tangent vector The tangent vector or (Cartan notation) (notation of this book) (11.2) is constant along any given geodesic in this sense: the vector U at any point, trans- \u00a7 11.2. RELATIVE ACCELERATION OF NEIGHBORING GEODESICS 267 Figure 11.1. One-parameter family of geodesics. The \"selector parameter\" n tells which geodesic. 1 he affine parameter A tells where on a given geodesic. The two tangent vectors indicated in the diagram are u = alaA (Cartan: agolaA) and n = alan (Cartan: 2&'Ian). ported parallel to itself along the geodesic, arrives at a second point coincident in direction and length with the u already existing at that point. The \"separation vector\" Separation vector (Cartan notation) or a n= -an (notation of this book) (11.3) measures the separation between the geodesic n, regarded as the fiducial geodesic, and the typical nearby geodesic, n + LIn (for small LIn), in the sense that LI LIn a~1'\" 1 ( n)n =lLIn ~ J change in on measures the J an jposition1 an~ function (1104) brought about", " by transfer of attention from the one geodesic to the other at a fixed value of the affine parameter A. This vector is represented by the arrow ~'Jll!:2 in the first diagram in Box 11.2. (continued on page 270) 268 11. GEODESIC DEVIATION AND SPACETIME CURVATURE Box 11.2 GEODESIC DEVIATION REPRESENTED AS AN ARROW \"Fiducial geodesic\" n. Separation vector n LIn = '31lf2 leads from point ~\")/l on it, to point f! with same value of affine parameter.\\ (timelike quan tity) on neighboring \"test geodesic\" n + LIn. n +.:in n Parallel transport of '31lf2 by \"Schild's ladder con struction\" (Box 10.2) to ~~ and Ed. If the test geodesic n + LIn had kept a constant separation from the fiducial geodesic n, its tracer point would have arrived at d at the value (.\\.1.\\) of the affine parameter, and at ~ at (.\\ +.1.\\). Actual location of tracer point of test geodesic at values of the timelike affine parameter (.\\.1.\\),.\\, and (.\\ +.1.\\). Confrontation between actual course of tracer point on test geodesic and \"canonical course\": course it would have had to take to keep constant separation from the tracer point moving along the fiducial geodesic. \u00a7 11.2. RELATIVE ACCELERATION OF NEIGHBORING GEODESICS 269 Test geodesic same as before, except for uniform stretchout in scale of affine parameter. Any meas ure of departure of the actual course of geodesic from the canonical course (Cl2~), to be useful, should be independent of this stretchout. Hence, take as measure of geodesic deviation, not the vector ~'!il alone, nor the vector Cl'3', but the combination 6 2 = (~'!il) + stretch-independent (Cl'3'). Here the sign of addition implies that the two vectors have been transported parallel to themselves, before addition, to a common point (2 in the diagram; '31l in the differential calculus limit L1n -+ 0, Lt\\ -+ 0). Alternative courses that the test geodesic of D could have taken through 2 (families of geodesics characterized by different degrees of divergence from the left or convergence towards the right). Tilt changes values of Cl<J' (to Cl~) and Q3'!il (to ;1j(yl) individually, but not value of the sum 6 2. = (~13'!il) + (Cl'3') (\"lever principle\"). Note that arrow Q/3qz is of first order in Lt\\ and of first order in L1n; similarly for Cl<J'; hence the combination 6 2 is of second order in Lt\\ and first order in L1n. Conclude that the arrow 6 2/(Lt\\)2(L1n) is the desired measure of geodesic deviation in the sense that: size of mesh (ultimately to go to zero) cancels out; parameterization of test geodesic cancels out; slope of test geodesic cancels out. Give this arrow the name \"relative-acceleration vector':' and by examining it more closely (Box 11.3), discover the formula 6 2 /(Lt\\)2(L1n) = V u Vun for it. Relative-acceleration vector 270 11. GEODESIC DEVIATION AND SPACETIME CURVATURE Box 11.2 illustrates what it means to speak of geodesic deviation. One transports the separation n L1n = VIl!!! parallel to itself along the fiducial geodesic. The tip of this vector traces out the canonical course that the nearby tracer point would have to pursue if it were to maintain constant separation from the fiducial tracer point. The actual course of the test geodesic deviates from this \"canonical\" course. The deviation, a vector (Cl'!! of Box 11.2), changes with the affine parameter CCl'!! at Cl, Oat 2, :'i3('il at :13). The first derivative of this vector with respect to the affine parameter is sensitive to the scale of parameterization along the test geodesic, and to its slope (Box 11.2, F). Not so the second derivative. It depends only on the tangent vector u of the fiducial geodesic, and on the separation vector n L1n. Divide this second derivative of the deviation by L1", "n and give it a name: the \"relative-acceleration vector\". Discover (Box 11.3) a simple formula for it (relative-acceleration vector) = Vu Vun. (11.5) \u00a711.3. TIDAL GRAVITATIONAL FORCES AND RIEMANN CURVATURE TENSOR With \"relative acceleration\" now defined, tum to the \"tidal gravitational force\" (i.e., \"spacetime curvature\") that produces it. Use a Newtonian analysis of tidal forces Box 11.3 GEODESIC DEVIATION: ARROW CORRELATED WITH SECOND COVARIANT DERIVATIVE The arrow 6 2 in Box 11.2 measures, not the rate of change of the separation of the test geodesic n + L1n from the \"canonical course\" Cl2ri3 as baseline, but the second derivative: ( first derivative at.\\ + 21.1.\\) = Vun = 'J{/!il - 'Jl:13 =~..1.\\ L1n '.1.\\ L1n ( first derivative at.\\ _1.1.\\) = V n = \u00a3Cl - \u00a3'!! = -Cl'!!.1.\\ L1n..1.\\ L1n 2 u Transpose to common location.\\, take difference, and divide it by L1.\\ to obtain the second covariant derivative with respect to the vector u; thus v V = (Vunh,+!.::Ix.1.\\ u un (Vunh,_!.::IX (:13&l + Cl'!!)vectorstransportedto common location = = \"relative acceleration vector\" for neighboring geodesics. (.1.\\)2 L1n = 6 2 (.1.\\)2 L1n \u00a7 11.3. TIDAL FORCES AND RIEMANN TENSOR 271 (left half of Box llA) to motivate the geometric analysis (right half of same box). Thereby arrive at the remarkable equation (11.6) Tide-producing gravitational forces expressed in terms of a commutator \"relative I acceleration\", \"tide-producing, wavitational forces\", This equation is remarkable, because at first sight it seems crazy. The term [Vn, Vu]u involves second derivatives of u, and a first derivative of Vn : (11.7) It thus must depend on how u and n vary from point to point. But the relative acceleration it produces, Vu Vun, is known to depend only on the values of u and n at the fiducial point, not on how u and n vary (see Box 11.2, F). How is this possible? Somehow all derivatives must drop out of the tidal-force quantity [Vn, Vu]u. One must be able to regard ['V..., 'V... ]... as a purely local, algebraic machine with three slots, whose output is a vector. If it is purely local and not differential, then it is even linear (as one sees from the additivity properties of V), so it must be a tensor. Give this tensor the name Riemann, and give it a fourth slot for inputting a I-form: Riemann (..., C,A, B) Riemann \u00ab(1, C,A, B) [VA' VB ]C; \u00ab(1, [VA, VB ]C). This is only a tentative definition of Riemann. Before accepting it, one should verify that it is, indeed, a tensor. Does it really depend on only the values of A, B, C at the point of evaluation, and not on how they are changing there? The answer (derived in Box 11.5) is \"almost.\" It fails the test, but with a slight modification it will pass. The modification is to replace the commutator [VA, VB] by the \"curvature operator\" (11.8) Curvature operator defined where V[A,Bl is the derivative along the vector [A, B] (commutator of A and B). [VA,VB ] for the fields A = nand B = u of the geodesic-deviation (!1l(A, B) problem, because [n, u] = 0.) Then the modified and acceptable definition of the Riemann curvature tensor is Riemann curvature tensor defined Riemann (..., C, A, B) Riemann \u00ab(1, C,A, B) _ \u00ab(1,!:ii'(A, B)C).!Yi'(A, B)C; (11.9) To define Riemann thus, and to verify its tensorial character (exercise 11.2", "), does not by any means teach one what curvature is all about. To understand curvature, one must scrutinize Riemann from all viewpoints. That is the task of the rest of this chapter. (continlled all page 275) 272 11. GEODESIC DEVIATION AND SPACETIME CURVATURE Box 11.4 RELATIVE ACCELERATION OF TEST PARTICLES- GEOMETRIC ANALYSIS PATTERNED ON NEWTONIAN ANALYSIS Newtonian Analysis Geometric Analysis 1. Consider a family of test-particle trajectories three-dimensional space: xi(t, n) in ordinary, \"t\" is time measured by particle's clock, or any clock; \"n\" is \"selector parameter.\" 1. Consider a family of test-particle trajectories (geodesics), g'(A, n), in spacetime: \"A\" is affine parameter, time measured by particle's clock; \"n\" is \"selector parameter.\" i.e., 2. Equation of motion for each trajectory: 2. Geodesic equation for each trajectory: aZx i atZ n ) + al/J. = 0 ax''( where l/J is Newtonian potential. 3. Take difference between equations of motion for neighboring trajectories, nand n + LIn, and take limit as LIn -+ O-i.e., take derivative (~) [( aZx an t i atZ n ) +~] _0 ax; -. Vuu = O. [Looks like first-order equation; is actually second-order because the \"u\" being differen tiated is itself a derivative, u = (a&'/aA)n'] 3. Take difference between geodesic equations for neighboring geodesics nand n + LIn, and take limit as LIn -+ O-i.e., take covariant deriva tive 4. When a/an acts on second term, rewrite it as 4. There is no second term, so leave equation in k a) ) a;; t = a;;- t ax k = n ax k ; ( ax k a a ( form Thereby obtain (~) (~) (ax i an t at n at ) + aZl/J nk - 0 ax; ax k -. n 5. To obtain equation for relative acceleration, move (a/an)t through both of the (a/at)n terms (permissible because partial derivatives com mute!): 5. To obtain equation for relative acceleration, V u Vun, move Vn through Vu and through the afaA. of U = a'3'/aA.: a. First step: In Vn Vuu = 0, move Vn through (~) (~) (ax J at n at n an taX; axk 2 lP nk = o. ) + a This is equivalent to a2nJ ) - - + ( at2 11'relative acceleration\"J [ a2lP ax J axk n k = 0. 1 tr'tide-producing 1 gravitational forces] Vu. The result: (Vu Vn + [Vn, Vuj}u = o. r--t---:--~ Icommutator; must be included I as protection against possibility,that Vu Vn i- Vn Vu. b. Second step: Move Vn U = a'3'/aA.; i.e., write a'3' Vn af = Vnu = Vun = Vu - through a/aA. of ~ef. ofu~ a'3' ~def :: nJ Why? Because symmetry of covariant derivative says Vnu - Vun = [n, u] 02 = a;;'aI =~-~=o; [ a a] a2 c. Result: Vu Vun + [Vm Vu]u = 0 ~~----lt--, I \"relative t I I\"tide-producing acceleration\"l gravitational forces\"; Le.. \"spacetime curvature\" I Box 11.5 RIEMANN CURVATURE TENSOR A. Definition of Riemann Motivated by Tidal Gravitational Forces: 1. Tidal forces (spacetime curvature) produce relative acceleration of test particles (geod~sics) given by (1) 274 11. GEODESIC DEVIATION AND SPACETIME CURVATURE Box 11.5 (continued) 2. This motivates the definition Riemann (..., C,A, B) = [VA' Vs]C. (2) 4empty slot for inserting a one-form] B. Failure of this Definition 1. Definition acceptable only if Riemann (..., C, A, B) is a linear machine, inde pendent of how A, B", ", C vary from point to point. 2. Check, in part: change variations of C, but not C itself, at event <;10 : CNEWW) = fW)CoLDW), '4arbitrary function except fWo) = 1] 3. Does this change [VA' Vs]C? Yes! Exercise 11.1 shows C. Modified Definition of Riemann: 1. The term causing trouble, COLD V[A,SJ!, can be disposed of by subtracting a \"correction term\" resembling it from Riemann-i.e., by redefining Riemann (..., C,A, B) &leA, B)C, &leA, B) [VA' Vs ] - V[A.Sj (3) (4) 2. The above calculation then gives a result independent of the \"modifying func tion\" f D. Is Modified Definition Compatible with Equation for Tidal Gravitational Forces? 1. One would like to write Vu Vun + Riemann (..., u, n, u) = O. 2. This works just as well for modified definition of Riemann as for original defini tion, because \u00a711.3. TIDAL FORCES AND RIEMANN TENSOR 275 ('ff(n, u) = [Vn, Vul- V[n,ul = [Vn, Vul\u00b7 t 1= 0 because n = (a/anh. and] LU = (a /a.\\)n commute Geodesic deviation and tidal forces cannot tell the difference between ('ff(n, u) and [Vno Vul, nor consequently between old and new definitions of Riemann. E. Is Modified Definition Acceptable? I.e., is Riemann (..., C, A, B) ent of how A, B, C vary near point of evaluation? YES! (See exercise 11.2.) ('ff(A, B)C a linear machine with output independ- Take stock, first, of what one knows already about the Riemann curvature tensor. (1) Riemann is a tensor; despite the appearance of V in its definition (11.9), no derivatives actually act on the input vectors A, B, and C. (2) Riemann is a mtensor; its first slot accepts a I-form; the others, vectors. (3) Riemann is determined entirely by V, or equivalently by the geodesics of spacetime, or equivalently by spacetime's parallel transport law; nothing but V and the input vectors and I-form are required to fix Riemann's output. (4) Riemann produces the tidal gravitational forces that pry geodesics (test-particle trajectories) apart or push them together; i.e., it charac terizes the \"curvature of spacetime\": Vu Vun + Riemann (..., u, n, u) = O. (11.10) (This \"equation of geodesic deviation\" follows from equations 11.6, 11.8, and 11.9, and the relation [n, ul = 0.) All these facets of Riemann are pictorial (e.g., geodesic deviation; see Boxes 11.2 and 11.3) or abstract (e.g., equations 11.8 and 11.9 for Riemann in terms of V). Riemann's component facet, (11.11) is related to the component facet of V by the following equation, valid in any coordinate basis {ea } = {a/ax a }: (11.12) (See exercise 11.3 for derivation, and exercise 11.4 for the extension to noncoordinate bases,) These components of Riemann, with no sign of any derivative operator anywhere, may leave one with a better feeling in one's stomach than the definition (11.8) with its non differentiating derivatives! Tide-producing gravitational forces expressed i'n terms of Riemann Components of Riemann expressed in terms of connection coefficients 276 11. GEODESIC DEVIATION AND SPACETIME CURVATURE EXERCISES Exercise 11.1. (Based on Box 11.5.) Let C:-;EW and COLD be vector fields related by [VA' VslC DEPENDS ON DERIVATIVES OF C CNEW(~Y~) =f(~1')COLDU1~). ~arbitrary function, exceptf(:ro) = I) Show that Exercise 11.2. PROOF THAT Riemann IS A TENSOR Show from its definition (11.8, 11.9) that Riemann is a tensor. Hint: Use the following procedure. (a) If f(", "P) is an arbitrary function, show that ~-H(A,B)fC =fij/(A,B)C. (b) Similarly show that ~'Il(jA, B)C =j!'ll(A, B)C and ;'Il(A,jB)C =fM(A, B)C. (c) Show that ~(A, B)C is linear; Le., ~(A + a,B)C =?il(A,B)C + (1l(a,B)C;?ileA, B + b)C =?ileA, B)C +!Yl(A, b)C;?ileA, B)(C + c) =!ileA, B)C +?ileA, B)c. (d) Now use the above properties to prove the most crucial feature ofij/(A, B)C: Modify the variations (gradients) ofA. B, and C in an arbitrary manner, but leave A, B, C unchanged at Po: A --+ A + a\"e\" B --+ B + bae\" C --+ C + caea 1 a\"W), b\"W), c\"W) arbitrary except they all vanish at,:,P = Po. Show that this modification leaves?ileA, B)C unchanged at '!Po. (e) From these facts, conclude that Riemann is a tensor. Exercise 11.3. COMPONENTS OF Riemann IN COORDINATE BASIS [ standard way to calculate components Derive equation (11.12) for the components of the Riemann tensor in a coordinate basis. [Solution: R\"I3Y8 = Riemann (w\", e 13, e y, e8) = (w a, ij/(ey, e 8)e13 ) = (w a, (Vy V 8 - V 8 V y)e13 ) =(w a, eP.Fp.13 8,y + (e v FVp.y)FP.138 - eP.Fp.l3y,8 = (FP.13 8,y - FP.l3y,8)(W a, ep.) + (FVp.yFP.138 - FVp.8FP.l3y)(wa, e v ), which reduces (upon using (w a, ep.) = 0\\) to (11.12).) by definition (11.8) plus.. ley, e 8) = 0 In coord. basIs [by definition (11.9\u00bb) (e v FVp.8)Fp.I3Y) [ ] ] \u00a7 11.4. PARALLEL TRANSPORT AROUND A CLOSED CURVE 277 Exercise 11.4. COM PONENTS 0 F RIEMANN IN NONCOORDINATE BASIS In a noncoordinate basis with commutation coefficients cap Y defined by equation (9.22), derive the following equation for the components of Riemann: (11.13) \u00a711.4. PARALLEL TRANSPORT AROUND A CLOSED CURVE What are the effects of spacetime curvature, and how can one quantify them? One effect is geodesic deviation (relative acceleration of test bodies), quantified by equa tion (11.10). Another effect, almost as important, is the change in a vector caused by parallel transport around a closed curve. This effect shows up most clearly in the same problem, geodesic deviation, that motivated curvature in the first place. The relative acceleration vector V u Vun is also the change 8u in the vector u caused by parallel transport around the curve whose legs are the vectors nand u: Change in a vector due to parallel transport around a closed curve: Related to geodesic deviation (See Box 11.6 for proof.) Hence, in this special case one can write 8u + Riemann (..., u, n, u) = O. The expected generalization is obvious: pick a closed quadrilateral with legs u..1a and v..1b (Figure 11.2;..1a and..1b are small parameters, to go to zero at end of discussion). Parallel transport the vector A around this quadrilateral. The resultant change in A should satisfy the equation 8A + Riemann (...,A, u..1a, v..1b) = 0; (11.14) Equation for change or, equivalently, or, more precisely, 8A +..1a..1b tJl(u, v)A = 0; Lim ( 8A b) + Riemann (...,A, u, v) = O..1a_O", "..1a..1.1b-O (11.14') (11.14\") The proof is enlightening, for it reveals the geometric origin of the correction term V[u.vl in the curvature operator. The circuit of transport (Figure 11.2) is to be made from two arbitrary vector fields u..1a and v..1b. However, a circuit made only of these fields has a gap in it, for a simple reason. The magnitude of u varies the wrong way from place to place. The displacement u..1a that reaches. across at the bottom of the quadrilateral from Derivation of equation for change y [v. u]Ja.1b v.1b uJa A b~forf' transport Figure 11.2. The change SA in a vector A as a result of parallel transport around a closed curve. The edges of the curve are the vector fields u.Ja and v Jb. plus the \"c1oser of the quadrilateral\" [v.Jb, u.Ja] == [v, u].Ja Jb (see Box 9.2). one line of v's to another cannot make the connection at the top of the quadrilateral. Similarly the v's vary the wrong way from place to place to connect the u's. To close the gap and complete the circuit, insert the \"closer of quadrilaterals\" [v Lib, u Lla] = [v, u]Lla Lib. (See Box 9.2 for why this vector closes the gap.) With the route now specified, the vector A is to be transported around it. One way to do this, \"geometrical construction\" by the method of Schild's ladder applied over and over, is the foundation for planning a possible experiment. For planning an abstract and coordinate-free calculation (the present line of action), introduce a \"fiducial field,\" only to take it away at the end of the calculation. Plan: Conceive of A, not as a localized vector defined solely at the start of the trip, but as a vector field (defined throughout the trip). Purpose: To provide a standard of reference (comparison of A transported from the origin with A at the place in question). Principle: The standard of reference will cancel out in the end. Procedure: -8A = - Net change made in taking the vector A, originally localized at the) start of the circuit, and transporting it parallel to itself (\"mobile A\") around the closed circuit. This quantity cannot be evaluated until completion of circuit because there is no preexisting standard of reference along the way. ( A quantity subject to analysis for each leg of circuit individually. This new quantity is defined by introducing throughout the whole region = + a vector field A (field), smoothly varying, and in agreement at starting point with the original localized A, but otherwise arbitrary. This new quantity is then given by A (field) at starting point (same as A (localized) at starting point) minus A(mobile) at finish point (after transit). Change in A (field) relative to A (mobile) in the course of transport along) specified leg. Value for anyone leg depends on the arbitrary choice ofA (field), but this arbitrariness cancels out in end because ofclosure = 2: ( ~frcu~~ of circuit. \u00a7 11.4. PARALLEL TRANSPORT AROUN D A CLOS ED CU RVE 279 Change in A (field) relative to the parallel-transported A (mobile) as standard of reference, made up of contributions along following legs of Figure 11.2: v Lib, giving V vA (field) Lib (on line displaced u Lla from start) - v Lib, giving - VvA (field) Lib (on line through starting point) - u Lla, giving - VuA(field) Lla (on line displaced v Lib from start) + u Lla, giving + [v, uJLla Lib, giving V[v.ujA(field) Lla Lib VuA(field) Lla (on line through starting point) = {Vu V v - V v V u + V[v.uj} A (field) Lla Lib = Riemann (...,A(field),u,v)LlaLib == 0'l(u,v)A(field)LlaLib. (IU5) Profit: The curvature operator Riemann (...,..., u, v) == '!il(u, v) = [V\", VvJ- V[u,vj' Box 11.6 GEODESIC DEVIATION AND PARALLEL TRANSPORT AROUND CLOSED CURVE: TWO ASPECTS OF SAME CONSTRUCTION Geodesic Deviation V V n = Lim {Clfj'", " + ~''!il}. (Lli\\)2 Lin u u &.._0.:In _ 0 (See Boxes 11.2 and 11.3) Geodesic Deviation Same result; different construction. To simplify the connection with closed-curve transport, change the tilt and dilate the parametrization of geodesic '!i:2ijl in A. The result: B, where (j and Cl coincide. From F of Box 11.2 one knows Cl~y~ + ~'B:il = Cl9 + ~\u00b7B~-i.e. Vu Vun the is Box 11.6 (continued) same for this family of geodesics as for the original family v V n = Lim { u u.:3,\\ _ 0.In _ 0!f!(.i (..::I\"W..::In }. Also, to simplify discussion set..::In =..::Ii\\ = 1, and assume nand u are small enough that one can evaluate Vu Vun without taking the limit: Parallel Transport Around Closed Curve, Performed by Same Construction Plan: Parallel transport the vector u..::Ii\\ = 2(.i counterclockwise around the curve!2 ---+ &> ---+ E ---+ ':.~ ---+ 2. Execution: (1) Call transported vector u(m) (\"m\" for \"mobile\"). (2) At f2, u(m) = 2(.i. (3) At &>, u(m) = &>2 because &>2(.i is a geodesic and u(m) is its tangent vector. (4) At E, u(m) = E'!JlZ according to Schild's ladder of the picture. (5) At '!JlZ, u(m) = '!JlZ'!Jl because E'!JlZ'!Jl is a geodesic and u(m) is now its tangent vector. (6) At 2, u(m) = 2q] according to Schild's ladder. Result: The change in u(m) is -q](.i. Had the curve been circuited in opposite direction (E ---+ &> ---+ 2 ---+ '!JlZ ---+ E), the change would have been +q](.i: (8U)duetoparalleltransportupn, outu, down-n, and = q](.i = V u Vun. back a1ong-uto starting point applied to the vector field A (field), gives the negative of the change in the localized vector A (localized) (called A (mobile) during the phase of travel) on parallel transport around the closed circuit. It does not give the change in A (field) on traversal of that circuit, for A (field) has the same value at the end of the journey as at the beginning. Equation (11.14') expresses that change in terms of the conveniently calculated differential operator, &leu, v) = [Vu' V.,] - V[u,.,j' Paradox: Neither wanted nor evaluated is the change in the quantity A (field) acted on by this operato~. Payoff: Ostensibly differential in the character of its action on A, the operator Riemann (\u2022.\u2022,..., u, v) = &leu, v) is actually local. Thus, replace the proposed smoothly varying vector field A (field) by a quite different but also smoothly varying vector field A (field, new). Then the two fields need agree only at the one point in question for them to give the same output Riemann (...,A, u, v) = &leu, v)A at that point. This one \u00a7 11.4. PARALLEL TRANSPORT AROUND A CLOSED CURVE 281 knows from the fact that <'lA, the quantity calculated, has an existence and value independent of the choice of A (field). This one can also verify by detailed calculation (exercise 11.2). Power: Although they cancel out in their response to any change of A with location, the several differentiations in the curvature operator respond directly to the \"rate of change of geometry with location\" (\"geodesic deviation\"). Prolongation: The closed curve need not be a quadrilateral. The curvature operator tells how a vector changes on parallel trarisport about small curves of arbitrary shape (Box 11.7). \u2022 Exercise 11.5. COPLANARITY OF CLOSED CURVES Let f 1 and f 2 be the bivectors (see Box 11.7) for two small closed curves at the same event. Show that the curves are coplanar if and only if f 1", " = af2 for some number a. EXERCISE Box 11.7 THE LAW FOR PARALLEL TRANSPORT ABOUT A CLOSED CURVE A. Special Case Curve is closed quadrilateral formed by vector fields u and v. 1. Law says (in component form) (I) 2. On what characteristics of the closed curve does this depend? a. Notice that RlX/3Y8 = -RlX/38Y (antisymmetry in last two indices; obvious in equation 11.12 for components; also obvious because reversing the direction the curve is traversed-i.e., interchanging u and v-should reverse sign of <'lA). two indices. The symmetric part of u \u00ae v must give zero. Only the antisymmetric part, u /\\ v = u \u00ae v - v \u00ae u can contribute: b. Equation (1) contracts u \u00ae v into these antisymmetric, last 3. This antisymmetric part is a \"bivector.\" It is independent of the curve's shape; it depends only on (a) the plane the curve lies in, and (b) the area enclosed by the curve. [Although without metric \"area\" is meaningless, \"relative areas at an event in a given plane\" have just as much meaning as \"relative lengths at an (2) Box 11.7 (continued) event along a given direction.\" Two vectors at the same event lie on the same line if they are multiples of each other; their relative length in that case is their ratio. Similarly, two small closed curves at the same event lie in the same plane if their bivectors are multiples of each other (exercise 11.5); their relative area in that case is the ratio of their bivectors.] B. General Case Arbitrary but small closed curve. 1. Break the curve down into a number of quadrilaterals, all lying in the same plane as the curve. 2. Traverse each quadrilateral once in the same sense as the curve is to be tra versed. Result: all interior edges get traversed twice in opposite directions (no net traversal); the outer edge (the curve itself) gets traversed once. 3. Thus,8A due to traversing curve is the sum of the 8A's from traversal of each quadrilateral: 8Aa = 2 2: quadrilaterals Ra!3Y8A!3( U /\\ Vfor given qUadrilateral)Y8. Define the bivector f for the curve as the sum of the bivectors for its component quadrilaterals: f 2: quadrilaterals (u /\\ V)qUadrilateral (add \"areas\"; keep plane the same). 4. Then 8Aa + 1. Ra 2 A!3I'Y8 = 8Aa + Ra!3y8 J -!3ly81 A!31'Y8 - 0 J -. C. Warning This is valid only for closed curves of small compass: 8A doubles when the area doubles; but the error increases by a factor _23/2 [SA ex: Lla Lib in calculation of \u00a711.4; but error ex: (Lla)2Lib or Lla(Llb)2]. \u00a7 11.5. FLATNESS MEANS ZERO RIEMANN CURVATURE 283 \u00a711.5. FLATNESS IS EQUIVALENT TO ZERO RIEMANN CURVATURE To say that space or spacetime or any other manifold is fiat is to say that there exists a coordinate system {xa(g')} in which all geodesics appear straight: Flatness of a manifold defined xa(i\\) == fiX + bai\\. (11.16) (Example: Lorentz spacetime of special relativity, where test bodies move on such straight lines.) They can appear sO if and only if the connection coefficients in the geodesic equation d 2x f3 di\\2 + r f3 dx JL dxv di\\ JLV di\\ _ - 0, expressed in the same coordinate system, all vanish: r f3 JLV == o. (11.17) (11.18) From the vanishing of these connection coefficients, it follows immediately (equation 11.12) that all the components of the curvature tensor are zero: Flatness implies Riemann = 0 R f3 - 0. \u2022 YJLV - (11.19) Proof that Riemann = 0 implies flatness [Geometric restatement of (I 1.16) ---+ (11.18) ---+ (11.19): For all geodesics to be straight in a given coordinate system means that initially parallel geodesics preserve their separation; the geodesic deviation is zero; and therefore the curvature vanishes", ".] Is the converse true? Does zero Riemann curvature imply the existence of a coordinate system in which all geodesics appear straight? Yes, as one sees by the following construction. Transport a vector parallel to itself from?f0 to!!!., and then back from!!!. to?f 0 along a slightly different route. It returns to its starting point with no alteration in magnitude or direction, because Riemann everywhere vanishes. Therefore parallel transport of a base vector e JL from q'0 to!!!. yields at!!!. a base vector e JL that is independent, both in magnitude and in direction, of the route of transportation (for routes obtainable one from the other by any contiimous sequence of deformations). As for!!!., so for all points of the manifold; and as for the one base vector e JL, so for a complete set of base vectors (p. == 0, 1,2,3): Parallel transport of a basis {ea(g'o)} yields everywhere a field of frames (\"frame field\"), each base vector of which suffers zero change (relative to the frame field) on parallel transport from any point to any nearby point: thus, or Vep' == 0; (11.20) (11.21) With the vanishing of these individual derivatives, there also vanishes the commuta tor of any two basis-vector fields: (11.22) 284 11. GEODESIC DEVIATION AND SPACETIME CURVATURE The gap in the quadrilateral of Figure 11.2 (there read \"e,/' for \"u,\" \"e.\" for \"v\") closes up completely. Thereupon one can introduce coordinates xl-', each of which increases with a motion in the direction of the corresponding vector field; and with appropriate scaling of these coordinates, one can write a e =-axl-' I-' (11.23) (see exercise 9.9). With this coordinate basis in hand, one can employ the formula (11.24) to calculate the connection coefficients. From the vanishing of the quantities on the left, one concludes that all the connection coefficients on the right (\"bending of geodesics\") must be zero; so spacetime is indeed flat. Summary: Spacetime is flat -i.e., there exist \"flat coordinates\" in which r I-'a/3 = 0 lines, xa(i\\) = an + bai\\-if and only if everywhere and geodesics are straight Riemann = O. Lorentz coordinates exist if and only if Riemann = 0 Flatness does not imply Euclidean topology Note: In the spacetime of Einstein, which has a metric, one can choose {el-'Wo)} in the above argument to be orthonormal, el-' \u2022 e v = 1/I-'V at?fo. The resulting field of frames will then be orthonormal everywhere, and the resulting coordinate system will be Lorentz. Thus, in Einsteinian gravity the above summary can be rewritten: spacetime isflat (there exists a Lorentz coordinate system) ifand only ifRiemann = O. Warning: Flatness does not necessarily imply Euclidean topology. Take a sheet of paper. It is flat. Roll it up into a cylinder. It is still flat, intrinsically. The tracks of geodesics over it have not changed. Distances between neighboring points have not changed. Only the topology has changed, so far as an observer confined forever to the sheet is concerned. (The \"extrinsic geometry\"-the way the sheet is embedded in the surrounding three-dimensional space-has also changed; but an observer on the sheet knows nothing of this, and it is not the subject of the present chapter. See, instead \u00a721.5.) Take this cylinder. Bend it around and glue its two ends together, without changing its flat intrinsic geometry. Doing so is impossible if the cylinder remains embedded in flat, three-dimensional Euclidean space; perfectly possible if it is embedded in a Euclidean space of 4 dimensions. However, embedding is unimportant to observers confined to the cylinder, since all they ever measure is intrinsic geometry; so all that matters to them is the topological identification of the two ends of the cylinder with each other. The result is topologically a torus; but the tracks of geodesics are still unchanged; the intrinsic geometry is flat; Riemann vanishes. By analogy, take flat Minkowskii spacetime. Pick some'Lorentz frame, and in it pick a cube 10 10 light years on each side (0 < x < 1010 light years; similarly for y and z). Identify opposite faces of the cube so that a geodesic exiting across one face enters across the other. The result is topologically a three-torus: a \"closed universe\" with finite volume,", " with flat, Minkowskii geometry, and with a form that changes not at all as Lorentz time t passes (no expansion, no contraction). \u00a7 11.6. RIEMANN NORMAL COORDINATES 285 \u00a711.6. RIEMANN NORMAL COORDINATES In curved spacetime one can never find a coordinate system with r a /3y = 0 every where. But one can always construct local inertial frames at a given event?f0; and as viewed in such frames, free particles must move along straight lines, at least locally-which means r a /3y must vanish, at least locally. A very special and useful realization of such a local inertial frame is a Riemann normal coordinate system. Pick an event?f0 and a set of basis vectors {ea(?f o)} to be used there by an inertial observer. Fill spacetime, near?f0' with geodesics radiating out from?f0 like the quills of a hedgehog or porcupine. Each geodesic is determined by its tangent vector v at?f0; and the general point on it can be denoted \u2022 Riemann normal coordinates: a realization of local frames inertial Geometric construction of Riemann normal coordinates?f = \u00a7(i\\; v). affine parameter; tells \"where\" on geodesicr 1 [ t tr tangent vector at?f0; } Ltells \"which geodesic\" (11.25) Actually, this gives more geodesics than are needed. One reaches the same point tangent vector is 2v, as one reaches after after parameter length ~i\\ if the initial i\\ if the tangent vector is v: Thus, by fixing i\\ = 1 and varying v in all possible ways, one can reach every point in some neighborhood of?fo' This is the foundation for constructing Riemann normal coordinates. Choose an event?f. Find that tangent vector v at?f0 for which?f = \u00a7 (1; v). Expand that v in terms of the chosen basis and give its components the names x a :?f = r:] (1; xaea). (11.26) The point?f determines x a uniquely (if?f is near enough to?f0 that spacetime curvature has not caused geodesics to cross each other). Similarly, x a determines?f uniquely. Hence, x a can be chosen as the coordinates ofg'-its \"Riemann-normal coordinates, based on the event?f0 and basis {eaWo)}.\" Equation (11.26) summarizes Riemann-normal coordinates concisely. Other equations, derived in exercise 11.9, summarize their powerful properties: eaWo) = (a /axa);i'o; r a/3y(?fo) = 0; /3Y,Il('.f'o) = - ~ (Ra/3YIl + R\\/3,,), r a (11.27) (11.28) (11.29) If spacetime has a metric (as it does in actuality), and if the observer's frame at?f0 has been chosen orthonormal (ea' e /3 = 7Ia/3)' then Mathematical properties of Riemann normal coordinates 286 11. GEODESIC DEVIATION AND SPACETIME CURVATURE galA?f~0) = 1/a/3' ga/3,/J.(':J~o) = 0, I g,,/3,/J.v(?fo) = - \"3 (Ra/l/3,' + R\"v/3/J.) 2= - 3\" Ja/3/lV' (11.30) (11.31) (11.32) (11.32') Other mathematical realizations of a local frame inertial Here Ja/3/lV are components of the Jacobi curvature tensor (see exercise 11.7). Is this the only coordinate system that is locally inertial at. '!?0 (Le., has r a /3y = 0 there) and is tied to the basis vectors ea there (Le., has a/axa = e a there)? No. But all such coordinate systems (called \"normal coordinates\") will be the same to second order: Moreover, only those the same to third order, XNEw(?f) = xgLD(?f) + corrections of order (x(lLD)4, will preserve the beautiful ties (11.29) and (11.32) to the Riemann curvature tensor. EXERCISES Exercise 11.6. SYMMETRIES OF Riemann (To be discussed in Chapter 13). Show that Riemann has the following symmetries: R", "\"/3Y8 = R\"/3ly81 R\"I/3y81 = 0 (antisymmetric on last 2 indices) (11.33a) (vanishing of completely antisymmetric part) (11.33b) Exercise 11.7. GEODESIC DEVIATION MEASURES ALL CURVATURE COMPONENTS The equation of geodesic deviation, written up to now in the form or T u Tun + Riemann (..., u, n, u) = 0 also lets itself be written in the Jacobi form T u Tun + j(u, u)n = O. Here j (u, v), the \"Jacobi curvature operator,\" is defined by j(u, v)n =t [&len, u)v + &len, v)u], and is related to the \"Jacobi curvature tensor\" by Jacobi (..., n, u, v) =feu, v)n, (11.34) (11.35) \u00a7 11.6, RIEMANN NORMAL COORDINATES which implies 287 (11.36) (a) Show that j!J.CapYl = 0 follows from RIJ.apy = RIJ.a[py]' (b) Show that by studying geodesic deviation (allowing arbitrary u and n in T u Tun + j(u, u)n = 0) one can measure all components of Jacobi. (c) Show that Jacobi contains precisely the same information as Riemann, [Hint: show that -1. R avp - 3 (J vap IJ. IJ. _ IJ.'J pav), (11.37) this plus equation (11.36) for jIJ. vap proves \"same information content\",) Hence, by studying geodesic deviation one can also measure all the components of Riemann. (d) Show that the symmetry of RIJ.[vaP] = 0 is essential in the equivalence between Jacobi and Riemann by exhibiting proposed values for RlJ. vap = -RlJ.vPa for which RIJ.[vap] :j:. 0, and from which one would find jIJ.vap = 0, Exercise 11,8, GEODESIC DEVIATION IN GORY DETAIL Write out the equation of geodesic deviatio.n in component form in a coordinate system, Expand all covariant derivatives (semicolon notation) in terms of ordinary (comma) deriva tives and in terms of F's to show all r and a terms explicitly, Exercise 11,g, RIEMANN NORMAL COORQINATES IN GENERAL Derive properties (11.27), (11.28), (11.29), (11.31), (11.32), and (11.32') of Riemann normal coordinates, Hint: Proceed as follows, (a) From definition (11.26), derive (ap /aX a )9 (b) Similarly, from definition (11.26), show that each of the curves x a = val.. (where the o = e a, va are constants) is a geodesic through Po, with affine parameter A. (c) Show that rapy(po) = 0 by SUbstituting x a = val.. into the geodesic equation. (d) Since the curves x a = val.. are geodesics for every choice of the parameters va, they provide not only a geodesic tangent u =(a/aA)V\"' but also several deviation vectors NCal =(a/avah, Compute the components of these vectors in the Riemann normal coordi nate system, and substitute into the geodesic deviation equation as written in exercise 1l,8. (e) Equate to zero the coefficients of the zeroth and first powers of A in the geodesic deviation equation of part (d), using rapYI,.=,., = AvlJ.rapyjPo) + 0(1..2 ), which is a Taylor series for r, In this way arrive at equation (11.29) for rapy,lJ. in terms of the Riemann tensor, (f) From equations (l1.28), (11.29), and (8,24) for the connection coefficients in terms of the metric, derive equations (l1.31), (l1.32), and (11.32'), Exercise 11,10, BIANCHI Show that the Riemann curvature tensor satisfies the following \"Bianchi identities\" IDENTITIES Ra p[y8;<l = O. (l1.38) The geometric meaning of these identities will be discussed in Chapter 15, [Hint: Perform", " the calculation at the origin of a Riemann normal coordinate system,) y 288 11. GEODESIC DEVIATION AND SPACETIME CURVATURE Exercise 11.11. CURVATURE OPERATOR ACTS ON 1-FORMS Let ~il(u. v) be the operator ~il(u, v) = [Vu. Vv ) - V[u. vI when acting on I-forms 0' (or other tensors) as well as on tangent vectors. Show that (tJl(u, v)O', w) = -(0', M(u, v)w). Exercise 11.12. ROTATION GROUP: RIEMANN CURVATURE [Continuation of exercises 9.13,9.14, and 10.17.) Calculate the components of the Riemann curvature tensor for the rotation group's manifold SO(3); use the basis of generators {e,,}. [Answer: R \" _ I.\"p \"2 UY8'Py8 - (l1.39) where 8~~ is the permutation symbol defined in equation (3.501): 8~~ =(8\"y8 P8 - 8\"88Py). Note that this answer is independent of location tJ' in the group manifold.) CHAPTER 12 NEWTONIAN GRAVITY IN THE LANGUAGE OF CURVED SPACETIME The longest period of time for which a modern painting has hung upside down in a public gallery unnoticed is 47 days. This occurred to Le Bateau by Matisse in the Museum of Modern Art New York City. In this time 116,000 people had passed through the gallery. McWHIRTER AND McWHIRTER (1971) \u00a712.1. NEWTONIAN GRAVITY IN BRIEF The equivalence principle is not unique to Einstein's description of the facts of gravity. What is unique to Einstein is the combination of the equivalence principle and local Lorentz geometry. To return to the world of Newton, forget everything discovered in the last century about special relativity, light cones, the limiting speed of light, and proper time. Return to the \"universal time\" t of earlier centuries. In terms of that universal time, and of rectangular, \"Galilean\" space coordinates, Newtonian theory gives for the trajectories of neutral test particles This chapter is entirely Track 2. Chapters 9-11 are necessary preparation for it. It is not needed for any later chapter, but it will be helpful (1) Chapter 17 (Einstein field in equations) and (2) Chapters 38 and 39 (experimental tests and other theories of gravity). l/J (sometimes denoted - U) = Newtonian potential. (l2.l ) (12.2) Newtonian gravity: original formulation Customarily one interprets these equations as describing the \"curved paths\" xi(t) along which test particles fall in Euclidean space (not spacetime). These curved paths include circular orbits about the Earth and the parabolic trajectory of a baseball. Cartan (1923, 1924) asks one to abandon this viewpoint. Instead, he says, regard these trajectories as geodesics [t(i\\), xi(i\\)] in curved spacetime. (This change of viewpoint was embodied in Figures Band C of Box 1.6.) Since the \"affinely ticking\" Newtonian gravity: translation into language of curved spacetime 290 12. NEWTONIAN GRAVITY IN LANGUAGE OF CURVED SPACETIME Newtonian clocks carried by test particles read universal time (or some multiple, 'A = at + b, thereof), the equation of motion (12.1) can be rewritten (12.3) By comparing with the geodesic equation one can read off the values of the connection coefficients: all other r a /3y vanish. (12.4) And by inserting these into the standard equation (11.12) for the components of the Riemann tensor, one learns (exercise 12.1). R' OkO - -.. R' OOk - axi ax k'a2l/J - all other Ra/3ya vanish. (12.5) Finally, the source equation for the Newtonian potential V 2l/J 2: l/J,ii = 47TP i one can rewrite with the help of the \"Ricci curvature tensor\" R a/3 R /la/l/3 (contraction of Riemann) in the geometric form (exercise 12.2) Roo = 47TP; all other R a /3 vanish. (12.6) (12.7) (12.8) Equation (12.4) for r a/3y' equation (12.5) for Ra/3ya, equation (12.", "8) for R a/3' plus the law of geodesic motion are the full content of Newtonian gravity, rewritten in geometric language. It is one thing to pass quickly through these component manipulations. It is quite another to understand fully, in abstract and pictorial terms, the meanings of these equations and the structure of Newtonian spacetime. To produce such understanding, and to compare Newtonian spacetime with Einsteinian spacetime, are the goals of this chapter, which is based on the work of Cartan (1923, 1924), Trautman (1965), and Misner (1969a). EXERCISES Exercise 12.1. RIEMANN CURVATURE OF NEWTONIAN SPACETIME Derive equation (12.5) for R\"py8 from equation (12.4) for r\"py. Exercise 12.2. NEWTONIAN FIELD EQUATION Derive the geometric form (12.8) of the Newtonian field equation from (12.5) through (12.7). \u00a7 12.2. STRATIFICATION OF NEWTONIAN SPACETIME 291 \u00a7 12.2. STRATIFICATION OF NEWTONIAN SPACETIME Galileo and Newton spoke of a fiat, Euclidean \"absolute space\" and of an \"absolute time,\" two concepts distinct and unlinked. In absolute space Newtonian physics took place; and as it took place, absolute time marched on. No hint was there that space and time might be two aspects of a single entity, a curved \"spacetime\"-until Einstein made the unification in relativity physics, and Cartan (1923) followed suit in New tonian physics in order to provide clearer insight into Einstein's ideas. How do the absolute space of Galileo and Newton, and their absolute time, fit into Cartan's \"Newtonian spacetime\"? The key to the fit is stratification; stratification produced by the universal time coordinate t. Regard t as a function (scalar field) defined once and for all in Newtonian space time t = t(!7'). (12.9) Without it, spacetime could not be Newtonian, for \"t\" is every bit as intrinsic to Newtonian spacetime as the metric \"g\" is to Lorentz spaceJime. The layers of spacetime are the slices of constant t-the \"space slices\"-each of which has an identical geometric structure: the old \"absolute space.\" Adopting Cartan's viewpoint, ask what kind of geometry is induced onto each space slice by the surrounding geometry of spacetime. A given space slice is endowed, by the Galilean coordinates of\u00a712.l, with basis vectors e i = a/ax i ; and this basis has vanishing connection coefficients, r 4 ki = 0 [cf. equation (12.4)]. Consequently, the geometry of each space slice is completely flat. \"Absolute space\" is Euclidean in its geometry, according to the old viewpoint, and the Galilean coordinates are Cartesian. Translated into Cartan's language, this says: not only is each space slice (t = constant) fiat, and not only do its Galilean coordinates have vanishing connection coefficients, but also each space slice is en dowed with a three-dimensional metric, and its Galilean coordinate basis is orthonormal, (l2.l 0) The geometry of Newtonian spacetime: \"Universal time\" as a scalar field Space slices with Euclidean geometry If the space slices are really so fiat, where do curvature and geodesic deviation enter in? They are properties of spacetime. Parallel transport a vector around a closed curve lying entirely in a space slice; it will return to its starting point unchanged. But transport it forward in time by..1t, northerly in space by..1x k, back in time by -..1t, and southerly by -..1x k to its starting point; it will return changed by Curvature acts in spacetime, not in space slices 8A = -fJl.(..1tl.at'..1x k _ a_. )A''ax\" i.e., Geodesics of a space slice (Euclid's straight lines) that are initially parallel remain (12.11) y 292 12. NEWTONIAN GRAVITY IN LANGUAGE OF CURVED SPACETIME always parallel. But geodesics of spacetime (trajectories of freely falling particles) initially parallel get pried apart or pushed together by spacetime curvature, or equivalently in Galilean coordinates: nO = dno/dt = 0 initially =- nO = 0 always; a2l/J. ax} ax k d 2n i - - + dt 2 n k = 0 (12.l2a) (l2.l2b) (see Box 12.1 and", " exercise 12.3). EXERCISE Exercise 12.3. GEODESIC DEVIATION DERIVED Produce a third column for Box 11.4, one that carries out the \"geometric analysis\" in component notation using the Galilean connection coefficients (12.4) of Newtonian spacetime. Thereby achieve a deeper understanding of how the geometric analysis parallels the old Newtonian analysis. Galilean coordinates defined \u00a712.3. GALILEAN COORDINATE SYSTEMS The Lorentz spacetime of special relativity has an existence and structure completely independent of any coordinate system. But a special property of its geometry (zero curvature) allows the introduction of a special class of coordinates (Lorentz coordi nates), which cling to spacetime in a special way (o/ax\"). (a /ax/3) = 1/,,/3 everywhere. By studying these special coordinate systems and the relationships between them (Lorentz transformations), one learns much about the structure of spacetime itself (breakdown in simultaneity; Lorentz contraction; time dilatation;...). Similarly for Newtonian spacetime. Special properties of its geometry (explored in abstract later; Box 12.4) permit the introduction of special coordinates (Galilean coordinates), which cling to spacetime in a special way XO(?f) = t(?f); (a/ax i ). (a/ax k ) = 8jk ; r ioo = l/J,i for some scalar field P, and all other r\"/3Y vanish. To understand Newtonian spacetime more deeply, study the relations between these Galilean coordinate systems. Box 12.1 GEODESIC DEVIATION IN NEWTONIAN SPACETIME Coordinate system for calculation: Galilean space coordinates xi and universal time coordinate t. General component form of equation: d!3 __n_ + Ra ~ nY ---.:!- = 0 D 2 a cfA2!3y8 d\"A. d 8 d\"A.. Special conditions for this calculation: let the par ticles' clocks (affine parameters) all be normalized to read universal time, \"A. = t. This means that the separation vector na = (oxajon).;.. between geodesics has zero time component, nO = 0; Le., in abstract language, (dt, n) = t,ana = nO = 0; i.e., in geometric language, n lies in a space slice (surface of constant t). Evaluation of covariant derivative: )--y X 1= 15 min. Evaluation of tidal accelerations: since RiokO and RioOk are only nonzero components. Ri ~ d\"A. t - dx!3 nY dx 8 _ Ri d\"A..!!!...- nk.!!i. - Ri OkO d\"A. d\"A. 40 unless y is space index] for y a space index: 0 unless f3 = /) = 0) OkO l nk _ o2l/J - oxi oxk nk Resultant equation of geodesic deviation: agrees with result nO = 0 always, which followed from choice \"A. = t for all particles agrees with Newton-type calculatiOn) in Box 11.4; see also exercise 12.3. ( ( ) 294 12. NEWTONIAN GRAVITY IN LANGUAGE OF CURVED SPACETIME Point of principle: how can one write down the laws of gravity and properties of spacetime in Galilean coordinates first (\u00a712.1), and only afterward (here) come to grip with the nature of the coordinate system and its nonuniqueness? Answer: (a quotation from \u00a73.1, slightly modified): \"Here and elsewhere in science, as emphasized not least by Henri Poincare, that view is out of date which used to say 'Define your terms before you proceed.' All the laws and theories of physics, includ ing Newton's laws of gravity, have this deep and subtle character, that they both define the concepts they use (here Galilean coordinates) and make statements about these concepts.\" The Newtonian laws of gravity, written in a Galilean coordinate system make the statement \"Fioo = lP,i and all other Fapy = 0\" about the geometry of spacetime. This statement in turn gives information about the relationships between different Galilean systems. Let one Galilean system {xa(g')} be given, and seek the most general coordinate transformation leading to another, {xa'(g')}. The following constraints exist: (1) xO' = XO = t (both time coordinates must be universal time); (2) at fixed t (i.e., in a fixed space slice) both sets of space coordinates must be Euclidean, so they must be related by a", " rotation and a translation: x f = Aj'kXk + at L ~translatiOn] rotation matrix, i.e., AflAk'l = <'lrk'] X = Aj'kX' k., 'h k - a,wIt a =Aj'ka. k j'- (12.13a) (12.13b) The rotation and translation might, a priori, be different on different slices, Ark = Aj'k(t) and ai = ai(t); but (3) they must be constrained by the required special form of the connection coefficients. Calculate the connection coefficients in the new coordinate system, given their form in the old. The result (exercise 12.4) is: (produces \"Coriolis forces\"); ~ Fl 0,0' = ax\" + A\"k(A1'kx - t); alP.. k.. l' (12.14) [\"centrifugal forces\"J 4\"inertial forces\"] all other Fa'p'y' vanish (\"Euclidean\" index conventions; repeated space indices to be summed even if both are down; dot denotes time derivative). These have the standard Galilean form (12.4) if and only if 0 A\u00b7 rk =, Newtonian potential in1 t new coordinate systemr [ no.f 'P = 'P - a x + constant. \"k k no. t rNewtonian potential in] L old coordinate system (12.15) \u00a7 12.3. GALILEAN COORDINATE SYSTEMS 295 These results can be restated in words: any two Galilean coordinate systems are related by (1) a time-independent rotation of the space grid (same rotation on each space slice), and (2) a time-dependent translation of the space grid (translation possibly different on different slices) Transformations.Iinking Galilean coordinate systems x f = AfkXk + af (t). [constant] t Wtime-dependent] (12.16) The Newtonian potential is not a function defined in spacetime with existence independent of all coordinate systems. (There is no coordinate-free way to measure it.) Rather, it depends for its existence on a particular choice of Galilean coordinates; and if the choice is changed via equation (12.16), then l/J is changed: Newtonian potential depends on choice of Galilean coordinate system (By contrast, an existence independent of all coordinates is granted to the universal time tW) and the covariant derivative V.) (12.17) Were all the matter in the universe concentrated' in a finite region of space and then one could impose the global surrounded by emptiness (\"island universe\"), boundary condition \u2022 (12.18) Absolute Galilean coordinates defined This would single out a subclass of Galilean coordinates (\"absolute\" Galilean coor dinates), with a unique, common Newtonian potential. The transformation from one absolute Galilean coordinate system to any other would be Transformations linking absolute Galilean coordinate systems rcon'ta::~~ ti~kXk+ at +4vrt ~otatlOnJ con'tant] velocity (12.19) constant Idisplacemenq (\"Galilean transformation\"). But, (1) by no local measurements could one ever distinguish these absolute Galilean coordinate systems from the broader class of Galilean systems (to distinguish, one must integrate the locally measurable quantity l/J,i = r ioo out to infinity); and (2) astronomical data deny that the real universe is an island of matter surrounded by emptiness. It is instructive to compare Galilean coordinates and Newtonian spacetime as described above with Lorentz coordinates and the Minkowskii spacetime of special relativity, and with the general coordinates and Einstein spacetime of general rela tivity; see Boxes 12.2 and 12.3. (continued on page 298) ~ r Box 12.2 NEWTONIAN SPACETIME, MINKOWSKIIAN SPACETIME, AND EINSTEINIAN SPACETIME: COMPARISON AND CONTRAST Query Newtonian spacetime Minkowskiian spacetime (special relativity) Einsteinian spacetime (general relativity) What a priori geometric structures does spacetime possess? What preferred coordinate systems are present? What is required to select OUt a particular preferred coordinate system? (I) Universal time function t (2) Covariant derivative V (3) Spatial metric \"'''; but spacetime metric can not be defined (exercise 12.10) (I) Galilean coordinates in general (2) Absolute Galilean coordinates in an island universe (this case not considered here) (I) A single spatial orientation, the same throughout all spacetime (three Euler angles) (2) The arbitrary world line of the origin of space coordinates (three functions of time)", " A spacetime metric that is flat (vanishing Riemann curv,lture) A spacetime metric In general, every coordinate systcm is equally preferred (t~ough in special cases with symmetry there are special preferred coordinates) All four functions of position x n(!l') Lorentz coordinates (I) A single spatial orientation, the same throughout all spacetime (three Euler angles) (2) The loclltion of the origin of coordinates (four numbers) (3) The velocity of the origin of space coordinates (thrce numbers) Under what conditions is \"~1' and 2 are simullaneous\" well-defined? In general; it is a coordinate-free geometric concept Only after a choice of Lorentz frame has been made; \"simultaneity\" depends on the frame's velocity Only after arbitrary choice of limc coordinate has been madc Only after choice of Galilean coordinates has been made Only after choice of Lorentz coordinates has been made Only after arbitrary choice of space coordinates has been made Under what conditions is \"<jJ and 2 occur at same point in space\" well defined? Under what conditions is \"u and v, at different events, point in same direction\" well-defined? Always Only if u and v are both spatial vectors (dt, u) = (dt, v) = 0); or if they lie in the same space slice and are arbitrary vectors; or if there exists a preferred route connecting their locations, along which to compare them by parallel transport Under what conditions is \"the invariant distance between ~i' and 2\" well-defined? Only if ~1' and 2 lie in the same space slice Always Only if u and v lie at events infinitesimally close together: or if there exists a preferred route (e.g., a unique geodesic) connecting their locations, along which to conlpare thcm by parallel transport Only if ~1' and!! are sutliciently close together; or if there exists a unique preferred world line (e.g., a geodesic) linking them, along which to measure the distance \u00a7 12.3. GALILEAN COORDINATE SYSTEMS 297 Box 12.3 NEWTONIAN GRAVITY A LA CARTAN. AND EINSTEINIAN GRAVITY: COMPARISON AND CONTRAST Property Newton-Cartan Einstein Idea in brief (formulations of the equivalence principle of very different scope) Laws of motion of free parti cles in a local, freely falling, nonrotating frame are identical to Newton's laws of motion as expressed in a gravity-free Galilean frame Laws of physics in a local, freely falling, nonrotating frame are identical with the laws of physics as formulated in special relativity in a Lorentz frame Idea even more briefly stated Point mechanics simple in a local inertial frame Everything simple in a local inertial frame Consequence (tested to one part in lO\" by Roll-Krotkov Dicke experiment) Test particles of diverse composition started with same initial position and same initial velocity follow the same world line (\"definition of geodesic\"). Test particles of diverse composition started with same initial position and same initial velocity follow the same world line (\"definition of geodesic\") Another consequence Consequence of way light rays travel in real physical world? Summary of spacetime structure In every local region, there exists a local frame (\"freely faIling frame\") in which all geodesics appear straight (all r a #> = 0) Disregarded or evaded. All light rays have same velocity? Speed depend on motion of source? Speed depend on motion of observer? possible to move fast enough to catch up with a light ray? No satisfactory position on any of these issues Stratified into spacelike slices; geometry in each slice Euclidean; each slice charac terized by value of universal time (geodesic parameter): displacement of one slice with respect to another not spec ified; no such thing as a spacetime interval In every local region there exists a local frame (\"freely falling frame\") in which all geodesics appear straight (all r a #> = 0) Spacetime always and every where has local Lorentz character No stratification. Welldefined interval between every event and every nearby event; spacetime has everywhere local Lorentz character, with one local frame (specific space and time axes) as good as another (other space and time axes); \"homogeneous\" rather than stratified This structure expressed in mathematical language r a.,.'s. yes: spacetime metric gu,\" no: r i - 21' ('-123)' 00 - - -. 1_ t \u2022 \u2022 2x' all other r a#,. vanish r a u,.'s have no independent existence; all derived from ra = gaP ~ (. 2gB> ax# 2 #> + agB# ax\" _ 2g#\u00bb", " 2x P (\"metric theory of gravity\") 298 12. NEWTONIAN GRAVITY IN LANGUAGE OF CURVED SPACETIME EXERCISES x' ________ y' 1= 10 1 = I\" + ';; '2\", Figure 12.1. The coordinate system carried by an orbital laboratory as it moves in a circular orbit about the Earth. Exercise 12.4. CONNECTION COEFFICIENTS FOR ROTATING, ACCELERATING COORDINATES Beginning with equation (12.4) for the connection coefficients of a Galilean coordinate system (xa(P)}, derive the connection coefficients (12.14) of the coordinate system (xa'(P)} of equations (12.13). From this, verify that (12.15) are necessary and sufficient for (xa'(P)} to be Galilean. Exercise 12.5. EINSTEIN'S ELEVATOR Use the formalism of this chapter to discuss \"Einstein's elevator\" -i.e., the equivalence of \"gravity\" to an acceleration of one's reference frame. Which aspects of \"gravity\" are equiva lent to an acceleration, and which are not? Exercise 12.6. GEODESIC DEVIATION ABOVE THE EARTH A manned orbital laboratory is put into a circular orbit about the Earth [radius of orbit = ro, angular velocity = w = (M/ro 3)1!2-why?]. An astronaut jetisons a bag of garbage and watches it move along its geodesic path. He observes its motion relative to (non-Galilean) space coordinates (xf'(p)} which-see Figure 12.1-(1) are Euclidean at each moment of universal time [(a/ax!') \u2022(a/ax k') = B!k), (2) have origin at the laboratory's center, (3) have a/ax' pointing away from the Earth, (4) have a/ax' and a/ay' in the plane of orbit. Use the equation of geodesic deviation to calculate tile motion of the garbage bag in this coordi nate system. Verify the answer by examining the Keplerian orbits of laboratory and garbage. Hints: (I) Calculate Ra'/3'Y'8' in this coordinate system by a trivial transformation of tensorial components. (2) Use equation (12.14) to calculate r a'/3'y' at the center of the laboratory (i.e., on the fiducial geodesic). \u00a712.4. GEOMETRIC, COORDINATE-FREE FORMULATION OF NEWTONIAN GRAVITY To restate Newton's theory of gravity in coordinate-independent, geometric language is the principal goal of this chapter. It has been achieved, thus far, with extensive assistance from a special class of coordinate systems, the Galilean coordinates. To \u00a712.4. COORDINATE-FREE FORMULATION OF NEWTONIAN GRAVITY 299 climb out of Galilean coordinates and into completely coordinate-free language is straightforward in principle. One merely passes from index notation to abstract notation. Example: Restate in coordinate-free language the condition rOa /3 = 0 of Galilean coordinates. Solution: Write rOa /3 = -(V/3 dt, ea>; the vanishing of this for all a means V/3 dt = 0 for all /3, which in turn means V u dt = 0 for all u. In words: the gradient of universal time is covariantly constant. By this process_ one can construct a set of coordinate-free statements about New tonian spacetime (Box 12.4) that are completely equivalent to the standard, non geometric version of Newton's gravitation theory. From standard Newtonian theory, one can deduce these geometric statements (exercise 12.7); from these geometric statements, regarded as axioms, one can deduce standard Newtonian theory (exercise 12.8). Coordinate-free. geometric axioms for Newton's theory of gravity Exercise 12.7. FROM NEWTON TO CARTAN From the standard axioms of Newtonian theory (last part of Box 12.4) derive the geometric axioms (first part of Box 12.4). Suggested procedure: Verify each of the geometric axioms by a calculation in the Galilean coordinate system. Make free use of the calculations and results in \u00a7 12.1. EXERCISES \u2022 Exercise 12.8. FROM CARTAN TO NEWTON From the geometric axioms of Newtonian theory (first part of Box 12.4) derive the standard axioms (last part of Box 12.4). Suggested procedure: (I) Pick three orthonormal, spatial basis vectors (e j with e j'e k = Bjk ) at some event", " Po' Parallel transport each of them by arbitrary routes to all other events in spacetime. (2) Use the condition '!il(u, n)e j = 0 for all u and n [axiom (3)) and an argument like that in \u00a7 11.5 to conclude: (a) the resultant vector fields e j are independent of the arbitrary transport routes. (b) Vel = 0 for the resultant fields, and (c) [e j, e k ) = O. (3) Pick an arbitrary \"time line\", which passes through each space slice (slice of constant t) once and only once. Parametrize it by t and select its tangent vector as the basis vector eo at each event along it. Parallel transport each of these eo's throughout its respective space slice by arbitrary routes. (4) From axiom (4) conclude that the resultant field is independent of the transport routes; also show that the above construction process guarantees ~eo = VOej = O. (5) Show that [e a, ep) = 0 for all pairs of the four basis-vector fields, and conclude from this that there exists a coordinate system (\"Galilean coordinates\") in which e a = (J/(Jx a (see \u00a711.5 and exercise 9.9). (6) Show that in this coordinate system e j'e k = Bik ~verywhere (space coordinates are Euclidean). and the only nonzero components of the connection coefficient are r joo: here axioms (6) and (2) will be helpful. (7) From the self-adjoint property of the Jacobi curvature operator (axiom 7) show that RjOkO = R\\Jjo= show that in terms of the connection coefficients this reads r j and from this conclude that there exists a potential r/J such that r j on = r/J.i\" oO \u2022k = r k oo \u2022j : (8) Show that the geometric field equation (axiom 5) reduces to Poisson's equation V 2r/J = 4rrp. (9) Show that the geodesic equation for free fall (axiom 8) reduces to the Newtonian equation of motion d 2x j /dt 2 + r/J j = O. (continued on page 302) 300 12. NEWTONIAN GRAVITY IN LANGUAGE OF CURVED SPACETIME Box 12.4 NEWTONIAN GRAVITY: GEOMETRIC FORMULATION CONTRASTED WITH STANDARD FORMULATION Geometric Formulation Newton's theory of gravi ty and the properties ofNewtonian spacetime can be derived from the following axioms. (For derivation see exercise 12.8.) (l) There exists a function t called \"universal time\", and a symmetric covariant derivative V (with associated geodesics, parallel transport law, curvature opera tor, etc.). (2) The I-form dt is covariantly constant; i.e., V u dt = 0 for all u. [Consequence: if w is a spatial vector field (Le., w lies everywhere in a surface of constant t; i.e. (dt, w) = 0 everywhere), then Vuw is also spatial for every u, (dt, Vuw) = V u (dt, w) - (Vu dt, w) = 0.]'-. -.'' -. -.'[0 alWay,J to always] (3) Spatial vectors are unchanged by parallel transport around infinitesimal closed curves; i.e., &/(u, n)w = 0 if w is spatial, for every u and n. (4) All vectors are unchanged by parallel transport around infinitesimal, spatial, closed curves; i.e., &l(v, w) = 0 for every spatial v and w. (5) The Ricci curvature tensor, Ra {3 RP.ap.{3' has the form Ricci = 4'lTp dt \u00ae dt, where p is the density of mass. (6) There exists a metric\"\u00b7\" defined on spatial vectors only, which is compatible with the covariant derivative in this sense: for any spatial w and v, and for any u whatsoever, Vu(w\u00b7 v) = (Vuw)\u00b7 v + w\u00b7 (Vuv). [Note: axioms (1), (2), and (3) guarantee that such a spatial metric can exist; see exercise 12.9.] \u00a7 12.4. COORDINATE-FREE FORMULATION OF NEWTONIAN GRAVITY 301 (7) The Jacobi curvature operator j(u, n), defined for any vectors u, n, p by $(u,n)p =", "2[~(p,n)u + ~(p,u)n], 1 is \"self-adjoint\" when operating on spatial vectors; i.e., v' [$ (u, n)w] = w\u00b7 [$ (u, n)v] for all spatial v, w; and for any u, n. (8) \"Ideal rods\" measure the lengths that are calculated with the spatial metric; \"ideal clocks\" measure universal time t (or some multiple thereof); and \"freely falling particles\" move along geodesics of V. [Note: this can be regarded as a definition of \"ideal rods,\" \"ideal clocks,\" and \"freely falling particles.\" A more complete theory (e.g., general relativity; see \u00a7 16.4) would predict in advance whether a given physical rod or clock is ideal, and whether a given real particle is freely falling.] Note: For an alternative but equivalent set of axioms, see pp. 106-107 of Trautman (1965). Standard Formulation The following standard axioms are equivalent to the above. (1) There exist a universal time t, a set of Cartesian space coordinates xi (called \"Galilean coordinates\"), and a Newtonian gravitational potential t/J. (2) The density of mass p generates the Newtonian potential by Poisson's equation, (3) The equation of motion for a freely falling particle is d?x i at/J dt?' + ox i = O. (4) \"Ideal rods\" mea~ure the Galilean coordinate lengths; \"ideal clocks\" measure universal time. 302 12. NEWTONIAN GRAVITY IN LANGUAGE OF CURVED SPACETIME Exercise 12.9. SPATIAL METRIC ALLOWED BY OTHER AXIOMS Show that the geometric axioms (I), (2), and (3) of Box 12.4 permit one to introduce a spatial metric satisfying axiom (6). Hint: Pick an arbitrary spatial basis {e;} at some event. Define it to be orthonormal. e;' el: =Bjk\" Extend this basis through all spacetime by the method used in (I) of exercise 12.8. Define e j \u2022 e k =Bjk everywhere in spacetime for this basis. Then prove that the resulting metric satisfies the compatibility condition of axiom (6). Exercise 12.10. SPACETIME METRIC FORBIDDEN BY OTHER AXIOMS Show that in Newtonian spacetime it is impossible to construct a nondegenerate spacetime metric g. defined on all vectors, that is compatible with the covariant derivative in the sense that (12.20) Note: to prove this requires mastery of the material in Chapter 8 or 13; so study either 8 or 13 before tackling it. Hint: Assume that such a 9 exists. Show, by the methods of exercise 12.8, that in a Galilean coordinate system the spatial components gil: are independent of position in spacetime. Then use this and the form of RQ/3-;8 in Galilean coordinates to prove R iOkO and - RO;kO are not identical, a result that conflicts with the symmetries of the Riemann tensor [eq. (8.45)) in a manifold with compatible metric and covariant derivative. \u00a712.5. THE GEOMETRIC VIEW OF PHYSICS: A CRITIQUE An important digression is in order. \"Every physical quantity must be describable by a (coordinate-free) geometric object, and the laws of physics must all be expressible as geometric relationships between these geometric objects.\" This view of physics, sometimes known as the \"principle of general covariance,\" pervades twentieth-century thinking. But does it have any forcible content? No, not at all, according to one viewpoint that dates back to Kretschmann (1917). Any physical theory originally written in a special coordinate system can be recast in geometric, coordinate-free language. Newtonian theory is a good example, with its equivalent geometric and standard formulations (Box 12.4). Hence, as a sieve for separating viable theories from nonviable theories, the principle of general covariance is useless. But another viewpoint is cogent. It constructs a powerful sieve in the form of a slightly altered and slightly more nebulous principle: \"Nature likes theories that are simple when stated in coordinate-free, geometric language.\"* According to this principle, Nature must love general relativity, and it must hate Newtonian theory. Of all theories ever conceived by physicists, general relativity has the simplest, most elegant geometric foundation [three axioms: (1) there is a metric; (2) the metric is governed by the Einstein field equation G = 8'lTT;", " (3) all special relativistic laws of physics are valid in local Lorentz frames of metric]. By contrast, what diabolically *Admittedly, this principle is anthropomorphic: twentieth-century physicists like such theories and even find them effective in correlating observational data. Therefore, Nature must like them too! The principle of general covariance has no forcible content Twentieth-century viewpoint judges a theory by simplicity of its geometric formulation Einstein's theory of gravity is simple; Newton's is complex \u00a7 12.5. GEOMETRIC VIEW OF PHYSICS: A CRITIQUE 303 clever physicist would ever foist on man a theory with such-a complicated geometric foundation as Newtonian theory? Of course, from the nineteenth-century viewpoint, the roles are reversed. It judged simplicity of theories by examining their coordinate formulations. In Galilean coordinates, Newtonian theory is beautifully simple. Expressed as differential equa tions for the metric coefficients in a specific coordinate system, Einstein's field equations (ten of them now!) are horrendously complex. The geometric, twentieth-century view prevails because it accords best with exper imental data (see Chapters 38-40). In Chapter 17 it will be applied ruthlessly to make Einstein's field equation seem compelling. CHAPTER 13 RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF ALL Philosophy is written in this great book (by which I mean the universe) which stands always open to our view. but it cannot be understood unless one first learns how to comprehend the language and interpret the symbols in which it is written, and its symbols are triangles, circlas, and other geometric figures, without which it is not humanly possible to comprehend even one word of it; without these one wanders in a dark labyrinth. GAll LEO GAll LEI (1623) \u00a713.1. NEW FEATURES IMPOSED ON GEOMETRY BY LOCAL VALIDITY OF SPECIAL RELATIVITY This chapter is entirely Track 2. Chapters 9-11 are necessary preparation for it. It will be needed as preparation for (11 the second half of Chapter 14 (calculation of curvaturel. and (21 the details. but not the message, of Chapter 15 (Bianchi identitiesl. \u00a7 13.6 (proper reference framel will be useful throughout the applications of gravitation theory (Chapters 18-401\u00b7 Constraints imposed on spacetime by special relativity Freely falling particles (geodesics) define and probe the structure of spacetime. This spacetime is curved. Gravitation is a manifestation of its curvature. So far, so good, in terms of Newton's theory of gravity as translated into geometric language by Cartan. What is absolutely unacceptable, however, is the further consequence of the Cartan-Newton viewpoint (Chapter 12): stratification of spacetime into slidable slices, with no meaning for the spacetime separation between an event in one slice and an event in another. Of all the foundations of physics, none is more firmly established than special relativity; and of all the lessons of special relativity none stand out with greater force than these. (I) Spacetime, far from being stratified, is homogeneous and isotropic throughout any region small enough (\"local region\") that gravitational tide-producing effects (\"spacetime curvatures\") are negligible. (2) No local experi ment whatsoever can distinguish one local inertial frame from another. (3) The speed of light is the same in every local inertial frame. (4) It is not possible to give frame-independent meaning to the separation in time (\"no Newtonian stratifica- Metric: the instrument which imposes the constraints Metric described in three languages \u00a7 13.2. METRIC 305 there exists a frame tion\"). (5) Between every event and every nearby event independent, coordinate-independent spacetime interval (\"Riemannian geometry\"). (6) Spacetime is always and everywhere locally Lorentz in character (\"local Lorentz character of this Riemannian geometry\"). What mathematics gives all these physical properties? A metric: a metric that is locally Lorentz (\u00a7\u00a7 13.2 and 13.6). All else then follows. In particular, the metric destroys the stratified structure of Newtonian spacetime, as well as its gravitational potential and universal time coordinate. But not destroyed are the deepest features of Newtonian gravity: (1) the equivalence principle (as embodied in geodesic de scription of free-fall motion, \u00a7\u00a713.3 and 13.4); and (2) spacetime curvature (as measured by tidal effects, \u00a7 13.5). The skyscraper of vectors, forms, tensors (Chapter 9), geodesics, parallel transport, covariant derivative (Chapter 10), and curvature (Chapter 11) has rested on crum bling foundations-Newtonian physics and a geodesic law based on Newtonian physics", ". But with metric now on the scene, the whole skyscraper can be transferred to new foundations without a crack appearing. Only one change is necessary: the geodesic law must be selected in a new, relativistic way; a way based on metric (\u00a7\u00a7 13.3 and 13.4). Resting on metric foundations, spacetime curvature acquires additional and stronger properties (the skyscraper is redecorated and extended), which are studied in \u00a713.5 and in Chapters 14 and 15, and which lead almost inexorably to Einstein's field equation. \u00a713.2. METRIC A spacetime metric: a curved spacetime metric; a locally Lorentz, curved spacetime metric. This is the foundation of spacetime geometry in the real, physical world. Therefore take a moment to recall what \"metric\" is in three contrasting languages. In the language of elementary geometry, \"metric\" is a table giving the interval between every event and every other event (Box 13.1 and Figure 13.1). In the language of coordinates, \"metric\" is a set of ten functions of position, g/lv(x\"'), such that the expression (13.1 ) gives the interval between any event x'\" and any nearby event x'\" + Llx\"'. In the a bilinear machine, language of abstract differential geometry, metric is (u' v)\"] out of two 9 =-( tangent vectors. u and v. ). to produce anum ber [\"scalar product g(u, v) The link between the abstract. machine viewpoint and the concrete coordinate viewpoint is readily exhibited. Let the tangent vector represent the displacement between two neighboring events. The abstract viewpoint gives (, olllilll/<:d Oil page 310) y 306 13. RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF ALL Box 13.1 METRIC DISTILLED FROM DISTANCES Raw Data on Distances Let the shape of the earth be described as in Figure 13.1, by giving distances between some of the principal identifiable points: buoys. ships, icebergs, lighthouses, peaks, and flags: points to a total of n = 2 X 10 7 \u2022 The total number of distances to be given is l1(n - 1)/2 = 2 X 10 14. With 200 distances per page of printout, this means Firsl point 9,316,434 9.316,434 9,316,434 9' I r A\"A Second point Distance (Nautical miles) First point Seco' p\" 14.117,103 14.117.104 14,117,105 1410.316 1812.717 1629.291 9.316,434 9,316,416 9.~, 10 12 pages weighing 6 g each, or 6 X 106 metric tons of data. With 6 tons per truck this means 106 truckloads of data; or with one truck passing by every 5 seconds,.~ ~ C idS ; : ; f night and day t,.!lie to get in the data. ~~~~c First Distillation: Distances to Nearby Points Only Get distances between faraway points by adding distances covered on the elementary short legs of the trip. Boil down the table of distances to give only the distance between each point and the hundred nearest points. Now have 100 n = 2 X 109 distances, or 2 X 109/200 = 10 7 pages ofdata, or 60 tons of records, or 10 truckloads. Second Distillation: Distances Between Nearby Points in Terms of Metric Idealize the surface of the earth as smooth. Then in any sufficiently limited region the geometry is Euclidean. This circumstance has a happy consequence. It is.enough to know a few distances between the nearby points to be able to determine all the \u00a7 13.2. METRIC 307 (10) 2 distances between the nearby points. Locate point 2 so that (102) is a right triangle; thus, (12)2 = (10)2 + (20)2. Consider a point 3 close to O. Define and x(3) = (13) - (10) y(3) = (23) - (20). Then the distance (03) does not have to be supplied independently; calculated from the formula* it can be (03)2 = [x(3)]2 + [y(3)]2. Similarly for a point 4 and its distance (04) from the local origin O. Similarly for the distance (mn) between any two points m and n that are close to 0: (mn)2 = [x(m) - x(n)j2 + [y(m) - y(n)j2. Thus it is only needful to have the distance (1m) (from point 1) and (2m) (from point 2) for each", "imated by a polyhedron built out of triangles (\"skeleton geometry\"). The approximation can be made arbitrarily -good by makfrig- the number of triangles sufficiently great and the size of each sufficiently small. Upper right: The geometry in each triangle is Euclidean: giving the three edge lengths fixes all the features of the figure, including the indicated angle. Lower right: The triangles that belong to a given vertex, laid out on a flat surface, fail to meet. The deficit angle measures the amount of curvature concentrated at that vertex on the tear-drop earth. The sum of these deficit angles for all vertices of the tear drop equals 4'1T. This \"Gauss-Bonnet theorem\" is valid for any figure with the topology of the 2-sphere; for the simplest figure of all, a tetrahedron, four vertices with a deficit angle at each of 180' are needed-3 triangles X 60' per triangle available = 180' deficit. In brief. the shape of the tear drop, in the given skeleton-geometry approximation, is determined by its 50 visible edge lengths plus. say, 32 more edge lengths hidden behind the figure, or a total of82 edge lengths, and by nothing more (\"distances determine geom<:lry\"k\"Meu-.ic\u00b7' 1etIs the distance between every point and every nearby point. If volcanic action the distances between that Icelandic capital and nearby points increase accordingly: that there is not a great bump on the earth in the vicinity distances again reveal shape. Conversely, of Iceland, and that the earth does not now have a tear-drop shape, can be unambiguously established by analyzing the pattern of distances from point to point in a sufficiently well-distributed network of points, with no call for any observations other than measurements of distance. _ ------raises Rejkjavik, _ Polygon laid out flat fails to close -Gap measures curvature concentrated at Cairo v Covariant components of metric \"Line element\" compared with \"metric as bilinear machine\" 310 13. RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF ALL for the interval between those events; comparison with the coordinate viewpoint [equation (13.1)] reveals (13.2) (standard equation for calculating components of a tensor). Just as modern differential geometry replaces the old style \"differential\" df by the \"differential form\" df (Box 2.3, page 63), so it also replaces the old-style \"line element\" ds 2 = gil' dx ll dx' = (\"interval between x'\" and x'\" + dx\"''') (13.3) by the bilinear machine (\"metric tensor\") -d 2 - 9 = s = gil' dll!O.d'\u00b7 '01 X. X (13.4) The output g\u00ab(, () of this machine, for given displacement-vector input, is identical to the old-style interval. Hence, ds 2 = gil' dxll I8l dx' represents the interval of an unspecified displacement; and the act of inserting ( into the slots of ds 2 is the act of making explicit the interval g\u00ab(, () = gil\".::lx ll.::lx'\u00b7 of an explicit displacement. In curved spacetime with metric, just as in flat spacetime with metric (\u00a72.5), a Metric produces a correspondence between '-forms and tangent vectors particular I-form ii corresponds to any given tangent vector u: ii is defined by \"(ii, v) = g(u, v) for all v\" (13.5) (\"representation of the same physical quantity in the two alternative versions of vector and I-form\"; \"corresponding representations\" as (A)-tensor and as (~)-tensor). the I-form ii corresponding to a basis vector u = e\", has components Example: standard way ~- t J t - to compute upJ Up = (ii, ep) = g(u, ep) = g(e\"\" ep) = g\",P; '--iby u = e\",] ~equation (13.2)] [definition (13.5) [ thus g\",pwP is the I-form e\", corresponding to e\",. (13.6) Also as in flat spacetime (\u00a73.2), a tensor can accept either a vector or a I-form into any given slot S(ii, G, v) = S(u, G, v). (13.7) Lowering indices Equivalently, in component language, the indices of a tensor can be lowered with the covariant components of the metric S/y = S(e\"\" w P, e y) = S(e\"\" w P, e y) = S(g\"'", "llwlJ., w P, e y) = g\"'IlSIJ.Py- (13.8) l[definition of S/y] l[by equation (13.6)] \u00a7 13.2. METRIC 311 The basis vectors {e a } can be chosen arbitrarily at each event. Therefore the corresponding components gap of the metric are quite arbitrary (though symmetric: gaP = gpa)' But the mixed components gap are not arbitrary. In particular, equations (13.5) and (13.7) imply g(ii, v) =g(u, v) =(ii, v). (13.9) Therefore one concludes that the metric tensor in mixed representation is identical with the unit matrix: Mixed and contravariant components of metric This feature of the metric in tum fixes the contravariant components of the metric: g a~g _ ga _ ~a. u 13' 13 - ~p - (13.11) l[\"lowering an index\" of gall] (13.10) l.e., IlgaPl1 is the matrix inverse of Ilgapii. (13.12) This reciprocity enables one to undo the lowering of tensor indices (i.e., raise indices) Raising indices with gap: s~P - 81l sap - gllVg sap - gllVS 13 y - y - y - va a v yo (13.13) The last two paragraphs may be summarized in brief: p; (1) gap = 8a (2) IlgaPl1 = Ilgap il-l; (3) tensor indices are lowered with gap; (4) tensor indices are raised with gap. In this formalism of metric and index shuffling, a big question demands attention: how can one tell whether the metric is locally Lorentz rather than locally Euclidean or locally something else? Of course, one criterion (necessary; not sufficient!) is dimensionality-a locally Lorentz spacetime must have four dimensions. (Recall the method of \u00a71.2 to determine dimensionality.) Confine attention, to four dimensional manifolds. What else must one demand? One must demand that at every event tj' there exist an orthonormal frame (orthonormal set of basis vectors {eli}) in which the components of the metric have their flat-spacetime form then, gliP = eli' ep = 'TIap = diagonal (-1,1,1,1). (13.14) To test for this is straightforward (exercise 13.1). (1) Search for a timelike vector u (u' u < 0). If none exist, spacetime is not locally Lorentz. If one is found, then (2) examine all non-zero vectors v perpendicular to u. If they are all spacelike (v' v > 0), then spacetime is locally Lorentz. Otherwise it is not. Metric must be locally Lorentz y EXERCISES 312 13. RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF ALL Exercise 13.1. TEST WHETHER SPACETIME IS LOCAL LORENTZ Prove that the above two-step procedure for testing whether spacetime is locally Lorentz is valid: i.e.. prove that if the procedure says \"yes,\" then there exists an orthonormal basis with gap = 1)all at the event in question; if it says \"no.\" then no such basis exists. Exercise 13.2. PRACTICE WITH METRIC A four-dimensional manifold with coordinates v, r, O. ep has line element (old-style notation) corresponding to metric (new-style notation) ds 2 = -(I - 2M/r) dv I8i dv + dv I8i dr + dr I8i dv + r2(dO I8i dO + sin20 dep I8i d\u00a2), where M is a constant. (a) Find the \"covariant\" components gall and \"contravariant\" components gall of the metric, gc/>e:> = r2 in this coordinate system. [Answer: gvv = -(I - 2M/r), gVT = grv = I, gee = r2 sin2 0; all other gall vanish; gVT = gTV = I, gTT = (I - 2M/r), gee = r- 2, gc/>c/> = r- 2 sin- 20, all other gall vanish.] (b) Define a scalar field t by t= v - r - 2Mln[(r/2M) - I]. ) of the I-form Ii =dt? What are the covariant and contravariant components (u a and u a What", " is the squared length u 2 =u\u00b7 u, of the corresponding vector? Show that u is timelike in the region r > 2M. [Answer: Uv = I, uT= -1/(1 - 2M/r), ue = Uc/> = 0; UV = -1/(1 2M/r), uT= 0, u e = uc/> = 0; u 2 = -1/(1 - 2M/r).] (c) Find the most general non-zero vector worthogonal to II in the region r > 2M. and show that it is spacelike. Thereby conclude that spacetime is locally Lorentz in the region r> 2M. [Answer: Since w' u = wau a = -wv/(I - 2M/r), Wv must vanish, but wT' we' we:> are arbitrary, and w 2 = (I - 2M/r)w/ + r-2w/ + r- 2 sin-20w/ > 0.] (d) Let t, r, 0, ep be new coordinates for spacetime. Find the line element in this coordinate system. [Answer: This is the \"Schwarzschild\" line element (e) Find an orthonormal basis, for which gaP = 1)all in the region r > 2M. [Answer: eo = (I - 2M/r)-1/2 a/at, e; = (I - 2M/r)1/2 a/or, eiJ = r-1 0/00, e\u00a2 = (r sin 0)-1 a/oep.] \u00a713.3. CONCORD BETWEEN GEODESICS OF CURVED SPACETIME GEOMETRY AND STRAIGHT LINES OF LOCAL LORENTZ GEOMETRY More could be said about the mathematical machinery and physical implications of \"metric,\" but an issue of greater urgency presses for attention. What has metric (or spacetime interval) to do with geodesic (or world line of test particle)1 Answer: \u00a7 13.3 GEODESIC AS STRAIGHT LINE OF LOCAL LORENTZ GEOMETRY 313 Two mathematical objects (\"straight line iii a local Lorentz frame\" and \"geodesic of the over-all global curved spacetime geometry\") equal to the same physical object (\"world line of test particle\") must be equal to each other (\"condition of consist ency\"). As a first method to spell out this consistency requirement,examine the two mathematical representations of the world line ofa test particle in the neighborhood of a given event tJ'0. The local-Lorentz representation says: \"Pick a local Lorentz frame at tJ'o. [As spelled out in exercise 13.3, such a local Lorentz frame is the closest thing there is to a global Lorentz frame at tJ'0; i.e., it is a coordinate system in which gaP(tJ'0) = 1Iap (flat-spacetime metric), gaP,y(tJ'O) = 0, gap,ya(tJ'o) \u00a5= 0 except in special cases, such as flat space.] (13.15a) (13.15b) (13.15c) The world line in that frame has zero acceleration, d 2x a / dr 2 = 0 at tJ'0 (\"straight-line equation\"), (13.16) Local-Lorentz description of straight lines where 7' is proper time as measured by the particle's clock.\" The geodesic representation says \"In the local Lorentz frame, as in any coordinate frame, the world line satisfies the geodesic equation Geodesic description of straight lines (13.17) (7' is an affine parameter because it is time as measured by the test particle's clock).\" Consistency of the two representations for any and every choice of test particle (any and every choice of dx a / d7' at tJ'0) demands rapy{tJ'o) = 0 in any local Lorentz frame [coordinate system satisfying equations (13.15) at tJ'0]; (13.18) i.e., it demands that everyJocal Lorentz frame is a local inertial frame. (On local inertial frames see \u00a711.6.) In such a frame, all local effects of \"gravitation\" disappear. That is the physical shorthand for (13.18). One does not have to speak in the language of a specific coordinate system when one demands identity between the geodesic (derived from the r apy) and the straight line of the local Lorentz geometry (gp..)' The local Lorentz specialization of coordi", " nates may be the most immediate way to see the physics (\"no local effects of gravitation\"), but it is not the right way to formulate the basic mathematical re quirement in its full generality and power. The right way is to demand Vg = 0 (\"compatibility of 9 and V\"). (13.19) Stated in the language of an arbitrary coordinate system, this requirement reads (13.19') Condition of consistency: r a py = 0 in local Lorentz frame Consistency reformulated: Vg = O. y 314 13. RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF ALL That this covariant requirement is fulfilled in every coordinate system follows from its validity in one coordinate system: a local Lorentz frame. (The first term in this equation, and the last two terms, are separately required to vanish in the local Lorentz frame at point &'o-and required to vanish by the physics.) From Vg = 0, one can derive both the abstract chain rule (13.20) f3\"y expressed in terms of r a metric (Exercise 13.4) and the following equations for the connection coefficients in any frame in terms of (1) the metric coefficients, ga{3 = 8 a'8{3' and (2) the covariant commutation coefficients of that frame: (13.21) (definition of rp.{3Y)' (13.22) r a{3y = gap. rp.{3y 1 =\u00b1(gp.{3,y + gp.y,{3 - rp.{3y = \"2 (gp.{3,y + cp.{3y + gp.y,{3 + cp.y{3 - g{3Y,p. - c{3yp.) (13.23) g{3Y,p.) in any coordinate frame. (See Exercise 13.4). Equations (13.23) are the connection coefficients required to make the geodesics of curved spacetime coincide with the straight lines of the local Lorentz geometry. And they are fixed uniquely; no other choice of connection coefficients will do the job! Summary: in curved spacetime with a local Lorentz metric, the following seemingly different statements are actually equivalent: (1) the geodesics of curved spacetime coincide with the straight lines of the local Lorentz geometry; (2) every local Lorentz frame [coordinates with ga{3('3'o) = 1Ia{3' ga{3,r<'3'o) = 0) is a local inertial frame [r a{3Y('3'o) = 0); (3) the metric and covariant derivative satisfy the compatibility condition Vg = 0; (4) the covariant derivative obeys the chain rule (13.20); (5) the connection coefficients are determined by the metric in the manner of equations (13.23). A sixth equivalent statement, derived in the next section, says (6) the geodesics of curved spacetime coincide with world lines of extremal proper time. EXERCISES Exercise 13.3. MATHEMATICAL REPRESENTATION OF LOCAL LORENTZ FRAME By definition, a local Lorentz frame at a given event Po is the closest thing there to a global Lorentz frame. Thus, it should be a coordinate system with gp.vWo) = TJp.v' and with as many derivatives of gp.v as possible vanishing at Po. Prove that there exist coordinates in which gp.vWo) = TJp.v and gp.v,pWo) = 0, but that gp.v,p\"Wo) cannot vanish in general. Hence, such coordinates are the mathematical representation of a local Lorentz frame. [Hint: Let {x'\" W)} be an arbitrary but specific coordinate system, and {xP.(P)} be a local Lorentz frame, both \u00a7 13.4. GEODESICS AS WORLD LINES OF EXTREMAL PROPER TIME 315 with origins at '!i' o' Expand the coordinate transformation between the two in powers of xp. and use the transformation matrix La'p. =oxa'/oxp. to get gp.v('!i'0)' gp.vj'!i'0), and gp.v,pCl ('!i'0) in terms of ga'/3' and its derivatives and the constants Ma Po' Na p.v' pap.vp' Show that whatever ga'(j' may be (so long as it is nonsingular, so ga'f3' exists!), one can choose the 16 constants Map. to make gp.v = TJp.v (ten conditions); one can choose the 4 X 10 = 40 constants NYp.v to make", " the 10 X 4 = 40 gp.v,p('!i'0) vanish; but one cannot in general choose the 4 X 20 = 80 pap.vp to make the lOX 10 = 100 gp.v,pCl vanish.] Exercise 13.4. CONSEQUENCES OF COMPATIBILITY BETWEEN 9 AND V (a) From the condition of compatibility V9 = 0, derive the chain rule (13.20). (b) From the condition of compatibility Vg = 0 and definitions (13.21) a.nd (13.22), derive equation (13.23) for the connection coefficients. [Answer: See exercise 8.15, p. 216.] \u00a713.4. GEODESICS AS WORLD LINES OF EXTREMAL PROPER TIME In a local Lorentz frame, it is easy to distinguish a world line that is straight from one that is not. Position the Lorentz frame and so orient it that the starting point of the world line, {f, lies at the origin and the end point, ~E, lies at x = 0, y = 0, z = 0, t = T As an example ofa nonstraight world line, consider passage at uniform velocity from {f to point tj' with coordinates (~T; 0, 0, ~R) and from there again with uniform velocity to point qa. The lapse of proper time from start to finish (\"length of world line\") is Thus the lapse of proper time is diminished from its straight-line value, and dimin ished moreover for any choice of R whatsoever, except for the zero or straight-line value R = 0. As for this simple nonstraight curve, so also for any other nonstraight curve: the lapse of proper time between {f and \u00a313 is less than the straight-line lapse (Exercise 6.3). Thus, in flat spacetime, extremal length of world line is an indicator of straightness. Any local region of the curved spacetime of the real, physical world is Lorentz in character. In this local Lorentz geometry, it is easy to set up Lorentz coordinates and carry out the extremal-length analysis just sketched to distinguish between a straight line and a nonstraight line: In flat spacetime. straight length lines have extremal Extremal spacetime length in curved 7' = f 'N 'N d7' = f' (-'I)p.v dxp. dx v)1/2 {/ {/ a maxim urn for straight line = as compared to any variant of ). (13.24 ) ( the straight line y 316 13. RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF ALL Such a test for straightness can be carried out separately in each local Lorentz region along the world line, or, with greater efficiency, it can be carried out over many local Lorentz regions simultaneously, i.e., over a region with endpoints {f and ~1j so widely separated that no single Lorentz frame can possibly contain them both. To carry out the analysis, one must abandon local Lorentz coordinates. Therefore introduce a general curvilinear coordinate system and find 'N 7' = J d7' = J (- gill' dx ll dXV '11 )I/2 {/ {/ an extremum for timelike world line that is straight in each local Lorentz frame along its path, as compared to any \"nearby\". variant of this world line ) ( (13.25) In the real world, the path of extremal 7', being straight in every local Lorentz frame, must be a geodesic of spacetime. Notice that the word \"maximum\" in equation (13.24) has been replaced by \"extremum\" in the statement (13.25). When {f and ('8 are widely separated, they may be connected by several different geodesics with differing lapses of proper time (Figure 13.2). Each timelike geodesic extremizes 7' with respect to nearby de formations of itself, but the extremum need not be a maximum. When several distinct geodesics connect two events, the typical one is not a local maximum (\"moun tain peak\") but a saddle point (\"mountain pass\") in such a diagram as Figure 13.2 or 13.3. Concord between locally straight lines (lines of extremal 7') and geodesics of curved spacetime demands that timelike geodesics have extremal proper length. If so, then any curve x ll (;\\) between {f (where ;\\ = 0) and qa (where ;\\ = 1) that extremizes 7' should satisfy the geodesic equation. To test for an extremal by comparing times, pick a curve suspected to be a geodesic, and deform it slightly but", ". <\u00b7\u00b7\u00b7\u00b7K\u00b7 -- :~::.......\u2022.... /'1 -r--,~+--+-f4-+-+-\"\"-+---j\"--t+-~I-+\"\"'-I-+-+-\"\"'++-H\"'~oJ Figure 13.3. Extremizing lapse of proper time by suitable choice of world line. Left: Spacetime; and world line F that extremizes the lapse of proper time 'T from t! to!l3 compared to other world lines. The specific world lines depicte1 in the diagram happen to be distinguished from fiducial world line G by two \"Fourier amplitudes\" oJ and \u00b02:.sO~(A) = OJ sin ('ITA) + 02 sin (2'ITA)~ _ where the arbitrary scaling of A, and its zero, are so adjusted that A(t!) = 0, A(!l3) = l. Right: \"Path space.\" The coordinates in this space are the Fourier amplitudes oJ and 02' Only these two amplitudes (\"two dimensions\") are shown out of what in principle are infinitely many amplitudes (\"infinite-dimensional path space\") required to represent the general timelike world line connecting d and {fl. Any given contour curve runs through all those points (in path space) for which the corresponding world lines (in spacetime) r.ack up the indicated lapse of proper time 'T. Foregoing description is classicaL according to quantum mechanics, all the timelike world lines connecting t! and!l3 occur with the same probability amplitude (\"principle of democracy of histories\") with the only difference from one to another being the phase of this complex probability amplitude exp ( - im'T/11) (m = mass of particle, 11 = quantum of angular momentum). In the sum over these probability amplitudes, however, destructive interference wipes out the contributions from all those histories which differ too much from the optimal or classical history (\"Fresnel wave zone\"; \"Feynman's principle of sum over histories\"; see Feynman and Hibbs, 1965). Capitalizing on this wave-mechanical background to show how the machinery of the physical world works, Box 25.3 spells out the Hamilton-Jacobi method (\"short-wavelength limit of quantum mechanics\") for determining geodesics, a method considerably more convenient for most applications than the usuai \"second-order differential equations for geodesics\" (equation 10.27). Focus on a specific world line, xlJ. = alJ.(;\\), with all deformations of it gone from view; one may replace alJ.(;\\) by xlJ.(;\\) everywhere. Then the differential equations (13.32) for an extremal world line reduce to (13.36) \u00a7 13.4. GEODESICS AS WORLD LINES OF EXTREMAL PROPER TIME 321 As an aside, note that the identity (13.34) now follows by one differentiation (with respect to 'T) of the equation (13.37) Thus the identity is to be interpreted as saying that 4-velocity and 4-acceleration are orthogonal for any world line, extremal or not. Now return to (13.36), raise an index with gf3 C1, and thereby bring the equation for a straight line of local Lorentz geometry into the form (13.38) Compare with the standard form of the equation for a geodesic in \"premetric geometry,\" (13.39) Conclude that the geodesics of the premetric geometry will agree with the straight. lines of the local Lorentz geometry if and only if two conditions are satisfied: (1) the 40 connection coefficients Ff3p.> that define geodesics, covariant derivatives, and parallel transport must be given in terms of the 10 metric coefficients gp.> (\"Einstein gravitation potentials\") by the equations (13.22) and (13.23) previously derived; and (2) the geodesic parameter ;\\ must agree with the proper time 'T up to an arbitrary normalization of zero point and an arbitrary but constant scale factor; thus ;\\ = a'T + b. (Nothing in the formalism has any resemblance whatsoever to the universal time t of Newton \"flowing everywhere uniformly\"; rather, there is a separate proper time 'T for each geodesic). See Box 13.3 for another variational principle, which gives in one step both the extremal world line and the right parametrization on that line. With this step, one has completed the transfer of the ideas of curved-space geometry from a foundation based on geodesics to a foundation based on metric. The resulting geometry always and everywhere anchors itself to the principle of \"local Lorentz character,\" as the geometry of Newton-Cartan never did", " and never could. Exercise 13.5. ONCE TIMELIKE. ALWAYS T1MELIKE Show that a geodesic of spacetime which is timelike at one event is everywhere timelike. Similarly, show that a geodesic initially spacelike is everywhere spacelike, and a geodesic initially null is everywhere nulL [Hint: This is the easiest exercise in the book!] EXERCISES (cominued on page 324) 322 13. RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF All Box 13.2 \"GEODESIC\" VERSUS \"EXTREMAL WORLD LINE\" Once the connection coefficients rap.I' have been expressed in terms of Einstein's gravitational po tentials gp.I' by the equations (13.22) and (13.23), as they are now and hereafter will be in this book (\"Riemannian or metric geometry\"), it is permis sible and appropriate to subsume under the one word \"geodesic\" two previously distinct ideas: (1) a parametrized world line that satisfies the geo desic equation d2x a d\"A.2 + rap.v d\"A. dxp. dx v d\"A. = 0; and (2) a world line that extremizes the proper time (or, if spacelike, a curve that extremizes the proper distance) between two events {f and qa. The one possible source of confusion is this, that (1) presupposes a properly parametrized curve (as was essential, for example, in the Schild's ladder con struction employed for parallel transport in Chap ter 10), whereas (2) cares only about the course of the world line through spacetime, being in different to what parametrization is used or whether any parametrization at all is introduced. This is not to deny the possibility of \"marking in afterward\" along the extremal curve the most nat ural and easily evaluated of all parameters, the proper time itself, whereupon the extremal curve of (2) satisfies the geodesic equation of (1). Am biguity is avoided by insisting on proper para metrization: henceforth the word \"curve\" means a parametrized curve, the word \"geodesic\" means a properly parametrized geodesic. Box 13.3 \"DYNAMIC\" VARIATIONAL PRINCIPLE FOR GEODESICS If the principle of extremal length f 'll [ T = d dxp. dxvJ1 d\"A. - gp.v d\"A. 2 / d\"A. = extremum (1 ) is indifferent to choice of parametrization [\"d\"A.\" canceling out in (1)] and if the geodesic equation finds the proper parametrization a matter of concern, it is appro priate to search for another extremal principle that yields in one package both the right curve and the right parameter. By analogy with elementary mechanics, one expects that an equation of motion [the geodesic equation whose leading term has the form \"x\" can be derived from a Lagrangian with leading term \"~X2\" (\"kinetic energy\"; \"dynamic\" term). The simplest coordinate invariant generalization of ~X2 is \u00a7 13.4. GEODESICS AS WORLD LINES OF EXTREMAL PROPER TIME 323 Thus one is led to try, in place of the \"geometric\" principle of extremal length, a new \"dynamic\" extremal principle: I f!1J. I =\"2 d dxlJ. dx' glJ.' dl\\. dA. dl\\. CT ( dX = d L x CT, dl\\. f!1J ) dl\\. = extremum (2) (replacement of square root in previous variational principle by first power). The condition for an extremum, here as before [equations (13.30) to (13.32)] is annulment of the so-called Euler-Lagrange \"functional derivative\" o =~ = (coefficient of oxCT in) the integrand of OJ ox CT oL CT = ox d oL - dl\\. o(~;); or, written out in full detail, d2x' gCT. dA.2 + 1 (og \"2 og CTIJ. oxlJ. + ax' CT' og) dxlJ. dx' dl\\. o. dA. =, IJ.' - oxCT or, after multiplication by the reciprocal metric, which translates into the geodesic equation (3) (4) (5) (6) Thus, the new \"dynamic\" expression (2) is indeed extremal for geodesic curves and, by contrast with proper", " length, (1), it is extremal when and only when the geodesic is affinely parametrized. [Its \"Euler-Lagrange equations\" (6) remain satisfied only under parameter changes l\\.new = al\\.Old + b, which keep the parameter affine; by contrast, the Euler-Lagrange equations (13.31) and (13.32) for the \"principle of extremal length\" (1) remain satisfied for any change of parameter whatsoever.] 324 13. RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF All Exercise 13.6. SPACELIKE GEODESICS HAVE EXTREMAL LENGTH Show that any spacelike curve linking two events (/ and q) is a geodesic if and only if it extremizes the proper length [Hint: This is almost as easy as exercise 13.5 if one has already proved the analogous theorem for timelike geodesics.] Exercise 13.7. METRIC TENSOR MEASURED BY LIGHT SIGNALS AND FREE PARTICLES [Kuchar] (a) Instead of parametrizing a timelike geodesic by the proper time T, parametrize it by an arbitrary parameter IL, T = F(IL). Write the geodesic equation in the IL-parametrization. (b) Use now the coordinate time t as a parameter. Throw out a cloud of free particles with different \"velocities\" Vi = dx i /dt and observe their \"accelerations\" a i = d2x i /dt 2\u2022 Discuss what combinations of the components of the affine connection r'K). one can measure in this way. (Assume that no standard clocks measuring 'T are available!) (c) Show that one can measure the conformal metric g\"\" i.e., the ratios of the components of the metric tensor g'K to a given component (say, goo) using only the light signals moving along the null geodesics g'K dx' dx K= O. (d) Combine now the results of (b) and (c). Assume that r'K). is generated by the metric tensor by (13.22), (13.23), in the coordinate frame x'. Show that one can determine A everywhere, if one prescribes it at one event (equivalent to fixing the unit of time). \u00a713.5. METRIC-INDUCED PROPERTIES OF RIEMANN Symmetries of Riemann in absence of metric In Newtonian spacetime, in the real, physical spacetime of Einstein-indeed, in any manifold with covariant derivative-the Riemann curvature tensor has these sym metries (exercise 11.6): Ra{3ya =Ra{3[yal (antisymmetry on last two indices) (13.40) Ra[{3yal =0 (vanishing of completely antisymmetric part). (13.41) In addition, it satisfies a differential identity (exercise 11.10): (\"Bianchi identity\") (13.42) New symmetries imposed by metric When metric is brought onto the scene, whether in Einstein spacetime or elsewhere, it impresses on Riemann the additional symmetry (exercise 13.8) (see Chapter 15 for geometric significance). \u00a7 13.5. METRIC-INDUCED PROPERTIES OF RIEMANN. 325 R af3ya = R[af3lya (antisymmetry o~ first tWo indices). This, together with (13.40) and (13.41), forms a complete set of symmetries for Riemann; other symmetries that follow from these (exercise 13.10) are (13.43) (symmetry under pair exchange), (13.44) and (vanishing of completely antisymmetric part). (13.45) These symmetries reduce the number of independent components of Riemann from 4 X 4 X 4 X 4 = 256 to 20 (exercise 13.9). With metric present, one can construct a variety of new curvature tensors from Riemann. Some that will play important roles later are as follows. (1) The double dual of Riemann, 6 *Riemann* (analog of Maxwell The curvature tensor 6 *Faraday), which has components L'af3 = 1- af3p. PR pCl 12 f ya - p.p U 2 fpClya - - _ 1-saf3p.p R pCl 4 pClya p.y (13.46) (exercise 13.11). (2) The Einstein curvature tensor, which is symmetric (exercise 13.11) Einstein tensor (13", ".47) (3) The Ricci curvature tensor, which is symmetric, and the curvature scalar Ricci tensor R =Rf3. f3' - (13.48) Curvature scalar which are related to the Einstein tensor by (exercise 13.12) (13.49) (4) The Weyl conformal tensor (exercise 13.13) Weyl conformal tensor C af3 - R af3 ya - ya - 2~[a Rf31 + 1- ~[a al U [y 3 U [yU al ~f31 R. (13.50) The Bianchi identity (13.42) takes a particularly simple form when rewritten in Bianchi identities terms of the double dual 6: (\"Bianchi identity\") (13.51) (exercise 13.11); and it has the obvious consequence (\"contracted Bianchi identity\"). (13.52) Chapter 15 will be devoted to the deep geometric significance of these Bianchi identities. 326 13. RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF All EXERCISES Exercise 13.8. RIEMANN ANTISYMMETRIC IN FIRST TWO INDICES (a) Derive the antisymmetry condition (13.43). [Hint: Prove by abstract calculations that any vector fields s, u, v, w satisfy 0 = '.il(u, v)(s' w) = s' [:~(u, v)w] + w\u00b7 Uff(U, v)s]. Then from this infer (13.43).] (b) Explain in geometric terms the meaning of this antisymmetry. Exercise 13.9. NUMBER OF INDEPENDENT COMPONENTS OF RIEMANN (a) In the absence of metric, a complete set of symmetry conditions for Riemann is Ra(3yS = Ra{3[ysJ and Ra[{3ysJ = O. Show that in four-dimensional spacetime these reduce the number of independent components from 4 X 4 X 4 X 4 = 256 to 4 X 4 X 6 - 4 X 4 = 96 16 = 80. (b) Show that in a manifold of n dimensions without metric, the number of independent components is n2(n - I)(n - 2) 6 (13.53) (c) In the presence of metric, a complete set of symmetries is R a {3Ys = R[a{3HYSJ' and R a [{3YSJ = O. Show that in four-dimensional spacetime, these reduce the number of inde pendent components to 6 X 6 - 4 X 4 = 36 - 16 = 20. (d) Show that in a manifold of n dimensions with metric, the number of independent components is n(n - 1)]2 _ n 2(n - 2 [ I)(n - 2) _ n 2(n 2 \u00b712 6 I). (13.54) Exercise 13.10. RIEMANN SYMMETRIC IN EXCHANGE OF PAIRS; COMPLETELY ANTISYMMETRIC PART VANISHES From the complete set of symmetries in the presence of a metric, R a {3YS = R[a{3HySJ and R a [{3YSJ = 0, derive: (a) symmetry under pair exchange, R a {3YS = R ysa{3' and (b) vanishing of completely antisymmetric part, R[a{3YSJ = O. Then (c) show that the following form a complete set of symmetries: (13.55) Exercise 13.11. DOUBLE DUAL OF RIEMANN; EINSTEIN (a) Show that 6 = *Riemann* contains precisely the same amount of information as Riemann, and satisfies precisely the same set of symmetries [(13.40), (13.41), (13.43) to (13.45)]. (b) From the symmetries of 6, show that Einstein [defined in (13.47)] is symmetric (c) Show that the Bianchi identities (13.42), when written in terms of 6, take the form (G[{3SJ = 0). (13.51) (\"vanishing divergence,\" V \u2022 6 = 0). (d) By contracting the Bianchi identities V\u00b7 6 = 0, show that G =Einstein has vanishing divergence [equation (13.52)]. Exercise 13.12. RICCI AND EINSTEIN RELATED (a) From the symmetries of Riemann, show that Ricci is symmetric (R[{3sJ = 0). (b) Show that Ricci is related to Einstein by equation (13.49).", " \u00a7 13.6. PROPER REFERENCE FRAME OF AN ACCELERATED OBSERVER 327 Exercise 13.13. THE WEYL CONFORMAL TENSOR (a) Show that the Weyl conformal tensor (13.50) possesses the same symmetries [(13.40), (13.41), (13.43) to (13.45)] as the Riemann tensor. (b) Show that the Weyl tensor is completely \"trace-free\"; i.e., that contraction of Caf3Y~ on any pair of slots vanishes. (13.56) Thus, Caf3Y~ can be regarded as the trace-free part of Riemann, and R af3 can be regarded as the trace of Riemann. Riemann is determined entirely by its trace-free part Caf3Y~ and its trace R af3 [see equation (13.50), and recall R = Raa]' (c) Show that in spacetime the Weyl tensor has 10 independent components. (d) Show that in an n-dimensional manifold the number of independent components of Weyl [defined by a modification of (13.50) that maintains (13.56)] is n 2(n 2 12 I) - n(n + I) fi 2 or n - > 3 ofor n :::; 3., (13.57) Thus, in manifolds of I, 2, or 3 dimensions, the Weyl tensor is identically zero, and the Ricci tensor completely determines the Riemann tensor. \u00a713.6. THE PROPER REFERENCE FRAME OF AN ACCELERATED OBSERVER A physicist performing an experiment in a jet airplane (e.g., an infrared astronomy experiment) may use several different coordinate systems at once. But a coordinate system of special utility is one at rest relative to all the apparatus bolted into the floor and walls of the airplane cabin. This \"proper reference frame\" has a rectangular \"i, y, j\" grid attached to the walls of the cabin, and one or more clocks at rest in the grid. That this proper reference frame is accelerated relative to the local Lorentz frames, the physicist knows from his own failure to float freely in the cabin, or, with greater precision, from accelerometer measurements. That his proper refer ence frame is rotating relative to local Lorentz frames he knows from the Coriolis forces he feels, or, with greater precision, from the rotation of inertial-guidance gyroscopes relative to the cabin walls. Exercise 6.8 gave a mathematical treatment of such an accelerated, rotating, but locally orthonormal reference frame in flat spacetime. This section does the same in curved spacetime. In the immediate vicinity of the spatial grid's origin xi = 0 (region of spatial extent so small that curvature effects are negligible), no aspect of the coordinate system can possibly reveal whether spacetime is curved or flat. Hence, all the details of exercise 6.8 must remain valid in curved spacetime. Never theless, it is instructive to rediscuss those details, and some new ones, using the powerful mathematics of the last few chapters. Begin by making more precise the coordinate grid to be used. The following is perhaps the most natural way to set up the grid. (1) Let T be proper time as measured by the accelerated observer's clock (clock at center of airplane cabin in above example). Let '3' ::: '3'O(T) be the observer's world line, as shown in Figure 13.4,a. Proper reference frame described physically Six-step construction of coordinate grid for proper frame 328 13. RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF All, / s=2./ / I I II s = 1 __~~~-=--\"\"\"- / /.... -_ /'i/ I -..-----.s=1'\\ '-,, '. s=2 \" '\\ /.s=3 \\ \\ / I.s=4 I \\ I ~ t \\ / s=3 Geodesic - _ - -....................... \" \" \" \" \"- '\" Geodesic 9' [4, ~ (el + e~), SJ (a) (b) Figure 1 3~4. The proper reference frame of an accelerated observer. Diagram (a) shows the observer's orthonormal tetrad {eu} being transported along his world line POC'T) [transport law (13.60)]. Diagram (b) shows geodesics bristling out perpendicularly from an arbitrary event ~o(4) on the observer's world line. Each geodesic is specified uniquely by (I) the proper time 'T at which it originates, and (2) the direction (unit tangent vector n = d", "(ds = n;e; along which it emanates). A given event on the geodesic is specified by 'T, n, and proper distance s from the geodesic's emanation point; hence the notation for the given event. The observer's proper reference frame attributes to this given event the coordinates P =.'i'['T, n, s] XO(.'i'['T, n, s]) = 'T, x;(.'i'['T, n, s]) = sn;. (2) The observer carries with himself an orthonormal tetrad {eoJ (Figure 13.4,a), with eo = u = dtJ'o/d-r = (4-velocity of observer) (13.58) (eo points along observer's \"time direction\"), and with ea' efj = Y/a/3 (13.59) (orthonormality). \u00a7 13.6. PROPER REFERENCE FRAME OF AN ACCELERATED OBSERVER 329 (3) The tetrad changes from point to point along the observer's world line, relative to parallel transport: Transport law for observer's tetrad Vue a = -0' e a, []IJ.' = alJ.u' - ulJ.a' + uaW/3f. a /3IJ.' = \"generator of infinitesimal Lorentz transformation.\" (13.60) (13.61) This transport law has the same form in curved spacetime as in flat (\u00a76.5 and exercise 6.8) because curvature can only be felt over finite distances, not over the infinitesimal distance involved in the \"first time-rate of change of a vector\" (equivalence principle). As in exercise 6.8, a = Vuu = (4-acceleration of observer), angular velocity of rotation of spatial ) w = basis vectors ej relative to Fermi-, ( Walker-transported vectors, i.e., relative to inertial-guidance gyroscopes (13.62) u'a = u'w = o. If w were zero, the observer would be Fermi-Walker-transporting his tetrad (gyroscope-type transport). Ifboth a and w were zero, he would be freely falling (geodesic motion) and would be parallel-transporting his tetrad, Vue a = O. (4) The observer constructs his proper reference frame (local coordinate system) in a manner analogous to the Riemann-nomal construction of \u00a711.6. From each event 9 0(T) on his world line, he sends out purely spatial geodesics (geodesics orthogonal to u = d90 /dT), with affine parameter equal to proper length. proper time; tells \"starting point\" of geodesic [ 9 = 9'[T, n, s]. t---, I tangent vector to geodesic at starting point; tells \"which\",geodesic (13.63) proper length along ~ geodesic from starting point; tells \"where\". on geodesic (See Figure 13.4,b.) The tangent vector has unit length, because the chosen affine parameter is proper length: n = (o\u00a7 /os).=o; nlJ. = (dxlJ./ds) along geodesic, n' n = glJ.' ( d;) ( d:s') = ~~ = 1. (13.64) (5) Each event near the observer's world line is intersected by precisely one of the geodesics \u00a7[T, n, s]. [Far away, this is not true; the geodesics may cross, either because of the observer's acceleration, as in Figure 6.3, or because of the curva ture of spacetime (\"geodesic deviation\").] y 330 13. RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF All (6) Pick an event,:P near the observer's world line. The geodesic through it originated on the observer's world line at a specific time T, had original direction n = njej, and needed to extend a distance s before reaching ':P. Hence, the four numbers (13.65) are a natural way of identifying the event 9. These are the coordinates of,:P in the observer's proper reference frame. (7) Restated more abstractly, XO(9'[T, n, s]) = T, Xj(9'[T, n, s]) = Sl1 j = snj = sn' ej' (13.65') In flat spacetime this construction process and the resulting coordinates x\"(~\") are identical to the", " process and resulting coordinates ~a'(9) of exercise 6.8. For use in calculations one wants not only the coordinate system, but also its metric coefficients and connection coefficients. Fortunately, g\"/3 and F\"/3y are needed only along the observer's world line, where they are especially simple. Only a foolish observer would try to use his own proper reference frame far from his world line, where its grid ceases to be orthonormal and its geodesic grid lines may even cross! (See \u00a76.3.) All along the observer's world line 9 0(T), the basis vectors of his coordinate grid are identical (by construction) to his orthonormal tetrad and therefore its metric coefficients are (13.66) (13.67) Connection coefficients along observer's world line Some of the connection coefficients are determined by the transport law (13.60) for the observer's orthonormal tetrad: V e' = Vre' = eIJF/3\u00b7\u00b7 aO a a u p 0 = -0' e\", = -e /3 []/3\",. Thus (13.68) Since 0 has the form (13.61) and the observer's 4-velocity and 4-acceleration have components Uo= -1, Uj = 0, 00 =\u00b0in the observer's own proper frame, these connection coefficients are (13.69a) \u00a7 13.6. PROPER REFERENCE FRAME OF AN ACCELERATED OBSERVER 331 The remaining connection coefficients can be read from the geodesic equation for the geodesics.9'[T, n, s] that emanate from the observer's world line. According to equation (13.65), the coordinate representation of each such geodesic is XO(s) = T = constant, hence, d 2x\"'jds 2 = 0 all along the geodesic, and the geodesic equation reads. d 2x'\" 0= - - + rafj. py ds ds 2., \u2022 = ra,\u00b7n'n\", k \u00b7 dx 4 dxY ds This equation is satisfied on the observer's world line for all spatial geodesics (all n;) if and only if r\"';\" = r\",;\" = 0 all along 9 0(T). (13.69b) The values (13.69) of the connection coefficients determine uniquely the partial derivatives of the metric coefficients [see equation (13.19')]: ga4,O = 0, goo,; = -2a;, (13.70) and these, plus the orthonormality condition g\"'4[tJ'0(T)] = 'YIa/3' imply that the line element near the observer's world line is ds 2 = - (1 + 2a;x;) dx02 - 2(f;/dX \"wI) dx\u00b0 dx; + O;k dx; dx\" + O(lx JI2) dx'\" dx4. (13.71) Metric of proper reference frame, and its physical interpretation Several features of this line element deserve notice, as follows. (1) On the observer's world line 9 0(T)-i.e., x; = 0-ds 2 = 'YIa/3 dx'\" dx 4. (2) The observer's acceleration shows up in a correction term to goo, ogoo = -2a-x, (13.72a) which is proportional to distance along the acceleration direction. For the flat-space time derivation of this correction term, see \u00a76.6. (3) The observer's rotation relative to inertial-guidance gyroscopes shows up in a correction term to go;, which can be rewritten in 3-vector notation (13.72b) (4) These first-order corrections to the line element are unaffected by spacetime curvature and contain no information about curvature. Only at second order, O(lxJI2), will curvature begin to show up. (5) In the special case of zero acceleration and zero rotation (a = w = 0), the observer's proper reference frame reduces to a local Lorentz frame (gaS = 'YIa/3' r a4r = 0) all along his geodesic world line! By contrast, the local Lorentz coordinate 332 13. RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF ALL systems constructed earlier in the book (\"general\" local Lorentz coordinates of\u00a78.6, \"Riemann normal coordinates\" of 111.6) are local Lorentz only at a single event. In the case of zero rotation and zero acceleration, one can derive the following expression for the metric, accurate to seoood order in Ixil: d", "s 2 = (- 1 - Ro~x!x.) dJZ - (: RoII-'x1x.) dt dx' + (at; - ~ RW.x1x.) dxedxl + O(lxJI3) dxd dx P (13.73) [see, e.g., Manasse and Misner (1963\u00bb), Here RaMl are the components of the Riemann tensor along the world line xl = O. Such coordinates are called \"Fermi Normal Coordinates.\" EXERCISES INERTIAL AND CO.UOLII fORCEI Ex.rci\u2022\u2022 13.14. An accelerated observer studies the path of a freely falling particle as it passes through the origin of his proper reference frame. If is the particle's ordinary velocity, show that its ordinary acceleration relative to the observer's proper reference frame is (13.74) d2 1 -:!... e; =-a - 2Iol X II + 2(a' II)'. [inertial accel~:ion~ i [Coriolis acceleration] L relativistic correction 1 L~o inertial accelerationJ (13.75).' Here a is the observer's own 4-acceleration, and ~ is the angular velocity with which his spatial basis vectors el are rotating [see equations (13.62)]. [Hint: Use the geodesic equation at the point xI = 0 of the particle's trajectory. Note; This result was derived in fiat spacetime in exercise 6.8 using a different method.] Exerci\u2022\u2022 13.15. ROTATION GROUP: METRIC (Continuation of exercises 9.13, 9.14, 10.17 and 11.l2). Show that for the manifold SO(3) of the rotation group, there exists a metric. that is compatible with the covariant derivative T. Prove existence by exhibiting the metric components explicitly in the noncoordinate basis of generators {ea }. [Answer; (13.76) Restated in words: If one postulates that: (I) the manifold of the rotation group is locally Euclidean; (2) the generators of infinitesimal rotations {eo} are orthonormal, eo' elJ = 8olJ ; and (3) {eo} obey the standard rotation-group commutation relations (13.77) then the resulting geodesics of S0(3) agree with the geodesics chosen in exercise 10.17.J --------------------------------....... ~ CHAPTER 14 CALCULATION OF CURVATURE \u00a714.1. CURVATURE AS A TOOL FOR UNDERSTANDING PHYSICS Elementary physics sometimes allows one to shortcircuit any systematized calculation of curvature (frequency of oscillation of test particle; tide-producing acceleration near a center of attraction; curvature of a closed 3-sphere model universe; effect of parallel transport on gyroscope or vector; see Figures 1.1, 1.10, and 1.12, and Boxes 1.6 and 1.7); but on other occasions a calculation of curvature is the quickest way into the physics. This chapter is designed for such occasions. It describes three ways to calculate curvature and gives the components of the Einstein curvature tensor for a plane gravitational wave (Box 14.4, equation 5), for the Friedmann geometry.. of the unIverse (Box 14.5), and for Schwarzschild geometry, both static (exercise 14.13) and dynamic (exercise 14.16). These and other calculations of curvature elsewhere are indexed under \"curvature tensors.\" It is enough to look at an expression for a 4-geometry as complicated as This chapter is entirely Track 2. Chapter 4 (differential forms) and Chapter 10, 11, and 13 (differential geometry) are necessary preparation for \u00a7\u00a714.5-14.6. This chapter is needed as preparation for Chapter 15 (Bianchi identities). It will be helpful in many applications of gravitation theory (Chapters 23-40). '\", ----\"-- - (14.1) Situations in which one must compute curvature \"Standard procedure\" for computing curvature Methods of displaying curvature formulas Computation of curvature using a computer EXERCISES 334 14. CALCULATION OF CURVATURE [Harrison (1959)] to realize that one might understand the physical situation better if one knew what the curvature is; similarly with any other complicated expressions for metrics that arise from solving Einstein's equations or that appear undigested in the literature. In any such case, the appropriate method often is: curvature first, understanding second. Curvature is the simplest local measure of geometric properties (see Box 14.1). Curvature is therefore a good first step toward a more comprehensive picture of the", " spacetime in question. One sometimes has an expression for a spacetime metric first, and then makes calculations of curvature to understand it. But more often one makes calculations of curvature, subject to specified conditions of symmetry in space and time, as an aid in arriving at an expression for a physically interesting metric (stars, Chapters 23 to 26; model cosmologies, Chapters 27 to 30; collapse and black holes, Chapters 31 to 34; and gravitational waves, Chapters 35 to 37). The basic \"standard procedure for computing curvature\" is illustrated in Box 14.2. Two formulas in Box 14.2, derived previously, are used in succession. The first (equations 1 and 2) has the form r -.. gog and provides the rlJ.a{3' The other (equation 3) has the form R -.. or + r 2 and gives the curvature components RlJ.va{3' After the curvature components have been computed, there are helpful ways to present the results. (1) Form the Ricci tensor RlJ.v = RalJ.av and the scalar curvature R = RIJ.IJ.' (2) Form other invariants such as RIJ.Va{3 Ra{3 IJ.V' (3) Form components R~j,ajj in a judiciously chosen orthonormal frame w a = La{3 dx{3, and (4) display R[~j,l[ajjl as a 6 X 6 matrix (in four dimensions; a 3 X 3 matrix in three dimensions) where [J!v] = [Of], [02], [03], [23], [31], [12] labels the rows and [a,8] labels the Columns (exercises 14.14 and 14.15). (5) Last, but by far the most important for general relativity, form the Einstein tensor G~j, as described in \u00a714.2. The method of computation outlined above and described in more detail in Box 14.2 is used wherever it is quicker to employ a standard method than to learn or invent a better method. The standard method is always preferable for the student in a short course where physical insight has higher priority than technical facility. It is, however, a dull method, better suited to computers than to people. E\"~the algebra can be handled by a computer (see Box 14.3). Exercise 14.1. CURVATURE OF A TWO-DIMENSIONAL HYPERBOLOID Compute the curvature of the hyperboloid 12 - x 2 - y2 == jZ == const in 2 + 1 Minkowski 2 == _d12 + dx2 + dy 2. First show that intervals within this two-dimen spacetime with dS3 sional surface can be expressed in.the form ds 2 == P(da2 + sinh2a dcp2) by a suitable choice of coordinates a, cp, on the hyperboloid. -~- Exercise 14.2. RIEMANNIAN CURVATURE EXPRESSIBLE IN TERMS OF RICCI CURVATURE IN TWO AND THREE DIMENSIONS In two dimensions, there is only one independent curvature component, R 1212. Evidently the single scalar quantity R must carry the same information. The two-dimensional identity Rp.vap == ~R(gp.agvp - gp.pgva) is established by noting that it is the only tensor formula giving (continued on page 343) \u00a7 14.1. CURVATURE AS A TOOL FOR UNDERSTANDING PHYSICS 335 Box 14.1 PERSPECTIVES ON CURVATURE 1. Historical point of departure: a curved line on a plane. There is no way to define the curvature of a line by measurements confined to (\"intrinsic to\") the line itself. One needs, for example, the azimuthal bearing 0 of the tangent vector relative to a fixed direction in the plane, as a function of proper distance s measured along the curve; thus, o= O(s). Then curvature K and its reciprocal, the radius of curvature p, are given by K(S) = l/p(s) = dO(s)/ds. Alternatively, one can examine departure, y, measured normally off from the tan gent line as a function of distance x measured along that tangent line; then K = 1/p = dZy/dxz. \u2022\\ \\ \\ p \\ \\ Reference \u2022 \u00bb\u00bb azimuth 2. This concept was later extended to a curved surface embedded in flat (Euclidean) 3-space. Departure, z, of the smooth curved surface from the flat surface tangent to it at a given point is described in the neighborhood of that point by the quadr", " Such calculations are sufficiently straightforward and deductive in logical structure that they can be handled by a computer. Since 1966, computers have been generally taking over such tasks. There are several computer languages in which the investigator can program his analytic calcula tions. The computer expert may wish to work in a machine-oriented language such as LISP [see, e.g., the work of Fletcher (1966) and of Hearn (1970)]. However, most appliers of relativity will prefer user-oriented languages such as REDUCE [created by Hearn (1970) and available for the IBM 360 and 370, and the PDP 10 computers], ALAM [created by D'Inverno (1969) and available on Atlas computers], CAMAL [created by Barton, Bourne, and Fitch (1970) and available on Atlas computers], and FORMAC [created by Tobey et al. (1967) and available on IBM 7090, 7094, 360, and 370]. For' a review of activity in this area, see Barton and Fitch (1971). Here we describe only FORMAC. It is the most widely available and widely used of the languages; but it is probably not the most powerful [see, e.g., D'Inverno (1969)]. FORMAC is to analytic work what the earliest and most primitive versions of FORTRAN were to numerical work. FORMAC manipulates algebraic expressions involving: numerical constants, such as Ij3; sym bolic constants, such as x or u; specific elementary functions, such as sin (u) or exp (x); and symbolic functions of several variables, such as f(x, u) or g(u). For example, it can add ax + bx2 to 2x + (3 + b)x2 and get (a + 2)x + (3 + 2b)x2 ; it can take the partial derivative of x 2uf(x, u) + cos (x) with respect to x and get 2xuf(x, u) + x 2u af(x, u)jax - sin (x). It can do any algebraic or differential-calculus computation that a human can do-but without making mistakes! Unfortunately, it cannot inte grate analytically; integration requires inductive logic rather than deductive logic. PLjl is a language that can be used simulta neously with FORMAC or independently of it. PLjl manipulates strings of characters-e.g., \"Zjl X29- + j.\" It knows symbolic logic; it can tell whether two strings are identical; it can insert new characters into a string or remove old ones; but it does not know the rules of algebra or differ ential calculus. Thus, its primary use is as an ad junct to FORMAC (though from the viewpoint of the computer system FORMAC is an adjunct of PLjl). FORMAC programs for evaluating Einstein's tensor in terms of given metric components and for doing other calculations are available from many past users [see, e.g., Fletcher, Clemens, Matzner, Thome, and Zimmerman (1967); Ernst (1968); Harrison (1970)]. However, programming in FORMAC is sufficiently simple that one ordi narily does not have difficulty creating one's own program to do a given task. If a difficulty does arise, it may be because the analytic computation exhausts the core of the computer. It is easy to create an expression too large to fit in the core of any existing computer by several differentiations of an expression half a page long! Users of FORMAC, confronted by core exhaustion, have devised several ways to solve their problems. One is to remove unneeded parts of the program and of the FORMAC system from the core. Routines called PURGE and KILL have been developed for this purpose by Clemens and Matzner (1967). Another is to create the answer to a given calculation in manageable-sized pieces and output those pieces from the computer's core onto its disk. One must then add all the pieces together-a using FORMAC alone, or even FORMAC plus PLjl, but a task that James Hartle has solved [see Hartle and Thome (1974)] by using a combination of FORMAC, PLjl, and IBM data-manipulation routines called SORT. impossible task that is \u00a714.2. FORMING THE EINSTEIN TENSOR 343 R/l.a/J as a linear function of R, constructed from R and the metric alone, and with the correct contracted value R/l'/l' == R. Establish a corresponding three-dimensional identity expressing Ri;kI in terms of the Ricci tensor R;k and the metric. Exercise 14.3. CURVATURE OF 3-SPHERE IN ORTHONORMAL FRAME Compute the curvature tensor for a 3-sphere or for a 3-hyper", "boloid (141) (14.3) Convert th: coordinate-based components Ri;kI to a corresponding orthonormal basis, RijU' Display RiiU == RCiiICfdl as a 3 X 3 matrix with appropriately labeled rows and columns. \u00a714.2. FORMING THE EINSTEIN TENSOR The distribution of matter in space does not immediately tell all details of the local curvature of space, according to Einstein. The stress-energy tensor provides infor mation only about a certain combination of components of the Riemann curvature tensor, the combination that makes up the Einstein tensor. Chapter 13 described two equivalent ways to calculate the Einstein tensor: (I) by successive contractions of the Riemann tensor Rp.. == Ro.p.o..' (14.4) [equations (13.48) and (13.49)]; (2) by forming the dual of the Riemann tensor and then contracting: G a - fi - {3 0.{3 o.a (14.5a) (l4.5b) [equations (13.46) and (13.47)]. A third method, usually superior to either of these, is discovered by combining equations (14.5a,b): (14.6) [Note: in any frame, orthonormal or not, the permutation tensor lJapu{3p.. has compo nents + I if lJpa is an even permutation of fJp.v, I if lJpa is an odd permutation of fJp.v, o otherwise; ( Three ways to compute the Einstein tensor from the Riemann tensor y 344 14. CALCUlATION OF CURVATURE to see this, simply evaluate S&pa{3J1.P using definition (3.S0h) and using the components (8.10) of (a{3J1.' and (paya.] Equation (14.6) for the Einstein tensor, written out explic itly, reads GOo = _(R12 23 + R3131 ), 12 + R23 G\\ = _(R02OZ + R0303 + R2323 ), G01 = ROZ 2 = R10 12 + R03 20 + R13 G1 13 23,, (14.7) Standard method of computing curvature is wasteful Ways to avoid \"waste\": (1) geodesic Lagrangian method (2) method of curvature 2-forms and every other component is given by a similar formula, obtainable by obvious permutations of indices. \u00a714.3. MORE EFFICIENT COMPUTATION If the answer to a problem or the result of a computation is not simple, then there is no simple way to obtain it. But when a long computation gives a short answer, then one looks for a better method. Many of the best-known applications of general relativity present one with metric forms in which many of the components gJl.P' F JI.a{3' and RJI. pa{3 are zero; for them the standard computation of the curvature (Box 14.2) involves much \"wasted\" effort. One computes many F JI.a{3 that tum out to be zero. One checks off many terms in a sum like -FJl. p{3FPa Jl. that are zero, or cancel with others to give zero. Two alternative procedures are available to eliminate some of this \"waste.\" The \"geodesic Lagrangian\" method provides an economical way to tabulate the r JI.a{3' The method of \"curvature 2-forms\" reorganizes the description from beginning to end, and computes both the connection and the curvature. The geodesic Lagrangian method is only a moderate improvement over the standard method, but it also demands only a modest investment in the calculus of variations, an investment that pays off in any case in other contexts in the world of mathematics and physics. In contrast, the method of curvature 2-forms is efficient, but demands a heavier investment in the mathematics of I-forms and 2-forms than anyone would normally find needful for any introductory survey ofrelativity. Anyone facing several days' work at computing curvatures, however, would do well to learn the algorithm of the curvature 2-forms. \u00a714.4. THE GEODESIC LAGRANGIAN METHOD One normally thinlcs that the connection coefficients FJI.a{3 must be known before one can write the geodesic equation (14.8) (Here and below dots denote derivative with respect to the affine parameter, A.) However, the argument can be reversed. Once the geodesic equations have been \u00a7 14.4. GE", "2e2/3i)\"13x dA. -----. The requirement that 51 = 0 for this variation (among others) gives Varying y, U, D, in-the same way gives 0= (Ue- 2/3j)\" = Ue- 2/3j + juac~, au 0= ii. Step 3: Rearrange to get XJl. leading terms. If this step is not straightforward, this method will not save time, and the technique of either Box 142 or Box 14.5 will be more suitable. In the example here, one quickly writes, using a prime for a/au, 0= x + 2(L-IL' + {J')iu, 0= j + 2(L-IL' - {J')ju, o = v + (L 2e2/3)(L-IL' + {J')i2 + (LZe-2/3)(L-IL' - {J')yz, o = ii. (3x) (3y) (3 v) (3u) t \u00a714.4. GEODESIC LAGRANGIAN METHOD 347 Step 3': Interpret these equations as a tabulation of rJl.a{3' Equations (3) are the standard equations for a geodesic, llu + r Z llu must appear in the coefficient (r Z Therefore it is enough to scan them to find the value of any desired r. For instance r z llu of the jli term in the equation for x. But no jli term appears in equation (3x). Therefore r z llu is zero in this example. Note that equations (3) are simple, in the sense that they contain few terms; therefore most of the r JI.a{3 must be zero. For instance, it follows from equation (3u) that all ten r U uz = (L-IL' + [1') from equation (3x), r ll [1') from equation (3y), and r vzz and r v a{3 are zero. The only non-zero r's are r z 1111 from equation (3u). ull = (L-IL' - ulI ) = 2r z llu = r ll zu = r z Step 4: Compute each RJI. va{3' etc. There is little relief from routine in systematically applying equation (3) from Box 14.2. One must list 21 components RJl. va{3 that are not related by any of the symmetries RJl.va{3 = R a{3J1.v = - RJl.v{3a, and compute each. In the example here, one notes that r ua{3 = 0 implies RUa{3y = - R Va{3y = O. There fore 15 of the list of 21 vanish at one swat. The list then is: R Va{3y = - RUa{3y = 0, R uzuz = -Rvzuz = - ( rV zz )' + rvzzrzzu _= -(Ue2{3) (~' + [1\" + 2 f [1' + [1'2), R uzzII = -RvZ1:11 = 0, R uzllu = -Rvzllu = 0, R ullull = -RvIIUY = _(r V IlIl = -(Ue-2{3) (~' - R ullzlI = -RvllzlI = 0, R zllzlI = (L 2e2{3)RZllzlI = O. )' + rVllyrYllu [1\" - 2 f [1' + [1'2), (4) One can now calculate the Einstein tensor via equation (14.7). In the example here, however, it is equally simple to form first the Ricci tensor by the straightforward contraction RJI.aJl.{3\" Only p. = x and p. =y give any contribution, because no super script index can be a u, and no subscript a u. Thus one finds and Ruu = -2[L-1L\" + [1'2], all other Ra {3 = 0, R =0. (5) (6) From this last result, it follows that here the desired Einstein tensor is identical with the Ricci tensor. 348 14. CALCULATION OF CURVATURE (a) Set up the variational integral (14.10) for a geodesic in this metric. then successively vary t. x. 8. and </> to obtain, after some rearrangement. four equations 0 = i\" + \"., 0 = X", "+ \"', etc. displaying the T's in the form of equation (14.8). (b) Use this display as a table of r's to compute R\\I'P and RX9I'p, of which only R\\tx and RX9X9 are non\u00b7zero (consequence of the complete equivalence of all directions tangent to the X8</> sphere). (c) Convert to an orthonormal frame with wi = dt. W X = a dX. w 9 =? w\u00a2 =?, and list X9' Explain why all other components are known by symmetry in terms of Rixix and RX9 these two. (d) Calculate, using equations (14.7), all independent components of the Einstein ten sor GPj,. [Answer: See Box 14.5.] \u00a714.5. CURVATURE 2-FORMS In electrodynamics the abstract notation saves space compared to the explicit notation F=dA...,etc. (six equations); Concepts needed for method of curvature 2-forms there is no reason to shun similar economies in dealing with the dynamics of geometry. Cartan introduced the decisive ideas, seen above, of differential forms (where a simple object replaces a listing of four components; thus, q = (JIJ. dxIJ.), and of the exterior derivative d. He went on (1928, 1946) to package the 21 compo nents Rp.va{3 of the curvature tensor into six curvature 2-forms, Regarded purely as notation, these 2-forms automatically produce a profit. They cut down the weight of paper work required to list one's answer after one has it. They also provide a route into deeper insight on \"curvature as a geometric object,\" although that is not the objective of immediate concern in this chapter. Cartan's exterior derivative d automatically effects many cancelations in the calculation of curvature. It often cancels terms before they ever need to be evaluated. Extension of Cartan's calculus from electromagnetism and other applications (Chapter 4) to the analysis of curvature (this chapter) requires two minor additions to the armament of forms and exterior derivative: (1) the idea of a vector-valued (or tensor-valued) exterior differential form; and (2) a corresponding generalization \u00a714.5. CURVATURE 2-FORMS 349 of the exterior derivative d. This section uses both these tools in deriving the key formulas (14.18), (14.25), (14.31), and (14.32). Once derived, however, these formulas demand no more than the standard exterior derivative for all applications and for all calculations of curvature (\u00a714.6 and Box 14.5). The extended exterior derivative leads to nothing new in the first two contexts Extended exterior derivative: to which one applies it: a scalar function (''O-form'') and a vector field (\"vector-valued O-form''). Thus, take any function f Its derivative in an unspecified direction is a I-form; or, to make a new distinction that will soon become meaningful, a \"scalarvalued I-form.\" Specify the direction in which differentiation is to occur (\"fill in the slot in the I-form\"). Thereby obtain the ordinary derivative as it applies to a function (14.11) (1) acting on a scalar Next, take any vector field v. Its covariant derivative in an unspecified direction is a \"vector-valued I-form.\" Specify the direction u in which differentiation is to occur (\"fill in the slot in the I-form\"). Thereby obtain the covariant derivative (2) acting on a vector ( 14.l2a) This object too is not new; it is the covariant derivative of the vector v taken in the direction of the vector u. When one abstracts away from any special choice of the direction of differentiation u, one finds an expression that one has encountered before, though not under its new name of \"vector-valued I-form.\" This expression measures the covariant derivative of the vector v in an unspecified direction (\"slot for direction not yet filled in\"). From a look at (14.l2a), one sees that this extended exterior derivative is applied to v, without reference to u, is dv = Tv. (14.l2b) Similarly, for any \"tensor-valued O-form\" [i.e. (~) tensor] S, dS = TS. Before proceeding further with the exterior (soon to be marked as \"antisymmetric\") differentiation of tensors, write down a formula (see exercise 14.5) for the exterior (antisymmetric) derivative of a product of forms: d(a /\\ P) = (Ila) /\\ P + (-l)Pa /\\ dp, (14.l3a)", " where a is a p-form and P is a q-form. Now extend the exterior derivative from elementary forms to the exterior product (3) defined in general of a tensor-valued p-form S with any ordinary q-form,!J; thus, deS /\\ P) = dS /\\ P + (-I)PS /\\ d!J. (14.l3b) This equation can be regarded as a general definition of the extended exterior derivative. For example, ifS is a-tensor-valued 2-form, S = SIlPlY31BIlB P dx Y /\\ dx\", then equation (14.13b) says dS == d(BIlBpSIlPIY3IXdxY /\\ dx 3\u00bb) = d(BIlBpSIlPjyal) /\\ (dxY /\\ dx\"). y 350 14. CALCULATION OF CURVATURE As another example, use (14.13b) to calculate d(uCT), where u is a vector-valued O-form (vector) and CT is a scalar-valued I-form (I-form): d(uCT) = (du) 1\\ CT + u dCT. If one were following the practice of earlier chapters, one would have written u \u00ae CT where UCT appears here, u \u00ae dCT instead of u dCT, and e a \u00ae e p instead of eae p' However, to avoid overcomplication in the notation, all such tensor product symbols are omitted here and hereafter. Equations (14.12) and (14.13) do more than define the (extended) exterior deriva tive d and provide a way to use it in computations. They also allow one to define and calculate the antisymmetrized second derivatives, e.g., d 2v. The relation where v is a vector will then introduce the \"operator-valued\" or \"(D-tensor valued\" curvature 2-form qil. The notation of the extended exterior derivative puts a new look on the old apparatus of base vectors and parallel transport, and opens a way to calculate the curvature 2-form qil. Let the vector field v be expanded in terms of some field of basis vectors e 1'; thus Then the exterior derivative of this vector is Definition of wJJ. p (14.14) Expand the typical vector-valued I-form de I' in the form Here the \"components\" w\u00b7I' in the expansion of de I' are I-forms. Recall from equation (10.13) that the typical w\u00b7I' is related to the connection coefficients by Therefore the expansion of the \"vector\" (really, \"vector-valued I-form\") is dv = e idul' + wI'.u\u00b7). Now differentiate once again to find (14.15) (14.16) d 2v = dea 1\\ (du a + w a.u\u00b7) + e/d 2 ul' + dwl'.u\u00b7 - wI'. 1\\ du\u00b7) = el'(Wl'a 1\\ dua + wl'a 1\\ wa.u\u00b7 + d 2ul' + dwl'.u\u00b7 - wI'a 1\\ dua). The simplifications made here use (l) the equation (14.14), for a second time; and (2) the product rule (14.13a), which introduced the minus sign in the last term, ready \u00a7 14.5. CURVATURE 2\u00b7FORMS 351 to cancel the first term. Now consider the term d 2vJl.. Recall that any given compo nent, for example, v3, is an ordinary scalar function of position (as contrasted to vor 83 or 8 3v3). Therefore the standard exterior derivative (Chapter 4) as applied to a scalar function is all that d can mean in d 2vJl.. But for the standard exterior derivative applied twice, one has automatically d 2vJl. = 0 (Box 4.1, B; Box 4.4). This circumstance reduces the expansion for d 2v to the form Curvature 2-forms ~JJ..: where the qilJl. v are abbreviations for the curvature 2-forms (14.17) (1) in terms of d 2v (14.18) (2) in terms of wJJ.. Ordinarily, equation (14.18) surpasses in efficiency every other known method for calculating the curvature 2-forms. The remarkable form of equation (14.17) deserves comment. On the left appear two d's, reminders that one has twice differentiated the vector field v. But on the right, as the result of the differentiation, one has only the vector field vat the point in question,", " undifferentiated. How v varies from place to place enters not one whit in the answer. All that matters is how the geometry varies from place to place. Here is curvature coming into evidence. It comes into evidence free of any special features of the vector field v, because the operation d 2 is an antisymmetrized covariant derivative [compare equation (11.8) for this antisymmetrized covariant derivative in the previously developed abstract language, and see Boxes 11.2 and 11.6 for what is going on behind the scene expressed in the form of pictures]. In brief, the result of operating on v twice with d is an algebraic linear operation on v; thus, Here qil is an abbreviation for the \"CD-tensor valued 2-form,\" (14.19) Tensor-valued curvature 2-form ~ (14.20) If d is a derivative with a \"slot in it\" in which to insert the vector saying in what direction the differentiation is to proceed, then the d 2 w of d 2 w = qilw has two slots and calls for two vectors, say, u and v. These two vectors define the plane in which the antisymmetrized exterior derivative of (14.19) is to be evaluated (change in w upon going around the elementary route defined by u and v and coming back to its starting point; Boxes 11.6 and 11.7). To spell out explicitly this insertion of vectors into slots, return first to a simpler context, and see the exterior derivative of a I-form (itself a 2-form) \"evaluated\" for a bivector u 1\\ v (\"count of honeycomblike cells of the 2-form over the parallelogram-shaped domain defined by the two vectors u and v\"), and see the result of the evaluation (exercise 14.6) expressed as a com mutator, (da, u 1\\ v) = 0u(a, v) - 0v(a, u) - (a, [u, v]). (14.21) ) / / 352 14. CALCULATION OF CURVATURE This result generalizes itself to a tensor-valued I-form S of any rank in an obvious way; thus, (dS, u 1\\ v) = V,,( S, v) - Vv( S, u) - (S, [u, v]). (14.22) Apply this result to the vector-valued I-form S = dw. Recall the expression for a directional derivative, (dw, u) = V\"w. Thus find the result (d 2w, U 1\\ v) = V\" Vvw - Vv V\"w - Vlu,vjW = q[(u, v)w, (14.23) Relation of curvature 2-form to curvature -oper.ator '.il tJl where q[(u, v) is the curvature operator defined already in Chapter 11 [equation (11.8)]. The conclusion is simple: the (i)-tensor-valued 2-form fJl of(14.19), evaluated on the bivector (\"parallelogram\") u 1\\ v, is identical with the curvature operator '!fl(u, v) introduced previously; thus Relation of tJl to components of Riemann ('!fl, u 1\\ v) = q[ (u, v). (14.24) Now go from the language of abstract operators to a language that begins to make components show up. Substitute on the left the expression (14.20) and on the right the value of the curvature operator from (11.11); and rewrite (14.24) in the form Compare and conclude that the typical individual curvature 2-form is given by the formula (14.25) (sum over a, [1, restricted to a < [1; so each index pair occurs only once). Equation (14.25) provides the promised packaging of 21 curvature components into six curvature 2-forms; and equation (14.18) provides the quick means to calcu late these curvature 2-forms. It is not necessary to take the key calculational equations (14.18) on faith, or to master the extended exterior derivative to prove or use them. Not one mention of any d do they make except the standard exterior d of Chapter 4. These key equations, moreover, can be verified in detail (exercise 14.8) by working in a coordinate frame. One adopts basis I-forms w d = dx d \u2022 One goes on to use UJ/l. p = FJl. ph dx h from equation (14.15). In this way one obtains the \"standard formula for the curvature\" [equation (11", ".12) and equation (3) of Box 14.2] by standard methods. In summary, the calculus of forms and exterior derivatives reduces the calculation to the computation. Now look at the other link in the chain that leads from metric to curvature. It used to be \u00a7 14.5. CURVATURE 2-FORMS 353 It now reduces to the calculation of \"connection I.forms\"; thus. Two principles master this first step in the curvature computation: (I) the symmetry of the covariant derivative; and (2) its compatibility with the metric. Condition (I), symmetry, appears in hidden guise in the principle Symmetry of covariant derivative: d'lfi' = O. (14.26) (1) expressed 8S d2~ =0 Here the notation \":1' for point\" comes straight out of Cartan. He thought of a vector as defined by the movement of one point to another point infinitesimally close to it. To write tIP was therefore to take the \"derivative of a point\" [make a construction with a \"point deleted\" (tail of vector) and \"point reinserted nearby\" (tip of vector)]. The direction of the derivative d in tIP is indefinite. In other words, tIP contains a \"slot.\" Only when one inserts into this slot a definite vector v does tIP give a definite answer for Cartan's vector. What is that vectO\u00a3 that tIP then gives? It is v itself. \"The movement that is v tells the point :1' to reproduce the movement that is v\"; or in concrete notation, <tIP, v) = v. (14.27) Put the content of this equation into more formalistic terms. The quantity tIP is a (i)-tensor (14.28) It is distinguished from the generic cD-tensor by the special value of its components In this sense it deserves the name of \"unit tensor.\" Insert this tensor in place of S into equation (14.22) and obtain the result (14.29) The zero on the right is a restatement of equation (1O.2a) or of \"the closing of the vector diagram\" in the picture called \"symmetry of covariant differentiation\" in Box 102. The vanishing of the righthand side for arbitrary u and v demands the vanish ing of d 2<j' on the left; and conversely, the vanishing of d 2<j' demands the symmetry of the covariant derivative. The other principle basic to the forthcoming computa tions is \"compatibility of covariant derivative with metric,\" as expressed in the form d(u' v) = (du). v + u\u00b7 (dv). (14.30) It is essential here to ascribe to the metric (the \"dot\") a vanishing covariant deriva tive; thus d(') = O. 354 14. CALCULATION OF CURVATURE Capitalize on the symmetry and compatibility of the covariant derivative by using basis vectors (and where appropriate the basis I-forms dual to these basis vectors) in equations (14.26) and (14.30). Thus from compute.d9=ep.wP. and conclude that the coefficient of e p. must vanish; or (\"symmetry\"). (14.31a) Next, into (14.30) in place of the general u and v insert the specific ep' and e v, respectively, and find (\"compatibility\"), (14.31b) where In equations (14.31) one has the connection between metric and connection forms expressed in the most compact way. (14.31 c) \u00a714.6. COMPUTATION OF CURVATURE USING EXTERIOR DIFFERENTIAL FORMS (2) expressed as dwP. + wp. p /\\ w p = 0 Compatibility of 9 and V expressed as dgp. p = wp. p + wpp. Method of curvature 2-forms in 4 steps: (1) select metric and frame (2) calculate connection 1-forms wp. p The use of differential forms for the computation of curvature is illustrated in Box 14.5. This section outlines the method. There are three main steps: compute wp.v; compute q[ p. v; and compute Gp. v' More particularly, first select a metric and a frame. Thereby fix the basis forms wp. = LP.a' dx a' and the metric components gp.v in ds 2 = gp.vwp. \u00ae w v. Then determine the connection forms wp.v' and determine them uniquely, as solutions of the equations 0= dwP. + wP.v /\\ w V, dgp.v = Wp.v + wvp.\u00b7 (14.", "31a) (14.31b) The \"guess and check\" method of finding a solution to these equations (described and illustrated in Box 14.5) is often quick and easy. [Exercise (14.7) shows that a solution always exists by showing that the Christoffel formula (14.36) is the unique solution in coordinate frames.] It is usually most convenient to use an orthonormal frame with gp.v = 1/p.v (or some other simple frame where gp.v = const, e.g., a null frame). Then dgp.v = 0 and equation (14.31b) shows that Wp.v = -wvp.' Therefore there are only six W p.v for which to solve in four dimensions. (continued on page 358) \u00a7 14.6. METHOD OF EXTERIOR DIFFERENTIAL FORMS 355 Box 14.5 CURVATURE COMPUTED USING EXTERIOR DIFFERENTIAL FORMS (METRIC FOR FRIEDMANN COSMOLOGY) The Friedmann metric (Box 27.1) represents a spacetime where each constant-t hypersurface is a three dimensional hypersphere of proper circumference 27Ta(t). An orthonormal basis is easily found in this spacetime; thus, where w t = dt, wx. = adX, w 8 = a sin XdO, w\u00a2 = a sin X sin 0 dcf;. A. Connection Computation Equation (l4.3Ib) gives, since dgJl.v = cfrIJl.v = 0, just (1) (2) so there are only six I-forms w Jl.V to be found. Turn to the second basic equation (l4.3la). The game now is to guess a solution (because this is so often quicker than using systematic methods) to the equations\u00b0= dwJl. + wJl.v /\\ WVin which the WV and thus also dwJl. are known, and wJl.v are unknown. The solution wJl.v is known to be unique; so guessing (if it leads to any answer) can only give the right answer. Proceed from the simplest such equation. From w t = dt, compute dw t = 0. Compare this with dw t = -wtJl./\\ wJl.or(sincewt t = -w\u00a3\u00a3 = 0, bywJl.v = -WVJl.) dw t = -w t k /\\ w k = 0. This equation could be satisfied by having w t with cancelations among different terms, or more simply by w t k ex:: w k, or in more complicated ways k = 0. Proceed, not 356 14. CALCULATION OF CURVATURE Box 14.5 (continued) looking for trouble, until some non-zero wJl.v is required. From cpx = a dX, find dw X = adt /\\ dX =(aja)w i /\\ W X,; -(aja)wX /\\ wi. Compare this with..- _ '......, dw x = -wxJI./\\ wJl. If:, = -.W Xi /\\ wi - WXiJ /\\ w 8 - w x\u00a2 /\\ w\u00a2. Guess that wXi = (aja)w x from the first term; and hope the other terms vanish. this allows w~ /\\ w X = -wix /\\ w x = wxi /\\ w x = 0 in the dw i (Note that equation.) Look at wiJ = a sin X dfJ, and write - dwo = (aja)w t /\\ W O+ a-I cot XW X /\\ W O = -Wei /\\ wi - w o x /\\ W X - w 8if, /\\ w\u00a2. Guess, consistent with previously written equations, that Wei = w i w 8 o = (aja)wO, x = -wxiJ = a-I cot Xwe. Finally from dW\u00a2 = (aja)w i /\\ w\u00a2 + a-I cot XWX /\\ w\u00a2 + (a sin xt l cot 0 WO /\\ w\u00a2 = -w\u00a2\u00b7/\\ wi - w\u00a2\u00b7 A W X - w\u00a2\u00b7 /\\ W O X 1\\ 8't deduce values of w\u00a2i' w\u00a2x' and w\u00a2o. These are not inconsistent with previous assumptions that terms like w 8if, /\\ w\u00a2 vanish (in the dWo equation); so one has in fact solved dwJl. = -wJl.v /\\ W V for a set of connection forms wJl.", " v, as follows: wki = w\\ = (aja)w k, wOx = -w x o = a-I cot X WO = cos X dfJ, w\u00a2x = -w x\u00a2 = a-I cot X w\u00a2 = cos X sin 0 d</>, w\u00a2o = -w\u00b0if, = (asinxtIcotOw\u00a2 = cosO d</>. (3) Of course, if these hit-or-miss methods of finding wJl. v do not work easily in some problem, one may simply use equations (14.32) and (14.33). \u00a7 14.6. METHOD OF EXTERIOR DIFFERENTIAL FORMS 357 B. Curvature Computation The curvature computation is a straightforward substitution of wIJ. v 'from equations (3) above into equation (14.34), which is This equation is short enough that one can write out the sum qzl. = dWi. + WI. /\\ wO. + WI. /\\ w\u00a2. 8 X \u00a2 X x x in contrast to the ten terms in the corresponding R = ar + r 2 equation [equation. (3) of Box 14.2]. Warning!: From w l X = (a/a)' w l /\\ w x. Missing is the term (a/a) dWx. Instead write w l X = (a/a)wX = X = ii dt /\\ dX = (ii/a)w l /\\ w x. With elementary care, a dx, and then find dw I then, in correctly substituting from (3) for the wIJ. v in the formula for qzIJ. v' one finds X = (a/a)w X, do not compute dw I and qzix= (ii/a)w l /\\ w X, This completes the computation of the RIJ.vap, since in this isotopic model universe, all space directions in the orthonormal frame wIJ. are algebraically equivalent. One can therefore write qzik = (ii/a)w i /\\ w k, qzk 1 = a-2(1 + a2)w k /\\ Wi, for the complete list of qzIJ. v' Specific components, such as or Ri... = ii/a xtx'R O... - a-2(1 + a2 ) \u00a28\u00a2 -, are easily read out of this display of qzIJ. v' C. Contraction From equations (14.7), find Gtt = +3a-2(1 + a2 ), and (4) (5a) (5b) (5c) (6) 358 14. CALCULATION OF CURVATURE If guessing is not easy, there is a systematic way to solve equations (14.31) in an orthonormal frame or in any other frame in which dgllv = O. Compute the dWIl and arrange them in the format (14.32) In this way display the 24 \"commutation coefficients\" ell.'\" These quantities enter into the formula (14.33) (3) calculate curvature 2-forms!'ilJJ.. Once the w llV are known, one computes the curvature forms tjlllv (again only six in four dimensions, since tjlllV = _tjlVIl) by use of the formula to provide the six Wllv (exercise 14.12). (14.34) (4) calculate components of curvature tensors Out of this tabulation, one reads the individual components of the curvature tensor by using the identification scheme The Einstein tensor Gilv is computed by scanning the tjlllV display to find the appropriate RIlVafJ components for use in formulas (14.7). (14.35) EXERCISES Exercise 14.5. EXTERIOR DERIVATIVE OF A PRODUCT OF FORMS Establish equation (l4.13a) by working up recursively from forms of lower order to forms of higher order. [Hints: Recall from equation (4.27) that for a p-form the exterior derivative is defined by Applied to the product a /\\ P of two I-forms, this formula gives d(a /\\ P) = d[(ax dx x) /\\ (f3JJ. dXJJ.)] = d[(axf3JJ.)(dxX /\\ dxJJ.)] = o(a xf3p) dx K /\\ dx x /\\ dxJJ. ox K oxK = (oa x dxK /\\ dXX) /\\ f3 dxJJ. _ (a, dxX) /\\ (of3JJ. dx K /\\ dXJJ.) JJ", ".'=(da) /\\ p - a /\\ dp. oxK Extend the reasoning to forms of higher order.] \u00a7 14.6. METHOD OF EXTERIOR DIFFERENTIAL FORMS 359 Exercise 14.6. RELATIONSHIP BETWEEN EXTERIOR DERIVATIVE AND COMMUTATOR Establish formula (14.21) by showing (a) that the righthand side is an algebraic linear function of u and an algebraic linear function of v, and (b) that the equation holds when u and v are coordinate basis vectors u = %x k, V = %xl. Exercise 14.7. CHRISTOFFEL FORMULA DERIVED FROM CONNECTION FORMS In a coordinate frame wJJ. = dxJJ., show that equation (l4.3la) requires rJJ.ap = rJJ.pa, and that, with this symmetry established, equation (14.31 b) gives an expression for ogJJ.,/ox a which can be solved to give the Christoffel formula (14.36) Exercise 14.8. RIEMANN-CHRISTOFFEL CURVATURE FORMULA RELATED TO CURVATURE FORMS Substitute wJJ., = rJJ.,x dx x into equation (14.18), and from the result read out, according to equation (14.25), the classical formula (3) of Box 14.2 for the components RJJ.,a/3\" Exercise 14.9. MATRIX NOTATION FOR REVIEW OF CARTAN STRUCTURE EQUATIONS Let e=(61,...,6n ) be a row matrix whose entries are the basis vectors, and let w be a column of basis I-forms WJJ.. Similarly let il = IlwJJ..I1 and ~ = 11~JJ..I1 be square matrices with I-form and 2-form entries. This gives a compact notation in which d6 JJ. = 6,w'JJ. and d!P = 6 JJ.WJJ. read de = eil and dP = ew, (14.37) respectively. (a) From equations (14.37) and d 2P = 0, derive equation (l4.3la) in the form \u00b0= dw + il /\\ w. [Solution: d 2P = de /\\ w + e dw = e(il 1\\ w + dw).] ( 14.38) (b) Compute d 2e as motivation for definition (14.18), which reads ~ = dil + il 1\\ il. (c) From d 2w = 0, deduce ~ /\\ w =\u00b0and then decompress the notation to get antisymmetry relation RJJ.[a/3Y] = 0. (14.39) the (d) Compute d~ from equation (14.39), and relate it to the Bianchi identity RJJ.,[a/3;Y] = 0. (e) Let u = {uJJ.} be a column of functions; so v = eu = 6 JJ.uJJ. is a vector field. Compute, in compact notation, dvand d 2v to show d 2v = e!'Ru (which is equation 14.17). Exercise 14.10. TRANSFORMATION RULES FOR CONNECTION FORMS IN COMPACT NOTATION Using the notation of the previous exercise, write e' = eA in place of 6 JJ.' = 6,A\"JJ.\" and similarly w' = A-1w, to represent a change of frame. Show that dP =ew = e'w'. Substitute e' = eA in de' = e'il' to deduce the transformation law Rewrite this in decompressed notation for coordinate frames with A'JJ.' = ox' j(: x JJ.' as a formula of the form rJJ.'a'/3' = (?). (14.40) 360 14. CALCULATION OF CURVATURE Exercise 14.11. SPACE IS FLAT IF THE CURVATURE VANISHES (see \u00a711.5) If coordinates exist in which all straight lines (d 2xJJ./d}..2 = 0) are geodesics, then one says the space is flat. Evidently all rJJ.OI.{3 and RJJ.YOI.{3 vanish in this case, by equation (14.8) and equation (3) in Box 14.2. Show conversely that, if (~ =0, then such coordinates exist. Use the results of the previous probl,em to find differential equations for a transformation A to a basis e' where il' = O. What are the conditions for complete integrability of these equations? [Note that dfK = FK(x,f) is", " completely integrable if d 2fK = 0 modulo the original equations.] Why will the basis forms wJJ.' in this new frame be coordinate differentials wJJ.' ::: dxJJ.'? Exercise 14.12. SYSTEMATIC COMPUTATION OF CONNECTION FORMS IN ORTHONORMAL FRAMES Deduce equation (14.32) by applying equation (14.21) to basis vectors, using equations (8.14) to define cJJ.yOI.. Then show that, in an orthonormal frame (or any frames with gJJ.Y ::: const), equation (14.33) provides a solution of equations (14.31), which define wJJ.Y' [Compare also equation (8.24b).] Exercise 14.13. SCHWARZSCHILD CURVATURE FORMS Use the obvious orthonormal frame wi::: e'\" dt, w;. = eA dr, w 9 = r dO, wJ> = r sin 8 d<j> for the Schwarzschild metric in which i/J and A are functions of r only; and compute the curvature forms eil,P.p and the Einstein tensor GiL p by the methods of Box 14.5. [Answer; tJli;' = Ew i /\\ w;., tJli9 = Ew i /\\ w 9, eiliJ> = Ew i /\\ wJ>, (:il9J> = Fw 9 /\\ wJ>, tJlJ>;' = FwJ> /\\ w;., 91,;'9 = Fw;' /\\ w 9, with (14.41 ) E = - e- 2A(i/J\" + i/J'2 E = _.!. e-2Ai/J', r i/J'A'), and then Gi = -(F + 2F), i ; G;. = -(F + 2E), J> 9 - G9 = GJ> = -(E + E + F), i i i 9 G; = G9 = GJ> = 0 = G9 = GJ> = G4>'] ;. ; (14.42) (14.43) Exercise 14.14. MATRIX DISPLAY OF THE RIEMANN-TENSOR COMPONENTS Use the symmetries of the Riemann tensor to justify displaying its components in an ortho normal frame in the form W\"\" OI.~ 01 02 _ 03 - 23 31 12 E I I I I I -------+-------, H I I _HT I I, F (14.44) \u00a7 14.6. METHOD OF EXTERIOR DIFFERENTIAL FORMS 361 where the rows are labeled by index pairs fiii = 01, 02, etc., as shown; and the columns tiP, similarly. Here E, F, and H are each 3 X 3 matrices with (why?) trace H = 0, (14.45) where E T means the transpose of E. Exercise 14.15. RIEMANN MATRIX WITH VANISHING Show that the empty-space Einstein equations GiL. =\u00b0allow the matrix in equation (14.44) EINSTEIN TENSOR to be simplified to the form where now, in addition to the equality E = F that this form implies, the further conditions trace E = 0, H = HT (14.47) hold. (14.46) Exercise 14.16. COMPUTATION OF CURVATURE FOR A PULSATING OR COLLAPSING STAR Spherically symmetric motions of self-gravitating bodies are discussed in Chapters 26 and 32. A metric form often adopted in this situation is (14.48) where now i/J, A, and r -are each functions of the two coordinates Rand T. Compute the curvature 2-forms and the Einstein tensor for this metric, using the methods of Box 14.5. In the guessing of the wJJ. p' most of the terms will already be evident from the corresponding calculation in exercise 14.13. [Answer, in the obvious orthonormal frame w T = e'\" dT, w R = eA dR, w 9 = rd8, wJ> = rsin8 d<j>: ~ilJ'R = Ew t 1\\ w R, 9 = Ew t 1\\ w 9 + Hw R 1\\ w 9, tJl t tJlTJ> = Ew t 1\\ wJ> + Hw R 1\\ wJ>, tJl9J> = Fw 9 1\\ wJ>, tJlR 9 = Fw R 1\\ w 9 - Hw t 1\\ w 9, tJl", "an, replies, \"Through the principle that 'the boundary of a boundary is zero'\" (Box 15.1). Actually, two features of the curvature are automatically conserved; or, otherwise stated, the curvature satisfies two Bianchi identities, the subject of this chapter. Both features of the curvature, both \"geometric objects,\" lend themselves to representation in diagrams, moreover, diagrams that show in action the principle that \"the boundary of a boundary is zero.\" In this respect, the geometry of spacetime shows a striking analogy to the field of Maxwell electrodynamics. In electrodynamics there are four potentials that are united in the I-form A = A,.. dx\"'. Out of this quantity by differentiation follows the Faraday, F = dA. This \u00a7 15.1. BIANCHI IDENTITIES IN BRIEF 365 field satisfies the identity dF = \u00b0(identity, yes; identity lending itself to the definition of a conserved source, no). In gravitation there are ten potentials (metric coefficients g/.I') that are united in the metric tensor 9 = g/.l' dx/.l \u00ae dx\u00b7. Out of this quantity by two differentiations follows the curvature operator dF = 0 qz = 1. e 4 /.I /\\ e R/.I\u00b7 \u2022 afJ dx a /\\ dx fJ \u2022 This curvature operator satisfies the Bianchi identity cMl = 0, where now \"d\" is a generalization of Cartan's exterior derivative, described more fully in Chapter 14 (again an identity, but again one that does not lend itself to the definition of a conserved source). In electromagnetism, one has to go to the dual, *F, to have any feature of the field that offers a handle to the source, d*F = 4'1T *J. The conservation of the source, d*J = 0, appears as a consequence of the identity dd*F = 0; or, by a rewording of the reasoning (Box 15.1), as a consequence of the vanishing of the boundary of a boundary. (continued on page 370) dd*F =0 plus Maxwell equations ~ d*J = 0 Box 15.1 THE BOUNDARY OF A BOUNDARY IS ZERO A. The Idea in Its 1-2-3-Dimensional Form Begin with an oriented cube or approximation to a cube (3-dimensional). Its boundary is composed of six oriented faces, each two-dimensional. Orientation of each face is indicated by an arrow. Boundary of anyone oriented face consists of four oriented edges or arrows, each one-dimen sional. Every edge unites one face with another. No edge stands by itself in isolation. \"Sum\" over all these edges, with due regard to sign. Find that any given edge is counted twice, once going one way, once going the other. Conclude that the one-dimensional boundary of the two-dimensional boundary of the three-di mensional cube is identically zero. ~\u00b7r.r. 366 15. BIANCHI IDENTITIES AND THE BOUNDARY OF A BOUNDARY Box 15.1 (continued) B. The Idea in Its 2-3-4-Dimensional Form Begin with an oriented four-dimensional cube or approximation thereto. The coordinates of the typical corner of the four-cube may be taken to be (to -I-! Lit, X o -I-! L1x, Yo -I-! L1y, =0 -I-! Liz); and, accordingly, a sample corner itself, in an obvious abbreviation, is conveniently abbreviated + - - +. There are 16 of these corners. Less com plicated in appearance than the 4-cube itself are its three-dimensional faces, which are \"exploded off of it\" into the surrounding area of the diagram, where they can be inspected in detail. The boundary of the 4-cube is composed of eight oriented hyperfaces, each of them three-dimen sional (top hyperface with extension L1x L1y Liz, for example; a \"front\" hyperface with extension Lit L1y Liz; etc.) ~Y x +--- ++++ ~z x - - - - - --.---- r--~-+++ 1:= -1/2411 \u00a7 15.1. BIANCHI IDENTITIES IN BRIEF 367 Boundary of anyone hyperface (\"cube\") consists of six oriented faces, each two-dimensional. Every face (for example, the hatched face.:1x L1y in the lower lefthand corner) unites one hypersurface with another (the \"3-cube side face\" L1t.:1x L1y in the lower lefthand corner with the \"3-cube top face\".:1x L1", "y LIz, in this example). No face stands by itself in isolation. The three-dimensional boundary of the 4-cube exposes no 2-surface to the outside world. It is faceless. \"Sum\" over all these faces, with due regard to orientation. Find any given face is counted twice, once with one orientation, once with the opposite orientation. Conclude that the two-dimensional boundary of the three-dimensional boundary of the four-dimensional cube is identically zero. C. The Idea in Its General Abstract Form 00 = 0 (the boundary of a boundary is zero). D. Idea Behind Application to Gravitation and Electromagnetism. The one central point is a law of conservation (conservation of charge; conservation of momentum-energy). The other central point is \"automatic fulfillment\" of this conservation law. \"Automatic conservation\" requires that source not be an agent free to vary arbitrarily from place to place and instant to instant. Source needs a tie to something that, while having degrees of freedom of its own, will cut down the otherwise arbitrary degrees of freedom of the source sufficiently to guarantee that the source automatically fulfills the conservation law. Give the name \"field\" to this something. Define this field and \"wire it up\" to the source in such a way that the conservation of the source shall be an automatic consequence ofthe \"zero boundary ofa boundary. \" Or, more explicitly: Conservation demands no creation or destruction of source inside the four-dimensional cube shown in the diagram. Equivalently, integral of \"creation events\" (integral of d*J for electric charge; integral of d*T for energy-momentum) over this four-dimensional region is required to be zero. Integral of creation over this four-dimensional region translates into integral of source density-current (*J or *T) over three-dimensional boundary of this region. This boundary consists of eight hyperfaces, each taken with due regard to orientation. Integral over upper hyperface (\"L1x L1y Liz)\" gives amount of source present at later moment; over lower hyperface gives amount of souee present at earlier moment; over such hyperfaces as \"Lit L1x L1y\" gives outflow of source over intervening period of time. Conservation demands that sum of these eight three-dimensional integrals shall be zero (details in Chapter 5). 368 15. BIANCHI IDENTITIES AND THE BOUNDARY OF A BOUNDARY Box 15.1 (continued) VaUishing of this sum of three-dimensional integrals states the conservation requirement, but does not provide the machinery for \"automatically\" (or, in mathe matical terms, \"identically\") meeting this requirement. For that, turn to principle that \"boundary of a boundary is zero.\" Demand that integral of source density-current over any oriented hyperface '1/ (three-dimensional region; \"cube\") shall equal integral of field over faces of this \"cube\" (each face being taken with the appropriate orientation and the cube being infinitesimal): 4'1Tf *J =f *F; 'TO ilT 8 f '1T f (moment Of) rotation. 'TO *T = il'T' Sum over the six faces of this cube and continue summing until the faces of all eight cubes are covered. Find that any given face (as, for example, the hatched face in the diagram) is counted twice, once with one orientation, once with the other (\"boundary of a boundary is zero\"). Thus is guaranteed the conservation of source: integral of source density-current over three-dimensional boundary of four-dimen sional region is automatically zero, making integral of creation over interior of that four-dimensional region also identically zero. Repeat calculation with boundary of that four-dimensional region slightly dis placed in one locality [the \"bubble differentiation\" ofTomonaga (1946) and Schwin ger (1948)], and conclude that conservation is guaranteed, notonly in the four-di mensional region as a whole,\u00b7 but at every point within it, and, by extension, everywhere in spacetime. E. Relation of Source to Field One view: Source is primary. Field may have other duties, but its prime duty is to serve as \"slave\" of source. Conservation of source comes first; field has to adjust itself accordingly. Alternative view: Field is primary. Field takes the responsibility of seeing to it that the source obeys the conservation law. Source would not know what to do in absence of the field, and would not even exist. Source is \"built\" from field. Conser vation of source is consequence of this construction. One model illustrating this view in an elementary context: Concept of \"classical\" electric charge as nothing but \"electric lines of force trapped in the topology of a multiply connected space\" [Weyl (1924b); Wheeler (1955); Mis", "ner and Wheeler (1957)]. On any view: Integral of source density-current over any three-dimensional region (a \"cube\" in simplified analysis above) equals integral of field over boundary of this region (the six faces of the cube above). No one has ever found any other way to understand the correlation between field law and conservation law. \u00a7 15.1. BIANCHI IDENTITIES IN BRIEF 369 F. Electromagnetism as a Model: How to \"Wire Up\" Source to Field to Give Automatic Conservation of Source Via \"00 = 0\" in Its 2-3-4-Dimensional Form Conservation means zero creation of charge (zero creation in four-dimensional region g). Conservation therefore demands zero value for integral of charge density-current over three-dimensional boundary of this volume; thus, in the Track-l language of Chapters 3 and 5. Equivalently, in the coordinate-free abstract language of \u00a7\u00a74.3-4.6, one has o=f d *J =I *J, il ail where *J = *J123 dx 1 1\\ dx 2 1\\ dx 3 + *J023 dxo 1\\ dx 2 1\\ dx 3 + *J031 dxo 1\\ dx 3 1\\ dx 1 + *J0l2 dxo 1\\ dx 1 1\\ dx 2 (\"eggcrate-like structure\" of the 3-form of charge-density and current-density). Fulfill this conservation requirement automatically (\"identically\") through the principle that \"the boundary of a boundary is zero\" by writing 4'1T *J = d*F; thus, 4'1TI *J = I d*F = I ail ail aaJ](zero!) *F = 0 or, in Track-l language, write 4'1TJIl = F/lV;v, and have In other words, half of Maxwell's equations in their familiar flat-space form, div E = V. E = 4'1Tp, curl B = V X B = E + 4'1TJ, \"wire up\" the source to the field in such a way that the law of conservation of source follows directly from \"aag = 0.\" G. Electromagnetism Also Employs \"00 = 0\" in its 1-2-3-Dimensional Form (\"No Magnetic Charge\") Magnetic charge is linked with field via 4...J rnag = dF (see point F above for transla terms). Absence of tion of this compact Track-2 language into equivalent Track-l 370 15. BIANCHI IDENTITIES AND THE BOUNDARY OF A BOUNDARY Box 15.1 (continued) any magnetic charge says that integral of J mag over any 3-volume '1' is necessarily zero; or (\"integration by parts,\" generalized Stokes theorem) o= r dF = r F = (to~~l magnetic flUX). eXItmg through O'Y JeW J'l' In order to satisfy this requirement \"automatically,\" via principle that \"the boundary of a boundary is zero,\" write F = dA (\"expression of field in terms of 4-potential\"), and have i F =i eW d'i' dA =i A = O. dd'i' (zero!) H. Structure of Electrodynamics in Outline Form A (potential) +F (field; Faraday) = dA t dF = 0 (identity based on aa = 0) - *F (dual field; Maxwell) ~. d*F = 41T*J +d*J = 0 (expressed as an identity based on aa = 0) d*J=Oor'lf'J=O (\"automatic\" conservation of source) In gravitation physics, one has to go to the \"double dual\" (two pairs of alternating indices, two places to take the dual) G = *R* of Riemann to have a feature of the field that offers a handle to the source: G = TrG = Einstein = 8'1TT = 8'1T X (density of energy-momentum). The conservation of the source T = e /I.T!J..w' can be stated V\u00b7 T = O. But better suited for the present purpose is the form (see Chapter 14 and exercise 14.18) \u00a7 15.1. BIANCHI IDENTITIES IN BRIEF 371 I. Structure of Geometrodynamics in Outline Form 9 (metric) (parallel transport; covariant derivative; ~ '\" = d! generalized exterior derivative) t operator) tJl = d 2 (curvature --- 6 = *R* (double dual) dtJl = 0 (full Bianchi identity; based on aa = 0) d*G = 0", " (contracted Bianchi identity based on aa = 0) or \"'\u00b7T = 0 (\"automatic conservation of source) where d*T = 0, d*G =0 plus Einstein field equation =- d*T = 0 This conservation law arises as a consequence of the \"contracted Bianchi identity\", d*G = 0, again interpretable in terms of the vanishing of the boundary of a boundary. 372 15. BIANCHI IDENTITIES AND THE BOUNDARY OF A BOUNDARY = Aatend IlA x Figure 15.1. Combine rotations associated with each of the six faces of the illus trated 3-volume and end up with zero net rotation (\"full Bianchi identity\"). Reason: Contribution of any face is measured by change in a test vector A carried in parallel transport around the perimeter of that face. Combine contributions of all faces and end up with each edge traversed twice, once in one direction, once in the other direction [boundary (here one-dimensional) of boundary (two-dimensional) of indicated three-dimensional figure is zero]. Detail: The vector A, residing at the indicated site, is transported parallel to itself over to the indicated face, then carried around the perimeter of that face by parallel transport, experiencing in the process a rotation measured by the spacetime curvature associated with that face, then transported parallel to itself back to the original site. To the lowest relevant order of small quantities one can write (change in A) = - 41)' 41z!il(ev' e z ) A in operator notation; or in coordinate language, -SA\" = R\"/lv.<at x + 41x)A/l 41)' 41z. \u00a715.2. BIANCHI IDENTITY dtJl = 0 AS A MANIFESTATION OF \"BOUNDARY OF BOUNDARY = 0\" Bianchi identity, d(>i =0, interpreted in terms of parallel transport around the six faces of a cube. Such is the story of the two Bianchi identities in outline form; it is now appropriate to fill in the details. Figure 15.1 illustrates the full Bianchi identity, d&l = 0 (see exercise 14.17), saying in brief, \"The sum of the curvature-induced rotations associ ated with the six faces of any elementary cube is zero.\" The change in a vector A associated with transport around the perimeter of the indicated face evaluated to the lowest relevant order of small quantities is given by -oA'\" = R\"'f3l1z (at x +.::lx)Af3 L1yLlz. (15.1) The opposite face gives a similar contribution, except that now the sign is reversed and the evaluation takes place at x rather than at x +.::lx. The combination of the contributions from the two faces gives (15.2) \u00a7 15.3. MOMENT OF ROTATION: KEY TO BIANCHI IDENTITY 373 when Riemann normal coordinates are in use. In such coordinates, the vanishing of the total - oA'\" contributed by all six faces implies R'\" f3l1z ;z + R'\" f3zz;1I + R'\" f3zlI;z = 0. (15.3) Here semicolons (covariant derivatives) can be and have been inserted instead of commas (ordinary derivatives), because the two are identical in the context of Riemann normal coordinates; and the covariant version (15.3) generalizes itself to arbitrary curvilinear coordinates. Tum from an xyz cube to a cube defined by any set of coordinate axes, and write Bianchi's identity in the form (15.4) (See exercise 14.17 for one reexpression of this identity in the abstract coordinate independent form, d&l = 0, and \u00a715.3 for another.) This identity occupies much the same place in gravitation physics as that occupied by the identity dF = ddA \u00b0 in electromagnetism: (15.5) \u00a715.3. MOMENT OF ROTATION: KEY TO CONTRACTED BIANCHI IDENTITY The contracted Bianchi identity, the identity that offers a \"handle to couple to the source,\" was shown by Elie Cartan to deal with \"moments of rotation\" [Canan (1928); Wheeler (1964b); Misner and Wheeler (1972)]. Moments are familiar in elementary mechanics. A rigid body will not remain at rest unless all the forces acting on it sum to zero: (15.6) Although necessary, this condition is not sufficient. The sum of the moments of these forces about some point qp must also be zero: (15.7) Exactly what point these moments are taken about happily does not matter, and this for a simple reason. The arbitrary point in the vector product (15.7) has for", " coefficient the quantity IiF(i), which already has been required to vanish. The situation is similar in the elementary cu be of Figure 15.1. Here the rotation associated with a given face is the analog of the force F(i) in mechanics. That the sum of these rotations vanishes when extended over all six faces of the cube is the analog of the vanishing of the sum of the forces F(i). What is the analog for curvature of the moment of the force that one encounters in mechanics? It is the moment of the rotation associated with a given face of the Net moment of rotation over all six faces of a cube: (1) descri bed (2) equated to integral of source, f *T. over interior of cube 374 15. BIANCHI IDENTITIES AND THE BOUNDARY OF A BOUNDARY cube. The value of any individual moment depends on the reference point 'cPo However, the sum of these moments taken over all six faces of the cube will have a value independent of the reference point qp, for the same reason as in mechanics. Therefore <jJ can be taken where one pleases, inside the elementary cube or outside it. Moreover, the cube may be viewed as a bit of a hypersurface sliced through spacetime. Therefore <j-' can as well be off the slice as on it. It is only required that all distances involved be short enough that one obtains the required precision by calculating the moments and the sum of moments in a local Riemann-normal coordinate system. One thus arrives at a P-independent totalized moment of rotation (not necessarily zero; gravitation is not mechanics!) associated with the cube in question. Now comes the magic of \"the boundary of the boundary is zero.\" Identify this net moment of rotation of the cube, evaluated by summing individual moments of rotation associated with individual faces, with the integral of the source density current (energy-momentum tensor *T) over the interior of the 3-cube. Make this identification not only for the one 3-cube, but for all eight 3-cubes (hyperfaces) that bound the four-dimensional cube in Box 15.1. Sum the integrated source density current *Tnot only for the one hyperface of the 4-cube, but for all eight hyperfaces. Thus have f 4-cube (:::~:n) = f d * T 3\u00b7boundary of this 4-cube source current-) density, *T ( = 2: these eight bounding 3-cubes net moment of rotatiOn) associated with specifi d e b cu e ( moment of rotation associated with specified. face of specified cube ) ( (15.8) = 2: eight bounding 3\u00b7cubes 2: six faces bounding given 3\u00b7cube (zero!) (3) conserved Let the moments of rotation, not only for the six faces of one cube, but for all the faces of all the cubes, be taken with respect to one and the same point P. Recall (Box 15.1) that any given face joins two cubes or hyperfaces. It therefore appears twice in the count of faces, once with one orientation (\"sense of circumnavigation in parallel transport to evaluate rotation\") and once with the opposite orientation. Therefore the double sum vanishes identically (boundary of a boundary is zero!) This identity establishes existence of a new geometric object, a feature of the curva ture, that is conserved, and therefore provides a handle to which to couple a source. The desired result has been achieved. Now to translate it into standard mathematics! \u00a7 15.4. CALCULATION OF THE MOMENT OF ROTATION 375 \u00a715.4. CALCULATION OF THE MOMENT OF ROTATION It remains to find the tensorial character and value of this conserved Cartan moment of rotation that appertains to any elementary 3-volume. The rotation associated with the front face L1y Llze ll /\\ e z of the cube in Figure 15.1 will be represented by the bivector (4) evaluated rotation associated with front L1y LIz face ( ) = e /\\ e RIAI'I I' A LI Liz liZ ~ (15.9) \"-located -at Pfro~t = (t =-! Lit, x + L1x, Y +! L1y, z +! LIz). This equation uses Rie mann normal coordinates; indices enclosed by strokes, as in IA!-tl, are summed with the restriction A <!-t. The moment of this rotation with respect to the point '!l will be represented by the trivector moment of rotation) associated with f ront L1y Liz laCe., ( = (q>center of front face - q\u00bb 1\\ e A 1\\ el'RIAI'!lI", "z L1y Liz. (15.10) Here neither q>centerfront nor q> has any well-defined meaning whatsoever as a vec tor, but their difference is a vector in the limit of infinitesimal separation, L1q> = q>. With the back face a similar moment of rotation is associated, q>center front with the opposite sign, and with q>centerfront replaced by Pcenterback' In the difference between the two terms, the factor q> is of no interest. because one is already assured it will cancel out [Bianchi identity (15.4); analog of IF(i) = 0 in mechanics]. The difference P center front q>center back has the value L1xez\u00b7 Summing over all six faces, one has ~:t~~:::~~~~ted ) = with cube or hyper face L1x L1y LIz ( e z /\\ e A /\\ e I'RIAI'IIIZ L1x L1y Liz (front and back). + ell /\\ e A /\\ el'RIAl'lzzLlyLlzLlx (sides) + e z /\\ e A /\\ el'RIAl'lzlI Liz L1x L1y (top and bottom). (I 5.1 1) This sum one recognizes as the value (on the volume element e z /\\ ell /\\ e z L1x L1y Liz) of the 3-form e \" /\\ e /\\ e RIAI'I I' A la(31 dx\" /\\ dx a /\\ dx(3. Moreover this 3-form is defined, and precisely defined, at a point, whereas (I 5.1 1), applying as it does to an extended region, does not lend itself to an analysis that is at the same time brief and precise. Therefore forego (I 5.1 1) in favor of the 3-form. Only remember, when it comes down to interpretation, that this 3-form is to be y 376 15. BIANCHI IDENTITIES AND THE BOUNDARY OF A BOUNDARY evaluated for the \"cube\" e z /\\ ell /\\ e z..1x L1y Liz. Now note that the \"trivector valued moment-of-rotation 3-form\" can also be written as (5) abstracted to give d'3' 1\\ ~t moment of) rotation ( _. = dqiJ /\\ <Jl = e v /\\ e I /\\ e R AI' I' I A laf31 f3 dx v /\\ dx a /\\ dx. (15.12) Here (15.13) is Cartan's (D unit tensor. Also tJl is the curvature operator. treated as a bivector valued 2-form: (15.14) Using the language of components as in (15.11), or the abstract language intro duced in (15.12), one finds oneself dealing with a trivector. A trivector can be left a trivector, as, in quite another context, an element of 3-volume on a hypersurface in 4-space can be left as a trivector. However, there it is more convenient to take the dual representation, and speak of the element of volume as a vector. Denote by * a duality operation that acts only on contravariant vectors, trivectors, etc. (but not on forms). Then in a Lorentz frame one has *(e 1 /\\ e 2 /\\ e 3 ) = eo; but *(dx3 ) = dx3. More generally, (15.15) (6) abstracted to give *(dtJ> 1\\ ~) = fI\"G\"Td 32 T In this notation, the \"vector-valued moment-of-rotation 3-form\" is moment of rotatIOn ) (. = *(d?l /\\ tJl) = e e \"RIAI'I dx v /\\ dx a /\\ dx f3 laf31 = e,,(* R)v\" laf31 dx V /\\ dx f3, \" VAP. / \\ dx a or, in one more step, ( momen~ of rotatIon ) = *(d?l /\\ tJl) = e (*R*) \"VT d 3I \" v Here d 3I T is a notation for basis 3-forms, as in Box 5.4; thus,. T (15.16) (15.17) (In a local Lorentz frame, dx 1 /\\ dx 2 /\\ dx3 = d 3I o.) Nothing is more central to the analysis of curvature than the formula (15.16). It starts with an element of 3", "-volume and ends up giving the moment of rotation in that 3-volume. The tensor that connects the starting volume with the final moment, the \"contracted double-dual\" of Riemann, is so important that it deserves and receives a name of its own, G Einstein; thus This tensor received attention in \u00a7\u00a713.5 and 14.2, and also in the examples at the (15.18) \u00a7 15.5 CONSERVATION OF MOMENT OF ROTATION FROM \"00 = 0\" 377 end of Chapter 14. In terms of Einstein, the connection between element of 3-vol ume and \"vector-valued moment of rotation\" is (momen~ ) = *(d?f/\\ qil) = e GUT d 3:E. of rotatIon U T (15.19) The amount of \"vector-valued moment of rotation\" contained in the element of 3-volume d3:E IJ. is identified by general relativity with the amount of energy-mo mentum contained in that 3-volume. However, defer this identification for now. Concentrate instead on the conservation properties of this moment of rotation. See them once in the formulation of integral calculus, as a consequence of the principle 0.\" See them then a second time, in differential formulation, as a consequence \"00 of \"dd 0.\" \u00a715.5. CONSERVATION OF MOMENT OF ROTATION SEEN FROM \"BOUNDARY OF A BOUNDARY IS ZERO\" The moment of rotation defines an automatically conserved quantity. In other words, the value of the moment of rotation for an elementary 3-volume.J.x.Jy.Jz after the lapse of a time.Jt is equal to the value of the moment of rotation for the same 3-volume at the beginning of that time, corrected by the inflow of moment of rotation over the six faces of the 3-volume in that time interval (quantities proportional to.Jy.Jz.Jt, etc.) Now verify this conservation of moment of rotation in the language of \"the boundary of a boundary.\" Follow the pattern of equation (15.8), but translate the words into formulas, item by item. Evaluate the amount of moment of rotation created in the elementary 4-cube [}, and find Conservation of net moment of rotation: (1) derived from \"00 = 0\" \"creation\" t ( \"creation of moment of f rotation\" in the elementary =f d*G; ) 4-cube of spacetime [} t [J definition definition f d*G =f *G =f *(tJ6.J 1\\ ~il) = 1 i\"O i M o 2: the eight 3-cubes that bound [J * moment of rotation f 3-cube (cit 1\\ tJl) associated with specified 3-cube step 1 step 2 step 3 step 4 moment of rotation (~1J 1\\ (off) f face associated with specified face of specified cube 2: six fa(\"e~ hounding' specified 3\u00b7cube * o. (15.20) 1step 5 2: eight bonnding 3-cubes = 1 step 4 378 15. BIANCHI IDENTITIES AND THE BOUNDARY OF A BOUNDARY Here step 1 is the theorem of Stokes. Step 2 is the identification established by (15.19) between the Einstein tensor and the moment of rotation. Step 3 breaks down the integral over the entire boundary ail into integrals over the individual 3-cubes that constitute this boundary. Moreover, in all these integrals, the star * is treated as a constant and taken outside the sign of integration. The reason for such treatment is simple: the duality operation * involves only the metric, and the metric is locally constant throughout the infinitesimal 4-cube over the boundary of which the inte gration extends. Step 4 uses the formula d(':i' 1\\!'Jl) = di' 1\\ qz + ~J 1\\ dtJl = diP 1\\ (oil (15.21 ) and the theorem of Stokes to express each 3-cube integral as an integral of q> 1\\ tJl over the two-dimensional boundary of that cube. The culminating step is 5. It has nothing to do with the integrand. It depends solely on the principle aa o. In brief, the conservation of moment of rotation follows from two circumstances. (1) The moment of rotation associated with any elementary 3-cube is by definition a net value, obtained by adding the six moments of rotation associated with the six faces of that cube. (2) When one sums these net values for all eight 3-cubes in (15.20), which are the boundary of the elementary 4-cube il, one counts the contribution of a given 2-face twice, once with", " one sign and once with the opposite sign. In virtue of the principle that \"the boundary of a boundary is zero,\" the conservation of moment of rotation is thus an identity. \u00a715.6. CONSERVATION OF MOMENT OF ROTATION EXPRESSED IN DIFFERENTIAL FORM (2) derived from.. dd = 0\u00b7\u00b7 Every conservation law stated in integral form lends itself to restatement in differ ential form, and conservation of moment of rotation is no exception. The calculation is brief. Evaluate the generalized exterior derivative of the moment of rotation in three steps, and find that it vanishes; thus: d*G = d[*(dq> 1\\ tJl)] = *[d(d?P 1\\ tJl)] = *[d 2q> 1\\ tJl _ d?P 1\\ dtJl] = 0 jstep 1 step 2 step 3 Step 1 uses the relation d* = *d. The star duality and the generalized exterior derivative commute because when d is applied to a contravariant vector, it acts as a covariant derivative, and when * is applied to a covariant vector or I-form, it is without effect. Step 2 applies the standard rule for the action of d on a product of tensor-valued forms [see equation (14.l3b)]. Step 3 deals with two terms. The first term vanishes because the first factor in it vanishes; thus, d 2q> = 0 [Cartan's equation of structure; expresses the \"vanishing torsion\" of the covariant derivative; see equation (1426)]. The second term also vanishes, in this case, because the second factor in it vanishes; thus, dtJl = 0 (the full Bianchi identity). Thus briefly is conser vation of moment of rotation established. Box 15.2 THE SOURCE OF GRAVITATION AND THE MOMENT OF ROTATION: THE TWO KEY QUANTITIES AND THE MOST USEFUL MATHEMATICAL REPRESENTATIONS FOR THEM Energy-momentum as source of gravitation (curvature of space time) Moment of rotation as automati cally conserved feature of the geometry Representation as a vector-valued 3-form, a coordinate-independent geometric object Machine to tell how much energy momentum is contained in an elementary 3-volume: \"T = e.T\"TtFIT (\"dual of stress-energy tensor\") Machine to tell how much net moment of rotation-expressed as a vector-is obtained by add\u00b7 ing the six moments of rotation associated with the six faces of the elementary 3-cube: *(d9/\\ t;il) =\"G = e.GaTrPIT ('\"dual of Einstein\") Representation as a (~)-tensor (also a coordinate independent geometric object) Representation in language of components (values depend on choice of coordinate system) Conservation law in language of components Conservation in abstract lan guage. for the (~)-tensor Conservation in abstract lan guage. as translated into exterior derivative of the dual tensor (vec tor-valued 3-form) Same conservation law expressed in integral form for an element of 4-volume fJ Stress-energy tensor itself: Einstein itself: T = e.TaTeT TaT;T = 0 \"7'T=O d'T= 0 f \"T= 0 an GaT;T = 0 \"7'G:=O d\"G=O or d*(cf}! /\\ (~) = 0 iln J \"G =0 or *f (d: 1' /\\ 'il) = 0 or *f (:1' /\\ ~~)=O iln ailn \u00a715.7. FROM CONSERVATION OF MOMENT OF ROTATION TO EINSTEIN'S GEOMETRODYNAMICS: A PREVIEW Mass, or mass-energy, is the source of gravitation. Mass-energy is one component of the energy-momentum 4-vector. Energy and momentum are conserved. The amount of energy-momentum in the element of 3-volume d 3E is (see Box 15.2). Conservation of energy-momentum for an elementary 4-cube [} expresses itself in the form f *T = O. ail (15.23) (15.22) Einstein field equation \"derived\" from demand that (conservation of net moment of rotation) => (conservation of source) 380 15. BIANCHI IDENTITIES AND THE BOUNDARY OF A BOUNDARY This conservation is not an accident. According to Einstein and Cartan, it is \"auto matic\"; and automatic, moreover, as a consequence of exact equality between energy-momentum and an automatically conserved feature of the geometry. What is", " this feature? It is the moment of rotation, which satisfies the law of automatic conservation, i *G = o. au (15.24) In other words, the conservation of momentum-energy is to be made geometric in character and automatic in action by the following prescription: Identify the stress energy tensor (up to a factor 8'lT, or 8'lTG/c4, or other factor that depends on choice of units) with the moment of rotation; thus, *(dq> 1\\ 91) = *G = 8'lT *T; (15.25) or equivalently (still in the language of vector-valued 3-forms) moment Of). rotatIOn ( = *(dP 1\\ fJl) = e GUT d 3:E = 8'lTe TUT d 3:E ; U T U T or, in the language of tensors, or, in the language of components, (15.26) (15.27) (15.28) (Einstein's field equation; more detail, and more on the question of uniqueness, will be found in Chapter 17; see also Box 15.3). Thus simply is all of general relativity tied to the principle that the boundary of a boundary is zero. No one has ever discovered a more compelling foundation for the principle of conservation of mo mentum and energy. No one has ever seen more deeply into that action of matter on space, and space on matter, which one calls gravitation. In summary, the Einstein theory realizes the conservation of energy-momentum as the identity, \"the boundary of a boundary is zero. \" EXERCISES Exercise 15.1. THE BOUNDARY OF THE BOUNDARY OF A 4-SIMPLEX In the analysis of the development in time of a geometry lacking all symmetry, when one is compelled to resort to a computer, one can, as one option, break up the 4-geometry into simplexes [four-dimensional analog of two-dimensional triangle, three-dimensional tetrahe dron; vertices of \"central simplex\" conveniently considered to be at (t, x,y, z) = (0, 1, 1, 1), (0,1, -1, -1), (0, -1, 1, -1), (0, -1, -1, 1), (5 112,0,0,0), for example], sufficiently nu merous, and each sufficiently small, that the geometry inside each can be idealized as fiat (Lorentzian), with all the curvature concentrated at the join between simplices (see discussion of dynamics of geometry via Regge calculus in Chapter 42). Determine (\"give a mathematical \u00a715.7. FROM MOMENT OF ROTATION TO GEOMETRODYNAMICS: A PREVIEW 381.-----------------------------........ ~ Box 15.3 OTHER IDENTITIES SATISFIED BY THE CURVATURE (1) The source of gravitation is energy-momentum. (2) Energy-momentum is expressed by stress energy tensor (or by its dual) as a vector-valued 3-form (\"energy-momentum per unit 3-volume\"). (3) This source is conserved (no creation in an elementary spacetime 4-cube). These principles form the background for the probe in this chapter of the Bianchi identities. That is why two otherwise most interesting identities [Allendoerfer and Weil (1943); Chern (1955,1962)] are dropped from attention. One deals with the 4-form II - 1 \"y (JaN) 24'lT 2 g g :n\"(J 1\\ N) :nya, (1) and the other with the 4-form r = 8'lT 2[det gl'vl 1/ 2 (qil12 1\\ qil30 + qil13 1\\ qil02 1 + qillO 1\\ qil23)\u00b7 (2) Both quantities are built from the tensorial \"cur vature 2-forms\" spacetime geometry interior to that surface (pro vided that these changes neither abandon the con tinuity nor change the connectivity of the 4-geom region). Property (1) kills any etry in that possibility of identifying the integral, a scalar, with energy-momentum, a 4-vector. Property (2) kills it for the purpose of a conservation law, because it implies a non-zero creation in Q. Also omitted here is the Bel-Robinson tensor (see exercise 15.2), built bilinearly out of the cur vature tensor, and other tensors for which see, e.g., Synge (1962). One or all of these quantities may be found someday to have important physical content. The integral of the 4-form r of equation (2", ") over the entire manifold gives a number, an inte ger, the so-called Euler-Poincare characteristic of the manifold, whenever the integral and the inte ger are well-defined. This result is the four-dimen sional generalization of the Gauss-Bonnet integral, widely known in the context of two-dimensional geometry: Riemannian scalar curvature) f invariant (value 2/a2 for a sphere of radius a) ( gl/2 d 2x. The four-dimensional integral of either quantity over a four-dimensional region Q has a value that (1) is a scalar, (2) is not identically equal to zero, (3) depends on the boundary of the region of spacetime over which the integral is extended, but (4) is independent of any changes made in the This integral has the value 8'lT for any closed, oriented, two-dimensional manifold with the to pology of a 2-sphere, no matter how badly dis torted; and the value 0 for any 2-torus, again no matter how rippled and twisted; and other equally specific values for other topologies. description of\") the boundary (three-dimensional) of such a simplex. Take one piece of this boundary and determine its boundary (two-dimensional). For one piece of this two-dimen sional boundary. verify that there is at exactly one other place, and no more, in the book keeping on the boundary of a boundary. another two-dimensional piece that cancels it (\"facelessness\" of the 3-boundary of the simplex). 382 15. BIANCHI IDENTITIES AND THE BOUNDARY OF A BOUNDARY Exercise 15.2. THE BEL-ROBINSON TENSOR [Bel (1958, 1959, 1962), Robinson (1959b), Sejnowski (1973); see also Pirani (1957) and Lichnerowicz (1962)]. Define the Bel-Robinson tensor by Show that in empty spacetime this tensor can be rewritten as Show also that in empty spacetime Ta{3ya:a = 0, T a{3ya is symmetric and traceless on all pairs of indices. (15.29) (l5.30a) (l5.30b) (15.30c) Discussion: It turns out that Einstein's \"canonical energy-momentum pseudotensor\" (\u00a720.3) for the gravitational field in empty spacetime has a second derivative which, in a Riemann normal coordinate system, is (l5.3la) Here T a{3ya is the completely symmetric Bel-Robinson tensor, and Sa{3ya is defined by (l5.3Ib) SI1{3ya appears in the empty-space covariant wave equation where L1 is a variant of the Lichnerowicz-de Rham wave operator [Lichnerowicz (1964)], when one rewrites this wave equation as (15.3Id) PART IV EINSTEIN'S GEOMETRIC THEORY OF GRAVITY Wherein the reader is seduced into marriage with the most elegant temptress of all-Geometrodynamics-and learns from her the magic potions and incantations that control the universe. --------....~ CHAPTER 16 EQUIVALENCE PRINCIPLE AND MEASUREMENT OF THE IIGRAVITATIONAL FIELD\" Rather than have one global frame with gravitational forces we have many local frames without gravitational forces. STEPH EN SCH UTZ (1966) \u00a716.1. OVERVIEW With the mathematics of curved spacetime now firmly in hand, one is tempted to rush headlong into a detailed study of Einstein's field equations. But such temptation must be resisted for a shon time more. To grasp the field equations fully, one must first understand how the classical laws of physics change, or do not change, in the transition from flat spacetime to curved (\u00a7\u00a716.2 and 16.3); and one must understand how the \"gravitational field\" (metric; covariant derivative; spacetime curvature;...) can be \"measured\" (\u00a7\u00a7 16.4 and 16.5). Purpose of this chapter \u00a716.2. THE LAWS OF PHYSICS IN CURVED SPACETIME Wherever one is and whenever one probes, one finds that then and there one can introduce a local inertial frame in which all test particles move along straight lines. Moreover, this local inertial frame is also locally Lorentz: in it the velocity of light has its standard value, and light rays, like world lines of test particles, are straight. But physics is more, and the analysis of physics demands more than an account solely of the motions of test particles and light rays. What happens", " to Maxwell's equations, the laws of hydrodynamics, the principles of atomic structure, and all the rest of physics under the influence of \"powerful gravitational fields\"? Einstein's equivalence principle Equivalence principle as tool to mesh nongravitational laws with gravity 386 16. EQUIVALENCE PRINCIPLE AND MEASUREMENT OF GRAVITATIONAL FIELD The answer is simple: in any and every local Lorentz frame, anywhere and anytime in the universe, all the (nongravitational) laws of physics must take on their familiar special-relativistic forms. Equivalently: there is no way, by experiments confined to infinitesimally small regions of spacetime, to distinguish one local Lorentz frame in one region of spacetime from any other local Lorentz frame in the same or any other region. This is Einstein's principle of equivalence in its strongest form-a principle that is compelling both philosophically and experimentally. (For the relevant experimental tests, see \u00a738.6.) The principle of equivalence has great power. With it one can generalize all the special relativistic laws of physics to curved spacetime. And the curvature need not be small. It may be as large as that in the center of a neutron star; as large as that at the edge of a black hole; arbitrarily large, in fact-or almost so. Only at the endpoint of gravitational collapse and in the initial instant of the \"big bang,\" i.e., only at \"singularities of spacetime,\" will there be a breakdown in the conditions needed for direct application of the equivalence principle (see \u00a7\u00a728.3, 34.6, 43.3, 43.4, and chapter 44). Everywhere else the equivalence principle acts as a tool to mesh all the nongravitational laws of physics with gravity. Example: Mesh the \"law of local energy-momentum conservation,\" V \u2022 T = 0, with gravity. Solution: (l) The law in flat spacetime, written in abstract geometric form, reads (2) Rewritten in a global Lorentz frame of flat spacetime, it reads V\u00b7T=O. TJJ.P,p = O. (l6.la) (l6.l b) (3) Application of equivalence principle gives same equation in local Lorentz frame of curved spacetime: Tit.,. = 0 at origin of local Lorentz frame. (l6.l c) Because the connection coefficients vanish at the origin of the local Lorentz frame, this can be rewritten as Tit.;. = 0 at origin of local Lorentz frame. (l6.ld) (4) The geometric law in curved spacetime, of which these are the local-Lorentz components, is and its component formulation in any reference frame reads V'T= 0; TJJ.P;p = O. (16.1 e) (16.lf) Compare the abstract geometric law (l6.le) in curved spacetime with the corre sponding law (l6.l a) in flat spacetime. They are identical! That this is not an accident one can readily see by tracing out the above four-step argument for any other law \u00a7 16.2. THE LAWS OF PHYSICS IN CURVED SPACETIME 387 of physics (e.g., Maxwell's equation V \u2022 F = 47TJ). The laws of physics, written in abstract geometric form, differ in no way whatsoever between curved spacetime and flat spacetime; this is guaranteed by, and in fact is a mere rewording of, the equiva lence principle. Compare the component version of the law V. T = 0, as written in an arbitrary frame in curved spacetime [equation (l6.l f)], with the component version in a global Lorentz frame of flat spacetime [equation (l6.lb)]. They differ in only one way: the comma (partial derivative; flat-spacetime gradient) is replaced by a semicolon (covariant derivative; curved-spacetime gradient). This procedure for rewriting the equations has universal application. The laws ofphysics, written in component form, change on passage from flat spacetime to curved spacetime by a mere replacement of all commas by semicolons (no change at all physically or geometrically; change due only to switch in reference frame from Lorentz to non-Lorentz!). This statement, like the nonchanging of abstract geometric laws, is nothing but a rephrased version of the equivalence principle. The transition in formalism from flat spacetime to curved spacetime is a trivial process when performed as outlined above. But it is nontrivial in its implications. It meshes gravity with all the laws of physics. Gravity enters in an essential way through the", " covariant derivative of curved spacetime, as one sees clearly in the following exercise. \"Com ma-goes-to-sem icolon\" rule Exercise 16.1. HYDRODYNAMICS IN A WEAK GRAVITATIONAL FIELD (a) In \u00a718.4 it will be shown that for a nearly Newtonian system, analyzed in an appropriate nearly global Lorentz coordinate system, the metric has the form EXERCISES ds 2 = -(I + 2tP) dt 2 + (I - 2tP)(dx2 + df + dz 2 ) (l6.2a) where tP is the Newtonian pOtential (-I ~ tP < 0). Consider a nearly Newtonian perfect fluid [stress-energy tensor raP = (p + p)uauP + pgap, (l6.2b) see Box 5.1 and \u00a75.1O] moving in such a spacetime with ordinary velocity Vi =dxildt ~ I. Show that the equations p.P;p = 0 for this system reduce to the familiar Newtonian law of mass conservation, and the Newtonian equation of motion for a fluid in a gravitational field: (l6.2c) avi dp Cit = -p axi'where dldt is the time derivative comoving with the matter. a a d -=-+v'dt - at Oxi'(l6.3a) (l6.3b) 388 16. EQUIVALENCE PRINCIPLE AND MEASUREMENT OF GRAVITATIONAL FIELD (b) Use these equations to calculate the pressure gradient in the Earth's atmosphere as a function of temperature and pressure. In the calculation, use the nonrelativistic relation p = nM J1.M' where nM is the number density of molecules and J1.M is the mean rest mass per molecule; use the ideal-gas equation of state p = nMkT (k = Boltzmann's constant); and use the spherically symmetric form, t/J = -M/r, for the Earth's Newtonian potential. If the pressure at sea level is l.01 X 106 dynes/cm2, what, approximately, is the pressure on top of Mount Everest (altitude 8,840 meters)? (Make a reasonable assumption about the temperature distribution of the atmosphere.) Exercise 16.2. WORLD LINES OF PHOTONS Show that in flat spacetime the conservation law for the 4-momentum of a freely moving photon can be written (l6.4a) According to the equivalence principle, this equation must be true also in curved spacetime. Show that this means photons move along null geodesics of curved spacetime with affine parameter>.. related to 4-momentum by p = d/d>.. (l6.4b) In exercise 18.6 this result will be used to calculate the deflection of light by the sun. \u00a716.3. FACTOR-ORDERING PROBLEMS IN THE EQUIVALENCE PRINCIPLE Factor-ordering problems and coupling to curvature On occasion in applying the equivalence principle to get from physics in flat space time to physics in curved spacetime one encounters \"factor-ordering problems\" analogous to those that beset the transition from classical mechanics to quantum mechanics.* Example: How is the equation (3.56) for the vector potential of electro dynamics to be translated into curved spacetime? If the flat-spacetime equation is written then its transition (\"comma goes to semicolon\") reads However, if the flat-spacetime equation is written with two of its partial derivatives interchanged (16.5) \u2022 For a discussion of quantum-mechanical factor-ordering problems, see, e.g., Merzbacher (1961), pp. 138-39 and 334-35; also Pauli (1934). \u00a7 16.3. FACTOR-ORDERING PROBLEMS IN EQUIVALENCE PRINCIPLE 389 then its translation reads which can be rewritten (Ricci tensor appears as result of interchanging covariant derivatives; see exercise 16.3.) Which equation is correct-(16.5) or (16.5')? This question is nontrivial, just as the analogous factor-ordering problems of quantum theory are nontrivial. For rules-of-th umb that resolve this and most factor-ordering problems, see Box 16.1. These rules tell one that (16.5') is correct and (16.5) is wrong (see Box 16.1 and \u00a722.4). (16.5') Exercise 16.3. NONCOMMUTATION OF COVARIANT DERIVATIVES Let B be a vector field and S be a second-rank tensor", " field. Show that EXERCISES BIJ.;aP = BIJ.;Pa + RlJ.ppaBP SW;ap = SW;pa + Ri'-pPaSPP + RPpPaSIJ.P. From equation (16.6a), show that (l6.6a) (l6.6b) (16.6c) [Hint for Track-l calculation: Work in a local Lorentz frame, where Fa py = 0 but Fapy,a #- 0; expand the lefthand side in_t.erms of Christoffel symbols and partial derivatives; and use equation (8.44) for the Riemann tensor. An alternative Track-2 calculation notices that Vp VaB is not linear in ea' and that BIJ.;ap are not its components; but, rather, that BIJ.;aP = VV B(wlJ., ea' e p). 4Third-rank tensor] (16.7) The calculation then proceeds as follows: (wlJ., Vp VaB) = (wlJ., Vp(e a'V B\u00bb = (wlJ., (Vpe a)' VB + ea' (Vp V B\u00bb = (wlJ.,FP apep'VB + VVB(...,ea,ep\u00bb = BIJ.;pFPaP + BIJ.;a{3' Consequently BIJ.:ap - BIJ.: Pa = (wlJ., [Vp, Va]B) - BlJ.jFPap - FP pa) (wlJ., V(vpe.- v.ep)B) = (wlJ., [Vp, Va]B) = (wlJ., ([Vp, Va] - V[ep.e)B) = (WI', ~il(ep. ea)B) = RlJ.ppaBP, in agreement with (l6.6a). Note: because of slight ambiguity in the abstract notation, one must think carefully about each step in the above calculation. Component notation, by contrast, is completely unambiguous.] (collfilll/ed on page 392) 390 16. EQUIVALENCE PRINCIPLE AND MEASUREMENT OF GRAVITATIONAL FIELD Box 16.1 FACTOR ORDERING AND CURVATURE COUPLING IN APPLICATIONS OF THE EQUIVALENCE PRINCIPLE The Problem In what order should derivatives be written when applying the \"comma-goes-to semicolon rule\"? Interchanging derivatives makes no difference in flat spacetime, but in curved spacetime it produces terms that couple to curvature, e.g., 2B\";[Y.81 B\";y.8 - B\";.8Y = R\"It.8yBIt for any vector field (see exercise 16.3). Hence, the problem can be restated: When must the comma-goes-to-semicolon rule be augmented by terms that couple to curvature? The Solution There is no solution in general, but in most cases the following types of mathematical and physical reasoning resolve the problem unambiguously. A. Mathematically, curvature terms almost always arise from the noncommutation of covariant derivatives. Consequently, one needs to worry about curvature terms in any equation that contains a double covariant derivative (e.g., -A\",1tIt + AIt,/ = laws 4'lTJ\"); or in any equation whose derivation from more fundamental involves double covariant derivatives (e.g. VuS = 0 in Example B(3) below). But one can ignore curvature coupling everywhere else (e.g., in Maxwell's first-order equations). B. Coupling to curvature can surely not occur without some physical reason. Therefore, if one applies the comma-goes-to-semicolon rule only to physically measurable quantities (e.g., to the electromagnetic field, but not to the vector potential), one can \"intuit\" whether coupling to curvature is likely. Examples: (1) Local energy-momentum conservation. A coupling to curvature in the equa tions T\".8;/3 = O-e.g., replacing them by T\".8;.8 = R\".BY8T.8Yu8-would not make sense at all. In a local inertial frame such terms as R\",BY8T.8Yu8 would be interpreted as forces produced at a single point by curvature. But it should not be possible to feel curvature except over finite regions (geodesic deviation, etc.)! Put differently, the second derivatives of the gravitational potential (metric) can hardly produce net forces at a point; they should only produce tidal forces! \u00a716.3. FACTOR-ORDERING PROBLEMS IN", " EQUIVALENCE PRINCIPLE 391 (2) Maxwell's equations for the electromagnetic field tensor. Here it would also be unnatural to introduce curvature terms. They would cause a breakdown in charge conservation, in the sense of termination of ~lectric and magnetic field lines at points where there is curvature but no charge. To maintain charge conservation, one omits curvature coupling when one translates Maxwell's equations (3.32) and (3.36) into curved spacetime: Moreover, one continues to regard Fit' as arising from a vector potential by the curved-spacetime translation of (3.54') These points granted, one can verify that the second of Maxwell's equations is automatically satisfied, and verify also that the first is satisfied if and only if (See \u00a722.4 for fuller discussion and derivation.) (3) Transport law for Earth's angular-momentum vector. If the Earth were in fiat spacetime, like any other isolated body it would parallel-transport its angu lar-momentum vector 5 along the straight world line of its center of mass, Vu 5 = 0 (\"conservation of angular momentum\"). When translating this transport law into curved spacetime (where the Earth actually resides!), can one ignore curvature coupling? No! Spacetime curvatures due to the moon and sun produce tidal gravitational forces in the Earth; and because the Earth has an equatorial bulge, the tidal forces produce a nonzero net torque about the Earth's center ofmass. (In Newtonian language: the piece of bulge nearest the Moon gets pulled with greater force, and hence greater torque, than the piece of bulge farthest from the Moon.) Thus, in curved spacetime one expects a transport law of the form VuS = (Riemann tensor) X (Earth's quadrupole moment). This curvature-coupling torque produces a precession of the Earth's rotation axis through a full circle in the plane of the ecliptic once every 26,000 years (\"general precession\"; \"precession of the equinoxes\"; discovered by Hip parchus about 150 B.C.). The precise form of the curvature-coupling term is derived in exercise 16.4. 392 16. EQUIVALENCE PRINCIPLE AND MEASUREMENT OF GRAVITATIONAL FIELD Exercise 16.4. PRECESSION OF THE EQUINOXES (a) Show that the transport law for the Earth's intrinsic angular momentum vector sa in curved spacetime is (16.8) Here dldT = U is 4-velocity along the Earth's world line; f{3~ is the Earth's \"reduced quadru pole moment\" (trace-free part of second moment of mass distribution), defined in the Earth's local Lorentz frame by fOfJ = fo; = 0, (16.9) and R~,.)'?; is the Riemann curvature produced at the Earth's location by the moon, sun, and planets. [Hint: Derive this result in the Earth's local Lorentz frame, ignoring the spacetime curvature due to the Earth. (In this essentially Newtonian situation, curvature components R'OkO due to the Earth, sun, moon, and planets superpose linearly; \"gravity too weak to be nonlinear\"). Integrate up the torque produced about the Earth's center of mass by tidal gravitational forces (\"geodesic deviation\"): acceleration at xi, relative to center of mass (x' = 0), ) produced by tidal gravitational forces but counterbalanced\u00b7 in part by Earth's internal stresses ( - (d 2 k X dr2 - ) geodesic deviation = _RkolOXi [see equation (1.8')]; force per unit volume due to thiS)k acceleration, relative to center of mass ( torque per unit volume relative) to center of mass ; ( total torque about center) of mass i (. k O\"k = \u00a3\u00b7\".xi(-pR 'i' x )' i... = [\u00a3'''k X '( -pRk..\u00b7 X f 0 0 I, 010 0\" )] d 3,i. Put this expression into a form involving fJI, equate it to dSr/dT, and then reexpress it in frame-independent, component notation. The result should be equation (16.8).] (b) Rewrite equation (16.8) in the Earth's local Lorentz frame, using the equation RiOko = a2tPlax; ax k for the components of Riemann in terms of the Newtonian gravitational potential. (Newto nian apprOXimation to Einstein theory. Track-2 readers have met this equation in Chapter 12; track-one readers will meet it in \u00a717.4.) (c) Calculate dSlldf", " using Newton's theory of gravity from the beginning. The answer should be identical to that obtained in part (b) using Einstein's theory. (d) Idealizing the moon.and sun as point masses, calculate the long-term effect of the spacetime curvatures that they produce upon the Earth's rotation axis. Use the result of part (b), together with moderately accurate numerical values for the relevant SOlar-system param eters. [Answer: The Earth's rotation axis precesses relative to the axes of its local Lorentz frame (\"precession of the equinoxes\"; \"general precession\"); the precession period is 26,000 years. The details of the calculation will be found in any textbook on celestial meChanics.] \u00a716.4. RODS AND CLOCKS USED TO MEASURE SPACETIME INTERVALS 393 \u00a716.4. THE RODS AND CLOCKS USED TO MEASURE SPACE AND TIME INTERVALS Turn attention now from the laws of physics in the presence of gravity to the nature of the rods and clocks that must be used for measuring the length and time intervals appearing in those laws. One need not-and indeed must not!-postulate that proper length s is measured by a certain type of rod (e.g., platinum meter stick), or that proper time T is measured by a certain type of clock (e.g., hydrogen-maser clock). Rather, one must ask the laws of physics themselves what types of rods and clocks will do the job. Put differently, one defines an \"ideal\" rod or clock to be one which measures proper length as given by ds = (g\"/3 dx\" dX/3)1/2 or proper time as given by dT = defined ( - g\"/3 dx\" dX/3)1/2 (the kind of clock to which one was led by physical arguments in \u00a71.5). One must then determine the accuracy to which a given rod or clock is How ideal are real clocks? ideal under given circumstances by using the laws of physics to analyze its behavior. As an obvious example, consider a pendulum clock. If it is placed at rest on the Earth's surface, if it is tiny enough that redshift effects from one end to the other and time dilation effects due to its swinging velocity are negligible, and if the accuracy one demands is small enough that time variations in the local gravitational acceleration due to Earth tides can be ignored, then the laws of physics report (Box 16.2) that the pendulum clock is \"ideal.\" However, in any other context (e.g., on a rocket journey to the moon), a pendulum clock should be far from ideal. Wildly changing accelerations, or no acceleration at all, will make it worthless! \"Ideal\" rods and clocks (1) pendUlum clocks Of greater interest are atomic and nuclear clocks of various sorts. Such a clock is analyzed most easily if it is freely falling. One can then study it in its local Lorentz rest frame, using the standard equations of quantum theory; and, of course, one will find that it measures proper time to within the'precision (.::Jt/t -- 10-9 to 10- 14) of the technology used in its construction. However, one rarely permits his atomic clock to fall freely. (The impact with the Earth's surface can be expensive!) Nevertheless, even when accelerated at \"1 g\" = 980 cm/sec2 on the Earth's surface, and even when accelerated at \"2 g\" in an airliner trying to avoid a midair collision (Box 16.3), an atomic clock-if built solidly-will still measure proper time dT = ( - g\"/3 dx\" dX/3)1/2 along its world line to nearly the same accuracy as if it were freely falling. To discover this one can perform an experiment. Alternatively, one can analyze the clock in its own \"proper reference frame\" (\u00a713.6), with Fermi-Walkertransported basis vectors, using the standard local Lorentz laws of quantum mechanics as adapted to accelerated frames (local Lorentz laws plus an \"inertial force,\" which can be treated as due to a potential with a uniform gradient. Of course, any clock has a \"breaking point,\" beyond which it will cease to function properly (Box 16.3). But that breaking point depends entirely on the construction of the clock-and not at all on any \"universal influence of acceleration on the march of time.\" Velocity produces a universal time dilation; acceleration does not. The aging of the human body is governed by the same electromagnetic and quantum-mechanical laws as govern the periodicities and level transitions in atoms and molecules. Consequently, aging, like atomic processes, is tied to proper time (continued on page 396) (2) atomic clocks (3) human clocks", " 394 16. EQUIVALENCE PRINCIPLE AND MEASUREMENT OF GRAVITATIONAL FIELD Box 16.2 PROOF THAT A PENDULUM CLOCK AT REST ON THE EARTH'S SURFACE IS IDEAL That is, a proof that it measures the interval dT ::::: ( - g\"/3 dx\" dx/3 )1/2. A. Constraint on the Pendulum It must be so small that it cannot couple to the spacetime curvature-Le., so small that the Earth's gravitational field looks uniform in its neighborhood-and that the velocity of its ball is totally negligible compared to the speed of light. B. Coordinate System and Metric (I) General coordinate system: because the Earth's field is nearly Newtonian, one can introduce the coordinates of \"linearized theory\" (\u00a7 18.4; one must take this on faith until one reaches that point) in which ds 2 ::::: - (l + 2tP) dt,2 + (I - 2tP)(dx'2 + dy'2 + dZ'2), z' where tP is the Newtonian potential. I I I I -----+---_ x' I I m (2) Put the origin of coordinates at the pendulum's equilibrium position, and orient the x',z'.plane so the pendulum swings in it. (3) Renormalize the coordinates so they measure proper length and proper time at the equilibrium position t ::::: [1 + 2tP(O)]l/2t', Then near the pendulum (inhomogeneities in the field neglected!) \u00a7 16.4. RODS AND CLOCKS USED TO MEASURE SPACETIME INTERVALS 395 W= W(O) + gz, g = \"acceleration of gravity,\" ds 2 = - (l + 2gz) dt 2 + (l - 2gz)(dx 2 + dy 2 + dz2). (1) (2) C. Analysis of Pendulum Motion (l) Put the total mass m of the pendulum in its ball (negligible mass in its rod). Let its rod have proper length I. (2) Calculate the 4-acceleration a = Vuu of the pendulum's ball in terms of d 2x\"/dt 2, using the velocity condition v\u00ab< 1 and dt/dT:=::: 1: aX = d 2x/dT2 + r X oo(dt/dT)2 = d 2x/dt 2 + r x a Z = d 2z/dT2 + r Zoo(dt/dT)2 = d 2z/dt 2 + r z oo = d 2x/dt 2, oo = d 2z/dt 2 + g. (3) (3) This 4-acceleration must be produced by the forces in the rod, and must be directed up the rod so that (for x ~ I so g ~ d 2z/dt 2) d 2x/dt 2 = aX = -(x/l)aZ = -(g/l)x. (4) Solve this differential equation to obtain x = X o cos (tViii>. (4) (5) (5) Thus conclude that the pendulum is periodic in t, which is proper time at the ball's equilibrium position (see equation 2). This means that the pendulum is an ideal clock when it is at rest on the Earth's surface. Note: The above analysis ignores the Earth's rotation; for an alternative analysis including rotation, one can perform a similar calculation at the origin of the pendu lum's \"proper reference frame\" [\u00a7 13.6; line element (13.71 )]. The answer is the same; but now \"g\" is a superposition of the \"gravitational acceleration,\" and the \"centrifu gal acceleration produced by Earth's rotation.\" 396 16. EQUIVALENCE PRINCIPLE AND MEASUREMENT OF GRAVITATIONAL FIELD Box 16.3 RESPONSE OF CLOCKS TO ACCELERATION AND TO TIDAL GRAVITATIONAL FORCES Consider an atomic clock with frequency stabilized by some atomic or molecular process-for exam ple, fixed by the \"umbrella vibrations\" of ammo nia molecules [see Feynman et. al. (1964)]. When subjected to sufficiently strong accelerations or tidal forces, such a clock will cease to measure proper time with its normal precision. Two types of effects could lead to such departures from \"ideality\": A. Influence ofthe acceleration or tidal force on the atomic process that provides the frequency stability. Example: If tidal forces are significant over dis tances of a few angstroms (e.g., near a spacetime \"singularity", "\" terminating gravitational collapse), then they can and will deform an ammonia mole cule and destroy the regularity of its umbrella vibrations, thereby making useless any ammonia atomic clock, no matter how constructed. Simi larly, if an ammonia molecule is subjected to ac celerations of magnitude comparable to its internal atomic (a - 1012\"g\" -. 1015 cml sec2), which change in times of the order of the \"umbrella\" vibration period, then it must cease to vibrate regularly, and any clock based on its vibra tions must fail. Such limits of principle on the ideality of a clock will vary from one atomic proc ess to another. However, they are far from being a limiting factor on clock construction in 1973. Much more important today is: accelerations which produces the periodic signal output, must be locked to the regulating atomic process in some way. The lock will be disturbed by moderate ac celerations. The toughest task for the manufac turer of aircraft clocks is to guarantee that precise locking will be maintained, even when the aircraft is maneuvering desperately to avoid collision with another aircraft or with a missile. In 1972 a solidly built rubidium clock will maintain its lock, with no apparent degradation of stability [.Jtlt -. 10-12(1 seclt)I/2 for 1 sec ~ t ~ 103 sec] under steady-state accelerations up to 50 \"g\" or more. But, because of the finite bandwidth of the lock loop (typically.Jp -. 20 to 50 Hz), sudden changes in acceleration will temporarily break the lock, degrading the clock stability to that of the unlocked crystal oscillator-for which an accelera tion a produces a change in frequency of about (all \"g\") X 10-9. But the lock to the rubidium standard is restored quickly (ot -. 1l.Jp), bring ing the clock back to its normal highly stable performance.* Tidal forces are so small in the solar system that the clock manufacturer can ignore them. However, a 1973 atomic clock, subjected to the tidal acceler ations near a spacetime singularity, should break the \"lock\" to its atomic process long before the tidal forces become strong enough to influence the atomic process itself. B. Influence ofthe acceleration or tidal force on the macroscopic structure of the clock-a structure dic tated by current technology. The crystal oscillator, \u00b7For this information on the response of rubidium clocks to acceleration, we thank H. P. Stratemeyer of General Radio Company, Concord, Massachusetts. Ideal rods and clocks constructed from geodesic world lines as governed by the metric-though, of course, it is also tied to other things, such as cigarette smoking. In principle, one can build ideal rods and clocks from the geodesic world lines of freely falling test particles and photons. (See Box 16.4.) In other words, spacetime has its own rods and clocks built into itself, even when matter and nongravitational fields are absent! \u00a716.4. RODS AND CLOCKS USED TO MEASURE SPACETIME INTERVALS 397 Box 16.4 IDEAL RODS AND CLOCKS BUILT FROM GEODESIC WORLD' LINES\u00b7 The Standard Interval. A specific timelike inter val-the interval between two particular neigh boring events (f and!i3-is chosen as the standard interval, and is assigned unit length. It is used to calibrate a huge set of geodesic clocks that pass through (f. Each geodesic clock is constructed and calibrated as follows: (l) A timelike geodesic (fe (path of freely falling particle) passes through (f. (2) A neighborin\"g world line, everywhere parallel to (fe (and thus not a geodesic), is constructed by the method of Schild's ladder (Box 10.2), which relies only on geodesics. (3) Light rays (null geodesics) bounce back and forth between these parallel world lines; each round trip constitutes one \"tick.\" (4) The proper time lapse, TO' between ticks is related to the interval (f!i3 by -1 ((f~3)2 = -(N1T O)(N2T O)' where N1 and N2 are the number of ticks be tween the events shown in the diagrams. [Proof see diagram at right.] to point e '13 Spacetime is filled with such geodesic clocks. Those that pass through (f are calibrated as above against the standard interval ([':13, and are used subsequently to calibrate all other clocks they meet. \u2022 Based on Marzke and Wheeler (\\ 964). {/ x In local Lorentz rest frame of geodesic clock: (N1TO)(N2TO) = (I - X)(I + x) =", " 12 _ x2 = _({/~tJ)2 398 16. EQUIVALENCE PRINCIPLE AND MEASUREMENT OF GRAVITATIONAL FIELD Box 16.4 (continued) Any interval 92 along the world line of a geodesic clock can be measured by the same method as was used in calibration. The interval 92 can be time like, spacelike, or null; its squared length in all three cases will be To achieve a precision of measurement good to one part in N, where N is some large number, take two precautions: (1) Demand that the intervals (fqJ and 92 be sufficiently small compared to the scale of cur vature of spacetime; or specifically, and R<AB)((fqJf ~ liN RlPQ)(92)2 ~ liN, where R(AB) and R<PQ) are the largest relevant components of the curvature tensor in the two regions in question. (2) Demand that the time scale, TO, of the geodesic clocks employed be small compared to (fqJ and 92 individually; thus, TO ~ (fqJIN, TO ~ '!l'21N. The Einstein principle that spacetime is de scribed by Riemannian geometry exposes itself to destruction by a \"thousand\" tests. Thus, from the fiducial interval, (fqJ, to the interval under meas urement, 92, there are a \"score\" of routes of in tercomparison, all of which must give the same value for the ratio 921(fqJ. Moreover, one can easily select out \"fifty\" intervals 92 to which the same kind of test can be applied. Such tests are not all items for the future. Some 5 X 109 years ago, electrons arrived by different routes at a common location, a given atom of iron in the core of the earth. This iron atom does not collapse. The Pauli principle of \u00a716.5. MEASUREMENT OF THE GRAVITATIONAL FIELD 399 exclusion keeps the electrons from all falling into the K-orbit. The Pauli principle would not apply if the electrons were not identical or nearly so. From this circumstance it would appear possible to draw an important conclusion (Marzke and Wheeler). With each electron is associated a standard length, its Compton wavelength, h/me. If these lengths had started different, or changed by different amounts along the different routes, and if the resulting difference in properties were as great as one part in -(5 X 109 yr) X (3 X 107 sec/yr) X (5 X 1018 rev/sec) - 1036, by now this difference would have shown up, the varied electrons would have fallen into the K orbit, and the earth would have collapsed, contrary to observation. The Marzke-Wheeler construction expresses an arbitrary small interval 9f2, anywhere in space time, in terms of the fiducial interval (N3, an inter val which itself may be taken for definiteness to be the \"geometrodynamic standard centimeter\" of \u00a7 1.5. This construction thus gives a vivid meaning to the idea of Riemannian geometry. The M-W construction makes no appeal what- soever to rods and clocks of atomic constitution. This circumstance is significant for the following reasons. The length of the usual platinum meter stick is some multiple,N1(h 2/me 2), of the Bohr atomic radius. Similarly, the wavelength of the Kr86 line is some multiple, N 2(he/ e2)(h2/me2), of a second basic length that depends on the atomic constants in quite a different way. Thus, if there is any change with time in the dimensionless ratio he/e2 = 137.038, one or the other or both of these atomic standards of length must get out of kilter with the geometrodynamic standard centimeter. In this case, general relativity says, \"Stick to the geo metrodynamic standard centimeter.\" Hermann Weyl at first thought that one could carry out the comparison of lengths by light rays alone, but H. A. Lorentz pointed out that one can dispense with the geodesics neither of test particles nor of light rays in the measurement process, the construction for which, however, neither Weyl nor Lorentz supplied [literature and Wheeler (1964)]. Ehlers, Pirani, and Schild (1972) have given a deeper analysis of the separate parts played in the measurement process by the affine connection, by the conformal part of the metric, and by the full metric. in Marzke. \u00a716.5. THE MEASUREMENT OF THE GRAVITATIONAL FIELD \"I know how to measure the electromagnetic field using test charges; what is the analogous procedure for measuring the gravitational field?\" This question- has,", " at the same time, many answers and none. It has no answers because nowhere has a precise definition of the term \"gravita tional field\" been given-nor will one be given. Many different mathematical entities are associated with gravitation: the metric, the Riemann curvature tensor, the Ricci curvature tensor, the connection coefficients, etc. Each of these plays an important role in gravitation theory, and none is so much more central than the others that it deserves the name \"gravitational field.\" Thus it is that throughout this book the terms \"gravitational field\" and \"gravity\" refer in a vague, collective sort of way to all of these entities. Another, equivalent term used for them is the \"geometry of spacetime.\" the covariant derivative, the curvature scalar, The many faces of gravity, and how one measures them 400 16. EQUIVALENCE PRINCIPLE AND MEASUREMENT OF GRAVITATIONAL FIELD To \"measure the gravitational field,\" then, means to \"explore experimentally various properties of the spacetime geometry.\" One makes different kinds of meas urements, depending on which geometric property of spacetime one is interested in. However, all such measurements must involve a scrutiny of the effects of the spacetime geometry (i.e., of gravity) on particles, on matter, or on nongravitational fields. For example, to \"measure\" the metric near a given event, one typically lays out a latticework of rods and clocks (local orthonormal frame, small enough that curva ture effects are negligible), and uses it to determine the interval between neighboring events. To measure the Riemann curvature tensor near an event, one typically studies the geodesic deviation (relative accelerations) that curvature' produces between the world lines of a variety of neighboring test particles; alternatively, one makes measurements with a \"gravity gradiometer\" (Box 16.5) if the curvature is static or slowly varying; or with a gravitational wave antenna (Chapter 37) if the curvature fluctuates rapidly. To study the large-scale curvature of spacetime, one examines large-scale effects of gravity, such as the orbits of planets and satellites, or the bending of ligh t by the sun's gravitational field. But whatever aspect of gravity one measures, and however one measures it, one is studying the geometry of spacetime. EXERCISE Exercise 16.5. GRAVITY GRADIOMETER The gravity gradiometer of Box 16.5 moves through curved spacetime along an accelerated world line. Calculate the amplitude and phase of oscillation of one arm of the gradiometer relative to the other. [Hint: Perform the calculation in the gradiometer's \"proper reference frame\" (\u00a713.6), with Fermi-Walker-transported basis vectors. Use, as the equation for the relative angular acceleration of the two arms, 2m!2(ii + a/T + W2a ) = (Drivin.g torque produced by), o 0 Riemann curvature where 2m!2 = (moment of inertia of one arm), a = (angular displacement of one arm from equilibrium), -I + 2a = (angular separation of the two arms), 2m! 2w02 = (torsional spring constant), W o = (angular frequency of free vibrations), TO =(decay time for free vibrations to damp out due to internal frictional forces). If ( is the vector from the center of mass of the gradiometer to mass 1, then one has (continued on page 403) \u00a7 16.5. MEASUREMENT OF THE GRAVITATIONAL FIELD 401 Box 16.5 GRAVITY GRADIOMETER FOR MEASURING THE RIEMANN CURVATURE OF SPACETIME This gravity gradiometer was designed and built by Robert M. Forward and his colleagues at Hughes Research Laboratories, Malibu, California. It measures the Riemann curvature of spacetime produced by nearby masses. By flying a more advanced version of such a gradiometer in an airplane above the Earth's surface, one should be able to measure subsurface mass variations due to varying geological structure. In an Earth-orbiting satellite, such a gradiometer could measure the gravitational multipole moments of the Earth. Technical details of the gradiometer are spelled out in the papers of Forward (1972), and Bell, Forward, and Williams (1970). The principles of its operation are outlined below. The gradiometer consists of two orthogonal arms with masses m on their ends, connected at their centers by a torsional spring. When the arms are twisted out of orthogonal alignment, they oscillate. A piezoelectric strain transducer is used to meas ure the oscillation amplitude. m", " m m m 402, 6. EQUIVALENCE PRINCIPLE AND MEASUREMENT OF GRAVITATIONAL FIELD Box 16.5 (continued) When placed near an external mass, M, the gra diometer experiences a torque: because of the gra dient in the gravitational field of M (i.e., because of the spacetime curvature produced by M), the Newtonian forces Fl and F2 are greater than F3 and F4 ; so a net torque pulls masses 1 and 2 to ward each other, and 3 and 4 toward each other. [Note: the forces Fl, F2, F3, F4 depend on whether the gradiometer is in free fall (geodesic motion; Vuu = 0) or is moving on a'n accelerated world line. But the net torque is unaffected by accelera tion; equal Newtonian forces on all four masses, with zero net torque.] acceleration produces Net torque 2 3 wI = 0 wI = 17/4 wI = 17/2 When in operation the gradiometer rotates with angular velocity w about its center. As it rotates, the torques on its arms oscillate: at wt = 0 net torque pushes 1 and 2 toward each other; at wt = '17/4 net torque is zero; at wt = '17/2 net torque pushes 1 and 2 away from each other. The angular frequency of the oscillating torque is 2w. If 2w is set equal to Wo = (natural oscillation frequency of the arms), the oscillating torque drives the arms into resonant oscillation. The re sulting oscillation amplitude, in the 1970 prototype of the gradiometer, was easily detectable for grav ity gradients (Riemann curvatures) of magnitude 2(mass of earth) ] ~ 0.0002 (radius of earth)3 [ -- 1 X 10-30 cm- 2 --.01 g/cm3 ~Riemann curvature produced by a tWO_kilometer] high mountain, idealized as a two-kilometer high cube, at a distance of 15 kilometers. (Neglected in this idealization are isostacy and any lowering of density of Earth's crust in regions of mountain uplift.) For a mathematical analysis of the gradiometer, see exercise 16.5. \u00a716.5. MEASUREMENT OF THE GRAVITATIONAL FIELD 403 torque acting on mass 1) rela~ve to center of gradlOmeter (. i =(r,TcE,( -mRTcOloEl)' The torque on mass 4 is identical to this (replace ( by -n, so the total torque on arm 1-4 is twice this. The components Rf<{)to of Riemann can be regarded as components of a 3 X 3 symmetric matrix. By appropriate orientation of the reference frame's spatial axes (orientation along \"principal axes\" of RTcot{)), one can make RTco!o diagonal at some initial moment of time R;o;o ::p 0, R;,o;,o ::P 0, R zozo ::P 0, all others vanish. Assume that Riemann changes sufficiently slowly along the gradiometer's world line that throughout the experiment R,oTco remains diagonal and constant. For simplicity, place the gradiometer in the X, j-plane, so it rotates about the i axis with angular velocity w :::::!wo: Angle of arm 1:4) = wt. ( relative to X axiS Show that the resultant equation of oscillation is and that the steady-state oscillations are a -1m - (1 ] -2 (R;o;o - R;,o;,o) 2wo(wo - 2w + i/2TO) Thus, for fixed w (e.g., 2w = wo), by measuring the amplitude and phase of the oscillations, one can learn the magnitude and sign of R;o;{) - R;,o;,o. The other differences, R yoyo R zozo and R.ozo - R;o;o can be measured by placing the gradiometer's rotation axis along the x and j axes, respectively.] ei2wt \u2022 ~\".f,;1 j,; CHAPTER 17 HOW MASS-ENERGY GENERATES CURVATURE the simplest laws which such a metric can satisfy, The physical world is represented as a four-dimensional continuum. If in this I adopt a Riemannian metric, and look for I arrive at the relativistic gravitation theory of empty space. If I adopt in this space a vector field, or the antisymmetrical tensor field derived from it, and if I look for the simplest laws which such a field I arrive at the Maxwell equations for free space", ". can satisfy,... at any given moment, out of all conceivable constructions, a single one has always proved itself absolutely superior to all the rest... ALBERT EINSTEIN (1934, p. 18) \u00a717.1. AUTOMATIC CONSERVATION OF THE SOURCE AS THE CENTRAL IDEA IN THE FORMULATION OF THE FIELD EQUATION This section derives the \u00b7\u00b7Einstein field equation\" Turn now from the response of matter to geometry (motion of a neutral test particle on a geodesic; \"comma-goes-to-semicolon rule\" for the dynamics of matter and fields), and analyze the response of geometry to matter. Mass is the source of gravity. The density of mass-energy as measured by any observer with 4-velocity u is Therefore the stress-energy tensor T is the frame-independent \"geometric object\" that must act as the source of gravity. (17.1) Equation describing how matter generates gravity must have form G = KT, where T is stress-energy tensor Properties that the tensor G must have \u00a717.1. AUTOMATIC CONSERVATION OF SOURCE AS CENTRAL IDEA 405 This source, this geometric object, is not an arbitrary symmetric tensor. It must have zero divergence V\u00b7T=O, (172) because only so can the law of conservation of momentum-energy be upheld. Place this source, T, on the righthand side of the equation for the generation of gravity. On the lefthand side will stand a geometric object that characterizes gravity. That object, like T, must be a symmetric, divergence-free tensor; and if it is to characterize gravity, it must be built out of the geometry of spacetime and nothing but that geometry. Give this object the name \"Einstein tensor\" and denote it by G, so that the equation for the generation of gravity reads G = KT (17.3) t[proportionality factor;] to be evaluated later (Do not assume that G is the same Einstein tensor as was encountered in Chapters 8, 13, 14, and 15; that will be proved below!) The vanishing of the divergence V \u2022 G is not to be regarded as a consequence of V \u2022 T = O. Rather, the obedience of all matter and fields to the conservation law V \u2022 T = 0 is to be regarded (1) as a consequence of the way [equation (17.3)] they are wired into the geometry of spacetime, and therefore (2) as required and enforced by an automatic conservation law, or identity, that holds for any smooth Riemannian spacetime whatsoever, physical or not: V\u00b7 G = O. (See Chapter 15 for a fuller discussion and \u00a7 17.2 below for a fuller justification.) Accordingly, look for a symme tric tensor G that is an \"automatically conserved measure of the curvature of spacetime\" in the following sense: (1) G vanishes when spacetime is flat. (2) G is constructed from the Riemann curvature tensor and the metric, and from nothing else. (3) G is distinguished from other tensors which can be built from Riemann and 9 by the demands (i) that it be linear in Riemann, as befits any natural measure of curvature; (ii) that, like T, it b~ symmetric and of second rank; and (iii) that it have an automatically vanishing divergence, (17.4) Apart from a multiplicative constant, there is only one tensor (exercise 17.1) that satisfies these requirements of being an automatically conserved, second-rank tensor, linear in the curvature, and of vanishing when spacetime is flat. It is the Einstein curvature tensor, G, expressed in Chapter 8 in terms of the Ricci curvature tensor: Proof that G must be the Einstein curvature tensor of Chapter 8 (17.5) Evaluation of K (in G = KT) by comparing with Newtonian theory of gravity 406 17. HOW MASS-ENERGY GENERATES CURVATURE This quantity was given vivid meaning in Chapter 15 as the \"moment of rotation of the curvature\" or, more simply, the \"moment of rotation,\" constructed by taking the double-dual of the Riemann curvature tensor, and then contracting this double dual, 6 = \u2022Riemann\u00b7 (17.6a) (17.6b) In Chapter 15 the vanishing of V. G was shown to follow as a consequence of the elementary principle of topology that \"the boundary of a boundary is zero.\" To evaluate the proportionality constant K in the \"Einstein field equation\" G = KT, one can compare with the well-tested Newtonian theory of gravity. To facilitate the comparison,", " examine the relative acceleration (geodesic deviation) of particles that fall down a pipe inserted into an idealized Earth of uniform density p (Figure 1.12). According to Newton, the relative acceleration is governed by the density; according to Einstein, it is governed by the Riemann curvature of spacetime. Direct comparison of the Newtonian and Einstein predictions using Newtonian coordinates (where g /lV :::::: 11 /lv) reveals the relation Roo = RQoQO = 4'7Tp. (17.7) (See \u00a7 1.7 for details of the derivation; see Chapter 12 for extensive discussion of Newtonian gravity using this equation.) When applied to the Earth's interior, the Einstein field equation G = KT must thus reduce to Roo = 4'7Tp. In component form, the Einstein field equation reads Its trace reads In consequence, it predicts -R = R - 2R = KT Roo =\"2gooR + KToo =\"2 K(2Too - ~T) 1 1 1 = \"2 K[2Too + (T 0 + 1\";)] o \u00b7 -1 1 = \"2K(Too + 1\";),. which red uces to (17.8) \u00a717.1. AUTOMATIC CONSERVATION OF SOURCE AS CENTRAL IDEA 407 when one recalls that for the Earth-as for any nearly Newtonian system-the stresses T;k are very small compared to the density of mass-energy Too = p: IT I --l!!..- __ Too pressure density. __ - dp dp -- (velocity of sound)2 ~ 1.. The equation Roo = 4'ITp (derived by comparing relative accelerations in the Newton and Einstein theories) and the equation Roo =!Kp (derived directly from the Einstein field equation) can agree only if the proportionality constant K is 8'IT. Thus, the Einstein field equation, describing the generation of curvature by mass-energy, must read G = 8'ITT (17.9) The lefthand side (\"curvature\") has units cm-2, since a curvature tensor is a linear machine into which one inserts a displacement (units: cm) and from which one gets a relative acceleration (units: cm/sec2 - cm/cm2 - cm- I ). The right-hand side also has dimensions cm-2, since it is a linear machine into which one inserts 4-velocity (dimensionless) and from which one gets mass density [units: g/cm3 -- cm/cm3 - cm-2; recall from equation (1.12) and Box 1.8 that Ig = (1 g) X (G/c 2) = (1g) X (0.742 X 10-28 cm/g) = 0.742 X 10-28 cm]. This concludes the simplest derivation of Einstein's field equation that has come to hand, and establishes its correspondence with the Newtonian theory of gravity under Newtonian conditions. That correspondence had to be worked out to deter mine the factor K = 8'IT on the righthand side of (17.9). Apart from this factor, the central point in the derivation was the demand for, and the existence of, a unique tensorial measure of curvature G with an identically vanishing divergence. Result: \"Einstein field equation\" G = 8'ITT Exercise 17.1. UNIQUENESS OF THE EINSTEIN TENSOR (a) Show that the most general second-rank, symmetric tensor constructable from Riemann and g, and linear in Riemann, is EXERCISES aR\",[3 + bRg\",[3 + Ag\",[3 = aRIL\"'IL[3 + bRILVp.Pg\",[3 + Ag\",[3' (17.10) where a, b, and A are constants. (b) Show that this tensor has an automatically vanishing divergence if and only if b = -~a. (c) Show that, in addition, this tensor vanishes in flat spacetime, if and only if A = O-i.e., if and only if it is a multiple of the Einstein tensor Ga [3 = Raf3 to prove that V' G =0; assume it as a result from Chapter 13.)!Rga [3' (Do not bother Exercise 17.2. NO TENSOR CONSTRUCTABLE FROM FIRST DERIVATIVES OF METRIC Show that there exists no tensor with components constructable from the ten metric coeffi cients g\"'[3 and their 40 first derivatives ga[3,IL-except the metric tensor g, and products of it with itself; e.g., g ~ g. [Hint: Assume there exists some other", " such tensor, and examine its hypothesized components in a local inertial frame.] 408 17. HOW MASS-ENERGY GENERATES CURVATURE Exercise 17.3. RIEMANN AS THE ONLY TENSOR CONSTRUCTABLE FROM, AND LINEAR IN SECOND DERIVATIVES OF METRIC Show that (1) Riemann, (2) g, and (3) tensors (e.g., Ricci) formed from Riemann and g but linear in Riemann, are the only tensors that (a) are constructable from the teI! gap, _. the 40 gnP.!,' and the 100 gaP.!,'\" and (b) are linear in the gap,!'p' [Hint: Assume there exists some other such tensor, and examine its hypothesized components in an orthonormal, Riemann-normal coordinate system. Use equations (11.30) to (11.32).] Exercise 17.4. UNIQUENESS OF THE EINSTEIN TENSOR (a) Show that the Einstein tensor, Gap = Rap!Rgap, is the only second-rank, symmetric tensor that (1) has components constructable solely from gap, gaP.!\" gaP.!'p: (2) has components linear in gaP.!'.': (3) has an automatically vanishing divergence, V\u00b7 G = 0; and (4) vanishes in flat spacetime. This provides added motivation for choosing the Einstein tensor as the left side of the field equation G = 817T. (b) Show that, when condition (4) is dropped, the most general tensor is G + Ag, where A is a constant. (See \u00a717.3 for the significance of this.) \u00a717.2. AUTOMATIC CONSERVATION OF THE SOURCE: A DYNAMIC NECESSITY The answer G = 8'7TT is now on hand; but what is the question? An equation has been derived that connects the Einstein-Cartan \"moment of rotation\" G with the stress-energy tensor T, but what is the purpose for which one wants this equation in the first place? If geometry tells matter how to move, and matter tells geometry how to curve, does one not have in one's hands a Gordian knot? And how then can one ever untie it? The story is no different in character for the dynamics of geometry than it is for other branches of dynamics. To predict the future, one must first specify, on an \"initial\" hypersurface of \"simultaneity,\" the position and velocity of every particle, and the amplitude and time-rate of change of every field that obeys a second-order wave equation. One can then evolve the particles and fields forward in time by means of their dynamic equations. Similarly, one must give information about the geometry and its first time-rate of change on the \"initial\" hypersurface if the Einstein field equation is to be able to predict completely and deterministically the future time development of the entire system, particles plus fields plus geometry. (See Chap ter 21 for details.) If a prediction is to be made of the geometry, how much information has to be supplied for this purpose? The geometry of spacetime is described by the metric that is, by the ten functions gafj of location ':f in spacetime. It might then seem that ten functions must be predicted; and, if so, that one would need for the task ten Einstein field equation governs the evolution of spacetime geometry \u00a717.3. COSMOLOGICAL CONSTANT 409 equations. Not so. Introduce a new set of coordinates xli by way of the coordinate transformations and find the same spacetime geometry, with all the same bumps, rills, and waves, described by an entirely new set of metric coefficients gap('3'). It would transgress the power as well as the duty of Einstein's \"geometrodynamic law\" G = 8'7TT if, out of the appropriate data on the \"initial-value hypersurface,\" it were to provide a way to calculate, on out into the future, values for all ten functions ga/3(':f). To predict all ten functions would presuppose a choice of the coordinates; and to make a choice among coordinate systems is exactly what the geometrodynamic law cannot and must not have the power to do. That choice resides of necessity in the man who studies the geometry, not in the Nature that makes the geometry. The geometry in and by itself, like an automobile fender in and by itself, is free of coordinates. The coordinates are the work of man. It follows that the ten components Ga/3 = 8'7TTa/3 of the field equation must not determine completely and uniquely all ten components gil' of the metric. On the contrary, Ga/3 =", " 8'7TTa/3 must place only six independent constraints on the ten g/l'(':f), leaving four arbitrary functions to be adjusted by man's specialization of the four coordinate functions xa(':f). How can this be so? How can the ten equations Ga/3 = 8'7TTa/3 be in reality only six? Answer: by virtue of the \"automatic conservation of the source.\" More specifi cally, the identity Ga/3;/3 = 0 guarantees that the ten equations Ga/3 = 8'7TTa/3 contain the four \"conservation laws\" Ta/3;/3 = O. These four conservation laws-along with other equations-govern the evolution of the source. They do not constrain in any way the evolution of the geometry. The geometry is constrained only by the six remaining, independent equations in Ga/3 = 8'7TTa/3' When viewed in this way, the \"automatic conservation of the source\" is not merely a philosophically attractive principle. It is, in fact, an absolute dynamic necessity. Without \"automatic conservation of the source,\" the ten Ga/3 = 8'7TTa/3 would place ten constraints on the ten ga/3' thus fixing the coordinate system as well as the geometry. With \"automatic conservation,\" the ten Ga/3 = 8'7TTa/3 place four con straints (local conservation of energy and momentum) on the source, and six con straints on the ten ga/3' leaving four of the ga/3 to be adjusted by adjustment of the coordinate system. G = 8'ITT must determine only six metric components; the other four are adjustable by changes of.coordinates G = 817T leaves four components of metric free because it satisfies four identities o = V. G = 817V' T (\"automatic conser-vation of source\") \u00a717.3. COSMOLOGICAL CONSTANT In 1915, when Einstein developed his general relativity theory, the permanence of the universe was a fixed item of belief in Western philosophy. \"The heavens endure from everlasting to everlasting.\" Thus, it disturbed Einstein greatly to discover (Chapter 27) that his geometrodynamic law G = 8'7TT predicts a nonpermanent universe; a dynamic universe; a universe that originated in a \"big-bang\" explosion, Einstein's motivation for introducing a cosmological constant 410 17. HOW MASS-ENERGY GENERATES CURVATURE or will be destroyed eventually by contraction to infinite density, or both. Faced with this contradiction between his theory and the firm philosophical beliefs of the day, Einstein weakened; he modified his theory. The only conceivable modification that does not alter vastly the structure of the theory is to change the lefthand side of the geometrodynamic law G = 8'7TT. Recall that the lefthand side is forced to be the Einstein tensor, Ga /3 = Ra /3!Rga /3' by three assumptions: (1) G vanishes when spacetime is flat. (2) G is constructed from the Riemann curvature tensor and the metric and nothing else. (3) G is distinguished from other tensors that can be built from Riemann and 9 by the demands (1) that it be linear in Riemann, as befits any natural measure of curvature; (2) that, like T, it be symmetric and of second rank; and (3) that it have an automatically vanishing divergence, V. G = 0, Denote a new, modified lefthand side by \"G\", with quotation marks to avoid confus.ion with the standard Einstein tensor. To abandon V. \"G\" = 0 is impossible on dynamic grounds (see \u00a7 17.2). To change the symmetry or rank of \"G\" is impossi ble on mathematical grounds, since \"G\" must be equated to T. To let \"G\" be nonlinear in Riemann would vastly complicate the theory. To construct \"G\" from anything except Riemann and 9 would make \"G\" no longer a measure of spacetime geometry and would thus violate the spirit of the theory. After much anguish, one concludes that the assumption which one might drop with least damage to the beauty and spirit of the theory is assumption (1), that \"G\" _vanish when spacetime is flat. But even dropping this assumption is painful: (1) although \"G\" might still be in some sense a measure of geometry, it can no longer be a measure of curvature; and (2) flat, empty spacetime will no longer be compatible with the geometrodynamic law (G \u00a5- 0 in flat, empty space, where T = 0). Nevertheless, these consequences were less painful to Einstein than a dynamic universe. The only tensor that satisfies conditions (2) and (3) [", "with (1) abandoned] is the Einstein tensor plus a multiple of the metric: Einstein's field equation with the cosmological constant Why Einstein abandoned the cosmological constant (exercise 17.1; see also exercise 17.4). Thus was Einstein (1917) led to his modified field equation G + Ag = 8'7TT. (17.11) The constant A he called,the \"cosmological constant\"; it has dimensions cm-2\u2022 The modified field equation, by contrast with the original, admits a static, un changing universe as one particular solution (see Box 27.5). For this reason, Einstein in 1917 was inclined to place his faith in the modified equation. But thirteen years later Hubble discovered t~e expansion of the universe. No longer was the cosmo logical constant necessary. Einstein, calling the cosmological constant \"the biggest \u00a717.3. COSMOLOGICAL CONSTANT 411 blunder of my life,\" abandoned it and returned to his original geometrodynamic law, G = 8'7TT [Einstein (1970)]. A great mistake A was indeed!-not least because, had Einstein stuck by his original equation, he could have claimed the expansion of the universe as the most triumphant prediction of his theory of gravity. A mischievous genie, once let out of a bottle, is not easily reconfined. Many workers in cosmology are unwilling to abandon the cosmological constant. They insist that it be abandoned only after cosmological observations reveal it to be negligibly small. As a modern-day motivation for retaining the cosmological constant, one sometimes rewrites the modified field equation in the form G = 8'7T[T + TlVAC)], r<VAC) = -(A/8'7T)g (17.l2a) (17.l2b) and interprets r<VAC) as a stress-energy tensor associated with the vacuum. This viewpoint speculates [Zel'dovich (1967)] that the vacuum polarization of quantum field theory endows the vacuum with the nonzero stress-energy tensor (17.12 b), which is completely unobservable except by its gravitational effects. Unfortunately, today's quantum field theory is too primitive to allow a calculation of r<VAC) from first principles. (See, however, exercise 17.5.) The mass-energy density that the cosmological constant attributes to the vacuum is (17.13) A modern-day motivation for the cosmological constant: vacuum polarization. If A \u00a5- 0, it must at least be so small that plVAC) has negligible gravitational effects [Ip(VAC)1 < p(MATTER>] wher-ever Newton's theory of gravity gives a successful account ofobservations. The systems oflowest density to which one applies Newtonian theory with some (though not great) success are small clusters of galaxies. Hence, one can place the limit Observational limit on the cosmological constant on the value of the cosmological constant. Evidently, even if A \u00a5- 0, A is so small that it is totally unimportant on the scale of a galaxy or a star or a planet or a man or an atom. Consequently it is reasonable to stick with Einstein's original geometrodynamic law (G = 8'7TT; A = 0) everywhere, except occasionally when discussing cosmology (Chapters 27-30). Why one ignores the cosmological constant everywhere except in cosmology (17.14) Exercise 17.5. MAGNITUDE OF COSMOLOGICAL CONSTANT (a) What is the order of magnitude of the influence of the cosmological constant on the celestial mechanics of the solar system if A - 10-57 cm-2? EXERCISE 1~..\u2022.\u2022.'.-..<\u2022\u2022.\u2022. i; \u2022\u2022 ~,. 412 17. HOW MASS-ENERGY GENERATES CURVATURE (b) Show that the mass-energy density of the vacuum plYAe) = A/81':\" - 10-29 g/cm3, corresponding to the maximum possible value of A, agrees in very rough magnitude with rest mass of an elementary particle 1 C 0 partlcle) ( ompton wave engt h f. \" 3 X (gravitatIOnal fine-structure constant) m m 6 = (tl/m)3 I i =h1 m 2 [Zel'dovich (1967, 1968)]. This numerology is suggestive, but has not led to any believable derivation of a stress-energy tensor for the vacuum. \u00a717.4. THE NEWTONIAN LIMIT Just as quantum mechanics reduces to classical mechanics in the \"correspondence limit\" of large actions, I ~ 11, so general relativity reduces to Newtonian theory in the \"correspondence limit", "\" of weak gravity and low velocities. (On \"correspondence limits,\" see B6x 17.1.) This section elucidates, in some mathematical detail, the correspondence between general relativity and Newtonian theory. It begins with \"passive\" aspects of gravitation (response of matter to gravity) and then turns to \"active\" aspects (generation of gravity by matter). Consider an isolated system-e.g., the solar system-in which Newtonian theory is highly accurate. In order that special relativistic effects not be noticeable, all Box 17.1 CORRESPONDENCE PRINCIPLES A. General Remarks and Specific Examples through which simpler 1. As physics develops and expands, its unity is maintained by a network of correspondence principles, theories maintain their vitality by links to more sophis ticated but more accurate ones. a. Physical optics, with all the new diffraction and interference phenomena for which it accounted, nevertheless also had to account, and did account, for the old, elementary, geometric optics of mirrors and lenses. Geo metric optics is recovered from physical optics in the mathematical \"correspondence principle limit\" in which the wavelength is made indefinitely small in comparison with all other relevant dimensions of the physical system. b. Newtonian mechanics is recovered from the mechanics of special relativity in the mathe matical \"correspondence principle limit\" in which all relevant velocities are negligibly small compared to the speed of light. c. Thermodynamics is recovered from its suc cessor theory, statistical mechanics, in the principle mathematical limit\" in which so many particles are taken that fluctuations in pressure, into account \"correspondence \u00a717.4. THE NEWTONIAN LIMIT 413 particle number, and other physical quanti ties are negligible compared to the average values of these parameters of the system. d. Classical mechanics is recovered from quan tum mechanics in the \"correspondence prin ciple limit\" in which the quantum numbers of the quantum states in question are so large, or the quantities of action that come into play are so great compared to 11, that wave and diffraction phenomena make neg ligible changes in the predictions of standard deterministic classical mechanics. Niels Bohr formulated and took advantage of this cor any respondence principle proper quantum theory existed. He used it to predict approximate values of atomic en ergy levels and of intensities of spectral lines. He also expounded it as a guide to all physi cists, first in searching for a proper version of the quantum theory, and then in elucidat ing the content of this theory after it was found. even before 2. In all these examples and others, the newer, more sophisticated theory is \"better\" than its predecessor because it gives a good description of a more extended domain of physics, or a more accurate description of the same domain, or both. the older 3. The correspondence between the newer theory and its predecessor (a) gives one the power to recover theory from the newer; (b) can be exhibited by straightforward mathe matics; and (c), according to the historical rec ord, often guided the development of the newer theory. B. Correspondence Structure of General Relativity 1. Einstein's theory of gravity has as distinct limit ing cases (a) special relativity; (b) the \"linear- ized theory of gravity\"; (c) Newton's theory of gravity; and (d) the post-Newtonian theory of gravity. Thus, it has a particularly rich corre spondence structure. a. Correspondence with special relativity: Gen eral relativity has two distinct kinds of corre spondence with special relativity. The first is the limit of vanishing gravitational field everywhere (vanishing curvature); in this inertial limit one can introduce a global frame, set g/lV = 71/l v, and recover completely and precisely the theory of special relativity. The second is local rather than global; it is the demand (\"correspondence principle\"; \"equivalence principle\") that in a local iner tial frame all the laws of physics take on their special relativistic forms. As was seen in Chapter 16, this puts no restrictions on the metric (except that g/lV = 71/l v and g/lv,a = 0 in local inertial frames); but it places severe constraints on the behavior of matter and fields in the presence of gravity. b. Correspondence with Newtonian theory: In the limit of weak gravitational fields, low velocities, and small pressures, general rela tivity reduces to Newton's theory of gravity. The correspondence structure is explored mathematically in the text of \u00a7 17.4. c. Correspondence with post-Newtonian theory: When Newtonian theory is nearly valid, but \"first-order relativistic corrections\" might be important, one often uses the \"post-Newton ian theory of gravity.\" Chapter 39 expounds the post-Newtonian theory and its corre spond", "ence with both general relativity and Newtonian theory. d. Correspondence with linearized theory: In the limit of weak gravitational fields, but possi 1, bly large velocities and pressures (v T;k - Too) general relativity reduces to the \"linearized theory of gravity\". This corre spondence is explored in Chapter 18. Conditions which a system must satisfy for Newton's theory of gravity to be accurate \"Newtonian coordinates\" defined 414 17. HOW MASS-ENERGY GENERATES CURVATURE velocities in the system, relative to its center of mass and also relative to the Newto nian coordinates, must be small compared to the speed of light v ~ 1. (17.15a) As a particle falls from the outer region of the system to the inner region, grav~ty Im<Pl max. [Here <P < 0 is Newton's gravita accelerates it to a kinetic energy 2mv2 tional potential, so normalized that <P((0) = 0.] The resulting velocity will be small only if (17.15b) Internal stresses in the system also produce motion-e.g., sound waves. Such waves have characteristic velocities of the order of ITii/Tooll/2-for example, the speed of sound in a perfect fluid is v = (dp/dp)1/2 _ (p/p)1/2 _ITif/Tooll/2. In order that these velocities be small compared to the speed of light, all stresses must be small compared to the density of mass-energy (17.15c) When, and only when conditions (17.15) hold, one can expect Newtonian theory to describe accurately the system being studied. Correspondence of general relativity with Newtonian theory for gravity in a passive role then demands that the geodesic world lines of freely falling particles reduce to the Newtonian world lines (17.16) Moreover, they must reduce to this form in any relativistic coordinate system where the source and test particles have low velocities v ~ 1, and where coordinate lengths and times agree very nearly with the lengths and times of the Newtonian coor dinates-which in turn are proper lengths and times as measured by rods and clocks. Thus, the relevant coordinates (called \"Galilean\" or \"Newtonian\" coordinates) are ones in which (17.17) (weak gravitational field; nearly inertial coordinates; coordinate system, the geodesic world lines of test particles have the form low velocities). In such a dx a dx{3 a{3 dT dT d 2x i dT2 __ Fi = -Fioo = -FiOO 1= \"2 hOO,i - (since dt/dT::::: 1 when IhjLpl ~ 1 and Ivil ~ 1) (geodesic equation) (since dt/dT::::: 1 and Idxf/dTI ~ 1) (since gjLP ::::: 1'JjLp) hOi,o (equation for Fa{3Y in terms of ga{3,y) 1=-hooi'2 all velocities small compared to c implies time ) de~ivatives small compared to space derivatives -l.e., ha{3,o - Vha{3,i (. \u00a717.4. THE NEWTONIAN LIMIT 415 These geodesic world Iffies do, indeed, reduce to those of Newtonian theory [equation (17.16)] if one makes the identification (17.18) Together with the boundary conditions <P(r = 00) = 0 and hp.p(r = 00) = 0 (coor dinates Lorentz far from the source), this identification implies hoo = -2<P; i.e., goo = -1 - 2<P for nearly Newtonian systems in Newtonian coordinates. (17.19) Einstein gravity reduces to Newton gravity only if. in Newtonian coordinates. goo = -1 - 2I/J Note that the correspondence tells one the form of hoo for nearly Newtonian systems, but not the forms of the other components of the metric perturbation. In fact, the other hp.pcould perfectly well be of the same order of magnitude as hoo <P, without influencing the world lines of slowly moving particles, because they always enter the geodesic equation multiplied by the small numbers vor v 2, or differentiated by t rather than by Xi. The forms of the other hp.p and their small corrections to the Newtonian motion will be explored in Chapters 18,39, and 40. The relation goo = -1 - 2<P is the mathematical embodiment ofthe correspondence between general relativity theory and Newtonian theory for passive aspects of gravity. Together with the \"validity conditions\" (17.15", ", 17.17), it is a foundation from which one can derive all other aspects of the correspondence for \"passive gravity,\" including the relation (17.20) (exercise 17.6). Alternatively. all other aspects of this correspondence can be derived by direct comparison of Newton's predictions with Einstein's. For example, to derive equation (17.20), examine the relative acceleration of two test particles, one at Xi + gi and the other at Xi. According to Newton The correspondence between Einstein theory and Newton theory for all \"passive\" aspects of gravity d 2gi dt2 = d 2(X i + gi) dt2 - = _H!-I i QX at\",i+ ~i d2xi dt2 +~I i QX = at\",i For comparison, Einstein predicts (equation of geodesic deviation) D 2gi.. - - = -- = -R'o\u00b7og'\u00b7 dT2, d 2gi dt2 t[bY conditions (17.15) and (17.17)] Direct comparison gives relation (17.20). Turn now from correspondence for passive aspects of gravity to correspondence for active aspects. According to Einstein, mass generates gravity (spacetime curva ture) by the geometrodynamic law G = 8'i7T. Apply this law to a nearly Newtonian system, and by the chain of reasoning that preceeds equation (17.8) derive the relation Roo = 4'i7p. (17.21) The Newtonian limit of the Einstein field equation is V 21/J = 4'i7p 416 17. HOW MASS-ENERGY GENERATES CURVATURE Combine with the contraction of (17.20), Roo = RiaiD + RO000 = o2ifJ /ox i oxi = V 2ifJ, t O and thereby obtain Newton's equation for the generation of gravity by mass (17.22) Thus, Einstein's field equation reduces to Newton's field equation in the Newtonian limit. The correspondence between Newton and Einstein, although clear and straight forward as outlined above, is even more clear and straightforward when Newton's theory of gravity is rewritten in Einstein's language of curved spacetime (Chapter 12; exercise 17.7). EXERCISES Exercise 17.6. RAMIFICATIONS OF CORRESPONDENCE FOR GRAVITY IN A PASSIVE ROLE From the correspondence relation goo = -1 - 21/J, and from conditions (17.15) and (17.17) for Newtonian physics, derive the correspondence relations Exercise 17.7. CORRESPONDENCE IN THE LANGUAGE OF CURVED SPACETIME [Track 2] Exhibit the correspondence between the Einstein theory and Cartan's curved-spacetime formulation of Newtonian theory (Chapter 12). \u00a717.5. AXIOMATIZE EINSTEIN'S THEORY? Find the most compact and reasonable axiomatic structure one can for general relativity? Then from the axioms derive Einstein's field equation, That approach would follow tradition. However, it may be out of date today. More than half a century has gone by since November 25, 1915. For all that time the equation has stood unchanged, if one ignores Einstein's temporary \"aberration\" of adding the cosmological constant. In contrast the derivations have evolved and become more numerous and more varied. In the beginning axioms told what equa tion is acceptable. By now the equation tells what axioms are acceptable. Box 17.2 sketches a variety of sets of axioms, and the resulting derivations of Einstein's equation. (continued on page 429) There are many ways (Box 17.2) to derive the Einstein field equation \u00a717.5. AXIOMATIZE EINSTEIN'S THEORY? 417 Box 17.2 SIX ROUTES TO EINSTEIN'S GEOMETRODYNAMIC LAW OF THE EQUALITY OF CURVATURE AND ENERGY DENSITY (\"EINSTEIN'S FIELD EQUATION\") [Recommended to the attention of Track-l readers are only route 1 (automatic conservation of the source, plus correspondence with Newtonian theory) and route 2 (Hilbert's variational principle); and even Track-2 readers are advised to finish the rest of this chapter before they study route 3 (physics on a spacelike slice), route 4 (going from superspace to Einstein's equation), route 5 (field of spin 2 in an \"unobservable fiat spacetime\" background), and route 6 (gravitation as an elasticity of space that arises from particle physics).] 1. Model geometrodynamics after electrodynamics and treat \"automatic conserva tion of the source\" and correspondence with the Newtonian theory of gravity as", " the central considerations. a. Particle responds in electrodynamics to field; in general relativity, to geometry. b. The potential for the electromagnetic field is the 4-vector A (components A/.L)' The potential for the geometrj is tht! metric tensor 9 (components g/.Lv)' c. The electromagnetic potential satisfies a wave equation with source term (4-current) on the right, aA v _ aAfL);V = 471', ax V r.J/.L ox/.L ( (1) so constructed that conservation of the source, j/.L;/.L = 0, is automatic (conse quence of an identity fulfilled by the lefthand side). By analogy, the geometro dynamic potential must also satisfy a wave equation with source term (stress energy tensor) on the right, so constructed that conservation of the source, T/.L/ =\u00b0(Chapter 16) is \"automatic.\" This conservation is automatic here because the lefthand side of the equation is a tensor (the Einstein tensor; see Box 8.6 or Chapter 15), built from the metric components and their second derivatives, that fulfills the identity G/.L/ = 0. (2) d. No other tensor which (1) is linear in the second derivatives of the metric components, (2) is free of higher derivatives, and (3) vanishes in fiat spacetime, satisfies such an identity. e. The constant of proportionality (871) is fixed by the choice of units [here geometric; see Box 1.8] and by the requirement (\"correspondence with Newto nian theory\") that a test particle shall oscillate back and forth through a collection of matter of density p, or revolve in circular orbit around that collection of matter, at a circular frequency given by \",2 = (4'1T/3)p (Figure 418 17. HOW MASS-ENERGY GENERATES CURVATURE Box 17.2 (continued) 1.12). The foregoing oversimplifies; and omits Einstein's temporary false~turns,. but otherwise summarizes the reasoning he pursued in arriving at his field equation. This reasoning is spelled out in more detail in the text of Chap ter 17. 2. Take variational principle as central. a. Construct out of the metric components the only scalar that exists that (1) is linear in the second derivatives of the metric tensor, (2) contains no higher derivatives, and (3) vanishes in flat spacetime: namely, the Riemann scalar curvature invariant, R. b. Construct the invariant integral, 1= _1_1R( - g)1/2 d 4x. c. Make small variations, l)gIJ. P, in the metric coefficients glJ. P in the interior of the four-dimensional region 0, and find that this integral changes by the amount 1671!l (3) (4) d. Demand that I should be an extremum with respect to the choice of geometry in the region interior to 0 (l>I = 0 for arbitrary l)gIJ. P; \"principle of extremal action\"). e. Thus arrive at the Einstein field equation for empty space, GIJ.P = O. (5) f. The continuation of the reasoning leads to the identity Chapter 21, on the variational principle, gives more detail and takes up the additional term that appears on the righthand side of(5) when matter or fields or both are present. g. This approach goes back to David Hilbert (1915). No route to the field equa tions is quicker. Moreover, it connects immediately (see the following section here, 2') with the quantum principle of the \"democracy of all histories\" [Feyn man (1942); Feynman and Hibbs (1965)]. The variational principle is spelled out in more detail in Chapter 21. 2'. An aside on the meaning of the classical action integral for the real world of quantum physics. a. A \"history of geometry,\" H, is a spacetime, that is to say, a four-dimensional manifold with four-dimensional - + + + Riemann metric that (1) reduces on one spacelike hypersurface (\"hypersurface of simultaneity\") to a specified \"initial value 3-geometry,\" A, with positive definite metric and (2) reduces on The rest of this chapter is Track 2. No previous track-2 material is needed as preparation for it. nor is it necessary prep aration for any later chapter. but it will be helpful in Chap ter 21 (initial value equations and variational principle) and in Chapter 39 (other theories of gravity). \u00a717.5. AXIOMATIZE EINSTEIN'S THEORY] 419 another spacelike hypersurface to", " a specified \"final value 3-geometry,\" B, also with positive definite metric. b. The classical variational principle of Hilbert, as reformulated by Arnowitt, Oeser, and Misner, provides a prescription for the dynamical path length, I H, of any conceivable history H, classically allowed or not, that connects A and B (see Chapter 21 for a fuller statement for what can and must be specified on the initial hypersurface of simultaneity, and on the final one, and for alternative choices of the integrand in the action principle). c. Classical physics says that a history H is allowed only if it extremizes the dynamic path length I as compared to all nearby histories. Quantum physics says that all histories occur with equal probability amplitude, in the following sense. The probability amplitude for \"the dynamic geometry of space to transit from A to B\" by way of the history H with action integral IH, and by way of histories that lie within a specified infinitesimal range, 6j)H, of the history H, is given by the expression probability amplitude to transit from A to B by way of history H - exp (iIH /h)N6j)H. and histories lying within the range oJ)H about H (6) Here the normalization factor, N, is the same for all conceivable histories H, allowed or not; that lead from A to B (\"principle of democracy of histories\"). The quantum of angular momentum, h = h/27T, expressed in geometric units, has the value (7) where L * is the Planck length, L* = 1.6 X 10-33 em. d. The classically allowed history receives \"preference without preference.\" That history, and histories H that differ from it so little that 8I = IH ICZa\u2022\u2022 is only of the order h and less, give contributions to the probability amplitude that interfere constructively. In contrast, destructive interference effectively wipes out the contribution (to the probability amplitude for a transition) that comes from histories that differ more from the classically allowed history. Thus there are quantum fluctuations in the geometry, but they are fluctuations of limited magnitude. The smallness of h ensures that the scale of these fluctuations is unnoticeable at everyday distances (see the further discussion in Chapters 43 and 44). In this sense classical geometrodynamics is a good approximation to the geometrodynamics of the real world of quantum physics. 3. \"Physics on a spacelike slice or hypersurface of simultaneity,\" again with electro magnetism as the model. a. Say over and over \"lines of magnetic force never end\" and come out with half of Maxwell's equations. Say over and over \"lines of electric force end y 420 17. HOW MASS-ENERGY GENERATES CURVATURE Box 17.2 (continued) only on charge\" and arrive at the other half of Maxwell's equations. Similarly, say over and over - intrinsic ) curvature scalar ( extrinSiC) (lOCal denSity) + curvature = 16.... of massenergy scalar ( (8) and end up with all ten components of Einstein's equation. To \"say over and over\" is an abbreviation for demanding that the stated principles hold on every spacelike slice through every event of spacetime. b. Spell out explicitly this \"spacelike-slice formulation\" of the equations of Maxwell and Einstein. Consider an arbitrary point of spacetime, 9 (\"event\"), and an arbitrary \"simultaneity\" S through 9 (hypersurface of simultaneity; spacelike slice through spacetime). Magnetic lines offorce run about through out S, but nowhere is even a single one of them permitted to end. Recall (\u00a73.4) that the demand \"lines of magnetic force never end,\" when imposed on all reference frames at 9 (for all choices of the \"simultaneity\" S), guarantees not only V. B = 0, but also V X E + oBjot = 0. Similarly (\u00a73.4) the demand that \"electric lines of force never end except on electric charge,\" V' E = 47TJO, when imposed on all \"simultaneities\" through 9, guarantees the remaining Maxwell equation V X B = oEjot + 47TJ. c. Each simultaneity S through 9 has its own slope and curvature. The possibility of different slopes (different local Lorentz frames at 9) is essential for deriving all of Maxwell's equations from the requirements of conservation of flux. Relevant though the slope thus is, the curvature of the hypersurface S never matters for the analysis of electromagnetism. It does matter, however, for any analysis of gravitation modeled on the foregoing treatment of electromagne tism. tTime Lspace-... s \"Simultaneity\" S (spacel", "ike hypersurface or \"slice through spacetime\") that cuts through event 9. The \"simultaneity\" may be considered to be de fined by a set of \"observers\" a,b,c,.... Their world lines cross the simultaneity orthogonally, and their clocks all read the same proper time at the instant of crossing. Another simultaneity through 9 may have at 9 a different curvature or a different slope or both; and it is defined by a different band of observers, with other wrist watches. \u00a717.5. AXIOMATIZE EINSTEIN'S THEORY] 421 d. \"Mass-energy curves space\" is the central principle of gravitation. To spell out this principle requires one to examine in succession the terms \"space\" and \"curvature of space\" and \"density of mass-energy in a given region of space.\" \"Space\" means spacelike hypersurface; or, more specifically, a hypersurface of simultaneity S that includes the point 9 where the physics is under exami nation. e. Denote by u the 4-vector nonpal to S at 9. Then the density of mass-energy in the spacelike hypersurface' S at 9 is (9) in accordance with the definition of the stress-energy tensor given in Chap ter 5. f. This density is a single number, dependent on the inclination of the slice one cuts through spacetime, but independent of how curved one cuts this slice. If it is to be equated to \"curvature of space,\" that curvature must also be independent of how curved one cuts the slice. g. Conclude that the geometric quantity, \"curvature of space,\" must (l) be a single number (a scalar) that (2) depends on the inclination u of the cut one makes through spacetime at 9 in constructing the hypersurface S, but (3) must be unaffected by how one curves his cut. The demand made here appears para doxical. One seems to be asking for a measure of curvature that is independent of curvature! h. A closer look discloses that three distinct ideas come into consideration here. One is the scalar curvature invariant (3)R of the 3-geometry intrinsic to the hypersurface S at 9: \"intrinsic\" in the sense that it is defined by, and depends exclusiv.ely on, measurements of distance made within the hypersurface. The second is the \"extrinsic curvature\" of this 3-geometry relative to the 4-geometry of the enveloping spacetime (\"how curved one cuts his slice\"; see Box 14.1 for more on the distinction between extrinsic and intrinsic curvature). The third is the curvature of the four-dimensional spacetime itself, \"normal to u,\" in some sense yet to be more closely defined. This is the quantity that is inde pendent of how curved one cuts his slice. It is the quantity that is to be identified, up to a factor that depends on the choice of units, with the density of mass-energy. 422 17. HOW MASS-ENERGY GENERATES CURVATURE Box 17.2 (continued) 1. These three quantities are related in the following way: scalar curvature invariant, (3) R, of the 3-geometry intrinsic to the spacelike hypersurface S, a quantity dependent on \"how curved one cuts the slice\" + a correction term that (a) depends only on the \"ex trinsic curvature\" K af3 (Box 14.1 and Chapter 21) of the hypersurface relative to the four-dimensional geometry in which it is imbedded, and (b) is so calculated (a uniquely determinate calcu lation) that the sum of this correction term and (3)R is independent of \"how curved one cuts his slice,\" and (c) has the precise value (Tr K)2 - Tr K2 = (Kaa)Z - Kaf3Kaf3 a measure of the curvature of spacetime that depends on the 4-geemetry of the spacetime and on the inclination u of the spacelike slice S cut through spacetime, but is independent, by construc tion, of \"how curved one cuts the slice\" 2uaGaf3Uf3, where Gaf3 is) = the Einstein curvature = 2 tensor of equation 8.49 and Box 8.6 ( a scalar quantity that (a) is completely defined by what has just been said and (b) can there fore be calculated in all completeness by standard differential geometry (details in Chapter 21) a quantity interpreted in Track 2, Chapter 15, as the \"moment of rotation\" asso ciated with a unit element of 3-volume located at 9 in the hypers", "urface orth ogonal to u (10) j. Conclude that the central principle, \"mass-energy curves space,\" translates to the formula (3)R + (Tr K)Z - Tr K2 = 16r.p, (II) \u00a717.5. AXIOMATIZE EINSTEIN'S THEORY] 423 or, in shorthand form, moment of) rotation ( ( intrinsic) ( extrinsiC) (density of ) = curvature + curvature = mass-energy, (12) valid for every spacelike slice through spacetime at any arbitrary point 9. k. All of Einstein's geometrodynarnics is contained in this statement as truly as all of Maxwell's electrodynamics is contained in the statement that the number of lines of force that end in an element of volume is equal to 471 times the amount of charge in that element of volume. The factor 1671 is appropriate for the geometric system of units in use in this book (density p in cm-2 given by G/c 2 = 0.742 X 10-28 cm/g multiplied by the density Peony expressed in the conventional units of g/cm3 ). 1. Reexpress the principle that \"mass-energy curves space\" in the form (13) Demand that this equation should hold for every simultaneity that cuts through 9, whatever its \"inclination\" u. m. Conclude that the coefficients on the two sides of (13) must agree; thus, (14) Einstein's equation in the language of components; or, in the language of abstract geometric quantities, G = 8'1TT. (15) 4. Going from superspace to Einstein's equation rather than from Einstein's equa tion to superspace. a. A fourth route to Einstein's equation starts with the advanced view of geome the trodynamics that dynamics of geometry unfolds in superspace. Superspace has an infinite number of dimensions. Anyone point in superspace describes a complete 3-geometry, (3)~, with all its bumps and curvatures. The dynamics of geom etry leads from point to point in superspace. is spelled out in Chapter 43. One notes there that b. Like the dynamics of a particle, the dynamics of geometry lends itself to distinct but equivalent mathematical formulations, associated with the names of Lagrange, of Hamilton, and of Hamilton and Jacobi. Of these the most convenient for the present analysis is the last (\"H-J\"). c. In the problem of one particle moving in one dimension under the influence of a potential V(x), the H-J equation reads at _as = _1_(aS)2 + V(x). 2m ax kin~tid'- - -'i total i (16) 424 17. HOW MASS-ENERGY GENERATES CURVATURE Box 17.2 (continued) It has the solution SE(X,t) = -Et +f Z [2m(E - V)]1/2 dx. (17) Out of this solution one reads the motion by applying the \"condition of constructive interference,\" aSE(x,t) aE = 0 (18) (one equation connecting the two quantities x and t; for more on the condition of constructive interference and the H-J method in general, see Boxes 25.3 and 25.4). d. In the corresponding equation for the dynamics of geometry, one deals with a function S = S\u00ab3)~) of the 3-geometry. It depends on the 3-geometry itself, and not on the vagaries of one's choice of coordinates or on the corresponding vagaries in the metric coefficients of the 3-geometry, \u00ab3) to indicate 3-geometry omitted hereafter for simplicityj;-This function obeys the H-J equation [the analog of (16)] (19) (20) f. e. Out of this equation for the dynamics of geometry in superspace one can deduce the Einstein field equation by reasoning similar to that employed in going from (17) to (18) (Gerlach 1969). It would appear that one must break new ground, and establish new founda tions, if one is to find out how to regard the \"Einstein-Hamilton-Jacobi equa tion\" (20) as more basic than the Einstein field equation that one derives from it. [Since done, by Hojman, Kuchar, and Teitelboim (1973 preprint).] 5. Einstein's geometrodynamics viewed as the standard field theory for a field of spin 2 in an \"unobservable fiat spacetime\" background. a. This approach to Einstein's field equation has a long history, references to which will be found in \u00a77.1 and \u00a7I8.1. (Further discussion of this approach will be found in", " those two sections and in Box 7.1, exercise 7.3, and Box 18.1). b. The following summary is quoted from Oeser (1970): \"We wish to give a simple physical derivation of the nonlinearity..., using a now familiar argument :.. leading from the linear, massless, spin-2 field to the full Einstein equa tions.... \u00a7 17.5. AXIOMATIZE EINSTEIN'S THEORY] 425 c. \"The Einstein equations may be derived nongeometrically by noting that the free, massless, spin-2 field equations, RLp.p(</\u00bb - ~ RLaa(</>Jrlp.p = GLp.p(</\u00bb = [(1Jp.a1JPf3 -1Jp.p1Ja(3)O + 1Jp.poaof3 + 1Jaf30p.op - 1Jp.aopof3 -1Jpf30p.0al<Paf3 = 0, (21) whose source is the matter stress-tensor Tp. p, must actually be coupled to the total stress-tensor, including that of the </>-field itself. That is, while the free-field equations (21) are of course quite consistent as they stand, [they are not] when there is a dynamic system's Tp.p as a source. For then the left side, which is identically divergenceless, is inconsistent with the right, since the coupling implies that TP.P,p, as computed from the matter equations of motion, is no longer conserved. d. \"To remedy this [violation of the principle of conservation of momentum and the stress tensor lZ)8p.p arising from the quadratic Lagrangian lZ)L energy], responsible for equation (21) is then inserted on the right. e. \"But the Lagrangian (3)L leading to these modified equations is then cubic, and itself contributes a cubic (3)8p. p\u2022 f. \"This series continues indefinitely, and sums (if properly derived!) to the full nonlinear Einstein equations, Gp.p ([calculated from] 1Jaf3 + </>a(3) = -KTp.p [+ 8'1TTp.p in the geometric units and sign conventions of this book], which are an infinite series in the deviation </>p.p of the metric gp.p from its Minkowskian value 1Jp.p' g. Once the iteration is begun (whether or not a Tp.p is actually present), it must be continued to all orders, since conservation only holds for the full series 2: In)8p.p, Thus, the theory is either left in its (physically irrelevant) free linear n=Z form (21), or it must be an infinite series.\" h. For details, see Oeser (1970); the paper goes on (1) to take advantage of a well-chosen formalism (2) to rearrange the calculation, and thus (3) to \"derive the full Einstein equations, on the basis of the same self-coupling requirement, but with the advantages that the full theory emerges in closed form with just one added (cubic) term, rather than as an infinite series.\" 1. Oeser summarizes the analysis at the end thus: \"Consistency has therefore led us to universal coupling, which implies the equivalence principle. It is at this point that the geometric interpretation of general relativity arises, since all matter now moves in an effective Riemann space of metric gp.p = 1Jp.p + hP. p.... [The] initial fiat 'background' space is no longer observable.\" In other words, this approach to Einstein's field equation can be summarized as \"cur vature without curvature\" or-equally well-as \"fiat spacetime without fiat spacetime\"! 426 17. HOW MASS-ENERGY GENERATES CURVATURE Box 17.2 (continued) 6. Sakharov's view of gravitation as an elasticity of space that arises from particle physics. a. The resistance of a homogeneous isotropic solid to deformation is described by two elastic constants, Young's modulus and Poisson's ratio. b. The resistance of space to deformation is described by one elastic constant, the Newtonian constant of gravity. It makes its appearance in the action principle of Hilbert I = _1_f (4)R( - g)1/2 d 4x l6r.G + f (Lmatter + L fields)( - g)1/2 d 4x = extremum. (22) it was first c. According to the historical records, learned how many elastic constants it takes to", " describe a solid from microscopic molecular models of matter (Newton, Laplace, Navier, Cauchy, Poisson, Voigt, Kelvin, Born), not from macroscopic considerations of symmetry and invariance. Thus, count the energy stored up in molecular bonds that are deformed from natural length or natural angle or both. Arrive at an expression for the energy of deformation per unit volume of the elastic material of the form e = A(Tr s)Z + B Tr(S2). Here the strain tensor (23) (24) measures the strain produced in the elastic medium by motion of the typical point that was at the location x m to the location x m + ~m(x). The constants A and B are derived out of microscopic physics. They fix the values of the two elastic constants of the macroscopic theory of elasticity. d. Andrei Sakharov (1967) (the Andrei Sakharov) has proposed a similar micro scopic foundation for gravitation or, as he calls it, the \"metric elasticity of space.\" He identifies the action term of Einstein's geometrodynamics [the first term in (22)] \"with the change in the action of quantum fluctuations of the vacuum [associated with the physics of particles and fields and brought about] when space is curved.\". e. Sakharov notes that present-day quantum field theory \"gets rid by a renor malization process\" of an energy density in the vacuum that would formally be infinite if not removed by this renormalization. Thus, in the standard analysis of the degrees of freedom of the electromagnetic field in flat space, one counts the number of modes of vibration per unit volume in the range \u00a717.5. AXIOMATIZE EINSTEIN'S THEORY? 427 of circular wave numbers from k to k + dk as (2 \u00b74'17\"18w3)k 2 dk. Each mode of oscillation, even at the absolute zero of temperature, has an absolute irre ducible minimum of \"zero-point energy of oscillation,\"!hv =!hck [the fluctu ating electric field associated with which is among the most firmly established of all physical effects. It acts on the electron in the hydrogen atom in supple ment to the electric field caused by the proton alone, and thereby produces most of the famous Lamb-Retherford shift in the energy levels of the hydrogen atom, as made especially clear by Welton (1948) and Dyson (1954)]. The totalized density of zero-point energy of the electromagnetic field per unit volume of spacetime (units: cm4 ) formally diverges as (hlh 2 ) i '\"k 3 dk. o (25) Equally formally this divergence is \"removed\" by \"renormalization\" [for more on renormalization see, for example, Hepp (1969)]. f. Similar divergences appear when one counts up formally the energy associated with other fields and with vacuum fluctuations in number of pairs of electrons, /L-mesons, and other particles in the limit of quantum energies large in com parison with the rest mass of any of these particles. Again these divergences in formal calculations are \"removed by renormalization.\" g. Removed by renormalization is a contribution not only to the energy density, and therefore to the stress-energy tensor, but also to the total Lagrange function e of the variational principle for all these fields and particles, I =f e d 4x = extremum. (26) h. Curving spacetime alters all these energies, Sakharov points out, extending an argument of Zel'dovich (1967). Therefore the process of\"renormalization\" or \"subtraction\" no longer gives zero. Instead, the contribution of zero-point energies to the Lagrangian, expanded as a power series in powers of the curvature, with numerical coefficients A, B,... of the order of magnitude of unity, takes a form simplified by Ruzmaikina and Ruzmaikin (1969) to the following: e(R) = Ah f k 3 dk + Bh(4)R f k dk + h[C\u00ab4)R)Z + DRa/3R<>/3]f k-1dk + (higher-order terms). (27) [For the alteration in the number of standing waves per unit frequency in a curved manifold, see also Berger (1966), Sakharov (1967), Hill in De Witt (1967c), Polievktov-Nikoladze (1969), and Berger, Gauduchon, and Mazet (1971).] 1. Renormalization physics argues that the first term in (27) is to be dropped. The second term, Sakharov notes, is identical in form to the Hilbert action 428 17. HOW MASS", "-EN ERGY GENERATES CURVATURE Box 17.2 [continued) principle, equation (3) above, with the exception that there the constant that multiplies the Riemann scalar curvature invariant is -c3/16...G (in con ventional units), whereas here it is Bhfk dk (in the same conventional units). The higher order terms in (27) lead to what Sakharov calls \"corrections... to Einstein's equations.\" j. Overlooking these corrections, one evidently obtains the action principle of Einstein's theory when one insists on the equality G = ( Newtonian constant of gravity ) = c3 l67TBhfk dk. (28) With B a dimensionless numerical factor of the order of unity, it follows, Sakharov argues, that the effective upper limit or \"cutoff\" in the formally divergent integral in (28) is to be taken to be of the order of magnitude of the reciprocal Planck length [see equation (7)], kcutorr-(c3/hG)1/2 = l/L* = 1/1.6 X 1O-33 cm. (29) In effect Sakharov is saying (1) that field physics suffers a sea change into something new and strange for wavelengths less than the Planck length, and 1028 eV or 10-5 g or more; for quantum energies of the order of hckeutorr (2) that in consequence the integral fk dk is cut off; and (3) that the value of this cutoff, arising purely out of the physics of fields and particles, governs the value of the Newtonian constant of gravity, G. k. In this sense, Sakharov's analysis suggests that gravitation is to particle physics as elasticity is to chemical physics: merely a statistical measure of residual energies. In the one case, molecular bindings depend on departures of mole cule-molecule bond lengths from standard values. In the other case, particle energies are affected by curvatures of the geometry. 1. Elasticity, which looks simple, gets its explanation from molecular bindings, which are complicated; but molecular bindings, which are complicated, receive their explanation in terms of SchrOdinger's wave equation and Coulomb's law of force between charged point-masses, which are even simpler than elasticity. m. Einstein's geometrodynamics, which looks simple, is interpreted by Sakharov as a correction term in particle physics, which is complicated. Is particle physics, which is complicated, destined some day in its turn to unravel into something simple-something far deeper and far simpler than geometry (\"pregeometry\"; Chapter 44)? \u00a717.6. EINSTEIN'S THEORY DISTINGUISHED BY \"NO PRIOR GEOMETRY\" 429 \u00a717.6. \"NO PRIOR GEOMETRY\": A FEATURE DISTINGUISHING EINSTEIN'S THEORY FROM OTHER THEORIES OF GRAVITY Whereas Einstein's theory of gravity is exceedingly compelling, one can readily construct less compelling and less elegant alternative theories. The physics literature is replete with examples [see Ni (1972), and Thorne, Ni, and Will (1971) for reviews]. However, when placed among its competitors, Einstein's theory stands out sharp and clear; it agrees with experiment; most of its competitors do not (Chapters 38-40). It describes gravity entirely in terms of geometry; most of its competitors do not. It is free of any \"prior geometry\"; most of its competitors are not. Set aside, until Chapter 38, the issue of agreement with experiment. Einstein's theory remains unique. Every other theory either introduces auxiliary gravitational fields [e.g., the scalar field of Brans and Dicke (1961 )], or involves \"prior geometry,\" or both. Thus, every other theory is more complicated conceptually than Einstein's theory. Every other theory contains elements of complexity for which there is no experimental motivation. The concept of \"prior geometry\" requires elucidation, not least because the rejection of prior geometry played a key role in the reasoning that originally led Einstein to his geometrodynamic equation G = 8r.T By \"prior geometry\" one means any aspect of the geometry of spacetime that is fixed immutably, i.e., that cannot be changed by changing the distribution of gravitating sources. Thus, prior geometry is not generated by or affected by matter; it is not dynamic. Example: Nordstwm (1913) formulated a theory in which the physical metric of spacetime 9 (the metric that enters into the equivalence principle) is generated by a \"background\" flat-spacetime metric '1, and by a scalar gravitational field tf>: Einstein's theory compared with other theories of gravity All other theories introduce auxiliary gravitational fields or prior geometry.. Prior geometry\"", " defi ned Nordstrom's theory as an illustration of prior geometry generation of tf> by), stress-energy construction of g). from tf> and '1 ( ( (17.23a) (17.23b) In this theory, the physical metric 9 (governor of rods and clocks and of test-particle motion) has but one changeable degree of freedom-the freedom in tf>. The rest of 9 is fixed by the flat spacetime metric (\"prior geometry\") '1. One does not remove the prior geometry by rewriting Nordstwm's equations (17.23) in a form rcurvature scalar l t Lconstructed from 9J R = 24r.T, t. 4 a /3 Ta~ IJ.P - - 0 c a /3 t rweyl tensor Lconstructed from 9 J (17.24) devoid of reference to '1 and tf> [Einstein and Fokker (1914); exercise 17.8]. Mass can still influence only one degree of freedom in the spacetime geometry. The other degrees of freedom are fixed a priori-they are prior geometry. And this prior geometry can perfectly well (in principle) be detected by physical experiments that make no reference to any equations (Box 17.3). 430 17. HOW MASS-ENERGY GENERATES CURVATURE Box 17.3 AN EXPERIMENT TO DETECT OR EXCLUDE CERTAIN TYPES OF PRIOR GEOMETRY (Based on December 1970 discussions between Alfred Schild and Charles W. Misner) Choose a momentarily static universe populated with a large supply of suitable pulsars. The pulsars should be absolutely regular, periodically emitting characteristic pulses of both gravitational and electromagnetic waves. Two fleets of spaceships containing receivers are sent out \"on station\" to collect the experimental data. Admiral Weber's fleet carries gravitational wave receivers; Admiral Hertz's fleet, electromag netic receivers. The captain of each spaceship holds himself \"on station\" by monitoring three suitably chosen pulsars (of identical frequency) and maneuvering so that their pulses always arrive in coincidence. The experimental data he collects consist of the pulses received from all other pul sars, which he is not using for station keeping, each registered as coincident with or interlaced among the reference (stationary) pulses. [For display pur poses, the pattern produced by any single pulsar can be converted to acoustic form. The reference pulses can be played acoustically (by the data processing computer) on one drum at a fixed rate, and the pulses from other pulsars can be played on a second drum. A pattern of rythmic beats will result.] When the data fleet is checked out and tuned up, each captain reports stationary patterns. Now the experiment begins. One or more massive stars are towed in among the fleet. The fleet reacts to stay on station, and reports changes in the data patterns. The spaceships on the outside edges of the fleet verify that no detectable changes occur at their stations; so the incident radiation from the distant pulsars can be regarded as unaffected by the newly placed stars. Data stations nearer the movable stars report the interesting data. What are the results? In a universe governed by the laws of special relativity (spacetime always flat), no patterns change. (Weber's fleet was unable to get checked out in the first place, as no gravitational waves were ever detected from the pulsars). Neither stars, nor anything else, can produce gravitational fields. All aspects of the spacetime geometry are fixed a priori (complete prior geometry!). There is no gravity; and no light deflection takes place to make Hertz's captains adjust their positions. In a universe governed by Nordstr0m's theory of gravity (see text) both fleets get checked out i.e., both see waves. But neither fleet sees any changes in the rhythmic pattern of beats. The stars being towed about have no influence on either gravitational waves or electromagnetic waves. The prior geometry ('1) present in the theory precludes any light deflection or any gravitational-wave deflection. In a universe governed by Whitehead's (1922) theory of gravity [see Will (1971 b) and references cited therein], radio waves propagate along geo desics of the \"physical metric\" g, and get deflected by the gravitational fields ofihe stars. But gravita tional waves propagate along geodesics of a flat background metric '1, and are thus unaffected by the stars. Consequently, Hertz's captains must ma neuver to keep on station; and they hear a chang ing beat pattern between the reference pulsars and the other pulsars. But Weber's fleet remains on station and records no changes in the beat pattern. The prior geometry ('1) shows itself clearly in the experimental result. In a universe governed by Einstein's theory, both fleets see effects (no sign of prior geometry because Einstein's theory has no prior", " geometry). Moreover, if the fleets were originally paired, one Weber ship and one Hertz at each station, they remain paired. No differences exist between the propagation of high-frequency light waves and high-frequency gravitational waves. Both propa gate along geodesics of g. \u00a717.7. A TASTE OF THE HISTORY OF EINSTEIN'S EQUATION 431 Mathematics was not sufficiently refined in 1917 to cleave apart the demands for \"no prior geometry\" and for a \"geometric, coordinate-independent formulation of physics.\" Einstein described both demands by a single phrase, \"general covariance.\" The \"no-prior-geometry\" demand actually fathered general relativity, but by doing so anonymously, disguised as \"general covariance,\" it also fathered half a century of confusion. [See, e.g., Kretschmann (1917).].A systematic treatment of the distinction between prior geometry (\"absolute objects\") and dynamic fields (\"dynamic objects\") is a notable feature of Anderson's (1967) relativity text. \"No prior geometry\" as a part of Einstein's principle of.. general covariance\" Exercise 17.8. EINSTEIN-FOKKER REDUCES TO NORDSTR~M The vanishing of the Weyl tensor [equation (13.50)] for a spacetime metric 9 guarantees that the metric is conformally flat-i.e., that there exists a scalar field </> such that 9 = </>211, where 11 is a flat-spacetime metric. [See, e.g., Schouten (1954) for proof.] Thus, the Einstein Fokker equation (17.24), C a /3 P' = 0, is equivalent to the Nordstrom equation (l7.23b). With this fact in hand, show that the Einstein-Fokker field equation R = 247fT reduces to the Nordstrom field equation (l7.23a). EXERCISE ~--- \u00a717.7. A TASTE OF THE HISTORY OF EINSTEIN'S EQUATION Nothing shows better what an idea is and means today than the battles and changes it has undergone on its way to its present form. A complete history of general relativity would demand a book. Here let a few key quotes from a few of the great papers give a little taste of what a proper history might encompass, Einstein (1908): \"We... will therefore in the following assume the complete physical equivalence of a gravitational field and of a corresponding acceleration of the reference system.... the clock at a point P for an observer anywhere in space runs (I + <P/c 2 ) times faster than the clock at the coordinate origin.... it follows that light rays are curved -by the gravitational field.,.. an amount of energy E has a mass E/cz.\" Einstein and Grossmann (1913): \"The theory described here originates from the conviction that the proportionality between the inertial and the gravitational mass of a body is an exact law of nature that must be expressed as a foundation principle of theoretical physics.... An observer enclosed in an elevator has no way to decide whether the elevator is at rest in a static gravitational field or whether the elevator is located in gravitation-free space in an accelerated motion that is maintained by forces acting on the elevator (equivalence hypothesis).... In the decay of radium, for example, that decrease [of mass] amounts to 1/10,000 of the total mass. If those changes in inertial mass did not correspond to changes in gravitational mass, then deviations of inertial from gravitational masses would arise that are far larger than the Eotvos experiments allow. It must therefore be considered as very probable that the identity of gravitational and inertial mass is exact. 432 17. HOW MASS-ENERGY GENERATES CURVATURE \"The sought for generalization will surely be of the form tensor gp.,' where\" is a constant and r p.v is a contravariant tensor of the second rank that arises through differential operations.... it proved out of the fundamental impossible to find a differential expression for rp.,. that is a generalization of[Poisson's).1<1>, and that is a tensor with respect to arbitrary transformations.... It seems most natural to demand that the system of equations should be covariant against arbitrary the equations of the transformations. That stands in conflict with the result gravitational field do not possess this property.\" that Einstein and Grossman (1914): \"In a 1913 treatment... we could not show general covariance for these gravitational equations. [Origin of their difficulty: part of the two-index curvature tensor was put on the left, to constitute the second-order part of the field equation, and part was put on the right with Tp.v and was called gravitational stress-energy. It was asked that lefthand and righthand sides transform as", " tensors, which they cannot do under general coordinate transformations.] Einstein (l915a): \"In recent years I had been trying to found a general theory of relativity on the assumption of the relativity even of nonuniform motions. I believed in fact that I had found the only law of gravitation that corresponds to a reasonably formulated postulate of general relativity, and I sought to establish the necessity of exactly this solution in a paper that appeared last year in these proceedings. \"A renewed analysis showed me that that necessity absolutely was not shown in the approach adopted there; that it nevertheless appeared to be shown rested on an error. \"For these reasons, I lost all confidence in the field equations I had set up, and I sought for an approach that would limit the possibilities in a natural way. In this way I was led back to the demand for the general covariance of the field equations, from which I had departed three years ago, while working with my friend Grossmann, only with a heavy heart. In fact we had already at that time come quite near to the solution of the problem that is given in what follows. \"According to what has been said, it is natural to postulate the field equations of gravitation in the form since we already know that these equations are covariant with respect to arbitrary transformations of determinant I. In fact, these equations satisfy aUcondit~aL_ we have to impose on them. [Here Rp.v is a piece of the Ricci tensor that Einstein regarded as covariant.]... \"Equations (22a) give in the first approximation o2ga f3 -\"o-x-\"a':-ox-f3\"'\" = O. By this [condition] the coordinate system is still not determined, in the sense that for this determination four equations are necessary.\" (Session of Nov. 4, 1915, published Nov. II.) Einstein (l915b): \"In a recently published investigation, I have shown how a theory of the gravitational field can be founded on Riemann's covariant theory of many-di- \u00a717.7. A TASTE OF THE HISTORY OF EINSTEIN'S EQUATION 433 mensional manifolds. Here it will now be proved that, by introducing a surely bold additional hypothesis on the structure of matter, a still tighter logical structure of the theory can be achieved.... it may very well be possible that in the matter to which the given expression refers, gravitational fields play an essential part. Then P'I' can appear to be positive for the entire structure, although in reality only TIJ.I' + tI'l' is positive, and TI'l' vanishes everywhere. We assume in the following that in fact the condition PI' = 0 is fulfilled [quite] generally. \"Whoever does not from the beginning reject the hypothesis that molecular [small scale] gravitational fields constitute an essential part of matter will see in the following a strong support for this point of view. \"Our hypothesis makes it possible... to give the field equations of gravitation in a generally covariant form... [where GI\" is the Ricci tensor].\" (Session of Nov. II, 1915; published Nov. 18.) Einstein (l915c): \"I have shown that no objection of principle stands in the way of this hypothesis [the field equations], by which space and time are deprived of the last trace of objective reality. In the present work I find an important confirmation of this most radical theory of relativity: it turns out that it explains qualitatively and quantitatively the secular precession of the orbit of Mercury in the direction of the orbital motion, as discovered by Leverrier, which amounts to about 45\" per century, without calling on any special hypothesis whatsoever.\" Einstein (l915d; session of Nov. 25,1915; published Dec. 2): \"More recently I have found that one can proceed without hypotheses about the energy tensor of matter when one introduces the energy tensor of matter in a somewhat different way than was done in my two earlier communications. The field equations for the motion of the perihelion of Mercury are undisturbed by this modification.... \"Let us put [where Gim is the Ricci tensor].\"...... these equations, in contrast to (9), contain no new condition, so that no other assumption has to be made about the energy tensor of matter than obedience to the energy-momentum [conservation] laws. \"With this step, general relativity is finally completed as a logical structure. The postulate of relativity in its most general formulation, which makes the spacetime coordinates into physically meaningless parameters, leads compellingly to a completely determinate theory of gravitation that explains the perihelion motion of Mercury. In contrast, the general-relativity postulate is able to open up to us nothing about the nature of the other processes of nature that special relativity has not already taught. The opinion on this point that I recently expressed in these proceedings", " was erroneous. Every physical theory compatible with special relativity can be aligned into the system of general relativity by means of the absolute differential calculus, without [general relativity] supplying any criterion for the acceptability of that theory.\" Hilbert (1915): \"Axiom I [notation changed to conform to usage in this book]. The I) I 434 17. HOW MASS-ENERGY GENERATES CURVATURE law of physical events is determined through a world function [Mie's terminology; better known today as \"Lagrangian\"] L, that contains the following arguments: og\"p 02g\"p gIL\" ax'\" ax\" ox/3'and specifically the variation of the integral f L( - g)1/2 d 4x must vanish for [changes in] every one of the 14 potentials gup' Au'... \"Axiom II (axiom of general invariance). The world function L is invariant with respect to arbitrary transformations of the world parameters [coordinates] x\".... \"For the world function L, still further axioms are needed to make its choice unambiguous. If the gravitation equations are to contain only second derivatives of the potentials gUP, then L must have the form L = R + L e1ec, where R is the invariant built from the Riemann tensor (curvature of the four-dimen sional manifold.\" (Session of Nov. 20, 1915.) Einstein (l916c): \"Recently H. A. Lorentz and D. Hilbert have succeeded in giving general relativity an especially transparent form in deriving its equations from a single variation principle. This will be done also in the following treatment. There it is my aim to present the basic relations as transparently as possible and in a way as general as general relativity allows.\" Einstein (1916b): \"From this it follows, first of all, that gravitational fields spread out with the speed of light.... [plane] waves transport-energy.... One thus gets... the radiation of the system per unit time.... ~ '\" (03J\"/3)2 \". 2477 L.J \",/3 0(3 Hilbert (1917): \"As for the principle of causality, the physical quantities and their time-rates of change may be known at the present time in any given coordinate system; a prediction will then have a physical meaning only when it is invariant with respect to all those transformations for which exactly those coordinates used for the present time remain unchanged. I declare that predictions of this kind for the future are all uniquely determined; that is, that the causality principle holds in this formulation: in the present, all predictions about the same quantities in the future follow necessarily and uniquely insofar as they have physical meaning.\" \"From the knowledge of the 14 physical potentials gILP' A\", ----------.....~ CHAPTER 18 WEAK GRAVITATIONAL FIELDS The way that can be walked on is not the perfect way. The word that can be said is not the perfect word. LAD-TZU (-3rd century B.C.) \u00a7 18.1. THE LINEARIZED THEORY OF GRAVITY Because of the geometric language and abbreviations used in writing them, Einstein's field equations, Gp. p = 87TTp. p, hardly seem to be differential equations at all, much less ones with many familiar properties. The best way to see that they are is to apply them to weak-field situations (18.1 ) e.g., to the solar system, where Ihp.pl tional wave propagating through interstellar space. IIPI ::s M0 / R0 - 10-6 ; or to a weak gravita In a weak-field situation, one can expand the field equations in powers of hp. p using a coordinate frame where (18.1) holds; and without much loss of accuracy, one can keep only linear terms. The resulting formalism is often called \"the linearized theory of gravity,\" because it is an important theory in its own right. In fact, it is precisely this \"linearized theory\" that one obtains when one asks for the classical field corresponding to quantum-mechanical particles of (1) zero rest mass and (2) spin two in (3) flat spacetime [see Fierz and Pauli (1939)]. Track-2 readers have already explored linearized theory somewhat in \u00a77.l, exercise 7.3, and Box 7.1. There it went under the alternative name, \"tensor-field theory of gravity in flat spacetime.\", \"linearized theory of gravity\": (1) as weak-field limit of general relativity (2) as standard \"field-theory\" description of gravity in \"flat spacetime\" (3) as a foundation for \"deriving\" general relativity Details of linearized theory: (1) connection coefficients 436", " 18. WEAK GRAVITATIONAL FIELDS Just as one can \"descend\" from general relativity to linearized theory by linearizing about flat spacetime (see below), so one can \"bootstrap\" one's way back up from linearized theory to general relativity by imposing consistency between the linearized field equations and the equations of motion. or, equivalently, by asking about: (1) the stress-energy carried by the linearized gravitational field hlJ.v; (2) the influence of this stress-energy acting as a source for corrections hUllJ.v to the field; (3) the stress-energy carried by the corrections HIllJ.v; (4) the influence of this stress-energy acting as a source for corrections H2ll' I' to the corrections hUllJ.v; (5) the stress-energy carried by the corrections to the corrections; and so on. This altern(;1tive w'ilj'!o derive general relativity has been developed and explored- by Gupta (1954, 1957, 1962), Kraichnan (1955), Thirring (1961), Feynman (1963a), Weinberg (1965), and Oeser (1970). But because the outlook is far from geometric (see Box 18.1), the details of the derivation are not presented here. (But see part 5 of Box 17.2.) Here attention focuses on deriving linearized theory from general relativity. Adopt the form (18.1) for the metric components. The resulting connection coefficients [equations (8.24b)], when linearized in the metric perturbation hlJ.v' read r I' - 1 IJ.V(h a{3 - 2 TJ aV,{3 + {3v,a h - h ) a{3,v =1 hI' h I' - 2 ( a,{3 + {3,a _ h,I' a/3)' (18.2) The second line here introduces the convention, used routinely whenever one ex pands in powers of hlJ.v' that indices of hlJ.v are raised and lowered using TJI'V and TJIJ.V' not glJ.Vand glJ.v' A similar linearization of the Ricci tensor [equation (8.47)] yields RlJ.v = ralJ.v,a - rap.a,v = ~ (h/,va + hva,p.a - hlJ.v,aa - h,lJ.v)' (18.3) where After a further contraction to form R =glJ.VRIJ.~ ;::::: TJIJ.V RlJ.v' one finds that the Einstein equations, 2GI'V= l6'1TTlJ.v' read h =ha a - -,,{3h 1/ a{3' (18.4) hlJ.a,.a + h va,lJ.a - hlJ.v,aa - h,IJ.V - TJIJ..(ha{3,a{3 - h,l) = l6'1TTlJ.v\u00b7 The number of terms has increased in passing from RlJ.v (18.3) to GIJ.V = RlJ.v (18.5), but this annoyance can be counteracted by defining (18.5) ~glJ.vR (18.6) and using a bar to imply a corresponding operation on any other symmetric tensor. (2) \"gravitational potentials\" hI\" \u00a7 18.1. THE LINEARIZED THEORY OF GRAVITY 437 Box 18.1 DERIVATIONS OF GENERAL RELATIVITY FROM GEOMETRIC VIEWPOINT AND FROM SPIN-TWO VIEWPOINT, COMPARED AND CONTRASTED Einstein derivation Spin\u00b72 derivation Nature of primordial spacetime geometry? Not primordial; geometry is a dynamic participant in physics \"God-given\" flat Lorentz spacetime manifold Topology (multiple connected ness) of spacetime? Vision of physics? Laws of physics are local; they do not specify the topology Dynamic geometry is the \"master field\" of physics Starting points for this deri vation of general relativity I. Equivalence principle (world lines of photons and test particles are geodesics of the spacetime geometry) 2. That tensorial conserved quantity which is derived from the curvature (Cartan's moment of rotation) is to he identified with the tensor of stress-momentum-energy (see Chapter 15). Simply connected Euclidean topology This field, that field, and the other field all execute their dynamics in a f1at spacetime manifold", " I. Begin with field of spin two and zero rest mass in flat spacetime. 2. Stress-energy tensor built from this field serves as a source for this field. Resulting equations Einstein's field equations Einstein's field equations Resulting assessment of the spacetime geometry from which derivation started Fundamental dynamic participant in physics None. Resulting theory eradi cates original flat geometry from all equations, showing it to be unobservable View about the greatest single crisis of physics to emerge from these equations: complete gravitational collapse Central to understanding the nature of matter and the universe Unimportant or at most peripheral Thus Gp.. = Rp.v to first order in the hp.., and hI'\" = hp..; i.e., hI\" = hI\" this notation the linearized field equations become - ~l1p..h. With The first term in these linearized equations is the usual flat-space d'Alembertian, and the other terms serve merely to keep the equations \"gauge-invariant\" (see Box (l8.7) (3) linearized field equations (4) gauge conditions (5) field equations and metric in Lorentz gauge 438 18. WEAK GRAVITATIONAL FIELDS 18.2). In Box 18.2 it is shown that, without loss of generality, one can impose the \"gauge conditions\" h-p.a - 0,a:-. (I8.8a) These gauge conditions are the tensor analog of the Lorentz gauge Aa,a = 0 of electromagnetic theory. The field equations (I8.7) then become (l8.8b) The gauge conditions (I8.8a), the field equations (I8.8b), and the definition of the metric (I8.8c) are the fundamental equations of the linearized theory of gravity written in Lorentz gauge. EXERCISES Exercise 18.1. GAUGE INVARIANCE OF THE RIEMANN CURVATURE Show that in linearized theory the components of the Riemann tensor are (18.9) Then show that these components are left unchanged by a gauge transformation of the form discussed in Box 18.2 [equation (4b)]. Since the Einstein tensor is a contraction of the Riemann tensor, this shows that it is also gauge-invariant. Exercise 18.2. JUSTIFICATION OF LORENTZ GAUGE Let a particular solution to the field equations (18.7) of linearized theory be given, in an arbitrary gauge. Show that there necessarily exist four generating functions ~/l(t, xi) whose gauge transformation [Box 18.2, eq. (4b)] makes hneW/la.Q = 0 (Lorentz gauge). Also show that a subsequent gauge transformation leaves this Lorentz gauge condition unaffected if and only if its generating functions satisfy the sourceless wave equation ~Q./3/3 = O. Exercise 18.3. EXTERNAL FIELD OF A STATIC. SPHERICAL BODY Consider the external gravitational field of a static spherical body, as described in the body's (nearly) Lorentz frame-i.e., in a nearly rectangular coordinate system Ih/lvl ~ I, in which the body is located at x = y = z = 0 for an t. By fiat, adopt Lorentz gauge. (a) Show that the field equations (l8.8b) and gauge conditions (l8.8a) imply hoo = 4M/(x2 + y2 + Z2)1I2, hoo = h,.,. = huu = hzz = 2M/(x2 + y2 + Z2)1I2, hOi = hik = 0, ha /3 = 0 if IX :;:. p, where M is a constant (the mass of the body; see \u00a719.3). \u00a7 18.1. THE LINEARIZED THEORY OF GRAVITY 439 / Box 18.2 GAUGE TRANSFORMATIONS AND COORDINATE TRANSFORMATIONS IN LINEARIZED THEORY A. The Basic Equations of Linearized Theory, written in any coordinate system that is nearly globally Lorentz, are (18.1) and (18.7): gp.. = TJp.. + hp..' Ihp..1 ~ 1; -hp..,aa - TJp..ha/3,a/3 + hp.a,a. + h.a,ap. = 16'JTTp.\u2022. (1) (2) Two different types of coordinate transformations connect nearly globally Lorentz systems to each other: global Lorentz transformations, and infinitesimal coordinate transformations. 1. Global Lorentz Transformations: (3a) These transform the metric", " coefficients via oxp. ox' TJa'/3' + ha'/3' = ga'/3' = ox a' ox/3' gp.. = Alla,A'/3'(TJp.. + hp..) = TJa'/3' + AP.a' A'/3,hll\u2022\u00b7 Thus, hp..-and likewise hp..-transform like components of a tensor in flat spacetime (3b) 2. Infinitesimal Coordinate Transformations (creation of \"ripples\" in the coordi nate system): (4a) where gP.W) are four arbitrary functions small enough to leave Ihp.'.,1 ~ 1. Infinitesimal transformations of this sort make tiny changes in the functional forms of all scalar, vector, and tensor fields. Example: the temperature T is a unique function of position, TW); so when written as a function of coordinates it changes T(x ll' = all) = T(xP. + gp. = aP.) = T(xP. = aP. = T(xP. = aP.) - T,p.gp.; ), and if T = cos2(xO), then i.e., if $!! = 0.001 sin(x 1 gp.) T = cos2(XO) + 0.002 sin(x 1 ) cos(xO) sin(xO} 440 18. WEAK GRAVITATIONAL FIELDS Box 18.2 (continued) These tiny changes can be ignored in all quantities except the metric, where tiny deviations from TJlLv contain all the information about gravity. The usual tensor transformation law for the metric when combined with the transformation law (4a) and with reveals that gp\"Axa' = aa) = TJ pC1 + hpC1(x a = a\") - gp,C1 + negligible corrections -- hpC1,aga and -- hpa ga,C1' gC1,P Hence, the metric perturbation functions in the new (x lL') and old (x lL ) coordinate systems are related by (4b) whereas thefunctionalforms ofall other scalars, vectors, and tensors are unaltered, to within the precision of linearized theory. B. Gauge Transformations and Gauge Invariance. regards equation (4b) as gauge transformations, analogous to those In linearized theory one usually of electromagnetic theory. The fact that gravitational gauge transformations do not affect the functional forms of scalars, vectors, or tensors (i.e., observables) is called \"gauge invariance.\" Just as a straightforward calculation reveals the gauge invariance of the electromagnetic field, (5a) Fnew = A new _ A new = AOld + '!' \",J1. J1.,\" \",p. J1.V _ Aold - p.,v,vp. '!' = Fold,,J1.V J1.V (5b) so a straightforward calculation (exercise 18.1) reveals the gauge invariance of the Riemann tensor R new - ROld ILva{3 - ILva{3' (6) Such gauge invariance was already guaranteed by the fact that R lLva{3 are the components of a tensor, and are thus essentially the same whether calculated in an orthonormal frame g~. = TJlLv, in the old coordinates where glL v = TJ~v + h~~, or in the new coordinates where glLv = TJlLv + h~~. \u00a718.1. THE LINEARIZED THEORY OF GRAVITY 441 Like the Riemann tensor, the Einstein tensor and the stress-energy tensor are unaffected by gauge transformations. Hence, if one knows a specific solution hll \u2022 to the linearized field equations (2) for a given Til', one can obtain another solution that describes precisely the samep hysical situation (all observables unchanged) by the change of gauge (4), in which gil are four arbitrary but small functions. C. Lorentz Gauge. One can show (exercise 18.2) that for any physical situation, one can specialize the gauge (i.e., the coordinates) so that hila,a = O. This is the Lorentz gauge introduced in \u00a718.1. The Lorentz gauge is not fixed uniquely. The gauge condition hlla,a = 0 is left unaffected by any gauge transformation for which (See exercise 18.2.) ta,/3 c;; - 0 /3 - \u2022 D. Curvilinear Coordinate Systems. Once the gauge has been fixed by fiat for a given system (e.g., the solar system), one can regard hll \u2022 and hll \u2022 as components of tensors in flat spacetime; and one can regard", " the field equations (2) and the chosen gauge conditions as geometric, coordinate-independent equations in flat spacetime. This viewpoint allows one to use curvilinear coordinates (e.g., spherical coordinates centered on the sun), if one wishes. But in doing so, one must everywhere replace the Lorentz components of the metric, TIll\" by the metric's components gil' flat in the flat-spacetime curvilinear coordinate system; and one must replace all ordinary derivatives (\"commas\") in the field equations and gauge conditions by covariant derivatives whose connection coefficients come from g.... See exercise 18.3 for an example. flat r (b) Adopt spherical polar coordinates, x = rsinll cos</>, y = r sin II sin <1>, z=rcosll. By regarding h and by using the usual form IL,. and hILv as components of tensors in fiat spacetime (see end of Box 18.2), tensor transformation laws, put the solution found in (a) into the 2M hoo = -,r 2M hjk = -g j k r nat where gall nat are the components of the fiat-spacetime metric in the spherical coordinate system goonat = -I, g\u00a2\u00a2 =r 2 sin211. nat grrnat = \\, gall nat goo nat = r2 \u2022 = 0 when a f:. fi. 442 18. WEAK GRAVITATIONAL FIELDS Thereby conclude that the general relativistic line element, accurate to linearized order, is (c) Derive this general, static, spherical1y symmetric, Lorentz-gauge, vacuum solution to. the linearized field equations from scratch, working entirely in spherical coordinates. [Hint: As discussed at the end of Box 18.2, 1)llv in equation (18.8c) must be replaced by gil' : and in the field equations and gauge conditions (l8.8a, b), al1 commas (partial derivativ~a~) must be replaced by covariant derivatives, whose connection coefficients come from gil v oar] (d) Calculate the Riemann curvature tensor for this gravitational field. The answer should agree with equation (1.14). \u00a718.2. GRAVITATIONAL WAVES Linearized theory and electromagnetic theory compared The gauge conditions and field equations (I8.8a, b) oflinearized theory bear a close resemblance to the equations of electromagnetic theory in Lorentz gauge and flat spacetime, A\"',,,, = 0, -AIL,,,,'\" = 4'JTJIL. (I8.lOa) (18.lOb) They differ only in the added index (hlL' versus AIL, P' versus JIL). Consequently, from past experience with electromagnetic theory, one can infer much about linear ized gravitation theory. For example, the field equations (I8.8b) must have gravitational-wave solutions. The analog of the electromagnetic plane wave All = AII(t - z), AO = 0, Plane gravitational waves will be the gravitational plane wave Jizz = Jizz(t - z), JiZII = JiZII(t - z), JiILO= JiJlZ = 0 for all p.. Jillll = Jillll(t - z), (I 8.1 1) Although a detailed study of such waves will be delayed until Chapters 35-37, some properties of these waves are explored in the exercises at the end of the next section. \u00a718.3. EFFECT OF GRAVITY ON MATTER How to analyze effects of weak gravity on matter The effects of weak gravitational fields on matter can be computed by using the linearized metric (I8.1) and Christoffel symbols (I 8.2) in the appropriate equations of motion-i.e., in the geodesic equation (for the motion of particles or light rays), in the hydrodynamic equations (for fluid matter), in Maxwell's equations (for elec tromagnetic waves), or in the equation V' T = 0 for the total stress-energy tensor \u00a7 18.3. EFFECT OF GRAVITY ON MATTER 443 of whatever fields and matter may be present. Exercises 18.5, 18.6 and 18.7 provide examples, as do the Newtonian-limit calculations in exercises 16.1 and 16.4, and in \u00a717.4. If, however, the lowest-order (linearized) gravitational \"forces\" (Chris toffel-symbol terms) have a significant influence on the motion of the sources of the gravitational field, one finds that the linearized field equation (I8.7) is inadequate, and better approximations to Einstein's equations must be considered. [", "Thus emission of gravitational waves by a mechanically or electrically driven oscillator falls within the scope of linearized theory, but emission by a double-star system, or by stellar oscillations that gravitational forces maintain, will require discussion of nonlinear terms (gravitational \"stress-energy\") in the Einstein equations; see \u00a7\u00a736.9 to 36.11.] The above conclusions follow from a consideration of conservation laws associated with the linearized field equation. Just as the electromagnetic equations (I 8.1 Oa, b) guarantee charge conservation jp. = 0,,p. f all space jO(t, x) dx dy dz=Q = const, so the gravitational equations (I8.8a, b) guarantee conservation of the total 4-mo mentum and angular momentum of any body bounded by vacuum: Conservation of 4-momentum and angular momentum in linearized theory f body TP.O(t, x) dx dy dz = pp. = const; (x aT!3P. - x!3 pP.) = 0,,p. f (x a_T!3 0 _ x!3 TaO) dx dy dz = ja!3 = const. body (I8.12a) (I8.12b) (I8.l3a) (I8.l3b) (See \u00a75.11 for the basic properties of angular momentum in special relativity. The angular momentum here is calculated relative to the origin of the coordinate system.) Now it is important that the stress-energy components TIL', which appear in the linearized field equations (18.7) and in these conservation laws, are precisely the components one would calculate using special relativity (with gp.. = Tlp..)' As a result, the energy-momentum conservation formulated here contains no contributions or effects of gra vity! From this one sees that linearized theory assumes that gravitational forces do no significant work. For example, energy losses due to gravitational radiation-damping forces are neglected by linearized theory. Similarly, conservation ~enttrnl-Pi'fOr eaCh oTllie oodies acting as sources of hp.. means that each body moves along a geodesic of TIp.. (straight lines in the nearly Lorentz coordinate system) rather than along a geodesic of gp.. = TIp.. + hp.\u2022. Thus, linearized theory can be used to calculate the motion of test particles and fields, using gp.. = Tlp.v + hp..; but to include gravitational corrections to the motion of the sources themselves-to allow them to satisfy Tp.v;. = 0 rather than p.v\", = O-one must reinsert into the field equations the nonlinear terms that linearized theory discards. (See, e.g., Chapter 20 on conservation laws; \u00a7\u00a736.9-36.11 on the generation of gravitational waves and radiation reaction; and Chapter 39 on the post-Newtonian approximation.) Limit on validity of linearized theory: gravity must not affect motions of sources significantly 444 18. WEAK GRAVITATIONAL FIELDS The energy, momentum, and angular momentum radiated by gravitational waves in linearized theory can be calculated by special-relativistic methods analogous to those used in electromagnetic theory for electromagnetic waves [Fiertz and Pauli (1939)], but it will be more informative and powerful to use a fully gravitational approach (Chapters 35 and 36). EXERCISES Exercise 18.4. SPACETIME CURVATURE FOR A PLANE GRAVITATIONAL WAVE Calculate the components of the Riemann curvature tensor [equations (18.9)] for the gravita tional plane wave (18.11). [Answer: R - -R.1'OyO -.1'01lZ - + zzyz - R - R zzyO - - - I Ii I. XlI,lt' an other components vanish except those obtainable from the above by the symmetries Ro:/3y5 = R[o:/3Hy51 = Ry5o:/3' Exercise 18.5. A PRIMITIVE GRAVITATIONAL-WAVE DETECTOR (see Figure 18.1) Two beads slide almost freely on a smooth stick; only slight friction impedes their sliding. The stick fans freely through spacetime, with its center moving along a geodesic and its ends attached to gyroscopes, so they do not rotate. The beads are positioned equidistant (distance ~ 1) from the stick's center. Plane gravitational waves [equation (18.11) and exercise 18.4]. impinging on the stick, push the beads back and forth (\"geodesic deviation\"; \"tidal gravita tional forces\"). The resultant friction of beads on stick heats the stick; and the", " passage of the waves is detected by measuring the rise in stick temperature.* (Of course, this is not the best of all conceivable designs!) Neglecting the effect of friction on the beads' motion, calculate the proper distance separating them as a function of time. [Hints: Let ( be the separation between the beads; and let n = (/1(1 be a unit vector that points along the stick in the stick's own rest frame. Then their separation has magnitude 1 = (. n. The fact that the stick is nonrotating is em bodied in a paranel-transport law for n. Vun = O. (\"Fermi Walker transport\" of\u00a7\u00a76.5, 6.6, and 13.6 reduces to parallel transport, because the stick moves along a geodesic with a = Vuu = 0.) Thus, d1/dr = V u \u00ab(' n) = (V,,()\u00b7 n_ d 21/dr 2 = V u V u \u00ab(' n) = (Vu Vu ( )'n, where r is the stick's proper time. But Vu Vu ( the wave (geodesic deviation); is produced by the Riemann curvature of Vu Vu ( = projection along n of [-Riemann (..., u, (, u)]. (The geodesic-deviation forces perpendicular to the stick, Le., perpendicular to n, are coun- *This thought experiment was devised by Bondi [1957, 1965; Bondi and McCrea (1960\u00bb) as a means for convincing skeptics of the reality of gravitational waves. \u00a7 18.4. NEARLY NEWTONIAN GRAVITATIONAL FIELDS 445 r--t----..\\ +====~~_'~-~ Gyroscope \" / Bea Stick Bead ~ Gyroscope Figure 18.1. A primitive detector for gravitational waves, consisting of a beaded stick with gyroscopes on its ends [Bondi (1957\u00bb). See exercise 18.5 for discussion. terbalanced by the stick's pushing back on the beads to stop them from passing through it-no penetration of matter by matter!) Thus, d 21/d-r 2 = -Riemann (..., u, (, u)\u00b7 n = -Riemann (n, u, (, u). Evaluate this acceleration in the stick's local Lorentz frame. Orient the coordinates so the waves propagate in the z-direction and the stick's direction has components nZ = cos e, n Z = sin ecos <1>, n Y = sin esin <1>. Solve the resulting differential equation for 1(-r).] [Answer: where h;k are evaluated on the stick's world line (x = y = z = 0). Notice that, if the stick is oriented along the direction of wave propagation (if e = 0), the beads do not move. In this sense, the effect of the waves (geodesic deviation) is purely transverse. For further discussion, see \u00a7\u00a735.4 to 35.6.] \u00a718.4. NEARLY NEWTONIAN GRAVITATIONAL FIELDS The general solution to the linearized field equations in Lorentz gauge [equations (I8.8a, b)] lends itself to expression as a retarded integral of the form familiar from electromagnetic theory: h- ) - f 4 Tp.v(t - Ix - x'I, x') d3 I x. (I 8.14),x - p.v (t Ix - x'I Retarded-integral solution of linearized field equation The gravitational-wave aspects of this solution will be studied in Chapter 36. Here focus attention on a nearly Newtonian source: Too} ITo;l, Too} l1jk l, and velocities slow enough that retardation is negligible. In this case, (18.14) reduces to Newtonian gravity as a limit of linearized theory ifJ(t, x) = - f Too(t, x') d 3 Ix - x'I I \u2022. X' = NewtonIan potentia.. The corresponding metric (I8.8c) is ds'2 = -(1 + 2ifJ) dF + (1 - 2ifJ)(dx2 + dy 2 + d~2) ::::: -(1 - 2M/r) dt2 + (I + UvI/r)(dx 2 + df + d~'2) far from source. (18.15a) (18.15b) (18.15c) Bending of light and gravitational red shift predicted by linearized theory EXERCISES 446 18. WEAK GRAVITATIONAL FIELDS The errors in this metric are: (1) missing corrections of order (/J2 due to nonlinearities of which", " linearized theory is oblivious; (2) missing corrections due to setting hOi = 0 IToY Too is a typical velocity in the source); (these are o( order hOi (3) missing corrections\"due to setting ~k = 0 [these are of order ~k (/J(I1jkl/Too)]' In the solar system all these errors are _10- 12, whereas (/J - 10-6 \u2022 (/Ju, where u - Passive correspondence with Newtonian theory demanded only that goo = - (1 + 2(/J); see equation (17.19). However, linearized theory determines all the metric coefficients, up to errors of -(/Ju, _(/J2, and -(/J(I1jkl/Too)' This is sufficient accuracy to predict correctly (fractional errors _10-6 ) the bending of light and the gravitational redshift in the solar system, but not perihelion shifts. Exercise 18.6. BENDING OF LIGHT BY THE SUN To high precision, the sun is static and spherical, so its external line element is (18.15c) with rp ::: -M/r; i.e., ds 2 = -(I - 2M/r) dP + (I + 2M/r)(dx2 + df + dz 2 ) everywhere outside sun. (18.16) A photon moving in the equatorial plane (z ::: 0) of this curved spacetime gets deflected very slightly from the world line x = 1, y = b = \"impact parameter,\" z = O. (18.17) Calculate the amount of deflection as follows. (a) Write down the geodesic equation (16.4a) for the photon's world line, dp\" \" d>... + r i3YP p - o. i3 Y_ (18.18) [Herep = d/d>..\u00b7 = (4-momentum of photon) = (tangent vector to photon's null geodesic).] (b) By evaluating the connection coefficients in the equatorial plane, and by using the approximate values, Ipyl ~ pO =::: pz, of the 4-momentum components corresponding to the approximate world line (18.17), show that pz = po[ 1+ o( ~)] = const[ 1+ o( ~)]. (c) Integrate this equation for pU, assumingpY = 0 at x = - 00 (photon moving precisely in x-direction initially); thereby obtain (d) Show that this corresponds to deflection of light through the angle.:14> = 4M/b = 1\".75 (R 0 /b), (18.19) where R 0 is the radius of the sun. For a comparison of this prediction with experiment, see Box 40.1. \u00a718.4. NEARLY NEWTONIAN GRAVITATIONAL FIELDS 447 Exercise 18.7. GRAVITATIONAL REDSHIFT (a) Use the geodesic equation for a photon, written in the form I dp/dA* - T\"p./3p\"p/3 = 0, to prove that any photon moving freely in the sun's gravitational field [line element (18.16)] has dp~ 0; i.e-r-__ Po = constant along photon's world line. (18.20) (b) An atom at rest on the sun's surface emits a photon of wavelength A., as seen in its orthonormal frame. [Note: hll. = h/A. = (energy atom measures) = -p' u.' (18.21) where p is the photon's 4-momentum and u. is the emitter's 4-velocity.] An atom at rest far from the sun receives the photon, and measures its wavelength to be Ar \u2022 [Note: h/Ar = -p' u r.] Show tha.t the photon is redshifted by the amount [Hint: Ur = a/at; u. = (I - 2M/r)-1/2 a/at. Why?] For further discussion of the gravitational redshift and experimental results, see \u00a7\u00a77.4 and 38.5; also Figures 38.1 and 38.2. (18.22) MASS AND ANGULAR MOMENTUM OF A GRAVITATING SYSTEM \u00a719.1. EXTERNAL FIELD OF A WEAKLY GRAVITATING SOURCE Metric far from a weakly gravitating system, as a power series in 1If: Consider an isolated system with gravity so weak that in calculating its structure and motion one can completely ignore self-gravitational effects. (This is true of an asteroid, and of a nebula with high", "-energy electrons and protons spiraling in a magnetic field; it is not true of the Earth or the sun.) Assume nothing else about the system-for example, by contrast with Newtonian theory, allow velocities to be arbitrarily close to the speed oflight, and allow stresses Tik and momentum densities ]'Oi to be comparable to the mass-energy density ]'00. (1) derivation Calculate the weak gravitational field, gp.p = Tlp.p + hp. p, h = h =f 4 Tp.p(t - p.p - p.p Ix - x'l, x') d3x' Ix - x'i (19.1 ) (19.2)'produced by such a system [see \"baIred\" version of equation (18.14)]. Restrict attention to the spacetime region far outside the system, and expand hp. p in powers of x'/, = x'/Ixl, using the relations _ T (t p.p \" Ix - x'l, x') = \"\" L.. n! n=O 1 [on _ atn T (t p.p ] \" x') (, - Ix - x'l)n, (19.3a), - Ix - x'i = Xi ( 1 xix k (xi'x k x,.), 2, + - - -'-,2,'2 Q'k), +.. \" 1 ---=-+--+--Ix - X'I 1 xix k (3 Xi'x k',2 2,3 xi xi',2, 1,,'2 Q'k), + (19.3b) (19.3c) \u00a719.1. EXTERNAL FIELD OF A WEAKLY GRAVITATING SOURCE Perform the calculation in the system's rest frame, where pi=f TOi d 3x = 0, with origin of coordinates at the system's center of mass f xiTOo d 3x = O. The result, after a change of gauge to simplify hoo and hOi' is ds 2 = - [1 - 2~ + 0(/3)] dt 2 - [4\u00a3jklSk ;: + 0(/3)] dt dxi + [( 1 + 2M) Q'k + (gravit~tional radiation terms)] dxi dxk. r ) that die out as 0(1/r) 449 (19.4a) (19.4b) (2) resull (19.5) (see exercise 19.1 for derivation.) Here M and Sk are the body's mass and intrinsic angular momentum. M= f TOod 3x, S - f\u00a3 x l T mO d 3x kIm k -. (19.6a) (19.6b) The corresponding Newtonian potential is 1 (/J = - '2 (goo - M ( 1 ) 1)00) = - -;: + 0 73. (l9.6c) Conclusion: With an appropriate choice of gauge, (/J and goo far from any weak source are time-independent and are determined uniquely by the source's mass M; gOj is time-independent and is fixed by the source's intrinsic angular momentum Si; but gik has time-dependent terms (gravitational waves!) of O(1/r). The rest of this chapter focuses on the \"imprints\" of the mass and angular mo mentum in the gravitational field; the gravitational waves will be ignored, or almost so, until Chapter 35. How melric depends on syslem's mass M and angular momenlum S Exercise 19.1. DERIVATION OF METRIC FAR OUTSIDE A WEAKLY EXERCISE GRAVITATING BODY (a) Derive equalion (19.5). [Hints: (I) Follow Ihe procedure ouilined in Ihe lexl. (2) When calculating hoo' wrile oul expliciily Ihe n = 0 and n = I lerms of(19.2), 10 precision 0(I/r 2), and simplify Ihe n = 0 lerm using Ihe idenlilies ( 19.7a) (19.7b) 450 19. MASS AND ANGULAR MOMENTUM OF A GRAVITATING SYSTEM (Verify that these identities follow from T\"'/3'/3 = 0.) (3) When calculating hOm' write out explicitly the n = 0 term of (19.2), to precision 0(I/r2), and simplify it using the identity (Veri(v that this follows from T\"'/3./3 = 0.) (4) Simplify", " hoo and hOm by the gauge transforma tion generated by (19.7c) \" 10\"-1 J \"=2 ~m = - -3JToo'x' x m -, + 2: n! 01\"-1 2x i \" r (Too' + Tkk ') (r-lx-x'l)\" Ix _ x'i 3 d x', I 0\"-1 J (r - d 3x' + 4 2:,~ TOm' \"=In.u1 Ix - x'l)\" I _ 'I X X 3 d x' r + ~~ - 1(1) JT. I 0\"-2 J 2 r,m \" 0 - :?2 n! 01\"-2 T. (00 00 'r,2 d 3x' - (\u00a3) J(TOi'XfXk' - 1TOk'r'2) d3x' 2 r2,m'+ T.'kk) [(r - Ix - X'I)\"] Ix - x'i,m d 3x' Here Tp.: denotes Tp.v(t - r, x').] (b) Prove that the system's mass and angular momentum are conserved. [No Ie: Because T\"/3./3 = 0 (self-gravity has negligible influence), the proof is no different here than in flat spacetime (Chapter 5).] \u00a719.2. MEASUREMENT OF THE MASS AND ANGULAR MOMENTUM The values of a system's mass and angular momentum can be measured by probing the imprint they leave in its external gravitational field. Of all tools one might use to probe, the simplest is a test particle in a gravitationally bound orbit. If the particle is sufficiently far from the source, its motion is affected hardly at all by the source's angular momentum or by the gravitational waves; only the spherical, Newtonian part of the gravitational field has a significant influence. Hence, the particle moves in an elliptical Keplerian orbit. To determine the source's mass M, one need only apply Kepler's third law (perhaps better called \"Kepler's 1-2-3 law\"): M = ( 2'JT orbital period )2 (Semi-major aXis)3.'of ellipse (19.8) The source's angular momentum is not measured quite so easily. One must use a probe that is insensitive to Newtonian gravitational effects, but \"feels\" the off diagonal term, (19.9) For a weakly gravitating system: (1) total mass M can be measured by applying Kepler's\" 1-2-3\" law to orbiting particles \u00a719.3. FULLY RELATIVISTIC SOURCES 451 in the metric (19.5). One such probe is the precession of the perihelion of a corevolv ing satellite, relative to the precession for a counterrevolving satellite. A gyroscope is another such probe. Place a gyroscope at rest in the source's gravitational field. By a force applied to its center of mass, prevent it from falling. As time passes, the gOi term in the metric will force the gyroscope to precess relative to the basis vectors a/ax;; and since these basis vectors are \"tied\" to the coordinate system, which in turn is tied to the Lorentz frames at infinity, which in turn are tied to the \"fixed stars\" (cf. \u00a739.12), the precession is relative to the \"fixed stars.\" The angular velocity of precession, as derived in exercise 19.2, is (2) total angular momentum S can be measured by examining the precession of gyroscopes n ~~ - _ ~[-S 3(S' X)X] + 2 3 r r \u2022 (19.10) One sometimes says that the source's rotation \"drags the inertial frames near the source,\" thereby forcing the gyroscope to precess. For further discussion, see \u00a7\u00a721.12, 40.7, and 33.4. Exercise 19.2. GYROSCOPE PRECESSION Derive equation (19.10) for the angular velocity of gyroscope precession. [Hints: Place an orthonormal tetrad at the gyroscope's center of mass. Tie the tetrad rigidly to the coordinate system, and hence to the \"fixed stars\"; more particularly, choose the tetrad to be that basis {e,,} which is dual to the following I-form basis: EXERCISE The spatial legs of the tetrad,- eJ, rotate relative to the gyroscope with an angular velocity w given by [see equation (13.69)] wi = [I + (2M/r)p/2 dx i. (19.11) Consequently,", " the gyroscope's angular momentum vector L precesses relative to the tetrad with angular velocity n = -w: (19.12) Calculate Tilo for the given orthonormal frame, and thereby obtain equation (19.10) for n.] \u00a719.3. MASS AND ANGULAR MOMENTUM OF FULLY RELATIVISTIC SOURCES Abandon, now, the restriction to weakly gravitating sources. Consider an isolated, gravitating system inside which spacetime mayor may not be highly curved-a black hole, a neutron star, the Sun,... But refuse, for now, to analyze the system's interior or the \"strong-field region\" near the system. Instead, restrict attention to the weak 452 19. MASS AND ANGULAR MOMENTUM OF A GRAVITATING SYSTEM gravitational field far from the source, and analyze it using linearized theory in vacuum. Expand hl\"> in multipole moments and powers of l/r; and adjust the gauge, the Lorentz frame, and the origin of coordinates to simplify the resulting metric.. The outcome of such a calculation is a gravitational field identical to that for a weak source [equation (19.5)]! (Details of the calculation are not spelled out here because of their length; but see exercise 19.3.) But before accepting this as the distant field of an arbitrary source, one should examine the nonlinear effects in the vacuum field equations. Two types of nonlinear ities tum out to be important far from the source: (1) nonlinearities in the static, Newtonian part of the metric, which generate metric corrections (see exercise 19.3 and \u00a739.8), thereby putting the metric into the form [ - 1--+--+0 r_ dt2 2M 2M2 r2 + [(1 + 2M + 3M2) O. + (gravit~tional radiation terms)] dx j dxk; 4t:\u00b7 I S k -+0 ( 1 )] xl r 3 r3 ( 1 )] r 3. dtdx' Jk r [ (19.13) r 2r2 Jk that die out as O(1/r) (2) a gradual decrease in the source's mass, gradual changes in its angular momen tum, and gradual changes in its' \"rest frame\" to compensate for the mass, angular momentum, and linear momentum carried off by gravitational waves (see Box 19.1, which is best r~ad only after finishing this section). By measuring the distant spacetime geometry (19.13) ofa given source, one cannot discover whether that source has strong internal gravity, or weak. But when one expresses the constants M and Sj' which determine goo and go;, as integrals over the interior of the source, one discovers a crucial difference: if the internal gravity is weak, then linearized theory is valid throughout the source, and S - f' - j t:jklX k TfO d 3 \u2022 x, (19.14) but if the gravity is strong, these formulas fail. Does this failure prevent one, for strong gravity, from identifying the constants M and Sj of the metric (19.13) as the source's mass and angular momentum? Not at all, according to the following argument. Consider, first, the mass of the sun. For the sun one expects Newtonian theory to be highly accurate (fractional errors -- M0 / R 0 -- 10-6 ); so one can assert that the constant M appearing in the line element (19.13) is, indeed M = f Pd 3x = f Too d 3x = total mass. But might this assertion be wrong? To gain greater confidence and insight, adopt the viewpoint of \"controlled ignorance\"; i.e., do not pretend to know more than what is needed. (This style of physical argument goes back to Newton's famous \"Hypoth eses non fingo,\" i.e. \"I do not feign hypotheses.\") In evaluating the volume integral of Too (usual Newtonian definition of M), one needs a theory of the internal structure Metric far from any gravitating system, as a power series in 1/r Failure of volume integrals for M and S when source has strong internal gravity \u00a7 19.3. FULLY RELATIVISTIC SOURCES 453 of the sun. For example, one must know that the visible surface layers of the sun do not hide a massive central core, so dense and large that relativistic gravitational fields 111>1 -- 1 exist there. If one makes use in the analysis of a fluid-type stress-energy tensor p.v, one needs to know equations of state, opacities, and theories of energy generation and transport. One needs to justify the fluid description as an adequate approximation to the atomic constitution of matter. One needs to assume that an ultimate theory of", " matter explaining the rest masses of protons and electrons will not assign an important fraction of this mass to strong (nonlinear) gravitational fields on a submicroscopic scale. It is plausible that one could do all this, but it is also obvious that this is not the way the mass of the sun is, in fact, determined by astronomers! Theories of stellar structure are adjusted to give the observed mass; they are not constructed to let one deduce the mass from nongravitational observa tions. The mass of the sun is measured in practice by studying the orbits of planets in its external gravitational field, a procedure equivalent to reading the mass M off the line element (19.13), rather than evaluating the volume integral fTOo d 3x. To avoid all the above uncertainties, and to make theory correspond as closely as possible to experiment, one defines the \"total mass-energy\" M of the sun or any other body to be the constant that appears in the line element (19.13) for its distant external spacetime geometry. Similarly, one defines the body's intrinsic angular mo mentum as the constant 3-vector S appearing in its line element (19.13). Operationally, the total mass-energy M is measured via Kepler's third law; the angular momentum S is measured via its influence on the precession of a gyroscope or a planetary orbit. This is as true when the body is a black hole or a neutron star as when it is the sun. What kind of a geometric object is the intrinsic angular momentum S? It is defined by measurements made far from the source, where, with receding distance, spacetime is becoming flatter and flatter (asymptotically flat). Thus, it can be regarded as a 3-vector in the \"asymptotically flat spacetime\" that surrounds the source. But in what Lorentz frame is S a 3-vector? Clearly, in the asymptotic Lorentz frame where the line element (19.13) is valid; i.e., in the asymptotic Lorentz frame where the source's distant \"coulomb\" (\"M/r\") field is static; i.e., in the \"asymptotic rest frame\" of the source. Alternatively, one can regard S as a 4-vector, S, which is purely spatial (SO = 0) in the asymptotic rest frame. If one denotes the 4-velocity of the asymptotic rest frame by U, then the fact that S is purely spatial can be restated geometrically as S\u00b7 U = 0, or S\u00b7p = 0, p =MU =\"total 4-momentum of source\" (19.15) (19.16) where Definition of \"total mass-energy\" M and \"angular momentum\" S in terms of external gravitational field S as a geometric object in an asymptotically flat region far outside source \"Asymptotic rest frame\" and \"total 4-momentum\" is still another vector residing in the asymptotically flat region of spacetime. The total4-momentum P and intrinsic angular momentum S satisfy conservation laws that are summarized in Box 19.1. These conservation laws are valuable tools in gravitation theory and relativistic astrophysics, but the derivation of these laws (Chapter 20) does not compare in priority to topics such as neutron stars and basic cosmology; so most readers will wish to skip it on a first reading of this book. (conTinued on page 456) Conservation laws for total 4-momentum and angular momentum Box 19.1 TOTAL MASS-ENERGY, 4\u00b7MOMENTUM, AND ANGULAR MOMENTUM OF AN ISOLATED SYSTEM A. Spacetime is divided into (1) the source's inte~ rior; which is surrounded by (2) a strong-field vacuum region; which in turn is surrounded by (3) a weak-field, asymptotically flat, near-zone re gion; which in turn is surrounded by (4) a weak field, asymptotically flat, radiation-zone region. This box and this chapter treat only the asympto tically flat regions. The interior and strong-field regions are treated in the next chapter. I, 1, 1, I, 1, I,, \\ \"'-- J- y x I I I, :0 <:: \" 1 '..::;-.<> <> - c I..::;-.\u00a7 \";1.~ ~ B ~, - N -, <> I.~.g ~II g;; o Ic.~ E~ ~:;; 1 0: \"5 1 \",e.:: 1 <J;I <<:: I '.< I'\" '--'...../ t ---,--- Interior, ) --/", " f==1,.~ 1 ~ I \"0, 0:; 1 1 I 1 1 <C: if 2 <;; s 1 \"0 ;; I.9 I I.....L. / / - - I-----:.-.~y x Asymptotic rest frame B. The asymptotic rest frame of the source is that global, asymptotically Lorentz frame (coordinates t, x, y, z) in which the distant, \"coulomb\" part of the source's field is at rest (see diagram). The asymptotic rest frame does not extend into the strong-field region; any such extension of it would necessarily be forced by the curvature into a highly non-Lorentz, curvilinear form. The spatial origin of the asymptotic rest frame is so adjusted that the source is centered on it-Le., that the distant (j> = -M/(x2 + y2 + is Newtonian z2)1/2 + 0(1/ r 3); i.e., that (j> has no dipole term, D\u00b7 x/r3, such as would originate from an offset of the coordinates. potential C. To the source one can attribute a total mass-energy M, a 4-velocity U, a total 4-momentum P, and an intrinsic angular momentum vector, S. The 4-vectors U, P, and S reside in the asymptotically flat region of spacetime and can be moved about freely there (negligible curvature = parallel transport around closed curves does not change U, P, or S). The source's 4-velocity U is defined to equal the 4-velocity of the asymptotic rest frame (UO = 1, U = 0 in rest frame). The total mass-energy M is measured via Kepler's third (\"1-2-3\") law [equation (19.8)]. The total4-momentum is defined by P = MU. The intrinsic angular momentum S is orthogonal to the 4-velocity U, S' U = 0 (so So = 0; S # 0 in general in asymptotic rest frame); S is measured via gyroscope precession or differential perihelion precession (\u00a719.2). In the asymptotic rest frame, with an appropriate choice of gauge (i.e., of ripples in the coordinates), the slight deviations from fiat-spacetime geometry are described by the line element ds 2 = - + [1 - 2~+ 2:;2 + 0 (,13 )] dt2 [( 2) I + 2~ + 32~ - [ 4\u00a3ikl Sk ;; + 0 (,13 ) ] dt dxi 0ik + (gravitational radiation terms)] dx i dx k. (l) D. Conservation of 4-momentum and angular mo mentum: Suppose that particles fall into a source or are ejected from it; suppose that electromag netic waves flow in and out; suppose the source emits gravitational waves. All such processes break the source's isolation and can change its total 4 momentum P, its intrinsic angular momentum 5, and its asymptotic rest frame. Surround the source with a spherical shell S, which is far enough out to be in the asymptotically flat region. Keep this shell always at rest in the source's momentary asymptotic rest frame. By probing the source's gravitational field near S, measure its 4-momen tum P and intrinsic angular momentum 5 as func tions of the shell's proper time r. An analysis given in the next chapter reveals that the 4-momentum is conserved, in the sense that ~\\\\M\u00a2~,Ly.........J InterstelIar debris falIs into a black hole, and gravitational waves emerge. dpa -- = dr i'T aln. d(area) = s' - 1 (rate at whieh 4-momentum) flows inward through shell, (2) where n is the unit outward normal to S and the integral is performed in the shell's momentary rest frame. In words: the rate at which 4-momentum flows through the shell, as measured in the standard special relativistic manner, equals the rate of change of the source's gravitationally measured 4-momentum. Similarly, the angular momen tum is conserved in the sense that dS. dr' = - i (\u00a3lik xi Tkl)n 1 d(area) = momentum flows inward, (rate at which angular ) (3a) S through the shell dSo - - - -S dr a - dua dr - (Change required to keep 5 orthogonal to U;) \"Fermi-Walker-transport law\": cf. \u00a7\u00a76", ".5, 13.6. (3 b) In these conservation laws Ta{3 is the total stress-energy tensor at the shell, including contributions from matter, electromagnetic fields, and gravitational waves. The gravitational-wave contribution, called TWWla{3, is treated in Chapter 35. Note: The conservation laws in the form stated above contain fractional errors of order Mjr (contributions from \"gravitational potential energy\" of infalling mate rial), but such errors go to zero in the limit of a very large shell (r ~ 00). Note: The formulation of these conservation laws given in the next chapter is more precise and more rigorous, but less physically enlightening than the one here. [455] 456 19. MASS AND ANGULAR MOMENTUM OF A GRAVITATING SYSTEM EXERCISE Exercise 19.3. GRAVITATIONAL FIELD FAR FROM A STATIONARY, FULLY RELATIVISTIC SOURCE Derive the line element (19.13) for the special case of a source that is time-independent (glLl',t = 0). This can be a difficult problem, if one does not proceed careful1y along the fol1owing outlined route. (I) Initial1y ignore al1 nonlinearities in the Einstein field equations. The field is weak far from the source. These nonlinearities will be absent from the dominant terms. (2) Calculate the dominant terms using linearized theory in the Lorentz gauge [equa tions (18.8)]. (3) In particular, write the general solution to the vacuum, time-independent wave equation (l8.8b) in the fol1owing form involving ni=xi/r=(unit vector in radial direction): (19.17) (Round brackets denote symmetrization.) (4) Then impose the Lorentz gauge conditions ho:/3'/3 = 0 on this general solution, thereby learning Ai = 0, Aik = 0, Bik(Sik _ 3n in k) = 0, Bik!(Skl _ 3n knl ) = O. (19.18) (5) Write Bik as the sum of its trace 3B, its traceless symmetric part Sik, and its traceless antisymmetric part (these are its \"irreducible parts\"): Sii =-0. (19.19) Show that any tensor Bik can be put into such a form. Then show that the gauge conditions (19.18) imply Sik = O. (6) Similarly show that any tensor Bikl that is symmetric on its first two indices can be put into the form E km symmetric and traceless, i.e., Si k! symmetric and traceless, i.e., Then show that (7) Combining al1 these results, conclude that the gauge conditions (19.18) (19.20) Si k! = Slik!l, Siil = Sikk = Siki = O. imply Ci = - 2Ai and Ekm = Sikl = O. (19.21) \u00a719.4. A CLOSED UNIVERSE 457 Then use gauge transformations, which stay within Lorentz gauge, to eliminate B and Ai from hOi and hik ; so _ hoo - _ AO - + r (Bi + Ai)ni 2 r (-.!..) + 0 3'r (19.22) (8) Translate the origin of coordinates so xinew = xiold (Bi + A i)/AO; in the new coordinate system ho:f3 has the same form as (19.22), but with Bi + Ai removed. From the resultant ho:f3, construct the metric and redefine the constants AO and Fl to agree with equation (19.13). (9) All linear terms in the metric are now accounted for. The dominant nonlinear terms must be proportional to the square, (M/r)2, of the dominant linear term. The easiest way to get the proportionality constant is to take the Schwarzschild geometry for a fully relativistic, static, spherical source [equation (31.1)1, by a change of coordinates put it in the form ds 2 = _ (I - M/2r)2 dt2 + (I + M)4 (dx2 + dy2 + dz2) (19.23) 1+ M/2r 2r (exercise 25.8), and expand it in powers of M/r. \u00a719.4, MASS AND ANGULAR MOMENTUM OF A CLOSED UNIVERSE \"There are no snakes in Ireland. \" Statement of St. Patrick after driving the snakes out of Ireland (legend') There is no such thing as \"the energy (or angular", " momentum, or charge) of a closed universe,\" according to general relativity, and this for a simple reason, To weigh something one needs a platform on which to stand to do the weighing. To weigh the sun, one measures the periods and semimajor axes of planetary orbits, and applies Kepler's \"1-2-3\" law, M = w 2a 3. To measure the angular momentum, S, of the sun (a task for space technology in the 1970's or 1980's!), one measures the precession of a gyroscope in a near orbit about the sun, or one examines some other aspect of the \"dragging of inertial frames.\" To determine the electric charge \u2022 Stokes (1887) and other standard references deny this legend. In part I of Stokes the basic manuscript references are listed, including especially codex manuscript Rawlinson B.512 in 154 folios. in double columns, written by various hands in the fourteenth and fifteenth centuries (cf Ca/alogi codicum manuscrip/orum Biblio/hecae Bodleianae Par/is Quinlae Fasciculus Primus, Oxford, 1862, col. 728-732). In this manuscript, folio 97b.l, line 14, reads in the translation of Stokes. Part 1. p. xxx: \"as Paradise is without beasts, without a snake, without a lion, without a dragon, without a scorpion. without a mouse. without a frog, so is Ireland in the same manner without any harmful animal, save only the wolf...\" For a closed universe the total mass-energy M and angular momentum S are undefined and undefinable 458 19. MASS AND ANGULAR MOMENTUM OF A GRAVITATING SYSTEM of a body, one surrounds it by a large sphere, evaluates the electric field normal to the surface at each point on this sphere, integrates over the sphere, and applies the theorem of Gauss. But within any closed model universe with the topology of a 3-sphere, a Gaussian 2-sphere that is expanded widely enough from one point finds itself collapsing to nothingness at the antipodal point. Also collapsed to nothingness is the attempt to acquire useful information about the \"charge of the universe\": the charge is trivially zero. By the same token, every \"surface integral\" (see details in Chapter 20) to determine mass-energy or angular momentum collapses to nothingness. To make the same point in another way: around a closed universe there is no place to put a test object or gyroscope into Keplerian orbit to determine either any so-called \"total mass\" or \"rest frame\" or \"4-momentum\" or \"angular momentum\" of the system. These terms are undefined and undefinable. Words, yes; meanmg, no. Not having a defined 4-momentum for a closed universe may seem at first sight disturbing; but it would be far more disturbing to be given four numbers and to the components of some purported be told authoritatively that \"total energy-momentum 4-vector of the universe.\" Components with respect to what local Lorentz frame? At what point? And what about the change in this vector on parallel transport around a closed path leading back to that strangely preferred point? It is a happy salvation from these embarrassments that the issue does not and cannot arise! they represent Imagine a fancastically precise measurement of the energy of a y-ray. The experi menter wishes to know how much this y-ray contributes to the total mass-energy of the universe. Having measured its energy in the laboratory, he then corrects it for the negative gravitational energy by which it is boun-d to the Earth. The result, is the energy the photon will have after it climbs out of the Earth's gravitational field. But this is only the first in a long chain ofcorrections for energy losses (redshifts) as the photon climbs out of the gravitational fields of the solar system, the galaxy, the local cluster of galaxies, the supercluster, and then what? These corrections show no sigl!- of converging, unless to Ecorrected = O. Quite in contrast to the charge-energy-angular-momentum facelessness ofa closed universe are the attractive possibilities of defining and measuring all three quantities in any space that is asymptotically flat. One does not have to revolutionize present day views of cosmology to talk of asymptotically flat space. It is enough to note how small is the departure from flatness, as measured by the departure of ( - gOO)1/2 from unity, in cases of astronomical or astrophysical interest (Box 19.2). Surrounding a region where any dynamics, however complicated, is going on, whenever the geometry is asymptotically flat to some specified degree of precision, then to that degree", " of precision it makes sense to speak of the total energy-momentum 4-vector of the dynamic region, P, and its total intrinsic angular momentum, S. Parallel transport of either around any closed curve in the flat region brings it back to its Asymptotic flatness as the key to the definability of M and S \u00a7 19.4. A CLOSED UNIVERSE 459 Box 19.2 METRIC CORRECTION TERM NEAR SELECTED HEAVENLY BODIES m m r!!!... = 1 r (_goo)112 At shoulder of Venus de Milo 2 X 105 g = 1.5 X 10-23 em 30cm 5 X 10-25 ~ ~~At surface of Earth 6x1027 g 4 X 10-1 em 6.4 X 108 em 6 X 10-10 At Earth's distance from sun At sun's distance from center of galaxy At distance of galaxy from center of Virgo cluster of galaxies 2 X 1033 g 1.5 X 105 em 1.5 X 1013 em 1 X 10-8 2 X 1044 g 1.5 X 1016 em 2.5 X 1022 em 6 X 10-7 6 X 1047 g = 4 X 10 19 em 3 X 1025 em 1 X 10-6 starting point unchanged. Moreover, it makes no difference how enormous are the departures from flatness in the dynamic region (black holes, collapsing stars, intense gravitational waves, etc.); far away the curvature will be weak, and the 4-momentum and angular momentum will reveal themselves by their imprints on the spacetime geometry. CHAPTER 20 CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM We denote as energy of a material system in a certain state the contribution of all effects (measured in mechanical units of work) produced outside the system when it passes in an arbitrary manner from its stete to e reference state which has been defined ad hoc. WILUAM THOMPSON (later Lord Kelvin), as quoted by Max von Laue in Schilpp (1949), p. 5'14. All forms of energy possess inertia. ALBERT EINSTEIN, conclusion from his paper of September 26, 1905, as summarized by von Laue'in Schilpp (1949), p. 523. \"'\" \u00a720.1. OVERVIEW Chapter 5 (stress-energy tensor) is needed as preparation for this chapter, which in turn is needed as preparation for the Track-2 portion of Chapter 36 (generation of gravitational waves) and will be useful in understanding Chapter 35 (propagation of gravitational waves). Chapter 19 expounded the key features of total 4-momentum P and total angular momentum S for an arbitrary, gravitating system. But one crucial feature was left unproved: the conservation laws for P and S (Box 19.1). To prove those conservation laws is the chief purpose of this chapter. But other interesting, if less important, aspects of P and S will be encountered along the route to the proof-Gaussian flux integrals for 4-momentum and angular momentum; a stress-energy \"pseudotensor\" for the gravitational field, which is a tool in constructing volume integrals for P and S; and the nonlocalizability of the energy of the gravitational field. \u00a720.2. FLUX INTEGRALS FOR 4-MOMENTUM AND ANGULAR MOMENTUM 461 \u00a720.2. GAUSSIAN FLUX INTEGRALS FOR 4-MOMENTUM AND ANGULAR MOMENTUM In electromagnetic theory one can determine the conserved total charge of a source by adding up the number ofelectric field lines emanating from it-Le., by performing a Gaussian flux integral over a closed 2-surface surrounding it: Gaussian flux integrals for charge and Newtonian mass Q = -1-fEi d 2S. == -1-fFOi d 2S.. 1 1 4'lT 4'lT (20.1 ) Similarly, in Newtonian theory one can determine the mass ofa source by evaluating the Gaussian flux integral (20.2) These flux integrals work because the charge and mass of a source place indelible imprints on the electromagnetic and gravitational fields that envelop it. The external gravitational field (spacetime geometry) in general relativity possesses similar imprints, imprints not only of the source's total mass-energy M, but also of its total4-momentum P and its intrinsic angular momentum S (see Box 19.1). Hence, it is reasonable to search f?r Gaussian flux integrals that represent the 4-momentum and angular momentum of the source. To simplify the search, carry it out initially in linearized theory, and use Maxwell electrodynamics as a guide. In electrodynamics the Gaussian flux integral for charge follows", " from Maxwell's equations pt.'p,p = 4'lTP', plus the crucial fact that FP.P is antisymmetric, so that FOp.,p. = FOi,;: Q = fjO d 3x = -l-f FOp d 3x = -l-f poi. d 3x =-l-f poi d 2S\u00b7. 4'lT,1 4'lT 4'lT'p 1 [Gauss's theoremt--l To find analogous flux integrals in linearized theory, rewrite the linearized field equations (18.7) in an analogous form involving an entity with analogous crucial symmetries. The entity needed turns out to be (20.3) HJJ.av{l defined As one readily verifies from this expression, it has the same symmetries as the Riemann tensor Hp.aP{3 = HP{3p.a = J!lp.a][p{3], Hp.[ap{3] = O. (20.4) This entity, like!ip.p, transforms as a tensor under the Lorentz transformations of linearized theory; but it is not gauge-invariant, so it is not a tensor in the general relativistic sense. 462 20. CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM Linearized field equations in terms of HI\",ull In terms of H Wfl'{3, the linearized field equations (18.7) take on the much simplified form Gaussian flux integrals in linearized theory: (1) for 4-momentum (20.5) and from these, by antisymmetry of HIlQP{3 in v and /3, follow the source conservation laws of linearized theory, TIlP = _l_HIl(fP{3 16'ii tV,a{3v = 0, which were discussed back in \u00a718.3. The same antisymmetry as yields these equations of motion also produces a Gaussian flux integral for the source's tota14-momentum: d3x =-l-f Hp.aOj l6'lT. d3x,O:J,o:{3 pll =f Tllo d 3x =-l-f Hw,o{3 T1~'lT f Hp.aOj,o: d 2Sj. l6'lT l..{Gau:s's theorem] (20.6) Here the closed 2-surface of integration S must completely surround the source and must lie in a 3-surface of constant time xO. The integral (20.6) for the source's energy po, which is used more frequently than the integrals for pj, reduces to an especially simple form in terms of go:{3 = 'lJo:{3 + ho:{3: po = 1~'lT f (gjk,k - S gkk,j) d 2Sj (20.7) (see exercise 20.1). (2) for angular momentum integral for total angular momentum about the origin of coordinates: A calculation similar to (20.6), but more lengthy (exercise 20.2), yields a flux JIlP = f (x Il T\"0 - xPTIlO) d3x =-l-f (xllHPo:Oj l6'lT S -,0: x\"Hp.aOj + HlljOP - HpjOIl) d 2S,. (20.8),0: J Generalization of Gaussian flux integrals to full general relativity To evaluate the flux integrals in (20.6) to (20.8) (by contrast with the volume integrals), one need utilize only the gravitational field far outside the source. Since that gravitational field has the same form in full general relativity for strong sources as in linearized theory for weak sources, the flux integrals can be used to calculate pll and JIlP for any isolated source whatsoever, weak or strong: pll =-l-f Hp.aOj d2S. l6'lT S,0: J' JIlP =-l-f (xIJ.HPo:Oj l6'lT s,0: - xPHp.aOj,0: + HlljOP _ HpjOIl) d 2S \u2022 j in full general relativity theory, for any isolated source, when the closed surface of integration S is in the asymptotically flat region surrounding the source, and when asymptotically Minkowskian coordinates are used. (20.9) \u00a720.2", ". FLUX INTEGRALS FOR 4-MOMENTUM AND ANGULAR MOMENTUM 463 Knowing PIJ. and JIJ.v, one can calculate the source's total mass-energy M and intrinsic angular momentum SIJ. by the standard procedure of Box 5.6: Total mass-energy, center of mass, and intrinsic angular momentum M= (_PlJ.P~1/2, vector by which the source's asymptotic, P = -J/lPPp/M2 = \"M/r\", spherical field is displaced from ( being centered on the origin of coordinates S = -\u00a3 1 2 IJ.pup p (JIJ.P - ylJ.pP + YPPIJ.)pu/M. (20.10) (20.11) ) (20.12) Note especially that the integrands of the flux integrals (20.9) are not gauge-invari ant. In any local inertial frame at an event Cfo [glJ. p(9o) = 1)IJ.P' glJ.p,of!fo) = 0] they vanish, since g Il\",a. = h /lv,a = O=>HIJ.P<>{3 = o\u00b7,a'g = 1) IJ.P IJ.P =>HIJ.P<>{3 = 0. This is reasonable behavior; their Newtonian analog, the integrand fP,i = (gravita tional acceleration) of the Newtonian flux integral (20.2), similarly vanishes in local inertial frames. Although the integrands of the flux integrals are not gauge-invariant, the total integrals PIJ. (4-momentum) and JIJ.P (angular momentum) most assuredly are! They have meaning and significance independent of any coordinate system and gauge. They are tensors in the asymptotically flat region surrounding the source. The spacetime must be asymptotically flat if there is to be any possibility of deaning energy and angular momentum. Only then can linearized theory be applied; and only on the principle that linearized theory applies far away can one justify using the flux integrals (20.9) in the full nonlinear theory. Nobody can compel a physicist to move in close to define \"energy and angular momentum. He has no need to move in close; and he may have compelling motives not to: the internal structure of the sources may be inaccessible, incomprehensible, uninteresting, dangerous, expensively distant, or frightening. This requirement for far-away flatness is a remarkable feature of the flux integrals (20.9); it is also a decisive feature. Even the coordinates must be asymptotically Minkowskian; otherwise most formulas in this chapter fail or require modification. In particular, when evaluating the 4-momen tum and angular momentum of a localized system, one must apply the flux integrals (20.9) only in asymptotical(v Minkowskian coordinates. Ifsuch coordinates do not exist (spacetime not flat at infinity), one must completely abandon the flux integrals, and the quantities that rely on them for definition: the total mass, momentum, and angular momentum ofthe gravitating source. In this connection, recall the discussion of\u00a7 19.4. It described, in physical terms, why \"total mass-energy\" is a limited concept, useful only when one adopts a limited viewpoint that ignores cosmology. (Compare \"light ray\" or \"particle,\" concepts of enormous value, but concepts that break down when wave optics or wave mechanics enter significantly.) Summary: Attempts to use formulas (20.9) in ways that lose sight of the Minkowski boundary conditions (and especially simply adopting them unmodified in curvilinear coordinates) easily and unavoidably produce nonsense. Gaussian flux integrals valid only in asymptotically flat region of spacetime and in asymptotically Minkowskian coordinates 464 20. CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM EXERCISES Exercise 20.1. FLUX INTEGRAL FOR TOTAL MASS-ENERGY IN LINEARIZED THEORY Show that the flux integral (20.6) for po reduces to (20.7). Then show that. when applied to a nearly Newtonian source [line element (18.15c)]. it reduces further to the familiar Newtonian flux integral (20.2). Exercise 20.2. FLUX INTEGRAL FOR ANGULAR MOMENTUM IN LINEARIZED THEORY Derive the Gaussian flux integral (20.8) for JP.I'. [Hint: use the field equations (20.5) to show (20.13) and then use Gauss", "'s theorem to evaluate the volume integral of equation (20.8)]. Exercise 20.3. FLUX INTEGRALS FOR AN ARBITRARY STATIONARY SOURCE (a) Use the flux integrals (20.9) to calculate pp. and Jp.v for an arbitrary stationary source. For the asymptotically flat metric around the source, use (19.13), with the gravitational radiation terms set to zero. (b) Verify that the \"auxiliary equations\" (20.10) to (20.12) give the correct answer for this source's total mass-energy M and intrinsic angular momentum Sp.. \u00a720.3. VOLUME INTEGRALS FOR 4-MOMENTUM AND ANGULAR MOMENTUM It is easy, in linearized theory, to convert the surface integrals for pp. and Jp.v into volume integrals over the source; one can simply trace backward the steps that led to the surface integrals in the first place [equation (20.6); exercise 20.2]. How, in full general relativity, can one similarly convert from the surface integrals to volume integrals? The answer is rather easy, if one thinks in the right direction. One need only put the full Einstein field equations into the form Hp.av{3,a{3 - 16'lTTp.v eff (20.14) analogous to equations (20.5) of linearized theory. Here Hp.av{3 is to be defined in 1)p.v by equation (20.3), even deep inside the source where Ihp.vl terms of hp.v might be 2: I. This form of the Einstein equations then permits a conversion of the Gaussian flux integrals into volume integrals, just as in linearized theory: gp.v - Similarly, (20.16) The full Einstein field equations in terms of Hp.av/l Volume integrals for 4-momentum and angular momentum in full general relativity \u00a720.3. VOLUME INTEGRALS FOR 4-MOMENTUM AND ANGULAR MOMENTUM 465 [Crucial to the conversion is the use of partial derivatives rather than covariant derivatives in equations (20.14).] In these volume integrals, as throughout the pre ceeding discussion, the coordinates must become asymptotically Lorentz (g/lP ~ 1)/lp) far from the source. The form of T~ff can be calculated by recalling that HIJ.fX Pf3. a{3 is a linearized approximation to the Einstein curvature tensor (20.5). Define the nonlinear correc tions by (To calculate them in terms of g/lP or h/l p = g/lP 1)/lP is straightforward but lengthy. The precise form of these corrections will never be needed in this book.) Then Einstein's equations read (20.17) t/lV (\"stress-energy pseudotensor\") defined so that T~ff = pv + t/l v. (20.18) T:tr defined The quantity t/lV is sometimes called a \"stress-energy pseudotensor for the gravita tional field.\" The Einstein field equations (20.14) imply, because H/lav{3,a{3 is anti symmetric in v and /3, that (20.19) Conservation law for T:fr t/lV, and T:tr are H/lavll, coordinate-dependent objects to T/lV;v = 0, but they are written with partial These equations are equivalent derivatives rather than covariant derivatives-a fact that permits conversions back and forth between vohime integrals and surface integrals. All the quantities HlJ.fXv{3, T~ff' and t/lV depend for their definition and existence on the choice of coordinates; they have no existence independent of coordinates; they are not components of tensors or of any other geometric object. Correspond ingly, the equations (20.14) to (20.19) involving T~ff and t/l V have no geometric, coordinate-free significance; they are not \"covariant tensor equations.\" There is, nevertheless, adequate invariance under general coordinate transformations to give the values P/l and J/lV of the volume integrals (20.15) and (20.16) geometric, coor dinate-free significance in the asymptotically flat region far outside the source. Although this invariance is hard to see in the volume integrals themselves, it is clear from the surface-integral forms (20.9) that no coordinate transformation which changes the coordinates only inside some spatially bounded region can influence the values of the integrals. For coordinate changes in the distant, asymptotically flat regions, linearized theor", ")T\u00a3uarantees that under Lorentz transformations the -ifttegrats fofPTL anClJ1l\" will transform like special relativistic tensors, and that under infinitesimal coordinate transformations (gauge changes) they will be invariant. Because t/l Vare not tensor components, they can vanish at a point in one coordinate system but not in another. The resultant ambiguity in defining a localized energy density tOO for the gravitational field has a counterpart in ambiguities that exist in y Other, equally good versions tJJ-I', T::i: of HJJ-D.l./l, 466 20. CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM the formal definition of W. It is clear that any quantities H~~:! which agree with the original HJJ-<xv{3 in the asymptotic weak-field region will give the same values as HJJ-<rP{3 does for the pJJ- and JJJ-P surface integrals (20.9). One especially convenient choice has been given by Landau and Lifshitz (1962; \u00a7100), who define (20.20) where glJ-l' =(- g)1/2g JJ-p. Landau and Lifshitz show that Einstein's equations can be written in the form where the Landau-Lifshitz pseudotensor components (_g) tI!- = _l_{g<r{3 AgAJJ\"J.l L _ gaA Ag{3J.l + l.ga{3gA gAP gPJ.l 2,J.l,P l6'lT,,P _ (gaAg g{3P gJ.lP + g{3Ag gap gJ.lP ) + g + ~ (2g aAg{3J.l - ga{3gAJ.l)(2gpp gCTT - t A,A Ill' Il\",P,P J.l gPCTgPT)gPT,AgPCT,J.l} gPPgaA g{3J.l,\" tP All (20.21 ) (20.22) are precisely quadratic in the first derivatives of the metric. (Einstein also gave a pseudotensor tEJ.l Pwith this property, but it was not symmetric and so did not lead to an integral for JJ.lp.) Because Ht':f.ll has the same symmetries as HJ.lap{3 and equals HJ.l<rP{3 far from the source (exercise 20.4), and because the field equations (20.21) in terms of Ht':f.ll have the same form as in terms of HJ.lap{3, it follows that has all the properties of the T~ff introduced earlier in this section: TeLeff,p = 0, PJ.l =f Tt~Leff d3x, jJ.lP = f (xJ.lTL~Leff - xPTt~Leff) d3x. (20.23a) (20.23b) (20.23c) (20.23d) EXERCISE Exercise 20.4. FORM OF HeL/l FAR FROM SOURCE Show that the entities Hef! of equations (20.20) reduce to HJJ-D.P/l (20.3) in the weak-field region far outside the source. \u00a720.4. WHY THE ENERGY OF THE GRAVITATIONAL FIELD CANNOT BE LOCALIZED Consider an element of3-volume dI P and evaluate the contribution of the \"gravita tional field\" in that element of 3-volume to the energy-momentum 4-vector, using Why one cannot define a localized energy-momentum for the gravitational field \u00a720A. ENERGY OF GRAVITATIONAL FIELD CANNOT BE LOCALIZED 467 in the calculation either the pseudotensor t/lV or the pseudotensor teL discussed in the last section. Thereby obtain or Right? No, the question is wrong. The motivation is wrong. The result is wrong. The idea is wrong. To ask for the amount of electromagnetic energy and momentum in an element of3-volume makes sense. First, there is one and only one formula for this quantity. Second, and more important, this energy-momentum in principle \"has weight.\" It curves space. It serves as a source term on the righthand side of Einstein's field equations. It produces a relative geodesic deviation of twO nearby world lines that pass through the region of space in question. It is observable. Not one", " of these properties does \"local gravitational energy-momentum\" posses~. There is no unique formula for it, but a multitude of quite distinct formulas. The two cited are only twO among an infinity. Moreover, \"local gravitational energy-momentum\" has no weight. It does not curve space. It does not Serve as a source term on the righthand side of Einstein's field equations. It does not produce any relative geodesic deviation of two nearby world lines that pass through the region of space in question. It is not observable. Anybody who looks for a magic formula for \"local gravitational energy-momen tum\" is looking for the right answer to the wrong question. Unhappily, enormous time and effort were devoted in the past to trying to \"answer this question\" before investigators realized the futility of the enterprise. Toward the end, above all mathe matical arguments, one came to appreciate the quiet but rock-like strength of Einstein's equivalence principle. One can always find in any given locality a frame of reference in which all local \"gravitational fields\" (all Christoffel symbols; all r aI'P) disappear. No r's means no \"gravitational field\" and no local gravitational field means no \"local gravitational energy-momentum.\" Nobody can deny or wants to deny that gravitational forces make a contribution to the mass-energy of a gravitationally interacting system. The mass-energy of the Earth-moon system is less than the mass-energy that the system would have if the two objects were at infinite separation. The mass-energy of a neutron star is less than the mass-energy of the same number of baryons at infinite separation. Sur rounding a region of empty space where there is a concentration of gravitational waves, there is a net attraction, betokening a positive net mass-energy in that region of space (see Chapter 35). At issue is not the existence of gravitational energy, but the localizability of gravitational energy. It is not localizable. The equivalence principle forbids. Look at an old-fashioned potato, replete with warts and bumps. With an orange marking pen, mark on it a \"North Pole\" and an \"equator\". The length of the equator is very far from being equal to 2... times the distance from the North Pole to the 468 20. CONSERVATION LAWS FOR 4\u00b7MOMENTUM AND ANGULAR MOMENTUM equator. The explanation, \"curvature,\" is simple, just as the explanation, \"gravita tion\", for the deficit in mass of the earth-moon system (or deficit for the neutron star, or surplus for the region of space occupied by the gravitational waves) is simple. Yet it is not possible to ascribe the deficit in the length of the equator in the one case, Or in mass in the other case, in' any uniquely right way to different elements of the manifold (2-dimensional in the one case, 3-dimensional in the other). Look at a small region on the surface of the potato. The geometry there is locally flat. Look at any small region of space in any of the three gravitating systems. In an appropriate coordinate system it is free of gravitational field. The over-all effect one is looking at is a global effect, not a local effect. That is what the mathematics cries out. That is the lesson of the nonuniqueness of the t ltv! \u00a720.5. CONSERVATION LAWS FOR TOTAL 4-MOMENTUM AND ANGULAR MOMENTUM Consider a system such as our galaxy or the solar system, which is made up of many gravitating bodies. Some of the bodies may be highly relativistic (black holes; neutron stars), while others are not. However, insist that in the regions between the bodies spacetime be nearly flat (gravity be weak)-so flat, in fact, that one can cover the entire system with coordinates which are (almost) globally inertial, except in a small neighborhood of each body where gravity may be strong. Such coordinates can exist only if the Newtonian gravitational potential, (/J ::::: ~(1)oo - goo)' in the interbody region is small: (/JinterbOdy - (Mass of system)/(radius of system) ~ 1. The solar system certainly satisfies this condition ((/JinterbOdy Galaxy ((/Jinterbody - 10-6 universe as a whole does not ((/Jinterbody - ), as do clusters of galaxies ((/Jinterbody - 10-6 I)! In evaluating volume integrals for the system's totaI4-momentum, split its volume into a region containing each body (denoted \"A\") plus an interbody region; and neglect the pseudotensor contribution from the almost-flat interbody region: ), as does the ); but the 10-7 P~ystem = 2", " f T~& d 3x + 1 = 2 P'l + r A A mt.erbody regIOn po d 3x. A Jinu.rbody region T~f~ d 3x (20.24a) Because spacetime is asymptotically flat around each body, PAP. is the 4-momentum of body A as measured gravitationally by an experimenter near it. The integral of TP.o over the interbody region is the contribution of any gas, particles, or magnetic Total 4-momentum and angular momentum for a system of gravitating bodies \u00a720.5. CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM 469 fields out there to the total 4-momentum. A similar breakup of the angular momen tum reads Jp.p system = ~ Jp.p + r J, L.. A A int~rbody regIon (xP.po - xPPO) d 3x. (20.24b) In operational terms, these breakups show that the total 4-momentum and angular momentum of the system, as measured gravitationally by an experimenter outside it, are sums of pp. and Jp.p for each individual body, as measured gravitationally by an experimenter near it, plus contributions of the usual special-relativistic type from the interbody matter andfields. This is true even ifsome of the bodies are hurtling through the system with speeds near that of light; their gravitationally measured pp. and Jp.p contribute, on an equal footing with anyone else's, to the system's total pp. and Jp.P! Surround this asymptotically flat system by a two-dimensional surface S that is at rest in some asymptotic Lorentz frame. Then the 4-momentum and angular momentum inside S change at a rate (as measured in S's rest frame) given by Rates of change of total 4-momentum and angular momentum: d f dPp. - - = dt dt Tp.o d3x = eff f Tp.o d3x = - eff,O f. TP.I. d3x eff,l - f TP.j d 2S eff - - ;' and similarly dJp.p dt -f (XP.T~~f - XPT~'f) d 2Sj \u2022 8 2 (1) expressed as flux integrals of T~;r (20.25) (20.26) p Although the pseudotensor tp., in the interbody region and outside the system, contributes negligibly to the total 4-momentum and angular momentum (by as sumption), its contribution via gravitational waves to the time derivatives dPp./dt and dJp.p/dt can be important when added up over astronomical periods of time. Thus, one must not ignore it in the flux integrals (20.25), (20.26). In evaluating these flux integrals, it is especially convenient to use the Landau Lifshitz form of T~ff' since that form contains no second derivatives of the metric. Thus set where teL are given by equations (20.22). Only those portions of teL that die out as 1/r2 or 1/r3 at large r can contribute to the flux integrals (20.25), (20.26). For const. + O(1/r)]. teL dies out as l/r-l. Hence, the only contri static solutions [!Ip. p butions come from dynamic parts of the metric. which, at these large distances, are entirely in the form of gravitational waves. The study of gravitational waves in Chapter 35 will reveal it becomes a stress-energy tensor T(GW)P.P for the waves. which has all the properties one ever requires of any stress-energy tensor. (For example, via Einstein's equations that when teL is averaged over several wavelengths, 470 20. CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM G(B)ILP = 8'lTrGW)ILP, it contributes to the \"background\" curvature of the spacetime through which the waves propagate.) Moreover, averaging teL over several wave lengths before evaluating the flux integrals (20.25), (20.26) cannot affect the values of the integrals. Therefore', one can freely make in these integrals the replacement T~ff = TILP + rGW)ILP, thereby obtaining (2) expressed as flux integrals of T!'P + -rGWlp.v - dplL - T = f (Pi + y1GW)lLi) d 2Si, dJlLP", " =f [XIL(pj + T(GW)Pi) - XP(Pi + rGW)lLi)] d\"2Sj is dt s (20.27). (20.28) These are tensor equations in the asymptotically flat spacetime surrounding the system. All reference to pseudotensors and other nontensorial entities has disap peared. Equations (20.27) and (20.28) say that the rate of loss of 4-momentum and angular momentum from the system, as measured gravitationally, is precisely equal to the rate at which matter, fields, and gravitational waves carry off 4-momentum and angular momentum. This theorem is extremely useful in thought experiments where one imagines changing the 4-momentum or angular momentum of a highly relativistic body (e.g., a rotating neutron star) by throwing particles onto it from far away [see, e.g., Hartle (1970)]. EXERCISE Exercise 20.5. TOTAL MASS\u00b7ENERGY IN NEWTONIAN LIMIT (a) Calculate t't~L for the nearly Newtonian metric ds 2 = -(I + 2<P) dt 2 + (I - 2<P) 8ik dx i dx k (see \u00a718.4). Assume the source is slowly changing, so that time derivatives of <P can be neglected compared to space derivatives. [Answer: 00 t L - L - _ _.:Ln. n. 8'17 \"',i\"',i' tlj}'L = 0, t~_L =4~ (<P,i<P,k - \u00b18ik <P,t<P,t)\u00b7j (20.29) (Note: t~.L as given here is the \"stress tensor for a Newtonian gravitational field\"; cf exercises 39.5 and 39.6.) (b) Let the source of the gravitational field be a perfect fluid with p.P = (p + p)up-uP + pgP-P, pip - v2=(dxldt)2 - I<PI. \u00a720.6. EQUATIONS OF MOTION DERIVED FROM FIELD EQUATION 471 Let the Newtonian potential satisfy the source equation f/J,;; = 4'ITp. Show that the energy of the source is po = J(TOO + (00)( - g) d 3x =J[pI(! - V2)1/2 +t pv2 +t pf/J](gzzgllllgzz)l/2 dx dy dz t rpotentia~ t rproperJ 1~nergy J L~o!ume Lorentz. tJ fkineticl t LenergyJ'- -'\"'- -\".J,, contractIOn factor [ (20.30) + higher-order corrections. (c) Show that the \"equations of motion\" TeLaff,v = 0 reduce to the standard equations (16.3) of Newtonian hydrodynamics. \u00a720.6. EQUATIONS OF MOTION DERIVED FROM THE FIELD EQUATION Consider the Einstein field equation G = 8'lTT (20.31 ) under conditions where space is empty of everything except a source-free electro magnetic field: TIJ.v = _1 (Fwtg 4'17 a{3 p{3 _ -lgIJ.vF FUT) 4 UT (20.32) (ef the expression for stress-energy tensor of the electromagnetic field in \u00a75.6). To predict from (20.31) how the geometry changes with time, one has to know how the electromagnetic field changes with time. The field is expressed as the \"exterior derivative\" of the 4-potential, F = dA (language of forms) or F.,v = -_v - __IJ. (language of components), ~ oA OxIJ. oA ox v (20.33) and the time rate of change of the field is governed by the Maxwell equation or d*F = 0 (20.34) Vacuum Maxwell equations derived from Einstein field equation 472 20. CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM If it seems a fair division oflabor for the Maxwell equation to predict the develop ment in time of the Maxwell field and the Einstein equation to do the same for the Einstein field, then it may come as a fresh surprise to discover that the Einstein equation (20.31), plus expression (20.32) for the Maxwell stress-energy, can do both jobs. One does not have to be given the Maxwell \"equation of motion\" (20.34). One can derive it fresh from (20.31)", " plus (20.32). The proof proceeds in five steps (see also exercise 3.18 and \u00a75.1O). Step one: The Bianchi identity V 0 G 0 implies conservation of energy-momentum V 0 T = O. Step two: Conservation expresses itself in the language of components in the form Step three: Leaving the middle term unchanged, rearrange the first term so that, like the last term, it carries a factor F'YT. Thus in that first term let the indices v{3 of p{3 be replaced in turn by OT and by TO, to subdivide that term into (20.35) FIJ.a;ugaTFUT + FIJ.a;Tgau FTU = (FIJ.T;u - FIJ.U;T)FUT =gIJ.P(FPT;u + FUP;T)FuT. Step four: Combine the first and the last terms in (20.35) to give (20.36) (20.37) The indices on the derivatives of the field quantities stand in cyclic order. This circumstance annuls all the terms in the connection coefficients r a{3y when one writes out the covariant derivatives explicitly. Thus one can replace the covariant derivatives by ordinary derivatives. Moreover, these three derivatives annul one another identi cally when one substitutes for the fields their expressions (20.33) in terms of the potentials. Consequently, nothing remains in the conservation law (20.35) except the middle term, giving rise to four statements (fL = 0, 1,2, 3) about the four quantities ({3 = 0, 1,2,3) F IJ. F{3P = 0 ;P {3 (20.38) (20.39) Step five: The determinant of the coefficients in the four equations (20.38) for the four unknowns (20.39) has the value f'OoP\\F02 f'03.....,. \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 0 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 = -(Eo B)2 (20.40) \u00a720.6. EQUATIONS OF MOTION DERIVED FROM FIELD EQUATION 473 (see exercise 20.6, part i). In the generic case, this one function of the four variables (t, x, y, z) vanishes on one or more hypersurfaces; but off any such hypersurface (i.e., at \"normal points\" in spacetime) it differs from zero. At all normal points, the solution of the four linear equations (20.38) with their nonvanishing determinant that is to say, Maxwell's gives identically zero for the four unknowns (20.39); \"equations of motion\" are fulfilled and must be fulfilled as a straight consequence of Einstein's field equation (20.31 )-plus expression 20.32 for the stress-energy tensor. Special cases admit counterexamples (see exercise 20.8); but in the generic case one need not invoke Maxwell's equations of motion; one can deduce them from the Einstein field equation. Turn from the dynamics of the Maxwell field itself to the dynamics of a charged particle moving under the influence of the Maxwell field. Make no more appeal to outside providence for the Lorentz equation of motion than for the Maxwell equation of motion. Instead, to generate the Lorentz equation call once more on the Einstein field equation or, more directly, on its consequence, the principle of the local conservation of energy-momentum. Keep track of the world line of the particle from t = t to t = t +,jt (Figure 20.1). Generate a \"world tube\" around this world line. Thus, at each value of the time coordinate t, take the location of the particle as center; construct a sphere of radius \u00a3 around this center; and note how the successive spheres sweep out the desired world tube. Construct \"caps\" on this tube at times t and t +,jt. The two caps, together with the world tube proper, bound a region of spacetime in which energy and momentum can be neither created nor destroyed (\"no creation of moment of rotation,\" in the language of the Bianchi identities, Chapter 15). Therefore the energy-momentum emerging out of the \"top\" cap has to equal the energy-momen tum entering the \"bottom\" cap, supplemented by the amount of energy-momentum carried in across the world-tube by the Maxwell field. Out of such an analysis, as performed in flat spacetime, one ends up with the Lorentz equation of motion in its elementary form (see Chapters 3 and 4), Figure 20.1. \"World tube.\" The change in the 4-momentum of the particle is governed by the flow of 4-momentum across the boundary of the world tube", ". Lorentz force equation derived from the Einstein field equation 474 20. CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM dp/dT = e(F, u) (language of forms) or in curved spacetime, the Lorentz equation of motion in covariant form, (form language) or m[d 2 a + r a p dx x 2 dT dT JL JL P dx dT ] = eP{3 dx{3 dT (component language). (20.41) \"One ends up with the Lorentz equation of motion\"-but only after hurdling problems of principle along the way. One would understand what a particle is if one understood how to do the calculation of balance of energy-momentum with all rigor! Few calculations in all of physics have been done in so many ways by so many leading investigators, from Lorentz and his predecessors to Dirac and Rohrlich [see Teitelboim (1970, 1971) for still further insights]. Among the issues that develop are two that never cease to compel attention. (I) The particle responds according to the Lorentz force law (20.41) to a field. This field is the sum of a contribution from external sources and from the particle itself. How is the field exerted by the particle on itself to be calculated? Insofar as it is not already included in its effects in the \"experimental mass\" m in (20.41), this force is to be calculated as half the difference between the retarded field and the advanced field caused by that particle (see \u00a736.l1 for a more detailed discussion of the corresponding point for an emitter of gravitational radiation). This difference is singularity-free. On the world line, it has the following simple value [valid in general for point particles; valid for finite-sized particles when and only when the particle changes its velocity negligibly compared to the speed of light during the light-travel time across itself see, e.g., Burke (1970)] (20.42) Every acceptable line of reasoning has always led to expression (20.42). It also represents the field required to reproduce the long-known and thoroughly tested law of radiation damping. (2) \"Infinite self-energy.\" Around a particle at rest, or close to a particle in an arbitrary state of motion, the field is e/r2 and the field energy is (1/8'17)If (e/r 2)24'lTr2 dr = (e2/2)(r~n - C 1 ). (20.43) rmin This expression diverges as rmin is allowed to go to zero. To hurdle this difficulty, one arranges the calculation of energy balance in such a way that there always appears the sum of this \"self-energy\" and the \"bare mass.\" The two terms individ ually are envisaged as \"going to infinity\" as rmin goes to zero; but the sum is identified with the \"experimental mass\" and is required to remain finite. Of course, no particle is a classical object. A proper calculation of the energy has to be conducted at the quantum level. There it is easier to hide from sight the separate infinities-but they A particle acted on by its own electromagnetic field (\"radiation damping\") Infinite self-energy of a point particle Why one is justified to regard equations of motion as consequences of the Einstein field equation \u00a720.6. EQUATIONS OF MOTION DERIVED FROM FIELD EQUATION 475 are still present, and promise to remain until the structure of a particle is understood. Before one turns from the Maxwell and Lorentz equations of motion to a final example (deriving the geodesic equations of motion for an uncharged particle), is it not time to object to the whole program of \"deriving an equation of motion from Einstein's field equation\"? First, is it not a pretensious parade of pomposity to say it comes \"from Einstein's field equation\" (and even more, \"from Einstein's field equations\") when it really comes from a principle so elementary and long established as the law of conservation of 4-momentum? It cannot be contested that this conser vation principle, in historical fact, came before geometrodynamics, just as it came before electrodynamics and before the theories of all other established fields. How ever, in no theory but Einstein's is this principle incorporated as an identity. Only here does the conservation of energy-momentum appear as a fully automatic conse quence of the inner working of the machinery of the world (energy density tied to moment of rotation, and moment of rotation automatically conserved; see Chapter 17). Out of Einstein's theory one can derive the equation of motion of a particle. Out of Maxwell's one cannot. Thus, nothing prevents one from acting on", " a charge with an \"external\" force, over and above the Lorentz force, nor from tailoring this force in such a way that the charge follows some prescribed world line (\"engine driven source\"). It makes no difficulties whatsoever for Maxwell's equations that one has shifted attention from a world line that follows the Lorentz equation of motion to one that does not. Quite the contrary is true in general relativity. To shift from right world line (geodesic) to wrong world line makes the difference between satisfying Einstein's field equation in the vicinity of that world line and being unable to satisfy Einstein's field equation. The Maxwell field equations are so constructed that they automatically fulfill and demand the conservation of charge; but not everything has charge. The Einstein field equation is so constructed that it automatically fulfills and demands the conser vation of momentum-energy; and everything does have energy. The Maxwell field equations are indifferent to the interposition of an \"external\" force, because that force in no way threatens the principle of conservation of charge. The Einstein field equation cares about every force, because every force is a medium for the exchange of energy. Electromagnetism has the motto, \"I count all the electric charge that's here.\" All that bears no charge escapes its gaze. \"I weigh all that's here\" is the motto of spacetime curvature. No physical entity escapes this surveillance. Why, then, is the derivation of the geodesic equation of motion of an object said to be based on \"Einstein's geometrodynamic field equation\" rather than on \"the principle of conservation of 4-momentum\"? Because geometry responds by its curvature to mass-energy in every form. Most of all, because geometry outside tells about mass-energy inside, free of all concern about issues of internal structure (violent motions, unknown forces, tortuously curved and even multiply-connected geometry). If one objection to the plan to derive the equation of motion of a particle \"from the field equation\" has been disposed of, then the moment has come to deal with How one can avoid complexities of particle structure when deriving equations of motion: the \u00b7\u00b7external viewpoint\"\u00b7 Derivation of geodesic motion from Einstein field equation: (1) derivation in brief / (2) derivation with care Coupling of curvature to particle moments produces deviations from geodesic motion 476 20. CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM the other natural objection: Is there not an inner contradiction in trying to apply to a \"particle\" (implying idealization to a point) a field equation that deals with the continuum? Answer: There is a contradiction in dealing with a point. Therefore do not deal with a point. Do not deal with internal structure at all. Analyze the motion by looking at the geometry outside the object. That geometry provides all the handle one needs to follow the motion. Already here one sees the difference from the derivation of the Lorentz equation of motion as sketched out above. There (1) no advantage was taken of geometry outside as indicator of motion inside; (2) a detailed bookkeeping was envisaged of the localization in space of the electromagnetic energy; and (3) this bookkeeping brought up the issue of the internal structure of the particle, which could not be satisfactorily resolved. Now begin the analysis in the new geometrodynamic spirit. Surrounding \"the Schwarzschild zone of influence\" of the object, mark out a \"buffer zone\" (Figure 20.2) that extends out to the region where the \"background geometry\" begins to depart substantially from flatness. Idealize the geometry in the buffer zone as that of an unchanging source merging asymptotically (\"boundary!X1 of buffer zone\") into flat space. It suffices to recall the properties of the spacetime geometry far outside an unchanging (i.e., nonradiating) source (exercise 19.3) to draw the key conclusion: relative to this flat spacetime and regardless of its internal structure, the object remains at rest, or continues to move in a straight line at uniform velocity (conserva tion of total 4-momentum; \u00a720.5). In other words, it obeys the geodesic equation of motion. If this is the result in a flash, then it is appropriate to go back a step to review it, to find out what it means and what it demands. When the object is absent and the background geometry alone has to be consid ered, then the geodesic is a well-defined mathematical construct. Moreover, Fermi Walker transport along this geodesic gives a well-defined way to construct a comoving local inertial frame (see \u00a7 13.6). Relative to this frame, the representative point of the geodesic remains for all time at rest at the origin. In what way does the presence of the object change", " this picture? The object possesses an angular momentum, mass quadrupole moments, and higher multipole moments. They interact with the tide-producing accelerations (Riemann curvature) of the background geometry. Depending on the orientation in space of these mo ments, the interactions drive the object off its geodesic course in one direction or another (see \u00a740.9). These anomalies in the motion go hand in hand with anomalies in the geometry. On and near the ideal mathematical geodesic the metric is Min kowskian. At a point removed from this geodesic by a displacement with Riemann normal coordinates e, ~2, ~3 (see \u00a711.6), the metric components differ from their canonical values (-1, 1, 1,1) by amounts proportional (1) to the squares and products of the ~m and (2) to the components of the Riemann curvature tensor (tide-producing acceleration) of the background geometry. These second-order terms produce departures from ideality in the buffer zone, departures that may be described symbolically as of order 8(metric) _,2. R\u00b7 (spherical harmonic of order two). (20.44) \u00a720.6. EQUATIONS OF MOTION DERIVED FROM FIELD EQUATION 477 ~ t I Time I / / / / / / /,,,,,,,,,,,,,, /,,,,, fl\\. t- - I / f--.-.... I \" / 1 --____-...-:\"8 d,,, - - - -Yd \" Figure 20.2. \"Buffer zone\": the shell of space between d and ~3, where the geometry is appropriately idealized as the spherically symmetric \"Schwarzschild geometry\" of a localized center of attraction (the object under study) in an asymptotically flat space. Inside d: the \"zone of influence\" of the object. In the general case where this object lacks all symmetry, the metric is found to depart more and more from ideal \"Schwarzschild character\" as the exploration of the geometry is carried inward from d (effect of angular momentum of the object on the metric; effect of quadrupole moment; effect of higher moments). Outside ~: the \"background geometry.\" As this geometry is explored at greater and greater distances outside ~fl, it is found to depart more and more from flatness (effect of concentrations of mass, gravitational waves, and other geometrodynamics). Here r is the distance from the geodesic and R is the magnitude of the significant components of the curvature tensor. The object produces not only the standard \"Schwarzschild\" departure from flatness, - - - - - - - ----- ---------- --8(metric) -- m/r, (20.45) which by itself (in a flat background) would bring about no departure from geodesic motion, but also correction terms which may be symbolized as Coupling of spin to curvature 478 and 20. CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM 8(metric) - (Slr 2 ) (spherical harmonic of order one) (20.46) 8(metric)- (f/r 3 ) (spherical harmonic-of order two) (20.47) and higher-order terms. Here S(cm2) is a typical component of the angular momen tum vector or \"spin\"; f(cm3 ) is a representative component of the moment of inertia or quadrupole tensor (see Chapter 36 for details), and higher terms have higher-order coefficients. The tide-producing acceleration generated by the surroundings of the object (\"background geometry\") acts on the spin of the object with a force of order RS and pulls it away from geodesic motion with an acceleration of the order. acceleratIOn (cm- I ) R(cm- 2)S(cm2 ) - - - - - - m(cm) (20.48) (see exercise 40.8). Otherwise stated, the surrounding and the spin both put warps in the geometry, and these warps conspire together to push the object off track. The sum of the relevant two perturbations in the metric is qualitatively of the form The sum is least where r has a value of the order and there it has the magnitude r - (SI R)l/4, 8g - (SR)I/2. (20.49) (20.50) (20.51) To \"derive the geodesic equation of motion with some preassigned accuracy t:\" may be defined to mean that the metric in the buffer zone is Minkowskian within the latitude t:. In the illustrative example, this means that (SR)I/2 is required to be of", " the order of t: or less. Nothing can be done about the value of R because the background curvature R is a feature of the background geometry. One can meet the requirement only by imposing limits on the mass and moments of the object. In the example, where the dominating moment is the angular momentum, one must require that this parameter of the object be less in order of magnitude than the limit (20.52) Evidently this and similar conditions on the higher moments are most easily satisfied f = 0, higher by demanding that the object have spherical symmetry (S = 0, \u00a720.6. EQUATIONS OF MOTION DERIVED FROM FIELD EQUATION 479 moments = 0). Then the perturbation in the metric, again disregarding angle factors and indices, is qualitatively of the form og - r2R + mlr, and the buffer zone is best designed to bracket the minimizing value of r, ra ~ [r - (mIR)l/3] ~ r~. (20.53) (20.54) The departure of the metric from Minkowskian perfection in the buffer zone is of the order To achieve any preassigned accuracy t: for og, one must demand that the mass be less than a limit of the order (20.55) (20.56) No object of finite mass moving under the influence of a complex background will admit a buffer zone where the geometry approaches Minkowskian values with arbitrary precision. Therefore it is incorrect to say that such an object follows a geodesic world line. It is meaningless to say that an object of finite rest mass follows a geodesic world line. World line of what? If the object is a black hole, there is no point inside its \"horizon\" (capture surface; one-way membrane; see Chapters 33 and 34) that is relevant to the physics going on outside. Geodesic world line within what background geometry? It has no sense to speak of a geometry that \"lies behind\" or is \"background to\" a black hole. Turn from one motion of one object in one spacetime to a continuous one-param eter family of spacetimes, with the mass m of the object being the parameter that distinguishes one of these solutions of Einstein's field equation from another. Go to the limit m = O. Then the size of the buffer zone shrinks to zero and the departure of the metric from Minkowskian perfection in the buffer zone also goes to zero. In this limit (\"test particle\"), it makes sense to say that the object moves in a straight line with uniform velocity in the local inertial frame or, otherwise stated, it pursues a geodesic in the background geometry. Moreover, this background geometry is well-defined: it is the limit of the spacetime geometry as the parameter m goes to zero [see Infeld and Schild (1949)]. In this sense, the geodesic equation of motion follows as an inescapable consequence of Einstein's field equation. The concept of \"background\" as limit of a one-parameter family of spacetimes extends itself to the case where the object bears charge as well as mass, and where the surrounding space is endowed with an electromagnetic field. This time the one-parameter family consists of solutions of the combined Einstein-Maxwell equa tions. The charge-to-mass ratio elm is fixed. The mass m is again the adjustable parameter. In the limit when m goes to zero, one is left with (I) a background geometry, (2) a background electromagnetic field, and (3) a world line that obeys The sense in which no body can move on a geodesic of spacetime The sense in which test particles do move on geodesics of a background geometry Motion of a charged test particle in curved spacetime References on derivation of equations of motion from Einstein field equation Quantum mechanical limitations on the derivation 480 20. CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM the general-relativity version of the Lorentz equation of motion in this background as a consequence of the field equations [Chase (1954)]. In contrast, a so-called \"unified field theory of gravitation and electromagnetism\" that Einstein tentatively put forward at one stage of his thinking, as a conceivable alternative to the combina tion of his standard 1915 geometrodynamics with Maxwell's standard electrody namics, has been shown [Callaway (1953)] to lead to the wrong equation of motion for a charged particle. It moves as if uncharged no matter how much charge is piled on its back. If that theory were correct, no cyclotron could operate, no atom could exist, and life itself would be impossible. Thus the ability to yield the correct equation of motion of a particle has today become an added ace in the hand of general relativity. The idea", " for such a treatment dates back to Einstein and Grommer (1927). Corrections to the geodesic equation of motion arising from interaction between the spin of the object (when it has finite dimensions) and the curvature of the background geometry are treated by Papa petrou (1951) and more completely by Pirani (1956) (see exercise 40.8). A book on the subject exists [Infeld and Plebanski (1960)]. Section 40.9 describes how corrections to geodesic motion affect lunar and planetary orbits. Some of the problems that arise when the object under study fragments or emits a directional stream of radia tion, and unresolved issues of principle, are discussed by Wheeler (1961). When one turns from the limit of infinitesimal mass to an object of finite mass, no simpler situation presents itself then a system of uncharged black holes (Chapter 33). Everything about the motion of these objects follows from an application of the source-free Einstein equation G = 0 to the region of spacetime outside the horizons (see Chapter 34) of the several objects. The theory of motion is then geometrodynamics and nothing but geometrodynamics. - It has to be emphasized that all the considerations on motion in this section are carried out in the context of classical theory. In the real world of quantum physics, the geometry everywhere experiences unavoidable, natural, zero-point fluctuations (Chapter 43). The calculated local curvatures associated with these fluctuations at the Planck scale of distances [L = (hG/C 3)1/2 = 1.6 X 10-33 cm] are enormous I/L2 - 0.4 X 1066 cm-2] compared to the curvature produced on much larger [R scales by any familiar object (electron or star). No detailed analysis of the interaction of these two curvatures has ever been made. Such an analysis would define a smoothed-out average of the geometry over regions larger than the local quantum fluctuations. With respect to this average geometry, the object will follow geodesic motion: this is the expectation that no one has ever seen any reason to question-but that no one has proved.. EXERCISES Exercise 20.6. SIMPLE FEATURES OF THE ELECTROMAGNETIC FIELD AND ITS STRESS-ENERGY TENSOR (a) Show that the \"scalar\" - 1/2FaS FaB (invariant with respect to coordinate transforma tions) and the \"pseudoscalar\" 1/4Fa/FaB (reproduces itself under a coordinate transfor mation up to a \u00b1 sign, according as the sign of the Jacobian of the transformation is positive \u00a720.6. EQUATIONS OF MOTION DERIVED FROM FIELD EQUATION 481 or negative) have in any local inertial frame the values E2 - B2 and E\u00b7 B, respectively (\"the two Lorentz invariarlts\" of the electromagnetic field). (b) Show that the Poynting flux (E X B)/4rr is less in magnitude than the energy density (E2 + B2)/8rr, save for the exceptional case where both Lorentz invariants of the field vanish (case where the field is locally \"null\"). (c) A charged pith ball is located a small distance from the North Pole of a bar magnet. Draw the pattern of electric and magnetic lines of force, indicating where the electromagnetic field is \"null\" in character. Is it legitimate to say that a \"null field\" is a \"radiation field\"? (d) A plane wave is traveling in the z-direction. Show that the corresponding electromag netic field is everywhere null. (e) Show that the superposition of two monochromatic plane waves traveling in different directions is null on at most a set of points of measure zero. (f) In the \"generic case\" where the field (E, B) at the point of interest is not null, show that the Poynting flux is reduced to zero by viewing the field from a local inertial frame that is traveling in the direction of E X B with a velocity where the velocity parameter a is given by the formula v = tanh a, h2 (Poynting flux) 21E X BI tan a = (energy density) = E2 + B2. (20.57) (20.58) (g) Show that all components of the electric and magnetic field in this new frame can be taken to be zero except Ez and Bz \u2022 (h) Show that the 4 X 4 determinant built out of the components of the field in mixed representation, F/~, is invariant with respect to general coordinate transformations. (Hint Use the theorem that the determinant of the product of three matrices is equal to the product of the determinants of those three matrices.) (i) Show that this determinant", " has the value - (E\u00b7 B)2 by evaluating it in the special local inertial frame of (f). (j) Show that in this special frame the Maxwell stress-energy tensor has the form E/ + B/ llTi'v il = 8IT -1 0 0 0 0 -1 0 0 0 +1 0 0 0 0 0 +\\ (20.59) (Faraday tension along the lines of force; Faraday pressure at right angles to the lines of force). (k) In the other case, where the field is locally null, show that one can always find a local inertial frame in which the field has the form E = (0, F, 0), B = (0,0, F) and the stress-energy tensor has the value P IITIJ..I' = 417 / 1 -1 1 0 0 -IL for row, v for column). (20.60) (I) Regardless of whether the electromagnetic field is or is not nulL show that the Maxwell stress-energy tensor has zero trace, TIJ.IJ. = 0, and that its square is a multiple of the unit tensor, v a 8IJ. (8rr)2 TIJ. Ta = _V_[(E2 _ B2)2 + (2E. B)2] = (::)2 [(E2 + B2)2 - (2E X B)2]. (20.61 ) 482 20. CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM Exercise 20.7. THE STRESS-ENERGY TENSOR DETERMINES THE ELECTROMAGNETIC FIELD EXCEPT FOR ITS COMPLEXION (a) Given a non-zero symmetric 4 X 4 tensor Tp.v which has zero trace TP. p. = 0 and whose square is a multiple, }\\f4/(S.\".)2. of the unit matrix, show that, according as this multiple is\u00b7 zero (\"null case\") or positive, the tensor can be transformed to the form (20.60) or (20.59) by a suitable rotation in 3-space or by a suitable choice of local inertial frame, respectively. (b) In the generic (non-null) case in the frame in question, show that pv is the Maxwell tensor of the \"extremal electromagnetic field\" ~P.I' with components E(extremaO = (M, 0, 0), B(extremaO = (0,0,0). (20.62) Show that it is also the Maxwell tensor of the \"dual extremal field\" *~p.v with components *E(extremaO = (0,0,0), *B(extremaO = (M, 0, 0). (20.63) (c) Recalling that the duality operation * applied twice to an antisymmetric second-rank tensor (2-form) in four-dimensional space leads back to the negative of that tensor, show that the operator eOn (\"duality rotation\") has the value eOn = (cos a) + (sina)*. (20.64) (d) Show that the most general electromagnetic field which will reproduce the non-null tensor pv in the frame in question, and therefore in any coordinate system, is (20.65) (e) Derive a corresponding result for the null case. [The fieldFp.v defined in the one frame and therefore in every coordinate system by (d) and (e) is known as the \"Maxwell square root\" of Tp.v; ~p.v is known as the \"extremal Maxwell square root\" of Tp.v; and the angle a is called the \"complexion of the electromagnetic field.\" See Misner and Wheeler (1957); see also Boxes 20.1 and 20.2, adapted from that paper.] Box 20.1 CONTRAST BETWEEN PROPER LORENTZ TRANSFORMATION AND DUALITY ROTATION Quantity General proper Lorentz transformation Duality rotation Components of the Maxwell stress-energy tensor or the \"Maxwell square\" of the field F Transformed Unchanged The invariants \u00a32 - B2 and (E' B)2 Unchanged Transformed The combination [(\u00a32 - B2)2 + (2E' B)2] = [(\u00a32 + B2)2 - (2E X B)2] Unchanged Unchanged \u00a720.6. EQUATIONS OF MOTION DERIVED FROM FIELD EQUATION 483 Box 20.2 TRANSFORMATION OF THE GENERIC (NON-NULl) IN ELECTROMAGNETIC FIELD TENSOR F = (E. B) A LOCAL", " INERTIAL FRAME Field values At start At start E,B After simplifying Lorentz transfermation E and B parallel to each other and parallel to x -axis After simplifying duality rotation E and B perpendicular, and E greater than B E parallel to x-axis and B = 0 Exercise 20.8. THE MAXWELL EQUATIONS CANNOT BE DERIVED FROM THE LAW OF CONSERVATION OF STRESS-ENERGY WHEN (E\u00b7 B) = 0 OVER AN EXTENDED REGION Supply a counter-example to the idea that the Maxwell equations, follow from the Einstein equation; or, more precisely, show that (I) the condition that the Maxwell stress-energy tensor should have a vanishing divergence plus (2) the condition that this Maxwell field is the curl of a 4-potential AIL can both be satisfied, while yet the stated Maxwell equations are violated. [Hint: It simplifies the analysis without obscuring the main point to consider the problem in the context offiat spacetime. Refer to the paper of Teitelboim (1970) for the decomposition of the retarded field of an arbitrarily accelerated charge into two parts, of which the second, there called FILVII, meets the stated requirements, and has everywhere off the worldline (E' B) = 0, but does not satisfy the cited Maxwell equations.] Exercise 20.9. EQUATION OF MOTION OF A SCALAR FIELD AS CONSEQUENCE OF THE EINSTEIN FIELD EQUATION -~ stress-energy tensor of a massless scalar field is taken to be Derive the equation of motion of this scalar field from Einstein's field equation. (20.66) y,,----------------------------- CHAPTER 21 VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA Whenever any action occurs in nature. the quantity of action employed by this change is the least possible. PIERRE MOREAU DE MAUPERTUIS (1746) In the theory of gravitation. as in all other branches of theoretical physics. a mathematically correct statement of a problem must be determinate to the extent allowed by the nature it must ensure the of the problem; if possible. uniqueness of its solution. VLADIMIR ALEXANDRDVITCH FOCK (1959) Things are es they are because they were as they were. THOMAS GOLD (1972) Calculemus G. W. LEIBNIZ \u00a721.1. DYNAMICS REQUIRES INITIAL-VALUE DATA This chapter is entirely Track 2. No earlier Track\u00b72 material is needed as preparation for it. but Chapters 9-11 and 13-15 will be helpful. It is needed as preparation for Box 30.1 (mixmaster universe) and for Chapters 42 and 43. No plan for predicting the dynamics of geometry could be at the same time more mistaken and more right than this: \"Give the distribution of mass-energy; then solve Einstein's second-order equation, G = 8'1T1; (21.1 ) for the geometry.\" Give the distribution of mass-energy in spacetime and solve for t~e spacetime geometry? No. Give the fields that generate mass-energy, and their ~ To Karel Kuchar, Claudio Teitelboim, and James York go warm thanks for their collaboration in the preparation of this chapter, and for permission to draw on the lecture notes of K. K. and to quote results of K. K. [especially exercise 21.10] and of J. Y. [especially equations (2 \\.87), (21.88), and (21.152)] prior to publication elsewhere. \u00a721.1. DYNAMICS REQUIRES INITIAL-VALUE DATA 485 time-rates of change, and give 3-geometry of spa'ce and its time-rate of change, all at one time, and solve for the 4-geometry of spacetime at that one time? Yes, And only then let one's equations for geometrodynamics and field dynamics go on to predict for all time, in and by themselves, needing no further prescriptions from outside (needing only work!), both the spacetime geometry and the flow of mass energy throughout this spacetime, This, in brief, is the built-in \"plan\" of geometro dynamics, the plan spelled out in more detail in this chapter. Contest the plan. Point out that the art of solving any coupled set of equations lies in separating the unknowns from \u00b7what is known or to be presc:ribed. Insist that this separation is already made in (21.1). On the right already stands the source of curvature. On the left already stands the receptacle of curvature in the form of what one wants to know, the metric coefficients, twice differentiated. Claim therefore that one has nothing to do except to go ahead and solve", " these equations for the metric coefficients. However, in analyzing the structure of the equations to greater depth [see Cartan (l922a) for the rationale of analyzing a coupled set of partial differential equations], one discovers that one can only make the split between \"the source and the receptacle\" in the right way when one has first recognized the still more important split between \"the initial-value data and the future,\" Thus-to summarize the results before doing the analysis-four of the ten components of Einstein's law connect the curvature of space here and now with the distribution of mass-energy here and now, and the other six equations tell how the geometry as thus determined then proceeds to evolve. In determining what are appropriate initial-value data to give, one discovers no guide more useful than the Hilbert variational principle, I =JE d 4x =JL(- g)1/2 d 4x =JL d(proper 4-volume) = extremum (21.2) 4exercise 8.16] Give initial data, predict geometry Four of ten components of Einstein equation are conditions on initial-value data or the Arnowitt-Deser-Misner (\"ADM\") variant of it (\u00a72l.6) and generalizations thereof by Kuchar (\u00a721.9), Out of this principle one can recognize most directly what one must hold fixed at the limits (on an initial spacelike hypersurface and on a final spacelike hypersurface) as one varies the geometry (\u00a721.2) throughout the spacetime \"filling of this sandwich,\" if one is to have a well-defined extremum problem, The Lagrange function L (scalar function) or the Lagrangian density E = (- g)1/2L (quantity to be integrated over coordinate volume) is built of geometry alone, when one deals with curved empty space, but normally fields are present as well, and contribute also to the Lagrangian; thus, E = Egeom + Efie1d = (- g)1I2L; L = L geom + L fie1d. (21.3) The variation of the field Lagrangian with respect to the typical metric coefficient is, for proves to be, of all ways, calculating) the corresponding component of the symmetric stress-energy tensor of the field (\u00a721.3). the one most convenient for generating (that New view of stress-energy tensor 486 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA A computer, allowing for the effect of this field on the geometry and computing ahead from instant to instant the evolution of the metric with time, imposes its own ordering on the events of spacetime. In effect, it slices spacetime into a great number of spacelike slices. It finds it most convenient (\u00a721.4) to do separate bookkeeping on (I) the 3-geometry of the individual slices and (2) the relation between one such slice and the next, as expressed in a \"lapse function\" N and a 3-vector \"shift func tion\" Ni. The 3-geometry internal to the individual slice or \"simultaneity\" defines in and by itself the three-dimensional Riemannian curvature intrinsic to this hypersurface; but for a complete account of this hypersurface one must know also the extrinsic curvature (\u00a721.5) telling how this hypersurface is curved with respect to the envelop ing four-dimensional spacetime manifold. In terms of the space-plus-time split of the 4-geometry, the action principle of Hilbert takes a simple and useful form (\u00a721.6). In the most elementary example of the application of an action principle in mechanics, where one writes 1= Jz.t z',t' L(dxjdt, x, t) dt (21.4) and extremizes the integral, one already knows that the resultant \"dynamic path length\" or \"dynamic phase\" or \"action,\" S(x, t) = Iextremum' (21.5) is an important quantity, not least because it gives (up to a factor 11) the phase of the quantum-mechanical wave function. Moreover, the rate of change of this action function with position is what one calls momentum, p = oS(x, t)jox; (21.6) and the (negative of the) rate of change with time gives energy (Figure 21.1), E = - oS(x, t)jot; and the relation between these two features of a system of wave crests, Hamiltonian as a dispersion relation E = H(P,x), (21.7) (21.8) call it \"dispersion relation\" or call it what one will, is the central topic of mechanics.", " When dealing with the dynamics of geometry in the Arnowitt-Deser-Misner formulation, * one finds it convenient to think of the specified quantities as being *Historical remark. No one knew until recently what coordinate-free geometric-physical quantity really is fixed at limits in the Hilbert-Palatini variational principle. In his pioneering work on the Hamiltonian formulation of general relativity, Dirac paid no particular attention to any variational principle. He had to generalize the Hamiltonian formalism to accommodate it to general relativity, introducing \"first- and second-class constraints\" and generalizations of the Poisson brackets of classical mechanics. The work of Amowitl, Oeser, and Misner, by contrast, took the variational principle as the foundation for the whole treatment, even though they too did not ask what it is that is fixed at limits in the sense of Figure 21.1. Momentum and (the negative of the) energy viewed as rate of change of \"dynamic phase\" or \"action,\" S(x, t) = Iextremum(X, t) = al ( extremum)fz,t f v ue 0 z',t' L(x..i. t) dt, (1 ) with respect to position and time; thus, 8S = P 8x - E 8t. (2) The variation of the integral I with respect to changes of the history along the way, 8x(t), is already zero by reason of the optimization of the history; so the only change that takes place is 8S = 8Iextremum = L(x, X, t) 8t + f z+Jz.t 8L dt z+Jz,t (aL = L 8t + f = L8t+-.1x+ f z',t' aL ax ax z',t' z'.t' aL ) ---:- 8x + -'~x dt ax z+Jz,/ (aL ---- 8xdt. d aL) dt a.i,, ax (3) i rzero by reason l t 12f extremizationr When one contemplates only a change 8x in the coordinates (x, t) of the end point (change of history from 8P to 852), one has.1x = 8x. When one makes only a change 8t in the end point (change of history from 19P to 8S), one has.1x =(indicator of change from P to 6fl) = -x 8t. For the general variation of the final point, one thus has.1x = 8x - x 8t and - - - x ----1~~ One concludes that the \"dispersion relation\" is obtained by taking the relations [compare (2) and (4)] 8S = - aL ax 8x - (aL x ax ) - L 8t. (4) and rate of.change Of) dynamiC phase with position ( = (momentum) = p = 2L(x, x, t).?x - rate of change Of) dynamic ph;se with time ( = (energy) = E =.i ~ - L. 'L eX and eliminating x from them [solve (5) for.i and substitute that value of.i into (6)]: thus or E = H(p, x, t) _ as = H(as x t). ax\" ct (5) (6) (7) (8) Every feature of this elementary analysis has its analog in geometrodynamics. a coordinate-free geometric-physical quantity. The great payoff of this work Y,'as recognition of the lapse and shift functions of equation (21.40) as Lagrange multipliers. the coefficients of which gave directly and simply Dirac's constraints. They did not succeed in arriving at a natural and simple time-coordinate. but that goal has in the meantime been achieved in the \"extrinsic time\" of Kuchar and York (\u00a721.11). However, the Arnowitt-Deser Misner approach opened the door to the \"intrinsic time\" of Sharp. Baierlein. and Wheeler. where 3-geometry is fixed at limits. and 3-geometry is the carrier of information about time; and this led directly to Wheeler's \"superspace version\" of the treatment of Arnowitt. Oeser. and Misner. 488 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA the 3-geometry (3)~ of the initial spacelike hypersurface and the 3-geometry (3)~ of the final spacelike hypersurface. One envisages the action integral as extremized with respect to the choice of the spacetime that", " fills the \"sandwich\" between these two faces. If one has thus determined the spacetime, one has automatically by that very act determined the separation in proper time of the two hypersurfaces. There is no additional time-variable to be brought in or considered. The one concept (3)~ thus takes the place in geometrodynamics of the two quantities x, t of particle there depended on x and t here depends on the dynamics. The action S that 3-geometry of the face of the sandwich; thus, (21.9) A change in the 3-geometry changes the action. The amount of the change in action per elementary change in 3-geometry defines the \"field momentum\" 7TUue conjugate to the geometrodynamic field coordinate gij' according to the formula oS = J7TUue ogijd3x. (21.10) Comparing this equation out of the Arnowitt, Oeser, and Misner (ADM) canonical formulation of geometrodynamics (\u00a721.7) with the expression for change of action with change of endpoint in elementary mechanics, oS = pox - E ot, (21.11) one might at first think that something is awry, there being no obvious reference to time in (21.10). However, the 3-geometry is itself automatically the carrier of information about time; and (21.1 0) is complete. Moreover, with no \"time\" variable other than the information that (3)~ itself already carries about time, there is also no \"energy.\" Thus the \"dispersion relation\" that connects the rates of change of action with respect to the several changes that one can make in the \"field coordinates\" or 3-geometry takes the form (21.12) with the E-term of (21.8) equal to zero (details in \u00a721.7). All the content of Einstein's general relativity can be extracted from this one Hamiltonian, or \"super-Hamilton ian,\" to give it a more appropriate name [see DeWitt (1967 a), pp. 1113-1118, for an account of the contributions of Dirac, of Arnowitt, Oeser, and Misner, and of others to the Hamiltonian formulation of geometrodynamics; and see \u00a721.7 and subsequent sections of this chapter for the meaning and payoffs of this formulation]. The difference between a Hamiltonian and a super-Hamiltonian [see, for example, Kramers (1957)] shows nowhere more clearly than in the problem of a charged particle moving in flat space under the influence of the field derived from the electromagnetic 4-potential, A/xa ). The Hamiltonian treatment derives the equation of motion from the action principle,. Action viewed as dependent on 3-geometry Hamiltonian versus super-Hamiltonian \u00a721.1. DYNAMICS REQUIRES INITIAL-VALUE DATA 489 with.. ( H= -~<P+ m 2 +T/lJ Pi+~Ai pj+~Aj e [ e X e )]1/2. The super-Hamiltonian analysis gets the equations of motion from the action principle t) 0= oJ' = 0 J[ dx'\" P\", d\"A - X(Pa' x ) d\"A. f3 ] Here the super-Hamiltonian is given by the expression The variational principle gives Hamilton's equations for the rates of change and From these equations, one discovers that X itself must be a constant, independent of the time-like parameter \"A. The value of this constant has to be imposed as an initial condition, X = 0 (\"specification of particle mass\"), thereafter maintained by the Hamiltonian equations themselves. This vanishing of:JC in no way kills the partial derivatives, that enter Hamilton's equations for the rates of change, and -o:Jc;oxf3, and Whether derived in the one formalism or the other, the equations of motion are equivalent, but the covariance shows more clearly in the formalism of the super Hamiltonian, and similarly in general relativity. Granted values of the \"field coordinates\" gij(x,y, z) \u00ab3)~) and field momenta '1Tll-ue(x,y, z) = OSjOgij compatible with (21.12), one has what are called \"compatible initial-value data on an initial spacelike hypersurface.\" One can proceed as described in \u00a721.8 to integrate ahead in time step by step from one spacelike hypersurface Dynamic evolution of to another and another, and construct the whole 4-geometry. Here one is dealing with what in mathematical terminology are hyperbolic differential equations that have the character of a wave equation. geometry. In contrast, one deals with elliptic differential equations that have the character", " of a Poisson potential equation when one undertakes in the first place to construct the needed initial-value data (\u00a721.9). In the analysis of these elliptic equations, it Another choice of what to fix at boundary hypersurface: conformal part of 3-geometry plus extrinsic time Mach updated: mass-energy there governs inertia here 490 21. VARIATIONAL PRINCIPLE AND INITIAL\u00b7VALUE DATA proves helpful to distinguish in the 3-geometry between (I) the part of the metric that determines relative lengths at a point, which is to say angles (\"the conformal part of the metric\") and (2) the common multiplicative factor that enters all the components of the gij at a point to determine the absolute scale of lengths at that point. This breakdown of the 3-geometry into two parts provides a particularly simple way to deal with two special initial-value problems known as the time-symmetric and time-antisymmetric initial-value problems (\u00a721.1 0). The ADM formalism is today in course of development as summarized in \u00a721.11. In Wheeler's (l968a) \"superspace\" form, the ADM treatment takes the 3-geometry to be fixed on each of the bounding spacelike hypersurfaces. In contrast, York (\u00a721.11) goes back to the original Hilbert action principle, and discovers what it takes to be fixed on each of the bounding spacelike hypersurfaces. The appropriate data turn out to be the \"conformal part of the 3-geometry\" plus something closely related to what Kuchar (1971a and 1972) calls the \"extrinsic time.\" The contrast between Wheeler's approach and the Kuchar-York approach shows particularly clearly when one (I) deals with a flat spacetime manifold, (2) takes a flat spacelike section through this spacetime, and then (3) introduces a slight bump on this slice, of height t\". The 3-geometry intrinsic to this deformed slice differs from Euclidean geometry only to the second order in t\". Therefore to read back from the full 3-geom etry to the time (\"the forward advance of the bump\") requires in this case an the Kuchar-York operation something like extracting a square root. In contrast, treatment deals with the \"extrinsic curvature\" of the slice, something proportional to the first power of t\", and therefore provides what is in some ways a more convenient measure of time [see especially Kuchar (1971) for the construction of \"extrinsic time\" for arbitrarily strong cylindrical gravitational waves; see- also Box 30.1 on \"time\" as variously defined in \"mixmaster cosmology\"]. York shows that the time-variable is most conveniently identified with the variable \"dynamically conjugate to the conformal factor in the 3-geometry.\" The initial-value problem of geometrodynamics can be formulated either in the language of Wheeler or in the language of Kuchar and York. In either formulation (\u00a721.9 or \u00a721.l1) it throws light on what one ought properly today to understand by Mach's principle (\u00a721.12). That principle meant to Mach that the \"acceleration\" dealt with in Newtonian mechanics could have a meaning only ifit was acceleration with respect to the fixed stars or to something equally well-defined. It guided Einstein to general relativity. Today it is summarized in the principle that \"mass-energy there governs inertia here,\" and is given mathematical expression in the initial-value equations. The analysis of the initial-value problem connected past and future across a spacelike hypersurface. In contrast, one encounters a hypersurface that accommo dates a timelike vector when one deals (\u00a721.l3) with the junction conditions between one solution of Einstein's field equation (say, the Friedmann geometry interior to a spherical cloud of dust of uniform density) and another (say, the Schwarzschild geometry exterior to this cloud of dust). Section 21.13, and the chapter, terminate with notes on gravitational shock waves and the characteristic initial-value problem (the statement of initial-value data on a light cone, for example). \u00a721.2. HILBERT ACTION PRINCIPLE AND PALATINI METHOD OF VARIATION 491 \u00a721.2. THE HILBERT ACTION PRINCIPLE AND THE PALATINI METHOD OF VARIATION Five days before Einstein presented his geometrodynamic law in its final and now standard form, Hilbert, animated by Einstein's earlier work, independently discovered (l915a) how to formulate this law as the consequence of the simplest action principle of the form (21.2-21.3) that one can imagine: Variational principle the simplest route to Einstein's", " equation (21.13) (Replace 1/16'1T by c3/16'1TG when going from the present geometric units to conven tional units; or divide by 11 -- L\u00b72 to convert from dynamic phase, with the units of action, to actual phase of a wave function, with the units of radians). Here (4)R is the four-dimensional scalar curvature invariant, as spelled out in Box 8.4. This action principle contains second derivatives of the metric coefficients. In contrast, the action principle for mechanics contains only first derivatives of the dynamic variables; and similarly only derivatives of the type oA;./ox/3 appear in the action principle for electrodynamics. Therefore one might also have expected only first derivatives, of the form og!J.p/oxY, in the action principle here. However, no scalar invariant lets itself be constructed out of these first derivatives. Thus, to be an invariant, Lgeom has to have a value independent of the choice of coordinate system. But in the neighborhood of a point, one can always so choose a coordinate system that all first derivatives of the g!J.P vanish. Apart from a constant, there is no scalar invariant that can be built homogeneously out of the metric coefficients and their first derivatives. When one turns from first derivatives to second derivatives, one has all twenty distinct components of the curvature tensor to work with. Expressed in a local inertial frame, these twenty components are arbitrary to the extent of the six parameters of a local Lorentz transformation. There are thus 20 - 6 = 14 independent local features of the curvature (\"curvature invariants\") that are coordinate-independent, anyone of which one could imagine employing in the action principle. However, (4)R is the only one of these 14 quantities that is linear in the second derivatives of the metric coefficients. Any choice of invariant other than Hilbert's complicates the geometrodynamic law, and destroys the simple correspondence with the Newto nian theory of gravity (Chapter 17). Hilbert originally conceived of the independently adjustable functions of x, y, Z, t in the variational principle as being the ten distinct components of the metric tensor in contravariant representation, g!J.P. Later Palatini (1919) discovered a simpler and more instructive listing of the independently adjustable functions: not the ten g!J. P alone, but the ten g!J.P plus the forty r~p of the affine connection. To give up the standard formula for the connection r in terms of the metric g and let r \"flap in the breeze\" is not a new kind of enterprise in mathematical physics. Even in the simplest problem of mechanics, one can give up the standard formula for the momentum p in terms of a time-derivative of the coordinate x and also let Scalar curvature invariant the only natural choice 492 21. VARIATIONAL PRINCIPLE AND INITIAL\u00b7VALUE DATA Idea of varying coordinate and momentum independently p \"flap in the breeze.\" Then x(t) and p(t) become two independently adjustable functions in a new variational principle, 1= f r,t [ \"\",t' dx(t) p(t) -d- - H(p(t), x(t), t) dt = extremum. ] t (21.14) Happily, out of the extremization with respect to choice of the function p(t), one recovers the standard formula for the momentum in terms of the velocity. The extremization with respect to choice of the other function, x(t), gives the equation of motion just as does the more elementary variational analysis of Euler and La grange, where x(t) is the sole adjustable function, A further analysis of this equiva lence between the two kinds of variational principles in particle mechanics appears in Box 21.1, In that box, one also sees the two kinds of variational principle as applied to electrodynamics, To express the Hilbert variational principle in terms of the r~v and ga/3 regarded as the primordial functions of t, x, y, z, note that the Lagrangian density is Here, as in any spacetime manifold with an affine connection, one has (Chap ter 14) where (21.16) (21.17) and every r is given in advance (in a coordinate frame) as symmetric in its two lower indices, In order that the integral I of (21.2-21.3) should be an extremum, one requires that the variation in I caused by changes both in the g/LV and in the r's should vanish; thus, Consider now the variations of the individual factors in the first and second integrals in (21.18), The variation of the first factor is trivial, o", "ga/3. In the variation of the second factor, Ra/3' changes in the ga/3 play no part; only changes in the r's appear. Moreover, the variation or~/3 is a tensor even though r~/3 itself is not. Thus in the transformation formula (21.19) the last term destroys the tensor character of any set of r~T individually, but subtracts out in the difference or~T between two alternative sets of r's. Note that the variation oR\"\\/L/3 of the typical component of the curvature tensor consists of two terms of (continued on page 500) Variation of connection is a tensor \u00a721.2. HILBERT ACTION PRINCIPLE AND PALATINI METHOD OF VARIATION 493 Box 21.1 RATE OF CHANGE OF ACTI ON WITH DYNAM IC COORDINATE (= \"MOMENTUM\") AND WITH TIME, AND THE DISPERSION RELATION (= \"HAMILTONIAN\") THAT CONNECTS THEM IN PARTICLE MECHANICS AND IN ELECTRODYNAMICS A. PROLOG ON THE PARTICLE-MECHANICS ANALOG OF THE PALATINI METHOD In particle mechanics, one considers the history x = x(t) to be adjustable be tween the end points (x', t') and (x, t) and varies it to extremize the integral 1= f L(x, i, t) dt taken between these two limits. X\",!\" Expressed in terms of coordinates and momenta (see Figure 21.1), the integral has the form 1= f [pi - H(p, x, t)] dt, (I) where x(t) is again the function to be varied and p is only an abbreviation for a certain function of x and i; thus, p = oL(x, i, t)/'ai. Viewed in this way, the variation, op(t), of the momentum is gov erned by, and is only a reflection of, the variation ox(t). 1. Momentum Treated as Independently Variable There miraculously exists, however, quite another way to view the problem (see inset). One can re gard x(t) and p(t) as two quite uncorrelated and independently adjustable functions. One abandons the formula p = 'aL(x, i, t)/'ai, only to recover it, or the equivalent of it, from the new \"independ the ent-coordinate-and-momentum version\" of variation principle. The variation of (I), as defined and calculated in this new way, becomes OJ =p ox I :r\",t\" :r',t' + f :r',t' :r\",t\" [( i - OH) 'ap ( op + - p - - 'aH) J ox dt. ox (2) Demand that the coefficient of op vanish and have the sought-for new version,, 'aH(p, x, t) x = ---\"--- op Box 21.1 (continued) of the old relation, p = oL(x, i, t)/oi, between momentum and velocity. The Van ishing of the coefficient of ox gives the other Hamilton equation,, P = - oH(p, x, t) ox'equivalent in content to the original Lagrange equation of motion, (3) (4) That p(t) in this double variable conception is-before the extremization!-a function of time quite separate from and independent of the function x(t) shows nowhere more clearly than in the circumstance that p(t) has no end point conditions imposed on it, whereas x' and x\" are specified. Thus not only is the shape of the history subject to adjustment in x, p, t space in the course of achieving the extremum, but even the end points are subject to being slid along the two indicated lines in the inset, like beads on a wire. 2. Action as Tool for Finding Dispersion Relation Denote by S(x, t) the \"action,\" or extremal value of I, for the classical history that starts with (x', t' ) and ends at (x, t) (= 11 times phase of de Broglie wave). To change the end points to (x + ox, t) makes the change in action - oS = pox. Thus momentum is \"rate of change of action with dynamic coordinate.\" To change the end point to (x + ox, t + ot) = ([x + i ot] + [ox - i ot], t + ot) makes the change in action oS = p[ox - i ot] + Lot =P ox - Hot. (5) (6)", " (7) Thus the Hamiltonian is the negative of \"the rate of change of action with time.\" In terms of the Hamiltonian H = H(p, x), the \"dispersion relation\" for de Broglie waves becomes _oS=H(oS,x). ot ox (8) In the derivation of this dispersion relation, one Can profitably short-cut all talk of p(t) and x(t) as independently variable quantities, and derive the result in hardly more than one step from the definition 1= f L(x, X, t) dt. Similarly in electrody namics. The remainder of this box best follows a first perusal of Chapter 21. B. ANALOG OF THE PALATINI METHOD IN ELECTRODYNAMICS In source-free electrodynamics, one considers as given two spacelike hypersurfaces S' and S\", and the magnetic fields-as-a-function-of-position in each, B' and B\" (this second field will later be written without the\" superscript to simplify the notation). To be varied is an integral extended over the region of spacetime between the two hypersurfaces, IMaxwell =f.eMaxwell d 4x = - 1~'1T f Fp.v Fp.v( - g)1/2 d 4x. (9) 1. Variation of Field on Hypersurface and Variation of Location of Hypersurface are Cleanly Separated Concepts in Electromagnetism The electromagnetic field F is the physically relevant quantity in electromagnetism (compare the 3-geometry in geometrodynamics). By contrast, the 4-potential A has no direct physical significance. A change of gauge in the potentials, leaves unchanged the field components Fp.v = oAv/oxP. - oAp./oxv (compare the coordinate transformation that changes the gp.v while leaving un changed the (3)~). The variation of the fields within the body of the sandwich is nevertheless expressed most conveniently in terms of the effect of changes oAp. in the potentials. One also wants to see how the action integral is influenced by changes in the location of the upper spacelike hypersurface (\"many-fingered time\"). Think of the point of the hypersurface that is presently endowed with coordinates x, y, z, t(x, y, z) as being displaced to x, y, =, t + ot(x,y, z). Now renounce this use of a privi leged coordinate system. Describe the displacement of the simultaneity in terms of a 4-vector on (not a unit 4-vector) normal to the hypersurface X. The ele ment of 4-volume oil included between the original upper face of the sandwich and the new upper face, that had in the privileged coordinate system the form (- g)l/2 ot(x,y~~)d3X, in the notation of Chapter 20 becomes where the element of surface d 3X p. already includes the previously listed factor (_ g)1/2. (10) 496 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA Box 21.1 (continued) Counting together the influence of changes in the field values on the upper hypersurface and changes in the location of that hypersurface, one has oS = oIextremaI = -(l/16'1T)f upper..!' FP.I'Fp.p(on\u00b7 d 3E) + (I/4'1T)f upper..!' FP.P&d 3 I p t--,,__ (II) I I replace by its equivalent /OAp. - onaAp.;a\\ + (I/4'1T) f EP;p,OAp.( _g)1/2 d 4x. 4-volume t ---. I has to vanish because integral has I been extremized I I Simplify this expression by arranging the coordinates so that the hypersurface shall be a hypersurface of constant t, and so that lines of constant x, y, z shall be normal to this\u00b7 hypersurface. Then it follows that the element of volume on that hypersurface contains a single nonvanishing component, d 3I o = (_g)1/2d 3x. The antisymmetry of the field quantity FOP in its two indices requires that v be a spacelike label, i = 1, 2, 3. The variation of the action becomes, add and subtract {( )I/ZFiO - g4'1T Ao 01 },.,l 2. Meaning of Field \"Momentum\" in Electrodynamics Identify this expression with the quantity oS = f '", "1T1M OA i d 3x - f X oil, where (12) (13) oS (\"denSity of electromagnetic ) '1TkM =8 = momentum dynamically Canon- = ically conjugate to At Ai \u00a721.2. HILBERT ACTION PRINCIPLE AND PALATINI METHOD OF VARIATION 497 is a simple multiple of the electric field and where o X = - o~ = electromagnetic = (1/1617)[FIlVFllv + 4FiO(A i ;o - AO;i)] (15) (\"density of ) Hamiltonian\" = (1/817)(E2 + B2). The concept ofdynamic Hamiltonian density agrees with the usual concept ofdensity of electromagnetic energy, despite the very different context in which the two quantities are derived and used. However, the canonical momentum 17kM has nothing directly whatsoever to do with the density of electromagnetic momentum as defined, for example, by the Poynting vector, despite the confusing similarity in the standard names for the two quantities. Note that there is no term oAo in (13); that is, 17~M = O. 3. Bubble Differentiation to The \"bubble differentiation\" with respect \"many-fingered time\" that appears in (15) was first introduced by Tomonaga (1946). One thinks of a spacelike hypersurface I l' a magnetic field B de fined as a function of position on this hypersurface (by an observer on a world line normal to this hypersurface), and a prescription S that carries one from this information to a single number, the ac tion. (Divided by 1'1, this action gives the phase of the \"wave function\" or \"probability amplitude\" for the occurrence of this particular distribution of field values over this particular hypersurface.) One goes to a second hypersurface I 2 (see inset), which is identical with I l' except in the immediate vicinity of a given point. Take a distribution of field values over X 2 that is identical with the original distribution over I l' \"identity of location\" being defined by means of the normal. Evaluate the difference, oS, in the value of the dynamic phase or action in the two cases. Divide this difference by the amount of proper 4-volume oil = f(on' d 3E) contained in the \"bubble\" between the two hypersurfaces. Take the quotient, evaluate it in the limit in which the size of the bubble goes to zero, and in this way get the \"bubble-time derivative,\" oS/oil, of the action. 498 21. VARIATIONAL PRINCIPLE AND INITIAL\u00b7VALUE DATA Box 21.1 (continued) What does it mean to say that the action, S, besides depending on the hypersurface,.2, depends also on the distribution of the magnetic field, B, over that hypersurface? The action depends on the physical quantity, B = V X A, not on the prephysical quantity, A. Thus a change in gauge oA; = aAiaxi, cannot make any change in S. On the other hand, the calculated value of the change in S for this alteration in A is o(action) = oS =f ~S oA; d 3x - f oS aA d 3 oA; ax; x - - vA; - f (OS) oA; '(,;/\\ x,y,z ) d 3 x. (I6) In order that there shall be no dependence of action on gauge, it follows that this expression must vanish for arbitrary A(X,y, z), a result only possible if S(.2, B) = S(hypersurface, field on hypersurface) satisfies the identity (g;).= 7TkM,; = - (I/47T)G;,; = o. i \" (17) 4. Hamilton-Jacobi \"Propagation Law\" for Electrodynamics The \"dispersion relation\" or \"Hamilton-Jacobi equation\" Joe electromagnetism relates (I) the changes of the \"dynamic phase\" or \"action\" brought about by altera tions in the dynamic variables A; (the generalization of the x of particle dynamics) with (2) the changes brought about by alterations in many-fingered time (the gener alization of the single time t of particle dynamics); thus (I5) translates into (I8) C. DISPERSION RELATIONS FOR GEOMETRODYNAMICS AND ELECTRODYNAMICS COMPARED AND CONTRASTED Geometrodynamics possesses a direct analog of equation (I7) (\"action depends on no information carried by the vector potential A except", " the magnetic field B = V X A\"), in an equation that says the action depends on no information carried by the metric g;j on the \"upper face of the sandwich\" except the 3-geometry there, (3)~. It also possesses a direct analog of equation (I8) (\"dynamic equation for the propagation of the action\") with this one difference: in electrodynamics the field variable B and the many-fingered time are distinct in character, whereas in geo metrodynamics the \"field\" and the \"many-fingered time\" can be regarded as two aspects of one and the same (3l~: \u00a721.2. HILBERT ACTION PRINCIPLE AND PALATINI METHOD OF VARIATION 499 D. ACTION PRINCIPLE AND DISPERSION RELATION ARE ROOTED IN THE QUANTUM PRINCIPLE; FEYNMAN'S PRINCIPLE OF THE DEMOCRATIC EQUALITY OF ALL HISTORIES For more on action principles in physics, see for example Mercier (1953), Lanczos (1970), and Yourgrau and Mandelstam (1968). Newton (1687) in the first page of the preface to the first edition of his Principia notes that \"The description of right lines..., upon which geometry is founded, belongs to mechanics. Geometry does not teach us to draw these lines, but requires them to be drawn.\" Newton's remark is also a question. Mechanics moves a particle along a straight line, but what is the machinery by which mechanics accomplishes this miracle? The quantum principle gives the answer. The particle moves along the straight line only by not moving along the straight line. In effect it \"feels out\" every conceivable world line that leads from the start, (x', t'), to the point of detection, (x\", to), \"compares\" one with another, and takes the extremal world line. How does it accomplish this miracle? The particle is governed by a \"probability amplitude to transit from (x', t') to (x\", t\").\" This amplitude or \"propagator,\" (x\", t\"lx', t'), is the democratic sum with equal weight of contributions from every world line that leads from start to finish; thus, (IS) Here N is a normalization factor, the same for all histories.,vx is the \"volume element\" for the sum over histories. For a \"skeleton history\" defined by giving X n at tn = to + n LIt, one has oJ)x equal, up to a multiplicative constant, to dX 1 dX2... dx N. When the history is defined by the Fourier coefficients in such an expression as x'(t\" - t) + x\"(t (t\" _ t') t') x(t) =. (t - t') + ~ an SIn nTr (t\" _ t'), n (16) the volume element, again up to a multiplicative factor, is da1 da2.... Destructive interference in effect wipes out the contribution to the transItIOn probability from histories that differ significantly from the \"extremal history\" or \"classical history.\" Histories that are near that extremal history, on the other hand, contribute constructively, and for a simple reason: a small departure of the first order from the classical history brings about a change in phase which is only of the second order in the departure. In this elementary example, one sees illustrated why it is that extremal principles play such a large part in classical dynamics. They remind one that all classical physics rests on a foundation of quantum physics. The central ideas are (l) the principle 500 21. VARIATIONAL PRINCIPLE AND INITIAL\u00b7VALUE DATA Box 21.1 (continued) of superposition of probability amplitudes, (2) constructive and destructive interfer ence, (3) the \"democracy of all histories,\" and (4) the probability amplitude associ ated with a history His eilHlll, apart from a normalizing factor that is a multiplicative constant. For more on the democracy of histories and the sum over histories see Feynman (1942, 1948, 1949, 1951, and 1955), and the book of Feynman and Hibbs (1965); also Hibbs (1951), Morette (1951), Choquard (1955), Polkinghorne (1955), Fujiwara (1962), and the survey and literature references in Kursunoglu (1962); also reports of Dempster (1963) and Symanzik (1963). This outlook has been applied by many workers to discuss the quantum formulation of geometrodynamics, the first being Misner (1957) and one of the latest being Faddeev (1971).", "1/1617)f (gaP;A - 8~gaY;y) 8r~p d 4x. This integral is the only term in the action integral that contains the variations of the r's at the \"interior points\" of interest here. For the integral to be an extremum, the symmetrized coefficient of 8r~p must vanish, This set of forty equations for the forty covariant derivative gaP 'A has only the zero solution,, g\"'P;A=O. (21.25) Thus the \"density formed from the reciprocal metric tensor\" is covariantly constant. This simple result (1) brings many simple results in its train: the covariant con stancy of (2) (- g)1/2, (3) gap, (4) gaP' and (5) gap, Of these, (4) is of special interest here, and (2) is needed in proving it, as follows. Take definition (21.24) for the covariant derivative of ( - g)l/2, and calculate the ordinary derivative that appears in the first term from exercise 21.1. One encounters in this calculation terms of the form ogaPIOx A. Use (21.25) to evaluate them, and end up with the result From this result it follows that the covariant derivative of the G)-tensor density ( - g)1/2 8~ is also zero. But this tensor density is the product of the tensor density g\"'P by the ordinary metric tensor gpy' In the covariant derivative of this product by x\\ one already knows that the derivative of the first factor is zero. Therefore the first factor times the derivative of the second must be zero, and from this it follows that as was to be proven; or, explicitly, (21.26) Solve these equations for the r's, which up to now have been independent of the gpy, and end up with the standard equation for the connection coefficients, (21.27) as required for Riemannian geometry. \u00a721.2. HILBERT ACTION PRINCIPLE AND PALATINI METHOD OF VARIATION 503 Similarly, equate to zero the coefficient of oga{3 in the variation (21.23), and find all ten components of Einstein's field equation, in the form, Ga{3 = 817 (ga{3 L fie1d - 2 \u00b0OL~~d)., t ridentified in \u00a721.3 with L the stress-energy tensor Ta{3 g (2128) ] Among variations of the metric, one of the simplest is the change brought about by the infinitesimal coordinate transformation (21.29) (21.30) Although the metric changes, the 3-geometry does not. It does not matter whether the spacetime geometry that one is dealing with extremizes the action principle or not, whether it is a solution of Einstein's equations or not; the action integral I is a scalar invariant, a number, the value of which depends on the physics but not at all on the system of coordinates in which that physics is expressed, This invariance even obtains for both parts of the action principle individually (lgeom and l fields ), Therefore neither part will be affected in value by the variation (21.29), In other words, the quantity OIgeom = (1/1617)f Gap(~a;{3 + ~{3;a)( _g)l/2 d 4x _ =_-{1/817)f Ga{3:{3~a( - g)l/2 d 4x ~\"covariant integration by parts\"] (21.31) must vanish whatever the 4-geometry and whatever the change ~a, In this way, one sees from a new angle the contracted Bianchi identities of Chapter 15, The \"neutrality\" of the action principle with respect to a mere coordinate trans formation such as (21.29) shows once again that the variational principle-and with it Einstein's equation-cannot determine the coordinates or the metric, but only the 4-geometry itself, (21.32) Exercise 21.1. VARIATION OF THE DETERMINANT OF THE METRIC TENSOR Recalling that the change in the value of any determinant is given by multiplying the change in each element of that determinant by its cofactor and adding the resulting products (exercise 5.5) prove that c'J( - g)1/ 2 =t (- g)1/2g!'v c'Jg!'v and c'J( - g)1/2 = - t (- g)1!2g!,v c'Jg!'v. Also show that g = det 11!I!'v\" and Action unaffected by mere change in coordin", "atization EXERCISE 504 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA \u00a721.3. MAnER LAGRANGIAN AND STRESS-ENERGY TENSOR The derivation of Einstein's geometrodynamic law from Hilbert's action principle puts on the righthand side a source term that is derived from the field Lagrangian. In contrast, the derivation of Chapter 17 identified the source term with the stress energy tensor of the field. For the two derivations to be compatible, the stress-energy tensor must be given by the expression Lagrangian generates stress-energy tensOr or (2 I.33a) (2I.33b) What are the consequences of this identification? By the term \"Lagrange function of the field\" as employed here, one means the Lagrange function of the classical theory as formulated in flat spacetime, with the flat-spacetime metric replaced wherever it appears by the actual metric, and with the \"comma-goes-to-semicolon rule\" of Chapter 16 applied to all derivatives. Were one dealing with a general tensorial field, the comma-goes-to-semicolon rule would introduce, in addition to the derivative of the tensorial field with all its indices, a number of r's equal to the number of indices. The presence of these r's in the field Lagrangian would have unhappy consequences for the Palatini variational procedure described in \u00a721.2. No longer would the F's_e_nd up given in terms of the metric coefficients by the standard formula (21.27): No longer would the geom etry, as derived from the Hilbert-Palatini variation principle, be Riemannian. Then what? These troublesome issues do not arise in two well-known simple cases, a scalar field and an electromagnetic field. In the one case, the field Lagrangian becomes (21.34) No connection coefficient comes in; the quantity being differentiated is a scalar. In the other case, the field Lagrangian is built on first derivatives of the 4-potential A/l' Therefore r's should appear, according to the standard rules for covariant differentiation (Box 8.4). However, the derivatives of the A's appear, never alone, but always in an antisymmetric combination where the r's cancel, making covariant derivatives equivalent to ordinary derivatives: (21.35) Electromagnetism as an example Contrast to stress-energy tensor of \"canonical field theory\" In both cases, the differentiations of (21.33) to generate the stress-energy tensor are easily carried out (exercises 21.2 and 21.3) and give the standard expressions already seen [(5.22) and (5.23)] for T/l v in one of these two cases in an earlier chapter. Field theory provides a quite other method to generate a so-called canonical expression for the stress-energy tensor of a field [see, for example, Wentzel (1949)]. \u00a721.4. SPLITTING SPACETIME INTO SPACE AND TIME 505 the canonical in certain contexts. However, By the very manner of construction, such an expression is guaranteed also to satisfy the law of conservation of momentum and energy, and by this circumstance it too tensor is often not becomes useful symmetric in its two indices, and in such cases violates the law of conservation of angular momentum (see discussion in \u00a75.7). Even when symmetric, it may give a quite different localization of stress and energy than that given by (21.33). Field theory in and by itself is unable to decide between these different pictures of where the field energy is localized. However, direct measurements of the pull ofgravitation provide in principle [see, for. example, Feynman (1964)] a means to distinguish between alternative prescriptions for the localization of stress-energy, because gravi tation responds directly to density of mass-energy and momentum. It is therefore a happy circumstance that the theory of gravity in the variational formulation gives a unique prescription for fixing the stress-energy tensor, a prescription that, besides being symmetric, also automatically satisfies the laws of conservation of momentum and energy (exercises 212 and 21.3). [For an early discussion of the symmetrization of the stress-energy tensor, see Rosenfeld (1940) and Belinfante (1940). A more extensive discussion is given by Corson (1953) and Davis (1970), along with extensive references to the literature.] When one deals with a spinor field, one finds it convenient to take as the quantities to be varied, not the metric coefficients themselves, but the components of a tetrad of orthonormal vectors defined as a tetrad field over all space [see Davis (1970) for discussion and references]. Exercise 21.2.", " STRESS-ENERGY TENSOR FOR A SCALAR FIELD EXERCISES Given the Lagrange function (21.34) of a scalar field, derive the stress-energy tensor for this field. Also write down the field equation for the scalar field that one derives from this Lagrange function (in the general case where the field executes its dynamics within the arena of a curved spacetime). Show that as a consequence of this field equation, the stress-energy -tensor satisfies the conservation law, Ta /3;/3 = O. FARADAY-MAXWELL STRESS-ENERGY TENSOR Exercise 21.3. Given the Lagrangian density - FIJ.pFIJ.P/16'Tr, reexpress it in terms of the variables AIJ. and gIJ.P, and by use of (21.33) derive the stress-energy tensor as discussed in \u00a75.6. Also derive from the Lagrange variation principle the field equation Fa /3;/3 = 0 (curved spacetime, but-for simplicity-a charge-free region of space). As a consequence of this field equation, show that the Faraday-Maxwell stress-energy tensor satisfies the conservation law, Ta /3:/3 = O. For a more ambitious project, show that any stress-energy tensor derived from a field Lagrangian by the prescription of equation (21.33) will automatically satisfy the conservation law Ta /3;/3 =O. \u00a721.4. SPUrriNG SPACETIME INTO SPACE AND TIME There are many ways to \"push forward\" many-fingered time and explore spacetime faster here and slower there, or faster there and slower here. However, a computer is most efficiently programmed only when it follows one definite prescription. The y 506 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA Slice spacetime to compute spacetime Thin sandwich 4-geometry Figure 21.2. Building two 3-geometries into a thin sandwich 4-geometry, by interposing perpendicular connectors between the two, with preassigned lengths and shifts. What would otherwise be flexible thereupon becomes rigid. The flagged point illustrates equation (21.40). successive hypersurfaces on which it gives the geometry are most conveniently described by successive values of a time-parameter t. One treats on a different footing the 3-geometries ofthese hypersurfaces and the 4-geometry that fills in between these laminations. The slicing of spacetime into a one-parameter family of spacelike hypersurfaces is called for, not only by the analysis of the dynamics along the way, but also by the boundary conditions as they pose themselves in any action principle of the form, \"Give the 3-geometries on the two faces of a sandwich of ~pacetime, and adjust the 4-geometry in between to extremize the action.\" - ----. There is no simpler sandwich to consider than one of infinitesimal thickness (Figure 21.2). Choosing coordinates adapted to the (3 + I)-space-time split, designate the \"lower\" (earlier) hypersurface in the diagram as t = constant and the \"upper\" (later) one as t + dt = constant (names, only names; no direct measure whatsoever of proper time). Compare the two hypersurfaces with two ribbons of steel out of which one wants to construct a rigid structure. To give the geometry on the two ribbons by no means fixes this structure; for that purpose, one needs cross-connectors between the one ribbon and the other. It is not even enough (l) to specify that these connectors are to be welded on perpendicular to the lower ribbon; (2) to specify where each is to be welded; and (3) to give its length. One must in addition tell where each connector joins the upper surface. If the proper distances between tops of the connectors are everywhere shorter than the distances between the bases of the connectors, the double ribbon will have the curve ofthe cable of a suspension bridge; if everywhere longer, the curve of the arch of a masonry bridge. The data necessary for the construction of the sandwich are thus (l) the metric of the 3-geometry of the lower hypersurface, (21.36) telling the (distance)2 between one point in that hypersurface and another; (2) the metric on the upper hypersurface, \u00a721.4. SPLITTING SPACETIME INTO SPACE AND TIME gij(t + dt, x,y, z) dx i dx i; (3) a formula for the proper length, lapse of proper time between lower and upper hypersurface = fi ( \"lapse unctIOn ). \" dt = N(t, x,y, z) d", "t, 507 (21.37) (21.38) of the connector that is based on the point (x,y, z) of the lower hypersurface; and (4) a formula for the place on the upper hypersurface, (21.39) where this connector is to be welded. Omit part of this information, and find the structure deprived of rigidity. The rigidity of the structure of the thin sandwich is most immediately revealed in the definiteness of the 4-geometry of the spacetime filling of the sandwich. interval ds or dT between x'\" = (t, Xi) and x'\" +- dx'\" = Ask for (t + dt, Xi + dx i ). The Pythagorean theorem in its 4-dimensional form the proper Metric of 4-geometry depends on lapse and shift of connectors of the two 3-geometries ds 2 = (proper distance in base 3-geometry )2 _ (proper time from lower to upper 3-geometry )2 yields the result (see Figure 21.2). - ds 2 = gij(dx i + Ni dt)(dx i + Ni dt) - (N dt)2 (21.40) Here as in (21.36) the gij are the metric coefficients of the 3-geometry, distinguished by their Latin labels from the Greek-indexed components of the 4-metric, (21.41 ) labeled here with a suffix (4) to reduce the possibility of confusion. Comparing (21.41) and (21.40), one arrives at the following construction of the 4-metric out of the Details of the 4-geometry 3-metric and the lapse and shift functions [Arnowitt, Deser, and Misner (1962)]: (21.42) The welded connectors do the job! In (21.42), the quantities Nm are the components of the shift in its original primor dial contravariant form, whereas the N j = gimNm are the covariant components, as calculated within the 3-geometry with the 3-metric. To invert this relation, (21.43 ) y The components of the unit normal 508 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA is to deal with the reciprocaI3-metric, a quantity that has to be distinguished sharply from the reciprocal 4-metric. Thus, the reciprocal 4-metric is a result that one checks by calculating out the product wg (4)g{3Y = (4)8 Y a a{3 according to the standard rules for matrix multiplication. The volume element has the form (21.44) (21.45) Welding the connectors to the two steel ribbons, or adding the lapse and shift functions to the 3-metric, by rigidifying the 4-metric, also automatically determines the components of the unit timelike normal vector n. The condition of normalization on this 4-vector is most easily formulated by saying that there exists a I-form, also called n for the sake of convenience, dual to n, and such that the p~oduct of this vector by this I-form has the value (n,n) = -1. This I-form has the value n = n{3 dx{3 = -N dt + \u00b0+ 0+0. (21.46) (21.47) Only so can this I-form, this structure of layered surfaces, automatically yield a value of unity, one bong of the bell, when pierced as in Figure 2.4 by a vector that represents an advance of one unit in proper time, regardless of what x, y, and z displacements it also has. Thus the unit timelike normal vector in covariant I-form representation necessarily has the components n{3 = (-N,O,O,O) (21.48) Raise the indices via (21.44) to obtain the contravariant components of the same normal, represented as a tangent vector; thus, na = [(l/N), -(Nm/N)]. (21.49) This result receives a simple interpretation on inspection of Figure 21.2. Thus the typical \"perpendicular connector\" in the diagram can be said to have the components (dt, -Nm dt) and to have the proper length dT = N dt; so, ratioed down to a vector n of unit proper length, the components are precisely those given by (21.49). Scalar curvature as measure of area defic it \u00a721.5. INTRINSIC AND EXTRINSIC CURVATURE 509 \u00a721.5. INTRINSIC AND EXTRINSIC CURVATURE The central concept in Einstein's", " account of gravity is curvature, so it is appropriate to analyze curvature in the language of the (3 + l)-space-time split. The curvature intrinsic to the 3-geometry of a spacelike hypersurface may be defined and calculated by the same methods described and employed in the calculation offour-dimensional curvature in Chapter 14. Of all measures of the intrinsic curvature, one of the simplest is the Riemann scalar curvature invariant (3)R (written for simplicity of notation in what follows without the prefix, as R); and of all ways to define this invariant (see Chapter 14), one of the most compact uses the limit (see exercise 21.4) 4771,2 - ( proper area of a surface (approximatelY) a ~-sphere) defined ~s the locus of the pomts at a proper dIstance /; = Lim 18 - - - - - - - - - - - - - - - - - - 417 /;4 at point under study ) R (......0 (21.50) For a more detailed description of the curvature intrinsic to the 3-geometry, capitalize on differential geometry as already developed in Chapters 8 through 14, amending it only as required to distinguish what is three-dimensional from what is four-dimensional. Begin by considering a displacement (21.51) within the hypersurface. Here the e i are the basis tangent vectors e j = %xi (in one notation) or ei = o'!i'/}x i (in another notation) dual to the three coordinate I-forms dx i \u2022 Any field of tangent vectors A that happens to lie in the hypersurface lets itself be expressed in terms of the same basis vectors: The scalar product of this vector with the base vector e j is (21.52) (21.53) Now turn attention from a vector at one point to the parallel transport of the vector to a nearby point. A vector lying on the equator of the Earth and pointing toward the North Star, transported parallel to itself along a meridian to a point still on the Earth's surface, but 1,000 km to the north, will no longer lie in the 2-geometry of the surface of the Earth. A telescope located in the northern hemisphere has to raise its tube to see the North Star! The generalization to a three-dimensional hypersurface imbedded in a 4-geometry is immediate. Take vector A, lying in the hypersurface, and transport it along an elementary route lying in the hypersurface, and in the course of this transport displace it at each stage parallel to itself, where \"parallel\" means parallel with respect to the geometry of the enveloping 4-manifold. Then A will ordinarily 510 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA end up no longer lying in the hypersurface. Thus the \"covariant derivative\" of A in the direction of the i-th coordinate direction in the geometry of the enveloping spacetime (that is, the A at the new point diminished by the transported A) has the form (see \u00a7 lOA) (21.54) A special instance of this formula is the equation for the covariantly measured change of the base vector em itself, (21.55) In both (21.54) and (21.55) the presence of the \"out-of-the-hypersurface component\" (21.56) this component. Project (4)VA orthogonally onto the is quite evident. Now kill hypersurface. In this way arrive at a parallel transport and a covariant derivative that are intrinsic to the 3-geometry of the hypersurface. By rights this covariant derivative should be written (3)V; but for simplicity of notation it will be written as V in the rest of this chapter, except where ambiguity might arise. To get the value of the new covariant derivative, one has only to rewrite (21.54) with the suffix (4) replaced everywhere by a (3), or, better, dropped altogether and with the \"dummy index\" of summation Jl = (0,1,2,3) replaced by m = (1,2,3). However, it is more convenient, following Israel (1966), to turn from an expression dealing with contra variant components Ai of A to one dealing with covariant components Ai = (A\u00b7 ei)' Thus the covariant derivative of A in the direction of the i-th coordinate direction in the hypersurface, calculated with respect to the 3-geometry intrinsic to the hyper surface itself, has for its h-th covariant component ~he quantity [see equation (10.18)] Here the notation of the vertical stroke distinguishes this covariant derivative from the covariant derivative taken with respect to the 4-", "geometry, as, for example, in equations (l0.17ff). The connection coefficients here for three dimensions, like those dealt with earlier for four dimensions [see the equations leading from (14.14) through (14.15)], allow themselves to be expressed in terms of the metric coefficients and their first derivatives, and have the interpretation From parallel transport in 4-geometry to parallel transport in 3-geometry A new covariant derivative. taken with respect to the 3-geometry (3lr -r - e mhi = mhi - \u00b7Ve i h' m (21.58) From the connection coefficients in turn, one calculates as in Chapter 14 the full Riemann curvature tensor (3)Ri jmn of the 3-geometry intrinsic to the hypersurface. Over and above the curvature intrinsic to the simultaneity, one now encounters a concept not covered in previous chapters (except fleetingly in Box 14.1), the extrinsic curvature of the 3-geometry. This idea has no meaning for a 3-geometry \u00a721.5. INTRINSIC AND EXTRINSIC CURVATURE 511 8n = -K(89) 1; 9 I I 9+ 8':f Figure 21.3. Extrinsic curvature measures the fractional shrinkage and deformation of a figure lying in the spacelike hypersurface I that takes place when each point in the figure is carried forward a unit interval of proper time \"normal\"' to the hypersurface out into the enveloping spacetime. (No enveloping spacetime? No extrinsic curvature!) The extrinsic curvature tensor is a positive multiple of the unit tensor when elementary displace ments 89, in whatever direction within the surface they point, all experi enc~e same fractional_shrinkage. Thus the extrinsic curvature of the hypersurface illustrated in tlie figure is positive. The dashed arrow repre sents the normal vector n at the fiducial point ':f after parallel transport to the nearby point ':f + 8':f. conceived in and by itself. It depends for its existence on this 3-geometry's being imbedded as a well-defined slice in a well-defined enveloping spacetime. It measures the curvature of this slice relative to that enveloping 4-geometry (Figure 21.3). Take the normal that now stands at the point q> and, \"keeping its base in the hypersurface\" I, transport it parallel to itself as a \"fiducial vector\" to the point q> + oq>, and there subtract it from the normal vector that already stands at that point. The difference, on, may be regarded in the appropriate approximation as a \"vector,\" the value ofwhich is governed by and depends linearly on the \"vector\" of displacement oq>. To obviate any appeal to the notion of approximation, go from the finite displace ment oq> to the limiting concept of the vector-valued \"displacement I-form\" dq> [see equation 15.13]. Also replace the finite but not rigorously defined vector on by the limiting concept of a vector-valued I-form dn. This quantity, regarded as a vector, being the change in a vector n that does not change in length, must represent a change in direction and thus stand perpendicular to n. Therefore it can be regarded as lying in the hypersurface I. Depending linearly on dq>, it can be represented in the form dn = -K(cJC:J). (21.59) Extrinsic curvature as an operator Here the linear operator K is the extrinsic curvature presented as an abstract coor dinate-independent geometric object. The sign of K as defined here is positive when the tips of the normals in Figure 21.3 are closer than their bases, as they are, for example, during the recontraction of a model universe, in agreement with the conventions employed by Eisenhart (1926), Schouten (1954), and Arnowitt, Deser and Misner (1962), but opposite to the convention of Israel (1966). Into the slots in the I-forms that appear on the lefthand and righthand sides of (21.59), insert in place of the general tangent vector [which is to describe the general y 512 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA local displacement, so far left open, as in the discussion following (2.l2a)] a very special tangent vector, the basis vector ej, for a displacement in the i-th coordinate direction. Thus find (21.59) reading where the K/ are the com ponents of the linear operator K in a coordinate represen tation. Take the", " scalar product of both sides of (21.60) with the basis vector em' Recall (em' n) = O. Thus establish the symmetry of the tensor K jm, covariantly presented, in its two indices: (21.60) K. = K.ig. = K.i(e.\u00b7 e ) = -e. (4)V.n = n' (4)V.e ~m ~ 1. J Jm m m 1. ~ m = (n' e O)(4)r::'i = n' (4)Vm ej = K mj. Lsee (21.55)] (21.61 ) A knowledge of the tensor Kif of extrinsic curvature assists in revealing the changes of the four vectors n, e l, e z' e 3 under parallel transport. Equation (21.60) already tells how n changes under parallel transport. The change of em is to be read off from (21.55) as a vector. It is adequate identification of this vector to know its scalar product with each of four independent vectors: with the basis vectors e l, e z' and e 3, or, more briefly, with e s' in (21.58); and with the normal vector n in (21.61). Thus one arrives, following Israel (1966), at what are known as the equations of Gauss and Weingarten, in happy oversight of all change of notation in the interven ing century: (21.62) Knowing from this equation how each\u00b7 basis vector in I changes, one also knows how to rewrite (21.54) for the change in any vector field A that lies in I. The change in both cases is expressed relative to a fiducial vector transported from a fiducial nearby point. By the term \"parallel transport\" one now means \"parallel with respect to the geometry of the enveloping spacetime\": Gauss-Weingarten equation for 4-transport in terms of extri nsic curvatu re (4)VA = Ai. e + K Ai_n_. (n'n) if I. i i (21.63) --------- ------------ Of special importance is the evaluation of extrinsic curvature when spacetime is sliced up into spacelike slices according to the plan of Arnowitt, Deser, and Misner as described in \u00a721.4. The 4-geometry of the thin sandwich illustrated in Figure 21.2, rUdimentary though it is, is fully defined by the 3-metric on the two faces of the sandwich and by the lapse and shift functions Nand Ni. The normal in covariant representation according to (21.47) has the components The change in n relative to \"n transported parallel to itself in the enveloping 4-geometry,\" according to the definition of parallel transport, is (21.64) \u00a721.5. INTRINSIC AND EXTRINSIC CURVATURE 513 (21.65) Compare to the same change as expressed in terms of the extrinsic curvature tensor, Conelude that this tensor has the value K - -n - _N(4)FO ik - i;k - ik - (21.66) _N[(4)gOO<4lF + (4)gOp(4)F Oik ] pik, or, with the help of equations (21.42) and (21.44), Kik = (1/N)[(4)FOik - NP(3)Fpik] oNk 1 [ONi = 2N oxk + oxi - at - 2Fpik NP ogik ] O~k] = 2N Ni1k + Nk1i - at. 1 [ Extrinsic curvature in terms of shift and change of 3-metric (21.67) This is the extrinsic curvature expressed in terms of the ADM functions [Arnowitt, Deser, and Misner (1962)]. lapse and shift As an example, let I have the geometry of a 3-sphere ds 2 = a2[dx 2 + sin2X(d8 2 + sin28 dep2)]. (21.68) Extrinsic curvature of expanding 3-sphere Let the nearby spacelike slice in the one-parameter family of slices, the slice with the label t + dt (only a label!) have a 3-metric given by the same formula with the radius a replaced by a + da. The 4-geometry of the thin sandwich between these two slices is completely undetermined until one gives the lapse and shift functions. F or simplicity, take the shift vector Ni (see Figure 212) to be everywhere zero and the lapse function at every point on I to have the same value N", ". The separation in proper time between the two spheres is thus dT = N dt. Any geometric figure located in I expands with time. The fractional increase of any length in this figure per unit of proper time is the same in whatever direction that length is oriented, and has the value fractional inCrea~e) 1 1 d(a2) of length per umt =- - =----. 2N a2 dt of proper tIme a dT 1 da. ( (21.69) The negative of this quantity, multiplied by the <D unit tensor, 1 = e, gives the extrinsic curvature tensor in <D representation, 1 1 d(a2) K= - - - - -1. 2N a2 dt (21.70) Basic forms for calculating 4-curvature 514 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA One confirms this result (exercise 21.5) by direct calculation of the components K{ using the ADM formula (21.67) as the starting point. The Riemann curvature R\\cd = (3)R\\Cd intrinsic to the hypersurface I, together with the extrinsic curvature Kij' give one information on the Riemann and Einstein curvatures of the 4-geometry. In the calculation, it is not convenient to use the coordinate basis, basis vectors, basis I-forms eo = at, e i = ai, dt, dXi, because ordinarily the basis vector eo does not stand perpendicular to the hypersur face (see Figure 21.2). Adopt a different basis but one that is still self-dual, basis vectors, en = n = N-l(Ot - Nm am)' ei = ai' basis I-forms, w n = Ndt = (n \u00b7n)n Wi = dxi + Ni dt. (21.71) Also use Greek labels a = n, 1,2,3, instead of Greek labels a = 0, 1,2, 3, to list components. Recall that curvature is measured by the change in a vector on transport around a closed route; or, from equation (14.23), (21.72). Let the vector transported be e i and let the route be defined by ej and e k. The latter two vectors belong to a coordinate basis. Therefore the \"route closes automati cally\", [ej, e k] = 0, and the final term in (21.72) drops out of consideration. Call on (21.62) and (21.60) to find (4)V (4)V e. = (4)V [K _n__ + (3)Fm e e j,k (n \u2022 n) ej e., ik m ] = K',._n k.1 (n \u2022 n) KkK.me __1_ + (3)F1fl.e k '1 m (n \u2022 n) '.1 m (21.73) + (3)Fm [K._n_ + (3)F8. e ]. m1 (n \u2022 n) ik m1 8 Evaluate similarly the term with indices j and k reversed, subtract from (21.73), simplify, and find Gauss-Codazzi: 4-curvature in terms of intrinsic 3-geometry and extrinsic curvature &l(ej, ek)ei = (Kik1j - K Hlk )---!!.-) (n\u00b7 n + [(n\u00b7n)-l(KijKkm - KikKt) + (3)Rmijk]e m\u00b7 (21.74) The coefficients give directly the desired components of the curvature tensor <4)Rm ijk - and (3)Rm + (n \u2022 n)-l(K K m k ijk ij - K Km) ik j (21.75) (21.76) \u00a721.5. INTRINSIC AND EXTRINSIC CURVATURE\u00b7 515 Equations (21.75) and (21.76) are known as the equations of Gauss and Codazzi [for literature, see Eisenhart (1926)]. It follows from (21.75) that the components of the curvature of the 3-geometry will normally only then agree with the corre sponding components of the curvature of the 4-geometry when the imbedding happens to be accomplished at the point under study with a hypersurface free of extrinsic curvature. The directly opposite situation is illustrated by the familiar example of a 2-sphere imbedded in a flat 3-space, where the lefthand side of (", "21.75) (with dimensions lowered by one unit throughout!) is zero, and the extrinsic and intrinsic curvature on the right exactly cancel. Important components of the Einstein curvature let themselves be evaluated from the Gauss-Codazzi results. In doing the calculation, it is simplest to think of e i' e; tetrad, n being itself already normalized and and e k as being an orthonormal orthogonal to every vector in the hypersurface. Then, employing (14.7) and (21.75), one finds -Gg = (4)R12 = (3)R12 l2 + (4)R2323 + (4)R3l 23 + (3)R3l l2 + (3)R23 3l 3l Einstein curvature in terms of extrinsic curvature + (n' n)-l[(KiK~ - K~KD + (K~K1- K~~) + (KjK~ - K~ K~)] (21.77) = ; R - ; (n' n)-l[(Tr K)2 - Tr (K2)]. Here R is the 3-dimensional scalar curvature invariant and Tr stands for \"trace of\"; thus, Tr K = gi;Kj \u2022 = g.. Ki; = Ki,'I'and Tr K2 = (K2)i. = KmK ; = g. Ksmg.KH,m,S m I ', (21.78) (21.79) The result, though obtained in an orthonormal tetrad, plainly is covariant with respect to general coordinate transformations within the spacelike hypersurface; and it makes no explicit reference whatever to any time coordinate, in this respect providing a coordinate-free description of the Einstein curvature. The Einstein field equation equates (21.77) to 87TP, where p is the density of mass-energy. Expression (21.77) is the \"measure of curvature that is independent of how curved one cuts a spacelike slice.\" This measure of curvature is central to the derivation of Einstein's field equation that is sketched in Box 17.2, item 3, \"Physics on a Spacelike Slice.\" The other component of the Einstein curvature tensor that is easily evaluated by (14.7) from the results at hand has the form G~ = (4)Rn2 l2 + WR n3l3 = -(n' n)-1(Ki I2 - K~ll + K~13 - K~ll)' (21.80) when referred to an orthonormal frame. One immediately translates to a form valid for any frame e l, e 2, e 3 in the hypersurface, orthonormal or not, Equation (21. 77) is the central Einstein equation. \"mass-energy fixes curvature\" (21.81) The other initial-value equation 516 21. VARIATIONAL PRINCIPLE AND INITIAL\u00b7VALUE DATA The Einstein field equation equates this quantity to 817 times the i-th covariant component of the density of momentum carried by matter and fields other than gravity. The four components of the Einstein field equation so far written down will have a central place in what follows as \"initial-value equations\" of general relativity. The other six components will not be written out: (l) the dynamics lets itself be analyzed more simply by Hamiltonian methods; and (2) the calculation takes work. It demands that one evaluate the remaining type of object,!Jl(ej, n)e j \u2022 One step towards that calculation will be found in exercise 21.7. Sachs does the calculation (1964, equation 10) but only after specializing to Gaussian normal coordinates. These coordinates presuppose a very special slicing of spacetime: (1) geodesics issuing normally from the spacelike hypersurface n = 0 cut all subsequent simultaneities n = constant normally; and (2) the n coordinate directly measures lapse of proper time, or proper length, whichever is appropriate,* along these geodesics. In coordinates so special it is not surprising that the answer looks simple: (4)R n. = (n. n)-l --'- + K Km ) im k'oKk. an (,nk ( GaUSSia~ nOrmal) coordmates (21.82) Additional terms come into (21.82) when one uses, instead of the Gaussian normal coordinate system, the coordinate system of Arnowitt, Deser, and Misner. The ADM coordinates are employed here because they allow one to analyze the dynamics as one wants to analyze the dynamics, with freedom to push the spacelike hypersurface a~ead in time at different rates in different places (\"many-fingered", " time\"). Fischer (1971) shows how to evaluate and understand the geometric content ofsuch formulas in a coordinate-free way by using the concept of Lie derivative of a tensor field, an introduction to which is provided by exercise 21.8. * Here Sachs' equation (10) is generalized to the case where the unit normal n is not necessarily timelike. Sachs used n == a/at. EXERCISES Exercise 21.4. SCALAR CURVATURE INVARIANT IN TERMS OF AREA DEFICIT It being 10,000 km from North Pole to equator, one would have 62,832 km for the length of the \"equator\" if the earth were flat, as contrasted to the actual -40,000 km, a difference reflecting the fact that the surface is curved up into closure. Tum from this \"pre-problem\" to the actual problem, a 3-sphere Measure off from X = 0 a 2-sphere of proper radius E = ax. Determine the proper area of this 2-sphere as a function of X. Verify that relation (21.50) on the area deficit gives in the limit E---+-O the correct result R = 6/a2\u2022 For a more ambitious exercise: (1) take a general (smooth) 3-geometry; (2) express the metric near any chosen point in terms of Riemann's normal coordinates as given in \u00a711.6; (3) determine the locus of the set of points at the proper distance E to the lowest interesting power of E in terms of the spherical polar angles 8 and <I> (direction of start of geodesic of length E); (4) determine to the lowest interesting power of E the proper area of the figure defined by these points; and thereby establish (21.50) [for more on this topic see, for example, Cartan (1946), pp. 252-256]. \u00a721.5. INTRINSIC AND EXTRINSIC CURVATURE 517 Exercise 21.5. EXTRINSIC CURVATURE TENSOR FOR SLICE OF FRIEDMANN GEOMETRY Confirm the result (21.70) for the extrinsic curvature by direct calculation from formula (21.67). Exercise 21.6. EVALUATION OF!1l(e j\u2022 ek)n Evaluate this quantity along the model of (21.74) or otherwise. How can it be foreseen that the coefficient of n in the result must vanish identically? Comparing coefficients of em' find (4}Rmnjk and test for equivalence to equation (21.76). Exercise 21.7. EVALUATION OF THE COMMUTATOR [e j \u2022 n] The evaluation of this commutator is a first step toward the calculation of a quantity like!1l(ej' n)e;. Expressing ej as the differential operator %x j, use (21.49) to represent n also as a differential operator. In this way, show that the commutator in question has the value -(N./N)n -: (Nm~Nl~m'_ - - - Exercise 21.8. LIE DERIVATIVE OF A TENSOR (exercise provided byJ. W. York. Jr.) Define the Lie derivative of a tensor field and explore some of its properties. The Lie derivative along a vector field n is a differential operator that operates on tensor fields T of type G), converting them into tensors.1,nT, also of type G). The Lie differentiation process obeys the usual chain rule and has additivity properties [compare equations (lO.2b, lO.2c, lO.2d) for the covariant derivative]. For scalar functionsf, one has.1,nf= nUl =!p.np.. The Lie derivative of a vector field u along a vector field v was defined in exercise 9.11 by.1,uv= [u, v]. If the action of.1,n on I-forms is defined, the extension to tensors of general type will be simple, because the latter can always be decomposed into a sum of tensor products of vectors and I-forms. If\" is a I-form and v is a vector, then one defines.1,n\" to be that I-form satisfying (.1,n\", v) = n[(\", v)] - (\", [n, v]) for arbitrary v. (a) Show that in a coordinate basis (b) Show that in a coordinate basis where T is of type (~). (c) Show that in (a) and (b), all partial derivatives can be replaced by covariant derivatives. [Observe that Lie differentiation is defined independently of the existence of an", " affine connection. For more information, see, for example, Bishop and Goldberg (1968) and Schouten (1954)]. Exercise 21.9. EXPRESSION FOR DYNAMIC COMPONENTS OF THE CURVATURE TENSOR (exercise provided by J. W. York. Jr.) The Gauss-Codazzi equations can be viewed as giving 14 of the 20 algebraically independent c,om-p.onents of the spacetime curvature tensor in terms of the intrinsic and extrinsic geometry 'of three-dimensional (non-null) hypersurfaces. In order to accomplish a space-plus-time splitting of the Hilbert Lagrangian v=gwR, one must express, in addition. the remaining 518 21. VARIATIONAL PRINCIPLE AND INITIAL\u00b7VALUE DATA 6 components of the curvature tensor in an analogous manner. It is convenient for this purpose to express all tensors as spacetime tensors, and to use Lie derivation in the direction of the timelike unit normal field of the spacelike hypersurfaces as a generalized notion of time differentiation. A number of preliminary results must be proven: (a) (b).1,ugp.v = up.;v + uv;p.'.1,u(gp.v + up.u,.) =.1,u(Yp.v) =IIp.;v + uv;p. + up'a v + ap'u v' where Yp.v is the metric of the spacelike hypersurface, expressed in the spacetime coordinate basis, and aP.=u AvAup. is the curvature vector (4-acceleration) of the timelike normal curves whose tangent field is up.. (Recall that up.ap. = 0.) (c) Prove that the extrinsic curvature tensor is given by (d) The unit tensor of projection into the hypersurface is defined by In terms of.L show that one can write where and (e) From the fact that up. is the unit normal field for a family of spacelike hypersurfaces, show that w\",p = O. (f) The needed tools are now on hand. To obtain the result: (i) Write down.1,uKp.\" (see exercise 21.8); (ii) Insert this expression into the Ricci identity in the form (iii) Project the two remaining free indices into the hypersurface using.L, and show that one obtains -.L ~.L j\\ (4)Rp.vp\"u vu\" =.1,uK\",p + K\",yK~ + (3)V'(\",am+ a\",ap, where (3)V'\",ap =.L ~.L!3 V'p.a v can be shown to be the three-dimensional covariant derivative of ap. In Gaussian normal coordinates, show that one obtains from this result RojO! = ;t K j! + K jk K7\u00b7 (g) Finally, in the construction of (4)R, one needs to show that \u00a721.6. HILBERT ACTION PRINCIPLE AND ADM MODIFICATION THEREOF 519 Exercise 21.10. EXPRESSION OF (4)Rlnin IN TERMS OF EXTRINSIC CURVATURE, PLUS A COVARIANT DIVERGENCE (exercise provided by K. Kuchar) Let ex' be an arbitrary smooth set of four coordinates, not necessarily coordinated in any way with the choice of the I-parameter family of hypersurfaces. (a) Show that (4)RI 71171 - a:\"Y (3'( n g ) na:';(3'Y - na:':Y(3'. (b) Show that the covariant divergences (3' y'. (n n ;f3\");Y and (3' y' - n n ;Y');(3' ( can be removed from this expression in such a way that what is left behind contains only first derivatives of the unit normal vector n. (c) Noting that the basis vectors el and n form a complete set, justify the formula g(3'p.' = el'wiP.' + (n' n)-ln(3'nP.', where wi is the I-form dual to e i, (d) Noting that na:';f3\"n a:' = 0 and show that (4)Ri nin = (TrK)2 - TrK2 plus a covariant divergence. \u00a721.6. THE HilBERT ACTION PRINCIPLE AND THE ARNOWITT-DESER-MISNER MODIFICATION THEREOF IN", " THE SPACE-PLUS-TIME SPLIT For analyzing the dynamics, it happily proves unnecessary to possess the missing formula for (4) Rnink. It is essential, however, to have the Lagrangian density, 16'1T\u00a3 geom = (_(4)g)l/2(4)R, (21.83) in the Hilbert action principle as the heart of all the dynamic analysis. In the present ADM (1962) notation, this density has the form Kuchar (1971 b; see also exercise 21.10) shows how to calculate a sufficient part of this quantity without calculating all of it. The difference between the \"sufficient part\" and the \"whole\" is a time derivative plus a divergence, a quantity of the form Drop a complete derivative from the Hilbert action principle to get the ADM principle (21.85) 520 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA When one multiplies (21.83) by dt dx l dx 2 dx3 and integrates to obtain the action integral, the term (21.85) integrates out to a surface term. Variations of the geometry interior to this surface make no difference in the value of this surface term. Therefore it has no influence on the equations of motion to drop the term (21.85). The result of the calculation (exercise 21.10) is simple: what is left over after dropping the divergence merely changes the sign of the terms in Tr K2 and (Tr K)2 in (21.84). Thus the variation principle becomes (extremum) = Imodified =f t'modified d4x = (1/16'17)f [R + (n' n)((Tr K)2 - Tr K2)] Ng l/2 dt d 3x + f t'fields d4x. (21.86) This expression, rephrased, is the starting point for Arnowitt, Oeser, and Misner's analysis of the dynamics of geometry. Two supplements from a paper of York (1972b; see also exercise 21.9) enlarge one's geometric insight into what is going on in the foregoing analysis. First, the tensor of extrinsic curvature lets itself be defined [see also Fischer (1971)] most naturally in the form (21.87) where 9 is the metric tensor of the 3-geometry, n is the timelike unit normal field, and 4, is the Lie derivative as defined in exercise 21.8. Second, the divergence (21.85), which has to be added to the Lagrangian of (21.86) to obtain the full Hilbert Lagrangian, is -2[( -(4)g)l/2(n a'Tr K + aa')l,a., where the coordinates are general (see exercise 21.10), and a a' - n a' - ;f3' nf3' (21.88) (21.89) is the 4-acceleration of an observer traveling along the timelike normal n to the successive slices. \u00a721.7. THE ARNOWITT, OESER, AND MISNER FORMULATION OF THE DYNAMICS OF GEOMETRY Dirac (1959, 1964, and earlier references cited therein) formulated the dynamics of geometry in a (3 + I)-dimensional form, using generalizations of Poisson brackets and of Hamilton equations. Arnowitt, Oeser, and Misner instead made the Hilbert Palatini variational principle the foundation for this dynamics. Because of its sim plicity, this ADM (1962) approach is followed here. The gravitational part of the integrand in the Hilbert-Palatini action principle is rewritten in the condensed but standard form (after inserting a 16'17 that ADM avoid by other units) as I 67Tt'geom true = t'goomADM = - gij 07T ii/ot - NX - ~X i - 2 [ 7Tij~. - ~ N i Tr n + N1i(g)I/2],. (21.90) \" \u00a721.7. ADM FORMULATION OF DYNAMICS OF GEOMETRY 521 Here each item of abbreviation has its special meaning and will play its special part, a part foreshadowed by the name now given it: 7Tii true - \"geometrodynamic field momentum\" dyn 8(action) = amically conjugate to the \"geometrodynamic field coordinate\" gii 8gij ij 7T = 167T; 7T'J = g1l2(g'JTr K - K'J)...,.. Momenta conjugate to the dynamic gij (here the 7Tii of ADM is usually more convenient than 7T", "gue); and.'7Ctrue =.'7C(7Tgue' gij) = (\"super-Hamiltonian\") =.10167T;.'7C(7Tij, gij) = g-1I2 (Tr n 2 - ~ (Tr n)2) _ g1l2R; and (21.91) (21.92) Here the covariant derivative is formed treating 7Tik as a tensor density, as its to be (see \u00a721.2). The quantities to be varied to definition in (21.91) shows it extremize the action are the coefficients in the metric of the 4-geometry, as follows: the six gii and the lapse function N and shift function Ni ; and also the six \"geome trodynamic momenta,\" 7Tii. To vary these momenta as well as the metric is (1) to follow the pattern of elementary Hamiltonian dynamics (Box 21.1), where, by taking the momentum p to be as independently variable as the coordinate x, one arrives at two Hamilton equations of the first order instead of one Lagrange equation of the second order, and (2) to follow in some measure the lead of the Palatini variation principle of \u00a721.2. There, however, one had 40 connection coefficients to vary, whereas here one has come down to only six 7TH. To know these momenta and the 3-metric is to know the extrinsic curvature. Before carrying out the variation, drop 1.i from (21.90), since it gives rise only to surface integrals the divergence -2[ and therefore in no way affects the equations of motion that will come out of the variational principle. Also rewrite the first term in (21.90) in the form (21.94) and drop the complete time-derivative from the variation principle, again because it is irrelevant to the resulting equations of motion. The action principle now takes the form extremum = I true = I ADM/167T = (1/167T)f [7T ij ogi/ot - N.1C(rr ii, gij) - Nj:J('i(r. ii, gi)] d 4x + f Efield d 4x. (21.95) The action principle itself, here as always, tells one what must be fixed to make the action take on a well-defined value (if and when the action possesses an extre mum). Apart from appropriate potentials having to do with fields other than geom- y Action principle says. fix 3-geometry on each face of sandwich What a 3-geometry is Electromagnetism gives example of momentum conjugate to \"field coordinate\" 522 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA etry, the only quantities that have to be fixed appear at first sight to be the values of the six gij on the initial and final spacelike hypersurfaces. However, the ADM action principle is invariant with respect to any change of coordinates Xl, x 2, x 3 ~ xl, x 2, x3 within the successive spacelike slices. Therefore the quantities that really have to be fixed on the two faces of the sandwich are the 3-geometries ('3)~' (on the initial hypersurface) and (3)~ (on the final hypersurface) and nothing more. In mathematical terms, a 3-geometry (3)~ is the \"equivalence class\" of a set of differentiable manifolds that are isometrically equivalent to each other under diffeo morphisms. In the terms of the everyday physicist, a 3-geometry is the equivalence class of 3-metrics gi/x,y, z) that are equivalent to one another under coordinate transformations. In more homely terms, two automobile fenders have one and the same 2-geometry if they have the same shape, regardless of how much the coordinate rulings painted on the one may differ from the coordinate rulings painted on the other. To have in equation (21.95) an example of a field Lagrangian that is at the same time physically relevant and free of avoidable complications, take the case of a source-free electromagnetic field. It would be possible to take the field Lagrangian to have the standard Maxwell value, with (21.96) (21.97) The variation of the Lagrangian with respect to the independent dynamic variables of the field, the four potentials A a, would then immediately give the four second-order partial differential wave equations for these four potentials. However, to have instead a larger number of first-order equations is as convenient for electrodynamics as it is for geometrodynamics. One seeks for the analog of the Hamiltonian equations of particle dynamics", ", dx/dt = oH(x,p)/op, dp/dt = -\u00b7oH(x,p)/ox. (21.98) One gets those equations by replacing the Lagrange integral fLex, i) dt by the Hamilton integral f[pi - H(x,p)] dt. Likewise, here one replaces the action inte grand of (21.96) by what in flat spacetime would be (1/47T) [A FP.> + 1 F FP.>] 4 p.> p.,> \u2022 (21.99) In actuality, spacetime is to be regarded as not only curved but also sliced up into spacelike hypersurfaces. This (3 + I) split of the geometry made it desirable to split the ten geometrodynamic potentials into the six gij and the four lapse and shift functions. Here one similarly splits the four Ap into the three components Ai of the vector potential and the scalar potential Ao = -ep (with the sign so chosen that, in flat spacetime in a Minkowski coordinate system, ep = AO). In this notation, the \u00a721.7. ADM FORMULATION OF DYNAMICS OF GEOMETRY 523 Lagrange density function, including the standard density factor (_Wg )1/2 but dropping a complete time integral (0 /ot)(A i Gi) that has no influence on the equations of motion, is given by the formula 47Tt'field = _Gi oA;lot + epGi,i - ; Ng-1I2gii (GiGi + fi3if'Bi) + Ni[ijk]Gif'Bk. (2l.l00) Lagrange density for electromagnetism Here use is made of the alternating symbol [ijk], defined as changing sign on the interchange of any two labels, and normalized so that [123] = 1. Note that the 3-tensor eijk and the alternating symbol [ijk] are related much as are the correspond ing four-dimensional objects in equation (8.10), so that one can write (21.101) The quantities f'Bi are the components of the magnetic field in the spacelike slice. They are not regarded as independently variable. They are treated as fully fixed by the choice of the three potentials Ai' The converse is the case for the components Gi of the electric field: they are treated like momenta, and as independently variable. Extremizing the action with respect to the G i (exercise 2l.l1) gives the analog of the equation dx/dt = p/m in particle mechanics, and the analog of the equation of flat-spacetime electrodynamics; namely, (2l.l02) (2l.l03) The initial-value equation of electromagnetism Here the last term containing the shift functions Ni, arises from the obliquity of the coordinate system. ADM give the following additional but equivalent ways to state the result (21.103): G i = ; [ijk] * 0k = ; [ijk] g[jkJLPlC - (4)g) 112(4)glUl(4)gP,8Fa,8 }. (21.104 ) They note that Gi and!'i3 i are not directly the contravariant components of the fields in the simultaneity I, but the contravariant densities, (21.105) (21.106) Extremizing the action with respect to the three Ai (exercise 21.12) gives the curved-spacetime analog of the Ma~well equations, oE/ot = V X B. (21.107) 524 21. VARIATIONAL PRINCIPLE AND INITIAL\u00b7VALUE DATA Divergence relation by extremization with respect to \u00a2 The remaining potential, 9, enters the action principle at only one point. Extre mizing with respect to it gives immediately the divergence relation of source-free electromagnetism. Action principle tells what to fix at limits At limits, fix not potentials but magnetic field itself [,\"i. = O..J (21.108) If an action principle tells in and by itself what quantities are to be fixed at the limits, what lessons does (21.100) give on this score? One can go back to the example of particle mechanics in Hamiltonian form. as in Box 21.1, and note that there the momentum p could \"flap in the breeze.\" Only the coordinate x had to be fixed at the limits. Thus the variation of the action was 81 = 8f [pi - H(x,p)] dt = f {[i - oH/op]8p + (d/dt)(P 8x) + [-p - oH/ox]8x} dt.", " (21.109) To arrive at a well-defined extremum of the action integral I, it was not enough to annul the coefficients, in square brackets, of 8p and 8x; that is, to impose Hamil ton's equations of motion. It was necessary in addition to annul the quantities at limits, p 8x; that is, to specify x at the start and at the end of the motion. Similarly here. The quantities cp and [,\"i flap in the breeze, but the magnetic field has to be specified on the two faces of the sandwich to allow one to speak of a well-defined extremum of the action principle. Why the magnetic field, or the three quantities (21.110) why not the three Ai themselves? When one varies (21.100) with respect to the Ai' and integrates the variation of the first term by parts, as one must to arrive at the dynamic equations, one obtains a term at limits L [,\"i 8Ai d 3x - f [,\"i 8Ai d 3x. (21.111 ) \"\"\"initIal I final One demands that both these terms at limits must vanish in order to have a well defined variational problem. Go from the given vector potential to another vector potential, A inew'by the gauge transformation (21.112) The magnetic-field components given by the three Ainew differ in no way from those listed in (21.110). Moreover the \"variation at limits,\" (21.113) is automatically zero by virtue of the divergence condition (21.108), for any arbitrary choice of A. Therefore the quantities fixed at limits are not the three Ai themselves (mere potentials) but the physically significant quantities (21.110), the components of the magnetic field. Moreover, the divergence condition [,\"i i = 0 now becomes the initial-value equation for the determination of the potential\u00b7cp. \u00a721.7. ADM FORMULATION OF DYNAMICS OF GEOMETRY 525 In the preceding paragraph one need only replace \"the three At by \"the six gil' and \"the components of the magnetic field\" by \"the 3-geometry (3)~\" and \"the potential <p\" by \"the lapse and shift functions Nand Ni\" to pass from electro dynamics to geometrodynamics. With this parallelism in view, turn back to the variational principle (21.95) of general relativity in the ADM formulation. With the 3-geometry fixed on the two faces of the sandwich, vary conditions in between to extremize the action, varying in turn the 7Tii, the gij' and the lapse and shift functions. The geometrodynamic momenta appear everywhere only algebraically in the action principle, except in the term -2A'i7Tii /j\" Variation and integration by parts gives 2A'ili 87Tii. Collecting coefficients of 87Tii and annuling the sum of these coefficients, one arrives at one of the several conditions required for an extremum, ogi/ot = 2Ng-1I2 ( 7Tii - ; g;jTrn) + NiJj + Nili. (2l.l14) This result agrees with what one gets from equations (21.91) defining geometrody namic momentum in terms of extrinsic curvature, together with expression (21.67) for extrinsic curvature in terms of lapse and shift. The result (2l.l14) here is no less useful than the result dx/dt = oH(x,p)/op =p/m ADM principle reproduces formula for geometrodynamic momentum in the most elementary problem in mechanics: it marks the first step in splitting a second-order equation or equations into twice as many first-order equations. Now vary the action with respect to the gii and again, after appropriate integration by parts and rearrangement, find the remaining first-order dynamic equations of general relativity [simplified by use of equations (2l.l16) and (2l.l17)], Dynamic and initial-value equations out of ADM formalism O'l1 ii/ot = - Ng1l2 ( Rii _ ; giiR ) + ; Ng-1I 2gii (Tr n 2 - ; (Tr '11)2) - 2Ng-1I2 ('I1 im7Tmi + g1l2(Nlii - gijN1m 1m) + ('I1;i ~)Im 'I1 iiTr n) ; - Ni lm7Tmi - N 1m'l1 mi + other than geometry, omitted here for ~source terms arising from fields simplicity, but discussed by ADM (1962) (2l.l15) Jii. Finally extremize the action", " (21.95) with respect to the lapse function N and the shift functions Ni, and find the four so-called initial-value equations of general relativity, equivalent to (21.77) and (21.81) or to G~ = 8'11T~; thus, -(1/16\\7)T('I1ii, gi) = (1/8'11)Ng-1/2gij(GiGi +!'Bi~1i), -(1/16'11).'J(i(7Tii, g;i) = -(1/4'11)[ijk]Gi!fl k \u2022 (2l.l16) (2l.l17) 526 21. VARIATIONAL PRINCIPLE AND INITIAL\u00b7VALUE DATA EXERCISES Exercise 21.11. FIRST EXPLOITATION OF THE ADM VARIATIONAL PRINCIPLE FOR THE ELECTROMAGNETIC FIELD Extremize the action principle (11.100) with respect to the \u00a7i and derive the result (21.103). Exercise 21.12. SECOND EXPLOITATION OF THE ADM VARIATIONAL PRINCIPLE FOR THE ELECTROMAGNETIC FIELD Extremize (21.100) with respect to the Ai' and verify that the resulting equations in any Minkowski-flat region are equivalent to (21.107). Exercise 21,13. FARADAY-MAXWELL SOURCE TERM IN THE DYNAMIC EQUATIONS OF GENERAL RELATIVITY Evaluate the final indicated source terms in (21,115) from the Lagrangian (21, I00) of Maxwell electrodynamics, regarded as a function of the Ai and the gij' Exercise 21.14. THE CHOICE OF <P DOESN'T MAnER Prove the statement in the text that the dynamic development of the electric and magnetic fields themselves is independent of the choice made for the scalar potential \u00a2(t, x,y, z) in the analysis (a) in flat spacetime in Minkowski coordinates and (b) in general relativity, according to equations (21.103), and (21.107) as generalized in exercise 21.12, Exercise 21.15, THE CHOICE OF SLICING OF SPACETIME DOESN'T MAnER Given a metric (3)g;;(X, y, z) and an extrinsic curvature Ki;(X,y, z) on a spacelike hypersurface I, and given that these quantities satisfy the initial-value equations (21.116) and (21.117), and given two alternative choices for the lapse and shift functions (N, N j ) and (N + fiN, N; + fiNi)' show that the curvature itself (as distinguished from its components in these two distinct coordinate systems), as calculated at a point 'J' a \"little way\" (first order of small quantities) off the hypersurface, by way of the dynamic equations (21.114) and (21.115), is independent of this choice of lapse and shift. \u00a721.8. INTEGRATING FORWARD IN TIME In the Hamiltonian formalism of Arnowitt, Deser, and Misner [see also the many papers by many workers on the quantization of general relativity-primarily putting Einstein's theory into Hamiltonian form-cited, for example, in references 1 and 2 of Wheeler (1968)], the dynamics of geometry takes a form quite similar to the Hamiltonian dynamics of geometry. There one gives x and p at a starting time and integrates two first-order equations for dx/dt and dp/dt ahead in time to find these dynamically conjugate variables at all future times. Here one gives appropriate values of gij and '17'ii over an initial spacelike hypersurface and integrates the two first-order equations (21.114) and (21.115) ahead in time to find the geometry at future times. For example, one can rewrite the differential equations as difference equations according to the practice by now familiar in modern hydrodynamics, and then carry out the integration on an electronic.digital computer of substantial memory capacity. \u00a721.B. INTEGRATING FORWARD IN TIME 527 Time in general relativity has a many-fingered quality very different from the one-parameter nature of time in nonrelativistic particle mechanics [see, however, Dirac, Fock, and Podolsky (1932) for a many-time formalism for treating the relati vistic dynamics of a system of many interacting particles]. He who is studying the geometry is free to push ahead the spacelike hypersurface faster at one place than another, so long as he keeps it spacelike. This freedom expresses itself in the lapse function", " N(t, x,y, z) at each stage, t, of the integration. Equations (2UI4) and (2UI5) are not a conduit to feed out information on N to the analyst. They are a conduit for the analyst to feed in information on N. The choice of Nis to be made, not by nature, but by man. The dynamic equations cannot begin to fulfill their purpose until this choice is made. The \"time parameter\" t is only a label to distinguish one spacelike hypersurface from another in a one-parameter family of hypersurface; but N thus tells the spacing in proper time, as it varies from place to place, between the successive slices on which one chooses to record the time-evolution of the geometry. A cinema camera can record what happens only one frame at a time, but the operator can make a great difference in what that camera sees by his choice of angle for the filming of the scene. So here, with the choice of slicing. Another choice is of concern to the analyst, especially one doing his analysis on a digital computer. He is in the course of determining, via (2UI4-21.115) written as difference equations, what happens on a lattice work of points, typified by x =...,73,74, 75, 76, 77,..., etc. He finds that the curvatures are developing most strongly in a localized region in the range around x = 83 to x = 89. He wants to increase the density of coverage of his tracer points in this region. He does so by causing points at lesser and greater x values to drift into this region moment by moment as t increases: t =..., 122, 123, 124,.... He makes the tracer points at lesser x-values start to move to the right (Nl positive) and points at greater x-values move to the left (Nl negative). In other words, the choice of the three shift functions Ni(t, x,y, z) is just as much the responsibility of the analyst as is the choice of the lapse function N. The equations will never tell him what to pick. He has to tell the equations. These options, far from complicating dynamic equations (21.1 14-21.1 15), make them flexible and responsive to the wishes of the analyst in following the course of whatever geometrodynamic process is in his hands for study. The freedom that exists in general relativity in the choice of the four functions N, Ni, is illuminated from another side by comparing it with the freedom one has in electrodynamics to pick the one function <p(t, x,y, z), the scalar potential. In no way do the dynamic Maxwell equations (21.103) and (21.107), as generalized in exercise 21.12 determine <p. Instead they demand that it be determined (by the analyst) as the price for predicting the time-development of the vector potential Ai' An altered choice of <p(t, x,y, z) in its dependence on position and time means altered results from the dynamic equations for the development of the three Ai in time and space. However, the physically significant quantities, the electric and magnetic fields themselves on successive hypersurfaces, come out the same (exercise 21.14) regardless of this choice of <p. Similarly in geometrodynamics: an altered choice for the four Lapse and shift chosen to_ push forward the integration in time as one finds most convenient Same 4-geometry regardless of lapse and shift options 528 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA F t)r-------rE ~ c,---D=4) Figure 21.4. Some of the many ways to make distinct spacelike slices through one and the same (4)~, the complete Schwarzschild 4-geometry. functions N, ~, means (a) an altered laying down of coordinates in spacetime, and therefore (b) altered results for the intrinsic metric (3)gij and extrinsic curvature Kif of successive spacelike hypersurfaces, but yields the same 4-geometry (4)~ (Figure 21.4) regardless of this choice of coordinatization (exercise 2l.l5). \u00a721.9. THE INITIAL-VALUE PROBLEM IN THE THIN-SANDWICH FORMULATION Given appropriate initial-value data, one can integrate the dynamic equations ahead in time and determine the evolution of the geometry; but what are \"appropriate initial-value data\"? They are six functions (3)gi/X,y, z) plus six more functions ij (X,y, z) or Kij (x,y, z) that together satisfy the four initial-value equations (2l.l16) 7T and (2l.l", "17). To be required to give coordinates and momenta accords with the familiar plan of Hamiltonian mechanics; but to have consistency conditions or \"constraints\" imposed on such data is less familiar. A particle moving in two-dimen sional space is catalogued by coordinates x, y, and coordinates Pz' PI!; but a particle forced to remain on the circle x 2 + y2 = a2 satisfies the constraint xpz + YPI! = O. Thus the existence of a \"constraint\" is a signal that the system possesses fewer degrees Initial-value data: what is freely disposable? and what is thereby fixed? \u00a721.9. INITIAL-VALUE PROBLEM IN THE THIN-SANDWICH FORMULATION 529 of freedom than one would otherwise suppose. Fully to analyze the four \"initial value\" or \"constraint\" conditions (2l.l16) and (21.1l7) is thus to determine (I) how many dynamic degrees offreedom the geometry possesses and (2) what these degrees of freedom are; that is to say, precisely what \"handles\" one can freely adjust to govern completely the geometry and its evolution with time. The counting one can do today, with the conclusion that the geometry possesses the same count of true degrees of freedom as the electromagnetic field. The identification of the \"handles,\" or freely adjustable features of the dynamics, is less advanced for geometry than it is for electromagnetism (Box 212), but most instructive so far as it goes. By rights the identification of the degrees of freedom of the field, whether that of Einstein or that of Faraday and Maxwell, requires nothing more than knowing what must be fixed on initial and final spacelike hypersurfaces to make the appro priate variation principle well-defined. One then has the option whether (I) to give that quantity on both hypersurfaces or (2) to give that quantity and its dynamic conjugate on one hypersurface or (3) to give the quantity on both hypersurfaces, as in (I), but go to the limit of an infinitely thin sandwich, so that one ends up specifying the quantity and its time rate of change on one hypersurface. This third \"thin sandwich\" procedure is simplest for a quick analysis of the initial-value problem in both electrodynamics and geometrodynamics. Take electrodynamics first, as an illustration. Give the divergence-free magnetic field and its time-rate ofchange: on an arbitrary smooth spacelike hypersurface in curved spacetime in the general case; on the hypersurface t = \u00b0in Minkowski spacetime in the present illustrative treatment, In electromagnetism, give magnetic field and its rate of change as initial data eBi(O, x,y, z) given, q]3i(O, x,y, z) = at also given.. (OeB ) i (2l.l18) (2l.l19) i These quantities together contain four and only four independent data per space - _G i so that one can start point. How is one now to obtain the momenta '1T integrating the dynamic equations (2l.l03) and (2U07) forward in time? (1) Find a set of three functions Ai(O, x,y, z) such that their curl gives the three specified eB i. That this can be done at all is guaranteed by the vanishing of the divergence --~',i' H-owever, the choice of the Ai is not unique. The new set of potentials A inew = Ai + OAjox i with arbitrary smooth A, provide just as good a solution as the original Ai' No matter. Pick one solution and stick to it. (2) Similarly, find a set of three their curl gives the specified ~i(O,x,y,z), and resolve all Ai(O,x,y,z) such that arbitrariness of choice by fiat. (3) Recall that the electric field (negative of the field momentum) is given by (formula valid without amendment only in flat space). The initial-value or constraint equation Gi,i =\u00b0translates to the form (2I.l20) V\"2-f, = -nifA. _ \" l, }'y (2U21) ~---------..... Box 21.2 COUNTING THE DEGREES OF FREEDOM OF THE ELECTROMAGNETIC FIELD In curved spacetime, Fourier analysis is a less convenieTlt way of identifying the degrees of free dom of the electromagnetic field [for such a Fou rier analysis, see Misner and Wheeler (1957), es pecially their Table X and following text] than direct analysis in space, as above. C. Another Alternative: Analy", "ze\" Deformation of Structure\" Still a third way to get a handle on the degrees of freedom of a divergence-free field, whether 8 or (8, rests on the idea of deformation of structure the [diagram from Wheeler (1964)]. Represent A. First Approach: Number of \"Field Coordinates\" per Spacepoint 7T Superficial tally of the degrees of freedom of the source-free electromagnetic field gives three field coordinates Ai(x,y, z) per spacepoint on the initial simultaneity I, plus three field momenta 7Thue = i /417 [with 7T i = - $i(x,y, z)] per spacepoint. Closer inspection reveals that the number of coordinate degrees of freedom per spacepoint is not three but two. Thus the change in vector po tential Ai -+ Ai + OA/OX i makes no change in the actual physics, the magnetic field components, Bi = ~ [ijk](oAk/oxi - oA/oxk). Moreover, though those components are three in number, they satisfy one condition per spacepoint,?Bi,i = 0, thus reducing the effective net number of coordinate degrees of freedom per spacepoint to two. The momentum degrees of freedom per space poin.t are likewise reduced from three to two by the ~ne condition per spacepoint $i,i = 0. a. b. c. B. Alternative Approach: Count Fourier Coefficients In textbooks on field theory [see, for example, Wentzel (1949)], attention focuses on flat space time. The electromagnetic field is decomposed by Fourier analysis into individual running waves. Instead of counting degrees of freedom per point in coordinate space, one does the equivalent: counts up degrees of freedom per point in wave number space. Thus for each (kz'k ll, k z ), there are two independent states of polarization. Each state of polarization requires for its description an am plitude (\"coordinate\") and time-rate of change of amplitude (\"momentum\") at the initial time, t~. Thus the number of degrees of freedom per point in wave-number space is two for coordinates and two for momenta, in accord with what one gets by carrying out the count in coordinate space. i magnetic field by Faraday's picture of lines of force (a) continuing through space without ever ending, automatic guarantee that?Bi is every where zero. Insert \"knitting needles\" (b) into the spaghetti-like structure of the lines of for~ and move these needles as one will. Sliding the \"knit ting needles\" along a line of force causes no move ment of the line of force. (c) With the help of two knitting needles perpendicular to each other and to the line of force, one can give any given line of force any small displacement one pleases per pendicular to its length: again two degrees of free dom per spacepoint. Granted any non-zero field to begin with, no matter how small, one can build it up by a sequence of such small deformations to agree with any arbitrary field pattern of zero divergence, no matter what its complexity and strength may be. \u00a721.9. INITIAL-VALUE PROBLEM IN THE THIN-SANDWICH FORMULATION 531 i Solve for cp. Then (4) equation (2l.120) gives the initial-value electric field, or -- - $i, required (along with the field coordinate electrodynamic field momentum 7T Ai) for starting the integration of the dynamic equations (21.103) and (2l.l07). [Misner and Wheeler (1957) deal with the additional features that come in when the space is multiply connected. Each wormhole or handle of the geometry is able to trap electric lines of force. The flux trapped in anyone wormhole defines the classical electric charge qw associated with that wormhole. One has to specify all these charges once and for all in addition to the data (2l.l18) and (21.119) in order to determine fully the dynamic evolution of the electromagnetic field. There is no geometrodynamic analog to electric charge, according to Unruh (1971).] (5) In this integration, the scalar potential cp at each subsequent time step is not to be calculated; it is to be chosen. Only when one has made this free choice definite do the dynamic i or $i at these successive equations come out with definite results for the Ai and the 7T steps. In the thin-sandwich formulation of the initial-value problem of electrodynamics, to summarize, one gives ei3 i and ~i (equivalent to (B on two nearby hypersurfaces). One chooses the Ai and Ai with much arbitrariness to represent these initial-value data. The arbitrariness having been seized on to give the initial Ai and Ai' there", " is no arbitrariness left in the initial cp. However, at all subsequent times the situation is just the other way around. All the arbitrariness is sopped up in the choice of the cp, leaving no arbitrariness whatever in the three Ai (as given by the integration of the dynamic equation). The situation is quite similar in geometrodynamics. One gives the beginnings of a I-parameter family of spacelike hypersurfaces; namely, (3)~(O) given, (2l.122) (21.123) Then (I) one picks a definite set of coordinates Xi = (x,y, z) and in terms of those coordinates finds the unique metric coefficients gij(X, y, z) that describe that 3-geom etry. The existence of a solution is guaranteed by the circumstance that (3)~ is a Riemannian geometry. However, one could have started with different coordinates and ended up with different metric coefficients for the description of the same 3-geometry. No matter. Pick one set of coordinates, take the resulting metric co efficients, and stick to them as giving half the required initial-value data. (2) Simi larly, to describe the 3-geometry (3)~ + (3l-b dt at the value of the parameter t + dt, make use of coordinates xi + xi dt and arrive at the metric coefficients gif + gij dt. The arbitrariness in the Xi having thus been resolved by fiat, and the (3)~ being given as definite initial physical data, the glf are thereby completely fixed. (3) Recall that the components of the extrinsic curvature Kif or the momenta 7TH are given in terms of the gij and gij and the lapse and shift functions Nand N. by (21.67) or by (21.67) plus (21.91) or by (21.114). The four initial-value or \"constraint\" equations (21.116) and (2UI7) thus become four conditions for finding the four Scalar potential: fixed at start; freely disposable later In ADM treatment, give 3-geometry and its time-rate of change y Lapse and shift initially determinate; thereafter freely disposable Counting initial-value data 532 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA quantities N, Nl. One can shorten the writing of these conditions by introducing the abbreviations and Y = 2 \"shift anomaly\" ) ( = (Tr y)2 _ Tr y 2 (both for functions of x, y, z on the initial simultaneity). Then one has for the one initial-value equation; and for the other three, k Yi [ - 8i N k Tr y ] = - 87T7t\u00b7 Ik (2 I.l24) (21.125) (21.126) (21.127) In summary, one chooses the gij and gij with much arbitrariness (because of the arbitrariness in the coordinates, not by reason of any arbitrariness in the physics) to represent the given initial-value data, (3)~ and (3).b. The arbitrariness at the initial time all having been soaked up in this way, one expects no arbitrariness to be left in the initial Nand Ni as obtained by solving (21.126) and (2I.l27). However, on all later spacelike slices, the award of the arbitrariness is reversed. The lapse and shift functions are freely disposable, but, with them once chosen, there is no arbi trariness whatever in the six gij (and the six Kij or 7T ii)-as given by the integration of the dynamic equations (2I.l14) and (21.115). The analogy with electrodynamics is clear. There the one \"gauge-controlled\" function <p was fixed at the start by the elliptic equation (21.121), but was thereafter free. Here the four lapse and shift functions are fixed at the start by the four equations (21.126) and (2I.l27), but are thereafter free. Exercise 21.16 applies the initial-value equation (21.126) to analyze the whole evolution in time of any Friedmann universe in which one knows the equation p = p(p) connecting pressure with density. Exercise 21.17 looks for a variation principle on the spacelike hypersurface :\u00a3 equivalent in content to the elliptic initial-value equation (21.121) for the scalar potential <p. Exercises 21.18 and 21.19 look for similar variation principles to determine the lapse and shift functions. How many degrees of freedom, or how many \"handles,\" are there in the specifica tion of the", " 4-geometry that one will obtain? The metric coefficients of the initial 3-geometry provided six numbers per space point. However, they were arbitrary to the extent of a coordinate transformation, specified by three functions of position, x = x(x',y', z'), y =y(x',y', z'), z = z(x',y', z'). \u00a721.9. INITIAL-VALUE PROBLEM IN THE THIN-SANDWICH FORMULATION 533 The net number of quantities per space point with any physical information was therefore 6 - 3 = 3. One can visualize these three functions as the three diagonal components of the metric in a coordinate system in which gij has been transformed to diagonal form. Ordinarily it is not useful to go further and actually spell out the analysis in any such narrowly circumscribed coordinate system. Now think of the (3)~ in question as imbedded in the (4)~ that comes out of the integrations. Moreover, think of that (4)~ as endowed with the lumps, bumps, wiggles, and waves that distinguish it from other generic 4-geometries and that make Min kowski geometry and special cosmologies so unrepresentative. The (3)~ is a slice in that (41~. It partakes of the lumps, bumps, wiggles, and waves present in all those regions of the (4)~ that it intersects. To the extent that the (4)~ is generic, it does not allow the (3)~ to be moved to another location without becoming a different (3)~. If one tries to push the (3)~ \"forward in time\" a little in a certain locality, leaving it unchanged in location elsewhere, one necessarily changes the (3)~. By this circum stance, one sees that the (3)~ \"carries information about time\" [Sharp (1960); Baierl ein, Sharp, and Wheeler (1962)]. Moreover, this \"forward motion in time\" demands for its description one number per space point. It is possible to think of this number in concrete terms by imagining an arbitrary coordinate system t, x, y, z laid down in the (4)~. Then the hypersurface can be conceived as defined by the value t = l(x, y, Z) at which it cuts the typical line x, y, z. A forward movement carries it to l(x,y, Z) + 8l(x,y, Z), and changes shape and metric coefficients on (3)~ accordingly. It is usually better not to tie one's thinking down to such a concrete model, but rather to recognize as a general point of principle (1) that the location of the (3)~ in spacetime demands for its specification one datum per spacepoint, and (2) that this datum is already willy-nilly present in the three data per spacepoint that mark any (3)~. In conclusion, there are only two data per spacepoint in a (3)~ that really tell anything about the (4)~ in which it is imbedded, or to be imbedded (as distinguished from where the (3)~ slices through that (4)~). Similarly for the other (3)~ that defines the other \"face of the sandwich,\" whether thick or thin. Thus one concludes that the specification of(3)~ and (3).i:, actually gives four net pieces of dynamic information per spacepoint about the (4)~ (all the rest of the information being \"many-fingered time,\" telling where the 3-geometries are located in that (4)~). According to this line of reasoning, geometrodynamics has the same number of dynamic degrees of free dom as electrodynamics. One arrives at the same conclusion in quite another way through the weak-field analysis (\u00a735.3) of gravitational waves on a flat spacetime background: the same ranges of possible wave numbers as for Maxwell waves; and for each wave number two states of polarization; and for each polarization one amplitude and one phase (the equivalent of one coordinate and one momentum). In electrodynamics in a prescribed spacetime manifold, one has a clean separation between the one time-datum per spacepoint (when one deals with electromagnetism in the context of many-fingered time) and the two dynamic variables per spacepoint; but not so in the superspace formulation of geometrodynamics. There the two kinds of quantities are inextricably mixed together in the one concept of 3-geometry. Four pieces of geometrodynamic information per space point on initial sim ultaneity 534 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA Turn from initial- and final-value data to the action integral that is", " determined by (1) these data and (2) the principle that the action be an extremum, I = Iextremum = S. The action depends on the variables on the final hypersurface, according to the formula S = S(I,B) in electrodynamics, but according to the formula S = S(3)~) (21.128) (21.129) in geometrodynamics. In each case, there are three numbers per spacepoint in the argument of the functional (one in I; two in a divergence-free magnetic field; three in (3)~). This mixing of the one many-fingered time and the two dynamic variables in a 3-geometry makes it harder in general relativity than in Maxwell theory to know and give!'B and J3 when one has in hand appropriate initial value data. Give I on I: that was enough for electrodynamics. For geometrodynamics, to give the six gi;CX,y, z) and the six gij(X,y, z) is not necessarily enough. For example, let the time parameter t be a fake, so that dt, instead of leading forward from a given hypersur to a new hypersurface:\u00a3 + dI, merely recoordinatizes the present hypersur face I face: (21.130) A first inspection may make one think that one has adequate data in the six glj and the six glj = ~i1j + ~jli' (21.131) but in the end one sees that one has not both faces of the thin sandwich, as required, but only one. Thus one must reject, as improperly posed data in the generic problem of dynamics, any set of six glj that let themselves be expressed in the form (21.131) [Belasco and Ohanian (1969)]. Similar difficulties occur when the two faces of the thin sandwich, instead of coinciding everywhere, coincide in a limited region, be it three-dimensional, two dimensional, or even one-dimensional (\"crossover of one face from being earlier than the other to being later\"). Thus it is enough to have (21.131) obtaining even on only a curved line in I to reject the six gij as inappropriate initial-value data. That one can impose conditions on the gij and gij which will guarantee existence and uniqueness of the solution N(x, y, z), Nl(x, y, z) of the initial-value equations (21.126) and (21.127) is known as the \"thin-sandwich conjecture,\" a topic on which there has been much work by many investigators, but so far no decisive theorem. Problem in assuring completeness and consistency of initial data The \"thin sandwich conjecture\" \u00a721.10. TIME-SYMMETRIC AND -ANTISYMMETRIC INITIAL-VALUE PROBLEMS 535 To presuppose existence and uniqueness is to make the first step in giving mathe matical content to Mach's principle that the distribution of mass-energy throughout space determines inertia (\u00a721.12). \u00a721.10. THE TIME-SYMMETRIC AND TIME-ANTISYMMETRIC INITIAL-VALUE PROBLEMS Turn from the general initial-value problem to two special initial-value problems that lend themselves to detailed treatment, one known as the time-symmetric ini tial-value problem, the other as the time-antisymmetric problem. A 4-geometry is said to be time-symmetric when there exists a spacelike hypersur face :\u00a3 at all points of which the extrinsic curvature vanishes. In this case the three initial value equations (21.127) are automatically satisfied, and the fourth reduces to a simple requirement on the three-dimensional scalar curvature invariant, R = I 67Tp. (21.132) Still further simplifications result when one limits attention to empty space. Simplest of all is the case of spherical symmetry in which (21.132) yields at once the full Schwarzschild geometry at the moment of time symmetry (two asymptotically flat spaces connected by a throat), as developed in exercise 21.20. Consider a 3-geometry with metric Call it a \"base metric.\" Consider another 3-geometry with metric (21.133) (21.134) Angles are identical in the two geometries. On this account they are said to be conformally equivalent. The scalar curvature invariants of the two 3-geometries are related by the formula [Eisenhart (1926)] where (21.135) (21.136) Demand that the scalar curvature invariant R 2 vanish, and arrive [Brill (1959)] at the \"wave equation\" (21.137) for the", " conformal correction factor \"'. Brill takes the base metric to have the form suggested by Bondi, (21.138) 536 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA and takes the conformal correction factor I/; also to possess axial symmetry. In the application: Q1(P, z) measures the \"distribution of gravitational wave amplitude,\" assumed for A I/;(p, z) simplicity to vanish outside r = (p2 + Z2)1/2 = a; measures the \"amplitude of the distribution of gravitational wave ampli tude\"; is the conformal correction factor, which varies with position at large distances as I + (m/2r). The quantity m(cm) is uniquely determined by the condition that the geometry be asymtotically flat. It measures the mass-energy of the distribution of gravitational radiation. Wave amplitude determines mass-energy: m = m(A) \"Time-antisymmetric\" initial-value data The mass m of the gravitational radiation is proportional to A2 for small values of the amplitude A. It is inversely proportional to the reduced wavelength A' = (effective wavelength/2'17) that measures the scale of rapid variations in the gravitational wave amplitude Q1(P, z) in the \"active zone.\" Thus the metric is dominated by wiggles, proportional in amplitude to A, in the active zone, and at larger distances dominated by something close to a Schwarzschild (I + 2m/r) factor in the metric. When the amplitude A is increased, a critical value is attained, A = Acrit' at which m goes to infinity and the geometry curves up into closure (\"universe closed by its own c.ontent of gravitational-wave energy\"). Further analysis and examples will be found in Wheeler (l964a), pp. 399-451, also in Wheeler (l964c). Brill has carried out a similar analysis [Brill (1961)] for the vacuum case of what he calls time-antisymmetric initial-value conditions, sketched below as amended by York (1973). (I) The initial slice is maximal, Tr K = 0._(2) This slice is conformally flat, (21.139) (3) Work in the \"base space\" with metric 8ii and afterwards transform to the geometry (21.139). Three of the initial-value equations become K lj base,! - - 0. (21.140) To solve these equations, (I) take any localized trace-free symmetric tensor Bkm ; (2) solve the flat-space Laplace equation V2A = (3/2) o2Bkm /OX k ox m for A; (3) define the six potentials Akm = Bkm +!A 8km ; and (4) calculate (21.141) that automatically satisfy (21.140) and give Tr Kbase = O. Then Kij = 1/;-10K~ase also automatically satisfies these conditions, but now in the curved geometry (21.139). The final initial-value equation becomes a quasilinear elliptic equation, in the flat base space, for the conformal factor 1/;, 8V~ase I/; + r 7 2: (Kbase lj)2 = O. i,j (21.142) The asymptotic form of I/; reveals that the mass of the wave is positive. Finite dimensional dynamics for geometries endowed with high symmetry \u00a721.10. TIME-SYMMETRIC AND -ANTISYMMETRIC INITIAL-VALUE PROBLEMS 537 In addition to the time-symmetric and time-antisymmetric cases, there are at least two further cases where the initial-value problem possess special simplicity. One is the case ofa geometry endowed with a symmetry, as, for example, for the Friedmann universe of Chapter 27 or the mixmaster universe of Chapter 30 or cylindrical gravitational waves in the treatment of Kuchar (1971 a). One starts with a spacelike slice on which the glj and 'lflj have a special symmetry, and makes all future spacelike slices in a way that preserves this symmetry. The geometry on anyone of these simultaneities, though almost entirely governed by these symmetry considerations, still typically demands some countable number of parameters for its complete deter mination, such as the radius of the Friedmann universe, or the three principal radii of curvature of the mixmaster universe. These parameters and the momenta conju gate to them define a miniphase space. In this miniphase space, the dynamics runs its course as for any other problem of classical dynamics [see, for example, Box 30.1 and Misner (1969) for the mixmaster universe; Kuchar (1971 a) and (", "1972) for waves endowed with cylindrical symmetry; and Gowdy (1973) for waves with spherical symmetry]. Even the evidence for the existence of many-fingered time, most charac teristic feature of general relativity, is suppressed as the price for never having to give attention to any spacelike slice that departs from the prescribed symmetry. Exercise 21.16. POOR MAN'S WAY TO DO COSMOLOGY Consider a spacetime with the metric EXERCISES corresponding to a 3-geometry-with the form of a sphere of radius a(t) changing with time. the tensor of extrinsic curvature as expressed in a local Euclidean frame of Show that reference is K = -a- 1(da/dt) 1, where 1 is the unit tensor. Show that the initial value equation (21.77) reduces to (6/a 2)(da/dt)2 + (6/a 2 ) = 16'17p(a) [for the value of the second term on the left, see exercise 14.3 and Boxes 14.2 and 14.5J, and explain why it is appropriate to write the term on the right as 6ao/a3 for a \"dust-filled model universe.\" More generally, given any equation of state, p = p(p), explain how one can find p = p(a) from and how one can thus forecast the history of expansion and recontraction, a = a(t). Exercise 21.17. THIN-SANDWICH VARIATIONAL PRINCIPLE FOR THE SCALAR POTENTIAL IN ELECTRODYNAMICS (a) Choose the unknown u m in the expression a'\" 1 _ mn _Y_ _c_y + U m _'/'_ ax n 8'17 g a'\" ax m ax n 'A ~ / =;/ 538 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA in such a way that this expression, multiplied by the volume element g1l2 d 3x, and integrated over the simultaneity X, is extremized by a </>, and only by a </>. that satisfies the initial-value equation (21.108) of electrodynamics. (b) Show that the resulting variational principle, instead of having to be invented \"out of the blue,\" is none other than what foll.ows directly from the action principle build on the Lagrangian density (21.100) of electrodynamics (independent variation of \u00a2 and the three Ai everywhere between the two faces of a sandwich to extremize I, subject only to the prior specification of the Ai on the two faces of the sandwich, in the limit where the thickness of the sandwich goes to zero). Exercise 21.18. THIN-SANDWICH VARIATIONAL PRINCIPLE FOR THE LAPSE AND SHIFT FUNCTIONS IN GEOMETRODYNAMICS (a) Extremize the action integral 13 =f {[R - (TrK)2 + TrK 2 - 2T~nJN - 2T:kNk}g1l2 d 3x with respect to the lapse and shift functions, and show that one arrives in this way at the four initial-value equations of geometrodynamics. It is understood that one has given the six gij and the six ogi;l0t on the simultaneity where the analysis is being done. The extrinsic curvature is considered to be expressed as in (21.67) in terms of these quantities and the lapse and shift. The energy density and energy flow are referred to a unit normal vector n and three arbitrary coordinate basis vectors ei within the simultaneity, as earlier in this chapter, and the asterisk is an abbreviation for an omitted factor of 8'17. (b) Derive this variational principle from the ADM variational principle by going to the limit of an infinitesimally thin sandwich [see derivation in Wheeler (1964)J. Exercise 21.19. CONDENSED THIN-SANDWICH VARIATIONAL PRINCIPLE (a) Extremize the action 13 of the preceding exercise with respect to the lapse function N. (b) What is the relation between the result and the principle that \"3-geometry is a carrier of information about time\"? (c) By elimination of N, arrive at a \"condensed thin-sandwich variational principle\" in which the only quantities to be varied are the three shift functions Ni \u2022 Exercise 21.20. POOR MAN'S WAY TO SCHWARZSCHILD GEOMETRY On curved empty space evolving deterministically in time, impose the conditions (I) that it possess a moment of time-symmetry, a spacelike hypersurface, the extrinsic curvature of which, with", " respect to the enveloping spacetime, is everywhere zero, and (2) that this spacelike hypersurface be endowed with spherical symmetry. Write the metric of the 3-geometry in the form From the initial-value equation (21.127), show that the conformal factor 1ft up to a multiplica tive factor must have the form 1ft = (I + m/2r). Show that the proper circumference 2'17,,\u00a52(r) assumes a minimum value at a certain value ofr, thus defining the throat of the 3-geometry. Show that the 3-geometry is mirror-symmetric with respect to reflection in this throat in the sense that the metric is unchanged in form under the substitution r' = m 2/47. Find the transformation from the conformal coordinate r to the Schwarzschild coordinate r. \u00a721.11. YORK'S '\"HANDLES\" TO SPECIFY A 4-GEOMETRY 539 \u00a721.11. YORK'S \"HANDLES\" TO SPECIFY A 4-GEOMETRY On a simultaneity-or on the simultaneity-of extremal proper volume, give the conformal part of the 3-geometry and give the two inequivalent components of the dynamically conjugate momentum in order (l) to have freely specifiable, but also complete, initial-value data and thus (2) to determine completely the whole generic four-dimensional spacetime manifold. This in brief is York's extension (1971, I 972b) to the generic case of what Brill did for special cases (see the preceding section). York and Brill acknowledge earlier considerations of Lichnerowicz (1944) and Bruhat (1962 and earlier papers cited there on conformal geometry and the initial-value problem). But why conformal geometry, and why pick such a special spacelike hypersurface on which to give the four dynamic data per spacepoint? Few solutions of Maxwell's equations are simpler than an infinite plane mon ochromatic wave in Minkowski's flat spacetime, and few look more complex when examined on a spacelike slice cut through that spacetime in an arbitrary way, with local wiggles and waves, larger-scale lumps and bumps, and still larger-scale general curvatures. No one who wants to explore electrodynamics in its evolution with many-fingered time can avoid these complexities; and no one will accept these complexities of many-fingered time who wants to see the degrees of freedom of the electromagnetic field in and by themselves exhibited in their neatest form. He will pick the simplest kind of timelike slice he can find. On that simultaneity, there are two and only two field coordinates, and two and only two field momenta per spacepoint. Similarly in geometrodynamics. When one wants to untangle the degrees of freedom of the geometry, as distinct from analyzing the dynamics of the geometry, one therefore retreats from the three items of information perspacepoint that are contained in a 3-geometry [or in any other way of analyzing the geometrodynamics, as especially seen in the \"extrinsic time\" formulation of Kuchar (1971 band 1972)] and following York (1) picks the simultaneity to have maximal proper volume and (2) on this simultaneity specifies the two \"coordinate degrees of freedom per spacepoint\" that are contained in the conformal part of the 3-geometry. An element of proper volume gl/2 d 3x on the spacelike hypersurface I undergoes, in the next unit interval of proper time as measured normal to the hypersurface, a fractional increase of proper volume [see Figure 21.3 and equations 21.59 and 21.66] given by (21.143 ) The degrees of freedom of the geometry in brief Pick hypersurface of extremal proper volume For the volume to be extremal this quantity must vanish at every point of I. This condition is satisfied in a Friedmann universe (Chapter 27) and in a Taub universe (Chapter 30) at that value of the natural time-coordinate t at which the universe the same switches over from expansion to recontraction. It condition on the choice of simultaneity, I, lets itself be formulated in the same natural way, is remarkable that Tr K = \u00b0or Tr n = 0, (21.144) Case of open 3-geometry 540 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA for a closed universe altogether deprived of any symmetry whatsoever. Alternatively, one can deal with a spacetime that is topologically the product of an open 3-space by the real line (time). Then it is natural to think of specifying the location in it of a bounding spacelike 2-geometry S with the topology of a 2-sphere. Then one has many ways", " to fill in the interior of S with a spacelike 3-geometry I; but of all these I's, only the one that is extremal, or only the ones that are extremal, satisfy (21.144). Who is going to specify this 2-geometry with the topology of a 2-sphere? The choice of that 2-geometry is not a matter of indifference. In a given 4-geometry, distinct choices for the bounding 2-geometry will ordinarily give distinct results for the extremizing 3-geometry, and therefore different choices for the \"initial-value simultaneity,\" I. No consideration immediately thrusts itself forward that would give preference to one choice of2-geometry over another. However, no such infinity of options presents itself when one limits attention to a closed 3-geometry. Therefore it will give concreteness to the following analysis to consider it applied to a closed universe, even though the analysis surely lets itself be made well-defined in an open region by appropriate specification of boundary values on the closed 2-geometry that bounds that open region. In brief, by limiting attention to a closed 3-geometry, one lets the obvious condition of closure take the place of boundary conditions that are not obvious. York's analysis remains simple when his extrinsic time T = 19-1/2 Tr n =\u00b1 Tr K 3 3 Meaning of conformal 3-geometry has any constant value on the hypersurface, not only the value T = 0 appropriate for the hypersurface of extremal proper volume. On the simultaneity I specified by the condition of constant extrinsic time, T = constant, begin by giving the conformal 3-geometry, < = (3)< = he equivalence class of all those positive definite Riemannian three-dimensional metrics that are equivalent to each other under (l) diffeomorphism (smooth sliding of the points over the mainfold to new locations) or (2) changes of scale that vary smoothly from point to point, leaving fixed all local angles (ratios of local distances), but changing local distances themselves or (3) both. (21.145) The conformal 3-geometry is a geometric object that lends itself to definition and interpretation quite apart from the specific choice of coordinate system and even without need to use any coordinates at all. The conformal 3-geometry (on the hypersuiface I where T = constant) may be regarded much as one regards the magnetic field in electromagnetism. The case of conformally flat 3-geometry, (21.146) \u00a721.11. YORK'S \"HANDLES\" TO SPECIFY A 4-GEOMETRY 541 (with gi;base = 0i;)' is analogous to those initial-value situations in electromagnetism where the magnetic field is everywhere zero (the time-antisymmetric initial-value problem of Brill); but now we consider the case of general dS~ase' The six metric coefficients gi; of the conformal 3-geometry, subject to being changed by change of the three coordinates Xi, and undetermined at anyone point up to a common position-dependent multiplicative factor, carry 6 - 3 - 1 = 2 pieces of information per spacepoint. In this respect, they are like the components of the divergenceless magnetic field!lJ. The corresponding field momentum 'lT~M 0: Gi (Box 21.1, page 496) has its divergence specified by the charge density, and so also carries two pieces of information (in addition to the prescribed information about the density of charge) per spacepoint. (21.147) The comparison is a little faulty between the components of!lJ and the metric coefficients. They are more like potentials than like components of the physically relevant field. The appropriate measure of the \"field\" in geometrodynamics is the curvature tensor; but how can one possibly define a curvature tensor for a geometry that is as rudimentary as a conformal 3-geometry? York (1971) has raised and answered this question. The Weyl conformal-curvature tensor [equation (13.50) and exercise 13.13] is independent [in the proper @ representation], in spaces of higher dimen sionality, of the position-dependent factor If;4 with which one multiplies the metric coefficients, but vanishes identically in three-dimensional space (exercise 21.21). One arrives at a non-zero coflfoI'!fially invariant measure of the curvature only when one goes to one higher derivative (exercise 21.22). In this way, one comes to York's curvature lJab, here called yab, a tensor density with these properties: yab = yba (symmetric); y~ = 0 (traceless", "); yalJ 1b = 0 (transverse); yab invariant with respect to position-dependent changes in the conformal scale factor; yab = 0 when and only when the 3-geometry is conformally flat. (21.148) York's curvature tensor yab provides what York calls the pure spin-two representation of the 3-geometry intrinsic to I. It is the analog of the field q, of electrodynamics on the spacelike initial-value simultaneity. It directly carries physical information about the conformal 3-geometry. In addition to the conformal geometry (3) <, specified by the \"potentials\" gi;!g1l3, the _~nd l1!easured -by the ~'fidd components\" yi;, one must also specify on I corresponding conjugate momenta: y The associated momenta 542 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA ;;rab = ;rrab (symmetric); ;;r~ = 0 (traceless); ;;rab 1b = 0 (transverse) in case there is no flow of energy in space; otherwise 7T ab 1b = 8r. (density of flow of energy)a; two pieces of information (in addition to the prescribed information about the flow of energy) per spacepoint. (21.149) It might appear to be essential to specify with respect to which of the 3-geometries, distinguished from one another by different values of the conformal factor one calculates the covariant derivatives of tensor densities of weight 513 (see \u00a721.2) in (21.148) and (21.149). However, York has shown that the conditions (21.149) do not in any way depend on the value of the conformal factor If;4. These equations (21.149) for what York calls the \"momentum density of weight 5/3,\" (21.150) are linear, and therefore lend themselves to analysis by standard methods. It is a great help in this enterprise that York (l973a,b) has provided a \"conformally invari ant orthogonal decomposition of symmetric tensors on Riemannian manifolds\" that allows one to generate solutions of these req uirements (\"transverse traceless,\" \"con formal Killing,\" and \"trace\" parts, respectively, measure deformation of conformal part of geometry, mere recoordinatization, and change of scale). It is a further assistance, as York notes, that one has the same 7i- ab for an entire conformal equivalence class of metrics; that is, for a given -\" (21.151) Unique solution for conformal factor no matter how different the gab and If; themselves may be. The conformal 3-geometry and the \"momentum density of weight 5/3\" once the remaining initial-value equation (21.116) then becomes the \"scale\" picked, equation, (21.152) for the determination of the conformal factor If;. Here V 2 stands for the Laplacian V 2lf; =g-li2(O lox a)gl/2gab(olf;loxb). (21.153) It, like (3)R, M, and Q, refers to the base space. It is interesting that V2 _..!J3)R 8 is a conformally invariant wave operator, whereas V2 itself is not. The quantity M in York's analysis is an abbreviation for and (21.154ii) (21.154b) \u00a721.12. MACH'S PRINCIPLE AND THE ORIGIN OF INERTIA 543 One seeks a solution \\f; that is continuous over the closed manifold and everywhere real and positive. When does such a solution \\f; of the elliptic equation (21.152) exist? When is it unique? Always (when M> 0 and or \u00a5- 0), is the result of O'Murchadha and York (1973); see also earlier investigations of Choquet-Bruhat (1972). Some of the physical considerations that come into this kind of problem have been discussed by Wheeler (1964a, pp. 370-381). \u00a721.12. MACH'S PRINCIPLE AND THE ORIGIN OF INERTIA In my opinion the general theory of relativity can only solve this problem [of inertia] satisfactorily if it regards the world as spatially self-enclosed. ALBERT EINSTEIN (1934), p. 52. On June 25, 1913, two years before he had discovered the geometrodynamic law that bears his name, Einstein (1913b) wrote to Ernst Mach (Figure 21.5) to express his appreciation for the inspiration that he had derived", " for his endeavors from Mach's ideas. In his great book, The Science of Mechanics, Mach [(1912), Chapter 2, section 6] had reasoned that it could not make sense to speak of the acceleration of a masS relative to absolute space. Anyone trying to clear physics of mystical ideas would do better, he reasoned, to speak of acceleration relative to the distant stars. But how can a star at a distance of 109 light-years contribute to inertia in the here and the now? To make a long story short, one can say at once that Einstein's theory (1) identifies gravitation as the mechanism by which matter there influences inertia here; (2) says that this coupling takes place on a spacelike hypersurface [in what No violation of causality, one, without a closer examination, might mistakenly think to be a violation of the principle ofcausality; see Fermi (1932) for a discussion and clarification of the similar apparent paradox in electrodynamics; see also Einstein (1934), p. 84: \"Moreover I believed that I could show on general considerations a law of gravitation invariant in relation to any transformation of coordinates whatever was inconsistent with the principle of causation. These were errors of thought which cost me two years of excessively hard work, until I finally recognized them as such at the end of 1915\"]; (3) supplies in the initial-value equations of geometrodynamics a mathematical tool to describe this coupling; (4) demands closure of the geometry in space [one conjectures; see Wheeler (1959, 1964c) and Hanl (1962)], as a boundary condition on the initial-value equations if they are to yield a well-determined [and, we know now, a unique]4-geometry; and (5) identifies the collection of local Lorentz frames near any point in this resulting spacetime as what one meanS quantitatively by speaking of inertia at that point. This is how one ends up with inertia here determined by density and flow of mass-energy there. despite appearances There are many scores of papers in the literature on Mach's principle, including. An enormous literature many-even one by Lenin (English translation, 1927)-one could call anti-Machian; and many of them make interesting points [see especially the delightful dialog by Weyl (1924a) on \"inertia and the cosmos,\" and the article (1957) and book (1961) of Sciama]. However, most of them were written before one had anything like the understanding of the initial-value problem that one possesses today. Therefore no (continued on page 546) 544 Figure 21.5. Einstein's appreciation of Mach, written to Ernst Mach June 25, 1913, while Einstein was working hard at arriving at the final November 1915 formulation of standard general relativity. Regarding confirmation at a forthcoming eclipse: \"If so. then your happy investigations on the foundations of mechanics, Planck's unjustified criticism notwithstanding, will receive brilliant confirmation. For it necessarily turns out that inertia originates in a kind of interaction between bodies, quite in the sense of your considerations on Newton's pail experiment. The first consequence is on p. 6 of my paper. The following additional points emerge: (I) If one accelerates a hea\\'y shell of matter S, then a mass enclosed by that shell experiences an accelerative force. (2) If one rotates the shell relative to the fixed stars about an axis going through its center, a Coriolis force arises in the interior of the shell; that is, the plane of a Foucault pendulum is dragged around (with a practically unmeasurably small angular velocity).\" Following the death of Mach, Einstein (l916a) wrote a tribute to the man and his work. Reprinted with the kind permission of the estate of Albert Einstein, Helen Dukas and Otto Nathan, executors. y / Mach's principle updated and spelled out 546 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA attempt will be made to summarize or analyze the literature, which would demand a book in itself. Moreover, Mach's principle as presented here is more sharply formulated than Einstein ever put it in the literature [except for his considerations arguing that the universe must be closed; see Einstein's book (1950), pp. 107-108]; and Mach would surely have disowned it, for he could never bring himself to accept general (or even special) relativity. Nevertheless, it is a fact that Mach's principle that matter there governs inertia here-and Riemann's idea-that the geometry of space responds to physics and participates in physics-were the two great currents of thought which Einstein, by means of his powerful equivalence principle, brought together into the present-day geometric description of gravitation and motion. to be sure, \"Specify everywhere the distribution and flow of mass-energy and thereby deter times\". mine the inertial", " properties of every test particle everywhere and at all Spelled out, this prescription demands (1) a way of speaking about \"everywhere\": a spacelike hypersurface I. Let one insist-in conformity with Einstein-(2) that it be a closed 3-geometry, and for convenience, not out of necessity, (3) that or be independent of position on I. (4) Specify this 3-geometry to the extent of giving the conformal metric; without the specification of at least this much 3-geometry, there would be no evident way to say \"where\" the mass-energy is to be located. (5) Give density Pbase as a function of position in this conformal 3-geometry. (6) Recognize that giving the mass-energy only of fields other than gravity is an inadequate way to specify the distribution of mass-energy throughout space. For malistically, the gravitational fields does not and cannot make any contribution to the source term that stands on the righthand side of Einstein's field equation. However, the analysis of gravitational waves (Chapters 18 and 35) shows that perturbations in the geometry ofscale small compared t6 the scale ofobservation have to be regarded as carrying an effective content of mass-energy. Moreover, one has in a geon [Wheeler (1955); Brill and Hartle (1964); for more on gravitational-wave energy, see \u00a735.l4] an object built out of gravitational waves (or electromagnetic waves, or neutrinos, or any combination of the three) that holds itself together for a time that is long in' comparison to the characteristic period of vibration of the waves. It looks from a distance like any other mass, even though nowhere in its interior can one put a finger and say \"here is mass.\" Therefore it, like any other mass, must have \"its influence on inertia.\" But to specify this mass, one must give enough information to characterize completely the gravitational waves on the simul taneity I. For this, it is not enough merely to have given the two \"wave-coordinates\" per spacepoint that one possesses in (3) <. One must give in addition (7) the two \"wave-momenta\" per spacepoint that appear in York's \"momentum density ofweight 5/3,\" 7i'ab; and at the same time, as an inextricable part of this operation, one must (8) specify the density of flow of field energy. (9) Solve for the conformal factor 1/;. (10) Then one has complete initial-value data that satisfy the initial-value equa tions of general relativity. (11) These data now known, the remaining, dynamic, components of the field equation determine the 4-geometry into the past and the future. (12) In this way, the inertial properties of every test particle are determined everywhere and at all times, giving concrete realization to Mach's principle. Much must still be done to spell out the physics behind these equations and to \u00a721.12. MACH'S PRINCIPLE AND THE ORIGIN OF INERTIA 547 see this physics in action. Some significant progress had already been made in this direction before the present stage in one's understanding of the initial-value equa tions. Especially interesting are results of Thirring (1918) and (1921) and of Thirring and Lense (1918), discussed by Einstein (1950) in the third edition of his book, The Meaning of Relativity. Consider a bit of solid ground near the geographic pole, and a support erected there, and from it hanging a pendulum. Though the sky is cloudy, the observer watches the track of the Foucault pendulum as it slowly turns through. 360 0 \u2022 Then the sky clears and, miracle of miracles, the pendulum is found to be swinging all the time on an arc fixed relative to the far-away stars. If \"mass there governs inertia here,\" as envisaged by Mach, how can this be? Enlarge the question. By the democratic principle that equal masses are created equal, the mass of the earth must come into the bookkeeping of the Foucault pendulum. Its plane of rotation must be dragged around with a slight angular velocity, wdrag' relative to the so-called \"fixed stars.\" How much is wdrag? And how much would Wdrag be if the pendulum were surrounded by a rapidly spinning spherical shell of mass M and radius Rshell' turning at angular velocity Wshell? Einstein's theory says that inertia is a manifestation of the geometry of spacetime. It also says that geometry is affected by the presence of matter to an extent propor tional to the factor G/c 2 = 0.742 X 10-28 cm/g. Simple dimensional considerations leave no room except to say that the rate of drag", " is proportional to a expression of the form The Foucault pendulum The dragging of the inertial frame wdrag - - k!i.... mshell, cony W R c2 shell - k mshell W R shell - shell shell' (21.155) Here k is a numerical factor to be found only by detailed calculation. Lense and Thirring [(1918) and (1921)], starting with a flat background spacetime manifold, calculated in the weak-field approximation of Chapter 18 the effect of the moving current of mass on the metric. Expressed in polar coordinates, the metric acquires a non-zero coefficient g\u00a2t. Inserted into the equation of geodesic motion, this off diagonal metric coefficient gives rise to a precession. This precession (defined here about an axis parallel to the axis of rotation, not about the local vertical) is given by an expression of the form (21.155), where the precession factor k has the value k = 4/3. (21.156) There is a close parallelism between the magnetic component of the Maxwell field and the precession component of the Einstein field. In neither field does a source at rest produce the new kind of effect when acting on a test particle that is also at rest. One designs a circular current of charge to produce a magnetic field; and a test charge, in order to respond to this magnetic field, must also be in motion. Similarly here: no pendulum vibration means no pendulum precession. Moreover, the direction of the precession depends on where the pendulum is, relative to the rotating shell of mass. The precession factor k has the following values: y The \"sum for inertia\" 548 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA k = 4/3 for pendulum anywhere inside rotating shell of mass; k = 4/3 for pendulum at North or South pole; k = -2/3 for pendulum just outside the rotating shell at its equator. (21.157) This position-dependence of the drag, Wdrag, makes still more apparent the analogy with magnetism, where the field of a rotating charged sphere points North at the center of the sphere, and North at both poles, but South at the equator. Whether the Foucault pendulum is located in imagination at the center of the earth or in actuality at the North pole, the order of magnitude of the expected drag is m earth Wdrag - -R-- wearth - earth 0.44 cm I radian -6-X--1O-g-c-m- --13-7-0-0-se-c - 5 X 10-14 rad/sec, (21.158) too small to allow detection, let alone actual measurement, by any device so far built-but perhaps measurable by gyroscopes now under construction (\u00a740.7). By contrast, near a rapidly spinning neutron star or near a black hole endowed with substantial angular momentum, the calculated drag effect is not merely detectable; it is even important (see Chapter 33 on the physics of a rotating black hole). The distant stars must influence the natural plane of vibration of the Foucault pendulum as the nearby rotating shell of matter does, provided that the stars are not so far away (r radius of universe) that the curyature of space begins to introduce substantial corrections into the calculation of Thirring and Lense. In other words, no reason is apparent why all masses should not be treated on the same footing, so that (21.158) more appropriately, if also somewhat symbolically, reads m shell wplaneof - ~ w shell + L... - - - W\"star'\" vibration of Foucault pendulum far-away \"stars\" r\"star\" shell m\"star\" \"\" (21.159) Moreover, when there is no nearby shell of matter, or when it has negligible effects, the plane of vibration of the pendulum, if experience is any guide, cannot turn with respect to the frame defined by the far-away \"stars.\" In this event WFoucault must be identical with w stars ; or the \"sum for inertia,\" '\" L... - - - - m\"star\" far-away \"stars\" r\"star\" muniverse runiverse (21.160) must be of the order of unity. Just such a relation of approximate identity between the mass content of the universe and its radius at the phase of maximum expansion is a characteristic feature of the Friedman model and other simple models ofa closed universe (Chapters 27 and 30). In this respect, Einstein's theory of Mach's principle exhibits a satisfying degree of self-consistency. \u00a721.12. MACH'S PRINCIPLE AND THE ORIGIN OF INERTIA 549 At phases of the dynamics of the universe other than the stage of maximum expansion, runiverse can become arbitrarily", " small compared to muniver~e' Then the ratio (21.160) can depart by powers of ten from unity. Regardless of this circum stance, one has no option but to understand that the effective value of the \"sum for inertia\" is still unity after all corrections have been made for the dynamics of contraction or expansion, for retardation, etc. Only so can WFoucault retain its inescap able identity with wfar-away stars' Fortunately, one does not have to pursue the theology of the \"sum for inertia\" to the uttermost of these sophistications to have a proper account of inertia. Mach's idea that mass there determines inertia here has its complete msHhematical account in Einstein's geometrodynamic law, as already spelled ont. For-the first strong-field analysis of the dragging of the inertial reference system in the context of relativistic cosmology, see Brill and Cohen (1966) and Cohen and Brill (1967); see also \u00a733.4 for dragging by a rotating black hole. Still another clarification is required of what Mach's principle means and how it is used. The inertial properties of a test particle are perfectly well-determined when that particle is moving in ideal Minkowski space. \"Point out, please,\" the anti-Ma chian critic says, \"the masses that are responsible for this inertia.\" In answer, recall that Einstein's theory includes not only the geometrodynamic law, but also, in Einstein's view, the boundary condition that the universe be closed. Thus the section ofspacetime that is flat is to be viewed, not as infinite, but as part of a closed universe. (For a two-dimensional analog, fill a rubber balloon with water and set it on a glass tabletop and look at it from underneath). The part of the universe that is curved acquires its curvature by reason of its actual content of mass-energy or-if animated only by gravitational waves-by reason of its effective content of mass-energy. This mass-energy, real or effective, is to be viewed as responsible for the inertial properties of the test particle that at first sight looked all alone in the universe. It in no way changes the qualitative character of the result to turn attention to a model universe where the region of Minkowski flatness, and all the other linear dimensions of the universe, have been augmented tenfold (\"ten times larger balloon; ten times larger face\"). The curvature and density of the ~urved part of the model universe are down by a factor of 100, the volume is up by a factor of 1,000, the mass is up by a factor of 10; but the ratio of mass to radius, or the \"sum for inertia\" (the poor man's substitute for a complete initial-value calculation) is unchanged. Einstein acknowledged a debt of parentage for his theory to Mach's principle (Figure 21.5). It is therefore only justice that Mach's principle should in return today owe its elucidation to Einstein's theory. Minkowski geometry as limit of a closed 3-geometry \u00a3-xereise 21.21. WHY THE WEYL CONFORMAL CURVATURE EXERCISES TENSOR VANISHES How many independent components does the Riemann curvature tensor have in three dimensional space? How many does the Ricci curvature tensor have? Show that the two tensors are related by the formula y 550 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA Rdab<: = ogRac - o~Rab + gacRdb - gabR~ + tR(O~gab - oggac) with no need of any Weyl conformal-curvature tensor t~ specify (as in higher dimensions) the further details of the Riemann tensor. Show that the Weyl tensor, from an n-dimensional modification of equation (13.50) as in exercise 13.13, vanishes for n = 2. Exercise 21.22. YORK'S CURVATURE [York (1971)]. (a) Define the tensor [Eisenhart (1926)] (b) Show that a 3-geometry is conformally flat when and only when Rab<: = O. (c) Show that the following identities hold and reduce to five the number of independent components of Rab<:: ac = g\"bR bae = 0; Ra Rabe + Racb = 0; Rab<: + Reab + Rb<:a = O. (d) Show that Yorks' curvature yab = gl/3[aef] (Rl _1- olR) = - t gll3[aef]gbm R me' 4 Ie is conform ally invariant and has the properties listed in equations (21.148). Exercise 21.23. PULLING THE POYNTING", " FLUX VECTOR \"OUT OF THE AIR\" From the condition that the Hamilton-Jacobi functional S(gij, Am) (extremal of the action integral) for the combined Einstein and Maxwell fields, ostensibly dependent on the six metric coefficients gij(X,y, z) and the three potentials Am(x, y, z), shall actually depend only on the 3-geometry of the spacelike hypersurface and the distribution of magnetic field strength on this hypersurface, show that the geometrodynamic field momentum 'lT li = OS/ogij satisfies a condition of the form and evaluate the coefficient c in this equation [Wheeler (1968b)]. Hint: Note that the transfor mation in no way changes the 3-geometry itself, and therefore the corresponding induced change in S, oS =f[ oS ligij + ~OS OAm] d3x oglj uA m must vanish identically for arbitrary choice of the ~1(X, y, z), which measure the equivalent of the sliding of a ruled transparent rubber sheet over an automobile fender. \u00a721.13. JUNCTION CONDITIONS 551 Exercise 21.24. THE EXTREMAL ACTION ASSOCIATED WITH THE HILBERT ACTION PRINCIPLE DEPENDS ON CONFORMAL 3-GEOMETRY AND EXTRINSIC TIME [K. Kuchar (1972) and J. York (1972)] Show that the data demanded by the Hilbert action principle oJ!4)R( _(4)g)l/2 d4x = 0 on each of the two bounding spacelike hypersurfaces consist of (I) the conformal 3-geometry (3) < of the hypersurface plus (2) the extrinsic time variable defined by T = ~g-1/2 Tr n -! Tr K - 3'3 conveniently represented by the pictogram H, measured by one number per spacepoint, and independent of the conformal factor in the metric of the 3-geometry. This done, explain in a few words why in this formulation of geometrodynamics the Hamilton-Jacobi function (h times the phase of the wave function in the semiclassical or JWKB approximation) is appropriately expressed in the form S = S\u00ab3)<, H). \u00a721.13. JUNCTION CONDITIONS The intrinsic and extrinsic curvatures of a hypersurface, which played such funda mental roles in the initial-value formalism, are also powerful tools in the analysis of \"junction conditions.\" Recall the junction conditions of electrodynamics: across any surface (e.g., a capacitor plate), the tangential part of the electric field, ~I' and the normal part of the magnetic field, B1.' must be continuous; thus, Junction conditions for electrodynamics [~,l (discontinuity in ~I) (~I on \"+\" side of surface) surface) Elt - Eii = 0, [BJ..l B1- BJ: = 0; (~I on \"-\" side of (21.161 a) (21.161 b) while the \"jump\" in the parts EJ.. and BII must be related to the charge density (charge per unit area) (J, the current density (current per unit area) j, and the unit normal to the surface n by the formulas [E.d = E1 - EJ: = 4'i7(Jn, [Bill = Bit - Bii = 4'i7j X n. (21.161c) (21.161 d) Recall also that one derives these junction conditions by integrating Maxwell's equations over a \"pill box\" that is centered on the surface. Similar junction conditions, derivable in a similar manner, apply to the gravita tional field (spacetime curvature), and to the stress-energy that generates it. * Focus \"The original formulation of gravitational junction conditions stemmed from Lanczos (1922, 1924). The formulation given here. in terms of intrinsic and extrinsic curvature, was developed by Darmois (1927), Misner and Sharp (1964), and Israel (1966). For further references to the extensive literature, see Israel. 552 21. VARIATIONAL PRINCIPLE AND INITIAL VALUE DATA Figure 21.6. Gaussian normal coordinates in the neighborhood of a 3-surface I. The metric in Gaussian normal coordinates has the form ds 2 = (n' n)-l dn 2 + gij dx i dx; with n = %n, (n' n) = - I if the surface is spacelike, and (n\u00b7 n) = I if it is timelike. (See exercise 27.2.) The extrinsic", " curvature of the surfaces n = constant is Kij = -~ ogij/on, and the Einstein field equations written in \"3 + I\" form are (21.162). attention on a specific three-dimensional slice through spacetime-the 3-surface :s on Figure 21.6. Let the surface be either spacelike [unit normal n timelike; (n 0 n) = -1] or timelike [n spacelike; (n 0 n) = + 1]. The null case will be discussed later. As an aid in deriving junction conditions, introduce Gaussian normal coordi nates in the neighborhood of:S [see the paragraph preceeding equation (21.82)]. In terms of the intrinsic and extrinsic curvatures of :s and of neighboring 3-surfaces n = constant, the Einstein tensor and Einstein field equation have components Gnn = - ; (3)R + ; (n' n)-l{(Tr K)2 - Tr (K2)} = 8'iT1\"'n' (21.162a) Einstein equation in \"3 + 1\" form Gni = -(nonrl{Kimlm j = (3lGi; + (n 0 nr1 Gi { (K\\ - 8ij Tr K),n (TrK)li} = 8'iT1\"'i' (21.162b) Surface stress-energy tensor - (Tr K)Ki j + ; oi/Tr K)2 + ; Oij Tr (K2)} = 8'iTTij. (21.162c) [See equations (21.77), (21.81), (21.76), and (21.82).] Suppose that the stress-energy tensor 'Pp contains a \"delta-function singularity\" at :s-Le., suppose that :s is the \"world tube\" of a two-dimensional surface with finite 4-momentum per unit area (analog of surface charge and surface current in electrodynamics). Then define the surface stress-energy tensor on :s to be the integral of 'Pp with respect to proper distance (n), measured perpendicularly through :s: sap = ~~ [f+t 'Ppdn]. - t (21.163) \u00a721.13. JUNCTION CONDITIONS 553 To discover the effect of this surface layer on the spacetime geometry, perform a \"pill-box integration\" of the Einstein field equation (21.162) Derivation of junction conditions (21.164) Examine the integral of Gaf3\" If the 3-metric gi; were to contain a delta function or a discontinuity at :S, then :s would not have any well-defined 3-geometry-a physically inadmissible situation, even in the presence of surface layers. Absence of delta functions, Sen), in gi; means absence of delta functions in (3lR; absence of discontinuities in gi; means absence of delta functions in Ki; = -!gi;,n' Thus, equations (21.162) when integrated say f G\\ dn = 0 = 8'i7Sn f Gn i dn = 0 = 8'i7Sn f Gi;. dn = (n' n)(yi; - Si; Tr y)= 8'i7S i (2 1.1 65a) (21.165b) (21.165c) n' i, ;, where yi; is the \"jump\" in the components of the extrinsic curvature y = [K]- (K on \"n = +e side\" of :S) (K on \"n = -e side\" of :S) =K+ - K-. (21.166) In the absence of a delta-function surface layer, the above junction conditions say, simply, that y = [K] = O. In words: if one examines how :s is embedded in the spacetime above its \"upper\"face, and how it is embedded in the spacetime below its \"lower\" face, one must discover identical embeddings-Le., identical extrinsic curvatures K. Of course, the intrinsic curvature of:S must also be the same, whether viewed from above or below. More briefly: (absence of surface layers) ( ) (\"continuity\" of gi; and Ki;). (21.167) If a surface layer is present, then :s must be the world tube of a two-dimensional layer of matter, and the normal to :s must be spacelike, (n' n) = + 1. The junction conditions (21.165a,b) then have the simple physical meaning Sen,...) = 0 ( I the momentum flow is entirely in :s;", " ~ i.e., no momentum associated with the surface layer flows out of :S; i.e., :s is the world tube of the surface layer (,(21.168a) which tells one nothing new. The junction condition (21.165c) says that the surface stress-energy generates a discontin uity in the extrinsic curvature (different embedding in spacetime \"above\" :s than \"below\" :S), given by (21.168b) Of course, the intrinsic geometry of :s must be the same as seen from above and below, gij continuous across :So (21.169) Junction conditions in absence of surface layers Junction conditions for a surface layer Equation of motion for a surface layer Gravitational-wave shock fronts 554 21. VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA In analyzing surface layers, one uses not only the junction conditions (21.168a) to (21.169), but also the four-dimensional Einstein field equation applied on each side of the surface I s~parately, and also an equation of motion for the surface stress energy. The equation of motion is derived by examining the jump in the field equation G\"i = 8';71\"'; (equation 21.162b); thus [G n;] = 8';7[1\"';] says and when reexpressed in terms of Si m by means of the junction condition (2 1.1 68b), it says (21.170) [For intuition into this equation of motion, see Exercises 21.25 and 21.26. For applications of the \"surface-layer formalism\" see exercise 21.27; also Israel (1966). Kuchar (1968), Papapetrou and Hamoui (1968).] When one turns attention to junction conditions across a null surface I, one finds results rather different from those in the spacelike and timelike caseS. A \"pill-box\" integration of the field equations reveals that even in vacuum the extrinsic curvature may be discontinuous. A discontinuity in Ki ; across a null surface, without any stress-energy to produce it, is the geometric manifestation of a gravitational-wave shock/rant (analog of a shock-front in hydrodynamics). For quantitative details see, e.g., Pirani (1957), Papapetrou and Treder (1959, 1962), Treder (1962), and especially Choquet-Bruhat (1968b). That a discontinuity in the curvature tensor can propagate_with the speed of light is a reminder that all gravitational effects, like all electromagnetic effects, obey a causal law. The initial-value data on a spacelike initial-value hypersurface uniquely determine the resulting spacetime geometry [see the work of Cartan, Stellmacher, Lichnerowicz, and Bruhat (also under the names Foures-Bruhat and Choquet-Bru hat) and others cited and summarized in the article of Bruhat (1962)] but determine it in a way consistent with causality. Thus a change in these data throughout a limited region of the initial value 3-geometry makes itself felt on a slightly later hypersurface solely in a region that is also limited, and only a little larger than the original region. When one turns from classical dynamics to quantum dynamics, one sees new reason to focus attention on a spacelike initial-value hypersurface: the observables at different points on such a hypersurface commute with one another; i.e., are in principle simultaneously observable. Not every four-dimensional manifold admits a global singularity-free spacelike hypersurface. Those manifolds that do admit such a hypersurface have more to do with physics, it is possible to believe, than those that do not. Even in a manifold that does admit a spacelike hypersurface, attention has been given sometimes, in the context of classical theory, to initial-value data on a hyper surface that is not spacelike but \"characteristic,\" in the sense that it accomodates null geodesics [see, for example, Sachs (1964) and references cited there]. It is typical in such situations that one can predict the future but not the past, or predict the past but not the future. \u00a721.13. JUNCTION CONDITIONS 555 Children ofJight and children of darkness is the vision of physics that emerges from this chapter, as from other branches of physics. The children of light are the differential equations that predict the future from the present. The children of darkness are the factors that fix these initial conditions. Exercise 21.25. EQUATION OF MOTION FOR A SURFACE LAYER (a) Let u be the \"mean 4-velocity\" of the matter in a surface layer-so defined that an observer moving with 4-vel", "ocity u sees zero energy flux. Let a be the total mass-energy per unit proper surface area, as measured by such a \"co moving observer.\" Show that the surface stress-energy tensor can be expressed in the form EXERCISES S = au \u00ae u + t, where (t\u00b7 u) = 0, (21.171) and where t is a symmetric stress tensor. (b) Show that the component along u of the equation of motion (21.170) is where d/dT = u. Give a physical interpretation for each term. (c) Let OJ be that part of the 4-acceleration of the comoving observer which lies in the surface layer X. By projecting the equation of motion (21.170) perpendicular to u, show that (21.172) where Pja is the projection operator (21.173) (21.174) Give a physical interpretation for each term of equation (21.182). Exercise 21.26. THIN SHELLS OF DUST For a thin shell of dust surrounded by vacuum ([Tin] = 0, t = 0), derive equations the following da/dT = -au b 1b, a+ + a- = 0, a+ - a- = (4'1Ta)n y = 8'1Ta(u \u00ae u + i g ). (21.175a) (2 1.1 75b) (2 1.1 75c) (2 1.1 75d) Here a+ and a- are the 4-accelerations as measured by accelerometers that are fastened onto the outer and inner sides of the shell, and 9 is the 3-metric of the shell. Show that the first of these equations is the law of \"conservation of rest mass.\" Exercise 21.27. SPHERICAL SHELL OF DUST Apply the formalism of exercise 21.25 to a collapsing spherical shell of dust [Israel (I967b)]. For the metric inside and outside the shell. take the flat-spacetime and vacuum Schwarzschild expressions (Chapter 23), 556 21. VARIATIONAL PRINCIPLE AND INITIAL VALUE DATA ds2 = - dt2 + dr2 + r2(d(J2 + sin2(J d</>2) inside, ds 2 = - (I - 2~) dt2 + I _d;~/r + r2(d(J2 + sin2(J d\u00a22) outside. (2 1.1 76a) (2 1.176 b) Let the \"radius\" of the shell, as a function of proper time measured on the shell, be - R = -2 1 '17 X (proper circumference of shell) = R (T). (2 I. I76c) Show that the shell's mass density varies with time as J.L = constant = \"total rest mass\"; (2 1.1 76d) and derive and solve the equation of motion M = J.L { dR )2}1/2 I + d; ( J.L - TR\" (2 1.1 76e) --------------------------------........~ CHAPTER 22 THERMODYNAMICS, HYDRODYNAMICS, ELECTRODYNAMICS, GEOMETRIC OPTICS, AND KINETIC THEORY \u00a722.1. THE WHY OF THIS CHAPTER Astrophysical applications of gravitation theory are the focus of the rest ofthis book, except for Chapters 41-44. Each application-stars, star clusters, cosmology, collapse, black holes, gravitational waves, solar-system experiments-can be pursued by itself at an elementary level, without reference to the material in this chapter. But deep understanding of the applications requires a prior grasp of thermodynamics, hydro dynamics, electrodynamics, geometric optics, and kinetic theory, all in the context of curved spacetime. Hence, most Track-2 readers will want to probe these subjects at this point. \u00a722.2. THERMODYNAMICS IN CURVED SPACETIME* Consider, for concreteness and simplicity, the equilibrium thermodynamics of a perfect fluid with fixed chemical composition (\"simple perfect fluid\")-for example, the gaseous interior of a collapsing supermassive star. The thermodynamic state of a fluid element, as it passes through an event '3'0' can be characterized by various thermodynamic potentials, such as n, P, p, T, s, fL. The numerical value of each potential at '3'0 is measured in the proper reference frame (\u00a7 13.6) of an observer who moves with the fluid element-i.e., in the fluid element's \"rest frame.\" Despite \u2022 For more detailed treatments of this subject see, e.g.\u2022 Stueckelberg and Wanders (19", "53), Kluitenberg and de Groot (1954), Meixner and Reik (1959), and references cited therein: see also the references on hydrodynamics cited at the beginning of \u00a722.3, and the references on kinetic theory cited at the beginning of \u00a722.6. r This chapter is entirely Track 2. No earlier Track-2 material is needed as prepara~ion for it, but Chapter 5 (stress-energy tensor) will be helpful. \u00a722.5 (geometric optics) is needed as preparation for Chapter 34 (singularities and global methods). The rest of the chapter is not needed as preparation for any later chapter; but it will be in most extremely helpful applications of gravitation theory (Chapters 23-40). Thermodynamic potentials are defined in rest frame of fluid Definitions of thermodynamic potentials Definition of \"simple fluid\" Law of baryon conservation 558 22. THERMODYNAMICS. HYDRODYNAMICS,.... AND KINETIC THEORY this use of rest frame to measure the potentials. the potentials are frame-independent functions (scalar fields). At the chosen event ~J~o, a given potential (e.g., n) has a unique value 11(~fo): so n is a perfectly good frame-independent function. The values of 11, p, p, T, s, fL measure the following quantities in the rest frame of the fluid element: n, baryon number density: i.e., number of baryons per unit three-dimensional volume of rest frame, with antibaryons (if any) counted negatively. p, density of total mass-energy; i.e., total mass-energy (including rest mass, ther mal energy, compressional energy, etc.) contained in a unit three-dimensional volume of the rest frame. p, isotropic pressure in rest frame. T, temperature in rest frame. s, entropy per baryon in rest frame. (The entropy per unit volume is 11S.) fL, chemical potential of baryons in rest frame [see equation (22.8) below]. The chemical composition of the fluid (number density of hydrogen molecules, number density of hydrogen atoms, number density of free protons and electrons, number density of photons, number density of 238U nuclei, number density of A hyperons...) is assumed to be fixed uniquely by two thermodynamic variables-e.g., by the total number density of baryons n and the entropy per baryon s. In this sense the fluid is a \"simple fluid.\" Simple fluids occur whenever the chemical abundances are \"frozen\" (reaction rates too slow to be important on the time scales of interest; for example, in a supermassive star except during explosiv~ burning and except at temperatures high enough for e- - e+ pair produetion).- Simple fluids also occur in the opposite extreme of complete chemical equilibrium (reaction rates fast enough to maintain equilibrium despite changing density and entropy; for example, in neutron stars, where high pressures speed up all reactions). When one examines nuclear burning in a nonconvecting star, or explosive nuclear burning, or pair production and neutrino energy losses at high temperatures, one must usually treat the fluid as \"multicomponent\" Then one introduces a number density nJ and a chemical potential fLJ for each chemical species with abundance not fixed by n and s. For further details see, e.g., Zel'dovich and Novikov (1971). The most fundamental law of thermodynamics-even more fundamental than the \"first\" and \"second\" laws-is baryon conservation. Consider a fluid element whose moving walls are attached to the fluid so that no baryons flow in or out As the fluid element moves through spacetime, deforming along the way, its volume V changes. But the number of baryons in it must remain fixed, so d dr (nV) =0. (22.1 ) The changes in volume are produced by the flow of neighboring bits of fluid away from or toward each other-explicitly (exercise 22.1) dV/dr = (V' u)v, (22.2) \u00a722.2. THERMODYNAMICS IN CURVED SPACETIME 559 where u = d/dr is the 4-velocity of the fluid. Consequently, baryon conservation [equation (22.1)] can be reexpressed as o = dn +.E.. dV = V n + n(V. u) = u\u00b7 Vn + n(V' u) = V. (nu); dr Vdr U i.e., V\u00b7 S = 0, S = nu = baryon number-flux vector (22.3) (22.4) (see \u00a7", "5.4 and exercise 5.3.) Moreover, this abstract geometric version of the law must be just as valid in curved spacetime as in flat (equivalence principle). Note the analogy with the law of charge conservation, V\u00b7 J = 0, in electrody namics (exercise 3.16) and with the local law of energy-momentum conservation, V\u00b7 T = 0 (\u00a7\u00a75.9 and 16.2). In a very deep sense, the forms of these three laws are dictated by the theorem of Gauss (\u00a75.9, and Boxes 5.3, 5.4). The second law of thermodynamics states that, in flat spacetime or in curved, entropy can be generated but not destroyed. Apply this law to a fluid element of volume V containing a fixed number of baryons N. The entropy it contains is S = Ns = nsV. Entropy may flow in and out across the faces of the fluid element (\"heat flow\" between neighboring fluid elements); but for simplicity assume it does not; or if it does, assume that it flows too slowly to have any significance for the problem at hand. Then the entropy in the fluid element can only increase: d(nsV)/dr ~ 0 when negligible entropy is exchanged between neighboring fluid elements; i.e. [combine with equation (22.1)] ds/dr ~ 0 (no entropy exchange). (22.5) So long as the fluid element remains in thermodynamic equilibrium, its entropy will actually be conserved [\" =\" in equation (22.5)]; but at a shock wave, where equilibrium is momentarily broken, the entropy will increase (conversion of \"relative kinetic energy\" of neighboring fluid elements into heat). [For discussions of heat flow in special and general relativity, see Exercise 22.7. For discussion of shock waves, see Taub (1948), de Hoffman and Teller (1950), Israel (1960), May and White (1967), Zel'dovich and Rayzer (1967), Lichnerowicz (1967, 1971), and Thorne (l973a).] The first law of thermodynamics, in the proper reference frame of a fluid element, is identical to the first law in flat spacetime (\"principle of equivalence\"); and in flat spacetime the first law is merely the law of energy conservation: Second law of thermodynamics Shock waves and heat flow First law of thermodynamics d( energy in a volume element containing) a fixed number. A, of baryons ) ) = - p vO ume + U' en ropy ; T J( d( I t. 560 i.e., Le., 22. THERMODYNAMICS. HYDRODYNAMICS,..., AND KINETIC THEORY d(pA/n) = -pd(A/n) + Td(As); dp = - -dn + nTds. p+p n Query: what kind of a \"d\" appears here? For a simple fluid, the values of two potentials, e.g., nand s, fix all the others uniquely; so any change in p must be determined uniquely by the changes in nand s. It matters not whether the changes are measured along the world line of a given fluid element, or in some other direction. Thus, the \"d\" in the first law can 'be interpreted as an exterior derivative dp = - -d n + nTds; p+p n (22.6) and the changes along a given direction in the fluid (along a given tangent vec tor v) can be written p+p VvP = (dp, v) = - - (dn, v) + nT(ds, v) n Pressure and temperature calculated from p(n, s) Equation (22.6) lends itself to interpretation in two -opposite senses: as a way to deduce the density of mass-energy of the medium from information about pressure (as a function of nand s) and temperature (as a function of nand s); and conversely, as a way to deduce the two functions p(n, s) and T(n, s) from the one function p(n, s). It is natural to look at the second approach first; who does not like a strategy that makes an intellectual profit? Regarding p as a known (or calculable) function of nand s, one deduces from (22.6) p + P = (~), on 8 n nT= (:~t and thence pressure and temperature individually, p(n,s) = n(::)8 - p, I (OP) T(n,s) = AS n n (22.7a) (22.7b) (\"two equations of state from one\"). The analysis simplifies still further when the fluid, already", " assumed to be everywhere of the same composition, is also everywhere \u00a722.2. THERMODYNAMICS IN CURVED SPACETIME 561 endowed with the same entropy per baryon, s, and is in a state of adiabatic flow (no shocks or heat conduction). Then the density p == p(n, s) reduces to a function of one variable out of which one derives everything (p,p,p.) needed for the hydro dynamics and the gravitation physics of the system (next chapter). Other choices of the \"primary thermodynamlc potential\" are appropriate under other circum stances (see Box 22.1). If differentiation leads from p(n, s) to p(n, s) and T(n, s), it does not follow that one can take any two functions p(n, s) and T(n, s) and proceed \"backwards\" (by integration) to the \"primary function\", p(n, s). To be compatible with the first law of thermodynamics (22.6), the two functions must satisfy the consistency requirement [\"Maxwell relation\"; equality of second partial derivatives of p] (22.7 c) Maxwell relation Box 22.1 PRINCIPAL ALTERNATIVES FOR \"PRIMARY THERMODYNAMIC POTENTIAL\" TO DESCRIBE A FLUID Primary thermodynamic potential and quantities on which it is most appropriately envisaged to depend \"Secondary\" thermodynamic quanti ties obtained by differentiation of primary with or without use of Conditions under which convenient, appropriate, and relevant \"Density\"; total amount of mass energy (rest + thermal +...) per unit volume P = p(n, s) p(n, s) :;= n(~) - p T(n,s)=- - on \u2022 I (OP) n os n p + P (OP) on \u2022 n Il(n,s) = - - = - Conditions of adiabatic flow (no shocks or heat conduction), so\u00b7 that s stays constant along streamline \"Physical free energy\" o(n, T) =!!.. - Ts n p(n, T) = n2((0) on T s(n, T) = -(~~) n Know or can calculate 0 (or the \"sum over states\" of statistical me chanics) for conditions of specified volume per baryon and temperature p(n, T) = -nT2 - o(O/T) ] [ oT n \"Chemical free energy\" P+P n j(p, T) =- - - Ts \"Chemical potential\" (\"energy to in ject\" expressed on a \"per baryon\" basis) p+p Il(P,S) = - n I/n(p, T) = (oj/OP)T s(p, T) = -(ojloT)p p(p, T) = j - T(ojloT)p (oj/oP)T - P I/n(p, s) = (Oll/Op). T(p, s) = (OIl/OS)p p(p, s) = (Oll/OP). - P Il Relevant for determining equilibrium when pressure and temperature are specified When injection energy [= Fermi en ergy for an ideal Fermi gas, relativis tic or not; see exercise 22.3] is the center of attention y Chemical potential equals \"injection energy\" at fixed entropy per baryon and total volume 562 22. THERMODYNAMICS, HYDRODYNAMICS,.., AND KINETIC THEORY The chemical potential J.L is also a unique function of nand s. It is defined as follows. (l) Take a sample of the simple fluid in a fixed thermodynamic state (fixed nand s). (2) Take, separately, a much smaller sample of the same fluid, containing SA baryons in the same thermodynamic state as the large sample (same nand s). (3) Inject the smaller sample into the larger one, holding the volume of the large sample fixed during the injection process. (4) The total mass-energy injected, SMjnjected = p X (volume of injected fluid) = p(SA/n), plus the work required to perform the injection Sw.. = (WOrk done against pressure of large Sample) to open up space in it for the injected fluid lIljection is equal to J.L SA: =p(volume of injected fluid) = p(SA/n), J.L SA = SMjnjected + S Wjnjection = --- SA. p+p n Stated more briefly:' total mass-energy required, per baryon, to \"create\"", " and) J.L = inject a small additional amount of fluid into a given sample, without changing s or volume of the sample ( = p + P = (~). t on s n - - [by first law of thermodynamics (22.6)] (22.8) All the above laws and equations of thermodynamics are the same in curved spacetime as in flat spacetime; and the same in (relativistic) flat spacetime as in classical nonrelativistic thermodynamics-except for the inclusion of rest mass, together with all other forms of mass-energy, in p and J.L. The reason is simple: the laws are all formulated as scalar equations linking thermodynamic variables that one measures in the rest frame of the fluid. \u00a722.3. HYDRODYNAMICS IN CURVED SPACETIME* A simple perfect fluid flows through spacetime. It might be the Earth's atmosphere circulating in the Earth's gravitational field. It might be the gaseous interior of the Sun at rest in its own gravitational field. It might be interstellar gas accreting on to a black hole. But whatever and wherever the fluid may be, its motion will be governed by the curved-spacetime laws of thermodynamics (\u00a722.2) plus the local 'For more detailed treatments of this subject see, e.g., Ehlers (1961), Taub (1971), Ellis (1971), Lichnerowicz (1967), Cattaneo (1971), and references cited therein; see also the references on kinetic theory cited at the beginning of \u00a722.6. Laws of hydrodynamics for simple fluid without heat flow or viscosity: (1) Origins of laws \u00a722.3. HYDRODYNAMICS IN CURVED SPACETIME 563 law of energy-momentum conservation, V \u2022 i = O. The chief objective of this section is to reduce the equation V \u2022 T = 0 to usable form. The reduction will be performed in the text using abstract notation; the reader is encouraged to repeat the reduction using index notation. The stress-energy tensor for a perfect fluid, in curved spacetime as in flat (equival ence principle!), is T = (p + p)u \u00ae u + pg. (22.9) (See \u00a75.5.) Its divergence is readily calculated using the chain rule; using the \"com patibility relation between 9 and V,\" Vg = 0; using the identity (Vp)' 9 = Vp (which one readily verifies in index notation); and using 0= V\u00b7 T = [V(p + p)' uju + [(p + p) V\u00b7 uju + [(p + p)uj' Vu + (Vp)' 9 l[divergence on first slotj = [Vup + VuP + (p + p)V' uju + (p + p) Vuu + Vp. (22.10) The component of this equation along the 4-velocity is especially simple (recall that u' Vuu =! V uu 2 = 0 because u 2 -1): 0= u'(V' T) = -[Vup + VuP + (p + p)V'uj + VuP = - VuP - (p + p) V\u00b7 u. Combine this with the equation of baryon conservation (22.3); the result is dp _ (p + p) dn dr' dr- n (22.11a) (2) Local energy conservation: adiabaticity of flow Notice that this is identical to the first law of thermodynamics (22.6) applied along a flow line, plus the assumption that the entropy per baryon is conserved along a flow line ds/dr = O. (22.11 b) There is no reason for surprise at this result. To insist on thermodynamic equilibrium and to demand that the entropy remain constant is to require zero exchange of heat between one element of the fluid and another. But the stress-energy tensor (22.9) recognizes that heat exchange is absent. Any heat exchange would show up as an energy flux term in T (Ex. 22.7); but no such term is present. Consequently, when one studies local energy conservation by evaluating u' ('V. T) = 0, the stress-energy tensor reports that no heat flow is occurring-i.e. that ds/dr = O. Three components of V \u2022 T = 0 remain: the components orthogonal to the fluid's 4-velocity. One can pluck them out of V \u2022 T = 0, leaving behind the component along u, by use of the \"projection tensor\" P 9 + u \u00ae u. (22.12) ~-------...... Box 22.2 THERMODYNAMICS AND HYDRODY", "NAMICS FOR A SIMPLE PERFECT FLUID IN CURVED SPACETIME A. Ten Quantities Characterize the Fluid Equation for chemical potential Thermodynamic potentials all measured in rest frame n, baryon number density p, density of total mass-energy p, pressure T, temperature s, entropy per baryon fL, chemical potential per baryon Four components of the fluid 4-velocity B. Ten Equations Govern the Fluid's Motion fL = (p + p)/n, (4) which can be combined with p(n, s) and p(n, s) to give fL(n, s). Law of baryon conservation dn/dr Vun = -n V' u. (5) Conservation of energy along flow lines, which (assuming no energy exchange between adja cent fluid elements) means \"adiabatic flow\" ds/dT = 0 except in shock waves, where ds/dT > O. (6) Two equations of state p = p(n,s), T = T(n,s) (1), (2) subject to the compatibility constraint (\"Max well relation,\" which follows from first law of thermodynamics) (op/os)n = n 2(oT/on)s' First law of thermodynamics dp = - -dn + nTds, p+p n (3) [Shock waves are not treated in this book; see Taub (1948), de Hoffman and Teller (1950), Israel (1960), May and White (1967), Zel'do (1967); Lichnerowicz vich (1967, 1971); and Thorne (1973a).] and Rayzer Euler equations (p + p)Vuu = -(g + u \u00ae u)' Vp, (7), (8), (9) which determine the flow lines to which u is tangent. Normalization of 4-velocity which can be integrated to give p(n, s). u\u00b7u=-1. (10) (See exercise 22.4.) Contracting P with V. T = 0 [equation (22.10)] gives (3) Euler equation (p + p) Vuu = _po (Vp) -[Vp + (VuP)u]. (22.13) This is the \"Euler equation\" of relativistic hydrodynamics. It has precisely the same form as the corresponding flat-spacetime Euler equation: inertial mass. per umt volume X [exercise 5.4] ) ( ( 4 l'-acce eratlOn) f fl.d = - 0 Ul (pressure gradient. In the 3-surface orthogonal to 4-velocity ). (22.13') The pressure gradient, not \"gravity,\" is responsible for all deviation of flow lines from geodesics. Box 22.2 reorganizes and summarizes the above laws of thermodynamics and hydrodynamics. \u00a722.3. HYDRODYNAMICS IN CURVED SPACETIME 565 Exercise 22.1. DIVERGENCE OF FLOW LINES PRODUCES VOLUME CHANGES Derive the equation dV/dT = (V'. u) V [equation (22.2)] for the rate of change of volume of a fluid element. [Hint: Pick an event '3'0' and calculate in a local Lorentz frame at '3'0 which momentarily moves with the fluid (\"rest frame at '3'0\").] [Solution: At events near '3'0 the fluid has a very small ordinary velocity vi = dxi/dt. Consequently a cube of fluid at '3'0 with edges.:1x =.:1y =.:1z = L changes its edges, after time 8t, by the amounts EXERCISES 8(.:1x) = [(dx/dt) 8t]at\"frontface\" =(av Z /ax)L 8t, 8(.:1y) = (av Y/ay)L8t, 8(.:1z) = (av Z /az)L 8t. [(dx/dt) 8t]at\"backface\" The corresponding change in volume is so the rate of change of volume is av/at = v(avi/axi). But in the local Lorentz rest frame at and near '3'0 (where x a = 0), the metric coefficients are gl'v = Tll'v + 0(lxaI2), and the ordinary velocity is vi = O(lxal); so o _ dt _ u - d - (_ T gl'v dt X d I'd V)1/2 X _ I 0(1 a12) + X,", ". u 1 = - = v1 + (Ixal). 0 3. dxi dT Thus, the derivatives av/at and V(avi/axi) at '3'0 are av/at = uaav/axa = ua~a = dV/dT = V(aviiaxi) = V(aua/ax a) = Vua;a = V(V'\u00b7 u). Q.E.D.] [Note that by working in flat spacetime, one could have inferred more easily that aVfat = dV/dT and avi/axi = V'. u; one would then have concluded dV/dT = (V\" u) V; and one could have invoked the equivalence principle to move this law into curved spacetime.] Exercise 22.2. EQUATION OF CONTINUITY Show that in the nonrelativistic limit in flat spacetime the equation of baryon conservation (22.3) becomes the \"equation of continuity\" an a. ) = O. - + -. (nv 1 ex1 at Exercise 22.3. CHEMICAL POTENTIAL FOR IDEAL FERMI GAS Show that the chemical potential of an ideal Fermi gas, nonrelativistic or relativistic, is (at zero temperature) equal to the Fermi energy (energy of highest occupied momentum state) of that gas. Exercise 22.4. PROJECTION TENSORS Show that contraction of a tangent vector B with the \"projection tensor\" P =9 + u \u00ae u projects B into the 3-surface orthogonal to the 4-velocity vector u. [Hint: perform the 566 22. THERMODYNAMICS. HYDRODYNAMICS.... AND KINETIC THEORY calculation in an orthonormal frame with eo = u, and write B = Baea: then show that p. B = BJej.] If n is a unit spacelike vector, show that P =9 - n \u00ae n is the corresponding projection operator. Note: There is no unique concept of \"the projection orthogonal to a null vector.\" Why? [Hirlt:?r~w pictures in flat spacetime suppressing one spatial dimension.] Exercise 22.5. PRESSURE GRADIENT IN STATIONARY GRAVITATIONAL FIELD A perfect fluid is at rest (flow lines have xi = constant) in a stationary gravitational field (metric coefficients are independent of Xo). Show that the pressure gradient required to \"support the fluid against gravity\" (i.e., to make its flow lines be xi = constant instead of geodesics) is ap - -0 = 0, ax ap _ axi - ( - p + p ) aln ~ axi. (22.14) Evaluate this pressure gradient in the Newtonian limit, using the coordinate system and metric coefficients of equation (18.l5c). Exercise 22.6. EXPANSION, ROTATION, AND SHEAR Let a field of fluid 4-velocities u(q') be given. (a) Show that Vu can be decomposed in the following manner: where a is the 4-acceleration of the fluid (J is the \"expansion\" of the fluid world lines Pap is the projection tensor l1ap is the shear tensor of the fluid and wap is the rotation 210rm of the fluid (22.1 5a) (22J5b) \u2022 (22J5c) (22J5d) (22J5e) (22J5f) (b) Each of the component parts of this decomposition has a simple physical interpretation in the local rest frames of the fluid. The interpretation of the 4-acceleration a in terms of accelerometer readings should be familiar. Exercise 22.1 showed that the expansion (J = V \u2022 u describes the rate of increase of the volume of a fluid element, (J = (1/V)(dV/dT). (22J5g) Exercise 22.4 explored the meaning and use of the projection tensor P. Verify that in a local Lorentz frame (ga~ = TJap, ra~y = 0) momentarily' moving with the fluid (u a = sao), l1a~ and wa~ reduce to the classical (nonrelativistic) shear and rotation of the fluid. [See, e.g., \u00a7\u00a72.4 and 2.5 of Ellis (1971) for both classical and relativistic descriptions of shear and rotation.] \u00a722.3. HYDRODYNAMICS IN CURVED SPACETIME 567 Exercise 22.7. HYDRODYNAMICS WITH VISCOSITY AND HEAT FLOW: (a) In \u00a7 15 of", " Landau and Lifshitz (1959), one finds an analysis of viscous stresses for'a classical (nonrelativistic) fluid. By carrying that analysis over directly to the local Lorentz rest frame of a relativistic fluid, and by then generalizing to frame-independent language, show that the contribution of viscosity to the stress-energy tensor is T(v1sc) = - 2T/u - ~O P, (22.16a) where T/ ~ 0 is the \"coefficient ofdynamic viscosity\"; ~ ~ 0 is the \"coefficient ofbulk viscosity\"; and u, 0, P are the shear, expansion, and projection tensor of the fluid. (b) An idealized description of heat flow in a fluid introduces the heat-jiux 4-vector q with components in the local rest-frame of the fluid, q~ = 0, j = (energy per unit ~ime crossing unit). q surface perpendicular to ej (22.16b) By generalizing from the fluid rest frame to frame-independent language, show that the contribution of heat flux to the stress-energy tensor is T(heaO = u I8i q + q I8i u. (22.16c) Thereby conclude that, in this idealized picture, the stress-energy tensor for a viscous fluid with heat conduction is T al3 = puau13 + (p _ ~o)pal3 - 2T/oa13 + qau13 + uaql3. (22.16d) (c) Define the entropy 4-vector s by s _ nsu + q/T. By calculations in the local rest-frame of the fluid, show that v. S = (~ate of!ncrease of entropy) _ (rate at which ~eat and ~uid m a urnt volume carry entropy mto a umt volume = (rate at which entropy is being) generated in a unit volume. Thereby arrive at the following form of the second law of thermodynamics: (22.16e) ) (22.16f) V'S ~ O. (d) Calculate the law of local energy conservation, u' V. T = 0, for a viscous fluid with law of thermodynamics and with the law of baryon (22.16g) heat flow. Combine with the first conservation to obtain Interpret each term of this equation as a contribution to entropy generation (example: 2T/oal3oa13 describes entropy generation by viscous heating). [Note: The term qaaa is rela tivistic in origin. It is associated with the inertia of the flowing heaL] (e) When one takes account of the inertia of the flowing heat, one obtains the following generalization of the classical law of heat conduction: (22.16h) (22.16i) \u2022 Exercise supplied by John M. Stewart. 568 22. THERMODYNAMICS, HYDRODYNAMICS,.... AND KINETIC THEORY (Eckart 1940). Here K is the coefficient of thermal conductivity. Use this equation to show that, for a fluid at rest in a_stationary gravitational field (Exercise 22.5), (22.l6j) [Thus, thermal equilibrium corresponds not to constant temperature, but to the redshifted temperature distribution TV - goo = constant; Tolman (1934a), p. 313.] Also, use the ideal ized law of heat conduction (22.l6i) to reexpress the rate of entropy generation as TV's = ~02 + 2T/Oa /30a/3 + (K/T)pa/3(T,a + Taa)(T,/3 + Ta/3) ~ O. (22.l6k) [For further details about heat flow and for discussions of the limitations of the above idealized description, see e.g., \u00a74.l8 of Ehlers (1971); also MarIe (1969), Anderson (1970), Stewart (1971), and papers cited therein.] \u00a722.4. ELECTRODYNAMICS IN CURVED SPACETIME Electric and magnetic fields In a local Lorentz frame in the presence of gravity, an observer can measure the electric and magnetic fields E and B using the usual Lorentz force law for charged particles. As in special relativity, he can regard E and B as components of an electromagnetic field tensor, FO; = _F;o = E;, he can regard the charge and current densities as components of a 4-vector fa, and he can write Maxwell's equations and the Lorentz force equation in the special relativistic form, F'::r/J A = 47TJ\",,p,., maa == Fa", "/3qu A P pY,a F: A'+ FA' \u2022 + P.\" ap, y (m = mass of particle, q = charge, ) U\" = 4-velocity, a\" = 4-acceleration. ya,p A = 0, Maxwell equations and Lorentz force law In any other frame these equations will have the same form, but with commas replaced by semicolons Fa/3;/3 = 47TJa, Fa/3;Y + F/3y;a + Fya;/3 = 0, maa = Fa/3 quf3' (22.17a) (22.17b) (22.17 c) These are the basic equations of electrodynamics in the presence of gravity. From them follows everything else. For example, as in special relativity, so also here (exercise 22.9), they imply the equation of charge conservation Charge conservation (22.18a) \u00a722.4. ELECTRODYNAMICS IN CURVED SPACETIME 569 and for an electromagnetic field interacting with charged matter (exercise 22.10) they imply vanishing divergence for the sum of the stress-energy tensors (TIEMla/3 + TlMATTERla/3);/3 = O. (22.18b) Local conservation of energy-momentum As in special relativity, so also here, one can introduce a vector potential AIL. Vector potential Replacing commas by semicolons in the usual special-relativistic expression for FILin terms of AIL, one obtains (22.19a) If all is well, this equation should guarantee (as in special relativity) that the Maxwell equations (22.17b) are satisfied. Indeed, it does, as one sees in exercise 22.8. To derive the wave equation that governs the vector potential, insert expression (22.19a) into the remaining Maxwell equations (22.17a), obtaining (22.19b) then commute covariant derivatives in the first term using the identity (16.6c), to obtain (22.19b') Finally, adopting the standard approach of special relativity, impose the Lorentz gauge condition thereby bringing the wave equation (22.19b') into the form (Lld~A)a - Aa;/3 /3 + Ra/3A/3 = 4'TTJa. (22.19d) Wave equation for vector potential (22.19c) Lorentz gauge condition The \"de Rham vector wave operator\" Ll which appears here is, apart from sign, a generalized d'Alambertian for vectors in curved spacetime. Mathematically it is more powerful than -Aa;/3;/3' and than any other operator that reduces to (minus) the d'Alambertian in special relativity. [For a discussion, see de Rham (1955).] Although the electrodynamic equations (22.17a)-(22.19b) are all obtained from special relativity by the comma-goes-to-semicolon rule, the wave equation (22.19d) for the vector potential is not (\"curvature coupling\"; see Box 16.1). Nevertheless,.when spacetime is flat (so Ra /3 = 0), (22.19d) does reduce to the usual wave equation of special relativity. Exercise 22.8. THE VECTOR POTENTIAL FOR ELECTRODYNAMICS Show that in any coordinate frame the connection coefficients cancel out of both equations (22.19a) and (22.l7b), so they can be written EXERCISES Fp.v = Av.1' - AI'.v' Fa13 \u2022y + F13y,a + FYa,13 = O. (22.20a) (22.20b) 570 22. THERMODYNAMICS, HYDRODYNAMICS,..., AND KINETIC THEORY (In the language of differential forms these equations are F = dA, dF = 0.) Then use this form of the equations to show that equation (22.19a) implies equation (22.17b), as asserted in the text. Exercise 22.9. CHARGE CONSERVATION IN THE PRESENCE OF GRAVITY Show that Maxwell's equations (22.17a,b) imply the equation of charge conservation (22.18a) when gravity is present, just as they do in special relativity theory. [Hints: Use the antisym metry of FQ/3; and beware of the noncommutation of the covariant derivatives, which must be handled using equations (16.6). Alternatively, show that in coordinate frames, equation (22.17a) can be written as and (22.18a)", " as (22.17a') (22.18a') and carry out the demonstration in a coordinate frame.] Exercise 22.10. INTERACTING ELECTROMAGNETIC FIELD AND CHARGED MATTER As in special relativity, so also in the presence of gravity (\"equivalence principle\"), stress-energy tensor for an electromagnetic field is the Use Maxwell's equations (22.17a,b) in the presence of gravity to show that T(EMla/3;/3 = -Fa/3J/3' (22.21) (22.22) But Fa/3J/3 is just the Lorentz 4-force per unit volume with which the electromagnetic field acts on the charged matter [see the Lorentz force equation (22.17c); also equation (5.43)]; i.e., it is T(MATTER)a/3 ;/3' Consequently, the above equation can be rewritten in the form (22.18b) cited in the text. \u00a722.5. GEOMETRIC OPTICS IN CURVED SPACETIME* Radio waves from the quasar 3C279 pass near the sun and get deflected by its gravitational field. Light rays emitted by newborn galaxies long ago and far away propagate through the cosmologically curved spacetime of the universe, and get focused (and redshifted) producing curvature-enlarged (but dim) images of the galaxies on the Earth's sky. \u2022 Based in part on notes prepared by William L. Burke at Caltech in 1968. For more detailed treatments of geometric optics in curved spacetime, see, e.g., SachS (1961), Jordan, Ehlers, and Sachs (1961), and Robinson (1961); also references discussed and listed in \u00a741.11. \u00a722.5. GEOMETRIC OPTICS IN CURVED SPACETIME 571 These and most other instances of the propagation of light and radio waves are subject to the laws of geometric optics. This section derives those laws, in curved spacetime, from Maxwell's equations. The fundamental laws of geometric optics are: (1) light rays are null geodesics; (2) the polarization vector is perpendicular to the rays and is parallel-propagated along the rays; and (3) the amplitude is governed by an adiabatic invariant which, in quantum language, states that the number of photons is conserved. The conditions under which these laws hold are defined by conditions on three lengths: (1) the typical reduced wavelength of the waves, Overview of geometric optics Conditions for validity of geometric optics A = ~ = (\"classical distance of closest approach for ) a photon with one unit of angular momentum\", 2'IT - (22.23 a) as measured in a typical local Lorentz frame (e.g., a frame at rest relative to nearby galaxies); (2) the typical length e over which the amplitude, polarization, and wavelength of the waves vary, e.g., the radius of curvature of a wave front, or the length of a wave packet produced by a sudden outburst in a quasar; (3) the typical radius of curvature?it of the spacetime through which the waves propagate,?it = Itypical component of Riemann as measuredl-1I2 - in typical local Lorentz frame (22.23 b). Geometric optics is valid whenever the reduced wavelength is very short compared to each of the other scales present, and (22.23 c) so that the waves can be regarded locally as plane waves propagating through spacetime of negligible curvature. Mathematically one exploits the geometric-optics assumption, A ~ E and A ~?it, as follows. Focus attention on waves that are highly monochromatic over regions ~ E. (More complex spectra can be analyzed by superposition, Le., by Fourier analysis.) Split the vector potential ofelectromagnetic theory into a rapidly changing, real phase, 0- (distance propagated)1A, and a slowly changing, complex amplitude (i.e. one with real and imaginary parts), A = Real part of{amplitude X ei8 } R{amplitude X ei8 }. Imagine holding fixed the scale of the amplitude variation, E, and the scale of the spacetime curvature, 5l, while making the reduced wavelength, A, shorter and shorter. The phase will get larger and larger (0 ex: II A) at any fixed event in spacetime, but the amplitude as a function oflocation in spacetime can remain virtually unchanged, The' 'two-length-scale\" expansion underlying geometric optics A l't d P mIue= [dOminant part, independent of A ] + [small corrections (deviations from geometric optics) due to finite wavelength.. ] y 572 22. THERM", "ODYNAMICS, HYDRODYNAMICS,..., AND KINETIC THEORY This circumstance allows one to expand the amplitude in powers of A:* Amplitude = a + b + c + '\". independent,t lex: A] ~ ex: A2 ] Y of A [ [Actually, the expansion proceeds in powers of the dimensionless number AI(minimum of e and qz) AIL. (22.24) Applied mathematicians call this a \"two-length-scale expansion\"; see, e.g., Cole (1968). The basic short-wavelength approximation here has a long history; see, e.g., Liouville (1837), Rayleigh (1912). Following a suggestion of Debye, it was applied to Maxwell's equations by Sommerfeld and Runge (1911). It is familiar as the WKB approximation in quantum mechanics, and has many other applications as indicated by the bibliography in Keller, Lewis, and Seckler (1956). The contribution of higher order terms is considered by Kline (1954) and Lewis (1958). See especially the book of Froman and Froman (1965).] It is useful to introduce a parameter e that keeps track of how rapidly various terms approach zero (or infinity) as AIL approaches zero: The vector potential geometric optics in Basic concepts of geometric optics: A = i'{(a + eb + e2c +.. \u00b7)ei8/e} p. p. p. p.. (22.25) Any term with a factor en in front of it varies as (AI L)n in the limit of very small wavelengths [0 ex: (AI L)-l; cp. ex: (AIL)2; etc.]. By convention, e is a dummy expan sion parameter with eventual value unity; so it can be dropped from the calculations when it ceases to be useful. And by convention, all \"post-geometric-optics correc tions\" are put into the amplitude terms b, c,... ; none are put into O. Note that, while the phase 0 is a real function of position in spacetime, the amplitude and hence the vectors a, b, c,... are complex. For example, to describe monochromatic waves with righthand circular polarization, propagating in the z direction, one could set 0 = w(z - t) and a = 1/0a(e., + ie,) with a real; so A = i' {~ a(e., + ie,)ei.,<z-tl} = ~ a{cos [w(z - t)]e., - sin [w(z - t)]e ll } The assumed form (22.25) for the vector potential is the mathematical foundation of geometric optics. All the key equations of geometric optics result from inserting this vector potential into the source-free wave equation L1A = 0 [equation (22.l9d)] and into the Lorentz gauge condition V \u2022 A = 0 [equation (22.l9c)]. The resulting equations (derived below) take their simplest form only when expressed in terms of the following: 'The equations for A are linear. Therefore the analysis would proceed equally well assuming, instead of an amplitude independent of A:, a dominant term B ex: A:\", with b ex: A:n+ I, C ex: A:\"+2, etc. The results are independent of n. Choosing n = I would give field strengths Fp.. and energy densities Tp.. ex: F2 ex: A2/A:2 ex: constant as A: -+ O. \u00a722.5. GEOMETRIC OPTICS IN CURVED SPACETIME 573 \"wave vector,\" k VB; (22.26a) (1) wave vector \"scalar amplitude,\" a = (a \u2022 a)1I2 = (aIL Qj..)112; (22.26b) (2) scalar amplitude \"polarization vector,\" f ala = \"unit complex vector along a\". (22.26c) (3) polarization vector (Here a is the complex conjugate of a.) Light rays are defined to be the curves 9(A) normal to surfaces of constant phase B. Since k VBis the normal to these surfaces, the differential equation for a light ray is (4) light rays Derivation of laws of geometric optics (22.26d) Box 22.3, appropriate for study at this point, shows the polarization vector, wave vector, surfaces of constant phase, and light rays for a propagating wave; the scalar amplitude, not shown there, merely tells the length ofthe vector amplitude a. Insight into the complex polarization vector, if not familiar from electrodynamics, can be developed later", " in Exercise 22.12. So much for the foundations. Now for the calculations. First insert the geometric optics vector potential (22.25) into the Lorentz gauge condition: o = AIL;IL = 1\\ {[~ k/a lL + eblL +... ) + (aIL + eblL +... );ILJ eI8/e}. (22.27) The leading term (order lie) says k' a = 0 (amplitude is perpendicular to wave vector); (22.28) or, equivalently k' f = 0 (polarization is perpendicular to wave vector). (22.28') The post-geometric-optics breakdown in this orthogonality condition is governed by the higher-order terms [0(1), O(e), 0(e 2 ),..\u2022J in the gauge condition (22.27); for example, the 0(1) terms say k'b = iV 'a. Next insert the vector potential (22.25) into the source-free wave equation (22.19d): o= (LldRA)''' = -A\",;f3f3 + R'\"f3 Af3 = R{[J...kf3k (a'\" + eb'\" + e2c'\" +... ) - 2!...k f3(a'\" + eb'\" +...).f3. e2 e f3.; k f3 ;f3(a'\" + eb'\" +... ) - (a'\" +.,. yf3f3 + R'\"f3(a f3 +... )] ei8/e}. - (22.29) Collect terms of order l/e 2 and lie (terms of order higher than lie govern post geometric-optics corrections): (continued on page 576) 574 22. THERMODYNAMICS, HYDRODYNAMICS,.... AND KINETIC THEORY Box 22.3 GEOMETRY OF AN ELECTROMAGNETIC WAVE TRAIN =---_........... y direction of propagation as seen in local Lorentz frame ~-----~vr-------' one wavelength, 2,,-t, as seen in local Lorentz frame The drawing shows surfaces of constant phase, 0 = constant, emerging through the \"surface of simultaneity\", t = 0, of a local Lorentz frame. The surfaces shown are alternately \"crests\" (0 = I764'IT, 0 = 1766'IT,...) and \"troughs\" (0 = 1765'IT, 0 = 1767'IT,...) of the wave train. These surfaces make up a I-form, ii = dO. The \"corresponding vector\" k = VO is the \"wave vector.\" The wave vector is null, k \u00b0 k = 0, according to Maxwell's equations [equation (22.30)]. Therefore it lies in a surface of constant phase: number of surfaCeS) db k y pierce. ( =(dO,k)=(k,k)=kok=O. But not only does it lie in a surface of constant phase; it is also perpendicular to that surface! Any vector v in that surface must satisfy k \u00b0 v = (ii, v) = (dO, v) = 0 because it pierces no surfaces. Geometric optics assumes that the reduced wavelength A, as measured in a typical local Lorentz frame, is small compared to the scale e of inhomogeneities in the wave train and small compared to the radius of curvature of spacetime, qz. Thus, over regions much larger than A but smaller than e or qz, the waves are plane-fronted \u00a722.5. GEOMETRIC OPTICS IN CURVED SPACETIME 575 and monochromatic, and there exist Lorentz reference frames (Riemann normal coordinates). In one of these \"extended\" local Lorentz frames, the phase must be o= kax a + constant; no other expression will yield V 0 = k. The corresponding vector potential [equation (22.25)] will be AIL = R {aIL exp[i(k' x - k\u00b0t)]} + (\"post-geometric-optics corrections\"); hence, kO = 2'IT/(period of wave) = 2'ITv = w (angular frequency), Ikl = 2'IT/(wavelength of wave) = 1/A\" = w, k points along direction of propagation of wave. At each event in spacetime there is a wave vector; and these wave vectors, tacked end-on-end, form a family of curves-the \"light rays\" or simply \"rays\"-whose tangent vector is k. The rays, like their tangent vector,", " lie both in and perpendicular to the surfaces of constant phase. The affine parameter A of a ray (not to be confused with wavelength = 2'ITA\") satisfies k = d/dA; therefore it is given by A = t/ko + constant = t/w + constant, where t is proper time along the ray as measured, not by the ray itself (its proper time is zero!), but by the local Lorentz observer who sees angular frequency w. Thus, while w is a frame-dependent quantity and t is also a frame-dependent quantity, their quotient t/w when measured along the ray (not off the ray) is the frame-inde pendent affine parameter. For a particle it is possible and natural to identify the affine parameter A with proper time T. For a light ray this identification is unnatural and impossible. The lapse of proper time along the ray is identically zero. The springing up of A to take the place of the vanished T gives one a tool to do what one might not have suspected to be possible. Given a light ray shot out at event {f and passing through event gJ, one can give a third event e along the same null world line that is twice as \"far\" from {f as ~] is \"far,\" in a new sense of \"far\" that has nothing whatever directly to do with proper time (zero!), but is defined by equal increments of the affine parameter (Ae - A,,! = A,,; - Ad)' The \"affine parameter\" has a meaning for any null geodesic analyzed even in isolation. In this respect, it is to be distinguished from the so-called \"luminosity distance\" which is sometimes introduced in dealing with the propagation of radiation through curved spacetime, and which is defined by the spreading apart of two or more light rays coming from a common source. Maxwell's equations as explored in the text [equation (22.28')] guarantee that the complex polarization vector f is perpendicular to the wave vector k and that, therefore, it lies in a surface of constant phase (see drawing). Intuition into the polarization vector is developed in exercise 22.12. 22. THERMODYNAMICS, HYDRODYNAMICS,.... AND KINETIC THEORY 576 0(}2): ===>- k\u00b7 k = 0 (wave vector is null); ~kf3Jba - 2i( kf3aa;f3 + ~ kf3 ;f3a a) = 0 '4= OJ (22.30) - ~ (V. k)a (propagation equation for vector amplitude). (22.31) These equations (22.30, 22.31) together with equation (22.28) are the basis from which all subsequent results will follow. As a first consequence, one can obtain the geodesic law from equation (22.30). Form the gradient of k. k = 0, and use the fact that kf3 o;f3a = 0;af3 or 0. 13 is the gradient of a scalar to interchange indices, The main laws of geometric optics: The result is Vkk = 0 (propagation equation for wave vector). (22.32) (1) Light rays are null geodesics Notice that this is the geodesic equation! Combined with equation (22.30), it is the statement, derived from Maxwell's equations in curved spacetime, that light rays are null geodesics, the first main result of geometric optics. Turn now from the propagation vector k = VO to the wave amplitude a = af, and obtain separate equations for the magnitude a and polarization f. Use equation (22.31) to compute so aka = - ~ (V\u00b7 k)a (propagation equation for scalar amplitude). (22.33) Next write a = af in equation (22.31) to obtain \u00a722.5. GEOMETRIC OPTICS IN CURVED SPACETIME 577 or Vkf = 0 (propagation equation for polarization vector). (22.34) This together with equation (2228'), constitutes the second main result of geometric optics, that the polarization vector is perpendicular to the rays and is parallel-propa gated along the rays. It is now possible to see that these results, derived from equations (22.30) and (22.31) are consistent with the gauge condition (22.28). The vectors k and f, specified at one point, are fixed along the entire ray by their propagation equations. But because both propagation equations are parallel-tran sport laws, the conditions k\u00b7 k = 0, f\u00b7 i = 1, and k' f = 0, once imposed on the vectors at one point, will be satisfied along the entire ray. The equation (22.33) for the scalar amplitude can be reformulated", " as a conserva (k' V), one rewrites the equation as (k' V)a2 + a2V\u00b7 k = 0, tion law. Since Ok or (2) polarization vector is perpendicular to ray and is parallel propagated along ray V' (a 2k) = O. (22.35) (3) conservation of \"photon number\" Consequently the vector a 2k is a \"conserved current,\" and the integral f a2kIJ. d3I IJ. has a fixed, unchanging value for each 3-volume cutting a given tube formed of light rays. (The tube must be so formed of rays that an integral of a2k over the walls of the tube will give zero.) What is conserved? To remain purely classical, one could say it is the \"number of light rays\" and call a2k o the \"density of light rays\" on an X O = constant hypersurface. But the proper correspondence and more concrete physical interpretation make one prefer to call equation (22.35) the law of conserva tion ofphoton number. It is the third main result of geometric optics. Photon number, of course, is not always conserved; it is an adiabatic invariant, a quantity that is not changed by influences (e.g., spacetime curvature, -1/&/2) which change slowly (&/ ~ A) compared to the photon frequency. Box 22.4 summarizes the above equations of geometric optics, along with others derived in the exercises. Exercise 22.11. ELECTROMAGNETIC FIELD AND STRESS ENERGY EXERCISES Derive the equations given in part D of Box 22.4 for F, E, B, and T. POLARIZATION Exercise 22.12. At an event ':fo through which geometric-optics waves are passing, introduce a local Lorentz frame with :-axis along the direction of propagation. Then k = w(eo + ez)' Since the is f = fO(eo + ez) + fie, + f\"!.e y; and since polarization vector is orthogonal f\u00b7 j = I. it has If l 2 + If2i2 = I. to k. (a) Show that the componentfO of the polarization vector has no influence on the electric i.e., show that one can add a multiple and magnetic fields measured in the given frame; of k to f without affecting any physical measurements. it 1 (colllillued 011 page 581) 578 22. THERMODYNAMICS, HYDRODYNAMICS,..., AND KINETIC THEORY Box 22.4 GEOMETRIC OPTICS IN CURVED SPACETIME (Summary of Results Derived in Text and Exercises) A. Geometric Optics Assumption Electromagnetic waves propagating in a source-free region of spacetime are locally plane-fronted and monochromatic (reduced wavelength A ~ scale e over which amplitude, wavelength, or polarization vary; and A ~ ~'il ::: mean radius of curva ture of spacetime). B. Rays, Phase, and Wave Vector (see Box 22.3) Everything (amplitude, polarization, energy, etc.) is transported along rays; and the quantities on one ray do not influence the quantities on any other ray. The rays are null geodesics of curved spacetime, with tangent vectors (\"wave vectors\")k: The rays both lie in and are perpendicular to surfaces of constant phase, 0 ::: const.; and their tangent vectors are the gradient of 0: k ::: VO. In a local Lorentz frame, kO is the \"angular frequency\" and kO/2'77 is the ordinary frequency of the waves, and is a unit 3-vector pointing along their direction of propagation. C. Amplitude and Polarization Vector The waves are characterized by a real amplitude a and a complex polarization vector f of unit length, f\u00b7 j::: 1. (Of the fundamental quantities 0, k, a, f, all are real except f. See exercise 22.12 for deeper understanding of f.) The polarization vector is everywhere orthogonal to the rays, k\u00b7 f ::: 0; and is parallel-transported along them, Vkf::: O. The propagation law for the amplitude is \u00a722.5. GEOMETRIC OPTICS IN CURVED SPACETIME 579 This propagation law is equivalent to a law of conservation ofphotons (classically: of rays); a2k the \"conserved current\" satisfying V\u00b7 (a 2k) = 0; and (8'17n)-lfa2k\u00b0Vigld 3x is the number of photons (rays) in the 3-volume of integration", " on any XO = constant hypersurface, and is constant as this volume is carried along the rays. is The propagation law holds separately on each hypersurface of constant phase. There it can be interpreted as conservation of a a 2(f, where (f is a two-dimen sional cross-sectional area of a pulse of photons or rays. See exercise 22.13. D. Vector Potential, Electromagnetic Field, and Stress-Energy-Momentum At any event the vector potential in Lorentz gauge is where 1\\ denotes the real part. The electromagnetic field tensor is orthogonal to the rays, F\u00b7 k = 0, and is given by F = 1\\{iaei8k /\\ fl. The corresponding-electric and magnetic fields in any local Lorentz frame are E = 1\\{iakOei8(projection off perpendicular to k)}, B = n X E, where n k/ko. The stress-energy tensor, averaged over a wavelength, is corresponding to an energy density in a local Lorentz frame of and an energy flux of _~t~wSa~ = k/ko direction) with the speed of TO; = ToonL-~ - light. This is identical with the stress-energy tensor that would be produced by a beam of photons with 4-momenta p = nk. Conservation of energy-momentum V. T = 0 follows from the ray conservation law V \u2022 (a 2k) = 0 and the geodesic law Vkk (k\u00b7 V)k = 0: 8'17 V. T = V. (a 2k \u00ae k) = [V. (a2k)]k + a2(k\u00b7 V)k = O. 580 22. THERMODYNAMICS; HYDRODYNAMICS, ', \" AND KINETIC THEORY Box 22.4 (continued) The adiabatic (geometric optics) invariant \"ray number\" a2ko or \"photon number\" (8'17 Ii )-la2kO in a unit volume is proportional (8'17 )-la2(kO)2, divided by the frequency, kG-corresponding exactly to the harmonic oscillator adiabatic invariant E/w [Einstein (1912), Ehrenfest (1916), Landau and Lifshitz (1960)]. to the energy, E. Photon Reinterpretation of Geometric Optics The laws of geometric optics can be reinterpreted as follows. This reinterpretation becomes a foundation of the standard quantum theory of the electromagnetic field (see, e.g., Chapters 1 and 13 of Baym (1969)]; and the classical limit of that quantum theory is standard Maxwell electrodynamics. Photons are particles of zero rest mass that move along null geodesics of spacetime (the null rays). The 4-momentum of a photon is related to the tangent vector of the null ray (wave vector) by p = lik. A renormalization of the affine parameter, (new parameter) = (1/Ii) X (old parameter), makes p the tangent vector to the ray_ Each photon possesses a polarization vector, f, which is ()!1llOgonalto its 4-mo mentum (p \u2022 f = 0), and which it parallel-transports along its geodesic world line (Vpf = 0). A swarm of photons, all with nearly the same 4-momentum p and polarization vector f(as compared by parallel transport), make up a classical electromagnetic wave. The scalar amplitude a of the wave is determined by equating the stress-energy tensor of the wave- T = - a 2k \u00ae k = 8'17 1 8'17 1 (a)2 Ii p \u00ae P to the stress-energy tensor of a swarm of photons with number-flux vector S, [see equation (5.18)]. The result: T=p \u00aeS or, in any local Lorentz frame, a = (8'17li2S0/p 0)l/2 = (8'17) 1I21i ( number density of photons )112 energy of one photon \u00a722.5. GEOMETRIC OPTICS IN CURVED SPACETIME 581 (b) Show that the following polarization vectors correspond to the types of polarization listed: f = e \", linear polarization in x direction; f = e y, linear polarization in y direction; f = ~ (e\" + ie y), righthand circular polarization; f = ~ (e\" - ie y), lefthand circular polarization; f = ae\" + i(1 - ( 2)1/2ey, righthand elliptical polarization. (c) Show that the type", " of polarization (linear; circular; elliptical with given eccentricity of ellipse) is the same as viewed in any local Lorentz frame at any event along a given ray. [Hint: Use pictures and abstract calculations rather than Lorentz transformations and com ponent calculations.] Exercise 22.13. THE AREA OF A BUNDLE OF RAYS Write equation (22.31) in a coordinate system in which one of the coordinates is chosen to be X O =(), the phase (a retarded time coordinate). gation of a can be described within a single () = constant hypersurface. (a) Show that gOO = 0 and that no derivatives o;o(} appear in equation (22.33); so propa (b) Perform the following construction (see Figure 22.1). Pick a ray eo along which a is to be propagated. Pick a bundle of rays, with two-dimensional cross section, that (i) all lie in the same constant-phase surface as ($'0' and (ii) surround eo' (The surface is three-di- z k coo Bundle has area {f lying perpendicular to k. x (b) Figure 22.1. Geometric optics for a bundle of rays with two-dimensional cross section. all lying in a surface of constant phase. 0 = const. Sketch (a) shows the bundle, surrounding a central ray eo, in a spacetime diagram with one spatial dimension suppressed. Sketch (b) shows the bundle as viewed on a slice of simultaneity in a local Lorentz frame at the event ':1'0' Slicing the bundle turns each ray into a \"photon\"; so the bundk becomes a two-dimensional surface filled with photons. The area d of this photon-filled surface obeys the following laws (see exercises 22.13 and 22.14); (I) {f is independent of the choice of Lorentz frame; it depends only on location ':1'0 along the ray eo' (2) The amplitude a of the waves satisfies (\"conservation of photon flux\"). (3) d obeys the \"propagation equation\" (22.36). {fa2 = constant all along the ray eo 582 22. THERMODYNAMICS, HYDRODYNAMICS, '. \" AND KINETIC THEORY mensional, so any bundle filling it has a two-dimensional cross section.) At any event '3'0' in any local Lorentz frame there, on a \"slice of simultaneity\" X O = constant, measure the cross-sectional area {/ of the bundle. (Note: the area being measured is perpendicular to k in the three-dimensional Euclidean sense; it can be thought of as the region occupied momentarily by a group of photons propagating along, side by side, in the k direction.) Show that the area {/ is the same, at a given event ~i'o, regardless of what Lorentz frame is used to measure it; but the area changes from point to point along the ray eo as a result of the rays' divergence away from each other or convergence toward each other: (22.36) Then show that {/a 2 is a constant everywhere along the ray eo (\"conservation of photon flux\"). [Hints: (i) Any vector ( connecting adjacent rays in the bundle is perpendicular to k, because ( lies in a surface of constant 0 and k' ( = (ii, () = (dO, () = (change in 0 along 0 = O. (ii) Consider, for simplicity, a bundle with rectangular cross section as seen in a specific local Lorentz frame at a specific event '3'0 [edge vectors v and w with v\u00b7 w = 0 (edges perpendicular) and v\u00b7 eo = w\u00b7 eo = 0 (edges in surface of constant time) and v\u00b7 k = w\u00b7 k =0 (since edge vectors connect adjacent rays of the bundle)]. Show pictorially that in any other Lorentz frame at '3'0' the edge vectors are v' = v + exk and w' = w + 13k for some ex and {3. Conclude that in all Lorentz frames at '3'0 the cross section has identical shape and identical area, and is spatially perpendicular to the direction of propagation (k' v = k' W = 0). (iii) By a calculation in a local Lorentz frame show that 0k{/ = (V' k ){/. (iv) Conclude from aka = -!(V' k)a that 0k({/a2) = 0.] Exercise 22.14. FOCUSING THEOREM The cross-sectional area {/ of a bundle of rays all lying in the same surface of constant phase changes along the central", " ray of the bundle at the rate (22.36) (see Figure 22.1). (a) Derive the following equation (\"focusing equation\") for the second derivative of {/1/2 : d 2 {/1/2 = _ (1 0 12 + 1. R k\"'k(3){/1/2 d\"A 2 \"'f3 2'(22.37) where \"A is affine parameter along the central ray (k = did>\"), and the \"magnitude of the shear of the rays\", 101, is defined by the equation (22.38) [Hint: This is a vigorous exercise in index manipulations. The key equations needed in the manipulations are (/,,,,k'\" =(k\"';\",){/ [equation (22.36)]; k\"';f3kf3 =0 [geodesic equation (22.32) for rays]; k\"';f3 = kf3;\", [which follows from k\", = 0,,,,]; and the rule (16.6c) for interchanging covariant derivatives of a vector.] (b) Show that, in a local Lorentz frame where k = w(et + ez ) at the origin, (22.39) Thus, 101 2 is nonnegative, which justifies the use of the absolute value sign. (c) Discussion: The quantity 101 is called the shear of the bundle of rays because it measures the extent to which neighboring rays are sliding past each other [see, e.g., Sachs (1964)]. the focusing equation (22.37) says that shear focuses a bundle of rays (makes Hence, d 2{/1/ 2Id\"A 2 < 0); and spacetime curvature also focuses it if R\"'f3k\"'kf3 > 0, but defocuses it if R\"'f3 k\"'k f3 < O. (When a bundle of toothpicks, originally circular in cross section, is squeezed into an elliptic cross section, it is sheared.) \u00a722.6. KINETIC THEORY IN CURVED SPACETIME 583 (d) Assume that the energy density Too, as measured by any observer anywhere in space time, is nonnegative. By combining the focusing equation (22.37) with the Einstein field equation, conclude that 1/2 (fOr any bundle of rays, all in the same) d 2 ~2 :::;; 0 surfac~ of constant phase, anywhere in (22.40) spacetIme (focusing theorem). This theorem plays a crucial role in black-hole physics (\u00a734.5) and in the theory of singularities (\u00a734.6). \u00a722.6. KINETIC THEORY IN CURVED SPACETIME* The stars in a galaxy wander through spacetime, each on its own geodesic world line, each helping to produce the spacetime curvature felt by all the others. Photons, left over from the hot phases of the big bang, bathe the Earth, bringing with themselves data on the homogeneity and isotropy of the universe. Theoretical analyses of these and many other problems are unmanageable, if they attempt to keep track of the motion of every single star or photon. But a statistical description gives accurate results and is powerful. Moreover, for most problems in astrophysics and cosmology, the simplest of statistical descriptions-one ignoring collisions-is adequate. Usually collisions are unimportant for the large-scale behavior ofa system (e.g.; a galaxy), or they are so important that a fluid description is possible (e.g., in a stellar interior). _..-.... Consider, then, a swarm of particles (stars, or photons, or black holes, or...) that move through spacetime on geodesic world lines, without colliding. Assume, for simplicity, that the particles all have the same rest mass. Then all information of a statistical nature about the particles can be incorporated into a single function, the \"distribution junction\" or \"number density in phase space\", 91:. Define 91: in terms of measurements made by a specific local Lorentz observer at a specific event 90 in curved spacetime. Give the observer a box with 3-volume 0/\" (and with imaginary walls). Ask the observer to count how many particles, N, are inside the box and have local-Lorentz momentum components pi in the range Volume in phase space for a group of identical particles... pI - 2\" J.p' < p' < pI + 2\" J.p'... I I (He can ignore the particle energies po; since all particles have the Same rest mass m, energy \u00b7For more detailed and sophisticated treatments of this topic, see, e.g., Tauber and Weinberg (1961", "), and Lindquist (1966), Marie (1969), Ehlers (1971), Stewart (1971), Israel (1972). and references cited therein. Ehlers (1971) is a particularly good introductory review article. y 584 22, THERMODYNAMICS, HYDRODYNAMICS, ', \" AND KINETIC THEORY is fixed uniquely by momentum.) The volume in momentum space occupied by the N particles is 'Vp = Jpr Jp'J Jpz; and the volume in phase space is (22.41) Lorentz invariance of volume in phase space Other observers at ~,}'o, moving relative to the first, will disagree on how much spatial volume 'Yr and how much momentum volume 'Vp these same N particles occupy: 'Y Z and 0/p depend on the choice of Lorentz frame. (22.42) However, all observers will agree on the value of the product 0/ in phase space\"): 'Yr'Y p (\"volume The phase-space volume.y occupied by a given set of N identical particles at a given event in spacetime is independent of the local Lorentz frame in which it is measured. (22.43) (See Box 22.5 for proof.) Moreover, as the same N particles move through spacetime along their geodesic world lines (and through momentum space), the volume 0/ they span in phase space remains constant: Liouville's theorem (conservation of volume in phase space) The 0/ occupied by a given swarm of N particles is independent of location along the world line of the swarm (\"Liouville's theorem in curved spacetime\"). (22.44) Number density in phase space (distribution function) (See Box 22.6 for proof.) More convenient for applications than the volume!V in phase space occupied by a given set of N particles is the \"number density in phase space\" (\"distribution function\") in the neighborhood of one of these particles: 9l = Njo/. On what does this number density depend? It depends on the location in space time, 9, at which the measurements are made. It also depends on the 4-momentum p of the particle in whose neighborhood the measurements are made. But because the particles all have the same rest mass, p cannot take on any and every value in the tangent space at 9. Rather, p is confined to the \"forward mass hyperboloid\" at 9: (22.45) p lies inside future light cone. Thus, 9l = 9l. 9 (.ocatlon,. m spacetime ~ I (4-momentum p, which must lie )~,), on the forward mass hyperboloid. (22.46) of the tangent space at 9 Pick some one particle in the swarm, with geodesic world line 9(A.) [A. = (affine parameter) = (proper time, if particle has finite rest mass)], and with 4-momentum -----~~ \u00a72~6. KINETIC THEORY IN CURVED SPACETIME -. ~ 585 Box 22.5 VOLUME IN PHASE SPACE A. For Swarm of Identical Particles with Nonzero Rest Mass Pick an event 9 0, through which passes a particle named \"John\" with a 4-momentum named \"P\". In John's local Lorentz rest frame at 9 0 (\"barred frame\", S), select a small 3-volume, 'V\" Li.i Liy Liz, containing him. Also select a small \"3-volume in momentum space,\" 'Vp Lipi Lipii Lipz centered on John's momentum, which is pi = pii = pz = O. Focus attention on all particles whose world lines pass through 'Viand which have momenta pT in the range 'Vp surrounding pT = O. I x ----~x Examine this bundle in another local Lorentz frame (\"unbarred frame\", S) at 9 0, which moves with speed {3 relative to the rest frame. Orient axes so the relative motion of the frames is in the x and x directions. Then the space volume 'V-\" occu pied in the new frame has Liy = Liy, Liz = Liz (no effect of motion on transverse directions), and Lix = (1 - {32)1/2 Lix (Lorentz contraction in lon gitudinal direction). Hence 'V\" = (1 - {32)1/2'Vi (\"transformation law for space volumes\") or, equivalently [since po = ml(l - {32)1/2]: p0'V = m'l!_ = (constant, independent). \" \" o f Lorentz frame A momentum-space diagram, analogous to the spacetime diagram, depicts the momentum spread for particles in the", " area. Same is true of z _ p'. p(A). Examine the density in phase space in this particle's neighborhood at each point along its world line: 9l = 9l[9(A), p(A)]. Calculate 9l(A) as follows: (1) Pick an initial event 9(0) on the world line, and a phase-space volume 'V containing the particle. (2) Cover with red paint all the particles contained in 'Vat 9(0). (3) Watch the red particles move through spacetime alongside the initial particle. (4) As they move, the phase-space region they occupy changes shape extensively; but its volume 'V remains fixed (Liouville's theorem). Moreover, no particles can enter or leave that phase-space region (once in, always in; once out, always out; boundaries of phase-space region are attached to and move with the particles). (5) Hence, at any A along the initial particle's world line, the particle is in a phase-space region of unchanged volume 'V, unchanged number of particles N, and unchanged ratio 9l = NI'V: d91[9(A), p(A)] _ 0 -. dA (22.47) Collisionless Boltzmann equation (kinetic equation) This equation for the conservation of'3l along a particle's trajectory in phase space is called the \"collisionless Boltzmann equation,\" or the \"kinetic equation.\" Photons provide an important application of the Boltzmann equation. But when discussing photons one usually does not think in terms of the number density in phase space. Rather, one speaks of the \"specific intensity\" Iv of radiation at a given frequency v, flowing in a given direction. n, as measured in a specified local Lorentz frame: I = v - d(energy) d(time) d(area) d(frequency) d(solid angle) (22.48) Distribution function for photons expressed in terms of specific intensity, Iv Invariance and conservation of 1./1'3 588 22. THERMODYNAMICS, HYDRODYNAMICS,..., AND KINETIC THEORY (See Figure 22.2). A simple calculation in the local Lorentz frame reveals that (22.49) where h is Planck's constant (see Figure 22.2). Thus, if two different observers at the same or different events in spacetime look at the same photon (and neighboring photons) as it passes them, they will see different frequencies v (\"doppler shift,\" \"cosmological red shift,\" \"gravitational redshift\"), and different specific intensities I,,; but they will obtain identical values for the ratio I v/v 3 \u2022 Thus I v/v 3, like ~, is invariant from observer to observer and from event to event along a given photon's world line. EXERCISES Exercise 22.15. INVERSE SQUARE LAW FOR FLUX The specific flux of radiation entering a telescope from a given source is defined by (22.50) where integration is over the total solid angle (assumed ~ 4,,) sub tended by the source on the observer's sky. Use the Boltzmann equation (conservation of I v/v 3 to show that F. a: (distance from sourcet 2 for observers who are all at rest relative to each other in flat spacetime. ) Exercise 22.16. BRIGHTNESS OF THE SUN Does the surface of the sun look any brighter to an astronaut standing on Mercury than to a student standing on Earth?.- Exercise 22.17. BLACK BODY RADIATION An \"optically thick\" source of black-body radiation (e.g., the surface of a star, or the hot matter filling the universe shortly after the big bang) emits photons isotropically with a specific intensity, as seen by an observer at rest near the source, given (Planck radiation law) by 2hv3 Iv = ehv/kT _ 1. (22.51) Here T is the temperature of the source. Show that any observer, in any local Lorentz frame, anywhere in the universe, who examines this radiation as it flows past him, will also see a black-body spectrum. Show, further, that if he calculates a temperature by measuring the specific intensity Iv at anyone frequency, and if he calculates a temperature from the shape of the spectrum, those temperatures will agree..(Radiation remains black body rather than being \"diluted\" into \"grey-body.\") Finally, show that the temperature he measures is red shifted by precisely the same factor as the frequency of any given photon is redshifted, TObserved =(Vobserved) for a given photon. Temitted vemitted (22.52) [Note that the redshifts can be \"Doppler\"", "location on the world tube of a bundle of particles. (b) qJ[ can be regarded as a function oflocation '3' in spacetime and 4-momentum p inside the future light cone of the tangent space at '3': (22.56) (c) qJ[ satisfies the collisionless Boltzmann equation (kinetic equation) dqJ[['3'(A), p(A\u00bb) = 0 dA along geodesic trajectory of any particle. (22.57) (d) ~ can be rewritten in a local Lorentz frame as (22.58) (e) The stress-energy tensor at an event '3' can be written as an integral over the interior of the future light cone of momentum space p' =f (qJ[pl'p')m- 1 dpo dpl dp 2 dp 3 in a local Lorentz frame (Track-1 notation for integral; see Box 5.3); T =f (qJ[p \u00ae p)m- 1 *1 =f (qJ[p \u00ae p)m- 1 dpo 1\\ dpl 1\\ dp 2 1\\ dp 3 in frame-independent notation in a local Lorentz frame (Track-2 notation; see Box 5.4). (22.59) (22.59') PART V RELATIVISTIC STARS Wherein the reader, armed with the magic potions and powers of Geometrodynamics, conquers the stars. -----~ CHAPTER 23 SPHERICAL STARS \u00a723.1. PROLOG Beautiful though gravitation theory may be, it is a sterile subject until it touches the real physical world. Only the hard reality of experiments and of astronomical observations can bring gravitation theory to life. And only by building theoretical models of stars (Part V), of the universe (Part VI), of stellar collapse and black holes (Part VII), of gravitational waves and their sources (Part VIII), and of gravitational experiments (Part IX), can one understand clearly the contacts between gravitation theory and reality. The model-building in this book will follow the tradition of theoretical physics. Each Part (stars, universe, collapse,...) will begin with the most oversimplified model conceivable, and will subsequently add only those additional touches of realism necessary to make contact with the least complex of actual physical systems. The result will be a tested intellectual framework, ready to support and organize the additional complexities demanded by greater realism. Greater realism will not be attempted in this book. But the reader seeking it could start in no better place than the two-volume treatise on Relativistic Astrophysics by Zel'dovich and Novikov (1971, 1974). Begin, now, with models for relativistic stars. As a major simplification, insist (initially) that all stars studied be static. Thereby exclude not only exploding and pulsating stars, but even quiescent ones with stationary rotational motions. From the static assumption, plus a demand that the star be made of \"perfect fluid\" (no shear stresses allowed!), plus Einstein's field equations, it probably follows that the star is spherically symmetric. However, nobody has yet given a proof. [For proofs under more restricted assumptions, see Avez (1964) and Kunzle (1971).] In the absence of a proof, assume the result: insist that all starS studied be spherical as well as static. Preview of the rest of this book Static stars must be spherical 594 23. SPHERICAL STARS \u00a723.2. COORDINATES AND METRIC FOR A STATIC. SPHERICAL SYSTEM Metric for any static, spherical system: To deduce the gravitational field for a static spherical star-or for any other static. spherical system-begin with the metric of special relativity (no gravity) in the spherically symmetric form where (23.1 ) (23.2) (1) generalized from flat spacetime Try to modify this metric to allow for curvature due to the gravitational influence of the star, while preserving spherical symmetry. The simplest and most obvious guess is to allow those metric components that are already non-zero in equation (23.1) to assume different values: (23.3) where f/J, A, and R are functions of r only. (The static assumption demands ag/lvl at = 0.) To verify that this guess is good, use it in constructing stellar models, and check that the resulting models have the same generality (same set of quantities freely specifiable) as in Newtonian theory and as expected from general physical considerations. An apparently more general metric (23.4) actually is not more general in any physical sense. One can perform a coordinate transformation to a new time coordinate t' defined by ef/J dt' = a dt + b dr. (23.5) By inserting this in", " equation (23.4), and by defining e2A = b 2 + c2, one obtains the postulated line element (23.3), apart from a prime on the t.* The necessity to allow for arbitrary coordinates in general relativity may appear burdensome when one is formulating the theory; but it gives an added flexibility, something one should always try to turn to one's advantage when formulating and solving problems. The grt = 0 simplification (called a coordinate condition) in equa tion (23.3) results from an advantageous choice of the t coordinate. The r coordinate, however, is also at one's disposal (as long as one chooses it in a way that respects spherical symmetry; thus not r' = r + cos 8). One can turn this freedom to advan tage by introducing a new coordinate r'(r) defined by r' = R(r). (23.6) *Of course, equation (23.5) only succeeds in defining a new time coordinate t' if it is integrable as a differential equation for t'. By choosing the integrating factor e~ to be just e~ = a(r), one sees that t' = t + flb(r)/a(r\u00bb) dr is the integral of (23.5); thus the required t' coordinate always exists, no matter what the functions a(r). b(r). c(r), and R(r) in equation (23.4) may be. (2) specialized to \"Schwarzschild form\" \u00a723.3. PHYSICAL INTERPRETATION OF SCHWARZSCHILD COORDINATES 595 With this choice of the radial coordinate, and with the primes dropped, equation (23.3) reduces to (23.7) a line element with just two unknown functions, l/J(r) and A(r). This coordinate system and metric have been used in most theoretical models for relativistic stars since the pioneering work of Schwarzschild (l916b), Tolman (1939), and Oppen heimer and Volkoff (1939). These particular coordinates are sometimes called \"cur vature coordinates\" and sometimes \"Schwarzschild coordinates.\" The central idea of these coordinates, in a nutshell, is (Schwarzschild r-coordinate) = (proper circum ference)/2'7T. For a more rigorous proof that in any static spherical system Schwarzschild coordinates can be introduced, bringing the metric into the simple form (23.7), see Box 23.3 at the end of this chapter. (3) derived more rigorously ISOTROPIC COORDINATES AND NEWTONIAN LIMIT Exercise 23.1. An alternative set of coordinates sometimes used for static, spherical systems is the \"isotropic coordinate system\" (t, r, 8, cf\u00bb. The metric in isotropic coordinates has the form EXERCISE with l/J and p. being functions of r. (a) Exhibit the coordinate transformation connecting the Schwarzschild coordinates (23.7) to the isotropic coordinates (23.8). (b) From equation (16.2a) [or equivalently (18.l5c\u00bb), show that, in the Newtonian limit, the metric coefficient l/J of the isotropic line element becomes the Newtonian potential; and p. becomes equal to -l/J. By combining with part (a), discover that A = rdl/Jjdr in the Newtonian limit. (23.8) \u00a723.3. PHYSICAL INTERPRETATION OF SCHWARZSCHILD COORDINATES In general relativity, because the use of arbitrary coordinates is permitted, the physical significance of statements about tensor or vector components and other quantities is not always obvious. There are, however, some situations where the interpretation is almost as straightforward as in special relativity. The most obvious example is the center point of a local inertial coordinate system, where the principle of equivalence aliows one to treat all involving spacetime curvature) exactly as in special relativity. Schwarzschild coordinates for a spherical system turn out to be a second example. local quantities (quantities not One's first reaction when meeting a new metric should be to examine it, not in order to learn about the gravitational field, for which the curvature tensor is more The form of any metric can reveal the nature of the coordinates being used Geometric significance of the Schwarzschild coordinates: (1) 8, cf> are angles on sphere 596 23. SPHERICAL STARS directly informative, but to learn about the coordinates. (Are they, for instance, locally inertial at some point?) The names given to the coordinates have no intrinsic significance. A coordinate transformation t' = O. r' = 9, 0' = r, 9' = t is perfectly permissible, and has no influence on the physics or the mathematics", " of a relativistic problem. The only thing it affects is easy communication between the investigator who adopts it and his colleagues. Thus the names trOcp for the Schwarzschild coordinates (23.7) provide the geometric content of the coordinates.* In a mnemonic device pointing out particular, the names 0, \u00a2 are justified by the fact that on each two-dimensional surface of constant rand t, the distance between two nearby events is given by ds 2 = r 2 dil 2, as befits standard 0, cp coordinates on a sphere of radius r. The area of this two-dimensional sphere is clearly A = f(rdO)(rsinOdcp) = 4'7Tr 2 ; (23.9) (2) r measures surface area of sphere hence, the metric (23.7) tells how to measure the r coordinate that it employs. One can merely measure (in proper length units) the area A of the sphere, composed of all points rotationally equivalent to the point tj' for which the value r('!J) is desired; and one can then calculate (3) t has 3 special geometric properties (4) description of a \"machine\" to measure t r('!J) = (proper are~ of Spherej4'7T)l/2. through pomt tj' (23.9') The Schwarzschild coordinates have been picked for convenience, and not for the ease with which one could build a coordinate-measuring machine. This makes it more difficult to design a machine to measure t than mllchines to measure r, 0, cp. The geometric properties of t on which a measurIng device can be based are: (1) the time-independent distances (ogap/Ot = 0) between world lines of constant r, 0, cp; (2) the orthogonality (gtr = gtn = gt</> = 0) of these world lines to the t = constant hypersurfaces; and (3) a labeling of these hypersurfaces by Minkowski (special relativistic) coordinate time at spatial infinity, where spacetime becomes flat. This labeling produces a constraint cP(oo) = 0 (23.10) in the metric (23.7). [Mathematically, this constraint is imposed by a simple rescaling transformation t' = e<l>('X1t, and by then dropping the prime.] One \"machine\" design which constructs (mentally) such a t coordinate, and in the process measures it, is the following. Observers using radar sets arrange to move along the coordinate lines r, 0, cp = const. They do this by adjusting their velocities until each finds that the radar echos from his neighbors, or from \"benchmark\" reference points in the asymptotically flat space, require the same round-trip time at each repetition. Equivalently, each returning echo must show zero doppler shift; *For an example of misleading names. consider those in the equation ds2 = _ e2~(9') d<p'2 + e2A(I1') d!J'2 + 8'2 (dt'2 + sin2 t' dr'2), which is equivalent to equation (23.7). but employs the coordinates t' = 8, r' = <p, 8' = r, <p' = t. \u00a723.4. DESCRIPTION OF MAnER INSIDE A STAR 597 it must return with the same frequency at which it was sent out. Next a master clock is set up near spatial infinity (far from the star). It is constructed to measure proper time-which, for it, is Minkowski time \"at infinity\"-and to emit a standard one Hertz signal. Each observer adjusts the rate of his \"coordinate clock\" to beat in time with the signals he receives from the master clock. To set the zero of his \"coordinate clock,\" now that its rate is correct, he synchronizes with the master clock, taking account of the coordinate time ~t required for radar signals to travel from the master to him. [To compute the transit time, he assumes that for radar signals (treflection treflection) = ~t, so that the echo is obtained by time-inversion temission) = (treturn about the reflection event. This time-reversal invariance distinguishes the time t in the metric (23.7) from the more general t coordinates allowed by equation (23.4).] Each observer moving along a coordinate line (r, 8, cp = const.) now has a clock that measures coordinate time t in his neighborhood. The above discussion identifies the Schwarzschild coordinates of equation (23.7) by their intrinsic geometric properties. Not only are rand t radial and time variables, respectively (in that a/or and a/at are spacel", "ike and timelike, respectively, and are orthogonal also to the spheres defined by rotational symmetry), but they have particular properties [4'17r2 = surface area; ogJl.lot = 0; a/or' a/at = grt = 0; a/at\u00b7 a/at = gtt = -1 at r = 00] that distinguish them from other possible coordi nate choices [r' = f(r), t' = t + F(r)]. No claim is made that these are the only coordinates that might reasonably be called rand t; for an alternative choice (\"isotropic coordinates\"), see exercise 23.1. However, they provide a choice that is reasonable, unambiguous, useful, and often used. Other coordinates are possible, but Schwarzschild are particularly simple \u00a723.4. DESCRIPTION OF THE MATTER INSIDE A STAR To high precision, the matter inside any star is a perfect fluid. (Shear stresses are negligible, and energy transport is negligible on a \"hydrodynamic time scale.\") Thus, it is reasonable in model building to describe the matter by perfect-fluid parameters: Material inside star to be idealized as perfect fluid p = p(r) = density of mass-energy in rest-frame of fluid; p = p(r) = isotropic pressure in rest-frame of fluid; n = n(r) = nu,?ber density of baryons in rest-frame of fluid; uJl = uJl(r) = 4-velocity of fluid; TJlP = (p + p)uJluP + pgJlP = stress-energy tensor of fluid. (23.11 ) (23.12) Parameters describing perfect fluid: (1) p, p, n (For Track-l discussion, see Box 5.1; for greater Track-2 detail, see \u00a7\u00a722.2 and 22.3.) In order that the star be static, each element of fluid must remain always at rest in the static coordinate system; Le., each element must move along a world line of constant r, 8, cp; i.e., each element must have 4-velocity components u.p = d\u00a2/dT = O. ufJ = d8/dT = 0, uT = dr/dT = 0, (23.13a) (2) u 598 23. SPHERICAL STARS The normalization of 4-velocity, (3) T and this, together with the general form (23.12) of the stress-energy tensor and the form (23.7) of the metric, determines pI': (23.13b) Ta{3 = 0 if a f. p. T\u00a2.p = pr- 2 sin- 2 8, (23.14) Although these components of the stress-energy tensor in Schwarzschild coordi nates are useful for calculations, the normalization factors e- 2<fJ, e- 211, r- 2, r- 2 sin- 2 8 make them inconvenient for physical interpretations. More convenient are compo tetrads carried by the fluid elements (\"proper reference nents on orthonormal frames\"; see \u00a713.6): Proper reference frame of fluid Components of u and T in proper reference frame 1 0 e; =? or' w; = ell dr, 1 0 e O=;a1i' w O=rd8, 0 1 e\u00a2, = r sin 8 a;p; w\u00a2' = r sin 8 dep; ut = 1, ~i= Too = p, Tafi = 0 if a f. p. (23.15a) (23.15b) (23.15c) (23.15d) See exercise 23.2 below. The structure of a star-i.e., the set of functions ifJ(r), A(r), p(r), p(r), n(r)-is determined in part by the Einstein field equations, Gil' = 87iP', and in part by the law of local conservation of energy-momentum in the fluid, TJlV;. = O. However, these are not sufficient to fix the structure uniquely. Also necessary is the functional dependence of pressure p and density p on number density of baryons n: p = p(n), p = p(n). (23.16) Equation of state: (1) in general (2) idealized to.. on e-parameter form\" p = p(n). p = p(n) \u2022 Normally one cannot deduce p and p from a knowledge solely of n. One must know, in addition, then the laws of thermodynamics plus equations of state will determine all remaining thermodynamic variables: the temperature T or the entropy per baryon s; p", " = p(n, s), p = p(n, s),.... (See \u00a722.2 and Box 22.1 for full Track-2 discussions.) To pass from the given thermo dynamic knowledge, p(n, s) and p(n, s), to the desired knowledge, p(n) and p(n), one needs information about the star's thermal properties, and especially about the way in which energy\u00b7 generation plus heat flow have conspired to distribute the entropy, S = s(n): p(n) = p[n, s(n)], p(n) = p[n, s(n)]. Justification for idealized equation of state: (1) in neutron stars \u00a723.4. DESCRIPTION OF MATTER INSIDE A STAR 599 There exist three important applications of the theory of relativistic stars: neutron stars, white dwarfs, and supermassive stars (stars with M ~ 103M0, which may exist according to theory, but the existence of which has never yet been confirmed by observation). In all three cases, happily, the passage from p = p(n, s), p(n, s), to p = p(n), p = p(n), is trivial. Consider first a neutron star. Though hot by ordinary standards, a neutron star is so cold by any nuclear-matter scale of temperatures that essentially all its thermal degrees of freedom are frozen out (\"degenerate gas\"; \"quantum fluid\"). It is not important that a detailed treatment of the substance of a neutron star is beyond the capability of present theory (allowance for the interaction between baryon and baryon; production at sufficiently high pressures of hyperons and mesons). The simple fact is that one is dealing with matter at densities comparable to the density of matter in an atomic nucleus (2 X 1014 g/cm3) and higher. Everything one knows about nuclear matter [see, for example, Bohr and Mottelson (1969)] tells one that it is degenerate, and that one can estimate in order of magnitude its degeneracy temperature by treating it as though it were an ideal Fermi neutron gas. (In a normal atomic nucleus, a little more than 50 per cent of all baryons are neutrons, the rest are protons; in a neutron star, as many as 99 per cent are neutrons.) When approxi mating the neutron-star matter as an ideal Fermi neutron gas, one considers the neutrons to occupy free-particle quantum states, with two particles of opposite spin in each occupied state, and a sharp drop from 100 per cent occupancy of quantum states to empty states when the particle energy rises to the level of the \"Fermi energy\" [for more on such an ideal Fermi gas, see Kittel, Section 19 (1958); or at an intro ductory level, see Sears, Section 16-5 (1953)]. In matter at nuclear density, the Fermi energy is of the order EFermi - 30 MeV or 3 X 1011 K; and at higher density the temperature required to unfreeze the degeneracy is even greater. In other words, for matter at and above nuclear densities, already at zero temperature the kinetic energy of the particles (governed by the Pauli exclusion principle and by their Fermi energy) is a primary source of pressure. Nuclear forces make a large correction to this pressure, but for T ~ 30 MeV = 3 X 1011 K, energies of thermal agitation do not. A star, in collapsing from a normal state to a neutron-star state (see Chapter 24), temperatures ~ 1010 K, and thereby cools to emits a huge flux of neutrinos at T ~ 3 X lOll K within a few seconds after formation. Consequently, in all neutron stars older than a few seconds one can neglect thermal contributions to the pressure and density; i.e., one can set p(n, s) = p(n, S = 0) =p(n), p(n, s) = p(n, S = 0) = p(n). A white dwarf is similar, except that here electrons rather than neutrons are the source of Fermi gas pressure and degeneracy. Typical white-dwarf temperatures satisfy (2) in white dwarfs kT ~ EFermielectrons; (3) in supermassive stars 600 23. SPHERICAL STARS the Fermi kinetic energy (Pauli exclusion principle), and not random kT energy, is primarily responsible for the pressure and energy density; and one can set p(n, s) = p(n, S = 0) =p(n), \u00a31(n, s) = p(n, S = 0) = p(n). In a supermassive star (see Chapter 24), the situation is quite different. There temperature and entropy are almost the whole story", ", so far as pressure and energy density are concerned. However, convection keeps the star stirred up and produces a uniform entropy distribution S = const. independent of radius; so one can write p(n, s) =ps(n), l{functions depending on un.iform entropy per baryon, s, In the star p(n, s) = p.(n). J-1 In all three cases-neutron stars, white dwarfs, supermassive stars-one regards the relations p(n) and p(n) as \"equations of state\"; and having specified them, one can calculate the star's structure without further reference to its thermal properties. EXERCISE Exercise 23.2. PROPER REFERENCE FRAMES OF FLUID ELEMENTS (a) Verify that equations (23.l5a,b) define an orthonormaltetrad and its dual basis ofl-forms, at each event in spacetime. (b) Verify that the components of the fluid 4-ve1ocity relative to these tetrads are given by equations (23.l5c). Why do these components guarantee that the tetrads form \"proper reference frames\" for the fluid elements? (c) Verify equations (23.l5d) for the components of the stress-energy tensor. Five equations needed to determine 5 stellar-structure functions: CP, A, p, p, n \u00a723.5. EQUATIONS OF STRUCTURE The structure of a relativistic star is determined by five functions of radius r: the metric functions ifJ(r), A(r), the pressure p(r), the density of mass-energy p(r), and the number density of baryons, n(r). Hence, to determine the structure uniquely, one needs five equations of structure, plus boundary conditions. Two equations of structure, the equations of state p(n) and p(n), are already in hand. The remaining three must be the essential content of the Einstein field equations and of the law of local energy-momentum conservation, TP.\u00b7;. = O. One knows that the law of local energy-momentum conservation for the fluid follows as an identity from the Einstein field equations. Without loss of information, \u00a723.5. EQUATIONS OF STRUCTURE 601 one can therefore impose all ten field equations and ignore local energy-momentum conservation. But that is an inefficient way to proceed. Almost always the equations p,. '. = 0 can be reduced to usable form more easily than can the field equations. He~ce, the most efficient procedure is to: (1) evaluate the four equations p';. = 0; (2) evaluate enough field equations (six) to obtain a complete set (6 + 4 = 10); and (3) evaluate the remaining four field equations as checks of the results of (1) and (2). The Track-2 reader has learned (\u00a722.3) that the equations p';. = 0 for a perfect J!!iid take on an especially simple form when projected (1) on the 4-velocity U of the-fluid itself, ancf(2) orthogonal to II. Projection along u (Up.P\u00b7;. = 0) gives the local law of energy conservation (22.11 a), p + p dn dp dT = -(p +p)V'u = -n-dT' where u = djdT; i.e., T is proper time along the world line of any chosen element of the fluid. For a static star, or for any other static system, both sides of this equation must vanish identically (no fluid element ever sees any change in its own density). Projection of TP.\u00b7;. = 0 orthogonal to u gives the reasonable equation ------------------ --\"\"-\"-(mertiarmass. per umt volume Le., -)-- (4 - X -acce eratlOn = - 1--. ) (pressure gradient, prOjected). perpendicular to u The most efficient procedure for solving Einstein equations Equation of hydrostatic equilibrium derived [see equation (22.13)]. When applied to a static star, this equation tells how much pressure gradient is needec! to prevent a fluid element from falling. Only the radial component of this equation has content, since the pressure depends only on r. The radial component in. the Schwarzschild coordinate system says [see the line element (23.7) and the 4~velocity components (23.13)], -----~- (p + p)ur;.u\u00b7 = -(p + p)Far.uau\u00b7 = -(p + p)FOrOuouo = (p + p)tP,r = -P,r' (23.17) (Track-l readers can derive this from scratch at the", " end of the section, exercise 23.3.) (since goo = In the Newtonian limit, _e 2 </J ::::: -1 - 2tP), and the pressure becomes much smaller than the mass-energy density; consequently equation (23.17) becomes tP becomes the Newtonian potential ptP,r = -P,r' This is the Newtonian version of the equation describing the balance between gra vitational force and pressure gradient. (23.l7N) The pressure gradient that prevents a fluid element from falling appears in Einstein's theory as the source of an acceleration. This acceleration, by keeping the fluid element at a fixed r value, causes it to depart from geodesic motion (from \"fiducial world line\"; from motion of free fall into the center of the star). Newtonian Comparison of Newton and Einstein views of hydrostatic equilibrium Equation for A derived 602 23. SPHERICAL STARS theory, on the other hand, views as the fiducial world line the one that stays at a fixed r value. It regards the \"gravitational force\" as trying (without success, because balanced by the pressure gradient) to pull a particle from a fixed-r world line onto a geodesic world line. In the two theories the magnitudes of the acceleration, whether \"actually taking place\" (Einstein theory) or \"trying to take place\" (Newtonian theory), are the same to lowest order (but opposite in direction); so it is no surprise that (23.17) and (23.17N) differ only in detail. Turn next to the Einstein field equation. Here, as is often the case, the components of the field equation in the fluid's orthonormal frame [equations (23.l5a,b)] are simpler than the components in the coordinate basis. One already knows the stress energy tensor Tafi in the orthonormal frame [equation (23.15d)]; and Track-2 readers have already calculated the Einstein tensor Gafi (exercise 14.13; Track-l readers will face the task at the end of this section, exercise 23.4). All that remains is to equate Gafi to 8'lTTafi' Examine first the 00 component of the field equations: Goo = r- 2 - r- 2e-2A - r-1(d/dr)(e- 2A) = r- 2(d/dr)[r(l - e-2A)] = 8'lTToo = 8'lTp. This equation becomes easy to solve as soon as one notices that it is a differential equation linear in the quantity e-2A ; a bit of tidying up then focuses attentiorion the quantity r(l - e-2A). Give this quantity the name 2m(r) (so far only a name!); thus, In this notation the 00 component of the Einstein tensor becomes Integrate and find 2 dm(r) Goo =- - - = 8'lTp. r 2 dr m(r) = IT 4'lTr2p dr + m(O). o (23.18) (23.19) For the constant of integration m(O), a zero value means a space geometry smooth at the origin (physically acceptable); a non-zero value means a geometry with a singularity at the origin (physically unacceptable: no local Lorentz frame at r = 0): ds 2 = [l - 2m(0)/rr 1dr2 + r 2(d()2 + sin2() dep2) ::::: -[r/2m(0)] dr2 + r 2(d()2 + sin2() dep2) ds 2 = [l - (8'lT/3)Pcr2]-1 dr 2 + r 2(d()2 + sin2() dep2) at r::::: 0 ifm(O) f. 0; (23.20) ::::: dr2 + r 2(d()2 + sin2() dep2) at r::::: 0 ifm(O) = O. \"Mass-energy inside radius r,\" m(r). defined The quantity m(r), defined by equation (23.18) and calculated from equation (23.19) with m(O) = 0, is a relativistic analog of the \"mass-energy inside radius r.\" Box 23.1 spells out the analogy in detail. \u00a723.5. EQUATIONS OF STRUCTURE 603 Box 23.1 MASS-ENERGY INSIDE RADIUS r The total mass-energy M of an isolated star is well-defined (Chapter 19). But not well-defined, in general, is the distribution of that mass-energy from point to point inside the star and in its gravitational field (no unique \"gravitational stress-", "energy tensor\"). This was the crucial message of \u00a720A (Track 2). lJie-mes~ is lme--m general. But for the case of a spherical star-and only for that case-the message loses its bite. Spherical symmetry allows one to select a distribution of the total mass-energy that is physically reasonable. In Schwarzschild coordinates, it is defined by \"total mass-energy inside radius r\" = m(r) =I 4'17r 2p dr. r. 0 (1) The fully convincing argument for this definition is found only by considering a generalization of it to time-dependent spherically symmetric stars (pulsating, col lapsing, or exploding stars; see Chapters 26 and 32, and especially exercise 32.7). For them one finds that the mass-energy m associated with a given ball of matter (fixed baryon number) can change in time only to the extent that locally measurable energy fluxes can be detected at the boundary of the ball. [Such energy fluxes could be the power expended by pressure forces against the moving boundary surface, or heat fluxes, or radiation (photon or neutrino) fluxes. But since spherically symme tric gravitational waves do not exist (Chapters 35 and 36), neither physical intuition nor Einstein's equations require that problems oflocalizing gravitational-wave energy be faced.] Thus the energy m is localized, not by a mathematical convention, but by the circumstance that transfer of energy (with this definition of m) is detectable by local measurements. [For the mathematical details of m(r, t) in the time-depend ent case, see Misner and Sharp (1964), Misner (1965), and exercise 32.7.] In addition to the critical \"local energy flux\" property of m(r) described above, there are three further properties that verify its identification as mass-energy. They are: (1) Everywhere outside the star ( ) _ M = (total mass-energy of star as measured frOm).'- Kepler's third law for distant planets m r - (2) see \u00a723.6 for proof. (2) For a Newtonian star, where \"mass inside radius r\" has a unique meaning, m(r) is that mass. (3) For a relativistic star, m(r) splits nicely into \"rest mass-energy\" mo(r) plus \"internal energy\" U(r) plus \"gravitational potential energy\" il(r). To recognize and appreciate the split m(r) = mo(r) + U(r) + il(r), (3) proceed as follows. First split the total density of mass-energy, p, into a part fLon due to rest mass-where fLo is the average rest mass of the baryonic species pres- r,,\u00b7- -.J:. 604 23. SPHERICAL STARS Box 23.1 (continued) ent-and a part p Next notice that the proper volume of a shell of thickness dr is t-Lon due to internal thermal energy, compressional energy, etc. not 4'17r2dr. Consequently, the total rest mass inside radius r is mo =IT t-Lon d'V =IT 4'17r2(l - 2m/r)-lI2t-Lon dr, o 0 (4) (5) and the total internal energy is u =IT (p - o t-Lon) d'V =IT 4'17r2(l - 2m/r)-lI2(p - 0 t-Lon) dr. (6) Subtract these from the total mass-energy, m; the quantity that is left must be the gravitational potential energy, Sl = - IT p[(l - 2m/r)-lI2 - o 1]4'17r2 dr ::::: - IT (pm/r)4'17r2dr. l[Ne:tonian limit, m/r ~ 1] (7) (See exercise 23.7.) Equation for ep derived Turn next to the ;; component of the field equations: G;; = _r-2 + r-2e-2A + 2r- 1e-2A dtP/dr = 8'17T;; = 8'17p. Solving this equation for the derivative of tP, and replacing e-2A by 1 - 2m/r, one obtains an expression for the gradient of the potential tP: dtP dr - m + 4'17r3p r(r - 2m). This expression reduces to the familiar formula dtP/dr = m/r2 in the Newtonian limit. (23.21 ) (23.21N) \u00a723.5. EQUATIONS OF STRUCTURE 6", "05 In most studies of stellar structure, one replaces equation (23.17) by the equivalent equation obtained with the help of (23.21), dp dr = (p + p)(m + 4'lir3p) r(r - 2m) (23.22) Equation of hydrostatic equilibrium rewritten in \"OV\" form This is called the Oppenheimer-Volkoff (OV) equation of hydrostatic equilibrium. Its Newtonian limit, (23.22N) is familiar. Compare two stellar models, one relativistic and the other Newtonian. Suppose that at a given radius r [determined in both cases by (proper area) = 4'lir2 ], the two configurations have the same values of p, p, and m. Then in the relativistic model the pressure gradient is dp -:-:----~:-:--:-:---- = d(proper radial distance) dp ell dr = (p + p)(m + 4'lir3p) rZ(l - 2m/r)1/2 (23.23) In contrast, Newtonian theory gives for the pressure gradient dp d(proper radial distance) dp pm dr = -7' (23.23N) The relativistic expression for the gradient is larger than the Newtonian expression (1) because the numerator is larger (added pressure term in both factors) and (2) because the denominator is smaller [shrinkage factor (1 - 2m/r)1/2]. Therefore, as one proceeds deeper into the star, one finds pressure rising faster than Newtonian gravitation theory would predict. Moreover, this rise in pressure is in a certain sense \"self-regenerative.\" The more the pressure goes up, the larger the pressure-correction terms become in the numerator of (23.23); and the larger these terms become, the faster is the further rise of the pressure as one probes still deeper into the star. The geometric factor [l - 2m(r)/r]1/2 in the denominator of (23.23) further augments this regenerative rise of pressure towards the center. It is appropriate to summarize the situation in short-hand terms by saying that general relativity predicts stronger gravitational forces in a stationary body than does Newtonian theory. These forces, among their other important effects, can pull certain white-dwarf stars and super massive stars into gravitational collapse under circumstances (see Chapter 24) where Newtonian theory would have predicted stable hydrostatic equilibrium. As the most elementary indication that a new factor has surfaced in the analysis of stability, note that no star in hydrostatic equilibrium can ever have 2m(r)/r 2: 1 (see Box 23.2 for one illustration and \u00a723.8 for discussion), a phenomenon alien to Newtonian theory. Comparison of pressure gradients in Newtonian and relativistic stars Now in hand are five equations of structure [two equations of state (23.16); equation (23.19), expressing m(r) = ~r(1 - e-211 ) as a volume integral of p; the source Equations of stellar structure summarized 606 23. SPH ERICAL STARS equation (23.21) for tP; and the OV equation of hydrostatic equilibrium (23.22)] for the five structure functions p,p, n, tP, A. If the theory of relativistic stars as outlined then each of the remaining eight Einstein field equations above is well posed, - G;'IJ = 8'lTTaIJ must be either vacuous (\"0 = 0\"), or must be a consequence of the five equations of structure. This is, indeed, the case, as one can verify by straightfor ward but tedious computations. To construct a stellar model, one needs boundary conditions as well as structure equations. To facilitate the presentation of boundary conditions, the next section will examine the star's external gravitational field. EXERCISES Exercise 23.3. LAW OF LOCAL ENERGY-MOMENTUM CONSERVATION (for readers who have not studied Chapter 22) Evaluate the four components of the equation Ta/3;/3 = 0 for the stress-energy tensor (23.14) in the Schwarzschild coordinate system of equation (23.7). [Answer: only p/3;/3 = 0 gives a nonvacuous result; it gives equation (23.17).] Exercise 23.4. EINSTEIN CURVATURE TENSOR (for readers who have not studied Chapter 14) Calculate the components of the Einstein curvature tensor, G a/3' in Schwarzschild coordinates. Then perform a transformation to obtain GaP' the components in the orthonormal frame of equations (23.15a,b). [See Box 8.6, or Box 14.2 and equation (14.", "7).] Exercise 23.5. TOTAL NUMBER OF BARYONS IN A STAR Show that, if r = R is the location of the surface of a static star, then the total number of baryons inside the star is R A =f. o 4o;rr 2ne A dr. (23.24) [Hint: See the discussion of mo in Box 23.1.] Exercise 23.6. BUOYANT FORCE IN A STAR An observer at rest at some point inside a relativistic star measures the radial pressure-buoyant force, Fbuoy, on a small fluid element of volume V. Let him use the usual laboratory tech niques. Do not confuse him by telling him he is in a relativistic star. What value will he find for rt,UOy' in terms of p, p, m, V, and dp/dr? If he equates this buoyant force to an equal and opposite gravitational force, Fgrav' what will Fgrav be in terms of p, p, m, V, and r? (Use equation 23.22.) How do these results differ from the corresponding Newtonian results? Exercise 23.7. GRAVITATIONAL ENERGY OF A NEWTONIAN STAR Calculate in Newtonian theory the energy one would gain from gravity if one were to construct a star by adding one spherical shell of matter on top of another, working from the inside outward. Use Laplace's equation (r 2ep,r),r = 4'1Tr2p and the equation of hydrostatic equilibrium P,r = - pep,r to put the answer in the following equivalent forms: \u00a723.6. EXTERNAL GRAVITATIONAL FIELD 607 (energy gained from gravity) = - (gravitationai potential energy) R R o 1 =i (prcfl)4'17r 2dr =i (pm/r)4'17r 2dr = - -2i (pcfl)4'17r2dr = -8i (cfI)24'17r2dr R= 3i 4'17r2p dr23.6. EXTERNAL GRAVITATIONAL FIELD Outside a star the density and pressure vanish, so only the metric parameters tP and A = -~ In (l - 2m/r) need be considered. From equation (23.19) one sees that \"the mass inside radius r,\" m(r), stays constant for values of r greater than R (outside the star). Its constant value is denoted by M: m(r) = M for r > R (i.e., outside the star). (23.25) By integrating equation (23.21) with p = 0 and m = M, and by imposing the boundary condition (23.10) on tP at r = 00 (\"normalization of scale of time at r = 00 \"), one finds tP(r) = \"2 In (1 - 2M/r) 1 for r > R. (23.26) Consequently, outside the star the spacetime geometry (23.7) becomes ~------ ~-_(l _ 2M)dt 2 + --- ---r 2 dr (l - 2M/r) + r 2(d8 2 + sin28d,f,2). 't' (23.27) Spacetime outside star possesses \"Schwarzschild\" geometry This is called the \"Schwarzschild geometry\" or \"Schwarzschild gratitational field\" or \"Schwarzschild line element,\" because Karl Schwarzschild (l9l6a) discovered it as an exact solution to Einstein's field equations a few months after Einstein formu lated general relativity theory. In that region of spacetime, r ~ 2M, where the geometry is nearly flat, Newton's theory of gravity is valid, and the Newtonian potential is tP = -M/r for r > R, r ~ 2M. (23.26N) Consequently, M is the mass that governs the Keplerian motions of planets in the distant, Newtonian gravitational field-i.e., it is the star's \"total mass-energy\" (see Chapters 19 and 20). Since the metric (23.27) far outside the star is precisely diagonal (gtj =0), the star's total angular momentum must vanish. This result accords with the absence of internal fluid motions. Total mass-energy of star 608 23. SPHERICAL STARS \u00a723.7. HOW TO CONSTRUCT A STELLAR MODEL Equations of stellar structu re _~ collected together The equations of stellar structure (23.16), (23.19), (23.21), (23.22), and associated boundary conditions (to be discussed below), all gathered together along with the line element, read as follows. Line Element", " ds 2 = _e 2ifJ dt 2 + 2 dr 1 - 2m/r + r2(d()2 + sin2() dep2) (23.27') = -(1 - 2M)dt 2 + r 2 dr 1 - 2M/r + r2(d()2 + sin2()dep2) for r > R. Mass Equation m =IT 4'17r 2p dr, with m(r = 0) = O. \u00b0 (23.28a) OV Equation of Hydrostatic Equilibrium dp -d = r (p + p)(m + 4'17r 3p) ( 2 r r - m).'with p(r = 0) =Pc = central pressure. (23.28b) Equations of State p = p(n), p = p(n). (23.28c) (23.28d) Source Equation for f/J dtP d; - (m + 4'17r 3p) r(r - 2m), with tP(r = R) = ~ In (l - 2M/R). (23.28e) To construct a stellar model one can proceed as follows. First specify the equations of state (23.28c,d) and a value of the central pressure, Pc' Also specify an arbitrary tPo, for tP(r = 0). The boundary conditions (later to be renorrnalized) value, p(r = 0) =Pc' tP(r = 0) = tPo, m(r = 0) = 0 are sufficient to determine uniquely the. solution to the coupled equations (23.28). Integrate these coupled equations outward from r = 0 until the pressure vanishes. [The OV equation, (23.28b), guarantees that the pressure will decrease monotonically so long as the equations of state obey the How to solve the equations of stellar structu re \u00a723.7. HOW TO CONSTRUCT A STELLAR MODEL 609 reasonable restriction P ~ 0 for all p 2: 0.] The point at which the pressure reaches zero is the star's surface; the value of r there is the star's radius, R; and the value of m there is the star's total mass-energy, M. Having reached the surface, renormalize tP by adding a constant to it everywhere, so that it obeys the boundary condition (23.28e). The result is a relativistic stellar model whose structure functions tP, m, P, p, n satisfy the equations of structure. Notice that for any fixed choice of the equations of state p =p(n), p = p(n), the stellar models form a one-parameter sequence (parameter Pc)' Once the central pressure has been specified, the model is determined uniquely. The next chapter describes a variety of realistic stellar models constructed numer ically by the above prescription. For an idealized stellar model constructed analyti cally, see Box 23.2. Exercise 23.8. NEWTONIAN STARS OF UNIFORM DENSITY Calculate the structures of uniform-density configurations in Newtonian theory. Show that the relativistic configurations of Box 23.2 become identical to the Newtonian configurations in the weak-gravity limit. Also show that there are no mass or radius limits in Newtonian theory. (continued on page 612) EXERCISE Box 23.2 RELATIVISTIC MODEL STAR OF UNIFORM DENSITY For realistic equations of state (see next chapter), the equations of stellar structure (23.28) cannot be integrated analytically; numerical integration is necessary. How ever, analytic solutions exist for various idealized and ad hoc equations of state. One of the most useful analytic solutions [Karl Schwarzschild (1916b)] describes a star of uniform density, p = Po = constant for all p. (1) It is not necessary to indulge in the fiction of \"an incompressible fluid\" to accept this model as interesting. Incompressibility would imply a speed of sound, v = (dpjdp)1/2, of unlimited magnitude, therefore in excess of the speed of light, and therefore in contradiction with a central principle of special relativity (\"principle of causality\") that no physical effect can be propagated at a speed v > 1. (If a source could cause an effect so quickly in one local Lorentz frame, then there would exist another local Lorentz frame in which the effect would occur before the source had acted!) However, that the part of the fluid in the region of high pressure has the same density as the part of the fluid in the region of low pressure is an idea easy to admit, if only one thinks of the fluid having a composition that varies from one Box 23.2 (continued)", " r value to another (\"hand-tailored\"). Whether one thinks along this line, or simply has in mind a globe of water limited in size to a small fraction of the dimensions of the earth, one has in Schwarzschild's model an instructive example of hydrostatics done in the framework of Einstein's theory. The mass equation (23.28a) gives immediately for r < R}. for r > R from which follows the length-correction factor in the metric d(proper distance) dr II == e == [ I - 2 ( 111 r)/r] -1/2. (2) (3) When for ease of visualization the space geometry (r,~) of an equatorial slice through the star is viewed as embedded in a Euclidean 3-geometry (z, r,~) [see \u00a723.8], the \"lift\" out of the plane z == 0 is (I - 2Mr 2/ R3)1/2] (R3/2M)l/2[1 (R3/2M)l/2[1 _ (1 - 2M/R)1/2] + [8M(r - 2M)]1/2 - for r ~ R, { z(r) == [8M(R - 2M)]l/2 for r ;;::: R. (4) The knowledge of m(r) from (2) allows the equation of hydrostatic equilibrium (23.28b) to be integrated to give the pressure: (5) The pressure in turn leads via (23.28e) to the time-correction factor in the metric. d(proper time) - -=-.--:..----'-- - etP - dt [1(1 __2M_)1/2 _1(1 __2_M_r_ 2 2 R R3 2 (I - 2M/r)1/2 )l/2 for r < R]. (6) for r > R Several features of these uniform-density configurations are noteworthy. (1) For fixed energy density, Po' the central pressure (I - 2M/R)l/2 } Pc == Po 3(1 _ 2M/R)l/2 _ 1'1 - { (7) increases monotonically as the radius, R, increases-and, hence, also as the mass, M == (4'17/3)PoR3, and the ratio (\"strength of gravity\") (8) increase. This is natural, since, as more and more matter is added to the star, a greater and greater pressure is required to support it. (2) The central pressure becomes infinite when M, R, and 2M/R reach the limiting values R lim = (9/4)Mlim = (3'17Potl/2, (2M/ R)lim = 8/9. (9) (10) No star of uniform density can have a mass and radius exceeding these limits. These limits are purely relativistic phenomena; no such limits occur in Newtonian theory. (3) Inside the star the space geometry (geometry of a hypersurface t = constant) is that of a three-dimensional spherical surface with radius of curvature (11) [See equation (4), above.] Outside the star the (Schwarzschild) space geometry is that of a three-dimensional paraboloid of revolution. The interior and exterior geometries join together smoothly. All these details are shown in the following three diagrams. There all quantities are given in the following geometric units (to convert mass in g or density in g/cm3 into mass in cm or density in cm- 2, multiply by 0.742 X 10-28 cm/g): lengths, in units (3/8'17Po)l/2; pressure, in units Po; mass, in units (3/32'17Po)l/2. o 612 23. SPHERICAL STARS Box 23.2 (continued) The mass \"after assembly\" is what is called M. The mass of the same fluid, dispersed in droplets at infinite separation, is called Mbefore in the following table. J\\1bdore.tl Difference (binding): small 0.0882 0.894 1.0913 1.374 small 0.0828 0.636 0.729 0.838 (critical) ~lvt\u00b7,)/3 0.0054 0.258 0.362 0.536 Surface area of spheres, 4r.r 2 : (1) increases monotonically from center of star outward \u00a723.8. THE SPACETIME GEOMETRY FOR A STATIC STAR For a highly relativistic star, the spacetime geometry departs strongly from Euclid Lorentz flatness. Consequently, there is no", " a priori reason to expect that the surface area 4'17r 2, and hence also the radial coordinate r, will increase monotonically as one moves from the center of the star outward. Fortunately, the equations of stellar structure guarantee that r will increase monotonically from 0 at the star's center to 00 at an infinite distance away from the star, so long as p 2: 0 and so long as the star is static (equilibrium). The monotonicity of r can be seen as follows. Introduce as a new radial coordinate proper distance, 1, from the center of the star. By virtue of expression (23.27') for the metric, 1 and r are related by dr = +(1 - 2m/r)l/2 d1. (23.29) Note that r is zero at the center of the star (where m ex r3 ), and note that r is always nonnegative by definition. Therefore r must at first increase with 1 as one moves outward from 1 = 0; r(l) can later reach a maximum and start decreasing only at a point where 2m/r becomes unity [see equation (23.29)]. Such a behavior can and does happen in a closed model universe, a 3-sphere of uniform density and radius a, where r(l) = a sin (l/a) [see Chapter 27; especially the embedding diagram of Box 27.2(A)]. However, the field equations demand that such a system be dynamic. Here, on the contrary, attention is limited to a system where conditions are static. In such a system, the condition of hydrostatic equilibrium (23.28b) applies. Then the pressure gradient is given by an expression with the factor [l - 2m(r)/r] in its denominator. If 2m/r approaches unity with increasing 1 in some region of the star, the pressure gradient \u00a723.8. SPACETIME GEOMETRY FOR A STATIC STAR 613 there becomes so large that one comes to the point p = 0 (surface of the star) before one comes to any point where 2m(r)/r might attain unit value. Moreover, after the surface of the star is passed, m remains constant, m(r) = M, and 2m(r)/r decreases. Consequently, 2m/r is always less than unity; and r(l) cannot have a maximum, Q.E.D. (Details of the proof are left to the reader as exercise 23.9.) Although the radii of curvature, r, and corresponding spherical surface areas, 4'17r 2, increase monotonically from the center of a star outward, they do not increase at the same rate as they would in flat spacetime. In flat spacetime the rate of increase is given by dr/d(proper radial distance) = dr/dl = 1. In a star it is given by dr/dl = (1 - 2m/r)1/2 < 1. Consequently, if one were to climb a long ladder outward from the center of a relativistic star, measuring for each successive spher ical shell its Schwarzschild r-value (\"proper circumference\"/2'17), one would find these r-values to increase surprisingly slowly. This strange behavior is most easily visualized by means of an \"embedding diagram.\" It would be too much for any easy visualization if one were to attempt to embed the whole curved four-dimensional manifold in some higher-dimensional flat space. [See, however, Fronsdal (1959) and Clarke (1970) for a global embedding in 5 + 1 dimensions, and Kasner (1921 b) for a local embedding in 4 + 2 dimensions. One can never embed a non-flat, vacuum metric (GIlV = 0) in a flat space of 5 dimensions (Kasner, 1921c).] Therefore seek a simpler picture (Flamm 1916). Space at one time in the context of a static system has the same 3-geometry as space at t = constant. another time. Therefore, depict 3-space only as it is at one time, Moreover, at anyone time the space itself has spherical symmetry. Consequently, one slice through the center, r = 0, that divides the space symmetrically into two halves (for example, the equatorial slice, 8 = '17/2) has the same 2-geometry as any other such slice (any selected angle of tilt, at any azimuth) through the center. Therefore limit attention to the 2-geometry of the equatorial slice. The geometry on this slice is described by the line element (23.30) (2) but increases more slowly than in flat spacetime Embedding of spacetime in a flat space of higher dimensionality Now one may embed this two-dimensional curved-space geometry in the flat geom etry of a Eucl", "idean three-dimensional manifold. If the curvature of the two-dimensional slice is zero or negligible, the embedding is trivial. In this event, identify the 2-geometry with the slice z =:= 0 of the Euclidean 3-space. Moreover, introduce into that 3-space the familiar cylindrical coordinates z, r, ~, that one employs for any problem with axial symmetry (see Fig. 23.1 and Box 23.2 for more detail). Then one recognizes the flat two-dimensional slice as the - ------------set~pts o\u00a3.the Euclidean space with z = 0, with ~ running from 0 to 2'17, and r from 0 to 00. One has identified the r and ~ of the slice with the r and ~ of the Euclidean 3-space. If the 2-geometry is curved, as it is when the equatorial section is taken through a real star, then maintain the identification between the r, ~, of the slice and the r,~, of the Euclidean 3-geometry, but bend up the slice out of the plane z = 0 (except at the origin, r = 0). At the same time, insist that the bending be axially symmetric. In other words, require that the amount of the \"lift\" above the plane z = 0 shall Construction of \"embedding diagram\" for equatorial slice through star 614 23. SPHERICAL STARS ;; x Figure 23.1. Geometry within (grey) and around (white) a star of radius R = 2.66M, schematically displayed. The star is in hydrostatic equilibrium and has zero angular momentum (spherical symmetry). The two dimensional geometry ds2 = [1 - 2m(r)/rt 1 dr2 + r2 deJ>2 of an equatorial slice through the star (8 = 1T/2, I = constant) is represented as embedded in Euclidean 3-space, in such a way that distances between any two nearby points (r. eJ\u00bb and (r + dr, \u00a2 + d\u00a2) are correctly reproduced. Distances measured off the curved surface have no physical meaning; points off that surface have no physical meaning; and the Euclidean 3-space itself has no physical meaning. Only the curved 2-geometry has meaning. A circle of Schwarzschild coordinate radius r has proper circum ference 21Tr (attention limited to equatorial plane of star, 8 = 1T/2). Replace this circle by a sphere of proper area 41Tr2, similarly for all the other circles, in order to visualize.the entire 3-geometry in and around the star at any chosen moment of Schwarzschild coordinate time I. The factor [1 - 2m(r)/rt 1 develops no singularity as r decreases within r = 2M, because m(r) decreases sufficiently fast with decreas ing r. be independent of~, whatever may be its dependence on r. Thus the whole stOfY~~ of the embedding is summarized by the single function, the lift, z = z(r) (\"embedding formula\"). The geometry on this curved two-dimensional locus in Euclidean space (a made-up 3-space; it has nothing whatever to do with the real world) is to be identical with the geometry of the two-dimensional equatorial slice through the actual star; in other words, the line elements in the two cases are to be identical. To work out this requirement in mathematical terms, write the line element in three-dimensional Euclidean space in the form (23.31 ) Restrict to the chosen locus (\"lifted surface\") by writing z = z(r) or dz = (dzjdr) dr. Thus have (23.32) \u00a723.8. SPACETIME GEOMETRY FOR A STATIC STAR 615 on the two-dimensional locus in the 3-geometry, to be identified with in the actual star. Compare and conclude (d~~)r+ I = [1 - 2m(r)/r]-1. (23.33) This equation is information enough to find the lift as a function of r; thus, z() IT r = [ dr o _r__ 2m(r) everywhere, (23.34a) JU2 1 z(r) = [8M(r - 2M)]U2 + constant outside the star. (23.34b) Outside the star this embedded surface is a segment of a paraboloid of revolution. Its form inside the star depends on how the mass, m, varies as a function of r. Recall that m(r) varies as (4'17/3)Pcr3 near the center of the star. Conclude that the embedded surface there looks like a segment of a sphere of radius a = (3/8'", "17Pc)U2; thus, Description of embedded surface [a - z(r)j2 t r2 = a2 (23.34c) In the special case of a star with uniform density (Box 23.2), the entire interior is of the spherical form (23.34c); in the general case it is not. In all cases, because r> 2m(r), equation (23.34a) produces a surface with z and r as monotonically increasing functions of each other. This means that the embedded surface always opens upward and outward like a bowl; it always looks qualitatively like Figure 23.1; it never has a neck, and it never flattens out except asymptotically at r = 00. At the star's surface, even though the density may drop discontinuously to zero (p finite inside when p = 0; p zero outside), the interior and exterior geometries will join together smoothly [dz/dr, as given by equation (23.33), is continuous]. It must be emphasized that only points lying on the embedded 2-surface have physical significance so far as the stellar geometry is concerned: the three-dimen sional regions inside and outside the bowl of Figure 23.1 are physically meaningless. So is the Euclidean embedding space. It merely permits one to visualize the geometry of space around the star in a convenient manner. Exercise 23.9. GOOD BEHAVIOR OF r Carry out explicitly the full details of the proof, at the beginning of this section, that 2m/r is always less than unity and r is a monotonic function of 1. EXERCISES Exercise 23.10. CENTER OF STAR OCCUPIED BY IDEAL FERMI GAS AT EXTREME RELATIVISTIC LIMIT Opposite to the idealization of a star built from an incompressible fluid is the idealization in which it is built from an ideal Fermi gas [ideal neutron star; see Oppenheimer and Volkoff (1939)] at zero temperature, so highly compressed that the particles have relativistic energies, 616 23. SPHERICAL STARS in comparison with which any rest mass they possess is negligible. In this limit, with two particles per occupied cell of volume 11.1 in phase space, one has ( number density) = n = (2/17:1)4<7 rPF p2 dp = 8<7p.1/3173, of fermlOns density of mass-energy ) ( =P =( /17 )4<7 2 3 LPF 0 Jo cp' p. dp = 2\"cp 4/17 3 F? F'and finally d(energy P = - d(volume. per particle per partIcle ) d(p/II) ) = - d(1/II) = 2\"CPF / 4 3 3 17 = p/ 3, as if one were dealing with radiation instead of particles (PF = Fermi momentum; momen tum of highest occupied state). Box 23.3 RIGOROUS DERIVATION OF THE SPHERICALLY SYMMETRIC LINE ELEMENT Section 23.2 gave a heuristic derivation of the general spherically symmetric line element (23.7). This box attempts a more rigorous derivation, applicable to nonstatic systems, as well as static ones. Begin with a manifold M4 on which a metric ds 2 of Lorentz signature is defined. Assume M 4 to be spherically symmetric in the sense that to any 3 X 3 rotation matrix A there corresponds a mapping (rotation) of M4, also called A (A: M4 --+ M4: P -+ AP), that preserves the lengths of all curves. Further assumptions and constructions will be numbered (i), (ii), etc., so one can see what specializations are needed to get to the line ele ment (23.7). Daggers (t) indicate assumptions that are found inapplicable to some other physically interesting situations. For any point P, form the set s= S(P) = {AP E M 4 1A E SO(3)} of all points equivalent that s is to P under a two-dimensional for center points, where s is zero-dimensional), and (ii) that the metric on s is that of a standard 2-sphere. Then on s one will have rotations. Assume (i)t surface (except (1) where dSl 2 is the standard metric of a unit sphere (dSl 2 = d8 2 + sin28 d~2 for some 8, ~, defined on s), and where 2'17R is the circumference of s. If M2 is the set of all such surfaces s, then S: M4 -+ [the M2: P -+ s = S(P", ") allows one to obtain, from \"circumference\" R: M2 -+ (>jl: s -+ R(s) function on M2 as defined by equation (1)], a corresponding R: M4 -+ (>jl: q> -+ R(S(P\u00bb on M4 which in some cases can eventually be used as a coordinate on M4. (Note: (>jl denotes here the real numbers.) function Now assume (iii)t there is a spherically symme tric 4-velocity field u, defined so that if P = B(T) is one trajectory of u with u = d/dT, then each curve q> = AB(T) obtained by a rotation must also be a trajectory of u. The orthogonal projection of u onto any sphere s must then vanish, as there are no rotation invariant non-zero vector fields on 2-spheres. Thus u is orthogonal to each s. Also, if two trajectories of u start on some same sphere s, so B1(0) = AB2(0), then the same rotation A will always relate them, Bk) = AB2(T), since trajec tories are uniquely defined by anyone point on are both the them. Then S(B1(T\u00bb same curve in M2, whose tangent d/dT one can call also u; in this way one obtains a vector field u on M2. Give each trajectory of u on M2 a differ ent label r to define a function res) on M2. Denote by r = r(S(p\u00bb a corresponding function r on M4 with dr/dT = O. Since functions and their gradi ents on M4 define corresponding quantities on M2, inner products such as df\u00b7 dg can be defined on M2 by their values on M4; thus, from the metric on M4 one obtains a metric on M2. Then by equa- and S(B2(T\u00bb \u00a723.8. SPACETIME GEOMETRY FOR A STATIC STAR 617. (a) Write out the relativistic equation of hydrostatic equilibrium for a substance satisfying. the equation of state p = p13. (b) Show that there exists a well-defined analytic solution for the limiting case of infinite central density, in which m(r)/r has the value 3/14. (c) Find per), per), and nCr). (d) Show that the number of particles out to any finite r-value is finite, despite the fact that nCr) is infinite at the origin. (e) Show that the 3-geometry has a \"conical singUlarity\" at r = O. (f) Make an \"embedding diagram\" for this 3-geometry [\"lift\" z(r) as a function of r from (23.34)]. (Note that the conical singularity at r = 0, otherwise physically unreasonable, arises because the density of mass-energy goes to infinity at that point. Note also that the calculated mass of the system diverges to infinity as r -+ 00. In actuality with decreasing density the FermI momentum falls from relativistic to nonrelativistic values, the equation of state changes its mathematical form, and the total mass M converges to a finite value). tion (23.5) or equivalently by drawing curves in M2 orthogonal to the r = const. lines, and giving each a different label t, one obtains coordinates WIth grl- =: dr.~t-= O. Both rand t labels were assigned arbitrarily on the corresponding curves, so it is clear that transformations t' = t'(t) and r' = r'(r) are not excluded. On one 2-sphere s in M4, on the t = 0 hypersur face, choose a set of (), ~ coordinates by picking the pole (() = 0) and the prime meridian (~ = 0) arbitrarily. Then extend the definition of (),~, over the t = 0 hypersurface by requiring () and ~ to be constant on curves orthogonal to each 2-sphere s, i.e., by demanding that (a/ar)9</> be orthogonal to each s at t = O. Extend the definition of () and ~ to t i- 0 by requiring them to be constant on curves with tangent u, so (a/at)r9</> ex u. But each s is a surface of constant rand t; so (a/a()rt</> and to s, while u ex (a/at) is (a/a~)rt9 are tangent orthogonal to each s. Consequently, gt9 = (a/at)\u00b7 (a/a() = 0 gt</", "> = (a/at)\u00b7 (a/a~) = 0 in the tr()~ coordinate system just constructed. The vector (a/ar)t9</> does not depend on the arbitrary directions introduced in the original choice of (), ~ coordinates on one sphere s; it is invariant under transformations () = ()(()', ~'), ~ = ~(()', ~'). But nothing except () and ~ introduced nonrotationally invariant elements into the discussion; so (a/ar)t9</> must be a rotationally invariant vector field (un- and (3) (2) like, say, a/a~); so it is, like u, orthogonal to each 2-sphere s. This invariance then gives gr9 = (a/ar)' (a/a() = 0, gr</> = (a/ar)' (a/a~) = 0, (5) which, with gtr = 0 as previously established, gives gtr = O. The result is a line element of the form (23.3). Further specialization, a change of radial and time coordinates to Rand T, where R is de fined by (I) above and (4) dT = e\",[_I- aR dt __1_ aR dr] grr ar gtt at'\", _ (integrating) e factor, followed by a change of notation, to Schwarzschild coordinates and the line element (23.7)-though such a transformation is possible (i.e., nonsingular) only where dR A dT i- 0: leads (V R)2 = (aR/at)2 + (aR/ar)2 i- O. gtt grr If (iv)t spacetime is asymptotically flat, so r --+ 00 is a region where the metric can take on its special relativity values, then the arbitrariness in the t coordinate, t' = t'(t), can be eliminated byrequiringgtt = -I asr --+ 00. Then (a/at)r9</> is uniquely determined by natural requirements (independent of the arbitrary (),~, choices), and whenever it is desired to make the further physical assumption (v)t of a time-independent geometry, this can be appropriately restated as ag!\"./at = O. CHAPTER 24 PULSARS AND NEUTRON STARS; QUASARS AND SUPERMASSIVE STARS Go, wond'rous creature, mount where Science guides, Go, measure earth, weigh air, and state the tides: Instruct the planets in what orbs to run, Correct old time, and regulate the sun. ALEXANDER POPE (1733) Types of stellar configurations where relativity should be important \u00a724.1. OVERVIEW Five kinds of stellar configurations are recognized in which relativistic effects should be significant: white dwarfs, neutron stars, black holes, supermassive stars, and relativistic star clusters. The key facts about each type of configuration are summar ized in Box 24.1; and the most important details are described in the text of this chapter (white dwarfs in \u00a724.2; neutron stars and their connection to pulsars in \u00a7\u00a724.2 and 24.3; supermassive stars and their possible connection to quasars and galactic nuclei in \u00a7\u00a724.4 and 24.5; and relativistic star clusters in \u00a724.6; a detailed discussion of black holes is delayed until Chapter 33). The book Stars and Relativity by Zel'dovich and Novikov (1971) presents a clear and very complete treatment of all these astrophysical applications of relativistic stellar theory. In a sense, that book can be regarded as a companion volume to this one; it picks up, with astrophysical emphasis, all the topics that this book treats with gravitational emphasis. This chapter is meant only to give the reader a brief survey of the material to be found in Stars and Relativity. (continued on page 621) \u00a724.1. OVERVIEW 619 Box 24.1. STELLAR CONFIGURATIONS WHERE RELATIVISTIC EFFECTS ARE IMPORTANT [For detailed analyses and references on all these topics, see Zel'dovich and Novikov (1971).] A. White Dwarf Stars Are stars of about one solar mass, with radii about 5,000 kilometers and densities about 1 ton/cm3 ; support themselves 106 g/cm3 against gravity by the pressure of degenerate electrons; have stopped burning nuclear fuel, and are gradually cooling as they radiate away their remaining store of thermal energy. Were observed and studied astronomically long before they were understood theoretically. Key points in history: August 1926, Dirac (1926) formulated Fermi Dirac statistics, following Fermi (February). December 1926, R. H. Fowler (1926) used", " Fermi-Dirac statistics to explain the nature of white dwarfs; he invoked electron de generacy pressure to hold the star out against the inward pull of gravity. 1930, S. Chandrasekhar (1931a,b) calculated white-dwarf models taking account of spe cial relativistic effects in the electron-de generacy equation of state; he discovered that no white dwarf can be more massive than -1.2 solar masses (\"Chandrasekhar Limit\"). 1932, L. D. Landau (1932) gave an elemen tary explanation of the Chandrasekhar limit. 1949, S. A. Kaplan (1949) derived the effects of general relativity on the mass-radius curve for massive white dwarfs, and de duced that general relativity probably in duces an instability when the radius be comes smaller than 1.1 X 103 km. Role of general relativity in white dwarfs: negligible influence on structure; significant influence on stability, on pulsation frequencies, and on form of mass-radius curve near the Chandrasekhar limit (i.e., in massive white dwarfs). Electron capture also significant. See; e.g., Zel'dovich and Novi kov (1971); Faulkner and Gribbin (1968). B. Neutron Stars Are stars of about one solar mass, with radii about 10 km and densities about 1014 g/cm3 (same as density of an atomic nucleus); are supported against gravity by the pressure of degenerate neutrons and by nucleon-nucleon strong-interaction forces; are not burning nu clear fuel; the energy being radiated is the energy of rotation and the remaining store of internal thermal energy. Theoretical calculations predicted their 'exist ence in 1934, but they were not verified to exist observationally until 1968. Key points in history: 1932, neutron discovered by Chadwick (1932). 1933-34, Baade and Zwicky (1934a,b,c) (1) invented the concept of neutron star; (2) identified a new class of astronomical ob jects which they called \"supernovae\"; (3) suggested that supernovae might be created by the collapse of a normal star to form a neutron star. (See Figure 24.1.) 1939, Oppenheimer and Volkoff (1939) per formed the first detailed calculations of the structures of neutron stars; in the process, they laid the foundations of the general relativistic theory of stellar structure as pre sented in Chapter 23. (See Figure 24.1.) 1942, Duyvendak (1942) and Mayall and Oort (1942) deduced that the Crab nebula is a remnant of the supernova obseJ:Ved by Chi- 620 24. PULSARS. NEUTRON STARS; QUASARS, SUPERMASSIVE STARS Box 24.1 (continued) (1942) nese astronomers in A.D. 1054. Baade (1942) and Minkowskii the \"south preceding star,\" near the center of the Crab Nebula, as probably the (col lapsed) remnant of the star that exploded in 1054 (see frontispiece). identified 1967, Pulsars were discovered by Hewish et al. (1968). 1968, Gold (1968) advanced the idea that pulsars are rotating neutron stars; and sub sequent observations confirmed this sug gestion. 1969, Cocke, Disney, and Taylor (1969) dis covered that the \"south preceding star\" of the Crab nebula is a pulsar, thereby clinch ing the connection between supernovae, neutron stars, and pulsars. Role of general relativity in neutron stars: significant effects (as much as a factor of 2) on structure and vibration periods; gravitational radiation reaction may be the dominant force that damps nonradial vi brations. C. Black Holes Are objects created when a star collapses to a size smaller than twice its geometrized mass (R < 2M (M/ Mo) X 3 km), thereby creat ing such strong spacetime curvatures that it can no longer communicate with the external universe (detailed analysis of black holes in Chapters 33 and 34). No one who accepts general relativity has found any way to escape the prediction that black holes must exist in our galaxy. This prediction depends in no way on the complexity of the collapse that forms the black holes, or on unknown properties of matter at high density. However, the existence of black holes has not yet been verified observationally. Key points in history: 1795, Laplace (1795) noted that, according to Newtonian gravity and Newton's corpuscu lar theory of light, light cannot escape from a sufficiently massive object (Figure 24.1). 1939, Oppenheimer and Snyder (1939) calcu lated the collapse of a homogeneous sphere of", " pressure-free fluid, using general rela tivity, and discovered that the sphere cuts itself off from communication with the rest of the universe. This was the first calcula tion of how a black hole can form (Figure 24.1 ). 1965, Beginning of an era of intensive theo retical investigation of black-hole physics. Role of general relativity in black-hole physics: No sensible account of black holes possible in Newtonian theory. The physics of black holes calls on Einstein's description of grav ity from beginning to end. D. Supermassive Stars Are stars of mass between 103 and 109 solar masses, constructed from a hot plasma of density typically less than that in normal stars; are supported primarily by the pressure of photons, which are trapped in the plasma and are in thermal equilibrium with it; burn nuclear fuel (hydrogen) at some stages in their evolution. Theoretical calculations suggest (but not with complete confidence) that supermassive stars exist in the c~nters of galaxies and quasars, and perhaps elsewhere. Supermassive stars conceivably could be the energy sources for some quasars and galactic nuclei. However, astronomical not yet yielded definitive evidence about their exist ence or their roles in the universe if they do exist. observations have \u00a724.2. ENDPOINT OF STELLAR EVOLUTION 621 Key points in history: 1963, Hoyle and Fowler (1963a,b) conceived the idea of supermassive stars, calculated their properties, and suggested that they might be associated With galactic nuclei and quasars. (1964a,b) 1963-64, Chandrasekhar and Feynman (1964) developed the general relativistic theory of stellar pulsations; and Feynman used it to show that supermassive stars, although Newtonian in structure, are subject to a general-relativistic instability. 1964 and after, calculations by many workers have elaborated on and extended the ideas of Hoyle and Fowler, but have not pro duced any spectacular breakthrough. Role of general relativity in supermassive stars: negligible influence on structure, except in the extreme case ofa compact, rapidly rotating, disc-like configuration [see Bardeen and Wagoner (1971); Salpeter and Wagoner (1971 )]. significant influence on stability. E. Relativistic Star Clusters Are clusters of stars so dense that relativistic corrections to Newtonian theory modify their structure. Theoretical calculations suggest that relativistic star clusters might, but quite possibly do not, form in the nuclei of some galaxies and qua sars; if they do try to form, they might be destroyed during formation by star-star colli sions, which convert the cluster into super massive stars or into a dense conglomerate of stars and gas. Astronomical observations have yielded no definitive evidence, as yet, about the existence of relativistic clusters. Key points in history: 1965, Zel'dovich and Podurets (1965) con ceived the idea of relativistic star clusters, developed the theory of their structure using general relativity and kinetic theory (cf. \u00a725.7), and speculated about their sta bility. 1968, lpser (1969) developed the theory of star-cluster stability and showed (in agree ment with the Zel'dovich-Podurets specula tions) that, when it becomes too dense, a cluster begins to collapse to form a black hole. Role of general relativity in star clusters: significant effect on structure when gravita tional redshift from center to infinity ex ceeds Zc = &/;\\ - 0.05. induces collapse of cluster to form black hole when central redshift reaches Zc :::::; 0.50. \u00a724.2. THE ENDPOINT OF STELLAR EVOLUTION \" After the normal stages of evolution, stars \"die\" by a variety of processes. Some stars explode, scattering themselves into the interstellar medium; others contract into a white-dwarf state; and others-according to current theory-collapse to a neutron-star state, or beyond, into a black hole. Although one knows little at present about a star's dynamic evolution into its final state, much is known about the final states themselves. The final states include dispersed nebulae, which are of no interest here; cold stellar configurations, the subject of this section; and \"black holes,\" the subject of Part Vll. (continued on page 624) JANUARY IS, 1934 PHYSICAL REVIEW VOLUME 45 Proceedings of the American Physical Society MISUTES OF THE STANFORD MEETIl\\G, DECRMRER 15-16, 1933 38. Supernovae and Cosmic Rays. W. BAADE, Mt. Wilson Obseroatory. AND F. ZWICKY, California Institute of Technology.-Supernovae flare up in every stel1ar system (nebula) once in several centuries.", " at zero; the nuclear composition in principle is specified uniquely by the density; and therefore the pressure is also fixed uniquely once the density has been specified [equation of state p(p) for \"cold catalyzed matter\"]. The white dwarfs and neutron stars observed by astronomers are not really built of cold catalyzed matter. However, the matter in them is sufficiently near the end point of thermonuclear evolution and sufficiently cold that it can be idealized with fair accuracy as cold and catalyzed (see \u00a723.4). The equation of state, p(P), for cold catalyzed matter is shown graphically in Figure 24.2. This version of the equation of state was constructed by Harrison and Wheeler in 1958. Other versions constructed more recently [see Cameron (1970) and Baym, Bethe, and Pethick (1971) for references] are almost identical to the Harrison-Wheeler version at densities well below nuclear densities, p < 3 X 1013 g/cm3 \u2022 At nuclear and supernuclear densities, all versions differ because of differing assumptions about nucleon-nucleon interactions. Along with the equation of state, in Figure 24.2 are shown properties of the models of cold stars constructed from this equation of state by integrating numerically the equations of structure (23.28). The equation of state can be understood by following the transformations that occur as a sample of cold catalyzed matter is compressed to higher and higher densities. At each stage in the compression, each possible thermonuclear reaction is to be catalyzed to its endpoint and the resultant thermal energy is to be removed. When the sample is at zero pressure, it is a ball of pure, cold Fe56, since Fe56 is the most tightly bound of all nuclei. It has the density 7.86 g/cm3. As the sample is compressed, its internal pressure is provided at first by normal solid-state forces; but the atoms are soon squeezed so closely together that the electrons become quite oblivious of their nuclei, and begin to form a degenerate Fermi gas. By the time a density of p = 105 g/cm3 has been reached, valence forces are completely negligi ble, the degenerate electron pressure dominates, and the compressibility index, y (see legend for Figure 24.2), is 5/3, the value for a nonrelativistically degenerate Fermi gas. Between 105 and 107 g/cm3, the pressure-providing electrons gradually Equation of state ty o 2 4 6 8 10 12 14 16 18 - - - l o gIO p, g/cm3 \u2022 Stellar models H-W-W 1o ~ ~ I 1.0 0.5 17. o o - - - logIO R, km.. Figure 24.2. The Harrison-Wheeler equation of state for cold matter at the absolute end point of thermonuclear evolution, and the corresponding Harrison-Wakano-Wheeler stellar models. The equation of state is exhibited in the form of a plot of \"compressibility index,\" p + pdp y= - - -, dp P as a function of density of mass-energy, p. (Small y corresponds to easy compressibility.) The curve is parameterized by the logarithm of the pressure, logloP, in units of g/cm3 [same units as p; note that p(g/cm3) = (l/c 2) X p(dyne/cm2)]. The chemical composition of the matter as a function of density is indicated as follows: Fe, Fe56 nuclei; A, nuclei more neutron rich than Fe56 ; e, electrons; n, free neutrons; p, free protons. The firsU!lw of thef!Tl0ELnamics [equation (22.6)1, when applied to cold matter (zero entropy) says dp/(p + p) = dn/n; i.e., - p + P n = JLFe/56 exp - (fP dp ) p + P. 0 Here JLFe' the rest mass of an Fe56 atom, is the ratio between p + p ::::: p and n/56 in the limit of zero density. From this equation and a knowledge of p(p)-(see Figure)-one can calculate n(p). The equilibrium configurations are represented by curves of total mass-energy, M, versus radius, R. (R is defined such that 4'lTR2 is the star's surface area.) The M(R) curve is parameterized by the logarithm loglOPe' measured in g/cm3\u2022 Only configurations along two branches of the of the central density, curve are stable against small perturbations and can therefore exist in nature: the white dwarfs, with log lOPe < 8.38, and the neutron stars, with", " 13.43 < log lOPe < 15.78 (see Box 26.1). For greater detail on both the equation of state and the equilibrium configurations, see Harrison, Thorne, Wakano, and Wheeler (1965); also, for an updated table of the equation of state. see Hartle and Thorne (1968). 626 24. PULSARS, NEUTRON STARS; QUASARS. SUPER MASSIVE STARS become relativistically degenerate, and y approaches 4/3. Above P = 1.4 X 10' g/cm3, the rest mass of 62 Fe~~ nuclei, plus the rest mass of 44 electrons, plus the rather large Fermi kinetic energy of 44 electrons at the top of the Fermi sea, exceeds the rest mass of 56 Ni~\u00a7 nuclei. Consequently, as the catalyzed sample of matter is compressed past P = 1.4 X 107 g/cm3, the nuclear reaction 62 Fe56 (highly compressed) --+ 26 neutral atoms 56 Ni~2 (highly compressed) ~8 neutral atoms (24.1 ) goes to its end point, with a release of energy. As the compression continues beyond this point, the rising Fermi energy of the electrons induces new nuclear reactions similar to (24.1), but involving different nuclei. In these reactions more and more electrons are swallowed up to form new nuclei, which are more and more neutron rich. When the density reaches P = 3 X 1011 g/cm3, the nuclei are so highly neutron rich (Y \u00a7\u00a72) that neutrons begin to drip off them. The matter now becomes highly compressible for a short time (3 X 1011 :s P :s 4 X 1011), since most of the re maining electrons are swallowed up very rapidly by the dripping nuclei. Above P - 4 X 1011 g/cm3 free neutrons become plentiful and their degeneracy pressure 10 13 g/cm3 completely exceeds that of the electrons. Further compression to P disintegrates the remaining nuclei, leaving the sample almost pure neutrons with y = 5/3, the value for a nonrelativistically degenerate Fermi gas. Intermixed with the neutrons are just enough degenerate electrons to prevent the neutrons from decaying, and just enough protons to maintain charge neutrality. Compression 10 13 g/cm3 pushes the sample into the domain of nuclear densities where beyond P the physics of matter is only poorly understood. This Harrison-Wheeler version of the equation of state ignores all nucleon-nucleon interactions at and above nuclear densities; it idealizes matter as a noninteracting mixture of neutrons, protons, and electrons with neutrons dominating; and it shows a compressibility index of5/3 while the neutrons are nonrelativistic, but 4/3 after they attain relativistic Fermi energies. Other versions of the equation of state attempt to take into account the nucleon nucleon interactions in a variety of ways [see Cameron (1970), Baym, Bethe, and Pethick (197 I), and many references cited therein]. - - - ~ ~...- - Equilibrium configurations for cold, catalyzed matter: (') forms and stability Corresponding to each value of the central density, Pc' there is one stellar equilib rium configuration. Equilibrium, yes; but is the equilibrium stable? Stability studies (Chapter 26, especially Box 26.1) show that many of the models are unstable against small radial perturbations, which lead to gravitational collapse. Only white-dwarf stars in the range 10gIO Pc < 8.4 and neutron stars in the range 13.4 :s 10gIO Pc :s 15.8 are stable. Instability for the region of 10gIO Pc values between 8.4 and 13.4 is caused by a combination of (I) relativistic strengthening of the gravitational forces, and (2) high compressibility of the matter due to electron capture and neutron drip by \u00a724.3. PULSARS 627 the atomic nuclei. Neutron stars are stable for a simple reason. Neutron-dominated matter is so difficult to compress that even the relativistically strengthened gravita tional forces cannot overcome it. Above 10gIO Pc - 15.8, the gravitational forces become strong enough to win out over the pressure of the nuclear matter, and the stars are all unstable. [See Gerlach (1968) for the possibility-which, however, he rates as unlikely-that there might exist a third family of stable equilibrium con figurations, additional to white dwarfs and neutron stars.] The white-dwarf stars have masses below 1.2 Me and radii between -3000 and -20,000 km. They are supported almost entirely by the pressure of the degenerate electron gas. Relativistic deviations", " from Newtonian structure are only a fraction of a per cent, but relativistic effects on stability and pulsations are important from Pc::::: 108 g/cm3 to the upper limit of the white-dwarffamily at Pc = 108.4 g/cm3 [see, e.g., Faulkner and Gribbin (1968)]. The properties of white-dwarf models are fairly independent of whose version of the equation of state is used in the calculations. The properties of neutron stars are moderately dependent on the equation of state used. However, all versions lead to upper and lower limits on the mass and central density. The correct lower limits probably lie in the range 13.4 ::; 10gIO Pcmin ::; 14.0, 0.05 Me::; Mmin ::; 0.2 Me; the correct upper limits are probably in the range 15.0 ::; 10gIO Pc max ::; 16.0, 0.5 Me ::; Mmax ::; 3 Me (24.2) (24.3) (2) white-dwarf stars (3) neutron stars [see Rhoades (1971)]. Neui~on stars typically have radii between -6 km and -100 km. Relativistic deviations from Newtonian structure are great, sometimes more than 50 per cent. It appears certain that no cold stellar configuration can have a mass exceeding -5 Me [Rhoades (1971)] (1.2 Me according to the Harrison-Wheeler equation of state, Figure 24.2). Any star more massive than this must reduce its mass below this limit if it is to fade away into quiet obscurity, otherwise relativistic gravitational forces will eventually pull it into catastrophic gravitational collapse past white-dwarf radii, past neutron-star radii, and into a black hole a few kilometers in size (see Part Vll). (4) black holes \u00a724.3. PULSARS Theory predicts that, when a star more massive than the Chandrasekhar limit of 1.2 Me has exhausted the nuclear fuel in its core and has compressed its core to white-dwarf densities, an instability pushes the star into catastrophic collapse. The Birth of a neutron star by stellar collapse. Dynamics of a newborn neutron sta r Neutron sta r as a pu Isa r Pulsar radiation as a tool for studying neutron stars 628 24. PULSARS. NEUTRON STARS; QUASARS. SUPERMASSIVE STARS core implodes upon itself until nucleon-nucleon repulsion halts the implosion. The result is a neutron star, unless the core's mass is so great that gravity overcomes the nucleon-nucleon repulsion and pulls the star on in to form a black hole. Not all the star's mass should become part of the neutron star or black hole. Much of it, perhaps most, can be ejected into interstellar space by the violence that accom panies the collapse-violence due to flash nuclear burning, shock waves, and energy transport by neutrinos (\"stick of dynamite in center of star, ignited by collapse\"). The collapsed core holds more interest for gravitation theory than the ejected envelope. That core, granted a mass small enough to avoid the black-hole fate, will initially be a hot, wildly pulsating, rapidly rotating glob of nuclear matter with a strong, embedded magnetic field (see Figure 24.3). The pulsations must die out quickly. They emit a huge flux of gravitational radiation, and radiation reaction damps them in a characteristic time of --1 second [see Wheeler (1966); Thome (1969a)]. Moreover, the pulsations push and pull elementary particle reactions back and forth by raising and lowering the Fermi energies in the core's interior; these particle reactions can convert pulsation energy into heat at about the same rate as the pulsation energy is radiated by gravity. [See Langer and Cameron (1969); also \u00a71l.5 of Zel'dovich and Novikov (1971) for details and references.] The result, after a few seconds, is a rapidly rotating centrifugally flattened neutron star with a strong (perhaps 10 12 gauss) magnetic field; all the pulsations are gone. If the star is deformed from axial symmetry (e.g., by centrifugal forces or by a nonsymmetric magnetic field), its rotation produces a steady outgoing stream of gravitational waves, which act back on the star to remove rotational energy. Whether or not this occurs, the rotating magnetic field itself radiates electromagnetic waves. They slow the rotation and transport energy into the surrounding, exploding gas cloud (nebula). [See Pacini (1968), Goldreich and Julian (1968), and Ostriker and Gunn (1969) for basic considerations.] Somehow, but nobody understands in detail how, the rotating neutron star beams coherent radio waves and light out into space. Each", " time the beam sweeps past the Earth optical and radio telescopes see a pulse of radiation. The light is emitted synchronously with the radio waves, but the light pulses reach Earth earlier (-I second for the pulsar in the crab nebula) because of the retardation of the radio waves by the plasma along the way. This is the essence of the 1973 theory of pulsars, accepted by most astrophysicists. Although the mechanism of coherent emission is not understood, the pulsar radiation can nevertheless be a powerful tool in the experimental study of neutron stars. Anything that affects the stellar rotation rate, even minutely (fractional changes as small as 10-9 ) will produce measurable irregularities in the timing of the pulses at Earth. If the star's crust and mantle are crystalline, as 1973 theory predicts, they may be subject to cracking, faulting, or slippage (\"starquake\") that changes the moment of inertia, and thence the rotation rate. Debris falling into the star will also change its rotation. Whichever the cause, after such a disturbance the star may rotate differentially for awhile; and how it returns to rigid rotation may depend on such phenomena as superfluidity in its deep interior. Thus, pulsar-timing data may eventually give information about the interior and crust of the neutron star, and 629 ~~ A B' J--~---- - 104 km -----:r----I Figure 24.3. \"Collapse, pursuit, and plunge scenario\" [schematic from Ruffini and Wheeler (l97Ib)]. -------~.~_.~ \u2022 A star with white-dwarf core (A), slowly rotating, \u2022 evolves by straightforward astrophysics, \u2022 arrives at the point of gravitational instability, \u2022 collapses, and \u2022 ends up as a rapidly spinning neutron-star pancake (B,B'). \u2022 It then fragments (C) because it has too much angular momentum to collapse into a single stable object. If the substance of the neutron-star pancake were an incompressible fiuid, the fragmentation would have a close tie to well\"known and often observed phenomena (\"drop formation\"). However, the more massive a neutron star is, the smaller it is, so one's insight into this and subsequent stages of the scenario are of necessity subject to correction or amendment. One can not today guarantee that fragmentation takes place at all; nevertheless, fragmentation will be assumed in what follows. \u2022 The fragments dissipate energy and angular momentum via gravitational radiation. \u2022 One by one as they revolve they coalesce (\"pursuit and plunge scenario\"). \u2022 In each such plunge a pulse of gravitational radiation emerges. \u2022 Fragments of debris fall onto the coalesced objects (neutron stars or black holes, as the case may be), changing their angular momenta. \u2022 Eventually the distinct neutron stars or black holes or both unite into one such collapsed object with a final pulse of gravitational radiation. \u2022 The details of the complete scenario differ completely from one evolving star to another, depending on \u2022 the mass of its core, and \u2022 the angular momentum of this core. \u2022 An entirely different kind of picture therefore has to be drawn for altered values of these two parameters. \u2022 Even for the values of these parameters adopted in the drawing, the present picture can at best possess only qualitative validity. \u2022 Detailed computer analysis would seem essential for any firm prediction about the course of any selected scenario. 630 24. PULSARS, NEUTRON STARS; QUASARS, SUPERMASSIVE STARS thence (by combination with theory) about its mass and radius, These issues are discussed in detail in a review article by Ruderman (1972) as well as in Zel'dovich and Novikov (1971). \u00a724.4. SUPERMASSIVE STARS AND STELLAR INSTABILITIES Theory of the stability of Newtonian stars When a Newtonian star of mass M oscillates adiabatically in its fundamental mode, the change in its radius, oR, obeys a harmonic-oscillator equation, Mol? = -koR, (24.4) with a \"spring constant\" k that depends on the star's mean adiabatic index 1\\ [recall: r 1 (n/p)(ap/an)const.entropy]' on its gravitational potential energy [2, on the trace I = f pr2 dT of the second moment of its mass distribution, and on its mass M, k = 3M(l\\ - 4/3)1[21/1 (24.5) (See Box 24.2). If f 1 > 4/3 the Newtonian star is stable and oscillates; if f 1 < 4/3 the star is unstable and either collapses or explodes, depending on its initial condi tions and overall energetics. This result is a famous theorem in Newtonian stellar theory-but it is relevant only for adiabatic oscillations. Box 24.2 OSCILLATION OF A", " NEWTONIAN STAR The following is a volume-averaged analysis of the lowest mode of radial oscillation. Such analyses are useful in understanding the qualitative behavior and stability of a star. [See Zel'dovich and Novikov (1971) for an extensive exploitation of them.] However, for precise quantitative results, one must perform a more detailed analysis [see, e.g., Ledoux and Walraven (1958); also Chapter 26 of this book]. I. Let M = star's total mass R = star's radius p = mean density = (3/4'IT)M/ R3 p = mean pressure f 1 = mean adiabatic index = (n/p)(ap/ari)BJ.liabatiC = (p/p)(ap;ap)BJ.liabatic in Newtonian limit, where p = const. X n. 2. Then the mean pressure-buoyancy force ~UOY and the counterbalancing gravita tional force Fgrav in the equilibrium star are FbUOY = p/R = Fgrav = pM/R2 = (4'IT/3)p2R. \u00a724.4. SUPERMASSIVE STARS AND STELLAR INSTABILITIES 631 3. When the oscillating star has expanded or contracted so its radius is R + 8R, then its mean density will have changed to p + 8p = (3/4'IT)M[R-a + 8(R-a)] = p - 3(p/R) 8R, and its mean pressure will be p + 8p =P + (p/p) 1\\ 8p = p - 3(I'lf/R) 8R. The corresponding changes in the forces will be (8R) 8Fbuoy = R - RZ 8R = -(3r1 + I)RR = -(3r1 + I)FbUOY R'8p - - P 8R P 8F = (4'IT)(2-R0i5+-Z8R) = (4'IT_ ZR )(_s8R) = -SF (8R) gray R \u2022 gray R P P P 3 P 3 Consequently, the restoring force will be (recall: F bUOY = Fgrav) 4. This restoring force produces an acceleration, 8Fgray - 8FbUOY = - p 8R. Hence, the equation of motion for the oscillations is 8R = -3(I'1 - 4/3)(4'IT/3)p 8R, corresponding to a \"spring constant\" k and angular frequency of oscillation w, given by wZ = 4'17(1'1 - 4/3)p, and k = Mwz. S. A more nearly exact analysis (see exercise 39.7 for details, or Box 26.2 for an alternative derivation) yields the improved formula wZ= 3(I'1 - 4/3)lill/I, energy il =(star's self-gravitational) = ~f ~ d'Y = - ~f pp' I = (trace of seco~d ~on:ent Of) = f prz do/, 2 2 P Ix - x'i star's mass dlstnbutlOn do/ d'Y', for the square of the oscillation frequency. 6. Note that I'1 > 4/3 corresponds to stable oscillations; I'1 < 4/3 corresponds to exponentially developing collapse or explosion. Stability theory predicts \"engine-driven oscillations\" and quick death for stars of M > 60M0 Possible existence of supermassive stars Relativistic instabilities in a supermassive star 632 24. PULSARS, NEUTRON STARS; QUASARS, SUPERMASSIVE STARS In a real star no oscillation is precisely adiabatic. The oscillations in temperature cause corresponding oscillations in the stellar opacity and in nuclear burning rates. These insert energy into or extract energy from the gas vibrations. All main-sequence stars thus far observed and studied have masses below 60 Me. For such small masses, theory predicts low enough temperatures that gas pressure dominates over radiation pressure, and the adiabatic index is nearly that of nonrela tivistic gas, f 1 ::::: 5/3. Such stars vibrate stably. The net effect of the oscillating opacity and burning rate is usually to extract energy from the vibrations. Thus, they damp. (The vibrations of Cepheid variable stars are a notable exception.) the central No one has yet seen a main-sequence star with mass above about 60 Me. This is explained as follows. For masses above 60 Me' the temperature should be so high that radiation pressure dominates over gas pressure, and the adiabatic index f", " 1 is only slightly above the value 4/3 for pure radiation. Consequently the \"spring constant\" of the star, although positive, is very small. On the inward stroke of an temperature rises, and nuclear burning speeds up. (The oscillation, nuclear burning rate goes as a very high power of the central temperature; for example, in a massive star HCNO burning releases energy at a rate f ReNo ex 1;Y.) Because the spring constant is so small, the inward stroke lasts for a long time, and the enhanced nuclear burning produces a significant excess of thermal energy and pressure. Hence, on the outward stroke the star expands more vigorously than it contracted (\"engine\"). Successive vibrations are driven to higher and higher ampli tudes. Eventually, calculations suggest, the star either explodes, or it ejects enough mass by its vigorous vibrations to drop below the critical limit of M - 60 Me. Hence, stars of mass above 60 Me should not live long enough that astronomers could have a reasonable probability of discovering them. Of course, this \"engine action\" does not prevent massive stars from forming, living time, and then disrupting themselves. Such a possibility is particularly a short intriguing for supermassive stars [M between 103 Me and 109 Me - 0.01 X (mass of a galaxy)]. Although such stars may be exceedingly rare, by their huge masses and huge release of explosive energy they might play an important role in the universe. Moreover, it is conceivable that the oscillations of such stars, like those of Cepheid variables, might be sustained at large amplitudes for long times (a million years?), with nonlinear damping processes preventing their further growth. Theory predicts that general relativistic effects should strongly influence the oscillations of a supermassive star. The increase in \"gravitational force,\" oFgrav' acting on a shell of matter on the inward stroke is greater in general relativity than in Newtonian theory, and the decrease on the outward stroke is also greater. Conse quently the \"effective index\" rlcrit of gravitational forces is increased above the Newtonian value of 4/3; thus, fractional increase in \"pressure-like force of gravity\" per unit fractional change in baryon-number density r 1crit = (4/3) + a(M/R) + O(M2/R2), (24.6) \u00a724.4. SUPERMASSIVE STARS AND STELLAR INSTABILITIES 633 where a is a constant of the order of unity that depends on the structure of the star (see Box 26.2). To resist gravity, one has only the elasticity of the relativistic material of the star: ) fractional increase in \"pressure-like resisting force\" per unit fractional change in baryon number density ( _- _~P (OP) - r 1 - - \u2022 non 8 effective average over star (24.7) The effective spring constant for the vibrations of the star is governed by the delicate margin between these two indices: k = ( effective spring constant ) ContributiOn) = of \"elastic forces\" ( - contrib~tion ) of gravIty ( (24.8) (derivation in Chapter 26). The relativistic rise in the effective index of gravity above 4/3 [equation (24.6)] brings on the transition from stability (positive k; vibration) to instability (negative k; explosion or collapse) under conditions when one otherwise would have expected stability. For supermassive stars, Fowler and Hoyle (1964) show that where f is a constant of order unity. As a newly formed supermassive star contracts inward, heating up, but not yet hot enough to ignite its nuclear fuel, it approaches nearer and nearer to instability against collapse. Unless burning halts the contraction, collapse sets in at a radius Rcrit given by f 1 = 4/3 + nM/Me )-1I2 = rlcrit = 4/3 + aM/R; i.e., R = (a/2f)(Mj Me) 112 X (Schwarzschild Radius) - 104 X (Schwarzschild Radius) if M = 108 Me. The relativistic instability occurs far outside the Schwarzschild radius when the star is very massive. Relativity hardly modifies the star's structure at all; but because of the delicate balance between 8Fgrav and 8~uoy in the Newtonian oscillations (Box 24.2), tiny relativistic corrections to these forces can completely change the stability. In practice, the story of a supermassive star is far more complicated than has been indicated here. Rotation can stabilize it against relativistic collapse for a while. However, after the star has lost all angular momentum in excess of the critical value Temporary stabilization by rotation Possible scenarios for evolution and death of a supermassive star Supermassive stars as possible energy sources for quasars and galactic nuclei Other possible energy sources; dense star", " clusters; black holes Relativistic star clusters 634 24. PULSARS. NEUTRON STARS; QUASARS. SUPERMASSIVE STARS Jcrit = M2 (\"extreme Kerr limit\"; see Chapter 33), and after it has contracted to near the Schwarzschild radius, rotation is helpless to stave off implosion. Depending on its mass and angular momentum, the star may ignite its fuel before or after relativistic collapse begins, and before or after implosion through the Schwarzschild radius. When the fuel is ignited, it can wreak havoc, because even if the star is not then imploding, its adiabatic index will be very near the critical one, and the burning may drive oscillations to higher and higher amplitudes. These processes are so complex that in 1973 one is far from having satisfactory analyses of them, but for reviews of what is known and has been done, the reader can consult Fowler (1966). Thorne (1967), and Zel'dovich and Novikov (1971). The theory of stellar pulsations in general relativity is presented for Track-2 readers in Chapter 26 of this book. \u00a724.5. QUASARS AND EXPLOSIONS IN GALACTIC NUCLEI\u00b7 Supermassive stars were first conceived by Hoyle and Fowler (1963a,b) as an expla nation for explosions in the nuclei of galaxies. Shortly thereafter, when quasars were discovered, Hoyle and Fowler quite naturally appealed to their supermassive stars for an explanation of these puzzles as well. Whether galactic explosions or quasars are driven by supermassive stars remains a subject of debate in astronomical circles even as this book is being finished, in 1973. Hence, this book will avoid the issue except for the following remark. Whatever is responsible for quasars and galactic explosions must be a machine 106 to 1010 Me) and small radius (light-travel time across the of great mass (M machine, as deduced from light variations, is sometimes less than a day). The machine might be a coherent object, i.e., a supermassive star; or it might be a dense mixture of ordinary stars and much gas. Actually these two possibilities may not be distinct. Star-star collisions in a dense cluster can lead to stellar coalescence and the gradual building up of one or more supermassive stars [Sanders (1970); Spitzer (1971); Colgate (1967)]. Thus, at one stage in its life, a galactic nucleus or quasar might be driven by collisions in a dense star cluster; and at a later stage it might be driven by a supermassive star; and at a still later stage that star might collapse to leave behind a massive black hole (106-109 Me), but a black hole that is still \"live\" and active (Chapter 33). \u00a724.6. RELATIVISTIC STAR CLUSTERS The normal astrophysical evolution of a galactic nucleus is estimated [Sanders (1970); Spitzer (1971)] to lead under some circumstances to a star cluster so dense that general relativity influences its structure and evolution. The theory of relativistic star clusters is closely related to that of relativistic stars, as developed in Chapter 23. A star is a swarm of gas molecules that collide frequently; a star cluster is a swarm of stars that collide rarely. But the frequency of collisions is relatively unim- \u00a724.6. RELATIVISTIC STAR CLUSTERS 635 portant in a steady state. For the theory of relativistic star clusters, see: \u00a725.7 of this book; Zel'dovich and Podurets (1965); Fackerell, Ipser, and Thome (1969); Chapter 12 of Zel'dovich and Novikov (1971); and references cited there. A relativ istic star cluster is a latent volcano. No future is evident for it except to evolve with enormous energy release to a massive black hole, either by direct collapse (possibly a star at a time) or by first coalescing into a supermassive star that later collapses. CHAPTER 25 THE \"PIT IN THE POTENTIAL\" AS THE CENTRAL NEW FEATURE OF MOTION IN SCHWARZSCHILD GEOMETRY \"Eccentric, intervolved, yet regular Then most, when most irregular they seem; And in their motions harmony divine\" MILTON. 1665 This chapter is entirely Track 2. except for Figures 25.2 and 25.6. and Boxes 25.6 and 25.7 (pp. 639. 660. 674. and 677), which Track-1 readers should peruse for insight and flavor. No earlier Track-2 material is needed as preparation for it. \u00a725.2 (symmetries) is needed as preparation for Box 30.2 (Mixmaster cosmology). The rest of the chapter is not essential for any later chapter. but it will be helpful in understanding (1) Chapters 31-34 (gravitational", " collapse and black holes), and (2) Chapter 40 (solar-system experiments). Overview of this chapter \u00a725.1. FROM KEPLER'S LAWS TO THE EFFECTIVE POTENTIAL FOR MOTION IN SCHWARZSCHILD GEOMETRY No greater glory crowns Newton's theory of gravitation than the account it gives of the principal features of the solar system: a planet in its motion sweeps out equal areas in equal times; its orbit is an ellipse, with one focus at the sun; and the cube of the semimajor axis, a, of the ellipse, multiplied by the square of the average angular velocity of the planet in its orbit (w = 2'IT/period) gives a number with the dimensions of a length, the same number for all the planets (Box 25.1), equal to the mass of the sun: Exactly the same is true for the satellites of Jupiter (Figure 25.1), and of the Earth (Box 25.1), and true throughout the heavens. What more can one possibly expect of Einstein's theory of gravity when it in its tum grapples with this centuries-old theme of a test object moving under the influence of a spherically symmetric center of attraction? The principal new result can be stated in a single sentence: The particle is governed by an \"effective potential\" (Figure 25.2 and \u00a7\u00a725.5, 25.6) that possesses not only (1) the long distance - M/r attractive behavior and (2) the shorter distance \u00a725.1. FROM KEPLER'S LAWS TO EFFECTIVE POTENTIAL 637 (angular momentum)2jr2 repulsive behavior of Newtonian gravitational theory, but also (3) at still shorter distances a pit in the potential, which (1) captures a particle that comes too close; (2) establishes a critical distance of closest approach for this black-hole capture process; (3) for a particle that approaches this critical point without crossing it, lengthens the tum-around time as compared to Newtonian expectations; and thereby (4) makes the period for a radial excursion longer than the period of a revolution; (5) causes an otherwise Keplerian orbit to precess; and (6) deflects a fast particle and a photon through larger angles than Newtonian theory would predict. The pit in the potential being thus the central new feature of motion in Schwarzs child geometry and the source of major predictions (Box 25.2), it is appropriate to look for the most direct road into the concept of effective potential and its meaning and application. In this search no guide is closer to hand than Newtonian mechanics. Analytic mechanics offers several ways to deal with the problem of motion in a central field of force, and among them are two of central relevance here: (1) the world-line method, which includes second-order differential equations of motion, Lagrange's equations, search for constants of integration, reduction to first-order equations, and further integration in rather different ways according as one wants the shape of the orbit, () = (}(r), or the time to get to a given point on the world line, t = t(r); and (2) the wave-crest method, otherwise known as the \"eikonal method\" or \"Hamilton-Jacobi method,\" which gives the motion by the condition of \"constructive interference of wave crests,\" thus making a single leap from the Hamilton-Jacobi equation to the motion of the test object. Both methods are em(continued on page 641) West East West East 0\".0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 120 14.0 Figure 25.1. Jupiter's satellites, as followed from night to night with field glasses or telescope, provide an opportunity to check for oneself the central ideas of gravitation physics in the Newtonian approximation (dis tances large compared to Schwarzschild radius). For the practically circular orbits of these satellites, Kepler's law becomes Ml = w 2r3 (\"1-2-3 principle\") and the velocity in orbit isf3 = wr. Out of observa tions on any two of the quantities f3, M. w, r, one can find the other two. (In the opposite limiting case of two objects, each of mass M, going around their common center of gravity with separation r, one has M = w2r3/2, f3 = wr/2). The configurations of satellites I-IV of JUl,lier as given here for December 19-64 (days 0.0, 1.0. 2.0, etc. in \"universal time,\" for which see any good dictionary or encyclopedia) are taken from The American Ephemeris and Nalllical Almallac for 1964 [U.", "S. Government Printing Office (1962)]. y 638 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY Box 25.1 MASS FROM MEAN ANGULAR FREQUENCY AND SEMIMAJOR AXIS: M = w2a3 Appropriateness of Newtonian analysis shown by smallness of mass (or \"half Schwarzschild radius\" or \"extension of the pit in the potential\") as listed in last column compared to the semimajor axis a in the next-to-Iast column. Basic data from compilation of Allen (1963). Object PeriocJ& (days) w(cm- 1) a(cm) w2a3(cm) Planets Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto 87.9686 224.700 365.257 686.980 4332.587 i0759.20 30685 60188 90700 Major satellites of Jupiter 10 Europa Ganymede Callisto 1.769 138 3.551 181 7.154553 16.689018 Two satellites of Earth OS05 b Moon 95.6 min. 27.32 275.8 X 10-19 107.95 66.41 35.31 5.599 2.255 0.7905 0.4030 0.2674 X 10-19 13.711 X 10-16 6.831 3.391 1.454 X 10- 16 0.5791 X 1Q13 1.0821 1.4960 22794 7.783 14.27 28.69 44.98 59.00 X 1013 0.422 X 1011 0.671 1.070 1.883 X 1011 3.65 X 10-14 0.888 X 10- 16 6.916 X lOB 3.84 X 1010 1.477 X 10 5 1.477 1.477 1.477 1.478 1.477 1.476 1.478 1.469 X 105 141.3 141.0 140.8 141.1 0.442 0.446 'Sidereal period: time to make one revolution relative to fixed stars. bOrbiting scientific observatory launched Jan. 22, 1969, to observe x-rays and ultraviolet radiation from the sun. Perigee 531 km, apogee 560 km, above earth. SOME TYPICAL MASSES AND TIMES IN CONVENTIONAL AND GEOMETRIC UNITS. Conversion factor for mass. G/c 2 = 0.742 X 10-28 cm/g Mass Galaxy Sun Jupiter Earth 2.2 X 1044 1.6 X 1016 M eonv (g) M(em) Conversion factor for time, C = 2.998 X 1010 em/sec. One sidereal year = 365.256 days or 3.1558 X 107 sec. 1.989 X 1033 1.47 X 105 5.977 X 1027 0.444 1.899 X 1030 112 Period 1 sec 1 min 1 hr 1 day weonv (sec-I) w (em-I) 6.28 2.09 X 10-10 1.046 X 10-1 3.48 X 10-12 1.75 X 10- 3 5.80 X 10-14 7.28 X 10-5 2.42 X 10- 15 1 week - 1 monCfr 1 year 1.04 X 10-5 3.46 X 10-16 2.39 X 10-6 7.95 X 10-17 1.99 X 10-7 6.63 X 10-18 639 Figure 25.2. Effective potential for motion of a test particle in the Schwarzschild geometry of a concentrated mass M. Energy, in units of the rest mass 1\u00a3 of the particle, is denoted \u00a3 = \u00a3/1\u00a3; angular momentum, I = L/I\u00a3. The quantity r denotes the Schwarzschild r coordinate. The effective potential (also in units of 1\u00a3) is defined by equation (25.16) or, equiva lently, by the equation (:r+ V2(r) = \u00a32 (see also \u00a725.5) and has the value V= [(1- 2M/r)(1 + L2/r2)]1I2. It represents that value of \u00a3 at which the radial kinetic energy of the particle, at r, reduces to zero (\u00a3-value that makes r into a \"turning point\": V(r) = \u00a3. Note that one could equally well regard V2(r) as the effective potential, and define a turning point by the condition V2 = \u00a32. Which definition one chooses depends on convenience, on the intended application, on the tie to the archetypal differential equation!x2 + V(x) = E, and on the stress one wishes to put on correspondence with the effective potential of Newtonian theory). Stable circular orbits are possible", " (representative point sitting at mini mum of effective potential) only-for I values in excess of 2v'3 M. For any such fixed I value, the motion departs slightly from circu larity as the energy is raised above the potential minimum (see the two heavy horizontal lines for I = 3.75 M). In classical physics, the motion is limited to the region of positive kinetic energy. In quantum physics, the particle can tunnel through the region where the kinetic energy, as calculated classically, is negative (dashed prolongations of heavy horizontal lines) and head for the \"pit in the potential\" (capture by black hole). Such tunneling is absolutely negligible when the center of attraction has any macroscopic dimension, but in principle becomes important for a black hole of mass 1017 g (or 10-11 cm) if such an object can in principle exist. The diagram at the right gives values of the minimum and maxi mum of the potential as they depend on the angular momentum of the test particle. The roots of aV/or are given in terms of the \"reduced angular momentum parameter\" Lt = I/M = L/MI\u00a3 by r = -=1-+---:(\"\"I-_':\"::=12\"\"\"/~L-;:t2:::-):-;1/:;;-2' 6M \u00a32 = (U2 + 36) + (Lt2 - 12)(1 - 12/Lt2)112 54 [= 8/9 for Lt = (12)112; I for Lt = 4; (Lt2/27) + (1/3) + (I/U2) +... forU - - 00] (plus root for maximum of the effective potential; minus root for minimum: see exercise 25.18). 640 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY Box 25.2 MOTION IN SCHWARZSCHILD GEOMETRY REGARDED AS A CENTRAL POINT OF DEPARTURE FOR MAJOR APPLICATIONS OF EINSTEIN'S GEOMETRODYNAMICS 1. Newtonian effect of sun on planets and of earth on moon and man. 2. Bending of light by sun. 3. Red shift of light from sun. 4. Precession of the perihelion of Mercury around the sun. 5. Capture of a test object by a black hole as simple exemplar of gravitational collapse. 6. Dynamics of Friedmann universe derived from model of Schwarzschild \"lattice universe.\" Lat tice universe is constructed from 120 or some other magic number of concentrations of mass, each mass in an otherwise empty lattice cell of its own. Each lattice cell, though actually po lygonal, is idealized (see Wigner-Seitz approxi mation of solid-state physics) as spherical. A test object at the interface between two cells falls toward the center of each [standard radial motion in Schwarzschild geometry; see discus sion following equation (25.27). Therefore the two masses fall toward each other at a calcula ble rate. From this simple argument follows the entire dynamics of the closed 3-sphere lattice universe, in close concord with the predictions of the Friedmann model [see Lindquist and Wheeler (1957)]. 7. Perturbations of Schwarzschild geometry, I. Gravitational waves are incident on, scattered by, and captured into a black hole. Waves with wavelength short compared to the Schwarzs child radius can be analyzed to good approxi mation by the methods of geometric optics (ex ercises 35.15 and 35.16), as employed in this chapter to treat the motions of particles and photons. For longer wavelengths, there occur important physical-optics corrections to this geometric-optics idealization (see \u00a735.8 and ex ercises 32.10, 32.11). Similar considerations apply to electromagnetic and de Broglie waves. 8. Lepton number for an electron in its lowest quantum state in the geometry (\"gravitational field of force\") of a black hole is calculated to be transcended (capture of the electron!) or not according as the mass of this black hole is large or small compared to a certain critical mass M*e = M*2/me (_10 17 g or 10-11 cm) [Hartle (1971, 1972); Wheeler (197lb,c); Teitelboim (1972b,c)]. Similarly (with another value for the critical mass) for conservation of baryon num ber [Bekenstein (1972a,b), Teitelboim (1972a)]. To analyze \"transcendence or not\" one must solve quantum-mechanical wave equations, of which the Hamilton-Jacobi equation for parti cle and photon orbits is a classical limit. These quantum wave equations contain effective po tentials identical-aside from spin-", " in localizing the probability distribution that this dynamic phase obeys the simple Hamilton-Jacobi law of propagation, _oS = H(OS, x) =_1_(oS)2 + V(x). ox 2m ox 01 (2) 7. It is of no help in localizing the probability distribution that the solution of this equation for a particle of energy E is extraordinarily simple, S(x, I) = -EI + fX {2m[E - V(X)])l/2 dx + SE (3) Xo (with SE an arbitrary additive phase constant). The probability amplitude is still spread all over everywhere. There is not the slightest trace of anything like a localized world line, x = x(/). 8. To localize the particle, build a probability amplitude wave packet by superposing mono frequency (monoenergy) terms, according to a prescription qualitatively of the form tP(x, I) = tPE(x, I) + tPE+,jE(x, I) +.... (4) ( \\ 1\\ C\\.~~ { Monoenergy wave Superposition of monoenergy waves to give wave packet \u00a725.1. FROM KEPLER'S LAWS TO EFFECTIVE POTENTIAL 643 Destructive interference takes place almost everywhere. The wave packet is concentrated in the region of constructive interference. There the phases of the various waves agree; thus SE(X, t) = SE+i1E(X, t). (5) At last one has moved from a wave spread everywhere to a localized wave and thence, in the limit of indefinitely small wavelength, to a classical world line. This one equation of constructive interference ties together x and t (locus of world line in x, t, diagram). Smooth lines -20, -19, -18, etc. are wave crests of t/;E; dashed lines, wave crests for t/;E+ i1E' Shaded area is region of constructive interfer ence (wave packet). Black dots mark locus of classical world line, Lim SE+i1E(x, t) - SE(x, t) = O..JE i1E.....O it I 9. The Newtonian course of the world line through spacetime follows at once from this condition of constructive interference when one goes to the classicallirnit (Ii negligible compared to amounts of action involved; hence wavelength negli gibly short; hence spread of energies.JE required to build well-localized wave packet also negligible); thus reduces to SE+ i1E(x, t) - SE(x, t) = 0.JE CJSE(x, t) _ 0 CJE -. 644 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY Box 25.3 (continued) This condition in turn~ applied to expression (3), gives the time required to travel to the point x; thus, \" dx - + \"0 {(2/m)[E - V(X)]}1/ where 10 is an abbreviation for the quantity (\"difference in base value of dynamic phase per unit difference of energy\"). 10. Not one trace of the quantum of action comes into this final Newtonian result, for a simple reason: 1i has been treated as negligible and the wavelength has been treated as negligible. In this limit the location of the wave \"packet\" re duces to the location of the wave crest. The location of the wave crest is precisely what is governed by SE(X, I); and the condition of \"constructive interference\" OSE(X, 1)/oE = 0 gives without approximation the location of the sharply defined Newtonian world line x = x(/). 11. Relevance in the context of motion in a central field of force? Quickest known route to the concept of effective potential (Box 25.4). Box 25.4 MOTiON UNDER GRAVITATIONAL ATTRACTION OF A CENTRAL MASS ANALYZED BY HAMILTON-JACOBI METHOD A. Newtonian Theory of Gravitation Hamiltonian (1) (tildes over energy, momentum, etc., refer to test object of unit mass; test particle of mass JL follows same motion with energy E = JLE, momentum p = JLP, etc.). Equation of Hamilton-Jacobi for propagation of wave crests: _ oS =1(oS)2 + _1 (oS)2 + 1 01 2 or 2r2 00 2r 2 sin20 o</> (oS)2 _M. r (2) \u00a725.1. FROM KEPLER'S LAWS TO EFFECTIVE POTENTIAL 645 Box 25.4 (continued) Solve by \"method of separation of variables\" with convention that P = +0, 8", " ( S = - Et + P</>1> +f V - ::2(J ( M V)]1/ 2r 2 T[ - 2 )1/2 d(J dr + il- --. p\"\"L,E (3) (Check by substituting into Hamilton-Jacobi equation. Solution as sum of four terms corresponding to the four independent variables goes hand in hand with expression of probability amplitude in quantum mechanics as product of four factors, because is/ii = ilLS/ii is exponent in approximate expression for the probability amplitude.) Constructive interference of waves: (l) with slightly different E values (impose \"condition of constructive interfer ence\" oS-p l E(t, r, (J, 1\u00bb/oE = 0) tells when the particle arrives at a given r (that is, gives relation between t and r); \"',, (2) with slightly different values of the \"parameter of total angular momen tum per unit mass,\" L (impose condition of constructive interference OSj)\"\"l,E(t, r, (J, 1\u00bb/oL = 0) tells correlation between (J and r (a major feature of the shape of the orbit); (3) with slightly different values of the \"parameter of azimuthal angular mo mentum per unit mass,\" p</> (impose condition oS/op</> = 0) gives correlation between (J and 1>, (4) Planar character of the orbit. Puzzle out the value of this last integral with the help of a table of integrals? It is quicker and clearer to capture the content without calculation: the particle moves in a plane. The vector associated with the angular momentum L stands perpendicular to this plane. The projection of this angular momen tum along the z-axis is p</> = L cos a (definition of orbital inclination, a). Straight line connecting origin with particle cuts unit sphere in a point P. As time runs on, q> traces out a great circle on the unit sphere. The plane of this great circle cuts the equatorial plane in a \"line of nodes,\" at which \"hinge-line\" the two planes are separated by a dihedral angle, a. The orbit of the point P is described by x = r cos I/;,j = r sin 1/;, i = 0 in a Cartesian system of coordinates in which runs along the line of nodes and in which x lies in the plane of the orbit. j 646 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY Box 25.4 (continued) In a coordinate system in which y runs along the line of nodes and x lies in the plane of the equator, one has: r cos (J = z = i cos ex + x sin ex = r cos ~ sin ex; r sin (J cos \u00a2 = x = - i sin ex + x cos ex = r cos ~ cos ex; r sin (J sin \u00a2 = y = y = r sin ~. Eliminate reference to the Cartesian coordinates and, by taking ratios, also eliminate reference to r. Thus find the equation of the great circle route in parametric form, and tan 1> = tan~/cosex cos (J = cos ~ sin ex. Here increasing values of ~ spell out successive points on the great circle. Eliminate ~ via the relation to find or, more briefly, sin2ex - - cos2(J 2 tan 1> cos ex = I 2 sec 1> = tan ex tan (J. (5) traced out. Moreover, One verifies that 1> as calculated from (5) provides an integral of (4), thus confirming the physical argument just the arbitrary constant of integration that comes from (4), left out for the sake of simplicity from (5), is easily inserted by replacing\" 1> there by 1> - 1>0 (rotation of line of nodes to a new azimuth). The kind of physics just done in tracing out the relation between (J and 1> is evidently elementary solid geometry and nothing more. The same geometric relationships also show up, with no relativistic corrections whatsoever (how could there be any?!) for motion in Schwarzschild geometry. Therefore it is appropriate to drop this complication from attention here and hereafter. Let the particle move entirely in the direction of increasing (J, not at all in the direction of increasing 1>; that is, let it move in an orbit of zero angular momentum p</> (total angular momentum vector I inclined at angle ex = 1T/2 to z-axis). Consequently the dynamic phase S (to be divided by 11 to obtain phase of Schrodinger wave function when one turns from classical to quantum mechanics) becomes S = - Et + LO + - fr [ (TI M 2 L + -;:- - 2r 2 V )]1/2 dr + ill,", "E' (6) \u00a725.1. FROM KEPLER'S LAWS TO EFFECTIVE POTENTIAL 647 Shape of orbit: whence _ oS _ 0- ar - (J - fr Idrlr 2 [2(\u00a3 + Mlr _ V/2r2)p/2' r= VIM 1 + e cos(J Here e is an abbreviation for the eccentricity of the orb,it, e = (1 + 2EUI M2)1/2 (7) (8) (9) (greater than 1 for positive E, hyperbolic orbit; equal to 1 for zero E, parabolic orbit; less than 1 for negative E, elliptic orbit). A constant of integration has been omitted from (8) for simplicity. To reinstall it, replace (J by (J (Jo (rotation of direction of principal axis in the plane of the orbit). Other features of the orbit: semimajor axis Of) orbit when elliptic ( semiminor axis Of) orbit when elliptic ( _ VIM _ (1 _ e2)1/2 - b - I. (-2\u00a3)1/2''impact parameter\" ) for hyperbolic orbit, or \"distan~e of closest approach 1n absence of deflection' r actual distance Of) closest approach ( angle of defleCtion) in hyperbolic orbit ( b = (angular momentum per unit mass) (linear momentum per unit mass) _ -. I (2\u00a3)1/2' _ r min -. VIM (1 + 2EUIM2)1/2 + 1'8 = 'TT - 2 arc cos (1 Ie) = 2 arc tan [MI(2E}1/2I] = 2 arc tan [MI2Eb]; ( differenti~l scattering) cross sectlOn do dil 2'TTbdb 2'TT sin 8 de M2 (4E sin2812)2 (Rutherford). (10) (11) (12) (13) (14) (15) 648 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY Box 25.4 (continued) Time as correlated with position: oS o = OE = -t + fT dr [( _ M 2 \u00a3+ - - 2r 2 r 12 )]1/2\u00b7 Write r = M (-2\u00a3) _ (1 - e cos u) to simplify the integration. Get t = ~ ( _2\u00a3)3/2 (u - e sin u), (16) (17) (18) mean cirCUlar) = frequency ( 271\" (period) = W = (-2E)3/2 = (M)l/2. (19) M a3 Here u is the so-called \"mean eccentric anomaly\" (Bessel's time parameter). In terms of this quantity, one has also:. sm u - (1 - e 2)1/2 sin 0. 1 + e cosO'cos u = cos 0 + e ; 1 + e cos 0 cosO= cosu-e. l-ecosu'. smO = (1 - e 2)1/2 sin u l-ecosu ; x = rcosO = ~(cosu - e); (-2\u00a3). 0 y = rsm = (-2\u00a3)1/2 sm u.. I (20) (21) These expressions lend themselves to Fourier analysis into harmonic functions of the time, with coefficients that are standard Bessel functions: 1 f7r J (z) = -7r 271\" n.(. etzSll~u-nu du; ) x/a = - -e + ~ k- 1Jk _ 1(ke) cos kwt; 3 2 +'\" k =_00 k:t:O +00 y/a= (1- e 2)1/2 ~ k- 1Jk _ 1(ke) sin kwt k=-oo k:t: O (22) (23) (24) \u00a725.1. FROM KEPLER\u00b7S LAWS TO EFFECTIVE POTENTIAL 649 [for these and further formulas of this type, see, for example, Wintner (1941), Siegel (1956), and Siegel and Moser (1971)]. Via such Fourier analysis one is in a position to calculate the intensity of gravitational radiation emitted at the fundamental circular frequency wand at the overtone frequencies (see Chap ter 36). B. Einstein's Geometric Theory of Gravitation Connection between energy and momentum for a test particle of rest mass JL traveling in curved space, gafJpaPfJ + JL2 = O. (25) Gravitation shows up in no way other than in curvature of the geometry,", " in which the particle moves as free of all \"real\" force. Refer all quantities to basis of a test object of unit rest mass by dealing throughout with if == p/JL. Also write Pa = oS/oxa. Thus Hamilton-Jacobi equation for propagation of wave crests in Schwarzschild geometry (external field of a star; \u00a723.6) becomes _ 1 (1 - 2M/r) ( -)2 (-)2 oS + (1 _ 2M/r) oS + 1- oS + 01 ( -)2 r2 00 or ( -)2 oS r2 sin 20 o</> 1 + 1 == O. (26) Solve Hamilton-Jacobi equation. As in Newtonian problem, simplify by eliminat ing all motion in direction of increasing </>. Thus set 0 = ]it/> = oSlo</> (dynamic phase independent of </\u00bb and have S == -\u00a31 + l.JJ + IT [\u00a32 - (1 - 2M/r)(1 + 12/r2)]l/2 (1 _ ~M/r). (27) Find shape of orbit by \"principle of constructive interference\"; thus, oS ~T o == ar == 0 - J [\u00a32 _(1 _ 2M/r)(1 + V/r2)jl/2\u00b7 L dr/r 2 (28) [See equation (25.41) and associated discussion in text; also Figure 25.6.] Find time to get to given r by considering \"interference of wave crests\" belonging -----1-------- to slightl~' di~s: oS o == Eo == - I + IT [\u00a32 _ (1 _ 2M/r)(1 + V/r2)p/2 (1 _ 2M/r)\u00b7 (29) \u00a3 dr [See equation (25.32) and associated discussion in text; also Figure 25.5 and exercise 25.15.] 650 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY \u00a725.2. SYMMETRIES AND CONSERVATION LAWS From symmetries to conservation laws by: (1) Lagrangian or Hamiltonian approach (2) Killing-vector approach In analytic mechanics, one knows that symmetries of a Lagrangian or Hamiltonian result in conservation laws. Exercises 25.1 to 25.4 describe how these general prin:. ciples are used to deduce, from the symmetries of Schwarzschild spacetime, constants of motion for the trajectories (geodesics) offreely falling particles in the gravitational field outside a star. The same constants of motion are obtained in a different language in differential geometry, where a \"Killing vector\" is the standard tool for the de scription of symmetry. This section considers the general question of metric symme tries before proceeding to Schwarzschild spacetime. Killing vector, (, defined Let the metric components gp.. relative to some coordinate basis dx\" be independ ent of one of the coordinates XK, so ogp../ox\" = 0 for a = K. (25.1 ) This relation possesses a geometric interpretation. Any curve x\" = C\"'(A) can be translated in the x K direction by the coordinate shift.:1xK = e to form a \"congruent\" (equivalent) curve: x\" = C\"'(A) for a \u00a5- K and x K = CK(A) + e. Let the original curve run from A = A1 to A = A2 and have length Then the displaced curve has length L(e) = ~ 2 h [ { gp..(x(A\u00bb + e o~'K 0 } (dxP.ldA)(dx\u00b7 IdA) ]1/2 dA. 1 But the coefficient of e in the integrand is zero. Therefore the length of the new curve is identical to the length of the original curve: dL/de = O. The vector (25.2) provides an infinitesimal description of these length-preserving \"translations.\" It is called a \"Killing vector.\" It satisfies Killing's equation* Killing's equation derived (25.3) (condition on the vector field ( necessary and sufficient to ensure that all lengths are preserved by the displacement e(). This condition is expressed in covariant form. *Historical note: Wilhelm K. J. Killing, born May 10, 1847, in Burbach, Westphalia. died February II, 1923 in MUnster, Westphalia; Professor of Mathematics at the University of MUnster, 1892-1920. The key article that gives the name \"Killing vector\" to the kind of isometries considered here appeared almost a century ago [Killing (1892)]. \u00a725.2. SYMMETRIES", " as in (25.10) and show from [X, Pr] =\u00b0that {' is a Killing vectOr. Thus the commutator of two Killing vectors is itself a Killing vector. \u00a725.3. CONSERVED QUANTITIES FOR MOTION IN SCHWARZSCHILD GEOMETRY 655 Exercise 25.6. EIGENVALUE PROBLEM FOR KILLING VECTORS Show that any Killing vector satisfies ~p.;p. = 0, and is an eigenvector with eigenvalue K =\u00b0 of the equation Find a variational principle (Raleigh-Ritz type) for this eigenvalue equation. (25.11) \u00a725.3. CONSERVED QUANTITIES FOR MOTION IN SCHWARZSCHILD GEOMETRY Consider a test particle moving in the Schwarzschild geometry, described by the line element (25.12) This expression for the geometry applies outside any spherically symmetric center of attraction of total mass-energy M. It makes no difference, for the motion of the particle outside, what the geometry is inside, because the particle never gets there; before it can, it collides with the surface of the star-if the center of attraction is a star, that is to say, a fluid mass in hydrostatic equilibrium. At each point throughout such an equilibrium configuration, the Schwarzschild coordinate r exceeds the local value of the quantity 2m(r); see \u00a723.8. Therefore the Schwarzschild coordinate R of the surface exceeds 2M. Consequently, expression (25.12) applies outside any equilibrium configuration, no matter how compact (r > R > 2M implies that one need not face the issue of the \"singularity\" at r = 2M). The more compact the configuration, however, the more of the Schwarzschild geometry the test particle can explore. The ideal limit is not a star in hydrostatic equilibrium. It is a star that has undergone complete gravitational collapse to a black hole. Then (25.12) applies arbitrarily close to r = 2M. This idealization is assumed here (\"black hole\"), because the analysis can then cover the maximum range of possible situations. Why attention focuses on particle orbits around a black hole Wherever the test particle lies, and however fast it moves, project that point and project that 3-velocity radially onto a sphere of some fixed r value, say, the unit sphere r = 1. The point and the vector together define a point and a vector on the surface of the unit sphere; and they in tum mark the beginning and define the totality of a great circle. As the particle continues on its way, the radial projection of its ------Pp'00SlsiJ.ltiLCoIlLD will CQntinue_~Qn that great circle. To depart from the great circle on one side or the other would be to give preference to the one hemisphere or the other of the unit sphere, contrary to the symmetry of the situation. Orient the coordinate system so that the radial projection of the orbit coincides with the equator, () = 'TT/2, of the polar coordinates (see Box 25.4 for the spherical trigonometry of a more general orientation of the orbit, and for eventual specializa- Choice of coordinates to make particle orbit lie in \"equator,\" 8 = ':7/2 Conserved quantities for particle motion: (1) E (2) L (3) p. (4) E=Elp. (5) I =LIp. Effective potential V. and equations for orbit when p.;t:O 656 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY tion to a polar orbit, in contrast to the equatorial orbit considered here). In polar coordinates as so oriented, the particle has at the start, and continues to have, zero momentum in the B direction; thus, p 6 = dB/d'A = O. The expression (25.12) for the line element shows that the geometry is unaffected by the translations t ---+ t +..1t, 9 ---+ 1> +..11>. Thus the coordinates t and 1> are \"cyclic.\" The conjugate momenta Po=-E and p \u00a2 =-+- L (L 2:: 0) are therefore conserved. This circumstance allows one immediately to deduce the major features of the motion, as follows. The magnitude of the 4-vector of energy-momentum is given by the rest mass of the particle, or - (25.13) (25.14) (dr)2 L2 d'A + ~ + JL2 = O. E2 (l - 2M/r) + (l - 2M/r) I Moreover, one knows from the equivalence principle that test particles follow the same world line regardless of mass. Therefore what is relevant for the motion of particles is not the energy", " and angular momentum themselves, but the ratios \u00a3 = E/ = (energy per unit), JL rest mass I = L/JL = (angula~ momentum). per umt rest mass (25.15) Recall also 'A = 'T/ JL = (pr~per time per). umt rest mass Then (25.14) becomes an equation for the change of r-coordinate with proper time in which the rest mass makes no appearance: (:r = \u00a32 - (l - 2M/r)(l + 12/r2) = \u00a32 _ 0(r). Here (25.16a) (25.16b) is the \"effective potential\" mentioned in \u00a725.1 and Figure 25.2 and to be discussed \u00a725.3. CONSERVED QUANTITIES FOR MOTION IN SCHWARZSCHILD GEOMETRY 657 in \u00a725.5. For the rate of change of the other two relevant coordinates with proper time, one has, assuming a \"direct\" orbit (dcp/dr > 0; P\", = +L rather than -L), p'\" Jl dep -= - - -= - -= dr g\"''''L Jl 1 dep Jl d\"A (25.17) I r 2 and 1 - 2M/r (25.18) Knowing r as a function of 7\" from (25.16), one can find ep and t in their dependence on 7\" from (25.17) and (25.18). Symmetry considerations have in effect reduced the four coupled second-order differential equations plJ.;.p' = 0 of geodesic motion to the single first-order equation (25.16). For objects of zero rest mass, it makes no sense to refer to proper time, and a slightly different treatment is appropriate (\u00a725.6). Before looking, in \u00a725.5, at the motions predicted by equations (25.16) to (25.18), it is useful to analyze the physical significance of the constants Po and p \"\" and to identify other physically significant quantities whose values will be of interest in studying these orbits. One calls E = -Po the \"energy at infinity\"; and L = IP\",I, for equatorial orbits, the \"total angular momentum.\" To justify these names, compare them with standard quantities measured by an observer at rest on the equator of the Schwarzschild coordinate system as the test particle flies past him in its orbit. Let E 10Cal =po=(WO,p) =(lgool1/2 dt,p) = Igooll/2pO = Igooll/2dt/d\"A = (l - 2M/r)l/2 dt/d\"A (25.19) Interpretation of E as \u00b7\u00b7energyat infinity\u00b7\u00b7 and L as \u00b7\u00b7angular momentum\u00b7\u00b7 be the energy he measures in his proper reference frame, and let v\u00a2=p~ = (W\u00a2,p) = (lg\",,,,1 1/2 dep,d/d\"A) po E10Cal E10Cal = r(dep/ d\"A) =---.l!.:L rE loCal be the tangential component of the ordinary velocity he measures. [Note: wa are the basis one-forms of the observer's proper reference frame; see equations (23.15a,b).] In terms of these locally measured quantities, the energy-at-infinity is E10Cal (25.20) E = -Po = -goopO = Igool1/2Elocal = (l - 2M/r)l/2E10Cal = constant. (25.21) It therefore represents the locally measured energy E1oCal' corrected by a factor Igooll/ 2\u2022 For any particle that flies freely (geodesic motion) from this observer to r = 00, the correction factor reduces to unity, and E10Cal (as measured by a second observer, this time at infinity) becomes identical with E. Similarly the angular momentum from (25.20) is (25.22) 658 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY This, like E = -Po' represents a quantity that is conserved, and whose interpretation for r --+- 00 on any orbit is familiar. Finally, recall that the total 4-momentum of two colliding particles PI + P2 or (P/l)l + (p/lh is conserved in a point collision (at any r). Therefore the totals (E)1 + (\u00a3)2 = (-PO)1 + (-Poh and (P\u00a2)1 + (p\u00a2h are also conserved. One of the colliding particles may be on an orbit that could never reach out to r = 00, but this makes no difference.", " This conservation principle allows and forces one to take over the terms E = \"energy at infinity\" and L = \"angular momentum,\" valid for orbits that do reach to infinity, for an orbit that does not reach to infinity. EXERCISES Exercise 25.7. RADIAL VELOCITY OF A TEST PARTICLE Obtain a formula for the radial component of velocity v, that an observer at r would measure [see (25.20) for v\u00a2]. Express \u00a3local' vi, and v\u00a2 in terms of r and the constants \u00a3, P,p- Exercise 25.8. ROTATIONAL KILLING VECTORS FOR SCHWARZSCHILD GEOMETRY (a) Show that in the isotropic coordinates of exercise 23.1, the metric for the Schwarzschild geometry takes the form (b) Exhibit a coordinate transformation that brings this into the form with r = (x 2 + y2 + Z2)1I2. (c) Show that (x = y(o /oz) - z(o /0y) and similar vectors (y and (. are each Killing vectors, by verifying (see exercise 25.5c) that the Poisson brackets [X, Lx] vanish for each LK = P \u2022 (K' K = x, y, z. (d) Show that L = (2/oCP)t,r,0; and show that for orbits in the equatorial plane L. = P\u00a2' Lx = L y = O. Exercise 25.9. CONSERVATION OF TOTAL ANGULAR MOMENTUM OF A TEST PARTICLE Prove by a Poisson-bracket calculation that the total angular momentum squared, L2 = P0 2 + (sin 8)-2p\u00a22 is a constant of motion for any Schwarzschild geodesic. Exercise 25.10. SELECTING EQUATION BY SELECTING WHAT IS VARIED Write out the integral 1 that is varied in (25.8) for the special case of the Schwarzschild metric (25.12). What equation results from the demand 81 = 0 if only <1>(/\\) is varied? If only t(t..)? Exercise 25.11. MOTION DERIVED FROM SUPER-HAMILTONIAN Write out the super-Hamiltonian (25.10) for the special case of the Schwarzschild metric. Deduce from its form that Po and p\u00a2 are constants of motion. Derive (25.14), (25.17), and (25.18) from this super-Hamiltonian formalism. Law of \"energy redshift\"' (\"gravitational redshift\") \u00a725.5. ORBITS OF PARTICLES 659 \u00a725.4. GRAVITATIONAL REDSHIFT The conservation law Igool1/2ElOCa! = constant (equation 2521), which is valid in this form for any time-independent metric with go;= 0 and for particles with both zero and non-zero rest mass, is sometimes called the \"law of energy red-shift.\" It describes how the locally measured energy of any particle or photon changes (is \"red-shifted\" or \"blue-shifted\") as it climbs out of or falls into a static gravitational field. For a particle of zero rest mass (photon or neutrino), the locally measured energy ElOCa!' and wavelength Aloca! (not to be confused with affine parameter!), are related by ElOCa! = h/AlOCa!' where h is Planck's constant. Consequently, the law of energy red-shift can be rewritten as (25.25) A photon emitted by an atom at rest in the gravitational field at radius r, and received by an astronomer at rest at infinity is red-shifted by the amount z = LiA/A = (Arecelved - Aemitt~/Aemitted = Igoo(r)I-1/2 - 1, z = (l - 2M/r)-1/2 - 1, z::::; M/r in Newtonian limit. (25.26) (25.26N) Note that these expressions are valid whether the photon travels along a radial path or not. Exercise 25.12. REDSHIFT BY TIMED PULSES EXERCISE Derive expression (25.26) for the photon redshift by considering two pulses of light emitted successively by an atom at rest at radius r. [Hint: If..:I'Tem is the proper time between pulses as measured by the emitting atom, and..:I'Trec is the proper time separation as measured by the observer at r = 00, then one can idealize t.. em as..:I'Tem and t.. rec as..:I'Trec'] \u00a725.5. ORBITS OF PARTICLES Turn attention now from energy red-shift to the orbit of", " a particle in the Schwarzs child geometry. The position as a function of proper time follows upon solving (25.16 a), dr)2 dr. + V-(r) = E2, -? _ ( where V is the \"effective potential\" defined by (25.l6a) (25.l6b) Qualitative features of orbits diagnosed from effective-potential diagram 660 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY and illustrated in Figure 25.2 and Box 25.6. The first diagram in Box 25.6 gives V2(r) as a function of r. It is relevant even in the \"domain inside the black hole\" (r < 2M), where V2 is negative (see Chapter 31). It serves as a model for, and is closely related to, the \"effective potential\" B-2(r) used in \u00a725.6 to analyze photon orbits. The final diagram in Box 25.6 gives VCr) itself as a function of r. Energy levels in this diagram or in Figure 25.2 can be interpreted as in any conventional energy-level diagram. The difference in energy between two levels represents energy, as measured at infinity, of the photon given off in the transition from the one level to the other. Whether one plots VCr) or V2(r) as a function of r is largely a matter of convenience. The important point is this: a value of r where VCr) becomes equal to the available energy \u00a3, or V2(r) becomes equal to \u00a32, is a turning point. A particle that was moving to larger r values, Once arrived at a turning point, turns around and moves to smaller r values. Or when a particle moving to smaller r values comes to a turning point, it reverses its motion and proceeds to larger r values. A turning point is not a point of equilibrium. A stone thrown straight up does not sit at a point of equilibrium at the top of its flight. However, when E - VCr), or \u00a32 - per), instead of having a single root, has a double root, then one does deal with a point of equilibrium (only possible because of \"centrifugal force\" fighting against gravity). When this equilibrium Occurs at a minimum of VCr), it is a stable equilibrium; at a maximum, an unstable equilibrium. Thus all the major features of the motion in the r direction can be read from a plot of the effective potential as a function of r (plot depends on value of L) and from a knowledge of the \u00a3 value (Figure 25.2, with further details in Box 25.6). Box 25.6 QUALITATIVE FEATURES OF ORBITS OF A PARTICLE MOVING IN SCHWARZSCHILD GEOMETRY A. Equations Governing Orbit (see text for derivation) 1. Effective-potential equation for radial part of motion: (dr/d'T)2 + pel, r) = \u00a32, V2(L, r) = (l - 2M/r)(l + V/r 2), \u00a3 = (energy at infinity per unit rest mass), I = (angular momentum per unit rest mass). 2. Supplementary equations for angular and time motion for \"direct\" orbit, d\u00a2/d'T > 0: d\u00a2/d'T = L/r2, dt d'T \u00a3 =....,.---=~1 - 2M/r' \"Turning points\" of orbit occur where horizontal line of height \u00a32 crosses V2 \u00a725.5. ORBITS OF PARTICLES 661 B. Newtonian Limit, IE M/r<1,1/r<1 11 < 1, 1. Speak not about \"energy-at-infinity per unit rest mass,\" \u00a3 = EIIL = (l - V;,>-1/2, but instead about the \"nonrelativistic en ergy per unit rest mass,\" E = 1. (\u00a32 - 1) - \u00a3 _ 1 -1. v 2 -2 oc' -2 2. Speak not about V2(L, r) but instead about the Newtonian effective potential, - 1 - M L2 VN (L,r)=\"2(J72-1)::::;--;:+ 2r2'3. Rewrite effective-potential equation in the form 1 (dr)2 \"2 dr + VN(L, r) = E. - 4. From the effective-potential diagram and the subsidiary equation d\u00a2/dr = L/r2, conclude that: a. Particles with E ~ 0 (\u00a3 ~ 1) come in from r = 00 along hyperbolic or pa rabolic orbits, are reflected off the effective potential at E = VN[\u00a32 = V2; (dr/dr)2 = 0], and \"turn", "ing point\"; return to r = 00. b. Particles with E < 0 (\u00a3 < 1) move back and forth in an effective potential well between periastron (inner turning point of elliptic orbit) and apastron (outer turning point). C. Relativistic Orbits Use the effective-potenti~l dia~ram of part A - - - - - c-reprOCfiIcecrnereror va.rious L), in the same way one uses the Newtonian diagram of part B, to deduce the qualitative features of the orbits. The main conclusions are these. - -,. >0----.. o 662 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY Box 25.6 (continued) 1. Orbits with periastrons at r ~ Mare Kep lerian in form, except for the periastron shift (exercise 25.16; \u00a740.5) familiar for Mercury. 2. Orbits with periastrons at r ~ 10M differ markedly from Keplerian orbits. 3. For LIM:::; 20 there is no periastron; any incoming particle is necessarily pulled into r = 2M. 4. For 20 < II M < 4 there are bound orbits in which the particle moves in and out between periastron and apastron; but any particle coming in from r = 00 (un bound; \u00a32 ~ 1) necessarily gets pulled into r=2M. 5. For U = II M > 4, there are bound orbits; particles coming in from r = 00 with \u00a32 < Vmax 2 = (l - 2um)(l + U2 um2), 1 + VI - l21U2 6 um = reach periastrons and then return to r = 00; but particles from r = 00 with \u00a32 > Vmax2 get pulled into r = 2M. 6. There are stable circular orbits at the mini mum of the effective potential; the mini mum moves inward from r = 00 for I = 00 to r = 6M for U = II M = 20. The most tightly bound, stable circular orbit (II M = 20, r = 6M) has a fractional binding energy of Jl - E = 1 _ \u00a3 = 1 - v'879 = 0.0572. Jl 7. There are unstable circular orbits at the maximum of the effective potential; the maximum moves outward from r = 3Mfor liM = 20. A I = 00 particle in such a circular orbit, if per turbed inward, will spiral into r = 2M. If perturbed outward, and if it has \u00a32 > 1, it will escape to r = (lQ. If perturbed out- to r = 6M for 0.98 0.96 0.94 0.92 L.J..J........u..L..L.J....l...I...L..L.J...L...L.J...J...J...J-'u..L.l..J 25 10 20 15 5 o -r/M-. ward, and if it has \u00a32 < 1, it will either reach an apastron and then enter a spiral ing orbit that eventually falls into the star (e.g., if o\u00a3 > 0, with unchanged angular momentum); or it will move out and in between apastron and periastron, in a sta ble bound orbit (e.g., if o\u00a3 < 0, again with unchanged angular momentum). \u00a725.5. ORBITS OF PARTICLES 663 When one turns from qualitative features to quantitative results, one finds it appropriate to write down explicitly the proper time AT required for the particle to augment its Schwarzschild coordinate by the amount Ar; thus (with the convention that square roots may be negative or positive, # =-t-a) (25.27) The integration is especially simple for a particle falling straight in, or climbing Radial orbits: straight out, for then the angular momentum vanishes and the integral can be written in an elementary form that applies (with the change T --+- t) even in Newtonian mechanics, T =f dT =f dr [2M/r - 2M/R]1/2' (25.27') Here R =2M/(1 - \u00a32) is the radius at which the particle has zero velocity (\"apas tron\"). The motion follows the same \"cycloid principle\" that is so useful in nonrela tivistic mechanics (Figure 25.3). Thus, in parametric form, one has R r =T (1 + cos 1), R ( R )1/2 T = T 2M. (1) + SIn 1), (1) \"cycloidal\" form of r(or) for radial bound orbits (25.28) with the total", " proper time to fall from rest at r = R into r = 0 given by the expression T =!!. R (..B-)1/2 2 2M (25.29) (shorter by a factor 1/ Vi than the time for fall under pull of the same mass, distributed over a sphere of radius R; see dotted curve in Figure 25.3). What about the Schwarzschild-coordinate time taken for a given motion? Take equation (25.l6a) for general motion (radial or nonradial), and where dr/dT appears, replace it by dr dT = dr dt dt dT = \u00a3 dr dt 1 - 2M/r - dr\u00b7 = E-. dt Here r\u00b7 is an abbreviation for a new \"tortoise coordinate,\" (25.30) (25.31 ) which was introduced by Wheeler (1955) and popularized by Regge and Wheeler (1957). Thus find the equation ( Ed;; Y+ j/2 = \u00a32. (2) \"tortoise\" radial (25.32) coordinate as function of coordinate time, r\u00b7(t) 664 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY tr I -Time_.........~.~'\" Figure 25.3. A cycloid gives the relation between proper time and Schwarschild r coordinate for a test particle falling straight in toward center of gravitational attraction of negligible dimensions. The angle of turn of the wheel as it rolls on the base line and generates the cycloid is denoted by 11. In terms of this parameter, one has R r = 2\" (I + cos 11) T = - R ( R )112 2 2M (Schwarzschild r-coordinate) (11 + sin 11) (proper time) (note difference in scale factors in expressions for r and for T). The total lapse of proper time to fall from r = R to r = 0 is T = ('1T/2)(R3/2M)1I2. The same cycloid relation and the same expression for time to fall holds in Newton's nonrelativistic theory of gravitation, except that there the symbol T is to be replaced by the symbol t (ordinary time). Were one dealing in Newtonian theory with the same attracting mass M spread uniformly over a sphere of radius R, with a pipe thrust through it to make a channel for the motion of the test particle, then that particle would execute simple harmonic oscillations (dotted curve above). The angular frequency w of these vibrations would be identical with the angular frequency of revolution of the test particle in a circle just grazing the surface of the planet, a frequency given by Kepler's law M = w2R3. In this case, the time to fall to the center would be ('1T/2)(R3/M)1I2, longer by a factor 2112 than for a concentrated center of attraction (concentrated mass: stronger accelera tion and higher velocity in the later phases of the fall). The expression for the Schwarzschild-coordinate time t required to reach any point r in the fall under the influence of a concentrated center of attraction is complicated and is not shown here (see equation 25.37 and Figure 25.5). The same cycloidal relation that connects r with time for free fall of a particle also connects the radius of the \"Friedmann dust-filled universe\" with time (see Box 27.1), except that there the cycloid diagram applies directly, without any difference in scale between the two key variables: radius oJ -sp ere h 3 ( a = -2 (I - cos 11) \"\" -11 2 (for small 11), a 4 ) coordinate time identical with proper time as measured on dust particle ( = f (11 - sin 11) \"\" ~ 113 (for small 11). The starting point of 11 is renormalized to time of start of expansion; see Lindquist and Wheeler (1957) for more on correlation between fall of particle and expansion of universe. Here the effective potential is the same effective potential that one dealt with before, Moreover, the E on the righthand side is the same E that appeared in the ear~er equation for (dr/dr)2. Therefore the turning points and the qualitative description of the motion are both the same as before. \"A turning point is a turning point is (25.33) (3) details of the approach to the Schwarzschild radius (r = 2M) \u00a725.5. ORBITS OF PARTICLES 665 a turning point.\" Right? Right about turning points; wrong about the conclusion. The story has it that Achilles never could pass the tortoise. Whenever he caught up with where it had been, it had moved ahead to a new location; and when he got there, it was still further ahead; and so on ad infinitum", ". Imagine the race between Achilles and the tortoise as running to the left and the expected point of passing as lying at r = 2M. The r-coordinate has no inhibition about passing through the value r = 2M. Not so r\u00b7, the \"tortoise coordinate.\" It can go arbitrarily far in the direction of minus infinity (corresponding to the infinitely many times when Achilles catches up with where the tortoise was) and still r remains outside r = 2M:,/2M 1.0\u05e1\u05e1oo1 1.0001 1.01 1.278465,\u00b7/2M -12.8155 -8.2102 - 3.5952 0 2 2 5 10 10,000 6.386 12.303 10,009.210 It follows that there is a great difference between the description of the motion in terms of the proper time or of a clock on the falling particle (r goes all the way from r = R down to r = 0 in the finite proper time of 25.29) and a description of the motion in terms of the Schwarzschild-coordinate time t appropriate for the faraway observer (r\u00b7 goes all the way from r\u00b7 = R\u00b7 down to r\u00b7 = - 00; infinite t required for this; but even in infinite time, as r\u00b7 goes down to - 00, r is only brought asymptotically down to r - 2M). Thus the second description of the motion leaves out, and has no alternative but to leave out, the whole range of r values from r = 2M down to zero: perfectly good physics, and physics that the falling particle is going to see and explore, but physics that the faraway observer never will see and never can see. If the tortoise coordinate did not exist, it would have to be invented. It invests each factor ten of closer approach to r = 2M with the same interest as the last factor ten and the next to come. It proportions itself in accord with the amount of Schwarzschild-coordinate time available to the faraway observer to study these more and more microscopic amounts of motion in more and more detail. ~_ Figure 25.4 shows the effective potential Vof (25.33) and of Figure 25.2 replotted asa- function-~of the Tortoise coordinate. The approach of V to zero at r = 2M shows up as an exponential approach of V to zero as r\u00b7 goes to minus infinity. Thus in moving \"towards the black hole\" (r = 2M, r\u00b7 = - 00), the particle, as described in coordinate time t, soon casts off any effective influence of any potential, and moves essentially freely toward decreasing r\u00b7, in accordance with the equation E (..~ E2; dC12 dtJ (25.34) that is, \"with the speed of light\" (dr\u00b7 / dt ~ - I). This dependence of r\u00b7 on t implies at once an asymptotic dependence of r itself on Schwarzschild-coordinate time t, of the form r = 2M + (constant X e- tI2M ). (25.35) This result is independent of the angular momentum of the particle and independent also of the energy, provided only that the energy-per-unit-mass E is enough to 666 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY 1.2.-------r--------.;:---, tV I 0.8 0.6 0.4 0.2 -5 o 5 10 -r*/M-.. Figure 25.4. Effective potential for motion in Schwarzschild geom etry, expressed as a function of the tortoise coordinate, for selected values of the angular momentum of the test particle. The angular momentum L is expressed in units Mp., where M is the mass of the black hole and p. the mass of the test particle. The effective potential (includ thus, V == V/p.. ing rest mass) is expressed in units p.; The tortoise coordinate r\u00b7 = r + 2M In (r/2M I) is given in units M. surmount the barrier (Figure 25.4) of the effective potential-per-unit-mass V. (More will be said on the approach to r = 2M in Chapter 32, on gravitational collapse.) To replace the asymptotic formula (25.35) by a complete formula requires one to integrate (25.32); thus, t =f dt =f \u00a3 dr\u00b7 [\u00a32 _ 0]1/2 \u00a3 dr (25.36) [\u00a32 _ (l - 2M/r)(l + V /r2)]1/2 (l - 2M/r). f = The integration here is not easy, even for pure radial motion (L = 0), as is seen in the complication of the resulting expression (", "Khuri 1957): t = T + 2M 2M - 1 ) ( R [( R )1/2] R ( R 1) + T 2M - 1 (R/2M - 1)1/2 + tan (1)/2) I (R/2M - 1)1/2 - tan (1)/2). + 2Mln \\ )1/2. SIn 1) (25.37) Here 1) is the same cycloid parameter that appears in equation (25.28) and Figure 25.3 (see the detailed plot in Figure 25.5 of the correlation between rand t, illustrat- \u00a725.5. ORBITS OF PARTICLES 667 6 5 4 triM 3 I 2 o 5 10 15 20 25 30 -timeIM-. Figure 25.5. Fall toward a Schwanschild black hole as described (a) by a comoving observer (proper time T) and (b) by a faraway observer (Schwarzschild-coordinate time t). In the one description, the point r = 0 is attained, and quickly [see equation (25.28\u00bb). In the other description, r = 0 is never reached and even r = 2M is attained only asymptotically [equations (25.35) and (25.37\u00bb). The qualitative features of the motion in both cases are most easily deduced by inspection of the \"effective potential-per-unit-mass\" V in its dependence on r (Figure 25.2) when one is interested in proper time; or the same effective potential V in its dependence on the \"tortoise coordinate\" r\u00b7 [Figure 25.4 and equation (25.31\u00bb) when one is interested in Schwarzschild-coordinate time t. ing the asymptotic approach to r = 2M). The difficulty in the integration for t, as compared to the ease of the integration for or (25.28), has a simple origin. Only two r-values appear in (25.27a) as special points when L is zero: the starting point, r = R, where the velocity vanishes, and the point r = 0, where dr/dor becomes infinite. In contrast (25.36), rewritten as t - f dt- f [l - 2M/RF/2 dr [2M/r - 2M/R]1I2 (1 - 2M/r), (25.36') contains three special points: r = R, r = 0, and the added point with all the new physics, r = 2M. To admit angular momentum is to increase the number of special points still further, and to make the integral unmanageable except numerically or qualitatively (via the potential diagram of Figure 25.4), or in terms of elliptic functions [Hagihara (1931 )]. It is often convenient to abstract away from the precise value r = R at the start of the collapse. In this event, one deals with the limit R --+- 00. Then it is convenient to displace the zero of proper time to the instant of final catastrophe. In this limit, one has -or/2M =--(2/3)(r/2M)3/2, t/2M = -(2/3)(r/2M)3/2 _ 2(r/2M)1/2 + In (r/2M)~~: + 1. - 1 (r/2M) (25.38) ------ --~ ~-- At very large negative time, the particle is far away and approaching only very slowly. Then one can write (25.39a) (4) free-fall from r = 00 y Nonradial orbits: (1) Fourier analysis 668 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY whether one refers to coordinate time or to proper time. However, the final stages of infall are again very different, when expressed in terms of proper time (or --+- 0, r --+- 0), from what they are as expressed in terms of Schwarzschild-coordinate time, (25.39b) Turning from pure radial motion to motion endowed with angular momentum, one has a situation where one would like to express the principal quantities of the motion (components of displacement, velocity, and acceleration) in Fourier series (in Schwarzschild-coordinate time), these being so convenient in the Newtonian limit in analyzing radiation and perturbations of one orbit by another and tidal perturba tions of the moving particle itself by the tide-producing action of the center of attraction. Any exact evaluation of these coefficients would appear difficult. For the time being, the values of the Fourier amplitudes would seem best developed by successive approximations starting from the Newtonian analysis (see Box 25.4 and references cited", " there). In connection with any such Fourier analysis, it is appropriate to recall that the fundamental frequency alone appears, and all higher harmonics have zero amplitude, when the motion takes place in an exactly circular orbit (opposite extreme from the pure radial motion of I = 0). Therefore it is of interest to note (exercise 25.19) that the circular frequency w of this motion, as measured by a faraway observer, is correlated with the Schwarzschild r-value of the orbit by exactly the Keplerian formula of non-relativistic physics: (exact; general relativity). (25.40) (2) details of angular motion Turn now from the correlation between r and time to the correlation between r and angle of revolution (<I> in the analysis here; () in the Hamilton-Jacobi analysis of Box 25.4; this difference in name is irrelevant in what follows). Return to equation (25.16), and recall also equation (25.17) Solve the second equation for dor, and substitute into the first to find I dr)2 _ _ r 2 d<l> + V2(r) = \u00a32, ( or equivalently, with u = Mlr and U = II M = LIMil, dU)2 = \u00a32 - (l - 2u)(l + U2U2). d<l> U2 ( (25.41 ) (25.42) (3) n,,,'y 0;\"\"'\"O'b;\"~ periastron shift (4) qualitative features of angular motion Scattering of incoming particles: (1) Rutherford (nonrelativistic) cross section (2) new features due to relativistic gravity \u00a725.5. ORBITS OF PARTICLES 669 Exercise 25.16 presents an alternative differential equation derived from this for mula, and uses it to obtain the following expression for the angle swept out by the particle or planet, moving in a nearly circular orbit, between two successive points of closest approach: (25.43) The radial motion turns around from ingoing to outgoing, or from outgoing to ingoing, whenever the quantity \u00a32 - per), or \u00a3 - V(r), plotted as a function of r, undergoes a change of sign, and this as clearly here in the correlation between rand cI> as in the earlier correlation between r and time. Recall again the curves of Figure 25.2 for VCr) as a function of r for selected L values. From them one can read out, without any calculation at all, the principal features of typical orbits (Box 25.6) obtained by detailed numerical calculation. Characteristic features are (1) circular orbit when \u00a3 coincides with a minimum of the effective potential V(r), (2) precession when \u00a3 is a little more than Vmin' (3) temporary \"orbiting\" (many turns around the center of attraction) when \u00a3 is close to a maximum Vmax of the effective potential, (4) \"capture into the black hole\" when \u00a3 exceeds Vmax' A more detailed analysis appears in Box 25.6. [For explicit analytic calculation of orbits in the Schwarzschild geometry, see Hagihara (1931), Darwin (1959 and 1961), and Mielnik and Plebanski (1962).] For orbits of positive energy, no feature of the inverse-square force is better known than the Rutherford scattering formula. It gives the \"effective amount of target area\" presented by the center of attraction for throwing particles into a faraway receptor that picks up everything coming off into a unit solid angle at a specified angle of deflection e: da dO _ M2 2 [4(\u00a3 - 1) sin e/2] 2 (Rutherford; nonrelativistic) (25.44) (derivation in equations 8 to 15 of Box 25.4). When one turns from the Newtonian analysis to the general-relativity treatment, one finds two striking new features of the scattering associated with the phenomenon of orbiting. (1) The particles that come off at a given angle of deflection e now include not only those that have really been deflected by e (the only contribution in Rutherford scattering), but also those that have been deflected bye + 217, e + 417,... etc. (an infinite series of contribu tions). (2) These supplementary contributions, while finite in amount, and even finite in amount \"per unit range of e,\" are not finite in amount when expressed \"per unit of solid angle dO = 217 sin e de\" in either the forward direction (e = 0) or the backward direction (e = 17). This circumstance produces no spectacular change in the forward scattering, for that is already infinite in the nonrelativistic approxi mation (infinity in Rutherford value of da/dO as e = 0 is approached. arising from 670 25. PARTICLE MOTION IN SCHWARZS", "CHILD GEOMETRY particles flying past with large impact parameters and experiencing small deflections; see exercise 25.21). In contrast, the backward scattering, which was perfectly finite in the Rutherford analysis, acquires also an infinity: constant sine. (25.45) This concentration of scattering in the backward direction is known as a \"glory.\" The effect is most readily seen by looking at the brilliant illumination that surrounds the shadow of one's plane on clouds far below (180 0 scattering of light ray within waterdrop). It is also clearly seen in observations on the scattering of atoms by atoms near e = 180 0 \u2022 No dwarf star, not even any neutron star, is sufficiently compact to be out of the way of a high-speed particle trying to make such a 180 0 tum. Only a black hole is compact enough to produce this effect. Further interesting features of motion in Schwarzschild geometry appear in the exercises below. EXERCISES Exercise 25.13. QUALITATIVE FORMS OF PARTICLE ORBITS Verify the statements about particle orbits made in part C of Box 25.6. Exercise 25.14. For a scattering orbit (Le., unbound orbit), show that I parameter and v\", the asymptotic ordinary velocity; also show that IMPACT PARAMETER == -Ev~b, where b is the impact (25.46) Draw a picture illustrating the physical significance of the impact parameter. Exercise 25.15. TIME TO FALL TO r = 2M Show from equation (25.16) and the first picture in Box 25.6 that orbits (general I value!) which approach r = 2M do so in a finite proper time, but (equation 25.32) an infinite coordinate time t. For equilibrium stars, which must have radii R > 2M, the coordinate time t to fall to the surface is finite, of course. Exercise 25.16. PERIASTRON SHIFT FOR NEARLY CIRCULAR ORBITS Rewrite equation (25.42) in the form Express the constant Uo=Mlro in terms of LIM, and express Eo in terms of uo' Show for a nearly circular orbit of radius ro that the angle swept out between two successive periastra (points of closest approach to the star) is (25.47) Sketch the shape of the orbit for ro = 8M. (25.48) \u00a725.5. ORBITS OF PARTICLES 671 Exercise 25.17. ANGULAR MOTION DURING INFALL From equation (25.42), deduce that the total angle.jc/> swept out on a trajectory falling into r = 0 is finite. The computation is straightforward; but the interpretation, in view of the behavior of t(;\\) on the same trajectory (equation 25.32 and exercise 25.15), is not. The interpretation will be elucidated in Chapter 31. Exercise 25.18. MAXIMUM AND MINIMUM OF EFFECTIVE POTENTIAL Derive the expressions given in the caption of Figure 252 for the locations of the maximum and the minimum of the effective potential as a function of angular momentum. Determine also the limiting form of the dependence of barrier height on angular momentum in the limit in which r is very large compared to M. Exercise 25.19. KEPLER LAW VALID FOR CIRCULAR ORBITS From dc/>/dr of (25.17) and dt/dr of (25.18), deduce an expression for the circular frequency of revolution as seen by a faraway observer; and from the results of exercise 25.18 (or otherwise) show that it fulfills exactly the Kepler relation for any circular orbit of Schwarzschild revalue equal to r, whether stable (potential minimum) or unstable (potential maximum). Exercise 25.20. HAMILTON-JACOBI FUNCTION Construct the locus in the r, 8 diagram of points of constant dynamic phase S(t, r, 8) = 0 for t = 0 and for values L = 4M, E = 1 (or for L = 2 V3M, E = (8/9)112, or for some other equally simple set of values for these two parameters). Show that the whole set of surfaces of constant S can be obtained by rotating the foregoing locus through one angle, then another and another, and recopying or retracing. Interpret physically the principal features of the resulting pattern of curves. Exercise 25.21. DEFLECTION BY GRAVITY CONTRASTED WITH DEFLECTION BY ELECTRIC FORCE A test particle of arbitrary velocity 13 flies past a mass M at an impact parameter b so great that the defiection is small. Show that the defiection is 8 = ~;; (1 + 13 2 ). (25.49) Derive the defiection according to Newtonian mechanics for a particle moving with the speed of light", ". Show that (25.49) in the limit 13 -+- 1 is twice the Newtonian defiection. Derive also (flat-space analysis) the contrasting formula for the deflection of a fast particle of rest mass J.L and charge e by a nucleus of charge Ze, (25.50) How feasible is it to rule out a \"vector\" theory of gravitation [see, for example, Brillouin (1970)], patterned after electromagnetism, by observations on the bending of light by the sun? [Hint: To simplify the mathematical analysis, go back to (25.42). Differentiate once with respect to c/> to convert into a second-order equation. Rearrange to put on the left all those terms that would be there in the absence of gravity, and on the right all those that originate from the -2u term (gravitation) in the factor (1 - 2u). Neglect the right-hand side of the equation and solve exactly (straight-line motion). Evaluate the perturbing term 672 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY on the right as a function of ep by inserting in it the unperturbed expression for u(ep). Solve again and get the deflection.] Exercise 25.22. CAPTURE BY A BLACK HOLE Over and above any scattering of particles by a black hole, there is direct capture into the black hole. Show that the cross section for capture is 'Ub~rit' with the critical impact parameter bedt given by Lcritl(\u00a32 JL2)112. From the formulas in the caption of Fig. 25.2 or otherwise, show that for high-energy particles this cross section varies with energy as (J = 27'UM 2(1 + _2_ +... ) ea~ 3\u00a32 (photon limit for E --+- 00) and for low energies as where f3 is the velocity relative to the velocity of light [Bogorodsky (1962)]. (25.51 ) (25.52) \u00a725.6. ORBIT OF A PHOTON, NEUTRINO, OR GRAVITON IN SCHWARZSCHILD GEOMETRY Orbits for particles of zero rest mass: The concepts of \"energy per unit of rest mass\" and \"angular momentum per unit of rest mass\" make no sense for an object of zero rest mass (photon, neutrino, even the graviton of exercise 35.16). However, there is nothing about the motion of such an entity that cannot be discovered by considering the motion of a particle of finite rest mass JL and going to the limit JL --+- O. In this limit the quantities and \u00a3 = EIJL I = LIJL individually go to infinity; but the ratio (1) impact parameter defined ( angular impact parae) = b = momentum = _-::-..::L~::-::-:~ = (linear) meter momentum JL2)1I2 (E2 - ( ). I (\u00a32 - 1)11 2 (25.53) goes to the finite value Lim.\u00a3 = b. 1'-0 E (25.54) (2) shape of orbit In this limit, equation (25.41) for the shape of the orbit reduces at once to the simple form \u00a725.6. ORBIT OF A PHOTON, NEUTRINO, OR GRAVITON or or ( 1 dU)2 dq, + U (1 - 2u) - b =p. (M)2 _ _ 2 673 (25.55) (25.56) (25.57) Whichever way the differential equation for the orbit is written, one term in it depends on the choice of orbit (the term l/b 2) the other on the properties of the Schwarzschild geometry, but not on the choice of orbit. This second term defines a kind of effective potential, ) \"effective potential for = B-2(r) = 1 - r~Mlr. photon\" ( (25.58) (3) effective potential No attempt is made here to take the square root, as was done for a particle of finite rest mass. There one took the root in order to have a quantity that reduced to the Newtonian effective potential (plus the rest mass) in the nonrelativistic limit; but for light (D = 1) there is no nonrelativistic limit. Therefore the effective potential (25.58) is plotted directly in Box 25.7, and used there to analyze some of the principal features of the orbits of a photon in Schwarzschild geometry. On occasion it has proved useful to plot as a function of r, not the \"effective potential\" of (25.58), but the \"potential impact parameter B(r", ")\" calculated from that formula [see, for example, Power and Wheeler (1957), Zel'dovich and Novikov (1971)]. This potential impact parameter has the following interpretation: A ray, in order to real::h_ the point r, must have an impact parameter b that is equal to or less than B(r): b :::;; B(r) (\"condition of accessibility\"). (25.59) A ray with zero impact parameter (head-on impact), or any impact parameter less than bcrit = min[B(r)] = 3V3M, can get to any and all r values. The beautifully simple \"effective potential\" defined by (25.58) is used in (25.56) _ _ to det~rmine the shgIJe-Df an orbit; that is, the azimuth q, that the photOn has when it gets to a given r-value. In other connections, it can be equally interesting to know when, or at what Schwarzschild coordinate time, the photon gets to a given r value. \\fore broadly, the geodesic of a photon, for which proper time has no meaning, admits of analysis from first principles by way of an affine parameter A, as contrasted with the device of first considering a particle and then going to the limit p. --+- O. (continued on page 676) (4) critical impact parameter (5) affine parameter 674 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY Box 25.7 QUALITATIVE ANALYSIS OF ORBITS OF A PHOTON IN SCHWARZSCHILD GEOMETRY A. Equations Governing Orbit 1. Effective-potential equation for radial part of motion: (;~y + B- 2(r) = b- 2 ; B-2(r) = r- 2(1 - 2M/r); b = (impact parameter). 2. Supplementary equations to determine angular and time motion: d,pjd\"A = 1/r2 ; dt/d\"A = b-1(l - 2M/rt 1. 0.06M-2 b -2 1 0.04:11- 2 1/27 M2 b2 - 2 O t - - o b2 > 30 M ------1--- (dr/dA)2 B. Qualitative Features of Orbits (deduced from effective-potential diagram) 1. A zero-mass particle with b >3V3M, which falls in from r = 00, is \"reflected off the potential barrier\" (periastron; b = B; dr/d\"A = 0) and returns to infinity. a. For b \u00bb 3 V3M, the orbit is a straight line, except for a slight deflection of angle 4M/b (exercise 25.21; \u00a740.3). b. For 0 < b - 3 V3M ~ M, the particle circles the star many times (\"un stable circular orbit) at r ::::: 3M before flying back to r = 00. \u00a725.6. ORBIT OF A PHOTON, NEUTRINO, OR GRAVITON 675 2. A zero-mass particle with b < 3 \\13M, which falls in from r = 00, falls into r = 2M (no periastron). 3. A zero-mass particle emitted from near r = 2M escapes to infinity only if it has b < 3 \\13M; otherwise it reaches an apastron and then gets pulled back into r = 2M. C. Escape Versus Capture as a Function of Propagation Direction An observer at rest in the Schwarzschild gravitational field measures the ordi nary velocity of a zero-mass particle relative to his orthonormal frame [equa tions (23.15)]: vr= Igrrl 1l2 Igool1l 2 dt/d\"A dr/d\"A = +(1 _ b2/B2)1I 2.'Ig 1112 dtP/d\"A VJ, = 1~:11I2 dt/d\"A = biB; (vr? + (vJ,? = I; l3 = (angle between propagation direction and radial direction) = cos- I vr= sin- I vJ,' To be able to cross over the potential barrier, the particle must have b < 3 \\13M, or vJ,2B2 < 27M2, or sin2~ < 27M2/B2. This result, restated: 1. A particle ofzerorest mass at r < 3M will eventually escape to infinity, rather than be captured by a black hole at r = 2M if and only if vr is positive and sin l3 < 3 \\", "l3MB-I(r). 2. A particle ofzero rest mass at r > 3M will eventually escape to infinity if and only if: (1) vr is positive, or (2) vr is negative and sin l3 > 3 \\l3MB-I(r). White, escape; black, to black hole; directions in proper reference frarne e\u00a2. = r- 1 iJ/iJep.~.;k2MM''''/~ 10 o 676 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY Return to the statement of the conservation laws (25.17) and (25.18) in the form that makes reference to the affine parameter A but no reference to the rest mass p.; thus and dq, dA L -;:2 dt dA E = -:----::---c:-I - 2M/r' (25.60) (25.61 ) Recall that the course of a photon in a gravitational field is governed by its direction but not by its energy. Therefore neither E nor L individually are relevant but only the impact parameter b = L/E of (25.54) and exercise 25.14. This their ratio, circumstance leads one to replace the affine parameter A by a new affine parameter, (25.62) (6) equations for orbit that is equally constant along the world line of the photon. In this notation (drop the subscript \"new\" hereafter), the conservation laws take the form I dq, dA --;:2' dt dll. I = - - -'- - bel - 2M/r). (25.63) (25.64) The statement that the world line of the photon is a line of zero lapse of proper time, leads to the \"radial equation\" (25.65) (25.66) Here one encounters again the \"effective potential\" B-2(r) of (25.58). The present fuller set of equations for the geodesic of a photon have the advantage that they reach beyond space to a de~cription of the world line in spacetime. Return to spac;e! Figure 25.6 shows typical orbits for a photon in Schwarzschild geometry. Figure 25.7 shows angle of deflection as a function of impact parameter. From the information contained in this curve, one can evaluate the contributions to the differential scattering cross section (7) sClltterjng crClllS section da _ L I 2wbdb dO ''braIlches'' 2w sin e de I (25.67) from the various \"branches\" of the scattering curve of Figure 25.7 [one tum around the center of attraction, two turns, etc.; for more on these branches and the central \u20221 Figure 25.6. The orbit of a photon in the \"equatorial plane\" of a black hole, plotted in terms of the Schwarzschild coordinates rand cp, for selected values of the turning point of the orbit, rTP / M = 2.99, 3.00 (unstable circular orbit), 3.01, 3.5, 4, 5, 6, 7, 8, 9. The impact parameter is given by the formula b = rTP (1 - 2M/rTP )-1/2. In none of the cases shown, even for the inward plunging spiral, is the impact parameter less than berll = (27)1/2 M, nor are any of these orbits able to cross the circle r = 3 M. That only happens for orbits with b less than bcrit\u2022 For such orbits there is no turning point; the photon comes in from infinity and ends up at r = 0: straight in for b = 0 (head-on impact); only after many loops near r = 3M, when b/M = (27)1/2 - E, where E is a very small quantity. Appreciation is expressed to Prof. R. H. Dicke for permission to publish these curves, which he had a digital calcu lator compute and plot out directly from the formula d2u/dcp2 = 3u2 - u, where u = M/r. role of the deflection function 8 = 8(b) in the analysis ofscattering, see, for example, Ford and Wheeler (l959a,b)]. For small angles the \"Rutherford\" part of the scatter ing predominates.- The major\" part of the small-angle scattering, and in the limit 8 ~ 0 all of it, comes from large impact parameters, for which one has (25.68) (see exercises 25.21 and 25.24). It follows that the limiting form of the cross section is.f!!!- = (4M)2 8 2 dO (small 8). (25.69) Also, at 8 = '", "fT one has a singul~ in the differential scattering cross section, with the character of a glory [see discussion following equation (25.44)]. Writing down the contributions of the several branches of the scattering function to the differential cross section, and summing them, one has, near 8 = 'fT, ~~ = 'fT~28 (1.75 + 0.0029.+ 0.0000055 +... ) = 1.75 'fT~28' (25.70) Thus, in principle, if one shines a powerful source of light onto a black hole, one gets a direct return of a few photons from it. Equation (25.70) provides a means to calculate the strength of this return. See exercise 2_5.26. 678 25, PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY 2 -t8 (rad) I 1.0 \u2022lIlustrative point b = 5.40M 8 = 2.83 = I.80f = 162 0 3.517 317 2.517 2r. t I 8 7f 1.517 0.517 5 IO 0.1 0.01 0.001 o'--L--l.--L.-'----'----'---''--'----L---'--'----'---L--L---->-J l.---l.--L.-L...l-.L-I.-L..L-LJ 15 0.0001 Figure 25.7. Deflection of a photon by a Schwarzschild black hole, or by any spherically symmetric center of attraction small enough not to block the trajectory of the photon. The accurate calculations (smooth curves) are compared with formulas (dashed curves) valid asymptoticaily in the two limiting cases of an impact parameter, b: (I) very close to berll = 33/ 2M (many turns around the center of attraction); and (2) very large compared to bent (small deflection). The algorithm for the accurate calculation of the deflection proceeds as foilows (ail distances being given, for simplicity, in units of the mass value, M). (I) Choose a value, r = R, for the Schwarzschild coordinate of the point of closest approach. (2) Calculate the im pact parameter, b, from b2 = R3/(R - 2). (3) Calculate Q from Q2 = (R - 2)(R + 6). (4) Determine the modulus, k, of an \"elliptic integral of the first kind\" from sin2fJ = k 2 = (Q - R + 6)/2Q. (5) De termine the so-called amplitude eJ> = eJ>min of the same elliptic function from sn2umin = sin2eJ>min = (2 + Q - R)/(6 + Q - R). (6) Then the total deflection is (8) gravitational lens effect 8 = 4(R/Q)1/2[F(7f/2,fJ) - F(eJ>min,fJ)]- 7f. The values plotted here were kindly calculated by James A. Isenberg on the basis of the work of C. G. Darwin (1959, 1961). When the source of illumination, instead of being on the observer's side of the black hole, is on the opposite side, then in addition to the \"lens effect\" experienced by photons flying by with large impact parameter [literature too vast to summarize here, but see, e.g., Refsdal (1964)], and subsumed in equation (25.68), there is a glory type of illumination (intensity -l/sin 8, with now, however, 8 close to zero) received from photons that have experienced deflections 8 = 2?T, 4?T,.... This illumination comes from \"rings of brightness\" located at impact parameters given by blM - 33/ 2 = 0.0065,0.000012,.... Interesting though all these optical effects are as matters of principle, they are, among all the ways to observe a black hole, the worst; see part VI, C, of Box 33.3 for a detailed discussion. \u00a725.7. SPHERICAL STAR CLUSTERS 679 ------------------------------------------ ~ Exercise 25.23. QUALITATIVE FEATURES OF PHOTON ORBITS EXERCISES Verify all the statements about orbits for particles of zero rest mass made in Box 25.7. LIGHT DEFLECTION Exercise 25.24. Using the dimensionless variable u = Mjr in place of r itself, and ub = M/b in place of the impact parameter, transform (25.55) into the first-order equation (dU)2 dep + (1 - 2u)u2 = u~ and thence, by differentiation, into (", "25.71) (25.72) (a) In the large-impact-parameter or small-u approximation, in which the term on the right is neglected, show that the solution of (25.72) yields elementary rectilinear motion (zero deflection). (b) Insert this zero-order solution into the perturbation term 3u2 on the righthand side of (25.72), and solve anew for u (\"rectilinear motion plus first-order correction\"). In this way, verify the formula for the bending oflight by the sun given by putting (3 = 1 in equation (25.49). Exercise 25.25. CAPTURE OF LIGHT BY A BLACK HOLE Show that a Schwarzschild black hole presents a cross section (Jcapl = 27'lTM2 for capture of light. Exercise 25.26. RETURN OF LIGHT FROM A BLACK HOLE Show that fiashing a powerful pulse of light onto a black hole leads in principle to a return from rings of brightness located at bjM - 33/ 2 = 0.151,0.00028,.... How can one evaluate the difference in time delays of these distinct returns? Show that the intensity I of the return (ergjcm2) as a function of the energy Eo(ergjsteradian) of the original pulse, the mass M(cm) of the black hole, the distance R to it, and the lateral distance r from the \"fiashlight\" to the receptor of returned radiation is Eo 1= R3 r \"\" L. 8=(2N+1l.\". 1 2b db I de = ~1.75 + 0.0029 + 0.0000055 +... ) EoM2 under conditions where diffraction can be neglected. \u00a725.7. SPHERICAL STAR CLUSTERS By combining orbit theory, as developed in this chapter, with kinetic theory in curved spacetime as developed in \u00a722.6, one can formulate the theory of relativistic star clusters. Consider, for simplicity, a spherically symmetric cluster of stars (e.g., a globular cluster, but one so dense that relativistic gravitational effects might be important). Static, spherical star clusters: 680 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY Demand that the cluster be static, in the sense that the number density in phase space 9l is independent of time. (New stars, fiying along geodesic orbits, enter a fixed region in phase space at the same rate as \"old\" stars leave it.) Ignore collisions and close encounters between stars; i.e., treat each star's orbit as ageodesic in the spherically symmetric spacetime of the cluster as a whole. With these idealizations accepted, one can write down a manageable set of equa tions for the structure of the cluster.* Since the cluster is static and spherical, so must be its gravitational field. Consequently, one can introduce the same kind of coordinate system (\"Schwarzschild coordinates\") as was used for a static spherical star in Chapter 23: tP = tP(r), A = A(r). (25.73) In the tangent space at each event in spacetime reside the momentum vectors of the swarming stars. For coordinates in this tangent space (\"momentum space\"), it is convenient to use the physical components of 4-momentum, ill-i.e., components on the orthonormal frame we = rdB, w\u00a2' = r sin B dcp. (25.74) Then the number density of stars in phase space is a spherically symmetric, static function (25.75) [9l is independent of t because the cluster is static; and independent of B, cp, and angle 8 = tan- 1(p\u00a2'/pe) because of spherical symmetry.]\u00b7\u00b7.. The functions describing the structure of the cluster, tP, A, and 9l, are determined by the kinetic (also, in this context, called the Vlasoff) equation (\u00a722.6) d9l/dll. = 0, i.e., 9l conserved along orbit of each star in phase space; and by the Einstein field equations (25.76a) (25.76b) {The Vlasoff equation for Newtonian star clusters is treated by Ogorodnikov (1965). The above expression for the stress-energy tensor of a swarm of particles (stars) was derived in exercise 22.18. Here, as in exercise 22.18, the particles (stars) are assumed not all to have the same rest mass. Note that rest mass is here denoted p., but in Chapter 22 it was denoted m.] To solve the Vlasoff equation, one need only note that 9l is conserved along stellar orbits and therefore must be a function of the constants of the", " orbital motion. There is a constant of motion corresponding to each Killing vector in the cluster's static, spherical spacetime (see exercise 25.8): *These equations were first derived and explored by Zel'dovich and Podurets (1965). (1) foundations for analysis (2) solution of Vlasoff equation \u00a725.7. SPHERICAL STAR CLUSTERS 681 E = \"energy. at infinity\" = -p' (a/at) = -Po, Lz = \"z-component of angular momentum\" = p' ( z = p' (a/acp) =P\"\" LIJ = ''y-component of angular momentum\" =p' ( IJ' Lz = \"x-component of angular momentum\" = P \u2022 ( z' (25.77a) In addition, each star's rest mass is a constant of its motion. The general solution of the Vlasoff equation, then, has the form ~ = H(E, L z, LIJ' L z, p.). But this general solution is not spherically symmetric. For example, the distribution function (25.77b) in the equatorial plane (J = corresponds to a cluster of stars with orbits all 7T/2 (LIJ = L z = 0 for all stars in cluster). To be spherical the cluster's distribution function must depend only on the magnitude L = (L/ + L/ + L/)1I2 of the angular momentum, and not on its direction (not on the orientation of a star's orbital plane). Thus, the general spherical solution to the Vlasoff equation in a static, spherical spacetime must-have the form ~ = F(E, L, p.). (25.78) To use this general solution, one must reexpress the constants of the motion E, L, p., in terms of the agreed-on phase-space coordinates (t, r, (J, cp,po,l,pe,p\u00a2). The rest mass of a star is given by (25.77b). The energy-at-infinity is obtained by red shifting the locally measured energy (25.79a) For an orbit in the equatorial plane (Pe = pe = pe = 0; L z = LIJ = 0), the total angular momentum has the form L = ILzl = IF\",I = Irp\u00a21 = r X (\"tangential\" component of 4-momentum). By symmetry, the equation L = r X (\"tangential\" component of p) must hold true also for orbits in other planes; it must be perfectly general: L = rp1', p1'= (tangential component of 4-momentum) = [(P0)2 + (p\u00a2)2JlI2 (25.79b) (25.80) (see exercise'25.9). (3) \"smeared-out\" stress-energy tensor due to stars 682 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY Before solving the Einstein field equations, one finds it useful to reduce the stress-energy tensor to a more explicit form than (25.76b). The off-diagonal compo nents Toi and Til< (j i: k) all vanish because their integrands are odd functions of pi. The integrands for the diagonal.components Too, TIT, and!(T88 + T\u00a2\u00a2) are independent of angle 8 = tan- 1(p\u00a2jp 8) in the tangential momentum plane; so the momentum volume element can be rewritten as dpo di dp 8dp\u00a2 -+ 2?Tp1' dp1' di dpo. Changing variables from (p1',i,pO) to (p1', p.,pO) where p. = [(P0)2 _ (p,)2 _ (p1')2]1I2, and recognizing that two values of i brings the volume element into the form (-+-p,) correspond to each value of p., one The diagonal components of T [equation (25.76b)] then read p =Too = (total density of mass-energy) = 4?Tf F(eq,p 0, rp1', p.)(P02p 1'jp') dp1' dpo dp., I........ PT =\"2(Tee + T'I>\u00a2) = Tee = T\u00a2\u00a2 = (tangential pressure) ~bY spheric~_symmetry] = 2?Tf F(eq,po, rp1', p.)[(p'f)3ji] dp1' dpo dp., Pr = TIT = (radial pressure) = 4?Tf F(eq,po, r", "p1', p.)(prp'f) dp1' dpo dp... (25.8Ia) (25.8Ib) (25.8Ic) When performing these integrals, one must express i integration, in terms of the variables of (25.8Id) (4) solution of field equations The Einstein field equations for this stress-energy tensor and the metric (25.73), after use of expressions (14.43) for Gap and after manipulations analogous to those for a spherical star (\u00a723.5), reduce to dlP _ m + 4?Tr3Pr r(r - 2m) dr -. (25.82a) (25.82b) These equations, together with the assumed form F(E, L, p.) of the distribution \u00a725.7. SPHERICAL STAR CLUSTERS 683 function and the integrals (25.81) for p, P.\" and PT> determine the structure of the cluster. Box 25.8 gives an overview of these structure equations, and specializes them for an isotropic velocity distribution. Box 25.9 presents and discusses the solution to the equations for an isothermal star cluster (truncated Maxwellian velocity distri bution). Exercise 25.27. For a cluster with distribution function independent of angular momentum, derive properties B.l to B.6 of Box 25.8. ISOTROPIC STAR CLUSTER EXERCISES Exercise 25.28. SELF-SIMILAR CLUSTER [See Bisnovatyi.Kogan and Zel'dovich (1969), Bisnovatyi-Kogan and Thorne (1970).] (a) Find a solution to the equations of structure for a spherical star of infinite central density, with the equation of state P =yp, where y is a constant (0 < y < 1/3). (b) Find an isotropic distribution function F(E, /-I) that leads to a star cluster with the same distributions of p, P, m, and r[J as in the gas sphere of part (a). (See Box 25.8.) [Answer: P = yp = 1 y2 6 + y + y e 2A = (1 - 2m/r)-1 = (1 + 6y + y2)/(1 + y)2, 1 2 2 2''lTr B = const; A = const.] Exercise 25.29. CLUSTER WITH CIRCULAR ORBITS What must be the form of the distribution function to guarantee that all stars move in circular orbits? Specialize the equations of structure to this case. Analyze the stability of the orbits of individual stars in the cluster, using an effective-potential diagram. What conditions must the distribution function satisfy if all orbits are to be stable? [See Einstein (1939), Zapolsky (1968).] Box 25.8 EQUATIONS OF STRUCTURE FOR A SPHERICAL STAR CLUSTER A. To Build a Model for a Star Cluster, Proceed as Follows I. Specify the distribution function ~ = F(E, L, p.), where E = energy-at-infinity of a star, L = angular momentum of a star, p. = rest mass of a star. 2. Solve the following two integro-differential equations for the metric functions m = ~r(l - e- ZA ) and (jJ of the line element Box 25.8 (continued) ds 2 = _e2t/> dt2 + e2A dr2 + r 2dQ2: where m = I r 4'1Tr2p dr,, o m + 4'1Tr3Pr dlP dr = r(r - 2m) p = 4'1Tf F(et/>p 0, rp1', fL)[(P\u00b0)2p 1'ji] dp1' dpo dfL' PT = 2'1Tf F(et/>po, rp1',fL)[(p1)3jp r] dp1' dpo dfL' Pr = 4'1Tf F(et/>po, rp1', fL)(prp 1') dp1' dpo dfL' i = [(p0)2 - (p1')2 - fL2JlI2. The integrations for p, PT> and Pr go over all positive p1', pO, fL for which (P0)2 _ (p1)2 - fL2 ~ O. B. If the Distribution Function is Independent of Angular Momentum, Then 1. F = F(E, fL). 2. The distribution of stellar velocities at each point in the cluster is isotropic. 3. p = 4'1T", "fF(et/>p o, fL)[(p0)2 fL2]1I2(PO)2 dpo dfL. 4. The pressure is isotropic: Pr = PT = P = 4 3 '1T f F(et/>po, fL)(P\u00b02 - fL2)3/2 dpo dfL\u00b7 5. The total density of mass-energy p, the pressure P, and the metric functions lP and m = tr(1 - e- 2A) satisfy the equations of structure for a gas sphere (\"'star\"), dlP -= dr m = f 4'1Tr2p dr, m + 4'1Tr3p r(r - 2m) (p + P)(m + 4'1Tr3P) r(r - 2m), dP dr = 6. Thus, to every static, spherical star cluster with isotropic velocity distribution, there corresponds a unique gas sphere that has the same distributions of p, P, m, and lP. 7. Conversely [see Fackerell (1968)], given a gas sphere (solution to equations of stellar structure for p, P, m, and lP), one can always find a distribution function F(E, fL) that describes a cluster with the same p, P, m, and lP. But for some gas spheres F is necessarily negative in part of phase space, and is thus unphysical. \u00a725.7. SPHERICAL STAR CLUSTERS 685 Box 25.9 ISOTHERMAL STAR CLUSTERS A. Distribution Function 1. In any relativistic star cluster, one might expect that occasional close en counters between stars would \"thermalize\" the stellar distribution function. This suggests that one study isotropic, spherical clusters with the Boltzmann distribution function (tacitly assumed zero for pO = Ee-q, < f-Lo) ~ = F(E, L, f-L) = Ke-E T 8(f-L - f-Lo)' (I) / Here K is a normalization constant, T is a constant \"temperature,\" and for simplicity the stars are all assumed to have the same rest mass f-Lo. 2. In such a cluster, an observer at radius r sees a star of energy-at-infinity E to have locally measured energy pO = (rest mass-energy) + (kinetic energy) = f-Lo 2 1/2 = Ee-tf>(r). (I - v ) (2) Consequently, the stars in his neighborhood have a Boltzmann distribution with locally measured temperature (3) (4) Thus, the temperature of the cluster is subject to identically the same red shift-blueshift effects as photons, particles, and stars that move about in the cluster. (For a derivation of this same temperature-redshift law for a gas in thermal equilibrium, see part (e) of exercise 22.7.) 3. Actually, the Boltzmann distribution (I) can never be achieved. Stars with E > f-Lo are gravitationally unbound from the cluster and will escape. The Boltzmann distribution presumes that, as such stars go zooming off toward r = 00, an equal number of stars with the same energies come zooming in from r = 00 to maintain an unchanged distribution function. Such a situation is clearly unrealistic. Instead, one expects the escape of stars to truncate the distribution at some energy Emu slightly less than f-Lo' The result, in idealized form, is the \"truncated Boltzmann distribution\" ~ = F(E, L, f-L) = {Ke- E / T 8(f-L 0, f-Lo), E < Emu' E> Emu' (5) 686 25. PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY Box 25.9 (continued) B. Structure and Stability of Cluster Models 1. Models for star clusters with truncated Boltzmann distributions have been constructed by Zel'dovich and Podurets (1965), by Fackerell (1966), and by Ipser (1969), using the procedure of Box 25.8. Ipser has analyzed the colli sionless radial vibrations of such clusters. 2. In general, these clusters form a 4-parameter family (K, T, f-Lo, Emu)' Replace the parameter K by the total rest mass of the cluster, M o = f-LoN, where N is the total number of stars. Replace T by the temperature per unit rest mass, T = T/f-Lo' Replace Emu by the maximum energy per unit rest mass, Emu = Emu/f-Lo\u00b7 Then the clusters are parametrized by (Mo' T, f-Lo' Emu)' When", " one now doubles f-Lo' holding Mo' T, Emu fixed (and thus halving the total number of stars), all macroscopic features of the cluster remain un changed. In this sense f-Lo is a \"trivial parameter\" and can henceforth be ignored or changed at will. The total rest mass of the cluster Mo can be regarded as a \"scaling factor\"; all dimensionless features of the cluster are independent of it. For example, if Pc is the central density of mass-energy [equation (25.8Ia), evaluated at r = 0], then PcMoz is dimensionless and is thus independent of Mo' which means that Pc a: Mo- z. Only two nontrivial parameters remain: T and Emu' 3. Consider as an instructive special case [Zel'dovich and Podurets (1965)] the one-parameter sequence with Emu = I - fT. The following figure, com puted by Ipser (1969), plots for this sequence the fractional binding energy, (6) (here M is total mass-energy); the square of the angular frequency for collisionless vibrations (vibration amplitude a: e-iwt ) divided by central density of mass-energy, w Z / Pc; and the redshift, zc' of photons emitted from the center of the cluster and received at infinity. All these quantities are dimensionless, and thus depend only on the choice of T = T/f-Lo' 4. Notice that all models beyond the point of maximum binding energy (zc ~ 0.5) are unstable against collisionless radial perturbations (w imagi nary; amplitude of perturbation a: e1w1t ). When perturbed slightly, such clusters must collapse to form black holes. (See Chapter 26 for an analysis of the analogous instability in stars). 5. These results suggest an idealized story of the evolution of a spherical cluster [Zel'dovich and Podurets (1965); Fackerell, Ipser, and Thome (1969)]. The \u00a725.7. SPHERICAL STAR CLUSTERS 687 ~~ ~ \\ \\ \\ \\ \\ \\ \\ \\ \\ \"- \" \"- 1.0 t L Pc 0.5 I 0..,, - - - - - - - - - - - - - - - l0.0 \"\".............. ------- - 0.5 \" f:>e'''' 't = 0.27 :o\\e ~O.... U~\",\\\\l. ~ o=_;;;; Collapse 0.25 0.5 0.75 1.0 --Zc~ IE bind MO + +0.02 +0.04 0.0 cluster would evolve quasistatically along a sequence ofspherical equilibrium configurations such as those of the figure. The evolution would be driven by stellar collisions and by the evaporation of stars. When two stars collide and coalesce, they increase the cluster's rest mass and hence its fractional binding energy. When a star gains enough energy from such encounters to escape from the cluster, it carries away excess kinetic energy, leaving the cluster more tightly bound. Thus, both collisions and evaporation should drive the cluster toward states of tighter and tighter binding. When the cluster reaches the point, along its sequence, of maximum fractional binding energy, it can no longer evolve quasistatically. Relativistic gravitational collapse sets in: the stars spiral inward through the gravitational radius of the cluster toward its center, leaving behind a black hole with, perhaps, some remaining stars orbiting it. It is tempting to speculate that violent events in the nuclei of some galaxies and in quasars might be associated with the onset of such a collapse, or with encounters between an already collapsed cluster (black hole) and surrounding stars. CHAPTER 26 STELLAR PULSATIONS, \u00a726.1. MOTIVATION This chapter is entirely Track 2. but it neither depends on nor prepares for any other. chapter. The raison d' etre of this chapter In relativistic astrophysics, as elsewhere in physics, most problems of deep physical interest are too difficult and too complex to be solved exactly. They can be solved only by use of approximation techniques. And of all approximation techniques, the one that has the widest range of application is perturbation theory. Perturbation calculations are typically long, tedious, and filled with complicated mathematical expressions. Therefore, they are not appropriate for a textbook such as this. Nevertheless, because it is so important that aspiring astrophysicists know how to set up and carry out perturbation calculations in general relativity, the authors have chosen to present one example in detail. The example chosen is an analysis of adiabatic, radial pulsations of a nonrotating, relativistic star. Two features of this example are noteworthy: (l) it is sufficiently complex", " to be instructive, but sufficiently simple to be presentable; (2) in the results of the calculation one can discern and quantify the relativistic instability that is so important for modem astrophysics (see Chapter 24). The calculation presented here is patterned after that of Chandrasekhar (1964a,b), which first revealed the existence of the relativistic instability. For an alternative calculation, based on the concept of \"extremal energy,\" see Appendix B of Harrison, Thome, Wakano, and Wheeler (1965); and for a calculation based on extremal entropy, see Cocke (1965). The authors are deeply indebted to Mr. Carlton M: Caves, who found and corrected many errors in the equations of this chapter and of a dozen other chapters. \u00a726.2. SETIING UP THE PROBLEM 689 \u00a726.2. SETTING UP THE PROBLEM The system to be analyzed is a sphere of perfect fluid, pulsating radially with very small amplitude. To analyze the pulsations one needs (a) the exact equations gov erning the equilibrium configuration about which the sphere pulsates; (b) a coordi nate system for the vibrating sphere that reduces, for zero pulsation amplitude, to the standard Schwarzschild coordinates of the equilibrium sphere; (c) a set of small functions describing the pulsation (radial displacement and velocity, pressure and density perturbations, first-order changes in metric coefficients), in which to linearize; and (d) a set of equations governing the evolution of these \"perturbation functions.\" a. Equilibrium Configuration Setting up the analysis of stellar pulsations The equations of structure for the equilibrium sphere are those summarized in \u00a723.7. It will be useful to rewrite them here, with a few changes of notation (use of subscript \"0\" to denote \"unperturbed configuration\"; use of A = -i In (l - 2m/r) in place of m in all equations; use of a prime to denote derivatives with respect to r): Equilibrium configuration of star ds Z= - eZqJo dt Z + eZAo dr z + rZ(d(}Z + sinZ(} depZ), Ao' = ir (l - eZAo ) + 4?TrpoeZAo, Po' = - (Po + Po)lPo', lPo' = - ir (l - e2Ao) + 4?Trpoe2Ao. (26.1 a) (26.1 b) (26.1 c) (26.1 d) b. Coordinates for Perturbed Configuration -\"The gas spru:re pulsates in a radial, i.e., spherically symmetric, manner. Conse quently, its spacetime geometry must be spherical. In Box 23.3 it is shown that for any spherical spacetime, whether dynamic or static, one can introduce Schwarzschild coordinates with a line element lP = lP(t, r), A = A(t, r). (26.2) Coordinates for perturbed configuration One uses these coordinates for t~!L~here---because they reduce to the ------unpermrbgd t.:~ordinatesWhefithe pulsations have zero amplitude. c. Perturbation Functions When the pulsations have very small amplitude, the metric coefficients, lP and A, and the thermodynamic variables p, p, and n as measured in the fluid's rest frame 690 26. STELLAR PULSATIONS have very nearly their unperturbed values. Denote by 8f!), SA, Sp, Sp, and Sn the perturbations at fixed coordinate locations: iP(t, r) = iPo(r) + SiP(t, r), p(t, r) = po(r) + Sp(t, r), A(t, r) = Ao(r) + SA(t, r), p(t, r) = po(r) + Sp(t, r),. (263a) n(t, r) = no(r) + Sn(t, r). Besides SiP, SA, Sp, Sp, and Sn, one more perturbation function is needed to describe the pulsations: the radial displacement ~ of the fluid from its equilibrium position: A fluid element located at coordinate radius r in the unperturbed configuration is displaced to coordinate radius r + Hr, t) at coordinate time t in the vibrat ing configuration. (263b) To make the analysis of the pulsations tractable, all equations will be linearized in the perturbation functions ~, SiP, SA, Sp, Sp, and Sn. d. Equations of Evolution The evolution of the perturbation functions with time will be governed by the Einstein field equations, the local law of conservation of energy-momentum", " V. T = 0, and the laws of thermodynamics-all appropriately linearized. The analysis from here on is nothing but a reduction of those equations to \"manageable form.\" Of course, the reduction will proceed most efficiently if one knows in advance what form one seeks. The goal in this calculation and in most similar calculations is simple: (l) obtain a set of dynamic equations for the true dynamic degrees of freedom (only the fluid displacement ~ in this case; the fluid displacement and the amplitudes of the gravitational waves in a nonspherical case, where waves are possible); and (2) obtain a set of initial-value equations expressing the remaining perturbation functions (SiP, SA, Sp, Sp, and Sn in this case) in terms of the dynamic degrees of freedom (~). \u00a726.3. EULERIAN VERSUS LAGRANGIAN PERTURBATIONS -~ ----- Before deriving the dynamic and initial-value equations, it is useful to introduce a new concept: the \"Lagrangian perturbation\" in a thermodynamic variable. The perturbations Sp, Sp, and Sn of equations (263) are Eulerian perturbations in p, p, and n; Le., they are changes measured by an observer who sits forever at a fixed in the coordinate grid. By contrast, the Lagrangian perturbations point (t, r, (J, </\u00bb denoted Jp, Jp, and In-are changes measured by an observer who moves with Perturbation functions How to derive equations governing th~ perturbation. functions Eulerian perturbations defined lagrangian perturbations defined \u00a726.4. INITIAL-VALUE EQUATIONS 691 the fluid; i.e., by an observer who would sit at radius'r in the unperturbed configura tion, but sits at r + g(t, r) in the perturbed configuration: iJp(t, r) = p[t, r + g(t, r)] - po(r) :::::: Sp + Po't iJp(t, r) = p[t, r + g(t, r)] - po(r) :::::: Sp + Po'g; iJn(t, r) = n[t, r + g(t, r)] - no(r) :::::: Sn + no'g. (26.4a) Relation between Eulerian and Lagrangian perturbations (26.4b) (26.4c) \u00a726.4. INITIAL-VALUE EQUATIONS a. Baryon Conservation The law of baryon conservation, V' (nu) = 0 (\u00a722.2), governs the evolution of perturbations iJn and Sn in baryon number. By applying the chain rule to the divergence and using the relation u\u00b7 Vn = Vun = dn/dT, one can rewrite the con servation law as Derivation of initial value equations: (1) for baryon perturbations Lin and 13n dn/dT = -n(V' u). L-[derivative of n along fluid world line] In terms of iJn, the perturbation measured by an observer moving with the fluid, this equation can be rewritten as d iJn- -= -n(V\u00b7u). dT (26.5) To reduce this equation further, one needs an expression for the fluid's 4-velocity. It is readily derived from U: = (dr/dT) = (dr) dt dt/dT U (U t )2e2tJl _ along world line (u r )2e2A = 1. = ~ g, at The result to first order in g, SA, and 8f!) is (26.6) Using these components in equation (26.5), and using the relations a - = u = ua__, d ox a dT V' u = _~(V-gua),a, v-g together with the vibrating metric (26.2), one reduces equation (26.5) to a relation whose time integral is (26.7) 692 26. STELLAR PULSATIONS This is the initial value equation for Lin in terms of the dynamic variable g. The initial-value equation for Sn, which will not be needed later, one obtains by combin ing with equation (26.4c). b. Adiabaticity (2) for pressure perturbations Lip and l'Jp For adiabatic vibrations (negligible heat transfer between neighboring fluid ele ments), the Lagrangian changes in number density and pressure are related by aIn n s 41 aIn p ) _ r _ n J.p - p Lin. ( definition", " of adia-] batic index, r 1 (26.8) Combining this adiabatic relation with equation (26.7) for Lin, and equation (26.4a) for Sp in terms of Lip, one obtains the following initial-value equation for Sp: (26.9) c. Energy Conservation (3) for density perturbations Lip and l'Jp The local law of energy conservation [first law of thermodynamics; u' (V. T) = 0; see \u00a7\u00a722.2 and 22.3] says that dp dT = (p + p) dn dT' n Rewritten in terms of Lagrangian perturbations (recall: d/dT is a time derivative as measured by an observer moving with the fluid), this reads p + P d Lin d Lip dT -n-~' which has as its time integral (first-order analysis!) A LIp = Po + Po AO LIn. no (The constant of integration is zero, because, when Lin = 0, Lip must also vanish.) Combining this with equation (26.7) for Lin and equation (26.4b) for Sp in terms of Lip, one obtains the following initial-value equation for Sp: (26.10) (26.11 ) \u00a726.5. DYNAMIC EQUATION AND BOUNDARY CONDITIONS 693 d. Einstein Field Equations Two of the Einstein field equations, when linearized, reduce to initial-value equations for the metric perturbations SA and SiP. The equations needed, expressed in an orthonormal frame (4) for metric perturbations SA and Srp wi = eq, dt, (26.12) are Grt = 8'1TTrt, and G;.;. = 8'1TT;.;.. The components of the Einstein tensor in this orthonormal frame were evaluated in exercise 14.16: wi> = r sin () dep, w'=rd(}, r[linearized] GTi = 2(A/r)e-(A+q,> = 2,-le-(A.+q,.> SA; G;.;. = 2(iP'/r)e-2A + r-2(e-2A - 1) = (G;.;.)o + 2r-1e-2A\u2022 SiP' - 2e-2A.(2r-1iPo' + r-2) SA. L[linearized] (26. 13 a) (26.13b) The components of the stress-energy tensor, Ta~ = (p + p)uau~ + P\"1af3' as calcu lated using the 4-velocity (26.6) [transformed into the form u;; = -1, u;' = ge A\u2022 - q,.] and using expressions (26.3a) for p and p, reduce to Consequently, the field equation GTi = 8'1TTTi-after integration with respect to time and choice of the constant of integration, so that SA = 0 when ~ = O-reduces to T;.;. = Po + Sp. (26.14) (26.15) This is the initial-value equation for SA. The field equation G;.;. = 8'1TT;.;., after using (26.15) to remove SA and (26.9) to remove Sp, and (26.1c) to remove iPo', reduces to SiP' = -4'1Tr IPOr- 1e2A\u2022+ q,\u00b7(r2e-q,.~)' + [4'1Tpo'r - 4'1T(Po + po)]e2A.~. (26.16) This is the initial-value equation for SiP. \u00a726.5. DYNAMIC EQUATION AND BOUNDARY CONDITIONS The dynamic evolution of the fluid displacement Ht, r) is governed by the Euler equation (22.13): (p + p) X (4-acceleration) = -(projection of Vp orthogonal to u). (26.17) Derivation of equation of motion for fluid displacement ~ The 4-acceleration a = Vuu corresponding to the 4-velocity (26.6) in the metric (26.2) has as its only non-zero, linearized, covariant component: ar = iPo' + SiP' + e2(A.-q,.>{ 694 26. STELLAR PULSATIONS [The component at is trivial in the sense that it leads to an Euler equation that duplicates (26.1 c).] Combining this with P + P = Po + Po + Sp + Sp, with the", " radial component Po' + Sp' for the projection of Vp, and with the zero-order equation of hydrostatic equilibrium (26.1c), one obtains for the Euler equation (26.18) This equation of motion is put into its most useful form by using the initial-value equations (26.9), (26.11), and (26.16) to reexpress Sp, Sp, and 8f!)' in terms of ~, and by then manipulating terms extensively with the aid of the zero-order equations of structure (26.1). The result is wf = (pr)' + QK, where Kis a \"renormalized displacement function,\" and W; P, Q are functions of radius determined by the structure of the equilibrium star: (26.19) K r2e-tflo~; W (Po + po)r-2e3Ao+tflo; P r lPor-2eAo + 3tf1o; Q eAo+3tf1{ (po')2 r-2 _ 4po'r-3 - 8?T(Po + po)por-2e2Ao]. Po + Po (26.20) (26.21 a) (26.21 b) (26.21 c) Equation (26.19) is the dynamic equation governing the stellar pulsations. [This equation could be written in other forms; for instance, it could be multiplied by W-l or any other non-zero factor, and terms could be regrouped. The form given in equation (26.19) is preferred because it leads to a self-adjoint eigenvalue problem for the oscillation frequencies, as indicated in Box 26.1.] Not all solutions of the dynamic equation are acceptable. To be physically acceptthe displacement function must produce noninfinite density and pressure able, perturbations (Sp and Sp) at the center of the sphere, which means (~/r) finite or zero in limit as r -+ 0 (26.22a) [see (26.9) and (26.11 )]; also, it must leave the pressure equal to zero at the star's surface, which means iJp = -rlPOr-2etflo(r2e-tflo~)' -+ 0 as r -+ R [surface radius]---1 (26.22b) [see (26.8), (26.7), and (26.15)]. \u00a726.6. SUMMARY OF RESULTS Boundary conditions on fluid displacement Summary of theory of stellar pulsations If an initial displacement of the fluid, ~(t = 0, r), is specified subject to the boundary conditions (26.22), then its subsequent evolution ~(t, r) can be calculated by inte- \u00a726.6. SUMMARY OF RESULTS 695 grating the dynamic equation (26.19); and the form of the pressure, density, and metric perturbations can be calculated from g(t, r) using the initial-value equations (26.9), (26.11), (26.15), and (26.16). Several and 26.2. important consequences of these results are explored in Boxes 26.1 (continued on page 699) Box 26.1 EIGENVALUE PROBLEM AND VARIATIONAL PRINCIPLE FOR NORMAL-MODE PULSATIONS OF A STAR Assume that the renormalized displacement function (26.20) has a sinusoidal time dependence: Then the dynamic equation (26.19) and boundary conditions (26.22) reduce to an eigenvalue problem for the angular frequency w and amplitude K(r): (NT + QK + w 2WK = 0, K/r 3 finite or zero as r -+ 0, r lPor-2et/>ot -+ 0 as r -+ R. (1) (2a) (2b) Methods for solving this eigenvalue problem are catalogued and discussed by Bardeen, Thome, an~ Meltzer (1966). One method (but not the best for numerical calculations) is the variational principle: w 2 = extremal value of R I. [ 0 (PK,2 - QK 2 ) dr] R I. WK2 dr o'(3) where Kis varied over all functions satisfying the boundary conditions (2). [See e.g., \u00a712.3 of Mathews and Walker (1965) for discussion of the equivalence between this variational principle and the original eigenvalue problem.] The absolute minimum value of expression (3) is the squared frequency of the fundamental mode of pulsation. If it is negative, the star is unstable (e-i\"'t grows exponentially in time). If it is positive, the star is stable against adiabatic, radial perturbations", ". Therefore, since the denominator of expression (3) is positive definite, stability against] [R. adiabatic radial perturbations ~ (PK'2 - QK2) dr > 0 for all. fu~ctions KsatIsfymg (2) ] [. (4) 696 26. STELLAR PULSATIONS Box 26.1 (continued).-- -- By numerical solution of the eigenvalue equation (I), the pulsation frequencies have been calculated for a wide variety of models of neutron stars and supermassive stars. Example: The figure gives a plot of pulsation frequency as a function of central density for the lowest four normal modes of the Harrison-Wakano-Wheeler models at the endpoint of stellar evolution. (Make a detailed comparison with Figure 24.2.) These curves are based on calculations by Meltzer and Thorne (1966), with correc tions for the fundamental mode of massive white dwarfs by Faulkner and Gribbin (1968). 10 10 108 106 104 102 I 1.0 '\" I u 0 (I) or. \",' - 1.0 3 _10 2 _104 _10 6 _10 8 _10 10 _10 12 106 108 10 10 10 12 10 14 10 16 10 18 Pc. g/cm3 ~ 10-4 I 10-3 10-2 10- 1 1.0 10 00 1.0 10- 1 10-2 10-3 10-6 u (I) Of> ~.g c. (I) I I u (I) Of>.; E on c:a :2 \" \u2022 \u00a726.6. SUMMARY OF RESULTS 697 Box 26.2 THE CRITICAL ADIABATIC INDEX FOR NEARLY NEWTONIAN STARS A. Fully Newtonian Stars 1. For a Newtonian star that pulsates sinusoidally, g = g(r)e-iwt, the dynamic equation (26.19) reduces to 2. If F 1 = 4/3 throughout the star, the physically acceptable solution [solution satisfying boundary conditions (26.22)] for the fundamental mode of vibra tion (mode with lowest value of w2 ) is w2 = 0, g= a, ( = const. (2) Thus, for F 1 = 4/3 the fundamental mode is \"neutrally stable\" and has a \"homologous\" displacement function-independent of the star's equation of state or structure. If F 1 is allowed to differ slightly from 4/3 in an r-dependent way, then g(r) will differ slightly from the homologous form: 3. g= \u00a3r[l + r-dependent corrections of magnitude (F1 - 4/3)]. Consequently, if one uses the homologous expression g = a as a trial function in the variational principle of Box 26.1, one will obtain w2 accurate first-order errors in trial function produce to 0[(F1 second-order errors in value of variational expression.) The Newtonian limit of the variational expression [equation (3) of Box 26.1] becomes, with the homologous choice of trial function, - 4/3)2]. (Recall: Ri w2 = (31\\ - 4) 0 R i o 3por 2 dr + 0[(31\\ - 4)2], (3) por4 dr where l'l is the pressure-averaged adiabatic index ~ FIPo4'lTr2 dr _ F 1 = - : : - -R - - - i po4'lTr 2 dr R o (4) 698 26. STELLAR PULSATIONS Box 26.2 (continued) By use of the Newtonian virial theorem for the nonpulsating star [equation' (39.2lb) or exercise 23.7], one can convert equation (3) into the form (5) where g is the star's self-gravitational energy and I = f(p or2)4'1Tr2 dr is the trace of the second moment of its mass distribution (see Box 24.2 and exercise 39.6). B. Nearly Newtonian Stars 1. When one takes into account first-order relativistic corrections (corrections of order M/ R), but ignores higher-order corrections, one can rewrite the variational expression [equation (3) of Box 26.1] in the form R I po[F1r41),2 + (3F1 - 4)(r 31)2)'](1 + Ao + 34'0) dr - i'R I por4(1 + 3Ao + 4'0 + Po/Po)'l)2 dr o 0 R F',,1)2 dr, (6) 0 2 W = where 1) = f/", "r 3 = (~/r)(1 - 4'0)' (7) and mo(r) is the equilibrium mass inside radius r. 2. For a relativistic star with F 1 - 4/3 of order M/ R and with M/ R ~ 1, the homologous trial function ~ = f.r will still be highly accurate. Equally accu rate, and easier to work with, will be ~ = aeif>o ;::::: a(l + 4'0)' which corre sponds to 1) = f. = constant. Its fractional errors will be of order M/r; and the errors which it produces in w2 will be of order (M/ R)2. By inserting this trial function into the variational principle (6) and keeping only relativistic corrections of order M/ R, one obtains Here F1 is the pressure-averaged adiabatic index, and the critical value of the adiabatic index F 1crit is 4 Flcrit = 3\" + aM/R, (9) (8) \u00a726.6. SUMMARY OF RESULTS 699 with a a positive constant of order unity given by Expressions (8) and (9) for the pulsation frequency and the adiabatic index play an important role in the theory of supermassive stars (\u00a724.4). 3. For alternative derivations of the above result, see Chandrasekhar (1964a,b; 1965c), Fowler (1964, 1965), Wright (1964). (10) Exercise 26.1. DRAGGING OF INERTIAL FRAMES BY A SLOWLY EXERCISE ROTATING STAR A fluid sphere rotates very slowly. Analyze its rotation using perturbation theory; keep only effects and terms linear in the angular velocity of rotation. [Hints: (1) Centrifugal forces _are second-order in angular velocity. Therefort:, to first order the star is undeformed; its density and pressure distributions remain spherical and unperturbed. (2) Show, by symmetry and time-reversal arguments, that one can introduce coordinates in which ds 2 = _e2</> dt 2 + e2A dr2 + r 2[dB2 + sin2Bd\u00a22] (26.23) - 2(r2 sin2B)w dcj> dt, where tP = tP(r), A = A(r), and w = w(r, B). (26.24) Show that (3) Adopt the following precise definition of the angular velocity il(r, B): tP::::: tPo and A = Ao (no perturbations!) to first-order in angular velOcity. il =u\u00a2ju t = (dcj>jdt)movingwithlhenuid' Assuming u r = U 8 = 0 (i.e., rotation in the cj> direction), calculate the 4-velocity of the fluid. (4) Use the Einstein field equations to derive a differential equation for the metric perturba tion yJ in terms of the angular velocity il. (5) Solve that differential equation outside the star in terms of elementary functions, and express the solution for w(r, B) in terms of the star's total angular momentum S, as measured using distant gyroscopes (see Chapter 19).] For the original analyses of this problem and of related topics, see Gurovich (1965), Dorosh kevich, Zel'dovich, and Novikov (1965), Hartle and Sharp (1965), Brill and Cohen (1966), Hartle (1967), Krefetz (1967), Cohen and Brill (1968), Cohen (1968). (26.25) PART VI THE UNIVERSE Wherein the reader, flushed with joy at his conquest of the stars, seeks to control the entira universe, and is foiled by an un fathomed mystery: the Initial Singularity. CHAPTER 27 IDEALIZED COSMOLOGIES From my point of view one cannot arrive, by way of theory, at any at least somewhet relieble results in the field of cosmology, if one makes no use of the principle of general relativity. ALBERT EINSTEIN (1949b. p. 684) \u00a727.1. THE HOMOGENEITY AND ISOTROPY OF THE UNIVERSE Astronomical observations reveal that the universe is homogeneous and isotropic on scales of _108 light years and larger. Taking a \"fine-scale\" point of view, one sees the agglomeration of matter into stars, galaxies, and clusters of galaxies in regions of size -1 light year, _10 6 light years, and -3 X 107 light years, respec tively. But taking instead a\"large-scale\" viewpoint, one sees little difference", " between an elementary volume of the universe of the order of 108 light years on a side centered on the Earth and other elementary volumes of the same size located elsewhere. Cosmology, summarized in its simplest form in Box 27.1, takes the large-scale viewpoint as its first approximation; and as its second approximation, it treats the fine-scale structure as a perturbation on the smooth, large-scale background. This chapter (27) treats in detail the large-scale, homogeneous approximation. Chapter 28 considers such small-scale phenomena as the primordial formation of the ele ments, and the condensation of galaxies out of the primeval plasma during the expansion of the universe. Chapter 29 discusses observational cosmology. Evidence for the large-scale homogeneity and isotropy of the universe comes from several sources. (1) There is evidence in the distribution of galaxies on the sky and in the distribution of their apparent magnitudes and redshifts [see, e.g., Hubble (1934b, 1936); Sandage (1972a); Sandage, Tamman, and Hardy (1972); but note the papers claiming \"hierarchic\" deviations from homogeneity, which Sandage cites and attacks]. (2) There is evidence in the isotropy of the distribution of radio sources on the sky [see, e.g., Holden (1966), and Hughes and Longair (1967)]. (3) There is evidence in the remarkable isotropy of the cosmic microwave radiation [see, e.g., Boughn, Fram, and Partridge (1971 )]. For a review of most of the evidence, see Chapter 2 of Peebles (1971). (conlinued on page 711) The universe: fine-scale condensations contrasted with large-scale homogeneity Evidence for large-scale homogeneity and isotropy 704 27. IDEALIZED COSMOLOGIES Box 27.1 COSMOLOGY IN BRIEF Uniform density. Idealize the stars and atoms as scattered like dust through the heavens with an effective average density p of mass-energy everywhere the same. Geometry homogeneous and isotropic. Idealize the curvature of space to be every where the same. Closure. Accept the term, \"Einstein's geometric theory of gravity\" as including not only his field equation G = 8'1TT, but also his boundary condition of closure imposed on any solution of this equation.* A three-sphere satisfies the three requirements of homogeneity, isotropy, and closure, and is the natural generalization of the metric on a circle and a 2-sphere: Spheres of selected dimensionality Visualized as embedded in a Euclidean space of one higher dimensiona Transformation from Cartesian to polar coordinates Metric on sa expressed in terms of these polar coordinates x=acosq, y=asinq, x = asinl/cosq, y=asinl/sinq, z = a cos 1/ x = asinXsinl/cosq, y=asinXsinl/sinq, z=asinxcosl/ w=acosl/ ds2 = a2[dX 2 + sin2xCdl/2 + sin21/ dq,2\u00bb) \u2022 Excursion off the sphere is physically meaningless and is forbidden. The superfluous dimension is added to help the reason in reasoning, not to help the traveler in traveling. Least of all does it have anything whatsoever to do with time. The spacetime geometry is described by the metric (1) The dynamics of the geometry is known in full when one knows the radius a as a function of the time t. \u00b7\"Thus we may present the following arguments against the conception of a space-infinite, and for the conception of a space-bounded, universe: \"1. From the standpoint of the theory of relativity, the condition for a closed surface is very much simpler than the corresponding boundary condition at infinity of the quasi-Euclidean structure of the universe. \"2. The idea that Mach expressed, that inertia depends upon the mutual action of bodies, is contained, to a first approximation, in the equations of the theory of relativity;... But this idea of Mach's corresponds only to a finite universe, bounded in space, and not to a quasi-Euclidean, infinite universe\" [Einstein (1950), pp. 107-108). Many workers in cosmology are skeptical of Einstein's boundary condition of closure of the universe, and will remain so until astronomical observations confirm it. \u00a727.1. COSMOLOGY IN BRIEF 705 Eins~ein's field equation (doubled, for convenience), 2G = 16'lTT, has its whole force concentrated in its 00 (or tt) component, 6 6 (da)2 - +-= 16'lTp a 2 a 2 dt (2) [equation (Sa) of Box 14", ".5]. This component of Einstein's equation is as central as the component V\u00b7 E = 4'lTp of Maxwell's equations. It is described in the Track-2 Chapter 21 as the \"initial-value equation\" ofgeometrodynamics. There the two terms on the left receive separate names: the \"second invariant\" of the \"extrinsic curvature\" of a \"spacelike slice\" through the 4-geometry (tells how rapidly all linear dimensions are being stretched from instant to instant); and the \"intrinsic curvature\" or three dimensional scalar curvature invariant (31R of the \"spacelike slice\" (here a 3-sphere) at the given instant itself. The amount of mass-energy in the. universe changes from instant to instant in accordance with the work done by pressure during the expansion, d [ ( density of mass-energy ) X (volume) = -(pressure) d(volume). ] (3) Today the pressure of radiation is negligible compared to the density of mass-energy, and the righthand side of this equation (\"work done\") can be neglected. The same was true in the past, one estimates, back to a time when linear dimensions were about a thousand times smaller than they are today. During this \"matter-dominated phase\" of the expansion of the universe, the product ( density Of) ( I mass-energy X vo ume ) remained a constant, (4) Here the symbol M can look like mass in the form of matter, and can even be called mass; but one has to recall again (see \u00a719.4) that the concept of total mass-energy of a closed universe has absolutely no well-defined meaning whatsoever, not least because there is no \"platform\" outside the universe on which to stand to measure its attraction via periods of Keplerian orbits or in any other way. More convenient than M, because more significant in what follows, is the quantity amax (\"radius of universe at phase of maximum expansion\") defined by amax = 4M/3'lT. (5) 706 27. IDEALIZED COSMOLOGIES Box 27.1 (continued) The decisive component ofthe Einstein field equation, in the terms of this notation, becomes or (: y_a:ax _ -1 (6) The first term in (6) has the qualitative character of\"kinetic energy\" in an elementary problem in Newtonian mechanics. The second term has the qualitative character of a \"potential energy,\" V(a) = _ a max a -a~ Zero t I'(a) I (see diagram A), resulting from an inverse-square Newtonian force. Pursuing the analogy, one identifies the\" -1\" on the righthand side with the total energy in the Newtonian problem. The qualitative character of the dynamics shows up upon an inspection of diagram A. Values of the radius of the universe, a, greater than a max are not possible. If a were to become greater than am ax' the \"potential energy\" would exceed the total \"energy\" and the \"kinetic energy\" of expansion would have to become negative, which is impossible. Consequently the geometrodynamic system can never be in a state more expanded than a = amax' Starting in a state of small a, (a ~ amax) and expanding, the universe has for each a value a perfectly definite da/dt value. This velocity of expansion decreases as the expansion proceeds. It falls to zero at the turning point a = amax. Thereafter the system recontracts. \u00a727.1. COSMOLOGY IN BRIEF 707 Lack of option is the striking feature of the dynamics. Granted a specific amount of matter [specific M value in (5)], one has at his disposal no free parameter what soever. The value of amax is uniquely specified by the amount of matter present, and by nothing more. There is no such thing as an \"adjustable constant of energy,\" such as there would have been in a traditional problem of Newtonian dynamics. Where such an adjustable parameter might have appeared in equation (6), there appears instead the fixed number\" -I.\" This fixity is the decisive feature of a system bound up into closure. Were one dealing with a collection of rocks out in space, one would have a choice about the amount of dynamite one placed at their center. With a low charge of explosive, one would find the rocks flying out for only a limited distance before gravity halted their flight and brought them to collapse together again. With more propellant, they would fly out with escape velocity and never return. But no such options present themselves here, exactly because Einstein's condition of closure has been imposed; and once closed, always closed. Collapse of the universe is universal. This is simple cosmology in brief. Einstein's unhappiness at this result was great", ". At the time he developed general relativity, the permanence of the universe was a fixed item of belief in Western philosophy: \"The heavens endure from everlasting to everlasting.\" Yet the reasoning that led to the fixed equation left open no natural way to change that equation or its fantastic prediction. Therefore Einstein (1917), much against his will, introduced the least unnatural change he could imagine, a so-called cosmological term (\u00a727.ll), the whole purpose of which was to avoid the expansion of the universe. A decade later, Hubble (1929) verified the predicted expansion. Thereupon Einstein aban doned the cosmological term, calling it \"the biggest blunder of my life\" [Einstein (1970)]. Thus ended thenrst great cycle of apparent contradiction to general relativity, test, and dramatic vindication. Will one ever penetrate the mystery ofcreation? There is no more inspiring evidence that the answer will someday be \"yes\" than man's power to predict, and predict correctly, and predict against all expectations, so fantastic a phenomenon as the expansion of the universe. \"Newtonian cosmology\" provides an \"equation of energy\" similar to that of Einstein cosmology, but fails to provide any clean or decisive argument for closure or for the unique constant\" -I.\" It considers the mass in any elementary spherical region of space of momentary radius r, and the gravitational acceleration of a test particle at the boundary of this sphere toward the center of the sphere; thus, (mass) (distance)2 = (4'lT/3)pr 3 r2 4'lTp = --3-r. (7) Consider such imaginary spheres of varied radii drawn in the cosmological medium with the same center. Note that doubling the radius doubles the acceleration. This proportionality between acceleration and distance is compatible with a homogeneous 708 27. IDEALIZED COSMOLOGIES Box 27.1 (continued) deceleration of the expansion of the universe. Therefore define an expansion param eter a* as the ratio between the radius of anyone of these spheres now and the radius of the same sphere at some fiducial instant; thus, a* = r/ro is to be considered as independent of the particular sphere under consideration. Write P = por~/r3, where Po is the density at the fidu~ial instant. Insert this expression for p into the deceleration equation (7), multiply both sides of the equation through by dr/dt, integrate, and translate the result from an equation for dr/dt to an equation for da* / dt, finding da*)2 d t ( - (8r.Po/3)_ * - constant, a (8) in agreement with equation (6), except for (1) the trivial differences that arise because a* is a dimensionless expansion ratio, whereas a is an absolute radius with the dimensions of em, and (2) the all-important difference that here the constant is disposable, whereas in standard Einstein geometrodynamics it has the unique ca nonical value\" -I.\" For more on Newtonian insights into cosmology, see especially Bondi (1961). Free fall of a particle towards a Newtonian center of attraction according to Newtonian mechanics gives an equation of energy of the same form as (6), except that the \"radius of the universe,\" a, is replaced by distance, r, from the center of B attraction. The solution of this problem of free fall is described by a cycloid (diagram B; see also Figure 25.3 and Box 25.4), generated by rolling a circle of diameter amax on a line through an ever increasing angle 1); thus, 1 a = 2\" amax(1 - cos 1)), t = i arnax(1) - sin 1)). (9) \u00a727.1. COSMOLOGY IN BRIEF 709 Immediately observable today is the present rate of expansion of the universe, with every distance increasing at a rate directly proportional to the magnitude of that distance:'\" ( velocity of recession) of a galaxy (dIstance to a ga axy) 1. = (Hubble \"constant,\" Ho) - 55 lan/sec megaparsec _.. = 1 18 X 109 yr or 1 1.7 X 1028 cm ( rate of increase of the ) radius of the universe itself (radius of the universe) da/dt a (10) The Hubble time, H-;l - 18 X 109 yr (linearly extrapolated back to zero separation on the basis of the expansion rate observed today, as illustrated in the diagram) is predicted to be greater by a factor 1.5 or more (Box 27.3) than the actual time back to the start of the expansion as deduced from the rate of the development of stars (-10 X 109 yr). No such satisfactory concord between prediction and obser vation on this inequality existed in", " the 1940's. The scale of distances between galaxy and galaxy in use at that time was short by a factor more than five. The error arose from misidentifications of Cepheid variable stars and of HII regions, which are used as st~ndards of intensity to judge the distance of remote galaxies. The linearly extrapolated time,. (Hubb e tIme) = 1 (distance today) (recessIOn velocIty today).., back to the start of the expansion was correspondingly short by a factor more than five. The Hubble time came out to be only of the order of 3 X 109 yr. This number obviously violates the inequality ( ~3 X 109 yr HUbble) > 1.5 (-10 X 109 yr; actual t~me). tIme back to start of expansIon - It implies a curve for dimensions as a function of time not bending down, as in diagram B, but bending up. On some sides the proposal was made to regard the actual curve as rising exponentially. Thus began an era of \"theories of continuous creation of matter,\" all outside the context of Einstein's standard geometrodynamics. * H. is predicted to be independent of the choice of galaxy insofar as local motions are unimportant, and insofar as the difference between recession velocity now and recession velocity at the time when the light was emitted is unimportant. The latter condition is well fulfilled by galaxies close enough to admit of the necessary measurement of distance, for they have redshifts only of the order of z - 0.1 and less (little lapse of time between emission of light and its reception on earth: therefore little change in recession velocity between then and now; see \u00a729.3 and Box 29.4 for a fuller analysis). 710 27. IDEALIZED COSMOLOGIES Box 27.1 (continued) This eraertded when, for the first time, the distinction between stelIar populations of classes I and II was recognized and as a result Cepheid variables were correctly identified, by Baade (1952, 1956) and when Sandage (1958) discovered that Hubble had misidentified as bright stars the HII regions in distant galaxies. Then the scale of galactic distances was set straight. Thus ended the second great cycle of an apparent contradiction to general relativity, then test, and then dramatic vindication. The mystery of the missing matter marks a third cycle of doubt and test with the final decision yet to come. It follows from equation (2) that, if Einstein's closure boundary condition is correct, then the density of mass-energy must exceed a certain lower limit given by the equation (11) (\"critical amount of mass-energy required to curve up the geometry of the universe into closure\"). A Hubble expansion rate of Ho = 55 km/sec Megaparsec implies a lower limit to the density of 3 1 PH = 8'1T (1.7 X 1028 cm)2 or PH,conv = 5 X 10-30 g/cm3 (12) as compared to P -. 2 X 10-31 g/cm3 of \"luminous matter\" observed in galaxies (\u00a729.6) and more being searched for today in the space between the galaxies. A fuller treatment ofcosmology deals with conditions back in the past corresponding to redshifts of 10,000 or more and dimensions 10,000 times less than they are today, when radiation could not be neglected, and even dominated (\u00a727.1O). It also considers even earlier conditions, when anisotropy oscillations of the geometry of the universe as a whole (analogous to the transformation from a cigar to a pancake and back again) may conceivably have dominated (Chapter 30). More broadly, it takes. up the evolution of the universe into its present state (Chapter 28) and the present state and future evolution of the universe (Chapter 29). The present chapter examines the basic assumptions that underlie the simple standard cosmology thus traced out, and \u00a727.11 examines what kinds of qualitative changes would result if one or another of these assumptions were to be relaxed. \u00a727.2. STRESS-ENERGY CONTENT OF UNIVERSE-FLUID IDEALIZATION 711 \u00a727.2. STRESS-ENERGY CONTENT OF THE UNIVERSE THE FLUID IDEALIZATION By taking the large-scale viewpoint, one can treat galaxies as \"particles\" of a \"gas\" that fills the universe. These particles have internal structure (stars, globular clusters, etc.); but one ignores it. The \"particles\" cluster on a small scale (clusters of galaxies of size ~3 X 10 7 light years); but one ignores the clustering. To simplify calculations, one even ignores the particulate nature of the \"gas\" [though one can take it into account, if one wishes, by adopting a kinetic-theory description;", " see \u00a722.6 for kinetic theory, and Ehlers, Geren, and Sachs (1968) for its application to cosmology]. One removes the particulate structure of the gas from view by treating it in the perfect fluid approximation. Thus, one characterizes the gas by a 4-velocity, u (the 4-velocity of an observer who sees the galaxies in his neighborhood to have no mean motion), by a density of mass-energy, p (the smoothed-out density of mass-energy seen in the frame with 4-velocity u; this includes the rest mass plus kinetic energy of the galaxies in a unit volume, divided by the volume), and by a pressure p (the kinetic pressure of the galaxies). The stress-energy tensor for this \"fluid of galaxies\" is the familiar one T = (p + p)u ~ u + gp, (27.1) The rest of this chapter. except for Box 27.4. is Track 2. No earlier track-2 material is needed as preparation for it. but it is needed as preparation for Chapter 29 (Present state and future evolution of the universe). Idealization of matter in universe as a perfect fluid (\"fluid of galaxies\") where g is the metric tensor. Astronomical observations reveal that the rest-mass density of the galaxies is much greater than their density of kinetic energy. The typical ordinary velocities of the galaxies-and of stars in them-relative to each other are Large-scale conditions in universe today: (v) -.200 km/sec -. 10-3. (27.2) Consequently, the ratios of kinetic-energy density and of pressure to rest-mass den sity are (1) kinetic energy and pressure of stars and galaxies Ek1n/ Prm = ~ (v 2 ) ;::::: 10-6, P/Prm = j (V 2 ) ;::::: 10-6. (27.3) At least, these are the ratios today. Very early in the life of the universe, conditions must have been quite different. The total density of mass-energy, P, is thus very nearly the rest-mass density of (2) density of mass in the galaxies, Prm\u2022 Astronomical observations yield for Prm today galaxies Prm 2: 2 X 10-31 g/cm3 (27.4) (see \u00a729.6). 712 27. IDEALIZED COSMOLOGIES (3) cosmic-ray density Not all the matter in the universe is tied up in galaxies; there is also matter in cosmic rays, with an averaged-out density of mass-energy (4) density of intergalactic and, perhaps, gas in intergalactic space with gas Per ~ 10-33 g/cm3, (27.5) (27.6) [Delineating more sharply the value of Pig is one of the most important goals of current cosmological research. For a review of this question as of 1971, see \"The mean mass density of the universe,\" pp. 56-120 in Peebles (1971).] These sources of mass density, and the associated pressures, one can lump together with the galaxies into the \"cosmological fluid,\" with stress-energy tensor (27.1). Not all the stress-energy in the universe is in the form of matter. There are also magnetic fields, with mean energy density that almost certainly does not exceed the limit P < 10-35 g/cm3 mag _ (27.7) (5) magnetic fields (6) radiation density (corresponding to Bavg radiation, and perhaps gravitational radiation) totaling, one estimates, ~ 10-6 G), and radiation (electromagnetic radiation, neutrino The cosmic microwave radiation (27.8) The magnetic fields will be ignored in this chapter; they are unimportant for large scale cosmology, except perhaps very near the \"big-bang beginning\" of the uni verse-if they existed then. However, the radiation cannot be ignored, for it plays a crucial role. Most of the radiation density is in the form of \"cosmic microwave radiation,\" which was discovered by Penzias and Wilson (1965) [see also Dicke, Peebles, Roll, and Wilkinson (1965)], and has been studied extensively since then [for a review, see Partridge (1969)]. The evidence is very strong that this cosmic microwave radia tion is a remnant of the big-bang beginning of the universe. This interpretation will be accepted here. The cosmic microwave radiation has just the form one would expect if the earth were enclosed in a box (\"black-body cavity\") with temperature 2.7K. The spectrum is a Planck spectrum with this temperature, and the radiation is isotropic [Boughn, Fram, and Partridge (1971 )]. Consequently, its pressure and density of mass-energy are given by the formula, Pmlcrowave", " = 3Pmlcrowave = aT4 = 4 X 10-34 g/cm3 \u2022 (27.9) Thermodynamic considerations (\u00a727.1O) suggest that the universe should also be filled with neutrino radiation and perhaps gravitational radiation that have Planck spectra at approximately the same temperature (-3K). However, they are not detectable with today's technology. \u00a727.3. GEOMETRIC IMPLICATIONS OF HOMOGENEITY AND ISOTROPY 713 To high accuracy (~300 kIn/sec) the mean rest frame of the cosmic microwave radia tion near Earth is the same as the mean rest frame of the galaxies in the neighborhood of Earth [Boughn, Fram and Partridge (1971)]. Consequently, the radiation can be included, along with the matter, in the idealized cosmological fluid. Summary: From the large-scale viewpoint, the stress-energy of the universe can be idealized as a perfect fluid with 4-velocity u, density of mass-energy p, pressure P'_and _stress-energy tensor (27.10) The 4-velocity u at a given event 9 in spacetime is the mean 4-velocity of the galaxies near 9; it is also the 4-velocity with which one must move in order to measure an isotropic intensity for the cosmic microwave radiation. The density p is made up of material density (rest mass plus negligible kinetic energy of galaxies; rest mass plus kinetic energy of cosmic rays; rest mass plus thermal energy of intergalactic gas-all \"smeared out\" over a unit volume), and also of radiation energy density (electromagnetic radiation, neutrino radiation, gravitational radiation). The pressure p, like the density p, is due to both matter and radiation. Today the pressure of the-matteris Jl1u\u00a3h~less than its mass-energy density, Pmatter < Pmatter today, (27.1la) but this strong inequality cannot have held long ago. Always the pressure of the radiation is! its mass-energy density: PradJation = \"3 PradJation always. 1 (27.llb) \u00a727.3. GEOMETRIC IMPLICATIONS OF HOMOGENEITY AND ISOTROPY This chapter will idealize the universe to be completely homogeneous and isotropic. This idealization places tight constraints on the geometry of spacetime and on the motion of the cosmological fluid through it. In order to discover these constraints, one must first give precise mathematical meaning to the concepts of homogeneity and isotropy. Homogeneity means, roughly speaking, that the universe is the same everywhere at a given moment of time. A given moment of what time? Whose time? This is the crucial question that the investigator asks. In Newtonian theory there is no ambiguity about the concept \"a given moment of time.\" In special relativity there is some ambiguity because of the nonuniversality of simultaneity, but once an inertial reference frame has been specified, the concept becomes precise. In general relativity there are no global inertial frames (unless spacetime is flat); so the concept of \"a given moment of time\" is completely ambigu ous. However, another, more general concept replaces it: the concept of a three dimensional spacelike hypersurface. This hypersurface may impose itself on one's Summary of fluid idealization of matter in universe Spacelike hypersurface as generalization of \"moment of time\" 714 27. IDEALIZED COSMOLOGIES attention by reason of natural symmetries in the spacetime. Or it may be selected at the whim or convenience of the investigator. He may find it more convenient to explore spacetime here and there than elsewhere, and to push the hypersurface forward accordingly (\"many-fingered time\"; the dramatically new conception of time that is part of general relativity). At each event on a spacelike hypersurface, there is a local Lorentz frame whose surface of simultaneity coincides locally with the hypersurface. Of course, this Lorentz frame is the one whose 4-velocity is orthogonal to the hypersurface. These Lorentz frames at various events on the hypersurface do not mesh to form a global inertial frame, but their surfaces of simultaneity do mesh to form the spacelike hypersurface itself. The intuitive phrase \"at a given moment of time\" translates, in general relativity, into the precise phrase \"on a given spacelike hypersurface.\" The investigator can go further. He can \"slice up\" the entire spacetime geometry by means of a \"one parameter family\" of such spacelike surfaces. He can give the parameter that distinguishes one such slice from the next the name of \"time.\" Such a one-parameter family of slices through spacetime is not required in the Regge", " calculus of Chapter 42. However, such a \"slicing\" is a necessity in most other practical methods for analyzing the dynamics of the geometry of the universe (Chapters 21, 30, and 43~ The choice of slicing may dissolve away the difficulties of the dynamic analysis or may merely recognize those difficulties. The successive slices of \"moments of time\" may shine with simplicity or may only do a tortured legalistic bookkeeping for the dynamics. Which is the case depends on whether the typical spacelike hypersurface is distinguished by natural symmetries or, instead, is drawn arbitrarily. Homogeneity of the universe means, then, that through each event in the universe there passes a spacelike \"hypersurface of homogeneity\" (physical conditions identical at every event on this hypersurface). At each event on such a hypersurface the density, p, and pressure, p, must be the same; and the curvature of spacetime must be the same. The concept of isotropy must also be made precise. Clearly, the universe cannot look isotropic to all observers. For example, an observer riding on a 1020 eV cosmic ray will see the matter of the universe rushing toward him from one direction and receding in the opposite direction. Only an observer who is moving with the cosmo logical fluid can possibly see things as isotropic. One considers such observers in defining isotropy: Isotropy of the universe means that, at any event, an observer who is \"moving with the cosmological.fluid\" cannot distinguish one of his space directions from the others by any local physical measurement. Isotropy of the universe actually implies homogeneity; of this one can convince oneself by elementary reasoning (exercise 27.1). Isotropy guarantees that the world lines of the cosmological.fluid are orthogonal to each hypersurface of homogeneity. This one sees as follows. An observer \"moving with the fluid\" can discover by physical measurements on which hypersurface through a given event conditions are homogeneous. Moreover, he can measure his own ordinary velocity relative to that hypersurface.\u00b7 If that ordinary velocity is nonzero, it provides the observer with a way to distinguish one space direction in \"Homogeneity of universe\" defined in terms of spacelike hypersurfaces \"Isotropy of universe\" defined Isotropy implies fluid world lines orthogonal to homogeneous hypersurfaces \u00a727.4. COMOVING, SYNCHRONOUS COORDINATES FOR THE UNIVERSE 715 his rest frame from all others-in viola tion of isotropy. Thus in an isotropic universe, where the concept of \"observer moving with the fluid\" makes sense, each such observer must discover that he is at rest relative to the hypersurface of homogeneity. His world line is orthogonal to that hypersurface. ISOTROPY IMPLIES HOMOGENEITY Exercise 27.1. Use elementary thought experiments to show that isotropy of the universe implies homogeneity. EXERCISE Construction of a \"comoving, synchronous\" coordinate system for the universe \u00a727.4, COMOVING, SYNCHRONOUS COORDINATE SYSTEMS FOR THE UNIVERSE The results of the last section enable one to set up special coordinate systems in the spacetime manifold of an isotropic model universe (Figure 27.1). Choose a hypersurface of homogeneity S/. To all the events on it assign coordinate time, tI. Layout, in any manner desired, a grid of space coordinates (xl, x 2, x 3) on S[\" \"Propagate\" these coordinates off SI and throughout all spacetime by means of the world lines of the cosmological fluid. In particular, assign to every event on a given world line the space coordinates (xl, x 2, x 3 ) at which that world line intersects SI' This assignment has a simple consequence. The fluid is always at rest relative to the space coordinates. In-other words, the space coordinates are \"comoving':' they are merely labels for the world lines of the fluid. For the time coordinate t of a given event 9, use the lapse of proper time, f dr, of 9 from Sb as measured along the fluid world line that passes through 9, plus tI (\"standard of time\" on the initial hypersurface SI); thus, t(9) = tI + ( 'j' ~l ) dr along world llne of fiuld (27.12) The surfaces t = constant of such a coordinate system will coincide with the hyper surfaces of homogeneity of the universe. This one sees by focusing attention on observations made by two different observers, A and B, who move with the fluid along different world lines. At coordinate time tI (on SI) the universe looks the same to B as to A. Let A and B make observations again after their", " clocks have ticked away the same time interval Lh. Homogeneity of the initial hypersurface Sb plus the deterministic nature of Einstein's field equations, guarantees that A and B will again see identical physics. (Identical initial conditions on Sb plus identical lapses of proper time during which Einstein's equations govern the evolution of the universe near A and B, guarantee identical final conditions.) Therefore, after time lapse Lh, A and B are again on the same hypersurface of homogeneity-albeit a different 716 27. IDEALIZED COSMOLOGIES This fluid element carries the label (Xl, x 2, x 3) = (9, 2, 136) -'\\\"'\"-+-_..1-~~r~- A piece of the hyper surface t = t l + 5 World lines of particles of the fluid with spatial grid attached to them (\"comoving\") f- A piece of the initial hyper- surface 81' with arbitrary coordinate grid on it.....~ '10(. __L..--O-~ ~ \\\\.... \\\\...., u>.. \\I.... II ~ lJ' ;.... -'l' ;. - Figure 27.1. Comoving, synchronous coordinate system for the universe, as constructed in \u00a727.4 of the text. Key features of such a coordinate system are as follows (see \u00a7\u00a727.4 and 27.5). (I) The spatial coordinates move with the fluid, and the time coordinate is proper time along the fluid world lines; i.e., the coordinate description of a particular fluid world line is (Xl, x 2, x 3) = constant, XO == t = 'I' + constant. t r proper time measuredl J Lalong world line (2) Any surface of constant coordinate time is a hypersurface of homogeneity of the universe. Every such hypersurface is orthogonal to the world lines of all particles of the fluid. (3) The spatial grid on is the metric on the initial some initial hypersurface 5/ is completely arbitrary. (4) If Yu dx hypersurface in terms of its arbitrary coordinates (with Yu a function of Xl, x 2, x3), then the metric of spacetime in terms of the comoving, synchronous coordinate system is dxl'ds2 = -dt 2 + a2(t)yu dxl dx l \u2022 Thus, the entire dynamics of the geometry of the universe is embodied in a single function of time, aCt) = \"expansion factor\"; while the shape (but not size) of the hypersurfaces of homogeneity is embodied in the spatial 3-metric Yli dxl dxl \u2022 one from Sf' where they began. By virtue of definition (27.12) of coordinate time, the time coordinate at the intersection of B's world line with this new hypersurface is t = tf + LIT; and similarly for A. Moreover, observers A and B were arbitrary. Consequently the new hypersurface of homogeneity, like Sf' is a hypersurface of constant coordinate time. Q.E.D..Because the hypersurfaces of homogeneity are given by t =' constant, the basis vectors a/ax i at any given event 9 are tangent to the hypersurface of homogeneity that goes through that event. On the other hand, the time basis vector, a/at, is tangent to the world line of the fluid through 9, since that world line has xi = constant along it. Consequently, orthogonality of the world line to the hypersurface guarantees orthogonality of a/at to a/ax i : (a/at)\u00b7 (a/ax i ) = 0 for i = 1,2,3. (27.l3a) \u00a727.4. COMOVING. SYNCHRONOUS COORDINATES FOR THE UNIVERSE 717 The time coordinate has another special property: it measures lapse of proper time along the world lines of the fluid. Because of this, and because a/at is tangent to the world lines, one can write a/at = (d/dT)alongflUld'sworldtines = u, (27.13b) where u is the 4-velocity of the \"cosmological fluid.\" The 4-velocity always has unit length, (a/at)\u00b7 (a/at) = -1. (27.13c) Conditions (27.13a,c) reveal ~f3 =(%x a ). (%x f3 )], the line element for spacetime reads that, in the comoving coordinate frame [where (27.14) Form of the line element in this coordinate system Any coordinate system in which the line element has this form is said to be \"synchronous\" (l) because the coordinate time t", " measures proper time along the lines of constant Xi (i.e., gtt = -1), and (2) because the surfaces t = constant are (locally) surfaces of simultaneity for the observers who move with Xi = constant [i.e., gti = (a/at) \u2022 (%xi ) = 0]; it is also called a \"Gaussian normal coordinate system\" (cf Figure 21.6). A hypersurface of homogeneity, t = constant, has a spatial, three-dimensional geometry described by equation (27.14) with dt = 0: (dS2)onhypers~rfaceofhomogeneity = do 2 = [gii]t = const dx i dx i. (27.15) To know everything about the 3-geometry on each of these hypersurfaces is to know everything about the geometry of spacetime. Exercise 27.2. SYNCHRONOUS COORDINATES IN GENERAL In an arbitrary spacetime manifold (not necessarily homogeneous or isotropic), pick an initial spacelike hypersurface Sf> place an arbitrary coordinate grid on it, eject geodesic world lines orthogonal to it, and give these world lines the coordinates EXERCISE where T is proper time along the world line, beginning with T = 0 on Sf' Show that in this coordinate system the metric takes on the synchronous (Gaussian normal) form (27.14). Proof that. aside from an over-all \"expansion factor.\" all homogeneous hypersurfaces in the universe have the same 3-geometry 718 27. IDEALIZED COSMOLOGIES \u00a727.5. THE EXPANSION FACTOR To determine the 3-geometry, d0 2 = gij(t, x k) dx i dx i, of each of the hypersurfaces of homogeneity, split the problem into two parts: (l) the nature of the 3-geometry on an arbitrary initial hypersurface (dealt with in next section); and (2) the evolution of the 3-geometry as time passes, i.e., as attention moves from the initial hypersur face to a subsequent hypersurface, and another, and another,... (dealt with in this section). Assume that one knows the initial 3-geometry-i.e., the coefficients in the space part of the metric, (27.16) on the initial hypersurface Sf-in its arbitrary but explicitly chosen coordinate system. What form will the metric coefficients gik(t, x k) have on the other hypersurfaces of homogeneity? This question is easily answered by the following argument: Consider two adjacent world lines, {/ and!XJ, of the cosmological fluid, with coordinates (xl, x 2, x 3 ) and (Xl +.:!xl, x 2 +.:!x2, x 3 +.:!x3). At time t[ (on surface Sf) they are separated by the proper distance (27.17) At some later time t (on surface S), they will be separated by some other proper distance.1o(t). Isotropy of spacetime guarantees that the ratio of separations.1o(t)/.10(t[) will be independent of the direction from {/ to qJ (no shearing motion of the fluid). For any given direction, the additivity of small separations guarantees that.10(t)/.10(t[) will be independent of.1o(t[). Thus.1o(t)/.10(t[) must be the same for all pairs of world lines near a given world line. Finally, homogeneity guarantees that this scalar ratio will be independent of position on the initial surface S[-i.e., independent of xl, x 2, x 3 \u2022 Define a(t) to be this spatially constant ratio, a(t).10(t)/.1o(t[). (27.18) Thus, a(t) is the factor by which the separations of world lines expand between time t[ and time t. In other words, the function a(t) is a universal \"expansion factor,\" or \"scale factor.\" By combining equations (27.17) and (27.18), one obtains for the separation of adjacent world lines at time t.1o(t) = a(t)[Yij(xk).:!xi.:!xi]l/2. This corresponds to the spatial metric at time t, and to the spacetime metric, ds 2 = - dt2 + a2(t)yii(x k) dx i dx i. (27.19) (2720) \u00a727.5. THE EXPANSION FACTOR 719 Figure 27", ".2. Inflation of a balloon covered with pennies as a model for the expansion of the universe. Each penny A may well consider itself to be the center of the expansion because the distance from A to any neighbor B or C increases the more the more remote that neighbor was to begin with (\"the Hubble relation\"). The pennies themselves do not expand (constancy of sun-Earth distance, no expansion of a meter stick, no increase of atomic dimensions). The spacing today between galaxy and galaxy (- 106 lyr) is roughly ten times the typical dimension of a galaxy (- 105 lyr). k Notice that the coefficients \"Yij(X ) describe the shape not only of the initial hypersurface, but also of all other hypersurfaces of homogeneity. All that changes in the geometry from one hypersurface to the next is the scale of distances. All distances between spatial grid points (fluid world lines) expand by the same factor a(t), leaving the shape of the hypersurface unchanged. This is a consequence of homogeneity and isotropy; and it is precisely true only if the model universe is precisely homogeneous and isotropic. What expands in the universe. and what does not Of all the disturbing implications of \"the expansion of the universe,\" none is more upsetting to many a student on first encounter than the nonsense of this idea. The universe expands, the distance between one cluster of galaxies and another cluster expands, the distance between sun and earth expands, the length of a meter stick expands, the atom expands? Then how can it make any sense to speak of any expansion at all? Expansion relative to what? Expansion relative to nonsense! Only later does he realize that the atom does not expand, the meter stick does not expand, the distance between sun and earth does not expand. Only distances between clusters of galaxies and greater distances are subject to the expansion. Only at this gigantic scale of averaging does the notion of homogeneity make sense. Not so at smaller distances. No model more quickly illustrates the actual situation than a rubber balloon with pennies affixed to it, each by a drop of glue. As the balloon is inflated (Figure 27.2) the pennies increase their separation one from another but not a single one of them expands! [For mathematical detail see, e.g., Noerdlinger and Petrosian (1971).] 720 27. IDEALIZED COSMOLOGIES EXERCISE Exercise 27.3. ARBITRARINESS IN THE EXPANSION FACTOR How much arbitrariness is there in the definition of the expansion factor aCt)? Civilization A started long ago at time tAo For it, the expansion factor is proper distance between ) two particles of the \"cos mological fluid\" at time t proper distanc~ between) = aA(t). same two partIcles at time tA ( ( Subsequently men planted civilization B at time tB on a planet in a nearby galaxy. [At this time, the expansion factor aA had the value aA(tB)]. Civilization B defines the expansion factor relative to the time of its own beginning: proper distance between ) two particles of the \"cosmological fluid\" at time t proper distance between ) = aBet). the same two particles at time tB. ( ( At two subsequent events, C and D, of which both civilizations are aware, they assign to the universe in their bookkeeping by no means identical expansion factors, aA(tC) \"I: aB(tC)' aA(tD) \"I: aB(tD)' Show that the relative expansion of the model universe in passing from stage C to stage D in its evolution is nevertheless the same in the two systems of bookkeeping: aA(tD) = (relative expansion) = aB(tD). aB(tC) aA(tc) from C to D \u00a727.6. POSSIBLE 3-GEOMETRIES FOR A HYPERSURFACE OF HOMOGENEITY Tum now to the 3-geometry l'ii dx i dx i for the arbitrary initial hypersurface Sf. This 3-geometry must be homogeneous and isotropic. A close scrutiny of its three-dimen sional Riemann curvature must yield no \"handles\" to distinguish one point on Sf from any other, or to distinguish one direction at a given point from any other. \"No handles\" means that (3lRiemann must be constructed algebraically from pure numbers and from the only \"handle-free\" tensors that exist: the 3-metric l'ij and Riemann tensor for homogeneous. isotropic hypersurfaces \u00a727.6. 3-GEOMETRIES FOR HOMOGENEOUS HYPERSURFACES 721 the three-dimensional Levi-Civita tensor \u00a3iik' (All other tensors pick", " out preferred directions or locations.) One possible expression for (3lRiemann is (31~ikl = K('Yik'Yil - 'YiI'Yik); K = \"curvature parameter\" = constant. (2721). n na e onc:ubat this is the only expression that both has the correct symmetries for a curvature tensor and can be constructed solely from con stants, 'Yw and \u00a3iik' Hence, this must be the 3-curvature of 3/. [One says that any manifold with a curvature tensor of this form is a manifold of \"constant curvature. \"] As one might expect, the metric for 3/ is completely determined, up to coordinate transformations, by the form (27.21) ofits curvature tensor. (See exercise 27.4 below). With an appropriate choice ofcoordinates, the metric reads (see exercise 27.5 below), do 2 = 'Yii dx i dx i = K-l[dX 2 + sin2X(d8 2 + sin28 dcf>2)] if K > 0, do 2 = 'Yij dx i dx i = dX 2 + X2(d82 + sin28 dcf>2) if K = 0, do 2 = 'Yij dx i dx i = (-Kr 1[dX 2 + sinh2X(d82 + sin28 dcf>2)] if K < 0. (27.22) Metric for homogeneous, isotropic hypersurfaces: three possibilities-positive, zero, or negative spatial curvature Absorb the factor K-1/2 or (-Kr l12 into the expansion factor a(t) [see exercise 27.3], and define the function I =sin X, I =sinh X, I=X, if k = K/IKI = + 1 (\"positive spatial curvature\"), if k =K = \u00b0(\"zero spatial curvature\"), if k = K/IKI = -1 (\"negative spatial curvature\"). Thus write the full spacetime geometry in the form ds 2 = - dt2 + a2(thii dx i dx i, 'Yii dx i dx i = dX 2 + I2(d82 + sin28 dcf>2), (27.23) (27.24) and the 3-curvatures of the homogeneous hypersurfaces in the form (3lRijkl = [k/a2(t)][Yik'Yil - YilYik]' (27.25a) The curvature parameter K, after this renormalization, is evidently (27.25b) Why is the word \"renormalization\" appropriate? Previously a(t) was a scale factor describing expansion of linear dimensions relative to the linear dimensions as they stood at some arbitrarily chosen epoch; but the choice of that fiducial epoch was a matter of indifference. Now a(t) has lost that arbitrariness. It has been normalized so that its value here and now gives the curvature of a spacelike hypersurface of homogeneity here and now. Previously the factor a(t) was conceived as dimensionless. Now it has the dimensions of a length. This length is called the \"radius of the model universe\" when the curvature is positive. Even when the curvature is negative one sometimes speaks of a(t) as a \"radius.\" Only for zero curvature does the normaliza- Significance of normalization of the expansion factor 722 27. IDEALIZED COSMOLOGIES tion of a(t) still retain its former arbitrariness. Thus, for zero-curvature, consider two choices for a(t), one of them a(t), the other a(t) = 2a(t). Then with X= h, one can write proper distances in the three directions of interest with perfect indifference in either of two ways: proper distanCe) in ~he dir~ction = a(t) dX = a(t) dX, of increasing X proper distanCe) in the direction = a(t)x d() = a(t)x d(), of increasing () proper distanCe) in ~he dir~ction = a(t)x sin () de/> = a(t)x sin () de/>, of Increasmg \u00a2 ( ( ( No such freedom of choice is possible when the model universe is curved, because then the X's in the last two lines are replaced by a function, sin X or sinh X, that is not linear in its argument. Despite the feasibility in principle of determining the absolute value of the \"radius\" a(t) of a curved universe, in practice today's accuracy falls short of what is required to do so. Therefore it", " is appropriate in many contexts to continue to regard a(t) as a factor of relative expansion, the absolute value of which one tries to keep from entering into any equation exactly because it is difficult to determine. This motivation will account for the way much of the analysis of expansion is carried out in what follows, with calculations arranged to deal with ratios of a values rather than with absolute a values. Box 27.2 explores and elucidates the geometry of a hypersurface of homogeneity. EXERCISES Exercise 27.4. UNIQUENESS OF METRIC FOR 3-SURFACE OF CONSTANT CURVATURE Let Yij and Yi'j' be two sets of metric coefficients, in coordinate systems {Xi} and {Xi'}, that have Riemann curvature tensors [derived by equations (8.22) and (8.42)] of the constant curvature type (27.21). Let it be given in addition that the curvature parameters K and K' are equal. Show that Yij and Yi'J' are related by a coordinate transformation. [For a solution, see \u00a78.l0 of Robertson and Noonan (1968), or \u00a7\u00a710 and 27 of Eisenhart (1926).]. Exercise 27.5. METRIC FOR 3-SURFACE OF CONSTANT CURVATURE (a) Show that the following metric has expression (27.21) as its curvature tensor Yij = 1 + \"4 K 8kl X ( 1 )-2 k x l 8ij' (27.26)* *With this choice of spatial coordinates, the spacetime metric reads ds2 = -dt2 + (dx 2 + dy2 + dz2) [1 +!K(x2 + y2 + z2)j2 This is often called the \"Robertson- Walker line element,\" because Robertson (1935,1936) and Walker (1936) gave the first proofs that it describes the most general homogeneous and isotropic spacetime geometry. \u00a727.6. 3\u00b7GEOMETRIES FOR HOMOGENEOUS HYPERSURFACES 723 (b) By transforming to spherical coordinates (R, (J, ep) and then changing to a Schwarzs child radial coordinate (2'ITr = \"proper circumference\"), transform this metric into the form (2727) (c) Find a further change of radial coordinate that brings the metric into the form (2722). Exercise 27.6. PROPERTIES OF THE 3-SURFACES Verify all statements made in Box 27.2. ISOTROPY IMPLIES HOMOGENEITY Exercise 27.7. Use the contracted Bianchi identity (3)Glk 1k = 0 (where the stroke indicates a covariant derivative based on the 3-geometry alone) to show (1) that (3l\"'K = 0 in equation (27.21), and therefore to show (2) that direction-independence of the curvature [isotropy; curvature of form (27.21)] implies and demands homogeneity (K constant in space). (continued on page 726) Box 27.2 THE 3-GEOMETRY OF HYPERSURFACES OF HOMOGENEITY A. Universe with Positive Spatial Curvature '''Spatially Closed Universe\") Metric of each h'ypersurface is do 2 = a2[dx 2 + sin2x(d{12 + sin 2{1 d<l>2)]. (1) Typical point with </>::;: 'IT w To visualize this 3-geometry, imagine embed ding it in a four-dimensional Euclidean space (such embedding possible here; not possible for general three-dimensional manifold; only four freely disposable functions [w, x,y, z] of three vari ables [a, {3, y] are at one's disposal to try to repro duce six prescribed functions [gmn(a, (3, y)] of those same three variables). The embedding is achieved by /'/'w = a cos X, z = a sin X cos {I, x = a sin X sin {I cos <1>, y = a sin X sin (I sin <1>, (2) z -\u00a5----'-...+--- x /'l4 - -a I.. I \u2022 since it follows that do 2_ dw 2 + dx 2 + dyz + dz 2 =a2[dx 2 + sin2X(d{l2 + sin2{1 d\u00a22)]. (3) A 3-surface of positive curvature embedded in four-dimen sional Euclidean space. One rotational degree of freedom is suppressed by setting <I> ::;: 0 and 'Ir (\"slice through pole,\" 3", " sphere in 4-space looks like a 2-sphere in 3-space). Box 27.2 (continued) B. Universe with Zero Spatial Curvature (\"Spatially Flat Universe\") Metric of each hypersurface is Equations (2) for the embedded surface imply do 2 = a2[dx 2 + X2(d()2 + sin2() dcp2)]. (6) that (4) i.e., the surface is a 3-dimensional sphere in 4 dimensional Euclidean space. To verify homogeneity and isotropy, one need only notice that rotations in the four-dimensional embedding space can move any given point [any given (w, x,y, z) on the 3-sphere] and any given direction at that point into any other point and direction-while leaving unchanged the line ele ment do 2 = dw 2 + dx 2 + dy 2 + dz2. The above equations and the picture show that (1) The 2-surfaces of fixed X (which look like cir cles in the picture, because one rotational de gree of freedom is suppressed) are actually 2-spheres of surface area 4'ITa 2sin2x; and \u00ab(), cp) are standard spherical coordinates on these 2-spheres. (2) As X ranges from 0 to 'IT, one moves outward from the \"north pole\" of the hypersurface, through successive 2-spheres (\"shells\") of area 4'ITa2sin2X (2-spheres look like circles in pic ture). The area of these shells increases rapidly at first and then more slowly as one ap proaches the \"equator\" of the hypersurface, X = 'IT/2. Beyond the equator the area de creases slowly at first, and then more rapidly as one approaches the \"south pole\", <x = 'IT; area = 0). (3) The entire hypersurface is swept out by o ~ X~ 'IT, o ~ () ~ 'IT, o ~ cp ~ 2'IT (cp is cyclic; cp = 0 is same as cp = 2'IT); its 3-volume is 0/ =f (a dx)(a sin X dO)(a sin X sin 0 dcp) 1'11' = 4'ITa 2 sin2x(a dx) = 2'IT 2a3. 0- (5) This is a perfectly flat, three-dimensional, Eucli dean space described in spherical coordinates. In Cartesian coordinates x = ax sin 0 cos cp, y = ax sin () sin cp, z = aX cos (), the metric is do 2 = dx 2 + dy 2 + dz2. The entire hypersurface is swept out by o ~ X < 00, o ~ 0 ~ 'IT, o ~ cp ~ 2'IT; (7) (8) (9) and its volume is infinite. C. Universe with Negative Spatial Curvature (\"SpatiaUy open Universe\") Metric of each hypersurface is do 2 = a2[dx 2 + sinh2x(d()2 + sin2() dcp2)]. (10) This 3-geometry cannot be embedded in a four dimensional Euclidean space; but it can be em bedded in a flat Minkowski space do 2 = -dw 2 + dx 2 + dy 2 + dz 2. (11) To achieve the embedding, set w = a cosh X, z = a sinh X cos (), x = a sinh X sin 0 cos cp, Y = a sinh xsin 0 ~in cp; insert this into equation (11), and thereby obtain (10). (12) Equations (12) for the embedded surface imply that. (13) i.e., the surface is a three-dimensional hyperboloid in four-dimensional Minkowski space. (It has the w Typical point a z A 3-surface of negative curvature embedded in four-dimen sional Minkowski space. One rotational degree of freedom is suppressed by setting q, = 0 and 'IT (\"slice through pole\"; 3\u00b7 hyperboloid in 4-space looks like 2-hyperboloid in 3-space). same form as a mass hyperboloid in momentum space; see Box 22.5.) To verify homogeneity and isotropy, one need only notice that \"Lorentz transformations\" in the embedding space can move any given point on the 3-hyperboloid and any direction through that point into any other point and direction-while leaving unchanged the line element \u2022 do 2 = _dw 2 +- dx 2 + dy 2 + dz 2 The above equations and the picture show that (1) The 2-surfaces of fixed X (which look like cir cles in the picture because one rotational de gree", " of freedom is suppressed) are actually 2-spheres of surface area 4'ITa 2 sinh2X; and \u00ab(), ep) are standard spherical coordinates on these 2-spheres. (2) As X ranges from 0 to 00, one moves outward from the (arbitrarily chosen) \"pole\" of the hypersurface, through successive 2-spheres (\"shells\") of ever increasing area 4'ITa 2 sinh 2X. For large X, surface area increases far more ------i~____;r:=at:':i~d!Y than it would if the hypersurface were flat (proper surface area) 4'IT (proper distance? = 4'IT 12 A The entire hypersurface is swept out by o ~ X < 00, o ~ () ~ 'IT, o ~ ep ~ 2'IT (ep is cyclic; ep = 0 is same as ep = 2'IT). The volume of the hypersurface is infinite. D. Nonuniqueness of Topology (15) Warning: Although the demand for homogeneity and isotropy determines completely the local geo metric properties of a hypersurface of homogene ity up to the single disposable factor K, it leaves the global topology of the hypersurface undeter mined. The above choices of topology are the most straightforward. But other choices are possible. This arbitrariness shows most simply when the hypersurface is flat (k = 0). Write the full space time metric in Cartesian coordinates as ds 2 = -dt 2 + a 2(t)[dx 2 + dy 2 + dz 2]. (16) Then take a cube of coordinate edge L 0< x < L, 0< z < L, 0< Y < L, and identify opposite faces (process similar to roll ing up a sheet of paper into a tube and gluing its edges together; see last three paragraphs of \u00a711.5 for detailed discussion). The resulting geom etry is still described by the line element (16), but now all three spatial coordinates are \"cyclic,\" like the ep coordinate of a spherical coordinate system: (t, x,y, z) is the same event as (t, x + L,y + L, z + L). The homogeneous hypersurfaces are now \"3-tor uses\" of finite volume V = a3L3, analogous to the 3-toruses which one\u00b7 meets under the name \"periodic boundary conditions\" when analyzing electron waves and acoustic waves in solids and electromagnetic waves in space. Another example: The 3-sphere described in part A above (case of \"positive curvature\") has the same geometry, but not the same topology, as the manifold of the rotation group, SO(3) [see exer cises 9.12, 9.13, 10.16, and 11.12]. For detailed discussion, see for example Weyl (1946), Coxeter (1963), and Auslander and Markus.(1959).. y Euler equation is vacuous for a homogeneous universe 726 27. IDEALIZED COSMOLOGIES \u00a727.7. EQUATIONS OF MOTION FOR THE FLUID After the above analysis of anyone hypersurface of homogeneity, return to the dynamics of the universe. Examine, first, the evolution ofthe fluid, as governed by the law V. T = 0. Recall (\u00a722.3 and 23.5) that for a perfect fluid the equations of motion split into two parts. The component along the 4-velocity, u' (V. T) = 0, reproduces the first law of thermodynamics (d/dr)(pV) = -p(dVjdr), (27.28a) where V is the volume of any fluid element. The part orthogonal to the 4-velocity, (g + u @ u). (V. T) = 0, gives the force equation (\"Euler equation\") (p + p) X (4-acce1eration) = - (component of Vp orthogonal to u). (27.28b) For a static star (\u00a723.5) the first law of thermodynamics was vacuous, but the force equation was crucial. For a homogeneous universe, the converse is true; the force equation is vacuous (no accelerations), but the first law of thermodynamics is crucial. To see that the force equation is vacuous, notice that isotropy guarantees the vanishing of both sides of equation (27.28b). If either side were nonzero at any event '!l, it would distinguish a direction in the homogeneous hypersurface at '!l. In applying the first law of thermodynamics (27.28a) to cosmology, divide the density and pressure into contributions due to matter and contributions due to radiation: P = Pm +", " Pr' (27.29) \"Equations of state\" for matter and radiation First discuss the density of mass-energy. Today Pm(~ 10-31 g/cm3) dominates over Pr(_10- 33 g/cm3). Matter did not always dominate. Therefore, one cannot set Pr = 0. Now discuss the pressure. During that epoch of the universe when pressure was significant cosmologically, Pr dominated over Pm' Consequently, one can neglect Pm at all times, and one can use the \"equation of state\" for radiation, Pr = ~Pr' to write 1 P = \"3Pr' (27.30) When (27.30) is inserted into the first law of thermodynamics (27.28a), it yields the result (27.31 ) Energy exchange between matter and radiation is negligible One cannot integrate this equation until one knows how mass-energy is fed back and forth between matter and radiation-i.e., until one knows another relationship between Pm V and PrV. All estimates indicate that, except hi the first few seconds of the life of the universe, the energy exchanged between radiation and matter was \u00a727.7. EQUATIONS OF MOTION FOR THE FLUID 727 negligible compared to Pm Vand PrV individually (see \u00a728.l). Under these conditions, equation (2731) can be split into two parts: and The solutions are simple: (2732a) (2732b) Pm V = constant (conservation of matter) (2733a) and First law of thermodynamics used to express densities of radiation and matter in terms of expansion factor PrV4I3 = const = V-~/3 V (constancy of nUmber) (2733b) of photons P..---__t jenergy he/A of l one photon, up to a factor of jproportionalitYj Now what is V? It is the volume of any fluid element. It has the value for a fluid element with edges.1X,.18, &po Here X, 8, ep are constant along each world line of the fluid (comoving coordinates). Therefore the element ofhyperspher ical solid angle I2 sin 8.1X.18 &p (or pseudohyperspherical solid angle for the model of an open universe) is constant throughout all time for any fluid element. Therefore the volume of the fluid element grows in direct proportion to the cube of the expansion parameter a; thus, V/aJ = constant. Combining this result with the constancy of PmV and PrV4I3, one sees that pma3 = constant, p~4 = constant. (2734) Let Pmo be the density of matter today, Pro be the density of radiation today, and ao be the expansion factor for the universe today. Then, at any time in the past, and ~ a~ p(t) = Pmo a3(t) + Pro a4(t) 1 p(t) = '3 Pro a4(t). a 4 (2735a) (2735b) Evaluation of the Einstein field equation for a homogeneous universe: 728 27. IDEALIZED COSMOLOGIES These results were based on two key claims, which will be justified in detail later (Chapter 28): the claim that in the epoch when pressure was important Pm was much smaller than Pr; and the claim that exchange of mass-energy between radiation and matter was always negligible (except in the first few seconds after the \"creation\"). \u00a727.8. THE EINSTEIN FIELD EQUATION Once the time evolution of the expansion factor, a(t), is known, one can read off the time evolution of the density and pressure directly from equations (27.35). The density and pressure, in tum, determine how the expansion proceeds in time, via Einstein's field equations. Thus the field equations \"close the logic loop\" and give one a closed mathematical system from which to determine all three quantities, a(t), p(t) and p(t). One can readily calculate the components of the Einstein tensor for the model universe using the orthonormal basis one-forms, w 8= a(t)'s dO, W x=a(t)dX, wl=dt, The result [see equations (5) of Box 14.5] is 3k G\u00b7\u00b7 = 3...L +-, a2 a (a)2 tt G~. = 0 if JL 1= v. w\u00a2 = a(t)'s sin 0 dep. (27.36) (27.37 a) (27.37b) (27.37c) (With foresight, one will notice ahead of time that isotropy guarantees the equality Gxx = G88 = G\u00a2\u00a2' and similar equalities for the Riemann", " tensor; and one will calculate only Gxx' the component that is most easily calculated.) The basis one-forms, Wi, w X, w 8, w\u00a2, are the orthonormal basis carried along by an observer who moves with the \"cosmological fluid.\" Consequently, Ttl is the mass-energy density, p, that he measures; T,\"1 is the pressure,p; Tt; vanishes, because he sees no energy flux (no momentum density); and 1;; vanishes for i 1= j because he sees no shear stresses: Ttt = p, Txx = T88 = T\u00a2\u00a2 =p, T~. = 0 when JL 1= v. (27.38a) (27.38b) (27.38c) Equate the Einstein (\"moment of rotation\") tensor of equations (27.37) to the stress-energy tensor of equations (27.38). And if one insists, include the so-called \"A-term\" or \"cosmological term\" in the field equations [Einstein (1970): \"the biggest blunder of my life\"). Thus obtain two nonvacuous field equations. The first is an \u00a727.B. THE EINSTEIN FIELD EQUATION 729 \"initial value equation,\" which relates a,l to a and p at any initial moment of time: (::-f = k A - a2 +T+T P. 8'17 (27.39a) (1) initial value equation omit The second is a \"dynamic equation,\" which gives the second time-derivative of the expansion factor, and thereby governs the dynamic evolution away from the initial moment of time, 2!!.!! = - (!!.J!.)2 _ ~ + A _ 8'ITp. a a a2 ~ omit (27.39b) (2) dynamic equation If (27.39b) is to be compared with anything in Newtonian mechanics, it is to be compared with an equation for acceleration (equation of motion), and in the same spirit (27.39a) is to be compared with a first integral of the equation of motion; that is, an equation of energy. In accordance with this comparison, note that one only has to differentiate (27.39a) and combine it with the relation satisfied by the pressure, Why the dynamic equation is superfluous (\"law of conservation of energy\") to get the acceleration equation (27.39b). Without any loss of information, one can therefore ignore the \"acceleration equation\" or \"dynamic equation\" (27.39b) henceforth, and work with the analog of an energy equation or what is more properly known as an \"initial-value equation\" (details of initial-value problem for \"frack-2 readers in Chapter 21). What shows up here in the limited context of Friedmann cosmology is appropri ately viewed in the wider context of general geometrodynamics. Conservation of energy plus one field equation have just been seen to reproduce the other field equations. Conversely, by accepting both field equations, one can derive the law of conservation of energy in the form just stated. Thus, the very act of writing the field equation G = 8'ITT (or, if one insists upon the \"cosmological term,\" Side remarks about initial value equations, dynamic equations, and Bianchi identities in more general contexts G + Ag = 8'ITT) omit ~ was encouraged by and founded on the automatic vanishing of the divergence V \u2022 G (or the vanishing of the divergences of G and g), because one knew to begin with that energy and momentum are conserved, V \u2022 T = 0. It is not surprising, then, that there should be a redundancy between the conservation law, V\u00b7 T = 0, and the field equations. Neither is it surprising in the dynamics of the Friedmann universe that one can use what is here the one and only interesting component of the conser vation law, plus the one and only interesting initial value component (Gii component) of the field equations, to obtain the one and only interesting dynamic component (Gxx component) of the field equations. 730 27. IDEALIZED COSMOLOGIES In a similar way, in more general problems that lack symmetry, one can always eliminate some of the dynamic field equations, but when gravitational radiation is present, one cannot eliminate them all. The dynamic field equations that cannot. be eliminated, even in principle, govern the propagation of the gravitational waves. No gravitational waves are present in a perfectly homogeneous and isotropic cos mological model; its high degree of symmetry-in particular, its spherical (2-sphere!) symmetry about X = O-is incompatible with gravitational waves. Now turn back from general dynamics to Friedmann cosmology. To determine the time evolution of the expansion factor, a, insert into the initial-value equation", " (27.39a) the expression for the density of mass-energy given in (2735a), and arrive at an equation ready for integration, Differential equation for expansion factor a )2_,t = a ( omit (8 'IT /3)p(a) (27.40) When one has completed the integration of this equation for a = aCt), one turns back to equation (2735a,b) to get pet) and pet), and to expression (27.24) to get the geometry, (27.41 ) thus completing the solution of the problem. \u00a727.9. TIME PARAMETERS AND THE HUBBLE CONSTANT To the analysis of this dynamic problem, many investigators have contributed over the years, beginning with Friedmann himself in 1922. They discovered, among other results, that there are three natural choices of time variable, the one of greatest utility depending on the application that one has at hand. First is t, the original time variable. This quantity gives directly proper time elapsed since the start of the expansion. This is the time available for the formation of galaxies. It is also the time during which radioactive decay and other physical processes have been taking place. Second is aCt), the expansion factor, which grows with time, which therefore serves to distinguish one phase of the expansion from another, and which consequently can be regarded as a parametric measure of time in its own right. The ratio of aCt) at two times gives the ratio of the dimensions of the universe (cube root of volume) at those two times. It also gives the ratio (l + z) of wavelengths at those two times (see \u00a729.2). A knowledge of the red shift, z, experienced in time past by radiation received today is equivalent to a knowledge of a(t)/ao' where ao is the expansion factor today. Specifically, radiation coming in with z = 999 is radiation coming in from a time in the history of the universe when it had 10-3 of its present dimensions and 10-9 of its present volume. During the interval of time while the expansion Three choices of time parameter for universe: (1) proper time, t (2) expansion factor, a \u00a727.9. TIME PARAMETERS AND THE HUBBLE CONSTANT 731 parameter is increasing from a to a + da, the lapse of proper time, according to (27.40), is dt - da - [- k + (8'17 /3)a2p(a) + (A/3)a2]l/2' ------ omit (27.42) In terms of a as a new time parameter, it follows from this formula that the metric takes the form [Hughston (1969)] ds 2 = - (da)2 -k + (8'IT/3)a2p(a) + (A/3)a2 + a X + \"\". 2() J 2 UI + sm ucp )]. 27 3 (.4) 2[d 2 ~2(J.'f)2 ---...-omit Third is 1'/(t), the \"arc-parameter measure of time.\" During the interval of time dt, a photon traveling on a hypersphere of radius a(t) covers an arc measured in radians equal to (3) arc parameter. 1/ dt drj = a(t)' (27.44) When the model universe is open instead of closed, the same parameter lets itself be defined. Only the words \"hypersphere\" and \"are\" have to be replaced by the corresponding words for a flat hypersurface of homogeneity (k = 0) or a hyperbol oidal hypersurface (k = -1). In all three cases, the \"arc parameter\" is defined by the integral of this expression from the start of the expansion: rt dt a(t); 1'/ = Jo (27.45) thus small values of the \"arc parameter time,\" 1'/, mean early times; and larger values mean later times. In terms of this \"arc-parameter measure of time,\" the metric takes the form (27.46) Let a photon start at the \"North Pole\" of the 3-sphere (X = 0; any () and cp) at the \"arc parameter time\" 1'/ =1'/1' Then, by the \"arc parameter time\" 1'/ = 1'/2' the photon has traveled to a new point on the hypersphere and encountered a new set of particles of the \"cosmological fluid.\" They lie at the hyperpolar angle X = 1'/2 - 1'/1' When one makes a spacetime diagram on a piece of paper to show what is happening when an effect propagates from one point to another in the universe, one finds it most convenient to take (l) the space coordinate to be X (the life histories of", " distinct particles of the \"cosmological fluid\" thus being represented by distinct vertical lines), and (2) the time coordinate to be 1'/ (so that photons ate described by lines inclined at +45\u00b0). No time parameter is more natural to use than 1/ when one is tracing y 732 27. IDEALIZED COSMOLOGIES out the course of null geodesics. For an example, see the treatment of the cosmologi cal redshift in \u00a729.2. It also turns out that it is simpler analytically (when A is taken to be zero) to give a = a(7)) and t = t(7)) than to give a directly as a function of time. Thus one gets the connection between the dimension a and the \"arc-parameter time\" 7) from the formula 7) = drj = a(t) = [_ka2 + (817/3)a 4p(a) + (A/3)a4 ]l/2' f f dt da (27.47) f --------omit Hubble constant and Hubble time From a knowledge of the dimension a as a function of this time parameter, one immediately gets proper time itselfin terms of this time parameter, from the formula dt = a(7)) drj. (27.48) An equation (27.40) for the expansion factor and a choice of parameters for marking out time have now set the stage for a detailed analysis of idealized cosmol ogy, and some of the relevant questions have even been asked: How does the characteristic dimension, a, of the geometry (radius of3-sphere, in the case of closure) change with time? What is the spacetime geometry? How do geodesics, especially null geodesics, travel in this geometry? However, additional questions are equally important: Is the expansion of the universe decelerating and, if so, how fast? How do density and pressure of matter and radiation vary with time? And finally, for the simplest and most immediate tie between theory and observation, what is the expansion rate? In speaking of expansion rate, one refers to the \"Hubb1e constant,\" the fractional rate of increase of distances, H= ti(t), a(t) (27.49) which is normally evaluated today H(today) = Ho' but is in principle defined as a function of time for every phase of the history of the universe. The reciprocal of H is the \"Hubble time,\" H-l. This quantity represents the time it would have taken for the galaxies to attain their present separations, starting from a condition of infinite compaction, if they had maintained for all time their present velocities (\"time for expansion with dimensions linearly extrapolated back to the start\"). For the conversion from astrophysical to geometric units and to years, take the currently accepted value, Ho = 55 km/sec megaparsec (Box 29.4), as an illustration: H _ 55 km/sec \u00b0 - (299,793 km/sec)(3.0856 X 1024 em or 3.2615 X 106 yr of time) = 0.59 X 10-28 per em of light-travel time or 5.6 X 10-11 fractional expansion per yr, (27.50) H;;t = 1.7 X 10 28 em of light-travel time or 18 X 109 yr. \u00a727.10. FRIEDMANN COSMOLOGY OF A CLOSED UNIVERSE 733 \u00a727.10. THE ELEMENTARY FRIEDMANN COSMOLOGY OF A CLOSED UNIVERSE Take the simplest cosmological model, an isotropic homogeneous closed universe wi~h A = 0, and trace out its features in all detail in the two limiting cases where matter dotnlnates--and where-mdiation domina1es.--The term \"Friedmann universe\" is used here for both cases, although the matter-dominated model is sometimes referred to as the Friedmann universe and the radiation-dominated one as the Tolman universe. In this analysis, it will be appropriate to let the variable ti(t) represent the radius of the universe, as measured in cm, because only by reference to this radius does one have the tool in hand to discuss all the interesting geometric effects that in principle lend themselves to observation. After this discussion, it will be enough, in dealing with other models, to summarize their principal parts and comment on their differences from this simple model, without repeating the full investigation. Any reference to an open universe or any so-called \"cosmological constant\" or its effects will therefore be deferred to a brief final section, \u00a727.11. There the variable a(t) will sometimes be taken to represent only a parameter of relative expansion, as is appropriate for discussions reaching out only to, say, Z = 0.1, where global geometric issues are not taken", " up. Features of a closed Friedmann universe with A = 0: Rewrite the controlling component (27.40) of Einstein's field equation in the form (1) radius as function of time According as one neglects the radiation term or the matter term in this equation, the equation idealizes to _ (27.51) (da)2 _ amax = -1, dt a or (27.5~; matter) dommates (27.5~; radiation) dommates In both cases, the problem lends itself to comparison to the problem of particle motion in Newtonian mechanics with \"total energy\" - 1 and with an \"effective potential energy\" of the qualitative form shown in diagram A of Box 27.1-apart from minor differences in shape according as the potential goes as -l/a or as -1Ia2\u2022 The principal features of the solution are collected in Box 27.3. It is a striking feature of the radiation-dominated era of the early Friedmann universe that the density of the radiation depends on time according to a simple universal law, Pr = 3/32m 2 (27.53) (final line and final column of Box 27.3). This circumstance may someday provide (continued on page 736) (2) early era, when radiation dominates: types of radiation 734 27. IDEALIZED COSMOLOGIES Box 27.3 SOLUTIONS FOR THE ELEMENTARY FRIEDMANN COSMOLOGY OF A CLOSED UNIVERSE IN THE TWO LIMITING CASES IN WHICH (1) MATTER DOMINATES AND RADIATION IS NEGLIGIBLE. AND (2) RADIATION DOMINATES AND MATTER IS NEGLIGIBLE Idealization for dynamics of 3-sphere Model relevant when? Effective \"potential\" in da)2dt + V(a) = -I ( Value of constant in this \"potential\" in terms of conditions at some standard epoch Solution of dynamic equation ex pressed parametrically in terms of \"arc parameter\" 1) (radians of arc distance on 3-sphere covered by a photon travelling ever since start of expansion) Matter dominated Radiation dominated back into past to redshift z - 10,000; through today and through phase of maximum expansion, and re contraction down to dimensions -1O,ooo-fold smaller than today very early phase of expansion, for redshifts z - 1,000 and greater; and corresponding phase in late stages of recontraction; not di rectly relevant today. a max V(a) = - - a a a = ;ax (l - cos 1).2 a = -a p 8'17 4 3 0 ro a = a\u00b7 sin 1) t = a\u00b7(l - cos 1) Range of 1) from start of expansion 2'17 (one trip around the universe) '17 (gets only as far as antipodal point to end of recontraction Nature of curve relating radius a to cycloid timet Hubble time H-l = a (da/dt) a2 = - - (da/dr,) a max (l - cos 1)2 2 sin 1) of universe) semicircle \u2022 sin2 1)a - cos 1) \u00a727.10. FRIEDMANN COSMOLOGY OF A CLOSED UNIVERSE 735 Idealization for dynamics of 3-sphere Inequality between Hubble or \"ex trapolated\" time and actual time back to start of expansion Density of mass-energy Matter dominated Radiation dominated H-l ~ 1.51 H-l ~ 21 This density expressed in terms of Hubble expansion rate p = m 3H2 8'17 2 I + COS1j 3H2 p =---8'17 cos~ I r 3H2 P >- 8'17 r - \u00a729.5 and Figure 29.2 \u00a729.5 a = (2a*I)1/2 Inequality satisfied by density Analysis of magnification of distant galaxy by curvature of intervening space Limiting form of law of expansion for early times Other features of expansion at early times 736 27. IDEALIZED COSMOLOGIES a tool to tell how many kinds of radiation contributed to Pr in the early universe; or, in other words, to learn about field physics from observational cosmology. Express the density of radiation in the form (27.54) It would be surprising if electromagnetism made the sole contribution to the radiation density, since the following additional mechanisms are available to sop up thermal energy from a violently radiating source: electromagnetic radiation (already considered), gravitational black body radiation, neutrino plus antineutrino radiation of the electron neutrino type [its contribution depends on the chemical potential of the neutrinos, on which see Brill and Wheeler (1957); a zero value is assumed here for that potential], neutrino plus antineut", "rino radiation of the muon neutrino type [with the same assumptions as for \"e'S], pairs of positive and negative electrons produced out of the vacuum when temperatures are of the order of T = mc 2/k = 0.59 X 1010 K and higher, evaluated in the approximation in which these particles are treated as overwhelmingly more numerous than the unpaired electrons that one sees today, other particles such as mesons created out of the vac!em = 8; - 8' r, Jg - lep = 7; fp.p = 7;!e+e- = 14; uum when temperatures are two orders of magnitude higher (_10 12 K), and baryon-antibaryon pairs created out of the vacuum when temperatures are of the order of _10 13 K and higher, sum of these fvalues, fp.+ p.-' fr\"... ; f (27.55) As the expansion proceeds and temperatures drop below 1013 K, then 1012, then 1010, the various particle pairs presumably annihilate and disappear [see, however, Alfven and Klein (1962), Alfven (1971), Klein (1971), and Omnes (1969)]. One is left with the radiations of zero rest mass, and only these radiations, contributing to the specific heat of the vacuum. At the phases of baryon-antibaryon and electron positron annihilation, the thermal gravitational radiation present has already effec tively decoupled itself from the matter, according to all current estimates. Therefore the energy set free by annihilation of matter and antimatter is expected to pour at first into the other two carriers of energy: neutrinos and electromagnetic radiation. However, the neutrinos also decouple early (after baryon-antibaryon annihilation; before full electron-positron annihilation), because the mean free path for neutrinos \u00a727.10. FRIEDMANN COSMOLOGY OF A CLOSED UNIVERSE 737 rises rapidly with expansion. The energy of the subsequent annihilations goes almost exclusively into electromagnetic radiation. Thus the temperatures of the three radia tions at the present time are expected to stand in the order (27.56) Tem has been measured to be 2.7 K; Tv is calculated to be (4/11)1/3 Tem = 1.9 K, and Til has been calculated to be 1.5 K [Matzner (1968)] in a model where gravitons decouple during an early, quark-dominated era. Decoupled radiation, once in a Planck spectrum, remains in a Planck spectrum (see Box 29.2). Expansion leaves constant the product Pr,decoupled a4 or the product T,.~decoUP1ed a4. Compare the temperature of this particular radiation now to the temperature of the same radiation at any chosen fiducial time tfld after its era of decoupling. Find Tr,fld = -a- Tr,now = (1 + z)T,.,now\u00b7 Gnow fld (27.57) Here z represents the red shift of any \"tracer\" spectral line, given off at the fiducial time, and observed today, relative to the standard wavelength of the same transition as observed in the laboratory. If the three radiations could be catalyzed into thermodynamic equilibrium, then all radiations could be treated on the same footing during the radiation-domi nated era of cosmology. Their individual f values could be added directly to give f = 8 + 8 + 7 + 7 = 30. Temperature and time would then be connected by the formula. (1'/10 10 K)2(t/l sec) = 1.19. (27.58a) This formula together with (27.57) implies the relation T,.,now ) (1 + Z)]2(~) = 1.19. 1010 K 1 sec [( (27.58b) This relation concerns two radiations: (1) the actual electromagnetic radiation with Planck spectrum (a continuum); and (2) the redshift and time of emission of a \"tracer radiation\" (a line spectrum). A measured departure from this relation could serve as one potential (indirect) indication that, in accordance with standard theory,\u00b7 neutrinos and gravitational radiation today are cooler than electromagnetic radiation. Turn now from the radiation-dominated era ofcosmology to the matter-dominated era. Numbers sometimes elicit more response from the imagination than formulas. Therefore idealize to a matter-dominated cosmology, and for the moment arbitrarily adopt 20 X 109 yr and 10 X 109 yr as Hubble time and actual time, respectively, back to the start of the expansion. It is certain that future work will show both numbers to require revision, but probably not by more than a factor 2, in the opinion of observational cosmologists. Since any judgment on the best numbers is subject (3", ") later era, when matter dominates 738 27. IDEALIZED COSMOLOGIES to uncertainty, one can pick the numbers to be simple as well as reasonable. From Box 27.3, one then deduces the present value of the arc parameter time 1J, 20 X 109 1yr lOX 109 1yr amax (I - cos 1J )2 2 sin 1J or 1J = 1.975 (or 113.2\u00b0) (27.59) (27.60) (arc traveled by a photon on the 3-sphere from the start of the expansion to today.) This fixed, all other numbers emerge as shown in Box 27.4. Box 27.4 A TYPICAL COSMOLOGICAL MODEL COMPATIBLE WITH ASTRONOMICAL OBSERVATIONS AND WITH EINSTEIN'S CONCEPTION OF COSMOLOGY (A = 0; Universe Closed) Radius at phase of maximum expansion, Time from start to maximum, Time from start to final recontraction, Time from start to today (adopted value), Radius today, Hubble time today (adopted value), Hubble expansion rate today, Deceleration parameter today, qo [equation (29.1b)] Density today (3/8'1Tao 2) + (3H;/87T), Volume today, 27T 2ao3, Density at maximum (3/87Ta 2) + (3H2/87T), Volume at maximum, Rate of increase of radius today, Rate of increase of volume today, Amount of matter, Equivalent number of solar masses, Equivalent number of baryons, Fraction visible today 18.94 X 109 lyr; 29.76 X 109 yr; 59.52 X 109 yr; 10 X 109 yr; 13.19 X 109 lyr; 20 X 109 yr; 49.0 km/sec Megaparsec; 1.7 (7.67 + 3.33) X 10-58 cm-2 = 11.00 X 10-58 cm-2 or 14.8 X 10-30 g/cm3 ; 38.3 X 1084 cm3 ; (3.70 + 0.00) X 10-58 cm-2 = 5.0 X 10-30 g/cm3 ; 114 X 1084 cm3 ; 13.19 X 109 lyr/20 X 109 yr = 0.66 lyr/yr; 1.82 X 1068 cm3/sec; 5.68 X 1056 g; 2.86 X 1023 ; 3.39 X lOBO. 0.74 It must be emphasized that these numbers do not deserve the title of \"canonical,\" they can at most however convenient that adjective may be for describing them; be called illustrative. (4) \"Where is the new space created during expansion?\" -a meaningless question Figure 27.3. Many Schwarzschild zoneS are fitted together to make a closed uni verse. This universe is dynamic because a test particle at the interface between two zones rises up against the gravitational attraction of each and falls back under the gravitational attraction of each. Therefore the two centers themselves have to move apart and move back to gether again. The same being true for all other pairs of centers, it follows that the lattice universe itself expands and recontracts, even though each Schwarzschild geometry individually is viewed as static. This diagram is taken from Lindquist and Wheeler (1957). If every five seconds a volume of space is added to the universe, a volume equivalent to a cube 105 lyr (= 0.95 X 1023 em) on an edge, about equal to the volume occupied by the Milky Way, where does that volume make its entry? Rather than look for an answer, one had better reexamine the question. Space is not like water. The o'utpouring of fresh water beneath the ocean at the Jesuit Spring off Mount Desert Island can be detected and measured by surrounding the site with flowmeters. There is no such thing as a flowmeter to tell \"how fast space is streaming past.\" The very idea that \"space flows\" is mistaken. There is no way to define a flow of space, not least because there is no way to measure a flow of space. Water, yes; space, no. Life is very different for the flowmeter, according as it is stationary or moving with respect to the water. For a particle in empty space, however, physics is indistinguishable regardless of whether the particle is at rest or moves at high velocity relative to some chosen inertial frame. To try to pinpoint where those cubic kilometers of space get born is a mistaken idea, because it is a meaningless idea. One can get a fresh perspective on what is going on in expansion and recontraction by turning from a homogeneous isotropic closed universe to a Schwarzschild lattice closed universe. [Lindquist and Wheeler (1957)]. In the former case, the mass", " is idealized as distributed uniformly. In the latter, the mass is concentrated into 120 identical Schwarzschild black holes. Each is located at the center of its own cell, of dodecahedral shape, bounded by 12 faces, each approximately a pentagon; and space is empty. The dynamics is easy to analyze in the approximation in which each lattice cell is idealized as spherical, a type of treatment long familiar in solid-state physics as the \"Wigner-Seitz approximation\" (references in Lindquist and Wheeler). In this approximation, the geometry inside each lattice cell is treated as having exactly the Schwarzschild character (Figure 27.3); a test particle placed midway between black hole A and black hole B rises against the attraction of each, and ultimately falls back toward each, according to the law developed in Chapter 25 [equation (25.28) with a shift of 'IT in the starting point for defining 1j], r = ~ (I - COS1j), 1\" = - R ( R )1/2 2 2M (1j - sin1j). (27.61) Accordingly, the two masses in question must fall toward each other; and so it is with alI the masses. One comes out in this way with the conclusion that the lattice 740 27. IDEALIZED COSMOLOGIES universe follows the same law of expansion and recontraction as the Friedmann universe to an accuracy of better than 4 per cent [Lindquist and Wheeler; Wheeler (l964a), pp. 370-381]. Now ask again the same meaningless question about where the cubic kilometers of space pour into the universe while it is expanding, and where they pour out while it is recontracting. Receive a fuller picture why the question is meaningless. Surrounding each center of mass, the geometry is and remains the Schwarzschild geometry (until eventually the black holes come so close together that they coalesce). The situation inside each cell the interface between cell and cell is defined in imagination by a sprinkling of test particles so light that they have no influence on the geometry or its dynamics. The matchup between the geometry in one cell and the next is smooth (\"tangency between the two geometries\"). There is nothing abnormal whatsoever in the space time on and near the interface. One has as little right to say those cubic kilometers are \"created\".here as anywhere else. To speak of the \"creation\" of space is a bad way of speaking, and the original question is a bad question. The right way of speaking is to speak of a dynamic geometry. So much for one question! is therefore static. Moreover, In charting the dynamics of the geometry of a Friedmann universe, one often finds that it simplifies things to take as space coordinate the hyperpolar angle X, measured from some chosen world line (moving with the \"cosmological fluid\") as standard of reference; and to take as time coordinate the arc-parameter measure of time, 1J, as illustrated in Figure 27.4. Inspection of the (X, 1J)-diagram makes it clear that photons emitted from matter at one point cannot reach, in a limited time, any matter except that which is located in a limited fraction of the 3-sphere. In a short time t, according to Box 27.3, a photon can cover an arc distance on the 3-sphere equal only to 1J = (2tla*)1/2. Moreover, what is true of photons is true of other fields, forces, pressures, energies and influences: they cannot reach beyond this limit. Evidently the 3-sphere at time t is divided into a number of \"zones,\" (5) causal isolation of various regions of universe from each other N = (nUmber Of) = angle of entire 3-sphere = 211'2 == 311' (a*)3/2 ( hyperspherical solid ) \"zones\" (hyperspherical SOlid) angle of one zone 41Tx3/3 25 / 2 t'(27.62) effectively decoupled one from the other. As time goes on, there are fewer separate zones, and ultimately every particle has been subjected to influences from every other particle in the model universe. EXERCISES Exercise 27.8. MATTER-DOMINATED AND RADIATION-DOMINATED REGIMES OF FRIEDMANN COSMOLOGY Derive the results listed in the last two columns of Box 27.3, except for the focusing properties of the curved space. \u00a727.10. FRIEDMANN COSMOLOGY OF A CLOSED UNIVERSE 741 D C EO CD 21T 31T/2 + 11 I 1T 1T/2 Final collapse ~ '\"... '\"..c 0.. M.... 0.\", \"0 0.. * B CD A 0 0 1T ~ '\"..c 0...., '", "('.... 0 Final collapse 1T '\", \"0 + \u00b7 '\" 0 0.. 11 11T/2 ~.!!..c 00.. 0..'\" Z~ Maximum \u00b7.... expansion \u00b7 0 ~ '\"..c 0...., '('.... 0 '\", \"0 0.. 0 1T Figure 27.4. Use of \"arc parameter\" 11 as a time coordinate and hyperpolar angle Xas a space coordinate to describe travel of a photon (\u00b145' line) in a Friedmann universe that is matter-dominated (center) or radiation dominated (right). The burst of photons is emitted from the \"N-pole\" of the 3-sphere at a time very little after the big bang, and the locus of the cloud of photons at subsequent stages of the expansion and recontraction is indicated by sections of the 3-sphere in the diagrams at the left. The matter-dominated Friedmann universe appears to be a reasonable model for the physical universe, except when its dimensions have fallen to the order of one ten-thousandth of those at maximum expansion or less (\"radiation regime\"). Exercise 27.9. TRANSITION FROM RADIATION-DOMINATED REGIME TO MATTER-DOMINATED REGIME Including both the radiation and the matter terms in equation (27.51), restate the equation in terms of the arc parameter 1/ (with dr, = dt/a) as independent variable, and integrate to find a = (amax/2) - [(amax/2)Z + a\"2Jl/2 cos (1/ + 15), t = (amax/2)1/ - [(amax /2)2 + a\"2p/2[sin (1/ + 15) - sin 15], where 15 = arc tan [a\" /(amax/2)]. (27.63) (27.64) (27.65) (a) Verify that under appropriate conditions these expressions reduce at early times to a \"circle\" relation between radius and time and to a \"cycloid\" relation later. 742 27. IDEALIZED COSMOLOGIES (b) Assign to a*2 the value aoamax/iO,OOO (why?) and construct curves for the dimensionless measures of density, as a function of the dimensionless measure of time, What conclusions emerge from inspecting the logarithmic slope of these curves? Exercise 27.10. THE EXPANDING AND RECONTRACTING SPHERICAL WAVE FRONT An explosion takes place at the\" N-pole\" of the matter-dominated Friedmann model universe at the value of the \"arc parameter time\" 1/ = 'IT /3, when the radius of the universe has reached half its peak value. The photons from the explosion race out on a spherical wave front. Through what fraction of the \"cosmological fluid\" has this wave front penetrated at that instant when the wave front has its largest proper surface area? \u00a727.11. HOMOGENEOUS ISOTROPIC MODEL UNIVERSES THAT VIOLATE EINSTEIN'S CONCEPTION OF COSMOLOGY It violates Einstein's conception of cosmology (Box 27.1 )-though not the equations of his theory-to replace the closed 3-sphere of radius a by the open hyperboloidal geometry of equation (27.22) with the same scale length a. Even so, the results of Box 27.3 continue to apply in the two limiting regimes of matter-dominated and radiation-dominated dynamics when the following changes are made. (1) Change the constant - I on the righthand side of the analog of a \"Newtonian energy equation\" to + 1, thus going over from a bound system (maximum expansion) to I), sin 1/ an open system (forever expanding). (2) Replace (l - cos 1/) by (cosh 1/ by sinh 1/, cos 1/ by cosh 1/, and (1/ sin 1/) by (sinh 1/ - 1/). (3) The range of the \"arc parameter\" 1/ now extends from 0 to 00, and the curve relating \"radius\" a to time t changes from cycloid or circle to an ever-rising curve. (4) The listed inequalities on the Hubble time (as related to the actual time of expansion) and on the density (as related to 3H;/8'1T) no longer hold. (5) The formulas given in Box 27.3 for conditions at early times continue to hold, for a simple reason: at early times the curvature of spacetime \"in the direction of increasing time\" [the extrinsic curvature (6/a 2)(da/dt}2 as it appears in Box 27.1, equation (2)] is overwhelmingly more important than the curvature within any hypersurface of", " homogeneity, -+-6/a2 (the intrinsic curvature); therefore it makes no detectable difference at early times whether the sign is plus or minus, whether the space is closed or open, or whether the geometry of space is spherical or hyperboloidal. Why doesn't it make a difference? Not why mathematically, but why physically, doesn't it make a difference in early days whether the space is open or closed? Open Friedmann universe with A = 0: (1) expansion factor as function of time (2) early stage-same as for closed universe \u00a727.11. MODEL UNIVERSES WITH k ~ a or A ':j; 0 743 Because photons, signals, pressures, forces, and energies cannot get far enough to \"smell out\" the difference between closure and openness. The \"zones of influence\" of (27.62) are too small for anyone by itself to sense or to respond significantly to any difference between a negative space curvature -6/a2 and a positive space curvature +6/a2 \u2022 Therefore the simple power-law time-dependence of the density of the mass-energy of radiation given in Box 27.3 for a closed universe holds equally well in the earliest days of a radiation-dominated, open, isotropic model universe; thus, (27.66) Only at a later stage of the expansion, when the \"extrinsic curvature\" term [equation (2), Box 27.1], (6/a2)(da/dt)2 (initially varying as 1.5t-2, according to Box 27.3) has fallen to a value of the same order of magnitude as the \"intrinsic curvature\" term -+-6/a2 (initially varying as -+-3a*-lt- 1), does the sign of the intrinsic curvature begin to matter. Only then do the differences in rate of expansion begin to show up that distinguish the open model universe from the closed one. The open model goes on expanding forever. Therefore the density of mass-energy, whether matter-dominated and proportional to ama:x./a3, or radiation-dominated and proportional to a*2/ a4, or some combination of the two, (l) ultimately falls to a level that is negligible in comparison with the intrinsic curvature, -6/a2, and (2) thereafter can be neglected. Under these circumstances, the only term left to balance the intrinsic curvature is the extrinsic curvature. The important component of the field equation (after removal from all terms of a common factor 3) now reads (27.67) For a closed universe, the two terms (one sixth the extrinsic curvature and one sixth the intrinsic curvature) have the same sign, and any equation like (27.67) leads to an impossibility. Here, however, rather than impossibility, one has the remarkably simple solution and the corresponding metric Write a = t, r = t sinh x, tnew = t cosh X, (27.68) (27.69) (27.70) and find that (27.69), solution as it is of Einstein's empty-space field equation, is identical with the Lorentz-Minkowski metric of flat spacetime, (27.71) (see Box 27.2C). This geometry had acquired the flavor of an expanding universe (3) late stage-expansion forever 744 27. IDEALIZED COSMOLOGIES because the cosmological fluid, too thinly spread to influence the dynamics of the geometry, and serving only to provide marker points, was flying out in all directions [for a fuller discussion of this \"expanding Minkowski universe,\" see, for example, Chapter 16 of Robertson and Noonan (1968)]. The typical spacelike hypersurface of homogeneity looks to have a curved 3-space geometry, and does have a curved geometry (instrinsic curvature), because the slice (27.70) through flat spacetime is itself curved (extrinsic curvature). Turn now to a second violation of Einstein's conception of cosmology: a cosmolog- ical term in the field equation (27.39), Homogeneous cosmologies with A -:j; 0: (1) equation for evolution of expansion factor ( da/dt)2 A - - + 2' - -3 = -3 pea) = a k a 817 817PTMag/3 3 a + 817Pro~/3 4 a \u2022 (27.72) In analyzing the implications of this broadened equation, turn attention from the \"radius\" aCt) itself, which was the focus of interest in the previous section, \u00a727.1O, on Friedmann cosmology. Recognize that present measurements have not yet pro vided a good, direct handle on the absolute dimension aCt", ") of the universe. However, they do give good figures for the redshift z and therefore for the ratio between a at the time of emission and a = ao now, (27.73) For any comparison with observations designed to fix limits on k (Einstein value: k = + 1) and on A (expected to be zero), it is therefore appropriate to rewrite the foregoing equations so that they refer as much as possible only to ratios. Thus one rephrases (27.72) as the \"generalized Friedmann equation,\" [ddt (aa(ot) )J2 + k V(a/ao) = - -2 = -Ko' ao Here the quantity (27.74) (27.75) acts as an \"effective potential\" for the dynamics of the expansion. The constant term Ko represents one sixth of the intrinsic curvature of the model universe today. Its negative, - K o' plays the role of an \"effective energy\" in the generalized Friedmann equation (Box 27.5). All the qualitative features of the cosmology can be read off from the curve for the effective potential as a function of (a/ao) and from the value of K o' For a quantitative analysis, the log-log diagram of Figure 27.5 is often more useful than the straight linear plot of V against (a/ao) of Box 27.5. All the limiting features shown in the varied types of cosmology have been encountered before in the analysis of the elementary Friedmann cosmology (big bang out of a configuration of infinite compaction; reaching a maximum expansion at a turning point, or continued expansion to a Minkowski universe; recoIlapse to (2) qualitative features of evolution (3) de Sitter universe \u00a727.11. MODEL UNIVERSES WITH k ~ 0 or A -:j; 0 745 infinite density) or lend themselves to simple visualization (static but unstable Einstein universe; \"hesitation\" model; \"turnaround\" model), except for the even more rapid expansion that occurs when A is positive and the dimension a has surpassed a certain critical value. In this expansion, a eventually increases as exp [(A/3)1I2t] irrespective of the openness or closure of the universe (k = 0, -+-1). This expansion dominates every other feature of the cosmology. Therefore, in discussing it, it is appropriate to suppress every other feature of the cosmology, take the density of matter to be negligible, and take k = 0 (hypersurfaces of homogeneity endowed with flat 3-space geometry). In this limit, one has the following empty-space solution of Einstein's field equation with cosmological constant: (27.76) This \"de Sitter universe\" [de Sitter (1917a,b)] may be regarded as a four-dimensional surface, in a five-dimensional space endowed with the metric (27.77) (27.78) The correctness of this description may be checked directly by making the substitu tions ZO = (3/A)1/2 sinh [(A/3)1/2t] + 1- (A/3)1/2eW3)l12ta;X 2, 2 Z4 = (3/A) 11..2 cosh [(A/3)1/2t] _1- (A/3)1/2eW3l'/2ta;X2, Zl = a o Z2 = a o Z3 = a o e(A/3)1/2tx sin 0 cos ep, e(A/3)1/2tx sin 0 sin ep, e(A/3)1/2tx cos o. 2 (27.79) Because of its beautiful group-theoretical properties and invariance with respect to 5 X 4/2 = 10 independent rotations, the de Sitter geometry has been the subject of scores of mathematical investigations. The physical implications of a cosmology following the de Sitter model are described for example by Robertson and Noonan (1968, especially their \u00a716.2). The de Sitter model is the only model obeying Ein Stein's equations (with A =J. 0) which (1) continually expands and (2) looks the same to any observer who moves with the cosmological fluid, regardless of his position or his time. Any model of the universe satisfying condition (2) is said to obey the so-called \"perfect cosmological principle.\" This phrase arose in the past in studying models that lie outside the framework of general relativity, models in which matter is envisaged as continuously being created, and to which the name of \"steady-state universe\" has been given. Any such model has been abandoned by most investigators today, not least because it gives no satisfactory account of the 2.7 K background radiation. (continued on page 7", "48) Other non-Einsteinian cosmologies: (1) steady-state model 746 27. IDEALIZED COSMOLOGIES Box 27.5 EFFECT OF VALUE OF COSMOLOGICAL CONSTANT AND OF INTRINSIC CURVATURE OF MODEL UNIVERSE \"TODAY\". ON THE PREDICTED COURSE OF COSMOLOGY The \"effective potential\" V in the generalized Friedmann equation (27.74) is repre sented schematically here as a function of the expansion ratio a/ao' The diagram illustrates the influence on the cosmology of(1) the cosmological constant A (deter mines the behavior of the effective potential at large values of a/ao; see dashed curves) and (2) the value adopted for Ko = (one sixth of the intrinsic curvature of 3-space at the present epoch). The value of the quantity - K o determines the \"effec tive energy level\" and is shown in the diagram as a horizontal line. The difference between this horizontal line and the effective potential determines (a;;-l da/dt)2. Regions where this difference is negative are inaccessible. From the diagram one can read off the histories of 3-space on the facing page. (Open universe (will recontract if A < 0) ~ V I -(a/ao)~ 0l------=:::::::~====A level which giVes) \"hesitation\" solution for A > 0 Einstein static) universe; A > O. K > 0; unstable ( - K o level to give nOllsingular \"turn-around\" solution, A > 0, K o > 0 (, 7'M,conv 7'M,conv - = 10-30 g/cm3 or p -+-3 X 10-28 g/cm3 \u2022 p - The diagram is schematic, not quantitative. Representative values might be Aconv = 0 - 10-28 g/cm3 \u2022 and or (a;;-l da/dt)2 = H; = (1/20 X 109 yr)2 or 3.8 X 10-29 g/cm3 \u2022 At small values of a/ao the cosmological term - (A/3)(a/ao)2 is negligible. Not negligible at small values of (a/ao) is the difference between a model universe curved only by the density of matter (the dashed curve in the diagram) and one curved also by a density of radiation (the full curve). The different dependence of \"radius\" and density on time at early times in these two cases of a matter-dominated cosmology and a radiation dominated cosmology is spelled out in the last part of Box 27.3, giving in the one case p = 1/61Tt 2 and in the other p = 3/321Tt 2\u2022, \u00a727.11. MODEL UNIVERSES WITH k ~ 0 or A -:j; 0 747 ------------------------------------....... ~ Intrinsic curvature of space today Hyperbolic; Ko negative A negative Hyperbolic; Ko negative zero Closed; Ko positive Closed; Ko positive Closed; Ko positive Closed; Ko positive zero negative A more positive than a certain critical value: A > Aerit A positive and exactly equal to the\"critical value, A = Aeril, that puts the \"summit of the potential\" into coincidence with -Ko Closed; Ko positive A less positive than the critical value: o < A < Acrit(Ko ) Cosmology Universe starts in a condition of infinite density. It ex pands to a maximum extent (or minimum density) governed by the value of A. It then recontracts at an ever increasing rate to a condition of infinite density. Universe starts in a condition of infinite density. It ex pands. Ultimately the rate of expansion reaches a steady rate, da/dt = I. The 4-geometry is Minkowski flat spacetime. Only the curvature of the spacelike slices taken through this flat 4-geometry gives the 3-geometry its hyperbolic character [see equation (27.70)].. Standard Friedmann cosmology: expansion from infinite compaction to a finite radius and recontraction and collapse. Qualitatively same as foregoing. Quantitatively a slightly smaller radius at the phase of maximum expansion and a slightly shorter time from start to end. \"Summit\" of \"effective potential\" is reduced to a value slightly less than - Ko' The closed universe once again starts its expansion from a condition of infinite compaction. This expansion once again slows down as the expansion proceeds and then looks almost as if it is going to stop (\"moment of hesitation\"). However, the representative point slowly passes over the summit of the potential. Thereafter the expansion gathers more and more speed. It eventually follows the ex ponential law a =constant X exp [(A/3)1I2r", "J. Situation similar to that of a pencil with its tip dug into the table and provided with just enough energy to rise asymptotically in infinite time to the vertical position. Uni verse starts from a compact configuration and expanding approaches a certain radius (\"Einstein radius\", aE ) according to a law of the form a = aE - constant X exp ( - at). Or (Einstein's original proposal, when he thought that the universe is static, and added the \"cosmological term,\" against his will, to the field equation to permit a static uni verse) the representative point sits forever at the \"summit of the effective potential\" (Einstein universe). Aside from con tradicting present-day evidence on expansion, this configura tion has the same instability as does a pencil trying to stand on its tip. The least disturbance. will cause it to \"fall\" either way, toward collapse or toward accelerating expansion, in the expansion case ultimately approaching the law a = constant X exp [(A/3)1/2r]. Motion on the large a side of the \"potential barrier.\" Far back in the past the model universe has enormous dimen sions, but is also contracting with enormous rapidity, in approximate accord with the formula a = constant X exp [- (A/3)l/2 r]. The radius a reaches a minimum value and thereafter the universe reexparidS (\"turn-around solution\"). ultimately ap proaching the asymptotic law a = constant X exp [(A/3)1I2rJ. ~ / / 748 27. IDEALIZED COSMOLOGIES Figure 27.5. (facing page) Log-log plot of the effective potential vea) of equation (27.75) and Box 27.5 as it enters the generalized Friedmann equation d (a(I\u00bb)]2 dr 00 [ + I f( a) _ k _ - - - ~ 00 K o' - Horizontally is given the expansion ratio referred to (a,'ao)toda,' = I as standard of reference. Vertically is given the value of V(a;'ao ) in the geometric units of cm- 2. The' supplementary scale at the right translates to -(c2/G)(3/S?T)V(a/ao ) as an eqUivalent density, expressed in g/cm3. The contribution of radiation density to the effective potential is indicated by the wavy line in the diagram. It is normalized to a value of the radiation densily today of Pro = 10-33 g/cm3 and has a logarithmic slope of two. The contribution of matter density to the \"effective potential\" has a logarithmic slope of unity. Two choices are made for it. corresponding to a density of matter today of Pmo = 10-30 g/cm3 and Pm. = 10-28 g/cm3 (dashed lines in the diagram). The total effective potential in the two cases is also indicated in the diagram: a heavy line for the case Pmo = 10-30 g/cm 3 (no cosmological term included) and a light line for the case Pmo = 10-28 g/cm3\u2022 In this second case, a cosmological term is included. with the cosmological constant given by (3/Sr.)(A/3) = 10-29 g/cm 3\u2022 The line describing the contribution of this term has a negative slope of magnitude two (dashed line). The horizontal or \"level line\" is drawn for a value of the Hubble expansion rate today. Ho' equal to 1/(20 X 109 years). The vertical separation on the log plot between the potential curve and the level line gives the ratio - V/ Hf,. This ratio as evaluated at any time I has the value a2(1)/a; + KJt;2, where a== da/dl. As evaluated \"today\" (a/ao = I) this ratio has the value I + KoH;2. Knowing the Hubble expansion rate H; today, and knowing (or trying a certain set of parameters for) the potential curve, one can therefore deduce from the spread between the two the value of I + KoH;2, hence the value of KoH;2, hence the present value. Ko' of the curvature factor. As an example, consider the case of the low-density universe (heavy line) and read off \"today's\" I + K oH;2 = 0.223. From this follows Ko = -0.777 Hf, (open or hyperbolic universe), hence value, )1I2 = (1/0.777)1/ 2 20 X 109 yr = 22.7 X 109 yr. For the high-density model k = -I and ao = (k/ K o universe, with Pmo =", " 10-28 g/cm3, one similarly finds I + KoH;2 = 24.5, hence Ko = +23.5Hf\" hence k = + I (closed universe) and ao = (k/Ko )1/2 = (1/23.5)112 20 X 109 yr = 4.12 X 109 yr. Expansion stops, if and when it stops, at that stage when the ratio - V/Hf, between the potential and the level line, or a2(1)/a; + KoH;2, falls from its \"present value\" of I + KoH;2 to 0 + KoH;2; that is, from 0.223 to -0.777 in the one case, and from 24.5 to 23.5 in the other case. This log-log plot should be replaced by the linear plot of Box 27.5 when A < O.. (2) hierarchic model However great a departure it is from Einstein's concept of cosmology to give any heed to a cosmological constant or an open universe, it is a still greater departure to contemplate a \"hierarchic model\" of the universe, in which clusters of galaxies, and clusters of clusters of galaxies, in this part of the universe are envisaged to grade off in density with distance, with space at great distances becoming asymptotically flat [Alfven and Klein (1962), Alfven (1971), Klein (1971), Moritz (1969), de Vau couleurs (1971), Steigman (1971)]. The viewpoint adopted here is expressed by Oskar Klein in these words, \"Einstein's cosmology was adapted to the discovery by Hubble that the observed part is expanding; the so-called cosmological postulate has been used as a kind of an axiomatic background which, when analyzed, makes it appear that this expansion is shared by a very big, but still bounded system. This implies that our expanding metagalaxy is probably just one of a type of stellar objects in different phases of evolution, some expanding and some contracting.\" \u00a727.11. MODEL UNIVERSES WITH k:OS; 0 or A ;C 0 749 --alao----i~~ 10-4 10-2 1 T -I X 10- 59 -I X 10- 58 t -I X IO~\" V(a/ao) in cm- 2 -I X 10- 56 -I X 10-55 -I X 10-54 - 1 X 10- 53 ~~/,(;/. \"'-~~~~, ~/\"\\V /!/,,>'\" qq -I X 10- 52 ~ q \\S~ -I X 10- 51 102 1 10-33 10-29 \u202210- 28 10-27 10- 26 10- 25 10- 24 The contrast between the hierarchic cosmology and Einstein's cosmology [Einstein (1931) advocates a closed Friedmann cosmology] appears nowhere more strongly than here, that the one regards asymptotically flat spacetime as a requirement; the other, as an absurdity. \"Only the genius of Riemann, solitary and uncomprehended,\" Einstein (1934) puts it, \"had already won its way by the middle of the last century to a new conception of space, in which space was deprived of its rigidity, and in which its power to take part in physical events was recognized as possible.\" That statement epitomizes cosmology today. EXERCISES 750 27. IDEALIZED COSMOLOGIES But today's view of cosmology, as dominated by Einstein's boundary condition of closure (k = + 1) and his belief in A = 0, need not be accepted on faith forever. Einstein's predictions are clear and definite. They expose themselves to destruction. Observational cosmology will ultimately confirm or destroy them, as decisively as it has already destroyed the 1920 belief in a static universe and the 1948 steady-state models (see Box 27.7 on the history of cosmology). Exercise 27.11. ON SEEING THE BACK OF ONE'S HEAD Can a being at rest relative to the \"cosmological fluid\" ever see the back of his head by means of photons that travel all the way around a closed model universe that obeys the Friedmann cosmology and has a non-zero cosmological constant (see the entries in Box 27.3 for the case of a zero cosmological constant)? Exercise 27.12. DO THE CONSERVATION LAWS FORBID THE PRODUCTION OF PARTICLE-ANTIPARTICLE PAIRS OUT OF EMPTY SPACE BY TIDAL GRAVITATION FORCES? Find out what is wrong with the following argument: \"The classical equations Gaf3 = 8'lTTaf3 are not compatible with the production of pairs, since they lead to the identity Tl;f3 =", " (1965), Peebles (1971), Rindler (1969), and Sciema (1971); and especially see Peebles and Sciama for bibliographical references to modern developments listed below in abbrevieted form.] A. Before the Twentieth Century Concepts of very early Indian cosmology [summarized by Zimmer (1946)]: \"One thousand mahayugas-4,320,OOO,OOO years of human reckoning-constitute a single day of Brahmii., a single kalpa.... I have known the dreadful dissolution of the universe. I have seen all perish, again and again, at every cycle. At that terrible time, every single atom dissolves into the primal, pure waters of eternity, whence all originally arose.\" Plato, ca. 428 to ca. 348 B.C. [from the Timaeus, written late in his life, as translated by Cornford (1937)]: \"The world [universe] has been fashioned on the model of that which is comprehensible by rational discourse and understanding and is always in the same state.... this world came to be... a living creature with soul and reason.... its maker did not make two worlds nor yet an indefinite number; but this Heaven has come to be and is and shall be hereafter one and unique.... he fashioned it complete and free from age and sickness.... he turned its shape rounded and spherical... It had no need of eyes, for nothing visible was left outside; nor of hearing, for there was nothing outside to be heard.... in order that Time might be brought into being, Sun and Moon and five other stars-'wanderers,' as they are called-were made to define and preserve the numbers of Time.... the generation of this universe was a mixed result of the combination of Necessity and Reason... we must also bring in the Errant Cause.... that which is to receive in itself all kinds [all forms] must be free from all characters [all form].... For this reason, then, the mother and Receptacle of what has come to be visible and otherwise sensible must not be called earth or air or fire or water... but a nature invisible and characterless, all-receiving, partaking in some very puzzling way of the intelligible, and very hard to apprehend.\" Aristotle, 384-322 B.C. [from On the Heavens, as translated by Guthrie (1939)]: \"Throughout all past time, according to the records handed down from generation to generation, we find no trace of change either in the whole of the outermost heaven or in anyone of its proper parts.... the shape of the heaven must be spherical.... From these considerations [motion invariably in a straight line toward the center; regularity of rising and setting of stars; natural motion of earth toward the center of the universe] it is clear that the earth does not move, neither does it lie anywhere but at the center.... the earth... must have grown in the form of a sphere. This [shape of segments cut out of moon at time of eclipse of moon; and ability to see \u00a727.11. BRIEF HISTORY OF COSMOLOGY 753 in Egypt stars not visible in more northerly lands] proves both that the earth is spherical and that its periphery is not large... Mathematicians who try to calculate the circumference put it at 400,000 stades [I\u00b7 stade = 600 Greek feet = 606 English thus 24.24 X 107 ft/(60802 ft/nautical mile) = 39,900 nautical miles-the feet; oldest recorded calculation of the earth's circumference, and reportedly known to Columbus-85 per cent more than the true circumference, 60 X 360 = 21,600 nautical miles].\" Aristotle [from the Metaphysics, as translated by Warrington (1956)]: \"Euxodus [of Cnidos, 408-355 B.C.] supposed that the motion of the sun and moon involves, in each case, three spheres.... He further assumed that the motion of the planets involves, in each case, four spheres.... Calippus [of Cyzicus, flourished 330 B.C.]... considered that, in the light of observation, two more spheres should be added to the sun, two to the moon, and one more to each of the other planets.\" Eratosthenes, 276-194 B.C. [a calculation attributed to him by Claudius Ptolemy, who observed at Alexandria from 127 to 141 or 151 A.D., in his Almagest, I, \u00a7 12; see the translation by Taliaferro (1952)]: (Maximum distance of moon from earth) = (64i) (radius of earth); (Minimum distance of sun from earth) = (1,160) (radius of earth). Abu 'Ali ai-Husain ibn'AbdaIHih ibn Sina, otherwise known as Avicenna", ", 980-1037; physician, philosopher, codifier of Aristotle, and one of the most influential of those who preserved Greek learning and thereby made possible its transmission to me diaeval Europe [quoted in Nasr (1964), p. 225]: \"Time is the measure of motion.\" From the Rasa'il, a51-treatise encyclopedia, sometimes known as the Koran after the Koran, of the Ikhwan al-Safa' or Brothers of Sincerity, whose main center was at Basra, Iraq, roughly A.D. 950-1000 [as quoted by Nasr (1964), p. 64; see p. 78 for a list of distances to the planets (in terms of Earth radii) taken from the Rasa'il, as well as sizes of planets and the motions of rotation of the various Ptolemaic carrier-spheres]: Space is \"a form abstracted from matter and existing only in the consciousness.\" Abu RaiMn al-Biriini, 973-1030, a scholar, but concerned also with experiment, observation, and measurement, who calculated the circumference of the Earth from measurements he made in India as 80,780,039 cubits (about 4 per cent larger than the value accepted today), and gave a table of distances to the planets [as quoted in Nasr (1964), pp. 120 and 130]: \"Both [kinds of eclipses] do not happen together except at the time of the total collapse of the universe.\" Etiene Tempier, Bishop of Paris, in 1277, to settle a controversy then dividing much of the French theological community, ruled that one could not deny the power of God to create as many universes as He pleases. Roger Bacon, 1214-1294, in his Opus Majus (1268), gave the diameter of the sphere that carries the stars, on the authority of Alfargani, as 130,715,000 Roman miles 754 27. IDEALIZED COSMOLOGIES Box 27.7 (continued) [mile equal to 1,000 settings down of the right foot]; the volume of the sun, 170 times that of the Earth; first-magnitude star, 107 times; sixth-magnitude, 18 times Earth. Nicolas Cusanus, 1401-1464 [from Of Learned Ignorance (1440), as translated by Heron (1954)]: \"Necessa.rily all parts of the heavens are in movement.... It is evident from the foregoing that the Earth is in movement... the world [universe], its movement and form... will appear as a wheel in a wheel, a sphere in a sphere without a center or circumference anywhere.... It is now evident that this Earth really moves, though to us it seems stationary. In fact, it is only by reference to something fixed that we detect the movement of anything. How would a person know that a ship was in movement, if... the banks were invisible to him and he was ignorant of the fact that water flows?\" Nicolaus Copernicus, February 19, 1473, to May 24,1543 [from De Revolutionibus Orbium Coelestrum (1543), as translated by Dobson and Brodetsky (1947)]: \"I was induced to think of a method of computing the motions of the spheres by nothing less than the knowledge that the mathematicians are inconsistent in these investiga tions.... they cannot explain or observe the constant length of the seasonal year.... some use only concentric circles, while others eccentrics and epicycles.... Nor have they been able thereby to discern or deduce the principal thing-namely the shape of the universe and the unchangeable symmetry of its parts.... \"I found first in Cicero that Nicetas had realized that the Earth moved. Afterwards I found in Plutarch [~A.D. 46-120]... 'The rest hold the Earth to be stationary, but Philolaus the Pythagorean [born ~480 B.C.] says that she moves around the (central) fire on an oblique circle like the Sun and Moon. Heraclides of Pontus [flourished in 4th century B.C.] and Ecphantus the Pythagorean also make the Earth to move, not indeed through space but by rotating round her own center as a wheel on an axle from West to East.' Taking advantage of this I too began to think of the mobility of the Earth.... \"Should we not be more surprised if the vast Universe revolved in twenty-four hours, than that little Earth should do so?... Idle therefore is the fear of Ptolemy that Earth and all thereon would be disintegrated by a natural rotation.... That the Earth is not the center of all revolutions is proved by the apparently irregular motions of the planets and the variations in their distances from the Earth.... We therefore assert that the", " center of the Earth, carrying the Moon's path, passes in a great orbit among the other planets in an annual revolution round the Sun; that near the Sun is the center of the Universe; and that whereas the Sun is at rest, any apparent motion of the Sun can be better explained by motion of the Earth.... Particularly Mars, when he shines all night, appears to rival Jupiter in magnitude, being distinguishable only by his ruddy color; otherwise he is scarce equal to a star of the second magnitude, and can be recognized only when his movements are \u00a727.11. BRIEF HISTORY OF COSMOLOGY 755 carefully followed. All these phenomena proceed from the same cause, namely Earth's motion.... That there are no such phenomena for the fixed stars proves their immeasurable distance, compared to which even the size of the Earth's orbit is negligible and the parallactic effect unnoticeable.\" Thomas Digges, 1546-1595 [in a Perfit Description ofthe Caelestial! Orbes accord ing to the most aunciente doctrine of the Pythagoreans, /atelye reuiued by Copernicus and by Geometrical! Demonstrations approued (1576), the principal vehicle by which Copernicus reached England, as quoted in Johnson (1937)]: \"Of whiche lightes Celestiall it is to bee thoughte that we onely behoulde sutch as are in the inferioure partes of the same Orbe, and as they are hygher, so seeme they of lesse and lesser quantity, euen tyll our sighte beinge not able farder to reach or conceyue, the greatest part rest by reason of their wonderfull distance inuisible vnto vs.\" Giordano Bruno, born ca. 1548, burned at the stake in the Campo dei Fiori in Rome, February 17, 1600 [from On the Infinite Universe and Worlds, written on a visit to England in 1583-1585, as translated by Singer (1950)]: \"Thus let this surface be what it will, I must always put the question, what is beyond? If the reply is NOTHING, then I call that the VOID or emptiness. And such a Void or Emptiness hath no measure and no outer limit, though it hath an inner; and this is harder to imagine than is an infinite or immense universe.... There are then innumerable suns, and an infinite number of earths revolve around those suns, just as the seven we can observe revolve around this sun which is close to us.\" Johann Kepler established the laws of elliptic orbits and of equal areas (1609), and established the connection between planetary periods and semimajor axes (1619). GaIileo Galilei observed the satellites ofJupiter and realized they provided support for Copernican-theory, and interpreted the Milky Way as a collection of stars (1610). In 1638 he wrote: \"Salvati. Now what shall we do, Simplicio, with the fixed stars? Do we want to sprinkle them through the immense abyss of the universe, at various distances from any predetermined point, or place them on a spherical surface extending around a center of their own so that each of them will be at the same distance from that center? \"Simplicio. I had rather take a middle course, and assign to them an orb de-kribed around a definite center and included between two spherical surfaces...\" Isaac Newton (1687): \"Gravitation toward the sun is made up out of the gravita tions toward the several particles of which the body of the sun is composed, and in receding from the sun decreases accurately as the inverse square of the distances as far as the orbit of Saturn, as evidently appears from the quiescence of the aphelion of the planets.\" Isaac Newton [in a letter of Dec. 10, 1692, to Richard Bentley, quoted in Munitz (1957)]: \"If the matter of our sun and planets and all the matter of the universe were evenly scattered throughout all the heavens, and every particle had an innate gravity toward all the rest, and the whole space throughout which this matter was 756 27. IDEALIZED COSMOLOGIES Box 27.7 (continued) scattered was but finite, the matter on the outside of this space would, by its gravity, tend toward all the matter on the inside and, by consequence, fall down into the middle of the whole space and there compose one great spherical mass. But if the matter was evenly disposed throughout an infinite space, it could never convene into one mass; but some of it would convene into one mass and some into another, so as to make an infinite number of great masses scattered at great distances from one to another throughout all that infinite space. And thus might the sun and fixed stars be formed.\" Christiaan", " Huygens, 1629-1695 [in his posthumously published Cosmotheoros (1698)]: \"Seeing then that the stars... are so many suns, if we do but suppose one of them [Sirius, the Dog-star] equal to ours, it will follow [details, including telescope directed at sun; thin plate; hole in it; comparison with Sirius]... that his distance to the distance of the sun from us is as 27,664 to I.... Indeed it seems to me certain that the universe is infinitely extended.\" Edmund Halley (1720): \"If the number of the Fixt Stars were more than finite, the whole superficies of their apparent Sphere [i.e., the sky] would be luminous\" [by today's reasoning the same temperature as the surface of the average star; this is known today as Olber's paradox, or the paradox ofP. L. de Cheseaux (1744) and Heinrich Wilhelm Matthias Olbers (1826)]. Thomas Wright of Durham (1750): \"To... solve the Phaenomena of the Via granted... that the Milky Way is formed of an infinite number of small Lactea Stars imagine a vast infinite gulph, or medium, every way extended like a plane, and inclosed between two surfaces, nearly even on both sides.... Now in this space let us imagine all the Stars scattered promiscuously, but at such an adjusted distance from one another, as to fill up the whole medium with a kind of regular irregularity of objects. [Considering its appearance] \"to an eye situated... anywhere about the middle plane\"... all the irregularity we observe in it at the Earth, I judge to be entirely owing to our Sun's position... and the diversity of motion... amongst the stars themselves, which may here and there... occasion a cloudy knot of stars.\" Immanuel Kant, 1724-1804 (1755): \"It was reserved for an Englishman, Mr. Wright of Durham, to make a happy step... we will try to discover the cause that has made the positions of the fixed stars come to be in relation to a common plane.... granted... that the whole host of [the fixed stars] are striving to approach each other through their mutual attraction... ruin is prevented by the action of the centrifugal forces... the same cause [centrifugal force]... has also so directed their orbits that they are all related to one plane.... [The needed motion is calcu lated to be] one degree [or less] in four thousand years;... careful observers... will be required for it.... Mr. Bradley has observed almost imperceptible displace ments of the stars\" [known from later work to be caused by aberration (effect of observer velocity) rather than real parallax (effect of position of observer)]. \\ \u00a727.11. BRIEF HISTORY OF COSMOLOGY 757 Asks for the first time how a very remote galaxy would appear: \"circular if its plane is presented directly to the eye, and elliptical if it is seen from the side or obliquely. The feebleness of its light, its figure, and the apparent size of its diameter will dearly distinguish such a phenomenon when it is presented, from all the stars that are seen single.... this phenomenon... has been distinctly perceived by different observers [who]... have been astonished at its strangeness.... Analogy thus does not leave us to doubt that these systems [planets, stars, galaxies] have been formed and produced... out of the smallest particles of the elementary matter that filled empty space.\" Goes on to consider seriously \"the successive expansion of the creation [of planets, stars, galaxies] through the infinite regions of space that have the matter for it.... attraction is just that universal relation which unites all the parts of nature in one space. It reaches, therefore, to... all the distance of nature's infinitude.\" Johann Heinrich Lambert, 1728-1777 (1761): \"The fixed stars obeying central forces move in orbits. The Milky Way comprehends several systems offixed stars.... Each system has its center, and several systems taken together have a common center. Assemblages of their assemblages likewise have theirs. In fine, there is a universal center for the whole world round which all things revolve.\" [First spelling out of a \"hierarchical model\" for the universe, later taken up by C. V. I. Charlier and by H. Alfven and O. Klein (1962); see also O. Klein (1966 and 1971)]. Auguste Comte (1835) concluded that it is meaningless to speak of the chemical composition of distant stars because man will never be able to explore them; \"the field of positive philosophy lies wholly within the limits of our solar system, the study of the univ_~~e being inaccessible in any positive sense", ".\" The firstsuccessful determination of the parallax [1 second of parallax: 1 pc = 3.08 X 10 18 cm = 3.26 lyr] of any star was made in 1838 (for a Centauri by Hen derson, for a Lyrae by Struve, and for 61 Cygni by Bessel). B. The Twentieth Century Derivation by James Jeans in 1902 of the critical wavelength that separates short wavelength acoustical modes of vibration of a hot primordial gas and longer wave length modes of commencement of gravitational condensation of this gas. Applica tion of these considerations by P. J. E. Peebles and R. H. Dicke in 1968 to explain why globular star clusters have masses of the order of 105Mo. Investigations of cosmic rays from first observation by V. F. Hess and W. KolhOrster in 1911-1913 to date: determination that the energy density in interstellar space (in this galaxy) is about 1 eV/cm3 or 10- 12 erg/cm3, comparable to the density of energy of starlight, to the kinetic energy of clouds of ionized interstellar gas, averaged over the galaxy, and to the energy density of the interstellar 758 27. IDEALIZED COSMOLOGIES Box 27.7 (continued) magnetic field (-- 10- 5 gauss). E. Fermi (1949). In connection with this equality, see especially Discovery by Henrietta Leavitt in 19 I2 that there is a well-defined relation between the period of a Cepheid variable and its absolute luminosity. First determination of the radial velocity of a galaxy by V. M. Slipher in 19 12: Andromeda approaching at 200 km/sec. Thirteen galaxies investigated by him by 1915; all but two receding at roughly 300 km/sec. Albert Einstein (l915d): Interpreted gravitation as a manifestation of geometry; gave final formulation of the law that governs the dynamic development of the geometry of space with the passage of time. Albert Einstein (1917): Idealized the universe as a 3-sphere filled with matter at effectively uniform density; the radius of this 3-sphere could not be envisaged as static without altering his standard 1915 geometrodynamic law; for this reason Einstein introduced a so-called \"cosmological term,\" which he later dropped as \"the biggest blunder\" in his life [Gamow (1970)]. Formulation by W. de Sitter in 1917 of a cosmological model in which (I) the universe is everywhere isotropic (and therefore homogeneous) and (2) the universe does not change with time, so that the mean density of mass-energy and the mean curvature of space are constant, but in which perforce (3) a cosmological term (\"repulsion\") of the Einstein type has to be added to balance the attraction of the matter. Observation by de Sitter that he could obtain another static model by removing all the matter from the original model, but that the A-term would cause test particles to accelerate away from one another. From 1917 to 1920, debate about whether spiral nebulae are mere nebulous objects (Harlow Shapley) or are \"island universes\" or galaxies similar to but external to the Milky Way (H. D. Curtis). Discovery by Harlow Shapley in 19 18, by mapping distribution of about 100 globular clusters of this Galaxy (104 to 106 stars each) in space that center is in direction of Sagittarius (present value of distance from sun -- 10 kpc). Independent derivation of evolving homogeneous and isotropic cosmological models [also leading to the relation v = H' (distance)] by A. Friedmann in 1922 and G. Lemaitre in 1927, with Lemaitre tieing in his theoretical analysis with the then-ongoing Mt. Wilson work, to become the \"father of the big-bang cosmology\". (Universe, however, taken to expand smoothly away from Einstein's static A > 0 solution in Lemaitre's original paper). Remark by H. Weyl in 1923 that test particles in de Sitter model will separate at a rate given by a formula of the form v = H\u00b7 (distance). In 1924, resolution of debate about nature of spiral nebulae by Edwin P. Hubble with Mount Wilson 100-inch telescope; discovery of Cepheid variables in Andromeda and other spiral nebulae, and consequent determination ofdistances to these nebulae. \u00a7 27.11. BRIEF HISTORY OF COSMOLOGY 759 Determination by Jan Oort in 1927 of characteristic pattern of radial velocities of stars near sun, 8ur = Ar cos 2(8 - 8), showing that: (I) axis of rotation of stars in Milky Way is perpendicular to disc; (2) sun makes a complete revolution in ~ 108 yr", "; and (3) the effective mass pulling on the sun required to produce a revolution with this period is of the order ~ 1044 g or ~lOl1M0' Age of a uranium ore as established from lead-uranium ratio: greatest value found up to 1927, 1.3 X 109 yr (A. Holmes and R. W. Lawson). Age of the lead in the \"average\" surface rocks of the earth as calculated from time required to produce this lead from the uranium in the same surface rocks, 2 X 109 yr to 6 X 109 yr. Age of elemental uranium as estimated by Rutherford from time required for U 235 and U238 to decay from assumed roughly equal ratio in early days to known very unequal ratio today, ~3 X 109 yr. Establishment by Hubble in 1929 that out to 6 X 106 Iyr the velocity of recession of a galaxy is proportional to its distance. Note by A. S. Eddington in 1930 that Einstein A >\u00b0static universe is unstable against any small increase or decrease in the radius of curvature. Recommendation from Einstein in 193 I hereafter to drop the so-called cosmolog ical term. Proposal by Einstein and de Sitter in 1932 that one tentatively adopt the simplest assumption that 11= 0, that pressure is negligible, and that the reciprocal of the square of the radius of curvature of the universe is neither positive nor negative (spherical or hyperbolic universe) but zero (\"cosmologically flat\"), thus leading to the relation p = 3H2/87T (in geometric units). Evidence uncovered by Grote Reber in 1934 for the existence of a discrete radio source in Cygnus; evidence for this source, Cygnus A, firmed up by J. S. Hey, S. J. Parsons, and J. W. Phillips in 1946; six other discrete radio sources, including Taurus A and Centaurus A, discovered by J. G. Bolton in 1948. Discovery by E. A. Milne and W. H. McCrea in 1934 of close correspondence between Newtonian dynamics of a large gas cloud and Einstein theory of a dynamic universe, with the scale factor of the expansion satisfying the same equation in both theories, so long as pressure is negligible. Demonstration by H. P. Robertson and by A. G. Walker, independently, in 1935 the Lemaitre type of line element provides the most general Riemannian that geometry compatible with homogeneity and isotropy. Classification of nebulae as spiral, barred spiral, elliptical, and irregular by Hubble in 1936. First detailed theory of thermonuclear energy generation in the sun, H. A. Bethe, 1939. 760 27. IDEALIZED COSMOLOGIES Box 27.7 (continued) Reasoning by George Gamow in 1946 that matter in the early universe was dense enough and hot enough to undergo rapid thermonuclear reaction. and that energy densities were radiation-dominated. Proposal of so-called \"steady-state cosmology\" by H. Bondi, T. Gold, and F. Hoyle in 1948, lying outside the framework of Einstein's standard general rela tivity, with \"continuous creation of matter\" taking place throughout the universe, and the mean age of the matter present being equal to one third of the Hubble time. Prediction by R. A. Alpher. H. A. Bethe, and G. Gamow in 1948 that the black body radiation that originally filled the universe should today have a Planck spec trum corresponding to a temperature of25 K. Independent conception of same idea by R. H. Dicke in 1964 and start of an experimental search for this primordial cosmic-fireball radiation. Discovery of unwanted and unexpected 7 cm background radiation in 1965 by A. A. Penzias and R. W. Wilson with a temperature of about 3.5 K; immediate identification of this radiation by Dicke, P. J. E. Peebles, P. G. Roll, and D. T. Wilkinson as the expected relict radiation. Radio sources Taurus A, Virgo A, and Centaurus A tentatively and, as it later proved, correctly identified with the Crab Nebula and with the galaxies NGC 4486 and NGC 5128 by J. G. Bolton, G. J. Stanley, and O. B. Slee in 1949. Analysis by Lemaltre in 1950 of big-bang expansion approa\u00a3hing very closely the Einstein static universe (11 > 0) and then, at first slowly,' subsequently more and more rapidly, going into exponential expansion. Discovery by Walter Baade in 1952 that there are two types of Cepheid variables with different period-luminosity relations; consequent increase in Hubble distance scale by factor of about 2.6, and a corresponding increase in the original value (roughly 2 X 109 yr) of the Hubble time, H-;;l. Identification of radio source", " Cygnus A by W. Baade and R. Minkowski in 1954 with the brightest member of a faint cluster of galaxies, contrary to the then widely held view that the majority of radio sources lie within the Milky Way. Determination of redshift in the optical spectrum of 8A/A = z = 0.057 by Minkowski, implying for Cygnus A a distance of 170 Mpc and a radio luminosity of 1045 erg/sec, 107 times the radio power and ten times the optical power of a normal galaxy. Resolution of radio source Cygnus A in 1956 into two components symmetrically located on either side of the optical galaxy, the first indication that most radio sources are double. Still unsolved is the mystery of the explosion or other mechanism that caused this and other double sources. Calculation by G. R. Burbidge in 1956 of the kinetic energy in the electrons giving off synchrotron radiation in a radio galaxy and the energy of the magnetic field that holds these electrons in orbit; minimization of the sum of these two energies; \u00a727.11. BRIEF HISTORY OF COSMOLOGY 761 determination that this minimum is of the order of 1060 ergs (energy of annihilation of half a million suns) for Hercules A, for example. Solar system determined to have an age of 4.55 X 109 yr or more from relative abundances of Pb204,206,207 and U235,238 in meteorites and oceanic sediments by C. Patterson in 1956; and by others in 1965 and 1969 from evidence on the processes Rb87 -+ Sr87 and K40 -+ A40 in meteorites. Discovery by Allen Sandage in 1958 that what Hubble had identified in distant galaxies as bright stars were H II regions, clumps of hot stars surrounded by a plasma ionized by stars, and consequent upping of Hubble's distance scale by a further factor of about 2.2. Estimation by Jan Oort in 1958, from luminosity of other galaxies, that matter in galaxies contributes to the density of mass-energy in the universe roughly 3 X 10-31 g/cm3 [see Peebles (1971) for updated analysis], this being one or two orders of magnitude less than that called for by Einstein's concept that the universe is curved up into closure, and thereby giving rise to \"the mystery of the missing matter,\" the focus of much present-day research. Discovery of celestial (nonsolar) X-rays in 1962 by Giacconi, Gursky, Paolini, and Rossi. Majority ofsources in plane of the Milky Way, presumably local to this galaxy, as is the Crab nebula. Extragalactic sources include the radio galaxy Virgo A and the quasar 3C273. Revised \"3C-catalog\" of radio sources published in 1962 by A. S. Bennett, con taining 328 sources, nearly complete in coverage between declinations _50 and +90 0 for source~_~t:ighter than 9 flux units (9 X 10-26 watt/m2Hz) at 178 MHz. Identification\"of the first quasistellar object (QSO) by Maarten Schmidt at Mt. Palomar in 1963: radio-position determination of 3C273 to better than 1 second of arc by C. Hazard, M. B. Mackey, and A. J. Shimmins in 1962, followed by Schmidt's taking an optical spectrum of the star-like source and, despite all pre sumptions that it was a star in this galaxy, trying to fit it, and succeeding, with a redshift of the magnitude (unprecedented for a \"star\") of8A/A = z = 0.158. Distance implied by Hubble relation, 1.5 X 109 lyr; optical brightness, 100 times brightest known galaxy. Largest redshift of any QSO known in 1972, Z = 2.88 (4C05.34; C. R. Lynds). Such a source detectable even ifit had a redshift of3; but no QSO's known in 1972 with such redshifts. See Box 28.1. Reasoning by Dennis Sciama in 1964 [see also Sciama (1971)] that intergalactic hydrogen can best escape observation if at a temperature between 3 X 105 K and 106 K. With as many as 10-5 protons and 10-5 electrons per cm3 and a temperature lower than 3 X 105 K, the num ber density of neutral atoms would be great enough and the resulting absorption of Lyman a from a distant galaxy (z = 2) would be strong enough to show up, contrary to observation. In 1964 J. E. Gunn and B. A. Peterson, E. J. Wampler, and others determined that, at a temperature greater than 106 K, the intensity of 0.25 keV or 50 A x-rays 762 27. IDEAL", "IZED COSMOLOGIES Box 27.7 (continued) Emphasis by Wheeler (l964a) that from intergalactic space would be too high to be compatible with the observations. in Einstein's general relativity is 3-geometry, not 4-geometry, and that this dynamics, both classical and quantum, unrolls in the arena of superspace. the dynamic object Discovery by Sandage in 1965 of quasistellar galaxies (radio-quiet QSO's). Discovery by E. M. Burbidge, G. R. Burbidge, C. R. Lynds, and A. N. Stockton in 1965 of a QSO, 3C191, with numerous absorption lines, implying the coexistence of several redshifts in one spectrum. Fraction (by mass) of matter converted to helium in early few minutes of universe nearly independent of the relative numbers of photons and baryons, over a 106 range in values of this number ratio, so long as the universe at 10 10 K is still radiation dominated. Value of this plateau helium abundance (following earlier work of others) first accurately calculated as 27 per cent by P. J. E. Peebles in 1966 and by R. V. Wagoner, W. A. Fowler, and F. Hoyle in 1967. Proposal by C. W. Misner in 1968 to consider as an important part of early cosmology the anisotropy vibrations of the geometry of space previously brought to attention by E. Kasner and by 1. M. Khalatnikov and E. Lifshitz. [Misner's hope to account naturally in this way for the otherwise so puzzling homogeneity of the universe was later dashed.] Proof on the basis of standard general relativity by S. W. Hawking, G. F. R. Ellis, and R. Penrose in 1968 and 1969 [see also related work of earlier investigators cited in Chapter 44] that a model universe presently expanding and filled with matter and radiation obeying a physically acceptable equation of state must have been singular in the past, however wanting in symmetry it is today. Discovery of pulsars in 1968 by Hewish, Bell, Pilkington, Scott, and Collins, and their interpretation as spinning neutron stars (see Chapter 24). and artists. You cannot value him alone; you must set him, \"No poet, nor artist of any art, has his complete meaning alone. His significance, his appreciation, is the appreciation of his relation to the dead poets for contrast and comparison, among the dead... when a new work of art is created... something... happens simultaneously to all the works of art which preceded it. The existing monuments form an ideal order among themselves, which is modified by the introduction of the new (the really new) work of art among them.\" T. S. ELIOT (1920). CHAPTER 28 EVOLUTION OF THE UNIVERSE INTO ITS PRESENT STATE Cosmology... restrains the aberrations of the mere undisciplined imagination. ALFRED NORTH WHITEHEAD (1929, p. 21) \u00a728.1. THE \"STANDARD MODEL\" OF THE UNIVERSE Since the discovery of the cosmic microwave radiation in 1965, extensive theoretical research has produced a fairly detailed picture of how the universe probably evolved into its present state. This picture, called the \"standard hot big-bang model\" of the universe, is sketched in the present chapter, and its main features appear in Figure 28.1. Gravitation dominates the over-all expansion; but otherwise most details of the evolution are governed much less by gravitation than by the laws of thermody namics, hydrodynamics, atomic physics, nuclear physics, and high-energy physics. This fact, and the existence of three excellent recent books on the subject [Sciama (1971); Peebles (1972); Zel'dovich and Novikov (1974)], encourage brevity here. The past evolution of the universe is qualitatively independent of the nature of the homogeneous hypersurfaces (k = -1,0, or + 1) and qualitatively independent of the cosmological constant, since the contributions of k and A to the evolution are not important in early stages of the history (small a/ao) [see equation (27.40) and Figure 27.5]. One crucial assumption underlies the standard hot big-bang model: that the universe \"began\" in a state of rapid expansion from a very nearly homoge neous, isotropic condition of infinite (or near infinite) density and temperature. During the first second after the beginning, according to this analysis, the temper ature of the universe was so high that there was complete thermodynamic equilib- Evolution of universe according to \"standard hot big-bang model\": (1) initial state 764 28. EVOLUTION OF THE UNIVERSE INTO ITS PRESENT STATE Time I.", " sec :-';OW (12 X 109 yrs.\\ IO~ 104 10~ 10' 10 1\" 10'\" 10\" 10 16 10\" 10- 0 10- 1\" 10- 1 \" 10- 20 10- 2:; 10-30 10- 35 'E '-' \"e.o \"'~ \"r, C '-' Cl --'--,-_--' 10-40 10- 2 I 10 12 lO\"' ::<: ~ ::l E lOS uc.. E ~ 106 to.2.~ 't:l,. cr. 104 102 '---,-_-1..-,.-__-'--__--'10- 10, 10- 8 10-6 10-4 Expansion factor a/a o I I I I. Pairs recombine~ neutnnos decouple Transition from ~ t d' \" Primordial r----1 ra latlon dommance t-- t \u2022 Galaxies N form ow elements form to matter dommance Plasma. 1 - - - - - - - - - - - - - 41...-11 Matter and radiation in thermal equilibrium recombines Most distant quasar observed (z:::: 3\\ Figure 28.1. Evolution of the universe into its present state, according to the standard hot big-bang model. The curves are drawn assuming Pnw = 5 X 10-30 g/cm3, Pro = 10-33 g/cm3, k = 0; but for other values of Pnw' Pro' and k within the limits of observation, the curves are virtually the same (see exercise 28.1). See text and Box 28.1 for detailed discussion of the processes described at the bottom of the figure. [This figure is adapted from Dicke, Peebles, RoIl, and Wilkinson (1965).] (2) thermal equilibrium, decay of particles, recombination of pairs (0 < t ~ 10 sec.) (3) decoupling and free propagation of gravitons and neutrinos (t ~ 1 sec.) rium between photons, neutrinos, electrons, positrons, neutrons, protons, various hyperons and mesons, and perhaps even gravitons (gravitational waves) [see, e.g., Kundt (1971) and references cited therein]. However, by the time the universe was a few seconds old, its temperature had dropped to about 1010 K and its density was down to ~ 105 g/cm3 ; so all nucleon-antinucleon pairs had recombined, all hyperons and mesons had decayed, and all neutrinos and gravitons had decoupled from matter. The universe then consisted of freely propagating neutrinos, and perhaps gravitons, with black-body spectra at temperatures T ~ 1010 K, plus electron-posi tron pairs in the process of recombining, plus electrons, neutrons, protons, and photons all in thermal equilibrium at T ~ 1010 K. Since that early state, the gravitons (if present) and neutrinos have continued \u00a72B.1. THE \"STANDARD MODEL\" OF THE UNIVERSE 765 to propagate freely, maintaining black-body spectra; but their temperatures have been redshifted by the expansion of the universe in accordance with the law T ex: l/a (28.1 ) (Box 29.2). Consequently, today their temperatures should be roughly 3 K, and they should still fill the universe. Unfortunately, today's technology is far from being able to detect such a \"sea\" of neutrinos or gravitons. However, if and when they can be detected, they will provide direct observational information about the first one second of the life of the universe! 10 10 to _10 9 K, P - As the universe continued to expand after the first few seconds, it entered a period lasting from t - 2 seconds to t - 1,000 s~conds (T 10+5 to 10- 1 g/cm3), during which primordial element formation occurred. Before this period, there were so many high-energy protons around that they could blast apart any atomic nucleus (e.g., deuterium or tritium or He3 or He4) the moment it formed; after this period, the protons were too cold (had kinetic energies too low) to penetrate each others' coulomb barriers, and all the freely penetrating neutrons from the earlier, hotter stage had decayed into electrons plus protons. Only during the short, crucial period from t - 2 seconds to t - 1,000 seconds were conditions right for making elements. Calculations by Gamow (1948), by Alpher and Hermann (l948a, b; 1950), by Fermi and Turkevitch (1950), by Peebles (1966), and by Wagoner, Fowler, and Hoyle (1967) reveal that about 25 per", " cent of the baryons in the universe should have been converted into He4 (alpha particles) during this period, and about 75 per cent should have been left as HI (protons). Traces of deuterium, He3, and Li should have also been created, but essentially no heavy elements. All the heavy elements observed today must have been made later, in stars [see, e.g., Fowler (1967) or Clayton (1968)]. Current astronomical studies of the abundances of the elements give some support for these predictions; but the observational data are not yet very conclusive [see, e.g., Danziger (1968) and pp. 268-275 of Peebles (1971)]. After primordial element formation, the matter and radiation continued to interact thermally through frequent ionization and recombination of atoms, keeping each other at the same temperature. Were the temperatures of radiation and matter not locked together, the radiation would cool more slowly than the matter (for adiabatic expansion, 1;. ex: l/a, but Tm ex: 1/a 2 ). Thus thermal equilibrium was maintained only by a constant transfer of energy from radiation to matter. But the heat capacity of the radiation was far greater than that of the matter. Therefore the energy transfer had a negligible effect on Pr, Pr' and Tr. It held up the temperature of the matter (Tm = Tr) without significantly lowering the temperature of the radiation. On the other hand, the total mass-energy of matter was and is dominated by rest mass. Therefore the energy transfer had negligible influence on Pm' [This circumstance justifies the approximation of ignoring energy transfer when passing from equation (27.31) to (27.32).] When the falling temperature reached a few thousand degrees (a/ao - 10-3, P - 10-20 g/cm3, t - 105 years), two things ofinterest happened: the universe ceased to be radiation-dominated and became matter-dominated [Pm = pmo(ao/a)3 came to exceed Pr = Pro(ao/a)4]; and the photons ceased to be energetic enough to keep (4) primordial element formation (2 sec. ~ t ~ 1,000 sec.) (5) thermal interaction of matter and radiation (1,000 sec. < t < 105 years) - - (6) plasma recombination and transition to matter dominance (t - 105 yrs.) 766 28. EVOLUTION OF THE UNIVERSE INTO ITS PRESENT STATE hydrogen atoms ionized, so the electrons and protons quickly recombined. That these two events were roughly coincident is a result of the specific, nearly conserved value that the entropy per baryon has in our universe: s = entropy per - b aryon ~ (number of photons in universe) (number of baryons in universe) 108 ~. Why the universe began with this value of s, rather than some other value (e.g. unity), nobody has been able to explain. Recombination of the plasma at t ~ 105 years was crucial, because it brought an end to the interaction and thermal equilibrium between radiation and matter (\"de coupling\"). Thereafter, with very few free electrons off which to scatter, and with Rayleigh scattering off atoms and molecules unimportant, the photons propagated almost freely through space. Unless energy-releasing processes reionized the inter galactic medium sometime between a/ao ~ 0.1, the photons have been propagating freely ever since the plasma recombined. Even if reionization occurred, the photons have been propagating freely at least since a/ao ~ 0.1. ~ 10-3 and a/ao The expansion of the universe has redshifted the temperature of the freely propa gating photons in accordance with the equation T ex: l/a (see Box 29.2). As a consequence, today they have a black-body spectrum with a temperature of 2.7 K. They are identified with the cosmic microwave radiation that was discovered in 1965, and they give one direct information about the nature of the universe at the time ~ 10-3, t ~ 105 years if reionization did not they last interacted with matter (a/ao occur; a/ao ~ 0.1, t ~ 5 X 108 years if reionization did occur.) Return to the history of matter. Before plasma recombination, the photon pressure (\"elasticity of the cosmological fluid\") prevented the uniform matter (25 per cent He4, 75 per cent H) from condensing into stars, galaxies, or clusters of galaxies. However, after recombination, the photon pressure was gone, and condensation could begin. Small perturbations in the matter density, perhaps dating back to the begin ning of expansion, then began to grow larger and larger. Somewhere between ~ 1/10 (10 8 years :s t :s 109 years) these perturbations began a/ao developing into stars", ", galaxies, and clusters of galaxies. Slightly later, at a/ao ~ 1/4, quasars probably \"turned on,\" emitting light which astronomers now receive at Earth (see Box 28.1). ~ 1/30 and a/ao (7) subsequent propagation of photons (t 2: 105 yrs.) (8) condensation of stars. galaxies and clusters (10 8 yrs. ~ t ~ 109 yrs.) EXERCISE Exercise 28.1. UNCERTAINTY IN EVOLUTION Current observations, plus the assumption of complete homogeneity and isotropy at the beginning of expansion, plus the assumption that the excess of leptons over antileptons is less than or of the order of the excess of baryons over antibaryons, place the following limits on the cosmological parameters today: Matter density today == Pmo' between 10-28 and 2 X 10-31 g/cm3 k == 0 or + I or - I; temperature of electromagnetic radiation today == 2.7 \u00b1 0.1 K. Total radiation density [observed photons, plus neutrinos and gravitons that presumably originated in big bang in thermal equilibrium with photons] == Pro' between 0.7 X 10-33 and 1.2 X 10-33 g/cm3 \u2022 ; (continued on page 769) \u00a72B.1. THE \"STANDARD MODEL\" OF THE UNIVERSE 767 Box 28.1 EVOLUTION OF THE QUASAR POPULATION That the large-scale, average properties of the universe are changing markedly with time one can infer from quasar data. In brief, there appear to have been about 50 times more quasars in the universe at a redshift z ::::::: 2 than at z ::::::: 0.5; and there may well have been fewer, or none, at redshifts z > 3. (On the use of redshift to characterize time since the big bang, see Box 29.3.) In greater detail, Schmidt (1972) gives the following analysis of the data:* 1. Schmidt assumes from the outset that quasar redshifts are cosmological in origin [redshift = (Hubble constant) X (distance); \u00a729.2]. The evidence for this is a. Observational: Some quasars are located in clusters of galaxies [as evidenced both by position on sky and by quasar having same redshift as galaxies in cluster; see Gunn (1971 )]. Since the evidence for the cosmological distance redshift relation for galaxies is overwhelming (Boxes 29.4 and 29.5), the redshifts of these particular quasars must be cosmological. b. Theoretical: Observed quasar redshifts of z ~ 1 to 3 cannot be gravitational in origin; objects with gravitational redshifts larger than z ::::::: 0.5 are unstable against collapse (see Chapters 24 and 26 and Box 25.9). Nor are the quasar redshifts likely to be Doppler; how could so massive an object be accelerated to v::::::: 1 without complete disruption? The only remaining possibility is a cosmological redshift. For this reason, opponents of the cosmological hypothe sis usually feel pressed to invoke in the quasars a breakdown of the laws of physics as one~understands them today. [See, e.g., Arp (1971) and references cited therein. These references also describe evidence against the cosmological assumption, evidence that a few prominent investigators find compelling, but that most do not as of 1972.] 2. Schmidt then asks how many quasars, N, there were in the universe at a time corresponding to the redshift z, and with absolute luminosity per unit frequency, L.(2,500 A) at the wavelength 2500 A as measured in the quasar's local Lorentz frame. 3. The data on quasars available in 1972 are not at all sufficient to determine N[z, L.(2,500 A)] uniquely. But they are sufficient to show unequivocally that: a. There must have been evolution; N(z, L.) cannot be independent of z. b. The evolution cannot have resided primarily in the luminosities: the total number of quasars, Ntoiz ) L N(z, Lv) L,(2,500A) must have changed markedly with time (with z). \u2022 Our version of Schmidt's (1972) argument is oversimplified. The reader interested in greater precision should consult his original paper. Box 28.1 (continued) C. If the evolution was primarily in the total number, Ntot(z), Le., if the changes in the relative luminosity distribution at 2,500 A were negligible, and if the universe today is characterized by ao = qo = 1 (see Chapter 29 for notation), then the data show This steep increase in number as", " one goes backward in time-and all other basic features of the observed quasar redshift and magnitude distributions for z ~ 2-can be fit in a universe with ao = qo = 1 by either of the evolution laws Ntot(z) ex: (l + Z)6, Ntot[z(t)] ex: 105<10 -!l/Io\u2022 Here to is the current age of the universe and t was the age at redshift z. d. These evolution laws, when extrapolated beyond a redshift z :::::: 2 and when combined with the observed relative luminosity function!(z, Lv) for quasars near apparent magnitude 18, predict that an observer on Earth should see the following fractions of nineteenth and twentieth-magnitude quasars to have redshifts greater than 2.5: evolution law fraction with z > 2.5 m = 20 m = 19 (I + z)6 I OS(to- tilt. 29% 12% 49% 14% In 1972 about 30 quasars fainter than m = 18.5 are known, and of these only 1 (3%) has z > 2.5. This shows, in Schmidt's words, \"that the density law (l + Z)6 cannot persist beyond a redshift of around 2.5.\" Schmidt regards the 105<1 0 -!l/Io law (which becomes nearly constant at z> 2.5) to be also in apparent conflict with the observations, but he says that \"further spectroscopic work on faint quasars is needed to confirm this suspicion.\" One reason for caution is the difficult problem ofremoving \"observational selection effects\" from the data. Schmidt, Sandage, and others have independently searched for selection effects that might produce an artificial apparent decrease in the number of quasars at z > 2.5. None have been found. In the words of Sandage (l972d) \"The apparent cutoff in quasar redshifts near z = 2.8 [has been] examined for selection effects that could produce it artificially. If the cutoff is real, it may be the time of the birth of the first quasars, although the suggested redshift is unexpectedly small. At z = 3 in a qo = 1 universe, the look-back time is 89 per cent of the Friedmann age. Assessment of the 0 bservational selection effects shows that none are positively established that could produce the cutoff artificially.\" \u00a728.3. WHAT \"PRECEDED\" THE INITIAL SINGULARITY? 769 (The uncertainties taken into account in Pro are uncertainty about whether quadrupole moments at early times were sufficient to create gravitons at the full level corresponding to thermal equilibrium, and uncertainty about the number and statistical weights of particle species in equilibrium at the time gravitons decoupled.) Use the equations in \u00a727.10 to calculate the uncertainties in the evolutionary history (Figure 28.1) caused by these uncer tainties in the present state of the universe. \u00a728.2. STANDARD MODEL MODIFIED FOR PRIMORDIAL CHAOS The standard hot big-bang model is remarkably powerful and accords well with observations (primordial helium abundances; existence, temperature, and isotropy of cosmic microwave radiation; homogeneity and isotropy of universe in the large; close accord between age of universe as measured by expansion and ages of oldest stars;...). However, in 1972 it encounters apparent difficulty with one item: the origin of galaxies. In a universe that is initially homogeneous and isotropic it is not clear that random fluctuations will give rise (after plasma recombination) to pertur bations in the density of matter of sufficient amplitude to condense into galaxies. The perturbations that eventually form galaxies might have to reside in the initial, exploding state of the universe. [See Zel'dovich and Novikov (1974) for detailed review and discussion; see also references cited in \u00a730.1.] Is it reasonable to assume a small amount of initial inhomogeneity? Is it not much more reasonable to assume either perfect homogeneity (one extreme) or perfect chaos (the other extreme)? Thus, if perfect initiarliomogeneity turns out to be incompatible with the origin of galaxies, it is attractive to try \"perfect initial chaos\"-Le., completely random initial conditions, with a full spectrum of fluctuations in density, entropy; and local expansion rate [Misner (1968, 1969b)]. It is conceivable, but far from proved, that during its subsequent evolution such a model universe will quickly smooth itself out by natural processes (Chapter 30) such as \"Mixmaster oscillations,\" neutrino induced viscosity [see, e.g;, Matzner and Misner (1971)], and gravitational curva ture-induced creation of particle pairs [Zel'dovich (1972)]. Will one be left, after a", " few seconds or less, with a nearly homogeneous and isotropic, Friedmann uniyerse, containingjust enough remaining perturbations to condense eventually into galaxies? Theoretical calculations have not yet been carried far enough to give a clear answer. Of course, after the initial chaos subsides, if it subsides, such a model universe will evolve in accord with the standard big-bang model of the last section. \u00a728.3. WHAT \"PRECEDED\" THE INITIAL SINGULARITY? No problem of cosmology digs more deeply into the foundations of physics than the question ofwhat \"preceded\" the \"initial state\" ofinfinite (or near infinite) density, pressure, and temperature. And, unfortunately, no problem is farther from solution in 1973. What if the universe began chaotic? \u00b7 \u00b7 \u00b7. ~ :;-\" r \u00b7 \u00b7 \u00b7 \u00b7. \u00b7 \u00b7 The initial singularity and quantum gravitational effects Cosmologies that violate general relativity 770 28. EVOLUTION OF THE UNIVERSE INTO ITS PRESENT STATE General relativity predicts, inexorably, that even if the \"initial state\" was chaotic rather than smooth, it must have involved a spacetime \"singularity\" of some sort [see Hawking and Ellis (1968); also \u00a734.6 of this book]. And general relativity is incapable of projecting backward through the singularity to say what \"preceded\" it. Perhaps only by coming to grip with quantum gravitational effects (marriage of quantum theory with classical geometrodynamics) will one ever reach a clear under standing of the initial state and of what, if anything, \"preceded\" it [see Misner (1969c), Wheeler (197Ic)]. For further discussion of these deep issues, see \u00a7\u00a734.6, 43.4, the final section of Box 30.1, and Chapter 44. \u00a728.4. OTHER COSMOLOGICAL THEORIES This book confines attention to the cosmology of general relativity. If one were to abandon general relativity, one would have a much wider set of possibilities, includ ing (1) the steady-state theory [Hoyle (1948); Bondi and Gold (1948)], which has not succeeded in accounting for the cosmic microwave radiation or in explaining observed evolutionary effects in radio sources and quasars [Box 28.1]; (2) the Klein Alfven \"hierarchic cosmology\" of matter in an asymptotically flat spacetime [Alfven and Klein (1962), Alfven (1971), Klein (1971), Moritz (1969), de Vaucouleurs (1971)], which disagrees with cosmic-ray and gamma-ray observations [Steigman (1971 )]; and the Brans-Dicke cosmologies [Dicke (1968), Greenstein (1968a,b), Morganstern (1973)], which are qualitatively the same and quantitatively almost the same as the standard hot big-bang model. However, no motivation or justification is evident for abandoning general relativity. The experimental basis of general relativity has been strengthened substantially in the past decade (Chapters 38-40); and the standard big-bang model of the universe predicted by general relativity accords remarkably well with observations-far better than any other model ever proposed! CHAPTER 29 PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE \u00a729.1. PARAMETERS THAT DETERMINE THE FATE OF THE UNIVERSE Will the universe continue to expand forever; or will it slow to a halt, reverse into contraction, and implode back to a state ofinfinite (or near infinite) density, pressure, temperature, and curvature? The answer is not yet known for certain. To discover the answer is one of the central problems of cosmology today. The only known way to discover the answer is to measure, observationally, the present state of the univ~rse; and then to calculate the future evolution using Einstein's field equations. The field equations have already been solved in \u00a7\u00a727.l0 and 27.11. From those solutions one reads off the following correlation between the present state of the universe and its future. If A = 0 [in accord with Einstein's firmly held principle of simplicity]: Expansion forever ~ negative or zero spatial curvature for hypersurfaces of homogeneity, i.e., k/ao 2 ~ 0 (\"open\" or \"flat\"); This chapter is entirely Track 2. Chapter 27 (idealized cosmological models) is needed as preparation for it. but this chapter is not needed as prepa rat ion for any later chapter. '\" Expansion forever vs. recontraction of universe Recontraction ~ positive spatial curvature for homogeneous hypersurfaces, i.e., k/ao 2 > 0 (\"closed\"). If A \u00a5= 0:. ExpansIOn forever ~ A > ACrlt Recontraction ~ A < Ac", "rit'- {O 4 ( 7TPmoao )- 3 2'-, if k < 0 If k > 0; Evidently three parameters are required to predict the future: the cosmological constant, A; the curvature parameter today for the hypersurface of homogeneity, k/ ao2; and the density of matter today, Pmo' (To extrapolate into the past, as was done in the last chapter, one needs, besides these quantities, the radiation density today, Pro' But Pro is so small now and is getting smaller so fast (Pr ex: a-4 ; Pm ex: a-3 ) that it can have no influence on the decision between the possibilities just listed. Parameters required to predict future of universe: (1) \"relativity parameters\" \u2022 Pmo A. k/Bo 2 (2) \"observational parameters\" Ho \u2022 qo' ao 772 29. PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE The task of predicting the future, then, reduces to the task of measuring the \"rela tivity parameters\" A, k/ao2, and Pmo' In tackling this task, observational cosmologists prefer to replace the three \"rela tivity parameters,\" which have immediate significance for relativity theory, by parameters that are more directly observable. One parameter close to the observa tions is the Hubble expansion rate today, i.e., the \"Hubble constant,\" (29.1 a) Another is the dimensionless \"deceleration parameter\" today, qo' defined by qo - --.J.!. H 2 = a a I \u00b0 _(aa,~t). \u00b0 a,t And a third is the dimensionless \"density parameter,\" today, _ 4'ITpmo ao = 3H2. o (29.1 b) (29.1 c) (3) relationship between relativity parameters and observational parameters The relationships between these three \"observational parameters\" and the three 2, and Pmo (together making six \"cosmological param \"relativity parameters\" A, k/ao eters\") can be calculated by combining definitions (29.1) with the Einstein field equations (27.39), which, evaluated today, say 2 k Ho = - a o A 8'17' 2 +T+TPmo, (29.2) By combining these equations, one finds the relationships shown in Box 29.1, where the implications of several values of ao and qo are also shown. EXERCISE Exercise 29.1. Derive the results quoted in Box 29.1. IMPLICATIONS OF PARAMETER VALUES \u00a729.2. COSMOLOGICAL REDSHIFT Observed features of cosmological redshift One of the key pieces of observational data used in measurements of Ho' qo' and ao is the cosmological redshift: spectral lines emitted by galaxies far from Earth and received at Earth are found to be shifted in wavelength toward the red. For example, the [011];\\3727 line, when both emitted and observed in an Earth-bound laboratory, has a wavelength of 3727 A. However, when it is emitted by a star in the galaxy \u00a729.1. PARAMETERS THAT DETERMINE FATE OF UNIVERSE 773 Box 29.1 OBSERVATIONAL PARAMETERS COMPARED TO RELATIVITY PARAMETERS A. Relativity Parameters B. Observational Parameters 1. Matter density today, 1. Hubble constant (Hubble expansion rate 2. Curvature ofhypersurface of homogeneity Pmo today, k/ao 3. Cosmological constant, 2 A today), Ho (a,tla)o 2. Deceleration parameter, _ a tt 1 qo = --;- H2 \u00b0 3. Density parameter, 4. Radiation density today, Pro (unimportant for the present dynamics of the universe, and therefore ignored in this chapter) _ 4'11pm<> ao = 3H2 \u00b0 C. Observational Parameters in Terms of Relativity Parameters Ho2 = (8'17/3)pmo - k/ao 2 + A/3, (4'17/3 )pmo - A/3 D. Relativity Parameters in Terms of Observational Parameters k/ao Pmo = (3/4'11)Ho2ao' 2 = Ho2(3ao - qo - 1), A = 3Ho2(ao - qo)' (1) (2) (3) (4) (5) (6) E. Implications of Specific Parameter Values 1. A = 0 (in accord with Einstein's point of view) if and only if ao = qo' 2. Sign of A is same as sign of ao - qo' 774 29. PRESENT", " STATE AND FUTURE EVOLUTION OF THE UN/VERSE Box 29.1 (continued) 3. If A = 0 (a) qo > ; ~ Pmo > Perit S3'TTHoZ ~ k > 0 positive curvature;) \"closed\" universe ( ~ universe will eventually recontract; (b) qo = \"2 ~ Pmo- Perit = S'TT zero curvature;) \"flat\" universe ( ===> universe will expand forever; (C) qo < ; ~ Pmo < Perit ;'11 Hoz ~ k < 0 negative curvature;) \"open\" universe ( ===> universe will expand forever. 4. If A \u00a5= 0 (a) ao > ; (qo + I) ~ k > 0 positive curvature;) \"closed\" universe'( and in this case, I( Z ao - --3- ~ universe will expand forever, ao - qo ~ a o ao - qo < a:Z(ao - qo : 1r~ universe will eventually recontract; qo+I)3 zero curvature;) \"flat\" universe'( and in this case, ao ~ qo ~ universe will expand forever, ao < qo ~ universe will eventually recontract; negative curvature;) \"open\" universe'( and in this case, ao ~ qo ~ universe will expand forever, ao < qo ~ universe will eventually recontract. \u00a729.2. COSMOLOGICAL REDSHIFT 775 3C 295 (presumably with the same wavelength, Aem = 3727 A) and received at Earth, it is measured here to have the wavelength Arec = 5447 A. The fractional change in wavelength is Z (Arec - Aem)jAem = 0.4614 for 3C 295. (29.3) The cosmological redshift is observed to affect all spectral lines alike, and not only lines in the visible spectrum. Thus, the 21-cm line of hydrogen, with 400,000 times the wavelength of the central region of the visible, undergoes a redshift that agrees (within the errors of the measurements) with the redshifts oflines in the visible for recession velocities of the order of v - 0.005, according to observation of thirty objects by Dieter, Epstein, Lilley, and Roberts (1962) and further observations by Roberts (1965). No one has ever put forward a satisfactory explanation for the cosmological redshift other than the expansion of the universe (see below). The idea has been proposed at various times by various authors that some new process is at work (\"tired light\") in which photons interact with atoms or electrons on their way from source to receptor, and thereby lose bits and pieces of their energy. Va. B. Zel'dovich (1963) gives a penetrating analysis of the difficulties with any such ideas: (I) \"If the energy loss is caused by an interaction with the intergalactic matter, it is accompanied by a transfer of momentum; that is, there is a change of the direction of motion of the photon. There would then be a smearing out of images; a distant star would be seen as a disc, not a point, and that is not what is observed.\" (2) \"Let us suppose that the photon decays, y - - y' + k, giving up a small part of its energy to some particle, k. It follows from the conservation laws that k must move in the direction of the photon (this, by the way, avoids a smearing out), and must have zero rest mass. Because of the statistical nature of the process, however, some photons would lose more energy than others, and there would be a spectral broadening of the lines, which is also not observed.\" (3) If there does exist any such decay process, then simple arguments of special relativity that Zel'dovich attributes to M. P. Bronshtein, and spells out in detail, demand the relationship Why redshift cannot be due to \"tired light\" probability per) second of \"photon decay\" ( (a uni:ersal. constant With) = _t_h_e_d_I_m_e_n_s_1O_n_s_s_e_c_-_2 _ (frequency of photon in sec-I) \"Thus,\" Zel'dovich concludes, \"if the decay of photons is possible at all, those in radio waves must decay especially rapidly! This would mean that the Maxwell equation for a static electric field would have to be changed.... There is no experimental indication of such effects: the radio-frequency radiation from distant sources is transmitted to us not a bit more poorly than visible light, and the red shift measured in different parts of the spectrum is exactly the same.... Thus, suggestions", " that there is an explanation of the red shift other than Friedmann's fail completely.\" 776 29. PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE Emission: atom excites n-node standing wave: universe small. a(te) = aem : ~ wavelengths small. A(t.) = Aem. Reception: universe larger. a(t,) = arec : wavelengths larger. A(t,) = Arec ; number of nodes in standing wave unchanged; n = constant = - - - = - 217arec 2\".Arec aem Aem Figure 29.1. Redshift as an effect of standing waves. The ratio of wavelengths, AreclAem' is identical with the ratio of dimensions, a,eclaem in any closed spherically symmetrical (Friedmann) model universe. The atom excites an n-node standing wave in the universe. The number n stays constant during the expansion. Therefore wavelengths increase in the same proportion as the dimensions of the universe. One sees immediately in this way that the redshift is independent of all such details as (I) why the expansion came about (spherical symmetry, but arbitrary equation of state); (2) the rate-uniform or nonuni form-at which it came about; and (3) the distance between source and receptor at emission, at reception, or at any time in-between. The reasoning in the diagram appears to depend on the closure of the universe (standing waves; k = + I rather than 0 or -I). That closure is not required for this simple result is seen from the further analysis given in the text. Not the least among the considerations that lead one to accept the general recession of the galaxies as the explanation for the redshift is the circumstance that this general recession was predicted [Friedmann (1922)] before the redshift was observed [Hubble (1929)]. The cosmological redshift is easily understood (Figure 29.1) in terms of the standard big-bang model for the universe. A detailed analysis focuses attention on three processes: emission of the light, propagation of the light through curved spacetime from emitter to receiver, and reception of the light. Emission and reception I occur in the proper reference frames (orthonormal tetrads) of the emitter and they are special-relativistic phenomena. Propagation, by contrast, is a receiver; general-relativistic process; it is governed by the law of geodesic motion in curved spacetime. In calculating all three processes-emission, propagation, and absorption-one Derivation of red shift formula: A ex (expanSion) factor \u00a729.2. COSMOLOGICAL REDSHIFT 777 needs a coordinate system. Use the coordinates (t, X, B, cp) or (1j, x, B, cp) introduced in Chapter 27; and orient the space coordinates in such a way that the paths of the light rays through the coordinate system are simple. This is best done by putting the origin of the coordinate system (X = 0) at the Earth. Then the emitting galaxy will lie at some \"radius\" Xe and some angular position (Be' CPe). The cosmological line element ds 2 = -dt 2 + a2(t)[dX 2 + 2 2(dB2 + sin2Bdcp2)] = a2(1j)[ _d1j2 + dX 2 + 2 2(dB2 + sin2Bdcp2)], 2 = X ( sin X ~fk=+l, If k = 0, sinhX ifk= -1, (29.4a) (29.4b) is spherically symmetric about X = 0 (i.e., about the Earth) whether k = -I, 0, or + 1. Consequently, the geodesics (photon world lines) that pass through both Earth and the emitting galaxy must all be radial X = X(t). (29.5) (One who wishes to forego any appeal to symmetry can examine the geodesic if dB/d\"A = equation in the (t, X, B, cp) coordinate system, and discover dcp/d\"A = 0, then d2B/d\"A2 = d2cp/d\"A2 = O. Consequently a geodesic that is initially radial will always remain radial.) that Consider, now, emission. A galaxy at rest (moving with the \"cosmological fluid\") at (Xe, Be' CPe) emits two successive crests, A and B, of a wave train toward Earth at coordinate times tAe and tBe. It has been arranged that proper time as measured on the galaxy is the same as coordinate time (t = 'T + const. was part of the con struction process for the", " coordinate system in \u00a727.4). Consequently the period of the radiation as seen by the emitter is Pem = teB teA; and the wavelength is the same as the period when geometrized units are used: (29.6) Next examine propagation. Wave crests A and B propagate along null geodesics. This fact enables one to read the world lines of the wave crests, XA (t) and XB(t), directly from the line element (29.4): ds 2 = 0 guarantees that aCt) dX = -dt (-, not +, because the light propagates toward the Earth at X = 0). Consequently, the world lines are Xe - XA(t or 1j) = 1j - Xe - XB(t or 1j) = 1j - t 1jeA = f a-I dt, 1jeB = f a-I dt. t e\" t tes (29.7) Finally, examine reception. The receiver on Earth moves with the \"cosmological fluid,\" just as does the distant emitter. (Ignore the Earth's \"peculiar motion\" relative 778 29. PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE to the fluid-motion around the sun, motion around center of our Galaxy, etc.; it can be taken into account by an ordinary Doppler correction.) Thus, for receiver as for emitter, proper time is the same as coordinate time, and (29.8) where trB and trA are the times of reception of the successive wave crests. Now combine equations (29.6), (29.7), and (29.8) to obtain the redshift. The receiver is at X = O. Therefore equations (29.7) say Subtract these equations from each other to obtain and combine with (29.6) and (29.8) to discover I.e., (29.9) (29.10) (29.11) These redshift equations confirm the simple result of Figure 29.1: As the light ray propagates, its wavelength (as measured by observers moving with the \"fluid\") increases in direct proportion to the linear expansion of the universe. The ratio of the wavelength to the expansion factor, AIa, remains constant. For important applica tions of this result, see Boxes 29.2 and 29.3. EXERCISES Exercise 29.2. ALTERNATIVE DERIVATION OF REDSHIFT Notice that the only part of the line element that is relevant for the light ray is since dO = dcf> = 0 along its world line (spherical symmetry!). Regard the light ray as made....---------~ Box 29.2 COSMOLOGICAL REDSHIFT OF THE PRIMORDIAL RADIATION As an important application of the redshift for mula in the \"fluid,\" the temperature of the primordial radiation is redshifted A/a = constant (1) [equation (29.10)], consider the radiation emerging from the hoLbi-& bang. Because it is initially in thermal equilibriumWfth matter, this primordial radiation initially has a Planck black-body spec trum. Subsequent interactions with matter cannot change the spectrum, because the matter remains in thermal equilibrium with the radiation so long as interactions are occurring. The cosmological redshift can and does change the spectrum, how ever. It was shown in exercise 22.17, using kinetic theory, that radiation with a Planck spectrum as viewed by one observer has a Planck spectrum as viewed by all observers; but the observed temper ature is redshifted in precisely the same manner as the frequency of an individual photon is red shifted. Consequently, as seen by observers at rest T a: l/a. (2) This is true after plasma recombination, when the radiation and matter are decoupled, as well as before recombination, when they are interacting. And it is true not only for the primordial photons but also for thermalized neutrinos and gravitons emerging from the hot big bang. There is another way to derive the redshift equation (2). Combine the equation (3) for the energy density of black-body radiation in terms of temperature, with the equation Pr a: (volume)-4/3 a: (a 3)-4/3 a: a-4 (4) for the decrease of energy density with adiabatic expansion. Box 29.3 USE OFREDSHIFT TO CHARACTERIZE DISTANCES AND TIME Distance: When discussing objects within the Earth's cluster of galaxies, astronomers typically describe distances in units of lightyears or parsecs. But when dealing with more distant objects (gal axies, quasars, etc.), astronomers find it more con venient to describe distance in terms of what is actually observed: the", " statement \"the galaxy 3C 295 is at a redshift of 0.4614\" means that \"3C 295 is at that distance from Earth [given by equation (29.16)] which cor responds to a redshift of z = 0.4614.\" redshift. For example, Time: When discussing events that occurred dur ing the last few 109 years, astronomers usually measure time in units of years. Example: \"The solar system condensed out of interstellar gas 4.6 X 109 years ago\" [see Wasserburg and Burnett (1968)]. But when dealing with events much nearer the beginning of the universe, all of which have essentially the same age, of about 12 X 109 years, astronomers find it more convenient to describe time in terms of redshift. Example: \"The primor dial plasma recombined at a redshift of 1,000\" means that \"If a photon had been emitted at the time of plasma recombination, and had propa gated freely ever since, it would have experienced a total redshift between then and now of z = 1,000.\" Equivalently, since 1 + z = (ao/a) [see equation (29.11 )], \"the plasma recombined when the universe was a factor of 1 + z ;:::; 1,000 smaller than it is today.\" [Application: In Figure 28.1, where the past evolution of the universe is sum marized, one can freely replace the horizontal scale a/ao by 1/(1 + z), and thereby see that pri mordial element formation occurred at a redshift of z ;:::; 109.) The conversion from redshift units to time units is strongly dependent on the parameters 2 [see \u00a7\u00a727.10 and 27.11; also Pmo' Pro' and k/ao equation (29.15)]. 780 29. PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE of photons with 4-momenta p. From the geodesic equation (or, for the reader who has studied chapter 25, from arguments about Killing vectors), show that is conserved along the photon's world line. Use this fact, the fact that a photon's 4-momentum is null, p. p = 0, and the equation E = -p' u for the energy measured by an observer with 4-velocity u, to derive the redshift equation (29.11). _ - - Exercise 29.3. REDSHIFT OF PARTICLE DE BROGLIE WAVELENGTHS A particle of finite rest mass JL moves along a geodesic world line through the expanding cosmological fluid. Let _ ( = p' P = (I - V 2)1I2 )112 _ JLv P be the spatial 4-momentum of the particle as measured by observers at rest in the fluid. (The ordinary velocity they measure in their proper reference frames is v.) The associated \"de Broglie wavelength\" of the particle is A=hlp. (a) Show that this de Broglie wavelength is redshifted in precisely the same manner as a photon wavelength: AIa = constant. (b) Employing this result, show that, for the molecules of an ideal gas that fills the universe, their mean kinetic energy decreases in inverse proportion to a2 when the gas is nonrelativistic and (like photon energies) in inverse proportion to a when the gas is highly relativistic. \u00a729.3. THE DISTANCE-REDSHIFT RELATION; MEASUREMENT OF THE HUBBLE CONSTANT Equation (29.11) expresses the redshift in terms of the change in expansion factor between the event of emission and the event of reception. For \"nearby\" emitters (emitters at distances much less than 1/Ho' the \"Hubble length\") it is more conven ient to express the redshift in terms of the distance between the emitter and Earth. That distance (\"present distance\") is defined on the hypersurface of homogeneity that passes through Earth today, since that hypersurface agrees locally with the surface of simultaneity of the receiver today, and it is also, locally, a surface of simultaneity for any observer moving today with the \"cosmological fluid.\" The distance between emitter and observer today [the distance along the spatial geodesic of constant (t, (), cf\u00bb connecting (t.., 0, ()e' cf>e) and (t.., Xe, ()e' cf>e)] can be read directly from the line element (29.4): Using expression (29.9) for Xe, one finds (29.12) (29.12') Derivation of distance-redshift relation \u00a729.3. DISTANCE\u00b7REDSHIFT RELATION; MEASUREMENT OF H. 781 In the recent past, a(t) was given by a(t) =", " a(tr) + (a,t)tr(t - tr) + ~ (a,tt)tr(t - tr)2 + = a(tr)[1 + Ho(t - tr) - I '2qoHo 2(t - tr)2 +.. '], (29.13) where definitions (29.1) for the Hubble constant Ho and the deceleration parameter % have been used. Putting this expression into equation (29.12') and integrating, one finds for the distance the expression or, equivalently, 1 = (t r t r - - t) + -H (t 0 e I 2 - r t )2 +... e I t = 1 - -H 12 +.,. 2 e 0 \u2022 (29.14) The redshift [equation (29.11)] can be expressed as a power series in tr equation (29.13): te by using a(tr) - a(te) a(te) z=Ho(tr - a(tr)[Ho(tr - ---.:.~~e.,:-_~~~~....:.,.:-_~__--.:. - a(tr)[1 - Ho(tr - te) +... ] te) + ~qoHo2(tr - te)2 +... ] te) + Ho2( I + ~ qo)(tr - te)2 +.... (29.15) Combining this with equ~tion (29.14) for tr - te in terms of 1, one finally obtains (29.16) Result for distance-redshift relation This is the \"distance-redshift relation\" for the standard big-bang model of the uni verse. By comparing this distance-redshift relation with astronomical observations (see Measurement of Hubble Box 29.4, which is best read after the next section), Allan Sandage (1972a) obtains a Hubble constant of constant H. i.e., HO-l = (I8 -+- 2) X 109 years. (29.17) (29.18) one Megaparsec is 3.26 X 106 light years, or 3.08 X 1024 em.) The (Note: I Mpc uncertainty of -+-7 km sec-1 Mpc- 1 quoted here is the \"one-sigma\" statistical uncer tainty associated with the distance-redshift data. Systematic errors, not now under stood, might be somewhat larger; but the true value of Ho almost certainly is within a factor 2 of Sandage's value, 55 km sec- 1 Mpc- 1. 782 29. PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE Note that, if A = 0, then the \"critical density\" marking the dividing line between a \"closed\" universe and an \"open\" universe-i.e., between eventual recontraction and expansion forever-is Value of critical density 3 erit - -8 o P rr - HZ - 5 X 10-30 g/ cm3. (29.19) (As described in Box 29.1, P > Perit -<==>- \"closed\" -<==>- recontraction; P < Perit -<==> \"open\" -<==>- expansion forever.) Comparison with the actual density will be de layed until \u00a729.6. The distance measurements are not accurate enough to yield useful information about the deceleration parameter, %. \u00a729.4. THE MAGNITUDE-REDSHIFT RELATION; MEASUREMENT OF THE DECELERATION PARAMETER Apparent magnitude defined In astronomy one defines the apparent (bolometric) magnitude, m, of an object Information about % is best obtained by comparing the apparent magnitudes of galaxies with their redshifts. by the formula m = -2.5 loglO(S/2.52 X 10-5 erg cm-z sec-I) = -2.5 10glOS + constant, (29.20) where S is the flux of energy (energy per unit time per unit area) that arrives at Earth from the object. [Of course, one cannot measure the flux over the entire wavelength range 0 < A < 00; so one distinguishes various apparent magnitudes (m u, mB' m v,...) corresponding to fluxes in various wavelength ranges (\"U\" \"ultraviolet\"; \"B\" from gravitation physics to be treated here.] \"visual\"). However, these subtleties are too far \"blue\"; \"V\" Derivation of magnitude-redshift relation Calculate the apparent magnitude for a galaxy of intrinsic luminosity Land redshift z. To simplify the calculation, put the emitter at the origin of the space coordinates (Xe = 0); and put the Earth at (Xr, (),., CPr). (Note the reversal oflocations compared", " to redshift calculation of \u00a729.2.) On Earth, place a photographic\u00b7 plate of area A perpendicular to the incoming light. Then at time I r the plate is a tiny segment of a spherical two-dimensional surface (I = Ir, X = Xr ; () and cP vary) about the emitting galaxy. The total area of the 2-sphere surrounding the galaxy is (29.21) Therefore, the ratio of the area of the plate to the area of the 2-sphere is given by A A - 4rr[a(/rP:(xr)j2\u00b7 A (29.22) The plate catches a fraction A/A of the energy that pours out through the 2-sphere. \u00a729.4. MAGNITUDE-REDSHIFT RELATION; MEASUREMENT OF q. 783 If there were no redshift, the power crossing the entire 2-sphere at time tT would be precisely the luminosity of the emitter at time teo However, the redshift modifies this result in two ways. (1) The energy of each photon that crosses the 2-sphere is smaller, as measured in the local Lorentz frame of the fluid there, than it was as measured by the emitter: (29.23) (2) Two photons with the same () and ep, which are separated by a time.:itT as measured by an observer stationary with respect to the \"cosmological fluid\" at the 2-sphere, were separated by a shorter time.:ite as measured by the emitter: (29.24) The luminosity, L, as measured at the source, is the sum of the energies EemJ of the individual photons (labeled with the index J) emitted in a time interval.:ite, divided by.:ite: L = (1/.:ite) 2: EemJ\" J (29.25) The power that crosses the 2-sphere a time t T at the 2-sphere, is te later, as measured by the fluid P = (1/.:itT) 2: ETecJo J (29.26) where the summation runs over the same set of photons. Combining equations (29.23) to (29.26), one sees that the power crossing the 2-sphere is Of this, a fraction, P = L/(1 + z)2. crosses the photographic plate; so the flux measured at the Earth is where R is the \"radius of curvature\" of the 2-sphere surrounding the emitter and passing through the receiver at the time of reception, (29.27) R a02(XT - Xe) = aO[~T - xel ao sinh (XT - Xe ) [ ao Sin (XT - Xe ) if k = -I, if k = 0, if k = + 1 (29.28) 784 29. PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE [recall: Xe is 0 according to the present conventions, and ao = a(tT)]' The correspond ing apparent magnitude [equation (29.20)] is m = +5 10glO[(1 + z)R] - 2.5 10gloL + constant. (29.29) In order to relate the apparent magnitude to the redshift of the emitter, one must express the quantity R in terms of z. From equation (29.7) for the photon propagation (with sign reversed because positions of receiver and emitter have been reversed), one knows that _ XT Xe - _ It, t, -1 _ dt - a Ia<t)/a<t,) [_a ][~] [a(tT)] 1 d d () a tT a a, (29.30) and from equation (29.11) one knows that Hence _ =I l Xe 1 XT Z + [_a][~] d[a(tT)]. da a(tT) a (29.31) Equations (4) to (6) of Box 29.1, and (27.40), determine the function dt/da in terms of a/a(tT) and the constants Ho' %' ao. By inserting that result into equation (29.31) - Xe in terms of the redshift z and the cosmological and integrating, one obtains XT parameters Ho' %' ao : The 2-sphere radius of curvature R is obtained by inserting this expression into the equation (29.32b) [equation (29.28), with ao evaluated by equation (5) of Box 29.1]. Equations (29.29) and (29.32) determine the apparent magnitude, m, in terms of redshift,", " z. For the case of vanishing cosmological constant (ao = qo; A = 0), the integral (29.32a) can be expressed in terms of elementary functions, yielding B-1 R = 2(10 qo + z :::::: Ho- IZ [ I - so that ) [-qo + I + qoz + (qo - 1)(2qoz + 1)1/2], ~ (1 + qo)z + O(z2)J. (29.33) Result for magnitude-redshift relation m = 5 10glO [1 - qo + qoz + (qo - 1)(2qoz + 1)1/2] - 2.510glO L + const. :::::: 5 10glO z + 1.086(1 - qo)z + O(z2) - 2.5 10glO L + const. (29.34) for z ~ 1. \u00a729.4. MAGNITUDE-REDSHIFT RELATION; MEASUREMENT OF q. 785 (Note: the factor 1.086 is actually 2.5/In 10.) A power-series solution for nonzero A (for (Jo y!:. qo) reveals a dependence on (Jo only at O(z2) and higher: R;:::; Ho-1z[ 1 - ~ (1 + qo)z + (corrections of O(z2) depending on (Jo and qo)J. (29.35 a) tn ;:::; 5 10glO z + 1.086(1 - qo)z + O(z2) - 2.5 10glO L + const. (29.35b) Sheldon (1971) gives the exact solution for A y!:. 0 in terms of the Weierstrass elliptic function. Refsdal et al. (1967) tabulate and plot the exact solution. By comparing the theoretical magnitude-redshift relation (29.35b) with observa tions of the brightest galaxies in 82 clusters, Allan Sandage (1972a,c,d) obtains the following value for the deceleration parameter: % = 1.0 -+- 0.5, if (Jo = %(i.e. A = 0). (29.36) (Note: 0.5 is the \"one-sigma\" uncertainty. Sandage estimates with 68 per cent con fidence that 0.5 < qo < 1.5, and with 95 per cent confidence that 0 < qo < 2 providing unknown evolutionary effects are negligible.) The observations leading to this result and the uncertainties due to evolutionary effects are described in Box 29.4. Box 29.5 gives a glimpse of Edwin Hubble, the man who laid the foundations for such cosmological measurements. (continued on page 794) Measurement of deceleration parameter, q. Box 29.4 MEASUREMENT OF HUBBLE CONSTANT AND DECELERATION PARAMETER 1. Hubble Constant, Ho A. Objective: To measure the constant Ho by comparing observational data with the distance-redshift relation Here.\u00a3 is distance from Earth to source today; and z is redshift of source as measured at Earth. B. Key Difficulty: This distance-redshift relation does not apply to stars in our Galaxy: the Galaxy is gravitationally bound and therefore is impervious to the universal expansion. Nor does the distance-redshift relation apply to the separations between our Galaxy and nearby gahxies (the \"local group\"); gravitational attraction between our Galaxy and its neighbors is so great it perturbs their motions substantially away from universal expansion. Only on 786 29. PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE Box 29.4 (continued) - scales large enough to include many galaxies (scales where each galaxy or cluster of galaxies can be thought of as a \"grain of dust,\" with the grains distributed roughly homogeneously)-only on such large scales should the distance-redshift relation hold with good accuracy. But it is very difficult to obtain reliable measurements of the distances.\u00a3 to galaxies that are so far away! C. Procedure by which Ho has been measured [Sandage and Tamman, as summar ized in Sandage (l972a)1: 1. Cepheid variables are pulsating stars with pulsation periods (as measured by oscillations in light output) that are very closely correlated with their luminosities L-or, equivalently, with their absolute (bolometric) magni tudes, M: M= (apparent magnitude star would have were it at a) distance of 10 parsecs = 32.6 light years = -2.5 10glO (L/3.0 X 1035 erg sec-I) (", "1) [see equation (29.20).] By measurements within our Galaxy, astronomers have obtained the \"period-luminosity relation\" for cepheid variables. 2. Cepheid variables are clearly visible in galaxies as far away as -4 Mpc 4 X 106 parsecs). In each such galaxy one measures the (4 Megaparsecs periods of the cepheids; one then infers their absolute magnitudes M from the period-luminosity relation; one measures their apparent magnitudes m; and one then calculates their distances.\u00a3 from Earth using the relation m - M = 5 10glO (1/10 pc). (2) By this means one obtains the distances.\u00a3 to all galaxies within -4 Mpc ofour own. Unfortunately, such galaxies are too close to participate cleanly in the universal expansion. (They include only the \"local group,\" the \"M81 group,\" and the \"south polar group.\") Thus, one must push the distance scale out still farther before attempting to measure H o' 3. Galaxies of types Sc, Sd, Sm, and Ir within -4 Mpc contain huge clouds of ionized hydrogen, which shine brightly in \"Ha light.\" These clouds, called \"H II regions,\" exhibit a very tight correlation between diameter D of the H II region and luminosity L of the galaxy (or, equivalently, between D and absolute magnitude of galaxy, M). In fact, for a given galaxy luminosity L, the fractional spread in H II diameters is a(,1D/D) ~ 0.12. Using (a) the distances (~4 Mpc) to these galaxies as determined via cepheid variables, (b) the apparent magnitudes of the galaxies, and (c) the angular diameters of H II regions in the galaxies, one calculates \u00a729.4. MAGNITUDE\u00b7REDSHIFT RELATION; MEASUREMENT OF q. 787 the actual H II diameters D and galaxy luminosities L, and thereby obtains the \"diameter-luminosity relation\" D(L). 4. H II regions are large enough to be seen clearly in galaxies as far away as -60 Mpc. By measuring the H II angular diameters a = D/l and galaxy apparent (bolometric) magnitudes - -251 ( m -. oglO 2.52 X 10- 5 erg cm-2 secl 2 L/4'fT1 )'(3) and by combining with the diameter-luminosity relation, one obtains the distances 1 to all galaxies of types Sc, Sd, Sm, and Ir which possess H II regions and lie within -60 Mpc of Earth. Unfortunately, this is still not far enough away for local motions to be negligible compared with the universal expansion. 5. Within -60 Mpc reside enough galaxies of type Sc I for one to discover that their luminosities (absolute magnitudes) are rather constant (difference in L from one Sc I galaxy to another :s 50 per cent). Using the distances to such Sc I galaxies, as measured via H II regions, and using measurements of their apparent magnitudes, one calculates their universal absolute magnitude (measured photographically) to be M pg = -21.2. The Sc I galaxy M 101 at a distance 1 - 3 Mpc from Earth, as photographed with the 200-inch telescope. (Courtesy of Hale Observatories) 788 29. PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE Box 29.4 (continued) 6. One then examines all known Sc I galaxies with distances greater than -70 Mpc. For each, one measures the apparent magnitude and compares it with the universal absolute magnitude to obtain the distance! from Earth. And for each, one measures the redshift z = L1A/A of the spectral lines. From the resulting redshift-distance relation-and taking into ac count the statistical uncertainties in all steps leading up to it-Sandage and Tamman (work carried out in 1965-1972) obtain the value Ho = dz/d! = 55 -+- 7 (km/sec) Mpc- l = 1/[(18 -+- 2) X 109 years]. [For a review see Sandage (1972a).] The quoted error is purely statistical. Sys tematic errors are surely larger-but they almost surely do not exceed a factor 2 [i.e., 30 < Ho < 110 (km/sec) Mpc- l]...,..~...,.~:\"../,....\"~:'..:. \"~\"'-t\u00b7. \u2022. \u2022 \u2022 \u2022 \u2022\u2022 \u2022 \u2022 -1.2 -1.6 -2.0 -2.4 -2.8 '\" o'i5 0...J 8 9 10 II 12 13 14 15 Magnitude-", "redshift relation for Sc I galaxies at distances ~ 70 Mpc. Solid line is a least-squares fit to the data; dOlled line has the theoretical slope of 5. [From Sandage and Tamman.) II. Deceleration Parameter, qo' A. Objective: To measure the constant qo by comparing observational data with the magnitude-redshift relation: m = 5 loglo Z + 1.086(1 - %)z + O(z2) - 2.5 10glO L + const. (4) \u00a729.4. MAGNITUDE-REDSHIFT RELATION; MEASUREMENT OF q. 789 [Note: This relation is valid even if the cosmological constant is nonzero, Le., even if ao y!:. qo. Dependence on ao occurs only at O(z2) and higher.] B. Key Difficulty: One must use data for objects with the same absolute lumin osity L (\"standard candles\"). But one cannot measure L at distances great enough for the effects of qo to show up. C. The Search for a Standard Candle: One obvious choice for the standard candle would be Sc I galaxies, since they were found to all have nearly the same L (see above). But they are not bright enough to be seen at distances great enough for effects of qo to show up. An alternative choice, quasars, are bright enough to be seen at very large redshifts (z as large as -3). But their absolute luminosities have enormous scatter-or so one infers from the failure of quasars to falI on a straight line, even at small z, in the magnitude-redshift diagram. The best choice is the brightest type of object that has small scatter in L. Sandage (1972a,b,c) chooses the brightest galaxy in \"recognized regular clusters of galaxies.\" Such clusters are composed predominately of E-type galaxies, and the brightest members are remarkably similar from one cluster The E-type galaxy M87 at a distance 1 II :-'Ipc from Earth. as photographed with the 200-inch telescope. (Courtesy of Hale Observatories) to another (scatter in L is -25 per cent). The similarity shows up in their spectra and in the very precise straight lines they give when one plots angular diameter versus redshift (next page), or apparent magnitude versus redshift (next page), or angular diameter versus apparent magnitude. Box 29.4 (continued) 3\" 10\" 30\" 100\" 300\" 8,o~ -0.5 -1.0 \" 0'0 -1.5 0....J -2.0 -2.5 -0.4 -0.8 \"e 0'0 0....J -1.2 -1.6 -2.0 -2.4 \u2022 \u2022 8 0,333 0.100 :: 0.033 0.010 0.003 Angular diameter versus redshift for brightest galaxy in recognized regular clusters. From Sandage (1972a,b). [These data are not suffi ciently precise to yield useful in formation about qo and \"0; but improvements in 1973 may bring the needed precision; see \u00a729.5.J ~,.)~\\. \u2022\u2022\u2022.0.\u2022 tIP...,\"~\" o \u2022.!!. \u2022 \u2022 \u2022 \u20220\u2022\u2022 \u2022 \u2022 o 200 inch \u2022 48 inch 0.4 1.2 2.0 Log 10 8 (sec) Brightest galaxy in 82 clusters Magnitude versus redshift for brightest galaxy in recognized regular clusters. V26 - Kv - Av is the apparent magni tude with certain corrections taken into account. The line plotted corresponds to \". = q. = I line of slope 5). From Sandage (1972a). (straight 10 12 14 16 18 20 \u00a729.4. MAGNITUDE-REDSHIFT RELATION; MEASUREMENT OF q. 791 D. Procedure by which % has been measured [Sandage (l972a,c)]: 1. Data on magnitude versus redshift have been gathered for the brightest galaxy in 82 recognized regular clusters (see above). 2. The data, when fitted with a straight line, show a slope of dm/dloglOz = 5.150 + 0268 (rms), by comparison with a theoretical slope of 5. 3. The data, when fitted to the theoretical relation m = 5 10glO Z + 1.086(1 - qo)z + O(z2) + const, (5) (6) [with the correct O(z2) and higher terms included; see equations (29.29), (29.32), and (29.34)] yield qo = 1 + 0.5 (one-sigma)} if a = (A = 0). (two-sigma) 0", " qo = 1 + 1 (7) The data are inadequate to determine a 0 and qo simultaneously. [The O(z2) terms, which depend on ao, playa significant role in the fit to the data. For a graphical depiction of their theoretical effects see Figure 2 of Refsdal et. al (1967).] E. Evolutionary uncertainties 1. Sandage's fit of data to theory assumes that the luminosities of his \"stand ard candles\" are constant in time. If, because of evolution of old stars and formation of new ones, his galaxies were to dim by 0.09 magnitudes per 109 years, then galaxies 109 light-years away, which one sees as they were 109 years ago, would be 0.09 magnitudes brighter intrinsically than identical nearby galaxies. Correction for this effect would lower the most probable value of qo from 1 to 0 [Sandage (1972c)]. 2. Knowledge of the evolution of galaxies in 1972 is too rudimentary to confirm or rule out such an effect. [See references cited by Sandage (1972<:).] 792 29. PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE Box 29.5 EDWIN POWELL HUBBLE November 20, 1889, Marshfield Missouri September 28, 1953, Pasadena, California astronomy education, and three years later (1917) completed a Ph.D. thesis on \"Photographic Inves tigations of Faint Nebulae.\" When Hubble entered astronomy, it was sus pected that some nebulae lie outside the Galaxy, but the evidence was exceedingly weak. During the subsequent two decades, Hubble, more than anyone else, was responsible for opening to man's purview the extragalactic universe. Working with the 60-inch and IOO-inch telescopes at Mount Wil son, Hubble developed irrefutable evidence of the extragalactic nature of spiral nebulae, elliptical nebulae, and irregular nebulae (now called gal axies). He devised the classification scheme for galaxies which is still in use today. He systematized the entire subject of extragalactic research: deter mining distance scales, luminosities, star densities, and the peculiar motion of our Galaxy; and ob taining extensive evidence that the laws of physics outside the Galaxy are die same as near Earth (in Hubble's words: \"verifying the principle of the uniformity of nature\"). He discovered and quan tified the large-scale homogeneity of the universe. And-his greatest triumph of all!-he discovered the expansion of the universe. The details of Hubble's pioneering work are best sketched in his own words: Edwin Hubble, at age 24, earned a law degree from Oxford University and began practicing law in Louisville, Kentucky. After a year of practice he became fed up and, in his own words, \"chucked the law for astronomy, and I knew that even if I were second-rate or third-rate it was astronomy that mattered.\" He chose the University of Chi cago and Yerkes Observatory as the site for his \"Extremely little is known of the nature of nebulae; and no classification has yet been suggested.... The agreement [between the velocity of escape from a spiral nebula and that from our galaxy] is such as to lend some color to the hypothesis that the spirals are stellar systems at distances to be measured often in millions of light years. \" (1920; Ph.D. THESIS; PUBLICATION DELAYED 3 YEARS BY WORLD WAR I) This box is based largely on the biography of Hubble by Mayall (1970). \u00a729.4. MAGNITUDE-REDSHIFT RELATION; MEASUREMENT OF qo 793 \"The present investigation [using Cepheid variables for the first time as an indicator of distances beyond the Magellanic clouds] identifies NGC 6822 as an isolated system of stars and nebulae of the same type as the Magellanic clouds, although somewhat smaller and much more distant. A consistent structure is thus reared on the foundation of the Cepheid criterion. in which the dimensions, luminosities, and densities, both of the system [NGC 6822] as a whole and its separate members, are of orders of magnitude which are thoroughly familiar. The distance is the only quantity of a new order. The principle of the uniformity of nature thus seems to rule undisturbed in this remote region of space... (1925) \"Critical tests made with the 100-inch reflector, the highest resolving power available, show no difference between the photographic images of the so-called condensations in Messier 33 and the images of ordinary galactic stars.... The period-luminosity relation is conspicuous among the thirty-five Cepheids and indicates a distance about 8. 1 times that of the Small Magellanic Cloud. Using the distance of the spiral is about 263,000 par", "secs. \" Shapley's value for the latter, (1926a) \"[To the present paper (1926b)] is prefaced a general classification of nebulae... the various types [of extragalactic nebulae] are homogeneously distributed over the sky.... The data are now available for deriving a value for the order of the density of space. This is accomplished by means of the formulae for the numbers of nebulae to a given limiting magnitude and for the distance in terms of [The result is] the magnitude. p = 1.5 X 10-31 grams per cubic centimeter. This must be considered as a lower limit, for loose material scattered between the systems is entirely ignored. The mean density of space can be used to determine the dimensions of the finite but boundless universe of general reletivity.. R = __c__1_ =.., = 2.7 X 1010 parsecs.. y4'1Tk vp (1926b) individually, \"The data... indicate a linear correlation between distances and velocities [for extragalactic nebulae]. Two solutions have been made, one using the 24 nebulae the other combining them into 9 groups according the proximity in direction and distance. The results are... 24 objects: K = 465 \u00b1 50 km/sec per 106 parsecs: 9 groups: K = 513 \u00b1 60 km/sec per 106 parsecs.... The outstanding feature, however, is the possibility that the velocity-distance relation may represent the de Sitter effect, and hence that numerical data may be introduced into discussions of the general curvature of space... (1929)* * Hubble's value of K (the \"Hubble constant\") was later revised downward by the work of Baade and Sandage; see section titled The Hubble Time in Box 27.1. ~ Box 29.5 (continued) \"The velocity-distance relation is re-examined with the aid of 40 new velocities.... The new data extend out to about eighteen times the distance available in the first formulation of the velocity-distance relation, but the form of the relation remains unchanged except for [Shapley's 10 per cent] revision of the unit of distance. \" (1931). WITH M. L. HUMASON \"Many ways of producing such effects [redshifts in extragalactic nebulae] are known. but of them all, only one will produce large redshifts without introducing other effects which should be conspicuous but actually are not found. This one known permissible explanation interprets redshifts as due to actual motion away from the observer. \" (1934a) \"We now have a hasty sketch of some of the general features of the observable region as a unit. The next step will be to follow the reconnaissance with a survey-to repeat carefully the explorations with an eye to accuracy and completeness. The program, with its emphasis on methods, will be a tedious series of successive approximations. \" (1934b) Most of the remainder of Hubble's career was dedicated to this \"tedious series of successive ap proximations.\" Shortly before Hubble's death the 200-inch telescope went into operation at Palomar Mountain; and Hubble's student, Alan Sandage, began using it in a continuation of Hubble's quest into the true nature of the universe. (See Box 29.4). EXERCISES Exercise 29.4. m(z) DERIVED USING STATISTICAL PHYSICS Derive the magnitude-redshift relation using a statistical description of the photon distribution [cf. eq. (22.49) and associated discussion]. Exercise 29.5. DOPPLER SHIFT VERSUS COSMOLOGICAL REDSHIFT (a) Consider, in flat spacetime, a galaxy moving away from the Earth with velocity v, and is received at Earth. Let the distance between Earth and galaxy, as emitting light that measured in the Earth's Lorentz frame at some specific moment of emission, be r; and let the Doppler shift of the radiation when it is eventually received be z = &'/A. Show that the flux of energy S received at the Earth is related to the galaxy's intrinsic luminosity L by (29.37) [Track-2 readers will find it most convenient to use the statistical formalism of equation (22.49).] (b) Compare this formula for the flux with formula (29.27), where the redshift is of cosmological origin. Why is the number of factors of 1 + z different for the two formulas? [Mathematical answer: equation (6.28a) of Ellis (1971).] \u00a729.5. SEARCH FOR \"LENS EFFECT'\" OF THE UNIVERSE 795 \u00a729.5. SEARCH FOR \"LENS EFFECT\" OF THE UNIVERSE Curved", " space should act as a lens of great focal length. The curving of light rays has little effect on the apparent size of nearby objects. However, distant galaxies galaxies from a quarter of the way up to halfway around the universe-are expected to have greatly magnified angular diameters [Klauder, Wakano, Wheeler, and Willey (1959)]. To see a normal galaxy at such a distance by means of an optical telescope seems out of the question. However, radio telescopes resolve features in quasistellar sources and other radiogalaxies at redshifts of z = 2 or more. Moreover, paired radio telescopes at intercontinental distances (for example, Goldstone, California, and Woomera, Australia) resolve distant sources to better than 0\".001 or 4.8 X 10-9 radians or 15 lightyears for an object at a distance of 3 X 109 lightyears (Euclidean geometry temporarily being assumed). A radio telescope in space\u00b7 paired with a radio telescope on earth will be able to do even better on angular resolution. Will one be able to find any fiducial distance characteristic of anyone class of objects that will serve as a natural standard of length, for very great distances (z = 2 to z = 3) as well as for galaxies closer at hand? Perhaps not. However, it would seem unwise to discount this possibility, with all the advantages it would bring, in view of the demonstrated ability of skilled observers to find regularities elsewhere where one had no right to expect them in advance. Let L denote the actual length of a fiducial element (if any be found) in a galaxy; and let S() (radians!) denote the apparent length of the object, idealized as perpendi cular to the line of sight, as seen by the observer. The ratio of these two quantities defines the \"angle effective distance\" of the source, (angle effective distance) = raed = LIS(). (29.38a) In flat space and for objects with zero relative velocity, this distance is to be identified with the actual distance, r, to the source or with the actual time of flight, t, of light from source to observer. The situation is changed in an expanding universe. To calculate the angle effective distance as a function of redshift, place the Earth (receiver) at Xr = 0; and place the object under study (emitter) at Xe\u2022 Let the fiducial length L lie on the sphere at Xe (perpendicular to line of sight), and let it run from ()e to ()e + S() [one end of fiducial element at (Xe, ()e' cfJe); other at (Xe, ()e + 88, cfJe)]. Then and The hope for a fiducial in distant objects length Angle effective distance defined i.e. [see equation (29.28), with Xr and Xe reversed], r aed = RI(I + z). (29.38b) Angle effective distance as function of redshift Here R is given as a function of redshift of source, z, and cosmological parameters Ho' %' ao, by equations (29.32) in general, or by (29.33) if A = O. [Equation (29.38b) is modified if the beam preferentially traverses regions oflow mass density (\"vacuum between the galaxies\"); see equation (22.37) and Gunn (1967).] y Angle effective distance as a tool for determining whether universe is closed Measurements of mean mass density of universe: (1) luminous matter in galaxies 796 29. PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE -< '-. -< \"'J II '\" 2.0 1.8 1.6 1.4 1.2 1.0.8.6.4.2 0 0 2 4 6 8 rAED' 109 light years Figure 29.2. Angle effective distance versus redshift for twO typical cosmo logical models-one open (0 < ao = qo ~ I); the other closed (ao = qo = I); both with zero cosmological constant; both with Ho-l = 18 X 1091yr. Figure 29.2 shows angle effective distance as a function of redshift for a few selected choices of the relevant parameters. It is evident that the angle effective distance has a maximum for a redshift roughly of the order z -- I, provided that the universe is closed. However, there is a big difference if the universe is open (Figure 29.2). The rapid improvements taking place in radio astronomy make increasingly attractive the possibility it provides for testing whether the universe is closed, as Einstein argued it should be [Einstein (1950), pp. 107-108]. Moreover, even with optical telescopes, in 1973 one may be on the verge of measuring qo by studies of angle effective distance", ": preliminary studies [Sandage (1972b)] suggest that the optical size of the brightest E-type galaxies may be a usable fiducial length. \u00a729.6. DENSITY OF THE UNIVERSE TODAY It is exceedingly difficult to measure the mean density Pmo of the universe today. A large amount of matter may be in forms that astronomers have not yet managed to observe (intergalactic matter, black holes, etc.). Therefore, the best one can do is to add up all the luminous matter in galaxies and regard the resulting number as a lower limit on Pmo'Even adding up the luminous matter is a difficult and risky task, so difficult that even today no analysis is more definitive than the classic work of Oort (1958). [See, however, the very detailed review of the problem in Chapter 4 of Peebles (1971)]. Oort's result is (29.39) corresponding to (independent of the value of Ho)' (29.40) \u00a729.7. PRESENT KNOWLEDGE OF COSMOLOGICAL PARAMETERS 797 As an example (albeit an atypical one) of the danger inherent in any such estimate, Oort points to the Virgo cluster of galaxies. If the Virgo cluster is not gravitationally bound, then its -2,500 galaxies will go flying apart, destroying any semblance of a cluster, in about one billion years. If it is gravitationally bound, then the mean velocity of its galaxies relative to each other, when combined with the virial theorem, yields an estimate of the cluster's total mass. That estimate is 25 times larger than the value one gets by Oort's method of adding up the luminous mass of the cluster. Although one has no definitive evidence for or against large amounts of matter (enough to close the universe) in intergalactic space, one has tentative indirect limits: (1) If A =\u00b0(in accord with Einstein), then (10 = qo; so Sandage's value of qo ~ I-stretched to qo < 10 under the most wild of assumptions about galaxy evolu tion-implies Pig < 10-28 g/cm3 (2) Gott and Gunn (1971) point out that, if the density of gas in intergalactic space were? 10-30 g/cm3 (i.e., if (10 were?0.1), one would expect gas falling into the Coma cluster of galaxies to form a shock wave, which would emit large amounts of X-rays. From the current X-ray observations, one can place a limit on the amount of such infalling matter-and therefrom a limit p. < 10-30 g/cm3 'g - on the density of gas in intergalactic space. But these limits, like others obtained in other ways [see Chapter 4 of Peebles (1971) for a review] are far from definitive; they depend too much on theoretical calculations to make one feel fully comfortable. (2) matter in intergalactic space \u00a729.7. SUMMARY OF PRESENT KNOWLEDGE ABOUT COSMOLOGICAL PARAMETERS The best data available in 1973 [equations (29.18), (29.36), (29.40)] reveal Summary of observational parameters of universe Ho-1 = (18 -+- 2) X 109 years, qo = 1 -+- 0.5 (one-sigma) if (10 = %(A = 0), (29.41) (10? 0.02, for the observational parameters of the universe. These numbers are inadequate to reveal whether the universe is closed or open, and whether it will continue to expand forever or will eventually slow to a halt and recontract. If one is disappointed in this lack of knowledge, one can at least be consoled by Some quantitative triumphs the following. (1) There is excellent agreement between theory and observation for the linear (low-z) parts of the distance-redshift, magnitude-redshift, and angular diameter-redshift relations (Box 29.4). (2) There is remarkably good agreement between (a) the age of the universe (18 billion years if qo = (10 ~ 1; 12 billion years if qo = (10 = ~) as calculated from the measured value of Ho; (b) the ages of the of cosmology The bright prospects for observational cosmology EXERCISES 798 29. PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE oldest stars (-10 X 109 years) as calculated by comparing the theory of stellar evolution with the properties of the observed stars; (C) the time (-9 billion years) since nucleosynthesis of the uranium, thorium, and plutonium atoms that one finds on Earth, as calculated from the measured relative abundances of various nucleides; and (d) the ages (4.6 billion years) of the oldest meteorites", " and oldest lunar rock samples, as calculated from measured relative abundances of other nucleides. For further detail see, e.g., Sandage (1968, 1970), Wasserburg et al. (1969), Wasserburg and Burnett (1968), and Fowler (1972). (3) Observations of the cosmic microwave radiation and measurements of helium abundance are now capable of giving direct information about physical processes in the universe at redshifts z ~ 1 (Chapter 28). (4) One may yet find \"fiducial lengths\" in radio sources, visible out to z ~ 1, with which to measure qo and (10 by the angle-effective-distance method (\u00a729.5). (5) The enigmas of the nature of quasars and of their peculiar distribution with redshift (great congregation at z - 2; absence at z ~ 3) may yet be cracked and may yield, in the process, much new information about the origin of structure in the universe (Box 28.1). (6) The next decade may well bring as many great observational surprises, and corresponding new insights, as has the last decade. Exercise 29.6. SOURCE COUNTS Suppose that one could find (which one cannot) a family of light or radio sources that (1) are all identical with intrinsic luminosities L, (2) are distributed uniformly throughout the universe, and (3) are born at the same rate as they die so that the number in a unit comoving coordinate volume is forever fixed. (a) Show that the number of such sources N(z) with redshifts less than z, as observed from Earth today, would be N(z) = (constant)\u00b7 Z3[ I - t(l + qo): + 0(Z2)l (29.42) (b) Show that the number of sources N(S) with fluxes greater than S as observed at Earth today would be N(S) = (constant)\u00b7 ( (LH2)1/2 LH 2)3/2[ 4'1TS I - 3 4'1TS -.-L= z 2 + O(z 3) t (LH 2)] 4'1TS + 0 rfirst-order correction Lindependent of q. and (Jo (29.43) ] [Answer: See \u00a715.7 of Robertson and Noonan (1968).] Exercise 29.7. COSMIC-RAY DENSITY (Problem devised by Maarten Schmidt) Suppose the universe has contained the same number of galaxies indefinitely into the past. Suppose further that the cosmic rays in the universe were created in galaxies and that a negligible fraction of them have been degraded or lost since formation. Derive an expression for the average density of energy in cosmic rays in the universe today in terms of: (1) the number density of galaxies, No, today; and (2) the nonconstant rate, dEldz, at which the average galaxy created cosmic-ray energy during the past history of the universe. [At redshift z in range dz, the average galaxy liberates energy (dEldz) dz into cosmic rays.] \u00a729.7. PRESENT KNOWLEDGE OF COSMOLOGICAL PARAMETERS 799 Exercise 29.8. FRACTION OF SKY COVERED BY GALAXIES Assume that the redshifts of quasars are cosmological. Let the number of galaxies per unit physical volume in the universe today be No, and assume that no galaxies have been created or destroyed since a redshift of?. 7. Let D be the average angular diameter of a galaxy. Calculate the probability that the light from a quasar at redshift z, has passed through at least one intervening galaxy during its travel to Earth. [For a detailed discussion of this problem, see Wagoner (1967).] y CHAPTER 30 ANISOTROPIC AND INHOMOGENEOUS COSMOLOGIES '\" This chapter' is entirely Track 2. The main text requires no special preparation. althou9h Chapters 27-29 would be helpful. Box 30.1 contains more technical sections: ideal preparation for it would be Chapters 4.9-14.21. and 27-29. plus \u00a725.2; minimal preparation would be exercises 9.13. 9.14. and 25.2. Chapter 21 throu9h \u00a721.8. and \u00a7\u00a727.8. 27.11. and 29.2. Chapter 30 is not needed as preparation for any later chapter. Motivation for studying inhomogeneous and anisotropic cosmologies: Why is universe so uniform? \u00a730.1. WHY IS THE UNIVERSE SO HOMOGENEOUS AND ISOTROPIC? The last three chapters studied the Friedmann cosmological models and the relatively satisfactory picture they give of the universe and its evolution. This chapter describes", " less simplified cosmological models, and uses them to begin answering the question, \"Why are the very simple Friedmann models satisfactory?\" This question is intended to probe more deeply than the first, obvious answer-namely, that the models are satisfactory because they do not contradict observations. Accepting the agreement with observations, we want to understand why the laws of physics should demand (rather than merely permit) a universe that is homogeneous and isotropic to high accuracy on large scales. Because this question cannot be answered definitively in 1972, many readers will prefer to omit this chapter on the first reading and return to it only after they have surveyed the major results in other areas such as black holes (Chapter 33), gravitational waves (Chapters 35-37), and solar-system experi ments (Chapter 40). The approach described here to the question \"Why is the universe so highly symmetric?\" is to ask Einstein's equations to describe what would have happened if the universe had started out highly irregular. The first step in this approach is to ask what would have happened if the universe had started a little bit irregular. This problem can be tackled by analyzing small perturbations away from the high symmetry of the Friedmann models. Such an analysis is most fruitful in its discussion of the beginnings of galaxy formation, and \u00a730.2. KASNER MODEL FOR AN ANISOTROPIC UNIVERSE 801 in its ability to relate small upper limits on the present-day anisotropy bf the microwave background radiation to limits on density and temperature irregularities that might have existed ten billion years ago, when the radiation was emitted. These studies are described so well in the book by Zel'dovich and Novikov (1974) [see also Field (1973), Peebles (1969), Peebles and Yu (1970), Jones and Peebles (1972), and references cited therein] that we omit them here.. Another approach is to allow large deviations from the symmetry of the Friedmann universes, but to put the asymmetries into only a few degrees of freedom. \u00a730.2. THE KASNER MODEL FOR AN ANISOTROPIC UNIVERSE The prototype for cosmological models with great asymmetry in a few degrees of freedom is the Kasner (192Ia) metric, Kasner metric: an example of an anisotropic model universe (30.1 ) which was first studied as a cosmological model by Schiicking and Heckmann (1958). In this metric the Pi are constants satisfying (30.2) Each t = constant hypersurface of this cosmological model is a flat three-dimensional space. The world lines of constant x, y, z are timelike geodesics along which galaxies or other matter, treated as test particles, can be imagined to move. This model represents an expanding universe, since the volume element v=g=pg=t is constantly increasing. But it is an anisotropically expanding universe. The separa tion between two standard (constant x,y, z) observers is tPI Ax if only their x-coor dinates differ. Thus, distances parallel to the x-axis expand at one rate, 1. 1 ex: tP1, while those along the y-axis can expand at a different rate, 1.2 ex: tP2 \u2022 Most remarkable perhaps is the fact that along one of the axes distances contract rather than expand. This contraction shows up mathematically in the fact that equations (30.2) require one of the p's, say PI' to be nonpositive: I-\"3 SP1 SO. (30.3) As a consequence, in a universe of this sort, if black-body radiation were emitted at one time t and never subsequently scattered, later observers would see blue shifts near one pair of antIpodes on the sky and red shifts in most other directions. In terms of this example, the fundamental cosmological question is why the Friedmann metrics should be a more accurate approximation to the real universe than is this Kasner metric. Kasner model with matter becomes isotropic in \"old age\" Anisotropy energy Adiabatic cooling of anisotropy 802 30. ANISOTROPIC AND INHOMOGENEOUS COSMOLOGIES \u00a730.3. ADIABATIC COOLING OF ANISOTROPY In seeking an answer, ask a question. Ask, in particular, what would become of a universe that starts out near t = 0 with a form described by the Kasner metric oL equation (30.1). This metric is an exact solution of the vacuum Einstein equation G = O. It approximates a situation where the matter terms in the Einstein equations are negligible by comparison with typical non-zero components of the Riemann tensor. Schiicking and Heckmann (1958) give solutions with matter included as a pressureless fluid. In this situation,", " the curvature of empty spacetime dominates both the geometry and the expansion rate at early times, t -+ 0; but after some characteristic time tm the matter terms become more important, and the metric reduces asymptotically to the homogeneous, isotropic model with k = O. This example illustrates the possibility that the universe might achieve a measure of isotropy and homogeneity in old age, even if it were born in a highly irregular state. Whether the symmetry of our universe can be explained along these lines is not yet clear in 1972. The model universe just mentioned is only a hint, especially since the critical parameter tm can be given any value whatsoever. The standard Einstein general-relativity physics of this model can be described in other language (Misner, 1968) by ascribing to the anisotropic motions of empty spacetime an \"effective energy density\" Paniso' which enters the Goo component of the Einstein equation on an equal footing with the matter-energy density, and thereby helps to account for the expansion of the universe: (30.4) The anisotropy energy density is found to have an equation of state while P matter ex: (3)g-Y/2 - (volume)-Y. For pressureless matter y = 1; for a radiation fluid y = 4/3; for a nonrelativistic ideal gas y = 5/3). This arrangement of the Einstein equation allows one to think of the anisotropy motions as being adiabatically cooled by the expansion of the universe, just as the thermal motions of an ideal gas would be. Since the adiabatic index for homogeneous anisotropy is y = 2, the anisotropy will be the dominant source of \"effective energy\" in a highly compressed state, whereas the matter will dominate in an expanded state. \u00a730.4. VISCOUS DISSIPATION OF ANISTROPY The model universe sketched above can be further elaborated by introducing dissi pative mechanisms that convert anisotropy energy into thermal energy. Suppose that \u00a730.5. PARTICLE CREATION IN AN ANISOTROPIC UNIVERSE 803 such an anisotropic universe were filled at one time with thermal radiation. If the radiation were collisionless or nearly so, the quanta moving parallel to the contracting x-axis would get blueshifted and would develop an energy distribution corresponding to a high temperature. The quanta moving parallel to the other (expanding) axes would be redshifted to an energy distribution corresponding to a low temperature. Any collisions taking place between these two systems of particles would introduce a \"thermal contact\" between them, and would transfer energy from the hot system to the cold one, with a corresponding large production of entropy. This provides an irreversible dissipative process, which decreases Paniso and increases Pradiation relative to the values they would have had under conditions of adiabatic expansion. [For further details, see, e.g., Matzner and Misner (1972).] It is possible that both the adiabatic cooling of anisotropy and the dissipation of anisotropy by its action on a gas of almost collisionless quanta have played significant roles in the evolution of our universe. In particular, neutrinos above 1010 K may have undergone sufficient p-e scattering to have provided strong dissipation during the first few seconds of the life of the universe. \u00a730.5. PARTICLE CREATION IN AN ANISOTROPIC UNIVERSE Adiabatic cooling and viscous dissipation might not be the chief destroyers of anisotropy in an expanding universe. More powerful still might be another highly dissipative process, which might occur at still earlier times, very near the initial \"singularity.\" This is a process of particle creation which was first treated by DeWitt Creation of particles by (1953), then explored by-Parker (1966 and 1969) for isotropic cosmologies and finally anisotropy of expansion by Zel'dovich (1970) in the present context of anisotropic cosmologies. In this process one again turns to the Kasner metric for the simplest example, but now quantummechanical considerations enter the picture. One realizes that not only would real quanta propagating in different directions be subject to red shifts and blue shifts, but that virtual quanta must be considered as well. Vacuum fluctuations (zero-point oscillations) entail a certain minimum number of virtual quanta, which are subject to the redshifting and blueshifting action of the strong gravitational fields. Virtual quanta that are blueshifted sufficiently violently can materialize as real particles, thanks to their energy gain. In this context \"sufficiently violently\" means not adiabatically. In an adiabatic expansion, the number of particles does not change, although the energy of each one does. This adiabatic limit is just the", " geometric-optics approxima tion to wave equations, which was discussed in \u00a722.5. There one saw that, if spacetime were not flat on the scale of a wavelength, then the wave equation could not be replaced by a particle description with conserved particle numbers. Thus, the adia batic limit (geometric-optics approximation) is violated in the conditions of high curvature near the singularity at the beginning of the universe. By studying wave equations in the Kasner background metric, Zel'dovich and Starobinsky (1971) find quantitatively the consequences of the failure of the adia- 804 30. ANISOTROPIC AND INHOMOGENEOUS COSMOLOGIES batic approximation near the singularity. Classically, the amplitudes of waves at frequencies comparable to the Hubble constant for any given epoch increase faster than a simple blue-shift calculation would imply (amplification through parametric resonance). Quantum-mechanically, the same amplification, applied to zero-point oscillations, leads to the creation of particle-antiparticle pairs. The calculations indi cate that this effect is very strong at the characteristic time tq = VGfi/c5 ~ 10-43 sec. (All calculations performed thus far are inadequate when the effect becomes strong, thus for t ~ tq ). For the creation of massless particles, it is essential that an anisotropically ex panding universe be postulated (except for scalar particles, for which particle creation occurs already in the Friedmann universe, unless the particle satisfies the conformal invariant wave equation). The isotropic Friedmann universes are all conformally flat, so that solutions of the wave equation for a field of zero rest mass can be given in terms ofsolutions for flat-space wave equations where there is no particle creation. There is some particle creation even in the isotropic Friedmann universe when the particle has finite rest mass and low energy. However, the particle-creation process normally uses anisotropy energy as the energy supply that it converts into radiation energy. The pioneering work by Parker and Zel'dovich suggests that one should study in detail cosmological models in which the initial conditions are a singularity, and in which quantum effects near the time t = tq dissipate all anisotropies and simulta neously give rise to the matter content of the model. This program of research, which is in its infancy, seems to require extrapolating laws of physics down to the very natural looking but preposterously small dimension VGfi/c 5 ~ 10-43 sec, or equiv alently VGfi/c 3 - 10-33 cm, Anisotropy might have created the matter content of our universe, damping itself out in the process Inhomogeneous cosmological models: (1) with spherical symmetry (2) with (rather symmetric) gravitational waves (3) near a singularity, with few or no symmetries \u00a730.6. INHOMOGENEOUS COSMOLOGIES The model universes considered above were all homogeneous although anisotropic. It is also crucial to study inhomogeneous cosmological models, in which the metric has a nontrivial dependence on the space coordinates. One class of such models is spherically symmetric universes, where the matter density, expansion rate, and all other locally measurable physical quantities have spherical symmetry about some preferred origin. Models of this sort were first considered by Lemaitre (1933a,b), Tolman (1934b), and Datt (1938), and were also treated by Bondi in 1947. These models provide a means for studying density perturbations of large amplitude. A recent tool is making it possible to study large-amplitude, spatially varying curvature perturbations of other symmetries; this tool is the Gowdy (1971, 1973) metrics. These metrics, which are exact solutions of the Einstein equations, represent closed universes with various topologies (5 3,5 1 X 52, T3) containing gravitational waves. The wave form in these solutions is essentially arbitrary, but all the waves propagate along a single preferred direction and have a common polarization. A rather different approach to understanding the behavior of inhomogeneous and anisotropic solutions of the Einstein equations has been developed by Khalatnikov, \u00a730.7. THE MIXMASTER UNIVERSE 805 Lifshitz, and their colleagues. Rather than truncate the Einstein theory by limiting attention to specialized situations where exact solutions can be obtained, they have sought to study the widest possible class of solutions, but to describe their behavior only in the immediate neighborhood of the singularity. These studies give a greatly enhanced significance to some of the exact solutions, by showing that phenomena found in them are in fact typical of much broader classes of solutions. Thus, in the first large class of solutions studied [Lifshitz and Khalatnikov (1963)], it", " was found that near the singularity solutions containing matter showed no features not already found in the vacuum solutions. Furthermore, space derivatives in the Einstein equations became negligible near the singularity in these solutions, with the consequence that a metric of the Kasner form [equation (30.1)] described the local behavior of spacetime near the singularity, but with a different set ofPi values possible at each point of the singular hypersurface. Subsequently, broadened studies of solutions near a singularity [Belinsky and Khalatnikov (1970)] showed that the mixmaster universe [Misner (1969b); Belinsky, Khalatnikov, and Lifshitz (1970)] is a still better homogeneous prototype for singularity behavior than the Kasner metric. \u00a730.7. THE MIXMASTER UNIVERSE The simplest example of a mixmaster universe is described in Box 30.1. It shows how, near the singularity, the Kasner exponents Pi can become functions of time. The result is most simply described in terms of the Khalatnikov-Lifshitz param eter u: PI = -u/(l + u + u 2 ----P2 = (1 + u)/(1 + u + u 2 P3 = u(1 + u)/(l + u + u 2 ), ), ). (30.5) As one extrapolates backward in time toward the singularity, one finds that the expansion rates in the three principal directions correspond to those of the Kasner metric of equation (30.1), with Pi values corresponding to some fixed u parameter. In these mixmaster models, however, the metric is not independent of the space coordinates (the spacelike hypersurfaces can, for instance, have the same 3-sphere topology as the closed Friedmann universes). The Kasner-like behavior at fixed u can persist through many decades of volume expansion before effects of the spatial derivatives of the metric come into play. The role then played by the space curvature is brief and decisive. The expansion is converted from a type corresponding to a parameter value u = Uo to a type corre sponding to the value u = - Uo (which is equivalent, under a relabeling of the axes, to the value u = Uo - 1). Extrapolating still farther back toward the singularity, one finds a previous period with u = Uo - 2. Throughout an entire sequence u = uo, ~ 1, nearly the entire volume expansion is due Uo to expansion in the 3-direction, whereas the 1- and 2-directions change very little, alternating at each step between expansion and contraction. Sufficiently far in the past, however, such a sequence leads to a value of u between 0 and 1. This value 1, Uo - 2, Uo - 3,..., with Uo Mixmaster universe: (1) \"anisotropy oscillations\" explained in terms of Kasner model. (2) as a prototype for generic behavior near singularities Are there any other generic types of behavior near singularities? 806 30. ANISOTROPIC AND INHOMOGENEOUS COSMOLOGIES can be interpreted as the starting point for another, similar sequence, through the transformation u -+ l/u, which interchanges the names of axes 2 and 3. The extrapolation of the universe's evolution back toward the singularity at t = 0 therefore shows an extraordinarily complex behavior, in which similar but not precisely identical sequences of behavior are repeated infinitely many times. In terms of a time variable which is approximately log (log I-i), these behaviors are quasi periodic. In the generic example to which the Khalatnikov-Lifshitz methods lead, one has a metric whose asymptotic behavior near the singularity is at each point of the singular hypersurface described by a mixmaster-type behavior, but with the principal axes of expansion changing their directions as well as their roles (as characterized by the u parameter) at each step, and with the mixmaster parameters spatially variable. [For more details see Belinsky, Lifshitz, and Khalatnikov (1971), and Ryan (1971, 1972).] It is not yet (1972) known whether there are important solutions or classes of solutions relevant to the cosmological problem, with asymptotic singularity behavior no! described by the Khalatnikov-Lifshitz generic case. The difficulty in reaching a definitive assessment here is that Khalatnikov and Lifshitz use essentially local methods, confined to a single coordinate patch, whereas the desired assessment poses an essentially global question. The global approaches (described in Chapter 34) have not, however, provided any comparable description of the nature of the singularity whose necessity they prove. One attempt to bridge these differences in technique and content is the work by Eardley, Liang, and Sachs (1972). (continued 0/1 page 815", " d\"A, (29) where t =\"A has been written to label the specific time-coordinate choice that equation (28) implies. Mixmaster Dynamics If matter terms with no additional degrees of free dom are included, the super-Hamiltonian in equa tion (29) is modified simply. For an example, choose TOO = - TOo = (3/4)2(p.e- 3a + Fe-4a ) for. the energy density of matter in a frame with time-axis eo = N-I(Olot). The two terms represent a nonrelativistic perfect fluid (p ex: V-I) and a radiation fluid (p ex: V-4/3), respectively, and lead to (30) 2X = -Pa2 + p+2 + p_2 + e + p.&a + Fe2a. 4a (V - 1) 31 ( ) This Hamiltonian, with its simple quadratic mo mentum dependence, differs in only two ways from the Hamiltonians of elementary mechanics, namely, (1) in the sign of the Pa 2 term and (2) in the detailed shape of the \"potential\" term as func tion of ex and f3\u00b1, the study of which reduces to a study of the function V(f3). Hamilton's equa f3\u00b1> Pa' and P\u00b1 in equation tions, from varying ex, (29), yield Thus the sign of the Pa2 term causes ex to accelerate toward (rather than away from) higher values of the \"potential\" terms e4a( V-I) + p.e3a + Fe2a. When IVI ~ 1 (small anisotropy), equation (33) is identical to its form in the isotropic Friedmann model, and allows a deceleration only when ex is large enough that the positive curvature term (- e4a ) dominates over matter (p.&a) and radiation (Fe2a). Near the singularity (ex -+ -00), the positive always negligible compared to radiation and matter. curvature term is For studies of the singularity behavior, is sufficient to study the simplified super-Hamilton ian it since the other terms obviously vanish for ex -+ - 00. This form retains only the V term in 3RIx = ie-2a(1 - V), which dominates when the cur vature of this closed universe becomes negative, V ~ 1. If the term in V(f3) were also negligible, then X = -Pa2 + p/ + p_ 2 would make each Pa' p\u00b1 constant, giving the Kasner behavior with and IdlJldexl2 = 1 as expected (since matter and curvature have been neglected). To proceed fur ther, a study of V(f3) is required, based on equa tions (19) and (8), and their immediate conse quence: V(f3) = ~ e-8f3 + - ~ e-2f3 + cosh 2 V3f3. (35) 2 + 1 + 3\" e4f3+(cosh 4 V3f3_ - 1). d 2f3\u00b1 oX - -2 = - -= - - e of3\u00b1 d\"A 1 4a aV 2 of3\u00b1 (32) One finds that V(f3) is a positive definite \"potential well\" which has the same symmetries as an equi- \u00a730.7. THE MIXMASTER UNIVERSE 811 lateral triangle in the /3+/3- plane. Near the origin, /3\u00b1 = 0, the equipotentials are circles, since \"Bounce\" Interrupts Kasner-like Steps Toward the Singularity V(f3) = 8(/3/ + /3_ 2) + 0(/33). (36) For large /3 values, one finds and /3+ -+ - 00, (37) /3+ -+ + 00, 1/3-1 ~ 1. (38) The dominant feature of the V(f3) potential is evidently its steep (exponential) triangular walls, with equation (37) representing the typical one for study. Under the influence of this potential wall, the evolution of this model universe is governed by the super-Hamiltonian 2X- -p(/ + p/ + p_2 + ~ e4(a- 2fJ). (39) These two asymptotic forms, together with the triangular symmetry, give a complete asymptotic description of V(f3), as sketched in the figure, where on successive levels separated by jj/3 = 1, the potential V increases by a factor of e8 = 3 X 103\u2022 If a -+ - 00 with d/3+lda > 1/2", " X 1027 oscillations of a hyperfine transition in neutral Cesium. But again the critical early ticks of the clock (needed to locate the singularity in time by actual physical events) are missing. The time needed for stellar nucleo synthesis to produce the first Cesium disqualifies this clock on historical grounds, and the still earlier high temperatures nearer the singularity would have ionized all Cesium even if this element had predated stars. Thus proper time near the singularity is not a direct counting of simple and actual physical an elaborate mathematical phenomena, but extrapolation. Each actual clock has its \"ticks\" discounted by a suitable factor-3 X 10 7 seconds per orbit from the Earth-sun system, 1.1 X 10-10 seconds per oscillation for the Cesium transition, etc. Since no single clock (because of its finite size and strength) is conceivable all the way back to the singularity, a statement about the proper time since the singularity involves the\u00b7 concept of an infinite sequence of successively smaller and sturdier clocks with their ticks then discounted and added. \"Finite proper time,\" then, need not imply that any finite sequence of events was possible. It may describe -a necessarily infinite number of events (\"ticks\") in any physically conceivable his tory, converted by mathematics into a finite sum by the action of a non-local convergence factor, the \"discount\" applied to convert \"ticks\" into \"proper time.\" Here one has the conceptual inverse of Zeno's paradox. One rejects Zeno's suggestion that a sin gle swing ofa pendulum is infinitely complicated being composed of a half period, plus a quarter period, plus 2-n ad infinitum-because the terms in his infinite series are mathematical abstractions, not physically achieved discrete acts in a drama that must be played out. By a comparable stand ard, one should ignore as a mathematical abstrac tion the finite sum of the proper-time series for the age of the universe, if it can be proved that there must be an infinite number of discrete acts played out during its past history. In both cases, finiteness would be judged by counting the num ber of discrete ticks on realizable clocks, not by assessing the weight of unrealizable mathematical abstractions. Whether the universe is infinitely old by this standard remains to be determined. The quantum influences, in particular, remain to be calculated. The decisive question is whether each present epoch event is subject to the influence of infinitely many previous discrete events. In that case statisti cal assumptions (large numbers, random phases, etc.) could enter in stronger ways into theories of cosmology. The Mixmaster cosmological model does have an infinite past history in this sense, since each \"bounce\" from one Kasner-like motion to another is a recognizable cosmological event, of which infinitely many must be realized between any finite epoch and the singularity. \u00a730.8. HORIZONS AND THE ISOTROPY OF THE MICROWAVE BACKGROUND 815 \u00a730.8. HORIZONS AND THE ISOTROPY OF THE MICROWAVE BACKGROUND The fundamental cosmological question-\"Must a universe that is born chaotic necessarily become as homogeneous and isotropic as our universe is, and do so before. life evolves?\"-entails one further issue. This issue is horizons. As was discussed in \u00a727.l0, at any given epoch in the expansion ofa Friedmann universe (e.g., the present epoch), there may be significant portions of the universe from which no light signal or other causally propagating influence will have yet reached Earth in the time available since the initial singularity. \"If we should live so long,\" the question would arise, \"will the new portions of the universe which first come into view during the next ten billion years look statistically identical to the neighboring portions which are already being seen?\" Fortunately, this question need not be posed only for the future. It Can be asked as of some past time, and the answer then is yes. Microwave background radiation arrives at the earth from all directions in the sky with very nearly the same tempera ture. [The data of Boughn, Fram, and Partridge (1971) and of Conklin (1969) show zjT!-T~~ 0.004.] The plasma that emitted the microwave radiation coming to earth from one direction in the sky had not been able, before the epoch of emission, to communicate causally with the plasma emitting the radiation that arrives from other directions. If one adopts a Friedmann model of the universe, then different sectors of the microwave sky are disjoint from each other in this sense if they are separated from each other by more than 30\u00b0, even if the microwaves were emitted as recently as z = 7. (The critical angle is much smaller if the microwaves were last scattered at z = 1,000.) From this, one concludes that", " the foundations for the homogeneity and isotropy of the universe were laid long before the universe became approximately Friedmann, for ifstatistical homogeneity and isotropy of the universe had not already been achieved at the longest wavelengths earlier, these horizon limitations would have prevented any further synchronization of conditions over large scales while the universe was in a nearly Friedmann state, and small amplitude (10%) deviations from isotropy should be observed now. The mixmaster universe received its name from the hope that it could contribute to the solution of this problem. The very large u values that occur sporadically an infinite number of times near the singularity in a mixmaster universe give a geometry close to that of the Kasner model with PI = 1, pz = P3 = O. This model can be written in the form (30.6) where 1J = In t. If this metric is converted into a closed-universe model by inter preting x, y, z as angle coordinates each with period 4'iT, then one sees that a light ray can circumnavigate the universe in the x-direction in a time interval Lhj = 4'iT, which corresponds to a volume expansion by a factor V- gtl V- gz = e4 'IT. Unfor tunately, a quantitative analysis of the degree and frequency with which the mix master universe achieves this specific Kasner form suggests that the horizon breaking Horizons in a Friedmann universe Observed isotropy of microwave radiation proves foundations for homogeneity were laid before universe became Friedmann-like What made the universe homogeneous and isotropic? (1) Mixmaster oscillations?- probably not (2) particle creation near singularity? 816 30. ANISOTROPIC AND INHOMOGENEOUS COSMOLOGIES is inadequate to explain the present state of the universe [Doroshkevich, Lukash, and Novikov (1971); Chitre (1972)]. It may turn out that particle creation near the singularity can solve this horizon question, as well as provide for the dissipation of anisotropy. Hope is provided by the fact that particle creation, when described in purely classical terms, has some acausal appearances, even though it is a strictly causal process at the quantum level [Zel'dovich (1972)]. GRAVITATIONAL COLLAPSE AND BLACK HOLES Wherein the reader is transported to the land of black holes, and encounters colonies of static limits, ergospheres, and horizons-behind whose veils are hidden gaping, ferocious singularities. ---------------------------------. ~ CHAPTER 31 SCHWARZSCHILD GEOMETRY \u00a731.1. INEVITABILITY OF COLLAPSE FOR MASSIVE STARS There is no equilibrium state at the endpoint of thermonuclear evolution for a star containing more than about twice the number ofbaryons in the sun (A > A max -- 2A0 ). This is one of the most surprising-and disturbing-consequences of the discussion in Chapter 24. Stated differently: A star with A > Amax -- 2A0 must eject all but A max of its baryons-e.g., by nova or supernova explosions-before settling down into its final resting state;_otherwise there will be no final resting state for it to settle down into. What.is the fate of a star that fails to eject its excess baryons before nearing the endpoint of thermonuclear evolution? For example, after a very massive supernova explosion, what will become of the collapsed degenerate-neutron core when it contains more than A max baryons? Such a supercritical mass cannot explode, since it is gravitationally bound and it has no more thermonuclear energy to release. Nor can it reach a static equilibrium state, since there exists no such state for so large a mass. There remains only one alternative; the supercritical mass must collapse through its \"gravitational radius,\" r = 2M,leaving behind a gravitating \"black hole\" in space. The phenomenon of collapse through the gravitational radius, as described by classical general relativity, will be the subject of the next chapter. However, before tackling it, one must understand more fully t.han heretofore the Schwarzschild spacetime geometry, which surrounds black holes and collapsing stars as well as static stars. This chapter will concern itself with two topics that, at first sight, appear to be disconnected. One is theJaU of a test particle in. a preexisting Schwarzschild geom etry, which is iegarde((a-~-static,but can also be visualized as' all tnatremairis of a star that underweritcollapse some time ago. The second topic is the physical This chapter. on Schwarzschild geometry. is key preparation for understanding gravitational collapse (next chapter) and black holes (following chapter) 820 \u00b7", "31. SCHWARZSCHILD GEOMETRY character of this geometry, regarded in and by itself. For the exploration of this geometry, the test particle serves as the best of all explorers. But the test particle may also be regarded in another light. It can be viewed as a rag-tag johnny-come lately piece of the matter of the falling star. Regarded in this way, it provides the simplest of all illustrations of an aSY}1lmetryjn t_h~distribution_ofmass_ofa ~o!lapsing star. That this asymmetry irons itself out will therefore give one some preliminary insight into how more complicated asymmetries also iron themselves out. In brief, the motion of the test particle and the dynamics of the Schwarzschild geometry (for this geometry will prove to be dynamic), two apparently different problems, have the happy ability to throw light on each other. \u00a731.2. THE NONSINGULARITY OF THE GRAVITATIONAL RADIUS The Schwarzschild spacetime geometry (31.1 ) appears to behave badly near r = 2M; there gil becomes zero, and grr becomes infinite. However, one cannot be sure without careful study whether this pathology in the line element is due to a patl!~!~gy in the__~Lry_~ts~lf, or _~e.rely to ~ Rath0.!9gy of the (t, r, e, 1\u00bb coordinafesystem near r = 2M. (As an example of a coordinate:Iil(luce<tpaffiOIOgy, consider the neighborhood of e= 0 on one of the invariant spher~s, t = const and r = const. There g;; becomes zero because the coordinate system behaves badly; however, the intrinsic, coordinate-independent geometry of the sphere is well-behaved there. For another example, see Figure 1.4. The worrisome region of the Schwarzschild geometry, r = 2M, is called the \"grav itational radius,\" or the \"Schwarzschild radius,\" or the \"Schwarzschild surface,\" or the \"Schwarzschild horizon,\" or the \"Schwarzschild sphere.\" It is also called the \"Schwarzschild singularity\" in some of the older literature; but that is a misnomer, since, as will be shown, the spacetime geometry is not si~lgular there. To determine whether the spacetime geometry is singular -at the gravitational radius, send an explorer in from far away to chart it. For simplicity, let him fall freely and radially into the gravitational radius, carrying his orthonormal tetrad with him as he falls. His trajectory through spacetime [\"parabolic orbit\"; radial geodesic of metric (31.1)] is The Schwarzschild line element becomes singular at r = 2M (\"gravitational radius\") 2~ = ~(..!-)3/2 _ (-!-)1/2 3 2M 2 2M _t 2M ; C~Y/2 + constant, In I(rj2M)1/2 + 11 (rj2M)1/2 _ 1 + constant. + (31.2) [See \u00a725.5 and especially equation (25.38) for derivation and discussion.] One obtains the r coordinate of the explorer in terms of the proper time measured on a clock \u00a731.2. NONSINGULARITY OF THE GRAVITATIONAL RADIUS 821 he carries, r(7), by inverting the first equation; one finds his r coordinate in terms of coordinate time, r(t), by inverting the second equation. Of all the features of the traveler's trajectory, one stands out most dearly and disturbingly: to reach the gravitational radius, r = 2M, requires a finite lapse of proper time, but an' infinite lapse of coordinate time: r/2M = 1 - (7 + constant)/2M when near r = 2M; r/2M = 1 + constant X exp (-t/2M) in limit as t -+ 00. (313) (see Fig. 25.5.) Of course, proper time is the relevant quantity for the explorer's heart-beat and health. No coordinate system has the power to prevent him from reaching r = 2M. Only tIie-coordinate-i~d~p~de\"llt geometry of spacetime could possibly -dothat; and equation (31.3) shows it does not! An infalling observer reaches r = 2M in finite proper time but infinite coordinate time Let the explorer approach and reach r = 2M, then. What spacetime geometry does he measure there? Is it singular or nonsingular? Restated in terms ofmeal'urements, do infinite tidal gravitational forces tear the traveler apart and crush n", "im.: as he approache~J = ~M, or does he f;eY\"on!f finite tidal forces which in principle his body can witH-statld? 'L. h --\"\"\"\"\"li '. 4$ r ~... :, ~~V...., __,', ~. \", ) The tidal forces felt by the explorer as he passes a given radius r are measured by the components of the Riemann curvature tensor with respect to his orthonormal frame there (equation of geodesic deviation). To calculate those curvature compo nents at r, proceed in two steps. (1) Calculate the components, not in the traveler's frame, but rather in the \"static\" orthonormal frame. ( Wi = 1 - - 2M)1I2 r \u2022 dt w T ='dr (1 - 2M/r)1I2'. \u2022 w 8 = rdB wrf> = rsin 0 dr+- ~: 't', (31.4a) located at the event through which he is passing; the result [obtainable from equa tions (14.50) and (14.51) by setting e2t/J = e-2A = 1 - 2M/r] is -2M RE,.E,. = - - 3-, r 2M RfJif,fJif, = - 3'r (31.4b) all other Rafj-y~ vanish except those obtainable from the above by symmetries of Riemann. (2) Calculate the components in the explorer's frame by applying to the \"static frame\" components (31.4b) the appropriate transformation-forr >.2M, a Lorentz boost in the e,. direction with ordinary velocity v\"; for r < 2M, not a \"boost,\" but a transformation given by the standard boost formula (Box 2.4) with v\" > 1. Here v\" = (grr)1I2 dr = dr/dt = _ (2M)1I2. ( - gtt)1I2 dt 1 - 2M/r r (31.5) The amazing result (a consequence of special algebraic properties of the Schwarzs child geometry, and somewhat analogous to what happens-or, rather, does not hap- 822 31. SCHWARZSCHILD GEOMETRY pen-to the components of the electromagnetic field, E and B, when they are both parallel to a boost) is this: all the components of Riemann are left completely unaffected by the boost. If e p is the traveler's radial basis vector, and e;. = u is his time basis vector, then R;.p;:\" = -2Mjr3 RiJ;;'iJif, = 2Mjr3,, T8T8 R\u00b7.. \u00b7 = R\u00b7\u00b7\u00b7\u00b7 = Mjr 3 R\u00b7\u00b7\u00b7\u00b7 - R\u00b7\u00b7\u00b7\u00b7 - _Mjr3 P9P9 - p8p8 - T<;n\u00a2, (31.6) \u2022 The infalling observer does not feel r= 2M infinite tidal forces at the spacetime Thus, geometry is well behaved at r = 2M, but the coordinate system is pathological (See exercise 31.1.) The payoff of this calculation: according to equations (31.6), none of the compo nents of Riemann in the explorer's orthonormal frame become infinite at the gravitational radius. The tidal forces the traveler feels as he approaches r = 2M are finite; they do not tear him apart-at least not when the mass M is sufficiently great, because at r = 2M the typical non-zero component Ra!JY~ of the curvature tensor is of the order I j M2. The gravitational radius is a perfectly well-behaved, nonsingular region of spacetime, and nothing there can prevent the explorer from falling on inward. By contrast, deep inside the gravitational radius, at r = 0, the traveler must encounter infinite tidal forces, independently of the route he uses to reach there. One says that \"r = 0 is a physical singularity of spacetime.\" To see this, one need only calculate from equation (31.4b) or (31.6) the \"curvature invariant\": (31.7) Box 31.1 THE \"SCHWARZSCHILD SINGULARITY\": HISTORICAL REMARKS Although Eddington (1924) was the first to con struct a coordinate system that is nonsingular at r = 2M, he seems not to have recognized the sig nificance of his result. Lemaitre (1933c, especially p. 82) appears to have been the first to recognize that the so-called \"Schwarzschild singularity\"", " at r = 2M is not a singularity. He wrote, \"La singu larite du champ de Schwarzschild est done une singularite fictive, analogue a celIe qui se presen tait a l'horizon du centre dans la forme originale de l'univers de de Sitter\". He also provided a coordinate system to go through r = 2M. How ever, his coordinate system, like Eddington's, cov ered only half of the Schwarzschild geometry: regions I and II of Figure 31.3. Synge (1950) was the first to discover the incompleteness in the Ed dington and Lemaitre coordinate systems, and to provide coordinates that cover the entire geometry (regions I, II, III, IV of Figure 31.3). Fronsdal (1959), unaware of Synge's work, rediscovered the global structure of the Schwarzschild geometry by means of embedding diagrams and calculations. The coordinate system that provides maximum insight into the Schwarzschild geometry is the one generally known as the Kruskal-Szekeres coordi nate system. It was constructed independently by Kruskal (1960) and by Szekeres (1960). \u00a731.3. BEHAVIOR OF SCHWARZSCHILD COORDINATES AT, = 2M 823 In every local Lorentz frame this will be a sum of products of curvature components, and it will have the same value 48M2/r6. Thus, in every local Lorentz frame, including the traveler's, Riemann will have one or more infinite components as r -+ 0; i.e., tidal forces will become infinite. At, = 0 the curvature is infinite Exercise 31.1. TIDAL FORCES ON INFALLING EXPLORER (a) Carry out the details of the derivation of the Riemann tensor components (31.6). (b) Calculate, roughly, the critical mass Merit such that, if M > Merit the explorer's body (a human body made of normal flesh and bones) can withstand the tidal forces at r = 2M, but if M < Merit his body is mutilated by them. [Answer: Merit lOOOM0. Evidently, if M - M 0 the physicist should transform himself into an ant before taking the plunge! For details see \u00a732.6.] EXERCISE \u00a731.3. BEHAVIOR OF SCHWARZSCHILD COORDINATES AT r = 2M Since the spacetime geometry is well behaved at the gravitational radius, the singular Nature of the coordinate behavior there of the Schwarzschild metric components, gtt = - (1 - 2M/r) and grr = (1 - 2M/r)-I, must be due to a pathology there of the Schwarzschild coordinates t, r, 0, cp. Somehow one must find a way to get rid of that pathology-i.e., one must construct a new coordinate system from which the pathology is absent. Before doing this, it is helpful to understand better the precise nature of the pathology.., pathology at, = 2M: The most obvious pathology at r = 2M is the reversal there of the roles of t and \\(1) t and, reverse roles as timelike and spacelike coordinates r as timelike and spacelike coordinates. In the region r > 2M, the t direction, a/at, i is timelike (gtt < 0) and the r direction, a/or, is spacelike (grr > 0); but in the region r < 2M, a/at is spacelike (gtt > 0) and a/or is timelike (grr < 0). What does it mean for r to \"change in character from a spacelike coordinate to a timelike one\"? The explorer in his jet-powered spaceship prior to arrival at r = 2M always has the option to turn on his jets and change his motion from decreasing r (infall) to increasing r (escape). Quite the contrary is the situation when he has once allowed himself to fall inside r = 2M. Then the further decrease of r represents the passage of time. No command that the traveler can give to his jet engine will turn back time. That unseen power of the world which drags everyone forward willy-nilly from age twenty to forty and from forty to eighty also.drags the rocket in from time coordinate r = 2M to the later value of the fime coordinate r = O. No human act of will, no engine, no rocket, no force (see exercise 31.3) can make time stand still. As surely as cells die, as surely as the traveler's watch ticks away \"the unforgiving minutes,\" with equal certainty, and with never one hO~t~long the way", " at r = 5.2M and falls straight in, arriving in a finite proper time at the singularity r = 0 (zig-zag marking). The unhappiness of the Schwarzschild coordinate system shows in two ways: (I) in the fact that t goes to 00 partway through the motion; and (2) in the fact that t thereafter decreases as 7' (not shown) continues to increase. The course of the same trajectory prior to t = 0 may be constructed by reflecting the diagram in the horizontal axis (\"time inversion\"). The time-reversed image of F\" marks the ejection of the test particle from the singularity. AA'A\" is a timelike geodesic which comes in from r = + 00. BB'B\" is the null geodesic travelled by a photon that falls straight in (no summit; never at rest!). DD'D\" is a spacelike radial geodesic. So is ce, but E'E\" is timelike. Neither of the latter two ever succeed in crossing r = 2M. (Unanswered questions about these geodesics will answer themselves in Figure 31.4, where the same world lines are charted in a- \"Kruskal-Szekeres diagram\"). Described mathematically via equation (31.10), the geodesic F\"inverse F'inverse FF'F\" starts with ejection at r = 0 at t = -2\",M - - - rmax 2M ( )1/2( rmax ) - - + I 4M'I it flies outward with increasing proper time 7', but decreasing coordinate time, t, until it reaches the gravitational radius r=2Matt= -00, 7'= _(rmax3)1/2coS_l(4M -1)-rmax(l- 2M)1/\\ 8M rmax rmax it then continues to fly on outward, but with coordinate time now increasing from t = - 00, until it reaches its maximum radius it then falls inward, with t continuing to increase, until it crosses the gravitational radius again r = rmax at t = 0, 7' = 0 (event F in diagram); r=2},[att= +00, 7'= +(rmax3)1/2coS_l(4M -1)+rmax(l- 2Mt2 rmax (event F' in diagram); rmax 8M and it finally falls on in with decreasing t (but, of course, still increasing 7') to r = 0 at t = +2\"'M(~m;; - 7' = +~C;j~3t2 It 2c,; + I), (event F\" in diagram). Novikov coordinates: (1) how constructed 826 31. SCHWARZSCHILD GEOMETRY --( spread them out into a line in a new (rnew' tnew)-plane; and will squeeze the \"line\" (r == 2M, t from - 00 to + (0) in~l!..].i.t!gk-~oi.nLinJh~J!ne~-'ne.wl-l'.1~ane.One is the more prepared to accept this tentative conclusion and act on it because one has already seen (equation 31.8) that the region covering the (0, \u00a2) 2-sphere at r = 2M, and extending from t = - 00 to t = + 00, has zer<:>J'!.op-e.~_voluDle. What timelier indication could one want that the \"line\" r = 2if-= 00 < t < 00, is actually a point? \u00a731.4. SEVERAL WELL-BEHAVED COORDINATE SYSTEMS The well-behaved coordinate system that is easiest to visualize is one in which the radially moving test particles of equations (31.10) remain always at rest (\"comoving coordinates\"). Such coordinates were first used by Novikov (1963). Novikov attaches a specific value of his radial coordinate, R*, to each test particle as it emerges from the singularity of infinite tidal forces at r = 0, and insists that the particle carry that value of R* throughout its \"cycloidallife\"-up through r = 2M to r = rmax, then back down through r = 2M to r = O. For definiteness, Novikov expresses the R* value for each particle in terms of the peak point on its trajectory by (31.11) As a time coordinate, Novikov uses proper time T of the test particles, normalized so T = 0 at the peak of the or~it. Every particle in the swarm is ejected in such a manner that it arrives at the s\u00b7tiinmitof", " its trajectory (r = r max' T = 0) at one and the same value of the Schwarzschild coordinate time; namely, at t = O. Simple th~ugh-:they may be conceptually, the Novil<;v coordinates are related to the original Schwarzschild coordinates by a very complicated transformation: (1) combine equations (31.10b) and (31.11) to obtain 1/(T, R*); (2) combine 1/(T, R*) with (31.l0a) and (31.11) to obtain reT, R*); (3) combinq(T, R*) with (31.10c) and (31.11) to obtain t(T, R*). The resulting coordinate transformation, when applied to the Schwarzschild metric (31.1), yields the line element (2) line element ds 2 = -dT 2 + (R*~~ 1)(a~*Y dR*2 + r2(d02 + sin20dep2). (31.12a) (\"Schwarzschild geometry in Novikov coordinates\".) Here r is no longer a radial coordinate; it is now a metric function reT, R*) given implicitly by T -2M- = \u00b1 (R + 1) *2 [r -2M- - (r/2M)2 ]1/2 -'-R-*2-+-'-1 + (R*2 + 1)3/2 cos- 1 [ ( r/2M )1/2) R*2 + 1. (31.12b) Figure 31.2 shows the locations of several key regions of Schwarzschild spacetime in this coordinate system. The existence of two distinct regions with r = 0 (singulari ties) and two distinct regions with r -+ 00 (asymptotically flat regions; recall that 4?Tr2 = surface area!) will be discussed in \u00a731.5. T I I I I \\ \\ \\ \\ \\ \\ \\ \\ \\, I ~I / II \" I \\ \\ ~\\ \\cP \\~ \\ \\ \\ \\ \\ \\ I I R* Figure 31.2. The Novikov coordinate system for Schwarzschild spacetime (schematic). The dashed curves are curves of constant r (recall: 4.\".r2 == surface area about center of symmetry). The region shaded gray is not part of spacetime; it corresponds to r < 0, a region that cannot be reached because of the singularity of spacetime at - 00 < r == O. Notice t < + 00) of the Schwarzschild coordinate diagram (Figure 31.1) has been compressed into a point here, in accordance with the discussion at the end of \u00a731.3. (r == 2M, \"line\" that the Kruskal-Szekeres coordinates.lthough N~yikoy's cQ_Q!\"cii!).ate system i\u00a7_very _sil!!ple~2!1ceptually, ~h_e_ mathe matical -exp~~~ions fO_~Jhe metri~:~omponents in it are rath~!. ~ieldy. Simpler, more usable expressions have been obtained -in -a different coordinate system (\"Kruskal-Szekeres coordinates\") by Kruskal (1960), and independently by Szekeres (1960). Kruskal and Szekeres use a dimensionless radial coordinate it and a dimensionless time coordinate v related-to';he-SchwarzschiW';-;ndtby --------.----.. -..-. -_._---..;,-_. u = (r12M - 1)1I2er/4M cosh (tI4M) } v = (r12M - 1)1I2er/4M sinh (tI4M) u = (1 v = (1 - rI2M)1I2er/4M sinh (t14M) } rI2M)1I2er/4M cosh (tI4M) when r > 2M, (31.13a) when r < 2M. (31.13b) (Motivation for rntroduciiig\"SUCh coordinates is given in Box 31.2.) By making this change of coordinates in the Schwarzschild metric (31.1), one obtains the following line element: (31.14a) (\"Schwarzschild geometry in Kruskal-Szekeres coordinates\"). Here r is to be regarded as a function of u and v defined implicitly by (r12M - 1)er / 2M = u2 _ v2 (31.14 b) [cf. equations (31.13)]. (COlllillued on page 833) 8", "28 31. SCHWARZSCHILD GEOMETRY Box 31.2 MOTIVATION FOR KRUSKAL-SZEKERES COORDINATES\u00b7 A. EDDINGTON-FINKELSTEIN COORDINATES The motivation for the Kruskal-Szekeres system begins by introducing a different coordinate system, first devised by Eddington (1924) and rediscovered by Finkelstein (1958). Eddington and Finkelstein use as the foundation of their coordinate system, not freely falling particles as did Novikov, but freely falling photons. More par ticularly, they introduce co~rdinates D and V, which are labels for outgoing and ingoing, radial, null geodesics, The geodesics are given by ds 2 = 0 = -(1 - 2M/r) dt2 + (1 - 2M/r)-1 dr 2. Equivalently, outgoing geodesics are given by D= const, where and ingoing geodesics are given by V = const, where D t - r*; V t + r*, Here r* is the \"tortoise coordinate\" of \u00a725.5 and Figure 25.4: r* r + 2Mln Ir/2M - II, Ingoing Eddington-Finkelstein Coordinates-Adopt r and II as coordinates in place of rand t The Schwarzschild metric becomes. I \" / ds 2 = -(1 - 2M/r) dV2 + 2 dV dr + r 2 dil 2. The radial light cone, ds 2 = 0, has one leg and the other leg dV/dr = 0, dV dr 2 1 - 2M/r' From this, and this alone, one can infer all features of the drawing, -This box is based on Misner (l969a). (1 a) (1 b) (1 c) (2) (3 a) (3b) \u00a731.4. SEVERAL WELL-BEHAVED COORDINATE SYSTEMS 829 t, V r = 2M is world line of outgoing photons \\ \\-.. O~ ~\\o og.\\%~\\;: ~\\\" s.\\~ g~ \\ \\ \\ \\ \\ I t=3M \\ -=-------\\ t=2M I -r-=-M-----\\ Ingoing Eddington-Finkelstein coordinates (one rotational degree of freedom is suppressed; Le., (J is set equal to 'IT/2). Surfaces of constant Ii; being ingoing null surfaces, are plotted on a 45-degree slant, just as they would be in flat spacetime. Equivalently, surfaces of constant (= Ii' - r = t + 2Mln Ir/2M - II are plotted as horizontal surfaces. Outgoing Eddington-Finkelstein Coordinates-Adopt rand iJ as coordinates in place of rand t The Schwarzschild metric becomes ds 2 = -(1 - 2M/r) dU2 - 2 dU dr + r2 dil 2. (4) 830 31. SCHWARZSCHILD GEOMETRY Box 31.2 (continued) The radial light cone, ds 2 = 0, has one leg and the other leg dD/dr = 0, dD dr = 2 1 - 2M/,.\u00b7 (5a) (5b) From this, and this alone, one can infer all features of the drawing \\ \\ ~ ;'\\~ ~\\~ ;,:.\\ p:. '=' \\-;:. (JG \\ \"1,-,: 0\\_. --\\ ~\\ Outgoing Eddington-Finkelstein coordinates (one rotational degree of free dom is suppressed). (Surfaces of constant D, being outgoing null surfaces, are plotted on a 45-degree slant just as they would be in flat spacetime.) \u00a731.4. SEVERAL WELL-BEHAVED COORDINATE SYSTEMS 831 Notice that both Eddington-Finkelstein coordinate systems are better behaved at 1 the gravitational radius than is the Schwarzschild coo!dinate system;~ not fully well-behaved, The outgoing coordinates (U, r, 0, ep) describe in a non p;lthologkal mannefihe ejection of particles outward from r = 0 through r = 2M; but thei!..E-~scriptioJLQfJnfall through r - 2M has the~_~athQlog~ the <;Ie (Figure 31.1). Similarly, the ingoing scrip!i2jl..gill.en....b.y:...schwarzschil.~L~oordinates coordinates (r, r,", " 0, ep) describe wellthe--infa1f(i'fa particle through r = 2M, but they give a pathological description of outgoing trajectories. Moreover, the contrast between the t~o diagrams seems paradoxical: in one the gravitational radius is made up of world lines of outgoing photons; in the other it is made up of world lines of ingoing photons! To resolve the paradox, one must seek another, better-behaved coordinate system. [But note: because the ingoing Eddington-Finklestein coordinates \\ \\ describe infall so well, they are used extensively in discussions of gravitational ( \\, collapse (Chapter 32) and black holes (Chapters 33 and 34).] \\ B. TRANSITION FROM EDDINGTON-FINKELSTEIN TO KRUSKA~SZEKERES Perhaps one would obtain a fully well-behaved coordinate system by dropping r from view and using D, V, as the two coordinates in the radial-time plane. The resulting coordin~t~~stem is related to Schwarzschild coordinates by [see equations (1)] v- D = 2r*, V+ D= 2t; and the line element in terms of the new coordinates reads (6a) (6b) (7) Contrary to one's hopes, this coordinate system is pathological at r = 2M. Second thoughts about the construction reveal the trouble: the surfaces D= constant (outgoing null surfaces) used in constructing it are geometrically well defined, as are the surfaces V= constant (ingoing null surfaces); but the way of labeling them is not. Any relabeling, u = F( D) and v = G( V), will leave the surfaces unchanged physically. What one needs is a relabeling that will get rid of the singular factor 1 - 2M/r in the line element (7). A successful relabeling is suggested by the equation exp [( V - D)/4M] = exp (r*/2M) = (r/2M - 1) exp (r/2M), (8) Box 31.2 (continued) which follows from equations (6a) and (Ic). Experimenting with this relation quickly reveals that the relabeling Ii = _e- U/ 4.1f = -(r/2M - 1)l/:!eT/4Me-t/4M, V = e+V/ 4M = (r/2M _ 1)1/2eT/4Met/4M, (9a) (9b) will remove the offending 1 - 2M/r from the metric coefficients. In terms of these new coordinates, the line element reads Here r is still defined by 4\"r2 = surface area, but it must be regarded as a function ofv and Ii: (r/2M - 1)eT/2M = -liv. One can readily verify that this equation determines r uniquely (recall: r > O!) in terms of the product liv [details in Misner (I969a)]. (lOb) The coordinates, Ii, v, which label the ingoing and outgoing null surfaces, are null coordinates; i.e., a/eli, a/ali = gil;; = 0, a/avo a/av = g;; = 0 [see equation (lOa)]. If one is not accustomed to working with null coordinates, it is helpful to replace Ii and v by spacelike and timelike coordinates, u and v (Kruskal Szekeres coordinates!) defined by Ii) = (r/2M - u = ; (v v = ; (v + Ii) = (r/2M - 1)1/2eT/4M sinh (t/4M), I)l/2 eT/4M cosh (t/4M), so that dv 2 - du2 = dV dUo (Ila) (ll b) (I2) In terms of these coordinates, the line element has the Kruskal form (31.14), which is fully well-behaved at the gravitational radius. Although the Kruskal-Szekeres line element the transforma tion (ll) from Schwarzschild to Kruskal-Szekeres is not; it becomes meaningless (u and v \"imaginary\") when one moves from r > 2M to r < 2M. Of course, this is a manifestation of the pathologies of Schwarzschild coordinates. By trial and error, one readily finds a new transformation, to replace (II) at r < 2M, leading from Schwarzschild to Kruskal-Szekeres coordinates: is well behaved at r = 2M, u == (l - r/2M)1I2 eT/4M sinh (t", "/4M), v = (l - r/2M)1/2 eT/4M cosh (t/4M). (llc) (lId) \u00a731.5. KRUSKAL DIAGRAMS 833 \u00a731.5. RELATIONSHIP BETWEEN KRUSKAL-SZEKERES COORDINATES AND SCHWARZSCHILD COORDINATES In the Kruskal-Szekeres coordinate system, v2 - u2 = 1. Thus there are actually two singularities, not one; both the singularity r = 0 is located at v = +(1 + U2)1/2 and v = -(1 + U2)1/2 correspond to r = O! (31.15) Kruskal-Szekeres coordinates reveal that Schwarzschild spacetime has two \", = 0 singularities\" and two \", --+ 00 exterior regions\" This is not the only surprise that lies hidden in the Kruskal-Szekeres line element (31.14). Notice also that r > 2M(the region of spacetime far outside the gravitational radius) is given by u2 > v 2\u2022 Thus there are actually two exterior regions*; both 7':.( b(;.,~~'e. u > + Ivl and u ~ -Ivl correspond to r > 2M! (31.16) How can this be? When the geometry is charted in Schwarzschild coordinates, it contains one singularity and one exterior region; but when expressed in Kruskal Szekeres coordinates, it shows two of each. The answer must be that the Schwarzs child coordinates cover only part of the spacetimeriiariifofd; theymust be oniya l~al coordi;ate \"pa.tch-oll\"\u00b0the-fufl manifold.\u00b7Somehow:\"bY--~eans of the coordinate transformation that leads to Kruskal-Szekeres coordinates, one has analytically extended the limited Schwarzschild solution for the metric to cover all (or more nearly all) of the manifold. To understand this covering more clearly, transform back from Kruskal-Szekeres coordinates to Schwarzschild coordinates (see Figure 31.3). The transformation equations, as written down in. (31.13) were valid only for the quadrants u > Ivi [equation (31.13a)] and v\":>\" 'ul [equation (31.13b)] of Kruskal coordinates. Denote these quadrants by the numerals I and II; and denote the other quadrants by III and IV (see Figure 31.3). In the other quadrants, one can also transform the Kruskal Szekeres line element (31.14) into the Schwarzschild line element (31.1); but slightly different transformation equations are needed. One easily verifies that the following sets of transformations work: u = (r12M - 1)1/2er/4M cosh (tI4M) v = (r12M - 1)1/2er/4M sinh (tI4M)'u = (1 rI2M)1/2er/4M sinh (tI4M) v = (1 - rI2M)1/2er/4M cosh (tI4M) u = -(rI2M - 1)1/2er/4M cosh (tI4M) v = -(rI2M - 1)1/2er/4M sinh (tI4M) u = -(1 - rI2M)1/2er/4M sinh (tI4M) v = -(1 rI2M)1/2er/4M cosh (tI4M),,. (I) { (II) (III) (IV) { { { (31.l7a) (31.17b) (31.17c) (31.17d) Transformation between Schwarzschild coordinates and Kruskal-Szekeres coordinates *The global structure of the Schwarzschild geometry. including the existence of two singularities and two exterior regions. was first discovered by Synge (1950)0 See Box 31.1. 834 31. SCHWARZSCHILD GEOMETRY (\" ) (lJ / \"\" ~I ~I '-I \"Upper patch\" (IV) (III) r ~I ~I \"Lower '-I patch\" Schwarz~child patches v 0+ 1\\ I ::::. ~ II II II I ~... -v Kruskal-Szekeres \\\\ \\> 'J' ~ Figure 31.3. The transformation of the Schwarzschild vacuum geometry between Schwarzschild and Kruskal", "-Szekeres coordinates. Two Schwarzschild coordinate patches I, II. and III, IV (illustrated in the upper and lower portions of Figure 3J.5,a) are required to cover the complete Schwarzschild geometry, whereas a single Kruskal-Szekeres coordinate system suffices. The Schwarzschild geometry consists of four regions I, II, III, IV. Regions I and III represent two distinct, but identical, asymptotically flat universes in which r > 2M; while regions II and IV are two identical, but time-reversed, regions in which physical singulari. ties (r = 0) evolve. The transformation laws that relate the Schwarzschild and Kruskal-Szekeres coordi nate systems to each other are given by equations (3J.l7) and (3US). In the Kruskal-Szekeres u,v-plane, curves of constant r are hyperbolae with asymptotes u == \u00b1v, while curves of constant ( are straight lines through the origin. The inverse transformations are (r/2M - l)er/ 2M = u2 - v2 in I, II, III, IV; t = {4M tanh-l(v/u) ~n I and III, 4M tanh-leu/v) In II and IV. (31.18a) (3 l.l 8b) Two Schwarzschild coordinate patches are required to cover all of spacetime These coordinate transformations are exhibited graphically in Figure 31.3. Notice that two Schwarzschild coordinate patches, I, II, and III, IV, are required to cover the entire Schwarzschild geometry; but a single Kruskal coordinate system suffices. Schwarzschild patch I, II, is divided into two regions-region I, which is outside the gravitational radius (r > 2M), and region II, which is inside the gravitational radius (r < 2M). Similarly, Schwarzschild patch III, IV, consists of an exterior region (III) and an interior region (IV).,. ~ \u00a731.5. KRUSKAL DIAGRAMS 835 v - u.....t----.\",.....-,:;;;;.......=.---\"'---+--~u t M F\"/./ A',B',D',F' (att=oo) I J~ \\\\ \\\\ \\ \\\\ I \\ \\ \\ \\ \\ \\ I A\") ; \\ \\ \\. E\" Ei \\ \\ \\ \\ \\, --'~-.... C '\" C' _',-,,\"-~.D \"\",A,B F r o~---\u00a5'4-.+I---f-~\"...~. M 3 4 \\ 5 B,:..J/ /2 I.... (a) Schwarzschild -v (b) Kruskal-Szekeres Figure 31.4. (a) Typical radial timelike (A, E, F), lightlike (B), and spacelike (C, D) geodesics of the Schwarzschild geometry, as seen in the Schwarzschild coordinate system (schematic only). This is a reproduction of Figure 3J. J. (b) The same geodesics, as seen in the Kruskal-Szekeres coordinate system, and as extended either to infinite length or to the singularity of infinite curvature at r = 0 (schematic only). Equations (31.18) reveal that the regions of constant r (constant surface area) are hyperbolae with asymptotes u = -+-v in the Kruskal-Szekeres diagram, and that regions of constant t are straight lines through the origin. Several radial geodesics of the complete Schwarzschild geometry are depicted in Properties of the the Kruskal-Szekeres coordinate system in Figure 31.4. Notice how much more \\ Kruskal-Szekeres coordinate reasonable the geodesic curves look in Kruskal-Szekeres coordinates than in ( system Schwarzschild coordinates. Notice also that radial, lightlike geodesics (paths of radial \\ light rays) are 45-degree lines in the Kruskal-Szekeres coordinate system. This can be seen from the Kruskal-Szekeres line element (31.14), for which du = -+-dv guarantees ds = O. Because of this 45-degree property, Kruskal-Szekeres diagram has the same form as in the spa:ce-tirrie'd~ratii~o'fspecial relativity. Any radial curve that points \"generally upward\" (l.e., makes an angle of less- than 45 degrees with the vertical, v, axis) is timelike; and curves that point \"generally outward\" are spacelike. This property enables a Krusk", "al-Szekeres diagram to exhibit easily the causality relation between one event in spacetime and another (see exercises 31.2 to 31.4). -..- - - ---=---~ \\ light co~e..5n.a \\. the radial._-_.~ \\ 836 31. SCHWARZSCHILD GEOMETRY EXERCISES Exercise 31.2. NON RADIAL LIGHT CONES Show that the world line of a photon traveling non radially makes an angle less than 45 degrees with the vertical v-axis of a Kruskal-SzeKeres coordinate diagram. From this. infer that particles with finite rest mass, traveling nonradiaIly or radially. must always mO\\'e \"generally upward\" (angle less than 45 degrees with vertical v-axis). Exercise 31.3. THE CRACK OF DOOM Use a Kruskal diagram to show the following. (a) If a man aIlows himself to fall through the gravitational radius' = 2M, there is no way whatsoever for him to avoid hitting (and being killed in) the singularity at r = O. (b) Once a man has faIlen inward through r = 2M, there is no way whatsoever that he can send messages out to his friends at, > 2M, but he can still receive messages from them (e.g.. by radio waves, or laser beam, or infalling \"CARE packages\"). Exercise 31.4. HOW LONG TO LIVE? Show that once a man falling inward reaches the gravitational radius, no matter what he does subsequently (no matter in what directions, how long, and how hard he blasts his rocket engines), he wiIl be pulled into the singularity and killed in a proper time of 7\" < 7\"max = TTM = 1.54 X IO-5(M/Mo ) seconds. (31.19) [Hint: The trajectory of longest proper time lapse must be a geodesic. Use the mathematical tools of Chapter 25 to show that the geodesic of longest proper time lapse between, = 2M and, = 0 is the radial geodesic (31.10a), with'max = 2M, for which the time lapse is 7TM.] Exercise 31.5. EDDINGTON-FINKELSTEIN AND KRUSKAL-SZEKERES COMPARED Use coordinate diagrams to compare the ingoing and outgoing Eddington-Finkelstein coor dinates of Box 31.2 with the Kruskal-Szekeres coordinates. Pattern the comparison after that between Schwarzschild and Kruskal-Szekeres in Figures 31.3 and 31.4. Exercise 31.6. ANOTHER COORDINATE SYSTEM Construct a coordinate diagram for the D, V, 0, cf> coordinate system of Box 31.2 [equations (6) and (7)]. Show such features as (1) the relationship to Schwarzschild and to KruskaI Szekeres coordinates; (2) the location of r = 2M; and (3) radial geodesics. \u00a731.6, DYNAMICS OF THE SCHWARZSCHILD GEOMETRY What does the Schwarzschild geometry look like? This question is most readily answered by means of embedding diagrams analogous to those for an equilibrium star (\u00a723.8; Figure 23.1; and end of Box 23.2) and for Friedmann universes of positive and negative spatial curvature [equations (27.23) and (27.24) and Box 27.2]. Examine, first, the geometry of the spacelike hypersurface v = 0, which extends from u = + 00 (r = 00) into u = 0 (r = 2M) and then out to u = - 00 (r = 00). t = 0 [see In Schwarzschild coordinates this surface is a slice of constant equation (31.18b)]; it is precisely the surface for which an embedding diagram was calculated in equation (23.34b). The embedded surface, with one degree of rotational freedom suppressed, is described by the paraboloid of revolution time, (31.20) (a) ~_d (b) Figure 31.5. (a) The Schwarzschild space geometry at the \"moment of time\" t == v == 0, with one degree of rotational freedom suppressed (0 == 11\"/2). To restore that rotational freedom and obtain the full Schwarzschild 3-geometry, one mentally replaces the circles of constant r == (3:2 + r)1/2 with spherical surfaces of area 411\"-;:2. Note that the resultant 3-geometry becomes flat (Euclidean) far from the throat of the bridge in both directions (both \"universes\"). (b) An embedding of the Schwarzschild space geometry", " at \"time\" t == v == 0, which is geometrically identical to the embedding (a), but which is topologically different. Einstein's field equations fix the local geometry of spacetime, but they do not fix its topology; see the discussion at end of Box 27.2. Here the Schwarzschild \"wormhole\" connects two distant regions of a single, asymptotically flat universe. For a discussion of issues of causality associated with this choice of topology, see Fuller and Wheeler (1962). in the fiat Euclidean space with metric d(J2 = d12 + dz2 +,2 dP. (31.21 ) (See Figure 31.5.) Notice from the embedding diagram of Figure 31.5,a, that the Schwarzschild The 3-surface v = t = 0 is a \"wormhole\" connecting two asymptotically flat universes, or two different regions of one universe Schwarzschild geometry is dynamic in regions r < 2M 838 31. SCHWARZSCHILD GEOMETRY geometry on the spacelike hypersurface I == const consists of a bridge or \"wormhole\" connecting two distinct, but identical. asymptotically flat universes. This bridge is sometimes called the \"Einstein-Rosen bridge\" and sometimes the \"Schwarzschild throat\" or the \"Schwarzschild wormhole.\" If one so wishes, one can change the topology of the Schwarzschild geometry by connecting the two asymptotically flat universes together in a region distant from the Schwarzschild throat [Fuller and Wheeler (1962); Fig. 31.5b]. The single. unique universe then becomes multiply connected, with the Schwarzschild throat providing one spacelike path from point {/ to point!13, and the nearly fiat universe providing another. For concreteness, focus attention on the interpretation of the Schwarzschild geometry, not in terms of Wheeler's multiply connected single universe, but rather in terms of the Einstein Rosen double universe of Figure 31.5,a. I and III of One is usually accustomed to think of the Schwarzschild geometry as static. However, the static \"time translations,\" t ----+- t + Lit, which leave the Schwarzschild geometry unchanged, are time translations in the strict sense of the words only in regions II and IV, the Schwarzschild geometry. t ----+- t + Lit is a spacelike motion, not a timelike motion (see Fig. 31.3). Conse quently, a spacelike hypersurface, such as the surface t == const of Figure 31.5,a, which extends from region I through u == v == 0 into region III, is not static. As this spacelike hypersurface is pushed forward in time (in the + v direction of the Kruskal diagram), it enters region II, and its geometry begins to change. In regions In order to examine the time-development of the Schwarzschild geometry, one needs a sequence of embedding diagrams, each corresponding to the geometry of a spacelike hypersurface to the future of the preceding one. But how are the hyper surfaces to be chosen? In Newtonian theory or special relativity, one chooses hyper surfaces of constant time. But in dynamic regions of curved spacetime, no naturally preferred time coordinate exists. This situation forces one to make a totally arbitrary choice of hypersurfaces to use in visualizing the time-development of geometry, and to keep in mind how very arbitrary that choice was. Figure 31.6 uses two very different choices of hypersurfaces to depict the time development of the Schwarzschild geometry. (Still other choices are shown in Figure 21.4.) Notice that the precise geometry of the evolving bridge depends on the arbitrary choice of spacelike hypersurfaces, but that the qualitative nature of the evolution is independent of the choice of hypersurfaces. Qualitatively speaking, the two asymptotically fiat universes begin disconnected, with each one containing a singularity of infinite curvature (r == 0). As the two universes evolve in time, their singularities join each other and form a nonsingular bridge. The bridge enlarges, until it reaches a maximum radius at the throat of r == 2M (maximum circumference of 4'1TM; maximum surface area of 16'1TM2). It then contracts and pinches off, leaving the two universes disconnected and containing singularities (r == 0) once again. The formation, expansion, and collapse of the bridge occur so rapidly, that no particle or light ray can pass across the bridge from the faraway region of the one universe to the faraway region of the other without getting caught and crushed in the throat as it pinches off. (To verify this, examine the Kruskal-Szekeres diagram of Figure 31.3, where radial light rays move along 45-degree lines.) Time evolution of the wormhole", ": creation; expansion; recontraction; and pinch-off Communication through the wormhole is impossible: it pinches off too fast \u00a731.6. DYNAMICS OF THE SCHWARZSCHILD GEOMETRY 839 v E D - I I D' F' E' C' B' II -v Spacelike slices History A-B-C-D-E-F-G History A-W-X-D-Y-Z-G Figure 31.6. Dynamical evolution of the Einstein-Rosen bridge of the vacuum Schwarzschild geometry (schematic). Shown here are two sequences of embedding diagrams corresponding to two different ways of viewing the evolution of the bridge-History A-B-C-D-E-F-G, and History A- W-X-D- Y-Z-G. The embedding diagrams are skeletonized in that each diagram must be rotated about the appropriate vertical axis in order to become two-dimensional surfaces analogous to Figure 31.5,a. [Notice that the hypersurfaces of which embedding diagrams are given intersect the singularity only tangentially. Hypersurfaces that intersect the singularity at a finite angle in the u,v-plane are not shown because they cannot be embedded in a Euclidean space. Instead, a Minkowski space (indefinite metric) must be used, at least near r = O. For an example of an embedding in Minkowski space, see the discussion of a universe with constant negative spatial curvature in equations (27.23) and (27.24) and Box 27.2C.1 Figure 21.4 exhibits embed ding diagrams for other spacelike slices in the Schwarzschild geometry. From the Kruskal-Szekeres diagram and the 45-degree nature of its radial light rays, one sees that any particle that ever finds itself in region IV of spacetime must have been \"created\" in the earlier singularity; and any particle that ever Jalls into region II is doomed to be crushed in the later singularity. Only particles that stay forever in one of the asymptotically fiat universes I or III, outside the gravitational radius (r > 2M), are forever safe from the singularities. Some investigators, disturbed by the singularities at r = 0 or by the \"double-uni verse';-nature of the Schwarzschild geometry, have proposed modifications of its topology. One proposal is that the earlier and later singularities be identified with each other, so that a particle which falls into the singularity of region II, instead of being destroyed, will suddenly reemerge, being ejected, from the singularity of region IV. One cannot overstate the objections to this viewpoint: the region r = 0 is a physical singularity of infinite tidal gravitation forces and infinite Riemann curvature. Any particle that falls into that singularity must be destroyed by those Creation and destruction in the singularities Nonviable proposals for modifying the topology of Schwarzschild spacetime 840 31. SCHWARZSCHILD GEOMETRY forces. Any attempt to extrapolate its fate through the singularity using Einstein's field equations must fail; the equations lose their predictive power in the face of infinite curvature. Consequently, to postulate that the particle reemerges from the earlier singularity is to make up an ad hoc mathematical rule, one unrelated to physics. It is conceivable, but few believe it true. that any object of finite mass will modify the geometry of the singularity as it approaches r = 0 to such an extent that it can pass through and reemerge. However, whether such a speculation is correct must be answered not by ad hoc rules, but by concrete, difficult computations within the framework of general relativity theory (see Chapter 34). A second proposal for modifying the topology of the Schwarzschild geometry is this: one should avoid the existence of two different asymptotically fiat universes by identifying each point (u, u. B, 9) with its opposite point (- U, -II, B, <1\u00bb in the Kruskal-Szekeres coordinate system. Two objections to this proposal are: (1) it pro duces a sort of \"conical\" singularity (absence of local Lorentz frames) at (u, u) = (0,0), i.e., at the neck of the bridge at its moment of maximum expansion; and (2) it leads to causality violations in which a man can meet himself going backward in time. One good way for the reader to become conversant with the basic features of the Schwarzschild geometry is to reread \u00a7\u00a731.1-31.4 carefully, reinterpreting every thing said there in terms of the Kruskal-Szekeres diagram. EXERCISES Exercise 31.7. SCHWARZSCHILD METRIC IN ISOTROP", "IC COORDINATES (a) Show that, rewritten in the isotropic coordinates of Exercise 23.1, the Schwarzschild metric reads and derive the transformation (31.22) (31.23 ) between the two radial coordinates. (b) Which regions ofspacetime (I, II, III, IV; see Figure 31.3) are covered by the isotropic coordinate patch. and which are not? (c) Calculate and construct an embedding diagrarn:lOfThespacelike--t:ypersurface I e, O<r<oo. (d) Find a coordinate transformation that interchanges the region near r= 0 with the region near r = 00, while leaving the metric coefficients in their original form. Exercise 31.8. REISSNER-NORDSTR~M GEOMETRY (a) Solve the Einstein field equations for a spherically symmetric, static gravitational field ds2 = _e2<t>(r) dl2 + e2A(r) dr 2 + r 2(d(J2 + sin 2(J d\u00a22), \u00a731.6. DYNAMICS OF THE SCHWARZSCHILD GEOMETRY 841 with no matter present, but with a radial electric field B = 0, E = j(r)e, in the static orthonormal frame wi = e4> dt, w' = eA dr, w 8 =rdfJ, wJ> = rsin IJ d</>. Use as a source in the Einstein field equations the stress-energy of the electric field. [Answer: (3 1.24a) (31.24b) This is called the \"Reissner (1916)-Nordstwm (1918) metric\".] (b) Show that the constant Q is the total charge as measured by a distant observer (r ~ 2M and r ~ Q), who uses a Gaussian flux integral, or who studies the coulomb-force-dominated orbits of test charges with charge-to-mass ratio elll ~ MIQ. What is the charge-to-mass ratio, in dimensionless units, for an electron? Show that the constant M is the total mass as measured by a distant observer using the Keplerian orbits of electrically neutral particles. r = 00 down to r =0, where there is a physical singularity and infinite tidal forces. (c) Show that for Q > M, the Reissner-Nordstrem coordinate system is well-behaved from (d) Explore the nature of the spacetime geometry for Q < M, using all the techniques of this chapter (coordinate transformations, Kruskal-like coordinates, studies of particle orbits, embedding diagrams, etc.). [Solution: see Graves and Brill (1960); also Fig. 34.4 of this book.] (e) Similarly explore the spacetime geometry for Q = M. [Solution: see Carter (l966b).] (f) For the case of a large ratio of charge to mass [Q > M as in part (c)], show that the region near r = 0 is unphysical. More precisely, show that any spherically symmetric distri bution of charged stressed matter that gives rise to the fields (31.24) outside its boundary must modify these fields for r < ro = Q 2/2M. [Hint: Study the quantity mer) defined in equations (23.18) and (32.22h), noting its values deduced from equation (31.24), on the one hand, and from the appropriate Einstein equation within the matter distribution, on the other hand. See Figure 26 of Misner (l969a) for a similar argument.] CHAPTER 32 GRAVITATIONAL COLLAPSE Now, here, you see, it takes all the running you can do, to keep in the same place. If you want to get somewhere else, you must run twice as fast as that. The Red Queen. in Through the Looking Glass, LEWIS CARROLL (1871) \u00a732.1. RELEVANCE OF SCHWARZSCHILD GEOMETRY The story that unfolded in the preceeding chapter was fantastic! One began with the innocuous looking Schwarzschild line element (32.1 ) which was derived originally as the external field of a static star. One asked what happens if the star is absent; i.e., one probed the nature of the Schwarzschild geometry when no star is present to generate it. One might have expected the geometry to be that of a point mass sitting at r = O. But it was not. It turned out to represent a \"wormhole\" connecting two asymptotically fiat universes. Moreover, the wormhole was dynamic. It was created by the \"joining together\" of two \"r =", " 0\" singularities, one in each universe; it expanded to a maximum circumference of4wM; it then recontracted and pinched off, leaving the two universes disconnected once again, each with its own \"r = 0\" singularity. As a solution to Einstein's field equations, this expanding and recontracting wormhole must be taken seriously. It is an exact solution; and it is one of the simplest of all exact solutions. But there is no reason whatsoever to believe that such worm holes exist in the real universe! They can exist only if the expanding universe, -10 X 109 years ago, was \"born\" with the necessary initial conditions-with \"r = 0\" The roles and relevance of the Schwarzschild geometry \u00a732.2. BIRKHOFFS THEOREM 843 Schwarzschild singularities ready and waiting to blossom forth into wormholes. There is no reason at all to believe in such pathological initial conditions! Why, then, was so much time and effort spent in Chapter 31 on understanding illustrates clearly the highly non the Schwarzschild geometry? (1) Because it Euclidean character of spacetime geometry when gravity becomes strong; (2) because it illustrates many of the techniques one can use to analyze strong gravitational fields; and most importantly (3) because, when appropriately truncated, it is the spacetime geometry of a black hole and of a collapsing star-as well as of a wormhole. This chapter explores the role of the Schwarzschild geometry in gravitational collapse; the next chapter explores its role in black-hole physics. \u00a732.2. BIRKHOFF'S THEOREM That the Schwarzschild geometry is relevant to gravitational collapse follows from BirkhojJ's (1923) theorem: Let the geometry of a given region of spacetime (1) be spherically symmetric, and (2) be a solution to the Einstein field equations in vacuum. Then that geometry is necessarily a piece ofthe Schwarzschild geometry. The external field of any electrically neutral, spherical star satisfies the conditions of Birkhoff's theorem, whether the star is static, vibrating, or collapsing. Therefore the external -fietdmust be a piece of the Schwarzschild geometry. Birkhoff's theorem is easily understood on physical grounds. Consider an equilib rium configuration that is unstable against gravitational collapse and that, like all equilibrium configurations (see \u00a723.6), has the Schwarzschild geometry as its external gravitational field. Perturb this equilibrium configuration in a spherically symmetric way, so that it begins to collapse radially. The perturbation and subsequent collapse cannot affect the external gravitational field so long as exact spherical symmetry is maintained. Just as Maxwell's laws prohibit monopole electromagnetic waves, so Einstein's laws prohibit monopole gravitational waves. There is no possible way for any gravitational influence of the radial collapse to propagate outward. Not only is Birkhoff's theorem easy to understand, but it is also fairly easy to prove. Consider a spherical region of spacetime. Spherical symmetry alone is suffi cient to guarantee that conditions (i), (ii), and (iii) of Box 23.3 are satisfied, and thus to guarantee that one can introduce Schwarzschild coordinates ds 2 = _e 21> dt 2 + e 2A dr 2 + r 2(d8 2 + sin28 d\u00a22), (/J = (/J(t, r), and A = A(t, r). [See Box 23.3 for proof; and notice that: (1) for generality one must allow gtt = - e21> and gTT = e2A to be positive or negative (no constraint on sign!); (2) at events where the gradient of the \"circumference function\" r is zero or null, Schwarzschild coor dinates cannot be introduced. The special case (Vr)2 = 0 is treated in exercise 32.1.] (32.2) The uniqueness of the Schwarzschild geometry: Birkhoff's theorem The physics undMlying Birkhoff's theorem Proof of Birkhoff's theorem 844 32. GRAVITATIONAL COLLAPSE Impose Einstein's vacuum field equation on the metric (32.2), using the orthonormal components of the Einstein tensor as derived in exercise 14.16: Gii = r- 2(1 - e-2A ) + 2(A,,/r)e-2.1 = 0, Gi; = G;i = 2(A,t/r)e-1A + tPl = 0, G;; = 2(1/J,,/r)e-2A + r- 2(e- 2\u20221 GiJiJ = G;:,;:, = +(I/J'TT + 1/J,,2 - I/J\"A\" + I/J,,/r - A,,/r)e-2A", " (A,tt + A/ - A,tl/J,t)e-2tP = 0. I) = 0, (32.3a) (32.3c) (32.3b) (32.3d) Equation (32.3b) guarantees that A is a function of r only, and equation (32.3a) then guarantees that A has the same form as for the Schwarzschild metric: A = - \"2 In II - 2M/rl. I (32.4a) Equations (32.3c,d) then become two equivalent equations for I/J(t, r)-equivalent by virtue of the Bianchi identity, V' G = O-whose solution is I/J =t in II - 2M/rl + f(t). (32.4b) Here f is an arbitrary function. Put expressions (32.4) into the line element (32.2); thereby obtain ds 2 = _e2f1tl(1 _ 2M r ) dt2 + 2 dr I - 2M/r + r2(d8 2 + sin28 d\u00a22).. Then redefine the time coordinate t =f ef1t) dt new, and thereby bring the line element into the Schwarzschild form ds 2 = _(I _2M) dt2 + r dr'L I - 2M/r + r2(d8 2 + sin28 d\u00a22). Conclusion: When the spacetime surrounding any object has spherical symmetry and is free of charge, mass, and all fields other than gravity, then one can introduce coordinates in which the metric is that of Schwarzschild. Conclusion restated in coordinate-free language: the geometry of any spherically symmetric vacuum region of spacetime is a piece of the Schwarzschild geometry (Birkhoff's theorem). Q.E.D. EXERCISE Exercise 32.1. UNIQUENESS OF REISSNER-NORDSTRepM GEOMETRY [Track 2] Prove the following generalization of Birkhoff's theorem. Let the geometry ofa given region of spacetime (1) be spherically symmetric, and (2) be a solution to the Einstein field equations \u00a732.2. BIRKHOFFS THEOREM 845 with an electromagnetic field as source. Then tl.tat geometry is necessarily a piece of the Reissner-Nordstrem geometry [equation (3 I 24b)] with electric and magnetic fields, as meas ured in the standard static orthonormal frames [Hints: (1) First consider regions of spacetime in which (Vr)2 # 0, using the same methods as the text uses for Birkhoff's theorem. The result is the Reissner-Nordstrem solution. (2) Any region of dimensionality less than four, in which (Vr)2 = 0 (e.g., the Schwarzschild radius), can be treated as the join between four-dimensional regions with (Vr)2 # O. More over, the geometry of such a region is determined uniquely by the geometry of the adjoining four-dimensional regions (\"junction conditions\"; \u00a721.13). Since the adjoining regions are necessarily Reissner-Nordstrem (step I), then so are such \"sandwiched\" regions. (3) Next consider four-dimensional regions in which Vr = dr is null and nonzero. Show that in such regions there exist coordinate systems with where 'l' = 'l'(r, t). Show further that the Ricci tensor for this line element has an orthonor malized component whereas the stress-energy tensor for a spherically symmetric electromagnetic field has Q = const. These quantities, RfJfJ and 87TTfJfJ' must be equal (Einstein's field equation) but cannot be because of their different r-dependence. Thus, an electromagnetic field cannot generate regions with dr # 0, dr-dr= O. (4) Finally, consider four-dimensional regions in which dr = O. Denote the constant value of r by a, and show that any event can be chosen as the origin of a locally well-behaved coordinate system with A = ACT, z), A(1' = 0, z) = 0, ~(1' = 0, z) = O. [Novikov-type coordinate system; see \u00a731.4.] Show that, in the associated orthonormal frame, spherical symmetry demands and that the Einstein field equations then require Q = a and e A = cos 1', so that (5) This solution of the field equations [sometimes called the \"Bertotti (1959)-Robinson (1959a) Electromagnetic Universe,\" and explored in this coordinate system by Lind", "quist (1960)] is actually the throat of the Reissner-Nordstrom solution for the special case Q = M. Verify this claim by performing the following coordinate transformation on the Reissner-Nordstmm throat region [equation (31.24b) with Q = M and Ir - QI ~ Q]: r - Q = Qe-z cos 1', t = Qez tan 1'. 846 32. GRAVITATIONAL COLLAPSE (6) Thus, each possible case leads either to no solution at all, or to a segment of the Reissner-Nordstrvm geometry. Q.E.D.1 Note: The missing case. (Vr)2 = 0, in the text's proof of Birkholf's theorem. is resolved by noting that, for Q = 0, steps (3) and (4) above lead to no solutions at all. We thank G. F. R. Ellis for pointing out the omission of the case (Vrf =\u00b0in the preliminary version of this book. \u00a732.3. EXTERIOR GEOMETRY OF A COLLAPSING STAR Consider a star that is momentarily static, but will subsequently begin to collapse. Its space geometry at the initial moment of Schwarzschild coordinate time, t = 0, has two parts: in the exterior, vacuum region (r > R > 2M), it is the Schwarzschild geometry (Birkhoff's theorem!); but in the star's interior, it is some other, totally different geometry. Whatever the interior geometry may be, it has an embedding diagram at time t = 0 which is qualitatively like that of Figure 23.1. (For discussion and proof of this, see \u00a723.8.) Notice that the star's space geometry is obtained by discarding the lower universe of the full Schwarzschild geometry (Figure 31.5,a), and replacing it with a smooth \"bowl\" on which the matter of the star is contained. To follow the subsequent collapse of this star in the Schwarzschild coordinate system, or in the Kruskal-Szekeres coordinate system, or in an ingoing Eddington Finkelstein coordinate system, one can similarly discard that part of the coordinate diagram which lies inside the star's surface, and keep only the exterior Schwarzschild region. (See Figure 32.1.) In place of the discarded interior Schwarzschild region, one must introduce some other coordinate system, line element, and diagram that correctly describe the interior of the collapsing star. From truncated coordinate diagrams (such as Figures 32.l,a,b,c), one can readily discover and understand the various peculiar features of collapse through the gravi tational radius. (1) No matter how stiff may be the matter of which a (spherical) star is made, once its surface has collapsed within the gravitational radius, the star will continue to collapse until its surface gets crushed in the singularity at r = O. This one discovers by recalling that the star's surface cannot move faster than the speed of light, so its world line must always make an angle of less than 45 degrees with the v-axis of the Kruskal-Szekeres diagram. (2) No signal (e.g., photon) emitted from the star's surface after it collapses inside the gravitational radius can ever escape to an external observer. Rather, all signals emitted from inside the gravitational radius get caught and destroyed by the collapse of the surrounding geometry into the singularity at r = 0 as space \"pinches off\" around the star. (3) Consequently, an external observer can never see the star after it passes the gravitational radius; and he can never see the singularity that terminates its col lapse-unless he chooses to fall through the gravitational radius himself and pay the price of death for the knowledge gained. Gravitational collapse analyzed by examining the star's exterior, Schwarzschild geometry The gravitational radius as a point of no return, and the \"crushing\" at r = 0 The redshift of signals emitted just before passage through the gravitational radius Optical appearance of the collapsing star \u00a732.3. EXTERIOR GEOMETRY OF A COLLAPSING STAR 847 Does this mean that the collapsing star instantaneously and completely disappears from external view as it reaches the gravitational radius? No, not according to the analysis depicted in Figure 31.1,c: Place an astrophysicist on the surface of a collapsing star, and have him send a series of uniformly spaced signals to a distant astronomer, at rest at r ~ 2M, to inform him of the progress of the collapse. These signals propagate along null lines in the spacetime diagram of Figure 31.1 c. The signals originate on the world line of the stellar surface, and they are received by the distant astronomer when they intersect his world line, r = constant ~ M. As the star collapses closer and closer to its gravitational radius, R = 2M, the signals, which", " are sent at equally spaced intervals according to the astrophysicist's clock, are received by the astronomer at more and more widely spaced intervals. The astrono mer does not receive a signal emitted just before the gravitational radius is reached until after an infinite amount of time has elapsed; and he never receives signals emitted after the gravitational radius has been passed. Those signals, like the astro physicist who sends them, after brief runs get caught and destroyed by the collapsing geometry in the singularity, at r = O. It is not only the star that collapses. The geometry around the star collapses. Hence, to the distant astronomer, the collapsing star appears to slow down as it light from the star becomes more and more approaches its gravitational radius: red-shifted. Clocks on the star appear to run more and more slowly. It takes an infinite time for the star to reach its gravitational radius; and, as seen by the distant astronomer, the star never gets beyond there. The optical appearance of a collapsing star was first analyzed mathematically, giving main attention to radially propagating photons, by J. R. Oppenheimer and H. Snyder (1939). More recently a number of workers have reexamined the problem [see, e.g., Podurets (1964),- Ames and Thorne (1968) and Jaffe (1969)]. The most important quantitative results of these studies are as follows. In the late stages of collapse, when the distant astronomer sees the star to be very near its gravitational radius, he observes its total luminosity to decay exponentially in time L <X exp( __2__t_). 302M (32.5) Simultaneously, photons that travel exponentially increasing redshifts to him along radial trajectories arrive with z = LlA/A <X et / 4M\u2022 (32.6) However, the light from the star is dominated in these late stages, not by photons flying along radial trajectories from near the gravitational radius, but by photons that were deposited by the star in unstable circular orbits as its surface passed through r = 3M (see \u00a725.6 and Box 25.7). As time passes, these photons gradually leak out the diffuse spherical shell of trapped photons at r = 3M and fly off to the distant observer, who measures them to have redshift z :::: 2. Consequently, in the late stages of collapse the star's spectral lines are broadened enormously, but they are brightest at redshift z :::: 2. (colltinued 011 page 850) 49 48 47 46 45 t 43,\\1 42 I 41 40 39 38 37 36 35 3.0 2.5 2.0 1.5 0- ~-+'::-\"-t----I--- I = 42.8.\\1 I = 44.3.\\1 I = 43.5.\\1 1=42.1.\\-1 1= 41.3JI I = 40.2.M ----1.~ -2.5 o 0.5 1.0 1.5 2.0 2.5 (a) Schwarzschild (b) Kruskal-Szekeres Figure 32.1. The free-fall collapse of a star of initial radius Ri = 10 M as depicted alternatively in (a) Schwarzschild coordinates, (b) Kruskal-Szekeres coordinates, and (c) ingoing Eddington-Finkelstein coordinates (see Box 31.2). The region of spacetime inside the collapsing star is grey, is white. Only the geometry of the exterior region is that of Schwarzschild. The curve separating the grey and white regions is the geodesic world line of the surface of the collapsing star (equations [31.10) or [32.10) with rmax = R i = 10 M). This world line is parameterized by proper time,.,., as measured by an observer who sits on the surface of the star; the radial light cones, as calculated from ds 2 = 0, are attached to it. the same relative to Schwarzschild coordinates or relative to Eddington-Finkelstein coordinates, they are all the same relative to Kruskal Szekeres coordinates. This is because light rays travel along 45-degree lines in the u,v-plane (dv = \u00b1du), but they travel along curved paths in the r,l-plane and r, V-plane. Notice that, although the shapes of the light cones are not all that outside it The Kruskal-Szekeres spacetime diagram shown here is related to the Schwarzschild diagram by equations (31.13) pius a translation of Schwarzschild time: 1-+1 + 42.8 M. The Eddington-Finklestein diagram is related to the Schwarzschild diagram by V = I + r\u00b7 = 1+ r + 2 Min Ir/2 M - Ii (see Box 31.2). It is", " evident from these diagrams that the free-fall collapse is characterized by a constantly diminishing radius, which drops from R = 10 M to R = 0 in a finite and short comoving proper time interval,.11.,. = 35.1 M. The point R = 0 and the entire region r = 0 outside the star make up a physical \"singular ity\" at which infinite tidal gravitational forces-according to classical, unquantized general relativity-can and do crush matter to infinite density (see end of \u00a731.2; also \u00a732.6). 849./ II '! /t: /~o./:\u00a7 /t / \",,'\" (~ /~ /~ I~ ; r'j! II : <Ie!,g!~!~ J'\u00a7 :... i c: : 0 i 0 i~! l i i ::!; :: H i! :;;: ~!!: i!! i ~ '\"... E 0 <1:: \"0 ~ 'f '\"c: 0 \"0.<:: Q., VIM 68 66 64 62 60 54 52 50 48 46 :/ /~ :'.....,\"! II /j.../~ / /,,\"\"\" \" 0\" II T = 35.1 M :' ~ /\"1:1....!:,... I;?,''',: ~, 0! -0! Q...!;;...'...... ',:' l ;,t:::..... \"\" : : 0 T = 33.7M.. 0/~, I,/1,/ I I,/ /~ I try,/1 /1, /,/ : 1 I I I \u00b71... T = 32.4M --''--_........_----l__..\"... r/ M 32 I...;.._--'2 o 8 (c) Eddington-Finkelstein 6 10 The Eddington-Finkelstein diagram depicts a series of photons emitted radially from the surface of the collapsing star, and received by an observer at r = Rinitial = 10 M. The observer eventually-receives all photons emitted radially from outside the gravitational radius; all photons emitted after the star passes through its gravitational radius eventually get pulled into the singularity at r = 0; and any photon emitted radially at the gravitational radius stays at the gravitational radius forever. Non-free-fall collapse is similar to the collapse depicted here. When pressure gradients are present, only the detailed shape of the world line of the star's surface changes, 850 32. GRAVITATIONAL COLLAPSE Notice how short is the characteristic e-folding time for the decay of luminosity and for the radial redshift: 'Tchar = 2M;:::, I X 1O-5(M/M0 ) sec light-travel time across a flat-space distance equal to the gravitational radius. ) ( - (32.7) Here M0 denotes one solar mass. EXERCISE Exercise 32.2. REDSHIFTS DURING COLLAPSE The rest of this chapter is Track 2. No, previous Track-2 material is needed as preparation for it, but it is needed as preparation for (1) the Track-2 part of Chapter 33 (black holes), and (2) Chapter 34 (singularities and global methods). (a) Let a radio transmitter on the surface of a collapsing spherical star emit monochromatic waves of wavelength At: and let a distant observer. at the same 8, <1>. as the transmitter, receive the waves. Show that at late times the wavelength received varies as (32.8a) [equation (32.6)]. where f is proper time as measured by the distant observer. (b) [Track 2] Use kinetic theory for the outgoing photons (conservation of density in phase space: Liouville's theorem; \u00a722.6) to show that the energy flux of the radiation received (ergs/cm 2 sec) varies as (32.8b) (c) Suppose that nuclear reactions at the center of the collapsing star generate neutrinos of energy E., and that these neutrinos flow freely outward (negligible absorption in star). Show that the energy of the neutrinos received by a distant observer decreases at late times as (d) Show that the flux of neutrino energy dies out at late times as (32.9a) (32.9b) (e) Explain in elementary terms why the decay laws (32.8a) and (32.9a) for energy are the same, but the decay laws (32.8b) and (32.9b) for energy flux are different. (f) Let a collapsing star emit photons from its surface at the black-body rate dN ( - = 1.5 X 101 d'T 1 photonS) cm", " 2 sec K3 X (SUrfaCe area) of star X (temperature)3 of surface. Let a distant observer coUnt the photons as they pass through his sphere of radius r ~ M. Let him begin his count (time t = 0) when he sees (via photons traveling radially outward) the center of the star's surface pass through the radius r = 3M. Show that, in order of magnitude, the time he and his associates must wait, until the last photon that will ever get out has reached them, is t = (M/Mo)[8 X 10-4 + 5 X 1O-5 Iog10(Tll M/MdJ seconds, (32.9c) where Tll is the star's surface temperature in units of 1011 K. \u00a732.4. COLLAPSE WITH UNIFORM DENSITY AND ZERO PRESSURE 851 \u00a732.4. COLLAPSE OF A STAR WITH UNIFORM DENSITY AND ZERO PRESSURE When one turns attention to the interior of a collapsing star and to the precise world line that its surface follows in the Schwarzschild geometry, one encounters rather complicated mathematics. The simplest case to treat is that of a \"star\" with uniform density and zero pressure; and, indeed, until recently that was the only case which had been treated in detail. The original-and very complete-analysis of the collapse of such a uniform-density \"ball of dust\" was given in the classic paper of Oppen heimer and Snyder (1939). More recently, other workers have discussed it from slightly different points of view and using different coordinate systems. The ap proach taken here was devised by Beckedorff and Misner (1962). Because no pressure gradients are present to deflect their motion, the particles on the surface of any ball of dust must move along radial geodesics in the exterior Schwarzschild geometry. For a ball that begins at rest with finite radius, R = R i, at time t = 0, the subsequent geodesic motion of its surface is given by equations (31.10): The collapse, from rest, of a uniform-density ball of \" dust\": R = (RJ2)(1 + cos 1)), t = 2MlnI(RJ2M - 1)1/2 + tan (1)/2)I (Ri /2M - 1)112 - tan (1)/2) 1)112[1) + (R/4M)(1) + sin 1))]. + 2M(R/2M - (1) world line of ball's surface in exterior Schwarzschild coordinates (32.10a) (32.1Ob) Here R is the Schwarzschild radial coordinate (i.e., 4'ITR2 is the star's surface area) at Schwarzschild time t. This world line is plotted in Figure 32.1 for Ri = 10M, in terms of Schwarzschild coordinates, Kruskal-Szekeres coordinates, and Eddington Finkelstein coordinates. The proper time read by a clock on the surface of the collapsing star is given by equation (31.10b): (32.1Oc) Note that the collapse begins when the parameter 1) is zero (R = Ri, t = 'T = 0); and it terminates at the singularity (R = 0,1) = 'IT) after a lapse of proper time, as measured on any test particle falling with the dust, equal to It is interesting, though coincidental, that this is precisely the time-lapse required for free-fall collapse to infinite density in Newtonian theory [see equation (25.27'), -------BFrgule 25.3, and a~_iateddiS\u20acussion]. What is the behavior of the interior of the ball of dust as it collapses? A variety of different interiors for pressureless dust Can be conceived (exercise 32.8). But here attention focuses on the simplest of them: an interior that is homogeneous and isotropic everywhere, except at the surface-i.e., an interior locally identical to a dust-filled Friedmann cosmological model (Box 27.1). Is the Friedmann interior to be \"open\" (k = - 1), \"flat\" (k = 0), or \"closed\" (k = + I)? Only the closed case (2) interior of ball is identical to a portion of a closed Friedmann universe y (3) the join between Friedmann interior and Schwarzschild exterior 852 32. GRAVITATIONAL COLLAPSE is appropriate, since one has already demanded [equation (32.10)] that the star be at rest initially (initial rate of change of density equals zero; \"moment of maximum expansion\"). Using comoving hyperspherical coordinates, x, B, ep, for the star's interior, and putting the", " origin of coordinates at the star's center, one can write the line element in the interior in the familiar Friedmann form Here a(or) is given by the familiar cycloidal relation, and the density is given by (32.11 ) (32.12) (32.13) [see equations (1), (9), (4), and (5) of Box 27.1, with 1) replaced by 1) + 'IT]. There is one possible difficulty with this interior solution. In the cosmological case, the solution was homogeneous and isotropic everywhere. Here homogeneity and isotropy are broken at the star's surface-which lies at some radius (32.14) x = Xo for all or, as measured in terms of the hyperspherical polar angle x, a comoving coordinate (first picture in Box 27.2). At that surface (i.e., three-dimensional world tube enclosing the star's fluid) the interior Friedmann geometry must match smoothly onto the exterior Schwarzschild geometry. If the match cannot be achieved, then the Friedmann line element (32.11) cannot represent the interior of a collapsing star. An example of a case in which the matching could not be achieved is an interior of uniform and nonzero pressure, as well as uniform density. In that case there would be an infinite pressure gradient at the star's surface, which would blow off the outer layers of the star, and would send a rarefaction wave propagating inward toward its center. The uniform distribution of density and pressure would quickly be de stroyed. For the case of zero pressure, the match is possible. As a partial verification of the match, one can examine the separate and independent predictions made by the interior and exterior solutions for the star's circumference, C =lwR, as a function of proper time or at the star's surface. The external Schwarzschild solution predicts the cycloidal relation, C = 2'lTR = 2'IT(RJ2)(l + COS1)), or = (Ri 3j8M)l/2(1) + sin 1)) (32.15) [equations (32.10)]. The interior Friedmann solution predicts a similar cycloidal relation: \u00a732.4. COLLAPSE WITH UNIFORM DENSITY AND ZERO PRESSURE 853 c = 27TR = 27Ta sin XO = 27T(tam sin xo)(1 + cos 1)), 7\" = tam(1) + sin 1)). The two predictions agree perfectly for all time if and only if M - l - 2 amsm Xo'. 3 (32.16) (32.17a) (32.17b) A more complete verification of the match is given in exercise 32.4. For further insight into this idealized model of gravitational collapse, see Box 32.1. Exercise 32.3. EMBEDDING DIAGRAMS AND PHOTON PROPAGATION EXERCISES FOR COLLAPSING STAR Verify in detail the features of homogeneous collapse described in Box 32.1. Exercise 32.4. MATCH OF FRIEDMANN INTERIOR TO SCHWARZSCHILD EXTERiOR The Einstein field equations are satisfied on a star's surface if and only if the intrinsic and extrinsic geometries of the surface's three-dimensional world tube are the same, whether measured on its interior or on its exterior (see \u00a721.l3 for proof and discussion). Verify that for the collapsing star discussed above, the intrinsic and extrinsic geometries match at the join between the Friedmann ir;:terior and the Schwarzschild exterior. [Hints: (a) Use Tf, (), cf>, as coordinates on the world tube of the star's surface, and show that the intrinsic geometry has the same line element (32.18a) whether measured in the Schwarzschild exterior or in the Friedmann interior. (b) Show that the extrinsic curvature of the world tube has the same components K.\".\" ::: K.,,8 ::: K.,,\u00a2> ::: K8\u00a2> ::: 0, K88 ::: K\u00a2>\u00a2>/sin2 () ::: -a(Tf) sin XO cos Xo, (32.18b) whether measured in the Schwarzschild exterior or in the Friedmann interior.] Exercise 32.5. STARS THAT COLLAPSE FROM INFINITY (a) Patch together a truncated Schwarzschild geometry and the geometry ofa truncated \"flat\" (k ::: 0), dust-filled Friedmann universe to obtain a model of a star that collapses from rest at an infinite initial radius. [Hint: The world line of the star's surface in the Schwarzschild geometry is given by equations (31.2).] (b) Similarly patch together a truncated Schwarzschild geometry and the geometry of", " a truncated \"open\" (k ::: -1), dust-filled Friedmann universe to obtain a star which collapses from infinity with finite initial inward velocity. (continued on page 857) 854 32. GRAVITATIONAL COLLAPSE Box 32.1 AN IDEALIZED COLLAPSING STAR WITH FRIEDMANN INTERIOR AND SCHWARZSCHILD EXTERIOR (See \u00a732.4 and exercises 32.3 and 32.4 for justification of the results described here.) Initial State (I) Take a Friedmann universe ofradius a = am at its moment of maximum expansion, 1) = 0; and slice off and discard the region XO < X ::; 7T, where XO is some angle less than 7T/2. (2) Take a Schwarzschild geometry of mass M = (am/2) sin3 XO at the moment t = 0; and slice off and discard the region r < Ri = am sin Xo' (3) Glue the remaining pieces of Friedmann and Schwarzschild geometry together smoothly along their cut surfaces. The resultant object will be a momentarily static star of ), of mass M = (a m/2) sin3 xo, and of radius R, = uniform density Pi = 3/(87Tam am sin Xo' 2 Subsequent Evolution Release this star from its intial state, and let it collapse in accord with Einstein's field equations. The interior, truncated Friedmann universe and the exterior, truncated Schwarzschild geometry will evolve just as though they had never been cut up and patched together; and this evolution will preserve the smoothness of the match between interior and exterior! Details of the Collapse Probe the details of the collapse using sequences of embedding diagrams (histories ABeD and A WXY), and using photons that propagate radially outward (photons lX, {3, y, 8, t:). The example shown here has Xo = 0.96 and Rtf M = 2/sin2 Xo = 3. Exterior (Kruskal Szekeres) Interior (Friedmann) vt W' B' '1/ =0 A' 0.96 = Xo -X __ A'U- _ ---1I---1~. History of Collapse as Probed by Hypersurfaces ABCD: (1) Initial configuration, A - A' - A\", is constructed by cutting and sewing at 1) = t = O. that times (2) Each subsequent configuration has as its inte rior a slice of constant Friedmann time 1). (3) The interior remains always a spherical cup of half-angle Xo; but it contracts from radius R = Ri = am sin XO to R = 0 as time increases. (4) The matter in the star is all crushed simultane ously to infinite density when R reaches zero, and the external Schwarzschild \"funnel\" de velops a cusp-like singularity at that point. ------- ~-- --~--(5) As time increases further, this cusp pulls the region r < 2M of the funnel into r = 0 so fast that the outward-traveling photon 8 is gobbled up and crushed. A\" B\" A' A A\" D\" D These embedding diagrams must be rotated about the vertical axes in order to become 2-dimensional surfaces analogous to Figure 23.1. Box 32.1 (continued) A\" History as Probed by Hypersurfaces AWXY A' (l) Initial configuration, A - A' - A\", is again that constructed by cutting and sewing at 1) = t = O. (2) Subsequent hypersurfaces are very different from 1) = const. (3) As time passes, a neck develops in the geom etry just outside the surface of the star. X'\" (4) This neck becomes tighter and tighter and then pinches off, leaving the star completely iso lated from the rest of the universe, and leaving a deadly cusp-like singularity in the exterior geometry where the star used to be. (5) The isolated star, in its own little closed uni verse, continues to contract until it is crushed to infinite density, while the external geometry begins to develop another neck and the cusp quickly gobbles up photon 8. A\" H'\" Y'--- A IV Y The extreme difference between histories ABeD and AWXY dramatizes the \"many-fingered time\" of general relativity. The hypersurface on which one explores the geometry can be pushed ahead faster in time in one region, at the option of the party of explorers. Thus whether one region of the star collapses first, or another, is a function both of the spacetime or the entire star collapses simultaneously, geometry and of the choice of slicing. The party of explorers has this choice of slicing at their own control", ", and thus they themselves to this extent govern what kind of spacelike slices they will see as their exploration moves forward in time. The spacetime geometry that they slice, however, is in no way theirs to control or to change. To the extent that their masses are negligible and they serve merely as test objects, they have no influence whatsoever on the spacetime. It was fixed completely by the specification of the initial conditions for the collapse. In brief, spacetime is four-dimensional and slices are only three-dimensional (and in the - pictures here look only two-dimensional or one-dimensional). Anyone set of slices captures only a one-sided view of the whole story. To see the entire picture one must either examine the dynamics of the geometry as it reveals itself in varied choices of the slicing or become accustomed to visualizing the spacetime geometry as a whole. Spherical collapse with pressure is qualitatively the same as without pressure \u00a732.5. SPHERICAL COLLAPSE WITH PRESSURE 857 \u00a732.5. SPHERICALLY SYMMETRIC COLLAPSE WITH INTERNAL PRESSURE FORCES So far as the external gravitational field is concerned, the only difference between a freely collapsing star and a collapsing, spherically symmetric star with internal pressure is this: that the surfaces of the two stars move along different world lines in the exterior Schwarzschild geometry. Because the exterior geometry is the same in both cases, the qualitative aspects offreefall collapse as described in the last section can be carried over directly to the case of nonnegligible internal pressure. An important and fascinating question to ask is this: can large internal pressures in any way prevent a collapsing star from being crushed to infinite density by infinite tidal gravitational forces? From the Kruskal-Szekeres diagram of Figure 32.l,b, it is evident that, once a star has passed inside its gravitational radius (R < 2M), no internal pressures, regardless of how large they may be, can prevent the star's surface from being crushed in a singularity. The surface must move along a time-like world line, and all such world lines inside r = 2M hit r = O. Although there is no such theorem now available, one can reasonably conjecture that, if the surface of a spherical configuration is crushed in the r = 0 singularity, the entire interior must also be crushed. The details of the interior dynamics of a spherically symmetric collapsing star with pressure are not so well-understood as the exterior Schwarzschild dynamics. However, major advances in one's understanding of the interior dynamics are now being made by means of numerical computations and analytic analyses [see Misner (l969a) for a review]. In these computations and analyses, no new features (at least, no unexpected ones) have been encountered beyond those that occurred in the simple uniform-density, free-fall collapse of the last section. Exercise 32.6. GENERAL SPHERICAL COLLAPSE: METRIC EXERCISES IN COMOVING COORDINATES Consider an inhomogeneous star with pressure, undergoing spherical collapse. Spherical symmetry alone is enough to guarantee the existence of a Schwarzschild coordinate system (t, r, e, </\u00bb throughout the interior and exterior of the star [see equation (32.2) and preceding discussion]. Label each spherical shell of the star by a parameter a, which tells how many baryons are contained interior to that shell. Then rea, t) is the world line of the shell with label a. The expression for these world lines can be inverted to obtain a(t, r), the number of baryons interior to radius r at time t. Show that there exists a new time coordinate t(t, r), such that the' line element (32.2), rewritten in the coordinates (t; a, e, </\u00bb, has the form ~_ _ ------------- - -'-~-- - - \" --- ;p = ;p(t; a), r = r(t; a), r = r(t; a). These are \"comoving, synchronous coordinates\" for the stellar interior. (32.l9a) (32.l9b) 858 32. GRAVITATIONAL COLLAPSE Exercise 32.7. ADIABATIC SPHERICAL COLLAPSE: EQUATIONS OF EVOLUTION [Misner and Sharp (1964)] Describe the interior of a cOllapsing star by the co moving, synchronous metric (32.19). by the number density of baryons n. by the total density of mass-energy P. and by the pressure p. The 4-velocity of the star's fluid is since the fluid is at rest in the coordinate system. Let a dot denote a proper time derivative as seen by the fluid-e.g., (32.20) and let a prime denote a partial derivative with respect to b", "aryon number.-e.g. n' = (on/oa);. Denote by V the rate of change of (1/2'IT) X (circumference of shell), as measured by a man riding in a given shell: and denote by m(T, a) the \"total mass-energy interior to shell a at time T: V=;; m{i, a) =La 47Tr 2p(i, a)r' da. o (32.21a) (32.21 b) (See Box 23.1 for discussion of this method of localizing mass-energy.) Assume that the collapse is adiabatic (no energy flow between adjacent shells; stress-energy tensor entirely that of a perfect fluid). (a) Show that the equations of collapse [baryon conservation, (22.3); local energy conser vation, (22.11a); Euler equation, (22.13); and Einstein field -equations (ex. 14.16)] can be reduced to the following eight equations for the eight functions ;P, r, r, n, p, p, V, m: ; == V )' (nr 2 V' ---;,:z == - 7 P Ii --==p + p n (dynamic equation for r); (32.22a) (dynamic equation for n); (32.22b) except at a shock front, where adiabaticity breaks down (dynamic equation for p); (32.22c) V\u00b7 - r 2 p' _ m + 47Tr 3p p + P r' r 2 (d \" ynamlc equatIOn,or <' V) ; (32.22d) p == p(n, p) ;p, == -p'/(p + p), ;p == 0 at star's surface m' == 47Tr 2pr', r == sign (r')(l + V2 - 2m/r)1/2 m == 0 at a == 0, (equation of state); (source equation for ;p); (source equation for m); (32.22e) (32.22f) (32.22g) (algebraic equation for r). (32.22h) (b) The preceding equations are in a form useful for numerical calculations. [For particular numerical solutions and for the handling of shocks, see May and White (1966).] For analytic work it is often useful to replace (32.22b) by n == r /(47Tr 2r'), (32.22b') \u00a732.5. SPHERICAL COLLAPSE WITH PRESSURE and (3222d) by Derive these equations. 859 (32.22d') (c) Explain why equations (32.22g) and (32.22d') justify the remarks made in Box 23.1 about localizability of energy. Exercise 32.8. ANALYTIC SOLUTIONS FOR PRESSURE-FREE COLLAPSE [Tolman (1934b); Datt (1938)] Show that the general solution to equations (3222) in the case of zero pressure can be generated as follows. (a) Specify the mass inside shell a, mea); by equation (32.22d'), with p = 0, it will not change with time t. (b) Assume that all the dust particles have rest masses p. that depend upon radius, p.(a); so (c) Calculate r from the equation p = p.n. r = m'Ip.; (32.23a) (3223b) it will be independent of t. (d) Specify an initial distribution of circumference 27Tr as function of a, and solve the dynamic equation (~)2 _2m(a) = Pea) _ 1 at r (32.23c) to obtain the subsequent evolution of r(T, a). Notice that this equation has identically the same form as in Newtonian theory! (e) Calculate the remaining_quantities of interest from the algebraic equations ds 2 = -dt2 + (r'dalr)2 + r2 d[}2, p = p.n = m'/(47Tr2r'), U = or/at. ;p = 0, (32.23d) (32.23e) (32.23f) [Note: In this solution, successive \"shells\" may pass through each other, producing a surface 0 where m' \u00a5- 0), since there is no pressure built up of infinite density as they do (r' to stop shell crossing. When this happens, the coordinate system becomes pathological (a no longer increases monotonically outward), but spacetime remains well-behaved. The surface of infinite density (I) produces", " negligible tidal forces on neighboring dust particles; and (2) like the surface layers of \u00a721.13, it is an idealization that gets smeared down to finite density by finite pressure.] Exercise 32.9. COLLAPSE WITH UNIFORM DENSITY Recover the Friedtnann-Schwarzschild solution for collapse with uniform density and zero pressure by specifying appropriate forms for mea) and rea) in the prescription of exercise 32.8. In the interior of the star, give the dust particles nonzero rest masses, p. = constant \u00a5- 0; in the exterior give them zero rest masses, p. = 0 (\"imaginary dust particles\" in vacuum). Reduce the resulting metric (32.23d) to that of Friedmann inside the star, and to that of Novikov for the Schwarzschild geometry outside the star [equations (31.12)]. The effect of tidal forces on the body of a man falling into the r = 0 singularity: Stage 1: body resists deformation; stresses build up 860 32. GRAVITATIONAL COLLAPSE \u00a732.6. THE FATE OF A MAN WHO FALLS INTO THE SINGULARITY AT r = 0 Consider the plight of an experimental astrophysicist who stands on the surface of a freely falling star as it collapses to R = O. As the collapse proceeds toward R = 0, the various parts of the astrophysicist's body experience different gravitational forces. His feet, which.are on the surface of the star, are attracted toward the star's center by an infinitely mounting gravita tional force; while his head, which is farther away, is accelerated downward by a somewhat smaller, though ever rising force. The difference between the two acceler ations (tidal force) mounts higher and higher as the collapse proceeds, finally be coming infinite as R reaches zero. The astrophysicist's body, which cannot withstand such extreme forces, suffers unlimited stretching between head and foot as R drops to zero. But this is not all. Simultaneous with this head-to-foot stretching, the astrophysicist is pulled by the gravitational field into regions of spacetime with ever-decreasing circumferential area, 4'1Tr2. In order to accomplish this, tidal gravitational forces must compress the astrophysicist on all sides as they stretch him from head to foot. The circumferential compression is actually more extreme than the longitudinal stretch ing; so the astrophysicist, in the limit R -+ 0, is crushed to zero volume and indefinitely extended length. The above discussion can be put on a mathematical footing as follows. There are three stages in the killing of the astrophysicist: (l) the early stage, when his body successfully resists the tidal forces; (2) the intermediate stage, when it is gradually succumbing; and (3) the final stage, when it has been completely over whelmed. During the early stage, one can analyze the tidal forces by means of the equation of geodesic deviation, evaluated in the astrophysicist's orthonormal frame w T, w P, we, wJ, (see \u00a731.2). In this frame, the nonvanishing components of the Riemann curvature tensor are given by equations (31.6): RTPTP = -2M/r3, RliJ,liJ, = 2M/r3, R\u00b7\u00b7\u00b7\u00b7 - R\u00b7\u00b7\u00b7\u00b7 - -Mjr3 p</Jp</J -. pBpB - (32.24a) \u2022 The equation of geodesic deviation says that twofreefy moving particles, momentarily at rest in the astrophysicist's local inertial frame, and separated by the 3-vector must accelerate apart with a relative acceleration given by '.> D21J/d-r2 = _RL,.l:.k = _R,..\u2022 l:.k = _R\u00b7,\u00b7,l:.k,TkT'.> 1'''1''.> 1\"1''''.> \u2022 \u00a732.6. GORE AT THE SINGULARITY Using the components (32.24a) of the curvature tensor, one sees that D2~P/dT2 = +(2M/r3)~p, D2~8/dT 2 = _(M/r3)~D, D2~;P/dT 2 = -(M/r3)~;P. 861 (32.24b) To apply these equations to the astrophysicist's body, idealize it (for simplicity) as a homogeneous rectangular box of mass p.::::: 165 pounds::::: 75 kg, of length 1 ::::: 70 inches::::: 1.8 m in the e p direction, and of width and depth w::::: 10 inches ::::: 0.2 m in the eo and", " e;p directions. Then calculate the stresses that must be set up in this idealized body to prevent its particles from moving along diverging (and converging) geodesics. From the form of equations (32.24), it is evident that the principal directions of the stress will be e p, eo, and e;p (i.e., in the e p' eo, e;p basis, the stress tensor will be diagonal). The longitudinal component of the stress, at the astrophysicist's center of mass, can be evaluated as follows. A volume element of his body with mass dp., located at a height h above the center of mass (distance h measured along e p direction) would accelerate with a = (2M/r3)h away from the center of mass, if it were allowed to move freely. To prevent this acceleration, the astrophysicist's muscles must exert a force dF = a dp. = (2M/r3)h dp.. This force contributes to the stress across the horizontal plane (eo /\\ e;p plane) through the center of mass. The total force across that plane is the sum of the forces on all mass elements abo\\T~)t (which is also equal to the sum of the forces on the mass elements below it): F= f (region above plane) 1 p.Ml - 4\"-;:J' adp. =f. 0 1/2(2Mh)( P. -3 - - - 2 (w2dh) 1w r ) The stress is this force divided by the cross-sectional area w 2, with a minus sign because it is a tension rather than a pressure: T.. = _ ~ p.Ml ::::: -1.1 X 1015 M/Me;) dynes. PP 4 w 2r3 (r/l km)3 cm2 (32.25a) The components of the stress in the eo and e;p directions at the center of mass are, similarly, (32.25 b) (Recall that one atmosphere of pressure is 1.01 X 106 dynes/cm2.) y Stage 2: body gives way; man dies Stage 3: body gets crushed and distended 862 32. GRAVITATIONAL COLLAPSE The human body cannot withstand a tension or pressure of ~ 100 atmospheres ;:::: 10 8 dynes/cm2 without breaking. Consequently, an astrophysicist on a freely collapsing star of one solar mass will be killed by tidal forces when the star's radius is R - 200 km > 2M;:::: 3 km. By the time the star is much smaller than its gravitational radius, the baryons of the astrophysicist's body are moving along geodesics; his muscles and bones have completely given way. In this final stage of collapse, the timelike geodesics are curves along which the Schwarzschild \"time\"-coordinate, t, is almost constant [ef the nar rowing down of the light cones near r = 0 in Figure 32.l,a; also equation (31.2) in the limit r ~ 2M]. The astrophysicist's feet touch the star's surface at one value of t-say t = t,-while his head moves along the curve t = th > t,. Consequently, the length of the astrophysicist's body increases according to the formula lastroPh = [gtt(R)]l/2[t h - a: R-1/2 a: ( t,] = [2MjR]l/2[t h _ )-1/3 Tcollapse T. t,] (32.26a) Here T = [-f R lgrrl1/2 dr + constant] is proper time as it would be measured by the astrophysicist if he were still alive, and Tcollapse is the time at which he hits r = O. The gravitational field also constrains the baryons of the astrophysicist's body to fall along world lines of constant () and ep during the final stages of collapse. Conse quently, his cross-sectional area decreases according to the law t1 astroph = [gee(R)g<l>iR)]l/2 UJ,jep a: R2 a: (Tcollapse - T)4/3. (32.26b) By combining equations (32.26a,b), one sees that the volume of the astrophysicist's body decreases, during the last few moments of collapse, according to the law 'lIastroph = lastroPht1astroph a: R3/2 a: (Tcollapse - T). (32.26c) This crushing of matter to infinite density by infinitely large tidal gravitational forces can Occur not only on the surface of the collapsing star, but also at any", " other point along the r = 0 singularity outside the surface of the star. Hence, any foolish rocketeer who ventures below the radius r = 2M of the external gravitational field is doomed to destruction. For further discussion of spacetime singularities, and of the possibility that quan tum gravitational effects might force a reconsideration of the singularities predicted by classical gravitation theory, see Chapter 30, \u00a734.6, and Chapter 44. \u00a732.7. REALISTIC GRAVITATIONAL COLLAPSE AN OVERVIEW Review of spherical collapse Instability, implosion, horizon, and singularity; these are the key stages in the spherical collapse of any star. Instability: The star, having exhausted its nuclear fuel, and having contracted slowly inward, begins to squeeze its pressure-sustaining electrons or photons onto its atomic nuclei; this softens the equation of state, which induces an instability [see, e.g., \u00a7\u00a7lO.l5 and 11.4 ofZel'dovich and Novikov (1971) \u00a732.7. NON SPHERICAL GRAVITATIONAL COLLAPSE 863 for details]. Implosion: Within a fraction of a second the instability develops into a full-scale implosion; for realistic density distributions, the stellar core falls rapidly inward on itself, and the outer envelopes trail along behind [see, e.g., the numerical calculations of Colgate and White (1966), Arnett (1966, 1967), May and White (1966), and Ivanova, Imshennik, and Nadezhin (1969)]. Horizon: In the idealized spherical case, the star's surface falls through its gravitational radius (\"horizon\"; end of communication with the exterior; point of no return). From the star's vantage point this happens after a finite, short lapse of proper time. But from an external vantage point the star requires infinite time to reach the horizon, though it becomes black exponentially rapidly in the process [e-folding time - M - 10-5(M/M0 ) sec]. The result is a \"black hole\", whose boundary is the horizon (gravitational radius), and whose interior can never communicate with the exterior. Singularity: From the star's interior vantage point, within a short proper time interval.1T - M - 1O-5(M/Md sec after passing through the horizon, a singularity is reached (zero radius, infinite density, infinite tidal gravitational forces). Does this basic picture-instability, implosion, horizon, singularity-have any relevance for real stars? Might complications such as rotation, nonsphericity, mag netic fields, and neutrino fluxes alter the qualitative picture? No, not for small initial perturbations from sphericity. Perturbation theory analyses described in Box 32.2 and exercise 32.10 show that realistic, almost-spherically symmetric collapse, like implosion, horizon; and Penrose is characterized by instability, idealized collapse, (1965b; see \u00a734.6) proves that some type of singularity then follows. Highly nonspherical collapse is more poorly understood, of course. Nevertheless, a number of detailed calculations and precise theorems point with some confidence to two conclusions: (1) horizons (probably)form when and only when a mass M gets compacted into a region whose circumference in EVERY direction is e :s 4'1TM (Box 32.3); (2) the external gravitational field ofa horizon (black hole), after all the \"dust\" and gravitational waves have cleared away, is almost certainly the Kerr-Newman generalization of the Schwarzschild geometry (Chapter 33). If so, then the external field is determined uniquely by the mass, charge, and angular momentum that went \"down the hole.\" (This nearly proved theorem carries the colloquial title \"A black hole has no hair.\") The interior of the horizon, and the endpoint (if any) of the collapse are very poorly understood today. The various possibilities will be reviewed in Chapter 34. That a singularity occurs one can state with much certainty, thanks to theorems of Penrose, Hawking, and Geroch. But whether all, only some, or none of the collapsing matter and fields ultimately encounter the singularity one does not know. -~~--~~-------- - Summary of 1972 knowledge about realistic, nonspherical collapse: (1) horizon (2) black hole (3) singularity Exercise 32.10. PRICE'S THEOREM FOR A SCALAR FIELD EXERCISES [See Price (1971, 1972a), also Thorne (1972), for more details than are presented here.] A collapsing spherical star, with an arbitrary nonspherical \"scalar charge distribution,\" a = O. generates an external scalar field r/J. The vacuum field equation for r/J is Dr/J = r/J: a Ignore the", " back-reaction of the field's stress-energy on the geometry of spacetime. (continued on page 868) 864 32. GRAVITATIONAL COLLAPSE Box 32.2 COLLAPSE WITH SMALL NON-SPHERICAL PERTURBATIONS [based ondetailed calculations by Richard H. Price (1971, 1972a,b)]. A. Density Perturbations 1. When star begins to collapse, it possesses a small nonspherical \"lump\" in its density distri bution. 2. As collapse proceeds, lump grows larger and larger [instability of collapse against small per turbations-a phenomenon well known in Newtonian theory; see, e.g., Hunter (1967); Lin, Mestel, and Shu (1965)]. 3. The growing lump radiates gravitational waves. 4. Waves of short wavelength (A. ~ M), emitted from near horizon (r - 2M :S M), partly prop agate to infinity and partly get backscattered by the \"background\" Schwarzschild curvature of spacetime. Backscattered waves propagate into horizon (surface of black hole; gravita tional radius) formed by collapsing star. 5. Waves of long wavelength (A. > M), emitted from near horizon (r - 2M :S M), get fully backscattered by spacetime curvature; they never reach out beyond r - 3M; they end up propagating \"down the hole.\" 6. Is lump on star still there as star plunges through horizon, and does star thereby create a deformed (lumpy) horizon? Yes, according to calculations. 7. But external observers can only learn about existence of \"final lump\" by examining defor mation (quadrupole moment) in final gravita tional field. That final deformation in field does not and cannot propagate outward with infinite speed (no instantaneous \"action at a distance\"). It propagates with speed of light, in form of gravitational waves with near-infinite wave length (infinite redshift from edge of horizon to any external radius). Deformation in final field, like any other wave of long wavelength, gets fully backscattered by curvature of spacetime at r :S 3M; it cannot reach external observers. External observers can never learn of existence '~'\" \"... Col1apse depicted in ingoing Eddington-Finkelstein coordinates \u00a7 32.7. NONSPHERICAL GRAVITATIONAL COLLAPSE 865 C. Perturbations in Electromagnetic Field 1. Star possesses a magnetic field generated by currents in its interior, and an electric field due to an arbitrary internal charge distribution; and electromagnetic radiation is emitted by its hot matter. For simplicity, S is assumed zero. 2. Evolution of external electromagnetic field is similar to evolution of perturbations in external gravitational field. Distant observer can never learn \"final\" values of changeable quantities (magnetic dipole moment, electric dipole mo ment, quadrupole moments,...). Final values try to propagate out from horizon, carried by electromagnetic waves of near-infinite wave length. But they cannot get out: spacetime cur vature reflects them back down the hole; and they superpose destructively with their outgoing counterparts, to produce zero net field. 3. By contrast with all other quantities, which are changeable, the electric monopole moment (total flux of electric field; equal to 4'1T times~ It n~ver: total electric charge) changes from before star collapses, through the; collapse stage, into the quiescent black-hole stage. is conserved. 4. Hence, the final external electromagnetic field is a spherically symmetric coulomb field E = (Q/r 2)er} as measured by static B = 0 observer (r, 8, cp, constant); and the final spacetime geometry is that of Reissner and Nordstrom (charged black hole; see exercises 31.8 and 32.1): of final lump. Final external field is perfectly spherical, lumpjree, Schwarzschild geometry! 8. Even in region of backscatter (2M < r ~ 3M), final external field is lump-free. Backscattered waves, carrying information about existence of final lump, interfere destructively with outgoing waves carrying same information. Result is de struction of all deformation in external field and in horizon! 9. Final black hole is a Schwarzschild black hole! B. Perturbations in Angular Momentum 1. When star begins to collapse, it possesses a small, nonzero intrinsic angular momentum (\"spin\") S. 2. As collapse proceeds, S is conserved (except for a tiny, negligible change due to angular mo mentum carried off by waves; that change is proportional to square of amplitude of waves, i.e., to square of amplitude of perturbations in star, i.e., to S2). 3", ". Consequently, external field always and every where carries imprint of angular momentum S (on imprints, see Chapter 19). There is no need for that to propagate outward from near horizon. Moreover, it could not so propa gate even ifit tried, because of the conservation law for S (absence of dipole gravitational waves; see \u00a7\u00a736.1 and 36.10). imprint 4. Hence, the final external field is that of an undeformed, slowly rotating black hole: ds2 = _(1 _ 2M) dt2 + r 2 dr 1- 2M/r + r2 d{22.I, - Schwarzschild geometry 4S sin 8)( r2 (. 8 A ucp ) d t. r Sin rotational imprint, see exercise 26.1; also Chapter 19. Here the polar axis has been oriented along S. 866 32. GRAVITATIONAL COLLAPSE Box 32.2 (continued) D. Generalization; Price's Theorem 1. Let the star generate a \"zero-rest-mass, integer spin field.\" [\"Zero rest mass\" refers to the quantized particles associated with the classical field. Classically it means the field has a Cou lomb-law (1jr) fall off at large distances. The spin also is a property of the quantized parti cles; classically it is most easily visualized as describing the symmetries of a monochromatic plane wave under rotations about the direction of propagation; see \u00a735.6. A scalar field has spin zero; an electromagnetic field has spin one; Einstein's gravitational field has spin two;.... Of such fields, only gravitational (s = 2) and electromagnetic (s = 1) are known to exist in the real universe. See, e.g., Dirac (1936), Gard ing (1945), Bargmann and Wigner (1948), Pen rose (1965a), for further discussion.] 2. Let the spin-s field be sufficiently weak that its external, stress-energy perturbs Schwarzschild geometry only very slightly. star's the 3. Resolve the external field into spherical har monics (scalar spherical harmonics for s = 0; vector spherical harmonics for s = 1; tensor spherical harmonics for s 2 2); and label the spherical harmonics by the usual integer 1 (1 = o for monopole; 1 = 1 for dipole; 1 = 2 for quadrupole; etc.). 4. All multipole fields with 1 < s are conserved during the collapse (theorem from classical ra diation theory). A scalar field (s = 0) conserves nothing. The electromagnetic field (s = 1) con serves only its monopole parts (electric Cou lomb field, and vanishing magnetic Coulomb field). The gravitational field (s = 2) conserves its monopole part (with imprint equal to mass), and its dipole parts (with imprints measuring the angular momentum, and the standard grav itational dipole moment-which vanishes if coordinate system is centered on star). 5. For 1 2 s, and only for 1 2 s, radiation is pos sible (scalar waves can have any multipolarity; electromagnetic waves must be dipole and higher; gravitational waves must be quadrupole and higher; see \u00a736.1). 6. Price's theorem states that, as the nearly spheri cal star collapses to form a black hole, all things that can be radiated (all multi poles 1 2 s) get radiated completely away-in part \"off to in finity\"; in part \"down the hole\" (\"what is per mitted is compulsory\"). The final field is char acterized completely by its conserved quantities (multipole moments with 1 < s). 7. For proof of Price's theorem in the case of a scalar field, see exercise 32.1 O. E. Generalization to Nonclassical Fields See Hartle (1971, 1972) and Teitelboim (1972b,c) for neutrino fields; Bekenstein (1972a,b) and Teitelboim (1972a) for pion fields. \u00a732.7. NON SPHERICAL GRAVITATIONAL COLLAPSE 867 ------------------------------....... ~ Box 32.3 COLLAPSE IN ONE AND TWO DIMENSIONS A. The Question To produce a black hole (horizon from which nothing can emerge), must one compact matter strongly in all three spatial dimensions, to circum ferences 8.':S 4'17M (quasispherical compaction); or C. is it sufficient to compact only in one or two di mensions? B. The Answer for One Dimension theory, Consider, as an example readily generalized, the gravitational collapse of a spheroid of dust (zero pressure). Let the spheroid be highly Newtonian (r \u00bb>", " 2M) in its initial, momentary state of rest; and let it be slightly flattened (oblate). In Newto nian nonrotating any homogeneous, spheroid of dust remains homogeneous as it col lapses; but its deformations grow [see, e.g., Lin, Mestel, and Shu (1965) for details]. Hence, the spheroid of interest implodes to form a pancake of infinite density but finite mass per unit surface area. The final kinetic energy of the dust particles is roughly equal to their final potential energy: 1 2 IV - (812'17) M M = mass of spheroid, e = circumference of final pancake. Consequently, so long as 812'17 > 2M, the collapse velocities remain much smaller than light, and the gravitational energy remains much smaller than the rest mass-energy. This means that for 8/2'17 > 2M, the Newtonian analysis is an excellent ap proximation to general relativity all the way down to the pancake endpoint. Hence, no horizon can form, hardly any gravitational waves are emitted, and the whole story is exceedingly simple and fully Newtonian. However, since the pancake endpoint is not a singularity of spacetime (see the remarks at end of exercise 32.8), the evolution can proceed beyond it; and as e contracts to.':S4'17M, the evolu- tion will become very complicated and highly rel ativistic (see the \"collapse, pursuit, and plunge scenario\" of Figure 24.3). The Answer for Two Dimensions Consider, as an example not so readily general ized, the gravitational collapse of a homogeneous prolate spheroid of dust, initially highly Newto nian. Such a spheroid collapses to form a thin \"thread\" or \"spindle\" [see Lin, Mestel, and Shu (1965)]. Assume -that the spheroid is still Newto nian when its threadlike state is reached. It then has a length 1, a mass per unit length >-.. = Mil ~ 1, and a rapidly contracting equato rial radius R ~ 1. Subsequently, each segment of the thread collapses radially as though it were part of an infinite cylinder. [Ignore the instability of breakup into \"beads\"; see, e.g., Hunter (1967), Chandrasekhar (1968).] The radial collapse veloc ity approaches the speed of light and the gravita tional energy approaches the rest mass-energy only when the thread has become exceedingly thin, R.':S Rcrit - 1 exp (-1/4>-..). At this stage, relativistic deviations from Newtonian collapse come into play. Thorne (1972) and Morgan and Thorne (1973) have analyzed the relativistic effects using an idealized infinite-cylinder model. The results are very different from either the spherical case or the pancake case. The collapsing cylinder emits a large flux of gravitational waves; but they are powerless to halt the collapse. The collapse proceeds inward to a thread-like singularity, without the creation of any horizon (no black hole!). D. Objection to the Answer, a Reply, and a Conjecture One can object that the collapses of both pancake and cylinder can be halted short of their endpoints, especially that of the pancake. As the thickness of y 868 32. GRAVITATIONAL COLLAPSE Box 32.3 (continued) the pancake approaches zero, the vertical pull of gravity remains finite, but the pressure gradient caused by any finite pressure goes to infinity. Hence, pressure halts the collapse. Subsequently the rim of the pancake contracts toward the relativ istic regime ej2'1T ~ 2M. In the collapse of a cylinder according to Newtonian theory, with a pressure-density relation p a: pY, the gravitational acceleration ag and pressure-buoyancy accelera tion ap vary as ag = -2AjR, Hence, for y > I (the most common realistic case) pressure halts the collapse, but for y < I it does not. Whether this is true also after the relativistic domain is reached, one does not yet know. Actually, the ability of pressure to halt the col lapse is of no importance to the issue of black holes and horizons. The important thing is that in oblate collapse with final circumference e> 47TM, and also in prolate collapse with final thread length 1 > 2M, nO horizons are created. This, coupled with the omnipresent horizons in nearly spherical collapse (Box 32.2) suggests the following conjec ture [Thorne (1972)]: Black holes with horizons form when and only when a mass M gets compacted into a region whose circumference in EVER Y direc tion is e ~ 4'1TM. (Like most conject", "ures, this one is sufficiently vague to leave room for many differ ent mathematical formulations!) (a) Resolve the external field into scalar spherical harmonics, using Schwarzschild coor dinates for the external Schwarzschild geometry: Show that the vacuum field equation reduces to -'I't,tt + 'I'l,T'T' = 1 - -r- 7 ( 2M)(2M 1(1 + 1\u00bb) + r 2 (32.27a) 'I't. (32.27b) where\" is the \"tortoise coordinate\" of \u00a725.5 and Figure 25.4:,. = r + 2M In (r/2M - 1). (32.27c) Notice that (32.27b) is a flat-space, one-dimensional wave equation with effective potential V () _ (I _ 2M)(2M 1(1 + 1\u00bb) eff r - 3+ 2 r r r. (32.27d) Part of this effective potential [1(1 + 1)/r2 ) is the \"centrifugal barrier,\" and part [2M/r) is due to the curvature of spacetime. Notice the similarity of this effective potential for scalar waves, to the effective potentials for particles and photons moving in the Schwarzschild geometry, -2 (V )partlcles = (l - 2M/r)(l + L /r ), -2 2 (B-2 )Photons = (1 - 2M/r)r-2 (Boxes 25.6 and 25.7). The scalar-wave potentiaL like the photon potential, is positive for all r > 2M. It rises, from 0 at r = 2M, to a barrier summit; then falls back to 0 at r = 00. \u00a7 32.7. NON SPHERICAL GRAVITATIONAL COLLAPSE 869 (b) Show that there exist no physically acceptable, static scalar-wave perturbations of a Schwarzschild black hole. [More precisely, show that all static solutions to equation (32.27b) become infinite at either the horizon (r = 2M, r* = - 00) or at radial infinity.) This suggests that somehow the black hole formed by collapse must divest itself of the star's external scalar field before it can settle down into a quiescent state. (c) The general solution to the wave equation (32.27b) can be written in terms of a Fourier transform. For waves that begin near the horizon, propagate outward, and are partially transmitted and partially reflected (\"rightward-propagating waves\"), show that the general solution is where dZRk/dr*z = [_k Z + Veff(r\u00bb)RL Rt = eikT' + r~Rle-ikT' Rt = T't1eikr\" as r* --- - 00, as r* --- 00. (32.28a) (32.28b) (32.28c) Show that the \"reflection and transmission coefficients for rightward-propagating waves,\" r~Rl and T't>, have the following asymptotic forms for Ikl < 1/M (short wave number; long wavelength): r~Rl = -1 + a2Mik, r---:----:-----::---7t~--:---1 'produces complete reflection and complete' destructive interference in limit k --- 0; see Box 32.2 for detailed discussion of,consequences T (RI k - f3 (21 - (2Mik)l+ I (32.28d) I)!! t 'no transmission in limit k --- 0; see,Box 32.2 where a and f3 are constants of order unity. Give a similar analysis for waves that impinge on a Schwarzschild black hole from outside (\"leftward-propagating waves\"). (d) Show that, as the star collapses into the horizon, the world line of its surface in (t, r*) coordinates is r* = R*(t) - -t - Ro* exp (- t/2M) + const., (32.29a) where Ro* is related to the magnitude a of the surface's 4-acceleration (a > 0 for outward 4-acceleration) by Thus. the world line of the surface appears to become null near the horizon (t + r* = V == constant); of course, this is due to pathology of the coordinate system there. Show, further, that the scalar field on the star's surface (V = constant) must vary as (32.29b) '!' - Q + Q e-U14,1f 1 - lO 11'U= t - r*. (32.29c) when the star is approaching the horizon (t ---", " 00, r* --- -00, U --- 00), in order that the rate of change of'!' 1 be finite as measured on the star's surface. Notice that Q lO is the \"final value\" of the scalar field on the star's surface. It can be regarded as an outgoing wave with zero wave number (infinite wavelength); and, consequently, it gets completely and 870 32. GRAViTATIONAL COLLAPSE destructively reflected by (he e./feerive potential [see equation (32.28d): also Box 32.2).Conclu sion: All multipoles of the scalar field die out at finite r\u00b7 as ( --- 00. (Price's theorem for a scalar field.) For a more detailed analysis, including the rates at which the multipoles die out. see Price (1971, 1972a) or Thorne (1972). Exercise 32.11. NEWMAN-PENROSE \"CONSTANTS\" [See Press and Bardeen (1971), Bardeen and Press (1972), and Piir (1971) for more details than are presented here.] Wheeler (1955) showed that Maxwell's equations for an i-pole electromagnetic field residing in the Schwarzschild geometry can be reduced to the wave equation (32.30) [electromagnetic analogue of (32.27b\u00bb). After this equation has been solved. the components of the electromagnetic field can be obtained by applying certain differential operators to 'l'1(t. r*)Ylm(8,<;,\u00bb. (a) Show that the general solution to the electromagnetic wave equation (32.30) for dipole (i = 1) fields, with outgoing-wave boundary conditions at r\u00b7 --- + 00, has the form '1', ~ f:,(U-) -l-[1(D) -l- NU) -l-.... r 1 )0 -;-. \u2022 - 7 OJ r- -;- -;- where U= (- r* is \"retarded time\". and It' =/0, 12' = 0, In' = - (11 + li~n - 2)In_l + (n - 2)MIn_2' (32.31b) When spacetime is flat (M = 0), this solution becomes (32.31 N) [The l/r fall-off of the radiation field h'( U) has been factored out of '1'1: see the scalar-wave function (32.27a).] The terms.f2( [;)/r 2 +..., which are absent in flat spacetime, are attributable to backscatter of the outgoing waves by the curvature of spacetime. They are sometimes called the \"tail\" of the waves. (b) Show that the general static dipole field has the form (32.31a) with (/o)statlc = 0; (/l)statlc =D = dipole moment; (/2).tatlc =1MD;.... (32.32) (c) Consider a star (not a black hole!) with a dipole field that is initially static. At time ( = 0, let the star suddenly change its dipole moment to a new static value D '. Equations (32.31b) demand that.fz be conserved [\"Newman-Penrose (1965) constant\"). Hence,.fz will always exhibit a value, ~MD, corresponding (0 (he old dipole moment; i( can never change (0 ~MD '. This is a manifestation of the tail of the waves that are generated by the sudden change in dipole moment. To understand this tail effect more clearly, and to discover an important flaw in the above result, evaluate the solution (32.31) for retarded time U > 0, using the assumptions (1) field has static form (32.32) for U < 0,,(2) h = D' for U > o. (32.33) \u00a732.7. NONSPHERICAL GRAVITATIONAL COLLAPSE Put the answer in the form 3 \"2 MD _ D' - + --2- + 2: r r n=3 '1'1 - \"2M(D' _ D)( _1)R+1(n + I)U n - 2 (2 )n r 871 (32.34) (M2 M 2U) + 0 -3'-4 -. r r (The terms neglected are small compared to those kept for all Vir, so long as r ~ M.) Evidently, so long as the series converges the Newman-Penrose \"constant\" (coefficient of l/r 2) remembers the old D value and is conserved, as claimed above. Show, however, that the series diverges for U > 2r-i", ".e., it diverges inside a sphere that moves outward with asymptotically! the speed of light. Thus, the Newman-Penrose \"constant\" is well-defined and conserved only outside the \"!-speed-of-light cone.\" (d) Sum the series in (32.34) to obtain a solution valid even for U > 2r: '1' = D' + ~ MD' _ 2M(D' - D) (~+ 3r) + 0('\\12 ) r2. r (U + 2r)2 r3 \"tail term\" (32.35) = the series (32.34) for U < 2r (domain of convergence of that series) = D' + 1. MD' + o(,:!, M2) r2 for U ~ r ~ M. Ur r3 2 r From this result conclude that at fixed r and late times the electromagnetic field becomes asymptotically static, and the Newman-Penrose \"constant\" assumes the new value ~MD' appropriate to the new dipole moment., r 2 new static solution / CHAPTER 33 BLACK HOLES two hundred and fifty times larger than that of the Sun, would not, A luminous star, of the same density as the Earth, and whose diameter should be in consequence of its attraction, allow any of its rays to arrive at us; it is therefore through this cause, be invisible. possible that the largest luminous bodies in the universe may, P. S. LAPLACE (1798) A dialog explaining why black holes deserve their name \u00a733,1, WHY \"BLACK HOLE\"? Sagredus. What is all this talk about \"black holes\"? When an external observer watches a star collapse, he sees it implode with ever-increasing speed, until the relativistic stage is reached. Then it appears to slow down and become \"frozen,\" just outside its horizon (gravitational radius). However long the observer waits, he never sees the star proceed further. How can one reasonably give the name \"black hole\" to such a frozen object, which never disappears from sight? Salvatius. Let us take the name \"black hole\" apart. Consider first the blackness. Surely nothing can be blacker than a black hole. The very redshift that makes the collapsing star appear to freeze also makes it darken and become black In the continuum approximation, where one ignores the discreteness of photons, the intensity of the radiation received by distant observers decreases exponentially as time passes, L a: exp ( - 1/3 V3M), with an exceedingly short e-folding time T = 3 V3M = (2.6 X 10-5 sec)(MIM0 ). Within a fraction of a second, the star is essentially black Discreteness of photons makes it even blacker. The number of photons emitted before the star crosses its horizon is finite, so the exponential decay cannot continue For a more detailed exposition of the foundations of \"black-hole physics,\" see DeWitt and DeWitt (1973). \u00a733.1. WHY THE NAME \"BLACK HOLE\"? 873 forever. Eventually-only 1O-3(M/M0 ) seconds after the star begins to dim' (see exercise 32.2)-the last photon that will ever get out reaches the distant observers. Thereafter nothing emerges. The star is not merely \"essentially black\"; it is \"absolutely black.\" Sagredus. Agreed. But it is the word \"hole\" that concerns me, not \"black.\" How can one possibly regard the name \"hole\" as appropriate for an object that hovers forever just outside its horizon. True, absence of light makes the object invisible. But couldn't one always see it by shining a flashlight onto its surface? And couldn't one always fly down to its surface in a rocket ship and scoop up a few of the star's baryons? After all, as seen from outside the baryons at its surface will never, never, never manage to fall into the horizon! Salvatius. Your argument sounds persuasive. To test its validity, examine the collapse of a spherically symmetric system, using the ingoing Eddington-Finkelstein diagram of Figure 33.1. Let a family of external observers shine their flash lights onto the star's surface, as you have suggested. Let the surface of the star be carefully silvered so it reflects back all light that reaches it. Initially (low down in the spacetime diagram of Figure 33.1) the ingoing light beams Figure 33.1. Spherical gravitational collapse of a star to form a black hole. as viewed in ingoing Eddington-Finkelstein coordinates. The \"surface of last influence,\" :Jl, is an ingoing null surface that intersects the horizon in coincidence with the surface of the collapsing star. After an external observer,", " moving forward in time, has passed through the surface of last influence, he cannot interact with and influence the star before it plunges through the horizon. Thus, one can think of the surface of last influence as the \"birthpoint\" of the black hole. Before passing through this surface, the external observer can say his flashlight is probing the shape of a collapsing star; afterwards, he can regard his signals as probes of a black hole. For further discussion. see text. 874 33. BLACK HOLES reach the star's surface and are reflected back to the flashlights with no difficulty. But there is a critical point-an ingoing radial null surface :?l beyond which reflection is impossible. Photons emitted inward along '3l reach the star just as it is passing through its horizon. After reflection these photons.fly \"outward\" along the horizon, remaining forever at r = 2M. Other photons, emitted inward after the flashlight has passed through :?l, reach the surface of the star and are reflected only after the star is inside its horizon. Such photons can never return to the shining flashlights. Once inside the horizon, they can never escape. Thus, the total number of photons returned is finite and is subject to the same blackness decay law as is the intrinsic luminosity of the star. Moreover, if the observers do not turn on their flashlights until after they pass through the null surface:?l, they can never receive back any reflec ted photons! Evidently, flashlights are of no help in seeing the \"frozen star.\" Sagredus. I cannot escape the logic of your argument. Nevertheless, seeing is not the only means of interacting with the frozen star. I have already suggested swooping down in a rocket ship and stealing a few baryons from its surface. Similarly, one might let matter fall radially inward onto the frozen star. When the matter hits the star's surface, its great kinetic energy of infall will be converted into heat and into outpouring radiation. Salvatius. Thus it might seem at first sight. But examine again Figure 33.1. No swooping rocket ship and no infalling matter can move inward faster than a light ray. Thus, if the decision to swoop is made after the ship passes through the surface :?l, the rocket ship has no possibility of reaching the star before it plunges through the horizon; the rocket and pilot cannot touch the star, sweep up baryons, and return to tell their tale. Similarly, infalling matter to the future of:?l can never hit the star's surface before passing through the horizon. The surface:?l is, in effect, a \"surface oflast influence.\" Once anybody or anything has passed through :?l, he or it has no possibility whatever of influencing or interacting with the star in any way before it plunges through the horizon. Thus,from a \"causal\" or \"interaction\" standpoint, the collapsing star becomes a hole in space at the surface :?l. This hole is not black at first. Radiation from the collapsing star still emerges after :?l because of finite light-propagation times, just as radiation still reaches us today from the hot big-bang explosion of the universe. But if an observer at radius r ~ 2M waits for a time 2r after passing through :?l (time for :?l to reach horizon, plus time for rays emitted at R - 3M to get back to observer), then he will see the (10- 3 newly formed hole begin to turn black; and within a time.1t seconds)(M/M0 ) thereafter, it will be completely black. Sagredus. You have convinced me. For all practical purposes the phrase \"black hole\" is an excellent description. The alternative phrases \"frozen star\" and \"col lapsed star,\" which I find in the pre-1969 physics literature, emphasize an optical-illusion aspect of the phenomenon. Attention must be directed away from the star that created the black hole, because beyond the surface of last influence one has no means to interact with that star. The star is irrelevant \u00a733.2. GRAVITATIONAL AND ELECTROMAGNETIC FIELDS OF A HOLE 875 to the subsequent physics and astrophysics. Only the horizon and its' external spacetime geometry are relevant for the future. Let us agree to call that horizon the \"surface of a black hole,\" and its external geometry the \"gravita tional field of the black hole.\" Salvatius. Agreed. \u00a733.2. THE GRAVITATIONAL AND ELECTROMAGNETIC FIELDS OF A BLACK HOLE The collapse of an electrically neutral star endowed with spherical symmetry pro duces a spherical black hole with external gravitational field described by the Schwarzschild line element ds 2 = -(1 - 2M/r) dt2 + 2 dr 1 - 2M/r + r2(d()2 +", " sin2()dcj>2). (33.1 ) The surface of the black hole, i.e., the horizon, is located at r = 2M = (gravitational radius). Only the region on and outside the black hole's surface, r 22M, is relevant to external observers. Events inside the horizon can never influence the exterior. The gravitational collapse of a realistic star (nonspherical, collapse with small but nonzero net charge of one sign or the other) produces a black hole somewhat different from the simple Schwarzschild hole. For collapse with small charge and small asymmetries, perturbation-theory calculations (Box 32.2) predict a final black hole with external field determined entirely by the mass M, charge Q, and intrinsic angular momentum S of the collapsing star. For fully relativistic collapse, with large asym the final black hole (if one forms) is also metries and possibly a large charge, characterized uniquely by M, Q, and S. This is the conclusion that strongly suggests itself in 1972 from a set of powerful theorems described in Box 33.1. Why M, Q, and S should be the complete governors of the final external field of the black hole, one can understand heuristically as follows. Of all quantities intrinsic to any isolated source of gravity and electromagnetism, only M, Q, and S possess (and are defined in terms of) unique, conserved imprints in the distant external fields of the source (conserved Gaussian flux integrals; see Box 19.1 and \u00a720.2). When a star collapses to form a black hole, its distant external fields are fOrCed wmaintain unchanged the imprints of M, Q, and S. In effect, M, Q, and S provide anchors or constraints on the forms of the fields. Initially other constraints are produced by the distributions of mass, momentum, stress, charge, and current inside the star. But ultimately the star plunges through a horizon, cutting itself off causally from the external universe. (The nonpropagation of long-wavelength waves through curved spacetime plays a key role in this cutoff; see Box 32.2.) Subsequently, the only anchors remaining for the external fields are the conserved imprints of M, Q, and S. Consequently, the external fields quickly settle down into unique shapes corresponding to the given M, Q, and S. Of course, the settling down involves dynamic changes of the fields and an associated outflow of gravitational and electro- The structure of a black hole is d'etermined uniquely by its mass M, charge.Q, and intrinsic angular momentum, S Heuristic explanation of the M-Q-S uniqueness 876 33. BLACK HOLES Box 33.1 A BLACK HOLE HAS NO \"HAIR\" The following theorems come close to proving that the external gravitational and electromagnetic fields of a stationary black hole (a black hole that has settled down into its \"final\" state) are determined uniquely by the hole's mass M, charge Q, and intrin sic angular momentum S-i.e., the black hole can have no \"hair\" (no other independent characteris tics). For a detailed review, see Carter (1973). I. Stephen Hawking (1971 b, 1972a): A station ary black hole must have a horizon with spherical topology; and it must either be static (zero angular momentum), or axially sym metric, or both. II. Werner Israel (1967a, 1968): Any static black hole with event horizon of spherical topology has external fields determined uniquely by its mass M and charge Q; moreover, those exter n~l fields are the Schwarzschild solution if Q = 0, and the Reissner-Nordstmm solution (exercises 31.8 and 32.1) if Q f. 0 (both special cases of Kerr-Newman; see \u00a733.2). fields gravitational III. Brandon Carter (1970): \"All uncharged, sta tionary, axially symmetric black holes with event horizons of spherical topology fall into disjoint families not deformable into each other. The black holes in each family have determined external uniquely by two parameters: the mass M and the angular momentum S.\" (Note: the \"Kerr solutions\"-i.e., \"Kerr-Newman\" with Q = 0 -form one such family; is very likely that there are no others, but this has not been proved as of December 1972. It is also likely that Carter's theorem can be extended to the case with charge; but this has also not yet been done.) it IV. Conclusions made by combining all three the orems: (a) All stationary black holes are axially sym metric. (b) All static (nonrotating) black holes are characterized uniquely by M and Q, and hllve the Reissner-Nordstmm form.", " (c) All uncharged, rotating black holes fall. into distinct and disjoint families, with each black hole in a given family charac terized uniquely by M and S. The Kerr solutions form one such family. There may well be no other family. V. Remarks and Caveats: (a) The above statements of the theorems are all somewhat heuristic. Each theorem makes several highly technical assump tions, not stated here, about the global properties of spacetime. These assump tions seem physically reasonable and in nocuous, but they might not be. (b) Progress in black-hole physics is so rapid that, by the time this book is published, there may well exist theorems more pow erful than the above, which really prove that \"a black hole has no hair.\" (c) For insight into the techniques of \"global geometry\" used in proving the above the orems and others like them, see Chapter 34; for greater detail see the forthcoming book by Hawking and Ellis (1973). (d) For analyses which show that a black hole cannot exert any weak-interaction forces caused by the leptons which have gone down it, see Hartle (1971,1972) and Tei telboim (1972b,c). For similar analyses which show absence of strong-interaction forces from baryons that have gone down the hole, see Bekenstein (1972a,b) and Teitelboim (1972a). \u00a733.2. GRAVITATIONAL AND ELECTROMAGNETIC FIELDS OF A HOLE 877 magnetic waves. And, of course, the outflowing waves carry off mass and angular momentum (but not charge), thereby leaving M and S changed. And, of course, the external fields must then readjust themselves to the new M and S. But the process will quickly converge, producing a black hole with specific final values of M, Q, and S and with external fields determined uniquely by those values. The problem of calculating the external fields for given M, Q, and S and their given imprints, is analogous to the problem of Plateau-to calculate the shape of a soap film anchored to a wire of given shape.* One calculates the shape of the soap film by seeking a surface of minimum area spanning the bent wire. The condition of minimum area leads to a differential equation describing the soap film, which must be solved subject to the constraint imposed by the shape of the wire. To calculate the external fields of a black hole, one can extremize the \"action integral\" f(&l + E) -v:::gd 4x for interacting gravitational and electromagnetic fields (see Chapter 21) subject to the anchored-down imprints of M, Q, and S at radial infinity, and subject to the existence of a physically nonsingular horizon (no infinite curvature at horizon!). Extremizing the action is equivalent to solving the coupled Einstein-Maxwell field equations subject to the constraints imprinted by M, Q, and S. and the existence of the horizon. The derivation of the solution and the proof of its uniqueness are much too complex to be given here. (See references cited in Box 33.1.) However, the solution turns out to be the \"Kerr-Newman geometry\" and its associated electromagnetic field.t Variational principle for black-hole structure Written in the t, r, 8, </> coordinates of Boyer and Lindquist (1967) (generalization Details of black-hole structure: of Schwarzschild coordinates), the Kerr-Newman geometry has the form sin28 [dt - a SIn 8 d</>] + -? - [(r2 + a2) d</> - a dtj2 p.. 2 2 - L1 p2.. ~. ---2 - + ~ dr 2 + p2 d82, (1) metric (\"Kerl\"-Newman geometry\") (33.2) where L1 =r2 - 2Mr + a2 + Q2, p2 = r2 + a2 cos28, a = SjM= angular momentum per unit mass. (33.3a) (33.3b) (33.4) The corresponding electromagnetic field tensor, written as a 2-form (recall: dx a dx fJ = dx a \u00ae dx fJ - dx fJ \u00ae dx a ) is /\\ F = Qp-4(r2 - a2 cos28) dr /\\ [dr - a sin28 d</>] + 2Qp-4ar cos 8 sin 8 dO /\\ [(r2 + a2) d<j> - a dt]. (2) electromagnetic field (33.5) \u00b7On the problem of Plateau see, e.g., Courant (1937), Darboux (1941), or p. 157 of Lipman Bers (1952). tThe un", "charged (Q = 0) version was first found as a solution to Einstein's vacuum field equations by Kerr (1963). The charged generalization was first found as a solution to the Einstein-Maxwell field equations by Newman, Couch, Chinnapared, Exton, Prakash, and Torrence (1965). Only later was the connection to black holes discovered; see Box 33.1. 878 33. BLACK HOLES Expressions (33.2) for the metric and (33.5) for the electromagnetic field are sufficiently long to be somewhat frightening. Therefore, it is helpful to develop some qualitative insight into them and into their implications before attempting detailed _computations with them. Boxes 33.2, 33.3, and 33.4 develop qualitative insight by presenting, without derivation, a summary of the key features of the Kerr-Newman geometry and a summary of the physics and astrophysics of black holes. The re mainder of this chapter is a Track-2 justification and derivation of some, but not all, of the results cited in Boxes 33.2-33.4. (continued on page 891) Box 33.2 KERR-NEWMAN GEOMETRY AND ELECTROMAGNETIC FIELD I. Equations for metric and electromagnetic field A. Parameters appearing in equations: M = mass, Q = charge, a = SIM = angular momentum per unit mass, all as measured by their standard imprints on the distant fields. B. Constraint on parameters: The Kerr-Newman geometry has a horizon, and therefore describes a black hole, if and only if M2 ;;:: Q2 +' a2. It seems likely that in any collapsing body which violates this constraint, centrifugal forces and/or electrostatic repulsion will halt the collapse before a size -.. M is reached; see equation (33.56). C. Limiting cases: Q = 0, S= 0, Q = S = 0, M2 = Q2 + a2 Kerr (1963) geometry; Reissner-Nordstmm geometry and electromagnetic field (exercises 31.8 and 32.1); Schwarzschild geometry; \"Extreme Kerr-Newman geometry.\" D. Boyer-Lindquist (1967) coordinates (t, r, 8, <t>-generalization of Schwarzs child coordinates; black hole rotates in <t> direction): ds 2 = _(L1lp2)[dt - a sin28 d<t>J2 + (sin28I p2)[(r2 + a2) d<t> - a dt]2 (1) + (p2 I L1) dr2 + p2 d8 2; L1 =r2 - 2Mr + a2 + Q2, F = Qp-4(r2 - a2 cos28) dr /\\ [dt - a sin28 d<t>] p2 =r2 + a2 cos28. + 2Qp-4ar cos 8 sin 8 dO /\\ [(r 2 + a2) d<t> - a dt]. (2) (3) E. Kerr coordinates [v, r, 8, <;-generalization ofingoing Eddington-Finkelstein coordinates; (v, 8, <;) = constant is an ingoing, \"radial,\" null geodesic; black hole rotates in <; direction]: Relationship to Boyer-Lindquist: \u00a733.2. PRINCIPAL FEATURES OF HOLES 879 dV = dt + (r2 + a2)(dr/fj), d;P=d</>+a(dr/fj). ds2 = - [1 - p-2(2Mr - Q2)J dV2 + 2 dr dV + p2 d82 + p-2[(r2 + a2)2 - 2ap-2(2Mr - Q2) sin28 d;P dV. fja2sin28] sin28 d;P2 - 2a sin28 d;P dr (4) (5) F = Qp-4[(r2 - a2 cos28) dr A dV - 2a2r cos 8 sin 8 dO A dV (6) - a sin28(r 2 - a2 cos28) dr A d;P + 2ar(r2 + a2) cos 8 sin 8 d8 A d;P]~ II. Properties of spacetime geometry A. Symmetries (\u00a733.4): The metric coefficients in Boyer-Lindquist coordinates are independent of t and </>, and in Kerr coordinates are independent of V and ;Po Thus the spacetime geometry is \"time-independent\" (stationary) and ax- ially symmetric. The \"Killing vectors\" (", "\u00a725.2) associated with these two sym metries are (0/ot)r,8,\u00a2 = (0/0 V)r,8,\u00a2 and (0 /o</> )t,r, 8 = (0 /o;P)v,r, 8\u00b7 B. Dragging of inertial frames and static limit (\u00a733.4): 1. The \"dragging of inertial frames\" by the black hole's angular momentum produces a precession of gyroscopes relative to distant stars. By this preces sion one defines and measures the an gular momentum of the black hole (see \u00a7\u00a719.2 and 19.3). 2. The dragging becomes more and more extreme the nearer one approaches the horizon of the black hole. Before the horizon is reached, at a surface described by - - --~~ - r = ro(8)=M + YM2 _ Q2 _ a2 cos28, (7) the dragging becomes so extreme that no observer can possibly remain at rest there (i.e., be \"static\") relative to the distant stars. At and inside this surface (called the \"static limit\"), all observers with fixed rand 8 must orbit the black hole in the same direction in which the hole rotates: a sin8 - VL1 >----_---=._-(r 2 + a2) sin 8 - VL1a sin28 (~ 0 for a = S/M > 0 and r ::::;; ro). No matter how hard an observer, at fixed (r,8) inside the static limit, blasts his rocket engines, he can never lialt his angular motion relative to the distant stars. 3. The mathematical foundation for the above statement is this: world lines of the form (r, 8, </\u00bb = constant [tangent vector ex a/at = \"Killing vector in time direction\"] change from being timelike outside the static limit to being space like inside it. Therefore, on and inside the static limit, no observer can remain at rest. C. Horizon (\u00a733.4): 1. The horizon is located at r = r+ = M + yM2 - Q2 - a2. 2. As with the Schwarzschild horizon of a nonrotating black hole, so also here, particles and photons can fall inward through the horizon; but no particle or (8) 880 33. BLACK HOLES Box 33.2 (continued) photon can emerge outward through it. 3. The horizon is \"generated\" by outgoing null geodesics (outgoing photon world lines). D. Ergosphere (\u00a733.4): 1. The \"ergosphere\" is the region of space time between the horizon and the static limit. It plays a fundamental role in the physics of black holes (Box 33.3; \u00a733.7). 2. The static limit and the horizon touch the point where they are cut by at the axis of rotation of the black hole (8 ::: 0, '17); they are well-separated else where with the static limit outside the horizon, unless a::: \u00b0 (no rotation). When a ::: 0, the static limit and hori zon coincide; there is no dragging of inertial frames; there is no ergosphere. Qualitative representation of horizon, ergosphere, and static limit [adapted from Ruffini and Wheeler (l97Ib\u00bb). E. Singularity in Boyer-Lindquist coordi nates: 1. For a nonrotating black hole, the Schwarzschild coordinates become sin gular at the horizon. One manifestation the horizon, { -+ 00 of the singularity is the infinite amount of coordinate time required for any inward particle or photon to fall as through r -+ 2M. One way to remove the sin gularity (Eddington-Finkelstein way) is to replace t by a null coordinate V::: t + r + 2Mlnlr/2M attached [so infalling (0/or)V,8,\u00a2 is vector tangent to photon world lines]. 11 photons to 2. For a rotating black hole, the Boyer Lindquist coordinates, being generali zations of the Schwarzschild coordi nates, are also singular at the horizon. It requires an infinite coordinate time for any particle or photon to fall in ward through the horizon, t -+ 00 as r -+ r+. But that is not all. The drag ging of inertial frames forces particles and photons near the horizon to orbit the black hole with.[J = dcp/dt > 0. Consequently, for a particle falling through the horizon (r -+ r+), just as (infinite t -+ 00, so also cp -+ 00 twisting of world lines around horizon). 3. To remove the coordinate singularity, one must perform an infinite compres sion of coordinate time, and an infinite untwisting in the neighborhood", " of the horizon. Kerr coordinates achieve this by replacing t with a null coordinate V, and cp with an untwisted angular coor dinate ;p: dV::: dt + (r 2 + a2)(dr/.1), d;P ::: dcp + a(dr/.1). Both of the new coordinates are at tached to the world lines of a particular family of infalling photons; (0/or)V,8,\u00a2 is the field of vectors tangent to the world lines of this family of photons (ingoing principal null congruence; \u00a733.6). \u00a733.2. PRINCIPAL FEATURES OF HOLES 881 Kerr diagram for equatorial slice (6 = 'IT/2) through the spacetime of an \"extreme Kerr\" black hole (Q = O. a = M). F. Spacetime diagram: 1. A spacetime diagram in Kerr coordi nates looks much like an Eddington Finkelstein diagram for the Schwarzs child geometry. In both cases, one plots the surfaces of constant V not as hori zontal planes, but as \"backward light cones\" (\"45-degree surfaces\"), because they are generated by the world lines of ingoing photons. Equivalently, one plots surfaces of constant 7= V- r as horizontal planes. for are 2. The key differences between a Kerr diagram and an Eddington-Finkelstein diagram are: (a) Because the Kerr Newman geometry is not spherical, a Kerr diagram with one rotational de gree of freedom suppressed loses infor mation about the geometry. Kerr dia the grams usually made equatorial \"plane,\" 8 = 7T/2. (b) Just as the horizon pulls the light cones inward, so the dragging of inertial frames tilts the light cones in the direction of in creasing ;p, for a > 0 and r = constant. (c) The ingoing edge of a light cone (dr/dV-=--oo) does not tilt toward in the transformation from creasing ;P; Boyer-Lindquist coordinates to Kerr coordinates untwists the tilt with de creasing r, which would otherwise be produced by \"frame dragging.\" 3. The shapes of the light cones reveal the special features of the static limit and horizon. At the static limit, a vertical world line [r, 8,;P constant; (0/0 V)r,e,\u00a2\" = (%t)r.e,\u00a2 = tangent vector] lies on the light cone. At the horizon the light cones tilts fully inward, except for a single line of tangency to the horizon. the line of tangency has Notice that d<;/dV = a/(r+2 + a 2 i= O. Equiva lently, the outgoing null geodesics, which generate the horizon, twist about (\"barber-pole-twist\")-yet another it manifestation of the dragging of inertial frames. ) View from above showing the shapes of the light cones as a function of radius 882 33. BLACK HOLES Box 33.2 (continued) 4. The Kerr diagram, like the Eddington Finkelstein diagram,- describes infall through the horizon in a faithful, non singular way. 5. [The term \"Kerr diagram\" is a misno mer. Kerr has not published such dia grams himself, though nowadays others construct such diagrams using his coor dinate system. Penrose is the originator and greatest exploiter of such diagrams (see, e.g., Penrose, 1969). But several other types of diagrams bear Penrose's name, so it would be confusing to name them all after him.] G. Maximal analytic extension of Kerr-New man geometry: 1. When one abstracts the Schwarzschild geometry away from all sources (Chap ter 31), one discovers that it describes an expanding and recontracting bridge, connecting two different universes. But in the context of black holes, only half of the Schwarzschild geometry (regions I and II) is relevant. The other half (regions III and IV) gets fully replaced by the interior of the star that collapsed to form the black hole. Because only a Surface of collapsing star part of the Schwarzschild geometry ingoing Eddington comes into play, de Finkelstein scribe I and II well, but III and IV badly-are well-suited to black-hole physics. coordinates-which 2. Similarly, when one abstracts the Kerr Newman geometry away from all sour ces, one discovers that it describes a much larger, and more complex space time manifold than one might ever have suspected. This \"maximum ana lytic extension\" of the Kerr-Newman geometry has been analyzed in de tail by Boyer and Lindquist (1967) and by Carter (l966a, 1968a). But it is totally irrelevant to the subject of black holes,", " for two reasons. First, as with Schwarzschild, the star that collapsed to form the black hole replaces most of the inward extension of the Kerr-Newman manifold. Second, even outside the star, the Kerr-Newman geometry does not properly represent the true geometry at early times. At early times the star has not got far down the road to collapse. Gravitational moments of the star arise from mountains or prominences or tur bulence or other_ particularities that have not yet gone into the meat grinder. The geometry departs from flatness (1) by a term that varies for large distances as mass divided by distance, and (2) by another term that varies as angular mo mentum divided by the square of the distance and multiplied by a spherical harmonic of order one, but also (3) by higher-order to higher-order mass moments multiplied by higher spherical harmonics. These higher-order terms normally will devi ate at early times from the correspond ing terms in the mathematical analysis geometry of though the deviations will die out as time passes. For a system endowed with spherical symmetry, no such higher order terms do occur or can occur. the geometry outside is Therefore the Kerr-Newman proportional terms \u00a733.2. PRINCIPAL FEATURES OF HOLES 883 Schwarzschild in character at all stages of the collapse. However, when the sys tem lacks spherical symmetry, the ge ometry outside initially departs from Kerr-Newman character. Only well after the collapse occurs (asymptotic future), and in the region at and outside the hori zon, is the Kerr-Newman geometry a faithful descriptor of a black hole. This region is described in a nonsingular manner by Kerr coordinates and Kerr diagrams; and it is the only region that this book will explore. H. Test-particle orbits See \u00a7\u00a733.5-33.8 and Box 33.5. III. Properties of electromagnetic field (\u00a733.3): in the usual A. Far from the black hole, where spacetime spherical is nearly flat, frame (WI = dt, w.. = dr, orthonormal w O= r dO, w\u00a2 = r sin 0 d</\u00bb, the electric and magnetic fields have dominant com ponents E.. = ~; r 2Qa B.. = -3-COSO, Bo = -3 smO. r Qa. r These reveal that Q = charge of black hole, q)Jl = Qa = magnetic dipole moment of black hole. B. Notice that the gyromagnetic ratio, y = (magnetic moment)/(angular momentum), is equal to Q/M = (charge/mass), just as for an electron!. C. Notice that the value of the magnetic mo ment, like all other features of the black hole, is determined uniquely by the hole's mass, charge, and angular momentum: q)Jl = QS/M. This illustrates the theorem (Box 33.1) that a black hole has no \"hair.\" D. Other electric and magnetic moments are nonzero, but are determined uniquely by M, S, and Q. E. Near the black hole, the curvature of spacetime deforms the electric and mag netic fields produced by the charged, rota ting black hole. For a mathematical de scription of this deformed field, see Cohen and Wald (1971); for a diagrammatic rep resentation, Hanni and Ruffini (1973). Box 33.3 THE ASTROPHYSICS OF BLACK HOLES Black holes in nature should participate in astro physical processes that are as varied as those for stars. By searching for observable phenomena as sociated with these processes, astronomers have a good chance of discovering the first black hole sometime during the 1970's. This box lists some possible astrophysical processes, and a few rele vant references. I. Mechanisms of Formation A. \"Direct, in isolation\": A massive star(M > 3M0 ) collapses, almost spherically, pr; ducing a collapsed neutron-star core that is too massive to support itself against gravity. Gravity pulls the core on inward, producing a horizon and black hole. [May 884 33. BLACK HOLES Box 33.3 (continued) and White (1966, 1967); Chapter 32 of this book.] - II. How many black holes are there in our galaxy today? B. \"Indirect, in isolation\": \"Collapse, pursuit, and plunge scenario\" depicted in Figure 24.3 [Ruffini and Wheeler (1971b).] C. \"In the thick of things\": Stars collected into a dense cluster (e.g., the nucleus of a galaxy) exchange energy. Some acquire energy and move out into a halo. Others lose energy and make a more compact cluster. This process of segregation con tinues. The cluster becomes so compact that collisions ensue and gas is driven off. The gas moves", " toward the center of the gravitational potential well. Out of it new stars form. The process continues. Eventu ally star-star collisions may become suffi ciently energetic and inelastic that the cen ters of the colliding stars coalesce. In this way supermassive objects may be built up and may evolve. Ultimately (1) many \"small\" stars may collapse to form \"small\". black holes (M - M 0 ); (2) one or more supermassive stars may collapse to form huge black holes (M- 104M 0 to 109M 0 ); (3) the entire conglomerate of stars and gas and holes may become so dense that it collapses to form a single gigantic hole. [Sanders (1970), Spitzer (1971), Lynden Bell (1967, 1969), Colgate (1967), \u00a7\u00a724.5, 24.6, 25.7 of this book.] D. \"Primordially\": Perturbations in the initial density distribution of the expanding uni verse may produce collapse, resulting in \"primordial black holes.\" Those holes would subsequently grow by accretion of radiation and matter. By today all such holes might have grown into enormous 10 17 M0 ; Zel'dovich and objects Novikov (1966)]; but some of them might have avoided such growth and might be as small as 10-5 grams [Hawking (1971a)]. [M - Peebles (1972) has given an excellent review of this issue and of prospects for finding black holes in the near future. He says \"a good fraction of the mass of the disc of our galaxy was deposited [long ago] in stars capable of collapsing to black holes.... The indication is that the galaxy's disk may contain on the order of 109 black holes.\" III. \"Live\" black holes versus \"dead\" black holes A. A Schwarzschild black hole is \"dead\" in the sense that one can never extract from it any of its mass-energy. One aspect of this \"deadness\"-the fact that a Schwarzs child black hole is stable against small perturbations-is essential (1) to the iden tification of a black hole with the ultimate \"ground state\" of a large mass, and (2) to any assertion that general relativity theory predicts the possible existence of black holes. [For a proof of stability see Vish veshwara (1970). The problem was formu lated, and most of the necessary tech niques developed, by Regge and Wheeler (1957), with essential contributions also by Zerilli (1970a).] Thus a small pulse ofgravi tational (or other) radiation impinging on a Schwarzschild black hole does not initi ate a transition of the black hole into a very different object or state. B. A Kerr-Newman black hole-which is ro tating or charged or both-is not dead. The rotational and electromagnetic contri butions to the mass-energy can be extracted. (See \u00a7\u00a733.7 and 33.8 for mathematical de tails.) Thus, such black holes are \"live\"; they can inject energy into their sUITound -ings. By a suitable arrangement of external apparatus, one can trigger an exponen tially growing energy release [Press and Teukolsky (1972).] But for a perturbed \u00a733.2. PRINCIPAL FEATURES OF HOLES 885 black hole in isolation, the release is al ways \"controlled\" and damped; i.e., Kerr black holes are stable in any classical con text [Press and Teukolsky (1973)]. C. Most objects (massive stars; galactic nu clei;...) that can collapse to form black holes have so much angular momentum that the holes they produce should be \"very live\" (a nearly equal to M; S nearly equal to M2). [Bardeen (1970a).] D. By contrast, it is quite probable (but far from certain) that no black hole in the universe has substantial charge-i.e., that all black holes have Q < M. A black hole with Q - M (say, Q > 0 for concreteness) would exert attractive electrostatic forces on electrons, and repulsive electrostatic forces on protons, that are larger than the hole's gravitational pull by the factor (electrostatic force) = eQ _!.... _10 20 (gravitational force) JlM Jl. Here e is the electron charge and Jl is the electron (or proton) mass. Such huge differential forces are likely to pull in enough charge from outside the hole to neutralize it. E. But one has learned from the \"unipolar induction process\" for neutron stars [Gold reich and Julian (1968)] that charge neu tralization can sometimes be circum vented. Whether any black-hole process can possibly prevent neutralization one does not know in 1972. IV. Interaction of a black hole with its en", "viron ment A. Gravitational pull: A black hole exerts a gravitational pull on surrounding matter and stars. The pull is indistinguishable, at radii r ~ M, from the pull of a star with the same mass. B. Accretion and emission of x-rays and y rays: Gas surrounding a black hole gets pulled inward and is heated by adiabatic compression, by shock waves, by turbu lence, by viscosity, etc. Before it reaches the horizon, the gas may become so hot that it emits a large flux of x-rays and perhaps even y-rays. Thus, accreting mat ter can convert a black hole into a glowing \"white\" body [for a review of the litera ture, see Novikov and Thorne (1973)]. Accretion from a nonrotating gas cloud tends to decrease the angular momentum of a black hole [preferential accretion of particles with \"negative\" angular momen tum; Doroshkevich (1966), Godfrey (1970a)]. But the gas surrounding a hole is likely to be rotating in the same direc tion as the hole itself, and to maintain S - M2 [more precisely, S;:::; 0.998M2; Thorne (1973b)]. C. A lump of matter (an \"asteroid\" or a \"planet\" or a star) falling into a black hole should emit a burst of gravitational waves as it falls. The total energy radiated is E - 0.0 lJl(Jl/M), where Jl is the mass of the object. [Zerilli (1970b); Davis, Ruffini, Press, and Price (1971); Figure 36.2 of this book.] D. An object in a stable orbit around a black hole should spiral slowly inward because of loss of energy through gravitational ra diation, until it reaches the most tightly bound, stable circular orbit. It should then fall quickly into the hole, emitting a \"last gasp burst\" of waves. The total energy radiated during the slow inward spiral is equal to the binding energy of the last stable circular orbit: Eradiated = Jl - Elastorbit = (0.0572Jl for Schwarzschild hole, 0.4235Jl for Kerr hole with S = M2, Q = O. is the rest mass of the captured Here Jl object. [Box 33.5.] The total energy in the last-gasp burst is E - O.OlJl(Jl/M) if Jl < M. [Fig. 36.2.] 886 33. BLACK HOLES Box 33.3 (continued) E. When matter falls down a black hole, it can excite the hole's external spacetime geometry into vibration. The vibrations are gradually converted into gravitational waves, some of which escape, others go down the hole. [Press(197l, Goebel (1972).] These vibrations are analogous to an \"incipient gravitational geon\" [Wheeler (1962); Christodoulou (1971)]-except that for a vibrating black hole the background Kerr geometry holds the vibration energy from propagating together (prevents it away immediately), whereas in a geon it is curvature produced by the \"vibration energy\" itself that prevents disruption. F. By a non-Newtonian, induction-zone (i.e., nonradiative) gravitational interaction, a black hole gradually transfers its angular momentum to any non-axially-symmetric, nearby distribution of matter or fields. [Hawking (1972a); (1971), Press (1972), Hawking and Hartle (1972).] Ipser G. A star or planet falling into a large black hole will get torn apart by tidal gravita tional forces. If the tearing occurs near but outside the horizon, it may eject a blob of stellar matter that goes out with relativis tic velocity (\"tube-of-toothpaste effect\"). Moreover, the outgoing jet may extract a substantial amount of rotational energy from the hole's ergosphere-i.e., the hole might throw it off with a rest mass plus kinetic energy in excess of the rest mass of the original infalling object. [Wheeler (197ld); \u00a7\u00a733.7 and 33.8.] H. The magnetic field lines of a charged black hole may be anchored to surrounding plasma, may get wound up as the hole rotates, and may shake, twitch, and excite the plasma. V. Collisions between black holes A. Two black holes can collide and coalesce; but there is no way to blast a black hole apart into several black holes [Hawking (1972a); exercise 34.4]. B. When two black holes collide and coalesce, the surface area of the final black hole must exceed the sum of the surface areas of the two initial black holes (\"second law of black-hole dynamics\"; Hawking (197la", ",b); Box 33.4; \u00a734.5). This constraint places an upper limit on the amount of gravitational radiation emitted in the colli sion. For example, if all three holes are of the Schwarzschild variety and the two ini tial holes have equal masses M12, then 4'1T(2Mfinal)2 ~ 4'1T[2(MI2)j2 + 4'1T[2(MI2)]2, M final ~ M I Vi, so the energy radiated is Eradiated ::::;; M - MI Vi = 0.293M VI. Where and how to search for a black hole [For a detailed review, see Peebles (1971)]: A. When it forms, by the burst or bursts of gravitational radiation given off during formation [Figure 24.3]. B. In a binary star system: black-hole compo nent optically invisible, but may emit x-rays and y-rays due to accretion; visible component shows telltale Doppler shifts [Hoyle, Fowler, Burbidge, and Burbidge (1964); Zel'dovich and Guseynov (1965); Trimble and Thorne (1969); Pringle and Rees and Sunyaev Shakura (1973)]. The velocity of the visible compo nent and the period give information on the mass of the invisible component. If (1972); \u00a733.2. PRINCIPAL FEATURES OF HOLES 887 C. mass of this invisible component is four solar masses or more, it cannot be an ordi nary star, because an ordinary star of that mass would have (4)3 = 64 times the lu minosity of the sun. Neither can it be a white dwarf or a neutron star because ei ther object, so heavy, would instantly col lapse to a black hole. Therefore, it is attrac tive-though not necessarily compelling [see Trimble and Thorne (1969)]-to iden tify the invisible object as a black hole. [But one must not expect to see any no ticeable gravitational lens action from a black hole in a binary system: if it taxed the abilities of astronomers for decades to see the black disc of Mercury, 4,800 km in diameter, swim across the great face of the sun, little hope there is to see a black hole with an effective radius of only -3 km, enormously more remote, occult a companion star. Significant lens action re quires that the lens (black hole) be sepa interstellar distance rated by a normal from the star it focuses; whence the impact parameter of the focused rays is more than a stellar radius, so the lens,action is not more than that of a normal star. Moreover, even with 109 black holes in the galaxy, only one per year would pass directly be. tween the Earth and a more distant star, and action significant (Refsdal, 1964). Chance of watching the right spot on the sky at the right time with a sufficiently strong telescope: nil!] produce lens D. At the center of a globular cluster, where a black hole may settle down, attract nor mal stars to its vicinity, and thereby pro duce a cusp in the distribution of light from the cluster. [Cameron and Truran (1971), Peebles (1971).] E. In the nucleus of a galaxy, including even the Milky Way, where a single huge black hole (M - 104 to 108M0 ) might sit as an end-product of earlier activity of the ga lactic nucleus. Such a hole will emit gravi tational waves, light, and radio waves as it accretes matter. Much of the light may be converted into infrared radiation by surrounding dust. The black hole may also produce jets and other nuclear activity. (1969), Lynden-Bell' and [Lynden-Bell Rees (1971), Wheeler (197ld), Peebles (1971).] Box 33.4 THE LAWS OF BLACK-HOLE DYNAMICS The black-hole processes described in Box 33.3 are governed by the standard laws of physics: general relativity, plus Maxwell electrodynamics, plus hydrodynamic, quantum mechanical, and other laws for the physics of matter and radiation. From these standard laws of physics, one can derive certain \"rules\" or \"constraints,\" which all black-hole processes must satisfy. Those rules have a power, elegance, and simplicity that rival and resemble the power, elegance, and simplicity of the laws of thermodynamics. Therefore, they have been given the analogous name \"the laws of black-hole dynamics\" (Israel 1971). This box states two of the laws of black-hole 888 33. BLACK HOLES Box 33.4 (continued) dynamics and some of their ramifications. Two additional laws, not discussed here, have been formulated by Bardeen. Carter, and Hawking (1973). I. The First", " and Second Laws of Black-Hole Dynamics. A. The first law. 1. Like the first law of thermodynamics, the first law of black-hole dynamics is the standard law of conservation of total energy, supplemented by the laws of conservation of total momentum, angular momentum, and charge. For detailed discussions of these conservation laws, see Box 19.1 and Chapter 20. 2. Specialized to the case where matter falls down a black hole and gravita tional waves pour out, the first law takes the form depicted and discussed near the end of Box 19.1. 3. Specialized to the case of infalling electric charge, the first law says that the total charge Qof a black hole, as measured by the electric flux emerging from it, changes by an amount equal to the total charge that falls down the hole,.::1Q = qthatfalls in' 4. Specialized to the case where two black holes collide and coalesce (example given in Box 33.3), the first law says: (a) Let PI a.nd P2 be the 4-momenta of the two black holes as measured gravitationally, when they are so well-separated that they have negligible influence on each other. (PI and P2 are 4-vectors in the surrounding asymptotically flat spacetime.) Simi larly, let J I and J 2 be their total angular-momentum tensors (not intrinsic angular-momentum vectors!) relative to some arbitrarily chosen origin of coordinates, tj'0' in the surrounding asymptotically flat spacetime (J I and J 2 contain orbital angular momentum, as well as intrinsic angular mo mentum; see Box 5.6.). (b) Let P3 and J 3 be the similar total4-momentum and angular momentum of the final black hole. (c) Let Pr and J r be the total4-momentum and angular momentum radiated as gravitational waves during the collision and coalescence. Then [Note: to calculate the mass and intrinsic angular momentum of the final black hole from a knowledge of P3 and J 3, follow the prescription of Box 5.6. In that prescription, the world line of the final black hole is that world line, in the distant asymptotically Lorentz coordinates, on which the hole's distant spherical field is centered. \u00a733.2. PRINCIPAL FEATURES OF HOLES 889 B. The second law [expounded and applied by Hawking (197lb, 1972a)]. When anything falls down a black hole, or when several black holes collide and coalesce or collide and scatter, or in any other process whatsoever involv the sum of the surface areas (or squares of \"irreducible ing black holes, masses\"-see equation 3 below) ofall black holes involved can never decrease. (See \u00a734.5 for proof.) This is the second law of black-hole dynamics. II. Reversible and Irreversible Transformations; Irreducible Mass [Christodoulou (1970); Christodoulou and Ruffini (197l)-results derived inde pendently of and simultaneously with Hawking's discovery of the second law.] A. Consider a single Kerr-Newman black hole interacting with surrounding matter and fields. Its surface area, at any moment of time, is given in terms of its momentary mass M, charge Q, and intrinsic angular momentum per unit mass a _ SjM by A = 4'17[r+2 + a2] = 4'17[(M + yM2 - Q2 - a2)2 + a2] (1) (exercise 33.12). Interaction with matter and fields may change M, Q, and a in various ways; M can even be decreased-i.e., energy can be extracted from the black hole! [Penrose (1969); \u00a733.7.] But whatever may be the changes, they can never reduce the surface area A. Moreover, if any change in M, Q, and a ever increases the surface area, no future process can ever reduce it back to its initial value. B. Thus, one can classify black-hole processes into two groups. 1. Reversible transformations change M, Q, or a or any set thereof, while leaving the surface area ~hed. They can be reversed, bringing the black.JJ,ole back to its original state. 2. Irreve~~lble transformations change M, Q, or a or any set thereof, and increase the surface area in the process. Such a transformation can never be reversed. The black hole can never be brought back to its original state after an irreversible transformation. C. Examples of reversible transformations and of irreversible transformations induced by infalling particles are presented in \u00a7\u00a733.7 and 33.8. D. The r~ersible extraction of charge and angular momentum from a black hole ~----l--~--", "7Qeecc:nrease in Q and a holding A fixed) necessarily reduces the black hole's mass (energy extraction!). By the time all charge and angular momentum have been removed, the mass has dropped to a final \"irreducible value\" of M = (Aj16'17)1/2 = (mass of Schwarzschild IT black hole of surface area A ). (2) 890 33. BLACK HOLES Box 33.4 (continued) E. Expressed in terms of this fina~ irreducible mass, the initial mass-energy of the black hole (with charge Q and intrinsic angular momentum S) is M2 = ~r + 4M. ( Q2)2 IT S2 + 4M. 2 IT (3) [This formula, derived by Christodoulou and RUffini, may be obtained by combining equations (1), (2), and S = Ma]. E Thus, one can regard the total mass-energy of a black hole as made up of an irreducible mass, an electromagnetic mass-energy, and a rotational energy. But the temptation to think of these contributions as adding one must resist linearly. On the contrary, they combine in a way [equation (3)] analogous to the way rest mass and linear momentum combine to give energy, \u00a32 = m2 + p2. G. Contours of constant M/ MiT are depicted below in the \"charge-angular momentum plane.\" Black holes can exist only in the interior of the region depicted (Q2 + a2 :::;; M2). [Diagram adapted from Christodoulou (1971).] \u2022S \u00a733.3. MASS, ANGULAR MOMENTUM, CHARGE, AND MAGNETIC MOMENT 891 H. Since a black hole's irreducible mass is proportional to the square root of its surface area, one can restate the second law of black-hole dynamics as follows: In black-hole processes the sum of the squares of the irreducible masses of all black holes involved can never decrease. \u00a733.3. MASS, ANGULAR MOMENTUM, CHARGE, AND MAGNETIC MOMENT It is instructive to verify that the constants M, Q, and a, which appear in equations (33.2)-(33.5) for the Kerr-Newman geometry and electromagnetic field, are actually the black hole's mass, charge, and angular momentum per unit mass, as claimed above. Mass and angular momentum are defined by their imprints on the spacetime geometry far from the black hole. Therefore, to calculate the mass and angular momentum, one can expand the line element (33.2) in powers of l/r and examine the leading terms: ds2 = - [ 1 - _2~_~ 0 (}2)] dt2 - [4a M sin28 + 0 (}2)] dt d</> r + [ 1 + 0 ( ~ )] [dr2 + r2(d82 + sin28 d</>2)]. The examination is facilitated by transforming to asymptotically Lorentz coordi nates-x = r sin 8 cos </>, y = r sin 8 sin </>, z = r cos 8: ds 2 = - [1- 2~ + 0 (}2 )]dt2 - [4~~ + 0 (r14 ) ] [x dy - Y dx] + [1 + oe )][dx 2 + di + dz2]. (33.6') Direct comparison with the \"standard form\" [equation (19.13)] of the metric far from a stationary rotating source reveals that (1) the parameter M is, indeed, the mass of the black hole; and (2) the intrinsic angular momentum vector of the black hole IS S = (aM) %z = (aM). (unit vector. point~ng alon~ polar aXis). of Boyer-LmdqUIst coordmates (33.7) The charge is defined for the black hole, as for any source, by a Gaussian flux integral of its electric field over a closed surface surrounding the hole. The electric The rest of this chapter is Track 2. To be prepared for it, one needs to have covered the Track-2 part of Chapter 32 (gravitational collapse). reading it, one will be helped greatly by Chapter 25 (orbits in Schwarzschild geometry). The rest of this chapter is needed as preparation for Chapter 34 (singularities and global methods). In (33.6) The metric far outside a black hole: imprints of mass and angular momentum The electromagnetic field far outside a black hole: (1) electric field (2) magnetic field 892 33. BLACK HOLES field in the asymptotic rest frame of the black hole has as its orthonormal components E;.", " = Er = Frt = Q/r'.!. + 00/r3), EiJ = E 8/r = F8t /r = 00/r 4 E\u00a2 = Eqjr sin 8 = F8,/r sin 8 = O. ), _(33.8). Hence, the electric field is purely radial with a Gaussian flux integral of 4~Q, which reveals Q to be the black hole's charge. A similar calculation of the dominant components of the magnetic field reveals B\u00b7 = F\u00b7\u00b7 = 8\u00a2 r ~ r2 sin 8 \u00a2 = 2 - Qa r3 (1) cos 8 + 0 r4, BiJ = F\u00b7\u00b7 = ---!!- = - F r SIn 8 Qa. r3 \u00a2r SIn 8 + 0 - ( 1 ) r4, (33.9) (3) magnetic dipole moment q)fl = Qa = (Q/M)S = ~charge/mass~ X (angular momentum) (33.10) This is a dipole magnetic field, and from it one immediately reads off the value Nonspherical shape of hole's geometry. t...{\"gyromagnetic ratio\"] for the magnetic moment of the black hole. Just as the rotation of the black hole produces a magnetic field, so it also produces nonspherical deformations in the gravitational field of the black hole [see Hernandez (1967) for quantitative discussion]. But those deformations, like the magnetic mo ment, are not freely specifiable. They are determined uniquely by the mass, charge, and angular momentum of the black hole. \u00a733.4. SYMMETRIES AND FRAME DRAGGING The metric components (33.2) of a Kerr-Newman black hole are independent of the Boyer-Lindquist time coordinate t and angular coordinate <p. This means (see \u00a725.2) that Killing vectOrs for the Kerr-Newman geometry (33.11) are Killing vectors associated with the stationarity (time-translation invariance) and axial symmetry of the black hole. The scalar products of these Killing vectors with themselves and each other are \u00a733.4. SYMMETRIES AND FRAME DRAGGING 893 (33.l2a) (2Mr - Q2)a sin28 p2 (33.l2b) (33.l2c) Since Killing vectors are geometric propert'ies of spacetime, with existence inde pendent of any and all coordinate systems, their scalar products also have coor dinate-free meaning. It so happens (not by chance, but by careful choice of coordi nates!) that the Boyer-Lindquist metric components gtt' gt\u00a2' andg\u00a2\u00a2 are equal to these coordinate-independent scalar products. Thus gtt' gt\u00a2' and g\u00a2\u00a2 can be thought of as three scalar fields which embody information about the symmetries of spacetime. By contrast, the metric coefficients grr = p2l;:j and g88 = p2 carry no information at all about the symmetries.* They depend, for their existence and values, on the specific Boyer-Lindquist choice of coordinates. Any observer who moves along a world line of constant (r, 8) with uniform angular velocity sees an unchanging spacetime geometry in his neighborhood. Hence, such an observer can be thought of as \"stationary\" relative to the local geometry. If and only if his angular velocity is zero, that is, if and only if he moves along a world line of constant (r, 8, </\u00bb, will he also be \"static\" relative to the black hole's asymptotic Lorentz frame (i.e., relative to the \"distant stars\"). The precise definition of \"angular velocity relative to the asymptotic rest frame\" -or simply \"angular velocity\"-is (33.l3a) (see exercise 33.2). In terms of [J, the Killing vectors, and the scalar products of Killing vectors, the 4-velocity of a stationary observer is Stationary observers Static observers Angular velocity and 4-velocity of a stationary observer u = ut(o lot + [Jo lo</\u00bb = (t) + [J(p) I(t) + [J(\u00a2)I _ - (- gtt - 2[Jgt\u00a2 - (t) + [J(\u00a2) [J2g\u00a2\u00a2)1/2' (33.l3b) A stationary observer is static if and only if [J vanishes. The stationary observers at given r,8 cannot have any and every angular velocity. Only those values of [J are allowed for which the 4-velocity u lies inside the future light cone-i.e., for which \"This is not quite true. Kerr-Newman spacetime", " possesses, in addition to its two Killing vectors, also a \"Killing tensor\" which is closely linked to the Boyer-Lindquist coordinates rand (J. See Walker and Penrose (1970); also \u00a733.5. y 894 33. BLACK HOLES Frame dragging, static limit, and ergosphere Thus, the angular velocities of stationary observers are constrained by where Q min = W - Vw 2 Q max = W + Vw 2 ) = _k _ Q g\u00a2\u00a2 - w = I Q. - ~2 mm + max - gtt!g\u00a2\u00a2, - gtt!g\u00a2\u00a2, (2MI' - Q2)a - ( 2 + 2)2 I' a L1 2 \u2022 \"8' a sm- (33.14) (33.15a) (33.15b) (33.16) and it is assumed that S/M = a > O. The following features of these limits are noteworthy. (I) Far from the black hole, one has rQrnin = -I and rQ max = +1, corresponding to the standard limits imposed by the speed oflight in flat spacetime. (2) With decreasing radius, Q rnin increases (\"dragging of inertial frames\"). Finally, when gtt reaches zero, i.e., at (33.17) Q min becomes zero. At and inside this surface, all stationary observers must orbit the black hole with positive angular velocity. Thus, static observers exist outside and only outside I' = 1'0(8). For this reason I' = 1'0(8) is called the \"static limit\"; see Box 33.2. (3) As one moves through the static limit into the \"ergosphere,\" one sees the allowed range of angular velocities become ever more positive (ever more \"frame dragging\"). At the same time, one sees the allowed range narrow down, until finally, at the horizon (33.18) the limits Q min and Q max coalesce (w 2 = gtt/g</></\u00bb. Thus, at the horizon there are no stationary observers. All timelike world lines point inward. There is no escape from the black hole's \"pull.\" Further features of stationary observers and \"frame dragging\" are explored in the exercises. EXERCISES Exercise 33.1. KERR DESCRIPTION OF KILLING VECTORS (a) Use the transformation law from Boyer-Lindquist coordinates to Kerr coordinates [equation (4) of Box 33.2] to show that ((I) = (0/ot)r,9,</> = (0/0 V)r,9,\u00a2, (0/0</\u00bb1,.,9 = (0/0~)v,r,9' (</\u00bb - Verify explicitly by examining metric components that in accordance with equations (33.l9a,b). (33.l9a) (33.l9b) (33.l9c) \u00a733.4. SYMMETRIES AND FRAME DRAGGING 895 (b) Show that for a stationary observer (world line of constant r,O), the angular velocity expressed in terms of Kerr coordinates is II = dep/dt = d;P/dV = u\u00a2;/uv, so that the entire discussion of stationary observers in terms of Kerr coordinates is identical to the discussion in terms of Boyer-Lindquist coordinates. Differences between the coordinate systems show up only when one moves along world lines of changing r. Reconcile this fact with the fact that both coordinate systems use the same coordinates (r,O) but different time and aximuthal coordinates (t, ep versus Po ;p). Exercise 33.2. OBSERVATIONS OF ANGULAR VELOCITY An observer, far from a black hole and at rest in the hole's asymptotic Lorentz frame, watches (with his eyes) as a particle moves along a stationary (nongeodesic) orbit near the black hole. Let II = dep/dt be the particle's angular velocity, as defined and discussed above. The distant observer uses his stopwatch to measure the time required for the particle to make one complete circuit around the black hole (one complete circuit relative to the distant observer himself; i.e., relative to the hole's asymptotic Lorentz frame). (a) Show that the circuit time measured is 2'lT/ll. Thus, II can be regarded as the particle's \"angular velocity as measured from infinity.\" (b) Let the observer moving with the particle measure its circuit time relative to the asymptotic Lorentz frame, using his eyes and a stopwatch he carries. Show that his answer for the circuit time must be (33.20) ~\"redshift factor\"] Exercise 33.3. LOCALLY NONROT", "ATING OBSERVERS (Bardeen 1970b) (a) Place a rigid, circular mirror (\"ring mirror\") at fixed (r, 0) around a black hole. Let an observer at (r,O) with angular velocity II emit a flash of light. Some of the photons will get caught by the mirror and will skim along its surface, circumnavigating the black hole in the positive-ep direction. Others will get caught and will skim along in the negative-ep direction. Show that the observer will receive back the photons from both directions simulta neously only if his angular velocity is II = w(r, 0) = expression (33.16). Thus in this case, and only in this case, can the observer regard the +ep and -ep directions as equivalent in terms of local geometry. Put differently, in this case and only in this case is the observer \"nonrotating relative to the local spacetime geometry.\" Thus, it is appropriate to use the name \"locally nonrotating observer\" for an observer who moves with the angular velocity II = w(r,O). p\u00a2 =P' f(\u00a2), for geodesic motion. This quantity for any particle-whether it is moving along (b) Associated with the axial symmetry of a black hole is a conserved quantity, a geodesic or not-is called the \"component of angular momentum along the black hole's spin axis,\" or simply the particle's \"angular momentum.\" (See \u00a733.5 below.) Show that of all stationary observers at fixed (r, 0), only the \"locally nonrotating observer\" has zero angular momentum. [Note: Bardeen, Press, and Teukolsky (1972) have shown that the \"locally nonrotating observer\" can be a powerful tool in the analysis of physical processes near a black hole.] 896 33. BLACK HOLES Exercise 33.4. ORTHONORMAL FRAMES OF LOCALLY NONROTATING OBSERVERS (a) Let spacetime be filled with world lines of locally nonrotating observers, and let each such observer carry an orthonormal frame with himself. Show that the spatial orientations of these frames can be so chosen that their basis I-forms are Ig - w 2g 11/2 dt \u00a2\u00a2 It wi wi = (p/,1112) dr,, w\u00a2 = (g\u00a2\u00a2)1!2(d\u00a2 - w dt), w 8 = pd(J. (33.21 ) More specifically, show that these I-forms are orthonormal and that the dual basis has alai = u = 4-velocity of locally nonrotating observer. (33.22) Show that u = - wi is a rotation-free field of I-forms [dw i 1\\ wi = 0; exercise 4.4]. (b) One sometimes meets the mistaken notion that a \"locally nonrotating observer\" is in some sense locally inertial. To destroy this false impression, verify that: (i) such an observer has nonzero 4-acceleration,. a = rjiiej = \"2 'V In Igtt - w2g,p,pI, I (33.23) (ii) if such an observer carries gyroscopes with himself, applying the necessary accelerations at the gyroscope centers of mass, he sees the gyroscopes precess relative to his orthonormal frame (33.21) with angular velocity O(precess) = r8;j,ie; + r;j,TieiJ I = 2 IgII g 112,pp - w2g,p,p11/2 [w -J!!..e- - --J!..e- ],11/2W P P T 9 \u00b7 (33.24) [Hints: See exercise 19.2, equation (13.69), and associated discussions. The calculation of the connection coefficients is performed most easily using the methods of differential forms; see \u00a714.6.] Exercise 33.5. LOCAL LIGHT CONES Calculate the shapes of the light cones depicted in the Kerr diagram for an uncharged (Q = 0) Kerr black hole (part II.F of Box 33.2). In particular, introduce a new time coordinate 7= V- r (33.25) for which the slices of constant 7 are horizontal surfaces in the Kerr diagram. Then the Kerr diagram plots 7 vertically, r radially, and;P azimuthally, while holding (J = 'IT/2 (\"equatorial slice through black hole\"). (a) Show that the light cone emanating from given 7, r, ;p has the form dr (d;P) dt = a dt - 2M/r I + 2M/r \u00b1 j I (I + 2M/r)2 - r2(", "d;Pld\"?f I + 2M/r (b) Show that the light cone slices through the surface of constant radius along the curves (33.26b) where ilmJn and ilmu are given by expressions (33.l5a,b) (minimum and maximum allowed angular velocities for stationary observers). (c) Show that at the static limit, r = ro('IT/2), the light cone is tangent to a curve of constant r, (J, ;Po \u00a733.5. EQUATIONS OF MOTION FOR TEST PARTICLES (d) Show that the light cone slices the surface of constant ;p along the curves dr = _ I and I - 2M/r.. dt 1+ 2M/r 897 (33.26c) (e) Show that the light cone is tangent to the horizon. (f) Make pictures of the shapes of the light cone as a function of radius. (g) Describe qualitatively how the light cone must look near the horizon in Boyer-Lindquist coordinates. (Note: it will look \"crazy\" because the coordinates are singular at the horizon.) \u00a733.5. EQUATIONS OF MOTION FOR TEST PARTICLES [Carter (1968a)] Let a test particle with electric charge e and rest mass /L move in the external fields of a black hole. Were there no charge down the black hole, the test particle would move along a geodesic (zero 4-acceleration). But the charge produces an electromag netic field, which in turn produces a Lorentz force on the particle: p.a = eF' u. (Here u is the particle's 4-velocity, and a \"'uu is its 4-acceleration.) The geodesic equation, a = 0, for the uncharged case is equivalent to Hamilton's equations dp,/dA. = -oX/oxlJ., (33.27a) where A. is an affine parameter so normalized that d/dA. = p = 4-momentum, (33.27b) and where X \"super-Hamiltonian\" = ~ gIJ.\"p~\" (33.27c) (see exercise 25.2). Similarly (see exercise 33.6) the Lorentz-force equation, p.a = eF' u, for the charged case is equivalent to Hamilton's equations written in terms of position xIJ. and \"generalized momentum\" 7TIJ.: (33.28a) The form of the superhamiltonian :)(, in terms of the metric coefficients at the ), and the particle's charge e and generalized momentum particle's location, gIJ.\"(x a 7TIJ.' IS Superhamiltonian for a charged test particle in any electromagnetic field in curved spacetime (33.28b) [See \u00a773 of Goldstein (1959) for the analogous superhamiltonian in flat spacetime.] 898 33. BLACK HOLES The first of Hamilton's equations for this superhamiltonian reduces to pll (4-momentum) dXIl/d\"A. = ';TIL - eAIl (33.29a) (value of ';TIL in terms of pll, e, and All); the second, when combined with the first, reduces to the Lorentz-force equation (33.29b) For a Kerr-Newman black hole, the vector potential in Boyer-Lindquist coordi nates can be put in the form Vector potential for a charged black hole A = - - 2 (dt - a sm2() dcp),. Qr p (33.30) as one verifies by checking that reduces to the Faraday 2-form of equation (33.5). There is good reason for going through all this formalism, rather than tackling head-on the Lorentz-force equation in its most elementary coordinate version, \"Constants of motion\" for a charged test particle moving around a charged black hole: The Hamiltonian formalism enables one to discover immediately two constants of the motion; the elementary Lorentz-force equation does not. The key point is that the components All of A [equation (33.30)] and the components gil\" of the metric [inverse of gil\" of equation (33.2); see (33.35)] are independent of t and cp (stationarity and axial symmetry of both the electromagnetic field and the spacetime geometry). Consequently, the superhamiltonian is also independent of t and cp; and therefore Hamilton's equation (1) \"energy at infinity\" E (2) \"axial component of angular momentum\" Lz guarantees that TT t and TTq, are constants of the motion. Far from the black hole, where the vector potential vanishes and the metric becomes the constants of the motion become", " _ _ TT q, - Pq, - rp TTt = Pt = _pt = -energy, ;p _ (prOjection of angular momentum) along black hole's rotation axis -. \u00a733.5. EQUATIONS OF MOTION FOR TEST PARTICLES 899 Thus it is appropriate to adopt the names and notation E (\"energy at infinity\") -'1Tt = -(Pt + eAt), (333la) L z \"axial component of angUlar) momentum\", or simply \"angular momentum\" ( '1Tq, = Pq, + eAq, (3331 b) for the constants of the motion -'1Tt and '1Tq,' A third constant of the motion is the particle's rest mass (3) rest mass /l (333lc) In general, four constants of the motion are needed to determine uniquely the orbit of a particle through four-dimensional spacetime. If the black hole were to possess an additional symmetry-e.g., ifit were spherical, rather than merely axially symmetric-then automatically there would be a fourth constant of the motion. But in general, black holes are not spherical; so test-particle motion around a black hole possesses only three obvious constants. It is rather remarkable, then, that a constant turns out to exist. It was discovered by Carter (1968a), using Hamilton-Jacobi methods. As of 1973, nobody has given a cogent geometric explanation of why this fourth constant should exist-although hints of an explanation may be found in Carter (l968c) and Walker and Penrose (1970). Carter's \"fourth constant\" of the motion, as derived in exercise 33.7, is (4) \"52\" The constant of the motion (333ld) (3331 e) obtained by combining 2, Lz, and E, is often used in place of 2. Whereas 2 can be negative, X is always nonnegative: X = pl + (Lz - aEsin20)2/sin20 + a2p.2 cos20 2 0 everywhere = 0 only for case of photon (p. = 0) moving along polar axis (0 = 0, '1T). The contravariant components of the test particle's 4-momentum, pO: = dxO:/dA., are readily expressed in terms of the constants E, Lz'p., 2, by combining equations (3331) with the metric coefficients (33.2) and the components of the vector potential (3330). The result is p2 dO/dA. = ve: p2 dr/dA. = VR, p 2 dcf>/dA. = -(aE - Lz/sin 20) + (a/J)P, p2 dt/dA. = -a(aE sin20 - L z ) + (r 2 + a2)J-Ip. (3332a) (3332b) (3332c) (3332d) Equations of motion for charged test particles y EXERCISES 900 Here p2 = r2 + a2 cos2 are defined by () as defined in equation (33.3b), and the functions e, R, P 33. BLACK HOLES e =!!!. - cos2() [a2(p.2 - \u00a32) + Lz 2/sin2()], P = \u00a3(r 2 + a2 ) - Lza - eQr, R = p2 - J[p.2r 2 + (Lz - a\u00a3)'l +!!!']. (33.33a) (33.33b) (33.33c) When working in Kerr coordinates (to avoid the coordinate singularity at the horizon), one must replace equations (33.32c) and (33.32d) by p 2 dV/d'A. = -a(a\u00a3sin2() - L) + (r 2 + a2)J-l(VR + P), (33.32c') p2 d1>/d'A. = -(a\u00a3 - Lz/sin2 () + aJ-l(VR + P). (33.32d') [These follow from (33.32) and the transformation between the two coordinate systems-see equations (4) of Box 33.2.] In the above equations, the signs of VR and -vecan be chosen independently; but once chosen, they must be used consis tently everywhere. Applications of these equations of motion will playa key role in the rest of this chapter. Exercise 33.6. SUPERHAMILTONIAN FOR CHARGED-PARTICLE MOTION Show that Hamilton's equations (33.28a) for the Hamiltonian (33.28b) reduce to equation (33.29a", ") for the value of the generalized momentum, and to the Lorentz force equation (33.29b). [Hint: Use the relation (g\",f3gf3Y).1' = 0.] Exercise 33.7. HAMILTON-JACOBI DERIVATION OF EQUATIONS OF MOTION [Based on Carter (1968a)] Derive the first-order equations of motion (33.32) for a charged particle moving in the external fields of a Kerr-Newman black hole. Use the Hamilton-Jacobi method [Boxes 253 and 25.4 of this book; also Chapter 9 of Goldstein (l959)J, as follows. (a) Throughout the superhamiltonian.X of equation (33.28b), replace the generalized momentum 'IT,,, by the gradient as/ax'\" of the Hamilton-Jacobi function. (b) Write down the Hamilton-Jacobi equation [generalization of equation (2) of Box 25.4] in the form _ as = \",- [ '\" E-] _1..11. x, ax f3 aA - 2 g \"'f3 (E- _ )(as - A ) e f3 --axtJ ax'\" eA\",. (3334a) (c) Show that the metric components g\"'f3 for a Kerr-Newman black hole in Boyer-Lind quist coordinates are given by (33.35) \u00a733.6. PRINCIPAL NUll CONGRUENCES 901 (d) Use these metric components and the components (33.30) of the vector potential to bring the Hamilton-Jacobi equation (33.33) into the concrete form - as aA = - - - (r 2 + a2) - + a - - eQr as at as a</> ]2 33.34b [as + asin20 as]2 +.l.-A(aS)2 +.l_I_(aS)2 ( 2 p2 2 p2 00 Or at. ) I [ I 2,1p2 I 2 p2 sin20 a</> + I (e) Solve this Hamilton-Jacobi equation by separation of variables. [Hint: Because the equation has no explicit dependence on A, </>, or t, the solution must take the form where the values of the \"integration constants\" follow from as/aA = - :](, as/at = 'ITt' as/a</> = 'lTq,. Insert this assumed form into (33.35) and solve for S,(r) and S9(0) to obtain (33.36a) (33.36b) where R(r) and 8(0) are the functions defined in equation (33.33). Notice that the constant!2 arises naturally as a \"separation-of-variables constant\" in this procedure. It was in this way that Carter originally discovered!2, following Misner's suggestion that he seek analogies to a constant in Newtonian dipole fields (Corben and Stehle, 1960, p. 209).] (f)-By ~ive4c~etting-aS/a{2.+(Lz....,. aE)2], as/a/l-2, as/aE, and as/aLz obtain the following equations describing the test-particle orbits: f' _ f9 a2 cos20. (;5 A __ --v8 _ f9 -a(aE sin20 - Lz ) t - dO +. (;5 v8 </>= f - (aE sin20 - Lz ) sin20 ve r2. In dr, vR dO + f dO+ --dr. aP,1 VIi f' (r2 + a2)p. In,1vR dr, to zero, (3337a) (33.37b) (33.37c) (33.37d) (g) By differentiating these equations and combining them, obtain the equations of motion (33.32) cited in the text. (h) Derive equations (33.31) for E, Lz'/1-, and!2 by setting as/ax'\" = 'IT\", = P'\" + eA\",. \u00a733.6. PRINCIPAL NULL CONGRUENCES Two special families ofphoton trajectories \"mold themselves into\" the Kerr-Newman geometry in an especially harmonious way. They are called the \"principal null congruences\" of the geometry. (\"Congruence\" is an elegant word that means \"space- Principal null congruences for the spacetime geometry of a black hole 902 33. BLACK HOLES filling family of curves.\") These congru", "ences are the solutions to the test-particle equations of motion (33.32) with }.t = 0 (zero rest ma~s~photonl!-.. e = 0 (zero charge on photon), (33.38a) (33.38b) L z = aE sin20 a permissible value for L z) only because dO / dA. turns out to be zero!:2 = -(Lz - aE)2 = -a2E2cos4fJ.\u00b7 (, (33.38c) (33.38d) For these values of the constants of motion, the equations of motion (33.32) reduce to k r dr/dA. = +-E (\" +\" for outgoing photons, \" -\" for ingoing), k\u00a2 dtP/dA. = aE/.1, dt/dA. = (r 2 + a2 )E/J. k t (33.39a) (33.39b) (33.39c) (33.39d~. Significance of the principal null congruences In what sense are these photon trajectories more interesting than others? (1) They mold themselves to the spacetime curvature in such a way that, if CO:(JYB is the Weyl conformal tensor (\u00a713.5), and \u00b7CO:(JYB = f.O:(JIJ.pC1IJ.PIYB is its dual, then (33.40) [This relationship implies that the Kerr-Newman geometry is of \"Petrov-Pirani type D\" and that these photon trajectories are \"doubly degenerate, principal null con gruences.\" For details of the meanings and implications of these terms see, e.g., \u00a78 of Sachs (1964), or Ehlers and Kundt (1962), or the original papers by Petrov (1954, 1969) and Pirani (1957).] (2) By suitable changes of coordinates (exercise 33.8), one can bring the Kerr-Newman metric into the form where H is a scalar field and ko: are the components of the wave vector for one of the principal null congruences (either one; but not both!). [This was the property of the Kerr-Newman metric that led to its original discovery (Kerr, 1963). For further detail on metrics of this form, see Kerr and Schild (1965).] (3) In Kerr coordinates (Box 33.2), the ingoing principal null congruence is (33.41) r = - EA, 0 = const, ;p = const, v= const. (33.42a) ~arbitrary nOrmaliZatiOn] factor; ca~?e removed by redefinItIOn of A. \u00a733.6. PRINCIPAL NULL CONGRUENCES 903 These ingoing photon world lines are the generators ofthe conical surface V = const. in the Kerr diagram of Box 33.2. (4) The only kind of particle that can remain forever at the horizon is a photon with world line in the outgoing principal null congruence (exercise 33.9). Such photon world lines are \"generators\" of the horizon (dotted curves with a \"barber-pole twist\" in Kerr diagram of Box 332). They have angular velocity dep d;P a [J - - - - - - -,+2 + a2 - dt - dV - a 2M2 _ Q2 + 2M(M2 _ a2 _ Q2)1/2' (33.42b) Exercise 33.8. KERR-SCHILD COORDINATES (a) Show that in Kerr coordinates the ingoing null congruence (33.39) has the form (33.42a). Also show that the covariant components of the wave vector-after changing to a new affine parameter Anew = AoldE-are EXERCISES (b) Introduce new coordinates t, x, y, z, defined by x + iy = (r + ia)ei ;; sin 0, z=rcosO, t = V - r; (33.44a) and show that in this \"Kerr-Schild coordinate system\" the metric takes the form k~nl = -1. V (33.43) where H- Mr -!Q2 - r2 + a2(z/r) 2, k lin) dx a = _ r(x dx + y dy) - a(x dy - Y dx) _ z dz _ dt. r2 + a2 r a (33.44b) (33.44c) (33.44d) For the transformation to analogous coordinates in which see, e", ".g., Boyer and Lindquist (1967). Exercise 33.9. NULL GENERATORS OF HORIZON (a) Show that in Kerr coordinates the outgoing principle null congruence is described by the tangent vector dO dA = 0, dr dA = E, d~ E dA = 2a\"\"J\"' (33.45) (b) These components of the wave vector become singular at the horizon (41 = 0), not because of a singularity in the coordinate system-the coordinates are well-behaved!-but because of poor normalization of the affine parameter. For each outgoing geodesic, let 41 0 904 33. BLACK HOLES be a constant, defined as the value of.1 at the event where the geodesic slices the hypersurface j7 = 0. Then renorrnalize the affine parameter for each geodesic Anew = (E/.10 )A o1d \u2022 Show that the resulting wave vectors dO dA = 0, (33.46) (33.45') are well-behaved as one approaches the horizon; and show that the geodesics on the horizon have the form. o= const.. r = r+ = const., ;p = 2aA, (33.47) (c) Show that these are the only test-particle trajectories that remain forever on the horizon. [Hint: Examine the light cone.] \u00a733.7. STORAGE AND REMOVAL OF ENERGY FROM BLACK HOLES [Penrose (1969)] When an object falls into a black hole, it changes the hole's mass, charge, and intrinsic angular momentum (first law of black-hole dynamics; Box 33.4). If the infalling object is large, its fall produces much gravitational and electromagnetic radiation. To calculate the radiation emitted, and the energy and angular momentum it carries away-which are prerequisites to any calculation of the final state of the black hole-is an enormously difficult task. But if the object is very small (size of object ~ size of horizon; mass of object ~ mass of hole). and has sufficiently small charge, the radiation it emits in each circuit around the hole is negligible. For example, for gravitational radiation (energy emitted per circuit) (rest mass of object) (rest mass of object) (mass of hole) (33.48) [see \u00a736.5; also Bardeen, Press, and Teukolsky (1972)]. Because the energy emitted is negligible, radiation reaction is also negligible, and the object moves very nearly along a test-particle trajectory. In this case, application of the first law of black-hole dynamics is simple and straightforward. Consider, initially, a small object that falls directly into the black hole from far away. According to the first law, it produces the following changes in the mass, charge, and angular momentum of the black hole: JM = E = (\"energy at infinity\" of infalling object), JQ = e = (charge of infalling object), JI51 = L = (component of object's angular momentum). on black hole's rotation axis z (33.49a) (33.49b) (33.49c) LiS When a small object falls down a large hole: (1) energy radiated is negligible compared to object's rest mass (2) hole's mass. charge. and angular momentum change by 41M = E, 410 = e. 41S = Lz \u00a733.7. STORAGE AND REMOVAL OF ENERGY 905 The infalling object will also change the direction of S. In the black hole's original asymptotic Lorentz frame, its initial angular momentum vector points in the z-direc tion, Consequently, only the z-component of angular momentum of the infalling object can produce any significant change in the magnitude of S. But the x- and y-com ponents, L., and LIP can change the direction of S. If the object has negligible speed at infinity, then it prodll.ces the changes (exercise 33.10): JS., = L., = -(sinep\",)v'8: (cotO\",cosep\",)L.\" JS/I = L/I = (C05q>\"JvB: ::\"'lcotO~ sinep\",)L.\" L1(S.,2 + S/)l/2 =.\"jfi = (L2 - L.,2)1/2. (33.49d) (33.4ge) (33.49f) Here a subscript \"00\" means the value of a quantity at a point on the orbit far -from thLblack hole (at \"infinity\"). Consider, next, a morecomplkated process, first conceived of by Penrose (1969): (1) Shoot a small object A into", " the black hole from outside with energy-at-infinity EA, charge eA, and axial component of angular momentum L.,A' (2) When the object is deep down near the horizon, let it explode into two parts, Band C, each of which subsequently moves along a new test-particle trajectory, with new constants of the motion eB and ee, EB and Ee, L\"B and L.,e. (3) So design the explosion that object B falls down the hole and gets captured, but object C escapes back to radial infinity. What will be the change~~in mass, charge, and angular momentum of the black hole? According to the first law of black-hole dynamics, total energy that distant observers see) JM = fall inward past themselves minus total energy that they see reemerge ( =EA - E e\u00b7 ee and JS = L ZA - L.,e. Not unexpectedly, these changes can Similarly, JQ = eA be written more simply in terms of the constants of motion for object B, which went down the hole. View the explosion \"A ~ B + C\" in a local Lorentz frame down near the hole, which is centered on the explosive event. As viewed in that frame, the explosion must satisfy the special relativistic laws of physics (equivalence princi ple!). In particular, it must obey charge conservation (33.50a) and conservation of total 4-momentum (PA)immediatelybeforeexPloSion = (PB + Pe)immediately after explosion' 906 33. BLACK HOLES Moreover, conservation of 4-momentum P and charge e implies also conservation of generalized momentum n p - eA, nA = PA - eAA = PB + Pe - (eB + ee)A = n B + ne; and' hence also conservation of the components of generalized momentum along the vectors a/at and a/oet>, EA L ZA -'1TtA = -'1TtB '1T\u00a2B + '1Tq,e = LZB + L ze' '1T tC = EB + Ee, (33.50b) (33.50c) (conservation of \"energy-at-infinity\" and \"axial component of angular momentum\" in explosion). Combining these conservation laws with the expressions one obtains (33.51 ) This result restated in words: the changes in mass, charge, and angular momentum are equal to the \"energy-at-infinity,\" charge, and \"axial component of angular momentum\" that object B carries inward across the horizon, even though B may have ended up on a test-particle orbit that does not extend back to radial infinity! Straightforward extensions of the above thought experiment produce this general ization: In any complicated black-hole process that involves infalling, colliding, and exploding pieces ofmatter that emit negligible gravitational radiation, the total changes in mass, charge, and angular momentum of the black hole are sum of values of energy-at-infinity, E, JM = for all objects which cross the horizon-with ), (33.52a) ( E evaluated for each object at event of crossing JQ = (Similar sum, of charges, e, fOr) all objects crossing horizon'J (Similar sum of axial components of angular ) S = momentum, L z'for all objects crossing horizon. (33.52b) (33.52c) This result is not at all surprising. It is precisely what one might expect from the most naive of viewpoints. Not so expected, however, is the following consequence [Penrose (1969)]: By injecting matter into a black hole in a carefully chosen way, one can decrease the total mass-energy ofthe black hole-i.e., one can extract energy from the hole. For uncharged infalling objects, the key to energy extraction is the ergosphere [hence its name, coined by Ruffini and Wheeler (1971 a) from the Greek word \"Wyov\" a/at is timelike, as for \"work\"]. Outside the ergosphere, the Killing vector (t) is the 4-momentum P of every test particle; and therefore E = - P \u2022 (t) is necessarily positive. But inside the ergosphere (between the horizon and the static limit), (t> Changes in M. Q, S for any non radiative black-hole process Extraction of energy from a black hole by processes in the ergosphere \u00a733.8. REVERSIBLE AND IRREVERSIBLE PROCESSES 907 is spacelike, so for certain choices of timelike momentum vector (certain orbits of uncharged test particles), E = -P\" (t) is negative, whereas for others it is positive. The orbits of negative E are confined entirely to the ergosphere. Thus,", " to inject an uncharged object with negative E into the black hole-and thereby to extract energy from the hole-one must always change its E from positive to negative and therefore also change its orbit, after it penetrates into the ergosphere. Of course, this is not difficult in principle-and perhaps not even in practice; see Figure 33.2. For a charged object, electromagnetic forces alter the region where there exist orhits.of negatiYe energ)'-at-infinity. If the charges of object and hole have opposite sign, then the hole's electromagndiC field pulls inward on the object, giving it more kinetic energy when near the hole than one would otherwise expect. Thus, -P\" (t) becomes an overestimate of E, E = -(p - eA)\" ((t) = -P\"(t) +~; (33.53) i...[< 0 if eQ < 0] -----. - -~-and9r.bJts with E < 0 exist in a region somewhat larger than the ergosphere. If, on the other hand;-e-and Q have the same sign, then orbits with E < 0 are confined to a region smaller than the ergosphere. For given values e, Q, and rest mass p., the region where there exist orbits with E < 0 is called the \"effective ergosphere. \" The \"effective ergosphere\" for charged-particle processes Exercise 33.10. ANGULAR MOMENTUM VECTOR FOR INFALLING PARTICLE Derive equations (33.49d,e,f) for the components L\" and L y of the orbital angular momentum of a particle falling into a black hole. Assume negligible initial speed, \u00a32 - /1-2 ~ O. EXERCISE \u00a733.8. REVERSIBLE AND IRREVERSIBLE TRANSFORMATIONS [Christodoulou (1970), Christodoulou and Ruffini (1971)] Take a black hole of given mass M, charge Q, and angular momentum S. By injection ofsmall objects, make a variety of changes in M, Q, and S. Can one pick an arbitrary desired change, JM, JQ, and JS, and devise a process that achieves it? Or are there limitations? The second law of black-hole dynamics (nondecreasing surface area of black hole; Box 33.4; proof in \u00a734.5 of next chapter) provides a strict limitation. Then can all values within that limitation be achieved-and can that limitation be discovered by a direct examination of test-particle orbits? The answer is yes; and, in fact, the limitation was discovered by Christodoulou (1970) and Christodoulou and Ruffini (1971) from an examination of test-particle orbits, independently of and simultaneously with Hawking's (1971) discovery of the second law of black-hole dynamics. 908 Figure 33.2. An advanced civilization has constructed a rigid framework around a black hole, and has built a huge city on that framework. Each day trucks carry one million tons of garbage out of the city to the garbage dump. At the dump the garbage is shoveled into shuttle vehicles which are then, one after another, dropped toward the center of the black hole. Dragging of inertial frames whips each shuttle vehicle into a circling, inward-spiraling orbit near the horizon. When it reaches a certain \"ejection point,\" the vehicle ejects its load of garbage into an orbit of negative energy-at-infinity, Egarbage < O. As the garbage flies down the hole, changing the hole's total mass-energy by.JM = E garoage eieeted < 0, the shuttle vehicle recoils from the ejection and goes flying back out with more energy-at-infinity than it took down Evehicle out = Evehicle + garbage down - Egarbageejected > Eveblele + garbage down' The vehicle deposits its huge kinetic energy in a giant flywheel adjacent to the garbage dump; and the flywheel turns a generator, producing electricity for the city, while the shuttle vehicle goes back for another load of garbage. The total electrical energy generated with each round trip of the shuttle vehicle is (Energy per trip) = Eveblele out - (rest mass of vehicle) = (Eveblcle+garbagedown - Egarbageeieeted) = (rest mass of vehicle + rest mass of garbage = (rest mass of garbage) + (amount, - (rest mass of vehicle).JM) -.JM, by which hole's mass decreases). (rest mass of vehicle) Thus, not only can the inhabitants of the city use the black hole to convert the entire rest mass of their garbage into kinetic energy of the vehicle, and thence into electrical power, but they can also convert some of the mass of the black hole into electrical power! \u00a733.", "B. REVERSIBLE AND IRREVERSIBLE PROCESSES 909 To derive the limitation of nondecreasing surface area from properties of test- Properties of test-particle particle orbits, one must examine what values of energy-at-infinity, E, are allowed at a given location (r, () outside a black hole. Equations (33.32a,b), when combined, yield the value of E in terms of a test particle's location (r, (), rest mass p., charge e, axial component of angular momentum Lz' and momentapT = dr/d\"A.,p8 = d()/dA. in the rand () directions: orbits; aE2 - 2f3E + y = a; (33.54a) (1) E as function of p., e, Lz, r, 0, pT where a = (r 2 + a2)2 - Ja2 sin2 () > a everywhere outside horizon, f3 = (Lza + eQr)(r2 + a2) - Lza.1, y = (Lza + eQr)2 - J(Lz/sin ()2 - p.2 J p2 _ p4[(pT)2 + J(p8)2]. (33.54b) (33.54c) (33.54d) (One must take the positive square root, + Vf32 - ay, rather than the negative square root; positive square root corresponds to 4-momentum pointing toward future; while negative square root corresponds to past-pointing 4-momentum; see Figure 33.3.) Several features of the energy equation (33.54) are noteworthy. (1) For orbits in a, the energy equation yields an effective the equatorial \"plane,\" () = 7T/2 and p 8 potential for radial motion (Box 33.5). (2) Orbits of negative E must have f3 < a and y > a-which can be achieved only if Lza < aand/or eQ < a. Thus, one cannot decrease the mass of a black hole without simultaneously decreasing the magnitude ofits charge or angular momentum or both. (3) For an orbit at given (r, (), with given e and Lz' E is a minimum if pT = p8 = P. = a. Put differently, the rest mass and the r- and ()-components of momentum always contribute positively to E. By injecting an object into a black hole, produce small changes 8M=E, 8Q = e, (2) effective potential (3) negative E requires Lz8 < 0 and/or eO < 0 in its mass, charge, and angular momentum. For given changes in Q and S, what range of changes in M is possible? Clearly 8M can be made as large as one wishes by making the rest mass p. sufficiently large. But there will be a lower limit on 8M. That limit corresponds to the minimum value of E for given e and Lz \u2022 The orbit of minimum E crosses the horizon (otherwise no changes in M, Q, S would occur!), so one can evaluate E there. At the horizon, as anywhere, a minimum for E is achieved if p. = pT = p8 = a. Inserting these values and r = r+ (so.1 = a) into equations (33.54), one finds Changes in black-hole properties due to injection of particles: (33.55) 910 33. BLACK HOLES 1.25 tE~'~ 1.0 Positive-root states of positive energy 0.75 I 0.5 Negative-root states of positive energy Positive-root states of negative energy -0.75 Negative-root states of negative energy 1-1.0 -1.25 r = 3/',,1/2 Figure 33.3. Energy-at-infinity E allowed for a particle of angular momentum L z and rest mass /L, which is (I) in the \"equatorial plane\" (J = \",/2, (2) at radius r = 3M/2, (3) of an uncharged (Q = 0) extreme-Kerr (5 = M2) black hole. E is here plotted against Lz. \"Seas\" of \"positive and negative root\" states are shown. The positive root states have energies-at-infinity given by equations (33.54) E=f3+~ a and have 4-momentum vectors pointing into the future light cone. The negative root states (states of Dirac's \"negative energy sea\") have energies at infinity given by E= 13 - vi132 - a)' a, - and have 4-momentum vectors pointing into the past light cone. In the gap between the \"seas\" no orbits exist (for", "bidden region). The gap vanishes at the horizon r = M (infinite redshift of local energy gap, 2/L, gives zero gap in energy-at-infinity). [Figure adapted from Christodoulou (1971).] corresponding to changes in the black-hole properties of (1) limit on 8M for given 80 and 8S absolute ~inimum value Of). 8M for given 8S and 8Q ( (33.56) Notice an important consequence [Bardeen (l970a)]: if the black hole is initially of the \"extreme Kerr-Newman\" variety, with M2 = a2 + Q2, so that one might fear a change which makes M2 < a2 + Q2 and thereby destroys the horizon, one's fears are unfounded. Equation (33.56) then demands (since r+ = M and S = Ma) (2) preservation of the horizon so M2 remains greater than or equal to a2 + Q2, and the horizon is preserved. M 8M 2. a 8a + Q 8Q; \u00a733.8. REVERSI8LE AND IRREVERSIBLE PROCESSES 911 Box 33.5 ORBITS OF TEST PARTICLE IN \"EQUATORIAL PLANE\" OF KERR-NEWMAN BLACK HOLE Radial motion is governed by energy equation (33.54) with () = pe = 0: E = f3 + yf32 - ayo + ar4(pT)2 a a, f3, Yo are functions of r and of constants of motion, a = (r2 + a2)2 -.da2 > 0, f3 = (Lza + eQr)(r2 + a2) - Lza.d, Yo = (Lza + eQr)2.dL/ - p.2r 2.d; pT = (radial momentum) is (1) (2a) (2b) (2c) (3) Thus, equation (1) is an ordinary differential equation for dr/d\"A.. Qualitative features of the radial motion can be read off an effective-potential diagram. The effective potential V(r) is the minimum allowed value of E at radius r: V(r) = f3 + yf32 - a yo. a As in the Schwarzschild case (Figure 25.2), the allowed regions for a particle of energy-at-infinity E are the regions with V(r) S E; and the turning points (pT = dr/d\"A. = 0) occur where V(r) = E. Stable circular orbits occur at the minima of V(r). By examining V(r) closely, one finds that for uncharged black holes the innermost stable circular orbit (most tightly bound orbit) has the characteristics here tabulated [table adapted from Ruffini and Wheeler (197Ib)]. Newtonian (Figure 25.2) Schwarzschild (a = Q = 0) (Figure 25.2) Extreme Kerr (a 2 = M2, Q = 0) (see figure) [Bardeen (1970a)) if Lza > 0 if Lza < 0 0 -00 +00 0 6 20/3 0.0572 20 1 / 0 0.4226 2 / 0 9 5/(30) 0.0377 22/(30) Characteristic r/,\\4 E/IL (IL - E)/IL = \"fractional binding\" ILzl/ILM 912 33. BLACK HOLES Box 33.5 (continued) For a charged extreme Kerr-Newman black hole (M2 = Q2 + a2, Q i= 0 and a i= 0) stable circular orbits with 100 per cent binding (E = 0) are achieved in the limit e--+ -00, /-L ;-+ 0 (so a -+ M), [Christodoulou and Ruffini (1971 )]. The effective potential for an uncharged, extreme Kerr black hole (a = M) is shown in the figure [figure adapted from Ruffini and Wheeler (1971 b)]. For detailed diagrams of orbits in the equatorial plane, see de Felice (1968). For many interesting properties of orbits that are not confined to the equatorial plane, see Wilkins (1972). \u00a733.B. REVERSIBLE AND IRREVERSIBLE PROCESSES 913 The general limit (33.56) on the change in mass can be rewritten in an alternative form [Christodoulou (1970), Christodoulou and Ruffini (1971)]: where M - 1.r,. 2 + 2 = l[(M+./'12 _ Q2 _ a2\\2 + a2]1/2 ir=IY+ a ylvr 2 J (33.57) (33.58)'\" (3) irred", "ucible mass is the \"irreducible mass\" of the black hole. Equation (33.57) states that no black-hole transformation produced by the injection of small lumps of matter can ever reduce the irreducible mass ofa black hole. This result is actually a special case of the second law of black-hole dynamics, since the surface area of a black hole is (Exercise 33.12). Equation (33.58) can be combined with a = S/M and inverted to yield Q2 )2 S2 M2- M - - - lr+4M +4M2' [ir~ed~cible con- -,~el~ctr~mtgnetic con-'trr~tati~nal con- tnbutlOn to mass!,tnbutlOn to mass ] I l{tnbutlOn to mass - IT IT ( (33.59) (33.60) A black-hole transformation that holds fixed the irreducible mass is reversible; one that increases it is irreversible. The derivation of equation (33.56) revealed that the only injection processes that actually achieve the minimum possible value for 8M (and thus make 8Mir = 0) are those with}.t = pT = p8 = 0 at the horizon, r = r+. Restated in words: To produce a reversible transformation by injecting an object into a black hole, one must (1) give the object a rest mass}.t extremely small compared to its charge { or axial component of angular momentum L z'(4) reversible and irreversible transformations and (2) set the object down \"extremely gently\" (pT = p8 = 0), extremely close to the horizon (r = r+). This does not sound too difficult until one recalls that objects with pT = p8 = 0 at the horizon must be moving outward with the speed of light, and that the nearer one approaches the horizon as one sets down the object, the greater one's danger of \"slipping\" and getting swallowed! Clearly, any actual injection process will depart somewhat from irreversibili ty. Reversibility is an idealized limit, approachable but not attainable. y 914 33. BLACK HOLES EXERCISES Exercise 33.11. IRREDUCIBLE MASS IS IRREDUCIBLE Show that condition (33.56) is equivalent to BMir ~ O. Exercise 33.12. SURFACE AREA OF A BLACK HOLE Show that the surface area of the horizon of the Kerr-Newman geometry [area of surface r = r+ and I = const (Boyer-Lindquist coordinates) or j7 = const (Kerr coordinates)] is 16\\7Mir2. Exercise 33.13. ANGULAR VELOCITY OF A BLACK HOLE A general theorem [Hartle (1970) for relativistic case; Ostriker and Gunn (1969) for nonrela tivistic case] says that. if one injects angular momentum into a rotating star while holding fixed all other contributions to its total mass-energy (contributions from entropy and from baryonic rest mass), then the injection produces a change in total mass-energy given by B(mass-energy) = angular velocity of star). f'\" at POint 0 injection ( B(aneular momentum). (33.61 ) v By analogy, if one injects an angular momentum BS into a rotating black hole while holding fixed all other contributions to its total mass-energy (contributions from irreducible mass and from charge), one identifies the coefficient Q h in the equation as the angular velocity of the hole: (a) Show that the angular velocity of a black hole is equal to (33.62) (33.63) Notice that this is precisely the angular velocity of photons that live forever on the horizon [equation (33.42b); \"barber-pole twist\" of null generators of horizon]. (b) Show that any object falling into a black hole acquires an angular velocity (relative to Boyer-Lindquist coordinates) of Q = dep/dt = Q h in the late stages. as it approaches the horizon. (Recall that the horizon is a singularity of the Boyer-Lindquist coordinates. This is the reason that every object, regardless of its L., E, e, p., 2, can approach and does approach Q = Qh') Exercise 33.14. SEPARATION OF VARIABLES FOR WAVE EQUATIONS This chapter has studied extensively the motion of small objects in the external fields of black holes. Of almost equal importance, but not so well-understood yet because of its is the evolution of weak electromagnetic and gravitational perturbations complexity, (\"waves\") in the Kerr-Newman geometry. Just as one had no Ii priori reason to expect a \"fourth constant\" for test-particle motion", " in the Kerr-Newman geometry, so one had no reason to expect separability for Maxwell's equations. or for the wave equations describing gravitational perturbations-or even for the scalar wave equation O>f;=->f; aa = O. Thus it came as a great surprise when Carter (l968c) proved separability for the scalar wave equation, and later when Teukolsky (1972, 1973) separated both Maxwell's equations and the wave equations for gravitational perturbations. \u00a733.B. REVERSIBLE AND IRREVERSIBLE PROCESSES 915 Show that separation of variables for the scalar-wave equation in the (uncharged) Kerr geometry yields solutions of the form where m and 1 are integers with 0 :-:; Iml and Scharfke (1954)]; and u1m satisfies the differential equation :-:; 1; Sml is a spheroidal harmonic [see Meixner -d 2u/dr\u00b7 2 + Vu =o. (33.64b) In order to put the equation in this form, define a Regge-Wheeler (1957) \"tortoise\"-type radial coordinate r\u00b7 by (33.64a) and find an effective potential V(r\u00b7) given by dr\u00b7 = LI-\\r2 + a2) dr, V = - (w - ma r2 + a2 )2 + [(m _ wa)2 + ~](r2 + 02)-2 LI + 2(Mr - a2)(r2 + a2t 3 LI + 3a2(r 2 + a2t 4 Ll2. (33.64c) (33.64d) In this radial equation ~ is a constant (analog of Carter's constant for particle motion), given in terms of m and L by \", = \\ ~ - _ \"ml m, \"ml 2. \\ = [eigenfunction of spheroidal harmonic;] see Meixner and Scharfke (1954). (33.64e) [These details of the separated solution were derived by Brill et al. (1972). For studies of the interaction between fields and Kerr black holes-studies performed using the above solution. and using analogous solutions to the electromagnetic and gravitational wave equations-see Bardeen, Press, and Teukolsky (1972), Misner (I 972b), Teukolsky (1972), Ipser (1971), Press and Teukol~ky (1973), and Chrzanowski and Misner (1973).] CHAPTER 34 GLOBAL TECHNIQUES, HORIZONS, AND SINGULARITY THEOREMS This chapter is entirely Track 2. \u00a722.5 (geometric optics) and the Track-2 portions of Chapters 32 and 33 (collapse and black holes) are necessary preparation for it. It is not needed as preparation for any later chapter. Local techniques of analyzing spacetime physics contrasted with global techniques \u00a734.1. GLOBAL TECHNIQUES VERSUS LOCAL TECHNIQUES Until the 1960's, computations in gravitation theory used local techniques almost exclusively: the Einstein field equation describes how the stress-energy tensor Tat a given event generates curvature G at that same event (local physics). When reduced to differential equations for the metric coefficients, G = 8'7TT relates ga/3' aga/3/ax IL, and a2ga/3/ax ILax V at each given event to TYB at that same event (local equation). The solution of these differential equations is effected, on a computer or in any initial-value-type analysis, by integrating forward in time from event to event to event (local integration). The nongravitational laws of physics are obtained by invoking the equivalence principle in a local Lorentz frame at each individual event in spacetime. To build up an understanding of the global structure of spacetime, one performs local computations near each event, and then patches the local results together to form a global picture. Why this great reliance on local analyses? Because the laws of gravitation physics take on particularly simple forms when stated locally. That gravitation physics is also subject to powerful and simple global laws, physi cists did not realize until the mid 1960's. But since 1963, studies of black holes and of singularities have revealed global laws and global properties of spacetime that rival in their simplicity and elegance even the (local) equivalence principle. An example is the second law of black-hole dynamics: \"In an isolated system, the sum of the surface areas of all black holes can never decrease.\" As a result, there has developed a powerful body of knowledge and techniques for analyzing directly the global properties of spacetime. To give a full treatment of global techniques would require many chapters. Fortunately, a full treatment is being published, almost simultaneously with this book, by Hawking and Ellis (1973", "). Because Hawking and Ellis are much better qualified to write on this subject than are we (Misner, Thorne, and Wheeler), we have chosen to not write a \"competitive\" treatment. Instead, we give in this chapter only a brief taste of the subject-enough of a taste to make the reader acquainted with the types of techniques involved and several of the most important results, but References on global techniques Motivation for studying properties of spacetime near infinity Specific regions of infinity: /+, /0, /-. 1+, 1- \u00a734.2. \"INFINITY\" IN ASYMPTOTICALLY FLAT SPACETIMES 917 not enough to give him a working knowledge of the subject. The topics we have chosen to treat are those that contact most closely the rest of this book: properties of \"infinity\" in an asymptotically flat spacetime (\u00a734.2); causality and horizons (\u00a7\u00a734.3 and 34.4); a proof of the second law of black-hole dynamics (\u00a734.5); and theorems about the evolution of singularities in spacetime (\u00a734.6). For greater detail on global techniques, one can consult not only the book of Hawking and Ellis (1973), but also review articles by Geroch (1971), by Penrose (1968a, 1972), and by Hawking (1973), the thesis of Godfrey (1970b), and the more specialized papers cited in the body of this chapter. \u00a734.2.~NITY\" IN ASYMPTOTICALLY FLAT SPACETIMES When performing calculations in asymptotically flat spacetime, one often must examine the asymptotic forms of fields (e.g., the metric, or the curvature tensor, or the electromagnetic field) \"at infinity.\" For example, the mass and angular momen tum of an isolated system are determined by the asymptotic form of the metric (Chapter 19). It is rarely sufficient to examine asymptotic forms near \"spatial infin ity.\" For example, if one wishes to learn how much mass was carried away by gravitational and electromagnetic waves during a supernova explosion, one must examine the asymptotic form of the metric not just at \"spatial infinity,\" but at \"future null infinity\" (see Figure 34.1). Penrose (1964, 1965a) has developed a powerful body of mathematical technique for studying asymptotic properties of spacetime near \"infinity.\" The key to his technique is a \"conformal transformation\" of spacetime, which brings \"infinity\" in to a finite radius and thereby converts asymptotic calculations into calculations at \"finite points.\" Penrose's technique also provides rigorous definitions of several types of \"infinity\" that one encounters in asymptotically flat spacetimes. The details of Penrose's technique are not of importance to the rest of this chapter. However, this chapter will refer frequently to the various types of \"infinity\" defined by Penrose. In heuristic terms, they are as follows (see Figure 34.2a). 1+ \"future timelike infinity\": the region t ---+ + 00 at finite radius r (region toward which timelike lines extend). 1- \"past timelike infinity\": the region t ---+ - 00 at finite radius r (region from which timelike lines come). 1\u00b0 \"spacelike infinity\": the region r ---+ 00 at finite time t (region toward which spacelike slices extend). 1+ \"future null infinity\": the region t + r ---+ 00 at finite t (region toward which outgoing null lines extend). r 1- _ \"past null infinity\": the region t (region from which ingoing null lines come). r ---+ - 00 at finite t + r Note: 1 is a script I, and is sometimes given the name \"seri.\" 2 Afbrlorr goo = -I + - - r begin~ to collap~e x Figure 34.1. Measurement of the mass-energy radiated as gravitational and electromagnetic waves by a supernova explosion in asymptotically flat spacetime, The mass-energy radiated equals the mass (Mbefore) of the presupernova star, minus the mass (Malter) of the neutron star and nebula after the explosion: Mradiated = MOOfore - Malter' To measure MOOfore' one can examine the asymptotic form (in suitable coordinates) of goo at spatial infinity goo = - I + 2Mbefore + 0 (~) r2 r as r -+ 00, I = constant. But to measure Malter in the same way, one must wait, at any fixed r, until the radiation has flowed entirely past that point: alter + 0 (~) -1 + 2M r2 r goo = as r -", "+ 00 WI 'th t - r = (constant value sufficiently large) to be inside the burst of waves, Put differently, to measure Malter one must examine the asymptotic form of goo not at \"spatial infinity,\" but rather at \"future null infinity,\" Coordinate diagrams for eXhibiting structure of infinity It is often useful, in visualizing the asymptotic structure of spacetime, to introduce coordinates that attribute finite coordinate values to infinity. For example, in flat spacetime one can transform from the usual spherical coordinates t, r, 0, <P, with to new spherical coordinates 1/1, t 0, <p, with (34.1 ) \u00a734.2. \"INFINITY\" IN ASYMPTOTICALLY FLAT SPACETIMES 919 ~ J+ t t t tt 'l 1+ 'l/ r ~ ~1- +++++ j- '(i;' 'j:( <U '\":s = 2 os ~ s g \", 0 -- jO '\" I J+ \u2022 +11' 0 -11' j- I I (a) (b) Figure 34.2. Flat, \"Minkowski\" spacetime as depicted (a) in the usual spherical coordinates t, r, (J, <p of a global Lorentz frame, and (b) in the spherical coordinates of eq uations (34.2). The five regions of infinity-J+, j-, jO, 1+, 1--are shown in each coordinate diagram. In both coordinate systems, radial null lines make angles of 45 \u00b0 with the vertical axis, and nonradial null lines make angles less than 45\u00b0 [see equations (34.1) and (34.2c) I. See exercise 34.1 for further detail. t + r = tan 2(~ + n I t - r = tan l(~ - ~), 2 (34.2a) (34.2b) (34.2c) The resulting~, ~ coordinate diagram (Figure 34.2b) depicts /+, /-, /0, 1+, 1- more clearly than does the usual t, r, coordinate diagram. As another example, replace the Kruskal-Szekeres coordinates v, U, 0, <p for Schwarzschild spacetime by new coordinates ~, ~,O, <p: v + U = tan2(~ + ~), I v- U = tan ~ (~ - n (34.3 a) (34.3 b) (l - rj2M)e T / 2Jl = v 2 - u 2 = tan ~ (~ + ~) tan ~ (~ - ~), (34.3 c) 32M3 2 s= - d r 2M / ( _d~2 + d~2) e-T I 4 cos22(~ + ~) cos22(~ - ~) I 2(,J1J2' 20 d 2) +ruu+sm <p. (34.3d) The resulting coordinate diagram (Figure 34.3) depicts clearly the causal connections between the horizons, the singularities, and the various regions of infinity. 920 34. GLOBAL TECHNIQUES. HORIZONS. AND SINGULARITY THEOREMS J+ r = 0 singularity /+ /- r = 0 singularity /- Figure 34.3. Schwarzschild spacetime as depicted in the y. ~, n. <> coordinates of equations (34.3). This coordinate diagram should be compared with the Kruskal-Szekeres coordi nate diagram (Figure 31.3). In both diagrams, radial null geodesics are 45' lines. Each of the asymptolically flat regions (one on each side of the \"wormhole\" of Figure 31.5a) has its own set of infinities /+. /-, /0..'/+, and.'/-. See exercise 34.2 for justification of this diagram. EXERCISES Exercise 34.1. FLAT SPACETIME IN y, ~, e, ep COORDINATES (a) Derive equation (34.2c) from (34.1) and (34.2a,b). (b) Show that the regions [+, [-, [0, r, and 1- of flat spacetime are located at [+: y = 'IT, ~ = 0, [-; y = -'IT, ~ = 0, [0; y = 0, ~ = 'IT,!J+: y + ~ =:: 1-; y - 'ITIT, (34.4) ~ =:: -'IT, -'IT < y + ~ < 'IT. [see equations (34.2)]. These are the regions depicted in Figure 34.2. (c)", " Show that in flat spacetime, in a y, ~ coordinate diagram (Figure 34.2), radial null lines make angles of 45 \u00b0 with the vertical axis, and nonradial null lines make angles of less than 45\u00b0. Exercise 34.2. SCHWARZSCHILD SPACETIME IN y, ~, e, ep COORDINATES (a) Derive equations (34.3c,d) from (34.3a,b) and the Kruskal-Szekeres equations (3I.l4). (b) Use equations (34.3) to justify the precise form of the coordinate diagram in Fig ure 34.3. Exercise 34.3. REISSNER\u00b7NORDSTROM SPACETIME (a) Show that there exists a coordinate system in which the Reissner-Nordstrom geometry with 0 < IQI < M (exercises 31.8 and 32.1) has the form ds 2 = F2( _dy2 + d~2) + r 2(de 2 + sin2e dep2), r=r(y,n (34.5) and in which the horizons and infinities are as shown in Figure 34.4. [Note: This is a very difficult exercise unless one has in hand the solution to exercise 3 I.8(d). For solution, see Carter (l966b).] (b) Use Figure 34.4 to deduce that the Reissner-Nordstrom geometry describes a \"worm hole\" or bridge, connecting two asymptotically flat spacetimes, which: (i) expands to a state of maximum circumference; (ii) recontracts toward a state of minimum circumference, and in the process disconnects its outer regions from the two [O's (spatial infinity) and reconnects them to a pair of r = 0 singularities; (iii) bounces; (iv) reexpands, and in the process \u00a734.2. \"INFINITY\" IN ASYMPTOTICALLY FLAT SPACETIMES 921 Figure 34.4. Reissner-Nordstrem spacetime dsZ = - 1 - ( 2M, + - Q2) r 2 dt 2 + dr2 1 - 2M/r + Q2/r2 + r2(d(J2 + sin2(J d</>2) with 0 < IQI < M, as depicted in a new (>/!, ~, </\u00bb coordinate system where the line element (J, has the form ds2 = F2( - d1/;2 + d~) + r2(d(J2 + sin2(J d</>2). (see exercise 34.3.) This coordinate diagram re '\u00a2'eals the global structure of the geometry, in cluding its singularities at r = 0, its horizons at r = r+ = M + VM2 _ Q2 (which limit communication with 1+ and r), the null surfaces at r =,_ = M - VM2 - Q2 (which limit communication with the singulari ties), and the various asymptotically flat infinities, r, j-, [0,.1+, and r, From this diagram one can read off the \"causal structure\" of the geome try-i.e., the abilities of various regions to com\u00b7 municate with each other, For detailed discussion of the geometry, see Graves and Brill (1960) and Carter (1966b). For discussions of collaps ing charged stars, for which this geometry is field, see Novikov the external gravitational (1966a,b), de la Cruz and Israel (1967), and Bar deen (1968). in twO new asymptotically flat universes; disconnects its \"outer regions\" from the two singularities and reconnects them to a pair of its expansion to a halt; 10's (vi) recontracts toward a state of minimum circumference, and in the process disconnects its outer regions from the two 10's and reconnects them to a new pair of r = 0 singularities; etc. ad infinitum. slows (v) 922 34. GLOBAL TECHNIQUES, HORIZONS, AND SINGULARITY THEOREMS *- -jt [f) Note: ~I)<M, ~f{~. ~1)~'Jl,~){~, but ~ E j+(~f) Note: S is a spacelike slice which extends from [0 in to r = 2M, but does not include r = 2M. J+(S) does not include the leftmost horizon; but )+(8) is the leftmost horizon. (a) (b) Figure 34.5. Spacetime diagrams illustrating various causal relationships.", " Diagram (a) is a hypo thetical spacetime; diagram (b) is Schwarzschild spacetime (see Figure 34.3). In both diagrams, null lines have slopes of 45\u00b0. \u00a734.3. CAUSALITY AND HORIZONS Turn now to global* techniques for analyzing black holes. The goals ofthe discussion will be (1) to define the concept of horizon (this section), (2) to deduce global geometric properties of horizons (next section), and (3) to prove the second law of black-hole dynamics (following section). The entire discussion will be confined to spacetime manifolds that (I) contain at least one asymptotically flat region (\"the external universe\"; region \"outside black holes\"), and (2) are \"time-oriented.\" By \"time-oriented\" one means that at each event in spacetime a distinct choice has been made as to which light cone is the future cone and which is the past, and moreover that this choice is continuous from event to event throughout spacetime. The discussion begins with definitions of a variety of causal relationships between events and regions of spacetime (see Figure 34.5). Definition: q> ~!:2 or equivalently!:2 ~ q> (\"the event q> precedes the event 2\"; \"the event!:2 follows the event tJ''') means that there exists at least one smooth, future-directed timelike curve that extends from tJ' to 2. Definition: A causal curve 8(\\) is any smooth curve that is nowhere spacelike-i.e., that is timelike or null or \"zero\" [8(\\) = some fixed tJ', for all \\] or some admixture thereof. Definition: tJ' -<!:2 or equivalently 2 >- tJ' (\"the event tJ' causally precedes the event \u00b7Global, but not fully global; the \"universe\" of \u00a7\u00a734.3-34.5 is asymptotically flat; no account is taken here of possible closure or collapse of the universe or of their consequences. Restriction of discussion to asymptotically flat, time-oriented manifolds Definitions of several causality concepts \u00a734.3. CAUSALITY AND HORIZONS 923 ~ 52\"; the event 52 causally follows the event ~\") means that there exists at least one future-directed causal curve that extends from ~ to 52. Definition: J-W), called the causal past of~, is the set of all events that causally precede ~-i.e., J-W) = {521 52 -< ~}. Definition: rw), called the causal future of~, is the set of all events that causally follow ~-i.e., rw) = {52152 >- ~}. Definition: If S is a region of spacetime-e.g., a segment of a spacelike hyper surface-then J-(S) (the causal past of S) is the set of all events that causally precede at least one event in S-i.e., J-(S) = {52152 -< ~ for at least one ~ E S}. Definition: Similarly, r(S) (the causal future of S) is the set of all events that causally follow at least one event in S-i.e., r(s) = {521 52 >- ~ for at least one ~ E S}. Definition: j+(S) is the boundary of r(s), j-(S) is the boundary of J-(S). Definition: One defines the future of~, /+W); the past of~, /-W); the future of S, /+(S); the past of S, /-(S); the boundary of the future of S, j+(S); and the boundary of the past of S, j-( S) in precisely the same manner as above, except that the phrase \"causally precede\" is replaced by \"precede,\" and \"causally follow\" is replaced by \"follow.\" Example: /+(S) = {521 52 ~ ~ for at least one ~ E S}. Not all these definitions are needed in the following discussion; but the literature on global methods uses these concepts so extensively that the reader should be familiar with them. Focus attention on a specific spacetime manifold, and in that manifold select out a specific asymptotically flat region. [In the external field of a star, there is but one asymptotically flat region. In the vacuum Schwarzschild geometry without source (Figure 34.3), there are two. In the Reissner-Nordstr0m geometry without source (Figure 34.4), there are infinitely many different asymptotically flat regions.] The selected asymptotically flat region (\"external universe\") has", " one future timelike infinity /+, one past timelike infinity /-, one spacelike infinity /0, one future null infinity 1+, and one past null infinity 1-. It may also possess black holes, which form by stellar collapse, and which collide, coalesce, accrete matter, and generally wreak havoc in their immediate vicinities. The surfaces of all black holes (\"future horizons\") separate the external universe, which can send signals out to 1+, from the black-hole interiors, which cannot. One thus has the definition: Definition: The totality (or \"union\") of all future horizons (surfaces of all black holes) is the region j-(1+)-i.e., it is the boundary of the domain J-UI+) that can send future-directed causal curves out to future null infinity. Definition: surfaces of black holes; future horizons-j-(.'/+) 924 34. GLOBAL TECHNIQUES, HORIZONS. AND SINGULARITY THEOREMS }---, x Surface of a \"\"---'r>--black hole at \"time\" 52 _----....7 Figure 34.6. Black holes in an asymptotically flat spacetime (schematic spacetime diagram). J-(!/+) is the \"external universe\"-Le., the region which can send causal curves to future null infinity. j-(1+), the greyish region, is the boundary of the \"external universe\"-i.e., it is the union of all future horizons. At the \"time\" of spacelike there are no black holes in the universe. Between 5, and 52 tw~ stars slice 5\\, collapse to form black holes. The two closed 2-surfaces, in which 52 intersects J-(,1+) are the horizons of those black holes at \"time\" 52' Between 52 and 53' the two original black holes collide and coalesce, while a third black hole is being formed by stellar collapse. [Similarly, one can define the totality of all past horizons to be j+(1-). But past horizons are of little interest for astrophysics. Whereas gravitational collapse pro duces future horizons in a quite natural manner, past horizons must be primordial in origin-i.e., they must be postulated as initial structure in the origin ofthe universe [Novikov (1964), Ne'eman (1965)]. There is no good reason to believe that the universe began with or should have begun with such strange initial structure.] Any given spacelike slice S through spacetime will intersect j-(1+) in a number of disjoint, closed, two-dimensional surfaces. Each such 2-surface is the horizon of a single black hole at the \"moment of time\" S. See Figure 34.6. \u00a734.4. GLOBAL STRUCTURE OF HORIZONS The union of all future horizons, j-(1+), has an especially simple global geometric structure, as follows. Penrose's theorem on the structure of J-(!/+) (future horizons) THEOREM [Penrose (1968a)]: j-(1+) is generated by null geodesics that have no future end points. Stated more precisely (see Figure 34.7): (1) Definition: The \"generators\" of j-(1+) are null geodesics which (at least for some finite lapse of affine parameter) lie in j-(1+). \u00a734.4. GLOBAL STRUCTURE OF HORIZONS 925 Once it enters j-(!/+), a gen~rator never leaves.1(.'/+) nor passes through a caustic nor crosses any other generator )--y x Through each non-caustic event tj' passes one and only one generator :::\"--C:am;tic. Here generators enter j-(.'/+) from J-(.'/+) - Figure 34.7. The future horizon j-(.'f+) produced by the spherical gravitational collapse of a star. This horizon -illustrates the global geometric structure of j-(.'/+) as spelled out in Penrose's theorem (\u00a734.4 of text). In this special case, there is only one caustic. In general there will be many. (2) Theorem: When followed into the past, a generator may (but does not have to!) leave j-(.1+). Each event at which a generator leaves is called a \"caustic\" of j-(.1+). When a generator leaves, it goes into J-(!f+). (3) Once a generator, being followed into the future, enters j-(1+) from J-(1+) at a caustic, it can never thereafter leave j-(1+), nor can it ever intersect another generator. [Generators can intersect only at the \"caustics,\" where they enter j-(1", "+).] (4) Through each noncaustic event of j-(1+) there passes one and (aside from normalization of affine parameter) only one genera- - - - -___ --.tol:..-. ~. This theorem is proved in Box 34.1. For a Schwarzschild black hole, the generators of j-(!f+) are the world lines of radially outgoing photons at the gravitational radius [r = 2M, () and ep co~stant, y 926 34. GLOBAL TECHNIQUES, HORIZONS, AND SINGULARITY THEOREMS u = +v; dotted line on horizon in Figure 32.l(c)]. For a Kerr-Newman black hole, the generators of j-(.(j+) are the \"barber-pole-twist\" null geodesics of Box 33.2(F)-i.e., they are those members of the outgoing principal null congruence that lie on the horizon, r = T+ (\u00a733.6; exercise 33.9). But the theorem is more general. It refers to any black hole-dynamic or static; accreting matter, or coalescing with a neighboring black hole, or existing alone in isolation-in any time-oriented, asymptotically flat spacetime. (continued on page 931) Box 34.1 HORIZONS ARE GENERATED BY NONTERMINATING NULL GEODESICS (Penrose 1968a) A. Lemma: If(l) t\\(A) is a causal, future-directed curve from event ~.p to event f2, (2) ('2(A) is a causal, fu ture-directed curve from event :2 to event (>il, and (3) q> <\u00a2 ~# (?p is not in the past of (jl), then eland e2 are null geodesics, and their tangent vectors coin cide (aside from normalization) at event 2. Proof*: 1. Suppose that e1 were not a null geodesic, or ('2 were not a null geodesic, or both. Then some where along e1 u e2 there would be a timelike segment, or a nongeodesic null segment, or both. a. Ife 1 U e 2 contained a timelike segment, then a slight deformationt of e1 U e2 would pro duce a smooth curve e3 from,:P to (Off which is everywhere timeliket-comradicting the as sumption ':P <\u00a2 ~#. *The proof utilizes some elemen tary concepts of point-set topology: see, e.g., Wallace (1963) or Kelley (1955). tOne can always deform any curve in any spacetime manifold by a small amount in any direction one wishes, without running into singularities or into other boundaries of the manifold. This is possible because a manifold by definition is open. In physical terms, spacetime is open because each event in spacetime must possess a local Lorentz neighborhood which also lies in spacetime. tOne can convince oneself of this. and of similar claims made later in the proof, by arguments using local Lorentz frames. In the literature on global geometry, claims such as this are rarely substantiated though each author is always convinced that he could do so if forced to by a skeptic. Unfortunately, to substantiate such claims with rigor ous arguments would lengthen and complicate the discussion enor mously and would tend to obscure the simplicity of the underlying ideas. \u00a734.4. GLOBAL STRUCTURE OF HORIZONS 927 b. If 8 1 U 8 2 contained a nongeodesic null seg ment 8 reaching from event (/ to event 11, then, when compared to neighboring curves between (/ and 11, 8 would not have stationary length. This means that some curves from (/ to 11 would have larger squared length-i.e., would be spacelike-while others would have smaller squared length-i.e., would be time like. Thus, a slight deformation of 8 would produce a timelike segment from (/ to 11. Then a further deformation of81 U 8 2, as described in (a) above, would produce a smooth timelike curve from tJ' to qil, contradicting tJ' ~ qil. Thus, the supposition is wrong; i.e., both 8 1 and 8 2 must be null geodesics. 2. Suppose that the tangent vectors of 8 1 and 8 2 f!2. Then one did not agree at their join point, could \"round off the corner\" at f!2, producing a timelike segment there. One could then further deform 8 1 U 8 2 -as in (l a) above, to produce a smooth timelike", " curve from tJ' to qil-contradiet ing tJ' ~ qil. Thus, the supposition is wrong; i.e., the tangent vectors must agree at f!2. Q.E.D. I Timelike / I I e\\ 928 34. GLOBAL TECHNIQUES, HORIZONS. AND SINGULARITY THEOREMS Box 34.1 (continued) B. Lemma: If dE j-U/+) and ~1J E j-U/+), then d{ ~ij. Proof Assume d ~ ~yJ. 1. Then there exists a timelike curve from d to ~1J. 2. A slight deformation of that curve which keeps it still timelike will make it link an arbitrary event!2 in some sufficiently small neighborhood ~[d] to an arbitrary event ~il in some small ~[~1J]. 3. Pick!W to lie in J-(1+). Then join the timelike curve from!2 to f)l onto a causal curve from ~ to 1+. The resulting curve, when smoothed in a neighborhood of the join, becomes a causal curve from any arbitrary!2 E ~[dJ to 1+. 4. The existence of such curves implies that ~[d] c J-(1+), and hence that d ~ j-(1+)-in contra diction to the original hypotheses. Conclusion: d { YJ. Q.E.D. C. Lemma: Let t'(A) be a causal curve that intersects j-(1+) at some event!'E. Then when followed into the past from qs, 8(A) forever lies inj-(1+) U J-(1+). Proof 1. Pick an arbitrary event d on t'(A), with d -< 0'3. 2. Construct an arbitrarily small neighborhood :9[[(f]. 3. A small deformation of 8, between d and!'E, produces a timelike curve oj) from some event tJ' E ~[d] to ~. 4. Since qs E j-(1+), a slight deformation of (7), keeping it still timelike, produces a curve 0 from tJ' to some event!2 E J-(1+). 0 can then be pro longed, remaining causal, until it reaches 1+. The to 1+. Hence, result tJ' E J-(1+). is a causal curve from tJ' 5. But tJ' was in an arbitrarily small neighborhood :9[[(f]. Hence, (f must also be in J-(1+) or else in its boundary, j-(1+). Q.E.D. to </+f \" / / I / ~[(/l \u00a734.4. GLOBAL STRUCTURE OF HORIZONS 929 D. Theorem [Penrose (l968a)]: j-(!f+) is generated by null geodesics which have no future endpoints. [See text of \u00a734.4 for more detailed statement of theo rem.] Proof 1. Pick an arbitrary event ~ in j-(1+). Prove as follows that through ~ there passes a future-di rected null geodesic which lies in j-(1+): a. Construct an arbitrary neighborhood ~W]. [If j-(1+) happens somewhere to encounter a singularity of spacetime, then ~W] must be chosen small enough to keep the singularity outside it.] ing from ~i to 1+. b. In ~W] n J-(1+), construct a sequence of events {~i} which converges to the event ~. c. For each i, construct a causal curve ei extendd. Let f2 i be the intersection of ei with ~W], the boundary of~W]' Since 0l[~] is a compact set, the sequence f2 i must have a limit point, f2. e. Because there exist causal curves from events ~i arbitrarily near ~ to events f2 i arbitrarily near f2, there must be a causal curve from ~ to f2. Call that curve e.. /....--- / / / / / I I \\ \\ \\ \"-,................... I I f. Since f2 isa limit point of a sequence of events in J-(1+), f2 either lies in r(1+), or else lies ~- -~on-~ iffbourrdary j-(1+), or both. Suppose Toward 1+ f2 ~ J-(1+). 1. Then some small neighborhood ~[f2] is contained entirely in J-(1+). n. Construct a causal curve from ~ to 1+ by going from ~ to f2 along the causal curve e, then from f2 along a timelike curve to some event qil E ~[f", "2], and then from qil to 1+ along a causal curve-and by smooth ing at the join points f2 and qil. Ill. Since this curve from ~ to 1+ has a time like segment, it can be deformed smoothly, while being kept causal, so that it reaches any desired event S in some small neigh borhood 9['[9]. But that 9['[9] C J-(1+), hence that 9 ~ j-(1+) which contradicts the original definition of 9. this means Conclusion: 2, E j-UI+). Cannot occur! 930 34. GLOBAL TECHNIQUES, HORIZONS, AND SINGULARITY THEOREMS Box 34.1 (continued) since ':'F E J-(!f+) g. By Lemma. B, and 12 E j-(!f+), then Cj' { 2. But e is a future directed causal curve from tJ' to 2. Conse quently, by Lemma A, e is ~ null geodesic. h. Since the curve e intersects J-(!I+) at 2, be tween ':'F and!2 it must everywhere lie in j-(!I+) U J-(!f+) [Lemma C]. Apply the rea soning of (f) above, with 12 replaced by an arbitrary point on C' between?P and 2. Thereby conclude. that, everywhere between ':'F and!2, e lies in r(!f+). Summary: Through every event tJ' E j-(!f+) there passes a null geodesic e which, when followed into the future from tJ', lies in j-(!f+). This null geodesic is called a \"generator\" of J-(!f+). 2. Follow the generator e from tJ' to ~ and then onward still further. Can it ever leave J-(!f+)? No! tJ\" E ~or suppose it did leave, at some event J-(!f+). a. Repeat the entire construction of step I, with tc? conclude that there is a null the C\" C J-(!f+) replacing tJ', tJ\" geodesic causal future from tJ\" extending into to some event 2'. e cannot do this b. By Lemma B, since '3' E j-(!f+) and 2' E j-(!f+), tJ' ~ 2'. c. Then by Lemma A the null geodesic e from tJ' to tJ\" and the null geodesic e' from tJ\" to (aside 2' have tangents that coincide at from normalization). Thus, with a renormal ization of affine parameter, e' becomes the prolongati(:m of e-which means that e does not leave J-(!f+) at tJ\". tJ\" Conclusion: Once a generator, being followed into the future, enters j-(!f+), it can never thereafter leave j-(!f+). 3. Figure 34.7 provides an example of how a null geode~ic, being following into the future, can enter J-(!f+) and become a generator. Lemma C that, when a null geodesic enters guarantees j-(!f+), it enters from J-(1+). \u00a734.5. PROOF OF SECOND LAW OF BLACK-HOLE DYNAMICS 931 4. As indicated by the example of Fig. 34.?, at a \"caustic\" [entry point of generators onto J-(1+)] generators can cross each other. Follow a genera-- ~ - - - -tor t?-to the causaLfu1ur~ft:9m.jts entry point onto j-(1+). Can it ever again cross another generator? No. For suppose that at an event ~ the generator e were to croSS another generator 6iJ. a. To the causa~ future of ~, both generators always lie in J-(1+). Thus, events (/ and ~ of the picture are in J-(1+). b. Since ~ is to the causal future of the caustic where e enters j-(1+), there exists an event f!2 E j-(1+) n e to the causal past of~. c. Since f!2 E j-(1+) and ~ E j-(1+), f!2 1:: ~ [Lemma B]. d. Lemma A, applied to the curves e from f!2 to ~, and 6j) from ~ to ~, then guarantees that the tangent vectors u e and u6j) coincide at ~ (aside from nomialization), and that therefore (aside from normalization) the geodesics e and (ill are identical. This contradicts the supposition that e", " and (ill are different generators which crOSS at ~. Conclusion: Once a generator has entered J-(1+), it can never thereafter cross any other generator. \u00a734.5. PROOF OF SECOND LAW OF BLACK-HOLE DYNAMICS [Hawking (1971 b, 1972a, 1973)] All the tools are now in hand for a proof of the second law of black-hole dynamics. Consider the union of all future horizons, j-(1+), in an asymptotically flat space time, as depicted in Figure 34.8. Divide up the null-geodesic generators of j-(1+) into a large number of infinitesimal bundles, and give each bundle an identifying number, K. As one moves from past toward future along j-(1+), one occasionally sees new bundles of generators created in \"caustics\" of the 3-surface j-(1+). The caustic sources of new generators are created by such processes as the infall of matter through the horizon (example: bundle #42 in Figure 34.8), and the collision and coalescence of two black holes (example: bundle #29). But each bundle, once created, can never be destroyed (no termination of null generators as one moves from past toward future). Proof of second law of black-hole dynamics: 932 34. GLOBAL TECHNIQUES. HORIZONS. AND SINGULARITY THEOREMS )-y x Bundle ;=42 Bundle #17 Figure 34.8. Schematic spacetime diagram used in proving the second law of black-hole dynamics. See text for details of the proof, and see Figure 34.6 for physical interpretation of the diagram. Focus attention on a specific bundle of generators-bundle # K. At a specific event ':'F along that bundle, let various observers, moving with various velocities,'measure its (two-dimensional) cross-sectional area ClK('3'). As shown in Figure 22.1, exercise 22.13, and exercise 22.14: (l) the cross-sectional area ClK('3') is independent of the velocity of the observer who measures it-i.e., ClK('3') depends only on location tJ' along the bundle; and (2) ClK changes from event to event along the bundle in a manner governed by the \"focusing theorem\" if the energy density Too, as measured by all observers along the bundle, is nonnegative. (34.6) Proof assumes nonnegative energy density Here AK is affine parameter along the bundle. Assume-in accord with all physical experience and the best assessments of modern physics-that energy density Too can never be negative. (This assumption underlies the second law of black-hole dynamics. If it were ever found to be invalid, then one would have to abandon the second law.) Suppose that dCli/2/dAK were negative at some event tJ' along the bundle. Then, according to the focusing theorem, dCli/ 2/ dAK would always remain at least as \u00a734.5. PROOF OF SECOND LAW OF BLACK-HOLE DYNAMICS 933 negative as its value at &'-and, hence, after a lapse of affine parameter given by AA < K - ( Cl 1/2 K _ dCl 1/2/dA K ) K aU''(34.7) Cl//2 would go to zero. At the point where Cl//2 reaches zero, adjacent null geodesics in the bundle cross each other, giving rise to events in j-(1+) through which pass more than one null geodesic generator. But this violates Penrose's theorem on the global structure of horizons (\u00a734.4). Thus either the supposition of negative dCl//2/dAK is wrong; or else dClK l/2/dAK goes negative, but then, before the generators get a chance to cross [before the finite lapse (34.7) of affine parameter], the generators hit a singularity and cease to exist. To prove the second law of black-hole dynamics, one must assume that no singularity is hit by the horizon, and thereby conclude that dCl//2/ dAK never goes negative. Hawking (l971b, 1972a) makes an alternative assumption which implies dCl//2/dAK ;::: 0: Hawking assumes that spacetime is \"future asymptoticallypredictable. \" In essence this means that spacetime possesses no \"naked singularities\"-i.e., no singularities visible from 1+. (Naked singularities could influence the evolution of the external universe; and, therefore, unless one knew the laws of physics governing singularities which one does not-they would prevent one from predicting the future in the external universe.) Under either assumption (no naked singularities; or horizon never hits", " a singular ity), one concludes that dCl//2/dAK is nonnegative everywhere along bundle K. (34.8) This result says that the cross-sectional area ClK of each bundle can never decrease as one moves toward the future along j-(1+). Since new bundles can be created, but old ones can never be destroyed as one moves toward the future, the total cross-sectional area of j-(1+) cannot decrease toward the future. Equivalently, (see Figure 34.8), if Sl and S2 are spacelike hypersurfaces with S2 everywhere to the future of Sl' then the cross-sectional area of j-(1+) at its intersection with S2' Cl(S2)' cannot be less than the cross-sectional area at Sl' Cl(SI)' This is the second law of black-hole dynamics, reformulated in more precise language than that of Chapter 33, and finally proved. Proof assumes that horizon never hits a singularity (no naked singularities) Precise formulation of second law Exercise 34.4. A BLACK HOLE CAN NEVER BIFURCATE [Hawking (1972a)] Make plausible the theorem that no matter how hard one \"zaps\" a black hole, and no matter what one \"zaps\" it with, one can never make it bifurcate into two black holes. [Hint: By drawing pictures, make it plausible that, at any bifurcation point. some null geodesic genera tors of j-(.tf+) must leave j-(.</+) as one follows them into the future-in violation of Penrose's theorem (\u00a734.4). Assume that the surface of each hole is topologically a 2-sphere. Not~: The same argument, time-reversed, shows that if two black holes coalesce, generators enter J-(l/+) from J-(.'/+) at the coalescence point: and the surface area of the horizon increases.] EXERCISE Overview of theorems on singularities Singularity defined Trapped surface defined 934 34. GLOBAL TECHNIQUES, HORIZONS, AND SINGULARITY THEOREMS \u00a734.6. THEOREMS ON SINGULARITIES, AND THE \"ISSUE OF THE FINAL STATE\" Just as global techniques are powerful tools in the analysis of horizons. so they also are-powerful in the analysis of spacetime singularities. In fact, it was the proof of Penrose's (1965b) pioneering theorem on singularities that gave birth to global techniques for studying spacetime. For a detailed introduction to the global analysis of singularities, one can read the book of Hawking and Ellis (1973). Now that the reader has had a taste of global techniques, attention here will focus on a qualitative description of results: How does gravitational collapse terminate? Is the singularity at the end point of spherical collapse typical, or can asymmetries remove it? That singularities are very general phenomena, and cannot be wished away, has been known since 1965, thanks to theorems on singularities proved by Penrose, Hawking, and Geroch. [For a full list of references, see Hawking and Penrose (1969) or Hawking and Ellis (1973).] Before examining the theorems on singularities, one must make precise the concept of a singularity. This is not easy, as Geroch (1968) has emphasized in a long treatise on the wide variety of pathologies that can occur in spacetime manifolds. However, after vigorous efforts by many people, Schmidt (1970) finally produced a definition that appears to be satisfactory. Put in heuristic terms, Schmidt's highly technical definition goes something like this. In a spacetime manifold, consider all spacelike geodesics (paths of \"tachyons\"), all null geodesics (paths of photons), all timelike geodesics (paths of freely falling observers), and all timelike curves with bounded acceleration (paths along which observers are able, in principle, to move). Suppose that one of these curves terminates after the lapse of finite proper length (or finite affine parameter in the null-geodesic case). Suppose, further, that it is impossible to extend the spacetime manifold beyond that termination point-e.g., because of infinite curvature there. Then that termination point, together with all adjacent termination points, is called a \"singularity.\" (What could be more singular than the cessation of existence for the poor tachyon, photon, or observer who moves along the terminated curve?) Another concept needed in the singularity theorems is that of a trapped surface. This concept, devised by Penrose (1965b), is motivated by a close examination of the two-dimensional, spherical surfaces (r, t) = const. inside the horizon of the Schwarzschild geometry. These surfaces signal the nearness of a singularity (", "r = 0) by this property: light rays emitted from one of these surfaces in the perpendicular outward direction (i.e., outgoing, orthogonal, null geodesics) converge toward each other as they propagate; and inward light rays perpendicular to the 2-surface also converge. Penrose gives the name \"trapped surface\" to any closed 2-surface, spherical or not, that has this property. In Schwarzschild spacetime, the convergence of light rays, both outgoing and ingoing, can be attributed to the \"intense pull of gravity,\" which sucks the photons into the singularity. That this might also be true in asym metric spacetimes is suggested by the Hawking-Penrose (1969) theorem [the most powerful of a wide class; see Hawking and Penrose (1969) for references to others; and see Boxes 34.2 and 34.3 for introductions to Hawking and Penrose]: \u00a734.6. THEOREMS ON SINGULARITIES 935 A spacetime M necessarily contains incomplete, inextendable time/ike or null geo desics (and is thus singular in the Schmidt sense) if, in addition to Einstein's equations, the following four conditions hold: (1) M contains no closed timelike curves (reason able causality condition); (2) at each event in M and for each unit timelike vector u, the stress-energy tensor satisfies The Hawking-Penrose theorem on singularities (reasonable energy condition); (3) the manifold is \"general\" (i.e., not too highly symmetric) in the sense that every timelike or null geodesic with unit tangent u passes through at least one event where the curvature is not lined up with it in a specific way: (4) the manifold contains a trapped surface. All these conditions, except the trapped surface, seem eminently reasonable for any physically realistic spacetime! Note, especially, that the energy condition can be violated only if, as measured by some local observer in his proper frame, the total energy density E is negative or the principal pressures (eigenvalues of stress tensor) Pi are so negative that 2 Pi < -E. i The relevance of the Hawking-Penrose theorem for collapse follows from the general expectation that, in the real universe, trapped surfaces will always exist just below all future horizons,j-(1+). (Exceptions, such as the Kerr metric with a = M, are probably a \"set of measure zero.\") Since horizons and accompanying trapped surfaces are necessarily produced by slightly nonspherical collapse (Box 32.2), and since they probably also result from moderately deformed collapse (\u00a732.7), such collapse presumably produces singularities-or a violation of causality, which is also a rather singular occurrence! If the singularities are really such a general feature of collapse, then the exact nature of the singularity is oflife-and-death importance to anyone who falls through a horizon! Here one is on very shaky ground. Although the main results and conjectures described up to now in this section will probably survive all future research, opinions about the nature of the singularities are likely to change several times more before the whole story is in. Hence, it is safe only to describe the possibilities, not to attempt to judge them. Possibility 1 The singularity at the endpoint of a realistic collapse is a region of infinite tidal gravitational forces (infinite curvature), which crushes the collapsing matter to infinite density. Examples: the very special, homogeneous crushing of the Oppenheimer Snyder (1939) spherical collapse (\u00a732.4); also the very special inhomogeneous but spherical crushing described by Podurets (1966); also the special inhomogeneous, (continued on page 940) Relevance of the Hawking-Penrose theorem for gravitational collapse The nature of the singularity at the endpoint of realistic collapse: 4 possibilities y_---------9-3-6-- 3_4_._G_L_O_B_A_L_T_E_C_H_N_IQ_U_ES_'_H_O_R_I_Z_O_N_S_,_A_N_D_S_I_N_G_U_LA_RI_TY_T_H_E_O_R_E_M_S_ Box 34.2 ROGER PENROSE: Born August 8, 1931, Colchester, Essex, England Roger Penrose started' out as an algebraic geom eter. However, while at Cambridge from 1952-55 and again from 1957-60, his interest in general relativity was aroused by Hermann Bondi and Dennis Sciama. Because of his pure mathematical background, his approach to the subject was different from those which had been adopted hith erto. He was particularly interested in the global light-cone structure of spacetime and in the equ", "a tions of zero rest-mass fields, both of which are transformations, He preserved under conformal exploited this conformal invariance to give an treatment of gravitational elegant and powerful radiation in terms of a null surface 1+ at infinity. More recently this interest has led him to develop the theory of twistors, which are the spinors corre sponding to the conformal group of Minkowski space. These offer a new and very promising ap proach to the quantization of spacetime. His interest in conformal geometry also led him to study the properties of the causality relation ships between points of spacetime. These in turn led him to the theorems on the occurrence of sin- gularities in spacetime, which are probably the most important predictions of general relativity, since they seem to imply that spacetime has a beginning or an end. \"If spacetime is considered from the point of view of its conformal structure only, points at infinity can be treated on the same basis as finite points\" [PENROSE, IN INFELD (1964l] \"The argument will be to show that the existence of a trapped surface implies irrespective of symmetry-that singularities necessarily develop\" [PENROSE (1965bl] \"While the quantum effects of gravitation are normally thought to be significant only when curvatures approach 10 33 cm- l, all our local physics is based on the Poincare group being a good approximation of a local symmetry group at dimensions greater than 10-13 em. Thus, if curvatures ever even approach there can be little doubt but that extraordinary local effects are likely to take place\" 1013 cm- l, [HAWKING AND PENROSE (1969l] \u00a734.6. THEOREMS ON :-\" \u2022..:JULARITIES 937 \"We are thus presented with what is perhaps the most fundamental unanswered question of general-relativistic collapse theory, namely: does there exist a \"cosmic censor\" who forbids the appearance of naked singularities, clothing each one in an absolute event horizon?\" - [PENROSE (1969)] singiJlarity, \"Under normal circumstances, general relativity can, for practical purposes, remain remarkably apart-'-almost aloof-from the rest of physics. At a space-time the very reverse must surely be the case!\" \"I do not believe that a real understanding of the nature of elementary particles can ever be achieved without a simultaneous deeper understanding of the nature of spacetime itself. But if we are concerned with a level of phenomena for which such an understanding is not necessary-and this will cover almost all of present-day physics-then the smooth manifold picture presents an (unreasonably!) excellent framework for the discussion of phenomena. \" that space and time are not concepts that can be considered independently of one another but must be combined together to giv~ a four-dimensional picture of phenomena: the description in terms of spacetime\" \"The most important single lesson of relativity theory is, perhaps, [PENROSE (1968a)] will <:orrespond to the world as it actually is. spacetime as a pseudo-Riemannian differentiable manifold, \"If a formalism enables one to treat myriads of non-existent types of universe, then (effectively) it contains 'arbitrary parameters,, only special values of which In the ordinary approach to the dimension of the manifold and the signature of the metric are two such arbitrary parameters. \" \"As we localize the position of a particle, it jumps essentially along the null cone. Other particles are produced, which leap backward and forward essentially along null directions, without apparent regard for continuity, heeding only the positions of the-nul/cones themselves and \"topology\" only in the respect in which this term is applied to the structure of graphs\" [PENROSE (1966)] expression in terms of essentially combinatorial principles, \"My own view is that ultimately physical laws should find their most natural in terms of finite processes such as counting or other basically simple manipulative in accordance with such a view, should emerge some form of discrete or combinatorial spacetime\" procedures. Thus, that is to say, [PENROSE. IN KLAUDER (1972)] \"Complex numbers are... a very important constituent of the structure of physical/aws. The twistor theory carries this further in suggesting that complex numbers may also be very basically involved in defining the nature of spacetime itself. \" [PENROSE AND MACCALLUM (1973)] \"It is thus very tempting to believe that a link between spacetime curvature and quantum processes may be supplied by the use of twistors. Then, roughly it is the continual slight'shifting' of the interpretations of the quantum (twistor) operators which results in the curvature of spacetime\" speaking, [PENROSE (1968b)] / u g /_ - - - - 34.", " GLOBAL TECHNIQUES. HORIZONS. AND SINGULARITY THEOREMS 938 ~ Box 34.3 STEPHEN W. HAWKING: Born January 8, 1942, Oxford, England As a research student of Dennis Sciama's in Cam bridge, Stephen Hawking's early interest in rela tivity theory centered mainly on the question of spacetime singularities. With Ellis, he showed that a large class of homogeneous cosmological models must be singular. Then, encouraged by work of Penrose on the singularities arising in gravitational collapse, he developed new techniques which, in a series of papers in the Royal Society of London during 1966-67, established the important result that any plausible general-relativistic cosmology must be singular. The major portion of his later research has been concerned with black holes. He devised a series of arguments of great ingenuity which, together with the work of Israel and Carter, established to all intents and purposes the result that (vacuum) black holes in general relativity are described by Kerr metrics, that topologies other than spherical cannot occur, and that a certain limit on the en ergy emitted when two black holes congeal into one must be satisfied. Some of this work has had substantial pure mathematical interest (e.g., singularity theorems), some of it is concerned with astrophysics (e.g., work with Taylor on helium production in the big bang), some with observations (work with Gibbons on the possibility of black holes in binary star systems) and even experimental developments (with Gibbons on gravitational-wave detectors). In such scope is exhibited not only a considerable insight, depth, and versatility, but also the gift of an extraordinary determination to overcome se vere physical handicaps, to seek out and compre hend the truth. \"The observed isotropy of the microwave background indicates that the universe... This could possibly be regarded as an experimental verification of Mach's Principle\" is rotating very little if at all. [HAWKING (1969)] \"Undoubtedly, the most important results are the theorems... on the occurrence of singularities. These seem to imply either that the general theory of relativity breaks down or that there could be particles whose histories did not exist before (or after) a certain time. The author's own opinion is that the theory probably does break down but only when quantum gravitational effects become important. \" \"Although we have omitted the singular points from the definition of spacetime, we can still recognize the 'holes' left where they have been cut out by the existence of incomplete geodesics. \" \u00a734.6. THEOREMS ON SINGULARITIES 939 \"A good physical theory should not only correctly describe the currently experimental knowledge, but should also predict new results which can be tested by experiment the further the predictions from the original experiments, the greater the credit to the theory if they are found to be correct. Thus observations of whether or not singularities actually occurred, would provide a powerful test of the general theory of relativity in strong fields\" [HAWKING (1966a)} \"The construction of gravitational radiation detectors may open up a whole new field of 'gravitational astronomy' which could be as fruitful as radio astronomy has been in the last two decades.... Black hole collisions... would be much more effective in converting rest-mass energy into radiation than nuclear reactions, which can release only about 1 per cent of the rest-mass energy. In addition, black holes formed by collisions of smaller black holes can undergo further collisions, releasing more energy, whereas matter that has been fully processed by nuclear reactions cannot yield any more energy by the same means.... we are witnessing something really cataclysmic at the centre of our galaxy\" [HAWKING (1972b)] \"One might suggest that prior to the present expansion there was a collapsing phase. In this, local inhomogeneities grew large and isolated singularities occurred. Most of the matter avoided the singularities and reexpanded to give the present observed universe. \" \"It seems that we should draw a surface around regions where the radius of curvature is less than. say, 10-16 cm. On our side of this surface, a manifold picture of spacetime would be appropriate, but we have no idea what structure spacetime would have on the other side\" [HAWKING AND ELLIS (1968)] \"Presumably it would be necessary to consider quantum effects in very strong these would not become important until the radius of curvature became of the order of 10-14 cm, which for practical purposes is pretty singular. \" fields. However, reason for modifying them. However, \"The view has been expressed that singularities are so objectionable that if the Einstein equations were to predict their occurrence, this would be a compulsive the real test of a physical theory is not whether its predicted results are aesthetically attractive but whether they agree with observation. So far there are no observations which would show that singularities do not occur", "\" [HAWKING (1966b)] \"It is shown that a stationary black hole must have a topologically spherical boundary and must be axisymmetric if it is rotating. These results, together with those of Israel and Carter, go most of the way toward establishing the conjecture that any stationary black hole is a Kerr solution\" [HAWKING (1972a)] \"The fact that we have observed the universe to be isotropic is only a consequence of our existence. \" [COLLINS AND HAWKING (1973)] 940 34. GLOBAL TECHNIQUES. HORIZONS, AND SINGULARITY THEOREMS \"Kasner-like\" crushing of Lifschitz and Khalatnikov (1963a,b); also, most impor tantly, the very general \"mixmaster\" crushing (Chapter 30), discovered in the homo geneous case by Misner (1969b) and by Belinsky and Khalatnikov (1969a), and analyzed in the inhomogeneous case by Belinsky and Khalatnikov (1969b, 1970) and -by Khalatnikov and Lifschitz (1970). The mixmaster singularities-and only they among all explicitly known singularities-appear to be generic in this sense: if one perturbs slightly but arbitrarily the initial conditions of a spacetime that evolves a mixmaster singularity, then the resultant perturbed spacetime will also evolve a mix master singularity. Because of this, the prevalent opinion today (1973) is that realistic collapse probably produces, inside the horizon, a mixmaster singular ity. But that opinion might change tomorrow. Possibility 2 The singularity is a region of spacetime in which timelike or null geodesics terminate, not because of infinite tidal gravitational forces or infinite crushing, but because of other, more subtle pathologies. Example: \"Taub-NUT space\" [see Misner and Taub (1968)]. For other examples created specially to exhibit various pathologies, see Geroch (1968). Possibility 3 The singularity may be sufficiently limited in \"size\" and influence that all or most of the collapsing matter successfully avoids it. The matter cannot then explode back outward through the horizon that it went down; the horizon is a one-way membrane and will not let anything back out. Instead, the matter may reach a stage of maximum but finite contraction, and then reexplode into some other region of spacetime (multiply connected spacetime topology; \"wormhole\"). Analytical solutions for collapsing, charged spheres do reexplode in this manner [Novikov (1966); de la Cruz and Israel (1967); Bardeen (1968); see Figure 34.4]. Such a process requires that the \"exploding\" end of the wormhole be built into the initial conditions of the universe, with mass and angular momentum (as measured by Keplerian orbits and frame dragging) precisely equal to those that go down the black-hole end. This seems physically implausible. So does the \"explosion.\" Other Possibilities Various combinations of the above. If, as one suspects today, the singularities are of a very physical, infinite-curvature type, then one must face up to John Wheeler's (1964a) \"issue of the final state\" in its most raw and disturbing form. Wheeler, when faced with the issue, argues that infinite-curvature singularities signal a breakdown in classical general relativity-a breakdown forced by the onset of quantum gravitational phenomena (see Chapter 44). Whether quantization of gravity will actually save spacetime from such singu larities one cannot know until the \"fiery marriage of general relativity with quantum physics has been consummated\" [Wheeler (1964a); see also Misner (1969c), and the last section of Box 30.1]. Will quantization of spacetime save the universe from singularities? PART VIII GRAVITATIONAL WAVES Wherein the reader voyages on stormy seas of curvature ripples, searching for the ripple-generating storm gods, and battles through an electromagnetic and thermal fog that allows only uncertain visibility upon those seas. --------------------------------.......'\" CHAPTER 35 PROPAGATION OF GRAVITATIONAL WAVES Born: \"I should like to put to Herr Einstein a question, namely. how quickly the action of gravitation is propagated in your theory. That it happens with the speed of light does not elucidate it to me. There must be a very complicated connection between these ideas.., Einstein: \"It is extremely simple to write down the equations for the case when the perturbations that one introduces in the field are infinitely small. Then the g's differ only infinitesimally from those that would be present without the perturbation. The perturbations then propagate with the same velocity as light.., Born: \"But for great perturbations things are surely very complicated?\" Einstein: \"Yes, it", " is a mathematically complicated problem. It is especially difficult to find exact solutions of the equations. as the equations are nonlinear. \" Excerpts from discussion alter Einstein's Fall 1913 lecture in Vienna on \"The present position of the problem of 9ravitation:' already two years before he had the final field equations [EINSTEIN. 1913a] \u00a735.1. VIEWPOINTS Study one idealization after another. Build a catalog of idealizations, of their prop erties, of techniques for analyzing them. This is the only way to come to grips with so complicated a subject as general relativity! Spherical symmetry is the idealization that has dominated most of the last 12 chapters. Together with the idealization of matter as a perfect fluid, and of the universe as homogeneous, it has yielded insight into stars, into cosmology, into gravitational collapse. Turn attention now to an idealization of an entirely different type, one independent of any symmetry considerations at all: the idealization of a \"gravitational wave.\" Just as one identifies as \"water waves\" small ripples rolling across the ocean, so one gives the name \"gravitational waves\" to small ripples rolling across spacetime. We are deeply indebted to Mr. James M. Nester, who found and corrected many errors in the equations of this chapter and of a dozen others throughout the book. Gravitational waves compared to water waves on ocean: (1) approximate nature of a wave (2) local viewpoint vs. large-scale viewpoint 944 35. PROPAGATION OF GRAVITATIONAL WAVES Ripples of what? Ripples in the shape of the ocean's surface; ripples in the shape (i.e., curvature) of spacetime. Both types of waves are idealizations. One cannot, with infinite accuracy, delineate at any moment which drops of water are in the waves and which are in the underlying ocean: Similarly, one cannot delineate precisely which parts of the spacetime curvature are in the ripples and which are in the cosmological background. But one can almost do so; otherwise one would not speak of \"waves\"! Look at the ocean from a rowboat. Waves dominate the seascape. Changes in angle and level of the surface occur every 30 feet or less. These changes propagate. They obey a simple wave equation ( _1_\u00a3 + E + \u00a3) (height of surface) = O. g2 ot 4 0;2 ox2 Now get more sophisticated. Notice from a spaceship the large-scale curvature of the ocean's surface-curvature because the Earth is round, curvature because the sun and the moon pull on the water. As waves propagate long distances, this curvature bends their fronts and changes slightly their simple wave equation. Also important over large distance are nonlinear interactions between waves, interaction with the wind, Coriolis forces due to the Earth's rotation, etc. Spacetime is similar. Propagating through the universe, according to Einstein's theory, must be a complex pattern of small-scale ripples in the spacetime curvature, ripples produced by binary stars, by supernovae, by gravitational collapse, by explosions in galactic nuclei. Locally (\"rowboat viewpoint\") one can ignore the interaction of these ripples with the large-scale curvature. of spacetime and their nonlinear interaction with each other. One can pretend the waves propagate in fiat spacetime; and one can write down a simple wave equation for them. But globally one cannot. The large-scale curvature due to quiescent stars and galaxies will produce redshifts and will deform wave fronts; and the energy carried by the waves them selves will help to produce the large-scale curvature. This chapter treats both view points, the local (\u00a7\u00a735.2-6), and the global (\u00a7\u00a735.7-15). \u00a735.2. REVIEW OF \"LINEARIZED THEORY\" IN VACUUM Idealize, for awhile, the gravitational waves of our universe as propagating through fiat, empty spacetime (local viewpoint). Then they can be analyzed using the \"lin earized theory of gravity,\" which was introduced in Chapter 18. Linearized theory, one recalls, is a weak-field approximation to general relativity. The equations of linearized theory are written and solved as though spacetime were fiat (special-relativity viewpoint); but the connection to experiment is made through the curved-space formalism of general relativity. More specifically, linearized theory describes gravity by a symmetric, second-rank tensor field lip.v' In the standard (\"Lorentz,\" or Hilbert) gauge, this field satisfies the \"gauge\" or \"Subsidiary\" conditions (coordinate conditions) Linearized theory of gravitational waves: (1) Lorentz gauge condition h-p.a -0,a -. (35.1 a) \u00a735.3", ". PLANE-WAVE SOLUTIONS IN LINEARIZED THEORY 945 (Here, and throughout linearized theory, indices of hp.. are raised and lowered with the Minkowski metric 1'/a/3') In this gauge the propagation equations for vacuum gravitational fields are the familiar wave equations (35.1 b) (2) propagation equation Spacetime is really curved in linearized theory, although equations (35.1) are written and solved as though it were not. The global inertial frames of equations (35.1) are only almost inertial. In them the metric components are actually and the \"metric perturbation\" hp.. is related to the \"gravitational field\" hp.. by (35.2a)* (3) metric (35.2b) The metric (35.2a) governs the motion of test particles, the propagation of light, etc., in the usual general-relativistic manner. Recall the origin of the equations (35.1) that govern hp.\u2022. The subsidiary conditions h/.a = 0 were imposed by specializing th: coordinate system; and the Einstein field equations in vacuum then reduced to Ohp.. = O. Actually, as was shown in Box 18.2, the coordinates oflinearized theory are not fully fixed by the conditions h/.a = O. There remains an ambiguity embodied in further \"gauge changes\" (infinitesimal coordinate transformations), gp., which satisfy a restrictive condition (4) residual gauge. freedom in order to preserve conditions (35.1a). Then is the coordinate transformation and (35.3a) (35.3b) (35.3c) is the gauge change. All this was derived and discussed in Chapter 18. \u00a735.3. PLANE-WAVE SOLUTIONS IN LINEARIZED THEORY The simplest of all solutions to the linearized equations hp..,aa = h/.a = 0 is the monochromatic, plane-wave solution, Monochromatic, plane wave \u2022 A more nearly rigorous treatment defines h~. == g~p - 1/~p, and puts the small corrections O([h~pI2) into the field equations: (35.4a) 946 35. PROPAGATION OF GRAVITATIONAL WAVES Here l\\[.....] means that one must take the real part of the expression in brackets; while Af.l' (amplitude) and kf.l (wave vector) are constants satisfying (k a null vector), (A orthogonal to k) (35.4b) (35.4c) [consequences of hp.,.,aa = 0 and hf.l~a = 0, respectively; see (35.10) below for the true physics associated with this wave, the curvature tensor]. Clearly, this solution de scribes a wave with frequency (35.5) which propagates with the speed of light in the direction (l/kO)(kz, k y, kz ). At first sight the amplitude Ap.. of this plane wave appears to have six independent components (ten, less the four orthogonality constraints Ap.aka = 0). But this cannot be right! As Track-2 readers have learned in Chapter 21, the gravitational field in general relativity has two dynamic degrees of freedom, not six. Where has the analysis gone astray? One went astray by forgetting the arbitrariness tied up in the gauge freedom (35.3). The plane-wave vector with four arbitrary constants ep., generates a gauge transformation that can change arbitrarily four of the six independent components of Ap..' One gets rid of this arbitrariness by choosing a specific gauge. (35.6) Plane wave has two degrees of freedom in amplitude (two polarizations) \u00a735.4. THE TRANSVERSE TRACELESS (TT) GAUGE Transverse-traceless (TI) gauge: (1) for plane wave Select a 4-velocity u-not at just one event, but the same u throughout all of spacetime (special-relativistic viewpoint!). Bya specific gauge transformation (exer cise 35.1), impose the conditions (35.7a) These are only three constraints on Ap.., not four, because one of them-kP.(Ap..u\u00b7) = O-is already satisfied (35.4c). As a fourth constraint, use a gauge transformation (exercise 35.1) to set (35.7b) One now has eight constraints in all, Ap.aua = Ap.aka = Aaa = 0, on the ten components of the amplitude; and the coordinate system (gauge) is now fixed rigidly. Thus, the two remaining free components ofAp.. represent the two degrees of freedom (two polarizations) in the plane gravitational wave. \u00a735.4", ". THE TRANSVERSE TRACELESS GAUGE 947 It is useful io restate the eight constraints Ap.au a = Ap.aka = AI'Ii = 0 in a Lorentz frame where uO = 1, u; = 0, and in a form where k a does riot appear explicitly: Le., only the spatial components h;k are nonzero; Le., the spatial components are divergence-free; Le., the spatial components are trace-free. (35.8a) (35.8b) (35.8c) (Here and henceforth repeated spatial indices are to be summed, even if both are down; e.g., hkk = I~=lhkk.) Notice that, since h = h/ = hkk = 0, there is no dis tinction between hI'\" and hl'v in this gauge. Turn attention now away from plane waves to arbitrary gravitational waves in linearized theory. Any electromagnetic wave can be resolved into a superposition of plane waves, and so can any gravitational wave. For each plane wave in the superposition, introduce the special gauge (35.8). Note that the gauge conditions are all linear in hl'v' Therefore the arbitrary wave will also satisfy conditions (35.8). Thus arises the theorem: Pick a specific global Lorentz frame of linearized theory (i.e., pick a specific 4-velocity u). In that frame (where ua = 8a o), examine a specific gravitational wave ofarbitraryform. One can always find a gauge in which hl'v satisfies the constraints (35.8). Moreover, in this gauge only the h;k are nonzero. Therefore, one need only impose the six wave equations (35.9) Any symmetric tensor satisfying constraints (35.8) [but not necessarily the wave equations (35.9)] is called a transverse-traceless (TT) tensor-transverse because it is purely spatial (hoI' = 0) and, if thought of as a wave, is transverse to its own traceless because hkk = o. The most direction of propagation (hi;,; = hiik; = 0); general purely spatial tensor Sii can be decomposed [see Arnowitt, Deser, and Misner (1962) or Box 35.1] into a part Sf!, which is \"transverse and traceless\"; a part S~ = ~(8i;!.kk -!.ij), which is \"transverse\" (S~,j = 0) but is determined entirely by one functionfgiving the trace of S (Sfk = 'V 2f); and a part Sfi = Sf,; + Sr.i' which is \"longitudinal\" and is determined by the vector field Sf. In linearized theory hf; is a purely gauge part of hl'v' whereas h~ and hf! are gauge-invariant parts of hl'v\u00b7 The special gauge in which hl'v reduces to its transverse-traceless part is called the TT or transverse-traceless gauge. The conditions (35.8) defining this gauge can be summarized as (35.8d) As exercise 35.2 illustrates, only pure waves (and not more general solutions of the linearized field equations with source, Ohl'v = -16\\7Tl'v) can be reduced to TT gauge. (2) for any wave Decomposition of spatial tensors 948 35. PROPAGATION OF GRAVITATIONAL WAVES Curvature tensor in TT gauge In the TT gauge, the time-space components RiOkO = ROiOk = - RiOOk = - ROjkO of the Riemann curvature tensor have an especially simple form [see equation (18.9) and exercise 18.4]: R jOkO - 1 hTT -\"2 jk,OO' (35.10) Recall that the curvature tensor is gauge-invariant (exercise 18.1). It follows that hfJV cannot be reduced to still fewer components than it has in the TT gauge. Box 35.1 describes methods to calculate h'{J from a knowledge of hfJV in some other gauge. Box 35.1 METHODS TO CALCULATE \"TRANSVERSE-TRACELESS PART\" OF A WAVE Problem: Let a gravitational wave hfJv(t, xi) in an arbitrary gauge of linearized theory be known. How can one calculate the metric perturbation h'{J(t, xi) for this wave in the transverse-traceless gauge? Solution 1 (valid only for waves; i.e., when OhfJV = 0). Calculate the components R", "jOkO of Riemann in the initial gauge; then integrate equation (35.10) hJ'[,oo = -2RjOkO to obtain hft When the wave is monochromatic, hp.v = hp.v(xi)e-iwt ; then the solution of(1) has the simple form (1) is the unit vector in the direction of propagation. Verify that Pjk is a projection operator onto the transverse plane: Then the transverse part of hik is PjlhimPmk (or in matrix notation, PhP); and the TT part is this quantity diminished by its trace: hJ,'{ = PjlPmkhim - 1 \"2Pjk(Pmihim) (4) (index notation), h TT - p~p _l.p Tr (Ph) - 2 (2) (matrix notation). (4') Solution 2 (valid only for plane waves). \"Project out\" the TT components in an algebraic manner using the operator (3) The sequence of operations that gives hfT cuts two parts out of hii. The first part cut out is (5) Here which is transverse but is built from its own trace, \u00a735.4. THE TRANSVERSE TRACELESS GAUGE 949 Exercise 35.1. TRANSFORMATION OF PLANE WAVE TO TT GAUGE Let a plane wave of the form (35.4) be given, in some arbitrary gauge of linearized theory. Exhibit explicitly the transformation that puts it into the TT gauge. [Hint: Work in a Lorentz frame where the 4-velocity ul' of the TT gauge is UO = I, u; = O. Solve for the four constants el' of the generating function (35.6) by demanding that hI'> satisfy the TT constraints (35.7).] EXERCISES LIMITATION ON EXISTENCE OF TT GAUGE Exercise 35.2. Although the metric perturbation hI'> for any gravitational wave in linearized theory can be put into the TT form (35.8), nonradiative hl';s cannot. Consider, for example, the external field of a rotating, spherical star, which cannot be written as a superposition of plane waves: The second part cut out of hi; is the longitudinal part hL jk = hjk - P;lPmk hlm = n1nkh;1 + njn1h1k - n;nk(nlnmhlm); (6) or hL = h - PhP (6') Solution 3 (general case). Fourier analyze any symmetric array hi; = fhi;(k, t) exp (ikmx m) d3k, and apply the formulas (4) from solution 2 to each Fourier component individually. But note that in this case one can write the projection operator in the direction-independent form ~ or (7) (8) (provided the formulas are written with all h's standing on the right), since 01 = ik1 under the Fourier integraL Of course the operation I/V2 can be evaluated by other methods, e.g., by Green's functions, as well as by Fourier analysis. [The quantity l/; = V-2jstands for the solution l/; of the Poisson equation V2l/; =fl The advantage of this method is its power in certain analytic computa tions (see, e.g., below). Gauge Transformations. The change in hp.> 9ue to a gauge transformation is (9) The transverse part of this change is To verify this formula for a plane wave (solution 2), note that 01 = ilkln 1 and P;lnl = O. To verify the same result in general, use equation (7) to give the result P;101=0. (11) Thus both hf{ of equation (4), and h~ of equation (5) are gauge-invariant: ohf! = {jh~ = O. (12) In empty space (Tp.> = 0), both h~ and another gauge-invariant quantity hOk (discussed in exercise 35.4) vanish, by virtue of the field equations. 950 35. PROPAGATION OF GRAVITATIONAL WAVES 2M hoo = r [see equation (19.5)]. Here M is the star's mass and S is its angular momentum. Show that this CanilO( be put into TT gauge. [Hint: Calculate R jOkO and from it, by means of (35.10), infer hIt Then calculate R OZYZ in both the original gauge and the new gauge, and discover that they disagree-not only by virtue of the mass term, but", " to the special case of a plane wave. Suppose the test-particle separation lies in the direction of propagation of the wave. Then the wave cannot affect the separa tion; there is no oscillation: Transverse character of relative accelerations Only separations in the transverse direction oscillate; the wave is transverse not only in its mathematical description (hJ{), but also in its physical effects (geodesic deviation)! 952 35. PROPAGATION OF GRAVITATIONAL WAVES EXERCISE Exercise 35.5. ALTERNATIVE CALCULATION OF RELATIVE OSCILLATIONS Introduce a TT coordinate system in which, at time l = 0, the two particles are both at rest. Use the-geodesic equation to show that subsequently they both always remain at rest in the TT coordinates. despite the action of the wave. This means that the contravariant components of the separation vector are always constant in the TT coordinate frame: Call this constant xB<ol- Transform these components to the comoving orthonormal frame; the answer should be equation (35.15). \u00a735.6. POLARIZATION OF A PLANE WAVE Polarization of gravitational waves: (1) States of linear polarization, \"+\" and\" X\" Geodesic deviation in the transverse direction provides a means for studying and characterizing the polarizations of plane waves. Consider a plane, monochromatic wave propagating in the z direction. In the TT gauge the constraints h'g = 0, hflj = ikjhfT = 0, and hIT =\u00b0reveal that the only nonvanishing components of h'JJ are h TT - _FT - ~{A e-iw(t-zl} zz - yy - + h TT - h TT - ~{A e-iw(t-zl} yz - ~ x zy -.'(35.16) The amplitudes A+ and Ax represent two independent modes of polariza~ion. As for electromagnetic plane waves (Figure 35.1), so also for gravitational plane waves (Figure 35.2), one can resolve a given wave into two linearly polarized components, or, alternatively, into two circularly polarized components. Displacement, /lx, for polarization w(t - z) 2m! e z \u2022 (2n + i} --- (2n + I)\" (2n + t)\" \u2022..- e. \u2022 ; \u2022 i e R i --;..- e L ; ~ ~.- Figure 35.1. Plane Electromagnetic Waves. Polarization vector: ep Vector Potential A = R[Aoe-i\",<t-z)ep ) Acceleration of a test charge: a = (qlm)E = (qlm)( - aAlal) = R[iw(qlm)Aoe-i\",<t-')ep ) Displacement of charge relative to inertial frame: <'lx =R[q:: Aoe-i\",<t-Z)ep ] \u00a735.6. POLARIZATION OF A PLANE WAVE 953 For linearly polarized waves, the unit polarization vectors of electromagnetic theory are ex and e y. A test charge hit by a plane wave with polarization vector ex oscillates in the x-direction relative to an inertial frame; and similarly for ey. By analogy, the unit linear-polarization tensors for gravitational waves are e+ = ex \u00ae ex - e y \u00ae e y, ex = ex \u00ae e y + e y \u00ae ex. The plane wave (35.16), when Ax = 0, has polarization e+ and can be rewritten (35.17b) (35.17a) h - ~{A e-iw(t-z)e + ik - } +ik. (35.18) Its effect in altering the geodesic separation between two test particles depends on the direction of their separation. To see the effect in all directions at once, consider a circular ring of test particles in the transverse (x,y) plane, surrounding a central particle (Figure 35.2). As the plane wave (35.18) (polarization e+) passes, it deforms what was a ring as measured in the proper reference frame of the central particle into an ellipse with axes in the x and y directions that pulsate in and out: 000000000000000 00 etc. By contrast (Figure 35.2), a wave of polarization ex deforms the ring at a 45-degree angle to the x and y directions: 0 0 0 0 0 C) ~ C) 0 etc. For circularly polarized waves, the unit polarization vectors of electromagnetic theory are (35.19) (2) States of circular polarization Deformation of a ring of test particles w(t - z) e L e R e+ ex 2m", "! <;) e <;) ~ (2n + 1)\" 0 0 e ~ (2n + I)\" G ~ G G (2n + ~)\" 0 0 ~ e Figure 35.2. Plane Gravitational Waves. Polarization tensor: e p Metric perturbation: hik = R [Aoe-i..,<t-z)epikl Tidal acceleration between two test particles: D2n\u2022 __I = - RjOkOn\" =- _'_ nk Dr2 I o2h k 2 ot2 = 1\\ [ - +w2Aoe-i..,Il-:)epikn\" ] Separation between two test particles: n' = n JO) + 1\\ [..!.. A e-i..,<t-zle. n,(Ol] 2 0 \" P,k \u2022 Position of test particle B in proper reference frame of test particle A. (In drawing. A is the central particle and B is any peripheral particle): I A -i..,ll-zl ~ '2 oe mo) epjkXmO),,] Spin-2 character of gravitational field and its relation to symmetries of waves EXERCISES 954 35. PROPAGATION OF GRAVITATIONAL WAVES Similarly, the unit circular polarization tensors of gravitation theory are (35.20) A test charge hit by an electromagnetic wave of polarization eR moves around and around in a circle in the righthanded direction (counterclockwise for a wave propa gating toward the reader); for eL it circles in the lefthanded (clockwise) direction (see Figure 35.1). Similarly (Figure 352), a gravitational wave of polarization e R rotates the deformation of a test-particle ring in the righthanded direction, ooOOI:0~oo while a wave of e L rotates it in the lefthanded direction. The individual test particles in the ring rotate in small circles relative to the central particle. However, just as the drops in an ocean wave do not move along with the wave, so the particles on the ring do not move around the central particle with the rotating ellipse. Notice from Figure 35.2 that, at any moment of time, a gravitational wave is invariant under a rotation of 180\u00b0 about its direction of propagation. The analogous angle for electromagnetic waves (Figure 35.1) is 360\u00b0, and for neutrino waves it is 720\u00b0. This behavior is intimately related to the spin of the zero-mass particles associated with the quantum-mechanical versions of these waves: gravitons have spin 2, photons spin 1, and neutrinos spin 1/2. The classical radiation field of a spin-S particle is always invariant under a rotation of 360\u00b0 / S about its propagation direction. A radiation field of any spin S has precisely two orthogonal states of linear polarization. They are inclined to each other at an angle of 90\u00b0 IS; thus, for a neutrino field, with S = ~, the two states are distinguished as It> and 1+> (spin up and spin down; 180\u00b0 angle). For an electromagnetic wave S = 1 and two ortho gonal states of polarization are ex and ey (90\u00b0 angle). For a gravitational wave S = 2, and two orthogonal states are e+ and ex (45 \u00b0 angle). Exercise 35.6. ROTATIONAL TRANSFORMATIONS FOR POLARIZATION STATES Consider two Lorentz coordinate systems, one rotated by an angle 8 about the z direction relative to the other: t' = t, x' = x cos 8 + Y sin 8, y' =y cos 8 - x sin 8, z' = z. (35.21) Let It> and 1+> be quantum-mechanical states of a neutrino with spin-up and spin-down relative to the x direction; and similarly for It'> and It'>. Let ez, ey, ex\" ell be the unit polarization vectors in the two coordinate systems for an electromagnetic wave traveling in the z-direction; and similarly e+, ex' e+\" ex' for a gravitational wave in linearized theory. Derive the following transformation laws: 1+'> = It> cost8 + 1+> sint8; ex' = ez cos 8 + ey sin 8; e+, = e+ cos 28 + ex sin 28; ey, = -ex sin 8 + ey cos 8; ex' = - e+ sin 28 + ex cos 28. (35.22) \u00a735.7. STRESS-ENERGY CARRIED BY A GRAVITATIONAL WAVE 955 What is the generalization to the linear-polarization basis states for a radiation field of arbitrary spin S? Exercise 35.7. ELL", "IPTICAL POLARIZATION Discuss elliptically pOlarized gravitational waves in a manner analogous to the discussion of linearly and circularly polarized waves in Figure 35.2. \u00a735.7. THE STRESS-ENERGY CARRIED BY A GRAVITATIONAL WAVE Exercise 18.5 showed that in principle one can build detectors which extract energy from gravitational waves. Hence, it is clear that the waves must carry energy. Unfortunately, to derive and justify an expression for their energy requires a somewhat more sophisticated viewpoint than linearized theory. Such a viewpoint will be developed later in this chapter (\u00a7\u00a735.13 and 35.15). But for the benefit of Track-I readers, the key result is stated here. In accordance with the discussions in \u00a7\u00a719.4 and 20.4, the stress-energy carried by gravitational waves cannot be localized inside a wavelength. One cannot say whether the energy is carried by the crest of a wave, by its trough, or by its \"walls.\" However, one can say that a certain amount of stress-energy is contained in a given \"macroscopic\" region (region of several wavelengths' size), and one can thus talk about a tensor for an effective smeared-out stress-energy of gravitational waves, T~~W). In a (nearly) inertial frame of linearized theory, T~~Wi is given by Approximate localization of energy in a gravitational wave Effective stress-energy tensor for gravitational waves: T(GW) p.v - 3277 1 (hTT hTT > jk,v, jk,p. (35.23) (1) expressed in terms of metric perturbations where ( >denotes an average over several wavelengths and hTl means the (gauge invariant) transverse-traceless part of hp.v' which is simply hjk in the TT gauge. ~nother formula for T~~Wl, valid in any arbitrary gauge, with Ji f:. 0, Ji/,a f:. 0, and hop. f:. 0 is T(GW) = _1_ (Ji 3277 p.v Jia/3 _lJi Ji 2'P.,v,v - Jia/3 Ii,/3 ap.,v - Jia/3 ) Ii,/3 av,p. (35.23') a/3,p. This stress-energy tensor, like any other, is divergence-free in vacuum T(GWl v = o\u00b7 p.,v'(35.24) (2) subject to conservation law and it contributes to the large-scale background curvature (which linearized theory ignores) just as any other stress-energy does: G(B) = 8~(T(GW) + T(matter) + T(otherfields). Po\" IJ Po\" Po\" Po\" (35.25) (3) role as source of background curvature In writing here the term T~~W) for the effective smeared-out energy density of the gravitational wave, one is foregoing any further insertion of the gravitational wave into the Einstein equation. Otherwise one might end up counting twice over the (4) for a plane, monochromatic wave 956 35. PROPAGATION OF GRAVITATIONAL WAVES contribution of the same wave to the background curvature of space, even though expressed in very different formalisms. According to equation (35.23), the stress-energy tensor for the plane wave, h p.v - \" {(A X\"< +e+p.v + A xexp.v e ) -i\"'(t-z)}~, is T(GW) = T(GW) = _T(GW) = _1_w 2(IA 12 +IA 12 ) x'zz tz tt 32 'iT + (35.26) (35.27) Notice that the background radius of curvature ('jl~ (ignored by linearized theory), and the mean reduced wavelength A (= wavelength/2'iT) and amplitude (f of the gravitational waves satisfy ~Jl-2 - typical magnitude of components of R~JY8 - T(GW) _ pv ~ T(GW) _ p.v!7 (f2/ A2 if T(GW) is chief source of background curvature (/2/ A2 if T(GW) is not chief source pv, p.v. Consequently, the dimensionless numbers (f and A/qil are related by (35.28) Conditions for validity of gravitational-wave formalism Thus, the whole concept of a small-scale ripple propagating in a background of large-scale curvature breaks down, and the whole formalism of this chapter becomes meaningless, if the dimensionless amplitude of the wave approaches unity. One", " must always have (f ~ 1 as well as A ~ qil if the concept ofa gravitational wave is to make any sense! Nonlinear effects in gravitational waves: (1) radiation dam pi ng (2) refraction \u00a735.8. GRAVITATIONAL WAVES IN THE FULL THEORY OF GENERAL RELATIVITY Curving up of the background spacetime by the energy of the waves is but one of many new effects that enter, when one passes from linearized theory to the full, nonlinear general theory of relativity. In linearized theory one can consider a localized source of gravitational waves (e.g., a vibrating bar) in steady oscillation, radiating a strictly periodic wave. But the exact theory insists that the energy of the source decrease secularly, to counter balance the energy carried off by the radiation (energy conservation; gravitational radiation damping; see \u00a7\u00a736.8 and 36.11). This makes an exactly periodic wave impossible, though a very nearly periodic one can certainly be emitted [Papapetrou (1958); Arnowitt, Deser, and Misner as reported by Misner (1964b)]. In the real universe there are spacetime curvatures due not only to the energy of gravitational waves, but also, and more importantly, to the material content of the universe (planets, stars, gas, galaxies). As a gravitational wave propagates through these curvatures, its wave fronts change shape (\"refraction\"), its wavelength changes \u00a735.9. AN EXACT PLANE-WAVE SOLUTION 957 (gravitational redshift), and it backscatters off the curvatures to some extent. If the wave is a pulse, the backscatter will cause its shape and polarization to keep changing and will produce \"tails\" that spread out behind the moving pulse, traveling slower than light [see exercise 32.10; also Riesz (1949), DeWitt and Brehme (1960), DeWitt and DeWitt (1964a), Kundt and Newman (1968), Couch et. al. (1968)]. However, so long as Cl ~ I and A/&!' ~ I, these effects will be extremely small locally. They can only build up over distances of the order of &!'~and sometimes not even then. Thus, locally, linearized theory will remain highly accurate. Even in an idealized universe containing nothing but gravitational waves, back scatter and tails are produced by the interaction of the waves with the background curvature that they themselves produce. If the reduced wavelength A = \"A/2'iT and the mass-energy m of a pulse of waves satisfy A ~ m, it is possible (in principle) to focus the pulse into a region of size r < m, whereupon a part of the energy of the pulse will undergo gravitational collapse to a singularity, leaving behind a black hole [see Ruffini and Wheeler (1970), and pp. 7-24 of Christodoulou (1971)]. Short of a certain critical strength, no part of the pulse undergoes such a collapse. But it does undergo a time delay before reexpanding. This time delay is definable and measurable in the asymptotically flat space, far from the domain where the energy a little earlier underwent temporary focusing into dimensions of order A. All these effects can be analyzed in general relativity theory using approximation schemes which, in first order, are similar to or identical to linearized theory. Later in this chapter (\u00a7\u00a735.13-35.15), one such approximation scheme will be developed. But first it is helpful to study an exact solution that exhibits some of these effects. (3) redshift (4) backscatter (5) tails (6) self-gravitational attraction \u00a735.9. AN EXACT PLANE-WAVE SOLUTION Any exact gravitational-wave solution that can be given in closed mathematical form must be highly idealized; otherwise it could not begin to cope with the complexities outlined above. Consequently, mathematically exact solutions are useful for peda gogical purposes only. However, pedogogy should not be condemned: it is needed not only by students, but also by veteran workers in the field of relativity, who even today are only beginning to develop intuition into the nonlinear regime of geometrodynamics! From the extensive literature on exact solutions, we have chosen, as a compromise between realism and complexity, the following plane wave [Bondi et. al. (1959), Ehlers and Kundt (1962)]: The rest of this chapter is Track 2. No earlier Track-2 material is needed as preparation for it, but Chapter 20 (conservation laws) and \u00a722.5 (geometric optics) will be found to be helpful. It is not needed as preparation for any later chapter. Exact plane-wave solution of vacuum field equation: ds 2 = L 2(e 2/3 dx 2 + e- 2", "/3 d y 2) + dz 2 - dt 2 =L2(e 2/3 dx 2 + e-2/3 dy 2) - du du. (35029a) (1) form of metric Here u = t - z, u = t + z, L = L(u), 13 = f3(u). (3.5029b) 958 35. PROPAGATION OF GRAVITATIONAL WAVES The forms that the functions L(u) (\"background factor\") and f3(u) (\"wave factor\") can take are determined by the vacuum field equations. In the null coordinate system u, v. x, y, the only component of the Ricci tensor that does not vanish identically is (see Box 14.4, allowing for the difference in coordinates. 2Vthere = vhere) where the prime denotes dldu. Thus, Einstein's equations in vacuum read L\" + (f3')2L = O. (35.30) (3531) (\"effect of wave factor on background factor\") The linearized version of this equation is L\" = 0, since (13')2 is a second-order quantity. Therefore the solution corresponding to linearized theory is L = I, f3(u) arbitrary but small. The corresponding metric is ds 2 = (I + 213) dx 2 + (I - 213) dy 2 + dz 2 - dt 2, 13 = f3(t - z). (3532) Notice that this is a plane wave of polarization e+ propagating in the z-direction. (See exercise 35.10 at end of \u00a735.12 for the extension to a wave possessing both polarizations, e+ and ex') Return attention to the exact plane wave, and focus on the case where the \"wave factor\" f3(u) is a pulse of duration 2T, and 113'1 ~ liT throughout the pulse. Then the exact solution (Figure 35.3) is: (I) for u < - T (flat spacetime; pulse has not yet arrived), 13 = 0, L = 1; (3533a) (2) for - T < u < + T (interior of pulse), (2) generation of \"background factor\" L by \"wave factor\" f3 (3) linearized limit (4) special case: a plane-wave pulse 13 = f3(u) is arbitrary, except that 113'1 ~ liT, L(u) = I - i:{I: [f3'(u)f dU} dU + 0([137]4); (3) for u > T (after the pulse has passed), 13 = 0, u L = I -~' _ a = I T f (13')2 du -T + T 0([137]2) f (13')2 du -T (3533b) \u2022 (3533c) Before discussing the physical interpretation of this exact solution, one must come to grips with the singularity in the metric coefficients at u = a ~ T. (There L = 0, Is this a physical singularity like the region r = 0 of the so g:r:r = gyy = 0.) Schwarzschild geometry, or is it merely a coordinate singularity as r = 2M is in Schwarzschild coordinates (Chapters 31, 32, and 33)? The only nonzero compo nents of the Riemann tensor for the metric (35.29) are (see Box 14.4) \u00a735.9. AN EXACT PLANE-WAVE SOLUTION 959 Figure 35.3. Spacetime diagram and pulse profile for an exact plane-wave solution to Einstein's vacuum field equations. The metric has the form ds2 = L 2(t?f3 dx2 + e-2f3 dy 2) + dz 2 _ dt 2. The \"wave factor\" f3(u) == f3(t z) (short-scale ripples) and the \"background fac z) (large-scale bending of the background geometry by the tor\" L(u):= L(t effective mass-energy of the \"ripply\" gravitational wave) are shown in the drawing and are given analytically by equations (35.33). RZuzu == tRuu - {3\" - 2(L'IL){3', (35.34) R\"ullu == tRuu + {3\" + 2(L'1L){3'. Moreover, these components both vanish in any extended region where {3 == O. Thus, spacetime is completely flat in regions where the \"wave factor\" vanishes-which is everywhere outside the pulse! In particular, spacetime is flat near u == a, so the singularity there must be a coordinate singularity, not a physical singular", "ity. To eliminate this singularity, one can perform the coordinate transformation x.X = I _ Via' y y = I - Via' u = V, V = V + X2 + y2 a - V (35.35) throughout the region to the future of the pulse (u > T), where (3.5.36 a) (5) spacetime is flat outside the pulse y 960 35. PROPAGATION OF GRAVITATIONAL WAVES In the new X, Y, U, V. coordinates the metric has the explicitly flat form ds 2 = dX2 + dy2 - dUdV for U = u > T. (35.36b) EXERCISES Exercise 35.8. GLOBALLY WELL-BEHAVED COORDINATES FOR PLANE WAVE [based on Ehlers and Kundt (1962)] Find a coordinate transformation similar to (35.35), which puts the exact plane-wave solution (35.29a), (35.31), into the form ds 2 = dX2 + dy2 - dU dV + (X2 - YZ)F dU2, F = F( U) completely arbitrary. (35.37) This coordinate system has the advantage of no coordinate singularities anywhere; while the original coordinate system has the advantages of an easy transition to linearized theory, and easy interpretation of the action of the wave on test particles. Exercise 35.9. GEODESIC COMPLETENESS FOR PLANE-WAVE MANIFOLD [based on Ehlers and Kundt (1962)] Prove that the coordinate system (X, y, U, V) of exercise 35.8 completely covers its spacetime manifold. More specifically, show that every geodesic can be extended in both directions for an arbitrarily large affine-parameter length without leaving the X, Y, U, V coordinate system. (This property is called geodesic completeness.) [Hint: Choose an arbitrary event and an arbitrary tangent vector did>\" there, They determine an arbitrary geodesic. Perform a coordinate transformation that leaves the form of the metric unchanged and puts did>\" either in the (fJ, V) = constant 2-surface, or in the.Y, Y) = constant 2-surface. Verify that the two coordinate systems cover the same region of spacetime. Then analyze completeness of dld>..'s geodesic in (.Y, Y, fJ, V) coordinates.) \u00a735.10. PHYSICAL PROPERTIES OF THE EXACT PLANE WAVE Spacetime is completely flat both before the arrival of the plane-wave pulse (u < - T) and after it has passed (u > T). This is the message of the last paragraph. Complete flatness outside the pulse is very atypical for gravitational waves in the full, nonlinear general theory of relativity. It occurs in this example only because the wave fronts (surfaces of constant u and v, i.e\" constant z and t) are perfectly flat 2-surfaces. If the wave fronts were bent (e.g., spherical), the energy carried by the pulse would produce spacetime curvature outside it. To see nonlinear effects in action, turn from the flat geometry outside the pulse to the action of the pulse on freely falling test particles. Consider a family of particles that are all at rest in the original t, x,y, z coordinate system (world lines: [x,y, z] = constant) before the pulse arrives. Then even while the pulse is passing, and after it has gone, the particles remain at rest in the coordinate system. (This is true for any metric, such as (35.29a), with gOJL = _8\u00b0p.' for then r JLoo = 0, so x JL = 8JLo'T + const. satisfies the geodesic equation.) Flatness outside gravitational-wave pulses is unusual Action of exact gravitational-wave pulse on test pa rticles: (1) transverse character of relative accelerations (2) gravitational attraction due to energy in pulse \u00a735.11. ELECTROMAGNETIC AND GRAVITATIONAL PLANE WAVES COMPARED 961 Two particles whose separation is in the direction of propagation of the pulse (z-direction) have not only constant coordinate separation,..1x =..1y = 0 and Liz f:. 0; they also have constant proper separation, L1s = gzz1/Z Liz = Liz. Hence, the exact plane wave is completely transverse, like a plane wave of linearized theory. Neighboring particles transverse to the propagation direction, (..1x f:. 0,..1y f:. 0, Liz = 0) have a proper separation that wiggles as the pulse passes:..1s =", " L(I - z)[eZ,8(t-z>(..1x)Z + e-Z,8(t-z>(..1y)Z]1/Z ::::: L[(l + 2,8)(..1x)Z + (l - 2,8)(..1y)zjl/2. (35.38) Superimposed On the usual linearized-theory type of wiggling, due to the \"wave factor\",8, is a very small net acceleration of the particles toward each other, due to the \"background factor\" L [note the form of L(u) in Figure 35.3]. This is an acceleration of almost Newtonian type, caused by the gravitational attraction of the energy that the gravitational wave carries between the two particles. The total effect of all the energy that passes is to convert the particles from an initial state of relative rest, to a final state of relative motion with speed (35.39) where..1si = [(..1X)2 + (..1y)Zjl/2 = (initial particle separation). [Recall: Linitial = 1, Lfinal = 1 - u/a = 1 - (I - z)/a; equation (35.33).] Precisely the same effect can be produced by a pulse of electromagnetic waves (\u00a735.l1). \u00a735.11. COMPARISON OF AN EXACT ELECTROMAGNETIC PLANE WAVE WITH THE GRAVITATIONAL PLANE WAVE Consider the metric ds 2 = V(u)(dx Z + dy 2) - du dv, u = 1- Z} V=I+Z' { (35.40) which is always flat if it satisfies the vacuum Einstein equations (Rp.v = 0 or L\" = 0), and therefore cannot represent a gravitational wave. In this metric the electromag netic potential A = Ap. dxP. = A(u) dx (35.41) An electromagnetic plane-wave pulse satisfies Maxwell's equations for arbitrary A(u). It represents an electromagnetic plane wave analogous to the gravitational plane wave of the last few sections. The only nonzero field components of this wave are Fuz = A', Le., F;z = -~z = A'; (~5.42) 962 35. PROPAGATION OF GRAVITATIONAL WAVES so the electric vector oscillates back and forth in the x-direction, the magnetic vector oscillates in the y-direction, and the wave propagates in the.::-direction. The stress energy tensor in x,y, u, v, coordinates has only (35.43) nonzero. The Maxwell equations are already satisfied by the potential (35.41) in the back ground metric (35.40), as the reader can verify. In order to make that metric itself equally acceptable, one need only impose the Einstein equations G/LV = S7TT/Lv' They read [see equation (35.30) with 13 = 0] (35.44) Electromagnetic plane wave and gravitational plane wave produce same gravitational attractions This has exactly the form of the equation L\" + (f3')2L = 0 for the gravitational plane wave. Consequently, the relative motions of uncharged test particles produced by the \"background factor\" L(u) are the same whether L(u) f:. 1 is produced by the stress-energy of an electromagnetic wave, or by a corresponding gravitational wave with (35.45) The analogy can be made even closer. Decrease the wavelength of the waves, while holding (f3')2/4'iT and (A'l/4'iTU fixed:.1--+0. In the limit of very small wavelength (Le., from a viewpoint whose characteristic length is ~ A), the two solutions are completely indistinguishable. Their metrics are identical (A --+ 0 and \u00ab13')2) = const. imply 13 --+ 0), and their jigglings of test particles are too small to be seen. Only their curving up of spacetime (L f:. 1) and the associated gravitational pull of their energy are detectable. \u00a735.12. A NEW VIEWPOINT ON THE EXACT PLANE WAVE The above comparison suggests a viewpoint that was sketched briefly in the intro duction to this chapter and in \u00a735.S. Think of the exact gravitational plane-wave solution [Figure 35.3; equations (35.29) and (35.33)] as ripples in the spacetime curvature, described by f3(u), propagating on a very slightly curved background spacetime, characterized by L(u). The most striking difference between the back ground and the ripples is not in the magnitude of their spacetime curvatures, but in their characteristic lengths. The ripples have characteristic length A= (typical reduced wavelength", ", A/2'iT, of waves); (35.46) the background has characteristic length (\"radius of curvature of background geometry\") Exact gravitational plane waves reexamined in the language of \"short-wave approximation\": (1) ripples vs. background \u00a735.12. NEW VIEWPOINT ON EXACT PLANE WAVE &l - IL/L\"11I2insidewave - 1/1,8'1- 963 (35.47) that...t is somewhat smaller than the pulse length, 2T. Recall also that Recall 1,8'TI ~ 1. Conclude that the characteristic lengths of the \"wave factor\" and the \"background factor\" differ greatly: (35.48) This difference in scales enables one to separate out the background from the ripples. The ripples are very much smaller in scale (...t ~ &l) than the background. Never theless the local curvature in a ripple is correspondingly larger than the background curvature [equations (35.30), (35.34)]; thus, (RZuzu\\ackground = (RIIUVU)background = - L\"/L - 1/&l2, (RZuzu)waves = _(Rllullu)waves = -,8\" -1,8'I/...t l/(...t&l) (35.49) - (&l/...t)(RZuzu\\ackground' One is reminded of the mottled surface of an orange! The metric for the background of the gravitational plane wave is the same as for the electromagnetic one [equation (35.40)]: ds 2 = g~.! dxIJ. dx v = U(dx 2 + dy 2) - du du. (35.50) By comparison with equation (35.29a), one sees that the metric for the full spacetime (background plus ripple) is ds 2 = (g(Bl + h ) dxP. dx v ___ _ p.v p.v, h\"\"\" = -hllll = 2,8, all other hp.v = O. (35.51) (35.52) (Recall, in the region where,8 f:. 0, L is very nearly 1.) One can think of the ripples as a transverse, traceless, symmetric tensor field hp.v analogous to the electromagnetic field AIJ.' propagating in the background geometry. Just as the electromagnetic field produces the background curvature via (2) propagation of ripples in background Guu = -2L\"/L = 8'iTTuu' so the gravitational-wave ripples hp.v produce the background curvature via equation (35.31), which one can rewrite as (35.53) Here T(EFFl - I uu = 4'iT (,8')2 _ I h h - 32'iT jk,u jk,u (35.54) (3) effective stress-ene rgy tensor for ripples is the \"effective stress-energy tensor\" for the gravitational waves. Notice that it agrees, except for averaging, with the expression (35.23) that was written down without justification in \u00a735.7. ~ 9_6_4 3_5_'_P_R_O_PA_G_A_T_IO_N_O_F_G_R_A_V_IT_A_T_IO_N_A_L_W._A_VE_S EXERCISE Exercise 35.10. PLANE WAVE WITH TWO POLARIZATIONS PRESENT - The exact plane-wave solution (35.29) has polarization e+. Construct a similar solution. containing two arbitrary amplitudes, f1(u) and y(u), for polarizations e+ and ex' Extend the discussions of \u00a7\u00a735.9-35.l2 to this solution. Foundations for shortwave formalism: (1) 0/1,.t, and {j' defined (2) demand that {j' ~ 1 and.t/tJl ~ 1 (3) split of metric into background plus pertur.bation; \"steady coordinates\" \u00a735.13. THE SHORTWAVE APPROXIMATION The remainder of this chapter extends the above viewpoint in a rigorous manner to very general gravitational-wave solutions. This extension is called the \"shortwave formalism\"; it was largely devised by Isaacson (l968a,b), though it was built On foundations laid by Wheeler (l964a) and by Brill and Hartle (1964). Versions that are even more rigorous have been given in the W.K.B. or geometric-optics limit by Choquet-Bruhat (1969), and by MacCallum and", " Taub (1973). Consider gravitational waves propagating through a vacuum background space time. As in \u00a735.7, let tJl be the typical radius of curvature of the background; let A and d be the typical reduced wavelength (.\\j2'iT) and amplitude of the waves; and demand both d ~ 1 and.1/(>il ~ 1. The background curvature might be due entirely to the waves, or partly to waves and partly to nearby matter and nongravi tational fields. The analysis uses a coordinate system closely \"tuned\" to spacetime in the sense that the metric coefficients can be split into \"background\" coefficients plus perturba tions with these properties: (I) the amplitude of the perturbation is d hp.v ~ (typical value of g~,!) \u2022 d; (35.56a) (2) the scale on which gj,B,! varies is ~ (>il (35.55) g(B) < (typical value of g(B\u00bb/(>il' p.v,Cl. __ J.l.P, (35.56b) (3) the scale on which hp.v varies is -A hp.v,o:. - (typical value of hp.v)/.1. (35.56c) (4) Split of Ricci curvature A rather long computation (exercise 35.11) shows that the Ricci tensor for an tensor expanded metric of the form (35.55) is Such coordinates are called \"steady.\"!LV R = R(B) + RW(h) + R(2)(h) + error \u2022.12!LV d 2/.12 d /.12 d 3!LV!LV? / (35.57) \u00a735.13. SHORTWAVE APPROXIMATION 965 Here a marker (d/.12, etc.) has been placed under each term to show its typical order of magnitude; R~; is the Ricci tensor for the background metric g~j; and R~~ and R~~ are expressions defined by R~~(h)=t(-hlltv - hltvlaa + haltlVa + havl/)' (35.58a) R~~(h)=; [thaPllthaPIV + haP(haPIW + hltvlaP - haltlvp - havlltP) + hvalP(haltlP - hpltla) - (h aP I P - ~ hla)(haItIV + havllt - hItVla )]. (35.58b) In these expressions and everywhere below, indices are raised and lowered with g~;, and an upright line denotes a covariant derivative with respect to g~; (whereas in Chapter 21 it denoted covariant derivative with respect to.3-geometry). At the heart of the shortwave formalism is its method for solving the vacuum field equations Rltv = O. One begins by selecting out of expression (35.57) the part linear in the amplitude of the wave d, and setting it equal to zero. The action of the waves to curve up the background is a nonlinear phenomenon (linearized theory shows no sign of it); so R~B; cannot be linear in d. Hence, in expression (35.57), R~l~(h) is the only linear term, and it must vanish by itself (35.59a) [Of course hltv may contain nonlinear correction terms-call themjltv-of order d 2, which must not be constrained by this linear equation. They will be determined by (35.59c), below.]. One next splits the remainder of Rw into a part that is free of ripples-Le., that varies only on scales far larger than A (\"coarse-grain viewpoint\"), and a second part that contains the fluctuations. This split can be accomplished by averaging over several wavelengths (see exercise 35.14 for a precise treatment of the averaging process, also see Choquet-Bruhat (1969) for a class of solutions where such averaging is not required): R<:; + <R~Xh\u00bb + error = 0 Ismoothl LPart J (35.59b)? jfluctuatingl J LPart (35.59c) Split of vacuum field equations into \"wave part\" (exil) plus \"coarse-grain part\" (ex (/2 and smooth on scale A') plus \"fluctuational corrections\" (ex il 2 and ripply on scale A') That's all there is to it!-except for reducing the equations to manageable form, and a fuller interpretation of the physics. Begin with the interpretation. Physical interpretation of the three parts of field equations: ~ (1) propagation of waves", " 966 35. PROPAGATION OF GRAVITATIONAL WAVES Equation (35.59a) is an equation for the propagation of the gravitational waves hili\" (2) production of background curvature by energy of waves; T~<;'W) defined Equation (35.59b) shows how the stress-energy in the waves creates the background curvature. It can be rewritten in the more suggestive form G(B) = R(B) _l.R(B)g(B) = 8~T(GW) in vacuum p.v - p.v,i p.v lU' 2 where, (35.60) (35.61) (3) nonlinear self-interaction of waves is the stress-energy tensor for the gravitational waves. Now one sees the origin of the statement in \u00a735.7, that the stress-energy of gravitational waves is well-defined only in a smeared-out sense. Finally, equation (35.59c) shows how the gravitational waves h generate nonlinear corrections j to themselves (wave-wave scattering, harmonics of the fundamental frequency, etc.). These higher-order effects will not be investigated in this chapter. EXERCISE Exercise 35.11. CONNECTION COEFFICIENTS AND CURVATURE TENSORS FOR A PERTURBED METRIC In a specific coordinate frame of an arbitrary spacetime, write the metric coefficients in covariant representation in the form (35.62a) (At the end of the calculation, one can split hp.v into two parts, hp.v - - hp.\u2022. + jp.v; and out of this split obtain the formulas used in the text.) Assume that the typical components of hp.v are much less than those of g~; so one can expand Christoffel symbols and curvature tensors in hp.,.' Raise and lower indices of hp.v with g~; and denote by a \"I\" covariant derivatives relative to g~J and by a \";\" covariant derivatives relative to gp.v' (a) Here gp.v and g~J can be thought of as two different metrics coexisting in the spacetime manifold. Show that the difference between the corresponding covariant derivatives, V V(Bl = S-indeed, the difference between any two covariant derivatives!-is a tensor with components [Hint: See part B of Box 10.3.) (b) Show that and also that :(35.62b) (35.62c) (35.62c') \u00a735.14. EFFECT OF BACKGROUND CURVATURE ON WAVE PROPAGATION 967 (c) By calculating in a local Lorentz frame of g~; and then transforming back to the original frame, show that Sp.IJY = 2: gp.a(halJ Iy + haYl1J - hlJyla ), I R a lJy8 - R(Bla sa sa Sp. p.8 RIJ8 - R(B)1J8 = saIJ81a - saIJal8 + sap.asp.1J8 - sa/l8Sp.lJa\u00b7 - sa 1J81y - lJy8 - 1J8 - lJyl8 + sa Sp. p.y lJy' (d) Show that expression (35.62f) reduces to RIJ8 = RlB)1J8 + R~Hh) + R~1(h) +... where RW and R(2) are defined by equations (35.58). (35.62d) (35.62e) (35.62f) (35.62g) \u00a735.14. EFFECT OF BACKGROUND CURVATURE ON WAVE PROPAGATION Focus attention on the propagation equation R~l~(h) = O. As in linearized theory, so also here, the propagation is studied more simply in terms of h- - I h (B) p.p= /lP-2' gp.P' h than in terms of hp.p. Rewritten in terms of hp. p, R~~(h) = 0 says -h p.Pla + /lP a g(B)h-a/3 J/3a - 2h a + 2R(B) M/ll p) a/l/3p h-a/3 - 2R(B) h- a - 0 M/l p) - (35.63) hp., defined (35.64). Propagation equati'on for waves on curved background [To obtain this, invert equation (35.63) obtaining h/lp = h/l p igj,B,!h; insert this into (35.", "58a) and equate to zero; then commute covariant derivatives using the identity (l6.6b); finally contract to obtain an expression for hlaa and substitute that back in.] The propagation equation (35.64) can be simplified by a special choice of gauge. An infinitesimal coordinate transformation Specialization to \"lorentz gauge\" (35.65a) induces a first-order change in the functional forms of the metric coefficients given by (35.65b) [analog of the gauge transformation of linearized theory, equation (35.3c); see exercise 35.12]. By an appropriate choice of the four functions ~/l, one can enforce the four \"Lorentz gauge conditions\" (35.66) Coupling of waves to Ricci tensor can be ignored Propagation equation in lorentz gauge and its realm of validity lists of effects absent from and contained in propagation equation 968 35. PROPAGATION OF GRAVITATIONAL WAVES in the new coordinate system (exercise 35.13). This choice of gauge is analogous to that oflinearized theory. It makes the second and third terms in the propagation equation vanish. (For additional gauge conditions of the \"TT\" type. see exercise 35.13.) Toe last term of the propagation equation. -2R~/p.hl,)a. vanishes to within the precision of the analysis, for this reason: attention has been confined to vacuum; so the only source of a nonvanishing Ricci tensor is the stress-energy carried by the gravitational waves themselves [equation (35.60)]; hence R~ - {f2/A2 and (35.67) This is of the same order as R~:'(h). the third-order correction to the Ricci tensor, which is far below the precision of the analysis. For consistency in the analysis it will therefore be neglected. Summary of this section thus far: by choosing a gauge where lip.ala = 0, and by discarding terms of higher order than the precision of the analysis, one obtains the. vacuum propagation equation h- p.vla a + 2R(Bl h-a{3 - 0 ap.{3v -, (35.68) subject to the Lorentz gauge condition la = 0. li p.a Equation (35.68) is accurate to first order in the amplitude [corrections r:x{f2 are embodied in equation (35.59c)]; and its accuracy is independent of the ratio A/qJ., as one sees from equations (35.59). Thus, it can be applied whenever the waves are weak, even if the wavelength is large! All nonlinear interactions of the wave with itself are neglected in this first-order propagation equation. Absent is the mechanism for waves to scatter off each other and off the background curvature that they themselves produce. Also absent are any hints of a change in shape of pulse due to self-interaction as a pulse of waves propagates. There are no signs of the gravitational collapse that one knows must occur when a mass-energy m of gravitational waves gets compressed into a region of size ~m. To see all these effects, one must turn to corrections of second order in {f and higher [e.g., equations (35.59c) and (35.60)]. Actually contained in the propagation equation are all effects due to the linear action of the background curvature on the propagating wave. These effects are explored, for short wavelengths (A/~ ~ 1) and for nearly flat wave fronts, in exercises 35.15-35.17 at the end of the chapter. The effects considered include a gravitational redshift of gravitational radiation and gravitational deflection of the direction of propagation of gravitational radiation, identical to those for light; and also a rotation of the polarization tensor. When the wavelength is not small (A/~ not ~ 1), the propagation equation includes a back-scatter of the gravitational waves off the background curvature and a resultant pattern of wave \"tails\" analogous to that explored in exercise 32.10 [see, e.g., Couch et al. (1968), Price (1971a), Bardeen and Press (1972), Unt and Keres (1972)]. \u00a735.15. STRESS-ENERGY TENSOR FOR GRAVITATIONAL WAVES 969 Exercise 35.12. GAUGE TRANSFORMATIONS IN A CURVED BACKGROUND (a) Show that (35.65b) in the functional form of the metric perturbation. the infinitesimal coordinate transformation (35.65a) induces the change (b) Discuss the relationship between this gauge transformation and the concept of a Killing EXERCISES vector (\u00a725.2). Exercise 35.13. TRANSVERSE-TRACELESS GAUGE FOR GRAVITATIONAL WAVES PROPAGATING", " IN A CURVED BACKGROUND (a) Show that, in vacuum in a curved background spacetime, the gauge condition h/ la = 0 is preserved by transformations whose generator satisfies the wave equation ~I'laa = O. (b) Locally (over distances much smaller than gz) linearized theory is applicable; so there exists such a transformation which makes [see equations (35.7b) and (35.8a\u00bb) h = 0 + error, (35.69) ltJ = 0); i.e., Here u a is a vector field that is as nearly covariantly constant as possible (u a it is a constant vector in the inertial coordinates of linearized theory; and the errors are small over distances much less than gz. Show that h = 0 can be imposed globally along with hJLala = 0; i.e., show that, if it is imposed on an initial hypersurface, the propagation equation (35.68) preserves it. (c) Show that in general, the background curvature prevents any vector field from being covariantly constant (Ulrl~ - ulr/tJl at best); and from this show that hl'aua = 0 cannot be imposed globally along with hI'ala = O. \u00a735.15. STRESS-EN.ERGY TENSOR FOR GRAVITATIONAL WAVES Turn now to an evaluation of the effective stress-energy tensor T~~W) of equation (35.61). The evaluation requires averaging various quantities over several wave lengths. Useful rules for manipulatirig quantities inside the averaging brackets ( ) are the following (see exercise 35.14 for justification). (I) Covariant derivatives commute; e.g., (h hp.vla/3) = (h hp.vl/3a)' The fractional errors made by freely commuting are _(A/~)2, well below the inaccuracy of the computation. (2) Gradients average out to zero; e.g., \u00abh lah!Lv)I/3) = O. Fractional errors made The averaging process involved in \"coarse-grain\" viewpoint here are ~ A/~. (3) As a corollary, one can freely integrate by parts, flipping derivatives from one h to the other; e.g., (h hp.vla/3) = (-h l/3hl'vla)' A straightforward but long calculation using these rules, using equation (35.58b) for R~~(h), using definition (35.63) of lil'v' using the propagation equation (35.64), and using the definition (35.61) of T~<tW), yields (R(2)(h\u00bb = 0, and Evaluation of effective stress-energy tensor for gravitational waves, JiI'<;'W) T (GW)!LV - 1 (-h 32'iT a/311' lia/3 1 hIv -\"'2 II' -h Iv - 2lia/3 li 1/3 a(l'lv) ). (35.70) This is the result quoted in equation (35.23'), except that there one used an inertial 970 35. PROPAGATION OF GRAVITATIONAL WAVES coordinate system of linearized theory, where covariant derivatives and ordinary derivatives are the same. In a gauge where h/L\"Ia = 0, the last term vanishes. When, in addition, h/LV is traceless (see exercise 35.13), the second term vanishes; and there remains only T(GWl - _l_<h /LV - 32'iT al3ll' h(13) Iv (35.70') Accuracy of expression for T~<;'W) Properties of Ji/L<;'W) These expressions for the effective stress-energy of a gravitational wave have fractional errors of order Cl, due to the neglect of second-order corrections to hp.v; they also have fractional errors of order Aj:\"il, due to the averaging process, which makes no sense when A approaches qz in magnitude. Since Cl ~ A/gz (35.28), the dominant errors in T~~Wl are -A/(>il. To this accuracy, the stress-energy tensor for gravitational waves is on an equal footing with any other stress-energy tensor. It plays the same role in producing background curvature; and it enters into conservation laws in the same way. For example, one can show, either by direct calculation or from the identity G(Bl/Lv lv = 0, that TGWl/LV - 0 + error, Iv - (35.71) where the error _(A/(>il)(T(GWlp.v/(>il) is negligible in the", " shortwave approximation. Some of the properties of T~~W) have already been explored in \u00a735.7. Further properties are explored in exercises 35.18 and 35.19. EXERCISES Exercise 35.14. BRILL-HARTLE AVERAGE Isaacson (l968b) introduces the following averaging scheme, which he names \"Brill-Hartle averaging.\" (a) In the small region, of size several times..t, where the averaging occurs, there will be a unique geodesic of g~ connecting any two points '3\" and '3'; so given a tensor E('3\") at '3\", one can parallel transport it along this geodesic to '3', getting there a tensor E('3\")_p. (b) Let 1('3\", '3') be a weighting function that falls smoothly to zero when '3\" and '3' are separated by many wavelengths, and such that (35.72) (c) Then the average of the tensor field E('3\") over several wavelengths about t~e point '3' is (35.73) (i) Show that there exists an entity g~)a'('3', '3\"), whose primed index transforms as a tensor at '3\" and whose unprimed index transforms as a tensor at '3', such that (for E second rank) (35.74) This entity is called the \"bivector of geodesic parallel displacement\"; see DeWitt and Brehme (1960) or Synge (1960a). \u00a735.15. STRESS-ENERGY TENSOR FOR GRAVITATIONAL WAVES 971 (ii) Rewryting expression (35.73) in coordinate language as (Eap(x\u00bb =f g~~ll/L'(x, x')glflv' (x, x')E/L'p,(x')f(x, x')v - g(B)(X') d 4x', (35.73') derive the three averaging rules cited at the beginning of\u00a735.l5. [For solution, see Appendix of Isaacson (l968b).) Exercise 35.15. GEOMETRIC OPTICS Develop geometric optics for gravitational waves of small amplitude propagating in a curved background. Pattern the analysis after geometric optics for electromagnetic waves (\u00a722.5). In particular, let h/Lp have an amplitude that varies slowly (on a scale L ~ fJl) and a phase (J that varies rapidly (8,a - 1/..t). Expand the amplitude in powers of..tIL, so that (35.75) Here ( is a formal expansion parameter, actually equal to unity, which reminds one that the terms attached to (\" are proportional to (..tlfJl)\". Define the following quantities (with A:v denoting the complex conjugate of A/Lv): \"wave vector\": k a=8,a \"scalar amplitude\": A =(t A:vA/LVY/2 (35.76a) (35.76b) (35.76c) By inserting expression (35.75) into the gauge condition (35.66) and the propagation equation (35.68), derive the fundamental equations of geometrical optics as follows. (a) The rays (curves perpendicular to surfaces of constant phase) are null geodesics; i.e. kak a = 0, ka1pk P = 0. (35.77a) (35.77b) (b) The polarization is orthogonal to the rays and is parallel transported along them; e/Laka = 0, e/Lvlaka = 0. (35.77c) (35.77d) (c) The scalar amplitude decreases as the rays diverge away from each other in accordance with i.e., (A 2k a)la =\u00b0(\"conservation of gravitons\"). (d) The correction B/Lv to the amplitude obeys B/Laka = iAJLala, B /Lvla = -'2 la 1 ka B k a l'A a /LV + '2/ /Lvla + / 'RIBl Aap a/Lpv. In accordance with exercise 35.13, specialize the gauge so that Ii = 0, i.e., (35.77e) (35.77f) (35.77g) (35.77h) (35.77i) 972 35. PROPAGATION OF GRAVITATIONAL WAVES Then show that the stress-energy tensor (35.70') for the waves is T(GWl _ p.v - 32\" 1.Azk k v\u00b7 (35.77j) p. This has the same form", " as the stress-energy tensor for a beam of particles with zero rest mass (see \u00a75.4). Show explicitly that T(GWlp.v l,. = O. Exercise 35.16. GRAVITONS Show that geometric optics, as developed in the preceding exercise, is equivalent to the following: \"A graviton is postulated to be a particle of zero rest mass and 4-momentum p, which moves along a null geodesic (Vpi' = 0). It parallel transports with iIself (Vpe = 0) a transverse (e. p = 0) traceless (ea a = 0) polarization tensor e. Geometric optics is the theory of a stream of such gravitons moving through spacetime.\" Exhibit the relationship between the quantities in this version of geometric optics and the quantities in the preceding version (e.g., p = nk, where n is Planck's reduced constant h/27T). Exercise 35.17. GRAVITATIONAL DEFLECTION OF GRAVITATIONAL WAVES Show that gravitational waves of short wavelength passing through the solar system experi ence the same redshift and gravitational deflection as does light. (One should be able to infer this directly from exercise 35.15.) Exercise 35.18. GAUGE INVARIANCE OF T'p.<;'W) Show that the stress-energy tensor T<;\"W) of equation (35.70) is invariant under gauge trans formations of the form (35.65). Exercise 35.19. T'2W ) EXPRESSED AS THE AVERAGE OF A STRESS-ENERGY PSEUDOTENSOR Calculate the average over several wavelengths of the Landau-Lifshitz stress-energy pseu dotensor [equation (20.22)] for gravitational waves with A/(-il ~ 1. The result should be equal to T~<;'W). [Hint: Work in a gauge where h/ 1a = h = 0, to simplify the calculation.] Exercise 35.20. SHORTWAVE APPROXIMATION FROM A VARIATIONAL VIEWPOINT Readers who have studied the variational approach to gravitation theory in Chapter 21 may - find attractive the following derivation of the basic equations of the shortwave approximation. It was devised, independently, by Sandor Kovacs and Bernard Schutz, and by Bryce DeWitt (unpublished, 1971). MacCallum and Taub (1973) give a \"non-Palatini\" version. (a) Define.(35.78a) (35.78b) Raise and lower indices on hp.v and wp. py with the background metric. Using the results of exercise 35.11, derive the following expression for the Lagrangian of the gravitational field: \u00a3 = _1_( _ )1/2R = \u00a3' g 167T - (perfect divergence) + (corrections of order + of form o::za/cxa {/3/A2, R<:j{/, and smaller ), (35.78c) \u00a735.15. STRESS-ENERGY TENSOR FOR GRAVITATIONAL WAVES 973 where \u00a3,=_I_(_gUll)1/2[R<Bl _ hl'V(Wa - 16'1T I'vla _ Wa ) /La Iv [Hint: recall that + glBll'v(wapa WP I'v - W a pv w P/La\u00bb)' (35.78d) for any Ba.) Drop the corrections of order {/3j;t2 from E; and, knowing in advance that the field equations will demand R~.! - (/2j;t2, drop also the corrections of order Rr;!{/. Knowing that a perfect divergence contributes nothing in an extremization calculation, drop the divergence term from \u00a3. Then \u00a3' is the only remaining part of E. (b) Extremize 1=j\u00a3' d4x by the Palatini method (\u00a721.2); i.e., abandon (temporarily) definition (35.78b) of WI'PY' and extremize 1 with respect to independent variations of Wl'py = Wl'yp, hl'v = hVl', and giRl = grltl' Show that extremization with respect to Wl'py lea~s back to equation (35.78b) for Wl'py in terms of hl'v' Show that extremization with respect to hl'v, when combined with equations (35.78a,b), leads to the propagation equation for gravitational waves (35.64). Show that extremization with respect to g<Bll'v, when combined with equations (35.78a", ",b) and with the propagation equation (35.64), and when averaged over several wavelengths, leads to where T~<;'Wl is given by equation (35.70). [Warning: The amount of algebra in this exercise lp = 0 from the is enormous, unless one chooses to impose the gauge conditions h = ha outset.) P CHAPTER 36 GENERATION OF GRAVITATIONAL WAVES Matter is represented by curvature, but not every curvature does represent matter; there may be curvature \"in vacuo:' G. lEMAITRE in Schilpp (1949). p. 440 Generation of gravitational waves analyzed by electromagnetic analog \u00a736.1. THE QUADRUPOLE NATURE OF GRAVITATIONAL WAVES Masses in an isolated, nearly Newtonian system move about each other. How much gravitational radiation do they 'emit? For an order-of-magnitu'de estimate, one can apply the familiar radiation formulas of electromagnetic theory, with the replacement e2 -+ _m 2, which converts the static coulomb law into Newton's law of attraction. This procedure treats gravity as though it were a spin-one (vector) field, rather than a spin-two (tensor) field; consequently, it introduces moderate errors in numerical factors and changes angular distributions. But it gives an adequate estimate of the total power radiated. In electromagnetic theory, electric-dipole radiation dominates, with a power output or \"luminosity,\" L, given (see \u00a74.4 and Figure 4.6) by Lelectric dipole = (213 )e2a2 for a single particle with acceleration a and dipole moment changing as d = eX = ea; Lelectricdipole = (2/3) d2 for a general system with dipole moment d. [Geometric units: luminosity in cm of mass-energy per cm of light travel time; charge in em, e = (Gl/2Ic2)econv = (2.87 X 10- 25 cm/esu) X (4.8 X 10-10 esu) = 1.38 X 10-34 cm, acceleration in cm of distance per cm of time per em of time. For conventional units, with e in esu or (g cm3/sec2)l/2, \u00a736.1. THE QUADRUPOLE NATURE OF GRAVITATIONAL WAVES 975 insert a factor c-3 on the right and get L in erg/sec]. The gravitational analog of the electric dipole moment is the mass dipole moment Why gravitational waves cannot be dipolar d= L mAxA particles A Its first time-rate of change is the total momentum of the system, d= L mAxA =p. particles A \u2022 The second time-rate of change of the mass dipole moment has to vanish because of the law of conservation of momentum, d =Ii = O. Therefore there can be no mass dipole radiation in gravitation physics. The next strongest types of electromagnetic radiation are magnetic-dipole and electric-quadrupole. Magnetic-dipole radiation is generated by the second time derivative of the magnetic moment, ii. Here again the gravitational analog is a constant of the motion, the angular momentum, II- = L (position of A) X (current due to A) = L rA X (mvA ) = J; A A so it cannot radiate. Thus, there can be no gravitational dipole radiation of any sort. When one turns to qua_drupole radiation, one finally gets a nonzero result (see Figure 36.1). The power output predicted by electromagnetic theory, Leleetricqua<!ruPole - _ 1 \"'2 _ 1 20 Q = 20 QjkQjk'. (Qjkhere = QjkinmUchotherliterature)' has as its gravitational counterpart (36.1 ) (36.2) Gravitational-wave power output expressed in terms of \"reduced quadrupole moment\" of source Formula (36.1) contains the correct factor of 1/5, which comes from tensor calcula tions (see \u00a736.10), instead of the incorrect factor 1;20 suggested by the electromag netic analog; and the righthand side of (36.1) has been averaged (\"( )\") over several characteristic periods of the source to accord with one's inability to localize the energy of gravitational radiation inside a wavelength. Receptor. r ---------J ea sinB r Source )!e.,:~1~~ ):e.,:::~~----- )fm (a) (b) (c) F~il----~lml (d) A \"~~----\u00ad mofi'ao.. \\ \\ \\ / /,", " / \" \\ (e),. / '~,,,''eml I' I \" I \\ __ _._----..1.-m~ 2 \\, L cos B = difference in time of travel to receptor; gives rise to difference in phase 8(J = wLcosB Conclusion: (amplitude)net = (amplitude)lor2 8(J mlal=--l)B r (wLcos B ) - \" mw2LSinB) ( r w3mUsin2B r - fir (luminosity) - (/)2 Figure 36.1. Why gravitational radiation is ordinarily weak. In brief, contributions to the amplitude of the outgoing wave from the mass dipole moments of the separate masses cancel, (mlal + m2a2)/r = 0 (principle that action equals reaction). (a) Radiation from an accelerated charge (see \u00a74.4 and Figure 4.6). (b) Representation of the field at the great distance r in terms of the typical rotating-vector diagram of electrical engineering; however, here, for ease of visualization, the vertical projection of the rotating vector gives the observed field (usual dipole-radiation field produced by a charge in circular orbit). (c) Corresponding rotating-vector diagram for gravitational radiation, based on the simplified model of the gravitation field as a spin-one or vector field (to be contrasted with its true tensor charader; hence details of angular distribution and total radiation as given by this simple diagram are not correct; but order of magnitude of luminosity is correct). (d) The two masses ml and m 2 that hold each other in orbit give equal and opposite contributions to the amplitude of the outgoing wave because of the principle that action equals reaction. (In electro magnetic radiation from a hydrogen atom, the corresponding radiation amplitudes do not cancel: eeleeaelec + eprotaprot - eelecaelec :j. 0). (e) In a better approximation, one has to allow for the difference in time of arrival at the receptor of the effects from the two masses. The two vectors that formerly opposed each other exactly are now drawn inclined, at the phase angle M. The amplitude of the resulting field goes as T, where f is the reduced quadrupole moment; and the luminosity is proportional to 12\u2022 \u00a736.1. THE QUADRUPOLE NATURE OF GRAVITATIONAL WAVES 977 Notation: There is no ambiguity about the definition of the \"second moment of the mass distribution\" as it appears throughout the physics and mathematics literature Nor is there any ambiguity about how one constructs the moment of inertia tensor 1 jk from this second moment of the mass distribution: The moments that characterize a source radiating quadrupole gravitational radiation to be the \"tracejree part of the second are here taken, equally unambiguously, moment of the mass distribution\": This notation is adopted because it simplifies formulas, it simplifies calculations, it meshes well with much of the literature of gravitational-wave theory [e.g. Peters (1964), Peres and Rosen (1964)], and it is easy to remember. Another name for the quantities f jk is reduced quadrupole moment. This terminology makes clear the distinction between the quantities used here and the three-times-larger quantities that are called quadrupole moments in the standard text of Landau and Lifshitz (1962) and in the literature on nuclear quadrupole moments, and the 3/2-times-larger quantities used in the th_eoryof spherical harmonics: Q (LandaU and Lifshitz; also zz nuclear quadrupole moments ) = Jp(3z2 _r2) d 3x, Qzz (theory of spherical harmonics) =Jp(fZ2- ~r2) d 3x, f zz ( reduced quadrupole moment;) unambiguous measure of source strength adopted here = Jp(Z2 - tr2)d3x. Thus the f jk notation has the merit of circumventing the existing ambiguity in the literature. That electromagnetic radiation is predominantly dipolar (spherical-harmonic index 1= 1), and gravitational radiation is quadrupolar (I = 2) are consequences of a general theorem. Consider a classical radiation field, whose associated quantum mechanical particles have integer spin S, and zero rest mass. Resolve that radiation field into spherical harmonics-i.e., into multipole moments. All components with I < S will vanish; in general those with I ~ S will not; and this is independent of the nature of the source! [See, e.g., Couch and Newman (1972).] Since the lowest nonvanishing multipoles generally dominate for a slowly moving source (speeds ~ c), electromagnetic radiation (S = 1) is ordinarily dipolar (I =", " S = 1), while g~avita- Why gravitational waves are ordinarily quadrupolar 978 36. GENERATION OF GRAVITATIONAL WAVES tional radiation (S = 2) is ordinarily quadrupolar (I = S = 2). Closely connected theorem\" [e.g., Lifshitz (1949)], with this theorem is the \"topological fixed-point which distinguishes between scalar, vector, and tensor fields. For a scalar disturbance, such as a pressure wave, there is no difficulty in having a spherically symmetric source. Thus, over a sphere of a great radius r, there is no difficulty in having a pressure field that everywhere, at anyone time, takes on the same value p. In contrast, there is no way to lay down on the surface of a 2-sphere a continuous vector field, the magnitude of which is non-zero and everywhere the same (\"no way to comb smooth the hair on the surface of a billiard ball\"). Likewise, there is no way to lay down on the surface of a 2-sphere a continuous non-zero transverse-traceless 2 X 2 matrix field that differs from one point to another at most by a rotation. Topology thus excludes the possibility of any spherically symmetric source of gravitational radiation whatsoever. \u00a736.2. POWER RADIATED IN TERMS OF INTERNAL POWER FLOW Expression (36.1) for the power output can be rewritten in a form that is easier to use in order-of-magnitude estimates. Notice that the reduced quadrupole moment is i. __ ( )k mass of that part Of) system which moves X system (time for masses to move from)3 (Size Of)2 one side of system to other M(R/T)2 T i. __ L )k mternal ( nonspherical part) of kinetic energy T (power flowing from one): side of system to other (36.4) Gravitationa I-wave power output in terms of internal power flow of source Consequently, equation (36.1) says that the power output in gravitational waves (\"luminosity\") is roughly the square of the internal power flow (36.5) If this equation seems crazy (who but a fool would equate a power to the square of a power?), recall that in geometrized units power is dimensionless. The conversion factor to conventional units is One may freely insert this factor of La = 1 wherever one wishes in order to feel more comfortable with the appearance of the equations. For example, one can rewrite equation (36.5) in the form (36.7) (36.6) \u00a736.3. LABORATORY GENERATORS OF GRAVITATIONAL WAVES 979 In applying the equation L GW,;.,.. (Linternal)2, one must be careful to ignore those internal power flows that cannot radiate at all, i.e., those that do not accompany a time-changing quadrupole moment. For example, in a star the internal power flows associated with spherical pulsation and axially symmetric rotation must be ignored. Conservation of energy guarantees that radiation reaction forces will pull down the internal energy of the system at the same rate as gravitational waves carry energy away (see Box 19.1). The characteristic time-scale for radiation reaction to change the system markedly is 'Treact -- [l/(rate at which energy is lost)] X [energy in motions that radiate] -- [1/L Gw] X [(Linternal) X (characteristic period T of internal motions)] -- (Linternal/LGw)T -- (Lo/ Linternal)T (36.8) Consequently, radiation reaction is important in one characteristic period only if the system achieves the enormous internal power flow Linternal ~ La = 3.63 X 1059 ergs/sec = I! \u00a736.3. LABORATORY GENERATORS OF GRAVITATIONAL WAVES As a laboratory generator of gravitational waves, consider a massive steel beam of radius r = 1 meter, length 1= 20 meters. density p = 7.8 g/cm3, mass M = 4.9 X 108 g (490 tons), and tensile strength t = 40,000 pounds per square inch or 3 X 109 dyne/cm2\u2022 Let the beam rotate about its middle (so it rotates end over end), with an angular velocity w limited by the balance between centrifugal force and tensile strength w = (8t/pI2)l/2 = 28 radians/sec. The internal power flow is Characteristic time-scale for radiation-reaction effects Power output from a rotating steel beam - (1. T 2) __1_ MI2 3 2 ~w w - 24 w internal - L :::::: 2 X 1018 erg/sec:::::: 10-41La' The", " order of magnitude of the power radiated is LGW -- (l0-41)2La -- 10-23 erg/sec. (36.9) (An exact calculation using equation (36.1) gives 2.2 X 10-22 erg/sec; see Exercise 36.1.) Evidently the construction of a laboratory generator of gravitational radiation ----------~Ism uilaftractive enterprise in the absence of new engineering or a new idea or both. To rely on an astrophysical source and to build a laboratory or solar-system detector is a more natural policy to consider. Detection will be discussed in the next chapter. Here attention focuses on astrophysical sources. 980 36. GENERATION OF GRAVITATIONAL WAVES EXERCISE Exercise 36.1. GRAVITATIONAL WAVES FROM ROTATING BEAM A long steel beam of length I and mass M rotates end over end with angular velocity\"'. Show that the power it radiates as gravitational waves is Use this formula to verify that the rod described in the text radiates 2.2 X 10-22 ergs/sec. (36.10) \u00a736.4. ASTROPHYSICAL SOURCES OF GRAVITATIONAL WAVES; GENERAL DISCUSSION Consider a highly dynamic astrophysical system (a star pulsating and rotating wildly, or a collapsing star, or an exploding star, or a chaotic system of many stars). If its mass is M and its size is R, then according to the virial theorem (exercise 39.6) its kinetic energy is --M2/R. The characteristic time-scale for mass to move from one side of the system to the other, T, is T -- (mean velocity -- (M/R)l/2 = M R (R3)1I2 R (36.11 a) (-- time of free fall; -- time to tum one radian in Kepler orbit; see Chapter 25). Consequently, the internal power flow is. Lmternal __ (kinetic energy) __ (M2) (M)1/2 __ (M)5/2 R R3 R T (36.11 b) Power output from violent astrophysical sources, in terms of mass and radius Upper limit on power output The gravitational-wave output or \"luminosity\" is the square of this quantity, or (36.llc) (If the system is rather symmetric, or if only a small portion of its mass is in motion, then its quadrupole moment does not change much, and the estimate of L GW must be reduced accordingly. The wave amplitude goes down in proportion to the fraction of the mass in motion, and the power is reduced in proportion to the square of that fraction.) Clearly, the maximum power output occurs when the system is near its gravitational radius; and because nothing, not even gravitational waves, can escape from inside the gravitational radius, the maximum value of the output is -- L o = 3.63x 1059 ergs/second, regardless of the nature of the system! Actually, the above derivation of this limit and of equation (36.11 c) uses approxi mations to general relativity that break down near the gravitational radius. [Velocities small compared to light are required in deriving the standard formula (36.1) for \u00a736.5. GENERATION BY COLLAPSE, BLACK HOLES, SUPERNOVAE, PULSARS 981 L GW (see \u00a736.7); nearly Newtonian fields are required for the virial theorem argu ments of (36.l1a), as well as for the L GW formula.] Nevertheless, in rough order of magnitude, equation (36.l1c) is valid to quite near the Schwarzschild'radius, say, R - 3M; and inside that point gravity is so strong that no system can resist collapse for an effective length of time much longer than T - M. The time required for radiation-reaction forces to affect a system substantially [equation (36.8)] is of the order (36.l1d) where T is the characteristic time (36.l1a) of rotation or free fall. (Note how one inserts and removes the factor L o = 1 at will!) Consequently, the effect of radiation reaction, as integrated over one period, is unimportant except when the system is near its gravitational radius. When a system such as a pulsating star is settling down into an equilibrium state, the radiation reaction will damp its internal motions. On the other hand, when the system, like a binary star system, is far from any state of equilibrium, then loss of energy (and angular momentum) to radiation under certain circumstances may speed up the angular velocity or speed up the internal motions and augment the radiation. \u00a736.5. GRAVITATIONAL COLLAPSE, BLACK HOLES, SUPERNOVAE,", " AND PULSARS AS SOURCES (Mj R)5Lo' the most intense gravitational waves reaching Earth must Since L GW come from a dynamic, deformed system near its gravitational radius (LGW drops by a factor 100,000 with each increase by 10 of R!). The scenario of Figure 24.3 gives an impression of some of the dynamic processes that not only may happen but probably must happen. The sequence ofevents sketched out there includes pulses of gravitational radiation interspersed with intervals of continuous radiation of gradually increasing frequency. Pulse number one comes at the time of the original collapse of a star with white-dwarf core to a pancake-shaped neutron star. The details of what then goes on will differ enormously depending on the original mass and angular momentum of this \"pancake.\" In the illustration, this pancake fragments into a constellation of corevolving neutron stars, which then one by one undergo \"pursuit and plunge.\" Whether in this kind of scenario or otherWise, perhaps the most favorable source of gravitational radiation is a star (the original very temporary \"pancake\" or one of the fragments therefrom) collapsing through its gravitational radius in a highly nonspherical manner. Such a star should terminate life with a last blast of gravita tional waves, which carry off a sizeable fraction of its rest mass. Thus an order-of magnitude estimate gives Radiation reaction in astrophysical SOurces Gravitational waves from: (1) stellar collapse and formation of a black hole (energy radIated) = L GW dt - L o \u2022. J (time during which k I.. ummoslty occurs pea ) (36.12) -LoM= M. (2) the fall of debris into a black hole 982 36. GENERATION OF GRAVITATIONAL WAVES (Whether the energy radiated is 0.9M, or O.IM, or O.OIM is not known for certain today; but it must lie in this range of orders of magnitude.) The radiation should be weak at low frequencies; it should rise to a peak at a frequency a little smaller than I!M; and it should cut off sharply for circular frequencies above w -- 1/M. Matter (\"debris\"; see Figure 24.3) falling into a black hole can also be a significant source of gravitational waves. The infalling matter will radiate only weakly when it is far from the gravitational radius; but as it falls through the gravitational radius (between r -- 4M and r = 2M), it should emit a strong burst. If m is the mass of an infalling lump of matter and M is the total mass of the black hole, then the total energy in the final burst is (36.13) and it comes off in a time --M with a power output of L GW -- (m/ M)2 La' (See exercise 36.2.) Actually, this is an extremely rough estimate of the energy output. In the limit where the infalling lump is small in both size and mass [(size oflump) ~ (gravitational radius of black hole); m ~ M; \"delta-function lump\"], one can perform an exact calculation of the spectrum and energy radiated by treating the lump and the waves as small perturbations on the Schwarzschild geometry of the black hole. The foundations for such a treatment were given by Zerilli (1970b). Zerilli's formula was corrected and applied to the case of head-on impact by Davis, Ruffini, Press, and Price (1971). They predict the spectrum of Figure 36.2 and the total energy output Eradiated = 0.0104m 2/M (36.14) (3) collisions of black holes A collision between black holes should also produce a strong burst of gravitational for m ~ M and (size of lump) ~ M. (4) supernova explosions waves-through such collisions are probably very rare! Not quite so rare, but still not common, are supernova explosions (about one per galaxy per 100 years). According to current theory as verified by pulsar observations, a supernova is triggered by the collapse of the core of a highly evolved star (see \u00a724.3). The collapse itself and the subsequent wild gyrations of the collapsed core (neutron star) should produce a short, powerful burst of gravitational waves. The characteristics of the burst, as estimated with formulas (36.11), and assuming large departures from sphericity, are (energy radiated) -- (neutron-star binding energy) -- M2/R -- O.1M -- 1053 ergs, (mean frequency) -- liT -- (M/ R3)1/2 -- 0.03M-l -- 3000 Hz, (36.15) (power output) -- (M/ R)5La -- 10-5La -- 3 X 1054 ergs/sec,", " time for gravitational radiation to damp the motion if turbulence, heat conduction, and other effects do not damp it sooner =: 7' -- M(M/ R)-4 -- 0.1 sec -- 300 periods. \u00a736.5. GENERATION BY COLLAPSE, BLACK HOLES, SUPERNOVAE, PULSARS 983 10-1 10-2 1=2 (quadrupole) ~ 10-3 '\":: \u00b7c :=.5 ~ 10-4 \"-~ 10-5 o 0.2 0.4 0.6 0.8 1.0 1.2 1.4 w, in units 1/ M Figure 36.2. Spectrum of the gravitational waves emitted by a \"delta-function\" lump of matter of mass m, falling head-on into a nonrotating (Schwarzschild) black hole of mass M ~ m. The total energy radiated is distributed among multipoles according to the empirical law (energy in I-pole waves)::::: (0.44 m2/M)e-21, and the total spectrum peaks at angular frequency w max = 0.32/M. These results were calculated by treating the infalling lump and the gravitational waves as small perturbations on the Schwarzschild geometry of the black hole. The relevant perturbation-theory equations were derived by Zerilli (1970), and were solved numerically to give these results by Davis, Ruffini, Press, and Price (1971). In the last stages of the stellar pulsations, when the amplitude ~ = Sr has dropped to Sr/r ~ 1, one can calculate the pulsation frequencies and damping times exactly by treating the fluid motions and gravitational waves as small perturbations of an equilibrium stellar model. The results of such a calculation, which are in good agreement with the above rough estimates, are shown in Box 36.1. Long after the pulsations of the neutron star have been damped out by gravita tional radiation reaction and by other forces, the star will continue to rotate; and as it rotates, carrying along with its rotation an off-axis-pointing magnetic moment, it will beam out the radio waves, light, and x-rays that astronomers identify as \"pulsar radiation.\" In this pulsar phase, gravitational radiation is important only if the star is somewhat deformed from axial symmetry (axial symmetry ==>- constant quadru pole moment ==>- no gravitational waves). According to estimates in exercise 36.3, a deformation that contains only 1).001 of the star's mass could radiate 1038 ergs per secon~!JQLthey'OungeSt-known pulsar (Crab nebula); and the accompanying radiaTion reaction could be a significant source of the pulsar's slowdown. However, it is not at all clear today (1973)-indeed, it seems unlikely-that the neutron star could support even so small a deformation. (continued on page 986) (5) young pulsars Box 36.1 GRAVITATIONAL WAVES FROM PULSATING NEUTRON STARS The table given here, taken from Thorne (I 969a), shows vari ous characteristics of the quadrupole oscillations of several typical neutron-star models. Note that the gravitational waves emitted by the most massive models (I) have frequencies p = I/Tn - 3,000 Hz, (2) last for a time of -~ second, (3) damp out the stellar vibrations after only -1,000 oscillations, and (4) carry off a total energy of _(1054 ergs) X (15R/ R)2, where 15R/R is the initial fractional am plitudc of vibration of the star's surface. These results are not based on the nearly Newtonian slow- motion formalism of this chapter [equation (36.1), \u00a7\u00a736.7 and 36.8], because that formalism is invalid here: the reduced wave length of the radiation, A 15 km for waves from the most massive star, is not large compared to the star's gravitational radius, 2M - 6 km; and the star's internal gravitational field is not weak (M/ R a'i large as 0.29). Consequently, these results were derived using an alternative technique, which is valid for rapid motions and strong internal ficlds, but which assumes small perturbations away from the equilibrium stcllar model. See Thorne (1969a) and papers cited therein for details. QUADRUPOLE PULSATIONS OF NEUTRON STARS Equation of state H-W H-W Vy Vy Pc (g cm-J ) M/Mo 3 X 1014 0.405 6 X 1015 0.682 2M/R 0.0574 0.240 5.15 X 1014 0.677", " 0.159 3 X 1015 1.954 0.580 Tn (msec) 1.197 0.3109 0.1713 0.1179 0.0938 0.6991 0.2358 0.3777 0.1556 0.1026 'Tn (sec) 13. 0.19 0.28 1.3 24. 1.7 II. 0.22 1.6 2.6 'Tn/Tn 11000 610 1600 11000 25\u05e1\u05e1oo 2400 47000 600 1\u05e1\u05e1oo 25000 Elf \u00abfJR/R)2) (ergs) Power- - \u00ab(FJR/R)2) (ergs sec-I) 7.8 X 10,,(1 1.2 X 10',(1 2.8 X 10\"2 3.6 X 10\"1 2.6 X 10\"\" 8.9 X 1048 - 5.7 X 10,\u00b72 6.0 X 10.'>0 1.7 X 10M 1.5 X 1054 5.2 X 10,\u00b71 2.9 X 10,\u00b7:1 2.6 X 10\"2 3.9 X 10511 7. X 1047 7. X 10,\u00b72 1.1 X 105(1 1.6 X 10\"5 1.9 X 1054 4.0 X 1O~.1 (flR/R) (flr/rl c + 7.4 + 5.2 14. + 55. -350. + 1.4 38. - + 1.9 2.1 + 2.9 flO./flO, + 3.1 + 3.7 3.3 + 5.9 -24. + 1.3 4.7 - + 3.1 0.66 + 0.40 The columns in the table have the following meanings. Equation of state: the equation of state pep) used in con structing the equilibrium stellar model and in calculating the adiabatic index from y = [(p + p)/p] dp/dp; H-W is the Harrison-Wheeler equation of state in the tabular form given by Hartle and Thorne (1968), Table I; Vy is the Levinger Simmons-Tsuruta-Cameron Vy equation of state in the tabu lar form given by Hartle and Thorne (1968), Table 2.. Pc: central density of total mass-energy for the equilibrium stellar model. M/ Mo: total mass-energy of the equilibrium model (i.e., the mass that governs distant Keplerian orbits), in units of the sun's mass. 2M/ R = 2GM/ Rc 2 : ratio of the gravitational radius of the equilibrium model to its actual radius (radii are defined by 4'lTR2 = surface area). n: the \"order\" of the pulsational normal mode under study (for all models given here, n is also the number of nodes in the radial relative eigenfunction, 8r/r.). Note: n = 0 is the fundamental (quadrupole) mode. [985] Tn = 2'lT/wn : the pulsation period of the quasinormal mode measured in milliseconds. Tn: the damping time for the amplitude of the normal mode measured in seconds. Tn/Tn = wnTn/2'lT: the number of pulsation periods required for the amplitude to drop by a factor of I/e. E'!Jd\u00abliR/R)2): energy of pulsation of the star, divided by the square of the relative amplitude of radial motion of the star's surface averaged over its surface. Power/ \u00aboR/R)2): radiated as gravitational waves, divided by the averaged square of the relative ampli tude at the star's surface. the power (oR/R)(or/r)e- 1 : relative amplitude of radial motion at the star's surface divided by relative amplitude at the star's center. 8o% oe = &t>%<pe : amplitude of the angular displacement of the star's fluid at its surface divided by the same amplitude at its center. ~ 986 36. GENERATION OF GRAVITATIONAL WAVES Of the sources discussed in this section, most are \"impulsive\" rather than continu ous (star collapsing through gravitational radius; debris falling into a black hole; collision between black holes; supernova explosion). They give rise to bursts of gravitational waves. An order-of-magnitude method of analyzing such bursts is spelled out in Box 36.2. It is difficult and risky to pass from the above description of processes that should generate gravitational waves to an estimate of the characteristics of the waves that actually bathe the earth. For such an estimate, made in 1972 and subject to extensive revision as one's understanding", " of the universe improves, see Press and Thorne (1972). EXERCISES Exercise 36.2. GRAVITATIONAL WAVES FROM MATTER FALLING INTO A BLACK HOLE A lump of matter with mass m falls into a black hole of mass M. Show that a burst of gravitational waves is emitted with duration -M and power L GW (m/M)2Lo' so that the total energy radiated is given in crude order of magnitude by equation (36.13). Exercise 36.3. GRAVITATIONAL WAVES FROM A VIBRATING NEUTRON STAR Idealize a neutron star as a sphere of incompressible fluid of mass M and radius R, with structure governed by Newton's laws of gravity. Let the star pulsate in its fundamental quadrupole mode. Using Newtonian theory, calculate: the angular frequency of pulsation, the energy of pulsation Epu1s ; the quantity g( j 2), which, according to equation (36.1), w; is the power radiated in gravitational waves, L GW ; and the e-folding time, 'T = E pu1s / L GW' for radiating away the energy of the pulsations. Compare the answers with equations (36.15) which are based on a much cruder approximation-and with the results in Box 36.1, which are based on much better approximations. [For solution, see Table 13 of Wheeler (1966).] Exercise 36.4. PULSAR SLOWDOWN The pulsar NP0532 in the Crab Nebula has a period of 0.033 seconds and is slowing down the rate dP/ dt = 1.35 X 10-5 sec/yr. Assuming the pulsar is a typical neutron star, at calculate the rate at which it is losing rotational energy. If this energy loss is due primarily to gravitational radiation reaction, what is the magnitude of the star's nonaxial deformation? [For solution, see Ferrari and Ruffini (1969); for a rigorous strong-field analysis, see Ipser (1970).] \u00a736.6. BINARY STARS AS SOURCES Binary stars as sources of gravitational waves: The most numerous sources of weak gravitational waves are binary star systems. Moreover, roughly half of all stars are in binary or multiple systems [see, for example, the compilation of Allen (1962)]. According to Kepler's laws, two stars of masses m1 and m2 that circle each other have angular frequency w and separation a coupled to each other by the formula -------~ Box 36.2 ANALYSIS OF BURSTS OF RADIATION FROM IMPULSE EVENTS' Typical moment relevant for radiation Its Fourier transform Name for this quantity Time decomposition of total radiative energy loss..:1E Decomposition of..:1E according to circular frequency Integrand nearly constant with respect to w from w = 0 up to a critical value of w, beyond which radiation falls off very fast -d..:1E/dw for w < werit Electromagnetism Gravitation fzz(t) (2'1T)-1/2f d, exp [iwt] dt (2'1T)-1/2ff zz exp [iwt] dt werit - 1/..:1t Zero frequency moment that enters this formula..:1(\u00abKinetic Energy\u00bb)zz Rewrite of -d..:1E/dw Total energy of pulse \u2022 Box adapted from pp. 113 and 114 of Wheeler (1962). -This/..:1t -This/..:1t As sample applications of this analysis, Wheeler (1962) cites the following: Parameter Mass Velocity Energy Fraction assumed relevant to radiative moment Time integral of this moment = \u00abK\u00b7E.\u00bbzz \u00abK\u00b7E.\u00bbzz/c2 dE _~(<<K.E,\u00bbzzr dw..:11..:1w..:1Eradtated Assumed distance 1/..:11 c2 C to detector..:1E/4'1Tr2 One atomic-nucleus fission of 180 MeV Fission bomb yield 1 7 kilotons at 10% efficiency Meteorite striking earth at escape velocity Explosion pf star when 10-4 of mass is released 4 X 10-22 g 1.2 X 109 cm/s 2.9 X 10-4 erg 2.9 X 10-4 erg 3.2 X 10-25 g 2.3 X 1O-\u00a37~ rad/s 10-21 s 1021 rad/s 10-46 erg 103 cm 10-53 erg!cm2 104 g 4 X 108 cm/s 7 X 1020 erg 0.1 109 g II X 105 cm / s 6 X 1020 erg 7 X 1019 erg 6 X 1020 erg 0.08 g 1", ".4 X 1O-2O~ rad/s 10-8 S 108 radjs 10-12 erg 0.67 g 1.0 X 1O-18~ rad/s 10-3 s 103 rad/s 10-15 erg 103 cm 10-19 erg/cm2 109 cm 10-34 erg/cm 2 2XI033 g 4 X 108 cm/s 1.8 X 1050 erg 0.1 1.8 X 1049 erg 2XI028 g 9 X 1038~ rad!s 104 s 10-4 rad/s 1030 erg 1023 cm 10-12 erg/ cm2 The reader might find it informative to extend this table to the bursts of waves emitted by (I) debris falling into a black hole, (2) collisions between two black holes, and (3) a supernova explosion in which a star of two solar masses collapses to nuclear densities, ejecting half its mass in the process. 988 36. GENERATION OF GRAVITATIONAL WAVES In this motion the kinetic energy is. I m 1m2 I. I (kmetic energy) = - 2\" (potentIa energy) =\"2-a-\u00b7 The power that they radiate as gravitational waves can be estimated roughly as the square of the circulating power, L -- w X (kinetic energy); thus, p.2M3 LGw--~Lo' where p. = m 1m2/ M is the familiar reduced mass, and M = m 1 + m2 is the total mass of this binary system. An exact calculation based on equation (36.1) gives a result larger than this by a factor --30: for a binary system of semimajor axis a and eccentricity e, the power output averaged over an orbital period is (1) power output 32 p.2M3 L GW = 5\" ~ f(e)Lo' where f(e) is the dimensionless \"correction function,\" f(e) = [1 + 1l e2 + lZ.e 4 ] [l - e2r 7/ 2. 24 96 (36.l6a) (36.l6b) (2) effects of radiation reaction [See exercise 36.6 at end of \u00a736.8; also Peters and Mathews (1963).] As the binary system loses energy by gravitational radiation, the stars spiral in tightening of gravitational binding). For toward each other (decrease of energy; circular orbits the energy, E = -~mlm2/a = -~M/a, decreases as dE/dt = 1/2(p.M/a2)(da/dt) 32 p.2M3 = -LGw = -5\"~. Consequently, the evolution of the orbital radius is given by the formula where ao = atoday and (36.l7a) (36.l7b) Thus, unless nongravitational forces intervene, the two stars will spiral together in a time To (spiral time). For an elliptical orbit, the eccentricity also evolves. Radiation is emitted primarily at periastron. Therefore the braking forces of radiation reaction act there with greatest force. This effect deprives the stars of some of the kinetic energy of the excursions in their separation (\"radial kinetic energy\"). In consequence, the orbit becomes more nearly circular. [See Peters and Mathews (1963) for detailed calculations.] (3) particular binaries observed by astronomers \u00a736.7. FORMULAS FOR RADIATION FROM SLOW-MOTION SOURCES 989 The calculated power output, flux at Earth, and damping times are shown in Box 36.3 for several known binary stars and several interesting hypothetical cases. Notice that in the most favorable known cases the period is a few hours; the damping time is the age of the universe (could the absence of better cases be due to radiation reaction's having destroyed them?); the output of power in the form of gravitational waves is _1030 to 1032 ergs/sec (approaching the light output of the sun, 3.9 X 1033 ergs/sec); and the calculated flux at the Earth is _10- 10 to 10-12 ergs/sec (too small too small several decades hence; see Chap to detect in 1973, but perhaps not ter 37). The hypothetical cases in Box 36.3 illustrate the general relations for astrophysical systems that were derived in \u00a736.4-namely, that only as the system approaches its gravitational radius can L GW approach L o, and only then can damping remove nearly the whole energy in a single period. \u00a736.7. FORMULAS FOR RADIATION FROM NEARLY NEWTONIAN SLOW-MOTION SOURCES Turn now from illustrative astrophysical sources to rigorous formulas valid for a wide variety of sources. One such formula has already been", " written down, (36.1 ) but it has not yet been derived, nor has its realm of validity been discussed. This formula for the p()wer output is actually valid for any \"nearly Newtonian, slow-motion source\"-more particularly, for any source in which (size of source)/(reduced wavelength of waves) ~ 1, (36.l8a) INewtonian potential I ~ (size of source)/(reduced wavelength), (36.18b) The\" nearly Newtonian, slow-motion approximation\" for analyzing sources of gravitational waves Itypical stressesl (mass density). ~ (size of source) d (re uce wave ength) d I. (36.18c) It is not valid, except perhaps approximately, for fast-motion or strong-field sources. Moreover, there is no formalism available today which can handle effectively and in genera-t the fast-motion case or the strong-field case. - The rest of this chapter is devoted to a detailed analysis of gravitational waves from nearly Newtonian, slow-motion sources. But the analysis (Track 2; \u00a7\u00a736.9 36.11) will be preceded by a Track-I summary in this section and the next. For any source of size R and mean internai velocity v, the characteristic reduced wavelength (A\" = Al2'ii) of the radiation emitted is (amplitude of mo tions)/v ~ R/v. Consequently the demand (36.l8a) that R/A\" be ~ 1 [Le., that the source be confined to a small region deep inside the near (nonradiation) zone] enforces the slow-motion constraint A\" - v ~ 1. Box 36.3 GRAVITATIONAL RADIATION FROM SEVERAL BINARY STAR SYSTEMS\" Type of system Name Period Solar System (Sun + Jupiter) Typical resolved binaries from compilation of Van de Kamp (1958) Typical eclipsing binaries from compilation of Gaposhkin (1958) Favorable cases from compilation of Braginsky (1965) Hypothetical binaries (neu tron stars or black holes) Solar System 11 Cas EBoo Sirius Fu 46 f3 Lyr UWCMa f3 Per WUMa UV Leo V Pup i Boo YY Eri SW Lac WZ Sge 104 km 103 km 102 km 10 km 11.86 yr. 480 yr. 149.95 yr. 49.94 yr. 13.12 yr. 12.925 day 4.395 day 2.867 day 0.33 day 0.6 day 1.45 day 0.268 day 0,321 day 0,321 day 81 min 12.2 sec 0.39 sec 12.2 msec 0.39 msec m l Mo 1.0 0.94 0.85 2.28 0,31 19.48 40.0 4.70 0.76 1.36 16.6 1.35 0.76 0.97 0.6 1.0 1.0 1.0 1.0 m 2 Mo 9.56 X 10-4 0.58 0.75 0.98 0.25 9.74 31.0 0.94 0.57 1.25 9.8 0.68 0.50 0.83 0,03 1.0 1.0 1.0 1.0 Distance from earth (pc) Earth is in near zone 5.9 6.7 2.6 6.5 330 1470 30 110 68 390 12 42 75 100 1000 1000 1000 1000 Spiral timeb Low (ergs/sec) Flux at earth (erg/sec cm 2) 2.5 X 1023 yr 5.2 X 1010 9.5 X 1024 3.8 X 1023 7.2 X 1021 3.2 X 1021 7.0 X 1011 8.2 X 109 3.2 X lO\" 6.2 X 109 1.0 X 1010 2,3 X 109 2.0 X 109 6.6 X 109 3.5 X 109 1.1 X 109 yr 3.2 yr 2.8 hr 1.0 sec 0.10 msec 5.6 3.6 1.1 3.6 X 1010 X 10 12 X 101 \" X 1014 0.057 X 10all X 1030 49. 0.014 X lOall 0.47 X 10-10 65. 0.63 X 103<1 X 1030 X lOall 3.2 0.42 X 103<1 X 103<1 1.5 X 10-10 0.5 3.25 X 1041 3.25 X 1046 3.25 X 1051 3.25 X 1056 1.4 6.7 1.3 7.1 X 10-29 X 10-2 \" X 10-24 X 10- 26 0.0004 X 10- 11 0.019 X 10- 11 0.013", " X 10-11 0.032 X 10- 11 0.012 X IO- l l X 10-11 0.36 X IO- l l 18. X 10- 11 0.20 X 10- 11 0.21 X IO- l l 0.04 2.7 2.7 2.7 2.7 X 10-3 X 102 X 10 7 X 1012 aBased on tables by Braginsky (1965) and by Ruffini and Wheeler (197Ib). \"The spiral time. TO' as given by equ!ltion (36.l7b) is the time for the two slars to spiral into each other if no nongravitalional forces intervene. \u00a736.7. FORMULAS FOR RADIATION FROM SLOW-MOTION SOURCES 991 These related conditions, v ~ I and R ~ A\", are satisfied by all presently conceived laboratory generators of gravitational waves. No one has seen how to bring a macroscopic mass up to a speed v 1. These conditions are also satisfied by every gravitationally bound, nearly Newtonian system. Thus, for such a system of mass M, the condition for gravitational binding, ~Mv2 ~ M2/ R guarantees that v ~ (M/ R)l/2 ~ 1. The conditions M/R ~ R/A\" and ITikl/TOo ~ R/A\" are satisfied by all nearly Newtonian sources of conceivable interest. Typical sources (e.g. binary stars) have (virial theorem). In those rare cases where (M/R or ITikl/TOO) ~ R/A\" (e.g., a mar ginally stable, slowly vibrating star), the motion is so very slow that the radiation will be too weak to be interesting. For any nearly Newtonian slow-motion system, there is a spacetime region deep inside the near zone (r ~ A\"), but outside the boundary of the source (r > R), in which vacuum Newtonian gravitation theory is nearly valid. An observer in this Newtonian region can measure the Newtonian potential fJj and can expand it in powers of I / r: Definitions of mass, dipole moment, and reduced quadrupole moment for a slow-motion source where ni = xi/r. (36.l9a) He can then give names to the coefficients in this expansion: M \"total mass~energy\" = \"active gravitational mass\"; di \"dipole moment\" [if he chooses the origin of coordinates carefully, he can make di = 0]; (36.l9b) f ik \"reduced quadrupole moment\" {because the system is nearly Newtonian, f ik is given by expression (36.3)]. As this Newtonian potential reaches out into the radiation zone, the static portions of it (-M/r - di ni/r 2 ) maintain their Newtonian form, unchanged. But the dynamic part (-~fiknink/r 3) ceases to be describable in Newtonian terms. As retardation effects become noticeable (at increasing r values), it gradually changes over into outgoing gravitational waves, which must be described in the full general theory of relativity, or in linearized theory, or in the \"shortwave\" approximation of \u00a735.13. If one chooses to use linearized theory in the radiation zone, and if one imposes the transverse-traceless gauge there (hf{;. = 0, hT! = 0, hf,{,k = 0), then the gravita tional waves take the form [derived later as equation (36.47)] Properties of gravitational waves in terms of reduced quadrupole moment: 2.. hJ,T = -;: fTt(t - r) + corrections of order [I. r2 fJ,T(t - ]. r) (36.20) (1) the wave field hIJ 992 36. GENERATION OF GRAVITATIONAL WAVES Here ff[ is the second time-derivative of the transverse-traceless part of the quad rupole moment (transverse to the radial direction; see \u00a735.4); thus, Pab = (cSab - nanb) (projection operator), (36.21 ) na = x a Ir (unit radial vector). The effective stress-energy tensor for these outgoing waves (\u00a735.7) has the same form as for a swarm of zero-mass particles traveling radially outward with the speed of light; at large distances its components of lowest nonvanishing order are (2) effective stress-energy tensor T(GW) = _T(GW) = T(GW) = _1_(h'fT hTT) = _1_(f'TTf''fT) 32'17 Ik jk 00 rr Or = 8;r", "2 C.j~Jik - Ik,O jk,O 8'ITr2 2nif~/j~krik + ~ (nif~knk)2), (36.22) ) denotes an average over several wavelengths. (Recall that one cannot where ( localize the energy more closely than a wavelength!) The total power crossing a sphere of radius rat time t is (3) total power radiated (36.23) (See exercise 36.9.) This is the formula with which this chapter began: equation (36.1). The wave fronts are not precisely spherical. For example, for a binary star system the wave fronts in the equatorial plane must be spirals. This means that there is a tiny nonradial component of the momentum flux, which decreases in strength as I/r 3. Associated with this nonradial momentum is an angular momentum density (angular momentum relative to the system's center, r = 0), which drops off as 1/r2 [Peters (1964), as corrected by DeWitt (1971), p. 286]: (4) density of angular momentum (36.24) The integral of this quantity over a sphere is the total angular momentum being transported outward per unit time, (5) total angular momentum radiated (36.25) (See exercise 36.9.) \u00a736.8. RADIATION REACTION IN SLOW-MOTION SOURCES 993 \u00a736.8. RADIATION REACTION IN SLOW-MOTION SOURCES\u00b7 The conservation laws discussed in Box 19.1 and derived in \u00a720.5 guarantee that the source must lose energy and angular momentum at the same rate as the gravita tional waves carry them off. The agent that produces these losses is a tiny component of the spacetime curvature inside the source, which reverses sign if one changes from a (realistic) outgoing-wave boundary condition at infinity to the opposite (unrealistic) ingoing-wave condition. These \"radiation-reaction\" pieces of the curvature can be described in Newtonian language when the source obeys the nearly Newtonian, slow-motion conditions (36.18). The dynamical part of the Newtonian potential, in its \"standard form\" Outgoing-wave boundary condition gives rise to a Newtonian-type radiation-reaction potential has no retardation in it. (Newtonian theory demands action at a distance!) Conse quently, there is no way whatsoever for the standard potential to decide, at large radii, whether to join onto outgoing waves or onto ingoing waves. Being undecided, it takes the middle track of joining onto standing waves (half outgoing, plus half ingoing). But this is not what one wants. It turns out (see \u00a736.11) that the join can be made to purely outgoing waves if and only if fJj is augmented by a tiny \"radia tion-reaction\" potential fJj _ fJj - standard Newtonian theory + fJj(react), (36.26a) (36.26b) Form and magnitude of the radiation-reaction po!ential If, instead, one sets fJj = fJjstandard waves. fJj(react), the potential will join onto purely ingoing In order of magnitude, the radiation-reaction potential is (36.27) Consequently, near the source it is tiny compared to the standard Newtonian potential [a factor (Rj tl)5....... v5 smaller!]. However, at the inner boundary of the radiation zone (r....... tl), it is of the same order of magnitude as the dynamic, quadru pole part of the standard potential. The radiation-reaction part of the Newtonian potential plays the same role as a producer of accelerations that any other part of the Newtonian potential does. Any particle in the Newtonian region experiences a gravitational acceleration given by a\u00b7 == -fJj. = -fJj t,1 1 d d' - s an ar,1 fJj(~eact),1. (36.28) Effects of the potential: (1) radiation-reaction accelerations \"The ideas and formalism described in this section were devised by Burke (1970). Thome (1969b). and Chandrasekhar and Esposito (1970). Among the forerunners of these ideas were the papers of Peters (1964). and Peres and Rosen (1964). 994 36. GENERATION OF GRAVITATIONAL WAVES Inside the source, this acceleration leads to energy and angular momentum losses given by and. (36.29a) (36.29b) (2) loss of energy and angular momentum (Here p is the density, vi is the velocity, and ai as above is the acceleration of the matter in the source.) Standard Newtonian theory conserves the energy and angular", " momentum. Therefore only the reaction part of the potential can produce losses: (36.30) A straightforward calculation (exercise 36.5) using expression (36.26b) for the reac tion potential yields, for the time-averaged losses, dE/dt = - s Uikfik)' 1...... 2 dJ/dt = - S\u00a3iklUkafal)' \"\". (36.3 I) Notice that these results agree with the energy and angular momentum carried off by the radiation as given by equations (36.1) and (36.25). The agreement is an absolute imperative. The laws of conservation of total energy and angular momen tum demand it. A slow-motion electromagnetic system emitting electric dipole radiation has a radiation-reaction potential Radiation-reaction potential for electromagnetic waves Aireact) - 1 -, 0 A (react) o _ \"\"(react) - -Yo'l -3 i d'\" Xi'(36.32) which is completely analogous to fJj(react) of gravitation theory [see, e.g., Burke (1971 )]. However, attention does not usually focus on this potential and the reaction forces it produces. Instead, it focuses on the reaction force in a special case: that of an isolated charge being accelerated by nonelectromagnetic forces. For such a charge, the reaction force is '\" F ea = -e2 x. <Rct) 2 3 (36.33) No such formula is relevant to gravitation theory, because there is no such thing as a gravitationally isolated, radiating particle (i.e., one accelerated by forces that have no coupling to gravity). \u00a736.9. FOUNDATIONS FOR DERIVATION OF RADIATION FORMULAS 995 ------------------------------------- ~ Exercise 36.5. ENERGY AND ANGULAR MOMENTUM LOSSES DUE TO EXERCISES RADIATION REACTION Derive equations (36.31) for the rate at which gravitational radiation damping saps energy and angular momentum from a slow-motion source. Base the derivation on equations (36.26b) and (36.30). Exercise 36.6. GRAVITATIONAL WAVES FROM BINARY STAR SYSTEMS Apply the full formalism of \u00a7\u00a736.7 and 36.8 to a binary star system with circular orbits. Calculate the angular distribution of the gravitational waves; the total power radiated; the total angular momentum radiated; the radiation-reaction forces; and the loss of energy and angular momentum due to radiation reaction. Compare the answers with the results quoted in \u00a736.6. [For further details of the solution, see Peters and Mathews (1963).] \u00a736.9. FOUNDATIONS FOR DERIVATION OF RADIATION FORMULAS Tum now from the formulas for radiation from a nearly Newtonian system in slow motion to a derivation of these formulas. Initially (this section) work in the full general theory of relativity without any approximations-not even that of slow motion. Impose only the constraint that the source be isolated, and that spacetime become asymptotically flat far away from it..Use a coordinate system that becomes asymptotically Lorentz as rapidly as space time curvature permits, when one moves radially outward from the source toward infinity. Everywhere in this coordinate system, even inside the source, which may be relativistic, define (36.34) The hJ.Lp are clearly not the components of a tensor. Neither is 11J.Lp the true metric tensor. Nevertheless, one is free to raise and lower indices on hllP with 11J.Lp and to define r The rest of this chapter is Track 2. Chapter 20 (conservation laws) is needed as preparation for it. It will be helpful in Chapter 39 (post-N ewtonian formalism). but is not needed as preparation for any other chapters. Derivation of formula for the gravitational-wave field produced by a slow-motion source: (36.35) (1) definition of hllP Moreover, one can always specialize the coordinates so that the four conditions (36.36) are exactly satisfied everywhere, including the interior of the source. With these definitions and conventions, hJ.Lp becomes the gravitational field of linearized theory far from the source, and also inside the source if gravity is weak there. But jf!he interior gravity is strong (lhJ.Lpl not ~ I), hJ.Lp in the interior has no ~ connection whatsoever to linearized theory. 996 36. GENERATION OF GRAVITATIONAL WAVES The exact Einstein field equations can be written in terms of hIL\" as [cf. \u00a720.3; in particular, combine equations (20.14), (20.18), and (20.", "3); and impose the coordi nate condition (36.36)] -- h J.LV. /31)a/3 = - 16'17(TJ.LV + t ILV ),,a (36.37) where TJ.LV are the components of stress-energy tensor, and tlLV are quantities (components of the \"stress-energy pseudo tensor for the gravitational field\") that are of quadratic order and higher in hJ.LV. Recall the \"philosophy of controlled ignorance\" expounded in \u00a719.3. One is so ignorant that nowhere does one ever write down an explicit expression for tlLV in terms of Jia/3; and this ignorance is so controlled that one will never need such an expression in the calculations to follow! More specifi cally, the strength of the outgoing wave is proportional to the integral of a compli cated expression over the interior of a system where \"gravitational stresses\" may ITiki. No matter. All that will count for be comparable to material stresses, Itikl the radiation is the quadrupole part of the field. Moreover, that quadrupole moment is empirically definable by purely Newtonian measurements in the Newtonian region (I) well inside the wave zone, but (2) well outside the surface of the source. One does not have to know the inner workings of a star to define its mass (influence on Kepler orbits outside) nor does one have to know those inner workings to define its quadrupole moment as sensed externally. Einstein's equations (36.37), augmented by an outgoing-wave boundary condition, are equivalent to the integral equations _ hJ.Lv(t, xi) = 4J all space [~v+tJ.LV] ret d 3x', I Ix - X I._ (36.38) (2) fiel~ equations in terms of hlLv (3) philosophy of controlled ignorance (4) integral formulation of field equations where Ix - x'I ~ (xi - Xi')2 ] [ 1/2, 1 and the subscript \"ret\" means the quantity is to be evaluated at spacetime point the retarded (t' = t - Ix - x'I, xi) These are integral equations because the unknowns, hJ.LV, appear both outside and inside the integral (inside they are contained in tJ.LV). Notice that in passing from the wave equations (36.37) to the integral equations (36.38), one has cavalierly behaved as though hJ.LV were fields in flat spacetime. This is certainly not true; but the mathematical manipulations are valid nevertheless!-and the integral equations (36.38) are valid for any field point (t, xi), even inside the source. \u00a736.10. EVALUATION OF THE RADIATION FIELD IN THE SLOW-MOTION APPROXIMATION (5) specialization to slow motion Thus far the analysis has been exact. Now it is necessary to introduce the slow-moTIoh assumption of \u00a736.7: R ~ tt. \u00a736.10. DERIVATION OF RADIATION FIELD 997 z \"Gravitational source\" [Region of size L, where I~' gives significant contribution to integral (36.38\u00bb) Material source [Region of size R, where T~' # 0) --~~-----------~ y Radiation zone (r} A) x Figure 36.3. A slow-motion source radiating gravitational waves. The origin of spatial coordinates is located inside the source. The size of the source, R, is very small compared to a reduced wavelength, R <li; A. Significant contributions to the retarded integral (36.38) for Ji~' come only from a region of size L - R <li; A surrounding the source, because outside the source-but in the near zone (R <li; r <li; A)-the \"stress-energy pseudotensor\" I~' dies out as l/r4 (see exercise 36.7). In the radiation zone, I~' ceases to die out as I/r\\ and begins to die out as l/r2; it is trying to describe (but cannot, really, without appropriate averaging) the stress-energy carried by the gravitational waves. If the source has been emitting waves long enough, contributions from the radiation zone to the retarded integral (36.38) may be nonnegligible: [wI ret t r'2 -...!..==>f[I~') d3x'-f...!..r'2 dil'dr'. -.::....r!.,.'.;..2~__-'. [for r> Aj--l [ma~ have significant contri- 1 t J bullons from large r' ret Such contributions are ignored in the text, in", " calculations of the radiated waves, because they have nothing whatsoever to do with the emission process itself. Rather, they are part of the propagation process treated in the last chapter. They include the background curvature produced by the stress-energy of the waves, scattering of waves off the background curvature, wave-wave scattering, etc.; and they are totally negligible in the neighborhood of the source itself (r ~ 1,000 A, for example) because a slow-motion source radiates so very weakly. Place the origin of spatial coordinates inside the source, as shown in Figure 36.3. For slow-motion systems, the only significant contributions to the retarded integrals (36.38) come from deep inside the near zone (from a region of size L -- R ~..t; see Figure 36.3). Confine attention to \"field points\" (points of observation). xi far outside this \"source region,\". 998 36, GENERATION OF GRAVITATIONAL WAVES Ixl r ~ L ~ lx'I, (36.39a) and expand the retarded integral (36.38) in powers of x'lr-in just the same manner as was done in \u00a7I?.1. (Such an expansion is justified by and requires the slow-motion assumption, AIR........ AIL ~ 1.) The result is Jill\"\u00ab(, x) = ~ f [P\"(x', ( - r) + (1l\"(X', ( - r)] d 3x'+ 0 {,~~ f xi'[PV(x', ( - r) + (1l\"(X', ( - r)] d3x} (36.40) Of the ten components of Ji IlV, only the six spa tial ones, Jiik, are of interest, since only they are needed in projecting out the transverse-traceless radiation field Jill'. The spatial components are expressed by equations (36.40) in terms of integrals over the \"stress distribution\" Tik + (ik. It will be convenient, in making comparisons with Newtonian theory, to reexpress Jiik in terms of integrals over the \"energy distribu tion\" TOO + (00. One can make the conversion with the help of the exact equations of motion Tllv;v = 0, which have the special form (36.41 ) (6) calculation of!iik in radiation zone in the coordinate system being used [see equations (36.36) and (36.37); also the discussion in \u00a720.3]. Applying these relations twice in succession, one obtains the identity (TOO + (00),00 = - (TO I + (01),10 = - (TLO + (LO),OI = +(TIm + (Im),ml'.. From this and the elementary chain rule for differentiation, it follows that [(TOO + (OO)xixk] = (TIm + (1m),00,m! XiXk = [(TIm + (Im)xixk],m! _ 2[(TIi + (li)X k + (TIk + (lk)Xi],I + 2(Tik + (ik), whence where (36.42a) (36.42b) (7) specialization to nearly Newtonian sources Now introduce the nearly Newtonian assumption. It guarantees that gravitation contributes only a small fraction of the total energy: hence (36.42b/) The quantity I ik thus represents the second moment of the mass distribution. \u00a736.10. DERIVATION OF RADIATION FIELD 999 By combining equations (36.42) and (36.40), and by noting that inside the source Itjkl - liP,jiP,kl - TOOliPl, one obtains 2Ijk(t = 2 d2Ijk(t - lijk(t, x) = ~ d dt2 r r dt2 r) + 0 [~(ITjkl + liPl)!!'\" M] TOO r r) {I + 0 [ITjkl TOO + R R A\" M]~}. (36.43) [negligible by assumptions (36.I8)rl Actually, what one wants are hTl', not lijk. They can be obtained by first lowering indices, using 111m = 81m, and then projecting out the TT part using the projection operator for radially traveling waves: (8) conversion, by projection, to hJ[ (36.44) (see Box 35.1). (Because h;k and hjk differ only in the trace, they have the same TT parts). The result is hTT(t jk,x - ) _ ~ d r 2 ITl'(t d 2 t r)'where (36", ".45 a) (36.45 b) This is not the best form in which to write the answer, because an external observer cannot measure directly the second moment of the mass distribution, I jk. Fortunately, one can replace I jk by the reduced quadrupole moment, ; 8jkI =f (TOO + tOO) (xjx k - ; 8jkr2 ) d 3x, (36.46) I jk - f jk (9) reexpression of hT[ in terms of reduced quadrupole moment and write h TT jk (t, X - ) _ 2 d 2 fTl'(t d 2 t r r) \u2022 (36.47) This is allowed because the TT parts of I jk and f jk are identical (exercise 36.8). The reduced quadrupole moment f jk has a well-defined, elementary physical significance for an observer confined to the exterior of the source. In the near zone (r ~ A\"), but outside the source so that vacuum Newtonian theory is very nearly valid, the Newtonian potential is =-fall space [see equation (36.38)]. Any nearly Newtonian, slow-motion source satisfies 1000 36. GENERATION OF GRAVITATIONAL WAVES [recall: t\u00b0{3 - (<P)2 - Tool<PI]. Hence, one can write m( Y' X, t) - - f [TOO(x' I x-x t)] 'd 3, 'I x. (36.48) Expanding Ix - x'1- 1 in powers of 1/r, one obtains m _ Y' - j (M djx - + - - + r 3 r k 3fikx iX 2r 5 +... ) fi or r ~ A\", but r neverthe-) less large enough that vacuum Newtonian theory is valid (, (36.49a) where M = (total mass-energy of source) =f TOO d 3x, (dipole moment of source) = f TOOx i dx 3, di (36.49b) f ik (reduced quadrupole moment of source) = expression (36.46). Thus, the quantities f jk, whose second time-derivatives determine the radiation field by equation (36.47), are precisely the components of the star's reduced quadrupole moment, as measured by an observer who explores its Newtonian potential <P deep inside the near zone (r ~ A\") (\"empirical quadrupole moment.\"). The final answer (36.47) for the radiation field interms of fJ{ was quoted in the summary of results given in \u00a736.7. Also quoted there were expressions for the effective stress-energy tensor of the radiation and for the energy and angular mo mentum radiated [equations (36.22) to (36.25)]. Those expressions can be derived using the formalism of the shortwave approximation. (See exercise 36.9.) Exercise 36.7. MAGNITUDE OF til\" Consider a slow-motion source of gravitational waves. Show that far from the source, but in the near zone (R ~ r ~ A) the components of the \"stress-energy pseudotensor\" til\" die out as l/r4, but in the radiation zone (r ~ A) they die out only as l/r 2\u2022 Exercise 36.8. PROOF THAT THE TRANVERSE TRACELESS PARTS OF.: f jk AND I jk ARE IDENTICAL Prove by direct computation that the TT parts of Ijk (36.42b) and f jk (36.46) are identical, no matter where the observer is who does the IT projection (i.e., no matter what the unit vector n in the projection operator may be). Exercise 36.9. ENERGY AND ANGULAR MOMENTUM RADIATED (a) For the gravitational waves in asymptotically flat spacetime described by equation (36.47), calculate the smeared-out stress-energy tensor T~~W) of equation (35.23). [Answer: equation (36.22).] EXERCISES \u00a7 36.11. DERIVATION OF RADIATION-REACTION POTENTIAL 1001 (b) Perform the integrals of equations (36.23) and (36.25) to obtain the total power and angular momentum radiated. [Hint: Derive and use the following averages over a sphere 4~ Jni dO = 0, Here n =x/lxl is the unit radial vector.] \u00a736.11. DERIVATION OF THE RADIATION-REACTION POTENTIAL Turn, finally, to a derivation of the radiation-reaction results quoted in", " \u00a736.8. The analysis starts with the solution (36.43) for the spatial part of the radiation field in the original (i.e., not IT) gauge: Derivation of formula for the radiation-reaction potential: - k h' (t, x) = - Iik(t - 2.. r r). (36.50) Although this solution was originally derived by discarding all terms that die out faster than l/r, it is in fact an exact solution to the vacuum field equations liik a a = 0 of linearized theory. This means that it is valid in the intermediate and near zones (r ~..t, but r > R) as well as in the radiation zone. Were one to replace the outgoing-wave condition by an ingoing-wave condition at infinity, the exact solution (36.50) for liik would get replaced by 2.. hik(t, x) = - Iik(t + r). r Thus, in order to delineate the effects of the outgoing-wave boundary condition, one can write the exact solution in the form f = \u00b1l, (36.51 ) (1) form ula for hjk anywhere outside source, with either outgoing or ingoing waves and then focus attention on the effects of the sign of f. In the near zone (r ~..t), but outside the nearly Newtonian source, this solution for lijk, as expanded in J?owers of r, becomes... J. (36.52a) (2) hjk specialized to near zone where 1002 36. GENERATION OF GRAVITATIONAL WAVES The corresponding forms of hOi and hoo can be generated from this by the gauge conditions h/,/3 = 0; i.e., by hiO,o = h;k,k and hoo,o = hOi,i' The results are: (3) hoo and hOi in near zone calculated by gauge conditions _ [/(I)x k /(3)X k - h. = 2 _ _,_k_ + _,_k_ 2!,,3 0, 2/(4)X k _ \u00a3 _,_k_ + 3! 3/(5)X k,,k 4! 4/(6)X k,2] _ \u00a3 _',-,k__ 5! + (static terms not associated with radiation); h - 2 00 - [(3Xixk,5,2 8ik) - / ik (XiXk _,28ik) 2!,3 /(2) ik - \u00a3 2 3! /(3) jj (36.52b) + 3(XiXk +,28ik). 4! \"k 1(4) - E -.. - 4(2xiXk +,28ik ) 5! -.... ] /(5) + ik ~.52c) + (static and time-linear terms not associated with radiation). The leading term in these expressions rises as 1/,3 when one approaches the source: It is precisely the leading term in the dynamic, quadrupole part of the Newtonian potential, <P = -~hoo = -1hoo' All other terms without \u00a3's in front of them are corrections to the Newtonian potential. They produce effects like the perihelion shift of Mercury that in no way deplete the energy and angular momentum of the system. The terms with \u00a3's are associated with radiation reaction. Pluck the leading ones out and call them \"reaction potentials\": (4) plucking out the radiation-reaction potenti?ls from ha13 h- (reactl 00 _ ~ /(3) __1_ (2xix k +,2 8ik)/(5). 3,i 15,k The corresponding metric perturbations ha/3 = ha/3 -~h1)a/3 are hlreactl - ik -2/(3) + ~ /(3) 8. + 0(/(5),2),k'3 11,k,k (36.53) (36.54) h(reactl = _ i /(3) __1_ (Xixk + 3,28. )/(5) 15 3 11 00,k,k' These reaction potentials in the near zone are understood most clearly by a change of gauge that brings them into Newtonian form. Set \u00a736.11. DERIVATION OF RADIATION-REACTION POTENTIAL 1003 with h IJ.Jln8W =h IJ.Jl old -~ -~ p.,JI JI,p. (5) conversion of radiation-reaction potentials to Newtonian gauge ----~ --- - - Then in the new gauge h lreacfJ - 0(/(", "5)r 2) jk'jk - h(react) - 0(/(61r3) jk'OJ - (36.55) (36.56) This gauge is ideally suited to a Newtonian interpretation, since in it the geodesic equation for slowly moving particles has the form. d 2 _ / t'\"\"lreact) - - '\u00a5. +. (terms not sensitive to.. outgomg-wave condItIOn,I ), (36.57) d 2 x with \"\"lreact) __ 1 h(react) - 1 f(5)XjXk 2 00 5 jk '\u00a5 - - \u2022 (36.58) Thus, the leading radiation-reaction effects (with fractional errors - [..t/r]2) can be described in the ne~r _zQne of a nearly Newtonian source by appending the term M~~xjxk to the Newtonian potential. The resulting formalism and a qualitative version of the above derivation were presented in \u00a736.8. CHAPTER 37 DETECTION OF GRAVITATIONAL WAVES I often say that when you can measure what you are speaking about. and express it in numbers, you know something about it; but when you cannot measure it. when you cannot express it in numbers. your knowledge is of a meagre and unsatisfactory kind_' it may be the beginning of knowledge. but you in your thoughts. advanced to the stage of science. whatever the matter may be. have scarcely. WILLIAM THOMSON, LORD KELviN [(1889), p. 73] -- - ---- \u00a737.1. COORDINATE SYSTEMS AND IMPINGING WAVES The detector is even easier to analyze than the generator or the transmission when one deals with gravitational waves within the framework of general relativity. Man's potential detectors all lie in the solar system, where gravity is so weak and spacetime so nearly flat that a plane gravitational wave coming in remains for all practical purposes a plane gravitational wave. (Angle of deflection of wave front passing limb of sun is only 1:' 75.) Moreover, the nearest source of significant waves is so far away.that, for all practical purposes, one can consider the waves as plane-fronted when they reach the Earth. Consequently, as they propagate in the z-direction past a detector, they can be described to high accuracy by the following transverseatraceless linearized expressions Metric perturbation: h;; = -h~ = A+(t - z), h;c = h~ = Ax(t - z), (37.1 a) Linearized description of gravitational waves propagating past Earth. Riemann tensor: R zozo = -RyOIIO = - 2\"A+(t - z), - I (37.1 b) \u00a737.1. COORDINATE SYSTEMS AND IMPINGING WAVES 1005 Figure 37.1. The proper reference frame of a vibrating-bar detector. The bar hangs by'a wire from a cross beam, which is supported by vertical posts (not shown) that are embedded in the Earth. Consequently, the bar experiences a 4-acceleration given, at the moment when this diagram is drawn, by a :;:: g(a/ai), where g is the \"local acceleration of gravity\" (g - 980 cm/sec2). Later, the spatial axes will have rotated relative to the bar (\"Foucoult\u00b7pendulum effect\" produced by Earth's rotation), so the components of a but not its magnitude will have changed. The proper reference frame relies on an imaginary clock and three imaginary gyroscopes located at the bar's center of mass (and shown above in a cut-away view). Coordinate time is equal to proper time as measured by the clock, and the directions of the spatial axes a/axi are attached to the gyroscopes. The forces that prevent the gyroscopes from falling in the Earth's field must be applied at the centers of mass of the individual gyroscopes (no torque!). Stress-energy: Toow = Tzzw = - (G) (G) T(GW) _ Oz 1 <A'z + A'Z) 16'17 + x timeavg.\u00b7 (37.1 c) (See exercise 37.1.) To analyze most easily the response of the detector to these impinging waves, use not the TT coordinate system {x a } (which is specially \"tuned\" to the waves), but rather use coordinates {x\"'} specially \"tuned\" to the experimenter and his detector. The detector might be a vibrating bar, or the vibrating Earth, or a loop it will of tubing filled with fluid (see Figures 37.1 and 37.2). But whatever it is, have a center of mass.", " Attach the spatial origin, xi = 0, to this center of mass; and attach orthonormal spatial axes, ajaxi, to (possibly imaginary) gyroscopes located at this spatial origin (Figure 37.1). If the detector is accelerating (Le., not falling freely), make the gyroscopes accelerate with it by applying the necessary forces at their centers of mass (no torque!). Use, as time coordinate, the proper time X O= T measured by a clock at the spatial origin. Finally, extend these locally defined coordinates x'\" throughout all spacetime in the \"straightest\" manner possibl~. (See Proper reference frame of a detector EXERCISES 1006 37. DETECTION OF GRAVITATIONAL WAVES Track 2's \u00a713.6 for full details.) The metric in this \"proper reference frame of the detector\" will have the following form [equation (13.71) with wI = 0]. Here aj are the spatial components of the detector's 4-acceleration. (Since a must be orthogonal to the detector's 4-velocity, aovanishes.) Notice that, except for the acceleration term in goo (\"gravitational redshift term\"; see \u00a738.5 and exercise 6.6), this reference frame is locally Lorentz. (31.2) _. Exercise 37.1. GENERAL PLANE WAVE IN TT GAUGE Show that the most general linearized plane wave can be described in the transverse-traceless gauge of linearized theory by expressions (37.1). [Hint: Express the plane wave as a superpo sition (Fourier integral) of monochromatic plane waves, and describe each monochromatic plane wave by expressions (35.16). Use equations (35.10) and (35.23) to calculate ROIf3Y~ and T~~W).l Exercise 37.2. TEST-PARTICLE MOTION IN PROPER REFERENCE FRAME Show that a slowly moving test particle, falling freely through the proper reference frame of equation (37.2), obeys the equation of motion (geodesic equation) tJ2xJjdi2 = -aJ + O(lxi:I). Thus, one can interpret -aJ as the \"local acceleration of gravity\" (see caption of Fig ure 37.1). Equations of motion for a mechanical detector \u00a737.2. ACCELERATIONS IN MECHANICAL DETECTORS lo The proper reference frame of equation (37.2) is the closest thing that exists to the reference frame a Newtonian physicist would use in analyzing the detector. In fact, it is so nearly Newtonian that (according to the analysis of Box 37.1) the equations of motion for a mechanical detector, when written in this proper reference frame, take their standard Newtonian form and can be viewed and dealt with in a fully Newtonian manner, with one exception: the gravitational waves produce a driving force of non Newtonian origin, given by the familiar expression for geodesic deviation (d 2 I) force per unit mass (i.e., aCCeleratiOn)) o a partlc e at x' re atlve to etector s = -_\u2022 l'x f m2 \u2022 center of mass at x' = 0 '1 d'( (37.3) due to waves To use this equation, and to calculate detector cross sections later, one must know the components of the curvature tensor RapyS, and of the waves' stress-energy tensor, T1!vW), in the detector's proper reference frame. One cannot calculate RapyS directly - -(R.-r.-) 10\"0 due to waves X k. \u00a737.2. ACCELERATIONS IN MECHANICAL DETECTORS 1007 Box 37.1 DERIVATION OF EQUATIONS OF MOTION FOR A MECHANICAL DETECTOR Consider a \"mass element\" in a mechanical detector (e.g., a cube of aluminum one millimeter on each edge if the detector is the bar of Figure 37.1; or an element of fluid with volume I mm3 if the detector is the tube filled with fluid shown in part h of Figure 37.2). This mass element gets pushed and pulled by adjacent matter and electromagnetic fields, as the medium of the detector vibrates or flows or does whatever it is supposed to do. Let f = (4-fOrCe per unit mass exerted on mass-element ) (I) - by adjacent matter and by electromagnetic fields. This 4-force per unit mass gives the mass element a 4-acceleration Vuu = f; or, in terms of components in the detector's proper reference frame, ji = Du ijdT. Assume that the mass element has a very small velocity (v ~ I) in the", " detector's proper reference frame (i.e., relative to the detector's center of mass). Then, ignoring ), O(lxiI2), and O(lxiiv), one has [see equation (37.2)] terms of O(v 2 (2) and Exercise 37.3 calculates riM to precision of O(lxJI). Inserting its result and rearrang ing terms, one finds that (4) (\"equation of motion for mass element \"). Examine this equation, first from the viewpoint of an Einsteinian physicist, and then from the viewpoint of a Newtonian physicist. The Einsteinian physicist recognizes d 2x ijii2 as the \"coordinate acceleration\" of the mass element-but he keeps in mind that, to precision of O(lxiI2), coordinate lengths and proper lengths are the same [see equation (37.2)]. The coordinate acceleration d 2x ijii2 has three causes: (I) the externally applied force, (I + 2a\u00b7 x)fi = (d2x ijdt 2 )externalfOrCe = (I + 2a\u00b7 x)(d2xijdT 2 )externalfOrCe (5a) (the origin of the a\u00b7 x correction is simply the conversion between coordinate time 1008 37. DETECTION OF GRAVITATIONAL WAVES Box 37.1 (continued) and prope~ time); (2) the \"inertial force\" due to the acceleration of the reference frame, (5b) (see exercise 37.4 for explanation of the a\u00b7 x correction); and (3) a \"Riemann curvature force,\" which will include Riemann curvature due to local, Newtonian gravitational fields (fields of Earth, sun, moon, etc.), plus Riemann curvature due to the impinging gravitational waves, -(RJ61co)wavesx1c - (Ri01cO)Newtonfieldsx1c = (d 2x J/ii 2)curvature (5c) (linear superposition because all gravitational fields in the solar system are so weak). This \"Riemann curvature force\" is not, of course, \"felt\" by the mass element; it does not produce any 4-acceleration. Rather, like the inertial force, it originates in the choice of reference frame: The spatial coordinates xi measure proper distance and direction away from the detector's center of mass; and Riemann curvature tries to change this proper distance and direction (\"relative acceleration;\" \"geodesic deviation\"). A Newtonian physicist views the equation of motion (4) in a rather different manner. Having been told that the spatial coordinates xi measure proper distance and direction away from the detector's center of mass, he thinks of them as the standard Euclidean spatial coordinates of Newtonian theory. He then rewrites equation (4) in the form where Fi=(Ne~tonian force per unit mass) actmg on mass element = (l + 2a\u00b7 x)i - ai(l + a\u00b7 x) - (RJo1co)Newtonfieldsx1c. (6) (7) The Newtonian physicist is free to express Fi in a form more familiar than t~is. He can ignore the subtleties of the a \u2022 x \"redshift effects\" because (l) they are small and (2) they are steady in time, and therefore-by contrast with the equally small wave-induced forces-they cannot excite resonant motions of the detector. Also, he (8) \u00a7 37.2. ACCELERATIONS IN MECHANICAL DETECTORS 1009 can separate the \"inertial acceleration,\" - aJ, into a contribution from the local acceleration of gravity at - (o<P jOX;)zi= 0' plus a the detector's center of mass, contribution - aJabsolute due to acceleration of the detector relative to the \"absolute space\" of Newtonian theory. Finally, he can rewrite the Riemann curvature due to Newtonian gravity in the familiar form R;oko = o2<Pjox; ox k, The net result is total Ne,:\"tonian force per unit] mass actmg on mass element Newtonian force per unit mass exerted by adjacent matter and by electromagnetic fields ] [ [ [ - a j absolute -(::Jt mass clement inertial force per unit mass due to acceleration ] of detector relative to Newtonian absolute space = -(o<Pjox;)zJ=o (o2<Pjox; OXk)X k ] = Newtonian gravitational acceleration. (9) [ Conclusion: The equation of motion for a mass element of a mechanical detector, when written in the detector's proper reference frame, has the standard Newtonian form (6), with standard", " to the center of mass of the detector (origin of spatial coordinates) they are -2 1.. d 2x/dt = '2 (A+x + AxY),.. \"'2 (-A+y + Axx),.. -2 _ 1 - 2 d y/dt d2i/ii2 = O. (1) j Notice that these accelerations are divergence-free. Consequently they can be represented by \"lines of force,\" analogous to those of a vacuum electric field. At a value of i - i where i( = 0 (so polari zation is entirely e+), the lines of force are the hyperbolas shown here [sketch (a)]. The direction of the acceleration at any point is the direction of the arrow there; the magnitude of the acceleration is the density of force lines. Since acceleration is proportional to distance from center of mass, the force lines get twice as close together when one moves twice as far away from the origin in a given direction. When A+ is positive, the arrows on the force lines are as show~ in (a); when it is negative, they are reversed. As IA+I increases, the force lines move in \"toward the origin so their density goes up; as IA+\\ decreases, they move out toward in finity so their density goes down. For polarization ex the force lines are rotated by 45 0 from the above diagram. For intermediate polarization (values of i - i where A+ and Ax are both nonzero), the diagram is rotated by an inter mediate angle [sketch (b)] (a) Force lines for A'x = 0,.4.'+ > \u00b0 j ---I--+--+-+-+----L-x 1 tPo = '2 arc tan (Ax! A+)..... (2) (b) 1012 37. DETECTION OF GRAVITATIONAL WAVES Box 37.2 (continued) B. Three-Dimensional Diagram At each value of i - i, the wave-produced accel erations have a specific polarization [orientation angle 90 of sketch (b)] and a specific amplitude (density of lines of force). Draw the lines of force ~n a three-dimensional (i,y, i) diagram for fixed t. Then as time passes the over-all diagram will remain unchanged in form, but will propagate with the speed of light in the i direction. Sketch (c) shows such a diagram for righthand circularly polarized waves of unchanging ampli tude. Note: The authors are not aware of diagrams such as these [(a), (b), (c) above] and their use in analyzing detector response prior to William H. Press (1970). i Exercise 37.4. WHY THE a' x? Explain the origin of the a' x correction in equation (5b) of Box 37.1. [Hint: Take the viewpoint of an observer at rest at the spatial origin who watches two freely falling particles respond to the inertial force. At time i = 0, put one particle at the origin and the other at xi. As time passes, the separation of the particles in their common Lorentz frame remains fixed; so there develops a Lorentz contraction from the viewpoint of the observer at x' = 0.] Exercise 37.5. ORIENTATION OF POLARIZATION DIAGRAM Derive equation (2) of Box 37.2. \u00a737.3. TYPES OF MECHANICAL DETECTORS Eight types of mechanical detectors: Figure 37.2 shows eight different types of mechanical detectors for gravitational waves. (By \"mechanical detector\" is meant a detector that relies on the relative..)\".. \\ ( Moon Laser beam Earth (a) Oscillations in Earth-moon separation (see exercise 37.7) (c) Oscillations in Earth's crust [see Dyson (1969\u00bb) (b) Normal-mode vibrations of earth and moon [see Weber (1968\u00bb) (d) Normal-mode vibrations of an elastic bar [see Weber (1969) and references cited therein) (e) Normal-mode vibrations of an elas.tic square, or hoop, or tuning fork [see Douglass (1971\u00bb) (f) Angular acce lerations of rotating bars [\"Heterodyne detector\": see Braginsky, Zel'dovich, and Rudenko (1969)) / - - -'\" '\\ \\ \\ I / / I A (g) Angular accelerations of driven oscillators [Sakharov (1969\u00bb) Figure 37.2. Various types of gravitational-wave detectors. (h) Pumping of fluid in a rotating loop of - pipe [Press (1970)). The pipe rotates with the same angular velocity as the waves: so the position of the pipe in the righthand polarized lines of force remains forever fixed (1) freely falling bodies 1014 37. DETECTION OF G", "RAVITATIONAL WAVES motions of matter. Nonmechanical detectors are described in \u00a737.9, near end of this chapter.) These eight detectors, and others, can be analyzed easily using the force-line diagrams of Box 37.2. A qualitative discussion of each of the eight detectors is given below. (A full quantitative analysis for each one would entail experimental techni calities' for which general relativity is irrelevant, and which are beyond the scope of this book. However, some quantitative details are spelled out in \u00a7\u00a737.5-37.8.) 1. The Relative Motions of Two Freely Falling Bodies As a gravitational wave passes two freely falling bodies, their proper separation oscillates (Figure 37.3). This produces corresponding oscillations in the redshift and round-trip travel times for electromagnetic signals propagating back and forth between the two bodies. Either effect, oscillating redshift or oscillating travel time, could be used in principle to detect the passage of the waves. Examples of such detectors are the Earth-Moon separation, as monitored by laser ranging [Fig. 37.2(a)]; Earth-spacecraft separations as monitored by radio ranging; and the separation between two test masses in an Earth-orbiting laboratory, as monitored by redshift measurements or by laser interferometry. Several features of such detectors are explored in exercises 37.6 and 37.7. As shown in exercise 37.7, such detectors have so Iowa sensitivity that they are of little experimental interest. Figure 37.3. Time of round-trip travel between two geodesics responds to oscillations in the curvature of spacetime (diagram is schematic only; symbolic of a laser pulse sent from the Earth to a comer reflector on the Moon and back at a time when a very powerful, long-wavelength gravitational wave passes by; the wave would have to be powerful because a direct measure of distance to better than 10 cm is difficult, and such precision produces a much less sensitive indicator of waves than the vibrations in length [10- 14 cm or less) of a Weber bar; see exercise 37.7). The geodesics are curved toward each other in regions where the relevant component of the Riemann curvature tensor, call it Rzoz6' has one sign, and curved away from each other in regions where it has the opposite sign. The diagram allows one to see at a glance the answer to an often expressed puzzlement: Is not any change in round\u00b7 trip travel time mere trumpery flummery? The metric perturbation, 8h pp, of the wave changes the scale of distances slightly but also correspondingly changes the scale of time. Therefore does not every possibility of any really meaningful and measurable effect cancel out? Answer: (I) The widened separation between the geodesics is not a local effect but a cumulative one. It does not arise from the local value of 8hpp directly or eve~ from the local value of the curvature. It arises from an accumulation of the bending process over an entire half\u00b7period of the gravitational wave. (2) The change in separation of the geodesics is a true change in proper distance, and shows up in a true change in proper time (see \"ticks\" on the world line of one of the particles). See exercise 37.6. Note: When one investi\u00b7 gates the separation between the geodesics, not over a single period, as here, but over a large number of periods, he finds a cumulative, systematic, net slow bending of the rapidly wiggling geodesics toward each other. This small, attractive acceler ation is evidence in gravitation physics of the effective mass-energy carried by the gravitational waves (see Chapter 35). (2) Earth and Moon (3) Earth's crust \u00a737.3. TYPES OF MECHANICAL DETECTORS 1015 2. Normal-Mode Vibrations of the Earth and Moon A gravitational wave sweeping over the Earth will excite its quadrupole modes of vibration, since the driving forces in the wave have quadrupole spatial distributions [see Fig. 37.2(b)). The fundamental quadrupole mode of the Earth has a period of 54 minutes, while that of the moon has a period of 15 minutes. Thus, the Earth and Moon should selectively pick out the 54-minute and 15-minute components of any passing wave train. Section 37.7 will analyze quantitatively the interaction between the wave and solid-body vibrations. By comparing that analysis with seismometer studies of the Earth's vibrations, Weber (1967) put the first observational limit ever on the cosmic flux of gravitational waves: I. d / flux requency.' < 3 X 107 erg cm- 2 sec-1 HZ-1 at v = 3.1 X 10-4 Hz. (37.8) 3. Oscillations in the Earth's Crust If the neutron star in a pulsar is slightly", " deformed from axial symmetry, its rotation will produce gravitational waves. The period of the waves is half the period of the pulsar (rotation of star through 180 0 produces one period ofwaves)-i.e., it should range from 0.017 sec for NP0532 (Crab Pulsar) to 1.87 sec. for NP0527. Such a wave train cannot excite the 54-minute quadrupole vibration or any of the other normal, low-frequency modes of vibration of the Earth. The kind of vibrations it can excite allow themselves in principle to be described in the language of normal modes. However, they are clearly and more conveniently envisaged as vibrations of localized regions of the Earth; or, more particularly, vibrations of the Earth's crust. Dyson (1969) has analyzed the response of an elastic solid, such as the Earth, to an incident, off-resonance gravitational wave. He shows that the response depends on irregularities in the elastic modulus for shear waves, and that it is strongest at a free surface [Figure 37.2(c)). For the fraction of gravitational-wave energy crossing a flat surface that is converted into energy of elastic motion of the solid, he finds the expression (fraction) = (8'ITGp/w 2)(s/C)3 X sin20lcosOI-1[1 + cos20 + (s/v) sin20). (37.9) Here s and v are the velocities of shear waves and compressive waves, respectively, and 0 is the angle between the direction of propagation of the waves and the normal to the surface. Considering a flux of 2 X 10-5 erg/cm2 sec (an optimistic but con ceivable value for waves from a pulsar) incident horizontally (0 = 'IT/2; \"divergent\" factor Icos 01-1 cancels out in calculation!), and taking s to be 4.5 X 105 cm/sec and w to be 6 rad/sec, he calculates that the I-Hz horizontal displacement produced in the surface has an amplitude of ~o -- 2 X 10-17 cm, too small by a factor of the order of 105 to be detected against background seismic noise. He points to the possibilities of improvements, especially via resonance (elastic waves reflected back and forth between two surfaces; Antarctic ice sheet). (4) elastic bar (5) elastic bodies of other shapes 1016 37. DETECTION OF GRAVITATIONAL WAVES 4. Normal-Mode Vibrations of an Elastic Bar As of 1972, the most often-discussed type of detector is the aluminum bar invented by Jos.epn Weber (1960, 1961) [see Figures 37.1 and 37.2(d)]. Weber's bars are cylindrical in shape, with length 153 cm, diameter 66 cm, and weight 1.4 X 106 g. Each bar is suspended by a wire in vacuum and is mechanically decoupled from its surroundings. Around its middle are attached piezoelectric strain transducers, which couple into electronic circuits that are sensitive to the bar's fundamental end-to-end mode ofoscillation (frequency v = 1,660 Hz). When a gravitational wave hi ts the bar broadside, as shown in Figure 37.2(d), the rela tive accelerations carried by the wave will excite the fundamental mode of the bar. As of 1972, Weber observes sudden, simultaneous excitations in two such bars, one at the University of Maryland, near Washington, D.C.; the other at Argonne National Laboratory, near Chicago [see Weber (1969, 1970a,b)]. No one has yet come forward with a workable explana tion for Weber's coincidences other than gravitational waves from outer space. However, the history of physics is rich with instances where supposedly new effects had to be attributed in the end to long familiar phenomena. Therefore it would seem difficult to rate the observed events as \"battie-tested.\" To achieve that con fidence rating would seem to require confirmation with different equipment, or under different circumstances, or both; experiments to provide that confirmation are now (1972) underway. If one makes this tentative assessment, one can be excused for expressing at the same time the greatest admiration for the experimental ingenuity, energy, and magnificent persistence that Joseph Weber has shown in his more than decade-long search for the most elusive radiation on the books of physics. Mechanical detectors of the above four types represent systems on which meas urements have been made; so practical difficulties and realizable noise levels can be estimated properly. In the continuing search for improved methods, more elaborate detectors are being studied, and in 1972 one can list a number of interesting pro posals, as below. For these it is hard to know how JP.uch development would be required in order to achieve the desired performance. 5. Normal-Mode Vibrations of Elastic Bodies of Other", " Shapes The \"bar\" of a Weber detector need not be cylindrical in shape. For a discussion of a detector with the shape of a hollow square, a hoop, or a tuning fork, see Douglass (1971); such a detector might allow its fundamental frequency to be adjdted for the most favorable response, with given mass, or given maximum dimension, or both. Sections 37.4 and 37.7 and exercises 37.9 to 37.12 analyze in detail the operation of a \"vibrating-bar\" detector of arbitrary shape. See also Douglass and Tyson (1971). 6. Angular Accelerations of Rotating Bars (6) rotating bars (\"heterodyne detector\") All the potential detectors described thus far respond in the most obvious of manners to the tidal accelerations of a gravitational wave: relative distances oscillate in and \u00a737.3. TYPES OF MECHANICAL DETECTORS 1017 out. But the tidal accelerations contain, in addition to a length-changing component, also a tangential, rotation-producing component. In picture (a) of Box 37.2, the length-changing component dominates near the x and j axes, whereas the rotation producing component dominates half-way between the axes. Vladimir B. Braginsky was the first to propose a detector that responds to the rotation-producing accelera tions [see Braginsky, Zel'dovich, and Rudenko (1969); Braginsky and Nazarenko (1971)]. It consists of two metal rods, oriented perpendicular to each other, and rotating freely with angular velocity Wo in their common plane [see Fig. 37.2(0]. (The rotation is relative to the gyroscopes of the proper reference frame of the detector; equivalently, it is relative to the Lorentz frame local to the detector.) Let monochromatic gravitational waves of angular frequency w = 2wo (change of phase per unit of time equals twice the angular velocity at which the pattern of lines of force turns) impinge broadside on the rotating rods. The righthand circularly polar ized component of the waves will then rotate with the rods; so their orientation in its lines-of-force diagram will remain forever fixed. With the orientation of Fig. 37.2(0, rod A will undergo angular acceleration, while rod B will decelerate. The experimenter can search for the constant relative angular acceleration of the two rods (constant so long as the angle between them does not depart significantly from 90\u00b0). Better yet, the experimenter can (all too easily) adjust the rotation rate Wo so it does not quite match the waves' frequency w. Then for iwo/lw - 2wol rotations, rod A will accelerate and B will decelerate; then will follow!wo/lw - 2wol rotations in which A decelerates and B accelerates, and so on (frequency beating). The experimenter can search for oscillations in the relative orientation of the rods. One need not worry about the lefthand polarized waves marring the experiment. Since they do not rotate with the rods, their angular accelerations average out over one cycle. Such a device is called a \"heterodyne detector\" by Braginsky. He envisages that such detectors might be placed in free-fall orbits about the Earth late in the 1970's. Heterodyne detectors would work most efficiently for long monochromatic wave trains such as those from pulsars; but even for short bursts of waves they may be more sensitive than vibrating bars [see Braginsky and Nazarenko (1971 )]. 7. Angular Accelerations of Driven Oscillators Andrei D. Sakharov (1969) has proposed a different type of detector for the angular accelerations of a gravitational wave. Instead of two rotating bars, it consists of two identical, driven oscillators, initially parallel and nonrotating, but oscillating out of phase with each other. Each oscillator experiences angular accelerations in one direction at one phase of a passing wave, and in the opposite direction at the next phase, but the torques do not cancel out. When the oscillator is maximally distended, it experiences a greater torque (acceleration ex length; torque ex length2) than when it is maximally contracted. Consequently, if the driven oscillations have the same angular frequency as a passing, monochromatic wave, and if the phases are as shown in Figure 37.2(g), then oscillator A will receive an angular acceleration in the right hand direction, while B receives an angular acceleration in the lefthand direction. (7) rotation of driven oscillators (8) fluid in pipe EXERCISES 1018 37. DETECTION OF GRAVITATIONAL WAVES 8. Pumping of Fluid in a Rotating Loop of Pipe A third type of detector that responds to angular accelerations has been described by William Press", " (1970). This detector would presumably be far less sensitive than others, \"and therefore not worth constructing; but it is intriguing in its novel design; and it illustrates features of gravitational waves ignored by other detectors. Press's detector consists of a loop of rotating pipe, containing a superfluid. The shape of the pipe and its constant rotation rate are chosen so that the gravitational waves will pump the fluid around inside the pipe. One conceivable pipe design (a bad one to build in practice, but an easy one to analyze) is shown in Fig. 37.2(h). Note that use is made of the variation in tidal acceleration along the direction ofpropaga tion of the wave as well as perpendicular to that direction. To analyze the response of the fluid to a righthand circularly polarized wave, one can mentally place the rotating pipe in the three-dimensional line-of-force diagram of Box 37.2(c). Exercise 37.6. RELATIVE MOTION OF FREELY FALLING BODIES AS A DETECTOR OF GRAVITATIONAL WAVES [see Figures 37.2(a) and 37.3.] Consider two test bodies initially at rest with respect to each other in flat, empty spacetime. (The case where other, gravitating bodies are nearby can be treated without too much more difficulty; but this exercise concerns only the simplest example!) A plane, nearly monochro matic gravitational wave, with angular frequency wand polarization e+, impinges on the bodies, coming from the -z direction. As shown in exercise 35.5, the bodies remain forever at rest in those TT coordinates that constituted the bodies' glo~bal inertial frame before the wave arrived. Calculate, for arbitrary separations (Ax, Ay, Az) of the test bodies, the redshift and the round-trip travel time of photons going back and forth between them. Compare the answer, for large Ax, Ay, Az, with the answer one would have obtained by using (without justification!) the equation of geodesic deviation. Physically, why does the correct answer oscillate with increasing separation? Discuss the feasibility and the potential sensitivity of such a detector using current technology. ~ Exercise 37.7. EARTH-MOON SEPARATION AS A GRAVITATIONAL-WAVE DETECTOR In the early 1970's one can monitor the Earth-moon separation using laser ranging to a precision of 10 cm, with successive observations separated by at least one round-trip travel time. Suppose that no oscillations in round-trip travel time are observed except those (of rather long periods) to be expected from the Earth-moon-sun-planets gravitational interac tion. What limits can one then place on the energy flux of gravitational waves that pass the Earth? The mathematical formula for the answer should yield numerically ~ Flux ~ 1018 erg/cm2 sec for OJ cycle/sec ~ v ~ I cycle/day, (37. lOa) corresponding to a limit on the mass density in gravitational waves of Density ~ 10-13 g/crn3. (37. lOb) Why is this an uninteresting limit? \u00a737.4. VIBRATING DETECTORS: INTRODUCTORY REMARKS 1019 \u00a737.4. VIBRATING, MECHANICAL DETECTORS: INTRODUCTORY REMARKS The remainder of this chapter (except for \u00a737.9) gives a detailed analysis of vibrating, mechanical detectors (Earth; Weber bar; \"bars\" with complex shapes; and so on). The details of the analysis and its applications depend in a crucial way on the values of two dimensionless numbers: (1) the ratio Taw/To, where r ) _ (Characteristic time scale for changes in Taw = gravitational-wave amplitude and spectrum'1'0 e-fOlding time for detector vibrations (in) normal mode of interest) to die out as a result of internal damping ( ; (37.l1a) (37.l1b) The rest of this chapter is Track 2. No earlier track-2 material is needed as preparation for it, nor is it needed for any later chapter. and (2) the ratio Evibration/kT, where _ Evibration \u2022 (mean value of detector's vibration energy (in) normal mode of interest) while waves are passing and driving detector, (37.l2a) kT (Boltzman's) X (detector's ) constant temperature Mean energy in normal mOde) of interest when g:~vit~tional. = - waves are not eXCItmg It ( (37.12b) When 'Taw ~ 'To, the detector views the radiation as having a \"steady flux,\" and it responds with steady-state vibrations; when Taw ~ 1'0 (short burst of waves), the waves deal a \"hammer blow\" to", "AVES Figure 37,4, An idealized detector (vibrator) responding to linearly polarized gravitational waves. of the antenna, and on other, more detailed characteristics of its response, on coupling of the antenna to the displacement sensor, on response characteristics of the sensor, on antenna noise, on sensor noise, and so on. For an overview of these issues, and for discussions of detectors for which the concept of cross section is useless, see, e.g., Press and Thorne (1972). \u00a737.5. IDEALIZED WAVE-DOMINATED DETECTOR, EXCITED BY STEADY FLUX OF MONOCHROMATIC WAVES Idealized detector: oscillator driven by a steady flux of monochromatic waves: (1) derivation of equation of motion Begin with the case of a wave-dominated detector (Evibration ~ kT) being driven by a steady flux of radiation ('Taw ~ 'To)' Deal at first, not with a solid bar of arbitrary shape, but rather with the idealized detector of Figure 37.4: an oscillator made of two masses M on the ends of a spring of equilibrium length 2L. Let the detector have a natural frequency of vibration Wo and a damping time 'To ~ l/wo' so that its equation of motion (in the detector's proper reference frame) is ~ + tI'To + w6~ = driving acceleration. (37.l3) Let gravitational waves of polarization e+ and angular frequency w impinge on the detector from the - i direction; and let the polar angles of the detector relative to. the wave-determined x,y, i-axes be () and cp. The incoming waves are described by equations (37.1) with the amplitude Ax = 0, (37.14) (Here and throughout one must take the real part of all complex expressions.) \u00a737.5. IDEALIZED DETECTOR, DRIVEN BY STRONG, STEADY FLUX 1023 Assume that the detector is much smaller than a wavelength, so that one can set z :::::: i = 0 throughout it. Then the tidal accelerations produced by the wave = -R-A'-X] - -lw 2d e-iwtx zUJO - + 2'(ti) dt 2 2y ) d di 2 ( due to wave due to wave have as their component along the oscillator Consequently, the equation of motion for the oscillator is (37.15) The driving force varies as cos 2et> because of the \"spin-2\" nature of gravitational waves: a rotation through 180 0 in the transverse plane leaves the waves unchanged; a rotation through 90 0 reverses the phase. The sin2(1 term results from the transverse nature of the waves [one factor of sin (I to account for projection onto the detector's direction], plus their tidal-force nature [another factor of sin (I to account for (relative force) ex (distance in transverse plane)]. The straightforward -steady-state solution of the equation of motion (37.15) is ~ = ~ w 2d +L sin2(1 cos 2et> 2 + iwlTo w2 - w0 _ e-iwt. (37.16) (2) oscillator amplitude as function of frequency and orientation Iw -+ wol ~ liTo, When the incoming waves are near resonance with the detector, the oscillator is excited to large amplitude. Otherwise the excitation is small. Focus attention henceforth on near-resonance excitations; then equation (37.16) can be simplified (note: W o is positive, but w may be negative or positive): lwod+L sin2(1 cos 2et> ~ = 4 _ e-iwt. Iwl - Wo + ; sgn(w)iITo (37.16') One measure of the detector's usefulness is its cross section for absorbing gravita tional-wave energy. The steady-state vibrational energy in a detector with the above amplitude and with 2 masses Mis (37.17) v 1024 37. DETECTION OF GRAVITATIONAL WAVES This energy is being dissipated internally at a rate Evibration/'TO. If one ignores reradiation of energy as gravitational waves (a negligible process!), one can equate the dissipation rate to the rate at which the detector absorbs energy from the incomi!1g waves-which in turn equals the \"cross section\" a times the incoming flux: (3) cross sections for polarized radiation Consequently, near resonance (Iw -+- wol ~ wo), the cross section for interception of gravitational-wave energy is for polarized radiation. (37.18) This expression applies to monochromatic radiation. However, experience with many other kinds of waves has taught that one often has to deal with a broad", " continuum of frequencies, with the \"bandwidth\" of the incident radiation far greater than the width of the detector resonance (see Box 37.3). Under these conditions, the relevant quantity is not the cross section itself, but the \"resonance integral\" of the cross section, f resonance a dv =: f a(dw/27i) =: 27iMUw0 2sin4(1 cos22ep, (37.19) for polarized radiation. Before examining the magnitude of this cross section, scrutinize its directionality (the \"antenna-beam pattern\"). The factor of sin4(1 cos22ep refers to linearly polarized, e+ radiation (see Figure 37.4). For the orthogonal mode\"of polarization, ex' cos22ep is to be replaced by sin22ep; and for unpolarized (incoherent mixture) radiation or circularly polarized radiation, the cross section is the average of these two expres sions; thus for unpolarized radiation. (37.20) Notice that this unpolarized cross section is peaked, with half-width 33 0, about the equatorial plane of the detector. Averaged over all possible directions of incoming waves, the cross section is (4) crosS sections for unpolarized radiation (a)a1ldirections =:\"2 a Sill (I d(l = 15amax 8 1 f'lr\u00b7 o = I I (87i/15)ML2(w0 (w - wo) + (1 / 2 2/'To) 2 'To) 2 for unpolanzed radiation.... (37.21) One can rewrite the above cross sections in several suggestive forms. For example, On resonance, the cross section (37.21) reads \u00a737.5. IDEALIZED DETECTOR, DRIVEN BY STRONG, STEADY FLUX 1025 Recall that Wo'T'o defines the \"Q\" of a detector, l/Q = (fraction of energy dissipated per radian of oscillation). Note that 27T/ Wo is the wavelength AO of resonant radiation. Finally, denote by rg = 4Mthe gravitational radius of the detector. In terms of these three' familiar quantities, find for the cross section the formula (a)a1ldirections _ (cross section for absorbing waves on resonance) (2L)2 (\"geometric\" cross section of detector) Magnitude of cross sections for any resonant detector = (47T 2/15)(r/Ao)Q for unpolarized radiation on resonance. (3722) This relation holds in order of magnitude for any resonant detector. It shows starkly that gravitational-wave astronomy must be a difficult enterprise. How large could you make the factor rgiAo' given a reasonable budget? Weber's 1970 detectors have 2Leffecti~e:::::: 1 meter, r g:::::: (0.74 X 10-28 em/g) X (106 g):::::: 10-22 em, Po = W O/27T = 1,660 Hz, AO :::::: 200 km, rgiAo :::::: ~ X 10-29, TO:::::: 20 sec, Q:::::: 2 X 105 ; so that aWeber :::::: 3 X 10-20 cm2 on resonance. (3723) What flux of gravitational-wave energy would have to be incident to excite a cold detector (-0\u00b0 K) into roughly steady-state vibrations with a vibration energy of (Boltzmann's constant) X (room temperature) - 4 X 10-14 erg? The vibrator, if cooled enough to be wave-dominated, dissipates its energy at the rate Evibration/To - 2 X 10-15 erg/sec. The incident flux has to make up this loss, at the rate Flux required to excite a Weber-type detector Tb~W)a - 2 X 10-15 erg/sec, (37.24a) implying an incident flux of the order of 2 X 10-15/3 X 10-20 Moreover, this flux has to be concentrated in the narrow range of resonance 105 erg/cm2 sec. - P:::::: Po -+ 1/47TTo = (1660 -+ 0.004) Hz. (37.24b) By anybody's standards, this is a very high flux of gravitational radiation for such a small bandwidth (-107 erg/cm2 sec Hz, as compared to the flux of blackbody gravitational radiation, 87Tp 2kT/c 2 = 3 X 10-27 erg/cm2 sec Hz, that would corre spond to Planck equilibrium at the same temperature; the large factor of difference is a direct reflection of the difference in rate of damping of the oscillator by friction and by gravitational radiation). Equation (37.", "/cm2) = f r<o~W) dt = f_l_A+2 dt = f 2 w21A 1 161T + dw = 161T f w21A 1 8 2 + dv (37.30) (Parseval's theorem again!). Consequently, the energy per unit frequency interval, per unit area carried by the waves is [for - 00 < v < + 00; double this for 0 < v < + 00, a convention we use for the rest of this chapter]. This is 21T times the second term in (37.29). Combining equations (37.18), (37.29), and (37.31), then, one finds (37.31) total ~nergy) = fa(v):f(v) dv. deposIted \u2022 ( (37.32) (2) answer( energ~ deposited ) =Jo~. dp How One can measure energy deposited 1028 37. DETECTION OF GRAVITATIONAL WAVES This is the total energy deposited, regardless ofthe spectrum ofthe waves, and regardless of whether they come in a steady flux for a long time, or in a short burst, or in any other form. It is perfectly general-so long as the detector is wave-dominated (Evibration ~ kT) while the waves are driving it. How can an experimenter measure the total energy deposited? He cannot measure it directly, in general, but he can measure a quantity equal to it: the total energy that goes into internal damping, i.e., into \"friction.\" Energy is removed by \"friction\" at a rate Evibration/TO' when the vibration energy is much greater than kT (during period of wave-dominance). Therefore, the experimenter can measure ( total ~nergy) =.If Evibration dt, deposIted TO in general. 4integrate over the period that Evibration ~ kTf (37.33) In the special case of \"hammer-blow waves\" (TGW = duration of waves ~ TO)' the vibration energy is driven \"instantaneously\" from -kT to a peak value, E~~a~tion ~ kT, and then decays exponentially back to -kT; thus total energy) =.If\"\" Epeak. vibration deposited To 0 ( e-tlTo dt = Epeak. vibration (37.34) for hammer-blow waves. When the waves are steady for a long period of time (TGW ~ TO)' with specific flux (ergs/cm2 sec Hz), then the energy will be deposited at a constant rate (dEjdt) = (total energy deposited)/TGW; and equation (37.32) can be rewritten rate of deposit) = fa(v)F. dv, of energy ( for steady waves (TGW ~ TO)' (37.35) Equations (37.32) and (37.35) are the key equations for application of the concept of cross section to realistic situations. They are applicable not only to polarized radiation, but also to unpolarized radiation and to radiation coming in froin all directions, if one merely makes sure to use the appropriate cross section!equation (37.20) or (37.21) instead of(37.18)]. For examples of their application, see Box 37.3. \u00a737.7. GENERAL WAVE-DOMINATED DETECTOR, EXCITED BY ARBITRARY FLUX OF RADIATION The cross sections of the idealized spring-plus-mass detector can be put into a form more elegant than equations (37.18) to (37.21)-a form that makes contact with many \u00a737.7. GENERAL DETECTOR DRIVEN BY STRONG, VARYING FLUX 1029 branches of physics, and is valid for any vibrating resonant detector whatsoever. Introduce the \"Einstein A-coefficients,\" which describe the rate at which a unit amount of detector energy is lost to internal damping and to reradiation of gravita tional waves: Cross sections reexpressed in terms of \"Einstein A-coefficients\" = (rate at which energy is dissipated internally) _ ~ 'To' - energy in oscillations of detector - Adiss = (rate at which energy is reradiated) - energy in oscillations. A GW (37.36a) (37.36b) For the idealized detector of Figure 37.4, the standard fonnula (36.1) for the emission of gravitational waves yields (see exercise 37.8). Consequently A -liML2 4 w. 15 GW - (37.37) (37.38) One can use these relations,to rewrite the detector cross sections in terms of AdiSS' AGW'and the reduced wavelength A= l/w (37", ".39) of the radiation. For example, the cross section (37.2 I)-now with w;::: O-is for unpolarized radiation (37.40) (recall the assumption Iw - wol ~ W o in all cross-section formulas) and the corre sponding integral over the resonance is f (a)a1ldirections dv = ~ 7TA 02AGW for polarized radiation. (37.41) These expressions for the cross section are comprehensive in their application. They apply to any vibrating, resonant, gravitational-wave detector whatsoever, as one sees from the \"detailed balance\" calculation of exercise 37.9, and from the dynamic calculations of exercise 37.10. They also apply, with obvious changes in statistical factors and notation, to compound-nucleus reactions in nuclear physics (\"Breit-Wig ner formula\"; see Blatt and Weisskopf, pp. 392-94,408-10,555-59), to the absorption of photons by atoms and molecules, to reception of electromagnetic waves by a television antenna, etc. Equation (37.41) says in effect, \"Calculate the rate at which the oscillator is damped by emission of gravitational radiation; multiply that rate by the geometric factor familiar in all work with antennas, ~I7A02, and immediately obtain the resonance integral of the cross section. The result is expressed in geometric Generality of the A-coefficient formalism Scattering of radiation by detector EXERCISES 1030 37. DETECTION OF GRAVITATIONAL WAVES units (cm). To get conversion factor c = 3 X 10 10 cm Hz. the resonance integral in conventional units, multiply by the The 'dynamic analysis' of the idealized masses-on-spring detector, as developed in the last section, is readily extended to a vibrating detector of arbitrary shape (Earth; Weber's bar; an automobile fender; and so on). The extension is carried out in exercise 37.10 and its main results are summarized in Box 37.4. Part of the energy that goes into a detector is reradiated as scattered gravitational radiation. For any detector of laboratory dimensions with laboratory damping coefficients, this fraction is fantastically small. However, in principle one can envisage a larger system and conditions where the reradiation is not at all negligible. In such an instance one is dealing with scattering. No attempt is made here to analyze such scattering processes. For a simple order-of-magnitude treatment, one can use the same type of Breit-Wigner scattering formula that one employs to calculate the scattering of neutrons at a nuclear resonance or photons at an optical resonance. A still more detailed account will analyze the correlation between the polarization of the scattered radiation and the polarization of the incident radiation. The kind of formalism useful here for gravitational radiation with its tensor character will be very much like that now used to treat polarization of radiation with a spin-l character. Here notice especially the \"Madison Convention\" [Barschall and Haeberli (1971)] developed by the collaborative efforts of many workers after experience during many years with a variety of conflicting notations. Considering the way in which the best notation that is available today for spin-l radiation was evolved, one can only feel that it is too early to canonize anyone notation for describing the scattering parameters for an object that is scattering gravitational radiation. Exercise 37.8. POWER RERADIATED The idealized gravitational wave detector of Figure 37kvibrates with angular frequency w. Show that the power it radiates as gravitational waves is given by equation (37.37). Exercise 37.9. CROSS SECTIONS CALCULATED BY DETAILED BALANCE Use the principle of detailed balance to derive the cross sections (37.41) for a vibrating, resonant detector of any size, shape, or mass (e.g., for the vibrating Earth, or Weber's vibrating cylinder, or the idealized detector of Figure 37.4). [Hints: Let the detector be in thermal equilibrium with a bath of blackbody gravitational waves. Then it must be losing energy by reradiation as rapidly as it is absorbing it from the waves. (Internal damping can be ignored because, in true thermal equilibrium, energy loss by internal damping will match energy gain from random internal Brownian forces.) In detail, the balance of energy in and out reads [with Iv = \"specific intensity,\" equation (22.48)] [4?TIv(v = VO)]blackbOdy X J<o)alldirectlons dv = AGW X (Energy in normal mode of detector). Solve for J<o) dv, using the familiar form of the Planck spectrum and the fact that gravita tional waves have two independent states of polarization.] Note: Because detailed balance \u00a737.7.", " GENERAL DETECTOR DRIVEN BY STRONG, VARYING FLUX 1031 Box 37.4 VIBRATING. RESONANT DETECTOR OF ARBITRARY SHAPE A. Physical Characteristics of Detector 1. Detector is a solid object (Earth, Weber bar, automobile fender,...) with density distribution p(x) and total mass M = fp d3x. 2. Detector has normal modes of vibration. The nth normal mode is character ized by: W n = angular frequency; e-fOlding time for vibration energy ) 'Tn = to decay as result of internal damping ~ l/wn ; ( un(x) = eigenfunction (defined here to be dimensionless and real). The eigenfunctions un are orthonormalized, so that (1) (2) 3. During a nOIlllal-mode vibration with Evibration ~ kT, a mass element origi nally at x receives the displacement l)x = ~ = un(x)g'Jne-iw.t- tIT., 4constant amplitude] the densiiy at fixed x changes by and the moment of inertia tensor oscillates Here llnljk is the \"moment of inertia factor for the nth normal mode\": -f' llnljk = - (PUn ),tXIX d X. k 3 t (3a) (3b) (3c) (4) [dimensions: mass X length, multiply by ~i1n (length) to get lid. The corresponding \"reduced quadrupole factor for the nth normal mode\" is (5) /?) / 1032 37. DETECTION OF GRAVITATIONAL WAVES Box 37.4 (continued) B. Cross. Sections for Detector (exercise 37.10) 1. For polarized radiation with propagation direction n and polarization ten sor e: hjk = A(t - n \u2022 x)ejk, (6) (7a) (7b) 2. For unpolarized radiation (random mixture of polarizations) with propaga tion direction n, cross sections are (8a) (8b) Here fGJjk is the transverse-traceless part of f(nljk (transverse and traceless relative to the propagation direction n): (9) (See Box 35.1) 3. Cross sections for unpolarized radiation, averaged over all directions, are (an (v)aIIdirections - 2 TTA (jwj - w _ ~ 2 AGWAdiss }2 + (A n /2}2' diss f resonance (an) all directions dv = ~?TA 2AGW' where the Einstein A coefficients are Adiss = 1/'Tn, 1 (f(nljk)2 A GW = 5 M wn \u2022 4 (lOa) ~ (lOb) (l1) (l2) \u00a737.7. GENERAL DETECTOR DRIVEN BY STRONG. VARYING FLUX 1033 C. Spectrum Radiated by an Aperiodic Source (exercise 37.11) It is instructive to compare these formulas with expressions for the radiation emitted by an aperiodic source. 1. Fourier-analyze the reduced quadrupole factor of the source fjk(t) = (27T)-1/2f +00 f;k(w)e- iwt dw. -00 2. Then the total energy per unit frequency (v ;::: 0) radiated over all time, into a unit solid angle about the direction n, and with polarization tensor e, is [compare with equations (7)]. Summed over polarizations, this is (13a) (13b) [compare with equations (8)]. Here v ;::: 0. 3. The total energy radiated per unit frequency, integrated over all directions, still with v ;::: 0, is (14) [compare with equations (10)-(12)]. can be applied to any kind of resonant system in interaction with any kind of thermal bath of radiation or particles, equations (37.40) and (37.41), :.vith appropriate changes of statistical factors, have wide generality. Exercise 37.10. NORMAL-MODE ANALYSIS OF VIBRATING. RESONANT DETECTORS Derive all the results for vibrating, resonant detectors quoted in Box 37.4. Pattern the derivation after the treatment of the idealized detector in \u00a737.6. (Guidelines: (a) Let the detector be driven by the polarized waves of equation (6), Box 37.4; and let it be wave-dom inated (Evibration ;p kT). Show that the displ", "acements ox = ~(x, t) of its mass elements are described by (37.42a) y 1034 37. DETECTION OF GRAVITATIONAL WAVES where the time-dependent amplitude for the nth mode satisfies the driven-oscillator equation and where the curvature-induced driving term is \u00a5 \u2022 (See Box 37.4 for notation.) (b) Fourier-analyze the amplitudes of the detector and waves, A(t) = (2~)-1!2l\", A(w)e-iwt dw, +'\" (37.42b) (37.42c) (37.42d) (37.42e) and solve the equation of motion (37.42b,c) to obtain, in the neighborhood of resonance, (c) Calculate the total energy deposited in the detector by integrating (denerg~ d) =J(Force per ~nit volume)' (Velocity) d3x dr. eposlte - Thereby obtain ( energy deposited in) =1. (l nth normal mode 4. e ) JAB dr. (n)Jk]k n (d) Apply Parseval's theorem and combine with expression (37.42f) to obtain energy deposited in) =J0 (v)\u00a7' (v) dv, ( nth normal mode n P where On is given by equation (7a) of Box 37.4, and (for - 00 < W < + 00) (37.42f) (37.43) ) (37.44) (e) Show that :fp(v) is the total energy per unit area per unit frequency carried by the waves past the detector. (f) Obtain all the remaining cross sections quoted in Box 37.4 by appropriate manipula tions of this cross section. Use the mathematical tools for projecting out and integrating \"transverse-traceless parts,\" which were developed in Box 35.1 and exercise 36.9. Exercise 37.11. SPECTRUM OF ENERGY RADIATED BY A SOURCE Derive the results quoted in the last section of Box 37.4. \u00a737.7. GENERAL DETECTOR DRIVEN BY STRONG. VARYING FLUX 1035 Exercise 37.12. PATTERNS OF EMISSION AND ABSORPTION The elementary dumbbell oscillator of Figure 37.4, initially unexcited, has a cross section for absorption of unpolarized gravitational radiation proportional to sin48, and when excited radiates with an intensity also proportional to sin48 (Chapter 36). The patterns of emission and absorption are identical. Any other dumbbell oscillator gives the same pattern, apart from a possible difference of orientation. Consider a nonrotating oscillator of general shape undergoing free vibrations in a single nondegenerate (and therefore nonrotatory) mode, or excited from outside by unpolarized radiation. (a) Show that its pattern ofemission is identical with its pattern of absorption. [Hint: Make the comparisons suggested in the last few parts of Box 37.4.] (b) Show that this emission pattern (= absorption pattern), apart from three Euler angles that describe the orientation of this pattern in space, and apart from a fourth parameter that determines total intensity, is uniquely fixed by a single (\"fifth\") parameter. (c) Construct diagrams for the pattern of intensity for the two extreme values of this parameter and for a natural choice of parameter intermediate between these two extremes. (d) -Define the parameter in question in terms of a certain dimensionless combination of the principal moments of the reduced quadrupole tensor. Exercise 37.13. MULTIMODE DETECTOR Consider a cylindrical bar of length very long compared to its diameter. Designate the fundamental mode of end-to-end vibration of the bar as \"n = I,\" and call the mode with n - 1 nodes in its eigenfunction the \"nth mode.\" Show that the cross section for the inter ception of unpolarized gravitational waves at the nth resonance, integrated over that reso nance, and averaged over direction, is given by the formula [Ruffini and Wheeler (1971b\u00bb) ~ 1 nth resonance; random o(v)dv = -15-2-2 for n odd (zero for even n), 32 v 2 M 'iT C n (37.45) where v is the speed of sound in the bar expressed in the same units as the speed of light, c; and M is the mass of the bar (geometric units; multiply the righthand side by the factor G/c = 2.22 X 10- 18 cm2 Hz/g when employing conventional units). Show that this expression gives fo dv = 1.0 X 10-21 cm2 Hz for the lowest mode of Weber's bar. Multimode detectors are", " (1973) under construction by William Fairbank and William Hamilton, and by David Douglass and John A. Tyson. Exercise 37.14. CROSS SECTION OF IDEALIZED MODEL OF EARTH FOR ABSORPTION OF GRAVITATIONAL RADIATION The observed period of quadrupole vibration of the earth is 54 minutes [see, e.g., Bolt (1964) or Press (1965) for survey and bibliography]. To analyze that mode of vibration, with all due allowance for elasticity and the variation of density in the earth, is a major enterprise. Therefore, for a first estimate of the cross section of the 'earth for the absorption of quadrupole radiation, treat it as a globe of fluid of uniform density held in the shape of a sphere by gravitational forces alone (zero rigidity). Let the surface be displaced from r = a to r = a + aaP2(cos8), (37.46a) where 8 is polar angle measured from the North Pole and a is the fractional elongation of the principal axis. The motion of lowest energy compatible with this change of shape is described by the velocity field e = -'2 ax, I e = az (37.46b) (zero divergence. zero curl). 1036 37. DETECTION OF GRAVITATIONAL WAVES (a) Show that the sum of the kinetic energy and the gravitational potential energy is (b) Show that the angular frequency of the free quadrupole vibration is (c) Show that the reduced quadrupole moments are (37.46c) (37.46d) (37.46e) (d) Show that the rate of emission of vibrational energy, averaged over a period, is (37.46f) (e) Show that the exponential rate of decay of energy by reason of gravitational wave damping, or \"gravitational radiation line broadening,\" is (37.46g) (f) Show that the resonance integral of the absorption cross section for radiation incident from random directions with random polarization is J reSOnance (o(v\u00bb d,' = ('1T/2)..t 2A GW = (2'1Tj25)Ma 2/..t 2 \u2022 (37.46h) (g) Evaluate this resonance integral. Note: This model of a globe of fluid or\" uniform density would imply for the earth, with average density 5.517 g/cm3, a quadrupole vibration period of 94 min, as compared to the observed 54 min; and a moment of inertia (2/5)Ma 2 as compared to the observed O.33Ma 2\u2022 Ruffini and Wheeler (1971 b) have estimated correction factors for both effects and give for the final resonance integral -5 cm2 Hz. \u00a737.8. NOISY DETECTORS When the bandwidth of the incoming waves is large compared to the resonance width of the detector, the waves deposit a total energy in the detector given by (total energy deposited) = fog'\". dv = g'\".(v = Vo) f 0 dv. [ergst-l [erg cm-2Hz-l~ [cm2Hzr1 Extraction of small signal from large noise-general remarks At least, this is so if the detector is wave-dominated (Le., if Evibration ~ k'1 while waves act; i.e., if initial amplitude of oscillation, produced by Brownian forces, is too small to interfere constructively or destructively with the amplitude due to waves). Unfortunately, all experiments today (1973) are faced with noisy detectors. Nobody has yet found waves so strong, or constructed a detector so sensitive, that the detector is wave-dominated. Consequently, a key experimental task today is to pick a small signal out of large noise. Many techniques for doing this have been developed and used in a variety of fields of physics, as well as in astronomy, psychology and engineering [see, e.g., Davenport and Root (1958), Blackman and Tukey (1959), and \u00a737.B. NOISY DETECTORS 1037 Figure 37.5. Detection of hammer-blow gravitational waves with a noisy detector. Detection of even a weak pulse is possible if the time of the pulse is short enough. The amplitude g,. of the detector's vibrations changes by an amount _~~ms (Jllr.)112 during a time interval Jl, due to thermal fluctuations (random-walk, Brownian-noise forces). Depicted in the inset is a change in amplitude produced by a burst of waves of duration 'TGW arriving out of phase with the detector's thermal motions (energy extracted by waves!). The waves are detectable because even though JI~.I -{ ~~ms. JI~.ldu.towav\u2022\u2022 }> ~~ms ('TGw/'Tn)1I2. ~", " references given there]. The key point is always to find some feature of the signal that is statistically more prominent than the same feature of the noise, plus a correlation to show that it arises from the expected signal source and not from elsewhere (\"protection from systematic error\"). Thus to detect steady gravitational waves from a pulsar, one might seek to define very precisely two numbers (N2) and \u00abN + S)2) = (N2) + (S2), where Nand S are the noise and signal amplitudes respectively. A long series of observations (with the pulsar out of the antenna beam) gives one value of (N2). Another equally long series of observations, interspersed with the first series, will be expected in zeroth approximation to give the same value of (N2). In the next approximation one recognizes and calculates the influence of normal statistical fluctuations. In an illustrative example, theory, confirmed by statistical tests of other parameters drawn from the same data, guarantees that the fluctuations are less than 1O-5(N2) with 95 per cent confidence (only 5 per cent chance of exceeding 1O-5(N2); this limit is set by time and money, not by absolute limitations of physics). Let the second series of observations be carried out only at times when the pulsar is in the antenna beam. Let it give ____________Then in first approximation one can say that (S2) lies with 95 per cent confidence ~ in the limits (7.3 -+- 1.0) X 1O-5(N2). Many conceivable sources of gravitational radiation produce bursts rather than a steady signal strength (Figure 37.5). Thus one is led to ask in what features \"hammer-blow radiation\" (-Tow ~ 'To) differs from noise. The \"Brownian motion\" noise in the detector may be thought of as arising from large numbers of small Rate-of-change of detector amplitude as a tool for extracting burst signals from thermal noise y Description of thermal noise in resonant detector 1038 37. DETECTION OF GRAVITATIONAL WAVES (molecular) energy exchanges with a heat bath. The calculations below estimate the typical rate of change of amplitude that a series of such molecular \"knocks\" can produce in a detector, and compare it with the rapid amplitude change produced by a \"hammer-blow\" pulse of radiation. The calculations show that sudden thermally induced changes, even of very small amplitude, are rare. Thus sudden changes are a suitable feature for the observations to focus on. The actual detection of pulses requires a more extended analysis, however, which goes beyond the estimates made below. Such an analysis would calculate the probabilities that rare events (sudden changes in amplitude) occur by chance (i.e., due to thermal fluctuations) in specified periods of time, the still smaller probabilities that they occur in coincidence between two or more detectors, and the correlations with postulated sources. Consider a realistic detector of the type described in Box 37.4. But examine it at a time when it is not radiation-dominated. Then its motions are being driven by internal Brownian forces (thermal fluctuations), and perhaps also by an occasional burst of gravitational waves. Focus attention on a particular normal mode (mode \"n\"), and describe that mode's contribution to the vibration of the detector by the vector field (37.47) 2 d 3x = M), the complex Since un is dimensionless with mean value unity (Jpun number ~:Bn(t) is the mass-weighted average of the amplitudes of motion of the detector's mass elements. This amplitude changes slowly with time (rate ~ wn ) as a result of driving by Brownian forces; but averaged over time it has a magnitude corresponding to a vibration energy of kT: l.e., (37.48) (37.49) Example: for Weber's detector (M -- 103 kg, W o at room temperature has 104/sec), the fundamental mode!t3rms = o ( 2 X 1.38 X 10-16 X 300 erg )1/2 106 g X 108 sec 2 = 3 X 10-14 em. (37.50) ~ One's hope for detecting weak hammer-blow radiation lies not in an examination of the detector's vibration amplitude (or energy), but in an examination of its rate of change (Figure 37.5). The time-scale for large Brownian fluctuations in amplitude!t3~ms), when the detector is noisy, is the same as the time scale Tn for internal (1.:i!t3nl forces to damp the detector, when it is driven to Evibration ~ kT. Thus, the amplitude!t3n does a \"random walk\" under the influence of Brownian forces, with the mean time g'1~ms) being.:it ;:::: Tn\" The change in g'1n over shorter times for \"large walks\" (1.:i!", "t3nl.:it is smaller by the \"1/VN factor,\" which always enters into random-walk processes: \u00a737.8. NOISY DETECTORS -IN = ( number of vibration cycles in time Tn )1/2 =(.2...)1/2..:it' <1.:ig'1~thermaJ)l) :::: g'1bms (~)1/2 = ( 2k~ )1/2 (.:it )1/2 ~uring. tlme.:it number of vibration cycles in time.:it Tn Tn MWn 1039 (37.51) (37.52)* Effect of a burst of waves on a noisy. resonant detector Now suppose that \"hammer-blow\" radiation (burst of duration.:it = TOW ~ Tn) strikes the detector, producing a change.:ig'1~ow) in the detector's amplitude. This change in amplitude, because it comes so quickly, (1) superposes linearly on any change in amplitude produced in the same time interval by the action of Brownian motion forces; and (2) is therefore independent in value of the presence or absence of Brownian-motion forces, Le., independent of all thermal agitation. Therefore.1.q'1~OW) (a quantity with both magnitude and phase!) is identical to what it would have been if the detector were at zero temperature: +Mw; I.:iB~OW) 12 = f an(vYf.(v) dv = :1.(wn/2'iT) f an(v) dv; t energy that would be deposited if detector were at zero temperature iFor hammer-blow radiation, bandwidth of radiation is always ~ bandwidth of detector; see Box 37.4 i.e., (37.53). This wave-induced change in amplitude will be distinguishable from thermal changes only if it is significantly bigger than the thermal changes (37.52) expected during the same length of time TOW: 1.:i~t'~OW) 1 ~ <1.jg'1~thermaJ) I) during time TOWI k equivalently: F.(wn /2'iT) ~ ( criteria for T \\ C~:) detectability fan dv1 Criteria for detectability of burst (37.54) Of course, if one is equipped only to measure the magnitude of the detector's amplitude or energy, and not its phase, these criteria for detectability are not quite sufficient. The wave-induced change in squared amplitude (proportional to change in energy) will depend on the relative phases of the initial amplitude and amplitude change * For a fuller derivation and discussion of this formula. see, e.g.. Braginsky (1970). Two key points covered there are: (I) a statistical version of the formula, which d~scribes the probability that in time J( the amplitude will change by a given amount. from a given initial value; and (2) quantum-mechanical corrections, which come into play in the limit as.,.. -+ oc:. but which are unimportant for detectors of the early 1970's. y Ways to improve sensitivity of detector Non-mechanical detectors The future of gravitational-wave astronomy 1040 37. DETECTION OF GRAVITATIONAL WAVES.j1~BnI2 = Igj~nitiall +.jg,~GW)12 _ 1~'il~nitiaIl12 :::: 21!11~nitiallll.j!1,~GW)1 :::: 0 :::: _21!1,~nitiaIlIW1,~GW)1 if in phase if phase difference is -+-'iT/2 if phase difference is 'iT. (37.55) Thus, only a burst that arrives in phase with the initial motion of the detector or with reversed phase will be measurable. But for such a burst, the criteria (37.54) are sufficient. Equations (37.54) make it clear that there are three ways to improve the sensitivity of vibratory detectors to hammer-blow radiation: (l) increase the detector's integrated cross-section [which can be done only by increasing the rate AGW at which it reradiates gravitational waves; see equations (lOb) and (l1 b) of Box 37.4]; (2) cool the detector; (3) increase the detector's damping time. Box 37.5 applies the above detectability criteria to some detectors that seem feasible in the 1970's, and to some bursts of waves predicted by theory. The conclu sions of that comparison give one hope! To be complete, the above discussion should have analyzed not only noise in the detector, but also the noise in the sensor which one uses", " to measure the amplitude of the detector's displacements. However, the theory of displacement sensors is beyond the scope of this book. For a brief discussion and for references, see Press and Thorne (1972). \u00a737.9. NON-MECHANICAL DETECTORS When gravitational waves flow through matter, they excite it into motion. Such excitations are the basis for all detectors described thus far. But gravitational waves interact not only with matter; they also interact with electromagnetic fields; and those interactions can also be exploited in detectors. One of the most promising detectors that may be built in the future, one designed by Braginsky and Menskii (1971), relies on a resonant interaction between gravitational waves and electro magnetic waves. It is described in Box 37.6. \u00a737.10. LOOKING TOWARD THE FUTURE As this book is being written, it is not at all clear whether the experimental results of Joseph Weber constitute a genuine detection of gravitational waves. (See \u00a737.4, part 4.) But whether they do or not, gravitational-wave astronomy has beglJn, and seems to have a bright future. The technology of 1973 appears sufficient for the construction of detectors that will register waves from a star that collapses to form a black hole anywhere in our galaxy (Box 37.5); and detectors of the late 1970's and early 1980's may well register waves from pulsars and from supernovae in other galaxies. The technical difficulties to be surmounted in constructing such detectors are enormous. But physicists are ingenious; and with the impetus provided by Joseph Weber's pioneering work, and with the support of a broad lay public sincerely interested in pioneering in science, all obstacles will surely be overcome. \u00a737.10. LOOKING TOWARD THE FUTURE 1041 Box 37.5 DETECTABILITY OF HAMMER-BLOW WAVES FROM ASTROPHYSICAL SOURCES: TWO EXAMPLES (The following calculations are accurate only to within an order of magnitude or so) A. Waves from a Star of Ten Solar Masses Collapsing to Form a Black Hole; 1972 Detector with 1975 (?) Sensor 1. Predicted characteristics of radiation: (intensity at Earth) = ~ -.M 0 2 4'iT(dIstance) v -(2 X 105 ergs/cm2 Hz) [(distance to center of galaxy)/(distance)j2, (frequency of waves) = v (bandwidth of waves) =.1v - 103 Hz, 103 Hz, (duration of burst) = 'Tow - 10-3 sec to 10-1 sec. 2. Detector properties: A Weber bar, vibrating in its fundamental mode, with, M = 106 g, fa dv = 10-21 cm2 Hz (exercise 37.13), T = 3 K (liquid Helium temperature), Vo = wo/2'iT = 1,660 Hz, 'T0.== 20 seconds, rms _ (2 X 1.37 X 10-16 X 3 erg )1/2 _ f'B o - 108 I.1YR~hermalll = (3 X 10-15 cm)(10- 3/20)112 = 2 X 10-17 cm, - 3 X 10 g X 106 sec -2 -15 cm, during.1{ = 10- 3 sec, 1.1~:i3~hermalll = 2 X 10-16 cm, during.1{ = 0.1 sec. 3. Effect of waves [equation (37.53)]:.1't'<OW) = (2 X 2 X 10 I. 0 I X ;0- 21 ergS)1/2 (center of Galax) -} distance 106 X 108 sec- 2 5 distance to ) _ 2 - 10-15 X ~. \\center of Galaxy dIstance. 4. Conclusion: Gravitational waves from a massive star collapsing to form a black hole anywhere in our galaxy are readily detectable, ifone can construct a \"sensor\" to measure changes in vibration amplitudes of magnitude < 10-15 cm on time scales <0.1 seconds. This does appear to be feasible ;ith 1972 technology; see Press and Thorne (1972).. Box 37.5 (continued) B. Waves from a Supernova Explosion in the Virgo Cluster of Galaxies; a Detector that might be constructable by late 1970's or early 1980's 1. Predicted characteristics of radiation:. (mtenslty at Earth) = F\" - 4 (11. 'iT 0.03Mo 2 megaparsecs) v - 4 X IO- J ergs; cm:! Hz, (frequency of waves) = t' (bandwidth of waves) v - 103 Hz, 103 Hz, (duration of burst) = 'Tow - OJ sec, or 'Tow -", " 2 X 10- 3 sec.* 2. Detector: A Weber-type bar made not of metal, but ofa 1,000-kg monocrys tal of quartz, cooled to a temperature of 3 X 10-3 K. (For such a monocrys tal, it is thought that the damping time would increase in inverse proportion to temperature, 'TO ex 1IT.) Estimated properties of such a detector: M - 106 g, Vo = Wo/2'iT - 1,500 Hz, f a dv = 10-21 cm2 Hz (same as for Weber bar), T = 3 X 10-3 K;.. TO - 106 sec, ~wms = o ( 2 X 1.37 X 10- 16 X 3 X 10- 3 erg )1/2 106 g X 108 sec- 2 = 1 X 10-16 cm, l.jqj~herma1l1 = (1 X 10-16 cm) (OJ or 2 :: 10-3)1/2 = 10 6 X 10-20 cm, or 5 X 10-21 cm. { 3. Effect of waves [equation (37.53)]: 1&%OWll = ( 2 X 4 X 10- 3 X 10-21 ergs )1/2 106 X 108 sec 2 = 3 X 10-19 cm. 4. Conclusion: Gravitational waves are detectable from a supernova in the Virgo cluster, if one can construct a sensor to measure changes in vibration amplitudes of magnitude ~ 10-19 cm on time scales of ~O.l seconds; and if one can construct a detector with the above characteristics. * For the duration of waves from a supernova explosion, two time scales appear to be relevant: (I) the time required for the final stages of the collapse of the white-dwarf core to a neutron star or a neutron-star (dimensions of neutron star)/(speed of sound in nuclear matter) - 2 X 10-3 sec (\"pulse pancake,.,. of gravitational radiation\"); and (2) the time required for a vibrating neutron star to lose its energy of vibration by gravitational radiation (\"damped train of waves\"),.,. - 0.3 sec. \u00a737.10. LOOKING TOWARD THE FUTURE 1043 ~---------------------------------- ~ Box 37.6 A NONMECHANICAL DETECTOR OF GRAVITATIONAL WAVES [Braginsky and Menskii (1971)] The Idea in Brief (see diagram at right) A toroidal waveguide contains a monochromatic train of electromagnetic waves, traveling around and around it. Gravitational waves propagate per pendicular to the plane of the torus. If the circuit time for the EM waves is twice the period of the gravitational waves, then one circularly polarized component of the gravitational waves will stay always in phase with the traveling EM waves. Result: a resonance develops. In one region of the EM wave train, gravitational tidal forces always \"push\" the waves forward (blue shift!) in another forces \"push\" backward (red region the tidal shift!). An EM frequency difference builds up linearly with time; a phase difference builds up quadratically. Outline of Quantitative Analysis 1. Let waveguide fall freely in an Earth orbit. Orient axes of waveguide's proper reference frame (= local Lorentz frame) so (1) wave guide lies in X, ji-plane, and (2) gravitational waves propagate in i direction. 2. Let gravitational waves have amplitudes A+ - iA x = (f e- iw(! - z) (1) [Recall: i:::: t, i:::: z; i.e., proper frame and TT coordinates almost agree.] Then in plane of waveguide (z = 0), R--\" = -R\"\u00b7~ = lw2Ll~~uJtr \",0\",0 ~_~. w =(angular frequency of gravitational waves) =(rate of change of phase of waves with time) = (two times angular velocity ~ith which pattern of \"lines of force\" rotates) r = (radius of torus), is adjusted so the speed of propa gation of EM waves in waveguide is v =!wr. ewl=1t/2 A (redshift) B (blueshift) [EM waves propagate counterclockwise; gravitational line-of-force diagram rotates counterclockwise; they stay in phase.] -----I---~--~ R--\" = R\"\" = lw 2{f sin (wt) \",0 yO yOzO 2 (2) 3. Consider two neighboring the EM wave, one at \u00a2 = 0: +!wt; the other at \u00a2 = 0: + 00: + ~wt. Treat them as photons. Each moves along a null ~eodesic, except for - ------ parts of - - -", " - \" - - -.... -- ~ 1044 37. DETECTION OF GRAVITATIONAL WAVES Box 37.6 (continued) the deflective guidance of the wave guide. Thus, their wave vectors k satisfy V k = (deflective \"acceleration\").'f 'd o wavegUl e k (3) and the difference ok = Vnk between the wave vectors of the two parts of the wave (difference measured via parallel transport) satisfies the equation Vkbk = Vk Vnk = [Vk, Vnlk + VnVkk = Riemann (..., k, k, n) + V nVkk (4) ~ [deflective acceleration of wave guidet! the direction of The waveguide influences propagation of the waves, but not their fre quency. Thus only Riemann enters into the 0 component of the above equation: kaoko,a = ROa!3Ykak PnY. (5) 4. Let kO = We be the angular frequency of the electromagnetic wave, The direction of the space component k of the propagation 4-vector is along the purely spatial vector n; so kO = We' k = (vwe/roa)n, Use these relations to rewrite equation (5) as nO = O. (6) t Spacetime diagram (d owe/di)movingWlthPhotons = (vwe/roa)RoWJnin J. (7) 5. Combine the expression for n in the spacetime diagram with equations (2) and (7), and with the world line \u00a2 = a + ~wt for the photons, to obtain (d owe/dt)movingWlthPhotons n = (vector connecting) photons = rtia (-sin <!> e z + cos<!> e.) = _1. vwew2tlr (cos 20:) 00:. (8) 2. 6. Integrate over time and over a to obtain PART IX EXPERIMENTAL TESTS OF GENERAL RELATIVITY Wherein the reader is tempted by a harem of charming gravitation theories (and some not so charming), is saved from his foolish passions by an army of experiments, cleaves unto his faithful spouse, Geometrodynamics, vows to lead an honest life hereafter, and becomes a True Believer. --------------------------------....... ~ CHAPTER 38 TESTING THE FOUNDATIONS OF RELATIVITY Provando e riprovando (Verify the one and disprove the other) GAll LEO \u00a738.1. TESTING IS EASIER IN THE SOLAR SYSTEM THAN IN REMOTE SPACE For the first half-century of its life, general relativity was a theorist's paradise, but an experimentalist's hell. No theory was thought more beautiful, and none was more difficult to test. The situation has changed. In the last few years general relativity has become one of the most exciting and fruitful branches ofexperimental physics. A half-century late, the march of technology has finally caught up with Einstein's genius-not only on the astronomical front, but also in laboratory experiments. On the astronomical front, observers search for phenomena in which relativity is important, and study them: cosmology, pulsars, quasars, gravitational waves, black holes. Unfortunately, in pulsars and quasars, and in the sources of cosmological radiation and gravity waves, gravitational effects are tightly interwoven with the local hydrodynamics and local plasma physics. There is little hope ofseparating the several effects sufficiently sharply to get clean tests of the nature of gravity. Instead, astro physicists must put the laws of gravity into their calculations along with all the other laws of physics and the observational data; and they must then seek, as output, information about the doings of matter and fields \"way out there.\" Thus, for clean tests of general relativity one turns to the laboratory-but to a laboratory that is much larger today than formerly: a laboratory that includes the entire solar system. Clean tests of general relativity are currently confined to solar system 1048 38. TESTING THE FOUNDATIONS OF RELATIVITY Capabilities of technology in 1970's In the solar system all relativistic effects are tiny. Nonetheless, some of them are in the 1970's, of one part in 1,000 of their whole measurable with a precision, magnitude or better (see Box 38.1). \u00a738.2. THEORETICAL FRAMEWORKS FOR ANALYZING TESTS OF GENERAL RELATIVITY There are now possible many experiments for testing general relativity. But most of them are expensive; very expensive. They involve atomic clocks flown on space- Box 38.1 TECHNOLOGY OF THE 1970's CONFRONTED WITH RELATIVISTIC PHENOMENA Quantity to be measured Magnitude of relativistic effects Precision", " of a one-day measurement in the early 1970's Angular separation of two sources on the sky Solar deflection of starlight (I) if light ray grazes edge of Sun, (a) With optical telescope, -I\" (b) Angular separation of two quasars with 1\".75 (2) if light ray comes in perpendicular to Earth-sun line, radio telescope (differential measurement from day to day, not absolute measurement) 0\".004 in 1970, -0\".1 in mid 1970's, -0\".001 Distance between two bodies in solar system (a) Perihelion shift per Earth year (I) for Mercury, 120 km IS km (2) for Mars, (b) Relativistic time delay for radio waves from Earth, past limb of sun, to Venus (one way), I X 10'-4 sec = 30 km (c) Periodic relativistic effects in Earth-moon separation (I) in general relativity, 100 cm (2) in Jordan\u00b7Brans-Dicke theory, (840 cm)/(2 + w) 100 cm; (a) Separation of another planet (Mercury, Venus, Mars) from Earth, by bouncing radar signals off it, -0.3 km (b) Separation of a radio transponder (on another planet or in a space craft) from Earth, by measuring round-trip radio travel time, -3 X 10-8 sec = 10 m = 0.01 km (c) Earth-moon separation by laser ranging, -lOcm Difference in lapse of proper time between two world lines in solar system (a) Clock on Earth vs. clock in synchronous Earth orbit, Jt/t - 6 X 10-10 Stability of a hydrogen maser clock, Jt/t - 10- 13 for t up to one year (b) Clock on Earth vs. cl()ck in orbit about sun, Jt/t - 10-8 Theoretical frameworks for analyzing gravitation experiments: (1) Dicke framework (2) PPN framework \u00a738.2. THEORETICAL FRAMEWORKS FOR ANALYZING TESTS 1049 craft; radar signals bounced off planets; radio beacons and transponders landed on planets or orbited about them; etc. Because of the expense, it is crucial to have as good a theoretical framework as possible for comparing the relative values of various experiments-and for proposing new ones, which might have been overlooked. Such a framework must lie outside general relativity. It must scrutinize the foun dations of Einstein's theory. It must compare Einstein's theory with other viable theories of gravity to see which experiments can distinguish between them. It must be a \"theory of theories.\" At present, in 1973, there are two different frameworks in broad use. One, devised largely by Dicke (1964b),* assumes almost nothing about the nature of gravity. It is used to design and discuss experiments for testing, at a very fundamental level, the nature of spacetime and gravity. Within it, one asks such questions as: Do all bodies respond to gravity with the same acceleration? Is space locally isotropic in its intrinsic properties? What are the theoretical implications of local isotropy? What types of fields, if any, are associated with gravity: scalar fields, vector fields, tensor fields, affine fields? Although some of the experiments that tackle these questions will be discussed below, this book will not attempt a detailed exposition of the Dicke framework. The second framework in broad use is the \"parametrized post-Newtonian (PPN) formalism.\" It has been developed to higher and higher levels of sophistication by Eddington (1922), Robertson.(1962), Schiff (1962, 1967), Nordtvedt (1968b, 1969), Will (1971c), and Will and Nordtvedt (1972). The PPN formalism is an approximation to general relativity, and also to a variety of other contemporary theories of gravity, called \"metric theories.\" It is a good approximation whenever, as in the solar system, the sources of the field gravitate weakly (14'11/c2 ~ 1) and move slowly (v2 / c2 ~ 1). The PPN formalism contains a set of ten parameters whose values differ from one theory to another. Solar-system experiments (measurements of perihelion shift, light deflection, etc.) can be regarded as attempts to measure some of these PPN parameters, and thereby to determine which metric theory of gravity is correct-general relativity, Brans-Dicke (1961) Jordan (1959) theory, one of Bergmann's (1968) scalar-tensor theories, one of Nordstrom's theories, Whitehead's (1922) theory, or something else. [For reviews of Nordstrom and Whitehead, see Whitrow", " and Morduch (1965), Will (1971 b), and Ni (1972). For a significant nonmetric theory, see Cartan (1920) and Trautman (1972).] Chapter 39 will discuss the concept of a metric theory of gravity and will construct the PPN formalism; and then Chapter 40 will use the PPN formalism to analyze the systematics of the solar system, and to discuss a variety of past and future experiments that distinguish between various metric theories of gravity. But first, as a prelude to those topics, this chapter will examine experiments that test the foundations of general relativity-foundations on which most other metric theories also rest. For a more detailed discussion of most of these experiments, see Dicke (1964b). \u00b7See Thome and Will (1971), or Will (1972), for expositions of both frameworks and a comparison of them. y The rest of this chapter is Track 2. No earlier Track-2 material is needed as preparation for it. but Chapter 7 (incompatibility of gravity and special relativity) will be helpful. This chapter is not needed as preparation for any later chapter. but it will be helpful in Chapters 39 and 40 (other theories; PPN formalism; experimental tests). Eotvos-Dicke experiment' to test uniqueness of free fall 1050 38. TESTING THE FOUNDATIONS OF RELATIVITY \u00a738.3. TESTS OF THE PRINCIPLE OF THE UNIQUENESS OF FREE FALL: EOTVOS-DICKE EXPERIMENT One fundamental building block common to Einstein's theory of gravity and to almost all other modern theories is the principle of \"uniqueness ojJree Jail\": * \"The world line oj a Jreely Jalling test body is independent oj its composition or structure. \" By \"test body\" is meant an electrically neutral body, small enough that (1) its self-gravitational energy, as calculated using standard Newtonian theory, can be neglected compared to its rest mass (M/ R ~ 1), and (2) the coupling of its multipole moments to inhomogeneities of the gravitational field can be neglected.t The uniqueness of free fall permits one to regard spacetime as filled with a set of curves, the test-body trajectories, which are unique aside from parametrization. Through each event, along each timelike or null direction in spacetime, there passes one and only one test-body trajectory. Describe these trajectories mathematically: that is a central imperative of any theory of gravity. When translated into Newtonian language, the uniqueness of free fall states that any two test bodies must fall with the same acceleration in a given external gravita tional field. Experimental tests of this principle search for differences in acceleration from one body to another. The most precise experiments to date are of a type devised by Baron Lorand von Eotvos (Box 38.2), redesigned and pushed to much higher precision by the Princeton group of Robert H. Dicke (Box 38.3), and extended with modifications by the Moscow group of Vladimir B. Braginsky. (See Figure 1.6 and Box 1.2 for experimental details.) These E5tvos-Dicke experiments are \"null experiments.\" They balance the accel eration of one body against the acceleration of another, and look for tiny departures from equilibrium. The reason is simple. Null experiments typically have much higher precision than experiments measuring the value of a nonzero quantity. Eotvos, Pekar, and Fekete (1922) checked to an accuracy of 5 parts in 109 that the Earth imparts the same acceleration to wood, platinum, copper, asbestos, water, magnalium (90% AI, 10% Mg), copper sulphate, and tallow. Renner (1935) checked, to 7 parts in 1010, the Earth's acceleration of platinum, copper, bizmuth, brass, glass, ammonium fluoride, and an alloy of 30% Mg, 70% Cu. Dicke, and later Braginsky, chose to use the sun's gravitational acceleration rather than the Earth's, since the alternation in the direction of the sun's pull every 12 hours lends itself to amplifica tion by resonance. (See Figure 1.6.) Roll, Krotkov, and Dicke (1964) reported an *R. H. Dicke calls this principle \"The weak equivalence principle.\" We prefer to avoidJconfusion with the equivalence principle (Chapter 16). tIn general relativity, one often uses an alternative definition of test body, which places no constraint on the self-gravitational energy [abandon condition (I) while retaining (2\u00bb). Such a definition is preferable, in principle, because the theory of matter has not been developed sufficiently to decide whether (and no objective test has ever been proposed to", " decide whether), gravitational energy at the subnuclear scale is a small fraction, a large fraction, or the entirety of the rest mass. But for present purposes a definition constraining test bodies to have M/R < I is preferable for two reasons. First, most theories of gravity that currently \"compete\" with Einstein's (a) agree with the principle of uniqueness of free fall when the macroscopic, Newtonian, self-gravitational energy is neglected (M/ R < I), but (b) violate that principle when macroscopic, Newtonian self-gravitational energy is taken into account See \u00a740.9 for the test bodies used in the EOtvos-Dicke experiment have M/R so small that their details. Second, macroscopic, Newtonian, self-gravitational energies are, in fact, negligible (M/ R - Egrav/ M _ 10-27). \u00a738.3. EOrvOS-DICKE EXPERIMENT 1051 agreement of 1 part in 1011 between the sun's acceleration of aluminum and gold, while Braginsky and Panov (1971) reported agreement to 1 part in 1012 for aluminum and platinum. From this agreement, one can infer the response of neutrons, protons, electrons, virtual electron-positron pairs, \u00b7nuclear binding energy, and electrostatic energy to the sun's gravity. Gold is 60% neutrons, while aluminum is only 50% neutrons. Therefore even from the 1964 results one could conclude that neutrons and protons must have the same acceleration to within [0.6 - 0.5 = 0.1]-1 parts in 1011 = 1 part in 1010. Similarly, electrons must accelerate the same as nucleons to 2 parts in 107 ; virtual pairs (being more abundant in gold than in aluminum) must accelerate the same to 1 part in 104; nuclear binding energy, to 1 part in 107 ; and electrostatic energy to 3 parts in 109 \u2022 This accuracy of testing gives one confidence in the principle of the uniqueness ojJreejall. (continued on page 1054) Theoretical Eotvos-Dicke experiment implications of Box 38.2 BARON LORAND VON EOTVOS Budapest, JUly 27, 1848-Budapest, April 8, 1919 Eotvos (pronounced ut'rush) studied at Heidelberg with Kirchhoff, Helmholtz, and Bunsen and at Konigsberg with Neumann and Richelot. His 1870 Heidelberg Ph.D. thesis dealt with an issue of relativity: can the motion of a light source relative to an \"ether\" be detected by comparing the light intensities in the direction of the motion and in the opposite direction? Studies of his at the same time resulted in the Eotvos law of capillarity, (surface tension) ;::::; 2.12 (Tcrit - T)j(specific volume)2/3. Eotvos, made professor of physics at Budapest in 1872, concentrated on gravity from 1886 onward. He developed and extended the original Michell-Cavendish torsion balance, which measured not only f/J,xx and f/J,XY (where f/J is the gravitational potential) but also f/J,xz and f/J,yZ' all to a precision destined to be unexcelled for decades. He showed that the so-called \"ratio between gravitational mass and inertial mass\" cannot vary from material to material by more than 5 parts in 109 \u2022 He investigated the paleomagnetism of bricks and other ceramic objects, and studied the shape of the earth. He served (June 1894-January 1895) as minister of public instruction and r~ligious affairs (a cabinet position held in earlier years by his father). He founded a school which trained high-school teachers, to whose leavening influence one can give some of the credit for such outstanding scientists as von Karman, von Neuman, Teller, and Wigner. He served one year as rector of the University of Budapest. \"I can never forget the moment when my train rushed into the railroad station of Heidelberg along the banks of the Neckar... I cannot forget my happiness that now I could breathe the same air as those men of science whose fame attracted me there.\" [EOTVOS IN 1887. AS QUOTED IN FEJER AND MIKOLA (1918). P. 259.] 1052 38. TESTING THE FOUNDATIONS OF RELATIVITY Box 38.2 (continued) Photograph by A. Szekely 1913 that is, declare that p., intervening compensator quadrants. \"Insofar as it is permitted on the basis of a few experiments, we can therefore the weakening of the Earth's attraction through the is less than one part in 5", " X 1010.. \u2022 \u2022 the absorption (of gravity) by the entire earth along a diameter is less than about one part in 800. \"We have carried out a series of observations which surpassed all previous ones in precision, but in no case could we discover any detectable deviation from the law of proportionality of gravitation and inertia.\" [EOTVOS. PEKAR. AND FEKETE (1922).] \"Science shall never find that formula by which its necessary character could be proved. Actually science itself might cease if we were to find the clue to the secret. \" [EOTVOS, PRESIDENTIAL ADDRESS TO THE HUNGARIAN ACADEMY OF SCIENCES, 1890, AS QUOTED IN FEJER AND MIKOLA (1918). P. 280.] \"We should consider it as one of the most astonishing errors of the present ';Jge thet so many people listen to the words of pseudoprophets who, in place of the dogmas of religion, offer scientific dogmas with medieval impatience but without historical ;ustification.\" [EOTVOS, 1877, AS QUOTED IN FEJER AND MIKOLA (1918), P. 280.] \u00a738.3. EOrvOS-DICKE EXPERIMENT 1053.-----------------------------------------. ~ Box 38.3 ROBERT HENRY DICKE May 6. 1916. St. Louis. Missouri Cyrus Fogg Brackett Professor of Physics at Princeton University During 1941-1946, Dicke was a leader in replacing the outmoded concept of lumped circuit elements by a new microwave analysis based on symmetry considerations, conservation laws, reciprocity rela tions, and the scattering matrix-concepts that led, among others, to the lock-in amplifier and the microwave radiometer. Searching for a means to lines for reduce the Doppler width of spectral precision measurements, Dicke discovered recoil less radiation in atomic systems held in a box or in a buffer gas. This development led to (1) the discovery of the basic idea of the gas-cell atomic clock and (2) a much more precise measurement of the gyromagnetic ratio of electrons in the Is and 2s levels of hydrogen and of the hyperfine struc ture of atomic hydrogen. A fundamental paper by Dicke in 1954 set forth the theory of coherent fadiation processes and of the superradiant state, and laid the foundation for the future development of the laser and the maser, to which he also contributed. His patent no. 2,851,652 (filed May 21,1956) was the first disclo sure of a device for- the generation of infrared radiation by a coherent process, and supplied the first suggestion for combining the use of an etalon resonator with an amplifying gas. Beginning in the 1960's, Dicke brought his talent for precision measurement to the service of experi mental cosmology, and with his collaborators: (1) checked the equivalence principle with the up to-then unprecedented accuracy of 1 part in 1011 ; (2) determined the solar oblateness; and (3) sug gested that the primordial cosmic-fireball radia tion, a tool for seeing deeper into the past history of the universe than has ever before been possible, should be observable, and therefore should be hunted down and found. \"For want of a better term, a gas which is radiating strongly because of coherence will be called'superradiant. '... As the system radiates it passes to states of lower m with r unchanged-to the'superradiant' region m - 0\" (1954) \"Possibilities are examined for the excitation of optical'superradiant' states of gas\" (1957) 1054 38. TESTING THE FOUNDATIONS OF RELATIVITY Box 38.3 (continued) ')I 'gravitational oblateness' of [the sun of] 5 X 10-5 would require the abandonment of Einstein's purely geometrical theory of gravitation.... Such a flattening [of the sun] could be understood as the effect of a rather rapidly rotating the solar corona with its magnetic field has acted as a brake on the surface of the sun\" interior.... The answer appears to be that in the past and to this day, (1964a) \"New measurements of the solar oblateness have given a value for the fractional difference of equatorial and polar radii of (5.0 \u00b1 O. 7) X 10- 5\" [DICKE AND GOLDENBERG (1967)] '1The universe must] have aged sufficiently for there to exist elements other than hydrogen. It is well-known that carbon is required to make physicists\" (1961) \"The question of the constancy of such dimensionless numbers is to be settled not", " by definition but by measurements\" [BRANS AND DICKE (1961)] \"The geophysical data lead to an upper limit of 3 parts in 10 13 per year on the rate of change of the fine-structure constant\" [DICKE AND PEEBLES (1962)] \u00a738.4. TESTS FOR THE EXISTENCE OF A METRIC GOVERNING LENGTH AND TIME MEASUREMENTS, AND PARTICLE KINEMATICS Special relativity, general relativity, and all other metric theories of gravity assume the existence of a metric field and predict that this field determines the rates of ticking of atomic clocks and the lengths of laboratory rods by the familiar relation - dr2 = ds 2 = gaP dx a dx p. Experimental evidence for existence of a metric The experimental evidence for a metric comes largely from elementary particle physics. It is of two types: first, experiments that measure time intervals directly, e.g., measurements of the time dilation of the decay times of unstable P51rticles;* second, expenments that reveal the fundamental ro e played by the Lorentz group in particle kinematics and elsewhere in particle physics.t To cast out the metric tensor entirely would leave one with no theoretical framework adequate for interpreting such experiments. 1. t *For a 2 per cent test of time dilation with muons of (I - zr)-1/2 - 12 in a storage ring, see Farley, Bailey, Brown, Giesch, Jostlein, van der Meer, Picasso, and Tannenbaum (1966). For earlier time-dilation experiments see Frisch and Smith (1963); Durbin, Loar, and Havens (1952); and Rossi and Hall (1941). tSee p. 18 of Lichtenberg (1965) for a discussion of Lorentz invariance, spin and statistics, the TCP theorem, and relevant experiments. Particle experiments do not guarantee existence of global Lorentz frames. or geodesic motion for test particles Particle experiments do suggest proper time is independent of acceleration \u00a738.5. GRAVITATIONAL REDSHIFT EXPERIMENTS 1055 Notice what particle-physics experiments do and do not tell one about the metric tensor, g. First, they do not guarantee that there exist global Lorentz frames, i.e., coordinate systems extending throughout all of spacetime, in which gap = T/ap. However, they do suggest that at each event tJ' there exist orthonormal frames with eci:(tJ')\u00b7 ep(tJ') = T/ap, which are related to each other by Lorentz transformations. These orthonormal frames provide one with a definition of the inner product between any two vectors at a given event-and, thereby, they define the metric field. Second, particle experiments do not guarantee that freely falling particles move along geodesics of the metric field, i.e., along straight lines in local Lorentz frames. (Here, in \u00a7\u00a738.4 and 38.5, the phrase \"local Lorentz frame\" means a \"normal\" co ordinate system at an event tJ', in which gaP(tJ') = T/ap and gaP,itJ') = O. The term \"inertial frame\" is avoided because no assertions are made, yet, about test-body motion.) In particular, one does not know from elementary-particle experiments wheth~r the local Lorentz frames in the laboratory are freely falling (so they fly up from the center of the earth and then fall back with Newtonian acceleration g = 980 cm/sec2), whether they are forever at rest relative to the laboratory walls, or whether they undergo some other type of motion. All one is led to believe is that a metric determines the nature of the spacetime intervals (dr 2 = - gp.p dxP. dx P measured by atomic clocks, that the various local Lorentz frames in the laboratory therefore move with uniform velocity relative to each other (they are connected by Lorentz transformations), and that electric and magnetic fields and the energies and momenta of particles undergo Lorentz transformations in the passage from one local Lorentz frame to another. ) Third, elementary particle experiments do suggest that the times measured by atomic clocks depend only on velocity, not on acceleration. The measured squared interval is ds 2 = gap dx a dX P, -independently of acceleration (until the acceleration becomes so great it disturbs the structure of the clock; see \u00a716.4 and Box 16.3). Equivalently, but more physically, the time interval measured by a clock moving with velocity vj relative to a local Lorentz frame is independently of the clock's acceleration d 2x j / dt2\u2022 If this were not so, then particles moving in circular orbits in strong magnetic fields would exhibit different decay", " rates than freely moving particles-which they do not [Farley et al. (1966)].* (38.1 ) \u00a738.5. TESTS OF GEODESIC MOTION: GRAVITATIONAL REDSHIFT EXPERIMENTS The uniqueness of free fall, as tested by the Dicke-Eotvos experiments, implies that spacetime is filled with a family of preferred curves, the test-body trajectories. There \u2022 The experiment of Farley et al. is a 2 percent check of acceleration-independence of the muon decay 12 and for accelerations, as measured in the muon rest frame, rate for energies Elm = (I - v2r '12 of a = 5 X 1020 cm/sec2 = 0.6 em-I. y Physical meaning of a comparison between test-body trajectories and geodesics of metric Pound-Rebka-Snider redshift experiment as a test of geodesic motion 1056 38. TESTING THE FOUNDATIONS OF RELATIVITY is also another family of preferred curves, the geodesics of the metric g. It is tempting to identify these geodesics with the test-body trajectories. Einstein's geometric theory of gravity makes this identification (\"equivalence principle\"). One might conceive of theories that reject this identification. What is the experimental evidence on this point? In order to see what kinds of experiments are relevant, it is helpful to elucidate the physical significance of the geodesics. A geodesic of 9 is most readily identified locally by the fact that it is a straight line in the local Lorentz frames. Put differently, a body's motion is unaccelerated as measured in a local Lorentz frame if and only if the body moves along a geodesic of g. Hence, to determine whether test-body trajectories are geodesics, one must compare experimentally the motion of the spatial origin of a local Lorentz frame (as defined by atomic-clock readings) with the motion of a test body (material particle). It is easy to study experimentally the motions of test bodies; relative to an earth bound laboratory, they accelerate downward with g = 980 cm/sec2 ; and this accel eration can be measured at a given location on the Earth to a precision of I part in. 106. Unfortunately, it is much more difficult to measure the motion of a local Lorentz frame, once again as defined by atomic-clock readings. The only direct experimental handle one has on this today, with sufficient precision to be interesting, is gravita tional redshift experiments. (See \u00a7\u00a77.2-7.5 and \u00a725.4 for theoretical discussions of the gravitational redshift in the framework of general relativity.) The redshift experiment of highest precision is that of Pound and Rebka (1960), as improved by Pound and Snider (1965); see Figure 38.1. It used the Mossbauer effect to measure the redshift of 14.4 keY gamma rays from Fe57 \u2022 The emitter and absorber of the gamma rays were placed at rest at the bottom and top of a tower at Harvard University, separated by a height h = 74 feet =22.5 meters. The meas ured redshift agreed, to I percent precision, with the general relativistic prediction of L1A/A = gh = 2.5 X 10-15\u2022 (38.2) This result tells one that the local Lorentz frames are not at rest relative to the Earth's surface; rather, they are accelerating downward with the same acceleration, g, as acts on a free particle (to within 1 percent precision). To arrive at this conclusion, one analyzes the experiment in the laboratory reference frame, where ev~rything (the experimental apparatus, the Earth's gravitational field) is static. Relative to the laboratory a local Lorentz frame, momentarily at rest, accelerates downward (horizontal accelerations being ruled out by symmetry) with some un known acceleration a. Equivalently, the laboratory accelerates upward (in +z direc tion) with acceleration a relative to the local Lorentz frame. Consequently, the spacetime metric in the laboratory frame has the standard form the Earth, (38.3) 4th fl. gr. 69.55' Counts, N t Counter for inverted operation Source velocity (V.II - v J ) Figu re 38.1. The experiment of Pound and Rebka (1959) and Pound and Snider (1965) on the gravitational redshift of photons rising 22.5 meters against gravity through a helium-filled tube in a shaft in the Jefferson Physical Laboratory of Harvard University. The source of C0 57 had an initial strength greater than a curie. The 14.4 keY gamma rays had to pass in through an absorber enriched in Fe'>; to reach the large-window proportional counters. Both source and absorber were placed in temperature-regulated", " ovens. The velocity of the source consisted of two parts: one steady (v.II), to put the center of the emission line on the part of the transmission curve that is nearly straight; and the other alternating between + vJ and -vJ \u2022 to sweep the transmission curve in this straight region; similarly when the steady velocity was - vii' The departure from symmetry between the two cases + v.1I and - vJI allows one to determine the offset vD (effect of gravitational redshift) from the zero-gravity case of stationary emitter and stationary absorber. The final result for the redshift was (0.9990 :::':: 0.0076) times the value 4.905 X 1O- 15 0f2gh/c 2 predicted from the principle of equivalence (difference between \"up\" experiment and \"down\" experi ment). Diagrams adapted from Pound and Snider (1965). 1058 38. TESTING THE FOUNDATIONS OF RELATIVITY which Track-2 readers have met in \u00a7\u00a76.6 and 13.6; and Track-l readers have met and used in Box 16.2. Moreover, in the laboratory frame the metric is static, gravity is static, and the experimental apparatus is static. Therefore the crest of each electro mag~etic wave that climbs upward must follow a world line t(z) identical in form to the world lines of the crests before and after it; thus, wave crest #0: t = to(z), wave crest # 1: t = to(z) + LIt, wave crest #n: t = to(z) + n LIt. [Here, as in Schild's argument (\u00a77.3) that redshift implies spacetime curvature, no assumption is made about the form of the wave-crest world lines to(z); see Figure 7.1.] Hence, expressed in coordinate time, the interval between reception of successive wave crests is the same as the interval between emission. Both are LIt. But the atomic clocks of the experiment (Fe57 nuclei) are assumed to measure proper time LIT = ( - gaf3 Llxa LlX (3 )l/2, not coordinate time. Thus Areceived _ LITreceived _ Aemitted - LITemitted (1 + aZreceiVed) LIt (1 + aZemitted) LIt = 1 + a(zreceived - zemitted); i.e., L1A -=ah A theoretical prediction based on assumptions (i) that atomic clocks measure LIT = (- gaf3 L1xa L1x(3)1/2; (ii) that electromagnetic radiation has the form of a wave train; (iii) that local Lorentz frames accelerate downward with acceleration a relative to the laboratory. (38.4) Direct comparison with the experimental result (38.2) reveals that local Lorentz frames in an Earthbound laboratory accelerate downward with the same acceleration g as acts on a test particle (to within 1 per cent precision). [The above discussion is basically a reworked version of Schild's proof (\u00a77.2) that the redshift experiment implies spacetime is curved. After all, how could spacetime possibly be flat if Lorentz frames in Washington, Moscow, and Peking all accelerate toward the Earth's center with g = 980 crn/sec2?] Of all redshift experiments, the Pound-Rebka-Snider experiment is the easiest to interpret theoretically, because it was performed in a uniform gravitational field. Complementary to it is the experiment by Brault (1962), which measured the redshift of the sodium D1 line emitted on the surface of the sun and received at Earth (Figure 38.2). To a precision of5 per cent, he found a redshift ofGM0 /R0 c2, where M 0 and R0 are the mass and radius of the sun. This is just the redshift to be expected if Other redshift experiments Movable It output slit Photo detector Amplifier and meter SOlar~ radiation 0.8 0.6 Plane.grating Photo current Redshift of sodium D 1 coming from the sun o Center 8. Limb f 0. 50 100 ISO ----Slit oscillation amplitude in rnA ~ Figure 38.2. The measurement by Brault (1962) of the redshift of the D 1 line of sodium gives 1.05 \u00b1 0.05 of the gravitational redshift predicted by general relativity. This strong line. in contrast to the weak lines used by earlier investigators (I) is emitted high in the sun's atmosphere, above the regions strongly disturbed by the pressure and convective shifts, and yet lower than the chromosphere. and (2) comes closer to standing up cleanly above the background than any other line in the visible spectrum. Brault built a new photoelectric spect", "rometer (upper diagram), with its slit vibrated mechanically back and forth across a narrow region of the spectrum, to define the position of the line peak (I) electronically, (2) independently of subjective judgment, and (3) with a precision greater by a factor of the order of ten than that afforded by conventional visual methods. The slit is considered set on a line when its mean position is such that the photomultiplier current contains no signal at the frequency of the modulation. The redshift measured in this way is corrected for orbital motion and for rotation of the sun and the Earth to give the points in circles and triangles in the lower diagram. Extrapolation to zero vibration of the slit gives the cited number for the redshift. Figure adapted from thesis of Brault (1962). 1060 38. TESTING THE FOUNDATIONS OF RELATIVITY the local Lorentz frames, at each point along the photon trajectory, fall in step with freely falling test bodies.* In summary, redshift experiments reveal that, to a precision of several percent, the local Lorentz frames at the Earth's-surface and near the sun are unaccelerated relative to freely falling test bodies. Equivalently, test bodies move along straight lines in the local Lorentz frames. Equivalently, the test-body trajectories are geodesics of the metric g. \u00a738.6. TESTS OF THE EQUIVALENCE PRINCIPLE Of all the principles at work in gravitation, none is more central than the equivalence principle. As enunciated in \u00a7 16.2, it states: \"In any and every local Lorentz frame, anywhere and anytime in the universe, all the (nongravitational) laws ofphysics must take on their familiar special-relativistic forms. \" That test bodies move along straight lines in local Lorentz frames (geodesic motion) is one aspect of the equivalence principle. Other aspects are the universality of Maxwell's equations (38.5) in all local Lorentz frames; the universality of the law of local energy-momentum conservation T a f3 - o\u00b7,.f3 - (38.6) and the universality of the values of the dimensionless constants that enter into the local laws of physics: (l' = \u00a3 = nc e - 1 137.0360... = (electromagnetic fine-); structure constant mneutron = 1.00138..., mproton melectron = -:-::c:-::\"--:l,...\".-_ 1836.12...'mproton etc. (38.7) (Attention here is confined to dimensionless constants, since only they are independ ent of one's arbitrary choice of units of measure.) If one focuses attention on a given event and asks about invariance of the form of the physical laws [equations (38.5), (38.6), etc.] from one Lorentz frame to another, one is then in the province of special relativity. Here a multitude of exp!eriments verify the equivalence principle (see \u00a738.4). If one asks about variations in the form of the laws from one event to another, one opens up a Pandora's box of possibilities that one hardly dares to contemplate. However, no experimental evidence has ever given the slightest warrant to consider any such \"departure from democracy\" in the action of the laws of physics. Moreover, astronomical observations provide strong evidence that the laws of physics are the *For a review of other, less-precise redshift experiments, see Bertotti, Brill, and Krotkov (1962). Tests of the equivalence principle: (1) geodesic motion (2) physical laws are locally Lorentz-invariant (3) laws do not vary from event to event \u00a738.6. TESTS OF THE EQUIVALENCE PRINCIPLE 1061 same in distant stellar systems as in the solar system, and the same in distant galaxies as in our own Galaxy. (See, in Box 29.5, Edwin Hubble's expressions of joy upon discovering this.) Constancy of the dimensionless \"constants\" from event to event can be tested to high precision, if one assumes constancy of the physical laws. Dirac (1937, 1938), Teller (1948), Jordan (1955, 1959), Gamow (1967), and others have proposed that the fine-structure \"constant\" ae might be a slowly varying scalar field, perhaps governed by a cosmological equation. However, rather stringent limits on such variations follow from data on the fine-structure splitting of the spectral lines of quasars and radio galaxies. For the quasar 3C 191 with redshift z = 1.95, Bahcall, Sargent, and Schmidt (1967) find ai3C 19l", ")/aiEarth) = 0.97 -+- 0.5. With a cos this corresponds to a limit mological (l/ae)(dae/dt) ~ 1/1011 years. An even tighter limit has been obtained from radio galaXy data, where there is no question about the interpretation of the redshift. Bahcall and Schmidt (1967) measured fine-structure splitting in five radio gal axies with z z 0.20, corresponding to an emission of light 2 X 109 years ago. They obtained aiz = 0.20)/aiEarth) = 1.001 -+- 0.002, which yields the limit I(1/a e)(dae/dt)1 ~ 1/1012 years. interpretation of the quasar redshift, Dyson (1972) points out that comparison of the rate of beta decay of Re 187 in times past (via osmium-rhenium abundance ratios in old ores) with the rate of beta-decay today provides.. means to check on any possible variation of a e with time more sensitive than redshift data and more sensitive than any changes in rates of alpha decay and fission between early times and now. He summarizes the available data on Re187 and arrives at the limit For further evidence of the constancy of the fundamental constants see Minkowski and Wilson (1956), Dicke (1959a,b), Dicke and Peebles (1962b). Spatial variations of ae' mneutron/mproton' and other \"constants\" in the solar system can be sought by means of Eotvos-type experiments. The reasoning [by Dicke (1969)] leading from such experiments to limits on any spatial variation of the constants is indirect. It recalls the reasoning used in standard treatises on polar molecules to deduce the acceleration of a polarizable molecule pulled on by an inhomogeneous electric field. It proceeds as follows. Suppose one of the dimensionless \"constants,\" \"a,\" depends on position. This will lead to a position-dependence of the total mass-energy of a laboratory test body. For example, if ae depends on position, then the coulomb energy of an atomic nucleus will also (ECOU1 ex: e4 ex: a/; oM/Ecou1 = 2oa e/a e ). One can calculate the change in a test body's mass-energy when it is moved from xll- to xll- + oxll- by assuming no change at all in the body's structure during its displacement: oM = (ilM/ila)fixedstructure(ila/ilxll-) oxll-. (38.8) After the displacement, a weakening of internal forces (due, e.g., to a decrease of a) (4) fundamental constants do not vary from event to event Eotvos-type experiments as tests for spatial variation of fundamental constants 1062 38. TESTING THE FOUNDATIONS OF RELATIVITY may cause a change in structure, but that change will be accompanied by a con version of internal potential energy into internal kinetic energy, which conserves M. Now consider the following thought experiment [an elaboration of the argument by which Einstein first derived the gravitational redshift (\u00a77.2)]: Take n particles, each with mass-energy JL. Make the particles with a structure such that a negligible fraction of JL is associated with the \"constant\" of interest, ex: (l/JL)(oJLloex) = o. (38.9) Place these particles at a height h in a (locally) uniform Newtonian field. Combine them together there, releasing binding energy EB(h), to form a composite body of mass which depends in a significant manner on the \"constant\" ex, (IIM)(oMloex) i O. (38.10) (38.11) Lower this body, and the released binding energy tied up in a little bag, a distance oh. The total force acting is (in Newtonian language) (38.12) Here g is acceleration experienced by the type of mass-energy that is independent of ex when it is in free fall. In contrast, \"free\" fall of the assembled body M is not really free fall, because of the supplementary \"polarization force\" pulling on this object. Hence the assembled body in \"free\" fall experiences an acceleration, a, a little different from g. However, the mass that is accelerated is precisely M, and therefore the force required to produce this acceleration is given by the product Ma. The energy gained in lowering the body and the bag is E(down) = F oh = Ma oh + EB(h)g oh. Put this energy in the bag. At h - oh use some of the energy from the bag to pull the body apart into its component particles. The", " energy required is ex EB(h - oh) = nJL - M(h - oh) = nJL - M(h) + aM dd h oex oh =E (h) + aM ~ oh''oex oh B so an energy is left in the bag. Use this energy to raise the n particles and the bag back up to (38.13) \u00a738.7. TESTS FOR EXISTENCE OF LONG-RANGE FIELDS 1063 height h. Assume total energy conservation, so that there will be no extra energy and no deficit when the n particles and bag have returned to the original state back at height h. This means that Ebag must be precisely the right amount of energy to do the raising: Combining expressions (38.13) and (38.14) for Ebag, discover that 1 'OM da a -g= - - -. M aa dh (38.14) (38.15) Thus, under the assumption of total energy conservation (no perpetual-motion machines!), a spatial dependence of a physical \"constant\" a will lead to the anomaly (38.15) in the acceleration of a body whose mass depends on a. Coulomb energy, which is proportional to a/, amounts in a gold nucleus to 0.4 per cent of the mass, and to 0.1 per cent in an aluroin urn nucleus. Hence, a spatial variation in a e should lead to a fractional difference in the gravitational accelerations of these two nuclei equal to i.e., ~ Idae I ~ 1 X 10-9 g::::::.; 1 X 10-9 cm/sec2 = 1 X 1O-30/cm ae dh (38.16) at the Earth due to the sun. Here use is made of the limit (1 X 10-11) from Dicke's experiment (\u00a738.3), and the acceleration g = 0.6 cm/sec2 due to the sun at Earth. Notice that this says the gradient of In ae is less than 1 X 10-9 the gradient of the Newtonian potential! \u00a738.7. TESTS FOR THE EXISTENCE OF UNKNOWN LONG-RANGE FIELDS Whether or not one accepts the assumption that test bodies move on geodesics of the metric, it remains conceivable that previously unknown long-range fields (fields with \"l/r\" fall-off at large distances) are somehow associated with gravity. If \"new\" long-range fields (not metric, not electromagnetic) do exist, waiting to be discovered, then there are two ways by which they could influence matter. First, they could couple directly to matter, producing, for example, slight deviations from geodesic motion (deviations smaller than the limits of \u00a738.5), or slight dependences of masses of particles on position (dependences smaller than the limits of \u00a738.6). Second (and harder to detect), they could couple indirectly to matter by being mere possible existence of new long-range fields associated with gravity Direct vs. indirect coupling y Experimental limits on direct-coupling fields: (1) Hughes-Drever experiment (2) ether-drift experiments 1064 38. TESTING THE FOUNDATIONS OF RELATIVITY participants in field equations that determine the geometry of spacetime. This section will describe tests for direct-coupling effects. Theories with fields that couple indi rectly will be described in Box 39.1, and tests for such fields will be discussed in Chapter 40. Dicke (l964b), using his framework for analyzing tests of gravitation theories (\u00a738.2), has shown that several null experiments place stringent limits on unknown, direct-coupling, long-range fields. One of these experiments is the \"Hughes-Drever Experiment\" [Hughes, Robinson, and Beltran-Lopez (1960); Drever (1961)]. It can be thought of as a search for a symmetric second-rank tensor field ha (3 that produces slight deviations of test-body trajectories from geodesics of the metric ga(3. Unless one's experiments happen to be made in a region of spacetime where ha (3 is a constant multiple of ga(3 (\"mere rescaling of all lengths and times by a constant factor\"), this tensor field must produce anisotropies in the properties of spacetime-which, in tum, will cause anisotropies in the inertial mass of a nucleon, and in tum will cause in an atomic nucleus relative shifts of degenerate energy levels with different magnetic quantum numbers. The Hughes-Drever experiment places stringent limits on such shifts, and thereby on a possible tensor field haW To quote Dicke (1964, p. 186), \"If two [tensor] fields are present with the one strongly anisotropic in a coordinate", " system chosen to make the other isotropic, the strength of [direct] coupling to one must be only of the order of 10-22 that of the other.... [Moreover], on the moving Earth with ever-changing velocity, anisotropy would be expected at some season.\" From the experiments of Hughes and Drever, then, one concludes that there is not the slightest evidence for the presence of a second tensor field. For further details see Dicke and Peebles (1962a). Another series of experiments, called \"ether-drift experiments,\" places stringent limits on any unknown, long-range vector field that couples directly to mass-energy. One can imagine such a field of cosmological origin. Being cosmological, the 4-vector would most naturally be expected to point in the same direction as the 4-vector u of the \"cosmological fluid\" (identical with the time direction eo of a frame in which the cosmic microwave radiation is isotropic). The 4-vector of the new field would then have spatial components in any other frame. In principle an observer could use them to discern his direction of motion and speed relative to the mean rest frame of the universe. The ether-drift experiments search for effects of such a field. For example, the experiment of Turner and Hill (1964) searches for a dependence of clock rates on such a vector field, by examining the transverse Doppler~shift as a function of direction for an emitter on the rim of a centrifuge and a receiver at its center (Figure 38.3). If there is any effect, it would most naturally be expected to have the form ( rate of clock moving relative) to universe with speed P rate ~f clock ~t rest) relatlVe to UOlverse ( _ 1 - 2 + yf3, y a small constant. (38.17) \u00a738.7. TESTS FOR EXISTENCE OF LONG-RANGE FIELDS 1065 Detector crystal ~A\"\"'b\" w Source, v Figure 38.3. The experiment of Turner and Hill (1964) looks for a dependence of proper clock rate (the clock being a C057 source placed near the rim of the centrifuge) on velocity relative to the distant matter of the universe; or, in operational terms, relative to a \"new local field\" described by a 4-vector. The 14.4 keY gamma rays from the CoS7 already experience a second-order Doppler shift of 1.3 parts in 1013\u2022 One searches for an additional shift yf32 where p = u + v(ez cos wI + eu sin wI) is the velocil)' relative to the frame in which the scalar field is purely timelike. The transmission of the gamma rays through the Fe57 absorber will drop linearly with any such additional shift, and will be noted as a drop in the counting rate of the NaI crystal. The source was 10 cm from the axis of rotation and the centrifuge turned at 15,000 rpm. The value of y deduced from the experiment was (I \u00b1 4) X 10-5 \u2022 A clock at the center of the centrifuge has {:J = u = uex'whereas one on the rim has (:J = u + v(ex cos wt + ey sin wt). Thus, the shift between rim and disk should vary with position L1A/A = -L1v/v = -2yuv cos wt + usual transverse shift. The data of Turner and Hill, using the Mossbauer effect, show that Iyl < 4 X 10-5. (38.18) Hence, a cosmological vector field, if present, has only a weak direct coupling to matter. For further discussion of these experiments and references on others like them, see Dicke (1964b). CHAPTER 39 OTHER THEORIES OF GRAVITY AND THE POST-NEWTONIAN APPROXIMATION \u00a739.1. OTHER THEORIES Among all bodies of physical law none has ever been found that is simpler or more beautiful than Einstein's geometric theory of gravity (Chapters 16 and 17); nor has any theory of gravity ever been discovered that is more compelling. As experiment after experiment has been performed, and one theory of gravity after another has fallen by the wayside a victim of the observations, Einstein's theory has stood firm. No purported inconsistency between experiment and Einstein's laws of gravity has ever surmounted the test of time. Query: Why then bother to examine alternative theories of gravity? Reply: To have \"foils\" against which to test Einstein's theory. To say that Einstein's geometrodynarnics is \"battle-tested\" is to say it has won every time it has been tried against a theory that makes a different prediction. How then does one select new antagonists for decisive new trials by combat? Not all", " theories of gravity are created equal. Very few, among the multitude in the literature, are sufficiently viable to be worth comparison with general relativity or with future experiments. The \"worthy\" theories are those which satisfy three criteria for viability: self-consistency, completeness, and agreement with past experi ment. Self-consistency is best illustrated by describing several theories that fail this test. The classic example of an internally inconsistent theory is the spin-two field theory of gravity [Fierz and Pauli (1939); Box 7.1 here], which is equivalent to linearized general relativity (Chapter 18). The field equations of the spin-two theory imply that all gravitating bodies move along straight lines in global Lorentz reference frames, whereas the equations of motion of the theory insist that gravity deflects Role of alternative gravitation theories as foils for experimental tests Criteria for viability of a theory: (1) self-consistency \u00a739.2. METRIC THEORIES OF GRAVITY 1067 bodies away from straight-line motion. (When one tries to remedy this inconsistency, one finds oneself being \"bootstrapped\" up to general relativity; see route 5 of Box 17.2.) Another self-inconsistent theory is that of Kustaanheimo (1966). It predicts zero gravitational redshift when the wave version of light (Maxwell theory) is used, and nonzero redshift when the particle version (photon) is used. Completeness: To be complete a theory of gravity must be capable of analyzing from \"first principles\" the outcome of every experiment of interest. It must therefore mesh with and incorporate a consistent set of laws for electromagnetism, quantum mechanics, and all other physics. No theory is complete if it postulates that atomic clocks measure the \"interval\" dr = (- gafJ dx a dx fJ )1/2 constructed from a particular metric. Atomic clocks are complex systems whose behavior must be calculated from the fundamental laws of quantum theory and electromagnetism. No theory is com plete if it postulates that planets move on geodesics. Planets are complex systems whose motion must be calculated from fundamental laws for the response of stressed matter to gravity. For further discussion see \u00a7\u00a716.4, 20.6, and 40.9. Agreement with past experiment: The necessity that a theory agree, to within several standard deviations, with the \"four standard tests\" (gravitational redshift, perihelion shift, electromagnetic-wave deflection, and radar time-delay) is obvious. Equally obvious but often forgotten is the need to agree with the expansion of the universe (historically the ace among all aces of general relativity) and with observations at the more everyday, Newtoni~n level. Example: Birkhoff's (1943) theory predicts the same redshift, perihelion shift, deflection, and time-delay as general relativity. But it requires that the pressure inside gravitating bodies equal the total density of mass-energy, p = p; and, as a consequence, it demands that sound waves travel with the speed of light. Of course, this prediction disagrees violently with experiment. Therefore, Birkhoff's -theory is not viable. Another example: Whitehead's (1922) theory of gravity was long considered a viable alternative to Einstein's theory, because it makes exactly the same prediction as Einstein for the \"four standard tests.\" Not until the work of Will (197lb) was it realized that Whitehead's theory predicts a time-dependence for the ebb and flow of ocean tides that is completely contradicted by everyday experience (see \u00a740.8). (2) completeness (3) agreement with past experiment \u00a739.2. METRIC THEORIES OF GRAVITY Two lines of argument narrow attention to a restricted class of gravitation theories, called metric theories. Why attention focuses on metric theories of gravity The first line of argument constitutes the theme of the preceding chapter. It examined experiment after experiment, and reached two conclusions: (1) spacetime possesses a metric; and (2) that metric satisfies the equivalence principle (the standard special relativistic laws of physics are valid in each local Lorentz frame). Theories of gravity that incorporate these two principles are called metric theories.* In brief, Chapter 38 says, \"For any adequate description of gravity, look to a metric theory.\" *For a slightly narrower definition of metric theories, see Thorne and will (1971). How metric theories differ 1068 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION Exception: Cartan's (1922b, 1923) theory [\"general relativity plus torsion\"; see Trautman (1972)] is nonmetric, but agrees with experiment and is experimentally indistinguishable from general relativity with the technology of the 1970's. The second line of argument pointing to metric", " theories begins with the issue of completeness (preceding section). To be complete, a theory must incorporate a self-consistent version of all the nongravitationallaws of physics. No one has found a way to incorporate the rest of physics with ease except to introduce a metric, and then invoke the principle of equivalence. Other approaches lead to dismaying complexity, and usually to failure of the theory on one of the three counts of self-consistency, completeness, and agreement with past experiment. All the theories known to be viable in 1973 are metric, except Cartan's. [See Ni (1972b); Will (1972).] In only one significant way do metric theories of gravity differ from each other: their laws for the generation of the metric. In general relativity theory, the metric is generated directly by the stress-energy of matter and of nongravitational fields. In Dicke-Brans-Jordan theory (Box 39.1, p. 1070), matter and nongravitational fields generate a scalar field 1>; then 1> acts together with the matter and other fields to generate the metric. Expressed in the language of \u00a738.7, 1> is a \"new long-range field\" that couples indirectly to matter. As another example, a theory devised by Ni (1970, 1972) (Box 39.1) possesses a flat-space metric 11 and a universal time coordinate t (\"prior geometry\"; see \u00a717.6); 11 acts together with matter and nongrav itational fields to generate a scalar field 1>; and then 11, t, and 1> combine to create the physical metric 9 that enters into the equivalence principle. All three of the above theories-Einstein, Dicke-Brans-Jordan, Ni-were viable in the summer of 1971, when this section was written. But in autumn 1971 Ni's theory, and many other theories that had been regarded as viable, were proved by Nordtvedt and Will (1972) to disagree with experiment. This is an example of the rapidity of current progress in experimental tests of gravitation theory! Henceforth, in this chapter and the next, attention will be confined to metric theories of gravity and their comparison with experiment. \u00a739.3. POST-NEWTONIAN LIMIT AND PPN FORMALISM Weak-field. slow-motion expansion of a metric theory The solar system, where experiments to distinguish between metric theories are performed, has weak gravity, 14'1 = INewtonian potential1~ 10-6 ;, (39.1 a) moreover, the matter that generates solar-system gravity moves slowly v 2 = (velocity relative to solar-system center of mass? ~ 10-7 (39.1b) and has small stress and internal energies l1jkl/Po = (stress divided by baryon \"mass\" density) ~ 10-6 II = ( _ )/ = (int.ernal energy density per) < 10-6. P Po Po umt baryon \"mass\" density -, (39.1 c) (39.1 d) \u00a739.3. POST-NEWTONIAN LIMIT AND PPN FORMALISM - 1069 [Here the baryon \"mass\" density Po' despite its name, and despite the fact it is sometimes even more misleadingly called \"density of rest mass-energy,\" is actually a measure of the number density of baryons n, and nothing more. It is defined as the product of n with some standard figure for the mass per baryon, JLo' in some well-defined standard state; thus, Po = nJLo'] Consequently, the analysis of solar-system experiments using any metric theory of gravity can be simplified, without significant loss of accuracy, by a simultaneous expansion in the small parameters 14'1, v 2 l1}kl/po' and II. Such a \"weak-field, slow-motion expansion\" gives: (1) flat, empty spacetime in \"zero order\"; (2) the Newtonian treatment of the solar system in \"first order\"; and (3) post-Newtonian corrections to the Newtonian treatment in \"second order\". (39.1 e), Post-Newtonian approximation PPN formalism The formalism of Newtonian theory plus post-Newtonian corrections is called the \"post-Newtonian approximation.\" Each metric theory has its own post-Newtonian approximation. Despite the great differences between metric theories themselves, their post-Newtonian approximations are very similar. They are so similar, in fact, that one can construct a single post-Newtonian theory of gravity, devoid of any reference to indirectly coupling fields (</> in Dicke-Brans-Jordan; 11, t, and </> in Ni; see Box 39.1), that contains the post-Newtonian approximation of", " every conceivable metric theory as a special qlse. This all-inclusive post-Newtonian theory is called the \"Parametrized Post-Newtonian (PPN) Formalism.\" It contains a set of parameters (called \"PPN parameters\") that can be specified arbitrarily. One set of values for these parameters makes the PPN formalism identical to the post-Newtonian limit of general relativity; another set of values makes it the post-Newtonian limit of Dicke-Brans-Jordan theory, etc. Subsequent sections of this chapter present a version of the PPN formalism devised by Clifford M. Will and Kenneth Nordtvedt, Jr. (1972). [See also Will (1972).] This version, containing ten PPN parameters, encompasses as special cases nearly every metric theory of gravity known to the authors. The few exceptions [Whitehead (1922) and theories reviewed by Will (1973)] all disagree with experiment. One can include them in the PPN formalism by adding additional terms and parameters. The ten parameters are described heuristically in Box 39.2, for the convenience of readers who would skip the full details of the formalism (\u00a7\u00a739.4-39.12). Accuracy of PPN formalism in solar system, l1}kl/Po' and II are all ~ 10-6 How accurate is the PPN formalism? Or, stated more precisely, how accurately does the post-Newtonian approXimation agree with the metric theory from which it comes? In the solar system, where 14'1, v 2, the post-Newtonian approximation makes fractional errors of ~ 10-6 in quantities of post-Newtonian order, and fractional errors of ;S 10-12 in quantities of Newtonian order. For example, it misrepresents the deflection of light by ;S 10-6 X (post-New tonian deflection) - 10-6 seconds of arc. And it ignores relativistic deformations of the Earth's orbit of magnitude <10- 12 X (one astronomical unit) 10 centimeters. Clearly, there is no need in the 1970's to use higher-order corrections to the post Newtonian approximation; and hence no need to construct a \"parametrized post post-Newtonian framework.\" However, in the words of Shapiro (1971 b): \"If one projects from the achievements in the last decade, it is not unreasonable to 'predict (continued all page 1072) 1070 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION Box 39.1 THE THEORIES OF DICKE-BRANS-JORDAN AND OF NI A. Dicke-Brans-Jordan References: Brans and Dicke (1961); Jordan (1959). [Notes: This is the special case 'I) = -1 of Jordan's theory. An alternative mathematical representation of the theory is given by Dicke (1962).] Fields associated with gravity: </>, a long-range scalar field; g, the metric of spacetime (from which are constructed the covariant derivative V and the curvature tensors, in the usual manner). Equations by which these fields are determined: The trace of the stress-energy tensor generates </> via the curved-spacetime wave equation O </> = cp,a = ;a 8'1T T. 3 + 2w'where w is the dimensionless \"Dicke coupling constant.\" The stress-energy tensor and </> together generate the metric (i.e., the spacetime curvature) via the field equations where Ga {3 is the Einstein tensor. Variational principle for these equations: where R is the scalar curvature and L is the matter Lagrangian. Equivalence principle is satisfied: The special-relativistic laws of physics are valid, without change, in the local Lorentz frames of the metric g. Consequence: the scalar field does not exert any direct influence on matter; its only role is that of participant in the field equations that determine the geometry of spacetime. It is an \"indirectly coupling field\" in the sense Qf\u00a738.7. This theory is self-consistent, complete, and for w > 5 in \"reasonable\" accord (two standard deviations or better) with all pre-1973 experiments. B. Ni References: Ni (1970, 1972) Fields associated with gravity:. 11, a flat \"background metric\" (\"prior geometry\" in sense of \u00a7 17.6). There exist, \u00a739.3. POST-NEWTONIAN LIMIT AND PPN FORMALISM 1071 by assumption, coordinate systems (\"background Lorentz frames\") in which everywhere at once 'lJ00 = - 1, 'lJOj = 0, and 'lJ", "jk = 0jk' t, a scalar field called the \"universal time coordinate\" (\"prior geometry\" in sense of \u00a7 17.6), which is so \"tuned\" to the background metric that where \",,\" denotes covariant derivative with respect to 11. This means there exists a background Lorentz frame (the \"rest frame of the :. universe\") in which x O = t. </>, a scalar field called the \"scalar gravitational field\". g, the metric of spacetime (from which are constructed the covariant derivative V and the curvature tensors, in the usual manner). Equations by which these fields are determined: The stress-energy of spacetime generates the scalar gravitational field </> via the wave equation D</> - </>,a;a = -271'Taf3 ogaf3/o</> = 471'Taf3['lJaf3e-z\u00a2 + (eZ\u00a2 + e-Z\u00a2)t,at,f31\u00b7 <fl, 11, and t together determine the metric of spacetime through the algebraic relation Note: In the \"rest frame of the universe\" that is presupposed in this theory, this metric reduces to ds Z = gaf3 dx a dx f3 = _ez\u00a2 dt Z + e-Z\u00a2(dx Z + dyz + dz Z). Variational principle for the field equation for </>: oJ(-2</>,a</>,a + 1671'L)( - g)l/Z d 4x = 0, where L is the matter Lagrangian. Equivalence principle is satisfied: The special-relativistic laws of physics are valid, without change, in the local Lorentz frames of the spacetime metric g. Consequence: <fl, 11, and t do not exert any direct influence on matter; they are \"indirectly coupling fields\" in the sense of \u00a738.7. This theory is self-consistent and complete. If the solar system were at rest in the \"rest frame of the universe\", the theory would agree with all experiments to date-except, possibly, the expansion of the universe. But the motion of the solar system through the universe leads to serious disagreement with experiment (Will and Nordtvedt 1972; \u00a740.8). 1072 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION Box 39.2 HEURISTIC DESCRIPTION OF THE TEN PPN PARAMETERS What it measures, relative to general relativity\" Value in General Relativity Value in Dicke-Brans Jordan Theoryb Value in Ni\"s Theoryb Para meter y f3 How much space curvature (gjk) is produced by unit rest mass? How much nonlinearity is there in the superposition law for gravity (goo)? How m?uch gravity (goo) is produced by unit kinetic energy (~pr)? How much gravity (goo) is produced by unit gravitational potential energy (Po U)? f3a How much gravity (goo) is produced by unit internal energy (Poll)? How much gravity (goo) is produced by unit pressure (p)? How much more gravity (goo) is produced by radial kinetic energy Hpo('\" i)2]-i.e., kinetic energy of motion toward obser ver-than by transverse kinetic energy? How much more gravity (goo) is produced by radial stress [i o t\u00b7 i] than by transverse stress? o o How much dragging of inertial frames (gOj) is produced by unit momentum (pou)? L12 How much easier is it for momentum (Pau) to drag inertial frames radially (toward the observer) than in a transverse direction? 1+\", 2+0: 3 + 20: 4 + 2\", 1+2\", 4 + 2\", 1+\", 2+\", o o 10 + 7\", 14 + 7\", o o I 7 \"These heuristic descriptions are based on equations (39.23). b For expositions of these theories see Box 39.1. For derivation of their PPN values and of PPN values for other theories, see Ni (1972). that in the 1980's techniques will be available to detect second-order effects of! general relativity. At that point the ratio of theoretical to experimental relativists may take a sharp tum downwards.\" Actually, there are a few exceptions to the claim that the post-Newtonian approxi mation suffices for the 1970's. These exceptions occur where the external universe impinges on and influences the solar system. For example, gravitational waves propagating into the solar system from distant sources (Chapters 35-37) are ignored by every post-Newtonian approximation and by the PPN framework. They must be treated using a full metric theory or a weak-field, \"fast", "-motion\" approximation \u00a739.4. PPN COORDINATE SYSTEM 1073 to such a theory. Similarly, time-dependence of the \"gravitational constant\" (\u00a740.8), induced in some theories by expansion of the universe, is beyond the scope of the PPN formalism, as is the expansion itself. The PPN formalism is used not only in interpreting experimental tests of gravita- Applications of PPN tion theories, but also as a powerful tool in theoretical astrophysics. By specializing all the PPN parameters to unity, except t = 'I) = 0, one obtains the post-Newtonian approximation to Einstein's theory of gravity. This post-Newtonian approximation can then be used (and has been used extensively) to calculate general relativistic corrections to such phenomena as the structure and stability of stars.* formalism to astrophysics History and notation of PPN formalism Historical and Notational Notes The earliest parametrizations of the post-Newtonian approximation were performed, and used in interpreting solar system experiments, by Eddington (1922), Robertson (1962), and Schiff (1962, 1967). However, they dealt solely with the vacuum gravita tional field outside an isolated, spherical body (the sun). Nordtvedt (1968b, 1969) devised the first full PPN fOrmalism, capable of treating all aspects of the solar -~~. --- system; he treated the sun, planets, and moon as made from \"gases\" of point-particles (atoms) that interact gravitationally and electromagnetically. Will (197lc) later used techniques devised by Chan<Jrasekhar (1965a) to modify Nordtvedt's formalism, so that it employs a stressed, continuous-matter description of celestial bodies. The version of the formalisIll Qresented here~-lkYise\\LbLWill al!.d Nordtvedt (1972), -----vgeeTrrerahzesall previous versions to acquire \"post-Galilean invarillIlce\" [seeThandra sekhar and Contopolous (1967)]. The most detailed and up-to-date review article on the PPN formalism is Will (1972). In the literature of post-Newtonian physics and the PPN formalism, the Newtonian potential is described traditionally not by tP, but by - JPo(x') d3x' u= -tP+ 1 '1.___ x - x _.. __ (39.2) To avoid confusion, this chapter and the next will use U, although the rest of the book uses tP. Turn now to a detailed, Track-2 exposition of,the PPN formalism. \u00a739.4. PPN COORDINATE--S-YSTEM ~-- The PPN formalism covers the solar system (or whatever system is being analyzed) with coordinates (t, Xj) = (t, xi) that are as nearly globally Lorentz as possible: EXPOSITION OF PPN FORMALISM: Coordinate system (39.3) \u2022 See. e.g.. a long series of papers by Chandrasekhar and his associates in the Astrophysical Journal. beginning with Chandrasekhar (1965a.b,c). r The rest of this chapter is Track 2. No earlier Track-2 material is needed as preparation fOr it, but the fOllowing will be helpful: (1) Chapter 7 (incompatibility of gravity and special relativity) (2) \u00a7 17.6 (no prior geometry); (3) \u00a7\u00a736.9-36.11 (generation of gravitational waves); and (4) Chapter 38 (tests of foundations). This chapter is not needed as preparation for any later chapter, but it will be helpful in Chapter 40 (solar-system tests) Description of matter 1074 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION (In this sense the PPN formalism is like linearized theory; see Chapter 18.) The velocity of the coordinate system (i.e., 4-velocity of its spatial origin) is so chosen that the solar system is approximately at rest in these coordinates. (Whether the center of mass of the solar system is precisely at rest, or is moving with some low velocity v ~ (M0 / R0 )l/2 - 10-3 - 300 km/sec, is a matter for the user of the formalism to decide. For more on the options, see \u00a7\u00a739.9 and 39.12.) The PPN coordinates provide one with a natural \"3 + 1\" split of spacetime into space plus time. That split is conveniently treated using the notation of three-dimen sional, flat-space vector analysis-even though spacetime and the three-", "dimensional hypersurfaces xO = constant are both curved. The resultant three-dimensional for malism will look more like Newtonian theory than like general relativity-as, indeed, one wishes it to; after all, one's goal is to study small relativistic corrections to Newtonian theory! \u00a739.5. DESCRIPTION OF THE MAnER IN THE SOLAR SYSTEM Relative to the PPN coordinates, the matter of the solar system (idealized as a stressed medium) has a coordinate-velocity field Vj = dx/dxo. Choose an event 9, and in its neighborhood transf~~k; an ortllonofffial frame that moves with the matter there. Orient the spatial axes ej of this comoving frame so that they coincide as accurately as possible with the PPN coordinate axes. (This requirement will be made more precise in \u00a739.10.) In the orthonormal comoving frame, define the following quantities, which describe the state of the matter: ---------- -- (39.4) --_.- - - (density of total mass-energy) (baryon \"mass\" density) = Po = (nUmber density) of baryons, n -. P; (standard re.st mass per baryon, p.o'); X for matter 10 some standard state (specific internal energy density)=II =(p - Po)/Po; (components of stress tensor) = tij = ei' T' ej; (pressure)=p =; (txx + tyy + tzz) (39.5a) (39.5b) (39.5c) (39.5d) ~ ------- (39.5e) = (average of stress over all directions). Anisotropies (i.e., shears) in the stress are important only in planets such as the Earth; and even there they are dominated by the isotropic pressure: tij = P 0ij + P X [corrections ~ 1]. (39.6) \u00a739.6. NATURE OF THE POST-NEWTONIAN EXPANSION 1075 For many purposes, especially inside the sun, one can ignore the anisotropies, thereby approximating the solar-system matter as a perfect fluid.* The isotropic part of the radiation field gives a significant contribution to the pressure, p, and the density of internal energy, Poll, inside the Sun. However, the anisotropic radiation flux is ignored in the stress-energy tensor. This approximation is allowable because in the sun the outward energy flux carried by radiation is less than 10-15 of the internal energy density poll; in planets it is even less. \u00a739.6. NATURE OF THE POST-NEWTONIAN EXPANSION For any gravitationally bound configuration such as the solar system, the Newtonian approximation imposes limits on the sizes of various dimensionless physical quanti ties (si:e exercise 39.1): f2 = maximum value of Newtonian potential U ---------- ~ues--anywhereof U, v2, p/po' Ilijl /Po' II. (39.7) Relative magnitudes of expansion parameters (The Newtonian potential at the center of the sun is f2 - 10-5. The values ofp/po' lij/Po' and II there are also -10- 5, and they are much smaller elsewhere. The orbital velocities of the planets are all less than 100 km/sec = 3 X 10-4, so v 2 < 10-7 Moreover, changes with tim~ of all quantities at fixed x j are due primarily to the motion of the matter. As a result, time derivatives are small by O(f) compared to space derivatives,.) aA/al I aA/axj - I Ivjl ~ f for any quantlty A,. (39.8) although nol in the radiation zone, where outgoing gravitational waves flow (dis tance ~ one light year from Sun). Consequently, the radiation zone must be excluded from the analysis when one makes a post-Newtonian expansion. To treat it requires different techniques, e.g., those of Chapter 36. Conditions 39.7 and 39.8 suggest that one expand the metric coefficients in powers Rules of the expansion of the small parameter f, treating U, v 2, p/po' Ii/Po' and II as though they were all of 0(f2) (often they are smaller!), and treating time derivatives as O(f) smaller than space derivatives. In this \"post-Newtonian\" expansion, terms odd in f (i.e., terms such as JPo(x', I)Vj(x', I) d3 Ix- x'i M X'- -V - f R 3 (39.9) whose total number of v's and (a/al)'S", " is odd) change sign under time reversal, \u2022 In the solar system, post-Newtonian corrections due to anisotropic stresses are so much smaller than other post-Newtonian corrections that there is no hope of measuring them in the 1970's. For this reason, elsewhere in the literature (but not in this book) the PPN formalism treats all stresses at the post-New tonian level as isotropic pressures, thereby setting to zero the PPN parameter 1) of \u00a7\u00a739.8-39.11. y Expanded form of metric 1076 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION whereas terms even in f do not. Time reversal (xu = _X O) also changes the sign of gOj(gOj = - gOj)' but leaves goo and gjk unchanged. Therefore, gOj must contain only terms odd in f; whereas goo and gjk must contain only even terms. (Actually, this ceases to be the case when radiation damping enters the picture. In the real world-one always insists on outgoing-wave boundary conditions. But time reversal converts outgoing waves to ingoing waves; so an extra sign change is required to convert back to out. Therefore, radiation damping terms in the near-zone metric are even in f for gOj' but odd for goo and gjk. However, radiation damping does not come into play until order f5 beyond Newtonian theory-see Chapter 36-so it will be ignored here.) The form of the expansion is already known through Newtonian order (see \u00a717.4, with t/J replaced by - U): Newtonian gravity is only obtained when one demands that goo = -1 + 2U + [terms ~ f4], gOj = [terms ~ f3], gij = Sij + [terms ~ f2]. (39.10) The stated limits on the higher-order corrections are dictated by demanding that the space components of the geodesic equation agree with the Newtonian equation of motion: d 2x j _ d 2 dt2 - d-r 2 - Xj _ - r j a dx fJ d-r dx afJ d-r = -Fioo - 2FiokVk - ri\"jVkVl ::::: -rja\" dx fJ dt a dx fJ dt (39.11) U,j + terms of order {fgok,j; f 2g kl,j}. One would get the wrong Newtonian limit if gOk were O(f) or greater, and if gkl - Ski were 0(1) or greater. The above pattern continues on to all orders in the expansion. Thus in the geodesic equation, and also in the law oflocal conservation of energy-momentum TafJ;fJ = 0, goo always goes hand-in-hand with fgok and f2 gjk (see exercise 39.2). Therefore, the post-Newtonian expansion has the form summarized in Box 39.3. EXERCISES Exercise 39.1. ORDERS OF MAGNITUDE IN GRAVITATIONALLY BOUND SYSTEMS Use Newtonian theory to derive conditions (39.7) for any gravitationally bound system. [Hint: Such concepts as orbital velocities, the speeds of sound and shear waves, the virial theorem, and hydrostatic equilibrium are relevant.] Exercise 39.2. PATTERN OF TERMS IN POST-NEWTONIAN EXPANSION Verify the statements in the paragraph following equation (39.11). In particular, suppose that one wishes to evaluate the coordinate acceleration, d 2xi/dt 2, to accuracy (2NU,i for some \u00a739.7. NEWTONIAN APPROXIMATION 1077 integer N. Show that this undertaking requires a knowledge of goo to accuracy (2N + 2, of gOk to (2N+l, and of gik to (2N. Also suppose that one knows Too to accuracy Po(2N, TOi to Po( 2N + 1, and Ti k to Po( 2N + 2 [see, e.g., equations (39.13) for N = 0 and (39.42) for N = 2J. Show that to calculate Toa;a with accuracy (2N +1po\u2022i and Tia;a with accuracy (2N + 2PO,j' one must know goo to (2N + 2, gOk to (2N + 1, and gik to (2N. This dictates the pattern of Box 39.3. \u00a739.7. NEWTONIAN APPROXIMATION At Newtonian order the metric has the form (39.10); and the 4", "-velocity and stress- Newtonian approximation energy tensor have components, relative to the PPN coordinate system, U O = + 1 + 0\u00ab(2), TOO = Po + 0(Po(2), Tjk = tjf< + PoVjVk + 0(Po(4) u j = vj + 0\u00ab(3); TOj = PoVj + 0(Po(3), (39.12) (39.13) (see exercise 39.3). Two sets of equations govern the structure and evolution of the solar system. (1) The Einstein field equations. As was shown in \u00a718.4, and also in \u00a717.4, in the Newtonian limit Einstein's equations reduce to Laplace's equation \u2022 which has the \"action-at-a-distance\" solution ~ U (x, t ) - JPo(x', t) d 3 Ix _ x'i -, x. (39.14a) (39.14b) Box 39.3 POST-NEWTONIAN EXPANSION OF THE METRIC COEFFICIENTS Level of approximation (and papers expanding general relativity to this level) flat. empty spacetime Newtonian approximation post-Newtonian approximation [(Fock (1959): Chandrasekhar (l965a)] post-post-Newtonian approximation [Chandrasekhar and Nutku (1969)] radiation damping [Chandrasekhar and Esposito (1970)] Order or value of terms go; 0 0 g;k O;k 0 goo -1 2U + terms _ (4 + terms _ (3 + terms _ (2 + terms _ (6 + terms _ (5 + terms _ (4 + terms _ (7 + terms _ (6 + terms _ (5 1078 39. OTHER THEORIES OF GRAVITY AND POST\u00b7NEWTONIAN APPROXIMATION (2) The law of local energy-momentum conservation, T a nent of this law reduces to the conservation of rest mass (3;{3 = O. The time compo (39.l5a) and the space components reduce to Newton's second law of motion, \"F = rna\": Po du/dt = po(au/axj ) d/dt =(time derivative following the matter) =a/at + Uk a/axk - atj,,/axk + fractional errors of 0\u00ab(2), (39.l5b) (39.16) (see exercise 39.3). Equations (39.14)-(39.16), together with equations of state describing the planetary and solar matter, are the foundations for all Newtonian calculations of the structure and motion of the sun and planets. Notice that the internal energy density Poll nowhere enters into these equations. It is of no importance to Newtonian hydrody namics. It matters for the sun's thermal-energy balance; but that is irrelevant here. EXERCISES Exercise 39.3. NEWTONIAN APPROXIMATION (a) Derive equations (39.13) for the components of the stress-energy tensor in the PPN coordinate frame. [Hint: In the rest frame of the matter (\"comoving orthonormal frame\") Too = P = Po + 0\u00ab(2), Tol = 0, 1}k = t'k; see equations (39.5). Lorentz-transform these components by a pure boost with ordinary velocity - vJ to obtain TaW] (b) Show that, in the PPN coordinate frame, TOa;a = 0 reduces to equation (39.15a), and Tja;a = 0, when combined with (39.15a), reduces to equation (39.15b).] Exercise 39.4. A USEFUL FORMULA Derive from equations (39.15) the following useful formula, valid for any function f(x, t): d f dt Po(x, t )ji( x, t) d 3 x = Po(x, t) -d-t- f + fractional errors of 0\u00ab(2). df(x, t) d 3 x (39.17) Here both integrals are extended over all of space; and df/dt is the derivative following the matter (39.16). Exercise 39.5. STRESS TENSOR FOR NEWTONIAN GRAVITATIONAL FIELD Define a \"stress tensor for the Newtonian gravitational field U\" as follows: ~ (39.18) Show that the equations of motion for the matter (39.15b) can be rewritten in the forms Po!!!!L = - -;. (tik + t'k) + fractional errors of 0\u00ab(2), (PoVi),! + (tik + tlk + PoVj", "Vk),k = 0 + fractional errors of 0\u00ab(2). ax dt (39.19) (39.19') \u00a739.7. NEWTONIAN APPROXIMATION 1079 Exercise 39.6. NEWTONIAN VIRIAL THEOREMS (a) From equation (39.19') show that where Ilk is the second moment of the system's mass distribution, This is called the \"time-dependent tensor virial theorem.\" (b) From this infer that, if < )longtime denotes an average over a long period of time, then (39.2Gb) This is called the \"tensor virial theorem.\" (c) By contraction of indices and use of equations (39.18), (39.14a), and (39.5e), derive the (ordinary) virial theorems: where I is the trace of the seco~d moment of the mass distribution I = f.. = I p r 2 d 3x' 11 0'and (I pov2 d 3x, J, - +J poUd 3x + I 3Ip d3x)... I. = O(IpO\u00a34 d3X). longtime (39.21b) 2 X (kinetic) + (gravitational) + 3 X (?ressure) Integral energy energy Exercise 39.7. PULSATION FREQUENCY FOR NEWTONIAN STAR Use the ordinary, time-dependent virial theorem (39.21a) to derive the following equation for the fundamental angular frequency of pulsation of a nonrotating, Newtonian star: w2 = (31' _ 4) I. Istar's self-gravitational energyl (trace of second moment of star's mass distribution), (39.22a) f I - (pressUre-weighted average) _ SFIP d3x = SP d3x - of adiabatic index. (39.22b) In the derivation assume that the pulsations are \"homologous\"-i.e., that a fluid element with equilibrium position xl (relative to center of mass xl = G) gets displaced tox l + ~I(x, I), where ~I = (small constant)xle-iwt. Assume nothing else. Notes: (1) The result (39.22) was derived differently in Box 26.2 and used in \u00a724.4. (2) The assumption of homologous pulsation is fully justified if iFl - 4/3! = constant <; 1: see Box 26.2. (3) The result (39.22) is readily generalized to slowly Post-Newtonian corrections to metric: (1) rules governing forms 1080 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION It can also be rotating Newtonian stars; see, e.g., Chandrasekhar and Lebovitz (1968). generalized to non rotating post-Newtonian stars using general relativity (Box 26.2), or using the PPN formalism for any metric theory [Ni (l973)J. And it can be generalized to slowly rotating, post-Newtonian stars [see, e.g., Chandrasekhar and Lebovitz (l968)J. \u00a739.8. PPN METRIC COEFFICIENTS (3, kij - (4, kOi - The post-Newtonian corrections kafJ to the metric coefficients gafJ are calculated, in any metric theory of gravity, by lengthy manipulations of the field equations. (See, e.g., exercise 39.14 near the end of this chapter for general relativity.) But without ever picking some one theory, and without ever writing down any set of field equations, one can infer theforms of the post-Newtonian corrections k afJ \u2022 Their forms are fixed by the following constraints: (1) They must be of post-Newtonian order (koo (2). (2) They must be dimensionless. (3) koo must be a scalar under rotations, k Oj must be components of a 3-vector, and k ik must be components of a 3-tensor. (4) The corrections must die out at least as fast as l/r far from the solar system, so that the coordinates become globally Lorentz and spacetime becomes flat at r = 00. (5) For simplicity, one can assume that the metric components are generated only by Po, Poll, tii' p, products of these with the velocity Vi' and time-derivatives of such quantities;* but not by their spatial gradients. [This assumption of simplicity is satisfied by all metric theories examined up to 1973, except Whitehead (1922) and theories reviewed", " by Will (1973)-which disagree with experiment.] Note the further justification for this assumption in exercise 39.8. (2) construction of corrections Begin with the corrections to the spatial components, k ii - (2. There are only two functionals of PO' p, II, tik' Vi' that die out at least as fast as l/r, are dimensionless, are 0\u00ab(2), and are second-rank, symmetric 3-tensors; they are u..( t) X, t) - JPo(x', t)(xi - xi')(Xi I x-x '1 3 - x;') d 3, X \u2022 (39.23a) Thus, k ii must be kij = 2\"1 8ij U + 2r Uij> for some constant \"PPN parameters\" \"I and r. By an infinitesimal coordinate transformation [Xi NEW = XiOLD + r axlaxil with X(x, t) = - fpo(x', t)lx - x'i d 3x'] one can set r = 0, thereby obtaining ~39.23b) 'One allows for time derivatives because retarded integrals contain such terms when expanded to post-Newtonian order; thus, f po(x', 1 - Ix - x'i Ix - x'D iJ3x' = f [Po(X', I) _ oPo(x', I) +... ] iJ3x'. Ix - x'i 01 However, it turns out that, with a suitable choice of coordinates (\"gauge\"), all time-odd retarded terms [e.g., f(oP%l} iJ3x] vanish, except at \"the post5/ 2-Newtonian order\" and at higher orders of approxima tion; there they lead to radiation damping (see Box 39.3). For example, f(oPo/OI} iJ3x = (d/dt)fpo iJ3x vanishes by virtue of the conservation of baryon number. \u00a739.8. PPN METRIC COEFFICIENTS 1081 Next consider k Oj - (3. Trial and error yield only two vector functionals that die out as l/r or faster, are dimensionless, and are 0\u00ab(3); They are 1 V(x t) =I Po(X', t)vj(x', t) d 3x' Ix - x'i _ I Po(X', t)[(x - x')\u00b7 v(x', t)](X j - x;') d3x' I x-x '1 3'' W \u00b7(x t)'1 (39.23c) (39.23d) Thus, k Oj must be a linear combination of these, involving unknown constants (PPN parameters)..:11 and..:1 2 : (39.23e) Finally consider koo - (4. Trial and error yields a variety of terms, which can all be combined together :with the Newtonian part of goo to give goo = -I + 2U + koo = -I + 2U - 2f3U2 + 4'!' - ~{j' -1j6f}, (39.23f) where''I '!'(x, t) - \u00b7 _I Po(X', t)1/;(x', t) d3x' I x-x 2 + f3 2U + ; f3 3Il + ~ f34p/PO' --- I po(x', t)[(x - x')\u00b7 v(x', t)F 3'Ix- x'i d x, 3 1/; = f3 1V (j'(x, t) = (39.23g)* (39.23h) (:D(x t) =I, [t]k(X', t) - ~ 8jktll(X', t) ] (xj - x;')(x k - x/) Ix - x1 3 d 3x'. (39.23i) Also, f3, in the literature the term -1j6f} in goo is ignored (see footnote on p. 1075). f34, t 1j are unknown constants (PPN parameters). Elsewhere f3 2, f3 3, f3 1, Yet another term is possible: one could have ~et goo = [value in equation (39.23f)] _IIf Po(x', t)po(x\", t)[(x - x')\u00b7 (x' - x\")]d3x' d 3x\", Ix - x'llx' - x\"1 3 (39.24) where I is another P", "PN parameter. [It can be shown, using the Newtonian equations (39.14)-(39.16), that this expression dies out as l/r far from the solar system.] If 'WARNING: Throughout the literature the notation (/J is used where we use'!' for the functional (39.23g), and \u00a2 is used for our y. We are forced to violate the standard notation to avoid confusion with the Newtonian potential (/J = - U. However, we urge that nobody else violate the standard notation! 1082 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION such a I infinitesimal coordinate transformation term had been included, then one could have removed it by making the o ~ 0 X new - X old _ ~ I JPo(x', t)[(x - x') \u2022 ~'(x', t)] 3\u00b7' Ix _ x'i d X 2 (39.25) Rigidity of coordinate system The absence of the I term from goo means that the time coordinate has been (see exercise 39.9). Thus, there is no necessity to include the I term. fixed rigidly up through post-Newtonian order: O has uncertainties only of O(Rd 5 X ) - 10-14 seconds. (39.26a) The space coordinates are also fixed rigidly through post-Newtonian order: Xi has uncertainties only of O(Rd 4) - 0.1 em, (39.26b) because any transformation of the form Summary of PPN metric and parameters would destroy the form (39.23b) of the space part of the metric. In summary, for almost every metric theory of gravity yet invented, accurate through post-Newtonian order the metric coefficients have the form (39.23). One theory is distinguished from another by the values of its ten \"post-Newtonian parameters\" f3, f3 1, f3 2, f33, f3 4, \"I, t 1),..:11 and..:12'These are determined by comparing the field equations of the given theory with the form (39.23) of the post-Newtonian metric. The parameter values for general relativity and for several other theories are given in Box 39.2, along with a heuristic description of each parameter. EXERCISES Exercise 39.8. ABSENCE OF \"METRIC-GENERATES-METRIC\" TERMS IN POST-NEWTONIAN LIMIT In writing down the post-Newtonian metric corrections, one might be tempted to include terms that are generated by the Newtonian potential acting alone, without any direct aid from the matter. After all, general relativity and other metric theories are nonlinear; so the two-step process (matter) ---+ U ---+ (post-Newtonian metric corrections) seems quite natural. Show that such terms are not needed, because the equations (39.14)-(39.16) of the Newtonian approximation enable one to reexpress them in terms of direct integrals over the matter distribution. In particular, show that 02 U(X', t)/ox/ at d3, - 2 [U( --'---'--'---;:,-'--Ix - x I X - 'IT Y i x, t f ) W;( - iX, t )J (3927) where V; and Uj are defined by equations (3923c,d); also show that [a U(x', t)/ox;'][o U(x', t)/ox/J d 3x' Ix - x'i f = -2'lT[U(x, t)J2 + 4'lTf Po(x', t)U(x', t) d 3x'. Ix - x'i (39.28) \u00a739.9. VELOCITY OF PPN COORDINATE FRAME 1083 Note that the terms on the righthand sides of (3927) and (3928) are already included in the expressions (39.23e,f) for go; and goo' Exercise 39.9. REMOVAL OF 2 TERM FROM 900 Show that the coordinate transformation (39.25) removes the 2 term from the metric coefficient goo of equation (3924), as claimed in the text. Exercise 39.10. VERIFICATION OF FORMS OF POST-NEWTONIAN CORRECTIONS Verify the claims in the text immediately preceding equations (39.23a,b,c,f). \u00a739.9. VELOCITY OF PPN COORDINATES RELATIVE TO \" \"UNIVERSAL REST FRAME\" Thus far it has been assumed tacitly that the center of mass of the solar system is at rest in the PPN coordinate system. Is this really a permissible assumption? Put", " differently, can one always so adjust the PPN coordinate system that its origin moves with any desired velocity (e.g., that of the solar system); or is the PPN coordinate system rigidly and irrevocably attached to some \"universal rest frame\"? In general relativity, the geometry of curved spacetime picks out no preferred coordinate frames (except in cases with special symmetry). Therefore, one expects the velocity of the PPN coordinate frame to be freely specifiable. Put differently, one expects the entire PPN formalism, for general relativity, to be invariant under Lorentz transformations of the PPN coordinates [combined, perhaps, with \"in finitesimal coordinate transformations\" to maintain the gauge conditions that the \"I\" and \"U;k\" terms of (39.24) and (39.23a) be absent]. By contrast, in Ni's theory of gravity (Box 39.1) the geometry of spacetime always picks out a preferred coordi nate frame: the \"rest frame of the universe.\" One would not be surprised, in this case, to find the PPN coordinate frame rigidly attached to the universal rest frame. The above intuition is correct, according to calculations by Will (1971d) and by Will and Nordtvedt (1972). When dealing with general relativity and other theories with little or no \"prior geometry,\" one can freely specify the velocity of the PPN coordinate system (at some initial instant of time). But for theories like Ni's, with a preferred \"universal rest frame\" (\"preferred-frame theories\"), only in the preferred frame can the post-Newtonian metric assume the form derived in the last section [equations (39.23)]. This restriction on the PPN metric does not mean that one is confined, in preferred-frame theories, to perform all calculations in the universal rest frame. Rather, it means that for such theories the PPN metric requires generali zation to take account of coordinate-frame motion relative to the universal rest frame. The required generalization can be achieved by subjecting the PPN metric (39.23) to (1) a Lorentz boost from the preferred frame {x'OLD} to a new PPN frame {x~lEw}' which moves with velocity w, plus (2) a change of gauge designed to keep the metric coefficients as simple as possible. The boost-plus-gauge-change is [Will and Nordt vedt (1972)] Preferred-frame theories of gravity Generalization of PPN metric to moving frames ~ / / 1084 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION X OLD = X NEW + ; (XNEW \u2022 W)W + (1 + ; 11'2 )WtNEW (39.29a) + 0\u00ab(5tNEW + (4XNEW )' tOLD = tNEW (1 + ; 11'2 + ~ 11'4) + (1 + ; 11'2 )XNEW \u2022 II'!' ( 1 A ax + -2 \"-12 + ~ - 1 wj -, -,- + 0\u00ab( tNEw + ( x NEW), ax NEW -l-.{gauge change] 5 6 ) (39.29b) X(tNEW' x NEW) = - JPo(tNEW' xNEw)lxNEw - x;'/Ewl d3x NEW' (39.29c) [Note: One insists, in the spirit of the post-Newtonian approximation, that the velocity II' of the new PPN frame relative to the universal rest frame be no larger than the characteristic internal velocities of the system: 111'1 ~ c] (39.30) This change of coordinates produces corresponding changes in the velocity of the matter VOLD = dt dXOLD OLD 1 2) ( = VNEW 1 - w\u00b7 VNEW + II' (1 - ; w\u00b7 VNEW ) + 0\u00ab(5). 2\"11' (39.31) A long but straightforward calculation (exercise 39.11) yields the following compo nents for the metric in the new PPN coordinates. [Note: The subscripts NEW are here and hereafter dropped from the notation.] Final form of metric (39.32a) goo = -1 + 2U - 2f3U2 + 4'!' - + (0: 2 + 0:3 - 2 0:1)W ~(f U + (20: 3 - O:I)Wj~ - 1J6j) 0:2WjWkUjk + 0\u00ab(6). (39.32c) Here 0: 1, 0: 2, and 0: 3 are certain combinations of PPN parameters 0: 1 = 7.11 +.1 2 - 4'1 -", " 4, 0: 2 =.1 2 + ~ - 1, 0: 3 = 4f3 1 - 2'1 - 2 - ~. (39.33a) (39.33b) (39.33c) The \"gravitational potentials\" U, ~, Uj, '!', (f, and 6j) appearing here are to be calculated in the new, \"moving\" PPN coordinate system by the same prescriptions \u00a739.9. VELOCITY OF PPN COORDINATE FRAME 1085 as one used in the universal rest frame. Thus, their functional forms are the same as previously, but their values at any given event are different (see exercise 39.11): ', Ix- x'i U(x t) =JPo(x', t) d 3x\" V(x t) =JPo(x', t)vlx', t) d 3x'\u00b7'Ix - X'I _ JPO(X', t)[(X - X')\u00b7 V(X', t)](Xj - X/) d 3x'. I x-x '!'(X, t) =JPO(X', t)1/;(X', t) d 3x', '1 3 1'' IX- x'i w j(X, t) - 1/; = f3 1V2 + f3 2U + ~ f33II + ~ f34P/PO ; JP (x', t)[(X - x') \u2022 v(X', t)F Ix - x'1 3 Cl(x t) =, 0 d 3x\"'(39.34a) (39.34b) (39.34c) (39.34d) (39.34e) q)(X, t) = J [t,,,(X', t) - ; 8jktll(X', t) ] (X j - X;')(X k - Xk'), Ix - x'1 3 d 3x'. (39.34f) The quantity Ujk is the gravitational potential defined in equation (39.23a): U. ( )k x, t) - JPo(x', t)(xj I x-x - xi)(xk - x;J d3'X. '1 3 (39.34g) Notice that the velocity;'; of the PPN coordinate system relative to the universal rest frame appears explicitly in the PPN metric only if one or more of the coefficients 0: 1,0: 2,0: 3, is nonzero. Thus, theories with 0: 1 = 0: 2 = 0: 3 = 0 (e.g., general relativity) possess no preferred universal rest frame in the post-Newtonian limit; all their PPN frames are \"created equal.\" By contrast, theories with at least one of 0: 1, 0: 2, 0: 3, nonzero (e.g., Ni's theory) do possess a preferred frame. The generalized form (39.32) of the PPN metric, by virtue of the process used to construct it, is invariant under a Lorentz boost plus a gauge adjustment [\"PostGalilean transformation\"; see Chandrasekhar and Contopolous (1967)]: Post-Galilean invariance XOLD = xNEW + T(XNEW'mJ3 + I + Tf3 2 J3tNEW ( I ). I + O(f 5tNEW + f 4 XNEW)' (39.35) tOLD = (I + ~ f32 + ~ (34) tNEw + (I + ~ (32) xNEW \u2022 13 + (~.1 2 + ~1 - 1)13' VNEWX + O(f 6 tNEW + f 5 X}IEW)' Of course, it is also invariant under spatial rotations. 1086 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION EXERCISE Exercise 39.11. TRANSFORMATION TO MOVING FRAME Show that the change of coordinates (39.29) changes the PPN metric coefficients from the form (39.23) to the form (39.32). [Hints: (I) Keep firmly in mind the fact that the potentials U, JIf, J-J!f, {/, and 'v are not scalar fields. Each coordinate system possesses its own potentials. For example, by using equations (39.29) in the integral for UOLD'one finds U ( OLD XOLD' tOLD = I ) _ / f Pa(X~LD' tOLD) d 3 / = [ UNEW - W;(JlfNEW - X OLD X OLD I X OLD J-J!f", "NEW) +t w; WkX,;k] + 0\u00ab(6)..l'NEW.!SEW (39.36) (2) The law of baryon conservation (39.44) may be useful.] \u00a739.10. PPN STRESS-ENERGY TENSOR The motion of the solar system is governed by the equations TafJ;fJ = O. Before studying them, one must calculate the post-Newtonian corrections to the stress-energy tensor in the PPN coordinate frame. This requires a transformation from the comov ing, orthonormal frame w a, where TOO = Pa(l + II), Toi = 0, TiT< - h - jK' (39.37) to the coordinate frame. One can effect this transformation in two stages: stage 2 is a transformation WO = [l - U + 0\u00ab(4)] dt + [~.11Vf + ~.12 fV; + G0: 1 - 0:2)W;U + 0:2Wk U ki + 0\u00ab(5)]dX i, (39.38a) (39.38b). between the coordinate frame and an orthonormal frame attached to it; stage I is a pure Lorentz transformation (boost) between the two orthonormal frames w a and w a. The 4-velocity of this boost is minus the 4-velocity of the matter, which has components ui = v\u00b7uo, ) in coord. frame; (39.39) uT= VjU O, U O= I + 1- v 2 + 0\u00ab(4),] 2 in w a frame. (39.40) Vj = vi[l + (I + y)U] Combining the boost, which has ordinary velocity f3j = -Uj, with the transformation (39.38), and then inverting, one obtains the result (exercise 39.12) \u00a739.11. PPN EQUATIONS OF MOTION 1087 W P = orthonormal comoving basis, dx a = PPN coordinate basis; { Transformation from rest frame of matter to PPN coordinate frame ~~ ~ AOo\u00ab(4), AOj = Vi[l + ; v 2 + (2 + Y)UJ - ~.11 Vf - ;.12 UJ + (0: 2 - ; O:l)WjU - 0:2Wk Ukj + 0\u00ab(5), (39.41) Ajo = Vj [I + ; v 2 + UJ+ 0\u00ab(5),. A'\" = ( I I - yU) 8jk + 2\" VjVk + 0\u00ab(4). This transformation, when applied to the stress-energy tensor (39.37) yields, in the PPN coordinate frame, Stress-energy tensor in coordinate frame TOO = Po(l + II + v 2 + 2U) + 0(Po(4), TOj = Po(l + II + v 2 + 2U)Vj + (;.;;,um + O(Po( 5), Tjk = (j,,(l - 2yU) + Po(l + II + V2 + 2U)VjVk I+ 2\" (V;!k';;'Vm + Vk(;.;;.Vm) + O(Po( 6)., (39.42a) (39.42b) (39.42c) Exercise 39.12. THE TRANSFORMATION BETWEEN COMOVING FRAME EXERCISE AND PPN FRAME Carry out the details of the derivation of the transformation matrix (39.41); and in the process calculate the correction of 0\u00ab(4) to AOo. \u00a739.11. PPN EQUATIONS OF MOTION The post-Newtonian corrections to the Newtonian equations of motion (39.15) and (39.16) are derived from the law of conservation of baryon number (Poua);a = 0, and from the law of conservation of local energy-momentum, TafJ;fJ = 0. The simplest of the equations of motion is the conservation of baryon number. Its exact expression is (Poua);a = (I/M)(V-gPoua).a = 0. Define a new quantity P*=Po( 1+ ; v 2 + 3yU) = Pouo M + O(Po( 4) (39.43) 1088 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION [see (39.39) for uo, and (39.32) for the metric].", " Then rest-mass conservation takes on the same form as at the Newtonian order (39.15a), except now it is more accurate: - Law of baryon conservation P*,t + (p*vj).j = 0 + errors of O(Pa,jf 5). (39.44) The next simplest equation of motion is TOa'a = O. Straightforward evaluation, using the metric ofequations (39.32) and the stres~-energytensor of equations (39.42), yields [Pa(l + II + v 2 + 2U)l,t + [Pa(l + II + v 2 + 2U)vj + t;,nvm].j + (3\"1 - 2)PaU. t + (3\"1 - 3)PaVkU,k = O(Pa,jf 5 ). (39.45) By subtracting equation (39.44) from this, and using the Newtonian equations of motion (39.15) and (39.16) to simplify several terms where the Newtonian approxi mation is adequate, one obtains Law of energy conservation (39.46) Notice that this is nothing but the first law of thermodynamics (local energy conser vation) with energy flow through the matter being neglected. (Neglecting energy flow was justified in \u00a739.5.) This first law of thermodynamics is actually a post-New tonian equation in the context of hydrodynamics, rather than a Newtonian equation, because II does not affect the hydrodynamic motion at Newtonian order (see \u00a739.7). The last of the equations of motion, T;a'a = 0, reduces to the post-Newtonian Euler equation, --- -.' Post-Newtonian Euler equation dv\u00b7 p*-t - p*U,j + [tjk(l + 3yU)],k - ( I - T tjk,k \"2 v2 + II ~ (7.1 1 +.12)Vj - ~O:IUWj] - ) t'fJfd + p* ;[(2\"1 + 2)Uv; - vjp*U,t + Vktkl,t + ~.12p*(Vj - ffj),t + ~ p*[(7.11 +.12)Vk + (0: 1 - 20: 3)Wk]~,J - p*[2,!, - ~ ~(j' - ~ 1J6f) - ~ 0:2WiWkUik + 0:2Wi(V'i - Wi)] (39.47) - p* U,j[ yv 2 - ~ 0: 1W\u00b7 v + ~ (0: 2 + 0: 3 - 0:1)W2 -,J (2fJ - 2)U + iyP/ p*] I + \"2(V;,k tk,nVm - t;,nVm,kVk) + \"2[Vm(t,njVk),k - Vj(tUVk),L] = o. I Partial derivatives are denoted by commas; d/dt is the time-derivative following the matter [equation (39.16)]. Equations (39.44), (39.46), and (39.47) are a complete set of equations' of motion at the post-Newtonian order. \u00a739.11. PPN EQUATIONS OF MOTION 1089 ------------------------------------------ ~ Exercise 39.13. EQUATIONS OF MOTION Carry out the details of the derivation of the equations of motion (39.44), (39.46), and (39.47). As part of the derivation, calculate the following values of the Christoffel symbols in the PPN coordinate frame: EXERCISES rooo = - U,t + O(U,i(3), rOik = yU,tSik + tL1lV(i,kl + tL1ZJt(i,kl + (tal - aZ)w<;U,kl rOOi = - U,i + O(U,lZ), + aZw;U;(i,kl + O(U,i(3). r ioo = -U,i + [({1 + y)UZ - 2'1' + ttl? + t7J<.v + t(a 1 - az - a3)wZU I + \"2 (a l - 2a3)w;f~ + \"2 aZw;wk U;k I ] 7 \"2 L1 1 Vi,t -.,J I \"2", " L1z u-j,t (39.48) + (a z - tal) wiU,t - aZw;Uii,t + O(U,l4), r iOk = yU,t 8ik - (tL11 + t L1z) V(i,kl - talW\u00a3jU,kl + O(U,i(3), rik! = -y(U,i8kl - 2U,(k81li) + O(U,i(Z). Here square brackets on tensor indices denote antisymmetrization, and round brackets denote symmetrization. As part of the tlerivation, it may be useful to prove and use the relations x(t, x) = - f Po(t, x')lx - x'i d 3x', Here X is the function originally defined in equation (39.29c). U1k,il = V(k,il' (39.49a) (39.49b) (39.49c) (39.50) Exercise 39.14. POST-NEWTONIAN APPROXIMATION TO GENERAL RELATIVITY Perform a post-Newtonian expansion of Einstein's field equations, thereby obtaining the values cited in Box 39.2 for the PPN parameters of general relativity. The calculations might best follow the approach of Chandrasekhar (1965a): Set ga/3 = 7Ja /3 + ha/3' and assume Choose the space and time coordinates so that the four \"gauge conditions\" hjk,k - t h.i = O\u00ab(4/Ro ) 1 hOk,k - t hkk\u2022O= 0\u00ab( 5/Ro ) with h = ha/37J a/3 = -hoo + hu (39.51 ) (39.52) are satisfied. (a) Show that the spatial gauge conditions are the post-Newtonian approximations to those (35.la) used in the study of weak gravitational waves, but that the temporal gauge condition is not. 1090 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION (b) Use these gauge conditions and the post-Newtonian limit in equations (8.24) and (8.47) to obtain for the Ricci tensor, accurate to linearized order. _ I ROj - -\"2 hOi,mm - I 2 '4 hoo,ol + 0\u00ab( /R O ). 5 (39.53b) (c) Combine these with the Newtonian form (39.13) of the stress-energy tensor, and with equation (39.27), to obtain the following metric coefficients, accurate to linearized order: h - 2 00 - U + koo + 0\u00ab( ), 6 I W O( 5) hOI - -\"2 V; -\"2 j + (, 7 - Lunknown post-Newtonian correctionJ (39.54) hjk = 2U 8jk + 0\u00ab(4). Here U, V;, and UJ are to be regarded as defined by equations (39.34a,b,c). By comparing these metric coefficients with equations (39.32), discover that y = I, Ll 1 = I, (39.55) for general relativity. (d) With this knowledge of the metric in linearized order, one can carry out the analysis of\u00a739.10 (using y = Ll1 = Ll 2 = I throughout), to obtain the post-Newtonian corrections to the stress-energy tensor [equation (39.42) with y = IJ. (e) Calculate, similarly, the post-Newtonian corrections to-the Ricci tensor component Roo' using gaP = 7Ja p + haP' using haP as given in equations (39.54), and using the gauge conditions (39.52). The answer should be Roo = (-U - tkoo - U2) + 4UU,mm + O\u00ab(6/Ro2). (39.56),mm (f) Evaluate the Einstein equation Roo = 8'IT(Too - ~gooT), accurate to post-Newtonian order, and solve it to obtain the post-Newtonian metric correction koo = -2U2 + 4'1', (39.57) where 'I' is given by equation (39.43d) with f31 = f32 = f33 = f34 = I. By comparing with equations (39.32c) and (39.34d), discover that for general relativity. (g) Knowing the full post-Newtonian metric, and the full post-Newtonian stress-energy tensor,", " one can carry out the calculations of \u00a739.11 (using y = f3 = f31 = f32 = f33 = f34 = Ll 1 = Ll 2 = I, ~ = 7J = 0) to obtain the post-Newtonian equations of motion for the matter [equations (39.44), (39.46), and (39.47)]. (39.58) \u00a739.13. SUMMARY OF PPN FORMALISM 1091 \u00a739.12. RELATION OF PPN COORDINATES TO SURROUNDING UNIVERSE One crucial issue remains to be clarified: What is the orientation of the PPN coordi nate system relative to the surrounding universe? More particularly: Does the PPN coordinate system rotate relative to the \"fixed stars on the sky;\" or is it \"rigidly attached\" to them, in some sense? In order to answer this question, imagine using the PPN formalism to analyze the solar system. Make no assumptions about the solar system's velocity through the PPN coordinate frame. Then, as one moves outward from the Sun, past the Earth's orbit, past Pluto's orbit, and on out toward interstellar space, one sees the PPN coordinate frame become more and more Lorentz in its global properties [ga,8 = 1Ja,8 + O(M0 /r)]. Thus, far from the solar system the PPN coordinates become a \"Lorentz frame moving through the galaxy.\" This means, of course, that the spatial axes of the PPN coordinate frame behave as though they were attached to gyroscopes far outside the solar system. Equivalently: The PPN coordinate system Fermi-Walker-transports its spatial axes through the spacetime geometry of the galaxy and universe. \u00a739.13. SUMMARY OF PPN FORMALISM, The PPN formalism, as constructed in this chapter, is summarized in Box 39.4. Much of the recent literature uses a different set of PPN parameters than are used in this book; for a translation from one parameter set to the other, see Box 39.5. Solar system's PPN coordinate frame is attached to a local Lorentz frame of Galaxy Exercise 39.15. MANY-BODY SYSTEM IN POST-NEWTONIAN LIMIT EXERCISE OF GENERAL RELATIVITY Consider, in the post-Newtonian limit of general relativity, a system made up of many gravitationally interacting bodies with separations large compared to their sizes (example: the solar system). Idealize each body to be spherically symmetric, to be free of internal motions, and to have isotropic internal stresses, tii< = 8jkP. Let the world line of the center of body A, in some chosen PPN coordinate frame, be xA(t); and let the (coordinate) velocity of the center of body A be (39.59a) The total mass-energy of body A as measured in its neighborhood (rest mass-energy plus internal energy plus self-gravitational energy) is given by MA = L(1 + n - i Uself)d(rest mass) + O(MA(4), A (39.59b) where Uself is the body's own Newtonian potential (no contributions from other bodies), and 'VA is the interior of the body. (continued on page 1094) 1092 39. OTHER THEORIES OF GRAVITY AND POST\u00b7NEWTONIAN APPROXIMATION Box 39.4 SUMMARY OF THE PPN FORMALISM I. Variables Po(x, f): baryon \"mass\" density (\u00a739.3), as measured in rest frame II(x, f): specific internal energy (dimensionless; \u00a739.3), as measured in rest frame fjf\u00abX' f): components of stress referred to orthonormal axes of rest frame vj(x, f): coordinate velocity of matter (i.e., rest frame) relative to PPN coordinates Iij(x, f), u-j(x, f), Ujk(x, f): gravitational potentials U(x, f), '!'(x, f), tl(x, f), <i>(x, f),.1 2't 1): parameters whose values distinguish one theory y, f3, f31' f3 4,.11, f3 2, f3 3, from another (see Box 39.2) w: velocity of PPN coordinate frame relative to \"universal rest frame\" [relevant only for theories with nonzero 0: 1, 0: 2, or 0: 3 ; see eq. (39.33)]. II. Equations governing evolution of these variables Po: conservation of rest mass, equation (39.44) II: first law of thermodynamics, equation (39.46) fjk", ": determined in terms of Po' II, and other material variables (chemical composi tion, strains, etc.) by equations of state and the usual theory of a stressed medium-which is not discussed here Vj: equations of motion (\"F = rna\"), equations (39.47) U, '!', tl, 6iJ, Iij, u-j, Ujk : source equations (39.34) III. Quantities to be calculated from these variables goo(x, f), gOj(x, f), gjk(x, f): these components of metric in PPN coordinate frame are expressed in terms of gravitational potentials by equations (39.32) UO(x, f), uj(x, f): these components of matter 4-velocity in PPN coordinate frame are given by equations (39.39) TOO(x, f), TOj(x, f), Tjk(x, f): these components of stress-energy tensor in PPN coordinate frame are given by equations (39.42) IV. Relation between rest frame, PPN coordinates, and the universe 1. Orthonormal basis we. of rest frame, where fjf< are defined, is related to P.PN coordinate basis dx a by equations (39.41) 2. Far from the sun, the PPN coordinates become asymptotically Lorentz; i.e., they form an inertial frame moving through the spacetime geometry of the galaxy and the universe. 3. Gives no account of expansion of universe or of cosmic gravitational waves impinging on solar system. \u00a739.13. SUMMARY OF PPN FORMALISM 1093 Box 39.5 PPN PARAMETERS USED IN LITERATURE: A TRANSLATOR'S GUIDE The original \"point-particle version\" of the PPN formalism [Nordtvedt (1968b)], and the original \"perfect-fluid version\" [Will (l971c)] used different sets of PPN param eters. This book has adopted Will's set, and has added the parameter 1) character izing effects of anisotropic stresses. More recently, Will and Nordtvedt have jointly adopted a revised set of parameters, described below. A. Translation Table Will-Nordtvedt revised parameters' Revised parameters in notation of this bookb Revised parameters in notation of Nordtvedt (1968b)C y f3 a1 \"'z a3 ~ 1 ~z ~3 ~4 y f3 7 Ll1 + Liz - 4y - 4 Liz + ~ - 1, 4f31 - 2y - 2 - ~ ~ 2f3 + 2f3z - 3y - 1 f33 - 1 {34 - y y f3 8 LI - 4y - 4 a'\" - 1 4a\" - a'\" - 2y - 1 a'\" - X 2f3 - a' - 1 absent absent \"Revised parameters are used by Will and Nordtvedt (1972), Nordtvedt and Will (1972), Will (1972), and Ni (1973). bNotation of this book is used by Will (l97Ia,b,c,d). Ni (1972), and Thome, Ni, and Will (1971). 'Nordtvedt's original \"point-particle\" parameters were used by Nordtvedt (l968b, 1970, 197Ia,b). B. Significance of Revised Parameters ai' a 2, a 3 measure the extent of and nature of \"preferred-frame effects\"; see \u00a739.9. Any theory of gravity with at least one a nonzero is called a preferred-frame theory. ~ I' ~ 2' ~ 3' ~ 4' a 3 measure the extent of and nature of breakdowns in global conser vation laws. A theory of gravity possesses, at the post-Newtonian level, all 10 global conservation laws (4 for energy-momentum, 6 for angular momentum; see Chapters 19 and 20) if and only if ~I = ~2 = ~3 = ~4 = a 3 = O. See Will (1971d), Will and Nordtvedt (1972), Will (1972), for proofs and discussion. Any theory with ~ I = ~ 2 = ~3 = ~4 = a 3 = 0 is called a conservative theory. In general relativity and the Dicke-Brans-Jordan theory, all a's and ts vanish. Thus, general relativity and Dicke-Brans-Jordan are conservative theories with no preferred-frame effects. 1094 39. OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION (a) Show that, when written in the chosen PPN coordinate frame, this expression for MA becomes MA =", " \u00a3. Po(L+ Jl+ ~l'A2 + 3U- ~ Uself )d 3x + O(MA(4). (39.59c) A Use equations (39.43), (39.44), and (39.46) to show that MA is conserved as the bodies move about, dMA/dt = O. (b) Pick an event (t, x) outside all the bodies, and at time t denote (39.59d) Show that the general-relativistic, post-Newtonian metric (39.32) at the chosen event has the form (39.60a) (39.60b) (39.60c) [Hint: From the Newtonian virial theorem (39.2Ia), applied to body A by itself in its own rest frame, conclude that where the integral is performed in the PPN frame.] (c) Perform an infinitesimal coordinate transformation, (39.61) (39.62) to bring the metric (39.60) into the standard form originally devised by Einstein, Infeld, and Hoffman (1938), and by Eddington and Clark (1938):, (39.63a) (39.63b) (39.63c) \u00a739.13. SUMMARY OF PPN FORMALISM 1095 where X [equation (39.49a)] is given by (d) The equations of motion for the bodies can be obtained in either of two ways: by performing a volume integral of the Euler equation (39.48) over the interior of each body; or by invoking the general arguments of\u00a720.6. The latter way is the easier. Use it to conclude that any chosen body K moves along a geodesic of the metric obtained by omitting the terms A = K from the sums in (39.63). Show that the geodesic equation for body K reduces to d2~K =dVK = 2: rAK M~ [I _42: MB _ 2: MC(I _ r,AK'rgA ) 2 - 4vA'VK - ~ (VAr~:AKYJ + vi + 2vA B #K rBK C#A rCA A #K 2rCA rAK dt dt (39.64) Equations (39.63) and (39.64) are caned the Einstein-Infeld-Hoffman (\"EIH\") equations for the geometry and evolution otfa many-body system. They are used widely in analyses of planetary orbits in the solar system. For example, the Caltech Jet Propulsion Laboratory uses them, in modified form, to calculate ephemerides for high-precision tracking of planets and spacecraft. The above method of deriving the EIH equations and metric was devised by Fock (1959). For a similar calculation in the Dicke-Brans-Jordan theory, see Estabrook (1969); and for a derivation of the analogous many-body equations in the fun PPN formalism, see Will (1972). CHAPTER 40 SOLAR-SYSTEM EXPERIMENTS \u00a740.1. MANY EXPERIMENTS OPEN TO DISTINGUISH GENERAL RELATIVITY FROM PROPOSED METRIC THEORIES OF GRAVITY No audience will show up for a fight if in everyone's eyes the challenger has zero chance to win. No battle-hungry promoter desperately trying to finance the fight can afford to put into the ring against the champion-any but the best contender that he can find. Against Einstein's metric theory of gravity, the judgment of the day (as \u00a739.2 showed) leaves one no option except to put up another theory ofgravity that is also metric (or metric plus torsion). To put on a contest, then, is to design and perform an experiment that distinguishes general relativity from some not completely implausible metric theory of gravity. This chapter describes such experiments--some already performed; some to be performed in the future-and analyses their significance using the PPN formalism of Chapter 39. In most of the experiments to be described, one investigates the motion of the moon, planets, spacecraft, light rays, or gyroscopes through the spacetime geometry of the solar system. That spacetime geometry is very complicated. It includes the spherical fields of the sun and all the planets, nonspherical fields due ~o their quadrupolar and higher-order deformations, and fields due to their momentum and angular momentum. Moreover, the spacetime geometry results-or at least in the post-Newtonian formalism it is viewed as resulting-from a nonlinear superposition of all these fields. * * Of course, from the point of view of Einstein's full general relativity theory, all that legalistically counts is the one and only curved-spacetime", " geometry of the real physical world. All these \"individual fields\" are mere bookkeepers' discourse, and they are best abandoned (they cease to be useful) when one passes from the post-Newtonian limit to the full Einstein theory. This chapter analyzes experiments using PPN formalism Complexity of solar system's spacetime geometry \u00a740.1. EXTERNAL FIELD OF SUN 1097 Fortunately for this discussion, several of the most important experiments are free of almost all these complications. The effects they measure are associated entirely with the spherical part of the sun's gravitational field. A description of these experi ments will come first (\u00a7\u00a740.2-40.5), and then attention will tum to experiments that are more complex in principle. To discuss central-field experiments, one needs an expression for the external gravitational field of an idealized, isolated, static, spherical sun. In general relativity, such a gravitational field is described by the Schwarzschild line element, Idealization of geometry to that of isolated. static. spherical sun: (1) in Schwarzschild coordinates r But this line element is not what one wants, for two reasons: (1) it is \"too accurate\"; (2) it is written in the \"wrong\" coordinate system. Why too accurate? Because it is simple only when unperturbed and unmodified; whereas some modified theories show up new effects that are so complex they are tractable only in the post-Newtonian approximation. Why wrong coordinate system? Because physicists, astronomers, and other celestial mechanics have adopted the fairly standard convention of using \"isotropic coordinates\" rather than \"Schwarzs child coordinates\" when analyzing the solar system. Example: post-Newtonian expansions, including the PPN formalism of Chapter 39, almost always use isotropic coordinates. Another example: the relativistic ephemeris for the solar system, pre pared by the Caltech Jet Propulsion Laboratory [Ohandley et al. (1969); Anderson (1973)] and used extensively throughout the world, employs isotropic coordinates. Modify the Schwarzschild'line element, then. First transform to isotropic coordi nates (Exercise 31.7); then expand the metric coefficients in powers of M0 /r, to post-Newtonian accuracy:-Thereby obtain ds 2 = - [1 - 2 ~0 + 2 (~0fJ dt2 + [1 + 2 ~0J[dr2 + r2(d02 + sin20 dIf>2)] = - [1- 2 ~0 + 2 (~0) Jdt2 + [1 + 2 ~0J[dx2 + dyZ + dz2]. 2 (40.1) (2) in isotropic coordinates Here r, 0, If> are related to x, y, z in the usual manner: \u00a2 = tan-ley/x); (40.2) and r is the new, \"isotropic\" radial coordinate, not to be confused with the Schwarzschild r. (The reader who has not studied \u00a739.6 will discover in the next section why one keeps terms of order M02/r 2 in goo but not in gjk') Note: this post Newtonian expression for the metric is a special case of the result derived in exercise 19.3. If one calculates the gravitational field of the same source (the sun) in the same post-Newtonian approximation in other metric theories of gravity, one obtains a very similar result: 1098 40. SOLAR\u00b7 SYSTEM EXPERIMENTS (3) in PPN formalism (see exercise 40.1). Here y and f3 are two of the ten PPN parameters described in Box 39.2. Recall from that box that y measures \"the amount of space curvature produced by unit rest mass,\" while f3 measures \"the amount of nonlinearity in the superposition law for goo'\" These heuristic descriptions find their mathematical counterparts in the above form for the idealized metric surrounding a spherically symmetric center of attraction. By measuring the parameter y to high precision, one can distinguish between general relativity (y = 1) and the Dicke-Brans-Jordan theory [y = (1 + w)j(2 + w), where w is the \"Dicke coupling constant\"]; see Box 39.2. But general relativity and Dicke-Brans-Jordan predict the same value for f3 (f3 = 1). This identity does not mean that f3 is unworthy of measurement. A value f3 ::j: 1 is predicted by other theories [see Ni (1972)]; so measurements of f3 are useful in distinguishing such theories from general relativity. Actually, the above form (40.3) for the sun's metric is not fully general. In any theory with a preferred \"universal rest frame\" (e.g., Ni's theory; Box 39.1", "), there are additional terms in the metric due to motion of the sun relative to that preferred frame (exercise 40.1): (4) including preferred-frame effects ds 2 = (expression\u00b740.3) + (0: 2 + 0: 3 -0:2[~0 x;x k - ~~ (x;xk - M 0: 1) ----f w2 dt 2 +2 0: 2 - 2\" 0: 1 ---!- w;dx; dt (1 )M ~ r 2 8;k) ] w; dt(2 dx k + wk dt). (40.3') In these \"preferred-frame terms\" 10 = I;; = fpr 2 d 3x is the trace of the second moment of the sun's mass distribution; 0: 1 = 7.11 +.12 - 4y - 4, 0: 2 =.12 + K- 1, 0: 3 = 4f31 - 2y - 2 - K are combinations of PPN parameters; and w is the sun's velocity (= velocity of coordinate system) relative to the preferred frame. (Theories such as general relativity and Dicke-Brans-Jordan, which possess no preferred frame, have 0: 1 = 0:2 = 0:3 = 0, and therefore have no preferred-frame terms in the metric.) For ease of exposition, all equations and calculations in this chapter will ignore the preferred-frame terms; but the consequences of those terms will be discussed and references analyzing them will be cited. \u2022 \u00a740.2. USE OF LIGHT RAYS AND RADIO WAVES TO TEST GRAVITY 1099 Exercise 40.1. PPN METRIC FOR IDEALIZED SUN [Track 2] Show that for an isolated, static, spherical sun at rest at the origin of the PPN coordinate system, the PPN metric (39.32) reduces to expressions (40.3), (40.3'). As part of the reduction, show that the sun's total mass-energy is given by EXERCISES M0 = LRo o Po(l + 2/32 U + /33ll + 3/34P/po)4wr2dr. (40.4) [Warning: One must not look at this formula and immediately think: \"The contribution of the contribution of gravitational energy is J2/32PoU4wr 2dr, etc.\" rest mass is Jpo4wr2dr, Rather, in making any such interpretation one must remember that (I) spacetime is curved, so 4wr2dr is not proper volume as measured by physical meter sticks; also (2) virial theorems (exercise 39.6) and other integral theorems can be used to change the form of the integrand. For further discussion see exercises 40.9 and 40.10 below.) \u00a740.2. THE USE OF LIGHT RAYS AND RADIO WAVES TO TEST GRAVITY In the Newtonian limit, pla;J.etary and spacecraft orbits are strongly influenced by gravity; but light propagation and radio-wave propagation (at \"infinite\" velocity) are not influenced at all. For this reason, experimental studies of orbits are beset by the problem of separating the relativistic effects from much larger standard Newtonian effects. By contrast; experimental studies of light and radio-wave propa gation do not contend with ~my such overpowering Newtonian background. Not surprisingly, they are to date (1973) the clearest and most definitive of the solar system experiments. Light rays and raaio waves give\" clean\" tests of relativity Mathematically, the parameter that distinguishes a light ray from a planet is its high speed. In the geodesic equation, the magnitude of the velocity determines which metric coefficients can influence the motion. Consider, for example, a weak, static field gaf3 = 1Jaf3 + haf3, and a particle at (x, y, z) = (r, 0, 0) moving with velocity (v\"\" Vy' vz) = (0, v,O); see Figure 40.1. Here the effect of gravity on the trajectory of the particle can be characterized by the quantity curvature of trajectory in 3-dimensional,) = (radius of curvature)-l nearly Euclidean, space of trajectory ( d2x = dyZ = dy dr (1 - v2) v2 dr d (dr dX) dy dr =;; dr -;;; = (uY)Z I du'\" dr I d (U\"') dx a dx f3 dr dx a dx f3 r'\" - - - -= --r'\" ---dt - _ = -r\"'oov- 2 = ~ hOO,,,,v-2 + (hoy,,,, - hO\"\"y)v- 1 + (; hyy,,,,", " - h\",y,y). - 2r\"'Oyv- 1 af3 dr af3 dt r\"'yy I v2 - 1100 y Here the trajectory as described in isotropic coordinates has (radius of curvature)-l = hOM + (h\u00a2\u00a2/ r2 l.. 2v 2 2 40. SOLAR-SYSTEM EXPERIMENTS = _ ~,0(~ + r). r- v\u00b7 L---------F'---------i~x Figure 40.1. The bending of the trajectory of a test body at its point of closest approach to the sun, as a function of its 3-velocity. (See text for computa tion and discussion.) Reexpressed in spherical coordinates, in the terminology of the idealized solar line element (40.3), this formula says ( ~urvature of trajectory) = 1. h ill 3-space 2 OO,r v-2 + 1. (h 2 /r2),r \u00a2\u00a2 ;:::: -(M0 /r 2)(v- 2 + y) (40.5) for a particle at its point of closest approach to th_e ~un. (Compare with exercise 25.21.) Note that here y is a PPN parameter; it is not (1 - V~-l/2. - (post-Newtonian expansion parameter \u20ac2) Notice what happens as one boosts the velocity of the particle. For slow velocities [v 2 ;:::: M0 /R0 ], the Newtonian part of hoo dominates completely; and the tiny post-Newtonian corrections come equally from the \u20ac4 part of hoo, the \u20ac3 part of hOi' and the \u20ac2 part of hik. [This was the justification for expanding hoo to 0(\u20ac4), in the post Newtonian limit; see \u00a739.6.] But as v increases, the ordering of the terms changes. the bending of the trajectory has become almost In the high-v regime (v imperceptible because of the high forward momentum of the particle and the short time it receives transverse momentum from the sun. What bending is left is due to the \u20ac2 (Newtonian) part of hoo' and the \u20ac2 (post-Newtonian) part of hik. Nothing else can have a significant influence. Notice, moreover, that-even when onf allows for \"preferred-frame\" effects-these dominant terms, and hik to 0(\u20ac2) hOi to 0(\u20ac3), I ~ \u20ac2), Light rays are governed solely by Newtonian potential and PPN parameter y depend only on the Newtonian potential U= -iP and the PPN parameter y. This is a special case of a more general result: Aside from fractional corrections of \u20ac2 ~ 10-6, relativistic effects on light and radio-wave propagation are governed entirely by the Newtonian potential U and the PPN parameter y. These relativistic effects include the gravitational redshift (discussed in the last chapter; independent \u00a740.3. LIGHT DEFLECTION 1101 of y), the gravitational deflection oflight and radio waves (discussed below; depend ent on y), and the \"relativistic time-delay\" (discussed below; dependent on y). \u00a740.3. \"L1GHT\" DEFLECTION Consider a light or radio ray coming into a telescope on Earth from a distant star or quasar. Do not assume, as in the usual discussion (exercises 18.6 and 25.24), that the ray passes near the sun. The deflection by the sun's gravitational field will probably be measurable, in the middle or late 1970's, even when the ray passes far from the sun! [The calculation that follows is due to Ward (1970), but Shapiro (1967) first derived the answer.] Orient the PPN spherical coordinates of equation (40.3) so that the ray lies in the \"ptane\" 0 = 'fT/2. By symmetry, if it starts out in this plane far from the Earth, it must lie in this plane always. Let the incoming ray enter the solar system along the line <p = 0; and let the Earth be located at r = rE, <p = <PE when the ray reaches it. (See Figure 40.2.) One wishes to calculate the angle a between the incoming light ray and the center of the sun, ~s measured in the orthonormal frame (e;, e:;,) of an observer on Earth. If the sun had zero mass (flat, Euclidean space), a would be 'fT <PE + Sa. The deflection angle Sa is the true objective of the calculation. <P-E (see Figure 40.2). However, the sun produces a deflection: a = 'fT - In the calculation, ignore the", " Earth's orbital and rotational motions. They lead to aberration, for which correction can be made by the usual formula of special relativity (Lorentz transformation in the neighborhood of the telescope.) Also ignore deflection of the light ray due to the Earth's gravitational field (deflection angle - 2ME/ RE -- 0\".0003), which might be detectable in the late 1970's. Light deflection: (1) derivation.......II----b---~ t ~ \u00b01-;;; III \"'2,.Ii -e- Figure 40.2. Coordinates used in the text for calculating the deflec tion of light. Notice that in this diagram \u00a2 increases in the clockwise direction. 1102 40. SOLAR-SYSTEM EXPERIMENTS As the first step in calculating the deflection angle, determine the trajectory of the ray in the r, <1>-plane. This can be calculated either using the geodesic equation, or using the eikonal method of geometric optics (Hamilton-Jacobi method; \u00a722.5 and Box 25.4). The result of such a calculation (exercise 40.2) is an equation con necting r with <1>; thus, b. -; = Sin <1> + (1 + y)M b 0 (1 - cos <1\u00bb. (40.6) Notice that b has a simple geometric interpretation: far from the sun, the ray trajectory is <1> = b/r + O(M0 b/r2). Consequently, b is the impact parameter in the usual sense of classical scattering theory (see Figure 40.2). The ray makes its closest approach to the sun (assuming it is not intercepted by the Earth first) at the PPN coordinate radius = b[I _(1 + Y)M0 ] - b _. b r m1n (40.7) Thus, b can also be thought of as the radius of the ray's \"perihelion.\" Notice that the ray returns to r = 00, not at an angle <1> = 'fT, but rather at <1>(r = 00) = 'fT + 2(1 + y)Mdb. (40.8a) Thus, the total deflection angle is (angle of total deflection) = 2(1 + y)M0 /b I. = \"2 (1 + y)l \".75 for a ray that just grazes the sun. (40.8b) But this is not the quantity of primary interest. Rather, one seeks the position of the star as seen by an astronomer on Earth. The angle a = 'fT <1>E + Sa between the sun and the star as measured by the astronomer is given by (see Figure 40.2) tan('fT - <1>E + Sa) = -tan <1>E + Sa/cos2<1>E = ~ = [(1 + yM0 /r)rd<1>/dA] = [rd<1>] (1 + yM0 /r) dr/dA E dr E ur (40.9) = _ [(b/r) d<1>], d(b/r) E where u f3 = dx f3 IdA are the components of a tangent to the ray at the Earth. By inserting into this equation expression (40.6) for the trajectory of the ray, one obtains sin <1>E + [(1 + y)M0 /b](l - cos <1>E) tan <1>E - - - - = ---'-\"=--'-;\";~~'--\":::\"'-\"\":\";'-:-:--:-----''''::;''' cos <1>E + [(1 + y)M0 /b] sin <1>E Sa COS2<1>E = tan <1>E - [(1 + y)M0 /b](1 - cos <1>E)/COS2<1>E' (40.10) \u00a740A. TIME-DELAY IN RADAR PROPAGATION 1103 Thus, the deflection angle measured at the Earth is Sa = (I + y)M0 (I + cos a) = (I + y)~ (1 + cos a )1/2. b rE 1 - cos a (40.11 ) (2) formula for deflection angle It ranges from zero when the ray comes in opposite to the sun's direction (a = 'IT), through the value (40.12) when the ray comes in perpendicular to the Earth-Sun line (a = 'IT/2), to the \"classi cal value\" of i(l + y) X 1\".75 when the ray comes in grazing the sun's limb. All experiments to date (1972) have examined", " the case of grazing passage. The experiinental results are stated and discussed in Box 40.1. They show that the PPN parameter y has its general relativistic value of 1 to within an uncertainty of about 20 percent. By the middle or late 1970's, measurements of the deflection of radio waves from quasars should determine y to much better than 1 percent. Also, by that time radio astronomers may be making progress toward setting up high-precision coordinates on the sky using very long baseline interferometry. If so, they will have to use equation (40.11) to compensate for the \"warping\" of the coordinates caused by the sun's deflection of radio waves in all regions of the sky, not just near the solar limb. Experimental measurements of deflection Exercise 40.2. TRAJECTORY OF LIGHT RAY IN SUN'S GRAVITATIONAL FIELD EXERCISE Derive equation (40.6) for the path of a light ray in isotropic coordinates (40.3) in the sun's \"equatorial plane.\" Use one or more of three alternative approaches: (1) direct integration of the geodesic equation (the hardest approach!); (2) computation based on the three integrals of the motion k\u00b7k = 0, k\u00b7 (a/at) = ko, k =d/d>.. = tangent vector to geodesic (see \u00a7\u00a725.2 and 25.3); (3) computation based on the Hamilton-Jacobi method (Box 25.4), which for photons (zero rest mass) reduces to the \"eikonal method\" of geometric optics (see \u00a722.5).. \u00a740A. TIME-DELAY IN RADAR PROPAGATION Another effect of spacetime curvature on electromagnetic waves is a relativistic delay in the round-trip travel time for radar signals. It was first pointed out by Shapiro (1964); see also Muhleman and Reichley (1964, 1965). (continued on page l106j Radar time delay: Box 40.1 DEFLECTION OF LIGHT AND RADIO WAVES BY SUN: EXPERIMENTAL RESULTS Eclipse Measurements Until 1968 every experiment measured the deflection of star light during total eclipse of the sun. The measurements were beset by difficulties such as poor weather, optical distortions due to temperature changes, and the strange propensity of eclipses to attain maximum time of totality in jungles, in the middles of oceans, in deserts, and in arctic tundras. Lists of all the results and references are given by Bertotti, Brill and Krotkov (1962), and by Kliiber (1960). Dicke (l964b) sum marizes the results as follows: \"The analyses [of the experimental data] scatter from a deflection at the limb of the sun of 1.43 seconds of arc to 2.7 seconds [compared to a general relativistic value of 1.75 seconds]. The scatter would not be too bad if one could be lieve that the technique was free of systematic errors. It ap pears that one must consider this observation uncertain to at least 10 percent, and perhaps as much as 20 percent.\" This result corresponds to an uncertainty in 'I of 20 to 40 percent. Measurements on the Deflection of Radio Waves Each October 8 the sun, as seen from the Earth, passes in front of the quasar 3C279. By monitoring the angular separa tion between 3C279 and a nearby quasar 3C273, radio ~s\u00ad tronomers can measure the deflection by the sun of the 3C279 radio waves. The monitoring uses radio interferometers. [See references cited in table for discussion of the technique.] Technology of the early 1970's should permit measurements to a precision 0.001 seconds of arc or better, if the two ends of the interferometer are separated by several thousand kilo meters (\"transcontinental\" or \"transworld\" baseline). But as of 1971 the only succ;;sful experiments were less ambitious: they used baselines of less than 10 kilometers. A summary of these pre-1971, short-baseline results is shown in the table..,., J 1\u00b0 2\u00b0..... o 0':5 1':0 1\"\"1,,,,1 Scale of light deflections Observed light deflections (mean of two instruments) of the 15 best measured stars within 2\u00b0.5 of the sun's center in the total solar eclipse of September 21, 1922 at WalIal, Western Australia, as determined by Campbell and Trumpler (1928). The arrows represent in size and direction the observed light deflections relative to the reference stars (5 \u00b0 to 10\u00b0 from the sun's center). (See Box 1.6 for Einstein's description of the deflection in terms of the curvature of geometry near the sun). - The 9O-foot (", "background) and \\30-foot (foreground) radio interferometer system at CaItech's Owens VaHey Radio Observatory. These were used by Seielstadt, Sramek, and Weiler (1970) in their pioneering measurement of the deflection of quasar radio waves by the sun. During the experiment the two antennas were separated by 1.07 kilometers. (Photo by Alan Moffel.) Experimenters and reference Seielstadt, Sramek, Weiler (1970) Muhleman, Ekers, Fomalont (1970) Sramek (1971) Number of telescopes and separations 2, 1.07 km 2, 21.56 km 3, 0.80 km, 1.90 km, 2.70 km Wave lengths 3.1 cm 12.5 cm I I.l cm, 3.7 cm Hill (1971) 3, 0.66 km, 1.41 km Il.l cm, 6.0 cm Experimental results& 1 -(1 + y) = 2 ( Observed) deflection ( Einstein ) prediction 1.01 1.04 0.90 1.07 Formal standard error \u00b1O.l2 \u00b10.05 \u00b10.05 Onesigma error \u00b1O.l2 +0.15 -0.10 \u00b10.05 \u00b10.I7 \u00b1O.l7 Dates of observation Observatory Sept. 30-0cl. 15 Owens VaHey 1969 (CaItech) Oct. 2-0ct. 10 1969 Goldstone (Caltech-JPL) Oct. 2-0ct. 12 1970 National Radio Astronomy Observatory (USA) Sept. 30-0ct. 15 MuHard Radio 1970 Astronomy Observatory (Cambridge Univ.) \"Here (observed detlection)/(Einstein prediction) is the number ~(I + y) obtained by tilling the observalional data to the PPN prediction (40.11). [For these experiments the ray passes near Ihe solar limb; so (40.11) reduces 108,. = ~(I + y)(Me/b).] The \"formal standard error\" is obtained from the data by standard statistical techniques. However, it is not usually a good measure of Ihe ceflainty of the resull, because it fails to take account of systemalic errors. The quoted \"one-sigma error\" is the experimenters' best estimate of the combined statistical and syslematic uncertain lies. The experimeDlers estimate a probability of 68 percenl that the true result is wilhin \"117\" of their measured value; a probability of 95 percent that it is within \"217\"; etc. '- [1105] ~J~ ~ 1106 40. SOLAR-SYSTEM EXPERIMENTS y __ _ i - _Beam path used in computation (y = b = consl.) Actual beam path Tran!;miller - - - - --t--..,::.:::=::=~- - - I- - - aT Reflector -----=--=-::.~-='=\"=.-=--\u00b71 a R (1) foundations for calculation; Fermat's principle (2) details of calculation -----------:--f~--II-----------------~x Figure 40.3. Diagram, in the PPN coordinate system, for the calculation of the relativistic time delay. Sun Let a radar transmitter on Earth send a radio wave out to a reflector elsewhere in the solar system, and let the reflector return the wave to Earth. Calculate the round-trip travel time, as measured by a clock on Earth. For simplicity of calculation, idealize both Earth and reflector as nonrotating and as at rest in the static, spherical gravitational field of the Sun. At the end of the calculation, the effects of rotation and motion will be discussed separately. Also ignore time dilation of the transmitter's clock due to the Earth's gravitational field; it is easily corrected for, and it is so small that it will not come into play in these radar experiments before the middle or late 1970's. The gravitational effects of the other planets on the radio waves are too small to be discernible in the foreseeable future, unless the beam grazes the limb of one of them. However, the effects of dispersion in the solar wind and corona are discernible and must be corrected for. These corrections will not be discussed here, since they are free of any general-relativistic influence. The calculation of the round-trip travel time can be simplified by using a general relativistic version of Fermat's principle: In any static field (goi = 0, gap,o = 0) consider all null curves between two points in space, xi = ai and xi = bi. Each such null curve, xi(t), requires a particular coordinate time interval Lit to get from ai to bi. The curves of extremal Lit", " are the null geodesics of spacetime. The proof of this theorem is outlined in exercise 40.3. Because of Fermat's principle, the lapse of coordinate time between transmission of the radar beam and reflection at the reflector, tTR, is the same for a straigft path in the PPN coordinates as for the slightly curved path which the beam actually (angle of deflec follows. tion? ~ 10-12, which is far from discernable.) Hence, in the computation one can ignore the gravitational bending of the beam. (The two differ by a fractional amount LltTR/tTR - Adopt Cartesian PPN coordinates with the sun at the origin; the transmitter, sun, and reflector in the z = 0 \"plane\"; and the transmitter-reflector line along the x direction (see Figure 40.3). The transmitter is at (x,y) = (-a1\" b) in the PPN coordi nates, and the reflector is at (x,y) = (aR, b). Recall that for a null ray ds 2 = 0 = \u00a740.4. TIME-DELAY IN RADAR PROPAGATION 1107 goo dt2 - gzz dx2. It follows that the lapse of coordinate time between transmission and reflection is The lapse of coordinate time in round-trip travel has twice this magnitude. The lapse of proper time measured by an Earth-based clock is (40.13) LIT = Igool~:th2tTR' LIT = 2(aR + aT) (1 _ M0 ) V4+ b2 2(1 + +y 0 n )M 1 [caR + y\"\"'aR\"'\"\"2-+-\"b'\"\"'2)(aT + V4 + b2)] b2 (3) formula for delay (40.14) (4) comparison with experiment This is the lapse oftime on an Earth-based clock, aside from corrections for the orbital and rotational motion of the Earth, for the orbital motion of the reflector, for dispersion of radiation traversing the solar wind and corona, and for time dilation in the Earth's gravitational field. Any reader is reasonable who objects to the form (40.14) in which the time-delay has been written. The quantities aR, a1\" and b are coordinate positions in the PPN coordinate system, rather than numbers the astronomer can measure directly. They differ from coordinate ~~itigns in other, equally good coordinate systems by amounts of the order ofM0 1.5 km. The objection is not mathematical in its origin. The quantities aR, a1\" and b are perfectly well-defined [with post-post-Newtonian uncertainties of order b(M0 /b)Z ~ 10-6 kmj, because the PPN coordinate system is perfectly well-defined. But they are not quantities which the experimenter can measure directly, with precision anywhere near that required to test the relativistic terms in the time-delay formula (40.14). In practice, fortunately, the experimenter does not need to measure aR, a1\" or b with high precision. Instead, he checks the time-delay formula by measuring the changes in LIT as the Earth and reflector move in their orbits about the Sun; Le., he measures LIT as a function of Earth-based time T. Notice that when the beam is passing near the sun (b ~ aR, b ~ aT; but db/dT ~ daR/dT and db/dT ~ daT/dT because the Earth's and reflector's orbits are nearly circular), the change of b in the In term of (40.14) dominates all other relativistic corrections to the Newtonian delay; consequently (using db/dT - 10 km/sec for typical experiments) (Constant Newtonian) _ dLiT - ~ ~rt 4 I ) M0 db --( +y - b ~ (40.15) _ 4( I + )( 1.5 km) ( 10 km) _ 30 p'sec. y 106 km sec day 1108 40. SOLAR-SYSTEM EXPERIMENTS Such differential shifts in round-trip travel time-which rise as the Earth-reflector line moves toward the Sun and falls as it moves away-are readily observable. In practice, in order to obtain precisions better than about 20 percent in the determination of the parameter y by time-delay measurements, one must carefully collecCand analyze data for a large fraction of a year-from a time when the beam 108 km), to the time of superior conjunction (b - R0 is far from the sun (b - aT 106 km), and on around to a time of distant beam again. Such a long \"arc\" of data is needed to determine the reflector", "04 'Here (observed delay)/(Einstein prediction) is lhe value oq(l + y) obtained by fining the observational data,.:IT(T), to a more sophisticated version of the PPj\\; prediction (40.14). This more sophisticated version includes the gravilational influences of all the planets on the orbits of reflector and Earth: also lhe effect of the moon on the Earth's orbil and the effect of the Earth's rotation on the lravel lime; also, to as good an extent as possible. the delay due to dispersion in the solar corona and wind. \"Formal standard error\" and \"one-sigma error\" are defined in the lable in Box 40.1. Perihelion shift for geodesic orbits around spherical sun, ignoring preferred-frame effects 1110 40. SOLAR-SYSTEM EXPERIMENTS \u00a740.5. PERIHELION SHIFT AND PERIODIC PERTURBATIONS IN GEODESIC ORBITS The light-deflection and time-delay experiments both measured y. To measure other PPN parameters, one must examine the effects of gravity on slowly moving bodies; this was the message of \u00a740.2. Begin with the simplest of cases: the geodesic orbit of a test body in the sun's spherical gravitational field, ignoring all gravitational effects of the planets, of solar oblateness, and of motion relative to any preferred frame. The PPN metric then has the form (40.3): (40.3) Orient the coordinates so the test body moves in the equatorial \"plane\" 0 = 'fT/2; and calculate the shape r(<p) of its nearly Keplerian, nearly elliptical geodesic orbit. The result, accurate to order M0 /r beyond Newtonian theory, is r= (1 - e2)a 1 + e cos [(1 - 8<po/2'fT)<P], where a and e are constants of integration, and 8<po is defined by _ (2 - 8 <Po - f3 + 2y) 3 6'fTM0 a(1 _ e2) \u00b7\u00b7 6'fTM0 ---,,-:,- in general relativity. a(1 - e2) (40.17) (40.18) (For derivation, see exercise 40.4.) Notice that, if 8<po were zero-as it is in the Newtonian limit-then the orbit (40.18) would be an ellipse with semimajor axis a and eccentricity e (see Box 25.4). The constant 8<po merely makes the ellipse precess: for r to go through a complete circuit, from perihelion to aphelion to perihelion again, (1 - 8<po/2'fT)<p must change by 2'fT; so <p must change by 2'fT + 8<po. Thus, the perihelion shifts forward by an angle 8<po with each circuit around the ellipse. Relative to what does the perihelion shift? (I) Relative to the PPN cooidinate system; (2) relative to inertial frames at the outskirts of the solar system (since the PPN coordinates are tied to those frames; see \u00a739.12); (3) relative to a frame determined by the \"fixed stars in the sky\" (since the inertial frames at the outskirts of the solar system, inertial frames elsewhere in our galaxy, and inertial frames in our cluster of galaxies should not rotate significantly relative to each other); (4) relative to the perihelia of (other) planets, which themselves are shifting at calculable rates that decrease as one moves outward in the solar system from Mercury to Venus to Earth to.... Periodic perturbations in geodesic orbits Comparison of theory with planetary orbits \u00a740.5. PERIHELION SHIFT IN GEODESIC ORBITS 1111 The perihelion shift is not the only rela tivistic effect contained in the orbital motion for a test body. There are other effects, but they are all periodic rather than cumula tive with time; so, with the limited technology of the pre-space-age era, it was impossible to detect them. But the technology of the 1970's is bringing them within reach. Moreover, many space-age experiments are necessarily of short duration (~ one orbit)-particularly those involving spacecraft and transponders landed on planets. For these, the periodic perturbations in an orbit are of almost as much experimental valu_e-1l~ the Cu_ml!l\u00a3itive perihelion shift. The periodic effects are not obvious in the PPN orbital equation (40.17); it looks like the simplest of precessing ellips", "es. But the quantities the observer measures directly are not a, e, and 8<po. Instead, he measures the time evolution of round-trip radar travel times,.1r(T), and of angular positions on the sky [Oo(T), <Po(T )]. To compute these quantities is perfectly straightforward in principle, but in practice is a very complex task. The calculations predi! relativistic effects periodic with the frequency of the orbit and all its har monics. The amplitudes of these effects, for the lower harmonics, must obviously be of the order of M0 10 p.sec - 0\".01 arc on the sky. (The distance M0 = 1.48 km is the characteristic length for all relativistic effects in the sun's spherical field!) I km - The most favorable orbits for experimental tests of the perihelion shift and of periodic effects are those that go nearest the sun and have the highest eccentricity [see equation (40.18)]-the orbits of Mercury, Venus, Earth, Mars, and the asteroid Icarus. But how does one know that these orbits are geodesics? After all, planets are not \"test bodies\"; they themselves produce nonnegligible curvature in spacetime. It turns out (see \u00a740.9 for full discussion) that there should exist tiny deviations from geodesic motion, but they are too small to compete with the perihelion shift or with the periodic effects disciiSs\"e<fabove, at least for these five bodies. Extensive astronomical observations of planetary orbits, dating back to the mid nineteenth century and aided by radar since 1966, have yielded accurate values for planetary perihelion shifts (accurate to \u00b10.4 seconds of arc per century for Mercury). From the data, which are summarized and discussed in Box 40.3, one obtains the value I \"3 (2 - {+0.01 f3 + 2y) = 1.00 -0.10 (40.19a) for the ratio ofobserved relativistic shift to general relativistic prediction. Combining this result with the radar-delay value for y (40.16), one obtains a value (40.19b) Experimental result for f3 for the PPN parameter f3. (Recall: f3 measures the \"amount of nonlinearity in the superposition law for goo.\") The periodic effects in the planetary orbits have not yet (1973) been studied experimentally. The above discussion and Box 40.3 have ignored the motion of the solar system relative to the preferred frame (if one exists); i.e., they have ignored the terms.(40.3') 1112 40. SOLAR\u00b7SYSTEM EXPERIMENTS Box 40.3 PERIHELION SHIFTS; EXPERIMENTAL RESULTS Relativistic corrections to Newtonian theory are not the only cause of shift in the perihelion of a planetary orbit. Any departure of the Newtonian gravitational field from its idealized, spherical, inverse-square-Iaw form also produces a shift. Such nonsphericities and resulting shifts are brought about by (1) the gravitational pulls of other planets, and (2) deformation of the sun (\"solar oblateness\"; \"quadrupole moment\"). In addition, when the primary data are optical posi tions of planets on the sky (right ascension and declination as functions of time), there is an ap parent perihelion shift caused by the precession of the Earth's axis (\"general precession\"; observer not on a \"stable platform\"; see exercise 16.4). The perihelion shifts due to a general precession and to the gravitational pulls of other planets can be calculated with high precision. But in 1973 there is no fully reliable way to determine the solar quadrupole moment. It is conventional to quantify the sun's quadrupole moment by a dimensionless parameter J2, which appears in the following ex pression for the Newtonian potential, U = ~0 [I _ J2 ~~2 Ccos~ - I )l If the sun were rotating near breakup velocity, J2 would be near 1. Very careful measurements of the optical shape of the sun [Dicke and Golden berg (1967)] show a flattening, which suggests J2 may be near 3 X 10-5\u2022 The total perihelion shift produced by relativity is (see exercise plus solar quadrupole moment 40.5) The Haystack radar antenna, which Irwin Shapiro and his group have used to collect extensive data on the systematics of the inner part of the solar system. Those data are rapidly becoming the most important source of information about perihelion shifts. (Picture courtesy of Lincoln Laboratories, MIT.) Note that relativistic and quadrupole shifts have different dependences on the semimajor axis a and ecentricity e of", " the orbit. This difference in depen dence allows one to obtain values for both tte quadrupole moment parameter J2, and the PPN f3 + 2y) by combining measure parameter i(2 ments of 8<p for more than one planet. The experimental results, as of 1972, are as follows. \u00a740.5. PERIHELION SHIFT IN GEODESIC ORBITS 1113 I. Data for Mercury from optical studies [Clemence (1943, 1947)]* (general relativity with no solar oblateness predicts 43.03\"/century) Quantity (a) Total observed shift per century (b) Contribution to shift caused by observer not being in an inertial frame far from the sun (\"general precession\" as evaluated in 1947) (c) Shift per century produced by Newtonian gravitation of other planets (d) Residual shift per century to be attributed to general relativity plus solar oblateness (e) Residual shift if one uses the 1973 value for the \"general precession\" (I) Corresponding value of Ap (see above) If. 1970 Results of Shapiro (1970, 197Ia,b), Shapiro et al. (1972) Value 5599\".74 \u00b1 0\".41 5025\".645 \u00b1 0\".50 531\".54 \u00b1 0\".68 42\".56 \u00b1 0\".94 41\".4 \u00b1 0\".90 Ap = 0.96 \u00b1 0.02 (a) Values of Ap obtained by reanalyzing all the world's collection of optical data, and combining it with radar data (b) Value of /2 obtained by comparing the observed shifts for Mercury and Mars ( (Ap)MercUry : (Ap)Mars /2 :S 3 X 10-5 1.00 ~ O.ol 1.07 - 0.10 III. Theoretical implications of Shapiro's results (a) Value of (2 - f3 + 2y)/3 (b) Value of f3 obtained by combining with y from time delay experiments [equation (40.16)] +0.01 I 00 {. -0.10 1.0 { ~g:i \u2022 Clemence (1947) notes, \"The observations cannot be made in a Newtonian frame of reference. They are referred to the moving equinox, that is, they are affected by the precession of the equinoxes, and the determination of the precessional motion is one of the most difficult problems of observational astronomy, if not the most difficult. In the light of all these hazards, it is not surprising that a difference of opinion could exist regarding the closeness of agreement between the observed and theoretical motions:'. in the sun's metric. When one takes account of these terms, one finds an additional contribution to the perihelion shift, given for small eccentricities e ~ I by Perihelion shift due to preferred-frame forces (40.20) [see Nordtvedt and Will (1972)]. Here M 0, ilo and W0 are the sun's mass, self-gravi tational energy, and rotational angular velocity; lV is the sun's velocity relative to the preferred frame; a and e are the semimajor axis and eccentricity of the orbit; P is the unit vector pointing from the sun to the perihelion; and Q is a unit vector orthogonal to P and lying in the orbital plane. Comparison with observations for w....\u00b7 \" Experimental result for 0:3 The future of orbital experiments EXERCISES 1114 40. SOLAR-SYSTEM EXPERIMENTS Mercury-and combination with limits on a 1 and a z discussed below [equations (40.46b) and (40.48)]-yields the limit w\u00b7 Q I ~ 2 X 10-5. 3200 km/sec Ia. (40.21a) Since the velocity of the sun around the Galaxy is -200 km/sec, and the peculiar motion of the Galaxy relative to other nearby galaxies is -200 km/sec, a value W - 200 km/sec is reasonable. Moreover, there is no reason to believe that wand Q are orthogonal, so one is fairly safe in concluding la31= 1413 1 - 2y - 2 - KI ~ 2 X 10-5 (40.21 b) This is a stringent limit on theories that possess universal rest frames. For example, with great certainty it rules out a theory devised by Coleman (1971), which has 13 = y = 1, but a 3 = -4; see Ni(1972). Looking toward the future, one cannot expect data on orbits of spacecraft to give decisive tests of general relativity, despite the high precision (-10 meters in 1972) with which spacecraft can be tracked. Spacecraft are buffeted by the solar wind. They respond to fluctuations in this wind and in the pressure of solar radiation, and respond also", " to \"outgassing\" from leaky jets. Unless one can develop excellent \"drag-free\" or \"conscience-guided\" spacecraft, one must therefore continue to rely on planets as the source of data on geodesics. However, planetary data themselves can be greatly improved in the future by placing radar transponders on the surfaces of planets or in orbit about them, by improvements in radar technology, and by the continued accumulation of more and more observations. Exercise 40.4. DERIVATION OF PERIHELION SHIFT IN PPN FORMALISM [See exercise 25.16 for a derivation in general relativity, accurate when gravity is strong (2M/r as large as!) but the orbital eccentricity is small. The present exercise applies to any \"metric theory\" and to any eccentricity, but it assumes gravity is weak (2M/r -< 1) and ignores motion relative to any universal rest frame.] Derive equation (40.17) for the shape of any bound orbit of a test particle moving in the equatorial plane of the PPN gravitational field (40.3). Keep only \"first-order\" corrections beyond Newtonian theory (first order in powers of M0 /r). [Sketch ofsolution using Hamilton-Jacobi theory (Box 25.4): (1) Hamilton-Jacobi equation, referred to a test body of unit mass, is (2) Solution to Hamilton-Jacobi equation is -.. 1\".< S = -1:.t + L<p \u00b1 rz [ - -;z 1 - fr{ 2M 2 r~ (2 - - -(1 - \u00a32) + -r-' [1 - 2M0 ]}1/2 dr, f3 + 2y) (1 + y)(l - \u00a32)] - (40.22) \u00a740.5. PERIHELION SHIFT IN GEODESIC ORBITS 1115 where post-post-Newtonian corrections have been discarded. In discarding post-post-New tonian corrections, recall that \u00a3 is the conserved energy per unit rest mass and r is the angular momentum per unit rest mass (see Box 25.4). Consequently one has the order-of-magnitude relations I - \u00a32 - (velocity of test body)2 - M0 /r and (3) The shape of the orbit is determined by the \"condition of constructive interference,\" as/or = 0: { I - \u00a32 L2 2M. $= \u00b1f --_-+~[I-(l + y)(l- \u00a32)] L2r 2M 2 I - ~(2 L2 ]}-112 f3 + 2y) I r2 - [ d(l/r). (4) This integral is readily evaluated in terms of trigonometric functions. For a bound orbit (\u00a3 < I), the integral is $= + - cos ( I 8$0 ) 2'1T where -1 [ (1 - e2)a er I J - e (40.23) (5) Straightforward manipulations bring this result into the form of equations (40.17) and (40.18).] Exercise 40.5. PERIHELION SHIFT FOR OBLATE SUN (a) The Newtonian potential for an oblate sun has the form u = ~0 ( I _ J 2 ~~2 3 cos~ - I ), (40.24) where J2 is the \"quadrupole-moment parameter.\" One knows that J2. Show that if an oblate sun is at rest at the origin of the PPN coordinate system, the metric of the surrounding spacetime [equations (39.32)] can be put into the form ~ 3 X 10-5 ds2 = _ [I _2~0 _ 2J2(M~~02)C cos~ - I) + 2f3(~0 YJdt2 + [ I + 2y ~0 J[dr2 + r2(d8 2 + sin28 d\u00a22)] + corrections of post-post-Newtonian magnitude. (40.25) 1116 40. SOLAR\u00b7SYSTEM EXPERIMENTS (b) Let a test particle move in a bound orbit in the equatorial plane. Use Hamilton-Jacobi theory to show that its orbit is a precessing ellipse [equation (40.17)] with a precession per orbit given by (40.26) For the significance of this result, see Box 40.3. The rest of this chapter is Track 2. No earlier Track-2 material preparation for it. but the following will be helpful: (1) Chapter 6 (accelerated is needed as observers) ; (2) \u00a7 17.6 (no prior geometry); and (3) Chapters 38 and 39 (tests", " of foundations; other theories; PPN formalism). It is not needed as prepara tion for any later chapter. 3-body effects in lunar orbit: (1) theory (2) prospects for measurement \u00a740.6. THREE-BODY EFFECTS IN THE LUNAR ORBIT The relativistic effects discussed thus far all involve the spherical part of the sun's external gravitational field, and thus they can probe only the PPN parameters f3 and y plus the \"preferred-frame\" parameters aI' a 2, and a 3. Attempts to measure other PPN parameters can focus on three-body interactions (discussed here), the dragging of inertial frames by a rotating body (\u00a740.7), anomalies in the locally measured gravitational constant (\u00a740.8), and deviations of planetary and lunar orbits from geodesics (\u00a740.9). There is no better place to study three-body interactions than the Earth-moon orbit. The pulls of the Earth, the moon, and the sun all contribute. Perturbations in the motion of Earth and moon about their common center of gravity can be measured with high precision using laser ranging (earth~m6on separation measured to -10 cm in early 1970's) and using a radio beacon on the moon's surface (angular position on sky potentially measurable to better than 0\".001 of arc). Over and above any Newtonian three-body interactions, the Earth and the sun, acting together in a nonlinear manner, should produce relativistic perturbations in the lunar orbit that are barely within the range of this technology. These effects depend on the familiar parameters y (measuring space curvature) and f3 [measuring amount of nonlinear superposition, (UEarth + Usun)2, in goo]. In addition, they depend on f3 2, which regulates the extent to which the sun's potential, Usun' acting inside the Earth, affects the strength of the Earth's gravitational pull, causing it to vary as the Earth moves nearer and farther from the sun. These effects are expected to depend also on r,.11' and.12, which regulate the extent to which the Earth's orbital momentum and anisotropies in kinetic energy (caused by the sun) gravitate. Bromberg (1958), Baierlein (1967), and Krogh and Baierlein (1968) have calculated the three dominant three-body effects in the Earth-moon orbit using general relativity and the Dicke-Brans-Jordan theory. These effects are noncumulative and have amplitudes of -100 cm, -20 em, and -10 cm. The 100-cm effect [which was originally discovered by de Sitter (1916)] is known to depend only on y. The precise dependence of the other effects on the PPN parameters is not known. The prospects for measuring these effects in the 1970's are dim; they are masked by peculiarities in the orbit of the moon that have nothing to do with relativity. \u00a740. 7. THE DRAGGING OF INERTIAL fRAMES 1117 \u00a740.7. THE DRAGGING OF INERTIAL FRAMES The experiments discussed thus far study the motion of electromagnetic waves, spacecraft, planets, and asteroids through the solar system. An entirely different type Gyroscope precession: of experiment measures changes in the orientation of a gyroscope moving in the gravitational field of the Earth. This experiment is particularly important because it can measure directly the \"dragging of inertial frames\" by the angular momentum of the Earth. (1) general analysis It is useful, before specializing to a rotating Earth, to derive a general expression for the precession of a gyroscope in the post-Newtonian limit. (Track-l readers, and others who have not studied Chapters 6 and 39, may have difficulty following the derivation. No matter. It is the answer that counts!) Let sa be the spin of the gyroscope (i.e., its angular momentum vector), and let ua be i!s 4-velocity. The spin is always orthogonal to the 4-velocity, saua = 0 (see Box 5.6). Assume that any nongravitationalforces acting on the gyroscope are applied at its center of mass, so that there is no torque in its proper reference frame. Then the gyroscope will \"Fermi-Walker transport\" its spin along its world line (see \u00a76.5): a = VuU = 4-acceleration. (40.27) The objective of the calculation is to write down and analyze this transport equation in the post-Newtonian limit. The gyroscope moves relative to the PPN coordinate grid with a velocity vj = dx j Idt = dx;Jdt. Assume that vj :S f, where f is the post-Newtonian expansion parameter (f 2 :", "::: Mol Ro)' As the gyroscope moves, it carries with itself an orthonor mal frame e \"\" which is related to the PPN coordinate frame by a pure Lorentz boost, plus a renormalization of the lengths of the basis vectors [transformation (39.41 )]. The spin is a purely spatial vector (SO = 0) in this comoving frame; its length (SjSj)1/2 remains fixed (conservation of angular momentum); and its direction is regulated by the Fermi-Walker transport law. The basis vectors e\", of the comoving frame are not Fermi-Walker transported, by contrast with the spin. Rather, they are tied by a pure boost (no rotation!) to the PPN coordinate grid, which in turn is tied to an inertial frame far from the solar system, which in turn one expects to be fixed relative to the \"distant stars.\" Thus, by calculating the precession of the spin relative to the comoving frame, one is in effect evaluating the spin's angular velocity of precession, il j, relative to a frame fixed on the sky by the distant stars. Calculate dSj/dT: (40.28) dS, _J - V (5' e,) = (V 5)' e, + 5\u00b7 (V e,) = 5' V e, dT -u uJ' UJ U J J (40.29) Here use is made of the fact that ~ 5 is in the u direction [equation (40.27)] and 1118 40. SOLAR-SYSTEM EXPERIMENTS is thus orthogonal to ej' The quantity S\u00b7 Vuej is readily evaluated in the PPN coordinate frame. In the evaluation, one uses as metric coefficients [equations (39.32)] the expressions ~fln = -1 + 2U + 0\u00ab(4), go\u00b7 = - 2..11 1 2 2 V; - 1.12 H-j + (\"preferred- frame terms\" gjk = 8jk (1 + 2yU) + 0\u00ab(4), ) + 0\u00ab(5); (40.30) one takes as the components of e] and S [obtained via the transformation (39.41)] the expressions so = VjS] + 0\u00ab(3Sj ), 1 Sk = (1 - yV)S\" + \"2 VkVjS] + 0\u00ab(4Sj ); and one uses the relation (40.31 ) (40.32) where aj (assumed::; V,j) are the components of the 4-acceleration. One finds (see exercise 40.6) for the precession of the spin the result (2) general PPN formula for precession Rewritten in three-dimensional vector form this result becomes dSjdT = 0 X S + Dv X V V, (40.33a) (40.33b) (40.33c) (3) specialization: Thomas precession In this final answer it does not matter whether the 3-vectors entering into 0 are evaluated in the coordinate frame or in the comoving orthonormal frame, since ej and ajaxj differ only by corrections of order (2. Equations (40.33) describe in complete generality at the post-Newtonianilevel of approximation the precession of the gyroscope spin S relative to the comoving ortho normal frame that is rotationally tied to the distant stars. For an electron with spin S in orbit around a proton, only the first term, -!v X a, is present (no gravity). This term leads to the Thomas precession, which plays an important role in the fine structure of atomic spectra [see, e.g., Ruark and Urey (1930)]. For other ways of deriving the Thomas precession, see exercise 6.9 and \u00a741.4. The Thomas precession comes into play for a gyroscope on the surface of the Earth (a = Newtonian acceleration of gravity), but not for a gyroscope in a freely moving satellite. \u00a740. 7. THE DRAGGING OF INERTIAL FRAMES 1119 If one ignores the rotation of the Earth and preferred-frame effects, and puts the PPN coordinate frame at rest relative to the center of the Earth, then gOj vanishes and 0 is given by (4) specialization: precessions due to acceleration and Earth's Newtonian potential o=vx[-~a+(y+~)Vu] =yv X VUfor gyroscope on Earth's surface = (y + ~)v X V U for gyroscope in orbit. (40.34) The general-relativistic term (y +!)v X V U is caused by", " the motion ofthe gyroscope through the Earth's curved, static spacetime geometry. Notice that it depends solely on the same parameter y as is tested by electromagnetic-wave experiments. In order of magpitude, for a gyroscope in a near-Earth, polar orbit, 3(M )l/2(M ) R~::::: 8 seconds of arc per year. {] :::::\"2 R; (40.35) The general-relativistic precession ~v X V U was derived by W. de Sitter (1916) for the \"Earth-moon gyroscope\" orbiting the sun. Eleven years later L. H. Thomas (1927) derived the special relativistic precession -~v X a for application to atomic physics. The Earth's rotation produces off-diagonal terms, gOj' in the PPN metric (exer- cise 40.7): (5) specialization: precession due to Earth's rotation (40.36) Here J is the Earth's angular momentum. These off-diagonal terms contribute an amount (40.37) -to the precession of the gyroscope. Notice that this contribution, unlike the others, is independent of the linear velocity of the gyroscope. One can think of it in the following way. The gyroscope is rotationally at rest relative to the inertial frames in its neighbor hood. It and the local inertial frames rotate relative to the distant galaxies with the angular velocity 0 because the Earth's rotation \"drags\" the local inertial frames along with it. Notice that near the north and south poles the local inertial frames rotate in the same direction as the Earth does (0 parallel to J), but near the equator they rotate in the opposite direction (0 antiparallel to J; compare 0 with the magnetic field of the Earth!). Although this might seem paradoxical at first, an analogy devised by Schiffmakes it seem more reasonable.* Consider a rotating, solid sphere immersed in a viscous fluid. As it rotates, the sphere will drag the fluid along with it. At various points in the fluid, set down little rods, and watch how the fluid 'This analogy can be made mathematically rigorous; see footnote on p. 255 of Thome (1971); see also, \u00a721.12 on Mach's principle. (6) prospects for measuring precession EXERCISES 1120 40. SOLAR-SYSTEM EXPERIMENTS rotates them as it flows past. Near the poles the fluid will clearly rotate the rods in the same direction as the star rotates. But near the equator, because the fluid is dragged more rapidly at small radii than at large, the end of a rod closest to the sphere is dragged by the fluid more rapidly -than the far end of the rod. Conse quenfly, the rod rotates in the direction opposite to the rotation of the sphere. In order of magnitude, the precessional angular velocity caused by the Earth's rotation is {} - lEIR1- 0.1 seconds of arc per year. (40.37') Both this precession, and the larger one [equation (40.35)] due to motion through the Earth's static field, may be detectable in the 1970's. Equipment aimed at detecting them via a satellite experiment is now (1973) under construction at Stanford Univer sity; see Everitt, Fairbank, and Hamilton (1970); also O'Connell (1972).* The gyroscope precession produced by motion of the Earth relative to the preferred frame (if any) is too small to be of much interest. \u2022 The dragging of inertial frames by a rotating body plays important roles elsewhere in gravitation physics, e.g., in the definition of angular momentum for a gravitating body (\u00a719.2), and in black-hole physics (Chapter 33). The effect was first discussed and calculated by Thirring and Lense (1918). More recent calculations by Brill and Cohen (1966) of idealized situations where the effect may be large give insight into the mechanism of the effect. See also the discussion of Mach's principle in \u00a721.12. Exercise 40.6. PRECESSIONAL ANGULAR VELOCITYDerive equations (40.33) for the precession of a gyroscope in the post-Newtonian limit. Base the derivation on equations (40.29)-(40.32). Exercise 40.7. OFF\u00b7DIAGONAL TERMS IN METRIC ABOUT THE EARTH Idealize the Earth as an isolated, rigidly rotating sphere with angular momentum J. Use equations (39.34b,c) and (39.27) to show that (in three-dimensional vector notation) (40.38) outside the Earth, in the Earth's PPN rest frame. From this, infer equation (40.36). Exercise 40.B.", " SPIN-CURVATURE COUPLING Consider a spinning body (e.g., the Earth or a gyroscope or an electron) movin& through curved spacetime. Tidal gravitational forces produced by the curvature of spacetime act on the elementary pieces of the spinning body. These forces should depend not only on the positions of the pieces relative to the center of the object, but also on their relative velocities. Moreover, the spin of the body, s =J(pr X v) d(volume) in comoving orthonormal frame, is a measure of the relative positions and velocities of its pieces. Therefore one expects the spin to couple to the tidal gravitational forces-i.e., to the curvature of spacetime-producing \u00a740.8. IS THE GRAVITATIONAL CONSTANT CONSTANT? 1121 deviations from geodesic motion. Careful solution of the PPN equations of Chapter 39 for general relativity reveals [Papapetrou (1951), Pirani (1956)] that such coupling occurs and causes a deviation of the worldline from the course that it would otherwise take; thus, (40.39) Evaluate, in order of magnitude, the effects of the supplementary term on planetary orbits in the solar system. [Answer: They are much too small to be detected. However, there are two other material places to look for the effect: (I) when a rapidly spinning neutron star, or a black hole endowed with substantial angular momentum enters the powerful tidal field of another neutron star or black hole; and (2) when an individual electron, or the totality of electrons in the \"Dirac sea of negative energy states,\" enter a still more powerful tidal field (late phase of gravita tional collapse). Such a tidal field, or curvature, pulls oppositelY on electrons with the two opposiit directions of spin [Pirani (1956); DeWitt (1962), p. 338; Schwinger (1963a,b)] just as an electric field pulls oppositely on electrons with the two opposite signs of charge [\"vacuum polarization\"; see especially Heisenberg and Euler (1936)]. In principle, the tidal field pulling on the spin of an electron need not be due to \"background\" spacetime curvature; it might be due to a nearby massive spinning object, such as a \"live\" black hole (chapter 33) [\"gravitational spin-spin coupling\"; O'Connell (1972)]. \u00a740.8. IS THE GRAVITATIONAL CONSTANT CONSTANT? The title and subject of this section are likely to arouse confusion. Throughout this book one has used geometrized units, in which G = c = 1. Therefore, one has locked oneself into a viewpoInt that forbids asking whether the gravitational constant changes from event to event. False! One can perfectly well ask the question in the context of G = c = 1, if one makes clear what is meant by the question. In \u00a7\u00a7 1.5 and 1.6, c was defined to be a certain conversion factor between centi meters and seconds; and G/c 2 was defined to be a certain conversion factor between grams and centimeters. These definitions by fiat do not guarantee, however, that a Cavendish experiment* to measure the attraction between two bodies will yield If general relativity correctly describes classical gravity, and if the values of the conversion factors G and c have been chosen precisely right, then any Cavendish experiment, anywhere in the universe, will yield \"Force = -m1m2/r 2 \". But if the \u2022 See any standard textbook for a description of Cavendish experiments. By his original version of the experiment, with two separated spheres suspended by fine wires, Henry Cavendish (1798) inferred the mass and hence the density of the Earth. He reported: \"By a mean of the experiments made with the wire first used, the density of the Earth comes out 5.48 times greater than that of water; and by a mean of those made with the latter wire it comes out the same; and... the extreme results do not differ from the mean more than 0.38, or 1/14 of the whole.\" The most precise method of measuring G today [Rose et af. (1969)J gives Gc = (6.674 \u00b1.004) X 10-8 cm3/g sec~ (one standard deviati~n). \"Cavendish gravitational constant.\" Gc' defined Changes of Gc with time Spatial variations in Gc 1122 40, SOLAR\u00b7SYSTEM EXPERIMENTS Dicke-Brans-Jordan theory, or almost any other metric theory gives the correct description of gravity, the force in the Cavendish experiment will depend on where and when the experiment is performed, as well as on m 1, m2, and r. To discuss Cavendish experiments as tests of gravitation theory, then,", " one must introduce a new 'proportionality factor Gc = Gcaven~sh = (\"Cavendish gravitational constant\"), (40.40) which enters into the Newtonian force law (40.41) This Cavendish constant will be unity in general relativity, but in most other metric theories it will vary from event to event in spacetime. In some theories, such as Dicke-Brans-Jordan, the Cavendish constant is deter mined by the distribution of matter in the universe. As a result, the expansion of the universe changes its value: I dGc Gc --;]'t -- - (0.1 to I ) - I age of universe -- 1010 or 1011 years [see, e.g. Brans and Dicke (1961 )]. A variety of observations place limits on such time variations. Big time changes in Gc during the last 4.6 billion years would have produced marked effects on the Earth, the sun, and the entire solar system. The expected geophysical effects have been summarized and compared with observations by Dicke and Peebles (1965). It is hard to draw firm conclusions because of the complexity of the geophysics involved, but a fairly certain limit is (I/GC>(dGc/dt) ~ 1/1010 years (geophysical). (40.42a) Eventually, high-precision measurements of the orbital motions of planets will yield a better limit. For the present, planetary observations show (l/GC>(dGc/dt) ~ 4/1010 years (planetary orbits) (40.42b) [Shapiro, Smith, et al. (1971 )]. These limits are tight enough to begin to be interesting, but not yet tight enough to disprove any otherwise viable theories of gravity. If Gc is determined by the distribution of matter in the universe, then it should depend on where in the universe one is, as well as when. In particular, as one moves from point to point in the solar system, closer to the Sun and then fartlfer away, one should see Gc change. Indeed this is the case in most metric theories of gravity, though not in general relativity. Analyses of Cavendish experiments using the PPN formalism reveal spatial variation in Gc given by L1Gc = -2Gc(f3 + y - f3 2 - l)U (40.43) [Nordtvedt (1970, 197Ia); Will (197Ib)]. The amplitude of these variations along the Earth's elliptical orbit is L1Gc/Gc - f3 2 - 1 -- 1. This is far too small to measure directly in the, if f3 + y - 10-10 \u00a740.8. IS THE GRAVITATIONAL CONSTANT CONSTANT? 1123 1970's. Despite great ingenuity and effort, the most accurate experiments measuring the value of Gc have precisions in 1972 no better than I part in 104 [see Beams (1971)]. Experiments to search for yearly variations in Gc on Earth without measuring the actual value (\"null-type experiments\") can surely be performed with better precision than 1 in 104-but not with precisions approaching I in 1010\u2022 On the other ~aI!d,j!!di[~C9nSe~~nc~sof a spatialvariation of Gc in the solar system are almost certainly measurable (see \u00a740.9 below).. In Ni's theory of gravity (Box 39.1), and other two-tensor or vector-tensor theories like it, where the prior geometry picks out a preferred \"universal rest frame,\" the Cavendish constant Gc can depend on velocity relative to the pr)!'erred frame. For Cavendish experiments with two equal masses separated by distances large compared to their sizes, Gc varies as Dependence of Gc on velocity [Will (1971 b)]. Here v is the velocity of the Cavendish apparatus relative to the preferred frame, and n is the unit vector between the two masses. For experiments where one body is a massive sphere (e.g., the Earth), and the other is a small object on the sphere's surface, Gc varies as (40.44) J1Gcl Gc = \"2 [(a 3 - I ( 1) + a 2(1 - i ail - 3IIMR2)(v o n)2 II MR2)]V 2 (40.44') [Nordtvedt and Will (1972)]. Here M and R are the mass and radius of the sphere, and is the trace of the second moment of its mass distribution. Consequences of these effects for planetary orbits have not yet been spelled out, but consequences for Earthbound experiments have. Think of a Cavendish experiment in which one mass is the Earth, and the other is a gravimeter on the Earth's surface. The gravimeter", " gives a reading for the \"local acceleration of gravity,\" (40.45) As the Earth turns, so the unit vector n between its center and the gravimeter rotates, Gc and hence g will fluctuate with a period of 12 sidereal hours and an amplitude Here 8m is the minimum, as the Earth rotates, of the angle between v (constant vector) and n (rotating vector). (Note: we have used the value II MR2 ~ 0.5 for the Earth.) These fluctuations will produce tides in the Earth of the same type as are Anomalies in Earth tides due to anisotropies in Gc : (1) experimental value of \u00b02 (2) experimental disproof of Whitehead theory 1124 40. SOLAR-SYSTEM EXPERIMENTS produced by the moon and sun. As of 1972, gravimeter measurements near the Earth's equator show no sign of any anomalous 12-sidereal-hour effects down to an amplitude of _10- 9 [Will (l97lb)]. Consequently, _ Icx 21112u cos 8m = 1.12 + r - 111/2U COS 8m ~ 6 X 10-5 - 20 km/sec.(40.46a) Using a rough estimate of u - 200 kIn/sec forthe--Barth's veleeity-TeUuiY~-to-the universal rest frame, and 8m ~ 60\u00b0 for the angle between v and the Earth's equatorial plane, one obtains the rough limit (40.46b) [This limit does not affect the three theories in Box 39.1; of them, only Ni's theory has prior geometry and a universal rest frame; and it predicts isotropic effects in L1Gc/Gc [equation (40.44)], but no anisotropic effects. However, other theories with universal rest frames-e.g. Papapetrou's (1954a,b,c) theory-are ruled out by this limit; see Ni (1972), Nordtvedt and Will (1972).] Whitehead's theory of gravity (which is a two-tensor theory with a rather different type of prior geometry from Ni's) predicts that the galaxy should produce velocity independent anisotropies in Gc. These, in turn, would produce Earth tides with periods of 12 sidereal hours and amplitudes of.1 / _ 2 X 10-7 _ 100 X (experimen~al limit on) g g such amphtudes [Will (197lb)]. The absence of such tides proves Whitehead's theory to be incor rect-a feat of disproof beyond the power of all redshift~light-defiection, time-delay, and perihelion-shift measurements. (For all these \"standard experiments,\" the predictions of Whitehead and Einstein are identical!) Equation (40.44') predicts a periodic annual variation of the Cavendish constant on Earth, as the Earth moves around the sun: Anomalies in Earth rotation rate due to dependence of Gc on velocity _ (Velocity of Earth) relatIve to sun v -. (velocity of sun relative) _ +.. d f to prelerre. = vE + W, rame cx 1 ) (w 2 + vi + 2w\u00b7 vE ). (40.47) (L1GC/GC>averagedover = ~ (; cx 2 + cx 3 - Surface of Earth t [varies sinusoidally with period of one year}-1 This annual variation, assuming all PPN parameters are of order unity, is l,doo times larger than the one produced by the Earth's motion in and out through the sun's gravitational potential [equation (40.43)]. In response to this changing Cavendish constant, the Earth's self-gravitational pull should change, and the Earth should \"breathe\" inward (greater pull) and outward (relaxed pull). The resulting annual variations in the Earth's moment of inertia should produce annual changes in its rotation rate w (changes in \"length of day\" as measured by atomic clocks): \u00a740.8. IS THE GRAVITATIONAL CONSTANT CONSTANT? 1125 [Nordtvedt and Will (1972)]. Comparison with the measured annual variations of rotation rate (all of which geophysicists attribute to seasonal changes in the Earth's atmosphere) yields the following limit I; a 2 + as - all = <0.2. (40.48) Experimental value of 2 + \u00b03 - \u00b01 i 0 [See Nordtvedt and Will (1972)]. This limit rules out several preferred-frame theories of gravity, including that of Ni (Boxes 39.1 and 392). The experimental results (40.21), (40.46), and (40.48), when combined, place the following very rough limits", " on any theory that possesses a Universal rest frame: la 11= 17.11 +.12 - 4y la 2 1 = 1.12 + r - 11 ~ 0.03, la 31= 14/31 - 2y - 2 - rl ~ 2 X 10-5 41 ~ 0.2,. (40.49) These limits completely disprove all theories with preferred frames that have been examined to date except one devised by Will and Nordtvedt [see Ni (1972); Nordtvedt and Will (1972)]. In some theories of gravity, the result of a Cavendish experiment depends on the Dependence of Gc on chemical composition chemical composition and internal structure of the test bodies (exercises 40.9 and 40.10). Kruezer (1968) has performed the most accurate search for such effects to date. He finds that Gc is the same for fluorine and bromine to a precision of IGc(bromine)G~ Gc(fiuorine) I~ 5 X 10-5. (40.50) Exercise 40.9. CAVENDISH CONSTANT FOR IDEALIZED SUN Idealize the sun as a static sphere of perfect fluid at rest at the origin of the PPN coordinates. Then its external gravitational field has the form (40.3), with Mo given by (40.4). Consequently, a test body of mass m, located far away at radius r, is accelerated by a gravitational force EXERCISES (40.51a) (a) Calculate the mass of the sun, M, in the sense of the amount of energy required to construct it by adding one spherical shell of matter on top of another, working from the inside outward. [Answer: M = I I Ro '2, 0 Po(1 + II + 3yU)4'iir2 dr - I Ro rest mass + internal energy =~Ro Po[1 + II + (3y - ~) u] 4'iir2 dr.],,0 poU4'iir 2 dr, gravitational potential energy (40.51b) (b) Use the virial theorem [equation (39.21b)] to rewrite equation (40.4) in the form (40.51c) 1126 40. SOLAR-SYSTEM EXPERIMENTS (c) Combine the above equations with the definition Force = - GCrnM/r 2 of the Cavendish constant for r far outside the sun, to obtain ( mass of sun as defined by its effect in bending world line of a faraway test particle Gc = (mass-energy as defined by applying law Of) conservation of energy to the steps in the construction of the sun ) (40.51 d) (40.52) Unless f33 = 1, and 4f32 + f34 - 6y + 1 = 0 (as they are, of course, in Einstein's theory), Gc will depend on the sun's internal structure! Specialize equation (40.52) to \"conservative theories of gravity (Box 39.5), and explain why the result is what one would expect from equation (40.43). Exercise 40.10. CAVENDISH CONSTANT FOR ANY BODY Extend the analysis of exercise 40.9 to a source that is arbitrarily stressed and has arbitrarv shape and internal velocities (subject to the constraints v2 ~ 1, Itif< I/Po ~ 1, U ~ 1, II ~ 1, of the post-Newtonian approximation). Assume that the body is at rest relative to the universal rest frame. Show that Gc depends on the internal structure of the source unless ~ =0, Of course, these PPN constraints are all satisfied by Einstein's theory. 11=0. (40.53) The sense in which general relativity predicts geodesic motion for planets and sun \u00a740.9. DO PLANETS AND THE SUN MOVE ON GEODESICS? Crucial to solar-system experiments is the question of whether the sun and the planets move on geodesics of spacetime. This question is complicated by the contributions to the spacetime curvature made by the moving body itself. To elucidate the question-and to obtain an answer in the framework of general relativity-consider an \"Einstein elevator\" type of argument. The astronomit:al object under consideration has an outer boundary, and each point on this boundary describes a world line. These world lines define a world tube. Some distance outside of this world tube construct a \"buffer zone\" as in \u00a720.6. Tailor its inner and outer dimensions, according to the mass and moments of the object and the curvature of the enveloping space (\"strength of the tide-producing force of the external gravi tational field\"), in such a way that the departure ( (cf. \u00a720.6) of the metric from flatness in this", " buffer zone takes on values equal at most to twice the extremal achievable value (extrem (a minimum with respect to variations in r, a maximum \u00a740.9. DO PLANETS AND THE SUN MOVE ON GEODESICS? 1127 with respect to variations in direction; in other words, a minimax). Then, apart from errors of order fextrem' the object can be regarded as moving in an asymptotically flat space. The law of conservation of total4-momentum applies. It assures one that the object moves in a (locally) straight line with uniform velocity. Consider, next, a \"background geometry\" that agrees just outside the buffer zone with the actual geometry to accuracy f extrem or better, but that inside is a source-free solution of Einstein's field equation. Then, to an accuracy governed by the magnitude of f extrem, the locally straight line along which the astronomical object moves will be a geodesic of this background geometry. Insofar as one can give any well-defined meaning to the departure of the actual motion from this geodesic (a task complicated by the fact that the background geometry does not actually exist!), one can calculate this departure by making use of the fPN formalism or some other approximation scheme [see, e.g., Taub (1965)]. This deviation springs ordinarily in substantial measure, and sometimes almost wholly, from a coupling between the Riemann curvature tensor of the external field and the multipole moments of the astronomical object (angular momentum associ ated with rotation; quadrupole and higher moments associated with deformation; see, e.g., exercises 40.8 and 16.4). This coupling is important for the Earth-moon system, but one need not use relativity to calculate it; Newtonian theory does the job to far greater accuracy than needed-or would, if one understood the interiors of the Earth and the moon well enough! For the planets and sun, the effect is negligible. (Exercise: use Newtonian theory to prove so!). Thus, in general relativity as applied to the solar system, one can approximate the orbit of the sun, the Earth-moon mass center, and each other planet, as a geodesic of that \"background spac~time geometry\" which would exist if its own curvature effects were absent. This is the approach used to analyze the perihelion shift for planets in \u00a740.5 in the context of general relativity, and to derive in exercise 39.15 the post-Newtonian \"many-body equations of motion.\" In most other metric theories of gravity, including the Dicke-Brans-Jordan theory, there are substantial departures from geodesic motion. The \"Einstein elevator\" argument fails in these theories because spacetime is endowed not only with a metric, but also with a long-range field that couples indirectly (cf. \u00a7\u00a738.7 and 39.2) to massive, gravitating bodies. This phenomenon is best understood in terms of Dicke's argument about the influence of spatial variations of the fundamental constants on experiments of the Eotvos-Dicke type (see \u00a738.6). In a theory where the Cavendish gravitational constant Gc depends on position (as it does not and cannot in general relativity), a body with significant self-gravitational energy Egrav must fall, in a perfectly uniform external Newtonian gravitational field, with an anomalous acceleration: Deviations from geodesic motion: (1) due to curvature coupling (2) due to spatial dependence of gravitational constant (Nordtvedt effect) acceleration Of) _ (aCCeleration Of) = ~(aEgrav) VG c massive body test body M aGc ( (40.54) = Egrav VG MGc c 1128 40. SOLAR-SYSTEM EXPERIMENTS [see equation (38.15)]. In Dicke-Brans-Jordan theory, Gc is essentially the reciprocal of the scalar field; and it contains a small part that is proportional to the Newtonian potential, U [equation (40.43) with the appropriate values of the parameters from Box 39.2]. As a result, the sun falls with an acceleration smaller by one part in 106 than the acceleration of a test body; Jupiter falls with an acceleration one part in 109 smaller; and the Earth, one part in 1010 smaller. Translated into relativistic language: the scalar field, by influencing the gravitational self-energy of a massive body, produces deviations from geodesic motion. One can use the full PPN formalism of Chapter 29 to calculate the motion of massive bodies in any metric theory of gravity. Nordtvedt (l968b) and Will (l971a) have done this. They find that a massive body at rest in a uniform external field experiences a (Newtonian-type) PPN coordinate acceleration given by where E", "jk is a quantity depending on the body's structure: E jk = 8jk { 1 - (7.11 - 30y - 4/3) Emav }- (2/3 + 2/32 - 30y +.12 - 2) ~k, g. = _ ~f Pop~(Xj - xj)(xk - xk) d 3 d 3 /X-X'/3 X Jk 2 (40.55)'x, E gray = \"\" goo L.\" 11' Here m is the body's total mass-energy, gjk is the \"Chandrasekhar potential-energy tensor,\" and Egrav is the body's self-gravitational energy. [Note: Dicke's method of calculating the anomalous acceleration (40.54) breaks down!n theories that are not \"conservative\" (Box 39.5).] In general relativity, the combinations of PPN coefficients appearing in Ejk vanish; so Ejk = 8jk, and the body falls with the usual acceleration-i.e., it moves along a geodesic. But in most other theories of gravity Ejk ::P 8jk ; the body does not move on a geodesic; and its acceleration may even be in a different direction than the gradient of the Newtonian potential! This predicted departure from geodesic motion is called the UNordtuedt effect.\" - The possibility of such an effect was first noticed in passing by Dicke (1961c), but was discovered independently and explored in great detail by Nordtvedt (1968a,b). The Nordtvedt effect in a theory other than general relativity produces a number of phenomena in the solar system that are potentially observable. [See Nordtvedt (1971 b) for an enumeration and references.] The effect most suitable for a test is a \"polarization\" of the Earth-moon orbit due to the fact that the moon should fall toward the sun with a greater acceleration than does the Earth. This \"polarization\" results in an eccentricity in the orbit that points always toward the Sun and has the amplitude Nordtvedt effect in Earth-moon orbit \u00a740.9. DO PLANETS AND THE SUN MOVE ON GEODESICS? 8r = 840 [3oy + 4[3 - 7.11 - j (2[3 + 2[32 - 30y +.12 - 2)] em = 67 meters =~ meters 2+\", in Ni's theory (Boxes 39.1 and 392) in Dieke-Brans-Jordan theory (Boxes 39.1 and 39.2) = 0 in Einstein's theory. 1129 (40.56) Box 40.4 CATALOG OF EXPERIMENTS Type of experiment Description of experiment Where discussed 1. Tests of uniqueness of free fall (Eotvos-Dicke-Braginsky \u00a738.3; Figure 1.6; 1. Tests of foundations of general relativity II. Post-Newtonian (\"solar-system\")_ _ experiments experiments 2. Tests for existence of metric (time dilation of particle de cays; role of Lorentz group in particle kinematics; etc.) 3. Searches for new, direct-coupling, long-range fields (Hughes Drever experiment; ether-drift experiments) 4. Gravitational redshift experiments 5. Constancy, in space and time, of the nongravitational phys- ical constants 1. Deflection of light and radio waves by Sun 2. Relativistic delay in round-trip travel time for radar beams passing near Sun 3. Perihelion shifts and periodic perturbations in planetary orbits 4. Three-body effects in the Lunar orbit 5. Precession of gyroscopes (\"geodetic precession\" and preces sion due to dragging of inertial frames by Earth's rotation) 6. Spatial variation of the Cavendish gravitational constant in the solar system 7. Dependence of the Cavendish gravitational constant on the chemical composition of the gravitating body 8. Earth tides with sidereal periods 9. Annual variations in Earth rotation rate 10. Periodicities in Earth-Moon separation due to breakdown of geodesic motion Box 1.1 \u00a738.4 \u00a738.7; Figure 38.3 \u00a738.5; Figures 38.1 and 38.2; \u00a7\u00a77.2, 7.3, and 7.4 \u00a738.6 \u00a740.3; Box 40.1. \u00a740.4; Box 40.2 \u00a740.5; Box 40.3 \u00a740.6 \u00a740.7 \u00a7\u00a74O.8 and 40.9 \u00a74O.8 \u00a74O.8 \u00a740.8 \u00a740.9 \u00a74O.8 III. Cosmological observations 1. Change of Cavendish gravitational constant with time", " in solar system 2. Large-scale features of universe (expansion, isotropy, homo geneity; existence and properties of cosmic microwave radiation;...) Chapters 27-30; especially Chapter 29 3. Agreement of various measures of age of universe (age from expansion; ages of oldest stars: age of solar system) \u00a729.7 IV. Gravitational-Wave experiments Existence of waves; propagation speed; polarization properties;... Chapters 35-37; especially Chapter 37 \u00a740.10. SUMMARY OF EXPERIMENTAL TESTS OF GENERAL RELATIVITY 1131 Figure 40.4. (facing page) Measuring the separation between earth and moon by determining the time-delay (about 2.5 sec) between the emission of light from a laser on the earth and the return of this light to the earth. A key element in the program is a comer reflector, the first of which was landed on the moon July 20, 1969, by the Apollo 11 flight crew. In November 1971, there were three such refiectors on the moon: two American, and one French-built and Soviet-landed. A pulsed ruby laser projects a beam out of the 107-inch reflecting telescope of the McDonald Observatory of the University of Texas, on Mount\"Locke, 119 miles east of El Paso. This beam makes a spot of light on the moon's surface about 3.2 km in diameter. Laser light is bounced straight back to the earth by the \"laser ranging retroreflectors\" (LR3). Each consists of an aluminum panel of 46 cm by 46 cm with 100 fused silica comer cubes each 3.8 cm in diameter. The first reflector ever set up appears in the first inset, near the lunar landing mOdule. It is tilted with respect to the landscape of the moon. The photograph was made shortly before astronauts Neil A. Armstrong and Edwin E. Aldrin, Jr., took off for the earth. The second inset is a photograph made by D. G. Currie of the field of view in the guiding eyepiece of the McDonald 107-inch telescope in an interval when the laser was not firing at the Apollo 11 site. One guides the telescope to Tranquility Base (small circle) by aligning flducial marks on more visible moonscape features. In November 1971, the LR~ experiment and continuing time-of-fiight measurements were the responsibility of the National Aeronautics and Space Administration and a Lunar Retroreflecting Ranging Team of representatives from several centers of research. One of the members of this team, Carroll Alley, of the University of Maryland, is hereby thanked for his kindness in providing the photographs used in this montage. Thanks to this NASA work, the distance between the laser source on the earth and the reflectors on the moon is known with an accuracy now better than half a meter. The astronauts left behind on the moon not only LR3 and a seismometer and other equipment, but also a plaque: \"We came in peace for all mankind.\" By the mid 1970's, lunar laser-ranging data will probably be able to determine the amplitudes of this polarization to a precision of one meter or better [see Bender et al. (1971); also Figure 40.4]. \u00a740.10. SUMMARY OF EXPERIMENTAL TESTS OF GENERAL RELATIVITY No longer is general relativity \"a theorist's Paradise, but an experimentalist's Hell.\" It is now a Paradise for all-as one can see quickly by perusing the catalog of experiments given in Box 40.4 on page 1129. Moreover, general relativity has emerged from each of its tests unscathed-a remarkable 1973 tribute to the 1915 genius of Albert Einstein. t PART X FRONTIERS mathematician, four times enticed (once by an old friend), Wherein the reader-who, during a life of continued variety for forty chapters (besides the Preface), was eight chapters a four chapters a cosmologist, and four chapters a transported astrophysicist in the land of black holes, and who at last lived honest- and became a True Believer-now ventures forth in search of new frontiers to conquer. inherited a wealth of experiments, - -------- - ~----------- CHAPTER 41 SPINORS \u00a741.1. REFLECTIONS, ROTATIONS, AND THE COMBINATION OF ROTATIONS Spinors and their applications in relativity grew out of the analysis of \"rotations,\" r first in space, then in spacetime. Take a cube (Figure 41.1). Rotate it about one axis through 90\u00b0. Then pick another axis at right angles to the first. About it rotate the cube again through 90\u00b0. In this way the cube is carried from the orientation marked \"Initial\" to that marked \"Final.\" How can one make this net transformation in a single step, with a single rotation?JIL other ~ds,__\\\\", "'l1~!js)he law for the This chapter is entirely Track 2. No earlier Track-2 material is needed as preparation for it, nor is it needed as preparation for any later chapter. \" The problem of combining rotations _------eombinatiun -or-rotations? - Were rotations described by vectors, then one could apply the law of combination of vectors. The resultant of two vectors of the same magnitude (90\u00b0) separated by a right angle, is a single vector that (1) lies in the same plane and (2) has the magnitude 2 1/ 2 X 90\u00b0 = 127.28\u00b0. Both predictions are wrong. To turn the cube from initial to final orientation in a single turn, (1) take an axis running from the center through the vertex A and (2) rotate through 120\u00b0. __ __ _~_.-Wha.t-.eQm~tfiiD-can ever reproduce a law of combination of rotations apparently so strange? On the evening of October 16, 1843, William Rowan Hamilton was walking with his wife along the Royal Canal in Dublin when the answer leaped to his mind, the fruit of years of reflectiorr. With his knife he then and there carved on a stone on Brougham Bridge the formulas* -~ ;2 =P = k 2 = ijk = - 1, *In the same city on June 21, 1972 President Eamon de Valera told one of the authors that, while in jail one evening in 1916, scheduled to be shot the next morning, he wrote down the formula of which he was so fond, ;2 =j2 = k 2 = ijk = - 1. y Initial Final Figure 41.1. Rotation about the vertical axis through 90\u00b7. followed by rotation about the horizontal axi. through 90\u00b7. gives a net change in orientation that can be acl1ieved by a single rotation through 120\u00b7 about an axis emergent from the center through the comer A. which in today's notation, Uz =,,~ ~II =ii, take the form Rotation operators: (1) defined' To any rotation is associated a quantity (Hamilton's \"quaternion;\" today's \"spin matrix\" or \"spinor transformation\" or \"rotation operator\") R = cos (0/2) - i sin (0/2)(uz cos a + ullcos/3 + uzcosy), (2) as tools in combining rotations where 0 is the angle of rotation and a, /3, yare the angles between the axis of rotation and the coordinate axes. A rotation described by R 1 followed by a rotation described by Rz gives a net change in orientation described by the single rotation (41.2) This is Hamilton's formula for the combination of two rotations -(steps towarn--n- by Euler in 1776; obtained by Gauss in 1819 but never published by him). In the example in Figure 41.1, t (41.4) R1(rotation by 0 =-90\"-iibout z-axi~t::: (l - Rz(rotation by 0 = 90\u00b0 about x-axis) = (l - iuz)/2 11Z, iUz)j21/2, - and the product of the two is RzR1 = (l - iuz + iUlJ - iUz)/2 = cos 60\u00b0 - i sin 60\u00b0 (uz /3 1/Z - uy/31/Z + uz/3 1/Z). \u00a741.1. REFLECTIONS AND THE COMBINATION OF ROTATIONS 1137 Figure 41.2. Reflection in the plane MPQ carries A to B. Reflection in the plane NPQ carries B to C. The combination of the two reflections in the two planes separated by the angle 8/2 produces the same end result (transformation from A to C) as rotation through the angle 8 about the line PQ. c According to Hamilton's rule (41.3), this result implies a net rotation through 120\u00b0 abo\\lt a line that makes equal angles with the x-axis, the y-axis, and the z-axis, in conformity with what one already saw in Figure 41.1 (axis of rotation running from center of cube through the comer A). What one has just done in the special example one can do in the general case: obtain the parameters 0a, aa, f3a, Ya of the net rotation (four unknowns!) by identify iaz on both ing the four coefficients of the four Hamilton units 1, - iaz' sides of the equation Ra = RzR1\u2022 In this way one arrives at the four prequatemion formulas of Olinde Rodrigues (1840) for the combination of the two rotations. i", "all, - - Geometric reason that half angles appear in rotation operators Algebraic properties of rotation 0 perators Why do half-angles put in an appearance? And what is behind the law of combi nation of rotations? The answer to both questions is the same: a rotation through the angle 0 about a given axis may be visualized as the consequence of successive reflections in two planes that meet along that axis at the angle 0/2 (Figure 41.2). Two rotations therefore. imply four reflections. However, it can be arranged that reflections no. 2 and no. 3 take place in the same plane, the plane that includes the two axes of rotation. Then reflection no. 3 exactly undoes reflection no. 2. By now there remain only reflections no. 1 and no. 4, which together constitute one rotation: the net rotation that was desired (Figures 41.3 and 41.4). The rotation R = cos (0/2) - i sin (0/2)(az cos a + all cos f3 + az cos y) (41.3) is undone by the inverse rotation R-l = cos (0/2) + i sin (0/2)(az cos a + all cos f3 + az cos y). (41.3') Thus the product of the two rotation operators is an operator, the unit operator, that leaves unchanged everything that it acts on. The reciprocal R-l of the combination R = RzR1 of two rotations is (41.5) (41.5') (reverse order of factors!), as one verifies by substitution into (41.5). 1138 41. SPINORS 2 2 G ) D x G x Figure 41.3. Composition of two rotations seen in terms of reflections. The first rotation (for instance, 90\u00b7 about 02 in the example of Figure 41.1.) is represented in terms of reflection I followed by reflection 2 (the planes of the two reflections being separated by 90\u00b7/2 = 45\u00b7 in the example). The second reflection appears as the resultant of reflections 3 and 4. But the reflections 2 and 3 take place in the common plane 20X. Therefore one reflection undoes the other. Thus the sequence of four operations 1234 collapses to the two reflections I and 4. Their place in tum is taken by a single rotation about the axis OA. The conjugate transpose, M\u00b7, of a matrix M is obtained by taking the conjugate complex of every element in the matrix and then interchanging rows and columns. By direct inspection of matrix expressions (41.1) one sees that a/ = az, all\u00b7 = al/' a/ = az\u2022 Such matrices are said to be Hermitian. The conjugate transpose of the product M = PQ of two matrices is the product M\u00b7 = Q.P* of the individual conjugate transposed matrices taken in the reverse order. For the rotation matrix written down above, note that R\u00b7 = R-l. Such a matrix is said to be unitary. The t \u00a741.1. REFLECTIONS AND THE COMBINATION OF ROTATIONS 1139 Figu re 41.4. Law of composition of rotations epitomized by a spherical triangle in which each of the three important angles represents half an angle of ro tation. determinant of a unitary matrix may be seen to have absolute value unity from the following line of argument: I = det (unit matrix) = det (RR-I) = det (RR\u00b7) = det R det R\u00b7 = Idet R12. (41.6) In actuality the determinant of the rotation spin matrix is necessarily unity (\"uni modular matrix\") as shown in the following exercises Exercise 41.1. ELEMENTARY FEATURES OF THE ROTATION MATRIX Write equation (41.3) in the form EXERCISES R(O) = cos (0/2) - isin(0/2)(0-'n), and establish the following properties: (a) (b) (c) (0-' n)2 = I = unit matrix; tr (0-' n) = 0 (tr means \"trace,\" i.e., sum of diagonal elements); JR,(o-'n)],=R(o-'n) - (o-'n)R = 0; ~commutator] (d) dR i dii = - 2(0- \u2022 n)R. (41.7) [Note that if one thinks of 0 as increasing with angular velocity w, so dO/dt = w = constant, then this last equation reads i dR -= --(o-'w)R 2 dt (41.7') where w = wn.] 1140 41. SPINORS Exercise 41.2. ROTATION MATRIX HAS UNIT DETERMINANT Recall from", " exercise 5.5 that for any matrix M one has d[ln (det M)] == tr (M-l dM) and use this to show that det R in (41.7) is constant, and therefore equal to (det R)o = 0 = I. \u00a741.2. INFINITESIMAL ROTATIONS Infinitesimal rotations A given rotation can be obtained by performing in turn two rotations of half the magnitude, or four rotations of a fourth the magnitude, or eight of an eighth the magnitude, and so on. Thus one arrives in the limit at the concept of an infinitesimal rotation described by the spin matrix or Here the quantities R = 1 - (i dO/2)(0\" n). dOI/Z = -dOzl/ = nZdO = cos a dO, dOzz = -dOzz = nl/ dO = cos f3 dO, dOzl/ = -dOl/z = n Z dO = cos y dO, (41.8) (41.9) are the components of the infinitesimal rotation in the three indicated planes. An infinitesimal rotation in the (x,y)-plane through the angle dOzl/ transforms the vector x = (x,y, z) into a new vector with changed components x' and y' but with un changed component z' = z. More generally, the infinitesimal rotation (41.8) con sidered in this same \"active\" sense* produces the transformation with x-+x', (dOzz)z, x' = x (dOZl/)y y' = -(dOl/z)x + y z' = -(dOzz)x - (dOl/z)z, (dOzl/)y + z. (41.10) Representation of a 3-vector as a spin matrix Spinor calculus provides an alternative (and shorthand!) means to calculate the foregoing effect of a rotation on a vector. Associate with the vector x the spin matrix x = x(Jz + lUI/ + z(Jz = (x' 0'), (41.11) * An \"active\" transformation changes one vector into another, while leaving unchanged the underlying reference frame (if there is one). By contrast, a \"passive\" transformation leaves all vectors unchanged, but alters the reference frame. All transformations in previous chapters of this book were passive. \u00a741.2. INFINITESIMAL ROTATIONS 1141 and with the vector x' a corresponding spin matrix or quaternion X'. Then the effect of the rotation is summarized in the formula X --+ X' = RXR*. (41.12) Rotation of a 3-vector described in spin-matrix language Test this formula for the general infinitesimal rotation (41.10). It reads (x' 00-) = [1 - (i dO12)(0- 0n)](x 00-)[1 + (i dO12)(0- 0n)] o~t!:Uhe first order in the quantity dO, (x'oo-) = (xoo-) + (idOI2)[(x oo-)(o-on) - (41.13) The product of spin matrices A = (a 0 0-) and B = (b 00-) built from two distinct vectors a and b is (o-on)(xoo-)]. or, according to (41.2), AB = (a 0 b) + i(a X b) 0 0-. (41.14) Employ this formula to evaluate the right-hand side of(41.13). In the square brackets, the terms in (x 0 n) have opposite signs and cancel. In contrast, the terms in (n X x) have the same sign. They combine to cancel the factor 2 in (dOI2). End up with (x' 00-) = (x 00-) + dfJ(n X x) 00- or x' = [1 + (dO)n X] x (41.15) in agreement with (41.10), as was to be shown. A finite rotation about a given axis can be considered as the composition of infinitesimal rotations about that axis. To see this composition in simplest form, rewrite the spin matrix (41.8) associated with the general infinitesimal rotation as R(dO) = e-(i dO/2)(u\u00b7n) (41.16),-(exponential function defined by its power-series expansion). Note that (0- 0 n) commutes (a) with unity and (b) with itself, and in addition (c) has a unit square. Therefore the calculation of the exponential function proceeds no differently here, for spin", " matrices, than for everyday algebra. The composition of the spin matrices for infinitesimal rotations about an unchanging axis proceeds by adding exponents, to give R(O) = e- i (O/2)(u.n), (41.17) which can also be obtained immediately from equation (41.7). This expression can be put in another form by developing the power series; thus, Composition of finite rotation from infinitesimal rotations 1142 41. SPINORS R(O) = 2: (l/p!)(-iOCT'n/2)P p=O = 2: (l/p!)( -iO/2)P + (CT' n) 2: (l/p!)( - iO /2)P evenp oddp (41.18) = cos (0/2) - i sin (0/2)(CT' n) in agreement with the expression (41.3) originally given for a spinor transformation. The effect of one infinitesimal rotation after another after another... on a vector is given by X' = R(dO)... R(dO)XR*(dO)... R*(dO), with the consequence that even for a finite rotation R = R(O) one is correct in employing the formula X' = RXR*. (41.19) EXERCISE Exercise 41.3. MORE PROPERTIES OF THE ROTATION MATRIX Show that for X ::: x\u00b7 CT one has the commutation relation [(CT' n), X] ::: 2i(n X x). CT. Use this to obtain, from equation (41.19) in the form X = RXoR* [where Xo is constant, while R(O) is given by equation (41.17)], the formula d dO (x' CT) = (n X x). CT. Why is this equivalent to the standard definition dx (j(=WXx for the angular velocity? Reverse the argument to show that equation (41.7') correctly defines the rotation R(t) resulting from a time-dependent angular velocity Col(t), even though the simple solution R = exp [-!it(CT' Col)] of this equation can no longer be written when Col is not constant. i \u00a741.3. LORENTZ TRANSFORMATION VIA SPINOR ALGEBRA 4-vectors and Lorentz transformations in spin-matrix language Generate a rotation by two reflections in space? Then why not generate a Lorentz transformation by two reflections in spacetime? If for this purpose one has to turn from a real half-angle between the two planes of reflection to a complex half-angle, that development will come as no surprise; nor will it be a surprise that one can \u00a741.3. LORENTZ TRANSFORMATION VIA SPINOR ALGEBRA 1143 still represent the effect of the Lorentz transformation by a matrix multiplication of the form x --+ X' = LXL\u00b7. (41.20) Here the \"Lorentz spin transformation matrix\" L is a generalization of the rotation matrix, R. Also the \"coordinate-generating spin matrix\" X is now generalized from (41.11) to or X = t + (x\u00b7 0') X = II t + ~ x - iyll. X+lY t-z It is dJmanded that this matrix be Hermitian X=X\u00b7. (41.21) (41.22) (41.23) Then and only then are the coordinates (t, x,y, z) real. The conjugate transpose of the transformed spin matrix must also be Hermitian-and is: (X')\u00b7= (LXL\u00b7)\u00b7 = (L\u00b7)*(X)\u00b7(L)* = LXL\u00b7 = X'. (41.24) Therefore the new coordinates (t', x', y', z') are guaranteed to be real, as desired. This reality requirement is a rationale for the form of the spin-matrix transformation (41.20), with L appearing on one side of X and L\u00b7 on the other. A Lorentz transformation is defined by the circumstance that it leaves the interval invariant: (41.25) Note that the determinant of the matrix X as written out above has the value (41.26) Consequently the requirement for the preservation of the interval may be put in the form or det X' = det X (det L) (det X) (det L\u00b7) = det X. This requirement is fulfilled by demanding det L = 1 (41.27) (41.28) (41.29) [iLis-nola-useful ge-neraIization to multiply every element of L", " here by a common phase factor e ill, and therefore multiply det L by e 2ill, because the net effect of this phase factor is nil in the formula X' = LXL\u00b7]. Infinitesimal Lorentz transformations 1144 41. SPINORS The spin matrix associated with a rotation, whether finite or infinitesimal, already satisfied the condition det L = I [proved in exercise (41.2)]. This condition, being algebraic, will continue to hold when the real angles d(Jyz' d(Jzz' d(JZY' are replaced by complex angles, dfJ yZ + i da z' d(Jzz + i day, d(Jry + i da z. The spin-transformation matrix acquires in this way a total of six parameters, as needed to describe the general infinitesimal Lorentz transformation. Thus the spin matrix for the general infinitesi mal Lorentz transformation can be put in the form L = I - (i/2)(O\"r d(JyZ + O\"y d(Jzz + O\"z d(Jry) + (1/2)(O\"z da z + O\"y day + O\"z daz) (id(J/2)(u o n) + (u\u00b7da/2). = 1 - (41.30) The effect of this transformation upon the coordinates is to be read out from the formula x --+ X' = LXL* or t' + (u' x') = [1 - (i d(J/2)(u' n) + (u 0 da/2)] X [t + (u 0 x)][l + (i d(J/2)(u 0 n) + (u 0 da/2)] (41.31) Employ equation (41.14) for (u' A)(u. B) to reduce the right side to the form t + (u' x) + (u' da)t + d(J(n X x), u + (x 0 da). Now compare coefficients of 1, u z' u y and u z' respectively, on both sides of the equation, and find t' = t + (x 0 do.) x' = x + t do. + d(J(n X x), (41.32) in agreement with the conventional expression for an infinitesimal Lorentz transfor mation or \"boost\" of velocity do., in active form, as was to be shown. The composition of such infinitesimal Lorentz transformations gives a finite Lorentz transformation. The result, however, can be calculated easily only when all infinitesimal transformations commute. Thus assume that d(J and do. are in a fixed ratio, so Composition of finite Lorentz transformations from infinitesimal transformations d _do. w -n - an a - dT - d(J dT -, are constants, with T a parameter. Then integration with respect to T (composition of infinitesimal transformations) gives a finite transformation L = exp [-~iTu 0 (w + ia)]. For T = 1, so (In = WT, a = aT, this reads L = exp [(a - i(Jn) 0 u /2]. (41.33) In the special case of a pure boost (no rotation; (J = 0), the exponential function is evaluated along the lines indicated in (41.18), with the result L = cosh (a/2) + (na 0 u) sinh (a/2). (41.34) \u00a741.4. THOMAS PRECESSION VIA SPIN OR ALGEBRA 1145 Here na = a./a is a unit vector in the direction of the boost. The corresponding Lorentz transformation itself is evaluated from the formula or X' = LXL* t' + (x'\u00b7 (7) = [cosh a/2 + (na'(7) sinh a/2][t + (x' (7)] X [cosh a/2 + (na'(7) sinh a/2]. (41.35) Simplify with the help of the relations cosh2(a/2) + sinh2(a/2) = cosh a, 2 sinh (a/2) cosh (a/2) = sinh a, and and on both sides of the equation compare coefficients of I and (7, to find t' = (cosh a)t + (sinh a)(na \u2022 x), x' = [(sinh a)nat + (cosh a)(na \u2022 x)na ] (\"in-line part of transformation\") (x' na)na ] (\"perpendicular part of x unchanged\"). + [x", " - (41.36) In this way one verifies that connected with the velocity itself by the relations the quantity a is the usual \"velocity parameter,\" (l - f3(l - (32)-l/2 = cosh a, (32)-l/2 = sinh a, f3 = tanh a. (41.37) That velocity parameters add for successive boosts in the same direction shows nowhere more clearly than in the representation (41.33) of the spin-transformation matrix: L(a2)L(a l) = exp [a 2(na \u2022 (7)/2] exp [al(na \u2022 (7)/2] = exp [(a 2 + al)(na \u2022 0\")/2] = L(a 2 + a l). (41.38) Turn from this special case, and ask how to get the resultant of two arbitrary Lorentz transformations, each of which is a mixture of a rotation and a boost. No simpler method offers itself to answer this question than to use formula (41.33) together with the equation (41.39) \u00a741.4. THOMAS PRECESSION VIA SPINOR ALGEBRA torque, but undergoing acceleration, changes its A spinning object, free of all direction as this direction is recorded in an inertial frame of reference. This is the Origin of Thomas precession: composition of two boosts is not a pure boost Derivation of Thomas precession using spin matrices 1146 41. SPINORS Thomas precession [see exercise 6.9 and first term in equation (40.33b)]. This preces sion accounts for a factor two in the effective energy of coupling of spin and orbital angular momentum of an atomic electron. In a nucleus it contributes a little to the coupling of the spin and orbit ofa nucleon. The evaluation of the Thomas precession affords an illustration of spin-matrix methods in action. The precession in question can be discussed quite without reference either to angular momentum or to mass in motion. It is enough to consider a sequence of inertial frames of reference Set) with these two features. (1) To whatever point the motion has taken the mass at time t, at that point is located the origin of the frame Set). (2) The inertial frame Set + dt) at the next succeeding moment has undergone no rotation with respect to the inertial frame Set), as rotation is conceived by an observer in that inertial frame. However, it has undergone a rotation (\"Thomas precession\") as rotation is conceived and defined in the laboratory frame of reference. How is it possible for \"no rotation\" to appear as \"rotation\"? The answer is this: one pure boost, followed by another pure boost in another direction, does not have as net result a third pure boost; instead, the net result is a boost plus a rotation. This idea is not new in kind. Figure 41.1 illustrated how a rotation about the z-axis followed by a rotation about the x-axis had as resultant a rotation about an axis with not only an x-component and a z-component but also a y-component. What is true of rotations is true of boosts: they defy the law for the addition of vectors. Let the frame So coincide with the laboratory frame, and let the origin of this laboratory frame be where the moving frame is at time t. Let Set) be a Lorentz frame moving with this point at time t. Let one pure boost raise its velocity relative to the laboratory from 13 to 13 + df3. The resulting final configuration cannot be reached from So by a pure boost. Instead, first turn So relative to the laboratory frame (\"rotation R associated with the Thomas precession\") and then send it by a simple boost to the final configuration. Only one choice of this rotation will be right to produce match-up. Thus, distinguishing the spin matrices for pure boosts and pure rotations by the letters Band R, one has the relation B(f3 + df3)R(w dt) = \"B(df3)\"B(f3) (41.40) out of which to find the angular velocity w of the Thomas precession. The quotation marks in \"B(df3)\" carry a double warning: (1) the velocity of transformation that 13 = df3 (law of vector addition-or boosts Set) to Set + dt) is not (13 + df3) subtraction-not applicable to velocity), and (2) \"B(df3)\" does not appear as a pure boost in the laboratory frame. It appears as a pure boost only in the comovinrframe. Take care of the second difficulty first. It is only a difficulty because the fotmalism for combination of transformations, Ra = R 2R l, as developed in \u00a741.1 presupposes all operations R l, R 2, \u2022.\u2022, to be defined and", " carried out in the laboratory reference frame. In contrast, the quantity \"B(df3)\" is understood to imply a pure boost as defined and carried out in the comoving frame. Such an operation can be fitted into the formalism as follows. (1) Undo any velocity that the object already has. In other words apply the operator B-l(f3). Then the object is at rest in the laboratory frame. Then apply the necessary small pure boost, B(acomoving dT), where acomoving \u00a741.4. THOMAS PRECESSION VIA SPINOR ALGEBRA 1147 is the acceleration as it will be sensed by the object and dr is the lapse of proper time as it will be sensed by the object. At the commencement of this brief acceleration the object is at rest relative to the laboratory. What is a pure boost to it is a pure boost relative to the laboratory. It is also a pure boost in the spin-matrix formalism. Then transform back from laboratory to moving frame. Thus have the relation \"B(df3)\" = B(f3)B(acomoving dT)B-l(f3). The equation for the determination of the Thomas precession now reads B(f3 + df3)R(w dt) = B(f3)B(acomoving dT) or, with all unknowns put on the left, (41.41) (41.42) R(w dt)B-l(acomoving dT) = B-l(f3 + df3)B(f3). (41.43) The first task, to replace the erroneous value of the velocity change (df3) by a correct value (aComoving dT), is now made part of the problem along with the evaluation of the Thomas precession itself. Principles settled, the calculation proceeds by inserting the appropriate expressions for all four factors in (41.43), and evaluating both sides of the equation to the first order of small quantities, as follows: I - (i dtw + dra)' u /2 = [cosh (a'/2) - (na, \u2022 u) sinh (a '/2)] X [cosh (aI2) + (na'u) sinh (aI2)]. (41.44) Here a and na are the velocity parameter and unit vector that go with the velocity 13; a' = a + da, and na' =na + dna' go with 13 + df3. Develop the righthand side of (41.44) by the methods of calculus, writing a' = a + da and na. = na + dna' and applying the rule for the differentiation of a product. Equate coefficients of -u12 and -iul2 on both sides of the equation. Thus find acomoving d-r = (da)na + (sinh a) dna (41.45) and (41.46) Angular velocity of Thomas precession The one expression gives the change of velocity as seen in a comoving inertial frame. The other gives the precession as seen in the laboratory frame. For low velocities the expression for the Thomas precession reduces to w = a X /3/2. (41.47) Here a is the acceleration. Only the component perpendicular to the velocity 13 is relevall} for the precession. For an elementary account of the importance of the Thomas precession in atomic physics, see, for example, Ruark and Urey (1930). y Orientation-entanglement relation 1148 41. SPINORS \"-AJ-- Figure 41.5. \"Orientation-entanglement relation\" between a cube and the walls of a room. A 360 0 rotation of the cube entangles the threads. A 720 0 rotation might be thought to entangle them still more-but instead makes it possible completely to disen tangle them. \u00a741.5. SPINORS Paint each face of a cube a different color. Then connect each corner of the cube to the corresponding corner of the room with an elastic thread (Figure 41.5). Now rotate the cube through 2'lT = 360 0 \u2022 The threads become tangled. Nothing one can do will untangle them. It is impossible for every.thread to proceed on its way in a straight line. Now rotate the cube about the same axis by a further 2'lT. The threads become still mure tangled. However, a little work now completely straightens out the tangle (Figure 41.6). Every thread runs as it did in the beginning in a straight line from its corner of the cube to the corresponding corner of the room. More generally, rotations by 0, -+-4'lT, -+-8'lT,..., leave the cube in its standard \"orienta tion-ent", "anglement relation\" with its surroundings, whereas rotations by -+-2'lT, -+-6'lT, -+-1O'lT,..., restore to the cube only its orientation, not its orientation-entanglement relation with its surroundings. Evidently there is something about the geometry of orientation that is not fully taken into account in the usual concept of orientation; hence the concept of \"orientation-entanglement relation\" or (briefer term!) \"version\" (Latin versor, turn). Whether there is also a detectable difference in the physics (contact potential between a metallic object and its metallic surroundings, for exam ple) for two inequivalent versions of an object is not known [Aharonov and Susskind (1967)]. In keeping with the distinction between the two inequivalent versions of an object, the spin matrix associated with a rotation, R = cos (0/2) - i(n\u00b7 0\") sin (0/2), (41.48) t reverses sign on a rotation through an odd multiple of 2'lT. This sign change never shows up in the law of transformation of a vector, as summarized in the formula x --+ X' = RXR* (41.49) Spinor defined (two factors R; sign change in each!). The sign change does show up when one turns from a vector to a 2-component quantity that transforms according to the law (41.50) \u00a741.5. SPINORS 1149 2 5 8 3 6 9 4 7 Figure 41.6. An object is connected to its surroundings by elastic threads as in Figure 41.5. (Eight are shown here; any number could be used.) Rotating the object through 720 0 and then following the procedure outlined (Edward McDonald) in frames 2-8 (with the object remaining fixed), one finds that the connecting threads are left disentangled, as in frame 9 (lower right). Such a quantity is known as a spinor. A spinor reverses sign on a 360 0 rotation. It therefore provides a reasonable means to keep track of the difference between the two inequivalent versions of the cube. More generally, with each orientation entanglement relation between the cube and its surroundings one can associate a different value of the spinor f Moreover, there is nothing that limits the usefulness of the spinor concept to rotations. Also, for the general combination of boost and rotation, one can write (~1.5I) Lorentz transformation of a spinor 1150 41. SPINORS When the boost and rotation are both infinitesimal, the explicit form of this trans formation is simple: ~' = [1 - (i d(J/2)(n \u2022 0) + (df312)' O\"]~, or, according to (41.1), (~'l) = ~'2 1 + ~ (-i(Jzy + (3z) ~ (-i(JyZ + (Jzz + (3z + i{3y) (::) (41.52) For any combination of a boost in the z-direction of any magnitude and a finite rotation about the z-axis, one has (41.53) To keep track of the two components of the spinor, it is convenient and customary to introduce a label (capital Roman letter near beginning of alphabet) that takes on the values 1 and 2; thus (41.51) becomes (41.54) The spinor has acquired a significance of its own through one's having pulled out half of the transformation formula X' = LXL*. (41.55) Second type of spinor To be able to recover this formula, one requires the other half as well. It contains the conjugate complex of the Lorentz transformation. Therefore introduce another spinor 1/ that transforms according to the law [if = 1,2; V= i,2; dots and capital letters near the end of the alphabet are used to distinguish components that transform according to the conjugate complex (no transpose!) of the Lorentz spin matrix]. (41.56) \u00a741.6. CORRESPONDENCE BETWEEN VECTORS AND SPINORS Vector regarded as a Hermitian second-rank spinor To go back from spinors to vectors, note that the spin matrix X in (41.55) has the form _. _II (t + z) t + (x 0\") _ (x + 'y) X - 'y) II_II Xli XliII _.,. X21 X22 (x (t - z) (41.57) \u00a741.7. SPINOR ALGEBRA 1151 where the labels receive dots or no dots according as they are coupled in (41.55) to L* or to L. That equation of transformation becomes (41.58) (transpose obtained automatically by ordering of indices", "; thus LuYo not L*uv), The coefficients in this transformation are identical with the coefficients in the law for the transformation of a \"second-rank spinor with one index undotted and the other dotted:\" (41.59) In this sense one can say that \"a 4-vector transforms like a second-rank spinor.\" To be completely explicit about this connection between a 4-vector and a second-rank spinort note from (41.57) the relations Xli = X o + x3, X12 = Xl ix 2, X2i = Xl + ix 2, X22 = X o _ x3. (41.60) In a more compact form, one has XAU = [t + (x\u00b7 U)]AU = xf.La/u (41.61) where ao is the unit matrix. This equation tells immediately how to go from the components of a 4-vector, or \"I-index tensor,\" to the components of the correspond ing \"l,l-spinor\" (one undotted and one dotted index). With each rea14-vector x Q is associated a l,l-spinor that is Hermitian in the sense that (41.62) An example of a Hermitian l,l-spinor is provided by (41.61). The concept of Hermiticity can be stated in other words, and more generally. Associated with any N,N-spinors and Hermiticity N,N-spinor tP with components tPA,...A,A\u00b7\u00b7\u00b7~v is the conjugate complex spinor ;p with An N,N-spinor is said to be Hermitian when it is equal to its conjugate complex. (41.63) \u00a741.7. SPINOR ALGEBRA Equation (41.53) showed the component fl of a spinor rising exponentially with a boost in proportion to the factor e 1l2f3:, and the other component, ~'2 falling exponentially. If from two spinors ~ and ~, there is to be any quantity constructed Spinor algebra: which is unaffected in value by the boost, it must be formed out of such pr?ducts 1152 41. SPINORS (1) fAB, f AB defined as ~1~2 and ~2~1. One can restate this product prescnptlOn in other language. Introduce the alternating symbols fAB and f AB such that f 12 = f 12 = 1 and the only other nonvanishing components being f21 = f 21 = - 1. Define the lower label spinor ~A in terms of the upper-label spinor e by the equation (41.64) (2) raising and lowering spinor indices with the inverse Then the scalar product of one spinor by another is defined to be (3) scalar products of spinors (41.65) (41.66) (41.67) The value of this scalar product is unaffected by any boost or rotation or combination thereof: ~/ArA = fBfBA~'A =(LBD~D)fBA(LAC~C) = (det L)~DfDC~C =~c~c. (41.68) The proof uses the fact that the expression LBDfBA LAC (l) vanishes when D = C, and (2) reduces to the determinant of L (unity!) or its negative when D = 1, C = 2, or D = 2, C = 1. Note that the scalar product ~A~A is the negative of the scalar product ~A~A. The value of the scalar product of a spinor with itself is automatically zero (\"built-in null character of a spinor\"). The components of a vector with upper index have been expressed in terms of the components of a l,l-spinor with upper indices (41.69) (4) the mapping between vectors and 1, 1-spinors and a similar correlation holds between vector and l,l-spinor with lower indices; t us, h. i (5) Oil defined and related Here the \"associated basic spin matrices\" have the components to Oil (41.70) (JI'. AU -., nf.LP(J BYf \u2022 f \u00b7 \u00b7 BA VU' (41.71) \u00a741.7. SPINOR ALGEBRA or, explicitly, r~.a~1211 = a~:: a~22 1153 (41.72) ~ II for ~ = 0, ~ II for ~ = 1, -~ II for ~ = 2, -II~ -II~ + II~ -II~ _~ II for ~ = 3. The same type of multiplication law holds for these matrices, (a Z )2 =", "_ __ _ ~A new = ~A, TjA new = TjA + a~A, (41.85) with a any real or complex constant, as one checks at once by substitution into (41.81) or (41.84). The most general \"spinor mate\" to a given spinor e, satisfying the normalization condition (41.81), has this form (41.85). \u00a741.9. SPINOR VIEWED AS FLAGPOLE PLUS FLAG 1157 Figure 41.7. Spinor represented by (I) \"flagpole\" [Penrose terminology; track of pulse of light; null vector 19q>] plus (2) \"flag\" [arrow (q> - ) flashed onto surface of moon by laser pulse from earth or, in expanded view in the inset above, a flag itself, substi tuted for the arrow] plus (3) the orientation-entanglement relation between the flag and its surroundings [symbolized by strings drawn from corners of flag to surround ings]. When the spinor itself is mUltiplied by a factor peiu, the components of the null vector (flagpole) are multiplied by the factor p2 and the flag is rotated through the angle 20 about the flagpole. \u00a741.9, SPINOR VIEWED AS FLAGPOLE PLUS FLAG PLUS ORIENTATION-ENTANGLEMENT RELATION How can one visualize a spinor? Aim the laser, pull the trigger, and send a megajoule pulse from the here and now (event e) to the there and then (event '!i': center of the crater Aristarchus, 400,000 km from 19 in space, and 400,000 km from 19 in light-travel time). The laser has been designed to produce, not a mere spot of light, but an illuminated arrow. Following Roger Penrose, speak of the null vector 199 as a \"flagpole,\" and of the illuminated arrow as a \"flag.\" A spinor (Figure 41.7) consists of this combination of (l) null flagpole plus (2) flag plus (3) the orientation- Geometric representation of a spinor: 1158 41. SPINORS entanglement relation between the flag and its surroundings. \"Rotate the flag\" by repeatedly firing the laser, with a bit of rotation of the laser about its axis between one firing and the next. When the flag has turned through 360 0 and has come back to its original direction, the spinor has reversed sign. A rotation of the flag about the flagpole through any even multiple of 2'lT restores the spinor to its original value. One goes from a spinor ~, a mathematical object with two complex components e and ~2, to the geometric object of \"flagpole plus flag plus orientation-entanglement relation\" in two steps: first the pole, then the flag. Thus, go from the spinor ~A to the real null 4-vector of the \"pole\" by way of the formula (1) null vector (flagpole), plus or (t + z) (x + iy) II (x - 0')11 = II ~1!~ e!~ II. e~l e~2 (t - z) (41.86) (41.87) (2) bivectOr (flag) and its orientation-entanglement relation The matrix on the right has its first row identical up to a factor e/e with its second row. Therefore the determinant of the matrix on the right vanishes. So also for the left. Therefore the 4-vector eq> = (t, x,y, z) is a null vector. One \"stretches\" this vector by a factor p2 when one multiplies the spinor ~A by the nonzero complex number A = pe iu (p,o real); however, the vector is unchanged in direction. The 4-vector is also unaffected by the choice of the angle o. In other words, this null 4-vector is uniquely fixed by the spinor; but the spinor is not fixed with all uniqueness by the 4-vector. To a given 4-vector corresponds a whole family of spinors. They differ from one another by a multiplicative phase factor of the form e iu (\"flag factor\"). Looking further to see the influence of the flag factor showing up, turn from a real vector (four components) generated out of the spinor ~A to a real bivector (six components) generated out of the same spinor:. Jl---+ AU; v ---+ BV.. (41.88) That this quantity has no more than six distinct components (PL' = - PIl) follows from interchanging A with B and if with V, and noting the resultant change in sign on the righthand side of (41.88). To", " unfold the meaning of this bivector, look in (41.88) for every app~arance of the alternating factor fAB. Wherever such a factor appears, insert the expression (41.84) for this factor in terms of the starting spinor ~A and insert the additional spinor 1jA that is needed, along with ~A, to supply a basis for all spinors. In this way, find p' ---+ FABUV = ~A~BaiJ.rjV _ TiaF) + (~A1jB _ 1jA~B)fUfV (eTi u + 1jAfl)~Bfv = efU(~BTiV + 1jBfV) - (41.89) \u00a741.9. SPINOR VIEWED AS FLAGPOLE PLUS FLAG 1159 Thus the 2,2-spinor built from e represents a bivector constructed out of the two 4-vectors x and y. Of these, the first is the \"real null vector of the flagpole,\" already seen to be determined uniquely by the spinor ~A. The second vector, is also determined by ~A, but not uniquely, because the \"spinor mate,\" TjA, to ~A is not unique. Go from one choice of mate, TjA, to a new choice of mate (equation 41.85), (41.90) Then the real 4-vector yJ' goes to the new real 4-vector t yJ'new = yJ' + (a + a)xl'. (41.91) (41.92) Were the 4-vector y unique, there would project out from the flagpole, not a flag but an arrow. The range of values open for the real constant a + a makes one arrow into many arrows, all coplanar; hence the flag of Penrose. Otherwise stated, the choice of a spinor e fixes no individual arrow, but does fix the totality of the collection of arrows, and thus uniquely specifies the flag. The 4-vector y (and with it Ynew) is orthogonal to the null 4-vector x, x' Y = x!3y!3 = - ~ XAUyAU _. -. = - -~A~U(~A1jU + TjA~U) = 0, 1 2 and spacelike, (41.93) (41.94) (\"unit length of flag\"). Multiplication of the spinor ~A by the \"flag factor\" e iu rotates the flag about the Rotation of flag about ___________ -~le--by tirelfrtgle ~o,because the spinor mate, TjA, of ~A is multiplied by the factor e-iu [see the normalization c;ondition (41.81)]. These changes alter the vector Y to a rotated vector Yrot' with flagpole --+- y'\" cos 20 + z'\" sin 20. (41.95) Here the 4-vector z shares with the vector Y these properties: IS (1) real, (2) spacelike, (3) of unit magnitude, (4) orthogonal to the null 4-vector x of the flagpole, and (5) uniquely specified by the original spinor ~A up to the additive real It Equations relating spinor, flagpole, and flag 1160 41. SPINORS multiple (a + a) of x. In addition, z and yare orthogonal. Thus y and z provide basis vectors in the two-dimensional space in which-to overpictorialize-the \"tip of the flag\" undergoes its rotation. Recapitulate by returning to the laser pulse. Two numbers, such as the familiar polar \"angles 0 (angle with the z-axis) and \u00a2 (azimuth around z-axis from x-axis) tell the direction of its flight. A third number, r, gives the distance to the moon and also the travel time for light to reach the moon. A fourth number, an angle 1/;, tells the azimuth of the illuminated arrow shot onto the surface of the moon, this azimuth to be measured from the e(j direction (where I/; = 0), around the flagpole in a righthanded sense. Then the spinor associated with the flagpole plus flag (rotated arrow) is ( ~1) e - - (2r)l/2 (COS (0/2)e- i sin (0 /2)e i r/>/2+ i.;,l2 \u00a212 + i.;,l2) (41.96) according to the conventions adopted here [see (41.87)]. The mate 1/A to this spinor, unique up to an additive", " multiple of e, is 1/ 1) 1/2 ( = (2r)-1/2 (-Sin (0 /2)e- i r/>/2- i.;,l2) cos (0/2)e i r/>/2-i.;,l2. (41.97) The 4-vector of the flagpole determined by e is found from (41.87): ::) = (s~n;c~s 4\u00bb ( x 2 x 3 r sm 0 sm 4> rcosO. (41.98) To determine the flag itself, one requires, in addition to x\"', the unit spacelike 4-vector y\"', normal to x\"', and unique up to an additive real multiple of x\"'. This vector is evaluated by use of (41.90) and has the form YO) yl y2 y3 ( - ( cos 0 cos 4> cos I/; + sin 4> sin I/; ) 0 cos 0 sin 4> c~s I/; - cos 4> sin I/; -smO cos I/;. (41.99) From these expressions for xfJ. and yfJ., one calculates the components of the 1\u00a5vector (\"flag\") FfJ.\u00b7 = xfJ.y\" - yfJ.x\u00b7 by simple arithmetic. \u00a741.10. APPEARANCE OF THE NIGHT SKY; AN APPLICATION OF SPINORS Attention has gone here to extracting all four pieces of information contained in a spinor: separation in time (equal to separation in space), direction in space, and \u00a741.10. APPEARANCE OF NIGHT SKY ANALYZED BY SPINORS 1161 Figure 41.8. Representation of a direction in space (one of the stars of the Big Dipper, regarded as a point on the celestial sphere) as a point in the complex r plane (r = ratio (2/~1 of spinor components) by stereographic projection from the South Pole. rotation about that direction. Turn now to an application where not all that informa tion is needed. Look at the night sky and ask (1) how to describe its appearance and (2) how to change that appearance. As one way to describe its appearance, give the direction of each star. Abandon any concern about the distance of the star, and any concern about any rotation 1/J about the flagpole. In other words, the complex factor Spinors used to analyze \"Lorentz transformations\" of appearance of night sky common to ~1 and ~2 drops from attention. All that is left as significant is the ratio r of these spinor components: r = ~2/~1 = tan (O/2)e i r/>. (41.100) To give the one complex number r (\"stereographic coordinate;\" Figure 41.8) for each star in the sky is to catalog the pattern of the stars. Let the observer change his stance. The celestial sphere appears to rotate. Or let him rocket past his present location in the direction of the North Star with some substantial fraction of the velocity of light. To him all that portion of the celestial sphere is opened out, as if by a magnifying glass. To compensate, the remaining stars are packed into a smaller angular compass. Any such rotation or boost or combination of rotation and boost being described in spinor language by a transfor mation of the form (41.101) 1162 41. SPINORS implies a transformation of the complex stereographic coordinate of any given star of the form (41.102) In the special case of a boost in the z-direction with velocity parameter a (velocity f3 = tanh a), the off-diagonal components Liz and L21 vanish. The magnification of the overhead sky then expresses itself in the simple formula or 4>new = 9, tan (Onew/2) = eQ tan (0/2). (41.103) is Contrary to this prediction and false expectation, no magnification at all achieved of the regions around the North Star by moving with high velocity in that direction. On the contrary, any photon coming in from a star a little off that direction, with a little transverse momentum, keeps that transverse momentum in the new frame; but its longitudinal momentum against the observer is augmented by his motion. Thus the ratio of the momenta is decreased, and the observed angle relative to the North Star is also decreased. The consequence is not magnification, but diminution (\"looking through the wrong end of a telescope\"). The correct formula is not (41.103) but tan (Onew/2) = e-Q tan (0/2) (41.104) (reversal of the sign of a). The reason for this correction is not far to seek. The spinor analysis so far had dealt with an outgoing light pulse,", " and a 4-vector with positive time component. That feature was built into the formula adopted to tie the spinor to the 4-vector, (41.105) In contrast the 4-vector that reaches back to the origin of an incoming photon has a time component that is negative (or, alternatively, sign-reversed space compo nents)! For any null 4-vector with negative time component, one employs instead of (41.105) the formula (~1.106) It is enough to mention here this point of principle without going through the details that give the altered sign for a in (41.104). From now on, to preserve the previous arithmetic, change the problem. Deal, not with incoming photons, but with outgoing photons. Replace the receiving telescope by the projector of a planetarium. It projects out into space a separate beam of light for each star of the Big Dipper and also one for the North Star itself. Let an observer move in the positive z-direction with velocity parameter a. In his frame of reference the beams actually will be widened out in full accord with (41.103). \"The magnification process changes the size of the Big Dipper but not its shape.\" Lorentz transformations leave angles on sky unchanged invariance\") (\"conformal \u00a741.10. APPEARANCE OF NIGHT SKY ANALYZED BY SPINORS 1163 This statement is at the same time true and false. It is true of the Dipper and of any other constellation to the extent that the angular dimensions of that constellation can be idealized to be small compared to the entire compass of the sky. It is false in the sense that any well-rounded projected constellation, however small it may appear to an observer at rest with respect to the earth, can always be so \"opened out\" by the observer putting on any sufficiently high velocity, the observer still being near the earth, that the constellation encompasses a major fraction of the sky. That the \"Lorentz-transformation-induced magnification\" of a small object does not change its shape can be seen in three ways. (1) Stereographic projection (Figure 41.8) and \"fractional linear transformation\" (41.102) are both known to leave all angles unchanged [\"conformal invariance;\" see for example Penrose (1959)] and known even to turn every old circle into a new circle. (2) Consider a given star, M, in the constellation and immediate neighbors, Land N, just below it and just above\u00a3it in the count of the members of that constellation. Consider the flagpole pointed at M and the flag pointed first from M to L, then from M to N. The flag has turne<J....about the flagpole through an angle..p. The two corresponding spinors therefore differ by a phase factor e i if,/2. They differ in no other way. After an arbitrary Lorentz transformation they still differ by the phase factor e i if,/2, and in no other way. The angle between the arcs ML and MN on the celestial sphere therefore remains at its original value..p after the Lorentz transformation (again conformal invariance of patterns on the celestial sphere!). (3) An even more elementary calcu lation shows that infinitesimal arc lengths on the unit celestial sphere in the direction of increasing 0 and arc lengths in the direction of increasing 4> are magnified in the same proportion, thus leaving unchanged the angle between arc and arc (confor mal invariance). Thus, consider a photon shot out from the planetarium projector to a point on the celestial sphere (\"planetarium version of a Big-Dipper star\") with inclination 0 to the z-axis, as seen by an observer at rest relative to the earth. From the standard laws of transformation of angles in a Lorentz transformation (\"aberra tion\"; Box 2.4), one has for the sine of the transformed angle. sm 0new = 1 (l - f32)l/2. f3 cos 0 sm 0 (41.107) and (by differentiating the expression for the cosine of the transformed angle) dOnew = 1 (l - - f32)l/2 f3 cos 0 dO. (41.108) From these expressions it follows at once that the inclination, relative to a meridian line, on the transformed celestial sphere is identical to the direction, relative to the same meridian line, on the original celestial sphere: new tan inclination - ) _ sin 0new d9new dO ( = ) new original. mc mahon l' = tan. ( sinO de? dO (41.109) (again conformal invariance!). Spinor formalism in curved spacetime Spinors needed when analyzing fermions in gravitational fields Equivalence of spinor and tensor formalisms 1164 41. SPINORS So much for the elementary spinor and what it has to", " do with a null vector, with a \"flagpole\" pointed to the celestial sphere, and with rotation of a \"flag\" about such a flagpole. \u00a741.11. SPINORS AS A POWERFUL TOOL IN GRAVITATION THEORY Just as vectors, tensors, and differential forms are easily generalized from flat space time to curved, so are spinors. Each event 9 in curved spacetime possesses a tangent space. In that tangent space reside and operate all the vectors, tensors, and forms located at 9. The geometry of the tangent srace is Lorentzian (\"local Lorentz geometry at 9\"), since the scalar product of any two vectors u and v at 9, expressed in an orthonormal frame at '1', is u' v = g(u, v) = T/aPUav P. Thus, there is no mathematical difference between the tangent space at 9 on the one hand, and flat spacetime on the other. Whatever mathematical can be done in the one can also be done in the other. In particular, the entire formalism ofspinors, developed originally in flat spacetime, can be carried over without change to the tangent space at the arbitrary event 9 in curved spacetime. Let it be done. Now spinors reside at every event in curved spacetime; and at each event one can translate back and forth between spinor language and tensor language, using the equations (valid in orthonormal frames) of \u00a7\u00a741.6 and 41.7. Spinors in curved spacetime are an indispensible mathematical tool, when one wishes to study the influence of gravity on quantized particles of half-integral spin (neutrinos, electrons, protons,...). Consider, for example, Hartle's (1971) proof that a black hole cannot exert any long-range, weak-interaction forces on external matter (i.e., that a black hole has no \"weak-interaction hair\"). His proof could not function withQut a spinor description of neutrino fields in curved spacetime. Similarly for Wheeler's (1971 b) analysis of the quasibound states of an electron in the gravitational field of a small black hole (gravitational radius _10- 13 cm): it requires solving the Dirac equation for a spin-~ particle in the curved spacetime geometry of Schwarzs child. For a detailed discussion of the Dirac equation in curved spacetime see, e.g., Brill and Wheeler (1957). To use the mathematics of spinors, one need not be dealing with quantum theory or with particles of half-integral spin. The spinor formalism is perfectly applicable in situations where only integral-spin entities (scalars, vectors, tensors) are in view, and where in fact, the spinor formalism is fully equivalent to the tensor formalism that pervades earlier chapters of this book. Equations (41.77) and (41.78) provide the translation from one formalism to the other, once an orthonormal frame has been chosen at each event in spacetime. Certain types of problems in gravitation theory are far more tractable in the language of spinors than in the language of tensors. Examples are as follows. \u00a741.11. SPINORS AS A POWERFUL TOOL IN GRAVITATION THEORY 1165 (1) Geometric Optics (the theory of \"null congruences of geodesics\") Here spinors make almost trivial the lengthy tensor algebra needed in derivations of the \"focusing theorem\" [equation (22.37)]; and they yield an elegant, simple formalism for discussing and calculating how, with increasing affine parameter, a bundle of rays alters its size (\"expansion\"), its shape (\"shear\"), and its orientation (\"rotation\"). See, e.g., Sachs (1964), Pirani (1965), or Penrose (1968a) for a review and the original references. Applications of spinor formalism in classical gravitation theory (2) Radiation Theory in Curved Spacetime (both gravitational and electromagnetic) Spinors provide the most powerful of all formalisms for decomposing radiation fields into spherical harmonics and for manipulating their decomposed components. See, for example, Price's (1972a,b) analysis of how a perturbed Schwarzschild black hole radiates away all its radiatable perturbations, be they electromagnetic perturbations, gravitational perturbations, or perturbations in a fictitious field of spin 17; see, similarly, the analysis by Fackerell and Ipser (1972) and by Ipser (1971) of electro magnetic perturbations of a Kerr black hole, and the analysis by Teukolsky (1972) of gravitational perturbations of a Kerr hole. Spinors also yield an elegant and powerful analysis of the \"l/r\" expansion of a radiation field flowing out", " from a source into asymptotically flat space. Among its results is a \"peeling theorem,\" which describes the algebraic properties of the coefficients in a l/r expansion of the Riemann tensor. See, e.g., Sachs (1964) or Pirani (1965) for reviews and original references. (3) Algebraic Properties of Curvature Tensors The spinor formalism is a more powerful method than any other for deriving the \"Petrov-Pirani algebraic classification of the conformal curvature tensor,\" and for proving theorems about algebraic properties of curvature tensors. See, e.g., Sachs (1964) or Pirani (1965) or Penrose (1968a) for reviews and references. - - Of course, the spinor formalism, like any formalism, has its limitations. For ---example, ma.ny ofthe elementary problems ofgravitation theory, and a large fraction of the most difficult ones, would be more difficult in the language of spinors than in the language of tensors! But for certain classes of problems, especially those where null vectors playa central role, spinors are a most valuable tool. Cartan gave spinors to the world's physics and mathematics. His text (American edition, 1966) is an important reference to the subject. CHAPTER 42 REGGE CALCULUS This chapter is entirely Track 2. As preparation for it, Chapter 21 (variational principle and initial-value formalism) is needed. It is not needed as preparation for any later chapter, though it will be helpful (dynamics of geometry). in Chapter 43 The need for Regge calculus as a computational tool \u00a742.1. WHY THE REGGE CALCULUS? Gravitation theory is entering an era when situations of greater and greater com plexity must be analyzed. Before about 1965 the problems of central interest could mostly be handled by idealizations of special symmetry or special simplicity or both. The Schwarzschild geometry and its generalizations, the Friedmann cosmology and its generalizations, the joining together of the Schwarzschild geometry and the Friedmann geometry to describe the collapse of a bounded collection of matter, the vibrations of relativistic stars, weak gravitational waves propagating in an otherwise flat space: all these problems and others were solved by elementary means. But today one is pressed to understand situations devoid of symmetry and not amenable to perturbation theory: How do two black holes alter in shape, and how much gravitational radiation do they emit when they collide and coalesce? What are the structures and properties of the singularities at the endpoint of gravitational collapse, predicted by the theorems of Penrose, Hawking, and Geroch? Can a Universe that begins completely chaotic smooth itself out quickly by processes such as inhomogeneous mixmaster oscillations? To solve such problems, one needs new kinds of mathematical tools-and in response to this need, new tools are being developed. The \"global methods\" of Chapter 34 provide one set of tools. The Regge Calculus provides another' [Regge (1961); see also pp. 467-500 of Wheeler (1964a)]. \u00a742.2. REGGE CALCULUS IN BRIEF Approximation of smooth geometries by skeleton structures Consider the geodesic dome that covers a great auditorium, made of a multitude of flat triangles joined edge to edge and vertex to vertex. Similarly envisage space time, in the Regge calculus, as made of flat-space \"simplexes\" (four-dimensional Role of Einstein field equation in fixing the skeleton Structure \u00a742.3. SIMPLEXES AND DEFICIT ANGLES 1167 item in this progression: two dimensions, triangle; three dimensions, tetrahedron; four dimensions, simplex) joined face to face, edge to edge, and vertex to vertex. To specify the lengths of the edges is to give the engineer all he needs in order to know the shape of the roof, and the scientist all he needs in order to know the geometry of the spacetime under consideration. A smooth auditorium roof can be approximated arbitrarily closely by a geodesic dome constructed of sufficiently small triangles. A smooth spacetime manifold can be approximated arbitrarily closely by a locked-together assembly of sufficiently small simplexes. Thus the Regge calculus, reaching beyond ordinary algebraic expressions for the metric, provides a way to analyze physical situations deprived, as so many situations are, of spherical symme try, and systems even altogether lacking in symmetry. If the designer can give the roof any shape he pleases, he has more freedom than the an~yst who is charting out the geometry of spacetime. Given the geometry of spacetime up to some spacelike slice that, for want of a better name, one may call \"now,\" one has no freedom at all in the geometry from that instant on. Einstein's geometrodynamic law is fully deter", "ministic. Translated into the language of the Regge calculus, it provides a means to calculate the edge lengths of new simplexes from the dimensions of the simplexes that have gone before. Though the geometry is deterministically specified, how it will be approximated is not. The original spacelike hypersurface (\"now\") is approximated as a collection of tetrahedrons joined together face to face; but how many tetrahedrons there will be and where their vertices will be placed is the option of the analyst. He can endow the skeleton more densely with bones in a region of high curvature than in a region of low curvature to get the most \"accuracy profit\" from a specified number of points. Some of this freedom of choice for the lengths of the bones remains as one applies the geometro dynamic law in the form given by Regge (1961) to calculate the future from the past. This freedom would be disastrous to any computer program that one tried to write, unless the programmer removed all indefiniteness by adding supplementary conditions of his own choice, either tailored to give good \"accuracy profit,\" or otherwise fixed. Having determined the lengths of all the bones in the portion of skeletonized spacetime of interest, one can examine any chosen local cluster of bones in and by themselves. In this way one can find out all there is to be learned about the geometry in that region. Of course, the accuracy of one's findings will depend on the fineness with which the skeletonization has been carried out. But in principle that is no limit to the fineness, or therefore to the accuracy, so long as one is working in the context of classical physics. Thus one ends up with a catalog of all the bones, showing the lengths of each. Then one can examine the geometry of whatever spacelike surface one pleases, and look into many other questions besides. For this purpose one has only to pick out the relevant bones and see how they fit together. \u00a742.3. SIMPLEXES AND DEFICIT ANGLES Figure 42.1 recalls how a smoothly curved surface can be approximated by flat triangles. All the curvature is concentrated at the vertices. No curvature resides at 1168 42. REGGE CALCULUS Deficit angle as a skeletonized measure of curvature: (1) in two dimensions (2) in n (or four) dimensions Figure 42.1. A 2-geometry with continuously varying curvature can be approximated arbitrarily closely by a polyhedron built of triangles, provided only that the number of triangles is made sufficiently great and the size of each sufficiently small. The geometry in each triangle is Euclidean. The curvature of the surface shows up in the amount of deficit angle at each vertex (portion ABeD of polyhedron laid out above on a flat surface). the edge between one triangle and the next, despite one's first impression. A vector carried by parallel transport from A through Band C to D, and then carried back by another route through C and B to A returns to its starting point unchanged in direction, as one sees most easily by laying out this complex of triangles on a fiat surface. Only if the route is allowed to encircle the vertex common to A, B, C, and D does the vector experience a net rotation. The magnitude of the rotation is equal to the indicated deficit angle, 8, at the vertex. The sum of the deficit angles over all the vertices has the same value, 4'17, as does the half-integral of the continuously distributed scalar curvature (2)R = 2/a2 for a sphere of radius a) taken over the entirety of the original smooth figure, 2: skeleton geometry 8i = 1 J 2 actual smooth geometry (2)R d(surface) = 4'17. (42.1 ) Generalizing from the example of a 2-geometry, Regge calculus approximates a smoothly curved n-dimensional Riemannian manifold as a collection of n-dimen sional blocks, each free of any curvature at all,joined by (n - 2)-dimensional regions the curvature is concentrated (Box 42.1). For the four-dimensional in which all spacetime of general relativity, the \"hinge\" at which the curvature is concentrated has the shape of a triangle, as indicated schematically in the bottom row of Figure 42.2. In the example illustrated there, ten tetrahedrons have that triangle in common. Between one of these tetrahedrons and the next fits a four-dimensional simplex. Every feature of this simplex is determined by the lengths of its ten edges. One of the features is the angle a between one of the indicated tetrahedrons or \"faces\" of the simplex and the next. Thus a represents the angle subtended by this simplex \u00a742.4. SKELETON FORM OF FIELD EQUATIONS 1169 Box", " FIELD EQUATIONS 1171 Box 42.2 FLOW DIAGRAMS FOR REGGE CALCULUS A skeleton 4-geometry is completely determined by all its edge lengths. From the edge lengths one gets the integrated curvature by pursuing, for each hinge in the 4-geometry, the following flow diagram: cycle of blocks swinging on this hinge lone of these blocks I t the two tetrahedral \"faces\" that set this block off from the blocks before and after it in the cycle of blocks angle a between these two faces fixed by the block's n(n - 1)/2 edge lengths deficit angle at the given hinge is {) = 2'17 - '\" L..J blocks swinging on that hinge a contribution to integrated curvature (Box 42.1) is {) times area of hinge One finds it natural to apply this analysis in either of two ways. First, one can probe a given 4-geometry (given set of edge lengths!) in the sense edge lengths I t I curvature 1172 42. REGGE CALCULUS Box 42.2 (continued) Second-and this is the rationale of Regge calculus-one can use the skeleton calculus to deduce a previously unknown 4-geometry from Einstein's geometrody namic law, proceeding in the direction initial conditions translated into information about some of the edge lengths Einstein's equations expressed as con ditions on the curvature (deficit angle of each hinge) fix remainder of the edge lengths (apart from natural options in fineness of zoning) In the changes contemplated in this variational principle, certain edge lengths are thought of as being fixed. They have to do with the conditions specified at the boundaries of the region of spacetime under study. It is not necessary here to enter into the precise formulation of these boundary conditions, fortunately, since some questions of principle still remain to be clarified about the precise formulation of boundary conditions in general relativity (see \u00a721.12). Rather, what is important is the effect of changes in the lengths of the edges of the blocks in the interior of the region being analyzed, as they augment or decrease the deficit angles at the various hinges. In his basic paper on the subject, Regge (1961) notes that the typical deficit angle 8\" depends in a complicated trigonometric way on the values ofnumer ous edge lengths lp. However, he proves (Appendix of his paper) that \"quite reo markably, we can carry out the variation as if the 8\" were constants,\" thus reducing the variational principle to the form (1/8'17) 2: 8\" 8A\" = O. H hinges 1!=1 (42.5) Here the change in area of the h-th triangle-shaped hinge, according to elementary trigonometry, is (42.6) \u00a742.5. THE CHOICE OF LATIICE STRUCTURE 1173 In this equation 0ph is the angle opposite to the p-th edge in the triangle. Conse quently, Einstein's equations in empty space reduce in skeleton geometry to the form 8h cotan 0ph = 0, (p == 1,2,...), (42.7) Einstein field equation reduced to skeleton form 2: hinges that have the given edge pincommon one equation for each edge length in the interior of the region of spacetime being analyzed. \u00a742.5. THE CHOICE OF LATTICE STRUCTURE Two l4.uestions arise in the actual application of Regge calculus, and it is not clear that either has yet received the resolution which is most convenient for practical applications of this skeleton analysis: What kind of lattice to use? How best to capitalize on the freedom that exists in the choice of edge lengths? The first question is discussed in this section, the second in the next section. It might seem most natural to use a lattice made of small, nearly rectangular blocks, the departure of each from rectangularity being conditioned by the amount and directionality of the local curvature. However, such building blocks are \"floppy.\" One could give them rigidity by specifying certain angles as well as the edge lengths. But then one would lose the cleanness of Regge's prescription: give edge lengths, and give only edge lengths, and give each edge length freely and independently, in order to define a geometry. In addition one would have to rederive.the Regge equations, including new equations for the determination of the new angles. There fore one discards the quasirectangle in favor of the simplex with its 5\u00b7 4/2 == 10 edge lengths. This decided, one also concludes that even in flat spacetime the simplexes cannot all have identical edge lengths. Two-dimensional flat space can be filled with identical equilateral triangles, but already at three dimensions it ceases to be possible to fill out the manifold with identical equilateral tetrahedrons. One knows that a given carbon atom in", " diamond is joined to its nearest neighbors with tetrahedral bonds, but a little reflection shows that the cell assignable to the given atom is far from having the shape of an equilateral tetrahedron. Synthesis would appear to be a natural way to put together the building blocks: first make one-dimensional structures; assemble these into two-dimensional struc tures; these, into three-dimensional ones; and these, into the final four-dimensional structure. The one-dimensional structure is made of points, 1, 2, 3,..., alternating with line segments, 12, 23, 34,.... To start building a two-dimensional structure, pick up a second one-dimensional structure. It might seem natural to label its points 1', 2', 3',..., etc. However, that labeling would imply a cross-connection between 1 and 1', between 2 and 2', etc., after the fashion of a ladder. Then the elementary cells would be quasirectangles. They would have the \"floppiness\" that is to be excluded. Therefore relabel the points of the second one-dimensional structure as If, 2f, 3f, etc. The implication is that one cross-connects 2~' with points 2 and 3 of the original one-dimensional structure, etc. One ends up with something like the The choice of lattice structure: (1) avoiding floppiness (2) necessity for unequal edge lengths (3) construction of two dimensional structures y (4) 3-D structures built from 2-D structures by \"method of blocks\" (5) 3-D structures from 2-D by \"method of spheres\" 1174 42. REGGE CALCULUS girder structure of a bridge, fully rigid in the context of two dimensions, as desired. The same construction, extended, fills out the plane with triangles. One now has a simple, standard two-dimensional structure. One might mistakenly conclude that one is. ready to go ahead to build up a three-dimensional structure: the mistake lies in the tacit assumption that the flat-space topology is necessarily correct. Let it be the problem, for example, to determine the development in time of a 3-geometry that has the topology of a 3-sphere. This 3-sphere is perhaps strongly deformed from ideality by long-wavelength gravitational waves. A right arrangement of the points is the immediate desideratum. Therefore put aside for the present any consideration of the deformation of the geometry by the waves (alteration of edge lengths from ideality). Ask how to divide a perfect 3-sphere into two-dimensional sheets. Here each sheet is understood to be separated from the next by a certain distance. At this point two alternative approaches suggest themselves that one can call for brevity \"blocks\" and \"spheres.\" (1) Blocks. Note that a 3-sphere lets itself be decomposed into 5 identical, tetra hedron-like solid blocks (5 vertices; 5 ways to leave out anyone of these vertices!) Fix on one of these \"tetrahedrons.\" Select one vertex as summit and the face through the other three vertices as base. Give that base the two-dimensional lattice structure already described. Introduce a multitude of additional sheets piled above it as evenly spaced layers reaching to the summit. Each layer has fewer points than the layer before. The decomposition of the 3-geometry inside one \"tetrahedron\" is thereby accomplished. However, an unresolved question remains; not merely how to join on this layered structure in a regular way to the corresponding structure in the adjacent \"tetrahedrons\"; but even whether such a regular joinup is at all possible. The same question can be asked about the other two ways to break up the 3-sphere into identical \"tetrahedrons\" [Coxeter (1948), esp. pp. 292-293: 16 tetrahedrons defined by a total of 8 vertices or 600 tetrahedrons defined by a total of 120 verticesJ. One can eliminate the question of joinup of structure in a simple way, but at the price of putting a ceiling on the accuracy attainable: take the stated number of vertices (5 or 8 or 120) as the total number of points that will be employed in the skeletonization of the 3-geometry (no further subdivision required or admitted). Considering the boundedness of the memory capacity of any computer, it is hardly ridiculous to contemplate a limitation to 120 tracer points in exploratory calculations! (2) Spheres. An alternative approach to the \"atomization\" of the 3-sphere begins by introducing on the 3-sphere a North Pole and a South Pole and the hyperspherical angle X (X == 0 at the first pole, X == 'TT at the second, X == 'TT/2 at the equator; see Box 27.2). Let", " each two-dimensional layer lie on a surface of constant X (~ equal to some integer times some interval.1X). The structure of this 2-sphere is already to be regarded as skeletonized into elementary triangles (\"fully complete Buckminster Fuller geodesic dome\"). Therefore the number of \"faces\" or triangles F, the number of edge lengths E, and the number of vertices V must be connected by the relation of Euler: F _ E + V == (a topology-dependent ) == {2 for 2-sphere, (42.8) number or \"Euler character\" 0 for 2-torus. It follows from this relation that it is impossible for each vertex to sit at the center \u00a742.5. THE CHOICE OF LATIICE STRUCTURE 1175 of a hexagon (each vertex the point of convergence of 6 triangles). This being the case, one is not astonished that a close inspection of the pattern of a geodesic dome shows several vertices where only 5 triangles meet. It is enough to have 12 such 5-triangle vertices among what are otherwise all6-triangle vertices in order to meet the requirements of the Euler relation: n 5-triangle vertices V - n 6-triangle vertices F = (V - n)(6/3) + n(5/3) triangles E = (V - n)(6/2) + n(5/2) edges V = (V - n)(6/6) + n vertices Euler characteristic 2 = F - E + V = n/6 n = 12 (42.9) Among all figures with triangular faces, the icosahedron is the one with the smallest number of faces that meets this condition (5-triangle vertices exclusively!) If each 2-surface has the pattern of vertices of a geodesic dome, how is one dome to be joined to the next to make a rigid skeleton 3-geometry? Were the domes imbedded in a flat 3-geometry, rigidity would be no issue. Each dome would already be rigid in and by itself. However, the 3-geometry is not given to be flat. Only by a completely deterministic skeletonization of the space between the two 2-spheres will they be given rigidity in the context ofcurved space geometry. (1) Not by running a single connector from each vertex in one surface to the corresponding vertex in the next (\"floppy structure\"!) (2) Not by displacing one surface so each of its vertices comes above, or nearly ab6ve,- the center of a triangle in the surface \"below.\" First, the numbers of vertices and triangles ordinarily will not agree. Second, even when they do, it will not give the structure the necessary rigidity to connect the vertex of the surface above to the three vertices of the triangle below. The space between will contain some tetrahedrons, but it will not be throughout decomposed into tetrahedrons. (3) A natural and workable approach to the skeletonization of the 3-geometry is to run a connector from each vertex in the one surface to the corre sponding vertex in the next, but to flesh out this connection with additional structure intervening vertices and connectors as that will give rigidity to the 3-geometry: illustrated in Box 42.3. In working up from the skeletonization of a 3-geometry to the skeletonization ofa 4-geometry, it is natural to proceed similarly. (1) Use identical patterns of points in the two 3-geometries. (2) Tie corresponding points together by single connectors. (3) Halfway, or approximately halfway between the two 3-geometries insert a whole additional pattern of vertices. Each of these supplementary vertices is \"dual\" to and lies nearly \"below\" the center of a tetrahedron in the 3-geometry immediately above. (4) Connect each supplementary vertex to the vertices of the tetrahedron immediately above, to the vertices of the tetrahedron immediately below, and to those other supplementary vertices that are its immediate neighbors. (5) In this way get the edge lengths needed to divide the 4-geometry into simplexes, each of rigidly dt:fined dimensions. (6) 4-D structures built from 3-D structures 1176 42. REGGE CALCULUS Box 42.3 SYNTHESIS OF HIGHER-DIMENSIONAL SKELETON GEOMETRIES OUT OF LOWER-DIMENSIONAL SKELETON GEOMETRIES $' ''11' (1) One-dimensional structure as alternation of points and line segments. (2) Two-dimensional structure (a) \"floppy\" (unacceptable) and (b) rigid ified (angles of triangles fully", " determined by edge lengths). When this structure is extended, as at right, the \"normal\" vertex has six triangles hinging on it. However, at least twelve 5-triangle vertices of the type indicated at tl are to be interpolated if the 2-geometry is to be able to close up into a 2-sphere. (3) Skeleton 3-geometry obtained by filling in between the skeleton 2-geometry... and the similar structure tltJ3... ~ge... $6j) $'6j)'... as follows. (a) In... tl'tJ3' '.' ~'9'e' sert direct connectors such as 99' between corresponding points in the two 2-geometries. (b) Insert an intermediate layer of \"supplementary vertices\" such as SV~O/\"\u00a5.X.... Each of these supplementary vertices lies roughly halfway be tween the center of the triangle \"above\" it and the center of the corresponding triangle \"below\" it. (c) Connect each such \"supplementary vertex\" with its immediate neighbors above, below, and in the same plane. (d) Give all edge lengths. (e) Then the skeleton 3-geometry between the two 2-ge ometries is rigidly specified. It is made up of five (1) \"Right types of tetrahedrons, as follows. through blocks,\" such as 99'SV (six of these hinge on 99' when 9 is a normal vertex; five, when it is a 5-fold vertex, such as indicated by tl at the upper right). (2) \"Lower-facing blocks,\" such as tltJ39v. (3) \"Lower-packing blocks,\" such as tl9Sv. (4, 5) Corresponding \"upper-facing blocks\" and \"upper-packing blocks\" (not shown). The number of blocks of each kind is appropriately listed here for the two extreme cases of a 2-geom etry that consists (a) of a normal hexagonal lattice extending indefinitely in a plane and (b) of a lat tice consisting of the minimum number of 5-fold vertices (\"type tl vertices\") that will permit close up into a 2-sphere. \u00a742.6. THE CHOICE OF EDGE LENGTHS 1177 2-geometry of upper (or lower, face Hexagonal pattern of tdangles Icosahedron made of tdangles Its topology Vertices on upper face Nature of these vertices Edge lengths on upper fac\u00a5 Triangles on upper face Number of \"supplementary vertices\" Infinite 2-plane V 6-fold 3V 2V 2V Outer facing blocks Outer packing blocks Right through blocks Inner packing blocks Inner facing blocks t 2V 3V 6V 3V 2V 2-sphere 12 S-fold W=30 20 20 20 30 60 30 20 \u00a742.6. THE CHOICE OF EDGE LENGTHS So much for the lattice structure of the 4-geometry; now for the other issue, the freedom that exists in the choice of edge lengths. Why not make the simplest choice and let all edges be light rays? Because the 4-geometry would not then be fully determined. The geometry ga/3(x JL ) differs from the geometry ;\\.(x JL ) ga/3(x JL ), even though the same points that are connected by light rays in the one geometry are also connected by light rays in the other geometry. If none of the edges is null, it is nevertheless natural to take some of the edge lengths to be spacelike and some to be timelike. In consequence the area A of the triangle in some cases will be real, in other cases imaginary. In 3-space the parallelo gram (double triangle) spanned by two vectors Band C is described by a vector perpendicular to the two vectors. One obtains the magnitude of A from the formula 2A=BXC In 4-space, let Band C be two edges of the triangle. Then, as in three dimensions, 2A is dual to the bivector built from Band C. In other words, if B goes in the t direction and C in the z direction, then A is a bivector lying in the (x,y) plane. Consequently its magnitude A is to be thought of as a real quantity. Therefore the appropriate formula for the area A is (Tullio Regge) (42.10) The quantity A is real when the deficit angle S is real. Thus the geometrically important product AS is also real. The choice of edge lengths: (1) choose some timelike, others spacelike 1178 42. REGGE CALCULUS When the hinge lies in the (x,y) plane, on the other hand, the quantity A is purely imaginary.", " In that instance a test vector taken around the cycle of simplexes that swing on this hinge has undergone change only in its z and t components; that is, it has experienced a Lorentz boost; that is, the deficit angle Sis also purely imaginary. So again the product AS is a purely real quantity. Turn now from character of edge lengths to magnitude of edge lengths. It is desirable that the elementary building blocks sample the curvatures of space in different directions on a roughly equal basis. In other words, it is desirable not to have long needle-shaped building blocks nor pancake-shaped tetrahedrons and simplexes. This natural requirement means that the step forward in time should be comparable to the steps \"sidewise\" in space. The very fact that one should have to state such a requirement brings out one circumstance that should have been obvious before: the \"hinge equations\" Sh cotan () ph =\u00b0 (p = 1,2,...), (42.7) 2: hinges h that haveedgep in common though they are as numerous as the edges, cannot be regarded as adequate to determine all edge lengths. There are necessarily relations between these equations that keep them from being independent. The equations cannot determine all the details of the necessarily largely arbitrary skeletonization process. They cannot do so any more than the field equations of general relativity can determine the coordi nate system. With a given pattern of vertices (four-dimensional generalization of drawings in Box 42.3), one still has (a) the option how close together one will take successive layers of the structure and (b) how one will distribute a given number of points in space on a given layer to achieve the maximum payoff in accuracy (greater density of points in regions of greater curvature). To prepare a practical computer program founded on Regge calculus, one has to supply the machine not only with the hinge equations and initial conditions, but also with definite algorithms to remove all the arbitrariness that resides in options (a) and (b). Formulas from solid geometry and four-dimensional geometry, out of which to determine the necessary hyperdihedral angles a and the deficit angles S in terms of edge lengths and nothing but edge lengths, are summarized by Wheeler (1964a, pp. 469, 470, and 490) and by C. Y. Wong (1971). Regge (1961) also gives a formula for the Riemann curvature tensor itself in terms of deficit angles and number of edges running in a given direction [see also Wheeler (l964a, p. 471)]. (2) choose timelike lengths comparable to spacelike lengths (3) why some lengths must be chosen arbitrarily Deficit angles in terms of edge lengths \u00a742. 7. PAST APPLICATIONS OF REGGE CALCULUS Past applications of Regge calculus Wong (1971) has applied Regge calculus to a problem where no time development shows itself, where the geometry can therefore be treated as static, and where in addition it is spherically symmetric. He determined the Schwarzschild and Reissner Nordstrom geometries by the method ofske1etonization. Consider successive spheres \u00a742.8. THE FUTURE OF REGGE CALCULUS 1179 surrounding the center of attraction. Wong approximates each as an icosahedron. The condition (3)R = 16'17 (energy density) on the 3-space (\u00a72l.5) gives a recursion relation that determines the dimension of each icosahedron in terms of the two preceding icosahedra. Errors in the skeleton representation of the exact geometry range from roughly 10 percent to less than 1 percent, depending on the method of analysis, the quantity under analysis, and the fineness of the subdivision. Skeletonization of geometry is to be distinguished from mere rewriting of partial differential equations as difference equations. One has by now three illustrations Partial skeletonization that oIlf can capitalize on skeletonization without fragmenting spacetime all the way to the level of individual simplexes. The first illustration is the first part of Wong's work, where the time dimension never explicitly makes an appearance, so that the building blocks are three-dimensional only. The second is an alternative treatment, also given by Wong, that goes beyond the symmetry in t to take account of the symmetry in () and cp. It divides space into spherical shells, in each of which the geometry is \"pseudo-flat\" in much the same sense that the geometry of a paper cone is flat. The third is the numerical solution for the gravitational collapse of a spherical star by May and White (1966), in which there is symmetry in () and cp, but not in r or t. This zoning takes place exclusively in the r, t-plane. Each zone is a spherical shell. The difference as compared to Regge calculus (flat geometry within each building block) is the adjustable \"conicity\" given to each shell. The examples show that", " 1964c). Honl (1962); see also \u00a721.l2] or because that requirement simplifies the mathematical analysis, or for both reasons together. One can approximate a smooth, closed 3-geometry by a skeleton 3-geometry built out of tetrahedrons, as indicated schematically in Figure 43.1 (see Chapter 42 on the Regge calculus). Specify the 98 edge-lengths in this example to fix all the features of the geometry; and fix these 98 edge-lengths by giving the location of a single point in a space of 98 dimensions. This 98-dimensional manifold, this \"truncated superspace,\" goes over into superspace [Wheeler (1964a), pp. 453, 459, 463, 495] in the idealization in which the tracer points increase in density of coverage without limit. Accounts of superspace with more mathematical detail are given by DeWitt (1967a,b), Wheeler (1970), and Fischer (1970). Let the representative point move from one location to a nearby location, either in truncated superspace or in full superspace. Then all edge-lengths alter, and the 3-geometry of Figure 43.1 moves as if alive. No better illustration can one easily supply of what it means to speak of the \"dynamics of space.\" The term \"3-geometry\" makes sense as well in quantum geometrodynamics as in classical theory. So does superspace. But spacetime does not. Give a 3-geometry, and give its time rate of change. That is enough, under typical circumstances (see Chapter 21) to fix the whole time-evolution of the geometry; enough in other words, to determine the entire four-dimensional spacetime geometry, provided one is 3-geometry is the dynamic object Finite-dimensional \"truncated superspace\" y 1182 43. SUPERSPACE: ARENA FOR THE DYNAMICS OF GEOMETRY L, Superspace Figure 43.1. Superspace in the simplicial approximation. Upper left, space (depicted as two-dimensional but actually three-dimensional). Upper right, simplical approximation to space. The approximation can be made arbitrarily good by going to the limit of an arbitrarily fine decomposition. The curvature at a typical location is measured by a deficit angle. This angle is completely determined by the edge lengths (L l, [2'... L 8 in the figure) of the simplexes that meet at that location. When there are 98 edge lengths altogether in the simplicial representation of the geometry, then this geometry is completely specified by a single point in a 98-dimensional space (lower diagram; \"superspace\"). The concept of spacetime is incompatible with the quantum principle considering the problem in the context of classical physics. In the real *orld of quantum physics, however, one cannot give both a dynamic variable and its time-rate of change. The principle of complementarity forbids. Given the precise 3-geometry at one instant, one cannot also know at that instant the time-rate of change of the 3-geometry. In other words, given the geometrodynamic field coordinate, one cannot know the geometrodynamic field momentum. If one assigns the intrinsic 3-geometry, one cannot also specify the extrinsic curvature. The uncertainty principle thus deprives one of any way whatsoever to predict, or even to give meaning to, \"the deterministic classical history of space evolving \u00a743.1. SPACE, SUPERSPACE, AND SPACETIME DISTINGUISHED 1183 in time.\" No prediction ofspacetime, therefore no meaningfor spacetime, is the verdict of the quantum principle. That object which is central to all of classical general relativity, the four-dimensional spacetime geometry, simply does not exist, except in a classical approximation.,/ that These considerations reveal the concepts of spacetime and time are not primary but secondary ideas in the structure of physical theory. These concepts are valid in the classical approximation. However, they have neither meaning nor application under circumstances where quantum geometrodynamic effects become important. Then one has to forego that view of nature in which every event, past, present, or future, occupies its preordained position in a grand catalog called \"spacetime,\" with the Einstein interval from each event to its neighbor eternally established. There is no spacetime, there is no time, there is no before, there is no after. The question of what happens \"next\" is without meaning. Tha~spacetime is not the right way does not mean there is no right way to describe the dynamics of geometry consistent with the quantum principle. Superspace is the key to one right way to describe the dynamics (see Figure 43.2). Spacetime Space lme..... J ~.....,...., \"T.t., -I \\ __\\ I", ".... -~/ \\ ~ \\ _ _/~l \"..7.............. h \u2022 B' ctl / DJ I \"\"Superspace Figure 43.2. Space, spacetime, and superspace. Upper left: Five sample configurations, A, B, C, D, E, attained by space in the course of its expansion and recontraction. Below: Superspace and these five sample configurations, each represented by a point in superspace. Upper right: Spacetime. A spacelike cut, like A, through spacetime gives a momentary configuration of space. There is no compulsion to limit attention to a one-parameter family of slices, A, B, C, D, E through spacetime. The phrase \"many-fingered time\" is a slogan telling one not to so limit one's slices, and B' is an example of this freedom in action. The 3-geomelries B' andA, B, C, D, E, like a1l3-geometries obtained by all spacelike slices whatsoever through the given classical spacetime, lie on a single bent leaf of history, indicated in the diagram, and cutting its thin slice through superspace. A different spacetime, in other words, a different solution of Einstein's field equation, means a different leaf of history (not indicated) slicing through superspace. Spacetime is a classical of history slicing through superspace leaf 1184 \u00a743.2. 43. SUPERSPACE: ARENA FOR THE DYNAMICS OF GEOMETRY THE DYNAMICS OF GEOMETRY DESCRIBED IN THE LANGUAGE OF THE SUPERSPACE OF THE (3)J.j'S Given a spacetime, one can construct the corresponding leaf of history slicing through superspace. Conversely, given the leaf of history, one can reconstruct the spacetime. Consider the child's toy commonly known as \"Chinese boxes.\" One opens the outermost box only to reveal another box; when the second box is opened, there is another box, and so on, until eventually there are dozens of boxes scattered over the floor. Or conversely the boxes can be put back together, nested one inside the other, to reconstitute the original package. The packaging of(3)-b's into a (4)-b is much more sophisticated. Nature provides no monotonic ordering of the (3)-b'S. Two of the dynamically allowed (3)-b's, taken at random, will often cross each other one or more times. When one shakes the (4)-b apart, one therefore gets enormously more (3)-b's \"spread out over the floor\" than might have been imagined. Conversely, when one puts back together all of the (3)Hs lying on the leaf of history, one gets a structure with a rigidity that might not otherwise have been foreseen. This rigidity arises from the infinitely rich interleaving and intercrossing of clear-cut, well-defined (3)Hs one with another. In summary: (1) Classical geometrodynamics in principle constitutes a device, an algorithm, a rule for calculating and constructing a leaf of history that slices through superspace. (2) The (3)Hs that lie on this leaf of history are YES 3-geometries; the vastly more numerous (3)Hs that do not are NO 3-geometries. (3) The YES (3)Hs are the building blocks of the (4)-b that is classical spacetime. (4) The interweaving and interconnections of these building blocks give the (4)-b its existence, its dimen sionality, and its structure. (5) In this structure every (3)-b has a rigidly fixed location of its own. (6) In this sense one can say that the \"many-fingered time\" of each 3-geometry is specified by the very interlocking construction itself. Baierlein, Sharp and Wheeler (1962) say a little more on this concept of \"3-geometry as carrier of information about time.\" How different from the textbook concept of spacetime! There the geometry of spacetime is conceived as constructed out of elementary objects, or points, known as \"events.\" Here, by contrast, the primary concept is 3-geometry, in abstracto, and out of it is derived the idea of event. Thus, (1) the event lies at the intersection of such and such (3)Hs; and (2) it has a timelike relation to (earlier or later than, or synchronous with) some other (3)-b, which in turn (3) derives from the intercrossings of all the other (3)-b'S. t When one turns from classical theory to quantum theory, one gives up the concept of spacetime, except in the semiclass", "ical approximation. Therefore, one gives up any immediate possibility whatsoever of defining the concept, normally regarded as so elemental, of an \"event.\" The theory itself, however, here as always [Bohr and Rosenfeld (1933)], defines in and by itself, in its own natural way, the procedures in-principle for measuring all those quantities that are in principle measurable. Quantum theory upsets the sharp distinction between YES 3-geometries and NO \u00a743.3. THE EINSTEIN-HAMILTON-JACOBI EQUATION 1185 3-geometries. It assigns to each 3-geometry not a YES or a NO, but a probability amplitude, Probability amplitude for a 3-geometry (43.1 ) This probability amplitude is highest near the classically forecast leaf of history and falls off steeply outside a zone of finite thickness extending a little way on either side of the leaf. Were one to take, instead of a physically rele~ant probability amplitude function, a typical solution of the relevant wave equation, one would have to expect to see not one trace of anything like classical geometrodynamics. The typical probability amplitude function is spread all over superspace. No surprise! Already in classical theory one has to reckon with a Hamilton-Jacobi function,, S = S( 3l-b), (43.2) spread out over superspace. Moreover, this \"dynamic phase function\" of classical geometrodynamics gives at once the phase of!/;, according to the formula!/;( 3l-b) = (SIOWI~ varying. amplItude function ) e(i/1ilS~3l~l, (43.3) indication enough that!/; and Sare both unlocalized. Dynamics first clearly becomes recognizable when sufficiently many such spread out probability amplitude functions are superposed to build up a localized wave packet, as in the elementary examples of Boxes 25.3 and 25.4; thus, Wave packet recovers classical geometrodynamics _. (43.4) Constructive interference occurs where the phases of the several individual waves agree: (43.5) This is the condition that distinguishes YES 3-geometries from NO 3-geometries. It is the tool for constructing a leaf of history in superspace. It is the key to the dynamics of geometry. Moreover, it is an equation that says not one word about the quantum principle. It is not surprising that the equation of constructive interfer ence in (43.5) makes the leap from classical theory to quantum theory the shortest. \u00a743.3. THE EINSTEIN-HAMILTON-JACOBI EQUATION Should one write down a differential equation for the Hamilton-Jacobi function S( 3l-b), solve it, and then analyze the properties of the solution? The exact opposite is simpler: look at the properties of the solution, and from that inspection find out what equation the dynamic phase or action S must satisfy. 1186 43. SUPERSPACE: ARENA FOR THE DYNAMICS OF GEOMETRY Hilberts' principle of least action reads fHilbert = (1/16'IT)f l4lR( _g)I/2 d 4x = extremum. (43.6) After- one separates off complete derivatives in the integrand, what is left [see equations (21.13) and (21.95)] becomes (1/16'IT)fADM = f txue = (1/16'IT)f {'lT ii ogi;lot + Ng l/2R + Ng-l/2[~ (Trn)2 - Tr(n 2)] + 2Ni'ITiili} d 4x. (43.7) In (43.7), but not in (43.6), g stands for the determinant of the three-dimensional metric tensor, gii' and R for the scalar curvature invariant of the 3-geometry; the suffix (3) is omitted for simplicity. The integral is extended from (1) a spacelike hypersurface on which a 3-geometry is given with metric gi;'(X,y, z) to (2) a spacelike hypersurface on which a 3-geometry is given with metric gi/'(X,y, z). Whatever is adjustable in the chunk of spacetime between is now to be considered as having been adjusted to extremize the integral. Therefore the value of the integral f ADM becomes a functional of the metrics on the two hypersurfaces and nothing more. Next, holding fixed the metric g'ii(x,y, z) on the earlier hypersurface, change slightly or even more than slightly the metric on the later hypersurface. Solve the new variation problem and get", " a new value of f ADM' Proceeding further in this way, for each new gi;\" one gets a new value of f ADM\u2022 Call the functional f ADM of the metric defined in this way \"Hamilton's principal function,\" or the \"action\" or the \"dynamic path length,*\" S(gij(x,y, z)) of the \"history-of geometry\" that connects the two given 3-geometries. The double prime suffix is dropped from gi/' here and hereafter to simplify the notation. One knows from other branches of mechanics that the quantity defined in this way, S(gii)' when it exists, even though it is a special solution, nevertheless is always a solution of the Hamilton-Jacobi equation. Jacobi could look for more general solutions, but Hamilton already had one! For (43.7) to be an extremal with respect to variations of the lapse N and the shift components Ni, it was necessary (see Chapter 21) that the coefficients of these four quantities should vanish; thus, g-1/2U (Tr n)2 - Tr n 2] + gl/2R = 0 and 'lT ii - 0 Ii -. (43.8) (43.9) In the expression for the extremal value of the action, only one term, the first, is left: S(g(x,y, z)) = fADM, extremal =r {'lTi; ogi;lot} d 4x. gil. (43.10) g'l *Actually S == SADM == l6?TStrue = l6?T (true dynamic path length). Geometrodynamic momentum as rate of change of dynamic path length with respect to 3-geometry of terminal hypersurface Action depends on 3-geometry, not on metric coefficients individually \u00a743.3. THE EINSTEIN-HAMllTON-JACOBI EQUATION 1187 The effect of a slight change, 8gi ;, in the 3-metric at the upper limit is therefore easy to read off: (43.11) The language of \"functional derivative\" [see, for example, Bogoliubov and Shirkov (1959)] allows one to speak in terms of a derivative rather than an integral: 8S = 7TH. 8gH (43.12) The \"field momenta\" acquire a simple meaning: they give the rate of change of the action with respect to the continuous infinitude of \"field coordinates,\" gH(x,y, z). (Here tfe x, y, z, as well as the i and j, serve as mere labels.) Although the phase function S appears to depend on all six metric coefficients gij individually, it depends in actuality only on that combination of the gij which is locked to the 3-geometry. To verify this point, express a particular 3-geometry (3)-b throughout one local coordinate patch in terms of one set of coordinates x P by one set of metric coefficients gpq' Reexpress the same 3-geometry in terms of coordinates xP shifted by the small amount ~P, To keep the 3-geometry the same, that is, to keep unchanged the distance ds from one coordinate-independent point to another, the metric coefficients have to change: (43.13) Let the phase function S (or in quantum mechanics, let the probability amplitude!/J) be considered to be expressed as a functional of the metric coefficients gl1(x), g12(X),...,g33(X). Changes 8gpq(x) in these coefficients alter the H-J phase function and the probability amplitude by the amounts (43.14) (43.15) according to the standard definition of functional derivative. Therefore the coordi nate change produces an ostensible change in the dynamic path length or phase S given by 8S = f (8S/8gpq)(~plq + ~qlp) d 3x = -2 f (8S/8gpq)lq~p d3x. (43.16) This change must vanish if S is to depend on the 3-geometry alone, and not on 1188 43. SUPERSPACE: ARENA FOR THE DYNAMICS OF GEOMETRY the coordinates in terms of which that 3-geometry is expressed; and must vanish, moreover, for arbitrary choice of the ~P\" From this condition, one concludes -(~) - 0 8gpq Iq -. (43.17) Likewise, one finds three equations on the wave function I/; itself, as distinguished from its phase Sin; thus, 81/; ) 8gpq ( Iq =0. (43.18) But (43.17), by virtue of (43.12), is identical with (43.", "9). In this sense (43.9) merely verifies what one already knew had to be true: the classical Hamilton-Jacobi function S (like the probability amplitude function I/; of quantum theory) depends on 3-ge ometry, not on individual metric coefficients, and not on choice of coordinates. All the dynamic content of geometrodynamics is summarized in the sole remaining equation (43.8), which takes the form Law of propagation of wave crests in superspace -1/2[1 2 gpqgrs g _ gprgqs 8g ]~ 8S rs pq 8g + g 1/2R-0 -. (43.19) This is the Einstein-Hamilton-Jacobi equation, first given explicitly in the literature by Peres (1962) on the foundation of earlier work by himself and others on the Hamiltonian formulation of geometrodynamics. This equation tells how fronts of constant S (\"wave crests\") propagate in superspace. That the one EHJ equation (43.19) contains as much information as all ten components of Einstein's field equation has been demonstrated by Gerlach (1969). The central point in the analysis is the principle of constructive interference, and the main requirement for a proper treatment of this point is the concept of a completely parametrized solution of the EHJ equation. The problem of a particle moving in two-dimensional space, as treated by the Hamilton-Jacobi method in Boxes 25.3 and 25.4, required for complete analysis a solution containing two distinct and independently adjustable parameters, the energy per unit mass, E, and angular momentum per unit mass, l; thus S(r, (J, t; E, l) = -Et + l8 + fr [E2 - (1 - 2M/r)(1 + \u00a32/r 2)jl/2 1 ~ ( - M/r) + 8(E, l). \u2022(43.20) \u2022 Here the additive phase 8(E, l) is required if one is to be able to arrange for the particle to arrive at a given r-value at a specified t value and at a specified value of (J. One thinks of superposing four probability amplitudes, as in (43.4), with dynamic phases, S, given by (43.20) and the parameters taking on, respectively, the following four sets of values: (E, l); (E + LiE, l); (E, l + Lil); and (E + LiE, l + Lil). The principle of constructive interference leads to the conditions \u00a743.3. THE EINSTEIN-HAMILTON-JACOBI EQUATION as/aE = 0, as/al = 0. 1189 (4321) The points in the spacetime (r, (J, t) that satisfy these conditions are the YES points; they lie on the world line. The ones that don't are the NO points. The desired solution of the EHJ equation (43.19) contains not two parameters (plus an additive phase, 8, depending on these two parameters), but an infinity of parameters, and even a continuous infinity of parameters. Thus the parameters are not to be designated as ai' a2'...; f3 l, f32,... (parameters labeled by a discrete index), but as and a(u, D, w) f3(u, D, w) (two parameters \"labeled\" by three continuous indices u, D, w). Accidentally omit one of this infinitude of parameters? How could one ever hope to know that what purported to be a complete solution of the EHJ equation was not in actuality complete? Happily Gerlach provides a procedure to test the parameters for com pleteness. Granted completeness, Gerlach goes on to show that the \"leaf of history in superspace\" or collection of 3-geometries that satisfy the conditions of constructive interference, Condition of constructive interference gives classical \"leaf of history\" or spacetime 8S( 3l-b; a(u, D, w), f3(u, D, w)) = 0, 8S( 3l-b; a(u, D, w), f3(u, D, w)) = \u00b0 8a 8f3'(43.22) is identical with the leaf of history, or equivalent 4-geometry, given by the ten components of Einstein's geometrodynamic law. From the Hamilton-Jacobi equation for a problem in elementary mechanics, it is a short step to the corresponding Schroedinger equation; similarly in geometro dynamics. No one has done more than Bryce DeWitt to explore the meaning and consequences of this \"Einstein-Schroedinger equation\" [DeWitt (l967a,b)]. One of the most", " interesting consequences is the existence of a conserved current in super space, analogous to the conserved current that one encounters in the Klein-Gordon wave equation for a particle of spin zero. It is an unhappy feature of this Einstein-Schroedinger wave equation that it contains second derivatives; so one has to specify both the probability amplitude, and the normal derivative of the probability amplitude, on the appropriate \"super- 1190 43. SUPERSPACE: ARENA FOR THE DYNAMICS OF GEOMETRY hypersurface\" in superspace, in order to be able to predict the evolution of this state function elsewhere in superspace. One suggested way out of this situation-it is at least an inconvenience, possibly a real difficulty-has been proposed by Leutwyler. (1968): impose a natural boundary condition that reduces the number ofindependent solutions from the number characteristic of a second-order equation to the number characteristic of a first-order equation. Another way out is to formulate the dynamics quite differently, in the way proposed by Kuchar (see Chapter 21), in which the resulting equation is of first order in the variable analogous to time. The exploration of quantum geometrodynamics is simplified when one treats most of the degrees of freedom of the geometry as frozen out, by imposition of a high degree of symmetry. Then one is left with one, two. or three degrees of freedom (see Chapter 30, on mixmaster cosmology), or even infinitely many, and is led to manageable problems of quantum mechanics [Misner (1972a, 1973)]. \u00a743.4. FLUCTUATIONS IN GEOMETRY Of all the remarkable developments of physics since World War II, none is more impressive than the prediction and verification of the effects of the vacuum fluctua tions in the electromagnetic field on the motion of the electron in the hydrogen atom (Figure 43.3). That development made it impossible to overlook the effects of such fluctuations throughout all physics and, not least, in the geometry of spacetime itself. \u2022 Figure 43.3. Symbolic representation of motion of electron in hydrogen atom as affected by fluctuations in electric field in vacuum (\"vacuum\" or \"ground state\" or \"zero-point\" fluctuations). The electric field associated with the fluctuation, E.(t) = fEz(w)e- iwt dw, adds to the st~ic electric field provided by the nucleus itself. The additional field brings about in the most elementary approximation the displacement.dx = f(ejmw 2)E.(w)e-iwt dw. The average vanishes but the root mean square \u00ab(.dx)2) does not. In consequence the electron feels an effective atomic potential altered from the expected value V(x,y, z) by the amount.dV(x,y, z) = f \u00ab(.dx)2) v2 V(x,y, z). The average of this perturbation over the unperturbed motion accounts for the major part of the observed Lamb-Retherford shift.dE = (.d V(x,y, z\u00bb in the energy level. Con versely, the observation of the expected shift makes the reality of the vacuum fluctuations inescapably evident. Fluctuations for oscillator and for electromagnetic field \u00a743.4. FLUCTUATIONS IN GEOMETRY 1191 From the zero-point fluctuations of a single oscillator to the fluctuations of the electromagnetic field to geometrodynamic fluctuations is a natural order of progres sion. A harmonic oscillator in its ground state has a probability amplitude of I/;(x) = : ; ( ) 1/4 e-(mw/2ftlz2 (43.23) to be displaced by the distance x from its natural classical position of equilibrium. In this sense, it may be said to \"resonate\" or \"fluctuate\" between locations in space ranging over a region of extent (43.24) The~ electromagnetic field can be treated as an infinite collection of independent \"field oscillators,\" with amplitudes ~1' ~2'.... When the Maxwell field is in its state oflowest energy, the probability amplitude-for the first oscillator to have amplitude ~1' and simultaneously the second oscillator to have amplitude ~2' the third ~3' and so on-is the product of functions of the form (43.23), one for each oscillator. When the scale ofamplitudes for each oscillator is suitably normalized, the resulting infinite product takes the form This expression gives the probability amplitude I/; for a configuration B(x,y, z) of the magnetic field that is described by the Fourier coefficients ~1' ~2'.... One can forgo any mention of these Fourier coefficients if one desires, however, and rewrite (43.25) directly in terms of the magnetic field configuration itself [Wheel", "er (1962)J: (43.25) (43.26) One no longer speaks of \"the\" magnetic field, but instead of the probability of this, that, or the other configuration of the magnetic field, even under circumstances, as here, where the electromagnetic field is in its ground state. [See Kuchar (1970) for a similar expression for the \"ground state\" functional of the linearized gravitational field.J It is reasonable enough under these circumstances that the configuration of greatest probability is B(x,y, =) = O. Consider for comparison a configuration where the magnetic field is again everywhere zero except in a region of dimension L. There let the field. subject as always to the condition div B = 0, be of the order of magni tude LJB. The probability amplitude for this configuration will be reduced relative to the nil configuration by a factor exp ( - 1). Here the quantity J in the exponent is of the order (LJB f U It/c. Configurations for which J is large compared to I occur with negligible probability. Configurations for which I is small compared to I occur with practically the same probability as the nil configuration. In this sense, one can 1192 43. SUPERSPACE: ARENA FOR THE DYNAMICS OF GEOMETRY say that the fluctuations in the magnetic field in a region of extension L are of the order of magnitude (hC)1/2 LJ.B-- U \u00b7 (43.27) In other words, the field \"resonates\" between one configuration and another with the range of configurations of significance given by (43.27). Moreover, the smaller is the region of space under consideration, the larger are the field magnitudes that occur with appreciable probability. Still another familiar way of speaking about electromagnetic field fluctuations gives additional insight relevant to geometrodynamics. One considers a measuring device responsive in comparable measure to the magnetic field at all points in a region ofdimension L. One asks for the effect on this device of electromagnetic disturbances of various wavelengths. A disturbance of wavelength short compared to L will cause forces to act one way in some parts of the detector, and will give rise to nearly compensating forces in other parts of it. In contrast, a disturbance of a long wave length A. produces forces everywhere in the same direction, but of a magnitude too low to have much effect. Thus the field, estimated from the equation f ener gy I f 0 wave engt 0 one uantum h ~ 1\\ ) energy of electromagnetiC) wave of wavelength A. In a domam of volume A. 3.. ( ( or or Fluctuations in geometry dominate at the Planck scale of distances (hC)l/2 B - - - A. 2 (43.28) is very small if A. is large compared to the domain size L. The biggest effect is caused by a disturbance of wavelength A. comparable to L itself. This line of reasoning leads directly from (43.28) to the standard fluctuation formula (43.27). Similar considerations apply in geometrodynamics. Quantum fluctuations in the geometry are superposed on and coexist with the large-scale, slowly varying curvature predicted by classical deterministic general relativity. Thus, in a region of dimension L, where in a local Lorentz frame the normal values of the metric coefficierits will be - I, I, I, I, there will occur fluctuations in these coefficients of the order L* LJ.g - -L' fluctuations in the first derivatives of the gik'S of the order L* LJ.g LJ.F---L 2'L (43.29) (43.30) \u00a743.4. FLUCTUATIONS IN GEOMETRY and fluctuations in the curvature of space of the order Here L* = ( :~ Y/2 = 1.6 X 10-33 cm 1193 (43.31) (43.32) is the so-called Planck length [Planck (1899)). It is appropriate to look at orders of magnitude. The curvature of space within and near the earth, according to classical Einstein theory, is of the order R -- (~) p -- (0.7 X 10-28 cm/g)(5 g/cm3 ) -- 4 X 10- 28 cm-2. (43.33) This quantity has a very direct physical significance. It measures the \"tide-producing component of the gravitational field\" as sensed, for example, in a freely falling elevator or in a spaceship in free orbit around the earth. By comparison, the quantum fluctuations in the curvature of space are only LJR -- 10-33 cm- 2, (43.34) even in a domain of observation as small as 1 cm in extent. Thus the quantum fluctuations in the geometry of space are completely negligible under everyday circumstances. Even in atomic and nuclear physics the", " fluctuations in the metric, 10-33 cm __ 10- 25 LJg -- 10-8 cm and LJ __ g 10- 33 10- 13 cm cm __ 10-20'(43.35) are so small that it is completely in order to idealize the physics as taking place in a flat Lorentzian spacetime manifold. The quantum fluctuations in the geometry are nevertheless inescapable, if one is to believe the quantum principle and Einstein's theory. They coexist with the geometrodynamic development predicted by classical general relativity. The fluctua tions widen the narrow swathe cut through superspace by the classical history of the geometry. In other words, the geometry is not deterministic, even though it looks so at the everyday scale of observation. Instead, at a submicroscopic scale it \"reson ates\" between one configuration and another and another. This terminology means no more and no less than the following: (1) Each configuration (3)~ has its own probability amplitude y = y(3)~). (2) These probability amplitudes have comparable magnitudes for a whole range of 3-geometries included within the limits (43.29) on 1194 43. SUPERSPACE: ARENA FOR THE DYNAMICS OF GEOMETRY either side of the classical swathe through superspace. (3) This range of 3-geometries is far too variegated on the submicroscopic sale to fit into anyone 4-geometry, or anyone classical geometrodynamic history. (4) Only when one overlooks these small-scale fluctuations (_10- 33 em) and examines the larger-scale features of the 3-geof!1etries do they appearto fit into a single space-time manifold, such as comports with the classical field equations. These small-scale fluctuations tell one that something like gravitational collapse is taking place everywhere in space and all the time; that gravitational collapse is in effect perpetually being done and undone; that in addition to the gravitational collapse of the universe, and ofa star, one has also to deal with a third and, because it is constantly being undone, most significant level of gravitational collapse at the Planck scale of distances. Exercise 43.1. THE ACTION PRINCIPLE FOR A FREE PARTICLE IN NONRELATIVISTIC MECHANICS Taking as action principle 1= JL dt = extremum, with specified x', t' and x\", t\" at the two limits, and with L = ~m(dx/dt)2, find (I) the extremizing history x = x(t) and (2) the dynam ical path length or action S(x\", t\"; x', 1') = Iextremum in its dependence on the end points. Also (3) write down the Hamilton-Jacobi equation for this problem, and (4) verify that Sex, t; x', t') satisfies this equation. Finally, imagining the Hamilton-Jacobi equation not to be known, (5) derive it from the already known properties of the function S itself. Exercise 43.2. THE ACTION FOR THE HARMONIC OSCILLATOR The kinetic energy is ~m(dx/dt)2 and the potential energy is ~mw2x2. Carry through the analysis of parts (I), (2), (3), (4) of the preceding exercise. Partial answer: EXERCISES S= - mw (x2 + x'2) cos wet - 1') - 2xx' 2 sin wet - t'). Verify that as/ax gives momentum and -as/at gives energy. Exercise 43.3. QUANTUM PROPAGATOR FOR HARMONIC OSCILLATOR Show that the probability amplitude for a simple harmonic oscillator to transit from x', t' to x\", t\" is (x\", t\"; x', I') _ ( - mw 2r.ih sin w(t\" - t') )1/2 imw[(x\"2 + x'2) cos w(t\" - 1') - 2x\"x'] X exp 2h sin w(t\" - t') and that it reduces for the case of a free particle to,! X\" til. X',, ( m 2r.ih(t\" _ 1') ( )112 ex im(x\" _ x')2 t') p 2h(t\" - I -, t) - Note that one can derive all the harmonic-oscillator wave functions from the solution by use of the formula (x\", t\"; x', t') = ~ un(x\")un\u00b7(x') exp iEn(t' - t\")jh. n \u00a743.4. FLUCTUATIONS IN GEOMETRY 1195 Exercise 43.4. QUANTUM PROPAGATOR FOR FREE ELECTROM", "AGNETIC FIELD In flat.spacetime, one is given on the spacelike hypersurface t = t' the divergence-free magnet~c field B'(x,y, z) and on the spacelike hypersurface t = t\" the divergence-free magnetIc field B\"(x,y, z). By Fourier analysis (reducing this problem to the preceding problem) or otherwise, find the probability amplitude to transit from B' at t' to B\" at t\". Exercise 43.5. HAMILTON-JACOBI FORMULATION OF MAXWELL ELECTRODYNAMICS Regard the four components AI' of the electromagnetic 4-potential as the primary quantities; split them into a space part Ai and a scalar potential cp. (I) Derive from the action principle (in flat spacetime) by splitting off an appropriate divergence, an expression qualitatively similar in character to (43.7). (2) Show that the appropriate quantity to be fixed on the initial and final spacelike hypersurface is not really Ai itself, but the magnetic field, defined by B = curiA. (3) Derive the Hamilton-Jacobi equation for the dynamic phase or action S(B, S) in its dependence on the choice of hypersurface S, and the choice of magnetic field B on this hypersurface, The quantity on the left is Tomonaga's \"bubble time\" derivative [Tomonaga (1946); see also Box 21.1]. y CHAPTER-44 BEYOND THE END OF TIME \"Heaven wheels above you Displaying to you her eternal glories And still your eyes are on the ground\" DANTE The world \"stands before uS as a great, eternal riddle\" EINSTEIN (1949a) This chapter is entirely Track 2. No previous Track-2 material is needed as preparation for it, but Chapter 43 will be helpful. \u00a744.1. GRAVITATIONAL COLLAPSE AS THE GREATEST CRISIS IN PHYSICS OF ALL TIME The universe starts with a big bang, expands to a maximum dimension, then recon tracts and collapses: no more awe-inspiring prediction was ever made. It is prepos terous. Einstein himself could not believe his own prediction. It took Hubble's observations to force him and the scientific community to abandon the concept of a universe that endures from everlasting to everlasting. Later work of Tolman (1934a), Avez (1960), Geroch (1967), and Hawking and Penrose (1969) generalizes the conclusion. A model universe that is closed, that obeys Einstein's geometrodynamic law, and that contains a nowhere negative density of mass-energy, inevitably develops a singularity. No one sees any escape fr~m the density of mass-energy rising without limit. A computing machine calculating ahead step by step the dynamical evolution of the geometry comes to the point where it can not go on. Smoke, figuratively speaking, starts to pour out of the computer. Yet physics surely continues to go on if for no other reason than this: Physics is by definition that which does go on its eternal way despite all the shadowy changes in the surface appearance of reality. The Marchon lecture given by J. A. W. at the University of Newcastle-upon-Tyne, May 18, 1971, and the Nuffield lecture given at Cambridge University July 19, 1971, were based on the material presented in this chapter. \u00a74401. GRAVITATIONAL COLLAPSE: THE GREATEST CRISIS IN PHYSICS 1197 Some day a door will surely open and expose the glittering central mechanism of the world in its beauty and simplicity. Toward the arrival of that day, no develop ment holds out more hope than the paradox of gravitational collapse. Why paradox? Because Einstein's equation says \"this is the end\" and physics says \"there is no end.\" Why hope? Because among all paradigms for probing a puzzle, physics proffers none with more promise than a paradox. No previous period of physics brought a greater paradox than 1911 (Box 4401). Rutherford had just been forced to conclude that matter is made up of-localized positive and negative charges. Matter as so constituted should undergo electric collapse in _10- 17 sec, according to theory. Observation equally clearly proclaimed that matter is stable. No one took the paradox more seriously than Bohr. No one worked around and around the central mystery with more energy wherever work was possible. No one brought to bear a more judicious combination of daring and conseArativeness, nor a deeper feel for the harmony of physics. The direct opposite The paradox of collapse: physics stops, but physics must go on The 1911 crisis of electric collapse Box 44.1 COLLAPSE OF UNIVERSE PREDICTED BY CLASSICAL THEORY, COMPARED AND CONTR", "ASTED WITH CLASSICALLY PREDICTED COLLAPSE OF ATOM System Atom (1911) Universe (1970 Os) Dynamic entity System of electrons Geometry of space Nature of classi cally predicted collapse Electron headed toward point-center of attraction is driven in a finite time to infinite energy Not only matter but space itself arrives in a finite proper time at a condition of infinite compaction One rejected \"way Give up Coulomb law of force Give up Einstein's field equation out\" Another proposal \"Accelerated charge need not radiate\" \"Matter cannot be compressed beyond a certain density by any pressure, how ever high for a \"cheap way out\" that has to be rejected How this proposal violates principle of causality Coulomb field of point-charge cannot re adjust itself with infinite speed out to indefinitely great distances to sudden changes in velocity of charge Speed of sound cannot exceed speed of light; pressure cannot exceed density of mass-energy A major new consider ation introduced by recognizing quantum principle as overarching or ganizing principle of physics uncertainty principle; binding too close to center of attraction makes zero-point kinetic energy outbalance potential energy: consequent existence of a lowest quantum state: can't radiate because no lower state available to drop to Uncertainty principle; propagation of representative wave packet in super space does not lead deterministically to a singular configuration for the geometry of space; expect rather a probability distribution of outcomes, each outcome describing a universe with a different size, a different set of particle masses, a different number of particles, and a different length of time required for its expansion and recontractiono 1198 44. BEYOND THE END OF TIME of harmony, cacophony, is the impression that comes from sampling the literature of the 'teens on the structure of the atom. (1) Change the Coulomb law of force between electric charges? (2) Give up the principle that an accelerated charge must radiate? There was little inhibition against this and still wilder abandonings of the well-established. In contrast, Bohr held fast to these two principles. At the same time he insisted on the importance of a third principle, firmly established by Planck in quite another domain of physics, the quantum principle. With that key he opened the door to the world of the atom. Great as was the crisis of 1911, today gravitational collapse confronts physics with its greatest crisis ever. At issue is the fate, not of matter alone, but of the universe itself. The dynamics of collapse, or rather of its reverse, expansion, is evidenced not by theory alone, but also by observation; and not by one observation, but by observations many in number and carried out by astronomers of unsurpassed ability and integrity. Collapse, moreover, is not unique to the large-scale dynamics of the universe. A white dwarf star or a neutron star of more than critical mass is predicted to undergo gravitational collapse to a black hole (Chapters 32 and 33). Sufficiently many stars falling sufficiently close together at the center of the nucleus of a galaxy are expected to collapse to a black hole many powers of ten more massive than the sun. An active search is under way for evidence for a black hole in this Galaxy (Box 33.3). The process that makes a black hole is predicted to provide an experi mental model for the gravitational collapse of the universe itself, with one difference. For collapse to a black hole, the observer has his choice whether (1) to observe from a safe distance, in which case his observations will reveal nothing of what goes on inside the horizon; or (2) to follow the falling matter on in, in which case he sees the final stages of the collapse, not only of the matter itself, but of the geometry surrounding the matter, to indefinitely high compaction, but only at the cost of his own early demise. For the gravitational collapse of a closed model universe, no such choice is available to the observer. His fate is sealed. So too is the fate of matter and elementary particles, driven up to indefinitely high densities. The stakes in the crisis of collapse are hard to match: the dynamics of the largest object, space, and the smallest object, an elementary particle--and how both began. \u00a744.2. ASSESSMENT OF THE THEORY THAT PREDICTS COLLAPSE No one reflecting on the paradox of collapse (\"collapse ends physics\"; \"oollapse cannot end physics\") can fail to ask, \"What are the limits of validity of Einstein's geometric theory of gravity?\" A similar question posed itself in the crisis of 1911. The Coulomb law for the force acting between two charges had been tested at distances of meters and millimeters, but what warrant was there to believe that it holds down to the 10-8 cm of atomic dimensions? Of course, in the end it proves to hold not only at the level of the atom, and at the 10-13 cm level of the nucleus, but even down to 5 X 10", "-15 cm [Barber, Gittelman, O'Neill, and Richter, and Bailey et al. (1968), as reviewed by Farley (1969) and Brodsky and Drell (1970)], a striking \u00a744.2. ASSESSMENT OF THE THEORY THAT PREDICTS COLLAPSE 1199 example of what Wigner (1960) calls the \"unreasonable effectiveness of mathematics in the natural sciences.\" No theory more resembles Maxwell's electrodynamics in its simplicity, beauty, and scope than Einstein's geometrodynamics. Few principles in physics are more firmly established than those on which it rests; the local validity of special relativity (Chapters 2-7), the equivalence principle (Chapter 16), the conservation of momen tum and energy (Chapters 5, 15 and 16), and the prevalence of second-order field equations throughout physics (Chapter 17). Those principles and the demand for no \"extraneous fields\" (e.g., Dicke's scalar field) and \"no prior geometry\" (\u00a7 17.6) lead to the conclusion that the geometry of spacetime must be Riemannian and the geometrodynamic law must be Einstein's. To say that the geometry is Riemannian is to say that the interval between any two nearby events C and D, anywhere in spacetime, stated in terms of the interval AB between two nearby fiducial events, at quite another point in spacetime, has a value CD/ AB independent of the route of intercomparison (Chapter 13 and Box 16.4). There are a thousand routes. By this hydraheaded prediction, Einstein's theory thus exposes itself to destruction in a thousand ways (Box 16.4). Geometrodynamics lends itself to being disproven in other ways as well. The geometry has no option about the control it exerts on the dynamics of particles and fields (Chapter 20). The theory makes predictions about the equilibrium configura tions and pulsations of compact stars (Chapters 23-26). It gives formulas (Chapters 27-29) for the deceleration of the expansion of the universe, for the density of mass-energy, and for the magnifying power of the curvature of space, the tests of which are not far off. It predicts gravitational collapse, and the existence of black holes, and a wealth of physics associated with these objects (Chapters 31-34). It predicts gravitational waves (Chapters 35-37). In the appropriate approximation, it encompasses all the well-tested predictions of the Newtonian theory of gravity for the dynamics of the solar system, and predicts testable post-Newtonian corrections besides, including several already verified effects (Chapters 38-40). No inconsistency of principle has ever been found in Einstein's geometric theory of gravity. No purported observational evidence against the theory has ever stood the test of time. No other acceptable account of physics of comparable simplicity and scope has ever been put forward. Continue this assessment of general relativity a little further before returning to the central issue, the limits of validity of the theory and their bearing on the issue of gravitational collapse. What has Einstein's geometrodynamics contributed to the understanding of physics? First, it has dethroned spacetime from a post of preordained perfection high above the battles of matter and energy, and marked it as a new dynamic entity participating actively in this combat. Second, by tying energy and momentum to the curvature of spacetime, Einstein's theory has recognized the law of conservation of momentum and energy as an automatic consequence of the geometric identity that the boundary of a boundary is zero (Chapters 15 and 17). Third, it has recognized gravitation as a manifestation of the curvature of the Battle-tested theory of gravitation New view of nature flowing from Einstein's geometrodynamics 1200 44. BEYOND THE END OF TIME geometry of spacetime rather than as something foreign and \"physical\" imbedded in spacetime. Fourth, general relativity has reinforced the view that \"physics is local\"; that the _ analysis of physics becomes simple when it connects quantities at a given event with quanti-ties at immediately adjacent events. Fifth, obedient to the quantum principle, it recognizes that spacetime and time itself are ideas valid only at the classical level of approximation; that the proper arena for the Einstein dynamics of geometry is not spacetime, but superspace; and that this dynamics is described in accordance with the quantum principle by the propagation of a probability amplitude through superspace (Chapt~r 43). In conse quence, the geometry ofspace is subject to quantum fluctuations in metric coefficients of the order 5 g - (Planck length, L\u00b7 = (fiG/ C 3 )1/2 = 1.6 X 10-33 em) (linear extension of region under study). Electric charge as lines of force trapped in the topology of space Sixth, standard Einstein geometrodynamics is partial", " as little to Euclidean topology as to Euclidean geometry. A multiply connected topology provides a natural de scription for electric charge as electric lines of force trapped in the topology of a multiply connected space (Figure 44.1). Any other description ofelectricity postulates a breakdown in Maxwell's field equations for the vacuum at a site where charge Figure 44.1. Electric charge viewed as electric lines of force trapped in the topology of a multiply connected space [for the history of this concept see reference 36 of Wheeler (l968a)]. The wormhole or handle is envisaged as connecting two very different regions in the same space. One of the wormhole mouths, viewed by an observer with poor resolving power, appears to be the seat of an electric charge. Out of this region of 3-space he finds lines of force emerging over the whole 477\" solid angle. He may construct a boundary around this charge, determine the flux through this boundary, incorrectly apply the theorem of Gauss and \"prove\" that there is a charge \"inside the boundary.\" It isn't a boundary. Someone caught within it-to speak figuratively-can go into that mouth of the wormhole, through the throat, out tte other mouth, and return by way of the surrounding space to look at his \"prison\" from the outside. Lines of force nowhere end. Maxwell's equations nowhere fail. Nowhere can one place a finger and say, \"Here there is some charge.\" This classical type of electric charge has no direct relation whatsoever to quantized electric charge. There is a freedom of choice about the flux through the wormhole, and a specificity about the connection between one charge and another, which is quite foreign to the charges of elementary particle physics. For ease of visualization the number of space dimensions in the above diagram has been reduced from three to two. The third dimension, measured off the surface, has no physical meaning-it only provides an extra dimension in which to imbed the surface for more convenient diagrammatic representation. [For more detail see Misner and Wheeler (1957), reprinted in Wheeler (1962)]. \u00a744.2. ASSESSMENT OF THE THEORY THAT PREDICTS COLLAPSE 1201 is located, or postulates the existence of some foreign and \"physical\" electric jelly imbedded in space, or both. No one has ever found a way to describe electricity free of these unhappy features except to say that the quantum fluctuations in the geometry of space are so great at small distances that even the topology fluctuates, makes \"wormholes,\" and traps lines of force. These fluctuations have to be viewed, not as tied to particles, and endowed with the scale of distances associated with particle physics (_10- 13 em) but as pervading all space (\"foam-like structure of geometry\") and characterized by the Planck distance (_10-33 em). Thus a third type of gravitational collapse forces itself on one's attention, a collapse continually being done and being undone everywhere in space: surely a guide to the outcome of collapse at the level. of a star and at the level of the universe (Box 442). I Box 44.2 THREE LEVELS OF GRAVITATIONAL COLLAPSE 1. Universe 2. Black hole 3. Fluctuations at the Planck scale of distances Recontraction and collapse of the universe is a kind of mirror image of the \"big bang,\" on which one already has so much evidence. Collapse of matter to form a black hole is most natural at two distinct levels: (a) collapse of the dense white-dwarf core of an individual star (when that core exceeds the critical mass, -1 M0 or 2M0, at which a neutron star is no longer a possible stable end-point for collapse) and (b) coalescence one by one of the stars in a galactic nucleu~ to make a black hole of mass up to 106M0 or even 109M0. In either case: no feature of principle about matter falling into the black hole is more interesting than the option that the observer has (symbolized by the branching arrow in the inset). He can go along with the infall ing matter, in which case he sees the final stages of collapse, but only at the cost of his own demise. Or he in which case even after in can stay safely outside, definitely long time he sees only the first part of the collapse, with the infalling matter creeping up more and more slowly to the horizon. In the final stages of the collapse of a closed model universe, all black holes present are caught up and driven together, amalgamating one by one. No one has any way to look at the event from safely outside; one is inevitably caught up in it oneself. Collapse at the Planck scale of distances is taking place everywhere and all the time in quantum fluctuations in the geometry and, one believes, the topology of space. In this sense, collapse is continually being done", " and undone, modeling the undoing of the collapse of the universe itself, summarized in the term, \"the reproc essing of the universe\" (see text). Is matter built out of geometry? The richness of the physics of the vacuum 1202 44. BEYOND THE END OF TIME \u00a744.3. VACUUM FLUCTUATIONS: THEIR PREVALENCE AND FINAL DOMINANCE If Einstein's theory thus throws light on the rest of p'hysics, the rest of physics also. throws light on geometrodynamics. No point is more central than this,\u00b7 that em\"pty space'is not empty. It is the seat of the most violent physics. The electromagnetic field fluctuates (Chapter 43). Virtual pairs of positive and negative electrons, in effect, are continually being created and annihilated, and likewise pairs ofmu mesons, pairs of baryons, and pairs of other particles. All these fluctuations coexist with the quantum fluctuations in the geometry and topology of space. Are they additional to those geometrodynamic zero-point disturbances, or are they, in some sense not now well-understood, mere manifestations of them? Put the question in other words. Recall Clifford, inspired by Riemann, speaking to the Cambridge Philosophical Society on February 21,1870, \"On the Space Theory of Matter\" [Clifford (1879), pp. 244 and 322; and (1882), p. 21], and saying, \"I hold in fact (1) That small portions of space are in fact of a nature analogous to little hills on a surface which is on the average flat; namely, that the ordinary laws of geometry are not valid in them. (2) That this property of being curved or distorted is continually being passed on from one portion of space to another after the manner of a wave. (3) That this variation of the curvature of space is what really happens in that phenomenon which we call the motion of matter, whether ponderable or etherial. (4) That in the physical world nothing else takes place but this variation, subject (possibly) to the law of continuity.\" Ask if there is a sense in which one can speak of a particle as constructed out of geometry. Or rephrase the question in updated language: \"Is a particle a geometrodyna~ic ~xdton?\" What else is there out of which to build a particle except geometry itself? And what else is there to give discreteness to such an object except the quantum principle? The Clifford-Einstein space theory of matter has not been forgotten in recent years. \"In conclusion,\" one of the authors wrote a decade ago [Wheeler (1962)], \"the vision of Riemann, Clifford, and Einstein, of a purely geometric basis for physics, today has come to a higher state of development, and offers richer prospects-and presents deeper problems-than ever before. The quantum of action adds to this geometro dynamics new features, of which the most striking is the presence of fluctuations of the wormhole type throughout all space. If there is any correspondence at all between this virtual foam-like structure and the physical vacuum as it has come to be known through quantum electrodynamics, then there seems to be no escape from identifying these wormholes with 'undressed electrons.' Completely different from these 'undressed electrons,' according to all available evidence, are the el~ctrons and other particles of experimental physics. For these particles the geometrodynamic picture suggests the model of collective disturbances in a virtual foam-like vacuum, analogous to different kinds of phonons or excitons in a solid. \"The enormous factor from nuclear densities _10 14 g/cm3 to the density of field fluctuation energy in the vacuum, _1094 g/cm3, argues that elementary particles represent a percentage-wise almost completely negligible change in the locally violent conditions that characterize the vacuum. ['A particle (10 14 g/cm3) means as little \u00a744.4. PREGEOMETRY AS THE MAGIC BUILDING MATERIAL 1203 to the physics of the vacuum (1094 g/cm3) as a cloud (10-6 g / cm3) means to the physics of the sky (10-3 g/cm3).'] In other words, elementary particles do not form a really basic starting point for the description of nature. Instead, they represent a first-order correction to vacuum physics. That vacuum, that zero-order state of affairs, with its enormous densities of virtual photons and virtual positive-negative pairs and virtual wormholes, has to be described properly before one has a funda mental starting point for a proper perturbation-theoretic analysis.\" \"These conclusions about the energy density of the vacuum, its complicated topological character, and the richness of the physics which goes on in the vac uum, stand in no evident contradiction with what quantum electrodynamics", " has to say about the vacuum. Instead the conclusions from the'small distance' analysis (10-33 cm)-sketchy as it is-and from 'larger distance' analysis (10-11 em) would seem to [be able] to reinforce each other in a most natural way. \"The most evident shortcoming of the geometrodynarnic model as it stands is this, that it fails to supply any completely natural place for spin ~ in general and for the neutrino in particular.\" Attempts to find a natural place for spin l' in Einstein's standard geometrodynamics (Box 44.3) founder because there is no natural way for a change in connectivity to take place within the context of classical differential geometry. A uranium nucleus undergoing fission starts with one topology and nevertheless __~!1ds_!!p with an6ther topology. It makes this transition in a perfectly continuous way, classical differential geometry notwithstanding. There are reputed to be two kinds of lawyers. One tells the client what not to do. The other listens to what the client has to do and tells him how to do it. From the first lawyer, classical differential geometry, the client goes away disappointed, still searching for a natural way to describe quantum fluctuations in the connectivity of space. Only in this way can he hope to describe electric charge as lines of electric force trapped in the topology of space. Only in this way does he expect to be able to understand and analyze the final stages of gravitational collapse. Pondering his problems, he comes to the office of a second lawyer, with the name \"Pregeometry\" on the door. Full of hope, he knocks and enters. What is pregeometry to be and say? Born of a combination of hope and need, of philosophy and physics and mathematics and logic, pregeometry will tell a story unfinished at this writing, but full of incidents of evolution so far as it goes. \u00a744.4. NOT GEOMETRY, BUT PREGEOMETRY, AS THE MAGIC BUILDING MATERIAL An early survey (Box 44.4) asked whether geometry can be constructed with the help of the quantum principle out of more basic elements, that do not themselves have any specific dimensionality. The focus of attention in this 1964 discussion was \"dimensionality without dimen sionality.\" However, the prime pressures to ponder pregeometry were and remain (continued on page 12(6) No place in geometrodynamics for change of topology; therefore turn to \"pregeometry\" 1204 44. BEYOND THE END OF TIME Box 44.3 THE DIFFICULTIES WITH ATTEMPTS TO FIND A NATURAL PLACE FOR SPIN ~ IN EINSTEIN'S STANDARD GEOMETRODYNAMICS \"It is impossible\" [Wheeler (1962)] \"to accept any description of elementary particles that does not have a place for spin~. What, then, has any purely geometric description to offer in explanation of spin ~ in general? More particularly and more im portantly, what place is there in quantum geomet rodynamics for the neutrino-the only entity of half-integral spin that is a pure field in its own right, in the sense that it has zero rest mass and moves with the speed of light? No clear or satisfac tory answer is known to this question today. Un less and until an answer is forthcoming, pure geo metrodynamics must be judged deficient as a basis for elementary particle physics.\" A later publication [Wheeler (1968a)] takes up this issue again, noting that, \"A new world opens out for analysis in quantum geometrodynamics. The central new concept is space resonating be tween one foamlike structure and another. For this multiple-connectedness of space at submicroscopic distances no single feature of nature speaks more powerfully than electric charge. Yet at least as impressive as charge is the prevalence of spin ~ throughout the world of elementary particles.\" Repeating the statement that \"It is impossible to accept any description of elementary particles that does not have a place for spin 1,\" the article adds to the discussion a new note: \"Happily, the concept of spin manifold has come to light, not least through the work of John Milnor [see Lich nerowicz (1961a,b,c) and (1964); Milnor (1962), (1963), and (1965a,b); Hsiang and Anderson (1965); Anderson, Brown, and Peterson (1966a,b); and Penrose (1968a)]. This concept suggests a new and interesting interpretation of a spinor field within the context of the resonating microtopology of quantum geometrodynamics, as the nonclassical two-valuedness [Pauli's standard term for spin; see, for example, Pauli (1947)] that attaches to the prob ability amplitude for", " otherwise identical 3-geome tries endowed with alternative'spin structures.'\" More specifically: \"One does not classify the closed orientable 3-manifold of physics completely when one gives its topology, its differential struc tures, and its metric. One must tell which spin structure it has.\" [On a 3-geometry with the topol ogy of a 3-sphere, one can lay down a continuous field of triads (a triad consisting of three orthonor mal vectors). Any other continuous field of triads can be deformed into the first field by a continuous sequence of small readjustments. One says that the 3-sphere admits only one \"spin structure,\" a po tentially misleading standard word for what could just as well have been called a \"triad structure.\" In contrast, a 3-sphere with n handles or worm holes admits 2n \"spin structures\" (continuous fields of triads) inequivalent to one another under any continuous sequence of small readjustments what soever, and distinguished from one another in any convenient way by n \"descriptors\" wl' wz'...., wk'\u2022 \u2022 \u2022 \u2022 wn.] It is natural in quantum geometro dynamics to expect \"separate probability ampli tudes for a 3-geometry with descriptor W k = + I and for an otherwise identical 3-geometry with descriptor Wk = - 1. Does this circumstance imply that quantum geometrodynamics supplies all the machinery one needs to describe fields of spin 1 in general and the neutrino field in particular?... That is the only way that has ever turned up within the framework of Einstein's general relativity and Planck's quantum principle. Is this the right path? It is difficult to name any question more decisive than this in one's assessment of 'everything as geometry.''' Why not spell out these concepts, reduce them to practice, and compare them with what one knows about the behavior of fields of spin 1? There is a central difficulty in this enterprise. It assumet and demands on physical grounds that the topol ogy of the 3-geometry shall be free to change from one connectivity to another. In contrast, classical differential geometry says, in effect, \"Once one topology, always that topology.\" Try a question like this, \"When a new handle develops and the number of descriptors rises by one, what boundary condition in s\u00b7uperspace connects the probability amplitude 0/ for 3-geometries of the original topol ogy with the probability amplitudes 0/+ and 0/- for the two spin structures of the new topology?\" Classical differential geometry not only gives one no help in answering this question; it even forbids one to ask it. In other words, one cannot even get the enterprise \"on the road\" for want of a natural mathematical way to describe the required change in topology. The idea is therefore abandoned here and now that 3-geometry is \"the magic building material of the universe.\" In contrast, pregeometry (see text), far from being endowed with any frozen topology, is to be viewed as not even possessing any dimensionality. Box 44.4 \"A BUCKET OF DUST\"-AN EARLY ATTEMPT TO FORMULATE THE CONCEPT OF PREGEOMETRY [Wheeler (1964a)] i \"What line of thought could ever be imagined as leading to four dimensions-or any dimensionality at all-out of more primitive considerations? In the case of atoms one derives the yellow color of the sodium D-lines by analyzing the quantum dynamics of a system, no part of which is ever endowed with anything remotely resembling the attribute of color. Likewise any derivation of the four-dimensionality of spacetime can hardly start with the idea of dimensionality.\" \"Recall the notion of a Borel set. Loosely speak ing, a Borel set is l!_ collection of points (\"bucket of dust\") which have not yet been assembled into a manifold of any particular dimensionality.... Recalling the universal sway of the quantum prin ciple, one can imagine probability amplitudes for the points in a Borel set to be assembled into points with this, that, and the other dimensional ity.... More conditions have to be imposed on a given number of points-as to which has which for a nearest neighbor-when the points are put together in a five-dimensional array than when these same points are arranged in a two-dimen sional pattern. Thus one can think of each dimen sionalityas having a much higher statistical weight than the next higher dimensionality. On the other hand, for manifolds with one, two, and three di mensions, the geometry is too rudimentary-one can suppose-to give anything interesting. Thus Einstein's field equations, applied to a manifold of dimensionality so low, demand flat space; only when the dimensionality is as high as four do really interesting", " possibilities arise. Can four, therefore, be considered to be the unique dimen sionality which is at the same time high enough to give any real physics and yet low enough to have great statistical weight? \"It is too much to imagine that one has yet made enough mistakes in this domain of thought to explore such ideas with any degree of good judg ment.\" Consider a handle on the geometry. Let it thin halfway along its length to a point. In other words, let the handle dissolve into two bent prongs that touch at a point. Let these prongs separate and shorten. In this process two points part company that were once immediate neighbors. \"However sudden the change is in classical theory, in quan tum theory there is a probability amplitude func tion which falls off in the classically forbidden there is some residual domain. In other words, connection between points which are ostensibly very far away (travel from one 'tip' down one then through the larger space to which prong, these prongs are attached, and then up the other prong to the other tip). But there is nothing dis tinctive in principle about the two points that have happened to come into discussion. Thus it would seem that there must be a connection... between every point and every other point. Under these conditions the concept of nearest neighbor would appear no longer to make sense. Thus the tool disappears with the help of which one might oth erwise try to speak [un]ambiguously about dimen sionality.\" Sakharov: gravitation is the \"metric elasticity of space\" The stratification of space Comparison with everyday elasticity 1206 44. BEYOND THE END OF TIME two features of nature, spin ~ and charge, that speak out powerfully from every part of elementary particle physics. A fresh perspective on pregeometry comes from a fresh assessment of general relativity. \"Geometrodynamics is neither as important or as simple as it looks. Do not make it the point of departure in searching for underlying simplicity. Look deeper, at elementary particle physics.\" This is the tenor of interesting new consider ations put forward by Sakharov [the Sakharov] (1967) and summarized under the heading, \"Gravitation as the metric elasticity of space,\" in Box 17.2. In brief, as elasticity is to atomic physics, so-in Sakharov's view-gravitation is to elementary particle physics. The energy of an elastic deformation is nothing but energy put into the bonds between atom and atom by the deformation. The energy that it takes to curve space is nothing but perturbation in the vacuum energy of fields plus particles brought about by that curvature, according to Sakharov. The energy required for the deformation is governed in the one case by two elastic constants and in the other case by one elastic constant (the Newtonian constant of gravity) but in both cases, he reasons, the constants arise by combination of a multitude of complicated individual effects, not by a brave clean stroke on an empty slate. One gives all the more favorable reception to Sakhorov's view of gravity because one knows today, as one did not in 1915, how opulent in physics the vacuum is. In Einstein's day one had come in a single decade from the ideal God-given Lorentz perfection of flat spacetime to curved spacetime. It took courage to assign even one physical constant to that world of geometry that had always stood so far above physics. The vacuum looked for long as innocent of structure as a sheet of glass emerging from a rolling mill. With the discovery of the positive electron [Anderson (1933)], one came to recognize a little of the life that heat can unfreeze in \"empty\" space. Each new particle and radiation that was discovered brought a new accretion to the recognized richness of the vacuum. Macadam looks smooth, but a bulldozer has only to cut a single furrow through the roadway to disclose all the complications beneath the surface. Think of a particle as built out of the geometry of space; think of a particle as a \"geometrodynamic exciton\"? No model-it would seem to follow from Sakharov's assessment-could be less in harmony with nature, except to think of an atom as built out of elasticity! Elasticity did not explain atoms. Atoms explained elasticity. If, likewise, particles fix the constant in Einstein's geometrodynamic law (Sakharov), must it not be unreasonable to think of the geometrodynamic law as explaining particles? Carry the comparison between geometry and elasticity one stage deeper (Figi44.2). In a mixed solid there are hundreds of distinct bonds, all of which contribute to the elastic constants; some of them arise from Van der Waal's forces, some from ionic coupling, some from homopolar linkage; they have the greatest variety of -strengths; but all have their origin in something so fantastically simple as a", " system of positively and negatively charged masses moving in accordance with the laws of quantum mechanics. In no way was it required or right to meet each complication of the chemistry and physics ofa myriad of bonds with a corresponding complication of principle. By going to a level of analysis deeper than bond strengths, one had \u00a744.4. PREGEOMETRY AS THE MAGIC BUILDING MATERIAL 1207 Two elastic constants One gravitational constant Dynamic system of + and - charged Dynamic \"pregeometry\" masses Figure 44.2. Elasticity and geometrodynamics, as viewed at three levels of analysis. A hundred years of the study of elasticity did not reveal the existence of molecules, and a hundred years of the study of molecular chemistry did not reveal Schr6dinger's equation. Revelation moved upward in the diagram, not downward. emerged into a world of light, where nothing but simplicity and unity was to be seen. Compare with geometry. The vacuum is animated with the zero-point activity of distinct fields and scores of distinct particles, all of which, according to Sakharov, contribute to the Newtonian G, the \"elastic constant of the metric.\" Some interact via weak forces, some by way of electromagnetic forces, and some through strong forces. These interactions have the greatest variety of strengths. But must not all these particles and interactions have their origin in something fantastically simple? And must not this something, this \"pregeometry,\" be as far removed from geometry as the quantum mechanics of electrons is far removed from elasticity? Ifone once thought of general relativity as a guide to the discovery of pregeometry, nothing might seem more dismaying than this comparison with an older realm of physics. No one would dream of studying the laws of elasticity to uncover the principles of quantum mechanics. Neither would anyone investigate the work-hard ening of a metal to learn about atomic physics. The order of understanding ran not Work-hardening (1 em) ---+ dislocations (10-4 em) ---+ atoms (10-8 em), but the direct opposite, Atoms (10-8 em) ---+ dislocations (10-4 em) ---+ work-hardening (l em) Search for the central principle of pregeometry 1208 44. BEYOND THE END OF TIME One had to know about atoms to conceive of dislocations, and had to know about dislocations to understand work-hardening. Is it not likewise hopeless to go from the \"elasticity of geometry\" to an understanding of particle physics, and from particle physics to the uncovering of pregeometry? Must not the order of progress again be the direct opposite? And is not the source of any dismay the apparent loss of guidance that one experiences in giving up geometrodynamics-and not only geo metrodynamics, but geometry itself-as a crutch to lean on as one hobbles forward? Yet there is so much chance that this view of nature is right that one must take it seriously and explore its consequences. Never more than today does one have the incentive to explore pregeometry. \u00a744.5. PREGEOMETRY AS THE CALCULUS OF PROPOSITIONS Paper in white the floor of the room, and rule it off in one-foot squares. Down on one's hands and knees, write in the first square a set of equations conceived as able to govern the physics of the universe. Think more overnight. Next day put a better set of equations into square two. Invite one's most respected colleagues to contribute to other squares. At the end of these labors, one has worked oneself out into the door way. Stand up, look back on all those equations, some perhaps more hopeful than others, raise one's finger commandingly, and give the order \"Fly!\" Not one of those equations will put on wings, take off, or fly. Yet the universe \"flies.\" Some principle uniquely right and uniquely simple must, when one knows it, be also so compelling that it is clear the universe is built, and must be built, in such and such a way, and that it could not possibly be otherwise: But how can one discover that principle? If it was hopeless to learn atomic physics by studying work-hardening and dislocations, it may be equally hopeless to learn the basic operating principle of the universe, call it pregeometry or call it what one will, by any amount of work in general relativity and particle physics. Thomas Mann (1937), in his essay on Freud, utters what Niels Bohr would surely have called a great truth (\"A great truth is a truth whose opposite is also a great truth\") when he says, \"Science never makes an advance until philosophy authorizes and encourages it to do so.\" If the equivalence principle (Chapter 16) and Mach's principle (\u00a721.9) were the philosophical godfathers of general relativity, it is also true that what those principles", " do mean, and ought to mean, only becomes clear by study and restudy of Einstein's theory itself. Therefore it would seem reasonable to expect the primary guidance in the search for pregeometry to come ftom a principle both philosophical and powerful, but one also perhaps not destined to be wholly clear in its contents or its implications until some later day. Among all the principles that one can name out of the world ofscience, it is difficult to think of one more compelling than simplicity; and among all the simplicities of dynamics and life and movement, none is starker [Werner (1969)] than the binary choice yes-no or true-false. It in no way proves that this choice for a starting principle is correct, but it at least gives one some comfort in the choice, that Pauli's \"nonclassi cal two-valuedness\" or \"spin\" so dominates the world of particle physics. \u00a744.6. THE REPROCESSING OF THE UNIVERSE 1209 It is one thing to have a start, a tentative construction of pregeometry; but how does one go on? How not to go on is illustrated by Figure 44.3. The \"sewing machine\" builds objects of one or another definite dimensionality, or ofrnixed dimensionalities, according to the instructions that it receives on the input tape in yes-no binary code. Some of the difficulties of building up structure on the binary element according to this model, or anyone of a dozen other models, stand out at once. (I) Why N = 10,000 building units? Why not a different N? And if one feeds in one such arbitrary number at the start, why not fix more features \"by hand?\" No natural stopping point is evident, nor any principle that would fix such a stopping point. Such arbitrariness contradicts the principle of simplicity and rules out the model. (2) Quantum mechanics is added from outside, not generated from inside (from the model itself). On this point too the principle of simplicity speaks against the model. (3) Tl;J.e passage from pregeometry to geometry is made in a too-literal-minded way, with no appreciation of the need for particles and fields to appear along the way. The model, in the words used by Bohr on another occasion, is \"crazy, but not crazy enough to be right.\", Noting these difficulties, and fruitlessly trying model after model of pregeometry to see if it might be free of them, one suddenly realizes that a machinery for the combination of yes-no or true-false elements does not have to be invented. It already exists. What else can pregeometry be, one asks oneself, than the calculus of proposi tions? (Box 44.5.) \u00a744.6. THE BLACK BOX: THE REPROCESSING OF THE UNIVERSE No amount of searching has ever disclosed a \"cheap way\" out of gravitational collapse, any more than earlier it revealed a cheap way out of the collapse of the atom. Physicists in that earlier crisis found themselves in the end confronted with a revolutionary pistol, \"Understand nothing-or accept the quantum principle.\" Today's crisis can hardly force a lesser revolution. One sees no alternative except to say that geometry fails and pregeometry has to take its place to ferry physics through the final stages of gravitational collapse and on into what happens next. No guide is evident on this uncharted way except the principle of simplicity, applied to drastic lengths. Whether the whole universe is squeezed down to the Planck dimension, or more or less, before reexpansion can begin and dynamics can return to normal, may be irrelevant for some of the questions one wants to consider. Physics has long used the \"black box\" to symbolize situations where one wishes to concentrate on what goes in and what goes out, disregarding what takes place in between. At the beginning of the crisis of electric collapse one conceived of the electron as headed on a deterministic path toward a point-center of attraction, and unhappily destined to arrive at a condition of infinite kinetic energy in a finite time. After the advent of quantum mechanics, one learned to summarize the interaction between (continued on page 1213) A first try at a pregeometry built on the principle of binary choice A more reasonable picture: pregeometry is the calculus of propositions The role of the black box in physics 1210 Figure 44.3. \"Ten thousand rings\"; or an example of a way to think of the connection between pregeometry and geometry, wrong because it is too literal-minded, and for other reasons spelled out in the text. The vizier [story by Wheeler, as alluded to by Kilmister (1971)*] speaks: \"Take N = 10,000 brass rings. Take an automatic fastening device that will cut open a ring, loop it through another ring, and resolder the joint. Pour the brass rings into the hopper that feeds", " this machine. Take a strip of instruction paper that is long enough to contain N(N - 1)/2 binary digits. Look at the instruction in the (jk)-th location on this instruction tape (j,k = 1,2,..., N; j < k). When the binary digit at that location is 0, it is a signal to leave the j-th ring disconnected from the k-th ring. When it is I, it is lin instruction to connect that particular pair of rings. Thread the tape into the machine and press the start button. The clatter begins. Out comes a chain of rings 10,000 links long. It falls on the table and the machine stops. Pour in another 10,000 rings, feed in a new instruction tape. and push the button again. This time it is not a one-dimen sional structure that emerges, but a two-dimensional one: a Crusader's coat of mail, complete with neck opening and sleeves. Take still another tape from the library of tapes and repeat. Onto the table thuds a smaller version of the suit of mail, this time filled out internally with a solid network of rings, a three-dimensional structure. Now forego the library and make one's own instruction tape, a random string of O's and I's. Guided by it, the fastener produces a \"Christmas tree ornament,\" a collection of segments of one-dimensional chain, two-dimensional surfaces, and three-, four-, five-, and higher-di mensional entities, some joined together, some free-floating. Now turn from a structure deterministically fixed by a tape to a probability amplitude, a complex number, o/(tape) = 0/(n12, n13, n14' \u00b7 \u00b7 \u00b7 \u2022\u2022\u2022, nN-1,N) (n i; = 0, I), (I) defined over the entire range of possibilities for structures built of 10,000 rings. Let these probability amplitudes not be assigned randomly. Instead, couple together amplitudes, for structures that differ from \"each other by the breaking of a single ring, by linear form ulas that treat all rings on the same footing. The separate o/'s, no longer entirely independent, will still give non-zero probability amplitudes for \"Christmas tree ornaments.\" Of greater immediate interest than these \"unruly\" parts of the structures are the following questions about the smoother parts: (I) In what kinds of structures is the bulk of the probability concentrated? (2) What is the dominant dimensionality of these structures in an appropri ate correspondence principle limit? (3) In this semiclassicallimil, what is the form taken by the dynamic law of evolution of the geometry?\" No principle more clearly rules out this model for pregeometry than the principle of simplicity (see text). \u2022 Wheeler's story about the vizier and whal the vizier had to say about superspace was told at the May 18, 1970, Gwatt Seminar on the Bearings of Topology upon General Relativity. Kilmister's (1971) published article alludes to the unpublished slOry, but does not actual1y contain it. \u00a744.6. THE REPROCESSING OF THE UNIVERSE 1211 Box 44.5 \"PREGEOMETRY AS THE CALCULUS OF PROPOSITIONS\" A sample proposition taken out of a standard text on logic selected almost at random reads [Kneebone (1963), p. 40] [X ---+ \u00abX ---+ X) ---+ Y)] & (X ---+ Z) eq (X V Y V Z) & (X V Y V Z) & (X V Y V Z) & (X V Y V Z). The symbols have the following meaning: ;t, AV B, A &B, Not A; A or B or both (\"A vel B\"); A and B; A ---+ B, A implies B (\"if A, then B\"); A~B, B is equivalent to A (\"B if and only if A\"). Propositional formula ~ is said to be equivalent (\"eq\") to propositional formula ~ if and only if ~ +--+~ is a tautology. The letters A, B, etc., serve as connectors to \"wire together\" one proposition with another. Proceeding in this way, one can construct propositions of indefinitely great length. A switching circuit [see, for example, Shannon (1938) or Hohn (1966)] is isomor phic to a proposition. Compare a short_proposition or an elementary switching circuit to a molecular collision. No idea seemed more preposterous than that of Daniel Bernoulli (1733), that heat is a manifestation ofmolecular collisions. Moreover, a three-body encounter is difficult to treat, a four-body collision is more difficult, and a five- or more molecule system is essentially intractable.", " Nevertheless, mechanics acquires new elements of simplicity in the limit in which the number of molecules is very great and in which one can use the concept of density in phase space. Out of statistical mechanics in this limit come such concepts as temperature and entropy. When the temperature is well-defined, the energy of the system is not a well-defined idea; and when the energy is well-defined, the temperature is not. This complementarity is built ines capably into the principles of the subject. Thrust the finger into the flame of a match and experience a sensation like nothing else on heaven or earth; yet what happens is all a consequence of molecular collisions, early critics notwithstanding. Any individual proposition is difficult for the mind to apprehend when it is long; and still more difficult to grasp is the content of a cluster of propositions. Neverthe less, make a statistical analysis of the calculus of propositions in the limit where the number of propositions is great and most of them are long. Ask if parameters force themselves on one's attention in this analysis (l) analogous in some small measure to the temperature and entropy of statistical mechanics but (2) so much 1212 44. BEYOND THE END OF TIME Box 44.5 (continued) more numerous, and everyday dynamic in character, that they reproduce the contin uum of ev~iyday physics. Nothing could seem so preposterous at first glance as the thought that nature is built on a foundation as ethereal as the calculus of propositions. Yet, beyond the push to look in this direction provided by the principle of simplicity, there are two pulls. First, bare-bones quantum mechanics lends itself in a marvelously natural way to formulation in the language of the calculus of propositions, as witnesses not least the book of Jauch (1968). If the quantum principle were not in this way already automatically contained in one's proposed model for pregeometry, and if in contrast it had to be introduced from outside, by that very token one would conclude t~at the model violated the principle of simplicity, and would have to reject it. Second, the pursuit of reality seems always to take one away from reality. Who would have imagined describing something so much a part of the here and now as gravitation in terms of curvature of the geometry of spacetime? And when later this geometry came to be recognized as dynamic, who would have dreamed that geometrodynamics unfolds in an arena so ethereal as superspace? Little astonishment there should be, therefore, if the description of nature carries one in the end to logic, the ethereal eyrie at the center of mathematics. If, as one believes, all mathematics reduces to the mathematics of logic, and all physics reduces to mathematics, what alternative is there but for all physics to reduce to the mathematics of logic? Logic is the only branch of mathematics that can \"think about itself.\" \"An issue oflogic having nothing to do with physics\" was the assessment by many of a controversy of old about the axiom, \"parallel lines never meet.\" Does it follow from the other axioms of Euclidean geometry or is it independent? \"Independent,\" Bolyai and Lobachevsky proved. With this and the work of Gauss as a start, Riemann went on to create Riemannian geometry: Study nature, not Euclid, to find out about geometry, he advised; and Einstein went on to take that advice and to make geometry a part of physics. \"An issue oflogic having nothing to do with physics\" is one's natural first assess ment of the startling limitations on logic discovered by GOdel (1931), Cohen (1966), and others [for a review, see, for example, Kac and Ulam (1968)]. The exact opposite must be one's assessment if the real pregeometry of the real physical world indeed turns out to be identical with the calculus of propositions. \"Physics as manifestation oflogic\" or \"pregeometry as the calculus ofpropositioIlf\" is as yet [Wheeler (l971a)] not an idea, but an idea for an idea. It is put forward here only to make it a little clearer what it means to suggest that the order of progress may not be but physics --+ pregeometry pregeometry --+ physics. \u00a744.6. THE REPROCESSING OF THE UNIVERSE 1213 Figure 44.4. The \"black-box model\" applied (I) to the scattering of an electron by a center of attraction and (2) to the collapse of the universe itself. The deterministic electron world line of classical theory is replaced in quantum theory by a probability amplitude, the wave crests of which are illustrated schematically in the diagram. The catastrophe of classical theory is replaced in quantum theory by a probability distribution of outputs. The same diagram illustrates the \"black-box account\" of gravitational collapse mentioned in the text. The arena of the diagram is no longer spacetime, but", " superspace. The incident arrow marks no longer a classical world line of an electron through spacetime, but a classical \"leaf of history of geometry\" slicing through superspace (Chapter 43). The wave crests symbolize no longer the electron wave function propagating through spacetime, but the geometrodynamic wave function propagating through superspace. The cross-hatched region is no longer the region where the one-body potential goes to infinity, but the region of gravitational collapse where the curvature of space goes to infinity. The outgoing waves describe no longer alternative directions for the new course of the scattered electron, but the beginnings of alternative new histories for the universe itself after collapse and \"reprocessing\" end the present cycle. center of attraction and electron in a \"black box:\" fire in a wave-train of electrons traveling in one direction, and get electrons coming out in this, that, and the other direction with this, that, and the other well-determined probability amplitude (Figure 44.4). Moreover, to predict these probability amplitudes quantitatively and correctly, it was enough to translate the Hamiltonian of classical theory into the language of wave mechanics and solve the resulting wave equation, the key to the \"black box.\" A similar \"black box\" view of gravitational collapse leads one to expect a \"proba bility distribution of outcomes.\" Here, however, one outcome is distinguished from another, one must anticipate, not by a single parameter, such as the angle of scatter ing of the electron, but by many. They govern, one foresees, such quantities as the size of the system at its maximum of expansion, the time from the start of this new cycle to the moment it ends in collapse, the number of particles present, and a thousand other features. The \"probabilities\" of these outcomes will be governed by a dynamic law, analogous to (1) the Schrodinger wave equation for the electron, -or, to cite another black box problem, (2) the Maxwell equations that couple together, at a wave-guide junction, electromagnetic waves running in otherwise separate wave guides. However, it is hardly reasonable to expect the necessary dynamic law to spring forth as soon as one translates the Hamilton-Jacobi equation of general relativity (Chapter 43) into a Schrodinger equation, simply because geometrody namics, in both its classical and its quantum version, is built on standard differential geometry. That standard geometry leaves no room for any of those quantum fluc tuations in connectivity that seem inescapable at small distances and therefore also inescapable in the final stages of gravitational collapse. Not geometry, but pregeom etry, must fill the black box of gravitational collapse. Probability distribution of the outcomes lllf collapse \"Reprocessing\" the universe All conservation laws transcended in the collapse of the universe 1214 44. BEYOND THE END OF TIME Little as one knows the internal machinery of the black box, one sees no escape from this picture of what goes on: the universe transforms, or transmutes, or transits, or is reprocessed probabilistically from one cycle of history to another in the era of collapse. However straightforwardly and inescapably this picture of the reprocessing of the universe would seem to follow from the leading features of general relativity and the quantum principle, the two overarching principles of twentieth-century physics, it is nevertheless fantastic to contemplate. How can the dynamics of a system so incredibly gigantic be switched, and switched at the whim of probability, from one cycle that has lasted 1011 years to one that will last only 106 years? At first, only the circumstance that the system gets squeezed down in the course of this dynamics to incredibly small distances reconciles one to a transformation otherwise so unbe lievable. Then one looks at the upended strata of a mountain slope, or a bird not seen before, and marvels that the whole universe is incredible: mutation of a species, metamorphosis of a rock, chemical transformation, spontaneous transformation of a nucleus, radioactive decay of a particle, reprocessing of the universe itself. If it cast a new light on geology to know that rocks can be raised and lowered thousands of meters and hundreds of degrees, what doe~jt mean for physics to think of the universe as being from time to time \"squeezed through a knothole,\" drastically \"reprocessed,\" and started out on a fresh dynamic cycle? Three considerations above all press themselves on one's attention, prefigured in these compressed phrases: destruction of all constants of motion in collapse; particles, and the physical \"constants\" themselves, as the \"frozen-in part of the meteorology of collapse;\" \"the biological selection of physical constants.\" The gravitational collapse of a star, or a collection of stars, to a black hole extinguishes all details of the system (see Chapters 32 and33) except mass and charge and angular momentum. Whether made of matter or antimatter or radiation, whether endowed", " with much entropy or little entropy, whether in smooth motion orfhaotic turbulence, the collapsing system ends up as seen from outside, according to all indications, in the same standard state. The laws of conservation of baryon number and lepton number are transcended [Chapter 33; also Wheeler (1971 b\u00bb). No known means whatsoever will distinguish between black holes of the most different prove nance if only they have the same mass, charge, and angular momentum. But for a closed universe, even these constants vanish from the scene. Total charge is automatically zero because lines of force have nowhere to end except upon charge. Total mass and total angular momentum have absolutely no definable meaning whatsoever for a closed universe. This conclusion follows not least because there \u00a744.6. THE REPROCESSING OF THE UNIVERSE 1215 is no asymptotically flat space outside where one can put a test particle into Keplerian orbit to determine period and precession. Of all principles of physics, the laws of conservation of charge, lepton number, baryon number, mass, and angular momentum are among the most firmly estab lished. Yet with gravitational collapse the content of these conservation laws also collapses. The established is disestablished. No determinant of motion does one see left that could continue unchanged in value from cycle to cycle of the universe. Moreover, if particles are dynamic in construction, and if the spectrum of particle masses is therefore dynamic in origin, no option would seem left except to conclude that the mass spectrum is itself reprocessed at the time when \"the universe is squeezed through a knot hole.\" A molecule in this piece of paper is a \"fossil\" from photochemical synthesis in a tree a few years ago. A nucleus of the oxygen in this air is r fossil from thermonuclear combustion at a much higher temperature in a star a few 109 years ago. What else can a particle be but a fossil from the most violent event of all, gravitational collapse? That one geological stratum has one many-miles long slope, with marvelous linearity of structure, and another stratum has another slope, is either an everyday triteness, taken as for granted by every passerby, or a miracle, until one understands the mechanism. That an electron here has the same mass as an electron there is also a triviality or a miracle. It is a triviality in quantum electrodynamics because it is assumed rather than derived. However, it is a miracle on any view that regards the universe as being from time to time \"reprocessed.\" How can electrons at different times and places in the present cycle of the universe have the same mass if the spectrum of particle masses differs between one cycle of the universe and another? Inspect the interior of ~particle of one type, and magnify it up enormously, and in that interior see one view of the whole universe [compare the concept of monad of Leibniz (1714), \"The monads have no window through which anything can enter or depart\"]; and do likewise for another particle of the same type. Are particles of the same pattern identical in anyone cycle of the universe because they give identically patterned views of the same universe? No acceptable explanation for the miraculous identity of particles of the same type has ever been put forward. That identity must be regarded, not as a triviality, but as a central mystery of physics. Not the spectrum of particle masses alone, but the physical \"constants\" themselves, would seem most reasonably regarded as reprocessed from one cycle to another. Reprocessed relative to what? Relative, for example, to the Planck system of units, L * = (fiG/ C 3)l/2 = 1.6 X 10- 33 cm, T* = (fiG/ C5)l/2 = 5.4 X 10-44 sec, M* = (flC/G)l/2 = 2.2 X 10-5 g, the only system of units, Planck (1899) pointed out, free, like black-body radiation itself, of all complications of solid-state physics, molecular binding, atomic constitu tion, and elementary particle structure, and drawing for its background only on the simplest and most universal principles of physics, the laws of gravitation and black body radiation. Relative to the Planck units, every constant in every other part of physics is expressed as a pure number. Three hierarchies of fossils: molecules. nuclei. particles Reason for identity in mass of particles of the same species? Reprocessing of physical constants 1216 44. BEYOND THE END OF TIME No pure numbers in physics are more impressive than ne/e 2 = 137.0360 and the so-called \"big numbers\" [Eddington (1931,1936,1946); Dirac (1937,1938); Jordan (1955, 1959); Dicke (1959b, 1961, 1964b); Hayak", "awa (1965a,b); Carter (1968b\u00bb): _10 80 particles in the universe,* _10 40 _ 10\"8 em 10- 12 cm ( radiu.s of universe. at)* maximum expansIOn (\"SiZ~\" of an elementary), particle _10 40 - - - - e 2 GmM (electric forces) --'-.,---:------:---:----'-(gravitational forces), _10 20 _ 2 e me 2/ (nG/e 3)1/2 ( \"SiZ~\" of an elementary) particle (Planck length) _10 10 _ (?Umb.er of photons) III UnIverse (?um~er of baryonS). III UnIVerse Some understanding of the relationships between these numbers has been won [Carter (1968b\u00bb). Never has any explanation appeared for their enormous magnitude, nor will there ever, if the view is correct that reprocessing the universe reprocesses also the physical constants. These constants on that view are not part of the laws of physics. They are part of the initial-value data. Such numbers are freshly given for each fresh cycle of expansion of the universe. To look for a physical explanation for the \"big numbers\" would thus seem to be looking for the right answer to the wrong question. In the week between one storm and the next, most features of the weather are ever-changing, but some special patterns of the wind last the week. If the term \"frozen features of the meteorology\" is appropriate for them, much more so would it seem appropriate for the big numbers, the physical constants and the spectrum of particle masses in the cycle between one reprocessing of the universe and another. A per cent or so change one way in one of the \"constants,\" ne/e 2, will cause all stars to be red stars; and a comparable change the other way will make hll stars be blue stars, according to Carter (1968b). In neither case will any star like the sun be possible. He raises the question whether life could have developed if the deter minants of the physical constants had differed substantially from those that charac terize this cycle of the universe. Dicke (1961) has pointed out that the right order of ideas may not be, here is the universe, so what must man be; but here is man, so what must the universe \u00b7Values based on the \"typical cosmological model\" of Box 27.4; subject to much uncertainty, in the present state of astrophysical distance determinations, not least because the latitude in these numbers is even enough to be compatible with an open universe. Values of physical constants as related to the possibilities for life \u00a744.6. THE REPROCESSING OF THE UNIVERSE 1217 be? In other words: (1) What good is a universe without awareness of that upiverse? But: (2) Awareness demands life. (3) Life demands the presence of elements heavier than hydrogen. (4) The production of heavy elements demands thermonuclear combustion. (5) Thermonuclear combustion normally requires several 109 years of cooking time in a star. (6) Several 109 years of time will not and cannot be available in a closed universe, according to general relativity, unless the radius-at-maximum expansion of that universe is several 109 light years or more. So why on this view is the universe as big as it is? Because only so can man be here! In brief, the considerations of Carter and Dicke would seem to raise the idea of the \"biological selection of physical constants.\" However, to \"select\" is impossible unless there are options to select between. Exactly such options would seem for the first time to be held out by the only over-all picture of the gravitational collapse of the universe that one sees how to put forward today, the pregeometry black-box mode! of the reprocessing of the universe. Proceeding with all caution into uncharted territory, one must nevertheless be aware that the conclusions one is reaching and the questions one is asking at a given stage of the analysis may be only stepping stones on the way to still more penetrating questions and an even more remarkable picture. To speak of \"reprocessing and selection\" may only be a halfway point on the road toward thinking of the universe as Leibniz did, as a world of relationships, not a world of machinery. Far from being brought into its present condition by \"reprocessing\" from earlier cycles, may the uni verse in some strange sense be \"brought into being\" by the participation of those who participate? On this view the concept of \"cycles\" would even seem to be altogether wrong. Instead the vital act is the act of participation. \"Participator\" is the incontrovertible new concept given by quantum mechanics; it strikes down the term \"observer\" of classical theory, the man who stands safely", " behind the thick glass wall and watches what goes on without taking part. It can't be done, quantum mechanics says. Even with the lowly electron one must participate before one can give any meaning whatsoever to its position or its momentum. Is this firmly estab lished result the tiny tip of a giant iceberg? Does the universe also derive its meaning from \"participation\"? Are we destined to return to the great concept of Leibniz, of \"preestablished harmony\" (\"Leibniz logic loop\"), before we can make the next great advance? Rich prospects stand open for investigation in gravitation physics, from neutron stars to cosmology and from post-Newtonian celestial mechanics to gravitational waves. Einstein's geometrodynamics exposes itself to destruction on a dozen fronts and by a thousand predictions. No predictions subject to early test are mOi\"e en trancing than those on the formation and properties of a black hole, \"laboratory model\" for some of what is predicted for the universe itself. No field is more pregnant with the future than gravitational collapse. No more revolutionary views of man and the universe has one ever been driven to consider seriously than those that come out of pondering the paradox of collapse, the greatest crisis of physics of all time. All of these endeavors are based on the belief that existence should have a completely harmonious structure. Today we have less ground than ever before for allowing ourselves to be forced away from this wonderful belief. EINSTEIN (1934) Black hole as \"Iaboratory\" model for collapse of universe............................................, \u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022............................................... :~:!! ~: ~ ~ ~:!::!:::::::: ~:::::::: ~:::::::::: \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 f What is now proved was once only imagin'd. WILLIAM BLAKE (Sur I'air de \"J'ai rendez-vous avec vous,\" chanson de Georges Brasseus)................... ANON.. J. D'ALEMBERT We will first understand How simple the universe is When we realize How strange it is. Un argument qu'on revere Celui de Synge pour dire Ie tout Nous promet Ie quadrupolaire Tout est rlatif apres tout To.some one who could grasp the universe from a unified standpoint. the entire creation would appear as a unique truth and necessity.. Le Rayonnement dipolaire.On sait qu'il n'est pas pour nous C'est pour Maxwell, oui mais Maxwell on s'en fout Tout est rlatif apres tout Les sources quasi stellaires Disparaissent comme dans un trou Dans Ie Schwarzschild, oui mais Schwarzschild on s'en fout Tout est rlatif apres tout \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 :\u2022.... \u2022\u2022\u2022\u2022\u2022\u2022 \u2022.... \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022....:. \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022 \u2022\u2022....:..::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::.:::::::::::::::::::. \u2022\u2022~\u2022\u2022: :: : : : : : : : : e: : : : : e: : : : : e: : : : \u2022 e: : : \u2022 : : : ; \u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022l \u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022~ Les physiciens nucJeaires Comme ils nous aiment pas beaucoup Y gardent tout J'fric, oui mais Ie fric on s'en fout Tout est rlatif apres tout Probable-Possible, my black hen, She lays eggs in the Relative When. She doesn't lay eggs in the Positive Now Because she's unable to postulate How. Les experiences de Weber Le gyroscope, 9", "a coutent des sous Celles de pensees sont moins cheres Tout est rlatif apres tout Yo ho, it's hot... the sun is not A place where we could live But here on earth there'd be no life Without the light it gives Aux solutions singulieres On prefere et de beaucoup Une metrique partout reguliere Tout est rlatif apres tout From A Space Child's Mother Goose. \u00a9 1956. 1957. 1958 by Frederick Winsor and Marian Parry\u2022 by permission of Simon and Schuster. Reprinted with the kind permission of M. A. Tonnelat. M. A. TONNELAT F. WINSOR H. ZARET SHAKESPEARE GOTTFRIED WILHELM VON LEIBNIZ (Sur I'air de Aupres de rna blonde) PAT (Mrs. Hypatia Vourloumis at Knossos (1971\u00bb \"Omnibus ex nihil ducendis'sufficit unum!\" (One suffices to create Everything of nothing!) Tous les savants du monde Apportent leurs ecrits LoigravnadonneUe Sans tenseur d'energie Dans les 1'ardins d'Asnieres La science a refleuri Tous les savants du monde Apportent leurs ecrits When Arthur Evans began this excavation neither he nor anyone knew that he would uncover an unknown world. 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Le champ laisse des plumes Aux bosses de I'espace-temps En prendrons quelques unes Pour decrire Ie mouvement De ravissants modeJes..... Pour la cosmologie Pour moi ne m'en faut guere Car {en ai un 1'oli Pour moi ne m'en faut guere Car j'en ai un 1'oli \" est dans ma cerveUe Voici mon manuscrit Loi gravitationneUe Sans tenseur d'energie De ravissants modeles Pour la cosmologie '\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u20224 \u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022 C.", " CATTANEO, J. GEHENIAU M. MAVRIDES, and M. A. TONNELAT Appreciation and farewell to our patient reader. Reprinted with the kind permission of the authors. h' 1- - BIBLIOGRAPHY AND INDEX OF NAMES This bibliography, like that of Synge (1960b), merely provides points of entry into the literature, which is far too extensive for a comprehensive listing. For more extensive bibliographies of certain segments of the literature, see, for example, Lecat (1924), Boni, Russ, and Laurence (1960), Chick (1960, 1964), and Combridge (1965). Citations are sometimes not to the earliest publication, but to a later and more accessible source. For ease of reference, conference and summer-school proceedings are cited both under editors' names and under the more familiar place names (Brandeis, Les Houches, Varenna, etc.). Persons mentioned in the text without explicit biblio graphical reference are also indexed here. Most doctoral dissertations in the United States are available from University Microfilms, Inc., Ann Arbor, Mich. 48106. Appreciation is expressed to Gregory Cherlin for preparing the initial version of this bibliography, to Nigel Coote for numerous subsequent amendments, and to colleagues without whose help some of the most elusive, and most important, of these references would have escaped capture. Adler, R., M. Bazin, and M. Schiffer, 1965, Introduction to General Relativity, McGraw-Hill, New York. [endpapers] Aharonov, Y., and L. Susskind, 1967, \"Observability of the sign change of spinors under 2.\". rotations,\" Phys. Rev. 158, 1237-1238. [\u00a741.5] Ahrens, L. H., F. Press, and S. K. Runcorn, eds., 1964, Physics and Chemistry of the Earth, Pergamon Press, London. Cited under Bolt 1964. al Biruni, Abu Raihi'in. cited by name in Box 27.7. Aldrin, E. E., Jr., cited by name in Fig. 40.4. Alfven, H., 1971, \"Plasma physics applied to cosmology,\" Phys. Todav 24 (Feb. 1971),28-33. [\u00a7\u00a727.1O, 27.11, 28.4] Alfven, H., and O. Klein, 1962. \"Matter-antimatter annihila tion and cosmology.\" Ark. Fys. 23, 187-194. [\u00a7\u00a727.1O. 27.11, 28.4, Box 27.7] al Khazini, 1115, Book of the Balance of Wisdom, partially edited and translated by N. Khanikoff in J. Am. Oriental Soc. 6 (1860), 25-38. [\u00a71.7] Allen, C. W., 1963, Astrophysical Quantities, University of London, Athlone Press, London. [\u00a736.6, Box 25.1, end papers] Allendoerfer, c., and A. Wei!, 1943, \"The Gauss-Bonnet theo rem for Riemannian polyhedra,\" Trans. Am. Math. Soc. 53, 101-129. [Box 15.3] Alley, C. 0., cited by name in Fig. 40.4. See also Bender et al. 1971. Alpher, R. A., cited by name in Box 27.7. Alpher, R. A., and R. C. Herman 1948a, \"Evolution of the universe,\" Nature 162, 774-775. [\u00a728.1] Alpher, R. A., and R. C. Herman, 1948b, \"On the relative abundance of the elements,\" Phys. Rev. 74, 1737-1742. [\u00a728.1] Alpher, R. A., and R. C. Herman, 1950, \"Theory of the origin and relative-abundance distribution of the elements.\" Rev. Mod. Phys. 22. 153-213. [\u00a728.1] American Ephemeris and Nautical Almanac for 1964, 1962, U.S. Government Printing Office, Washington, D.C. [Fig. 25.1] Ames, W. L., and K. S. Thorne, 1968, \"The optical appearance of a star that is collapsing through its gravitational radius,\" Astrophys.1. 151, 659-670. [\u00a732.3] Anderson, C. D., 1933, \"The positive electron,\" Phys. Rev. 43, 491-494. [\u00a744.4] 1222 GRAVITATION Anderson, D. W., E. H. Brown,", " Jr., and F. P. Peterson, 1966a, \"Spin cobordism,\" Bul/. Am. Math. Soc. 72,256-260. [Box 44.3) Anderson, D. W., E. H. Brown, Jr., and F. P. Peterson, 1966b, and the \"SU-cobordism, KO:characteristic numbers, Kervaire invariant,\" Ann. MaJIl. 83. 54. [Box 44.3) Anderson. J. D., 1973. \"Lectures on physical and technical problems posed by precision radio tracking.\" in Bertotti 1973. [\u00a740.l) Anderson, J. D., P. B. Esposito, W. L. Martin. and D. O. Muhleman, 1971, \"Measurements of the general relativis tic time delay with data from Mariners 6 and 7,\" in Davies 1971, pp. 111-135. [Box 40.2) Anderson, J. L., 1967, Principles of Relativity P/(l'sics. Aca demic Press, New York. [\u00a717.6, endpapers) Anderson, J. L.. 1970, \"Relativistic Boltzmann theory and the Grad method of moments,\" in Carmeli, Fickler, and Witten 1970, pp. 109-124. [Ex. 22.7) Anonymous, source of quotation at end of Chapter 44: found on beach at Rincon, Puerto Rico, March 21, 1973. Aristotle, ca. 330 B.C., Physics, ed. by W. D. Ross, Clarendon Press, Oxford (1950). [Box 1.2) See also Guthrie 1939, Warrington 1956. Armstrong, N. A., cited by name in Fig. 40.4. Arnett, W. D., 1966, \"Gravitational collapse and weak inter actions,\" Call. J. Pllys. 44, 2553-2594. [\u00a732.7) Arnett, W. D., 1967, \"Mass dependence in gravitational col lapse of stellar cores,\" Can. J. Phys. 45, 1621-1641. [\u00a732.7) Arnold, V. I., and A. Avez. 1968, Ergodic Problems of Classical Mechanics, W. A. Benjamin, New York. [\u00a78.2) Arnowitt, R., cited by name in \u00a7\u00a721.1, 35.8, Box 17.2. Arnowilt, R., S. Oeser, and C. W. Misner, 1962, \"The dynamics of general relativity,\" in Witten 1962, pp. 227-265. [\u00a7\u00a721.1,. 21.4,21.5,21.6,21.7,21.8,35.4) Arp, H., 1971, \"Observational paradoxes in extragalactic as tronomy,\" Science 174, 1189-1200. [Box 28.1) Ash, M. E:: See Shapiro, Ash, et. al. 1971,. Shapiro, Smith, et. al. 1971, Shapiro et. al. 1972. Auslander, L., and R. E. MacKenzie, 1963, Introduction to Differemiable Manifolds, McGraw-Hill, New York. [Box 8.1) Auslander, L., and L. Markus, 1959, Flat Lorentz 3-Manifolds, American Mathematical Society, Providence, R.1. [Box 27.2) Avez, A., 1960, \"Proprietes globales des espace-temps period iques dos,\" Acad. des Sci., Paris, Comptes Rend. 250, 3583-3587. [\u00a744.I) Avez, A., 1964, \"Le ds2 de Schwarzschild parmi les ds2 stationnaires,\" Ann. Inst. Henri Poincare AI, 291-300. [\u00a723.I) Avez, A. See also Arnold and Avez 1968. Avicenna, cited by name in Box 27.7. Baade, W., 1942, \"The Crab NebUla,\" Astrophys. J. 96, 188 198. [Box 24.1) Baade, W., 1952, Report to Commission 28 of I.A.U., Rome, September 1952. [Box 27.1) Baade, W., 1956. \"The period-luminosity relation of the ceph eids,\" Pub. Astron. Soc. Pacific, 28, 5-16. [Box 27.1) Baade, W., cited by name in Box 27.7. Baade, W., and F. Zwicky, 1934a, \"Cosmic rays from super novae,\" Proc. Nat. Acad Sci. U.S. 20, 259-263. [Box 24.1", ") Baade. W.. and F. Zwicky. 1934b, \"On supernovae,\" Proc. Nat. Acad Sci. U.S. 20, 254-259. [Box 24.1) Baade, W., and F. Zwicky, 1934c. \"Supernovae and cosmic rays,\" Phys. Rev. 45, 138. Abstract of paper presented at Stanford meeting of Amer. Phys. Soc., Dec. 15-16, 1933. [Box 24.1) Bacon, R.. 1268, Opus Majus, J. H. Bridges, ed., Clarendon Press, Oxford, Eng. Vols. I and 2, 1897; vol. 3. 1900. [Box 27.7) Bahcall, J. N.. W. L. W. Sargent, and M. Schmidt, 1967, \"An analysis of the absorption spectrum of3C191,\" Astropllys. 1. 149, 11-15. [\u00a738.6) Bahcall, J. N., and M. Schmidt, 1967, \"Does the fine-structure constant vary with cosmic time?\" Pllys. Rev. Lett. 19, 1294-1295. [\u00a738.6) Baierlein, R. F.. 1967, \"Testing general relativity with laser ranging to the moon,\" Phys. Rev. 162, 1275-1288. [\u00a740.6) Baierlein, R. F., cited by name in \u00a721.1. See also Krogh and Baierlein 1968. Baierlein, R. F., D. H. Sharp, and J. A. Wheeler, 1962, \"Three-dimensional geometry as carrier of information about time,\" Pllys. Rev. 126, 1864-1865. [\u00a7\u00a721.9, 43.2) Bailey, J. See Farley et al. 1966. Bailey, J., W. Bartl, G. von Bochmann, R. C. A. Brown, F. J. M. Farley, H. Jostlein, E. Picasso, and R. W. Williams, 1968, \"Precision measurement of the anomalous magnetic moment of the muon,\" Phys. Lett. 28B, 287-289. [\u00a744.2) Bardeen, J. M., 1968, \"General relativistic collapse of charged dust (abstract),\" Bull. Am. Phys. Soc. 13, 41. [\u00a734.6, Fig. 34.4) Bardeen, J. M., 1970a, \"Kerr metric black holes,\" Nature 226, 64-65. [\u00a733.8, Boxes 33.3,-33.5) Bardeen, J. M., 1970b, \"A variational principle for rotating stars in general relativity,\" Astrophys. J. 162, 71-95. [Ex. 33.3) Bardeen, J. M. See also Press and Bardeen 1971. Bardeen, J. M., B. Carter, and S. W. Hawking, 1973, \"The four laws of black-hole mechanics,\" Commun. Math. Phys., in press. [Box 33.4) Bardeen, J. M., and W. H. Press, 1972, \"Radiation fields in the Schwarzschild background,\" J. Math. Phys. 14, 7-19. [\u00a735.14, Ex. 32.11) Bardeen, J. M., W. H. Press, and S. A. Teukolsky, 1972, \"Rota locally nonrotating frames, energy ex ting black holes: traction, and scalar synchrotron radiation,\" Astrophys. J. 178,347-369. [\u00a733.7, Ex. 33.3, 33.14) Bardeen, J. M., K. S. Thorne, and D. W. Meltzer, 1966, \"A catalogue of methods for studying the normal m~des of radial pulsation of general-relativistic stellar models,\" Astrophys. J. 145, 505-513. [Box 26.1) Bardeen, J. M., and R. V. Wagoner, 1971, \"Relativistic disks, I: Uniform rotation,\" Astrophys. J. 167, 359-423. [Box 24.1) Barger, R. L. See Evenson et. al. 1972. Bargmann, V., and E. P. Wigner, 1948, \"Group theoretical discussion of relativistic wave equations;\" Proc. Nat. Acad. Sci. U.S. 34, 211-222. [Box 32.2) Barnes, J. A., 1971, \"A non-mathematical discussion of some basic concepts of precise time measurement,\" in Tracor on Frequency 2, No. 2 (Tracor", " Industrial Instruments, Austin, Texas). [Box 1.4) BIBLIOGRAPHY AND INDEX OF NAMES 1223 Barschall, H. H. and W. Haeberli, 1971. \"The Madison con vention,\" pp. xxv-xxix in Polarization Phenomena in Nu clear Reactions, Univ. of Wisconsin Press, Madison. [\u00a737.7] Bartl, W. See Bailey et al. 1968. Barton, D., S. R. Bourne, and J. P. Fitch, 1970, \"An algebra system,\" Comput. J. 13, 32-39. [Box 14.3] Barton, D., and J. P. Fitch, 1971, \"General relativity and the application of algebraic manipulative systems,\" Comm. Assoc. Comput. Machinery 14, 542-547. [Box 14.3] Battelle Rencontres, Seattle 1967. See DeWitt and Wheeler 1968. Baym, G., 1969, Lectures on Quantum Mechanics, W. A. Benjamin, New York. [Box 22.4] Baym, G., H. A. Bethe, and C. J. Pethick, 1971, \"Neutron star matter,\" Nuclear Phys A. 175, 225-271. [\u00a724.2] Bazin, M. See Adler, Bazin, and Schiffer 1965. Beall, t. F., 1970, \"Measuring the gravitational interaction of elementary particles,\" Phys. Rev. D 1, 961-978. [\u00a71.3, Box 1.2] \"Oscillatory approach to a singular point in the relativistic cosmology,\" Usp. Fiz. Nauk 102, 463-500. English transla tion in Advances in Physics 19, 525-573. [\u00a730.6, Box 30.1] Belinsky, V. A., E. M. Lifshitz, and \\. M. Khalatnikov, 1971, \"Oscillatory mode of approach to a singularity in homo geneous cosmological models with rotating axes,\" 2h. Eksp. & Teor. Fit. 60. 1969-1979. English translation in Sov. Phys.-JETP 33, 1061-1066 (1971). [\u00a7\u00a730.7, 34.6] Bell, C. c.. R. L. Forward, L. R. Miller, T. D. Beard, and T. M. Baran, 1969, \"Static gravitational gradient field detec tion with a rotating torsional gradiometer,\" in NASL 1969. [Box 16.5] Bell, C. c., R. L. Forward, and H. P. Williams, 1970, \"Simu lated terrain mapping with the rotating gravity gradiome. ter, in W. T. Kattner, ed., Advances in Dynamic Gravime try, Instrument Society of America, Pittsburgh, 115-129. [Box 16.5] Bell, S. J., cited by name in Box 27.7. See also Hewish et al. Beams, J. W., 1971, \"Finding a better value for G,\" Phys. 1968. Today 24 (May 1971), 35-40. [\u00a740.8] Beams, J. W. See also Rose et al. 1969. Beckedorff, D. L., and C. W. Misner, 1962, unpublished; see D. L. Beckedorff's A. B. Senior Thesis, Princeton Univer sity. [\u00a732.4] Bekenstein, J., 1972a, \"Nonexistence of baryon number for static black holes,\" Phys. Rev. D 5, 1239-1246. [Boxes 25.2, 32.2, 33.1] Bekenstein. J., 1972b, \"Nonexistence of barvon number for static black holes. II,\" Phys. Rev. D 5, 2403-2412. [Boxes 25.2, 32.2, 33.1] Bel, L., 1958, \"Sur la radiation gravitationelle.\" Acad. Sci., Paris, Comptes Rend. )47,_1094-1096. [Ex. 15.2] Bel, L., 1959. \"Introduction d'un tenseur du quatrieme ordre,\" Acad. Sci., Paris, Comptes Rend. 248, 1297-1300. [Ex. 15.2] Bel; L., 1962, \"La radiation gravitationelle,\" in CNRS 1962, pp. 119-126. [Ex. 15.2] Belasco, E. Po, and H. C. Ohanian, 1969, \"Initial conditions relativity: Lapse and shift formulation,\" J. in general Mach. Phys. 10, 1503-1507.", " [\u00a721.9] Belinfante, F., 1940, \"On the current and the density of the electric charge, the energy, the linear momentum, and the angular momentum of arbitrary fields.\" Physica 7, 449 474. [\u00a721.3] Belinsky, V. A. See also Ryan 1972b. Belinsky. V. A.; and I. M. Khalatnikov, 1969a, \"On the nature of the singularities in the general solution of the gravita tional equations,\" 2h. Eksp. & Teor. Fl:. 56. 1700-1712. English translation in Sov. Phys.-JETP 29. 911-917 [\u00a734.6] Belinsky. V. A.. and I. M. Khalatnikov, 1969b. \"General solu tion of the gravitational equations with a physical singu larity:' 2h. Eksp. & Teor. Fit. 57, 2163-2175. English translation in Sov. Phys.-JETP 30. 1174-1180 (1970). [\u00a734.6] Belinsky. V. A.\u2022 and I. M. Khalatnikov, 1970. \"General solution of the gravitational equations with a physical oscillatory singularity:' 211. Eksp. & Teor. Fit. 59, 314-321. English translation in Sou. Phys.-JETP 31, 169-172. [\u00a7\u00a730.6. 34.6] Belinsky. V. A.. I. M. Khalatnikov. and E. M. Lif,hitz. 1970. Beltran-Lopez, V. B. See Hughes, Robinson, and Beltran Lopez 1960. Bender, P. L., R. H. Dicke, D. T. Wilkinson, C. O. Alley, D. G. Currie, J. E. Faller, J. D. Mulholland, E. C. Silverberg, H. E. Plotkin, W. M. Kaulla, and G. J. F. McDonald, 1971. \"The lunar laser-ranging experiment,\" in Davies 1971. [\u00a740.9] Bennett, A. S.\u2022 cited by name in Box 27.7. Bentley, R., cited by name in Box 27.7. Berger, B. K., D. M. Chitre, V. E. Moncrief, and Y. Nutku. 1972, \"Hamiltonian formulation of spherically symmetric gravitational fields,\" Phys. Rev. D 5, 2467-2470. [\u00a72\\.i0] Berger, M., 1966. \"Sur les spectres d'une variete Riemannienne,\" Acad. Sci Paris, Comptes Rendus 263, 13-16. [Box 17.2 part 6] Berger, M., P. Gauduchon, and E. Mazet. 1971, Le Spectre d'l/ne variere riemannienne, Springer. Berlin and New York. [Box 17.2 part 6] Bergmann, P. G., 1942. Introduction to the Theory of Relativi~v, Prentice-Hall, New York. [endpapers] Bergmann, P. G.\u2022 1968, \"Comments on the scalar-tensor the ory,\" Int. J. Theor. Phys. 1, 25-36. [\u00a738.2] Bern, Switzerland. 1955, Jubilee of Relativity Theory, confer ence held Julv 11-16. 1955; see Mercier and Kervaire 1956. - Bernoulli. D., 1733, Hydrodynamica. chapter 10. Archives of Academy of Science. Leningrad: summarized in C. C. Gillispie, ed.. Dictiollary ofSCientific Biograp/(v, Scribners, New York. vol. 2. 1970. [Box 44.5] Bers. L.. 1952. \"Singularities of minimal surfaces,\" in Proceed illgs of the International COllgress of Mathematicians, Cambridge. Mass., 1950, American Mathematical Society. Providence. R.\\. [\u00a733.2] Bers. L.. cited ~r name ill Box 8.5. Bertotti. B.\u2022 1959, \"Uniform electromagnetic field in the theory of general relativity.\" Phys. Rev. 116, 1331-1333. [Ex. 32.1] Bertotti. Boo ed.. 1973. proceedings of Course 56 of the Interna tional School of Physics Enrico Fermi. dealing with exper imental tests of relativjstic gra\\'itation theories. Academic Press. New York. in press. 1224 GRAVITATION Bertotti, B.. D. R. Brill, and R. Krotkov, 1962, \"Experiments Bondi, H., 1961,", " Cosmology, 2d ed., Cambridge Univ. Press. on gravitation,\" in Witten 1962. [\u00a738.5. Box 40.1] Cambridge. Eng. [Box 27.1] Bessel, F. W., cited by name in Box 27.7. Bethe, H. A.. cited ~V name in Box 27.7. See also Baym, Bethe. and Pethic k 1971. Birkhoff, G. D., 1923, Relatiuity and Modern Physics. Harvard University Press. Cambridge, Mass. [\u00a732.2] Birkhoff, G. D., 1943, \"Matter, electricity and gravitation in flat spacetime,\" Proc. Nat. Acad. Sci. U.S. 29, 231-239. [\u00a739.1] Bishop, R. L., and S. I. Goldberg, 1968, Tensor Analysis on Manifolds, Macmillan, New York. [Box 8.1, Ex. 21.8] Bisnovatyi-Kogan, G. S., and K. S. Thorne, 1970, \"Relativistic gas spheres and clusters of point masses with arbitrarily large central redshifts: Can they be stable?\" Astrophys. J. 160, 875-885. [Ex. 25.28] Bisnovatyi-Kogan, G. S., and Ya. B. Zel'dovich, 1969, \"Models of clusters of point masses with great central redshift,\" Astrojizika 5, 223-234. English translation in Astrophys. 5, 105-109. [Ex. 25.28] Blackman, R. B., and J. W. Tukey, 1959, The Measurement of Power Spectra, from the Point of View of Communications Engineering, Dover, New York. [\u00a737.8] Blake, W., quoted at end of Chapter 44 Blaton, J., 1950, \"On a geometrical interpretation of energy and momentum conservation in atomic collisions and disintegration processes,\" KgI. Danske Videnskab. Sels., Mat.-.fys. Medd. 24, no. 20. [\u00a72.1O] Blatt, J. M., and V. F. Weisskopf, 1952, Theoretical Nuclear Physic\" Wiley, New York. [\u00a737.7] Blumenthal, L. M., 1953, Theory and Applications of Distance Geometry, Clarendon Press, Oxford, England. [Box 13.1] Bochmann, G. von. See Bailey et al. 1968. Bogoliubov, N. N., and D. V. Shirkov, 1959, Introduction to the Theo~r of Quantized Fields, Interscience (now a divi sion of Wiley), New York. [\u00a743.3] Bogorodskii, A. F., 1962, Urauneniya Polya Einsteina i ikh Primeneniye u Astronomii (Einstein's Field Equations and Their Application in Astronomy), Kiev Univ. Press, USSR. [Ex. 25.22] Bohr, A., and B. R. Mottleson, 1969, Nuclear Structure, Vol. I: Single' Particle Motion, W. A. Benjamin, New York. [\u00a723.4] Bohr, N., 1918, \"On the quantum theory of line spectra,\" Mem Acad. R. Sci. & Lett. Danemark 4, no. I, parts 1-2; part I reprinted in Van der Waerden 1967, pp. 95-137. [Box 17.1] Bohr, N., cited by name in Preface, \u00a7\u00a744.I, 44.5, Box 17.1 Bohr, N., and L. Rosenfeld, 1933, \"Zur Frage der Messbarkeit elektromagnetischen Feldgrossen,\" KgI. Danske der Videnshab. Sels. Mat.-fys. Medd. 12, no. 8. [\u00a743.2] Boissonade, J. F. See Westerman and Boissonade 1878. Bolt, B. A., 1964, \"Recent information of the earth's interior from studies of mantle waves and eigenvibrations,\" in Ahrens, Press, and Runcorn 1964, pp. 55-119. [Ex. 37.14] Bolton, J. G., cited by name in Box 27.7. Bondi, H., 1947, \"Spherically Symmetrical Models in General Relativity,\" Mon. Not. R. Astron. Soc. 107,410-425. [\u00a730.6] Bondi, H., 1957, \"Plane gravitational waves in general relativ ity,\" Nature 179, \\072-\\073. [", "Ex. 18.5, Fig. 18.1] Bondi. H., 1965. \"Some special solutions of the Einstein equa tions,\" in Trautman, Pirani, and Bondi 1965. pp. 375-459. esp.p. 431. [Ex. 18.5] Bondi, H., cited bl' name in \u00a72I.IO, Boxes 27.7,34.2. See also Trautman, Pirani, and Bondi 1965. Bondi. H., and T. Gold, 1948, \"The steady-state theory of the expanding universe,\" Mon. Not. R. Astron. Soc. 108, 252 270. [\u00a728.4] Bondi, H., and W. H. McCrea, 1960, \"Energy transfer by gravitation in Newtonian theory,\" Proc. Cambridge Phil. Soc. 56, 410-413. [Ex. 18.5] Bondi, H., F. A. E. Pirani, and I. Robinson, 1959, \"Gravita tional waves in general relativity, Ill: Exact plane waves,\" Proc. R. Soc. London A 251, 519-533. [\u00a735.9] Boni, N., M. Russ, and D. H. Laurence, 1960, A Bibliographical Check List and Index to the Published Writings of Albert Einstein, Pageant Books. Paterson, N.J., cited in preface of present bibliography; based on earlier bibliography of M. C. Shields in Schilpp 1949. Born, M., 1971, The Born-Einstein letters: Correspondence Be tween Albert Einstein and Max and Hedwig Born from 1916 translated by to 1955 with Commentaries by Max Born, Irene Born, Walker, New York. [Box l.ll] Born, M., cited by name in Box 17.2 Boughn, S. P., D. M. Fram, and R. B. Partridge, 1971, \"Iso tropy of the microwave background at 8-millimeter wavelength,\" Astrophys. J. 165,439-444. [\u00a7\u00a727.1, 27.2, 30.8] Bourbaki, N., cited by name in Box 8.2; pseudonym for an organization of French mathematicians who, since 1939, have nearly completed the first ten Books of a projected larger work: Elements de Mathematique, Her mann, Paris. Bourne, S. R. See Barton, Bourne, and Fitch 1970. Boyer, R. H., and R. W. Lindquist, 1967, \"Maximal analytic extension of the Kerr metric,\" J. Math. Phys. 8, 265-281. [\u00a733.2, Box 33.2, Ex. 33.8] Braginsky, V. B., 1965, \"Gravitational radiation and the pros pect of its experimental discovery,\" Usp. Fiz. Nauk 86, 433-446. English translation in Sou. Phys.-Uspekhi 8, 513-521. [Box 36.3] Braginsky, V. B., 1970, Fizicheskie Eksperimenti s Probnimi Telami (Physics Experiments With Test Bodies), Nauka, Moscow, Russia. English translation published as NASA IT F-762, available from National Technical Information Service, Springfield, Va. [\u00a737.8] t Braginsky, V. B., A. B. Manukin, and M. Yu. Tikhonov, 1970, \"Investigation of dissipative ponderomotive effects of electromagnetic radiation,\" Zh. Eksp. & Teor. Fiz. 58, 1549-1552, English translation in Sou. Phys.-JETP 31, 829-830. [\u00a737.8] Braginsky, V. B., and M. B. Menskii, 1971, \"High-frequency detection of gravitational waves,\" Zh. Eks!!. & Teor. Fiz. Pis'ma 13, 585. English translation in Sou. Phys.-JETP Lett. 13,417-419. [\u00a737.9, Box 37.6] Braginsky, V. B., and V. S. Nazarenko, 1971, \"On the detection of gravitational radiation from some astronomical sources with the heterodyne detector,\" in Davies 1971, pp. 45-46. [\u00a737.3] BIBLIOGRAPHY AND INDEX OF NAMES 1225 Braginsky, V. B., and V. I. Panov, 1971, \"Verification of the equivalence of inertial and gravitational mass,\" Zh. Eksp. & Tear. Fiz. 61, 873-879. English translation in Sov. Phys.-JETP 34, 464-466.", " [\u00a7\u00a738.3, 40.9, Box 1.2) Braginsky, V. B., Ya. B. Zel'dovich, and V. N. Rudenko, 1969, \"Reception of gravitational radiation of extraterrestrial origin,\" Zh. Eksp. & Tear. Fiz. Pis'ma 10, 437-441. Eng lish translation in Sov. Phys.-JETP Lett. 10, 280-283. [\u00a737.3, Fig. 37.2) Brandeis Summer Institute in Theoretical Physics: 1964, see Trautman, Pirani, and Bondi 1965; 1968, see Chretian, Deser, and Goldstein 1969. Brans, C., cited by name in \u00a7\u00a740.1, 40.6, 40.9. Brans, C., and R. H. Dicke, 1961, \"Mach's principle and a relativistic theory of gravitation,\" Phys. Rev. 124,925-935. Reprinted in Dicke 1964b. [\u00a7\u00a717.6, 28.4, 38.2, 39.2, 40.8, Boxes 38.3, 39.1) BrasseJr, Georges, quoted in Fig. 44.5. Brault, J. W., 1962, The Gravitational Redshift in the Solar Spectrum, doctoral dissertation, Princeton Univ. Abstract, \"Gravitational Redshift of Solar Lines,\" in Bull. Amer. Phys. Soc. 8, 28 (1963). [\u00a738.5, Fig. 38.2) Brehme, R. W. See DeWitt and Brehme 1960. Bressanone, Italy: 1970 summer school sponsored by Centro Interzionale Matematico -Estivo. See Cattaneo 1971. Brill, D. R., 1959, \"On the positive definite mass of the Bondi time-symmetric gravitational waves,\" Weber-Wheeler Ann. Phys. (USA) 7, 466-483. [\u00a721.10) Brill, D. R., 1961, General Relativity: Selected Topics of Current Interest, report SUI61-4, Physics Department, State Uni versity of Iowa, Iowa City, Iowa (multilithed); summar ized in DeWitt and DeWitt 1964b, pp. 431-432; published as Brill 1964. [\u00a721.l0)_ Brill, D. R., 1964, \"General relativity: interest,\" Suppl. Nuovo Cimento 2, selected topics of I-56. See current Brill 1961. Brill, D. R. See also Bertotti, Brill, and Krotkov 1962, Cohen and Brill 1967 and 1968, and Graves and Brill 1960. Brill, D. R., P. L. Chrzanowski, C. M. Pereira, E. D. Fackerell, and J. R. Ipser, 1972, \"Solution of the scalar wave equa tion in a Kerr background by separation of variables,\" Phvs. Rev. D 5, 1913-1915. [Ex. 33.14) Brill, D. R., and J. M. Cohen, 1966, \"Rotating masses and their effect on inertial frames,\" Phys. Rev. 143, 1011-1015. [\u00a7\u00a721.12, 40.7, Ex. 26.1) Brill, D. R., and J. B. Hartle, 1964, \"Method of self-consistent field in general relativity and its application to the gravita tional geon.\" Phys. Rev. B 135, 271-278. [\u00a7\u00a721.12, 35.13) Brill, D. R., and J. A. Wheeler, 1957, \"Interaction of neutrinos and gravitational fields,\" Rev. Mod. Phys. 29, 465-479. [\u00a7\u00a727.I0,41.1I) Brillouin, L., 1970, Relativity Reexamined, Academic Press, New York. [Ex. 25.21) Brodetsky, S. See Dobson and Brodetsky 1947. Brodsky, S. J., and S. D. Drell, 1970, \"The present status of quantum ele~trodynamics,\" in Ann. Rev. Nllclear Sci. 20. 147-194. [\u00a744.2) Bromberg, V. A., 1958, \"Relativistic corrections in the theory of the motion of the moon,\" Bull. Inst. Theor. Astron., Akad Nauk SSSR 6, 733-756. [\u00a740.6) Bronshtein, M. P., cited by name in \u00a729.2. Brown, E. H., Jr. See Anderson, Brown, and Peterson 1966a and 1966b. Brown, R. C", ". A. See Bailey et al. 1968 and Farley et al. 1966. Bruhat, Y., 1962, \"The Cauchy problem,\" in Witten 1962. [\u00a7\u00a721.11, 21.13) Bruhat, Y. See also Choquet-Bruhat, Y., and Foures- Bruhat, Y. Bruno, G. See Singer 1950. Bunsen, R. W., cited by name in Box 38.2. Burbidge, E. M., cited by name in Box 27.7. See also Burbidge and Burbidge 1969a and 1969b and Hoyle et al. 1964. Burbidge, G. R., cited by name in Box 27.7. See also Hoyle et al. 1964. Burbidge, G. R., and E. M. Burbidge, 1969a, \"Red-shifts of quasi-stellar objects and related extragalactic systems,\" Nature 222, 735-741. [Ex. 27.14) Burke, W. L., 1970, \"Runaway solutions: Burbidge, G. R., and E. M. Burbidge, 1969b, \"Quasi-stellar objects-a progress report,\" Nature 224,21-24. [Ex. 27.14) remarks on the asymptotic theory of radiation damping,\" Phys. Rev. A 2, 1501-1505. [\u00a720.6) Burke, W. L., 1971, \"Gravitational radiation damping of slowly moving systems calculated using matched asymptotic ex pansions,\" J. Math. Phys. 12,402-418. [\u00a736.8) Burke, W. L., cited by name in \u00a722.5. Burnett, D. S. See Wasserburg and Burnett 1968. Cairns, S. S., ed., 1965, Differential and Combinatorial Topol ogy, Princeton Univ. Press, Princeton, N.J. Cairns, S. S. See also Morse and Cairns 1969. Callaway, J., 1953, \"The equations of motion in Einstein's new unified field theory,\" Phys. Rev. 92, 1567-1570. [\u00a720.6) Callippus, cited by name in Box 27.7. Caltech-JPL conference on experimental tests of gravitational theories, November 1970. See Davies 1971. Cameron, A. G. W., 1970, \"Neutron stars,\" Ann. Rev. Astron. & Astrophys. 8, 179-208. [\u00a724.2) Cameron, A. G. W., See also Langer and Cameron 1969, and Truran and Cameron 1971. Cameron, A. G. W., and J. Truran, 1971, work cited by Peebles 1971. [Box 33.3) Campbell, D. B. See Shapiro, Ash, et al. 1971 and Shapiro et al. 1972. Campbell, L., and Garnett. W., 1882, Life of James Clerk Maxwell, Macmillan, London. [Box 4.6) Campbell, W. W., and R. J. Trumpler, 1928. \"Observations made with a pair of five-foot cameras on the light-deflec tions in the sun's gravitational field at the total solar eclipse of September 21, 1922.\" Lick Obs. Bull. 13, 130 160. [Box 40.1) Carmeli, M., S. I. Fickler, and L. Witten, 1970, Relativity, Plenum, New York. [\u00a740.7) Carroll, L. See Dodgson, C. L. Cartan, E., 1901, \"Sur quelques quadratures dont l'element differentiel contient des fonctions arbitraires,\" Bill/. Soc.,Hath. France 29, 118-130. [Box 8.2) 1226 GRAVITATION Cartan, E.. 1922a. \"Sur les equations de la gravitation de Einstein,\" 1. Mach. Pures Appl. 1, 141-203. [\u00a72I.1) Cartan. E., I922b, \"Sur une generalisation de la notion de courbure de Riemann et les espaces a torsion.\" Acad. Sci.. Paris, Comptes Rend. 174. 593-595. [\u00a7\u00a738.2. 39.2) -Cartan. E., 1923,-\"Sur res varietes a connexion affine et la theorie de la relativite generallsee (premiere partie),\" Ann. Ecole Norm. Sup. 40. 325-412. [\u00a7\u00a78.2. 12.1, 12.2,38.2.39.2) la theorie de la relativite generalisee (suite", "),\" AI/n. Ecole Norm. Sup. 41, 1-25. [\u00a7\u00a78.2, 8.3, 12.1) Cartan, E., 1924. \"Sur les varietes a connexion affine et Cartan, E., 1925. La Geometrie des Espaces de Riemann. Me IX, des Sciences Mathematiques, Faseicule morial Gauthier- Villars, Paris. France. [\u00a714.5) Cartan, E., 1928 and 1946, Ler;ons sur la Geometrie des Espaces de Riemann. Gauthier-Villars. Paris. France. [\u00a7\u00a714.5, 15.3, Box 8.2, Ex. 21.4. endpapers) Cartan, E., 1935. La Methode du Repere Mobile. la Theorie des Groupes Cominus et les Espaces Generalisees, Hermann, Paris, France. [\u00a76.4) Cartan, E., 1945, \"Les systemes differentiels exterieurs et leurs applications geometriques,\" Exposes de Geometrie 14. Hermann, Paris, France. [\u00a74.I) Cartan, E., 1946. See Cartan 1928. Cartan, E., 1966, Theory of Spinors, M.LT. Press, Cambridge, Mass. [\u00a74I.1I) Cartan, E., cited by name in \u00a7\u00a79.l, 13.1, 13.4. 15.1, Boxes!O.l, 12.3, Ex. 12.7, 12.8. Cartan, H., 1970, Formes DifJeremiels, Hermann, Paris, France. [\u00a74.I) Carter, B., 1966a, \"Complete analytic extension of the sym metry axis of Kerr's solution of Einstein's equations,\" Phys. Rev. 141, 1242-1247. [Box 33.2) Carter, B., 1966b, \"The complete analytic extension of the Reissner-Nordstrom metric in the special case eZ = mZ,\" Phys. Lett. 21, 423-424. [Fig. 34.4, Ex. 31.8. 34.3) Carter, B.. 1968a, \"Global structure of ihe Kerr family of gravitational fields,\" Phys. Rev. 174, 1559-1571. [\u00a733.5, Box 33.2, Ex. 33.7) Carter, B., 1968b, \"Large numbers in astrophysics and cos mology,\" unpublished preprint, Institute of Theoretical Astronomy, Cambridge, Eng. [\u00a744.6) Carter, B., 1968c, \"Hamilton-Jacobi and Schrodinger separable solutions of Einstein's equations,\" Commun. Math. Phys. 10, 28()...310. [\u00a733.5) Carter, B., 1970, \"An axisymmetric black hole has only two degrees of freedom,\" Phys. Rev. Lett. 26, 331-333. [Box 33.1) Carter, B., 1973, \"Properties of the Kerr metric,\" in DeWitt and DeWitt 1973. [Box 33.1) Carter, B.. cited by name in Box 34.3. See also Bardeen, Carter, and Hawking 1973. Cattaneo, C., ed., 1971, Relativistic Fluid Dynamics, proceed ings of a summer school sponsored by Centro Inter nazionale Matematico Estivo, Bressanone, Italy, June 1970, Edizione Cremonese, Rome. [\u00a722.3) Cattaneo, C., quoted at end of Chap. 44. Cauchy, A. L., cited by name in Box 17.2. Ca\\\u00b7endish. H., 1798. \"Experiments to determine the density of the Earth\" (read June 21. 1798). Phil. TrailS. R. Soc. London. Part 11 (1798). pp. 469-526. [\u00a740.8) Caves, C. M., cited by name in \u00a726.1. Chadwick. J.. 1932. \"Possible existence of a neutron.\" Nature 129. 312. [Box 24.1) Chandrasekhar, S.. 1931a. \"The density of white dwarf stars.\" Phil. Mag. 11. 592-596. [Box 24.1) Chandrasekhar. S.. 1931 b, \"The maximum mass of ideal white dwarfs,\" Astrophys. J. 74, 81-82. [Box 24.1) Chandrasekhar. S.. 1964a. \"Dynamical instability of gaseous ma,ses approaching the Schwarzschild limit in general relativity.\" PhI'S. Rev. Lett. 12. 114-116 and 4", "37-438. [\u00a726.1. Boxes 24.1. 26.2) Chandrasekhar. S.. 1964b. \"The dynamical instability of gase ow; masses approaching the Schwarzschild limit in general relativity,\" Astropl~l\u00b7s. J. 140, 417-433. [\u00a726.I, Boxes 24.1, 26.2) Chandrasekhar, S.,_ 1965a, \"The post-Newtonian equations of hydrodynamics in general relativity,\" Astrophys. J. 142, 1488-1512. [\u00a739.3, Box 39.3, Ex. 39.14) Chandrasekhar, S., 1965b, \"The post-Newtonian effects of general relativity on the equilibrium of uniformly rotating bodies, I: The Maclaurin spheroids and the virial theo rem,\" Astrophys. J. 142, 1513-1518. [\u00a739.3, Box 39.3) Chandrasekhar, S., 1965c, \"The stability of gaseous masses for radial and non-radial oscillations in the post-Newtonian approximation of general relativity,\" Astrophys. J. 142, 1519-1540. [\u00a739.3, Box 26.2) Chandrasekhar, S., 1968, Hydrodynamic and Hydromagnetic Stability, Clarendon Press, Oxford. See especially \u00a7I08. [Box 32.3) appropriate transformation Chandrasekhar, S., and G. Contopolous, 1967, \"On a post post Galilean Newtonian theory of Einstein, Infeld, and Hoffmann,\" Proc. R. Soc. London A 298, 123-141. [\u00a739.3, 39.9) Chandrasekhar, S., and F. P. Esposito, 1970, \"The 2I1z-post Newtonian equations of hydrodynamics and radiation reaction in general relativity,\" Astrophys. J. 160, 153-180. [\u00a736.8, Box 39.3) the to Chandrasekhar, S., and N. R. Lebovitz, 1968, \"The pulsations and the dynamical stability of gaseous masses in uniform rotation,\" Astrophys. J., 152, 267-291. [Ex. 39.7) Chandrasekhar, S., and Y. Nutku, 1969, \"The second post Newtonian equations of hydrodynamics in general relativ ity,\" Astrophys. J. 158, 55-80. [Box 39.3) Chapel Hill 1957, Conference on the Role of Gravitation in Physics, Jan. 18-23, 1957. See DeWitt 1957. i Chase, D. M., 1954, \"The equations of motion of charged test particles in general relativity,\" Phys. Rev. 95, 243-246. [\u00a720.6) Cherlin, G., cited by name in imroduction to this bibliography. Chern, S.-S., 1955, \"On curvature and characteristic classes of a Riemann manifold,\" Hamburg Abh. 20, 117-126. [Box 15.3) Chern, S.-S., 1960, Geometric Structures on Manifolds, collo quium lectures, Amer. Math. Soc., Providence, R.L [\u00a72.2) Chern, S.-S., 1962, \"Geometry of a quadratic differential BIBLIOGRAPHY AND INDEX OF NAMES 1227 form,\" J. Soc. Indust. & Appl. Math. 10, 751-755, also in H. P. Robertson January 27, 1903-August 26, 1961, In Memoriam, Soc. Indust. and Applied Math., Philadelphia 1963, pp. 13-17. [Box 15.3) Chern, S.-S., 1966, \"The geometry of G-structures.\" Bull. Am. Math. Soc. 72, 167-219. [\u00a72.2) Chern, S.-S., 1971, \"Differential geometry: its past and its future,\" in Proceedings of the International Congress of Mathematicians, Nice, September 1970, Gauthier-Villars, Paris, pp. 41-53. [\u00a72.2) Chern, S.-S., and C. Chevalley, 1952, \"Elie Cartan and his mathematical work,\" Bull. Am. Math. Soc. 58, 217-250. [Box 8.2) Chevalley, C. See Chern and Chevalley 1952. Chick, H. J., 1960, Theory of Gravitation 192o-Present (Bibli ography), Report LMS-2328 of Los Alamos Scientific Laboratory, Los Alamos, New Mexico, cited in preface of ~resent bibliography. Chick, H. J., 1964, Theory of Gravitation", " 1959-1963 (Bibliog raphy), Report LMS-3099 of Los Alamos Scientific Labo ratory, Los Alamos, New Mexico, cited in preface of present bibliography. Chinnapared, R. See Newman et al. 1965. Chitre, D. M., 1972a, Investigations of the Vanishing of a Hori zon for Bianchi Type IX (Mixmaster) Universe, doctoral dissertation, Univ. of Maryland. [\u00a730.8) Chitre, D. M., 1972b, \"High-frequency sound waves to elimi nate a horizon in the mix master universe,\" Phys. Rev. D 6, 3390-3396. [\u00a730.8) Chitre, D. M. See also Berger et al. 1972. Chiu, H.-Y., and W. F. Hoffman, eds., 1964, Gravitation and Relativity, W. A. Benjamin, New York. Choquard, P., 1955, \"Traitement semi-classique des forces generales dans la repres~ntation de Feynman,\" Helv. Phys. Acta 28, 89-157. [Box 21.1, Fig. 13.2) Choquet-Bruhat, Y., 1968a, Geometrie difJerentielle et systemes exterieurs. Monographies Universitaires de Mathema tiques no. 28, Dunod, Paris, France. [\u00a74.I) Choquet-Bruhat, Y., \"Espace-temps Einsteiniens generaux, chocs gravitationels,\" Ann. Inst. Henri Poincare 8. 327-338. [\u00a721.13) 1968b, Choquet-Bruhat, Y., 1969, \"Construction de solutions radia tives approchees des equations d'Einstein,\" Commun. Math. Phys. 12, 16-35. [\u00a735.13) Choquet-Bruhat, Y., 1972, \"Probleme des contraintes sur une variete compacte,\" Acad. Sci. Paris, Comptes Rend. A 274, 682-684; fuller account in \"Global solutions of the equa tions of constraint in general relativity, on open and closed manifolds.\" Gen. Rei. & Grav. 1973. in press. [\u00a72I.I) Choquet-Bruhat, Y. See also Bruhat. Y, and Foures Bruhat, Y. Chretien, M., S. Deser, and J. Goldstein, eds.. 1969, Brandeis Slimmer Institute 1968, Astrophysics and General Relativity. Gordon and Breach, New York. Christodolou. D., 1970, \"Reversible and irreversible trans formations in black-hole physics,\" Phys. Rev. Lett. 25. 1596-1597. [\u00a733.8, Box 33.4) Christodolou, D., 1971. Investigations in Gravitational Collapse and the Physics of Black Holes. doctoral dissertation, Princeton Univ. [\u00a735.8. Boxes 33.3. 33.4, Fig. 33.3) Christodolou, D., and R. Ruffini, 1971, \"Reversible trans formations of a charged black hole,\" Phys. Rev. D 4, 3552-3555. [\u00a733.8, Boxes 33.4, 33.5) Chrzanowski, P. L., and C. W. Misner, 1973, \"Geodesic syn chrotron radiation in the Kerr geometry by the method of asymptotically factorized Green's functions,\" technical report of the Dept. of Physics and Astronomy, University of Maryland. [\u00a733.8) Chrzanowski, P. L. See also Brill et al. 1972. Cincinnati 1969, Relativity Conference in The Midwest, June 2-6, 1969, Cincinnati, Ohio. See Carmeli, Fickler, and Witten 1970. Clark, G. See Eddington and Clark 1938. Clarke, C. J. S., 1970, \"On the global isometric embedding of pseudo-Riemannian manifolds,\" Proc. Roy. Soc. London A 314, 417-428. [\u00a723.8) Clayton, D., 1968, PrinCiples of Stellar Evolution and Nucleo synthesis, McGraw-Hill, New York. [\u00a728.I) Clemence, G. M., 1943, \"The motion of Mercury, 1765-1937,\" Astron. Papers Am. Ephemeris 11, 1-221. [Box 40.3) Clemence, G. M., 1947, \"The relativity effect in planetary motions,\" Rev. Mod. Phys. 19, 361-364. [Box 40.3) Clemens, R. See", " also Fletcher et al. 1967. Clemens, R., and R. Matzner, 1967, Technical Report no. 635 of the Univ. of Maryland Department of Physics and Astronomy. [Box 14.3) Clifford, W. K., 1879, Lectures and Essays, L. Stephen and I. 244 and 322. F. Pollock, eds., Macmillan, London, [\u00a744.3) Clifford, W. K., 1882, Mathematical Papers, R. Tucker, ed., Macmillan, London. [\u00a744.3) CNRS, 1962, Les Theories relativistes de la gravitation, Editions du Centre National de la Recherche Scientifique, Paris, France. Same as Royaumont 1959. Cocke, W. J., 1965, \"A maximum-entropy principle in general relativity and the stability of fluid spheres,\" Ann. Inst. Henri Poincare A 2, 283-306. [\u00a726.I) Cocke, W. J., H. J. Disney and D. J. Taylor. 1969, \"Discovery of optical signals from pulsar NP0532.\" Nature 221, 525-527. [Box 24.1) Codazzi, D., cited by name in \u00a721.5. Cohen, J. M., 1968, \"Angular momentum and the Kerr metric,\" J. Math. Phys. 9, 905-906. [Ex. 26.1) Cohen, J. M. See also Brill and Cohen 1966. Cohen, J. M., and D. Brill, 1966, \"Rotating masses and their effect on inertial frames,\" Phys. Rev. 143, 1011-1015. [\u00a721.12) Cohen, J. M., and D. R. Brill, 1968, \"Further examples of 'Machian effects' of rotating bodies in general relativity,\" Nuovo Cim. 56B, 209-219. [Ex. 26.1) Cohen, J. Moo and R. M. Wald, 1971. \"Point charge in the vicinity of a Schwarzschild black hole.\" J. Math. Phys. 12, 1845-1849. [Box 33.1) Cohen, P., 1966, Set Theory and the Continuum H)pothesis, W. A. Benjamin, New York. [Box 44.5) Cole, J., 1968, Perturbation Methods in Applied Mathematics. Ginn-Blaisdell, Waltham. Mass. [\u00a722.5) Coleman, C. 1.. 1971. \"A simple relativistic theory of gravita tion,\" J. Phys. A 4. 611-616. [\u00a740.5) Coieman. J.. A., 1967, Ear~l' Theories of the Universe. New American Library. New York. [\u00a71.7) 1228 GRAVITATION Colgate, S. A., 1967, \"Stellar coalescence and the multiple supernova interpretation of quasi-stellar sources,\" Astro\u00b7 phys.1. 150, 163-192. [\u00a724.5. Box 33.3) Colgate, S. A., and R. H. White. 1966, \"The hydrodynamic beha vior of supernovae explosions:' Astrophys. J. 143, 626-681. [\u00a732.7) Collins, C. B., and S. W. Hawking, 1973, \"Why is the universe isotropic?\" Astrophys. J. 180, 317-334. [Box 34.3) Collins, R. A., cited by name in Box 27.7. See also Hewish et al. 1968. Combridge, J. T., 1965. Bibliography of Relativity and Gravita\u00b7 tion Theory 1921-1937, Kings College, London. cited in preface of present bibliography. Comte, A., 1835, Cours de philosophie positive. Paris. [Box 27.7). Conklin, E. K., 1969, \"Velocity of the Earth with respect to the cosmic background radiation,\" Nature 222, 971-972. [\u00a730.8) Contopolous, G. See Chandrasekhar and Contopolous 1967. Coote, N., cited by name in introduction to this bibliography. Copenhagen 1971, Sixth International Conference on General Relativity and Gravitation, held in Copenhagen, July 1971; portions of the proceedings appear as articles in Gen. Rel. Grav., especially vol. 3, 1972. Copernicus, N. See Dobson and Brodetsky 1947. Corben, H. C.. and P. Stehle, 1960, Classical Mechanics, 2d ed., Wiley, New York. [Ex. 33.7). Cornell University, 1965, Summer Seminar on Applied Mathe matics, arranged by the American", " Mathematical Society. Proceedings (vols. 8-10 in Lectures in Applied Mathe matics series) published as Ehlers 1967. Cornford, F. M., 1937, Plato's Cosmology: The Timaeus of Plato, Humanities Press, New York. [Box 27.7) Corson, E. M., 1953, Introduction to Tensors, Spinors, and Relativistic Wave-Equations. Hafner, New York. [\u00a721.3). Couch, w. E. See also Newman et al. 1965. Couch, w. E., and E. T. Newman, 1972, \"Generalized Lienard-Wiechart fields,\" J. Math. Phys. 13, 929-931. [\u00a736.l) Couch, w. E., R. J. Torrence, A. I. Janis, and E. T. Newman, 1968, \"Tail of a gravitational wave,\" J. Math. Phys. 9, 484-496. [\u00a7\u00a735.8, 35.14) Courant, R., 1937, \"Plateau's problem and Dirichlet's princi ple,\" Ann. Math. U.S. 38, 679-724. [\u00a733.2) Coxeter, H. S. M., 1963, Regular Polytopes. 2d ed., Macmillan, New York. [\u00a742.5, Box 27.2) Curott, D. R., 1966, \"Earth deceleration from ancient solar eclipses,\" Astron. J. 71, 264-269. [Fig. 1.8) Currie, D. G., cited by name in Fig. 40.4. See also Bender et al. 1971. Curtis, H. D., cited by name in Box 27.7. Cusanus, N. See Heron 1954. Dabbs, J. W. T\" J. A. Harvey, D. Paya, and H. Horstmann, 1965, \"Gravitational acceleration of free neutrons,\" Phys. Rev. B 139, 756-760. [\u00a71.3) D'Alembert, J., quoted at end of Chap. 44. Danielson, B. L. See Evenson et al. 1972. Dante Alighieri. Paradiso, translated by Philip H. Wicksteed, 1921, Dent, L\"ndon, England. [Epigraph to Chap. 44) Danziger, I. 1.. 1970, \"The cosmic abundance of helium,\" Ann. Rev. Astron. Astrophys.\u2022 8, 161-178. [\u00a728.I) Darboux, G., 1941. Le(ons sur la Theorie Ge/l(!rale des Surfaces, Part I, Gauthier-Villars, Paris, France. [\u00a733.2) Darmois, G., 1927. Les equations de la gravitation einsteinienne, Memorial des sciences mathematiques XXV, Gauthier Villars, Paris. [\u00a721.13) Darwin, c., 1959. \"The gravity field of a particle. I.\" Proc. R. Soc. London A 249. 180-194. [\u00a725.5, Fig. 25.7) Darwin, C., 1961. \"The gravity field of a particle. I/.\" Proc. R. Soc. London A 263. 39-50. [\u00a725.5, Fig. 25.7) Datt, B., 1938, \"Uber eine Klasse von Losungen der Gravita tionsgleichungen der Relativitlit,\" Z. Physik 108, 314-321. [\u00a730.6, Ex. 32.8) Davenport, W. B., and W. L. Root, 1958, An Introduction to the Theory of Random Signals and Noise. McGraw Hill, New York, [\u00a737.8) Davies, R. W., ed., 1971, Proceedings of the Conference on Experimental Tests of Gravitational Theories, November 11-13, 1970, California Institute of Technology, J.P.L. Technical Memorandum, 33-499. Davis, M., R. Ruffini, W. H. Press, and R. H. Price, 1971, \"Gravitational radiation from a particle falling radially into a Schwarzschild black hole,\" Phys. Rev. Lett. 27, 1466-1469. [\u00a7\u00a733.7, 36.5, Box 33.3, Fig. 36.2) Davis, W. R., 1970, Classical Theory of Particles and Fields and the Theory of Relativity, Gordon and Breach, New York. [\u00a721.3, endpapers) Day, G. W. See Evenson et al. 1972. de Broglie, M. See Einstein -1912. De", "bye, P. J. W., cited by name in \u00a722.5. De Cheseaux, P. L., 1744, Traite de la comete qui a paru en decembre 1743, Paris; the relevant material is in the Ap-. pendix. [Box 27.7) De Felice, F., 1968, \"Equatorial geodesic motion in the gravi tational field of a rotating source,\" Nuovo Cimento B 57, 351-388. [Box 33.5) Defoe, D., 1722, The Fortunes and Misfortunes of the Famous Moll Flanders, reprinted by Fine Editions Press, Cleveland, 1953. [epigraph to Part X) De Groot, S. R. See Kluitenberg and de Groot 1954. De Hoffmann, F., and E. Teller, 1950, \"Magneto-hydro dynamic shocks,\" Phys. Rev. 80, 692-703. [\u00a722.2, Box 22.2) De la Cruz, V., and W. Israel, 1967, \"Gravitational bounce,\" Nuovo Cimento A 51, 744-760. [\u00a734.6, Fig. 34.4) De Lange, F. G. See Refsdal, Stabel1, and de LangcJ 1967. Dempster, J. R. H., 1963, Feynman Quantization in Field The ory: Topology of Commutators and Functional Repre sentation of States, doctoral dissertation, Princeton Uni versi ty. [Box 21.1) De Rham, G., 1955, Varietes DifJerentiables, Herman, Paris. [\u00a7\u00a74.1,22.4) Deser. S., 1970, \"Self-interaction and gauge invariance,\" Gen. Rei. & Grav. 1, 9-18. [\u00a7\u00a77.I, 18.1, Box.17.2 part 5) BIBLIOGRAPHY AND INDEX OF NAMES 1229 Deser, S., cited by name in \u00a7\u00a721.1, 35.8, Box 17.2, part 2'. See also Arnowitl, Deser, and Misner 1962 and Chretien, Deser, and Goldstein 1969. Deser, S., and K. W. Ford, eds., 1965, Lectures on General Relativity; 1964 Brandeis Summer Institute in Theoretical Physics 1, Prentice-Hall, Englewood Cliffs, N.J. Same as Trautman, Pirani, and Bondi 1965. De Sitter, W., 1916, \"On Einstein's theory of gravitation and its astronomical consequences,\" Mon. Not. R. Astron. Soc. 77, 155-184 and 481. [\u00a7\u00a740.6, 40.7] De Sitter, W., 1917a, \"On the relativity of inertia: Remarks concerning Einstein's latest hypothesis,\" Proc. Kon. Ned. Akad. Wet. 19, 1217-1225. [\u00a727.11] De Sitter, W., 1917b, \"On the curvature of space,\" Proc. Kon. Ned. Akad Wet. 20, 229-243. [\u00a727.11] De Sitter, W., cited by name in \u00a730.3, Box 27.7. Deutsch, A. J., and W. B. Klemperer, eds., 1962, Space Age Aslronomy, Academic Press, New York. De Valera, E., cited by name in \u00a741.1. De Vaucouleurs, G., 1971, \"The large-scale distribution of galaxies and clusters of galaxies,\" Astron. Soc. Pacific Publ. 83, 113-143. [\u00a727.11] DeWitt, B. S., 1953, \"Pair production by a curved metric,\" Phys. Rev. 90, 357. [\u00a730.5] 1967 Lectures Benjamin, New York. in Mathematics and Physics, W. A. Dicke, R. H., 1954, \"Coherence in spontaneous radiation proc esses,\" Phys. Rev. 93,99-110. [Box 38.3] Dicke, R. H., 1957, \"Coherence and the quantum,\" J. Opt. Soc. Am. 47, 527. [Box 38.3] Dicke, R. H., 1959a, \"Dirac's cosmology and the dating of meteorites,\" Nature 183, 170-171. [\u00a738.6] Dicke, R. H., 1959b, \"New research on old gravitation,\" Sci\u00b7 ence 129, 621-624. [\u00a7\u00a738.6, 40.9, 44.6] Dicke, R. H., 1961, \"Dirac's", " cosmology and Mach's principle,\",Vature 192, 44()...441. [\u00a7\u00a738.6, 44.6, Box 38.3] Dicke, R. H., 1962, \"Mach's principle and invariance under transformation of units,\" Phys. Rev. 125, 2163-2167. Re printed in Dicke 1964b. [Box 39.1] Dicke, R. H., 1964a, \"The sun's rotation and relativity,\" Nature 202, 432-435. [Box 38.3] Dicke, R. H., I964b, The Theoretical Significance of Experi mental Relativity, Gordon and Breach, New York; also in DeWitt and DeWitt 1964a. [\u00a7\u00a738.2, 38.7, 44.6, Box 40.1] Dicke, R. H., I964c, \"Remarks on the observational basis of general relativity,\" in Chiu and Hoffman 1964, pp. 1-16. [\u00a740.9] Dicke, R. H., 1968, \"Scalar-tensor gravitation and the cosmic DeWitt, B. S., 1962, \"The quantization of geometry,\" in Witten fireball,\" Astrophys. J., 152, 1-24. [\u00a728.4] 1962. [Ex. 40.8] Dicke. R. H., 1969, Gravitation and the Universe, American DeWitt, B. S., 1967a, \"Quantum theory of gravity, I,\" Phys. Philosophical Society, Philadelphia. [\u00a7\u00a738.6, 40.9] Rev. 160, 1113-1148. [\u00a7\u00a721.1, 43.1, 43.3] DeWitt, B. S., 1967b. \"Quantum theory of gravity, II; The manifestly covariant theory.\" Phys. Rev. 162, 1195-1239. [\u00a7\u00a743.1,43.3] DeWitt, B. S., 1967c, \"Quantum theory of gravity, III; Appli cation of the covariant theory,\" Phys. Rev. 162, 1239-1256. [Box 17.2 part 6h ] \". DeWitt, B. S., 1971, Lectures on Relativity-Stanford, Fa/l1971 (unpublished manuscript). [\u00a736.7] DeWitt, B. S., cited by name in Ex. 35.20. See also DeWitt and DeWitt I964a, 1964b, and 1973. DeWitt, B. S., and R. W. Brehme, 1960, \"Radiation damping in a gravitational field,\" Ann. Phys. (U.S.A.) 9, 220-259. [\u00a735.8, Ex. 35.14] DeWitt, C., ed. 1957, Conference on the Role of Gravitation in Physics, Proceedings, W.A.D.C. Technical Report 57-216; ASTIA Document No. AD118180, Wright Air Development Center, Wright-Patterson Air Force Base, Ohio. DeWitt, C. See also Morette, C. DeWitt, C., and B. S. DeWitt, 1964a, \"Falling charges,\" Physics 1, 3-20. [\u00a735.8] DeWitt, C., and B. S. DeWitt, eds., 1964b, Relativity, Groups. and Topology, Gordon and Breach, New York. [\u00a738.2] DeWitt. c., and B. S. DeWitt, eds., 1973, Black Holes, Pro ceedings of 1972 session of Ecole d'ete de physique theo rique. Gordon and Breach, New York. [\u00a733.1] DeWitt, c., E. Schatzman, and P. Veron, eds., 1967, High. Energy Astrophysics. Vol. Ill. Gordon and Breach, New York. [\u00a724.4] DeWitt, C., and J. A. Wheeler, eds., 1968, Battelle Rencontres: Dicke, R. H., cited by name in \u00a7\u00a7IO.l, 38.3, 38.5, 40.1, 40.6, 40.9, Box 27.7. See also Bender et al. 1971, Brans and Dicke 1961, and Roll, Krotkov, and Dicke 1964. Dicke, R. H., and H. M. Goldenberg, 1967, \"Solar oblateness relativity,\" Phys. Rev. Lett. 18, 313-316. and general [Boxes 38.3, 40.3] Dicke, R. H., and P. J", ". E. Peebles, 1962a, \"Significance of spatial isotropy,\" Phys. Rev. 127, 629-631. Reprinted in Dicke I964b. [\u00a738.7] Dicke, R. H., and P. J. E. Peebles, 1962b, \"Cosmology and the radioactive decay ages of terrestrial rocks and meteor ites,\" Phys. Rev. 128, 2006-2011. [\u00a738.6, Box 38.3] Dicke. R. H., and P. J. E. Peebles, 1965, \"Gravitation and space science,\" Space Sci. Rev. 4, 419-460. [\u00a740.8] Dicke. R. H., P. J. E. Peebles, P. G. Roll, and D. T. Wilkinson, 1965, \"Cosmic-black-body radiation,\" Astrophys. J. 142, 414-419. [\u00a7\u00a727.2, 28.1] Dieter, N. H., E. E. Epstein, A. E. Lilley, and M. S. Roberts, 1962, \"A radio and optical investigation of extragalactic redshifts,\" Astron. J. 67, 270-271. [\u00a729.2] Dieudonne, J.. 1960, Foundations of Modern Ana~}'sis, Aca demic Press, New York. [\u00a72.6] Digges, T., cited by name in Box 27.7. d'inverno. R. A., 1969, \"ALAM-Atlas Lisp algebraic manipu lator.\" Comput. J. 12, 124-127. [Box 14.3] Dirac. P. A. M., 1926, \"On the theory of quantum mechanics,\" Proc. R. Soc. London A 112. 661-677. [Box 24.1] Dirac. P. A. M., 1\u00b7936, \"Relativistic wave equations,\" Proc. R. Soc. London A 155,447-459. [Box 32.2] Dirac. P. A. ~I., 1937, 'The cosmological constants,\" Nature 139. 323. [\u00a7\u00a738.6. 44.6] 1230 GRAVITATION Dirac. P. A. M., 1938, \"New basis for co~mologv.\" Proc. R. Soc. London A 165. 199-208. [\u00a7\u00a738.6, 44.6] Dirac, P. A. M., 1959, \"Fixation of coordinates in the Hamil tonian theory of gravitation,\" PhI's. Rev. 114, 924-930. [\u00a721.7]. Dirac, P. A. M. 1964, Lectures on Quantum.He~hanics. Be'lter Graduate School of Science' Monograph Series!\"umber Two. Yeshiva University, New York. [\u00a721.7] Dirac. Paul A. M., Festschrift on the occasion of his 70th birthday, Aug. 8, 1972. See Salam and Wigner 1972. Dirac, P. A. M.. cited ~r name in \u00a7\u00a720.6. 21.1. Box 4.4. Dirac, P. A. M., V. A. Fock. and B. Podolsky. 1932, \"On quantum electrodynamics,\" Phys. Z. SOlljetunion 2. 468-479. [\u00a721.8] Dobson. J. F., and S. Brodetsky. 1947. Nicolaus Copernicus's De Revolutionibus (translation of Preface and Book Il. Royal Astronomicai Society. London. [Box 27.7]. Dodgson, C. L., 1871, Through the Looking Glass. Macmillan, London. [Epigraph to Chap. 32] Doroshkevich, A. G., 1966, \"Change of the angular momentum of a star during accretion.\" Astron. 2hur. 43, 105-107. Terrible English translation in Sou. Astron.-A. J. 10, 83-84. [Box 33.3] Doroshkevich, A. G., v. N. Lukash, and I. D. Novikov, 1971, \"Impossibility of mixing in the Bianchi type IX cosmo logical model,\" 211. Eksp. & Teor. Fiz. 60, 1201-1205. English translation in Sov. Phys.-JETP 33, 649-651. [\u00a730.8] Doroshkevich, A. G., Ya. B. Zel'dovich, and I. D. Novikov, 1965, \"Gravitational collapse of nonsymmetric and rota ting masses", ". 1., 1969, \"Seismic response of the Earth to a gravita tional wave in the I-Hz band,\" Astrophys. 1. 156, 529-540. [\u00a737.3, Fig. 37.2] Dyson, F. J.. 1972. \"The fundamental constants and their time variation:' in Salam and Wigner 1972. [\u00a738.6] Eardley. D., E. Liang, and R. Sachs, 1972, \"Velocity-dominated singularities in irrotational dust cosmologies,\" J. Math. Phys. 13, 99-107. [\u00a730.7] Eckart, C., 1940, \"The thermodynamics of irreversible proc esses, III: Relativistic theory of the simple fluid,\" Phys. Rev., 58, 919-924. [Ex. 22.7] Ecphantus, cited by name in Box 27.7. Eddington, A. S., 1922, The Mathematical Theory of Relativity, Cambridge Univ. Press, Cambridge, Eng. [\u00a7\u00a738.2, 39.3] Eddington, A. S., 1924, \"A comparison of Whitehead's and Einstein's formulas,\" Nature 113, 192. [Boxes 31.1, 31.2] Eddington, A. S., 1931, \"Preliminary note on the masses of the electron, the proton, and the universe,\" Proc. Cambridge Phil. Soc. 27, 15-19. [\u00a744.6] Eddington, A. S., 1936, Relativity Theory of Protons and Elec trons, Cambridge Univ. Press, Cambridge, Eng. [\u00a744.6] Eddington, A. S., 1946, Fundamental Theory, Cambridge Univ. Press, Cambridge, Eng. [\u00a744.6] Eddington, A. S., cited by name in Box 27.7. Eddington, A., and G. Clark, 1938, \"The problem of n bodies in general relativity theory,\" Proc. Roy. Soc. A 166, 465 475. [Ex. 39.15] Ehlers, 1., 1961, \"Contributions to the relativistic mechanics of continuous media,\" Akad. Wiss. Lit. Mainz AM. Math.\u00b7 Nat. KI., 793-837. [\u00a722.3] Ehlers, J., ed., 1967, Relativity Theory and Astrophysics: I, Relativity and Cosmology; II, Galactic Structure; Ill, Stellar Structure, American Mathematical Society, Provi dence, R.I. Ehlers, J., 1971, \"General relativity and kinetic theory,\" In Sachs 1971. [\u00a722.6, Ex. 22.7, endpapers] Ehlers, J. See also Jordan, Ehlers, and Sachs 1961. Ehlers, J., P. Geren, and R. K. Sachs, 1968, \"Isotropic!i>lutions of the Einstein-Liouville equations,\" 1 Math. Phys. 9, 1344-1349. [\u00a727.2] Ehlers, J., and W. Kundt, 1962, \"Exact solutions of the gravita tional field equations,\" in Witten 1962. [\u00a7\u00a733.6, 35.9, Ex. 35.8, 35.9] Ehlers, J., F. A, E. Pirani, and A. Schild, 1972, \"The geometry of free-fall and light propagation,\" in O'Raifeartaigh 1972, pp. 63-84. [Boxes 10.2, 16.4] Ehrenfesl, P., 1916, \"On the adiabatic changes of a system in connection with the quantum theory,\" Amsterdam Acad., Proc. 19, 576-597. [Box 22.4] BIBLIOGRAPHY AND INDEX OF NAMES 1231 Einstein, A., 1905, \"Zur elektrodynamik bewegter Korper,\" Ann. Phys. (Germany) 17, 891-921. [Box 1.5] Einstein, A., 1908, \"Uber das Relativitatsprinzip und die aus demselben gezogenen Folgerungen,\" Jarb. Radioakt. 4, 411-462. [\u00a7\u00a77.4, 17.7] Einstein, A, 1911, \"Uber den Einfluss der Schwerkraft auf die Ausbreitung des Lichtes,\" Ann. Phys. (Germany) 35, 898 908, English translation in Lorentz et al. 1923. [\u00a7\u00a71.3, 7.2, 7.4, 10.1, 16.2] Einstein, A.' 1912, \"Si I'on modifie la longueur du pen dule...,\" p.", " 450 in P. Langevin and M. de Broglie, eds., La theorie du rayonnement et les quanta, Gauthier-Villars, Paris, 1912. [Box 22.4] Einstein, A., 1913a, \"Zum gegenwartigen Stand des Gravita tionsproblems,\" Phys. Z. 14, 1249-1262, discussion 1262 1266. [Epigraph to Chap. 35] Einstein, A., 1913b, letter to Ernst Mach dated June 26. [\u00a721.12, Fi~. 21.5] Einstein, A, 1915a, \"ZUr allgemeinen Relativitiitstheorie,\" Preuss. Akad. Wiss. Berlin, Sitzber., 778-786 (published November 11). [\u00a7I7.7] Einstein, A. 1915b, \"Ziir allgemeinen Relativitiitstheorie (Nachtrag),\" Preuss. Akad. Wiss. Berlin, Sitzber., 799-801 (published November 18). [\u00a7I7.7, Box 27.8] Einstein, A., 1915c. \"Erkliirung der Perihelhewegung des Merkur aus der allgemeinen Relativitiitstheorie,\" Preuss. Akad. Wiss. Berlin, Sitzber. 47, 831-839 (published No vember 25). [\u00a7I7.7] Einstein, A., 1915d, \"Die Feldgleichungen der Gravitation,\" Preuss. Akad. Wiss. Berlin, Sitzber., 844-847 (published December 2). [\u00a717.7, Box 27.7] 101-104. [Fig. 21.5] Einstein, A., 1916a, \"Emst Mach,\" Phys. Z. Sowjetunion 17, _ Einstein. A., 1916b, \"Niiherungsweise Integration der Feld gleichungen der Gravitation,\" Preuss, Akad. Wiss. Berlin, Sitzber. 688-696. [\u00a7I7.7] Einstein, A., 1916c, \"Hamiltonsches Prinzip und allgemeine Relativitlitstheorie,\" Preuss. Akad. Wiss. Berlin, Sitzber., 1111-1116. [\u00a717.7] Einstein, A., 1917, \"Kosmologische Betrachtungen zur allge meinen Relativitiitstheorie,\" Preuss. Akad. Wiss. Berlin, Sitzber. 142-152, reprinted in English in Lorentz et al. 1923. [\u00a717.3, Boxes 27.1, 27.7] Einstein, A., 1921, \"Geometry and Experience,\" address to Prussian Academy ofSeiences, Berlin, Jan. 27,1921; pub lished in Sidelights on Relativity, Mathuen, London, 1922. [Box 1.11] Einstein, A., 1931, \"Zum kosmologischen Problem der allge meinen Relativitiitstheorie,\" Preuss, Akad. Wiss. Berlin, Sitzber, 235-237. [\u00a727.II] Einstein. A., 1934. Essays in Science, Philosophical Library, New York. Translated from Mein Weltbild, Querido Ver lag. Amsterdam, 1933. [Epigraph to Chap. 17, \u00a7\u00a72 I.l 2. 27.11,27.12,43.1,44.6, Box 8.5] Einstein, A, 1939, \"On a stationary system with spherical symmetry consisting of many gravitating masses,\" Ann. Math. (U.S.A.) 40, 922-936. [Ex. 25.29] Einstein. Albert, Festschrift, in commemoration of his seven- tieth birthday, March 14, 1949; Rev. Mod. Phys. 21 (July 1949), 343-540. Einstein, A., 1949a, introduction to Schilpp 1949. [Preface, Epigraph to Chap. 44, \u00a71.2] Einstein, A, 1949b, \"Reply to criticisms,\" in Schilpp 1949. [Epigraph to Chap. 27] Einstein, A., 1950, The Meaning of Relativity, 3d ed., Princeton Vniv. Press, Princeton, N.J. [\u00a721.12, 29.5,43.1, Box 27.1, endpapers] Einstein, A., 1954, preface to Concepts of Space, M. Jammer, ed., Harper, New York, p. xv. [\u00a71.3] Einstein, A., 1970, quoted by G. Gamow in", " My World Line, Viking Press, New York (1970), p. 44. [\u00a7\u00a717.3, 27.8, Box 27.1] Einstein, A, cited by nam~ in \u00a7\u00a7I.I, 1.6, 7.1, 7.2, 7.4, 10.1, 12.1, 12.2, 17.3, 20.6, Boxes 12.2, 12.3, 27.7. See also Lorentz et al. 1923, Born 1971, Holton 1971, Klein 1971. Einstein, A, and A. D. Fokker, 1914, Ann. Phys. (Germany), 44, 321. [\u00a717.6] Einstein, A., and J. Grommer, 1927, \"Allgemeine Rela tivitatstheorie und Bewegungsgesetze,\" Sitzber. Deut. Akad. Wiss. Berlin, KI. Math. Phys. Tech. 2-13, 235-245. [\u00a720.6] Einstein, A, and M. Grossmann, 1913, \"Entwerf einer ver allgemeinerten Relativitiitstheorie und einer Theorie der Gravitation,\" Zeit. Math. Phys. 62, 225-261. [\u00a7I7.7] Einstein, A., and M. Grossmann, 1914, \"Kovarianzeigen schaften der Feldgleichungen der auf die verallgemeinerte Relativitlitstheorie gegrundeten Gravitationstheorie,\" Zeit. Math. Phys. 63, 215-225. [\u00a7I7.7] Einstein, A., L. Infeld, and B. Hoffmann, 1938, \"The gravita tional equations and the problem of motion,\" Ann. Math. 39, 65-100. [Ex. 39.15] Eisenhart, L. P., 1926, Riemannian Geometry. Princeton Vniv. Press, Princeton, N.J. [\u00a7\u00a721.5, 21.12, Ex. 27.4, endpapers] Ekers, R. D. See Muhleman, Ekers, and Fomalont 1970. Eliot, T. S., 1920, \"Tradition and individual talent,\" in The Sacred Wood; Essays on Poetry and Criticism, Methuen, London. [Box 27.7] Eliot, T. S., 1942, Little Gidding, Faber and Faber, London. [\u00a7L7] Ellis, G. F. R., 1971, \"Relativistic cosmology,\" in Sachs 1971. [\u00a722.3, Ex. 22.6, 29.5] Ellis., G. F. R., cited by name in Boxes 27.7, 34.3, Ex. 32.1. See also Hawking and Ellis 1968 and 1973, and Stewart and Ellis 1968. E6tv6s, R. V., 1889, \"Uber die Anziehung der Erde auf ver schiedene Substanzen,\",Hath. Naturw. Ber. aus Ungarn 8, 65-68. [Box 1.2] E6tv6s, R. V\" Cited by name in \u00a7\u00a7IO.I, 38.3, 38.5,40.9. E6tvds, R. V., D. Pekar. and E. Fekete, 1922, \"Beitriige zum Gesetze der Proportionalitiit von Triigheit und Gravilal,\" Ann. Ph.vs. (Germany) 68, 11-66. [\u00a7\u00a7IO.I, 38.3,40.9. Boxes 1.2,38.2] Epstein, E. E. See Dieter et al. 1962. Ernst, F. J., 1968, \"7090 FORMAC subroutines: Exterior cal culus.\" reporI number PORS lIT-9, Illinois Institute of Technology. [Box 14.3] 1232 GRAVITATION Esposito, F. P. See Chandrasekhar and Esposito 1970. Esposito, P. B. See Ander~on et al. 1971. ESRO Conference on Space Physics, Interlaken, Switzerland, September 4, 1969. See Ruffini and Wheeler 1971b. Estabrook, F. B.,I969, \"Post-Newtonian nobody equations of the Brans-Dicke theory,\" ASfrophys. J. 158, 81-83. [Ex. 39.15] Estermann, I., O. C. Simpson, and O. Stern, 1938, \"The free fall of molecules,\" Phys. Rev. 53, 947-9", "48. [\u00a71.3] Eudoxus, cited by name in Box 27.7. Euler, H. See Heisenberg and EtHer 1936. Euler, L., 1760, \"Demonstratio theorematis Fermatiani omnem numerum esse summam quatuor quadratorum,\" Novi Comment, Petrop. 5, 93, Opera 12, 369ff. [\u00a741.1] Evenson, K. M., J. S. Wells, F. R. Petersen, B. L. Danielson, G. W. Day. R. L. Barger, and J. L. Hall, 1972, \"Speed of light from direct frequency and wavelength measure ments of the methane-stabilized laser\", Phys. Rev. Letters 29, 1346-1349. [\u00a71.5] Everitt, C. W. F., 1970, \"Maxwell as a student,\" preprint, Stanford University, Stanford, Calif. [Box 14.6] Everitt, C. W. F., W. M. Fairbank, and W. O. Hamilton, 1970, \"General relativity experiments using low-temperature techniques,\" in Carmeli, Fickler, and Witten 1970, pp. 145-159. [\u00a740.7] Exton, A. See Newman et al. 1965. Fackerell, E. D., 1966, Relativistic Stellar Dynamics, unpub lished Ph.D. thesis, University of Sydney. [Box 25.9] Fackerell, E. D., 1968, \"Relativistic stellar dynamics,\" Astro phys. 1. 153, 643-660. [Box 25.8] Fackerell, E. D. See also Brill et al. 1972. Fackerell, E. D., and J. R. Ipser, 1972, \"Weak electromagnetic fields around a rotating black hole,\" Phys. Rev. D5,. 2455-2458. [\u00a74I.II] Fackerell, E. D., J. R. Ipser, and K. S. Thorne, 1969, \"Rela tivistic star clusters,\" Comments Astrophys. & Space Phys. 1, 134-139. [\u00a724.6, Box 25.9] Fadeev, L. D., 1971, \"Symplectic structure and quantization of the Einstein gravitation theory,\" in Actes du Congres International des Mathematiciens, Nice 1-10 Septembre 1970, Gauthier-Villars, Paris, III, 35-40. [Box 21.1] Fairbank, W. M. cited by name in Ex. 37.13. See also Everitt, Fairbank, and Hamilton 1970, Witteborn and Fairbank 1967, and Witteborn, Knight, and Fairbank 1967. Faller, J. E. See Bender et al. 1971. Faraday, M., cited by name in \u00a75.6. Farley, F. J. M., 1969, \"The status of quantum electrodynam ics,\" Nuovo Cimento 1, Numero Speciale, 59-86. [\u00a744.2] Farley, F. J. M. See also Bailey et al. 1968. Farley, F. J. M., J. Bailey, R. C. A. Brown, M. Giesch, H. Jostlein, S. van der Meer, E. Picasso, and M. Tannen baum, 1966, \"The anomolous magnetic moment of the negative muon,\" Nuovo Cimento 45, 281-286. [\u00a738.4] Farnsworth, D., J. Fink, J. Porter, and A. Thomson, eds., 1972, Methods of Local and Global Differential Geometry in Gen eral Relativity, Springer, Berlin. Faulkner, J., and J. R. Gribbin, 1968, \"Stability and radial vibration periods of the Hamada-Salpeter white-dwarf models,\" Nature 218, 734-735. [\u00a724.2, Boxes 24.1, 26.1] Fejer, L., and S. Mikola, eds., 1918, \"Baro Eotvos Lonind FUzet,\" Matem. Fiz. Lapok 27, Nos. 6-7, Oct.-Nov., kindly translated into English for use here by Rev. Stanley Jaki. [Box 38.2] Fekete, E. See Eotvos, Pekar, and Fekete 1922. Fermi, E., 1922, \"Sopra i fenomeni che avvengono in vicinanza di una linea oraria,\" Atti Accad Naz. Lincei CI.", " Sci. Fis. Mat. & Nat. 31, 184-187,306-309. [\u00a76.3] Fermi, E., 1932, \"Quantum theory of radiation,\" Rev. Mod. Phys. 4, 87-132. [\u00a721.12] Fermi, E., 1949, \"On the origin of the cosmic radiation,\" Phys. Rev. 75, 1169-1174. [Box 27.7]. Fermi, E., and A. Turkevich, 1950, unpublished work described in Alpher and Herman 1950, pp. I94ff. [\u00a728.1] Ferrari, A., and R. Ruffini, 1969, \"Theoretical implications of the second time derivative of the period of the pulsar NP 0532,\" Astrophys. J. Letters 158, L71-L75. [Ex. 36.4] Feynman, R. P., 1942. The Principle of Least Action in Quantum Mechanics, doctoral dissertation, Princeton University. [Boxes 17.2, 21.1] Feynman, R. P., 1948, \"Spacetime approach to nonrelativistic quantum mechanics,\" Rev. Mod. Phys. 20, 367-387. [Box 2I.l] Feynman, R. P., 1949, \"Spacetime approach to quantum elec trodynamics,\" Phys. Rev. 76, 769-789. [Box 21.1] Feynman, R. P., 1951, \"The concept of probability in quantum mechanics,\" in Neyman 1951, p. 533. [Box 2I.l] Feynman, R. P., 1955, \"Slow electrons in polar crystals,\" Phys. Rev. 97, 660-665. [Box 21.1] Feynman, R. P., 1963, Lectures on Gravitation, unpublished lecture notes prepared by F. B. Morinigo and W. G. Wagner, Calif. Institute.ofTechnology. [\u00a77.1, Box 7.1, \u00a718.1] Feynman, R. P., 1964, unpublished work cited in Fowler 1964. [Box 24.1] Feynman, R. P. See also Wheeler and Feynman 1945 and Wheeler and Feynman 1949. Feynman, R. P., and A. R. Hibbs, 1965, Quantum Mechanics and Path Integrals, McGraw-Hill, New York. [\u00a721.1, Fig. 13.3, Boxes 17.2, 21.1] Feynman, R. P., R. B. Leighton and M. Sands, 1964, The Feynman Lectures on Physics, Addison-Wesley, Reading, Mass. [\u00a721.3, Box 16.3] Fickler, S. I. See Carmeli, Fickler, and Witten 1970. Field, G., 1973, \"The origin of galaxies,\" in A. and M. Sand age, eds., Scars and Stellar Systems, IX: Galaxies and the Universe, Univ. of Chicago Press, Chicago (in press). [\u00a730.1] i Fierz, M., and W. Pauli, 1939, \"Relativistic wave equations for particles of arbitrary spin in an electromagnetic field,\" Proc. R. Soc. London A 173, 211-232. [\u00a7\u00a718.1, 18.3,39.1, Box 7.1] Fierz, M., and V. F. Weisskopf, 1960, Theoretical Physics in the 20th Century: A Memorial Volume to Wolfgang Pauli, Interscience, New York. Fink, J. See Farnsworth et al 1972. Finkelstein, D., 1958, \"Past-future asymmetry of the gravita tional field of a point particle,\" Phys. Rev. Il 0, 965-967. [Box 31.2] BIBLIOGRAPHY AND INDEX OF NAMES 1233 Fischer, A. E., and J. B. Marsden, 1972, \"The Einstein equa tions of evolution-A geometric approach,\" J. Math. Phys. 13, 546-568. [\u00a7\u00a721.5, 21.6] Fischer, A. E., 1970, \"The theory of superspace,\" in Carmeli, Fickler, and Witten 1970. [\u00a743.1]. Fitch, J. P. See Barton and Fitch 1971 and Barton, Bourne, and Fitch 1970. Flamm, L., 1916, \"Beitrage zur Einsteinschen Gravita tionstheorie,\" Physik. Z. 17, 448-454. [\u00a723.8] Flanders, H., 1953, \"Development of an extended exterior differential calculus,\" Trans. Am. Math. Soc. 75", ", 311-326. [\u00a714.5, Ex. 14.9, 14.10] Flanders, H., 1963, Differential Forms With Applications to the Physical Sciences, Academic Press, New York. [\u00a74.l, Boxes 4.4, 8.1] Fletcher, J. G., 1966, \"GRAD Assistant-a program for sym bolic algebraic manipulation and differentiation.\" Com mlin. Assoc. Comput. Machinery 9, 552-553. [Box 14.3] Fletcher, J. G., R. Clemens, R. Matzner, K. S. Thorne, and B. A. Zimmerman, 1967, \"Computer programs for calcu lating general-relativistic curvature tensors,\" Astrophys. J. Lett. 148, L91-L92. [Box 14.3] Fock, v. A., 1959, The Theory of Space, Time, and Gravitation, translated by N. Kemmer. [Box Pergamon, New York, 39.3. Ex. 39.15, Epigraph to Chap. 21, endpapers] Fock, V. A., chief editor, 1968. Abstracts, Fifth International Conference on Gravitation and the Theo~y of Relativity, Publishing House of Tbilisi University, Tbilisi, USSR. Fock. V. A. See also Dirac, Fock. and Podolsky 1932. Fokker, A. D., 1965, Time and Space, Weight and Inertia, Pergamon Press, London. [endpapers] Fokker. A. D. See Einstein and Fokker 1914. Fomalonl, E. B. See Muhleman. Ekers, and Fomalont 1970. Ford. K. W. See Oeser and Ford 1967. Ford, K. W., and J. A. Wheeler, 1959a, \"Application of semi classical scattering analysis,\" Ann. Phys. (U.S.A.) 7, 287 322. [\u00a725.6] Ford, K. W., and J. A. Wheeler, 1959b, \"Semiclassical descrip tion of scattering,\" Ann. Phys. (U.S.A.) 7, 259-286. [\u00a725.6] Forman, F. See Gittelman et al. 1968. Forward, R. L., 1972, \"Geodesy with orbiting gravity grad i ometers,\" in The Use of Artificial Satellites for Geodesy, Geophysical Monograph Series, vol. 15, American Geo physical Union, Washington, 239-243. [Box 16.5] Foures-Bruhat, Y. See Bruhat, Y, and Choquet-Bruhat, Y. Fowler, R. H., 1926. \"Dense matter,\" Mon. Not. R. Astron. Soc. 87,114-122. [Box 24.1] Fowler. W. A., 1964, \"Massive stars, relativistic polytropes, and gravitational radiation,\" Rev. Mod. Phys. 36, 545-555 and 1104. [Boxes 24.1, 26.2] Fowler, W. A.. 1966. \"The stability of supermassive stars.\" Fowler, W. A., and F. Hoyle, 1964, \"Neutrino processes and pair formation in massive stars and supernovae,\" Astro phys. J. Supp. 9, 201-319. [\u00a724.4] Fram. D. M. See Boughn, Fram. and Partridge 1971. Friedmann, A., 1922, \"Uber die Kriimmung des Raumes.\" Z. Phys. 10,377-386. [\u00a7\u00a714.1, 27.9, 29.2, Boxes 14.5,27.6] Friedmann, A., cited by name in \u00a7\u00a730.3, 42.1, Boxes 27.6, 27.7. Frisch, D. H., and J. H. Smith, 1963, \"Measurement of the relativistic time dilation using mu-mesons,\" Am. J. Phys. 31, 342-355. [\u00a738.4] Froman, N., and P. O. Froman, JWKB approximation: contri butions to the theory, North-Holland, Amsterdam, 1965. [\u00a722.5] Froman, P. O. See Froman and Froman 1965. Fronsdal, C., 1959, \"Completion and embedding of the Schwarzschild solution,\" Phys. Rev. 116, 778-781. [\u00a723.8, Box 3l.l] Fujiwara, I., 1962, \"The correspondence principle, I,\" in Pro ceedings of the Physics Seminar in Trondheim, no. 3, Nor", " dita, Copenhagen, Denmark. [Box 2l.l] FUller, R. W., and J. A. Wheeler, 1962, \"Causality and multi ply-connected space-time,\" Phys. Rev. 128, 919-929. [\u00a731.6, Fig. 31.5] Galilei, G., 1592, De Motu, unpublished work written at the University of Pisa, 1589-1592; translation in Drabkin and Drake 1961. p. 48. [Box 1.2] Galilei, G., 1610, Sidereus Nuncius, Translated by E. S. Carlos, The Sidereal messenger, Rivingtons, London, 1880. [Boxes 1.9, 27.7] Galilei. G., 1623, II Saggiatore, Accademia dei Lincei, Rome. [epigraph, Chap. 13] Galilei, G., 1632, Dialogo dei due massimi sistemi del mund,O, Landini, Florence. Translation by S. Drake, Galileo Gali lei-Dialogue Concerning the Two Chief World Systems Ptolemaic and Copernican, Univ. of California Press, Berkeley and Los Angeles, 1953. [Box 1.9] Galilei, G., 1638, Discorsi e dimostriazioni matematiche intorno a due nuove scienze, Elzevir, Leiden. Translation by H. Crew and A. de Salvio, Dialogues Concerning TIm New Sciences, Macmillan, New York, 1914; reprinted by Dover, New York, 1954. [Boxes 1.2, 1.9,27.7] Galilei, G., cited by name in \u00a712.2, quoted in epigraph of Chap. 38. Gamow, G., 1948, \"The evolution of the universe,\" Nature 162, 680-682. [\u00a728.1] Gamow, G., 1967, \"Electricity. gravity, and cosmology,\" Phys. Rev. Lett. 19, 759-761. [\u00a738.6] Gamow, Go, 1970, My World Line, Viking Press, New York. [Box 27.7] Gamow, George, March 4, 1904-August 20. 1968, Memorial volume. See Reines 1972. Astrophys. J. 144, 180-200. [\u00a724.4, Box 26.2] Gaposchkin. S\" 1958. \"The eclipsing binaries.\" Handbuch der Fowler. W. A., 1967, Nuclear Astrophysics. American Philo Phrsik 50. 225-242. [Box 36.3] sophical Society, Philadelphia, Penn. [\u00a728.1] Fowler. W. A.. 1972. \"New observations and old nucleocosmo chronologies.\" in Reines 1972. [\u00a729.7] Fowler. W. A. Cited by name in Box 27.7. See also Hoyle and Fowler 1963a and 1963b. Hoyle et al. 1964. and Wagoner. Fowler. and Hoyle 1967. Garding. L.. 1945. \"Relativistic wave equations for zero rest mass.\" Proc. Cambridge Phil. Soc. 41, 49-56. [Box 32.2] Garnett. W. See Campbell and Garnett 1882. Gauduchon. P. See Berger. Gauduchon, and Mazet 1971. Gasiorowizc. S.. 1966. Elementarl' PartiCle PhYSiCS, Wiley. l'ew York. [Box 7.1] 1234 GRAVITATION Gauss, K. F., 1817, letter to Olbers in his Werke, VIII, 177. [Epigraph to Chap. 8) of Static, Axisymmetric Vacuum Spacetimes. doctoral dis sertatiC'n, Princeton University. [\u00a734.l) Gauss. K. Fo, 1830, letter to Bessel in his Werke, VIII, 201. Goebel, C. J., 1972, \"Comments on the 'vibrations' of a black [Epigraph to Chap. 8) Gauss, K. F., 1827, \"Disquisitiones generales circa superficies IV, 217-258; curvas,\" in Karl Friedrich Gauss Werke, English translation by J. C. Morehead and A. M. Hille beitel, General Investigations of Curved Surfaces of 1827 and 1825, reprinted by Raven Press. New York. 1965. [\u00a7\u00a75.9. 22.2, Boxes 5.3, 5.4, 14.1) Gauss, K. F., cited by name in \u00a741.1. Geheniau, J., quoted", " at end of Chap. 44. Geren, P. See Ehlers, Geren, and Sachs 1968. Gerlach, U., 1968, \"The equation of state at supranuclear densities and the existence of a third family of superdense stars,\" Phys. Rev. 172, 1325-1330. [\u00a724.2) Gerlach, U., 1969, \"Derivation of the ten Einstein field equa tions from the semiclassical approximation to quantum geometrodynamics,\" Phys. Rev. 177, 1929-1941. [\u00a743.3, Box 17.2 part 4) Geroch, R. P., 1967, Singularities in the Spacetime of General Relativity: Their Definition, Existence, and Local Charac terization, doctoral dissertation, Princeton Universi ty. [\u00a7\u00a732.7, 44.1, Epigraph to Chap. 34) Geroch, R. P., 1968, \"What is a singularity in general relativ ity?\" Ann. Phys. (U.S.A.) 48, 526-540. [\u00a734.6) Geroch, R. P., 1971, \"Spacetime structure from a global view- point,\" in Sachs 1971. [\u00a734.1) Geroch, R. P., cited by name in \u00a7\u00a732.7, 42.1, Box 34.1. Giacconi, R., cited by name in Box 27.7. Gibbons, J. J., cited by name in Box 34.3. Giesch, M. See Farley et al. 1966. Gilbert, R. Po, and R. Newton, 1970, Analytic Methods in Mathematical Physics, Gordon and Breach, New York. [Box 8.5). Gillispie, C. C. See Bernoulli 1733. Ginzel, F. K.. 1882, \"Astronomische Untersuchungen tiber Finsternisse, I: Uber die zwischen 26 und 103 nach Chr. stattgefundenen Sonnenfinsternisse im Allgemeinen and die Finsterniss des Plutarch ins besonders,\" Sitzber. Kais. Akad. Wiss. Wien Mach.-Nat. KI. 85I1, 663-747. [Fig. 1.8) Ginzel, F. K., 1883, \"Astronomische Untersuchungen tiber Finsternisse, II: Grundlagen aus historischen Sonnenfin sternissen zur Ableitung empirischer Correctionen der Mondbahn,\" Sitzber. Kais. Akad. Wiss. Wien Math.-Nat. KI. 88I1, 629-755. [Fig. l.8) Ginzel, F. K., 1884, \"Astronomische Untersuchungen tiber Finsternisse, III: Ermittlung empirischer Correction der Bahn des Mondes,\" Sitzber. Kais. Akad. Wiss. Wien Math.-Nat. KI. 89I1. [Fig. 1.8) Ginzel, F. K., 1899, Spezieller Kanon der Sonnen- und Mond finsternisse jUr das Landergebiet der Klassischen Altertum wissenschaften und den Zeitraum von 900 vor Chr. bis 600 nach Chr., Mayer, Berlin. [Fig. 1.8) Godfrey, B. B., I970a, \"Mach's principle, the Kerr metric, and black-hole physics,\" Phys. Rev. D 1,2721-2725. [Box 33.3) Godfrey, B. B., I970b, A Survey of the Mathematical Properties hole,\" Astrophys. J. 172, L95-L96. [Box 33.3) Giidel. K., 1931, \"Uber formal unentscheidbare Satze der Principia Mathematica und I,\" Monatsh. Math. Phys. 38, 173-198, English translation by B. Meltzer, On Formally Undecidable Propositions, with an introduction by R. O. Braithwaite. Basic Books, New York, 1962: outlined and discussed in Nagel and Newman 1958. [Box 44.5) verwandte Systeme Gold, T., 1968. \"Rotating neutron stars as the origin of the pulsating radio sour;es,\" Nacure 218. 731-732. [Box 24.1) Gold, T., 1972, oral remark to J. A. Wheeler. [Epigraph to Chap. 21) Gold, T., cited by name in Box 27.7. See also Bondi and Gold 1948. Goldberg", ", S. I. See Bishop and Goldberg 1968. Goldenberg, H. M. See Dicke and Goldenberg 1967. Goldreich, P. and W. H. Julian, 1968, \"Pulsar elecirodynam- ics,\" Astrophys. J. 157, 869-880. [\u00a724.3, Box 33.3) Goldstein, H., 1959, Classical Mechanics, Addison-Wesley, Reading, Mass. [\u00a7\u00a72.I, 9.7,33.5, Ex. 33.7) Goldstein, J. See Chretien, Oeser, and Goldstein 1969. Gott, J. R., III, and J. E. Gunn, 1971, \"The Coma Cluster as an X-ray source: Some cosmological implications,\" Astro phys. J. Lett. 169, 13-15. [\u00a729.6) Gowdy, R. H., 1971, \"Gravitational waves in closed universes,\" Phys. Rev. Lett. 27, 826-829, and erratum p. 1102. [\u00a730.6) Gowdy, R. H., 1973, \"Vacuum spacetimes with two-parameter spacelike isometry groups and compact invariant hyper surfaces: Topologies and Boundary conditions,\" Univer sity of Maryland technical report 73-089 (unpublished). [\u00a7\u00a721.1O,30.6) Gratton, L., ed., 1966, High.Energy Astrophysics, Academic Press, New York. Graves, J. c., and D. R. Brill, 1960, \"Oscillatory character of Reissner-Nordstrom metric for an ideal charged worm hole,\" Phys. Rev. 120, 1507-1513. [Fig. 34.4, Ex. 31.8) Greenstein, G. S., I968a, \"Brans-Dicke Cosmology,\" Astro phys. Letters, 1, 139-141. [\u00a728.4) Greenstein, G. S., 1968b, \"Brans-Dicke Cosmology, II,\" Ap. Space Sci., 2, 157-165. [\u00a728.4) Gribbin, J. R. See Faulkner and Gribbin 1968. GRn, International Conferences on General Relativity and Gravitation. See Berne 1955 (GRO), Chapel Hill 1957 (GRI), Royaumont 1959 (GR2), Jablonna 1962 (GR3), London 1965 (GR4), Tbilisi 1968 (GRS), Copcrhagen 1971 (GR6). Grommer, J. See Einstein and Grommer 1927. Grossmann, M. See Einstein and Grossman 1913 and 1914. Gunn, J. E., 1967, \"On the propagation of light in inhomogeneous cosmologies: I, Mean effects,\" Astrophys. J. 150, 737-753. [\u00a729.5) Gunn, J. E., 1971, \"On the distances of the quasistellar ob jects,\" Astrophys. J. Letters, 164, Ll33-LI18. [Box 28.1) Gunn, J. E. Cited by name in Box 27.7. See also Gott and Gunn 1971, Ostriker and Gunn 1969. BIBLIOGRAPHY AND INDEX OF NAMES 1235 Gupta, S. N., 1954, \"Gravitation and electromagnetism,\" Phys. Rev. 96, 1683-1685. [\u00a7\u00a77.I, 18.1] Gupta, S. N., 1957, \"Einstein's and other theories of gravita tion,\" Rev. Mod. Phys. 29, 337-350. [\u00a7\u00a77.I, 18.1] Gupta, S. N., 1962, \"Quantum theory of gravitation,\" in Recent Developments in General Relativity, Pergamon, New York, pp. 251-258. [\u00a7\u00a77.1, 18.1] Gurovich, V. Ts., 1965, \"The gravitational field of rotating superdense configurations,\" Astron. Zhur. 42, 974-976. English translation in Sov. Astron.-A.J. 9, 75()...751. [Ex. 26.1] Gursky, H., cited by name in Box 27.7. Guseynov, O. H. See Zel'dovich and Guseynov 1965. Guthrie, W. K. C., trans., 1939, Aristotle's On the Heavens, Harvard Univ. Press, Cambridge, Mass. [Box 27.7] Haeberli, W. See Barschall and Haeberli 1971. Hagedorn, R., 1964, Relativistic Kinematics, W. A. Benjamin, r-tew York. [\u00a7", "2.IO] Hagihara, Y., 1931, \"Theory of the relativistic trajectories in a gravitational field of Schwarzschild,\" Japan. J. Astron. Geophys., 8, 67-175. [\u00a725.5] Hall, D. B. See Rossi and Hall 1941. Hall, J. L., 1972. See Evenson et al. 1972. Halley, E., 1693, \"Emendationes ac notal in vetustas Ablatenii observationes astronomicas cum restitutione tabularum lunisolarum,\" Roy. Soc., London, Phil. Trans. 17, 913. [\u00a71.5] Halley, E. 1720, \"Of the infinity of the sphere of fix'd stars\" and \"Of the number, order. and light of the fix'd stars,\" Roy. Soc. London, Phil. Trans. 31, 22-24 and 24-26. [Box 27.7] Hamilton, W.O., cited by name in Ex. 37.13. See also Everitt, Fairbank, and Hamilton 1970. Hamilton, W. R., cited by-name in \u00a741.1 Boxes 17.2, 25.3. Hamoui, A. See Papapetrou and Hamoui 1967 and 1968. Hanni, R. S., and R. Ruffini, 1973, paper in preparation. [Box 33.2] Hardy, E. See Sandage, Tamman, and Hardy 1972. Harrison, B. K., 1959, \"Exact three-variable solutions of the relativity,\" Phys. Rev. 116, field equations of general 1285-1296. [\u00a714.1] Harrison, B. K., K. S. Thorne, M. Wakano, and J. A. Wheeler, 1965, Gravitation Theory and Gravitational Collapse, Univ. of Chicago Press, Chicago. [\u00a726.1, Fig. 24.2] Harrison, B. K., and J. A. Wheeler, 1958, \"The Harrison Wheeler equation of state,\" first presented in Onzii?me Conseil de Physique Solvay: La Structure et!'evolution de l'univers, Editions Stoops, Brussels, 1959; presented in greater detail in Harrison et al. 1965. [\u00a724.2, Fig. 24.2] Harrison. G. H., 1970, \"A Compact Method for Symbolic Computation of the Riemann Tensor:' M. S. thesis. Tech. Report. Dept. of Physics and Astronomy. Univ. of Mary land. College Park. [Box 14.3] Hartle. J. B., 1967, \"Slowly rotating relativistic stars, I: Equa tions of structure,\" Astrophys. 1. 150. 1005-1029. [Ex. 26.1] Hartle. J. B., 1970, \"Slowly rotating relativistic stars, IV: Rota tional energy and moment of inertia for stars in differen tial rotation,\" Astrophys. 1. 161, 111-118. [\u00a720.5. Ex. 33.13] Hartle, J. B., 1971, \"Long-range neutrino forces exerted by Kerr black holes,\" Phys. Rev. D 3, 2938-2940. [\u00a74I.II, Boxes 25.2, 32.2, 33.1] Hartle, J. B., 1972, \"Can a Schwarzschild black hole exert long-range neutrino forces?\" in Klauder 1972. [Boxes 25.2, 32.2, 33.1] Hartle, J. B. See also Brill and Hartle 1964, Hawking and Hartle 1972. Hartle, J. B., and D. H. Sharp 1965, \"Variational principle for the hydrostatic equilibrium of a relativistic, rotating fluid,\" Phys. Rev. Lett. 15, 909-911. [Ex. 26.1] Hartle, J. B., and K. S. Thorne, 1968, \"Slowly rotating relativ istic stars, II: Models for neutron stars and superrnassive stars,\" Astrophys. J. 153, 807-834. [Box 36.1, Fig. 24.2] Hartle, J. B.. and K. S. Thorne, 1974. \"Slowly rotating rela tivistic stars, VII: Emission of gravitational waves due to quasiradial pulsation.\" Astrophys. 1., in preparation. [Box 14.3] Harvey, I. A. See Dabbs et al. 1965. Hauser, W., 1970, \"On the fundamental equations of electro magnetism,\" Am. j Phys. 38, 80-85. [\u00a74.I] Havens, W. W.,", " Ir. See Durbin, Loar, and Havens 1952. Hawking, S. W., 1966a, \"Singularities and the geometry of space-time\" Adams Prize Essay, Cambridge University, Cambridge, England. [Box 34.3] Hawking, S. W., 1966b, \"The occurrence of singularities in cosmology,\" Proc. R. Soc. London A 294, 511-521. [Box 34.3] Hawking, S. W.. 1969, \"On the rotation of the Universe,\" Observatory 89, 38-39. [Box 34.3] Hawking, S. W.. 1971a, \"Gravitationally collapsed objects of very low mass,\" Mon. Not. R. Astron. Soc. 152, 75-78. [!;lox 33.3] Hawking, S. W., 1971 b, \"Gravitational radiation from colliding black holes,\" Phys. Rev. Lett. 26, 1344-1346. [\u00a734.5. Boxes 33.1, 33.4, Ex. 34.4] Hawking, S. W., I972a, \"Black holes in general relativity.\" Commun. Math. Phys.25. 152-166. [\u00a734.5. Boxes 33.1. 33.3. 33.4, 34.3, Ex. 34.4] Hawking. S. W.. 1972b. \"Gravitational Radiation: The Theo retical Aspect.\" Concemp. Phys. 13. 273-282. [Box 34.3] Hawking, S. W., 1973, \"The Event Horizon,\" in DeWitt and DeWitt 1973. [\u00a7\u00a734.1, 34.5] Hawking, S. W.. cited by name in \u00a7\u00a732.7, 42.1, Box 27.7. See also Bardeen, Carter, and Hawking 1973 and Collins and Hawking 1973. Hawking, S. W.. and G. F. R. Ellis, 1968, \"The cosmic black body radiation and the existence of singularities in our universe,\" Astrophys. J. 152. 25-36. [\u00a728.3. Box 34.3] Hawking. S. W.. and G. F. R. Ellis. 1973. The Large Scale Structllre of Space-time. Cambridge Univ. Press. Cam bridge. Eng. [\u00a7\u00a734.1. 34.6. Box 33.1. endpapers] Hawking, S. W.. and J. B. Hartle. 1972. \"Energy and angular momentum flow into a black hole,\" COIllIllZtn. Math. Phys. 27, 283-290. [Box 33.3] Hawking. S. W.. and R. Penrose. 1969, \"The singularities of gravitational collapse and cosmology.. Proc. R. Soc. LOIl dOll A 314. 529-548. [\u00a7\u00a734.6. 44.1] 1236 GRAVITATION Hayakawa, S., I965a. \"Cosmological interpretation of the weak forces.\" Prog. Theor. Phys. 33, 538-539. [\u00a744.6) Hayakawa. S.. 1965b, \"Atomism and cosmology.\" Prog. Theor. Phys. Supp.. special number, Yukawa thirtieth anniversary issue, 532-541. [\u00a744.6) Hazard, C., cited by-name in Box 27.7. Hearn, A. e., 1970. \"A program lor symbolic algebraic com putation,\" in Proceedings of SHARE XXXIV. 4.470-4.481. [Box 14.3) Heckmann, O. See Schiicking and Heckmann 1958. Heisenberg, W., and H. Euler, 1936, \"Folgerungen aus der Diracschen Theorie des Positrons,\" Z. Phys. 98, 714-732. [Ex. 40.8) Helmholtz, H. L. F., cited by name in Box 38.2. Henderson, T., cited by name in Box 27.7. Hepp, K., 1969, Theorie de la renormalisation, Springer, Berlin. [Box 17.2 part 6) Herman, R. e. See Alpher and Herman 1948a.b and 1950. Hermann, R., 1966, Lie Groupsfor Physicists, W. A. Benjamin, New York. [\u00a78.2) Hernandez, W. C., Jr., 1967, \"Material sources for the Kerr metric,\" Phys. Rev. 159, 1070-1072. [\u00a733.3) Heron (Hero of Alexandria), 1899, Heronis Alexandrini Opera Omnia, Teubner, Leipzig; also in G. Sarton, A History of Science, Harvard Univ. Press", ", Cambridge, Mass., 1952, I, 73. [Box 13.1) Heron, G., trans., 1954, Nicolas Cusanus's Of Learned Ignorance, Yale Univ. Press, New Haven, Conn. [Box 27.7) Hertz, G. L., cited by name in Box 17.3. Hewish, A., cited by name in Box 27.7. Hewish, A., S. J. Bell, J. D. H. Pilkington, P. F. Scott, and R. A. Collins, 1968, \"Observation of a rapidly pulsating radio source,\" Nature 217, 709-713. [Box 24.1) Hey, J. S., cited by name in Box 27.7. Hibbs, A. R., 1951, \"Quantum mechanics,\" in M. Kac, Proba bility and Related Topics in Physical Science, Interscience, New York. pp. 205-221. [Box 21.1). Hibbs, A. R. See also Feynman and Hibbs 1965. Hicks, N. J., 1965, Notes on Differential Geometry, Van Nos Honl. H., 1962, \"Allgemeine Relativitlitstheorie und Machsches Prinzip.\" in P1(I'sikertagung /'Vien, Physik Verlag. Mos bach/Baden, Germany. [\u00a7\u00a72I.l2, 43.1) Hoffmann, B., ed., 1966, Perspectives in Geometry and Relativ Indiana Univ: ity: Essays in Honor of Vaclav Hlavat}. Press. Bloomington. Hoffman, B. See also Einstein, Infeld. and Hoffman 1938. Hoffman, W. F. See Chiu and Hoffman 1964. Hohn F. E., 1966, Applied Boolean Algebra: An Elementary Introduction, 2d ed., Macmillan, New York. [\u00a744.5) Holden, D. 1.. 1966, \"An investigation of the clustering of radio sources,\" Mon. Not. R. Astron. Soc., 133. 225-228. [\u00a727.I) Holdridge, D. B. See Ohandley et al., 1969. Holmes, A., cited ~I' nallle in Box 27.7. Holton, G., 1965, personal communication, London. [Preface) Holton. G., 1971, Einstein. The Life and Times, ~I' R. W. Clark, book review, New York Times Book Reviey,; September 5, 1-20. [Box 1.11) Holton, G. See also Kollros 1956. Horstmann, H: See Dabbs et al. 1965 Hoyle, F., 1948, \"A new model for the expanding universe,\" Mon. Not. R. Astron. Soc. 108, 372-382. [\u00a728.4) Hoyle, F., cited by name in Box 27.7. See also Fowler and Hoyle 1964 and Wagoner, Fowler, and Hoyle 1967. Hoyle, F., and W. A. Fowler, 1963a, \"On the nature of strong radio sources,\" Mon. Not. R. Astron. Soc. 125, 169-176. [\u00a724.5, Box 24.1) Hoyle, F., and W. A. Fowler, 1963b, \"Nature of strong radio sources,\" Nature 197, 533-535. [\u00a724.5, Box 24.1) Hoyle, F., W. A. Fowler, G. R. Burbidge, and E. M. Burbidge, 1964, \"On relativistic astrophysics,\" Astrophys. 1. 139, 909-928. [Box 33.3) Hsiang, w.-e., and B. J.- Sinderson, 1965, \"Twist-spinning spheres in spheres,\" I/linois 1. Math. 9, 651-659. [Box 44.3) Hubble, E. P., 1920, \"Photographic investigations of faint nebulae\" (Ph.D. thesis), Pub. Yerkes Obs., 4, 69-85. [Box 29.5) Hubble, E. P., 1925, \"NGC 6822, a remote stellar system,\" Astrophys. 1. 62, 409-433. [Box 29.5) trand, Princeton, N.J. [Box 8.1, endpapers) Hubble, E. P., 1926a, \"A spiral nebula as a stellar system: Hilbert, D., 1915, \"Die Grundlagen der Physik,\" Konigl. Gesell. d. Wiss. Gottingen, Nachr., Math.-Phys. KI., 395-407; see also Math. Annalen 92 (", "1924), 1-32. [\u00a7\u00a7I7.7, 21.2, Box 17.2) Hilbert, D., 1917, \"Die Grundlagen der Physik (Zweite Mit d. Wiss. Gottingen, Nachr., teilung),\" KonigL Gesell. Math.-Phys. KL, 53-76. [\u00a7I7.7) Hilbert, D., cited by name in \u00a721.1. Hill, H. A. See Turner and Hill 1964. Hill, J. M., 1971, \"A Measurement of the gravitatibnal deflec tion of radio waves by the sun,\" Mon. Not. R. Astron. Soc., 154, 7P-II P. [Box 40.1). Messier 33,\" Astrophys. J. 63, 236-274. [Box 29.5) Hubble, E. P., 1926b, \"Extra-galactic nebulae,\" Astrophys. J. 64, 321-369. [Box 29.5) Hubble, E. P., 1929, \"A relation between distance and radial velocity among extragalactic nebulae,\" Proc. Nat. Acad. Sci. U.S. 15, 169-173. [\u00a7\u00a727.9, 29.2, Boxes 27.1, 29.5) Hubble, E. P., I934a, Red-shifts in the spectra of nebulae (Halley Lecture), Clarendon Press, Oxford, En!; [Box 29.5) Hubble, E. P., 1934b, \"The distribution of extra-galactic nebu lae,\" Astrophys. 1. 79, 8-76. [\u00a727.1, Box 29.5) Hubble, E. P., 1936, \"Effects of redshifts on the distribution Hill, T., 1967, unpUblished work referred to in DeWitt I967c. of nebulae,\" Astrophys. J. 84, 517-554. [\u00a727.I) [Box 17.2 part 6) Hiltebeitel, A. M. See Gauss 1827 Hipparchus, cited by name in Box 16.1. Hlavaty, Vaclav, Festschrift on the occasion of his 70th birth day, Jan. 27, 1964. See Hoffmann 1966. Hubble, E. P., cited by name in \u00a717.3, Boxes 27.7,29.5. See also Mayall 1970. Hubble, E. P., and M. L. Humason, 1931, \"The velocity distance relation among extra-galactic nebulae,\" Astro phys. J. 74, 43-80. [Box 29.5) BIBLIOGRAPHY AND INDEX OF NAMES 1237 Hughes, R. G., and M. S. Longair, 1967, \"Evidence of the isotropy of faint radio sources,\" Mon. Not. R. Astron. Soc. 135, 131-137. [\u00a727.1) Hughes, V. W., H. G. Robinson, and V. B. Beltran-Lopez, 1960, \"Upper limit for the anisotropy of inertial mass from nuclear resonance experiments,\" Phys. Rev. Lett. 4, 342 344. [\u00a738.7) Hughston, L., 1969, \"MuItifiuid co~mologies,\" Astrophys. J. 158, 987-989. [\u00a727.9) Humason, M. L. See Hubble and Humason 1931. Huneke, J. C See Wasserburg, Schramm, and Huneke 1969. Hunter, C, 1967, \"Fragmentation,\" in Ehlers 1%7, pp. 169- 194. [Boxes 32.2, 32.3) Huygens, C, 1698, Cosmotheoros, The Hague. English transla tion, The Celestial Worlds Discovered or, Conjectures Con cerning the Inhabitants, Plants and Productions of the Worlds in the Planets, London, also 1698. [Box 27.7) Imshetnik, V. S. See Ivanova, Imshennik, and Nadezhin 1969. Infeld, Leopold, 1962, volume dedicated to him in connection with his 60th birthday, Feb. 8, 1958: Recent Developments in General Relativity, Pergamon Press, Oxford, and PWN-Polish Scientific Publishers, Warsaw. Infeld, L., ed., 1964, Conference Internationale sur les theories la gravitarion, Gaqthier-Villars, Paris, relativistes de France. Infeld, L. See also Einstein, Infeld, and Hoffman 1938. Infeld, L, and", " J. Plebanski, 1960, Motion and Relativity, Per gamon, New York. [\u00a720.6, endpapers) Infeld, L., and A. Schild, 1949, \"On the motion of test par ticles in general relativity,\" Rev. Mod. Phys. 21, 408-413. [\u00a720.6) Ingalls, R. P. See Shapiro, Ash et al. 1971, Shapiro, Smith et al. 1971, Shapiro et aI, 1972. Ipser, J. R., 1969, \"Relativistic. spherically symmetric star clus ters, III: Stability of compact isotropic models,\" Astrophys. J. 158, 17-43. [Boxes 24.1, 25.9) Ipser. J. R., 1970, \"Gravitational radiation from slowly rotat ing, fully relativistic stars,\" Asrrophys. J. 166, 175-196. [Ex. 36.4) Ipser, J. R., 1971, \"Electromagnetic test fields around a Kerr metric black hole,\" Phys. Rev. Lett. 27, 529-531. [\u00a741.1I, Ex. 33.14, Box 33.3) Ipser, J. R. See also Brill et al. 1972, Fackerell and Ipser 1972, and Fackerell, Ipser, and Thome 1969. Ipser, J. R., and K. S. Thome, 1968, \"Relativistic, spherically symmetric star clusters, I: Stability theory for radial per turbations,\" Astrophys. J. 154, 251-270. [Box 24.1) Isaacson, R. A., 1968a, \"Gravitational radiation in the limit of high frequency, I: TheJi~at16nand geometri c\"l epl;\"s.\"- P~. 166. 1263-1271. [\u00a735.13) Isaacson, R. A., 1968b, \"Gravitational radiation in the limit of high frequency. II: Nonlinear terms and the effective stress tensor.\" Phys. Rev. 166. 1172-1180. [\u00a735.13. Ex. 35.14) Isenberg, J. A.. cired by name in Figure 25.7. Israel, W.. 1960, \"Relativistic theory of shock waves:' Proc. Roj'. Soc. A 259, 129-143. [\u00a722.1, Box 21.2) Israel, W., 1966, \"Singular hypersurfaces and thin shells in general relativity,\" l\\'uovo Cimento 44 B, 1-14, and correc tion in 48 B, 463. [\u00a7\u00a721.5, 21.13) Israel, W., I967a, \"Event horizons in static vacuum space times,\" Phys. Rev. 164, 1776-1779. [Box 33.1] Israel, W., 1967b, \"Gravitational collapse and causality,\" Phys. Rev. 153, 1388-1393. [Ex. 21.27) Israel, W., 1968, \"Event horizons in static electrovac space times,\" Commun. Math. Phys. 8, 245-260. [Box 33.1) Israel, W., 1970, Differential Forms in General RelatiVity, Com munications of the Dublin Institute for Advanced Studies. series A, no. 17. [\u00a74.1) Israel, W., 1971, personal communication. [Box 33.4) Israel, W., 1972, \"The relativistic Boltzman Equation,\" in O'Raifeartaigh 1972, pp. 201-241. [\u00a722.6) Israel, W., cited by name in Box 34.3. See also de la Cruz and Israel 1967. Ithaca 1965. See Cornell University 1965 or Ehlers 1967. Ivanova, L. N., V. S. Imshennik, and D. K. Nadezhin, 1969, \"Investigations into the dynamics of supernova explo sions,\" Sci. In! Astron. Council USSR Acad. Sci. 13,3-93. [\u00a732.7) Jablonna 1962, Relativistic Theories ofGravitation, proceedings of a conference held in Warsaw and Jablonna, July 1962, edited by L. Infeld, Pergamon Press, Oxford. 1964. Jackson, J. D., 1962, Classical Electrodynamics, Wiley, New York. [\u00a72.1) Jacobi, K. G. J., cited by name in Box 17.2. Jaffe, 1.. 1969, \"Collapsing objects and the backward emission of light,\" Ann. Phys. U.S.A. 55. 374-390. [\u00a7", "32.3) Janis, A. I. See Couch et al. 1968. Jauch. 1., 1968, Foundations of Quantum Mechanics. Addison Wesley, Reading, Mass. [Box 44.5) Jeans, 1., cited by name in Box 27.7. Jones, B. T. J., and P. J. E. Peebles, 1972, \"Chaos in Cos mOlogy,\" Comm. Astrophys. Space Sci. 4, 121-128. [\u00a730.1) Jostlein, H. See Bailey et aL 1968 and Farley et al. 1966. Johnson, F. R., 1937, Astronomical Thought in Renaissance England, John Hopkins Press, Baltimore, [Box 27.7) Jordan, P., 1955, Schwerkraft und Weltall, Vieweg und Sohn. Braunschweig. [\u00a7\u00a738.6. 44.6) Jordan. P., 1959, \"Zum gegenwartigen Stand der Diracschen kosmologischen Hypothesen,\" Z. Phys. 157, 112-121. [\u00a7\u00a738.2, 38.6, 44.6, Box 39.1) Jordan, P., cited by name in \u00a7\u00a740.1, 40.6, 40.9. Jordan, P., J. Ehlers, and R. Sachs, 1961, \"Exact solutions of the field equations of general relativity, II: Contributions to the theory of pure gravitational radiation,\" (in German) Akad. Wiss. Lit. Main:: Abh. Math.-Nat. KI. 1, 3-60. (\u00a722.5) Julian, W. H. See Goldreich and Julian 1968. Jurgens. R. F. See Shapiro, Ash et al. 1971. Kac. M.. and S. Ulam. 1968, Mathemarics and Logic: Retrospecr and Prospects, Praeger. New York: abridged paperback edition by Pelican Books. Harmondsworth, Middlesex. 1971. [Box 44.5) Kant, I.. 1754. \"Whether the earth has undergone an alteration of its axial rotation,\" W6chemliche Frag- und An::eigungs Nachrichren. Kanter. Konigsberg. mid-June issue [\u00a7I.5) Kant, I.. 1755, Allgemeine Narurgeschichre und Theories des Himmels, oder Versuch von der Verfassung und dem i\\1ech anischen ljrsprunge des gan::en Welt gebiiudes nach New IOnischen Grundsiir:en- Abgehandelt. Petersen, Konigsberg and Leipzig. Translation by W. Hastie, with introduction 1238 GRAVITATION by 1\\1. K. :-.tunitz, L'niLwsal Satural History and Theory ofrhe Heavens. Univ. of Michigan Press. Ann Arbor. 1969.. [Box 27.7) Kaplan, S. A.. I949a, \"S,,~rkhplotnye' zvezdy ('Superdense' (Sci. Nores Cniv. LII\"oll\") 15. stars).\" Sal/kovy Zapiski 109-115. [Box 24.1) Kaplan. S. A.. 1949b. \"0 krugovykh orbitakh v teorii tyago teniya Ein_hteina (On circular orbits in Einslein's theory of gravitation),\" Zh. Ek.lp. & TeoI'. Fi:. 19, 951-952. [Box 24.1) Kasner. E.. 1921a. \"Geometrical theorems on Einstein's cos mological equations,\" Am. J. Marh. 43, 217-221. [\u00a730.2) Kasner, E.. 1921b. \"Finite representation of the solar gravita tional field in a space of six dimensions:' Am. J. Marh 43, 130-133. [\u00a723.8) Kasner. E., 1921c, \"The impossibility of Eimlein fields im mersed in flat space of five dimensions.\" Am. J. Marh 43, 126-129. [\u00a723.8) Kasner, E.\u2022 cited bv name in Boxes 27.7; 30.1. Kaula, W. M. See' Bender et al. 1971. Keller, J. B., R. M. Lewis, and B. D. Seckler, 1956, \"Asymp totic solution of some diffraction problems,\" Commun. Pure & Appl. Math. 9, 207-265. [\u00a722.5) Keller, L. V., cited by name in Box 27", ".6. Kelley, J. L.. 1955, General Topology, Van Nostrand, New York. [Box 34.2) Kelvin. See Thomson, W.. Lord Kelvin. Kemble, E. C.. 1937, The Fundamental Principles of Quanrum Mechanics, McGraw-Hill, New York. Later editions avail able in paperback. [Box 25.3) Keres, P. See Unt and Keres 1972. Kepler, J., 1609, Astronomia nova caTLOAOY'lTO~, seu Physica. coelestis tradira comenrariis de moribus stellae Martis, Prague. [Box 27.7) Kepler, J., 1619. De HarnlOnice Mundi; Augsburg. [Box 27.7) Kerr, R. P., 1963, \"Gravitational field of a spinning mass as an example of algebraically special metrics,\" Phys. Rev. Lett. 11, 237-238. [\u00a7\u00a733.2, 33.6, Box 33.2) Kerr, R. P., and A. Schild, 1965, \"A new class of vacuum solutions of the Einstein field equations,\" in Proceedings of the Gali/eo Galilei Centenary Meeting on General Rela tivity, Problems of Energy and Gravitational Waves, G. Barbera, ed., Comitato Nazionale per Ie Manifestazione Celebrative, Florence, pp. 222-233. [\u00a733.6) Khalatnikov, I. M., cited by name in Box 27,7. See also Belinsky and Khalatnikov 1969a, 1969b, and 1970, Belinsky, Khalatnikov, and Lifshitz 1970, Belinsky, Lifshitz, and Khalatnikov 1971, Lifshitz and Khalatnikov 1963a and 1963b, and Ryan I972b. Khalatnikov, I. M.. and E. M, Lifshitz, 1970, \"General cosmo logical solutions of the gravitational equations with a singularity in time,\" Phys. Rev. Lett, 24, 76-79. [\u00a734.6, Box 30.1) Khuri, N., 1957, unpublished calculation quoted in Lindquist and Wheeler 1957. [\u00a725,5) Killing, W., 1892, \"Uber die Grundlagen der Geometrie,\" J. Reine Angel<'. Math. 109, 121-186. [\u00a725.2) Ki!mister. C. W., 1971, \"More about the King and the Vizier,\" Gen. Relativi(l' & Grav. 2. 35-36. [Fig. 44.3) Kirchhoff, G.-R.. cited by n(Jme in Box 38.2. Kittel, c., 1958. Elemenral:l' Statisrical Physics, Wiley, New York. [\u00a723.4) Klauder, J.. ed., 1972, Magic Without Magic: John Archibald Wheeler. W. H. Freeman, San Francisco. Klauder, L. T., Jr., M. Wakano, J. A. Wheeler, and R. Willey, 1959. \"Kinematics of a spherical isotropic uniform uni verse,\" in On:ieme Comeil de Physique Solvay: La Struc rure et!'evolution de l'ullivers, Editions Stoops, Brussels, Belgium. [\u00a729.5) Klein, F., 1872, \"Erlangerprogram: Vergleichende Betrachtun gen iiber neuere geometrischen Forschungen,\" translation in Bull. Amer. Math. Suc. 2, 215-249, 1893. [\u00a72.2) Klein, M. J., 1971, Einstein: The Life and Times, by R. W. Clark, book review, Science, 174, 1315-1316. [Box l.ll) Klein, 0., 1966, \"Boundary conditions and general relativity\" in A. de-Shalit, H. Feshbach, and L. Van Hove, eds., Preludes in Theoretical Physics in Honor of V. F. Weiss kopf, North Holland, Amsterdam, pp. 23-26. [Box 27.7) Klein, 0., 1971, \"Arguments concerning relativity and cosmology,\" Science 171, 339-345. [\u00a7\u00a727.10, 27.11, Box 27.7) Klein, O. See also Alfven and Klein 1962. Klemperer, W. B. See Deutsch and Klemperer 1962. Kline, M., 1954, \"Asymptotic solution of linear hyperbolic partial-differential equations.\" J. Rat. Mech. & Anal. 3, 315-342. [\u00a722.5) K", "liiber, H. V., 1960, \"The determination of Einstein's light deflection in the gravitational field of the sun,\" Vistas in Astronomy 3, 47-77, London, Pergamon Press. [Box 40.1) Kluitenberg, G. A., and S.- R. -deGroot, 1954, \"Relativistic thermodynamics of irreversible processes, III.\" Physica 20, 199-209. [\u00a722.2) Kneebone, G. T., 1963, Mathematical Logic and the Founda tions of Mathematics: An Introductory Survey, Van Nos trand, Princeton, N. J. [Box 44.5) Knight, L. V. See Witteborn, Knight, and Fairbank 1967. Kollros, L., 1956, in Carl Seelig, ed. Helle Zeit- Dunkle Zeit-In Memoriam Albert Einstein, Europa, Zurich. We thank Professor G. Holton for this quotation. [preface) Kovacs, S., cited by name in ex. 35.20. Kraichnan, R. H., 1955, \"Special-relativistic derivation of gen erally covariant gravitation theory,\" Phys. Rev. 55, 1118 1122. [\u00a7\u00a77.1, 18.1) Kramers, H. A., 1957, Quanrum Mechanics. North Holland Pub. Co., Amsterdam, Holland. Translated from the German by D. ter Haar. [\u00a721.I) i Krefetz, E., 1967, \"The equilibrium of slowly rotating configu rations in the post-Newtonian approximatiofl: Corrections to Clairaut's equations,\" Astrophys. J. 148, 589-612. [Ex. 26,1) Kretschmann, E., 1917, \"Uber den physikalischen Sinn der Relativitlitspostulate, A, Einsteins neue und seine ur sprungliche Relativitll.tstheorie,\" Ann. Phys. (Germany) 53, 575-614. [\u00a7\u00a712.5, 17,6) Kreuzer, L. B\" 1968, \"Experimental measurement of the BIBLIOGRAPHY AND INDEX OF NAMES 1239 equivalence of active and passive gravitational mass,\" Phys. Rev. 169, 1007-1012. [\u00a740.8) Krogh, c., and R. Baierlein, 1968, \"Lunar laser ranging and the Brans-Dicke theory,\" Phys. Rev. 175, 1576-1579. [\u00a740.6) Krotkov, R. See Bertotti, Brill, and Krotkov 1962 and Roll, Krotkov, and Dicke 1964. Kruskal, M. D\" 1960, \"Maximal extension of Schwarzschild metric,\" Phys. Rev. 119, 1743-1745. [\u00a731.4, Box 31,1] Kuchar, K., 1968, \"Charged shells in general relativity and their gravitational collapse,\" Czech J. Phys, B 18, 435-463, [\u00a72I.l3] Kuchar, K., 1970, \"Ground state functional of the linearized gravitational field,\" J. Math. Phys. 11, 3322-3334. [\u00a743.4] Kuchar, K., 1971a, \"Canonical quantization of cylindrical gravitational waves,\" Phys. Rev. D 4, 955-986. [\u00a7\u00a72I.l, 2I.l0] KuchJi, K., 1971b, \"Gravitational action functional reex pressed by means of intrinsic and extrinsic curvature,\" unpublished lecture given in the course Relativity at Princeton University. [\u00a7\u00a721.6, 2I.lI, Ex. 21.10] Kuchar, K., 1972, \"A bubble-time canonical formalism for geometrodynamics,\" 1, Math. Phys. 13, 768-781. [\u00a7\u00a72I.l, 2I.l0, 2I.lI, Ex. 21.24] Kuchar, K\" cired by name in \u00a72I.l, Ex. 13.7.21.10. Kuhlthau, A. R. See Rose et al. 1969. Kundt, W., 1971, \"Cosmological density fluctuations during Hadron stage,\" in Sachs 1971, pp. 365-372. [\u00a728.1] Kundt, W. See also Ehlers and Kundt 1962. Kundt, W., and E. T. Newman, 1968, \"Hyperbolic differential equations in two dimensions,\" J. Math. Phys. 9, 2193-2210. [\u00a7\u00a735.8, 38.8] Kiinz", "le, H. P., 1971, \"On the Spherical Symmetry of a Static Perfect Fluid,\" Commun. Marh. Phys. 20. 85-100. [\u00a723.1] Kuper, C. G., and A. Peres, eds., 1971, Relariviry and Gravita tion, Gordon and Breach, New York, Kursunoglu, B., 1962, Modern Quanrum Theory, W. H. Free man, San Francisco. [Box 2I.l] Kustaanheimo, P., 1966, \"Route dependence of the gravita tional red shift,\" Phys. Lett. 23. 75-77. [\u00a739.1] Lagrange, J. L., cited by name in Box 17.2. Lamb, W. E., Jr., and R. C. Retherford, 1947, \"Fine structure of the hydrogen atom by a microwave method,\" Pllys. Rev. 72, 241-243. [Fig. 43.3, Box 17.2 part 6] Lambert, J. H., 1761, Kosmologischen Briefe, Augsburg; see his The Sysrem of the World, tranSlated, with modifications, from the French by J. Jacque, London, 1800; reprinted in part in Munitz 1957. [Box 27.7] Lanczos, K., 1922, \"Bemerkung zur de Sitterschen Welt.\" Phys. Z. 23, 539-543. [\u00a72I.l3] Lanczos, K., 1924. \"FHichenhafte Verteilung der Materie in der Einsteinschen Gravitationstheorie.\" Ann. Phys. (Gemlany) 74. 518-540. [\u00a72I.l3] Lanczos, K., 1970. The Variarional Principles of ilIechanics. Univ. of Toronto Press, Toronto, 4th ed. [Box 2l.I] Landau. L. D., 1932, \"On the theory of stars:' Phys. Z. Landau, L D., and E. M. Lifshitz, 1959, Pergamon Press, London, [Ex. 22.7) fluid Mechanics, Landau, L D., and E. M, Lifshitz, 1960, Mechanics, Addison Wesley, Reading, Mass. [Box 22,4) Landau, L. D., and E. M. Lifshitz, 1962, The Classical Theon' of Fields, translated by M, Hamermesh, Addison.Wesle~\u00b7 \u2022'Reading, Mass. [\u00a7\u00a720.3, 20.5, 36.1, endpapers) Landau, L. D. and E. M, Lifshitz, 1971, The Classical Theon' of Fields, 3d rev. English ed., Addison-Wesley, Reading, Massachusetts, and Pergamon, London. [endpapers) Langer, W, D., and A. G, W. Cameron, 1969, \"Effects of hyperons on the vibrations of neutron stars,\" Asrrophys. & Space Sci. 5, 213-253, [\u00a724.3] Lao-tzu, 3d century B.C. Tao Ie Ching, Poem I. Translated as The Way of Life, R. B. Blakney, New American Library. 1955; translation revised by Janet Ten Broeck, [Epigraph to Chap. 18] Langevin, P. See Einstein 1912. Laplace, P. S., 1795, Le Systeme du monde, vol. II, Paris. Pub lished in English as The System of the World, W. Flint. London, 1809. [Fig. 24,1, Box 24.1, Epigraph to chap. 33) Laplace, P. S., cired by name in Box 17,2. Lascoux, J., 1968, \"Perturbation theory in quantum field theory and homology,\" in DeWitt and Wheeler 1968, 354-419. [Ex. 4.5] Laurence, D. H. See Boni, Russ, and Laurence 1960, Lawson. R. W., cited by name in Box 27.7. Leavitt, H., cired by name in Box 27.7. Lebovitz, N. R. See Chandrasekhar and Lebovitz 1968, Lecat, M., 1924, \"Bibliographie de la relarivire,\" Lamertin.:, Bruxelles, cited in preface of present bibliography. Ledoux, P., and Th. Walraven, 1958, \"Variable stars,\" Hard. buch der Physik 51. S, Fliigge ed., Springer-Verlag", ". Berlin, pp. 353-604. [Box 24.2] Lefschetz, S. 1949, Introduction to Topology, Princeton lIni\\'. Press, Princeton, N.J. For a fuller account of fixed-point theorems, see Dunford and Schwartz 1958. [\u00a736, I] Leibniz, G. W., 1714, La Monadologie. English translation available in several books; parts included in P, P. Wiener. ed., Leibniz Selecrions. Scribners, New York, 1951. (~.6] Leibniz, G. W., 1961, untitled essay in his Philosophischen Schrifien, vol. 7, ed. by C. J. Gerhardt. alms Verlapbuch. handlung, Hildesheim, p. 200. [Epigraph to ChIlP, :!I] Leibniz, G. W., quored ar end of Cha p. 44. Leighton, R. B., 1959, Principles of Modern Physics. McGrll'l\" Hill, New York. [\u00a72.1] Leighton, R. B. See also Feynman, Leighton. and Sands 19M. Lemaitre, G.\u2022 1933a, \"Formation of nebulae in the exrandln\u00a3 /yo, 10ll( universe,\" Acad. Sci., Paris. COlllples Rend. 1087. [\u00a730.6] Lemaitre. G.. 1933b. \"Spherical condensations in lhe cxrand\u2022 ing universe,\" Acad. Sci.. Paris. Compres Rend /1J(j, 90) 904. [\u00a730.6] Lemaitre. G.. 1933c. \"L'univers en expansion:' Ann..'t\u00ab. So Bruxelles 1 A53. 51-85. [Box 3I.l] Lemaitre. G.. 1949. cited in Schilpp 1949, r. 440. (l\u00b7fUlTJrh.. SOI')enmion 1, 285-288. [Box 24.1] to Chap. 36] 1240 GRAVITATION Lemaitre. G., cited by name in Box 27.7. Lenin. V. I., 1927. Marerialism and Empiriocriricism: Crirical International Comments on a Reactionary Philosophy, Publications, New York, 2d English ed. of Materializm' i Empiriokriritsbn': Kriricheskia Zamerki ob' Odnoi Reaktsionnoi Filosofii, Zveno Press, Moscow, 1909. [\u00a72U2] Lense. J. See Thirring and Lense -1918. Les Houches. France. Ecole d'ete de physique theorique: 1963. see DeWitt and DeWitt 1964b; 1966, see DeWitt, Schatz man, and Veron 1967 (third volume of three): 1972, see DeWitt and DeWitt 1973. Leutwyler. H., 1968. \"Boundary conditions for the state func tional in the quantum theory of gravity,\" in DeWitt and Wheeler 1968. pp. 309-317. [\u00a743.3] Lewis. R. M., 1958, \"Asymptotic expansion of steady-state solutions of symmetric hyperbolic linear differential equa tions.\" J. Rat. Mech. & Math. 7, 593-641. [\u00a722.5] Lewis, R. M. See also Keller, Lewis, and Seckler 1956. Liang, E. See Eardley. Liang, and Sachs 1972. Lichnerowicz, A., 1944, \"L'inll:gration des equations de la gravitation relativiste et Ie probleme des n corps,\" J. Math. Pures & Appl. 23,37-63. [\u00a72UI] Lichnerowicz, A., 1955, Theories relativistes de la gravitation et de l'electromagnerisme, Masson, Paris. [endpapers] Lichnerowicz, A., 1961a, \"Anticommutateur du champ spino riel en relativite generale,\" Acad Sci., Paris, Comptes Rend 252, 3742-3744. [Box 44.3] Lichnerowicz, A., 1961b, \"Anticommutateur pour Ie champ correspondant au spin 3/2 sur un espace d'Einstein,\" Acad. Sd, Paris, Comptes Rend. 253, 940-942. [Box 44.3] Petiau-Duffin Kemmeren,\" Acad. Sci., Paris, Comptes Rend 253, 983 985.", " [Box 44.3] Lichnerowicz, A., 'Theorie 1961c, de Lichnerowicz, A., 1962, \"Radiations en relativite generale,\" especially \"Le tenseur de superenergie,\" in CNRS 1962, pp. 93-106. [Ex. 15.2] Lichnerowicz, A., 1964, \"Propagateurs, commutateurs, et anti in DeWitt and commutateurs en relativite generale,\" DeWitt 1964. [Box 44.3, Ex. 15.2] Lichnerowicz, A., 1967, Relativistic Hydrodynamics and Mag netohydrodynamics; Lectures on rhe Existence of Solutions, W. A. Benjamin, New York. [\u00a7\u00a722.2, 22.3, Box 22.2] Lichnerowicz, A., 1971, \"Ondes des choc, ondes infinitesimales et rayons en hydrodynamique et magnetohydrodynamique relativistes,\" in Cattaneo 1971, pp. 88-203. [\u00a722.2, Box 22.2] Lichtenberg, D. B., 1965, Meson and Baryon Spectroscopy, Springer-Yerlag, New York. [\u00a738.4] Lifshitz, E. M., cited by name in Box 27.7. See also Belinsky, Khalatnikov, and Lifshitz 1970, Belinsky, Lifshitz, and Khalatnikov 1971, Khalatnikov and Lifshitz 1970, and Landau and Lifshitz, 1959, 1960, 1962, and 1971. Lifshitz, E. M., and I. M. Khalatnikov, 1963a, \"Investigations in relativistic cosmology,\" in Advances in Physics 12, 185 249; translated from the Russian by J. L. Beeby. [\u00a7\u00a730.6, 34.6] relativistic cosmology,\" Usp. Fiz. Nauk 80, 391-438. Eng lish translation in Sov. Phys.-Uspekhi 6, 495-522. 1964. [\u00a7\u00a730.6. 34.6] Lifshitz, E. M., and I. M. Khalatnikov, 1970, \"Oscillatory approach to singular point in the open cosmological model,\" Zh. Eksp. & TeoI'. Fiz. Pis'ma., 11, 200-203. Eng lish translation in Sov. Phys.-JETP Lerr. 11, 123-125 (1971). LighthiIl, M. J., 1958, Introduction to Fourier Analysis and Generali;ed Functions, Cambridge University Press, Cam bridge, England. [\u00a74.7] Lille v, A. E. See Dieter et al. 1962. Lin. 'c. c.. L. Mestel, and F. H. Shu, 1965. \"The gravitational collapse of a uniform spheroid,\" Astrophys. J. 142, 1431 1446. [Boxes 32.2, 32.3] Lindquist, R. W.. 1960, The Two-Body Problem in Geomerro dynamics, doctoral dissertation, Princeton University, Princeton, N.J. [Ex. 32.1] Lindquist, R. W., 1966, \"Relativistic transport theory,\" Ann. Phys. (USA) 37, 487-518. [\u00a722.6] LindqUist, R. W. See also Boyer and Lindquist 1967. LindqUist, R. W., and J. A. Wheeler, 1957, \"Dynamics of a lattice universe by the Schwarzschild cell method,\" Rev. Mod Phys. 29, 432-433. [\u00a727.lO, Box 25.2, Fig. 25.3.27.3] Liouville, J., 1837, \"Second memoire sur Ie developpement des fonctions ou parties des fonctions en series....,\" liou ville's 16-35. [\u00a722.5] et Appliquees 2, de Math. Pures J. Loar, H. H. See Durbin, Loar, and Havens 1952. London 1965, International Conference on Relativistic Theo ries of Gravitation, King's College, University of London, july 1965. Proceedings multilithed in 2 volumes by King's College.. _. Longair, M. S. See Hughes and Longair 1967. Lorentz, H. A., cited by name in \u00a720.6 and Box 16.4. Lorentz, H. A., A. Einstein, H. Minkowski, and H. Weyl, 1923. The Principle of Relativity: A Collection of Original Mem oirs", ", Methuen, London. Paperback reprint, Dover, New York. Lowry, R. A. See Rose et al. 1969. Lukash, V. N. See Doroshkevich, Lukash, and Novikov 1971. Lurie, D., and N. Mukunda, eds., 1963, Proceedings of Seminar on Unified Theories of Elementary Particles, University of Rochester report URPA- II, Rochester, N.Y. Lynden-Bell, D., 1967, \"Statistical mechanics of violent relaxa tion in stellar systems,\" Mon. Not. R. Astron. Soc. 136, lOl-121. [Box 33.3] Lynden-Bell, D., 1969, \"Galactic nuclei as collapsed oJd qua- sars,\" Nature 223, 690-694. [Box 33.3] \u2022 Lynden-Bell, D., and M. Rees, 1971, \"On quasars, dust, and the galactic center,\" Mon. Not. R. Astron. Soc. 152, 461 475. [Box 33.3] Lynds, C. R., cited by name in Box 27.7. MacCallum, M. A. H. See Penrose and MacCallum 1973. MacDonald, G. J. F. See Bender et al. 1971 and Munk and MacDonald 1960. Lifshitz, E. M., and I. M. Khalatnikov, 1963b, \"Problems of Mach, E., 1912, Die Mechanik in Ihrer Entwicklung Historisch- BIBLIOGRAPHY AND INDEX OF NAMES 1241 Kritisch Dargestellt, Brockhaus, Leipzig. English transla tion by T. J. McCormack, published in 1960 as The Sci ence of Mechanics with an introduction by Karl Menger, Open Court, La Salle, Ill. [\u00a7\u00a721.1, 21.12] Mach, E., cited by name in Box 27.1. Machado, A., 1940, \"Proverbios y cantares XXIX,\" in his Obras, Editorial Seneca, Mexico, pp. 235-236. [Epigraph to Chap. 43] MacKenzie, R. E. See Auslander and MacKenzie 1963. Mackey, M. B., cited by name in Box 27.7. Manasse, F. K., and C. W. Misner, 1963, \"Fermi Normal Coordinates and Some Basic Concepts in Differential Geometry,\" J. Math. Phys. 4, 735-745. [\u00a713.6] Mandelstam, S. See Yourgrau and Mandelstam 1968. Mann, T., 1937, Freud, Goethe, Wagner, Knopf, New York. Essay on Freud published originally as Freud und die Zukunji (Vortrag Gehalten in Wien am 8 Mai 1936 zur Feie~ von Sigmund Freuds 80 Geburtstag), Bormann Fischer, Wien (1936). [\u00a744.5] Manukin, A B. See Braginsky, Manukin, and Tikhonov 1970. Marconi, G. W., cited by name in Preface. Marinus of Naples, cited by name in Fig. 1.8. Markus, L. See Auslander and Markus 1959. Marie, c., 1969, \"Sur I'etablissement des equations de I'hydro dynamique des fluides relativistes dissipatifs, I et II,\" Ann. 1nst. Henri Poincare, 10, 67-126 and 127-194. [\u00a722.6, Ex. 22.7] Marsden, J. B. See Fischer and Marsden 1972. Martin, W. L. See Anderson et al. 1971. Marzke, R. E, and J. A. Wheeler, 1964, in Chiu and Hoffman 1964. [Box 16.4] Mathews, 1. See also Peters and Mathews 1963. Mathews, J., and R. Walker, 19lj5, Mathematical Methods of Physics, W. A. Benjamin, New York. [Box 26.1] Matzner, R. A., 1968, \"The present temperature of primordial radiation,\" Astrophys. J. 154, black-body gravitational 1123-II25. [\u00a727.10] Matzner, R. A. See also Clemens and Matzner 1967 and Fletcher et al. 1967. Matzner, R. A., and C. W. Misner, 1972, \"Dissipative effects in the expansion of the universe,\" Astrophys. J. 171,415 432. [\u00a7\u00a728.2, 30.4] Maupertuis, P. L", ". M. de, 1746, \"Recherche des loix du mouvement,\" Acad. R. Sci. Berlin. [Epigraph to\u00b7Chap. 21] Mavrides, M., quoted at end of Chap. 44. Maxwell, J. c., cired by name in Preface, Box 4.6. May, M., and R. H. White, 1966, \"Hydrodynamic calculations of general-relativistic collapse,\" Phys. Rev. 141,1232-1241. [\u00a7\u00a732.7, 42.7, Box 33.3, Ex. 32.7] May, M., and R. H. White, 1967, \"Stellar dynamics and gravi tational COllapse,\" Meth. Computat. Phys. 73, 219-258. [\u00a722.2, Boxes 22.2, 33.3] Mayall, N. U., 1970, \"Edwin Powell Hubble,\" Biograph. Mem. Nat. Acad. Sci., 41, 176-214. [Box 29.5] Mayall, N. U., and Oort, J. H., 1942, \"Further data bearing on the identification of the Crab Nebula with the super nova of 1054 A.D., part II: The astronomical aspects.\" Proc. Asrron. Soc. Pacific 54,95-104. [Box 24.1] Mazet, E. See Berger, Gauduchon, and Mazet 1971. McCrea, W. H., cited by name in Box 27.7. See also Bondi and McCrea 1960. McDonald, E., cited by name in Fig. 41.6. McVittie, G. c., 1956, General Relativity and Cosmology, Chapman and Hall, London. [endpapers] McWhirter, N., and R. McWhirter, 1971, The Guinness Book of Records, Guinness Superlatives, Ltd., Enfield, Middle sex [Chap. 12 epigraph] Meer, S. van der. See Farley et al. 1966. Meixner, J., and H. G. Reik, 1959, \"Thermodynamik der irre versiblen Prozesse,\" Handbuch der Physik, 3, Teil 2, 413 505. [\u00a722.2] Meixner, J., and E W. Schafke, 1954, Mathieusche Funktionen und Sphiiroidjirnktionen, Springer-Verlag, Berlin, Ger many. [Ex. 33.14] Melbourne, W. G. See Ohandley et al. 1969. Meltzer, D. W. See Bardeen, Thorne, and Meltzer 1966. Meltzer, D. W., and K. S. Thorne, 1966, \"Normal modes of radial pUlsation of stars at the end point of thermonuclear evolution,\" Astrophys. J. 145, 514-543. [Box 26.1] Menger, Karl, 1949, see Schilpp 1949, pp. 466-467. [Epigraph to Chap. 2, Epigraph to Chap. 9] Menskii, M. B. See Braginsky and Menskii 197I. Mercier, A., 1953, Analytical and Canonical Formalism in Physics, North-Holland, Amsterdam, Holland. Paperback reprint, Dover, New York, 1963. [Box 2I.l] Mercier, A., and M. Kervaire, 1956, Jubilee of Relativity The ory, proceedings of a conference in Bern, Switzerland, July 11-16, 1955, Birkhauser, Bern. Merzbacher, E., 1961, Quantum Mechanics, Wiley. New York. [\u00a716.3] Messiah, A., 1961, Quantum Mechanics, translated from the French by G. M: Temmer, North-Holland, Amsterdam, Holland. [Ex. 3.12] Mestel, L. See Lin, Mestel, and Shu 1965. Mielnik, B., and J. Plebanski, 1962, \"A study of geodesic motion in the field of Schwarzschild's solution,\" Acra Phys. Polon. 21, 239-268. [\u00a725.5] Mikola, S. See Fejer and Mikola 1918. Milne, E. A., cited by name in Box 27.7. Milnor, J., 1962, \"A survey of cobordism theory.\" Enseigne ment Marh. 8, 16-23. [Box 44.3] Milnor, J., 1963, \"Spin structures on manifolds,\" Enseignemenr Math. 9, 198-203. [Box 44.3] Milnor, J., 1965a, \"On the Stiefel-Whit", "ney numbers of complex manifolds and of spin manifolds,\" Topology 3, 223-230. [Box 44.3] Milnor, 1., 1965b, \"Remarks concerning spin manifolds,\" in Cairns 1965, p. 55. [Box 44.3] Milton, J., 1665, Paradise Lost, Book V, Lines 623-625, in The Complere Poems of John Mihon, Union Library Associa tion of New York, 1936. [Epigraph to Chap. 25] Minkowski, H., 1908, address delivered at the 80th Assembly of German Natural Scientists and Physicians, Cologne. English translation, \"Space and time,\" in The Principle of Relarivi(l', Dover. New York. 1923. [\u00a71.5] Minkowski, H. See also Loremz er al. 1923. 1242 GRAVITATION l\\1inkowski. R., 1942, \"The Crab Nebula,\" Asrrophys. J. 96. M9111er. C, 1952, The Theory of RelarivirJ; Oxford Univ. Press, 199-213. [Box 24.1] Minkowski, R., cired ~r /lame in Box 27.7. Minkowski. Roo and O. C. Wilson. 1956. \"Proportionality of nebular redshifts to wave length,\" Asrrop/(I's. J. 113, 373 376. [\u00a738.6] Misner. C. Woo 1957. \"Feynman quantization of general rela tivity,\" Rev. Mod Ph,rs. 29. 497-509. [Box 2I.l] Misner. C. W., 1964a. \"Differential geometry and differential topology,\" in DeWitt and DeWitt I964b, pp. 881-929. [Box 8.1] Misner, C. W., 1964b. \"Waves, Newtonian fields, and coordi nate functions,\" in Infeld 1964. [\u00a735.8] Misner. C. W., 1965, \"Relativistic equations for spherical grav itational collapse with escaping neutrinos,\" Phys. Rev. B13l, 1360-1364. [Box 23.1] Misner. C. W., 1968, \"The isotropy of the universe,\" Astrophys. J. 151, 431-457. [\u00a7\u00a728.2, 30.3] Misner, C. W., 1969a, \"Gravitational collapse,\" in Chretien, Deser, and Goldstein 1969, vol I. [\u00a7\u00a712.I, 32.5, Boxes 8.1, 31.2, Ex. 31.8, endpapers] Misner, C. W., 1969b. \"Mixmaster universe,\" Phys. Rev. Lett. 22, 1071-1074. [\u00a7\u00a72I.IO, 28.2, 30.6, 34.6. Box 30.1] Misner, C. W., 1969c, \"Absolute zero of time,\" Phys. Rev. 186, 1328- 1333. [\u00a7\u00a728.3, 34.6] Misner, C. W., 1970, \"Classical and quantum dynamics of a closed universe,\" in Carmeli, Fickler, and Witten 1970, pp. 55-79. [Box 30.1] Misner, C. W., 1972a, \"Minisuperspace,\" in Klauder 1972. [\u00a743.3, Box 30.1] Misner, C. W., 1972b, \"Stability of Kerr black holes against scalar perturbations,\" Bull. Am. Phys. Soc. (II) 17, 472. [Ex. 33.14] Misner, C. W., 1973, \"A minsuperspace example: The Gowdy TJ cosmology,\" Caltech Orange Aid Preprint. Submitted to Phys. Rev. [\u00a743.3] Misner, C. W.\u2022 cired by name in \u00a7\u00a721.1, 35.8, Boxes 17.2 part 2', 17.3,27.7. See also Arnowitt, Deser, and Misner 1962, Beckedorff and Misner 1962, Chrzanowski and Misner 1973, Manasse and Misner 1963, and Matzner and Misner 1972. Misner, C. W., and D. H. Sharp, 1964, \"Relativistic equations for adiabatic, spherically symmetric gravitational col lapse,\" Phys. Rev. B 136,571-576. [\u00a7\u00a72I.l3, 22.2, Ex. 32.7, Box 23.1] Misner, C. W., and A. H. Taub", ", 1968, \"A singularity-free empty universe,\" Zh. Eksp. & Teor. Fiz. 55, 233. English original in Sov. Phys.-JETP 28, 122, 1969. [\u00a7\u00a732.5, 34.6] Misner, C. W., and J. A. Wheeler, 1957, \"Classical physics as geometry: Gravitation, electromagnetism, unquantized charge, and mass as properties of curved empty space,\" Ann. Phys. (U.S.A.) 2,525-603, reprinted in Wheeler 1962. [\u00a7\u00a720.6, 21.8, 21.9, Fig. 44.1, Boxes 4.1, 4.4, 15.1,21.2, Ex. 20.7] Misner, C. W., and J. A. Wheeler, 1972, \"Conservation laws and the boundary of a boundary,\" in Shelest 1972, pp. 338-351. [\u00a715.3] London: 2d ed.. 1972. [endpapers] l\\1\u00a2lIer, c., ed.. 1962, Evidence for Gravirarional Theories, Aca- demic Press. New York. Moncrief, V. E. See Berger et al.' 1972. Morduch, G. E. See Whitrow and Morduch 1965. Morehead, J. C. See Gauss 1827. Morette, c., 1951, \"On the definition and approximation of Feynman's path integrals,\" Phys. Rev. 81, 848-852. [Box 2I.l] Morette, C. See also DeWitt, C. Morgan. T., and K. S. Thorne, 1973, in preparation. [Box 32.3] Morganstern, R. E., 1973, \"Observational constraints imposed by Brans-Dicke cosmOlogies,\" Phys. Rev. D 7, 1570-1579. [\u00a728.4] Moritz, B. K., 1969, A Srudy of the General Relarivistic Dy namics of the Klein-Alfven Cosmology, doctoral disser tation, University of Maryland [\u00a727.11, \u00a728.4] Morse, M., 1934, The calculus ofvariations in the large, Ameri can Mathematical Society, Providence, R.I. [Fig. 13.2] Morse, M., and S. S. Cairns, 1969, Critical Point Theory in Global Analysis and Differential Topology: an Introduction, Academic Press, New York. [Fig. 13.2] Moser, J. K. See Siegel and Moser 1971. Mottleson, B. R. See Bohr and Mottleson 1969. Muhleman, D. O. See Anderson et al. 1971. Muhleman, D.O., R. D. Ekers, and E. B. Fomalont, 1970, \"Radio interferometric test of the general relativistic light bending near the sun,\" Phys. Rev. Lett. 24, 1377-1380. [Box 40.1] Muhlemann, D.O., and P. Reichley, 1964, \"Effects of general relativity on planetary radardistance measurements,\" JPL Space Programs Summary 4, no. 37-29, p. 239. [\u00a740.4] Muhlemann, D.O., and P. Reichley, 1965, \"General relativistic effects on radar propagation,\" JPL Space Programs Sum mary 4, no. 37-31, p. 342. [\u00a740.4] Mukunda, N. See Lurie and Mukunda 1963. Mulholland. J. D. See Bender et al. 1971 and Ohandley et al. 1969. Munitz, M. K., 1957, Theories ofthe Universe, Free Press, New York. [Box 27.7] Munk, W. H., and G. J. F. MacDonald, 1960, The Rotation of the Earth,- a Geophysical Discussion, Cambridge Uni versity Press, Cambridge, England. [Fig. 1.8] Nadezhin, D. K. See Ivanova, Imshennik, and Nadezhin 1969. Nagel, E., and J. R. Newman, 1958, GOdel's Proof, New York Univ. Press, New York. NASL, 1969, Proceedings of the 1969 Symposium on UnJonven tional Inertial Sensors (U) Vol. I, Naval Applied Science Laboratory, Brooklyn, New York. Nasr, H., 1964, Three Muslim sages: Avicenna, Suhrawardi, Ibn 'Arabi, Harvard University Press, Cambridge, Mass. [Box 27.7] Navier", ", L., cited by name in Box 17.2 part 6. Nazarenko, V. S. See Braginsky and Nazarenko 1971. Ne'eman, Y, 1965, \"Expansion as an energy source in quasistellar radio sources,\" Astrophys. J. 141, 1303-1305. [\u00a734.3] Moffet, A., cired by name in Box 40.1. Nester, J. M., cited by name in \u00a735.1, BIBLIOGRAPHY AND INDEX OF NAMES 1243 Neugebauer, 0., 1959, Greek Horoscopes, with H. B. van Hasen, American Philosophical Society, Philadelphia, Pa. [Fig. 1.8] Neugebauer, P. V., 1927, \"Zilr astronomischen Chronologie,\" Astron. Nachr. 230, 57-64. [Fig. 1.8] Neugebauer, P. V., 1929, Astronomische Chronologie, 2 vols, De Gruyter, Berlin. [Fig. 1.8] Neugebauer, P. V., 1930, \"Schoch Todesanzeige,\" Astron. Nachr. 237, 221-224. [Fig. 1.8] Neugebauer, P. V., and C. Schoch, 1930, \"Zur astronomischen Chronologie, II,\" Astronomische Abhandlungen. Ergan zungshefte zu den Astron. Nachrichten 8, no. 2,24-32. [Fig. 1.8] Newman, E. T. See also Couch and Newman 1972, Couch et al. 1968, and Kundt and Newman 1968. Newman, E. T., E. Couch, K. Chinnapared, A. Exton, A. Prakash. and R. Torrence, 1965, \"Metric of a rotating. char~d mass,\" J. Math. Phys. 6, 918-919. [\u00a733.2] Newman, E. T., and R. Penrose. 1965, \"10 exact gravita tionally-conserved quantities,\" Phys. Rev. Lett. 15, 231 233. [Ex. 32.11] Newman, J. R. See Nagel and Newman 1958. Newton, I., 1687, Philosophiae natllralis principia mathematica, 1st ed., Streater, London; final ed., in English translation by A. Motte, 1729, revised by A. Cajori, Sir Isaac Newron's Marhemarical Principles of Narural Philosophy and His Sysrem of rhe World, Univ. of California Press, Berkeley and Los Angeles, 1934; paperback, 1962. [Boxes 1.10,21.1, 27.7] Newton, I., 1730, Opricks, or a Treallse of the Reflection, Re fracrion, Inflecrion, & Colours of Light, 4th ed.; reprint, ed. by D. H. D. 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[Box 39.5, Ex. 39.7] - - ~on. H. K.,~", "encer, and N. E. Steenrod, 1959, Advanced Calculus, Van Nostrand, Princeton. New Jersey. [\u00a74.1J Nicetas, cired by name in Bo~ ---- Nijenhuis, A., 1952, \"Theory of the geometric object.\" disser tation, Amsterdam. [\u00a72.2] Nizer, 1.., 1972, quoted in the New York Times February 7, 1972. [Epigraph to Chap. 40] Noerdlinger, P. D., and V. Petrosian, 1971, \"The effect of cosmological expansion on self-gravitating ensembles of particles,\" Astrophys. J. 168, 1-9. [\u00a727.5] Noonan, T. W. See Robertson and Noonan 1968. Nordstr~m, G., 1913, \"Zur Theorie der Gravitation vom Standpunkt des Relativitatsprinzips,\" Ann. Phys. (Germany) 42, 533-554. [\u00a717.6] Nordstr\u00a2m, G., 1918, \"On the energy of the gravitational field in Einstein's theory,\" Proc. Kon. Ned. Akad. Wet. 20, 1238-1245. [Ex. 31.8] Nordstr~m, G., cired by name in \u00a7\u00a738.2, 42.7. Nordtvedt, K., 1968a, \"Equivalence principle for massive bod I: Phenomenology,\" Phys. Rev. 169, 1014-1016. ies, [\u00a7\u00a738.2, 40.9] Nordtvedt, K., 1968b, \"Equivalence principle for massive bod ies, II: Theory,\" Phys. Rev. 169, 1017-1025. [\u00a7\u00a738.2, 39.3, 40.9, Box 39.5] Nordtvedt, K., Jr., 1969, \"Equivalence principle for massive bodies including rotational energy and radiation pres sure,\" Phys. Rev., 180, 1293-1298. [\u00a7\u00a738.2, 39.3] Nordtvedt. K., 1970, \"Solar system Eatvos experiments,\" Icarus 12, 91-100. [\u00a740.8, Box 39.5] Nordtvedt, K., 1971a, \"Equivalence principle for massive bod ies, IV: Planetary bodies and modified Eatvos-type exper iments,\" Phys. Rev. D 3, 1683-1689. [\u00a7\u00a740.8, 40.9, Box 39.5] Nordtvedt, K., 1971b, in Davies 1971, pp. 32-37. [\u00a7\u00a739.5, 40.9. Box 39.5] Nordtvedt, K., Jr. See also Will and Nordtvedt 1972. Nordtvedt, K., Jr., and C. M. Will, 1972, \"Conservation laws II; Experi and preferred frames in relativistic gravity, mental evidence to rule out preferred-frame theories of' gravity,\" Astrophys. J. 177,775-792. [\u00a7\u00a739.2, 40.5, 40.8, Box 39.5] North, J. D., 1965, The Measure of the Universe, Clarendon Press, Oxford. [Box 27.7] Novikov, I. D., 1963, doctoral dissertation, Shternberg Astro nomical Institute, Moscow. [\u00a731.4] Novikov, I. D., 1964. \"Delayed explosion of a part of the Friedmann universe, and quasars,\" Asrron. Zh. 41, 1075 1083. English translation in Sov. Astron. \u00b7Astron. J. 8, 857 863. [\u00a734.3] Novikov, I. D., 1966a, \"Change of relativistic collapse into anticollapse and kinematics of a charged sphere,\" Zh. Eksp. & Teor. Fiz. Pis'rna 3, 223-227. English trans lation in Sov. Phys.-JETP Lett. 3, 142-144. [\u00a734.6, Fig. 34.4] Novikov, I. D., 1966b. \"The replacement of relativistic gravita tional contraction by expansion, and the physical singu larities during contraction,\" Asrrol!. Z/z. 43, 911-918. Eng lish translation in Sov. Asrron.-A. J. 10. 731-737. [\u00a734.6. Fig. 34.4] Novikov. I. D. See also Doroshkevich, Lukash. and Novikov 1971, Doroshkevich, Zel'dovich. and Novikov 1965, and Zel'd", "ovich and Novikov 1966. 1971, and 1974. Novikov. I. D.. and K. S. Thorne, 1973, \"Astrophysics of black holes.\" in De\\Vitt and DeWitt 1973, pp. 343-450. [Box 33.3] 1244 GRAVITATION Nutku, Y. See Berger et al. 1972 alld Chandrasekhar and Nutku 1969. O'Connell, R. E, 1972, \"Present status of the theory of the 3, relativistic gyroscope experiment,\" Gen. ReI. & Grav. 123-133. [\u00a740.7, Ex. 40.8] Ogorodnikov, K. E, 1965, Dynamics of Srellar Sysrems, ed. by A. Beer. Macmillan. New York. [\u00a725.7] Ohandley, D. A., D. B. Holdridge. W. G. Melbourne. and J. D. MUlholland, 1969,J.P.L. Development Ephemeris No. 69, J.P.L. Technical report 32-1465, Caltech Jet PropUlsion Laboratory, Pasadena. [\u00a740.l] Ohanian, H. C. See Belasco and Ohanian 1969. Olbers. H. W. M., 1826. \"Uber die Durchsichtigkcit des Welt raumes\" Bode Jahrbuch; 110; substantially the same article in Bibliorheque Universelle for February 1826, and in Edinburgh New Philosophical Journal for April-October 1826, 141. [Box 27.7] Omnes, R. L., 1969, \"Possibility of malter-antimatter separa tion at high temperature,\" Phys. Rev. Lett. 23, 38-40. [\u00a727.1O] see also O'Murchadha, N., O'Murchadha, N., and J. W. York, Jr., 1973, personal commu nication; \"Existence and uniqueness of solutions of the Hamiltonian constraint of general relativity,\" doctoral dissertation, Princeton U niv., February 5, 1973. [\u00a72l.lI] O'Neill, B., 1966, Elementary Differential Geomerry, Academic Press, New York. [Box 8.1] Oort, J. H., 1958, \"Distribution of galaxies and the density of the universe,\" in Onzieme Conseil de Physique Solvay: La Srructure et!'evolution de l'univers, Editions Stoops, Brus sels, Belgium. [\u00a729.6] Oort, J. H., 1970, \"Maller far from the galactic plane associated with spiral arms,\" in Spiral Structure ofour Galaxy, Basel, Switzerland, 29 Aug.-4 Sep. 1969, D. Reidel, Dordrecht, Netherlands. [\u00a728.2] Oort, J.' H., cited by name in Box 27.7. See also Mayall and Oort 1942. Oppenheimer, J. R., and H. Snyder, 1939, \"On continued gravitational contraction,\" Phys. Rev. 56, 455-459. [\u00a7\u00a732.3, 32.4,34.6, Fig. 24.1, Box 24.1] Oppenheimer, J. R., and G. Volkoff, 1939, \"On massive neu tron cores,\" Phys. Rev. 55, 374-381. [\u00a723.2, Fig. 24.1, Box 24.1, Ex. 23.10] Oppolzer, Th. V., 1887, \"Canon der Finsternisse,\" Math. Naturwiss. ct. d. Akad. der Wiss. Denkschriften 52, Wien. [Fig. 1.8] O'Raifeartaigh, L, ed., 1972, General Relativity, Papers in. Honor of J. L. Synge, Oxford Univ. Press, London. Ostriker, J. P., and 1. E. Gunn, 1969, \"On the nature of pUlsars, I: Theory,\" Astrophys. J. 157, 1395-1417. [\u00a724.3, Ex. 33.13] Pacini, F.. 1968, \"Rotating neutron stars, pUlsars, and super nova remnants,\" Nature 219, 145-146. [\u00a724.3] Page, L., 1941, lecture at Dec. 17 meeting of the American Institute of Electrical Engineers, New York. [Epigraph to Chap. 3] Palatini, A., 1919, \"Deduzione invariantiva delle equazioni gravitazionali dal principio di Hamilton,\" Rend.", " Circ. Mat. Palermo 43, 203-212. [\u00a721.2] Palatini, A., cited by name in \u00a7\u00a72l.l, 21.2. Panov, V. I. See Braginsky and Panov 1971. Paolini, E, R., cited by Ilame in Box 27.7. Papanastassiou. D. A.. and G. J. Wasserburg, 1969, \"Initial strontium isotopic abundances and the resolution of small time differences in the formation of planetary objects,\" Earth & Planetary Sci. Lett. 5, 361-376. [Box 29.3] Papapetrou, A.. 1951, \"Spinning test-particles in general rela tivity, I,\" Proc. R. Soc. London A 209, 248-258. [\u00a720.6, Ex. 40.8] Papapetrou, A., I954a, \"Eine neue Theorie des Gravita tionsfeldes, I.\" Math. Nacllr. 12, 129-141. [\u00a740.8] Papapetrou. A., 1954b, \"Eine neue Theorie des Gravita tionsfeldes, II,\" Marh. Nachr. 12, 143-154. [\u00a740.8] Papapetrou, A.. I954c, \"Eine Theorie des Gravitationsfeldes mit einer Feldfunktion,\" Z. Phys. 139, 518-532. [\u00a740.8] Papapetrou, A., 1958, \"Uber zeit abhangige Losungen der allgemeinen Relativitatstheorie.\" Feldgleichungen der Ann. Phys. (Germany) 2, 87-96. [\u00a735.8] Papapetrou, A., and A. Hamoui, 1967, \"Surfaces caustiques degenerees dans la solu tion de Tolman: -La singularite physique en relativite generale,\" Ann. Inst. Henri Poincare A 6, 343-364. [Ex. 32.7] Papapetrou, A., and A. Hamoui, 1968, \"Couches simples de matieres en relativite generale,\" Ann. Inst. Henri Poincare A 9, 179-211. [\u00a72l.l3] Papapetrou, A., and H. Treder, 1959, \"Das Sprungproblem erster Ordnung in der allgemeinen Relativitatstheorie,\" Math. Nachr. 20, 53-66. [\u00a72l.l3] Papapetrou, A., and H. Treder, 1962, \"Shock waves in general relativity,\" in Infeld 1962, pp. 351-360. [\u00a72l.l3] Paris 1967, Fluides et Champs Gravitationnel en relarivite ge nerale, Colloques Internationaux du CNRS, no. 170, orga nize au College de France,du 19 au 23 juin 1967, Editions du Centre National de la Recherche Scientifique, 1969. Parker, H. M. See Rose et al. 1969. Parker, L., 1966, Particle Creation in the Expanding Universe, Ph.D. thesis, Harvard University, unpublished. [\u00a730.5] Parker, L., 1969, \"Quantized fields and particle creation in I,\" Phys. Rev. 183, 1057-1068. expanding universes, [\u00a730.5] Parker, L., 1972, \"Particle Creation in Isotropic Cosmologies,\" Phys. Rev. Lett. 28, 705-708. [\u00a730.5] Parsons, S. J., cited by name in Box 27.7. Partridge, R. B., 1%9, \"The primevat fireball today,\" Am. Sci. 57, 37-74. [\u00a727.2] Partridge, R. B. See also Boughn, Fram, and Partridge 197 I. Pauli, W., 1921, \"Relativitatstheorie,\" in Enzyklopiidie der II, 539-775, Teubner, Mathematischen Wissenschaften, Leipzig, Germany. [Preface], Pauli, W., 1934, \"Die allgemeinen Prinzipien der Wellen I, 2d ed. mechanik,\" Handbuch der Physik 24, Part [\u00a716.3)\" Pauli, W., 1947, Exclusion Principle and QuanT1lmMl!r:hanics, 1945 Nobel prize lecture, Editions Grisson, Neuchatel, Switzerland. Related discussion in Fierz and Weiss", "kopf 1960. [Box 44.3] PaUli, W., ed., 1955, Niels Bohr and the Development ofPhysics, McGraw-Hill, New York. PaUli, W., 1958, Theory ofRelativity, Pergamon Press, London. English translation by Gerard Field of Pauli 1921 with supplementary notes (1956) by the author. [endpapers] BIBLIOGRAPHY AND INDEX OF NAMES 1245 Pauli, Wolfgang, Memorial volume. See Fierz and Weisskopf 1960. Pauli, W., cited by name in Preface. See also Fierz and Pauli 1939. Paya, D. See Dabbs et al. 1965. Peebles, P. J. E., 1966, \"Primordial He abundance and fireball, II,\" Astrophys. J. 146, 542-552. [\u00a728.l] Peebles, P. J. E., 1969, \"Cosmology,\" Proc. Roy. Astron. Soc. Canada, 63, 4-31. [\u00a730.1] Peebles, P. J. E., 1971, Physical Cosmology, Princeton Univ. Press, Princeton, N.J. [\u00a7\u00a727.I, 27.2, 28.1, 29.6, Box 27.7] Peebles, P. J. E., 1972, \"Gravitational collapse and related phenomena from an empirical point of view, or, black holes are where you find them,\" Gen. ReI. & Grav. 3, 63-82. [Box 33.3] Peebles, P. J. E. Cited by name in Box 27.7. See also Dicke and Peebles 1962a, 1962b, and 1965, Dicke er al. 1965, and Jones%nd Peebles 1972. Peebles, P. J. E., and J. T. Yu, 1970, \"Primeval adiababic perturbation in an expanding universe,\" Astrophys. J. 162, 815-836. [\u00a730.I] Pekar, D. See Eotvos, Pekar, and Fekete 1922. Penrose, R., 1959, \"The apparent shape of a relativistically moving sphere,\" Proc. Cambridge Phil. Soc. 55, 137-139. [\u00a741.10] Penrose, R., 1964, \"Conformal treatment of infinity,\" in DeWitt and DeWitt I964b. [\u00a734.2] Penrose, R., 1965a, \"Zero rest-mass fields including gravitation: Asymptotic behavior,\" Proc. R. Soc. London A 284, 159 203. [\u00a734.2, Box 32.2] Penrose, R., I965b, \"Gravitational collapse and space-time singularities,\" Phys. Rev. Lett. 14,57-59. [\u00a7\u00a732.7, 34.6, Box 34.2] Penrose, R., 1966, \"An analysis Orale structure of spacetime,\" Adams Prize Essay. [Box 34.2] Penrose, R., 1968a, \"Structure of spacetime,\" in DeWitt and Wheeler 1968. [\u00a7\u00a734.1, 34.4, 41.7, 4I.lI, Boxes 34.1, 34.2, 44.3] Penrose, R., 1968b, \"Twistor quantization and the curvature of spacetime,\" Int. J. Theor. Phys. 1, 61-99. [Box 34.2] Penrose, R., 1969, \"Gravitational collapse: The role of general relativity,\" Nuovo Cimento 1, special number, 252-276. [\u00a733.7, Boxes 33.2, 33.4, 34.2] ing of nongravitational motion,\" Ann. Phys. (U.S.A.) 10, 94-99. [\u00a7\u00a736.I, 36.8] Peters, P. C, 1964, \"Gravitational radiation and the motion of two point masses,\" Phys. Rev. B 136, 1224-1232. [\u00a7\u00a736.I, 36.7, Ex. 36.9] Peters, P. C, and J. Mathews, 1963,' \"Gravitational radiation' from point masses in a Keplerian orbit,\" Phys. Rev. 131, 435-440. [\u00a736.6, Ex. 36.6] Peterson, B. A., cited by name in Box 27.7. Peterson, F. P. See Anderson, Brown, and Peterson 1966a and I966b. Peterson, F. R. See Evenson et al. 1972. Pethick, C J. See Baym,", " Bethe, and Pethick 1971. Petrosian, V. See Noerdlinger and Petrosian 1971. Petrov, A. Z., 1954, Reports of the Stare Universiry of Kazan 114, book 8, 55. [\u00a733.6] Petrov, A. Z., 1969, Einstein Spaces, Pergamon, New York. [\u00a733.6] Petrov, Aleksei Zinoveivich, Memorial volume. See Shelest 1972. Pettengill, G. H. See Shapiro, Ash et al. 1971, Shapiro, Smith et al. 1971, and Shapiro et al. 1972. Philolaus, cited by name in Box 27.7. Phillips, J. W., cited by name in Box 27.7. Picasso, E. See Bailey et al. 1968 and Farley et al. 1966. Piir, I. I., 1971, \"Maxwell equations in the Schwarzschild grav itational field\" (in Russian), Izv. Akad Nauk Estonian S. S. R., Fiz. Mat., 20, 413-419. [Ex. 32.l1] Pilkington, J. D., cited by name in Box 27.7. See also Hewish et al. 1968. Pirani, F. A. E., 1956, \"On the physical significance of the Riemann tensor,\" Acta Phys. Pol. 15, 389-405. [\u00a720.6, Ex.. 40.8] Pirani, F. A. E., 1957, \"Invariant formulation of gravitational radiation theory,\" Phys. Rev. 105, 1089-1099. [\u00a7\u00a72I.l3, 33.6, Ex. 15.2] Pirani, F. A. E., 1965, \"Introduction to gravitational radiation theory,\" in Trautman, Pirani, and Bondi 1965, pp. 249 373. [\u00a7\u00a741.7, 4I.lI, endpapers] Pirani, F. A. E. See also Bondi, Pirani, and Robinson 1959, Ehlers, Pirani, and Schild 1972, and Trautman, Pirani, and Bondi 1965. Pittsburgh 1970, Regional Conference on Relativity, July Penrose, R., 1972, \"Techniques of differential topology in rela 13-17, 1970. See Farnsworth et al. 1972. tivity,\" in Farnsworth et al. (1972). [\u00a734.1] Penrose, R., cited by name in \u00a7\u00a732.7, 41.9, 42.1, Box 27.7, Fig. 41.7. See also Hawking and Penrose 1969, Infeld 1964, Klauder 1964, Newman and Penrose 1965. and Walker and Penrose 1970. Penrose, R., and M. A. H. MacCallum, 1973, \"Twistor Theory: An approach to the quantization of fields and spacetime,\" Physics Reports 6C, no. 4, 242-315. [Box 34.2] Penzias, A. A.. cited by name in Box 27.7. Penzias, A. A., and R. W. Wilson, 1965, \"A measurement of excess antenna temperature at 4080 Mc/s,\" Astrophys. J. 142, 419-421. [\u00a727.2] Pereira, C M. See Brill et al. 1972. Peres, A., 1962, \"On the Cauchy problem in general relativity, II,\" Xuovo Cimento 26, 53-62. [\u00a743.3] Peres, A.. and N. Rosen. 1964, \"Gravitational radiatio~damp- Planck, M., 1899, \"Uber irreversible Strahlungsvorgange,\" Sitzungsber. Deut. Akad Wiss. Berlin, Kl. Math.-Phys. Tech., 440-480. [\u00a7\u00a743.4, 44.6] Plato. See Cornford 1937. Plebanski, J. See Infeld and Plebanski 1960 and Mielnik and Plebanski 1962. Plotkin, H. E. See Bender et aL 1971. Plutarch, cited by name in Box 27.7. Podolsky, B. See Dirac, Fock, and Podolsky 1932. Podurets. M. A., 1964, \"Asymptotic behavior of the optical luminosity of a star in gravitational collapse in terms of general relativity theory,\" Astron. Zhur. 41. 1090-1096, English translation in Sov. Astron.-A. J. 8, 868-873 (1965). [\u00a732.3] Podurets, M. A.\u2022 1966, \"The c~aracter of the singularity in the gravitational collapse of a star,\" Dokl. Akad", ",Vauk SSSR 1246 GRAVITATION 167, 1025-1027. Engli,h translation in SOl'. PIz.-Dokla~l\u00b7 11, 275-276. [\u00a734.6. Ex. 32.3] Podurets. M. A. See also Zel'dovich and Podurets 1965. Poincare. H. cited ~)' l/Gme ill \u00a7\u00a73.1. 12.3. Poi\"on, S.. cited 6.;. flame in Box 17.2 part 6. Poli.:vktov-:\"ikoladze. N. M.. 1969. \"On the theory of the gravitational field,\" Zh. Eksp. & Teor. Fi:. 57. 2010-2020. English translation in Sov. Phys.-JETP 30 (1970). 1089 1095. [Box 17.2 part 6] Polkinghome. J. c., 1955. \"On the Feynman principle,\" Proc. R. Soc. London A 230, 272-276. [Box 21.1] Polubarinova-Kochina, P. Ya., 1963. \"Alexander Alexandro vitch Friedmann (In memory of A. A. Friedmann),\" Usp. Fi::. Nal/I.:. 80, 345-352. English translation in Sov. Phl's.-Uspekhi. XX, 467-472 (1964). [Box 27.6] Pope, A.. 1733, Essay on Mall, London [Epigraph to Chap. 24] Porter, J. See Farnsworth et al. 1972. Pound, R. V., and G. A. Rebka. 1960, \"Apparent weight of Photons.\" Phys. Rev. Lett. 4, 337-341. [\u00a7\u00a77.2, 7.4, 38.5, Fig. 38.1] Pound, R. V., and J. L. Snider, 1964, \"Effect of gravity on nuclear resonance,\" Phys. Rev. Lett. 13, 539-540. [\u00a7\u00a77.2, 7.4] Pound, R. V., and J. L. Snider, 1965, \"Effect of gravity on gamma radiation,\" Phys. Rev. B 140, 788-803. [\u00a7\u00a77.2. 38.5, Fig. 38.1] Power, E. A., and J. A. Wheeler, 1957, \"Thermal geons,\" Rev. Mod Phys. 29, 480-495. [\u00a725.6] Prakash, A. See Newman et al. 1965. Press, E, 1965, \"Resonant vibrations of the Earth,\" Sci. Ameri can 213, no. 5,28-37. [Ex. 37.14] Press, E See also Ahrens, Press. and Runcorn 1964. 'Press, W. H.. 1970, unpublished work described in part in Press and Thome 1972. [\u00a737.3, Fig. 37.2, Box 37.2] Press. W. H.. 1971. \"Long wave trains of gravitational waves from a vibrating black hole,\" Astrophys. 1. Lett. 170, 105 108. [Box 33.3] Press, W. H.. 1972, \"Time evolution of a rotating black hole immersed in a static scalar field,\" Astrophys. J. 175, 243 252. [Ex. 33.14, Box 33.3] Press, W. H. See also Bardeen and Press 1972, Bardeen, Press, and Teukolsky 1972, and Davis et al. 1971. Press, W. H., and J. M. Bardeen, 1971, \"Nonconservation of the Newman-Penrose conserved quantities,\" Phys. Rev. Lett. 27, 1303-1306. [Ex. 32.1 I] Press, W. H., and S. A. Teukolsky, 1972, \"Floating orbits, super radiant scattering and the black-hole bomb,\" Nature 238, 211-212. [Box 33.3] Press, W. H., and S. A. Teukolsky, 1973, \"Perturbations of a rotating black hole, II: Dynamical stability of the Kerr metric,\" Astrophys. J., in press. [Box 33.3, Ex. 33.14] Press, W. H., and K. S. Thorne, 1972, \"Gravitational-wave astronomy,\" Ann. Rev. Astron. Astrophys. 10, 335-374. [\u00a7\u00a736.5, 37.4, 37.8, Box 37.5] Price, R. H., 1971, Nonspherica1 Perturbations of Relativistic Gravitational Collapse, doctoral dissertation, California Institute of Technology.", " [Ex. 32.10, Box 32.2] Price. R. H.. 1972a. \"Nonspherical perturbations of relativistic gravitational collap,e. I: Scalar and gravitational perturba tions,\" Phys. Rev., D5, 2419-2438. [\u00a7\u00a735.l4. 41.11. Ex. 32.10. Box 32.2] Price. R. H.. 1972b. \"Nonspherical perturbations of relativistic zero-rest-ma,s gravitational II: fields.\" PI~\\\u00b7s. Rev.. D5, 2439-2454. [~41.l1. Box 32.2] Integer-spin. collapse. Price. R. H. See also Davis et al. 1971. Pringle. J. E., and M. J. Rees, 1972. \"Accretion disc models for compact X-ray sources,\" Astroll. Astrophys. 21, 1-9. [Box 33.3] Proclus of Athens, cited bl' name in Fig. 1.8. Ptolemy, C. See Taliaferro 1952. Ramsey. N. F.. 1972, \"History of atomic and molecular stand ards of frequency and time,\" L\u00a3.\u00a3.\u00a3. Trans. 1M-21. 90-98. [Box 16.3] Rayleigh. Lord. 1912, \"On the propagation of waves through a stratified medium, with special reference to the question of reflection,\" Proc. R. Soc. London A 86, 207-226. [\u00a722.5] Rayzer. Yu. P. See Zel'dovich and Rayzer 1967. Reber, G., cited by name in Box 27.7. Rebka, G. A. See Pound and Rebka 1960. Rees,,M. See Lynden-Bell and Rees 1971 and Pringle and Rees 1972. RefsdaI. S., 1964, \"The gravitational lens effect.\" Mon. Not. R. Astron. Soc. 128, 295-306. [\u00a725.6. Fig. 22.2. Box 33.3] Refsdal, S., R. Stabell, and F. G. de Lange, 1967, \"Numerical calculations on relativistic cosmological models,\" Mem. Roy. Astron. Soc., 71, 143-248. [\u00a729.4, Box 29.4] Regge, T., 1961, \"General relativity without coordinates.\" Nuovo Cimento 19, 558-571. [\u00a7\u00a742.1, 42.2, 42.4, 42.6] Regge, T., and J. A. Wheeler; 1957, \"Stability of a Schwarzs child singularity,\" Phys. Rev. 108, 1063-1069. [\u00a725.5, Box 33.3] Reichley, P. See Muhlemann and Reichley 1964 and 1965. Reik, H. G. See Meixner and Reik 1959. Reines, E, ed., 1972, Cosmolog}\\ FUSion, and Other Matters. Univ. of Colorado Press. Boulder. Reissner, H., 1916, \"Uber die Eigengravitation des elektrischen FeIdes nach der Einsteinschen Theorie,\" Ann. Phys. (Ger many) 50, 106-120. [Ex. 31.8] Reissner, H., cited by name in \u00a742.7. Retherford, R. C. See Lamb and Retherford 1947. Rhoades, C. E. J., 1971, Investigations in the Physics of Neutron Stars, doctoral dissertation, Princeton University. [\u00a724.2] Richelot, cited by name in Box 38.2. Riemann, G. F. B., achievements and picture in Box 8.5; f,)r collected works, see Weber 1953. i Riesz, M., 1949, \"L'integrale de Riemann-Liouville et Ie pn1 h. Ierne de Cauchy,\" Acta Math. 81, 1-223. [\u00a735.8] Rindler, W., 1969, Essential Relativity: Special, General, allli Cosmological, Van Nostrand, New York. [Box 27.7] Robb, A. A., 1914, A Theory of Time and Space, Cambridg.: Univ. Press, Cambridge, Eng. [Box 13.1] Robb, A. A., 1936, Geometry of Time and Space, Cambridge Univ. Press, Cambridge, Eng. Rev. ed. of Robb 1914. [Boxes 1.3, 13.1] BIBLIOGRAPHY AND INDEX OF NAMES 1247 Robb, A. A., quoted", ", epigraph of Chap. 4. Roberts, M. S., 1965, \"2 I-em. hydrogen measurements of NGC5668, a relatively distant galaxy,\" Astrophys. J. 142, 148-153. [\u00a729.2] Roberts, M. S. See also Dieter et al 1962. Robertson, H. P., 1935, \"Kinematics and world structure,\" Astrophys. J. 82, 248-301. [Ex. 27.5] Rudenko, V. N. See Braginsky, Zel'dovich, and Rudenko 1969. Ruderman, M., 1972, \"Pulsars: Structure and dynamics,\" Ann. Rev. Astron. Astrophys. 10, 427-476. [\u00a724.3] Ruffini, R. See Christodolou and Ruffini 1971; Davis et al. 1971; Ferrari and Ruffini 1969; and Hanni and Ruffini 1973. Ruffini, R., and J. A. Wheeler, 1970, \"Collapse of wave to Robertson, H. P., 1936, \"Kinematics and world structure,\" black hole,\" Bull. Am. Phys. Soc. 15, 76. [\u00a735.8] Astrophys. J. 83, 187-201, 257-271. [Ex. 27.5] Ruffini, R., and J. A. Wheeler, 1971a, \"Introducing the black Robertson, H. P., 1962, \"Relativity and cosmology,\" in Deutsch hole,\" Phys. Today 24, 30-36. [\u00a733.7] and Klemperer 1962, pp. 228-235. [\u00a7\u00a738.2, 39.3] Robertson, H. P., cited by name in \u00a730.3, Box 27.7. Robertson, H. P., and T. W. Noonan, 1968, Relativity and Cosmology, W. B. Saunders, Philadelphia. [\u00a727.11, Ex. 27.4, 29.6, endpapers] Robinson, H. G. See Hughes, Robinson, and Beltran-Lopez 1960. Robinscfu, I., 1959a, \"A solution of the Maxwell-Einstein equations,\" Bull. Acad Polon. Sci. 7, 351-352. [Ex. 32.1] Robinson, I., 1959b, unpublished report at Colloque Interna tionale sur les Theories Relativistes de la Gravitation, Abbaye de Royaumont. [Ex. 15.2] Robinson, I., 1961, \"Null Electromagnetic Fields,\" 1. Math. Phys., 2, 290-291. [\u00a722.5] Robinson, I., A. Schild, and E. Schiicking, eds., 1965, Quasi\u00b7 stellar Sources and Gravitational Collapse, The University of Chicago Press, Chicago. Robinson, I. See also Bondi, Pirani, and Robinson 1959, Douglas et al. 1969. Rohrlich, E, 1965, Classical Charged Particles; Foundations of Their Theory, Addison-Wesley, Reading, Mass. [\u00a74.7] Rohrlich, E, cited by name in \u00a720.6. Roll, P. G., cited by name in Box 27.7. See also Dicke et al 1965. - Roll, P. G., R. Krotkov, and R. H. Dicke, 1964, \"The equiva lence of inertial and passive gravitational mass,\" Ann. Phys. (U.S.A.) 26,442-517. [\u00a7\u00a71.3,!O.I, 38.3, Fig. 1.6, Box 1.2] Root, W. L. See Davenport and Root 1958. Rose, R. D., H. M. Parker, R. A. Lowry, A. R. KUh}thau, and J. W. Beams, 1969, \"Determination of the gravitational constant G, Phys. Rev. Lett. 23, 655-658. [\u00a740.8] Rosen, Nathan, Festschrift on The occasion of his 60th birth\u00b7 day, March 22, 1969. See Kuper and Peres 1971. Rosen, N. See also Peres and Rosen 1964. Rosenfeld, L., 1940, \"Sur Ie tenseur d'impulsion-energie,\" Mem. Acad R. Belg. Sci. 18, no. 6. [\u00a721.3] Rosenfeld, L. See also Bohr and Rosenfeld 1933. Rossi, B., cited by name in Box 27.7. Rossi, B., and D. B. Hall, 1941, \"Variation of the rate of decay of mesotrons with momentum,\" Phys. Rev. 59, 223-228. [\u00a738.4] Royaumont 1959, Les", " Theories relativistes de la gravitmion, Actes du Colloque International. organize,i Royaumont, du 21 au 27 juin 1959, Centre National de la Recherche Scientit1que. Paris, 1962. Ruffini, R., and J. A. Wheeler, 1971b, \"Relativistic cosmology and space platforms,\" in Proceedings of the Conference on Space Physics, European Space Research Organization, Paris, France. pp. 45-174. [Boxes 33.2, 33.3, 33.5, 36.3, Fig. 24.3, Ex. 37.13. 37.14] Runcom, S. K. See Ahrens, Press, and Runcorn 1964. Runge, J. See Sommerfeld and Runge 1911. Russ, N. See Boni, Russ, and Laurence 1960. Rutherford, E., cited by name in Preface, \u00a744.1. Ruzmaikin, A. A. See Ruzmaikina and Ruzmaikin 1969. Ruzmaikina, T. V., and A. A. Ruzmaikin, 1969, \"Quadratic corrections to the Lagrangian density of the gravitational field and the singularity,\" Zh. Eksp. & Teor. Fiz. 57, 680 685. English translation in Sov. Phys.-JETP 30, 372-374. [Box 17.2 part 6] Ryan, M. P., Jr., 1971, \"Qualitative cosmology: Diagrammatic solutions for Bianchi type IX universes with expansion, rotation, and shear, II: The general case,\" Ann. Phys. 68, 541-555. [\u00a730.7] Ryan. M. P., Jr., 1972a, Hamiltonian Cosmology, Springer, Berlin, Germany. [\u00a730.7, Box 30.1] Ryan, M. P., Jr., 1972b, \"The oscillatory regime near the singu larity in Bianchi type-IX universes,\" Ann. Phys. (U.S.A.) 70, 301-322. (Includes a report of work by V. A. Belinsky and I. M. Khalatnikov. [\u00a730.7, Box 30.1] Sachs. R. K., 1961, \"Gravitational waves in general relativity, VI: The outgoing radiation condition,\" Proc. R. Soc. Lon don A 264, 309-338. [\u00a722.5] Sachs, R. K., 1964, \"Gravitational radiation.\" in DeWitt and DeWitt 1964. [\u00a7\u00a72I.5, 2I.l3, 33.6, 4I.lI, Ex. 22.14, end papers] Sachs, R. K., ed., 1971, General Relativi(v and Cosmology, proceedings of course 47 of the International School of Physics \"Enrico Fermi\", Academic Press. [\u00a7\u00a727.2, 34.1, 40.7] Sachs, R. K. See also Eardley, Liang, and Sachs 1972, Ehlers, Geren, and Sachs 1968, and Jordan, Ehlers, and Sachs 1961. Sakharov. A. D., 1967, \"Vacuum quantum fluctuations in curved space and the theory of gravitation.\" Doklady Akad. Sauk S.S.S.R. 177, 70-71. English translation in Sov. Ph.rs. Doklad)' 12, 1040-1041 (1968). [\u00a744.4. Box 17.2 part 6] Sakharov, A. D.. 1969, unpublished work cited by Braginsky and Nazarenko 1971. [\u00a737.3. Fig. 37.2] Salam. A.. and E. P. Wigner, eds., 1972, Aspects of Quantum Ruark, A. E.. and H. C. Urey, 1930, Aloms. Jfolec/lles, and Theory. Cambridge Univ. Press. Cambridge, Eng. Qlul/lta. McGraw-Hill. New York. [\u00a7\u00a740.7. 41.4] Salpeter. E. E. and R. V. Wagoner. 1971, \"Structure of rela\u00b7 tivistic disks,\" ASlrophys.J. 164. 557-568. [Box 24.1] 1248 GRAVITATlONr Sandage, A., 1958. \"Current problems in the extragalactic distance scale.\" Astrophys. J. 127.513-526. [Box 27.1] Sandage. A.. 1968, \"The time scale for creation,\" in L. Woltjer, ed., Galaxies and the Universe. Columbia Univ. Press. New York, pp.", " 75-112. [\u00a729.7] Sandage. A., 1970. \"Main-,equence color magnitude diagrams. and ages for the globular clusters M3, M13, M15, and M92,\" Astrophys. J. 162, 841-870. [\u00a729.7] photometry. Sandage, A., 1972a, \"Distances to galaxies: the Hubble con stant, the Friedmann time. and the edge of the world.\" in Proceedings of the Symposium on the Gala.\\:1' and the Distance Scale, Essex. England. in press. [\u00a7\u00a727.1, 29.3. 29.4, Box 29.4] Sandage, A.\u2022 1972b. \"The redshift-distance relation, I. Angular diameter of first-ranked cluster galaxies as a function of the aperture correction to magnitudes,\" Astro redshift: phys. J. 173, 485-500. [\u00a729.5, Box 29.4] Sandage, A.. I972c, \"The redshift-distance relation, II. The Hubble diagram and its scatier for first-ranked cluster galaxies: a formal value for qo'\" Astrophys. J. 178, 1-24. [\u00a729.4, Box 29.4] Sandage, A., 1972d, \"The redshift-distance relation, Ill. Pho tometry and the Hubble diagram for radio sources and the possible turn-on time for QSO's,\" Astrophys. J. 178, 25-44. [\u00a729.4, Boxes 28.1, 29.4] Sandage, A., cited bl\" name in Boxes 27.7, 29.5. Sandage, A., G. A. Tamman, and E. Hardy, 1972, \"Limits on the local deviation of the universe from a homogeneous model,\" Astrophys. J. 172,253-263. [\u00a727.I] Sanders, R, 1970, \"The effects of stellar collisions in dense stellar systems,\" Astrophys. J. 162, 791-809. [\u00a7\u00a724.5, 24.6, Box 33.3] Sanderson, B. J. See Hsiang and Sanderson 1965. Sands, M. See Feynman, Leighton, and Sands 1964. Sard, R. D., 1970, Relativistic Mechanics, W. A. Benjamin, Ne\\Y York. [\u00a72.IO] Sargent, W. L. W. See Bahcall, Sargent, and Schmidt 1967. Schlifke, E W. See Meixner and Schlifke 1954. Schatzman, E. See DeWitt, Veron, and Schatzman 1967. Schiff, L. relativity: Theory and experiment,\" J. Izulust. & Appl. Math 10, 795-801. [\u00a7\u00a738.2, 39.3] I., 1962. \"General Schilpp, P. A.. ed.. 1949, Albert Einstein: Philosopher-Scielltist. Library of Living Philosophers, Evanston. III.: source of quotations frllm A. Einstein. K. Menger, and M. von Laue. [Box I.I I. epigraphs to \u00a71.2, Chaps. 2. 9] Schmidt, B. G.. 1971, \"A new definition of singular points in general relativity,\"Gen. ReI. Grav. 1, 269-280. [\u00a734.6] Schmidt. M., 1972. \"Statistical studies of the evolution of ex tragalactic radio sources, I. Quasars,\" Astrophys. J. 176, 273-287. [Box 28.1] Schmidt. M. Cited by name ill Box 27.7, Ex. 29.7. See also Bahcall, Sargent. and Schmidt 1967 and Bahcall and Schmidt 1967. Schoch, C. See Neugebauer and Schoch 1930. Schouten. J. A.. 1954, Ricci-calculus, Springer, Berlin, Ger many. [\u00a721.5, Ex. 17.8,21.8, endpapers] Schramm. D. N. See Wasserburg, Schramm. and Huneke 1969. Schrodinger, E., 1950, Space-Time Structure, Cambridge Univ. Press, England. [endpapers] Schiicking, E. See also Douglas et al. 1969 and Robinson. Schild, and Schiicking 1965. Schiicking, E., and O. Heckmann, 1958, \"World models,\" in Onzieme Conseil de Physique Solvay, Editions Stoops, Brussels, pp. 149-158. [\u00a7\u00a730.2, 30.3] Schutz, B.,", " cited by name in Ex. 35.20. Schutz, S., 1966, statement in January 1966 final examination in course in relativity, Princeton University. [Epigraph to Chap. 16] Schwartz, J. T. See Dunford and Schwartz 1958. Schwartz, L., 1950-1951, Theorie des distributions, vols. I and 2, Hermann, Paris. [\u00a74.7] Schwarzschild, K., 1916a, \"Uber das Gravitationsfeld eines Massenpunktes nach der Einsteinschen Theorie,\" Sitzber. Deut. Akad. Wiss. Berliiz;- Kl. Math.-Phys. Tech., 189-196. [Preface, \u00a7\u00a714.1, 23.6] Schwarzschild, K., 1916b, \"Uber das Gravitationsfeld einer Kugel aus inkompressibler Flussigkeit nach der Einstein schen Theorie,\" Sitzber. Deut. Akad. Wiss. Berlin, Kl. Math.-Phys. Tech., 424-434. [\u00a723.2, Box 23.2] Schwarzschild, K., cited by name in \u00a7\u00a742.l, 42.7. Schwinger, J., 1948, \"Quantum electrodynamics, I: A covariant formulation,\" Phys. Rev. 74, 1439-1461. [Box 15.1] Schwinger, J., ed., 1958, Selected Papers on Quantum Electro dynamics, Dover, New York. Schiff, L. I., 1967, \"Comparison of theory and observation in Schwinger, J., 1963a, \"Energy and momentum density in field general relativity,\" in Ehlers 1967. [\u00a7\u00a738.2, 39.3] theory,\" Phys. Rev. /30, 800-805. [Ex. 40.8] Schiffer, M. See Adler, Bazin, and Schiffer 1965. Schild, A., 1960, \"Time,\" Texas Quarterly, 3, no. 3, 42-62. [\u00a77.3] Schild, A., 1962, \"Gravitational theories of the Whitehead type and the principle of equivalence,\" in Mj6ller 1962. [\u00a7\u00a77.3, 38.5] Schi.ld, A., 1967, \"Lectures on General Relativity Theory,\" in Ehlers 1967, pp. 1-105. [\u00a77.3, endpapers] Schild, A., 1970, \"Tearing geometry to pieces: More on con formal geometry,\" unpublished lecture at Jan. 19 Prince ton University relativity seminar. [Box 10.2] Schild, A., cited by name in \u00a71O.3, Box 17.3, Ex. 10.6. See also Douglas et al. 1969, Ehlers, Pirani, and Schild 1972, Infeld and Schild 1949, Kerr and Schild 1965, and Robinson, Schild, and Schiicking 1965. Schwinger, J., 1963b, \"Quantized gravitational field,\" Phys. Rev. 130, 1253-1258. [Ex. 40.8] Sciama, D. W., 1957, \"Inertia,\" Sci. American 196, qo. 2, 99- 109. [\u00a721.12] \u2022 Sciama, D. W., 1961, The Unity of the Universe, Anchor, New York. [\u00a721.12] Sciama, D. W., 1971, Modern Cosmology, Cambridge Uni\\\u00b7. Press, Cambridge, Eng. [\u00a728.1, Box 27.7] Sciama, D. W., cited by name in Boxes 34.2, 34.3. Scott, P. E, cited by name in Box 27.7. See also Hewish et al. 1968. Sears, E W., 1953, An Introduction to Thermodynamics, the Kinetic Theory of Gases, and Statistical Mechanics, Addi son-Wesley, Reading, Mass. [\u00a723.4] BIBLIOGRAPHY AND INDEX OF NAMES 1249 Seattle 1967, Battelles Rencontres. See DeWitt and Wheeler 1968. Seckler, B. D. See Keller, Lewis, and Seckler 1956. Seelig, c., ed., 1956. See Kollros 1956. Seielstad, G. A., R. A. Sramek, and K. W. Weiler, 1970, \"Measurement of the deflection of 9.602-GHz radiation from 3C279 in the solar gravitational field,\" Phys. Rev. Lett. 24, 1373-1376. [Box 40.1] Sejnowski, T. J., 1973, personal communication. [Ex. 15.2] Shakespeare, W., quoted at end of Chap. 44. Shakura, N. I., and R. A", ". Sunyaev, 1973, \"Black holes in binary systems: Observational appearance,\" Astron. Astrophys. 24, 337-355. [Box 33.3] Shannon, c., 1938, \"A symbolic analysis of relay and switching circuits,\" Trans. Am. Inst. Elect. Eng. 57, 713-723. [\u00a744.5] Shapiro, I. I., 1964, \"Fourth test of general relativity,\" Phys. Rev. Lett. 13, 789-791. [\u00a740.4] Shapiro, Ii I., 1967, \"New method for the detection of light deflection by solar gravity,\" Science 157, 806-808. [\u00a740.3] Shapiro, I. I., 1968, \"Fourth test of general relativity: prelimi nary results,\" Phys. Rev. Lett. 20, 1265-1269. [\u00a740.4, Box 40.2] Shapiro, I. I., 1970, results reported at Third Cambridge Con ference on Experimental Relativity, New York City, June 1970. [Box 40.3] Shapiro, Shapiro, I. I., 1971a. \"Testing general I., 1971b. \"Testing general relativity: plOgress, problems, and prospects,\" in Davies, 1971, pp. 136-147. [Box 40.3] I. relativity: progress, problems, and prospects,\" lecture at Sixth International Conference on Gravitation and Relativity, published by NORDITA (Copenhagen) in multilith form for distribu tion at conference, also published in Gen. ReI. Grav. 3, 135-148 (1972). [\u00a739.3, Box 40.3] Shapiro, I. I., M. E. Ash, D. B. Campbell, R. B. Dyce, R. P. Ingalls, R. F. Jurgens, and G. H. Pettengill, 1971, \"Fourth test of general relativity: New radar result,\" Phys. Rev. Lett. 26, 1132-1135. [Box 40.2] Shapiro, I. I., W. B. Smith, M. E. Ash, R. P. Ingalls,. and G. H. Pettengill, 1971, \"Gravitational constant: Experimental bound on its time variation,\" Phys. Rev. Lett. 26, 27-30. [\u00a740.8] Shapiro, I. I., G. H. Pettengill, M. E. Ash, R. P. Ingalls, D. B. Campbell, and R. B. Dyce, 1972, \"Mercury's perihelion advance: determination by radar,\" Phys. Rev. Lett. 28, 1594-1597. [Box 40.3] Shapley, H., cited by name in Box 27.7. Sharp, D. H., 1960, \"One- and two-surface formulations of the boundary-value problem for Einstein-Maxwell theory and for 'already unified' field theory,\" Physics A. B. senior thesis, Princeton University, unpublished; deposited in Fine Library,. Princeton University. [\u00a721.9] Sharp, D. H., cited by name in \u00a721.1. See also Baierlein, Sharp. and Wheeler 1962, Hartle and Sharp 1965, alld Misner and Sharp 1964. Sheldon. S. E., 1971, \"A complete redshift-magnitude for mula,\" Astron. J. 76, 751-755. [\u00a729.4] Shelest, V. P., ed., 1972, Gravitats(ra: Problenzi i Perspektivi: pan~rati Alekseya Zinovievicha Petrova posvashaetsya, Naukova Dumka, Kiev. Shields, M. C. See Boni, Russ, and Laurence 1960. Shimmins, A. J., cited by name in Box 27.7. Shirkov, D. V. See Bogoliubov, N. N. Shu, F. H. See Lin, Mestel, and Shu 1965. Siegel, C. L., 1956, Vorlesungen uber Himmelsmechanik, Springer, Berlin. [Box 25.4] Siegel, C. L., and J. K. Moser, 1971, Lectures on Classical Mechanics, translated by C. I. Kalme, Springer, Berlin and New York, 1971. [Box 25.4] Silverberg, E. C. See Bender et al. 1971. Simpson, O. C. See Estermann, Simpson, and Stem 1938. Singer, D. W., 1950, Giordano Bruno, His Life and Thought with Annotated Translation of His Work, On the Infinite Universe and Worlds, Abelard-Schuman, New York.", " [Box 27.7] Slee, O. B., cited by name in Box 27.7. Slipher, V. M., cited by name in Box 27.7. Smith, J. H. See Frisch and Smith 1963. Smith, W. B. See Shapiro, Smith et al. 197I. Snider, J. L. See Pound and Snider 1964 and 1965. Snyder, H. See Oppenheimer and Snyder 1939. Soding, P., J. Bartels, A. Barbaro-Galtieri, J. E. Enstrom, T. A. Lasinski, A. Rittenberg, A. H. Rosenfeld, T. G. Trippe, N. Barash-Schmidt, C. Bricman, V. Chaloupka, and M. Roos, 1972, \"Review of particle properties,\" Physics Letters, 39B, 13. [endpapers] Solvay 1958, Onzieme Conseil de Physique Solvay, held at the Instituts Solvay, Brussels; proceedings: La Structure et revolution de l'univers, Editions Stoops, Brussels, Belgium. Sommerfeld, A., and J. Runge, 1911, \"Anwendung der Vektor rechnung auf die Grundlagen der geometrischen Optik,\" Ann. Phys. (Germany) 35, 277-298. [\u00a722.5] Spencer, D. C. See Nickerson, Spencer, and Steenrod 1959. Spitzer, L., Jr., 1971, \"Dynamical evolution of dense spherical star systems,\" in Semaine d'etude sur les noyeaux des gal axies 1970 (Pontificae Academiae Scientarum Scripta Varia no. 35). [\u00a7\u00a724.5, 24.6, Box 33.3] Spivak, M., 1965, Calculus on Manifolds, W. A. Benjamin, New York. [\u00a74.I, Box 4.l] Sramek, R. A., 1971, \"A measurement of the gravitational deflection of microwave radiation near the sun, 1970 Oc tober,\" Astrophys. J. Lett. 167, L55-L60. [Box 40.1] Sramek, R. A. See also Seielstad, Sramek, and Weiler 1970. St. Patrick. See Stokes 1887. Stabell, R. See Refsdal, Stabell, and de Lange 1967. Stanley, G. J., cited by name in Box 27.7. Starobinsky, A. A. See Zel'dovich and Starobinsky 1971. Steenrod, N: E. See Nickerson, Spencer, and Steenrod 1959. Stehle, P. See Corben and Stehle 1960. Steigman, G., 1971, \"Antimatter and cosmology,\" in Sachs 1971, pp. 373-382. [\u00a7\u00a727.l1, 28.4] Stephenson, F. R., contributed Fig. 1.8. Stern. O. See Estermann, Simpson, and Stern 1938. Sternberg, So, 1969, Celestial Mechanics, W. A. Benjamin, New York. [\u00a78.2] Steuckelberg, E. C. Goo and G. Wanders, 1953, \"Thermo dynamique en relativite general.\" Helv. Phys. Act., 26, 307-316. [\u00a722.2] Stewart, J. M., 1971..vall-Equilibrium Relalivistic Kinetic The 0':1', lecture notes in Phys!cs 10, Springer-Verlag, Berlin. [\u00a722.6, Ex. 22.7] 1250 GRAVITATION Stewart. J. M., cited hI' name in \u00a722.3. Stewart, J. M.. and G.\u00b7F. R. Ellis. 1968, \"Solutions of Einstein's equations for a fluid which exhibils local rotational sym metry.\" J. Math. Php. 9, 1072-1082. [Ex. 32.1] Stockton, A. N., cited by name in Box 27.7. Stokes, G. c.. cited bl' name in Box 4.6. Stokes, W., 1887, Th~ Tripartiie Life of St. Patrick With Other Documents Relating to That Saint, Her Majesty's Station ery Office, London. [\u00a719.4] Stratimyer, H., cited by name in Box 16.3. Struve, F. G. W., cited bl' name in Box 27.7. Sunyaev, R. A. See Shakura and Sunyaev 1973. Susskind, L. See Aharonov and Susskind 1967. Syman", "zik, K., 1963, \"Aspects of Euclidean field theory and Feynman path integrals,\" in Lurie and Mukunda 1963. [Box 21.1] Synge, J. L., 1950, \"The gravitational field of a particle,\" Proc. R. Irish Acad A 53, 83-114. [\u00a731.5, Box 31.1] Synge, J. L., I960a, \"On certain identities connected with the Irish Acad. A 61, 29-36. [Ex. Einstein tensor,\" Proc. R. 35.14] Synge, J. L., I960b, RelatiVity: The General Theory, North Holland, Amsterdam. [endpapers] Synge, J. L., 1962, \"Tensorial integral conservation laws in general relativity,\" in CNRS 1962, pp. 75-83. [Box 15.3] Synge, John Lighton, Festschrift on the occasion of his 75th birthday, March 23, 1972. See O'Raifeartaigh 1972. Szekeres, G., 1960, \"On the singularities of a Riemannian manifold,\" Publ. Mat. Debrecen 7, 285-301. [\u00a731.4, Box 3I.l] Taliaferro, R. c., trans. 1952, Ptolemy, The Almagest, in Great Books ofthe Western World, Encyclopedia Brittanica, Inc., Chicago, III., Volume 16. [Box 27.7] Tamman, G. A., cited by name in Box 29.4. See also Sandage, Tamman, and Hardy 1972. Tannenbaum, M. See Farley et al. 1966. Tartaglia, N. F., 1560, General Trattato dei Numeri et Misnre, Cited in Enzyklopiidie der Mathematischen Wissenschaften, vol. Ill, part i, p. 1056, Leipzig 1914. [Box 13.1] Taub, A. H., 1948, \"Relativistic Rankine-Hugonoit equations,\" Phys. Rev., 74, 328-334. [\u00a722.2, Box 22.2] Taub, A. H., 1965, \"Equations of motion of test particles,\" in London 1965. [\u00a740.9] Taub, A. H., 1971, \"Relativistic Hydrodynamics,\" Studies in Applied Mathematics-Math. Assoc. Amer. Studies in Math. 7, 150-180. [\u00a722.3] Taub, A. H. See also Misner and Taub 1968. Tauber, G. E., and J. W. Weinberg, 1961, \"Internal state of a gravitating gas,\" Phys. Rev. 122, 1342-1365. [\u00a722.6] Taylor, D. J., cited by name in Box 34.3. See also Cocke and Taylor 1969. Taylor, E. F., and J. A. Wheeler, 1966, Spacetime Physics, W. H. Freeman, San Francisco, Calif. [\u00a72.1, Fig. 2.7, Ex. 6.9] Tbilisi, USSR, 1968. Fifth International Conference on Gravi tation and the Theory of Relativity, TbiIisi State Univer sity, August 1968. See Fock 1968. Technion 1969, International Seminar on Relativity and Gravi- tation held at Technion City. Israel, in July 1969. in honor of the 60th birthday of Professor :-J'athan Rosell. St'e Kuper and Peres 1971. Teitelboim. c., 1970, \"Splitting of the t\\laxwell tensor: radia tion reaction without ad\\'anced fields.\" Pin's. Rev. D 1, 1572-1582. [\u00a720.6, Ex. 20.8]. Teitelboim, c., 1971, \"Radiation reaction as a retarded self interaction.\" P/~l's. Reu. D 4, 345-347. [\u00a720.6] Teitelboim, c., 1972a, \"Nonmeasurability of the baryon num II, no. 3, ber of a black hole,\" Nnovo Cimellto Lett., 326-328. [Box 25.2, Boxes 32.2, 33.1] Teitelboim, c., 1972b, \"Nonmeasurability of the lepton num ber of a black hole,\" /Vuovo Cimento, II 3, 397-400. [Boxes 25.2, 33.1] Teitelboim, C., I972c, \"Nonmeasurability of the Quantum Numbers of", " a Black Hole,\" Phys. Rev. D 5, 2941-2954 [Boxes 25.2, 32.2, 33.1] Teitelboim, C., cited by name in \u00a72I.l, Fig. 4.6. Teller, E., 1948, \"On the change of physical constants,\" Phys. Rev. 73, 801-802. [\u00a738.6] Teller, E., cited by name in Box 38.2. See also De Hoffman and Teller 1950. Tempier, E., cited ~l' name in Box 27.7. Teukolsky, S. A., 1972. \"Rotating black holes: Separable wave equations for gravitational and electromagnetic perturba tions,\" Phys. Rev. Lett. 29, 1114-1118. [\u00a74l.II, Box 33.3, Ex. 33.14] Teukolsky, S. A., 1973, \"Perturbations of a rotating black hole, electro I: Fundamental magnetic, and neutrino-field perturbations,\" Astrophys. J., in press. [Ex. 33.14] for gravitational, equations Teukolsky, S. A. See also Bardeen. Press, and Teukolsky 1972 and Press and Teukolsky 1972 and 1973. Texas Symposia on Relativistic Astrophysics: 1963, see Robin son, Schild, and Schiicking 1965 and Harrison et aL 1965: 1964, see Douglas et al. 1969; 1972, see Trans. N. Y. Acad. Sci. Thirring, H., 1918, \"Uber die Wirkung rotierender ferner Massen in der Einsteinschen Gravitationstheorie,\" Phys. Z. 19, 33-39. [\u00a72l.I2] Thirring, H., 1921, \"Berichtigung zu meiner Arbeit: \"Uber die Wirkung rotierender ferner Massen in der Einsteinschen Gravitationstheorie,\" Phys. A. 22, 29-30. [\u00a721.12] Thirring, H., and J. Lense, 1918, \"Uber den Einfluss der Eigenrotation der Zentralkorper auf die Bewegung der Planeten und Monde nach der Einsteinschen Gravita tionstheorie,\" Phys. Z. 19, 156-163. [\u00a7\u00a72l.I2, 40.7] Thirring, W. E., 1961, \"An alternative approach t~ the theory of gravitation,\" Ann. Phys. (U.S.A.) 16, 96-117. [\u00a7\u00a77.1. 18.1] Thomas, L. H., 1927, \"The kinematics of an electron with an axis,\" Phtl. Mag. 3, 1-22. [\u00a7\u00a740.7, 41.4, Ex. 6.9] Thomson, A. See Farnsworth et al. 1972. Thomson, J. J., 1907, Electricity and Matter, Archibald Con stable, London, Chapter III. [Fig. 4.6] Thomson, J. J., cited by name in Fig. 4.6. Thomson, W., Lord Kelvin, 1889, Popular Lectures and Ad dresses, Vol. I, Macmillan, London. [Epigraph to Chap. 37] Thomson, W., Lord Kelvin, 1904, Baltimore Lectures on Mo- BIBLIOGRAPHY AND INDEX OF NAMES 1251 lecular Dynamics and the Wave Theory ofLight, C. J. Clay, London. [Preface] Thomson, W., Lord Kelvin, cited by name in Box 17.2, quoted in epigraph of Chap. 20. Thome, K. S., 1967, \"The general relativistic theory of stellar structure and dynamics,\" in DeWitt, Schatzman, and Veron 1967, pp. 259-44 I. [\u00a724.4, endpapers] Thorne, K. S., 1969a, \"Nonradial pulsation of general Ill: Analytic and numerical relativistic stellar models, results for neutron stars,\" Astrophys. J. 158, 1-16. [\u00a724.3, Box 36.1] Thome, K. S., 1969b, \"Nonradial pulsation of general-relati vistic stellar models, IV: The weak-field limit,\" Astrophys. J. 158, 997-1019. [\u00a736.8, Box 36.1] Thorne, K. S., 1971, \"Relativistic stars, black holes, and gravi tational waves,\" in Sachs 1971, pp. 238-283. [\u00a740.7] Thorne, K. S., 1972, \"", "Nonspherical gravitational collapse, a short review,\" in Klauder 1972, pp. 231-258. [Box 32.3, Ex. 32.10] Thorne, K. S., 1973a. \"Relativistic shocks: The Taub adiabat,\" Astrophys. J. 179, 897-907. [\u00a722.2, Box 22.2] Thorne, K. S., 1973b, \"Black-hole models for compact x-ray sources,\" Astrophys. J., in press. [Box 33.3] Thorne, K. S. See also Ames and Thorne 1968, Bardeen, Thorne, and Meltzer 1966, Bisnovatyi-Kogan and Thome 1970, Fackerell, Ipser, and Thorne 1969, Fletcher et al. 1967, Harrison et al. 1965, Hartle and Thorne 1968 and 1974, Ipser and Thorne 1968, Meltzer and Thorne 1966, Morgan and Thome 1973, Novikov and Thome 1973, Press and Thorne 1972, and Trimble and Thorne 1969. Thorne, K. S., W. T. Ni, and C. M. Will, 1971, \"Theoretical frameworks for testing relativistic gravity: A review,\" in Davies 1971, pp. 10-31. [\u00a7H.6, Box 39.5] Thorne, K. S., and C. M. Will. 1971, \"Theoretical frameworks for texting relativistic gravity, I: Foundations,\" Astrophys. J. 163, 595-610. [\u00a7\u00a738.2, 39.2] Tikhonov, M. Yu. See Braginsky, Manukin. and Tikhonov 1970. Tobey, R. G., et al. 1967, PLj1 Fonnac Interpreter, Users Ref erence Manual IBM Controlled Program Library 360D 03.3004, Hawthorne, N.Y. [Box 14.3] Tolman, R. c., 1934a, RelatiVity, Thermodynamics, and Cos mology, Clarendon Press, Oxford. [\u00a744.1, Ex. 22.7, end papers] Tolman, R. c., 1934b, \"Effect of inhomogeneity on cosmOlogi cal models,\" Proc. Nat. Acad. Sci. U.S. 20, 169-176. [\u00a730.6, Ex. 32.8] Tolman, R. c., 1939, \"Static solutions of Einstein:s field equa tions for spheres of fluid,\" Ph),s. Rev. 55, 364-373. [\u00a723.2, Ex. 23.9] Tomonaga, S., 1946, \"On a relativistically invariant formulation of the quantum theory of wave fields,\" Prog. Theor. Phys. 1, 27-42, reprinted in Schwinger 1958. [Boxes 15.1, 2I.l, Ex. 43.5] Tonnelat, M. A., quoted at end of Chap. 44. Torrence, R. J. See Couch et a1 1968 and Newman et al. 1965. Trautman, A., 1965, \"Foundations andi current problems of general relativity theory,\" in Trautman, Pirani, and Bondi 1965. [\u00a712.1, Boxes 8.1. 12.4, endpapers] Trautman, A., 1972, \"On the Einstein-Carlan equations, I,\" Bulletin de I'Academie Polonaise des Sciences (math, astr., phys.) 20, 185-190. [\u00a7\u00a738.2, 39.2] Trautman, A., F. A. E. Pirani, and H. Bondi, 1965, Lectures on General Relativi~v, Brandeis 1964 Summer Institute on Theoretical Physics, vol. I, Prentice-Hall, Englewood Cliffs, N.J. Treder, H. J., 1962, Gravitative Stosswellen, Nichtanalytischen Wellenlosungen der Einsteinschen Gravitationsgleichungen, Akademie-Verlag, Berlin. [\u00a72I.l3] Treder, H. J. See also Papapetrou and Treder 1959 and 1962. Trimble, V. L., and K. S. Thome, 1969, \"Spectroscopic binaries and COllapsed stars,\" Astrophys. J. 156, 1OI3-1019. [Box 33.3] Trumpler, R. J. See Campbell and Trumpler 1928. Truran, J. See Cameron and Truran 1971. Truran, J., and A. G. W. Cameron, 1971", ", \"Evolutionary models of nucleosynthesis in the galaxy,\" Astrophys. & Space Sci. 14, 179-222. [Box 33.3] Tukey, J. W. See Blackman and Tukey 1959. Turkevich, A. See Fermi and Turkevich 1950. Turner, K. c., and H. A. Hill, 1964, \"New experimental limit on velocity-dependent interactions of clocks and distant matter,\" Phys. Rev. B 134, 252-256. 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R., 1887, \"Canon der Finsternisse.\" Sit:ber. /Vie/l Jlath.-SQI. K1. 52. i-xviii, 1-321. Kais. Akad. Wiss. [Fig. 1.8] von Neumann. 1.. cited by /lame i/l Box 38.2. Vourloumis, Hypatia, 1971, statement made at Knossos, May 1971, while serving as guide on Swan's Hellenic tour. Quoted at end of Chap. 44. Vyasa, about 400 A.D., The Mahabharata, translated from the Sanskrit by P. Lal, Writers Workshop, Calcutta. [Epigraph to \u00a71.5] Waage, H., cited by /lame in Fig. 1.7. Wagoner. R. V., 1967, \"Some effects of an intervening galaxy on the radiation from very distant objects,\" Astrophys. J. 149, 465-480, [Ex. 29.8] Wagoner, R. V., cited by name in Box 27.7. See also Bardeen and Wagoner 1971 and Salpeter and Wagoner 197I. Wagoner, R. V., W. A. Fowler, and F.", " Hoyle, 1967, \"On the synthesis of elements at very high temperatures,\" Astro phys. J. 148,3-49. [\u00a728.1] Wakano, M. See Harrison et al. 1965 and Klauder et al. 1959. Wald, R. M., 1971. See Cohen and Wald 1971. Walker, A. 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S. Burnett, 1968, \"The status of isotopic age determinations on iron and stone meteorites,\" in P. M. Millman, ed., Meteorite Research, Reidel, Dor drecht, Holland, pp. 467-479. [\u00a729.7, Box 29.3] Wasserburg, G. J., D. N. Schramm, and J. C. Huneke, 1969, \"Nuclear chronOlogies for the galaxy,\" Astrophys. J. 157, L91-L96. [\u00a729.7] Weber, H., ed., 1953, B. Riemann: Gesammelte Mathematische Werke, 2d ed., paperback reprint, Dover, New York. The Habilitationsvorlesung of 1854 was translated into English by W. K. Clifford, Nalure, 8, 14 (1873). [Box 8.5] Weber, J., 1960, \"Detection and generation of gravitational waves,\" Phys. Rev. 117, 307-313. [\u00a737.3] Weber, J., 1961, General Relativity and Gravitational Waves, Wiley-Interscience, New York. [\u00a737.3, endpapers] Weber. 1.. 1967, \"Gravitational radiation,\" Phys. Rev. Lett. 18, 498-50 I. [\u00a737.3, Fig. 37.2] Weber, 1.. 1969, \"Evidence for discovery of gravitational radia tion,\" Phys. Rev. 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[\u00a7\u00a721.8, 2l.l2, Ex. 21.23] Wheeler, J. A., 1970, \"Superspace,\" in Gilbert and Newton 1970, pp. 335-378. [\u00a743.1] Wheeler, J. A.. 1971a, notebook entry, \"Pregeometry and the calculus of propositions,\" 9: 10 A.M., April 10; seminar, De partment of Mathematics, Kings College, London, May 10; letter to L. Thomas, \"Pregeometry and propositions,\" June II, unpublished. [Box 44.5] Wheeler.]. A.. 1971 b, \"Transcending the law of conservation of leptons,\" in Atti del Convegno Intemazionale suI Tema: The Astrophysical Aspects of the Weak Interactions; Qua demo N. 157, Accademia Nazionale dei Lincei, Roma, pp. 133-164. [\u00a7\u00a74l.lI, 44.6, Box 25.2] Wheeler, J. A., 1971c, \"From Mendeleev's atom to the col lapsing star,\" 189-233 in A tti 7iel Convegno Mendeteeviano, Accademia delle Scienze, Torino; reprinted in Trans. New York Acad. Sci., 33, 745-779 (1971). See also Wyler 1971. [\u00a728.3, Box 25.2] Wheeler, J. A., 1971d, \"Mechanisms for jets,\" in Study week on nuclei of galaxies, Pontificiae Academiae Scientarum Scripta Varia no. 35, 539-567. [Box 33.3] Wheeler, John Archibald, Festschrift on the occasion of his sixtieth birthday, JUly 9, 1971. See Klauder 1972. Wheeler,]. A., cited by name in \u00a721.l, Box 8.5. See also Baier lein, Sharp, and Wheeler 1962, Brill and Wheeler 1957, DeWitt and Wheeler 1968, Douglas et al. 1969, Ford and Wheeler 1959a and 1959b, Fuller and Wheeler 1962, Har rison et al. 1965. Harrison and Wheeler 1958, Klauder et al. 1959, Lindquist and Wheeler 1957, Marzke and Wheeler 1964, Misner and Wheeler 1957 and 1972, Power and Wheeler 1957, Regge and Wheeler 1957, Ruffini and Wheeler 1970, 1971a. and 1971b, and Taylor and Wheeler 1966. Wheeler. 1. A., and R. P. Feynman, 1945. \"Interaction with the absorber as the mechanism of radiation,\" Rev. Mod. Phys.17, 157-181. [\u00a74.7, Box 2.1] Wheeler, J. A.. and R. P. Feynman. 1949, \"Classical electro inter-particle action,\" Rev. dynamics in terms of direct Mod. Phys. 21, 425-434. [\u00a74.7, Box 2.1] White, R. H. See COlgate and White 1966 and May and White 1966 alld 1967. Univ. Press, Cambridge, Eng. [\u00a7\u00a738.2, 39.1, 39.3, 39.8, Box 17.3] Whitehead, A. N., 1929, The Function of Reason, Princeton Univ. Press, Princeton, N.J. [Epigraph to Chap. 28] Whitehead, J. H. C. See Veblen and Whitehead 1932. Whitrow, G. J., and G. E. Morduch, 1965, \"Relativistic theories of gravitation, a comparative analysis with particular ref in Vistas in Astronomy 6, erence to astronomical 1-67. [\u00a738.2] tests, Whittaker, E. R., 1904, \"On an expression of the electro magnetic field due to electrons by means of two scalar potential functions,\" Proc. R. Soc. London 1, 367-372. [Ex. 4.10] Wigner. E. P., 1960, \"The unreasonable effectiveness of mathe matics in the natural sciences,\" Commun. Pure & Appl. Math. 13, 1-14, reprinted in Symmetries and Reflections: Scientific Essays of Eugene P. Wigner, W. J. Moore and M. Scriven, eds., Indiana Univ. Press, Bloomington, Ind., 1967. [\u00a744.2] Wigner, E. P., cited by name in Box 38.2. See also Bargmann and Wigner 1948 and Salam and Wigner 1972. Wilkins, D., 1972, \"Bound geodesics in the", " Kerr metric,\" Phys. Rev. D 5, 814-822. [Box 33.5] Wilkinson, D. T., cited by name in Box 27.7. See also Bender et al. 1971 and Dicke et al. 1965. Will, C. M., 1971a, \"Relativistic gravity in the solar system, I: Effect of an anisotropic gravitational mass on the earth-moon distance,\" Astrophys. 1. 165, 409-412. [Box 39.5] Will, C. M., 1971b, \"Relativistic gravity in the solar system, II: Anisotropy in the Newtonian gravitational constant,\" Astrophys.1. 169, 141-156. [\u00a7\u00a738.2. 39.1,40.8, Boxes 17.,3, 39.5] Will, C. M., 1971c, \"Theoretical frameworks for testing rela tivistic gravity, II: Parametrized post-Newtonian hydro dynamics, and the Nordtvedt effect,\" Astrophys. J. 163, 611-628. [\u00a7\u00a738.2. 39.3, 40.9, Box 39.5] Will, C. M., 1971d, \"Theoretical frameworks for testing rela tivistic gravity, III: Conservation laws, Lorentz invariance, and values of the PPN parameters.\" Astrophys. J. 169, 125-140. [\u00a739.9, Box 39.5, Ex. 39.15] Will, C. M., 1972. \"The theoretical tools of experimental gravi tation,\" Caltech Orange-Aid Preprint 289, to be published in Bertotti 1973. [\u00a7\u00a738.2, 39.3, Box 39.5] Will, C. M., 1973. \"Relativistic gravity in the solar system, III: Experimental disproof of a class of linear theories of gravitation.\" Astrophys. J., in press. [\u00a7\u00a739.3, 39.8] Will. C. M. See also Nordtvedt and Will 1972, Thorne, Ni, and Will 1971, alld Thorne and Will 1971. Will. C. M.. and K. Nordtvedt. Jr.. 1972. \"Conservation laws and preferred frames in relativistic gravity, I: Preferred frame theories and an extended PPN formalism.\" Astro phys. J. 177,757-774. [\u00a7\u00a738.2, 39.2. 39.3. 39.9.40.8. Boxes 39.1.39.5] Willey. R. See Klauder et al. 1959. Williams. R. W. See Bailey et al 1968. Willmore. T. 1.. 1959. An Imrodllctioll to Differential Geometry, Clarendon Press. OXford, [Box 8.1] Wilson. O. C. See Minkowski and Wilson 1956. Wilson. R. W.. cited by name in Box 27.7. See also Penzias and Whitehead. A. N.. 1922, The Principle of Relativity. Cambridge Wilson 1965. 1254 GRAVITATION \\Vim,or. F.. 'iuoted at end of Chap. 44. Wintner. A.. 1941. Ana~l'tical Foundations of Celestial.\\fe thanies. Princeton U niv. Press. Princeton. N.J. [Box 25.4] Witte born. F. c.. and W. M. Fairbank. 1967. '\"Experimental. comparison of the gravitational force on freely falling electrons and metallic electrons.\" Phs. Ret\u00b7. Lett. 19. 1049-1052. [\u00a71.3]. Witteborn. F. c.. L. V. Knight. and W. M. Fairbank. 1965. \"Use of low-temperature techniques to measure gra\\'ita tional forces on charged particles.\" in Proceedings of the 9th International Conference on Low-Temperature Physics, Columbus 1964. Plenum Press. New York, 1965, part B, pp. 1248-1252. [\u00a71.3} Witten. L., ed.. 1962. Gravitation: An 1ntroduction to Current Research. Wiley. New York. Witten. L. See also CarmeIi, Fickler. and Witten 1970. Wong, c.-Y.. 1971. \"Application of Regge calculus to the Schwarzschild and Reissner-Nordstrom geometries at the moment of time-symmetry,\" J. Math. Ph)'s. 12, 70-78. [\u00a7\u00a742.6, 42.7] Woolf, N. See Douglas et al. 1969. Wright, J. P., 1964, \"General relativistic instability,\" Phys", ". Rev. B 136, 288-289. [Box 26.2] Wright, T., 1750, An Original Theory or New fi.\\pothesis of the Universe, H. Chapelle, London. Reprinted in Munitz 1957. [Box 27.7] Wyler, J. A. (pseudonym), 1971, \"Raspulin, science, and the transmogrification of destiny,\" preprint bearing the Princeton University seal. York, J. W., Jr., 1971, \"Gravitational degrees of freedom and the initial-value problem,\" Phys. Rev. Lett. 26, 1656-1658. [\u00a7\u00a721.11, 21.12, Ex. 21.22} York, J. W., Jr., 1972a, personal communication. [Ex. 21.8, 2.1.9. 21.24} York. J. W., Jr., 1972b, \"The role of conformal 3-geometry in the dynamics of gravitation,\" Ph)'s. Rev. Letters 28, 1082 1085. [\u00a7\u00a721.6. 21.11. Ex. 21.24] York, J. W., Jr.. 1973a, \"Conformally invariant orthogonal decomposition of symmetric tensors on Riemannian man ifolds and the initial-value problem of general relativity,\" J. Math. Phys. 14, in press. [\u00a72I.lI] York, J. W., Jr., 1973b, personal communication. [\u00a7\u00a72 I. 10, 21.11] York, J. W., Jr., cited by name in \u00a72I.l, Ex. 21.8, 21.9. See also O'Murchadha and York 1973. Yourgrau, W., and S. Mandelstam, 1968, Variational Principles in Dynamics and Quantum Theory, W. B. Saunders, Phila delphia, Pa. [Box 2I.l} YU, J. T. See Peebles and Yu 1970. Zapolsky, H. S., 1968, \"Can the redshift of quasistellar objects be gravitational?\" Astrophys. J. Lett. 153, 163-169. [Ex. 25.29] Zaret, H., quoted at end of Chap. 44. Zeeman, E. c., 1964, \"Causality implies the Lorentz group,\" J. Math. Phys. 5, 190-493. [Box 2. I} ZeI'dovich, Va. B., 1963. \"The theory of the expanding universe as originated by A. A. Friedmann,\" Usp. Fiz. Nauk, 80, 357-390. English translation in SOl'. Phys.-Uspekhi. 6, 475-494 (1964). [\u00a729.2] ZeI'dovich, Va. B., 1967, \"Cosmological constant and elemen tary particles,\" Zh. Eksp. & Teor. Fiz. Pis'rna 6, 883-884. English translation in SOl'. Phn.-JETP Lett. 6. 316-317. [\u00a717.3. Ex. 17.5. Box 17.2 part 6] Zel'dovich, Va. B.. 1968. \"The cosmologi.:al constant and the theory of elementary particles.\" L'sp. Fi:..\"'auk 95. 209 230. English translation in Sou. Phys.-L'.\\pekhi 11, 381 393. [Ex. 17.5] Zel'dovich. Va. B.. 1970. \"Particle production in cosmology,\" Zh. Eksp. Teor. Fi:.-Pis\u00b7ma 12, 443-447. English transla tion in SOl'. Phys.-JETP Lett. 12. 307-311. [\u00a730.5. Ex. 27.12] Zel'dtlVkh, Va. B.. 1971. \"Particle creation by gravitational fields in collapse and singularities,\" Comlll. Astrophys. & Space Php. 3, 179-184. [Ex. 27.12] ZeI'dovich, Va. B.. 1972. \"Creation of particles and antiparti cles in an electric and gravitational field.\" in Klauder 1972, pp. 277-288. [\u00a730.8. Ex. 27.12} ZeI'dovich. Va. B. See also Bisnovatyi-Kogan and Zel'dovkh 1969, Braginsky. Zel'dovich. and Rudenko 1969, and Doroshkevich, Zel'dovich. and Novikov 1965. ZeI'dovich, Va. B.\u2022 and O. H. Guseynov, 1965. \"Collapsed stars in binaries,\" Astrophys. J. 144, 840-", "841. [Box 33.3] ZeI'dovich, Va. B., and I. D. Novikov, 1966, \"The hypothesis of cores retarded during expansion and the hot cosmo logical model,\" Astron. Zh. 43, 758-760. English transla tion in SOl'. Astron.-A. J. 10, 602-603. [Box 33.3] Zel'dovich, Va. B., and I. D. Novikov, 1971, Relativistic Astro physics, Vol. I: Stars and Relativity, Univ. of Chicago Press, Chicago, III. [\u00a7\u00a722.2, 23,1, 24.1, 24.3, 24.4, 24.6,25.6, 32.1. Boxes 24.1, 24.2] ZeI'dovich, Va. B., and I. D. Novikov, 1974, Relativistic Astro Ii: The Universe and Relativity, Univ. of physics. Vol. Chicago Press. Chicago. Ill.. in press. [\u00a7\u00a723.I, 28.1. 28.2, 30.1] ZeI'dovich, Va. B.. and M. A. Podurets, 1965, \"The evolution of a system of gravitationally interacting point masses,\" Astron. Zh. 42, 963-973. English translation in SOl'. As troll.-A. J. 9. 742-749 (1966). [\u00a7\u00a724.6, 25.7, Boxes 24.1. 25.9} ZeI'dovich, Va. B., and Yu. P. Rayzer, 1967, Physics of Shock Waves and High-temperature Hydrodynamic Phenomena, Academic Press, New York. [\u00a722.2, Box 22.2} ZeI'dovich, Va. B., and A. A. Starobinsky, 1971, \"Particle production and vacuum polarization in an anisotropic gravitational field,\" Zh. Eksp. Teor. Fiz. 61, 2161-2175. English translation in SOl'. Phys.-JETP 34, 1159-1166 (1972). [\u00a730.5] Zeno, cited by name in Box 30.1; see A. Griinbaum, 1967. Modern Science and Zeno's Paradoxes, We~eyan Uni\\'. Press, Middletown, Conn. \u2022 Zerilli, F. J., 1970a, \"Effective potential for even-parity Regge-Wheeler gravitational perturbation equations,\" Phys. Rev. Lett. 24, 737-738. [Box 33.3] Zerilli, F. J., 1970b, \"Gravitational field of a particle falling in Schwarzschild geometry analyzed in tensor harmon ics,\" Ph),s. Rev. D 2. 2141-2160. [\u00a736.5. Fig. 36.2, Box 33.3} Zimmer, H.\u2022 1946, Myths and Symbols in Indian Art and Civili\u00b7 zation, BoIlingen Foundation, Washington, D.C.; paper back reprint, Harper and Row. New York, 1962. [Box 27.7] Zimmerman, B. A. See Fletcher et al. 1967. Zwicky, F, See Baade and Zwicky 1934a and 1934b. SUBJECT INDEX Action at a distance, gravitational, 4 Newton's stricture against. 41 derived from local law, 120 Active vs. passive transformations, 1140 Adiabatic index defined. 692 Advanced fields, and radiation reaction, 474 Advanced potential. 121 After, undefined term in quantum geometrodynamics. 1183 Affine connection. See Connection, Covariant derivative Affine geometry characterized, 191, 242 in extenso. Chap. See also specific concepts. such as IO Covariant derivative, Connection coefficients Affine parameter, defined. 211, 244ff of geodesic, 244-246 effect of changing, on geodesic deviation, 269 variational principle adapted to, 322-323 in geometric optics, 575 no meaning of, for closed universe, 457ff contribution of interbody matter and fields to, 468 total unambiguous, despite contribution of pseudotensor to, 470 Gaussian flux integral for, 460-464 volume integral for. 460-466 measured by satellite-orbit precession, 451, 454, 457 measured by gyroscope pre~ession, 451, 454,457 measured by frame dragging, 451. 457 conservation laws for, 455. 468-471 for Kerr-Newman black hole, 891 Angular momentum in nat spacetime density of. 151, 1561' total, 156-159 decomposition of total into intrinsic and orbital, 1581' conservation of, 1561' intrinsic, sets lower limit to size, 162 parallel transport of,", ", 757 evolution of mean density of, 798 observations refute Klein-Alfven cosmological models, 770 Cosmological constant, 410ff Einstein's invention and retraction of, 41Of, 707, 758 influence on evolution of universe, 747, 771, 774 Cosmological models anisotropic, Chap. 30 Brans- Dicke, 770 closure of universe related to Mach's principle, 543, 549 as boundary condition, 1181 de Siller, 745, 758 Einstein static universe, 746f, 750, 758f flat, closed, static 3-torus model, 284 Friedmann discovery of, by Friedmann and Lemaitre. 751. 758 assumption of homogeneity and isotropy. 703, 713 assumption of perfect-fluid stress-energy tensor, 711f assumed equation of state, 713. 726 implications of homogeneity and isotropy. 714[, 720ff isotropy implies homogeneity, 715. 723 coordinate system constructed, 715ff expansion factor introduced, 718 arbitrariness in expansion factor, 720ff expansion factor renormalized, 721 f possible 3-geometries for homogeneous hypersurfaces. 720-725 curvature parameter K = k/a2 line element. various forms for, 72lff, \u2022 721 731. 759 embedding diagrams, 723, 725 topology not unique, 725 first law of thermodynamics for. 726ff assumption that mailer and radiation exchange negligible energy, 726ff, 765 assumption that pressure of maJter can always be neglected, 726, 728 density and pressure expressed in terms of expansion factor, 727 Einstein tensor for, 728 orthonormal frames allached to mailer. 728 initial-value equation (for a,?), 744 dynamic equation (for a,lt), 729 dynamic equation derivable from initial-value equation plus first law of thermodynamics, 729 time parameters: t, a, 1), 730-732 observer's parameters vs. relativity parameters, 77lff implications of parameter values for future of universe, 747, 771, 773f dynamics of early stage independent of k (unaffected by closure), 742f critical density for closure of universe if A = 0, 782 small perturbations of, 800f See also Hubble constant, Density parameter, Deceleration parameter Friedmann, closed (k = + I, A = 0) in extenso, 733-742 track-I overview, 704-711 Einstein's arguments favoring closure, 704, critical density for closure, 710, 782 geometry of 3-sphere hypersurfaces, 704,72I,723f radius of, defined, 704 radius of maximum expansion, 705 embedding diagram, 723f volume of, 724 topology not unique, 725 first law of thermodynamics applied to, 705, 726ff SUBJECT INDEX 1259 initial-value equation for, 537, 705f, was greatest prediction of Einstein's observed properties of universe 729, 733 effective potential for evolution of, 706 inevitability of recollapse, 707 solutions of field equations for, 734f radiation\u00b7 dominated era, 733-737, 740ff matter-dominated era, 733ff, 738-742 coordinate diagram for, 741 concrete numbers for a typical model, 738 theory, 411 what expandS and what does not, 719, 739 \"Where is the new space added?\" 719, 739 will Universe recontract? 747, 771, 774 See also Hubble expansion rate history of man's ideas and knOWledge of the universe, 752-762 homogeneity on large scales, 703, 815 isotropy on large scale, 703, 80 I, 815 rotation, observational limits on, 939 cosmological expansion, 772, 775f, 785-788, 793f age deduced from expansion rate, 709f, 797 ages of oldest stars, 709, 797f ages of rocks and meteorites, 759, 761, propagation of signals around universe, history of the universe according to the 798 741,750 causal isolation of various regions from each other, 740ff mocked up by Schwarzschild.lattice universe, 739f compared with Newtonian cosmological models, 707f Frieemann, flat and open (k = 0, k = -I; A = 0) geometry of homogeneous hypersurfaces, 721, 724f embedding diagram, 724f topology not unique, 725 Solutions of field equations for, 742 radiation.dominated era, 742f matter-dominated era, 743f J Friedmann,", " plus cosmological constant (k=O, \u00b1I; A;H) initial-value equation (for a,/), 744 effective potential for evolution of, 744, 746, 748f dynamical evolution of, 744-747 special cases of, 745ff, 750f hesitation universe, 750 hierarchic (island) universe, 748f, 770 inhomogeneous, Chap. )U- inhomogeneous Gowdy models, 804 inhomogeneous but spherical models, 804 Kasner model, 801, 805ff Klein-Alfven model, 748, 770 mixmaster, 805-814 Newtonian, 707f, 759 primordial chaos in big-bang models, 769, 802ff in extenso, Chap. 30 primordial black holes produced by, 884 See also Isotropy and homogeneity of universe, possible explanations of Schwarzschild lattice universe, 739f steady-state universe, 745, 750, 770 turnaround universe, 750f See also Cosmology: history of universe according to \"standard big. bang model\" Cosmology: expansion of universe prediction of by Friedmann, de Sitter, and Weyl, 758, 776 discovery of by Hubble, 759, 792-794 removed motive for cosmological term, 410-411 \"standard big-bang model\" in extenso, Chap. 28 initial singularity, 769f what \"preceded\" initial singularity? 769 possible roles of primordial chaos, 769, 803f, 816. See also Cosmological models: primordial chaos complete thermal equilibrium at t ~ I second, 736, 763f decoupling of gravitational waves and neutrinos, 736, 764 recombination of pairs, 736f, 764 thermal interaction of matter and radiation during expansion, 765f transition from matter dominance to radiation dominance, 741f, 765f condensation of stars, galaxies, and clusters of galaxies, 766, 769, 800 past history not much affected by k (by geometry of hypersurfaces), 742f, 763 expansion forever vs. recontraction, 747, 771, 774 deceleration parameter, 785, 788-791. density parameter, 796f mean density of luminous matter, 71Of, 761 mean density of cosmic rays, 712, 757, 798 mean density of intergalactic matter, 712, 761f, 797 mean density in electromagnetic radiation, 712 energy and pressure in kinetic motions of galaxies and stars, 711 abundances of elements, 765 entropy per baryon, 766 quasar population, evolution of, 767f, 770 \"fine\u00b7scale\" structure, 703 See also Cosmic microwave radiation, Hubble expansion rate speculations about initial and final states of universe, 707, 1209, 1213-1217 Coulomb field, \"pancaking\" of, for fast charged particle, 124 Coulomb force, from electromagnetic observational probes of standard model, 4'potential, 122 780-798 Coupling of fields to matter, direct vs. summary of, 797f distance-redshift relation, derivation of, indirect, 1063f Covariance, general. See General 780f distance-redshift relation, observational data, 781, 785-788, 792ff. See also Hubble expansion rate magnitude-redshift relation, derivations of, 782-785, 794 magnitude.redshift relation, observational data, 788-791 angle-effective distance vs. redshift (\"lens effect of universe\"), 795f source counts (number-flux relation), 798 mean mass density of universe, 71Off, 796f comparison of temperature, redshift, and emission times for cosmic background radiations, 737 abundances of elements, 765 comparison of ages deduced by various methods, 797f evolution of quasar population, 767f, 770 covariance Covariant components of a tensor, 76, 201-207, 312 Covariant derivative fundamental equations summarized, 223-224 defined by parallel transport, 208, 249 pictorialized, 209, 212 algebra of, 250-261 chain rule for, 214, 250, 252, 257f, 260f symmetry of (\"no torsion\"), 250, 252, 353f additivity of, 252 commutes with contraction, 214 compatibility with metric, 215f, 313ff, 353f noncommutation of two covariant derivatives, 389ff as a machine with slots, 253ff is not a tensor, 253, 255f connection coefficients as its components, 210, 256, 261f experimental tests of general relativity rotation I\u00b7forms constructed from, 349ff, using cosmological observations, 1047. 359f semicolon notation for, introduced, 210 1260 GRAV", "-star matter Degrees of freedom, counting of, for geometrodynamics and electrodynamics, 529-533 Delta, Kronecker, 22 Delta function, Dirac, 121 Democracy of histories, 418-419 Density of universe. See under Cosmology: observed properties of the universe \"Density parameter\" of universe, 772, 796f Derivative, covariant. See Covariant derivative Derivative, directional. See Directional derivative Derivative, following fluid, 153, 1078 Detailed balance, principle of, 1028ff, 1033, 1035f Determinant Directional derivative of a function along a vector, 59-60 operator for, 61 as a tangent vector, 227-230 Disks, rapidly rotating, in general relativity, 621 Dispersion relations obtained from Hamiltonians, 486f, 494, 498 Distance, proper. See Interval, Lorentz Distance-redshift relation. See under Cosmology: observational probes of standard model Distances, as raw material of metric, 306-309 Distant action. See Action at a distance Distant stars, inertial influence of. See Mach's principle Distribution. See Dirac delta function Distribution function, 583f, 590 Divergence of a vector or tensor, 82, 213, 222, 261 Double star. See Binary star Dragging of inertial frames in PPN formalism, 1117-1120 by Earth's rotation, 1119f by a slowly rotating star, 699 in Kerr-Newman geometry, 879ff, 893-896 prospects to measure, 1120 See also Mach's principle Dual bases, 60f, 119, 202, 232 Duality operation on forms, vectors, and tensors on forms, 88, 97f, 108, 119, lSI on simple forms, expressed in terms of perpendicularity, 98 application to electromagnetism, 88, 97f, 114 double dual of Riemann, 371, 376 not to be confused with duality of bases, 119. special star operation that does not act on forms, 376-380 Duality rotation of electromagnetic field, 108, 482f Dynamical path length in elementary mechanics, 486-487 as proportional to phase of wave function, 486 in superspace formulation of geometrodynamics, 419, 1186 See also Variational principle E Earth atmosphere and gravity, 388 crust, as detector of gravitational waves, 1013, 1015 general precession (precession of rotation axis), 391, 392, 1112, 1113 gravitational multipole moments, 401 mass, radius, density. See endpapers motion relative to cosmic microwave radiation, 713 particles oscillating in hole bored through,39 rotation of, drags inertial frames, 1119f. See also Day satellite orbits used to deduce mass, 638 shape as described by collection of distances, 306-309 subsurface mass variations, 40 I tides, as experimental test of general relativity, 1123f vibrations of, as detector for gravitational waves, 10 13, 10 15, 1035f Eccentricity of an elliptical orbit, 647 Eclipses, 24-26, 1104 Eddington-Finklestein coordinates, 828-831, 849 Eddington-Finklestein diagrams, 829, 830, 849, 864, 873 Effective potentials Divergences, in theory of particles and fields, 426-428 for test particles in Schwarzschild geometry, 639, 656, 659-662 1262 GRAVITATION Etl\"cctive potentials (continued) for charged test particles in equatorial plane of Kerr-Newman hole. 911 for waves in Schwarzsch ild geometry. 868, 870 for scalar waves in Kerr geometry. 915 for radius of Friedmann unh'erse. 706. 744. 746. 748f for oscillations of mixmaster universe. 809lf Einstein. See under Curvature, formalism of EIH equations of motion, 1091, 1094-1095 Eikonal method, 1102 Einstein A coefficients. 1029 Einstein's elevator. 298. See also Equivalence principle Einstein field equation. 43 1-434 elementary introduction to, 41lf integral equation equivalent to. 995-996 variational principles for. See under Variational principles derivations of in extenso. Chap. 17, esp. 406, 416-482 from automatic conservation of source, 379f,417 from Hilbert's action principle, 418 from physics on a spacelike slice, 419-423 from spin-2 field theory, 424f. 437 from superspace analysis, 423f from \"metric elasticity of space,\" 426lf modified by cosmological term, 410-412 correspondence with Newtonian theory, 412-416 and collapse, 1198-1199 See also Ge", "post-Newtonian) for spherical bodies, 1091, 1094-1095 deviations from geodesic motion, spinning, Thomas precession of, 175-176 1120-1121, 1128 Electron capture, in white-dwarf matter, Equations of state three-plus-one view versus geometric 619 view, 78-79 deduced from vector potential, 122 Elementary'particle experiments as tests of relativity theory, 1054f, 1060. See also for nuclear and white-dwarf matter,. 624-626 for \"cosmological fluid,\" 713, 726 SUBJECT INDEX 1263 Equinoxes, precession of, 391f, 1112f Equivalence principle enunciated, 386f, 1060 Einstein's 1911 formulation of, 17 bridge from special relativity to general relativity, 164, Chap. 16, 207 out of spin-2 field theory, 425 in Newton-Cartan theory, 297 basis for affine parameter, 211, 250 factor-ordering problems in, 388-391 role in metric theories, 1067f tests of, 187-190, 1054-1063 weak equivalence principle. See Uniqueness of free fall Ergosphere, 880 Ether, 1051, lO64f Euclidean geometry, 19-22 contrasted with Lorentz geometry, 51 Euler ilngles, 243 Euler relation, on vertices, edges, faces, 1175 Euler equation of hydrodynamics in flat spacetime, 152f in curved spacetime, 564 in PPN formalism, 1088 applied to a pulsating star, 693-694 applied to a collapsing star, 858 Eulerian perturbations, 690-691 Events, 6, 9f identifiability as key, 225 as classical, not valid quantum concept, 1184 Expansion of universe. See under Cosmology \"Expansion,\" of a bundle of null rays, 582, 1165 \"Expansion,\" of a congruericeo(world lines, 565f variations of. See under Cavendish gravitational constant gravitational waves, 1047, 1072 Hughes-Drever experiment, 1064 isotropy of space, 1064 Kreuzer experiment, 1125 laser ranging to moon, 1048, 113Q.-1I31 lunar orbit, 1048, 1116, 1119, 1127, 1128-1131 Newtonian experiments, 1067 Nordtvedt effect, 1128-1131 null experiments, 1050, 1064 perihelion shift, esp. 1110-1116 planetary orbits, deviations from geodesic motion, 1111, 1126-1131 planetary orbits, periodic effects in, 1069, 1111 Pound-Rebka-Snider experiment, 1056-1058 preferred-frame effects, 1098, 1113-1114 pulsars used for, 1047 quasars used for, 1047, 1048, 1061, 1101, 1103, 1104-1105 radar time delay, 1048, 1103, esp. 1106-1109 red shift, gravitational See Redshift, gravitational fedshift, due to \"ether drift,\" 1064-1065 singularities in spacetime, existence of, 939 Turner-Hill experiment, 1064-1065 See also Parametrized post-Newtonian formalism, Dicke's framework for analyzing experiments, Experimental tests of special relativity Experimental tests of special relativity, 1054-1055 Exterior calculus Extrinsic curvature. See under Curvature, formalism of Extrinsic time, of Kuchar and York, 487, 490 F Factor-ordering problems, 388-391 Faraday. See under Electromagnetic field Faraday stresses, 14Of, 481 Fast-motion approximation, 1072-1073 Fermat's principle in a static gravitational field, 1106, 1108 Fermi energy, in neutron stars and white dwarfs, 599-600 Fermi gas, ideal, 565, 599 Fermi normal coordinates, 332 Fermi-Walker transport, 165, 170f, 1117 Feynman's sum over histories, 320, 419, 499f Field equations. See Einstein field equations Fields, long range (i.e., zero rest mass) spin of, deduced from transformation laws for polarization of waves, 954 radiation fields must have I> S, 866, 977 role in slightly nonspherical ~ollapse of a star, 866 direct coupling vs. indirect coupling, 1063-1064 direct coupling, experimental searches for, 1063-1065 indirect coupling, 1068, 1069 Final state of stellar evolution, 624. See also White dwarfs, Neutron stars, Black holes Fine-structure constant, electromagnetic, constancy of, 399, 1061 introduction to and detailed summary of,", "ics, 364,,367-370 causal propagation of effects in, 554 Amowitt, Deser, Misner formulation of in brief, 486-490 action principle in, 521 geometrodynamic field momenta for, 521 3-geometry fixed at surfaces in, 522 split of variables made by, 525-526 electrodynamic analog, 522-524 Dirac formulation of, 520 subject to standard quantum indeterminism, 1182 illustrated in action, for Schwarzschild geometry, 528 See also Einstein field equation, General relativity, Initial value, Integrating forward in time Geometry. See specific rypes:--Spacetime, Euclidean, Lorentz, Differential, Affine, Riemannian, Prior. See also Curvature Geon, 886 Global techniques of analyzing spacetime structure in extenso, Chap. 34 basic references on, 916-917 examples of, 926-931 attempt to combine with local methods, 806 See also Infinity, regions of, in asymptotically flat spacetime; Causal structure of spacetime; Singularities in spacetim e Globular clusters, 757 black holes in, 887 \"Glory,\" in particle scattering, 670 Gowdy metrics, 804 Gradient of a scalar, in flat spacetime, 59f of a tensor, in flat spacetime, 81f in a curved manifold. 208-212, 259-261 See also Exterior. derivative Gradiometer, gravity. 400-403 Gravitation. 13. 163-164 local description in terms of tide-producing acceleration, 29-37 Gravitation-matter \"coupling loop,\" in brief, Gravitational potential. See under 5,37 Gravitation, theories of. catalogs of, 429 criteria for viability of, 1066-1067 Bergmann's scalar-tensor theories, 1049 Birkholf's, 1067 Cartan's (general relativity plus torsion), 1049, 1068 Cartan-Newton. See Newton~Cartan theory of gravity Coleman's, 1114 completeness of, 1067, 1068 conservative, 1093 Dicke-Brans-Jordan, l048f, 1068f, esp. 1070, 1093, 1098. 1122, 1127, 1129 cosmological models in. 770 general relativity, foundations of, Chaps. 16, 17 Kustaanheimo's, 1067 linearized. See Linearized theory of gravity metric. See Metric theories of gravity metric, not encompassed by the IO-parameter PPN formalism, 1069 Newton-Cartan theory, Post-Newtonian formalism Gravitational radiation reaction. See Gravitational waves; radiation reaction Gravitational radius, 820-826. See also Horizon, Black hole, Schwarzschild geometry, Kerr-Newman geometry Gravitational-wave detectors conceivable types of Earth-moon separation, 1013, 1014, 1018 normal-mode vibrations of Earth and moon, 1013, 1015 oscillations of Earth's crust, 10 13, 1015 normal-mode vibrations of an elastic bar, 1013, 1016, 1025, 1035, 1038 normal-mode vibrations of general elastic bodies, 1013, 1016, 1025, 1028-1035, 1041-1042 angular accelerations of rotating bars (\"heterodyne detector\"), lOB, 1016-1017 angular accelerations of driven oscillators, 10 13, 10 17 Newtonian. See Newton-Cartan theory of pumping of fluid in a rotating pipe, gravity 1013, 1018 Ni's, 1068f, esp. 1070f, 1083, 1098, 1123, idealized vibrator (2 masses on a 1129 Nordstrom's, 429ff, 1049 Papapetrou's, 1124 post-Newtonian. See Post-Newtonian approximation; Post-Newtonian formalism, parametrized preferred-frame, 1083, 1093, 1098, 1123-1125 prior-geometric, 429-431, 1068, 1070-1071 self-consistency of, 1066-1067 spin-O field, in flat spacetime, 178f spin-I field, in flat spacetime, 179 spin-2 field, in flat spacetime. See Linearized theory of gravity Whitehead's, 430, 1049, 1067, 1069, 1124 Gravitational collapse. See Collapse, gravitational Gravitational constant value of, 29, endpapers measurement of, 1121, 1123 as measure of \"metric elasticity of space,\" 426-428 See also Cavendish gravitational constant \"Gravitational field\" in general relativity theory as term with many meanings and none, 399f spacetime geometry as, 399-400 metric as, 399f covariant derivative and connection coefficients as. 387", ", 399-400 Riemann curvature as, 399-403 contribution of. to standard stress-energy tensor, specifically excluded. 131 Gravitational lens effect, 589, 887 Gravitational mass, 431 spring), 1022-1028 beads on stick, 444f nonmechanical detectors. 1040 electromagnetic waves in a toroidal wave guide, 1043-1044 methods of analyzing (for mechanical detectors small com paled to wavelength) proper reference frame of detector, 1005-1006, 1010, 1012 dynamic analysis: Newtonian equation of motion plus wave driving forces, 1006-1009 driving forces of waves, 1006, 1009, 1010 line-of-force diagram, 1011-10 12 method of detailed balance, 1028, 1029-1030, 1033 for noisy detector, 10 19, 1036-1040 detailed analysis of two freely falling bodies, 10 18 idealized vibrator (two masses on a spring), 1022-1028 any resonant vibrator. analyzed by detailed balance. 1030, 1033 any resonant vibrator. analyzed by dvnamic method, 1031-1034 noisy ~esonant vibrator (extraction of signal from noise). 1036-1040 Earth vibrating in quadrupole mode, 1035-1036 electromagnetic waves in a toroidal wave-guide, 1043-1044 cross sections limits on usefulness of concept of cross section. 10 19. 1022 1266 GRAVITATION Gr:lvitational-wave detectors (co/ltil/lIed) summary of ways to use. for wave-dominated detectors. 1020-1021 used to calculate total energy deposited in detector, 1027, 1028 use of. for noisy detectors. 1038-1039 related to emission patterns. 1032-1033. 1035 for idealized vibrator, 1024. 1025 for any resonant. mechanical detector. 1025. 1029. 1032 for a Weber bar. 1025 for a Weber bar in multimode operation. 1035 for Earth in fundamental quadrupole mode. 1036 thermally noisy detectors perturbations of Schwarzschild metric. 982. 983 vibrations of a relativistic star. by perturbations of equilibrium stellar structure, 984-985 rotation of a deformed relativistic star, by perturbations of spherical stellar structure. 986 initial-value solutions for, 536 intensity and spectrum of waves that bathe Earth. estimate of. 986 linearized theory of in extenso, 944-955 Lorentz gauge condition. 944-945 propagation equation. 945 gauge transformations that maintain Lorentz gauge. 945 plane-wave solutions, 945-946, 949, extraction of small signal from noise, 1004-1005 1036-1040 senSitivity of, to hammer-blow waves, 1039 ways to improve sensitivity. 1040 sensors for monitoring displacements. 1041, 1042 prospects for the future, 1040ff Gravitational waves exact solutions cylindrical wave, 950 plane waves with one state of polarization, 957-963. See also Plane gravitational waves plane waves with two polarization states, 964 transverse-traceless gauge 946-950 methods to calculate transverse-traceless part. 948-949 Riemann tensor, 948 geodesic deviation, 950-955, 1011-1Ol2 relative accelerations are purely transverse, 951 polarization, 952-955 Fourier analysis of, 1026, 1027 specific flux of, 1027 monopole and dipole waves absolutely forbidden. 977, 978 nonlinear interaction of waves with themselves nonexistence of precisely periodic experimental tests of general relativity waves, 956 using, 1047. 1072 generation by slow-motion, weak-field sources. nonexistence of monopole and dipole waves, 974-978 waves are predominantly quadrupolar, 975-978 assumptions underlying formulas. 989, 991 formula for metric perturbation, 991 formulas for emitted flux of energy and angular momentum, 992 formulas for total output of energy and angular momentum, 975, 992 formulas for radiation reaction in so urce, 993- 994 formulas for spectrum in various polarization states, 1033, 1035 formulas specialized to impulse events, 987 order-of-magnitude formulas for, 978-979, 980-981 derivation of formulas, 995-1003 role of \"gravitational stresses\" in generation, 996-998 generation by strong-field sources, techniques for calculating particle falling into black hole, by self-gravitational attraction, 957, 968 wave-wave scattering, 968 propagation through curved spacetime analogy with water waves on ocean, 993-994 refraction of wave fronts (deflection of rays) by background", " curvature, 956, 968, 972 gravitational redshift of frequency, 956-957, 968 backscatter off curvature, 957, 864-865, 869-871 tails due to interaction with background curvature, 957, 864-865, 869-871 shock fronts, 554 shortwave formalism for, 964-973. See also Gravitational waves: shortwave formalism propagation equation, 967-968 stress-energy tensor, 969-970. See also Gravitational waves: stress-energy tensor for geometric optics formalism, 971-972 propagation of polarization tensor, 968, 971 spinor formalism for, 1165 See also Gravitational waves: nonlinear interaction of waves with themselves radiation reaction order-of-magnitude formulas for. 979. 981 formalism for calculatin!!:. in weak field, slow-motion s'Ources. 993ff. lOOlff linked to outgoing-wave condition, 993. 1001-1002 forbids existence of exactly periodic waves, 956 damping of neutron-star vibrations by. 620, 628. 984f evolution of binary-star orbits due to. 988 shortwave formalism \"steady\" coordinates, 964 expansion parameters of, 964 assumptions underlying, 964 expansion of Ricci tensor, 964-965 coarse-grain viewpoint vs. fine-grain viewpoint, 965 propagation equation, 967-968 gauge freedom, 967-969 Lorentz gauge, 968 transverse-traceless gauge, 969 stress-energy tensor, 969-970. See also Gravitational waves: stress-energy tensor Brill-Hartle averaging process, 970 geometric optics specialization, 971-972 variational principle used to derive, 972-973 sources of astrophysical sources, order-of-magnitude formulas for, 980-981 big-bang origin of universe, 712, 736-737, 764-765 gravitational collapse of a star, 628, 629, 1041 supernova explosions, 982, 1040, 1042 explosion of a star, 987 collapses and explosions in Virgo cluster of galaxies, 1042 vibrations of neutron star, 982-986 rotation of a deformed neutron star (young pulsar), 628f, 983, 986, 1040 binary stars, 986, 988-990, 995 fall of matter into a black hole, 885, 982, 983, 986 collision and coalescence of black ~ \u2022 holes, 886, 939, 982 vibrations of a black hole, 886 rotating steel beam, 979-980 fission of an atomic nucleus, 987 atomic bomb, 987 meteorite striking earth, 987 stress-energy tensor for elementary summary of, 955-956 expressed in terms of metric perturbations, 969 SUBJECT INDEX 1267 expression for in traceless Lorentz Hamilton's principle for geodesic motion, gauge, 970 654 gauge invariance of, 972 expressed as an average of stress-energy Harrison-Wheeler equation of state, 625 Harrison-Wakano-Wheeler stellar models, pseudo tensor, 972 625lf, 696 divergence vanishes, 970 as source for background curvature of Hat product See Wedge product Heat flow in general relativity spacetime, 966, 973 for geometric-optics waves, 972 for waves in nearly flat spacetime, 955-956 for exact plane wave, 963 Gravitons, 972 Gravity gradiometer, 400-403 Group. See Rotation group: Lorentz group Group of motions, 652-653. See also Killing vector fields Gyroscopes employed in definition of Fermi-Walker hnsport, 165 employed in constructing proper reference frame, 327, 330f precession of, as experimental test of general relativity, 1117-1120 See also Dragging of inertial frames Gyromagnetic ratio, of Kerr-Newman black hole, 883, 892 H HII regions in galaxies, 710, 761, 786f Hair on a billiard ball, 978 Hair on a hole. See Kerr-Newman geometry, uniqueness of \"Hammer-blow waves\" defined, 1019 Hamilton-Jacobi theory, 486lf, 641-649 relation to quantum theory,Q.41-643 for harmonic oscillator, 1194 for free particle, 1194 for test-particle motion in Newtonian M/r potential, 644-649 in Schwarzschild gravitational field, 649 in Kerr-Newman gravitational and electromagnetic flelds, 900-901 deflection of light by sun, in PPN formalism, 1I02f, perihelion shift in P", " dissipation of anisotropy, 769, 802-803 pair creation by anisotropy energy, 769, 803-804 See also Cosmological models; Cosmology Isotropy implies homogeneity, 715, 723 J Jacobi identity, for commutators, 240 Jacobian, 93, 148, 160f Jacob's ladder. See Schild's ladder Jeans instability, 757 Junction conditions, 490 from electrodynamics as guide, 551 discovery by Kepler, 755 \"1-2-3\" law, 39, 450, 457 Kernel, of wave operator, 121 Kerr coordinates, 879f Kerr diagram, 881 Kerr geometry, as limiting case of Kerr-Newman, 878 Kerr-Newman geometry and electromagnetic field history of, 877n parameters of (M, Q, S, or a), 878 limiting cases (Schwarzschild, Reissner-Nordstrom, Kerr, extreme Kerr-Newman), 878 uniqueness as external field of a black hole heuristic explanation of uniqueness, 875, 877 theorems implying uniqueness, 876, 938, 939 implications for realistic gravitational collapse, 863 Boyer-Lindquist coordinates metric, 877, 878 electromagnetic field tensor, 877, 878 vector potential, 898 pathology of, at horizon, 880 Kerr coordinates electromagnetic field tensor, 879 metric, 879 transformation between Kerr and Boyer- ~ Lindquist coordinates, 879f Kerr-Schild coordinates, 903 stationary observers, 893-894 locally nonrotating observers, 895-896 Kerr diagram for, 881 maximal analytic extension of, 882 Killing vectors, 879, 892ff Killing tensor, 893 principal null congruences, 901-904 SUBJECT INDEX 1269 light cones, 891, 896-897 electromagnetic-field structure, 877ff, 883, 892 magnetic dipole moment, 883, 892 multipole moments of, 883, 892 horizon, 879ff null generators of, 903-904 area of, 889, 914 angular velocity of, 914 rotational properties collisionless Boltzmann equation (kinetic equation), 587, 590. specialized to photons, 587-589 basic concepts mass hyperboloid, 585 momentum space, 583ff, 590 ph ase space, 584f, 590 volume in phase space, 584-587, 590 distribution function (number density in phase space) defined, 583f, 590 intrinsic angular momentum vector, applications, 583 891 gyro magnetic ratio, 883, 892 static limit, 879ff, 894 ergosphere, 880 dragging of inertial frames, 879ff, 893-896 dynamic properties ro~e as endpoint of gravitational collapse, 882-883 stability against small perturbations, 884-885 change of M, Q, S when particles fall into horizon, 904-910, 913 reversible and irreversible transformations of, 889-890 rotational energy of, 890 electromagnetic energy of, 890 irreducible mass, 889-890, 913 test-particle motion in, super-Hamiltonian for, 897 energy-at-infinity, 898-899, 910 axial component of angular momentum, 898-899 rest-mass of particle, 899 Carter's fourth constant of the motion, f2 or X, 899 equations of motion in separated form, 899-900, 901 Hamilton-Jacobi derivation of equations of motion, 900-901 orbits in equatorial plane, 911-912 effective potential for equatorial motion, 911 binding energy of last stable circular orbit, 885, 911 wave propagation in, 914-915 Kerr-Schild coordinates, 903 Killing vector fields, 650-653 associated conservation laws for test-particle motion, 651 commutator of is Killing vector, 654 eigenvalue problem for finding, 654 for flat spacetime, 654 for spherically symmetric manifolds, 658 for Kerr-Newman geometry, 879, 892ff Killing's equation, 650 Killing tensor field, 893n Kinetic theory in curved spacetime in extenso, 583-590 basic laws elementary expression for pressure, 139-140 stress-energy tensor as in tegral over momentum space, 589f photons, 587ff relativistic star clusters, 679-687 computation of optical appearance of a collapsing star, 850 Klein-Alfven cosmology, 748, 770 Kronecker delta, 22 Kruskal diagrams, 528, 834f, 839, 848, 855 Kruskal-Szekeres coordinates for Schwarzschild geometry, 828-832 metric in, 827", " 398f curvature of, implied by gravitational red shift, 187ff stratification denied by locally Lorentz character of physics, 304f viewed as a \"gravitational field,\" 399f as dynamic participant in physics, 337 response to matter, as heart of general relativity, 404 as classical approximation, 1181 f as classical leaf slicing through superspace, 1184 See also General Relativity, Geometrodynamics, Curvature of spacetime Newtonian, Minkows'kiian, and Einsteinian, compared and contrasted, 296, 437 Special relativity briefly outlined, 47-48 spelled out, Chaps. 2-6 does not take in gravitation, Chap. 7 local validity as central feature of curved spacetime, 304f See also specific concepts, e.g., Electmmagnetic field, Lorentz transformations Specific intensity, defined, 587, 589 Specific flux, defined, 1025 Sphere, 2\u00b7dimensional (S2) two coordinate patches to cover, 12 topology of, 241f metric on, 340 Riemann tensor of, 341 Sphere, 3-dimensional (S3) volume of, 724 hyperspherical coordinates and metric for,723f Riemann curvature tensor, 721 embedding diagram, 723 compared with spheres of lower dimensionality, 704 isometric to manifold of rotation group, 725 Spherical symmetry, Killing vectors for, 658 Spherical systems, static Schwarzschild coordinates for, 594-597 isotropic coordinates for, 595 orthonormal frames for, 598 rigorous derivation of line element, 616f curvature tensors for, 360f curvature tensors for, 361f Birkhoff's theorem for, 883f Spin, as nonclassical tWQ-valuedness. 1204 'Spin matrices in law of combination of rotations, 1136 as quaternions or rotation operators, 1136 Hermitian conjugate of. 1138 alge braic properties, 1137-1142 and 3-vectors. 1140f and 4-vectors, 1142f associated spin matrices. 1152f multiplication law for, 1153 Spinning body equation of motion for, 1120f transport law for spin Fermi-Walker, in absence of curvature coupling, 165, 176f, 1117 modified by curvature coupling. 391f spin precessions \"general,\" 391f Thomas, 175f, 1118, 1145ff frame-dragging, 1119f due to space curvature (\"geodetic\"), 1119f Spindown of black holes, 886 Spinors general account, Chap. 41 and orientation-entanglement relation, 1148ff defined by their law of transformation, 1148ff conjugate complex, 1150 with dotted indices, 1150 correspondence with vectors, 1150ff of higher rank, 1151 algebra of, 1151-1155 spinor equivalent of tensors, 1153f spin space and basis spinors, 1156 flagpole plus flag plus orientation-entanglement relation, 1157-1160 in curved spacetime, 1164 analyze appearance of night sky, 1160-1164 as tool in gravitation theory, 1164f Standard candle, 789 Standard hot big-bang cosmological model. See under Cosmology Star clusters, relativistic, 621, 635 creation by evolution of a galactic nucleus, 634 analysis of structure, 679-683 equations of structure summarized, 683f relativistic instability in, 621, 686f collapse of, 884 possible roles in quasars and galactic nuclei, 634, 687 specific models with purely circular orbits, 683 self-similar, 683 isotropic, 683f isothermal, spherical, 685ff Spherical systems, dynamic Schwarzschild coordinates for, 616f, 689 Star operations. See Duality Starquake, 628 Stars, evolution into final state, 621, 624. 627-629 Stars, Newtonian equations of structure. 601-602, 605ff gravitational energy, 606-607 uniform-density model, 609 pulsation theory dynamical analysis, 697f virial-theorem analysis, 1079f volume-averaged analysis, 630f Stars, relativistic. nonrotating structure in extenso, Chaps. 23, 24 equations of structure summarized, 608-609, 689 must be spherical, 593 Schwarzschild coordinate system. 597 isotropic coordinate system, 595 curvature tensors for, 360f Newtonian limit of gravitational potentials, 595 parameters describing matter, 597-600 proper reference frame of fluid, 598 equations of structure derived, 600-606 Newtonian limit of equations of structure, 601-602, 605ff mass-energy inside radius r, 602ff must have 2mj", " a theory that apparently requires an in\ufb01nite number of particles. Even if we accept this, we still have not solved the problem of how to describe particles like photons or pions or alpha nuclei that do not obey Pauli exclusion. At this point, it is worthwhile to stop and re\ufb02ect on why it has proven to be so hard to \ufb01nd an acceptable relativistic wave equation for a single quantum particle. Perhaps there is something wrong with our basic approach. And there is. Recall the axiom of quantum mechanics that says that \u201cObservables are represented by hermitian operators.\u201d This is not entirely true. There is one observable in quantum mechanics that is not represented by a hermitian operator: time. Time enters into quantum mechanics only when we announce that the \u201cstate of the system\u201d depends on an extra parameter t. This parameter is not the eigenvalue of any operator. This is in sharp contrast to the particle\u2019s position x, which is the eigenvalue of an operator. Thus, space and time are treated very di\ufb00erently, a fact that is obscured by writing the Schr\u00a8odinger equation in terms of the position-space wave function \u03c8(x, t). Since space and time are treated asymmetrically, it is not surprising that we are having trouble incorporating a symmetry that mixes them up. So, what are we to do? In principle, the problem could be an intractable one: it might be impossible to combine quantum mechanics and relativity. In this case, there would have to be some meta-theory, one that reduces in the nonrelativistic limit to quantum mechanics, and in the classical limit to relativistic particle dynamics, but is actually neither. This, however, turns out not to be the case. We can solve our problem, but we must put space and time on an equal footing at the outset. There are two ways to do this. One is to demote position from its status as an operator, and render it as an extra label, like time. The other is to promote time to an operator. Let us discuss the second option \ufb01rst. If time becomes an operator, what do we use as the time parameter in the Schr\u00a8odinger equation? Happily, in relativistic theories, there is more than one notion of time. We can use the proper time \u03c4 of the particle (the time measured by a clock that moves with it) as the time parameter. The coordinate time T (the time measured by a stationary clock in an inertial frame) is then promoted to an operator. In the Heisenberg picture (where the state of the system is \ufb01xed, but the operators are functions of time that obey the classical equations of motion), we would have operators X \u00b5(\u03c4 ), where X 0 = T. Relativistic quantum mechanics can indeed be developed along these lines, but it is surprisingly 1: Attempts at relativistic quantum mechanics 26 complicated to do so. (The many times are the problem; any monotonic function of \u03c4 is just as good a candidate as \u03c4 itself for the proper time, and this in\ufb01nite redundancy of descriptions must be understood and accounted for.) One of the advantages of considering di\ufb00erent formalisms is that they may suggest di\ufb00erent directions for generalizations. For example, once we have X \u00b5(\u03c4 ), why not consider adding some more parameters? Then we would have, for example, X \u00b5(\u03c3, \u03c4 ). Classically, this would give us a continuous family of worldlines, what we might call a worldsheet, and so X \u00b5(\u03c3, \u03c4 ) would describe a propagating string. This is indeed the starting point for string theory. Thus, promoting time to an operator is a viable option, but is complicated in practice. Let us then turn to the other option, demoting position to a label. The \ufb01rst question is, label on what? The answer is, on operators. Thus, consider assigning an operator to each point x in space; call these operators \u03d5(x). A set of operators like this is called a quantum \ufb01eld. In the Heisenberg picture, the operators are also time dependent: \u03d5(x, t) = eiHt/\u00afh\u03d5(x, 0)e\u2212 iHt/\u00afh. (1.29) Thus, both position and (in the Heisenberg picture) time are now labels on operators; neither is itself the eigenvalue of an operator. So, now we have two di\ufb00erent approaches to relativistic quantum theory, approaches that might, in principle, yield di\ufb00erent results. This, however, is not the case: it turns out that any relativistic quantum physics that can be treated in one formalism", " can also be treated in the other. Which we use is a matter of convenience and taste. And, quantum \ufb01eld theory, the formalism in which position and time are both labels on operators, is much more convenient and e\ufb03cient for most problems. There is another useful equivalence: ordinary nonrelativistic quantum mechanics, for a \ufb01xed number of particles, can be rewritten as a quantum \ufb01eld theory. This is an informative exercise, since the corresponding physics is already familiar. Let us carry it out. Begin with the position-basis Schr\u00a8odinger equation for n particles, all with the same mass m, moving in an external potential U (x), and interacting with each other via an interparticle potential V (x1 \u2212 x2): i\u00afh \u2202 \u2202t \u03c8 = n \" Xj=1 \u2212 \u00afh2 2m \u2207 2 j + U (xj)! + n j 1 \u2212 Xj=1 Xk=1 V (xj \u2212 xk) # \u03c8, (1.30) where \u03c8 = \u03c8(x1,..., xn; t) is the position-space wave function. The quantum mechanics of this system can be rewritten in the abstract form of 1: Attempts at relativistic quantum mechanics 27 eq. (1.1) by \ufb01rst introducing (in, for now, the Schr\u00a8odinger picture) a quantum \ufb01eld a(x) and its hermitian conjugate a\u2020(x). We take these operators to have the commutation relations [a(x), a(x\u2032)] = 0, [a\u2020(x), a\u2020(x\u2032)] = 0, [a(x), a\u2020(x\u2032)] = \u03b43(x x\u2032), \u2212 (1.31) where \u03b43(x) is the three-dimensional Dirac delta function. Thus, a\u2020(x) and a(x) behave like harmonic-oscillator creation and annihilation operators that are labeled by a continuous index. In terms of them, we introduce the hamiltonian operator of our quantum \ufb01eld theory, H = \u00afh2 d3x a\u2020(x) 2m \u2207 \u2212 d3x d3y V (x) a(x) y)a\u2020(x)a\u2020(y)a(y)a(x). (1.32) Now consider a time-dependent state of the form \u03c8, t | i = Z d3x1... d3xn \u03c8(x1,..., xn; t)a\u2020(x1)... a\u2020(xn) 0 i |, (1.33) where \u03c8(x1,..., xn; t) is some function of the n particle positions and time, and is the vacuum state, the state that is annihilated by all the a\u2019s, 0 i | a(x) 0 i | It is now straightforward (though tedious) to verify that eq. (1.1), the abstract Schr\u00a8odinger equation, is obeyed if and only if the function \u03c8 satis\ufb01es eq. (1.30). (1.34) = 0. Thus we can interpret the state as a state of \u201cno particles\u201d, the state a\u2020(x1) as a state with one particle at position x1, the state a\u2020(x1)a\u2020(x2) 0 0 i | i | as a state with one particle at position x1 and another at position x2, and so on. The operator 0 i | N = d3x a\u2020(x)a(x) (1.35) counts the total number of particles. It commutes with the hamiltonian, as is easily checked; thus, if we start with a state of n particles, we remain with a state of n particles at all times. Z However, we can imagine generalizations of this version of the theory (generalizations that would not be possible without the \ufb01eld formalism) in which the number of particles is not conserved. For example, we could try adding to H a term like \u2206H \u221d Z d3x a\u2020(x)a2(x) + h.c. i h. (1.36) 1: Attempts at relativistic quantum mechanics 28 This term does not commute with N, and so the number of particles would not be conserved with this addition to H. Theories in which the number of particles can change as time evolves are a good thing: they are needed for correct phenomenology. We are already familiar with the notion that atoms can emit and absorb photons, and so", " we had better have a formalism that can incorporate this phenomenon. We are less familiar with emission and absorption (that is to say, creation and annihilation) of electrons, but this process also occurs in nature; it is less common because it must be accompanied by the emission or absorption of a positron, antiparticle to the electron. There are not a lot of positrons around to facilitate electron annihilation, while e+e\u2212 pair creation requires us to have on hand at least 2mc2 of energy available for the rest-mass energy of these two particles. The photon, on the other hand, is its own antiparticle, and has zero rest mass; thus photons are easily and copiously produced and destroyed. There is another important aspect of the quantum theory speci\ufb01ed by eqs. (1.32) and (1.33). Because the creation operators commute with each other, only the completely symmetric part of \u03c8 survives the integration in eq. (1.33). Therefore, without loss of generality, we can restrict our attention to \u03c8\u2019s of this type: \u03c8(... xi... xj... ; t) = +\u03c8(... xj... xi... ; t). (1.37) This means that we have a theory of bosons, particles that (like photons or pions or alpha nuclei) obey Bose-Einstein statistics. If we want Fermi-Dirac statistics instead, we must replace eq. (1.31) with a(x), a(x\u2032) } { a\u2020(x), a\u2020(x\u2032) } { a(x), a\u2020(x\u2032) } { = 0, = 0, = \u03b43(x x\u2032), \u2212 (1.38) = AB + BA is the anticommutator. Now only the fully where again antisymmetric part of \u03c8 survives the integration in eq. (1.33), and so we can restrict our attention to A, B { } \u03c8(... xi... xj... ; t) = \u2212 \u03c8(... xj... xi... ; t). (1.39) Thus we have a theory of fermions. It is straightforward to check that the abstract Schr\u00a8odinger equation, eq. (1.1), still implies that \u03c8 obeys the di\ufb00erential equation (1.30).1 Interestingly, there is no simple way to write 1Now, however, the ordering of the a and a\u2020 operators in the last term of eq. (1.32) becomes signi\ufb01cant, and must be as written. 1: Attempts at relativistic quantum mechanics 29 down a quantum \ufb01eld theory with particles that obey Boltzmann statistics, corresponding to a wave function with no particular symmetry. This is a hint of the spin-statistics theorem, which applies to relativistic quantum \ufb01eld theory. It says that interacting particles with integer spin must be bosons, and interacting particles with half-integer spin must be fermions. In our nonrelativistic example, the interacting particles clearly have spin zero (because their creation operators carry no labels that could be interpreted as corresponding to di\ufb00erent spin states), but can be either bosons or fermions, as we have seen. Now that we have seen how to rewrite the nonrelativistic quantum mechanics of multiple bosons or fermions as a quantum \ufb01eld theory, it is time to try to construct a relativistic version. Reference Notes The history of the physics of elementary particles is recounted in Pais. A brief overview can be found in Weinberg I. More details on quantum \ufb01eld theory for nonrelativistic particles can be found in Brown. Problems 1.1) Show that the Dirac matrices must be even dimensional. Hint: show 1, and that Tr \u03b2 = 0. To show that that the eigenvalues of \u03b2 are all Tr \u03b2 = 0, consider, e.g., Tr \u03b12 \u00b1 1\u03b2. Similarly, show that Tr \u03b1i = 0. 1.2) With the hamiltonian of eq. (1.32), show that the state de\ufb01ned in eq. (1.33) obeys the abstract Schr\u00a8odinger equation, eq. (1.1), if and only if the wave function obeys eq. (1.30). Your demonstration should apply both to the case of bosons, where the particle creation and annihilation operators obey the commutation relations of eq. (1.31), and to fermions, where the particle creation and annihilation operators obey the anticommutation relations of eq. (1.38). 1.", "3) Show explicitly that [N, H] = 0, where H is given by eq. (1.32) and N by eq. (1.35). 2 Lorentz Invariance Prerequisite: 1 A Lorentz transformation is a linear, homogeneous change of coordinates from x\u00b5 to \u00afx\u00b5, \u00afx\u00b5 = \u039b\u00b5 \u03bdx\u03bd, that preserves the interval x2 between x\u00b5 and the origin, where x2 \u2261 x\u00b5x\u00b5 = g\u00b5\u03bd x\u00b5x\u03bd = x2 c2t2. \u2212 This means that the matrix \u039b\u00b5 \u03bd must obey g\u00b5\u03bd \u039b\u00b5 \u03c1\u039b\u03bd \u03c3 = g\u03c1\u03c3, where 1 \u2212 +1 g\u00b5\u03bd = \uf8eb is the Minkowski metric. \uf8ec \uf8ec \uf8ed +1 \uf8f6. +1 \uf8f7 \uf8f7 \uf8f8 (2.1) (2.2) (2.3) (2.4) Note that this set of transformations includes ordinary spatial rotations: j = Rij, where R is an orthogonal 0 = 0, and \u039bi 0 = 1, \u039b0 i = \u039bi take \u039b0 rotation matrix. The set of all Lorentz transformations forms a group: the product of any two Lorentz transformations is another Lorentz transformation; the product is associative; there is an identity transformation, \u039b\u00b5 \u03bd ; and every Lorentz transformation has an inverse. It is easy to demonstrate these statements explicitly. For example, to \ufb01nd the inverse transformation \u03c3, (\u039b\u2212 \u03c3. On and that we can raise the \u03c1 index on both sides to get \u039b\u03bd the other hand, by de\ufb01nition, (\u039b\u2212 \u03bd, note that the left-hand side of eq. (2.3) can be written as \u039b\u03bd\u03c1\u039b\u03bd \u03c3. Therefore \u03bd = \u03b4\u00b5 \u03c3 = \u03b4\u03c1 \u03c3 = \u03b4\u03c1 \u03bd\u039b\u03bd \u03c1\u039b\u03bd 1)\u00b5 1)\u03c1 (\u039b\u2212 1)\u03c1 \u03bd = \u039b\u03bd \u03c1. Another useful version of eq. (2.3) is g\u00b5\u03bd \u039b\u03c1 \u00b5\u039b\u03c3 \u03bd = g\u03c1\u03c3. (2.5) (2.6) To get eq. (2.6), start with eq. (2.3), but with the inverse transformations 1)\u03bd \u03c3. Then use eq. (2.5), raise all down indices, and lower (\u039b\u2212 all up indices. The result is eq. (2.6). \u03c1 and (\u039b\u2212 1)\u00b5 For an in\ufb01nitesimal Lorentz transformation, we can write \u039b\u00b5 \u03bd = \u03b4\u00b5 \u03bd + \u03b4\u03c9\u00b5 \u03bd. (2.7) 2: Lorentz Invariance 31 Eq. (2.3) can be used to show that \u03b4\u03c9 with both indices down (or up) is antisymmetric: \u03b4\u03c9\u03c1\u03c3 = \u03b4\u03c9\u03c3\u03c1. (2.8) \u2212 Thus there are six independent in\ufb01nitesimal Lorentz transformations (in four spacetime dimensions). These can be divided into three rotations \u03b5ijk \u02c6nk\u03b4\u03b8 for a rotation by angle \u03b4\u03b8 about the unit vector \u02c6n) and (\u03b4\u03c9ij = three boosts (\u03b4\u03c9i0 = \u02c6ni\u03b4\u03b7 for a boost in the direction \u02c6n by rapidity \u03b4\u03b7). \u2212 Not all Lorentz transformations can be reached by compounding in1 = \ufb01nitesimal ones. If we take the determinant of eq. (2.5), we get (det \u039b)\u2212 1. Transformations with det \u039b = +1 are det \u039b, which implies det \u039b = proper, and transformations with det \u039b = 1 are improper. Note that the product of any two proper Lorentz transformations is proper, and that in\ufb01nitesimal transformations of the form \u039b = 1 + \u03b4\u03c9 are proper. Therefore, any transformation that can be reached by compounding in\ufb01nitesimal ones is proper. The proper transformations form a subgroup of the Lorentz group. \u00b1 \u2212 Another subgroup is that of the orthochronous Lorentz transformations: +1. Note that eq. (2.3) implies (\u039b0 those for which \u039b0 0 = 1; thus, either \u039b0 1. An in\ufb01nitesimal transformation is 0 \u2265", " clearly orthochronous, and it is straightforward to show that the product of two orthochronous transformations is also orthochronous. +1 or \u039b0 0 \u2264 \u2212 0 \u2265 0\u039bi 0)2 \u039bi \u2212 Thus, the Lorentz transformations that can be reached by compounding in\ufb01nitesimal ones are both proper and orthochronous, and they form a subgroup. We can introduce two discrete transformations that take us out of this subgroup: parity and time reversal. The parity transformation is \u00b5 P \u03bd = ( P 1)\u00b5 \u2212 \u03bd = \uf8eb +1 1 \u2212 \uf8f6. 1 (2.9) \u2212 \uf8ec \uf8ec \uf8ed It is orthochronous, but improper. The time-reversal transformation is \uf8f7 \uf8f7 \uf8f8 \u2212 1 \u00b5 \u03bd = ( T T 1)\u00b5 \u2212 \u03bd = \uf8eb 1 \u2212 +1 It is nonorthochronous and improper. \uf8ec \uf8ec \uf8ed +1 \uf8f6. +1 \uf8f7 \uf8f7 \uf8f8 (2.10) Generally, when a theory is said to be Lorentz invariant, this means under the proper orthochronous subgroup only. Parity and time reversal are treated separately. It is possible for a quantum \ufb01eld theory to be invariant under the proper orthochronous subgroup, but not under parity and/or time-reversal. 2: Lorentz Invariance 32 From here on, in this section, we will treat the proper orthochronous subgroup only. Parity and time reversal will be treated in section 23. In quantum theory, symmetries are represented by unitary (or antiunitary) operators. This means that we associate a unitary operator U (\u039b) to each proper, orthochronous Lorentz transformation \u039b. These operators must obey the composition rule U (\u039b\u2032\u039b) = U (\u039b\u2032)U (\u039b). (2.11) For an in\ufb01nitesimal transformation, we can write U (1+\u03b4\u03c9) = I + i 2\u00afh \u03b4\u03c9\u00b5\u03bd M \u00b5\u03bd, M \u03bd\u00b5 is a set of hermitian operators called the generators 1\u039b\u2032\u039b), let where M \u00b5\u03bd = 1U (\u039b\u2032)U (\u039b) = U (\u039b\u2212 of the Lorentz group. If we start with U (\u039b)\u2212 \u039b\u2032 = 1 + \u03b4\u03c9\u2032, and expand both sides to linear order in \u03b4\u03c9, we get (2.12) \u2212 \u03b4\u03c9\u00b5\u03bd U (\u039b)\u2212 1M \u00b5\u03bdU (\u039b) = \u03b4\u03c9\u00b5\u03bd \u039b\u00b5 \u03c1\u039b\u03bd \u03c3M \u03c1\u03c3. (2.13) Then, since \u03b4\u03c9\u00b5\u03bd is arbitrary (except for being antisymmetric), the antisymmetric part of its coe\ufb03cient on each side must be the same. In this case, because M \u00b5\u03bd is already antisymmetric (by de\ufb01nition), we have U (\u039b)\u2212 1M \u00b5\u03bdU (\u039b) = \u039b\u00b5 \u03c1\u039b\u03bd \u03c3M \u03c1\u03c3. (2.14) We see that each vector index on M \u00b5\u03bd undergoes its own Lorentz transformation. This is a general result: any operator carrying one or more vector indices should behave similarly. For example, consider the energymomentum four-vector P \u00b5, where P 0 is the hamiltonian H and P i are the components of the total three-momentum operator. We expect U (\u039b)\u2212 1P \u00b5U (\u039b) = \u039b\u00b5 \u03bdP \u03bd. (2.15) If we now let \u039b = 1 + \u03b4\u03c9 in eq. (2.14), expand to linear order in \u03b4\u03c9, and equate the antisymmetric part of the coe\ufb03cients of \u03b4\u03c9\u00b5\u03bd, we get the commutation relations [M \u00b5\u03bd, M \u03c1\u03c3] = i\u00afh g\u00b5\u03c1M \u03bd\u03c3 (\u00b5 \u2194 \u2212 \u03bd) (\u03c1 \u2194 \u2212 \u03c3). (2.16) These commutation relations specify the Lie algebra of the Lorentz group. We can identify the components of the angular momentum operator J as M i0. 2 \u03b5ijkM jk, and the components of the boost operator K as Ki \u2261 Ji \u2261 We then \ufb01nd from eq. (2.16) that 1 [Ji, Jj] =", " i\u00afh\u03b5ijkJk, [Ji, Kj] = i\u00afh\u03b5ijkKk, i\u00afh\u03b5ijkJk. [Ki, Kj] = \u2212 (2.17) 2: Lorentz Invariance 33 The \ufb01rst of these is the usual set of commutators for angular momentum, and the second says that K transforms as a three-vector under rotations. The third implies that a series of boosts can be equivalent to a rotation. Similarly, we can let \u039b = 1 + \u03b4\u03c9 in eq. (2.15) to get [P \u00b5, M \u03c1\u03c3] = i\u00afh g\u00b5\u03c3P \u03c1 (\u03c1 \u2194 \u2212 \u03c3), which becomes [Ji, H] = 0, [Ji, Pj] = i\u00afh\u03b5ijkPk, [Ki, H] = i\u00afhPi, [Ki, Pj] = i\u00afh\u03b4ijH, Also, the components of P \u00b5 should commute with each other: [Pi, Pj] = 0, [Pi, H] = 0. (2.18) (2.19) (2.20) Together, eqs. (2.17), (2.19), and (2.20) form the Lie algebra of the Poincar\u00b4e group. Let us now consider what should happen to a quantum scalar \ufb01eld \u03d5(x) under a Lorentz transformation. We begin by recalling how time evolution works in the Heisenberg picture: e+iHt/\u00afh\u03d5(x, 0)e\u2212 iHt/\u00afh = \u03d5(x, t). Obviously, this should have a relativistic generalization, e\u2212 iP x/\u00afh\u03d5(0)e+iP x/\u00afh = \u03d5(x), (2.21) (2.22) where P x = P \u00b5x\u00b5 = P \u00b7 de\ufb01ning the unitary spacetime translation operator \u2212 x Hct. We can make this a little fancier by Then we have T (a) exp( \u2212 \u2261 iP \u00b5a\u00b5/\u00afh). T (a)\u2212 1\u03d5(x)T (a) = \u03d5(x a). \u2212 (2.23) (2.24) For an in\ufb01nitesimal translation, \u2212 Comparing eqs. (2.12) and (2.25), we see that eq. (2.24) leads us to expect T (\u03b4a) = I i \u00afh \u03b4a\u00b5P \u00b5. (2.25) U (\u039b)\u2212 1\u03d5(x)U (\u039b) = \u03d5(\u039b\u2212 1x). (2.26) 2: Lorentz Invariance 34 Derivatives of \u03d5 then carry vector indices that transform in the appropriate way, e.g., U (\u039b)\u2212 1\u2202\u00b5\u03d5(x)U (\u039b) = \u039b\u00b5 \u03c1 \u00af\u2202\u03c1\u03d5(\u039b\u2212 1x), (2.27) where the bar on a derivative means that it is with respect to the argument \u00afx = \u039b\u2212 1x. Eq. (2.27) also implies U (\u039b)\u2212 1\u22022\u03d5(x)U (\u039b) = \u00af\u22022\u03d5(\u039b\u2212 1x), (2.28) so that the Klein-Gordon equation, ( invariant, as we saw in section 1. \u2212 \u22022 + m2/\u00afh2c2)\u03d5 = 0, is Lorentz Reference Notes A detailed discussion of quantum Lorentz transformations can be found in Weinberg I. Problems 2.1) Verify that eq. (2.8) follows from eq. (2.3). 2.2) Verify that eq. (2.14) follows from U (\u039b)\u2212 1U (\u039b\u2032)U (\u039b) = U (\u039b\u2212 1\u039b\u2032\u039b). 2.3) Verify that eq. (2.16) follows from eq. (2.14). 2.4) Verify that eq. (2.17) follows from eq. (2.16). 2.5) Verify that eq. (2.18) follows from eq. (2.15). 2.6) Verify that eq. (2.19) follows from eq. (2.18). 2.7) What property should be attributed to the translation operator T (a) that could be used to prove eq. (2.20)? 2.8) a) Let \u039b = 1 + \u03b4\u03c9", " in eq. (2.26), and show that [\u03d5(x), M \u00b5\u03bd ] = \u00b5\u03bd\u03d5(x), L where \u00b5\u03bd L \u2261 \u00afh i (x\u00b5\u2202\u03bd b) Show that [[\u03d5(x), M \u00b5\u03bd ], M \u03c1\u03c3] = \u03c1\u03c3\u03d5(x). x\u03bd \u2202\u00b5). \u2212 \u00b5\u03bd L L (2.29) (2.30) c) Prove the Jacobi identity, [[A, B], C] + [[B, C], A] + [[C, A], B] = 0. Hint: write out all the commutators. d) Use your results from parts (b) and (c) to show that [\u03d5(x), [M \u00b5\u03bd, M \u03c1\u03c3]] = ( L \u00b5\u03bd \u03c1\u03c3 \u03c1\u03c3 L \u2212 L L \u00b5\u03bd)\u03d5(x). (2.31) 2: Lorentz Invariance 35 e) Simplify the right-hand side of eq. (2.31) as much as possible. f) Use your results from part (e) to verify eq. (2.16), up to the possibility of a term on the right-hand side that commutes with \u03d5(x) and its derivatives. (Such a term, called a central charge, in fact does not arise for the Lorentz algebra.) 2.9) Let us write where \u039b\u03c1 \u03c4 = \u03b4\u03c1 \u03c4 + i 2\u00afh \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd V )\u03c1 \u03c4, (2.32) (S\u00b5\u03bd i (g\u00b5\u03c1\u03b4\u03bd are matrices which constitute the vector representation of the Lorentz generators. g\u03bd\u03c1\u03b4\u00b5 (2.33) V )\u03c1 \u03c4 \u2261 \u03c4 \u2212 \u03c4 ) \u00afh a) Let \u039b = 1 + \u03b4\u03c9 in eq. (2.27), and show that [\u2202\u03c1\u03d5(x), M \u00b5\u03bd ] = L \u00b5\u03bd\u2202\u03c1\u03d5(x) + (S\u00b5\u03bd V )\u03c1 \u03c4 \u2202\u03c4\u03d5(x). (2.34) b) Show that the matrices S\u00b5\u03bd relations as the operators M \u00b5\u03bd. Hint: see the previous problem. V must have the same commutation c) For a rotation by an angle \u03b8 about the z axis, we have 0 1 0 cos \u03b8 sin \u03b8 0 0 0 \u2212 0 sin \u03b8 cos \u03b8 0 0 0 0 1 \uf8f6. \uf8f7 \uf8f7 \uf8f8 \u039b\u00b5 \u03bd = \uf8eb \uf8ec \uf8ec \uf8ed \u039b = exp( \u2212 i\u03b8S12 V /\u00afh). Show that d) For a boost by rapidity \u03b7 in the z direction, we have cosh \u03b7 0 0 sinh \u03b7 0 0 1 0 0 1 0 0 sinh \u03b7 0 0 cosh \u03b7 \uf8f6. \uf8f7 \uf8f7 \uf8f8 \u039b\u00b5 \u03bd = \uf8eb \uf8ec \uf8ec \uf8ed \u039b = exp(+i\u03b7S30 V /\u00afh). Show that (2.35) (2.36) (2.37) (2.38) 3: Canonical Quantization of Scalar Fields 36 3 Canonical Quantization of Scalar Fields Prerequisite: 2 Let us go back and drastically simplify the hamiltonian we constructed in section 1, reducing it to the hamiltonian for free particles: H = = Z Z 2 a(x) 1 2m \u2207 d3x a\u2020(x) a\u2020(p) d3p 1 2m p2 \u2212 a(p), where e d3x (2\u03c0)3/2 e\u2212 e x a(x). ip \u00b7 a(p) = Z Here we have simpli\ufb01ed our notation by setting e \u00afh = 1. (3.1) (3.2) (3.3) The appropriate factors of \u00afh can always be restored in any of our formulas via dimensional analysis. The commutation (or anticommutation) relations of the a\u2020(p) operators are a(p) and e e [ a(p), a(p\u2032)] [ a\u2020(p), e a(p), [ e a\u2020(p\u2032)] e a\u2020(p\u2032)] e = 0, = 0, = \u03b43(p p\u2032), \u2212 \u2213 \u2213 \u2213 (3.4) \u2213 e e is either the commutator (if we want a theory of bosons) where [A, B] a\u2020(p) can or the anticommutator (if we want a theory", " of fermions). Thus be interpreted as creating a state of de\ufb01nite momentum p, and eq. (3.1) describes a theory of free particles. The ground state is the vacuum ; it 0 i | is annihilated by a(p), e e = 0, a(p) 0 i | and so its energy eigenvalue is zero. The other eigenstates of H are all of a\u2020(pn) the form, and the corresponding energy eigenvalue is 0 i | E(p1) +... + E(pn), where E(p) = 1 e 2m p2. It is easy to see how to generalize this theory to a relativistic one; all we need to do is use the relativistic energy formula E(p) = +(p2c2 + m2c4)1/2: a\u2020(p1)... (3.5) e e H = d3p (p2c2 + m2c4)1/2 a\u2020(p) a(p). (3.6) Z Now we have a theory of free relativistic spin-zero particles, and they can be either bosons or fermions. e e 3: Canonical Quantization of Scalar Fields 37 Is this theory really Lorentz invariant? We will answer this question (in the a\ufb03rmative) in a very roundabout way: by constructing it again, from a rather di\ufb00erent point of view, a point of view that emphasizes Lorentz invariance from the beginning. We will start with the classical physics of a real scalar \ufb01eld \u03d5(x). Real means that \u03d5(x) assigns a real number to every point in spacetime. Scalar means that Alice [who uses coordinates x\u00b5 and calls the \ufb01eld \u03d5(x)] and Bob \u03bdx\u03bd +a\u03bd, [who uses coordinates \u00afx\u00b5, related to Alice\u2019s coordinates by \u00afx\u00b5 = \u039b\u00b5 and calls the \ufb01eld \u00af\u03d5(\u00afx)], agree on the numerical value of the \ufb01eld: \u03d5(x) = \u00af\u03d5(\u00afx). This then implies that the equation of motion for \u03d5(x) must be the same as that for \u00af\u03d5(\u00afx). We have already met an equation of this type: the Klein-Gordon equation, \u22022 + m2)\u03d5(x) = 0. ( \u2212 Here we have simpli\ufb01ed our notation by setting c = 1 (3.7) (3.8) in addition to \u00afh = 1. As with \u00afh, factors of c can restored, if desired, by dimensional analysis. We will adopt eq. (3.7) as the equation of motion that we would like \u03d5(x) to obey. It should be emphasized at this point that we are doing classical physics of a real scalar \ufb01eld. We are not to think of \u03d5(x) as a quantum wave function. Thus, there should not be any factors of \u00afh in this version of the Klein-Gordon equation. This means that the parameter m must have dimensions of inverse length; m is not (yet) to be thought of as a mass. R : L = The equation of motion can be derived from variation of an action S = dt L, where L is the lagrangian. Since the Klein-Gordon equation is local, we expect that the lagrangian can be written as the space integral of. The integration a lagrangian density L \u03bdx\u03bd, measure d4x is Lorentz invariant: if we change to coordinates \u00afx\u00b5 = \u039b\u00b5 d4x = d4x. Thus, for the action to be Lorentz inwe have d4\u00afx = variant, the lagrangian density must be a Lorentz scalar: (\u00afx). Then we have \u00afS = d4x (x) = S. Any simple function of (\u00afx) = \u03d5 is a Lorentz scalar, and so are products of derivatives with all indices contracted, such as \u2202\u00b5\u03d5\u2202\u00b5\u03d5. We will take for det \u039b | | d4\u00afx \u00af L (x) = \u00af L. Thus, S = d3x d4x \u00b5\u03d5\u2202\u00b5\u03d5 L 2 m2\u03d52 + \u21260, 1 \u2212 where \u21260 is an arbitrary constant. We \ufb01nd the equation motion (also known as the Euler-Lagrange equation) by making an in\ufb01nitesimal variation \u03b4\u03d5(x) \u2212 L (3.", "9) 3: Canonical Quantization of Scalar Fields 38 in \u03d5(x), and requiring the corresponding variation of the action to vanish: 0 = \u03b4S d4x d4x = = Z Z 1 2 \u2202\u00b5\u03b4\u03d5\u2202\u00b5\u03d5 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03b4\u03d5 \u2212 h +\u2202\u00b5\u2202\u00b5\u03d5 h \u2212 m2\u03d5 i \u2212 \u03b4\u03d5. \u2212 m2\u03d5 \u03b4\u03d5 i (3.10) In the last line, we have integrated by parts in each of the \ufb01rst two terms, putting both derivatives on \u03d5. We assume \u03b4\u03d5(x) vanishes at in\ufb01nity in any direction (spatial or temporal), so that there is no surface term. Since \u03b4\u03d5 has an arbitrary x dependence, eq. (3.10) can be true if and only if ( \u2212 \u22022 + m2)\u03d5 = 0. One solution of the Klein-Gordon equation is a plane wave of the form exp(ik x \u00b7 \u00b1 i\u03c9t), where k is an arbitrary real wave-vector, and \u03c9 = +(k2 + m2)1/2. (3.11) The general solution (assuming boundary conditions that require \u03d5 to remain \ufb01nite at spatial in\ufb01nity) is then \u03d5(x, t) = d3k f (k) Z a(k)eik h x i\u03c9t + b(k)eik \u00b7 \u2212 \u00b7 x+i\u03c9t, (3.12) i where a(k) and b(k) are arbitrary functions of the wave vector k, and f (k) is a redundant function of the magnitude of k which we have inserted for later convenience. Note that, if we were attempting to interpret \u03d5(x) as a quantum wave function (which we most de\ufb01nitely are not), then the second term would constitute the \u201cnegative energy\u201d contributions to the wave function. This is because a plane-wave solution of the nonrelativistic Schr\u00a8odinger equation for a single particle looks like exp(ip iE(p)t), with E(p) = 1 2m p2; there is a minus sign in front of the positive energy. We are trying to interpret eq. (3.12) as a real classical \ufb01eld, but this formula does not generically result in \u03d5 being real. We must impose \u03d5\u2217(x) = \u03d5(x), where \u2212 x \u00b7 \u03d5\u2217(x, t) = Z Z = d3k f (k) d3k f (k) a\u2217(k)e\u2212 h ik x+i\u03c9t + b\u2217(k)e\u2212 \u00b7 ik \u00b7 x \u2212 i\u03c9t ik \u00b7 a\u2217(k)e\u2212 h x+i\u03c9t + b\u2217( \u2212 i k)e+ik \u00b7 x \u2212 i\u03c9t. (3.13) i In the second term on the second line, we have changed the dummy intek. Comparing eqs. (3.12) and (3.13), we see gration variable from k to \u2212 3: Canonical Quantization of Scalar Fields 39 that \u03d5\u2217(x) = \u03d5(x) requires b\u2217( can rewrite \u03d5 as \u2212 k) = a(k). Imposing this condition, we \u03d5(x, t) = Z Z Z = = d3k f (k) d3k f (k) d3k f (k) h h h a(k)eik \u00b7 x \u2212 i\u03c9t + a\u2217( \u2212 k)eik \u00b7 x+i\u03c9t a(k)eik \u00b7 x i\u03c9t + a\u2217(k)e\u2212 \u2212 ik \u00b7 x+i\u03c9t i i a(k)eikx + a\u2217(k)e\u2212 ikx, (3.14) i where kx = k x\u00b5 = (t, x) and k\u00b5 = (\u03c9, k): kx = k\u00b5x\u00b5 = g\u00b5\u03bd k\u00b5x\u03bd. Note that \u03c9t is the Lorentz-invariant product of the four-vectors \u2212 x \u00b7 k2 = k\u00b5k\u00b5 = k2 A four-momentum k\u00b5 that obeys k2 = or on shell for short. \u03c92 = m2. \u2212 (3.15) m2 is said to be on the mass shell, \u2212 \u2212 It is now convenient to choose f (k) so that d3k/f", " (k) is Lorentz invariant. An integration measure that is manifestly invariant under orthochronous Lorentz transformations is d4k \u03b4(k2+m2) \u03b8(k0), where \u03b4(x) is the Dirac delta function, \u03b8(x) is the unit step function, and k0 is treated as an independent integration variable. We then have + \u221e Z \u2212\u221e dk0 \u03b4(k2+m2) \u03b8(k0) = 1 2\u03c9. Here we have used the rule + \u221e Z \u2212\u221e dx \u03b4(g(x)) = 1 g\u2032(xi) | |, Xi (3.16) (3.17) where g(x) is any smooth function of x with simple zeros at x = xi; in our case, the only zero is at k0 = \u03c9. Thus we see that if we take f (k) \u03c9, then d3k/f (k) will be Lorentz invariant. We will take f (k) = (2\u03c0)32\u03c9. It is then convenient to give the corresponding Lorentz-invariant di\ufb00erential its own name: \u221d Thus we \ufb01nally have d3k (2\u03c0)32\u03c9. \u2261 dk f \u03d5(x) = dk Z f a(k)eikx + a\u2217(k)e\u2212 h ikx. i (3.18) (3.19) 3: Canonical Quantization of Scalar Fields 40 We can also invert this formula to get a(k) in terms of \u03d5(x). We have d3x e\u2212 ikx\u03d5(x) = 1 2\u03c9 a(k) + 1 2\u03c9 e2i\u03c9ta\u2217( k), \u2212 i 2 a(k) + i 2 e2i\u03c9ta\u2217( \u2212 k). (3.20) Z d3x e\u2212 ikx\u22020\u03d5(x) = Z We can combine these to get \u2212 a(k) = d3x e\u2212 ikx i\u22020\u03d5(x) + \u03c9\u03d5(x) i h d3x e\u2212 ikx \u2194\u22020 \u03d5(x), (3.21) Z = i Z where f \u2194\u2202\u00b5g time independent. \u2261 f (\u2202\u00b5g) (\u2202\u00b5f )g, and \u22020\u03d5 = \u2202\u03d5/\u2202t = \u02d9\u03d5. Note that a(k) is \u2212 Now that we have the lagrangian, we can construct the hamiltonian by the usual rules. Recall that, given a lagrangian L(qi, \u02d9qi) as a function of some coordinates qi and their time derivatives \u02d9qi, the conjugate momenta L. In are given by pi = \u2202L/\u2202 \u02d9qi, and the hamiltonian by H = our case, the role of qi(t) is played by \u03d5(x, t), with x playing the role of a P (continuous) index. The appropriate generalizations are then i pi \u02d9qi \u2212 and \u03a0(x) = \u2202 L \u2202 \u02d9\u03d5(x) = \u03a0 \u02d9\u03d5, \u2212 L H (3.22) (3.23) is the hamiltonian density, and the hamiltonian itself is H = where d3x H H. In our case, we have R and H \u03a0(x) = \u02d9\u03d5(x) = 1 2 \u03a02 + 1 2 ( \u2207 \u03d5)2 + 1 2 m2\u03d52 \u21260. \u2212 (3.24) (3.25) Using eq. (3.19), we can write H in terms of the a(k) and a\u2217(k) coe\ufb03cients: H = \u21260V + 1 2 \u2212 Z dk dk\u2032 d3x h i\u03c9 a(k)eikx + i\u03c9 a\u2217(k)e\u2212 f f ikx i\u03c9\u2032 a(k\u2032)eik\u2032x + i\u03c9\u2032 a\u2217(k\u2032)e\u2212 ik\u2032x \u2212 +ik a(k)eikx + + m2 \u2212 a(k)eikx + a\u2217(k)e\u2212 ikx ik a\u2217(k)e\u2212 ikx \u00b7 \u2212 +ik\u2032 a(k\u2032)eik\u2032x \u2212 a(k\u2032)eik\u2032x + a\u2217(k\u2032", ")e\u2212 ik\u2032x ik\u2032 a\u2217(k\u2032)e\u2212 ik\u2032x i 3: Canonical Quantization of Scalar Fields 41 \u21260V + 1 = \u2212 \u03b43(k dk dk\u2032 2 (2\u03c0)3 Z h k\u2032 + m2) k\u2032)(+\u03c9\u03c9\u2032 + k f \u00b7 \u03c9\u2032)t + a(k)a\u2217(k\u2032)e+i(\u03c9 i(\u03c9 \u2212 k\u2032 + m2) \u2212 a\u2217(k)a(k\u2032)e\u2212 \u03c9\u03c9\u2032 f k \u2212 a(k)a(k\u2032)e\u2212 \u00b7 \u2212 i(\u03c9+\u03c9\u2032)t + a\u2217(k)a\u2217(k\u2032)e+i(\u03c9+\u03c9\u2032)t \u03c9\u2032)t \u2212 \u00d7 + \u03b43(k + k\u2032)( \u00d7 \u21260V + 1 2 = \u2212 dk 1 2\u03c9 a\u2217(k)a(k) + a(k)a\u2217(k) Z h (+\u03c92 + k2 + m2) a(k)a( f + ( \u03c92 + k2 + m2) a\u2217(k)a(k) + a(k)a\u2217(k) k)e\u2212 dk \u03c9 \u2212 \u21260V + 1 2 \u2212 2i\u03c9t + a\u2217(k)a\u2217( = \u2212, Z f \u2212 k)e+2i\u03c9t i (3.26) where V is the volume of space. To get the second equality, we used d3x eiq \u00b7 x = (2\u03c0)3\u03b43(q). (3.27) Z dk\u2032 = d3k\u2032/(2\u03c0)32\u03c9\u2032. To get the third equality, we integrated over k\u2032, using The last equality then follows from \u03c9 = (k2+m2)1/2. Also, we were careful to keep the ordering of a(k) and a\u2217(k) unchanged throughout, in anticipation of passing to the quantum theory where these classical functions will become operators that may not commute. f Let us take up the quantum theory now. We can go from classical to quantum mechanics via canonical quantization. This means that we promote qi and pi to operators, with commutation relations [qi, qj] = 0, [pi, pj] = 0, and [qi, pj] = i\u00afh\u03b4ij. In the Heisenberg picture, these operators should be taken at equal times. In our case, where the \u201cindex\u201d is continuous (and we have set \u00afh = 1), we have [\u03d5(x, t), \u03d5(x\u2032, t)] = 0, [\u03a0(x, t), \u03a0(x\u2032, t)] = 0, [\u03d5(x, t), \u03a0(x\u2032, t)] = i\u03b43(x x\u2032). \u2212 (3.28) From these canonical commutation relations, and from eqs. (3.21) and (3.24), we can deduce [a(k), a(k\u2032)] = 0, [a\u2020(k), a\u2020(k\u2032)] = 0, [a(k), a\u2020(k\u2032)] = (2\u03c0)32\u03c9 \u03b43(k k\u2032). \u2212 (3.29) 3: Canonical Quantization of Scalar Fields 42 We are now denoting a\u2217(k) as a\u2020(k), since a\u2020(k) is now the hermitian conjugate (rather than the complex conjugate) of the operator a(k). We can now rewrite the hamiltonian as H = where Z f E0 = 1 2 (2\u03c0)\u2212 3 Z dk \u03c9 a\u2020(k)a(k) + ( \u21260)V, (3.30) E0 \u2212 d3k \u03c9 (3.31) is the total zero-point energy of all the oscillators per unit volume, and, using eq. (3.27), we have interpreted (2\u03c0)3\u03b43(0) as the volume of space V. If we integrate in eq. (3.31) over the whole range of k, the value of E0 is in\ufb01nite. If we integrate only up to a maximum value of \u039b, a number known as the ultraviolet cuto\ufb00, we \ufb01nd \u039b4 16\u03c02, E0 = m. This is physically justi\ufb01ed if, in the real where we have assumed \u039b world, the formalism of quantum \ufb01eld theory breaks down", " at some large energy scale. For now, we simply note that the value of \u21260 is arbitrary, and E0. With this choice, the ground state has so we are free to choose \u21260 = energy eigenvalue zero. Now, if we like, we can take the limit \u039b, with no further consequences. (We will meet more of these ultraviolet divergences after we introduce interactions.) (3.32) \u2192 \u221e \u226b The hamiltonian of eq. (3.30) is now the same as that of eq. (3.6), with a(k) = [(2\u03c0)32\u03c9]1/2 a(k). The commutation relations (3.4) and (3.29) are also equivalent, if we choose commutators (rather than anticommutators) in eq. (3.4). Thus, we have re-derived the hamiltonian of free relativistic bosons by quantization of a scalar \ufb01eld whose equation of motion is the Klein-Gordon equation. The parameter m in the lagrangian is now seen to be the mass of the particle in the quantum theory. (More precisely, since m has dimensions of inverse length, the particle mass is \u00afhcm.) e What if we want fermions? Then we should use anticommutators in eqs. (3.28) and (3.29). There is a problem, though; eq. (3.26) does not then become eq. (3.30). Instead, we get H = \u21260V, a simple constant. Clearly there is something wrong with using anticommutators. This is another hint of the spin-statistics theorem, which we will take up in section 4. \u2212 Next, we would like to add Lorentz-invariant interactions to our theory. With the formalism we have developed, this is easy to do. Any local function of \u03d5(x) is a Lorentz scalar, and so if we add a term like \u03d53 or \u03d54 to the lagrangian density, the resulting action will still be Lorentz invariant. Now, however, we will have interactions among the particles. Our next task is to deduce the consequences of these interactions. L 3: Canonical Quantization of Scalar Fields 43 However, we already have enough tools at our disposal to prove the spin-statistics theorem for spin-zero particles, and that is what we turn to next. Problems 3.1) Derive eq. (3.29) from eqs. (3.21), (3.24), and (3.28). 3.2) Use the commutation relations, eq. (3.29), to show explicitly that a state of the form k1... kni \u2261 | a\u2020(k1)... a\u2020(kn) 0 i | (3.33) is an eigenstate of the hamiltonian, eq. (3.30), with eigenvalue \u03c91 + = 0, and we... + \u03c9n. The vacuum take \u21260 = is annihilated by a(k), a(k) 0 i | 0 i | E0 in eq. (3.30). 1\u03d5(x)U (\u039b) = \u03d5(\u039b\u2212 3.3) Use U (\u039b)\u2212 1x) to show that U (\u039b)\u2212 1a(k)U (\u039b) = a(\u039b\u2212 1a\u2020(k)U (\u039b) = a\u2020(\u039b\u2212 1k), 1k), U (\u039b)\u2212 and hence that (3.34) = k1... kni U (\u039b) | \u039bk1... \u039bkni | = a\u2020(k1)... a\u2020(kn) 0 i |, (3.35) is a state of n particles with k1... kni where | momenta k1,..., kn. 3.4) Recall that T (a)\u2212 1\u03d5(x)T (a) = \u03d5(x iP \u00b5a\u00b5) is the spacetime translation operator, and P 0 is identi\ufb01ed as the hamiltonian H. a) Let a\u00b5 be in\ufb01nitesimal, and derive an expression for [\u03d5(x), P \u00b5]. a), where T (a) exp( \u2212 \u2212 \u2261 b) Show that the time component of your result is equivalent to the Heisenberg equation of motion i \u02d9\u03d5 = [\u03d5, H]. c) For a free \ufb01eld, use the Heisenberg equation to derive the KleinGordon equation.", " d) De\ufb01ne a spatial momentum operator P \u2261 \u2212 Z d3x \u03a0(x) \u2207 \u03d5(x). (3.36) Use the canonical commutation relations to show that P obeys the relation you derived in part (a). e) Express P in terms of a(k) and a\u2020(k). 3: Canonical Quantization of Scalar Fields 44 3.5) Consider a complex (that is, nonhermitian) scalar \ufb01eld \u03d5 with la- grangian density = L \u2212 \u2202\u00b5\u03d5\u2020\u2202\u00b5\u03d5 \u2212 m2\u03d5\u2020\u03d5 + \u21260. (3.37) a) Show that \u03d5 obeys the Klein-Gordon equation. b) Treat \u03d5 and \u03d5\u2020 as independent \ufb01elds, and \ufb01nd the conjugate momentum for each. Compute the hamiltonian density in terms of these conjugate momenta and the \ufb01elds themselves (but not their time derivatives). c) Write the mode expansion of \u03d5 as \u03d5(x) = dk Z a(k)eikx + b\u2020(k)e\u2212 h ikx. i (3.38) Express a(k) and b(k) in terms of \u03d5 and \u03d5\u2020 and their time derivatives. f d) Assuming canonical commutation relations for the \ufb01elds and their conjugate momenta, \ufb01nd the commutation relations obeyed by a(k) and b(k) and their hermitian conjugates. e) Express the hamiltonian in terms of a(k) and b(k) and their hermitian conjugates. What value must \u21260 have in order for the ground state to have zero energy? 4: The Spin-Statistics Theorem 45 4 The Spin-Statistics Theorem Prerequisite: 3 Let us consider a theory of free, spin-zero particles speci\ufb01ed by the hamiltonian H0 = dk \u03c9 a\u2020(k)a(k), (4.1) where \u03c9 = (k2 + m2)1/2, and either the commutation or anticommutation relations f Z [a(k), a(k\u2032)] [a\u2020(k), a\u2020(k\u2032)] [a(k), a\u2020(k\u2032)] \u2213 \u2213 \u2213 = 0, = 0, = (2\u03c0)32\u03c9 \u03b43(k k\u2032). \u2212 (4.2) Of course, if we want a theory of bosons, we should use commutators, and if we want fermions, we should use anticommutators. Now let us consider adding terms to the hamiltonian that will result in local, Lorentz invariant interactions. In order to do this, it is convenient to de\ufb01ne a nonhermitian \ufb01eld, \u03d5+(x, 0) \u2261 Z dk eik \u00b7 x a(k), and its hermitian conjugate f \u03d5\u2212(x, 0) \u2261 Z dk e\u2212 ik x a\u2020(k). \u00b7 f These are then time-evolved with H0: (4.3) (4.4) \u03d5+(x, t) = eiH0t\u03d5+(x, 0)e\u2212 iH0t = dk eikx a(k), \u03d5\u2212(x, t) = eiH0t\u03d5\u2212(x, 0)e\u2212 iH0t = Z Z f dk e\u2212 ikx a\u2020(k). (4.5) Note that the usual hermitian free \ufb01eld \u03d5(x) is just the sum of these: \u03d5(x) = \u03d5+(x) + \u03d5\u2212(x). f For a proper orthochronous Lorentz transformation \u039b, we have U (\u039b)\u2212 1\u03d5(x)U (\u039b) = \u03d5(\u039b\u2212 1x). (4.6) This implies that the particle creation and annihilation operators transform as U (\u039b)\u2212 U (\u039b)\u2212 1a(k)U (\u039b) = a(\u039b\u2212 1a\u2020(k)U (\u039b) = a\u2020(\u039b\u2212 1k), 1k). (4.7) 4: The Spin-Statistics Theorem 46 This, in turn, implies that \u03d5+(x) and \u03d5\u2212(x) are Lorentz scalars: U (\u039b)\u2212 1\u03d5\u00b1(x)U (\u039b) = \u03d5", "\u00b1(\u039b\u2212 1x). (4.8) We will then have local, Lorentz invariant interactions if we take the inL1 to be a hermitian function of \u03d5+(x) and teraction lagrangian density \u03d5\u2212(x). To proceed we need to recall some facts about time-dependent perturi to f | bation theory in quantum mechanics. The transition amplitude start with an initial state at time t = + \u2190 and end with a \ufb01nal state at time t = \u2212\u221e i i | Tf is i \u221e Tf i = f h T exp i dt HI (t) + \u221e \u2212 where HI(t) is the perturbing hamiltonian in the interaction picture, \u2212\u221e \u2190 Z |, i i | (4.9) HI(t) = exp(+iH0t) H1 exp( \u2212 iH0t), (4.10) R d3x H1(x, 0), and specify H1 is the perturbing hamiltonian in the Schr\u00a8odinger picture, and T is the time ordering symbol: a product of operators to its right is to be ordered, not as written, but with operators at later times to the left of those at earlier H1(x, 0) as a hermitian times. We write H1 = function of \u03d5+(x, 0) and \u03d5\u2212(x, 0). Then, using eqs. (4.5) and (4.10), we can see that, in the interaction picture, the perturbing hamiltonian density HI (x, t) is simply given by the same function of \u03d5+(x, t) and \u03d5\u2212(x, t). i to Now we come to the key point: for the transition amplitude Tf be Lorentz invariant, the time ordering must be frame independent. The time ordering of two spacetime points x and x\u2032 is frame independent if their separation is timelike; this means that the interval between them is x\u2032)2 < 0. Two spacetime points whose separation is spacelike, negative, (x x\u2032)2 > 0, can have di\ufb00erent temporal ordering in di\ufb00erent frames. In (x i being di\ufb00erent in di\ufb00erent frames, we must then require order to avoid \u2212 \u2212 \u2190 \u2190 Tf HI(x), [ Obviously, [\u03d5+(x), \u03d5+(x\u2032)] \u2213 HI (x\u2032)] = 0 whenever = [\u03d5\u2212(x), \u03d5\u2212(x\u2032)] x\u2032)2 > 0. (x \u2212 (4.11) = 0. However, \u2213 k\u2032x\u2032)[a(k), a\u2020(k\u2032)] \u2213 [\u03d5+(x), \u03d5\u2212(x\u2032)] \u2213 = = = \u2261 dk dk\u2032 ei(kx \u2212 Z x\u2032) \u2212 Z dk eik(x f f m f 4\u03c02r K1(mr) C(r). (4.12) 4: The Spin-Statistics Theorem 47 x\u2032)2 = r2 > 0, and K1(z) is In the next-to-last line, we have taken (x a modi\ufb01ed Bessel function. (This Lorentz-invariant integral is most easily evaluated in the frame where t\u2032 = t.) The function C(r) is not zero for any r > 0. (Not even when m = 0; in this case, C(r) = 1/4\u03c02r2.) On the other HI (x) must involve both \u03d5+(x) and \u03d5\u2212(x), by hermiticity. Thus, hand, generically, we will not be able to satisfy eq. (4.11). To resolve this problem, let us try using only particular linear combi- \u2212 nations of \u03d5+(x) and \u03d5\u2212(x). De\ufb01ne \u03d5\u03bb(x) \u03d5\u2020\u03bb(x) \u2261 \u2261 \u03d5+(x) + \u03bb\u03d5\u2212(x), \u03d5\u2212(x) + \u03bb\u2217\u03d5+(x), where \u03bb is an arbitrary complex number. We then have [\u03d5\u03bb(x), \u03d5\u2020\u03bb(x\u2032)] \u2213 = [\u03d5+(x), \u03d5\u2212(x\u2032)] \u2213 + 2[\u03d5\u2212(x), \u03d5+(x\u2032)] \u03bb | | \u2213 = (1 2) C(r) \u03bb | \u2213 | and [\u03d5\u03bb(x), \u03d5\u03bb(x\u2032)] \u2213 = \u03bb[\u03d5+(x), \u03d5\u2212(x\u2032", ")] \u2213 + \u03bb[\u03d5\u2212(x), \u03d5+(x\u2032)] \u2213 = \u03bb(1 \u2213 1) C(r). (4.13) (4.14) (4.15) \u03bb | | e+i\u03b1/2a(k) and a\u2020(k) i\u03b1/2\u03d5\u03bb(x) is hermitian. e\u2212 Thus, if we want \u03d5\u03bb(x) to either commute or anticommute with both \u03d5\u03bb(x\u2032) = 1, and we must and \u03d5\u2020\u03bb(x\u2032) at spacelike separations, we must choose choose commutators. Then (and only then), we can build a suitable HI (x) by making it a hermitian function of \u03d5\u03bb(x). But this has simply returned us to the theory of a real scalar \ufb01eld, because, for \u03bb = ei\u03b1, e\u2212 In fact, if we make the i\u03b1/2a\u2020(k), then the comreplacements a(k) i\u03b1/2\u03d5\u03bb(x) = \u03d5(x) = mutation relations of eq. (4.2) are unchanged, and e\u2212 \u03d5+(x) + \u03d5\u2212(x). Thus, our attempt to start with the creation and annihilation operators a\u2020(k) and a(k) as the fundamental objects has simply led us back to the real, commuting, scalar \ufb01eld \u03d5(x) as the fundamental object. Let us return to thinking of \u03d5(x) as fundamental, with a lagrangian density given by some function of the Lorentz scalars \u03d5(x) and \u2202\u00b5\u03d5(x)\u2202\u00b5\u03d5(x). Then, quantization will result in [\u03d5(x), \u03d5(x\u2032)] = 0 for t = t\u2032. If we choose anticommutators, then [\u03d5(x)]2 = 0 and [\u2202\u00b5\u03d5(x)]2 = 0, resulting in a trivial \u02d9\u03d5. This clearly does not that is at most linear in \u03d5, and independent of \u2192 \u2192 \u2213 L lead to the correct physics. This situation turns out to generalize to \ufb01elds of higher spin, in any number of spacetime dimensions. One choice of quantization (commuta, and so this choice tors or anticommutators) always leads to a trivial L 4: The Spin-Statistics Theorem 48 is disallowed. Furthermore, the allowed choice is always commutators for \ufb01elds of integer spin, and anticommutators for \ufb01elds of half-integer spin. If we try treating the particle creation and annihilation operators as fundamental, rather than the \ufb01elds, we \ufb01nd a situation similar to that of the spin-zero case, and are led to the reconstruction of a \ufb01eld that must obey the appropriate quantization scheme. Reference Notes This discussion of the spin-statistics theorem follows that of Weinberg I, which has more details. Problems 4.1) Verify eq. (4.12). Verify its limit as m 0. \u2192 5: The LSZ Reduction Formula 49 5 The LSZ Reduction Formula Prerequisite: 3 Let us now consider how to construct appropriate initial and \ufb01nal states for scattering experiments. In the free theory, we can create a state of one particle by acting on the vacuum state with a creation operator where The vacuum state 0 i | and has unit norm, k | i = a\u2020(k) 0 i |, a\u2020(k) = i \u2212 Z d3x eikx \u2194\u22020 \u03d5(x). is annihilated by every a(k), a(k5.1) (5.2) (5.3) (5.4) then has the Lorentz-invariant normalization k\u2032 | = (2\u03c0)3 2\u03c9 \u03b43(k i k\u2032), \u2212 (5.5) k | The one-particle state i k h where \u03c9 = (k2 + m2)1/2. Next, let us de\ufb01ne a time-independent operator that (in the free theory) creates a particle localized in momentum space near k1, and localized in position space near the origin: d3k f1(k)a\u2020(k), a\u20201 \u2261 Z (5.6) where f1(k) exp[ (k k1)2/4\u03c32] (5.7) \u2212 \u221d \u2212 is an appropriate wave packet, and \u03c3 is its width in momentum space. Consider the state a\u20201|. If we time evolve this", " state in the Schr\u00a8odinger 0 i picture, the wave packet will propagate (and spread out). The particle is thus localized far from the origin as t If we consider instead a = k2, then the two particles are widely, where k1 6 0 state of the form a\u20201a\u20202| i separated in the far past. \u2192 \u00b1\u221e. Let us guess that this still works in the interacting theory. One complication is that a\u2020(k) will no longer be time independent, and so a\u20201, eq. (5.6), becomes time dependent as well. Our guess for a suitable initial state of a scattering experiment is then i i | = lim t \u2192\u2212\u221e 0 a\u20201(t)a\u20202(t) i |. (5.8) 5: The LSZ Reduction Formula 50 By appropriately normalizing the wave packets, we can make = 1, and we will assume that this is the case. Similarly, we can consider a \ufb01nal state i i i | h f | f | = lim + t \u2192 \u221e i 0 a\u20201\u2032(t)a\u20202\u2032(t) i |, (5.9) i f h = k\u20322, and where k\u20321 6 = 1. This describes two widely separated particles in the far future. (We could also consider acting with more creation operators, if we are interested in the production of some extra particles in. i the collision of two.) Now the scattering amplitude is simply given by i |. To this end, let us i i | We need to \ufb01nd a more useful expression for f h f h note that a\u20201(+ ) \u221e \u2212 a\u20201( \u2212\u221e ) = = = = = = = + \u221e dt \u22020a\u20201(t) Z \u2212\u221e \u2212 \u2212 \u2212 \u2212 \u2212 d3k f1(k) d3k f1(k) d3k f1(k) d3k f1(k) d3k f1(k) d3k f1(k) Z Z Z Z Z d4x \u22020 eikx \u2194\u22020 \u03d5(x) d4x eikx(\u22022 0 + \u03c92)\u03d5(x) d4x eikx(\u22022 0 + k2 + m2)\u03d5(x) d4x eikx(\u22022 0 \u2212 \u2190 \u2207 2 + m2)\u03d5(x) d4x eikx(\u22022 d4x eikx( \u2212 Z \u2192 \u2207 2 + m2)\u03d5(x) 0 \u2212 \u22022 + m2)\u03d5(x). (5.10) The \ufb01rst equality is just the fundamental theorem of calculus. To get the second, we substituted the de\ufb01nition of a\u20201(t), and combined the d3x from this de\ufb01nition with the dt to get d4x. The third comes from straightforward evaluation of the time derivatives. The fourth uses \u03c92 = k2 + m2. The \ufb01fth x. The sixth uses integration by parts to writes k2 as 2 onto the \ufb01eld \u03d5(x); here the wave packet is needed to avoid a move the surface term. The seventh simply identi\ufb01es \u22022 2 acting on eik 2 as \u2212\u2207 \u22022. \u2207 \u00b7 In free-\ufb01eld theory, the right-hand side of eq. (5.10) is zero, since \u03d5(x) In an interacting theory, with (say) 2 g\u03d52. Thus the right-hand obeys the Klein-Gordon equation. L1 = 1 6 g\u03d53, we have instead ( side of eq. (5.10) is not zero in an interacting theory. \u22022 + m2)\u03d5 = 1 \u2212 0 \u2212 \u2207 \u2212 Rearranging eq. (5.10), we have a\u20201( \u2212\u221e ) = a\u20201(+ \u221e ) + i d3k f1(k) Z Z d4x eikx( \u2212 \u22022 + m2)\u03d5(x). (5.11) We will also need the hermitian conjugate of this formula, which (after a little more rearranging) reads a1(+ \u221e ) = a1( \u2212\u221e ) + i d3k f1(k) d4x e\u2212 Z Z ikx( \u2212 \u22022 + m2)\u03d5(x). (5.12) 5: The LSZ Reduction Formula 51 Let us return to the scattering amplitude, f h i i | =", " a1\u2032 (+ 0 | h \u221e )a2\u2032 (+ \u221e )a\u20201( \u2212\u221e )a\u20202( \u2212\u221e. 0 ) i | (5.13) Note that the operators are in time order. Thus, if we feel like it, we can put in a time-ordering symbol without changing anything: f h i i | = Ta1\u2032(+ 0 | h \u221e )a2\u2032(+ \u221e )a\u20201( \u2212\u221e )a\u20202( \u2212\u221e. 0 ) i | (5.14) The symbol T means the product of operators to its right is to be ordered, not as written, but with operators at later times to the left of those at earlier times. Now let us use eqs. (5.11) and (5.12) in eq. (5.14). The time-ordering )\u2019s to the right, where they anni)\u2019s move to the left, where they annihilate symbol automatically moves all ai\u2032( hilate. Similarly, all a\u2020i (+ 0 i |. 0 h | 0 limit in eq. (5.7), so that f1(k) = \u03b43(k k1). The initial and \ufb01nal states now have a delta-function normalization, the multiparticle generalization of eq. (5.5). We are left with The wave packets no longer play a key role, and we can take the \u03c3 \u2212\u221e \u2192 \u221e \u2212 = in+n\u2032 f h i i | Z ik\u2032 \u2212 1( d4x1 eik1x1( 1x\u2032 d4x\u20321 e\u2212 T\u03d5(x1)... \u03d5(x\u20321)... 0 | \u00d7 h 1 + m2)... \u22022 1\u2032 + m2)... \u22022 0 i | \u2212. (5.15) This formula has been written to apply to the more general case of n incoming particles and n\u2032 outgoing particles; the ellipses stand for similar factors for each of the other incoming and outgoing particles. Eq. (5.15) is the Lehmann-Symanzik-Zimmermann reduction formula, or LSZ formula for short. It is one of the key equations of quantum \ufb01eld theory. However, our derivation of the LSZ formula relied on the supposition that the creation operators of free \ufb01eld theory would work comparably in the interacting theory. This is a rather suspect assumption, and so we must review it. Let us consider what we can deduce about the energy and momentum eigenstates of the interacting theory on physical grounds. First, we assume that there is a unique ground state, with zero energy and momentum. 0 i | The \ufb01rst excited state is a state of a single particle with mass m. This state can have an arbitrary three-momentum k; its energy is then E = \u03c9 = (k2 + m2)1/2. The next excited state is that of two particles. These two particles could form a bound state with energy less than 2m (like the 5: The LSZ Reduction Formula 52 E 2m m 0 P Figure 5.1: The exact energy eigenstates in the (P, E) plane. The ground state is isolated at (0, 0), the one-particle states form an isolated hyperbola that passes through (0, m), and the multi-particle continuum lies at and above the hyperbola that passes through (0, 2m). hydrogen atom in quantum electrodynamics), but, to keep things simple, let us assume that there are no such bound states. Then the lowest possible energy of a two-particle state is 2m. However, a two-particle state with zero total three-momentum can have any energy above 2m, because the two particles could have some relative momentum that contributes to their total energy. Thus we are led to a picture of the states of theory as shown in \ufb01g. (5.1). Now let us consider what happens when we act on the ground state with the \ufb01eld operator \u03d5(x). To this end, it is helpful to write \u03d5(x) = exp( \u2212 iP \u00b5x\u00b5)\u03d5(0)exp(+iP \u00b5x\u00b5), (5.16) where P \u00b5 is the energy-momentum four-vector. (This equation, introduced in section 2, is just the relativistic generalization of the Heisenberg equation.) Now let us sandwich \u03d5(x) between the ground state (on the right), and other possible states (on the left). For example, let us put the ground state on the left as", " well. Then we have 0 \u03d5(x) 0 i | | h = = e\u2212 0 | h 0 \u03d5(0) 0 i | | h iP x\u03d5(0)e+iP x 0 i |. (5.17) 5: The LSZ Reduction Formula 53 To get the second line, we used P \u00b5 a Lorentz-invariant number. Since interacting theory, we have (in general) no idea what this number is. = 0. The \ufb01nal expression is just is the exact ground state of the 0 i | 0 i | We would like ), when acting on to be zero. This is because we would like 0 \u03d5(0) 0 i | | h, to create a single particle state. We do not 0 i | ) to create a linear combination of a single particle state and is not a\u20201( \u00b1\u221e want a\u20201( the ground state. But this is precisely what will happen if zero. 0 \u03d5(0) 0 i | | h \u00b1\u221e So, if v 0 \u03d5(0) 0 i | | \u2261 h is not zero, we will shift the \ufb01eld \u03d5(x) by the constant v. This means that we go back to the lagrangian, and replace \u03d5(x) everywhere by \u03d5(x) + v. This is just a change of the name of the operator of interest, and does not a\ufb00ect the physics. However, the shifted 0 \u03d5(x) 0 \u03d5(x) obeys, by construction, h i | |, where 0 \u03d5(x) Let us now consider i | | is a one-particle state with four-momentum p, normalized according to eq. (5.5). Again using eq. (5.16), we have = 0(x) i | | = e\u2212 p | h ipx = e\u2212 iP x\u03d5(0)e+iP x 0 i | p h 0 \u03d5(0) i | | is a Lorentz-invariant number. It is a function of p, but where the only Lorentz-invariant functions of p are functions of p2, and p2 is just the constant is just some number that depends on m 0 \u03d5(0) i | | and (presumably) the other parameters in the lagrangian. 0 \u03d5(0) i | | m2. So (5.18) p h p h \u2212, p h We would like 0 \u03d5(0) i | | and we know that, in free-\ufb01eld theory, a\u20201( ized one-particle state. Thus, for a\u20201( one-particle state in the interacting theory, we must have to be one. That is what it is in free-\ufb01eld theory, ) creates a correctly normal\u00b1\u221e ) to create a correctly normalized = 1. \u00b1\u221e p h So, if 0 \u03d5(0) i | | is not equal to one, we will rescale (or, one might say, renormalize) \u03d5(x) by a multiplicative constant. This is just a change of the name of the operator of interest, and does not a\ufb00ect the physics. However, the rescaled \u03d5(x) obeys, by construction, is a multiparticle state with, where 0 \u03d5(x) i | | total four-momentum p, and n is short for all other labels (such as relative momenta) needed to specify this state. We have 0 \u03d5(0) p h i | | p, n | Finally, consider p, n h = 1. i p h 0 \u03d5(0) i | | p, n h 0 \u03d5(x) i | | iP x\u03d5(0)e+iP x 0 i | = p, n h = e\u2212 e\u2212 | ipx 0 \u03d5(0) i | | p, n h ipxAn(p), = e\u2212 (5.19) where An(p) is a function of Lorentz invariant products of the various (relative and total) four-momenta needed to specify the state. Note that, 5: The LSZ Reduction Formula 54 from \ufb01g. (5.1), p0 = (p2 + M 2)1/2 with M \u2265 one of the parameters included in the set n. p, n ), h, to create a single particle state. We do not want 0 i | ) to create any multiparticle states. But this is precisely what may to be zero, because we would like a\u20201( 2m. The invariant mass M is We would like 0 \u03d5(x) i", " | | \u00b1\u221e when acting on a\u20201( \u00b1\u221e happen if p, n h 0 \u03d5(x) i | | is not zero. Actually, we are being a little too strict. We really need 0 \u03d5(x) i | | 0 ) p, n h i | is not. Also, we to be zero, and perhaps it will be zero even if really should test a\u20201( only against normalizable states. Mathematically, non-normalizable states cause all sorts of trouble; mathematicians don\u2019t consider them to be states at all. In physics, this usually doesn\u2019t bother us, but here we must be especially careful. So let us write p, n h a\u20201( | 0 ) i | \u00b1\u221e \u00b1\u221e = \u03c8 | i n Z X d3p \u03c8n(p) p, n | i, (5.20) where the \u03c8n(p)\u2019s are wave packets for the total three-momentum p. Note that eq. (5.20) is highly schematic; the sum over n includes integrals over continuous parameters like relative momenta. Now we want to examine \u03c8 h 0 a\u20201(t d3p \u03c8\u2217n(p) We will take the limit t becomes \u2192 \u00b1\u221e Z d3k f1(k) d3x eikx \u2194\u22020 h 0 \u03d5(x) | i | (5.21) in a moment. Using eq. (5.19), eq. (5.21) p, n Z. \u03c8 h a\u20201(t Next we use d3x ei(k R = \u03c8 h 0 a\u20201(t) i | | n Z X d3p \u03c8\u2217n(p) d3k f1(k) d3x eikx \u2194\u22020 e\u2212 ipx An(p) n Z X d3p \u03c8\u2217n(p) Z d3k f1(k) Z d3x (p0+k0)ei(k p)xAn(p). \u2212 Z Z (5.22) p) \u00b7 \u2212 x = (2\u03c0)3\u03b43(k p) to get \u2212 d3p (2\u03c0)3(p0+k0)\u03c8\u2217n(p)f1(p)An(p)ei(p0 k0)t, \u2212 (5.23) where p0 = (p2 + M 2)1/2 and k0 = (p2 + m2)1/2. Now comes the key point. Note that p0 is strictly greater than k0, 2m > m. Thus the integrand of eq. (5.23) contains a phase because M factor that oscillates more and more rapidly as t. Therefore, by the Riemann-Lebesgue lemma, the right-hand side of eq. (5.23) vanishes as t \u2192 \u00b1\u221e \u2265. \u2192 \u00b1\u221e 5: The LSZ Reduction Formula 55 Physically, this means that a one-particle wave packet spreads out differently than a multiparticle wave packet, and the overlap between them goes to zero as the elapsed time goes to in\ufb01nity. Thus, even though our operator a\u20201(t) creates some multiparticle states that we don\u2019t want, we can \u201cfollow\u201d the one-particle state that we do want by using an appropriate wave packet. By waiting long enough, we can make the multiparticle contribution to the scattering amplitude as small as we like. Let us recap. The basic formula for a scattering amplitude in terms of the \ufb01elds of an interacting quantum \ufb01eld theory is the LSZ formula, which is worth writing down again: = in+n\u2032 f h i i | d4x1 eik1x1( 1 + m2)... \u22022 \u2212 1( \u2212 Z d4x1\u2032 e\u2212 ik\u2032 1x\u2032 T\u03d5(x1)... \u03d5(x\u20321)... 0 | \u00d7 h 1\u2032 + m2)... \u22022 0 i |. (5.24) The LSZ formula is valid provided that the \ufb01eld obeys 0 \u03d5(x) 0 i | | h = 0 and k h 0 \u03d5(x) i | | = e\u2212 ikx. (5.25) These normalization conditions may con\ufb02ict with our original choice of \ufb01eld and parameter normalization in the lagrangian. Consider, for example, a lagrangian originally speci\ufb01ed as (5.26)", " = 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 1 2 m2\u03d52 + 1 6 g\u03d53. \u2212 After shifting and rescaling (and renaming some parameters), we will have instead \u2212 L = L \u2212 1 2 Z\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 1 2 Zmm2\u03d52 + 1 6 Zgg\u03d53 + Y \u03d5. (5.27) Here the three Z\u2019s and Y are as yet unknown constants. They must be chosen to ensure the validity of eq. (5.25); this gives us two conditions in four unknowns. We \ufb01x the parameter m by requiring it to be equal to the actual mass of the particle (equivalently, the energy of the \ufb01rst excited state relative to the ground state), and we \ufb01x the parameter g by requiring some particular scattering cross section to depend on g in some particular way. (For example, in quantum electrodynamics, the parameter analogous to g is the electron charge e. The low-energy Coulomb scattering cross section is proportional to e4, with a de\ufb01nite constant of proportionality and no higher-order corrections; this relationship de\ufb01nes e.) Thus we have four conditions in four unknowns, and it is possible to calculate Y and the three Z\u2019s order by order in powers of g. Next, we must develop the tools needed to compute the correlation functions T\u03d5(x1)... 0 | h 0 i | in an interacting quantum \ufb01eld theory. 5: The LSZ Reduction Formula 56 Reference Notes Useful discussions of the LSZ reduction formula can be found in Brown, Itzykson & Zuber, Peskin & Schroeder, and Weinberg I. Problems 5.1) Work out the LSZ reduction formula for the complex scalar \ufb01eld that was introduced in problem 3.5. Note that we must specify the type (a or b) of each incoming and outgoing particle. 6: Path Integrals in Quantum Mechanics 57 6 Path Integrals in Quantum Mechanics Prerequisite: none Consider the nonrelativistic quantum mechanics of one particle in one dimension; the hamiltonian is H(P, Q) = 1 2m P 2 + V (Q), (6.1) \u2212 t\u2032) and iH(t\u2032\u2032 q\u2032i | q\u2032\u2032i |, where where P and Q are operators obeying [Q, P ] = i. (We set \u00afh = 1 for notational convenience.) We wish to evaluate the probability amplitude for the particle to start at position q\u2032 at time t\u2032, and end at position q\u2032\u2032 at time t\u2032\u2032. This amplitude is are eigenstates q\u2032i e\u2212 q\u2032\u2032| h | of the position operator Q. We can also formulate this question in the Heisenberg picture, where operators are time dependent and the state of the system is time independent, as opposed to the more familiar Schr\u00a8odinger picture. In the Heisenberg piciHt. We can then de\ufb01ne an instantaneous ture, we write Q(t) = eiHtQe\u2212. These instantaneous eigenstates = q eigenstate of Q(t) via Q(t) q, t q, t | i i | = e+iHt can be expressed explicitly as. Then q q q, t | i | | our transition amplitude can be written as in the Heisenberg picture., where Q q | i q\u2032, t\u2032i q\u2032\u2032, t\u2032\u2032| h, we begin by dividing the time interval T To evaluate \u2261 t\u2032 into N + 1 equal pieces of duration \u03b4t = T /(N + 1). Then introduce t\u2032\u2032 \u2212 N complete sets of position eigenstates to get q\u2032\u2032, t\u2032\u2032| h q\u2032, t\u2032i = q i i q\u2032\u2032, t\u2032\u2032 h q\u2032, t\u2032 | i = N Z Yj=1 dqj h q\u2032\u2032 e\u2212 | iH\u03b4t qN ih | qN | e\u2212 iH\u03b4t qN | 1i \u2212... e\u2212 q1| h iH\u03b4t. q\u2032 | i (6.2) The integrals over the q\u2019s all run from q2| h Now consider q1i | exp(A + B) = exp(A) exp(B) exp( formula \u2212\u221e iH\u03b4t e\u2212 to +. \u221e. We can use the Campbell-Baker-Hausdorf 1 2 [A, B] +...) (6.3) \u2212 to write exp( \u2212 iH\u03b4t) = exp[ \u2212 i", "(\u03b4t/2m)P 2] exp[ \u2212 i\u03b4tV (Q)] exp[O(\u03b4t2)]. (6.4) Then, in the limit of small \u03b4t, we should be able to ignore the \ufb01nal exponential. Inserting a complete set of momentum states then gives e\u2212 q2| h iH\u03b4t q1i | = = Z Z q2| dp1 h dp1 e\u2212 e\u2212 i(\u03b4t/2m)P 2 p1ih | p1| e\u2212 i\u03b4tV (Q) i(\u03b4t/2m)p2 i\u03b4tV (q1) 1 e\u2212 q2| h p1ih p1| q1i | q1i 6: Path Integrals in Quantum Mechanics 58 dp1 2\u03c0 dp1 2\u03c0 = = Z Z e\u2212 i(\u03b4t/2m)p2 1 e\u2212 i\u03b4tV (q1) eip1(q2 q1). \u2212 e\u2212 iH(p1,q1)\u03b4t eip1(q2 q1). \u2212 (6.5) To get the third line, we used = (2\u03c0)\u22121/2 exp(ipq). If we happen to be interested in more general hamiltonians than eq. (6.1), then we must worry about the ordering of the P and Q operators in any term that contains both. If we adopt Weyl ordering, where the quantum hamiltonian H(P, Q) is given in terms of the classical hamiltonian H(p, q) by q h p | i H(P, Q) \u2261 Z dx 2\u03c0 dk 2\u03c0 Z eixP +ikQ dp dq e\u2212 ixp ikq H(p, q), \u2212 (6.6) in the last line, H(p1, q1) should be then eq. (6.5) is not quite correct; replaced with H(p1, \u00afq1), where \u00afq1 = 1 2 (q1 + q2). For the hamiltonian of eq. (6.1), which is Weyl ordered, this replacement makes no di\ufb00erence in the limit \u03b4t 0. Adopting Weyl ordering for the general case, we now have \u2192 q\u2032\u2032, t\u2032\u2032 h q\u2032, t\u2032 | i = N N dqk Z Yk=1 Yj=0 dpj 2\u03c0 eipj(qj+1\u2212 qj) e\u2212 iH(pj,\u00afqj)\u03b4t, (6.7) 2 (qj + qj+1), q0 = q\u2032, and qN +1 = q\u2032\u2032. If we now de\ufb01ne \u02d9qj \u2261 qj)/\u03b4t, and take the formal limit of \u03b4t 0, we get where \u00afqj = 1 (qj+1 \u2212 q\u2032\u2032, t\u2032\u2032 h q\u2032, t\u2032 | i = q D D i p exp t\u2032 Z Z dt p(t) \u02d9q(t) H(p(t), q(t)). (6.8) The integration is to be understood as over all paths in phase space that start at q(t\u2032) = q\u2032 (with an arbitrary value of the initial momentum) and end at q(t\u2032\u2032) = q\u2032\u2032 (with an arbitrary value of the \ufb01nal momentum). If H(p, q) is no more than quadratic in the momenta [as is the case for eq. (6.1)], then the integral over p is gaussian, and can be done in closed form. If the term that is quadratic in p is independent of q [as is the case for eq. (6.1)], then the prefactors generated by the gaussian integrals are all constants, and can be absorbed into the de\ufb01nition of q. The result of integrating out p is then D q\u2032\u2032, t\u2032\u2032 h q\u2032, t\u2032 | i = q exp i \" t\u2032 Z D Z dt L( \u02d9q(t), q(t)) #, (6.9) t\u2032\u2032 where L( \u02d9q, q) is computed by \ufb01rst \ufb01nding the stationary point of the p integral by solving 0 = \u2202 \u2202p p \u02d9q \u2212 H(p, q) = \u02d9q \u2202H(p, q) \u2202p \u2212 (6.10) \u2192 \u2212 t\u2032\u2032 6: Path Integ", "rals in Quantum Mechanics 59 for p in terms of \u02d9q and q, and then plugging this solution back into p \u02d9q H to get L. We recognize this procedure from classical mechanics: we are passing from the hamiltonian formulation to the lagrangian formulation. \u2212 Now that we have eqs. (6.8) and (6.9), what are we going to do with them? Let us begin by considering some generalizations; let us examine, for example,, where t\u2032 < t1 < t\u2032\u2032. This is given by q\u2032\u2032, t\u2032\u2032| h q\u2032, t\u2032 Q(t1) | | q\u2032, t\u2032i Q(t1) | q\u2032\u2032 = h i q\u2032\u2032, t\u2032\u2032 h iH(t\u2032\u2032 t1)Qe\u2212 \u2212 iH(t1 t\u2032) \u2212 e\u2212 |. q\u2032 | i (6.11) In the path integral formula, the extra operator Q inserted at time t1 will simply result in an extra factor of q(t1). Thus q\u2032\u2032, t\u2032\u2032 h q\u2032, t\u2032 Q(t1(t1) eiS, (6.12) where S = H). Now let us go in the other direction; consider q q(t1)q(t2)eiS. This clearly requires the operators Q(t1) and Q(t2), \u2212 but their order depends on whether t1 < t2 or t2 < t1. Thus we have R p D D t\u2032\u2032 t\u2032 dt (p \u02d9q R p D D Z q q(t1)q(t2) eiS = q\u2032\u2032, t\u2032\u2032 h q\u2032, t\u2032 TQ(t1)Q(t2) | | i. (6.13) where T is the time ordering symbol: a product of operators to its right is to be ordered, not as written, but with operators at later times to the left of those at earlier times. This is signi\ufb01cant, because time-ordered products enter into the LSZ formula for scattering amplitudes. To further develop these methods, we need another trick: functional derivatives. We de\ufb01ne the functional derivative \u03b4/\u03b4f (t) via \u03b4 \u03b4f (t1) f (t2) = \u03b4(t1 \u2212 t2), (6.14) where \u03b4(t) is the Dirac delta function. Also, functional derivatives are de\ufb01ned to satisfy all the usual rules of derivatives (product rule, chain rule, etc). Eq. (6.14) can be thought of as the continuous generalization of (\u2202/\u2202xi)xj = \u03b4ij. Now, consider modifying the lagrangian of our theory by including ex- ternal forces acting on the particle: H(p, q) H(p, q) \u2212 \u2192 f (t)q(t) \u2212 h(t)p(t), (6.15) where f (t) and h(t) are speci\ufb01ed functions. In this case we will write q\u2032\u2032, t\u2032\u2032 h q\u2032, t\u2032 | if,h = p D D q exp i \" t\u2032 Z Z dt p \u02d9q \u2212 H + f q + hp. (6.16) # t\u2032\u2032 6: Path Integrals in Quantum Mechanics 60 where H is the original hamiltonian. Then we have 1 i \u03b4 \u03b4f (t1) 1 i 1 i 1 i \u03b4 \u03b4f (t1) h q\u2032\u2032, t\u2032\u2032 q\u2032, t\u2032 | if,h = \u03b4 \u03b4f (t2) h q\u2032\u2032, t\u2032\u2032 q\u2032, t\u2032 | if,h = \u03b4 \u03b4h(t1) h q\u2032\u2032, t\u2032\u2032 q\u2032, t\u2032 | if,(t1) ei dt [p \u02d9q H+f q+hp], \u2212 R q q(t1)q(t2) ei dt [p \u02d9q H+f q+hp], \u2212 R q p(t1) ei dt [p \u02d9q H+f q+hp], \u2212 R (6.17) and so on. After we are done bringing down as many factors of q(ti) or p(ti) as we like, we can set f (t) = h(t) = 0, and return to the original hamiltonian. Thus, q\u2032\u2032, t\u2032\u2032 h TQ(t1)... P (tn)... | \u03b4 = \u03b4", "h(tn) \u03b4 \u03b4f (t1) 1 i 1 i... q\u2032, t\u2032 | i... q\u2032\u2032, t\u2032\u2032 h q\u2032, t\u2032 | if,h. (6.18) f =h=0 Suppose we are also interested in initial and \ufb01nal states other than position eigenstates. Then we must multiply by the wave functions for these states, and integrate. We will be interested, in particular, in the ground state as both the initial and \ufb01nal state. Also, we will take the limits t\u2032 \u2192 \u2212\u221e if,h = lim 0 0 | h. The object of our attention is then if,h \u03c80(q\u2032), dq\u2032\u2032 dq\u2032 \u03c8\u22170(q\u2032\u2032) and t\u2032\u2032 \u2192 q\u2032\u2032, t\u2032\u2032 h q\u2032, t\u2032 | (6.19) \u221e + t\u2032\u2192\u2212\u221e t\u2032\u2032\u2192+\u221e Z q h 0 i | where \u03c80(q) = is the ground-state wave function. Eq. (6.19) is a rather cumbersome formula, however. We will, therefore, employ a trick to simplify it. Let denote an eigenstate of H with eigenvalue En. We will suppose that E0 = 0; if this is not the case, we will shift H by an appropriate constant. Next we write n | i = eiHt\u2032 q\u2032, t\u2032 | i q\u2032 | i eiHt\u2032 n | n ih q\u2032 | i = = \u221e Xn=0 \u221e Xn=0 \u03c8\u2217n(q\u2032)eiEnt\u2032, n | i (6.20) where \u03c8n(q) = q n is the wave function of the nth eigenstate. Now, h | i i\u01eb)H in eq. (6.20), where \u01eb is a small positive in\ufb01nitesreplace H with (1 \u2212 of eq. (6.20) with \u01eb held \ufb01xed. Every imal. Then, take the limit t\u2032 \u2192 \u2212\u221e 6: Path Integrals in Quantum Mechanics 61 = 0. We then have a constant times state except the ground state is then multiplied by a vanishing exponential. Next, multiply by an arbi0 factor, and so the limit is simply \u03c8\u22170(q\u2032) i | trary function \u03c7(q\u2032), and integrate over q\u2032. The only requirement is that, and this constant can be 0 \u03c7 0 h i | | i 6 absorbed into the normalization of the path integral. A similar analysis of i\u01eb)H also picks shows that the replacement H q\u2032\u2032, t\u2032\u2032| h out the ground state as the \ufb01nal state in the t\u2032\u2032 \u2192 i\u01eb)H instead of H, we can be What all this means is that if we use (1 cavalier about the boundary conditions on the endpoints of the path. Any reasonable boundary conditions will result in the ground state as both the initial and \ufb01nal state. Thus we have (1 \u2212 limit. q\u2032\u2032| h \u2192 \u221e iHt\u2032\u2032 e\u2212 \u2212 + = if, exp i + \u221e dt p \u02d9q Z \u2212\u221e (1 \u2212 \u2212 i\u01eb)H + f q + hp. (6.21) Now let us suppose that H = H0 + H1, where we can solve for the eigenstates and eigenvalues of H0, and H1 can be treated as a perturbation. Suppressing the i\u01eb, eq. (6.21) can be written as if,h = 0 0 | h p i q exp D D Z Z \u2212\u221e H0(p, q) \u2212 \u2212 H1(p, q) + f q + hp + \u221e dt p \u02d9q \u03b4 \u03b4h(t) 1 i +, 1 i \u03b4 \u03b4f (t) = exp i \u2212 Z p D D \u00d7 Z + \u221e dt H1 \u2212\u221e q exp i Z \u221e dt p \u02d9q \u2212 H0(p, q) + f q + hp. (6.22) \u2212\u221e To understand the second line of this equation, take the exponential prefactor inside the path integral. Then the functional derivatives (that appear as the arguments of H1) just pull out appropriate factors of p(t) and q(t), generating the right-hand side of the \ufb01rst line. We assume that we can compute the functional integral in the second line, since it involves only the solvable hamiltonian H0. The exponential prefactor can", " then be expanded in powers of H1 to generate a perturbation series. If H1 depends only on q (and not on p), and if we are only interested in time-ordered products of Q\u2019s (and not P \u2019s), and if H is no more than quadratic in P, and if the term quadratic in P does not involve Q, then eq. (6.22) can be simpli\ufb01ed to if = exp 0 0 | h i + \u221e dt L1 Z \u2212\u221e q exp + i 1 i \u03b4 \u03b4f (t) \u221e dt L0( \u02d9q, q) + f q. (6.23) Z \u2212\u221e \u00d7 Z D where L1(q) = H1(q). \u2212 6: Path Integrals in Quantum Mechanics 62 Reference Notes Brown and Ramond I have especially clear treatments of various aspects of path integrals. For a careful derivation of the midpoint rule of eq. (6.7), see Berry & Mount. Problems 6.1) a) Find an explicit formula for be of the form should compute. D q = C Q q in eq. (6.9). Your formula should N j=1 dqj, where C is a constant that you D b) For the case of a free particle, V (Q) = 0, evaluate the path integral of eq. (6.9) explicitly. Hint: integrate over q1, then q2, etc, and look for a pattern. Express you \ufb01nal answer in terms of q\u2032, t\u2032, q\u2032\u2032, t\u2032\u2032, and m. Restore \u00afh by dimensional analysis. iH(t\u2032\u2032 c) Compute by inserting a complete set of momentum eigenstates, and performing the integral over the momentum. Compare with your result in part (b). q\u2032\u2032, t\u2032\u2032| h q\u2032, t\u2032i q\u2032\u2032| h q\u2032i | e\u2212 = t\u2032) \u2212 7: The Path Integral for the Harmonic Oscillator 63 7 The Path Integral for the Harmonic Oscillator Prerequisite: 6 Consider a harmonic oscillator with hamiltonian H(P, Q) = 1 2m P 2 + 1 2 m\u03c92Q2. (7.1) We begin with the formula from section 6 for the ground state to ground state transition amplitude in the presence of an external force, specialized to the case of a harmonic oscillator: if = 0 0 | h D Z p Dq exp i + \u221e dt p \u02d9q Z \u2212\u221e h (1 \u2212 \u2212 i\u01eb)H + f q. (7.2) i Looking at eq. (7.1), we see that multiplying H by 1 (1 the replacements m\u2212 m\u03c92 \u2212 1 [or, equivalently, m i\u01eb)m\u03c92. Passing to the lagrangian formulation then gives i\u01eb is equivalent to (1+i\u01eb)m] and i\u01eb)m\u2212 \u2192 \u2192 (1 \u2212 1 \u2192 \u2212 if = 0 0 | h Z Dq exp i + \u221e dt 1 2 (1+i\u01eb)m \u02d9q2 1 2 (1 \u2212 \u2212 i\u01eb)m\u03c92q2 + f q. (7.3) h From now on, we will simplify the notation by setting m = 1. \u2212\u221e i Z Next, let us use Fourier-transformed variables, q(E) = + \u221e Z \u2212\u221e dt eiEt q(t), q(t) = + \u221e Z \u2212\u221e dE 2\u03c0 iEt e\u2212 q(E). (7.4) The expression in square brackets in eq. (7.3) becomes e e = 1 2 + \u221e Z \u2212\u221e dE 2\u03c0 dE\u2032 2\u03c0 \u00b7 \u00b7 \u00b7 h i i(E+E\u2032)t e\u2212 (1+i\u01eb)EE\u2032 (1 \u2212 \u2212 \u2212 i\u01eb)\u03c92 q(E) q(E\u2032) f (E) q(E\u2032) + f (E\u2032) e (7.5) h +. q(E) e i Note that the only t dependence is now in the prefactor. Integrating over t then generates a factor of 2\u03c0\u03b4(E + E\u2032). Then we can easily integrate over E\u2032 to get e e e e S = + \u221e dt + \u221e Z \u2212\u221e 1 2 Z \u2212\u221e = \u00b7 \u00b7 \u00b7 h dE 2\u03c0 i (1+i\u01eb)E2 i\u01eb)\u03c92 (1 \u2212 \u2212 q(E) q( \u2212 E) h + f (E) q( \u2212", " E) + \u2212 f ( E) e q(E) e. (7.6) e e e e i 7: The Path Integral for the Harmonic Oscillator 64 The factor in large parentheses is equal to E2 can absorb the positive coe\ufb03cient into \u01eb to get E2 \u2212 \u03c92 + i(E2 + \u03c92)\u01eb, and we \u03c92 + i\u01eb. \u2212 Now it is convenient to change integration variables to x(E) = q(E) + f (E) E2 \u2212 e \u03c92 + i\u01eb. Then we get e e + \u221e 1 2 S = x(E)(E2 dE 2\u03c0 \" E) f ( \u2212 \u03c92 + i\u01eb # e Furthermore, because eq. (7.7) is just a shift by a constant, we have f (E) E2 e \u03c92 + i\u01eb) q = E) \u2212\u221e x7.7). (7.8) x. Now D if = exp \u2212\u221e x exp dE 2\u03c0 i 2 E) f ( f (E) \u2212 E2 + \u03c92 \u2212 e e dE \u221e x(E)(E2 2\u03c0 i\u01eb # \u2212 + Z \u2212\u221e \u03c92 + i\u01eb) x( \u2212 \u2212. (7.9) E) Now comes the key point. The path integral on the second line of if in the case f = 0. On the other hand, eq. (7.9) is what we get for 0 0 | h if there is no external force, a system in its ground state will remain in its ground state, and so if is given by the \ufb01rst line of 0 0 | h eq. (7.9), e e if =0 = 1. Thus 0 0 | h dE 2\u03c0 + \u221e i 2 f ( f (E) if = exp 0 0 \u2212 E2 + \u03c92 | h \u2212 e e if in terms of time-domain variables as 0 0 | h i\u01eb # E) \u2212\u221e \u2212 \" Z. We can also rewrite if = exp 0 0 | h i 2 + \u221e Z \u2212\u221e dt dt\u2032 f (t)G(t \u2212 t\u2032)f (t\u2032), where Note that G(t \u2212 G(t \u2212 t\u2032) = + \u221e Z \u2212\u221e dE 2\u03c0 t\u2032) iE(t e\u2212 \u2212 E2 + \u03c92. i\u01eb \u2212 \u2212 t\u2032) is a Green\u2019s function for the oscillator equation of motion: \u22022 \u2202t2 + \u03c92 G(t \u2212! t\u2032) = \u03b4(t t\u2032). \u2212 (7.13) This can be seen directly by plugging eq. (7.12) into eq. (7.13) and then taking the \u01eb t\u2032) explicitly by treating the integral over E on the right-hand side of eq. (7.12) as a contour integral 0 limit. We can also evaluate G(t \u2192 \u2212 (7.10) (7.11) (7.12) 7: The Path Integral for the Harmonic Oscillator 65 in the complex E plane, and then evaluating it via the residue theorem. The result is G(t t\u2032) = exp \u2212 i 2\u03c9 i\u03c9 t | \u2212 \u2212 t\u2032 |. (7.14) Consider now the formula from section 6 for the time-ordered product of operators. In the case of initial and \ufb01nal ground states, it becomes TQ(t1 \u03b4f (t1)... 0 0 if | h Using our explicit formula, eq. (7.11), we have (7.15). f =0 0 TQ(t1)Q(t2 \u03b4f (t1) 1 i \u03b4 0 0 if \u03b4f (t2) h | f =0 + \u221e \u2212\u221e = 1 i \u03b4 \u03b4f (t1) Z i G(t2 \u2212 h = 1 i G(t2 \u2212 = 1 t1). dt\u2032 G(t2 \u2212 t\u2032)f (t\u2032) 0 0 if | h f =0 0 0 t1) + (term with f \u2019s) if | h i f =0 (7.16) We can continue in this way to compute the ground-state expectation value of the time-ordered product of more Q(t)\u2019s. If the number of Q(t)\u2019s is odd, then there is always a left-over f (t) in the prefactor, and so the result is zero. If the number of Q(t)", "\u2019s is even, then we must pair up the functional derivatives in an appropriate way to get a nonzero result. Thus, for example, 0 TQ(t1)Q(t2)Q(t3)Q(t4) 0 i | | h = 1 G(t1\u2212 i2 h + G(t1\u2212 + G(t1\u2212 t2)G(t3\u2212 t3)G(t2\u2212 t4)G(t2\u2212 t4) t4). t3) i (7.17) (7.18) G(ti1 \u2212 ti2)... G(ti2n\u22121 \u2212 ti2n). Xpairings Problems More generally, 0 TQ(t1)... Q(t2n) 0 i | | h = 1 in 7.1) Starting with eq. (7.12), do the contour integral to verify eq. (7.14). 7.2) Starting with eq. (7.14), verify eq. (7.13). 7: The Path Integral for the Harmonic Oscillator 66 7.3) a) Use the Heisenberg equation of motion, \u02d9A = i[H, A], to \ufb01nd explicit expressions for \u02d9Q and \u02d9P. Solve these to get the Heisenberg-picture operators Q(t) and P (t) in terms of the Schr\u00a8odinger picture operators Q and P. b) Write the Schr\u00a8odinger picture operators Q and P in terms of the creation and annihilation operators a and a\u2020, where H = \u00afh\u03c9(a\u2020a+ 1 2 ). Then, using your result from part (a), write the Heisenberg-picture operators Q(t) and P (t) in terms of a and a\u2020. 0 c) Using your result from part (b), and a i | eqs. (7.16) and (7.17). = a\u2020 = 0, verify 0 | h 7.4) Consider a harmonic oscillator in its ground state at t = 2 that the oscillator is still in its ground state at t = +. It is then then subjected to an external force f (t). Compute the probability. \u221e Write your answer as a manifestly real expression, and in terms of + dt eiEtf (t). Your answer should the Fourier transform \u221e \u2212\u221e not involve any other unevaluated integrals. R 0 0 if | | |h f (E) = \u2212\u221e e 8: The Path Integral for Free Field Theory 67 8 The Path Integral for Free Field Theory Prerequisite: 3, 7 Our results for the harmonic oscillator can be straightforwardly generalized to a free \ufb01eld theory with hamiltonian density H0 = 1 2 \u03a02 + 1 2 ( \u2207 \u03d5)2 + 1 2 m2\u03d52. (8.1) The dictionary we need is q(t) Q(t) f (t) \u2212\u2192 \u2212\u2192 \u2212\u2192 \u03d5(x, t) (classical \ufb01eld) \u03d5(x, t) (operator \ufb01eld) J(x, t) (classical source) (8.2) The distinction between the classical \ufb01eld \u03d5(x) and the corresponding operator \ufb01eld should be clear from context. To employ the \u01eb trick, we multiply H0 with m2 alent to replacing m2 in simplicity, we will write m2 when we really mean m2 i\u01eb. The results are equivi\u01eb. From now on, for notational i\u01eb. H0 by 1 \u2212 \u2212 Let us write down the path integral (also called the functional integral) \u2212 for our free \ufb01eld theory: Z0(J) iJ = 0 0 | \u2261 h D Z \u03d5 ei d4x[ 0+J\u03d5], L R where L0 = is the lagrangian density, and \u2212 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 1 2 m2\u03d52 \u2212 \u03d5 D \u221d d\u03d5(x) x Y is the functional measure. Note that when we say path integral, we now mean a path in the space of \ufb01eld con\ufb01gurations. We can evaluate Z0(J) by mimicking what we did for the harmonic oscillator in section 7. We introduce four-dimensional Fourier transforms, \u03d5(k) = d4x e\u2212 ikx \u03d5(x), \u03d5(x) = d4k (2\u03c0)4 eikx Z \u03d5(k), (8.6) x, and k0", " is an integration variable. Then, starting e e where kx = with S0 = \u2212 d4x [ Z k0t + k \u00b7 S0 = R d4k (2\u03c0)4 1 2 Z \u2212 h L0 + J\u03d5], we get \u03d5(k)(k2 + m2) \u03d5( \u2212 k) + J(k) \u03d5( \u2212 k) + J( \u2212 k) \u03d5(k), (8.7) i e e e e e e (8.3) (8.4) (8.5) 68 (8.8) 8: The Path Integral for Free Field Theory where k2 = k2 \u2212 (k0)2. We now change path integration variables to J(k) k2 + m2. e Since this is merely a shift by a constant, we have e becomes \u03c7(k) = \u03d5(k) \u2212 e \u03d5 = D D \u03c7. The action S0 = 1 2 Z d4k (2\u03c0)4 \" J(k) J ( k) \u2212 k2 + m2 \u2212 e e \u03c7(k)(k2 + m2) \u03c7( \u2212. k) # (8.9) Just as for the harmonic oscillator, the integral over \u03c7 simply yields a factor of Z0(0) = e e iJ=0 = 1. Therefore 0 0 | h d4k (2\u03c0)4 Z0(J) = exp i 2 \" Z J ( J(k) k2 + m2 e e \u2212 \u2212 k) i\u01eb # = exp i 2 Z d4x d4x\u2032 J(x)\u2206(x \u2212 x\u2032)J(x\u2032). Here we have de\ufb01ned the Feynman propagator, \u2206(x \u2212 x\u2032) = Z d4k (2\u03c0)4 eik(x \u2212 k2 + m2 x\u2032) \u2212. i\u01eb (8.10) (8.11) The Feynman propagator is a Green\u2019s function for the Klein-Gordon equation, x + m2)\u2206(x \u22022 x\u2032) = \u03b44(x x\u2032). ( \u2212 \u2212 This can be seen directly by plugging eq. (8.11) into eq. (8.12) and then x\u2032) explicitly by taking the \u01eb 0 limit. We can also evaluate \u2206(x treating the k0 integral on the right-hand side of eq. (8.11) as a contour integral in the complex k0 plane, and then evaluating it via the residue theorem. The result is \u2192 \u2212 \u2212 (8.12) \u2206(x \u2212 x\u2032) = i Z dk eik \u00b7 (x \u2212 x\u2032) \u2212 i\u03c9 t\u2032 t \u2212 | | = i\u03b8(t t\u2032) f \u2212 Z dk eik(x x\u2032) + i\u03b8(t\u2032 \u2212 t) \u2212 Z dk e\u2212 ik(x x\u2032), \u2212 (8.13) f where \u03b8(t) is the unit step function. The integral over performed in terms of Bessel functions; see section 4. f dk can also be Now, by analogy with the formula for the ground-state expectation value of a time-ordered product of operators for the harmonic oscillator, we have f T\u03d5(x1 \u03b4J(x1). J=0... Z0(J) (8.14) 8: The Path Integral for Free Field Theory 69 Using our explicit formula, eq. (8.10), we have 0 T\u03d5(x1)\u03d5(x2 \u03b4J(x1) \u03b4 \u03b4J(x1) Z i \u2206(x2 \u2212 h = 1 i \u2206(x2 \u2212 = 1 J=0 \u03b4 \u03b4J(x2) d4x\u2032 \u2206(x2 \u2212 x1) + (term with J\u2019s) i Z0(J) x\u2032)J(x\u2032) Z0(J) Z0(J) J=0 x1). J=0 (8.15) We can continue in this way to compute the ground-state expectation value of the time-ordered product of more \u03d5\u2019s. If the number of \u03d5\u2019s is odd, then there is always a left-over J in the prefactor, and so the result is zero. If the number of \u03d5\u2019s is even, then we must pair up the functional derivatives in an appropriate way to get a nonzero result. Thus, for example, 0 T\u03d5(x1)\u03d5(x2)\u03d5(x3)\u03d5(x4", ") 0 i | | h = 1 \u2206(x1\u2212 i2 h + \u2206(x1\u2212 + \u2206(x1\u2212 x2)\u2206(x3\u2212 x3)\u2206(x2\u2212 x4)\u2206(x2\u2212 x4) x4). x3) i (8.16) \u2206(xi1\u2212 xi2)... \u2206(xi2n\u22121 \u2212 xi2n). (8.17) More generally, 0 T\u03d5(x1)... \u03d5(x2n) 0 i | | h = 1 in Xpairings This result is known as Wick\u2019s theorem. Problems 8.1) Starting with eq. (8.11), verify eq. (8.12). 8.2) Starting with eq. (8.11), verify eq. (8.13). 8.3) Starting with eq. (8.13), verify eq. (8.12). Note that the time derivatives in the Klein-Gordon wave operator can act on either the \ufb01eld (which obeys the Klein-Gordon equation) or the time-ordering step functions. 8.4) Use eqs. (3.19), (3.29), and (5.3) (and its hermitian conjugate) to verify the last line of eq. (8.15). 8.5) The retarded and advanced Green\u2019s functions for the Klein-Gordon y) = 0 y0. Find the pole prescriptions on the right-hand side of wave operator satisfy \u2206ret(x for x0 eq. (8.11) that yield these Green\u2019s functions. y0 and \u2206adv(x y) = 0 for x0 \u2265 \u2264 \u2212 \u2212 8: The Path Integral for Free Field Theory 70 8.6) Let Z0(J) = exp iW0(J), and evaluate the real and imaginary parts of W0(J). 8.7) Repeat the analysis of this section for the complex scalar \ufb01eld that was introduced in problem 3.5, and further studied in problem 5.1. Write your source term in the form J \u2020\u03d5 + J\u03d5\u2020, and \ufb01nd an explicit formula, analogous to eq. (8.10), for Z0(J \u2020, J). Write down the appropriate generalization of eq. (8.14), and use it to compute, 0 T\u03d5(x1)\u03d5(x2) 0 i | | h. Then verify your re0 T\u03d5\u2020(x1)\u03d5\u2020(x2) 0, and 0 T\u03d5\u2020(x1)\u03d5(x2) 0 h i | | h i | | sults by using the method of problem 8.4. Finally, give the appropriate generalization of eq. (8.17). 8.8) A harmonic oscillator (in units with m = \u00afh = 1) has a ground-state \u03c9q2/2. Now consider a real scalar \ufb01eld \u03d5(x), wave function and de\ufb01ne a \ufb01eld eigenstate 0 i \u221d | q h e\u2212 that obeys A i | \u03d5(x, 0) A i | = A(x) A i |, (8.18) where the function A(x) is everywhere real. For a free-\ufb01eld theory speci\ufb01ed by the hamiltonian of eq. (8.1), Show that the ground-state wave functional is 0 A i \u221d | h exp \"\u2212 1 2 Z d3k (2\u03c0)3 \u03c9(k) \u02dcA(k) \u02dcA( \u2212, k) # (8.19) where \u02dcA(k) \u2261 R d3x e\u2212 ik \u00b7 xA(x) and \u03c9(k) (k2 + m2)1/2. \u2261 9: The Path Integral for Interacting Field Theory 71 9 The Path Integral for Interacting Field Theory Prerequisite: 8 Let us consider an interacting quantum \ufb01eld theory speci\ufb01ed by a lagrangian of the form = L \u2212 1 2 Z\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 1 2 Zmm2\u03d52 + 1 6 Zgg\u03d53 + Y \u03d5. (9.1) As we discussed at the end of section 5, we \ufb01x the parameter m by requiring it to be equal to the actual mass of the particle (equivalently, the energy of the \ufb01rst excited state relative to the ground state), and we \ufb01x the parameter g by", " requiring some particular scattering cross section to depend on g in some particular way. (We will have more to say about this after we have learned to calculate cross sections.) We also assume that the \ufb01eld is normalized by = 0 0 \u03d5(x) 0 i | | h 0 \u03d5(x) i | | = 1, and 0 0 is the ground state, normalized via i | h Here one particle with four-momentum k\u00b5, where k2 = k\u00b5k\u00b5 = via = e\u2212 0 i | and k h \u2212 ikx. (9.2) is a state of k | m2, normalized i k\u2032 h k | i = (2\u03c0)32k0\u03b43(k\u2032 k). (9.3) \u2212, and 0 \u03d5 0 i | | h ), and we will use these four conditions to determine the values of 0 \u03d5 i | | Thus we have four conditions (the speci\ufb01ed values of m, g, k h the four remaining parameters (Y and the three Z\u2019s) that appear in. Before going further, we should note that this theory (known as \u03d53 theory, pronounced \u201cphi-cubed\u201d) actually has a fatal \ufb02aw. The hamiltonian density is L = 1 2 Z \u2212 \u03d5 \u03a02 1 Y \u03d5 + 1 2 Zmm2\u03d52 1 6 Zgg\u03d53. (9.4) H \u2212 \u2212 Classically, we can make this arbitrarily negative by choosing an arbitrarily large value for \u03d5. Quantum mechanically, this means that this hamiltonian has no ground state. If we start o\ufb00 near \u03d5 = 0, we can tunnel through the potential barrier to large \u03d5, and then \u201croll down the hill\u201d. However, this process is invisible in perturbation theory in g. The situation is exactly analogous to the problem of a harmonic oscillator perturbed by a q3 term. This system also has no ground state, but perturbation theory (both time dependent and time independent) does not \u201cknow\u201d this. We will be interested in eq. (9.1) only as an example of how to do perturbation expansions in a simple context, and so we will overlook this problem. We would like to evaluate the path integral for this theory, Z(J) iJ = 0 0 | \u2261 h D Z \u03d5 ei d4x[ 0+ 1+J\u03d5]. L L (9.5) R 9: The Path Integral for Interacting Field Theory 72 We can evaluate Z(J) by mimicking what we did for quantum mechanics at the end of section 6. Speci\ufb01cally, we can rewrite eq. (9.5) as Z(J) = ei d4x 1 L 1 i \u03b4 \u03b4J (x) R ei 1 i \u03b4 \u03b4J (x) d4x 1 L R \u221d \u03d5 ei d4x[ 0+J\u03d5]. L R D Z Z0(J), where Z0(J) is the result in free-\ufb01eld theory, Z0(J) = exp i 2 Z d4x d4x\u2032 J(x)\u2206(x \u2212 x\u2032)J(x\u2032). (9.6) (9.7) We have written Z(J) as proportional to (rather than equal to) the righthand side of eq. (9.6) because the \u01eb trick does not give us the correct overall normalization; instead, we must require Z(0) = 1, and enforce this by hand. Note that, in eq. (9.7), we have implicitly assumed that L0 = 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 1 2 m2\u03d52, \u2212 \u2212 L0 that gives us eq. (9.7). Therefore, the rest of (9.8) must be L since this is the included in L1. We write L1 = 1 Lct = 6 Zgg\u03d53 + 2 (Z\u03d5\u2212 1 Lct, 1)\u2202\u00b5\u03d5\u2202\u00b5\u03d5 1 2 (Zm\u2212 1)m2\u03d52 + Y \u03d5, (9.9) \u2212 where Y \u2212 0, Lct is called the counterterm lagrangian. We expect that, as g 1. In fact, as we will see, Y = O(g) and Zi = 1 + O(g2). 0 and Zi \u2192 \u2192 In order to make use of eq. (9.7), we will have to compute lots and lots of functional derivatives of Z0(J). Let us begin by ignoring", " the counterterms. We de\ufb01ne \u2192 Z1(J) exp \u221d i 6 \" Zgg d4x Z 1 i \u03b4 \u03b4J(x) 3 # Z0(J), (9.10) where the constant of proportionality is \ufb01xed by Z1(0) = 1. We now make a dual Taylor expansion in powers of g and J to get Z1(J) \u221d 1 V! \" iZgg 6 d4x Z 1 i \u03b4 \u03b4J(x) V 3 # \u221e XV =0 \u221e \u00d7 XP =0 1 P! i 2 Z d4y d4z J(y)\u2206(y P. z)J(z) \u2212 (9.11) If we focus on a term in eq. (9.11) with particular values of V and P, then the number of surviving sources (after we take all the functional derivatives) 9: The Path Integral for Interacting Field Theory 73 S = 2 3 S = 2 x 3! Figure 9.1: All connected diagrams with E = 0 and V = 2 = 24 S = 2 x 3! 3 S = 4! Figure 9.2: All connected diagrams with E = 0 and V = 4. \u2212 is E = 2P 3V. (Here E stands for external, a terminology that should become clear by the end of the next section; V stands for vertex and P for propagator.) The overall phase factor of such a term is then iV (1/i)3V iP = P, and the 3V functional derivatives can act on the 2P sources in iV +E \u2212 3V )! di\ufb00erent combinations. However, many of the resulting (2P )!/(2P expressions are algebraically identical. \u2212 To organize them, we introduce Feynman diagrams. In these diagrams, y), a i \u2206(x d4x J(x), and a d4x. Sets of diagrams with a line segment (straight or curved) stands for a propagator 1 \ufb01lled circle at one end of a line segment for a source i vertex joining three line segments for iZgg di\ufb00erent values of E and V are shown in \ufb01gs. (9.1\u20139.11). \u2212 R To count the number of terms on the right-hand side of eq. (9.11) that result in a particular diagram, we \ufb01rst note that, in each diagram, the number of lines is P and the number of vertices is V. We can rearrange the three functional derivatives from a particular vertex without changing the resulting diagram; this yields a counting factor of 3! for each vertex. Also, we can rearrange the vertices themselves; this yields a counting factor of V!. Similarly, we can rearrange the two sources at the ends of a particular propagator without changing the resulting diagram; this yields a counting R 9: The Path Integral for Interacting Field Theory 74 factor of 2! for each propagator. Also, we can rearrange the propagators themselves; this yields a counting factor of P!. All together, these counting factors neatly cancel the numbers from the dual Taylor expansions in eq. (9.11). However, this procedure generally results in an overcounting of the number of terms that give identical results. This happens when some rearrangement of derivatives gives the same match-up to sources as some rearrangement of sources. This possibility is always connected to some symmetry property of the diagram, and so the factor by which we have overcounted is called the symmetry factor. The \ufb01gures show the symmetry factor S of each diagram. Consider, for example, the second diagram of \ufb01g. (9.1). The three propagators can be rearranged in 3! ways, and all these rearrangements can be duplicated by exchanging the derivatives at the vertices. Furthermore the endpoints of each propagator can be simultaneously swapped, and the e\ufb00ect duplicated by swapping the two vertices. Thus, S = 2 3! = 12. Let us consider two more examples. In the \ufb01rst diagram of \ufb01g. (9.6), the exchange of the two external propagators (along with their attached sources) can be duplicated by exchanging all the derivatives at one vertex for those at the other, and simultaneously swapping the endpoints of each semicircular propagator. Also, the e\ufb00ect of swapping the top and bottom semicircular propagators can be duplicated by swapping the corresponding derivatives at each vertex. Thus, the symmetry factor is S = 2 2 = 4. In the diagram of \ufb01g. (9.10), we can exchange derivatives to match swaps of the top and bottom external propagators on the left, or the top", " and bottom external propagators on the right, or the set of external propagators on the left with the set of external propagators on the right. Thus, the symmetry factor is S = 2 2 = 8. 2 The diagrams in \ufb01gs. (9.1\u20139.11) are all connected: we can trace a path through the diagram between any two points on it. However, these are not the only contributions to Z(J). The most general diagram consists of a product of several connected diagrams. Let CI stand for a particular connected diagram, including its symmetry factor. A general diagram D can then be expressed as \u00d7 \u00d7 \u00d7 \u00d7 D = 1 SD (CI)nI, (9.12) YI where nI is an integer that counts the number of CI \u2019s in D, and SD is the additional symmetry factor for D (that is, the part of the symmetry factor that is not already accounted for by the symmetry factors already included in each of the connected diagrams). We now need to determine SD. 9: The Path Integral for Interacting Field Theory 75 S = 2 Figure 9.3: All connected diagrams with E = 1 and V = 1. S = 22 S = 22 S = 23 Figure 9.4: All connected diagrams with E = 1 and V = 3. S = 2 Figure 9.5: All connected diagrams with E = 2 and V = 0. S = 22 S = 22 Figure 9.6: All connected diagrams with E = 2 and V = 2. 9: The Path Integral for Interacting Field Theory 76 Since we have already accounted for propagator and vertex rearrangements within each CI, we need to consider only exchanges of propagators and vertices among di\ufb00erent connected diagrams. These can leave the total diagram D unchanged only if (1) the exchanges are made among di\ufb00erent but identical connected diagrams, and only if (2) the exchanges involve all of the propagators and vertices in a given connected diagram. If there are nI factors of CI in D, there are nI! ways to make these rearrangements. Overall, then, we have SD = nI!. (9.13) YI Now Z1(J) is given (up to an overall normalization) by summing all diagrams D, and each D is labeled by the integers nI. Therefore Z1(J) \u221d \u221d \u221d \u221d \u221d D nI } X{ nI } YI X{ \u221e 1 nI! (CI )nI 1 nI! (CI )nI YI XnI =0 exp (CI ) YI exp ( I CI ). (9.14) Thus we have a remarkable result: Z1(J) is given by the exponential of the sum of connected diagrams. This makes it easy to impose the normalization Z1(0) = 1: we simply omit the vacuum diagrams (those with no sources), like those of \ufb01gs. (9.1) and (9.2). We then have P where we have de\ufb01ned Z1(J) = exp[iW1(J)], iW1(J) CI, \u2261 0 = XI { } (9.15) (9.16) and the notation I from the sum, so that W1(0) = 0.1 0 } { = means that the vacuum diagrams are omitted Were it not for the counterterms in L1, we would have Z(J) = Z1(J). Let us see what we would get if this was, in fact, the case. In particular, let us compute the vacuum expectation value of the \ufb01eld \u03d5(x), which is given 1We have included a factor of i on the left-hand side of eq. (9.16) because then W1(J) is real in free-\ufb01eld theory; see problem 8.6. 6 6 9: The Path Integral for Interacting Field Theory 77 S = 2 3 S = 22 S = 23 = 24 S = 22 S = 22 Figure 9.7: All connected diagrams with E = 2 and V = 4. S = 3! Figure 9.8: All connected diagrams with E = 3 and V = 1. 9: The Path Integral for Interacting Field Theory 78 S = 3! S = 22 S = 22 Figure 9.9: All connected diagrams with E = 3 and V = 3. S = 23 Figure 9.10: All connected diagrams with E = 4 and V = 2. S = 24 S = 23 S = 24 S = 22 S = 22 S = 2 2 Figure 9.11: All connected diagrams with E = 4 and V = 4. 9: The Path Integral for Interacting Field Theory 79 Figure 9.12: All connected diagrams with E = 1, X number", " of one-point vertices from the linear counterterm), and V + X 1 (where X is the 3. \u2265 \u2264 by 0 \u03d5(x \u03b4J(x) = \u03b4 \u03b4J(x) Z1(J) W1(J) J=0 J=0. (9.17) This expression is then the sum of all diagrams [such as those in \ufb01gs. (9.3) and (9.4)] that have a single source, with the source removed: 0 \u03d5(x) 0 i | | h = 1 2 ig Z d4y 1 i \u2206(x \u2212 y) 1 i \u2206(y \u2212 y) + O(g3). (9.18) Here we have set Zg = 1 in the \ufb01rst term, since Zg = 1 + O(g2). We see the vacuum-expectation value of \u03d5(x) is not zero, as is required for the validity of the LSZ formula. To \ufb01x this, we must introduce the counterterm Y \u03d5. L1 introduces a new kind Including this term in the interaction lagrangian of vertex, one where a single line segment ends; the corresponding vertex d4y. The simplest diagrams including this new vertex are factor is iY shown in \ufb01g. (9.12), with a cross symbolizing the vertex. Assuming Y = O(g), only the \ufb01rst diagram in \ufb01g. (9.12) contributes at R O(g), and we have 0 \u03d5(x) 0 i | | h = iY + 1 2 (ig) 1 i \u2206(0) Z d4y 1 i \u2206(x \u2212 y) + O(g3). (9.19) Thus, in order to have = 0, we should choose 0 \u03d5(x) 0 | | i h 2 ig\u2206(0) + O(g3). Y = 1 (9.20) The factor of i is disturbing, because Y must be a real number: it is the coe\ufb03cient of a hermitian operator in the hamiltonian, as seen in eq. (9.4). Therefore, \u2206(0) must be purely imaginary, or we are in trouble. We have \u2206(0) = Z d4k (2\u03c0)4 1 k2 + m2. i\u01eb \u2212 (9.21) 9: The Path Integral for Interacting Field Theory 80 From eq. (9.21), it is not immediately obvious whether or not \u2206(0) is purely imaginary, but eq. (9.21) does reveal another problem: the integral diverges at large k. This is another example of an ultraviolet divergence, similar to the one we encountered in section 3 when we computed the zero-point energy of the \ufb01eld. To make some progress, we introduce an ultraviolet cuto\ufb00 \u039b, which we assume is much larger than m and any other energy of physical interest. Modi\ufb01cations to the propagator above some cuto\ufb00 may be well justi\ufb01ed physically; for example, quantum \ufb02uctuations in spacetime itself should become important above the Planck scale, which is given by the inverse square root of Newton\u2019s constant, and has the numerical value of 1019 GeV (compared to, say, the proton mass, which is 1 GeV). In order to retain the Lorentz-transformation properties of the propagator, we implement the ultraviolet cuto\ufb00 in a more subtle way than we did in section 3; spec\ufb01cally, we make the replacement \u2206(x y) \u2212 \u2192 Z d4k (2\u03c0)4 eik(x \u2212 k2 + m2 \u039b2 k2 + \u039b2 i\u01eb y) \u2212 2. i\u01eb! \u2212 (9.22) The integral is now convergent, and we can evaluate the modi\ufb01ed \u2206(0) with the methods of section 14; for \u039b m, the result is \u2206(0) = \u226b i 16\u03c02 \u039b2. (9.23) Thus Y is real, as required. If we like, we can now formally take the limit \u039b remains zero, at least to this order in g.. The parameter Y becomes in\ufb01nite, but 0 \u03d5(x) 0 i | | h \u2192 \u221e It may be disturbing to have a parameter in the lagrangian that is formally in\ufb01nite. However, such parameters are not directly measurable, and so need not obey our preconceptions about their magnitudes. Also, it is important to remember that", " Y includes a factor of g; this means that we can expand in powers of Y as part of our general expansion in powers of g. When we compute something measurable (like a scattering cross section), all the formally in\ufb01nite numbers will cancel in a well-de\ufb01ned way, leaving behind \ufb01nite coe\ufb03cients for the various powers of g. We will see how this works in detail in sections 14\u201320. As we go to higher orders in g, things become more complicated, but in principle the procedure is the same. Thus, at O(g3), we sum up the diagrams of \ufb01gs. (9.4) and (9.12), and then add to Y whatever O(g3) term = 0. In this way we can determine the is needed to maintain value of Y order by order in powers of g. 0 \u03d5(x) 0 i | | h Once this is done, there is a remarkable simpli\ufb01cation. Our adjustment = 0 means that the sum of all connected diagrams of Y to keep 0 \u03d5(x) 0 i | | h 9: The Path Integral for Interacting Field Theory 81 Figure 9.13: All connected diagrams without tadpoles with E V 4. \u2264 4 and \u2264 with a single source is zero. Consider now that same in\ufb01nite set of diagrams, but replace the single source in each of them with some other subdiagram. Here is the point: no matter what this replacement subdiagram is, the sum of all these diagrams is still zero. Therefore, we need not bother to compute any of them! The rule is this: ignore any diagram that, when a single line is cut, falls into two parts, one of which has no sources. All of these diagrams (known as tadpoles) are canceled by the Y counterterm, no matter what subdiagram they are attached to. The diagrams that remain (and need to be computed!) are shown in \ufb01g. (9.13). We turn next to the remaining two counterterms. For notational sim- plicity we de\ufb01ne A = Z\u03d5 \u2212 1, B = Zm \u2212 1, (9.24) 9: The Path Integral for Interacting Field Theory 82 and recall that we expect each of these to be O(g2). We now have Z(J) = exp i 2 \u2212 Z d4x 1 i \u03b4 \u03b4J(x) A\u22022 x + Bm2 \u2212 1 i \u03b4 \u03b4J(x) Z1(J). (9.25) We have integrated by parts to put both \u2202x\u2019s onto one \u03b4/\u03b4J(x). (Note that the time derivatives in this interaction should really be treated by including an extra source term for the conjugate momentum \u03a0 = \u02d9\u03d5. However, the space derivatives are correctly treated, and then the time derivatives must work out comparably by Lorentz invariance.) i) d4x ( Eq. (9.25) results in a new vertex at which two lines meet. The correA\u22022 sponding vertex factor is ( x acts on the x in \u2212 one or the other (but not both) propagators. (Which one does not matter, and can be changed via integration by parts.) Diagramatically, all we need do is sprinkle these new vertices onto the propagators in our existing diagrams. How many of these vertices we need to add depends on the order in g we are working to achieve. x + Bm2); the \u22022 \u2212 R This completes our calculation of Z(J) in \u03d53 theory. We express it as Z(J) = exp[iW (J)], (9.26) where W (J) is given by the sum of all connected diagrams with no tadpoles and at least two sources, and including the counterterm vertices just discussed. Now that we have Z(J), we must \ufb01nd out what we can do with it. Problems 9.1) Compute the symmetry factor for each diagram in \ufb01g. (9.13). (You can then check your answers by consulting the earlier \ufb01gures.) 9.2) Consider a real scalar \ufb01eld with L1, where 1 = L0 + L 2 m2\u03d52, 1 Lct, 1)\u2202\u00b5\u03d5\u2202\u00b5\u03d5 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 24 Z\u03bb\u03bb\u03d54 + 2 (Z\u03d5\u2212 \u2212 1 1 1 L0 = L1 = Lct = \u2212 \u2212 \u2212 2 (Zm\u2212 1)m2\u03d52. a) What kind of vertex appears in", " the diagrams for this theory (that is, how many line segments does it join?), and what is the associated vertex factor? b) Ignoring the counterterms, draw all the connected diagrams with 1 2, and \ufb01nd their symmetry factors. 4 and 0 E V \u2264 \u2264 \u2264 \u2264 c) Explain why we did not have to include a counterterm linear in \u03d5 to cancel tadpoles. 9: The Path Integral for Interacting Field Theory 83 9.3) Consider a complex scalar \ufb01eld (see problems 3.5, 5.1, and 8.7) with = L0 + L L1, where L0 = L1 = Lct = \u2212 \u2212 \u2212 \u2212 \u2202\u00b5\u03d5\u2020\u2202\u00b5\u03d5 4 Z\u03bb\u03bb(\u03d5\u2020\u03d5)2 + 1 (Z\u03d5\u2212 m2\u03d5\u2020\u03d5, Lct, \u2212 1)\u2202\u00b5\u03d5\u2020\u2202\u00b5\u03d5 1)m2\u03d5\u2020\u03d5. (Zm\u2212 This theory has two kinds of sources, J and J \u2020, and so we need a way to tell which is which when we draw the diagrams. Rather than labeling the source blobs with a J or J \u2020, we will indicate which is which by putting an arrow on the attached propagator that points towards the source if it is a J \u2020, and away from the source if it is a J. a) What kind of vertex appears in the diagrams for this theory, and what is the associated vertex factor? Hint: your answer should involve those arrows! b) Ignoring the counterterms, draw all the connected diagrams with 1 2, and \ufb01nd their symmetry factors. Hint: 4 and 0 \u2264 the arrows are important! \u2264 \u2264 \u2264 E V 9.4) Consider the integral exp W (g, J) \u2261 + \u221e dx exp 1 \u221a2\u03c0 Z \u2212\u221e 1 2 x2 + 1 6 gx3 + Jx i \u2212 h. (9.27) This integral does not converge, but it can be used to generate a joint power series in g and J, W (g, J) = \u221e \u221e CV,E gVJ E. (9.28) a) Show that XV =0 XE=0 CV,E = 1 SI, XI (9.29) where the sum is over all connected Feynman diagrams with E sources and V three-point vertices, and SI is the symmetry factor for each diagram. b) Use eqs. (9.27) and (9.28) to compute CV,E for V 5. (This is most easily done with a symbolic manipulation program like Mathematica.) Verify that the symmetry factors given in \ufb01gs. (9.1\u2013 9.11) satisfy the sum rule of eq. (9.29). 4 and E \u2264 \u2264 9: The Path Integral for Interacting Field Theory 84 c) Now consider W (g, J+Y ), with Y \ufb01xed by the \u201cno tadpole\u201d condition = 0. (9.30) \u2202 \u2202J W (g, J+Y ) \u221e \u221e J=0 XV =0 XE=0 1 SI XI CV,E = e W (g, J+Y ) = CV,E gVJ E. (9.31) e, (9.32) Then write Show that where the sum is over all connected Feynman diagrams with E sources and V three-point vertices and no tadpoles, and SI is the symmetry factor for each diagram. d) Let Y = a1g + a3g3 +..., and use eq. (9.30) to determine a1 and a3. Compute 4. Verify that the symmetry factors for the diagrams in \ufb01g. (9.13) satisfy the sum rule of eq. (9.32). CV,E for V 4 and E \u2264 \u2264 e = 9.5) The interaction picture. In this problem, we will derive a formula for without using path integrals. Suppose we have 0 T\u03d5(xn)... \u03d5(x1) 0 h i | | \u03d5)2 + H1, where a hamiltonian density 1 2 m2\u03d52, and H1 is a function of \u03a0(x, 0) and \u03d5(x, 0) and their spatial derivatives. (It should be chosen to preserve Lorentz invariance, but we will not be concerned with this issue.) We add a constant to H 0 so that H be the ground state of H0, with a constant i | |\u2205i H0.) The = 0. (H1 is then de\ufb01ned as H added to H0 so that", " H0|\u2205i Heisenberg-picture \ufb01eld is H0 = 1 2 \u03a02 + 1 2 ( H0 + = 0. Let H \u2207 \u2212 \u03d5(x, t) \u2261 eiHt\u03d5(x, 0)e\u2212 iHt. We now de\ufb01ne the interaction-picture \ufb01eld \u03d5I (x, t) \u2261 eiH0t\u03d5(x, 0)e\u2212 iH0t. (9.33) (9.34) a) Show that \u03d5I (x) obeys the Klein-Gordon equation, and hence is a free \ufb01eld. b) Show that \u03d5(x) = U \u2020(t)\u03d5I (x)U (t), where U (t) unitary. c) Show that U (t) obeys the di\ufb00erential equation i d where HI(t) = eiH0tH1e\u2212 teraction picture, and the boundary condition U (0) = 1. dt U (t) = HI(t)U (t), iH0t is the interaction hamiltonian in the in- \u2261 eiH0te\u2212 iHt is 9: The Path Integral for Interacting Field Theory 85 H1 is speci\ufb01ed by a particular function of the Schr\u00a8odinger-picture HI (t) is given by the same func- d) If \ufb01elds \u03a0(x, 0) and \u03d5(x, 0), show that tion of the interaction-picture \ufb01elds \u03a0I (x, t) and \u03d5I (x, t). e) Show that, for t > 0, t U (t) = T exp i \u2212 obeys the di\ufb00erential equation and boundary condition of part (c). What is the comparable expression for t < 0? Hint: you may need to de\ufb01ne a new ordering symbol. Z 0 dt\u2032 HI(t\u2032) (9.35) f) De\ufb01ne U (t2, t1) \u2261 U (t2)U \u2020(t1). Show that, for t2 > t1, U (t2, t1) = T exp t2 t1 Z i \u2212 dt\u2032 HI (t\u2032). (9.36) What is the comparable expression for t1 > t2? g) For any time ordering, show that U (t3, t1) = U (t3, t2)U (t2, t1) and that U \u2020(t1, t2) = U (t2, t1). h) Show that \u03d5(xn)... \u03d5(x1) = U \u2020(tn, 0)\u03d5I (xn)U (tn, tn 1)\u03d5I (xn... U (t2, t1)\u03d5I (x1)U (t1, 0). \u2212 1) \u2212 (9.37) i) Show that U \u2020(tn, 0) = U \u2020( U (t1,, 0). )U ( \u2212\u221e \u2212\u221e j) Replace H0 with (1 0, 0) and that U ( i | \u2212\u221e \u2212 = k) Show that. 0 i |\u2205ih\u2205|, 0)U ( \u221e, tn) and also that U (t1, 0) = \u221e i\u01eb)H0, and show that U \u2020( 0 | h \u221e, 0) = 0 |\u2205ih\u2205| h 0 \u03d5(xn)... \u03d5(x1) 0 i | | h = h\u2205|, tn)\u03d5I (xn)U (tn, tn U ( U (t2, t1)\u03d5I (x1)U (t1, \u221e 1)\u03d5I (xn \u2212 ) |\u2205i \u2212\u221e 1)... \u2212 (9.38) 0 i| \u00d7 |h\u2205| 2. l) Show that 0 T\u03d5(xn)... \u03d5(x1) 0 i | | h = m) Show that T\u03d5I (xn)... \u03d5I (x1)e\u2212 h\u2205| 0 i| \u00d7 |h\u2205| 2. i d4x HI (x) R |\u2205i (9.39) 0 i| |h\u2205| 2 = 1/ Te\u2212 h\u2205| i d4x HI (x) R. |\u2205i (9.40", ") 9: The Path Integral for Interacting Field Theory 86 Thus we have 0 T\u03d5(xn)... \u03d5(x1) 0 i | | h = h\u2205| T\u03d5I (xn)... \u03d5I (x1)e\u2212 i d4x HI (x) |\u2205i. Te\u2212 i d4x R HI (x) h\u2205| |\u2205i (9.41) We can now Taylor expand the exponentials on the right-hand side of eq. (9.41), and use free-\ufb01eld theory to compute the resulting correlation functions. R 10: Scattering Amplitudes and the Feynman Rules 87 10 Scattering Amplitudes and the Feynman Rules Prerequisite: 5, 9 Now that we have an expression for Z(J) = exp iW (J), we can take functional derivatives to compute vacuum expectation values of time-ordered products of \ufb01elds. Consider the case of two \ufb01elds; we de\ufb01ne the exact propagator via 0 T\u03d5(x1)\u03d5(x2) 0 i | | \u2261 h For notational simplicity let us de\ufb01ne i \u2206(x1 \u2212 x2) 1. (10.1) \u03b4j \u2261 1 i \u03b4 \u03b4J(xj ). (10.2) Then we have. J=0 J=0 J=0 J=0 J=0 \u2212 0 T\u03d5(x1)\u03d5(x2) 0 i | | h \u03b42iW (J) \u03b41iW (J) = \u03b41\u03b42Z(J) = \u03b41\u03b42iW (J) = \u03b41\u03b42iW (J) To get the last line we used \u03b4jW (J) = 0. Diagramat|J=0 = ically, \u03b41 removes a source, and labels the propagator endpoint x1. Thus 1 i \u2206(x1\u2212 x2) is given by the sum of diagrams with two sources, with those sources removed and the endpoints labeled x1 and x2. (The labels must be applied in both ways. If the diagram was originally symmetric on exchange of the two sources, the associated symmetry factor of 2 is then canceled by the double labeling.) At lowest order, the only contribution is the \u201cbarbell\u201d diagram of \ufb01g. (9.5) with the sources removed. Thus we recover the obvious fact that 1 x2) + O(g2). We will take up the subject of the O(g2) corrections in section 14. For now, let us go on to compute 0 \u03d5(xj ) 0 i | | h i \u2206(x1\u2212 i \u2206(x1\u2212 x2) = 1 (10.3) 0 T\u03d5(x1)\u03d5(x2)\u03d5(x3)\u03d5(x4) 0 i | | h = \u03b41\u03b42\u03b43\u03b44Z(J) = \u03b41\u03b42\u03b43\u03b44iW h + (\u03b41\u03b42iW )(\u03b43\u03b44iW ) + (\u03b41\u03b43iW )(\u03b42\u03b44iW ) J=0 i We have dropped terms that contain a factor of = 0. According to eq. (10.3), the last three terms in eq. (10.4) simply give products of the exact propagators. 0 \u03d5(x) 0 i | | h + (\u03b41\u03b44iW )(\u03b42\u03b43iW ). (10.4) 10: Scattering Amplitudes and the Feynman Rules 88 Let us see what happens when these terms are inserted into the LSZ formula for two incoming and two outgoing particles, = i4 f h i i | 1x\u2032 k\u2032 2x\u2032 k\u2032 2) Z \u2212 \u00d7 1\u2212 d4x1 d4x2 d4x\u20321 d4x\u20322 ei(k1x1+k2x2 \u22022 2 + m2)( \u22022 1 + m2)( \u22022 1\u2032 + m2)( ( \u2212 \u2212 \u2212. 0 T\u03d5(x1)\u03d5(x2)\u03d5(x\u20321)\u03d5(x\u20322) 0 | i | \u00d7h i \u2206(x2\u2212 x\u20321) 1 \u2212 \u22022 2\u2032 + m2) (10.5) If we consider, for example, 1 correlation function in eq. (10.5), we get from this term i \ufffd", "\ufffd(x1\u2212 x\u20322) as one term in the 1x\u2032 k\u2032 1\u2212 \u2212 2x\u2032 k\u2032 2)F (x11\u2032 )F (x22\u2032 ) k\u20322) F (\u00afk11\u2032 ) F (\u00afk22\u2032), (10.6) Z d4x1 d4x2 d4x\u20321 d4x\u20322 ei(k1x1+k2x2 = (2\u03c0)4\u03b44(k1\u2212 i +m2)( \u22022 ( \u2212 x\u2032j, and \u00afkij\u2032 k\u20321) (2\u03c0)4\u03b44(k2\u2212 j +m2)\u2206(xij ), \u22022 \u2261 \u2212 e e \u2261 \u2261 xi\u2212 F (k) is its Fourier transform, where F (xij) e (ki+k\u2032j)/2. The important point is the two delta xij\u2032 functions: these tell us that the four-momenta of the two outgoing particles (1\u2032 and 2\u2032) are equal to the four-momenta of the two incoming particles (1 and 2). In other words, no scattering has occurred. This is not the event whose probability we wish to compute! The other two similar terms in eq. (10.4) either contribute to \u201cno scattering\u201d events, or vanish due to factors like \u03b44(k1+k2) (which is zero because k0 2m > 0). In general, the diagrams that contribute to the scattering process of interest are only those that are fully connected: every endpoint can be reached from every other endpoint by tracing through the diagram. These are the diagrams that arise from all the \u03b4\u2019s acting on a single factor of W. Therefore, from here on, we restrict our attention to those diagrams alone. We de\ufb01ne the connected correlation functions via 1+k0 2 \u2265 0 T\u03d5(x1)... \u03d5(xE) 0 iC \u2261 | | h, J=0 (10.7) \u03b41... \u03b4EiW (J) and use these instead of 0 T\u03d5(x1)... \u03d5(xE ) 0 i | | h in the LSZ formula. Returning to eq. (10.4), we have iC = \u03b41\u03b42\u03b41\u2032 \u03b42\u2032iW 0 T\u03d5(x1)\u03d5(x2)\u03d5(x\u20321)\u03d5(x\u20322) 0 | | h. (10.8) The lowest-order (in g) nonzero contribution to this comes from the diagram of \ufb01g. (9.10), which has four sources and two vertices. The four \u03b4\u2019s remove the four sources; there are 4! ways of matching up the \u03b4\u2019s to the sources. These 24 diagrams can then be collected into 3 groups of 8 diagrams each; the 8 diagrams in each group are identical. The 3 distinct diagrams are shown in \ufb01g. (10.1). Note that the factor of 8 neatly cancels the symmetry factor S = 8 of the diagram with sources. J=0 10: Scattering Amplitudes and the Feynman Rules 89 Figure 10.1: The three tree-level Feynman diagrams that contribute to the connected correlation function iC. 0 T\u03d5(x1)\u03d5(x2)\u03d5(x\u20321)\u03d5(x\u20322) 0 | | h This is a general result for tree diagrams (those with no closed loops): once the sources have been stripped o\ufb00 and the endpoints labeled, each diagram with a distinct endpoint labeling has an overall symmetry factor of one. The tree diagrams for a given process represent the lowest-order (in g) nonzero contribution to that process. We now have 0 T\u03d5(x1)\u03d5(x2)\u03d5(x\u20321)\u03d5(x\u20322) 0 iC | | h 5 d4y d4z \u2206(y = (ig)2 1 i Z y)\u2206(x2\u2212 \u2206(x1\u2212 \u00d7 h y)\u2206(x\u20321\u2212 + \u2206(x1\u2212 y)\u2206(x\u20322\u2212 + \u2206(x1\u2212 + O(g4). z) \u2212 y)\u2206(x\u20321\u2212 y)\u2206(x2\u2212 y)\u2206(x2\u2212 z)\u2206(x\u20322\u2212 z)\u2206(x\u20322\u2212 z)\u2206(x\u20321\u2212 z) z) z) i (10.9) Next, we use eq. (10.9) in the LSZ formula, eq. (", "10.5). Each Klein-Gordon wave operator acts on a propagator to give ( \u2212 i + m2)\u2206(xi \u2212 \u22022 y) = \u03b44(xi \u2212 y). (10.10) The integrals over the external spacetime labels x1,2,1\u2032,2\u2032 are then trivial, and we get f h i i | = (ig)2 1 i d4y d4z \u2206(y Z \u2212 z) k\u2032 1z \u2212 \u2212 k\u2032 2z) ei(k1y+k2y + ei(k1y+k2z h + ei(k1y+k2z \u2212 \u2212 k\u2032 1y k\u2032 1z k\u2032 2z) k\u2032 2y) \u2212 \u2212 + O(g4). (10.11) i This can be simpli\ufb01ed by substituting \u2206(y \u2212 z) = Z d4k (2\u03c0)4 eik(y \u2212 k2 + m2 z) i\u01eb \u2212 (10.12) 10: Scattering Amplitudes and the Feynman Rules 90 into eq. (10.9). Then the spacetime arguments appear only in phase factors, and we can integrate them to get delta functions: = ig2 f h i i | Z d4k (2\u03c0)4 1 k2 + m2 i\u01eb \u2212 (2\u03c0)4\u03b44(k1+k2+k) (2\u03c0)4\u03b44(k\u20321+k\u20322+k) k2+k) h \u00d7 + (2\u03c0)4\u03b44(k1\u2212 + (2\u03c0)4\u03b44(k1\u2212 k\u20321\u2212 = ig2 (2\u03c0)4\u03b44(k1+k2\u2212 1 (k1+k2)2 + m2 + \u00d7 + O(g4). k\u20321+k) (2\u03c0)4\u03b44(k\u20322\u2212 k\u20322+k) (2\u03c0)4\u03b44(k\u20321\u2212 k\u20322) 1 k\u20321)2 + m2 + (k1\u2212 i 1 k2+k) + O(g4) (k1\u2212 k\u20322)2 + m2 (10.13) \u2212 In eq. (10.13), we have left out the i\u01eb\u2019s for notational convenience only; m2 is really m2 i\u01eb. The overall delta function in eq. (10.13) tells that that four-momentum is conserved in the scattering process, which we should, of course, expect. For a general scattering process, it is then convenient to de\ufb01ne a scattering matrix element T = (2\u03c0)4\u03b44(kin\u2212 where kin and kout are the total four-momenta of the incoming and outgoing particles, respectively. kout)i T (10.14) i i | f h via, Examining the calculation which led to eq. (10.13), we can take away some universal features that lead to a simple set of Feynman rules for computing contributions to i for a given scattering process. The Feynman rules are: T 1. Draw lines (called external lines) for each incoming and each outgoing particle. 2. Leave one end of each external line free, and attach the other to a vertex at which exactly three lines meet. Include extra internal lines in order to do this. In this way, draw all possible diagrams that are topologically inequivalent. 3. On each incoming line, draw an arrow pointing towards the vertex. On each outgoing line, draw an arrow pointing away from the vertex. On each internal line, draw an arrow with an arbitrary direction. 4. Assign each line its own four-momentum. The four-momentum of an external line should be the four-momentum of the corresponding particle. 10: Scattering Amplitudes and the Feynman Rules 91 k1 k2 k1 k2+ k1 k2 k1 k1 k1 k2 k 1 k 2 k1 k1 k 2 k2 k 1 k 2 Figure 10.2: The tree-level s-, t-, and u-channel diagrams contributing to i T for two particle scattering. 5. Think of the four-momenta as \ufb02owing along the arrows, and conserve four-momentum at each vertex. For a tree diagram, this \ufb01xes the momenta on all the internal lines. 6. The value of a diagram consists of the following factors: for each external line, 1; for each internal line with momentum k, for each vertex, iZgg. i/(k2 + m2 \u2212 i\u01eb); \u2212 7. A diagram with L closed loops will have L", " internal momenta that are not \ufb01xed by rule #5. Integrate over each of these momenta \u2113i with measure d4\u2113i/(2\u03c0)4. 8. A loop diagram may have some leftover symmetry factors if there are exchanges of internal propagators and vertices that leave the diagram unchanged; in this case, divide the value of the diagram by the symmetry factor associated with exchanges of internal propagators and vertices. 9. Include diagrams with the counterterm vertex that connects two propagators, each with the same four-momentum k. The value of this vertex is 1, and each is O(g2). i(Ak2 + Bm2), where A = Z\u03d5 \u2212 1 and B = Zm \u2212 \u2212 10. The value of i T is given by a sum over the values of all these diagrams. For the two-particle scattering process, the tree diagrams resulting from these rules are shown in \ufb01g. (10.2). Now that we have our procedure for computing the scattering amplitude, we must see how to relate it to a measurable cross section. T Problems 10: Scattering Amplitudes and the Feynman Rules 92 10.1) Use eq. (9.41) of problem 9.5 to rederive eq. (10.9). 10.2) Write down the Feynman rules for the complex scalar \ufb01eld of problem 9.3. Remember that there are two kinds of particles now (which we can think of as positively and negatively charged), and that your rules must have a way of distinguishing them. Hint: the most direct approach requires two kinds of arrows: momentum arrows (as discussed in this section) and what we might call \u201ccharge\u201d arrows (as discussed in problem 9.3). Try to \ufb01nd a more elegant approach that requires only one kind of arrow. 10.3) Consider a complex scalar \ufb01eld \u03d5 that interacts with a real scalar \ufb01eld \u03c7 via L1 = g\u03c7\u03d5\u2020\u03d5. Use a solid line for the \u03d5 propagator and a dashed line for the \u03c7 propagator. Draw the vertex (remember the arrows!), and \ufb01nd the associated vertex factor. 10.4) Consider a real scalar \ufb01eld with ated vertex factor. L1 = 1 2 g\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5. Find the associ- 10.5) The scattering amplitudes should be unchanged if we make a \ufb01eld rede\ufb01nition. Suppose, for example, we have = L \u2212 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 1 2 m2\u03d52, and we make the \ufb01eld rede\ufb01nition \u03d5 + \u03bb\u03d52. \u03d5 \u2192 (10.15) (10.16) Work out the lagrangian in terms of the rede\ufb01ned \ufb01eld, and the cor\u03d5\u03d5 responding Feynman rules. Compute (at tree level) the \u03d5\u03d5 scattering amplitude. You should get zero, because this is a free-\ufb01eld theory in disguise. (At the loop level, we also have to take into account the transformation of the functional measure \u03d5; see section 85.) \u2192 D 11: Cross Sections and Decay Rates 93 11 Cross Sections and Decay Rates Prerequisite: 10 Now that we have a method for computing the scattering amplitude, we must convert it into something that could be measured in an experiment. In practice, we are almost always concerned with one of two generic cases: one incoming particle, for which we compute a decay rate, or two incoming particles, for which we compute a cross section. We begin with the latter. T Let us also specialize, for now, to the case of two outgoing particles as well as two incoming particles. In \u03d53 theory, we found in section 10 that in this case we have + O(g4), = g2 T 1 (k1+k2)2 + m2 + 1 k\u20321)2 + m2 + (k1\u2212 1 (k1\u2212 k\u20322)2 + m2 (11.1) where k1 and k2 are the four-momenta of the two incoming particles, k\u20321 and k\u20322 are the four-momenta of the two outgoing particles, and k1+k2 = k\u20321+k\u20322. Also, these particles are all on shell: k2 i. (Here, for later use, we allow for the possibility that the particles all have di\ufb00erent masses.) i = m2 \u2212 Let us think about the kinematics of this process. In the center-ofmass frame, or CM frame", " for short, we take k1 + k2 = 0, and choose k1 to be in the +z direction. Now the only variable left to specify about the initial state is the magnitude of k1. Equivalently, we could specify the total energy in the CM frame, E1 + E2. However, it is even more convenient to (k1 + k2)2. In the CM frame, s reduces to de\ufb01ne a Lorentz scalar s (E1 + E2)2; s is therefore called the center-of-mass energy squared. Then, since E1 = (k2 in terms of s, with the result 2)1/2, we can solve for 1)1/2 and E2 = (k2 1 + m2 1 + m2 k1| | \u2261 \u2212 = k1| | 1 2\u221as s2 q \u2212 2(m2 1 + m2 2)s + (m2 1 \u2212 m2 2)2 (CM frame). (11.2) Now consider the two outgoing particles. Since momentum is conserved, we must have k\u20321 + k\u20322 = 0, and since energy is conserved, we must also have (E\u20321 + E\u20322)2 = s. Then we \ufb01nd = k\u20321| | 1 2\u221as s2 q \u2212 2(m2 1\u2032 + m2 2\u2032)s + (m2 1\u2032 m2 2\u2032)2 \u2212 (CM frame). (11.3) Now the only variable left to specify about the \ufb01nal state is the angle \u03b8 between k1 and k\u20321. However, it is often more convenient to work with the Lorentz scalar t k\u20321)2, which is related to \u03b8 by (k1 \u2212 \u2261 \u2212 t = m2 1 + m2 1\u2032 k1|| 2E1E\u20321 + 2 | k\u20321| \u2212 cos \u03b8. (11.4) 11: Cross Sections and Decay Rates 94 This formula is valid in any frame. The Lorentz scalars s and t are two of the three Mandelstam variables, de\ufb01ned as s t u (k1+k2)2 = k\u20321)2 = (k1\u2212 k\u20322)2 = (k1\u2212 (k\u20321+k\u20322)2, k\u20322)2, (k2\u2212 k\u20321)2. (k211.5) The three Mandelstam variables are not independent; they satisfy the linear relation s + t + u = m2 1 + m2 2 + m2 1\u2032 + m2 2\u2032. (11.6) In terms of s, t, and u, we can rewrite eq. (11.1) as = g2 T 1 m2 \u2212 + s 1 m2 \u2212 + t m2 1 u \u2212 + O(g4), (11.7) which demonstrates the notational utility of the Mandelstam variables. Now let us consider a di\ufb00erent frame, the \ufb01xed target or FT frame (also sometimes called the lab frame), in which particle #2 is initially at rest: k2 = 0. In this case we have = k1| | 1 2m2 q s2 \u2212 2(m2 1 + m2 2)s + (m2 1 \u2212 m2 2)2 (FT frame). (11.8) Note that, from eqs. (11.8) and (11.2), m2| k1|FT = \u221as k1|CM. | (11.9) This will be useful later. We would now like to derive a formula for the di\ufb00erential scattering cross section. In order to do so, we assume that the whole experiment is taking place in a big box of volume V, and lasts for a large time T. We should really think about wave packets coming together, but we will use some simple shortcuts instead. Also, to get a more general answer, we will let the number of outgoing particles be arbitrary. Recall from section 10 that the overlap between the initial and \ufb01nal states is given by f h i i | To get a probability, we must square initial and \ufb01nal states:. (11.10) kout)i T = (2\u03c0)4\u03b44(kin\u2212, and divide by the norms of the i i | 2 i i| | i i i | ih f h f P = |h f f | h (11.11). 11: Cross Sections and Decay Rates 95 The numerator of this expression is 2 = [(2\u03c0)4\u03b44(kin\u2212 i i| | We write the square of the delta function", " as f |h kout)]2 2. |T | (11.12) [(2\u03c0)4\u03b44(kin\u2212 kout)]2 = (2\u03c0)4\u03b44(kin\u2212 kout) \u00d7 (2\u03c0)4\u03b44(0), (11.13) and note that (2\u03c0)4\u03b44(0) = d4x ei0 \u00b7 x = V T. (11.14) Also, the norm of a single particle state is given by Z = (2\u03c0)32k0\u03b43(0) = 2k0V. (11.15) k h k | i Thus we have i i i | h = 4E1E2V 2, n\u2032 Yj=1 2k\u2032j 0V, f h f | i = (11.16) (11.17) where n\u2032 is the number of outgoing particles. If we now divide eq. (11.11) by the elapsed time T, we get a probability per unit time \u02d9P = (2\u03c0)4\u03b44(kin\u2212 4E1E2V 2 2 kout) V |T | n\u2032 j=1 2k\u20320 j V. (11.18) Q This is the probability per unit time to scatter into a set of outgoing particles with precise momenta. To get something measurable, we should sum each outgoing three-momentum k\u2032j over some small range. Due to the box, all three-momenta are quantized: k\u2032j = (2\u03c0/L)n\u2032j, where V = L3, and n\u2032j is a three-vector with integer entries. (Here we have assumed periodic boundary conditions, but this choice does not a\ufb00ect the \ufb01nal result.) In the limit of large L, we have V (2\u03c0)3 \u2192 Xn\u2032 j d3k\u2032j. Z (11.19) Thus we should multiply \u02d9P by a factor of V d3k\u2032j/(2\u03c0)3 for each outgoing particle. Then we get \u02d9P = (2\u03c0)4\u03b44(kin\u2212 4E1E2V kout) 2 |T | n\u2032 dk\u2032j, Yj=1 f (11.20) 11: Cross Sections and Decay Rates 96 where we have identi\ufb01ed the Lorentz-invariant phase-space di\ufb00erential d3k (2\u03c0)32k0 dk \u2261 (11.21) that we \ufb01rst introduced in section 3. f To convert \u02d9P to a di\ufb00erential cross section d\u03c3, we must divide by the incident \ufb02ux. Let us see how this works in the FT frame, where particle #2 is at rest. The incident \ufb02ux is the number of particles per unit volume that are striking the target particle (#2), times their speed. We have one k1| incident particle (#1) in a volume V with speed v = /E1, and so the | /E1V. Dividing eq. (11.20) by this \ufb02ux cancels the last incident \ufb02ux is k1| factor of V, and replaces E1 in the denominator with. We also set | m2 as a function of s; d\u03c3 will E2 = m2 and note that eq. (11.8) gives be Lorentz invariant if, in other frames, we simply use this function as the value of k1| | m2. Adopting this convention, and using eq. (11.9), we have k1| | k1| | d\u03c3 = 1 k1|CM\u221as |T | 4 | 2 dLIPSn\u2032(k1+k2), (11.22) where the n\u2032-body Lorentz-invariant phase-space measure k1|CM is given as a function of s by eq. (11.2), and we have de\ufb01ned | dLIPSn\u2032(k) \u2261 (2\u03c0)4\u03b44(k n\u2032 n\u2032 j=1 k\u2032i) dk\u2032j. Yj=1 f \u2212 P (11.23) Eq. (11.22) is our \ufb01nal result for the di\ufb00erential cross section for the scattering of two incoming particles into n\u2032 outgoing particles. Let us now specialize to the case of two outgoing particles. We need to evaluate dLIPS2(k) = (2\u03c0)4\u03b44(k k\u20321\u2212 \u2212 k\u20322) dk\u20321 dk\u20322, (11.24) where k = k1 + k", "2. Since dLIPS2(k) is Lorentz invariant, we can compute it in any convenient frame. Let us work in the CM frame, where k = k1 + k2 = 0 and k0 = E1 + E2 = \u221as; then we have f f dLIPS2(k) = 1 4(2\u03c0)2E\u20321E\u20322 \u03b4(E\u20321+E\u20322\u2212 \u221as ) \u03b43(k\u20321+k\u20322) d3k\u20321d3k\u20322. (11.25) We can use the spatial part of the delta function to integrate over d3k\u20322, with the result dLIPS2(k) = 1 4(2\u03c0)2E\u20321E\u20322 \u03b4(E\u20321+E\u20322\u2212 \u221as ) d3k\u20321, (11.26) 11: Cross Sections and Decay Rates 97 where now E\u20321 = k\u20321 2 + m2 1\u2032 and E\u20322 = k\u20321 2 + m2 2\u2032. (11.27) Next, let us write q q d3k\u20321 = k\u20321| | 2 d k\u20321| | d\u2126CM, (11.28) where d\u2126CM = sin \u03b8 d\u03b8 d\u03c6 is the di\ufb00erential solid angle, and \u03b8 is the angle between k1 and k\u20321 in the CM frame. We can carry out the integral over the 1, where xi magnitude of k\u20321 in eq. (11.26) using satis\ufb01es f (xi) = 0. In our case, the argument of the delta function vanishes R, the value given by eq. (11.3). Also, the derivative at just one value of of that argument with respect to dx \u03b4(f (x)) = f \u2032(xi) k\u20321| | |\u2212 i | P is k\u20321| | \u2202 k\u20321| | \u2202 E\u20321 + E\u20322 \u2212 \u221as = | k\u20321| E\u20321 + | k\u20321| E\u20322 E\u20321 + E\u20322 E\u20321 E\u20322 = | k\u20321| \u221as k\u20321| E\u20321E\u20322 = |. (11.29) Putting all of this together, we get dLIPS2(k) = | k\u20321| 16\u03c02\u221as d\u2126CM. Combining this with eq. (11.22), we have d\u03c3 d\u2126CM = 1 64\u03c02s k\u20321| | k1| | 2, |T | (11.30) (11.31) where and d\u2126CM is the di\ufb00erential solid angle in the CM frame. are the functions of s given by eqs. (11.2) and (11.3), k\u20321| | k1| | and The di\ufb00erential cross section can also be expressed in a frame-independent manner by noting that, in the CM frame, we can take the di\ufb00erential of eq. (11.4) at \ufb01xed s to get dt = 2 = 2 k1| | | k1| | | k\u20321| k\u20321| d cos \u03b8 d\u2126CM 2\u03c0. Now we can rewrite eq. (11.31) as d\u03c3 dt = 1 64\u03c0s 2, 2 |T | k1| | (11.32) (11.33) (11.34) 11: Cross Sections and Decay Rates 98 where is given as a function of s by eq. (11.2). k1| | We can now transform d\u03c3/dt into d\u03c3/d\u2126 in any frame we might like (such as the FT frame) by taking the di\ufb00erential of eq. (11.4) in that frame. depends on \u03b8 as well as s, so the result is more In general, though, complicated than it is in eq. (11.32) for the CM frame. k\u20321| | Returning to the general case of n\u2032 outgoing particles, we can de\ufb01ne a Lorentz invariant total cross section by integrating completely over all the outgoing momenta, and dividing by an appropriate symmetry factor S. If there are n\u2032i identical outgoing particles of type i, then and \u03c3 = S = n\u2032i!, Yi 1 S d\u03c3, (11.35) (11.36) Z where d\u03c3 is given by eq. (11.22). We need the symmetry factor because merely integrating over all the outgoing momenta in dLIPSn\u2032 treats the \ufb01", "nal state as being labeled by an ordered list of these momenta. But if some outgoing particles are identical, this is not correct; the momenta of the identical particles should be speci\ufb01ed by an unordered list (because, for ). The symmetry 0 is identical to the state a\u20202a\u20201| 0 example, the state a\u20201a\u20202| i i factor provides the appropriate correction. In the case of two outgoing particles, eq. (11.36) becomes \u03c3 = = 1 S Z 2\u03c0 S d\u2126CM d\u03c3 d\u2126CM +1 d cos \u03b8 d\u03c3 d\u2126CM, 1 Z \u2212 (11.37) (11.38) where S = 2 if the two outgoing particles are identical, and S = 1 if they are distinguishable. Equivalently, we can compute \u03c3 from eq. (11.34) via \u03c3 = 1 S tmax tmin Z dt d\u03c3 dt, (11.39) 1 where tmin and tmax are given by eq. (11.4) in the CM frame with cos \u03b8 = \u2212 and +1, respectively. To compute \u03c3 with eq. (11.38), we should \ufb01rst express t and u in terms of s and \u03b8 via eqs. (11.4) and (11.6), and then integrate over \u03b8 at \ufb01xed s. To compute \u03c3 with eq. (11.39), we should \ufb01rst express u in terms of s and t via eq. (11.6), and then integrate over t at \ufb01xed s. Let us see how all this works for the scattering amplitude of \u03d53 theory, eq. (11.7). In this case, all the masses are equal, and so, in the CM frame, 11: Cross Sections and Decay Rates 99 E = 1 eq. (11.4) becomes 2 \u221as for all four particles, and \u2212 From eq. (11.6), we also have t = 1 2 (s \u2212 k\u20321| | 4m2)(1 = 1 2 (s k1| | cos \u03b8). = \u2212 u = 1 2(s \u2212 \u2212 4m2)(1 + cos \u03b8). 4m2)1/2. Then \u2212 (11.40) (11.41) Thus limit, = T |T | k1| \u226a | 5g2 3m2 \" 2 is quite a complicated function of s and \u03b8. In the nonrelativistic m or equivalently s 4m2 \u2212 \u226a m2, we have s 8 15 4m2 \u2212 m2! + 5 18 1 \u2212 27 25 1 + cos2 \u03b8 s 2 4m2 \u2212 m2! +... # + O(g4). (11.42) Thus the di\ufb00erential cross section is nearly isotropic. In the extreme relativistic limit, m or equivalently s m2, we have k1| \u226b | g2 s sin2 \u03b8 \" = T 3 + cos2 \u03b8 \u2212 \u226b (3 + cos2 \u03b8)2 sin2 \u03b8 m2 s 16! \u2212 +... # + O(g4). (11.43) Now the di\ufb00erential cross section is sharply peaked in the forward (\u03b8 = 0) and backward (\u03b8 = \u03c0) directions. We can compute the total cross section \u03c3 from eq. (11.39). We have in 4m2) and tmax = 0. Since the two outgoing particles t, this case tmin = are identical, the symmetry factor is S = 2. Then setting u = 4m2 and performing the integral in eq. (11.39) over t at \ufb01xed s, we get (s \u2212 \u2212 \u2212 \u2212 s \u03c3 = g4 32\u03c0s(s 4m2) \" \u2212 + (s \u2212 s (s 2 m2 + 4m2 m2)(s \u2212 s 4m2 m2)2 \u2212 \u2212 \u2212 s ln 2m2) 2 3m2 \u2212 3m2 \u2212 m2!# + O(g6). (11.44) In the nonrelativistic limit, this becomes \u03c3 = 25g4 1152\u03c0m6 \" 1 \u2212 s 79 60 4m2 \u2212 m2! +... # + O(g6). (11.45) In the extreme relativistic limit, we get \u03c3 = g4 16\u03c0m2s2 \" 1 + 7 2 m2 s +... # + O(g6). (11.46) 11: Cross Sections and Decay Rates 100 These results illustrate how even a very", " simple quantum \ufb01eld theory can yield speci\ufb01c predictions for cross sections that could be tested experimentally. Let us now turn to the other basic problem mentioned at the beginning of this section: the case of a single incoming particle that decays to n\u2032 other particles. We have an immediate conceptual problem. According to our development of the LSZ formula in section 5, each incoming and outgoing particle should correspond to a single-particle state that is an exact eigenstate of the exact hamiltonian. This is clearly not the case for a particle that can decay. Referring to \ufb01g. (5.1), the hyperbola of such a particle must lie above the continuum threshold. Strictly speaking, then, the LSZ formula is not applicable. A proper understanding of this issue requires a study of loop corrections that we will undertake in section 25. For now, we will simply assume that the LSZ formula continues to hold for a single incoming particle. Then we can retrace the steps from eq. (11.11) to eq. (11.20); the only change is that the norm of the initial state is now i i i | h = 2E1V (11.47) instead of eq. (11.16). Identifying the di\ufb00erential decay rate d\u0393 with \u02d9P then gives 1 d\u0393 = 2E1 |T | 2 dLIPSn\u2032(k1), (11.48) 1 = m2 k2 1. In the CM frame (which is now the rest frame of where now s = the initial particle), we have E1 = m1; in other frames, the relative factor of E1/m1 in d\u0393 accounts for relativistic time dilation of the decay rate. \u2212 We can also de\ufb01ne a total decay rate by integrating over all the outgoing momenta, and dividing by the symmetry factor of eq. (11.35): \u0393 = 1 S Z d\u0393. (11.49) We will compute a decay rate in problem 11.1 Reference Notes For a derivation with wave packets, see Brown, Itzykson & Zuber, or Peskin & Schroeder. Problems 11: Cross Sections and Decay Rates 101 11.1) a) Consider a theory of a two real scalar \ufb01elds A and B with an L1 = gAB2. Assuming that mA > 2mB, compute the interaction total decay rate of the A particle at tree level. b) Consider a theory of a real scalar \ufb01eld \u03d5 and a complex scalar \ufb01eld \u03c7 with L1 = g\u03d5\u03c7\u2020\u03c7. Assuming that m\u03d5 > 2m\u03c7, compute the total decay rate of the \u03d5 particle at tree level. 11.2) Consider Compton scattering, in which a massless photon is scattered by an electron, initially at rest. (This is the FT frame.) In problem 2 for this process (summed over the possible 59.1, we will compute spin states of the scattered photon and electron, and averaged over the possible spin states of the initial photon and electron), with the result |T | 2 = 32\u03c02\u03b12 |T | \" m4 + m2(3s + u) s)2 (m2 su \u2212 + + \u2212 2m2(s + u + 2m2) s)(m2 (m2 u) # \u2212 \u2212 m4 + m2(3u + s) u)2 (m2 \u2212 su \u2212 + O(\u03b14) (11.50) where \u03b1 = 1/137.036 is the \ufb01ne-structure constant. a) Express the Mandelstam variables s and u in terms of the initial and \ufb01nal photon energies \u03c9 and \u03c9\u2032. b) Express the scattering angle \u03b8FT between the initial and \ufb01nal photon three-momenta in terms of \u03c9 and \u03c9\u2032. c) Express the di\ufb00erential scattering cross section d\u03c3/d\u2126FT in terms of \u03c9 and \u03c9\u2032. Show that your result is equivalent to the Klein-Nishina formula d\u03c3 d\u2126FT = \u03b12 2m2 2 \u03c9\u2032 \u03c92 \u03c9 \u03c9\u2032 + \u03c9\u2032 \u03c9 \u2212 sin2 \u03b8FT. (11.51) 11.3) Consider the process of muon decay, \u00b5\u2212 \u2192 |T | e\u2212\u03bde\u03bd\u00b5. In section 88, 2 for this process (summed over the possible spin we will compute states of the decay products, and averaged over the possible spin states of the initial muon), with the result 2 = 64G2 |T | F(k1 \u00b7 where GF is the Fermi", " constant, k1 is the four-momentum of the muon, and k\u20321,2,3 are the four-momenta of the \u03bde, \u03bd\u00b5, and e\u2212, respectively. In the rest frame of the muon, its decay rate is therefore k\u20322)(k\u20321 \u00b7 (11.52) k\u20323), \u0393 = 32G2 F m Z (k1 \u00b7 k\u20322)(k\u20321 \u00b7 k\u20323) dLIPS3(k1), (11.53) 11: Cross Sections and Decay Rates 102 where k1 = (m, 0), and m is the muon mass. The neutrinos are massless, and the electron mass is 200 times less than the muon mass, so we can take the electron to be massless as well. To evaluate \u0393, we perform the following analysis. a) Show that \u0393 = 32G2 F m Z dk\u20323 k1\u00b5k\u20323\u03bd Z k\u20322 \u00b5k\u20321 \u03bd dLIPS2(k1\u2212 k\u20323). b) Use Lorentz invariance to argue that, for m1\u2032 = m2\u2032 = 0, f k\u20321 \u00b5k\u20322 \u03bd dLIPS2(k) = Ak2g\u00b5\u03bd + Bk\u00b5k\u03bd, Z where A and B are numerical constants. c) Show that, for m1\u2032 = m2\u2032 = 0, dLIPS2(k) = Z 1 8\u03c0. (11.54) (11.55) (11.56) d) By contracting both sides of eq. (11.55) with g\u00b5\u03bd and with k\u00b5k\u03bd, and using eq. (11.56), evaluate A and B. e) Use the results of parts (b) and (d) in eq. (11.54). Set k1 = (m, 0), E\u20323 is the energy of the electron. and compute d\u0393/dEe; here Ee \u2261 Note that the maximum value of Ee is reached when the electron is emitted in one direction, and the two neutrinos in the opposite direction; what is this maximum value? f) Perform the integral over Ee to obtain the muon decay rate \u0393. 6 s. The muon g) The measured lifetime of the muon is 2.197 2. mass is 105.66 MeV. Determine the value of GF in GeV\u2212 (Your answer is too low by about 0.2%, due to loop corrections to the decay rate.) 10\u2212 \u00d7 1d\u0393/dEe. h) De\ufb01ne the energy spectrum of the electron P (Ee) Note that P (Ee)dEe is the probability for the electron to be emitted with energy between Ee and Ee + dEe. Draw a graph of P (Ee) vs. Ee/m\u00b5. \u0393\u2212 \u2261 11.4) Consider a theory of three real scalar \ufb01elds (A, B, and C) with = L \u2212 \u2212 1 1 2 \u2202\u00b5A\u2202\u00b5A 2 \u2202\u00b5B\u2202\u00b5B 2 \u2202\u00b5C\u2202\u00b5C 1 \u2212 + gABC. 1 1 2 m2 2 m2 2 m2 AA2 BB2 CC 2 1 \u2212 \u2212 \u2212 (11.57) 11: Cross Sections and Decay Rates 103 Write down the tree-level scattering amplitude (given by the sum of the contributing tree diagrams) for each of the following processes: AA AA AA AA AB AB \u2192 \u2192 \u2192 \u2192 \u2192 \u2192 AA, AB, BB, BC, AB, AC. (11.58) Your answers should take the form = g2 T cs m2 s \u2212 s + ct m2 t \u2212 + t cu m2 u \u2212 u, (11.59) where, in each case, each ci is a positive integer, and each m2 or m2 may be zero for some processes. B or m2 C. Hint: i is m2 A T 12: Dimensional Analysis with \u00afh = c = 1 104 12 Dimensional Analysis with \u00afh = c = 1 Prerequisite: 3 We have set \u00afh = c = 1. This allows us to convert a time T to a length L 1. via T = c\u2212 Thus any quantity A can be thought of as having units of mass to some power (positive, negative, or zero) that we will call [A]. For example, 1L, and a length L to an inverse mass M \u2212 1 via L = \u00afhc\u2212 1M \u2212 [m] = +1, [x\u00b5] = 1, [\u2202\u00b5] = +1, [ddx", "] = d. \u2212 \u2212 (12.1) (12.2) (12.3) (12.4) In the last line, we have generalized our considerations to theories in d spacetime dimensions. Let us now consider a scalar \ufb01eld in d spacetime dimensions with la- grangian density (12.5) (12.6) = L \u2212 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 1 2 m2\u03d52 N \u2212 Xn=3 1 n! gn\u03d5n. The action is and the path integral is Z(J) = S = Z ddx, L From eq. (12.7), we see that the action S must be dimensionless, because it appears as the argument of the exponential function. Therefore Z \u03d5 exp D i Z ddx ( L + J\u03d5). (12.7) Combining eqs. (12.4) and (12.8) yields [S] = 0. (12.8) [ L Then, from eqs. (12.9) and (12.3), and the fact that \u2202\u00b5\u03d5\u2202\u00b5\u03d5 is a term in we see that we must have ] = d. (12.9), L [\u03d5] = 1 2 (d 2). \u2212 Then, since gn\u03d5n is also a term in, we must have L [gn] = d 1 2 n(d 2). \u2212 \u2212 (12.10) (12.11) 12: Dimensional Analysis with \u00afh = c = 1 105 In particular, for the \u03d53 theory we have been working with, we have [g3] = 1 2 (6 d). \u2212 (12.12) Thus we see that the coupling constant of \u03d53 theory is dimensionless in d = 6 spacetime dimensions. Theories with dimensionless couplings tend to be more interesting than theories with dimensionful couplings. This is because any nontrivial dependence of a scattering amplitude on a coupling must be expressed as a function of a dimensionless parameter. If the coupling is itself dimensionful, this parameter must be the ratio of the coupling to the appropriate power of either the particle mass m (if it isn\u2019t zero) or, in the high-energy m2, the Mandelstam variable s. Thus the relevant parameregime s \u226b [g]/2. If [g] is negative [and it usually is: see eq. (12.11)], then ter is g s\u2212 [g]/2 blows up at high energies, and the perturbative expansion breaks g s\u2212 down. This behavior is connected to the nonrenormalizability of theories with couplings with negative mass dimension, a subject we will take up in section 18. It turns out that such theories require an in\ufb01nite number of input parameters to make sense; see section 29. In the opposite case, [g] [g]/2 goes positive, the theory becomes trivial at high energy, because g s\u2212 rapidly to zero. Thus the case of [g] = 0 is just right: scattering amplitudes can have a nontrivial dependence on g at all energies. Therefore, from here on, we will be primarily interested in \u03d53 theory in d = 6 spacetime dimensions, where [g3] = 0. Problems 12.1) Express \u00afhc in GeV fm, where 1 fm = 1 Fermi = 10\u2212 13 cm. 12.2) Express the masses of the proton, neutron, pion, electron, muon, and tau in GeV. 12.3) The proton is a strongly interacting blob of quarks and gluons. It has a nonzero charge radius rp, given by r2 d3x \u03c1(r)r2, where \u03c1(r) is the quantum expectation value of the electric charge distribution inside the proton. Estimate the value of rp, and then look up its measured value. How accurate was your estimate? p = R 13: The Lehmann-K\u00a8all\u00b4en Form of the Exact Propagator 106 13 The Lehmann-K\u00a8all\u00b4en Form of the Exact Propagator Prerequisite: 9 Before turning to the subject of loop corrections to scattering amplitudes, it will be helpful to consider what we can learn about the exact propagator \u2206(x y) from general principles. We de\ufb01ne the exact propagator via \u2212 \u2206(x y) 0 T\u03d5(x)\u03d5(y) 0 i i | | h. \u2261 \u2212 We take the \ufb01eld \u03d5(x) to be normalized so that 0 \u03d5(x) 0 i | | h = 0 and k h 0 \u03d5", "(x) i | | = e\u2212 ikx. (13.1) (13.2) In d spacetime dimensions, the one-particle state has the normalization k | i k h k\u2032 | i = (2\u03c0)d \u2212 1 2\u03c9 \u03b4d \u2212 1(k k\u2032), \u2212 (13.3) with \u03c9 = (k2 + m2)1/2. The corresponding completeness statement is f where I1 is the identity operator in the one-particle subspace, and dk k | k ih | = I1, Z dk \u2261 dd \u2212 (2\u03c0)d 1k 12\u03c9 \u2212 (13.4) (13.5) is the Lorentz invariant phase-space di\ufb00erential. We also de\ufb01ne the exact momentum-space propagator \u02dc\u2206(k2) via f \u2206(x y) \u2212 \u2261 Z ddk (2\u03c0)d eik(x y) \u02dc\u2206(k2). \u2212 In free-\ufb01eld theory, the momentum-space propagator is \u02dc\u2206(k2) = 1 k2 + m2. i\u01eb \u2212 (13.6) (13.7) It has an isolated pole at k2 = m2 with residue one; m is the actual, physical mass of the particle, the mass that enters into the energy-momentum relation. \u2212 We begin our analysis with eq. (13.1). We take x0 > y0, and insert a complete set of energy eigenstates between the two \ufb01elds. Recall from section 5 that there are three general classes of energy eigenstates: 13: The Lehmann-K\u00a8all\u00b4en Form of the Exact Propagator 107 1. The ground state or vacuum energy and momentum., which is a single state with zero 0 i | 2. The one particle states with energy \u03c9 = (k2 + m2)1/2. k |, speci\ufb01ed by a three-momentum k and i 3. States in the multiparticle continuum, speci\ufb01ed by a threei momentum k and other parameters (such as relative momenta among the di\ufb00erent particles) that we will collectively denote as n. The energy of one of these states is \u03c9 = (k2 + M 2)1/2, where M 2m; M is one of the parameters in the set n. k, n | \u2265 Thus we get 0 \u03d5(x)\u03d5(y) 0 i | | h = 0 \u03d5(y) 0 0 \u03d5(x) 0 i | | ih | | h dk k \u03d5(x) 0 | | h k ih 0 \u03d5(y dk k, n \u03d5(x) 0 | | h k, n ih 0 \u03d5(y) i | |. (13.8) f The sum over n is schematic, and includes integrals over continuous parameters like relative momenta. The \ufb01rst two terms in eq. (13.8) can be simpli\ufb01ed via eq. (13.2). Also, iP \u00b5x\u00b5)\u03d5(0)exp(+iP \u00b5x\u00b5), where P \u00b5 is the writing the \ufb01eld as \u03d5(x) = exp( energy-momentum operator, gives us \u2212 k, n h where k0 = (k2 + M 2)1/2. We now have 0 \u03d5(x) i | | = e\u2212 ikx k, n h 0 \u03d5(0) i | |, (13.9) 0 \u03d5(x)\u03d5(y) 0 i | | h = Z dk eik(x y) + \u2212 dk eik(x \u2212 y) k, n |h 0 \u03d5(0) i| | | 2. (13.10) Next, we de\ufb01ne the spectral density f n Z X f \u03c1(s) \u2261 k, n n |h X 0 \u03d5(0) i| | | 2 \u03b4(s \u2212 M 2). (13.11) 0 for s 4m2, and \u03c1(s) = 0 for s < 4m2. Now we have Obviously, \u03c1(s) \u2265 \u2265 dk eik(x 0 \u03d5(x)\u03d5(y) 0 i | | h = y) + \u2212 ds \u03c1(s) dk eik(x y). \u2212 (13.12) \u221e 4m2 Z In the \ufb01rst term, k0 = (k2+m2)1/2,", " and in the second term, k0 = (k2+s)1/2. Clearly we can also swap x and y to get f f Z Z 0 \u03d5(y)\u03d5(x) 0 i | | h = dk e\u2212 ik(x y) + \u2212 Z f \u221e 4m2 Z ds \u03c1(s) dk e\u2212 ik(x y) \u2212 (13.13) Z f 13: The Lehmann-K\u00a8all\u00b4en Form of the Exact Propagator 108 as well. We can then combine eqs. (13.12) and (13.13) into a formula for the time-ordered product 0 T\u03d5(x)\u03d5(y) 0 i | | h = \u03b8(x0 y0) 0 \u03d5(x)\u03d5(y) 0 i | | h \u2212 + \u03b8(y0 x0), 0 \u03d5(y)\u03d5(x) 0 i | | h (13.14) \u2212 where \u03b8(t) is the unit step function, by means of the identity ddk (2\u03c0)d eik(x \u2212 k2 + m2 y) i\u01eb \u2212 Z = i\u03b8(x0 \u2212 y0) dk eik(x \u2212 y) + i\u03b8(y0 Z x0) \u2212 f Z dk e\u2212 ik(x y) ; \u2212 (13.15) the derivation of eq. (13.15) was sketched in section 8. Combining eqs. (13.12\u2013 13.15), we get f 0 T\u03d5(x)\u03d5(y) 0 i i | | h = Z ddk (2\u03c0)d eik(x y) \u2212 \" 1 k2 + m2 i\u01eb \u2212 + \u221e 4m2 Z ds \u03c1(s) 1 k2 + s. i\u01eb # \u2212 Comparing eqs. (13.1), (13.6), and (13.16), we see that \u02dc\u2206(k2) = 1 k2 + m2 + \u221e 4m2 Z i\u01eb \u2212 ds \u03c1(s) 1 k2 + s. i\u01eb \u2212 (13.16) (13.17) This is the Lehmann-K\u00a8all\u00b4en form of the exact momentum-space propagator \u02dc\u2206(k2). We note in particular that \u02dc\u2206(k2) has an isolated pole at k2 = m2 with residue one, just like the propagator in free-\ufb01eld theory. \u2212 Problems 13.1) Consider an interacting scalar \ufb01eld theory in d spacetime dimensions, = L \u2212 1 2 Z\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 1 2 Zmm2\u03d52 \u2212 L1(\u03d5), (13.18) where L1(\u03d5) is a function of \u03d5 (and not its derivatives). The exact momentum-space propagator for \u03d5 can be expressed in LehmannK\u00a8all\u00b4en form by eq. (13.17). Find a formula for the renormalizing factor Z\u03d5 in terms of \u03c1(s). Hint: consider the commutator [\u03d5(x), \u02d9\u03d5(y)]. 14: Loop Corrections to the Propagator 109 14 Loop Corrections to the Propagator Prerequisite: 10, 12, 13 In section 10, we wrote the exact propagator as 1 i \u2206(x1\u2212 x2) 0 T\u03d5(x1)\u03d5(x2) 0 i | | \u2261 h, J=0 (14.1) = \u03b41\u03b42iW (J) where iW (J) is the sum of connected diagrams, and \u03b4i acts to remove a source from a diagram and label the corresponding propagator endpoint xi. In \u03d53 theory, the O(g2) corrections to 1 x2) come from the diagrams of \ufb01g. (14.1). To compute them, it is simplest to work directly in momentum space, following the Feynman rules of section 10. An appropriate assignment of momenta to the lines is shown in \ufb01g. (14.1); we then have i \u2206(x1\u2212 1 i \u02dc\u2206(k2) = 1 i \u02dc\u2206(k2) + 1 i \u02dc\u2206(k2) i\u03a0(k2) i h 1 i \u02dc\u2206(k2) + O(g4), (14.2) where \u02dc\u2206(k2) = 1 k2 + m2 i\u01eb \u2212 is the free-\ufb01", "eld propagator, and i\u03a0(k2) = 1 2 (ig)2 1 i 2 dd\u2113 (2\u03c0)d \u02dc\u2206((\u2113+k)2) \u02dc\u2206(\u21132) Z i(Ak2 + Bm2) + O(g4) \u2212 (14.3) (14.4) is the self-energy. Here we have written the integral appropriate for d spacetime dimensions; for now we will leave d arbitrary, but later we will want to focus on d = 6, where the coupling g is dimensionless. In the \ufb01rst term in eq. (14.4), the factor of one-half is the symmetry factor associated with exchanging the top and bottom semicircular propagators. Also, we have written the vertex factor as ig rather than iZgg because we expect Zg = 1 + O(g2), and so the Zg \u2212 1 contribution can be lumped into the O(g4) term. 1 and 1 are both expected to be O(g2). B = Zm \u2212 series It will prove convenient to de\ufb01ne \u03a0(k2) to all orders via the geometric In the second term, A = Z\u03d5 \u2212 1 i \u02dc\u2206(k2) = 1 i \u02dc\u2206(k2) + 1 i + 1 i +.... \u02dc\u2206(k2) h \u02dc\u2206(k2) h 1 i 1 i i\u03a0(k2) i i\u03a0(k2) i \u02dc\u2206(k2) \u02dc\u2206(k2) i\u03a0(k2) i h 1 i \u02dc\u2206(k2) (14.5) 14: Loop Corrections to the Propagator 110 k+l l k k k kk Figure 14.1: The O(g2) corrections to the propagator. Figure 14.2: The geometric series for the exact propagator. This is illustrated in \ufb01g. (14.2). The sum in eq. (14.5) will include all the diagrams that contribute to \u02dc\u2206(k2) if we take i\u03a0(k2) to be given by the sum of all diagrams that are one-particle irreducible, or 1PI for short. A diagram is 1PI if it is still connected after any one line is cut. The 1PI diagrams that make an O(g4) contribution to i\u03a0(k2) are shown in \ufb01g. (14.3). When writing down the value of one of these diagrams, we omit the two external propagators. If we sum up the series in eq. (14.5), we get \u02dc\u2206(k2) = k2 + m2 1 i\u01eb \u2212 \u2212. \u03a0(k2) (14.6) In section 13, we learned that the exact propagator has a pole at k2 = with residue one. This is consistent with eq. (14.6) if and only if m2 \u2212 m2) = 0, m2) = 0, \u03a0( \u03a0\u2032( \u2212 \u2212 (14.7) (14.8) where the prime denotes a derivative with respect to k2. We will use eqs. (14.7) and (14.8) to \ufb01x the values of A and B. Figure 14.3: The O(g4) contributions to i\u03a0(k2). 14: Loop Corrections to the Propagator 111 Next we turn to the evaluation of the O(g2) contribution to i\u03a0(k2) in eq. (14.4). We have the immediate problem that the integral on the righthand side diverges at large \u2113 for d 4. We faced a similar situation in \u2265 section 9 when we evaluated the lowest-order tadpole diagram. There we introduced an ultraviolet cuto\ufb00 \u039b that modi\ufb01ed the behavior of \u02dc\u2206(\u21132) at large \u21132. Here, for now, we will simply restrict our attention to d < 4, where the integral in eq. (14.4) is \ufb01nite. Later we will see what we can say about larger values of d. We will evaluate the integral in eq. (14.4) with a series of tricks. We \ufb01rst use Feynman\u2019s formula to combine denominators, 1 A1... An = dFn (x1A1 +... + xnAn)\u2212 n, (14.9) Z where the integration measure over the Feynman parameters xi is dFn = (n Z 1 1", ")! \u2212 0 Z dx1... dxn \u03b4(x1 +... + xn \u2212 1). (14.10) This measure is normalized so that dFn 1 = 1. Z We will prove eq. (14.9) in problem 14.1. In the case at hand, we have \u02dc\u2206((k+\u2113)2) \u02dc\u2206(\u21132) = 1 (\u21132 + m2)((\u2113 + k)2 + m2) = = = = 1 dx dx dx dx h h h h x((\u2113 + k)2 + m2) + (1 \u2212 x)(\u21132 + m2) \u21132 + 2x\u2113 \u00b7 k + xk2 + m2 2 \u2212 i x)k2 + m2 2 \u2212 i (\u2113 + xk)2 + x(1 \u2212 q2 + D 2 \u2212, i (14.11) 2 \u2212 i (14.12) where we have suppressed the i\u01eb\u2019s for notational convenience; they can be m2 restored via the replacement m2 i\u01eb. In the last line we have de\ufb01ned \u2192 \u2212 and \u2113 + xk q \u2261 D x(1 \u2212 \u2261 x)k2 + m2. (14.13) (14.14) 14: Loop Corrections to the Propagator 112 Im q 0 Re q 0 Figure 14.4: The q0 integration contour along the real axis can be rotated to the imaginary axis without passing through the poles at q0 = \u03c9 + i\u01eb and q0 = +\u03c9 i\u01eb. \u2212 \u2212 We then change the integration variable in eq. (14.4) from \u2113 to q; the jacobian is trivial, and we have dd\u2113 = ddq. \u2212\u221e to + Next, think of the integral over q0 from as a contour integral in the complex q0 plane. If the integrand vanishes fast enough as, | \u2192 \u221e we can rotate this contour clockwise by 90\u25e6, as shown in \ufb01g. (14.4), so that it runs from In making this Wick rotation, the contour does not pass over any poles. (The i\u01eb\u2019s are needed to make this statement unambiguous.) Thus the value of the integral is unchanged. It is now convenient to de\ufb01ne a euclidean d-dimensional vector \u00afq via q0 = i\u00afqd and qj = \u00afqj; then q2 = \u00afq2, where to +i i \u221e q0 | \u221e \u221e \u2212. Also, ddq = i dd\u00afq. Therefore, in general, \u00afq2 = \u00afq2 1 +... + \u00afq2 d. (14.15) ddq f (q2 i\u01eb) = i \u2212 Z Z 0 faster than 1/\u00afqd as \u00afq as long as f (\u00afq2) Now we can write \u2192 dd\u00afq f (\u00afq2) (14.16). \u2192 \u221e \u03a0(k2) = 1 2 g2I(k2) Ak2 \u2212 \u2212 Bm2 + O(g4), where I(k2) \u2261 1 dx 0 Z Z dd\u00afq (2\u03c0)d 1 (\u00afq2 + D)2. (14.17) (14.18) It is now straightforward to evaluate the d-dimensional integral over \u00afq in spherical coordinates. Before we perform this calculation, however, let us introduce another trick, one that can simplify the task of \ufb01xing A and B through the imposition of eqs. (14.7) and (14.8). Here is the trick: di\ufb00erentiate \u03a0(k2) twice with respect to k2 to get \u03a0\u2032\u2032(k2) = 1 2 g2I \u2032\u2032(k2) + O(g4), (14.19) 14: Loop Corrections to the Propagator 113 where, from eqs. (14.18) and (14.14), (14.20) I \u2032\u2032(k2) = dx 6x2(1 x)2 \u2212 1 0 Z dd\u00afq (2\u03c0)d 1 (\u00afq2 + D)4. Z Then, after we evaluate these integrals, we can get \u03a0(k2) by integrating with respect to k2, subject to the boundary conditions of eqs. (14.7) and (14.8). In this way we can construct \u03a0(k2) without ever explicitly computing A and B. Notice that this trick does something else for us as well. The integral over", " \u00afq in eq. (14.20) is \ufb01nite for any d < 8, whereas the original integral in eq. (14.18) is \ufb01nite only for d < 4. This expanded range of d now includes the value of greatest interest, d = 6. How did this happen? We can gain some insight by making a Taylor expansion of \u03a0(k2) about k2 = m2: \u2212 m2) + (A \u2212 m2) + A B)m2 i (k2 + m2) \u03a0(k2) = 1 2 g2I( h + \u2212 2 g2I \u2032( 1 h + 1 2! \u2212 2 g2I \u2032\u2032( 1 h + O(g4). \u2212 m2) i (k2 + m2)2 +... i (14.21) \u2212 \u2265 4, I \u2032( m2) is divergent for d it is straightforward to see that I( m2) From eqs. (14.18) and (14.14), 6, and, in general, is divergent for d \u2265 \u2212 4 + 2n. We can use the O(g2) terms in I (n)( m2) is divergent for d \u2265 A and B to cancel o\ufb00 the 1 m2) terms in \u03a0(k2), 2 g2I( whether or not they are divergent. But if we are to end up with a \ufb01nite \u03a0(k2), all of the remaining terms must be \ufb01nite, since we have no more free parameters left to adjust. This is the case for d < 8. m2) and 1 2 g2I \u2032( \u2212 \u2212 \u2212 \u2264 Of course, for 4 d < 8, the values of A and B (and hence the lagrangian coe\ufb03cients Z = 1 + A and Zm = 1 + B) are formally in\ufb01nite, and this may be disturbing. However, these coe\ufb03cients are not directly measurable, and so need not obey our preconceptions about their magnitudes. Also, it is important to remember that A and B each includes a factor of g2; this means that we can expand in powers of A and B as part of our general expansion in powers of g. When we compute \u03a0(k2) (which enters into observable cross sections), all the formally in\ufb01nite numbers cancel in a well-de\ufb01ned way, provided d < 8. \u2265 For d 8, this procedure breaks down, and we do not obtain a \ufb01nite expression for \u03a0(k2). In this case, we say that the theory is nonrenormalizable. We will discuss the criteria for renormalizability of a theory in detail in section 18. It turns out that \u03d53 theory is renormalizable for d 6. (The \u2264 14: Loop Corrections to the Propagator 114 problem with 6 < d < 8 arises from higher-order corrections, as we will see in section 18.) Now let us return to the calculation of \u03a0(k2). Rather than using the trick of \ufb01rst computing \u03a0\u2032\u2032(k2), we will instead evaluate \u03a0(k2) directly from eq. (14.18) as a function of d for d < 4. Then we will analytically continue the result to arbitrary d. This procedure is known as dimensional regularization. Then we will \ufb01x A and B by imposing eqs. (14.7) and (14.8), and \ufb01nally take the limit d 6. We could just as well use the method of section 9. Making the replace- \u2192 ment \u02dc\u2206(p2) 1 p2 + m2 \u2192 \u039b2 p2 + \u039b2 i\u01eb i\u01eb, (14.22) \u2212 \u2212 where \u039b is the ultraviolet cuto\ufb00, renders the O(g2) term in \u03a0(k2) \ufb01nite for d < 8; This procedure is known as Pauli\u2013Villars regularization. We then evaluate \u03a0(k2) as a function of \u039b, \ufb01x A and B by imposing eqs. (14.7) and (14.8), and take the \u039b limit. Calculations with Pauli-Villars regularization are generally much more cumbersome than they are with dimensional regularization. However, the \ufb01nal result for \u03a0(k2) is the same. Eq. (14.21) demonstrates that any regularization scheme will give the same result for d < 8, at least as long as it preserves the Lorentz invariance of the integrals. \u2192 \u221e We", " therefore turn to the evaluation of I(k2), eq. (14.18). The angular part of the integral over \u00afq yields the area \u2126d of the unit sphere in d dimensions, which is \u2126d = 2\u03c0d/2 \u0393( 1 2 d) ; (14.23) in this is most easily veri\ufb01ed by computing the gaussian integral both cartesian and spherical coordinates. Here \u0393(x) is the Euler gamma function; for a nonnegative integer n and small x, R dd\u00afq e\u2212 \u00afq2 \u0393(n+1) = n!, \u221a\u03c0, \u0393(n+ 1 2 ) = n+x) = \u0393( \u2212 (2n)! n!2n 1)n ( \u2212 n! 1 x \u2212 \u03b3 + X (14.24) (14.25) n k=1 k\u2212 1 + O(x), (14.26) where \u03b3 = 0.5772... is the Euler-Mascheroni constant. The radial part of the \u00afq integral can also be evaluated in terms of gamma functions. The overall result (generalized slightly) is dd\u00afq (2\u03c0)d (\u00afq2)a (\u00afq2 + D)b = Z \u0393(b 1 a 2 d)\u0393(a+ 1 (4\u03c0)d/2\u0393(b)\u0393( 1 2 d) \u2212 \u2212 2 d) D\u2212 (b \u2212 a \u2212 d/2). (14.27) 14: Loop Corrections to the Propagator 115 We will make frequent use of this formula throughout this book. In the case of interest, eq. (14.18), we have a = 0 and b = 2. There is one more complication to deal with. Recall that we want to focus on d = 6 because in that case g is dimensionless. However, for general d, g has mass dimension \u03b5/2, where \u03b5 6 \u2212 \u2261 d. (14.28) To account for this, we introduce a new parameter \u02dc\u00b5 with dimensions of mass, and make the replacement g \u02dc\u00b5\u03b5/2. g \u2192 (14.29) In this way g remains dimensionless for all \u03b5. Of course, \u02dc\u00b5 is not an actual parameter of the d = 6 theory. Therefore, nothing measurable (like a cross section) can depend on it. This seemingly innocuous statement is actually quite powerful, and will eventually serve as the foundation of the renormalization group. We now return to eq. (14.18), use eq. (14.26), and set d = 6 \u2212 I(k2) = \u0393( 1+ \u03b5 2 ) \u2212 (4\u03c0)3 1 dx D 4\u03c0 D \u03b5/2. 0 Z Hence, with the substitution of eq. (14.29), and de\ufb01ning g2 (4\u03c0)3 \u03b1 \u2261 for notational convenience, we have \u03b5; we get (14.30) (14.31) \u03a0(k2) = 1 2 \u03b1 \u0393( 1+ \u03b5 2 ) \u2212 Ak2 \u2212 \u2212 1 0 dx D \u03b5/2 4\u03c0 \u02dc\u00b52 D! Z Bm2 + O(\u03b12). Now we can take the \u03b5 0 limit, using eq. (14.26) and \u2192 A\u03b5/2 = 1 + \u03b5 2 ln A + O(\u03b52). The result is \u03a0(k2) = \u2212 1 2 \u03b1 \" Ak2 2 \u03b5 + 1 1 6 k2 + m2 + Bm2 + O(\u03b12). \u2212 1 0 Z dx D ln 4\u03c0 \u02dc\u00b52 e\u03b3D!# \u2212 (14.32) (14.33) (14.34) 14: Loop Corrections to the Propagator 116 Here we have used 6 k2 + m2. It is now convenient to de\ufb01ne \u221a4\u03c0 e\u2212 \u03b3/2 \u02dc\u00b5, (14.35) 1 0 dx D = 1 R \u00b5 \u2261 and rearrange things to get \u03a0(k2) = 1 2 \u03b1 Z 1 dx D ln(D/m2) 0 1 6 \u03b1 h \u03b1 1 \u03b5 + ln(\u00b5/m) + 1 \u03b5 + ln(\u00b5/m If we take A and B to have the form + A k2 i i o + B m2 + O(\u03b12). (14.36 + ln(\u00b5/m) + 1 \u03b5 + ln(\u00b5/m) + 1 1 2 + \u03baA 2 + \u03baB + O(\u03b12), i + O(\u03b12), (14.", "37) (14.38) where \u03baA and \u03baB are purely numerical constants, then we get h i 1 2 \u03b1 \u03a0(k2) = 1 dx D ln(D/m2) + \u03b1 Thus this choice of A and B renders \u03a0(k2) \ufb01nite and independent of \u00b5, as required. 6 \u03baAk2 + \u03baBm2 + O(\u03b12). (14.39) 0 Z 1 To \ufb01x \u03baA and \u03baB, we must still impose the conditions \u03a0( m2) = 0 and m2) = 0. The easiest way to do this is to \ufb01rst note that, schematically, \u2212 \u03a0\u2032( \u2212 \u03a0(k2) = 1 2 \u03b1 1 0 Z dx D ln D + linear in k2 and m2 + O(\u03b12). (14.40) We can then impose \u03a0( \u2212 m2) = 0 via \u03a0(k2) = 1 2 \u03b1 1 0 Z where dx D ln(D/D0) + linear in (k2 + m2) + O(\u03b12). (14.41) D0 \u2261 D = [1 \u2212 x(1 \u2212 m2 x)]m2. (14.42) Now it is straightforward to di\ufb00erentiate eq. (14.41) with respect to k2, and \ufb01nd that \u03a0\u2032( m2) vanishes for k2= \u2212 \u2212 \u03a0(k2) = 1 2 \u03b1 1 0 Z dx D ln(D/D0) \u2212 1 12 \u03b1(k2 + m2) + O(\u03b12). (14.43) The integral over x can be done in closed form; the result is \u03a0(k2) = 1 12 \u03b1 h c1k2 + c2m2 + 2k2f (r) i + O(\u03b12), (14.44) 14: Loop Corrections to the Propagator 117 \u0391 0.1 -30 -20 -10 10 20 30 k2 \u20ac\u20ac\u20ac\u20ac\u20ac\u20ac\u20ac m2 -0.1 -0.2 Figure 14.5: The real and imaginary parts of \u03a0(k2)/(k2 + m2) in units of \u03b1. where c1 = 3 \u2212 \u03c0\u221a3, c2 = 3 \u2212 2\u03c0\u221a3, and f (r) = r3 tanh\u2212 1(1/r), r = (1 + 4m2/k2)1/2. (14.45) (14.46) There is a branch point at k2 = part for k2 < \u2212 4m2; we will discuss this further in the next section. 4m2, and \u03a0(k2) acquires an imaginary We can write the exact propagator as \u2212 \u02dc\u2206(k2) = 1 1 \u2212 \u03a0(k2)/(k2 + m2) 1 k2 + m2. i\u01eb \u2212 (14.47) In \ufb01g. (14.5), we plot the real and imaginary parts of \u03a0(k2)/(k2 + m2) in units of \u03b1. We see that its values are quite modest for the plotted range. For much larger values of k2 |, we have | \u03a0(k2) k2 + m2 \u2243 1 12 \u03b1 h ln(k2/m2) + c1 + O(\u03b12). (14.48) i\u01eb) = ln i If we had kept track of the i\u01eb\u2019s, k2 would be k2 i\u01eb; when k2 is negative, \u2212 we have ln(k2 i\u03c0. The imaginary part of \u03a0(k2)/(k2 + m2) 1 12 \u03c0\u03b1 + O(\u03b12) when k2 is therefore approaches the asymptotic value of large and negative. The real part of \u03a0(k2)/(k2 + m2), however, continues to increase logarithmically with is large. We will begin to when address the meaning of this in section 26. k2 | k2 | k2 | | \u2212 \u2212 \u2212 | | 14: Loop Corrections to the Propagator 118 Problems 14.1) Derive a generalization of Feynman\u2019s formula, 1 1... A\u03b1n n A\u03b11 = \u0393( i \u03b1i) i \u0393(\u03b1i) P 1 (n 1)! \u2212 Z dFn ( 1 i x\u03b1i\u2212 i i xiAi) Q i\u03b1i. (14.49) Hint: start with Q \u0393(\u03b1) A\u03b1 =", " 0 Z P At, 1 e\u2212 \u221e dt t\u03b1 \u2212 P (14.50) which de\ufb01nes the gamma function. Put an index on A, \u03b1, and t, and take the product. Then multiply on the right-hand side by 1 = \u221e ds \u03b4(s 0 Z P iti). \u2212 (14.51) Make the change of variable ti = sxi, and carry out the integral over s. 14.2) Verify eq. (14.23). 14.3) a) Show that ddq q\u00b5f (q2) = 0, Z ddq q\u00b5q\u03bdf (q2) = C2 g\u00b5\u03bd Z ddq q2f (q2), Z (14.52) (14.53) and evaluate the constant C2 in terms of d. Hint: use Lorentz symmetry to argue for the general structure, and evaluate C2 by contracting with g\u00b5\u03bd. b) Similarly evaluate ddq q\u00b5q\u03bdq\u03c1q\u03c3f (q2). 14.4) Compute the values of \u03baA and \u03baB. R 14.5) Compute the O(\u03bb) correction to the propagator in \u03d54 theory (see \u03b5 spacetime dimensions, and compute the problem 9.2) in d = 4 \u2212 O(\u03bb) terms in A and B. 14.6) Repeat problem 14.5 for the theory of problem 9.3. 14.7) Renormalization of the anharmonic oscillator. Consider an anhar- monic oscillator, speci\ufb01ed by the lagrangian L = 1 2 Z \u02d9q2 1 2 Z\u03c9\u03c92q2 \u2212 \u2212 Z\u03bb\u03bb\u03c93q4. (14.54) We set \u00afh = 1 and m = 1; \u03bb is then dimensionless. 14: Loop Corrections to the Propagator 119 i i 0 i | and 2 P 2 + 1 2 \u03c92Q2, and [Q, P ] = i. 1 i | be the ground and \ufb01rst excited states of H. a) Find the hamiltonian H corresponding to L. Write it as H = H0 + H1, where H0 = 1 be the ground and \ufb01rst excited states of H0, and and b) Let let (We I \u2126 | | take all these eigenstates to have unit norm.) We de\ufb01ne \u03c9 to be the E\u2126. We normalize the position EI \u2212 excitation energy of H, \u03c9 1/2. Finally, to operator Q by setting Q 1 = I | h i h make things mathematically simpler, we set Z\u03bb equal to one, rather than using a more physically motivated de\ufb01nition. Write Z = 1 + A and Z\u03c9 = 1 + B, where A = \u03baA\u03bb + O(\u03bb2) and B = \u03baB\u03bb + O(\u03bb2). Use Rayleigh\u2013Schr\u00a8odinger perturbation theory to compute the O(\u03bb) corrections to the unperturbed energy eigenvalues and eigenstates. c) Find the numerical values of \u03baA and \u03baB that yield \u03c9 = EI \u2212 and = (2\u03c9)\u2212 \u2261 \u2126 | 0 i | Q | 1/2. E\u2126 = (2\u03c9)\u2212 I h Q | \u2126 | i d) Now think of the lagrangian of eq. (14.54) as specifying a quantum \ufb01eld theory in d = 1 dimensions. Compute the O(\u03bb) correction to the propagator. Fix \u03baA and \u03baB by requiring the propagator to have a pole at k2 = \u03c92 with residue one. Do your results agree with those \u2212 of part (c)? Should they? 15: The One-Loop Correction in Lehmann-K\u00a8all\u00b4en Form 120 15 The One-Loop Correction in Lehmann-K\u00a8all\u00b4en Form Prerequisite: 14 In section 13, we found that the exact propagator could be written in Lehmann-K\u00a8all\u00b4en form as \u02dc\u2206(k2) = 1 k2 + m2 + i\u01eb \u221e 4m2 ds \u03c1(s) 1 k2 + s i\u01eb, (15.1) Z where the spectral density \u03c1(s) is real and nonnegative. In section 14, on the other hand, we found that the exact propagator could be written as \u2212 \u2212 \u02dc\u2206(k2) = k2 + m2 1 i\u01eb \u2212 \u2212, \u03a0(k2) (15.2) and that, to O(g2) in \u03d53", " theory in six dimensions, \u03a0(k2) = 1 2 \u03b1 1 0 Z where dx D ln(D/D0) \u2212 1 12 \u03b1(k2 + m2) + O(\u03b12), (15.3) g2/(4\u03c0)3, \u03b1 \u2261 D = x(1 \u2212 D0 = [1 x)k2 + m2 i\u01eb, \u2212 x)]m2. x(1 \u2212 \u2212 (15.4) (15.5) (15.6) In this section, we will attempt to reconcile eqs. (15.2) and (15.3) with eq. (15.1). Let us begin by considering the imaginary part of the propagator. We will always take k2 and m2 to be real, and explicitly include the appropriate factors of i\u01eb whenever they are needed. We can use eq. (15.1) and the identity 1 \u2212 x = i\u01eb x x2 + \u01eb2 + i\u01eb x2 + \u01eb2 = P 1 x + i\u03c0\u03b4(x), where P means the principal part, to write Im \u02dc\u2206(k2) = \u03c0\u03b4(k2 + m2) + ds \u03c1(s) \u03c0\u03b4(k2 + s) \u221e 4m2 Z = \u03c0\u03b4(k2 + m2) + \u03c0\u03c1( \u2212 k2), (15.7) (15.8) 15: The One-Loop Correction in Lehmann-K\u00a8all\u00b4en Form 121 where \u03c1(s) \u2261 0 for s < 4m2. Thus we have \u03c0\u03c1(s) = Im \u02dc\u2206( \u2212 s) for s 4m2. \u2265 (15.9) Let us now suppose that Im \u03a0(k2) = 0 for some range of k2. (In section 14, we saw that the O(\u03b1) contribution to \u03a0(k2) is purely real for k2 > 4m2.) Then, from eqs. (15.2) and (15.7), we get \u2212 Im \u02dc\u2206(k2) = \u03c0\u03b4(k2 + m2 \u2212 \u03a0(k2)) for Im \u03a0(k2) = 0. (15.10) From \u03a0( ishes at k2 = this argument with respect to k2 equals one at k2 = m2) = 0, we know that the argument of the delta function vanm2) = 0, we know that the derivative of m2, and from \u03a0\u2032( \u2212 \u2212 \u2212 m2. Therefore \u2212 Im \u02dc\u2206(k2) = \u03c0\u03b4(k2 + m2) for Im \u03a0(k2) = 0. (15.11) Comparing this with eq. (15.8), we see that \u03c1( k2) = 0 if Im \u03a0(k2) = 0. Now suppose Im \u03a0(k2) is not zero for some range of k2. (In section 14, we saw that the O(\u03b1) contribution to \u03a0(k2) has a nonzero imaginary part for k2 < 4m2.) Then we can ignore the i\u01eb in eq. (15.2), and \u2212 \u2212 Im \u02dc\u2206(k2) = Im \u03a0(k2) (k2 + m2 + Re \u03a0(k2))2 + (Im \u03a0(k2))2 for Im \u03a0(k2) = 0. (15.12) (15.13) Comparing this with eq. (15.8) we see that \u03c0\u03c1(s) = s + m2 + Re \u03a0( ( \u2212 Im \u03a0( s) \u2212 s))2 + (Im \u03a0( s))2. \u2212 Since we know \u03c1(s) = 0 for s < 4m2, this tells us that we must also have 4m2. s) = 0 for s < 4m2, or equivalently Im \u03a0(k2) = 0 for k2 > Im \u03a0( This is just what we found for the O(\u03b1) contribution to \u03a0(k2) in section 14. \u2212 \u2212 \u2212 We can also see this directly from eq. (15.3), without doing the integral over x. The integrand in this formula is real as long as the argument of the logarithm is real and positive. From eq. (15.5), we see that D is real and positive if and only if x(1 x) \u2212 is 1/4, and so the argument of the logarithm is real and positive for the whole integration range 0 In this regime, Im \u03a0(k2) = 0", ". On the other hand, for k2 < 4m2, the argument of the logarithm becomes negative for some of the integration range, and 4m2. This is exactly what we need to reconcile = 0 for k2 < so Im \u03a0(k2) eqs. (15.2) and (15.3) with eq. (15.1). m2. The maximum value of x(1 1 if and only if k2 > x)k2 > 4m2. \u2212 \u2212 \u2212 \u2212 \u2264 \u2264 \u2212 x Problems 6 6 15: The One-Loop Correction in Lehmann-K\u00a8all\u00b4en Form 122 15.1) In this problem we will verify the result of problem 13.1 to O(\u03b1). a) Let \u03a0loop(k2) be given by the \ufb01rst line of eq. (14.32), with \u03b5 > 0. Show that, up to O(\u03b12) corrections, A = \u03a0\u2032loop( \u2212 m2). Then use Cauchy\u2019s integral formula to write this as dw 2\u03c0i \u03a0loop(w) (w + m2)2, A = I (15.14) (15.15) \u2212 m2 in the complex w plane. where the contour of integration is a small counterclockwise circle around b) By examining eq. (14.32), show that the only singularity of \u03a0loop(k2) is a branch point at k2 = 4m2. Take the cut to run along the negative real axis. \u2212 c) Distort the contour in eq. (15.15) to a circle at in\ufb01nity with a detour around the branch cut. Examine eq. (14.32) to show that, for \u03b5 > 0, the circle at in\ufb01nity does not contribute. The contour around the branch cut then yields 4m2 \u2212 dw 2\u03c0i 1 (w + m2)2 A = \u03a0loop(w+i\u01eb) h where \u01eb is in\ufb01nitesimal (and is not to be confused with \u03b5 = 6 \u03a0loop(w i\u01eb) \u2212\u221e \u2212 \u2212 Z i, (15.16) d). \u2212 d) Examine eq. (14.32) to show that the real part of \u03a0loop(w) is continuous across the branch cut, and that the imaginary part changes sign, so that \u03a0loop(w+i\u01eb) \u03a0loop(w i\u01eb) = \u2212 \u2212 \u2212 2i Im \u03a0loop(w i\u01eb). \u2212 (15.17) e) Let w = \u2212 s in eq. (15.16) and use eq. (15.17) to get A = 1 \u03c0 \u2212 \u221e 4m2 ds Im \u03a0loop( (s s \u2212 \u2212 m2)2 i\u01eb) Z Use this to verify the result of problem 13.1 to O(\u03b1). \u2212. (15.18) 15.2) Dispersion relations. Consider the exact \u03a0(k2), with \u03b5 = 0. Assume 4m2, that it obeys 2 at large k2. | | | that its only singularity is a branch point at k2 = \u2212 k2 eq. (15.17), and that \u03a0(k2) grows more slowly than | By recapitulating the analysis in the previous problem, show that \u03a0\u2032\u2032(k2) = 2 \u03c0 \u221e 4m2 Z ds Im \u03a0( s i\u03b5) \u2212 \u2212 (k2 + s)3. (15.19) 15: The One-Loop Correction in Lehmann-K\u00a8all\u00b4en Form 123 This is a twice subtracted dispersion relation. It gives \u03a0\u2032\u2032(k2) throughout the complex k2 plane in terms of the values of the imaginary part of \u03a0(k2) along the branch cut. 16: Loop Corrections to the Vertex 124 16 Loop Corrections to the Vertex Prerequisite: 14 Consider the O(g3) diagram of \ufb01g. (16.1), which corrects the \u03d53 vertex. In this section we will evaluate this diagram. We can de\ufb01ne an exact three-point vertex function iV3(k1, k2, k3) as the sum of one-particle irreducible diagrams with three external lines carrying momenta k1, k2, and k3, all incoming, with k1 + k2 + k3 = 0 by momentum conservation. (In adopting this convention, we allow k0 i to have either sign; if ki is the momentum of an external particle, then the sign of k0 i is positive if the particle is incoming, and", " negative if it is outgoing.) The original vertex iZgg is the \ufb01rst term in this sum, and the diagram of \ufb01g. (16.1) is the second. Thus we have iV3(k1, k2, k3) = iZgg + (ig)3 + O(g5). 3 1 i Z dd\u2113 (2\u03c0)d \u02dc\u2206((\u2113 \u2212 k1)2) \u02dc\u2206((\u2113+k2)2) \u02dc\u2206(\u21132) (16.1) In the second term, we have set Zg = 1 + O(g2). We proceed immediately to the evaluation of this integral, using the series of tricks from section 14. First we use Feynman\u2019s formula to write \u02dc\u2206((\u2113 \u2212 k1)2) \u02dc\u2206((\u2113+k2)2) \u02dc\u2206(\u21132) where = dF3 x1(\u2113 Z h \u2212 1 k1)2 + x2(\u2113+k2)2 + x3\u21132 + m2 3 \u2212, i (16.2) dx1 dx2 dx3 \u03b4(x1+x2+x3\u2212 0 Z We manipulate the right-hand side of eq. (16.2) to get dF3 = 2 Z 1). (16.3) \u02dc\u2206((\u2113 \u2212 k1)2) \u02dc\u2206((\u2113+k2)2) \u02dc\u2206(\u21132) (x1k1 \u2212 dF3 \u21132 2\u2113 \u2212 \u00b7 x2k2) + x1k2 1 + x2k2 2 + m2 3 \u2212 (\u2113 \u2212 x1k1 + x2k2)2 + x1(1 \u2212 x1)k2 1 + x2(1 q2 + D 3 \u2212. + 2x1x2k1 \u00b7 k2 + m2 i \u2212 3 \u2212 i In the last line, we have de\ufb01ned q \u2113 x1k1 + x2k2, and D x1(1 x1)k2 \u2212 \u2261 1 + x3x2k2 = x3x1k2 \u2261 1 + x2(1 \u2212 x2)k2 \u2212 2 + x1x2k2 2 + 2x1x2k1 \u00b7 3 + m2, k2 + m2 i x2)k2 2 (16.4) (16.5) = = Z h dF3 Z h = dF3 Z h 16: Loop Corrections to the Vertex 125 k1 l k l 1 k 3 k2 l + k2 Figure 16.1: The O(g3) correction to the vertex iV3(k1, k2, k3). where we used k2 line. 3 = (k1 + k2)2 and x1 + x2 + x3 = 1 to simplify the second After making a Wick rotation of the q0 contour, we have V3(k1, k2, k3)/g = Zg + g2 dF3 Z Z dd\u00afq (2\u03c0)d 1 (\u00afq2 + D)3 + O(g4), (16.6) where \u00afq is a euclidean vector. This integral diverges for d 6. We therefore evaluate it for d < 6, using the general formula from section 14; the result is \u2265 dd\u00afq (2\u03c0)d 1 (\u00afq2 + D)3 = 1 \u0393(3 2 d) 2(4\u03c0)d/2 D\u2212 \u2212 (3 \u2212 d/2). (16.7) \u03b5. To keep g dimensionless, we make the replacement Z Now we set d = 6 g g \u02dc\u00b5\u03b5/2. Then we have \u2212 \u2192 V3(k1, k2, k3)/g = Zg + 1 2 \u03b1 \u0393( \u03b5 2 ) \u03b5/2 4\u03c0 \u02dc\u00b52 D! dF3 Z + O(\u03b12), (16.8) where \u03b1 = g2/(4\u03c0)3. Now we can take the \u03b5 0 limit. The result is \u2192 V3(k1, k2, k3)/g = Zg + 1 2 \u03b1 \" 2 \u03b5 + dF3 ln Z 4\u03c0 \u02dc\u00b52 e\u03b3D!# + O(\u03b12), (16.9) where we have used dF3 = 1. We now let \u00b52 = 4\u03c0e\u2212 \u03b3 \u02dc\u00b52,", " set R Zg = 1 + C, (16.10) and rearrange to get V3(k1, k2, k3)/g = 1 + \u03b1 + C 1 \u03b5 + ln(\u00b5/m) i dF3 ln(D/m2) n h 1 2 \u03b1 \u2212 + O(\u03b12). Z o (16.11) 16: Loop Corrections to the Vertex 126 If we take C to have the form \u03b1 1 \u03b5 + ln(\u00b5/m) + \u03baC + O(\u03b12), (16.12) C = \u2212 h where \u03baC is a purely numerical constant, we get i V3(k1, k2, k3)/g = 1 1 2 \u03b1 \u2212 Z dF3 ln(D/m2) \u2212 \u03baC\u03b1 + O(\u03b12). (16.13) Thus this choice of C renders V3(k1, k2, k3) \ufb01nite and independent of \u00b5, as required. \u2212 \u2212 m2) = 0 and \u03a0\u2032( We now need a condition, analogous to \u03a0( m2) = 0, to \ufb01x the value of \u03baC. These conditions on \u03a0(k2) were mandated by known properties of the exact propagator, but there is nothing directly comparable for the vertex. Di\ufb00erent choices of \u03baC correspond to di\ufb00erent de\ufb01nitions of the coupling g. This is because, in order to measure g, we would measure a cross section that depends on g; these cross sections also depend on \u03baC. Thus we can use any value for \u03baC that we might fancy, as long as we all agree on that value when we compare our calculations with experimental measurements. It is then most convenient to simply set \u03baC = 0. This corresponds to the condition V3(0, 0, 0) = g. (16.14) This condition can then also be used to \ufb01x the higher-order (in g) terms in Zg. The integrals over the Feynman parameters in eq. (16.13) cannot be m2, done in closed form, but it is easy to see that if (for example) then k2 1| \u226b | V3(k1, k2, k3)/g 1 \u2212 \u2243 ln(k2 1 2 \u03b1 h 1/m2) + O(1) i + O(\u03b12). (16.15) Thus the magnitude of the one-loop correction to the vertex function inm2. This is the same behavior creases logarithmically with that we found for \u03a0(k2)/(k2 + m2) in section 14. k2 i | \u226b | k2 i | | when Problems 16.1) Compute the O(\u03bb2) correction to V4 in \u03d54 theory (see problem 9.2) in \u03b5 spacetime dimensions. Take V4 = \u03bb when all four external d = 4 momenta are on shell, and s = 4m2. What is the O(\u03bb) contribution to C? \u2212 16.2) Repeat problem 16.1 for the theory of problem 9.3. 17: Other 1PI Vertices 127 17 Other 1PI Vertices Prerequisite: 16 In section 16, we de\ufb01ned the three-point vertex function iV3(k1, k2, k3) as the sum of all one-particle irreducible diagrams with three external lines, with the external propagators removed. We can extend this de\ufb01nition to the n-point vertex iVn(k1,..., kn). There are two key di\ufb00erences between Vn>3 and V3 in \u03d53 theory. The \ufb01rst is that there is no tree-level contribution to Vn>3. The second is that the one-loop contribution to Vn>3 is \ufb01nite for d < 2n. In particular, the one-loop contribution to Vn>3 is \ufb01nite for d = 6. Let us see how this works for the case n = 4. We treat all the external momenta as incoming, so that k1 + k2 + k3 + k4 = 0. One of the three contributing one-loop diagrams is shown in \ufb01g. (17.1); in this diagram, the k3 vertex is opposite to the k1 vertex. Two other inequivalent diagrams are then obtained by swapping k3 \u2194 k2 and k3 \u2194 k4. We then have iV4 = g4 \u02dc\u2206((\u2113 k1)2) \u02dc\u2206", "((\u2113+k2)2) \u02dc\u2206((\u2113+k2+k3)2) \u02dc\u2206(\u21132) d6\u2113 (2\u03c0)6 Z + (k3 \u2194 + O(g6). \u2212 k2) + (k3 \u2194 k4) (17.1) Feynman\u2019s formula gives k1)2) \u02dc\u2206((\u2113+k2)2) \u02dc\u2206((\u2113+k2+k3)2) \u02dc\u2206(\u21132) dF4 dF4 h x1(\u2113 \u2212 q2 + D1234 k1)2 + x2(\u2113+k2)2 + x3(\u2113+k2+k3)2 + x4\u21132 + m2 4 \u2212 4 \u2212, i (17.2) \u02dc\u2206((\u2113 \u2212 = = Z Z h x1k1 + x2k2 + x3(k2+k3) and, after making repeated use of i where q = \u2113 x1+x2+x3+x4 = 1 and k1+k2+k3+k4 = 0, \u2212 D1234 = x1x4k2 1 + x2x4k2 2 + x2x3k2 + x1x2(k1+k2)2 + x3x4(k2+k3)2 + m2. 3 + x1x3k2 4 (17.3) We see that the integral over q is \ufb01nite for d < 8, and in particular for d = 6. After a Wick rotation of the q0 contour and applying the general formula of section 14, we \ufb01nd d6q (2\u03c0)6 1 (q2 + D)4 = i 6(4\u03c0)3D. Z (17.4) 17: Other 1PI Vertices 128 k1 l k4 l k1 l + k2 + k3 l + k2 k2 k3 Figure 17.1: One of the three one-loop Feynman diagrams contributing to the four-point vertex iV4(k1, k2, k3, k4); the other two are obtained by swapping k3 \u2194 k2 and k3 \u2194 k4. Thus we get V4 = g4 6(4\u03c0)3 dF4 Z 1 D1234 + 1 D1324 + 1 D1243 + O(g6). (17.5) This expression is \ufb01nite and well-de\ufb01ned; the same is true for the one-loop contribution to Vn for all n > 3. Problems 17.1) Verify eq. (17.3). 18: Higher-Order Corrections and Renormalizability 129 18 Higher-Order Corrections and Renormalizability Prerequisite: 17 In sections 14\u201317, we computed the one-loop diagrams with two, three, and four external lines for \u03d53 theory in six dimensions. We found that the \ufb01rst two involved divergent momentum integrals, but that these divergences could be absorbed into the coe\ufb03cients of terms in the lagrangian. If this is true for all higher-order (in g) contributions to the propagator and to the one-particle irreducible vertex functions (with n 3 external lines), then we say that the theory is renormalizable. If this is not the case, and further divergences arise, it may be possible to absorb them by adding some new terms to the lagrangian. If a \ufb01nite number of such new terms is required, the theory is still said to be renormalizable. However, if an in\ufb01nite number of new terms is required, then the theory is said to be nonrenormalizable. Despite the in\ufb01nite number of parameters needed to specify it, a nonrenormalizable theory is generally able to make useful predictions at energies below some ultraviolet cuto\ufb00 \u039b; we will discuss this in section 29. \u2265 In this section, we will deduce the necessary conditions for renormalizability. As an example, we will analyze a scalar \ufb01eld theory in d spacetime dimensions of the form = L \u2212 1 2 Z\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 1 2 Zmm2\u03d52 \u221e \u2212 Xn=3 1 n! Zngn\u03d5n. (18.1) Consider a Feynman diagram with E external lines, I internal lines, L closed loops, and Vn vertices that connect n lines", ". (Here Vn is just a number, not to be confused with the vertex function Vn.) Do the momentum integrals associated with this diagram diverge? We begin by noting that each closed loop gives a factor of dd\u2113i, and each internal propagator gives a factor of 1/(p2 + m2), where p is some linear combination of external momenta ki and loop momenta \u2113i. The diagram would then appear to have an ultraviolet divergence at large \u2113i if there are more \u2113\u2019s in the numerator than there are in the denominator. The number of \u2113\u2019s in the numerator minus the number of \u2113\u2019s in the denominator is the diagram\u2019s super\ufb01cial degree of divergence and the diagram appears to be divergent if D \u2261 dL \u2212 2I, D \u2265 0. (18.2) (18.3) 18: Higher-Order Corrections and Renormalizability 130 Next we derive a more useful formula for D. The diagram has E external lines, so another contributing diagram is the tree diagram where all the lines are joined by a single vertex, with vertex factor iZEgE; this is, in fact, the value of this entire diagram, which then has mass dimension [gE]. (The Z\u2019s are all dimensionless, by de\ufb01nition.) Therefore, the original diagram also has mass dimension [gE], since both are contributions to the same scattering amplitude: \u2212 [diagram] = [gE]. (18.4) On the other hand, the mass dimension of any diagram is given by the sum of the mass dimensions of its components, namely [diagram] = dL 2I + \u2212 \u221e Xn=3 Vn[gn]. From eqs. (18.2), (18.4), and (18.5), we get D = [gE] Vn[gn]. \u221e \u2212 Xn=3 (18.5) (18.6) This is the formula we need. From eq. (18.6), it is immediately clear that if any [gn] < 0, we expect uncontrollable divergences, since D increases with every added vertex of this type. Therefore, a theory with any [gn] < 0 is nonrenormalizable. According to our results in section 12, the coupling constants have mass dimension and so we have [gn] = d 1 2 n(d 2), \u2212 \u2212 [gn] < 0 if n > d 2d \u2212. 2 (18.7) (18.8) Thus we are limited to powers no higher than \u03d54 in four dimensions, and no higher than \u03d53 in six dimensions. The same criterion applies to more complicated theories as well: a theory is nonrenormalizable if any coe\ufb03cient of any term in the lagrangian has negative mass dimension. 0) are those for which [gE] What about theories with couplings with only positive or zero mass dimension? We see from eq. (18.6) that the only dangerous diagrams (those with D 0. But in this case, we can absorb the divergence simply by adjusting the value of ZE. This discussion also applies to the propagator; we can think of \u03a0(k2) as representing the loopcorrected counterterm vertex Ak2 + Bm2, with A and Bm2 playing the roles of two couplings. We have [A] = 0 and [Bm2] = 2, so the contributing \u2265 \u2265 18: Higher-Order Corrections and Renormalizability 131 Figure 18.1: The one-loop contribution to V4. Figure 18.2: A two-loop contribution to V4, and the corresponding counterterm insertion. diagrams are expected to be divergent (as we have already seen in detail), and the divergences must be absorbed into A and Bm2. D is called the super\ufb01cial degree of divergence because a diagram might diverge even if D < 0, or might be \ufb01nite even if D 0. The latter can happen if there are cancellations among \u2113\u2019s in the numerator. Quantum electrodynamics provides an example of this phenomenon that we will encounter in Part III; see problem 62.3. For now we turn our attention to the case of diagrams with D < 0 that nevertheless diverge. \u2265 Consider, for example, the diagrams of \ufb01gs. (18.1) and (18.2). The oneloop diagram of \ufb01g. (18.1) with E = 4 is \ufb01nite, but the two-loop correction from the \ufb01rst diagram of \ufb01g. (18.2) is not: the bubble on the upper propag", "ator diverges. This is an example of a divergent subdiagram. However, this is not a problem in this case, because this divergence is canceled by the second diagram of \ufb01g. (18.2), which has a counterterm vertex in place of the bubble. This is the generic situation: divergent subdiagrams are diagrams that, considered in isolation, have D 0. These are precisely the diagrams whose divergences can be canceled by adjusting the Z factor of the corresponding tree diagram (in theories where [gn] 0 for all nonzero gn). \u2265 Thus, we expect that theories with couplings whose mass dimensions are all positive or zero will be renormalizable. A detailed study of the \u2265 18: Higher-Order Corrections and Renormalizability 132 properties of the momentum integrals in Feynman diagrams is necessary to give a complete proof of this. It turns out to be true without further restrictions for theories that have spin-zero and spin-one-half \ufb01elds only. Theories with spin-one \ufb01elds are renormalizable for d = 4 if and only if these spin-one \ufb01elds are associated with a gauge symmetry. We will study this in Part III. Theories of \ufb01elds with spin greater than one are never renormalizable for d 4. \u2265 Reference Notes Explicit two-loop calculations in \u03d53 theory can be found in Collins, Muta, and Sterman. Problems 18.1) In any number d of spacetime dimensions, a Dirac \ufb01eld \u03a8\u03b1(x) carries a spin index \u03b1, and has a kinetic term of the form i\u03a8\u03b3\u00b5\u2202\u00b5\u03a8, where we have suppressed the spin indices; the gamma matrices \u03b3\u00b5 are dimensionless, and \u03a8 = \u03a8\u2020\u03b30. a) What is the mass dimension [\u03a8] of the \ufb01eld \u03a8? b) Consider interactions of the form gn(\u03a8\u03a8)n, where n integer. What is the mass dimension [gn] of gn? c) Consider interactions of the form gm,n\u03d5m(\u03a8\u03a8)n, where \u03d5 is a scalar \ufb01eld, and m 1 are integers. What is the mass dimension \u2265 [gm,n] of gm,n? 1 and n 2 is an \u2265 \u2265 d) In d = 4 spacetime dimensions, which of these interactions are allowed in a renormalizable theory? 19: Perturbation Theory to All Orders 133 19 Perturbation Theory to All Orders Prerequisite: 18 In section 18, we found that, generally, a theory is renormalizable if all of its lagrangian coe\ufb03cients have positive or zero mass dimension. In this section, using \u03d53 theory in six dimensions as our example, we will see how to construct a \ufb01nite expression for a scattering amplitude to arbitrarily high order in the \u03d53 coupling g. n \u2264 \u2264 \u2212 m2) = 0, and V3(0, 0, 0) = g. We begin by summing all one-particle irreducible diagrams with two external lines; this gives us the self-energy \u03a0(k2). We next sum all 1PI diagrams with three external lines; this gives us the three-point vertex function V3(k1, k2, k3). Order by order in g, we must adjust the value of the lagrangian coe\ufb03cients Z\u03d5, Zm, and Zg to maintain the conditions \u03a0( m2) = 0, \u03a0\u2032( \u2212 Next we will construct the n-point vertex functions Vn(k1,..., kn) with 4 E, where E is the number of external lines in the process of interest. We compute these using a skeleton expansion. This means that we draw all the contributing 1PI diagrams, but omit diagrams that include either propagator or three-point vertex corrections. That is, we include only diagrams that are not only 1PI, but also 2PI and 3PI: they remain connected when any one, two, or three lines are cut. (Cutting three lines may isolate a single tree-level vertex, but nothing more complicated.) Then we take the propagators and vertices in these diagrams to be given by the exact propagator \u02dc\u2206(k2) = (k2 + m2 1 and vertex V3(k1, k2, k3), rather than by the tree-level propagator \u02dc\u2206(k2) = (k2 + m2)\u2212 1 and vertex g. We then sum these skeleton diagrams to get Vn for 4 E. Order n by order in g, this procedure is equivalent to computing Vn by sum", "ming the usual set of contributing 1PI diagrams. \u03a0(k2))\u2212 \u2212 \u2264 \u2264 Next we draw all tree-level diagrams that contribute to the process of interest (which has E external lines), including not only three-point vertices, but also n-point vertices for n = 3, 4,..., E. Then we evaluate these diagrams using the exact propagator \u02dc\u2206(k2) for internal lines, and the exact 1PI vertices Vn; external lines are assigned a factor of one.1 We sum these tree diagrams to get the scattering amplitude; loop corrections have all been accounted for already in \u02dc\u2206(k2) and Vn. Order by order in g, this procedure is equivalent to computing the scattering amplitude by summing the usual set of contributing diagrams. Thus we now know how to compute an arbitrary scattering amplitude 1This is because, in the LSZ formula, each Klein-Gordon wave operator becomes (in i + m2 that multiplies each external propagator, leaving i = \u2212m2; by construction, momentum space) a factor of k2 behind only the residue of the pole in that propagator at k2 this residue is one. 19: Perturbation Theory to All Orders 134 to arbitrarily high order. The procedure is the same in any quantum \ufb01eld theory; only the form of the propagators and vertices change, depending on the spins of the \ufb01elds. The tree-level diagrams of the \ufb01nal step can be thought of as the Feynman diagrams of a quantum action (or e\ufb00ective action, or quantum e\ufb00ective action) \u0393(\u03d5). There is a simple and interesting relationship between the effective action \u0393(\u03d5) and the sum of connected diagrams with sources iW (J). We derive it in section 21. Reference Notes The detailed procedure for renormalization at higher orders is discussed in Coleman, Collins, Muta, and Sterman. 20: Two-Particle Elastic Scattering at One Loop 135 20 Two-Particle Elastic Scattering at One Loop Prerequisite: 19 We now illustrate the general rules of section 19 by computing the twoparticle elastic scattering amplitude, including all one-loop corrections, in \u03d53 theory in six dimensions. Elastic means that the number of outgoing particles (of each species, in more general contexts) is the same as the number of incoming particles (of each species). We computed the amplitude for this process at tree level in section 10, with the result i (ig)2 Ttree = 1 i \u02dc\u2206( h where \u02dc\u2206( i\u01eb) is the free-\ufb01eld propagator, and s, t, and u are the Mandelstam variables. Later we will need to remember that s is positive, that t and u are negative, and that s + t + u = 4m2. s) + \u02dc\u2206( t) + \u02dc\u2206( s) = 1/( s + m2 u) i (20.1) \u2212 \u2212 \u2212 \u2212 \u2212 \u2212, The exact scattering amplitude is given by the diagrams of \ufb01g. (20.1), with all propagators and vertices interpreted as exact propagators and vertices. (Recall, however, that each external propagator contributes only the m2, and that this residue is one; thus the facresidue of the pole at k2 = tor associated with each external line is simply one.) We get the one-loop approximation to the exact amplitude by using the one-loop expressions for the internal propagators and vertices. We thus have \u2212 i T1 \u2212 [iV3(s)]2 \u02dc\u2206( loop = 1 i + iV4(s, t, u), \u2212 s) + [iV3(t)]2 \u02dc\u2206( \u2212 t) + [iV3(u)]2 \u02dc\u2206( \u2212 where, suppressing the i\u01eb\u2019s, \u02dc\u2206( \u2212 s) = 1 s + m2 \u2212 \u2212, s) \u03a0( \u2212 u) (20.2) (20.3) 1 12 \u03b1( \u2212 \u2212 s + m2), (20.4) \u03a0( \u2212 s) = 1 2 \u03b1 1 0 Z V3(s)/g = 1 1 2 \u03b1 \u2212 Z dx D2(s) ln D2(s)/D0 D3(s)/m2, dF3 ln V4(s, t, u) = 1 6 g2\u03b1 dF4 Z 1 D4(s, t) 1 D4(t, u) + + 1 D4(u, s). Here \u03b1 = g2/(4\u03c0)3, the Feynman integration measure is dFn f (", "x) = (n Z 1 1)! \u2212 0 Z dx1... dxn \u03b4(x1+... +xn\u2212 1)f (x) (20.5) (20.6) 20: Two-Particle Elastic Scattering at One Loop 136 and we have de\ufb01ned D2(s) = 1 dx1 = (n 1)! \u2212 0 Z x1 1 \u2212 dx2... x1 1 \u2212 \u2212... \u2212 xn\u22122 dxn 1 \u2212 0 Z 0 Z f (x) xn=1 \u00d7, x1... \u2212 \u2212 xn\u22121 \u2212 x(1 \u2212 x)s + m2, x)]m2, \u2212 \u2212 x(1 x1x2s + [1 x1x2s \u2212 \u2212 \u2212 D0 = +[1 D3(s) = D4(s, t) = \u2212 \u2212 (x1+x2)x3]m2, x3x4t + [1 (x1+x2)(x3+x4)]m2. \u2212 (20.7) (20.8) (20.9) (20.10) (20.11) i to We obtain V3(s) from the general three-point function V3(k1, k2, k3) by setm2, and the third to ting two of the three k2 s. We obtain V4(s, t, u) from the general four-point function V4(k1,..., k4) by setting all four k2 i to u. (Recall that the vertex functions are de\ufb01ned with all momenta treated as incoming; here we have identi\ufb01ed k4 as the outgoing momenta.) t, and (k1 + k4)2 to m2, (k1 + k2)2 to s, (k1 + k3)2 to k3 and \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 Eqs. (20.2\u201320.11) are formidable expressions. To gain some intuition about them, let us consider the limit of high-energy, \ufb01xed angle scattering, all much larger than m2. Equivalently, we are where we take s, considering the amplitude in the limit of zero particle mass., and | u | | t | We can then set m2 = 0 in D2(s), D3(s), and D4(s, t). For the self- \u2212 \u2212 energy, we get s) = \u03a0( \u2212 = \u2212 \u2212 Thus, 1 2 \u03b1 s 1 0 Z 1 12 \u03b1 s ln( \u2212 h dx x(1 x) ln \u2212 s/m2) + 3 x(1 \u2212 x(1 x) x) \u2212 1 \u2212 + 1 12 \u03b1 s (20.12) s \u2212 m2 \u03c0\u221a3 + ln. i \u2212 \u02dc\u2206( \u2212 s( s) \u2212 \u2212 1 + 1 12 \u03b1 h s/m2) + 3 ln( \u2212 \u2212 \u03c0\u221a3 + O(\u03b12). (20.13) i The appropriate branch of the logarithm is found by replacing s by s + i\u01eb. For s real and positive, s lies just below the negative real axis, and so \u2212 For t (or u), which is negative, we have instead s) = ln s ln( \u2212 i\u03c0. \u2212 ln( t) = ln \u2212 ln t = ln, + i\u03c0. t | t | | | (20.14) (20.15) 20: Two-Particle Elastic Scattering at One Loop 137 k1 k2 k1 k1 k1 k2 k1 k2 k1 k2 k1 k2 k1 k1 k2+ k2 k1 k1 k2 k2 k 1 k2 Figure 20.1: The Feynman diagrams contributing to the two-particle elastic \u02dc\u2206(k), scattering amplitude; a double line stands for the exact propagator 1 i a circle for the exact three-point vertex V3(k1, k2, k3), and a square for the exact four-point vertex V4(k1, k2, k3, k4). An external line stands for the unit residue of the pole at k2 = m2. \u2212 20: Two-Particle Elastic Scattering at One Loop 138 For the three-point vertex, we get V3(s)/ dF3 ln( h s/m2) ln( \u2212 \u2212 i \u2212 s/m2) + ln(x1x2), i 3, (20.16) where the same comments about the appropriate branch apply. For the four-point vertex,", " the integral over the Feynman parameters can be done in closed form, with the result dF4 D4(s, t) = 3 s + t \u2212 Z \u03c02 + ln(s/t) 2 h i = + 3 u \u03c02 + ln(s/t) i h 2, (20.17) where the second line follows from s + t + u = 0. Putting all of this together, we have loop = g2 T1 \u2212 h where F (s, t, u) + F (t, u, s) + F (u, s, t), (20.18) i 1 s s/m2) + c 11 12 \u03b1 h \u2212 ln( \u2261 \u2212 F (s, t, u) 1 and c = (6\u03c02 + \u03c0\u221a3 39)/11 = 2.33. This is a typical result of a loop calculation: the original tree-level amplitude is corrected by powers of logarithms of kinematic variables. ln(t/u) i (20.19 Problems 20.1) Verify eq. (20.17). 20.2) Compute the O(\u03b1) correction to the two-particle scattering amplitude at threshold, that is, for s = 4m2 and t = u = 0, corresponding to zero three-momentum for both the incoming and outgoing particles. 21: The Quantum Action 139 21 The Quantum Action Prerequisite: 19 In section 19, we saw how to compute (in \u03d53 theory in d = 6 dimensions) the 1PI vertex functions Vn(k1,..., kn) for n 4 via the skeleton expansion: draw all Feynman diagrams with n external lines that are one-, two-, and three-particle irreducible, and compute them using the exact propagator \u02dc\u2206(k2) and three-point vertex function V3(k1, k2, k3). \u2265 We now de\ufb01ne the quantum action (or e\ufb00ective action, or quantum e\ufb00ective action) \u0393(\u03d5) \u2261 1 2 Z + ddk (2\u03c0)dx 1 n! Z \u221e Xn=3 \u2212 \u03a0(k2) \u03d5(k) ddkn (2\u03c0)d (2\u03c0)d\u03b4d(k1+... +kn) e \u03d5( k) k2 + m2 \u2212 ddk1 e (2\u03c0)d... Vn(k1,..., kn) \u00d7 \u03d5(k1)... \u03d5(kn), (21.1) \u03d5(k) = ddx e\u2212 ikx\u03d5(x). The quantum action has the property that where the tree-level Feynman diagrams it generates give the complete scattering amplitude of the original theory. e e e R In this section, we will determine the relationship between \u0393(\u03d5) and the sum of connected diagrams with sources, iW (J), introduced in section 9. Recall that W (J) is related to the path integral Z(J) = D Z \u03d5 exp iS(\u03d5) + i ddx J\u03d5, Z (21.2) where S = ddx L R is the action, via Z(J) = exp[iW (J)]. (21.3) Consider now the path integral Z\u0393(J) \u2261 Z D \u03d5 exp i\u0393(\u03d5) + i ddx J\u03d5 Z = exp[iW\u0393(J)]. (21.4) (21.5) W\u0393(J) is given by the sum of connected diagrams (with sources) in which each line represents the exact propagator, and each n-point vertex represents the exact 1PI vertex Vn. W\u0393(J) would be equal to W (J) if we included only tree diagrams in W\u0393(J). 21: The Quantum Action 140 We can isolate the tree-level contribution to a path integral by means of the following trick. Introduce a dimensionless parameter that we will call \u00afh, and the path integral Z\u0393,\u00afh(J) \u2261 Z D \u03d5 exp i \u00afh \u0393(\u03d5) + ddx J\u03d5 Z = exp[iW\u0393,\u00afh(J)]. (21.6) (21.7) E \u2212 \u2212 In a given connected diagram with sources, every propagator (including those that connect to sources) is multiplied by \u00afh, every source by 1/\u00afh, and every vertex by 1/\u00afh. The overall factor of \u00afh is then \u00afhP V, where V is the number of vertices, E is the number of sources (equivalently,", " the number of external lines after we remove the sources), and P is the number of propagators (external and internal). We next note that P V is equal to L 1, where L is the number of closed loops. This can be seen by counting the number of internal momenta and the constraints among them. SpecifE ically, assign an un\ufb01xed momentum to each internal line; there are P of these momenta. Then the V vertices provide V constraints. One linear combination of these constraints gives overall momentum conservation, and so does not constrain the internal momenta. Therefore, the number of 1), internal momenta left un\ufb01xed by the vertex constraints is (P and the number of un\ufb01xed momenta is the same as the number of loops L. E) (V \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 E So, W\u0393,\u00afh(J) can be expressed as a power series in \u00afh of the form W\u0393,\u00afh(J) = \u221e XL=0 \u00afhL \u2212 1 W\u0393,L(J). (21.8) If we take the formal limit of \u00afh 0, the dominant term is the one with L = 0, which is given by the sum of tree diagrams only. This is just what we want. We conclude that \u2192 W (J) = W\u0393,L=0(J). (21.9) Next we perform the path integral in eq. (21.6) by the method of stationary phase. We \ufb01nd the point (actually, the \ufb01eld con\ufb01guration) at which the exponent is stationary; this is given by the solution of the quantum equation of motion \u03b4 \u03b4\u03d5(x) \u0393(\u03d5) = J(x). \u2212 (21.10) Let \u03d5J (x) denote the solution of eq. (21.10) with a speci\ufb01ed source function J(x). Then the stationary-phase approximation to Z\u0393,\u00afh(J) is Z\u0393,\u00afh(J) = exp i \u00afh \u0393(\u03d5J ) + ddx J\u03d5J + O(\u00afh0). Z (21.11) 21: The Quantum Action 141 Combining the results of eqs. (21.7), (21.8), (21.9), and (21.11), we \ufb01nd W (J) = \u0393(\u03d5J ) + ddx J\u03d5J. (21.12) This is the main result of this section. Z Let us explore it further. Recall from section 9 that the vacuum expec- tation value of the \ufb01eld operator \u03d5(x) is given by 0 \u03d5(x) 0 i | | h = \u03b4 \u03b4J(x). J=0 (21.13) W (J) Now consider what we get if we do not set J = 0 after taking the derivative: 0 \u03d5(x) 0 iJ \u2261 | | h \u03b4 \u03b4J(x) W (J). (21.14) This is the vacuum expectation value of \u03d5(x) in the presence of a nonzero source function J(x). We can get some more information about it by using eq. (21.12) for W (J). Making use of the product rule for derivatives, we have iJ = 0 \u03d5(x) 0 | | h \u03b4 \u03b4J(x) \u0393(\u03d5J ) + \u03d5J (x) + d6y J(y) Z \u03b4\u03d5J (y) \u03b4J(x). (21.15) We can evaluate the \ufb01rst term on the right-hand side by using the chain rule, \u03b4 \u03b4J(x) \u0393(\u03d5J ) = Z d6y \u03b4\u0393(\u03d5J ) \u03b4\u03d5J (y) \u03b4\u03d5J (y) \u03b4J(x). (21.16) Then we can combine the \ufb01rst and third terms on the right-hand side of eq. (21.15) to get iJ = 0 \u03d5(x) 0 | | h d6y Z \u03b4\u0393(\u03d5J ) \u03b4\u03d5J (y) + J(y) \u03b4\u03d5J (y) \u03b4J(x) + \u03d5J (x). (21.17) Now we note from eq. (21.10) that the factor in large brackets on the righthand side of eq. (21.17) vanishes, and so iJ = \u03d5J (x). 0 \u03d5(x", ") 0 | | h (21.18) That is, the vacuum expectation value of the \ufb01eld operator \u03d5(x) in the presence of a nonzero source function is also the solution to the quantum equation of motion, eq. (21.10). We can also write the quantum action in terms of a derivative expansion, \u0393(\u03d5) = ddx Z h (\u03d5) 1 2 Z \u2212 \u2212 U (\u03d5)\u2202\u00b5\u03d5\u2202\u00b5\u03d5 +..., (21.19) i 21: The Quantum Action 142 U (\u03d5) and where the ellipses stand for an in\ufb01nite number of terms with more and more (\u03d5) are ordinary functions (not functionals) derivatives, and of \u03d5(x). (\u03d5) is called the quantum potential (or e\ufb00ective potential, or quantum e\ufb00ective potential), and it plays an important conceptual role in theories with spontaneous symmetry breaking; see section 31. However, it is rarely necessary to compute it explicitly, except in those cases where we are unable to do so. Z U Reference Notes Construction of the quantum action is discussed in Coleman, Itzykson & Zuber, Peskin & Schroeder, and Weinberg II. Problems 21.1) Show that \u0393(\u03d5) = W (J\u03d5) ddx J\u03d5\u03d5, \u2212 Z (21.20) where J\u03d5(x) is the solution of \u03b4 \u03b4J(x) W (J) = \u03d5(x) (21.21) for a speci\ufb01ed \u03d5(x). 21.2) Symmetries of the quantum action. Suppose that we have a set of \ufb01elds \u03d5a(x), and that both the classical action S(\u03d5) and the integration measure \u03d5 are invariant under D \u03d5a(x) \u2192 Z ddy Rab(x, y)\u03d5b(y) (21.22) for some particular function Rab(x, y). Typically Rab(x, y) is a constant matrix times \u03b4d(x y), or a \ufb01nite number of derivatives of \u03b4d(x y); see sections 22, 23, and 24 for some examples. \u2212 \u2212 a) Show that W (J) is invariant under Ja(x) \u2192 Z ddy Jb(y)Rba(y, x). (21.23) b) Use eqs. (21.20) and (21.23) to show that the quantum action \u0393(\u03d5) is invariant under eq. (21.22). This is an important result that we will use frequently. 21: The Quantum Action 143 21.3) Consider performing the path integral in the presence of a background \ufb01eld \u00af\u03d5(x); we de\ufb01ne exp[iW (J; \u00af\u03d5)] \u2261 Z D \u03d5 exp iS(\u03d5+ \u00af\u03d5) + i ddx J\u03d5. (21.24) Z Then W (J; 0) is the original W (J) of eq. (21.3). We also de\ufb01ne the quantum action in the presence of the background \ufb01eld, \u0393(\u03d5; \u00af\u03d5) \u2261 W (J\u03d5; \u00af\u03d5) \u2212 Z ddx J\u03d5\u03d5, (21.25) where J\u03d5(x) is the solution of \u03b4 \u03b4J(x) W (J; \u00af\u03d5) = \u03d5(x) (21.26) for a speci\ufb01ed \u03d5(x). Show that \u0393(\u03d5; \u00af\u03d5) = \u0393(\u03d5+ \u00af\u03d5; 0), (21.27) where \u0393(\u03d5; 0) is the original quantum action of eq. (21.1). 22: Continuous Symmetries and Conserved Currents 144 22 Continuous Symmetries and Conserved Currents Prerequisite: 8 Suppose we have a set of scalar \ufb01elds \u03d5a(x), and a lagrangian density (x) if we make \u03d5a(x) + \u03b4\u03d5a(x) in each \ufb01eld. We have (\u03d5a(x), \u2202\u00b5\u03d5a(x)). Consider what happens to (x) = L L L an in\ufb01nitesimal change \u03d5a(x) (x), where \u03b4 (x) + \u03b4 (x) \u2192 (x) is given by the chain rule, L \u2192 L L L (x) = \u03b4 L \u2202 L \u2202\u03d5a", "(x) \u03b4\u03d5a(x) + \u2202 L \u2202(\u2202\u00b5\u03d5a(x)) \u2202\u00b5\u03b4\u03d5a(x). (22.1) Next consider the classical equations of motion (also known as the Euler- Lagrange equations, or the \ufb01eld equations), given by the action principle \u03b4S \u03b4\u03d5a(x) = 0, (22.2) d4y where S = (y) is the action, and \u03b4/\u03b4\u03d5a(x) is a functional derivative. (For de\ufb01niteness, we work in four spacetime dimensions, though our results will apply in any number.) We have (with repeated indices implicitly summed) L R \u03b4S \u03b4\u03d5a(x) = d4y \u03b4 (y) L \u03b4\u03d5a(x) Z Z Z = = = d4y \u2202 (y) L \u2202\u03d5b(y) \u03b4\u03d5b(y) \u03b4\u03d5a(x) \" + \u2202 (y) L \u2202(\u2202\u00b5\u03d5b(y)) \u03b4(\u2202\u00b5\u03d5b(y)) \u03b4\u03d5a(x) # d4y \u2202 (y) L \u2202\u03d5b(y) \" \u03b4ba\u03b44(y x) + \u2212 \u2202 (y) L \u2202(\u2202\u00b5\u03d5b(y)) (x) \u2202 \u2202\u03d5a(x) \u2212 L \u2202\u00b5 \u2202 (x) L \u2202(\u2202\u00b5\u03d5a(x)). We can use this result to make the replacement \u2202 (x) \u2202\u03d5a(x) \u2192 L \u2202\u00b5 \u2202 (x) L \u2202(\u2202\u00b5\u03d5a(x)) + \u03b4S \u03b4\u03d5a(x) \u03b4ba\u2202\u00b5\u03b44(y x) # \u2212 (22.3) (22.4) in eq. (22.1). Then, combining two of the terms, we get (x) = \u2202\u00b5 \u03b4 L \u2202 (x) L \u2202(\u2202\u00b5\u03d5a(x)) \u03b4\u03d5a(x)! + \u03b4S \u03b4\u03d5a(x) \u03b4\u03d5a(x). (22.5) 22: Continuous Symmetries and Conserved Currents 145 Next we identify the object in large parentheses in eq. (22.5) as the Noether current \u03b4\u03d5a(x). (22.6) j\u00b5(x) \u2261 Eq. (22.5) then implies \u2202 (x) L \u2202(\u2202\u00b5\u03d5a(x)) \u2202\u00b5j\u00b5(x) = \u03b4 (x) L \u2212 \u03b4S \u03b4\u03d5a(x) \u03b4\u03d5a(x). (22.7) If the classical \ufb01eld equations are satis\ufb01ed, then the second term on the right-hand side of eq. (22.7) vanishes. The Noether current plays a special role if we can \ufb01nd a set of in\ufb01nitesimal \ufb01eld transformations that leaves the lagrangian unchanged, or = 0, and we say that the lagrangian has invariant. In this case, we have \u03b4 a continuous symmetry. From eq. (22.7), we then have \u2202\u00b5j\u00b5 = 0 whenever the \ufb01eld equations are satis\ufb01ed, and we say that the Noether current is conserved. In terms of its space and time components, this means that L \u2202 \u2202t j0(x) + \u2207 \u00b7 j(x) = 0. (22.8) If we interpret j0(x) as a charge density, and j(x) as the corresponding current density, then eq. (22.8) expresses the local conservation of this charge. Let us see an example of this. Consider a theory of a complex scalar \ufb01eld with lagrangian = L \u2212 \u2202\u00b5\u03d5\u2020\u2202\u00b5\u03d5 \u2212 m2\u03d5\u2020\u03d5 \u2212 1 4 \u03bb(\u03d5\u2020\u03d5)2. (22.9) We can also rewrite (\u03d51 + i\u03d52)/\u221a2 to get 2 \u2202\u00b5\u03d51\u2202\u00b5\u03d51 \u2212 = \u2212 L 1 in terms of two real scalar \ufb01elds by setting \u03d5 = L 1 2 \u2202\u00b5", "\u03d52\u2202\u00b5\u03d52 \u2212 1 2 m2(\u03d52 1 + \u03d52 2) \u2212 1 16 \u03bb(\u03d52 1 + \u03d52 2)2. (22.10) In the form of eq. (22.9), it is obvious that formation L is left invariant by the trans- \u03d5(x) \u2192 e\u2212 i\u03b1\u03d5(x), (22.11) where \u03b1 is a real number. This is called a U(1) transformation, a transformation by a unitary 1 1 matrix. In terms of \u03d51 and \u03d52, this transformation reads \u00d7 \u03d51(x) \u03d52(x)! \u2192 cos \u03b1 sin \u03b1 \u03d51(x) sin \u03b1 cos \u03b1! \u03d52(x)! \u2212. (22.12) If we think of (\u03d51, \u03d52) as a two-component vector, then eq. (22.12) is just a rotation of this vector in the plane by angle \u03b1. Eq. (22.12) is called an 22: Continuous Symmetries and Conserved Currents 146 2 matrix with SO(2) transformation, a transformation by an orthogonal 2 \u00d7 1, the only a special value of the determinant (namely +1, as opposed to other possibility for an orthogonal matrix). We have learned that a U(1) transformation can be mapped into an SO(2) transformation. \u2212 The in\ufb01nitesimal form of eq. (22.11) is \u03d5(x) \u03d5\u2020(x) \u2192 \u2192 \u03d5(x) \u2212 i\u03b1\u03d5(x), \u03d5\u2020(x) + i\u03b1\u03d5\u2020(x), (22.13) where \u03b1 is now in\ufb01nitesimal. In eq. (22.6), we should treat \u03d5 and \u03d5\u2020 as independent \ufb01elds. It is also conventional to scale the in\ufb01nitesimal parameter out of the current, so that we have \u03b1 j\u00b5 = \u2202 L \u2202(\u2202\u00b5\u03d5) \u03b4\u03d5 + \u2202 L \u2202(\u2202\u00b5\u03d5\u2020) \u03b4\u03d5\u2020 \u2202\u00b5\u03d5\u2020)( i\u03b1\u03d5) + ( = ( \u2212 \u2212 \u2194\u2202\u00b5\u03d5), = \u03b1 Im(\u03d5\u2020 \u2212 \u2202\u00b5\u03d5)(+i\u03b1\u03d5\u2020) (22.14) where A \u2194\u2202\u00b5B current is \u2261 A\u2202\u00b5B \u2212 (\u2202\u00b5A)B. Canceling out \u03b1, we \ufb01nd that the Noether j\u00b5 = Im(\u03d5\u2020 \u2194\u2202\u00b5\u03d5). (22.15) We can also repeat this exercise using the SO(2) form of the transformation. For in\ufb01nitesimal \u03b1, eq. (22.12) becomes \u03b4\u03d51 = +\u03b1\u03d52 and \u03b4\u03d52 = \u03b1\u03d51. Then the Noether current is given by \u2212 \u03b1 j\u00b5 = \u2202 L \u2202(\u2202\u00b5\u03d51) \u03b4\u03d51 + \u2202 L \u2202(\u2202\u00b5\u03d52) \u03b4\u03d52 = ( \u2212 \u2202\u00b5\u03d51)(+\u03b1\u03d52) + ( \u2212 \u2202\u00b5\u03d52)( \u03b1\u03d51) \u2212 = \u03b1 (\u03d51 \u2194\u2202\u00b5\u03d52), (22.16) which is (hearteningly) equivalent to eq. (22.14). Let us de\ufb01ne the Noether charge Q \u2261 d3x j0(x) = d3x Im(\u03d5\u2020 \u2194\u22020\u03d5), (22.17) Z Z and investigate its properties. If we integrate eq. (22.8) over d3x, use Gauss\u2019s j as a surface integral, and assume law to write the volume integral of \u2207\u00b7 that the boundary conditions at in\ufb01nity \ufb01x j(x) = 0 on that surface, then 22: Continuous Symmetries and Conserved Currents 147 we \ufb01nd that Q is constant in time. To get a better idea of the physical implications of this, let us rewrite Q using the free-\ufb01eld expansions \u03d5(x) = \u03d5\u2020(x) = dk f dk Z Z a(k)eikx + b\u2217(k)e\u2212 h b(k)eikx + a\u2217(k)e\u2212 h ikx ikx i i,. (22", ".18) f We have written a\u2217(k) and b\u2217(k) rather than a\u2020(k) and b\u2020(k) because so In a theory far our discussion has been about the classical \ufb01eld theory. with interactions, these formulae (and their \ufb01rst time derivatives) are valid at any one particular time (say, t = ). Then, we can plug them into eq. (22.17), and \ufb01nd (after some manipulation similar to what we did for the hamiltonian in section 3) \u2212\u221e Q = dk Z a\u2217(k)a(k) h \u2212 b(k)b\u2217(k)]. (22.19) f In the quantum theory, this becomes an operator that counts the number of a particles minus the number of b particles. This number is then timeindependent, and so the scattering amplitude vanishes identically for any process that changes the value of Q. This can be seen directly from the Feynman rules, which conserve Q at every vertex. To better understand the implications of the Noether current in the quantum theory, we begin by considering the path integral, Z(J) = D Z \u03d5 ei[S+ d4y Ja\u03d5a]. R (22.20) The value of Z(J) is unchanged if we make the change of variable \u03d5a(x) \u2192 \u03d5a(x) + \u03b4\u03d5a(x), with \u03b4\u03d5a(x) an arbitrary in\ufb01nitesimal shift that (we assume) leaves the measure \u03d5 invariant. Thus we have D 0 = \u03b4Z(J) = i D Z \u03d5 ei[S+ d4y Jb\u03d5b] d4x R Z \u03b4S \u03b4\u03d5a(x) + Ja(x) \u03b4\u03d5a(x). (22.21) We can now take n functional derivatives with respect to Jaj(xj), and then set J = 0, to get 0 = \u03d5 eiS Z D Z d4x i \" \u03b4S \u03b4\u03d5a(x) \u03d5a1(x1)... \u03d5an(xn) + n Xj=1 \u03d5a1(x1)... \u03b4aaj \u03b44(x \u2212 xj)... \u03d5an(xn) # \u03b4\u03d5a(x). (22.22) 22: Continuous Symmetries and Conserved Currents 148 Since \u03b4\u03d5a(x) is arbitrary, we can drop it (and the integral over d4x). Then, since the path integral computes the vacuum expectation value of the timeordered product, we have T 0 0 = i | h \u03b4S \u03b4\u03d5a(x) n \u03d5a1(x1)... \u03d5an(xn) 0 i | + T\u03d5a1(x1)... \u03b4aaj \u03b44(x 0 | Xj=1h \u2212 xj)... \u03d5an(xn) 0 i |. (22.23) These are the Schwinger-Dyson equations for the theory. To get a feel for them, let us look at free-\ufb01eld theory for a single real scalar \ufb01eld, for which \u03b4S/\u03b4\u03d5(x) = (\u22022 x \u2212 m2)\u03d5(x). For n = 1 we get ( \u2212 x + m2)i \u22022 0 T\u03d5(x)\u03d5(x1) 0 i | | h = \u03b44(x \u2212 x1). (22.24) That the Klein-Gordon wave operator should sit outside the time-ordered product (and hence act on the time-ordering step functions) is clear from the path integral form of eq. (22.22). We see from eq. (22.24) that the free\ufb01eld propagator, \u2206(x, is a Green\u2019s function for 0 T\u03d5(x)\u03d5(x1) 0 x1) = i i | | h the Klein-Gordon wave operator, a fact we \ufb01rst learned in section 8. \u2212 More generally, we can write T 0 | h \u03b4S \u03b4\u03d5a(x) \u03d5a1(x1)... \u03d5an(xn) 0 i | = 0 for x = x1,...,n. (22.25) We see that the classical equation of motion is satis\ufb01ed by a quantum \ufb01eld inside a correlation function, as long as its spac", "etime argument di\ufb00ers from those of all the other \ufb01elds. When this is not the case, we get extra contact terms. Let us now consider a theory that has a continuous symmetry and a corresponding Noether current. Take eq. (22.22), and set \u03b4\u03d5a(x) to be the in\ufb01nitesimal change in \u03d5a(x) that results in \u03b4 (x) = 0. Now sum over the index a, and use eq. (22.7). The result is the Ward (or Ward-Takahashi) identity L Tj\u00b5(x)\u03d5a1(x1)... \u03d5an(xn) 0 0 = \u2202\u00b5h 0 | i | n T\u03d5a1(x1)... \u03b4\u03d5aj(x)\u03b44(x 0 + i | Xj=1h \u2212 xj)... \u03d5an(xn) 0 i |. (22.26) Thus, conservation of the Noether current holds in the quantum theory, with the current inside a correlation function, up to contact terms with a speci\ufb01c form that depends on the details of the in\ufb01nitesimal transformation that leaves invariant. L 6 22: Continuous Symmetries and Conserved Currents 149 The Noether current is also useful in a slightly more general context. (x) is not zero, (x) = \u2202\u00b5K \u00b5(x) for some K \u00b5(x). Then Suppose we have a transformation of the \ufb01elds such that \u03b4 but instead is a total divergence: \u03b4 there is still a conserved current, now given by L L j\u00b5(x) = \u2202 (x) L \u2202(\u2202\u00b5\u03d5a(x)) \u03b4\u03d5a(x) \u2212 K \u00b5(x). (22.27) An example of this is provided by the symmetry of spacetime translations. a), where a\u00b5 is a constant fourWe transform the \ufb01elds via \u03d5a(x) a\u03bd\u2202\u03bd\u03d5a(x), and vector. The in\ufb01nitesimal version of this is \u03d5a(x) \u2192 a\u03bd\u2202\u03bd \u03d5a(x). Under this transformation, we obviously so we have \u03b4\u03d5a(x) = have (x)). Thus \u2212 in this case K \u03bd(x) = a\u03bd\u2202\u03bd L (x), and the conserved current is \u2212 a), and so \u03b4 \u2212 \u2202\u03bd(a\u03bd \u03d5a(x) (x) = (x) = \u03d5a(x \u2192 L (x) a\u03bd \u2192 (x \u2212 \u2212 \u2212 L L L \u2212 j\u00b5(x) = L \u2202 (x) L \u2202(\u2202\u00b5\u03d5a(x)) = a\u03bdT \u00b5\u03bd(x), a\u03bd\u2202\u03bd\u03d5a(x)) ( \u2212 a\u00b5 L \u2212 (x) (22.28) where we have de\ufb01ned the stress-energy or energy-momentum tensor T \u00b5\u03bd(x) \u2261 \u2212 \u2202 (x) L \u2202(\u2202\u00b5\u03d5a(x)) \u2202\u03bd\u03d5a(x) + g\u00b5\u03bd (x). L (22.29) For a renormalizable theory of a set of real scalar \ufb01elds \u03d5a(x), the lagrangian takes the form = 1 2 \u2202\u00b5\u03d5a\u2202\u00b5\u03d5a \u2212 \u2212 L V (\u03d5), where V (\u03d5) is a polynomial in the \u03d5a\u2019s. In this case T \u00b5\u03bd = \u2202\u00b5\u03d5a\u2202\u03bd\u03d5a + g\u00b5\u03bd. L In particular, (22.30) (22.31) 2 \u03a02 T 00 = 1 a + 1 2 ( \u2207 where \u03a0a = \u22020\u03d5a is the canonical momentum conjugate to the \ufb01eld \u03d5a. We recognize T 00 as the hamiltonian density that corresponds to the lagrangian density of eq. (22.30). Then, by Lorentz symmetry, T 0j must be the corresponding momentum density. We have \u03d5a)2 + V (\u03d5), (22.32) H T 0j = \u22020\u03d5a\u2202j\u03d5a = \u03a0a\u2207 \u2212 j\u03d5a. (22.33) To check that this is a sensible result, we use the free-\ufb01eld expansion for a set of real", " scalar \ufb01elds [the same as eq. (22.18) but with b(k) = a(k) for each \ufb01eld]; then we \ufb01nd that the momentum operator is given by P j = d3x T 0j(x) = dk kj a\u2020a(k)aa(k), (22.34) Z Z f 22: Continuous Symmetries and Conserved Currents 150 which is just what we would expect. We therefore identify the energymomentum four-vector as P \u00b5 = d3x T 0\u00b5(x). Z (22.35) Recall that in section 2 we de\ufb01ned the spacetime translation operator as and announced that it had the property that T (a) exp( \u2212 \u2261 iP \u00b5a\u00b5), T (a)\u2212 1\u03d5a(x)T (a) = \u03d5a(x a). \u2212 (22.36) (22.37) Now that we have an explicit formula for P \u00b5, we can check this. This is easiest to do for in\ufb01nitesimal a\u00b5; then eq. (22.37) becomes [\u03d5a(x), P \u00b5] = 1 i \u2202\u00b5\u03d5a(x). (22.38) This can indeed be veri\ufb01ed by using the canonical commutation relations for \u03d5a(x) and \u03a0a(x). One more symmetry we can investigate is Lorentz symmetry. If we make x), an in\ufb01nitesimal Lorentz transformation, we have \u03d5a(x) \u03c1x\u03c1. This case is very similar to that of where \u03b4\u03c9 spacetime translations; the only di\ufb00erence is that the translation parameter \u03c1x\u03c1. The resulting conserved current is a\u03bd is now x dependent, a\u03bd x is shorthand for \u03b4\u03c9\u03bd \u03d5a(x + \u03b4\u03c9 \u03b4\u03c9\u03bd \u2192 \u00b7 \u00b7 \u2192 \u2212 \u00b5\u03bd\u03c1(x) = x\u03bdT \u00b5\u03c1(x) x\u03c1T \u00b5\u03bd(x), (22.39) M \u00b5\u03bd\u03c1 = 0, with the derivative contracted with the \ufb01rst and it obeys \u2202\u00b5M \u00b5\u03bd\u03c1 is antisymmetric on its second two indices; this comes about index. M because \u03b4\u03c9\u03bd\u03c1 is antisymmetric. The conserved charges associated with this current are \u2212 M \u03bd\u03c1 = d3x 0\u03bd\u03c1(x), M (22.40) Z and these are the generators of the Lorentz group that were introduced in section 2. Again, we can use the canonical commutation relations for the \ufb01elds to check that the Lorentz generators have the right commutation relations, both with the \ufb01elds and with each other. Reference Notes The path-integral approach to Ward identities is treated in more detail in Peskin & Schroeder. An operator-based derivation can be found in Weinberg I. 22: Continuous Symmetries and Conserved Currents 151 Problems 22.1) For the Noether current of eq. (22.6), and assuming that \u03b4\u03d5a does not involve time derivatives, use the canonical commutation relations to show that [\u03d5a, Q] = i\u03b4\u03d5a, (22.41) where Q is the Noether charge. 22.2) Use the canonical commutation relations to verify eq. (22.38). 22.3) a) With T \u00b5\u03bd given by eq. (22.31), compute the equal-time (x0 = y0) commutators [T 00(x), T 00(y)], [T 0i(x), T 00(y)], and [T 0i(x), T 0j(y)]. b) Use your results to verify eqs. (2.17), (2.19), and (2.20). 23: Discrete Symmetries: P, T, C, and Z 152 23 Discrete Symmetries: P, T, C, and Z Prerequisite: 22 In section 2, we studied the proper orthochronous Lorentz transformations, which are continuously connected to the identity. In this section, we will consider the e\ufb00ects of parity, +1 \u00b5 P \u03bd = ( P 1) \uf8f7 \uf8f7 \uf8f8 \u2212 and time reversal, \u00b5 \u03bd = ( T T 1)1 +1 \uf8f6. +1 \uf8f7 \uf8f7 \uf8f8 (23.1) (23.2) We will also consider certain other discrete transformations such as charge conjugation. Recall from section 2 that for every proper", " orthochronous Lorentz trans\u03bd there is an associated unitary operator U (\u039b) with the prop- formation \u039b\u00b5 erty that U (\u039b)\u2212 1\u03d5(x)U (\u039b) = \u03d5(\u039b\u2212 1x). (23.3) Thus for parity and time-reversal, we expect that there are corresponding unitary operators P T \u2261 \u2261 such that ), ), U ( U ( P T P \u2212 T \u2212 1\u03d5(x)P = \u03d5( 1\u03d5(x)T = \u03d5( T P (23.4) (23.5) (23.6) (23.7) x), x). There is, however, an extra possible complication. Since the and matrices are their own inverses, a second parity or time-reversal transT formation should transform all observables back into themselves. Using eqs. (23.6) and (23.7), along with 2 = 1, we see that 2 = 1 and P P T 2\u03d5(x)P 2 = \u03d5(x), 2\u03d5(x)T 2 = \u03d5(x). P \u2212 T \u2212 (23.8) (23.9) 23: Discrete Symmetries: P, T, C, and Z 153 Since \u03d5(x) is a hermitian operator, it is in principle an observable, and so eqs. (23.8) and (23.9) are just what we expect. However, another possibility for the parity transformation of the \ufb01eld, di\ufb00erent from eqs. (23.6) and (23.7) but nevertheless consistent with eqs. (23.8) and (23.9), is P \u2212 T \u2212 1\u03d5(x)P = 1\u03d5(x)T = \u2212 \u2212 \u03d5( \u03d5( P T x), x). (23.10) (23.11) This possible extra minus sign cannot arise for proper orthochronous Lorentz transformations, because they are continuously connected to the identity, and for the identity transformation (that is, no transformation at all), we must obviously have the plus sign. If the minus sign appears on the right-hand side, we say that the \ufb01eld is odd under parity (or time reversal). If a scalar \ufb01eld is odd under parity, we sometimes say that it is a pseudoscalar.1 So, how do we know which is right, eqs. (23.6) and (23.7), or eqs. (23.10) and (23.11)? The general answer is that we get to choose, but there is a key principle to guide our choice: if at all possible, we want to de\ufb01ne P and T so that the lagrangian density is evenx)P = + (x)T = + x), x). ( P ( T L L (23.12) (23.13) Then, after we integrate over d4x to get the action S, the action will be invariant. This means that parity and time-reversal are conserved. For theories with spin-zero \ufb01elds only, it is clear that the choice of eqs. (23.6) and (23.7) always leads to eqs. (23.12) and (23.13), and so there is no reason to \ufb02irt with eqs. (23.10) and (23.11). For theories that also include spin-one-half \ufb01elds, certain scalar bilinears in these \ufb01elds are necessarily odd under parity and time reversal, as we will see in section 40. If a scalar \ufb01eld couples to such a bilinear, then eqs. (23.12) and (23.13) will hold if and only if we choose eqs. (23.10) and (23.11) for that scalar, and so that is what we must do. it is antiunitary, rather than unitary. Antiunitary means that T \u2212 There is one more interesting fact about the time-reversal operator T : i. To see why this must be the case, consider a Lorentz transformation of 1iT = \u2212 the energy-momentum four-vector, U (\u039b)\u2212 1P \u00b5U (\u039b) = \u039b\u00b5 \u03bdP \u03bd. (23.14) 1It is still a scalar under proper orthochronous Lorentz transformations; that is, eq. (23.3) still holds. Thus the appellation scalar often", " means eq. (23.3), and either eq. (23.6) or eq. (23.10), and that is how we will use the term. 23: Discrete Symmetries: P, T, C, and Z 154 For parity and time-reversal, we therefore expect P \u2212 T \u2212 1P \u00b5P = 1P \u00b5T = \u00b5 \u00b5 P \u03bdP \u03bd, \u03bdP \u03bd. (23.15) (23.16) T 1HP = +H and T \u2212 H. In particular, for \u00b5 = 0, we expect P \u2212 The \ufb01rst of these is \ufb01ne; it says the hamiltonian is invariant under parity, which is what we want.2 However, eq. (23.16) is a disaster: it says that the hamiltonian is invariant under time-reversal if and only if H = H, which is possible only if H = 0. 1HT = \u2212 \u2212 Can we just put an extra minus sign on the right-hand side of eq. (23.16), as we did for eq. (23.11)? The answer is no. We constructed P \u00b5 explicitly in terms of the \ufb01elds in section 22, and it is easy to check that choosing eq. (23.11) for the \ufb01elds does not yield an extra minus sign in eq. (23.16) for the energy-momentum four-vector. Let us reconsider the origin of eq. (23.14). We \ufb01rst recall that the space- time translation operator T (a) = exp( \u2212 a). iP \u00b7 (23.17) (which should not be confused with the time-reversal operator T ) transforms a scalar \ufb01eld according to T (a)\u2212 1\u03d5(x)T (a) = \u03d5(x a). \u2212 (23.18) The spacetime translation operator is a scalar with a spacetime coordinate as a label; by analogy with eq. (23.3), we should have U (\u039b)\u2212 1T (a)U (\u039b) = T (\u039b\u2212 1a). (23.19) Now, treat a\u00b5 as in\ufb01nitesimal in eq. (23.19) to get U (\u039b)\u2212 1(I \u2212 ia\u00b5P \u00b5)U (\u039b) = I = I \u2212 \u2212 For time-reversal, this becomes \u00b5a\u00b5P \u03bd i(\u039b\u2212 i\u039b\u00b5 1)\u03bd \u03bda\u00b5P \u03bd. T \u2212 1(I \u2212 ia\u00b5P \u00b5)T = I \u00b5 \u03bda\u00b5P \u03bd. i T \u2212 (23.20) (23.21) ia\u00b5 on each side, we get eq. (23.16), If we now identify the coe\ufb03cients of \u2212 which is bad. In order to get the extra minus sign that we need, we must impose the antiunitary condition T \u2212 1iT = i. \u2212 (23.22) 2When spin-one-half \ufb01elds are present, it may be that no operator exists that satis\ufb01es in this case we say that parity is either eq. (23.6) or eq. (23.10) and also eq. (23.15); explicitly broken. 23: Discrete Symmetries: P, T, C, and Z 155 We then \ufb01nd T \u2212 1P \u00b5T = \u00b5 \u03bdP \u03bd instead of eq. (23.16). This yields T \u2212 expression of time-reversal invariance. \u2212T 1HT = +H, which is the correct (23.23) We turn now to other unitary operators that change the signs of scalar \ufb01elds, but do nothing to their spacetime arguments. Suppose we have a theory with real scalar \ufb01elds \u03d5a(x), and a unitary operator Z that obeys Z \u2212 1\u03d5a(x)Z = \u03b7a\u03d5a(x), (23.24) \u2212 where \u03b7a is either +1 or 1 for each \ufb01eld. We will call Z a Z2 operator, because Z2 is the additive group of the integers modulo 2, which is equivalent 1. This also implies that Z 2 = 1, to the multiplicative group of +1 and 1 = Z. (For theories with spin-zero \ufb01elds only, the same is also and so Z \u2212 true of P and T, but things are more subtle for higher spin, as we will see in Part II.) \u2212", " Consider the theory of a complex scalar \ufb01eld \u03d5 = (\u03d51 + i\u03d52)/\u221a2 that was introduced in section 22, with lagrangian 1 L = = \u2212 \u2212 m2\u03d5\u2020\u03d5 \u2202\u00b5\u03d5\u2020\u2202\u00b5\u03d5 2 \u2202\u00b5\u03d51\u2202\u00b5\u03d51 \u2212 In the form of eq. (23.25), mation \u2212 1 L 1 4 \u03bb(\u03d5\u2020\u03d5)2 \u2212 2 \u2202\u00b5\u03d52\u2202\u00b5\u03d52 \u2212 (23.25) 1 2 m2(\u03d52 1 + \u03d52 2) 1 16 \u03bb(\u03d52 1 + \u03d52 2)2. (23.26) \u2212 is obviously invariant under the U(1) transfor- \u03d5(x) \u2192 e\u2212 i\u03b1\u03d5(x). (23.27) In the form of eq. (23.26), SO(2) transformation, L is obviously invariant under the equivalent \u03d51(x) \u03d52(x)! \u2192 \u2212 cos \u03b1 sin \u03b1 \u03d51(x) sin \u03b1 cos \u03b1! \u03d52(x)!. (23.28) However, it is also obvious that has an additional discrete symmetry, L \u03d5(x) \u03d5\u2020(x) \u2194 (23.29) in the form of eq. (23.25), or equivalently \u03d51(x) \u03d52(x)! \u2192 +1 0 \u03d51(x) 0 \u2212 1! \u03d52(x)!. (23.30) in the form of eq. (23.26). This discrete symmetry is called charge conjugation. It always occurs as a companion to a continuous U(1) symmetry. In terms of the two real \ufb01elds, it enlarges the group from SO(2) (the group 23: Discrete Symmetries: P, T, C, and Z 156 of 2 2 \u00d7 2 orthogonal matrices with determinant +1) to O(2) (the group of \u00d7 2 orthogonal matrices). We can implement charge conjugation by means of a particular Z2 op- (23.31) (23.32) (23.33) (23.34) (23.35) (23.36) erator C that obeys or equivalently C \u2212 1\u03d5(x)C = \u03d5\u2020(x), C \u2212 C \u2212 1\u03d51(x)C = +\u03d51(x), 1\u03d52(x)C = \u03d52(x). \u2212 We then have 1 C \u2212 (x)C = (x), L L and so charge conjugation is a symmetry of the theory. Physically, it implies that the scattering amplitudes are unchanged if we exchange all the a-type particles (which have charge +1) with all the b-type particles (which have charge 1). This means, in particular, that the a and b particles must have exactly the same mass. We say that b is a\u2019s antiparticle. \u2212 More generally, we can also have Z2 symmetries that are not related to antiparticles. Consider, for example, \u03d54 theory, where \u03d5 is a real scalar \ufb01eld with lagrangian 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 If we de\ufb01ne the Z2 operator Z via = \u2212 L 1 1 2 m2\u03d52 \u2212 \u2212 1 24 \u03bb\u03d54. Z \u2212 1\u03d5(x)Z = \u03d5(x), \u2212 L is obviously invariant. We therefore have Z \u2212 1HZ = H, or equivathen lently [Z, H] = 0, where H is the hamiltonian. If we assume that (as usual) the ground state is unique, then, since Z commutes with H, the ground state must also be an eigenstate of Z. We can \ufb01x the phase of Z [which is undetermined by eq. (23.36)] via 0 i | Then, using eqs. (23.36) and (23.37), we have = 23.37) 0 \u03d5(x) 0 i | | h = = ZZ \u2212 0 | h 0 \u03d5(x) 0 i | | \u2212h 1\u03d5(x)ZZ \u2212. 1 0 i | (23.38) 0 \u03d5(x) 0 i | | h Since is equal to minus itself, it must be zero. Thus, as long as the ground state is unique, the Z2 symmetry of \u03d54 theory guarantees that the \ufb01eld has", " zero vacuum expectation value. We therefore do not need to enforce this condition with a counterterm Y \u03d5, as we did in \u03d53 theory. (The assumption of a unique ground state does not necessarily hold, however, as we will see in section 30.) 24: Nonabelian Symmetries 157 24 Nonabelian Symmetries Prerequisite: 22 Consider the theory (introduced in section 22) of a two real scalar \ufb01elds \u03d51 and \u03d52 with = 1 2 \u2202\u00b5\u03d51\u2202\u00b5\u03d51 \u2212 1 2 \u2202\u00b5\u03d52\u2202\u00b5\u03d52 \u2212 \u2212 L 1 2 m2(\u03d52 1 + \u03d52 2) \u2212 1 16 \u03bb(\u03d52 1 + \u03d52 2)2. (24.1) We can generalize this to the case of N real scalar \ufb01elds \u03d5i with = 1 2 \u2202\u00b5\u03d5i\u2202\u00b5\u03d5i \u2212 1 2 m2\u03d5i\u03d5i \u2212 \u2212 L 1 16 \u03bb(\u03d5i\u03d5i)2, (24.2) where a repeated index is summed. This lagrangian is clearly invariant under the SO(N ) transformation \u03d5i(x) \u2192 Rij\u03d5j(x), (24.3) 1, where R is an orthogonal matrix with a positive determinant: RT = R\u2212 det R = +1. This largrangian is also clearly invariant under the Z2 transformation \u03d5i(x) \u03d5i(x), which enlarges SO(N ) to O(N ); see section 23. However, in this section we will be concerned only with the continuous SO(N ) part of the symmetry. \u2192 \u2212 Next we will need some results from group theory. Consider an in\ufb01nites- imal SO(N ) transformation, Rij = \u03b4ij + \u03b8ij + O(\u03b82). (24.4) Orthogonality of Rij implies that \u03b8ij is real and antisymmetric. It is convenient to express \u03b8ij in terms of a basis set of hermitian matrices (T a)ij. The index a runs from 1 to 1 1), the number of linearly independent, N matrices. We can, for example, choose hermitian, antisymmetric, N each T a to have a single nonzero entry i above the main diagonal, and a corresponding +i below the main diagonal. These matrices obey the normalization condition 2 N (N \u2212 \u00d7 \u2212 Tr(T aT b) = 2\u03b4ab. In terms of them, we can write \u03b8jk = \u2212 i\u03b8a(T a)jk, (24.5) (24.6) where \u03b8a is a set of 1 1) real, in\ufb01nitesimal parameters. The T a\u2019s are the generator matrices of SO(N ). The product of any two SO(N ) transformations is another SO(N ) transformation; this implies (see 2 N (N \u2212 24: Nonabelian Symmetries 158 problem 24.2) that the commutator of any two generator matrices must be a linear combination of generator matrices, [T a, T b] = if abcT c. (24.7) The numerical factors f abc are the structure coe\ufb03cients of the group, and eq. (24.7) speci\ufb01es its Lie algebra. If f abc = 0, the group is abelian. Otherwise, it is nonabelian. Thus, U(1) and SO(2) are abelian groups (since they each have only one generator that obviously must commute with itself), and SO(N ) for N 3 is nonabelian. If we multiply eq. (24.7) on the right by T d, take the trace, and use \u2265 eq. (24.5), we \ufb01nd. \u2212 1 2 i Tr f abd = [T a, T b]T d Using the cyclic property of the trace, we \ufb01nd that f abd must be completely antisymmetric. Taking the complex conjugate of eq. (24.8) (and remembering that the T a\u2019s are hermitian matrices), we \ufb01nd that f abd must be real. The simplest nonabelian group is SO(3). In this case, we can choose i\u03b5aij, where \u03b5ijk is the completely antisymmetric Levi-Civita (T a)ij = symbol, with \u03b5123 = +1. The commutation relations become (24", ".8) \u2212 [T a, T b] = i\u03b5abcT c. (24.9) That is, the structure coe\ufb03cients of SO(3) are given by f abc = \u03b5abc. Consider now a theory with N complex scalar \ufb01elds \u03d5i, and a lagrangian m2\u03d5\u2020i \u03d5i \u2212 \u2202\u00b5\u03d5\u2020i \u2202\u00b5\u03d5i \u2212 4 \u03bb(\u03d5\u2020i \u03d5i)2, (24.10) = \u2212 L 1 where a repeated index is summed. This lagrangian is clearly invariant under the U(N ) transformation \u03d5i(x) \u2192 Uij\u03d5j(x), (24.11) 1. We can write Uij = e\u2212 Uij, where U is a unitary matrix: U \u2020 = U \u2212 where \u03b8 is a real parameter and det Uij is called a special unitary matrix. Clearly the product of two special unitary matrices is another e special unitary matrix; the N N special unitary matrices form the group SU(N ). The group U(N ) is the direct product of the group U(1) and the group SU(N ); we write U(N ) = U(1) Uij = +1; SU(N ). \u00d7 e e i\u03b8 Consider an in\ufb01nitesimal SU(N ) transformation, \u00d7 i\u03b8a(T a)ij + O(\u03b82), (24.12) Uij = \u03b4ij \u2212 e where \u03b8a is a set of real, in\ufb01nitesimal parameters. Unitarity of that the generator matrices T a are hermitian, and det U implies U = +1 implies e e 24: Nonabelian Symmetries 159 that each T a is traceless. (This follows from the general matrix formula ln det A = Tr ln A.) The index a runs from 1 to N 2 1, the number of linearly independent, hermitian, traceless, N N matrices. We can choose these matrices to obey the normalization condition of eq. (24.5). For SU(2), the generators can be chosen to be the Pauli matrices; the structure coe\ufb03cients of SU(2) then turn out to be f abc = 2\u03b5abc, the same as those of SO(3), up to an irrelevant overall factor [which could be removed by changing the numerical factor on right-hand side of eq. (24.5) from 2 to 1 \u2212 \u00d7 For SU(N ), we can choose the T a\u2019s in the following way. First, there i above the main diagonal a correare the SO(N ) generators, with one \u2212 sponding +i below; there are 1 1) of these. Next, we get another set 2 N (N \u2212 by putting one +1 above the main diagonal and a corresponding +1 below; there are 1 1) of these. Finally, there are diagonal matrices with n n, followed by zeros 1\u2019s along the main diagonal, followed a single entry [times an overall normalization constant to enforce eq. (24.5)]; the are N 1 \u2212 of these. The total is N 2 1, as required. 2 N (N \u2212 \u2212 2 ]. However, if we examine the lagrangian of eq. (24.10) more closely, we \ufb01nd that it is actually invariant under a larger symmetry group, namely SO(2N ). To see this, write each complex scalar \ufb01eld in terms of two real scalar \ufb01elds, \u03d5j = (\u03d5j1 + i\u03d5j2)/\u221a2. Then \u2212 \u03d5\u2020j\u03d5j = 1 2 (\u03d52 11 + \u03d52 12 +... + \u03d52 N 1 + \u03d52 N 2). (24.13) Thus, we have 2N real scalar \ufb01elds that enter symmetrically, and so the actual symmetry group of eq. (24.10) is SO(2N ), rather than just the obvious subgroup U(N ). L We will, however, meet the SU(N ) groups again in Parts II and III, where they will play a more important role. Problems 24.1) Show that \u03b8ij in eq. (24.4) must be antisymmetric if R is orthogonal. 24.2) By considering the SO(N ) transformation R\u2032\u2212 1R\u2032R, where R and R\u2032 are independent in\ufb01nitesimal SO(N ) transformations, prove eq. (24.7). 1R\u2212", " 24.3) a) Find the Noether current ja\u00b5 for the transformation of eq. (24.6). b) Show that [\u03d5i, Qa] = (T a)ij\u03d5j, where Qa is the Noether charge. c) Use this result, eq. (24.7), and the Jacobi identity (see problem 2.8) to show that [Qa, Qb] = if abcQc. 24: Nonabelian Symmetries 24.4) The elements of the group SO(N ) can be de\ufb01ned as N R that satisfy Rii\u2032Rjj\u2032\u03b4i\u2032j\u2032 = \u03b4ij. 160 N matrices \u00d7 (24.14) The elements of the symplectic group Sp(2N ) can be de\ufb01ned as 2N 2N matrices S that satisfy \u00d7 Sii\u2032Sjj\u2032\u03b7i\u2032j\u2032 = \u03b7ij, (24.15) \u03b7ji, and where the symplectic metric \u03b7ij is antisymmetric, \u03b7ij = squares to minus the identity: \u03b72 = I. One way to write \u03b7 is \u2212 \u2212 24.16) where I is the N of Sp(2N ). \u00d7 N identity matrix. Find the number of generators 25: Unstable Particles and Resonances 161 25 Unstable Particles and Resonances Prerequisite: 14 Consider a theory of two real scalar \ufb01elds, \u03d5 and \u03c7, with lagrangian = L \u2212 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 1 2 m2 \u03d5\u03d52 1 2 \u2202\u00b5\u03c7\u2202\u00b5\u03c7 \u2212 \u2212 1 2 m2 \u03c7\u03c72 + 1 2 g\u03d5\u03c72 + 1 6 h\u03d53. (25.1) This theory is renormalizable in six dimensions, where g and h are dimensionless coupling constants. Let us assume that m\u03d5 > 2m\u03c7. Then it is kinematically possible for the \u03d5 particle to decay into two \u03c7 particles. The amplitude for this process is given at tree level by the Feynman diagram of \ufb01g. (25.1), and is simply = g. We can also choose to de\ufb01ne g as the value of the exact \u03d5\u03c72 vertex m2 \u03d5, T function V3(k, k\u20321, k\u20322) when all three particles are on shell: k2 = k\u20321 \u03c7. This implies that 2 = k\u20322 2 = m2 \u2212 \u2212 = g T (25.2) exactly. According to the formulae of section 11, the di\ufb00erential decay rate (in the rest frame of the initial \u03d5 particle) is d\u0393 = 1 2m\u03d5 dLIPS2 |T | 2, (25.3) where dLIPS2 is the Lorentz invariant phase space di\ufb00erential for two outgoing particles, introduced in section 11. We must make a slight adaptation for six dimensions: (2\u03c0)6\u03b46(k\u20321+k\u20322\u2212 k) Here k = (m\u03d5, 0) is the energy-momentum of the decaying particle, and dLIPS2 \u2261 dk\u20322. dk\u20321 f f (25.4) dk = d5k (2\u03c0)52\u03c9 (25.5) is the Lorentz-invariant phase-space di\ufb00erential for one particle. Recall that we can also write it as f dk = d6k (2\u03c0)6 2\u03c0\u03b4(k2 + m2 \u03c7) \u03b8(k0), (25.6) where \u03b8(x) is the unit step function. Performing the integral over k0 turns eq. (25.6) into eq. (25.5). f 25: Unstable Particles and Resonances 162 k k1 k2 Figure 25.1: The tree-level Feynman diagram for the decay of a \u03d5 particle (dashed line) into two \u03c7 particles (solid lines). Repeating for six dimensions what we did in section 11 for four dimen- sions, we \ufb01nd dLIPS2 = 3 k\u20321| | 4(2\u03c0)4m\u03d5 d\u2126, (25.7) = 1 \u03d5 \u2212 4m2 2 (m2 k\u20321| | d\u2126 = \u21265 = 2\u03c05/2/\u0393( 5 \u03c7)1/2 is the magnitude of the spatial momentum where of one of the outgoing particles. We can now plug this into eq. (25.3), and 3", " \u03c02. We also need to divide by a symmetry use factor of two, due to the presence of two identical particles in the \ufb01nal state. The result is 2m\u03d5 Z = 1 12 \u03c0\u03b1(1 \u2212 dLIPS2 |T | 2 4m2 \u03c7/m2 \u03d5)3/2 m\u03d5, (25.8) (25.9) where \u03b1 = g2/(4\u03c0)3. However, as we discussed in section 11, we have a conceptual problem. According to our development of the LSZ formula in section 5, each incoming and outgoing particle should correspond to a single-particle state that is an exact eigenstate of the exact hamiltonian. This is clearly not the case for a particle that can decay. Let us, then, compute something else instead: the correction to the \u03d5 propagator from a loop of \u03c7 particles, as shown in \ufb01g. (25.2). The diagram is the same as the one we already analyzed in section 14, except that the internal propagators contain m\u03c7 instead of m\u03d5. (There is also a contribution from a loop of \u03d5 particles, but we can ignore it if we assume that h g.) We have \u226a where \u03a0(k2) = 1 2 \u03b1 1 0 Z dx D ln D A\u2032k2 \u2212 \u2212 B\u2032m2 \u03d5, D = x(1 \u2212 x)k2 + m2 \u03c7 \u2212 i\u01eb, (25.10) (25.11) 25: Unstable Particles and Resonances 163 Figure 25.2: A loop of \u03c7 particles correcting the \u03d5 propagator. m2 But, we have a problem. For k2 = and A\u2032 and B\u2032 are the \ufb01nite counterterm coe\ufb03cients that remain after the in\ufb01nities have been absorbed. We now try to \ufb01x A\u2032 and B\u2032 by imposing the usual on-shell conditions \u03a0( \u03d5) = 0. \u03d5 and m\u03d5 > 2m\u03c7, D is negative for part of the range of x. Therefore ln D has an imaginary part. This imaginary part cannot be canceled by A\u2032 and B\u2032, since A\u2032 and B\u2032 must be real: they are coe\ufb03cients of hermitian operators in the lagrangian. The best we can do is Re \u03a0( \u03d5) = 0 and Re \u03a0\u2032( \u03d5) = 0. Imposing these gives \u03d5) = 0 and \u03a0\u2032( m2 m2 m2 m2 \u2212 \u2212 \u2212 \u2212 \u2212 (25.12) (25.13) \u03a0(k2) = 1 2 \u03b1 1 0 Z ) D0| dx D ln(D/ | \u2212 1 12 \u03b1(k2 + m2 \u03d5), where D0 = x(1 \u2212 \u2212 x)m2 \u03d5 + m2 \u03c7. Now let us compute the imaginary part of \u03a0(k2). This arises from the \u03c7/k2)1/2 are integration range x the roots of D = 0 when k2 < 4m2 i\u03c0; the minus sign arises because, according to eq. (25.11), D has a small negative imaginary part. Now we have 2 \u00b1 \u03c7. In this range, Im ln D = < x < x+, where x 1 2 (1 + m2 = 1 \u2212 \u2212 \u2212 \u00b1 Im \u03a0(k2) = 1 2 \u03c0\u03b1 \u2212 x+ dx D x\u2212 Z = 1 12 \u03c0\u03b1(1 + 4m2 \u03c7/k2)3/2 k2 \u2212 \u03c7. Evaluating eq. (25.14) at k2 = Im \u03a0( \u2212 m2 \u03d5) = 1 12 \u03c0\u03b1(1 \u2212 4m2 \u03c7/m2 m2 \u03d5, we get \u2212 \u03d5)3/2 m2 \u03d5. when k2 < \u2212 4m2 From this and eq. (25.9), we see that Im \u03a0( \u2212 m2 \u03d5) = m\u03d5\u0393. (25.14) (25.15) (25.16) This is not an accident. Instead, it is a general rule. We will argue this in two ways: \ufb01rst, from the mathematics of Feynman diagrams, and second, from the physics of resonant scattering in quantum mechanics. 25: Unstable Particles and Resonances 164 We begin with the mathematics of Feynman diagrams. Return to the diagrammatic expression for \u03a0(k2), before we evaluated any of the integrals: \u03a0(k2) = 1 2 ig2 \u2212 Z d6\u21131 (2\u03c0)6 d6\ufffd", "\ufffd2 (2\u03c0)6 (2\u03c0)6\u03b46(\u21131+\u21132\u2212 k) 1 \u21132 1 + m2 \u03c7 \u2212 1 \u21132 2 + m2 \u03c7 \u2212 i\u01eb i\u01eb \u00d7 (Ak2 + Bm2 \u03d5). \u2212 (25.17) Here, for later convenience, we have assigned the internal lines momenta \u21131 and \u21132, and explicitly included the momentum-conserving delta function that \ufb01xes one of them. We can take the imaginary part of \u03a0(k2) by using the identity 1 x where P means the principal part. We then get, in a shorthand notation, + i\u03c0\u03b4(x), (25.18) = P \u2212 i\u01eb x 1 Im \u03a0(k2) = 1 2 g2 \u2212 P1P2 \u2212 \u03c02\u03b41\u03b42. (25.19) Z Next, we notice that the integral in eq. (25.17) is the Fourier transform of [\u2206(x \u2212 y)]2, where \u2206(x \u2212 y) = Z d6k (2\u03c0)6 y) eik(x \u2212 k2 + m2 \u03c7 \u2212 i\u01eb (25.20) is the Feynman propagator. Recall (from problem 8.5) that we can get the retarded or advanced propagator (rather than the Feynman propagator) by replacing the \u01eb in eq. (25.20) with, respectively, \u2261 \u2212 sign(k0). Therefore, in eq. (25.19), replacing \u03b41 with s1\u03b41 and \u03b42 with +s2\u03b42 yields an integral that is the real part of the Fourier transform of y). But this product is zero, because the \ufb01rst factor y)\u2206adv(x \u2206ret(x vanishes when x0 y0. So we can subtract the modi\ufb01ed integrand from the original without changing the value of the integral. Thus we have y0, and the second when x0 s\u01eb or +s\u01eb, where s \u2212 \u2265 \u2212 \u2212 \u2264 Im \u03a0(k2) = 1 2 g2\u03c02 (1 + s1s2)\u03b41\u03b42. (25.21) Z 1 and \u21130 The factor of 1+ s1s2 vanishes if \u21130 2 have opposite signs, and equals 2 if they have the same sign. Because the delta function in eq. (25.17) enforces 1 + \u21130 \u21130 2 must be positive. 2 = k0, and k0 = m\u03d5 is positive, both \u21130 1 and \u21130 25: Unstable Particles and Resonances 165 So we can replace the factor of 1 + s1s2 in eq. (25.21) with 2\u03b8(\u21130 Rearranging the numerical factors, we have 1)\u03b8(\u21130 2). Im \u03a0(k2) = 1 4 g2 Z d6\u21131 (2\u03c0)6 d6\u21132 (2\u03c0)6 (2\u03c0)6\u03b46(\u21131+\u21132\u2212 1) 2\u03c0\u03b4(\u21132 \u03c7)\u03b8(\u21130 1 + m2 k) 2\u03c0\u03b4(\u21132 If we now set k2 = is the decay amplitude, we can rewrite eq. (25.22) as m2 \u2212 \u00d7 \u03d5, use eqs. (25.4) and (25.6), and recall that = g T 2 + m2 \u03c7)\u03b8(\u21130 2). (25.22) Im \u03a0( \u2212 m2 \u03d5) = 1 4 Z dLIPS2 |T | 2. Comparing eqs. (25.8) and (25.23), we see that we indeed have Im \u03a0( \u2212 m2 \u03d5) = m\u03d5\u0393. (25.23) (25.24) This relation persists at higher orders in perturbation theory. Our analysis can be generalized to give the Cutkosky rules for computing the imaginary part of any Feynman diagram, but this is beyond the scope of our current interest. To get a more physical understanding of this result, recall that in nonrelativistic quantum mechanics, a metastable state with energy E0 and angular momentum quantum number \u2113 shows up as a resonance in the partial-wave scattering amplitude, f\u2113(E) 1 \u223c E \u2212 E0 + i\u0393/2. (25.25) If we imagine conv", "olving this amplitude with a wave packet will \ufb01nd a time dependence \u03c8(E)e\u2212 iEt, we e 1 \u03c8(t) dE \u223c Z \u2212 \u0393t/2. iE0t e\u2212 \u2212 E E0 + i\u0393/2 \u03c8(E)e\u2212 iEt e \u223c \u0393t, and we identify \u0393 as the inverse lifetime of the (25.26) 2 Therefore \u03c8(t) | | metastable state. \u223c e\u2212 In the relativistic case, consider the scattering process \u03c7\u03c7 \u03c7\u03c7. The contributing diagrams from the e\ufb00ective action are those of \ufb01g. (20.1), where the exact internal propagator can be either \u03d5 or \u03c7. Suppose that the center-of-mass energy squared s is close to m2 \u03d5. Since the \u03d5 progator has a pole near s = m2 \u03d5, s-channel \u03d5 exchange, shown in \ufb01g. (25.3), makes the dominant contribution to the \u03c7\u03c7 scattering amplitude. We then have \u2192 g2 T \u2243 s + m2 \u03d5 \u2212 \u2212 s) \u03a0( \u2212. (25.27) 25: Unstable Particles and Resonances 166 Figure 25.3: For s near m2 s-channel \u03d5 exchange. \u03d5, the dominant contribution to \u03c7\u03c7 scattering is Here we have used the fact that the exact \u03d5\u03c7\u03c7 vertex has the value g when all three particles are on-shell. Now let us write s = (m\u03d5 + \u03b5)2 m2 \u03d5 + 2m\u03d5\u03b5, \u2243 (25.28) where \u03b5 o\ufb00 resonance. We \ufb01nd \u226a m\u03d5 is the amount of energy by which our incoming particles are T \u2243 \u2212 \u03b5 + \u03a0( g2/2m\u03d5 m2 \u2212 \u03d5)/2m\u03d5. (25.29) m2 Recalling that Re \u03a0( we should make the identi\ufb01cation of eq. (25.24). \u2212 \u03d5) = 0, and comparing with eq. (25.25), we see that Reference Notes The Cutkosky rules are discussed in more detail in Peskin & Schroeder. More details on resonances can be found in Weinberg I. 26: Infrared Divergences 167 26 Infrared Divergences Prerequisite: 20 In section 20, we computed the \u03d5\u03d5 in six dimensions in the high-energy limit (s, than m2). We found that \u03d5\u03d5 scattering amplitude in \u03d53 theory all much larger u | |, and | t | \u2192 T = T0 g2(s\u2212 1 \u2212 h 1 + t\u2212 11 12 \u03b1 1 + u\u2212 ln(s/m2) + O(m0) + O(\u03b12) i, (26.1) \u2192 T0 = 1) is the tree-level result, and the O(m0) where term includes everything without a large logarithm that blows up in the limit m 0.1 \u2212 Suppose we are interested in the limit of massless particles. The large log is then problematic, since it blows up in this limit. What does this mean? It means we have made a mistake. Actually, two mistakes. In this section, we will remedy one of them. Throughout the physical sciences, it is necessary to make various idealizations in order to make progress. (Recall the \u201cmassless springs\u201d and \u201cfrictionless planes\u201d of freshman mechanics.) Sometimes these idealizations can lead us into trouble, and that is one of the things that has gone wrong here. We have assumed that we can isolate individual particles. The reasoning behind this was explained in section 5, and it depends on the existence of an energy gap between the one-particle states and the multiparticle continuum. However, this gap vanishes if the theory includes massless particles. In this case, it is possible that the scattering process involved the creation of some extra very low energy (or soft) particles that escaped detection. Or, there may have been some extra soft particles hiding in the initial state that discreetly participated in the scattering process. Or, what was seen as a single high-energy particle may actually have been two or more particles that were moving colinearly and sharing the energy. Let us, then, correct our idealization of a perfect detector and account for these possibilities. We will work with \u03d53 theory, initially in d spacetime dimensions. Let be the amplitude for some scattering process in \u03d53 theory. Now consider the possibility that one of the outgoing particles in this process splits into two, as shown in \ufb01g. (26.1). The amplitude for this new process T 1In", " writing T in this form, we have traded factors of ln t and ln u for ln s by \ufb01rst using ln t = ln s + ln(t/s), and then hiding the ln(t/s) terms in the O(m0) catchall. 26: Infrared Divergences 168 k1 k2 k Figure 26.1: An outgoing particle splits into two. The gray circle stands for the sum of all diagrams contributing to the original amplitude i. T is given in terms of by T Tsplit = ig \u2212 i k2 + m2 T, (26.2) where k = k1 + k2, and k1 and k2 are the on-shell four-momenta of the two particles produced by the split. (For notational convenience, we drop our usual primes on the outgoing momenta.) The key point is this: in the massless limit, it is possible for 1/(k2 + m2) to diverge. To understand the physical consequences of this possibility, we should compute an appropriate cross-section. To get the cross section for the original process (without the split), we multiply dk (as well as by similar di\ufb00erentials for other outgoing particles, and by an overall energymomentum delta function). For the process with the split, we multiply dk. (The factor of one-half is for counting |Tsplit| of identical particles.) If we assume that (due to some imperfection) our detector cannot tell whether or not the one particle actually split into two, then we should (according to the usual rules of quantum mechanics) add the probabilities for the two events, which are distinguishable in principle. We can therefore de\ufb01ne an e\ufb00ectively observable squared-amplitude via dk2 instead of 2 by 1 2 2 by dk1 |T | f f f f 2 obs |T | dk = 2 |T | dk + 2 1 2 |Tsplit| dk1 dk2 +.... (26.3) Here the ellipses stand for all other similar processes involving emission of one or more extra particles in the \ufb01nal state, or absorption of one or more extra particles in the initial state. f f f f We can simplify eq. (26.3) by including a factor of 1 = (2\u03c0)d \u2212 1 2\u03c9 \u03b4d \u2212 1(k1+k2\u2212 k) dk (26.4) f 26: Infrared Divergences in the second term. Now all terms in eq. (26.3) include a factor of we can drop it. Then, using eq. (26.2), we get 169 dk, so f g2 2 2 obs \u2261 |T | |T | 1 + (k2 + m2)2 (2\u03c0)d 1 2\u03c9 \u03b4d \u2212 \u2212 1(k1+k2\u2212 k) 1 2 \" # (26.5) Now we come to the point: in the massless limit, the phase space integral in the second term in eq. (26.5) can diverge. This is because, for m = 0, f f dk1 dk2 +.... k2 = (k1 + k2)2 = 4\u03c91\u03c92 sin2(\u03b8/2), (26.6) \u2212 where \u03b8 is the angle between the spatial momenta k1 and k2, and \u03c91,2 = k1,2| |. Also, for m = 0, dk2 \u223c Therefore, for small \u03b8, f dk1 f 3 (\u03c9d \u2212 1 d\u03c91) (\u03c9d \u2212 2 3 d\u03c92) (sind \u2212 3 \u03b8 d\u03b8). (26.7) dk1 dk2 (k2)2 \u223c f f d\u03c91 d \u03c95 \u2212 1 d\u03c92 d \u03c95 \u2212 2 d\u03b8 \u03b87 \u2212 d. (26.8) 4, and the Thus the integral over each \u03c9 diverges at the low end for d integral over \u03b8 diverges at the low end for d 6. These divergent integrals would be cut o\ufb00 (and rendered \ufb01nite) if we kept the mass m nonzero, as we will see below. \u2264 \u2264 Our discussion leads us to expect that the m 0 divergence in the 0 divergence in the loop second term of eq. (26.5) should cancel the m 2. We will now see how this works (or fails to work) in detail correction to for the familiar case of two-particle scattering in six spacetime dimensions, where is given by eq. (26.1). For d = 6, there is no problem", " with soft particles (corresponding to the small-\u03c9 divergence), but there is a problem with collinear particles (corresponding to the small-\u03b8 divergence). |T | \u2192 \u2192 T Let us assume that our imperfect detector cannot tell one particle from two nearly collinear particles if the angle \u03b8 between their spatial momenta is less than some small angle \u03b4. Since we ultimately want to take the m 0 limit, we will evaluate eq. (26.5) with m2/k2 \u2192 \u03b42 1. We can immediately integrate over d5k2 using the delta function, which results in setting k2 = k k1 everywhere. Let \u03b2 then be the angle between k1 (which is still to be integrated over) and k (which is \ufb01xed). For twoparticle scattering, 0. We then have = 1 2 \u221as in the limit m \u2212 \u226a \u226a \u2192 k | | (2\u03c0)5 2\u03c9 \u03b45(k1+k2\u2212 k) 1 2 dk1 f dk2 \u2192 f \u21264 4(2\u03c0)5 \u03c9 \u03c91\u03c92 | k1| 4 d k1| | sin3\u03b2 d\u03b2, (26.9) 26: Infrared Divergences 170 where \u21264 = 2\u03c02 is the area of the unit four-sphere. Now let \u03b3 be the angle between k2 and k. The geometry of this trio of vectors implies \u03b8 = \u03b2 + \u03b3, k k1|. All three of the angles are = (sin \u03b2/sin \u03b8) | | | small and positive, and it then is useful to write \u03b2 = x\u03b8 and \u03b3 = (1 x)\u03b8, with 0 k, and = (sin \u03b3/sin \u03b8) | | k2| | 1. 1 and \u03b8 \u2212 x \u03b4 In the low mass limit, we can safely set m = 0 everywhere in eq. (26.5) except in the propagator, 1/(k2 + m2). Then, expanding to leading order in both \u03b8 and m, we \ufb01nd (after some algebra) \u2264 \u2264 \u2264 \u226a k2 + m2 x(1 \u2212 \u2243 \u2212 x)k2 \u03b82 + (m2/k2)f (x) i h, (26.10) where f (x) = (1 \u2212 k1| safely set \u03c91 = | integration variables in eq. (26.9) from x2)2. Everywhere else in eq. (26.5), we can x+x2)/(x \u2212 k k. Then, changing the = x x) = (1 | | | | k2| and \u03c92 = | k1| and \u03b2 to x and \u03b8, we get | \u2212 2 obs = |T | 2 |T | 1 + \" g2\u21264 4(2\u03c0)5 1 0 Z x(1 \u2212 x)dx \u03b4 0 Z Performing the integral over \u03b8 yields \u03b83 d\u03b8 [\u03b82 + (m2/k2)f (x)]2 +.... 1 2 ln(\u03b42k2/m2) 1 2 ln f (x) 1 2. \u2212 Then, performing the integral over x and using \u21264 = 2\u03c02 and \u03b1 = g2/(4\u03c0)3, we get \u2212 # (26.11) (26.12) 2 obs = |T | 2 |T | 1 + 1 h 12 \u03b1 4.11. \u2212 where c = (4 3\u221a3\u03c0)/3 = \u2212 ln(\u03b42k2/m226.13) The displayed correction term accounts for the possible splitting of one of the two outgoing particles. Obviously, there is an identical correction for the other outgoing particle. Less obviously (but still true), there is (A glib an identical correction for each of the two incoming particles. explanation is that we are computing an e\ufb00ective amplitude-squared, and this is the same for the reverse process, with in and outgoing particles switched. So in and out particles should be treated symmetrically.) Then, since we have a total of four in and out particles (before accounting for any splitting), 2 obs = |T | 2 |T | 1 + 4 h 12 \u03b1 ln(\u03b42k2/m2) + c + O(\u03b12) i. (26.14) We have now accounted for the O(\u03b1) corrections due to the failure of our detector to separate two particles whose spatial momenta are nearly 26: Infrared Divergences 171 parallel. Combining this with eq. (26.1), and recalling that k2 = 1 4 s, we get 2 obs = |", "T | 2 |T0| \u00d7 = 2 |T0 ln(s/m2) + O(m0) + O(\u03b12) 11 6 \u03b1 i ln(\u03b42s/m2) + O(m0) + O(\u03b12) i 3 ln(1/\u03b42) + O(m0) 3 \u03b1 2 ln(s/m2) + 1 3 h \u2212 + O(\u03b12) i. (26.15) We now have two kinds of large logs. One is ln(1/\u03b42); this factor depends on the properties of our detector. If we build a very good detector, one for which \u03b1 ln(1/\u03b42) is not small, then we will have to do more work, and calculate higher-order corrections to eq. (26.15). The other large log is our original nemesis ln(s/m2). This factor blows up in the massless limit. This means that there is still a mistake hidden somewhere in our analysis. Reference Notes Infrared divergences in quantum electrodynamics are discussed in Brown and Peskin & Schroeder. More general treatments can be found in Sterman and Weinberg I. 27: Other Renormalization Schemes 172 27 Other Renormalization Schemes Prerequisite: 26 To \ufb01nd the remaining mistake in eq. (26.15), we must review our renormalization procedure. Recall our result from section 14 for the one-loop correction to the propagator, \u03a0(k2 k2 \u03b5 + 1 6 \u03b1 1 dx D ln(D/\u00b52) + O(\u03b12 m2 i (27.1) where \u03b1 = g2/(4\u03c0)3 and D = x(1 respect to k2 is \u2212 x)k2+m2. The derivative of \u03a0(k2) with \u03a0\u2032(k2 dx x(1 2 i x) \u2212 0 Z ln(D/\u00b52) + 1 i h + O(\u03b12). (27.2) \u2212 m2) = 0 and We previously determined A and B via the requirements \u03a0( m2) = 0. The \ufb01rst condition ensures that the exact propagator \u02dc\u2206(k2) \u03a0\u2032( m2, and the second ensures that the residue of this has a pole at k2 = pole is one. Recall that the \ufb01eld must be normalized in this way for the validity of the LSZ formula. \u2212 \u2212 We now consider the massless limit. We have D = x(1 x)k2, and we should apparently try to impose \u03a0(0) = \u03a0\u2032(0) = 0. However, \u03a0(0) is now automatically zero for any values of A and B, while \u03a0\u2032(0) is ill de\ufb01ned. \u2212 Physically, the problem is that the one-particle states are no longer separated from the multiparticle continuum by a \ufb01nite gap in energy. Mathematically, the pole in \u02dc\u2206(k2) at k2 = m2 merges with the branch point at k2 = 4m2, and is no longer a simple pole. \u2212 The only way out of this di\ufb03culty is to change the renormalization = 0, where we know scheme. Let us \ufb01rst see what this means in the case m what we are doing. Let us try making a di\ufb00erent choice of A and B. Speci\ufb01cally, let A = B = 1 \u03b5 + O(\u03b12), 6 \u03b1 1 \u03b5 + O(\u03b12). \u03b1 1 \u2212 \u2212 (27.3) Here we have chosen A and B to cancel the in\ufb01nities, and nothing more; we say that A and B have no \ufb01nite parts. This choice represents a di\ufb00erent renormalization scheme. Our original choice (which, up until now, we have pretended was inescapable!) is called the on-shell or OS scheme. The choice \u2212 6 27: Other Renormalization Schemes 173 of eq. (27.3) is called the modi\ufb01ed minimal-subtraction or MS (pronounced \u201cemm-ess-bar\u201d) scheme. g \u02dc\u00b5\u03b5/2, with \u00b52 = 4\u03c0e\u2212 just plain minimal subtraction or MS.] Now we have [\u201cModi\ufb01ed\u201d because we introduced \u00b5 via g \u2192 \u03b3 \u02dc\u00b52; had we set \u00b5 = \u02dc\u00b5 instead, the scheme would be \u03a0MS(k2) = \u2212 1 12 \u03b1(k2 + 6", "m2) + 1 2 \u03b1 dx D ln(D/\u00b52) + O(\u03b12), (27.4) 1 0 Z as compared to our old result in the on-shell scheme, \u03a0OS(k2) = \u2212 1 12 \u03b1(k2 + m2) + 1 2 \u03b1 dx D ln(D/D0) + O(\u03b12), (27.5) 1 0 Z x)k2 + m2, and D0 = [ x)+1]m2. Notice that where again D = x(1 0 limit, whereas \u03a0OS(k2) does not. On the \u03a0MS(k2) has a well-de\ufb01ned m other hand, \u03a0MS(k2) depends explicitly on the fake parameter \u00b5, whereas \u03a0OS(k2) does not. x(1 \u2192 \u2212 \u2212 \u2212 What does this all mean? First, in the MS scheme, the propagator \u2206MS(k2) will no longer have a pole at k2 = m2. The pole will be somewhere else. However, by de\ufb01nition, the actual physical mass mph of the particle is determined by the location of this pole: k2 = ph. Thus, the lagrangian parameter m is no longer the same as mph. m2 \u2212 \u2212 Furthermore, the residue of this pole is no longer one. Let us call the residue R. The LSZ formula must now be corrected by multiplying its 1/2 for each external particle (incoming right-hand side by a factor of R\u2212 1/2\u03d5(x) that now has unit or outgoing). This is because it is the \ufb01eld R\u2212 amplitude to create a one-particle state. \u22022+m2 Note also that, in the LSZ formula, each Klein-Gordon wave operator \u22022+m2; also, each external four-momentum should be m2. A review of the derivation of the should square to LSZ formula clearly shows that each of these mass parameters must be the actual particle mass, and not the parameter in the lagrangian. ph, and not ph, and not m2 \u2212 \u2212 \u2212 \u2212 Finally, in the LSZ formula, each external line will contribute a factor of R when the associated Klein-Gordon wave operator hits the external propagator and cancels its momentum-space pole, leaving behind the residue R. 1/2 for each \ufb01eld, we get a net Combined with the correction factor of R\u2212 factor of R1/2 for each external line when using the MS scheme. Internal i)/(k2 + m2), where m is the lagrangianlines each contribute a factor of ( parameter mass, and each vertex contributes a factor of iZgg, where g is the lagrangian-parameter coupling. \u2212 Let us now compute the relation between m and mph, and then compute R. We have \u2206MS(k2)\u2212 1 = k2 + m2 \u03a0MS(k2), \u2212 (27.6) 27: Other Renormalization Schemes and, by de\ufb01nition, \u2206MS( \u2212 m2 ph)\u2212 1 = 0. 174 (27.7) Setting k2 = \u2212 m2 ph in eq. (27.6), using eq. (27.7), and rearranging, we \ufb01nd m2 ph = m2 \u03a0MS( \u2212 \u2212 m2 ph). (27.8) Since \u03a0MS(k2) is O(\u03b1), we see that the di\ufb00erence between m2 O(\u03b1). Therefore, on the right-hand side, we can replace m2 only make an error of O(\u03b12). Thus ph and m2 is ph with m2, and m2 ph = m2 \u03a0MS( \u2212 \u2212 m2) + O(\u03b12). (27.9) Working this out, we get m2 ph = m2 1 2 \u03b1 \u2212 where D0 = [1 x(1 1 1 \u2212 6 m2 m2 + dx D0 ln(D0/\u00b52) x)]m2. Doing the integrals yields 0 Z + O(\u03b12), (27.10) \u2212 \u2212 ph = m2 m2 1 + 5 ln(\u00b52/m2) + c\u2032 12 \u03b1 h + O(\u03b12) i. (27.11) where c\u2032 = (34 3\u03c0\u221a3)/15 = 1.18. \u2212 Now, physics should be independent of the fake parameter \u00b5. However, the right-hand side of eq. (27.11) depends explicitly on \u00b5. It must, be, then, that m and \u03b1 take on di\ufb00erent numerical values as \u00b5", " is varied, in just the right way to leave physical quantities (like mph) unchanged. We can use this information to \ufb01nd di\ufb00erential equations that tell us how m and \u03b1 change with \u00b5. For example, take the logarithm of eq. (27.11) and divide by two to get ln mph = ln m + 5 ln(\u00b5/m) + 1 2 c\u2032 + O(\u03b12). (27.12) 12 \u03b1 Now di\ufb00erentiate with respect to ln \u00b5 and require mph to remain \ufb01xed: 0 = d d ln \u00b5 ln mph = 1 m dm d ln \u00b5 + 5 12 \u03b1 + O(\u03b12). (27.13) To get the second line, we had to assume that d\u03b1/d ln \u00b5 = O(\u03b12), which we will verify shortly. Then, rearranging eq. (27.13) gives dm d ln \u00b5 = 5 12 \u03b1 + O(\u03b12) \u2212 m. (27.14) 27: Other Renormalization Schemes 175 The factor in large parentheses on the right is called the anomalous dimension of the mass parameter, and it is often given the name \u03b3m(\u03b1). Turning now to the residue R, we have R\u2212 1 = d dk2 1 \u2206MS(k2)\u2212 h. k2= m2 ph \u2212 i (27.15) Using eq. (27.6), we get R\u2212 1 = 1 \u03a0\u2032MS( \u2212 \u2212 m2 ph) \u03a0\u2032MS( = 1 \u2212 = 1 + 1 m2) + O(\u03b12) \u2212 ln(\u00b52/m2) + c\u2032\u2032 12 \u03b1 + O(\u03b12), (27.16) (27.17) (27.18) where c\u2032\u2032 = (17 3\u03c0\u221a3)/3 = 0.23. \u2212 We can also use MS to de\ufb01ne the vertex function. We take and so C = \u2212 \u03b5 + O(\u03b12), \u03b1 1 V3,MS(k1, k2, k3) = g 1 where D = xyk2 \u2212 Z 3 + m2. 2 + zxk2 Let us now compute the \u03d5\u03d5 1 + yzk2 1 2 \u03b1 dF3 ln(D/\u00b52) + O(\u03b12) \u03d5\u03d5 scattering amplitude in our fancy In the low-mass limit, repeating the steps new renormalization scheme. that led to eq. (26.1), and including the LSZ correction factor (R1/2)4, we get \u2192 = R2 T0 T g2(s\u2212 where T0 = eq. (27.16), we \ufb01nd \u2212 1 \u2212 h 1 + t\u2212 11 12 \u03b1 1 + u\u2212 ln(s/\u00b52) + O(m0) + O(\u03b12) i 1) is the tree-level result. Now using R from, (27.19) = T0 T 1 \u03b1 \u2212 h 11 12 ln(s/\u00b52) + 1 6 ln(\u00b52/m2) + O(m0) + O(\u03b12) i To get an observable amplitude-squared with an imperfect detector, we must square eq. (27.20) and multiply it by the correction factor we derived in section 26,. (27.20) 2 obs = |T | 2 |T | 1 + 1 h 3 \u03b1 ln(\u03b42s/m2) + O(m0) + O(\u03b12) i, (27.21) where \u03b4 is the angular resolution of the detector. Combining this with eq. (27.20), we get 2 obs = |T | 2 |T0| 1 \u2212 \u03b1 h 3 2 ln(s/\u00b52) + 1 3 ln(1/\u03b42) + O(m0) + O(\u03b12) (27.22) i. 27: Other Renormalization Schemes 176 All factors of ln m2 have disappeared! Finally, we have obtained an expres0 limit. sion that has a well-de\ufb01ned m 2 Of course, \u00b5 is still a fake parameter, and so obs cannot depend on |T | it. It must be, then, that the explicit dependence on \u00b5 in eq. (27.22) is canceled by the implicit \u00b5 dependence of \u03b1. We can use this information 2 = O(g4) = O(\u03b12), to \ufb01gure out how \u03b1 must vary with \u00b5. Noting that we have |T0| \u2192 ln 2 obs = C1 + 2 ln \u03b1", " + 3\u03b1 |T | where C1 and C2 are independent of \u00b5 and \u03b1 (but depend on the Mandelstam variables). Di\ufb00erentiating with respect to ln \u00b5 then gives + O(\u03b12), ln \u00b5 + C2 (27.23) or, after rearranging, 0 = d d ln \u00b5 ln 2 obs |T | = 2 \u03b1 d\u03b1 d ln \u00b5 + 3\u03b1 + O(\u03b12), d\u03b1 d ln \u00b5 = \u2212 3 2 \u03b12 + O(\u03b13). (27.24) (27.25) The right-hand side of this equation is called the beta function. Returning to eq. (27.22), we are free to choose any convenient value of \u00b5 that we might like. To avoid introducing unnecessary large logs, we should choose \u00b52 s. \u223c To compare the results at di\ufb00erent values of s, we need to solve eq. (27.25). Keeping only the leading term in the beta function, the solution is \u03b1(\u00b52) = \u03b1(\u00b51) 1 + 3 2 \u03b1(\u00b51) ln(\u00b52/\u00b51). (27.26) Thus, as \u00b5 increases, \u03b1(\u00b5) decreases. A theory with this property is said to be asymptotically free. In this case, the tree-level approximation (in the MS scheme with \u00b52 s) becomes better and better at higher and higher energies. \u223c Of course, the opposite is true as well: as \u00b5 decreases, \u03b1(\u00b5) increases. As we go to lower and lower energies, the theory becomes more and more strongly coupled. If the particle mass is nonzero, this process stops at \u00b5 m. This is because the minimum value of s is 4m2, and so the factor of ln(s/\u00b52) m. We should therefore not use becomes an unwanted large log for \u00b5 \u223c \u226a 27: Other Renormalization Schemes 177 \u226a values of \u00b5 below m. Perturbation theory is still good at these low energies if \u03b1(m) 1. If the particle mass is zero, \u03b1(\u00b5) continues to increase at lower and lower energies, and eventually perturbation theory breaks down. This is a signal that the low-energy physics may be quite di\ufb00erent from what we expect on the basis of a perturbative analysis. In the case of \u03d53 theory, we know what the correct low-energy physics is: the perturbative ground state is unstable against tunneling through the potential barrier, and there is no true ground state. Asymptotic freedom is, in this case, a signal of this impending disaster. Much more interesting is asymptotic freedom in a theory that does have a true ground state, such as quantum chromodynamics. In this example, the particle excitations are colorless hadrons, rather than the quarks and gluons we would expect from examining the lagrangian. If the sign of the beta function is positive, then the theory is infrared free. The coupling increases as \u00b5 increases, and, at su\ufb03ciently high energy, perturbation theory breaks down. On the other hand, the coupling decreases as we go to lower energies. Once again, though, we should stop this process at \u00b5 m if the particles have nonzero mass. Quantum electrodynamics with massive electrons (but, of course, massless photons) is in this category. \u223c Still more complicated behaviors are possible if the beta function has a zero at a nonzero value of \u03b1. We brie\ufb02y consider this case in the next section. Reference Notes Minimal subtraction is treated in more detail in Brown, Collins, and Ramond I. Problems 27.1) Suppose that we have a theory with \u03b2(\u03b1) = b1\u03b12 + O(\u03b13), \u03b3m(\u03b1) = c1\u03b1 + O(\u03b12). (27.27) (27.28) Neglecting the higher-order terms, show that m(\u00b52) = c1/b1 \u03b1(\u00b52) \u03b1(\u00b51) m(\u00b51). (27.29) 28: The Renormalization Group 178 28 The Renormalization Group Prerequisite: 27 In section 27 we introduced the MS renormalization scheme, and used the fact that physical observables must be independent of the fake parameter \u00b5 to \ufb01gure out how the lagrangian parameters m and g must change with \u00b5. In this section we re-derive these results from a much more formal (but calculationally simpler) point of view, and see how they extend to all orders of perturbation theory. Equations that tells us how the lagrangian parameters (and other objects that are not", " directly measurable, like correlation functions) vary with \u00b5 are collectively called the equations of the renormalization group. Let us recall the lagrangian of our theory, and write it in two di\ufb00erent ways. In d = 6 \u03b5 dimensions, we have \u2212 2 Z\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5 1 = L \u2212 and 1 2 Zmm2\u03d52 + 1 6 Zgg \u02dc\u00b5\u03b5/2\u03d53 + Y \u03d5 \u2212 (28.1) 1 1 L = \u2212 0\u03d52 0 + 1 2 \u2202\u00b5\u03d50\u2202\u00b5\u03d50 \u2212 2 m2 The \ufb01elds and parameters in eq. (28.1) are the renormalized \ufb01elds and parameters. (And in particular, they are renormalized using the MS scheme, with \u00b52 = 4\u03c0e\u2212 \u03b3 \u02dc\u00b52.) The \ufb01elds and parameters in eq. (28.2) are the bare \ufb01elds and parameters. Comparing eqs. (28.1) and (28.2) gives us the relationships between them: 0 + Y0\u03d50. 6 g0\u03d53 (28.2) \u03d50(x) = Z 1/2 1/2 m0 = Z \u2212 \u03d5 g0 = Z \u2212 \u03d5 Y0 = Z \u2212 \u03d5 \u03d5 \u03d5(x), Z 1/2 m m, Zgg \u02dc\u00b5\u03b5/2, Y. 1/2 3/2 (28.3) (28.4) (28.5) (28.6) Recall that, after using dimensional regularization, the in\ufb01nities coming from loop integrals take the form of inverse powers of \u03b5 = 6 d. In the MS renormalization scheme, we choose the Z\u2019s to cancel o\ufb00 these powers of 1/\u03b5, and nothing more. Therefore the Z\u2019s can be written as \u2212 Z\u03d5 = 1 + Zm = 1 + Zg = 1 + an(\u03b1) \u03b5n, bn(\u03b1) \u03b5n cn(\u03b1) \u03b5n,, \u221e Xn=1 \u221e Xn=1 \u221e Xn=1 (28.7) (28.8) (28.9) 28: The Renormalization Group 179 where \u03b1 = g2/(4\u03c0)3. Computing \u03a0(k2) and V3(k1, k2, k3) in perturbation theory in the MS scheme gives us Taylor series in \u03b1 for an(\u03b1), bn(\u03b1), and cn(\u03b1). So far we have found a1(\u03b1) = b1(\u03b1) = c1(\u03b1(\u03b12), \u03b1 + O(\u03b12), \u03b1 + O(\u03b12), and that an(\u03b1), bn(\u03b1), and cn(\u03b1) are all at least O(\u03b12) for n (28.10) (28.11) (28.12) 2. \u2265 Next we turn to the trick that we will employ to compute the beta function for \u03b1, the anomalous dimension of m, and other useful things. This is the trick: bare \ufb01elds and parameters must be independent of \u00b5. Why is this so? Recall that we introduced \u00b5 when we found that we had to regularize the theory to avoid in\ufb01nities in the loop integrals of Feynman diagrams. We argued at the time (and ever since) that physical quantities had to be independent of \u00b5. Thus \u00b5 is not really a parameter of the theory, but just a crutch that we had to introduce at an intermediate stage of the calculation. In principle, the theory is completely speci\ufb01ed by the values of the bare parameters, and, if we were smart enough, we would be able to compute the exact scattering amplitudes in terms of them, without ever introducing \u00b5. The point is this: since the exact scattering amplitudes are independent of \u00b5, the bare parameters must be as well. Let us start with g0. It is convenient to de\ufb01ne \u03b10 \u2261 and also 0/(4\u03c0)3 = Z 2 g2 \u03d5 \u02dc\u00b5\u03b5\u03b1, g Z \u2212 3 G(\u03b1, \u03b5) \u2261 ln(Z 2 g Z \u2212 3 \u03d5 ). From the general structure of eqs. (28.7) and (28.9), we have G(\u03b1, \u03b5) = Gn(\u03b1) \u03b5n, \u221e Xn=1 where, in particular, G1(\u03b1) = 2c1(\u03b1) 3a1(\u03b1) \u2212 2 \u03b1 + O(", "\u03b12). 3 = \u2212 (28.13) (28.14) (28.15) (28.16) The logarithm of eq. (28.13) can now be written as ln \u03b10 = G(\u03b1, \u03b5) + ln \u03b1 + \u03b5 ln \u02dc\u00b5. (28.17) 28: The Renormalization Group 180 Next, di\ufb00erentiate eq. (28.17) with respect to ln \u00b5, and require \u03b10 to be independent of it: 0 = d d ln \u00b5 ln \u03b10 = \u2202G(\u03b1, \u03b5) \u2202\u03b1 d\u03b1 d ln \u00b5 + 1 \u03b1 d\u03b1 d ln \u00b5 + \u03b5. (28.18) Now regroup the terms, multiply by \u03b1, and use eq. (28.15) to get 0 = 1 + \u03b1G\u20321(\u03b1) \u03b5 + \u03b1G\u20322(\u03b1) \u03b52 +... d\u03b1 d ln \u00b5 + \u03b5\u03b1. (28.19) Next we use some physical reasoning: d\u03b1/d ln \u00b5 is the rate at which \u03b1 must change to compensate for a small change in ln \u00b5. If compensation is possible at all, this rate should be \ufb01nite in the \u03b5 0 limit. Therefore, in a renormalizable theory, we should have \u2192 d\u03b1 d ln \u00b5 = \u2212 \u03b5\u03b1 + \u03b2(\u03b1). (28.20) \u03b5\u03b1, is \ufb01xed by matching the O(\u03b5) terms in eq. (28.19). The The \ufb01rst term, second term, the beta function \u03b2(\u03b1), is similarly determined by matching the O(\u03b50) terms; the result is \u2212 \u03b2(\u03b1) = \u03b12G\u20321(\u03b1). (28.21) Terms that are higher-order in 1/\u03b5 must also cancel, and this determines all the other G\u2032n(\u03b1)\u2019s in terms of G\u20321(\u03b1). Thus, for example, cancellation 1) terms \ufb01xes G\u20322(\u03b1) = \u03b1G\u20321(\u03b1)2. These relations among the of the O(\u03b5\u2212 G\u2032n(\u03b1)\u2019s can of course be checked order by order in perturbation theory. From eq. (28.21) and eq. (28.16), we \ufb01nd that the beta function is \u03b2(\u03b1) = \u2212 3 2 \u03b12 + O(\u03b13). (28.22) Hearteningly, this is the same result we found in section 27 by requiring the 2 observed scattering cross section obs to be independent of \u00b5. However, |T | simply as a matter of practical calculation, it is much easier to compute G1(\u03b1) than it is to compute 2 obs. |T | Next consider the invariance of m0. We begin by de\ufb01ning M (\u03b1, \u03b5) \u2261 = ln(Z 1/2 m Z \u2212 \u03d5 1/2 ) Mn(\u03b1) \u03b5n. \u221e Xn=1 (28.23) 28: The Renormalization Group 181 From eqs. (28.10) and (28.12) we have M1(\u03b1) = 1 2 b1(\u03b1) 5 1 2 a1(\u03b1) \u2212 12 \u03b1 + O(\u03b12). Then, from eq. (28.4), we have = \u2212 (28.24) ln m0 = M (\u03b1, \u03b5) + ln m. (28.25) Take the derivative with respect to ln \u00b5 and require m0 to be unchanged: 0 = d d ln \u00b5 ln m0 = = \u2202M (\u03b1, \u03b5) \u2202\u03b1 d\u03b1 d ln \u00b5 + 1 m dm d ln \u00b5. \u2202M (\u03b1, \u03b5) \u2202\u03b1 \u2212 \u03b5\u03b1 + \u03b2(\u03b1) + 1 m dm d ln \u00b5. (28.26) Rearranging, we \ufb01nd 1 m dm d ln \u00b5 \u221e \u03b2(\u03b1) = \u03b5\u03b1 \u2212 Xn=1 = \u03b1M \u20321(\u03b1) +..., M \u2032n(\u03b1) \u03b5n (28.27) where the ellipses stand for terms with powers of 1/\u03b5. In a renormalizable theory, dm/d ln \u00b5 should be \ufb01nite in the \u03b5 0 limit, and so these terms must actually all be zero. Therefore, the anomalous dimension of the mass, de\ufb01ned via \u2192, (28.28) is given by \u03b3m(\u03b1) \u2261 1 m dm d ln \u00b5 \u03b3m", "(\u03b1) = \u03b1M \u20321(\u03b1) = \u2212 5 12 \u03b1 + O(\u03b12). (28.29) Comfortingly, this is just what we found in section 27. Let us now consider the propagator in the MS renormalization scheme, \u02dc\u2206(k2) = i d6x eikx Z 0 T\u03d5(x)\u03d5(0) 0 i | | h. (28.30) The bare propagator, \u02dc\u22060(k2) = i d6x eikx Z 0 T\u03d50(x)\u03d50(0) 0 i | | h, (28.31) 28: The Renormalization Group 182 should be (by the now-familiar argument) independent of \u00b5. The bare and renormalized propagators are related by \u02dc\u22060(k2) = Z\u03d5 \u02dc\u2206(k2). (28.32) Taking the logarithm and di\ufb00erentiating with respect to ln \u00b5, we get 0 = d d ln \u00b5 ln \u02dc\u22060(k2) = = d ln Z\u03d5 d ln \u00b5 d ln Z\u03d5 d ln \u00b5 + + d d ln \u00b5 ln \u02dc\u2206(k2) 1 \u02dc\u2206(k2) \u2202 \u2202 ln \u00b5 + d\u03b1 d ln \u00b5 \u2202 \u2202\u03b1 + dm d ln \u00b5 \u2202 \u2202m \u02dc\u2206(k2). (28.33) We can write Then we have ln Z\u03d5 = a2(\u03b1) a1(\u03b1) \u03b5 + \u2212 \u03b52 1 2 a2 1(\u03b1) +.... (28.34) d ln Z\u03d5 d ln \u00b5 = = = \u2202 ln Z\u03d5 \u2202\u03b1 d\u03b1 d ln \u00b5 a\u20321(\u03b1) \u03b5 +... \u03b1a\u20321(\u03b1) +..., \u2212 \u2212 \u03b5\u03b1 + \u03b2(\u03b1) (28.35) where the ellipses in the last line stand for terms with powers of 1/\u03b5. Since \u02dc\u2206(k2) should vary smoothly with \u00b5 in the \u03b5 0 limit, these must all be zero. We then de\ufb01ne the anomalous dimension of the \ufb01eld \u2192 \u03b3\u03d5(\u03b1) 1 2 d ln Z\u03d5 d ln \u00b5. \u2261 From eq. (28.35) we \ufb01nd \u03b3\u03d5(\u03b1) = \u2212 = + 1 1 2 \u03b1a\u20321(\u03b1) 12 \u03b1 + O(\u03b12). (28.36) (28.37) Eq. (28.33) can now be written as \u2202 \u2202 ln \u00b5 + \u03b2(\u03b1) \u2202 \u2202\u03b1 + \u03b3m(\u03b1)m \u2202 \u2202m + 2\u03b3\u03d5(\u03b1) \u02dc\u2206(k2) = 0 (28.38) 28: The Renormalization Group 183 \u2192 in the \u03b5 0 limit. This is the Callan-Symanzik equation for the propagator. The Callan-Symanzik equation is most interesting in the massless limit, and for a theory with a zero of the beta function at a nonzero value of \u03b1. So, let us suppose that \u03b2(\u03b1 and ) = 0 for some \u03b1 \u2217 6 \u2217 m = 0, the Callan-Symanzik equation becomes = 0. Then, for \u03b1 = \u03b1 \u2217 The solution is \u2202 \u2202 ln \u00b5 + 2\u03b3\u03d5(\u03b1 \u2217 ) \u02dc\u2206(k2) = 0. \u02dc\u2206(k2) = C(\u03b1 ) \u2217 k2 \u00b52 k2! \u2212 \u03b3\u03d5(\u03b1\u2217), (28.39) (28.40) ) is an integration constant. where C(\u03b1 \u2217 has mass dimension dependence.) Thus the naive scaling law \u02dc\u2206(k2) \u02dc\u2206(k2) nomena, which is beyond the scope of this book. (We used the fact that \u02dc\u2206(k2) 2 to get the k2 dependence in addition to the \u00b5 2 is changed to \u03b3\u03d5(\u03b1\u2217)]. This has applications in the theory of critical phe- k\u2212 k\u2212 \u223c \u2212 \u223c 2[1 \u2212 Reference Notes The formal development of the renormalization group is explored in more detail in Brown, Collins, and Ramond I. 28.1) Consider \u03d54 theory, Problems = 1 2 Z\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5 1 2 Zmm2\u03d52 1 24 Z\u03bb", "\u03bb\u02dc\u00b5\u03b5\u03d54, \u2212 \u2212 \u2212 L \u03b5 dimensions. Compute the beta function to O(\u03bb2), the in d = 4 anomalous dimension of m to O(\u03bb), and the anomalous dimension of \u03d5 to O(\u03bb). \u2212 (28.41) 28.2) Repeat problem 28.1 for the theory of problem 9.3. 28.3) Consider the lagrangian density = L \u2212 1 1 1 2 Z\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 2 Z\u03c7\u2202\u00b5\u03c7\u2202\u00b5\u03c7 \u2212 6 Zgg \u02dc\u00b5\u03b5/2\u03d53 + 1 2 Zmm2\u03d52 + Y \u03d5 2 ZM M 2\u03c72 2 Zhh\u02dc\u00b5\u03b5/2\u03d5\u03c72 1 \u2212 + 1 (28.42) in d = 6 \u2212 is adjusted to make \u03c7?) 0 \u03d5(x) 0 i | | h \u03b5 dimensions, where \u03d5 and \u03c7 are real scalar \ufb01elds, and Y = 0. (Why is no such term needed for 28: The Renormalization Group 184 a) Compute the one-loop contributions to each of the Z\u2019s in the MS renormalization scheme. b) The bare couplings are related to the renormalized ones via 3/2 g0 = Z \u2212 \u03d5 1 \u03d5 Z \u2212 h0 = Z \u2212 \u03c7 Zgg \u02dc\u00b5\u03b5/2, 1/2 Zhh\u02dc\u00b5\u03b5/2. De\ufb01ne G(g, h, \u03b5) = H(g, h, \u03b5) = \u221en=1 Gn(g, h)\u03b5\u2212 \u221en=1 Hn(g, h)\u03b5\u2212 P n n \u2261 \u2261 3/2 ln(Z \u2212 \u03d5 1 \u03d5 Z \u2212 \u03c7 ln(Z \u2212 Zg), 1/2 Zh). (28.43) (28.44) (28.45) (28.46) By requiring g0 and h0 to be independent of \u00b5, and by assuming that dg/d\u00b5 and dh/d\u00b5 are \ufb01nite as \u03b5 0, show that P \u2192 \u00b5 \u00b5 dg d\u00b5 dh d\u00b5 = = \u2212 \u2212 1 2 \u03b5g + 1 2 g 2 \u03b5h + 1 2 h 1 g g \u2202G1 \u2202g \u2202H1 \u2202g + h + h \u2202G1 \u2202h \u2202H1 \u2202h,. (28.47) (28.48) \u2192 \u2261 \u2261 lim\u03b5 lim\u03b5 0 \u00b5 dg/d\u00b5 and \u03b2h(g, h) c) Use your results from part (a) to compute the beta functions \u03b2g(g, h) 0 \u00b5 dh/d\u00b5. You should \ufb01nd terms of order g3, gh2, and h3 in \u03b2g, and terms of order g2h, gh2, and h3 in \u03b2h. d) Without loss of generality, we can choose g to be positive; h can then be positive or negative, and the di\ufb00erence is physically signi\ufb01cant. (You should understand why this is true.) For what numerical range(s) of h/g are \u03b2g and \u03b2h/h both negative? Why is this an interesting question? \u2192 29: E\ufb00ective Field Theory 185 29 Effective Field Theory Prerequisite: 28 So far we have been discussing only renormalizable theories. In this section, we investigate what meaning can be assigned to nonrenormalizable theories, following an approach pioneered by Ken Wilson. We will begin by analyzing a renormalizable theory from a new point of view. Consider, as an example, \u03d54 theory in four spacetime dimensions: = L \u2212 1 2 Z\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 1 2 Zmm2 ph\u03d52 1 24 Z\u03bb\u03bbph\u03d54. \u2212 (29.1) (This example is actually problematic, because this theory is trivial, a technical term that we will exlain later. For now we proceed with a perturbative analysis.) We take the renormalizing Z factors to be de\ufb01ned in an on-shell scheme, and have emphasized this by writing the particle mass as mph and the coupling constant as \u03bbph. We de\ufb01ne \u03bbph as the value of the exact 1PI four-point vertex with zero external four-momenta: The path integral is given by \u03bbph \u2261 V4(0, 0, 0, 0). Z(J) = D Z J\u03d5, \u03d5 eiS+i R d", "4x J\u03d5. where S = d4x and J\u03d5 is short for L Our \ufb01rst step in analyzing this theory will be to perform the Wick rotation (applied to loop integrals in section 14) directly on the action. We it. Then we have de\ufb01ne a euclidean time \u03c4 R R R \u2261 Z(J) = Z D LE, d4x = d3x d\u03c4, 2 Z\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5 + 1 \u03d5 e\u2212 SE\u2212 J\u03d5, R 2 Zmm2 ph\u03d52 + 1 24 Z\u03bb\u03bbph\u03d54, where SE = d4x R and LE = 1 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 = (\u2202\u03d5/\u2202\u03c4 )2 + ( \u2207 \u03d5)2. Note that each term in SE is always positive (or zero) for any \ufb01eld con\ufb01guration \u03d5(x). This is the advantage of working in euclidean space: eq. (29.4), the euclidean path integral, is strongly damped (rather than rapidly oscillating) at large values of the \ufb01eld and/or its derivatives, and this makes its convergence properties more obvious. (29.2) (29.3) (29.4) (29.5) (29.6) 186 (29.7) (29.8) 29: E\ufb00ective Field Theory Next, we Fourier transform to (euclidean) momentum space via \u03d5(x) = Z d4k (2\u03c0)4 eikx \u03d5(k). The euclidean action becomes e SE = 1 2 \u03d5( \u2212 d4k (2\u03c0)4 1 24 e Z\u03bb\u03bbph Z + Z d4k1 (2\u03c0)4... \u03d5(k1) \u00d7 0. e \u03c4 \u2265 k) Z\u03d5k2 + Zmm2 ph \u03d5(k) d4k4 (2\u03c0)4 (2\u03c0)4\u03b44(k1+k2+k3+k4) e \u03d5(k2) \u03d5(k3) \u03d5(k4). Note that k2 = k2 + k2 e e e We now introduce an ultraviolet cuto\ufb00 \u039b. It should be much larger than the particle mass mph, or any other energy scale of practical interest. Then we perform the path integral over all > \u039b. We also take J(k) = 0 for > \u039b. Then we \ufb01nd \u03d5(k) with k | | k | | Z(J\u039b e\u2212 Se\ufb00 (\u03d5;\u039b) \u2212 J\u03d5, R (29.9) e where e\u2212 Se\ufb00 (\u03d5;\u039b) = >\u039b e\u2212 SE(\u03d5). \u03d5 | k D Se\ufb00 (\u03d5; \u039b) is called the Wilsonian e\ufb00ective action. We can write the corresponding lagrangian density as Z | (29.10) 2 Z(\u039b)\u2202\u00b5\u03d5\u2202\u00b5\u03d5 + 1 Le\ufb00(\u03d5; \u039b) = 1 + cd,i(\u039b) 2 m2(\u039b)\u03d52 + 1 Od,i, 24 \u03bb(\u039b)\u03d54 Xd \u2265 where the Fourier components of \u03d5(x) are now cut o\ufb00 at 6 Xi > \u039b: k | | (29.11) (29.12) \u03d5(x) = \u039b d4k (2\u03c0)4 eikx 0 Z \u03d5(k). \u2265 6 and that are even under \u03d5 e The operators Od,i in eq. (29.11) consist of all terms that have mass dimension d \u03d5; i is an index that distinguishes operators of the same dimension that are inequivalent after integrations by parts of any derivatives that act on the \ufb01elds. (The operators must be even \u03d5 symmetry of the original under \u03d5 lagrangian.) \u03d5 in order to respect the \u03d5 \u2194 \u2212 \u2194 \u2212 \u2194 \u2212 The coe\ufb03cients Z(\u039b), m2(\u039b), \u03bb(\u039b), and cd,i(\u039b) in eq. (29.11) are all \ufb01nite functions of \u039b. This is established by the following argument. We can 29: E\ufb00ective Field Theory 187... Figure 29.1: A one-loop 1PI diagram with 2n external lines. Each external line represents a \ufffd", "\ufffd\ufffdeld with < \u039b. The internal (dashed) line represents a \ufb01eld with > \u039b. k | | k | | di\ufb00erentiate eq. (29.9) with respect to J(x) to compute correlation functions of the renormalized \ufb01eld \u03d5(x), and correlation functions of renormalized \ufb01elds are \ufb01nite. Using eq. (29.9), we can compute these correlation functions Le\ufb00. These as a series of Feynman diagrams, with Feynman rules based on rules include an ultraviolet cuto\ufb00 \u039b on the loop momenta, since the \ufb01elds with higher momenta have already been integrated out. Thus all of the loop integrals in these diagrams are \ufb01nite. Therefore the other parameters that enter the diagrams\u2014Z(\u039b), m2(\u039b), \u03bb(\u039b), and cd,i(\u039b)\u2014must be \ufb01nite as well, in order to end up with \ufb01nite correlation functions and those with < \u039b \ufb01elds), n vertices, and a 1, the dominant contribution to cd,i(\u039b) for an operator To compute these parameters, we can think of eq. (29.8) as the action > \u039b. Then | < \u039b \ufb01elds only. For Od,i with 2n 2n derivatives is then given by a one-loop diagram with 2n > \u039b \ufb01eld k | for two kinds of \ufb01elds, those with we draw all 1PI diagrams with external lines for \u03bbph \u226a \ufb01elds and d \u2212 external lines (representing circulating in the loop; see \ufb01g. (29.1). The simplest case to consider is \u03d52n. With 2n external lines, O2n,1 \u2261 there are (2n)! ways of assigning the external momenta to the lines, but 2 of these give the same diagram: 2n for exchanging the two external 2n lines that meet at any one vertex; n for rotations of the diagram; and 2 for re\ufb02ection of the diagram. Since there are no derivatives on the external \ufb01elds, we can set all of the external momenta to zero; then all (2n)!/(2n2n) diagrams have the same value. With a euclidean action, each internal line contributes a factor of 1/(k2 + m2 ph), and each vertex contributes a factor of ph). The vertex factor associated with the term \u03bbph + O(\u03bb2 Z\u03bb\u03bbph = \u00d7 \u00d7 n | \u2212 \u2212 29: E\ufb00ective Field Theory 188 c2n,1(\u039b)\u03d52n in Le\ufb00 is \u2212 (2n)! c2n,1(\u039b). Thus we have (2n)! c2n,1(\u039b) = \u2212 ( \u2212 \u03bbph)n(2n)! 2n2n + O(\u03bbn+1 Z ph ). d4k (2\u03c0)4 \u221e \u039b 1 k2 + m2 ph! n For 2n \u2265 6, the integral converges, and we \ufb01nd c2n,1(\u039b) = \u2212 ( \u2212 32\u03c02n(n \u03bbph/2)n 2) \u2212 1 \u039b2n \u2212 4 + O(\u03bbn+1 ph ). (29.13) (29.14) We have taken \u039b mph, and dropped terms down by powers of mph/\u039b. For 2n = 4, we have to include the tree-level vertex; in this case, we \u226b have \u03bb(\u039b) = \u2212 Z\u03bb\u03bbph + \u2212 3 2 ( \u2212 \u03bbph)2 + O(\u03bb3 ph). \u221e d4k (2\u03c0)4 1 k2 + m2 ph! 2 \u039b Z (29.15) This integral diverges. To evaluate it, we note that the one-loop contribution to the exact four-point vertex is given by the same diagram, but with \ufb01elds of all momenta circulating in the loop. Thus we have V4(0, 0, 0, 0) = \u2212 Z\u03bb\u03bbph + \u2212 3 2 ( \u2212 \u03bbph)2 + O(\u03bb3 ph). \u221e d4k (2\u03c0)4 0 Z 2 1 k2 + m2 ph! (29.16) Then, using V4(0, 0, 0, 0) = \u03bbph and subtracting eq. (29.15) from eq. (29.16), we get", " \u03bbph + \u03bb(\u039b) = \u2212 3 2 ( \u2212 \u03bbph)2 d4k \u039b (2\u03c0)4 0 Z 1 k2 + m2 ph! 2 + O(\u03bb3 ph). (29.17) Evaluating the (now \ufb01nite!) integral and rearranging, we have \u03bb(\u039b) = \u03bbph + 3 16\u03c02 \u03bb2 ph h ln(\u039b/mph) \u2212 1 2 i + O(\u03bb3 ph). (29.18) Note that this result has the problem of a large log; the second term is 1. To cure this problem, we smaller than the \ufb01rst only if \u03bbph ln(\u039b/mph) must change the renormalization scheme. We will take up this issue shortly, but \ufb01rst let us examine the case of two external lines while continuing to use the on-shell scheme. \u226a 29: E\ufb00ective Field Theory 189 For the case of two external lines, the one-loop diagram has just one vertex, and by momentum conservation, the loop integral is completely indepedent of the external momentum. This implies that the one-loop contribution to Z(\u039b) vanishes, and so we have Z(\u039b) = 1 + O(\u03bb2 ph). (29.19) The one-loop diagram does, however, give a nonzero contribution to m2(\u039b); after including the tree-level term, we \ufb01nd m2(\u039b) = \u2212 Zmm2 ph + \u2212 1 2 ( \u2212 \u221e d4k (2\u03c0)4 \u03bbph) Z \u039b 1 k2 + m2 ph + O(\u03bb2 ph). (29.20) This integral diverges. To evaluate it, recall that the one-loop contribution to the exact particle mass-squared is given by the same diagram, but with \ufb01elds of all momenta circulating in the loop. Thus we have m2 ph = \u2212 \u2212 Zmm2 ph + 1 2 ( \u2212 \u03bbph) Z 0 \u221e d4k (2\u03c0)4 1 k2 + m2 ph + O(\u03bb2 ph). (29.21) Then, subtracting eq. (29.20) from eq. (29.21), we get m2 ph + m2(\u039b) = \u2212 1 2 ( \u2212 \u03bbph) 0 Z \u039b d4k (2\u03c0)4 1 k2 + m2 ph + O(\u03bb2 ph). (29.22) Evaluating the (now \ufb01nite!) integral and rearranging, we have m2(\u039b) = m2 ph \u2212 \u03bbph 16\u03c02 \u039b2 h \u2212 m2 ph ln(\u039b2/m2 + O(\u03bb2 ph). (29.23) ph) i We see that we now have an even worse situation than we did with the large log in \u03bb(\u039b): the correction term is quadratically divergent. As already noted, to \ufb01x these problems we must change the renormalization scheme. In the context of an e\ufb00ective action with a speci\ufb01c value of the cuto\ufb00 \u039b0, there is a simple way to do so: we simply treat this e\ufb00ective action as the fundamental starting point, with Z(\u039b0), m2(\u039b0), \u03bb(\u039b0), and cd,i(\u039b0) as input parameters. We then see what physics emerges at energy scales well below \u039b0. We can set Z(\u039b0) = 1, with the understanding that the \ufb01eld no longer has the LSZ normalization (and that we will have to correct the LSZ formula to account for this). We will also assume that the parameters \u03bb(\u039b0), m2(\u039b0), and cd,i(\u039b0) are all small when measured in units of the cuto\ufb00: \u03bb(\u039b0) 1, \u226a m2(\u039b0) | | \u226a \u039b2 0, cd,i(\u039b0) \u226a 4) (d \u2212. \u039b\u2212 0 (29.24) (29.25) (29.26) 29: E\ufb00ective Field Theory 190 The proposal to treat the e\ufb00ective action as the fundamental starting point may not seem very appealing. For one thing, we now have an in\ufb01nite number of parameters to specify, rather than two! Also, we now have an explicit cuto\ufb00 in place, rather than trying to have a theory that works at all energy scales", ". On the other hand, it may well be that quantum \ufb01eld theory does not work at arbitrarily high energies. For example, quantum \ufb02uctuations in spacetime itself should become important above the Planck scale, which is given by the inverse square root of Newton\u2019s constant, and has a numerical 1019 GeV (compared to, say, the proton mass, which is 1 GeV). value of So, let us leave the cuto\ufb00 \u039b0 in place for now. We will then make a twopronged analysis. First, we will see what happens at much lower energies. Then, we will see what happens if we try to take the limit \u039b0 \u2192 \u221e we will set We begin by examining lower energies. To make things more tractable, \u223c \u223c. later we will examine the e\ufb00ects of a more general choice. cd,i(\u039b0) = 0 ; (29.27) A nice way to see what happens at lower energies is to integrate out some more high-energy degrees of freedom. Let us, then, perform the functional integral over Fourier modes \u03d5(k) with \u039b < < \u039b0; we have e\u2212 Se\ufb00 (\u03d5;\u039b) = e \u03d5\u039b< k | | D Z Se\ufb00 (\u03d5;\u039b0). (29.28) We can do this calculation in perturbation theory, mimicking the procedure that we used earlier. We \ufb01nd m2(\u039b) = m2(\u039b0) + \u03bb(\u039b) = \u03bb(\u039b0) 3 2 \u2212 c2n,1(\u039b) = 1)n ( \u2212 2n2n \u2212 1 2 \u03bb(\u039b0) Z \u039b0 \u039b d4k (2\u03c0)4 1 k2 + m2(\u039b0) \u03bb2(\u039b0) Z \u039b0 \u039b \u03bbn(\u039b0) Z \u039b \u039b0 d4k (2\u03c0)4 1 k2 + m2(\u039b0) d4k (2\u03c0)4 1 k2 + m2(\u039b0) n +..., (29.29) 2 +..., (29.30) +..., (29.31) where the ellipses stand for higher-order corrections. For \u039b not too much less than \u039b0 (and, in particular, for \u039b2), we \ufb01nd | k | <\u039b0 e\u2212 m2(\u039b) = m2(\u039b0) + +..., \u039b2 m2(\u039b0) | 1 16\u03c02 \u03bb(\u039b0) 3 16\u03c02 \u03bb2(\u039b0) ln 1)n ( \u2212 32\u03c022nn(n 0 \u2212 \u039b0 \u039b \u03bbn(\u039b0) 1 \u039b2n 2) \u2212 | \u226a \u039b2 +..., \u2212 \u03bb(\u039b) = \u03bb(\u039b0) c2n,1(\u039b) = \u2212 (29.32) (29.33) 1 \u039b2n 29.34) 29: E\ufb00ective Field Theory 191 Figure 29.2: A one-loop contribution to the \u03d54 vertex for \ufb01elds with The internal (dashed) line represents a \ufb01eld with > \u039b. k | | < \u039b. k | | We see from this that the corrections to m2(\u039b), which is the only coe\ufb03cient with positive mass dimension, are dominated by contributions from the high end of the integral. On the other hand, the corrections to cd,i(\u039b), coe\ufb03cients with negative mass dimension, are dominated by contributions from the low end of the integral. And the corrections to \u03bb(\u039b), which is dimensionless, come equally from all portions of the range of integration. For the cd,i(\u039b), this means that their starting values at \u039b0 were not very important, as long as eq. (29.26) is obeyed. Nonzero starting values would contribute another term of order 1/\u039b2n to c2n,1(\u039b), but all such 0 terms are less important than the one of order 1/\u039b2n 4 that comes from doing the integral down to = \u039b. \u2212 \u2212 4 Similarly, nonzero values of cd,i(\u039b0) would make subdominant contributions to \u03bb(\u039b). As an example, consider the contribution of the diagram in \ufb01g. (29.2). Ignoring numerical", " factors, the vertex factor is c6,1(\u039b0), and the loop integral is the same as the one that enters into m2(\u039b); it yields a factor of \u039b2 0. Thus the contribution of this diagram to \u03bb(\u039b) is \u223c of order c6,1(\u039b0)\u039b2 0. This is a pure number that, according to eq. (29.26), is small. This contribution is missing the logarithmic enhancement factor ln(\u039b0/\u039b) that we see in eq. (29.33). 0 \u2212 \u039b2 \u039b2 0 when cd,i(\u039b0) On the other hand, for m2(\u039b), there are in\ufb01nitely many contributions of order \u039b2 = 0. These must add up to give m2(\u039b) a value that is much smaller. Indeed, we want to continue the process down to lower and lower values of \u039b, with m2(\u039b) dropping until it becomes of order m2 ph mph. For this to happen, there must be very precise cancellations at \u039b 0 that contribute to m2(\u039b). In some sense, it among all the terms of order \u039b2 is more \u201cnatural\u201d to have m2 \u03bb(\u039b0)\u039b2 0, rather than to arrange for these very precise cancellations. This philosophical issue is called the \ufb01ne-tuning problem, and it generically arises in theories with spin-zero \ufb01elds. ph \u223c \u223c k | | In theories with higher-spin \ufb01elds only, the action typically has more 6 29: E\ufb00ective Field Theory 192 symmetry when these \ufb01elds are massless, and this typically prevents divergences that are worse than logarithmic. These theories are said to be technically natural, while theories with spin-zero \ufb01elds (with physical masses well below the cuto\ufb00) generally are not. (The only exceptions are theories where supersymmetry relates spin-zero and spin-one-half \ufb01elds; the spinzero \ufb01elds then inherent the technical naturalness of their spin-one-half partners.) For now, in \u03d54 theory, we will simply accept the necessity of \u039b. \ufb01ne-tuning in order to have mph \u226a Returning to eqs. (29.32\u201329.34), we can recast them as di\ufb00erential equations that tell us how these parameters change with the vaule of the cuto\ufb00 \u039b. In particular, let us do this for \u03bb(\u039b). We take the derivative of eq. (29.33) with respect to \u039b, multiply by \u039b, and then set \u039b0 = \u039b to get d d ln \u039b \u03bb(\u039b) = 3 16\u03c02 \u03bb2(\u039b) +.... (29.35) \u2261 Notice that the right-hand side of eq. (29.33) is apparently the same as the d\u03bb/d ln \u00b5 that we calculated in problem 28.1, where beta function \u03b2(\u03bb) it represented the rate of change in the MS parameter \u03bb that was need to compensate for a change in the MS renormalization scale \u00b5. Eq. (29.33) gives us a new physical interpretation of the beta function: it is the rate of change in the coe\ufb03cient of the \u03d54 term in the e\ufb00ective action as we vary the ultraviolet cuto\ufb00 in that action. Actually, though, there is a technical detail: 2\u03bb(\u039b) that is most closely analogous to the MS parameter \u03bb. This is because, if we rescale \u03d5 so that it has a canonical kinetic term of 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5, then the coe\ufb03cient of the \u03d54 term is Z(\u039b)\u2212 2\u03bb(\u039b). Since Z(\u039b) = 1 + O(\u03bb2(\u039b)), this has no e\ufb00ect at the one-loop level, but it does matter at two loops. We can account for the e\ufb00ect of this wave function renormalization (in all the couplings) by writing, instead of eq. (29.11), it is really Z(\u039b)\u2212 2 Z(\u039b)\u2202\u00b5\u03d5\u2202\u00b5\u03d5 + 1 Le\ufb00(\u03d5; \u039b) = 1 + Z nd,i/2(\u039b)cd,i(\u039b) 2 Z(\u039b)m2(\ufffd", "\ufffd)\u03d52 + 1 Od,i, 24 Z 2(\u039b)\u03bb(\u039b)\u03d54 Xd \u2265 6 Xi Od,i. Now the beta where nd,i is the number of \ufb01elds in the operator function for \u03bb is universal up through two loops; see problem 29.1. At three and higher loops, di\ufb00erences with the MS beta function can arise, due to the di\ufb00erent underlying de\ufb01nitions of the coupling \u03bb in the cuto\ufb00 scheme and the MS scheme. We now have the overall picture of Wilson\u2019s approach to quantum \ufb01eld theory. First, de\ufb01ne a quantum \ufb01eld theory via an action with an explicit (29.36) 29: E\ufb00ective Field Theory 193 momentum cuto\ufb00 in place.1 Then, lower the cuto\ufb00 by integrating out higher-momentum degrees of freedom. As a result, the coe\ufb03cients in the e\ufb00ective action will change. If the \ufb01eld theory is weakly coupled\u2014which in practice means eqs. (29.24\u201329.26) are obeyed\u2014then the coe\ufb03cients of the operators with negative mass dimension will start to take on the values we would have computed for them in perturbation theory, regardless of their precise initial values. If we continuously rescale the \ufb01elds to have canonical kinetic terms, then the dimensionless coupling constant(s) will change according to their beta functions. The \ufb01nal results, at an energy scale E well below the initial cuto\ufb00 \u039b0, are the same as we would predict via renormalized perturbation theory, up to small corrections by powers of E/\u039b0. The advantage of the Wilson scheme is that it gives a nonperturbative de\ufb01nition of the theory which is applicable even if the theory is not weakly coupled. With a spacetime lattice providing the cuto\ufb00, other techniques (typically requiring large-scale computer calculations) can be brought to bear on strongly-coupled theories. The Wilson scheme also allows us to give physical meaning to nonrenormalizable theories. Given an action for a nonrenormalizable theory, we can regard it as an e\ufb00ective action. We should then impose a momentum cuto\ufb00 \u039b0, where \u039b0 can be de\ufb01ned by saying that the coe\ufb03cient of every operator Oi with mass dimension Di > 4 is given by ci/\u039bDi\u2212 1. Then we can use this theory for physics at energies below \u039b0. At energies E far below \u039b0, the e\ufb00ective theory will look like a renormalizable one, up to corrections by powers of E/\u039b0. (This renormalizable theory might simply be a free-\ufb01eld theory with no interactions, or no theory at all if there are no particles with physical masses well below \u039b0.) with ci \u2264 0 4 We now turn to the \ufb01nal issue: can we remove the cuto\ufb00 completely? Returning to the example of \u03d54 theory, let us suppose that we are somehow able to compute the exact beta function. Then we can integrate the renormalization-group equation d\u03bb/d ln \u039b = \u03b2(\u03bb) from \u039b = mph to \u039b = \u039b0 to get \u03bb(\u039b0) \u03bb(mph) Z d\u03bb \u03b2(\u03bb) = ln \u039b0 mph. (29.37). Obviously, the right-hand side of We would like to take the limit \u039b0 \u2192 \u221e eq. (29.37) becomes in\ufb01nite in this limit, and so the left-hand side must as well. However, it may not. Recall that, for small \u03bb, \u03b2(\u03bb) is positive, and it 1This can be done in various ways: for example, we could replace continuous spacetime with a discrete lattice of points with lattice spacing a; then there is an e\ufb00ective largest momentum of order 1/a. 29: E\ufb00ective Field Theory 194 increases faster than \u03bb. If this is true for all \u03bb, then the left-hand side of eq. (29.37) will approach a \ufb01xed, \ufb01nite value as we take the upper limit of integration to in\ufb01nity. This yields a maximum possible value for the initial cuto\ufb00, given by ln \u039bmax mph \u2261 \u221e \u03bb(mph) d\u03bb \u03b2(\u03bb). (", "29.38) Z If we approximate the exact beta function with its leading term, 3\u03bb2/16\u03c02, and use the leading term in eq. (29.18) to get \u03bb(mph) = \u03bbph, then we \ufb01nd \u039bmax \u2243 mph e16\u03c02/3\u03bbph. (29.39) The existence of a maximum possible value for the cuto\ufb00 means that we cannot take the limit as the cuto\ufb00 goes to in\ufb01nity; we must use an e\ufb00ective action with a cuto\ufb00 as our starting point. If we insist on taking the cuto\ufb00 to in\ufb01nity, then the only possible value of \u03bbph is \u03bbph = 0. Thus, \u03d54 theory is trivial in the limit of in\ufb01nite cuto\ufb00: there are no interactions. (There is much evidence for this, but as yet no rigorous proof. The same is true of quantum electrodynamics, as was \ufb01rst conjectured by Landau; in this case, \u039bmax is known as the location of the Landau pole.) However, the cuto\ufb00 can be removed if the beta function grows no faster than \u03bb at large \u03bb; then the left-hand side of eq. (29.37) would diverge as we take the upper limit of integration to in\ufb01nity. Or, \u03b2(\u03bb) could drop to zero (and then become negative) at some \ufb01nite value \u03bb, the left-hand side of eq. (29.37) would diverge as the upper limit of integration In this case, the e\ufb00ective coupling at higher and higher approaches \u03bb energies would remain \ufb01xed at \u03bb is called an ultravioldet \ufb01xed point of the renormalization group.. Then, if \u03bbph < \u03bb \u2217, and \u03bb = \u03bb \u2217 \u2217 \u2217 \u2217. If the beta function is negative for \u03bb = \u03bb(mph), the theory is said to In be asymptotically free, and \u03bb(\u039b) decreases as the cuto\ufb00 is increased. this case, there is no barrier to taking the limit \u039b. In four spacetime dimensions, the only asymptotically free theories are nonabelian gauge theories; see section 69. \u2192 \u221e Reference Notes E\ufb00ective \ufb01eld theory is discussed in Georgi, Peskin & Schroeder, and Weinberg I. An introduction to lattice theory can be found in Smit. Problems 29.1) Consider a theory with a single dimensionless coupling g whose beta function takes the form \u03b2(g) = b1g2 + b2g3 +.... Now consider a new de\ufb01nition of the coupling of the form \u02dcg = g + c2g2 +.... 29: E\ufb00ective Field Theory 195 a) Show that \u03b2(\u02dcg) = b1\u02dcg2 + b2\u02dcg3 +.... b) Generalize this result to the case of multiple dimensionless couplings. 29.2) Consider \u03d53 theory in six euclidean spacetime dimensions, with a cuto\ufb00 \u039b0 and lagrangian = 1 2 Z(\u039b0)\u2202\u00b5\u03d5\u2202\u00b5\u03d5 + 1 6 Z 3/2(\u039b0)g(\u039b0)\u03d53. (29.40) L We assume that we have \ufb01ne-tuned to keep m2(\u039b) neglect the mass term. \u226a \u039b2, and so we a) Show that Z(\u039b) = Z(\u039b0) 1 1 2 \u2212 g2(\u039b0) \u039b0 d dk2 \"Z \u039b d6\u2113 (2\u03c0)6 1 (k+\u2113)2\u21132 #k2=0 +...,! g(\u039b) = Z 3/2(\u039b0) Z 3/2(\u039b) g(\u039b0) \u039b0 1 + g2(\u039b0) Z \u039b d6\u2113 (2\u03c0)6 1 (\u21132)3 +....! Hint: note that the tree-level propagator is \u02dc\u2206(k) = [Z(\u039b0)k2]\u2212 b) Use your results to compute the beta function 1. \u03b2(g(\u039b)) d d ln \u039b \u2261 g(\u039b), (29.41) and", " compare with the result in section 27. 30: Spontaneous Symmetry Breaking 196 30 Spontaneous Symmetry Breaking Prerequisite: 21 Consider \u03d54 theory, where \u03d5 is a real scalar \ufb01eld with lagrangian = L \u2212 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 1 2 m2\u03d52 \u2212 \u2212 1 24 \u03bb\u03d54. (30.1) As we discussed in section 23, this theory has a Z2 symmetry: ant under \u03d5(x) implements this: is invari\u03d5(x), and we can de\ufb01ne a unitary operator Z that \u2192 \u2212 L Z \u2212 1\u03d5(x)Z = \u03d5(x). \u2212 We also have Z 2 = 1, and so Z \u2212 this makes Z hermitian as well as unitary. 1 = Z. Since unitarity implies Z \u2212 (30.2) 1 = Z \u2020, Now suppose that the parameter m2 is, in spite of its name, negative rather than positive. We can write in the form = L \u2212 where the potential is L 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 1 V (\u03d5), \u2212 V (\u03d5) = 1 = 1 2 m2\u03d52 + 1 24 \u03bb(\u03d52 24 \u03bb\u03d54 v2)2 \u2212 \u2212 1 24 \u03bbv4. In the second line, we have de\ufb01ned \u2261 We can (and will) drop the last, constant, term in eq. (30.4). v m2 +(6 | /\u03bb)1/2. | From eq. (30.4) it is clear that there are two classical \ufb01eld con\ufb01gurations that minimize the energy: \u03d5(x) = +v and \u03d5(x) = v. This is in contrast to the usual case of positive m2, for which the minimum-energy classical \ufb01eld con\ufb01guration is \u03d5(x) = 0. \u2212 We can expect that the quantum theory will follow suit. For m2 < 0, and, with the property that there will be two ground states, 0 \u2212i | = +v, 0+ i | 0+ \u03d5(x) | | 0 \u03d5(x) \u2212i | i 0+ h 0 \u2212| h = v, \u2212 (30.6) up to quantum corrections from loop diagrams that we will treat in detail in section 30. These two ground states are exchanged by the operator Z, and they are orthogonal: 0+ h 0 \u2212i | Z 0+ | = i = 0. 0 \u2212i |, (30.7) (30.3) (30.4) (30.5) 30: Spontaneous Symmetry Breaking 197 This last claim requires some comment. Consider a similar problem in quantum mechanics, H = 1 2 p2 + 1 24 \u03bb(x2 v2)2. \u2212 (30.8) There are two approximate ground states in this case, speci\ufb01ed by the approximate wave functions (x) = \u03c8 \u00b1 0 x \u00b1i \u223c | h exp[ \u2212 \u03c9(x \u2213 v)2/2], (30.9) where \u03c9 = (\u03bbv2/3)1/2 is the frequency of small oscillations about the minimum. However, the true ground state is a symmetric linear combination of these. The antisymmetric linear combination has a slightly higher energy, due to the e\ufb00ects of quantum tunneling. \u2207 We can regard a \ufb01eld theory as an in\ufb01nite set of oscillators, one for each point in space, each with a hamiltonian like eq. (30.8), and coupled together \u03d5)2 term in the \ufb01eld-theory hamiltonian. There is a tunneling by the ( into amplitude for each oscillator, but to turn the \ufb01eld-theoretic state, all the oscillators have to tunnel, and so the tunneling amplitude gets 0 | \u2212i raised to the power of the number of oscillators, that is, to the power of in\ufb01nity (more precisely, to a power that scales like the volume of space). Therefore, in the limit of in\ufb01nite volume, 0+ | i 0+ h to use as the ground state. Let vanishes. 0 \u2212i | Thus we can pick either i. Then we can de\ufb01ne a shifted \ufb01eld, i 0+ | or 0 \u2212i | us choose 0+ | \u03c1(x) = \u03d5(x) v, \u2212 (30.10) which obeys which we will do at the", " end of this section.) The potential becomes = 0. (We must still worry about loop corrections, 0+ \u03c1(x) | | 0+ h i V (\u03d5) = 1 24 \u03bb[(\u03c1 + v)2 6 \u03bbv2\u03c12 + 1 v2]2 \u2212 6 \u03bbv\u03c13 + 1 = 1 24 \u03bb\u03c14, (30.11) (30.12) and so the lagrangian is now = L \u2212 1 2 \u2202\u00b5\u03c1\u2202\u00b5\u03c1 \u2212 1 6 \u03bbv2\u03c12 \u2212 1 6 \u03bbv\u03c13 1 24 \u03bb\u03c14. \u2212 6 \u03bbv2 = We see that the coe\ufb03cient of the \u03c12 term is 1. This coe\ufb03cient | should be identi\ufb01ed as 1 \u03c1, where m\u03c1 is the mass of the corresponding \u03c1 particle. Also, we see that the shifted \ufb01eld now has a cubic as well as a quartic interaction. m2 | 2 m2 Eq. (30.12) speci\ufb01es a perfectly sensible, renormalizable quantum \ufb01eld theory, but it no longer has an obvious Z2 symmetry. We say that the Z2 symmetry is spontaneously broken. 30: Spontaneous Symmetry Breaking 198 This leads to a question about renormalization. If we include renormal- izing Z factors in the original lagrangian, we get = L \u2212 1 2 Z\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 1 2 Zmm2\u03d52 \u2212 1 24 Z\u03bb\u03bb\u03d54. (30.13) \u2264 For positive m2, these three Z factors are su\ufb03cient to absorb in\ufb01nities for d 4, where the mass dimension of \u03bb is positive or zero. On the other hand, looking at the lagrangian for negative m2 after the shift, eq. (30.12), we would seem to need an extra Z factor for the \u03c13 term. Also, once we have a \u03c13 term, we would expect to need to add a \u03c1 term to cancel tadpoles. So, the question is, are the original three Z factors su\ufb03cient to absorb all the divergences in the Feynman diagrams derived from eq. (30.13)? The answer is yes. To see why, consider the quantum action (introduced in section 21) \u0393(\u03d5) = 1 2 Z + d4k (2\u03c0)4 1 n! Z \u221e Xn=3 \u03d5(k) \u03a0(k2) d4kn (2\u03c0)4 (2\u03c0)4\u03b44(k1+... +kn) e \u03d5( k) k2 + m2 \u2212 \u2212 d4k1 e (2\u03c0)4... Vn(k1,..., kn) \u00d7 \u03d5(k1)... \u03d5(kn), (30.14) e e computed with m2 > 0. The ingredients of \u0393(\u03d5)\u2014the self-energy \u03a0(k2) and the exact 1PI vertices Vn\u2014are all made \ufb01nite and well-de\ufb01ned (in, say, the MS renormalization scheme) by adjusting the three Z factors in eq. (30.13). Furthermore, for m2 > 0, the quantum action inherits the Z2 symmetry of the classical action. To see this directly, we note that Vn must zero for odd n, simply because there is no way to draw a 1PI diagram with an odd number of external lines using only a four-point vertex. Thus \u0393(\u03d5) also has the Z2 symmetry. This is a simple example of a more general result that we proved in problem 21.2: the quantum action inherits any linear symmetry of the classical action, provided that it is also a symmetry \u03d5. (Linear means that the transformed \ufb01elds of the integration measure are linear functions of the original ones.) The integration measure is almost always invariant; when it is not, the symmetry is said to be anomalous. We will meet an anomalous symmetry in section 75. D Once we have computed the quantum action for m2 > 0, we can go ahead and consider the case of m2 < 0. Recall from section 21 that the quantum equation of motion in the presence of a source is \u03b4\u0393/\u03b4\u03d5(x) = J(x), and that the solution of this equation is also the vacuum expectation \u2212 value of \u03d5(x). Now set J(x) = 0, and look for a translationally invariant (that is, constant) solution \ufffd", "\ufffd(x) = v. If there is more than one such solution, we want the one(s) with the lowest energy. This is equivalent to 30: Spontaneous Symmetry Breaking 199 minimizing the quantum potential (\u03d5), where U \u0393(\u03d5) = d4x Z h (\u03d5) 1 2 Z \u2212 \u2212 U (\u03d5)\u2202\u00b5\u03d5\u2202\u00b5\u03d5 +..., (30.15) i and the ellipses stand for terms with more derivatives. In a weakly coupled (\u03d5) to be qualitatively theory, we can expect the loop-corrected potential similar to the classical potential V (\u03d5). Therefore, for m2 < 0, we expect that there are two minima of v, where v = \u00b1 is the exact vacuum expectation value of the \ufb01eld. (\u03d5) with equal energy, located at \u03d5(x) = U U Thus we have a description of spontaneous symmetry breaking in the quantum theory based on the quantum action, and the quantum action is made \ufb01nite by adjusting only the three Z factors that appear in the original, symmetric form of the lagrangian. 0 \u03d5(x) 0 i | | h In the next section, we will see how this works in explicit calculations. 31: Broken Symmetry and Loop Corrections 200 31 Broken Symmetry and Loop Corrections Prerequisite: 30 Consider \u03d54 theory, where \u03d5 is a real scalar \ufb01eld with lagrangian = L \u2212 1 2 Z\u03d5\u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 1 2 Zmm2\u03d52 \u2212 1 24 Z\u03bb\u03bb\u03d54. (31.1) In d = 4 spacetime dimensions, the coupling \u03bb is dimensionless. We begin by considering the case m2 > 0, where the Z2 symmetry of L \u03d5 is manifest. We wish to compute the three renormalizing under \u03d5 \u03bb\u02dc\u00b5\u03b5 (where \u02dc\u00b5 Z factors. We work in d = 4 \u03b5 dimensions, and take \u03bb has dimensions of mass) so that \u03bb remains dimensionless. \u2192 \u2212 \u2192 \u2212 The propagator correction \u03a0(k2) is given by the diagrams of \ufb01g. (31.1), which yield i\u03bb\u02dc\u00b5\u03b5) 1 i i\u03a0(k2) = 1 2 ( 1 and B = Zm \u2212 \u2212 1, and where A = Z\u03d5 \u2212 \u02dc\u2206(0) \u2212 i(Ak2 + Bm2), (31.2) \u02dc\u2206(0) = Z dd\u2113 (2\u03c0)d 1 \u21132 + m2. Using the usual bag of tricks from section 14, we \ufb01nd \u02dc\u00b5\u03b5 \u02dc\u2206(0) = \u2212 i (4\u03c0)2 2 \u03b5 + 1 + ln(\u00b52/m2) m2, where \u00b52 = 4\u03c0e\u2212 \u03b3 \u02dc\u00b52. Thus \u03a0(k2) = \u03bb 2(4\u03c0)2 2 \u03b5 + 1 + ln(\u00b52/m2) m2 \u2212 Ak2 \u2212 Bm2. From eq. (31.5) we see that we must have A = O(\u03bb2), B = \u03bb 16\u03c02 1 \u03b5 + \u03baB + O(\u03bb2), (31.3) (31.4) (31.5) (31.6) (31.7) where \u03baB is a \ufb01nite constant (that may depend on \u00b5). In the MS renormalization scheme, we take \u03baB = 0, but we will leave \u03baB arbitrary for now. Next we turn to the vertex correction, given by the diagram of \ufb01g. (31.2), k4; all momenta are treated as plus two others with k2 \u2194 incoming. We have k3 and k2 \u2194 iV4(k1, k2, k3, k4) = iZ\u03bb\u03bb + 1 2 ( \u2212 + O(\u03bb3). \u2212 i\u03bb)2 2 1 i h iF ( s) + iF ( \u2212 \u2212 t) + iF ( u) \u2212 i (31.8) 31: Broken Symmetry and Loop Corrections 201 l k k k k Figure 31.1: O(\u03bb) corrections to \u03a0(k2). k4 1k k2 Figure 31.2: The O(\u03bb2) correction to V4(k1, k2, k3, k4). Two other diagrams, obtained from this one via k2 \u2194 k3 and k2 \u2194 k4, also contribute. Here we have de", "\ufb01ned s = and \u2212 (k1 + k2)2, t = (k1 + k3)2, u = (k1 + k4)2, \u2212 \u2212 iF (k2) \u02dc\u00b5\u03b5 Z i 16\u03c02 \u2261 = dd\u2113 (2\u03c0)d 1 ((\u2113+k)2 + m2)(\u21132 + m2) 1 2 \u03b5 + 0 Z dx ln(\u00b52/D), (31.9) x)k2 + m2. Setting Z\u03bb = 1 + C in eq. (31.8), we see that where D = x(1 we need \u2212 C = 3\u03bb 16\u03c02 1 \u03b5 + \u03baC + O(\u03bb2), (31.10) where \u03baC is a \ufb01nite constant. We may as well pause to compute the beta function, \u03b2(\u03bb) = d\u03bb/d ln \u00b5, where the derivative is taken with the bare coupling \u03bb0 held \ufb01xed, and the \ufb01nite parts of the counterterms set to zero, in accord with the MS prescription. We have \u03bb0 = Z\u03bbZ \u2212 2 \u03d5 \u03bb\u02dc\u00b5\u03b5, (31.11) with 2 ln 3\u03bb 16\u03c02 Let L1(\u03bb) be the coe\ufb03cient of 1/\u03b5 in eq. (31.12). Our analysis in section 28 shows that the beta function is then given by \u03b2(\u03bb) = \u03bb2L\u20321(\u03bb). Thus we \ufb01nd Z\u03bbZ \u2212 \u03d5 + O(\u03bb2). (31.12) 1 \u03b5 = \u03b2(\u03bb) = 3\u03bb2 16\u03c02 + O(\u03bb3). (31.13) 31: Broken Symmetry and Loop Corrections 202 k l k Figure 31.3: The O(\u03bb) correction to the vacuum expectation value of the \u03c1 \ufb01eld. The beta function is positive, which means that the theory becomes more and more strongly coupled at higher and higher energies. Now we consider the more interesting case of m2 < 0, which results in the spontaneous breakdown of the Z2 symmetry. Following the procedure of section 30, we set \u03d5(x) = \u03c1(x) + v, where /\u03bb)1/2 minimizes the potential (without Z factors). Then the | m2 v = (6 | lagrangian becomes (with Z factors) = L 1 1 1 \u2212 4 Z\u03bb\u2212 2 Z\u03d5\u2202\u00b5\u03c1\u2202\u00b5\u03c1 \u2212 + 1 2 (Zm\u2212 6 Z\u03bb(3\u03bb\u02dc\u00b5\u03b5)1/2m\u03c1\u03c13 4 Zm)m2 2 ( 3 Z\u03bb)(3/\u03bb\u02dc\u00b5\u03b5)1/2m3 \u03c1\u03c1 24 Z\u03bb\u03bb\u02dc\u00b5\u03b5\u03c14, \u03c1\u03c12 1 1 \u2212 \u2212 (31.14) where m2 m2 \u03c1 = 2 |. Now we can compute various one-loop corrections. | We begin with the vacuum expectation value of \u03c1. The O(\u03bb) correction is given by the diagrams of \ufb01g. (31.3). The three-point vertex factor is \u2212 iZ\u03bbg3, where g3 can be read o\ufb00 of eq. (31.14): g3 = (3\u03bb\u02dc\u00b5\u03b5)1/2m\u03c1. (31.15) The one-point vertex factor is iY, where Y can also be read o\ufb00 of eq. (31.14): Y = 1 2 (Zm\u2212 Z\u03bb)(3/\u03bb\u02dc\u00b5\u03b5)1/2m3 \u03c1. (31.16) Following the discussion of section 9, we then \ufb01nd that 0 \u03c1(x) 0 i | | h = iY + 1 2 ( \u2212 iZ\u03bbg3) 1 i \u02dc\u2206(0) d4y 1 i \u2206(x \u2212 y), (31.17) Z plus higher-order corrections. Using eqs. (31.15) and (31.16), and eq. (31.4) with m2 \u03c1, the factor in large parentheses in eq. (31.17) becomes m2 \u2192 i 2 (3/\u03bb)1/2m3 \u03c1 Z\u03bb + Zm\u2212 \u03bb 16\u03c02 2 \u03b5 + 1 + ln(\u00b52/m2 \u03c1) + O(\u03bb2). (31.18) Using Zm = 1 + B and Z\u03bb = 1 + C, with B and C from eqs. (31.7) and (31.10), the factor in large parentheses in eq.", " (31.18) becomes \u03bb 16\u03c02 \u03baB \u2212 h \u03baC + 1 + ln(\u00b52/m2 \u03c1) i. (31.19) 31: Broken Symmetry and Loop Corrections 203 All the 1/\u03b5\u2019s have canceled. The remaining \ufb01nite vacuum expectation value for \u03c1(x) can now be removed by choosing \u03baB \u2212 \u03baC = 1 \u2212 \u2212 ln(\u00b52/m2 \u03c1). (31.20) This will also cancel all diagrams with one-loop tadpoles. Next we consider the \u03c1 propagator. The diagrams contributing to the iX, O(\u03bb) correction are shown in \ufb01g. (31.4). The counterterm insertion is where, again reading o\ufb00 of eq. (31.14), \u2212 X = Ak2 + ( 3 4 C 1 4 B)m2 \u03c1. \u2212 (31.21) Putting together the results of eq. (31.2) for the \ufb01rst diagram (with m2 m2 \u2192 \u03c1), eq. (31.9) for the second (ditto), and eq. (31.21) for the third, we get \u03a0(k2) = = \u02dc\u2206(0) + 1 2 g2 3F (k2) X + O(\u03bb2) \u2212 i 1 \u03c1 2 \u03b5 2 (\u03bb\u02dc\u00b5\u03b5) 1 \u2212 \u03bb 32\u03c02 m2 \u03bb 32\u03c02 m2 Ak2 ( 3 4 C + \u03c1 + 1 + ln(\u00b52/m2 \u03c1) 1 + 2 \u03b5 dx ln(\u00b52/D) \u03c1 + O(\u03bb2). 0 Z 4 B)m2 1 \u2212 Again using eqs. (31.7) and (31.10) for B and C, we see that all the 1/\u03b5\u2019s cancel, and we\u2019re left with \u2212 \u2212 (31.22) \u03a0(k2) = \u03bb 32\u03c02 m2 + O(\u03bb2). \u03c1 1 + ln(\u00b52/m2 \u03c1) + 1 0 Z dx ln(\u00b52/D) + 1 2 (9\u03baC \u2212 \u03baB) (31.23) \u2212 m2 We can now choose to work in an OS scheme, where we require \u03a0( and \u03a0\u2032( k2. Thus, we automatically have \u03a0\u2032( to \ufb01x \u03a0( \u03baB and \u03baC to this order in \u03bb. \u03c1) = 0 \u03c1) = 0. We see that, to this order in \u03bb, \u03a0(k2) is independent of \u03baB \u03c1) = 0. Together with eq. (31.20), this completely determines \u03c1) = 0, and we can choose 9\u03baC \u2212 m2 m2 \u2212 \u2212 \u2212 m2 Next we consider the one-loop correction to the three-point vertex, given by the diagrams of \ufb01g. (31.5). We wish to show that the in\ufb01nities are canceled by the value of Z\u03bb = 1 + C that we have already determined. The \ufb01rst diagram in \ufb01g. (31.5) is \ufb01nite, and so for our purposes we can ignore it. The remaining three, plus the original vertex, sum up to give iV3(k1, k2, k3)div = \u2212 iZ\u03bbg3 + 1 2 ( i\u03bb)( ig3) 2 1 i \u2212 \u2212 2) + iF (k2 1) + iF (k2 iF (k2 3) i h (31.24) \u00d7 + O(\u03bb5/2), 31: Broken Symmetry and Loop Corrections 204 Figure 31.4: O(\u03bb) corrections to the \u03c1 propagator. k1 k1 k2 k3 k3 k2 k3 k2 k1 k3 k2 k1 Figure 31.5: O(\u03bb) corrections to the vertex for three \u03c1 \ufb01elds. where the subscript div means that we are keeping only the divergent part. Using eq. (31.9), we have V3(k1, k2, k3)div = g3 1 + C \u2212 3\u03bb 16\u03c02 1 \u03b5 \u2212 + O(\u03bb2). (31.25) From eq. (31.10), we see that the divergent terms do indeed cancel to this order in \u03bb. Finally, we have the correction to the four-point vertex. In this case, the divergent diagrams are just those of \ufb01g. (31.1), and", " so the calculation of the divergent part of V4 is exactly the same as it is when m2 > 0 (but with m\u03c1 in place of m). Since we have already done that calculation (it was how we determined C in the \ufb01rst place), we need not repeat it. We have thus seen how we can compute the divergent parts of the counterterms in the simpler case of m2 > 0, where the Z2 symmetry is unbroken, and that these counterterms will also serve to cancel the divergences in the more complicated case of m2 < 0, where the Z2 symmetry is spontaneously broken. This a general rule for renormalizable theories with spontaneous symmetry breaking, regardless of the nature of the symmetry group. Reference Notes Another example of renormalization of a spontaneously broken theory is worked out in Peskin & Schroeder. 32: Spontaneous Breaking of Continuous Symmetries 205 32 Spontaneous Breaking of Continuous Symmetries Prerequisite: 22, 30 Consider the theory (introduced in section 22) of a complex scalar \ufb01eld \u03d5 with \u2212 This lagrangian is obviously invariant under the U(1) transformation \u2212 \u2212 L = \u2202\u00b5\u03d5\u2020\u2202\u00b5\u03d5 m2\u03d5\u2020\u03d5 1 4 \u03bb(\u03d5\u2020\u03d5)2. (32.1) \u03d5(x) \u2192 e\u2212 i\u03b1\u03d5(x), (32.2) where \u03b1 is a real number. Now suppose that m2 is negative. The minimum of the potential of eq. (32.1) is achieved for where \u03d5(x) = 1 \u221a2 ve\u2212 i\u03b8, m2 v = (4 | /\u03bb)1/2, | (32.3) (32.4) and the phase \u03b8 is arbitrary; the factor of root-two in eq. (32.3) is conventional. Thus we have a continuous family of minima of the potential, parameterized by \u03b8. Under the U(1) transformation of eq. (32.2), \u03b8 changes to \u03b8 + \u03b1; thus the di\ufb00erent minimum-energy \ufb01eld con\ufb01gurations are all related to each other by the symmetry. In the quantum theory, we therefore expect to \ufb01nd a continuous family of ground states, labeled by \u03b8, with the property that \u03b8 h \u03b8 \u03d5(x) | | i = 1 \u221a2 ve\u2212 i\u03b8. (32.5) = \u03b8. Also, according to the discussion in section 30, we expect \u03b8\u2032 6 Returning to classical language, there is a \ufb02at direction in \ufb01eld space that we can move along without changing the energy. The physical consequence of this is the existence of a massless particle called a Goldstone boson. = 0 for \u03b8 \u03b8\u2032| h i Let us see how this works in more detail. We \ufb01rst choose the phase \u03b8 = 0, and then write \u03d5(x) = 1 \u221a2 [v + a(x) + ib(x)], (32.6) where a and b are real scalar \ufb01elds. Substituting eq. (32.6) into eq. (32.1), we \ufb01nd = L \u2212 1 2 \u2202\u00b5a\u2202\u00b5a a2 m2 | 1 2 \u2202\u00b5b\u2202\u00b5b \u2212 2 \u03bb1/2 1 m | \u2212 \u2212 | a(a2 + b2) | 1 16 \u03bb(a2 + b2)2. \u2212 (32.7) 32: Spontaneous Breaking of Continuous Symmetries 206 We see from this that the a \ufb01eld has a mass given by 1. The b | \ufb01eld, on the other hand, is massless, and we identify it as the Goldstone boson. m2 | 2 m2 a = A di\ufb00erent parameterization brings out the role of the massless \ufb01eld more clearly. We write \u03d5(x) = 1 \u221a2 (v + \u03c1(x))e\u2212 i\u03c7(x)/v, (32.8) where \u03c1 and \u03c7 are real scalar \ufb01elds. Substituting eq. (32.8) into eq. (32.1), we get = L \u2212 1 + 1 2 \u2202\u00b5\u03c1\u2202\u00b5\u03c1 \u03c12 m2 | \u2212 | \u2212 1 2 \u2212 1 2 \u03bb1/2 2 \u03c1 v \u03c13 m | | \u2202\u00b5\u03c7\u2202\u00b5\u03c7 1 16 \u03bb\u03c14. \u2212 (32.9)", " We see from this that the \u03c1 \ufb01eld has a mass given by 1, and that | the \u03c7 \ufb01eld is massless. These are the same particle masses we found using the parameterization of eq. (32.6). This is not an accident: the particle masses and scattering amplitudes are independent of \ufb01eld rede\ufb01nitions. m2 | 2 m2 \u03c1 = Note that the \u03c7 \ufb01eld does not appear in the potential at all. Thus it parameterizes the \ufb02at direction. In terms of the \u03c1 and \u03c7 \ufb01elds, the U(1) transformation takes the simple form \u03c7(x) \u03c7(x) + \u03b1. Does the masslessness of the \u03c7 \ufb01eld survive loop corrections? It does. To see this, we note that if the \u03c7 \ufb01eld remains massless, its exact propagator \u02dc\u2206\u03c7(k2) should have a pole at k2 = 0; equivalently, the self-energy \u03a0\u03c7(k2), related to the propagator by \u02dc\u2206\u03c7(k2) = 1/[k2 \u03a0\u03c7(k2)], should satisfy \u03a0(0) = 0. \u2212 \u2192 We can evaluate \u03a0\u03c7(0) by summing all 1PI diagrams with two external \u03c7 lines, each with four-momentum k = 0. We note from eq. (32.9) that the derivatives acting on the \u03c7 \ufb01elds imply that the vertex factors for the \u03c1\u03c7\u03c7 k2, where k1 and k2 are the and \u03c1\u03c1\u03c7\u03c7 vertices are each proportional to k1 \u00b7 momenta of the two \u03c7 lines that meet at that vertex. Since the external lines have zero momentum, the attached vertices vanish; hence, \u03a0\u03c7(0) = 0, and the \u03c7 particle remains massless. The same conclusion can be reached by considering the quantum action \u0393(\u03d5), which includes all loop corrections. According to our discussion in section 29, the quantum action has the same symmetries as the classical action. Therefore, in the case at hand, \u0393(\u03d5) = \u0393(e\u2212 i\u03b1\u03d5). (32.10) Spontaneous symmetry breaking occurs if the minimum of \u0393(\u03d5) is at a constant, nonzero value of \u03d5. Because of eq. (32.10), the phase of this 32: Spontaneous Breaking of Continuous Symmetries 207 constant is arbitrary. Therefore, there must be a \ufb02at direction in \ufb01eld space, corresponding to the phase of \u03d5(x). The physical consequence of this \ufb02at direction is a massless particle, the Goldstone boson. All of this has a straightforward extension to the nonabelian case. Con- sider = 1 2 \u2202\u00b5\u03d5i\u2202\u00b5\u03d5i \u2212 1 2 m2\u03d5i\u03d5i \u2212 \u2212 L 1 16 \u03bb(\u03d5i\u03d5i)2, (32.11) where a repeated index is summed. This lagrangian is invariant under the in\ufb01nitesimal SO(N ) transformation \u03b4\u03d5i = \u2212 i\u03b8a(T a)ij\u03d5j, (32.12) where runs from 1 to 1 1) real, in\ufb01nitesimal parameters, and each antisymmetric generator matrix T a has a single nonzero entry i above the main diagonal, and a corresponding +i below the main diagonal. 1), \u03b8a is a set of 1 2 N (N 2 N (N \u2212 \u2212 \u2212 Now let us take m2 < 0 in eq. (32.11). The minimum of the potential m2 is achieved for \u03d5i(x) = vi, where v2 = vivi = 4 /\u03bb, and the direction | | in which the N -component vector ~v points is arbitrary. In the quantum theory, we interpret vi as the vacuum expectation value (VEV for short) of the quantum \ufb01eld \u03d5i(x). We can choose our coordinate system so that vi = v\u03b4iN ; that is, the VEV lies entirely in the last component. Now consider making an in\ufb01nitesimal SO(N ) transformation. This changes the VEV; we have i\u03b8a(T a)ijvj vi \u2192 vi \u2212 = v\u03b4iN \u2212 i\u03b8a(T a)iN v. (32.13) For some choices of \u03b8a, the second term on the right-hand side of eq. (32", ".13) vanishes. This happens if the corresponding T a has no nonzero entry in the 1 T a\u2019s with a nonzero entry in the last column: last column. There are N i in the \ufb01rst row and last column, in the second row and those with the \u2212 1th row and last column. These T a\u2019s are last column, etc, down to the N \u2212 said to be broken generators. A generator is broken if (T a)ijvj 6 = 0, and unbroken if (T a)ijvj = 0. \u2212 An in\ufb01nitesimal SO(N ) transformation that involves a broken generator changes the VEV of the \ufb01eld, but not the energy. Thus, each broken generator corresponds to a \ufb02at direction in \ufb01eld space. Each \ufb02at direction implies the existence of a corresponding massless particle. This is Goldstone\u2019s theorem: there is one massless Goldstone boson for each broken generator. The unbroken generators, on the other hand, do not change the VEV of the \ufb01eld. Therefore, after rewriting the lagrangian in terms of shifted 32: Spontaneous Breaking of Continuous Symmetries 208 \ufb01elds (each with zero VEV), there should still be a manifest symmetry corresponding to the set of unbroken generators. In the present case, the number of unbroken generators is 1 2). This 2 N (N \u2212 is the number of generators of SO(N 1) \u2212 to be an obvious symmetry of the lagrangian after it is written in terms of shifted \ufb01elds. 1) 2 (N 1). Therefore, we expect SO(N 1) = 1 1)(N (N \u2212 \u2212 \u2212 \u2212 \u2212 Let us see how this works in the present case. We can rewrite eq. (32.11) as with = 1 2 \u2202\u00b5\u03d5i\u2202\u00b5\u03d5i \u2212 \u2212 L V (\u03d5), V (\u03d5) = 1 16 \u03bb(\u03d5i\u03d5i \u2212 v2)2, (32.14) (32.15) /\u03bb)1/2, and the repeated index i is implicitly summed m2 where v = (4 | | from 1 to N. Now let \u03d5N (x) = v + \u03c1(x), and plug this into eq. (32.14). 1, we have With the repeated index i now implicitly summed from 1 to N \u2212 = 1 2 \u2202\u00b5\u03d5i\u2202\u00b5\u03d5i \u2212 \u2212 L 1 2 \u2202\u00b5\u03c1\u2202\u00b5\u03c1 \u2212 V (\u03c1, \u03d5), (32.16) where V (\u03c1, \u03d5) = 1 v2]2 16 \u03bb[(v+\u03c1)2 + \u03d5i\u03d5i \u2212 16 \u03bb(2v\u03c1 + \u03c12 + \u03d5i\u03d5i)2 4 \u03bbv2\u03c12 + 1 = 1 = 1 4 \u03bbv\u03c1(\u03c12 + \u03d5i\u03d5i) + 1 16 \u03bb(\u03c12 + \u03d5i\u03d5i)2. (32.17) There is indeed a manifest SO(N Also, the N bosons. \u2212 1 \u03d5i \ufb01elds are massless: they are the expected N 1) symmetry in eqs. (32.16) and (32.17). 1 Goldstone \u2212 \u2212 Reference Notes Further discussion of Goldstone\u2019s theorem can be found in Georgi, Peskin & Schroeder, and Weinberg II. Problems 32.1) Consider the Noether current j\u00b5 for the U(1) symmetry of eq. (32.1), and the corresponding charge Q. a) Show that e\u2212 i\u03b1Q \u03d5 e+i\u03b1Q = e+i\u03b1\u03d5. i\u03b1Q b) Use eq. (32.5) to show that e) Show that Q = 0; that is, the charge does not annihilate the \u03b8 = 0 vacuum. Contrast this with the case of an unbroken symmetry. 0 i 6 | 32: Spontaneous Breaking of Continuous Symmetries 209 32.2) In problem 24.3, we showed that [\u03d5i, Qa] = (T a)ij\u03d5j, where Qa is the Noether charge in the SO(N ) symmetric theory. Use this result to show that Qa = 0 if and only if Qa is broken. 0 i 6 | 32.3) We de\ufb01ne the decay constant f of the Goldstone boson via k h j\u00b5(x) 0 i | | = if k\u00b5e\u2212 ikx, (32.18)", " i k | 0 i | is the state of a single Goldstone boson with four-momentum is the \u03b8 = 0 vacuum, and j\u00b5(x) is where k, normalized in the usual way, the Noether current. a) Compute f at tree level. (That is, express j\u00b5 in terms of the \u03c1 and \u03c7 \ufb01elds, and then use free \ufb01eld theory to compute the matrix element.) A nonvanishing value of f indicates that the corresponding current is spontaneously broken. b) Discuss how your result would be modi\ufb01ed by higher-order corrections. Part II Spin One Half 33: Representations of the Lorentz Group 211 33 Representations of the Lorentz Group Prerequisite: 2 In section 2, we saw that we could de\ufb01ne a unitary operator U (\u039b) that implemented a Lorentz transformation on a scalar \ufb01eld \u03d5(x) via U (\u039b)\u2212 1\u03d5(x)U (\u039b) = \u03d5(\u039b\u2212 1x). (33.1) As shown in section 2, this implies that the derivative of the \ufb01eld transforms as U (\u039b)\u2212 1\u2202\u00b5\u03d5(x)U (\u039b) = \u039b\u00b5 \u03c1 \u00af\u2202\u03c1\u03d5(\u039b\u2212 1x), (33.2) where the bar on he derivative means that it is with respect to the argument \u00afx = \u039b\u2212 1x. Eq. (33.2) suggests that we could de\ufb01ne a vector \ufb01eld A\u00b5(x) that would transform as U (\u039b)\u2212 1A\u03c1(x)U (\u039b) = \u039b\u00b5 \u03c1A\u03c1(\u039b\u2212 1x), (33.3) or a tensor \ufb01eld B\u00b5\u03bd(x) that would transform as U (\u039b)\u2212 1B\u00b5\u03bd(x)U (\u039b) = \u039b\u00b5 \u03c1\u039b\u03bd \u03c3B\u03c1\u03c3(\u039b\u2212 1x). (33.4) Note that if B\u00b5\u03bd is either symmetric, B\u00b5\u03bd(x) = B\u03bd\u00b5(x), or antisymmetric, B\u03bd\u00b5(x), then the symmetry is preserved by the Lorentz transB\u00b5\u03bd(x) = g\u00b5\u03bd B\u00b5\u03bd(x), then, using formation. Also, if we take the trace to get T (x) g\u00b5\u03bd \u039b\u00b5 \u03c3 = g\u03c1\u03c3, we \ufb01nd that T (x) transforms like a scalar \ufb01eld, \u03c1\u039b\u03bd \u2261 \u2212 U (\u039b)\u2212 1T (x)U (\u039b) = T (\u039b\u2212 1x). (33.5) Thus, given a tensor \ufb01eld B\u00b5\u03bd(x) with no particular symmetry, we can write B\u00b5\u03bd(x) = A\u00b5\u03bd (x) + S\u00b5\u03bd (x) + 1 4 g\u00b5\u03bd T (x), (33.6) A\u03bd\u00b5) and S\u00b5\u03bd is symmetric (S\u00b5\u03bd = where A\u00b5\u03bd is antisymmetric (A\u00b5\u03bd = \u2212 S\u03bd\u00b5) and traceless (g\u00b5\u03bd S\u00b5\u03bd = 0). The key point is that the \ufb01elds A\u00b5\u03bd, S\u00b5\u03bd, and T do not mix with each other under Lorentz transformations. Is it possible to further break apart these \ufb01elds into still smaller sets that do not mix under Lorentz transformations? How do we make this decomposition into irreducible representations of the Lorentz group for a \ufb01eld carrying n vector indices? Are there any other kinds of indices we could consistently assign to a \ufb01eld? If so, how do these behave under a Lorentz transformation? The answers to these questions are to be found in the theory of group representations. Let us see how this works for the Lorentz group (in four spacetime dimensions). 33: Representations of the Lorentz Group 212 Consider a \ufb01eld (not necessarily hermitian) that carries a generic Lorentz index, \u03d5A(x). Under a Lorentz transformation, we have U (\u039b)\u2212 1\u03d5A(x)U (\u039b) = LA B(\u039b)\u03d5B(\u039b\u2212 1x), (33.7) where LA matrices must obey the group composition rule", " B(\u039b) is a matrix that depends on \u039b. These \ufb01nite-dimensional LA B(\u039b\u2032)LB C (\u039b) = LA C(\u039b\u2032\u039b). (33.8) We say that the matrices LA group. B(\u039b) form a representation of the Lorentz For an in\ufb01nitesimal transformation \u039b\u00b5 \u03bd = \u03b4\u00b5 \u03bd + \u03b4\u03c9\u00b5 \u03bd, we can write U (1+\u03b4\u03c9) = I + i 2 \u03b4\u03c9\u00b5\u03bd M \u00b5\u03bd, (33.9) where the operators M \u00b5\u03bd are the generators of the Lorentz group. As shown in section 2, the generators obey the commutation relations [M \u00b5\u03bd, M \u03c1\u03c3] = i g\u00b5\u03c1M \u03bd\u03c3 (\u00b5 \u2194 \u2212 \u03bd) (\u03c1 \u2194 \u2212 \u03c3), (33.10) which specify the Lie algebra of the Lorentz group. We can identify the components of the angular momentum operator ~J as 2 \u03b5ijkM jk and the components of the boost operator ~K as Ki \u2261 M i0. Ji \u2261 We then \ufb01nd from eq. (33.10) that 1 [Ji, Jj] = +i\u03b5ijkJk, [Ji, Kj] = +i\u03b5ijkKk, i\u03b5ijkJk. [Ki, Kj] = \u2212 (33.11) (33.12) (33.13) For an in\ufb01nitesimal transformation, we also have LA B(1+\u03b4\u03c9) = \u03b4A B + i 2 \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd)A B, (33.14) Eq. (33.7) then becomes [\u03d5A(x), M \u00b5\u03bd ] = L \u00b5\u03bd \u03d5A(x) + (S\u00b5\u03bd )A B\u03d5B(x), (33.15) \u00b5\u03bd 1 i (x\u00b5\u2202\u03bd where representation matrices (S\u00b5\u03bd )A tion relations as the generators themselves; see problems 2.8 and 2.9. \u00b5\u03bd and the B must separately obey the same commuta- x\u03bd \u2202\u00b5). Both the di\ufb00erential operators \u2261 \u2212 L L Our problem now is to \ufb01nd all possible sets of \ufb01nite-dimensional matrices that obey eq. (33.10), or equivalently eqs. (33.11\u201333.13). Although the operators M \u00b5\u03bd must be hermitian, the matrices (S\u00b5\u03bd )A B need not be. 33: Representations of the Lorentz Group 213 (2j+1) hermitian matrices If we restrict our attention to eq. (33.11) alone, we know (from standard results in the quantum mechanics of angular momentum) that we J1, can \ufb01nd three (2j+1) J3 that J2, and \u00d7 obey eq. (33.11), and that the eigenvalues of (say) j+1,..., +j, j, J3 are \u2212 \u2212 where j has the possible values 0, 1 2, 1,.... We further know that these matrices constitute all of the inequivalent, irreducible representations of the Lie algebra of SO(3), the rotation group in three dimensions. Inequivalent means not related by a unitary transformation; irreducible means cannot be made block-diagonal by a unitary transformation. (The standard derivation assumes that the matrices are hermitian, but allowing nonhermitian matrices does not enlarge the set of solutions.) Also, when j is a half integer, a rotation by 2\u03c0 results in an overall minus sign; these representations of the Lie algebra of SO(3) are therefore actually not represenations of the group SO(3), since a 2\u03c0 rotation should be equivalent to no rotation. As we saw in section 24, the Lie algebra of SO(3) is the same as the Lie algebra of SU(2); the half-integer representations of this Lie algebra do qualify as representations of the group SU(2). We would like to extend these conclusions to encompass the full set of eqs. (33.11\u201333.13). In order to do so, it is helpful to de\ufb01ne some nonhermitian operators whose physical signi\ufb01cance is obscure, but which simplify the commutation relations. These are Ni \u2261 N \u2020i \u2261 1 2 (Ji \u2212 iKi), 1 2 (Ji + iKi). In terms of Ni and N \u2020i, eqs. (", "33.11\u201333.13) become [Ni, Nj] = i\u03b5ijkNk, [N \u2020i, N \u2020j ] = i\u03b5ijkN \u2020k, [Ni, N \u2020j ] = 0. (33.16) (33.17) (33.18) (33.19) (33.20) We see that we have two di\ufb00erent SU(2) Lie algebras that are exchanged by hermitian conjugation. As we just discussed, a representation of the SU(2) Lie algebra is speci\ufb01ed by an integer or half integer; we therefore conclude that a representation of the Lie algebra of the Lorentz group in four spacetime dimensions is speci\ufb01ed by two integers or half-integers n and n\u2032. We will label these representations as (2n+1, 2n\u2032+1); the number of components of a representation is then (2n+1)(2n\u2032+1). Di\ufb00erent components within a representation can also be labeled by their angular momentum representations. To do this, we \ufb01rst note that, from eqs. (33.16) and (33.17), we have Ji = Ni + N \u2020i. Thus, deducing the allowed values 33: Representations of the Lorentz Group 214 of j given n and n\u2032 becomes a standard problem in the addition of angular momenta. The general result is that the allowed values of j are +1,..., n+n\u2032, and each of these values appears exactly once. n, n\u2032| | The four simplest and most often encountered representations are (1, 1), n\u2032| n | \u2212 \u2212 (2, 1), (1, 2), and (2, 2). These are given special names: (1, 1) = Scalar or singlet (2, 1) = Left-handed spinor (1, 2) = Right-handed spinor (2, 2) = Vector (33.21) It may seem a little surprising that (2, 2) is to be identi\ufb01ed as the vector representation. To see that this must be the case, we \ufb01rst note that the vector representation is irreducible: all the components of a four-vector mix with each other under a general Lorentz transformation. Secondly, the vector representation has four components. The only candidate irreducible representations are (4, 1), (1, 4), and (2, 2). The \ufb01rst two of these contain angular momenta j = 3 2 only, whereas (2, 2) contains j = 0 and j = 1. This is just right for a four-vector, whose time component is a scalar under spatial rotations, and whose space components are a three-vector. In order to gain a better understanding of what it means for (2, 2) to be the vector representation, we must \ufb01rst investigate the spinor representations (1, 2) and (2, 1), which contain angular momenta j = 1 2 only. Reference Notes An extended treatment of representations of the Lorentz group in four dimensions can be found in Weinberg I. Problems 33.1) Express A\u00b5\u03bd(x), S\u00b5\u03bd(x), and T (x) in terms of B\u00b5\u03bd(x). 33.2) Verify that eqs. (33.18\u201333.20) follow from eqs. (33.11\u201333.13). 34: Left- and Right-Handed Spinor Fields 215 34 Left- and Right-Handed Spinor Fields Prerequisite: 3, 33 Consider a left-handed spinor \ufb01eld \u03c8a(x), also known as a left-handed Weyl \ufb01eld, which is in the (2, 1) representation of the Lie algebra of the Lorentz group. Here the index a is a left-handed spinor index that takes on two possible values. Under a Lorentz transformation, we have U (\u039b)\u2212 1\u03c8a(x)U (\u039b) = La b(\u039b)\u03c8b(\u039b\u2212 1x), (34.1) b(\u039b) is a matrix in the (2, 1) representation. These matrices satisfy where La the group composition rule La b(\u039b\u2032)Lb c(\u039b) = La c(\u039b\u2032\u039b). (34.2) For an in\ufb01nitesimal transformation \u039b\u00b5 \u03bd = \u03b4\u00b5 \u03bd + \u03b4\u03c9\u00b5 \u03bd, we can write La b(1+\u03b4\u03c9) = \u03b4a b + i 2 \u03b4", "\u03c9\u00b5\u03bd (S\u00b5\u03bd L )a b, (34.3) where (S\u00b5\u03bd commutation relations as the generators M \u00b5\u03bd, namely b is a set of 2 (S\u03bd\u00b5 L )a L )a b = \u00d7 \u2212 2 matrices that obey the same [S\u00b5\u03bd L, S\u03c1\u03c3 L ] = i g\u00b5\u03c1S\u03bd\u03c3 L \u2212 (\u00b5 \u2194 \u03bd) (\u03c1 \u2194 \u2212 \u03c3). U (1+\u03b4\u03c9) = I + i 2 \u03b4\u03c9\u00b5\u03bd M \u00b5\u03bd, Using eq. (34.1) becomes [\u03c8a(x), M \u00b5\u03bd ] = L \u00b5\u03bd\u03c8a(x) + (S\u00b5\u03bd L )a b\u03c8b(x), (34.4) (34.5) (34.6) L \u2212 \u00b5\u03bd = 1 i (x\u00b5\u2202\u03bd x\u03bd\u2202\u00b5). The \u00b5\u03bd term in eq. (34.6) would also be where L present for a scalar \ufb01eld, and is not the focus of our current interest; we will suppress it by evaluating the \ufb01elds at the spacetime origin, x\u00b5 = 0. Recalling that M ij = \u03b5ijkJk, where Jk is the angular momentum operator, we have \u03b5ijk[\u03c8a(0), Jk] = (Sij L )a b\u03c8b(0). (34.7) Recall that the (2, 1) representation of the Lorentz group includes angular momentum j = 1 2 only. For a spin-one-half operator, the standard convention is that the matrix on the right-hand side of eq. (34.7) is 1 2 \u03b5ijk\u03c3k, where we have suppressed the row index a and the column index b, and where \u03c3k is a Pauli matrix: \u03c31 = 0 1 1 0, \u03c32 = 0 i i \u2212 0, \u03c33 = 1 0 0 1 \u2212. (34.8) 34: Left- and Right-Handed Spinor Fields We therefore conclude that (Sij L )a b = 1 2 \u03b5ijk\u03c3k, 216 (34.9) 2 = L )1 1 = + 1 2, (S12 Thus, for example, setting i=1 and j=2 yields (S12 and so (S12 L )2 1 2, and (S12 Once we have the (2, 1) representation matrices for the angular momentum operator Ji, we can easily get them for the boost operator Kk = M k0. This is because Jk = Nk + N \u2020k and Kk = i(Nk \u2212 N \u2020k), and, acting on a \ufb01eld in the (2, 1) representation, N \u2020k is zero. Therefore, the representation matrices for Kk are simply i times those for Jk, and so 2 \u03b512k\u03c3k = 1 1 = 0. L )a 2 = (S12 b = 1 L )2 2 \u03c33, L )1 \u2212 (Sk0 L )a b = 1 2 i\u03c3k. (34.10) Now consider taking the hermitian conjugate of the left-handed spinor \ufb01eld \u03c8a(x). Recall that hermitian conjugation swaps the two SU(2) Lie algebras that comprise the Lie algebra of the Lorentz group. Therefore, the hermitian conjugate of a \ufb01eld in the (2, 1) representation should be a \ufb01eld in the (1, 2) representation; such a \ufb01eld is called a right-handed spinor \ufb01eld or a right-handed Weyl \ufb01eld. We will distinguish the indices of the (1, 2) representation from those of the (2, 1) representation by putting dots over them. Thus, we write Under a Lorentz transformation, we have [\u03c8a(x)]\u2020 = \u03c8\u2020\u02d9a(x). U (\u039b)\u2212 1\u03c8\u2020\u02d9a(x)U (\u039b) = R \u02d9a \u02d9b(\u039b)\u03c8\u2020\u02d9b (\u039b\u2212 1x), (34.11) (34.12) \u02d9b(\u039b) is a matrix in the (1, 2) representation. These matrices satisfy where R \u02d9a the group composition rule R \u02d9a \u02d9b(\u039b\u2032)R\u02d9b \u02d9c(\u039b) = R \u02d9a \u02d9c", "(\u039b\u2032\u039b). (34.13) For an in\ufb01nitesimal transformation \u039b\u00b5 \u03bd = \u03b4\u00b5 \u03bd + \u03b4\u03c9\u00b5 \u03bd, we can write R \u02d9a \u02d9b(1+\u03b4\u03c9) = \u03b4 \u02d9a \u02d9b + i 2 \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd R ) \u02d9a \u02d9b, (34.14) where (S\u00b5\u03bd commutation relations as the generators M \u00b5\u03bd. We then have \u02d9b is a set of 2 (S\u03bd\u00b5 \u02d9b = R ) \u02d9a R ) \u02d9a \u2212 \u00d7 2 matrices that obey the same [\u03c8\u2020\u02d9a(0), M \u00b5\u03bd ] = (S\u00b5\u03bd R ) \u02d9a \u02d9b\u03c8\u2020\u02d9b (0). Taking the hermitian conjugate of this equation, we get [M \u00b5\u03bd, \u03c8a(0)] = [(S\u00b5\u03bd R ) \u02d9a \u02d9b]\u2217\u03c8b(0). (34.15) (34.16) 34: Left- and Right-Handed Spinor Fields 217 Comparing this with eq. (34.6), we see that (S\u00b5\u03bd R ) \u02d9a \u02d9b = [(S\u00b5\u03bd L )a b]\u2217. \u2212 (34.17) In the previous section, we examined the Lorentz-transformation properties of a \ufb01eld carrying two vector indices. To help us get better acquainted with the properties of spinor indices, let us now do the same for a \ufb01eld that carries two (2, 1) indices. Call this \ufb01eld Cab(x). Under a Lorentz transformation, we have U (\u039b)\u2212 1Cab(x)U (\u039b) = La c(\u039b)Lb d(\u039b)Ccd(\u039b\u2212 1x). (34.18) The question we wish to address is whether or not the four components of Cab can be grouped into smaller sets that do not mix with each other under Lorentz transformations. To answer this question, recall from quantum mechanics that two spinone-half particles can be in a state of total spin zero, or total spin one. Furthermore, the single spin-zero state is the unique antisymmetric combination of the two spin-one-half states, and the three spin-one states are the three symmetric combinations of the two spin-one-half states. We can write this schematically as 2 3S, where we label the representa2 = 1A \u2295 tion of SU(2) by the number of its components, and the subscripts S and A indicate whether that representation appears in the symmetric or antisymmetric combination of the two 2\u2019s. For the Lorentz group, the relevant relation is (2, 1) (3, 1)S. This implies that we should be able to write (2, 1) = (1, 1)A \u2295 \u2297 \u2297 Cab(x) = \u03b5abD(x) + Gab(x), (34.19) where D(x) is a scalar \ufb01eld, \u03b5ab = \u03b5ba is an antisymmetric set of constants, and Gab(x) = Gba(x). The symbol \u03b5ab is uniquely determined by its sym\u03b512 = +1. metry properties up to an overall constant; we will choose \u03b521 = Since D(x) is a Lorentz scalar, eq. (34.19) is consistent with eq. (34.18) \u2212 \u2212 only if La c(\u039b)Lb d(\u039b)\u03b5cd = \u03b5ab. (34.20) This means that \u03b5ab is an invariant symbol of the Lorentz group: it does not change under a Lorentz transformation that acts on all of its indices. In this way, \u03b5ab is analogous to the metric g\u00b5\u03bd, which is also an invariant symbol, since \u03c3g\u03c1\u03c3 = g\u00b5\u03bd. (34.21) We use g\u00b5\u03bd and its inverse g\u00b5\u03bd to raise and lower vector indices, and we can use \u03b5ab and and its inverse \u03b5ab to raise and lower left-handed spinor indices. Here we de\ufb01ne \u03b5ab via \u03c1\u039b\u03bd \u039b\u00b5 \u03b512 = \u03b521 = +1, \u03b521 = \u03b512 = 1. \u2212 (34.22) 34: Left-", " and Right-Handed Spinor Fields 218 With this de\ufb01nition, we have \u03b5ab\u03b5bc = \u03b4a c, \u03b5ab\u03b5bc = \u03b4a c. We can then de\ufb01ne \u03c8a(x) \u2261 \u03b5ab\u03c8b(x). We also have (suppressing the spacetime argument of the \ufb01eld) \u03c8a = \u03b5ab\u03c8b = \u03b5ab\u03b5bc\u03c8c = \u03b4a c\u03c8c, (34.23) (34.24) (34.25) as we would expect. However, the antisymmetry of \u03b5ab means that we must be careful with minus signs; for example, eq. (34.24) can be written in various ways, such as \u03c8a = \u03b5ab\u03c8b = \u03b5ba\u03c8b = \u2212 \u2212 \u03c8b\u03b5ba = \u03c8b\u03b5ab. (34.26) We must also be careful about signs when we contract indices, since \u03c8a\u03c7a = \u03b5ab\u03c8b\u03c7a = \u03b5ba\u03c8b\u03c7a = \u03c8b\u03c7b. (34.27) \u2212 In section 35, we will (mercifully) develop an index-free notation that automatically keeps track of these essential (but annoying) minus signs. \u2212 An exactly analogous discussion applies to the second SU(2) factor; (1, 3)S, we can from the group-theoretic relation (1, 2) deduce the existence of an invariant symbol \u03b5 \u02d9a\u02d9b = \u03b5\u02d9b \u02d9a. We will normalize \u03b5 \u02d9a\u02d9b according to eq. (34.22). Then eqs. (34.23\u201334.27) hold if all the undotted indices are replaced by dotted indices. (1, 2) = (1, 1)A \u2295 \u2212 \u2297 Now consider a \ufb01eld carrying one undotted and one dotted index, Aa \u02d9a(x). Such a \ufb01eld is in the (2, 2) representation, and in section 33 we concluded that the (2, 2) representation was the vector representation. We would more naturally write a \ufb01eld in the vector representation as A\u00b5(x). There must, then, be a dictionary that gives us the components of Aa \u02d9a(x) in terms of the components of A\u00b5(x); we can write this as Aa \u02d9a(x) = \u03c3\u00b5 a \u02d9aA\u00b5(x), (34.28) where \u03c3\u00b5 be deduced from the group-theoretic relation a \u02d9a is another invariant symbol. That such a symbol must exist can (2, 1) (1, 2) \u2297 \u2297 (2, 2) = (1, 1).... \u2295 (34.29) As we will see in section 35, it turns out to be consistent with our already established conventions for S\u00b5\u03bd L and S\u00b5\u03bd R to choose \u03c3\u00b5 a \u02d9a = (I, ~\u03c3). (34.30) 34: Left- and Right-Handed Spinor Fields 219 Thus, for example, \u03c33 1 \u02d91 = +1, \u03c33 2 \u02d92 = 1, \u03c33 1 \u02d92 = \u03c33 2 \u02d91 = 0. \u2212 In general, whenever the product of a set of representations includes the singlet, there is a corresponding invariant symbol. For example, we can deduce the existence of g\u00b5\u03bd = g\u03bd\u00b5 from (2, 2) (2, 2) = (1, 1)S \u2295 (1, 3)A \u2295 (3, 1)A \u2295 \u2297 (3, 3)S. (34.31) Another invariant symbol, the Levi-Civita symbol, follows from (2, 2) (2, 2) (2, 2) (2, 2) = (1, 1)34.32) where the subscript A denotes the completely antisymmetric part. The Levi-Civita symbol is \u03b5\u00b5\u03bd\u03c1\u03c3, which is antisymmetric on exchange of any pair of its indices, and is normalized via \u03b50123 = +1. To see that \u03b5\u00b5\u03bd\u03c1\u03c3 is \u03b4\u03b5\u03b1\u03b2\u03b3\u03b4 is antisymmetric on exchange \u03b3\u039b\u03c3 invariant, we note that \u039b\u00b5 of any two of its uncontracted indices, and therefore must", " be proportional to \u03b5\u00b5\u03bd\u03c1\u03c3. The constant of proportionality works out to be det \u039b, which is +1 for a proper Lorentz transformation. \u03b2\u039b\u03c1 \u03b1\u039b\u03bd We are \ufb01nally ready to answer a question we posed at the beginning of section 33. There we considered a \ufb01eld B\u00b5\u03bd(x) carrying two vector indices, and we decomposed it as B\u00b5\u03bd(x) = A\u00b5\u03bd (x) + S\u00b5\u03bd (x) + 1 4 g\u00b5\u03bd T (x), (34.33) A\u03bd\u00b5) and S\u00b5\u03bd is symmetric (S\u00b5\u03bd = where A\u00b5\u03bd is antisymmetric (A\u00b5\u03bd = \u2212 S\u03bd\u00b5) and traceless (g\u00b5\u03bd S\u00b5\u03bd = 0). We asked whether further decomposition into still smaller irreducible representations was possible. The answer to this question can be found in eq. (34.31). Obviously, T (x) corresponds to (1, 1), and S\u00b5\u03bd (x) to (3, 3).1 But, according to eq. (34.31), the antisymmetric \ufb01eld A\u00b5\u03bd (x) should correspond to (3, 1) (1, 3). A \ufb01eld in the (3, 1) representation carries a symmetric pair of left-handed (undotted) spinor indices; its hermitian conjugate is a \ufb01eld in the (1, 3) representation that carries a symmetric pair of right-handed (dotted) spinor indices. We should, then, be able to \ufb01nd a mapping, analogous to eq. (34.28), that gives A\u00b5\u03bd (x) in terms of a \ufb01eld Gab(x) and its hermitian conjugate G\u2020 \u02d9a\u02d9b This mapping is provided by the generator matrices S\u00b5\u03bd R. We \ufb01rst note that the Pauli matrices are traceless, and so eqs. (34.9) and (34.10) imply that (S\u00b5\u03bd a = 0. Using eq. (34.24), we can rewrite this as \u03b5ab(S\u00b5\u03bd L )ab must be symmetric on exchange of its two spinor indices. An identical argument shows that L )ab = 0. Since \u03b5ab is antisymmetric, (S\u00b5\u03bd (x). L and S\u00b5\u03bd L )a \u2295 1Note that a symmetric traceless tensor has three independent diagonal components, and six independent o\ufb00-diagonal components, for a total of nine, the number of components of the (3, 3) representation. 34: Left- and Right-Handed Spinor Fields 220 R ) \u02d9a\u02d9b must be symmetric on exchange of its two spinor indices. Further- (S\u00b5\u03bd more, according to eqs. (34.9) and (34.10), we have (S10 L )a b = i(S23 L )a b. \u2212 This can be written covariantly with the Levi-Civita symbol as Similarly, (S\u00b5\u03bd L )a b = i 2 \u03b5\u00b5\u03bd\u03c1\u03c3(SL \u03c1\u03c3)a b. \u2212 (S\u00b5\u03bd R ) \u02d9a \u02d9b = + i 2 \u03b5\u00b5\u03bd\u03c1\u03c3(SR \u03c1\u03c3) \u02d9a \u02d9b. (34.34) (34.35) (34.36) Eq. (34.36) follows from taking the complex conjugate of eq. (34.35) and using eq. (34.17). Now, given a \ufb01eld Gab(x) in the (3, 1) representation, we can map it into a self-dual antisymmetric tensor G\u00b5\u03bd (x) via G\u00b5\u03bd (x) \u2261 (S\u00b5\u03bd L )abGab(x). By self-dual, we mean that G\u00b5\u03bd (x) obeys G\u00b5\u03bd (x) = i 2 \u03b5\u00b5\u03bd\u03c1\u03c3G\u03c1\u03c3(x). \u2212 (34.37) (34.38) Taking the hermitian conjugate of eq. (34.37), and using eq. (34.17), we get G\u2020 \u00b5\u03bd (x) = (S\u00b5\u03bd R ) \u02d9a\u02d9bG\u2020 \u02d9a\u02d9b (x), \u2212 which is anti-self-dual, G\u2020 \u00b5\u03bd (", "x) = + i 2 \u03b5\u00b5\u03bd\u03c1\u03c3G\u2020\u03c1\u03c3(x). (34.39) (34.40) Given a hermitian antisymmetric tensor \ufb01eld A\u00b5\u03bd(x), we can extract its self-dual and anti-self-dual parts via G\u00b5\u03bd (x) = 1 G\u2020 \u00b5\u03bd (x) = 1 i 2 A\u00b5\u03bd (x) \u2212 2 A\u00b5\u03bd (x) + i 4 \u03b5\u00b5\u03bd\u03c1\u03c3A\u03c1\u03c3(x), 4 \u03b5\u00b5\u03bd\u03c1\u03c3A\u03c1\u03c3(x). Then we have A\u00b5\u03bd (x) = G\u00b5\u03bd (x) + G\u2020 \u00b5\u03bd (x). (34.41) (34.42) (34.43) The \ufb01eld G\u00b5\u03bd (x) is in the (3, 1) representation, and the \ufb01eld G\u2020 \u00b5\u03bd (x) is in the (1, 3) representation; these do not mix under Lorentz transformations. Problems 34.1) Verify that eq. (34.6) follows from eq. (34.1). 34: Left- and Right-Handed Spinor Fields 221 34.2) Verify that eqs. (34.9) and (34.10) obey eq. (34.4). 34.3) Show that the Levi-Civita symbol obeys \u03b5\u00b5\u03bd\u03c1\u03c3\u03b5\u03b1\u03b2\u03b3\u03c3 = \u03b4\u00b5 \u2212 + \u03b4\u00b5 \u03b1\u03b4\u03bd \u03b2\u03b4\u03bd \u03b2\u03b4\u03c1 \u03b1\u03b4\u03c1 \u03b4\u00b5 \u03b3 \u2212 \u03b3 + \u03b4\u00b5 \u03b2\u03b4\u03bd \u03b1\u03b4\u03bd \u03b3\u03b4\u03c1 \u03b3\u03b4\u03c1 \u03b4\u00b5 \u03b1 \u2212 \u03b2 + \u03b4\u00b5 \u03b3\u03b4\u03bd \u03b3\u03b4\u03bd \u03b1\u03b4\u03c1 \u03b2\u03b4\u03c1 \u03b2 \u03b1, \u03b5\u00b5\u03bd\u03c1\u03c3\u03b5\u03b1\u03b2\u03c1\u03c3 = \u03b5\u00b5\u03bd\u03c1\u03c3\u03b5\u03b1\u03bd\u03c1\u03c3 = \u2212 \u2212 2(\u03b4\u00b5 \u03b1\u03b4\u03bd \u03b2 \u2212 \u03b4\u00b5 \u03b2\u03b4\u03bd \u03b1), 6 \u03b4\u00b5 \u03b1. (34.44) (34.45) (34.46) 34.4) Consider a \ufb01eld Ca...c \u02d9a... \u02d9c(x), with N undotted indices and M dotted indices, that is furthermore symmetric on exchange of any pair of undotted indices, and also symmetric on exchange of any pair of dotted indices. Show that this \ufb01eld corresponds to a single irreducible representation (2n+1, 2n\u2032+1) of the Lorentz group, and identify n and n\u2032. 35: Manipulating Spinor Indices 222 35 Manipulating Spinor Indices Prerequisite: 34 In section 34 we introduced the invariant symbols \u03b5ab, \u03b5ab, \u03b5 \u02d9a\u02d9b, and \u03b5 \u02d9a\u02d9b, where \u03b512 = \u03b5 \u02d91 \u02d92 = \u03b521 = \u03b5 \u02d92 \u02d91 = +1, \u03b521 = \u03b5 \u02d92 \u02d91 = \u03b512 = \u03b5 \u02d91 \u02d92 = 1. \u2212 (35.1) We use the \u03b5 symbols to raise and lower spinor indices, contracting the second index on the \u03b5. (If we contract the \ufb01rst index instead, then there is an extra minus sign). Another invariant symbol is \u03c3\u00b5 a \u02d9a = (I, ~\u03c3), where I is the 2 2 identity matrix, and \u00d7 0 1 1 0 \u03c31 =, \u03c32 = are the Pauli matrices. 0 i i \u2212 0, \u03c33 = 1 0 0 1 \u2212 (35.2) (35.3) Now let us consider some combinations of invariant symbols with some indices contracted, such as g\u00b5\u03bd \u03c3\u00b5 a \u02d9a\u03c3\u03bd. This object must also be invariant. b\u02d9b Then, since it carries two undotted and two dotted spinor indices, it must be proportional to \u03b5ab\u03b5 \u02d9a\u02d9b. Using eqs. (35.1) and (35.2), we can laboriously check this; it turns out to be correct.1 The proportionality constant works out to be minus two: \u03c3\u00b5 a \u02d9a\u03c3\u00b5b\u02d9b = 2\u03b5ab\u03b5 \u02d9a\u02d9b. \u2212 (35.4) Similarly, \u03b5ab\u03b5 \u02d9a\u02d9b\u03c3\ufffd", "\ufffd constant is again minus two: a \u02d9a\u03c3\u03bd b\u02d9b must be proportional to g\u00b5\u03bd, and the proportionality b\u02d9b = Next, let\u2019s see what we can learn about the generator matrices (S\u00b5\u03bd \u2212 (35.5) 2g\u00b5\u03bd. \u03b5ab\u03b5 \u02d9a\u02d9b\u03c3\u00b5 a \u02d9a\u03c3\u03bd L )a a \u02d9a are all invariant symbols. b \u02d9b from the fact that \u03b5ab, \u03b5 \u02d9a\u02d9b, and \u03c3\u00b5 and (S\u00b5\u03bd R ) \u02d9a Begin with \u03b5ab = L(\u039b)a cL(\u039b)b d\u03b5cd, (35.6) which expresses the Lorentz invariance of \u03b5ab. For an in\ufb01nitesimal transformation \u039b\u00b5 \u03bd, we have \u03bd + \u03b4\u03c9\u00b5 \u03bd = \u03b4\u00b5 La b(1+\u03b4\u03c9) = \u03b4a b + i 2 \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd L )a b, (35.7) 1If it did not turn out to be correct, then eq. (35.2) would not be a viable choice of numerical values for this symbol. 35: Manipulating Spinor Indices 223 and eq. (35.6) becomes \u03b5ab = \u03b5ab + i 2 \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd L )a c\u03b5cb + (S\u00b5\u03bd L )b d\u03b5ad + O(\u03b4\u03c92) h = \u03b5ab + i 2 \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd L )ab + (S\u00b5\u03bd L )ba \u2212 h i i + O(\u03b4\u03c92). (35.8) Since eq. (35.8) holds for any choice of \u03b4\u03c9\u00b5\u03bd, it must be that the factor in square brackets vanishes. Thus we conclude that (S\u00b5\u03bd L )ba, which we had already deduced in section 34 by a di\ufb00erent method. Similarly, starting from the Lorentz invariance of \u03b5 \u02d9a\u02d9b, we can show that (S\u00b5\u03bd R ) \u02d9a\u02d9b = (S\u00b5\u03bd L )ab = (S\u00b5\u03bd R )\u02d9b \u02d9a. Next, start from \u03c3\u03c1 a \u02d9a = \u039b\u03c1 \u03c4 L(\u039b)a bR(\u039b) \u02d9a \u02d9b\u03c3\u03c4 b\u02d9b, (35.9) which expresses the Lorentz invariance of \u03c3\u03c1 formation, we have a \u02d9a. For an in\ufb01nitesimal trans- \u039b\u03c1 \u03c4 = \u03b4\u03c1 La b(1+\u03b4\u03c9) = \u03b4a R \u02d9a \u02d9b(1+\u03b4\u03c9) = \u03b4 \u02d9a \u03c4 + i b + i V )\u03c1 \u03c4, 2 \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd 2 \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd L )a \u02d9b + i 2 \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd R ) \u02d9a b, \u02d9b, (35.10) (35.11) (35.12) where (S\u00b5\u03bd i (g\u00b5\u03c1\u03b4\u03bd Substituting eqs. (35.10\u201335.13) into eq. (35.9) and isolating the coe\ufb03cient of \u03b4\u03c9\u00b5\u03bd yields (35.13) g\u03bd\u03c1\u03b4\u00b5 V )g\u00b5\u03c1\u03b4\u03bd \u03c4 \u2212 Now multiply by \u03c3\u03c1c \u02d9c to get g\u03bd\u03c1\u03b4\u00b5 \u03c4 )\u03c3\u03c4 a \u02d9a + i(S\u00b5\u03bd L )a b\u03c3\u03c1 b \u02d9a + i(S\u00b5\u03bd R ) \u02d9a \u02d9b\u03c3\u03c1 a\u02d9b = 0. (35.14) \u03c3\u00b5 c \u02d9c\u03c3\u03bd a \u02d9a \u2212 c \u02d9c\u03c3\u00b5 \u03c3\u03bd a \u02d9a + i(S\u00b5\u03bd L )a b\u03c3\u03c1 b \u02d9a\u03c3\u03c1c \u02d9c + i(S\u00b5\u03bd R ) \u02d9a \u02d9b\u03c3\u03c1 a\u02d9b\u03c3\u03c1c \u02d9c = 0. (35.15) Next use eq. (35.4) in each of the last two terms to get \u03c3\u00b5 c \u02d9c\u03c3\u03bd a", " \u02d9a \u2212 c \u02d9c\u03c3\u00b5 \u03c3\u03bd a \u02d9a + 2i(S\u00b5\u03bd L )ac\u03b5 \u02d9a \u02d9c + 2i(S\u00b5\u03bd R ) \u02d9a \u02d9c\u03b5ac = 0. (35.16) If we multiply eq. (35.16) by \u03b5 \u02d9a \u02d9c, and remember that \u03b5 \u02d9a \u02d9c(S\u00b5\u03bd that \u03b5 \u02d9a \u02d9c\u03b5 \u02d9a \u02d9c = 2, we get a formula for (S\u00b5\u03bd L )ac, namely R ) \u02d9a \u02d9c = 0 and \u2212 (S\u00b5\u03bd L )ac = i 4 \u03b5 \u02d9a \u02d9c(\u03c3\u00b5 a \u02d9a\u03c3\u03bd c \u02d9c \u2212 a \u02d9a\u03c3\u00b5 \u03c3\u03bd c \u02d9c). (35.17) 35: Manipulating Spinor Indices Similarly, if we multiply eq. (35.16) by \u03b5ac, we get (S\u00b5\u03bd R ) \u02d9a \u02d9c = i 4 \u03b5ac(\u03c3\u00b5 a \u02d9a\u03c3\u03bd c \u02d9c \u2212 a \u02d9a\u03c3\u00b5 \u03c3\u03bd c \u02d9c). These formulae can be made to look a little nicer if we de\ufb01ne \u00af\u03c3\u00b5 \u02d9aa \u03b5ab\u03b5 \u02d9a\u02d9b\u03c3\u00b5 b\u02d9b. \u2261 Numerically, it turns out that Using \u00af\u03c3\u00b5, we can write eqs. (35.17) and (35.18) as \u00af\u03c3\u00b5 \u02d9aa = (I, ~\u03c3). \u2212 (S\u00b5\u03bd L )a b = + i (S\u00b5\u03bd R ) \u02d9a \u02d9b = \u2212 4 (\u03c3\u00b5 \u00af\u03c3\u03bd 4 (\u00af\u03c3\u00b5\u03c3\u03bd i \u03c3\u03bd \u00af\u03c3\u00b5)a b, \u00af\u03c3\u03bd \u03c3\u00b5) \u02d9a \u02d9b. \u2212 \u2212 224 (35.18) (35.19) (35.20) (35.21) (35.22) In eq. (35.22), we have suppressed a contracted pair of undotted indices arranged as c c, and in eq. (35.21), we have suppressed a contracted pair of dotted indices arranged as \u02d9c \u02d9c. We will adopt this as a general convention: a missing pair of contracted, c, and a missing pair of \u02d9c. Thus, if \u03c7 and undotted indices is understood to be written as c contracted, dotted indices is understood to be written as \u02d9c \u03c8 are two left-handed Weyl \ufb01elds, we have \u03c7\u03c8 = \u03c7a\u03c8a and \u03c7\u2020\u03c8\u2020 = \u03c7\u2020\u02d9a\u03c8\u2020 \u02d9a. (35.23) We expect Weyl \ufb01elds to describe spin-one-half particles, and (by the spinstatistics theorem) these particles must be fermions. Therefore the correspoding \ufb01elds must anticommute, rather than commute. That is, we should have Thus we can rewrite eq. (35.23) as \u03c7a(x)\u03c8b(y) = \u03c8b(y)\u03c7a(x). \u2212 \u03c7\u03c8 = \u03c7a\u03c8a = \u2212 \u03c8a\u03c7a = \u03c8a\u03c7a = \u03c8\u03c7. (35.24) (35.25) The second equality follows from anticommutation of the \ufb01elds, and the a (which introduces an extra minus sign). third from switching a Eq. (35.25) tells us that \u03c7\u03c8 = \u03c8\u03c7, which is a nice feature of this notation. Furthermore, if we take the hermitian conjugate of \u03c7\u03c8, we get a to a (\u03c7\u03c8)\u2020 = (\u03c7a\u03c8a)\u2020 = (\u03c8a)\u2020(\u03c7a)\u2020 = \u03c8\u2020\u02d9a\u03c7\u2020 \u02d9a = \u03c8\u2020\u03c7\u2020. (35.26) That (\u03c7\u03c8)\u2020 = \u03c8\u2020\u03c7\u2020 is just what we would expect if we ignored the indices completely. Of course, by analogy with eq. (35.25), we also have \u03c8\u2020\u03c7\u2020 = \u03c7\u2020\u03c8\u2020. 35: Manipulating Spinor Indices 225 In order to", " tell whether a spinor \ufb01eld is left-handed or right-handed when its spinor index is suppressed, we will adopt the convention that a right-handed \ufb01eld is always written as the hermitian conjugate of a lefthanded \ufb01eld. Thus, a right-handed \ufb01eld is always written with a dagger, and a left-handed \ufb01eld is always written without a dagger. Let\u2019s try computing the hermitian conjugate of something a little more complicated: \u03c8\u2020 \u00af\u03c3\u00b5\u03c7 = \u03c8\u2020\u02d9a \u00af\u03c3\u00b5 \u02d9ac\u03c7c. (35.27) This behaves like a vector \ufb01eld under Lorentz transformations, U (\u039b)\u2212 1[\u03c8\u2020 \u00af\u03c3\u00b5\u03c7]U (\u039b) = \u039b\u00b5 \u03bd[\u03c8\u2020 \u00af\u03c3\u03bd \u03c7]. (35.28) (To avoid clutter, we suppressed the spacetime argument of the \ufb01elds; as 1x on the right.) The hermitian usual, it is x on the left-hand side and \u039b\u2212 conjugate of eq. (35.27) is [\u03c8\u2020 \u00af\u03c3\u00b5\u03c7]\u2020 = [\u03c8\u2020\u02d9a \u00af\u03c3\u00b5 \u02d9ac\u03c7c]\u2020 = \u03c7\u2020\u02d9c (\u00af\u03c3\u00b5a \u02d9c)\u2217\u03c8a = \u03c7\u2020\u02d9c \u00af\u03c3\u00b5 \u02d9ca\u03c8a = \u03c7\u2020 \u00af\u03c3\u00b5\u03c8. (35.29) \u2212 In the third line, we used the hermiticity of the matrices \u00af\u03c3\u00b5 = (I, ~\u03c3). We will get considerably more practice with this notation in the follow- ing sections. Problems 35.1) Verify that eq. (35.20) follows from eqs. (35.2) and (35.19). Hint: write everything in \u201cmatrix multiplication\u201d order, and note that, numerically, \u03b5ab = \u03b5ab = i\u03c32. Then make use of the properties of the Pauli matrices. \u2212 35.2) Verify that eq. (35.21) is consistent with eqs. (34.9) and (34.10). 35.3) Verify that eq. (35.22) is consistent with eq. (34.17). 35.4) Verify eq. (35.5). 36: Lagrangians for Spinor Fields 226 36 Lagrangians for Spinor Fields Prerequisite: 22, 35 Suppose we have a left-handed spinor \ufb01eld \u03c8a. We would like to \ufb01nd a suitable lagrangian for it. This lagrangian must be Lorentz invariant, and it must be hermitian. We would also like it to be quadratic in \u03c8 and its hermitian conjugate \u03c8\u2020\u02d9a, because this will lead to a linear equation of motion, with plane-wave solutions. We want plane-wave solutions because these describe free particles, the starting point for a theory of interacting particles. Let us begin with terms with no derivatives. The only possibility is \u03c8\u03c8 = \u03c8a\u03c8a = \u03b5ab\u03c8b\u03c8a, plus its hermitian conjugate. Because of anticom\u03c8a\u03c8b), this expression does not vanish (as mutation of the \ufb01elds (\u03c8b\u03c8a = it would if the \ufb01elds commuted), and so we can use it as a term in \u2212. Next we need a term with derivatives. The obvious choice is \u2202\u00b5\u03c8\u2202\u00b5\u03c8, plus its hermitian conjugate. This, however, yields a hamiltonian that is unbounded below, which is unacceptable. To get a bounded hamiltonian, the kinetic term must involve both \u03c8 and \u03c8\u2020. A candidate is i\u03c8\u2020 \u00af\u03c3\u00b5\u2202\u00b5\u03c8. This not hermitian, but L (i\u03c8\u2020 \u00af\u03c3\u00b5\u2202\u00b5\u03c8)\u2020 = (i\u03c8\u2020\u02d9a \u00af\u03c3\u00b5 \u02d9ac\u2202\u00b5\u03c8c)\u2020 = \u2212 i\u2202\u00b5\u03c8\u2020\u02d9c (\u00af\u03c3\u00b5a \u02d9c)\u2217\u03c8a i\u2202\u00b5\u03c8\u2020\u02d9c \u00af\u03c3\u00b5 \u02d9ca\u03c8a = = i\u03c8\u2020\u02d9c \u00af\u03c3\u00b5 \u02d9ca\u2202\u00b5\u03c8a \u2212 = i\u03c8\u2020 \u00af\u03c3", "\u00b5\u2202\u00b5\u03c8 \u2212 \u2212 i\u2202\u00b5(\u03c8\u2020\u02d9c \u00af\u03c3\u00b5 \u02d9ca\u03c8a). i\u2202\u00b5(\u03c8\u2020 \u00af\u03c3\u00b5\u03c8). (36.1) In the third line, we used the hermiticity of the matrices \u00af\u03c3\u00b5 = (I, ~\u03c3). In the fourth line, we used \u2202(AB). In the last line, the second term is a total divergence, and vanishes (with suitable boundary conditions on the \ufb01elds at in\ufb01nity) when we integrate it over d4x to get the action S. Thus i\u03c8\u2020 \u00af\u03c3\u00b5\u2202\u00b5\u03c8 has the hermiticity properties necessary for a term in (\u2202A)B = A\u2202B \u2212 \u2212 \u2212. L Our complete lagrangian for \u03c8 is then = i\u03c8\u2020 \u00af\u03c3\u00b5\u2202\u00b5\u03c8 1 2 m\u03c8\u03c8 \u2212 \u2212 1 2 m\u2217\u03c8\u2020\u03c8\u2020, L (36.2) where m is a complex parameter with dimensions of mass. The phase of m i\u03b1/2 \u02dc\u03c8 in eq. (36.2); is actually irrelevant: if m = then we get a lagrangian for \u02dc\u03c8 that is identical to eq. (36.2), but with m. So we can, without loss of generality, take m to be real replaced by | ei\u03b1, we can set \u03c8 = e\u2212 | m | m | 36: Lagrangians for Spinor Fields 227 and positive in the \ufb01rst place, and that is what we will do, setting m\u2217 = m in eq. (36.2). The equation of motion for \u03c8 is then 0 = \u03b4S \u03b4\u03c8\u2020 \u2212 = \u2212 i\u00af\u03c3\u00b5\u2202\u00b5\u03c8 + m\u03c8\u2020, Restoring the spinor indices, this reads 0 = i\u00af\u03c3\u00b5 \u02d9ac\u2202\u00b5\u03c8c + m\u03c8\u2020 \u02d9a. \u2212 Taking the hermitian conjugate (or, equivalently, computing get (36.3) (36.4) \u03b4S/\u03b4\u03c8), we \u2212 0 = +i(\u00af\u03c3\u00b5a \u02d9c)\u2217 \u2202\u00b5\u03c8\u2020\u02d9c + m\u03c8a = +i\u00af\u03c3\u00b5 \u02d9ca\u2202\u00b5\u03c8\u2020\u02d9c + m\u03c8a \u02d9c + m\u03c8a. a \u02d9c \u2202\u00b5\u03c8\u2020 = In the second line, we used the hermiticity of the matrices \u00af\u03c3\u00b5 = (I, In the third, we lowered the undotted index, and switched \u02d9c gives an extra minus sign. \u02d9c to \u02d9c i\u03c3\u00b5 \u2212 (36.5) ~\u03c3). \u2212 \u02d9c, which Eqs. (36.5) and (36.4) can be combined to read c m\u03b4a i\u00af\u03c3\u00b5 \u02d9ac \u2202\u00b5 \u2212 \u2212 a \u02d9c \u2202\u00b5 i\u03c3\u00b5 m\u03b4 \u02d9a \u03c8c \u02d9c! \u03c8\u2020 \u02d9c! = 0. (36.6) We can write this more compactly by introducing the 4 \u03b3\u00b5 0 \u03c3\u00b5 a \u02d9c \u2261 \u00af\u03c3\u00b5 \u02d9ac 0!. Using the sigma-matrix relations, (\u03c3\u00b5 \u00af\u03c3\u03bd + \u03c3\u03bd \u00af\u03c3\u00b5)a (\u00af\u03c3\u00b5\u03c3\u03bd + \u00af\u03c3\u03bd\u03c3\u00b5) \u02d9a c = \u02d9c = 2g\u00b5\u03bd \u03b4a 2g\u00b5\u03bd \u03b4 \u02d9a c, \u02d9c, \u2212 \u2212 4 gamma matrices \u00d7 (36.7) (36.8) which are most easily derived from the numerical formulae \u03c3\u00b5 \u00af\u03c3\u00b5 \u02d9aa = (I, ~\u03c3), we see that the gamma matrices obey a \u02d9a = (I, ~\u03c3) and \u2212 \u03b3\u00b5, \u03b3\u03bd { } = \u2212 2g\u00b5\u03bd, (36.9) where A, B { understood 4 a four-component Majorana \ufb01eld } \u2261 \u00d7 AB + BA denotes the anticommutator, and there is an 4 identity matrix on the right-hand side. We also introduce \u03a8 \u03c8c \u2261 \u03c8\u2020 \u02d9c!. (36.10) 36: Lagrangians for Spinor Fields Then eq. (", "36.6) becomes i\u03b3\u00b5\u2202\u00b5 + m)\u03a8 = 0. ( \u2212 228 (36.11) This is the Dirac equation. We \ufb01rst encountered it in section 1, where the gamma matrices were given di\ufb00erent names (\u03b2 = \u03b30 and \u03b1k = \u03b30\u03b3k). Also, in section 1 we were trying (and failing) to interpret \u03a8 as a wave function, rather than as a quantum \ufb01eld. Now consider a theory of two left-handed spinor \ufb01elds with an SO(2) symmetry, = i\u03c8\u2020i \u00af\u03c3\u00b5\u2202\u00b5\u03c8i \u2212 1 2 m\u03c8i\u03c8i \u2212 L 1 2 m\u03c8\u2020i \u03c8\u2020i, (36.12) where the spinor indices are suppressed and i = 1, 2 is implicitly summed. As in the analogous case of two scalar \ufb01elds discussed in sections 22 and 23, this lagrangian is invariant under the SO(2) transformation \u03c81 \u03c82! \u2192 \u2212 cos \u03b1 sin \u03b1 \u03c81 sin \u03b1 cos \u03b1! \u03c82!. (36.13) We can write the lagrangian so that the SO(2) symmetry appears as a U(1) symmetry instead; let \u03c7 = 1 \u221a2 \u03be = 1 \u221a2 (\u03c81 + i\u03c82), (\u03c81 \u2212 i\u03c82). (36.14) (36.15) In terms of these \ufb01elds, we have = i\u03c7\u2020\u00af\u03c3\u00b5\u2202\u00b5\u03c7 + i\u03be\u2020 \u00af\u03c3\u00b5\u2202\u00b5\u03be L m\u03c7\u03be \u2212 \u2212 m\u03be\u2020\u03c7\u2020. (36.16) Eq. (36.16) is invariant under the U(1) version of eq. (36.13), i\u03b1\u03c7, e\u2212 e+i\u03b1\u03be. \u03c7 \u03be \u2192 \u2192 (36.17) Next, let us derive the equations of motion that we get from eq. (36.16), following the same procedure that ultimately led to eq. (36.6). The result is c m\u03b4a i\u00af\u03c3\u00b5 \u02d9ac \u2202\u00b5 \u2212 a \u02d9c \u2202\u00b5 i\u03c3\u00b5 m\u03b4 \u02d9a \u02d9c! \u02d9c! \u03be\u2020 \u03c7c \u2212 = 0. (36.18) We can now de\ufb01ne a four-component Dirac \ufb01eld \u03a8 \u03c7c \u2261 \u02d9c! \u03be\u2020, (36.19) 36: Lagrangians for Spinor Fields 229 which obeys the Dirac equation, eq. (36.11). (We have annoyingly used the same symbol \u03a8 to denote both a Majorana \ufb01eld and a Dirac \ufb01eld; these are di\ufb00erent objects, and so we must always announce which is meant when we write \u03a8.) We can also write the lagrangian, eq. (36.16), in terms of the Dirac \ufb01eld \u03a8, eq. (36.19). First we take the hermitian conjugate of \u03a8 to get Introduce the matrix \u03a8\u2020 = (\u03c7\u2020\u02d9a, \u03bea). \u03b2 0 \u03b4 \u02d9a \u02d9c \u2261 c \u03b4a 0!. (36.20) (36.21) Numerically, \u03b2 = \u03b30. However, the spinor index structure of \u03b2 and \u03b30 is di\ufb00erent, and so we will distinguish them. Given \u03b2, we de\ufb01ne \u03a8 \u2261 \u03a8\u2020\u03b2 = (\u03bea, \u03c7\u2020\u02d9a). \u03a8\u03a8 = \u03bea\u03c7a + \u03c7\u2020\u02d9a\u03be\u2020 \u02d9a. (36.22) (36.23) Then we have Also, \u03a8\u03b3\u00b5\u2202\u00b5\u03a8 = \u03bea\u03c3\u00b5 (\u2202A)B + \u2202(AB), the \ufb01rst term on the right-hand side of \u02d9c + \u03c7\u2020\u02d9a \u00af\u03c3\u00b5 \u02d9ac \u2202\u00b5\u03c7c. a \u02d9c \u2202\u00b5\u03be\u2020 (36.24) Using A\u2202B = \u2212 eq. (36.24) can be rewritten as \u03bea\u03c3\u00b5 a \u02d9c \u2202\u00b5\u03be\u2020 \u02d9c = (\u2202\u00b5\u03bea)\u03c3", "\u00b5 a \u02d9c \u03be\u2020 \u02d9c + \u2202\u00b5(\u03bea\u03c3\u00b5 a \u02d9c \u03be\u2020 \u02d9c). (36.25) \u2212 Then the \ufb01rst term on the right-hand side of eq. (36.25) can be rewritten as (\u2202\u00b5\u03bea)\u03c3\u00b5 a \u02d9c \u03be\u2020 \u02d9c = +\u03be\u2020 \u02d9c\u03c3\u00b5 a \u02d9c \u2202\u00b5\u03bea = +\u03be\u2020\u02d9c \u00af\u03c3\u00b5 \u02d9ca\u2202\u00b5\u03bea. (36.26) \u2212 Here we used anticommutation of the \ufb01elds to get the \ufb01rst equality, and switched \u02d9c a (thus generating two minus signs) to get the second. Combining eqs. (36.24\u201336.26), we get \u02d9c and a a to a \u02d9c to \u02d9c \u03a8\u03b3\u00b5\u2202\u00b5\u03a8 = \u03c7\u2020\u00af\u03c3\u00b5\u2202\u00b5\u03c7 + \u03be\u2020\u00af\u03c3\u00b5\u2202\u00b5\u03be + \u2202\u00b5(\u03be\u03c3\u00b5\u03be\u2020). (36.27) Therefore, up to an irrelevant total divergence, we have = i\u03a8\u03b3\u00b5\u2202\u00b5\u03a8 L m\u03a8\u03a8. \u2212 (36.28) This form of the lagrangian is invariant under the U(1) transformation i\u03b1 \u03a8, e\u2212 e+i\u03b1 \u03a8, \u03a8 \u03a8 \u2192 \u2192 (36.29) 36: Lagrangians for Spinor Fields 230 which, given eq. (36.19), is the same as eq. (36.17). The Noether current associated with this symmetry is j\u00b5 = \u03a8\u03b3\u00b5\u03a8 = \u03c7\u2020 \u00af\u03c3\u00b5\u03c7 \u03be\u2020\u00af\u03c3\u00b5\u03be. \u2212 (36.30) In quantum electrodynamics, the electromagnetic current is e\u03a8\u03b3\u00b5\u03a8, where e is the charge of the electron. As in the case of a complex scalar \ufb01eld with a U(1) symmetry, there is an additional discrete symmetry, called charge conjugation, that enlarges SO(2) to O(2). Charge conjugation simply exchanges \u03c7 and \u03be. We can de\ufb01ne a unitary charge conjugation operator C that implements this, C \u2212 C \u2212 1\u03c7a(x)C = \u03bea(x), 1\u03bea(x)C = \u03c7a(x), (36.31) where, for the sake of precision, we have restored the spinor index and spacetime argument. We then have C \u2212 (x)C = (x). 1 To express eq. (36.31) in terms of the Dirac \ufb01eld, eq. (36.19), we \ufb01rst L L introduce the charge conjugation matrix \u03b5ac 0 C \u2261 0 \u03b5 \u02d9a \u02d9c!. (36.32) Next we notice that, if we take the transpose of \u03a8, eq. (36.22), we get \u03a8 T = \u03bea \u03c7\u2020\u02d9a!. (36.33) Then, if we multiply by conjugate of \u03a8, C, we get a \ufb01eld that we will call \u03a8C, the charge \u03a8C \u2261 C \u03a8 T = \u03bea \u03c7\u2020 \u02d9a!. (36.34) We see that \u03a8C is the same as the original \ufb01eld \u03a8, eq. (36.19), except that the roles of \u03c7 and \u03be have been switched. We therefore have C \u2212 1\u03a8(x)C = \u03a8C(x) (36.35) for a Dirac \ufb01eld. The charge conjugation matrix has a number of useful properties. As a numerical matrix, it obeys C (36.36) 36: Lagrangians for Spinor Fields and we can also write it as = C \u03b5ac \u2212 0 0 \u03b5 \u02d9a \u02d9c! \u2212. A result that we will need later is \u03c3\u00b5 b \u02d9c \u03b5ce 0 0! 0 \u03b5 \u02d9c \u02d9e! 1\u03b3\u00b5 \u2212 = C C \u03b5ab 0 0 \u03b5 \u02d9a\u02d9b! 0 \u00af\u03c3\u00b5\u02d9bc \u03b5ab\u03c3\u00b5 0 \u03b5 \u02d9a\u02d9b\u00af\u03c3\u00b5\u02d9", "bc\u03b5ce 0 \u03c3\u00b5 \u02d9ae \u2212 \u00af\u03c3\u00b5a \u02d9e \u2212 0! b \u02d9c\u03b5 \u02d9c \u02d9e! 0. = = 231 (36.37) (36.38) The minus signs in the last line come from raising or lowering an index by contracting with the \ufb01rst (rather than the second) index of an \u03b5 symbol. Comparing with 0 \u03c3\u00b5 e \u02d9a \u00af\u03c3\u00b5 \u02d9ea 0!, (36.39) \u03b3\u00b5 = we see that 1\u03b3\u00b5 \u2212 = (\u03b3\u00b5)T. (36.40) C C \u2212 Now let us return to the Majorana \ufb01eld, eq. (36.10). It is obvious that a Majorana \ufb01eld is its own charge conjugate, that is, \u03a8C = \u03a8. This condition is analogous to the condition \u03d5\u2020 = \u03d5 that is satis\ufb01ed by a real scalar \ufb01eld. A Dirac \ufb01eld, with its U(1) symmetry, is analogous to a complex scalar \ufb01eld, while a Majorana \ufb01eld, which has no U(1) symmetry, is analogous to a real scalar \ufb01eld. We can write our original lagrangian for a single left-handed spinor \ufb01eld, eq. (36.2), in terms of a Majorana \ufb01eld, eq. (36.10), by retracing eqs. (36.20\u2013 \u03c8. The result is 36.28) with \u03c7 \u03c8 and \u03be \u2192 \u2192 = i L 2 \u03a8\u03b3\u00b5\u2202\u00b5\u03a8 1 2 m\u03a8\u03a8. \u2212 (36.41) However, we cannot yet derive the equation of motion from eq. (36.41) because it does not yet incorporate the Majorana condition \u03a8C = \u03a8. To \u03a8 T remedy this, we use eq. (36.36) to write the Majorana condition \u03a8 =. Then we can replace \u03a8 in eq. (36.41) by \u03a8T as \u03a8 = \u03a8T to get C C C = i 2 \u03a8T C L \u03b3\u00b5\u2202\u00b5\u03a8 1 2 m\u03a8T \u03a8. C \u2212 (36.42) The equation of motion that follows from this lagrangian is once again the Dirac equation. 36: Lagrangians for Spinor Fields 232 We can also recover the Weyl components of a Dirac or Majorana \ufb01eld by means of a suitable projection matrix. De\ufb01ne \u03b35 \u2261 c \u03b4a \u2212 0 0 +\u03b4 \u02d9a \u02d9c!, (36.43) where the subscript 5 is simply part of the traditional name of this matrix, rather than the value of some index. Then we can de\ufb01ne left and right projection matrices PL \u2261 1 2 (1 \u2212 \u03b35) = PR \u2261 1 2 (1 + \u03b35) = c \u03b4a 0 0 0 0! 0 0 \u03b4 \u02d9a \u02d9c!,. Thus we have, for a Dirac \ufb01eld, PL\u03a8 = PR\u03a8 = \u03c7c 0!, 0 \u02d9c! \u03be\u2020. The matrix \u03b35 can also be expressed as \u03b35 = i\u03b30\u03b31\u03b32\u03b33 i = \u2212 24 \u03b5\u00b5\u03bd\u03c1\u03c3\u03b3\u00b5\u03b3\u03bd\u03b3\u03c1\u03b3\u03c3, (36.44) (36.45) (36.46) where \u03b50123 = 1. \u2212 Finally, let us consider the behavior of a Dirac or Majorana \ufb01eld under a Lorentz transformation. Recall that left- and right-handed spinor \ufb01elds transform according to U (\u039b)\u2212 1\u03c8a(x)U (\u039b) = L(\u039b)a c \u03c8c(\u039b\u2212 1x), U (\u039b)\u2212 1\u03c8\u2020\u02d9a(x)U (\u039b) = R(\u039b) \u02d9a \u02d9c \u03c8\u2020\u02d9c(\u039b\u2212 1x), where, for an in\ufb01nitesimal transformation \u039b\u00b5 \u03bd = \u03b4\u00b5 \u03bd + \u03b4\u03c9\u00b5 \u03bd, L(1+\u03b4\u03c9)a c = \u03b4a R(1+\u03b4\u03c9) \u02d9a \u02d9c = \u03b4 \u02d9a c + i \u02d9c + i L )", "a 2 \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd 2 \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd R ) \u02d9a c, \u02d9c, (36.47) (36.48) (36.49) (36.50) 36: Lagrangians for Spinor Fields and where (S\u00b5\u03bd L )a c = + i 4 (\u03c3\u00b5 \u00af\u03c3\u03bd 4 (\u00af\u03c3\u00b5\u03c3\u03bd i (S\u00b5\u03bd R ) \u02d9a \u02d9c = \u2212 233 (36.51) (36.52) \u03c3\u03bd \u00af\u03c3\u00b5)a c, \u00af\u03c3\u03bd \u03c3\u00b5) \u02d9a \u02d9c. \u2212 \u2212 From these formulae, and the de\ufb01nition of \u03b3\u00b5, eq. (36.7), we can see that i 4 [\u03b3\u00b5, \u03b3\u03bd ] = +(S\u00b5\u03bd L )a c 0 0 (S\u00b5\u03bd R ) \u02d9a \u2212 S\u00b5\u03bd. \u02d9c! \u2261 (36.53) Then, for either a Dirac or Majorana \ufb01eld \u03a8, we can write U (\u039b)\u2212 1\u03a8(x)U (\u039b) = D(\u039b)\u03a8(\u039b\u2212 1x), (36.54) where, for an in\ufb01nitesimal transformation, the 4 D(1+\u03b4\u03c9) = 1 + i \u00d7 2 \u03b4\u03c9\u00b5\u03bd S\u00b5\u03bd, 4 matrix D(\u039b) is (36.55) with S\u00b5\u03bd given by eq. (36.53). The minus sign in front of S\u00b5\u03bd is compensated by the switch from a \u02d9c contraction in eq. (36.54). \u02d9c contraction in eq. (36.50) to a \u02d9c R in eq. (36.53) \u02d9c Problems 36.1) Using the results of problem 2.9, show that, for a rotation by an angle \u03b8 about the z axis, we have D(\u039b) = exp( \u2212 i\u03b8S12), (36.56) and that, for a boost by rapidity \u03b7 in the z direction, we have D(\u039b) = exp(+i\u03b7S30). (36.57) 36.2) Verify that eq. (36.46) is consistent with eq. (36.43). 36.3) a) Prove the Fierz identities (\u03c7\u20201 \u00af\u03c3\u00b5\u03c72)(\u03c7\u20203 \u00af\u03c3\u00b5\u03c74) = 2(\u03c7\u20201\u03c7\u20203)(\u03c72\u03c74), \u2212 (\u03c7\u20201 \u00af\u03c3\u00b5\u03c72)(\u03c7\u20203 \u00af\u03c3\u00b5\u03c74) = (\u03c7\u20201 \u00af\u03c3\u00b5\u03c74)(\u03c7\u20203 \u00af\u03c3\u00b5\u03c72). (36.58) (36.59) b) De\ufb01ne the Dirac \ufb01elds \u03a8i \u2261 \u03c7i \u03be\u2020i!, \u03a8C i \u2261 \u03bei \u03c7\u2020i!. (36.60) 36: Lagrangians for Spinor Fields 234 Use eqs. (36.58) and (36.59) to prove the Dirac form of the Fierz identities, (\u03a81\u03b3\u00b5PL\u03a82)(\u03a83\u03b3\u00b5PL\u03a84) = \u2212 2(\u03a81PR\u03a8C 3 )(\u03a8C 4 PL\u03a82), (36.61) (\u03a81\u03b3\u00b5PL\u03a82)(\u03a83\u03b3\u00b5PL\u03a84) = (\u03a81\u03b3\u00b5PL\u03a84)(\u03a83\u03b3\u00b5PL\u03a82). (36.62) c) By writing both sides out in terms of Weyl \ufb01elds, show that \u03a81\u03b3\u00b5PR\u03a82 = \u03a8C 2 \u03b3\u00b5PL\u03a8C 1, \u2212 \u03a81PL\u03a82 = +\u03a8C 2 PL\u03a8C 1, \u03a81PR\u03a82 = +\u03a8C 2 PR\u03a8C 1. (36.63) (36.64) (36.65) Combining eqs. (36.63\u201336.65) with eqs. (36.61\u201336.62) yields more useful forms of the Fierz identities. 36.4) Consider a \ufb01eld \u03d5A(x) in an unspeci\ufb01ed representation of the Lorent", "z group, indexed by A, that obeys U (\u039b)\u2212 1\u03d5A(x)U (\u039b) = LA B(\u039b)\u03d5B(\u039b\u2212 1x). (36.66) For an in\ufb01nitesimal transformation, LA B(1+\u03b4\u03c9) = \u03b4A B + i 2 \u03b4\u03c9\u00b5\u03bd (S\u00b5\u03bd )A B. (36.67) a) Following the procedure of section 22, show that the energy-momentum tensor is T \u00b5\u03bd = g\u00b5\u03bd L \u2212 \u2202 L \u2202(\u2202\u00b5\u03d5A) \u2202\u03bd\u03d5A. (36.68) b) Show that the Noether current corresponding to a Lorentz transformation is \u00b5\u03bd\u03c1 = x\u03bd T \u00b5\u03c1 M \u2212 x\u03c1T \u00b5\u03bd + B\u00b5\u03bd\u03c1, where B\u00b5\u03bd\u03c1 i \u2202 L \u2202(\u2202\u00b5\u03d5A) (S\u03bd\u03c1)A B\u03d5B. \u2261 \u2212 c) Use the conservation laws \u2202\u00b5T \u00b5\u03bd = 0 and \u2202\u00b5M that (36.69) (36.70) \u00b5\u03bd\u03c1 = 0 to show T \u03bd\u03c1 \u2212 T \u03c1\u03bd + \u2202\u00b5B\u00b5\u03bd\u03c1 = 0. (36.71) d) De\ufb01ne the improved energy-momentum tensor or Belinfante tensor \u0398\u00b5\u03bd \u2261 T \u00b5\u03bd + 1 2 \u2202\u03c1(B\u03c1\u00b5\u03bd B\u00b5\u03c1\u03bd \u2212 \u2212 B\u03bd\u03c1\u00b5). (36.72) 36: Lagrangians for Spinor Fields 235 Show that \u0398\u00b5\u03bd is symmetric: \u0398\u00b5\u03bd = \u0398\u03bd\u00b5. Also show that \u0398\u00b5\u03bd is d3x T 0\u03bd = P \u03bd, where conserved, \u2202\u00b5\u0398\u00b5\u03bd = 0, and that P \u03bd is the energy-momentum four-vector. (In general relativity, it is R the Belinfante tensor that couples to gravity.) d3x \u03980\u03bd = R e) Show that the improved tensor \u039e\u00b5\u03bd\u03c1 \u2261 x\u03bd\u0398\u00b5\u03c1 \u2212 x\u03c1\u0398\u00b5\u03bd (36.73) obeys \u2202\u00b5\u039e\u00b5\u03bd\u03c1 = 0, and that M \u03bd\u03c1 are the Lorentz generators. f) Compute \u0398\u00b5\u03bd for a left-handed Weyl \ufb01eld with given by eq. (36.28). and for a Dirac \ufb01eld with d3x \u039e0\u03bd\u03c1 = d3x R R L M L 0\u03bd\u03c1 = M \u03bd\u03c1, where given by eq. (36.2), 36.5) Symmetries of fermion \ufb01elds. (Prerequisite: 24.) Consider a theory with N massless Weyl \ufb01elds \u03c8j, = i\u03c8\u2020j \u03c3\u00b5\u2202\u00b5\u03c8j, L (36.74) where the repeated index j is summed. This lagrangian is clearly invariant under the U(N ) transformation, \u03c8j \u2192 Ujk\u03c8k, (36.75) where U is a unitary matrix. State the invariance group for the following cases: a) N Weyl \ufb01elds with a common mass m, = i\u03c8\u2020j \u03c3\u00b5\u2202\u00b5\u03c8j \u2212 L 1 2 m(\u03c8j\u03c8j + \u03c8\u2020j \u03c8\u2020j ). (36.76) b) N massless Majorana \ufb01elds, = i 2 \u03a8T j C L \u03b3\u00b5\u2202\u00b5\u03a8j. c) N Majorana \ufb01elds with a common mass m, = i 2 \u03a8T j C \u03b3\u00b5\u2202\u00b5\u03a8j \u2212 1 2 m\u03a8T j C L \u03a8j. d) N massless Dirac \ufb01elds, = i\u03a8j\u03b3\u00b5\u2202\u00b5\u03a8j. L e) N Dirac \ufb01elds with a common mass m, = i\u03a8j\u03b3\u00b5\u2202\u00b5\u03a8j \u2212 L m\u03a8j\u03a8j. (36.77) (36.78) (36.79) (36.80) 37: Canonical Quantization of Spinor Fields I 236 37 Canonical Quantization of Spinor Fields I Prerequisite: 36 Consider a left-handed Weyl \ufb01eld \u03c8 with lagrangian = i\ufffd", "\ufffd\u2020 \u00af\u03c3\u00b5\u2202\u00b5\u03c8 L \u2212 1 2 m(\u03c8\u03c8 + \u03c8\u2020\u03c8\u2020). (37.1) The canonically conjugate momentum to the \ufb01eld \u03c8a(x) is then1 The hamiltonian is H \u03c0a(x) \u2202 L \u2202(\u22020\u03c8a(x)) \u2261 = i\u03c8\u2020\u02d9a(x)\u00af\u03c30 \u02d9aa. = \u03c0a\u22020\u03c8a \u2212 L = i\u03c8\u2020\u02d9a \u00af\u03c30 \u02d9aa\u03c8a \u2212 L i\u03c8\u2020 \u00af\u03c3i\u2202i\u03c8 + 1 = \u2212 2 m(\u03c8\u03c8 + \u03c8\u2020\u03c8\u2020). The appropriate canonical anticommutation relations are \u03c8a(x, t), \u03c8c(y, t) } { \u03c8a(x, t), \u03c0c(y, t) } { Substituting in eq. (37.2) for \u03c0c, we get = 0, = i\u03b4a c \u03b43(x y). \u2212 \u00af\u03c30 \u02d9cc = \u03b4a \u03c8a(x, t), \u03c8\u2020\u02d9c (y, t) } { Then, using \u00af\u03c30 = \u03c30 = I, we have c \u03b43(x y). \u2212 or, equivalently, \u03c8a(x, t), \u03c8\u2020\u02d9c (y, t) } { = \u03c30 a \u02d9c \u03b43(x y), \u2212 \u03c8a(x, t), \u03c8\u2020 { \u02d9c(y, t) } = \u00af\u03c30 \u02d9ca \u03b43(x y). \u2212 (37.2) (37.3) (37.4) (37.5) (37.6) (37.7) (37.8) We can also translate this into four-component notation for either a Dirac or a Majorana \ufb01eld. A Dirac \ufb01eld is de\ufb01ned in terms of two lefthanded Weyl \ufb01elds \u03c7 and \u03be via \u03a8 \u03c7c \u2261 \u02d9c! \u03be\u2020. (37.9) 1Here we gloss over a subtlety about di\ufb00erentiating with respect to an anticommuting object; we will take up this topic in section 44, and for now simply assume that eq. (37.2) is correct. 37: Canonical Quantization of Spinor Fields I We also de\ufb01ne where The lagrangian is \u03a8 \u2261 \u03a8\u2020\u03b2 = (\u03bea, \u03c7\u2020\u02d9a), \u03b2 0 \u03b4 \u02d9a \u02d9c \u2261 c \u03b4a 0!. = i\u03c7\u2020\u00af\u03c3\u00b5\u2202\u00b5\u03c7 + i\u03be\u2020 \u00af\u03c3\u00b5\u2202\u00b5\u03be L m(\u03c7\u03be + \u03be\u2020\u03c7\u2020) \u2212 = i\u03a8\u03b3\u00b5\u2202\u00b5\u03a8 m\u03a8\u03a8. \u2212 237 (37.10) (37.11) (37.12) The \ufb01elds \u03c7 and \u03be each obey the canonical anticommutation relations of eq. (37.5). This translates into \u03a8\u03b1(x, t), \u03a8\u03b2 (y, t) { } \u03a8\u03b1(x, t), \u03a8\u03b2(y, t) } { = 0, = (\u03b30)\u03b1\u03b2 \u03b43(x y), \u2212 where \u03b1 and \u03b2 are four-component spinor indices, and \u03b3\u00b5 0 \u03c3\u00b5 a \u02d9c \u2261 \u00af\u03c3\u00b5 \u02d9ac 0!. (37.13) (37.14) (37.15) Eqs. (37.13) and (37.14) can also be derived directly from the four-component form of the lagrangian, eq. (37.12), by noting that the canonically conjugate momentum to the \ufb01eld \u03a8 is \u2202 /\u2202(\u22020\u03a8) = i\u03a8\u03b30, and that (\u03b30)2 = 1. A Majorana \ufb01eld is de\ufb01ned in terms of a single left-handed Weyl \ufb01eld L \u03c8 via We also de\ufb01ne \u03a8 \u03c8c \u2261 \u03c8\u2020 \u02d9c!. \u03a8\u2020\u03b2 = (\u03c8a, \u03c8\u2020\u02d9a). A Majorana \ufb01eld obeys the Majorana condition \u2261 \u03a8 \u03a8 =", " \u03a8T, C where \u03b5ac \u2212 0 C \u2261 0 \u03b5 \u02d9a \u02d9c! \u2212 is the charge conjugation matrix. The lagrangian is L \u2212 = i = i\u03c8\u2020 \u00af\u03c3\u00b5\u2202\u00b5\u03c8 2 \u03a8\u03b3\u00b5\u2202\u00b5\u03a8 2 \u03a8T = i \u2212 \u03b3\u00b5\u2202\u00b5\u03a8 C 1 2 m(\u03c8\u03c8 + \u03c8\u2020\u03c8\u2020) 1 2 m\u03a8\u03a8 1 2 m\u03a8T \u03a8. \u2212 C (37.16) (37.17) (37.18) (37.19) (37.20) 37: Canonical Quantization of Spinor Fields I 238 The \ufb01eld \u03c8 obeys the canonical anticommutation relations of eq. (37.5). This translates into \u03a8\u03b1(x, t), \u03a8\u03b2 (y, t) { } \u03a8\u03b1(x, t), \u03a8\u03b2(y, t) } { \u03b30)\u03b1\u03b2 \u03b43(x = ( C = (\u03b30)\u03b1\u03b2 \u03b43(x y), \u2212 y), \u2212 (37.21) (37.22) where \u03b1 and \u03b2 are four-component spinor indices. To derive eqs. (37.21) and (37.22) directly from the four-component form of the lagrangian, eq. (37.20), requires new formalism for the quantization of constrained systems. This is because the canonically conjugate momentum to the \ufb01eld \u03a8 is \u2202 /\u2202(\u22020\u03a8) = i \u03b30, and this is linearly related to \u03a8 itself; this relation constitutes a 2 \u03a8T constraint that must be solved before imposition of the anticommutation relations. In this case, solving the constraint simply returns us to the Weyl formalism with which we began. L C The equation of motion that follows from either eq. (37.12) or eq. (37.20) is the Dirac equation, i/\u2202 + m)\u03a8 = 0. (37.23) ( \u2212 Here we have introduced the Feynman slash: given any four-vector a\u00b5, we de\ufb01ne To solve the Dirac equation, we \ufb01rst note that if we act on it with a\u00b5\u03b3\u00b5. /a \u2261 (37.24) i/\u2202 + m, we get 0 = (i/\u2202 + m)( i/\u2202 + m)\u03a8 \u2212 = (/\u2202 /\u2202 + m2)\u03a8 = ( \u22022 + m2)\u03a8. \u2212 Here we have used /a/a = a\u00b5a\u03bd\u03b3\u00b5\u03b3\u03bd = a\u00b5a\u03bd 1 2 { 2 [\u03b3\u00b5, \u03b3\u03bd ] \u03b3\u00b5, \u03b3\u03bd + 1 } 2 [\u03b3\u00b5, \u03b3\u03bd ] g\u00b5\u03bd + 1 \u2212 a\u00b5a\u03bdg\u00b5\u03bd + 0 a2. = a\u00b5a\u03bd = = \u2212 \u2212 (37.25) (37.26) From eq. (37.25), we see that \u03a8 obeys the Klein-Gordon equation. Therefore, the Dirac equation has plane-wave solutions. Let us consider a speci\ufb01c solution of the form \u03a8(x) = u(p)eipx + v(p)e\u2212 ipx. (37.27) 37: Canonical Quantization of Spinor Fields I 239 where p0 = \u03c9 constant spinors. Plugging eq. (37.27) into the eq. (37.23), we get (p2 + m2)1/2, and u(p) and v(p) are four-component \u2261 (/p + m)u(p)eipx + ( \u2212 /p + m)v(p)e\u2212 ipx = 0. (37.28) Thus we require (/p + m)u(p) = 0, /p + m)v(p) = 0. ( \u2212 (37.29) Each of these equations has two linearly independent solutions that we (p); their detailed properties will be worked out in will call u the next section. The general solution of the Dirac equation can then be written as (p) and v \u00b1 \u00b1 \u03a8(x) = dp s= X \u00b1 Z f bs(p)us(p)eipx + d\u2020s(p)vs(p)e\u2212 h ipx, i (37.30) where the integration measure is as usual d3p (2\u03c0)32\u03c9. dp \u2261 f Problems (37.31)", " 37.1) Verify that eqs. (37.13) and (37.14) follow from eqs. (37.4) and (37.5). 38: Spinor Technology 240 38 Spinor Technology Prerequisite: 37 The four-component spinors us(p) and vs(p) obey the equations (/p + m)us(p) = 0, /p + m)vs(p) = 0. ( \u2212 (38.1) Each of these equations has two solutions, which we label via s = + and = 0, we can go to the rest frame, p = 0. We will then s = distinguish the two solutions by the eigenvalue of the spin matrix. For m \u2212 Sz = i 4 [\u03b31, \u03b32] = i 2 \u03b31\u03b32 = Speci\ufb01cally, we will require 1 2 \u03c33 0 0 1 2 \u03c33!. (0), Szu Szv \u00b1 \u00b1 (0) = (00). (38.2) (38.3) The reason for the opposite sign for the v spinor is that this choice results in [Jz, b\u2020 \u00b1 [Jz, d\u2020 \u00b1 (0)] = (0)] = \u00b1 \u00b1 1 2 b\u2020 \u00b1 1 2 d\u2020 \u00b1 (0), (0), (38.4) where Jz is the z component of the angular momentum operator. Eq. (38.4) implies that b\u2020+(0) and d\u2020+(0) each creates a particle with spin up along the z axis. We will verify eq. (38.4) in problem 39.2. For p = 0, we have /p = \u2212 m\u03b30, where \u03b30 = 0 I I 0!. (38.5) Eqs. (38.1) and (38.3) are then easy to solve. Choosing (for later conve(0), we nience) a speci\ufb01c normalization and phase for each of u get (0) and v \u00b1 \u00b1 u+(0) = \u221am \uf8eb v+(0) = \u221a0) = \u221am \uf8eb u \u2212 \uf8ec \uf8ec \uf8ed (0) = \u221a38.6) \uf8ec \uf8ec \uf8ed 6 38: Spinor Technology For later use we also compute the barred spinors us(p) vs(p) \u2261 \u2261 u\u2020s(p)\u03b2, v\u2020s(p)\u03b2, \u03b2 = 0 I I 0! \u03b2T = \u03b2\u2020 = \u03b2\u2212 1 = \u03b2. where satis\ufb01es We get u+(0) = \u221am (1, 0, 1, 0), (0) = \u221am (0, 1, 0, 1), u \u2212 v+(0) = \u221am (0, v \u2212 (0) = \u221am (1, 0, 1, 0, 1), 1, 0). \u2212 \u2212 241 (38.7) (38.8) (38.9) (38.10) We can now \ufb01nd the spinors corresponding to an arbitrary three-momentum p by applying to us(0) and vs(0) the matrix D(\u039b) that corresponds to an appropriate boost. This is given by D(\u039b) = exp(i\u03b7 \u02c6p where \u02c6p is a unit vector in the p direction, K j = i the boost matrix, and \u03b7 Thus we have (38.11) 2 \u03b3j\u03b30 is 1( p /m) is the rapidity (see problem 2.9). | | 4 [\u03b3j, \u03b30] = i sinh\u2212 K), \u2261 \u00b7 us(p) = exp(i\u03b7 \u02c6p vs(p) = exp(i\u03b7 \u02c6p K)us(0), K)vs(0). \u00b7 \u00b7 We also have us(p) = us(0) exp( vs(p) = vs(0) exp( i\u03b7 \u02c6p i\u03b7 \u02c6p K), \u00b7 K). \u2212 \u2212 \u00b7 (38.12) (38.13) This follows from K j = K j, where for any general combination of gamma matrices, In particular, it turns out that A \u2261 \u03b2A\u2020\u03b2. (38.14) \u03b3\u00b5 = \u03b3\u00b5, S\u00b5\u03bd = S\u00b5\u03bd, i\u03b35 = i\u03b35, \u03b3\u00b5\u03b35 = \u03b3\u00b5\u03b35, i\u03b35S\u00b5\u03bd = i\u03b35S\u00b5\u03bd. (38.15) 38: Spinor Technology The barred spinors satisfy the equations us", "(p)(/p + m) = 0, /p + m) = 0. vs(p)( \u2212 242 (38.16) It is not very hard to work out us(p) and vs(p) from eq. (38.12), but it is even easier to use various tricks that will sidestep any need for the explicit formulae. Consider, for example, us\u2032(p)us(p); from eqs. (38.12) and (38.13), we see that us\u2032(p)us(p) = us\u2032(0)us(0), and this is easy to compute from eqs. (38.6) and (38.10). We \ufb01nd us\u2032(p)us(p) = +2m \u03b4s\u2032s, vs\u2032(p)vs(p) = 2m \u03b4s\u2032s, \u2212 us\u2032(p)vs(p) = 0, vs\u2032(p)us(p) = 0. (38.17) Also useful are the Gordon identities, 2m us\u2032(p\u2032)\u03b3\u00b5us(p) = us\u2032(p\u2032) h (p\u2032 + p)\u00b5 2m vs\u2032(p\u2032)\u03b3\u00b5vs(p) = vs\u2032(p\u2032) h (p\u2032 + p)\u00b5 2iS\u00b5\u03bd (p\u2032 2iS\u00b5\u03bd (p\u2032 \u2212 \u2212 \u2212 \u2212 p)\u03bd p)\u03bd us(p), i vs(p). (38.18) i \u2212 To derive them, start with \u03b3\u00b5 /p = 1 2 { /p \u2032\u03b3\u00b5 = 1 2 { \u03b3\u00b5, /p + 1 2 [\u03b3\u00b5, /p ] = 2 [\u03b3\u00b5, /p \u2032 ] = 1 p\u00b5 p\u2032 2iS\u00b5\u03bd p\u03bd, \u2212 \u00b5 + 2iS\u00b5\u03bd p\u2032\u03bd. \u2212 \u2212 (38.19) (38.20) } } \u2212 \u03b3\u00b5, /p \u2032 Add eqs. (38.19) and (38.20), sandwich them between u \u2032 and u spinors (or v \u2032 and v spinors), and use eqs. (38.1) and (38.16). An important special case is p\u2032 = p; then, using eq. (38.17), we \ufb01nd us\u2032(p)\u03b3\u00b5us(p) = 2p\u00b5\u03b4s\u2032s, vs\u2032(p)\u03b3\u00b5vs(p) = 2p\u00b5\u03b4s\u2032s. With a little more e\ufb00ort, we can also show us\u2032(p)\u03b30vs( vs\u2032(p)\u03b30us( \u2212 \u2212 p) = 0, p) = 0. (38.21) (38.22) Consider now the spin sums We will need eqs. (38.21) and (38.22) in the next section. us(p)us(p) and vs(p)vs(p), each of which is a 4 4 matrix. The sum over eigenstates of Sz should remove any memory of the spin-quantization axis, and so the result should P P \u00d7 s= s= \u00b1 \u00b1 38: Spinor Technology 243 be expressible in terms of the four-vector p\u00b5 and various gamma matrices, m\u03b30, and it is with all vector indices contracted. In the rest frame, /p = vs(0)vs(0) = easy to check that m\u03b30 s= m. We therefore conclude that us(0)us(0) = m\u03b30 + m and \u2212 s= \u00b1 \u00b1 \u2212 P P us(p)us(p) = vs(p)vs(p) = /p + m, \u2212 /p \u2212 \u2212 m. s= X \u00b1 s= X \u00b1 (38.23) We will make extensive use of eq. (38.23) when we calculate scattering cross sections for spin-one-half particles. From eq. (38.23), we can get u+(p)u+(p), etc, by applying appropriate spin projection matrices. In the rest frame, we have 1 2 (1 + 2sSz)us\u2032(0) = \u03b4ss\u2032 us\u2032(0), 1 2sSz)vs\u2032(0) = \u03b4ss\u2032 vs\u2032(0). 2 (1 \u2212 (38.24) In order to boost these projection matrices to a more general frame, we \ufb01rst recall that I 0 \u03b35 \u2261 i\u03b30\u03b31\u03b32\u03b33 = \u2212 0 2 \u03b31\u03b32 as Sz = I! 1 2 \u03b35\u03b3", "3\u03b30. In the rest frame, \u2212 /p/m, and \u03b33 as /z, where z\u00b5 = (0, \u02c6z); thus we have (38.25). This allows us to write Sz = i we can write \u03b30 as \u2212 Sz = 1 2m \u03b35/z /p. (38.26) Now we can boost Sz to any other frame simply by replacing /z and /p with their values in that frame. (Note that, in any frame, z\u00b5 satis\ufb01es z2 = 1 and z p = 0.) Boosting eq. (38.24) then yields \u00b7 1 2 (1 1 2 (1 \u2212 \u2212 s\u03b35/z)us\u2032(p) = \u03b4ss\u2032 us\u2032(p), s\u03b35/z)vs\u2032(p) = \u03b4ss\u2032 vs\u2032(p), (38.27) where we have used eq. (38.1) to eliminate /p. Combining eqs. (38.23) and (38.27) we get us(p)us(p) = 1 vs(p)vs(p) = 1 2 (1 2 (1 \u2212 \u2212 s\u03b35/z)( s\u03b35/z)( \u2212 \u2212 /p + m), m). /p \u2212 (38.28) It is interesting to consider the extreme relativistic limit of this formula. Let us take the three-momentum to be in the z direction, so that it is parallel to the spin-quantization axis. The component of the spin in the 38: Spinor Technology 244 direction of the three-momentum is called the helicity. A fermion with 1/2 helicity +1/2 is said to be right-handed, and a fermion with helicity is said to be left-handed. For rapidity \u03b7, we have \u2212 1 m p\u00b5 = (cosh \u03b7, 0, 0, sinh \u03b7), z\u00b5 = (sinh \u03b7, 0, 0, cosh \u03b7). (38.29) The \ufb01rst equation is simply the de\ufb01nition of \u03b7, and the second follows from z2 = 1 and p z = 0 (along with the knowledge that a boost of a four-vector \u00b7 in the z direction does not change its x and y components). In the limit of large \u03b7, we see that z\u00b5 = 1 m p\u00b5 + O(e\u2212 Hence, in eq. (38.28), we can replace /z with /p/m, and then use the matrix m2. For relation (/p/m)( consistency, we should then also drop the m relative to /p, since it is down by a factor of O(e\u2212 m), which holds for p2 = \u03b7). We get (38.30) m) = ( \u2212 \u03b7). \u2213 \u2212 \u00b1 \u00b1 \u2212 /p /p \u2212 \u2212 \u2212 \u2192 \u2192 /p), /p). us(p)us(p) vs(p)vs(p) 1 2 (1 + s\u03b35)( 1 s\u03b35)( 2 (1 The spinor corresponding to a right-handed fermion (helicity +1/2) is (p) for a d-type particle. According u+(p) for a b-type particle and v to eq. (38.31), either of these is projected by 1 2 (1 + \u03b35) = diag(0, 0, 1, 1) onto the lower two components only. In terms of the Dirac \ufb01eld \u03a8(x), this is the part that corresponds to the right-handed Weyl \ufb01eld. Similarly, lefthanded fermions are projected (in the extreme relativistic limit) onto the upper two spinor components only, corresponding to the left-handed Weyl \ufb01eld. (38.31) \u2212 The case of a massless particle follows from the extreme relativistic limit In particular, eqs. (38.1), (38.16), (38.17), (38.21), of a massive particle. (38.22), and (38.23) are all valid with m = 0, and eq. (38.31) becomes exact. Finally, for our discussion of parity, time reversal, and charge conjugation in section 40, we will need a number of relationships among the u and v spinors. First, note that \u03b2us(0) = +us(0) and \u03b2vs(0) = vs(0). Also, \u03b2K j = K j\u03b2. We then have \u2212 \u2212 us( vs( \u2212 \u2212 p) =", " +\u03b2us(p), \u03b2vs(p). p) = \u2212 Next, we need the charge conjugation matrix 0 +38.32) (38.33) 245 (38.34) (38.35) (38.36) us(0)T = vs(0) (K j)T. From \u2212 38: Spinor Technology which obeys C T = \u2020 =, \u2212 C \u2212C \u2212C 1 = \u03b2, (\u03b3\u00b5)T. = = C \u03b2 C 1\u03b3\u00b5 \u2212 C Using eqs. (38.6), (38.10), and (38.33), we can show that and this we can conclude that vs(0)T = us(0). Also, eq. (38.36) implies 1K j C\u2212 \u2212 C C C C = C us(p)T = vs(p), vs(p)T = us(p). C Taking the complex conjugate of eq. (38.37), and using uT get (38.37) \u2217 = u\u2020 = \u03b2u, we u\u2217s(p) = v\u2217s (p) = \u03b2vs(p), \u03b2us(p). C C Next, note that \u03b35us(0) = +s v \u03b35K j = K j\u03b35. Therefore \u2212 s(0) and \u03b35vs(0) = \u03b35us(p) = +s v \u03b35vs(p) = s u s(p), s(p). \u2212 \u2212 \u2212 Combining eqs. (38.32), (38.38), and (38.39) results in (38.38) s u \u2212 \u2212 s(0), and that (38.39) s( s( u\u2217 \u2212 v\u2217 \u2212 \u2212 p) = p) = \u2212 C s s \u03b35us(p), \u03b35vs(p). \u2212 We will need eq. (38.32) in our discussion of parity, eq. (38.37) in our discussion of charge conjugation, and eq. (38.40) in our discussion of time reversal. \u2212 C (38.40) Problems 38.1) Use eq. (38.12) to compute us(p) and vs(p) explicity. Hint: show 1, and that, for any matrix 1, ecA = (cosh c) + (sinh c)A, where c is an K has eigenvalues \u00b1 that the matrix 2i\u02c6p A with eigenvalues arbitrary complex number. \u00b1 \u00b7 38.2) Verify eq. (38.15). 38.3) Verify eq. (38.22). 38.4) Derive the Gordon identities (p\u2032 + p)\u00b5 (p\u2032 + p)\u00b5 us\u2032(p\u2032) h vs\u2032(p\u2032) h 2iS\u00b5\u03bd (p\u2032 2iS\u00b5\u03bd (p\u2032 \u2212 \u2212 \u2212 \u2212 p)\u03bd p)\u03bd \u03b35us(p) = 0, i \u03b35vs(p) = 0. i (38.41) 39: Canonical Quantization of Spinor Fields II 246 39 Canonical Quantization of Spinor Fields II Prerequisite: 38 A Dirac \ufb01eld \u03a8 with lagrangian obeys the canonical anticommutation relations = i\u03a8/\u2202\u03a8 m\u03a8\u03a8 \u2212 L \u03a8\u03b1(x, t), \u03a8\u03b2 (y, t) { } \u03a8\u03b1(x, t), \u03a8\u03b2(y, t) } { = 0, = (\u03b30)\u03b1\u03b2 \u03b43(x y), \u2212 and has the Dirac equation as its equation of motion. The general solution is i/\u2202 + m)\u03a8 = 0 ( \u2212 (39.1) (39.2) (39.3) (39.4) (39.5) \u03a8(x) = dp s= X \u00b1 Z f bs(p)us(p)eipx + d\u2020s(p)vs(p)e\u2212 h ipx, i where bs(p) and d\u2020s(p) are operators; the properties of the four-component spinors us(p) and vs(p) were belabored in the previous section. Let us express bs(p) and d\u2020s(p) in terms of \u03a8(x) and \u03a8(x). We begin with d3x e\u2212 ipx\u03a8(x) = 1 2\u03c9 bs\u2032(p)us\u2032(p) + 1 2\u03c9 e2i\u03c9td\u2020s\u2032( p)vs\u2032( p). \u2212 \u2212 Z", " i (39.6) Next, multiply on the left by us(p)\u03b30, and use us(p)\u03b30us\u2032(p) = 2\u03c9\u03b4ss\u2032 and us(p)\u03b30vs\u2032( p) = 0 from section 38. The result is \u00b1 h Xs\u2032= \u2212 bs(p) = d3x e\u2212 ipx us(p)\u03b30\u03a8(x). (39.7) Note that bs(p) is time independent. Z To get b\u2020s(p), take the hermitian conjugate of eq. (39.7), using us(p)\u03b30\u03a8(x) i h \u2020 = us(p)\u03b30\u03a8(x) = \u03a8(x)\u03b30us(p) = \u03a8(x)\u03b30us(p), where, for any general combination of gamma matrices A, A \u2261 \u03b2A\u2020\u03b2. (39.8) (39.9) 39: Canonical Quantization of Spinor Fields II 247 Thus we \ufb01nd b\u2020s(p) = d3x eipx \u03a8(x)\u03b30us(p). (39.10) To extract d\u2020s(p) from \u03a8(x), we start with Z Z 1 2\u03c9 e\u2212 d3x eipx\u03a8(x) = 2\u03c9 d\u2020s\u2032(p)vs\u2032(p) i (39.11) Next, multiply on the left by vs(p)\u03b30, and use vs(p)\u03b30vs\u2032(p) = 2\u03c9\u03b4ss\u2032 and vs(p)\u03b30us\u2032( p) = 0 from section 38. The result is p)us\u2032( Xs\u2032= \u00b1 h 2i\u03c9tbs\u2032( p) + 1 \u2212 \u2212. \u2212 d\u2020s(p) = d3x eipx vs(p)\u03b30\u03a8(x). (39.12) Z To get ds(p), take the hermitian conjugate of eq. (39.12), which yields ds(p) = d3x e\u2212 ipx \u03a8(x)\u03b30vs(p). (39.13) Z Next, let us work out the anticommutation relations of the b and d operators (and their hermitian conjugates). From eq. (39.2), it is immediately clear that bs(p), bs\u2032(p\u2032) } { ds(p), ds\u2032 (p\u2032) } { bs(p), d\u2020s\u2032 (p\u2032) } { = 0, = 0, = 0, (39.14) because these involve only the anticommutator of \u03a8 with itself, and this vanishes. Of course, hermitian conjugation also yields Now consider bs(p), b\u2020s\u2032(p\u2032) } { b\u2020s(p), b\u2020s\u2032(p\u2032) } { d\u2020s(p), d\u2020s\u2032 (p\u2032) } { b\u2020s(p), ds\u2032 (p\u2032) } { = 0, = 0, = 0. (39.15) = = Z d3x d3y e\u2212 ipx+ip\u2032y us(p)\u03b30 \u03b30us\u2032(p\u2032) \u03a8(x), \u03a8(y) } { d3x e\u2212 i(p p\u2032)x us(p)\u03b30\u03b30\u03b30us\u2032(p\u2032) \u2212 Z = (2\u03c0)3\u03b43(p = (2\u03c0)3\u03b43(p \u2212 \u2212 p\u2032) us(p)\u03b30us\u2032(p) p\u2032) 2\u03c9\u03b4ss\u2032. (39.16) In the \ufb01rst line, we are free to set x0 = y0 because bs(p) and b\u2020s\u2032(p\u2032) are In the third, we used (\u03b30)2 = 1, and in the actually time independent. fourth, us(p)\u03b30us\u2032(p) = 2\u03c9\u03b4ss\u2032. 39: Canonical Quantization of Spinor Fields II 248 Similarly, d\u2020s(p), ds\u2032(p\u2032) } { = = Z d3x d3y eipx ip\u2032y vs(p)\u03b30 \u2212 \u03b30vs\u2032(p\u2032) \u03a8(x), \u03a8(y) } { d3x ei(p \u2212 p\u2032)x vs(p)\u03b30\u03b30\u03b30vs\u2032(p\u2032) Z = (2\u03c0)3\u03b43(p = (2\u03c0)3\u03b43(p \u2212 \u2212", " p\u2032) vs(p)\u03b30vs\u2032(p) p\u2032) 2\u03c9\u03b4ss\u2032. (39.17) And \ufb01nally, bs(p), ds\u2032(p\u2032) } { d3x d3y e\u2212 ipx ip\u2032y us(p)\u03b30 \u2212 \u03b30vs\u2032(p\u2032) \u03a8(x), \u03a8(y) } { d3x e\u2212 i(p+p\u2032)x us(p)\u03b30\u03b30\u03b30vs\u2032(p\u2032) = = Z Z = (2\u03c0)3\u03b43(p + p\u2032) us(p)\u03b30vs\u2032( \u2212 p) = 0. (39.18) According to the discussion in section 3, eqs. (39.14\u201339.18) are exactly what we need to describe the creation and annihilation of fermions. In this case, we have two di\ufb00erent kinds: b-type and d-type, each with two possible spin states, s = + and s =. \u2212 Next, let us evaluate the hamiltonian H = Z d3x \u03a8( \u2212 i\u03b3i\u2202i + m)\u03a8 (39.19) in terms of the b and d operators. We have i\u03b3i\u2202i + m)\u03a8 = ( \u2212 s= X \u00b1 Z dp f \u2212 i\u03b3i\u2202i + m bs(p)us(p)eipx + d\u2020s(p)vs(p)e\u2212 ipx = = s= X \u00b1 Z s= X \u00b1 Z Therefore dp bs(p)(+\u03b3ipi + m)us(p)eipx f + d\u2020s(p)( \u03b3ipi + m)vs(p)e\u2212 ipx \u2212 dp bs(p)(\u03b30\u03c9)us(p)eipx h h i f + d\u2020s(p)( \u2212 \u03b30\u03c9)vs(p)e\u2212 ipx. (39.20) i H = dp dp \u2032 d3x b\u2020s\u2032(p\u2032)us\u2032(p\u2032)e\u2212 ip\u2032x + ds\u2032(p\u2032)vs\u2032(p\u2032)eip\u2032x Xs,s\u2032 Z f f 39: Canonical Quantization of Spinor Fields II 249 = dp Xs,s\u2032 Z f f \u00d7 dp \u2032 d3x \u03c9 \u03c9 bs(p)\u03b30us(p)eipx d\u2020s(p)\u03b30vs(p)e\u2212 ipx \u2212 p)x i(p\u2032 \u2212 b\u2020s\u2032(p\u2032)bs(p) us\u2032(p\u2032)\u03b30us(p) e\u2212 b\u2020s\u2032(p\u2032)d\u2020s(p) us\u2032(p\u2032)\u03b30vs(p) e\u2212 h \u2212 + ds\u2032(p\u2032)bs(p) vs\u2032(p\u2032)\u03b30us(p) e+i(p\u2032+p)x ds\u2032(p\u2032)d\u2020s(p) vs\u2032(p\u2032)\u03b30vs(p) e+i(p\u2032 p)x i(p\u2032+p)x \u2212 \u2212 = Xs,s\u2032 Z dp 1 2 b\u2020s\u2032(p)bs(p) us\u2032(p)\u03b30us(p) f b\u2020s\u2032( h \u2212 + ds\u2032( p)d\u2020s(p) us\u2032( \u2212 p)\u03b30vs(p) e+2i\u03c9t \u2212 \u2212 p)bs(p) vs\u2032( ds\u2032(p)d\u2020s(p) vs\u2032(p)\u03b30vs(p) \u2212 p)\u03b30us(p) e\u2212 2i\u03c9t = dp \u03c9 b\u2020s(p)bs(p) s Z X f Using eq. (39.17), we can rewrite this as ds(p)d\u2020s(p) \u2212 i. i \u2212 h i (39.21) H = dp \u03c9 b\u2020s(p)bs(p) + d\u2020s(p)ds(p) E0V, 4 \u2212 i (39.22) s2\u03c0)\u2212 E0 = 1 d3k \u03c9 is the zero-point energy per unit volume that where we found for a real scalar \ufb01eld in section 3, and V = (2\u03c0)3\u03b43(0) = d3x is the volume of space. That the zero-point energy is negative rather than positive is characteristic of fermions; that it is larger in magnitude by a factor of four is due to the four types", " of particles that are associated with a Dirac \ufb01eld. We can cancel o\ufb00 this constant energy by including a constant term \u21260 = E0 in the original lagrangian density; from here on, we will 4 assume that this has been done. 0 i | The ground state of the hamiltonian (39.22) is the vacuum state that is annihilated by every bs(p) and ds(p), \u2212 R bs(p) 0 i | = ds(p) 0 i | = 0. (39.23) Then, we can interpret the b\u2020s(p) operator as creating a b-type particle with momentum p, energy \u03c9 = (p2 + m2)1/2, and spin Sz = 1 2 s, and the d\u2020s(p) operator as creating a d-type particle with the same properties. The b-type and d-type particles are distinguished by the value of the charge d3x j0, where j\u00b5 = \u03a8\u03b3\u00b5\u03a8 is the Noether current associated with the Q = e+i\u03b1\u03a8. invariance of R under the U(1) transformation \u03a8 i\u03b1\u03a8, \u03a8 e\u2212 \u2192 \u2192 L 39: Canonical Quantization of Spinor Fields II 250 Following the same procedure that we used for the hamiltonian, we can show that Q = d3x \u03a8\u03b30\u03a8 = = Z s= X \u00b1 Z s= X \u00b1 Z f dp f dp b\u2020s(p)bs(p) + ds(p)d\u2020s(p) h h b\u2020s(p)bs(p) \u2212 d\u2020s(p)ds(p) i i + constant. (39.24) Thus the conserved charge Q counts the total number of b-type particles minus the total number of d-type particles. (We are free to shift the overall value of Q to remove the constant term, and so we shall.) In quantum electrodynamics, we will identify the b-type particles as electrons and the d-type particles as positrons. Now consider a Majorana \ufb01eld \u03a8 with lagrangian = i 2 \u03a8T /\u2202\u03a8 C \u2212 1 2 m\u03a8T \u03a8. C L (39.25) The equation of motion for \u03a8 is once again the Dirac equation, and so the general solution is once again given by eq. (39.5). However, \u03a8 must also \u03a8 T. Starting from the barred form of obey the Majorana condition \u03a8 = eq. (39.5), C \u03a8(x) = s= X \u00b1 Z we have \u03a8 T(x) = C s= X \u00b1 Z dp f dp f From section 38, we have b\u2020s(p)us(p)e\u2212 h ipx + ds(p)vs(p)eipx, (39.26) i b\u2020s(p) h C uT s (p)e\u2212 ipx + ds(p) C s (p)eipx vT. (39.27) i us(p)T = vs(p), vs(p)T = us(p), C C (39.28) and so \u03a8 T(x) = C s= X \u00b1 Z b\u2020s(p)vs(p)e\u2212 h dp f ipx + ds(p)us(p)eipx. (39.29) i Comparing eqs. (39.5) and (39.29), we see that we will have \u03a8 = ds(p) = bs(p). \u03a8 T if C (39.30) 39: Canonical Quantization of Spinor Fields II 251 Thus a free Majorana \ufb01eld can be written as \u03a8(x) = dp bs(p)us(p)eipx + b\u2020s(p)vs(p)e\u2212 ipx. (39.31) s= X \u00b1 Z f h i The anticommutation relations for a Majorana \ufb01eld, \u03a8\u03b1(x, t), \u03a8\u03b2 (y, t) { } \u03a8\u03b1(x, t), \u03a8\u03b2(y, t) } { can be used to show that \u03b30)\u03b1\u03b2 \u03b43(x = ( C = (\u03b30)\u03b1\u03b2 \u03b43(x y), \u2212 y), \u2212 (39.32) (39.33) bs(p), bs\u2032(p\u2032) } { bs(p), b\u2020s\u2032(p\u2032) } { = 0, = (2\u03c0)3\ufffd", "\ufffd3(p p\u2032) 2\u03c9\u03b4ss\u2032, \u2212 (39.34) as we would expect. The hamiltonian for the Majorana \ufb01eld \u03a8 is H = 1 2 = 1 2 Z Z d3x \u03a8T C d3x \u03a8( i\u03b3i\u2202i + m)\u03a8 ( \u2212 i\u03b3i\u2202i + m)\u03a8, \u2212 (39.35) and we can work through the same manipulations that led to eq. (39.21); the only di\ufb00erences are an extra overall factor of one-half, and ds(p) = bs(p). Thus we get H = 1 2 s= X \u00b1 Z dp \u03c9 b\u2020s(p)bs(p) \u2212 bs(p)b\u2020s(p). (39.36) i h f Note that this would reduce to a constant if we tried to use commutators rather than anticommutators in eq. (39.34), a re\ufb02ection of the spin-statistics theorem. Using eq. (39.34) as it is, we \ufb01nd H = s= X \u00b1 Z f dp \u03c9 b\u2020s(p)bs(p) E0V. 2 \u2212 (39.37) Again, we can (and will) cancel o\ufb00 the zero-point energy by including a term \u21260 = E0 in the original lagrangian density. 2 The Majorana lagrangian has no U(1) symmetry. Thus there is no associated charge, and only one kind of particle (with two possible spin states). \u2212 Problems 39: Canonical Quantization of Spinor Fields II 252 39.1) Verify eq. (39.24). 39.2) Use [\u03a8(x), M \u00b5\u03bd ] = i(x\u00b5\u2202\u03bd x\u03bd\u2202\u00b5)\u03a8(x) + S\u00b5\u03bd \u03a8(x), plus whatever \u2212 spinor identities you need, to show that \u2212 Jz b\u2020s(p\u02c6z) 0 i | Jz d\u2020s(p\u02c6z\u2020s(p\u02c6z) 0 i | 2 s d\u2020s(p\u02c6z) 0 i |,, (39.38) where p = p\u02c6z is the three-momentum, and \u02c6z is a unit vector in the z direction. 39.3) Show that U (\u039b)\u2212 U (\u039b)\u2212 1b\u2020s(p)U (\u039b) = b\u2020s(\u039b\u2212 1d\u2020s(p)U (\u039b) = d\u2020s(\u039b\u2212 1p), 1p), (39.39) and hence that where p, s, q U (\u039b) | i = \u039bp, s, q | i, (39.40) p, s, + | p, s, | i \u2261 \u2212i \u2261, b\u2020s(p) 0 i | d\u2020s(p) 0 i | (39.41) are single-particle states. 39.4) The spin-statistics theorem for spin-one-half particles. We will follow the proof for spin-zero particles in section 4. We start with bs(p) and b\u2020s(p) as the fundamental objects; we take them to have either commutation ( ) or anticommutation (+) relations of the form \u2212 [bs(p), bs\u2032 (p\u2032)] \u2213 [b\u2020s(p), b\u2020s\u2032 (p\u2032)] \u2213 [bs(p), b\u2020s\u2032 (p\u2032)] \u2213 = 0, = 0, = (2\u03c0)32\u03c9\u03b43(p p\u2032)\u03b4ss\u2032. \u2212 (39.42) De\ufb01ne \u03a8+(x) \u03a8\u2212(x) \u2261 \u2261 dp bs(p)us(p)eipx, s= X \u00b1 Z f dp b\u2020s(p)vs(p)e\u2212 ipx. (39.43) a) Show that U (\u039b)\u2212 s= X \u00b1 Z f 1\u03a8\u00b1(x)U (\u039b) = D(\u039b)\u03a8\u00b1(\u039b\u2212 1x). 39: Canonical Quantization of Spinor Fields II 253 b) Show that [\u03a8+(x)]\u2020 = [\u03a8\u2212(x)]T term in the lagrangian must involve both \u03a8+(x) and \u03a8\u2212(x). c) Show that [\u03a8+ y)2 > 0. C \u03b2. Thus a hermitian interaction \u03b1 (x), \u03a8\u2212\u03b2 (y)] \u03b1 (x), \ufffd", "\ufffd\u2212\u03b2 (y)] \u2213 \u2213 6 = 0 for (x [\u03a8+ \u2212 = d) Show that [\u03a8+ y)2 > 0. \u03b2 (y), \u03a8\u2212\u03b1 (x)] \u2213 \u03a8+(x)+\u03bb\u03a8\u2212(x), where \u03bb is an arbitrary complex for \u2213 = 1 and and [\u03a8\u03b1(x), \u03a8\u03b2(y)] e) Consider \u03a8(x) number, and evaluate both [\u03a8\u03b1(x), \u03a8\u03b2(y)] (x we use anticommutators. y)2 > 0. Show these can both vanish if and only if for (x \u03bb | | \u2261 \u2212 \u2212 \u2212 \u2213 40: Parity, Time Reversal, and Charge Conjugation 254 40 Parity, Time Reversal, and Charge Conjugation Prerequisite: 23, 39 Recall that, under a Lorentz transformation \u039b implemented by the unitary operator U (\u039b), a Dirac (or Majorana) \ufb01eld transforms as U (\u039b)\u2212 1\u03a8(x)U (\u039b) = D(\u039b)\u03a8(\u039b\u2212 1x). (40.1) For an in\ufb01nitesimal transformation \u039b\u00b5 given by \u03bd = \u03b4\u00b5 \u03bd + \u03b4\u03c9\u00b5 \u03bd, the matrix D(\u039b) is D(1+\u03b4\u03c9) = I + i 2 \u03b4\u03c9\u00b5\u03bd S\u00b5\u03bd, where the Lorentz generator matrices are S\u00b5\u03bd = i 4 [\u03b3\u00b5, \u03b3\u03bd ]. (40.2) (40.3) In this section, we will consider the two Lorentz transformations that cannot be reached via a sequence of in\ufb01nitesimal transformations away from the identity: parity and time reversal. We begin with parity. De\ufb01ne the parity transformation \u00b5 P \u03bd = ( P 1)\u00b5 \u2212 \u03bd = \uf8eb \uf8ec \uf8ec \uf8ed + \u2212 and the corresponding unitary operator Now we have P ). U ( P \u2261 P \u2212 1\u03a8(x)P = D( P )\u03a8( P x). The question we wish to answer is, what is the matrix D( )? First of all, if we make a second parity transformation, we get P P \u2212 2\u03a8(x)P 2 = D( P )2\u03a8(x), (40.4) (40.5) (40.6) (40.7) )2 = 1, so that we and it is tempting to conclude that we should have D( return to the original \ufb01eld. This is correct for scalar \ufb01elds, since they are themselves observable. With fermions, however, it takes an even number of \ufb01elds to construct an observable. Therefore we need only require the weaker condition D( )2 = 1. P P \u00b1 40: Parity, Time Reversal, and Charge Conjugation 255 We will also require the particle creation and annihilation operators to transform in a simple way. Because P \u2212 P \u2212 1PP = P, 1JP = +J, \u2212 (40.8) (40.9) where P is the total three-momentum operator and J is the total angular momentum operator, a parity transformation should reverse the threemomentum while leaving the spin direction unchanged. We therefore require P \u2212 P \u2212 1b\u2020s(p)P = \u03b7 b\u2020s( \u2212 1d\u2020s(p)P = \u03b7 d\u2020s( p), \u2212 p), (40.10) where \u03b7 is a possible phase factor that (by the previous argument about observables) should satisfy \u03b72 = 1. We could in principle assign di\ufb00erent phase factors to the b and d operators, but we choose them to be the same so that the parity transformation is compatible with the Majorana condition ds(p) = bs(p). Writing the mode expansion of the free \ufb01eld \u00b1 \u03a8(x) = dp s= X \u00b1 Z f bs(p)us(p)eipx + d\u2020s(p)vs(p)e\u2212 h ipx, i (40.11) the parity transformation reads P \u2212 1\u03a8(x)P = = = s= X \u00b1 Z s= X \u00b1 Z s= X \u00b1 Z dp f dp f dp f P \u2212 1bs(p)P us(p)eipx + P \u2212 1d\u2020s(p)P vs(p)e\u2212 ipx p)us(p)eipx + \u03b7d\u2020s(", " \u2212 ipx p)vs(p)e\u2212 i p)eip P x + \u03b7d\u2020s(p)vs( \u2212 \u2212 i ip P x p)e\u2212 i. (40.12) \u2212 h \u03b7\u2217bs( h \u03b7\u2217bs(p)us( h In the last line, we have changed the integration variable from p to We now use a result from section 38, namely that p. \u2212 us( vs( \u2212 \u2212 p) = +\u03b2us(p), \u03b2vs(p), p) = \u2212 where \u03b2 = 0 I I 0!. (40.13) (40.14) 40: Parity, Time Reversal, and Charge Conjugation 256 Then, if we choose \u03b7 = i, eq. (40.12) becomes \u2212 P \u2212 1\u03a8(x)P = dp ibs(p)\u03b2us(p)eip P x + id\u2020s(p)\u03b2vs(p)e\u2212 ip P x \u00b1 Z s= X = i\u03b2 \u03a8( h f x). P i (40.15) Thus we see that D( in D( ) = P P \u2212 i\u03b2; either choice is acceptable.) ) = i\u03b2. (We could also have chosen \u03b7 = i, resulting The factor of i has an interesting physical consequence. Consider a state of an electron and positron with zero center-of-mass momentum, = \u03c6 i | Z dp \u03c6(p)b\u2020s(p)d\u2020s\u2032 ( \u2212 p) 0 i | ; (40.16) f here \u03c6(p) is the momentum-space wave function. Let us assume that the. Let us also assume that = vacuum is parity invariant\u2113\u03c6(p). Then, applying p) = ( the wave function has de\ufb01nite parity: \u03c6( \u2212, we get \u03c6 the inverse parity operator on i | 1b\u2020s(p)P dp \u03c6(p) 0 i | (P \u2212 p)P P \u2212 P \u2212 P \u2212 = \u2212 1d\u2020s\u2032( 1 1 1. \u2212 0 i | dp \u03c6(p)b\u2020s( p)d\u2020s\u2032(p) 0 i | \u2212 \u2212 p)b\u2020s(p)d\u2020s\u2032( \u2212 p) 0 i | (40.17) \u03c6 i | Z = ( \u2212 i)2 f Z i)2 f dp \u03c6\u2113 f. \u03c6 i | Thus, the parity of this state is opposite to that of its wave function; an 1. This also applies to a electron-positron pair has an intrinsic parity of pair of Majorana fermions. This in\ufb02uences the selection rules for fermion pair annihilation in theories that conserve parity. (A pair of electrons also has negative intrinsic parity, but this is less interesting because the electrons are prevented from annihilating by charge conservation.) \u2212 Let us see what eq. (40.15) implies for the two Weyl \ufb01elds that comprise the Dirac \ufb01eld. Recalling that \u03a8 = \u03c7a \u02d9a! \u03be\u2020, we see from eqs. (40.14) and (40.15) that P \u2212 1\u03c7a(x)P = i\u03be\u2020 \u02d9a( P x), P \u2212 1\u03be\u2020 \u02d9a(x)P = i\u03c7a( P x). (40.18) (40.19) 40: Parity, Time Reversal, and Charge Conjugation 257 Thus a parity transformation exchanges a left-handed \ufb01eld for a righthanded one. If we take the hermitian conjugate of eq. (40.19), then raise the index on one side while lowering it on the other (and remember that this introduces a relative minus sign!), we get P \u2212 1\u03c7\u2020 \u02d9a(x)P = i\u03bea( x), P \u02d9a( P P \u2212 1\u03bea(x)P = i\u03c7\u2020 x). (40.20) Comparing eqs. (40.19) and (40.20), we see that they are compatible with the Majorana condition \u03c7a(x) = \u03bea(x). Next we take up time reversal. De\ufb01ne the time-reversal transformation \u00b5 \u03bd = ( T T 1)1 +1 \uf8f6 \uf8f7 \uf8f7 \uf8f8 +1 and the corresponding operator Now we have T U ( T \u2261 ). T \u2212 1\u03a8(x)T = D( T )\u03a8( T x). The question we wish", " to answer is, what is the matrix D( (40.21) (40.22) (40.23) )? As with parity, we can conclude that D( 1, and we will require the particle creation and annihilation operators to transform in a simple way. Because \u00b1 T )2 = T T \u2212 T \u2212 1PT = 1JT = P, J, \u2212 \u2212 (40.24) (40.25) where P is the total three-momentum operator and J is the total angular momentum operator, a time-reversal transformation should reverse the direction of both the three-momentum and the spin. We therefore require T \u2212 T \u2212 1b\u2020s(p)T = \u03b6s b\u2020 \u2212 1d\u2020s(p)T = \u03b6s d\u2020 \u2212 p), s( \u2212 p). s( \u2212 (40.26) This time we allow for possible s-dependence of the phase factor. Also, we recall from section 23 that T must be an antiunitary operator, so that 40: Parity, Time Reversal, and Charge Conjugation 258 T \u2212 1iT = i. Then we have \u2212 1\u03a8(x)T T \u2212 = = = s= X \u00b1 Z s= X \u00b1 Z s= X \u00b1 Z dp f dp f dp f h \u03b6 \u2217s b h \u03b6 \u2217 \u2212 h T \u2212 1bs(p)T u\u2217s(p)e\u2212 ipx + T \u2212 1d\u2020s(p)T v\u2217s (p)eipx p)u\u2217s(p)e\u2212 ipx + \u03b6sd\u2020 \u2212 s( \u2212 s( \u2212 \u2212 i p)v\u2217s (p)eipx sbs(p)u\u2217 \u2212 p)eip T x + \u03b6 s( \u2212 sd\u2020s(p)v\u2217 \u2212 \u2212 s( \u2212 i p)e\u2212 x ip T. i (40.27) In the last line, we have changed the integration variable from p to the summation variable from s to 38, namely that p, and s. We now use a result from section \u2212 \u2212 s( s( u\u2217 \u2212 v\u2217 \u2212 \u2212 \u2212 p) = p) = s s \u2212 \u2212 C C \u03b35us(p), \u03b35vs(p). Then, if we choose \u03b6s = s, eq. (40.27) becomes T \u2212 1\u03a8(x)T = \u03b35\u03a8( T C x). (40.28) (40.29) Thus we see that D( resulting in D( ) = \u03b35; either choice is acceptable.) As with parity, we can consider the e\ufb00ect of time reversal on the Weyl (We could also have chosen \u03b6s = T \u2212C ) = \u03b35. \u2212 s, T C \ufb01elds. Using eqs. (40.18), (40.29), and we see that = C \u03b5ab \u2212 0 0 \u03b5 \u02d9a\u02d9b! \u2212, \u03b35 = c \u03b4a \u2212 0 0 +\u03b4 \u02d9a \u02d9c!, T \u2212 1\u03c7a(x)T = +\u03c7a( T x), T \u2212 1\u03be\u2020 \u02d9a(x)T = \u03be\u2020\u02d9a( T \u2212 x). (40.30) (40.31) (40.32) Thus left-handed Weyl \ufb01elds transform into left-handed Weyl \ufb01elds (and right-handed into right-handed) under time reversal. 40: Parity, Time Reversal, and Charge Conjugation 259 If we take the hermitian conjugate of eq. (40.32), then raise the index on one side while lowering it on the other (and remember that this introduces a relative minus sign!), we get T \u2212 1\u03c7\u2020 \u02d9a(x)T = \u03c7\u2020\u02d9a( T 1\u03bea(x)T = +\u03bea( T \u2212 T \u2212 x), x). (40.33) Comparing eqs. (40.32) and (40.33), we see that they are compatible with the Majorana condition \u03c7a(x) = \u03bea(x). It is interesting and important to evaluate the transformation properties of fermion bilinears of the form \u03a8A\u03a8, where A is some combination of gamma matrices. We will consider A\u2019s that satisfy A = A, where A \u03b2A\u2020\u03b2; in this case, \u03a8A\u03a8 is hermitian. \u2261 Let us begin with", " parity transformations. From \u03a8 = \u03a8\u2020\u03b2 and eq. (40.15) we get Combining eqs. (40.15) and (40.34) we \ufb01nd P \u2212 1\u03a8(x)P = i\u03a8( P \u2212 x)\u03b2, 1 P \u2212 \u03a8A\u03a8 P = \u03a8 \u03b2A\u03b2 \u03a8, (40.34) (40.35) where we have suppressed the spacetime arguments (which transform in the obvious way). For various particular choices of A we have \u03b21\u03b2 = +1, \u2212 \u03b2i\u03b35\u03b2 = i\u03b35, \u03b2\u03b30\u03b2 = +\u03b30, \u03b3i, \u03b2\u03b3i\u03b2 = \u03b2\u03b30\u03b35\u03b2 = \u03b30\u03b35, \u03b2\u03b3i\u03b35\u03b2 = +\u03b3i\u03b35. \u2212 \u2212 (40.36) Therefore, the corresponding hermitian bilinears transform as 1 P \u2212 \u03a8\u03a8 P = + \u03a8\u03a8, 1 P \u2212 \u03a8i\u03b35\u03a8 P \u2212 1 P \u2212 1 \u03a8\u03b3\u00b5\u03a8 \u03a8\u03b3\u00b5\u03b35\u03a8 P = \u2212 P = + P = \u03a8i\u03b35\u03a8, \u00b5 \u03bd\u03a8\u03b3\u03bd\u03a8, P \u00b5 \u03bd\u03a8\u03b3\u03bd\u03b35\u03a8, \u2212 P (40.37) Thus we see that \u03a8\u03a8 and \u03a8\u03b3\u00b5\u03a8 are even under a parity transformation, while \u03a8i\u03b35\u03a8 and \u03a8\u03b3\u00b5\u03b35\u03a8 are odd. We say that \u03a8\u03a8 is a scalar, \u03a8\u03b3\u00b5\u03a8 40: Parity, Time Reversal, and Charge Conjugation 260 is a vector or polar vector, \u03a8i\u03b35\u03a8 is a pseudoscalar, and \u03a8\u03b3\u00b5\u03b35\u03a8 is a pseudovector or axial vector. Turning to time reversal, from eq. (40.29) we get T \u2212 1\u03a8(x)T = \u03a8( T x)\u03b35C 1. \u2212 (40.38) Combining eqs. (40.29) and (40.38), along with T \u2212 1AT = A\u2217, we \ufb01nd 1 T \u2212 \u03a8A\u03a8 T = \u03a8 1A\u2217 \u2212 \u03b35C C \u03b35 \u03a8, (40.39) where we have suppressed the spacetime arguments (which transform in the 1\u03b35C obvious way). Recall that = \u03b35. Also, = \u2212 \u03b30 and \u03b35 are real, hermitian, and square to one, while \u03b3i is antihermitian. Finally, \u03b35 anticommutes with \u03b3\u00b5. Using all of this info, we \ufb01nd (\u03b3\u00b5)T and that 1\u03b3\u00b5 C\u2212 C\u2212 C \u2212 \u2212 11\u2217 \u03b35C \u2212 1(i\u03b35)\u2217 1(\u03b30)\u2217 1(\u03b3i)\u2217 \u2212 1(\u03b30\u03b35)\u2217 1(\u03b3i\u03b35)\u2217 \u03b35C \u03b35C \u03b35C \u03b35C \u03b35C \u2212 \u2212 \u03b35 = +1, i\u03b35, \u03b35 = \u2212 \u03b35 = +\u03b30, \u03b3i, \u03b35 = \u2212 \u03b35 = +\u03b30\u03b35, \u03b3i\u03b35. \u03b35 = \u2212 C C C C C C Therefore, 1 T \u2212 \u03a8\u03a8 T = + \u03a8\u03a8, 1 T \u2212 \u03a8i\u03b35\u03a8 T = T \u2212 1 T \u2212 1 \u03a8\u03b3\u00b5\u03a8 \u03a8\u03b3\u00b5\u03b35\u03a8 T = T = \u03a8i\u03b35\u03a8, \u2212 \u00b5 \u03bd\u03a8\u03b3\u03bd\u03a8, \u00b5 \u03bd\u03a8\u03b3\u03bd\u03b35\u03a8. \u2212 T \u2212 T (40.40) (40.41) Thus we see that \u03a8\u03a8 is even under time reversal, while \u03a8i\u03b35\u03a8, \u03a8\u03b3\u00b5\u03a8, and \u03a8\u03b3\u00b5\u03b35\u03a8 are odd. For completeness we will also consider the transformation properties of bilinears under charge conjugation. Recall that C \u2212 1\u03a8(x)C = \u03a8 T(x), C C \u2212 1\u03a8(x)C = \u03a8T(x) C. The bilinear \u03a8A\u03a8 therefore transforms as 1 C \u2212 \u03a8A\u03a8 C = \u03a8T \u03a8 T. A C C (40.42) (40.43) 40: Parity, Time Reversal, and Charge Conjugation 261 Since all indices are contracted, we can rewrite the right-hand", " side as its transpose, with an extra minus sign for exchanging the order of the two fermion \ufb01elds. We get Recalling that T = C TAT \u03a8 C \u2212 C T\u03a8. 1 C \u2212 \u03a8A\u03a8 C =, we have 1 = C\u2212 C \u2212 1 \u2212C \u03a8A\u03a8 C = \u03a8 1AT \u2212 \u03a8. C C Once again we can go through the list: \u2212 11T \u2212 C 1(i\u03b35)T 1(\u03b3\u00b5)T \u2212 C 1(\u03b3\u00b5\u03b35)T \u2212 C = +1, = +i\u03b35, \u03b3\u00b5, = = +\u03b3\u00b5\u03b35. \u2212 C C C C Therefore, C 1 C \u2212 \u03a8\u03a8 C = + \u03a8\u03a8, 1 C \u2212 \u03a8i\u03b35\u03a8 1 C \u2212 \u03a8\u03b3\u00b5\u03a8 C = + \u03a8i\u03b35\u03a8, \u03a8\u03b3\u00b5\u03a8, C = \u2212 1 C \u2212 \u03a8\u03b3\u00b5\u03b35\u03a8 C = + \u03a8\u03b3\u00b5\u03b35\u03a8. (40.44) (40.45) (40.46) (40.47) Thus we see that \u03a8\u03a8, \u03a8i\u03b35\u03a8, and \u03a8\u03b3\u00b5\u03b35\u03a8 are even under charge conjugation, while \u03a8\u03b3\u00b5\u03a8 is odd. For a Majorana \ufb01eld, we have C \u2212 1\u03a8C = \u03a8; this 1(\u03a8A\u03a8)C = \u03a8A\u03a8 for any combination of gamma matrices A. \u03a8\u03b3\u00b5\u03a8 for either a Dirac or implies C \u2212 Since eq. (40.47) tells that C \u2212 Majorana \ufb01eld, it must be that \u03a8\u03b3\u00b5\u03a8 = 0 for a Majorana \ufb01eld. 1\u03a8C = \u03a8 and C \u2212 1(\u03a8\u03b3\u00b5\u03a8)C = \u2212 Let us consider the combined e\ufb00ects of the three transformations (C, P, and T ) on the bilinears. From eqs. (40.37), (40.41), and (40.47), we have (CP T )\u2212 1 \u03a8\u03a8 CP T = + \u03a8\u03a8, (CP T )\u2212 1 \u03a8i\u03b35\u03a8 CP T = + \u03a8i\u03b35\u03a8, (CP T )\u2212 1 \u03a8\u03b3\u00b5\u03a8 CP T = (CP T )\u2212 1 \u03a8\u03b3\u00b5\u03b35\u03a8 CP T = \u03a8\u03b3\u00b5\u03a8, \u2212 \u2212 \u03a8\u03b3\u00b5\u03b35\u03a8, (40.48) \u03c1. We see that \u03a8\u03a8 and \u03a8i\u03b35\u03a8 are both where we have used even under CP T, while \u03a8\u03b3\u00b5\u03a8 and \u03a8\u03b3\u00b5\u03b35\u03a8 are both odd. These are (it \u03bd \u03c1 = \u03bdT \u03b4\u00b5 \u2212 P \u00b5 40: Parity, Time Reversal, and Charge Conjugation 262 turns out) examples of a more general rule: a fermion bilinear with n vector indices (and no uncontracted spinor indices) is even (odd) under CP T if n is even (odd). This also applies if we allow derivatives acting on the \ufb01elds, since each component of \u2202\u00b5 is odd under the combination P T and even under C. For scalar and vector \ufb01elds, it is always possible to choose the phase factors in the C, P, and T transformations so that, overall, they obey the same rule: a hermitian combination of \ufb01elds and derivatives is even or odd depending on the total number of uncontracted vector indices. Putting this together with our result for fermion bilinears, we see that any hermitian combination of any set of \ufb01elds (scalar, vector, Dirac, Majorana) and their derivatives that is a Lorentz scalar (and so carries no indices) is even under CP T. Since the lagrangian must be formed out of such combinations, we have is invariant. x) under CP T, and so the action S = \u2192 L This is the CP T theorem. ( \u2212 d4x (x) L L R Reference Notes A detailed treatment of CP T for \ufb01elds of any spin is given in Weinberg I. Problems 40.1) Find the transformation properties of \u03a8S\u00b5\u03bd \u03a8 and \u03a8iS\u00b5\u03bd \u03b35\u03a8 under P, T, and C. Verify that they are both even under CP T, as", " claimed. Do either or both vanish if \u03a8 is a Majorana \ufb01eld? 41: LSZ Reduction for Spin-One-Half Particles 263 41 LSZ Reduction for Spin-One-Half Particles Prerequisite: 5, 39 Let us now consider how to construct appropriate initial and \ufb01nal states for scattering experiments. We will \ufb01rst consider the case of a Dirac \ufb01eld \u03a8, and assume that its interactions respect the U(1) symmetry that gives rise to the conserved current j\u00b5 = \u03a8\u03b3\u00b5\u03a8 and its associated charge Q. In the free theory, we can create a state of one particle by acting on the vacuum state with a creation operator: where the label \u00b1, i p, s, + | p, s, | = b\u2020s(p) 0 i | = d\u2020s(p) 0 i | on the ket indicates the value of the U(1) charge Q, and (41.2) (41.1) \u2212i, b\u2020s(p) = d3x eipx \u03a8(x)\u03b30us(p), d\u2020s(p) = Z Z d3x eipx vs(p)\u03b30\u03a8(x). (41.3) (41.4) Recall that b\u2020s(p) and d\u2020s(p) are time independent in the free theory. The states have the Lorentz-invariant normalization p, s, | \u00b1i p, s, q h p\u2032, s\u2032, q\u2032 | i = (2\u03c0)3 2\u03c9 \u03b43(p p\u2032) \u03b4ss\u2032 \u03b4qq\u2032, \u2212 (41.5) where \u03c9 = (p2 + m2)1/2. Let us consider an operator that (in the free theory) creates a particle with de\ufb01nite spin and charge, localized in momentum space near p1, and localized in position space near the origin: d3p f1(p)b\u2020s1(p), b\u20201 \u2261 Z (41.6) where f1(p) exp[ (p p1)2/4\u03c32] (41.7) \u2212 \u221d \u2212 is an appropriate wave packet, and \u03c3 is its width in momentum space. If we time evolve (in the Schr\u00a8odinger picture) the state created by this timeindependent operator, then the wave packet will propagate (and spread. If out). The particle will thus be localized far from the origin as t = p2,, where p1 6 0 = b\u20201b\u20202| we consider instead an initial state of the form i then we have two particles that are widely separated in the far past. \u2192 \u00b1\u221e i i | Let us guess that this still works in the interacting theory. One complication is that b\u2020s(p) will no longer be time independent, and so b\u20201, eq. (41.6), 41: LSZ Reduction for Spin-One-Half Particles 264 becomes time dependent as well. Our guess for a suitable initial state for a scattering experiment is then i i | = lim t \u2192\u2212\u221e 0 b\u20201(t)b\u20202(t) i |. (41.8) By appropriately normalizing the wave packets, we can make = 1, and we will assume that this is the case. Similarly, we can consider a \ufb01nal state i i i | h f | f | = lim + t \u2192 \u221e i 0 b\u20201\u2032(t)b\u20202\u2032(t) i |, (41.9) i f h = p\u20322, and where p\u20321 6 = 1. This describes two widely separated particles in the far future. (We could also consider acting with more creation operators, if we are interested in the production of some extra particles in the collision of two, or using d\u2020 operators instead of b\u2020 operators for some or all of the initial and \ufb01nal particles.) Now the scattering amplitude is simply given by We need to \ufb01nd a more useful expression for f h. To this end, let us \u20201(+ \u221e + \u221e note that b\u20201( \u2212\u221e ) \u2212 = = = = = = \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 = i Z dt \u22020b\u20201(t) Z \u2212\u221e d3p f1(p) d4x \u22020. eipx \u03a8(x)\u03b30us1(p) us1(p)eipx d4x \u03a8(x) d4x \u03a8(x) d4x \u03a8(x) d4x \u03a8(x) i\u03b30p0 \u03b30\u2190\u2202 0", " \u2212 i\u03b3ipi \u2212 \u03b30\u2190\u2202 0 \u2212 \u03b3i\u2192\u2202 i \u2212 \u03b30\u2190\u2202 0 \u2212 \u03b30\u2190\u2202 0 + \u03b3i\u2190\u2202 i \u2212 im us1(p)eipx us1(p)eipx d4x \u03a8(x)(+i\u2190/\u2202 + m)us1(p)eipx. im im us1(p)eipx (41.10) Z Z Z Z Z d3p f1(p) d3p f1(p) d3p f1(p) d3p f1(p) d3p f1(p) Z Z Z Z Z Z The \ufb01rst equality is just the fundamental theorem of calculus. To get the second, we substituted the de\ufb01nition of b\u20201(t), and combined the d3x from this de\ufb01nition with the dt to get d4x. The third comes from straightforward evaluation of the time derivatives. The fourth uses (/p + m)us(p) = 0. The \ufb01fth writes ipi as \u2202i acting on eipx. The sixth uses integration by parts to move the \u2202i onto the \ufb01eld \u03a8(x); here the wave packet is needed to avoid a surface term. The seventh simply identi\ufb01es \u03b30\u22020 + \u03b3i\u2202i as /\u2202. In free-\ufb01eld theory, the right-hand side of eq. (41.10) is zero, since \u03a8(x) obeys the Dirac equation, which, after barring it, reads \u03a8(x)(+i\u2190/\u2202 + m) = 0. (41.11) 41: LSZ Reduction for Spin-One-Half Particles 265 In an interacting theory, however, the right-hand side of eq. (41.10) will not be zero. We will also need the hermitian conjugate of eq. (41.10), which (after some slight rearranging) reads b1(+ ) \u221e b1( ) \u2212 = i \u2212\u221e d3p f1(p) Z Z d4x e\u2212 ipx us1(p)( \u2212 i/\u2202 + m)\u03a8(x), (41.12) and the analogous formulae for the d operators, d\u20201( \u2212\u221e ) d1(+ ) \u221e \u2212 = \u2212 = d\u20201(+ i \u2212 Z d1( i \u2212 Z ) \u221e d3p f1(p) ) \u2212\u221e d3p f1(p) d4x eipx vs1(p)( \u2212 i/\u2202 + m)\u03a8(x), (41.13) d4x \u03a8(x)(+i\u2190/\u2202 + m)vs1(p)e\u2212 ipx. (41.14) Z Z Let us now return to the scattering amplitude we were considering, f h i i | = b2\u2032 (+ 0 | h \u221e )b1\u2032(+ \u221e )b\u20201( \u2212\u221e )b\u20202( \u2212\u221e. 0 ) i | (41.15) Note that the operators are in time order. Thus, if we feel like it, we can put in a time-ordering symbol without changing anything b2\u2032 (+ )b1\u2032(+ )b\u20201( \u2212\u221e \u221e \u221e )b\u20202( \u2212\u221e. 0 ) i | (41.16) The symbol T means the product of operators to its right is to be ordered, not as written, but with operators at later times to the left of those at earlier times. However, there is an extra minus sign if this rearrangement involves an odd number of exchanges of these anticommuting operators. \u2212\u221e Now let us use eqs. (41.10) and (41.12) in eq. (41.16). The time-ordering )\u2019s to the right, where they anni)\u2019s move to the left, where they annihilate symbol automatically moves all bi\u2032( hilate. Similarly, all b\u2020i (+ 0 i |. 0 h | 0 limit in eq. (41.7), so that f1(p) = \u03b43(p p1). The initial and \ufb01nal states now have a delta-function normalization, the multiparticle generalization of eq. (41.5). We are left with the Lehmann-Symanzik-Zimmermann reduction formula for spin-one-half particles, The wave packets no longer play a key role, and", " we can take the \u03c3 \u2192 \u221e \u2212 f h i i | = i4 d4x1 d4x2 d4x1\u2032 d4x2\u2032 Z e\u2212 \u00d7 ip\u2032 1x\u2032 1 [us1\u2032 (p1\u2032)( i/\u22021\u2032 + m)]\u03b11\u2032 \u2212 41: LSZ Reduction for Spin-One-Half Particles 266 ip\u2032 2x\u2032 e\u2212 \u00d7 2 [us2\u2032 (p2\u2032)( i/\u22022\u2032 + m)]\u03b12\u2032 \u2212 0 | \u00d7 h T \u03a8\u03b12\u2032 (x2\u2032)\u03a8\u03b11\u2032 (x1\u2032 )\u03a8\u03b11(x1)\u03a8\u03b12 (x2) 0 i | [(+i\u2190/\u2202 1 + m)us1(p1)]\u03b11 eip1x1 [(+i\u2190/\u2202 2 + m)us2(p2)]\u03b12 eip2x2. (41.17) \u00d7 \u00d7 The generalization of the LSZ formula to other processes should be clear; insert a time-ordering symbol, and make the following replacements: b\u2020s(p)in \u2192 +i bs(p)out \u2192 +i d\u2020s(p)in \u2192 \u2212 i ds(p)out \u2192 \u2212 i Z Z Z Z d4x \u03a8(x)(+i\u2190/\u2202 + m)us(p) e+ipx, (41.18) d4x e\u2212 ipx us(p)( \u2212 i/\u2202 + m)\u03a8(x), (41.19) d4x e+ipx vs(p)( \u2212 i/\u2202 + m)\u03a8(x), (41.20) d4x \u03a8(x)(+i\u2190/\u2202 + m)vs(p) e\u2212 ipx, (41.21) \u2192 \u221e, respectively. where we have used the subscripts \u201cin\u201d and \u201cout\u201d to denote t t + All of this holds for a Majorana \ufb01eld as well. In that case, ds(p) = bs(p), and we can use either eq. (41.18) or eq. (41.20) for the incoming particles, and either eq. (41.19) or eq. (41.21) for the outgoing particles, whichever is more convenient. The Majorana condition \u03a8 = \u03a8T guarantees that the results will be equivalent. \u2192 \u2212\u221e and C As in the case of a scalar \ufb01eld, we cheated a little in our derivation of the LSZ formula, because we assumed that the creation operators of free \ufb01eld theory would work comparably in the interacting theory. After performing an analysis that is entirely analogous to what we did for the scalar in section 5, we come to the same conclusion: the LSZ formula holds provided the \ufb01eld is properly normalized. For a Dirac \ufb01eld, we must require 0 \u03a8(x) 0 i | h | 0 \u03a8(x) p, s, + i | | h 0 \u03a8(x) p, s, | h i \u2212| p, s, + 0 \u03a8(x) | h i | 0 \u03a8(x) p, s, i | h \u2212| = 0, = 0, = vs(p)e\u2212 = us(p)e\u2212 ipx, ipx, = 0, (41.22) (41.23) (41.24) (41.25) (41.26) where 0 0 i | h eq. (41.5). = 1, and the one-particle states are normalized according to 41: LSZ Reduction for Spin-One-Half Particles 267 = \u00b1| \u00b1i p, s, 0 i |. \u00b1i p, s, |. If 0 i | = 0 and Q The zeros on the right-hand sides of eqs. (41.23) and (41.26) are required by charge conservation. To see this, start with [\u03a8(x), Q] = +\u03a8(x), take the matrix elements indicated, and use Q 0 i | The zero on the right-hand side of eq. (41.22) is required by Lorentz invariance. To see this, start with [\u03a8(0), M \u00b5\u03bd ] = S\u00b5\u03bd \u03a8(0), and take the is Lorentz invariant (as we expectation value in the vacuum state will assume), then it is annihilated by the Lorentz generators M \u00b5\u03bd, which means that we must have S\u00b5\u03bd = 0; this is possible for all \u00b5 and \u03a8(0)", " \u03bd only if = 0, which (by translation invariance) is possible only 0 0 i | | h = 0. 0 \u03a8(x) 0 if | i | h The right-hand sides of eqs. (41.24) and (41.25) are similarly \ufb01xed by Lorentz invariance: only the overall scale might be di\ufb00erent in an interacting theory. However, the LSZ formula is correctly normalized if and only if eqs. (41.24) and (41.25) hold as written. We will enforce this by rescaling (or, one might say, renormalizing) \u03a8(x) by an overall constant. This is just a change of the name of the operator of interest, and does not a\ufb00ect the physics. However, the rescaled \u03a8(x) will obey eqs. (41.24) and (41.25). (These two equations are related by charge conjugation, and so actually constitute only one condition on \u03a8.) 0 \u03a8(0) 0 i | | h For a Majorana \ufb01eld, there is no conserved charge, and we have 0 \u03a8(x) 0 i | h | 0 \u03a8(x) p, s | | h i 0 \u03a8(x) p, s i | | h instead of eqs. (41.22\u201341.26). = 0, = vs(p)e\u2212 = us(p)e\u2212 ipx, ipx, (41.27) (41.28) (41.29) The renormalization of \u03a8 necessitates including appropriate Z factors in the lagrangian. Consider, for example, = iZ\u03a8/\u2202\u03a8 Zmm\u03a8\u03a8 \u2212 \u2212 L 1 4 Zgg(\u03a8\u03a8)2, (41.30) where \u03a8 is a Dirac \ufb01eld, and g is a coupling constant. We choose the three constants Z, Zm, and Zg so that the following three conditions are satis\ufb01ed: m is the mass of a single particle; g is \ufb01xed by some appropriate scattering cross section; and eq. (41.24) and is obeyed. [Eq. (41.25) then follows by charge conjugation.] Next, we must develop the tools needed to compute the correlation in an interacting quantum \ufb01eld 0 i | T\u03a8\u03b11\u2032 (x1\u2032)... \u03a8\u03b11(x1)... 0 | h functions theory. Problems 41.1) Assuming that eq. (39.40) holds for the exact single-particle states, verify eqs. (41.23) and (41.26), up to overall scale. 42: The Free Fermion Propagator 268 42 The Free Fermion Propagator Prerequisite: 39 Consider a free Dirac \ufb01eld \u03a8(x) = \u03a8(y) = s= X \u00b1 Z dp bs(p)us(p)eipx + d\u2020s(p)vs(p)e\u2212 h f dp \u2032 ipx, i ip\u2032y + ds\u2032(p\u2032)vs\u2032(p\u2032)eip\u2032y b\u2020s\u2032(p\u2032)us\u2032(p\u2032)e\u2212 h Xs\u2032= \u00b1 Z f where and bs(p) 0 i | = ds(p) 0 i | = 0, bs(p), b\u2020s\u2032(p\u2032) { } ds(p), d\u2020s\u2032 (p\u2032) { } = (2\u03c0)3\u03b43(p = (2\u03c0)3\u03b43(p p\u2032) 2\u03c9\u03b4ss\u2032, p\u2032) 2\u03c9\u03b4ss\u2032, \u2212 \u2212 (42.1), (42.2) i (42.3) (42.4) (42.5) and all the other possible anticommutators between b and d operators (and their hermitian conjugates) vanish. We wish to compute the Feynman propagator S(x y)\u03b1\u03b2 \u2261 0 T\u03a8\u03b1(x)\u03a8\u03b2(y42.6) where T denotes the time-ordered product, T\u03a8\u03b1(x)\u03a8\u03b2(y) \u03b8(x0 \u2212 \u2261 y0)\u03a8\u03b1(x)\u03a8\u03b2(y) \u03b8(y0 \u2212 \u2212 x0)\u03a8\u03b2(y)\u03a8\u03b1(x), (42.7) and \u03b8(t) is the unit step function. Note the minus sign in the second term; this is needed because \u03a8\u03b1(x)\u03a8\u03b2", "(y) = We can now compute 0 \u03a8\u03b1(x)\u03a8\u03b2(y) 0 i | | h serting eqs. (42.1) and (42.2), and then using eqs. (42.3\u201342.5). We get by in- \u2212 = y0. \u03a8\u03b2(y)\u03a8\u03b1(x) when x0 0 \u03a8\u03b2(y)\u03a8\u03b1(x) 0 i | | h and \u03a8\u03b1(x)\u03a8\u03b2(y) 0 0 | h i | dp \u2032 eipx e\u2212 dp = ip\u2032y us(p)\u03b1us\u2032(p\u2032)\u03b2 h ip\u2032y us(p)\u03b1us\u2032(p\u2032)\u03b2 (2\u03c0)3\u03b43(p bs(p)b\u2020s\u2032(p\u2032) 0 0 i | | f dp f dp \u2032 eipx e\u2212 Xs,s\u2032 Z Xs,s\u2032 Z = = = f f dp eip(x y) us(p)\u03b1us(p)\u03b2 \u2212 X s Z f dp eip(x \u2212 Z f y) ( \u2212 /p + m)\u03b1\u03b2. p\u2032) 2\u03c9\u03b4ss\u2032 \u2212 (42.8) 6 42: The Free Fermion Propagator 269 To get the last line, we used a result from section 38. Similarly, 0 \u03a8\u03b2(y)\u03a8\u03b1(x) 0 i | | h = = = = Xs,s\u2032 Z Xs,s\u2032 Z dp dp \u2032 e\u2212 f dp f dp\u2032 e\u2212 ipx eip\u2032y vs(p)\u03b1vs\u2032(p\u2032)\u03b2 h ipx eip\u2032y vs(p)\u03b1vs\u2032(p\u2032)\u03b2 (2\u03c0)3\u03b43(p ds\u2032 (p\u2032)d\u2020s(p) 0 0 i | | p\u2032) 2\u03c9\u03b4ss\u2032 \u2212 f f dp e\u2212 ip(x y) vs(p)\u03b1vs(p)\u03b2 \u2212 s Z X f dp e\u2212 ip(x Z \u2212 y) ( \u2212 /p \u2212 m)\u03b1\u03b2. (42.9) We can combine eqs. (42.8) and (42.9) into a compact formula for the timeordered product by means of the identity f d4p (2\u03c0)4 Z \u2212 eip(x p2 + m2 y)f (p) i\u01eb \u2212 = i\u03b8(x0 y0) dp eip(x y) f (p) \u2212 \u2212 + i\u03b8(y0 Z x0) \u2212 f Z dp e\u2212 ip(x \u2212 y) f ( \u2212 p), (42.10) where f (p) is a polynomial in p; the derivation of eq. (42.10) was sketched in section 8. We get f T\u03a8\u03b1(x)\u03a8\u03b2(y) 0 0 | h i | = d4p (2\u03c0)4 eip(x \u2212 1 i Z y) ( /p + m)\u03b1\u03b2 i\u01eb \u2212 p2 + m2 \u2212, (42.11) and so S(x \u2212 y)\u03b1\u03b2 = Note that S(x \u2212 d4p (2\u03c0)4 eip(x \u2212 y) ( /p + m)\u03b1\u03b2 i\u01eb \u2212 p2 + m2 \u2212 Z. (42.12) y) is a Green\u2019s function for the Dirac wave operator: i/\u2202x + m)\u03b1\u03b2S(x ( \u2212 y)\u03b2\u03b3 = \u2212 Z = Similarly, S(x \u2212 y)\u03b1\u03b2(+i \u2190/\u2202y + m)\u03b2\u03b3 = Z = \u2212 d4p (2\u03c0)4 eip(x d4p (2\u03c0)4 eip(x y)\u03b4\u03b1\u03b3. \u2212 \u2212 d4p (2\u03c0)4 eip(x d4p (2\u03c0)4 eip(x y)\u03b4\u03b1\u03b3. \u2212 \u2212 /p + m)\u03b2\u03b3 i\u01eb \u2212 y) (/p + m)\u03b1\u03b2( \u2212 p2 + m2 y) (p2 + m2)\u03b4\u03b1\u03b3 p2 + m2 i\u01eb \u2212 (42.13) /p + m)\u03b1\u03b2(/p + m)\u03b2\u03b3 y) ( \u2212 \u2212 p2 + m2 y) (p2 + m2)\u03b4\u03b1\u03b3 p2 + m2 i\u01eb \u2212 i\u01eb \u2212 (42.14) Z = \u03b44(x Z = \u03b44(x 42: The Free F", "ermion Propagator 270, but it is 0 T\u03a8\u03b1(x)\u03a8\u03b2(y) We can also consider 0 0 T\u03a8\u03b1(x)\u03a8\u03b2(y easy to see that now there is no way to pair up a b with a b\u2020 or a d with a d\u2020, and so and 0 T\u03a8\u03b1(x)\u03a8\u03b2(y) 0 i | | h T\u03a8\u03b1(x)\u03a8\u03b2(y. Next, consider a Majorana \ufb01eld \u03a8(x) = \u03a8(y) = s= X \u00b1 Z dp bs(p)us(p)eipx + b\u2020s(p)vs(p)e\u2212 h f dp \u2032 b\u2020s\u2032(p\u2032)us\u2032(p\u2032)e\u2212 ipx, i ip\u2032y + bs\u2032(p\u2032)vs\u2032(p\u2032)eip\u2032y (42.15) (42.16) (42.17). (42.18) Xs\u2032= \u00b1 Z h i f It is easy to see that is the same as it is in the Dirac case; the only di\ufb00erence in the calculation is that we would have b and b\u2020 in place of d and d\u2020 in the second line of eq. (42.9), and this does not change the \ufb01nal result. Thus, T\u03a8\u03b1(x)\u03a8\u03b2(y) 0 0 | h i | 0 T\u03a8\u03b1(x)\u03a8\u03b2(y) 0 i i | | h = S(x \u2212 y)\u03b1\u03b2, (42.19) y) is given by eq. (42.12). However, eqs. (42.15) and (42.16) no longer hold for a Majorana \ufb01eld., which can be rewritten as Instead, the Majorana condition \u03a8 = \u03a8T \u03a8T = \u03a8 1, implies C where S(x \u2212 C\u2212 0 T\u03a8\u03b1(x)\u03a8\u03b2(y) 0 i i | | h ( 0 T\u03a8\u03b1(x)\u03a8\u03b3(y]\u03b1\u03b2. = [S(x y) \u2212 \u2212 C 1)\u03b3\u03b2 \u2212 (42.20) Similarly, using 1\u03a8, and so C\u2212 T = C C\u2212 1, we can write the Majorana condition as \u03a8 T = T\u03a8\u03b1(x)\u03a8\u03b2(y( C \u2212 1)\u03b1\u03b3h 0 T\u03a8\u03b3(x)\u03a8\u03b2(y) 0 | i | 1S(x y)]\u03b1\u03b2. = [ C \u2212 \u2212 (42.21) Of course, C\u2212 and (42.21) as they are. \u2212C 1 =, but it will prove more convenient to leave eqs. (42.20) We can also consider the vacuum expectation value of a time-ordered product of more than two \ufb01elds. In the Dirac case, we must have an equal number of \u03a8\u2019s and \u03a8\u2019s to get a nonzero result; and then, the \u03a8\u2019s and \u03a8\u2019s must pair up to form propagators. There is an extra minus sign if the 42: The Free Fermion Propagator 271 ordering of the \ufb01elds in their pairs is an odd permutation of the original ordering. For example, i2 0 T\u03a8\u03b1(x)\u03a8\u03b2(y)\u03a8\u03b3(z)\u03a8\u03b4(w) 0 i | | h = + S(x S(x \u2212 y)\u03b1\u03b2 S(z w)\u03b1\u03b4 S(z \u2212 \u2212 \u2212 \u2212 w)\u03b3\u03b4 y)\u03b3\u03b2. (42.22) In the Majorana case, we may as well let all the \ufb01elds be \u03a8\u2019s (since we can always replace a \u03a8 with \u03a8T ). Then we must pair them up in all possible ways. There is an extra minus sign if the ordering of the \ufb01elds in their pairs is an odd permutation of the original ordering. For example, C i2 0 T\u03a8\u03b1(x)\u03a8\u03b2(y)\u03a8\u03b3(z)\u03a8\u03b4(w) 0 i | | h = + [S(x [S(x \u2212 + [S(x \u2212 \u2212 \u2212 y) C z) C w) 1]\u03b1\u03b2 [S(z \u2212 1]\u03b1\u03b3 [S(y 1]\u03b1\u03b4 [S(y \u2212 \u2212 C w) w) C C \u2212 \u2212 z) C \u2212 1]\u03b3\u03b4 \u2212 1]\u03b2\u03b4 \u2212 1]\u03b2\u03b3. \u2212 (42", ".23) Note that the ordering within a pair does not matter, since [S(x y) C \u2212 1]\u03b1\u03b2 = \u2212 [S(y x) C \u2212 \u2212 1]\u03b2\u03b1. \u2212 (42.24) This follows from anticommutation of the \ufb01elds and eq. (42.20). Problems 42.1) Prove eq. (42.24) directly, using properties of the matrix. C 43: The Path Integral for Fermion Fields 272 43 The Path Integral for Fermion Fields Prerequisite: 9, 42 We would like to write down a path integral formula for the vacuumexpectation value of a time-ordered product of free Dirac or Majorana \ufb01elds. Recall that for a real scalar \ufb01eld with L0 = = \u2212 \u2212 1 2 m2\u03d52 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 1 2 \u03d5( \u2212 \u22022 + m2)\u03d5 \u2212 1 2 \u2202\u00b5(\u03d5\u2202\u00b5\u03d5), \u2212 (43.1) we have where T\u03d5(x1 \u03b4J(x1) Z0(J) = \u03d5 exp i d4x (. (43.3) In this formula, we use the epsilon trick (see section 6) of replacing m2 with m2 i\u01eb to construct the vacuum as the initial and \ufb01nal state. Then we get Z Z \u2212, (43.2) J=0... Z0(J) L0 + J\u03d5) Z0(J) = exp Z where the Feynman propagator d4x d4y J(x)\u2206(x \u2212 y)J(y), D i 2 \u2206(x \u2212 y) = Z d4k (2\u03c0)4 eik(x \u2212 k2 + m2 y) i\u01eb \u2212 is the inverse of the Klein-Gordon wave operator: x + m2)\u2206(x \u22022 ( \u2212 \u2212 y) = \u03b44(x y). \u2212 For a complex scalar \ufb01eld with L0 = = \u2212 \u2212 \u03d5\u2020( \u2212 \u2202\u00b5\u03d5\u2020\u2202\u00b5\u03d5 m2\u03d5\u2020\u03d5 \u2212 \u22022 + m2)\u03d5 \u2202\u00b5(\u03d5\u2020\u2202\u00b5\u03d5), \u2212 we have instead T\u03d5(x1)... \u03d5\u2020(y1 \u03b4J \u2020(x1)... 1 i \u03b4 \u03b4J(y1)... Z0(J \u2020, J), J=J \u2020=0 (43.8) where Z0(J \u2020, J) = \u03d5\u2020 D Z D \u03d5 exp i d4x ( Z d4x d4y J \u2020(x)\u2206(x L0 + J \u2020\u03d5 + \u03d5\u2020J) \u2212 y)J(y). (43.9) = exp i Z (43.4) (43.5) (43.6) (43.7) 43: The Path Integral for Fermion Fields 273 We treat J and J \u2020 as independent variables when evaluating eq. (43.8). In the case of a fermion \ufb01eld, we should have something similar, except that we need to account for the extra minus signs from anticommutation. For this to work out, a functional derivative with respect to an anticommuting variable must itself be treated as anticommuting. Thus if we de\ufb01ne an anticommuting source \u03b7(x) for a Dirac \ufb01eld, we can write \u03b4 \u03b4\u03b7(x) \u03b4 \u03b4\u03b7(x) Z Z d4y d4y \u03b7(y)\u03a8(y) + \u03a8(y)\u03b7(y) h i \u03b7(y)\u03a8(y) + \u03a8(y)\u03b7(y) i h \u03a8(x), = \u2212 (43.10) = +\u03a8(x). (43.11) The minus sign in eq. (43.10) arises because the \u03b4/\u03b4\u03b7 must pass through \u03a8 before reaching \u03b7. Thus, consider a free Dirac \ufb01eld with L0 = i\u03a8/\u2202\u03a8 = \u03a8( m\u03a8\u03a8 \u2212 i/\u2202 + m)\u03a8. \u2212 \u2212 (43.12) A natural guess for the appropriate path-integral formula, based on analogy with eq. (43.9), is T\u03a8\u03b11 (x1)", "... \u03a8\u03b21(y1)... 0 | h \u03b4 \u03b4\u03b7\u03b11(x1) 0 i | \u03b4 \u03b4\u03b7\u03b21(y143.13) \u03b7=\u03b7=0... Z0(\u03b7, \u03b7) where Z0(\u03b7, \u03b7) = \u03a8 D Z Z and the Feynman propagator i = exp D \u03a8 exp i d4x ( Z d4x d4y \u03b7(x)S(x L0 + \u03b7\u03a8 + \u03a8\u03b7) \u2212, y)\u03b7(y) (43.14) S(x \u2212 y) = Z d4p (2\u03c0)4 ( \u2212 /p + m)eip(x \u2212 p2 + m2 i\u01eb \u2212 is the inverse of the Dirac wave operator: i/\u2202x + m)S(x ( \u2212 \u2212 y) = \u03b44(x y). \u2212 y) (43.15) (43.16) Note that each \u03b4/\u03b4\u03b7 in eq. (43.13) comes with a factor of i rather than the usual 1/i; this re\ufb02ects the extra minus sign of eq. (43.10). We treat \u03b7 and \u03b7 43: The Path Integral for Fermion Fields 274 as independent variables when evaluating eq. (43.13). It is straightforward to check (by working out a few examples) that eqs. (43.13\u201343.16) do indeed reproduce the result of section 42 for the vacuum expectation value of a time-ordered product of Dirac \ufb01elds. This is really all we need to know. Recall that, for a complex scalar \ufb01eld with interactions speci\ufb01ed by Z(J \u2020, J) \u221d exp i d4x Z 1 i L1 L1(\u03d5\u2020, \u03d5), we have \u03b4 \u03b4J(x) \u03b4 \u03b4J \u2020(x) 1 i, Z0(J \u2020, J), (43.17) where the overall normalization is \ufb01xed by Z(0, 0) = 1. Thus, for a Dirac \ufb01eld with interactions speci\ufb01ed by Z(\u03b7, \u03b7) \u221d exp i d4x Z i L1 L1(\u03a8, \u03a8), we have \u03b4 \u03b4 1 \u03b4\u03b7(x) i \u03b4\u03b7(x), Z0(\u03b7, \u03b7), (43.18) where again the overall normalization is \ufb01xed by Z(0, 0) = 1. Vacuum expectation values of time-ordered products of Dirac \ufb01elds in an interacting theory will now be given by eq. (43.13), but with Z0(\u03b7, \u03b7) replaced by Z(\u03b7, \u03b7). Then, just as for a scalar \ufb01eld, this will lead to a Feynman-diagram expansion for Z(\u03b7, \u03b7). There are two extra complications: we must keep track of the spinor indices, and we must keep track of the extra minus signs from anticommutation. Both tasks are straightforward; we will take them up in section 45. Next, let us consider a Majorana \ufb01eld with L0 = i = 2 \u03a8T C 1 2 \u03a8T \u2212 i/\u2202 + m)\u03a8. 1 2 m\u03a8T /\u2202\u03a8 \u03a8 C ( \u2212 C A natural guess for the appropriate path-integral formula, based on analogy with eq. (43.2), is \u2212 (43.19) T\u03a8\u03b11(x1 \u03b4\u03b7\u03b11 (x1), \u03b7=0... Z0(\u03b7) (43.20) where Z0(\u03b7) = \u03a8 exp i Z D Z d4x ( L0 + \u03b7T\u03a8) = exp i 2 \u2212 Z The Feynman propagator S(x operator i/\u2202 + m): ( \u2212 C d4x d4y \u03b7T(x)S(x y) C \u2212 \u2212 1\u03b7(y). (43.21) y) C\u2212 \u2212 1 is the inverse of the Majorana wave ( \u2212 C i/\u2202x + m)S(x y) C \u2212 1 = \u03b44(x \u2212 y). \u2212 (43.22) 43: The Path Integral for Fermion Fields 275 The extra minus sign in eq. (43.21), as compared with eq. (43.14), arises because all", " functional derivatives in eq. (43.20) are accompanied by 1/i, rather than half by 1/i and half by i, as in eq. (43.13). It is now straightforward to check (by working out a few examples) that eqs. (43.20\u201343.22) do indeed reproduce the result of section 42 for the vacuum expectation value of a time-ordered product of Majorana \ufb01elds. 44: Formal Development of Fermionic Path Integrals 276 44 Formal Development of Fermionic Path Integrals Prerequisite: 43 In 43, we formally de\ufb01ned the fermionic path integral for a free Dirac \ufb01eld \u03a8 via Z0(\u03b7, \u03b7) = \u03a8 D D Z = exp i Z \u03a8 exp i d4x \u03a8(i/\u2202 Z d4x d4y \u03b7(x)S(x m)\u03a8 + \u03b7\u03a8 + \u03a8\u03b7 \u2212 \u2212 y)\u03b7(y), (44.1) where the Feynman propagator S(x operator: \u2212 y) is the inverse of the Dirac wave i/\u2202x + m)S(x ( \u2212 y) = \u03b44(x y). (44.2) \u2212 We would like to \ufb01nd a mathematical framework that allows us to derive this formula, rather than postulating it by analogy. \u2212 Consider a set of anticommuting numbers or Grassmann variables \u03c8i that obey \u03c8i, \u03c8j } { = 0, (44.3) where i = 1,..., n. Let us begin with the very simplest case of n = 1, and thus a single anticommuting number \u03c8 that obeys \u03c82 = 0. We can de\ufb01ne a function f (\u03c8) of such an object via a Taylor expansion; because \u03c82 = 0, this expansion ends with the second term: f (\u03c8) = a + \u03c8b. (44.4) The reason for writing the coe\ufb03cient b to the right of the variable \u03c8 will become clear in a moment. Next we would like to de\ufb01ne the derivative of f (\u03c8) with respect to \u03c8. Before we can do so, we must decide if f (\u03c8) itself is to be commuting or anticommuting; generally we will be interested in functions that are themselves commuting. In this case, a in eq. (44.4) should be treated as an ordinary commuting number, but b should be treated as an anticommuting number: = 0. In this case, f (\u03c8) = a + \u03c8b = a b\u03c8. = Now we can de\ufb01ne two kinds of derivatives. The left derivative of f (\u03c8) with respect to \u03c8 is given by the coe\ufb03cient of \u03c8 when f (\u03c8) is written with the \u03c8 always on the far left: b, b } { b, \u03c8 { } \u2212 \u2202\u03c8f (\u03c8) = +b. (44.5) 44: Formal Development of Fermionic Path Integrals 277 Similarly, the right derivative of f (\u03c8) with respect to \u03c8 is given by the coe\ufb03cient of \u03c8 when f (\u03c8) is written with the \u03c8 always on the far right: f (\u03c8)\u2190\u2202 \u03c8 = b. \u2212 (44.6) Generally, when we write a derivative with respect to a Grassmann variable, we mean the left derivative. However, in section 37, when we wrote the canonical momentum for a fermionic \ufb01eld \u03c8 as \u03c0 = \u2202 /\u2202(\u22020\u03c8), we actually meant the right derivative. (This is a standard, though rarely stated, convention.) Correspondingly, we wrote the hamiltonian density as L H \u2212 L, with \u22020\u03c8 to the right of \u03c0. = \u03c0\u22020\u03c8 Finally, we would like to de\ufb01ne a de\ufb01nite integral, analogous to integrating a real variable x from minus to plus in\ufb01nity. The key features of such an integral over x (when it converges) are linearity, + \u221e Z \u2212\u221e dx cf (x) = c + \u221e Z \u2212\u221e dx f (x), (44.7) and invariance under shifts of the dependent variable x by a constant: + \u221e Z \u2212\u221e dx f (x + a) = + \u221e Z \u2212\u221e dx f (x). (44.8) Up to an overall numerical factor that is the same for every f (", "\u03c8), the only possible nontrivial de\ufb01nition of d\u03c8 f (\u03c8) that is both linear and shift invariant is R d\u03c8 f (\u03c8) = b. Now let us generalize this to n > 1. We have Z f (\u03c8) = a + \u03c8ibi + 1 2 \u03c8i1\u03c8i2ci1i2 +... + 1 n! \u03c8i1... \u03c8indi1...in, (44.9) (44.10) where the indices are implicitly summed. Here we have written the coef\ufb01cients to the right of the variables to facilitate left-di\ufb00erentiation. These coe\ufb03cients are completely antisymmetric on exchange of any two indices. The left derivative of f (\u03c8) with respect to \u03c8j is \u2202 \u2202\u03c8j f (\u03c8) = bj + \u03c8icji +... + 1 (n \u2212 Next we would like to \ufb01nd a linear, shift-invariant de\ufb01nition of the integral of f (\u03c8). Note that the antisymmetry of the coe\ufb03cients implies that 1)! \u03c8i2... \u03c8indji2...in. (44.11) di1...in = d \u03b5i1...in. (44.12) where d is a just a number (ordinary if f is commuting and n is even, Grassmann if f is commuting and n is odd, etc.), and \u03b5i1...in is the completely 44: Formal Development of Fermionic Path Integrals 278 antisymmetric Levi-Civita symbol with \u03b51...n = +1. This number d is a candidate (in fact, up to an overall numerical factor, the only candidate!) for the integral of f (\u03c8): dn\u03c8 f (\u03c8) = d. (44.13) Z dn\u03c8, Although eq. (44.13) really tells us everything we need to know about we can, if we like, write dn\u03c8 = d\u03c8n... d\u03c81 (note the backwards ordering), and treat the individual di\ufb00erentials as anticommuting: = 0, d\u03c8i \u03c8j = \u03b4ij as our basic d\u03c8i, \u03c8j} { formulae, and use them to derive eq. (44.13). R d\u03c8i, d\u03c8j} { = 0. Then we take d\u03c8i = 0 and R R Let us work out some consequences of eq. (44.13). Consider what hap- pens if we make a linear change of variable, \u03c8i = Jij\u03c8\u2032j, (44.14) where Jji is a matrix of commuting numbers (and therefore can be written on either the left or right of \u03c8\u2032j). We now have f (\u03c8! (Ji1j1\u03c8\u2032j1)... (Jinjn\u03c8\u2032jn)\u03b5i1...ind. (44.15) Next we use \u03b5i1...inJi1j1... Jinjn = (det J)\u03b5j1...jn, (44.16) which holds for any n n matrix J, to get \u00d7 f (\u03c8! \u03c8\u2032i1... \u03c8\u2032in\u03b5i1...in(det J)d. (44.17) If we now integrate f (\u03c8) over dn\u03c8\u2032, eq. (44.13) tells us that the result is (det J)d. Thus, dn\u03c8 f (\u03c8) = (det J)\u2212 1 dn\u03c8\u2032 f (\u03c8). (44.18) Z Z Recall that, for integrals over commuting real numbers xi with xi = Jijx\u2032j, we have instead dnx f (x) = (det J)+1 dnx\u2032 f (x). (44.19) Z Z Note the opposite sign on the power of the determinant. Now consider a quadratic form \u03c8TM \u03c8 = \u03c8iMij\u03c8j, where M is an antisymmetric matrix of commuting numbers (possibly complex). Let\u2019s evalu2 \u03c8TM \u03c8). For example, for n = 2, we ate the gaussian integral have dn\u03c8 exp( 1 R M = 0 +m m 0! \u2212, (44.20) 44: Formal Development of Fermionic Path Integrals 279 and \u03c8TM \u03c8 = 2m\u03c81\u03c82. Thus exp", "( 1 2 \u03c8TM \u03c8) = 1 + m\u03c81\u03c82, and so dn\u03c8 exp( 1 2 \u03c8TM \u03c8) = m. Z (44.21) For larger n, we use the fact that a complex antisymmetric matrix can be brought to a block-diagonal form via 0 +m1 m1 \u2212 0... \uf8f7 \uf8f7 \uf8f7 \uf8f8 U TM 44.22) where U is a unitary matrix, and each mI is real and positive. (If n is odd there is a \ufb01nal row and column of all zeroes; from here on, we assume n is even.) We can now let \u03c8i = Uij\u03c8\u2032j; then, we have dn\u03c8 exp( 1 2 \u03c8TM \u03c8) = (det U )\u2212 1 Z n/2 d2\u03c8I exp( 1 2 \u03c8TMI \u03c8), (44.23) YI=1 Z 2 blocks in eq. (44.22). Each of these where MI represents one of the 2 two-dimensional integrals can be evaluated using eq. (44.21), and so \u00d7 dn\u03c8 exp( 1 2 \u03c8TM \u03c8) = (det U )\u2212 1 Z n/2 YI=1 mI. (44.24) Taking the determinant of eq. (44.22), we get (det U )2(det M ) = m2 I. n/2 YI=1 (44.25) We can therefore rewrite the right-hand side of eq. (44.24) as dn\u03c8 exp( 1 2 \u03c8TM \u03c8) = (det M )1/2. (44.26) Z In this form, there is a sign ambiguity associated with the square root; it is resolved by eq. (44.24). However, the overall sign (more generally, any overall numerical factor) will never be of concern to us, so we can use eq. (44.26) without worrying about the correct branch of the square root. It is instructive to compare eq. (44.26) with the corresponding gaussian integral for commuting real numbers, dnx exp( \u2212 Z 1 2 xTM x) = (2\u03c0)n/2(det M )\u2212 1/2. (44.27) 44: Formal Development of Fermionic Path Integrals 280 Here M is a complex symmetric matrix. Again, note the opposite sign on the power of the determinant. Now let us introduce the notion of complex Grassmann variables via \u03c7 \u00af\u03c7 \u2261 \u2261 1 \u221a2 1 \u221a2 (\u03c81 + i\u03c82), (\u03c81 \u2212 i\u03c82). We can invert this to get \u03c81 \u03c82! = 1 \u221a2 1 i 1 \u00af\u03c7 i! \u03c7! \u2212. The determinant of this transformation matrix is i, and so d2\u03c8 = d\u03c82d\u03c81 = ( \u2212 i)\u2212 \u2212 1d\u03c7 d \u00af\u03c7. (44.28) (44.29) (44.30) Also, \u03c81\u03c82 = \u2212 i \u00af\u03c7\u03c7. Thus we have d\u03c7 d \u00af\u03c7 \u00af\u03c7\u03c7 = ( \u2212 1 i)\u2212 i)( \u2212 Z Z Thus, if we have a function d\u03c82d\u03c81 \u03c81\u03c82 = 1. (44.31) its integral is In particular, f (\u03c7, \u00af\u03c7) = a + \u03c7b + \u00af\u03c7c + \u00af\u03c7\u03c7d, (44.32) d\u03c7 d \u00af\u03c7 f (\u03c7, \u00af\u03c7) = d. (44.33) Z d\u03c7 d \u00af\u03c7 exp(m \u00af\u03c7\u03c7) = m. (44.34) Let us now consider n complex Grassmann variables \u03c7i and their com- Z plex conjugates, \u00af\u03c7i. We de\ufb01ne dn\u03c7 dn\u00af\u03c7 \u2261 d\u03c7nd \u00af\u03c7n... d\u03c71d \u00af\u03c71. (44.35) Then under a change of variable, \u03c7i = Jij\u03c7\u2032j and \u00af\u03c7i = Kij \u00af\u03c7\u2032j, we have dn\u03c7 dn\u00af\u03c7 = (det J)\u2212 1(det K)\u2212 1 dn\u03c7\u2032 dn\u00af\u03c7\u2032. (44.36) Note that we need not require Kij = J \u2217ij, because, as far as the integral is concerned, it is does not matter whether or not \u00af\ufffd", "\ufffdi is the complex conjugate of \u03c7i. We now have enough information to evaluate dn\u03c7 dn\u00af\u03c7 exp(\u03c7\u2020M \u03c7), where M is a general complex matrix. We make the change of variable R 44: Formal Development of Fermionic Path Integrals 281 \u03c7 = U \u03c7\u2032 and \u03c7\u2020 = \u03c7\u2032\u2020V, where U and V are unitary matrices with the property that V M U is diagonal with positive real entries mi. Then we get dn\u03c7 dn\u00af\u03c7 exp(\u03c7\u2020M \u03c7) = (det U )\u2212 1(det V )\u2212 1 d\u03c7id \u00af\u03c7i exp(mi \u00af\u03c7i\u03c7i) n Z = (det U )\u2212 1(det V )\u2212 1 Yi=1 Z n mi Yi=1 = det M. (44.37) This can be compared to the analogous integral for commuting complex iyi)/\u221a2, with dnz dn\u00afz = dnx dny, variables zi = (xi + iyi)/\u221a2 and \u00afz = (xi \u2212 namely dnz dn\u00afz exp( \u2212 Z z\u2020M z) = (2\u03c0)n(det M )\u2212 1. (44.38) We can now generalize eqs. (44.26) and (44.37) by shifting the integration variables, and using shift invariance of the integrals. Thus, by making the replacement \u03c8 1\u03b7 in eq. (44.26), we get \u03c8 M \u2212 \u2192 \u2212 dn\u03c8 exp( 1 2 \u03c8TM \u03c8 + \u03b7T\u03c8) = (det M )1/2 exp( 1 2 \u03b7TM \u2212 1\u03b7). (44.39) Z (In verifying this, remember that M and its inverse are both antisym1\u03b7 and metric.) Similarly, by making the replacements \u03c7 \u03c7\u2020 \u2192 \u03c7\u2020 \u2212 dn\u03c7 dn\u00af\u03c7 exp(\u03c7\u2020M \u03c7 + \u03b7\u2020\u03c7 + \u03c7\u2020\u03b7) = (det M ) exp( 1 in eq. (44.37), we get \u03b7\u2020M \u2212 1\u03b7). (44.40) \u03b7\u2020M \u2212 M \u2212 \u2192 \u2212 \u03c7 Z We can now see that eq. (44.1) is simply a particular case of eq. (44.40), with the index on the complex Grassmann variable generalized to include both the ordinary spin index \u03b1 and the continuous spacetime argument x of the \ufb01eld \u03a8\u03b1(x). Similarly, eq. (43.21) for the path integral for a free Majorana \ufb01eld is simply a particular case of eq. (44.39). In both cases, the determinant factors are constants (that is, independent of the \ufb01elds and sources) that we simply absorb into the overall normalization of the path integral. We will meet determinants that cannot be so neatly absorbed in sections 53 and 71. \u2212 45: The Feynman Rules for Dirac Fields 282 45 The Feynman Rules for Dirac Fields Prerequisite: 10, 12, 41, 43 In this section we will derive the Feynman rules for Yukawa theory, a theory with a Dirac \ufb01eld \u03a8 (with mass m) and a real scalar \ufb01eld \u03d5 (with mass M ), interacting via L1 = g\u03d5\u03a8\u03a8, where g is a coupling constant. In this section, we will be concerned with tree-level processes only, and so we omit renormalizing Z factors. (45.1) In four spacetime dimensions, \u03d5 has mass dimension [\u03d5] = 1 and \u03a8 has mass dimension [\u03a8] = 3 2 ; thus the coupling constant g is dimensionless: [g] = 0. As discussed in section 12, this is generally the most interesting situation. Note that i\u03b1\u03a8, L1 is invariant under the U(1) transformation \u03a8 as is the free Dirac lagrangian. Thus, the corresponding Noether current \u03a8\u03b3\u00b5\u03a8 is still conserved, and the associated charge Q (which counts the number of b-type particles minus the number of d-type particles) is constant in time. e\u2212 \u2192 We can think of Q as electric charge, and identify the b-type particle as the electron e\u2212, and the d-type particle as the positron e+. The scalar particle is electrically neutral (and could, for example, be thought of as the Higgs boson; see section 88). We now use the general result of sections 9 and 43 to write Z(\u03b7, \u03b7, J) \u221d exp ig", " d4x Z 1 i \u03b4 \u03b4J(x) \u03b4 \u03b4\u03b7\u03b1(x) i 1 i \u03b4 \u03b4\u03b7\u03b1(x) Z0(\u03b7, \u03b7, J), (45.2) where and Z0(\u03b7, \u03b7, J) = exp Z exp i \u00d7 i 2 Z d4x d4y \u03b7(x)S(x \u2212 d4x d4y J(x)\u2206(x y)\u03b7(y) y)J(y) \u2212 S(x \u2206(x \u2212 \u2212 y) = y) = Z Z d4p (2\u03c0)4 ( \u2212 /p + m)eip(x \u2212 p2 + m2 i\u01eb \u2212 y), d4k (2\u03c0)4 eik(x \u2212 k2 + M 2 y) i\u01eb \u2212 are the appropriate Feynman propagators for the corresponding free \ufb01elds. We impose the normalization Z(0, 0, 0) = 1, and write Z(\u03b7, \u03b7, J) = exp[iW (\u03b7, \u03b7, J)]. (45.6), (45.3) (45.4) (45.5) 45: The Feynman Rules for Dirac Fields 283 (a) (c) (b) (d) Figure 45.1: Tree contributions to iW (\u03b7, \u03b7, J) with four or fewer sources. Then iW (\u03b7, \u03b7, J) can be expressed as a series of connected Feynman diagrams with sources. \u2212 \u2212 R R i S(x i \u2206(x We use a dashed line to stand for the scalar propagator 1 y), and a solid line to stand for the fermion propagator 1 y). The only allowed vertex joins two solid lines and one dashed line; the associated vertex factor is ig. The blob at the end of a dashed line stands for the \u03d5 d4x J(x), and the blob at the end of a solid line for either the \u03a8 source i d4x \u03b7(x). To tell which is which, we d4x \u03b7(x), or the \u03a8 source i source i d4x \u03b7(x) if the adopt the \u201carrow rule\u201d of problem 9.3: the blob stands for i R arrow on the attached line points away from the blob, and the blob stands d4x \u03b7(x) if the arrow on the attached line points towards the blob. for i L1 involves one \u03a8 and one \u03a8, we also have the rule that, at each Because vertex, one arrow must point towards the vertex, and one away. The \ufb01rst few tree diagrams that contribute to iW (\u03b7, \u03b7, J) are shown in \ufb01g. (45.1). We omit tadpole diagrams; as in \u03d53 theory, these can be cancelled by shifting the \u03d5 \ufb01eld, or, equivalently, adding a term linear in \u03d5 to. The LSZ formula is valid only after all tadpole diagrams have been cancelled in this way. L R R The spin indices on the fermionic sources and propagators are all contracted in the obvious way. For example, the complete expression corresponding to \ufb01g. (45.1)(b) is Fig. (45.1)(b) = i3 3 1 i (ig) d4x d4y d4z d4w Z \u00d7 \u00d7 \u03b7(x)S(x y)S(y \u2212 w)J(w). h \u2206(y \u2212 \u2212 z)\u03b7(z) i (45.7) Our main purpose in this section is to compute the tree-level amplitudes 45: The Feynman Rules for Dirac Fields 284 y w1 ww 2 x y w1 ww 2 x z1 z2 z 2 z 1 Figure 45.2: Diagrams corresponding to eq. (45.8). e\u2212\u03d5 and for various two-body elastic scattering processes, such as e\u2212\u03d5 e+e\u2212 \u2192 \u03d5\u03d5; for these, we will need to evaluate the tree-level contributions iC. Other pro0 T\u03a8\u03a8\u03d5\u03d5 0 to connected correlation functions of the form | | h e+e\u2212; for these, we e\u2212e\u2212 and e+e\u2212 \u2192 cesses of interest include e\u2212e\u2212 \u2192 will need to evaluate the tree-level contributions to connected correlation iC. 0 T\u03a8\u03a8\u03a8\u03a8 functions of the form 0 | h | T\u03a8\u03a8\u03d5\u03d5 0 iC, the relevant tree-level contribution to iW (\ufffd", "\ufffd, \u03b7, J) 0 | h | For \u2192 is given by \ufb01g. (45.1)(c). We have 0 T\u03a8\u03b1(x)\u03a8\u03b2(y)\u03d5(z1)\u03d5(z2) 0 iC | | h \u03b4 \u03b4J(z1) \u03b4 \u03b4\u03b7\u03b2(y) \u03b4 \u03b4\u03b7\u03b1(x \u03b4J(z2) = 1 i 5 (ig)2 Z + z1 \u2194 d4w1 d4w2 [S(x \u2212 \u2206(z1\u2212 + O(g4). \u00d7 \u00d7 z2 w1)S(w1\u2212 w1)\u2206(z2\u2212 iW (\u03b7, \u03b7, J) \u03b7=\u03b7=J=0 w2)S(w2\u2212 w2) y)]\u03b1\u03b2 (45.8) The corresponding diagrams, with sources removed, are shown in \ufb01g. (45.2). iC, the relevant tree-level contribution to iW (\u03b7, \u03b7, J) 0 | is given by \ufb01g. (45.1)(d), which has a symmetry factor S = 2. We have T\u03a8\u03a8\u03a8\u03a8 0 | h For T\u03a8\u03b11(x1)\u03a8\u03b21(y1)\u03a8\u03b12(x2)\u03a8\u03b22(y2) 0 0 iC | h | = 1 i \u03b4 \u03b4\u03b7\u03b11(x1) i \u03b4 \u03b4\u03b7\u03b21(y1) 1 i \u03b4 \u03b4\u03b7\u03b12(x2) i \u03b4 \u03b4\u03b7\u03b22 (y2) iW (\u03b7, \u03b7, J). \u03b7=\u03b7=J=0 (45.9) The two \u03b7 derivatives can act on the two \u03b7\u2019s in the diagram in two di\ufb00erent ways; ditto for the two \u03b7 derivatives. This results in four di\ufb00erent terms, but two of them are algebraic duplicates of the other two; this duplication cancels the symmetry factor (which is a general result for tree diagrams). 45: The Feynman Rules for Dirac Fields 285 Figure 45.3: Diagrams corresponding to eq. (45.10). We get 0 T\u03a8\u03b11 (x1)\u03a8\u03b21(y1)\u03a8\u03b12 (x2)\u03a8\u03b22(y2) 0 iC | | h = 5 1 i (ig)2 Z d4w1 d4w2 [S(x1\u2212 \u2206(w1\u2212 [S(x2\u2212 \u00d7 \u00d7 w1)S(w1\u2212 w2) w2)S(w2\u2212 y1)]\u03b11\u03b21 y2)]\u03b12\u03b22 + O(g4). (y2, \u03b22) \u00d7 (y1, \u03b21) (45.10) \u2212 \u2194 The corresponding diagrams, with sources removed, are shown in \ufb01g. (45.3). Note that we now have a relative minus sign between the two diagrams, due to the anticommutation of the derivatives with respect to \u03b7. In general, the overall sign for a diagram can be determined by the following procedure. First, draw each diagram with all the fermion lines horizontal, with their arrows pointing from left to right, and with the left endpoints labeled in the same \ufb01xed order (from top to bottom). Next, in each diagram, note the ordering (from top to bottom) of the labels on the right endpoints of the fermion lines. If this ordering is an even permutation of an arbitrarily chosen \ufb01xed ordering, then the sign of that diagram is positive, and if it is an odd permutation, the sign is negative. (This rule arises because endpoints with arrows pointing away from the vertex come from derivatives with respect to \u03b7 that anticommute. Of course, we could equally well put the right endpoints in a \ufb01xed order, and get the sign from the permutation of the left endpoints, which come from derivatives with respect to \u03b7 that anticommute.) Also, in loop diagrams, a closed fermion loop yields an extra minus sign; we will discuss this rule in section 51. Let us now consider a particular scattering process: e\u2212\u03d5 scattering amplitude is e\u2212\u03d5. The \u2192 (k\u2032)outbs\u2032(p\u2032)outb\u2020s(p)ina\u2020(k)in | 0 i. (45.11) Next we make the replacements b\u2020s(p)in \u2192 i Z d4y \u03a8(y)(+i\u2190", "/\u2202 + m)us(p) e+ipy, (45.12) 45: The Feynman Rules for Dirac Fields 286 p k p+k p k p k p k p k Figure 45.4: Diagrams for e\u2212\u03d5 \u2192 e\u2212\u03d5, corresponding to eq. (45.16). bs\u2032(p\u2032)out \u2192 a\u2020(k)in \u2192 a(k\u2032)out \u2192 i i i Z Z Z d4x e\u2212 ip\u2032x us\u2032(p\u2032)( \u2212 i/\u2202 + m)\u03a8(x), (45.13) d4z1 e+ikz1( \u22022 + m2)\u03d5(z1), d4z2 e\u2212 \u22022 + m2)\u03d5(z2) \u2212 ik\u2032z2( \u2212 (45.14) (45.15) in eq. (45.11), and then use eq. (45.8). The wave operators (either KleinGordon or Dirac) act on the external propagators, and convert them to delta functions. After using eqs. (45.4) and (45.5) for the internal propagators, all dependence on the various spacetime coordinates is in the form Integrating over the internal coof plane-wave factors, as in section 10. ordinates then generates delta functions that conserve four-momentum at each vertex. The only new feature arises from the spinor factors us(p) and us\u2032(p\u2032). We \ufb01nd that us(p) is associated with the external fermion line whose arrow points towards the vertex, and that us\u2032(p\u2032) is associated with the external fermion line whose arrow points away from the vertex. We can therefore draw the momentum-space diagrams of \ufb01g. (45.4). Since there is only one fermion line in each diagram, the relative sign is positive. The tree-level e\u2212\u03d5 e\u2212\u03d5 scattering amplitude is then given by \u2192 i Te\u2212\u03d5 \u2192 e\u2212\u03d5 = 1 i (ig)2 us\u2032(p\u2032) \" /p \u2212 \u2212 /k + m s + m2 + \u2212 \u2212 \u2212 /p + /k\u2032 + m u + m2 # us(p), (45.16) (p+k)2 and u = k\u2032)2. (We can safely ignore the i\u01eb\u2019s in the where s = propagators, because their denominators cannot vanish for any physically allowed values of s and u.) (p \u2212 \u2212 \u2212 Next consider the process e+\u03d5 e+\u03d5. We now have (k\u2032)outds\u2032(p\u2032)outd\u2020s(p)ina\u2020(k)in | 0 i. (45.17) The relevant replacements are d\u2020s(p)in \u2192 \u2212 i Z d4x e+ipx vs(p)( \u2212 i/\u2202 + m)\u03a8(x), (45.18) 45: The Feynman Rules for Dirac Fields 287 p k p+k p k p k p k p k Figure 45.5: Diagrams for e+\u03d5 \u2192 e+\u03d5, corresponding to eq. (45.22). ds\u2032(p\u2032)out \u2192 \u2212 i a\u2020(k)in \u2192 a(k\u2032)out \u2192 i i Z Z d4y \u03a8(y)(+i\u2190/\u2202 + m)vs\u2032(p\u2032) e\u2212 ipy, (45.19) Z d4z1 e+ikz1( \u2212 \u22022 + m2)\u03d5(z1), d4z2 e\u2212 ikz2( \u2212 \u22022 + m2)\u03d5(z2). (45.20) (45.21) We substitute these into eq. (45.17), and then use eq. (45.8). This ultimately leads to the momentum-space Feynman diagrams of \ufb01g. (45.5). Note that we must now label the external fermion lines with minus their fourmomenta; this is characteristic of d-type particles. (The same phenomenon occurs for a complex scalar \ufb01eld; see problem 10.2.) Regarding the spinor vs(p) is associated with the external fermion line factors, we \ufb01nd that \u2212 vs\u2032(p\u2032) with the external whose arrow points away from the vertex, and fermion line whose arrow points towards the vertex. The minus signs attached to each v and v can be consistently dropped, however, as they only a\ufb00ect the overall sign of the amplitude (and not the relative signs among contributing diagrams). The tree-level expression for the", " e+\u03d5 e+\u03d5 amplitude is then \u2192 \u2212 i Te+\u03d5 \u2192 e+\u03d5 = 1 i (ig)2 vs(p) \" /p /k\u2032 + m u + m2 + \u2212 \u2212 /p + /k + m s + m2 # vs\u2032(p\u2032), (45.22) (p + k)2 and u = where again s = (p After working out a few more of these (you might try your hand at some of them before reading ahead), we can abstract the following set of Feynman rules. \u2212 \u2212 \u2212 \u2212 k\u2032)2. 1. For each incoming electron, draw a solid line with an arrow pointed towards the vertex, and label it with the electron\u2019s four-momentum, pi. 2. For each outgoing electron, draw a solid line with an arrow pointed away from the vertex, and label it with the electron\u2019s four-momentum, p\u2032i. 45: The Feynman Rules for Dirac Fields 288 3. For each incoming positron, draw a solid line with an arrow pointed away from the vertex, and label it with minus the positron\u2019s fourmomentum, pi. \u2212 4. For each outgoing positron, draw a solid line with an arrow pointed towards the vertex, and label it with minus the positron\u2019s four-momentum, p\u2032i. \u2212 5. For each incoming scalar, draw a dashed line with an arrow pointed towards the vertex, and label it with the scalar\u2019s four-momentum, ki. 6. For each outgoing scalar, draw a dashed line with an arrow pointed away from the vertex, and label it with the scalar\u2019s four-momentum, k\u2032i. 7. The only allowed vertex joins two solid lines, one with an arrow pointing towards it and one with an arrow pointing away from it, and one dashed line (whose arrow can point in either direction). Using this vertex, join up all the external lines, including extra internal lines as needed. In this way, draw all possible diagrams that are topologically inequivalent. 8. Assign each internal line its own four-momentum. Think of the fourmomenta as \ufb02owing along the arrows, and conserve four-momentum at each vertex. For a tree diagram, this \ufb01xes the momenta on all the internal lines. 9. The value of a diagram consists of the following factors: for each incoming or outgoing scalar, 1; for each incoming electron, usi(pi); (p\u2032i); for each outgoing electron, us\u2032 for each incoming positron, vsi(pi); (p\u2032i); for each outgoing positron, vs\u2032 for each vertex, ig; i i for each internal scalar, \u2212 for each internal fermion, i/(k2 + M 2 i\u01eb); \u2212 /p + m)/(p2 + m2 i( \u2212 \u2212 i\u01eb). \u2212 10. Spinor indices are contracted by starting at one end of a fermion line: speci\ufb01cally, the end that has the arrow pointing away from the vertex. The factor associated with the external line is either u or v. Go along the complete fermion line, following the arrows backwards, and write down (in order from left to right) the factors associated with the 45: The Feynman Rules for Dirac Fields 289 1p 1p k 1 k 1 1p p2 p2 k2 k2 p2 k 1 Figure 45.6: Diagrams for e+e\u2212 \u2192 \u03d5\u03d5, corresponding to eq. (45.23). vertices and propagators that you encounter. The last factor is either a u or v. Repeat this procedure for the other fermion lines, if any. 11. The overall sign of a tree diagram is determined by drawing all contributing diagrams in a standard form: all fermion lines horizontal, with their arrows pointing from left to right, and with the left endpoints labeled in the same \ufb01xed order (from top to bottom); if the ordering of the labels on the right endpoints of the fermion lines in a given diagram is an even (odd) permutation of an arbitrarily chosen \ufb01xed ordering, then the sign of that diagram is positive (negative). 12. The value of i (at tree level) is given by a sum over the values of the contributing diagrams. T There are additional rules for counterterms and loops; in particular, each closed fermion loop contributes an extra minus sign. We will postpone discussion of loop corrections to section 51. Let us apply these rules to e+e\u2212 \u2192 \u03d5\u03d5. Let the initial electron and positron have four-momenta p1 and p2, respectively, and the two", " \ufb01nal scalars have four-momenta k\u20321 and k\u20322. The relevant diagrams are shown in \ufb01g. (45.6); there is only one fermion line, and so the relative sign is positive. The result is i Te+e\u2212 \u2192 \u03d5\u03d5 = 1 i (ig)2 vs2(p2) \" \u2212 /p1 + /k\u20321 + m t + m2 + \u2212 /p1 + /k\u20322 + m u + m2 us1(p1), # \u2212 where t = k\u20321)2 and u = (p1 \u2212 Next, consider e\u2212e\u2212 \u2192 \u2212 (p1 \u2212 e\u2212e\u2212. Let the initial electrons have fourmomenta p1 and p2, and the \ufb01nal electrons have four-momenta p\u20321 and p\u20322. The relevant diagrams are shown in \ufb01g. (45.7), and according to rule #11 the relative sign is negative. Thus the result is (45.23) k\u20322)2. \u2212 \u2212 i Te\u2212e\u2212 \u2192 e\u2212e\u2212 = 1 i (ig)2 (u \u20321u1)(u \u20322u2) t + M 2 \u2212 \u2212 (u \u20322u1)(u \u20321u2) u + M 2 \u2212, (45.24) 45: The Feynman Rules for Dirac Fields 290 p1 p 2 p1 p 1 p2 p1 p1 p1 p 2 p 2 p 2 p1 Figure 45.7: Diagrams for e\u2212e\u2212 \u2192 e\u2212e\u2212, corresponding to eq. (45.24). p1 p1 p1 p1 p2 p2 p1 p1 p1 p2+ p2 p2 Figure 45.8: Diagrams for e+e\u2212 \u2192 e+e\u2212, corresponding to eq. (45.25). One more: e+e\u2212 \u2192 where u1 is short for us1(p1), etc., and t = p\u20322)2. (p1 \u2212 e+e\u2212. Let the initial electron and positron have four-momenta p1 and p2, respectively, and the \ufb01nal electron and positron have four-momenta p\u20321 and p\u20322, respectively. The relevant diagrams are shown in \ufb01g. (45.8). If we redraw them in the the standard form of rule #11, as shown in \ufb01g. (45.9), we see that the relative sign is negative. Thus the result is p\u20321)2, u = (p1 \u2212 \u2212 \u2212 i Te+e\u2212 \u2192 e+e\u2212 = 1 i (ig)2 p1 (u \u20321u1)(v2v\u20322) \u2212 p1 t + M 2 \u2212 p1 (v2u1)(u \u20321v\u20322) u + M 2 \u2212, p2 (45.25) p1 p1 p1 p2 p2 p2 p2 p1 Figure 45.9: Same as \ufb01g. (45.8), but with the diagrams redrawn in the standard form given in rule #11. 45: The Feynman Rules for Dirac Fields 291 where s = \u2212 (p1 + p2)2 and t = p\u20321)2. \u2212 (p1 \u2212 Problems 45.1) a) Determine how \u03d5(x) must transform under parity, time reversal, and charge conjugation in order for these to all be symmetries of the theory. (Prerequisite: 40) b) Same question, but with the interaction given by instead of eq. (45.1). L1 = ig\u03d5\u03a8\u03b35\u03a8 45.2) Use the Feynman rules to write down (at tree level) i for the pro- T cesses e+e+ e+e+ and \u03d5\u03d5 e+e\u2212. \u2192 \u2192 46: Spin Sums 292 46 Spin Sums Prerequisite: 45 In the last section, we calculated various tree-level scattering amplitudes in Yukawa theory. For example, for e\u2212\u03d5 e\u2212\u03d5 we found \u2192 = g2 us\u2032(p\u2032) \" T /p \u2212 \u2212 /k + m s + m2 + \u2212 \u2212 \u2212 /p + /k\u2032 + m u + m2 # us(p), (46.1) \u2212 (p + k)2 and u = where s = responding cross section, we must evaluate simplifying eq. (46.1) a little; we use (/p + m)us(p) = 0 to replace the each numerator with m. We then abbreviate eq. (46.1) as k\u2032)2", ". In order to compute the corT T \u2217. We begin by /p in 2 = |T | (p \u2212 \u2212 \u2212 = u \u2032Au, T (46.2) where Then we have where in general A Thus we have \u2261 g2 A \u2261 \" /k + 2m \u2212 m2 s \u2212 + /k\u2032 + 2m m2 u # \u2212. (46.3) \u2217 = T T = u \u2032Au = uAu\u2032, (46.4) \u03b2A\u2020\u03b2, and, for the particular A of eq. (46.3), A = A. 2 = (u \u2032Au)(uAu\u2032) |T | u \u2032\u03b1A\u03b1\u03b2u\u03b2u\u03b3A\u03b3\u03b4u\u2032\u03b4 u\u2032\u03b4u \u2032\u03b1A\u03b1\u03b2u\u03b2u\u03b3A\u03b3\u03b4 = = X\u03b1\u03b2\u03b3\u03b4 X\u03b1\u03b2\u03b3\u03b4 Next, we use a result from section 38: = Tr (u\u2032u \u2032)A(uu)A i h. (46.5) us(p)us(p) = 1 2 (1 s\u03b35/z)( \u2212 /p + m), (46.6) \u2212 where s = tization axis z. We then have \u00b1 tells us whether the spin is up or down along the spin quan- |T | 2 = 1 4 Tr h (1 \u2212 s\u2032\u03b35/z\u2032)( \u2212 /p \u2032 + m)A(1 \u2212 s\u03b35/z)( \u2212 /p + m)A. (46.7) i We now simply need to take traces of products of gamma matrices; we will work out the technology for this in the next section. 46: Spin Sums 293 However, in practice, we are often not interested in (or are unable to easily measure or prepare) the spin states of the scattering particles. Thus, if we know that an electron with momentum p\u2032 landed in our detector, but 2 over the two possible spin know nothing about its spin, we should sum states of this outgoing electron. Similarly, if the spin state of the initial electron is not specially prepared for each scattering event, then we should 2 over the two possible spin states of this initial electron. Then average |T | we can use |T | us(p)us(p) = /p + m \u2212 s= X \u00b1 (46.8) in place of eq. (46.6). Let us, then, take initial spins, and call the result |T | 2, sum over all \ufb01nal spins, and average over all. In the present case, we have i 2 h|T | 2 2 h|T | i \u2261 1 2 Xs,s\u2032 |T | ( 2 Tr \u2212 h = 1 /p \u2032 + m)A( /p + m)A, (46.9) \u2212 i which is much less cumbersome than eq. (46.7). Next let\u2019s try something a little harder, namely e+e\u2212 \u2192 found in section 45 that e+e\u2212. We We then have = g2 T (u \u20321u1)(v2v\u20322) (v2u1)(u \u20321v\u20322) M 2 \u2212 t \u2212 M 2 s \u2212 = g2 T (u1u\u20321)(v \u20322v2) (u1v2)(v \u20322u\u2032146.10) (46.11) When we multiply, we will get four terms. We want to arrange the factors in each of them so that every u and every v stands just to the left of the corresponding u and v. In this way, we get by T T 2 = + |T | + g4 (M 2 \u2212 g4 (M 2 \u2212 \u2212 (M 2 \u2212 (M 2 \u2212 \u2212 t)2 Tr h u1u1u\u20321u \u20321 Tr v\u20322v \u20322v2v2 h i u1u1v2v2 s)2 Tr h g4 t)(M 2 Tr s) \u2212 i i Tr v\u20322v \u20322u\u20321u \u20321 h i u1u1v2v2v\u20322v \u20322u\u20321u \u20321 h i g4 s)(M 2 Tr u1u1u\u20321u \u20321v\u20322v \u20322v2v2 h i. t) \u2212 (46.12) 46: Spin Sums 294 Then we average over initial spins and sum over \ufb01nal spins, and use eq. (46.8) and vs(p)vs(p) = /p \u2212 \u2212 m. s= X \u00b1 (46.13) We then must evaluate traces of products of up to four gamma matrices", ". 47: Gamma Matrix Technology 295 47 Gamma Matrix Technology Prerequisite: 36 In this section, we will learn some tricks for handling gamma matrices. We need the following information as a starting point: 2g\u00b5\u03bd, = \u03b3\u00b5, \u03b3\u03bd } \u2212 { \u03b32 5 = 1, \u03b3\u00b5, \u03b35} = 0, { Tr 1 = 4. (47.1) (47.2) (47.3) (47.4) Now consider the trace of the product of n gamma matrices. We have Tr[\u03b3\u00b51... \u03b3\u00b5n] = Tr[\u03b32 5 \u03b3\u00b5n] 5 \u03b3\u00b51\u03b32 5... \u03b32 = Tr[(\u03b35\u03b3\u00b51\u03b35)... (\u03b35\u03b3\u00b5n \u03b35)] 5 \u03b3\u00b51)... ( \u03b32 5\u03b3\u00b5n )] \u03b32 \u2212 1)n Tr[\u03b3\u00b51... \u03b3\u00b5n]. = Tr[( = ( \u2212 \u2212 (47.5) We used eq. (47.2) to get the \ufb01rst equality, the cyclic property of the trace for the second, eq. (47.3) for the third, and eq. (47.2) again for the fourth. If n is odd, eq. (47.5) tells us that this trace is equal to minus itself, and must therefore be zero: Tr[ odd # of \u03b3\u00b5\u2019s ] = 0. Similarly, Tr[ \u03b35 ( odd # of \u03b3\u00b5\u2019s ) ] = 0. Next, consider Tr[\u03b3\u00b5\u03b3\u03bd]. We have Tr[\u03b3\u00b5\u03b3\u03bd] = Tr[\u03b3\u03bd \u03b3\u00b5] = 1 2 Tr[\u03b3\u00b5\u03b3\u03bd + \u03b3\u03bd\u03b3\u00b5] g\u00b5\u03bd Tr 1 4g\u00b5\u03bd. \u2212 = = \u2212 (47.6) (47.7) (47.8) The \ufb01rst equality follows from the cyclic property of the trace, the second averages the left- and right-hand sides of the \ufb01rst, the third uses eq. (47.1), and the fourth uses eq. (47.4). A slightly nicer way of expressing eq. (47.8) is to introduce two arbitrary four-vectors a\u00b5 and b\u00b5, and write Tr[/a/b] = 4(ab), \u2212 (47.9) 47: Gamma Matrix Technology 296 where /a = a\u00b5\u03b3\u00b5, /b = b\u00b5\u03b3\u00b5, and (ab) = a\u00b5b\u00b5. Next consider Tr[/a/b/c/d]. We evaluate this by moving /a to the right, using eq. (47.1), which is now more usefully written as Using this repeatedly, we have /a/b = /b/a \u2212 \u2212 2(ab). (47.10) Tr[/a/b/c/d] = Tr[/b/a/c/d] 2(ab)Tr[/c/d] \u2212 \u2212 = +Tr[/b/c/a/d] + 2(ac)Tr[/b/d] 2(ab)Tr[/c/d] \u2212 2(ad)Tr[/b/c] + 2(ac)Tr[/b/d] Tr[/b/c/d/a] = \u2212 \u2212 2(ab)Tr[/c/d]. (47.11) \u2212 Now we note that the \ufb01rst term on the right-hand side of the last line is, by the cyclic property of the trace, equal to minus the left-hand side. We can then move this term to the left-hand side to get 2 Tr[/a/b/c/d] = \u2212 2(ad)Tr[/b/c] + 2(ac)Tr[/b/d] \u2212 2(ab)Tr[/c/d]. (47.12) Finally, we evaluate each Tr[/a/b] with eq. (47.9), and divide by two: Tr[/a/b/c/d] = 4 (ad)(bc) \u2212 (ac)(bd) + (ab)(cd). (47.13) i h This is our \ufb01nal result for this trace. Clearly, we can use the same technique to evaluate the trace of the product of any even number of gamma matrices. Next, let\u2019s consider traces that involve \u03b35\u2019s and \u03b3\u00b5\u2019s. Since = 0, we can always bring all the \u03b35\u2019s together by moving", " them through the \u03b3\u00b5\u2019s (generating minus signs as we go). Then, since \u03b32 5 = 1, we end up with either one \u03b35 or none. So we need only consider Tr[\u03b35\u03b3\u00b51... \u03b3\u00b5n]. And, according to eq. (47.7), we need only be concerned with even n. \u03b35, \u03b3\u00b5 { } Recall that an explicit formula for \u03b35 is Eq. (47.13) then implies Similarly, we can show that \u03b35 = i\u03b30\u03b31\u03b32\u03b33. Tr \u03b35 = 0. Tr[\u03b35\u03b3\u00b5\u03b3\u03bd] = 0. (47.14) (47.15) (47.16) Finally, consider Tr[\u03b35\u03b3\u00b5\u03b3\u03bd\u03b3\u03c1\u03b3\u03c3]. The only way to get a nonzero result is to have the four vector indices take on four di\ufb00erent values. If we consider the 297 1 \u2212 47: Gamma Matrix Technology special case Tr[\u03b35\u03b33\u03b32\u03b31\u03b30], plug in eq. (47.14), and then use (\u03b3i)2 = and (\u03b30)2 = 1, we get i( 4i, or equivalently 1)3 Tr 1 = \u2212 Tr[\u03b35\u03b3\u00b5\u03b3\u03bd \u03b3\u03c1\u03b3\u03c3] = \u2212 4i\u03b5\u00b5\u03bd\u03c1\u03c3, \u2212 (47.17) where \u03b50123 = \u03b53210 = +1. Another category of gamma matrix combinations that we will eventually encounter is \u03b3\u00b5/a... \u03b3\u00b5. The simplest of these is \u03b3\u00b5\u03b3\u00b5 = g\u00b5\u03bd \u03b3\u00b5\u03b3\u03bd = 1 \u03b3\u00b5, \u03b3\u03bd 2 g\u00b5\u03bd { g\u00b5\u03bd g\u00b5\u03bd \u2212 } = = d. \u2212 (47.18) To get the second equality, we used the fact that g\u00b5\u03bd is symmetric, and so only the symmetric part of \u03b3\u00b5\u03b3\u03bd contributes. In the last line, d is the number of spacetime dimensions. Of course, our entire spinor formalism has been built around d = 4, but we will need formal results for d = 4 \u03b5 when we dimensionally regulate loop diagrams involving fermions. \u2212 We move on to evaluate \u03b3\u00b5/a\u03b3\u00b5 = \u03b3\u00b5( 2a\u00b5) \u2212 2/a \u2212 \u03b3\u00b5/a \u2212 \u03b3\u00b5\u03b3\u00b5/a 2)/a. \u2212 = \u2212 = (d We continue with and \u03b3\u00b5/a/b\u03b3\u00b5 = 4(ab) (d \u2212 \u2212 4)/a/b \u03b3\u00b5/a/b/c\u03b3\u00b5 = 2/c/b/a + (d \u2212 4)/a/b/c ; the derivations are left as an exercise. Problems 47.1) Verify eq. (47.16). 47.2) Verify eqs. (47.20) and (47.21). (47.19) (47.20) (47.21) 47.3) Show that the most general 4 4 matrix can be written as a linear combination (with complex coe\ufb03cients) of 1, \u03b3\u00b5, S\u00b5\u03bd, \u03b3\u00b5\u03b35, and \u03b35, where 1 is the identity matrix and S\u00b5\u03bd = i 4 [\u03b3\u00b5, \u03b3\u03bd ]. Hint: if A and B are two di\ufb00erent members of this set, prove linear independence by showing that Tr A\u2020B = 0 vanishes. Then count. \u00d7 48: Spin-Averaged Cross Sections 298 48 Spin-Averaged Cross Sections Prerequisite: 46, 47 |T | 2 for (among other processes) e+e\u2212 \u2192 e+e\u2212. In section 46, we computed We take the incoming and outgoing electrons to have momenta p1 and p\u20321, respectively, and the incoming and outgoing positrons to have momenta p2 m2, where m is the electron 2 and p\u20322, respectively. We have p2 i = i = p\u2032 (and positron) mass. The Mandelstam variables are (p\u20321 + p\u20322)2, p\u20322)2, (p2 \u2212 p\u20321)2, u = (p2 \u2212 and they obey s + t + u = 4m2. Our result was (p1 + p2)2 = p\u20321)2 = (p1 \u2212 p\u20322)2 = (p1 \u2212 (48.1 = g4 \u03a6ss |T | (M 2 where M is the scalar mass, and s)2 \u2212", " (M 2 \u2212 \u03a6st + \u03a6ts s)(M 2 \u2212 t) \u2212 + \u03a6tt (M 2 t)2 \u2212, (48.2) \u03a6ss = Tr \u03a6tt = Tr \u03a6st = Tr \u03a6ts = Tr,, i i v\u20322v \u20322u\u20321u \u20321 v\u20322v \u20322v2v2 h Tr h Tr u1u1v2v2 h i u1u1u\u20321u \u20321 h i u1u1u\u20321u \u20321v\u20322v \u20322v2v2 h u1u1v2v2v\u20322v \u20322u\u20321u \u20321 h, i i. (48.3) Next, we average over the two initial spins and sum over the two \ufb01nal spins to get Then we use 2 h|T | i = 1 4 X s1,s2,s\u2032 1,s\u2032 2 |T | 2. us(p)us(p) = vs(p)vs(p) = /p + m, \u2212 /p \u2212 \u2212 m, s= X \u00b1 s= X \u00b1 to get \u03a6ssi h \u03a6tti h \u03a6sti h \u03a6tsi h = 1 4 Tr = 1 4 Tr = 1 4 Tr = 1 4 Tr ( \u2212 p1+m)( \u2212 /p1+m)( \u2212 /p1+m)( \u2212 i m) /p2\u2212 /p \u20321+m) i /p \u20321+m)( /p1+m)( \u2212 /p2\u2212 m)( \u2212 /p \u20321+m) i m) Tr Tr ( /p \u20322\u2212 \u2212 /p \u20322\u2212 m)( h ( \u2212 h /p \u20322\u2212 /p \u20322\u2212 m)( m)( m)( \u2212 /p2\u2212 \u2212, m) /p2\u2212 i /p \u20321+m) i. \u2212 \u2212 \u2212 i (48.4) (48.5) (48.6) (48.7) (48.8) (48.9),, 48: Spin-Averaged Cross Sections 299 It is now merely tedious to evaluate these traces with the technology of section 47. For example, Tr ( \u2212 h /p1+m)( \u2212 /p2\u2212 m) i m2 Tr 1 = Tr[/p1 /p2] \u2212 4(p1p2) \u2212 = \u2212 4m2, (48.10) It is convenient to write four-vector products in terms of the Mandelstam variables. We have p1p2 = p\u20321p\u20322 = \u2212 p1p\u20321 = p2p\u20322 = + 1 p1p\u20322 = p\u20321p2 = + 1 1 2(s 2(t 2(u 2m2), 2m2), 2m2), \u2212 \u2212 \u2212 and so Tr Thus, we can easily work out eqs. (48.6) and (48.7): h i ( \u2212 /p1+m)( \u2212 /p2\u2212 m) = 2s \u2212 8m2. (48.11) (48.12) \u2212 \u2212 = (t = (s (48.14) (48.13) 4m2)2, 4m2)2. and make the swap s \u03a6ssi h \u03a6tti h Obviously, if we start with. \u03a6tti \u03a6ssi h h We could have anticipated this from eqs. (48.6) and (48.7): if we start with p\u20321, we get the right-hand side of eq. (48.6) and make the swap p2 \u2194 \u2212 the right-hand side of eq. (48.7). But from eq. (48.11), we see that this momentum swap is equivalent to s t. \u2194. These two are also related by p2 \u2194 \u03a6tsi h p\u20321, and so we only need to compute one of them. We have \u03a6sti h Let\u2019s move on to t, we get \u03a6sti h = 1 and \u2194 \u2212 /p \u20321 /p2 + /p \u20322 /p2] + 1 4 m4 Tr 1 4 Tr[/p1 /p \u20321 /p \u20322 /p2] + 1 /p1 /p \u20322 \u2212 4 m2 Tr[/p1 /p \u20321 \u2212 /p1 /p2 \u2212 = (p1p\u20321)(p2p\u20322) (p1p\u20322)(p2p\u20321) + (p1p2)(p\u20321p\u20322) \u2212 m2[", "p1p\u20321 \u2212 p1p\u20322 \u2212 \u2212 2 st + 2m2u. 1 p1p2 \u2212 p\u20321p\u20322 \u2212 /p \u20321 /p \u20322 \u2212 = \u2212 p\u20321p2 + p2p\u20322] + m4 (48.15) To get the last line, we used eq. (48.11), and then simpli\ufb01ed it as much as possible via s + t + u = 4m2. Since our result is symmetric on s t, we have \u2194 = \u03a6tsi h. \u03a6sti h Putting all of this together, we get 2 h|T | i = g4 \" (s \u2212 (M 2 4m2)2 s)2 + (M 2 \u2212 st 4m2u \u2212 s)(M 2 \u2212 \u2212 + t) (t \u2212 (M 2 4m2)2 t)2 # \u2212. (48.16) 48: Spin-Averaged Cross Sections 300 This can then be converted to a di\ufb00erential cross section (in any frame) via the formulae of section 11. Let\u2019s do one more: e\u2212\u03d5 e\u2212\u03d5. We take the incoming and outgoing electrons to have momenta p and p\u2032, respectively, and the incoming and outgoing scalars to have momenta k and k\u2032, respectively. We then have p2 = p\u2032 M 2. The Mandelstam variables are m2 and k2 = k\u2032 2 = 2 = \u2192 \u2212 \u2212 (p + k)2 = p\u2032)2 = k\u2032)2 = (p (p\u2032 + k\u2032)2, k\u2032)2, p\u2032)2, (k (k \u2212 \u2212 \u2212 \u2212 \u2212 (48.17) and they obey s + t + u = 2m2 + 2M 2. Our result in section 46 was 2 h|T | i = 1 2 Tr A( \u2212 /p + m)A( \u2212, /p \u2032 + m) i (48.18) h A = g2 /k + 2m \u2212 m2 s \u2212 \" + /k\u2032 + 2m m2 u # \u2212. (48.19) \u03a6ssi h (m2 \u2212 s)2 + h (m2 \u03a6sui \u2212 \u03a6usi + h s)(m2 \u2212 + h (m2 u) \u03a6uui \u2212 u)2 (48.20), where Thus we have 2 h|T | i = g4 where now \u03a6ssi h \u03a6uui h \u03a6sui h \u03a6usi h = 1 2 Tr = 1 2 Tr = 1 2 Tr = 1 2 Tr p \u2032+m)( \u2212 /k+2m)( \u2212 /p+m)( /k+2m), (48.21) \u2212 i /p+m)(+/k\u2032+2m) /p \u2032+m)(+/k\u2032+2m)( \u2212 /p \u2032+m)( \u2212 /k+2m)( \u2212 /p+m)(+/k\u2032+2m) /p \u2032+m)(+/k\u2032+2m)( \u2212 /p+m)( \u2212 i /k+2m) i i, (48.22),. (48.23) (48.24) We can evaluate these in terms of the Mandelstam variables by using our trace technology, along with M 2), pk = p\u2032k\u2032 = \u2212 pp\u2032 = + 1 kk\u2032 = + 1 pk\u2032 = p\u2032k = + 1 1 2 (s 2 (t 2 (t 2 (u \u2212 \u2212 \u2212 m2 \u2212 2m2), 2M 2), m2 \u2212 Examining eqs. (48.21) and (48.22), we see that formed into each other by k \u2212 are transk\u2032. Examining eqs. (48.23) and (48.24), we \u03a6uui h \u03a6ssi h and M 2). (48.25) \u2194 \u2212 48: Spin-Averaged Cross Sections 301 are also transformed into each other by k and see that \u03a6sui h From eq. (48.25), we see that this is equivalent to s only compute \u03a6sui \u2194 h This is, again, merely tedious, and the results are \u03a6usi h and \u03a6ssi h, and then take s \u2194 u to get k\u2032. \u2194 \u2212 u. Thus we need. \u03a6usi and \u03a6uui h h \u03a6ssi h \u03a6uui h \u03a6sui h \u03a6usi h su + m2(9s + u) + 7m4 su + m2(", "9u + s) + 7m4 = = \u2212 \u2212 = +su + 3m2(s + u) + 9m4 = +su + 3m2(s + u) + 9m4 8m2M 2 + M 4, (48.26) 8m2M 2 + M 4, (48.27) 8m2M 2 8m2M 48.28) (48.29) \u2212 \u2212 \u2212 \u2212 Problems \u00d7 48.1) The tedium of these calculations is greatly alleviated by making use of a symbolic manipulation program like Mathematica or Maple. One 4 matrices like /p in the CM approach is brute force: compute 4 frame, and take their products and traces. If you are familiar with a symbolic-manipulation program, write one that does this. See if you can verify eqs. (48.26\u201348.29). for e+e\u2212 \u2192 \u03d5\u03d5. You should \ufb01nd that your result is the same as that for e\u2212\u03d5 t, and an extra overall minus sign. This relationship is known as crossing symmetry. There is an overall minus sign for each fermion that is moved from the initial to the \ufb01nal state. e\u2212\u03d5, but with s 48.2) Compute 2 h|T | \u2194 \u2192 i 48.3) Compute 2 h|T | for e\u2212e\u2212 \u2192 i the same as that for e+e\u2212 \u2192 example of crossing symmetry. e\u2212e\u2212. You should \ufb01nd that your result is u. This is another e+e\u2212, but with s \u2194 48.4) Suppose that M > 2m, so that the scalar can decay to an electron- positron pair. a) Compute the decay rate, summed over \ufb01nal spins. |T | 2 = 0 if s1 = 2 for decay into an electron with spin s1 and a positron b) Compute with spin s2. Take the fermion three-momenta to be along the z axis, and let the x-axis be the spin-quantization axis. You should \ufb01nd that s2, or if M = 2m (so that the outgoing three-momentum of each fermion is zero). Discuss this in light of conservation of angular momentum and of parity. (Prerequisite: 40.) 2 for decay into an electron with helicity s1 and a c) Compute positron with helicity s2. (See section 38 for the de\ufb01nition of helicity.) |T | |T | \u2212 48: Spin-Averaged Cross Sections 302 You should \ufb01nd that the decay rate is zero if s1 = in light of conservation of angular momentum and of parity. \u2212 s2. Discuss this L1 = ig\u03d5\u03a8\u03b35\u03a8, and comd) Now consider changing the interaction to pute the spin-summed decay rate. Explain (in light of conservation of angular momentum and of parity) why the decay rate is larger than it was without the i\u03b35 in the interaction. e) Repeat parts (b) and (c) for the new form of the interaction, and explain any di\ufb00erences in the results. 48.5) The charged pion \u03c0\u2212 is represented by a complex scalar \ufb01eld \u03d5, the, and the muon neutrino \u03bd\u00b5 by a spin\u03b35). The charged pion muon \u00b5\u2212 by a Dirac \ufb01eld M, where PL = 1 projected Dirac \ufb01eld PL N can decay to a muon and a muon antineutrino via the interaction 2 (1 \u2212 L1 = 2c1GFf\u03c0\u2202\u00b5\u03d5 M \u03b3\u00b5PL N + h.c., (48.30) where c1 is the cosine of the Cabibbo angle, GF is the Fermi constant, and f\u03c0 is the pion decay constant. a) Compute the charged pion decay rate \u0393. b) The charged pion mass is m\u03c0 = 139.6 MeV, the muon mass is m\u00b5 = 105.7 MeV, and the muon neutrino is massless. The Fermi 2, constant is measured in muon decay to be GF = 1.166 and the cosine of the Cabibbo angle is measured in nuclear beta decays to be c1 = 0.974. The measured value of the charged pion lifetime is 8 s. Determine the value of f\u03c0 in MeV. Your result is too 2.603 large by 0.8%, due to neglect of electromagnetic loop corrections. 5 GeV\u2212 10\u2212 10\u2212 \u00d7 \u00d7 49: The Feynman Rules for", " Majorana Fields 303 49 The Feynman Rules for Majorana Fields Prerequisite: 45 In this section we will deduce the Feynman rules for Yukawa theory, but with a Majorana \ufb01eld instead of a Dirac \ufb01eld. We can think of the particles associated with the Majorana \ufb01eld as massive neutrinos. We have 2 g\u03d5\u03a8\u03a8 L1 = 1 = 1 2 g\u03d5\u03a8T \u03a8, C (49.1) where \u03a8 be a Majorana \ufb01eld (with mass m), \u03d5 is a real scalar \ufb01eld (with mass M ), and g is a coupling constant. In this section, we will be concerned with tree-level processes only, and so we omit renormalizing Z factors. From section 41, we have the LSZ rules appropriate for a Majorana \ufb01eld, b\u2020s(p)in \u2192 \u2212 i Z d4x e+ipx vs(p)( \u2212 i/\u2202 + m)\u03a8(x) = +i bs\u2032(p\u2032)out \u2192 +i Z Z d4x e\u2212 d4x \u03a8T(x) C (+i\u2190/\u2202 + m)us(p)e+ipx, ip\u2032x us\u2032(p\u2032)( \u2212 i/\u2202 + m)\u03a8(x), (49.2) (49.3) (49.4) = i \u2212 Z d4x e\u2212 ip\u2032x \u03a8T(x) C (+i\u2190/\u2202 + m)vs\u2032(p\u2032)e\u2212 ip\u2032x. (49.5) Eq. (49.3) follows from eq. (49.2) by taking the transpose of the right-hand 1; side, and using vs\u2032(p\u2032)T = us\u2032(p\u2032) and ( similarly, eq. (49.5) follows from eq. (49.4). Which form we use depends on convenience, and is best chosen on a diagram-by-diagram basis, as we will see shortly. i/\u2202 + m)T = (+i/\u2202 + m) C\u2212 \u2212C \u2212 C y) C\u2212 2 in i S(x 1, and the \u03d5\u03a8\u03a8 vertex is ig Eqs. (49.2\u201349.5) lead us to compute correlation functions containing \u03a8\u2019s, In position space, this leads to Feynman rules where the but not \u03a8\u2019s. fermion propagator is 1 ; the factor \u2212 of 1 L1 is canceled by a symmetry factor of 2! that arises from having In a particular diagram, as we move along two identical \u03a8 \ufb01elds in L1. in the a fermion line, the 1 can be canceled by 1 at one end. This vertex, leaving over a \ufb01nal a from eq. (49.3) (for an incoming particle) or eq. (49.5) (for an outgoing particle). On the other hand, for the other end of the same line, we should use either eq. (49.2) (for an incoming particle) or eq. (49.4) (for an outgoing 1 in the propagator will cancel against the C\u2212 C\u2212 C\u2212 C C C 49: The Feynman Rules for Majorana Fields 304 particle) to avoid introducing an extra avoid ever having explicit factors of at that end. In this way, we can C in our Feynman rules.1 Using this approach, the Feynman rules for this theory are as follows. C 1. The total number of incoming and outgoing neutrinos is always even; call this number 2n. Draw n solid lines. Connect them with internal dashed lines, using a vertex that joins one dashed and two solid lines. Also, attach an external dashed line for each incoming or outgoing scalar. In this way, draw all possible diagrams that are topologically inequivalent. 2. Draw arrows on each segment of each solid line; keep the arrow di- rection continuous along each line. 3. Label each external dashed line with the momentum of an incoming or outgoing scalar. If the particle is incoming, draw an arrow on the dashed line that points towards the vertex; If the particle is outgoing, draw an arrow on the dashed line that points away from the vertex. 4. Label each external solid line with the momentum of an incoming or outgoing neutrino, but include a minus sign with the momentum if (a) the particle is incoming and the arrow points away from the vertex, or (b) the particle is outgoing and the arrow points towards the vertex. Do this labeling of external lines in all possible inequivalent ways. Two diagrams are considered equivalent if they can", " be transformed into each other by reversing all the arrows on one or more fermion lines, and correspondingly changing the signs of the external momenta on each arrow-reversed line. 5. Assign each internal line its own four-momentum. Think of the fourmomenta as \ufb02owing along the arrows, and conserve four-momentum at each vertex. For a tree diagram, this \ufb01xes the momenta on all the internal lines. 6. The value of a diagram consists of the following factors: for each incoming or outgoing scalar, 1; for each incoming neutrino labeled with +pi, usi(pi); pi, vsi(pi); for each incoming neutrino labeled with (p\u2032i); (p\u2032i); for each outgoing neutrino labeled with +p\u2032i, us\u2032 p\u2032i, vs\u2032 for each outgoing neutrino labeled with \u2212 i \u2212 i 1This is not always possible if the Majorana \ufb01elds interact with Dirac \ufb01elds, and we use the usual rules for the Dirac \ufb01elds. 49: The Feynman Rules for Majorana Fields 305 k p1 p2 k p1 p2 Figure 49.1: Two equivalent diagrams for \u03d5 \u03bd\u03bd. \u2192 for each vertex, ig; for each internal scalar, \u2212 for each internal fermion, i/(k2 + M 2 i\u01eb); \u2212 /p + m)/(p2 + m2 i( \u2212 \u2212 i\u01eb). \u2212 7. Spinor indices are contracted by starting at one end of a fermion line: speci\ufb01cally, the end that has the arrow pointing away from the vertex. The factor associated with the external line is either u or v. Go along the complete fermion line, following the arrows backwards, and writing down (in order from left to right) the factors associated with the vertices and propagators that you encounter. The last factor is either a u or v. Repeat this procedure for the other fermion lines, if any. 8. The overall sign of a tree diagram is determined by drawing all contributing diagrams in a standard form: all fermion lines horizontal, with their arrows pointing from left to right, and with the left endpoints labeled in the same \ufb01xed order (from top to bottom); if the ordering of the labels on the right endpoints of the fermion lines in a given diagram is an even (odd) permutation of an arbitrarily chosen \ufb01xed ordering, then the sign of that diagram is positive (negative). To compare two diagrams, it may be necessary to use the arrow-reversing equivalence relation of rule #4; there is then an extra minus sign for each arrow-reversed line. 9. The value of i T is given by a sum over the values of all these diagrams. There are additional rules for counterterms and loops, but we will postpone those to section 51. Let\u2019s look at the simplest process, \u03d5 \u03bd\u03bd. There are two possible diagrams for this, shown in \ufb01g. (49.1). However, according to rule #4, these two diagrams are equivalent, and we should keep only one of them. T1 = ig v \u20322u\u20321 and the second iT2 = ig v \u20321u\u20322. Rule The \ufb01rst diagram yields i #8 then implies that we should have \u2212T2. To check this, we note that (after dropping primes to simplify the notation) T1 = \u2192 v1u2 = [v1u2]T 49: The Feynman Rules for Majorana Fields 306 p1 p2 p1 p1 p1 p2 p1 p2 p2 p1 p2 p1 p 2 p1 p1 p2 p2 p1 Figure 49.2: Diagrams for \u03bd\u03bd \u2192 \u03bd\u03bd, corresponding to eq. (49.7). = uT 2 vT 1 1 = v2C = \u2212 C v2u1, \u2212 \u2212 1u1 (49.6) as required. \u2192 \u03bd\u03d5 or \u03bd\u03bd In general, for processes with a total of just two incoming and outgo\u03d5\u03d5, these rules give (up to an ing neutrinos, such as \u03bd\u03d5 \u2192 as we would get for the corirrelevant overall sign) the same result for i T e\u2212\u03d5 or e+e\u2212 \u2192 \u03d5\u03d5. (Note, responding process in the Dirac case, e\u2212\u03d5 \u2192 L1 = g\u03d5\u03a8\u03a8, as compared with however, that in the Dirac case, we have L1 = 1 2 g\u03d5\u03a8\u03a8 in the Majorana case.) The", " di\ufb00erences between Dirac and Majorana fermions become more pronounced for \u03bd\u03bd \u03bd\u03bd. Now there are three inequivalent contributing diagrams, shown in \ufb01g. (49.2). The corresponding amplitude can be written as \u2192, (49.7) = 1 i (ig)2 i T (u \u20321u1)(u \u20322u2) t + M 2 \u2212 (u \u20322u1)(u \u20321u2) u + M 2 + (v2u1)(u \u20321v\u20322p1\u2212 (p1+p2)2, t = p\u20321)2 and u = \u2212 p\u20322)2. After arbitrarily (p1\u2212 where s = assigning the \ufb01rst diagram a plus sign, the minus sign of the second diagram follows from rule #8. To get the sign of the third diagram, we compare it with the \ufb01rst. To do so, we reverse the arrow direction on the lower line of the \ufb01rst diagram (which yields an extra minus sign), and then redraw it in standard form. Comparing this modi\ufb01ed \ufb01rst diagram with the third diagram (and invoking rule #8) reveals a relative minus sign. Since the modi\ufb01ed \ufb01rst diagram has a minus sign from the arrow reversal, we conclude that the third diagram has an overall plus sign. After taking the absolute square of eq. (49.7), we can use relations like eq. (49.6) on a term-by-term basis to put everything into a form that allows the spin sums to be performed in the standard way. In fact, we have already done all the necessary work in the Dirac case. The s-s, s-t, and t-t terms in e+e\u2212, while the t-t, \u03bd\u03bd are the same as those for e+e\u2212 \u2192 2 h|T | for \u03bd\u03bd \u2192 i 49: The Feynman Rules for Majorana Fields 307 t-u, and u-u terms are the same as those for the crossing-related process e\u2212e\u2212. Finally, the s-u terms can be obtained from the s-t terms e\u2212e\u2212 \u2192 s. Thus the result is u, or equivalently from the t-u terms via t via t \u2194 2 h|T | i = g4 \" + 4m2)2 \u2212 4m2)2 (s \u2212 (M 2 (t \u2212 (M 2 s)2 + (M 2 t)2 + (M 2 + (u \u2212 (M 2 \u2212 4m2)2 u)2 + \u2212 (M 2 \u2212 st \u2212 tu \u2212 us \u2194 4m2u \u2212 s)(M 2 4m2s \u2212 t)(M 2 \u2212 4m2t \u2212 u)(M 2 \u2212 t) \u2212 u), s) # (49.8) which is neatly symmetric on permutations of s, t, and u. Problems 49.1) Let \u03a8 be a Dirac \ufb01eld (representing the electron and positron), X be a Majorana \ufb01eld (represeting the photino, the hypothetical supersymmetric partner of the photon, with mass m\u02dc\u03b3), and EL and ER be two di\ufb00erent complex scalar \ufb01elds (representing the two selectrons, the hypothetical supersymmetric partners of the left-handed electron and the right-handed electron, with masses ML, and MR; note that the subscripts L and R are just part of their names, and do not signify anything about their Lorentz transformation properties). They interact via L1 = \u221a2eE\u2020LXPL\u03a8 + \u221a2eE\u2020RXPR\u03a8 + h.c., (49.9) \u2213 \u2243 1/137 is the \ufb01ne-structure constant, and PL,R = where \u03b1 = e2/4\u03c0 1 \u03b35). 2 (1 a) Write down the hermitian conjugate term explicitly. b) Find the tree-level scattering amplitude for e+e\u2212 \u2192 \u02dc\u03b3\u02dc\u03b3. Hint: there are four contributing diagrams, two each in the t and u channels, with exchange of either EL or ER. c) Compute the spin-averaged di\ufb00erential cross section for this process in the case that me (the electron mass) can be neglected, and | \u226a ML = MR. Express it as a function of s and the center-of-mass scattering angle \u03b8. u | t |, | 50: Massless Particles and Spinor Helicity 308 50 Massless Particles and Spinor Helicity Prerequisite: 48 Scattering", " amplitudes often simplify greatly if the particles are massless (or can be approximated as massless because the Mandelstam variables all have magnitudes much larger than the particle masses squared). In this section we will explore this phenomenon for spin-one-half (and spin-zero) particles. We will begin developing the technology of spinor helicity, which will prove to be of indispensible utility in Part III. Recall from section 38 that the u spinors for a massless spin-one-half particle obey us(p)us(p) = 1 2 (1 + s\u03b35)( \u2212 /p), (50.1) speci\ufb01es the helicity, the component of the particle\u2019s spin where s = measured along the axis speci\ufb01ed by its three-momentum; in this notation the helicity is 1 2 s. The v spinors obey a similar relation, \u00b1 vs(p)vs(p) = 1 2 (1 In fact, in the massless case, with the phase conventions of section 38, we have vs(p) = u s(p). Thus we can con\ufb01ne our discussion to u-type spinors only, since we need merely change the sign of s to accomodate v-type spinors. (50.2) s\u03b35)( /p). \u2212 \u2212 \u2212 Consider a u spinor for a particle of negative helicity. We have (p)u u \u2212 \u2212 (p) = 1 2 (1 \u03b35)( \u2212 \u2212 /p). pa \u02d9a \u2261 p\u00b5\u03c3\u00b5 a \u02d9a. p \u02d9aa = \u03b5ac\u03b5 \u02d9a \u02d9cpc \u02d9c = p\u00b5 \u00af\u03c3\u00b5 \u02d9aa. Let us de\ufb01ne Then we also have Then, using \u03b3\u00b5 = 0 \u00af\u03c3\u00b5 \u03c3\u00b5 0!, 1 2 (1 \u2212 \u03b35) = 1 0 0 0! in eq. (50.3), we \ufb01nd (p)u u \u2212 \u2212 (p) = 0 0 pa \u02d9a \u2212 0!. (50.3) (50.4) (50.5) (50.6) (50.7) On the other hand, we know that the lower two components of u vanish, and so we can write \u2212 (p) (p) = u \u2212 \u03c6a 0!. (50.8) 50: Massless Particles and Spinor Helicity 309 Here \u03c6a is a two-component numerical spinor; it is not an anticommuting object. Such a commuting spinor is sometimes called a twistor. An explicit numerical formula for it (veri\ufb01ed in problem 50.2) is \u03c6a = \u221a2\u03c9 sin( 1 \u2212 + cos( 1 2 \u03b8)e\u2212 2 \u03b8) i\u03c6,! (50.9) where \u03b8 and \u03c6 are the polar and azimuthal angles that specify the direction p of the three-momentum p, and \u03c9 =. Barring eq. (50.8) yields | | (p) = ( 0, \u03c6\u2217\u02d9a ), (50.10) u \u2212 where \u03c6\u2217\u02d9a = (\u03c6a)\u2217. Now, combining eqs. (50.8) and (50.10), we get (p)u u \u2212 \u2212 (p) = 0 \u03c6a\u03c6\u2217\u02d9a 0 0!. (50.11) Comparing with eq. (50.7), we see that pa \u02d9a = \u03c6a\u03c6\u2217\u02d9a. (50.12) \u2212 This expresses the four-momentum of the particle neatly in terms of the twistor that describes its spin state. The essence of the spinor helicity method is to treat \u03c6a as the fundamental object, and to express the particle\u2019s four-momentum in terms of it, via eq. (50.12). Given eq. (50.8), and the phase conventions of section 38, the positive- helicity spinor is 0 u+(p) =, (50.13) \u02d9a! \u02d9a = \u03b5 \u02d9a \u02d9c\u03c6\u2217\u02d9c. Barring eq. (50.13) yields \u03c6\u2217 where \u03c6\u2217 u+(p) = ( \u03c6a, 0 ). (50.14) Computation of u+(p)u+(p) via eqs. (50.13) and (50.14), followed by comparison with eq. (50.1) with s = +, then reproduces", " eq. (50.12), but with the indices raised. In fact, the decomposition of pa \u02d9a into the direct product of a twistor and its complex conjugate is unique (up to an overall phase for the twistor). To see this, use \u03c3\u00b5 = (I, ~\u03c3) to write pa \u02d9a = p0 + p3 \u2212 p1 + ip2 p1 ip2 \u2212 p0 \u2212 \u2212 p3!. (50.15) (p0)2 + p2, and this vanishes because The determinant of this matrix is the particle is (by assumption) massless. Thus pa \u02d9a has a zero eigenvalue. \u2212 50: Massless Particles and Spinor Helicity 310 Therefore, it can be written as a projection onto the eigenvector corresponding to the nonzero eigenvalue. That is what eq. (50.12) represents, with the nonzero eigenvalue absorbed into the normalization of the eigenvector \u03c6a. Let us now introduce some useful notation. Let p and k be two fourmomenta, and \u03c6a and \u03baa the corresponding twistors. We de\ufb01ne the twistor product Because \u03c6a\u03baa = \u03b5ac\u03c6c\u03baa, and the twistors commute, we have [p k] \u03c6a\u03baa. \u2261 From eqs. (50.8) and (50.14), we can see that [k p] = [p k]. \u2212 Similarly, let us de\ufb01ne u+(p)u \u2212 (k) = [p k]. p k h i \u2261 \u03c6\u2217\u02d9a\u03ba\u2217 \u02d9a. Comparing with eq. (50.16) we see that p k h i = [k p]\u2217, which implies that this product is also antisymmetric, Also, from eqs. (50.10) and (50.13), we have k p h i = p k \u2212h. i Note that the other two possible spinor products vanish: (p)u+(k+(p)u+(k) = u (p)u \u2212 \u2212 (k) = 0. (50.16) (50.17) (50.18) (50.19) (50.20) (50.21) (50.22) (50.23) The twistor products p k h i and [p k] satisfy another important relation, p k h [k p] = (\u03c6\u2217\u02d9a\u03ba\u2217 i \u02d9a)(\u03baa\u03c6a) \u02d9a) = (\u03c6\u2217\u02d9a\u03c6a)(\u03baa\u03ba\u2217 = p \u02d9aaka \u02d9a \u2212 where the last line follows from \u00af\u03c3\u00b5 \u02d9aa\u03c3\u03bd = 2p\u00b5k\u00b5, a \u02d9a = 2g\u00b5\u03bd. \u2212 (50.24) 50: Massless Particles and Spinor Helicity 311 Let us apply this notation to the tree-level scattering amplitude for e\u2212\u03d5 in Yukawa theory, which we \ufb01rst computed in Section 44, and e\u2212\u03d5 \u2192 which reads Ts\u2032s = g2 us\u2032(p\u2032) For a massless fermion, \u02dcS(p) = (p + k)2 = 2p k\u2032)2 = \u00b7 propagator numerators in eq. (50.25), because /pus(p) = 0. Thus we have /p/p2. If the scalar is also massless, then k\u2032. Also, we can remove the /p\u2019s in the 2p \u02dcS(p+k) + \u02dcS(p h k\u2032) i k and (p us(p). (50.25) \u2212 \u2212 \u2212 \u2212 \u00b7 Ts\u2032s = g2 us\u2032(p\u2032) \" /k k \u2212 2p \u00b7 /k\u2032 k\u2032 # + \u2212 2p \u00b7 us(p). Now consider the case s\u2032 = s = +. From eqs. (50.13), (50.14), and /k = \u2212 0 \u03baa\u03ba\u2217\u02d9a \u03ba\u2217 \u02d9a\u03baa 0!, (50.26) (50.27) we get u+(p\u2032)( \u2212 /k)u+(p) = \u03c6\u2032 \u02d9a a\u03baa\u03ba\u2217\u02d9a\u03c6\u2217 = [p\u2032 k] k p h i. (50.28) Similarly, for s\u2032 = s =, we \ufb01nd \u2212 u \u2212 (p\u2032)( \u2212 /k)u \u2212 (p) = \u03c6\u2032\u2217\u02d9a", " \u03ba\u2217 \u02d9a\u03baa\u03c6a = p\u2032 k h i [k p], (50.29) while for s\u2032 6 = s, the amplitude vanishes: u \u2212 (p\u2032)( \u2212 /k)u+(p) = u+(p\u2032)( \u2212 /k)u \u2212 (p) = 0. (50.30) Then, using eq. (50.24) on the denominators in eq. (50.26), we \ufb01nd T++ = \u2212 g2 = g2 \u2212 T\u2212\u2212 [p\u2032 k] [p k] p\u2032 k h p k h i i while +, [p\u2032 k\u2032] [p k\u2032] p\u2032 k\u2032i + h p k\u2032i h, (50.31) T\u2212 Thus we have rather simple expressions for the \ufb01xed-helicity scattering amplitudes in terms of twistor products. \u2212 = + = 0. T+ (50.32) 50: Massless Particles and Spinor Helicity 312 Reference Notes Spinor-helicity methods are discussed by Siegel. Problems 50.1) Consider a bra-ket notation for twistors, p] = u | p | i [p | (p) = v+(p), \u2212 = u+(p) = v (p), = u+(p) = v \u2212 (pp) = v+(p). (50.33) We then have k h [] = [k p], | | k h [k | | p | | p] = 0, = 0. i a) Show that p p] h | where p is any massless four-momentum. [p i /p = p | \u2212 + | (50.34) (50.35), | b) Use this notation to rederive eqs. (50.28\u201350.30). 50.2) a) Use eqs. (50.9) and (50.15) to verify eq. (50.12). b) Let the three-momentum p be in the +\u02c6z direction. Use eq. (38.12) (p) explicitly in the massless limit (corresponding to to compute u p the limit \u03b7 /m). Verify that, when \u03b8 = 0, | | your results agree with eqs. (50.8), (50.9), and (50.13)., where sinh \u03b7 = \u00b1 \u2192 \u221e 50.3) Prove the Schouten identity, p q h r s ih + p r h i s q ih + p s h i q r ih i = 0. (50.36) Hint: note that the left-hand side is completely antisymmetric in the three labels q, r, and s, and that each corresponding twistor has only two components. 50: Massless Particles and Spinor Helicity 313 50.4) Show that p q h and evaluate the right-hand side. r s h [q r] i i [s p] = Tr 1 2 (1 \u2212 \u03b35)/p/q/r/s, (50.37) 50.5) a) Prove the useful identities, p | \u03b3\u00b5 | i p], | p h p h \u03b3\u00b5 k] = [k | | \u03b3\u00b5 k]\u2217 = | | \u03b3\u00b5 p | h \u03b3\u00b5 p h | [p \u03b3\u00b5 | k h p] = 2p\u00b5, | k | \u03b3\u00b5 | i k] = 0. | = 0, (50.38) (50.39) (50.40) (50.41) (50.42) b) Extend the last two identies of part (a): show that the product p of an odd number of gamma matrices sandwiched between either | h k] vanishes. Also show that the product of an even and | number of gamma matrices between either k | vanishes. k] or [p | or [p and and and ) Prove the Fierz identities, q]\u03b3\u00b5 = 1 p 2 h 1 2 [p \u2212 \u03b3\u00b5| | q \u03b3\u00b5| | Now take the matrix element of eq. (50.44) between another useful form of the Fierz identity, p q] h | [p q i | [q p i | q p] h | \u03b3\u00b5 = 50.43) (50.44) r h | and s] to get | [p \u03b3\u00b5 | q | r i h s] = 2 [p s] \u03b3\u00b5| | q r h i. (50.45) 51: Loop Corrections in Yukawa", " Theory 314 51 Loop Corrections in Yukawa Theory Prerequisite: 19, 40, 48 In this section we will compute the one-loop corrections in Yukawa theory with a Dirac \ufb01eld. The basic concepts are all the same as for a scalar \ufb01eld, and so we will mainly be concerned with the extra technicalities arising from spin indices and anticommutation. First let us note that the general discussion of sections 18 and 29 leads us to expect that we will need to add to the lagrangian all possible terms whose coe\ufb03cients have positive or zero mass dimension, and that respect the symmetries of the original lagrangian. These include Lorentz symmetry, the U(1) phase symmetry of the Dirac \ufb01eld, and the discrete symmetries of parity, time reversal, and charge conjugation. The mass dimensions of the \ufb01elds (in four spacetime dimensions) are [\u03d5] = 1 and [\u03a8] = 3 2. Thus any power of \u03d5 up to \u03d54 is allowed. But there are no additional required terms involving \u03a8: the only candidates contain \u03a8) and either \u03b35 (e.g., i\u03a8\u03b35\u03a8) and are forbidden by parity, or are forbidden by the U(1) symmetry. (e.g, \u03a8T C C Nevertheless, having to deal with the addition of three new terms (\u03d5, \u03d53, \u03d54) is annoying enough to prompt us to look for a simpler example. Consider, then, a modi\ufb01ed form of the Yukawa interaction, LYuk = ig\u03d5\u03a8\u03b35\u03a8. (51.1) This interaction will conserve parity if and only if \u03d5 is a pseudoscalar: P \u2212 1\u03d5(x, t)P = \u03d5( \u2212 \u2212 x, t). (51.2) Then, \u03d5 and \u03d53 are odd under parity, and so we will not need to add them to. The one term we will need to add is \u03d54. L Therefore, the theory we will consider is \u2212 = m\u03a8\u03a8 L1, L0 + L L0 = i\u03a8/\u2202\u03a8 L1 = iZgg\u03d5\u03a8\u03b35\u03a8 \u2212 1)\u03a8/\u2202\u03a8 Lct = i(Z\u03a8\u2212 \u2212 2 (Z\u03d5\u2212 1 1 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 24 Z\u03bb\u03bb\u03d54 + 1 (Zm\u2212 \u2212 \u2212 Lct, 1)m\u03a8\u03a8 1 2 (ZM \u2212 \u2212 1)\u2202\u00b5\u03d5\u2202\u00b5\u03d5 2 M 2\u03d52, 1)M 2\u03d52 (51.3) (51.4) (51.5) (51.6) where \u03bb is a new coupling constant. We will use an on-shell renormalization scheme. The lagrangian parameter m is then the actual mass of the electron. We will de\ufb01ne the couplings g and \u03bb as the values of appropriate vertex functions when the external four-momenta vanish. Finally, the 51: Loop Corrections in Yukawa Theory 315 \ufb01elds are normalized according to the requirements of the LSZ formula. In practice, this means that the scalar and fermion propagators must have appropriate poles with unit residue. We will assume that M < 2m, so that the scalar is stable against decay into an electron-positron pair. The exact scalar propagator (in momentum space) can be then written in Lehmann-K\u00a8all\u00b4en form as \u02dc\u2206(k2) = 1 k2 + M 2 i\u01eb \u2212 + \u221e M 2 th Z ds \u03c1(s) k2 + s \u2212, i\u01eb (51.7) where the spectral density \u03c1(s) is real and nonnegative. The threshold mass Mth is either 2m (corresponding to the contribution of an electron-positron pair) or 3M (corresponding to the contribution of three scalars; by parity, there is no contribution from two scalars), whichever is less. We can also write \u02dc\u2206(k2)\u2212 1 = k2 + M 2 i\u01eb \u2212 \u2212 \u03a0(k2), (51.8) where i\u03a0(k2) is given by the sum of one-particle irreducible (1PI for short; see section 14) diagrams with two external scalar lines, and the external propagators removed. The fact that \u02dc\u2206(k2) has a pole at k2 = M 2 with M 2) = 0; this \ufffd", "\ufffd\ufffdxes the residue one implies that \u03a0( coe\ufb03cients Z\u03d5 and ZM. M 2) = 0 and \u03a0\u2032( \u2212 \u2212 \u2212 All of this is mimicked for the Dirac \ufb01eld. When parity is conserved, the exact propagator (in momentum space) can be written in Lehmann-K\u00a8all\u00b4en form as /p + m \u02dcS(/p) = \u2212 p2 + m2 i\u01eb \u2212 + \u221e m2 th Z ds \u2212 /p\u03c11(s) + \u221as\u03c12(s) p2 + s i\u01eb \u2212, (51.9) where the spectral densities \u03c11(s) and \u03c12(s) are both real, and \u03c11(s) is nonnegative and greater than \u03c12(s). The threshold mass mth is m + M (corresponding to the contribution of a fermion and a scalar), which, by assumption, is less than 3m (corresponding to the contribution of three fermions; by Lorentz invariance, there is no contribution from two fermions). Since p2 = \u2212 /p/p, we can rewrite eq. (51.9) as \u02dcS(/p) = 1 /p + m + \u221e m2 th Z ds ( \u2212 i\u01eb \u2212 /p\u03c11(s) + \u221as\u03c12(s) i\u01eb)(/p + \u221as \u2212 /p + \u221as \u2212 \u2212, i\u01eb) (51.10) with the understanding that 1/(...) refers to the matrix inverse. However, since /p is the only matrix involved, we can think of \u02dcS(/p) as an analytic function of the single variable /p. With this idea in mind, we see that \u02dcS(/p) has an isolated pole at /p = m with residue one. This residue corresponds to the \ufb01eld normalization that is needed for the validity of the LSZ formula. \u2212 51: Loop Corrections in Yukawa Theory 316 k+l l k k k k k k l Figure 51.1: The one-loop and counterterm corrections to the scalar propagator in Yukawa theory. We can also write the exact fermion propagator in the form \u02dcS(/p)\u2212 1 = /p + m i\u01eb \u2212 \u2212 \u03a3(/p), (51.11) where i\u03a3(/p) is given by the sum of 1PI diagrams with two external fermion lines, and the external propagators removed. The fact that \u02dcS(/p) has a pole at /p = m) = 0; this \ufb01xes the coe\ufb03cients Z\u03a8 and Zm. m with residue one implies that \u03a3( m) = 0 and \u03a3\u2032( \u2212 \u2212 \u2212 We proceed to the diagrams. The Yukawa vertex carries a factor of Zgg\u03b35. Since Zg = 1 + O(g2), we can set Zg = 1 in the i(iZgg)\u03b35 = one-loop diagrams. \u2212 Consider \ufb01rst \u03a0(k2), which receives the one-loop (and counterterm) corrections shown in \ufb01g. (51.1). The \ufb01rst diagram has a closed fermion loop. As we will see in problem 51.1 (and section 53), anticommutation of the fermion \ufb01elds results in an extra factor of minus one for each closed fermion loop. The spin indices on the propagators and vertices are contracted in the usual way, following the arrows backwards. Since the loop closes on itself, we end up with a trace over the spin indices. Thus we have i\u03a0\u03a8 loop(k2) = ( \u2212 1)( \u2212 2 g)2 1 i Z d4\u2113 (2\u03c0)4 Tr \u02dcS(/\u2113+/k)\u03b35 \u02dcS(/\u2113)\u03b35 h i, where \u02dcS(/p) = \u2212 /p + m p2 + m2 i\u01eb \u2212 is the free fermion propagator in momentum space. We now proceed to evaluate eq. (51.12). We have (51.12) (51.13) Tr[( /\u2113 \u2212 \u2212 \u2212 /k + m)\u03b35( /\u2113 + m)\u03b35] = Tr[( /k + m)(+/\u2113 + m)] /\u2113 \u2212 \u2212 = 4[(\u2113 + k)\u2113 + m2] 4N. \u2261 (51.14) 51: Loop", " Corrections in Yukawa Theory 317 The \ufb01rst equality follows from \u03b32 5 = 1 and \u03b35 /p\u03b35 = /p. \u2212 Next we combine the denominators with Feynman\u2019s formula. Suppress- ing the i\u01eb\u2019s, we have 1 (\u2113+k)2 + m2 1 \u21132 + m2 = 1 dx 1 (q2 + D)2, 0 Z (51.15) where q = \u2113 + xk and D = x(1 x)k2 + m2. We then change the integration variable in eq. (51.12) from \u2113 to q; the \u2212 result is 1 d4q (2\u03c0)4 dx i\u03a0\u03a8 loop(k2) = 4g2 N (q2 + D)2, 0 Z xk)+m2. The integral diverges, and so we \u03b5 spacetime dimensions. (Here we ignore a 5 = 1 where now N = (q +(1 x)k)(q analytically continue it to d = 4 subtlety with the de\ufb01nition of \u03b35 in d dimensions, and assume that \u03b32 and \u03b35 /p\u03b35 = /p continue to hold.) We also make the replacement g g \u02dc\u00b5\u03b5/2, where \u02dc\u00b5 has dimensions of mass, so that g remains dimensionless. (51.16) \u2212 \u2212 \u2192 \u2212 \u2212 Z Expanding out the numerator, we have N = q2 x(1 \u2212 \u2212 x)k2 + m2 + (1 \u2212 2x)kq. (51.17) The term linear in q integrates to zero. For the rest, we use the general result of section 14 to get Z \u02dc\u00b5\u03b5 \u02dc\u00b5\u03b5 ddq (2\u03c0)d ddq (2\u03c0)d 1 (q2 + D)2 = q2 (q2 + D)2 = i 16\u03c02 2 \u03b5 \u2212 ln(D/\u00b52), (51.18) i 16\u03c02 2 \u03b5 + 1 2 \u2212 ln(D/\u00b52) ( \u2212 2D), (51.19) Z where \u00b52 = 4\u03c0e\u2212 eqs. (51.18) and (51.19) into eq. (51.16) yields \u03b3 \u02dc\u00b52, and we have dropped terms of order \u03b5. Plugging \u03a0\u03a8 loop(k2) = g2 4\u03c02 \" 1 \u03b5 \u2212 (k2 + 2m2) + 1 6 k2 + m2 1 dx \u2212 0 Z 3x(1 \u2212 x)k2 + m2 ln(D/\u00b52) #. (51.20) We see that the divergent term has (as expected) a form that permits cancellation by the counterterms. We evaluated the second diagram of \ufb01g. (51.1) in section 31, with the result \u03a0\u03d5 loop(k2) = \u03bb (4\u03c0)2 1 \u03b5 + 1 2 \u2212 1 2 ln(M 2/\u00b52) M 2. (51.21) 51: Loop Corrections in Yukawa Theory 318 l p p+l p p p Figure 51.2: The one-loop and counterterm corrections to the fermion propagator in Yukawa theory. The third diagram gives the contribution of the counterterms, (Z\u03d5\u2212 Adding up eqs. (51.20\u201351.22), we see that \ufb01niteness of \u03a0(k2) requires (ZM \u2212 \u2212 \u2212 \u03a0ct(k2) = 1)M 2. 1)k2 (51.22) Z\u03d5 = 1 \u2212 ZM = 1 + 1 \u03b5 g2 4\u03c02 \u03bb 16\u03c02 \u2212 + \ufb01nite, (51.23) m2 M 2! 1 \u03b5 g2 2\u03c02 + \ufb01nite, (51.24) plus higher-order (in g and/or \u03bb) corrections. Note that, although there is an O(\u03bb) correction to ZM, there is not an O(\u03bb) correction to Z\u03d5. M 2) = 0 by writing We can impose \u03a0( \u2212 \u03a0(k2) = g2 4\u03c02 1 dx 0 Z 3x(1 x)k2 + m2 \u2212 ln(D/D0) + \u03ba\u03d5(k2 + M 2), (51.25) x)M 2 + m2, and \u03ba\u03d5 is a constant to be determined. We where D0 = \ufb01x \u03ba\u03d5 by imposing \u03a0\u2032( x(", "1 \u2212 \u2212 M 2) = 0, which yields \u03ba\u03d5 = 0 Z dx x(1 \u2212 x)[3x(1 \u2212 x)M 2 \u2212 m2]/D0. (51.26) Note that, in this on-shell renormalization scheme, there is no O(\u03bb) correction to \u03a0(k2). Next we turn to the \u03a8 propagator, which receives the one-loop (and counterterm) corrections shown in \ufb01g. (51.2). The spin indices are contracted in the usual way, following the arrows backwards. We have i\u03a31 loop(/p) = ( \u2212 2 g)2 1 i Z d4\u2113 (2\u03c0)4 h \u03b35 \u02dcS(/p + /\u2113)\u03b35 \u02dc\u2206(\u21132), i where \u02dcS(/p) is given by eq. (51.13), and \u02dc\u2206(\u21132) = 1 \u21132 + M 2 i\u01eb \u2212 (51.27) (51.28) \u2212 1 51: Loop Corrections in Yukawa Theory 319 is the free scalar propagator in momentum space. We evaluate eq. (51.27) with the usual bag of tricks. The result is i\u03a31 loop(/p) = g2 \u2212 1 dx 0 Z Z d4q (2\u03c0)4 N (q2 + D)2, where q = \u2113 + xp and N = /q + (1 x)/p + m, D = x(1 \u2212 \u2212 x)p2 + xm2 + (1 x)M 2. \u2212 (51.29) (51.30) (51.31) \u03b5 g \u02dc\u00b5\u03b5/2, and take the limit 0. The term linear in q in eq. (51.30) integrates to zero. Using The integral diverges, and so we analytically continue it to d = 4 spacetime dimensions, make the replacement g as \u03b5 eq. (51.18), we get \u2192 \u2192 \u2212 \u03a31 loop(/p) = g2 16\u03c02 \" 1 \u03b5 \u2212 (/p + 2m) 1 dx (1 \u2212 \u2212 0 Z x)/p + m ln(D/\u00b52) #. (51.32) We see that the divergent term has (as expected) a form that permits cancellation by the counterterms, which give \u03a3ct(/p) = (Z\u03a8\u2212 \u2212 1)/p (Zm\u2212 \u2212 1)m. (51.33) Adding up eqs. (51.32) and (51.33), we see that \ufb01niteness of \u03a3(/p) requires Z\u03a8 = 1 Zm = 1 \u2212 \u2212 g2 16\u03c02 1 \u03b5 1 \u03b5 g2 8\u03c02 + \ufb01nite, + \ufb01nite, (51.34) (51.35) plus higher-order corrections. We can impose \u03a3( \u2212 m) = 0 by writing \u03a3(/p) = g2 16\u03c02 1 dx (1 x)/p + m \u2212 0 Z where D0 is D evaluated at p2 = mined. We \ufb01x \u03ba\u03a8 by imposing \u03a3\u2032( to /p, we take the p2 in D, eq. (51.31), to be \u2212 \u2212, ln(D/D0) + \u03ba\u03a8(/p + m) m2, and \u03ba\u03a8 is a constant to be determ) = 0. In di\ufb00erentiating with respect (51.36) /p 2; we \ufb01nd \u2212 \u03ba\u03a8 = 1 2 \u2212 0 Z dx x2(1 \u2212 x)m2/D0. (51.37) 51: Loop Corrections in Yukawa Theory 320 l p p+l p +l p k Figure 51.3: The one-loop correction to the scalar-fermion-fermion vertex in Yukawa theory. Next we turn to the correction to the Yukawa vertex. We de\ufb01ne the vertex function iVY (p\u2032, p) as the sum of one-particle irreducible diagrams with one incoming fermion with momentum p, one outgoing fermion with p. The momentum p\u2032, and one incoming scalar with momentum k = p\u2032 \u2212 original vertex Zgg\u03b35 is the \ufb01rst term in this sum, and the diagram of \ufb01g. (51.3) is the second. Thus we have \u2212 iVY (p\u2032, p) = \u2212 Zgg\u03b35 + iVY, 1 loop(p\u2032, p) + O(g", "5), (51.38) where iVY, 1 loop(p\u2032, p) = ( \u2212 3 g)3 1 i Z dd\u2113 (2\u03c0)d h \u03b35 \u02dcS(/p \u2032+/\u2113)\u03b35 \u02dcS(/p+/\u2113)\u03b35 \u02dc\u2206(\u21132). i (51.39) The numerator can be written as N = (/p \u2032 + /\u2113 + m)( \u2212 /\u2113 + m)\u03b35, /p \u2212 and the denominators combined in the usual way. We then get iVY, 1 loop(p\u2032, p) = ig3 \u2212 dF3 Z Z d4q (2\u03c0)4 N (q2 + D)3, (51.40) (51.41) where the integral over Feynman parameters was de\ufb01ned in section 16, and now q = \u2113 + x1p + x2p\u2032, N = [/q x1 /p + (1 x2)/p \u2032 + m][ \u2212 D = x1(1 \u2212 \u2212 x1)p2 + x2(1 2 x2)p\u2032 + (x1+x2)m2 + x3M 2. \u2212 \u2212 \u2212 /q (1 x1)/p + x2 /p \u2032 + m]\u03b35, \u2212 \u2212 2x1x2p p\u2032 \u00b7 (51.42) (51.43) (51.44) Using /q/q = \u2212 q2, we can write N as N = q2\u03b35 + N + (linear in q), (51.45) e 51: Loop Corrections in Yukawa Theory 321 k 2 k 3 l k 1 k 4 Figure 51.4: One of six diagrams with a closed fermion loop and four external scalar lines; the other \ufb01ve are obtained by permuting the external momenta in all possible inequivalent ways. where N = [ \u2212 x1 /p + (1 \u2212 x2)/p \u2032 + m][ \u2212 (1 \u2212 x1)/p + x2 /p \u2032 + m]\u03b35. (51.46) e The terms linear in q in eq. (51.45) integrate to zero, and only the \ufb01rst term is divergent. Performing the usual manipulations, we \ufb01nd iVY, 1 loop(p\u2032, p) = g3 8\u03c02 \" 1 \u03b5 \u2212 1 4 \u2212 1 2 Z dF3 ln(D/\u00b52) \u03b35 + 1 4 Z From eq. (51.38), we see that \ufb01niteness of VY (p\u2032, p) requires dF3. N D # e (51.47) Zg = 1 + g2 8\u03c02 1 \u03b5 + \ufb01nite, (51.48) plus higher-order corrections. To \ufb01x the \ufb01nite part of Zg, we need a condition to impose on VY (p\u2032, p). We will mimic what we did in \u03d53 theory in section 16, and require VY (0, 0) to have the tree-level value ig\u03b35. We leave the details to problem 51.2. Next we turn to the corrections to the \u03d54 vertex iV4(k1, k2, k3, k4); the iZ\u03bb\u03bb. There are diagrams with a closed fermion tree-level contribution is loop, as shown in \ufb01g. (51.4), plus one-loop diagrams with \u03d5 particles only that we evaluated in section 31. We have \u2212 iV4, \u03a8 loop = ( \u2212 1)( \u2212 4 g)4 1 i Z d4\u2113 (2\u03c0)4 Tr \u02dcS(/\u2113+/k2+/k3)\u03b35 \u02dcS(/\u2113+/k2)\u03b35 \u02dcS(/\u2113)\u03b35 \u02dcS(/\u2113 h /k1)\u03b35 \u2212 \u00d7 + 5 permutations of (k2, k3, k4). i (51.49) Again we can employ the standard methods; there are no unfamiliar aspects. This being the case, let us concentrate on obtaining the divergent 51: Loop Corrections in Yukawa Theory 322 part; this will give us enough information to calculate the one-loop contributions to the beta functions for g and \u03bb. To obtain the divergent part of eq. (51.49), it is su\ufb03cient to set ki = 0. The term in the numerator that contributes to the divergent part is Tr (/\u2113\u03b3", "5)4 = 4(\u21132)2, and the denominator is (\u21132 + m2)4. Then we \ufb01nd, after including identical contributions from the other \ufb01ve permutations of the external momenta, V4, \u03a8 loop = 3g4 \u03c02 \u2212 1 \u03b5 + \ufb01nite. From section 31, we have V4, \u03d5 loop = 3\u03bb 16\u03c02 1 \u03b5 + \ufb01nite. Then, using (51.50) (51.51) V4 = \u2212 Z\u03bb\u03bb + V4, \u03a8 loop + V4, \u03d5 loop +..., (51.52) we see that \ufb01niteness of V4 requires Z\u03bb = 1 + 3\u03bb 16\u03c02 \u2212 3g4 \u03c02\u03bb! 1 \u03b5 + \ufb01nite, (51.53) plus higher-order corrections. Reference Notes A detailed derivation of the Lehmann-K\u00a8all\u00b4en form of the fermion propagator can be found in Itzykson & Zuber. Problems 51.1) Derive the fermion-loop correction to the scalar propagator by working through eq. (45.2), and show that it has an extra minus sign relative to the case of a scalar loop. 51.2) Finish the computation of VY (p\u2032, p), imposing the condition VY (0, 0) = ig\u03b35. (51.54) 51.3) Consider making \u03d5 a scalar rather than a pseudoscalar, so that the LYuk = g\u03d5\u03a8\u03a8. In this case, renormalizability Yukawa interaction is 6 Z\u03ba\u03ba\u03d53, as well as term linear in \u03d5 to requires us to add a term cancel tadpoles. Find the one-loop contributions to the renormalizing Z factors for this theory in the MS scheme. L\u03d53 = 1 52: Beta Functions in Yukawa Theory 323 52 Beta Functions in Yukawa Theory Prerequisite: 28, 51 In this section we will compute the beta functions for the Yukawa coupling g and the \u03d54 coupling \u03bb in Yukawa theory, using the methods of section 28. The relations between the bare and renormalized couplings are g0 = Z \u2212 \u03d5 \u03bb0 = Z \u2212 1 1/2 \u03a8 Zg \u02dc\u00b5\u03b5/2g, Z \u2212 \u03d5 Z\u03bb \u02dc\u00b5\u03b5\u03bb. 2 Let us de\ufb01ne ln Z \u2212 \u03d5 1/2 Z \u2212 1 \u03a8 Zg = = ln Z \u2212 2 \u03d5 Z\u03bb Gn(g, \u03bb) \u03b5n Ln(g, \u03bb) \u03b5n,. \u221e Xn=1 \u221e Xn=1 From our results in section 51, we have (52.1) (52.2) (52.3) (52.4) G1(g, \u03bb) = 5g2 16\u03c02 +..., g2 3\u03bb 16\u03c02 + 2\u03c02 \u2212 where the ellipses stand for higher-order (in g2 and/or \u03bb) corrections. L1(g, \u03bb) = 3g4 \u03c02\u03bb +..., (52.5) (52.6) Taking the logarithm of eqs. (52.1) and (52.2), and using eqs. (52.3) and (52.4), we get ln g0 = ln \u03bb0 = Gn(g, \u03bb) \u03b5n Ln(g, \u03bb) \u03b5n \u221e Xn=1 \u221e Xn=1 + ln g + 1 2 \u03b5 ln \u02dc\u00b5, (52.7) + ln \u03bb + \u03b5 ln \u02dc\u00b5. (52.8) We now use the fact that g0 and \u03bb0 must be independent of \u00b5. We di\ufb00erentiate eqs. (52.7) and (52.8) with respect to ln \u00b5; the left-hand sides vanish, and we multiply the right-hand sides by g and \u03bb, respectively. The result is 0 = 0 = \u221e g Xn=1 \u221e \u03bb Xn=1 \u2202Gn \u2202g dg d ln \u00b5 + g \u2202Gn \u2202\u03bb d\u03bb d ln \u00b5 \u2202Ln \u2202g dg d ln \u00b5 + \u03bb \u2202Ln \u2202\u03bb d\u03bb d ln \u00b5 1 \u03b5n + dg d ln \u00b5 1 \u03b5n + d\u03bb d ln \u00b5 + 1 2 \u03b5g, (52", ".9) + \u03b5\u03bb. (52.10) 52: Beta Functions in Yukawa Theory 324 In a renormalizable theory, dg/d ln \u00b5 and d\u03bb/d ln \u00b5 must be \ufb01nite in the \u03b5 0 limit. Thus we can write \u2192 dg d ln \u00b5 d\u03bb d ln \u00b5 = = \u2212 \u2212 1 2 \u03b5g + \u03b2g(g, \u03bb), \u03b5\u03bb + \u03b2\u03bb(g, \u03bb). (52.11) (52.12) Substituting these into eqs. (52.9) and (52.10), and matching powers of \u03b5, we \ufb01nd \u03b2g(g, \u03bb) = g \u03b2\u03bb(g, \u03bbg \u2202 \u2202g + \u03bb + \u03bb \u2202 \u2202\u03bb \u2202 \u2202\u03bb G1, L1. (52.13) (52.14) The coe\ufb03cients of all higher powers of 1/\u03b5 must also vanish, but this gives us no more information about the beta functions. Using eqs. (52.5) and (52.6) in eqs. (52.13) and (52.14), we get \u03b2g(g, \u03bb) = 5g3 16\u03c02 +..., 1 16\u03c02 3\u03bb2 + 8\u03bbg2 The higher-order corrections have extra factors of g2 and/or \u03bb. \u03b2\u03bb(g, \u03bb) = +.... 48g4 \u2212 (52.15) (52.16) Problems 52.1) Compute the one-loop contributions to the anomalous dimensions of m, M, \u03a8, and \u03d5. 52.2) Consider the theory of problem 51.3. Compute the one-loop contributions to the beta functions for g, \u03bb, and \u03ba, and to the anomalous dimensions of m, M, \u03a8, and \u03d5. 52.3) Consider the beta functions of eqs. (52.15) and (52.16). \u03bb/g2, and compute d\u03c1/d ln \u00b5. Express your answer in a) Let \u03c1 terms of g and \u03c1. Explain why it is better to work with g and \u03c1 rather than g and \u03bb. Hint: the answer is mathematical, not physical. \u2261 b) Show that there are two \ufb01xed points, \u03c1\u2217+ and \u03c1\u2217 \u2212 0, and \ufb01nd their values., where d\u03c1/d ln \u00b5 = c) Suppose that, for some particular value of the renormalization scale 1. What happens to \u03c1 at much higher \u00b5, we have \u03c1 = 0 and g \u226a\u226a 52: Beta Functions in Yukawa Theory 325 values of \u00b5 (but still low enough to keep g values of \u00b5? \u226a 1)? At much lower d) Same question, but with an initial value of \u03c1 = 5. e) Same question, but with an initial value of \u03c1 = 5. \u2212 f) Find the trajectory in the (\u03c1, g) plane that is followed for each of the three starting points as \u00b5 is varied up and down. Hint: you should \ufb01nd that the trajectories take the form g = g0 \u03bd \u03c1 \u03c1 \u03c1\u2217+ \u03c1\u2217 \u2212 \u2212 \u2212 for some particular exponent \u03bd. Put arrows on the trajectories that point in the direction of increasing \u00b5. g) Explain why \u03c1\u2217 \u2212 \u03c1\u2217+ is called an infrared stable \ufb01xed point. is called an ultraviolet stable \ufb01xed point, and why 53: Functional Determinants 326 53 Functional Determinants Prerequisite: 44, 45 In the section we will explore the meaning of the functional determinants that arise when doing gaussian path integrals, either bosonic or fermionic. We will be interested in situations where the path integral over one particular \ufb01eld is gaussian, but generates a functional determinant that depends on some other \ufb01eld. We will see how to relate this functional determinant to a certain in\ufb01nite set of Feynman diagrams. We will need the technology we develop here to compute the path integral for nonabelian gauge theory in section 70. We begin by considering a theory of a complex scalar \ufb01eld \u03c7 with = \u2202\u00b5\u03c7\u2020\u2202\u00b5\u03c7 m2\u03c7\u2020\u03c7 + g\u03d5\u03c7\u2020\u03c7, (53.1) \u2212 where \u03d5 is a real scalar background \ufb01eld. That is, \u03d5(x) is treated as a \ufb01xed function of spacetime. Next we de\ufffd", "\ufffd\ufffdne the path integral \u2212 L Z(\u03d5) = \u03c7\u2020 D D Z \u03c7 ei d4x L, R (53.2) where we use the \u01eb trick of section 6 to impose vacuum boundary conditions, and the normalization Z(0) = 1 is \ufb01xed by hand. Recall from section 44 that if we have n complex variables zi, then we can evaluate gaussian integrals by the general formula dnz dn \u00afz exp ( \u2212 i\u00afziMijzj) \u221d (det M )\u2212 1. Z (53.3) In the case of the functional integral in eq. (53.2), the index i on the integration variable is replaced by the continuous spacetime label x, and the \u201cmatrix\u201d M becomes M (x, y) = [ \u2212 x + m2 \u22022 \u2212 g\u03d5(x)]\u03b44(x y). \u2212 (53.4) In order to apply eq. (53.3), we have to understand what it means to compute the determinant of this expression. To this end, let us \ufb01rst note that we can write M = M0 M, which is shorthand for where M (x, z) = d4y M0(x, y) M (y, z), f (53.5) Z M0(x, y) = ( \u2212 x + m2)\u03b44(x \u22022 f \u2212 y), M (y, z) = \u03b44(y z) \u2212 \u2212 g\u2206(y \u2212 z)\u03d5(z). f (53.6) (53.7) (53.10) (53.11) 53: Functional Determinants Here \u2206(y \u2212 z) is the Feynman propagator, which obeys \u22022 y + m2)\u2206(y ( \u2212 \u2212 z) = \u03b44(y z). \u2212 327 (53.8) After various integrations by parts, it is easy to see that eqs. (53.5\u201353.7) reproduce eq. (53.4). Now we can use the general matrix relation det AB = det A det B to conclude that det M = det M0 det M. (53.9) The advantage of this decomposition is that M0 is independent of the 1 in Z(\u03d5) can background \ufb01eld \u03d5, and so the resulting factor of (det M0)\u2212 simply be absorbed into the overall normalization. Furthermore, we have M = I G, where f \u2212 f is the identity matrix, and I(x, y) = \u03b44(x y) \u2212 G(x, y) = g\u2206(x y)\u03d5(y). \u2212 Thus, for \u03d5(x) = 0, we have M = 1. Then, using eq. (53.3) and the normalization condition Z(0) = 1, we see that for nonzero \u03d5(x) we must have simply M = I and so det f f Z(\u03d5) = (det M )\u2212 1. (53.12) Next, we need the general matrix relation det A = exp Tr ln A, which is most easily proved by working in a basis where A is in Jordan form (that is, all entries below the main diagonal are zero). Thus we can write f det M = exp Tr ln M f = exp Tr ln(I f \u2212 G) = exp Tr \"\u2212 \u221e Xn=1 1 n Gn. # Combining eqs. (53.12) and (53.13) we get Z(\u03d5) = exp Tr Gn, 1 n \u221e Xn=1 where (53.13) (53.14) Tr Gn = gn Z d4x1... d4xn \u2206(x1\u2212 x2)\u03d5(x2)... \u2206(xn\u2212 x1)\u03d5(x1). (53.15) 53: Functional Determinants 328... Figure 53.1: All connected diagrams with \u03d5(x) treated as an external \ufb01eld. Each of the n dots represents a factor of ig\u03d5(x), and each solid line is a \u03c7 or \u03a8 propagator. This is our \ufb01nal result for Z(\u03d5). L To better understand what it means, we will rederive it in a di\ufb00erent way. Consider treating the g\u03d5\u03c7\u2020\u03c7 term in as an interaction. This leads to a vertex that connects two \u03c7 propagators; the associated vertex factor is ig\u03d5(x). According to the general analysis of section 9, we have", " Z(\u03d5) = exp i\u0393(\u03d5), where i\u0393(\u03d5) is given by a sum of connected diagrams. (We have called the exponent \u0393 rather than W because it is naturally interpreted as a quantum action for \u03d5 after \u03c7 has been integrated out.) The only connected diagrams we can draw with these Feynman rules are those of \ufb01g. (53.1), 1. The diagram with n vertices with n insertions of the vertex, where n has an n-fold cyclic symmetry, leading to a symmetry factor of S = n. The factor of i associated with each vertex is canceled by the factor of 1/i associated with each propagator. Thus the value of the n-vertex diagram is \u2265 1 n gn Z d4x1... d4xn \u2206(x1\u2212 x2)\u03d5(x2)... \u2206(xn\u2212 x1)\u03d5(x1). (53.16) Summing up these diagrams, and using eq. (53.15), we \ufb01nd i\u0393(\u03d5) = \u221e Tr Gn. (53.17) 1 n Xn=1 This neatly reproduces eq. (53.14). Thus we see that a functional determinant can be represented as an in\ufb01nite sum of Feynman diagrams. Next we consider a theory of a Dirac fermion \u03a8 with = i\u03a8/\u2202\u03a8 \u2212 L m\u03a8\u03a8 + g\u03d5\u03a8\u03a8, (53.18) where \u03d5 is again a real scalar background \ufb01eld. We de\ufb01ne the path integral Z(\u03d5) = \u03a8 D D Z \u03a8 ei d4x L, R (53.19) 53: Functional Determinants 329 where we again use the \u01eb trick to impose vacuum boundary conditions, and the normalization Z(0) = 1 is \ufb01xed by hand. Recall from section 44 that if we have n complex Grassmann variables \u03c8i, then we can evaluate gaussian integrals by the general formula dn\u00af\u03c8 dn\u03c8 exp Z \u2212 i \u00af\u03c8iMij\u03c8j det M. \u221d (53.20) In the case of the functional integral in eq. (53.19), the index i on the integration variable is replaced by the continuous spacetime label x plus the spinor index \u03b1, and the \u201cmatrix\u201d M becomes M\u03b1\u03b2(x, y) = [ \u2212 i/\u2202x + m \u2212 g\u03d5(x)]\u03b1\u03b2 \u03b44(x y). \u2212 (53.21) In order to apply eq. (53.20), we have to understand what it means to compute the determinant of this expression. To this end, let us \ufb01rst note that we can write M = M0 M, which is shorthand for where M\u03b1\u03b3(x, z) = d4y M0\u03b1\u03b2(x, y) M\u03b2\u03b3(y, z), Z f M0\u03b1\u03b2(x, y) = ( \u2212 i/\u2202x + m)\u03b1\u03b2\u03b44(x y), \u2212 gS\u03b2\u03b3(y M\u03b2\u03b3(y, z) = \u03b4\u03b2\u03b3\u03b44(y z) \u2212 \u2212 z)\u03d5(z). (53.24) \u2212 Here S\u03b2\u03b3(y \u2212 z) is the Feynman propagator, which obeys f z) = \u03b4\u03b1\u03b3\u03b44(y z). (53.25) i/\u2202y + m)\u03b1\u03b2S\u03b2\u03b3(y ( \u2212 \u2212 After various integrations by parts, it is easy to see that eqs. (53.22\u201353.24) reproduce eq. (53.21). \u2212 Now we can use eq. (53.9). The advantage of this decomposition is that M0 is independent of the background \ufb01eld \u03d5, and so the resulting factor of det M0 in Z(\u03d5) can simply be absorbed into the overall normalization. G, where Furthermore, we have M = I \u2212 I\u03b1\u03b2(x, y) = \u03b4\u03b1\u03b2\u03b44(x f y) \u2212 is the identity matrix, and G\u03b1\u03b2(x, y) = gS\u03b1\u03b2(x y)\u03d5(y). \u2212 (53.26) (53.27) M = 1. Then, using Thus, for \u03d5(x) = 0, we have eq. (53.20) and the normalization condition Z(0) = 1, we see that for f nonzero \u03d5(x) we must have simply M =", " I and so det f Z(\u03d5) = det M. (53.28) f f (53.22) (53.23) 53: Functional Determinants Next, we use eqs. (53.13) and (53.28) to get Z(\u03d5) = exp 1 n \u221e \u2212 Xn=1 Tr Gn, 330 (53.29) where now Tr Gn = gn Z d4x1... d4xn tr S(x1\u2212 x2)\u03d5(x2)... S(xn\u2212 x1)\u03d5(x1), (53.30) and \u201ctr\u201d denotes a trace over spinor indices. This is our \ufb01nal result for Z(\u03d5). L To better understand what it means, we will rederive it in a di\ufb00erent way. Consider treating the g\u03d5\u03a8\u03a8 term in as an interaction. This leads to a vertex that connects two \u03a8 propagators; the associated vertex factor is ig\u03d5(x). According to the general analysis of section 9, we have Z(\u03d5) = exp i\u0393(\u03d5), where i\u0393(\u03d5) is given by a sum of connected diagrams. (We have called the exponent \u0393 rather than W because it is naturally interpreted as a quantum action for \u03d5 after \u03a8 has been integrated out.) The only connected diagrams we can draw with these Feynman rules are those of \ufb01g. (53.1), with n insertions of the vertex, where n 1. The diagram with n vertices has an n-fold cyclic symmetry, leading to a symmetry factor of S = n. The factor of i associated with each vertex is canceled by the factor of 1/i associated with each propagator. The closed fermion loop implies a trace over the spinor indices. Thus the value of the n-vertex diagram is \u2265 1 n gn Z d4x1... d4xn tr S(x1\u2212 x2)\u03d5(x2)... S(xn\u2212 x1)\u03d5(x1). (53.31) Summing up these diagrams, we \ufb01nd that we are missing the overall minus sign in eq. (53.29). The appropriate conclusion is that we must associate an extra minus sign with each closed fermion loop. Part III Spin One 54: Maxwell\u2019s Equations 332 54 Maxwell\u2019s Equations Prerequisite: 3 The most common (and important) spin-one particle is the photon. Emission and absorption of photons by matter is an important phenomenon in many areas of physics, and so that is the context in which most physicists \ufb01rst encounter a serious treatment of photons. We will use a brief review of this subject (in this section and the next) as our entry point into the theory of quantum electrodynamics. Let us begin with classical electrodynamics. Maxwell\u2019s equations are E = \u03c1, \u02d9E = J, \u2207\u00b7 \u2212 B E + \u02d954.1) (54.2) (54.3) (54.4) \u2207\u00b7 where E is the electric \ufb01eld, B is the magnetic \ufb01eld, \u03c1 is the charge density, and J is the current density. We have written Maxwell\u2019s equations in Heaviside-Lorentz units, and also set c = 1. In these units, the magnitude of the force between two charges of magnitude Q is Q2/4\u03c0r2. Maxwell\u2019s equations must be supplemented by formulae that give us the dynamics of the charges and currents (such as the Lorentz force law for point particles). For now, however, we will treat the charges and currents as speci\ufb01ed sources, and focus on the dynamics of the electromagnetic \ufb01elds. The last two of Maxwell\u2019s equations, the ones with no sources on the right-hand side, can be solved by writing the E and B \ufb01elds in terms of a scalar potential \u03d5 and a vector potential A, \u02d9A, (54.5) E = \u2212\u2207 \u03d5 B = \u2212 A. (54.6) \u2207 \u00d7 The potentials uniquely determine the \ufb01elds, but the \ufb01elds do not uniquely determine the potentials. Given a particular \u03d5 and A that result in a particular E and B, we will get the same E and B from any other potentials \u03d5\u2032 and A\u2032 that are related by \u03d5\u2032 = \u03d5 + \u02d9\u0393, A\u2032 = A \u0393, \u2212 \u2207 (54.7) (54", ".8) where \u0393 is an arbitrary function of spacetime. A change of potentials that does not change the \ufb01elds is called a gauge transformation. The E and B \ufb01elds are gauge invariant. 54: Maxwell\u2019s Equations 333 All this becomes more compact and elegant in a relativistic notation. We de\ufb01ne the four-vector potential or gauge \ufb01eld We also de\ufb01ne the \ufb01eld strength A\u00b5 \u2261 (\u03d5, A). F \u00b5\u03bd \u2202\u00b5A\u03bd \u2202\u03bdA\u00b5. \u2212 Obviously, F \u00b5\u03bd is antisymmetric: F \u00b5\u03bd = (54.6) with eqs. (54.9) and (54.10), we see that \u2261 \u2212 (54.9) (54.10) F \u03bd\u00b5. Comparing eqs. (54.5) and F 0i = Ei, F ij = \u03b5ijkBk. The \ufb01rst two of Maxwell\u2019s equations can now be written as where \u2202\u03bdF \u00b5\u03bd = J \u00b5, J \u00b5 \u2261 (\u03c1, J) (54.11) (54.12) (54.13) (54.14) is the charge-current density four-vector. If we take the four-divergence of eq. (54.13), we get \u2202\u00b5\u2202\u03bd F \u00b5\u03bd = \u2202\u00b5J \u00b5. The left-hand side of this equation vanishes, because \u2202\u00b5\u2202\u03bd is symmetric on exchange of \u00b5 and \u03bd, while F \u00b5\u03bd is antisymmetric. We conclude that we must have or equivalently \u2202\u00b5J \u00b5 = 0, \u02d9\u03c1 + \u2207\u00b7 J = 0 ; that is, the electromagnetic current must be conserved. The last two of Maxwell\u2019s equations can be written as \u03b5\u00b5\u03bd\u03c1\u03c3\u2202\u03c1F \u00b5\u03bd = 0, (54.15) (54.16) (54.17) where \u03b5\u00b5\u03bd\u03c1\u03c3 is the completely antisymmetric Levi-Civita tensor; see section 34. Plugging in eq. (54.10), we see that eq. (54.17) is automatically satis\ufb01ed, since the antisymmetric combination of two derivatives vanishes. Eqs. (54.7) and (54.8) can be combined into A\u2032 \u00b5 = A\u00b5 \u2202\u00b5\u0393. \u2212 Setting F \u2032 \u03bd \u00b5\u03bd = \u2202\u00b5A\u2032 \u2202\u03bdA\u2032 \u00b5 and using eq. (54.18), we get \u2212 F \u2032 \u00b5\u03bd = F \u00b5\u03bd (\u2202\u00b5\u2202\u03bd \u2212 \u2212 \u2202\u03bd\u2202\u00b5)\u0393. (54.18) (54.19) 54: Maxwell\u2019s Equations 334 The last term vanishes because derivatives commute; thus the \ufb01eld strength is gauge invariant, F \u2032 \u00b5\u03bd = F \u00b5\u03bd. (54.20) Next we will \ufb01nd an action that results in Maxwell\u2019s equations as the equations of motion. We will treat the current as an external source. The action we seek should be Lorentz invariant, gauge invariant, parity and time-reversal invariant, and no more than second order in derivatives. The only candidate is S =, where d4x L R = L \u2212 1 4 F \u00b5\u03bdF\u00b5\u03bd + J \u00b5A\u00b5. (54.21) The \ufb01rst term is obviously gauge invariant, because F \u00b5\u03bd is. After a gauge transformation, eq. (54.18), the second term becomes J \u00b5A\u2032\u00b5, and the di\ufb00erence is J \u00b5(A\u2032\u00b5 \u2212 A\u00b5) = = J \u00b5\u2202\u00b5\u0393 (\u2202\u00b5J \u00b5)\u0393 \u2212 \u2212 \u2212 \u2202\u00b5(J \u00b5\u0393). (54.22) The \ufb01rst term in eq. (54.22) vanishes because the current is conserved. The second term is a total divergence, and its integral over d4x vanishes (assuming suitable boundary conditions at in\ufb01nity). Thus the action speci\ufb01ed by eq. (54.21) is gauge invariant. Setting F \u00b5\u03bd = \u2202\u00b5A\u03bd \u2202\u03bdA\u00b5 and multiplying out the terms, eq. (54.21) \u2212 becomes L = \u2212 = + 1 1 2 \u2202\u00b5A\u03bd\u2202\u00b5A\u03bd + 1 2 A\u00b5(g\u00b5\u03bd \u22022 2 \u2202\u00b5A\u03bd", "\u2202\u03bdA\u00b5 + J \u00b5A\u00b5 \u2202\u00b5\u2202\u03bd )A\u03bd + J \u00b5A\u00b5 \u2212 \u2212 \u2202\u00b5K\u00b5, (54.23) (54.24) where K\u00b5 = 1 \u2202\u03bdA\u00b5). The last term is a total divergence, and can be dropped. From eq. (54.24), we can see that varying A\u00b5 while requiring S to be unchanged yields the equation of motion 2 A\u03bd(\u2202\u00b5A\u03bd \u2212 (g\u00b5\u03bd \u22022 \u2202\u00b5\u2202\u03bd )A\u03bd + J \u00b5 = 0. (54.25) \u2212 g\u00b5\u03bd \u22022)A\u03bd, we see that Noting that \u2202\u03bdF \u00b5\u03bd = \u2202\u03bd(\u2202\u00b5A\u03bd eq. (54.25) is equivalent to eq. (54.13), and hence to Maxwell\u2019s equations. \u2202\u03bdA\u00b5) = (\u2202\u00b5\u2202\u03bd \u2212 \u2212 55: Electrodynamics in Coulomb Gauge 335 55 Electrodynamics in Coulomb Gauge Prerequisite: 54 Next we would like to construct the hamiltonian, and quantize the electromagnetic \ufb01eld. There is an immediate di\ufb03culty, caused by the gauge invariance: we have too many degrees of freedom. This problem manifests itself in several ways. For example, the lagrangian L = = \u2212 \u2212 1 4F \u00b5\u03bdF\u00b5\u03bd + J \u00b5A\u00b5 2\u2202\u00b5A\u03bd \u2202\u00b5A\u03bd + 1 1 2 \u2202\u00b5A\u03bd\u2202\u03bd A\u00b5 + J \u00b5A\u00b5 (55.1) (55.2) does not contain the time derivative of A0. Thus, this \ufb01eld has no canonically conjugate momentum and no dynamics. To deal with this problem, we must eliminate the gauge freedom. We do this by choosing a gauge. We choose a gauge by imposing a gauge condition. This is a condition that we require A\u00b5(x) to satisfy. The idea is that there should be only one A\u00b5(x) that results in a given F \u00b5\u03bd (x) and that also satis\ufb01es the gauge condition. One possible class of gauge conditions is n\u00b5A\u00b5(x) = 0, where n\u00b5 is a constant four-vector. If n is spacelike (n2 > 0), then we have chosen axial gauge; if n is lightlike, (n2 = 0), it is lightcone gauge; and if n is timelike, (n2 < 0), it is temporal gauge. Another gauge is Lorenz gauge, where the condition is \u2202\u00b5A\u00b5 = 0. We will meet a family of closely related gauges in section 62. In this section, we will work in Coulomb gauge, also known as radiation gauge or transverse gauge. The condition for Coulomb gauge is We can impose eq. (55.3) by acting on Ai(x) with a projection operator, A(x) = 0. \u2207\u00b7 (55.3) Ai(x) \u03b4ij \u2212 \u2192 \u2207i\u2207j 2 \u2207 Aj(x). (55.4) Ai(k), multiplying We construct the right-hand side of eq. (55.4) by Fourier-transforming Ai(x) kikj/k2, and then Fourierto transforming back to position space. From now on, whenever we write Ai, we will implicitly mean the right-hand side of eq. (55.4). Ai(k) by the matrix \u03b4ij \u2212 e e Now let us write out the lagrangian in terms of the scalar and vector potentials, \u03d5 = A0 and Ai, with Ai obeying the Coulomb gauge condition. Starting from eq. (55.2), we get = 1 2 \u02d9Ai \u02d9Ai \u2212 1 2 \u2207jAi\u2207jAi + JiAi L 336 (55.5) 55: Electrodynamics in Coulomb Gauge \u03c1\u03d5. + 1 + 1 \u2207i\u03d5 2 \u2207iAj\u2207jAi + \u02d9Ai\u2207i\u03d5 2 \u2207i\u03d5 \u2212 In the second line of eq. (55.5), the parts; in the \ufb01rst term, we will then get a factor of second term, we will get a factor of of the gauge condition dropped. \u2207i in each term can be integrated by \u2207iAi), and in the \u2207j( \u2207i \u02d9Ai. Both of these vanish by virtue \u2207iAi =", " 0, and so both of these terms can simply be to be stationary), we \ufb01nd If we now vary \u03d5 (and require S = d4x that \u03d5 obeys Poisson\u2019s equation, R 2\u03d5 = \u03c1. \u2212\u2207 L The solution is \u03d5(x, t) = d3y Z \u03c1(y, t) y x 4\u03c0 | \u2212 | (55.6) (55.7). This solution is unique if we impose the boundary conditions that \u03d5 and \u03c1 both vanish at spatial in\ufb01nity. Eq. (55.7) tells us that \u03d5(x, t) is given entirely in terms of the charge density at the same time, and so has no dynamics of its own. It is therefore legitimate to plug eq. (55.7) back into the lagrangian. After an integration by parts to turn 2\u03d5 = \u03d5\u03c1, the result is \u03d5 \u2207i\u03d5 = 1 2 \u2207i\u03d5 into \u02d9Ai \u02d9Ai \u2212 L \u2207 \u2212 1 2 \u2207jAi\u2207jAi + JiAi + where Lcoul = \u2212 d3y 1 2 Z \u03c1(x, t)\u03c1(y, t) x | 4\u03c0 \u2212 y | We can now vary Ai; keeping proper track of the implicit projection operator in eq. (55.4), we \ufb01nd that Ai obeys the massless Klein-Gordon equation with the projected current as a source, Lcoul, (55.8). (55.9) \u22022Ai(x) = \u2212 \u03b4ij \u2212 \u2207i\u2207j 2 \u2207 Jj(x). (55.10) For a free \ufb01eld (Ji = 0), the general solution is A(x) = dk \u03b5\u2217\u03bb(k)a\u03bb(k)eikx + \u03b5\u03bb(k)a\u2020\u03bb(k)e\u2212 ikx, (55.11) X\u03bb= \u00b1 Z h i f where k0 = \u03c9 = k (k) are po, | | larization vectors. In order to satisfy the Coulomb gauge condition, the polarization vectors must be orthogonal to the wave vector k. We will dk = d3k/(2\u03c0)32\u03c9, and \u03b5+(k) and \u03b5 \u2212 f 55: Electrodynamics in Coulomb Gauge 337 choose them to correspond to right- and left-handed circular polarizations; for k = (0, 0, k), we then have \u03b5+(k) = 1 \u221a2 (k) = 1 \u221a2 \u03b5 \u2212 (1, i, 0), \u2212 (1, +i, 0). (55.12) More generally, the two polarization vectors along with the unit vector in the k direction form an orthonormal and complete set, k \u03b5\u03bb\u2032(k) \u03b5\u03bb(k) = 0, \u03b5\u2217\u03bb(k) = \u03b4\u03bb\u2032\u03bb, \u00b7 \u00b7 \u03b5\u2217i\u03bb(k)\u03b5j\u03bb(k) = \u03b4ij \u2212 kikj k2. X\u03bb= \u00b1 (55.13) (55.14) (55.15) The coe\ufb03cients a\u03bb(k) and a\u2020\u03bb(k) will become operators after quantization, which is why we have used the dagger symbol for conjugation. In complete analogy with the procedure used for a scalar \ufb01eld in section 3, we can invert eq. (55.11) and its time derivative to get a\u03bb(k) = +i \u03b5\u03bb(k) a\u2020\u03bb(k) = i \u03b5\u2217\u03bb(k) \u2212 \u00b7 \u00b7 Z Z d3x e\u2212 ikx \u2194\u22020 A(x), d3x e+ikx \u2194\u22020 A(x), (55.16) (55.17) where f \u2194\u2202\u00b5g = f (\u2202\u00b5g) (\u2202\u00b5f )g. \u2212 Now we can proceed to the hamiltonian formalism. First, we compute the canonically conjugate momentum to Ai, \u03a0i = \u2202 L \u2202 \u02d9Ai = \u02d9Ai. (55.18) Note that \u2207iAi = 0 implies \u2207i\u03a0i = 0. The hamiltonian density is then H = \u03a0i \u02d9Ai \u2212 L = 1 2 \u03a0i\u03a0i + 1 2 \u2207jAi\u2207jAi \u2212 JiAi + Hcoul, (55.19) where H", "coul = \u2212Lcoul. To quantize the \ufb01eld, we impose the canonical commutation relations. Keeping proper track of the implicit projection operator in eq. (55.4), we have [Ai(x, t), \u03a0j (y, t)] = i = i Z \u2207i\u2207j \u03b4ij \u2212 2 \u2207 d3k (2\u03c0)3 eik \u00b7 \u03b43(x y) \u2212 (x y) \u2212 \u03b4ij \u2212 kikj k2. (55.20) 55: Electrodynamics in Coulomb Gauge 338 The commutation relations of the a\u03bb(k) and a\u2020\u03bb(k) operators follow from eq. (55.20) and [Ai, Aj] = [\u03a0i, \u03a0j ] = 0 (at equal times). The result is [a\u03bb(k), a\u03bb\u2032 (k\u2032)] = 0, [a\u2020\u03bb(k), a\u2020\u03bb\u2032 (k\u2032)] = 0, [a\u03bb(k), a\u2020\u03bb\u2032 (k\u2032)] = (2\u03c0)32\u03c9 \u03b43(k\u2032 k)\u03b4\u03bb\u03bb\u2032. \u2212 (55.21) (55.22) (55.23) We interpret a\u2020\u03bb(k) and a\u03bb(k) as creation and annihilation operators for photons of de\ufb01nite helicity, with helicity +1 corresponding to right-circular polarization and helicity 1 to left-circular polarization. It is now straightfoward to write the hamiltonian explicitly in terms of \u2212 these operators. We \ufb01nd H = dk \u03c9 a\u2020\u03bb(k)a\u03bb(k) + 2 A(x) + Hcoul, (55.24) Z X\u03bb= \u00b1 E0 = 1 d3k \u03c9 is the zero-point energy per unit volume that where we found for a real scalar \ufb01eld in section 3, V is the volume of space, the Coulomb hamiltonian is f 2 (2\u03c0)\u2212 R 3 E0V \u2212 Z d3x J(x) \u00b7 Hcoul = d3x d3y 1 2 Z \u03c1(x, t)\u03c1(y, t) x | 4\u03c0 \u2212 y |, (55.25) and we use eq. (55.11) to express Ai(x) in terms of a\u03bb(k) and a\u2020\u03bb(k) at any one particular time (say, t = 0). This is su\ufb03cient, because H itself is time independent. This form of the hamiltonian of electrodynamics is often used as the starting point for calculations of atomic transition rates, with the charges and currents treated via the nonrelativistic Schr\u00a8odinger equation. The Coulomb interaction appears explicitly, and the J A term allows for the creation and annihilation of photons of de\ufb01nite polarization. \u00b7 Reference Notes A more rigorous treatment of quantization in Coulomb gauge can be found in Weinberg I. Problems 55.1) Use eqs. (55.13\u201355.20) and [Ai, Aj] = [\u03a0i, \u03a0j] = 0 (at equal times) to verify eqs. (55.21\u201355.23). 55.2) Use eqs. (55.11), (55.14), (55.19), and (55.21\u201355.23) to verify eq. (55.24). 56: LSZ Reduction for Photons 339 56 LSZ Reduction for Photons Prerequisite: 5, 55 In section 55, we found that the creation and annihilation operators for free photons could be written as i \u03b5\u2217\u03bb(k) a\u2020\u03bb(k) = \u2212 a\u03bb(k) = +i \u03b5\u03bb(k) \u00b7 Z \u00b7 d3x e+ikx \u2194\u22020 A(x), d3x e\u2212 ikx \u2194\u22020 A(x), (56.1) (56.2) Z where \u03b5\u03bb(k) is a polarization vector. From here, we can follow the analysis of section 5 line by line to deduce the LSZ reduction formula for photons. The result is that the creation operator for each incoming photon should be replaced by a\u2020\u03bb(k)in \u2192 i \u03b5\u00b5 \u2217\u03bb (k) Z d4x e+ikx( \u2212 \u22022)A\u00b5(x), (56.3) and the destruction operator for each outgoing photon should be replaced by a\u03bb(k)out \u2192 i \u03b5\u00b5 \u03bb(k) d4x e\u2212 ikx( \u2212 \u22022)A\u00b5(x), (56.4) Z", " and then we should take the vacuum expectation value of the time-ordered product. Note that, in writing eqs. (56.3) and (56.4), we have made them look nicer by introducing \u03b50 0, and then using four-vector dot products rather than three-vector dot products. \u03bb(k) \u2261 The LSZ formula is valid provided the \ufb01eld is normalized according to the free-\ufb01eld formulae Ai(x) 0 0 | | h i Ai(x) 0 k, \u03bb i | | h = 0, = \u03b5i \u03bb(k)eikx, where k, \u03bb i | is a single photon state, normalized according to k\u2032, \u03bb\u2032 h k, \u03bb i | = (2\u03c0)32\u03c9 \u03b43(k\u2032 k)\u03b4\u03bb\u03bb\u2032. \u2212 (56.5) (56.6) (56.7) The zero on the right-hand side of eq. (56.5) is required by rotation invariance, and only the overall scale of the right-hand side of eq. (56.6) might be di\ufb00erent in an interacting theory. The renormalization of Ai necessitates including appropriate Z factors in the lagrangian, = L \u2212 1 4Z3F \u00b5\u03bdF\u00b5\u03bd + Z1J \u00b5A\u00b5. (56.8) 56: LSZ Reduction for Photons 340 Here Z3 and Z1 are the traditional names; we will meet Z2 in section 62. We must choose Z3 so that eq. (56.6) holds. We will \ufb01x Z1 by requiring the corresponding vertex function to take on a certain value for a particular set of external momenta. Next we must compute the correlation functions. As 0 i | usual, we begin by working with free \ufb01eld theory. The analysis is again almost identical to the case of a scalar \ufb01eld; see problem 8.4. We \ufb01nd that, in free \ufb01eld theory, TAi(x)... 0 | h TAi(x)Aj(yij(x y), \u2212 (56.9) where the propagator is \u2206ij(x y) = \u2212 Z d4k (2\u03c0)4 eik(x k2 y) \u2212 i\u01eb \u2212 P \u2217\u03bb (k)\u03b5j \u03b5i \u03bb(k). \u03bb= \u00b1 (56.10) As with a free scalar \ufb01eld, correlations of an odd number of \ufb01elds vanish, and correlations of an even number of \ufb01elds are given in terms of the propagator by Wick\u2019s theorem; see section 8. We would now like to evaluate the path integral for the free electromag- netic \ufb01eld Z0(J) iJ = 0 0 | \u2261 h D A ei d4x [ 1 4 F \u00b5\u03bd F\u00b5\u03bd +J \u00b5A\u00b5]. \u2212 (56.11) Z Here we treat the current J \u00b5(x) as an external source. R We will evaluate Z0(J) in Coulomb gauge. This means that we will integrate over only those \ufb01eld con\ufb01gurations that satisfy A = 0. We begin by integrating over A0. Because the action is quadratic in A\u00b5, this is equivalent to solving the variational equation for A0, and then substituting the solution back into the lagrangian. The result is that we have the Coulomb term in the action, \u2207\u00b7 Scoul = 1 2 \u2212 Z d4x d4y \u03b4(x0 y0) \u2212 J 0(x)J 0(y) x 4\u03c0 | \u2212 y |. (56.12) Since this term does not depend on the vector potential, we simply get a factor of exp(iScoul) in front of the remaining path integral over Ai. We wll perform this integral formally (as we did for fermion \ufb01elds in section 43) by requiring it to yield the correct results for the correlation functions of Ai when we take functional derivatives with respect to Ji. In this way we \ufb01nd that Z0(J) = exp iScoul + i 2 Z d4x d4y Ji(x)\u2206ij(x \u2212 y)Jj(y). (56.13) 56: LSZ Reduction for Photons 341 We can make Z0(J) look prettier by writing it as Z0(J) = exp where we have de\ufb01ned i 2 Z d4x d4y J\u00b5(x)\u2206\u00b5\u03bd (x", " \u2212 y)J\u03bd (y), (56.14) \u2206\u00b5\u03bd(x y) \u2212 \u02dc\u2206\u00b5\u03bd(k) \u2212 d4k (2\u03c0)4 eik(x 1 k2 \u03b4\u00b50\u03b4\u03bd0 + \u2261 Z \u2261 \u2212 y) \u02dc\u2206\u00b5\u03bd(k), (56.15) 1 \u2212 k2 P i\u01eb \u03bb= \u00b1 \u03b5\u00b5 \u2217\u03bb (k)\u03b5\u03bd \u03bb(k). (56.16) The \ufb01rst term on the right-hand side of eq. (56.16) reproduces the Coulomb term in eq. (56.13) by virtue of the facts that + \u221e Z \u2212\u221e Z dk0 2\u03c0 e\u2212 ik0(x0 y0) = \u03b4(x0 \u2212 y0), \u2212 d3k (2\u03c0)3 eik y) (x \u00b7 \u2212 k2 = 4\u03c0 1 x | \u2212 y |. (56.17) (56.18) The second term on the right-hand side of eq. (56.16) reproduces the second term in eq. (56.13) by virtue of the fact that \u03b50 \u03bb(k) = 0. Next we will simplify eq. (56.16). We begin by introducing a unit vector in the time direction, \u02c6t\u00b5 = (1, 0). (56.19) Next we need a unit vector in the k direction, which we will call \u02c6z\u00b5. We \ufb01rst note that \u02c6t \u00b7 k0, and so we can write k = \u2212 (0, k) = k\u00b5 + (\u02c6t \u00b7 k)\u02c6t\u00b5. The square of this four-vector is k2 = k2 + (\u02c6t \u00b7 k)2, where we have used \u02c6t 2 = \u2212 1. Thus the unit vector that we want is \u02c6z\u00b5 = k)\u02c6t\u00b5 k\u00b5 + (\u02c6t \u00b7 [k2 + (\u02c6t k)2]1/2 \u00b7. Now we recall from section 55 that \u2217\u03bb (k)\u03b5j \u03b5i \u03bb(k) = \u03b4ij \u2212 kikj k2. X\u03bb= \u00b1 (56.20) (56.21) (56.22) (56.23) 56: LSZ Reduction for Photons 342 This can be extended to i \u2192 \u03b5\u00b5 \u2217\u03bb (k)\u03b5\u03bd \u00b5 and j \u03bd by writing \u2192 \u03bb(k) = g\u00b5\u03bd + \u02c6t\u00b5\u02c6t\u03bd \u02c6z\u00b5 \u02c6z\u03bd. \u2212 (56.24) X\u03bb= \u00b1 It is not hard to check that the right-hand side of eq. (56.24) vanishes if \u00b5 = 0 or \u03bd = 0, and agrees with eq. (56.23) for \u00b5 = i and \u03bd = j. Putting all this together, we can now write eq. (56.16) as \u02dc\u2206\u00b5\u03bd(k) = \u2212 \u02c6t\u00b5\u02c6t\u03bd k2 + (\u02c6t \u00b7 + k)2 g\u00b5\u03bd + \u02c6t\u00b5\u02c6t\u03bd k2 \u2212 \u02c6z\u00b5 \u02c6z\u03bd. \u2212 i\u01eb (56.25) The next step is to consider the terms in this expression that contain factors of k\u00b5 or k\u03bd ; from eq. (56.22), we see that these will arise from the \u02c6z\u00b5 \u02c6z\u03bd term. In eq. (56.15), a factor of k\u00b5 can be written as a derivative with respect to x\u00b5 acting on eik(x y). This derivative can then be integrated by parts in eq. (56.14) to give a factor of \u2202\u00b5J\u00b5(x). But \u2202\u00b5J\u00b5(x) vanishes, because the current must be conserved. Similarly, a factor of k\u03bd can be turned into \u2202\u03bdJ\u03bd(y), and also leads to a vanishing contribution. Therefore, we can ignore any terms in \u02dc\u2206\u00b5\u03bd (k) that contain factors of k\u00b5 or k\u03bd. \u2212 From eq. (56.22), we see that this means we can make the substitution \u02c6z\u00b5 \u2192 Then eq. (56.25) becomes k)\u02c6t\u00b5 (\u02c6t \u00b7 [k2 + (\u02c6t \u00b7 k)2]1/2. (56.26) \u02dc\u2206\u00b5\u03bd (", "k) = 1 g\u00b5\u03bd + k2 k2 + (\u02c6t \u00b7 + 1 \u2212 k)2 (\u02c6t k)2 \u00b7 k2 + (\u02c6t \u00b7 k2 i\u01eb \" \u2212 \u2212 # (56.27) where the three coe\ufb03cients of \u02c6t\u00b5\u02c6t\u03bd come from the Coulomb term, the \u02c6t\u00b5\u02c6t\u03bd term in the polarization sum, and the \u02c6z\u00b5 \u02c6z\u03bd term, respectively. A bit of algebra now reveals that the net coe\ufb03cient of \u02c6t\u00b5\u02c6t\u03bd vanishes, leaving us with the elegant expression k)2! \u02c6t\u00b5\u02c6t\u03bd, \u02dc\u2206\u00b5\u03bd(k) = g\u00b5\u03bd k2 \u2212. i\u01eb (56.28) Written in this way, the photon propagator is said to be in Feynman gauge. (It would still be in Coulomb gauge if we had retained the k\u00b5 and k\u03bd terms that we previously dropped.) In the next section, we will rederive eq. (56.28) from a more explicit path-integral point of view. 56.1) Use eqs. (55.11) and (55.21\u201355.23) to verify eqs. (56.9\u201356.10). Problems 57: The Path Integral for Photons 343 57 The Path Integral for Photons Prerequisite: 8, 56 In this section, in order to get a better understanding of the photon path integral, we will evaluate it directly, using the methods of section 8. We begin with Z0(J) = S0 = Z Z A eiS0, D d4x \u2212 h 1 4 F \u00b5\u03bd F\u00b5\u03bd + J \u00b5A\u00b5. i (57.1) (57.2) Following section 8, we Fourier-transform to momentum space, where we \ufb01nd S0 = d4k (2\u03c0)4 1 2 Z k2g\u00b5\u03bd \u2212 h A\u00b5(k) J \u00b5(k) e A\u00b5( + \u2212 k) \u2212 k\u00b5k\u03bd A\u03bd( J \u00b5( e k) \u2212 A\u00b5(k) i e A so as to \u201ccomplete the The next step is to shift the integration variable k\u00b5k\u03bd. However, 4 matrix k2g\u00b5\u03bd square\u201d. This involves inverting the 4 \u2212 this matrix has a zero eigenvalue, and cannot be inverted. (57.3) k) + \u2212 \u00d7 e e e e. To see this, let us write where we have de\ufb01ned k2g\u00b5\u03bd \u2212 k\u00b5k\u03bd = k2P \u00b5\u03bd (k), P \u00b5\u03bd (k) g\u00b5\u03bd \u2261 \u2212 k\u00b5k\u03bd/k2. This is a projection matrix because, as is easily checked, P \u00b5\u03bd (k)P\u03bd \u03bb(k) = P \u00b5\u03bb(k). (57.4) (57.5) (57.6) Thus the only allowed eigenvalues of P are zero and one. There is at least one zero eigenvalue, because P \u00b5\u03bd(k)k\u03bd = 0. On the other hand, the sum of the eigenvalues is given by the trace g\u00b5\u03bd P \u00b5\u03bd(k) = 3. Thus the remaining three eigenvalues must all be one. (57.7) (57.8) 57: The Path Integral for Photons 344 e Now let us imagine carrying out the path integral of eq. (57.1), with A\u00b5(k) into components S0 given by eq. (57.3). Let us decompose the \ufb01eld aligned along a set of linearly independent four-vectors, one of which is k\u00b5. (It will not matter whether or not this basis set is orthonormal.) Because A\u00b5 involves the matrix k2P \u00b5\u03bd (k), and P \u00b5\u03bd(k)k\u03bd = 0, the term quadratic in A\u00b5(k) that lies along k\u00b5 does not contribute to this the component of quadratic term. Furthermore, it does not contribute to the linear term e either, because \u2202\u00b5J\u00b5(x) = 0 implies k\u00b5 J\u00b5(k) = 0. Thus this component does not appear in the path integal at all! It then makes no sense to A to mean integration over only integrate over it. We therefore de\ufb01ne those components that are spanned by the remaining three basis vectors, and therefore satisfy k\u00b5 A\u00b5(k) = 0. This is", " equivalent to imposing Lorenz gauge, \u2202\u00b5A\u00b5(x) = 0. e D e R The matrix P \u00b5\u03bd(k) is simply the matrix that projects a four-vector into the subspace orthogonal to k\u00b5. Within the subspace, P \u00b5\u03bd(k) is equivalent to the identity matrix. Therefore, within the subspace, the inverse of k2P \u00b5\u03bd (k) is (1/k2)P \u00b5\u03bd (k). Employing the \u01eb trick to pick out vacuum boundary conditions replaces k2 with k2 i\u01eb. e We can now continue following the procedure of section 8, with the \u2212 result that Z0(J) = exp = exp i 2 i 2 \" Z Z d4k (2\u03c0)4 J\u00b5(k) e P \u00b5\u03bd (k) k2 i\u01eb \u2212 d4x d4y J\u00b5(x)\u2206\u00b5\u03bd (x J\u03bd( \u2212 k) # e y)J\u03bd (y) \u2212, where \u2206\u00b5\u03bd(x y) = \u2212 Z d4k (2\u03c0)4 eik(x \u2212 y) P \u00b5\u03bd (k) k2 i\u01eb \u2212 (57.9) (57.10) is the photon propagator in Lorenz gauge (also known as Landau gauge). Of course, because the current is conserved, the k\u00b5k\u03bd term in P \u00b5\u03bd (k) does not contribute, and so the result is equivalent to that of Feynman gauge, where P \u00b5\u03bd (k) is replaced by g\u00b5\u03bd. 58: Spinor Electrodynamics 345 58 Spinor Electrodynamics Prerequisite: 45, 57 In the section, we will study spinor electrodynamics: the theory of photons interacting with the electrons and positrons of a Dirac \ufb01eld. (We will use the term quantum electrodynamics to denote any theory of photons, irrespective of the kinds of particles with which they interact.) We construct spinor electrodynamics by taking the electromagnetic current j\u00b5(x) to be proportional to the Noether current corresponding to the U(1) symmetry of a Dirac \ufb01eld; see section 36. Speci\ufb01cally, j\u00b5(x) = e\u03a8(x)\u03b3\u00b5\u03a8(x). (58.1) 0.302822 is the charge of the electron in Heaviside-Lorentz Here e = \u2212 (We will rely on context to distinguish this e units, with \u00afh = c = 1. from the base of natural logarithms.) In these units, the \ufb01ne-structure constant is \u03b1 = e2/4\u03c0 = 1/137.036. With the normalization of eq. (58.1), Q = d3x j0(x) is the electric charge operator. R Of course, when we specify a number in quantum \ufb01eld theory, we must always have a renormalization scheme in mind; e = 0.302822 corresponds to a speci\ufb01c version of on-shell renormalization that we will explore in sections 62 and 63. The value of e is di\ufb00erent in other renormalization schemes, such as MS, as we will see in section 66. The complete lagrangian of our theory is thus \u2212 = L \u2212 1 4 F \u00b5\u03bdF\u00b5\u03bd + i\u03a8/\u2202\u03a8 \u2212 m\u03a8\u03a8 + e\u03a8\u03b3\u00b5\u03a8A\u00b5. (58.2) In this section, we will be concerned with tree-level processes only, and so we omit renormalizing Z factors. We have a problem, though. A Noether current is conserved only when the \ufb01elds obey the equations of motion, or, equivalently, only at points in \ufb01eld space where the action is stationary. On the other hand, in our development of photon path integrals in sections 56 and 57, we assumed that the current was always conserved. This issue is resolved by enlarging the de\ufb01nition of a gauge transformation to include a transformation on the Dirac \ufb01eld as well as the electromagnetic \ufb01eld. Speci\ufb01cally, we de\ufb01ne a gauge transformation to consist of A\u00b5(x) \u03a8(x) \u03a8(x) \u2192 \u2192 \u2192 A\u00b5(x) exp[ \u2212 \u2202\u00b5\u0393(x), \u2212 ie\u0393(x)]\u03a8(x), exp[+ie\u0393(x)]\u03a8(x). (58.3) (58.4) (58.5", ") 58: Spinor Electrodynamics 346 It is not hard to check that (x) is invariant under this transformation, whether or not the \ufb01elds obey their equations of motion. To perform this as check most easily, we \ufb01rst rewrite L L 4 F \u00b5\u03bdF\u00b5\u03bd + i\u03a8 /D\u03a8 1 = L \u2212 m\u03a8\u03a8, \u2212 (58.6) where we have de\ufb01ned the gauge covariant derivative (or just covariant derivative for short) ieA\u00b5. D\u00b5 \u2261 \u2202\u00b5 \u2212 In the last section, we found that F \u00b5\u03bd is invariant under eq. (58.3), and so the F F term in It is also obvious L that the m\u03a8\u03a8 term in is invariant under eqs. (58.4) and (58.5). This leaves the \u03a8 /D\u03a8 term. This term will also be invariant if, under the gauge transformation, the covariant derivative of \u03a8 transforms as is obviously invariant as well. (58.7) L D\u00b5\u03a8(x) exp[ \u2212 \u2192 ie\u0393(x)]D\u00b5\u03a8(x). (58.8) To see if this is true, we note that D\u00b5\u03a8 \u2192 \u2202\u00b5 \u2212 = exp[ ie[A\u00b5 \u2212 \u2202\u00b5\u03a8 ie\u0393] \u2202\u00b5\u0393] exp[ \u2212 ie(\u2202\u00b5\u0393)\u03a8 \u2212 \u2212 ie\u0393]\u03a8 ie[A\u00b5 \u2212 \u2212 \u2202\u00b5\u0393]\u03a8 = exp[ \u2212 = exp[ \u2212 ie\u0393] \u2202\u00b5 \u2212 ie\u0393]D\u00b5\u03a8. ieA\u00b5 \u03a8 (58.9) So eq. (58.8) holds, and \u03a8 /D\u03a8 is gauge invariant. We can also write the transformation rule for D\u00b5 a little more abstractly as D\u00b5 \u2192 e\u2212 ie\u0393D\u00b5 e+ie\u0393, (58.10) where the ordinary derivative in D\u00b5 is de\ufb01ned to act on anything to its right, including any \ufb01elds that are left unwritten in eq. (58.10). Thus we have D\u00b5\u03a8 \u2192 e\u2212 ie\u0393D\u00b5e+ie\u0393 e\u2212 ie\u0393\u03a8 = e\u2212 ie\u0393D\u00b5\u03a8, (58.11) which is, of course, the same as eq. (58.9). We can also express the \ufb01eld strength in terms of the covariant derivative by noting that [D\u00b5, D\u03bd]\u03a8(x) = ieF \u00b5\u03bd (x)\u03a8(x). \u2212 (58.12) 58: Spinor Electrodynamics We can write this more abstractly as F \u00b5\u03bd = i e [D\u00b5, D\u03bd], 347 (58.13) where, again, the ordinary derivative in each covariant derivative acts on anything to its right. From eqs. (58.10) and (58.13), we see that, under a gauge transformation, F \u00b5\u03bd ie\u0393D\u00b5 e+ie\u0393, e\u2212 ie\u0393D\u03bd e+ie\u0393 i e \u2192 e\u2212 h ie\u0393 = e\u2212 i e [D\u00b5, D\u03bd ] = e\u2212 ie\u0393F \u00b5\u03bd e+ie\u0393 i e+ie\u0393 = F \u00b5\u03bd. (58.14) ie\u0393 factors against each other In the last line, we are able to cancel the e\u00b1 because no derivatives act on them. Eq. (58.14) shows us that (as we already knew) F \u00b5\u03bd is gauge invariant. It is interesting to note that the gauge transformation on the fermion \ufb01elds, eqs. (58.4\u201358.5), is a generalization of the U(1) transformation i\u03b1\u03a8, e\u2212 e+i\u03b1\u03a8, \u03a8 \u03a8 \u2192 \u2192 (58.15) (58.16) that is a symmetry of the free Dirac lagrangian. The di\ufb00erence is that, in the gauge transformation, the phase factor is allowed to be a function of spacetime, rather than a constant that is the same everywhere. Thus, the gauge transformation is also called a local U(1) transformation, while eqs. (58.15\u201358.16) correspond to a global U(1) transformation. We say that, in a gauge theory, the global U(1) symmetry is promoted to a local U(1) symmetry, or that we have gauged the U(1", ") symmetry. In section 57, we argued that the path integral over A\u00b5 should be restricted to those components of A\u00b5(k) that are orthogonal to k\u00b5, because the component parallel to k\u00b5 did not appear in the integrand. Now we must make a slightly more subtle argument. We argue that the path integral over the parallel component is redundant, because the fermionic path integral over \u03a8 and \u03a8 already includes all possible values of \u0393(x). Therefore, as in section 57, we should not integrate over the parallel component. (We will make a more precise and careful version of this argument when we discuss the quantization of nonabelian gauge theories in section 71.) e 58: Spinor Electrodynamics 348 By the standard procedure, this leads us to the following form of the path integral for spinor electrodynamics: Z(\u03b7, \u03b7, J) exp ie \" \u221d d4x Z 1 i \u03b4 \u03b4J \u00b5(x) \u03b4 \u03b4\u03b7\u03b1(x) i (\u03b3\u00b5)\u03b1\u03b2 1 i \u03b4 \u03b4\u03b7\u03b2(x)!# Z0(\u03b7, \u03b7, J), \u00d7 (58.17) where and Z0(\u03b7, \u03b7, J) = exp i d4x d4y \u03b7(x)S(x \u2212 y)\u03b7(y) Z exp \u00d7 i 2 Z d4x d4y J \u00b5(x)\u2206\u00b5\u03bd (x y)J \u03bd(y) \u2212, (58.18) S(x \u2206\u00b5\u03bd(x y) = y) = \u2212 \u2212 Z Z d4p (2\u03c0)4 d4k (2\u03c0)4 /p + m) ( \u2212 p2 + m2 i\u01eb \u2212 eip(x y), \u2212 (58.19) g\u00b5\u03bd k2 i\u01eb \u2212 eik(x \u2212 y) (58.20) are the appropriate Feynman propagators for the corresponding free \ufb01elds, with the photon propagator in Feynman gauge. We impose the normalization Z(0, 0, 0) = 1, and write Z(\u03b7, \u03b7, J) = exp[iW (\u03b7, \u03b7, J)]. (58.21) Then iW (\u03b7, \u03b7, J) can be expressed as a series of connected Feynman diagrams with sources. The rules for internal and external Dirac fermions were worked out in the context of Yukawa theory in section 45, and they follow here with no change. For external photons, the LSZ analysis of section 56 implies that each external photon line carries a factor of the polarization vector \u03b5\u00b5(k). Putting everything together, we get the following set of Feynman rules for tree-level processes in spinor electrodynamics. 1. For each incoming electron, draw a solid line with an arrow pointed towards the vertex, and label it with the electron\u2019s four-momentum, pi. 2. For each outgoing electron, draw a solid line with an arrow pointed away from the vertex, and label it with the electron\u2019s four-momentum, p\u2032i. 58: Spinor Electrodynamics 349 3. For each incoming positron, draw a solid line with an arrow pointed away from the vertex, and label it with minus the positron\u2019s fourmomentum, pi. \u2212 4. For each outgoing positron, draw a solid line with an arrow pointed towards the vertex, and label it with minus the positron\u2019s four-momentum, p\u2032i. \u2212 5. For each incoming photon, draw a wavy line with an arrow pointed towards the vertex, and label it with the photon\u2019s four-momentum, ki. (Wavy lines for photons is a standard convention.) 6. For each outgoing photon, draw a wavy line with an arrow pointed away from the vertex, and label it with the photon\u2019s four-momentum, k\u2032i. 7. The only allowed vertex joins two solid lines, one with an arrow pointing towards it and one with an arrow pointing away from it, and one wavy line (whose arrow can point in either direction). Using this vertex, join up all the external lines, including extra internal lines as needed. In this way, draw all possible diagrams that are topologically inequivalent. 8. Assign each internal line its own four-momentum. Think of the fourmomenta as \ufb02owing along the arrows, and conserve four-momentum at each vertex. For a tree diagram, this \ufb01xes the momenta on all the internal lines. 9. The value of a diagram consists of the following factors", ": i for each incoming photon, \u03b5\u00b5 (ki); \u2217\u03bbi for each outgoing photon, \u03b5\u00b5 (k\u2032i); \u03bb\u2032 for each incoming electron, usi(pi); (p\u2032i); for each outgoing electron, us\u2032 for each incoming positron, vsi(pi); (p\u2032i); for each outgoing positron, vs\u2032 for each vertex, ie\u03b3\u00b5; i i for each internal photon, for each internal fermion, \u2212 ig\u00b5\u03bd /(k2 i\u01eb); /p + m)/(p2 + m2 \u2212 i( \u2212 \u2212 i\u01eb). \u2212 10. Spinor indices are contracted by starting at one end of a fermion line: speci\ufb01cally, the end that has the arrow pointing away from the vertex. The factor associated with the external line is either u or v. 58: Spinor Electrodynamics 350 Go along the complete fermion line, following the arrows backwards, and write down (in order from left to right) the factors associated with the vertices and propagators that you encounter. The last factor is either a u or v. Repeat this procedure for the other fermion lines, if any. The vector index on each vertex is contracted with the vector index on either the photon propagator (if the attached photon line is internal) or the photon polarization vector (if the attached photon line is external). 11. The overall sign of a tree diagram is determined by drawing all contributing diagrams in a standard form: all fermion lines horizontal, with their arrows pointing from left to right, and with the left endpoints labeled in the same \ufb01xed order (from top to bottom); if the ordering of the labels on the right endpoints of the fermion lines in a given diagram is an even (odd) permutation of an arbitrarily chosen \ufb01xed ordering, then the sign of that diagram is positive (negative). 12. The value of i (at tree level) is given by a sum over the values of all T the contributing diagrams. In the next section, we will do a sample calculation. Problems 58.1) Compute P \u2212 1A\u00b5(x, t)P, T \u2212 1A\u00b5(x, t)C, assuming that P, T, and C are symmetries of the lagrangian. (Prerequisite: 40.) 1A\u00b5(x, t)T, and C \u2212 58.2) Furry\u2019s theorem. Show that any scattering amplitude with no external fermions, and an odd number of external photons, is zero. 59: Scattering in Spinor Electrodynamics 351 59 Scattering in Spinor Electrodynamics Prerequisite: 48, 58 In the last section, we wrote down the Feynman rules for spinor electrodynamics. In this section, we will compute the scattering amplitude (and its spin-averaged square) at tree level for the process of electron-positron annihilation into a pair of photons, e+e\u2212 \u2192 expression for the scattering amplitude is The contributing diagrams are shown in \ufb01g. (59.1), and the associated \u03b3\u03b3. = e2 \u03b5\u00b5 1\u2032\u03b5\u03bd 2\u2032 v2 \u03b3\u03bd \u2212 T \u2212 where \u03b5\u00b5 1\u2032 is shorthand for \u03b5\u00b5 \u03bb\u2032 1 The Mandelstam variables are /p1 + /k\u20321 + m t + m2 \u03b3\u00b5 + \u03b3\u00b5 \u2212 /p1 + /k\u20322 + m u + m2 \u03b3\u03bd u1, \u2212 (59.1) (k\u20321), v2 is shorthand for vs2(p2), and so on. \u2212 s = (p1 + p2)2 = k\u20321)2 = (p1 \u2212 k\u20322)2 = (p1 \u2212 and they obey s + t + u = 2m2. u = t = \u2212 \u2212 (k\u20321 + k\u20322)2, k\u20322)2, (p2 \u2212 k\u20321)2, (p2 \u2212 \u2212 \u2212 \u2212 Following the procedure of section 46, we write eq. (59.1) as = \u03b5\u00b5 1\u2032\u03b5\u03bd 2\u2032 v2A\u00b5\u03bdu1, T where e2 A\u00b5\u03bd \u2261 We also have \u03b3\u03bd \u2212 /p1 + /k\u20321 + m t + m2 \u2212 \u03b3\u00b5 + \u03b3\u00b5 \u2212 /p1 + /k\u20322 + m u + m2 \u2212 \u03b3\u03bd. T Using /a/b... =... /b/a, we see from eq. (59.4) that T \u2217 = = \u03b5\u03c1 \u22171\u2032 \u03b5\u03c3 \u22172", "\u2032 u1 A\u03c1\u03c3 v2. Thus we have A\u03c1\u03c3 = A\u03c3\u03c1. 2 = \u03b5\u00b5 1\u2032\u03b5\u03bd 2\u2032 \u03b5\u03c1 \u22171\u2032 \u03b5\u03c3 \u22172\u2032 (v2A\u00b5\u03bd u1)(u1A\u03c3\u03c1v2). |T | (59.2) (59.3) (59.4) (59.5) (59.6) (59.7) Next, we will average over the initial electron and positron spins, using the technology of section 46; the result is 2 = 1 4 \u03b5\u00b5 1\u2032\u03b5\u03bd 2\u2032\u03b5\u03c1 \u22171\u2032 \u03b5\u03c3 1 4 s1,s2 X |T | A\u00b5\u03bd ( \u22172\u2032 Tr h \u2212 /p1+m)A\u03c3\u03c1( \u2212 /p2\u2212. m) i (59.8) 59: Scattering in Spinor Electrodynamics 352 1p 1k 1p k2 p 1 k 1 p 1 k 2 p2 k2 p2 1k Figure 59.1: Diagrams for e+e\u2212 \u2192 \u03b3\u03b3, corresponding to eq. (59.1). We would also like to sum over the \ufb01nal photon polarizations. From eq. (59.8), we see that we must evaluate \u03bb(k)\u03b5\u03c1 \u03b5\u00b5 \u2217\u03bb (k). X\u03bb= \u00b1 (59.9) We did this polarization sum in Coulomb gauge in section 56, with the result that \u03b5\u00b5 \u03bb(k)\u03b5\u03c1 \u2217\u03bb (k) = g\u00b5\u03c1 + \u02c6t\u00b5\u02c6t\u03c1 \u02c6z\u00b5 \u02c6z\u03c1, \u2212 (59.10) X\u03bb= \u00b1 where \u02c6t\u00b5 is a unit vector in the time direction, and \u02c6z\u00b5 is a unit vector in the k direction that can be expressed as \u02c6z\u00b5 = k)\u02c6t\u00b5 k\u00b5 + (\u02c6t \u00b7 [k2 + (\u02c6t k)2]1/2 \u00b7. (59.11) It is tempting to drop the k\u00b5 and k\u03c1 terms in eq. (59.10), on the grounds that the photons couple to a conserved current, and so these terms should not contribute. (We indeed used this argument to drop the analogous terms in the photon propagator.) This also follows from the notion that the scattering amplitude should be invariant under a gauge transformation, as represented by a transformation of the external polarization vectors of the form \u03b5\u00b5 \u03bb(k) \u03b5\u00b5 \u03bb(k) i\u02dc\u0393(k)k\u00b5. \u2212 Thus, if we write a scattering amplitude particular outgoing photon with four-momentum k\u00b5 as \u2192 T (59.12) for a process that includes a T or a particular incoming photon with four-momentum k\u00b5 as = \u03b5\u00b5 \u03bb(k) M\u00b5, = \u03b5\u00b5 \u2217\u03bb (k) M\u00b5, T (59.13) (59.14) 59: Scattering in Spinor Electrodynamics 353 then in either case we should have k\u00b5 M\u00b5 = 0. Eq. (59.15) is in fact valid; we will give a proof of it, based on the Ward identity for the electromagnetic current, in section 67. For now, we will take eq. (59.15) as given, and so drop the k\u00b5 and k\u03c1 terms in eq. (59.10). (59.15) This leaves us with \u03bb(k)\u03b5\u03c1 \u03b5\u00b5 \u2217\u03bb (k) \u2192 g\u00b5\u03c1 + \u02c6t\u00b5\u02c6t\u03c1 X\u03bb= \u00b1 (\u02c6t k)2 \u00b7 k2 + (\u02c6t \u00b7 k)2 \u2212 \u02c6t\u00b5\u02c6t\u03c1. (59.16) But, for an external photon, k2 = 0. Thus the second and third terms in eq. (59.16) cancel, leaving us with the beautifully simple substitution rule \u03bb(k)\u03b5\u03c1 \u03b5\u00b5 \u2217\u03bb (k) g\u00b5\u03c1. \u2192 X\u03bb= \u00b1 (59.17) Using eq. (59.17), we can sum 2 over the polarizations of the outgoing photons, in addition to averaging over the spins of the incoming fermions; the result is |T | 2 h|T | i \u2261 1 4 2 |T | 1,\u03bb\u2032 X\u03bb\u2032 s1,s2 2 X = 1 4 Tr = e4 A\u00b5\u03bd ( h \u2212 \u03a6tti h (m2 \u2212 /p1+m)A\u03bd", "\u00b5( t)2 + h (m2 m) /p2\u2212 i + \u03a6uti h t)(m2 \u2212 \u2212 \u03a6tui \u2212 + h (m2 u) \u03a6uui \u2212 u)2 (59.18), where \u03a6tti h \u03a6uui h \u03a6tui h \u03a6uti h = 1 4 Tr = 1 4 Tr = 1 4 Tr = 1 4 Tr \u03b3\u03bd( h \u03b3\u00b5( h \u03b3\u03bd( h \u03b3\u00b5( h /p1+/k\u20321+m)\u03b3\u00b5( \u2212 /p1+m)\u03b3\u00b5( \u2212 /p1+/k\u20321+m)\u03b3\u03bd( \u2212 \u2212 /p1+/k\u20322+m)\u03b3\u03bd ( \u2212 \u2212 /p1+/k\u20321+m)\u03b3\u00b5( \u2212 \u2212 /p1+m)\u03b3\u03bd ( \u2212 /p1+m)\u03b3\u03bd ( \u2212 /p1+/k\u20322+m)\u03b3\u00b5( \u2212 /p1+/k\u20322+m)\u03b3\u00b5( \u2212 /p1+/k\u20322+m)\u03b3\u03bd ( \u2212 /p1+m)\u03b3\u00b5( \u2212 /p1+/k\u20321+m)\u03b3\u03bd( \u2212 \u2212 m) i m) i m) i m),,,. /p2\u2212 /p2\u2212 /p2\u2212 /p2\u2212 i (59.19) and \u03a6tti Examinging \u03a6uui h h k\u20322, which is equivalent to t other by k\u20321 \u2194. Thus we need only compute and \u03a6uti h. \u03a6uti u to get t h, we see that they are transformed into each \u03a6tui h, and then take u. The same is true of \u03a6tui h \u2194 \u03a6tti h \u03a6uui h and and \u2194 59: Scattering in Spinor Electrodynamics 354 Now we can apply the gamma-matrix technology of section 47. In particular, we will need the d = 4 relations \u2212 \u03b3\u00b5\u03b3\u00b5 = 4, \u03b3\u00b5/a\u03b3\u00b5 = 2/a, \u03b3\u00b5/a/b\u03b3\u00b5 = 4(ab), \u03b3\u00b5/a/b/c\u03b3\u00b5 = 2/c/b/a, in addition to the trace formulae. We also need p1p2 = \u2212 k\u20321k\u20322 = \u2212 p1k\u20321 = p2k\u20322 = + 1 p1k\u20322 = p2k\u20321 = + 1 1 2 (s 1 2 s, 2 (t 2 (u 2m2), \u2212 m2), m2). \u2212 \u2212 (59.20) (59.21) 1 = p2 which follow from eq. (59.2) plus the mass-shell conditions p2 2 2 2 = 0. After a lengthy and tedious calculation, we \ufb01nd 1 = k\u2032 and k\u2032 2 = m2 \u2212 which then implies \u03a6tti h \u03a6tui h = 2[tu \u2212 = 2m2(s m2(3t + u) 4m2), \u2212 m4], \u2212 \u03a6uui h \u03a6uti h = 2[tu \u2212 = 2m2(s m2(3u + t) 4m2). \u2212 m4], \u2212 (59.22) (59.23) (59.24) (59.25) This completes our calculation. Other tree-level scattering processes in spinor electrodynamics pose no new calculational di\ufb03culties, and are left to the problems. In the high-energy limit, where the electron can be treated as massless, we can reduce our labor with the method of spinor helicity, which was introduced in section 50. We take this up in the next section. Problems 59.1) Compute 2 h|T | i for Compton scattering, e\u2212\u03b3 e\u2212\u03b3. You should \ufb01nd that your result is the same as that for e+e\u2212 \u2192 \u03b3\u03b3, but with s t, and an extra overall minus sign. This is an example of crossing \u2194 symmetry; there is an overall minus sign for each fermion that is moved from the initial to the \ufb01nal state. \u2192 59: Scattering in Spinor Electrodynamics 355 59.2) Compute 59.3) Compute 2 h|T | i 2 h|T | i for Bhabha scattering, e+e\u2212 \u2192 for M\u00f8ller scattering, e\u2212e\u2212 \u2192 e+e\u2212. \ufb01nd that your result is the same as that for e+e\u2212 \u2192", " u. This is another example of crossing symmetry. s \u2194 e\u2212e\u2212. You should e+e\u2212, but with 60: Spinor Helicity for Spinor Electrodynamics 356 60 Spinor Helicity for Spinor Electrodynamics Prerequisite: 50, 59 In section 50, we introduced a special notation for u and v spinors of de\ufb01nite helicity for massless electrons and positrons. This notation greatly simpli\ufb01es calculations in the high-energy limit (s, all much greater than m2)., and | u | | t | We de\ufb01ne the twistors p] | p | [p i \u2261 | \u2261 u \u2212 \u2261 (p) = v+(p), u+(p) = v \u2212 (p), u+(p) = v (p), \u2212 We then have p h | \u2261 u \u2212 (p) = v+(p). (60.1) p] = [k p], [k k h [] = 0,, i (60.2) where the twistor products [k p] and are antisymmetric, k p h i [p k], [k ph and related by complex conjugation, i\u2217 = [k p]. They can be expressed p k h explicitly in terms of the components of the massless four-momenta k and p. However, more useful are the relations i (60.3) and [q rp k] = Tr 1 2 (1 \u03b35)/k /p \u2212 p 2k \u00b7 (k + p)2 = = \u2212 \u2212 [s p] = Tr 1 2 (1 \u03b35)/p/q/r/s \u2212 q)(r = 2[(p s) r)(q \u00b7 \u00b7 \u2212 + i\u03b5\u00b5\u03bd\u03c1\u03c3p\u00b5q\u03bdr\u03c1s\u03c3]. (p \u00b7 \u00b7 (60.4) s) + (p s)(q r) \u00b7 \u00b7 (60.5) 60: Spinor Helicity for Spinor Electrodynamics Finally, for any massless four-momentum p we can write /p = p | [p i | + p p] h | |. \u2212 357 (60.6) We will quote other results from section 50 as we need them. To apply this formalism to spinor electrodynamics, we need to write photon polarization vectors in terms of twistors. The formulae we need are \u03b5\u00b5 +(k) = (k) = \u03b5\u00b5 \u2212 \u2212 \u2212 \u03b3\u00b5 q h | \u221a2 k] | q k h \u03b3\u00b5 [q k i | | \u221a2 [q k] i,, (60.7) (60.8) where q is an arbitrary massless reference momentum. We will verify eqs. (60.7) and (60.8) for a speci\ufb01c choice of k, and then rely on the Lorentz transformation properties of twistors to conclude that the result must hold in any frame (and therefore for any massless fourmomentum k). We will choose k\u00b5 = (\u03c9, \u03c9\u02c6z) = \u03c9(1, 0, 0, 1). Then, the most general form of \u03b5\u00b5 +(k) is \u03b5\u00b5 +(k) = ei\u03c6 1 \u221a2 (0, 1, \u2212 i, 0) + Ck\u00b5. (60.9) Here ei\u03c6 is an arbitrary phase factor, and C is an arbitrary complex number; the freedom to add a multiple of k comes from the underlying gauge invariance. To verify that eq. (60.7) reproduces eq. (60.9), we need the explicit form when the three-momentum is in the z direction. of the twistors Using results in section 50 we \ufb01nd k] and | k | i 0 1 0 0 k] = \u221a2\u03c9 \uf8eb | \uf8ec \uf8ec \uf8ed For any value of q, the twistor = \u221a2\u03c9 \uf8eb \uf8f6. \uf8f7 \uf8f7 \uf8f8 takes the form \uf8f6, \uf8f7 \uf8f7 \uf8f8 q h | = (0, 0, \u03b1, \u03b2), q h | (60.10) (60.11) where \u03b1 and \u03b2 are complex numbers. Plugging eqs. (60.10) and (60.11) into eq. (60.7), and using \u03b3\u00b5 = 0 \u00af\u03c3\u00b5 \u03c3\u00b5 0! (60.12) (60.13) (60.14) (60.15) (60.16) 60: Spinor Helicity for Spinor Electrod", "ynamics 358 (k) = [q k] along with another result from section 50, along with \u03c3\u00b5 = (I, ~\u03c3) and \u00af\u03c3\u00b5 = (I, eq. (60.9) with ei\u03c6 = 1 and C = eq. (60.8), because \u03b5\u00b5 \u2212 ~\u03c3), we \ufb01nd that we reproduce \u03b2/(\u221a2\u03b1\u03c9). There is now no need to check \u2212 [\u03b5\u00b5 q k +(k)]\u2217, as can be seen by using i\u2217 = h \u03b3\u00b5 q]. | | In spinor electodynamics, the vector index on a photon polarization vector is always contracted with the vector index on a gamma matrix. We can get a convenient formula for /\u03b5 (k) by using the Fierz identities k]\u2217 = | \u03b3\u00b5 | k h q h \u2212 \u2212 \u2212 \u2212 \u2212 We then have 1 2\u03b3\u00b5 q h 2\u03b3\u00b5[q 1 \u03b3\u00b5| | k \u03b3\u00b5| | \u00b1 k] = = i q k] h | [q k i | | | + + [k q i | k q] h | | |,. /\u03b5+(k;q) = q k] h | | + q | [k i | (k;q) = /\u03b5 \u2212 k | [q i | + k q] h | | \u221a2 q k h \u221a2 [q k] i,, where we have added the reference momentum as an explicit argument on the left-hand sides. Now we have all the tools we need for doing calculations. However, we can simplify things even further by making maximal use of crossing symmetry. \u2212 Note from eq. (60.1) that u p], while u+ (an outgoing electron) and v | (which is the factor associated with an incoming electron) and v+ (an outgoing positron) are both represented by the (an incoming positron) twistor \u2212 are both represented by [p. Thus the square-bracket twistors correspond | to outgoing fermions with positive helicity, and incoming fermions with negative helicity. Similarly, the angle-bracket twistors correspond to outgoing fermions with negative helicity, and incoming fermions with positive helicity. Let us adopt a convention in which all particles are assigned fourmomenta that are treated as outgoing. A particle that has an assigned fourmomentum p then has physical four-momentum \u01ebpp, where \u01ebp = sign(p0) = +1 if the particle is physically outgoing, and \u01ebp = sign(p0) = 1 if the particle is physically incoming. \u2212 Since the physical three-momentum of an incoming particle is opposite to its assigned three-momentum, a particle with negative helicity relative to its physical three-momentum has positive helicity relative to its assigned three-momentum. From now on, we will refer to the helicity of a particle relative to its assigned momentum. Thus a particle that we say has \u201cpositive helicity\u201d actually has negative physical helicity if it is incoming, and positive physical helicity if it is outgoing. 60: Spinor Helicity for Spinor Electrodynamics 359 p1 p2 p3 3 p4 p1 p p1 p2 p4 4 p3 p1 p Figure 60.1: Diagrams for fermion-fermion scattering, with all momenta treated as outgoing. With this convention, the square-bracket twistors represent p] and [p | | positive-helicity fermions, and the angle-bracket twistors reprep and p | h sent negative-helicity fermions. When \u01ebp = sign(p0) = 1, we analytically continue the twistors by replacing each \u03c91/2 in eq. (60.10) with i 1/2. Then \u03c9 | | all of our formulae for twistors and polarizations hold without change, with the exception of the rule for complex conjugation of a twistor product, which becomes \u2212 i | p k h \u2217 = \u01ebp\u01ebk[k p]. i (60.17) Now we are ready to calculate some amplitudes. Consider \ufb01rst the process of fermion-fermion scattering. The contributing tree-level diagrams are shown in \ufb01g. (60.1). p h \u03b3\u00b5 | The \ufb01rst thing to notice is that a diagram is zero if two external fermion lines that meet at a vertex have the same helicity. This is because (as shown in section 50) we get zero if we sandwich the product of an odd number of gamma matrices", " between two twistors of the same helicity. In particular, k] = 0. Thus, we will get a nonzero result we have | for the tree-level amplitude only if two of the helicities are positive, and two are negative. This means that, of the 24 = 16 possible combinations of helicities, only six give a nonzero tree-level amplitude:,, T+ T++ Ts1s2s3s4. FurT+ thermore, the last three of these are related to the \ufb01rst three by complex conjugation, so we only have three amplitudes to compute., where the notation is = 0 and [p +, and \u03b3\u00b5 | T\u2212\u2212 ++, k | T\u2212 T\u2212 +, \u2212\u2212 \u2212\u2212 ++ \u2212 \u2212 \u2212 \u2212 + + i T+ Let us begin with +. Only the \ufb01rst diagram of \ufb01g. (60.1) contributes, because the second has two postive-helicity lines meeting at a vertex. To evaluate the \ufb01rst diagram, we note that the two vertices contribute a factor of (ie)2 = e2, and the internal photon line contributes a factor of ig\u00b5\u03bd /s13, where we have de\ufb01ned the Mandelstam variable \u2212\u2212 \u2212 Following the charge arrows backwards on each fermion line, and dividing sij \u2261 \u2212 (pi + pj)2. (60.18) T+ \u2212 T++ e2 \u2212 \u2212\u2212 \u03b3\u00b5 3 | h T++ \u2212\u2212 = = 60: Spinor Helicity for Spinor Electrodynamics 360 by i to get T (rather than i ), we \ufb01nd T + = T+ \u2212\u2212 e2 \u03b3\u00b5 3 h | = +2e2 [1 4] \u2212 /s13 2 \u03b3\u00b5| 1] [4 | i | /s13, 2 3 i h (60.19) where Fierz identity to get the second line. is short for p3| h 3 | h, etc, and we have used yet another form of the The computation of is exactly analogous, except that now it is only the second diagram of \ufb01g. (60.1) that contributes. According to the Feynman rules, this diagram comes with a relative minus sign, and so we have T+ + \u2212 \u2212 Finally, we turn to. Now both diagrams contribute, and we have + = \u2212 \u2212 2e2 [1 3] 2 4 i h /s14. (60.20) 1] 2] \u03b3\u00b5| 4 | h | s13 \u2212 \u03b3\u00b5 4 | h 1] 2] \u03b3\u00b5| 3 | h | s14 2e2 [1 2] \u2212 3 4 h i 1 s13 + 1 s14 = +2e2 [1 2] 3 4 h s12 s13s14, (60.21) i where we used the Mandelstam relation s12 + s13 + s14 = 0 to get the last line. To get the cross section for a particular set of helicities, we must take the absolute squares of the amplitudes. These follow from eqs. (60.4), (60.17), and (60.18) : 1 2 2 = |h i| [1 2] | | 2 = \u01eb1\u01eb2s12 = s12| |. (60.22) We can then compute the spin-averaged cross section by summing the absolute squares of eqs. (60.19\u201360.21), multiplying by two to account for the processes in which all helicities are opposite (and which have amplitudes that are related by complex conjugation), and then dividing by four to average over the initial helicities. Making use of s34 = s12 and its permutations, we \ufb01nd 2 h|T | i = 2e4 s2 14 s2 13 + s2 13 s2 14 + s4 12 13s2 s2 14! = 2e4 s4 12 + s4 13 + s4 14 13s2 s2 14.! (60.23) For the processes of e\u2212e\u2212 \u2192 s13 = t, and s14 = u; for e+e\u2212 \u2192 s12 = u. e\u2212e\u2212 and e+e+ e+e+, we have s12 = s, e+e\u2212, we have s13 = s, s14 = t, and \u2192 60: Spinor Helicity for Spinor Electrodynamics 361 p1 p2 p1 k 3 k 3 k4 p1 p2 k 4 p1 k 4 k 3 Figure 60.2: Diagrams for fermion-photon scattering, with all momenta treated as outgoing. Now", " we turn to processes with two external fermions and two external photons, as shown in \ufb01g. (60.2). The \ufb01rst thing to notice is that a diagram is zero if the two external fermion lines have the same helicity. This is because the corresponding twistors sandwich an odd number of gamma matrices: one from each vertex, and one from the massless fermion propagator \u02dcS(p) = +\u03bb3\u03bb4 is related by /p/p2. Thus we need only compute \u03bb3\u03bb4 since T+ \u2212 T\u2212 \u2212 complex conjugation. Next we use eqs. (60.15\u201360.16) and (60.2\u201360.3) to get p] = 0, (k;p) | (k;p) = 0. /\u03b5 \u2212 /\u03b5 [p | \u2212 p /\u03b5+(k;p) | = 0, i /\u03b5+(k;p) = 0, | p h (60.24) (60.25) (60.26) (60.27) Thus we can get some amplitudes to vanish with appropriate choices of the reference momenta in the photon polarizations. So, let us consider \u03bb3\u03bb4 = T+ \u2212 e2 e2 \u2212 \u2212 /\u03b5\u03bb4 2 | h /\u03b5\u03bb3 2 | h (k4;q4)(/p1 + /k3)/\u03b5\u03bb3 (k3;q3)(/p1 + /k4)/\u03b5\u03bb4 1] /s13 (k3;q3) | 1] /s14. (k4;q4) | (60.28) If we take \u03bb3 = \u03bb4 =, then we can get both terms in eq. (60.28) to vanish by choosing q3 = q4 = p1, and using eq. (60.24). If we take \u03bb3 = \u03bb4 = +, then we can get both terms in eq. (60.28) to vanish by choosing q3 = q4 = p2, and using eq. (60.27). \u2212 Thus, we need only compute, we can \u2212\u2212 get the second term in eq. (60.28) to vanish by choosing q3 = p2, and using eq. (60.27). Then we have + and. For T+ T+ T+ + + \u2212 \u2212 \u2212 \u2212 + T+ \u2212 \u2212 = e2 \u2212 /\u03b5 2 | h \u2212 1] /s13 (k4;q4)(/p1 + /k3)/\u03b5+(k3;p2) | 60: Spinor Helicity for Spinor Electrodynamics 362 e2 \u221a2 [q44] h 2 4 i = \u2212 [q4| 2 (/p1 + /k3) i | [3 1] \u221a2 2 3 i h 1 s13. (60.29) Next we note that [p /p = 0, and so it is useful to choose either q4 = p1 | or q4 = k3. There is no obvious advantage in one choice over the other, and they must give equivalent results, so let us take q4 = k3. Then, using eq. (60.6) for /k3, we get + T+ \u2212 \u2212 = 2e2 h [3 1] [3 1] 2 4 i [3 4] 1 2 i h s13 2 3 i h (60.30) 1 2 Now we use [3 1] in the numerator (see problem 60.2), h i and set s13 = [3 1] in the denominator. Canceling common factors and 1 3 i h using antisymmetry of the twistor product then yields 4 2 i h [3 4] \u2212 = + T+ = 2e2 ih h + simply by exchanging the labels 3 and 4, \u2212 \u2212. (60.31) We can now get T+ \u2212\u2212 \u2212\u2212 T+ + = 2e2 ih h We can compute the spin-averaged cross section by summing the absolute squares of eqs. (60.31) and (60.32), multiplying by two to account for the processes in which all helicities are opposite (and which have amplitudes that are related by complex conjugation), and then dividing by four to average over the initial helicities. The result is (60.32). 2 h|T | i + = 2e4 s14 s13 s14 s13 e\u2212\u03b3 and e+\u03b3 e+\u03b3, we have s13 = s, s12 = t, e+e\u2212 we have s12 = s, s13 = t, and (60.33) \u2192. \u2192 \u03b3\u03b3 and \u03b3\u03b3 For the processes of e\u2212\u03b3", " and s14 = u; for e+e\u2212 \u2192 s14 = u. \u2192 Problems 60.1) a) Show that p \u00b7 \u03b5+(k;q) = h (k;q) = p \u03b5 \u00b7 \u2212 q p \u221a2 [p k] i q k i h [q p] p k i h \u221a2 [q k],. (60.34) (60.35) 60: Spinor Helicity for Spinor Electrodynamics 363 Use this result to show that which is required by gauge invariance, and also that (k;q) = 0, k \u03b5 \u00b7 \u00b1 (k;q) = 0. q \u03b5 \u00b7 \u00b1 b) Show that \u03b5+(k;q) (k;q) \u03b5 \u2212 \u03b5+(k;q) \u00b7 \u00b7 \u00b7 \u03b5 (k\u2032;q\u2032) = q q\u2032i [k k\u2032] \u03b5+(k\u2032;q\u2032) = h q k q\u2032 k\u2032i h i h k k\u2032i [q q\u2032] h [q k] [q\u2032 k\u2032] q k\u2032i (k\u2032;q\u2032) = h q k i h [k q\u2032] [q\u2032 k\u2032] \u03b5 \u2212 \u2212,,. (60.36) (60.37) (60.38) (60.39) (60.40) Note that the right-hand sides of eqs. (60.38) and (60.39) vanish if q\u2032 = q, and that the right-hand side of eq. (60.40) vanishes if q = k\u2032 or q\u2032 = k. 60.2) a) For a process with n external particles, and all momenta treated as outgoing, show that n i j Xj=1 h i [j k] = 0 and n Xj=1 [i j] j k h i = 0. (60.41) Hint: make use of eq. (60.6). b) For n = 4, show that [3 1] 1 2 i h = \u2212 [3 4]. 4 2 i h 60.3) Use various identities to show that eq. (60.31) can also be written as + T+ \u2212 \u2212 2e2 = \u2212 [1 3]2 [1 4] [2 4]. (60.42) 60.4) a) Show explicitly that you would get the same result as eq. (60.31) if you set q4 = p1 in eq. (60.29). b) Show explicitly that you would get the same result as eq. (60.31) if you set q4 = p2 in eq. (60.29). 61: Scalar Electrodynamics 364 61 Scalar Electrodynamics Prerequisite: 58 In this section, we will consider how charged spin-zero particles interact with photons. We begin with the lagrangian for a complex scalar \ufb01eld with a quartic interaction, \u2212 This lagrangian is obviously invariant under the global U(1) symmetry \u2212 \u2212 L = \u2202\u00b5\u03d5\u2020\u2202\u00b5\u03d5 m2\u03d5\u2020\u03d5 1 4 \u03bb(\u03d5\u2020\u03d5)2. (61.1) \u03d5(x) \u2192 e\u2212 i\u03b1\u03d5(x), \u03d5\u2020(x) e+i\u03b1\u03d5\u2020(x). \u2192 We would like to promote this global symmetry to a local symmetry, (61.2) \u03d5(x) \u03d5\u2020(x) \u2192 \u2192 exp[ \u2212 ie\u0393(x)]\u03d5(x), exp[+ie\u0393(x)]\u03d5\u2020(x). (61.3) (61.4) To do so, we must replace each ordinary derivative in eq. (61.1) with a covariant derivative where A\u00b5 transforms as D\u00b5 \u2261 \u2202\u00b5 \u2212 ieA\u00b5, A\u00b5(x) A\u00b5(x) \u2212 \u2192 \u2202\u00b5\u0393(x), which implies that D\u00b5 transforms as \u2212 Our complete lagrangian for scalar electrodynamics is then exp[ ie\u0393(x)]D\u00b5 exp[+ie\u0393(x)]. D\u00b5 \u2192 (61.5) (61.6) (61.7) (61.8) = (D\u00b5\u03d5)\u2020D\u00b5\u03d5 m2\u03d5\u2020\u03d5 1 4 \u03bb(\u03d5\u2020\u03d5)2 1 4 F \u00b5\u03bdF\u00b5\u03bd. L \u2212 \u2212 \u2212 \u2212 We have added the usual gauge-invariant kinetic term for the gauge \ufb01eld. The quartic interaction term has a dimensionless coe\ufb03", "cient, and so is necessary for renormalizability. For now, we omit the renormalizing Z factors. Of course, eq. (61.8) is invariant under a global U(1) transformation as well as a local U(1) transformation: we simply set \u0393(x) to a constant. Then we can \ufb01nd the conserved Noether current corresponding to this symmetry, following the procedure of section 22. In the case of spinor electrodynamics, this current is same as it is for a free Dirac \ufb01eld, j\u00b5 = \u03a8\u03b3\u00b5\u03a8. In the case of a complex scalar \ufb01eld, we \ufb01nd j\u00b5 = \u2212 i[\u03d5\u2020D\u00b5\u03d5 \u2212 (D\u00b5\u03d5)\u2020\u03d5]. (61.9) 61: Scalar Electrodynamics 365 k k Figure 61.1: The three vertices of scalar electrodynamics; the corresponding 2ie2g\u00b5\u03bd, and vertex factors are ie(k + k\u2032)\u00b5, i\u03bb. \u2212 \u2212 With a factor of e, this current should be identi\ufb01ed as the electromagnetic current. Because the covariant derivative appears in eq. (61.9), the electromagnetic current depends explicitly on the gauge \ufb01eld. We had not previously contemplated this possibility, but in scalar electrodynamics it arises naturally, and is essential for gauge invariance. It also poses no special problem in the quantum theory. We will make the same assumption that we did for spinor electrodynamics: namely, that the correct procedure is to omit integration over the component of \u02dcA\u00b5(k) that is parallel to k\u00b5, on the grounds that this integration is redundant. This leads to the same Feynman rules for internal and external photons as in section 58. The Feyman rules for internal and external scalars are the same as those of problem 10.2. We will call the spin-zero particle with electric charge +e a scalar electron or selectron (recall that our convention is that e is negative), and the spin-zero particle with electric charge e \u2212 a scalar positron or spositron. Scalar lines (traditionally drawn as dashed in scalar electrodynamics) carry a charge arrow whose direction must be preserved when lines are joined by vertices. To determine the kinds of vertices we have, we \ufb01rst write out the inter- action terms in the lagrangian of eq. (61.8): L1 = ieA\u00b5[(\u2202\u00b5\u03d5\u2020)\u03d5 \u2212 \u03d5\u2020\u2202\u00b5\u03d5] \u2212 e2A\u00b5A\u00b5\u03d5\u2020\u03d5 1 4 \u03bb(\u03d5\u2020\u03d5)2. \u2212 (61.10) This leads to the vertices shown in \ufb01g. (61.1). The vertex factors associated 2ie2g\u00b5\u03bd and i\u03bb. To get the vertex factor with the last two terms are is an incoming selectron state, then k for the \ufb01rst term, we note that if | = eikx and = 0; and if 0 is an outgoing selectron k \u03d5\u2020(x) k \u03d5(x) 0 k\u2032| h h h i | | i | | ik\u2032x and state, then = 0. Therefore, in free \ufb01eld k \u03d5(x) 0 = e\u2212 0 \u03d5\u2020(x) k\u2032| i | | h i | h theory, \u2212 \u2212 i k\u2032 h k (\u2202\u00b5\u03d5\u2020)\u03d5 | | i = \u2212 ik\u2032\u00b5e\u2212 i(k\u2032 k)x, \u2212 k\u2032 h k \u03d5\u2020\u2202\u00b5\u03d5 | | i = +ik\u00b5e\u2212 i(k\u2032 k)x. \u2212 (61.11) (61.12) 61: Scalar Electrodynamics 366 This implies that the vertex factor for the \ufb01rst term in eq. (61.10) is given by i(ie)[( (ik\u00b5)] = ie(k + k\u2032)\u00b5. ik\u2032\u00b5) Putting everything together, we get the following set of Feynman rules \u2212 \u2212 for tree-level processes in scalar electrodynamics. 1. For each incoming selectron, draw a dashed line with an arrow pointed towards the vertex, and label it with the selectron\u2019s four-momentum, ki. 2. For each outgoing selectron, draw a dashed line with an arrow pointed away from the vertex, and label it with the selectron\u2019s four-momentum, k\u2032i. 3", ". For each incoming spositron, draw a dashed line with an arrow pointed away from the vertex, and label it with minus the spositron\u2019s fourmomentum, ki. \u2212 4. For each outgoing spositron, draw a dashed line with an arrow pointed towards the vertex, and label it with minus the spositron\u2019s fourmomentum, k\u2032i. \u2212 5. For each incoming photon, draw a wavy line with an arrow pointed towards the vertex, and label it with the photon\u2019s four-momentum, ki. 6. For each outgoing photon, draw a wavy line with an arrow pointed away from the vertex, and label it with the photon\u2019s four-momentum, k\u2032i. 7. There are three allowed vertices, shown in \ufb01g. (61.1). Using these vertices, join up all the external lines, including extra internal lines as needed. In this way, draw all possible diagrams that are topologically inequivalent. 8. Assign each internal line its own four-momentum. Think of the fourmomenta as \ufb02owing along the arrows, and conserve four-momentum at each vertex. For a tree diagram, this \ufb01xes the momenta on all the internal lines. 9. The value of a diagram consists of the following factors: for each incoming photon, \u03b5\u00b5 \u2217\u03bbi for each outgoing photon, \u03b5\u00b5 \u03bbi for each incoming or outgoing selectron or spositron, 1; (ki); (ki); for each scalar-scalar-photon vertex, ie(k + k\u2032)\u00b5; 61: Scalar Electrodynamics 367 1k k 1 1k 2k k 2 2k k2 Figure 61.2: Diagrams for e+ k 1 k 2 1k 2k \u03b3\u03b3. e\u2212 \u2192 e \u2212 e for each scalar-scalar-photon-photon vertex, 2ie2g\u00b5\u03bd ; for each four-scalar vertex, for each internal photon, for each internal scalar, i\u03bb; \u2212 ig\u00b5\u03bd /(k2 \u2212 \u2212 i/(k2 + m2 \u2212 i\u01eb); i\u01eb). \u2212 10. The vector index on each vertex is contracted with the vector index on either the photon propagator (if the attached photon line is internal) or the photon polarization vector (if the attached photon line is external). 11. The value of i (at tree level) is given by a sum over the values of all T the contributing diagrams. Let us compute the scattering amplitude for a particular process, e\u2212 \u2192 e\u2212 denotes a selectron. We have the diagrams of \ufb01g. (61.2). The e e e+ \u03b3\u03b3, where amplitude is e = (ie)2 1 i (2k1\u2212 i T k\u20321)\u00b5\u03b5\u00b5 1\u2032(k1\u2212 m2 t \u2212 k2)\u03bd \u03b5\u03bd 2\u2032 k\u20321\u2212 + (1\u2032 2\u2032) \u2194 2ie2g\u00b5\u03bd \u03b5\u00b5 1\u2032\u03b5\u03bd 2\u2032, \u2212 (k1 \u2212 where t = simpli\ufb01ed by noting that k1 \u2212 we have k\u20321)2 and u = k\u20321 \u2212 \u2212 (61.13) (k1 \u2212 \u2212 k2 = k\u20322 \u2212 k\u20322)2. This expression can be \u03b5\u2032i = 0. Then 2k2, and that k\u2032i\u00b7 = e2 \u2212 T 4(k1 \u00b7 \u03b51\u2032)(k2 \u00b7 m2 t \u2212 \u03b52\u2032) + \u03b51\u2032) 4(k1 \u00b7 \u03b52\u2032)(k2 \u00b7 m2 u \u2212 + 2(\u03b51\u2032. \u03b52\u2032) \u00b7 (61.14) To get the polarization-summed cross section, we take the absolute square of eq. (61.14), and use the substitution rule \u03bb(k)\u03b5\u03c1 \u03b5\u00b5 \u2217\u03bb (k) g\u00b5\u03c1. \u2192 X\u03bb= \u00b1 (61.15) This is a straightforward calculation, which we leave to the problems. 61: Scalar Electrodynamics 368 Problems 61.1) Compute 2 h|T | e\u2212 \u2192 the Mandelstam variables. e e+ for e i 61.2) Compute for the process e\u2212\u03b3 2 h|T | i your result is the same as that for example of crossing symmetry. e \u03b3\u03b3, and express your answer in terms of e\u2212\u03b3. You should \ufb01nd that t, an \u03b3\u03b3, but with s \u2194 \u2192 e\u2212 \u2192 e e e+ e 62: Loop Corrections in Spinor Electrodynamics 369 62 Loop Corrections", " in Spinor Electrodynamics Prerequisite: 51, 59 In this section we will compute the one-loop corrections in spinor electrodynamics. First let us note that the general discussion of sections 18 and 29 leads us to expect that we will need to add to the free lagrangian L0 = i\u03a8/\u2202\u03a8 \u2212 m\u03a8\u03a8 \u2212 1 4 F \u00b5\u03bd F\u00b5\u03bd (62.1) all possible terms whose coe\ufb03cients have positive or zero mass dimension, and that respect the symmetries of the original lagrangian. These include Lorentz symmetry, the U(1) gauge symmetry, and the discrete symmetries of parity, time reversal, and charge conjugation. The mass dimensions of the \ufb01elds (in four spacetime dimensions) are [A\u00b5] = 1 and [\u03a8] = 3 2. Gauge invariance requires that A\u00b5 appear only in the form of a covariant derivative D\u00b5. (Recall that the \ufb01eld strength F \u00b5\u03bd can be expressed as the commutator of two covariant derivatives.) Thus, L0 that does not involve the \u03a8 the only possible term we could add to \ufb01eld, and that has mass dimension four or less, is \u03b5\u00b5\u03bd\u03c1\u03c3F \u00b5\u03bd F \u03c1\u03c3. This term, however, is odd under parity and time reversal. Similarly, there are no terms meeting all the requirements that involve \u03a8: the only candidates contain \u03a8) and either \u03b35 (e.g., i\u03a8\u03b35\u03a8) and are forbidden by parity, or are forbidden by the U(1) symmetry. (e.g, \u03a8T C Therefore, the theory we will consider is speci\ufb01ed by L0 is given by eq. (62.1), and = L0 + L C L1, where L1 = Z1e\u03a8 /A\u03a8 + 1)\u03a8/\u2202\u03a8 Lct = i(Z2\u2212 Lct, \u2212 We will use an on-shell renormalization scheme. \u2212 (Zm\u2212 1)m\u03a8\u03a8 (62.2) 1 4 (Z3\u2212 1)F \u00b5\u03bd F\u00b5\u03bd. (62.3) We can write the exact photon propagator (in momentum space) as a geometric series of the form \u02dc\u2206\u00b5\u03bd (k) = \u02dc\u2206\u00b5\u03bd(k) + \u02dc\u2206\u00b5\u03c1(k)\u03a0\u03c1\u03c3(k) \u02dc\u2206\u03c3\u03bd (k) +..., (62.4) where i\u03a0\u00b5\u03bd (k) is given by a sum of one-particle irreducible (1PI for short; see section 14) diagrams with two external photon lines (and the external propagators removed), and \u02dc\u2206\u00b5\u03bd(k) is the free photon propagator, \u02dc\u2206\u00b5\u03bd(k) = 1 \u2212 k2 g\u00b5\u03bd \u2212 i\u01eb \u03be) (1 \u2212 k\u00b5k\u03bd k2. (62.5) 62: Loop Corrections in Spinor Electrodynamics 370 Here we have used the freedom to add k\u00b5 or k\u03bd terms to put the propagator into generalized Feynman gauge or R\u03be gauge. (The name R\u03be gauge has historically been used only in the context of spontaneous symmetry breaking\u2014see section 85\u2014but we will use it here as well. R stands for renormalizable and \u03be stands for \u03be.) Setting \u03be = 1 gives Feynman gauge, and setting \u03be = 0 gives Lorenz gauge (also known as Landau gauge). Observable squared amplitudes should not depend on the value of \u03be. This suggests that \u03a0\u00b5\u03bd(k) should be transverse, k\u00b5\u03a0\u00b5\u03bd (k) = k\u03bd \u03a0\u00b5\u03bd(k) = 0, (62.6) so that the \u03be dependent term in \u02dc\u2206\u00b5\u03bd (k) vanishes when an internal photon line is attached to \u03a0\u00b5\u03bd (k). Eq. (62.6) is in fact valid; we will give a proof of it, based on the Ward identity for the electromagnetic current, in problem 68.1. For now, we will take eq. (62.6) as given. This implies that we can write \u03a0\u00b5\u03bd (k) = \u03a0(k2) = k2 \u03a0(k2)P \u00b5\u03bd (k), k2g\u00b5\u03bd \u2212 k\u00b5k\u03bd (62.7) (62.8) where \u03a0(k2) is a scalar function, and", " P \u00b5\u03bd(k) = g\u00b5\u03bd jection matrix introduced in section 57. \u2212 k\u00b5k\u03bd/k2 is the pro- Note that we can also write \u02dc\u2206\u00b5\u03bd(k) = 1 \u2212 i\u01eb k2 P\u00b5\u03bd (k) + \u03be k\u00b5k\u03bd k2. (62.9) Then, using eqs. (62.8) and (62.9) in eq. (62.4), and summing the geometric series, we \ufb01nd \u02dc\u2206\u00b5\u03bd(k) = P\u00b5\u03bd (k) \u03a0(k2)] \u2212 i\u01eb \u2212 + \u03be k\u00b5k\u03bd /k2 k2 i\u01eb \u2212. k2[1 (62.10) The \u03be dependent term should be physically irrelevant (and can be set to zero by the gauge choice \u03be = 0, corresponding to Lorenz gauge). The remaining term has a pole at k2 = 0 with residue P\u00b5\u03bd (k)/[1 \u03a0(0)]. In our on-shell renormalization scheme, we should have \u2212 \u03a0(0) = 0. (62.11) This corresponds to the \ufb01eld normalization that is needed for validity of the LSZ formula. 62: Loop Corrections in Spinor Electrodynamics 371 k+l l k k Figure 62.1: The one-loop and counterterm corrections to the photon propagator in spinor electrodynamics. Let us now turn to the calculation of \u03a0\u00b5\u03bd (k). The one-loop and coun- terterm contributions are shown in \ufb01g. (62.1). We have i\u03a0\u00b5\u03bd (k) = ( \u2212 1)(iZ1e)2 2 1 i Z 1)(k2g\u00b5\u03bd d4\u2113 \u02dcS(/\u2113+/k)\u03b3\u00b5 \u02dcS(/\u2113)\u03b3\u03bd (2\u03c0)4 Tr h k\u00b5k\u03bd ) + O(e4), i (62.12) i(Z3\u2212 \u2212 \u2212 where the factor of minus one is for the closed fermion loop, and \u02dcS(/p) = ( i\u01eb) is the free fermion propagator in momentum space. \u2212 Anticipating that Z1 = 1 + O(e2), we set Z1 = 1 in the \ufb01rst term. /p+m)/(p2+m2 \u2212 We can write Tr \u02dcS(/\u2113+/k)\u03b3\u00b5 \u02dcS(/\u2113)\u03b3\u03bd h 1 dx = 0 Z 4N \u00b5\u03bd (q2 + D)2, i where we have combined denominators in the usual way: q = \u2113 + xk and (62.13) D = x(1 \u2212 x)k2 + m2 i\u01eb. \u2212 The numerator is 4N \u00b5\u03bd = Tr h Completing the trace, we get /\u2113 ( \u2212 \u2212 /k+m)\u03b3\u00b5( /\u2113+m)\u03b3\u03bd \u2212 (62.14) (62.15) i N \u00b5\u03bd = (\u2113+k)\u00b5\u2113\u03bd + \u2113\u00b5(\u2113+k)\u03bd \u2212 [\u2113(\u2113+k) + m2]g\u00b5\u03bd. (62.16) Setting \u2113 = q grate to zero), we \ufb01nd \u2212 xk and and dropping terms linear in q (because they inte- N \u00b5\u03bd 2q\u00b5q\u03bd 2x(1 \u2212 \u2212 x)k\u00b5k\u03bd [q2 \u2212 x(1 \u2212 \u2212 \u2192 x)k2 + m2]g\u00b5\u03bd. (62.17) The integrals diverge, and so we analytically continue to d = 4 \u03b5 dimensions, and replace e with e\u02dc\u00b5\u03b5/2 (so that e remains dimensionless for any d). \u2212 62: Loop Corrections in Spinor Electrodynamics 372 Next we recall a result from problem 14.3, ddq q\u00b5q\u03bdf (q2) = 1 d Z g\u00b5\u03bd ddq q2f (q2). (62.18) Z This allows the replacement N \u00b5\u03bd 2x(1 x)k\u00b5k\u03bd + \u2192 \u2212 \u2212 2 d \u2212 1 q2 + x(1 x)k2 m2 \u2212 \u2212 Using the results of section 14, along with a little manipulation of gamma functions, we can show that h g\u00b5\u03bd. i (62.19) 2 d \u2212 1 Z ddq (2\u03c0)d q2 (q2 +", " D)2 = D Z ddq (2\u03c0)d 1 (q2 + D)2. (62.20) (62.21) Thus we can make the replacement ( 2 \ufb01nd N \u00b5\u03bd 2x(1 \u2212 \u2192 d \u2212 x)(k2g\u00b5\u03bd \u2192 k\u00b5k\u03bd ). \u2212 1)q2 D in eq. (62.19), and we This guarantees that the one-loop contribution to \u03a0\u00b5\u03bd (k) is transverse (as we expected) in any number of spacetime dimensions. Now we evaluate the integral over q, using ddq (2\u03c0)d 1 (q2 + D)2 = \u02dc\u00b5\u03b5 Z = \u03b5/2 i 4\u03c0 \u02dc\u00b52/D 16\u03c02 \u0393( \u03b5 2 ) 2 ln(D/\u00b52) 1 \u03b5 \u2212 i 8\u03c02 1, (62.22) where \u00b52 = 4\u03c0e\u2212 Combining eqs. (62.7), (62.12), (62.13), (62.21), and (62.22), we get \u03b3 \u02dc\u00b52, and we have dropped terms of order \u03b5 in the last line. \u03a0(k2) = e2 \u03c02 \u2212 1 dx x(1 \u2212 x) 0 Z 1 \u03b5 \u2212 Imposing \u03a0(0) = 0 \ufb01xes 1 2 ln(D/\u00b52) \u2212 1) + O(e4). (62.23) (Z3\u2212 Z3 = 1 e2 6\u03c02 1 \u03b5 \u2212 \u2212 ln(m/\u00b5) + O(e4) (62.24) and \u03a0(k2) = e2 2\u03c02 Z 1 0 dx x(1 \u2212 x) ln(D/m2) + O(e4). (62.25) Next we turn to the fermion propagator. The exact propagator can be written in Lehmann-K\u00a8all\u00b4en form as \u02dcS(/p) = 1 /p + m + \u221e m2 th Z ds i\u01eb \u2212 \u03c1\u03a8(s) /p + \u221as \u2212. i\u01eb (62.26) 62: Loop Corrections in Spinor Electrodynamics 373 We see that the \ufb01rst term has a pole at /p = m with residue one. This residue corresponds to the \ufb01eld normalization that is needed for the validity of the LSZ formula. \u2212 There is a problem, however: in quantum electrodynamics, the threshold mass mth is m, corresponding to the contribution of a fermion and a m. zero-energy photon. Thus the second term has a branch point at /p = The pole in the \ufb01rst term is therefore not isolated, and its residue is ill de\ufb01ned. \u2212 This is a re\ufb02ection of an underlying infrared divergence, associated with the massless photon. To deal with it, we must impose an infrared cuto\ufb00 that moves the branch point away from the pole. The most direct method is to change the denominator of the photon propagator from k2 to k2 + m2 \u03b3, where m\u03b3 is a \ufb01ctitious photon mass. Ultimately, as in section 26, we must deal with this issue by computing cross-sections that take into account detector ine\ufb03ciencies. In quantum electrodynamics, we must specify the lowest photon energy \u03c9min that can be detected. Only after computing cross sections with extra undectable photons, and then summing over them, is 0. It turns out that it is not also necessary it safe to take the limit m\u03b3 \u2192 to abandon the on-shell shell renormalization scheme (as we were forced to do in massless \u03d53 theory in section 27), as long as the electron is massive. An alternative is to use dimensional regularization for the infrared divergences as well as the ultraviolet ones. As discussed in section 25, there are no soft-particle infrared divergences for d > 4 (and no colinear divergences at all in quantum electrodynamics with massive charged particles). In practice, infrared-divergent integrals are \ufb01nite away from even-integer dimensions, just like ultraviolet-divergent integrals. Thus we simply keep d = 4 0 limit \u03b5 all the way through to the very end, taking the \u03b5 only after summing over cross sections with extra undetectable photons, all computed in 4 \u03b5 dimensions. This method is calculationally the simplest, but requires careful bookkeeping to segregate the infrared and ultraviolet singularities. For that reason, we will not pursue it further. We can write the", " exact fermion propagator in the form \u2192 \u2212 \u2212 \u02dcS(/p)\u2212 1 = /p + m i\u01eb \u2212 \u2212 \u03a3(/p), (62.27) where i\u03a3(/p) is given by the sum of 1PI diagrams with two external fermion lines (and the external propagators removed). The fact that \u02dcS(/p) has a pole at /p = m) = 0; this \ufb01xes the coe\ufb03cients Z2 and Zm. As we will see, we must have an infrared cuto\ufb00 in place in order to have a \ufb01nite value for \u03a3\u2032( m with residue one implies that \u03a3( m) = 0 and \u03a3\u2032( m). \u2212 \u2212 \u2212 Let us now turn to the calculation of \u03a3(/p). The one-loop and counter- \u2212 62: Loop Corrections in Spinor Electrodynamics 374 l p p+l p p p Figure 62.2: The one-loop and counterterm corrections to the fermion propagator in spinor electrodynamics. term contributions are shown in \ufb01g. (62.2). We have i\u03a3(/p) = (iZ1e)2 2 1 i i(Z2\u2212 It is simplest to work in Feynman gauge, where we take 1)/p \u2212 \u2212 Z d4\u2113 (2\u03c0)4 \u03b3\u03bd \u02dcS(/p + /\u2113)\u03b3\u00b5 \u02dc\u2206\u00b5\u03bd (\u2113) h i 1)m + O(e4). i(Zm\u2212 \u02dc\u2206\u00b5\u03bd(\u2113) = g\u00b5\u03bd \u21132 + m2 \u03b3 \u2212 ; i\u01eb (62.28) (62.29) here we have included the \ufb01ctitious photon mass m\u03b3 as an infrared cuto\ufb00. We now apply the usual bag of tricks to get i\u03a3(/p) = e2 \u02dc\u00b5\u03b5 1 dx ddq (2\u03c0)d N (q2 + D)2 Z 0 Z i(Z2\u2212 \u2212 where q = \u2113 + xp and 1)/p i(Zm\u2212 \u2212 1)m + O(e4), (62.30) D = x(1 \u2212 x)p2 + xm2 + (1 \u2212 x)m2 \u03b3, (62.31) N = \u03b3\u00b5( \u2212 /p = (d \u2212 \u2212 /\u2113 + m)\u03b3\u00b5 \u2212 2)(/p + /\u2113) dm \u2212 = (d 2)[/q + (1 x)/p] dm, (62.32) \u2212 where we have used (from section 47) \u03b3\u00b5\u03b3\u00b5 = 2)/p. The term linear in q integrates to zero, and then, using eq. (62.22), we get d and \u03b3\u00b5 /p\u03b3\u00b5 = (d \u2212 \u2212 \u2212 \u2212 \u2212 \u03a3(/p) = \u2212 \u2212 e2 8\u03c02 1 dx 0 Z 1)/p (Z2\u2212 (Zm\u2212 \u2212 (2 \u03b5)(1 \u2212 x)/p + (4 \u2212 \u03b5)m \u2212 1)m + O(e4). 1 \u03b5 \u2212 1 2 ln(D/\u00b52) (62.33) 62: Loop Corrections in Spinor Electrodynamics 375 l p p+l p +l p Figure 62.3: The one-loop correction to the photon-fermion-fermion vertex in spinor electrodynamics. We see that \ufb01niteness of \u03a3(/p) requires (62.34) (62.35) + O(e4), (62.36) Z2 = 1 Zm = 1 e2 8\u03c02 e2 2\u03c02 1 \u03b5 1 \u03b5 \u2212 \u2212 + \ufb01nite + O(e4), + \ufb01nite + O(e4). We can impose \u03a3( m) = 0 by writing \u2212 1 e2 8\u03c02 \u03a3(/p) = dx (1 x)/p + 2m 0 Z where D0 is D evaluated at p2 = \u2212 m2, ln(D/D0) + \u03ba2(/p + m) \u2212 D0 = x2m2 + (1 \u2212 x)m2 \u03b3, (62.37) and \u03ba2 is a constant to be determined. We \ufb01x \u03ba2 by imposing \u03a3\u2032( m) = 0. In di\ufb00erentiating with respect to /p, we take the p2 in D", ", eq. (62.31), to be \u2212 /p 2; we \ufb01nd \u2212 \u03ba2 = 1 2 \u2212 0 Z dx x(1 \u2212 x2)m2/D0 = \u2212 2 ln(m/m\u03b3) + 1, (62.38) where we have dropped terms that go to zero with the infrared cuto\ufb00 m\u03b3. Next we turn to the loop correction to the vertex. We de\ufb01ne the vertex function iV\u00b5(p\u2032, p) as the sum of one-particle irreducible diagrams with one incoming fermion with momentum p, one outgoing fermion with momentum p\u2032, and one incoming photon with momentum k = p\u2032\u2212 p. The original vertex iZ1e\u03b3\u00b5 is the \ufb01rst term in this sum, and the diagram of \ufb01g. (62.3) is the second. Thus we have iV\u00b5(p\u2032, p) = iZ1e\u03b3\u00b5 + iV\u00b5 1 loop(p\u2032, p) + O(e5), (62.39) 62: Loop Corrections in Spinor Electrodynamics 376 where iV\u00b5 1 loop(p\u2032, p) = (ie)3 1 i 3 d4\u2113 (2\u03c0)4 Z \u03b3\u03c1 \u02dcS(/p \u2032+/\u2113)\u03b3\u00b5 \u02dcS(/p+/\u2113)\u03b3\u03bd h \u02dc\u2206\u03bd\u03c1(\u2113). i (62.40) We again use eq. (62.29) for the photon propagator, and combine denominators in the usual way. We then get iV\u00b5 1 loop(p\u2032, p) = e3 d4q (2\u03c0)4 N \u00b5 (q2 + D)3, dF3 Z Z (62.41) where the integral over Feynman parameters is dF3 \u2261 2 Z 1 0 Z dx1dx2dx3 \u03b4(x1+x2+x3\u2212 1), (62.42) and q = \u2113 + x1p + x2p\u2032, D = x1(1 x1)p2 + x2(1 2 x2)p\u2032 + (x1+x2)m2 + x3m2 \u03b3, \u2212 \u2212 2x1x2p p\u2032 \u00b7 \u2212 (62.43) (62.44) N \u00b5 = \u03b3\u03bd( /p \u2032 /\u2113 + m)\u03b3\u00b5( /p /\u2113 + m)\u03b3\u03bd = \u03b3\u03bd[ \u2212 \u2212 /q + x1 /p \u2212 \u2212 x2)/p \u2032 + m]\u03b3\u00b5[ (1 \u2212 = \u03b3\u03bd/q\u03b3\u00b5/q\u03b3\u03bd + \u2212 \u2212 N \u00b5 + (linear in q), \u2212 /q (1 \u2212 \u2212 x1)/p + x2 /p \u2032 + m]\u03b3\u03bd (62.45) where e N \u00b5 = \u03b3\u03bd[x1 /p (1 x2)/p \u2032 + m]\u03b3\u00b5[ (1 x1)/p + x2 /p \u2032 + m]\u03b3\u03bd. e \u2212 \u2212 \u2212 The terms linear in q in eq. (62.45) integrate to zero, and only the \ufb01rst term is divergent. After continuing to d dimensions, we can use eq. (62.18) to make the replacement \u2212 (62.46) \u03b3\u03bd/q\u03b3\u00b5/q\u03b3\u03bd q2 \u03b3\u03bd\u03b3\u03c1\u03b3\u00b5\u03b3\u03c1\u03b3\u03bd. 1 d 2)\u03b3\u00b5 twice to get \u2192 Then we use \u03b3\u03c1\u03b3\u00b5\u03b3\u03c1 = (d \u2212 \u03b3\u03bd/q\u03b3\u00b5/q\u03b3\u03bd (d 2)2 \u2212 d \u2192 q2 \u03b3\u00b5. Performing the usual manipulations, we \ufb01nd V\u00b5 1 loop(p\u2032, p) = e3 8\u03c02 \" 1 \u03b5 \u2212 1 1 2 \u2212 Z dF3 ln(D/\u00b52) \u03b3\u00b5 + 1 4 dF3 Z (62.47) (62.48). N \u00b5 D # e (62.49) 62: Loop Corrections in Spinor Electrodynamics 377 From eq. (62.39), we see that \ufb01niteness of V\u00b5(p\u2032, p) requires Z1 = 1 e2 8\u03c02 1 \u03b5 \u2212 + \ufb01nite + O(e4). (62.50) To completely \ufb01x V\u00b5(p\u2032, p), we need a suitable condition to impose on it. We take this up in the", " next section. Problems 62.1) Show that adding a gauge \ufb01xing term results in eq. (62.9) as the photon propagator. Explain why \u03be = 0 corresponds to Lorenz gauge, \u2202\u00b5A\u00b5 = 0. 1(\u2202\u00b5A\u00b5)2 to 1 2 \u03be\u2212 \u2212 L 62.2) Find the coe\ufb03cients of e2/\u03b5 in Z1,2,3,m in R\u03be gauge. In particular, show that Z1 = Z2 = 1 + O(e4) in Lorenz gauge. 62.3) Consider the six one-loop diagrams with four external photons (and no external fermions). Show that, even though each diagram is logarithmically divergent, their sum is \ufb01nite. Use gauge invariance to explain why this must be the case. 63: The Vertex Function in Spinor Electrodynamics 378 63 The Vertex Function in Spinor Electrodynamics Prerequisite: 62 In the last section, we computed the one-loop contribution to the vertex function V\u00b5(p\u2032, p) in spinor electrodynamics, where p is the four-momentum of an incoming electron (or outgoing positron), and p\u2032 is the four-momentum of an outgoing electron (or incoming positron). We left open the issue of the renormalization condition we wish to impose on V\u00b5(p\u2032, p). For the theories we have studied previously, we have usually made the mathematically convenient (but physically obscure) choice to de\ufb01ne the coupling constant as the value of the vertex function when all external fourmomenta are set to zero. However, in the case of spinor electrodynamics, the masslessness of the photon gives us the opportunity to do something more physically meaningful: we can de\ufb01ne the coupling constant as the 2 = value of the vertex function when all three particles are on shell: p2 = p\u2032 p is the photon four-momentum. Because \u2212 the photon is massless, these three on-shell conditions are compatible with momentum conservation. m2, and q2 = 0, where q p\u2032 \u2212 \u2261 To be more precise, let us sandwich V\u00b5(p\u2032, p) between the spinor factors that are appropriate for an incoming electron with momentum p and an outgoing electron with momentum p\u2032, impose the on-shell conditions, and de\ufb01ne the electron charge e via us\u2032(p\u2032)V\u00b5(p\u2032, p)us(p) p2=p\u20322=\u2212m2 (p\u2032\u2212p)2=0 = e us\u2032(p\u2032)\u03b3\u00b5us(p). (63.1) p2=p\u20322=\u2212m2 (p\u2032\u2212p)2=0 This de\ufb01nition is in accord with the usual one provided by Coulomb\u2019s law. To see why, consider the process of electron-electron scattering. According to the general discussion in section 19, we compute the exact amplitude for this process by using tree diagrams with exact internal propagators and vertices, as shown in \ufb01g. (63.1). In the last section, we renormalized the photon propagator so that it approaches its tree-level value \u02dc\u2206\u00b5\u03bd(q) when q2 0. And we have just chosen to renormalize the electron-photon vertex function by requiring it to approach its tree-level value e\u03b3\u00b5 when q2 0, and when sandwiched between external spinors for on-shell incoming and outgoing electrons. Therefore, as q2 0, the \ufb01rst two diagrams in \ufb01g. (63.1) approach the tree-level scattering amplitude, with the electron charge equal to e. Furthermore, the third diagram does not have a pole at q2 = 0, and so can be neglected in this limit. Physically, q2 0 means that the electron\u2019s momentum changes very little during the scattering. Measuring a slight de\ufb02ection in the trajectory of one charged particle (due \u2192 \u2192 \u2192 \u2192 63: The Vertex Function in Spinor Electrodynamics 379 p1 p2 p1 p1 p2 p1 p1 p2 p2 p1 p2 p1 p1 p2 p1 p2 Figure 63.1: Diagrams for the exact electron-electron scattering amplitude. The vertices and photon propagator are exact; external lines stand for the m2. usual u and u spinor factors, times the unit residue of the pole at p2 = \u2212 to the presence of another) is how we measure the coe\ufb03cient in Coulomb\u2019s law. Thus, eq. (", "63.1) corresponds to this traditional de\ufb01nition of the charge of the electron. We can simplify eq. (63.1) by noting that the on-shell conditions actually enforce p\u2032 = p. So we can rewrite eq. (63.1) as us(p)V\u00b5(p, p)us(p) = e us(p)\u03b3\u00b5us(p) = 2ep\u00b5, (63.2) where p2 = right-hand side vanishes (and hence does not specify a value for e). m2 is implicit. We have taken s\u2032 = s, because otherwise the \u2212 Now we can use eq. (63.2) to completely determine V\u00b5(p\u2032, p). Using the freedom to choose the \ufb01nite part of Z1, we write V\u00b5(p\u2032, p) = e\u03b3\u00b5 e3 16\u03c02 \u2212 dF3 \" Z ln(D/D0)+2\u03ba1 \u03b3\u00b5 \u2212 N \u00b5 2D # +O(e5), (63.3) where D = x1(1 x1)p2 + x2(1 + (x1+x2)m2 + x3m2 2 x2)p\u2032 \u03b3, \u2212 \u2212 2x1x2p p\u2032 \u00b7 \u2212 D0 is D evaluated at p\u2032 = p and p2 = m2, \u2212 D0 = (x1+x2)2m2 + x3m2 \u03b3 x3)2m2 + x3m2 \u03b3, = (1 \u2212 (63.4) (63.5) and N \u00b5 = \u03b3\u03bd[x1 /p (1 \u2212 \u2212 x2)/p \u2032 + m]\u03b3\u00b5[ \u2212 (1 \u2212 x1)/p + x2 /p \u2032m]\u03b3\u03bd ; (63.6) 63: The Vertex Function in Spinor Electrodynamics 380 N \u00b5 was called convenience. N \u00b5 in section 62, but we have dropped the tilde for notational We \ufb01x the constant \u03ba1 in eq. (63.3) by imposing eq. (63.2). This yields e 4\u03ba1p\u00b5 = dF3 Z us(p)N \u00b5 2D0 0 us(p), (63.7) where N \u00b5 0 is N \u00b5 with p\u2032 = p and p2 = p\u2032 2 = m2. So now we must evaluate uN \u00b5 \u2212 0 u. To do so, we \ufb01rst write N \u00b5 = \u03b3\u03bd(/a1+m)\u03b3\u00b5(/a2+m)\u03b3\u03bd, (63.8) where a1 = x1p a2 = x2p\u2032 (1 \u2212 \u2212 x2)p\u2032, (1 \u2212 \u2212 x1)p. (63.9) Now we use the gamma matrix contraction identities to get N \u00b5 = 2/a2\u03b3\u00b5/a1 + 4m(a1+a2)\u00b5 + 2m2\u03b3\u00b5. (63.10) Here we have set d = 4, because we have already removed the divergence and taken the limit \u03b5 mu and mu, along with u\u03b3\u00b5u = 2p\u00b5 and uu = 2m, and recalling that u/p = x1+x2+x3 = 1, we \ufb01nd 0. Setting p\u2032 = p, and using /pu = \u2192 \u2212 \u2212 uN \u00b5 0 u = 4(1 \u2212 4x3+x2 3)m2p\u00b5. (63.11) Using eqs. (63.5), (63.7), and (63.11), we get 1 2 Z 1 0 Z 4x3+x2 1 3 \u2212 x3)2 + x3m2 \u03b3/m2 dF3 (1 \u2212 dx3 (1 \u2212 x3) (1 \u2212 4x3+x2 1 3 \u2212 \u03b3/m2 x3)2 + x3m2 \u03ba1 = = = 2 ln(m/m\u03b3) + 5 2 \u2212 0. We see that an infrared regulator is necessary for (63.12) in the limit of m\u03b3 \u2192 the vertex function as well as the fermion propagator. Now that we have V\u00b5(p\u2032, p), we can extract some physics from it. Consider again the process of electron-electron scattering, shown in \ufb01g. (63.1). In order to compute the contributions of these diagrams, we must evaluate us\u2032(p\u2032)V\u00b5(p\u2032, p)us(p) with p2 = p)2 arbitrary. m2, but with q2 = (p\u2032 \u2212", " 2 = \u2212 \u2212 p\u2032 63: The Vertex Function in Spinor Electrodynamics 381 To evaluate u \u2032N \u00b5u, we start with eq. (63.10), and use the anticommutation relations of the gamma matrices to move all the /p\u2019s in N \u00b5 to the far right (where we can use /pu = mu) and all the /p \u2032\u2019s to the far left (where we can use u \u2032 /p \u2032 = \u2212 mu \u2032). This results in \u2212 N \u00b5 \u2192 x1\u2212 [4(1 \u2212 + 4m(x2 1\u2212 p\u2032 + 2(2x1\u2212 x2+x1x2)p \u00b7 x2+x1x2)p\u00b5 + 4m(x2 2\u2212 x2 1+2x2\u2212 x1+x1x2)p\u2032 x2 2)m2]\u03b3\u00b5 \u00b5. (63.13) \u00b5 terms into p\u2032 + p p combinations, and make use of x1+x2+x3 = 1 to simplify some m2, group the p\u00b5 and p\u2032 1 2 q2 \u2212 \u2212 p\u2032 with Next, replace p and p\u2032 \u2212 coe\ufb03cients. The result is \u00b7 (x3+x1x2)q2]\u03b3\u00b5 N \u00b5 \u2192 2[(1 3)m2 x2 2x3\u2212 \u2212 \u2212 3)(p\u2032 + p)\u00b5 x2 2m(x3\u2212 2m[(x1+x2 1) \u2212 \u2212 (x2+x2 2)](p\u2032 p)\u00b5. \u2212 (63.14) \u2212 m2 and p p\u2032 = 1 2 q2 m2 to get \u2212 (63.15) \u2212 x1x2q2 + (1 \u2212 \u00b7 x3)2m2 + x3m2 \u03b3. \u2212 D \u2192 In the denominator, set p2 = p\u2032 2 = Note that the right-hand side of eq. (63.15) is symmetric under x1 \u2194 x2. Thus the last line of eq. (63.14) will vanish when we integrate u \u2032N \u00b5u/D over the Feynman parameters. Finally, we use the Gordon identity from section 38, u \u2032(p\u2032 + p)\u00b5u = u \u2032[2m\u03b3\u00b5 + 2iS\u00b5\u03bdq\u03bd]u, (63.16) where S\u00b5\u03bd = i 4 [\u03b3\u00b5, \u03b3\u03bd ], to get N \u00b5 \u2192 2[(1 4x3+x2 3)m2 \u2212 \u2212 3)S\u00b5\u03bdq\u03bd. x2 4im(x3\u2212 (x3+x1x2)q2]\u03b3\u00b5 So now we have \u2212 (63.17) us\u2032(p\u2032)V\u00b5(p\u2032, p)us(p) = eu \u2032 F1(q2)\u03b3\u00b5 h i m F2(q2)S\u00b5\u03bdq\u03bd u, (63.18) i \u2212 where we have de\ufb01ned the form factors F1(q2) = 1 e2 16\u03c02 \u2212 + Z \" ln 1 + dF3 x1x2q2/m2 (1 (x3+x1x2)q2/m2 x1x2q2/m2 + (1 x3)2! (1 \u2212 \u2212 x3)2 + x3m2 (1 4x3+x2 3) \u03b3/m2 # \u2212 + \u2212 \u2212 4x3+x2 1 3 \u2212 x3)2 + x3m2 \u03b3/m2 + O(e4), (63.19) F2(q2) = e2 8\u03c02 Z dF3 x3\u2212 x1x2q2/m2 + (1 x2 3 \u2212 x3)2 + O(e4). (63.20) 63: The Vertex Function in Spinor Electrodynamics 382 We have set m\u03b3 = 0 in eq. (63.20), and in the logarithm term in eq. (63.19), because these terms do not su\ufb00er from infrared divergences. We can simplify F2(q2) by using the delta function in dF3 to do the x1), making the change of x3\u2212 x3)y, and performing the integral over x3 from zero to integral over x2 (which replaces x2 with 1 variable x1 = (1 one; the result is \u2212 \u2212 F2(q2) = e2 8\u03c02 1 0 Z 1 \u2212 dy y)q2/m2 + O(", "e4). y(1 \u2212 (63.21) This last integral can also be done in closed form, but we will be mostly interested in its value at q2 = 0, corresponding to an on-shell photon: F2(0) = \u03b1 2\u03c0 + O(\u03b12), (63.22) where \u03b1 = e2/4\u03c0 = 1/137.036 is the \ufb01ne-structure constant. We will explore the physical consequences of eq. (63.22) in the next section. Problems 63.1) The most general possible form of u \u2032V \u00b5(p\u2032, p)u is a linear combination \u00b5 sandwiched between u \u2032 and u, with coe\ufb03cients that of \u03b3\u00b5, p\u00b5, and p\u2032 depend on q2. (The only other possibility is to include terms with \u03b35, but \u03b35 does not appear in the tree-level propagators or vertex, and so it cannot be generated in any Feynman diagram; this is a consequence of parity conservation.) Thus we can write us\u2032(p\u2032)V\u00b5(p\u2032, p)us(p) = eu \u2032[A(q2)\u03b3\u00b5 + B(q2)(p\u2032 + p)\u00b5 + C(q2)(p\u2032 \u2212 p)\u00b5]u. (63.23) a) Use gauge invariance to show that q\u00b5u \u2032V\u00b5(p\u2032, p)u = 0, and determine the consequences for A, B, and C. b) Express F1 and F2 in terms of A, B, and C. 64: The Magnetic Moment of the Electron 383 64 The Magnetic Moment of the Electron Prerequisite: 63 In the last section, we computed the one-loop contribution to the vertex function V\u00b5(p\u2032, p) in spinor electrodynamics, where p is the four-momentum of an incoming electron, and p\u2032 is the four-momentum of an outgoing electron. We found us\u2032(p\u2032)V\u00b5(p\u2032, p)us(p) = eu \u2032 F1(q2)\u03b3\u00b5 h \u2212 i m F2(q2)S\u00b5\u03bdq\u03bd u, (64.1) i where q = p\u2032 \u2212 p is the four-momentum of the photon (treated as incoming), and with complicated expressions for the form factors F1(q2) and F2(q2). For our purposes in this section, all we will need to know is that F1(0) = 1 exactly, F2(0) = \u03b1 2\u03c0 + O(\u03b12). (64.2) Eq. (64.1) follows from a quantum action of the form \u0393 = d4x Z eF1(0)\u03a8 /A\u03a8 + e 2m h F2(0)F\u00b5\u03bd \u03a8S\u00b5\u03bd \u03a8 +..., (64.3) i where the ellipses stand for terms with more derivatives. The displayed terms yield the vertex factor of eq. (64.1) with q2 = 0. To see this, recall \u03b5\u2217\u00b5eiqx, and therethat an incoming photon translates into a factor of A\u00b5 \u223c q\u03bd\u03b5\u2217\u00b5)eiqx; the two terms in F\u00b5\u03bd cancel the extra factor fore of F\u00b5\u03bd \u223c of one half in the second term in eq. (64.3). i(q\u00b5\u03b5\u2217\u03bd\u2212 Now we will see what eq. (64.3) predicts for the magnetic moment of the electron. We de\ufb01ne the magnetic moment by the following procedure. We take the photon \ufb01eld A\u00b5 to be a classical \ufb01eld that corresponds to a constant magnetic \ufb01eld in the z direction: A0 = 0 and A = (0, Bx, 0). This F21 = B, with all other components of F\u00b5\u03bd vanishing. Then yields F12 = we de\ufb01ne a normalized state of an electron at rest, with spin up along the z axis: \u2212 e i \u2261 | Z dp f (p)b\u2020+(p) 0 i |, (64.4) where the wave packet is rotationally invariant (so that there is no orbital f angular momentum) and sharply peaked at p = 0, something like f (p) exp( \u2212 \u223c a2p2/2) (64.5) with a \u226a e e have i | h 1/m. We normalize the wave packet by = 1. dp 2 = 1; then we f (p) | | R f 64", ": The Magnetic Moment of the Electron 384 Now we de\ufb01ne the interaction hamiltonian as what we get from the two displayed terms in eq. (64.3), using our speci\ufb01ed \ufb01eld A\u00b5, and with the form-factor values of eq. (64.2): H1 \u2261 \u2212 eB Z d3x \u03a8 x\u03b32 + \u03b1 2\u03c0m h S12 \u03a8. i Then the electron\u2019s magnetic moment \u00b5 is speci\ufb01ed by \u00b5B e H1| e i | \u2261 \u2212h. (64.6) (64.7) In quantum mechanics in general, if we identify H1 as the piece of the hamiltonian that is linear in the external magnetic \ufb01eld, then eq. (64.7) de\ufb01nes the magnetic moment of a normalized quantum state with de\ufb01nite angular momentum in the B direction. Now we turn to the computation. We need to evaluate Using the usual plane-wave expansions, we have. e \u03a8\u03b1(x)\u03a8\u03b2(x) e i | | h b+(p\u2032)\u03a8\u03b1(x)\u03a8\u03b2(x)b\u2020+(p) 0 0 | h i | = u+(p\u2032)\u03b1u+(p)\u03b2 ei(p \u2212 p\u2032)x. (64.8) Thus we get e H1| e i | h = \u2212 eB Z dp dp\u2032 d3x ei(p \u2212 p\u2032)x \u00d7 f x\u03b32 + \u03b1 2\u03c0m u+(p)f (p). f \u2217(p\u2032)u+(p\u2032) f i h i\u2202p1 acting on ei(p S12 \u2212 p\u2032)x, and integrate We can write the factor of x as by parts to put this derivative onto u+(p)f (p); the wave packets kill any surface terms. Then we can complete the integral over d3x to get a factor of (2\u03c0)3\u03b43(p\u2032 \u2212 p), and do the integral over dp\u2032. The result is \u2212 (64.9) e H1| e i | h = \u2212 eB dp 2\u03c9 Z f f \u2217(p)u+(p) h f i\u03b32\u2202p1 + \u03b1 2\u03c0m S12 u+(p)f (p). i (64.10) Suppose the \u2202p1 acts on f (p). Since f (p) is rotationally invariant, the result is odd in p1. We then use u+(p)\u03b3iu+(p) = 2pi to conclude that this term is odd in both p1 and p2, and hence integrates to zero. The remaining contribution from the \ufb01rst term has the \u2202p1 acting on u+(p). Recall from section 38 that us(p) = exp(i\u03b7 \u02c6p K)us(0), \u00b7 (64.11) where K j = Sj0 = i direction, and \u03b7 = sinh\u2212 2 \u03b3j\u03b30 is the boost matrix, \u02c6p is a unit vector in the p 1( p /m) is the rapidity. Since the wave packet is | | 64: The Magnetic Moment of the Electron 385 sharply peaked at p = 0, we can expand eq. (64.11) to linear order in p, take the derivative with respect to p1, and then set p = 0; the result is = i \u2202p1u+(p) p=0 m K 1u+(0) 1 2m 1 2m \u2212 \u2212 \u03b31\u03b30u+(0) \u03b31u+(0), = = (64.12) (64.13) (64.14) where we used \u03b30us(0) = us(0) to get the last line. Then we have u+(p)i\u03b32\u2202p1u+(p) p=0 \u03b32\u03b31u+(0) i = u+(0) \u2212 2m = 1 m u+(0)S12u+(0) Plugging this into eq. (64.10) yields e H1| e i | h = \u2212 eB = \u2212 eB 2m2 Next we use S12u \u00b1 (0) = dp 2\u03c9 | 2 f (p) | Z f 1 + \u03b1 2\u03c0 1 2 u (0) and u \u00b1 \u00b1 e H1| e i | h = eB 2m \u2212 1 m u+(0)S12u+(0) 1 + \u03b1 2\u03c0 u+(0)S12u+(0). (0)u (0) = 2m to get \u00b1 \u00b1 1 + \u03b1 2\u03c0.", " (64.15) Comparing with eq. (64.7), we see that the magnetic moment of the electron is 1 2 where e/2m is the Bohr magneton, the extra factor of one-half is for the electron\u2019s spin (a classical spinning ball of charge would have a magnetic moment equal to the Bohr magneton times its angular momemtum), and g is the Land\u00b4e g factor, given by e 2m (64.16 2\u03c0 Since g can be measured to high precision, calculations of \u00b5 provide a stringent test of spinor electrodynamics. Corrections up through the \u03b14 term have been computed; the result is currently in good agreement with experiment. + O(\u03b12) (64.17). Problems 64.1) Let the wave packet be f (p) a2p2/2)Y\u2113m(\u02c6p), where Y\u2113m(\u02c6p) exp( is a spherical harmonic. Find the contribution of the orbital angular momentum to the magnetic moment. \u2212 \u223c 65: Loop Corrections in Scalar Electrodynamics 386 65 Loop Corrections in Scalar Electrodynamics Prerequisite: 61, 62 In this section we will compute the one-loop corrections in scalar electrodynamics. We will concentrate on the divergent parts of the diagrams, enabling us to compute the renormalizing Z factors in the MS scheme, and hence the beta functions. This gives us the most important qualitative information about the theory: whether it becomes strongly coupled at high or low energies. Our lagrangian for scalar electrodynamics in section 61 already includes all possible terms whose coe\ufb03cients have positive or zero mass dimension, and that respect Lorentz symmetry, the U(1) gauge symmetry, parity, time reversal, and charge conjugation. Therefore, the theory we will consider is (65.1) (65.2) (65.3) = L1, L0 + L \u2202\u00b5\u03d5\u2020\u2202\u00b5\u03d5 L0 = \u2212 \u2212 L1 = iZ1e[\u03d5\u2020\u2202\u00b5\u03d5 \u2212 (Z2\u2212 Lct = 4 Z\u03bb\u03bb(\u03d5\u2020\u03d5)2 + 1)\u2202\u00b5\u03d5\u2020\u2202\u00b5\u03d5 \u2212 \u2212 1 4 F \u00b5\u03bdF\u00b5\u03bd, 1 m2\u03d5\u2020\u03d5 \u2212 (\u2202\u00b5\u03d5\u2020)\u03d5]A\u00b5 \u2212 Lct, Z4e2\u03d5\u2020\u03d5A\u00b5A\u00b5 (Zm\u2212 \u2212 1)m2\u03d5\u2020\u03d5 1 4 (Z3\u2212 \u2212 1)F \u00b5\u03bd F\u00b5\u03bd. (65.4) We will use the MS renormalization scheme to \ufb01x the values of the Z\u2019s. We begin with the photon self-energy, \u03a0\u00b5\u03bd(k). The one-loop and coun- terterm contributions are shown in \ufb01g. (65.1). We have i\u03a0\u00b5\u03bd (k) = (iZ1e)2 2 1 i Z d4\u2113 (2\u03c0)4 (2\u2113 + k)\u00b5(2\u2113 + k)\u03bd ((\u2113+k)2 + m2)(\u21132 + m2) 2iZ4)e2g\u00b5\u03bd + ( \u2212 Z 1)(k2g\u00b5\u03bd i(Z3\u2212 \u2212 d4\u2113 (2\u03c0)4 1 \u21132 + m2 k\u00b5k\u00b5) +..., \u2212 (65.5) where the ellipses stand for higher-order (in e2 and/or \u03bb) terms. We can set Zi = 1 + O(e2,\u03bb) in the \ufb01rst two terms. It will prove convenient to combine these \ufb01rst two terms into d4\u2113 (2\u03c0)4 N \u00b5\u03bd ((\u2113+k)2 + m2)(\u21132 + m2) i\u03a0\u00b5\u03bd (k) = e2 \u2212 Z i(Z3\u2212 1)(k2g\u00b5\u03bd k\u00b5k\u00b5) +..., \u2212 (65.6) 65: Loop Corrections in Scalar Electrodynamics 387 where N \u00b5\u03bd = (2\u2113 + k)\u00b5(2\u2113 + k)\u03bd (65.7) Then we continue to d dimensions, replace e with e\u02dc\u00b5\u03b5/2, and combine the denominators with Feynman\u2019s formula; the result is 2[(\u2113+k)2 + m2]g\u00b5\u03bd. \u2212 i\u03a0\u00b5\u03bd (k)", " = e2 \u02dc\u00b5\u03b5 1 dx ddq (2\u03c0)d N \u00b5\u03bd (q2 + D)2 0 Z i(Z3\u2212 \u2212 Z 1)(k2g\u00b5\u03bd k\u00b5k\u00b5) +..., \u2212 (65.8) where q = \u2113 + xk and D = x(1 \u2212 x)k2 + m2. The numerator is N \u00b5\u03bd = (2q + (1 2x)k)\u00b5(2q + (1 \u2212 = 4q\u00b5q\u03bd + (1 + (linear in q) 2x)2k\u00b5k\u03bd \u2212 \u2212 2x)k)\u03bd \u2212 2[q2 + (1 \u2212 2[(q + (1 \u2212 \u2212 x)k)2 + m2]g\u00b5\u03bd x)2k2 + m2]g\u00b5\u03bd 4d\u2212 \u2192 1g\u00b5\u03bd q2 + (1 \u2212 2x)2k\u00b5k\u03bd 2[q2 + (1 \u2212 \u2212 x)2k2 + m2]g\u00b5\u03bd, (65.9) where we used the symmetric-integration identity from problem 14.3 to get the last line. We can rearrange eq. (65.9) into N \u00b5\u03bd = 2 2 d \u2212 1 g\u00b5\u03bd q2 + (1 \u2212 2x)2k\u00b5k\u03bd 2[(1 \u2212 \u2212 x)2k2 + m2]g\u00b5\u03bd. (65.10) Now we recall from section 62 that, when q2 is integrated against (q2+D)\u2212 we can make the replacement ( 2 1)q2 D; thus we have 2, N \u00b5\u03bd 2Dg\u00b5\u03bd + (1 \u2192 = (1 \u2212 2x)2k\u00b5k\u03bd 2[(1 x)2k2 + m2]g\u00b5\u03bd \u2212 2x)(1 \u2212 x)k2g\u00b5\u03bd. 2(1 (65.11) \u2212 \u2212 Next we note that if we make the change of variable x = y + 1 1 4 y2)k2 + m2, and y is integrated from have D = (1 any term in N \u00b5\u03bd that is even in y will integrate to zero. We then get 2 to + 1 2, then we 2. Therefore, \u2212 \u2212 \u2212 \u2212 1 \u2192 d \u2212 2x)2k\u00b5k\u03bd N \u00b5\u03bd = 4y2k\u00b5k\u03bd 2(2y2 y)k2g\u00b5\u03bd \u2212 4y2(k2g\u00b5\u03bd \u2212 k\u00b5k\u03bd ). \u2192 \u2212 \u2212 (65.12) Thus we see that \u03a0\u00b5\u03bd (k) is transverse, as expected. Performing the integral over q in eq. (65.8), and focusing on the diver- gent part, we get ddq (2\u03c0)d 1 (q2 + D)2 = i 8\u03c02 1 \u03b5 \u02dc\u00b5\u03b5 Z + O(\u03b50). (65.13) 65: Loop Corrections in Scalar Electrodynamics 388 l k+l k l k k k k k Figure 65.1: The one-loop and counterterm corrections to the photon propagator in scalar electrodynamics. l l k k+l k k k l k k k k Figure 65.2: The one-loop and counterterm corrections to the scalar propagator in scalar electrodynamics. Then performing the integral over y yields 1/2 1/2 Z \u2212 dy N \u00b5\u03bd = 1 3 (k2g\u00b5\u03bd \u2212 \u2212 k\u00b5k\u03bd ). Combining eqs. (65.8), (65.13), and (65.14), we get \u03a0\u00b5\u03bd (k) = \u03a0(k2)(k2g\u00b5\u03bd k\u00b5k\u03bd), \u2212 (65.14) (65.15) where \u03a0(k2) = e2 24\u03c02 1 \u03b5 \u2212 + \ufb01nite (Z3 \u2212 \u2212 1) +.... (65.16) Thus we \ufb01nd, in the MS scheme, Z3 = 1 e2 24\u03c02 1 \u03b5 \u2212 +.... (65.17) Now we turn to the one-loop corrections to the scalar propagator, shown in \ufb01g. (65.2). It will prove very convenient to work in Lorenz gauge, where 65: Loop Corrections in Scalar Electrodynamics 389 the photon propagator is \u02dc\u2206\u00b5\u03bd (\u2113) = P\u00b5\u03bd (\u2113) \u21132 i\u01eb \u2212, (65.18) with P\u00b5\u03bd (", "\u2113) = g\u00b5\u03bd \u2212 \u2113\u00b5\u2113\u03bd /\u21132. The diagrams in \ufb01g. (65.2) then yield i\u03a0\u03d5(k2) = (iZ1e)2 2 1 i Z d4\u2113 (2\u03c0)4 P\u00b5\u03bd (\u2113)(\u2113 + 2k)\u00b5(\u2113 + 2k)\u03bd \u21132((\u2113+k)2 + m2) d4\u2113 (2\u03c0)4 P\u00b5\u03bd (\u2113) \u21132 + m2 \u03b3 1 \u21132 + m2 + ( \u2212 1 i 2iZ4e2g\u00b5\u03bd ) d4\u2113 (2\u03c0)4 i\u03bb) 1 i Z + ( \u2212 Z 1)k2 i(Z2\u2212 \u2212 i(Zm\u2212 1)m2 +.... (65.19) \u2212 In the second line, m\u03b3 is a \ufb01ctitious photon mass; it appears here as an infrared regulator. We can set Zi = 1 + O(e2,\u03bb) in the \ufb01rst three lines. Continuing to d \u03bb\u02dc\u00b5\u03b5, and using 1, we get e\u02dc\u00b5\u03b5/2 and \u03bb dimensions, making the replacements e the relations \u2113\u00b5P\u00b5\u03bd (\u2113) = \u2113\u03bdP\u00b5\u03bd (\u2113) = 0 and g\u00b5\u03bd P\u00b5\u03bd (\u2113) = d \u2192 \u2192 \u2212 i\u03a0\u03d5(k2) = 4e2 \u02dc\u00b5\u03b5 dd\u2113 (2\u03c0)d P\u00b5\u03bd (\u2113)k\u00b5k\u03bd \u21132((\u2113+k)2 + m2) Z 1)e2 \u02dc\u00b5\u03b5 2(d \u2212 \u2212 Z dd\u2113 (2\u03c0)d 1 \u21132 + m2 \u03b3 \u03bb\u02dc\u00b5\u03b5 Z i(Z2\u2212 \u2212 \u2212 d4\u2113 (2\u03c0)4 1 \u21132 + m2 1)k2 \u2212 i(Zm\u2212 1)m2 +.... (65.20) We evaluate the second and third lines via dd\u2113 (2\u03c0)d 1 \u21132 + m2 = i 8\u03c02 1 \u03b5 \u2212 \u02dc\u00b5\u03b5 Z m2 + O(\u03b50). (65.21) Then, taking the limit m2 0 (with \u03b5 \ufb01xed) in eq. (65.21) shows that the second line of eq. (65.20) vanishes when the infrared regulator is removed. To evaluate the \ufb01rst line of eq. (65.20), we multiply the numerator and \u2192 denominator by \u21132 and use Feynman\u2019s formula to get dd\u2113 (2\u03c0)d \u21132P\u00b5\u03bd (\u2113)k\u00b5k\u03bd \u21132\u21132((\u2113+k)2 + m2) Z = dF3 Z Z ddq (2\u03c0)d N (q2 + D)3, (65.22) 65: Loop Corrections in Scalar Electrodynamics 390 l l l+k k l l+k k l l l k k l+k Figure 65.3: The one-loop corrections to the three-point vertex in scalar electrodynamics. where q = \u2113 + x3k, D = x3(1 x3)k2 + x3m2, and \u2212 N = \u21132k2 (\u2113 k)2 \u00b7 \u2212 x3k)2k2 = (q \u2212 = q2k2 q2k2 \u2192 \u2212 \u2212 (q k x3k2)2 \u2212 \u00b7 k)2 + (linear in q) \u2212 (q \u00b7 1q2k2. d\u2212 (65.23) Now we use q2 (q2 + D)3 = Combining eqs. (65.20\u201365.24), and requiring \u03a0\u03d5(k2) to be \ufb01nite, we \ufb01nd ddq (2\u03c0)d + O(\u03b50). i 8\u03c02 \u02dc\u00b5\u03b5 1 \u03b5 Z (65.24) Z2 = 1 + Zm = 1 + 3e2 8\u03c02 \u03bb 8\u03c02 65.25) (65.26) in the MS scheme. Now we turn to the one-loop corrections to the three-point (scalar\u2013 scalar\u2013photon) vertex, shown in", " \ufb01g. (65.3). In order to simplify the calculation of the divergent terms as much as possible, we have chosen a special set of external momenta. (If we wanted the complete vertex function, including the \ufb01nite terms, we would need to use a general set of external momenta.) We take the incoming scalar to have zero four-momentum, and 65: Loop Corrections in Scalar Electrodynamics 391 the photon (treated as incoming) to have four-momentum k; then, by momentum conservation, the outgoing scalar also has four-momentum k. We take the internal photon to have four-momentum \u2113. Now comes the magic of Lorenz gauge: in the second and third diagrams of \ufb01g. (65.3), the vertex factor for the leftmost vertex is ie\u2113\u00b5, and this is zero when contracted with the P\u00b5\u03bd (\u2113) of the internal photon propagator. Thus the second and third diagrams vanish. Alas, we will have to do some more work to evaluate the \ufb01rst and fourth diagrams. We have iV\u00b5 3 (k, 0) = ieZ1k\u00b5 + (iZ1e)( \u2212 2iZ4e2g\u00b5\u03bd ) 1 i + ( \u2212 iZ\u03bb\u03bb)(iZ1e d4\u2113 (2\u03c0)4 d4\u2113 (2\u03c0)4 P\u03bd\u03c1(\u2113)(\u2113 + 2k)\u03c1 \u21132((\u2113+k)2 + m2) (2\u2113 + k)\u00b5 (\u21132 + m2)((\u2113+k)2 + m2) (65.27) We can set Zi = 1 + O(e2,\u03bb) in the second and third lines. We then do the usual manipulations; the integral in the third line becomes 1 dx 0 Z Z ddq (2\u03c0)d 2q\u00b5 + (1 2x)k\u00b5 \u2212 (q2 + D)2, (65.28) x)k2 + m2. The term linear in q vanishes where q = \u2113 + xk and D = x(1 upon integration over q, and the term linear in k vanishes upon integration over x. Thus the third line of eq. (65.27) evaluates to zero. \u2212 To evaluate the second line of eq. (65.27), we note that, since P\u03bd\u03c1(\u2113)\u2113\u03c1 = 0, it already has an overall factor of k. We can then treat k as in\ufb01nitesimal, and set k = 0 in the denominator. We can then use the symmetricintegration identity to make the replacement \u2113\u03bd\u2113\u03c1/\u21132 1g\u03bd\u03c1 in the numerator. Putting all of this together, and using eq. (65.13), we \ufb01nd d\u2212 \u2192 V\u00b5 3 (k, 0)/e = Z1k\u00b5 3e2 8\u03c02 1 \u03b5 \u2212 k\u00b5 + O(\u03b50) +..., (65.29) and so in the MS scheme. Z1 = 1 + 3e2 8\u03c02 1 \u03b5 +..., (65.30) Next up is the four-point, scalar\u2013scalar\u2013photon\u2013photon vertex. Because 2iZ4e2g\u00b5\u03bd, does not depend on the external the tree-level vertex factor, four-momenta, we can simply set them all to zero. Then, whenever an \u2212 65: Loop Corrections in Scalar Electrodynamics 392 l l l l Figure 65.4: The nonvanishing one-loop corrections to the scalar\u2013scalar\u2013 photon\u2013photon vertex in scalar electrodynamics (in Lorenz gauge with vanishing external momenta). internal photon line attaches to an external scalar with a three-point vertex, the diagram is zero, for the same reason that the second and third diagrams of \ufb01g. (65.3) were zero. This kills a lot of diagrams; the survivors are shown in \ufb01g. (65.4). We have iV\u00b5\u03bd 4 (0, 0, 0) = 2iZ4e2g\u00b5\u03bd \u2212 2iZ4e2)2 + ( \u2212 2 1 i Z d4\u2113 (2\u03c0)4 g\u00b5\u03c1P\u03c1\u03c3(\u2113)g\u03c3\u03bd \u21132(\u21132 + m2) + (\u00b5 \u03bd) \u2194 + (iZ1e)2( \u2212 1 i iZ\u03bb", "\u03bb) 3 Z d4\u2113 (2\u03c0)4 + ( \u2212 iZ\u03bb\u03bb)( \u2212 2iZ4e2g\u00b5\u03bd ) ) \u2194 (2\u2113)\u00b5(2\u2113)\u03bd (\u21132 + m2)3 + (\u00b5 d4\u2113 (2\u03c0)4 1 (\u21132 + m2)2 (65.31) The notation +(\u00b5 \u03bd) in the second and third lines means that we must add the same expression with these indices swapped; this is because the original and swapped versions of each diagram are topologically distinct, and contribute separately to the vertex function. \u2194 As usual, we set Zi = 1 + O(e2,\u03bb) in the second through fourth lines. After using the symmetric-integration identity, along with eqs. (65.13) and (65.24), we can see that the divergent parts of the third and fourth lines cancel each other. The \ufb01rst line is easily evaluated with symmetric integration and eq. (65.13). Then we have V\u00b5\u03bd 4 (0, 0, 0)/e2 = 2Z4g\u00b5\u03bd + \u2212 3e2 4\u03c02 1 \u03b5 g\u00b5\u03bd + O(\u03b50) +..., (65.32) and so Z4 = 1 + 3e2 8\u03c02 1 \u03b5 +... (65.33) 65: Loop Corrections in Scalar Electrodynamics 393 Figure 65.5: The nonvanishing one-loop corrections to the four-scalar vertex in scalar electrodynamics (in Lorenz gauge with vanishing external momenta). in the MS scheme. Finally, we have the one-loop corrections to the four-scalar vertex. Once again, because the tree-level vertex factor, iZ\u03bb\u03bb, does not depend on the external four-momenta, we can set them all to zero. Then, whenever an internal photon line attaches to an external scalar with a three-point vertex, the diagram is zero. The remaining diagrams are shown in \ufb01g. (65.5). \u2212 Even though we have set the external momenta to zero, we still have to keep track of which particle is which, in order to count the diagrams correctly; thus the external lines are labelled 1 through 4. Lines 1 and 2 have arrows pointing towards their vertices, and 3 and 4 have arrows pointing away from their vertices. The symmetry factor for each of the \ufb01rst three diagrams is S = 2; for each of the last two, it is S = 1. The di\ufb00erence arises because the last two diagrams have the charge arrows pointing in opposite directions on the two internal propagators, and so these propagators cannot be exchanged. It is clear that the \ufb01rst two diagrams will yield identical contributions to the vertex function (when the external momenta are all zero). Similarly, except for symmetry factors, the contributions of the last three diagrams are also identical. Thus we have iV4\u03d5(0, 0, 0) = iZ\u03bb\u03bb \u2212 + 1 2 + 1 2 2iZ4e2)2 ( \u2212 2 1 i Z d4\u2113 (2\u03c0)4 g\u00b5\u03bd P\u03bd\u03c1(\u2113)g\u03c1\u03c3P\u03c1\u00b5(\u2113) (\u21132 + m2 \u03b3)2 65: Loop Corrections in Scalar Electrodynamics 394 + 1 2 + 1 + 1 +.... iZ\u03bb\u03bb)2 ( \u2212 2 1 i Z d4\u2113 (2\u03c0)4 1 (\u21132 + m2)2 (65.34) Using the familiar techniques, we \ufb01nd V4\u03d5(0, 0, 0) = iZ\u03bb\u03bb + \u2212 3e4 2\u03c02 1 \u03b5 + 5\u03bb2 16\u03c02 1 \u03b5 + O(\u03b50) +..., (65.35) and so in the MS scheme. Z\u03bb = 1 + 3e4 2\u03c02\u03bb + 5\u03bb 16\u03c02! 1 \u03b5 +... (65.36) Problems 65.1) What conditions should be imposed on V\u00b5 4 (k, p\u2032, p) in the OS scheme? (Here k is the incoming four-momentum of the photon at the \u00b5 vertex, and p\u2032 and p are the four-momenta of the outgoing and incoming scalars, respectively.) 3 (p\u2032, p) and V\u00b5\u03bd 65.2) Consider a gauge transformaton A\u00b5 \u2202\u00b5\u0393. Show that there is a transformation of \u03d5 that leaves the lagrangian of eqs. (65.", "1\u201365.4) invariant if and only if Z4 = Z 2 1 /Z2. A\u00b5 \u2192 \u2212 66: Beta Functions in Quantum Electrodynamics 395 66 Beta Functions in Quantum Electrodynamics Prerequisite: 52, 62 In this section we will compute the beta function for the electromagnetic coupling e in spinor electrodynamics and scalar electrodynamics. We will also compute the beta function for the \u03d54 coupling \u03bb in scalar electrodynamics. In spinor electrodynamics, the relation between the bare and renormal- ized couplings is e0 = Z \u2212 3 1/2 1 2 Z1 \u02dc\u00b5\u03b5/2e. Z \u2212 (66.1) It is convenient to recast this formula in terms of the \ufb01ne-structure constant \u03b1 = e2/4\u03c0 and its bare counterpart \u03b10 = e2 2 Z 2 2 0/4\u03c0, 1 \u02dc\u00b5\u03b5\u03b1. \u03b10 = Z \u2212 1 3 Z \u2212 (66.2) From section 62, we have Z1 = 1 Z2 = 1 Z3 = 1 \u03b1 2\u03c0 \u03b1 2\u03c0 2\u03b1 3\u03c0 (\u03b12), + O(\u03b12), + O(\u03b12), in the MS scheme. Let us write ln \u221e = Xn=1 En(\u03b1) \u03b5n. Then we have \u221e ln \u03b10 = From eqs. (66.3\u201366.5), we get Xn=1 En(\u03b1) \u03b5n + ln \u03b1 + \u03b5 ln \u02dc\u00b5. E1(\u03b1) = 2\u03b1 3\u03c0 + O(\u03b12). Then, the general analysis of section 28 yields \u03b2(\u03b1) = \u03b12E\u20321(\u03b1), (66.3) (66.4) (66.5) (66.6) (66.7) (66.8) (66.9) where the prime denotes di\ufb00erentiation with respect to \u03b1. Thus we \ufb01nd \u03b2(\u03b1) = 2\u03b12 3\u03c0 + O(\u03b13) (66.10) 66: Beta Functions in Quantum Electrodynamics 396 in spinor electrodynamics, We can, if we like, restate this in terms of e as \u03b2(e) = e3 12\u03c02 + O(e5). (66.11) To go from eq. (66.10) to eq. (66.11), we use \u03b1 = e2/4\u03c0 and \u02d9\u03b1 = e \u02d9e/2\u03c0, where the dot denotes d/d ln \u00b5. The most important feature of either eq. (66.10) or eq. (66.11) is that the beta function is positive: the electromagnetic coupling in spinor electrodynamics gets stronger at high energies, and weaker at low energies. It is easy to generalize eqs. (66.10) and (66.11) to the case of N Dirac \ufb01elds with electric charges Qie. There is now a factor of Z2i for each \ufb01eld, and of Z1i for each interaction. These are found by replacing \u03b1 in eqs. (66.3) i \u03b1. Then we \ufb01nd Z1i/Z2i = 1 + O(\u03b12), so that this ratio and (66.4) with Q2 is universal, at least through O(\u03b1). In fact, as we will see in section 67, Z1i/Z2i is always exactly equal to one, and so it always cancels in eq. (66.6). As for Z3, now each Dirac \ufb01eld contributes separately to the fermion loop in the photon self-energy, and so we should replace \u03b1 in eq. (66.5) with i Q2 i \u03b1. Thus we \ufb01nd that the generalization of eq. (66.11) is P \u03b2(e) = N i=1 Q2 i 12\u03c02 P e3 + O(e5). (66.12) Now we turn to scalar electrodynamics. (Prerequisite: 65.) The rela- tions between the bare and renormalized couplings are 1 \u03d5 Z1 \u02dc\u00b5\u03b5/2e. Z \u2212 \u03d5 Z4 \u02dc\u00b5\u03b5e2. 2 e0 = Z \u2212 3 1/2 e2 0 = Z \u2212 \u03bb0 = Z \u2212 2 3 Z \u2212 \u03d5 Z\u03bb \u02dc\u00b5\u03b5\u03bb. 2 (66.13) (66.14) (66.15) We have two di\ufb00erent relations between e and e0, coming from the two types of", " vertices. We can guess (and will demonstrate in section 67) that these two renormalizations must work out to give the same answer. Indeed, from section 65, we have Z1 = 1 + Z2 = 1 + 3e2 8\u03c02 3e2 8\u03c02 1 \u03b5 1 \u03b5 +..., +..., Z3 = 1 e2 24\u03c02 1 \u03b5 \u2212 +..., (66.16) (66.17) (66.18) 66: Beta Functions in Quantum Electrodynamics 397 (66.19) Z4 = 1 + 3e2 8\u03c02 1 \u03b5 +..., Z\u03bb = 1 + 3e4 2\u03c02\u03bb + 5\u03bb 16\u03c02! 1 \u03b5 +..., (66.20) in Lorenz gauge in the MS scheme; the ellipses stand for higher powers of e2 and/or \u03bb. We see that Z1 = Z2 = Z4, at least through O(e2,\u03bb). The correct guess is that this is true exactly. Thus eqs. (66.13) and (66.14) both collapse to e0 = Z \u2212 3 e, just as in spinor electrodynamics. 1/2 Thus we can write 1/2 ln Z \u2212 3 = ln Z \u2212 2 2 Z\u03bb = En(e, \u03bb) \u03b5n Ln(e, \u03bb) \u03b5n,. \u221e Xn=1 \u221e Xn=1 (66.21) (66.22) Then we have ln e0 = ln \u03bb0 = En(e, \u03bb) \u03b5n Ln(e, \u03bb) \u03b5n \u221e Xn=1 \u221e Xn=1 + ln e + 1 2 e ln \u02dc\u00b5, (66.23) + ln \u03bb + e ln \u02dc\u00b5. (66.24) Using eqs. (66.17), (66.18) and (66.20), we have E1(e, \u03bb) = L1(e, \u03bb) = e2 24\u03c02 +..., 1 16\u03c02 5\u03bb + 24e4/\u03bb \u2212 (66.25) 12e2 +..., (66.26) Now applying the general analysis of section 52 yields \u03b2e(e, \u03bb) = \u03b2\u03bb(e, \u03bb) = e3 48\u03c02 +..., 1 16\u03c02 5\u03bb2 \u2212 (66.27) 6\u03bbe2 + 24e4 +.... (66.28) Both right-hand sides are strictly positive, and so both e and \u03bb become large at high energies, and small at low energies. Generalizing eq. (66.25) to the case of several complex scalar \ufb01elds with charges Qie works in the same way as it does in spinor electrodynamics. 66: Beta Functions in Quantum Electrodynamics 398 For a theory with both Dirac \ufb01elds and complex scalar \ufb01elds, the one-loop contributions to Z3 simply add, and so the beta function for e is \u03b2e(e, \u03bb) = 1 12\u03c02 \u03a8 Q2 \u03a8 + 1 4 \u03d5 Q2 \u03d5 e3 +.... (66.29) P Problems P 66.1) Compute the one-loop contributions to the anomalous dimensions of m, \u03a8, and A\u00b5 in spinor electrodynamics in Feynman gauge. 66.2) Compute the one-loop contributions to the anomalous dimensions of m, \u03d5, and A\u00b5 in scalar electrodynamics in Lorenz gauge. 66.3) Use the results of problem 62.2 to compute the anomalous dimension of m and the beta function for e in spinor electrodynamics in R\u03be gauge. You should \ufb01nd that the results are independent of \u03be. 66.4) The value of \u03b1(MW). The solution of eq. (66.12) is 1 \u03b1(MW) = 1 \u03b1(\u00b5) \u2212 2 3\u03c0 Xi Q2 i ln(MW/\u00b5), (66.30) where the sum is over all quarks and leptons (each color of quark counts separately), and we have chosen the W \u00b1 boson mass MW as a reference scale. We can de\ufb01ne a di\ufb00erent renormalization scheme, modi\ufb01ed decoupling subtraction or DS, where we imagine integrating out a \ufb01eld when \u00b5 is below its mass. In this scheme, eq", ". (66.30) becomes 1 \u03b1(MW) = 1 \u03b1(\u00b5) \u2212 2 3\u03c0 Xi Q2 i ln[MW/min(mi, \u00b5)], (66.31) where the sum is now over all quarks and leptons with mass less than MW. For \u00b5 < me, the DS scheme coincides with the OS scheme, and we have 1 \u03b1(MW) = 1 \u03b1 \u2212 2 3\u03c0 Q2 i ln(MW/mi), (66.32) Xi where \u03b1 = 1/137.036 is the \ufb01ne-structure constant in the OS scheme. 300 MeV for the light quark masses (because Using mu = md = ms \u223c quarks should be replaced by hadrons at lower energies), and other quark and lepton masses from sections 83 and 88, compute \u03b1(MW). 67: Ward Identities in Quantum Electrodynamics I 399 67 Ward Identities in Quantum Electrodynamics I Prerequisite: 22, 59 In section 59, we assumed that scattering amplitudes would be gauge invariant, in the sense that they would be unchanged if we replaced any photon polarization vector \u03b5\u00b5 with \u03b5\u00b5 + ck\u00b5, where k\u00b5 is the photon\u2019s four-momentum and c is an arbitrary constant. Thus, if we write a scatfor a process that includes an external photon with tering amplitude four-momentum k\u00b5 as T = \u03b5\u00b5 M\u00b5, T (67.1) then we should have M\u00b5 = 0. In this section, we will use the Ward identity for the electromagnetic current to prove eq. (67.2). (67.2) k\u00b5 We begin by recalling the LSZ formula for scalar \ufb01elds, f h i i | = i Z d4x1 e\u2212 ik1x1( \u2212 1 + m2)... \u22022 T\u03d5(x167.3) We have treated all external particles as outgoing; an incoming particle has k0 i < 0. We can rewrite eq. (67.3) as f h i i | = lim k2 i \u2192\u2212 m2 (k2 1 + m2)... T \u02dc\u03d5(k167.4) d4x e\u2212 Here \u02dc\u03d5(k) = i extra factor of i), and we do not \ufb01x k2 = R m2. \u2212 ikx\u03d5(x) is the \ufb01eld in momentum space (with an We know that the right-hand side of eq. (67.4) must include an overall energy-momentum delta function, so let us write = (2\u03c0)4\u03b44( i, ki \u00b7 T \u02dc\u03d5(k1)... 0 | h kj) is a function of the Lorentz scalars k2 i, ki\u00b7 i ki) 0 i | (k2 P F (k2 kj), i and ki\u00b7 (67.5) kj. Then, where since F f h = i(2\u03c0)4\u03b44( i i ki) | i T should have a multivariable pole as each k2 i apis the residue of this pole. That is, near (67.6) P, eq. (67.4) tells us that proaches k2 i = m2, and that i T takes the form \u2212 m2, F \u2212 F (k2 i, ki \u00b7 F kj) = i T (k2 1 + m2)... (k2 n + m2) + nonsingular. (67.7) 67: Ward Identities in Quantum Electrodynamics I 400 T The key point is this: contributions to pole do not contribute to. that do not have this multivariable F We have framed this discussion in terms of scalar \ufb01elds in order to keep the notation as simple as possible, but the general point holds for \ufb01elds of any spin. In section 22, we analyzed how various classical \ufb01eld equations apply to quantum correlation functions. For example, we derived the SchwingerDyson equations T 0 | h \u03b4S \u03b4\u03c6a(x) \u03c6a1(x1)... \u03c6an(xn) 0 i | n = i T\u03c6a1 (x1)... \u03b4aaj \u03b44(x 0 | Xj=1h \u2212 xj)... \u03c6an(xn) 0 i |. (67.8) Here we have used \u03c6a(x) to denote any kind of \ufb01eld, not necessarily a scalar \ufb01eld,", " carrying any kind of index or indices. The classical equation of motion for the \ufb01eld \u03c6a(x) is \u03b4S/\u03b4\u03c6a(x) = 0. Thus, eq. (67.8) tells us that the classical equation of motion holds for a \ufb01eld inside a quantum correlation function, as long as its spacetime argument and indices do not match up exactly with those of any other \ufb01eld in the correlation function. These matches, which constitute the right-hand side of eq. (67.8), are called contact terms. Suppose we have a correlation function that, for whatever reason, includes a contact term with a factor of, say, \u03b44(x1\u2212 x2). After Fouriertransforming to momentum space, this contact term is a function of k1 +k2, k2; hence it cannot take the form of the singular but is independent of k1 \u2212 term in eq. (67.7). Therefore, contact terms in a correlation function do not contribute to the scattering amplitude F. Now let us consider a scattering process in quantum electrodyamics that involves an external photon with four-momentum k. In Lorenz gauge (the simplest for this analysis), the LSZ formula reads T f h i i | = i\u03b5\u00b5 d4x e\u2212 Z ikx( \u2212 \u22022)... TA\u00b5(x67.9) and the classical equation of motion for A\u00b5 is Z3 \u22022A\u00b5 = \u2212 \u2202 \u2202A\u00b5. L (67.10) In spinor electrodynamics, the right-hand side of eq. (67.10) is Z1j\u00b5, where j\u00b5 is the electromagnetic current. (This is also true for scalar electrodynamics if Z4 = Z 2 1 /Z2; we saw in problem 65.2 that this condition is necessary 67: Ward Identities in Quantum Electrodynamics I 401 for gauge invariance.) We therefore have 1 3 Z1\u03b5\u00b5 d4x e\u2212 ikx.... Z f h i i | 0 i | = iZ \u2212 + contact terms Tj\u00b5(x)... 0 | h h (67.11) The contact terms arise because, as we saw in eq. (67.8), the classical equations of motion hold inside quantum correlation functions only up to contact terms. However, the contact terms cannot generate singularities in the k2\u2019s of the other particles, and so they do not contribute to the left-hand side. (Remember that, for each of the other particles, there is still an appropriate wave operator, such as the Klein-Gordon wave operator for a scalar, acting on the correlation function. These wave operators kill any term that does not have an appropriate singularity.) i \u2202\u00b5 acting on the e\u2212 Now let us try replacing \u03b5\u00b5 in eq. (67.11) with k\u00b5. We are attempting to prove that the result is zero, and we are almost there. We can write the factor of ik\u00b5 as ikx, and then we can integrate by parts to get \u2202\u00b5 acting on the correlation function. (Strictly speaking, we need a wave packet for the external photon to kill surface terms.) Then we have \u2202\u00b5 on the right-hand side. Now we use another result from section 22, namely that a Noether current for an exact symmetry obeys \u2202\u00b5j\u00b5 = 0 classically, and Tj\u00b5(x\u00b5 Tj\u00b5(x)... 0 | h 0 i | = contact terms (67.12) quantum mechanically; this is the Ward (or Ward-Takahashi) identity. But once again, the contact terms do not have the right singularities to contribute to vanishes if we replace an external photon\u2019s polarization vector \u03b5\u00b5 with its four-momentum k\u00b5, quo erat demonstratum.. Thus we conclude that Reference Notes Diagrammatic proofs of the Ward identity in spinor electrodynamics can be found in Peskin & Schroeder and Zee. Problems 67.1) Show explicitly that the tree-level \u03b3\u03b3 scattering amplitude in scalar electrodynamics, = \u2212 T e2 vanishes if \u03b5\u00b5 \u03b52\u2032) 4(k1 \u00b7 \u03b51\u2032)(k2 \u00b7 m2 t \u2212 \u00b5 1. 1\u2032 is replaced with k\u2032 + e+ e\u2212 \u2192 e e 4(k1 \u00b7 \u03b52\u2032)(k2 \u00b7 m2 u \u2212 \u03b51\u2032) + 2(\u03b51\u2032, \u03b52\u2032) \u00b7 67: Ward Identities in Quantum Electrodynamics I 402 67.2", ") Show explicitly that the tree-level e+e\u2212 \u2192 in spinor electrodynamics, \u03b3\u03b3 scattering amplitude = e2 v2 /\u03b52\u2032 \u2212 T /p1 + /k\u20321 + m m2 t vanishes if \u03b5\u00b5 \u00b5 1. 1\u2032 is replaced with k\u2032 \u2212 /\u03b51\u2032 + /\u03b51\u2032 \u2212 /p1 + /k\u20322 + m m2 u \u2212 /\u03b52\u2032 u1, 68: Ward Identities in Quantum Electrodynamics II 403 68 Ward Identities in Quantum Electrodynamics II Prerequisite: 63, 67 In this section, we will show that Z1 = Z2 in spinor electrodynamics, and that Z1 = Z2 = Z4 in scalar electrodyanmics (in the OS and MS renormalization schemes). Let us specialize to the case of spinor electrodynamics with a single Dirac \ufb01eld, and consider the correlation function ip\u2032y+ipz C \u00b5 \u2212 Z \u2261 iZ1 \u03b1\u03b2(k, p\u2032, p) d4x d4y d4z eikx Tj\u00b5(x)\u03a8\u03b1(y)\u03a8\u03b2(z) 0 0 | h i | (68.1) where j\u00b5 = e\u03a8\u03b3\u00b5\u03a8 is the electromagnetic current. As we saw in section 67, including Z1j\u00b5(x) inside a correlation function adds a vertex for an external photon; the factor of Z1 provides the necessary renormalization of this vertex. The explicit fermion \ufb01elds on the right-hand side of eq. (68.1) combine with the fermion \ufb01elds in the current to generate propagators. Thus we have, C \u00b5 \u03b1\u03b2(k, p\u2032, p) = (2\u03c0)4\u03b44(k+p p\u2032) h where \u02dcS(p) is the exact fermion propagator, and V\u00b5(p\u2032, p) is the exact 1PI photon\u2013fermion\u2013fermion vertex function. \u02dcS(p\u2032)iV\u00b5(p\u2032, p) 1 i \u02dcS(p) i (68.2) \u2212 \u03b1\u03b2 1 i, Now let us consider k\u00b5C \u00b5 \u03b1\u03b2(k, p\u2032, p). Using eq. (68.1), we can write the factor of ik\u00b5 on the right-hand side as \u2202\u00b5 acting on eikx, and then integrate \u2202\u00b5 acting on j\u00b5(x). (Strictly speaking, we need a wave by parts to get packet for the external photon to kill surface terms.) Thus we have \u2212 k\u00b5C \u00b5 \u03b1\u03b2(k, p\u2032, p) = \u2212 Z d4x d4y d4z eikx \u2212 ip\u2032y+ipz \u2202\u00b5h Tj\u00b5(x)\u03a8\u03b1(y)\u03a8\u03b2(z) 0 0 i | | (68.3). Now we use the Ward identity from section 22, which in general reads TJ \u00b5(x)\u03c6a1 (x1)... \u03c6an(xn) 0 \u2202\u00b5h 0 | i | n \u2212 = i T\u03c6a1 (x1)... \u03b4\u03c6aj (x)\u03b44(x 0 | Xj=1h \u2212 xj)... \u03c6an(xn) 0 i |. (68.4) Here \u03b4\u03c6a(x) is the change in a \ufb01eld \u03c6a(x) under an in\ufb01nitesimal transformation that leaves the action invariant, and J \u00b5 = \u2202 L \u2202(\u2202\u00b5\u03c6a) \u03b4\u03c6a (68.5) 68: Ward Identities in Quantum Electrodynamics II 404 is the corresponding Noether current. In the case of spinor electrodynamics, \u03b4\u03a8(x) = ie\u03a8(x), \u2212 \u03b4\u03a8(x) = +ie\u03a8(x), (68.6) where we have dropped an in\ufb01nitesimal parameter on the right-hand sides, /\u2202(\u2202\u00b5\u03a8) = iZ2\u03a8\u03b3\u00b5 but included a factor of the electron charge e. Using \u2202 /\u2202(\u2202\u00b5\u03a8) = 0 we \ufb01nd that, with these conventions, the Noether and \u2202 current is J \u00b5 = Z2e\u03a8\u03b3\u00b5\u03a8 = Z2j\u00b5. Thus the Ward identity becomes L L Tj\u00b5(", "x)\u03a8\u03b1(y)\u03a8\u03b2(z) 0 0 Z2\u2202\u00b5h i | | \u2212 Recall that = +e \u03b44(x e \u03b44(x \u2212 0 T\u03a8\u03b1(y)\u03a8\u03b2(z) 0 y) | | h i 0 T\u03a8\u03b1(y)\u03a8\u03b2(z) 0 z) i | | h. (68.7) \u2212 \u2212 T\u03a8\u03b1(y)\u03a8\u03b2(z d4q (2\u03c0)4 eiq(y z) \u02dcS(q)\u03b1\u03b2. \u2212 (68.8) Using eqs. (68.7) and (68.8) in eq. (68.3), and carrying out the coordinate integrals, we get k\u00b5C \u00b5 \u03b1\u03b2(k, p\u2032, p) = \u2212 iZ \u2212 2 Z1(2\u03c0)4\u03b44(k+p 1 e\u02dcS(p) \u2212 p\u2032) h \u2212 e\u02dcS(p\u2032) \u03b1\u03b2 i. (68.9) On the other hand, from eq. (68.2) we have k\u00b5C \u00b5 \u03b1\u03b2(k, p\u2032, p) = \u2212 i(2\u03c0)4\u03b44(k+p \u2212 p\u2032) h \u02dcS(p\u2032)k\u00b5V\u00b5(p\u2032, p)\u02dcS(p) \u03b1\u03b2 i. (68.10) Comparing eqs. (68.9) and (68.10) shows that p)\u00b5\u02dcS(p\u2032)V\u00b5(p\u2032, p)\u02dcS(p) = Z \u2212 (p\u2032 \u2212 1 2 Z1e h \u02dcS(p) \u2212, \u02dcS(p\u2032) i (68.11) where we have dropped the spin indices. We can simplify eq. (68.11) by multiplying on the left by \u02dcS(p\u2032)\u2212 1, and on the right by \u02dcS(p)\u2212 1, to get 1 \u2212 (p\u2032 p)\u00b5V\u00b5(p\u2032, p) = Z \u2212 1 2 Z1e h Thus we \ufb01nd a relation between the exact photon\u2013fermion\u2013fermion vertex function V\u00b5(p\u2032, p) and the exact fermion propagator \u02dcS(p). (68.12) \u02dcS(p\u2032)\u2212 \u02dcS(p)\u2212 Since both \u02dcS(p) and V\u00b5(p\u2032, p) are \ufb01nite, eq. (68.12) implies that Z1/Z2 must be \ufb01nite as well. In the MS scheme (where all corrections to Zi = 1 are divergent), this immediately implies that \u2212 i 1. Z1 = Z2. (68.13) 68: Ward Identities in Quantum Electrodynamics II 405 In the OS scheme, we recall that near the on-shell point p2 = p\u2032 m2 p)2 = 0 we have \u02dcS(p) = /p + m and V\u00b5(p\u2032, p) = e\u03b3\u00b5. Plugging and (p\u2032 \u2212 these expressions into eq. (68.12) then yields Z1 = Z2 for the OS scheme. 2 = \u2212 To better understand this result, we note that when Z1 = Z2, we can combine the fermion kinetic term iZ2\u03a8/\u2202\u03a8 and the interaction term ieA\u00b5 is the covariant derivative. Z1e\u03a8 /A\u03a8 into iZ2\u03a8 /D\u03a8, where D\u00b5 = \u2202\u00b5 Recall that it is D\u00b5 that has a simple gauge transformation, and so we might expect the lagrangian, written in terms of renormalized \ufb01elds, to include \u2202\u00b5 and A\u00b5 only in the combination D\u00b5. It is still necessary to go through the analysis that led to eq. (68.12), however, because quantization requires \ufb01xing a gauge, and this renders suspect any naive arguments based on gauge invariance. Still, in this case, those arguments yield the correct result. \u2212 We can make a similar analysis in scalar electrodynamics. We leave the details to the problems. Reference Notes BRST symmetry (see section 74) can be used to derive the Ward identities; see Ramond I. Problems 68.1) Consider the current correlation function electrodynamics. Tj\u00b5(x)j\u03bd (y) 0 0 i | | h in spinor a) Show that its Fourier transform is proportional to \u03a0\u00b5\u03bd (k) + \u03a0\u00b5\u03c1(k) \u02dc", "\u2206\u03c1\u03c3(k)\u03a0\u03c3\u03bd (k) +.... (68.14) b) Use this to prove that \u03a0\u00b5\u03bd (k) is transverse: k\u00b5\u03a0\u00b5\u03bd (k) = 0. 68.2) Verify that eq. (68.12) holds at the one-loop level in any renormaliza- tion scheme with Z1 = Z2. 68.3) Scalar electrodynamics. (Prerequisite: 65.) a) Consider the Fourier transform of TJ \u00b5(x)\u03d5(y)\u03d5\u2020(z), where 0 0 i | | h (\u2202\u00b5\u03d5\u2020)\u03d5] 2Z1e2A\u00b5\u03d5\u2020\u03d5 (68.15) J \u00b5 = ieZ2[\u03d5\u2020\u2202\u00b5\u03d5 \u2212 \u2212 is the Noether current. You may assume that Z4 = Z 2 necessary for gauge invariance). Show that \u2212 1 /Z2 (which is (p\u2032 \u2212 p)\u00b5V\u00b5 3 (p\u2032, p) = Z \u2212 1 2 Z1e h \u02dc\u2206(p\u2032)\u2212 1 \u2212 \u02dc\u2206(p)\u2212 1, (68.16) i 68: Ward Identities in Quantum Electrodynamics II 406 where V\u00b5 \u02dc\u2206(p) is the exact scalar propagator. 3 (p\u2032, p) is the exact scalar-scalar-photon vertex function, and b) Use this result to show that Z1 = Z2 in both the MS and OS renormalization schemes. c) Consider the Fourier transform of Show that TJ \u00b5(x)A\u03bd (w)\u03d5(y)\u03d5\u2020(z). 0 0 i | | h k\u00b5V\u00b5\u03bd 4 (k, p\u2032, p) = Z \u2212 1 V\u03bd 1 Z4e h 3 (p\u2032 k, p) \u2212 \u2212 V\u03bd 3(p\u2032, p+k) i, (68.17) where V\u00b5\u03bd 4 (k, p\u2032, p) is the exact scalar-scalar-photon-photon vertex function, with k the incoming momentum of the photon at the \u00b5 vertex. 69: Nonabelian Gauge Theory 407 69 Nonabelian Gauge Theory Prerequisite: 24, 58 Consider a lagrangian with N scalar or spinor \ufb01elds \u03c6i(x) that is invariant under a continuous SU(N ) or SO(N ) symmetry, \u03c6i(x) \u2192 Uij\u03c6j(x), (69.1) \u00d7 \u00d7 where Uij is an N N special unitary matrix in the case of SU(N ), or an N special orthogonal matrix in the case of SO(N ). (Special means N that the determinant of U is one.) Eq. (69.1) is called a global symmetry transformation, because the matrix U does not depend on the spacetime label x. In section 58, we saw that quantum electrodynamics could be under- stood as having a local U(1) symmetry, \u03c6(x) \u2192 U (x)\u03c6(x), (69.2) \u2212 where U (x) = exp[ 1 unitary matrix ie\u0393(x)] can be thought of as a 1 that does depend on the spacetime label x. Eq. (69.2) can be a symmetry of the lagrangian only if we include a U(1) gauge \ufb01eld A\u00b5(x), and promote ieA\u00b5. ordinary derivatives \u2202\u00b5 of \u03c6(x) to covariant derivatives D\u00b5 = \u2202\u00b5 \u2212 Under the transformation of eq. (69.2), we have \u00d7 U (x)D\u00b5U \u2020(x). D\u00b5 \u2192 (D\u00b5\u03d5)\u2020D\u00b5\u03d5, or a With this transformation rule, a scalar kinetic term like fermion kinetic term like i\u03a8 /D\u03a8, is invariant, as are mass terms like m2\u03d5\u2020\u03d5 and m\u03a8\u03a8. We call eq. (69.3) a gauge transformation, and say that the lagrangian is gauge invariant. (69.3) \u2212 Eq. (69.3) implies that the gauge \ufb01eld transforms as A\u00b5(x) \u2192 U (x)A\u00b5(x)U \u2020(x) + i e U (x)\u2202\u00b5U \u2020(x). If we use U (x) = exp[ \u2212 ie\u0393(x)], then eq. (69.4) simpl", "i\ufb01es to A\u00b5(x) A\u00b5(x) \u2212 \u2192 \u2202\u00b5\u0393(x), (69.4) (69.5) which is what we originally had in section 54. We can now easily generalize this construction of U(1) gauge theory to SU(N ) or SO(N ). (We will consider other possibilities later.) To be concrete, let us consider SU(N ). Recall from section 24 that we can write an in\ufb01nitesimal SU(N ) transformation as Ujk(x) = \u03b4jk \u2212 ig\u03b8a(x)(T a)jk + O(\u03b82), (69.6) 69: Nonabelian Gauge Theory 408 where we have inserted a coupling constant g for later convenience. The indices j and k run from 1 to N, the index a runs from 1 to N 2 1 (and is implicitly summed), and the generator matrices T a are hermitian and traceless. (These properties of T a follow immediately from the special unitarity of U.) The generator matrices obey commutation relations of the form \u2212 [T a, T b] = if abcT c, (69.7) where the real numerical factors f abc are called the structure coe\ufb03cients of the group. If f abc does not vanish, the group is nonabelian. We can choose the generator matrices so that they obey the normaliza- tion condition Tr(T aT b) = 1 2 \u03b4ab ; (69.8) then eqs. (69.7) and (69.8) can be used to show that f abc is completely antisymmetric. For SU(2), we have T a = 1 2 \u03c3a, where \u03c3a is a Pauli matrix, and f abc = \u03b5abc, where \u03b5abc is the completely antisymmetric Levi-Civita symbol. Now we de\ufb01ne an SU(N ) gauge \ufb01eld A\u00b5(x) as a traceless hermitian N \u00d7 N matrix of \ufb01elds with the gauge transformation property A\u00b5(x) \u2192 U (x)A\u00b5(x)U \u2020(x) + i g U (x)\u2202\u00b5U \u2020(x). (69.9) Note that this is identical to eq. (69.4), except that now U (x) is a special unitary matrix (rather than a phase factor), and A\u00b5(x) is a traceless hermitian matrix (rather than a real number). (Also, the electromagnetic coupling e has been replaced by g.) We can write U (x) in terms of the generator matrices as U (x) = exp[ \u2212 ig\u0393a(x)T a], where the real parameters \u0393a(x) are no longer in\ufb01nitesimal. The covariant derivative is D\u00b5 = \u2202\u00b5 \u2212 igA\u00b5(x), (69.10) (69.11) N identity matrix multiplying \u2202\u00b5. Acting where there is an understood N on the set of N \ufb01elds \u03c6i(x) that transform according to eq. (69.2), the covariant derivative can be written more explicitly as \u00d7 (D\u00b5\u03c6)j(x) = \u2202\u00b5\u03c6j(x) \u2212 igA\u00b5(x)jk\u03c6k(x), (69.12) with an understood sum over k. The covariant derivative transforms acwith covariant cording to eq. (69.3). Replacing all ordinary derivatives in L 69: Nonabelian Gauge Theory derivatives renders had a global SU(N ) symmetry). L gauge invariant (assuming, of course, that 409 originally L We still need a kinetic term for A\u00b5(x). Let us de\ufb01ne the \ufb01eld strength F\u00b5\u03bd (x) i g [D\u00b5, D\u03bd ] \u2261 = \u2202\u00b5A\u03bd \u2212 \u2202\u03bdA\u00b5 \u2212 ig[A\u00b5, A\u03bd ]. (69.13) (69.14) Because A\u00b5 is a matrix, the \ufb01nal term in eq. (69.14) does not vanish, as it does in U(1) gauge theory. Eqs. (69.3) and (69.13) imply that, under a gauge transformation, Therefore, F\u00b5\u03bd (x) \u2192 U (x)F\u00b5\u03bd (x)U \u2020(x). Lkin = \u2212 1 2 Tr(F \u00b5\u03bd F\u00b5\u03bd ) (69.", "15) (69.16) is gauge invariant, and can serve as a kinetic term for the SU(N ) gauge \ufb01eld. (Note, however, that the \ufb01eld strength itself is not gauge invariant, in contrast to the situation in U(1) gauge theory.) Since we have taken A\u00b5(x) to be hermitian and traceless, we can expand it in terms of the generator matrices: A\u00b5(x) = Aa \u00b5(x)T a. Then we can use eq. (69.8) to invert eq. (69.17): Similarly, we have Aa \u00b5(x) = 2 Tr A\u00b5(x)T a. F\u00b5\u03bd (x) = F a \u00b5\u03bd T a, \u00b5\u03bd (x) = 2 Tr F\u00b5\u03bd T a. F a Using eq. (69.19) in eq. (69.14), we get \u00b5\u03bd T c = (\u2202\u00b5Ac F c \u03bd \u2212 = (\u2202\u00b5Ac \u03bd \u2212 \u2202\u03bdAc \u00b5)T c igAa \u00b5Ab \u03bd[T a, T b] \u2212 \u00b5 + gf abcAa \u2202\u03bdAc \u00b5Ab \u03bd)T c. Then using eqs. (69.20) and (69.8) yields \u00b5\u03bd = \u2202\u00b5Ac F c \u03bd \u2212 \u2202\u03bdAc \u00b5 + gf abcAa \u00b5Ab \u03bd. Also, using eqs. (69.19) and (69.8) in eq. (69.16), we get Lkin = \u2212 1 4 F c\u00b5\u03bdF c \u00b5\u03bd. (69.17) (69.18) (69.19) (69.20) (69.21) (69.22) (69.23) 69: Nonabelian Gauge Theory 410 From eq. (69.22), we see that Lkin includes interactions among the gauge \ufb01elds. A theory of this type, with nonzero f abc, is called nonabelian gauge theory or Yang\u2013Mills theory. Everything we have just said about SU(N ) also goes through for SO(N ), with unitary replaced by orthogonal, and traceless replaced by antisymmetric. There is also another class of compact nonabelian groups called Sp(2N ), and \ufb01ve exceptional compact groups: G(2), F(4), E(6), E(7), and E(8). Compact means that Tr(T aT b) is a positive de\ufb01nite matrix. Nonabelian gauge theory must be based on a compact group, because otherwise Lkin would have the wrong sign, leading to a hamilsome of the terms in tonian that is unbounded below. As a speci\ufb01c example, let us consider quantum chromodynamics, or QCD, which is based on the gauge group SU(3). There are several Dirac \ufb01elds corresponding to quarks. Each quark comes in three colors; these are the values of the SU(3) index. (These colors have nothing to do with ordinary color.) There are also six \ufb02avors: up, down, strange, charm, bottom (or beauty) and top (or truth). Thus we consider the Dirac \ufb01eld \u03a8iI (x), where i is the color index and I is the \ufb02avor index. The lagrangian is 1 L 2 Tr(F \u00b5\u03bd F\u00b5\u03bd ), = i\u03a8iI /Dij\u03a8jI \u2212 mI \u03a8I \u03a8I \u2212 where all indices are summed. The di\ufb00erent quark \ufb02avors have di\ufb00erent masses, ranging from a few MeV for the up and down quarks to 178 GeV for the top quark. (The quarks also have electric charges: + 2 for the u, e 3 | | for the d, s, and b quarks. For now, however, we c, and t quarks, and omit the appropriate coupling to the electromagnetic \ufb01eld.) The covariant derivative in eq. (69.24) is (69.24) 1 e 3 | | \u2212 (D\u00b5)ij = \u03b4ij\u2202\u00b5 \u2212 igAa \u00b5T a ij. (69.25) The index a on Aa one particles are the eight gluons. \u00b5 runs from 1 to 8, and the corresponding massless spin- In a nonabelian gauge theory in general, we can consider scalar or spinor \ufb01elds in di\ufb00erent representations of the group. A representation of a compact nonabel", "ian group is a set of \ufb01nite-dimensional hermitian matrices T a R (the R is part of the name, not an index) that obey that same commutation relations as the original generator matrices T a. Given such a set of D(R) D(R) matrices (where D(R) is the dimension of the representation), and a \ufb01eld \u03c6(x) with D(R) components, we can write its covariant \u00b5T a D(R) identity derivative as D\u00b5 = \u2202\u00b5 \u2212 \u00d7 matrix multiplying \u2202\u00b5. Under a gauge transformation, \u03c6(x) UR(x)\u03c6(x), \u2192 where UR(x) is given by eq. (69.10) with T a replaced by T a R. The theory will be gauge invariant provided that the transformation rule for Aa \u00b5 is in- R, with an understood D(R) igAa \u00d7 69: Nonabelian Gauge Theory 411 depedent of the representation used in eq. (69.9); we show in problem 69.1 that it is. We will not need to know a lot of representation theory, but we collect some useful facts in the next section. Problems 69.1) Show that eq. (69.9) implies a transformation rule for Aa \u00b5 that is independent of the representation used in eq. (69.9). Hint: consider an in\ufb01nitesimal transformation. 69.2) Show that [T aT a, T b] = 0. 70: Group Representations 412 70 Group Representations Prerequisite: 69 Given the structure coe\ufb03cients f abc of a compact nonabelian group, a representation of that group is speci\ufb01ed by a set of D(R) D(R) traceless hermitian matrices T a R (the R is part of the name, not an index) that obey that same commutation relations as the original generators matrices T a, namely \u00d7 [T a R, T b R] = if abcT c R. (70.1) The number D(R) is the dimension of the representation. The original T a\u2019s correspond to the fundamental or de\ufb01ning representation. (T a R )\u2217 also obey these commutation relations. If Consider taking the complex conjugate of the commutation relations, eq. (70.1). Since the structure coe\ufb03cients are real, we see that the matrices R )\u2217 = T a R, or if we can \u2212 R )\u2217 = T a (T a \ufb01nd a unitary transformation T a R for U \u2212 every a, then the representation R is real. If such a unitary transformation (T a R )\u2217 = does not exist, but we can \ufb01nd a unitary matrix V \u2212 V \u2212 If such a unitary matrix also does not exist, then the representation R is complex. In this case, the complex conjugate representation R is speci\ufb01ed by = I such that R V for every a, then the representation R is pseudoreal. \u2212 R U that makes R \u2192 1T a 1T a (T a \u2212 T a R = (T a R )\u2217. \u2212 (70.2) One way to prove that a representation is complex is to show that at least one generator matrix T a R (or a real linear combination of them) has eigenvalues that do not come in plus-minus pairs. This is the case 3. For SU(2), for the fundamental representation of SU(N ) with N ( 1 2 \u03c3a, 2 \u03c3a)\u2217 6 the fundamental representation is pseudoreal, because 2 \u03c3a)V with V = \u03c32. For SO(N ), the fundamental but representation is real, because the generator matrices are antisymmetric, and every antisymmetric hermitian matrix is equal to minus its complex conjugate. ( 1 2 \u03c3a)\u2217 = V \u2212 \u2265 \u2212 = 1 1( 1 \u2212 An important representation for any compact nonabelian group is the adjoint representation A. This is given by (T a A )bc = if abc. \u2212 (70.3) Because f abc is real and completely antisymmetric, T a A is manifestly hermitian, and also satis\ufb01es eq. (70.2); thus the adjoint representation is real. The dimension of the adjoint representation D(A) is equal to the number of generators of the group; this number is also called the dimension of the group. 6 70: Group Representations 413 To see that the T a A \u2019s satisfy the commutation relations, we use the Jacobi identity f", " abdf dce + f bcdf dae + f cadf dbe = 0, (70.4) which holds for the structure coe\ufb03cients of any group. To prove the Jacobi identity, we note that Tr T e [[T a, T b], T c] + [[T b, T c], T a] + [[T c, T a], T b] = 0, (70.5) where the T a\u2019s are the original generator matrices. That the left-hand side of eq. (70.5) vanishes can be seen by writing out all the commutators as matrix products, and noting that they cancel in pairs. Employing the commutation relations twice in each term, followed by Tr(T aT b) = 1 2 \u03b4ab, (70.6) ultimately yields eq. (70.4). Then, using the antisymmetry of the structure coe\ufb03cients, inserting some judicious factors of i, and moving the last term of eq. (70.4) to the right-hand side, we can rewrite it as ( \u2212 if abd)( \u2212 if cde) ( \u2212 \u2212 if cbd)( \u2212 if ade) = if acd( \u2212 if dbe). (70.7) Now we use eq. (70.3) in eq. (70.7) to get (T a A )bd(T c A)de (T c A)bd(T a A )de = if acd(T d A)be, (70.8) \u2212 or equivalently [T a A, T c commutation relations. A] = if acdT d A. Thus the T a A \u2019s satisfy the appropriate Two related numbers usefully characterize a representation: the index T (R) and the quadratic Casimir C(R). The index is de\ufb01ned via Tr(T a R T b R) = T (R)\u03b4ab. (70.9) Next we recall from problem 69.2 that the matrix T a R commutes with every generator, and so must be a number times the identity matrix; this number is the quadratic Casimir C(R). It is easy to show that R T a T (R)D(A) = C(R)D(R). (70.10) With the standard normalization conventions for the generators, we have T (N) = 1 2 for the fundamental representation of SU(N ) and T (N) = 2 for the fundamental representation of SO(N ). We show in problem 70.2 that T (A) = N for the adjoint representation of SU(N ), and in problem 70.3 that T (A) = 2N 4 for the adjoint representation of SO(N ). A representation R is reducible if there is a unitary transformation T a 1T a R \u2192 R U that puts all the nonzero entries into the same diagonal blocks for U \u2212 \u2212 70: Group Representations 414 each a; otherwise it is irreducible. Consider a reducible representation R whose generators can be put into (for example) two blocks, with the blocks forming the generators of the irreducible representations R1 and R2. Then R is the direct sum representation R = R1 \u2295 R2, and we have D(R1\u2295 T (R1\u2295 R2) = D(R1) + D(R2), R2) = T (R1) + T (R2). (70.11) (70.12) Suppose we have a \ufb01eld \u03d5iI (x) that carries two group indices, one for the representation R1 and one for the representation R2, denoted by i and R2. I respectively. This \ufb01eld is in the direct product representation R1 \u2297 The corresponding generator matrix is (T a R1\u2297R2 )iI,jJ = (T a R1 )ij\u03b4IJ + \u03b4ij(T a R2 )IJ, (70.13) where i and I together constitute the row index, and j and J together constitute the column index. We then have D(R1\u2297 T (R1\u2297 R2) = D(R1)D(R2), R2) = T (R1)D(R2) + D(R1)T (R2). (70.14) (70.15) To get eq. (70.15), we need to use the fact that the generator matrices are traceless, (T a R )ii = 0. At this point it is helpful to introduce a slightly more re\ufb01ned notation for the indices", " of a complex representation. Consider a \ufb01eld \u03d5 in the complex representation R. We will adopt the convention that such a \ufb01eld carries a \u201cdown\u201d index: \u03d5i, where i = 1, 2,..., D(R). Hermitian conjugation changes the representation from R to R, and we will adopt the convention that this also raises the index on the \ufb01eld, (\u03d5i)\u2020 = \u03d5\u2020 i. (70.16) Thus a down index corresponds to the representation R, and an up index to R. Indices can be contracted only if one is up and one is down. Generator matrices for R are then written with the \ufb01rst index down and the second index up: (T a j. An in\ufb01nitesimal group transformation of \u03d5i takes the form R )i \u03d5i \u2192 (1 \u2212 = \u03d5i \u2212 i\u03b8aT a R )i j\u03d5j i\u03b8a(T a R )i j\u03d5j. The generator matrices for R are then given by (T a R )i j = (T a R )j i, \u2212 (70.17) (70.18) 70: Group Representations 415 where we have used the hermiticity to trade complex conjugation for transi takes position of the indices. An in\ufb01nitesimal group transformation of \u03d5\u2020 the form i \u03d5\u2020 (1 \u2212 \u2192 i = \u03d5\u2020 j j\u03d5\u2020 i\u03b8aT a R )i i\u03b8a(T a R )i \u2212 i + i\u03b8a(T a R )j = \u03d5\u2020 j j\u03d5\u2020 i\u03d5\u2020 j, (70.19) where we used eq. (70.18) to get the last line. Note that eqs. (70.17) and (70.19) together imply that \u03d5\u2020 i\u03d5i is invariant, as expected. Consider the Kronecker delta symbol with one index down and one up: \u03b4i j. Under a group transformation, we have j \u03b4i (1 + i\u03b8aT a \u2192 = (1 + i\u03b8aT a R )i R )i k(1 + i\u03b8aT a l\u03b4k R )j i\u03b8aT a R )l l j k\u03b4k l(1 \u2212 = \u03b4i j + O(\u03b82). (70.20) j is an invariant symbol of the group. This exisEq. (70.20) shows that \u03b4i tence of this invariant symbol, which carries one index for R and one for R, tells us that the product of the representations R and R must contain the singlet representation 1, speci\ufb01ed by T a 1 = 0. We therefore can write R R = 1.... (70.21) The generator matrix (T a \u2297 j, which carries one index for R, one for R, and one for the adjoint representation A, is also an invariant symbol. To see this, we make a simultaneous in\ufb01nitesimal group transformation on each of these indices, R )i \u2295 (T b R)i j i\u03b8aT a R )i k(1 i\u03b8aT a R )j l(1 i\u03b8aT a A )bc(T c R)k l (1 \u2212 \u2192 = (T b R)i j \u2212 + O(\u03b82). \u2212 R )i i\u03b8a[(T a k(T b R)k \u2212 R )j j + (T a l(T b R)i l + (T a A )bc(T c R)i j] (70.22) The factor in square brackets should vanish if (as we claim) the generator matrix is an invariant symbol. Using eqs. (70.3) and (70.18), we have [... ] = (T a R )i k(T b R)k j (T a R )l j(T b R)i l if abc(T c R)i j = (T a R T b R)i j \u2212 \u2212 RT a (T b R )i j \u2212 \u2212 if abc(T c j R)i = 0, (70.23) 70: Group Representations 416 where the last line follows from the commutation relations. The fact that (T a j is an invariant symbol implies that R )70.24) If we now multiply both sides of eq. (70.24) by A, and use A [which follows from eq. (70.21) and", " the reality of A], we \ufb01nd R Combining this with eq. (70.21), we have \u2297 \u2297 70.25) That is, the product of a representation with its complex conjugate is always reducible into a sum that includes (at least) the singlet and adjoint representations. For the fundamental representation N of SU(N ), we have N = 1 N \u2297 A, \u2295 (70.26) with no other representations on the right-hand side. To see this, recall that D(1) = 1, D(N) = D(N) = N, and, as shown in section 24, D(A) = N 2 1. From eq. (70.14), we see that there is no room for anything else on the right-hand side of eq. (70.26). \u2212 Consider now a real representation R. From eq. (70.25), with R = R, we have 70.27) \u2295 The singlet on the right-hand side implies the existence of an invariant symbol with two R indices; this symbol is the Kronecker delta \u03b4ij. It is invariant because \u2295 \u2297 \u03b4ij \u2192 (1 \u2212 = \u03b4ij \u2212 i\u03b8aT a R )i k(1 i\u03b8aT a R )j l\u03b4kl \u2212 R )ij + (T a i\u03b8a[(T a R )ji] + O(\u03b82). (70.28) The term in square brackets vanishes by hermiticity and eq. (70.18). The fact that \u03b4ij = \u03b4ji implies that the singlet on the right-hand side of eq. (70.28) appears in the symmetric part of this product of two identical representations. The fundamental representation N of SO(N ) is real, and we have N N = 1S \u2295 AA \u2295 \u2297 SS. (70.29) The subscripts tell whether the representation appears in the symmetric or antisymmetric part of the product. The representation S corresponds to a \ufb01eld with a symmetric traceless pair of fundamental indices: \u03d5ij = \u03d5ji, 70: Group Representations 417 \u03d5ii = 0, where the repeated index is summed. We have D(1) = 1, D(N) = N, and, as shown in section 24, D(A) = 1 1). Also, a traceless symmetric tensor has D(S) = 1 1 independent components; thus eq. (70.14) is ful\ufb01lled. 2 N (N +1) 2 N (N \u2212 \u2212 Consider now a pseudoreal representation R. Since R is equivalent to its complex conjugate, up to a change of basis, eq. (70.27) still holds. However, we cannot identify \u03b4ij as the corresponding invariant symbol, because then eq. (70.28) shows that R would have to be real, rather than pseudoreal. From the perspective of the direct product, the only alternative is to have the singlet appear in the antisymmetric part of the product, rather than the symmetric part. The corresponding invariant symbol must then be antisymmetric on exchange of its two R indices. An example (the only one that will be of interest to us) is the fundamental representation of SU(2). For SU(N ) in general, another invariant symbol is the Levi-Civita tensor \u03b5i1...iN, which carries N fundamental indices and is completely antisymmetric. It is invariant because, under an SU(N ) transformation, \u03b5i1...iN \u2192 Ui1 j1... UiN jN \u03b5j1...jN = (det U )\u03b5i1...iN. (70.30) Since det U = 1 for SU(N ), we see that the Levi-Civita symbol is invariant. We can similarly consider \u03b5i1...iN, which carries N completely antisymmetric \u03b5ji; antifundamental indices. For SU(2), the Levi-Civita symbol is \u03b5ij = this is the two-index invariant symbol that corresponds to the singlet in the product \u2212 2 2 = 1A \u2295 \u2297 3S, (70.31) where 3 is the adjoint representation. We can use \u03b5ij and \u03b5ij to raise and lower SU(2) indices. This is another way to see that there is no distinction between the fundamental representation 2 and its complex conjugate 2. That is, if we have a \ufb01eld \u03d5i in the representation 2, we can get a \ufb01", "eld in the representation 2 by raising the index: \u03d5i = \u03b5ij\u03d5j. The structure constants f abc are another invariant symbol. This follows if abc, since we have seen that generator matrices (in any from (T a representation) are invariant. Alternatively, given the generator matrices in a representation R, we can write A )bc = \u2212 T (R)f abc = \u2212 i Tr(T a R [T b R, T c R]). (70.32) Since the right-hand side is invariant, the left-hand side must be as well. 70: Group Representations 418 If we use an anticommutator in place of the commutator in eq. (70.32), we get another invariant symbol, A(R)dabc \u2261 1 2 Tr(T a R { R, T c T b R} ), (70.33) where A(R) is the anomaly coe\ufb03cient of the representation. The cyclic property of the trace implies that A(R)dabc is symmetric on exchange of any pair of indices. Using eq. (70.18), we can see that Thus, if R is real or pseudoreal, A(R) = 0. We also have A(R) = A(R). \u2212 A(R1\u2295 A(R1\u2297 R2) = A(R1) + A(R2), R2) = A(R1)D(R2) + D(R1)A(R2). (70.34) (70.35) (70.36) We normalize the anomaly coe\ufb03cient so that it equals one for the smallest complex representation. In particular, for SU(N ) with N 3, the smallest complex representation is the fundamental, and A(N) = 1. For SU(2), all representations are real or pseudoreal, and A(R) = 0 for all of them. \u2265 Reference Notes More group and representation theory can be found in Ramond II. Problems 70.1) Verify eq. (70.10). 70.2) a) Use eqs. (70.12) and (70.26) to compute T (A) for SU(N ). b) For SU(2), the adjoint representation is speci\ufb01ed by (T a A)bc = i\u03b5abc. Use this to compute T (A) explicitly for SU(2). Does your \u2212 result agree with part (a)? c) Consider the SU(2) subgroup of SU(N ) that acts on the \ufb01rst two components of the fundamental representation of SU(N ). Under this 2)1\u2019s. UsSU(2) subgroup, the N of SU(N ) transforms as 2 ing eq. (70.26), \ufb01gure out how the adjoint representation of SU(N ) transforms under this SU(2) subgroup. (N \u2212 \u2295 d) Use your results from parts (b) and (c) to compute T (A) for SU(N ). Does your result agree with part (a)? 70: Group Representations 419 70.3) a) Consider the SO(3) subgroup of SO(N ) that acts on the \ufb01rst three components of the fundamental representation of SO(N ). Under this SO(3) subgroup, the N of SO(N ) transforms as 3 3)1\u2019s. Using eq. (70.29), \ufb01gure out how the adjoint representation of SO(N ) transforms under this SO(3) subgroup. (N \u2212 \u2295 b) Use your results from part (a) and from problem 70.2 to compute T (A) for SO(N ). 70.4) a) For SU(N ), we have N \u2297 N = AA \u2295 SS, corresponds to a \ufb01eld with two antisymmetric fundamental corresponds to a \ufb01eld with two ) where SU(N ) indices, \u03d5ij = symmetric fundamental SU(N ) indices, \u03d5ij = +\u03d5ji. Compute D( and D( \u03d5ji, and (70.37) A A \u2212 ). S b) By considering an SU(2) subgroup of SU(N ), compute T ( T ( ). A c) For SU(3), show that = 3. A d) By considering an SU(3) subgroup of SU(N ), compute A( A( ). A ) and ) and S S S 70.5) Consider a \ufb01eld \u03d5i in the representation R1 and a \ufb01eld \u03c7I in the representation R2. Their", " product \u03d5i\u03c7I is then in the direct product R2, with generator matrices given by eq. (70.13). representation R1 \u2297 a) Prove the distribution rule for the covariant derivative, [D\u00b5(\u03d5\u03c7)]iI = (D\u00b5\u03d5)i\u03c7I + \u03d5i(D\u00b5\u03c7)I. (70.38) b) Consider a \ufb01eld \u03d5i in the complex representation R. Show that \u2202\u00b5(\u03d5\u2020 i\u03d5i) = (D\u00b5\u03d5\u2020)i\u03d5i + \u03d5\u2020 i(D\u00b5\u03d5)i. (70.39) Explain why this is a special case of eq. (70.38). 70.6) The \ufb01eld strength in Yang-Mills theory is in the adjoint representa- tion, and so its covariant derivative is (D\u03c1F\u00b5\u03bd )a = \u2202\u03c1F a \u00b5\u03bd \u2212 Prove the Bianchi identity, igAc \u03c1(T c A)abF b \u00b5\u03bd. (70.40) (D\u00b5F\u03bd\u03c1)a + (D\u03bd F\u03c1\u00b5)a + (D\u03c1F\u00b5\u03bd )a = 0. (70.41) 71: The Path Integral for Nonabelian Gauge Theory 420 71 The Path Integral for Nonabelian Gauge Theory Prerequisite: 53, 69 We wish to evaluate the path integral for nonabelian gauge theory (also known as Yang\u2013Mills theory), Z(J) \u221d SYM(A, J) = Z Z A eiSYM(A,J), D 1 4 F a\u00b5\u03bd F a \u00b5\u03bd + J a\u00b5Aa \u00b5 d4x \u2212 h (71.1) (71.2). i In section 57, we evaluated the path integral for U(1) gauge theory by arguing that, in momentum space, the component of the U(1) gauge \ufb01eld parallel to the four-momentum k\u00b5 did not appear in the action, and hence should not be integrated over. This argument relied on the form of the U(1) gauge transformation, In the nonabelian case, however, the gauge transformation is nonlinear, A\u00b5(x) A\u00b5(x) \u2212 \u2192 \u2202\u00b5\u0393(x). (71.3) A\u00b5(x) \u2192 U (x)A\u00b5(x)U \u2020(x) + i g U (x)\u2202\u00b5U \u2020(x), (71.4) where A\u00b5(x) = Aa \u00b5(x)T a. For an in\ufb01nitesimal transformation, U (x) = I = I \u2212 \u2212 ig\u03b8(x) + O(\u03b82) ig\u03b8a(x)T a + O(\u03b82), we have or equivalently A\u00b5(x) \u2192 A\u00b5(x) + ig[A\u00b5(x), \u03b8(x)] \u2202\u00b5\u03b8(x), \u2212 Aa \u00b5(x) Aa \u00b5(x) gf abcAb \u00b5(x)\u03b8c(x) \u2192 = Aa \u00b5(x) = Aa \u00b5(x) = Aa \u00b5(x) = Aa \u00b5(x) \u2212 \u2212 \u2212 \u2212 \u2212 \u2202\u00b5\u03b8a(x) \u2212 \u00b5(x)]\u03b8c(x) if bac)]\u03b8c(x) \u2212 A)ac]\u03b8c(x) \u00b5(T b igAb \u00b5( [\u03b4ac\u2202\u00b5 + gf abcAb [\u03b4ac\u2202\u00b5 \u2212 [\u03b4ac\u2202\u00b5 \u2212 \u00b5 \u03b8c(x), Dac igAb (71.5) (71.6) (71.7) where Dac \u00b5 is the covariant derviative in the adjoint representation. We see the similarity with the abelian case, eq. (71.3). However, the fact that 71: The Path Integral for Nonabelian Gauge Theory 421 it is D\u00b5 that appears in eq. (71.7), rather than \u2202\u00b5, means that we cannot account for gauge redundancy in the path integral by simply excluding the components of Aa \u00b5 that are parallel to k\u00b5. We will have to do something more clever. Consider an ordinary integral of the form dx dy eiS(x), Z \u221d Z (71.8) where both x and y are integrated from minus to plus in\ufb01nity. Because y does not appear in S(x", "), the integral over y is redundant. We can then de\ufb01ne Z by simply dropping the integral over y, dx eiS(x). Z \u2261 Z (71.9) This is how we dealt with gauge redundancy in the abelian case. We could get the same answer by inserting a delta function, rather than by dropping the y integral: Z = dx dy \u03b4(y) eiS(x). (71.10) Z Furthermore, the argument of the delta function can be shifted by an arbitrary function of x, without changing the result: Z = dx dy \u03b4(y Z f (x)) eiS(x). \u2212 (71.11) Suppose we are not given f (x) explicitly, but rather are told that y = f (x) is the unique solution, for \ufb01xed x, of G(x, y) = 0. Then we can write \u03b4(G(x, y)) = \u03b4(y f (x)) \u2212 \u2202G/\u2202y | |, (71.12) where we have used a standard rule for delta functions. We can drop the absoute-value signs if we assume that \u2202G/\u2202y is positive when evaluated at y = f (x). Then we have Z = dx dy Z \u2202G \u2202y \u03b4(G) eiS. (71.13) Now let us generalize this result to an integral over dnx dny. We will need n functions Gi(x, y) to \ufb01x all n components of y. The generalization of eq. (71.13) is Z = dnx dny det Z \u2202Gi \u2202yj! i\u03b4(Gi) eiS. (71.14) Q 71: The Path Integral for Nonabelian Gauge Theory 422 Now we are ready to translate these results to path integrals over nonabelian gauge \ufb01elds. The role of the redundant integration variable y is played by the set of all gauge transformations \u03b8a(x). The role of the integration variables x and y together is played by the gauge \ufb01eld Aa \u00b5(x). The role of G is played by a gauge-\ufb01xing function. We will use the gauge-\ufb01xing function appropriate for R\u03be gauge, which is Ga(x) \u2202\u00b5Aa \u00b5(x) \u2212 \u2261 \u03c9a(x), (71.15) where \u03c9a(x) is a \ufb01xed, arbitrarily chosen function of x. (We will see how the parameter \u03be enters later.) In eq. (71.15), the spacetime argument x and the index a play the role of the index i in eq. (71.14). Our path integral becomes Z(J) \u221d Z A det D where SYM is given by eq. (71.2). \u03b4G \u03b4\u03b8 x,a\u03b4(G) eiSYM, (71.16) Q Now we have to evaluate the functional derivative \u03b4Ga(x)/\u03b4\u03b8b(y), and then its functional determinant. From eqs. (71.7) and (71.15), we \ufb01nd that, under an in\ufb01nitesimal gauge transformation, Thus we have Ga(x) Ga(x) \u2212 \u2192 \u2202\u00b5Dab \u00b5 \u03b8b(x). \u03b4Ga(x) \u03b4\u03b8b(y) = \u2212 \u2202\u00b5Dab \u00b5 \u03b44(x y), \u2212 (71.17) (71.18) where the derivatives are with respect to x. Now we need to compute the functional determinant of eq. (71.18). Luckily, we learned how to do this in section 53. A functional determinant can be written as a path integral over complex Grassmann variables. So let us introduce the complex Grassmann \ufb01eld ca(x), and its hermitian conjugate \u00afca(x). (We use a bar rather than a dagger to keep the notation a little less cluttered.) These \ufb01elds are called Faddeev-Popov ghosts. Then we can write det c \u03b4Ga(x) \u03b4\u03b8b(y) \u221d D Z d4x Lgh, and the ghost lagrangian is \u00afc eiSgh, D (71.19) where the ghost action is Sgh = R \u00b5 cb Lgh = \u00afca\u2202\u00b5Dab \u2202\u00b5\u00afcaDab = \u00b5 cb \u2212 \u2202\u00b5\u00afca\u2202\u00b5ca + ig\u2202\u00b5\u00afcaAc \u00b5(T c A)ab", "cb \u2202\u00b5\u00afca\u2202\u00b5ca + gf abcAc \u00b5\u2202\u00b5\u00afcacb. (71.20) = = \u2212 \u2212 71: The Path Integral for Nonabelian Gauge Theory 423 We dropped a total divergence in the second line. We see that ca(x) has the standard kinetic term for a complex scalar \ufb01eld. (We need the factor of i in front of Sgh in eq. (71.19) for this to work out; this factor a\ufb00ects only the overall phase of Z(J), and so we can choose it at will.) The ghost \ufb01eld is also a Grassmann \ufb01eld, and so a closed loop of ghost lines in a Feynman diagram carries an extra factor of minus one. We see from eq. (71.20) that the ghost \ufb01eld interacts with the gauge \ufb01eld, and so we will have such loops. Since the particles associated with the ghost \ufb01eld do not in fact exist (and would violate the spin-statistics theorem if they did), it must be that the amplitude to produce them in any scattering process is zero. This is indeed the case, as we will discuss in section 74. We note that in abelian gauge theory, where f abc = 0, there is no interaction term for the ghost \ufb01eld. In that case, it is simply an extra free \ufb01eld, and we can absorb its path integral into the overall normalization. We have one \ufb01nal trick to perform. Our gauge-\ufb01xing function, Ga(x) contains an arbitrary function \u03c9a(x). The path integral Z(J) is, however, independent of \u03c9a(x). So, we can multiply Z(J) by an arbitrary functional of \u03c9, and then perform a path integral over \u03c9; the result can change only the overall normalization of Z(J). In particular, let us multiply Z(J) by exp i 2\u03be Z \u2212 d4x \u03c9a\u03c9a. (71.21) Because of the delta-functional in eq. (71.16), it is easy to integrate over \u03c9. The \ufb01nal result for Z(J) is Z(J) A D D \u00afc D \u221d Z c exp iSYM + iSgh + iSgf, (71.22) where SYM is given by eq. (71.2), Sgh is given by the integral over d4x of eq. (71.20), and Sgf (gf stands for gauge \ufb01xing) is given by the integral over d4x of Lgf = \u2212 1 2 \u03be\u2212 1\u2202\u00b5Aa \u00b5\u2202\u03bdAa \u03bd. (71.23) In the next section, we will derive the Feynman rules that follow from this path integral. 72: The Feynman Rules for Nonabelian Gauge Theory 424 72 The Feynman Rules for Nonabelian Gauge Theory Prerequisite: 71 Let us begin by considering nonabelian gauge theory without any scalar or spinor \ufb01elds. The lagrangian is \u2202\u03bdAe\u00b5 + gf abeAa\u00b5Ab\u03bd )(\u2202\u00b5Ae \u2202\u03bdAe \u00b5 + gf cdeAc \u00b5Ad \u03bd ) \u03bd \u2212 LYM = = = \u2212 \u2212 \u2212 1 1 \u00b5\u03bd 4F e\u00b5\u03bdF e 4(\u2202\u00b5Ae\u03bd 2\u2202\u00b5Ae\u03bd\u2202\u00b5Ae \u03bd + 1 gf abeAa\u00b5Ab\u03bd \u2202\u00b5Ae \u2212 1 2 \u2202\u00b5Ae\u03bd\u2202\u03bdAe \u00b5 \u03bd \u2212 To this we should add the gauge-\ufb01xing term for R\u03be gauge, \u2212 1 4 g2f abef cdeAa\u00b5Ab\u03bdAc \u00b5Ad \u03bd. (72.1) (72.2) Lgf = \u2212 1 2 \u03be\u2212 1\u2202\u00b5Ae \u00b5\u2202\u03bdAe \u03bd. Adding eqs. (72.1) and (72.2), and doing some integrations-by-parts in the quadratic terms, we \ufb01nd LYM + Lgf = 1 \u2212 2 Ae\u00b5(g\u00b5\u03bd \u22022 \u2202\u00b5\u2202\u03bd )Ae\u03bd + 1 \u2212 gf abcAa\u00b5Ab\u03bd\u2202\u00b5Ac \u03bd \u2212 2 \u03be\u2212 1Ae\u00b5\u2202\u00b5\u2202\u03bdAe\u03bd 4 g2f abef c", "deAa\u00b5Ab\u03bdAc 1 \u00b5Ad \u03bd. (72.3) The \ufb01rst line of eq. (72.3) yields the gluon propagator in R\u03be gauge, \u03b4ab \u02dc\u2206ab \u00b5\u03bd(k) = g\u00b5\u03bd \u2212 The second line of eq. (72.3) yields three- and four-gluon vertices, shown in \ufb01g. (72.1). The three-gluon vertex factor is (72.4) i\u01eb k2 \u2212. k\u00b5k\u03bd k2 + \u03be k\u00b5k\u03bd k2 iVabc \u00b5\u03bd\u03c1(p, q, r) = i( gf abc)( ir\u00b5g\u03bd\u03c1) \u2212 + [ 5 permutations of (a,\u00b5,p), (b,\u03bd,q), (c,\u03c1,r) ] \u2212 = gf abc[(q r)\u00b5g\u03bd\u03c1 + (r \u2212 \u2212 p)\u03bdg\u03c1\u00b5 + (p \u2212 q)\u03c1g\u00b5\u03bd ]. (72.5) The four-gluon vertex factor is iVabcd \u00b5\u03bd\u03c1\u03c3 = = ig2f abef cdeg\u00b5\u03c1g\u03bd\u03c3 \u2212 + [ 5 permutations of (b,\u03bd), (c,\u03c1), (d,\u03c3) ] ig2 [ f abef cde(g\u00b5\u03c1g\u03bd\u03c3 \u2212 + f acef dbe(g\u00b5\u03c3g\u03c1\u03bd \u2212 + f adef bce(g\u00b5\u03bd g\u03c3\u03c1 \u2212 g\u00b5\u03c3g\u03bd\u03c1) g\u00b5\u03bd g\u03c1\u03c3) g\u00b5\u03c1g\u03c3\u03bd ) ]. \u2212 (72.6) 72: The Feynman Rules for Nonabelian Gauge Theory 425 Figure 72.1: The three-gluon and four-gluon vertices in nonabelian gauge theory. These vertex factors are quite a bit more complicated that the ones we are used to, and they lead to rather involved formulae for scattering cross sections. For example, the tree-level gg gg cross section (where g is a gluon), averaged over initial spins and colors and summed over \ufb01nal spins and colors, has 12,996 terms! Of course, many are identical and the \ufb01nal result can be expressed much more simply, but this is no help to us at the initial stages of computation. For this reason, we postpone any attempt at tree-level calculations until section 81, where we will make use of some techniques (color ordering and Gervais-Neveu gauge) that, combined with spinor-helicity methods, greatly reduce the necessary labor. \u2192 For loop calculations, we need to include the ghosts. The ghost la- grangian is Lgh = = = \u2212 \u2212 \u2212 \u2202\u00b5\u00afcbDbc \u00b5 cc \u2202\u00b5\u00afcc\u2202\u00b5cc + ig\u2202\u00b5\u00afcbAa \u00b5(T a A )bccc \u2202\u00b5\u00afcc\u2202\u00b5cc + gf abcAa \u00b5\u2202\u00b5\u00afcbcc. (72.7) The ghost propagator is \u02dc\u2206ab(k2) = \u03b4ab k2 \u2212. i\u01eb (72.8) Because the ghosts are complex scalars, their propagators carry a charge arrow. The ghost-ghost-gluon vertex shown in \ufb01g. (72.2); the associated vertex factor is iVabc \u00b5 (q, r) = i(gf abc)( \u2212 iq\u00b5) = gf abcq\u00b5. (72.9) 72: The Feynman Rules for Nonabelian Gauge Theory 426 c r b q a \u00b5 Figure 72.2: The ghost-ghost-gluon vertex in nonabelian gauge theory. j i a \u00b5 Figure 72.3: The quark-quark-gluon vertex in nonabelian gauge theory. If we include a quark coupled to the gluons, we have the quark la- grangian m\u03a8i\u03a8i Lq = i\u03a8i /Dij\u03a8j \u2212 = i\u03a8i/\u2202\u03a8i \u2212 m\u03a8i\u03a8i + gAa \u00b5\u03a8i\u03b3\u00b5T a ij\u03a8j. (72.10) The quark propagator is \u02dcSij(p) = ( \u2212 p2 + m2 /p + m)\u03b4ij i\u01eb \u2212. (72.11) The quark-quark-gluon vertex shown in \ufb01g. (72.3); the associated vertex factor is ij = ig\u03b3", "\u00b5T a ij. If the quark is in a representation R other than the fundamental, then T a ij becomes (T a (72.12) iV\u00b5a R )ij. Problems 72.1) Consider a complex scalar \ufb01eld \u03d5i in a representation R of the gauge group. Find the vertices that involve this \ufb01eld, and the associated vertex factors. 73: The Beta Function in Nonabelian Gauge Theory 427 73 The Beta Function in Nonabelian Gauge Theory Prerequisite: 70, 72 In this section, we will do enough loop calculations to compute the beta function for the Yang\u2013Mills coupling g. We can write the complete lagrangian, including Z factors, as = 1 2 Z3Aa\u00b5(g\u00b5\u03bd \u22022 \u2202\u00b5\u2202\u03bd )Aa\u03bd + 1 \u2212 L \u2212 Z3ggf abcAa\u00b5Ab\u03bd \u2202\u00b5Ac Z2\u2032\u2202\u00b5 \u00afC a\u2202\u00b5C a + Z1\u2032gf abcAc \u03bd \u2212 1 \u00b5\u2202\u00b5 \u00afC aC b 2 \u03be\u2212 1Aa\u00b5\u2202\u00b5\u2202\u03bdAa\u03bd 4 Z4gg2f abef cdeAa\u00b5Ab\u03bdAc \u00b5Ad \u03bd \u2212 + iZ2\u03a8i/\u2202\u03a8i \u2212 Zmm\u03a8i\u03a8i + Z1gAa \u00b5\u03a8i\u03b3\u00b5T a ij\u03a8j. (73.1) Note that the gauge-\ufb01xing term in the \ufb01rst line does not need a Z factor; we saw in section 62 that the \u03be-dependent term in the propagator is not renormalized. We see that g appears in several places in, and gauge invariance leads If in terms of bare \ufb01elds and parameters, and compare with us to expect that it will renormalize in the same way in each place. we rewrite eq. (73.1), we \ufb01nd that L L g2 0 = Z 2 1 Z 2 2 Z3 g2 \u02dc\u00b5\u03b5 = Z 2 1\u2032 Z 2 2\u2032Z3 g2 \u02dc\u00b5\u03b5 = Z 2 3g Z 3 3 g2 \u02dc\u00b5\u03b5 = g2 \u02dc\u00b5\u03b5, Z4g Z 2 3 (73.2) \u2212 where d = 4 \u03b5 is the number of spacetime dimensions. To prove eq. (73.2), we have to derive the nonabelian analogs of the Ward identities, known as Slavnov-Taylor identities. For now, we simply assume that eq. (73.2) holds; we will return to this issue in section 74. The simplest computation to perform is the renormalization of the quark-quark-gluon vertex. This is partly because much of the calculation is the same as it is in spinor electrodynamics, and so we can make use of our results in section 62. We then must compute Z1, Z2, and Z3. We will work in Feynman gauge, and use the MS renormalization scheme. We begin with Z2. The O(g2) corrections to the fermion propagator are shown in \ufb01g. (73.1). These diagrams are the same as in spinor electrodynamics, except for the factors related to the color indices. The loop diagram has a color-factor of (T aT a)ij = C(R)\u03b4ij. (Here we have allowed the quark to be in an arbitrary representation R; for notational simplicity, we will continue to omit the label R on the generator matrices.) In section 62, we found that, in spinor electrodynamics, the divergent part of this (e2/8\u03c02\u03b5)/p to the electron self-energy \u03a3(/p). Thus diagram contributes \u2212 73: The Beta Function in Nonabelian Gauge Theory 428 l p+ Figure 73.1: The one-loop and counterterm corrections to the quark propagator in quantum chromodynamics Figure 73.2: The one-loop corrections to the quark\u2013quark\u2013gluon vertex in in quantum chromodynamics. in Yang\u2013Mills gauge theory, the divergent part of this diagram contributes (g2/8\u03c02\u03b5)C(R)\u03b4ij /p to the quark self-energy \u03a3ij(/p). This divergent term \u2212 must be cancelled by the counterterm contribution of 1)\u03b4ij /p. There(Z2\u2212 fore, in Yang\u2013Mills theory, with a quark in the representation R, using Feynman gauge and the MS renormalization scheme,", " we have \u2212 C(R) Z2 = 1 g2 8\u03c02 Moving on to the quark-quark-gluon vertex, we the contributing oneloop diagrams are shown in \ufb01g. (73.2). The \ufb01rst diagram is again the same as it is in spinor electrodynamics, except for the color factor of (T bT aT b)ij. We can simplify this via + O(g4). (73.3) 1 \u03b5 \u2212 T bT aT b = T b T bT a + if abcT c = C(R)T a + 1 = C(R)T a + 1 = C(R)T a 2 if abc[T b, T c] 2 (if abc)(if bcd)T d A)cbT d A )bc(T d 2 (A) i 1 = C(R) h \u2212 In the second line, we used the complete antisymmetry of f abc to replace T bT c with 1 A ) = T (A)\u03b4ad. In 2 [T b, T c]. To get the last line, we used Tr(T a A T d (73.4) 73: The Beta Function in Nonabelian Gauge Theory 429 section 62, we found that, in spinor electrodynamics, the divergent part of this diagram contributes (e2/8\u03c02\u03b5)ie\u03b3\u00b5 to the vertex function iV\u00b5(p\u2032, p). Thus in Yang\u2013Mills theory, the divergent part of this diagram contributes C(R) \u2212 1 2 T (A) i g2 8\u03c02\u03b5 igT a ij\u03b3\u00b5 (73.5) h to the quark-quark-gluon vertex function iVa\u00b5 ij (p\u2032, p). This divergent term, along with any divergent term from the second diagram of \ufb01g. (73.2), must be cancelled by the tree-level vertex iZ1g\u03b3\u00b5T a ij. Now we must evaluate the second diagram of \ufb01g. (73.2). The divergent part is independent of the external momenta, and so we can set them to zero. Then we get a contribution to iVa\u00b5 ij (0, 0) of (ig)2gf abc(T cT b)ij 3 d4\u2113 1 i (2\u03c0)4 Z \u2113))\u00b5g\u03bd\u03c1 + ( ( \u2212 \u2212 \u03b3\u03c1( /\u2113+m)\u03b3\u03bd \u21132\u21132(\u21132+m2) 0)\u03bd g\u03c1\u00b5 + (0 \u2113 \u2212 \u2212 \u2212 [(\u2113 \u2212 \u00d7 \u2113)\u03c1g\u00b5\u03bd ]. (73.6) We can simplify the color factor with the manipulations of eq. (73.4), f abcT cT b = 1 = 1 2 f abc[T c, T b] 2 if abcf cbdT d 2 iT (A)T a. 1 \u2212 The numerator in eq. (73.6) is = (73.7) N \u00b5 = \u03b3\u03c1( \u2212 \u03b3\u03c3\u2113\u03c3 + m)\u03b3\u03bd(2\u2113\u00b5g\u03bd\u03c1 \u2113\u03bdg\u03c1\u00b5 \u2212 \u2212 \u2113\u03c1g\u00b5\u03bd ). (73.8) We can drop the terms linear in \u2113, and make the replacement \u2113\u03c3\u2113\u00b5 d\u2212 1\u21132g\u03c3\u00b5. Thus we have \u2192 N \u00b5 d\u2212 d\u2212 d\u2212 \u2192 \u2212 \u2192 \u2212 \u2192 \u2212 1\u21132 (\u03b3\u03c1\u03b3\u03c3\u03b3\u03bd)(2g\u03c3\u00b5g\u03bd\u03c1 1\u21132 (2\u03b3\u03bd \u03b3\u00b5\u03b3\u03bd \u2212 1\u21132 (2(d \u2212 \u03b3\u00b5\u03b3\u03bd\u03b3\u03bd \u2212 2) + d + d)\u03b3\u00b5. \u2212 g\u03c3\u03bd g\u03c1\u00b5 g\u03c3\u03c1g\u00b5\u03bd ) \u2212 \u03b3\u03c1\u03b3\u03c1\u03b3\u00b5) (73.9) Because we are only keeping track of the divergent term, we are free to set d = 4, which yields Using eqs. (73.7) and (73.10) in eq. (73.6), we get N \u00b5 \u2192 \u2212 3\u21132\u03b3\u00b5. 3 2 T (A) g3 T a ij\u03b3\u00b5 d4\u2113 (2\u03c0)4 1 \u21132(\u21132+m2). Z (73.10) (73.11) 73: The Beta Function in Nonabelian", " Gauge Theory 430 +l l c \u03c1 k+l l \u03c3d + Figure 73.3: The one-loop and counterterm corrections to the gluon propagator in quantum chromodynamics. After continuing to d dimensions, the integral becomes i/8\u03c02\u03b5 + O(\u03b50). Combining eqs. (73.5) and (73.6), we \ufb01nd that the divergent part of the quark-quark-gluon vertex function is h \u2212 Va\u00b5 ij (0, 0)div = Z1 + C(R) g2 8\u03c02\u03b5 1 2 T (A) i \u2212 + 3 2 T (A) g2 8\u03c02\u03b5! gT a ij\u03b3\u00b5. (73.12) Requiring Va\u00b5 ij (0, 0) to be \ufb01nite yields g2 8\u03c02 1 \u03b5 Z1 = 1 C(R) + T (A) + O(g4) (73.13) h in Feynman gauge and the MS renormalization scheme. i Note that we have found that Z1 does not equal Z2. In electrodynamics, we argued that gauge invariance requires all derivatives in the lagrangian ieA\u00b5 should to be covariant derivatives, and that both pieces of D\u00b5 = \u2202\u00b5 \u2212 therefore be renormalized by the same factor; this then implies that Z1 must equal Z2. In Yang\u2013Mills theory, however, this argument fails. This failure is due to the introduction of the ordinary derivative in the gauge\ufb01xing function for R\u03be gauge: once we have added LYM, we \ufb01nd that both ordinary and covariant derivatives appear. (This is especially obvious for Lgh.) Therefore, to be certain of what gauge invariance does and does not imply, we must derive the appropriate Slavnov-Taylor identities, a subject we will take up in section 74. Lgf and Lgh to Next we turn to the calculation of Z3. The O(g2) corrections to the gluon propagator are shown in \ufb01g. (73.3). The \ufb01rst diagram is proportional d4\u2113/\u21132; as we saw in section 65, this integral vanishes after dimensional to regularization. R 73: The Beta Function in Nonabelian Gauge Theory 431 The second diagram yields a contribution to i\u03a0\u00b5\u03bdab(k) of 1 2 g2f acdf bcd 1 i 2 Z d4\u2113 (2\u03c0)4 N \u00b5\u03bd \u21132(\u2113+k)2, (73.14) where the one-half is a symmetry factor, and N \u00b5\u03bd = [(k+\u2113) k [( \u2212 ( \u2212 \u2113) \u2113))\u00b5g\u03c1\u03c3 + ( \u2212 \u2113)\u03bdg\u03c1\u03c3 + (\u2113 \u2113 ( \u2212 \u2212 k)\u03c1\u03b4\u03c3 k))\u03c1g\u03c3\u00b5 + (( \u2212 \u03bd + (k k k) \u2212 \u2113))\u03c3\u03b4\u03bd (k+\u2113))\u03c3g\u00b5\u03c1] \u00d7 = \u2212 \u00d7 \u2212 \u2212 \u2212 [(2\u2113+k)\u00b5g\u03c1\u03c3 \u2212 [(2\u2113+k)\u03bd g\u03c1\u03c3 \u2212 (\u2113 (\u2113 \u2212 \u2212 \u2212 k)\u03c1g\u03c3\u00b5 \u03bd k)\u03c1\u03b4\u03c3 \u2212 \u2212 ( \u2212 \u2212 \u2212 (\u2113+2k)\u03c3g\u00b5\u03c1] (\u2113+2k)\u03c3\u03b4\u03bd \u03c1] \u03c1] (73.15) The color factor can be simpli\ufb01ed via f acdf bcd = T (A)\u03b4ab. We combine denominators with Feynman\u2019s formula, and continue to d = 4 \u03b5 dimensions; we now have \u2212 1 2 g2 T (A)\u03b4ab \u02dc\u00b5\u03b5 \u2212 1 dx 0 Z Z ddq (2\u03c0)d N \u00b5\u03bd (q2 + D)2, (73.16) where D = x(1 \u2212 x)k2 and q = \u2113 + xk. The numerator is N \u00b5\u03bd = \u2212 \u00d7 [(2q+(1 [(2q+(1 2x)k)\u00b5g\u03c1\u03c3 \u2212 2x)k)\u03bd g\u03c1\u03c3 \u2212 \u2212 \u2212 (1+x)k)\u03c1g\u03c3\u00b5 \u03bd (1+x)k)\u03c1\u03b4\u03c3 (q (q \u2212 \u2212 \u2212 \u2212 (q+(2 (q+(2 \u2212 \u2212 x)k)\u03c3g\u00b5\u03c1] x)k)\u03c3\u03b4\u03bd \u03c1]. (73.17) Terms", " linear in q will integrate to zero, and so we have N \u00b5\u03bd \u2192 \u2212 \u2212 \u2212 6)q\u00b5q\u03bd x)2]k2g\u00b5\u03bd \u2212 2q2g\u00b5\u03bd (4d [(1+x)2 + (2 [d(1 \u2212 \u2212 2x)2 + 2(1 x)(1+x) \u2212 2(2 \u2212 \u2212 \u2212 2x)(1+x) 2(2 \u2212 x)(1 \u2212 2x)]k\u00b5k\u03bd. (73.18) \u2212 Since we are only interested in the divergent part, we can go ahead and set 1 4 q2g\u00b5\u03bd. d = 4 in the numerator. We can also make the replacement q\u00b5q\u03bd Then we \ufb01nd \u2192 N \u00b5\u03bd 9 2 q2g\u00b5\u03bd (5 \u2212 \u2212 \u2192 \u2212 2x+2x2)k2g\u00b5\u03bd + (2+10x \u2212 10x2)k\u00b5k\u03bd. (73.19) We saw in section 62 that, when integrated against (q2 + D)\u2212 replaced with ( 2 1D; in our case this is 2x(1 x)k2. This yields 2, q2 can be 1)\u2212 d \u2212 (5 \u2212 11x+11x2)k2g\u00b5\u03bd + (2+10x \u2212 10x2)k\u00b5k\u03bd. (73.20) N \u00b5\u03bd \u2192 \u2212 \u2212 \u2212 73: The Beta Function in Nonabelian Gauge Theory 432 We now use ddq (2\u03c0)d 1 (q2 + D)2 = i 8\u03c02\u03b5 + O(\u03b50) \u02dc\u00b5\u03b5 Z (73.21) in eq. (73.16) to get ig2 16\u03c02 T (A)\u03b4ab 1 \u03b5 \u2212 0 Z Performing the integral over x yields 1 dx N \u00b5\u03bd + O(\u03b50). (73.22) ig2 16\u03c02 T (A)\u03b4ab 1 \u03b5 \u2212 19 6 k2g\u00b5\u03bd + 11 3 k\u00b5k\u03bd \u2212 (73.23) as the divergent contribution of the second diagram to i\u03a0\u00b5\u03bdab(k). Next we have the third diagram of \ufb01g. (73.3), which makes a contribu- tion to i\u03a0\u00b5\u03bdab(k) of 1)g2f acdf bdc ( \u2212 2 1 i Z d4\u2113 (2\u03c0)4 (\u2113+k)\u00b5\u2113\u03bd \u21132(\u2113+k)2, (73.24) where the factor of minus one is from the closed ghost loop. The color factor T (A)\u03b4ab. After combining denominators, the numerator is f acdf bdc = becomes \u2212 (\u2113+k)\u00b5\u2113\u03bd = (q + (1 \u2212 x)k)\u00b5(q xk)\u03bd \u2212 x)k\u00b5k\u03bd 1 4 q2g\u00b5\u03bd \u2192 1 2 x(1 \u2212 \u2192 \u2212 x(1 \u2212 \u2212 x)k2g\u00b5\u03bd x(1 \u2212 \u2212 x)k\u00b5k\u03bd. (73.25) We then use eq. (73.21) in eq. (73.24); performing the integral over x yields ig2 8\u03c02 T (A)\u03b4ab 1 \u03b5 \u2212 1 12 k2g\u00b5\u03bd 1 6 k\u00b5k\u03bd \u2212 \u2212 (73.26) as the divergent contribution of the third diagram to i\u03a0\u00b5\u03bdab(k). Finally, we have the fourth diagram. This is the same as it is in spinor electrodynamics, except for the color factor of Tr(T aT b) = T (R)\u03b4ab. If there is more than one \ufb02avor of quark, each contributes separately, leading to a factor of the number of \ufb02avors nF. Then, using our results in section 62, we \ufb01nd ig2 6\u03c02 nFT (R)\u03b4ab 1 \u03b5 \u2212 k2g\u00b5\u03bd k\u00b5k\u03bd \u2212 (73.27) as the divergent contribution of the fourth diagram to i\u03a0\u00b5\u03bdab(k). 73: The Beta Function in Nonabelian Gauge Theory 433 Adding up eqs. (73.23), (73.26), and (73.27), as well as the counterterm contriubtion, we \ufb01nd that the gluon self-energy is transverse, \u03a0\u00b5\u03bdab(k) = \u03a0(k2)(k2g\u00b5\u03bd k\u00b5k\u03bd)\u03b4ab, \u2212", " (73.28) and that \u03a0(k2)div = (Z3\u2212 1) + \u2212 Thus we \ufb01nd 5 3 T (A) \u2212 4 3 nFT (R) i h g2 8\u03c02 1 \u03b5 + O(g4). (73.29) Z3 = 1 + 5 3 T (A) h \u2212 4 3 nFT (R) i g2 8\u03c02 1 \u03b5 + O(g4) (73.30) in Feynman gauge and the MS renormalization scheme. Let us collect our results: Z1 = 1 Z2 = 1 \u2212 \u2212 g2 8\u03c02 1 \u03b5 + O(g4), C(R) + T (A) i h C(R) g2 8\u03c02 1 \u03b5 + O(g4), (73.31) (73.32) Z3 = 1 + 5 3 T (A) h \u2212 4 3 nFT (R) i g2 8\u03c02 1 \u03b5 + O(g4), (73.33) in Feynman gauge and the MS renormalization scheme. We de\ufb01ne g2 4\u03c0. \u03b1 \u2261 \u03b10 = Z 2 1 Z 2 2 Z3 \u03b1\u02dc\u00b5\u03b5. Then we have Let us write ln Then we have \u221e ln \u03b10 = Xn=1 From eqs. (73.31\u201373.33), we get \u221e = Xn=1 Gn(\u03b1) \u03b5n. Gn(\u03b1) \u03b5n + ln \u03b1 + \u03b5 ln \u02dc\u00b5. (73.34) (73.35) (73.36) (73.37) 4 3 nFT (R) i Then, the general analysis of section 28 yields 11 3 T (A) G1(\u03b1) = \u2212 \u2212 h \u03b1 2\u03c0 + O(\u03b12). (73.38) \u03b2(\u03b1) = \u03b12G\u20321(\u03b1), (73.39) 73: The Beta Function in Nonabelian Gauge Theory 434 where the prime denotes di\ufb00erentiation with respect to \u03b1. Thus we \ufb01nd \u03b2(\u03b1) = 11 3 T (A) \u2212 4 3 nFT (R) i \u2212 h \u03b12 2\u03c0 + O(\u03b13). (73.40) in nonabelian gauge theory with nF Dirac fermions in the representation R of the gauge group. We can, if we like, restate eq. (73.40) in terms of g as \u03b2(g) = 11 3 T (A) \u2212 4 3 nFT (R) i \u2212 h g3 16\u03c02 + O(g5). (73.41) To go from eq. (73.40) to eq. (73.41), we use \u03b1 = g2/4\u03c0 and \u02d9\u03b1 = g \u02d9g/2\u03c0, where the dot denotes d/d ln \u00b5. In quantum chromodynamics, the gauge group is SU(3), and the quarks are in the fundamental representation. Thus T (A) = 3 and T (R) = 1 2, and 16, the the factor in square brackets in eq. (73.41) is 11 beta function is negative: the gauge coupling in quantum chromodynamics gets weaker at high energies, and stronger at low energies. 3 nF. So for nF \u2264 \u2212 2 This has dramatic physical consequences. Perturbation theory cannot serve as a reliable guide to the low-energy physics. And indeed, in nature we do not see isolated quarks or gluons. (Quarks, in particular, have fractional electric charges and would be easy to discover.) The appropriate conclusion is that color is con\ufb01ned: all \ufb01nite-energy states are invariant under a global SU(3) transformation. This has not yet been rigorously proven, but it is the only hypothesis that is consistent with all of the available theoretical and experimental information. Problems 73.1) Compute the beta function for g in Yang\u2013Mills theory with a complex scalar \ufb01eld in the representation R of the gauge group. Hint: all the real work has been done already in this section, problem 72.1, and section 66. 73.2) Write down the beta function for the gauge coupling in Yang\u2013Mills theory with several Dirac fermions in the representations Ri, and several complex scalars in the representations R\u2032j. 73.3) Compute the one-loop contributions to the anomalous dimensions of m, \u03a8, and A\u00b5. 74: BRST Symmetry 435 74 BRST Symmetry Prerequisite: 70, 71 Lgf + In this section we will rederive the gauge-\ufffd", "\ufffd\ufffdxed path integral for nonabelian gauge theory from a di\ufb00erent point of view. We will discover that the Lgh, still has a residual L \u2261 LYM + complete gauge-\ufb01xed lagrangian, form of the gauge symmetry, known as Becchi-Rouet-Stora-Tyutin symmetry, or BRST symmetry for short. BRST symmetry can be used to derive the Slavnov-Taylor identities that, among other useful things, show that the coupling constant is renormalized by the same factor at each of its appearances in. Also, we can use BRST symmetry to show that gluons whose polarizations are not both spacelike and transverse (perpendicular to the four-momentum) decouple from physical scattering amplitudes (as do particles that are created by the ghost \ufb01eld). L Consider a nonabelian gauge theory with a gauge \ufb01eld Aa \u00b5(x), and a scalar or spinor \ufb01eld \u03c6i(x) in the representation R. Then, under an in\ufb01nitesimal gauge transformation parameterized by \u03b8a(x), we have \u03b4Aa \u00b5(x) = \u03b4\u03c6i(x) = \u2212 \u2212 Dab\u03b8b(x), ig\u03b8a(x)(T a R )ij\u03c6j(x). (74.1) (74.2) We now introduce a scalar Grassmann \ufb01eld ca(x) in the adjoint representation; this \ufb01eld will turn out to be the ghost \ufb01eld that we introduced in section 71. We de\ufb01ne an in\ufb01nitesimal BRST transformation via \u03b4BAa \u00b5(x) Dab \u00b5 cb(x) \u2261 = \u2202\u00b5ca(x) \u03b4B\u03c6i(x) \u2261 gf abcAc \u00b5(x)cb(x), \u2212 igca(x)(T a R )ij\u03c6j(x). (74.3) (74.4) (74.5) This is simply an in\ufb01nitesimal gauge transformation, with the ghost \ufb01eld \u03b8a(x). Therefore, any combica(x) in place of the in\ufb01nitesimal parameter nation of \ufb01elds that is gauge invariant is also BRST invariant. In particular, the Yang\u2013Mills lagrangian LYM (including the appropriate lagrangian for the scalar or spinor \ufb01eld \u03c6i) is BRST invariant, \u2212 \u03b4BLYM = 0. We now place a further restriction on the BRST transformation: we require a BRST variation of a BRST variation to be zero. This requirement will determine the BRST transformation of the ghost \ufb01eld. Consider (74.6) \u03b4B(\u03b4B\u03c6i) = ig(\u03b4Bca)(T a R )ij\u03c6j \u2212 igca(T a R )ij\u03b4B\u03c6j. (74.7) 74: BRST Symmetry 436 There is a minus sign in front of the second term because \u03b4B acts as an anticommuting object, and it generates a minus sign when it passes through another anticommuting object, in this case ca. Using eq. (74.5), we have \u03b4B(\u03b4B\u03c6i) = ig(\u03b4Bca)(T a R )ij\u03c6j \u2212 cacb in the second term to replace T a \u2212 R, T b 2 [T a R with its antiR. Then, after relabeling some dummy R)ik\u03c6k. R T b 2 f abcT c R] = i (74.8) g2cacb(T a R T b We now use cbca = symmetric part, 1 indices, we have \u03b4B(\u03b4B\u03c6i) = ig(\u03b4Bcc + 1 2 gf abccacb)(T c R)ij\u03c6j. (74.9) The right-hand side of eq. (74.9) will vanish for all \u03c6j(x) if and only if \u03b4Bcc(x) = \u2212 1 2 gf abcca(x)cb(x). (74.10) We therefore adopt eq. (74.10) as the BRST variation of the ghost \ufb01eld. Let us now check to see that the BRST variation of the BRST variation of the gauge \ufb01eld also vanishes. From eq. (74.4) we have \u03b4B(\u03b4BAa \u00b5", ") = (\u03b4ab\u2202\u00b5 \u2212 \u00b5 (\u03b4Bcb) = Dab gf abc(Dcd \u2212 \u00b5 cd)cb gf abcAc \u00b5)(\u03b4Bcb) gf abc(\u03b4BAc \u00b5)cb \u2212 \u2212 = Dab \u00b5 (\u03b4Bcb) gf abc(\u2202\u00b5cc)cb + g2f abcf cdeAe \u00b5cdcb. (74.11) We now use the antisymmetry of f abc in the second term to replace (\u2202\u00b5cc)cb with its antisymmetric part, 1 1 (\u2202\u00b5c[c)cb] 2 (\u2202\u00b5cc)cb \u2212 2 (\u2202\u00b5cc)cb + 1 2 \u2202\u00b5(cccb). Similarly, we use the antisymmetry of cdcb in the third term to replace f abcf cde with its antisymmetric part, 2 (\u2202\u00b5cb)cc 2 cc(\u2202\u00b5cb) \u2261 = 1 (74.12) = 1 A )ce 1 1 1 \u2212 \u2212 = \u2212 A)ac(T d f adcf cbe) = 2 (f abcf cde 2 [(T b A )ae 2 if bdh(T h 2 f bdhf hae, which is just the Jacobi identity. Now we have 2 gf abc(\u2202\u00b5cccb) (\u03b4ah\u2202\u00b5 \u2212 \u2212 \u00b5 (\u03b4Bch + 1 2 gf bchcbcc). \u00b5) = Dab \u00b5 (\u03b4Bch) \u00b5 (\u03b4Bcb) \u03b4B(\u03b4BAa = Dah = Dah \u2212 = \u2212 1 1 \u2212 gf aheAe (T d A)ac(T b A)ce] \u2212 \u00b5cdcb 1 2 g2f bdhf haeAe 2 gf bchcccb \u00b5) 1 (74.13) (74.14) 74: BRST Symmetry 437 We see that this vanishes if the BRST variation of the ghost \ufb01eld is given by eq. (74.10). Now we introduce the antighost \ufb01eld \u00afca(x). We take its BRST trans- formation to be \u03b4B\u00afca(x) = Ba(x), where Ba(x) is a commuting (as opposed to Grassmann) scalar \ufb01eld, the Lautrup-Nakanishi auxiliary \ufb01eld. Because Ba(x) is itself a BRST variation, we have (74.15) \u03b4BBa(x) = 0. (74.16) Note that eq. (74.15) is in apparent contradiction with eq. (74.10). However, there is actually no need to identify \u00afca(x) as the hermitian conjugate of ca(x). The role of these \ufb01elds (in producing the functional determinant that must accompany the gauge-\ufb01xing delta functional) is ful\ufb01lled as long as ca(x) and \u00afca(x) are treated as independent when we integrate them; whether or not they are hermitian conjugates of each other is irrelevant. We identi\ufb01ed them as hermitian conjugates in section 71 only for the sake of familiarity in deriving the associated Feynman rules. Now, however, we must abandon this notion. In fact, it will be most convenient to treat ca(x) and \u00afca(x) as two real Grassmann \ufb01elds. Now that we have introduced a collection of new \ufb01elds\u2014ca(x), \u00afca(x), and Ba(x)\u2014what are we to do with them? Consider adding to object, O LYM a new term that is the BRST variation of some (74.17) =. LYM + \u03b4BO L is BRST invariant, because ) = 0. corresponds to \ufb01xing a gauge; which gauge we LYM is, and because \u03b4B(\u03b4BO Clearly We will see that adding \u03b4BO get depends on L. O We will choose O (x) = \u00afca(x) h 1 2 \u03beBa(x) \u2212 Ga(x) i, (74.18) where Ga(x) is a gauge-\ufb01xing function, and \u03be is a parameter. If we further choose Ga(x) = \u2202\u00b5Aa \u00b5(x), (74.19) then we end up with R\u03be gauge. Let us see how this works. We have \u03b4BO = (\u03b4B\u00afca)", " h 1 2 \u03beBa \u2212 \u2202\u00b5Aa \u00b5 \u2212 i h \u00afca 1 2 \u03be(\u03b4BBa) \u2212 \u2202\u00b5(\u03b4BAa \u00b5) i. (74.20) There is a minus sign in front of the second set of terms because \u03b4B acts as an anticommuting object, and so it generates a minus sign when it 74: BRST Symmetry 438 passes through another anticommuting object, in this case \u00afca. Now using eqs. (74.3), (74.15), and (74.16), we get \u03b4BO = 1 2 \u03beBaBa \u2212 Ba\u2202\u00b5Aa \u00b5 + \u00afca\u2202\u00b5Dab \u00b5 cb. (74.21) Lgh that we found in We see that the last term is the ghost lagrangian section 71. If we like, we can integrate the ordinary derivative by parts, so that it acts on the antighost \ufb01eld, \u03b4BO \u2192 1 2 \u03beBaBa Ba\u2202\u00b5Aa \u00b5 \u2212 \u2212 \u2202\u00b5\u00afcaDab \u00b5 cb. (74.22) Examining the \ufb01rst two terms in eq. (74.22), we see that no derivatives act on the auxiliary \ufb01eld Ba(x). Furthermore, it appears only quadratically. We can, therefore, perform the path integral over it; and linearly in \u03b4BO the result is equivalent to solving the classical equation of motion ) \u2202(\u03b4BO \u2202Ba(x) = \u03beBa(x) \u2212 \u2202\u00b5Aa \u00b5(x) = 0, (74.23) and substituting the result back into \u03b4BO \u00b5\u2202\u03bdAa \u03b4BO \u2192 \u2212 1\u2202\u00b5Aa 1 2 \u03be\u2212 \u03bd \u2212. This yields \u2202\u00b5\u00afcaDab \u00b5 cb. (74.24) = We see that the \ufb01rst term is the gauge-\ufb01xing lagrangian in section 71. We now take note of all the symmetries of the action S = LYM +\u03b4BO Lgf that we found d4x, where, they are: (1) Lorentz invariance; (2) L O the discrete symmetries of parity, time reversal, and charge conjugation; (3) global gauge invariance (that is, invariance under a gauge transformation with a spacetime-independent parameter \u03b8a); (4) BRST invariance; (5) ghost number conservation; and (6) antighost translation invariance.. With our choice of L R Global gauge invariance simply requires every term in to have all the group indices contracted in a group-invariant manner. Ghost number conservation corresponds to assigning ghost number +1 to ca, 1 to \u00afca, and to have ghost number zero to all other \ufb01elds, and requiring every term in zero. Antighost translation invariance corresponds to \u00afca(x) \u00afca(x) + \u03c7, \u2192 invariant because, in the where \u03c7 is a Grassmann constant. This leaves form of eq. (74.22), contains only a derivative of \u00afcb(x). \u2212 L L L L We now claim that already includes all terms consistent with these symmetries that have coe\ufb03cients with positive or zero mass dimension. This means that we will not encounter any divergences in perturbation theory that cannot be absorbed by including a Z factor for each term in. Furthermore, loop corrections should respect the symmetries, and BRST symmetry requires that g renormalize in the same way at each of L L 74: BRST Symmetry 439 its appearances. (Filling in the mathematical details of these claims is a lengthy project that we will not undertake.) We can regard a BRST transformation as in\ufb01nitesimal, and hence con- struct the associated Noether current via the standard formula j\u00b5 B (x) = XI \u2202 L \u2202(\u2202\u00b5\u03a6I(x)) \u03b4B\u03a6I (x), (74.25) where \u03a6I(x) stands for all the \ufb01elds, including the matter (scalar and/or spinor), gauge, ghost, antighost, and auxiliary \ufb01elds. We can then de\ufb01ne the BRST charge QB = d3x j0 B(x). (74.26) If we think of ca(x) and \u00afca(x) as independent hermitian \ufb01elds, then QB is hermitian. The BRST charge generates a BRST transformation, Z i[QB", ", Aa \u00b5(x)] = Dab \u00b5 cb(x), 1 2 gf abccb(x)cc(x), = \u2212 QB, ca(x) i { } QB, \u00afca(x) i } { i[QB, Ba(x)] = 0, = Ba(x), i[QB, \u03c6i(x)] \u00b1 = igca(x)(T a R )ij\u03c6j(x). (74.27) (74.28) (74.29) (74.30) (74.31) } A, B { is the commu= AB + BA is the anticommutator, and [, ] where tator if \u03c6i is a scalar \ufb01eld, and the anticommutator if \u03c6i is a spinor \ufb01eld. Also, since the BRST transformation of a BRST transformation is zero, QB must be nilpotent, \u00b1 Q2 B = 0. (74.32) i \u03c8 | Eq. (74.32) has far-reaching consequences. In order for it to be satis\ufb01ed, many states must be annihilated by QB; such states are said to be in the kernel of QB. A state which is annihilated by QB may take the form of QB acting on some other state; such states are said to be in the image of QB. There may be some states in the kernel of QB that are not in the image; such states are said to be in the cohomology of QB. Two states in the cohomology of QB are identi\ufb01ed if their di\ufb00erence is in the image; that, and if \u03c7 \u03c7 = QB| \u03b6 + QB| \u03c8 is, if QB| = | i i 6 i for some state as a single element of the \u03c8\u2032i, then we identify | i cohomology of QB. = 0 but \u03b6 | for any state \u03c8\u2032i | and Note any state in the image of QB has zero norm, since if, i = 0. (Here we have used the hermiticity of QB to \u03c7 = QB| \u03c8 | = i \u03c8 h \u03c7 then \u03c8 \u03c8 QB| | i h | = 0 implies \u03c8 conclude that QB| i i \u03c8 h QB = 0.) | 74: BRST Symmetry 440 i \u03c8 | \u03c8 h Now consider starting at some initial time with a normalized state \u03c8 i |. (This last equation \u03c7 = QB| \u03c8 = 1, QB| in the cohomology: \u03c8 = 0, i i 6 | is actually redundant, because if \u03c8 for some state \u03c7 = QB| \u03c8 i | i | is BRST invariant, the hamiltonian that we derive has zero norm.) Since from it must commute with the BRST charge: [H, QB] = 0. Thus, an that is annihilated by QB must still be annihilated by it at \u03c8 initial state | iHtQB| later times, since QBe\u2212 = 0. Also, since unitary time \u03c8 evolution does not change the norm of a state, the time-evolved state must still be in the cohomology., then i = e\u2212 \u03c8 | \u03c7 | iHt i i L i i i We now claim that the physical states of the theory correspond to the cohomology of QB. We have already shown that if we start with a state in the cohomology, it remains in the cohomology under time evolution. Consider, then, an initial state of widely separated wave packets of incoming particles. According to our discussion in section 5, we can treat these states as being created by the appropriate Fourier modes of the \ufb01elds, and ignore interactions. We will suppress the group index (because it plays no essential role when interactions can be neglected) and write the mode expansions A\u00b5(x) = \u03bb=>,<, X +,\u2212 Z dk f \u2217\u03bb (k)a\u03bb(k)eikx + \u03b5\u00b5 \u03b5\u00b5 h \u03bb(k)a\u2020\u03bb(k)e\u2212 ikx, (74.33) i c(x) = \u00afc(x) = \u03c6(x) = dk f dk f dk Z Z Z ikx c(k)eikx + c\u2020(k)e\u2212 h b(k)eikx + b\u2020(k)e\u2212 h a\u03c6(k)eikx + a\u2020\u03c6(k)e\u2212 h ikx i,, i ikx, i (74.34) (74.35) (74.36) f Here", ", for maximum simplicity, we have taken \u03c6(x) to be a real scalar \ufb01eld. (This is possible if R is a real representation.) In eq. (74.33), we have included four polarization vectors that span four-dimensional spacetime. For k\u00b5 = (\u03c9, k) = \u03c9(1, 0, 0, 1), we choose these four polarization vectors to be \u03b5\u00b5 >(k) = 1 \u221a2 \u03b5\u00b5 <(k) = 1 \u221a2 \u03b5\u00b5 +(k) = 1 \u221a2 \u03b5\u00b5 \u2212 (k) = 1 \u221a2 (1, 0, 0, 1), (1, 0, 0, (0, 1, \u2212 1), \u2212 i, 0), (0, 1, +i, 0). (74.37) The \ufb01rst two of these, > and <, are lightlike vectors; \u03b5\u00b5 k\u00b5, and \u03b5\u00b5 <(k) is spatially opposite. The latter two, + and >(k) is parallel to, are spacelike \u2212 74: BRST Symmetry 441 and transverse: they correspond to physical photon polarizations of de\ufb01nite helicity. We set g = 0, plug eqs. (74.33\u201374.36) into eqs. (74.27\u201374.31), and use ikx, eq. (74.23) to eliminate the auxiliary \ufb01eld. Matching coe\ufb03cients of e\u2212 we \ufb01nd [QB, a\u2020\u03bb(k)] = \u221a2\u03c9 \u03b4\u03bb> c\u2020(k), = 0, QB, c\u2020(k) } { QB, b\u2020(k) { } [QB, a\u2020\u03c6(k)] = 0. = \u03be\u2212 1\u221a2\u03c9 a\u2020<(k), (74.38) (74.39) (74.40) (74.41, QB| in the cohomology: = 0. Consider a normalized state Eq. (74.38) tells us that if we add a photon with the unphysical polarization with a\u2020>(k), then this state is not annihilated by QB; > by acting on hence the state a\u2020>(k) is not in the cohomology. Eq. (74.40) tells us that \u03c8 | the state a\u2020<(k) ; hence the state a\u2020<(k) is proportional to QBb\u2020(k) \u03c8 \u03c8 \u03c8 | i i | | is also not in the cohomology. On the other hand, the states a\u2020+(k) and \u03c8 | a\u2020 are annihilated by QB, but they cannot be written as QB acting \u2212 on some other state; hence these states are in the cohomology. Also, by similar reasoning, the state with one extra \u03c6 particle, a\u2020\u03c6(k), is in the \u03c8 | cohomology. (k) \u03c8 | i i i i i i Eq. (74.38) tells us that if we add a ghost particle by acting on \u03c8 i | with c\u2020(k), then this state is proportional to QBa\u2020>(k) \u03c8 ; hence the state | c\u2020(k) \u03c8 is not in the cohomology. Eq. (74.40) tells us that if we add an | with b\u2020(k), then this state is not anniantighost particle by acting on hilated by QB; hence the state b\u2020(k) is also not in the cohomology. \u03c8 | i We conclude that the only particle creation operators that do not take (k). Of course, it a state out of the cohomology are a\u2020\u03c6(k), a\u2020+(k), and a\u2020 \u2212 is precisely these operators that create the expected physical particles. \u03c8 | i i Finally, we note that the vacuum must be in the cohomology, because it is the unique state with zero energy and positive norm. Thus we can conclude that we can build an initial state of widely separated particles that is in the cohomology only if we do not include any ghost or antighost particles, or photons with polarizations other than + and. Since a state in the cohomology must evolve to another state in the cohomology, no ghosts, antighosts, or unphysically polarized photons can be produced in the scattering process. \u2212 Reference Notes A detailed treatment of BRST symmetry can be found in Weinberg II. 0 i | 74: BRST Symmetry 442 Problems 74.1) The creation operator for a photon of positive helicity can be written as a\u2020+(k)", " = i \u03b5\u00b5 \u2217+ (k) \u2212 Z d3x e+ikx \u2194\u22020 A\u00b5(x). (74.42) Consider the state a\u2020+(k) \u03c8 | De\ufb01ne a gauge-transformed polarization vector, where i \u03c8 | i is in the BRST cohomology. +(k) = \u03b5\u00b5 \u02dc\u03b5\u00b5 +(k) + ck\u00b5, (74.43) where c is a constant, and a corresponding creation operator \u02dca\u2020+(k). Show that = a\u2020+(k) \u02dca\u2020+(k) \u03c8 \u03c8 \u03c7 + QB| i | | and a\u2020+(k) which implies that \u02dca\u2020+(k) represent the same element \u03c8 \u03c8 | | of the cohomology, and hence are physically equivalent. Find the state \u03c7 | (74.44). i i i i i, 75: Chiral Gauge Theories and Anomalies 443 75 Chiral Gauge Theories and Anomalies Prerequisite: 70, 72 So far, we have only discussed gauge theories with Dirac fermion \ufb01elds. Recall that a Dirac \ufb01eld \u03a8 can be written in terms of two left-handed Weyl \ufb01elds \u03c7 and \u03be as \u03a8 = \u03c7 \u03be\u2020!. (75.1) If \u03a8 is in a representation R of the gauge group, then \u03c7 and \u03be\u2020 must be as well. Equivalently, \u03c7 must be in the representation R, and \u03be must be in the complex conjugate representation R. (For an abelian theory, this means Q.) that if \u03a8 has charge +Q, then \u03c7 has charge +Q and \u03be has charge Thus a Dirac \ufb01eld in a representation R is equivalent to two left-handed Weyl \ufb01elds, one in R and one in R. \u2212 If the representation R is real, then we can have a Majorana \ufb01eld \u03a8 = \u03c8 \u03c8\u2020! (75.2) instead of a Dirac \ufb01eld; the left-handed Weyl \ufb01eld \u03c8 and and its hermitian conjugate \u03c8\u2020 are both in the representation R. Thus a Majorana \ufb01eld in a real representation R is equivalent to a single left-handed Weyl \ufb01eld in R. Now suppose that we have a single left-handed Weyl \ufb01eld \u03c8 in a complex representation R. Such a gauge theory is automatically parity violating (because the right-handed hermitian conjugate of the left-handed Weyl \ufb01eld is in an inequivalent represetnation of the gauge group), and is said to be chiral. The lagrangian is = i\u03c8\u2020 \u00af\u03c3\u00b5D\u00b5\u03c8 L R. Since T a \u00b5T a 1 4 F a\u00b5\u03bdF a \u00b5\u03bd, \u2212 (75.3) igAa R is a hermitian matrix (even when R where D\u00b5 = \u2202\u00b5 \u2212 is a complex representation), i\u03c8\u2020 \u00af\u03c3\u00b5D\u00b5\u03c8 is hermitian (up to a total divergence, as usual). We cannot include a mass term for \u03c8, though, because \u03c8\u03c8 transforms as R R does not contain a singlet if R is complex. Thus, \u03c8\u03c8 is not gauge invariant. But without a mass term, this lagrangian would appear to possess all the required properties: Lorentz invariance, gauge invariance, and no terms with coe\ufb03cients with negative mass dimension. R, and R \u2297 \u2297 However, it turns out that most chiral gauge theories do not exist as quantum \ufb01eld theories; they are anomalous. The problem can ultimately be traced back to the functional measure for the fermion \ufb01eld; it turns out 75: Chiral Gauge Theories and Anomalies 444 that this measure is, in general, not gauge invariant. We will explore this surprising fact in section 77. For now we will content ourselves with analyzing Feynman diagrams. We will \ufb01nd an insuperable problem with gauge invariance at the one-loop level that a\ufb04icts most chiral gauge theories. We will work with the simplest possible example: a U(1) theory with a single Weyl \ufb01eld \u03c8 with charge +1. The lagrangian is = i\u03c8\u2020 \u00af\u03c3\u00b5(\u2202\u00b5 \u2212 4 F \u00b5\u03bdF\u00b5\u03bd. igA\u00b5)\u03c8 \u2212", " L 1 (75.4) We can use the following trick to write this theory in terms of a Dirac \ufb01eld \u03a8. We note that PL\u03a8 =, (75.5) \u03c8 0! where PL = 1 the right-handed components of \u03a8. Then we can write eq. (75.4) as \u03b35) is the left-handed projection matrix, does not involve 2 (1 \u2212 = i\u03a8\u03b3\u00b5(\u2202\u00b5 \u2212 L igA\u00b5)PL\u03a8 1 4 F \u00b5\u03bdF\u00b5\u03bd, \u2212 (75.6) and treat \u03a8 as a Dirac \ufb01eld when we derive the Feynman rules. To better understand the physical consequences of eq. (75.5), consider the case of a free \ufb01eld. The mode expansion is PL\u03a8(x) = dp s= X \u00b1 Z f bs(p)PLus(p)eipx + d\u2020s(p)PLvs(p)e\u2212 h ipx i. (75.7) For a massless \ufb01eld, we learned in section 38 that PLu+(p) = 0 and PLv (p) = 0. Thus we can write eq. (75.7) as \u2212 PL\u03a8(x) = dp b Z h f (p)u \u2212 \u2212 (p)eipx + d\u2020+(p)v+(p)e\u2212 ipx. (75.8) i \u2212 Eq. (75.8) shows us that there are only two kinds of particles associated with (p) creates a particle this \ufb01eld (as opposed to four with a Dirac \ufb01eld): b\u2020 \u2212 1/2, and d\u2020+(p) creates a particle with charge with charge +1 and helicity 1 and helicity +1/2. In this theory, charge and spin are correlated. \u2212 We can easily read the Feynman rules o\ufb00 of eq. (75.6). In particular, PL /p/p2, and the fermion\u2013 the fermion propagator in momentum space is fermion\u2013photon vertex is ig\u03b3\u00b5PL. \u2212 When we go to evaluate loop diagrams, we need a method of regulating the divergent integrals. However, our usual choice, dimensional regularization, is problematic, due to the close connection between \u03b35 and four-dimensional spacetime. In particular, in four dimensions we have Tr[\u03b35\u03b3\u00b5\u03b3\u03bd\u03b3\u03c1\u03b3\u03c3] = 4i\u03b5\u00b5\u03bd\u03c1\u03c3, \u2212 (75.9) 75: Chiral Gauge Theories and Anomalies 445 k+l l k k Figure 75.1: The one-loop and counterterm corrections to the photon propagator. where \u03b50123 = +1. It is not obvious what should be done with this formula i\u03b30\u03b31\u03b32\u03b33. in d dimensions. One possibility is to take d > 4 and de\ufb01ne \u03b35 \u2261 Then eq. (75.9) holds, but with each of the four vector indices restricted to = 0 for \u00b5 = 0, 1, 2, 3, but [\u03b3\u00b5, \u03b35] = 0 \u03b3\u00b5, \u03b35} span 0, 1, 2, 3. We also have { for \u00b5 > 3. This approach is workable, but cumbersome in practice. It is therefore tempting to abandon dimensional regularization in favor of, say, Pauli\u2013Villars regularization, which involves the replacement PL /p p2 \u2192 PL /p \u2212 p2 \u2212 /p + \u039b \u2212 p2 + \u039b2. (75.10) Pauli\u2013Villars regularization is equivalent to adding an extra fermion \ufb01eld with mass \u039b, and a propagator with the wrong sign (corresponding to changing the signs of the kinetic and mass terms in the lagrangian). But, a Dirac \ufb01eld with a chiral coupling to the gauge \ufb01eld cannot have a mass, since the mass term would not be gauge invariant. So, in a chiral gauge theory, Pauli\u2013Villars regularization violates gauge invariance, and hence is unacceptable. Given the di\ufb03culty with regulating chiral gauge theories (which is a hint that they may not make sense), we will sidestep the issue for now, and see what we can deduce about loop diagrams without a regulator in place. Consider the correction to the photon propagator, shown in \ufb01g. (75.1). We have i\u03a0\u00b5\u03bd (k) = ( \u2212 \u2212 1)(ig)2 2 1 i Z 1)(k2g\u00b5\u03bd i", "(Z3\u2212 d4\u2113 (2\u03c0)4 N \u00b5\u03bd (\u2113+k)2\u21132 k\u00b5k\u03bd ) + O(g4), (75.11) \u2212 where the numerator is N \u00b5\u03bd = Tr[PL(/\u2113+/k)\u03b3\u00b5PLPL/\u2113\u03b3\u03bd PL]. (75.12) We have P 2 L = PL and PL\u03b3\u00b5 = \u03b3\u00b5PR (and hence PL\u03b3\u00b5\u03b3\u03bd = \u03b3\u00b5\u03b3\u03bdPL), and so all the PL\u2019s in eq. (75.12) can be collapsed into just one; this is generically 75: Chiral Gauge Theories and Anomalies 446 true along any fermion line. Thus we have N \u00b5\u03bd = Tr[(/\u2113+/k)\u03b3\u00b5/\u2113\u03b3\u03bd PL]. (75.13) The term in eq. (75.13) with PL \u2192 get in spinor electrodynamics with a Dirac \ufb01eld. The term in eq. (75.13) with PL \u2192 \u2212 vanishing contribution to \u03a0\u00b5\u03bd (k). To see this, \ufb01rst note that 1 2 \u03b35, on the other hand, yields a 1 2 simply yields half the result that we N \u00b5\u03bd Thus we have 1 2 Tr[(/\u2113+/k)\u03b3\u00b5/\u2113\u03b3\u03bd \u03b35] \u2192 \u2212 = 2i\u03b5\u03b1\u00b5\u03b2\u03bd(\u2113+k)\u03b1\u2113\u03b2 = 2i\u03b5\u03b1\u00b5\u03b2\u03bdk\u03b1\u2113\u03b2. (75.14) \u2113\u03b2 (\u2113+k)2\u21132 (75.15) \u03a0\u00b5\u03bd (k) = 1 2 \u03a0\u00b5\u03bd (k)Dirac \u2212 1)(k2g\u00b5\u03bd (Z3\u2212 2g2\u03b5\u03b1\u00b5\u03b2\u03bd k\u03b1 d4\u2113 (2\u03c0)4 Z k\u00b5k\u03bd) + O(g4). \u2212 \u2212 The integral is logarithmically divergent. But, it carries a single vector index \u03b2, and the only vector it depends on is k. Therefore, any Lorentzinvariant regularization must yield a result that is proportional to k\u03b2. This then vanishes when contracted with \u03b5\u03b1\u00b5\u03b2\u03bd k\u03b1. We therefore conclude that, at the one-loop level, the contribution to \u03a0\u00b5\u03bd (k) of a single charged Weyl \ufb01eld is half that of a Dirac \ufb01eld. This is physically reasonable, since a Dirac \ufb01eld is equivalent to two charged Weyl \ufb01elds. Nothing interesting happens in the one-loop corrections to the fermion propagator, or the fermion\u2013fermion\u2013photon vertex. There is simply an extra factor of PL along the fermion line, which can be moved to the far right. Except for this factor, the results exactly duplicate those of spinor electrodynamics. All of this implies that a single Weyl \ufb01eld makes half the contribution of a Dirac \ufb01eld to the leading term in the beta function for the gauge coupling. Next we turn to diagrams with three external photons, and no external fermions, shown in \ufb01g. (75.2). In spinor electrodynamics, the fact that the vector potential is odd under charge conjugation implies that the sum of these diagrams must vanish; see problem 58.2. For the present case of a single Weyl \ufb01eld, there is no charge-conjugation symmetry, and so we must evaluate these diagrams. The second diagram in \ufb01g. (75.2) is the same as the \ufb01rst, with p and \u00b5 \u2194 \u03bd. Thus we have iV\u00b5\u03bd\u03c1(p, q, r) = ( \u2212 1)(ig)3 3 1 i Z d4\u2113 (2\u03c0)4 N \u00b5\u03bd\u03c1 p)2\u21132(\u2113+q)2 (\u2113 \u2212 q \u2194 75: Chiral Gauge Theories and Anomalies 447 \u03c1 r l p l+q \u00b5 p \u03bd q l \u03c1 r l+p l q l \u00b5 p \u03bd q Figure 75.2: One-loop contributions to the three-photon vertex. + (p, \u00b5 \u2194 q, \u03bd) + O(g5), (75.16) where N \u00b5\u03bd\u03c1 = Tr[( /\u2113+/p)\u03b3", "\u00b5( /\u2113)\u03b3\u03bd( /q)\u03b3\u03c1PL]. (75.17) \u2212 \u2212 /\u2113 \u2212 \u2212 1 2 simply yields half the result that we The term in eq. (75.17) with PL \u2192 get in spinor electrodynamics with a Dirac \ufb01eld, which gives a vanishing contribution to V\u00b5\u03bd\u03c1(p, q, r). Hence, we can make the replacement PL \u2192 1 2 \u03b35 in eq. (75.17). Then, after cancelling some minus signs, we have \u2212 N \u00b5\u03bd\u03c1 1 2 Tr[(/\u2113 \u2212 \u2192 /p)\u03b3\u00b5/\u2113\u03b3\u03bd(/\u2113+/q)\u03b3\u03c1\u03b35]. (75.18) We would now like to verify that V\u00b5\u03bd\u03c1(p, q, r) is gauge invariant. We should have p\u00b5V\u00b5\u03bd\u03c1(p, q, r) = 0, q\u03bdV\u00b5\u03bd\u03c1(p, q, r) = 0, r\u03c1V\u00b5\u03bd\u03c1(p, q, r) = 0. (75.19) (75.20) (75.21) Let us \ufb01rst check the last of these. From eq. (75.16) we \ufb01nd r\u03c1V\u00b5\u03bd\u03c1(p, q, r) = ig3 d4\u2113 (2\u03c0)4 (\u2113 Z r\u03c1N \u00b5\u03bd\u03c1 p)2\u21132(\u2113+q)2 \u2212 + (p, \u00b5 \u2194 q, \u03bd) + O(g5), (75.22) where r\u03c1N \u00b5\u03bd\u03c1 = 1 2 Tr[(/\u2113 \u2212 /p)\u03b3\u00b5/\u2113\u03b3\u03bd(/\u2113+/q)r\u03c1\u03b3\u03c1\u03b35]. (75.23) It will be convenient to use the cyclic property of the trace to rewrite eq. (75.23) as r\u03c1N \u00b5\u03bd\u03c1 = 1 2 Tr[/\u2113\u03b3\u03bd(/\u2113+/q)r\u03c1\u03b3\u03c1(/\u2113 \u2212 /p)\u03b3\u00b5\u03b35]. (75.24) 75: Chiral Gauge Theories and Anomalies 448 To simplify eq. (75.24), we write r\u03c1\u03b3\u03c1 = /r = Then \u2212 (/q+/p) = \u2212 (/\u2113+/q) + (/\u2113 /p). \u2212 (/\u2113+/q)r\u03c1\u03b3\u03c1(/\u2113 \u2212 /p) = (/\u2113+/q)[ (/\u2113+/q) + (/\u2113 /p)](/\u2113 /p) \u2212 = (\u2113+q)2(/\u2113 \u2212 \u2212 p)2(/\u2113+/q). /p) (\u2113 \u2212 \u2212 \u2212 (75.25) Now we have r\u03c1N \u00b5\u03bd\u03c1 = 1 2 (\u2113+q)2 Tr[/\u2113\u03b3\u03bd(/\u2113 \u2212 2i\u03b5\u03b1\u03bd\u03b2\u00b5 = \u2212 = +2i\u03b5\u03b1\u03bd\u03b2\u00b5 /p)\u03b3\u00b5\u03b35] 1 2 (\u2113 p)2 Tr[/\u2113\u03b3\u03bd(/\u2113+/q)\u03b3\u00b5\u03b35] \u2212 p)2\u2113\u03b1(\u2113+q)\u03b2 \u2212 p)\u03b2 \u2212 \u2212 (\u2113+q)2\u2113\u03b1(\u2113 h (\u2113+q)2\u2113\u03b1p\u03b2 + (\u2113 h \u2212 (\u2113 \u2212 p)2\u2113\u03b1q\u03b2 i (75.26) i Putting eq. (75.26) into eq. (75.22), we get r\u03c1V\u00b5\u03bd\u03c1(p, q, r) = \u2212 2g3 \u03b5\u03b1\u03bd\u03b2\u00b5 + (p, \u00b5 \u2194 d4\u2113 (2\u03c0)4 \u2113\u03b1p\u03b2 \u21132(\u2113 p)2 + Z q, \u03bd) + O(g5), \u2212 \u2113\u03b1q\u03b2 \u21132(\u2113+q)2 (75.27) Consider the \ufb01rst term in the integrand. Because the only four-vector that it depends on is p, any Lorentz-invariant regularization of its integral must yield a result proportional to p\u03b1p\u03b2. Similarly, any Lorentz-invariant regularization of the integral of the second term must yield a result proportional to q\u03b1q\u03b2. Both", " p\u03b1p\u03b2 and q\u03b1q\u03b2 vanish when contracted with \u03b5\u00b5\u03b1\u03bd\u03b2. Therefore, we have shown that r\u03c1V\u00b5\u03bd\u03c1(p, q, r) = 0, (75.28) as required by gauge invariance. It might seem that now we are done: we can invoke symmetry among the external lines to conclude that we must also have p\u00b5V\u00b5\u03bd\u03c1(p, q, r) = 0 and q\u03bdV\u00b5\u03bd\u03c1(p, q, r) = 0. However, eq. (75.16) is not manifestly symmetric on the exchanges (p, \u00b5 r, \u03c1). So it still behooves us to \u2194 compute either p\u00b5V\u00b5\u03bd\u03c1(p, q, r) or q\u03bdV\u00b5\u03bd\u03c1(p, q, r). r, \u03c1) and (q, \u03bd \u2194 From eq. (75.16) we \ufb01nd p\u00b5V\u00b5\u03bd\u03c1(p, q, r) = ig3 d4\u2113 (2\u03c0)4 Z (\u2113 \u2212 p\u00b5N \u00b5\u03bd\u03c1 p)2\u21132(\u2113+q)2 + (p, \u00b5 \u2194 q, \u03bd) + O(g5), (75.29) where p\u00b5N \u00b5\u03bd\u03c1 = 1 2 Tr[(/\u2113 \u2212 /p)p\u00b5\u03b3\u00b5/\u2113\u03b3\u03bd(/\u2113+/q)\u03b3\u03c1\u03b35]. (75.30) 75: Chiral Gauge Theories and Anomalies 449 To simplify eq. (75.30), we write p\u00b5\u03b3\u00b5 = /p = (/\u2113 \u2212 \u2212 /p) + /\u2113. Then (/\u2113 \u2212 /p)p\u00b5\u03b3\u00b5/\u2113 = (/\u2113 \u2212 = (\u2113 \u2212 /p)[ \u2212 p)2/\u2113 Now we have (/\u2113 /p) + /\u2113]/\u2113 \u2212 \u21132(/\u2113 \u2212 /p). \u2212 (75.31) /p)\u03b3\u03bd (/\u2113+/q)\u03b3\u03c1\u03b35] p\u00b5N \u00b5\u03bd\u03c1 = 1 2 (\u2113 p)2 Tr[/\u2113\u03b3\u03bd (/\u2113+/q)\u03b3\u03c1\u03b35] 1 2 \u21132 Tr[(/\u2113 \u2212 \u2212 2i\u03b5\u03b1\u03bd\u03b2\u03c1 2i\u03b5\u03b1\u03bd\u03b2\u03c1 2i\u03b5\u03b1\u03bd\u03b2\u03c1 = = = \u2212 \u2212 \u2212 p)2\u2113\u03b1q\u03b2 \u2212 p)2\u2113\u03b1q\u03b2 \u2212 p)2\u2113\u03b1q\u03b2 \u2212 \u2212 \u2212 \u2212 (\u2113 h (\u2113 h (\u2113 h \u2212 \u21132(\u2113 \u2212 \u21132(\u2113 \u21132(\u2113 \u2212 \u2212 p)\u03b1(\u2113+q)\u03b2 i p)\u03b1(\u2113 \u2212 p + p+q)\u03b2 i p)\u03b1(p+q)\u03b2. (75.32) i Putting eq. (75.32) into eq. (75.29), we get p\u00b5V\u00b5\u03bd\u03c1(p, q, r) = 2g3 \u03b5\u03b1\u03bd\u03b2\u03c1 + (p, \u00b5 Z \u2194 d4\u2113 (2\u03c0)4 \u2113\u03b1q\u03b2 \u21132(\u2113+q)2 \u2212 q, \u03bd) + O(g5), p)\u03b1(p+q)\u03b2 p)2(\u2113+q)2 (\u2113 (\u2113 \u2212 \u2212 (75.33) The \ufb01rst term on the right-hand side of eq. (75.33) must vanish, because any Lorentz-invariant regularization of the integral must yield a result proportional to q\u03b1q\u03b2, and this vanishes when contracted with \u03b5\u03b1\u03bd\u03b2\u03c1. As for the second term, we can shift the loop momentum from \u2113 to \u2113+p, which results in p)\u03b1(p+q)\u03b2 p)2(\u2113+q)2 \u2192 \u2113\u03b1(p+q)\u03b2 \u21132(\u2113+p+q)2. (75.34) (\u2113 (\u2113 \u2212 \u2212 We can now use Lorentz invariance to argue that the integral of the righthand side of eq. (75.34) must yield something proportional to (p+q)\u03b1(p+q)\u03b2; this vanishes when contracted with", " with \u03b5\u03b1\u03bd\u03b2\u03c1. Thus, we have shown that p\u00b5V\u00b5\u03bd\u03c1(p, q, r) = 0, as required by gauge invariance, provided that the shift of the loop momentum did not change the value of the integral. This would of course be true if the integral was convergent. Instead, however, the integral is linearly divergent, and so we must be more careful. Consider a one-dimensional example of a linearly divergent integral: let + \u221e dx f (x+a), (75.35) I(a) \u2261 Z, with c+ and c \u2212\u221e where f ( two \ufb01nite constants. If the integral converged, then I(a) would be independent of a. In the present case, however, ) = c \u00b1\u221e \u2212 \u00b1 75: Chiral Gauge Theories and Anomalies 450 we can Taylor expand f (x+a) in powers of a, and note that f ( implies that every derivative of f (x) vanishes at x = ) = c \u00b1\u221e. Thus we have \u00b1 \u00b1\u221e I(a) = + \u221e Z \u2212\u221e dx f (x) + af \u2032(x) + 1 2 a2f \u2032\u2032(x) +... h i = I(0) + a(c+ \u2212 c \u2212 We see that I(a) is not independent of a. Furthermore, even if we cannot assign a de\ufb01nite value to I(0) (because the integral is divergent), we can assign a de\ufb01nite value to the di\ufb00erence (75.36) ). \u2212 Now let us return to eqs. (75.33) and (75.34). De\ufb01ne I(0) = a(c+ \u2212 c \u2212 ). I(a) f\u03b1(\u2113) \u2113\u03b1 \u21132(\u2113+p+q)2. \u2261 Using Lorentz invariance, we can argue that (75.37) (75.38) d4\u2113 (2\u03c0)4 f\u03b1(\u2113) = A(p+q)\u03b1, where A is a scalar that will depend on the regularization scheme. Now consider (75.39) Z f\u03b1(\u2113 \u2212 p) = f\u03b1(\u2113) \u2212 \u2202\u2113\u03b2 f\u03b1(\u2113) +.... p\u03b2 \u2202 (75.40) The integral of the \ufb01rst term on the right-hand side of eq. (75.40) is given by eq. (75.39). The integrals of the remaining terms can be converted to surface integrals at in\ufb01nity. Only the second term in eq. (75.40) falls o\ufb00 slowly enough to contribute. To determine the value of its integral, we make a Wick rotation to euclidean space, which yields a factor of i as usual; then we have d4\u2113 (2\u03c0)4 Z \u2202 \u2202\u2113\u03b2 f\u03b1(\u2113) = i lim \u2113 \u2192\u221e Z dS\u03b2 (2\u03c0)4 f\u03b1(\u2113), (75.41) where dS\u03b2 = \u21132\u2113\u03b2d\u2126 is a surface-area element, and d\u2126 is the di\ufb00erential solid angle in four dimensions. We thus \ufb01nd d4\u2113 (2\u03c0)4 Z \u2202 \u2202\u2113\u03b2 f\u03b1(\u2113) = i lim \u2192\u221e Z \u21264 (2\u03c0)4 = i \u2113 d\u2126 (2\u03c0)4 \u2113\u03b2\u2113\u03b1 (\u2113+p+q)2 1 4 g\u03b1\u03b2 = i 32\u03c02 g\u03b1\u03b2, (75.42) (75.44) (75.45) 75: Chiral Gauge Theories and Anomalies 451 where we used \u21264 = 2\u03c02. Combining eqs. (75.38\u201375.42), we \ufb01nd d4\u2113 (2\u03c0)4 (\u2113 Z p)\u03b1 (\u2113 p)2(\u2113+q)2 = A(p+q)\u03b1 \u2212 \u2212 \u2212 i 32\u03c02 p\u03b1. (75.43) Using this in eq. (75.33), we \ufb01nd p\u00b5V\u00b5\u03bd\u03c1(p, q, r) = = ig3 16\u03c02 \u03b5\u03b1\u03bd\u03b2\u03c1p\u03b1(p+q)\u03b2 + (p, \u00b5 ig3 8\u03c0", "2 \u03b5\u03b1\u03bd\u03b2\u03c1p\u03b1q\u03b2 + O(g5). An exactly analogous calculation results in q, \u03bd) + O(g5) \u2194 q\u03bdV\u00b5\u03bd\u03c1(p, q, r) = ig3 8\u03c02 \u03b5\u03b1\u03c1\u03b2\u00b5q\u03b1p\u03b2 + O(g5). Eqs. (75.44) and (75.45) show that the three-photon vertex is not gauge invariant. Since r\u03c1V\u00b5\u03bd\u03c1(p, q, r) = 0, eqs. (75.44) and (75.45) also show that the three-photon vertex does not exhibit the expected symmetry among the external lines. This is a puzzle, because the only asymmetric aspects of the diagrams in \ufb01g. (75.2) are the momentum labels on the internal lines. The resolution of the puzzle lies in the fact that the integral in eq. (75.16) is linearly divergent, and so shifting the loop momentum changes its value. To account for this, let us write V\u00b5\u03bd\u03c1(p, q, r) with \u2113 replaced with \u2113 + a, where a is an arbitrary linear combination of p and q. We de\ufb01ne V\u00b5\u03bd\u03c1(p, q, r; a) 1 2 ig3 \u2261 Z Tr[(/\u2113+a /p)\u03b3\u00b5(/\u2113+a)\u03b3\u03bd(/\u2113+a+/q)\u03b3\u03c1\u03b35] p)2(\u2113+a)2(\u2113+a+q)2 d4\u2113 (2\u03c0)4 \u2212 (\u2113+a \u2212 q, \u03bd) + O(g5). (75.46) + (p, \u00b5 \u2194 Our previous expression, eq. (75.16), corresponds to a = 0. The integral in eq. (75.46) is linearly divergent, and so we can express the di\ufb00erence between V\u00b5\u03bd\u03c1(p, q, r; a) and V\u00b5\u03bd\u03c1(p, q, r; 0) as a surface integral. Let us write V\u00b5\u03bd\u03c1(p, q, r; a) = 1 2 ig3I\u03b1\u03b2\u03b3(a) Tr[\u03b3\u03b1\u03b3\u00b5\u03b3\u03b2\u03b3\u03bd \u03b3\u03b3\u03b3\u03c1\u03b35] q, \u03bd) + O(g5), + (p, \u00b5 \u2194 where I \u03b1\u03b2\u03b3(a) d4\u2113 (2\u03c0)4 \u2261 Z (\u2113+a (\u2113+a \u2212 \u2212 p)\u03b1(\u2113+a)\u03b2(\u2113+a q)\u03b3 p)2(\u2113+a)2(\u2113+a+q)2. \u2212 (75.47) (75.48) 75: Chiral Gauge Theories and Anomalies 452 Then we have I\u03b1\u03b2\u03b3(a) \u2212 I\u03b1\u03b2\u03b3(0) = a\u03b4 d4\u2113 (2\u03c0)4 \u2202 \u2202\u2113\u03b4 Z (\u2113 (\u2113 \u2212 \u2212 \u2113\u03b4(\u2113 (\u2113 q)\u03b3 p)\u03b1\u2113\u03b2(\u2113 p)2\u21132(\u2113+q)2 \u2212 p)\u03b1\u2113\u03b2(\u2113 \u2212 p)2(\u2113+q)2 \u2212 \u2212 q)\u03b3 d\u2126 (2\u03c0)4 1 24 g\u03b4\u03b1g\u03b2\u03b3 + g\u03b4\u03b2 g\u03b3\u03b1 + g\u03b4\u03b3g\u03b1\u03b2 = ia\u03b4 lim \u2113 \u2192\u221e Z = ia\u03b4 \u21264 (2\u03c0)4 = i 192\u03c02 a\u03b1g\u03b2\u03b3 + a\u03b2g\u03b3\u03b1 + a\u03b3g\u03b1\u03b2. (75.49) Using this in eq. (75.47), we get contractions of the form g\u03b1\u03b2\u03b3\u03b1\u03b3\u00b5\u03b3\u03b2 = 2\u03b3\u00b5. The three terms in eq. (75.49) all end up contributing equally, and after using eq. (75.9) to compute the trace, we \ufb01nd V\u00b5\u03bd\u03c1(p, q, r; a) \u2212 V\u00b5\u03bd\u03c1(p, q, r; 0) = \u2212 ig3 16\u03c02 \u03b5\u00b5\u03bd\u03c1\u03b2a\u03b2 + (p, \u00b5 \u2194 q, \u03bd) + O(g5). (75.50) Since the Levi-Civita symbol is antisymmetric on \u00b5 a that is antisymmetric on p we will", " set a = c(p \u03bd, only the part of \u2194 q contributes to V\u00b5\u03bd\u03c1(p, q, r; a). Therefore q), where c is a numerical constant. Then we have \u2194 \u2212 V\u00b5\u03bd\u03c1(p, q, r; a) \u2212 V\u00b5\u03bd\u03c1(p, q, r; 0) = ig3 8\u03c02 c \u03b5\u00b5\u03bd\u03c1\u03b2(p \u2212 \u2212 q)\u03b2 + O(g5). (75.51) Using this, along with eqs. (75.28), (75.44), and (75.45), and making some simplifying rearrangements of the indices and momenta on the right-hand sides (using p+q+r = 0), we \ufb01nd p\u00b5V\u00b5\u03bd\u03c1(p, q, r; a) = q\u03bdV\u00b5\u03bd\u03c1(p, q, r; a) = r\u03c1V\u00b5\u03bd\u03c1(p, q, r; a) = \u2212 \u2212 \u2212 c)\u03b5\u03bd\u03c1\u03b1\u03b2q\u03b1r\u03b2 + O(g5), ig3 8\u03c02 (1 \u2212 ig3 8\u03c02 (1 \u2212 ig3 8\u03c02 (2c)\u03b5\u00b5\u03bd\u03b1\u03b2p\u03b1q\u03b2 + O(g5). c)\u03b5\u03c1\u00b5\u03b1\u03b2 r\u03b1p\u03b2 + O(g5), (75.52) (75.53) (75.54) We see that choosing c = 1 removes the anomalous right-hand side from eqs. (75.52) and (75.53), but it then necessarily appears in eq. (75.54). Chosing c = 1 3 restores symmetry among the external lines, but now all three right-hand sides are anomalous. (This is what results from dimensional regularization of this theory with \u03b35 = i\u03b30\u03b31\u03b32\u03b33.) We have therefore failed to construct a gauge-invariant U(1) theory with a single charged Weyl \ufb01eld. 75: Chiral Gauge Theories and Anomalies 453 i Q3 Consider now a U(1) gauge theory with several left-handed Weyl \ufb01elds igQiA\u00b5)\u03c8i. \u03c8i, with charges Qi, so that the covariant derivative of \u03c8i is (\u2202\u00b5\u2212 Then each of these \ufb01elds circulates in the loop in \ufb01g. (75.2), and each vertex has an extra factor of Qi. The right-hand sides of eqs. (75.52\u201375.54) are i happens to be zero, then gauge now multiplied by invariance is restored! The simplest possibility is to have the \u03c8\u2019s come in P pairs with equal and opposite charges. (In this case, they can be assembled into Dirac \ufb01elds.) But there are other possibilities as well: for example, one \ufb01eld with charge +2 and eight with charge 1. Such a gauge theory is still chiral, but it is anomaly free. (It could be that further obstacles to gauge invariance arise with more external photons and/or more loops, but this turns out not to be the case. We will discuss this in section 77.) i. And if i Q3 P \u2212 RT c R T b R) to the \ufb01rst diagram in \ufb01g. (75.2), and a factor of Tr(T a All of this has a straightforward generalization to nonabelian gauge theories. Suppose we have a single Weyl \ufb01eld in a (possibly reducible) representation R of the gauge group. Then we must attach an extra factor of Tr(T a RT b R) to the second; here the group indicies a, b, c go along with the momenta p, q, r, respectively. Repeating our analysis shows that the diagrams with 2 T (R)f abcT c R; PL \u2192 these contribute to the renormalization of the tree-level three-gluon vertex. Diagrams with PL \u2192 \u2212 2 come with an extra factor of 1 1 2 \u03b35 come with an extra factor of 2 Tr([T a R) = i R, T b R] Tr( R} { R) = A(R)dabc. T c (75.55) Here dabc is a completely symmetric tensor that is independent of the representation, and A(R) is the anomaly coe\ufb03cient of R, introduced in secIn order for this theory to exist, we must have A(R) = 0. As tion 70. shown in section 70, A(R) = A(R); thus a theory whose left-", "handed Weyl R pairs is automatically anomaly free (as is one whose \ufb01elds come in R Weyl \ufb01elds are all in real representations). Otherwise, we must arrange the cancellation by hand. For SU(2) and SO(N ), all representations have A(R) = 0. For SU(N ) with N > 3, the fundamental representation has A(N) = 1, and most complex SU(N ) representations R have A(R) = 0. So the cancellation is nontrivial. \u2295 \u2212 We mention in passing two other kinds of anomalies: if we couple our theory to gravity, we can draw a triangle diagram with two gravitons and one gauge boson. This diagram violates general coordinate invariance (the gauge symmetry of gravity). If the gauge boson is from a nonabelian group, the diagram is accompanied by a factor of Tr T a R = 0, and so there is no anomaly. If the gauge boson is from a U(1) group, the diagram is accomi Qi, and this must vanish to cancel the anomaly. panied by a factor of There is also a global anomaly that a\ufb04icts theories with an odd number of Weyl fermions in a pseudoreal representation, such as the fundamental P 6 75: Chiral Gauge Theories and Anomalies 454 representation of SU(2). The global anomaly cannot be seen in perturbation theory; we will discuss it brie\ufb02y in section 77. Reference Notes Discussions of anomalies emphasizing di\ufb00erent aspects can be found in Georgi, Peskin & Schroeder, and Weinberg I. Problems 75.1) Consider a theory with a nonabelian gauge symmetry, and also a U(1) gauge symmetry. The theory contains left-handed Weyl \ufb01elds in the representations (Ri, Qi), where Ri is the representation of the nonabelian group, and Qi is the U(1) charge. Find the conditions for this theory to be anomaly free. 76: Anomalies in Global Symmetries 455 76 Anomalies in Global Symmetries Prerequisite: 75 In this section we will study anomalies in global symmetries that can arise in gauge theories that are free of anomalies in the local symmetries (and are therefore consistent quantum \ufb01eld theories). A phenomenological application will be discussed in section 90. The simplest example is electrodynamics with a massless Dirac \ufb01eld \u03a8 with charge Q = +1. The lagrangian is = i\u03a8 /D\u03a8 \u2212 L 1 4 F a\u00b5\u03bdF a \u00b5\u03bd, (76.1) where /D = \u03b3\u00b5D\u00b5 and D\u00b5 = \u2202\u00b5 \u2212 igA\u00b5. (We call the coupling constant g rather than e because we are using this theory as a formal example rather than a physical model.) We can write \u03a8 in terms of two left-handed Weyl \ufb01elds \u03c7 and \u03be via \u03a8 = \u03c7 \u03be\u2020!, (76.2) where \u03c7 has charge Q = +1 and \u03be has charge Q = \u03be, the lagrangian is \u2212 1. In terms of \u03c7 and = i\u03c7\u2020\u00af\u03c3\u00b5(\u2202\u00b5 \u2212 L igA\u00b5)\u03c7 + i\u03be\u2020\u00af\u03c3\u00b5(\u2202\u00b5 + igA\u00b5)\u03be \u2212 1 4 F a\u00b5\u03bdF a \u00b5\u03bd. The lagrangian is invariant under a U(1) gauge transformation \u03a8(x) \u03a8(x) A\u00b5(x) \u2192 \u2192 \u2192 e\u2212 ig\u0393(x)\u03a8(x), e+ig\u0393(x)\u03a8(x), A\u00b5(x) \u2212 \u2202\u00b5\u0393(x). In terms of the Weyl \ufb01elds, eqs. (76.4) and (76.5) become \u03c7(x) \u03be(x) \u2192 \u2192 e\u2212 ig\u0393(x)\u03c7(x), e+ig\u0393(x)\u03be(x). (76.3) (76.4) (76.5) (76.6) (76.7) (76.8) Because the fermion \ufb01eld is massless, the lagrangian is also invariant under a global symmetry in which \u03c7 and \u03be transform with the same phase, \u03c7(x) \u03be(x) \u2192 \u2192 e+i\u03b1\u03c7(x), e+i\u03b1\u03be(x). (76.9) (76.10) 76: Anomalies in Global Symmetries", " In terms of \u03a8, this is \u03a8(x) \u03a8(x) \u2192 \u2192 e\u2212 i\u03b1\u03b35 \u03a8(x), \u03a8(x)e\u2212 i\u03b1\u03b35. 456 (76.11) (76.12) This is called axial U(1) symmetry, because the associated Noether current j\u00b5 A(x) \u2261 \u03a8(x)\u03b3\u00b5\u03b35\u03a8(x) (76.13) is an axial vector (that is, its spatial part is odd under parity). Noether\u2019s theorem leads us to expect that this current is conserved: \u2202\u00b5j\u00b5 A = 0. However, in this section we will show that the axial current actually has an anomalous divergence, \u2202\u00b5j\u00b5 A = g2 16\u03c02 \u03b5\u00b5\u03bd\u03c1\u03c3F\u00b5\u03bd F\u03c1\u03c3. \u2212 (76.14) We will see in section 77 that eq. (76.14) is exact; there are no higher-order corrections. We will demonstrate eq. (76.14) by making use of our results in section j\u03c1 75. Consider the matrix element is a state of two, where 0 A(z) i | | outgoing photons with four-momenta p and q, and polarization vectors \u03b5\u00b5 and \u03b5\u2032\u03bd, respectively. (We omit the helicity label, which will play no essential role.) Using the LSZ formula for photons (see section 67), we have p,q h p,q h | p,q h j\u03c1 0 A(z) i | | where = (ig)2 \u03b5\u00b5\u03b5\u2032\u03bd d4x d4y e\u2212 i(px+qy) Tj\u00b5(x)j\u03bd (y)j\u03c1 0 | h, 0 A(z) i | (76.15) \u03a8(x)\u03b3\u00b5\u03a8(x) (76.16) Z j\u00b5(x) \u2261 is the Noether current corresponding to the U(1) gauge symmetry. Since both j\u00b5(x) and j\u00b5 A(x) are Noether currents, we expect the Ward identities \u2202 Tj\u00b5(x)j\u03bd (y)j\u03c1 0 \u2202x\u00b5 h | \u2202 Tj\u00b5(x)j\u03bd (y)j\u03c1 0 \u2202y\u03bd h | 0 A(z) i | 0 A(z) i | = 0, = 0, \u2202 Tj\u00b5(x)j\u03bd (y)j\u03c1 0 \u2202z\u03c1 h | 0 A(z) i | = 0, (76.17) (76.18) (76.19) to be satis\ufb01ed. Note that there are no contact terms in eqs. (76.17\u201376.19), because both j\u00b5(x) and j\u00b5 A(x) are invariant under both U(1) transformations. If we use eq. (76.19) in eq. (76.15), we see that we expect \u2202 \u2202z\u03c1 h p,q j\u03c1 0 A(z) i | | = 0. (76.20) 76: Anomalies in Global Symmetries 457 However, our experience in section 75 leads us to proceed more cautiously. Let us de\ufb01ne C \u00b5\u03bd\u03c1(p, q, r) via (2\u03c0)4\u03b44(p+q+r)C \u00b5\u03bd\u03c1(p, q, r) d4x d4y d4z e\u2212 i(px+qy+rz) Tj\u00b5(x)j\u00b5(y)j\u03c1 0 | h 0 A(z) i |. (76.21) \u2261 Z Then we can rewrite eq. (76.15) as p,q h j\u03c1 0 A(z) i | | = \u2212 g2\u03b5\u00b5\u03b5\u2032\u03bdC \u00b5\u03bd\u03c1(p, q, r)eirz Taking the divergence of the current yields. r= p q \u2212 \u2212 p,q h \u2202\u03c1j\u03c1 | 0 A(z) i | = \u2212 ig2\u03b5\u00b5\u03b5\u2032\u03bdr\u03c1C \u00b5\u03bd\u03c1(p, q, r)eirz The expected Ward identities become. p q \u2212 \u2212 r= p\u00b5C \u00b5\u03bd\u03c1(p, q, r) = 0, q\u03bdC \u00b5\u03bd\u03c1(p, q, r) = 0, r\u03c1C \u00b5\u03bd\u03c1(p, q, r) = 0. (76.22) (76.23) (76.24) (76.25) (76.26) To check eqs. (76.24\u201376.26", "), we compute C \u00b5\u03bd\u03c1(p, q, r) with Feynman diagrams. At the one-loop level, the contributing diagrams are exactly those we computed in section 75, except that the three vertex factors are now \u03b3\u00b5, \u03b3\u03bd, and \u03b3\u03c1\u03b35, instead of ig\u03b3\u00b5PL, ig\u03b3\u03bd PL, and ig\u03b3\u03c1PL. But, as we saw, the three PL\u2019s can be combined into just one at the last vertex, and then 1 2 \u03b35. Thus, the vertex function iV\u00b5\u03bd\u03c1(p, q, r) this one can be replaced by of section 75 is related to C \u00b5\u03bd\u03c1(p, q, r) by \u2212 iV\u00b5\u03bd\u03c1(p, q, r) = \u2212 1 2 (ig)3C \u00b5\u03bd\u03c1(p, q, r) + O(g5). (76.27) In section 75, we saw that we could choose a regularization scheme that preserved eqs. (76.24) and (76.25), but not also (76.26). For the theory of this section, we de\ufb01nitely want to preserve eqs. (76.24) and (76.25), because these imply conservation of the current coupled to the gauge \ufb01eld, which is necessary for gauge invariance. On the other hand, we are less enamored of eq. (76.26), because it implies conservation of the current for a mere global symmetry. Using eq. (76.27) and our results from section 75, we \ufb01nd that preserving eqs. (76.24) and (76.25) results in r\u03c1C \u00b5\u03bd\u03c1(p, q, r) = i 2\u03c02 \u03b5\u00b5\u03bd\u03b1\u03b2p\u03b1q\u03b2 + O(g2) \u2212 (76.28) 76: Anomalies in Global Symmetries 458 in place of eq. (76.26). Using this in eq. (76.23), we \ufb01nd p,q h \u2202\u03c1j\u03c1 | 0 A(z) i | = \u2212 g2 2\u03c02 \u03b5\u00b5\u03bd\u03b1\u03b2p\u03b1q\u03b2\u03b5\u00b5\u03b5\u2032\u03bde\u2212 i(p+q)z + O(g4). (76.29) Now we come to the point. The right-hand side of eq. (76.29) is exactly what we get in free-\ufb01eld theory for the matrix element of the right-hand side of eq. (76.14). We conclude that eq. (76.14) is correct, up to possible higher-order corrections. In the next section, we will see that eq. (76.14) is exact. Problems 76.1) Verify that the right-hand side of eq. (76.29) is exactly what we get in free-\ufb01eld theory for the matrix element of the right-hand side of eq. (76.14). 77: Anomalies and the Path Integral for Fermions 459 77 Anomalies and the Path Integral for Fermions Prerequisite: 76 In the last section, we saw that in a U(1) gauge theory with a massless Dirac \ufb01eld \u03a8 with charge Q = +1, the axial vector current j\u00b5 A = \u03a8\u03b3\u00b5\u03b35\u03a8, (77.1) which should (according to Noether\u2019s theorem) be conserved, actually has an anomalous divergence, \u2202\u00b5j\u00b5 A = g2 16\u03c02 \u03b5\u00b5\u03bd\u03c1\u03c3F\u00b5\u03bd F\u03c1\u03c3. \u2212 (77.2) In this section, we will derive eq. (77.2) directly from the path integral, using the Fujikawa method. We will see that eq. (77.2) is exact; there are no higher-order corrections. We can also consider a nonabelian gauge theory with a massless Dirac \ufb01eld \u03a8 in a (possibly reducible) representation R of the gauge group. In this case, the triangle diagrams that we analyzed in the last section carry an extra factor of Tr(T a \u2202\u00b5j\u00b5 A = R) = T (R)\u03b4ab, and we have R T b g2 16\u03c02 T (R)\u03b5\u00b5\u03bd\u03c1\u03c3\u2202[\u00b5Aa \u2202\u03bdAa \u03bd]\u2202[\u03c1Aa \u2212 \u03c3] + O(g3), (77.3) \u2202\u00b5Aa where \u2202[\u00b5Aa \u00b5. We expect the right-hand side of eq. (77.3) to be gauge invari", "ant (since this theory is free of anomalies in the currents coupled to the gauge \ufb01elds); this suggests that we should have \u03bd] \u2261 \u03bd \u2212 (77.4) \u2202\u00b5j\u00b5 A = \u2212 g2 16\u03c02 T (R)\u03b5\u00b5\u03bd\u03c1\u03c3F a \u00b5Ac \u00b5\u03bd F a \u03c1\u03c3, \u00b5\u03bd = \u2202\u00b5Aa where F a \u03bd is the nonabelian \ufb01eld strength. We will see that eq. (77.4) is correct, and that there are no higher-order corrections. \u00b5 + gf abcAb \u2202\u03bd Aa \u03bd \u2212 We can write eq. (77.4) more compactly by using the matrix-valued gauge \ufb01eld and \ufb01eld strength A\u00b5 \u2261 R Aa T a \u00b5 F\u00b5\u03bd = \u2202\u00b5A\u03bd \u2212 Then eq. (77.4) can be written as \u2202\u03bd A\u00b5 \u2212 ig[A\u00b5, A\u03bd]. \u2202\u00b5j\u00b5 A = g2 16\u03c02 \u03b5\u00b5\u03bd\u03c1\u03c3 Tr F\u00b5\u03bd F\u03c1\u03c3. \u2212 (77.5) (77.6) (77.7) 77: Anomalies and the Path Integral for Fermions 460 We now turn to the derivation of eqs. (77.2) and (77.7). We begin with the path integral over the Dirac \ufb01eld, with the gauge \ufb01eld treated as a \ufb01xed background, to be integrated later. We have Z(A) \u2261 Z \u03a8 eiS(A), \u03a8 D D where \u2261 is the Dirac action, i /D = i\u03b3\u00b5D\u00b5 is the Dirac wave operator, and Z S(A) d4x \u03a8i /D\u03a8 D\u00b5 = \u2202\u00b5 \u2212 igA\u00b5 (77.8) (77.9) (77.10) is the covariant derivative. Here A\u00b5 is either the U(1) gauge \ufb01eld, or the matrix-valued nonabelian gauge \ufb01eld of eq. (77.5), depending on the theory under consideration. Our notation allows us to treat both cases simultaneously. We can formally evaluate eq. (77.8) as a functional determinant, Z(A) = det(i /D). (77.11) However, this expression is not useful without some form of regularization. We will take up this issue shortly. Now consider an axial U(1) transformation of the Dirac \ufb01eld, but with a spacetime dependent parameter \u03b1(x): \u03a8(x) \u03a8(x) \u2192 \u2192 e\u2212 i\u03b1(x)\u03b35 \u03a8(x), \u03a8(x)e\u2212 i\u03b1(x)\u03b35. (77.12) (77.13) We can think of eqs. (77.12) and (77.13) as a change of integration variable in eq. (77.8); then Z(A) should be independent of \u03b1(x). The corresponding change in the action is S(A) \u2192 S(A) + Z d4x j\u00b5 A(x)\u2202\u00b5\u03b1(x). (77.14) We can integrate by parts to write this as S(A) S(A) \u2192 \u2212 d4x \u03b1(x)\u2202\u00b5j\u00b5 A(x). (77.15) \u03a8 is invariant under the axial U(1) If we assume that the measure transformation, then we have D Z \u03a8 D Z(A) \u03a8 D D \u2192 Z \u03a8 eiS(A)e\u2212 i d4x \u03b1(x)\u2202\u00b5j\u00b5 A(x). (77.16) R 77: Anomalies and the Path Integral for Fermions 461 This must be equal to the original expression for Z(A), eq. (77.8). This implies that \u2202\u00b5j\u00b5 A(x) = 0 holds inside quantum correlation functions, up to contact terms, as discussed in section 22. However, the assumption that the measure \u03a8 is invariant under the axial U(1) transformation must be examined more closely. The change of variable in eqs. (77.12) and (77.13) is implemented by the functional matrix \u03a8 D D J(x, y) = \u03b44(x \u2212 y)e\u2212 i\u03b1(x)\u03b35. (77.17) Because the path integral is over fermionic variables (rather than bosonic), 1 (rather than det J", ") for each of the we get a jacobian factor of (det J)\u2212 transformations in eqs. (77.12) and (77.13), so that we have \u03a8 \u03a8 D D \u2192 (det J)\u2212 2 \u03a8 D D \u03a8. Using log det J = Tr log J, we can write (det J)\u2212 2 = exp 2i Z d4x \u03b1(x) Tr \u03b44(x \u2212 x)\u03b35, (77.18) (77.19) where the explicit trace is over spin and group indices. Like eq. (77.11), this expression is not useful without some form of regularization. We could try to replace the delta-function with a gaussian; this is equiv- alent to \u03b44(x \u2212 y) \u2192 x/M 2 e\u22022 \u03b44(x \u2212 y), (77.20) where M is a regulator mass that we would take to in\ufb01nity at the end of the calculation. However, the appearance of the ordinary derivative \u2202, rather than the covariant derivative D, implies that eq. (77.20) is not properly gauge invariant. So, another possibility is \u03b44(x \u2212 y) \u2192 eD2 x/M 2 \u03b44(x \u2212 y). (77.21) However, eq. (77.21) presents us with a more subtle problem. Our regularization scheme for eq. (77.19) should be compatible with our regularization scheme for eq. (77.11). It is not obvious whether or not eq. (77.21) meets this criterion, because D2 has no simple relation to i /D. To resolve this issue, we use \u03b44(x \u2212 y) \u2192 e(i /Dx)2/M 2 \u03b44(x \u2212 y) (77.22) to regulate the delta function in eq. (77.19). To evaluate eq. (77.22), we write the delta function on the right-hand side of eq. (77.22) as a Fourier integral, \u03b44(x \u2212 y) \u2192 Z d4k (2\u03c0)4 e(i /Dx)2/M 2 eik(x y). \u2212 (77.23) 77: Anomalies and the Path Integral for Fermions 462 Then we use f (\u2202)eikx = eikxf (\u2202 + ik); eq. (77.23) becomes \u03b44(x \u2212 y) \u2192 Z d4k (2\u03c0)4 eik(x y) e(i /D \u2212 \u2212 /k)2/M 2, (77.24) where a derivative acting on the far right now yields zero. We have /k)2 = /k2 (i /D \u2212 /D2 \u2212 k2 /k, /D i } \u2212 { \u03b3\u00b5, \u03b3\u03bd i { \u2212 = \u2212 Next we use \u03b3\u00b5\u03b3\u03bd = 1 \u03b3\u00b5, \u03b3\u03bd 2 ( { } (i /D \u2212 /k)2 = \u2212 k2 + 2ik \u00b7 \u03b3\u00b5\u03b3\u03bd D\u00b5D\u03bd. k\u00b5D\u03bd \u2212 } g\u00b5\u03bd + [\u03b3\u00b5, \u03b3\u03bd ]) = 2iS\u00b5\u03bd to get \u2212 D + D2 + 2iS\u00b5\u03bd D\u00b5D\u03bd. \u2212 (77.25) (77.26) In the last term, we can use the antisymmetry of S\u00b5\u03bd to replace D\u00b5D\u03bd with 1 2 [D\u00b5, D\u03bd ] = 1 2 igF\u00b5\u03bd, which yields \u2212 (i /D \u2212 /k)2 = \u2212 k2 + 2ik \u00b7 D + D2 + gS\u00b5\u03bdF\u00b5\u03bd. (77.27) We use eq. (77.27) in eq. (77.24), and then rescale k by M ; the result is \u03b44(x \u2212 M 4 y) \u2192 Z d4k (2\u03c0)4 eiM k(x y) e\u2212 \u2212 k2 e2ik \u00b7 D/M +D2/M 2+gS\u00b5\u03bdF\u00b5\u03bd /M 2. (77.28) Thus we have Tr \u03b44(x \u2212 x)\u03b35 \u2192 M 4 Z d4k (2\u03c0)4 e\u2212 k2 Tr e2ik \u00b7 D/M +D2/M 2+gS\u00b5\u03bdF\u00b5\u03bd /M 2 \u03b35. 4 will survive the M (77.29) We can now expand the exponential in inverse powers of M ; only terms up to M \u2212 limit. Furthermore, the trace over", " spin indices will vanish unless there are four or more gamma matrices multiplying \u03b35. Together, these considerations imply that the only term that can make a nonzero contribution is 1 2 (gS\u00b5\u03bdF\u00b5\u03bd )2/M 4. Thus we \ufb01nd \u2192 \u221e Tr \u03b44(x \u2212 x)\u03b35 \u2192 1 2 g2 d4k (2\u03c0)4 e\u2212 k2 Z (Tr F\u00b5\u03bd F\u03c1\u03c3)(Tr S\u00b5\u03bd S\u03c1\u03c3\u03b35), (77.30) where the \ufb01rst trace is over group indices (in the nonabelian case), and the second trace is over spin indices. The spin trace is Tr S\u00b5\u03bd S\u03c1\u03c3\u03b35 = Tr ( i 2 \u03b3\u03c1\u03b3\u03c3)\u03b35 2 \u03b3\u00b5\u03b3\u03bd)( i 4 Tr \u03b3\u00b5\u03b3\u03bd\u03b3\u03c1\u03b3\u03c3\u03b35 1 = \u2212 = i\u03b5\u00b5\u03bd\u03c1\u03c3. (77.31) 77: Anomalies and the Path Integral for Fermions 463 To evaluate the integral over k in eq. (77.30), we analytically continue to euclidean spacetime; this results in an overall factor of i, as usual. Then each of the four gaussian integrals gives a factor of \u03c01/2. So we \ufb01nd Tr \u03b44(x \u2212 x)\u03b35 \u2192 \u2212 g2 32\u03c02 \u03b5\u00b5\u03bd\u03c1\u03c3 Tr F\u00b5\u03bd F\u03c1\u03c3. (77.32) Using this in eq. (77.19), we get (det J)\u2212 2 = exp ig2 16\u03c02 \"\u2212 Z d4x \u03b1(x) \u03b5\u00b5\u03bd\u03c1\u03c3 Tr F\u00b5\u03bd (x)F\u03c1\u03c3(x) #. (77.33) Including the transformation of the measure, eq. (77.18), in the transformation of the path integral, eq. (77.16), then yields Z(A) \u03a8 D D \u2192 Z \u03a8 eiS(A)e\u2212 i d4x \u03b1(x)[(g2/16\u03c02)\u03b5\u00b5\u03bd\u03c1\u03c3Tr F\u00b5\u03bd (x)F\u03c1\u03c3(x)+\u2202\u00b5j\u00b5 A(x)] (77.34) in place of eq. (77.16). This must be equal to the original expression for Z(A), eq. (77.8). This implies that eq. (77.7) holds inside quantum correlation functions, up to possible contact terms. R Note that this derivation of eq. (77.7) did not rely on an expansion in powers of g, and so eq. (77.7) is exact; there are no higher-order corrections. This result is known as the Adler-Bardeen theorem. It can also be (and originally was) established by a careful study of Feynman diagrams. The Fujikawa method can be used to \ufb01nd the anomaly in the chiral gauge theories that we studied in section 75, but the analysis is more involved. Here we will quote only the \ufb01nal result. Consider a left-handed Weyl \ufb01eld in a (possibly reducible) represen- tation R of the gauge group. We de\ufb01ne the chiral gauge current ja\u00b5 \u03a8T a Noether\u2019s theorem) is given by \u2261 R \u03b3\u00b5PL\u03a8. Its covariant divergence (which should be zero, according to Dab \u00b5 jb\u00b5 = g2 24\u03c02 \u03b5\u00b5\u03bd\u03c1\u03c3\u2202\u00b5Tr T a R (A\u03bd\u2202\u03c1A\u03c3 \u2212 h 1 2 igA\u03bd A\u03c1A\u03c3) i. (77.35) Note that the right-hand side of eq. (77.35) is not gauge invariant. The anomaly spoils gauge invariance in chiral gauge theories, unless this righthand side happens to vanish for group-theoretic reasons. We show in problem 77.1 that this occurs if and only if A(R) = 0, where A(R) is the anomaly coe\ufb03cient of the representation R. For comparison, note that eq. (77.7) can be written as \u2202\u00b5j\u00b5 A = \u2212 g2 4\u03c02 \u03b5\u00b5\u03bd\u03c1\u03c3\u2202\u00b5Tr A\u03bd \u2202\u03c1A\u03c3 \u2212 h 2 3 igA\u03bd A\u03c1A\u03c3. i (77.36) 77: Anomalies and the Path Integral for Fermions 464 The relative value of the overall numerical prefactor in eqs. (77.35) and (77.36) is easy to understand: there is a minus one-", "half in eq. (77.35) from 1 2 \u03b35, and a one-third from regularizing to preserve symmetry among PL \u2192 \u2212 the three external lines in the triangle diagram. (The relative coe\ufb03cients of the second terms have no comparably simple explanation.) Finally, a related but more subtle problem, known as a global anomaly, arises for theories with an odd number of Weyl \ufb01elds in a pseudoreal representation, such as the fundamental representation of SU(2). In this case, every gauge \ufb01eld con\ufb01guration A\u00b5 can be smoothly deformed into another gauge \ufb01eld con\ufb01guration A\u2032\u00b5 that has the same action, but has Z(A\u2032) = Z(A). Thus, when we integrate over A, the contribution from A\u2032 cancels the contribution from A, and the result is zero. Since its path integral is trivial, this theory does not exist. \u2212 Problems 77.1) Show that the right-hand side of eq. (77.35) vanishes if and only if A(R) = 0. 77.2) Show that the right-hand side of eq. (77.36) equals the right-hand side of eq. (77.7). 78: Background Field Gauge 465 78 Background Field Gauge Prerequisite: 73 In the section, we will introduce a clever choice of gauge, background \ufb01eld gauge, that greatly simpli\ufb01es the calculation of the beta function for Yang\u2013 Mills theory, especially at the one-loop level. We begin with the lagrangian for Yang\u2013Mills theory, where the \ufb01eld strength is LYM = \u2212 1 4 F a\u00b5\u03bdF a \u00b5\u03bd, F a \u00b5\u03bd = \u2202\u00b5Aa \u03bd \u2212 \u2202\u03bdAa \u00b5 + gf abcAb \u00b5Ac \u03bd. (78.1) (78.2) Lgf + To evaluate the path integral, we must choose a gauge. As we saw in section 71, one large class of gauges corresponds to choosing a gauge-\ufb01xing function Ga(x), and adding Lgh to LYM, where 1 2\u03be\u2212 Lgf = Lgh = \u00afca \u2202Ga Here Dbc \u00b5 = \u03b4bc\u2202\u00b5 \u2212 \u00b5 = \u03b4bc\u2202\u00b5 + gf bacAa \u00b5 is the covariant derivative in the adjoint representation, and c and \u00afc are the ghost and antighost \ufb01elds. The notation \u2202Ga/\u2202Ab \u00b5 in Ga now act to the right in eq. (78.4). \u00b5 means that any derivatives that act on Ab 1GaGa, A )bcAa \u00b5 cc. ig(T a (78.3) (78.4) \u2202Ab \u00b5 Dbc \u2212 We get R\u03be gauge by choosing Ga = \u2202\u00b5Aa gauge, we \ufb01rst introduce a \ufb01xed, classical background \ufb01eld \u00afAa corresponding background covariant derivative, \u00b5. To get background \ufb01eld \u00b5(x), and the \u00afD\u00b5 \u2261 \u2202\u00b5 \u2212 igT a A \u00afAa \u00b5. (78.5) Then we choose The ghost lagrangian becomes by parts, Ga = ( \u00afD\u00b5)ab(A \u2212 \u00afA)b \u00b5. (78.6) Lgh = \u00afca \u00afD\u00b5abDbc \u00b5 cc, or, after an integration ( \u00afD\u00b5\u00afc)a(D\u00b5c)a, \u00b5 cc. (78.7) \u2212 Lgh = where ( \u00afD\u00b5\u00afc)a = \u00afD\u00b5ab\u00afcb and (D\u00b5c)a = Dac Under an in\ufb01nitesimal gauge transformation, the change in the \ufb01elds is \u03b4GAa \u00b5(x) = Dac \u00b5 \u03b8c(x), \u2212 ig\u03b8a(x)(T a A )bccc(x), \u03b4Gcb(x) = \u2212 \u00b5(x) = 0. \u03b4G \u00afAa (78.8) (78.9) (78.10) 78: Background Field Gauge 466 The antighost \u00afc transforms in the same way as c (since the adjoint representation is real). The background \ufb01eld \u00afA is \ufb01xed, and so does not change under a gauge transformation. Of course, this means that Lgh are not gauge invariant; their role is to \ufb01x the gauge. We can, however, de\ufb01ne a background \ufb01eld gauge transformation, under Lg", "f and which only the background \ufb01eld transforms, \u03b4BG \u00afAa \u03b4BGAa \u00b5(x) = \u2212 \u00b5(x) = 0, \u00afDac \u00b5 \u03b8c(x), \u03b4BGcb(x) = 0. (78.11) (78.12) (78.13) Obviously, LYM is invariant under this transformation (since it does not involve the background \ufb01eld at all), but Lgh are not. However, Lgf and Lgh are invariant under the combined transformation \u03b4G+BG. For Lgh, as given by eq. (78.7), this follows immediately from the fact that D\u00b5 and \u00afD\u00b5 have the same transformation property under the combined transformation, and that using covariant derivatives with all group indices contracted always yields a gauge-invariant expression. Lgf and To use this argument on Lgf, as given by eqs. (78.3) and (78.6), we need to show that (A \u00b5 transforms under the combined transformation in the same way as does an ordinary \ufb01eld in the adjoint representation, such as the ghost \ufb01eld in eq. (78.9). To show this, we write \u00afA)a \u2212 \u03b4G+BG(A \u00afA)b \u00b5 = \u2212 (D \u2212 \u2212 \u00afD)ba \u00b5 \u03b8a = +ig(A \u00afA)c \u00b5(T c A)ba\u03b8a \u2212 A )bc(A ig\u03b8a(T a = \u00afA)c \u00b5. (78.14) \u2212 \u2212 We used the complete antisymmetry of (T c if cba to get the last line. \u00afA)a We see that (A \u00b5 transforms like an ordinary \ufb01eld in the adjoint representation, and so any expression that involves only covariant derivatives (either \u00afD\u00b5 or D\u00b5) acting on this \ufb01eld, with all group indices contracted, is invariant under the combined transformation. A)ba = \u2212 \u2212 L = Lgf + LYM + Therefore, the complete lagrangian, under the combined transformation. Lgh, is invariant Now consider constructing the quantum action \u0393(A, c, \u00afc; \u00afA). Recall from section 21 that the quantum action can be expressed as the sum of all 1PI diagrams, with the external propagators replaced by the corresponding \ufb01elds. In a gauge theory, the quantum action is in general not gauge invariant, because we had to \ufb01x a gauge in order to carry out the path integral. The quantum action thus depends on the choice of gauge, and 78: Background Field Gauge 467 hence (in the case of background \ufb01eld gauge) on the background \ufb01eld \u00afA. This is why we have written \u00afA as an argument of \u0393, but separated by a semicolon to indicate its special role. An important property of the quantum action is that it inherits all linear symmetries of the classical action; see problem 21.2. In the present case, these symmetries include the combined gauge transformation \u03b4G+BG. Therefore, the quantum action is also invariant under the combined transformation. The quantum action takes its simplest form if we set the external \ufb01eld A equal to the background \ufb01eld \u00afA. Then, \u0393( \u00afA, c, \u00afc; \u00afA) is invariant under a gauge transformation of the form \u03b4G+BG \u00afAa \u00b5(x) = \u03b4G+BGcb(x) = \u2212 \u2212 \u00afDac \u00b5 \u03b8c(x), ig\u03b8a(x)(T a A )bccc(x). (78.15) (78.16) This is now simply an ordinary gauge transformation, with \u00afA as the gauge \ufb01eld. The quantum action can be expressed as the classical action, plus loop corrections. For A = \u00afA, we have \u0393( \u00afA, c, \u00afc; \u00afA) = d4x Z \u00afF a\u00b5\u03bd \u00afF a \u00b5\u03bd \u2212 1 4 \u2212 h ( \u00afD\u00b5\u00afc)a( \u00afD\u00b5c)a +..., (78.17) i where the ellipses stand for the loop corrections. Note that Lgf has disappeared [because we set A = \u00afA in eq. (78.6)], and Lgh has the form of a kinetic term for a complex scalar \ufb01eld in the adjoint representation. This term is therefore manifestly gauge invariant, as is the \u00afF \u00afF term. The gauge invariance of the quantum action has an important consequence for the loop corrections.", " In background \ufb01eld gauge, the renormalizing Z factors must respect the gauge invariance of the quantum action. Therefore, using the notation of section 73, we must have Z1 = Z2, Z1\u2032 = Z2\u2032, Z3 = Z3g = Z4g. (78.18) (78.19) (78.20) Thus the relation between the bare and renormalized gauge couplings becomes g2 0 = Z \u2212 1 3 g2 \u02dc\u00b5\u03b5. (78.21) This relation now involves only Z3. We can therefore compute the beta function from Z3 alone. This is the major advantage of background \ufb01eld gauge. 78: Background Field Gauge 468 To compute the loop corrections, we need to evaluate 1PI diagrams in background-\ufb01eld gauge with the external propagators removed and replaced with external \ufb01elds; the external gauge \ufb01eld should be set equal to the background \ufb01eld. The easiest way to do this is to set A = \u00afA + A (78.22) at the beginning, and to write the path integral in terms of A \ufb01eld appears only on internal lines, and the \u00afA \ufb01eld only on external lines. The gauge-\ufb01xing term now reads. Then the A Lgf = \u2212 1 2 \u03be\u2212 1( \u00afD\u00b5 A\u00b5)a( \u00afD\u03bd A\u03bd)a, (78.23) and the ghost term is given by eq. (78.7). The Feynman rules that follow from LYM + Lgh are closely related to those we found in R\u03be gauge in section 72. The ghost and gluon propagators are the same, and vertices involving all internal lines are also the same. But if one or more gluon lines are external, then there are additional Lgh. We leave the details to Lgf and contributions to the vertices from problem 78.1. Further simpli\ufb01cations arise at the one-loop level. Using eq. (78.22) in Lgf + eq. (78.2), we \ufb01nd \u00b5\u03bd = \u2202\u00b5 \u00afAa F a \u2202\u03bd \u00afAa \u00afAc \u03bd \u00b5 + gf abc( \u00afAb a \u00b5 + gf abc \u00afAb \u03bd \u2212 \u00b5 a + \u2202\u00b5A \u2202\u03bd A \u03bd \u2212 \u00b5\u03bd + ( \u00afD\u00b5A\u03bd)a c \u03bd + \u00b5A A ( \u00afD\u03bd A\u00b5)a + gf abc A \u2212 b \u00b5 \u00afAc b \u00b5A = \u00afF a \u03bd) + gf abc A c \u03bd. c \u03bd b \u00b5A (78.24) We then have LYM = 1 1 4 \u2212 \u2212 \u03bd)a( \u00afD\u00b5A\u03bd)a + 1 \u00afF a\u00b5\u03bd \u00afF a \u00b5\u03bd \u2212 2 gf abc \u00afF a\u00b5\u03bd 1 A 2 ( \u00afD\u00b5 A c \u03bd +...,. where the ellipses stand for terms that are linear, cubic, or quartic in A Vertices arising from terms linear in cannot appear in a 1PI diagram, and A the cubic and quartic vertices do not appear in the one-loop contribution to the \u00afA propagator. \u03bd)a( \u00afD\u03bd A\u00b5)a 2 ( \u00afD\u00b5 A (78.25) b \u00b5A The last term on the \ufb01rst line of eq. (78.25) can be usefully manipulated A ( \u00afD\u00b5 \u03bd)a( \u00afD\u03bd A\u00b5)a = ( \u00afD\u03bd \u2212A with some dummy-index relabelings and integrations by parts; we have A\u00b5)c( \u00afD\u00b5 A\u03bd)c \u00b5( \u00afD\u03bd \u00afD\u00b5)bc b c \u03bd A \u00b5( \u00afD\u00b5 \u00afD\u03bd [ \u00afD\u00b5, \u00afD\u03bd])bc b \u2212 \u00b5( \u00afD\u00b5 \u00afD\u03bd)bc b c \u03bd \u2212 A A\u00b5)c( \u00afD\u03bd A\u03bd)c c \u03bd A A )bc \u00afF a\u00b5\u03bd gf abc \u00afF a\u00b5\u03bd \u2212A = +( \u00afD\u00b5 b \u00b5A c \u03bd. A b \u00b5A ig(T a \u2212A A \u2212 = = = c \u03bd (78.26) 78: Background Field Gauge 469 Figure 78.1: The one-loop contributions to the \u00afA propagator in background gauge \ufb01elds. \ufb01eld gauge; the dashed lines can be either ghosts or internal The dot denotes the \u00afF vertex.", " A AA Now the \ufb01rst term on the right-hand side of eq. (78.26) has the same form as the gauge-\ufb01xing term. If we choose \u03be = 1, these two terms will cancel. Setting \u03be = 1, and including a renormalizing factor of Z3, the terms of interest in the complete lagrangian become 4 Z3 \u00afF a\u00b5\u03bd \u00afF a \u00b5\u03bd \u2212 Z3gf abc \u00afF a\u00b5\u03bd 2 Z3( \u00afD\u00b5 )a( \u00afD\u00b5A\u03bd)a b \u00b5A A \u2212 ( \u00afD\u00b5\u00afc)a( \u00afD\u00b5c)a \u2212 (78.27) In the ghost term, we have replaced D\u00b5 with \u00afD\u00b5; the vertex corresponding term does not appear in the one-loop contribution to to the dropped the \u00afA propagator. Also, we can rescale to absorb Z3 in all terms except the \ufb01rst; since never appears on an external line, its normalization is irrelevant, and always cancels among propagators and vertices. (The same is true of the ghost \ufb01eld.) A A A The one-loop diagrams that contribute to the \u00afA propagator are shown in \ufb01g. (78.1). The dashed lines in the \ufb01rst two diagrams represent either the A \ufb01eld or the ghost \ufb01elds. In either case, the second diagram vanishes, because d4\u2113/\u21132, which is zero after dimensional regularization. it is proportional to Note that the ghost term in eq. (78.27) has the form of a kinetic term for a complex scalar \ufb01eld in the adjoint representation. In problem 73.1, we found the contribution of a complex scalar \ufb01eld in a representation RCS to \u03a0(k2) is R \u03a0CS(k2) = g2 24\u03c02 T (RCS) 1 \u03b5 \u2212 + \ufb01nite. (78.28) The ghost contribution is minus this, with RCS \u2192 from the closed ghost loop.) Thus we have A. (The minus sign is \u03a0gh(k2) = + g2 24\u03c02 T (A) 1 \u03b5 + \ufb01nite + O(g4). (78.29) For reference we recall that the counterterm contribution is \u03a0ct(k2) = (Z3\u2212 \u2212 1). (78.30) 78: Background Field Gauge 470 Next we consider the diagrams with \ufb01elds in the loop. If the \u00afF A A A \u00afD AA interaction term was absent, the calculation would again be a familiar one; the \u00afD term in eq. (78.27) has the form of a kinetic term for a real scalar \ufb01eld that carries an extra index \u03bd. That this index is a Lorentz vector index is immaterial for the diagrammatic calculation; the index is simply summed around the loop, yielding an extra factor of d = 4. There is also an extra factor of one-half (relative to the case of a complex scalar) because is real rather than complex. (Equivalently, the diagram has a symmetry factor of S = 2 from exchange of the top and bottom internal propagators when they do not carry charge arrows.) We thus have A \u03a0 \u00afD \u00afD A A (k2) = g2 12\u03c02 T (A) 1 \u03b5 \u2212 + \ufb01nite + O(g4). (78.31) AA If we now include the \u00afF interaction, we can think of \u00afF a \u00b5\u03bd as a constant external \ufb01eld. We can then draw the third diagram of \ufb01g. (78.1), where and each dot denotes a vertex factor of \u00b5\u03bd. This vacuum diagram has a symmetry factor of S = 2 2: one factor of two for exchanging the top and bottom propagators, and one for exchanging the left and right sources. Its contribution to the quantum action is 2igf abc \u00afF a \u00d7 \u2212 i\u0393 \u00afF AA /V T = 1 4 ( \u2212 2igf acd \u00afF a \u00b5\u03bd )( \u2212 2igf beg \u00afF b \u03c1\u03c3) 2 \u02dc\u00b5\u03b5 1 i Z dd\u2113 (2\u03c0)d g\u00b5\u03c1\u03b4ce \u21132 g\u03bd\u03c3\u03b4dg \u21132 = g2 T (A) \u00afF a\u00b5\u03bd \u00afF a \u00b5\u03bd i 8\u03c02\u03b5 + \ufb01nite, (78.32) where V T is the volume of spacetime. Comparing this with the tree", "-level 4 Z3 \u00afF \u00afF, and recalling eq. (78.30), we see that eq. (78.32) is lagrangian equivalent to a contribution to \u03a0(k2) of \u2212 1 (k2) = + \u03a0 \u00afF AA g2 2\u03c02 T (A) 1 \u03b5 + \ufb01nite. (78.33) There is also a one-loop diagram with one \u00afD AA vertex; however, contracting the vector indices on the \ufb01elds around the loop leads to a factor of \u00afF \u00b5\u03bdg\u00b5\u03bd = 0. Similarly, a one-loop diagram with a single \u00afF vertex and one \u00afF vertex vanishes. \u00afD A A A We could also couple the gauge \ufb01eld to a Dirac fermion in the representation RDF, and a complex scalar in the representation RCS. The corresponding contributions to \u03a0(k2) were computed in section 73, and are given by eq. (78.28) and AA \u03a0DF(k2) = g2 6\u03c02 T (RDF) 1 \u03b5 \u2212 + \ufb01nite, (78.34) 78: Background Field Gauge 471 Adding up eqs. (78.28), (78.29), (78.30), (78.31), (78.33), and (78.34), we \ufb01nd that \ufb01niteness of \u03a0(k2) requires g2 24\u03c02 Z3 = 1 + +1 2 + 12 T (A) h in the MS renormalization scheme. \u2212 4T (RDF) \u2212 \u2212 T (RCS) i 1 \u03b5 + O(g4) (78.35) The analysis of section 28 now results in a beta function of \u03b2(g) = g3 48\u03c02 \u2212 11T (A) h 4T (RDF) \u2212 \u2212 T (RCS) i + O(g5). (78.36) A Majorana fermion or a Weyl fermion makes half the contribution of a Dirac fermion in the same representation; a real scalar \ufb01eld makes half the contribution of a complex scalar \ufb01eld. (Majorana fermions and real scalars must be in real representations of the gauge group.) In quantum chromodynamics, the gauge group is SU(3), and there are nF = 6 \ufb02avors of quarks (which are Dirac fermions) in the fundamental representation. We therefore have T (A) = 3, T (RDF) = 1 2 nF, and T (RCS) = 0; therefore \u03b2(g) = 11 2 3 nF + O(g5). \u2212 We see that the beta function is negative for nF \u2264 asymptotically free. \u2212 g3 16\u03c02 (78.37) 16, and so QCD is Problems 78.1) Compute the tree-level vertex factors in background \ufb01eld gauge for all vertices that connect one or more external gluons with two or more internal lines (ghost or gluon). 78.2) Our one-loop corrections can be interpreted as functional determi- nants. De\ufb01ne \u00afD2 + gT a R \u00b5\u03bd S\u00b5\u03bd \u00afF a (a,b), (78.38) R,(a,b) \u2261 R ) \u00afAa ig(T a where \u00afD\u00b5 = \u2202\u00b5 \u2212 \u00b5 is the background-covariant derivative in the representation R, implicitly multiplied by the indentity matrix for the (a, b) representation of the Lorentz group, and S\u00b5\u03bd (a,b) are the Lorentz generators for that representation; in particular, S\u00b5\u03bd (1,1) = 0, (1,2) = i S\u00b5\u03bd (2,1) \u2295 (S\u00b5\u03bd (2,2))\u03b1\u03b2 = 4 [\u03b3\u00b5, \u03b3\u03bd ], \u03b1\u03b4\u03bd i(\u03b4\u00b5 \u2212 \u03b2 \u2212 (78.39) (78.40) (78.41) \u03b4\u03bd \u03b1\u03b4\u00b5 \u03b2). 78: Background Field Gauge 472 Show that the one-loop contribution to the terms in the quantum action that do not depend on the ghost \ufb01elds is given by exp i\u03931 \u2212 loop( \u00afA, 0, 0; \u00afA) \u221d \u00d7 (det A,(1,1))+1 (det A,(2,2))\u2212 (det RDF,(2,1) \u2295 (det RCS,(1,1))\u2212 \u00d7 1/2 \u00d7 (1", ",2))+1/2 1. (78.42) Verify that this expression agrees with the diagrammatic analysis in this section. 79: Gervais\u2013Neveu Gauge 473 79 Gervais\u2013Neveu Gauge Prerequisite: 78 In section 78, we used background \ufb01eld gauge to set up the computation of a quantum action that is gauge invariant. Given this quantum action, we can use it to compute scattering amplitudes via the corresponding tree diagrams, as discussed in section 19. Since the ghost \ufb01elds in the quantum action do not contribute to tree diagrams, we can simply drop all the ghost terms in the quantum action. Because the quantum action computed in background \ufb01eld gauge is itself gauge invariant, it requires further gauge \ufb01xing to specify the gluon propagator and vertices. We can choose whatever gauge is most convenient; for example, R\u03be gauge. In principle, this gauge \ufb01xing involves introducing new ghost \ufb01elds, but, once again, these do not contribute to tree diagrams, and so we can ignore them. If we start with the tree-level approximation to the quantum action, then, in R\u03be gauge, we simply get the gluon propagator and vertices of section 72. As we noted there, the complexity of the three- and four-gluon vertices in R\u03be gauge leads to long, involved computations of even simple processes like gluon-gluon scattering. In this section, we will introduce another gauge, Gervais\u2013Neveu gauge, that simpli\ufb01es these tree-level computations. We begin by specializing to the gauge group SU(N ), and working with \u00b5T a. For later convenience, we will normal- the matrix-valued \ufb01eld A\u00b5 = Aa ize the generators via Tr T aT b = \u03b4ab. With this choice, their commutation relations become [T a, T b] = i\u221a2f abcT c. The tree-level action is speci\ufb01ed by the Yang\u2013Mills lagrangian, where the matrix-valued \ufb01eld strength is LYM = \u2212 1 4 Tr F \u00b5\u03bdF\u00b5\u03bd, F\u00b5\u03bd = \u2202\u00b5A\u03bd \u2212 Let us introduce the matrix-valued complex tensor \u2202\u03bdA\u00b5 \u2212 [A\u00b5,A\u03bd ]. ig \u221a2 H\u00b5\u03bd \u2261 \u2202\u00b5A\u03bd \u2212 ig \u221a2 A\u00b5A\u03bd. Then F\u00b5\u03bd is the antisymmetric part of H\u00b5\u03bd, F\u00b5\u03bd = H\u00b5\u03bd \u2212 H\u03bd\u00b5. (79.1) (79.2) (79.3) (79.4) (79.5) (79.6) 79: Gervais\u2013Neveu Gauge 474 The Yang\u2013Mills lagrangian can now be written as H \u00b5\u03bdH\u00b5\u03bd \u2212 To \ufb01x the gauge, we choose a matrix-valued gauge-\ufb01xing function G(x), LYM = H \u00b5\u03bdH\u03bd\u00b5 (79.7) 1 2 Tr \u2212. and add (79.8) to LYM. Here we have set the gauge parameter \u03be to one, and ignored the ghost lagrangian (since, as we have already discussed, it does not a\ufb00ect tree diagrams). The choice of G that yields Gervais\u2013Neveu gauge is Lgf = \u2212 1 2 Tr GG G = H \u00b5 \u00b5. (79.9) At \ufb01rst glance, this choice seems untenable, because we see from eq. (79.5) Lgf ) is not hermitian. However, because the role of that this G (and hence Lgf to be nonhermitian. Lgf is merely to \ufb01x the gauge, it is acceptable for Combining eqs. (79.7) and (79.8), we get a total, gauge-\ufb01xed lagrangian L = \u2212 1 2 Tr H \u00b5\u03bdH\u00b5\u03bd \u2212 with two derivatives. After some integrations by Consider the terms in parts, those from the third term in eq. (79.10) cancel those from the second, leading to H \u00b5\u03bdH\u03bd\u00b5 + H \u00b5 (79.10) \u00b5H \u03bd L \u03bd. L2\u2202 = \u2212 1 2 Tr \u2202\u00b5A\u03bd \u2202\u00b5A\u03bd, (79.11) just as in R\u03be gauge with \u03be = 1. Now consider the terms with no derivatives. Once again, those from the third term in eq. (79", ".10) cancel those from the second (after using the cyclic property of the trace), leading to L0\u2202 = + 1 4 g2 Tr A\u00b5A\u03bdA\u00b5A\u03bd. (79.12) Finally, we have the terms with one derivative, L1\u2202 = + ig \u221a2 Tr \u2202\u00b5A\u03bdA\u00b5A\u03bd \u2212 \u2202\u00b5A\u03bdA\u03bdA\u00b5 + \u2202\u00b5A\u00b5A\u03bdA\u03bd. (79.13) Each derivative acts only on the \ufb01eld to its immediate right. If we integrate by parts in the last term in eq. (79.13), we generate two terms; one of these cancels the \ufb01rst term in eq. (79.13), and the other duplicates the second. Thus we have L1\u2202 = Combining eqs. (79.11), (79.12), and (79.14), we \ufb01nd i\u221a2g Tr \u2202\u00b5A\u03bdA\u03bdA\u00b5. \u2212 (79.14) = Tr L \u2212 1 2 \u2202\u00b5A\u03bd\u2202\u00b5A\u03bd \u2212 i\u221a2g \u2202\u00b5A\u03bdA\u03bdA\u00b5 + 1 4 g2A\u00b5A\u03bdA\u00b5A\u03bd. (79.15) 79: Gervais\u2013Neveu Gauge 475 Because this lagrangian has a rather simple structure in terms of the matrixvalued \ufb01eld A\u00b5, it is helpful to stick with this notation, rather than trying \u00b5 = Tr(T aA\u00b5). In the next section, we explore to reexpress the Feynman rules for a matrix-valued \ufb01eld in a simpli\ufb01ed context. in terms of Aa L Reference Notes Gervais\u2013Neveu gauge and some interesting variations are discussed in Siegel. 80: The Feynman Rules for N N Matrix Fields \u00d7 476 80 The Feynman Rules for N Fields Prerequisite: 10 N Matrix \u00d7 In section 79, we found that the lagrangian for SU(N ) Yang\u2013Mills theory in Gervais\u2013Neveu gauge is 2 \u2202\u00b5A\u03bd\u2202\u00b5A\u03bd \u2212 i\u221a2g \u2202\u00b5A\u03bdA\u03bdA\u00b5 + 1 4 g2A\u00b5A\u03bdA\u00b5A\u03bd (80.1) = Tr \u2212 L 1, where A\u00b5(x) is a traceless hermitian N N matrix. In this section, we will work out the Feynman rules for a simpli\ufb01ed model of a scalar \ufb01eld that keeps the essence of the matrix structure. \u00d7 Let B(x) be a hermitian N \u00d7 a complete set of N 2 hermitian N N matrix that is not traceless. Let T a be N matrices normalized according to \u00d7 Tr T aT b = \u03b4ab. (80.2) We will take one of these matrices, T N 2, to be proportional to the identity matrix; then eq. (80.2) requires the rest of the T a\u2019s to be traceless. We can expand B(x) in the T a\u2019s, with coe\ufb03cient \ufb01elds Ba(x), B(x) = Ba(x)T a, Ba(x) = Tr T aB(x), (80.3) (80.4) where the repeated index in eq. (80.3) is implicitly summed over a = 1 to N 2. Consider a lagrangian for B(x) of the form = Tr 1 2\u2202\u00b5B\u2202\u00b5B + 1 3 gB3 1 4 \u03bbB4 L \u2212 Using eqs. (80.2) and (80.3), we \ufb01nd an expression for coe\ufb03cient \ufb01elds, \u2212 (80.5) in terms of the. L = L \u2212 1 2 \u2202\u00b5Ba\u2202\u00b5Ba + 1 3 g Tr(T aT bT c)BaBbBc 1 4 \u03bb Tr(T aT bT cT d)BaBbBcBd. \u2212 (80.6) It is easy to read o\ufb00 the Feynman rules from this form of for the coe\ufb03cient \ufb01eld Ba is L. The propagator \u02dc\u2206ab(k2) = \u03b4ab \u2212 k2. i\u01eb (80.7) There is a three-point vertex with vertex factor 2ig Tr(T aT bT c), and a four6i\u03bb Tr(", "T aT bT cT d). This clearly leads to point vertex with vertex factor messy and complicated formulae for scattering amplitudes. \u2212 80: The Feynman Rules for N N Matrix Fields \u00d7 477 i j l k Figure 80.1: The double-line notation for the propagator of a hermitian matrix \ufb01eld. Figure 80.2: 3- and 4-point vertices in the double-line notation. Instead, let us work with in the form of eq. (80.5). Writing the matrix indices explicitly, with one up and one down (and employing the rule that two indices can be contracted only if one is up and one is down), we have B(x)i j. This implies that the propagator for Bi j = Ba(x)(T a)i j is L j \u02dc\u2206i k l(k2) = (T a)i k2 j(T a)k i\u01eb \u2212 l. (80.8) Since the T a matrices form a complete set, there is a completeness relation of the form (T a)i l j. To get the constant of proportionality, set \u221d j=k and l=i to turn the left-hand side into (T a)i i = Tr(T aT a), and k(T a)k k = N 2. From eq. (80.2) we have Tr(T aT a) = i\u03b4k the right-hand side into \u03b4i \u03b4aa = N 2. So the constant of proportionality is one, and j(T a)k l\u03b4k \u03b4i (T a)i j(T a)k l = \u03b4i l\u03b4k j. (80.9) We can represent the B propagator with a double-line notation, as shown in \ufb01g. (80.1). The arrow on each line points from an up index to a down index. Since the interactions are simple matrix products, with an up index from one \ufb01eld contracted with a down index from an adjacent \ufb01eld, the vertices follow the pattern shown in \ufb01g. (80.2). Since an n-point vertex of this type has only an n-fold cyclic symmetry (rather than an n!fold permutation symmetry), the vertex factor is i times the coe\ufb03cient of Tr(Bn) in times n (rather than n!). Thus, for the lagrangian of eq. (80.5), the 3- and 4-point vertex factors are ig and i\u03bb. L Now consider a scattering process. Particles corresponding to the coef\ufb01cient \ufb01elds labeled by the indices a1 and a2 (and with four-momenta k1 \u2212 80: The Feynman Rules for N 1 2 4 3 \u00d7 1 2 N Matrix Fields 478 4 3 1 2 4 3 Figure 80.3: Tree diagrams with four external lines. Five more diagrams of each of these three types, with the external labels 2, 3, and 4 permuted, also contribute. and k2) scatter into particles corresponding to the coe\ufb03cient \ufb01elds labeled by the indices a3 and a4 (and with four-momenta k3 and k4). We wish to compute the scattering amplitude for this process, at tree level. There are 18 contributing Feynman diagrams. Three are shown in \ufb01g. (80.3); the remaining 15 are obtained by making noncylic permutations of the labels 1, 2, 3, 4 (equivalent to making unrestricted permutations of 2, 3, 4). For simplicity, we will treat all external momenta as outgoing; then k0 2 are negative, and k1 + k2 + k3 + k4 = 0. Each external line carries a factor of T ai, with its matrix indices contracted by following the arrows backward through the diagrams. Omitting the i\u01eb\u2019s in the propagators (which are not relevant for tree diagrams), the resulting tree-level amplitude is 1 and k0 = Tr(T a1T a2T a3T a4) i T i) (ig)2( (k1+k2)2 + \u2212 i) (ig)2( \u2212 (k1+k4)2 \u2212 i\u03bb! + (234) (342), (423), (243), (432), (324). (80.10) \u2192 More generally, we can see that the value of any tree-level diagram with n external lines is proportional to Tr(T ai1... T ain ). If the diagram is drawn in planar fashion (that is, with no crossed lines), then the ordering of the ai indices in the trace", " is determined by the cyclic ordering of the labels on the external lines (which we take to be couterclockwise). Then, each i/k2, each 3-point vertex a factor of internal line contributes a factor of ig, and each four-point vertex a factor of i\u03bb. These are the color-ordered Feynman rules for this theory. \u2212 \u2212 Return now to i as given by eq. (80.10). Suppose that we wish to square this amplitude, and sum over all possible particle types for each incoming or outgoing particle. We then have to evaluate expressions like T Tr(T a1T a2T a3T a4)[Tr(T a1T a2T a4T a3)]\u2217, (80.11) 80: The Feynman Rules for N N Matrix Fields \u00d7 479 1 2 3 4 1 2 4 3 Figure 80.4: Evaluation of Tr(T a1T a2T a3T a4)[Tr(T a1 T a2T a4T a3)]\u2217, with all repeated indices summed. Each of the two closed single-line loops yields a factor of \u03b4i i = N. with all repeated indices summed. Using the hermiticity of the T a matrices, we have [Tr(T a1... T an)]\u2217 = Tr(T an... T a1), (80.12) It is then easiest to evaluate eq. (80.11) diagrammatically, as shown in i = N. The \ufb01g. (80.4). Each closed single-line loop yields a factor of \u03b4i result is that the absolute square of any particular trace yields a factor of N 4, and the product of any trace times the complex conjugate of any other di\ufb00erent trace yields a factor of N 2. The coe\ufb03cient of both Tr(T a1T a2T a3T a4) and Tr(T a1T a4T a3T a2) in eq. (80.10) is A3 \u2261 g2 (k1+k2)2 + g2 (k1+k4)2 \u2212 \u03bb. (80.13) Similarly, the coe\ufb03cient of both Tr(T a1 T a3T a4T a2) and Tr(T a1T a2T a4T a3) is A4 \u2261 g2 (k1+k3)2 + g2 (k1+k2)2 \u2212 \u03bb, (80.14) and of both Tr(T a1T a4T a2T a3) and Tr(T a1T a3T a2T a4) is A2 \u2261 g2 (k1+k4)2 + g2 (k1+k3)2 \u2212 \u03bb. (80.15) Thus we have |T | a1,a2,a3,a4 X 2 = (2N 4 + 2N 2) = (2N 4 2N 2) \u2212 Xj Xj 2 + 4N 2 Aj| | A\u2217j Ak =k Xj 2 + 4N 2( Aj| | A\u2217j )( Ak), (80.16) Xj Xk where j and k are summed over 2, 3, 4. 6 80: The Feynman Rules for N i j \u00d7 l k N Matrix Fields 480 i 1 N j l k Figure 80.5: The propagator for a traceless hermitian \ufb01eld. Now suppose we wish to impose the condition that the matrix \ufb01eld B is traceless: Tr B = 0. This means that we eliminate the component \ufb01eld with a=N 2, corresponding to the matrix T N 2 1/2I. We must also eliminate T N 2 from the sum in eq. (80.9), leading to = N \u2212 (T a)i j(T a)k l = \u03b4i l\u03b4k j 1 N \u03b4i j\u03b4k l. \u2212 (80.17) This can all be done diagrammatically by replacing the propagator in i/k2, is un\ufb01g. (80.1) with the one in \ufb01g. (80.5). (The kinematic factor, changed.) Fig. (80.5) must now be used as the internal propagator in the diagrams of \ufb01g. (80.3). Also, when we multiply one diagram by the complex 2, we must use the conjugate of another in the computation of propagator of \ufb01g. (80.5) to connect the external line of one diagram with the matching external line of the complex conjug", "ate diagram. Although these computations are still straightforward, they can become considerably more involved. a1...an |T | P \u2212 Problems 80.1) Show that the color-ordered Feynman rules, and the rules for compo- nent \ufb01elds given after eq. (80.6), agree in the case N = 1. 80.2) Verify the results quoted after eq. (80.12). 80.3) Compute T a\u2019s. a1,a2,a3,a4 | Tr(T a1T a2T a3T a4) 2 for the case of traceless | P 80.4) The large-N limit. Let \u03bb = cg2, where c is a number of order one. Now consider evaluating the path integral, without sources, as a function of g and N, Z(g, N ) = eiW (g,N ) = B ei ddx L, R D Z (80.18) where W (g, N ) is normalized by W (0, N ) = 0. As usual, W can be expressed as a sum of connected vacuum diagrams, which we draw in the double-line notation. Consider a diagram with V3 three-point vertices, V4 four-point vertices, E propagators or edges, and F closed single-line loops or faces. 80: The Feynman Rules for N N Matrix Fields \u00d7 481 a) Find the dependence on g and N of a diagram speci\ufb01ed by the values of V3, V4, E, and F. b) Express E for a vacuum diagram in terms of V3 and V4. c) Recall, derive, or look up the formula for the Euler character \u03c7 of the two-dimensional surface of a polyhedron in terms of the values of V3 + V4, E, and F. The Euler character is related to the genus V counts the number of handles, so \u2261 of the surface by \u03c7 = 2 G \u2212 that a sphere has genus zero, a torus has genus one, etc. G 2 G ; with the \u2019t Hooft coupling d) Consider the limit g \u00af\u03bb = g2N held \ufb01xed (and not necessarily small). Show that W (\u00af\u03bb, N ) has a topological expansion of the form 0 and N \u2192 \u221e \u2192 W (\u00af\u03bb (\u00af\u03bb), G \u221e =0 XG (80.19) (\u00af\u03bb) is given by a sum over diagrams that form polyhedra where W G In particular, the leading term, W0(\u00af\u03bb), is given by with genus a sum over diagrams with spherical topology, also known as planar diagrams. G. 81: Scattering in Quantum Chromodynamics 482 81 Scattering in Quantum Chromodynamics Prerequisite: 60, 79, 80 In section 79, we found that the lagrangian for SU(N ) Yang\u2013Mills theory in Gervais\u2013Neveu gauge is = Tr L \u2212 1 2 \u2202\u00b5A\u03bd\u2202\u00b5A\u03bd \u2212 i\u221a2g \u2202\u00b5A\u03bdA\u03bdA\u00b5 + 1 4 g2A\u00b5A\u03bdA\u00b5A\u03bd, (81.1) \u00d7 N matrix. For quantum chromowhere A\u00b5(x) is a traceless hermitian N dynamics, N = 3, but we will leave N unspeci\ufb01ed in our calculations. In section 80, we worked out the color-ordered Feynman rules for a scalar matrix \ufb01eld; the same technology applies here as well. In particular, we draw each tree diagram in planar fashion (that is, with no crossed lines). Then the cyclic, counterclockwise ordering i1... in of the external lines \ufb01xes the color factor as Tr(T ai1... T ain), where the generator matrices are normalized via Tr(T aT b) = \u03b4ab. The tree-level n-gluon scattering amplitude is then written as T = gn \u2212 2 Tr(T a1... T an)A(1,..., n), (81.2) Xnoncyclic perms where we have pulled out the coupling constant dependence, and A(1,..., n) is a partial amplitude that we compute with the color-ordered Feynman rules. The partial amplitudes are cyclically symmetric, A(2,..., n, 1) = A(1, 2,..., n). (81.3) The sum in eq. (81.2) is over all noncyclic permutations of 1... n, which is equivalent to a sum over", " all permutations of 2... n. From the \ufb01rst term in eq. (81.1), we see that the gluon propagator is simply \u02dc\u2206\u00b5\u03bd(k) = g\u00b5\u03bd k2 \u2212. i\u01eb (81.4) Here we have left out the matrix indices since we have already accounted for them with the color factor in eq. (81.2). The second and third terms in eq. (81.1) yield three- and four-gluon vertices. The three-gluon vertex factor (again without the matrix indices) is iV\u00b5\u03bd\u03c1(p, q, r) = i( i\u221a2g)( ip\u03c1g\u00b5\u03bd ) \u2212 + [ 2 cyclic permutations of (\u00b5,p), (\u03bd,q), (\u03c1,r) ] \u2212 i\u221a2g(p\u03c1g\u00b5\u03bd + q\u00b5g\u03bd\u03c1 + r\u03bdg\u03c1\u00b5), = \u2212 (81.5) 81: Scattering in Quantum Chromodynamics 483 where the four-momenta p, q, and r are all taken to be outgoing. The four-gluon vertex factor is simply iV\u00b5\u03bd\u03c1\u03c3 = ig2g\u00b5\u03c1g\u03bd\u03c3. (81.6) However, in the context of the color-ordered rules, it is simpler to designate the outgoing four-momentum on each external line as ki, and contract the vector index with the corresponding polarization vector \u03b5i. (For now we suppress the helicity label \u03bb = In this notation, the vertex factors become \u00b1.) iV123 = i\u221a2g \u2212 (\u03b51\u03b52)(k1\u03b53) + (\u03b52\u03b53)(k2\u03b51) + (\u03b53\u03b51)(k3\u03b52) i h iV1234 = +ig2(\u03b51\u03b53)(\u03b52\u03b54),, (81.7) (81.8) where the external lines are numbered sequentially, counterclockwise around the vertex. (Of course, if an attached line is internal, the corresponding polarization vector is simply a placeholder for an internal propagator.) The color-ordered three-point vertex, eq. (81.7), is antisymmetric on the 321, while the four-point vertex, eq. (81.8), is symmetric on reversal 123 the reversal 1234 \u2194 4321. This implies the re\ufb02ection identity, \u2194 A(n,..., 2, 1) = ( \u2212 1)nA(1, 2,..., n), (81.9) which will be useful later. It is clear from eqs. (81.7) and (81.8) that every term in any tree-level scattering amplitude is proportional to products of polarization vectors with each other, or with external momenta. (Actually, this follows directly from Lorentz invariance, and the fact that the scattering amplitude is linear in each polarization.) We get one momentum factor from each three-point vertex. Since every tree diagram with n external lines has no more than n 2 momenta to contract with the n polarizations. Therefore, every term in every tree-level amplitude must include at least one product of two polarization vectors. Then, if the product of every possible pair of polarization vectors vanishes, the tree-level amplitude for that process is zero. 2 vertices, there are no more than n \u2212 \u2212 We will now show that this is indeed the case if all, or all but one, of the external gluons have the same helicity. (Here we are using the semantic convention of section 60: the helicity of an external gluon is speci\ufb01ed relative to the outgoing four-momentum ki that labels the corresponding external line. If that gluon is actually incoming\u2014as indicated by a negative value of k0 i \u2014then its physical helicity is opposite to its labeled helicity.) 81: Scattering in Quantum Chromodynamics 484 To proceed, we recall from section 60 some formulae for products of polarization vectors in the spinor-helicity formalism, \u03b5+(k;q) (k;q) \u03b5 \u2212 \u03b5+(k;q) \u00b7 \u00b7 \u00b7 \u03b5 (k\u2032;q\u2032) = q q\u2032i [k k\u2032] \u03b5+(k\u2032;q\u2032) = h q k q\u2032 k\u2032i h i h k k\u2032i [q q\u2032] h [q k] [q\u2032 k\u2032] q k\u2032i (k\u2032;q\u2032) = h q k i h [k q\u2032] [q\u2032 k\u2032] \u03b5 \u2212 \u2212,,. (81.10) (81.11) (", "81.12) i q k h and [q k] are antisymmetric, and hence The \ufb01rst argument of each \u03b5 is the momentum of the corresponding line; the second argument is an arbitrary reference momentum. Recall that the = [q q] = twistor producs 0. Using this fact in eq. (81.10), we see that choosing the same reference momentum q for all positive-helicity polarizations results in a vanishing product for any pair of them. Furthermore, if we choose this q equal to the momentum k\u2032 of a negative-helicity gluon, eq. (81.12) tells us that the product of its polarization with that of any positive-helicity gluon also vanishes. Thus, if all, or all but one, of the external gluons have positive helicity, all possible polarization products are zero, and hence the tree-level scattering amplitude is also zero. Thus we have shown that q q h i A(1\u00b1, 2+,..., n+) = 0, (81.13) where the superscripts are the helicities. Of course, the same is true if all, or all but one, of the helicities are negative, A(1\u00b1, 2\u2212,..., n\u2212) = 0. (81.14) Now we turn to the calculation of some nonzero tree-level partial amplitudes, beginning with A(1\u2212, 2\u2212, 3+, 4+). The contributing color-ordered Feynman diagrams are shown in \ufb01g. (81.1). We choose the reference momenta to be q1 = q2 = k3 and q3 = q4 = k2. Then all polarization products vanish, with the exception of \u03b51\u00b7 \u03b54 = \u03b5 \u2212 (k1, q1) \u00b7 (k1, k3) \u00b7 \u03b5+(k4, q4) \u03b5+(k4, k24 3] [3 1]. (81.15) With this choice of the reference momenta, the third diagram in \ufb01g. (81.1) \u03b53 = 0 (and also, for good obviously vanishes, because it has a factor of \u03b51\u00b7 81: Scattering in Quantum Chromodynamics 485 Figure 81.1: Diagrams for the partial amplitude A(1, 2, 3, 4). measure, \u03b52 \u00b7 we have \u03b54 = 0). Now consider the 235 vertex in the second diagram; V235 \u221d (\u03b52\u03b53)(k2\u03b55) + (\u03b53\u03b55)(k3\u03b52) + (\u03b55\u03b52)(k5\u03b53), (81.16) where \u03b55 is a placeholder for an internal propagator. The \ufb01rst term in eq. (81.16) vanishes because \u03b52 \u00b7 \u03b53 = 0. The second term vanishes because \u03b52 = 0. Finally, the third term vanishes because k5 = k3 = q2 and q2 \u00b7 \u03b53 = 0. Hence the 235 vertex vanishes, k3, and q3\u00b7 q3\u2212 k3 = k2\u2212 \u2212 and therefore so does the second diagram. That leaves only the \ufb01rst diagram. We then have \u03b53 = k3 \u00b7 \u2212 ig2A(1\u2212, 2\u2212, 3+, 4+) = (iV125)(iV345\u2032 ) where 5\u2032 means the momentum is k5 rather than k5, and \u03b5\u00b5 5 \u03b5\u03bd 5 \u2192 ig\u00b5\u03bd/s12, (81.17) \u2212 (k1 + k2)2 = [2 1]. 1 2 i h (81.18) s12 \u2261 \u2212 We have i\u221a2g iV125 = (\u03b51\u03b52)(k1\u03b55) + (\u03b52\u03b55)(k2\u03b51) + (\u03b55\u03b51)(k5\u03b52) h i but the \ufb01rst term vanishes because \u03b51 \u00b7 \u2212 \u03b52 = 0. Similarly, the \ufb01rst term of, (81.19) iV345\u2032 = i\u221a2g \u2212 (\u03b53\u03b54)(k3\u03b55) + (\u03b54\u03b55)(k4\u03b53) + (\u03b55\u03b53)( h \u2212 k5\u03b54) i (81.20) also vanishes. When we take the product of these two vertices, and replace the internal polarizations with the propagator, as indicated in eq. (81.17), only the product of the third term of eq. (81.19) with the second term of eq. (81.20) is nonzero; all other terms include a vanishing product of polarizations. We get", " ig2A(1\u2212, 2\u2212, 3+, 4+) = ( \u2212 i\u221a2g)2(i/s12)(\u03b51\u03b54)(k5\u03b52)(k4\u03b53). (81.21) 81: Scattering in Quantum Chromodynamics 486 Since k5 = k1 \u00b7 \u2212 \u03b52 and k4 \u00b7 k2, and k2 \u00b7 k1\u2212 \u03b53 via the general formulae \u03b52 = 0, we have k5 \u00b7 \u03b5+(k;q) = h p \u00b7 q p \u221a2 [p k] i q k h i p k [q p] h i \u221a2 [q k] (k;q) = p \u03b5 \u2212 \u00b7 Setting q2 = k3 and q3 = k2, we get \u03b52 = k1 \u00b7 [3 1] 1 2 i h \u221a2 [3 2] k4 \u00b7 \u03b53 = h 2 4 [4 3] i \u221a2 2 3 i h,. \u03b52 = k1 \u00b7 \u2212 \u03b52. We evaluate,. (81.22) (81.23) (81.24) (81.25) Using eqs. (81.15), (81.18), (81.24), and (81.25) in eq. (81.21), and using antisymmetry of the twistor products to cancel common factors, we get A(1\u2212, 2\u2212, 3+, 4+) = h 2 1 i [2 1] [3 2] [4 3]2 2 3 i h. (81.26) We can make our result look nicer by multiplying the numerator and denominator by. In the numerator, we use 3 4 i h 3 4 i h [4 3] = s34 = s12 = [2 1], 1 2 i h (81.27) and cancel the [2 1] with the one in the denominator. Now multiply the nu, and use the momentum-conservation merator and denominator by 4 1 i h in the numerator with identity (see problem 60.2) to replace 4 1 i h [2 3], and cancel the [2 3] with the [3 2] in the denominator (which \u2212h yields a minus sign). Finally, multiply the numerator and denominator by 1 2 i h 2 1 i to get [4 3 ih This is our \ufb01nal result for A(1\u2212, 2\u2212, 3+, 4+). A(1\u2212, 2\u2212, 3+, 4+) = 1 2 h ih. i (81.28) 4 1 ih Now, using cyclic symmetry, we can get any partial amplitude where the two negative helicities are adjacent; for example, A(1+, 2\u2212, 3\u2212, 4+) =. (81.29) 1 2 h We must still calculate one partial amplitude where the negative helicities are not adjacent, such as A(1\u2212, 2+, 3\u2212, 4+). Once we have it, we can use cyclic symmetry to get all the remaining partial amplitudes. 4 1 ih ih ih 81: Scattering in Quantum Chromodynamics 487 Before turning to this calculation, let us consider the problem of squaring the total amplitude and summing over colors. Because the generator matrices are traceless, we should (as we discussed in section 80) use the completeness relation (T a)i j(T a)k l = \u03b4i l\u03b4k j 1 N \u03b4i j\u03b4k l. \u2212 (81.30) However, recall that the Yang\u2013Mills \ufb01eld strength is ig \u221a2 (81.31) [A\u00b5,A\u03bd ]. F\u00b5\u03bd = \u2202\u00b5A\u03bd \u2212 \u2202\u03bdA\u00b5 \u2212 If we allow a generator matrix proportional to the identity, which corresponds to a gauge group of U(N ) rather than SU(N ), then this extra U(1) generator commutes with every other generator. Thus the U(1) \ufb01eld does not appear in the commutator term in eq. (81.31). Since it is this commutator term that is responsible for the interaction of the gluons, the U(1) \ufb01eld is a free \ufb01eld. Therefore, any scattering amplitude involving the associated particle (which we will call the \ufb01ctitious photon) must be zero. Thus, if we write a scattering amplitude in the form of eq. (81.2), and replace one of the T a\u2019s with the identity matrix, the result must be zero. This decoupling of the \ufb01ctitious photon allows us to use the much simpler completeness relation (T a)i j", "(T a)k l = \u03b4i l\u03b4k j (81.32) in place of eq. (81.30). There is no need to subtract the U(1) generator from the sum over the generators, as we did in eq. (81.30), because the terms involving it vanish anyway. The decoupling of the \ufb01ctitious photon is useful in another way. Let us I in eq. (81.2). Then we have apply it to the case of n = 4, and set T a4 \u221d 0 = Tr(T a1 T a2T a3) h + Tr(T a1 T a3T a2) h A(1, 2, 3, 4) + A(1, 2, 4, 3) + A(1, 4, 2, 3) A(1, 3, 2, 4) + A(1, 3, 4, 2) + A(1, 4, 3, 2). (81.33) i i The contents of each square bracket must vanish. Requiring this of the \ufb01rst term yields A(1, 2, 3, 4) = A(1, 2, 4, 3) \u2212 \u2212 A(1, 4, 2, 3). (81.34) Assigning some helicities, this reads A(1\u2212, 2+, 3\u2212, 4+) = A(1\u2212, 2+, 4+, 3\u2212) \u2212 \u2212 A(1\u2212, 4+, 2+, 3\u2212). (81.35) Note that we have now expressed a partial amplitude with nonadjacent negative helicities in terms of partial amplitudes with adjacent negative 81: Scattering in Quantum Chromodynamics 488 helicities, which we have already calculated. Thus we have A(1\u2212, 2+, 3\u2212, 4+) = \u2212 \" 4 3 1 i h 2 4 1 2 ih + 4 3 i ih 3 1 h ih 4 3 1 i h 4 2 1 4 ih 2 3 i # ih 3 1 ih ih ih + ih 1 ih 2 3 i ih 2 3 3 4 i i ih 1 2 h ih 1 3 \u2212 h ih 3 4 1 2 h ih ih 1 4 4 2 i 2 3 ih i, (81.36) where the last line follows from the Schouten identity (see problem 50.3). A \ufb01nal clean-up yields (1\u2212, 2+, 3\u2212, 4+) =. (81.37) 1 2 h Now that we have all the partial amplitudes, we can compute the color2. There are only three partial amplitudes that are not related summed by either cyclic permutations, eq. (81.3), or re\ufb02ections, eq. (81.9); we can take these to be |T | 4 1 ih ih ih i A3 \u2261 A4 \u2261 A2 \u2261 A(1, 2, 3, 4), A(1, 3, 4, 2), A(1, 4, 2, 3), (81.38) (81.39) (81.40) where the subscript on the left-hand side is the third argument on the right-hand side. (Switching the second and fourth arguments is equivalent to a re\ufb02ection and a cyclic permutation, and so leaves the partial amplitude unchanged.) This mimics the notation we used at the end of section 80, and we can apply our result from there to the color sum, 2 = 2N 2(N 2 |T | Xcolors 1)g4 \u2212 Xj Aj| | 2 + 4N 2g4( A\u2217j )( Ak), (81.41) Xj Xk where j and k are summed over 2, 3, 4. eq. (81.34) is equivalent to vanishes. Our result for the color-summed squared amplitude is then In the present case, however, jAj = 0, so the second term in eq. (81.41) P 2 = 2N 2(N 2 |T | Xcolors 1)g4 \u2212 2 + A2| | 2 + A3| | 2 A4| |. (81.42) 81: Scattering in Quantum Chromodynamics 489 Figure 81.2: Color-ordered diagrams for \u00afqqgg scattering. 1 2 4 3 1 2 4 3 Figure 81.3: The double-line version of \ufb01g. (81.2); the associated color factor is (T a3 T a4)i2 i1. For the case where 1 and 2 are the incoming gluons, and 3 and 4 are the outgoing gluons, we can write this in terms of the usual Mandelstam", " variables s = s12 = s34, t = s13 = s24, and u = s14 = s23 by recalling that, etc. Let us take the case where gluons 1 and 2 |h have negative helicity, and 3 and 4 have positive helicity. In this case, we see from eqs. (81.28) and (81.37) that the numerator in every nonvanishing partial amplitude is 2 = [1 2] | | s12| | 2 = 1 2 i| 4. Then we get 1 2 i h 2 1\u22122\u22123+4+ = 2N 2(N 2 |T | Xcolors 1)g4s4 \u2212 1 s2t2 + 1 t2u2 + 1 u2s2. (81.43) We can also sum over helicities. There are six patterns of two positive + ++ and ++ and two negative helicities; \u2212\u2212 + yield u4. The helicity sum is and + and + + \u2212 therefore yield a factor of s4, yield t4, and \u2212\u2212 ++ \u2212\u2212 \u2212 \u2212 \u2212 + \u2212 \u2212 2 = 4N 2(N 2 |T | Xcolors helicities 1)g4(s4 + t4 + u4) \u2212 1 s2t2 + 1 t2u2 + 1 u2s2. (81.44) Of course, we really want to average (rather than sum) over the initial colors and helicities; to do so we must divide eq. (81.44) by 4(N 2 1)2. Next we turn to scattering of quarks and gluons. We consider a single type of massless quark: a Dirac \ufb01eld in the N representation of SU(N ). \u2212 81: Scattering in Quantum Chromodynamics 490 The lagrangian for this \ufb01eld is is D\u00b5 = \u2202\u00b5 \u2212 L (ig/\u221a2)A\u00b5. Thus the color-ordered vertex factor is = i\u03a8 /D\u03a8, where the covariant derivative iV\u00b5 = (ig/\u221a2)\u03b3\u00b5. (81.45) To get the color factor, we use the double-line notation, with a single line for the quark. As an example, consider the process of \u00afqq gg (and its crossing-related cousins). The contributing color-ordered tree diagrams are shown in \ufb01g. (81.2). The corresponding double-line diagrams are shown in \ufb01g. (81.3); the quark is represented by a single line, with an arrow direction that matches its charge arrow. To get the color factor, we start with line 2, and follow the arrows backwards; the result is (T a3 T a4)i2 i1. The complete amplitude can then be written as \u2192 = g2 T (T a3 T a4)i2 h i1A(1\u00afq, 2q, 3, 4) + (T a4T a3)i2 i1A(1\u00afq, 2q, 4, 3) i, (81.46) where A(1\u00afq, 2q, 3, 4) is the appropriate partial amplitude. The subscripts q and \u00afq indicate the labels that correspond to an outgoing quark and outgoing antiquark, respectively. From our results for spinor electrodynamics in section 60, we know that a nonzero amplitude requires opposite helicities on the two ends of any fermion line. Consider, then, the case of +\u03bb3\u03bb4. The partial amplitude T\u2212 corresponding to the diagrams of \ufb01g. (81.2) is ig2A(1\u2212\u00afq, 2+ q, 3, 4) = (ig/\u221a2)2(1/i)[2 /\u03b53( | \u2212 + (ig/\u221a2)[2 iV345 1 /\u03b55| i | /p5/p2 1 5)/\u03b54| i \u03b5\u00b5 5 \u03b5\u03bd 5 \u2192 Suppose both gluons have positive helicity. Then using ig\u00b5\u03bd/s12. (81.47) /\u03b5+(k;q) = q k] h | | + q | [k i |, (81.48) i we can get both lines of eq. (81.47) to vanish by choosing q3 = q4 = p1. Similarly, if both gluons have negative helicity, then using \u221a2 q k h (k;q) = /\u03b5 \u2212 \u221a2 [q k] k | [q i | + k q] h | |, (81.49) we can get both lines of eq. (81.47) to vanish by choosing q3 = q4 = p", "2. So the gluons must have opposite helicities to get a nonzero amplitude. Consider, then, the case of \u03bb3 = + and \u03bb4 =. We can get V345 to vanish by choosing q3 = k4 and q4 = k3. The partial amplitude is then given by just the \ufb01rst line of eq. (81.47), \u2212 A(1\u2212\u00afq, 2+ q, 3+, 4\u2212) = 1 /\u03b53+(/p1+/k4)/\u03b54\u2212 2 [2 | /( 1 i | \u2212 s14). (81.50) 81: Scattering in Quantum Chromodynamics 491 With q3 = k4 and q4 = k3, we have /\u03b53+ = /\u03b54\u2212 = \u221a2 4 3 h \u221a2 [3 4] [3 4 | i | +, 4 3] | h | i [3 4 | i | +. 4 3] | h | /p = p \u2212 | [p i p p] h | \u2212 |, | Using the identity eq. (81.50) becomes (81.51) (81.52) (81.53). (81.54) A(1\u2212\u00afq, 2+ q, 3+, 4\u2212) = [2 3] 4 1 [1 3] 4 1 i h i h [3 4] s14 4 3 i h [2 3] In the numerator, we use [1 3] set s14 = s23 = 2 3 i h tor by, and use [2 3] 1 2 i h multiply both by = 4 1 i h. In the denominator, we 4 2 i h [3 2]. Then we multiply the numerator and denominain the numerator. Finally we = \u2212 \u2212 [4 3] 1 2 i h 1 4 i h, and rearrange to get 1 4 h i A(1\u2212\u00afq, 2+ q, 3+, 4\u2212) = ih ih ih ih ih ih 81.55) (81.56) An analogous calculation yields A(1\u2212\u00afq, 2+ q, 3\u2212, 4+) = The remaining nonzero amplitudes are related by complex conjugation. Now that we have all the partial amplitudes, we can compute the color2. To do so, we multiply eq. (81.46) by its complex conjugate, summed and use hermiticity of the generator matrices to get |T | 2 = g4 |T | Xcolors 2 2 + A3| | Tr(T aT bT bT a) h + Tr(T aT bT aT b) A\u22173A4 + A\u22174A3 A4| | i, (81.57) A(1\u00afq, 2q, 3, 4) and A4 \u2261 A(1\u00afq, 2q, 4, 3). The traces are easily where A3 \u2261 evaluated with the double-line technique of section 80; because the \ufb01ctitious photon couples to the quark, we must use eq. (81.30) to project it out. The traces in eq. (81.57) are also easily evaluated with the grouptheoretic methods of section 70, with the normalization that the index of the fundamental representation is one: T (N) = 1. Either way, the results are Tr(T aT bT bT a) = +(N 2 Tr(T aT bT aT b) = (N 2 \u2212 1)2/N, 1)/N. \u2212 \u2212 (81.58) (81.59) 81: Scattering in Quantum Chromodynamics 492 \u2212 \u2212 The sum over the four possible helicity patterns ( + ) is left as an exercise. + Now that we have calculated these scattering amplitudes for quarks and gluons, an important questions arises: why did we bother to do it? Quarks and gluons are con\ufb01ned inside colorless bound states, the hadrons, and so apparently cannot appear as incoming and outgoing particles in a scattering event. +, + \u2212\u2212 ++ +, \u2212 \u2212 \u2212 \u2212 +, \u2261 To answer this question, suppose we collide two hadrons with a centerof-mass energy E = \u221as large enough so that the QCD coupling g is small when renormalized in the MS scheme with \u00b5 = E. (In the real world, we g2/4\u03c0 = 0.12 for \u00b5 = MZ = 91 GeV.) Then we can think of have \u03b1 each hadron as being made up of a loose collection of quarks and gluons, and these parts of a hadron, or partons, can be treated as independent participants in scattering processes. In order to extract quantitative results for", " hadron scattering (a project beyond the scope of this book), we need to know how each hadron\u2019s energy and momentum is shared among its partons. This is described by parton distribution functions. At present, these cannot be calculated from \ufb01rst principles, but they have to satisfy a variety of consistency conditions that can be derived from perturbation theory, and that relate their values at di\ufb00erent energies. These conditions are well satis\ufb01ed by current experimental data. Reference Notes More detail on how hadron scattering experiments can be compared with parton scaterring amplitudes can be found in Peskin & Schroeder, Muta, Quigg, and Sterman. Problems 81.1) Compute the four-gluon partial amplitude A(1\u2212, 2+, 3\u2212, 4+) directly from the Feynman diagrams, and verify eq. (81.37). 81.2) Compute the \u00afqqgg partial amplitude A(1\u2212\u00afq, 2+ q, 3\u2212, 4+) with q4 = p1 and q3 = k4. Show that, with this choice of the reference momenta, the \ufb01rst line of eq. (81.47) vanishes. Evaluate the second line, and veri\ufb01y eq. (81.55). 81.3) Compute A(1\u2212\u00afq, 2+ q, 3\u2212, 4+), and verify eq. (81.56). 81.4) a) Verify eqs. (81.58) and (81.59) using the double-line notation of section 80. b) Compute Tr(T a R) in terms of the index T (R) and dimension D(R) of the representation R, and the index R ) and Tr(T a RT a RT RT b 81: Scattering in Quantum Chromodynamics 493 T (A) and dimension D(A) of the adjoint representation. Verify that your results reproduce eqs. (81.58) and (81.59). 81.5) Compute the sum over helicities of eq. (81.57). Express your answer in terms of s, t, and u for the process \u00afqq gg. \u2192 81.6) Consider the partial amplitude A(1\u2212\u00afq, 2+ q, 3\u2212, 4+, 5+). Show that, with the choice q3 = k2 and q4 = q5 = k1, there are just two contributing diagrams. Evaluate them. After some manipulations, you should be able to put your result in the form A(1\u2212\u00afq, 2+ q, 3\u2212, 4+, 5+) = ih ih 1 2 h ih. (81.60) 5 1 ih i 82: Wilson Loops, Lattice Theory, and Con\ufb01nement 494 82 Wilson Loops, Lattice Theory, and Confinement Prerequisite: 29, 73 In this section, we will contruct a gauge-invariant operator, the Wilson loop, whose vacuum expectation value (VEV for short) can diagnose whether or not a gauge theory exhibits con\ufb01nement. A theory is con\ufb01ning if all \ufb01nite-energy states are invariant under a global gauge transformation. U(1) gauge theory\u2014quantum electrodynamics\u2014is not con\ufb01ning, because there are \ufb01nite-energy states (such as the state of a single electron) that have nonzero electric charge, and hence change by a phase under a global gauge transformation. Con\ufb01nement is a nonperturbative phenomenon; it cannot be seen at any \ufb01nite order in the kind of weak-coupling perturbation theory that we have been doing. (This is why we had no trouble calculating quark and gluon scattering amplitudes.) In this section, we will introduce lattice gauge theory, in which spacetime is replaced by a discrete set of points; the inverse lattice spacing 1/a then acts as an ultraviolet cuto\ufb00 (see section 29). This cuto\ufb00 theory can be analyzed at strong coupling, and, as we will see, in this regime the VEV of the Wilson loop is indicative of con\ufb01nement. The outstanding question is whether this phenomenon persists as we simultaneously lower the coupling and increase the ultraviolet cuto\ufb00 (with the relationship between the two governed by the beta function), or whether we encounter a phase transition, signalled by a sudden change in the behavior of the Wilson loop VEV. We take the gauge group to be SU(N ). Consider two spacetime points x\u00b5 and x\u00b5 + \u03b5\u00b5, where \u03b5\u00b5 is in\ufb01nitesimal. De\ufb01", "ne the Wilson link W (x+\u03b5, x) \u2261 exp[ig\u03b5\u00b5A\u00b5(x)], (82.1) where A\u00b5(x) is an N Since \u03b5 is in\ufb01nitesimal, we also have \u00d7 N matrix-valued traceless hermitian gauge \ufb01eld. W (x+\u03b5, x) = I + ig\u03b5\u00b5A\u00b5(x) + O(\u03b52). (82.2) Let us determine the behavior of the Wilson link under a gauge transformation. Using the gauge transformation of A\u00b5(x) from section 69, we \ufb01nd W (x+\u03b5, x) \u2192 1 + ig\u03b5\u00b5U (x)A\u00b5(x)U \u2020(x) \u2212 \u03b5\u00b5U (x)\u2202\u00b5U \u2020(x), (82.3) where U (x) is a spacetime-dependent special unitary matrix. Since U U \u2020 = 1, we have U \u2202\u00b5U \u2020 = +(\u2202\u00b5U )U \u2020; thus we can rewrite eq. (82.3) as \u2212 W (x+\u03b5, x) \u2192 (1 + \u03b5\u00b5\u2202\u00b5)U (x) U \u2020(x) + ig\u03b5\u00b5U (x)A\u00b5(x)U \u2020(x). (82.4) 82: Wilson Loops, Lattice Theory, and Con\ufb01nement 495 In the \ufb01rst term, we can use (1 + \u03b5\u00b5\u2202\u00b5)U (x) = U (x+\u03b5) + O(\u03b52). In the second term, which already contains an explicit factor of \u03b5\u00b5, we can replace U (x) with U (x+\u03b5) at the cost of an O(\u03b52) error. Then we get W (x+\u03b5, x) \u2192 which is equivalent to U (x+\u03b5) 1 + ig\u03b5\u00b5A\u00b5(x) U \u2020(x), (82.5) W (x+\u03b5, x) U (x+\u03b5)W (x+\u03b5, x)U \u2020(x). \u2192 Note also that eq. (82.1) implies W \u2020(x+\u03b5, x) = W (x shift x to x + \u03b5 at the cost of an O(\u03b52) error, and so (82.6) \u03b5, x). We can \u2212 W \u2020(x+\u03b5, x) = W (x, x+\u03b5), (82.7) which is consistent with eq. (82.6). Now consider mutiplying together a string of Wilson links, speci\ufb01ed by a starting point x and n sequential in\ufb01nitesimal displacement vectors \u03b5j. The ordered set of \u03b5\u2019s de\ufb01nes a path P through spacetime that starts at x and ends at y = x + \u03b51 +... + \u03b5n. The Wilson line for this path is WP (y, x) W (y, y \u2261 \u03b5n)... W (x+\u03b51+\u03b52, x+\u03b51)W (x+\u03b51, x). \u2212 (82.8) Using eq. (82.6) and the unitarity of U (x), we see that, under a gauge transformation, the Wilson line transforms as WP (y, x) \u2192 U (y)WP (y, x)U \u2020(x). (82.9) Also, since hermitian conjugation reverses the order of the product in eq. (82.8), using eq. (82.7) yields where \u2212 W \u2020P (y, x) = W P (x, y), \u2212 (82.10) P denotes the reverse of the path P. Now consider a path that returns to its starting point, forming a closed, oriented curve C in spacetime. The Wilson loop is the trace of the Wilson line for this path, WC \u2261 Using eq. (82.9), we see that the Wilson loop is gauge invariant, Tr WC(x, x). Also, eq. (82.10) implies WC \u2192 WC. W \u2020C = W C, \u2212 (82.11) (82.12) (82.13) 82: Wilson Loops, Lattice Theory, and Con\ufb01nement 496 \u2212 where C denotes the curve C traversed in the opposite direction. To gain some intuition, we will calculate for U(1) gauge theory, without charged \ufb01elds. This is simply a", " free-\ufb01eld theory, and the calculation can be done exactly. 0 WC | 0 i | h In order to avoid dealing with i\u01eb issues, it is convenient to make a Wick rotation to euclidean spacetime (see section 29). The action is then S = d4x 1 4 F\u00b5\u03bd F\u00b5\u03bd, Z (82.14) \u2202\u03bd A\u00b5. The VEV of the Wilson loop is now given by where F\u00b5\u03bd = \u2202\u00b5A\u03bd \u2212 the path integral 0 WC | 0 i | h = A eig D Z dx\u00b5A\u00b5e\u2212 S. C H (82.15) C dx\u00b5 as a current J\u00b5(x), we can apply our results If we formally identify g for the path integral from section 57. After including a factor of i from the Wick rotation, we get H 0 WC | 0 i | h = exp 1 2 g2 \u2212 IC dx\u00b5 IC dy\u03bd \u2206\u00b5\u03bd (x \u2212, y) (82.16) where \u2206\u00b5\u03bd(x man gauge, we have \u2212 y) is the photon propagator in euclidean spacetime. In Feyn- \u2206\u00b5\u03bd(x \u2212 y) = \u03b4\u00b5\u03bd Z d4k (2\u03c0)4 eik y) (x \u00b7 \u2212 k2 = \u03b4\u00b5\u03bd = \u03b4\u00b5\u03bd 4\u03c0 (2\u03c0)4 4\u03c0 (2\u03c0)4 \u03b4\u00b5\u03bd = = 4\u03c02(x \u2212 \u03b4\u00b5\u03bd 4\u03c02(x \u2212 y)2 0 Z y)2, x y | \u2212 | cos \u03b8 k3 dk k2 k3 dk k2 \u221e \u221e 0 Z 0 Z \u03c0 d\u03b8 sin2\u03b8 eik 0 Z x \u03c0J1( ) | \u221e du J1(u) (82.17) where J1(u) is a Bessel function. Since \u2206\u00b5\u03bd (x y, the double line integral in eq. (82.16) will yield a factor of the perimeter P of the curve C. There is also an ultraviolet divergence as x approaches y; we will cut this o\ufb00 at a length scale a. The result is then y) depends only on x \u2212 \u2212 0 WC | 0 i | h = exp[ \u2212 (\u02dccg2/a)P ], (82.18) 82: Wilson Loops, Lattice Theory, and Con\ufb01nement 497 R \u226a T. 0 WC | 0 i | h where \u02dcc is a numerical constant that depends on the shape of C and the details of the cuto\ufb00 procedure. This behavior of the Wilson loop in euclidean spacetime\u2014exponential decay with the length of the perimeter\u2014is called the perimeter law. It is indicative of uncon\ufb01ned charges. We can gain more insight into the meaning of by taking C to be a rectangle, with length T in the time direction and R in a space direction, where a (Of course, in euclidean spacetime, the choice of the time direction is an arbitrary convention.) The reason for C dx\u00b5 corresponds to a point this particular shape is that the current g charge moving along the curve C. When the particle is moving backwards H in time, the associated minus sign is equivalent to a change in the sign of its charge. So when we compute, we are doing the path integral 0 WC | 0 i | h in the presence of a pair of point charges with opposite sign, separated by a distance R, that exists for a time T. On general principles (see section 6), this path integral is proportional to exp( EpairT ), where Epair is the ground-state energy of the quantum electrodynamic \ufb01eld in the presence of the charged particle pair. \u226a \u2212 We now turn to the calculation. If both x and y are on the same side of the rectangle, we \ufb01nd L L dx dy 0 0 Z Z (x \u2212 y)2 = 2L/a 2 ln(L/a) + O(1), (82.19) \u2212 where L is the length of the side (either R or T ), and the O(1) term is a numerical constant that depends on the details of the short-distance cuto\ufb00. If x and y are on perpendicular sides, the double line integral is zero, because then dx dy = 0. If x is on one short side and y on the other, the integral evaluates to R2/T 2, and this we can neglect. Finally, if x is on one long side and y is on the other, we have \u00b7 T T 0 0 Z Z (x dx dy y)2 + R", "2 = \u03c0T /R \u2212 2 ln(T /R) \u2212 \u2212 2 + O(R2/T 2). (82.20) Adding up all these contributions, we \ufb01nd in the limit of large T that dx (x \u00b7 \u2212 dy y)2 = 4/a \u2212 IC IC 2\u03c0/R T + O(ln T ). (82.21) Combining this with eqs. (82.16) and (82.17), and setting \u03b1 = g2/4\u03c0, we \ufb01nd 0 WC | 0 i | h = exp \u2212 2\u03b1 \u03c0a \u2212 \u03b1 R T. (82.22) Comparing this with the general expectation EpairT ), we \ufb01nd a cuto\ufb00 dependent contribution to Epair that represents a divergent exp( \u2212 0 WC | 0 i \u221d | h 82: Wilson Loops, Lattice Theory, and Con\ufb01nement 498 2 x 1 Figure 82.1: The minimal Wilson loop on a hypercubic lattice goes around an elementary plaquette; this one lies in the 1-2 plane. self-energy for each point particle, plus the Coulomb potential energy for the pair, V (R) = \u03b1/R. In the nonabelian case, where there are interactions among the gluons, \u2212 we must expand everything in powers of g. Then we \ufb01nd 0 WC | 0 i | h 1 = Tr \u2212 1 2 g2T aT a dx\u00b5 IC IC dy\u03bd \u2206\u00b5\u03bd(x \u2212 y) + O(g4). (82.23) Since T aT a equals the quadratic Casimir C(N) for the fundamental representation (times an identity matrix), we see that to leading order in g2 we simply reproduce the results of the abelian case, but with g2 g2C(N). We can also consider a Wilson loop in a di\ufb00erent representation by setting A\u00b5(x) = Aa R. Then, at leading order, we get a factor of C(R) instead of C(N). Perturbative corrections can be computed via standard Feynman diagrams with gluon lines that, in position space, have at least one end on the curve C. \u00b5(x)T a \u2192 Next we turn to a strong coupling analysis. We begin by constructing a lattice action for nonabelian gauge theory. Consider a hypercubic lattice of points in four-dimensional euclidean spacetime, with a lattice spacing a between nearest-neighbor points. The smallest Wilson loop we can make on this lattice goes around an elementary square or plaquette, as shown in \ufb01g. (82.1). Let \u03b51 and \u03b52 be vectors of length a in the 1 and 2 directions, and let x be the point at the center of the plaquette. Using the center of each link as the argument of the gauge \ufb01eld, and using the lower-left corner as the starting point, we have (multiplying the Wilson links from right to left along the path) Wplaq = Tr e\u2212 igaA2(x \u03b51/2)e\u2212 \u2212 igaA1(x+\u03b52/2)e+igaA2(x+\u03b51/2)e+igaA1(x \u2212 \u03b52/2). (82.24) If we now treat the gauge \ufb01eld as smooth and expand in a, we get Wplaq = Tr e\u2212 igaA2(x)+iga2\u22021A2(x)/2+... e\u2212 \u2212 e+igaA2(x)+iga2\u22021A2(x)/2+... e+igaA1(x) \u2212 igaA1(x) iga2\u22022A1(x)/2+... iga2\u22022A1(x)/2+.... (82.25) \u00d7 82: Wilson Loops, Lattice Theory, and Con\ufb01nement 499 Next we use eAeB = eA+B+[A,B]/2+... to combine the two exponential factors on the \ufb01rst line of eq. (82.25), and also the two exponential factors on the second line. Then we use this formula once again to combine the two results. We get Wplaq = Tr e+iga2(\u22021A2 \u22022A1 ig[A1,A2])+..., \u2212 \u2212 where all \ufb01elds are evaluated at x. expand the exponentials, we \ufb01nd If we now take Wplaq + W (82.", "26) plaq and \u2212 plaq = 2N g2a4 TrF 2 12 +..., (82.27) \u2212 ig[A1, A2] is the Yang\u2013Mills \ufb01eld strength. From where F12 = \u22021A2\u2212 eq. (82.27), we conclude that an appropriate action for Yang\u2013Mills theory on a euclidean spacetime lattice is \u2212 Wplaq + W \u22022A1\u2212 S = 1 2g2 \u2212 Wplaq, Xplaq (82.28) where the sum includes both orientations of all plaquettes. Each Wplaq is N matrices, expressed as the trace of the product of four special unitary N one for each oriented link in the plaquette. If U is the matrix associated with one orientation of a particular link, then U \u2020 is the matrix associated with the opposite orientation of that link. The path integral for this lattice gauge theory is \u00d7 where U e\u2212 S, Z = D Z U = dUlink, D (82.29) (82.30) Ylinks and dU is the Haar measure for a special unitary matrix. The Haar measure V U, where V is a constant speis invariant under the transformation U \u2192 cial unitary matrix, and is normalized via dU = 1; this \ufb01xes it uniquely. For N 3, it obeys R \u2265 dU Uij = 0, Z dU UijUkl = 0, Z dU UijU \u2217kl = 1 N \u03b4ik\u03b4jl, (82.31) (82.32) (82.33) which is all we will need to know. Z Now consider a Wilson loop, expressed as the trace of the product of the U \u2019s associated with the oriented links that form a closed curve C. For simplicity, we take this curve to lie in a plane. We have 0 WC | WC e\u2212 S. (82.34) 82: Wilson Loops, Lattice Theory, and Con\ufb01nement 500 S \u2192 We will evaluate eq. (82.34) in the strong coupling limit by expanding e\u2212 in powers of 1/g2. At zeroth order, e\u2212 1; then eq. (82.31) tells us that the integral over every link in C vanishes. To get a nonzero result, we need S. This can only come to have a corresponding U \u2020 from the expansion of e\u2212 from a plaquette containing that link. But then the integral over the other links of this plaquette will vanish, unless there is a compensating U \u2020 for requires us each of them. We conclude that a nonzero result for 0 WC | 0 i | h S. Since to \ufb01ll the interior of C with plaquettes from the expansion of e\u2212 each plaquette is accompanied by a factor of 1/g2, we have S 0 WC | 0 i \u223c | h (1/g2)A/a2, (82.35) where A is the area of the surface bounded by C, and A/a2 is the number of plaquettes in this surface. Eq. (82.35) yields the area law for a Wilson loop, where 0 WC | 0 i \u221d | h e\u2212 \u03c4 A, (82.36) \u03c4 = c(g)/a2 is the string tension. In the strong coupling limit, c(g) = ln(g2) + O(1). (82.37) The area law for the Wilson loop implies con\ufb01nement. To see why, let us again consider a rectangular loop with area A = RT. Comparing eq. (82.36) with the general expectation EpairT ), we see that Epair = V (R) = \u03c4 R. This corresponds to a linear potential between nonabelian point charges in the fundamental representation. It takes an in\ufb01nite amount of energy to separate these charges by an in\ufb01nite distance; the charges are therefore con\ufb01ned. The coe\ufb03cient \u03c4 of R in V (R) is called the string tension because a linear potential is what we get from two points joined by a string with a \ufb01xed energy per unit length; the energy per unit length of a string is its tension. 0 WC | 0 i \u221d | h exp( \u2212 The string tension \u03c4 is a physical quantity that should remain \ufb01xed as we remove the cuto\ufb00 by lowering a. Thus lowering a requires us to lower g. The outstanding question is whether c(g) reaches zero at a \ufb01nite, nonzero value of g. If so, at this point there is a phase transition to an", " uncon\ufb01ned phase with zero string tension. This has been proven to be the case for abelian gauge theory (which also exhibits an area law at strong coupling, by the identical argument). In nonabelian gauge theory, on the other hand, analytic and numerical evidence strongly suggests that c(g) remains nonzero for all nonzero values of g. At small a and small g, the behavior of g as a function of a is governed a dg/da. (The minus sign arises because the by the beta function, \u03b2(g) = ultraviolet cuto\ufb00 is a\u2212 \u2212 1.) Requiring \u03c4 to be independent of a yields c\u2032(g) = 1 2 \u03b2(g)c(g). \u2212 (82.38) 82: Wilson Loops, Lattice Theory, and Con\ufb01nement 501 At small g, we have \u03b2(g) = b1g3 + O(g5), (82.39) \u2212 where b1 = 11N/48\u03c02 for SU(N ) gauge theory without quarks. Solving for c(g) yields c(g) = C exp(1/b1g2), (82.40) where C is an integration constant, which is nonzero if there is no phase transition. In this case, at small g, the string tension has the form \u03c4 = C exp(1/b1g2)/a2. (82.41) We then want to take the continuum limit of a \ufb01xed. \u2192 0 and g \u2192 0 with \u03c4 held Note that eq. (82.41) shows that the string tension, at weak coupling, is not analytic in g, and so cannot be computed via the Taylor expansion in g that is provided by conventional weak-coupling perturbation theory. Instead, the path integrals of eqs. (82.29) and (82.34) can be performed on a \ufb01nite-size lattice via numerical integration. The limiting factor in such a calculation is computer resources. Reference Notes An introduction to lattice theory is given in Smit. Problems 82.1) Let C be a circle of radius R. Evaluate the constant \u02dcc in eq. (82.18), y)2 where P = 2\u03c0R is the circumference of the circle. Replace 1/(x with zero when < a. Assume a R. \u2212 y x | \u2212 | \u226a 83: Chiral Symmetry Breaking 502 83 Chiral Symmetry Breaking Prerequisite: 76, 82 In the previous section, we discussed con\ufb01nement in Yang\u2013Mills theory without quarks. In the real world, there are six di\ufb00erent \ufb02avors of quark; see Table 1. Each \ufb02avor has a di\ufb00erent mass, and is represented by a Dirac \ufb01eld in the fundamental or 3 representation of the color group SU(3). Such a Dirac \ufb01eld is equivalent to two left-handed Weyl \ufb01elds, one in the fundamental representation, and one in the antifundamental or 3 representation. The lightest quarks are the up and down quarks, with masses of a few MeV. These masses are small, in the following sense. The gauge coupling g of QCD becomes large at low energies. If we truncate the beta function after some number of terms (in practice, four or fewer), and integrate it, we \ufb01nd that g becomes in\ufb01nite at some \ufb01nite, nonzero value of the MS parameter \u00b5; this value is called \u039bQCD. Measurements of the strength of 0.2 GeV. The up and the gauge coupling at high energies imply \u039bQCD \u223c down quark masses are much less than \u039bQCD. We can therefore begin with the approximation that the up and down quarks are massless. The mass of the strange quark is also somewhat less than \u039bQCD. It is sometimes useful (though clearly less justi\ufb01ed) to treat the strange quark as massless as well. 1 GeV, we can ignore the charm, bottom, and top quarks entirely; we will also ignore the strange quark for now. Let us, then, consider QCD with nF = 2 \ufb02avors of massless quarks. We then have left-handed Weyl \ufb01elds \u03c7\u03b1i, where \u03b1 = 1, 2, 3 is a color index for the 3 representation, and i = 1, 2 is a \ufb02avor index, and left-handed Weyl \ufb01elds \u03be\u03b1\u00af\u0131, where \u03b1 = 1, 2, 3 is a color index for the 3 representation, and \u00af\u0131 =", " 1, 2 is a \ufb02avor index; we distinguish this \ufb02avor index from the one for the \u03c7\u2019s by putting a bar over it, and we write it as a superscript for later notational convenience. We suppress the undotted spinor index carried by both \u03c7 and \u03be. The lagrangian is If we are interested in hadron physics at energies below \u223c = i\u03c7\u2020 \u03b1i \u00af\u03c3\u00b5(D\u00b5)\u03b1 L \u03b1. where D\u00b5 = \u2202\u00b5 \u2212 In addition to the SU(3) color gauge symmetry, this lagrangian has a global U(2) U(2) \ufb02avor symmetry: is invariant under \u00b5, with (T a 3 \u03b2 \u03c7\u03b2i + i\u03be\u2020\u00af\u0131\u03b1 \u00af\u03c3\u00b5( \u00afD\u00b5)\u03b1 \u00b5 and \u00afD\u00b5 = \u2202\u00b5 \u2212 4 F a\u00b5\u03bdF a igT a 3 \u03b2 \u03be\u03b2\u00af\u0131 (83.1) 3 Aa igT a \u00b5\u03bd, (T a \u03b2 = 3 )\u03b2 Aa )\u03b1 \u2212 \u2212 1 \u00d7 L \u03c7\u03b1i \u2192 \u03be\u03b1\u00af\u0131 \u2192 Li j\u03c7\u03b1j, (R\u2217)\u00af\u0131 \u00af\uf6be \u03be\u03b1\u00af\uf6be, (83.2) (83.3) where L and R\u2217 are independent 2 2 constant unitary matrices. (The \u00d7 complex conjugation of R is a notational convention that turns out to be 83: Chiral Symmetry Breaking 503 name up down strange charm bottom top Q symbol mass (GeV) 0.0017 +2/3 1/3 0.0039 \u2212 1/3 0.076 \u2212 +2/3 1.3 4.3 1/3 \u2212 +2/3 178 u d s c b t Table 1: The six \ufb02avors of quark. Each \ufb02avor is represented by a Dirac \ufb01eld in the 3 representation of the color group SU(3). Q is the electric charge in units of the proton charge. Masses are approximate, and are MS parameters. For the u, d, and s quarks, the MS scale \u00b5 is taken to be 2 GeV. For the c, b, and t quarks, \u00b5 is taken to be equal to the corresponding mass; e.g., the bottom quark mass is 4.3 GeV when \u00b5 = 4.3 GeV. convenient.) In terms of the Dirac \ufb01eld eqs. (83.2) and (83.3) read \u03a8\u03b1i = \u03c7\u03b1i \u03be\u2020\u03b1\u00af\u0131!, PL\u03a8\u03b1i \u2192 PR\u03a8\u03b1\u00af\u0131 \u2192 Li jPL\u03a8\u03b1i, R\u00af\u0131 \u00af\uf6bePR\u03a8\u03b1\u00af\uf6be, (83.4) (83.5) (83.6) where PL,R = 1 U(2)L \u00d7 a Dirac \ufb01eld di\ufb00erently is said to be chiral. \u03b35). Thus the global \ufb02avor symmetry is often called U(2)R. A symmetry that treats the left- and right-handed parts of 2 (1 \u2213 \u2192 e\u2212 However, there is an anomaly in the axial U(1) symmetry corresponding i\u03b1\u03b35 \u03a8 for the Dirac \ufb01eld). to L = R\u2217 = ei\u03b1I (which is equivalent to \u03a8 U(1)V, SU(2)R\u00d7 Thus the nonanomalous global \ufb02avor symmetry is SU(2)L\u00d7 where V stands for vector. The U(1)V transformation corresponds to i\u03b1\u03a8. The corresponding conL = R = e\u2212 served charge is quark number, the number of quarks minus the number of antiquarks; this is one third of the baryon number, the number of baryons minus the number of antibaryons. (Baryons are color-singlet bound states of three quarks; the proton and neuton are baryons. Mesons are colorsinglet bound states of a quark and an antiquark; pions are mesons.) i\u03b1I, or equivalently \u03a8 e\u2212 \u2192 Thus, U(1)V results in classi\ufb01cation of hadrons by their baryon numSU(2)R symmetry realized in nature? The vector ber. How is the SU(2)L \u00d7 83: Chiral Symmetry Breaking 504 subgroup SU(2)V, obtained by setting R = L in eq. (83.3), is known as isotopic spin or isospin symmetry. Hadrons clearly come in representations of SU(2)V: the", " lightest spin-one-half hadrons (the proton, mass 0.938 GeV, and the neutron, mass 0.940 GeV) form a doublet or 2 representation, while the lightest spin-zero hadrons (the \u03c00, mass 0.135 GeV, and the \u03c0\u00b1, mass 0.140 GeV) form a triplet or 3 representation. Isospin is not an exact symmetry; it is violated by the small mass di\ufb00erence between the up and down quarks, and by electromagnetism. Thus we see small di\ufb00erences in the masses of the hadrons assigned to a particular isotopic multiplet. The role of the axial part of the SU(2)L \u00d7 SU(2)R symmetry, obtained by setting R = L\u2020 in eq. (83.3), is harder to identify. The hadrons do not appear to be classi\ufb01ed into multiplets by a second SU(2) symmetry group. In particular, there is no evidence for a classi\ufb01cation that distinguishes the left- and right-handed components of spin-one-half hadrons like the proton and neutron. Reconciliation of these observations with the SU(2)L\u00d7 SU(2)R symmetry of the underlying lagrangian is only possible if the axial generators are spontaneously broken. The three pions (which have spin zero, odd parity, and are by far the lightest hadrons) are then identi\ufb01ed as the corresponding Goldstone bosons. They are not exactly massless (and hence are sometimes SU(2)R symmetry is, called pseudogoldstone bosons) because the SU(2)L \u00d7 as we just discussed, not exact. To spontaneously break the axial part of the SU(2)L \u00d7 SU(2)R, some operator that transforms nontrivially under it must acquire a nonzero vacuum expectation value, or VEV for short. To avoid spontaneous breakdown of Lorentz invariance, this operator must be a Lorentz scalar, and to avoid spontaneous breakdown of the SU(3) gauge symmetry, it must be a color singlet. Since we have no fundamental scalar \ufb01elds that could acquire a nonzero VEV, we must turn to composite \ufb01elds instead. The simplest cana = \u03a8\u03b1\u00af\uf6bePL\u03a8\u03b1i, where a is an undotted spinor index. (The didate is \u03c7a product of two \ufb01elds is generically singular, and a renormalization scheme must be speci\ufb01ed to de\ufb01ne it.) We assume that \u03b1i\u03be\u03b1\u00af\uf6be \u03b1i\u03be\u03b1\u00af\uf6be \u03c7a 0 0 a | i | h where v is a parameter with dimensions of mass. depends on the renormalization scheme; for MS with \u00b5 = 2 GeV, v 0.23 GeV. v3\u03b4i \u00af\uf6be, Its numerical value (83.7) \u2212 = \u2243 To see that this fermion condensate does the job of breaking the axial SU(2)R while preserving the vector generators, we generators of SU(2)L \u00d7 note that, under the transformation of eqs. (83.2) and (83.3), \u03b1i\u03be\u03b1\u00af\uf6be \u03c7a 0 0 a | i \u2192 | h Li k(R\u2217)\u00af\uf6be \u03b1k\u03be\u03b1\u00afn \u03c7a 0 0 \u00afnh a | i | 83: Chiral Symmetry Breaking v3(LR\u2020)i \u00af\uf6be, \u2192 \u2212 505 (83.8) where we used eq. (83.7) to get the second line. If we take R = L, corresponding to an SU(2)V transformation, the right-hand side of eq. (83.8) is unchanged from its value in eq. (83.7). This signi\ufb01es that SU(2)V [and also U(1)V] is unbroken. However, for a more general transformation with R = L, the right-hand side of eq. (83.8) does not match that of eq. (83.7), signifying the spontaneous breakdown of the axial generators. Eq. (83.7) is nonperturbative: \u03b1i\u03be\u03b1\u00af\uf6be \u03c7a vanishes at tree level. Pertur0 0 a | i | h bative corrections then also vanish, because of the chiral \ufb02avor symmetry of the lagrangian. Thus the value of v is not accessible in perturbation theory. On general grounds, we expect v \u039bQCD, since \u039bQ", "CD is the only mass scale in the theory when the quarks are massless. Similarly, \u039bQCD sets the scale for the masses of all the hadrons that are not pseudogoldstone bosons, including the proton and neutron. \u223c We can construct a low-energy e\ufb00ective lagrangian for the three pseudogoldstone bosons (to be identi\ufb01ed as the pions) in the following way. We allow the orientation in \ufb02avor space of the VEV of \u03c7a a to vary slowly as a function of spacetime. That is, in place of eq. (83.7), we write \u03b1i\u03be\u03b1\u00af\uf6be \u03b1i(x)\u03be\u03b1\u00af\uf6be \u03c7a 0 | h 0 a (x) i | = \u2212 v3Ui \u00af\uf6be(x), (83.9) where U (x) is a spacetime dependent unitary matrix. We can write it as U (x) = exp[2i\u03c0a(x)T a/f\u03c0], (83.10) where T a = 1 2 \u03c3a with a = 1, 2, 3 are the generator matrices of SU(2), \u03c0a(x) are three real scalar \ufb01elds to be identi\ufb01ed with the pions, and f\u03c0 is a parameter with dimensions of mass, the pion decay constant. We do not include a fourth generator matrix proportional to the identity, since the corresponding \ufb01eld would be the Goldstone boson for the U(1)A symmetry that is eliminated by the anomaly. Equivalently, we require det U (x) = 1. We will think of U (x) as an e\ufb00ective, low energy \ufb01eld. Its lagrangian should be the most general one that is consistent with the underlying SU(2)R transSU(2)L \u00d7 formation, we have SU(2)R symmetry.1 Under a general SU(2)L \u00d7 U (x) \u2192 LU (x)R\u2020, (83.11) where L and R are independent special unitary matrices. We can organize the terms in the e\ufb00ective lagrangian for U (x) (also known as the chiral lagrangian) by the number of derivatives they contain. Because U \u2020U = 1, 1U(1)V acts trivially on U (x), and so we need not be concerned with it. 6 83: Chiral Symmetry Breaking 506 there are no terms with no derivatives. There is one term with two (all others being equivalent after integrations by parts), = L \u2212 1 4 f 2 \u03c0 Tr \u2202\u00b5U \u2020\u2202\u00b5U. (83.12) If we substitute in eq. (83.10) for U, and expand in inverse powers of f\u03c0, we \ufb01nd = L \u2212 1 2 \u2202\u00b5\u03c0a\u2202\u00b5\u03c0a + 1 6 f \u2212 2 \u03c0 (\u03c0a\u03c0a\u2202\u00b5\u03c0b\u2202\u00b5\u03c0b \u03c0a\u03c0b\u2202\u00b5\u03c0b\u2202\u00b5\u03c0a) +.... (83.13) \u2212 Thus the pion \ufb01elds are conventionally normalized, and they have interactions that are dictated by the general form of eq. (83.12). These interactions lead to Feynman vertices that contain factors of momenta p divided by f\u03c0. Therefore, we can think of p/f\u03c0 as an expansion parameter. Of course, we should also add to all possible inequivalent terms with four or more derivatives, with coe\ufb03cients that include inverse powers of f\u03c0. These will lead to more vertices, but their e\ufb00ects will be suppressed by additional powers of p/f\u03c0. Comparison with experiment then yields f\u03c0 = 92.4 MeV. (In practice, the value of f\u03c0 is more readily determined from the decay rate of the pion via the weak interaction; see section 90 and problem 48.5.) L is accompanied by a factor of (4\u03c0f\u03c0)\u2212 This value for f\u03c0 may seem low; it is, for example, less than the mass of the \u201calmost masless\u201d pions. However, it turns out that tree and loop diagrams contribute roughly equally to any particular process if each extra derivative in \u03c0, and each loop momentum is cut o\ufb00 at 4\u03c0f\u03c0. Thus it is 4\u03c0f\u03c0 \u223c 1 GeV that sets the scale of the interactions, rather than f\u03c0 \u223c Now let us consider the e\ufb00ect of including the small masses for the up and down quarks. The most general mass term", " we can add to the lagrangian is 1 rather that f \u2212 100 MeV. L 1 Lmass = = = \u2212 \u2212 \u2212 \u03be\u03b1\u00af\uf6beM\u00af\uf6be i\u03c7\u03b1i + h.c. M\u00af\uf6be i\u03c7\u03b1i\u03be\u03b1\u00af\uf6be + h.c. Tr M \u03c7\u03b1\u03be\u03b1 + h.c., where M is a complex 2 transformation, we can bring M to the form 2 matrix. By making an SU(2)L \u00d7 \u00d7 M = mu 0 0 md! e\u2212 i\u03b8/2, (83.14) SU(2)R (83.15) where mu and md are real and positive. We cannot remove the overall phase \u03b8, however, without making a forbidden U(1)A transformation. A 83: Chiral Symmetry Breaking 507 nonzero value of \u03b8 has physical consequences, as we will discuss in section 9, and so 94. For now, we note that experimental observations \ufb01x we will set \u03b8 = 0 on this phenomenological basis. < 10\u2212 \u03b8 | | Next, we replace \u03c7\u03b1\u03be\u03b1 in eq. (83.14) with its spacetime dependent VEV, eq. (83.7). The result is a term in the chiral lagrangian that incorporates the leading e\ufb00ect of the quark masses, Lmass = v3 Tr(M U + M \u2020U \u2020). Here we continue to distinguish M and M \u2020, even though, with \u03b8 = 0, they are the same matrix. If we think of M as transforming as M RM L\u2020, while U transforms as U LU R\u2020, then Tr M U is formally invariant. We then require all terms in the chiral lagrangian to exhibit this formal invariance. (83.16) \u2192 \u2192 If we expand Lmass = = 4(v3/f 2 Lmass in inverse powers of f\u03c0, and use M \u2020 = M, we \ufb01nd \u03c0) Tr(M T aT b)\u03c0a\u03c0b +... T a, T b \u03c0) Tr(M { \u03c0)(Tr M )\u03c0a\u03c0a +.... )\u03c0a\u03c0b +... } 2(v3/f 2 (v3/f 2 \u2212 = \u2212 (83.17) \u2212 2 \u03b4ab to get the last line. From We used the SU(2) relation } eq. (83.17), we see that all three pions have the same mass, given by the Gell-Mann\u2013Oakes\u2013Renner relation, T a, T b { = 1 m2 \u03c0 = 2(mu + md)v3/f 2 \u03c0. (83.18) On the right-hand side, the quark masses and v3 depend on the renormalization scheme, but their product does not. In the real world, electromagnetic interactions raise the mass of the \u03c0\u00b1 slightly above that of the \u03c00. This framework is easily expanded to include the strange quark. The three pions (\u03c0+, \u03c0\u2212, mass 0.140 GeV; \u03c00, mass 0.135 GeV), the four kaons (K +, K \u2212, mass 0.494 GeV; K 0, K 0, mass 0.498 GeV), and the eta (\u03b7, mass 0.548 GeV) are identi\ufb01ed as the eight expected Goldstone bosons. We can assemble them into the hermitian matrix \u03c0aT a/f\u03c0 = \u03a0 \u2261 1 2f\u03c0 \u03b7 \u221a2\u03c0+ \u221a2K + \u03c00 + 1 \u221a3 \u221a2\u03c0\u2212 \u2212 \u221a2K \u2212 \u03c00 + 1 \u221a3 \u221a2 K 0 2 \u221a2K 0 \uf8f6. (83.19) The second line of eq. (83.17) still applies, but now the T a\u2019s are the generators of SU(3), and M includes a third diagonal entry for the strange quark mass. We leave the details to the problems. 83: Chiral Symmetry Breaking 508 Next we turn to the coupling of the pions to the nucleons (the proton and neutron). We de\ufb01ne a Dirac \ufb01eld Ni, where N1 = p (the proton) SU(2)R and N2 = n (the neutron). We assume that, under an SU(2)L \u00d7 transformation, PLNi \u2192 PRN\u00af\u0131 \u2192 Li jPLNj, R\u00af\u0131 \u00af\uf6bePRN\u00af\uf6be. (83.20) (83.21) SU(2)R invari", "ant, The standard Dirac kinetic term iN /\u2202N is then SU(2)L \u00d7 but the standard mass term mN NN is not. (Here mN is the value of the nucleon mass in the limit of zero up and down quark masses.) However, we can construct an invariant mass term by including appropriate factors of U and U \u2020, \u2212 There is one other parity, time-reversal, and SU(2)L \u00d7 term with one derivative. Including this term, we have mN N (U \u2020PL + U PR)N. Lmass = (83.22) SU(2)R invariant L = iN /\u2202N 1 \u2212 2 (gA\u2212 \u2212 mN N (U \u2020PL + U PR)N 1)iN \u03b3\u00b5(U \u2202\u00b5U \u2020PL + U \u2020\u2202\u00b5U PR)N, (83.23) where gA = 1.27 is the axial vector coupling. Its value is determined from the decay rate of the neutron via the weak interaction; see section 90. The form of the lagrangian in eq. (83.23) is somewhat awkward. It can be simpli\ufb01ed by \ufb01rst de\ufb01ning u(x) \u2261 exp[ i\u03c0a(x)T a/f\u03c0], so that U (x) \u2261 u2(x). Then we de\ufb01ne a new nucleon \ufb01eld (u\u2020PL + uPR)N. N \u2261 (83.24) (83.25) (This is a \ufb01eld rede\ufb01nition in the sense of problem 11.5.) Equivalently, using the unitarity of u, we have N = (uPL + u\u2020PR) N. (83.26) Using eq. (83.26) in eq. (83.23), along with the identities \u2202\u00b5U = (\u2202\u00b5u)u + u\u2020(\u2202\u00b5u), etc., we ultimately \ufb01nd u(\u2202\u00b5u), (\u2202\u00b5u\u2020)u = L /\u2202 N \u2212 mN NN + N /v N \u2212 gAN /a\u03b35N, (83.27) N where we have de\ufb01ned the hermitian vector \ufb01elds v\u00b5 \u2261 a\u00b5 \u2261 1 2 i[u\u2020(\u2202\u00b5u) + u(\u2202\u00b5u\u2020)], 1 2 i[u\u2020(\u2202\u00b5u) \u2212 u(\u2202\u00b5u\u2020)]. (83.28) (83.29) \u2212 = i 83: Chiral Symmetry Breaking 509 If we now expand u and u\u2020 in inverse powers of f\u03c0, we get = i N /\u2202 L N \u2212 mN NN + (gA/f\u03c0)\u2202\u00b5\u03c0a T a\u03b3\u00b5\u03b35 N N +.... (83.30) N and We can integrate by parts in the interaction term to put the derivative on the \ufb01elds. Then, if we consider a process where an o\ufb00-shell pion is scattered by an on-shell nucleon, we can use the Dirac equation to replace the derivatives of with factors of mN. We then \ufb01nd a coupling and of the pion to an on-shell nucleon of the form N N N L\u03c0NN = \u2212 ig\u03c0NN \u03c0a \u03c3a\u03b35 N, N (83.31) where we have set T a = 1 constant, 2 \u03c3a, and identi\ufb01ed the pion-nucleon coupling g\u03c0NN = gAmN /f\u03c0. The value of g\u03c0NN can be determined from measurements of the neutronproton scattering cross section, assuming that it is dominated by pion exchange; the result is g\u03c0NN = 13.5. Eq. (83.32), known as the Goldberger\u2013 Treiman relation, is then satis\ufb01ed to within about 5%. (83.32) Reference Notes The chiral lagrangian is treated in Georgi, Ramond II, and Weinberg II. Light quark masses are taken from MILC. Problems 83.1) Suppose that the color group is SO(3) rather than SU(3), and that each quark \ufb02avor is represented by a Dirac \ufb01eld in the 3 representation of SO(3). a) With nF \ufb02avors of massless quarks, what is the nonanomalous \ufb02avor symmetry group? b) Assume the formation of a color-", "singlet, Lorentz scalar, fermion condensate. Assume that it preserves the largest possible unbroken subgroup of the \ufb02avor symmetry. What is this unbroken subgroup? c) For the case nF = 2, how many massless Goldstone bosons are there? d) Now suppose that the color group is SU(2) rather than SU(3), and that each quark \ufb02avor is represented by a Dirac \ufb01eld in the 2 representation of SU(2). Repeat parts (a), (b), and (c) for this case. Hint: at least one of the answers is di\ufb00erent! 83: Chiral Symmetry Breaking 510 83.2) Why is there a minus sign on the right-hand side of eq. (83.7)? 83.3) Verify that eq. (83.13) follows from eq. (83.12). 83.4) Use eqs. (83.12) and (83.16) to compute the tree-level contribution to \u03c0c\u03c0d. Work in the isospin limit, m. Express your answer in terms of the Mandelstam the scattering amplitude for \u03c0a\u03c0b mu = md \u2261 variables and the pion mass m\u03c0. \u2192 83.5) Verify that eq. (83.27) follows from eqs. (83.26) and eq. (83.23). 83.6) Consider the case of three light quark \ufb02avors, with masses mu, md, and ms. u,d/ms. ms, and drop terms that are of order m2 \u03c0\u00b1 and m2 a) Find the masses-squared of the eight pseudogoldstone bosons. Take the limit mu,d \u226a b) Assume that m2 K \u00b1 each receive an electromagnetic contribution; to zeroth order in the quark masses, this contribution is the same for both, but the comparatively large strange quark mass results in an electromagnetic contribution to m2 K \u00b1 that is roughly twice as large as the electromagnetic contribution \u2206m2 \u03c0\u00b1. Use the observed masses of the \u03c0\u00b1, \u03c00, K \u00b1, and K 0 to compute muv3/f 2 \u03c0, mdv3/f 2 c) Compute the quark mass ratios mu/md and ms/md. d) Use your results from part (b) to predict the \u03b7 mass. How good is your prediction? \u03c0, and \u2206m2 \u03c0, msv3/f 2 EM to m2 EM. 83.7) Suppose that the U(1)A symmetry is not anomalous, so that we must include a ninth Goldstone boson. We can write U (x) = exp[ 2i\u03c0a(x)T a/f\u03c0 + i\u03c09(x)/f9]. (83.33) The ninth Goldstone boson is given its own decay constant f9, since there is no symmetry that forces it to be equal to f\u03c0. We write the two-derivative terms in the lagrangian as = L \u2212 1 4 f 2 \u03c0 Tr \u2202\u00b5U \u2020\u2202\u00b5U 1 4 F 2\u2202\u00b5(det U \u2020)\u2202\u00b5(det U ). (83.34) \u2212 a) By requiring all nine Goldstone \ufb01elds to have canonical kinetic terms, determine F in terms of f\u03c0 and f9. ms. Find the masses b) To simplify the analysis, let mu = md \u2261 of the nine pseudogoldstone bosons. Identify the three lightest as the pions, and call their mass m\u03c0. Show that another one of the nine has a mass less than or equal to \u221a3 m\u03c0. (The nonexistence of such a \u226a m 83: Chiral Symmetry Breaking 511 particle in nature is the U(1) problem; the axial anomaly solves this problem.) 83.8) a) Write down all possible parity and time-reversal invariant terms with no derviatives that are bilinear in the nucleon \ufb01eld N and that have one factor of the quark mass matrix M. b) Reexpress your result in terms of the nucleon \ufb01eld mp = c) Use the observed neutron-proton mass di\ufb00erence, mn \u2212 1.293 MeV, and the mu/md ratio you found in problem 83.6, to detemine as much as you can about the coe\ufb03cients of the terms wrote down. (Ignore the mass di\ufb00erence due to electromagnetism.) N. 84: Spontaneous Breaking of Gauge Symmetries 512 84", " Spontaneous Breaking of Gauge Symmetries Prerequisite: 32, 70 Consider scalar electrodynamics, speci\ufb01ed by the lagrangian = (D\u00b5\u03d5)\u2020D\u00b5\u03d5 V (\u03d5) 1 4 F \u00b5\u03bdF\u00b5\u03bd, \u2212 \u2212 igA\u00b5, and where \u03d5 is a complex scalar \ufb01eld, D\u00b5 = \u2202\u00b5 \u2212 \u2212 L V (\u03d5) = m2\u03d5\u2020\u03d5 + 1 4 \u03bb(\u03d5\u2020\u03d5)2. (84.1) (84.2) (We call the gauge coupling constant g rather than e because we are using this theory as a formal example rather than a physical model.) So far we have always taken m2 > 0, but now let us consider m2 < 0. We analyzed this model in the absence of the gauge \ufb01eld in section 32. Classically, the \ufb01eld has a nonzero vacuum expectation value (VEV for short), given by 0 \u03d5(x) 0 i | | h = 1 \u221a2 v, (84.3) where we have made a global U(1) transformation to set the phase of the VEV to zero, and We therefore write m2 v = (4 | /\u03bb)1/2. | \u03d5(x) = 1 \u221a2 (v + \u03c1(x))e\u2212 i\u03c7(x)/v, (84.4) (84.5) where \u03c1(x) and \u03c7(x) are real scalar \ufb01elds. The scalar potential depends only on \u03c1, and is given by V (\u03d5) = 1 4 \u03bbv2\u03c12 + 1 4 \u03bbv\u03c13 + 1 16 \u03bb\u03c14. (84.6) Since \u03c7 does not appear in the potential, it is massless; it is the Goldstone boson for the spontaneously broken U(1) symmetry. The big di\ufb00erence in the gauge theory is that we can make a gauge transformation that shifts the phase of \u03d5(x) by an arbitrary spacetime function. We can use this gauge freedom to set \u03c7(x) = 0; this choice is called unitary gauge. Using eq. (84.5) with \u03c7(x) = 0 in eq. (84.1), we have (D\u00b5\u03d5)\u2020D\u00b5\u03d5 = \u2212 \u2212 1 2(\u2202\u00b5\u03c1 + ig(v + \u03c1)A\u00b5)(\u2202\u00b5\u03c1 2\u2202\u00b5\u03c1\u2202\u00b5\u03c1 \u2212 2 g2(v + \u03c1)2A\u00b5A\u00b5. 1 1 = ig(v + \u03c1)A\u00b5) \u2212 Expanding out the last term, we see that the gauge \ufb01eld now has a mass \u2212 (84.7) M = gv. (84.8) 84: Spontaneous Breaking of Gauge Symmetries 513 This is the Higgs mechanism: the Goldstone boson disappears, and the gauge \ufb01eld acquires a mass. Note that this leaves the counting of particle spin states unchanged: a massless spin-one particle has two spin states, but a massive one has three. The Goldstone boson has become the third or longitudinal state of the now-massive gauge \ufb01eld. A scalar \ufb01eld whose VEV breaks a gauge symmetry is generically called a Higgs \ufb01eld. This generalizes in a straightforward way to a nonabelian gauge theory. Consider a complex scalar \ufb01eld \u03d5 in a representation R of the gauge group. (D\u00b5\u03d5)\u2020D\u00b5\u03d5, where the covariant derivative is The kinetic term for \u03d5 is \u00b5(T a j\u03d5j, and the indices i and j run from 1 to d(R). (D\u00b5\u03d5)i = \u2202\u00b5\u03d5i \u2212 We assume that \u03d5 acquires a VEV igaAa \u2212 R )i 0 \u03d5i(x) 0 i | | h where the value of vi is determined (up to a global gauge transformation) (D\u00b5\u03d5)\u2020D\u00b5\u03d5, by minimizing the potential. If we replace \u03d5 by its VEV in we \ufb01nd a mass term for the gauge \ufb01elds, (84.9) vi, \u2212 = 1 \u221a2 Lmass = \u2212 where the mass-squared matrix is 1 2 (M 2)abAa\u00b5Ab \u00b5, (84.10) (M 2)ab = 1 2 g2v\u2217i { R, T b", " T a R}ijvj. \u00b5 is symmetric on a (84.11) b, and \u2194 The anticommutator appears because Aa\u00b5Ab R with 1 so we replaced } R T b If the \ufb01eld \u03d5 is real rather than complex (which is possible only if R is a real representation), then we remove the factor of root-two from the right-hand side of eq. (84.9), but this is compensated by an extra factor of one-half from the kinetic term for a real scalar \ufb01eld; thus eq. (84.11) holds as written. If there is more than one gauge group, then the g2 in eq. (84.11) is replaced by gagb, where ga is the coupling constant that goes along with the generator T a, and all generators of all gauge groups are included in the mass-squared matrix. Recall from section 32 that a generator T a is spontaneously broken if (T a = 0. From eq. (84.11), we see that gauge \ufb01elds corresponding R )ijvj 6 to broken generators get a mass, while those corresponding to unbroken generators do not. The unbroken generators (if any) form a gauge group with massless gauge \ufb01elds. The massive gauge \ufb01elds (and all other \ufb01elds) form representations of this unbroken group. Let us work out some simple examples. Consider the gauge group SU(N ), with a complex scalar \ufb01eld \u03d5 in the fundamental representation. We can make a global SU(N ) transformation to bring the VEV entirely into the last component, and furthermore 84: Spontaneous Breaking of Gauge Symmetries 514 \u2212 \u2212 N = 1 2 for i = N (there are N make it real. Any generator (T a)i j that does not have a nonzero entry in the last column will remain unbroken. These generators form an unbro1) gauge group. There are three classes of broken generators: ken SU(N those with (T a)i 1 of these); those with 1 (T a)i N = 1 of these), and \ufb01nally the sin2 i for i gle generator T N 2 1)). The gauge \u2212 \ufb01elds corresponding to the generators in the \ufb01rst two classes get a mass M = 1 2 gv; we can group them into a complex vector \ufb01eld that transforms in 1) subgroup. The the fundamental representation of the unbroken SU(N gauge \ufb01eld corresponding to T N 2 1)/2N ]1/2gv; it is a singlet of SU(N = N (there are also N 1 = [2N (N \u2212 1 gets a mass M = [(N 1/2 diag(1,..., 1, 1)]\u2212 (N 1). \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 Consider the gauge group SO(N ), with a real scalar \ufb01eld in the fundamental representation. We can make a global SO(N ) transformation to bring the VEV entirely into the last component. Any generator (T a)i j that does not have a nonzero entry in the last column will remain unbroken. These generators form an unbroken SO(N 1 \u2212 broken generators, those with (T a)i N = = N. The corresponding gauge \ufb01elds get a mass M = gv; they form a fundamental representation of the unbroken SO(N 1) subgroup. In the case N = 3, this subgroup is SO(2), which is equivalent to U(1). 1) subgroup. There are N \u2212 i for i \u2212 \u2212 \u2212 igAa 1 2 g2 Tr \u00b5[T a, \u03a6], and the VEV of \u03d5 is a traceless hermitian N Consider the gauge group SU(N ), with a real scalar \ufb01eld \u03a6a in the adjoint representation. It will prove more convenient to work with the matrix-valued \ufb01eld \u03a6 = \u03d5aT a; the covariant derivative of \u03a6 is D\u00b5\u03a6 = \u2202\u00b5\u03a6 N matrix V. Thus the mass-squared matrix for the gauge \ufb01elds is (M 2)ab = [T a, V ], [T b, V ]. We can make a global SU(N ) transformation to } { \u2212 bring V into diagonal form. Suppose the diagonal entries consist of N1 v1\u2019s, i Nivi = 0. Then all followed by N2 v2\u2019s, etc., where v1 < v2 <..., and generators whose nonzero entries lie entirely within the", " ith block commute with V, and hence form an unbroken SU(Ni) subgroup. Furthermore, the linear combination of diagonal generators that is proportional to V also commutes with V, and forms a U(1) subgroup. Thus the unbroken gauge group is SU(N1) U(1). The gauge coupling constants for the di\ufb00erent groups are all the same, and equal to the original SU(N ) gauge coupling constant. SU(N2)... P \u00d7 \u00d7 \u00d7 \u00d7 As a speci\ufb01c example, consider the case of SU(5), which has 24 gener3, + 1 2 )v. \u2212 U(1). The number of 1 = 12. The generator of the U(1) sub3 \u2212 \u2212 3 c, + 1 1 1 2 c), where c2 = 3/5. Under the 3 c, U(1) subgroup, the 5 representation of SU(5) ators. Let the diagonal entries of V be given by ( The unbroken subgroup is then SU(3) broken generators is 24 \u2212 1 group is T 24 = diag( 3 c, \u2212 SU(2) unbroken SU(3) \u00d7 2 c, + 1 2, + 1 SU(2 84: Spontaneous Breaking of Gauge Symmetries 515 transforms as 1 3 ) Here the last entry is the value of T 24/c. The 5 of SU(5) then transforms as (1, 2, + 1 (84.12) (3, 1, 2 ). \u2192 \u2212 \u2295 5 (\u00af3, 1, + 1 3 ) To \ufb01nd out how the adjoint or 24 representation of SU(5) transforms under the SU(3) U(1) subgroup, we use the SU(5) relation (84.13) SU(2) 1 2 ). (1, 2, \u2192 \u2212 \u2295 5 \u00d7 \u00d7 5 = 24 5 \u2297 1. \u2295 (84.14) From eqs. (84.12) and (84.13), we have 5 5 \u2297 \u2192 [(3, 1, 1 3 ) \u2212 \u2295 (1, 2, + 1 2 )] \u2297 [(\u00af3, 1, + 1 3 ) (1, 2, 1 2 )]. \u2212 \u2295 (84.15) If we expand this out, and compare with eq. (84.14), we see that 24 \u2192 (8, 1, 0) (3, 2, \u2295 \u2295 \u2212 (1, 3, 0) 5 6 ) \u2295 (\u00af3, 2, + 5 (1, 1, 0) 6 ). \u2295 (84.16) \u00d7 \u00d7 SU(2) The \ufb01rst line on the right-hand side of eq. (84.16) is the adjoint repreU(1); the corresponding gauge \ufb01elds remain sentation of SU(3) massless. The second line shows us that the gauge \ufb01elds corresponding to the twelve broken generators can be grouped into a complex vector \ufb01eld 5 6 ). Since it is an irreducible representation of in the representation (3, 2, the unbroken subgroup, all twelve vectors \ufb01elds must have the same mass. g2 Tr([T 4, V ][T 4, V ]), This mass is most easily computed from (M 2)44 = where we have de\ufb01ned (T 4)i gv. 1); the result is M = 5 6\u221a2 j = 1 4 + \u03b4i 2 (\u03b4i 4\u03b4j 1\u03b4j \u2212 \u2212 Problems 84.1) Conside a theory with gauge group SU(N ), with a real scalar \ufb01eld \u03a6 in the adjoint representation, and potential V (\u03a6) = 1 2 m2 Tr \u03a62 + 1 4 \u03bb1 Tr \u03a64 + 1 4 \u03bb2 (Tr \u03a62)2. (84.17) This is the most general potential consistent with SU(N ) symmetry \u03a6, which we impose to keep things simand a Z2 symmetry \u03a6 ple. We assume m2 < 0. We can work in a basis in which \u03a6 = v diag(\u03b11,..., \u03b1N ), with the constraints a) Extremize V (\u03a6) with respect to v. Solve for v, and plug your result back into V (\u03a6). You should \ufb01nd i \u03b1i = 0 and i = 1. \u2194 \u2212 i \u03b12 P P V (\u03a6) = 1 4(m2)2 \u03bb1A(\u03b1) + \u03bb2B(\u03b1) \u2212, (", "84.18) 84: Spontaneous Breaking of Gauge Symmetries 516 where A(\u03b1) and B(\u03b1) are functions of \u03b1i. b) Show that \u03bb1A(\u03b1) + \u03bb2B(\u03b1) must be everywhere positive in order for the potential to be bounded below. c) Show that the absolute mimimum of the potential (assuming that it is bounded below) occurs at the absolute minimum of \u03bb1A(\u03b1) + \u03bb2B(\u03b1). d) Show that, at any extremum of the potential, the \u03b1i take on at most three di\ufb00erent values, and that these three values sum to zero. Hint: impose the constraints with Lagrange multipliers. e) Show that, for \u03bb1 > 0 and \u03bb2 > 0, at the absolute minimum of V (\u03a6) the unbroken symmetry group is SU(N+) U(1), \u00d7 = 1 1) if N is odd. where N+ = N 2 N if N is even, and N \u00d7 2 (N SU(N = 1 \u2212 ) \u00b1 \u00b1 \u2212 85: Spontaneously Broken Abelian Gauge Theory 517 85 Spontaneously Broken Abelian Gauge Theory Prerequisite: 61, 84 Consider scalar electrodynamics, speci\ufb01ed by the lagrangian = (D\u00b5\u03d5)\u2020D\u00b5\u03d5 V (\u03d5) 1 4 F \u00b5\u03bdF\u00b5\u03bd, L \u2212 where \u03d5 is a complex scalar \ufb01eld and D\u00b5 = \u2202\u00b5 \u2212 2 v2)2, V (\u03d5) = 1 4 \u03bb(\u03d5\u2020\u03d5 \u2212 \u2212 1 \u2212 igA\u00b5. We choose which yields a nonzero VEV for \u03d5. We therefore write \u03d5(x) = 1 \u221a2 (v + \u03c1(x))e\u2212 i\u03c7(x)/v, (85.1) (85.2) (85.3) where \u03c1(x) and \u03c7(x) are real scalar \ufb01elds. The scalar potential depends only on \u03c1, and is given by V (\u03d5) = 1 4 \u03bbv2\u03c12 + 1 4 \u03bbv\u03c13 + 1 16 \u03bb\u03c14. We can now make a gauge transformation to set \u03c7(x) = 0. (85.4) (85.5) This is unitary gauge. The kinetic term for \u03d5 becomes (D\u00b5\u03d5)\u2020D\u00b5\u03d5 = 1 2 \u2202\u00b5\u03c1\u2202\u00b5\u03c1 \u2212 \u2212 1 2 g2(v + \u03c1)2A\u00b5A\u00b5. \u2212 (85.6) We see that the gauge \ufb01eld has acquired a mass M = gv. The terms in L that are quadratic in A\u00b5 are 4 F \u00b5\u03bdF\u00b5\u03bd \u2212 L0 = \u2212 1 1 2 M 2A\u00b5A\u00b5. The equation of motion that follows from eq. (85.8) is \u22022 + M 2)g\u00b5\u03bd + \u2202\u00b5\u2202\u03bd]A\u03bd = 0. [( \u2212 If we act with \u2202\u00b5 on this equation, we get (85.7) (85.8) (85.9) M 2\u2202\u03bdA\u03bd = 0. (85.10) 85: Spontaneously Broken Abelian Gauge Theory 518 If we now use eq. (85.10) in eq. (85.9), we \ufb01nd that each component of A\u03bd obeys the Klein-Gordon equation, The general solution of eqs. (85.9) and (85.10) is \u22022 + M 2)A\u03bd = 0. ( \u2212 (85.11) A\u00b5(x) = dk X\u03bb= \u2212,0,+ Z \u2217\u03bb (k)a\u03bb(k)eikx + \u03b5\u00b5 \u03b5\u00b5 h \u03bb(k)a\u2020\u03bb(k)e\u2212 ikx, (85.12) i f where the polarization vectors must satisfy k\u00b5\u03b5\u00b5 \u03bb(k) = 0. In the rest frame, where k = (M, 0, 0, 0), we choose the polarization vectors to correspond to de\ufb01nite spin along the \u02c6z axis, (0, 1, i, 0), \u03b5+(0) = 1 \u221a2 (0) = 1 \u03b5 \u221a2 \u2212 \u03b50(0) = (0, 0, 0, 1). \u2212 (0, 1, +i, 0), (85.13) More generally, the three polarization vectors along with the tim", "elike unit vector k\u00b5/M form an orthonormal and complete set, \u03b5\u00b5 \u03bb(k) = 0, k \u00b7 \u03b5\u2217\u03bb(k) = \u03b4\u03bb\u2032\u03bb, \u03bb(k) = g\u00b5\u03bd + \u03b5\u03bb\u2032(k) \u00b7 \u03b5\u00b5 \u2217\u03bb (k)\u03b5\u03bd X\u03bb= \u2212,0,+ (85.14) (85.15) (85.16) k\u00b5k\u03bd M 2. Since the lagrangian of eq. (85.8) has no manifest gauge invariance, quantization is straightforward. The coe\ufb03cients a\u2020\u03bb(k) and a\u03bb(k) become particle creation and annihilation operators in the usual way, and the propagator of the A\u00b5 \ufb01eld is given by TA\u00b5(x)A\u03bd (y) 0 0 i i | | h = = d4k (2\u03c0)4 eik(x \u2212 k2 + M 2 d4k (2\u03c0)4 eik(x \u2212 k2 + M 2 y) \u2212 y) \u2212 Z Z \u03b5\u00b5 \u2217\u03bb (k)\u03b5\u03bd \u03bb(k) i\u01eb \u03bb X g\u00b5\u03bd + i\u01eb k\u00b5k\u03bd M 2. (85.17) The interactions of the massive vector \ufb01eld A\u00b5 with the real scalar \ufb01eld \u03c1 can be read o\ufb00 of eq. (85.6). The self-interactions of the \u03c1 \ufb01eld can be read o\ufb00 of eq. (85.4). The resulting Feynman rules can be used for tree-level calculations. Loop calculations are more subtle. We have imposed the gauge condition \u03c7(x) = 0, which corresponds to inserting a functional delta function 85: Spontaneously Broken Abelian Gauge Theory 519 x \u03b4(\u03c7(x)) into the path integral. In order to integrate over \u03c7, we must make a change of integration variables from Re \u03d5 and Im \u03d5 to \u03c1 and \u03c7; this Q is simply a transformation from cartesian to polar coordinates, analogous to dx dy = r dr d\u03c6. So we must include a factor analogous to r in the functional measure; this factor is v + \u03c1(x) = det(v + \u03c1) x Y \u221d \u221d det(1 + v\u2212 1\u03c1) \u00afc D D Z im2 gh c e\u2212 d4x \u00afc(1+v\u22121\u03c1)c. (85.18) R In the last line, we have written the functional determinant as an integral over ghost \ufb01elds. We see that they have no kinetic term, and we have chosen the overall nomalization of their action so that their mass is mgh, where mgh is an arbitrary mass parameter. Thus the momentum-space propagator for the ghosts is simply \u02dc\u2206(k2) = 1/m2 gh. We also see that there 1, but there is no is a ghost-ghost-scalar vertex, with vertex factor interaction between the ghosts and the vector \ufb01eld. ghv\u2212 im2 \u2212 This seems like a fairly convenient gauge for loop calculations, but there is a complication. The fact that the ghost propagator is independent of the momentum means that additional internal ghost propagators do not help the convergence of loop-momentum integrals. The same is true of vector\ufb01eld propagators; from eq. (85.17) we see that, in momentum space, the propagator scales like 1/M 2 in the limit that all components of k become large. Thus, in unitary gauge, loop diagrams with arbitrarily many external lines diverge. This makes it di\ufb03cult to establish renormalizability. A gauge that does not su\ufb00er from this problem is a generalization of R\u03be gauge (and in fact this name has traditionally been applied only to this generalization). We begin by using a cartesian basis for \u03d5, \u03d5 = 1 \u221a2 (v + h + ib), (85.19) where h and b are real scalar \ufb01elds. In terms of h and b, the potential is V (\u03d5) = 1 4 \u03bbv2h2 + 1 4 \u03bbvh(h2 + b2) + 1 16 \u03bb(h2 + b2)2, and the covariant derivative of \u03d5 is D\u00b5\u03d5 = 1 \u221a2 (\u2202\u00b5h + gbA\u00b5) + i(\u2202\u00b5b h \u2212 g(v+h)A\u00b5) i. Thus the kinetic term", " for \u03d5 becomes (85.20) (85.21) (D\u00b5\u03d5)\u2020D\u00b5\u03d5 = 1 2 (\u2202\u00b5h + gbA\u00b5)2 1 2 (\u2202\u00b5b \u2212 \u2212 \u2212 \u2212 g(v+h)A\u00b5)2. (85.22) 85: Spontaneously Broken Abelian Gauge Theory 520 Expanding this out, and rearranging, we get (D\u00b5\u03d5)\u2020D\u00b5\u03d5 = \u2212 2 g2v2A\u00b5A\u00b5 + gvA\u00b5\u2202\u00b5b 1 2 \u2202\u00b5h\u2202\u00b5h \u2212 \u2212 + gA\u00b5(h\u2202\u00b5b \u2212 gvhA\u00b5A\u00b5 \u2212 \u2212 1 \u2212 1 2 \u2202\u00b5b\u2202\u00b5b b\u2202\u00b5h) 2 g2(h2 + b2)A\u00b5A\u00b5. 1 (85.23) The \ufb01rst line on the right-hand side of eq. (85.23) contains all the terms that are quadratic in the \ufb01elds. The \ufb01rst two are the kinetic terms for the h and b \ufb01elds. The third is the mass term for the vector \ufb01eld. The fourth is an annoying cross term between the vector \ufb01eld and the derivative of b. In abelian gauge theory, in the absence of spontaneous symmetry break- ing, we \ufb01x R\u03be gauge by adding to L the gauge-\ufb01xing and ghost terms Lgf + Lgh = \u2212 1 2 \u03be\u2212 1G2 \u00afc \u03b4G \u03b4\u03b8 \u2212 c, (85.24) where G = \u2202\u00b5A\u00b5, and \u03b8(x) parameterizes an in\ufb01nitesimal gauge transformation, A\u00b5 \u2192 \u03d5 \u2192 A\u00b5 \u2212 \u03d5 \u2212 \u2202\u00b5\u03b8, ig\u03b8\u03d5. (85.25) (85.26) With G = \u2202\u00b5A\u00b5, we have \u03b4G/\u03b4\u03b8 = interactions, and can be ignored. \u2212 \u22022. Thus the ghost \ufb01elds have no In the presence of spontaneous symmetry breaking, we choose instead G = \u2202\u00b5A\u00b5 \u2212 which reduces to \u2202\u00b5A\u00b5 when v = 0. Multiplying out G2, we have \u03begvb, (85.27) Lgf = = 1 2 \u03be\u2212 1 2 \u03be\u2212 1\u2202\u00b5A\u00b5\u2202\u03bdA\u03bd + gvb\u2202\u00b5A\u00b5 \u2212 gvA\u00b5\u2202\u00b5b 1\u2202\u00b5A\u03bd\u2202\u03bdA\u00b5 \u2212 \u2212 \u2212 \u2212 1 2 \u03beg2v2b2 2 \u03beg2v2b2, 1 (85.28) where we integrated by parts in the \ufb01rst two terms to get the second line. Note that the second term on the second line of eq. (85.28) cancels the annoying last term on the \ufb01rst line of eq. (85.23). Also, the last term on the second line of eq. (85.28) gives a mass \u03be1/2M to the b \ufb01eld. We must still evaluate Lgh. To do so, we \ufb01rst translate eq. (85.26) into (85.29) h h + g\u03b8b, \u2192 Then we have b b \u2212 \u2192 g\u03b8(v + h). \u03b4G \u03b4\u03b8 = \u2212 \u22022 + \u03beg2v(v + h). (85.30) (85.31) 85: Spontaneously Broken Abelian Gauge Theory 521 From eq. (85.24) we see that the ghost lagrangian is Lgh = = \u2212 \u2212 \u00afc \u22022 + \u03beg2v(v + h) i \u03beg2vh\u00afcc. \u03beg2v2\u00afcc \u2212 h \u2202\u00b5\u00afc\u2202\u00b5c c \u2212 \u2212 (85.32) We see from the second term that the ghost has acquired the same mass as the b \ufb01eld, \u03be1/2M. Now let us examine the vector \ufb01eld. Including are quadratic in the vector \ufb01eld can be written as Lgf, the terms in L that L0 = \u2212 1 2 A\u00b5 g\u00b5\u03bd ( \u2212 \u22022 + M 2) + (1 \u2212 In momentum space, this reads", " h \u03be\u2212 1)\u2202\u00b5\u2202\u03bd A\u03bd. (85.33) i \u02dc L0 = \u2212 \u02dcA\u00b5( 1 2 \u2212 h The kinematic matrix is k) (k2 + M 2)g\u00b5\u03bd + (1 \u2212 \u03be\u2212 1)k\u00b5k\u03bd \u02dcA\u03bd(k). i (85.34)... h i = (k2 + M 2)g\u00b5\u03bd + (1 \u03be\u2212 1)k\u00b5k\u03bd \u2212 P \u00b5\u03bd (k) + k\u00b5k\u03bd/k2 = (k2 + M 2) = (k2 + M 2)P \u00b5\u03bd (k) + \u03be\u2212 \u03be\u2212 1)k\u00b5k\u03bd + (1 \u2212 1(k2 + \u03beM 2)k\u00b5k\u03bd/k2, (85.35) k\u00b5k\u03bd/k2 projects onto the transverse subspace; where P \u00b5\u03bd (k) = g\u00b5\u03bd P \u00b5\u03bd (k) and k\u00b5k\u03bd/k2 are orthogonal projection matrices. Using this fact, it is easy to invert eq. (85.35) to get the propagator for the massive vector \ufb01eld in R\u03be gauge, \u2212 \u02dc\u2206\u00b5\u03bd (k) = P \u00b5\u03bd(k) \u03be k\u00b5k\u03bd/k2 k2 + M 2 i\u01eb k2 + \u03beM 2 i\u01eb +. (85.36) \u2212 We see that the transverse components of the vector \ufb01eld propagate with mass M, while the longitudinal component propagates with the same mass as the b and ghost \ufb01elds, \u03be1/2M. \u2212 Eq. (85.36) simpli\ufb01es greatly if we choose \u03be = 1; then we have \u02dc\u2206\u00b5\u03bd (k) = g\u00b5\u03bd k2 + M 2 i\u01eb \u2212 (\u03be = 1). (85.37) On the other hand, leaving \u03be as a free parameter allows us to check that all \u03be dependence cancels out of any physical scattering amplitude. Since their masses depend on \u03be, the ghosts, the b \ufb01eld, and the longitudinal component of the vector \ufb01eld must all represent unphysical particles that do not appear in incoming or outgoing states. 85: Spontaneously Broken Abelian Gauge Theory 522 h = 1 2 \u03bbv2 and propagator 1/(k2 + m2 To summarize, in R\u03be gauge we have the physical h \ufb01eld with masssquared m2 h), the unphysical b \ufb01eld with propagator 1/(k2 + \u03beM 2), the ghost \ufb01elds \u00afc and c with propagator 1/(k2 + \u03beM 2), and the vector \ufb01eld with the propagator of eq. (85.36). For external vectors, the polarizations are still given by eq. (85.13), and obey the sum rules of eq. (85.16). The mass parameter M is given by M = gv. The interactions of these \ufb01elds are governed by L1 = 1 4 \u03bbvh(h2 + b2) \u2212 + gA\u00b5(h\u2202\u00b5b \u03beg2vh\u00afcc. \u2212 \u2212 \u2212 b\u2202\u00b5h) 1 16 \u03bb(h2 + b2)2 gvhA\u00b5A\u00b5 \u2212 \u2212 1 2 g2(h2 + b2)A\u00b5A\u00b5 (85.38) It is interesting to consider the limit \u03be. In this limit, the vector propagator in R\u03be gauge, eq. (85.36), turns into the massive vector propagator of eq. (85.17), \u2192 \u221e \u02dc\u2206\u00b5\u03bd (k) = g\u00b5\u03bd + k\u00b5k\u03bd/M 2 k2 + M 2 i\u01eb \u2212 (\u03be = ). \u221e (85.39) \u2212 \u2212 i\u03beg2v = i(\u03beM 2)v\u2212 The b \ufb01eld becomes in\ufb01nitely heavy, and we can drop it. (Equivalently, its propagator goes to zero.) The ghost \ufb01elds also become in\ufb01nitely heavy, but we must be more careful with them because their interaction term, the last line of eq. (85.38), also contains a factor of \u03be. The vertex factor for this 1. Note that this is the same vertex interaction is factor that we found in unitary gauge for the in", "eraction between the \u03c1 \ufb01eld and the ghost \ufb01elds; see eq. (85.18) and take m2 gh = \u03beM 2. Thus we cannot drop the ghost \ufb01elds, but we can take their propagator to be 1/m2 gh rather m2 than 1/(k2 + m2. This is the ghost propagator that we found in unitary gauge. We conclude that R\u03be gauge in the limit \u03be is equivalent to unitary gauge. Of course, in this limit, we reencounter the problems with divergent diagrams that led us to consider alternative gauge choices in the \ufb01rst place. For practical loop calculations, R\u03be gauge with \u03be = 1 is typically the most convenient. gh = \u03beM 2 in the limit \u03be gh), since k2 \u2192 \u221e \u2192 \u221e \u226a In the next section, we consider R\u03be gauge for nonabelian theories. 86: Spontaneously Broken Nonabelian Gauge Theory 523 86 Spontaneously Broken Nonabelian Gauge Theory Prerequisite: 85 In the previous section, we worked out the lagrangian for a U(1) gauge theory with spontaneous symmetry breaking in R\u03be gauge. In this section, we extend this analysis to a general nonabelian gauge theory. As in section 85, it will be convenient to work with real scalar \ufb01elds. We therefore decompose any complex scalar \ufb01elds into pairs of real ones, and organize all the real scalar \ufb01elds into a big list \u03c6i, i = 1,..., N. These real scalar \ufb01elds form a (possibly reducible) representation R of the gauge a be the gauge-group generator matrices that act on \u03c6; they group. Let are linear combinations of the generators of the SO(N ) group that rotates all components of \u03c6i into each other. Because these SO(N ) generators a)ij is a real, are hermitian and antisymmetric, so are the antisymmetric matrix. a\u2019s. Thus i( T T T The lagrangian for our theory can now be written as = L \u2212 1 2 D\u00b5\u03c6D\u00b5\u03c6 V (\u03c6) \u2212 \u2212 1 4 F a\u00b5\u03bdF a \u00b5\u03bd, where (D\u00b5\u03c6)i = \u2202\u00b5\u03c6i \u2212 igaAa \u00b5( T a)ij\u03c6j (86.1) (86.2) is the covariant derivative, and the adjoint index a runs over all generators a)ij is a real of all gauge groups. Because \u03c6i and Aa matrix, (D\u00b5\u03c6)i is real. \u00b5 are real \ufb01elds, and i( T Now we suppose that the potential V (\u03c6) is minimized when \u03c6 has a VEV A generator 0 \u03c6i(x) 0 i | | h a is unbroken if ( T T = vi. a)ijvj = 0, and broken if ( T a)ijvj 6 (86.3) = 0. Each broken generator results in a massless Goldstone boson. To see this, we note that the potential must be invariant under a global gauge transformation, Expanding to linear order in the in\ufb01nitesimal parameter \u03b8, we \ufb01nd V ((1 \u2212 i\u03b8a T a)\u03c6) = V (\u03c6). \u2202V \u2202\u03c6j ( T a)jk\u03c6k = 0. We di\ufb00erentiate eq. (86.5) with respect to \u03c6k to get \u22022V \u2202\u03c6i\u2202\u03c6j ( T a)jk\u03c6k + \u2202V \u2202\u03c6j ( T a)jk = 0. (86.4) (86.5) (86.6) 86: Spontaneously Broken Nonabelian Gauge Theory 524 Now set \u03c6i = vi; then \u2202V /\u2202\u03c6i vanishes, because \u03c6i = vi minimizes V (\u03c6). Also, we can identify (m2)ij = \u22022V \u2202\u03c6i\u2202\u03c6j \u03c6i=vi (86.7) as the mass-squared matrix for the scalars (after spontaneous symmetry breaking). Thus eq. (86.6) becomes (m2)ij( T av)j = 0. (86.8) av is an eigenvector of the mass-squared We see that if matrix with eigenvalue zero. So there is a zero eigenvalue for every linearly", " independent broken generator. = 0, then av T T Let us write where \u03c7i is a real scalar \ufb01eld. The covariant derivative of \u03c6 becomes \u03c6i(x) = vi + \u03c7i(x), (86.9) (D\u00b5\u03c6)i = \u2202\u00b5\u03c7i \u2212 igaAa \u00b5( T a)ij(v + \u03c7)j It is now convenient to de\ufb01ne a set of real antisymmetric matrices (\u03c4 a)ij \u2261 iga( T a)ij, and the real rectangular matrix F a i \u2261 (\u03c4 a)ijvj. We can now write (D\u00b5\u03c6)i = \u2202\u00b5\u03c7i \u2212 Aa \u00b5(F a + \u03c4 a\u03c7)i. The kinetic term for \u03c6 becomes (86.10) (86.11) (86.12) (86.13) 1 2D\u00b5\u03c6D\u00b5\u03c6 = \u2212 1 1 2 \u2202\u00b5\u03c7i\u2202\u00b5\u03c7i \u2212 \u2212 + Aa 2 (F a \u00b5\u03c7i(\u03c4 a)ij\u2202\u00b5\u03c7j \u2212 2 Aa\u00b5Ab \u00b5\u03c7i(\u03c4 a\u03c4 b)ij\u03c7j. 1 \u2212 iF b Aa\u00b5Ab i)Aa\u00b5Ab \u00b5F a \u00b5 + F a i(\u03c4 b)ij\u03c7j iAa \u00b5\u2202\u00b5\u03c7i (86.14) We see (from the second term on the right-hand side) that the mass-squared matrix for the vector \ufb01elds is (M 2)ab = F a iF b i = (FF T)ab. (86.15) A theorem of linear algebra states that every real rectangular matrix can be written as F a i = Sab(M b\u03b4b j)Rji, (86.16) 6 86: Spontaneously Broken Nonabelian Gauge Theory 525 where S and R are orthogonal matrices, and the diagonal entries M a are real and nonnegative. From eq. (86.15) we see that these diagonal entries are the masses of the vector \ufb01elds. The vector \ufb01elds of de\ufb01nite mass are then given by \u02dcAa \u00b5 = SbaAb \u00b5. Now we are ready to \ufb01x R\u03be gauge. To do so, we add to \ufb01xing and ghost terms the gauge- L Lgf + Lgh = \u2212 1 2 \u03be\u2212 1GaGa \u00afca \u03b4Ga \u03b4\u03b8b cb, \u2212 where we choose Then we have Lgf = = 1 2 \u03be\u2212 1 2 \u03be\u2212 \u2212 \u2212 Ga = \u2202\u00b5Aa \u00b5 \u2212 \u03beF a i\u03c7i. 1\u2202\u00b5Aa \u00b5\u2202\u03bdAa \u03bd + F a 1\u2202\u00b5Aa \u03bd \u2202\u03bdAa \u00b5 \u2212 F a i\u03c7i\u2202\u00b5Aa \u00b5 \u2212 \u00b5\u2202\u00b5\u03c7i \u2212 iAa (86.17) (86.18) j)\u03c7i\u03c7j iF a 1 2 \u03be(F a 2 \u03be(F a 1 iF a j)\u03c7i\u03c7j. (86.19) We integrated by parts in the \ufb01rst two terms to get the second line. Note that the second term on the second line of eq. (86.19) cancels the annoying last term on the \ufb01rst line of eq. (86.14). Also, the last term on the second line of eq. (86.19) makes a contribution to the mass-squared matrix for the \u03c7 \ufb01elds, j = \u03be(F TF )ij. \u03be(M 2)ij = \u03beF a iF a (86.20) Eq. (86.16) tells us that the eigenvalues of this matrix are \u03be1/2M a, where M a are the vector-boson masses. The mass-squared matrix \u03beM 2 should be added to the mass-squared matrix m2 that we get from the potential, eq. (86.7). Note that eqs. (86.8) and (86.12) imply that (m2)ijF a j = 0; eq. (86.20) then yields (m2)ij(\u03beM 2)jk = 0. Thus these two contributions to the mass-squared matrix of the scalar \ufb01elds live in orthogonal subspaces. The scalar \ufffd", "\ufffd\ufffdelds of de\ufb01nite mass are \u02dc\u03c7i = Rij\u03c7j, where the block of R in the m2 subspace is chosen to diagonalize m2. The m2 subspace consists of the physical, massive scalars, and the \u03beM 2 subspace consists of the unphysical Goldstone bosons; these are the \ufb01elds that would be set to zero in unitary gauge. We must still evaluate Lgh. To do so, we recall that \u03b8a(x) parameterizes an in\ufb01nitesimal gauge transformation, Aa Aa \u00b5 \u2192 \u03c7i \u2192 \u2212 Dab \u00b5 \u03b8b, \u00b5 \u2212 \u03b8a(\u03c4 a)ij(v + \u03c7)j. (86.21) (86.22) Thus we have \u03b4Ga \u03b4\u03b8b = \u2212 \u2202\u00b5Dab \u00b5 + \u03beF a j(\u03c4 b)jk(v + \u03c7)k 86: Spontaneously Broken Nonabelian Gauge Theory 526 \u2202\u00b5Dab \u00b5 + \u03beF a jF b j + \u03beF a j(\u03c4 b)jk\u03c7k \u2202\u00b5Dab \u00b5 + \u03be(M 2)ab + \u03beF a j(\u03c4 b)jk\u03c7k, (86.23) = = \u2212 \u2212 and so the ghost lagrangian is Lgh = \u2212 \u2202\u00b5\u00afcaDab \u00b5 cb \u2212 \u03be(M 2)ab\u00afcacb \u2212 \u03beF a j(\u03c4 b)jk\u03c7k\u00afcacb. (86.24) The ghost \ufb01elds of de\ufb01nite mass are \u02dcca = Sbacb and \u02dc\u00afca = Sba\u00afcb. The complete gauge-\ufb01xed lagrangian is now given by eqs. (86.1), (86.14) (86.19), and (86.24). We can rewrite it in terms of the \ufb01elds of de\ufb01nite mass. This results in the replacements F a i \u2192 (\u03c4 a)ij \u2192 f abc \u2192 M a\u03b4a i, Sab(RT\u03c4 bR)ij, SadSbeScgf deg (86.25) (86.26) (86.27) throughout L. The Feynman rules then follow in the usual way. Problems i\u03b8a(T a 86.1) Let \u03d5i be a complex scalar \ufb01eld in a complex representation R of the gauge group. Under an in\ufb01nitesimal gauge transformation, we have (\u03c6i+i\u03c6i+d(R)), where \u03c6i is a \u03b4\u03d5i = real scalar \ufb01eld with the index i running from 1 to 2d(R). Then, under a)ij\u03c6j. i\u03b8a( an in\ufb01nitesimal gauge transformation, we have \u03b4\u03c6i = T a in terms of the real and imaginary parts of T a R. j\u03d5j. Let us write \u03d5i = 1 \u221a2 a) Express R )i \u2212 \u2212 T b) Show that the relations. a matrices satisfy the appropriate commutation T 87: The Standard Model: Gauge and Higgs Sector 527 87 The Standard Model: Gauge and Higgs Sector Prerequisite: 84 We now turn to the construction of the Standard Model of elementary particles, also called the Glashow\u2013Weinberg\u2013Salam model. This is the complete (except for gravity) quantum \ufb01eld theory that appears to describe our world. It can be succinctly speci\ufb01ed as a gauge theory with gauge group U(1), with left-handed Weyl \ufb01elds in three copies of the SU(3) (\u00af3, 1, (\u00af3, 1, + 1 (3, 2, + 1 3 ), and 6 ) representation (1, 2, \u2295 \u2295 1 a complex scalar \ufb01eld in the representation (1, 2, 2 ). Here the last entry of each triplet gives the value of the U(1) charge, known as hypercharge. The lagrangian includes all terms of mass dimension four or less that are allowed by the gauge symmetries and Lorentz invariance. (1, 1, +1) SU(2) 1 2 ) 2 3 ) \u2295 \u2212 \u2212 \u00d7 \u2295 \u2212 \u00d7 We will construct the Standard Model over several sections. We begin U(1), and the complex 1 2 ). The U(1) with the electroweak part of the gauge group, SU(2", ") scalar \ufb01eld \u03d5, known as the Higgs \ufb01eld, in the representation (2, Higgs \ufb01eld acquires a nonzero VEV that spontaneously breaks SU(2) to U(1); the unbroken U(1) is identi\ufb01ed as electromagnetism. \u00d7 \u00d7 \u2212 We begin with the covariant derivative of the Higgs \ufb01eld \u03d5, (D\u00b5\u03d5)i = \u2202\u00b5\u03d5i \u2212 i[g2Aa \u00b5T a + g1B\u00b5Y ]i j\u03d5j, (87.1) 2 \u03c3a and Y = 1 where T a = 1 2 I; Y is the hypercharge generator. It will \u2212 \u00b5T a + g1B\u00b5Y in matrix form, prove useful to write out g2Aa iA2 g2A3 \u00b5) g2(A1 Now suppose that \u03d5 has a potential g1B\u00b5 \u00b5 \u2212 \u00b5 + iA2 \u00b5) \u00b5T a + g1B\u00b5Y = g2(A1 g2A3 \u00b5 \u2212 \u00b5 \u2212 1 2 g1B\u00b5! g2Aa (87.2) \u2212. V (\u03d5) = 1 4 \u03bb(\u03d5\u2020\u03d5 1 2 v2)2. \u2212 (87.3) This potential gives \u03d5 a nonzero VEV. We can make a global gauge transformation to bring this VEV entirely into the \ufb01rst component, and furthermore make it real, so that 0 \u03d5(x) 0 i | | h = 1 \u221a2 v 0!. (87.4) The kinetic term for \u03d5 is \ufb01nd a mass term for the gauge \ufb01elds, \u2212 (D\u00b5\u03d5)\u2020D\u00b5\u03d5. After replacing \u03d5 by its VEV, we Lmass = \u2212 1 8 v2 ( 1, 0 ) g2A3 g2(A1 g1B\u00b5 \u00b5 \u2212 \u00b5 + iA2 \u00b5) g2(A1 g2A3 \u00b5 \u2212 \u00b5 \u2212 \u2212 iA2 \u00b5) 2 1 g1B\u00b5! 0!. (87.5) 87: The Standard Model: Gauge and Higgs Sector 528 To diagonalize this mass-squared matrix, we \ufb01rst de\ufb01ne the weak mixing angle \u03b8W \u2261 tan\u2212 1(g1/g2), (87.6) and the \ufb01elds iA2 (A1 1 \u221a2 \u00b5), \u00b5 \u2213 (87.7) cWA3 W \u00b1\u00b5 \u2261 Z\u00b5 \u2261 A\u00b5 \u2261 cos \u03b8W. In terms of these \ufb01elds, eq. (87.5) becomes \u00b5 \u2212 \u00b5 + cWB\u00b5, sWB\u00b5, sWA3 (87.8) (87.9) where sW \u2261 sin \u03b8W, cW \u2261 Lmass = \u2212 = = \u2212 \u2212 1 8 g2 2v2 ( 1, 0 ) 1 cW \u221a2 W \u2212\u00b5 1 Z\u00b5 \u221ag2v/2cW)2 Z \u00b5Z\u00b5 0 1 (g2v/2)2 W +\u00b5W \u2212\u00b5 \u2212 WW +\u00b5W \u2212 \u00b5Z\u00b5, where we have identi\ufb01ed MW = g2v/2, MZ = MW/cos \u03b8W. \uf8f6 \uf8f8 (87.10) (87.11) (87.12) The observed masses of the W \u00b1 and Z 0 particles are MW = 80.4 GeV and MZ = 91.2 GeV. Eq. (87.12) then implies cos \u03b8W = 0.882, or, as it is more usually expressed, sin2\u03b8W = 0.223.1 Note that the A\u00b5 \ufb01eld remains massless; this signi\ufb01es that there is an unbroken U(1) subgroup. We will identify this unbroken U(1) with the gauge group of electromagnetism. Before introducing leptons and quarks (which we do in sections 87 and 88), let us work out the complete lagrangian for the gauge and Higgs \ufb01elds, in unitary gauge. This is su\ufb03cient for tree-level calculations. The two complex components of the \u03d5 \ufb01eld yield four real scalar \ufb01elds; three of these become the longitudinal components of the W \u00b1 and Z 0. The 1Of course, this number is only meaningful once a renormalization scheme has been speci\ufb01", "ed. We are implicitly using an on-shell scheme in which \u03b8W is de\ufb01ned by the relation cos \u03b8W = MW/MZ, where MW and MZ are the actual particle masses. The relation g1 = g2 tan \u03b8W is then subject to loop corrections that depend on the precise de\ufb01nitions adopted for g1 and g2. In the MS scheme, on the other hand, \u03b8W is de\ufb01ned by eq. (87.6), and for \u00b5 = MZ, we have sin2 \u03b8W = 0.231. 87: The Standard Model: Gauge and Higgs Sector 529 remaining scalar \ufb01eld must be able to account for shifts in the overall scale of \u03d5. Thus we can write, in unitary gauge, \u03d5(x) = 1 \u221a2 v + H(x) 0,! (87.13) where H is a real scalar \ufb01eld; the corresponding particle is the Higgs boson. The potential now reads V (\u03d5) = 1 4 \u03bbv2H 2 + 1 4 \u03bbvH 3 + 1 16 \u03bbH 4. (87.14) 2 \u03bbv2. (As of We see that the mass of the Higgs boson is given by m2 this writing, the Higgs boson has not been observed; the lower limit on its mass is mH > 115 GeV.) The kinetic term for H comes from the kinetic 1 2 \u2202\u00b5H\u2202\u00b5H. Finally, term for \u03d5, and is the usual one for a real scalar \ufb01eld, recall that the mass term for the gauge \ufb01elds, eq. (87.10), is proportional to v2. Hence it should be multiplied by a factor of (1 + v\u2212 1H)2. H = 1 \u2212 Now we have to work out the kinetic terms for the gauge \ufb01elds. We have where = L \u2212 1 4 F a\u00b5\u03bdF a \u00b5\u03bd \u2212 1 4 B\u00b5\u03bdB\u00b5\u03bd, \u00b5\u03bd = \u2202\u00b5A1 F 1 \u03bd \u2212 \u00b5\u03bd = \u2202\u00b5A2 F 2 \u03bd \u2212 \u2202\u03bdA1 \u00b5 + g2(A2 \u00b5A3 \u03bd \u2212 A2 \u03bdA3 \u00b5), \u2202\u03bdA2 \u00b5 + g2(A3 \u00b5A1 \u03bd \u2212 \u00b5\u03bd = \u2202\u00b5A3 F 3 \u2202\u03bdA3 \u00b5 + g2(A1 \u00b5A2 A3 \u03bdA1 \u00b5), A1 \u03bdA2 \u00b5), \u03bd \u2212 \u03bd \u2212 \u2202\u00b5B\u03bd \u2212 Next, form the combinations F 1 B\u00b5\u03bd \u2261 \u2202\u03bdB\u00b5. iF 2 \u00b5\u03bd. Using eq. (87.7), we \ufb01nd 1 \u221a2 1 \u221a2 (F 1 (F 1 iF 2 \u00b5\u03bd \u2212 \u00b5\u03bd + iF 2 \u00b5\u03bd \u00b1 \u00b5\u03bd ) = D\u00b5W + \u03bd \u2212 \u00b5\u03bd ) = D\u2020\u00b5W \u2212\u03bd \u2212 D\u03bd W + \u00b5, D\u2020\u03bd W \u2212\u00b5, where we have de\ufb01ned a covariant derivative that acts on W + \u00b5, D\u00b5 \u2261 \u2202\u00b5 \u2212 = \u2202\u00b5 \u2212 ig2A3 \u00b5 ig2(sWA\u00b5 + cWZ\u00b5). (87.22) If we identify A\u00b5 as the electromagnetic vector potential, and assign electric charge Q = +1 (in units of the proton charge) to the W +, then we see from (87.15) (87.16) (87.17) (87.18) (87.19) (87.20) (87.21) 87: The Standard Model: Gauge and Higgs Sector 530 eq. (87.22) that we must identify the electromagnetic coupling constant e as e = g2 sin \u03b8W. (87.23) Here we are adopting the convention that e is positive. (In our treatment of quantum electrodynamics, we used the convention that e is negative, but that is less convenient in the present context.) We also have \u00b5\u03bd = \u2202\u00b5A3 F 3 \u2202\u03bdA3 \u00b5 \u2212 \u03bd \u2212 = sWF\u00b5\u03bd + cWZ\u00b5\u03bd \u2212 sWZ\u00b5\u03bd, B\u00b5\u03bd = cWF\u00b5\u03bd \u2212 ig2(W + \u00b5 W \u2212\u03bd \u2212 W + ig2(W + \u00b5 W \u2212\u03bd \u2212 \u03bd W \u2212\u00b5 ) W + \u03bd W \u2212\u00b5 ), (87.24) (87.25) where F\u00b5\u03bd = \u2202\u00b5", "A\u03bd \u2212 \u2202\u03bdA\u00b5 is the usual electromagnetic \ufb01eld strength, and Z\u00b5\u03bd \u2261 \u2202\u00b5Z\u03bd \u2212 \u2202\u03bd Z\u00b5 (87.26) is the abelian \ufb01eld strength associated with the Z\u00b5 \ufb01eld. Now we can assemble all of this into the complete lagrangian for the electroweak gauge \ufb01elds and the Higgs boson in unitary gauge. We will express g2 in terms of e and \u03b8W via g2 = e/sin \u03b8W, and \u03bb in terms of mH and v via \u03bb = 2m2 H/v2. We ultimately get 4 Z \u00b5\u03bdZ\u00b5\u03bd \u2212 4 F \u00b5\u03bdF\u00b5\u03bd \u2212 \u00b5W \u2212 D\u2020 \u2212 + ie(F \u00b5\u03bd + cot \u03b8WZ \u00b5\u03bd)W + \u00b5 W \u2212\u03bd 1 1 = L \u03bdD\u00b5W + \u03bd + D\u2020 \u00b5W \u2212 \u03bdD\u03bdW + \u00b5 1 2 (e2/sin2\u03b8W)(W +\u00b5W \u2212\u00b5 W +\u03bdW \u2212\u03bd \u2212 Z Z \u00b5Z\u00b5)(1 + v\u2212 2 M 2 (M 2 1H 3 HH 2 2 \u2202\u00b5H\u2202\u00b5H 1 Hv\u2212 WW +\u00b5W \u2212\u00b5 + 1 2 m2 2 m2 W +\u00b5W + 1H)2 8 m2 1 1 1 \u2212 \u2212 \u2212 \u00b5 W \u2212 \u03bdW \u2212\u03bd ) Hv\u2212 2H 4, (87.27) \u2212 \u2212 \u2212 where D\u00b5 = \u2202\u00b5 \u2212 ie(A\u00b5 + cot \u03b8WZ\u00b5). (87.28) \u00b5 \ufb01eld assigned electric charge Q = +1, this lagrangian exhibits With the W + manifest electromagnetic gauge invariance. The full underlying SU(2) \u00d7 U(1) gauge invariance is not manifest, however, because we have \ufb01xed unitary gauge. Reference Notes Discussions of the Standard Model in R\u03be gauge can be found in Cheng & Li and Ramond II. Problems 87: The Standard Model: Gauge and Higgs Sector 531 87.1) Find the generator Q of the unbroken U(1) subroup as a linear com- bination of the T a\u2019s and Y. 87.2) a) Ignoring loop corrections, \ufb01nd the numerical values of v, g1, and g2. Take e2/4\u03c0 = \u03b1(MZ) = 1/127.9 and sin2 \u03b8W = 0.231. b) The Fermi constant is de\ufb01ned (at tree level) as GF \u2261 e2 4\u221a2 sin2\u03b8WM 2 W. (87.29) Find its numerical value in GeV\u2212 2. c) Express GF in terms of v. 87.3) In this problem we will work out the generator matrices introduced in section 86 for the case of the Standard Model. a) Write the Higgs \ufb01eld as \u03d5 = 1 \u221a2 \u03c61 + i\u03c63 \u03c62 + i\u03c64!. (87.30) where \u03c6i is a real scalar \ufb01eld. Express the SU(2) generators T a and the hypercharge generator Y as 4 that act on \u03c6i. Hint: see problem 86.1. b) Compute the matrix F a c) Compute the mass-squared matrix for the vector \ufb01elds, (M 2)ab = F a i, de\ufb01ned in eq. (86.12). i, and \ufb01nd its eigenvalues. 4 matrices a and iF b \u00d7 Y T 87.4) Work out the Feynman rules for the lagrangian of eq. (87.27). 87.5) Assume that mH > 2MZ, and compute (at tree level) the decay rate of the Higgs boson into W +W \u2212 and Z 0Z 0 pairs. Express your answer in GeV for mH = 200 GeV. 88: The Standard Model: Lepton Sector 532 88 The Standard Model: Lepton Sector Prerequisite: 75, 87 Leptons are spin-one-half particles that are singlets of the color group. There are six di\ufb00erent \ufb02avors of lepton; see Table 2. The six \ufb02avors are naturally grouped into three families or generations: e and \u03bde, \u00b5 and \u03bd\u00b5, \u03c4 and \u03bd\u03c4. Let us begin by describing a single lepton family, the electron and its neutrino. We introduce left", "-handed Weyl \ufb01elds \u2113 and \u00afe in the representaU(1). Here the bar over the e in the tions (2, \ufb01eld \u00afe is part of the name of the \ufb01eld, and does not denote any sort of conjugation. The covariant derivatives of these \ufb01elds are 1 2 ) and (1, +1) of SU(2) \u00d7 \u2212 (88.1) (88.2) (88.3) (D\u00b5\u2113)i = \u2202\u00b5\u2113i \u2212 D\u00b5\u00afe = \u2202\u00b5\u00afe \u2212 ig2Aa \u00b5(T a)i j\u2113j \u2212 ig1( \u2212 1 2 )B\u00b5\u2113i, ig1(+1)B\u00b5\u00afe, and their kinetic terms are i \u00af\u03c3\u00b5(D\u00b5\u2113)i + i\u00afe\u2020 \u00af\u03c3\u00b5D\u00b5\u00afe. Lkin = i\u2113\u2020 1 2 ) The representation (2, plex; hence the gauge theory is chiral, and therefore parity violating. (1, +1) for the left-handed Weyl \ufb01elds is com- \u2295 \u2212 We cannot write down a mass term involving \u2113 and/or \u00afe because there is no gauge-group singlet contained in any of the products (21, +1) (2, (2, 1 2 ), \u2212 (1, +1), (1, +1). \u2297 \u2297 \u2297 (88.4) However, we are able to write down a Yukawa coupling of the form (88.5) LYuk = y\u03b5ij\u03d5i\u2113j \u00afe + h.c., \u2212 1 where \u03d5 is the Higgs \ufb01eld in the (2, 2 ) representation that we introduced in the last section, and y is the Yukawa coupling constant. A gauge-invariant Yukawa coupling is possible because there is a singlet on the right-hand side of \u2212 (2, 1 2 ) \u2212 \u2297 (2, 1 2 ) \u2212 \u2297 (1, +1) = (1, 0) (3, 0). \u2295 (88.6) There are no other gauge-invariant terms involving \u2113 or \u00afe that have mass dimension four or less. Hence there are no other terms that we could add to while preserving renormalizability. L We add eqs. (88.3) and (88.5) to the lagrangian for \u03d5 and the gauge \ufb01elds that we worked out in the last section. In unitary gauge, we replace 88: The Standard Model: Lepton Sector 533 name electron electron neutrino muon muon neutrino tau tau neutrino symbol mass (MeV) 0.511 0 105.7 0 1777 0 e \u03bde \u00b5 \u03bd\u00b5 \u03c4 \u03bd\u03c4 Table 2: The six \ufb02avors of lepton. Q is the electric charge in units of the proton charge. Each charged \ufb02avor is represented by a Dirac \ufb01eld, each neutral \ufb02avor by a Majorana \ufb01eld (or, equivalently, a left-haned Weyl \ufb01eld). Neutrino masses are exactly zero in the Standard Model. \u03d51 with 1 \u221a2 Higgs boson, and \u03d52 with zero. The Yukawa term becomes (v+H), where H is the real scalar \ufb01eld representing the physical LYuk = \u2212 1 \u221a2 y(v + H)(\u21132\u00afe + h.c.). (88.7) It is now convenient to assign new names to the SU(2) components of \u2113, \u2113 = \u03bd e!. (88.8) (We will rely on context to distinguish the \ufb01eld e from the electromagnetic coupling constant e.) Then eq. (88.7) becomes LYuk = = 1 \u221a2 1 \u221a2 \u2212 \u2212 y(v + H)(e\u00afe + \u00afe\u2020e\u2020) y(v + H) EE where we have de\ufb01ned a Dirac \ufb01eld for the electron, e E \u2261 \u00afe\u2020!. We see that the electron has acquired a mass me = yv \u221a2. The neutrino has remained massless. (88.9) (88.10) (88.11) 88: The Standard Model: Lepton Sector 534 We can describe the neutrino", " with a Majorana \ufb01eld \u03bd N \u2261 \u03bd\u2020!. However, it is often more convenient to work with NL \u2261 PL N = \u03bd 0!, (88.12) (88.13) \u2212 2 (1 \u03b35). We can think of where PL = 1 neutrino kinetic term i\u03bd\u2020\u00af\u03c3\u00b5\u2202\u00b5\u03bd can be written as i NL as a Dirac \ufb01eld; for example, the NL. Now we return to eqs. (88.1) and (88.2), and express the covariant derivatives in the terms of the W \u00b1\u00b5, Z\u00b5, and A\u00b5 \ufb01elds. From our results in section 87, we have NL/\u2202 g2A1 \u00b5T 1 + g2A2 \u00b5T 2 = g2 \u221a2 0 W + \u00b5 W \u2212\u00b5 0! (88.14) and g2A3 \u00b5T 3 + g1B\u00b5Y = e sW (sWA\u00b5 + cWZ\u00b5)T 3 + e cW (cWA\u00b5 \u2212 sWZ\u00b5)Y = e(A\u00b5 + cot \u03b8WZ\u00b5)T 3 + e(A\u00b5 \u2212 = e(T 3 + Y )A\u00b5 + e(cot \u03b8WT 3 tan \u03b8WZ\u00b5)Y tan \u03b8WY )Z\u00b5. (88.15) \u2212 Since we identify A\u00b5 as the electromagnetic \ufb01eld and e as the electromagnetic coupling constant (with the convention that e is positive), we identify Q = T 3 + Y as the generator of electric charge. Then, since 2 \u03bd, T 3e = T 3\u03bd = + \u00afe = 0, 1 Y \u00afe = +\u00afe, 2 e, \u2212 \u2212 (88.16) (88.17) (88.18) we see from eq. (88.16) that Q\u03bd = 0, Qe = e, \u2212 Q\u00afe = +\u00afe. (88.19) This is just the set of electric charge assignments that we expect for the electron and the neutrino. Then (since the action of Q on the \ufb01elds is more 88: The Standard Model: Lepton Sector 535 familiar than the action of Y ) it is convenient to replace Y in eq. (88.15) with Q T 3. We \ufb01nd \u2212 \u00b5T 3 + g1B\u00b5Y = eQA\u00b5 + e[(cot \u03b8W + tan \u03b8W)T 3 g2A3 tan \u03b8WQ]Z\u00b5 \u2212 (88.20) \u2212 In terms of the four-component \ufb01elds, we have = eQA\u00b5 + e sWcW (T 3 s2 WQ)Z\u00b5. (g2A3 (g2A3 \u00b5T 3 + g1B\u00b5Y ) E \u00b5T 3 + g1B\u00b5Y ) = eA\u00b5 + e \u2212 h sWcW NL = e ( sWcW \u2212 2 )Z\u00b5 NL. (+ 1 1 2 PL + s2 W)Z\u00b5 E i, (88.21) (88.22) Using eqs. (88.14) and (88.21\u201388.22) in eqs. (88.1\u201388.3), we \ufb01nd the coupings of the gauge \ufb01elds to the leptons, Lint = 1 \u221a2 g2W + \u00b5 J \u2212 \u00b5 + 1 \u221a2 g2W \u2212\u00b5 J +\u00b5 + e sWcW Z\u00b5J \u00b5 Z + eA\u00b5J \u00b5 EM, J +\u00b5 where we have de\ufb01ned the currents \u2261 E L\u03b3\u00b5 NL, \u2261 NL\u03b3\u00b5 EL, J \u00b5 WJ \u00b5 s2 3 \u2212 2 NL\u03b3\u00b5 \u03b3\u00b5 EM \u2261 \u2212E \u00b5 1 NL \u2212. EM, 1 2 E L\u03b3\u00b5 EL, (88.23) (88.24) (88.25) (88.26) (88.27) (88.28) In many cases, we are interested in scattering amplitudes for leptons whose momenta are all well below the W \u00b1 and Z 0 masses. In this case, we can integrate the W \u00b1\u00b5 and Z\u00b5 \ufb01elds out of the path integral, as discussed in section 29. We get the leading term (in a double expansion in powers of the gauge couplings and inverse powers of MW and MZ) by ignoring the kinetic energy and other interactions of the W \u00b1\u00b5 and Z\u00b5 \ufffd", "\ufffd\ufffdelds, solving the equations of motion for them that follow from Lint is given by eq. (88.23) and Lint, where Lmass + M 2 (88.29) Lmass = and \ufb01nally substituting the solutions back into Lint. This is equivalent to evaluating tree-level Feynman diagrams with a single W \u00b1 or Z 0 exchanged, with the propagator g\u00b5\u03bd /M 2 Z Z \u00b5Z\u00b5, Lmass + W,Z. The result is WW +\u00b5W \u2212\u00b5 \u2212 2 M 2 \u2212 1 Le\ufb00 = = g2 2 2M 2 W e2 WM 2 2s2 W J +\u00b5J \u2212\u00b5 + e2 WM 2 Wc2 Z 2s2 J \u00b5 Z JZ\u00b5 (J +\u00b5J \u2212\u00b5 + J \u00b5 Z JZ\u00b5) = 2\u221a2 GF(J +\u00b5J \u2212\u00b5 + J \u00b5 Z JZ\u00b5). (88.30) 88: The Standard Model: Lepton Sector 536 p1 \u00b5 p 1 \u03bd\u00b5 p 2 \u03bde p 3 e Figure 88.1: Feynman diagram for muon decay. The wavy line is a W propagator. We used e = g2 sin \u03b8W and MW = MZ cos \u03b8W to get the second line, and we de\ufb01ned the Fermi constant GF \u2261 e2 4\u221a2 sin2\u03b8WM 2 W (88.31) Le\ufb00 to compute the tree-level scattering in the third line. We can use \u03bdee\u2212; we leave this to the problems. amplitude for processes like \u03bdee\u2212 \u2192 Having worked out the interactions of a single lepton generation, we now examine what happens when there is more than one of them. Let us consider the \ufb01elds \u2113iI and \u00afeI, where I = 1, 2, 3 is a generation index. The kinetic term for all these \ufb01elds is Lkin = i\u2113\u2020 i I \u00af\u03c3\u00b5(D\u00b5)i j\u2113jI + i\u00afe\u2020I \u00af\u03c3\u00b5D\u00b5\u00afeI, (88.32) where the repeated generation index is summed. The most general Yukawa term we can write down now reads LYuk = \u2212 \u03b5ij\u03d5i\u2113jIyIJ \u00afeJ + h.c., (88.33) LIJ\u2113J and \u00afeI \u2192 where yIJ is a complex 3 3 matrix, and the generation indices are summed. \u00d7 We can make unitary transformations in generation space on the \ufb01elds: \u00afEIJ \u00afeJ, where L and \u00afE are independent unitary \u2113I \u2192 matrices. The kinetic terms are unchanged, and the Yukawa matrix y is replaced with LTy \u00afE. We can choose L and \u00afE so that LTy \u00afE is diagonal with EI then have masses meI = positive real entries yI. The charged leptons yIv/\u221a2, and the neutrinos remain massless. In the currents, eqs. (88.24\u2013 88.28), we simply add a generation index I to each \ufb01eld, and sum over it. Let us work out the details for one process of particular importance: muon decay, \u00b5\u2212 \u2192 for the electron, Nm for the M muon neutrino. Only the charged currents J \u00b1\u00b5 are relevant; the neutral e\u2212\u03bde\u03bd\u00b5. Let the four-component \ufb01elds be Ne for the electron neutrino, and for the muon, E 88: The Standard Model: Lepton Sector 537 current J \u00b5 Z and the electromagnetic current J \u00b5 the \u03c4 terms in the charged currents, we have EM do not contribute. Ignoring J +\u00b5 = ML\u03b3\u00b5 NmL\u03b3\u00b5 The relevant term in the e\ufb00ective interaction is E L\u03b3\u00b5 NeL\u03b3\u00b5 NeL + EL + \u00b5 = J \u2212 NmL, ML. (88.34) (88.35) E L\u03b3\u00b5 This can be simpli\ufb01ed by means of a Fierz identity (see problem 36.3), Le\ufb00 = 2\u221a2 GF( NmL\u03b3\u00b5 ML). NeL)( (88.36) Le\ufb00 = \u2212 4\u221a2 GF( CPL Ne)( PR N E C m). M (88.37) Assigning momenta as shown in \ufb01g. (88.1), and using the usual Feynman rules for incoming and outgoing particles and antiparticles, the scattering amplitude", " is T = = \u2212 \u2212 4\u221a2 GF(uT 1 C PLv\u20322)(u \u20323PR C 4\u221a2 GF(v1PLv\u20322)(u \u20323PRv\u20321). u \u2032 T 1 ) Taking the complex conjugate, and using PL = PR, we \ufb01nd \u2217 = T \u2212 4\u221a2 GF(v \u20322PRv1)(v \u20321PLu\u20323). (88.38) (88.39) Multiplying eqs. (88.38) and (88.39), summing over \ufb01nal spins and averaging over the initial spin, we get 2 h|T | i = 1 2 (4\u221a2)2G2 F Tr[( Tr[( \u2212 /p1\u2212 m\u00b5)PL( \u2212 /p \u20323+me)PR( \u2212 \u2212 /p \u20322)PR] /p \u20321)PL]. \u00d7 The traces are easily evaluated, with the result 2 h|T | i = 64G2 F(p1p\u20322)(p\u20321p\u20323). (88.40) (88.41) We get the decay rate \u0393 by multiplying ing over p\u20321,2,3. We worked out the result (in the limit me \u226a 11.3, by dLIPS3(p1) and integratm\u00b5) in problem i 2 h|T | \u0393 = G2 Fm5 \u00b5 192\u03c03. (88.42) After including one-loop corrections from electromagnetism, and accounting for the nonzero electron mass, the measured muon decay rate is used to determine the value of GF, with the result GF = 1.166 5 GeV\u2212 10\u2212 2. \u00d7 88: The Standard Model: Lepton Sector 538 Reference Notes Lepton phenomonology is covered in more detail in in Cheng & Li, Georgi, Peskin & Schroeder, Quigg, and Ramond II. Problems 88.1) Verify the claim made immediately after eq. (88.6). 88.2) Show that a neutrino always has negative helicity, and that an an- tineutrino always has positive helicity. Hint: see section 75. 88.3) Show that the sum of eqs. (88.32) and (88.33), when rewritten in terms of \ufb01elds of de\ufb01nite mass, has a global symmetry U(1) U(1). The corresponding charges are called electron number, muon number, and tau number; the sum of the charges is the lepton number. List the value of each charge for each Dirac \ufb01eld U(1) \u00d7 \u00d7 EI and NLI. 88.4) Compute for muon decay using eq. (88.36), without making the Fierz transformation to eq. (88.37), and verify eq. (88.41). 2 h|T | i 88.5) a) Write down the term in Express your answer in the form Le\ufb00 that is relevant for and \u03bd\u00b5e\u2212 \u2192 \u03bd\u00b5e\u2212. \u221a2 \u03b3\u00b5(1 Le\ufb00 = 1 \u03b3\u00b5(CV\u2212 is the muon neutrino \ufb01eld, and determine the values of CV CA\u03b35) E GF N (88.43) N E \u03b35) \u2212, where N and CA. b) Repeat part (a) for \u03bdee\u2212 \u2192 2 c) Compute h|T | electron mass, and CV and CA. i \u03bdee\u2212. as a function of the Mandelstam variables, the 88.6) Compute the rates for the decay processes W + e+e\u2212, \u03bde\u03bde. Neglect the electron mass. Express your results in e+\u03bde, Z 0 \u2192 \u2192 and Z 0 GeV. \u2192 88.7) Anomalous dimension of the Fermi constant. The coe\ufb03cient of the e\ufb00ective interaction for muon decay, eq. (88.36), is subject to renormalization by quantum electrodynamic processes. In particular, we can compute its anomalous dimension \u03b3G, de\ufb01ned via \u00b5 d d\u00b5 GF(\u00b5) = \u03b3G(\u03b1)GF(\u00b5), (88.44) where \u03b1 = e2/4\u03c0 is the \ufb01ne-structure constant in the MS scheme with renormalization scale \u00b5. 88: The Standard Model: Lepton Sector 539 a) Argue that it is GF(MW) that is given by eq. (88.31). b)", " Multiply eq. (88.36) by a renormalzing factor ZG, and de\ufb01ne ln(ZG/Z2) = Gn(\u03b1) \u03b5n, \u221e Xn=1 (88.45) where Z2 is the renormalizing factor for a \ufb01eld of unit charge in spinor electrodynamics. Show that \u03b3G(\u03b1) = \u03b1 \u20321(\u03b1). G (88.46) c) If \u03b3G(\u03b1) = c1\u03b1 + O(\u03b12) and \u03b2(\u03b1) = b1\u03b12 + O(\u03b13), show that GF(\u00b5) = c1/b1 \u03b1(\u00b5) \u03b1(MW) GF(MW) (88.47) (For \u00b5 > MW, we should not be using an e\ufb00ective for \u00b5 < MW. interaction.) d) If \u03b1(\u00b5) ln(MW/\u00b5) \u226a 1, show that eq. (88.47) becomes GF(\u00b5) = 1 h \u2212 c1\u03b1(\u00b5) ln(MW/\u00b5) i GF(MW). (88.48) e) Use a Fierz identity to rewrite eq. (88.36) in charge retention form, Le\ufb00 = 2\u221a2 ZGGF( E L\u03b3\u00b5 ML)( NmL\u03b3\u00b5 NeL). (88.49) f) Consider the process of muon decay with an extra photon connecting the \u00b5 and e lines. Work in Lorenz gauge, and with the fourfermion vertex provided by eq. (88.49). Use your results from problem 62.2 to show that, in this gauge, there is no O(\u03b1) contribution to ZG in the MS scheme. g) Use your result from part (d), and your result for Z2 in Lorenz gauge from problem 62.2, to show that c1 = 0, and hence that GF(\u00b5) = GF(MW) at the one-loop level. 89: The Standard Model: Quark Sector 540 89 The Standard Model: Quark Sector Prerequisite: 88 Quarks are spin-one-half particles that are triplets of the color group. There are six di\ufb00erent \ufb02avors of quark; see Table 1 in section 83. The six \ufb02avors are naturally grouped into three families or generations: u and d, c and s, t and b. Let us begin by describing a single quark family, the up and down quarks. We introduce left-handed Weyl \ufb01elds q, \u00afu, and \u00afd in the representations (3, 2, + 1 U(1). Here the bar over the letter in the \ufb01elds \u00afu and \u00afd is part of the name of the \ufb01eld, and does not denote any sort of conjugation. The covariant derivatives of these \ufb01elds are 3 ), and (\u00af3, 1, + 1 3 ) of SU(3) 6 ), (\u00af3, 1, SU(2) \u00d7 \u2212 \u00d7 2 (D\u00b5q)\u03b1i = \u2202\u00b5q\u03b1i \u2212 (D\u00b5 \u00afu)\u03b1 = \u2202\u00b5 \u00afu\u03b1 (D\u00b5 \u00afd)\u03b1 = \u2202\u00b5 \u00afd\u03b1 \u2212 \u2212 ig3Aa \u00b5(T a 3 )\u03b1 ig2Aa ig1(+ 1 \u03b2q\u03b2i \u2212 \u2212 ig1( \u03b2 \u00afu\u03b2 \u2212 ig1(+ 1 \u2212 \u2212 \u00b5(T a jq\u03b2j 2 )i 6 )B\u00b5q\u03b1i, 3 )B\u00b5 \u00afu\u03b1, 3 )B\u00b5 \u00afd\u03b1. 2 \u00af3 )\u03b1 3 )\u03b1 ig3Aa \u00b5(T a ig3Aa \u00b5(T a \u03b2 \u00afd\u03b2 (89.1) (89.2) (89.3) We rely on context to distinguish the SU(3) gauge \ufb01elds from the SU(2) gauge \ufb01elds. The kinetic terms for q, \u00afu, and \u00afd are Lkin = iq\u2020 \u03b1i \u00af\u03c3\u00b5(D\u00b5q)\u03b1i + i\u00afu\u2020\u03b1 \u00af\u03c3\u00b5(D\u00b5 \u00afu)\u03b1 + i \u00afd\u2020\u03b1 \u00af\u03c3\u00b5(D\u00b5 \u00afd)\u03b1. (89.4) The representation (3, 2, + 1 3 ) for the left-handed 6 ) Weyl \ufb01elds is complex; hence the gauge theory is chiral, and therefore parity violating. (\u00af3, 1, + 1 (\u00af3, 1, 2 3 ) \u2295", " \u2295 \u2212 We cannot write down a mass term involving q, \u00afu, and/or \u00afd because there is no gauge-group singlet contained in any of the products of their representations. But we are able to write down Yukawa couplings of the form \u2212 y\u2032\u03b5ij\u03d5iq\u03b1j \u00afd\u03b1 LYuk = \u2212 1 2 ) representation that we introduced where \u03d5 is the Higgs \ufb01eld in the (1, 2, \u2212 in section 87, and y\u2032 and y\u2032\u2032 are the Yukawa coupling constants. These gauge-invariant Yukawa couplings are possible because there are singlets on the right-hand sides of (89.5) y\u2032\u2032\u03d5\u2020 iq\u03b1i \u00afu\u03b1 + h.c., (1, 2, 1 2 ) \u2212 (1, 2, + 1 2 ) (3, 2, + 1 6 ) (3, 2, + 1 6 ) \u2297 \u2297 \u2297 \u2297 (\u00af3, 1, + 1 (\u00af3, 1, \u2212 3 ) = (1, 1, 0) 2 3 ) = (1, 1, 089.6) (89.7) 89: The Standard Model: Quark Sector 541 There are no other gauge-invariant terms involving q, \u00afu, or \u00afd that have mass dimension four or less. Hence there are no other terms that we could add to while preserving renormalizability. In unitary gauge, we replace \u03d51 with 1 \u221a2 (v + H), where H is the real scalar \ufb01eld representing the physical Higgs boson, and \u03d52 with zero. The Yukawa term becomes L LYuk = \u2212 1 \u221a2 y\u2032(v + H)q\u03b12 \u00afd\u03b1 1 \u221a2 \u2212 y\u2032\u2032(v + H)q\u03b11 \u00afu\u03b1 + h.c.. (89.8) It is now convenient to assign new names to the SU(2) components of q, q = u d!. (89.9) Then eq. (89.8) becomes LYuk = = 1 \u221a2 \u2212 1 \u221a2 y\u2032(v + H)(d\u03b1 \u00afd\u03b1 + \u00afd\u2020\u03b1d\u2020 \u03b1 y\u2032(v + H) 1 \u221a2 D\u03b1 \u2212 where we have de\ufb01ned Dirac \ufb01elds for the down and up quarks, \u2212 D U y\u2032\u2032(v + H) \u03b1 U\u03b1, \u03b1) 1 \u221a2 \u2212 y\u2032\u2032(v + H)(u\u03b1 \u00afu\u03b1 + \u00afu\u2020\u03b1u\u2020 \u03b1) (89.10) (89.11) D\u03b1 \u2261 d\u03b1 \u00afd \u2020\u03b1!, u\u03b1 U\u03b1 \u2261 \u00afu\u2020\u03b1!. We see from eq. (89.10) that the up and down quarks have acquired masses md = y\u2032v \u221a2, mu = y\u2032\u2032v \u221a2. (89.12) Now we return to eqs. (89.1\u201389.3), and express the covariant derivatives in the terms of the W \u00b1\u00b5, Z\u00b5, and A\u00b5 \ufb01elds. From our results in section 88, we have g2A1 \u00b5T 1 + g2A2 \u00b5T 2 = g2 \u221a2 0 W + \u00b5 W \u2212\u00b5 0!, (89.13) g2A3 \u00b5T 3 + g1B\u00b5Y = eQA\u00b5 + e sWcW (T 3 \u2212 s2 WQ)Z\u00b5, (89.14) where Q = T 3 + Y (89.15) is the generator of electric charge. Then, since 2 d, T 3 \u00afu = 0, 2 u, T 3d = Y \u00afu = 6 d, 6 u, T 3u = + u, \u2212 we see from eq. (89.15) that T 3 \u00afd = 0, Y \u00afd = + 1 3 \u00afd, (89.16) (89.17) 89: The Standard Model: Quark Sector 542 Qu = + 2 3 u Qd = \u2212 1 3 d, Q\u00afu = \u2212 2 3 \u00afu, Q \u00afd = + 1 3 \u00afd. (89.18) This is just the set of electric charge assignments that we expect for the up and down quarks. In terms of the four-component \ufb01elds, we have (g2A3 \u00b5T 3 + g1B\u00b5Y ) U (g2A3 \u00b5T 3 + g1B\u00b5Y ) D = = 3 eA\u00b5 + e sWcW (+ 1 1 3 eA\u00b5", " + e sWcW ( \u2212 2 PL \u2212 2 PL + 1 1 2 3 s2 W)Z\u00b5 3 s2 W)Z\u00b5 + 2 h \u2212 h, (89.19) U i, (89.20) D i Using eqs. (89.13) and (89.19\u201389.20) in eqs. (89.1\u201389.4), we \ufb01nd the coupings of the electroweak gauge \ufb01elds to the quarks, Lint = 1 \u221a2 g2W + \u00b5 J \u2212 \u00b5 + 1 \u221a2 g2W \u2212\u00b5 J +\u00b5 + e sWcW where we have de\ufb01ned the currents Z\u00b5J \u00b5 Z + eA\u00b5J \u00b5 EM, J +\u00b5 \u00b5 J \u2212 \u2261 DL\u03b3\u00b5 U L, \u2261 U L\u03b3\u00b5 DL, J \u00b5 WJ \u00b5 s2 3 \u2212 2 U L\u03b3\u00b5 \u03b3\u00b5 + EM \u2261 EM, 1 2 DL\u03b3\u00b5 \u03b3\u00b5 1 3 D DL,. (89.21) (89.22) (89.23) (89.24) (89.25) (89.26) U \u2212 Having worked out the interactions of a single quark generation, we now examine what happens when there is more than one of them. Let us consider the \ufb01elds q\u03b1iI, \u00afuI, and \u00afdI, where I = 1, 2, 3 is a generation index. The kinetic term for all these \ufb01elds is D \u03b1iI \u00af\u03c3\u00b5(D\u00b5)\u03b1i Lkin = iq\u2020 \u03b2 \u00afd\u03b2 I, (89.27) where the repeated generation index is summed. The most general Yukawa term we can write down now reads \u03b2jq\u03b2jI + i\u00afu\u2020\u03b1I \u00af\u03c3\u00b5(D\u00b5)\u03b1 I + i \u00afd\u2020\u03b1I \u00af\u03c3\u00b5(D\u00b5)\u03b1 \u03b2 \u00afu\u03b2 LYuk = \u2212 \u03b5ij\u03d5iq\u03b1jIy\u2032IJ \u00afd\u03b1 J \u2212 \u03d5\u2020 iq\u03b1iIy\u2032\u2032IJ \u00afu\u03b1 J + h.c., (89.28) where y\u2032IJ and y\u2032\u2032IJ are complex 3 are summed. In unitary gauge, this becomes \u00d7 3 matrices, and the generation indices LYuk = \u2212 1 \u221a2 (v + H)d\u03b1Iy\u2032IJ \u00afd\u03b1 J \u2212 1 \u221a2 (v + H)u\u03b1Iy\u2032\u2032IJ \u00afu\u03b1 J + h.c.. (89.29) We can make unitary transformations in generation space on the \ufb01elds: \u00afUIJ \u00afuJ, where U, D, \u00afU dI \u2192 and \u00afD are independent unitary matrices. The kinetic terms are unchanged UIJuJ, and \u00afuI \u2192 DIJ dJ, \u00afdI \u2192 \u00afdJ, uI \u2192 \u00afDIJ 89: The Standard Model: Quark Sector 543 (except for the couplings to the W \u00b1, as we will discuss momentarily), and the Yukawa matrices y\u2032 and y\u2032\u2032 are replaced with DTy\u2032 \u00afD and U Ty\u2032\u2032 \u00afU. We can choose D, \u00afD, U, and \u00afU so that DTy\u2032 \u00afD and U Ty\u2032\u2032 \u00afU are diagonal with positive real entries y\u2032I and y\u2032\u2032I. The down quarks DI then have masses UI have masses muI = y\u2032\u2032I v/\u221a2. mdI = y\u2032Iv/\u221a2, and the up quarks In the neutral currents J \u00b5 EM, we simply add a generation index I to each \ufb01eld, and sum over it. The charged currents are more complicated, however; they become 3 and J \u00b5 J +\u00b5 = \u00b5 = DLI(V \u2020)IJ \u03b3\u00b5 U LIVIJ\u03b3\u00b5 ULJ, (89.30) J \u2212 DLK, U \u2020D is the Cabibbo\u2013Kobayashi\u2013Maskawa matrix (or CKM mawhere V trix for short). Note that we did not have this complication in the lepton sector, because there we had only one Yukawa term. (89.31) \u2261 \u00d7 A 3 3 unitary matrix has 9 real parameters. However, we are still free UI, to make the independent phase rotations as these leave the kinetic and mass terms invariant. These phase changes allow us to make the \ufb01rst row and column of VIJ real, eliminating 5 of the 9 parameters. The remaining four can be chosen as \u03b81 (the Cabibbo angle), \u03b82, \u03b8", "3, and \u03b4, where DI and DI \u2192 UI \u2192 ei\u03b1I ei\u03b2I c1 +s1c3 V = \uf8eb s1c2 \u2212 s1s2 \uf8ec \uf8ec \uf8ed \u2212 s2s3ei\u03b4 c1c2c3 \u2212 c1s2c3 + c2s3ei\u03b4 +s1s3 c1c2s3 + s2c3ei\u03b4 c2c3ei\u03b4 c1s2s3 \u2212 \uf8f6, \uf8f7 \uf8f7 \uf8f8 (89.32) and ci = cos \u03b8i and si = sin \u03b8i. The measured values of these angles are s1 = 0.224, s2 = 0.041, s3 = 0.016, and \u03b4 = 40\u25e6. Note that the charged currents now have some terms with a phase factor ei\u03b4, and some without. Since the time-reversal operator T is antiunitary (T \u2212 i), the charged currents do not transform in a simple way under time reversal. This implies that the charged current terms in Lint are is not time-reversal invariant; hence the electroweak interactions violate time-reversal symmetry. Since CP T is always a good symmetry, time-reversal violation is equivalent to CP violation; \u03b4 is therefore sometimes called the CP violating phase. 1iT = \u2212 At high energies, we can use our results to compute electroweak contributions to scattering amplitudes involving quarks. This is because, at high energies, the SU(3) coupling g3 is weak; we can, for example, reliably compute the decay rates of the W \u00b1 and Z 0 into quarks, because g2 3(MZ)/4\u03c0 = 0.12 is small enough to make QCD loop correc\u03b13(MZ) tions a few-percent e\ufb00ect. \u2261 89: The Standard Model: Quark Sector 544 To understand low-energy processes such as neutron decay, we must \ufb01rst write the currents in terms of hadron \ufb01elds. We take this up in the next section. For now, we simply note that the terms in the charged currents that involve only up and down quarks are J +\u00b5 = c1DL\u03b3\u00b5 \u00b5 = c1 U L\u03b3\u00b5 J \u2212 where c1 is the cosine of the Cabibbo angle. UL, DL, (89.33) (89.34) Reference Notes Electroweak interactions of quarks are discussed in more detail in Cheng & Li, Georgi, Peskin & Schroeder, Quigg, and Ramond II. 89.1) Verify the claims made immediately after eqs. (89.4) and (89.7). Problems \u2192 89.2) Compute the rates for the decay processes W + \u00afuu, \u00afdd. Neglect the quark masses. Express your results in and Z 0 GeV. Combine your answers with those of problem 88.6, and sum over generations to get the total decay rates for the W \u00b1 and Z 0. You can neglect the masses of all quarks and leptons except the top quark, and take \u03b82 = \u03b83 = 0. \u2192 \u2192 u \u00afd, Z 0 89.3) Show that the Standard Model is anomaly free. Hint: you must consider 3\u20133\u20133, 2\u20132\u20132, 3\u20133\u20131, 2\u20132\u20131, and 1\u20131\u20131 anomalies, where the number denotes the gauge group of one of the external gauge \ufb01elds in the triangle diagram. Why do we not need to worry about the unlisted combinations? 89.4) Compute the leading term in the beta function for each of the three gauge couplings of the Standard Model. 89.5) After integrating out the W \u00b1 \ufb01elds, we get an e\ufb00ective interaction between the hadron and lepton currents that includes Le\ufb00 = 2\u221a2 ZCC( E L\u03b3\u00b5 NeL)( U L\u03b3\u00b5 DL), (89.35) where we have de\ufb01ned C c1GF at a renormalization scale \u00b5 = MW, and ZC is a renormalizing factor. This interaction contributes to neutron decay; see section 90. In this problem, following the analysis of problem 88.7, we will compute the anomalous dimension \u03b3C of C due to one-loop photon and gluon exchange. \u2261 89: The Standard Model: Quark Sector 545 a) Use Fierz identities to show that eq. (89.35) can be rewritten as Le\ufb00 = 2\u221a2 ZCC( E L\u03b3\u00b5 DL)( U L\u03b3", "\u00b5 NeL), (89.36) and also as Le\ufb00 = \u2212 4\u221a2 ZCC( CPL Ne)( PR U E D C). (89.37) b) Working in Lorenz gauge and using the results of problem 88.7, show that gluon exchange does not make a one-loop contribution to ZC. c) Show that only a photon connecting the e and u lines makes a one-loop contribution to ZC. E PR U C = e\u2020u\u2020, and compare this with d) Note that = e\u2020\u00afe\u2020 + h.c.. Argue that the photon-exchange contribution to ZC is given by the one-loop contribution to Zm in spinor electrodynamics in Lorenz gauge, with the replacement ( e) Let \u03b3C(\u03b1) = c1\u03b1 +..., where \u03b1 = e2/4\u03c0, and \ufb01nd c1. (This c1 should not be confused with the cosine of the Cabibbo angle.) 1)(+1)e2 1)(+ 2 3 )e2. ( \u2212 EE \u2192 \u2212 90: Electroweak Interactions of Hadrons 546 90 Electroweak Interactions of Hadrons Prerequisite: 83, 89 Now that we know how quarks couple to the electroweak gauge \ufb01elds, we can use this information to obtain amplitudes for various processes involving hadrons. We will focus on three of the most important: neutron decay, n \u00b5\u2212 \u00af\u03bd\u00b5; and neutral pion decay, \u03c00 \u03b3\u03b3. pe\u2212 \u00af\u03bd; charged pion decay, \u03c0\u2212 \u2192 \u2192 Recall from section 83 the chiral lagrangian for pions and nucleons, \u2192 = L 1 4 f 2 \u2212 + iN /\u2202N 2 (gA\u2212 \u2212 1 \u03c0 Tr \u2202\u00b5U \u2020\u2202\u00b5U + v3 Tr(M U + M \u2020U \u2020) mN N (U \u2020PL + U PR)N \u2212 1)iN \u03b3\u00b5(U \u2202\u00b5U \u2020PL + U \u2020\u2202\u00b5U PR)N, (90.1) where U (x) = exp[2i\u03c0a(x)T a/f\u03c0], \u03c0a is the pion \ufb01eld, N is the nucleon \ufb01eld, f\u03c0 is the pion decay constant, M is the quark mass matrix, v3 is the value of the quark condensate, mN is the nucleon mass, and gA is the axial vector coupling. The electroweak gauge group SU(2) U(1) is a subgroup of the \u00d7 U(1)V \ufb02avor group that we have in the limit of zero SU(2)L \u00d7 quark mass. It will prove convenient to go through the formal procedure of gauging the full \ufb02avor group, and only later identifying the electroweak subgroup. We therefore de\ufb01ne matrix-valued gauge \ufb01elds l\u00b5(x) and r\u00b5(x) that transform as SU(2)R \u00d7 l\u00b5 \u2192 r\u00b5 \u2192 Ll\u00b5L\u2020 + iL\u2202\u00b5L\u2020, Rr\u00b5R\u2020 + iR\u2202\u00b5R\u2020. (90.2) (90.3) SU(2)R \u00d7 2 unitary matrices that correspond to a general Here L(x) and R(x) are 2 U(1)V gauge transformation; we restrict the U(1) part SU(2)L \u00d7 of the transformation to the vector subgroup by requiring det L = det R and Tr l\u00b5 = Tr r\u00b5. The transformation rules for the pion and nucleon \ufb01elds are \u00d7 (90.4) U LUR\u2020, LNL, \u2192 NL \u2192 NR \u2192 PRN are the left- and right-handed parts of RNR, (90.6) (90.5) where NL \u2261 the nucleon \ufb01eld. PLN and NR \u2261 We can make the chiral lagrangian gauge invariant (except for terms involving the quark masses) by replacing ordinary derivatives with appropriate covariant derivatives. We determine the covariant derivative of each 90: Electroweak Interactions of Hadrons 547 \ufb01eld by requiring it to transform in the same way as the \ufb01eld itself; for example, D\u00b5U L(D\u00b5U )R\u2020. We thus \ufb01nd \u2192 D\u00b5U = \u2202\u00b5U \u2212 il\u00b5U + iU r\u00b5, ir\u00b5U \u2020, D\u00b5U \u2020 = \ufffd", "\ufffd\u00b5U \u2020 + iU \u2020l\u00b5 \u2212 D\u00b5NL = (\u2202\u00b5 \u2212 il\u00b5)NL, D\u00b5NR = (\u2202\u00b5 \u2212 D in ir\u00b5)NR. (90.7) (90.8) (90.9) (90.10) Making the substitution \u2202 \u2192 couple to these gauge \ufb01elds. L, we learn how the pions and nucleons As in section 83, it is more convenient to work with the nucleon \ufb01eld, de\ufb01ned via N where u2 = U. Making this transformation, we ultimately \ufb01nd N = (uPL + u\u2020PR), (90.11 Tr(\u2202\u00b5U \u2020\u2202\u00b5U il\u00b5U \u2194\u2202\u00b5U \u2020 ir\u00b5U \u2020 \u2194\u2202\u00b5U \u2212 \u2212 2l\u00b5U r\u00b5U \u2020) + l\u00b5l\u00b5 + r\u00b5r\u00b5 \u2212 + v3 Tr(M U + M \u2020U \u2020) + i N 2/\u02dc\u2113 + 1 gAN + (/v + 1 N \u2212 2 /\u02dcr ) N /\u2202 mN N N N \u2212 2/\u02dc\u2113 (/a + 1 \u2212 2 /\u02dcr )\u03b35N 1 where v\u00b5 \u2261 a\u00b5 \u2261 \u02dcl\u00b5 \u2261 \u02dcr\u00b5 \u2261 It is now convenient to set 1 2 i[u\u2020(\u2202\u00b5u) + u(\u2202\u00b5u\u2020)], 1 2 i[u\u2020(\u2202\u00b5u) \u2212 u(\u2202\u00b5u\u2020)], u\u2020l\u00b5u, ur\u00b5u\u2020. l\u00b5 = la r\u00b5 = ra \u00b5T a + b\u00b5, \u00b5T a + b\u00b5., (90.12) (90.13) (90.14) (90.15) (90.16) (90.17) (90.18) We have normalized b\u00b5 so that the corresponding charge is baryon number. The SU(2) gauge \ufb01elds of the Standard Model can now be identi\ufb01ed as and the electromagnetic gauge \ufb01eld as g2Aa \u00b5 = la \u00b5, eA\u00b5 = l3 \u00b5 + r3 \u00b5 + 1 2 b\u00b5. (90.19) (90.20) 90: Electroweak Interactions of Hadrons 548 Eq. (90.20) follows from reconciling eqs. (90.9) and (90.10) with the requirement that the electromagnetic covariant derivatives of the proton \ufb01eld p = N1 and the neutron \ufb01eld n = N2 be given by (\u2202\u00b5 \u2212 We can now \ufb01nd the hadronic parts of the currents that couple to the with respect to them, and then setting ieA\u00b5)p and \u2202\u00b5n. gauge \ufb01elds by di\ufb00erentiating them to zero. We \ufb01nd L J a\u00b5 L = (\u2202 = 1 L l=r=0 \u03c0 Tr T aU \u2194\u2202\u00b5U \u2020 + 1 4 if 2 = + 1 /\u2202la \u00b5) 2 f\u03c0\u2202\u00b5\u03c0a 1 2 N 2 \u03b5abc\u03c0b\u2202\u00b5\u03c0c + 1 \u2212 u\u2020T a\u03b3\u00b5(1 gA\u03b35)u \u2212 T a\u03b3\u00b5(1 N gA\u03b3590.21) J a\u00b5 R = (\u2202 l=r=0 /\u2202ra \u00b5) L \u03c0 Tr T aU \u2020 4 if 2 2 f\u03c0\u2202\u00b5\u03c0a 1 \u2212 \u2212 = 1 = uT a\u03b3\u00b5(1+gA\u03b35)u\u2020 N \u2194\u2202\u00b5U + 1 2 N 2 \u03b5abc\u03c0b\u2202\u00b5\u03c0c + 1 1 T a\u03b3\u00b5(1+gA\u03b35) N +..., (90.22) 2 N J \u00b5 B = (\u2202 = N /\u2202b\u00b5). L \u03b3\u00b5 N l=r=0 (90.23) In the third lines of eqs. (90.21) and (90.22), we have expanded in inverse powers of f\u03c0. We can now identify the currents that couple to the physical W \u00b1\u00b5, Z\u00b5, and A\u00b5 \ufb01elds as J +\u00b5 = c1(J 1\u00b5 = 1 \u221a2 iJ 2\u00b5 L ) L \u2212 c1(", "f\u03c0\u2202\u00b5\u03c0+ + i\u03c00 \u2194\u2202\u00b5\u03c0+) + 1 2 c1 n\u03b3\u00b5(1 \u2212 gA\u03b35)p +..., (90.24) J \u2212 \u00b5 = c1(J 1\u00b5 = 1 \u221a2 L + iJ 2\u00b5 L ) c1(f\u03c0\u2202\u00b5\u03c0\u2212 \u2212 WJ \u00b5 s2 EM, 3\u00b5 J \u00b5 = 1 L i\u03c00 \u2194\u2202\u00b5\u03c0\u2212) + 1 2 c1 p\u03b3\u00b5(1 \u2212 gA\u03b35)n +..., (90.25) (90.26) gA\u03b35)p 2 (f\u03c0\u2202\u00b5\u03c00 + i\u03c0+ \u2194\u2202\u00b5\u03c0\u2212) + 1 4 p\u03b3\u00b5(1 L + J 3\u00b5 \u2212 R + 1 = i\u03c0+ \u2194\u2202\u00b5\u03c0\u2212 + p\u03b3\u00b5p +..., 2 J \u00b5 \u2212 B EM = J 3\u00b5 J \u00b5 1 4 n\u03b3\u00b5(1 \u2212 gA\u03b35)n +..., (90.27) (90.28) 90: Electroweak Interactions of Hadrons 549 where c1 is the cosine of the Cabibbo angle, and the interactions are speci\ufb01ed by \u221a2 \u00b5 J \u2212 g2W + \u00b5 + 1 \u221a2 Lint = 1 g2W \u2212\u00b5 J +\u00b5 + e For low-energy processes involving W \u00b1 or Z 0 exchange, we can use the e\ufb00ective current-current interaction that we derived in section 88, Le\ufb00 = 2\u221a2 GF(J +\u00b5J \u2212\u00b5 + J \u00b5 Z + eA\u00b5J \u00b5 Z JZ\u00b5). Z\u00b5J \u00b5 (90.29) (90.30) sWcW EM. We should include both hadronic and leptonic contributions to the currents. \u00b5\u2212\u03bd\u00b5. The relevant terms in the Consider charged pion decay, \u03c0\u2212 \u2192 charged currents (neutral currents do not contribute) are J \u2212 \u00b5 = 1 \u221a2 c1f\u03c0\u2202\u00b5\u03c0\u2212, (90.31) J +\u00b5 = 1 \u03b3\u00b5(1 \u03b35) (90.32) 2 M Nm is the muon neutrino \ufb01eld. The relevant Nm, \u2212 M where is the muon \ufb01eld and term in the e\ufb00ective interaction is then Le\ufb00 = GFc1f\u03c0\u2202\u00b5\u03c0\u2212 The corresponding decay amplitude is \u03b3\u00b5(1 \u2212 \u03b35) Nm. M = GFc1f\u03c0k\u00b5 u1\u03b3\u00b5(1 \u2212 T \u03b35)v2, where the four-momenta of the pion, muon, and antineutrino are k, p1, and p2. Eq. (90.34) can be simpli\ufb01ed by using /k = /p1 + /p2 along with u1 /p1 = m\u00b5u1 and /p2v2 = 0; we get \u2212 We see that \u00b5\u2212 \u00af\u03bd\u00b5 is preferred over decay to e\u2212 \u00af\u03bde. T Squaring T T \u2212 = \u2212 \u03b35)v2. GFc1f\u03c0m\u00b5u1(1 is proportional to the muon mass; since m\u00b5 \u226b and summing over \ufb01nal spins, we \ufb01nd = (GFc1f\u03c0m\u00b5)2( 8p1 \u00b7 = 4(GFc1f\u03c0m\u00b5)2(m2 m2 \u00b5). \u03c0 \u2212 \u00b5 + 0 + m2 m2 (p1+p2)2 = 2 h|T | p2 = p2 1 + p2 p2) \u2212 i \u2212 (90.35) me, decay to (90.36) \u03c0 to get the second We used \u2212 line. We now have 2p1\u00b7 \u0393 = dLIPS2(k) 2 h|T | i 2 \u2212 1 2m\u03c0 Z p1| 8\u03c0m2 = | 2 \u03c0 h|T | i = G2 Fc2 \u00b5m\u03c0 1f 2 \u03c0m2 4\u03c0 2 m2 \u00b5 m2 \u03c0! 1 \u2212, (90.37) (90.33) (90.34) 90: Electroweak Interactions of Hadrons 550 where we used \u00b5)/2m\u03c0 to get the last line. Since we determine the value of GF from the decay rate of the muon (see section 88), the charged pion decay rate allows us", " to \ufb01x the value of c1f\u03c0. p1| | \u03c0 \u2212 = (m2 m2 The value of c1 can be determined from the rate for the decay pro\u03c00e\u2212 \u00af\u03bde, which we will calculate in problem 90.6. The relevant cess \u03c0\u2212 \u2192 hadronic term in the charged current is J \u2212 \u00b5 = 1 \u221a2 \u2212 ic1\u03c00 \u2194\u2202\u00b5\u03c0\u2212, (90.38) which depends on c1 but not f\u03c0. Comparison with experiment then yields c1 = 0.974. We note that the key feature of eq. (90.38) is that it involves spin-zero hadrons that are members of an isospin triplet; eq. (90.38) applies to any such hadrons, including nuclei. Thus c1 can also be measured in superallowed Fermi decays, which take a nucleus from one spin-zero state to another spin-zero state with the same parity in the same isospin multiplet. Having thus determined c1, the charged pion decay rate yields f\u03c0 = 92.4 MeV. Next we consider neutron decay, n charged currents (neutral currents do not contribute) are pe\u2212 \u00af\u03bde. The relevant terms in the \u2192 (90.39) (90.40) J \u2212 \u00b5 = 1 J +\u00b5 = 1 2 c1 p\u03b3\u00b5(1 \u03b3\u00b5(1 \u2212 \u03b35) \u2212 Ne, gA\u03b35)n, 2 E is the electron \ufb01eld and where relevant term in the e\ufb00ective interaction is then Ne is the electron neutrino \ufb01eld. The E Le\ufb00 = 1 \u221a2 GFc1 p\u03b3\u00b5(1 \u2212 gA\u03b35)n \u03b3\u00b5(1 \u2212 \u03b35) Ne. E (90.41) Consider a neutron with four-momentum pn = (mn, 0), and spin up along the z axis; the decay amplitude is = 1 \u221a2 T GFc1[up\u03b3\u00b5(1 \u2212 gA\u03b35)un][ue\u03b3\u00b5(1 \u2212 \u03b35)v\u00af\u03bd ], (90.42) \u2212 \u2212 2 (1 \u03b35/z)( /pn+mn). We take the absolute square of where unun = 1 and sum over the \ufb01nal spins. Since the maximum available kinetic energy is mn \u2212 mp, the proton is nonrelativistic, and we me = 0.782 MeV \u226a mpE\u00af\u03bd in addition p\u00af\u03bd \u2243 \u2212 mpEe and pp\u00b7 pe \u2243 \u2212 can use the approximations pp\u00b7 mnE\u00af\u03bd. After a tedious p\u00af\u03bd = mnEe and pn\u00b7 pe = to the exact formulae pn\u00b7 \u2212 \u2212 but straightforward calculation, we \ufb01nd mp \u2212 T 2 h|T | i = 16 G2 Fc2 1 (1 + 3g2 p\u00af\u03bd pe \u00b7 EeE\u00af\u03bd A)mnmpEeE\u00af\u03bd \u02c6z pe \u00b7 Ee + A + B 1 + a \u00d7 \u02c6z p\u00af\u03bd \u00b7 E\u00af\u03bd, (90.43) 90: Electroweak Interactions of Hadrons 551 l k 1 k l++ Figure 90.1: One-loop diagrams contributing to \u03c00 a proton. \u2192 \u03b3\u03b3. The solid line is where the correlation coe\ufb03cients are given by a = g2 1 A \u2212 1 + 3g2 A, A = 2gA(1 gA) \u2212 1 + 3g2 A, B = 2gA(1 + gA) 1 + 3g2 A. (90.44) When we integrate over the \ufb01nal momenta to get the total decay rate, the correlation terms vanish, and so the rate is proportional to G2 Fc2 A). \u03c00e\u2212 \u00af\u03bde. the neutron Since we get the value of G2 decay rate allows us to determine 1 + 3g2 A. To get the sign of gA, we need a measurement of either A or B. (The antineutrino three-momentum can be determined from the electron and proton three-momenta.) The result is that gA = +1.27. The measured values of the three correlation coe\ufb03cients are all consistent with eq. (90.44). 1 from the rate for \u03c0\u2212 \u2192 1(1 + 3g2 Fc2 \u2192 Finally, we consider the decay of the neutral pion into two photons, \u03c00 \u03b3\u03b3. None of the terms in our chiral lagrangian, eq. (90.12), couple a", " single \u03c00 to two photons. Therefore, without adding more terms, this process does not occur at tree level. However, at the one-loop level, we have the diagrams of \ufb01g. (90.1); a proton circulates in the loop. Let us evaluate these diagrams. In section 83, we found that the coupling of the \u03c00 to the nucleons is given by 1 2 (gA/f\u03c0)\u2202\u00b5\u03c00(p\u03b3\u00b5\u03b35p L\u03c00N N = This leads to a \u03c00pp vertex factor of 1 2 (gA/f\u03c0)k\u03c1\u03b3\u03c1\u03b35. The diagrams in \ufb01g. (90.1) are then identical to the diagrams we evaluated in section 76, and so the one-loop decay amplitude is n\u03b3\u00b5\u03b35n). (90.45) \u2212 \u2212 loop = 1 2 (gA/f\u03c0)(ie)2\u03b51\u00b5\u03b52\u03bdk\u03c1C \u00b5\u03bd\u03c1(k1, k2, k), (90.46) i T1 \u2212 where k\u03c1C \u00b5\u03bd\u03c1(k1, k2, k) = i 2\u03c02 \u03b5\u00b5\u03bd\u03b1\u03b2k1\u03b1k2\u03b2. \u2212 Here we have chosen to renormalize so as to have k1\u00b5C \u00b5\u03bd\u03c1(k1, k2, k) = 0 and k2\u03bd C \u00b5\u03bd\u03c1(k1, k2, k) = 0; this is required by electromagnetic gauge invariance. (90.47) 90: Electroweak Interactions of Hadrons 552 Combining eqs. (90.46) and (90.47), we get loop = T1 \u2212 gAe2 4\u03c02f\u03c0 \u2212 \u03b5\u03b1\u00b5\u03b2\u03bd k1\u03b1\u03b51\u00b5k2\u03b2\u03b52\u03bd. (90.48) This result is subject to higher-loop corrections. Note that diagrams with extra internal pion lines attached to the nucleon loop are not suppressed by any small expansion parameter. Thus we cannot trust the overall coe\ufb03cient in eq. (90.48). Note that this amplitude would arise at tree level from an interaction \u03c00\u03b5\u03b1\u00b5\u03b2\u03bdF\u00b5\u03b1F\u03bd\u03b2. If we integrate out the nucleon \ufb01elds of the form to get an e\ufb00ective lagrangian for the pions and photons alone, such a term should appear. L\u03c00\u03b3\u03b3 \u221d There is a problem, however. The SU(2)L \u00d7 U(1)V symmetry of the e\ufb00ective lagrangian implies that a pion \ufb01eld that has no derivatives acting on it must be accompanied by at least one factor of a quark mass. M \u2020U \u2020)\u03b5\u03b1\u00b5\u03b2\u03bdF\u00b5\u03b1F\u03bd\u03b2. The For example, we could have problem is that there are no quark-mass factors in eq. (90.48). So we have an apparent contradiction between our explicit one-loop result, and a general argument based on symmetry. SU(2)R \u00d7 L\u03c00\u03b3\u03b3 \u221d i Tr(M U \u2212 This contradiction is resolved by noting that the electromagnetic gauge J 3\u00b5 R. In terms \ufb01eld results in an anomaly in the axial current J 3\u00b5 of the quark doublet J 3\u00b5 L \u2212 A \u2261 this current is = Q,! U D (90.49) J 3\u00b5 A = T 3\u03b3\u00b5\u03b35Q \u03b3\u00b5\u03b35 U \u2212 2 U where we have suppressed the color indices. Using our results in sections 76 and 77, the anomalous divergence of this current is given by \u03b3\u00b5\u03b35D Q = 1 (90.50) 1 2 D, \u2202\u00b5J 3\u00b5 A = e2 16\u03c02 Tr(T 3Q2)\u03b5\u00b5\u03bd\u03c1\u03c3F\u00b5\u03bd F\u03c1\u03c3, \u2212 where Q = + 2 3 0 0 1 3! \u2212 is the electric charge matrix acting on the quark \ufb01elds, and the trace includes a factor of three for color; we thus have Tr(T 3Q2) = 3 1 2 (+ 2 3 )2 1 2 ( \u2212 1 3 )2 \u2212 = + 1 2, (90.53) (90.51) (90.52) 90: Electroweak Interactions of Hadrons and so e2 32\u03c02 \u03b5\u00b5\u03bd\u03c1\u03c3F\u00b5\u03bd F\u03c1\u03c3. This formula is exact in the limit of zero quark mass. \u2202\u00b5J 3\u00b5 A = \u2212 553 (90.54) Now using eqs. (90.21) and (90.22", "), we can write the axial current in terms of the pion \ufb01elds as J 3\u00b5 A \u2261 J 3\u00b5 L \u2212 J 3\u00b5 R = f\u03c0\u2202\u00b5\u03c00 +.... (90.55) (We do not include the nucleon contribution because we are considering the e\ufb00ective lagrangian for pions and photons after integrating out the nucleons.) From eq. (90.55) we have \u2202\u00b5J 3\u00b5 A = f\u03c0\u22022\u03c00 +... ; Combining this with eq. (90.54), we get \u22022\u03c00 = \u2212 e2 32\u03c02f\u03c0 \u03b5\u00b5\u03bd\u03c1\u03c3F\u00b5\u03bd F\u03c1\u03c3 + O(f \u2212 2 \u03c0 ). (90.56) This equation of motion would follow from an e\ufb00ective lagrangian that included an interaction term of the form L\u03c00\u03b3\u03b3 = This interaction leads to a \u03c00 e2 32\u03c02f\u03c0 \u03c00\u03b5\u00b5\u03bd\u03c1\u03c3F\u00b5\u03bd F\u03c1\u03c3. (90.57) \u03b3\u03b3 decay amplitude of \u2192 e2 4\u03c02f\u03c0 = \u2212 T \u03b5\u00b5\u03bd\u03c1\u03c3k1\u00b5\u03b51\u03bdk2\u03c1\u03b52\u03c3. (90.58) This amplitude receives no higher-order corrections in e2, but is subject \u03c0/(4\u03c0f\u03c0)2. to quark-mass corrections; these are suppressed by powers of m2 Comparing eq. (90.58) with eq. (90.48), we see the one-loop result (which receives unsuppressed corrections) is too large by a factor of gA = 1.27. T Squaring, summing over \ufb01nal spins, integrating over dLIPS2(k), and multiplying by a symmetry factor of one half (because there are two identical particles in the \ufb01nal state), we ultimately \ufb01nd that the decay rate is \u0393 =. (90.59) \u03b12m3 \u03c0 64\u03c03f 2 \u03c0 This prediction is in agreement with the experimental result, which has an uncertainty of about 7%. Reference Notes Electoweak interactions of hadrons are treated in Georgi and Ramond II. 90: Electroweak Interactions of Hadrons 554 Problems 90.1) Verify that the covariant derivatives in eqs. (90.7\u201390.10) transform appropriately. 90.2) Verify that substituting eq. (90.11) into eq. (90.1) yields eq. (90.12). 90.3) Compute the rate for the decay process \u03c4 \u2212 \u2192 measured value and compare with your result. \u03c0\u2212\u03bd\u03c4. Look up the 90.4) a) Verify eq. (90.43). b) Compute the total neutron decay rate. Given the measured neu2 and tron lifetime \u03c4 = 886 s, and using GF = 1.166 c1 = 0.974, compute gA. Your answer is about 4% too high, because we neglected loop corrections, and the Coulomb interaction between the outgoing electron and proton. 5 GeV\u2212 10\u2212 \u00d7 90.5) Use your results from problems 88.7 and 89.5 to show that the neutron \u03c0 \u03b1 ln(MW/mp). How much decay rate is enhanced by a factor of 1 + 2 of the 4% discrepancy is accounted for by this e\ufb00ect? 90.6) Compute the rate for the decay process \u03c0\u2212 \u2192 m\u03c00 = 4.594 MeV \u03c00e\u2212 \u00af\u03bde. Note that, m\u03c00, the outgoing \u03c00 is nonrelasince m\u03c0+ \u226a tivistic. Compare your calculated rate with the measured value of 1 to determine c1. Your answer is about 1% too low, due to 0.397 s\u2212 neglect of loop corrections. \u2212 90.7) Verify eq. (90.59). Express \u0393 in eV. 91: Neutrino Masses 555 91 Neutrino Masses Prerequisite: 89 Recall from sections 88 and 89 that a single generation of quarks and leptons consists of left-handed Weyl \ufb01elds q\u03b1i, \u00afu\u03b1, \u00afd\u03b1, \u2113i, and \u00afe in the representa3 ), (\u00af3, 1, + 1 6 ), (\u00af3, 1, 1 tions (3, 2, + 1 2 ), and (1, 1, +1) of the gauge 3 ), (1, 2, \u2212 group SU(3) U(1). The Higgs \ufb01eld is a complex scalar \u03d5i in the SU(2) \u00d7 \u00d7 1 representation", " (1, 2, 2 ). The Yukawa couplings among these \ufb01elds that are allowed by the gauge symmetry are \u2212 \u2212 2 LYuk = \u2212 y\u03b5ij\u03d5i\u2113j \u00afe \u2212 y\u2032\u03b5ij\u03d5iq\u03b1j \u00afd\u03b1 \u2212 y\u2032\u2032\u03d5\u2020 iq\u03b1i \u00afu\u03b1 + h.c.. (91.1) After the Higgs \ufb01eld acquires its VEV, these three terms give masses to the electron, down quark, and up quark, respectively. The neutrino remains massless. Thus, massless neutrinos are a prediction of the Standard Model. However, there is now good experimental evidence that the three neutrinos actually have small masses. The data implies that mass of the heaviest neutrino is in the range from 0.04 eV to 0.5 eV. To account for this, we must extend the Standard Model. Let us continue to consider a single generation. We introduce a new left-handed Weyl \ufb01eld \u00af\u03bd in the representation (1, 1, 0); this \ufb01eld does not couple to the gauge \ufb01elds at all, and its kinetic term is simply i\u00af\u03bd\u2020\u00af\u03c3\u00b5\u2202\u00b5 \u00af\u03bd. (The bar over the \u03bd in the \ufb01eld \u00af\u03bd is part of the name of the \ufb01eld, and does not denote any sort of conjugation.) With this new \ufb01eld, we can introduce a new Yukawa coupling of the form L\u03bd Yuk = In unitary gauge, this becomes \u2212 \u02dcy\u03d5\u2020 i\u2113i \u00af\u03bd + h.c.. L\u03bd Yuk = We see that the neutrino mass is \u02dcm = \u02dcyv/\u221a2. \u2212 1 \u221a2 \u02dcy(v + H)(\u03bd \u00af\u03bd + \u00af\u03bd\u2020\u03bd\u2020). (91.2) (91.3) If this was the end of the story, we would have no understanding of why the neutrino mass is so much less than the other \ufb01rst-generation quark and lepton masses; we would simply have to take \u02dcy much less than y, y\u2032, and y\u2032\u2032. However, because \u00af\u03bd is in a real representation of the gauge group, we are allowed by the gauge symmetry to add a mass term of the form L\u00af\u03bd mass = \u2212 1 2 M (\u00af\u03bd \u00af\u03bd + \u00af\u03bd\u2020 \u00af\u03bd\u2020). (91.4) Here M is an arbitrary mass parameter. In particular, it could be quite large. 91: Neutrino Masses 556 Adding eqs. (91.3) and (91.4), we \ufb01nd a mass matrix of the form L\u03bd \u00af\u03bd mass = \u2212 1 2 ( \u03bd \u00af\u03bd ) 0 \u02dcm \u03bd \u02dcm M! \u00af\u03bd! + h.c.. (91.5) \u226b If we take M \u02dcm2/M. \u02dcm, then the eigenvalues of this mass matrix are M and (The sign of the smaller eigenvalue can be absorbed into the \u2212 phase of the corresponding eigen\ufb01eld.) Thus, if \u02dcm is of the order of the electron mass, then \u02dcm2/M is less than 1 eV if M is greater than 103 GeV. So \u02dcy can be of the same order as the other Yukawa couplings, provided M is large. This is called the seesaw mechanism for getting small neutrino masses. The eigen\ufb01eld corresponding to the smaller eigenvalue is mostly \u03bd, and the eigen\ufb01eld corresponding to the larger eigenvalue is mostly \u00af\u03bd. Another way to get this result is to integrate out the heavy \u00af\u03bd \ufb01eld at the beginning of our analysis. We get the leading term (in an expansion in inverse powers of M ) by ignoring the kinetic energy of the \u00af\u03bd \ufb01eld, solving the equation of motion for it that follows from L\u03bd Yuk, and \ufb01nally L\u00af\u03bd mass + substituting the solution back into i\u2113i)(\u03d5\u2020 L\u00af\u03bd mass + L\u03bd Yuk. The result is j\u2113j) + h.c. i (\u03d5\u2020 h \u02dcy 2 2M L\u03bdYuk+mass = = 1 2 m\u03bd(\u03bd\u03bd + \u03bd\u2020\u03bd\u2020)(1 + H/v)2, \u2212 where m\u03bd \u2261 \u2212 \u02dcm2 M = \u02dcy2v2 2M \u2212. \u2192 Again, we can absorb the minus sign in eq. (91.7) by making the \ufffd", "\ufffdeld rede\ufb01nition \u03bd i\u03bd. The seesaw mechanism has a straightforward extension to three generations. Let us consider the \ufb01elds \u2113iI, \u00afeI, and \u00af\u03bdI, where I = 1, 2, 3 is a generation index. The most general Yukawa and mass terms we can write down now read \u2212 LYuk+mass = i\u2113iI \u02dcyIJ \u00af\u03bdJ \u2212 3 matrices, MIJ is a complex symmetric 3 matrix, and the generation indices are summed. In unitary gauge, \u03b5ij\u03d5i\u2113jIyIJ \u00afeJ \u2212 where yIJ and \u02dcyIJ are complex 3 3 this becomes 1 2 MIJ \u00af\u03bdI \u00af\u03bdJ + h.c., (91.8) \u03d5\u2020 \u00d7 \u00d7 LYuk+mass = \u2212 1 \u221a2 (v+H)eIyIJ \u00afeJ \u2212 1 \u221a2 (v+H)\u03bdI \u02dcyIJ \u00af\u03bdJ \u2212 1 2 MIJ \u00af\u03bdI \u00af\u03bdJ + h.c.. (91.9) We can now integrate out the \u00af\u03bdI \ufb01elds; eq. (91.9) is then replaced with LYuk+mass = \u2212 1 \u221a2 (v+H)eIyIJ \u00afeJ \u2212 1 2 (m\u03bd )IJ(\u03bdI\u03bdJ +\u03bd\u2020I \u03bd\u2020J)(1+H/v)2, (91.10) (91.6) (91.7) 91: Neutrino Masses 557 where we have de\ufb01ned the complex symmetric neutrino mass matrix (m\u03bd )IJ \u2261 \u2212 1 2 v2(\u02dcyTM \u2212 1 \u02dcy)IJ. (91.11) EIJeJ, \u00afeI \u2192 \u00afEIJ \u00afeJ, and \u03bdI \u2192 We can make unitary transformations in generation space on the \ufb01elds: NIJ\u03bdJ, where E, \u00afE, and N are indeeI \u2192 pendent unitary matrices. The kinetic terms are unchanged (except for the couplings to the W \u00b1, as we will discuss momentarily), and the matrices y and m\u03bd are replaced with ETy \u00afE and N Tm\u03bdN. We can choose the unitary matrices E, \u00afE, and N so that ETy \u00afE and N Tm\u03bdN are diagonal with positive real entries yI and m\u03bdI. The neutrinos NI then have masses m\u03bdI, and EI have masses meI = yIv/\u221a2. In the neutral currents the charged leptons J \u00b5 3 and J \u00b5 EM, we simply add a generation index I to each \ufb01eld, and sum over it. The charged currents are more complicated, however; they become J +\u00b5 = E LI(X \u2020)IJ\u03b3\u00b5 N LIXIJ\u03b3\u00b5 \u00b5 = J \u2212 ELK, N \u2020E is the analog in the lepton sector of the CKM matrix V (91.13) NLJ, (91.12) where X \u2261 in the quark sector. \u2212 One di\ufb00erence, though, between X and V is that the phases of the Majorana NI \ufb01elds are \ufb01xed by the requirement that the neutrino masses are real and positive. Thus we cannot change these phases to make the \ufb01rst column of X real, as we did with V. We are allowed to change the phases EI \ufb01elds, so we can make the \ufb01rst row of X real. Thus X has of the Dirac 3 = 6 parameters, two more than the CKM matrix V. 9 The presence of X in the charged currents leads to the phenomenon of neutrino oscillations. A neutrino that is produced by scattering an electron o\ufb00 a target will be a linear combination XIJ \u03bdJ of the neutrinos of de\ufb01nite mass. The di\ufb00erent mass eigenstates propagate at di\ufb00erent speeds, and then (in a subsequent scattering) may become (if there is enough energy) It is the observation of neutrino muons or taus rather than electrons. oscillations that leads us to believe that neutrinos do, in fact, have mass. Reference Notes Neutrino masses are discussed in detail in Ramond II. Problems 91.1) Show that introducing neutrino masses via the seesaw mechanism results in lepton number no longer being conserved. 92: Solitons and Monopoles 558 92 Solitons and Monopoles Prerequisite: 84 Consider a real scalar \ufffd", "\ufffdeld \u03d5 with lagrangian = L \u2212 1 2 \u2202\u00b5\u03d5\u2202\u00b5\u03d5 \u2212 V (\u03d5), (92.1) with V (\u03d5) = 1 8 \u03bb(\u03d52 As we discussed in section 30, this potential yields two ground states or vacua, corresponding to the classical \ufb01eld con\ufb01gurations \u03d5(x) = +v and \u03d5(x) = v), we \ufb01nd that the particle mass is m = \u03bb1/2v. v. After shifting the \ufb01eld by its VEV (either +v or v2)2. (92.2) \u2212 \u2212 \u2212 Let us consider this theory in two spacetime dimensions (one space dimension x and time t). In this case, \u03d5 and v are dimensionless, and \u03bb has dimensions of mass squared. In the quantum theory, the coupling is weak if \u03bb \u226a The case of one space dimension is interesting for the following reason. m2. \u221e The boundary of one-dimensional space consists of two points, x = \u2212\u221e and x = +. This topology of the spatial boundary is mirrored by the topology of the space of vacuum \ufb01eld con\ufb01gurations, which also consists v and \u03d5(x) = +v. In each vacuum, both spatial of two points, \u03d5(x) = boundary points (x = ) are mapped to the same \ufb01eld v or +v). This is a trivial map. More interesting is the value (either identity map, where x = is mapped to \u03d5 = +v. This map does not correspond to a vacuum; the \ufb01eld must smoothly interpolate between \u03d5 = and \u03d5 = +v at, and this requires energy. The interesting question is whether it x = + can be done at the cost of a \ufb01nite amount of energy. is mapped to \u03d5 = v, and x = + and x = + \u2212 \u2212\u221e v at x = \u2212\u221e \u2212\u221e \u221e \u221e \u221e \u2212 \u2212 \u2212 To make these notions more precise, we will look for a minimum energy, time-independent solution of the classical \ufb01eld equations, with the boundary conditions The total energy is given by lim \u2192\u00b1\u221e x \u03d5(x) = v. \u00b1 E = + \u221e dx Z \u2212\u221e h 1 2 \u02d9\u03d52 + 1 2 \u03d5\u2032 2 + V (\u03d5) i. (92.3) (92.4) The solution of interest is time independent, so we can set \u02d9\u03d5 = 0. We can also rewrite the remaining terms in E as E = + \u221e Z \u2212\u221e dx 1 2 \u03d5\u2032 h \u2212 \u221a2V (\u03d5) 2 + \u221a2V (\u03d5) \u03d5\u2032 i 92: Solitons and Monopoles 559 = = + \u221e Z \u2212\u221e + \u221e Z \u2212\u221e dx 1 2 \u03d5\u2032 dx 1 2 \u03d5\u2032 \u2212 \u2212 \u221a2V (\u03d5) 2 + \u221a2V (\u03d5) 2 +v v Z \u2212 \u221a2V (\u03d5) d\u03d5 + 2 3 (m2/\u03bb)m. (92.5) Since the \ufb01rst term in eq. (92.5) is positive, the minimum possible energy is 2 3 (m2/\u03bb)m; this is much larger than the particle mass m if the theory M m2). Requiring the \ufb01rst term in eq. (92.5) to vanish is weakly coupled (\u03bb yields \u03d5\u2032 = \u221a2V (\u03d5), which is easily integrated to get \u226a \u2261 \u03d5(x) = v tanh 1 2 m(x, (92.6) \u2212 x0) where x0 is a constant of integration. The energy density is localized near x = x0, and goes to zero exponentially fast for > 1/m. This solution is a soliton, a solution of the classical \ufb01eld equations with an energy density that is localized in space, and that does not dissipate or change its shape with time. In this case (and in all cases of interest to us), its existence is related to the topology of the boundary of space and the topology of the set of vacua, and the existence of a nontrivial map from the boundary of space to the set of vacua. x | \u2212 x0| Given eq. (92.6), we can get other soliton solutions by making a Lorentz boost; these solutions take the form 1 2 \u03b3m(x \u03d5(x, t) = v tanh x0 \u2212", " 1/2; their energy is E = \u03b3M = (p2 + M 2)1/2, where where \u03b3 = (1 M = 2 3 (m2/\u03bb)m is the energy of the soliton at rest, and p = \u03b3\u03b2M is the momentum of the soliton, found by integrating the momentum density T 01 = \u03d5\u2032\u22020\u03d5. \u03b22)\u2212 (92.7) \u03b2t) \u2212 \u2212, We see that the soliton behaves very much like a particle. We may expect that, in the quantum theory, the soliton will correspond to a new species of particle with mass M, in addition to the elementary \ufb01eld excitation with mass m. The soliton solution is still interesting if there is more than one spatial dimension. In that case, eq. (92.6) describes a domain wall, a structure that is localized in one particular spatial direction, but extended in the others. The wall has a surface tension (energy per unit transverse area) given by \u03c3 = 2 3 m3/\u03bb. Having found a theory that has a soliton that is localized in one spatial direction, let us try to \ufb01nd a theory that has a soliton that is localized in two spatial directions. In two space dimensions, the spatial boundary has the topology of a circle, denoted by the symbol S1. There is no smooth 92: Solitons and Monopoles 560 nontrivial map from a circle to two points; continuity of the map requires the entire circle to be mapped into one of the two points. But there do exist smooth nontrivial maps from one circle to another circle, as we will discuss momentarily. So, we would like to \ufb01nd a theory whose vacua have the topology of a circle. To this end, let us consider a complex scalar \ufb01eld \u03d5(x), with lagrangian where = L \u2212 \u2202\u00b5\u03d5\u2020\u2202\u00b5\u03d5 V (\u03d5), \u2212 V (\u03d5) = 1 4 \u03bb(\u03d5\u2020\u03d5 v2)2. \u2212 The vacuum \ufb01eld con\ufb01gurations are \u03d5(x) = vei\u03b1, (92.8) (92.9) (92.10) where \u03b1 is an arbitrary angle. This angle speci\ufb01es a point on a circle, and so the space of vacua does indeed have the topology of S1. Let us write x = r(cos \u03c6, sin \u03c6); then the angle \u03c6 speci\ufb01es a point on the spatial circle at in\ufb01nity. We can specify a map from the spatial circle to the vacuum circle by giving \u03b1 as a function of \u03c6. In order for \u03d5(x) to be single valued, this function must obey \u03b1(\u03c6 + 2\u03c0) = \u03b1(\u03c6) + 2\u03c0n, where the integer n is the winding number of the map: we wind around the vacuum circle n times for every one time that we wind around the spatial circle. (If n is negative, the vacuum winding is opposite in direction to the spatial winding.) An example of a map with winding number n is U (\u03c6) = ein\u03c6. Setting n = 0 then yields the trival map, n = 1 the identity map, and n = 1 the inverse of the identity map. \u2212 Given a smooth map U (\u03c6), its winding number can be written as n = 2\u03c0 i 2\u03c0 0 Z d\u03c6 U \u2202\u03c6U \u2020, (92.11) where U \u2020 is the complex conjugate of U. To verify that eq. (92.11) agrees with our previous de\ufb01nition, we \ufb01rst check that plugging in our example map indeed yields the correct value of the winding number. We then show that the right-hand side of eq. (92.11) is invariant under smooth deformations of U (\u03c6); see problem 92.2. Thus any U (\u03c6) that can be smoothly deformed to ein\u03c6 has winding number n. Next, we want to look for a \ufb01nite-energy solution of the classical \ufb01eld equations for the theory speci\ufb01ed by eqs. (92.8) and (92.9), with the boundary condition \u03d5(r, \u03c6) = vU (\u03c6), lim r \u2192\u221e (92.12) 92: Solitons and Monopoles 561 with U (\u03c6) corresponding to a map with nonzero winding number. We therefore make the ansatz \u03d5(r, \u03c6) = vf (r)ein\u03c6, (92.13", ") where f (r) is a real function that obeys f ( f (0) = 0 so that \u03d5(r, \u03c6) is well de\ufb01ned at r = 0. \u221e ) = 1. We must also have Alas, it is easy to see that there is no \ufb01nite-energy solution of this form. \u2207 The gradient of the \ufb01eld is \u03d5 = v \u2207 and the gradient energy density is f \u2032(r)\u02c6r + inr\u2212 h 1f (r) \u02c6\u03c6 ein\u03c6, i (92.14) 2 = v2 \u03d5 | |\u2207 h f \u2032(r)2 + n2r\u2212 2f (r)2. (92.15) i At large r, f (r) must approach one; then the integral over the second term in eq. (92.15) diverges logarithmically, d2x 2 \u03d5 | |\u2207 \u223c 2\u03c0n2v2 Z \u221e dr r. Z (92.16) So the energy is in\ufb01nite. This is, in fact, a very general result, known as Derrick\u2019s theorem: with scalar \ufb01elds only, there are no \ufb01nite-energy, timeindependent solitons that are localized in more than one dimension. The problem is that the gradient energy always diverges at large distances from the putative soliton\u2019s core. To get solitons that are localized in more than one dimension, we must introduce gauge \ufb01elds. Note that the lagrangian of eq. (92.8) has a global U(1) symmetry. Let us gauge this U(1) symmetry, so that the lagrangian becomes where = L \u2212 (D\u00b5\u03d5)\u2020D\u00b5\u03d5 V (\u03d5) \u2212 \u2212 1 4 F \u00b5\u03bdF\u00b5\u03bd, D\u00b5\u03d5 = \u2202\u00b5\u03d5 ieA\u00b5\u03d5, \u2212 (92.17) (92.18) and V (\u03d5) still given by eq. (92.9). The gauge symmetry is therefore spontaneously broken, and the mass of the vector particle is mV = ev. The mass of the scalar particle is mS = \u03bb1/2v. The gradient energy density of the scalar \ufb01eld is now ~D\u03d5 2 = | | ( \u2207 \u2212 | 2. ieA)\u03d5 | (92.19) Thus we have the opportunity to choose A so as to partially cancel the badly behaved second term in eq. (92.14). To see how to do this, recall that 92: Solitons and Monopoles a gauge transformation in this theory takes the form \u03d5 \u2192 A\u00b5 \u2192 U \u03d5, UA\u00b5U \u2020 + i e U \u2202\u00b5U \u2020, 562 (92.20) (92.21) \u00d7 1 unitary matrix that is a function of spacetime. As r where U is a 1, \u2192 \u221e our ansatz for \u03d5, eq. (92.13), corresponds to a gauge transformation of a vacuum, \u03d5 = v, by U = ein\u03c6. The corresponding transformation of A\u00b5 = 0 is lim r \u2192\u221e A(r, \u03c6) = i U \u2020 \u2207 e U n er = \u02c6\u03c6. (92.22) \u2207 \u2212 For n Before making the gauge transformation, we have \u03d5 = v and A\u00b5 = 0, and so D\u00b5\u03d5 = 0; by gauge invariance, this must be true after the transformation as well. Indeed, it is easy to check that, with A given by eq. (92.22), we have ( ieA)vei\u03c6 = 0. = 0, the gauge transformation U = ein\u03c6 is large: it cannot be smoothly deformed to U = 1. This implies that we cannot extend it from r = into the interior of space without meeting an obstruction, a point where U (r, \u03c6) is ill de\ufb01ned. For example, the simplest attempt at such an extension, U (r, \u03c6) = ein\u03c6, is ill de\ufb01ned at r = 0. Near the obstruction, the \ufb01elds \u03d5 and A must deviate from a gauge transformation of a vacuum. This deviation costs energy, and results in a soliton. \u221e Our ansatz for a soliton in the theory speci\ufb01ed by eq. (92.17) is then \u03d5(r, \u03c6) = vf (r)U (\u03c6), A(r, \u03c6", ") = i e a(r)U (\u03c6) \u2207 U \u2020(\u03c6), (92.23) (92.24) where U (\u03c6) = ein\u03c6, and we require f ( approaches a large gauge transformation of a vacuum as r f (0) = a(0) = 0 (so that A and this soliton is a Nielsen\u2013Olesen vortex. ) = 1 (so that the solution ) and \u03d5 are well de\ufb01ned at r = 0). For n = 1, ) = a( \u2192 \u221e \u221e \u221e \u2207 The nonzero vector potential results in a perpendicular magnetic \ufb01eld B = = = A \u2202 \u2202r \u2207 \u00d7 1 r n e a\u2032(r) r (rA\u03c6) \u2202 \u2202\u03c6 \u2212 Ar \u02c6z \u02c6z. (92.25) 6 92: Solitons and Monopoles 563 The corresponding magnetic \ufb02ux is \u03a6 = dS \u00b7 Z = lim r \u2192\u221e Z i lim e r \u2192\u221e 2\u03c0n e = =. B d\u2113 A \u00b7 a(r) 0 Z 2\u03c0 d\u03c6 U \u2202\u03c6U \u2020 (92.26) Here the second line follows from Stokes\u2019 theorem, the third from eq. (92.24), and the fourth from eq. (92.11). The energy of the soliton is E = d2x \u03d5) + 1 ieA)\u03d5 | 2 B2. i (92.27) Substituting in our ansatz, eqs. (92.23) and (92.24), we get E = 2\u03c0v2 \u221e dr r 0 Z n2 r2 (a \u2212 2 + f \u2032 1)2f 2 + 1 4 \u03bbv2(f 2 1)2 + \u2212 n2 e2v2r2 a\u2032 2. (92.28) It is convenient to de\ufb01ne a dimensionless radial coordinate \u03c1 \u03bb/e2 = m2 Let us also de\ufb01ne \u03b22 V. Then eq. (92.28) becomes S/m2 \u2261 evr = mVr. \u2261 E = 2\u03c0v2 \u221e d\u03c1 \u03c1 f \u2032 2 + 0 Z n2 \u03c12 (a \u2212 1)2f 2 + 1 4 \u03b22(f 2 1)2 + \u2212 n2 \u03c12 a\u2032 2, (92.29) where a prime now denotes a derivative with respect to \u03c1. We can \ufb01nd the equations obeyed by f (\u03c1) and a(\u03c1) either by substituting the ansatz into the equations of motion, or by applying the variational principle directly to eq. (92.29). Either way, the result is f \u2032\u2032 + f \u2032 \u03c1 \u2212 n2f \u03c12 (1 \u2212 a)2 + 1 2 \u03b22(1 a\u2032 \u03c1 + (1 f 2)f = 0, (92.30) a)f 2 = 0, (92.31) \u2212 \u2212 a\u2032\u2032 \u2212 with the boundary conditions a(0) = f (0) = 0 and a( ) = f ( ) = 1. e\u2212 \u226a \u2212 1, we can show that a(\u03c1) \u03c1 and 1 f (\u03c1) a(\u03c1) Eqs. (92.30) and (92.31) have no closed-form solution. However, for \u03c1 1, that c\u03c1, where c = min(\u03b2, 2); see problem 1 92.4. For n and \u03b2 of order one, the integral in eq. (92.29) also results in a 2\u03c0v2. For \u03b2 > 1, it is possible number of order one, and so we have E to prove a Bogomolny bound, E > 2\u03c0v2 In this case, a soliton with. | \u03c12 and f (\u03c1) e\u2212 \u03c1n; and for \u03c1 \u223c \u223c \u223c n | \u226b \u2212 \u223c \u223c \u221e \u221e 92: Solitons and Monopoles 564 winding number n is unstable against breaking up into n | with winding number one (or minus one, if n is negative). | solitons, each Once we have our soliton solution, we can translate and/or boost it; thus we expect the soliton to behave like a particle in two space dimensions. In three space dimensions, the soliton becomes a Nielsen-Olesen string (also called a gauge string), a structure that is localized in two directions, but extended in the third. Such strings can bend, and even form closed loops. In certain uni\ufb01ed theories (see section 97), gauge strings may have", " formed in the early universe; they are then called cosmic strings. Now let us try to \ufb01nd a soliton that is localized in three spatial directions. In three space dimensions, the spatial boundary has the topology of a two-dimensional sphere S2. There are smooth nontrivial maps from S2 to S2, as we will discuss momentarily, so let us look for a theory whose vacua have the topology of S2. Consider three real scalar \ufb01elds \u03d5a, a = 1, 2, 3, with lagrangian where = L \u2212 1 2 \u2202\u00b5\u03d5a\u2202\u00b5\u03d5a V (\u03d5) = 1 8 \u03bb(\u03d5a\u03d5a V (\u03d5), v2)2. \u2212 \u2212 The vacuum \ufb01eld con\ufb01gurations are \u03d5a(x) = v \u02c6\u03d5a, (92.32) (92.33) (92.34) where \u02c6\u03d5 is an arbitrary unit vector. This unit vector speci\ufb01es a point on a two-sphere, and so the space of vacua does indeed have the topology of S2. Let us write x = r(sin \u03b8 cos \u03c6, sin \u03b8 sin \u03c6, cos \u03b8); then the polar and azimuthal angles \u03b8 and \u03c6 specify a point on the spatial two-sphere at in\ufb01nity. We can specify a map from the spatial two-sphere to the vacuum twosphere by giving \u02c6\u03d5 as an (appropriately periodic) function of \u03b8 and \u03c6. We can de\ufb01ne a winding number n that counts the number of times the vacuum two-sphere covers the spatial two-sphere, with n negative if the orientation is reversed. An example of a map with winding number n can be constructed by taking the polar angle of \u02c6\u03d5 to be \u03b8, and the azimuthal angle to be n\u03c6. Setting n = 1 then yields the identity map, and n = 1 the inverse of the identity map. \u2212 Given a smooth map \u02c6\u03d5a(\u03b8, \u03c6), its winding number can be written as n = 1 8\u03c0 d2\u03b8 \u03b5abc\u03b5ij \u02c6\u03d5a\u2202i \u02c6\u03d5b\u2202j \u02c6\u03d5c, (92.35) Z where d2\u03b8 = d\u03b8 d\u03c6, \u22021 = \u2202/\u2202\u03b8, \u22022 = \u2202/\u2202\u03c6, and \u03b512 = \u03b521 = +1. To verify that eq. (92.35) agrees with our previous de\ufb01nition, we \ufb01rst check \u2212 92: Solitons and Monopoles 565 that plugging in our example map indeed yields the correct value of the winding number; see problem 92.5. We then show that the right-hand side of eq. (92.35) is invariant under smooth deformations of \u02c6\u03d5a(\u03b8, \u03c6); see problem 92.6. It is also worthwhile to note that the right-hand side of eq. (92.35) is invariant under a change of coordinates, because the jacobian for d2\u03b8 is cancelled by the jacobian for \u22021\u22022. This is of course closely related to the invariance under smooth deformations, since one way to make such a deformation is via a coordinate change. Next, we want to look for a \ufb01nite-energy solution of the classical \ufb01eld equations with nonzero winding number, but we already know that these will not exist unless we introduce gauge \ufb01elds. We therefore take \u03d5a to be in the adjoint representation of an SU(2) gauge group. The lagrangian is now where = L \u2212 1 2 (D\u00b5\u03d5)a(D\u00b5\u03d5)a V (\u03d5) \u2212 \u2212 (D\u00b5\u03d5)a = \u2202\u00b5\u03d5a + e\u03b5abcAb \u00b5\u03d5c, 1 4 F a\u00b5\u03bdF a \u00b5\u03bd, \u00b5\u03bd = \u2202\u00b5Aa F a \u03bd \u2212 \u2202\u03bdAa \u00b5 + e\u03b5abcAb \u00b5Ac \u03bd, (92.36) (92.37) (92.38) and V (\u03d5) is given by eq. (92.33). We have called the gauge coupling e for reasons that will become clear in a moment. The gauge symmetry is spontaneously broken to U(1). If we take the vacuum \ufb01eld con\ufb01guration to be \ufffd", "\ufffda = v\u03b4a3, then the A3 \u00b5 \ufb01eld remains massless; we will think of it as the electromagnetic \ufb01eld. The complex vector \ufb01elds W \u00b1\u00b5 = (A1 \u00b5)/\u221a2 get a mass mW = ev, and have electric charge (This is the reason for calling the gauge coupling e.) This e. theory, known as the Georgi-Glashow model, was once considered as an alternative to the Standard Model of electroweak interactions (but is now ruled out, because it does not have a Z 0 boson). \u00b5 \u2213 iA2 \u00b1 When the vacuum \ufb01eld con\ufb01guration is \u03d5a = v\u03b4a3, the electromagnetic \u00b5. We can write down a gauge-invariant \ufb01eld strength is F\u00b5\u03bd = \u2202\u00b5A3 expression that reduces to F\u00b5\u03bd when we set \u03d5a = v\u03b4a3; this expression is \u2202\u03bd A3 \u03bd \u2212 F\u00b5\u03bd = \u02c6\u03d5aF a e\u2212 1\u03b5abc \u02c6\u03d5a(D\u00b5 \u02c6\u03d5)b(D\u03bd \u02c6\u03d5)c. \u03d5 | | \u00b5\u03bd \u2212 Here \u02c6\u03d5a = \u03d5a/ = (\u03d5a\u03d5a)1/2. We can, in fact, use eq. (92.39) as, where \u03d5 | | the de\ufb01nition of the electromagnetic \ufb01eld strength at any spacetime point = 0, the SU(2) symmetry is unbroken, and there is no where gauge-invariant way to pick out a particular component of the nonabelian \ufb01eld strength F a \u00b5\u03bd.) If we substitute in eqs. (92.37) and (92.38), and make \u03b4be\u03b4cd, it is repeated use of \u02c6\u03d5a \u02c6\u03d5a = 1 and the identity \u03b5abc\u03b5ade = \u03b4bd\u03b4ce = 0. (If \u03d5 | | \u03d5 | | 6 (92.39) \u2212 92: Solitons and Monopoles 566 possible to rewrite eq. (92.39) as F\u00b5\u03bd = \u2202\u00b5( \u02c6\u03d5aAa \u03bd) \u2202\u03bd ( \u02c6\u03d5aAa \u00b5) \u2212 \u2212 e\u2212 1\u03b5abc \u02c6\u03d5a\u2202\u00b5 \u02c6\u03d5b\u2202\u03bd \u02c6\u03d5c. (92.40) In particular, the magnetic \ufb01eld is Bi = 1 2 \u03b5ijkFjk = \u03b5ijk\u2202j( \u02c6\u03d5aAa k) \u2212 (2e)\u2212 1\u03b5ijk\u03b5abc \u02c6\u03d5a\u2202j \u02c6\u03d5b\u2202k \u02c6\u03d5c. (92.41) dS Let us consider the magnetic \ufb02ux through a sphere at spatial in\ufb01nity; B, where dSk = r2 sin \u03b8 d\u03b8 d\u03c6 \u02c6xk, and \u02c6x = x/r this is given by \u03a6 = is a radially outward unit vector. The \ufb01rst term in eq. (92.41) for B is ( \u02c6\u03d5aAa); since this is a curl, it has zero divergence, and therefore zero \u2207 \u00d7 surface integral. From eq. (92.35), we see that the second term in eq. (92.41) results in R \u00b7 \u03a6 = 4\u03c0n e. (92.42) \u2212 This \ufb02ux implies that any soliton with nonzero winding number is a magnetic monopole with magnetic charge QM = \u03a6. (In Heaviside-Lorentz units, the Coulomb \ufb01eld of an electric point charge QE is Ei = QE \u02c6xi/4\u03c0r2, and so the total electric \ufb02ux is QE. We adopt the same convention for magnetic charge.) If we add a \ufb01eld in the fundamental representation of SU(2), then the 1 2 e. This is the smallest electric component \ufb01elds have electric charges charge we can get, and all possible electric charges are integer multiples of it. Eq. (92.42) tells us that all possible magnetic charges are integer multiples of 4\u03c0/e. Thus the possible electric and magnetic charges obey the Dirac charge quantization condition, which is \u00b1 QEQM = 2\u03c0k, (92.43) where k is an integer. This condition", " can be derived from general considerations of the quantum properties of monopoles. Now let us turn to the explicit construction of a soliton solution. This simplest case to consider is provided by the identity map (which has winding number n = 1); the soliton we will \ufb01nd is the \u2019t Hooft-Polyakov monopole. The boundary condition on the scalar \ufb01eld is \u03d5a(x) = vxa/r. lim r \u2192\u221e (92.44) We can \ufb01nd the appropriate boundary condition on the gauge \ufb01eld by requiring (D\u00b5\u03d5)a = 0 in the limit of large r. This condition yields \u2202i(xa/r) + e\u03b5abcAb i xc/r = 0. (92.45) 92: Solitons and Monopoles 567 We have \u2202i(xa/r) = (r2\u03b4ai \u2212 use the identity \u03b5jda\u03b5abc = \u03b4jb\u03b4dc \u03b4jc\u03b4db to get \u2212 xaxi)/r3. Next we multiply by rxj\u03b5jda, and \u03b5dijxj + e(xdxjAj i \u2212 r2Ad i ) = 0. (92.46) If we ingore the \ufb01rst term in the parentheses, we \ufb01nd Ad then xdAd have found the needed asymptotic behavior of Aa i = \u03b5dijxj/er2. But i = 0, and so the \ufb01rst term in the parentheses vanishes. Thus we i. Our ansatz is therefore \u03d5a(x) = vf (r)xa/r, Aa i (x) = a(r)\u03b5aijxj/er2. (92.47) (92.48) We require f ( \u221e totic limits) and f (0) = a(0) = 0 (so that Aa r = 0). ) = 1 (so that Aa ) = a( \u221e i and \u03d5a have the desired asympi and \u03d5a are well de\ufb01ned at The total energy of the soliton (which we will call M, because it is the mass of the monopole) is given by M = d3x 1 2 Ba i Ba i + 1 Z The nonabelian magnetic \ufb01eld is h 2 (Di\u03d5)a(Di\u03d5)a + V (\u03d5) i Ba i = 1 jk 2 \u03b5ijkF a = \u03b5ijk\u2202jAa k + 1 2 e\u03b5ijk\u03b5abcAb jAc k.. (92.49) (92.50) If we write eq. (92.48) as Aa \ufb01nd that eq. (92.50) becomes Ba Ki = a(r)xi/er2 then yields i = \u03b5aijKj, then after some manipulation we \u03b4ai\u2202jK j + KaKi. Plugging in i = \u2202aKi \u2212 Ba i = 1 e \" a\u2032 r \u2212 \u02c6xa \u02c6xi + \u03b4ai \u2212 The magnetic \ufb01eld energy then becomes a2 2a \u2212 r2 \u02c6xa \u02c6xi. # 1 2 Ba i Ba i = 1 2e2r4 2r2a\u2032 h 2 + (2a \u2212 a2)2. i The covariant derivative of the scalar \ufb01eld is (92.51) (92.52) (Di\u03d5)a = v (1 a)f \u2212 r \u03b4ai \u2212 The scalar gradient energy density then becomes \u02c6xa \u02c6xi + f \u2032 \u02c6xa \u02c6xi. (92.53) 1 2 (Di\u03d5)a(Di\u03d5)a = v2 2r2 2(1 h \u2212 a)2f 2 + r2f \u2032 2. (92.54) i 92: Solitons and Monopoles The scalar potential energy density is V (\u03d5) = 1 8 \u03bbv4(f 2 1)2. \u2212 568 (92.55) We can plug eqs. (92.52), (92.54), and (92.55) into eq. (92.49), and then use the variational principle to get the second-order di\ufb00erential equations obeyed by f (r) and a(r). We can get a lower bound on M by performing a trick analogous to the one we used in eq. (92.5). We write 1 2 Ba i Ba i + 1 2 (Di\u03d5)a(Di\u03d5)a = 1 2 [Ba", " i + (Di\u03d5)a]2 Ba i (Di\u03d5)a. \u2212 (92.56) We can apply the distribution rule for covariant derivatives (see problem 70.5) to rewrite the last term as Ba (DiBi)a\u03d5a. Then we note that the Bianchi identity (see problem 70.6) implies (DiBi)a = 0. i \u03d5a), and this is a total divergence. Then Gauss\u2019s Thus Ba theorem yields i (Di\u03d5)a = \u2202i(Ba i (Di\u03d5)a = \u2202i(Ba i \u03d5a) \u2212 d3x \u2202i(Ba i \u03d5a) = dSi Ba i \u03d5a, (92.57) where the integral is over the surface at spatial in\ufb01nity. On this surface, we have \u03d5a = v\u02c6xa. Z Z Next we use eq. (92.39). At spatial in\ufb01nity, the covariant derivatives i \u03d5a = vBi, where Bi is the magnetic \ufb01eld of of \u03d5 vanish; thus we have Ba electromagnetism. We can now see that the right-hand side of eq. (92.57) evaluates to v\u03a6, where \u03a6 = QM = 4\u03c0n/e is the magnetic charge of the monopole. \u2212 In our case, n = 1 and QM is negative; thus the last term in eq. (92.56) integrates to v. (For the case of positive QM, we can swap the plus and minus signs in eq. (92.56) to get the same result.) Thus the mass of the monopole, eq. (92.49), can be written as QM| | M = 4\u03c0 v | n | e + d3x Z h 1 2 [Ba i + (sign n)(Di\u03d5)a]2 + V (\u03d5) i, (92.58) Both terms in the integrand of eq. (92.58) are positive, and so we have a Bogomolny bound on the mass of the monopole. For \u03bb > 0, a monopole with winding number n is unstable against breaking up into monopoles, n | | each with winding number one (or minus one, if n is negative). Using mW = ev and \u03b1 = e2/4\u03c0, we can write the Bogomolny bound as mW \u03b1 | (92.59) M \u2265 n. | Since \u03b1 1, the monopole is much heavier than the W boson. Alas, the Georgi-Glashow model, which has monopole solutions, is not in accord with nature, while the Standard Model, which is in accord with \u226a 92: Solitons and Monopoles 569 nature, does not have monopole solutions. This is because, in the Standard Model, electric charge is a linear combination of an SU(2) generator and the U(1) hypercharge generator. Nothing prevents us from introducing an SU(2) singlet \ufb01eld with an arbitrarily small hypercharge. Such a \ufb01eld would have an arbitrarily small electric charge (in units of e), and then the Dirac charge quantization condition would preclude the existence of magnetic monopoles. 1015 GeV. This disappointing situation is remedied in uni\ufb01ed theories (see section 97), where the gauge group has a single nonabelian factor like SU(5). In uni\ufb01ed theories, the monopole mass is of order mX/\u03b1, where mX is the mass of a superheavy vector boson; typically mX \u223c Returning to the Georgi-Glashow model, we can saturate the Bogo0; then V (\u03d5) vanishes. molny bound if we consider the formal limit of \u03bb (This limit is formal because we need a nonzero potential to \ufb01x the magni(sign n)(Di\u03d5)a. tude of \u03d5 at in\ufb01nity.) Then we saturate the bound if Ba In the case of the \u2019t Hooft-Polyakov monopole, we have n = 1, and Ba i and (Di\u03d5)a are given by eqs. (92.51) and (92.53). Matching the coe\ufb03\u02c6xa \u02c6xi and \u02c6xa \u02c6xi yields a pair of \ufb01rst-order di\ufb00erential equacients of \u03b4ai \u2212 tions. These look nicer if we introduce the dimensionless radial coordinate \u03c1 evr = mWr; then we \ufb01nd i = \u2192 \u2212 \u2261 a)f, a\u2032 = (1 \u2212 f \u2032 = (2a", " a2)/\u03c12, \u2212 (92.60) (92.61) where a prime now denotes a derivative with respect to \u03c1. These equations have a closed-form solution, a(\u03c1) = 1 \u03c1 sinh \u03c1, \u2212 f (\u03c1) = coth \u03c1 1 \u03c1. \u2212 (92.62) (92.63) This is the Bogomolny-Prasad-Sommerfeld (or BPS for short) solution. A soliton that saturates a Bogomolny bound is generically called a BPS soliton. Reference Notes Discussions of solitons, and their relation to the theory of homotopy groups, can be found in Coleman and Weinberg II. Problems 92: Solitons and Monopoles 570 92.1) Derrick\u2019s theorem says that, in a theory with scalar \ufb01elds only, there are no solitons localized in more than one dimension. To prove this, consider a theory in D space dimensions with a set of real scalar \ufb01elds \u03d5i; any complex scalar \ufb01elds are written as a pair of real ones. The 0. Suppose lagrangian is we have a soliton solution \u03d5i(x); its energy is E = T + U, where T = 1 2 a) Now consider \u03d5i(x/\u03b1), where \u03b1 is a positive real number. Show 2T + \u03b1DU. that, for this \ufb01eld con\ufb01guration, the energy is E(\u03b1) = \u03b1D L \u03d5i)2 and U = 2 \u2202\u00b5\u03d5i\u2202\u00b5\u03d5i \u2212 V (\u03d5i), with V (\u03d5i) dDx V (\u03d5i). dDx ( \u2207 \u2265 \u2212 = R R 1 \u2212 b) Argue that we must have E\u2032(1) = 0. c) Use this to prove the theorem. 92.2) The winding number n for a map from S1 S1 is given by eq. (92.11), where U \u2020U = 1. We will prove that n is invariant under an in\ufb01nitesimal deformation of U. Since any smooth deformation can be made by compounding in\ufb01nitesimal ones, this will prove that n is invariant under any smooth deformation. \u2192 a) Consider an in\ufb01nitesimal deformation of U, U that \u03b4U \u2020 = b) Use this result to show that \u03b4(U \u2202\u03c6U \u2020) = c) Use this to show that \u03b4n = 0. 2\u03b4U. U \u2020 \u2212 \u2212 \u2202\u03c6(U \u2020\u03b4U ). U + \u03b4U. Show \u2192 92.3) Show that if Un(\u03c6) and Uk(\u03c6) are maps from S1 S1 with winding numbers n and k, then Un(\u03c6)Uk(\u03c6) is a map with winding number n + k. Hint: consider smoothly deforming Un(\u03c6) to equal one for 0 \u03c0. How should Uk(\u03c6) be deformed? \u2192 \u03c6 \u2264 \u2264 92.4) Verify the statements made about the solutions to eqs. (92.30) and (92.31) in the limit of large and small \u03c1. 92.5) Use eq. (92.35) to compute the winding number for the map speci\ufb01ed by \u02c6\u03d5 = (sin \u03b8 cos n\u03c6, sin \u03b8 sin n\u03c6, cos \u03b8). 92.6) The winding number n for a map from S2 S2 is given by eq. (92.35), where \u02c6\u03d5a \u02c6\u03d5a = 1. We will prove that n is invariant under an in\ufb01nitesimal deformation of \u02c6\u03d5. Since any smooth deformation can be made by compounding in\ufb01nitesimal ones, this will prove that n is invariant under any smooth deformation. \u2192 \u03b4 \u02c6\u03d5 = 0 and that \u02c6\u03d5 a) Consider an in\ufb01nitesimal deformation of \u02c6\u03d5, \u02c6\u03d5 that \u02c6\u03d5 \u00b7 b) Use these results to show that \u03b5abc\u03b4 \u02c6\u03d5a\u2202i \u02c6\u03d5b\u2202j \u02c6\u03d5c = 0. c) Use this to show that \u03b4n = 0. \u2202i \u02c6\u03d5 = 0. \u2192 \u00b7 \u02c6\u03d5 + \u03b4 \u02c6\u03d5. Show 93: Instantons and Theta Vac", "ua 571 93 Instantons and Theta Vacua Prerequisite: 92 Consider SU(2) gauge theory, with gauge \ufb01elds only. The classical \ufb01eld con\ufb01guration corresponding to the ground state is F a \u00b5\u03bd = 0. This implies that \u00b5T a = the vector potential Aa i g U \u2202\u00b5U \u2020. \u00b5 is a gauge transformation of zero, A\u00b5 = Aa Let us restrict our attention to gauge transformations that are time independent, U = U (x). This \ufb01xes temporal gauge, A0 = 0. We will also impose the boundary condition that U (x) approaches a particular constant matrix as, independent of direction. This is equivalent to adding a spatial \u201cpoint at in\ufb01nity\u201d where U has a de\ufb01nite value; space then has the topology of a three-dimensional sphere S3. | \u2192 \u221e x | Can every U (x) be smoothly deformed into every other U (x)? If the answer is yes, then all these \ufb01eld con\ufb01gurations are gauge equivalent, and they correspond to a single quantum vacuum state. If the answer is no, then there must be more than one quantum vacuum state. To see why, suppose that U (x) and \u02dcU (x) cannot be smoothly deformed into each other. The g \u02dcU \u2202\u00b5 \u02dcU \u2020, are both associated vector potentials, A\u00b5 = i gauge transformations of zero, and so both F\u00b5\u03bd and \u02dcF\u00b5\u03bd vanish. However, if we try to smoothly deform A\u00b5 into \u02dcA\u00b5, we must pass through vector potentials that are not gauge transformations of zero, and whose \ufb01eld strengths therefore do not vanish. These nonzero \ufb01eld strengths imply nonzero energy: there is an energy barrier between the \ufb01eld con\ufb01gurations A\u00b5 and \u02dcA\u00b5. Therefore, they represent two di\ufb00erent minima of the hamiltonian in the space of classical \ufb01eld con\ufb01gurations. Di\ufb00erent minima in the space of classical \ufb01eld con\ufb01gurations correspond to di\ufb00erent vacuum states in the quantum theory. g U \u2202\u00b5U \u2020 and \u02dcA\u00b5 = i It turns out that every U (x) can not be smoothly deformed into every other U (x); the \ufb01eld con\ufb01gurations speci\ufb01ed by U (x) are classi\ufb01ed by a 2 special unitary winding number. To see this, we \ufb01rst note that any 2 matrix U can be written in the form \u00d7 where a4 and the three-vector ~a are real, and U = a4 + i~a ~\u03c3, \u00b7 ~a 2 + a2 4 = 1 ; (93.1) (93.2) (~a, a4) speci\ufb01es a euclidean four-vector of unit see problem 93.1. Thus a\u00b5 \u2261 length, a\u00b5a\u00b5 = 1, and hence a point on a three-sphere. We will call this the vacuum three-sphere. Since our boundary conditions give space the 93: Instantons and Theta Vacua 572 topology of a three-sphere, U (x) provides a map from the spatial threesphere to the vacuum three-sphere. We can de\ufb01ne a winding number n that counts the number of times the vacuum three-sphere covers the spatial three-sphere, with n negative if the orientation is reversed. It is convenient to specify the spatial three-sphere by a euclidean four(~z, z4) of unit length, z\u00b5z\u00b5 = 1. An explicit relation between vector z\u00b5 \u2261 z\u00b5 and x can be constructed by (for example) stereographic projection: r2)/(1+r2), where we take \u02c6z = ~z/ ~z |. Then we can construct an example of a map from the spatial S3 x r = | | to the vacuum S3 with winding number n by taking the two polar angles of a\u00b5 to be equal to the two polar angles of z\u00b5, and the azimuthal angle of a\u00b5 to be equal to n times the azimuthal angle of z\u00b5. (The polar angles run from 0 to \u03c0, and the azimuthal angle from 0 to 2\u03c0.) = 2r/(1+r2), z4 = (1 = \u02c6x, and ~z | \u2212 |", " | Given a smooth map U (x), its winding number can be written as n = \u2212 1 24\u03c02 Z d3x \u03b5ijk Tr[(U \u2202iU \u2020)(U \u2202j U \u2020)(U \u2202kU \u2020)]. (93.3) Here we have used the original x coordinates, but we could also use the angles that specify z\u00b5: the integral in eq. (93.3) is invariant under a change of coordinates, because the jacobian for d3x is cancelled by the jacobian for \u22021\u22022\u22023. To verify that eq. (93.3) agrees with our previous de\ufb01nition, we \ufb01rst check that plugging in our example map indeed yields the correct value of the winding number; see problem 93.5. We then show that the righthand side of eq. (93.3) is invariant under smooth deformations of U (x); see problem 93.3. So, we have concluded that SU(2) gauge theory has an in\ufb01nite number of classical \ufb01eld con\ufb01gurations of zero energy, distinguished by an integer n, and separated by energy barriers. This is analogous to a scalar \ufb01eld theory with a potential V (\u03d5) = \u03bbv4[1 cos(2\u03c0\u03d5/v)]. \u2212 (93.4) This potential has minima at \u03d5 = nv, where n is an integer. Let be the quantum state corresponding to the minimum at \u03d5 = nv. Generically, between two quantum states that are separated by an energy barrier, there is a tunneling amplitude of the form n\u2032i | and n | n | i i n\u2032 h H | n | i \u223c e\u2212 S, (93.5) where H is the hamltonian, and S is the euclidean action for a classical solution of the euclidean \ufb01eld equations that mediates between the \ufb01eld con\ufb01guration corresponding to n at t =, and the \ufb01eld con\ufb01guration. In the scalar \ufb01eld theory, this solution corresponding to n\u2032 at t = + \u2212\u221e \u221e 93: Instantons and Theta Vacua 573 is independent of x. Thus, S scales like the volume of space V, and so vanishes in the in\ufb01nite volume limit. The minima of eq. (93.4) H n\u2032| h therefore remain exactly degenerate in the quantum theory. n | i Things are di\ufb00erent in the SU(2) gauge theory. In this case, there is a classical solution of the euclidean \ufb01eld equations that mediates between states with winding numbers n and n\u2032, and that has an action that stays \ufb01xed and \ufb01nite in the in\ufb01nite-volume limit. The value of this action is S1, where S1 = 8\u03c02/g2, and g is the Yang-Mills coupling conS = | stant. For n\u2032 = n + 1, this solution is the instanton. The instanton is localized in all four euclidean directions. For n\u2032 = n 1, the solution is the antiinstanton. For n n\u2032 \u2212 | | n is negative) distributed throughout instantons (or antiinstantons, if n\u2032 \u2212 euclidean spacetime. > 1, the solution is a dilute gas of n\u2032 \u2212 | n\u2032\u2212 | \u2212 n n | We will shortly construct the instanton and examine its properties, but \ufb01rst we study the consequences of its existence. For SU(2) gauge theory, eq. (93.5) reads n\u2032 h H | n | i \u223c e\u2212| n\u2032 n S1. | \u2212 (93.6) These matrix elements depend only on n\u2032 \u2212 by theta vacua of the form n, and so H can be diagonalized = \u03b8 | i + \u221e n= X \u2212\u221e in\u03b8 e\u2212 ; n | i (93.7) 1, and so we can neglect all see problem 93.2. For weak coupling, S1 \u226b 1. Then we \ufb01nd that the matrix elements of H except those with n\u2032 = n \u00b1 energy of a theta vacuum is proportional to cos \u03b8. (We are of course free to add a constant to H so that the lowest lying state, the theta vacuum with \u03b8 = 0, has energy zero.) \u2212 We have derived these results in the weak-coupling regime. However, we are discussing properties of low-energy states, and", " the gauge coupling becomes large at low energies. Therefore we must consider the theory to be in the strong-coupling regime. How does this a\ufb00ect our conclusions? n n\u2032 \u2212 | The topological properties of the gauge \ufb01elds are independent of the value of the coupling constant, so we still expect vacuum states labeled by will depend the winding number n to exist. We also expect that only on. To see this, consider making a gauge transformation by | Uk(x), where Uk(x) has winding number k. The product of two maps with winding numbers n and k is a map with winding number n + k; see problem 93.4. Thus, making a gauge transformation by Uk(x) converts a \ufb01eld con\ufb01guration with winding number n to one with winding number n + k. In the quantum theory, the gauge transformation is implemented by H n\u2032| h n | i 93: Instantons and Theta Vacua 574 a unitary operator Uk, and we should have n+k | On the other hand, the hamiltonian, which is built out of \ufb01eld strengths, must be invariant under time-independent gauge transformations: n Uk| (93.8) = i i. UkH U \u2020k = H. Inserting factors of I = eqs. (93.8) and (93.9), we \ufb01nd U \u2020k Uk on either side of H in (93.9) H n\u2032| h n |, and using 93.10) n+k | n\u2032+k h n\u2032 h H n\u2032| We conclude that h winding number is reversed by parity, P i Mills hamiltonian is parity invariant, P HP \u2212 H n\u2032| that h\u2212 H n\u2032| h n i | n. We can also note that depends only on n\u2032 \u2212 =, and that the Yangi |\u2212 1 = H, to conclude similarly depends only on i n\u2032 \u2212 | n\u2032 \u2212 H n\u2032| | h tells us that the theta vacua are still eigenstates of H. Furthermore, their energies must be a periodic, even function of \u03b8. Of course, the eigenvalues of H should scale with the volume of space V. Then, on dimensional grounds, we have n. Thus n | i depends only on n H n\u2032| | h The fact that |\u2212 \u039b4 \u03b8 QCDf (\u03b8) |, i (93.11) where \u039bQCD is the scale where the gauge coupling becomes strong. The function f (\u03b8) must obey f (\u03b8 + 2\u03c0) = f (\u03b8) and f ( \u03b8) = f (\u03b8). We expect the minimum of f (\u03b8) to be at \u03b8 = 0. \u2212 \u2212 T, we set A\u00b5(x) = i We turn now to the solutions of the euclidean \ufb01eld equations. At eu(x). Similarly, at euclidean time x4 = +T, we set g U+(x)\u2202\u00b5U \u2020+(x), where U+(x) has winding number n+. At T, we set the boundary condition A\u00b5 = 0. This g U \u2202\u00b5U \u2020 with \u2202\u00b5U \u2020 = 0; we therefore set U (x) to a = R. We want to take T and R to in\ufb01nity at the clidean time x4 = has winding number n A\u00b5(x) = i x x4 \u2264 = R, for | | is equivalent to A\u00b5 = i x constant matrix at | | end of the calculation. (x)\u2202\u00b5U \u2020 \u2212 (x), where We have now speci\ufb01ed U (x, x4) on the cylindrical boundary of fourdimensional spacetime shown in \ufb01g. (93.1). This boundary is topologically a three-sphere. The winding number of the map on this three-sphere is. We see this by using eq. (93.3), and noting that the cylindrical wall n+\u2212 makes no contribution (because \u2202\u00b5U \u2020 = 0 there), the upper cap contributes n+, and the lower cap contributes ; the sign is negative because the n \u2212 orientation of the cap as part of the boundary is reversed from its original orientation. \u2212 \u2212 93: Instantons and Theta Vacua 575 +T x 4 \u2212T R Figure 93.1: The boundary in euclidean spacetime. We have a \ufb01eld conT, and one with \ufb01guration with winding number n winding number n+ on the cap at x4 = +T. On the cylindrical surface", " at x | | = R, the \ufb01eld vanishes. on the cap at x4 = \u2212 \u2212 Since we are interested in large R and T, and since the shape of the boundary should not matter in this limit, we instead consider the boundary. On this boundary, we have a to be a three-sphere at \u03c1. map U (\u02c6x), where \u02c6x\u00b5 = x\u00b5/\u03c1; this map has winding number n Our \ufb01rst task will be to construct a Bogomolny bound on the euclidean (x\u00b5x\u00b5)1/2 = n+ \u2212 \u221e \u2261 \u2261 n \u2212 action of a \ufb01eld that obeys the boundary condition Z S = 1 2 d4x Tr(F \u00b5\u03bdF\u00b5\u03bd ) A\u00b5(x) = i g U (\u02c6x)\u2202\u00b5U \u2020(\u02c6x), lim \u03c1 \u2192\u221e (93.12) (93.13) where U (\u02c6x) is a map with winding number n. The \ufb01eld strength is given in terms of the vector potential by \u2202\u03bd A\u00b5 \u2212 We begin by de\ufb01ning the polar angles \u03c7 and \u03c8, and the azimuthal angle F\u00b5\u03bd = \u2202\u00b5A\u03bd \u2212 ig[A\u00b5, A\u03bd]. (93.14) \u03c6, via \u02c6x\u00b5 = (sin \u03c7 sin \u03c8 cos \u03c6, sin \u03c7 sin \u03c8 sin \u03c6, sin \u03c7 cos \u03c8, cos \u03c7). (93.15) (We use \u03c8 rather than \u03b8 in order to avoid any possible confusion with the vaccum angle.) Next we write the winding number in terms of these angles, n = \u2212 1 24\u03c02 \u03c0 d\u03c7 \u03c0 d\u03c8 0 Z 0 Z 0 Z 2\u03c0 d\u03c6 \u03b5\u03b1\u03b2\u03b3 Tr[(U \u2202\u03b1U \u2020)(U \u2202\u03b2U \u2020)(U \u2202\u03b3U \u2020)], (93.16) 93: Instantons and Theta Vacua 576 where \u03b1, \u03b2, \u03b3 run over \u03c7, \u03c8, \u03c6; \u2202\u03c6 = \u2202/\u2202\u03c6, etc.; and \u03b5\u03c7\u03c8\u03c6 = +1. Now we write eq. (93.16) as a surface integral over a surface at in\ufb01nity in fourdimensional euclidean space, n = 1 24\u03c02 Z dS\u00b5 \u03b5\u00b5\u03bd\u03c3\u03c4 Tr[(U \u2202\u03bd U \u2020)(U \u2202\u03c3U \u2020)(U \u2202\u03c4 U \u2020)], (93.17) where \u2202\u00b5 = \u2202/\u2202x\u00b5, and \u03b51234 = +1. (This implies that \u03b5\u03c1\u03c7\u03c8\u03c6 = 1, as can be checked by computing the jacobian for this change of coordinates; that is why the overall minus sign disappeared.) Now we use eq. (93.13) to write the winding number in terms of the vector potential, \u2212 n = ig3 24\u03c02 dS\u00b5 \u03b5\u00b5\u03bd\u03c3\u03c4 Tr(A\u03bd A\u03c3A\u03c4 ). (93.18) Z Next, we will write this surface integral as a volume integral. To do so, we \ufb01rst de\ufb01ne the Chern-Simons current, J \u00b5 CS \u2261 2 \u03b5\u00b5\u03bd\u03c3\u03c4 Tr(A\u03bd F\u03c3\u03c4 + 2 3 igA\u03bd A\u03c3A\u03c4 ). (93.19) This current is not gauge invariant, but the relative coe\ufb03cient of its two terms has been chosen so that its divergence is gauge invariant, where \u2202\u00b5J \u00b5 CS = \u03b5\u00b5\u03bd\u03c3\u03c4 Tr(F\u00b5\u03bd F\u03c3\u03c4 ) = 2 Tr( \u02dcF \u00b5\u03bdF\u00b5\u03bd ), \u02dcF \u00b5\u03bd \u2261 1 2 \u03b5\u00b5\u03bd\u03c3\u03c4F\u03c3\u03c4 (93.20) (93.21) is the dual \ufb01eld strength. On the surface at in\ufb01nity, the vector potential is a gauge transformation of zero, and so the \ufb01eld strength F\u00b5\u03bd vanishes there. Thus we can use eq. (93.19) to write eq. (93.18) as n = g2 32\u03c02 Z Using Gauss\u2019s theorem, this becomes dS\u00b5 J \u00b5 CS. (93.22) n = g2 32\u03c02 Z d4x \u2202\u00b5J \u00b5 CS. (93.23) Finally, we use", " eq. (93.20) to get n = g2 16\u03c02 Z d4x Tr( \u02dcF \u00b5\u03bdF\u00b5\u03bd ). (93.24) 93: Instantons and Theta Vacua 577 Thus we have expressed the winding number as a (four-dimensional) volume integral of a gauge-invariant expression. Now it is easy to construct a Bogomolny bound. We \ufb01rst note that \u02dcF \u00b5\u03bd \u02dcF\u00b5\u03bd = F \u00b5\u03bdF\u00b5\u03bd, and hence 1 2 Tr( \u02dcF\u00b5\u03bd \u00b1 F\u00b5\u03bd )2 = Tr(F \u00b5\u03bdF\u00b5\u03bd ) Tr( \u02dcF \u00b5\u03bdF\u00b5\u03bd ). \u00b1 (93.25) The left-hand side of eq. (93.25) is nonnegative, and so we have d4x Tr(F \u00b5\u03bdF\u00b5\u03bd ) \u2265 The left-hand side of eq. (93.26) is twice the euclidean action, while the right-hand side is, according to eq. (93.24), 16\u03c02 Z Z. d4x Tr( \u02dcF \u00b5\u03bdF\u00b5\u03bd ) (93.26) n | /g2. Thus we have | 8\u03c02 /g2. | S \u2265 n | Eq. (93.27) gives us the minimum value of the euclidean action for a solution of the euclidean \ufb01eld equations that mediates between a vacuum con\ufb01guration with winding number n and a vacuum con\ufb01guration at x4 = with winding number n+ = n \u2212\u221e \u2212 + n at x4 = + From eq. (93.25) we see that we can saturate the bound in eq. (93.27) if (93.27) \u221e \u2212. and only if \u02dcF\u00b5\u03bd = (sign n)F\u00b5\u03bd. (93.28) We can \ufb01nd an explicit solution of eq. (93.28) for a map with winding number n = 1; this solution is the instanton. We take the map on the boundary to be the identity map, U (\u02c6x) = ~\u03c3 x4 + i~x \u03c1 \u00b7 = cos \u03c7 + i sin \u03c7 cos \u03c8 i sin \u03c7 sin \u03c8 e+i\u03c6 i sin \u03c7 sin \u03c8 e\u2212 i\u03c6 cos \u03c7 \u2212 i sin \u03c7 cos \u03c8!, (93.29) which has n = 1. We then make the ansatz A\u00b5(x) = i g f (\u03c1)U (\u02c6x)\u2202\u00b5U \u2020(\u02c6x), (93.30) where f ( ) = 1 (so that the solution obeys the boundary condition) and f (0) = 0 (so that A\u00b5 is well de\ufb01ned at \u03c1 = 0). Using eq. (93.14), we \ufb01nd that the \ufb01eld strength is \u221e F\u00b5\u03bd = i g (\u2202\u00b5f )U \u2202\u03bdU \u2020 + f \u2202\u00b5U \u2202\u03bd U \u2020 + f 2(U \u2202\u00b5U \u2020)(U \u2202\u03bd U \u2020) h (\u00b5 \u03bd). \u2212 \u2194 i (93.31) 93: Instantons and Theta Vacua 578 Using (\u2202\u00b5U \u2020)U = to F\u00b5\u03bd = i g U \u2020\u2202\u00b5U in the third term, eq. (93.31) can be simpli\ufb01ed \u2212 (\u2202\u00b5f )U \u2202\u03bdU \u2020 + f (1 h \u2212 f )\u2202\u00b5U \u2202\u03bdU \u2020 (\u00b5 \u2194 \u2212 \u03bd). i (93.32) In three-spherical coordinates, we have \u2202 = \u02c6\u03c1 \u2202\u03c1 + \u02c6\u03c7 \u03c1\u2212 1\u2202\u03c7 + \u02c6\u03c8 (\u03c1 sin \u03c7)\u2212 1\u2202\u03c8 + \u02c6\u03c6 (\u03c1 sin \u03c7 sin \u03c8)\u2212 1\u2202\u03c6, (93.33) where \u2202\u03c6 = \u2202/\u2202\u03c6, etc., and \u02c6\u03c1 is the same radial unit vector as \u02c6x. Note that f is a function of \u03c1 only, while U is a function of the angles only. We therefore have F\u03c1\u03c7 = i g f \u2032\u03c1\u2212 1U \u2202\u03c7U \u2020, (93.34) F\u03c8\u03c6 = i g f (1 f )(\u03c12 sin", "2\u03c7 sin \u03c8)\u2212 1(\u2202\u03c8U \u2202\u03c6U \u2020 \u2202\u03c6U \u2202\u03c8U \u2020). (93.35) \u2212 \u2212 Next we note that that eq. (93.21) implies \u02dcF\u03c1\u03c7 = 1), \u2212 and since \u02dcF\u00b5\u03bd = F\u00b5\u03bd for the instanton solution, we have F\u03c1\u03c7 = F\u03c8\u03c6. Thus the right-hand side of eq. (93.34) equals minus the right-hand side of eq. (93.35). We then use separation of variables to conclude that F\u03c8\u03c6 (since \u03b5\u03c1\u03c7\u03c8\u03c6 = \u2212 \u2212 \u03c1f \u2032 = cf (1 f ), \u2212 1(sin2\u03c7 sin \u03c8)\u2212 (93.36) 1(\u2202\u03c8U \u2202\u03c6U \u2020 \u2212 \u2202\u03c6U \u2202\u03c8U \u2020), (93.37) U \u2202\u03c7U \u2020 = c\u2212 \u2212 where c is the separation constant. If we plug eq. (93.29) into eq. (93.37), we \ufb01nd that it is satis\ufb01ed if c = 2. The solution of eq. (93.36) is then f (\u03c1) = \u03c12 \u03c12 + a2, (93.38) where a, the size of the instanton, is a constant of integration. The instanton solution is also parameterized by the location of its center; we have used the spacetime origin, but translation invariance allows us to displace it. If we consider initial and \ufb01nal states whose winding numbers di\ufb00er by more than one, we can construct a mediating solution by patching together instantons (or antiinstantons) whose centers are widely separated on the scale set by their sizes. Each instanton (or antiinstanton) contributes S1 = 8\u03c02/g2 to the action, and so the minimum total action is n S1. To better understand the role of the \u03b8 parameter, let us consider the euclidean path integral, with the boundary condition that we start with a, and end with a state with winding state of winding number n at x4 = n+\u2212 | \u2212| \u2212 \u2212\u221e Z\u03b8\u2032 \u03b8(J) = \u2190 n\u2212,n+ X + n, so that n+\u03b8\u2032 \u2212 93: Instantons and Theta Vacua 579 number n+ at x4 = + those with winding number n+ \u2212 n\u2212(J) = Zn+ \u221e \u2190 \u2212. The only \ufb01eld con\ufb01gurations that contribute are. We can therefore write n An+ D \u2212 Z n\u2212 e\u2212 S+JA, (93.39) R n\u2212(J) depends only on n+ \u2212 d4x Tr(J \u00b5A\u00b5), and the subscript on the \ufb01eld di\ufb00erwhere JA is short for ential means that we integrate only over \ufb01elds with that winding number., and not separately on We see that Zn+ n+ and n n | depends only on n\u2032 \u2212 vacuum \u03b8 | we see from eq. (93.7) that the corresponding path integral is Suppose now that we are interested in starting with a particular theta. Then,, and ending with a (possibly di\ufb00erent) theta vacuum i. This in accord with our previous conclusion that H n\u2032| h \u03b8\u2032i | n. n \u2190 \u2212 \u2212 i ei(n+\u03b8\u2032 in\u2212\u03b8)Zn+ \u2212 n\u2212(J). \u2190 (93.40) n n\u2212(J) Let n+ = n \u2212 appears in eq. (93.40) only through a factor of depends only on n, n ein\u2212(\u03b8\u2032 \u03b8), which implies that the value of \u03b8 is time indpendent. (Of course, we already knew this, because \u03b8 labels energy eigenstates.) We now have then generates \u03b4(\u03b8\u2032 \u2212 \u03b8). Summing over n \u03b8) + n\u03b8\u2032. Since Zn+ (\u03b8\u2032 \u2212 \u03b8 = n \u2190 \u2212 \u2212 \u2212 \u2212 \u2212 Z\u03b8\u2032 \u2190 \u03b8(J) = \u03b4(\u03b8\u2032 \u03b8) \u2212 ein\u03b8 n X Z An e\u2212 D S+JA. (93.41) We can drop the delta function, and just de\ufb01ne Z\u03b8(J) \u2261 D ein\u03b8 An e\u2212 S+JA. (93.42) Z Next, we combine the sum over n and the", " integral over An into an integral over all A. To account for the factor of ein\u03b8, we use eq. (93.24); we get n X Z\u03b8(J) = A exp D Z Z d4x Tr 1 2 \"\u2212 F \u00b5\u03bdF\u00b5\u03bd + ig2\u03b8 16\u03c02 \u02dcF \u00b5\u03bdF\u00b5\u03bd + J \u00b5A\u00b5. (93.43) # The vacuum angle \u03b8 now appears as the coe\ufb03cient of an extra term in the Yang-Mills lagrangian. We can write the path integral in Minkowski space by setting x4 = it. The extra term contains one derivative with respect to x4, and thus picks 1. Putting all this up a factor of together, we get i. Also, \u03b51234 = +1 but \u03b51230 = \u2212 \u2212 Z\u03b8(J) = Z D A exp i Z d4x Tr \"\u2212 1 2 F \u00b5\u03bdF\u00b5\u03bd \u2212 g2\u03b8 16\u03c02 \u02dcF \u00b5\u03bdF\u00b5\u03bd + J \u00b5A\u00b5 (93.44) # 93: Instantons and Theta Vacua 580 in Minkowski space. We see that the extra term is gauge invariant, Lorentz invariant, hermitian, and has a dimensionless coe\ufb03cient. We therefore could have included it when we \ufb01rst considered Yang-Mills theory. We did not do so because this term is a total divergence; see eq. (93.20). We have always dropped total divergences from the lagrangian, because they do not a\ufb00ect the equations of motion or the Feynman rules. In this case, however, the new term does change the quantum physics, as we have seen. We will explore this in more detail in the next section. So far, we have only discussed SU(2) gauge theory, without scalar or fermion \ufb01elds. What can we say more generally? Adding scalar \ufb01elds has no e\ufb00ect on our analysis. Changing from SU(2) to another simple nonabelian group also has no e\ufb00ect; instanton solutions always reside in an SU(2) subgroup. If the gauge group is U(1), there are no instantons, and hence no vacuum angle. If the gauge group includes more than one nonabelian factor, then there is an independent vacuum angle for each of these factors. On the other hand, adding fermions can signi\ufb01cantly change the physics. We take this up in the next section. Reference Notes Instantons are treated in more detail in Coleman and Weinberg II. Problems 93.1) Show that any 2 \u00d7 form U = a4 + i~a \u00b7 ~a 2 + a2 4 = 1. 2 special unitary matrix U can be written in the ~\u03c3, where a4 and the three-vector ~a are real, and 93.2) Verify that a state of the form of eq. (93.7) is an eigenstate of any n). hamiltonian with matrix elements of the form H n\u2032| h n | i = f (n\u2032 \u2212 93.3) The winding number n for a map from S3 S3 is given by eq. (93.3), where U \u2020U = 1. We will prove that n is invariant under an in\ufb01nitesimal deformation of U. Since any smooth deformation can be made by compounding in\ufb01nitesimal ones, this will prove that n is invariant under any smooth deformation. \u2192 a) Consider an in\ufb01nitesimal deformation of U, U that \u03b4U \u2020 = U \u2020\u03b4U U \u2020, and hence that \u03b4(U \u2202kU \u2020) = U + \u03b4U. Show U \u2202k(U \u2020\u03b4U )U \u2020. \u2192 \u2212 \u2212 b) Show that \u03b4n = \u2212 3 24\u03c02 Z d3x \u03b5ijk Tr[(U \u2202iU \u2020)(U \u2202j U \u2020)\u03b4(U \u2202kU \u2020)]. (93.45) 93: Instantons and Theta Vacua 581 Plug in your result from part (a), and integrate \u2202k by parts. Show that the resulting integrand vanishes. Hint: make repeated use of U \u2202iU \u2020 = \u2202iU U \u2020, and the antisymmetry of \u03b5ijk. \u2212 93.4) Show that if Un(x) and Uk(x) are maps from S3 S3 with winding numbers n and k, then Un(x)Uk(x", ") is a map with winding number n + k. Hint: consider smoothly deforming Un(x) to equal one for x3 < 0. How should Uk(x) be deformed? \u2192 93.5) Use eq. (93.16) to compute the winding number for the map given in eq. (93.29), and for a generalization where \u03c6 is replaced by n\u03c6. 93.6) Use eq. (93.16) to compute the winding number for U n, where U is the map given in eq. (93.29). Hint: \ufb01rst show that U can be written in the form U = exp[i~\u03c7 ~\u03c3], where ~\u03c7 is a three-vector that you should \u00b7 specify. Is your result in accord with the theorem of problem 93.4? 94: Quarks and Theta Vacua 582 94 Quarks and Theta Vacua Prerequisite: 77, 83, 93 Consider quantum chromodynamics with one \ufb02avor of massless quark, represented by a Dirac \ufb01eld \u03a8 in the fundamental representation of the gauge group SU(3). The path integral is Z = Z A D D \u03a8 D \u03a8 exp i Z i\u03a8 /D\u03a8 d4x \" 1 4 \u2212 F a\u00b5\u03bdF a \u00b5\u03bd \u2212 g2\u03b8 32\u03c02 \u02dcF a\u00b5\u03bdF a \u00b5\u03bd # ; (94.1) for the sake of brevity, we have not written the source terms explicitly. U(1)A In addition to the SU(3) gauge symmetry, there is a U(1)V \u00d7 global symmetry of the quark action. However, the U(1)A symmetry is anomalous. As we saw in section 77, under a U(1)A transformation e\u2212 i\u03b1\u03b35 \u03a8, \u03a8e\u2212 i\u03b1\u03b35, \u03a8 \u03a8 \u2192 \u2192 the integration measure picks up a phase factor, \u03a8 \u03a8 D D \u2192 exp \"\u2212 i Z d4x g2\u03b1 16\u03c02 \u02dcF a\u00b5\u03bdF a \u00b5\u03bd \u03a8. \u03a8 D #D (94.2) (94.3) (94.4) Using this in eq. (94.1), we see that the e\ufb00ect of a U(1)A transformation is to change the value of the theta angle from \u03b8 to \u03b8 +2\u03b1. Since the value of \u03b8 can be changed by making a U(1)A transformation (which is simply a change of the dummy integration variable in the path integral), we must conclude that Z does not depend on \u03b8. Apparently (and surprisingly), adding a massless quark has turned the theta angle into a physically irrelevant, unobservable parameter. How do we reconcile this with our analysis in the previous section, where we concluded that instanton-mediated tunneling amplitudes make the vacuum energy density depend on \u03b8? The answer is that when we perform the integral over the quark \ufb01eld in eq. (94.1), we get a functional determinant; the path integral becomes Z = Z A det(i /D)eiSein\u03b8, D (94.5) where S is the Yang-Mills action and n is the winding number. We see from eq. (94.5) that Z would be independent of \u03b8 if gauge \ufb01elds with nonzero winding number did not contribute. This will be the case if det(i /D) van= 0. We conclude that i /D must have a zero ishes for gauge \ufb01elds with n 6 94: Quarks and Theta Vacua 583 eigenvalue or zero mode whenever the gauge \ufb01eld has nonzero winding number. This renders Z independent of \u03b8. Now consider adding a mass term for the quark. If we write the Dirac \ufb01eld \u03a8 in terms of two left-handed Weyl \ufb01elds \u03c7 and \u03be, \u03a8 = \u03c7 \u03be\u2020!, (94.6) the mass term reads \u2212 We have allowed m to be complex: m = can be written as \u2212 Lmass = m\u03c7\u03be m\u2217\u03be\u2020\u03c7\u2020. (94.7) m | ei\u03c6. In terms of \u03a8, eq. (94.7) | i\u03c6\u03b35\u03a8. (94.8) Lmass = m \u03a8e\u2212 | A U(1)A transformation changes \u03c6 to \u03c6 + 2\u03b1. Since \u03b8 simultaneously i\u03b8, is unchanged. changes to \u03b8 + 2\u03b1, we see that \u03c6 i\u03b8, but not on m and Thus, the path integral can (", "and does) depend on me\u2212 \u03b8 separately. \u03b8, or equivalently me\u2212 \u2212| \u2212 With more quark \ufb01elds, the mass term is Mij\u03c7i\u03bej + h.c.; a U(1)A transformation changes the phase of every \u03c7i and \u03bei by ei\u03b1, and so every matrix element of M picks up a factor of e2i\u03b1. Simultaneously, \u03b8 changes i\u03b8 is to \u03b8 + 2N \u03b1, where N is the number of quark \ufb01elds. Thus (det M )e\u2212 invariant under a U(1)A transformation. \u2212 = L To understand the e\ufb00ects of the theta angle on hadronic physics, we turn to the e\ufb00ective lagranagian (for the case of two light \ufb02avors) that we developed in section 83, = L 1 1 \u2212 \u2212 1)iN \u03b3\u00b5(U \u2202\u00b5U \u2020PL + U \u2020\u2202\u00b5U PR)N \u03c0 Tr \u2202\u00b5U \u2020\u2202\u00b5U + v3 Tr(M U + M \u2020U \u2020) mN N (U \u2020PL + U PR)N 4 f 2 \u2212 + iN /\u2202N 2 (gA\u2212 c1N (M PL + M \u2020PR)N c3 Tr(M U + M \u2020U \u2020) N (U \u2020PL + U PR)N c4 Tr(M U U PR)N, M \u2020U \u2020) N (U \u2020PL \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 c2N (U \u2020M \u2020U \u2020PL + U M U PR)N (94.9) where U is a 2 \u00d7 the \ufb01eld for the nucleon doublet, 2 special unitary matrix \ufb01eld representing the pions, N is M = mu 0 0 md! i\u03b8/2 e\u2212 (94.10) is the quark mass matrix, v3 is the value of the quark condensate, and gA and ci are numerical constants. (The terms in the last three lines were introduced in problem 83.8.) 94: Quarks and Theta Vacua For the case of \u03b8 = 0, the potential V (U ) = \u2212 v3 Tr M U + h.c. 584 (94.11) is minimized by U = I, and we can expand about this point in powers of the pion \ufb01elds, as we did in section 83. However, for nonzero \u03b8, the minimum of V (U ) occurs at U = U0, where U0 is diagonal (because M is) and has unit determinant (because U is required to have unit determinant). We can therefore write e+i\u03c6 0 0 e\u2212 i\u03c6!. (94.12) U0 = We determine \u03c6 by minimizing V (U0) = \u2212 the result is 2v3 mu cos(\u03c6 h \u2212 1 2 \u03b8) + md cos(\u03c6 + 1 2 \u03b8) i ; (94.13) tan \u03c6 = md mu \u2212 mu + md tan( 1 2 \u03b8). (94.14) As we will see shortly, experimental results require the value of than 10\u2212 of \u03b8. For \u03b8 to be less 9; therefore, we can work to \ufb01rst order in an expansion in powers 1, eqs. (94.12) and (94.14) can be written in the elegant form \u03b8 | | \u226a where M U0 = M0 \u2212 i\u03b8 \u02dcmI + O(\u03b82), M0 = mu 0 0 md! (94.15) (94.16) is the quark mass matrix with \u03b8 set to zero, I is the identity matrix, and \u02dcm = mumd mu + md (94.17) is the reduced mass of the up and down quarks. We can now expand in powers of the pion \ufb01elds. Though it is not at all obvious, it turns out that the most convenient way to de\ufb01ne the pion \ufb01elds is by writing (94.18) 0 = U0 and u(x) = exp[i\u03c0a(x)T a/f\u03c0]. We also de\ufb01ne \u02dcU(x) = u2(x), U (x) = u0u2(x)u0, where u2 and a new nucleon \ufb01eld via N N = (u0uPL + u\u20200u\u2020PR) N. (94.19", ") 94: Quarks and Theta Vacua 585 Substituting eqs. (94.18) and (94.19) into eq. (94.9), and using u0M u0 = M U0 (which follows because u0 and M are both diagonal, and hence commute), we ultimately get = L 1 N \u2212 mN NN /a\u03b35 N gAN + [u(M U0)u + u\u2020(M U0)\u2020u\u2020] N [u(M U0)u \u03c0 Tr \u2202\u00b5 \u02dcU \u2020\u2202\u00b5 \u02dcU + v3 Tr[(M U0) \u02dcU + (M U0)\u2020 \u02dcU \u2020] 4 f 2 \u2212 + i /\u2202 /v N 1 2 c+ N \u2212 + 1 2 c \u2212 N c3 Tr[(M U0) \u02dcU + (M U0)\u2020 \u02dcU \u2020] (M U0)\u2020 \u02dcU \u2020] \u2212 + c4 Tr[(M U0) \u02dcU u\u2020(M U0)\u2020u\u2020]\u03b35 N N \u2212 NN N \u2212, \u2212 \u03b35 N N 2 i[u\u2020(\u2202\u00b5u) (94.20) 2 i[u\u2020(\u2202\u00b5u) + u(\u2202\u00b5u\u2020)], a\u00b5 = 1 = c2. Eq. (94.20) is exactly what we found in section 83, except that the where v\u00b5 = 1 c1 \u00b1 quark mass matrix M has been replaced everywhere by M U0. u(\u2202\u00b5u\u2020)], and c \u2212 \u00b1 We can now use eq. (94.15) to get the O(\u03b8) contributions to. Using \u02dcU \u2020) vanishes in the case of two light \ufb02avors, we \ufb01nd the fact that Tr( \u02dcU L \u2212 1 \u2212 i\u03b8 \u02dcm L\u03b8 = + 1 2 c N \u03b35 N Expanding in powers of the pion \ufb01elds yields 2 c+ N \u2212 h + c4 Tr( \u02dcU + \u02dcU \u2020) N \u02dcU \u2020) ( \u02dcU \u2212 i \u2212 N. ( \u02dcU + \u02dcU \u2020)\u03b35 N (94.21) (\u03b8c+ \u02dcm/f\u03c0)\u03c0a \u03c3a +..., (94.22) L\u03b8 = i\u03b8 \u02dcm(c +4c4) \u03b35 N \u2212 \u2212 N N N \u2212 where we used T a = 1 2 \u03c3a. We can eliminate the \ufb01rst term with a \ufb01eld rede\ufb01nition of the form. This generates some new terms in eq. (94.20), but all have at least two factors of quark masses, and hence can be neglected. The second term in eq. (94.22) provides a pion-nucleon coupling that violates both parity P and time-reversal T (equivalently, CP ). N \u2192 i\u03b1\u03b35 e\u2212 N mn = The value of c+ can be \ufb01xed by baryon mass di\ufb00erences. The c+ md) to the protonterm in eq. (94.20) makes a contribution of c+(mu \u2212 1.3 MeV. Using mu = 1.7 MeV neutron mass di\ufb00erence, mp \u2212 and md = 3.9 MeV yields c+ = 0.6. However, there is a comparable electromagnetic contribution to the proton-neutron mass di\ufb00erence. We get a better estimate from the masses of baryons with strange quarks, m\u03a30 = 122 MeV; using ms = 76 MeV yields c+(ms \u2212 c+ = 1.7. (All these values assume the MS renormalization scheme with \u00b5 = 2 GeV.) 1 2 md) = m\u039e0 2 mu \u2212 \u2212 \u2212 1 For comparison with the interaction of eq. (94.22), the dominant (P and CP conserving) pion-nucleon interaction comes from the last term in the 94: Quarks and Theta Vacua 586 Figure 94.1: Diagrams contributing to the electric dipole moment of the neutron that are enhanced by a chiral log. The CP violating vertex is denoted with a cross. second line of eq. (94.20), and is L\u03c0NN = (gA/f\u03c0)\u2202\u00b5\u03c0a N T a\u03b3\u00b5\u03b35 N, (94.23) where gA = 1.27. As we did in section 83", ", we can integrate by parts to put the derivative on the nucleon \ufb01elds, and then (for on-shell nucleons) use the Dirac equation to get L\u03c0NN = \u2212 i(gAmN /f\u03c0)\u03c0a \u03c3a\u03b35 N N. (94.24) The strongest limit on a CP violating pion-nucleon coupling comes from measurements of the electric dipole moment of the neutron. The Feynman diagrams of \ufb01g. (94.1) contribute to an ampitude of the form \u2212 T p. In the q where q = p\u2032 \u2212 e\ufb00ective lagrangian of = 2iD(q2)\u03b5\u2217\u00b5(q)us\u2032(p\u2032)S\u00b5\u03bdq\u03bdi\u03b35us(p), (94.25) 0 limit, this corresponds to a term in the \u2192 = D(0)F\u00b5\u03bd nS\u00b5\u03bd i\u03b35n, L (94.26) where n is the neutron \ufb01eld; see section 64. If the factor of i\u03b35 was absent, this would represent a contribution of D(0) to the magnetic dipole moment of the neutron. To account for the factor of i\u03b35, we use \u2212 to see that eq. (94.26) is equivalent to S\u00b5\u03bd i\u03b35 = 1 2 \u03b5\u00b5\u03bd\u03c1\u03c3S\u03c1\u03c3 = L \u2212 D(0) \u02dcF\u00b5\u03bd nS\u00b5\u03bd n, (94.27) (94.28) where \u02dcF\u00b5\u03bd = 1 2 \u03b5\u00b5\u03bd\u03c1\u03c3F\u03c1\u03c3 is the dual \ufb01eld strength. Since \u02dcB = E, eq. (94.26) represents a contribution of D(0) to the electric dipole moment of the neutron dn. \u2212 94: Quarks and Theta Vacua 587 q l+q/2 l\u2212q/2 p l+(p+p )/2 p Figure 94.2: Momentum \ufb02ow in the diagrams of \ufb01g. (94.1). \u223c 4.2, where \u039b The salient feature of the diagrams shown in \ufb01g. (94.1) is that the photon line attaches to the pion line. These diagrams are enhanced by a chiral log ln(\u039b2/m2 4\u03c0f\u03c0 is the ultraviolet cuto\ufb00 in the e\ufb00ec\u03c0) tive theory. No other contributing diagrams have this enhancement; it is an infrared e\ufb00ect, due to the light pion. Of course, 4.2 is not an impressively large number, and so we cannot be certain that the remaining contributions are not signi\ufb01cant. These contributions depend on coe\ufb03cients in the e\ufb00ective lagrangian that are not well determined by other experimental results. \u223c Using \u03c0a\u03c3a = \u221a2\u03c0+ \u03c00 \u221a2\u03c0\u2212 \u2212 \u03c00!, (94.29) we can write the charged-pion terms in eqs. (94.24) and (94.21) as L\u03c0NN = L\u03b8\u03c0NN = \u2212 \u2212 i\u221a2(gAmN /f\u03c0)(\u03c0+ p\u03b35n + \u03c0\u2212 n\u03b35p), \u221a2(\u03b8c+ \u02dcm/f\u03c0)(\u03c0+ pn + \u03c0\u2212 np). (94.30) (94.31) From these we read o\ufb00 the pion-nucleon vertex factors. We label the internal momenta as shown in \ufb01g. (94.2); for small q, \u2113 is a pion momentum 4\u03c0f\u03c0. Because the terms of interest have a that should be cut o\ufb00 at \u039b chiral log produced by an infrared divergence at small \u2113, we can treat \u2113 as much less than p and p\u2032. Thus the internal proton is nearly on-shell, which justi\ufb01es the use of eq. (94.24) for the CP conserving interaction. \u223c The diagrams of \ufb01g. (94.1) yield an amplitude of = i T 3 1 i (ie)(\u221a2gAmN /f\u03c0)( i\u221a2\u03b8c+ \u02dcm/f\u03c0)\u03b5\u2217\u00b5 \u2212 \u00d7 (2\u2113\u00b5) u \u2032[( /\u2113 \u2212 \u2212 N )((\u2113+ 1 ((\u2113+\u00afp)2 + m2 /\u00afp + mN )\u03b35 + \u03b35( 2 q)2 + m2 /\ufffd", "\ufffd \u2212 \u2212 \u03c0)((\u2113 \u2212 \u039b d4\u2113 (2\u03c0)4 0 Z /\u00afp + mN )]u 1 2 q)2 + m2 \u03c0), (94.32) 94: Quarks and Theta Vacua 588 where \u00afp = 1 simplify to 2 (p\u2032+p). Using \u03b3\u00b5, \u03b35} { u \u2032[...]u = 2mN u \u2032\u03b35u. = 0, the spinor factors in the numerator (94.33) From the spinor properties established in section 38, it is easy to check that u \u2032\u03b35u vanishes when p\u2032 = p; thus u \u2032\u03b35u must be O(q), and so we can set q = 0 everywhere else. Also, taking \u2113 \u2113 in the denominator of eq. (94.32). We now have p, we can set (\u2113+\u00afp)2 + m2 N = 2p \u226a \u00b7 = 4(e\u03b8gAc+ \u02dcmm2 N /f 2 \u03c0)\u03b5\u2217\u00b5 T \u039b d4\u2113 (2\u03c0)4 0 Z (2\u2113\u00b5) u \u2032\u03b35u \u2113)(\u21132 + m2 \u03c0)2. (2p \u00b7 Integrating over the direction of \u2113 results in \u2113\u00b5 p \u00b7 \u2113 \u2192 p\u00b5 p2 = p\u00b5 m2 N \u2212. (94.34) (94.35) Next we use the Gordon identity (see problem 38.4) p\u00b5 u \u2032\u03b35u = u \u2032S\u00b5\u03bdq\u03bdi\u03b35u + O(q2), (94.36) which veri\ufb01es that u \u2032\u03b35u is linear in q. We now have = \u2212 T 4(e\u03b8gAc+ \u02dcm/f 2 \u03c0)\u03b5\u2217\u00b5 u \u2032S\u00b5\u03bdq\u03bdi\u03b35u \u039b d4\u2113 (2\u03c0)4 1 (\u21132 + m2 \u03c0)2. 0 Z (94.37) 0, the integral diverges at small \u2113, generating a chiral In the limit m\u03c0 \u2192 log. This infrared divergence can only arise from diagrams with two pion propagators, which is why it appears only if the photon is attached to the pion. After a Wick rotation, the integral evaluates to (i/16\u03c02) ln(\u039b2/m2 \u03c0). Comparing with eq. (94.25), we see that the electric dipole moment of the neutron is e\u03b8gAc+ \u02dcm 8\u03c02f 2 \u03c0 Putting in numbers (gA = 1.27, c+ = 1.7, \u02dcm = 1.2 MeV), we \ufb01nd \u03c0) + O(1) i ln(\u039b2/m2 dn = h. (94.38) dn = 3.2 10\u2212 16 \u03b8 e cm. (94.39) \u00d7 dn| | < 6.3 \u00d7 10\u2212 26 e cm, and so we must The experimental upper limit is 10\u2212 have < 2 10. \u03b8 | | \u00d7 Such a small value for a fundamental parameter cries out for an explanation; this is the strong CP problem. Several solutions have been proposed. (1) The up quark mass may actually be zero, since a massless quark renders \u03b8 unobservable (and e\ufb00ectively zero). This requires higher-order 94: Quarks and Theta Vacua 589 corrections in the quark masses to account for the masses of the pseudogoldstone bosons. (2) The fundamental lagrangian may be CP invariant, and the observed CP violation in weak interactions due to spontaneous breaking of CP symmetry. (3) The theta parameter may be promoted to a \ufb01eld, the axion, which would minimize its energy by rolling to \u03b8 = 0; see problem 94.2. All of these solutions have interesting physical consequences. Reference Notes Quarks and theta vacua are discussed in Coleman, Ramond II, and Weinberg II. Problems 94.1) Carry out the \ufb01eld rede\ufb01nition discussed after eq. (94.22), and verify that all new terms generated in the lagrangian are suppressed by at least two powers of quark masses. 94.2) Consider adding to the Standard Model a massless quark, represented by a pair of Weyl fermions \u03c7 and \u03be in the 3 and 3 representations of SU(3). Also add a complex scal", "ar \u03a6 in the singlet representation. Assume that these \ufb01elds have a Yukawa interaction of the form LYuk = y\u03a6\u03c7\u03be + h.c., where y is the Yukawa coupling constant. Assume that the scalar potential V (\u03a6) depends only on \u03a6\u2020\u03a6. a) Show that the lagrangian is invariant under a Peccei-Quinn trans2i\u03b1\u03a6, all other \ufb01elds unformation \u03c7 changed. ei\u03b1\u03be, \u03a6 ei\u03b1\u03c7, \u03be e\u2212 \u2192 \u2192 \u2192 b) Show that this global U(1)PQ symmetry is anomalous, and that \u03b8 \u03b8 + 2\u03b1 under a U(1)PQ transformation. c) Suppose that V (\u03a6) has its minimum at Show that this gives a mass to the quark we introduced. \u03a6 | | = f /\u221a2, with f = 0. \u2192 1/2(f + \u03c1)eia/f, where \u03c1 and a are \ufb01elds. Argue that, d) Write \u03a6 = 2\u2212 in eq. (94.10), we should replace \u03b8 with \u03b8 + a/f, and add to eq. (94.9) a kinetic term 1 2 \u2202\u00b5a\u2202\u00b5a for the a \ufb01eld. d) Show that the minimum of V (U ), de\ufb01ned in eq. (94.11), is at U = I f \u03b8. Show that P and CP are conserved at this minimum. and a = \u2212 e) The particle corresponding to the a \ufb01eld is the axion; compute its mass, assuming f f\u03c0. f) Note that if f is large, the extra quark becomes very heavy, and the axion becomes very light. Show that couplings of the axion to the hadrons are all suppressed by a factor of 1/f. \u226b \u2212 6 95: Supersymmetry 590 95 Supersymmetry Prerequisite: 69 Supersymmetry is a continuous symmetry that mixes up bosonic and fermionic degrees of freedom. A supersymmetric theory (in four spacetime dimensions) has a set of supercharges QaA, where a is a left-handed spinor index, and A is an internal index that runs from 1 to, where the allowed values of are 1, 2, and 4. The supercharges can be obtained as integrals over d3x of the time component of a supercurrent. The supercurrent is found via the Noether procedure, once we have identi\ufb01ed the set of supersymmetry transformations that leaves the action invariant. N N The supercharges QaA and their hermitian conjugates Q\u2020\u02d9aA, together with the generators of the Poincare group P \u00b5 and M \u00b5\u03bd, obey a supersymmetry algebra [QaA, P \u00b5] = 0, [Q\u2020\u02d9aA, P \u00b5] = 0, [QaA, M \u00b5\u03bd ] = (S\u00b5\u03bd [Q\u2020\u02d9aA, M \u00b5\u03bd ] = (S\u00b5\u03bd QaA, QbB} { QaA, Q\u2020\u02d9aB} { = \u2212 = ZAB\u03b5ab, 2\u03b4AB\u03c3\u00b5 L )a R ) \u02d9a cQcA, \u02d9cQ\u2020\u02d9cA, a \u02d9aP\u00b5. (95.1) (95.2) (95.3) (95.4) (95.5) (95.6) Eqs. (95.1) and (95.2) simply say that the supercharges are conserved, and eqs. (95.3) and (95.4) simply say that their spinor indices are indeed spinor ZBA must commute with QaA, P \u00b5, and M \u00b5\u03bd, indices. In eq. (95.5), ZAB = and so represents a central charge in the supersymmetry algebra. We will be concerned only with the case of = 1 supersymmetry: the index A then takes on only one value (and so can be dropped), and ZAB = 0. N \u2212 N = 1 supersymmetric theories are most easily formulated in superspace, where we augment the usual spacetime coordinate x\u00b5 with an anticommuting left-handed spinor coordinate \u03b8a and its right-handed complex conjugate \u03b8\u2217\u02d9a. We de\ufb01ne super\ufb01elds \u03a6(x, \u03b8, \u03b8\u2217) that are functions of all these coordinates. The energy-momentum vector generates", " translations of the usual space- time coordinate x\u00b5 in the usual way, [\u03a6(x, \u03b8, \u03b8\u2217), P \u00b5] = i\u2202\u00b5\u03a6(x, \u03b8, \u03b8\u2217). \u2212 By analogy, we would expect [\u03a6(x, \u03b8, \u03b8\u2217), Qa] = Qa\u03a6(x, \u03b8, \u03b8\u2217), i \u2212 (95.7) (95.8) 95: Supersymmetry [\u03a6(x, \u03b8, \u03b8\u2217), Q\u2020\u02d9a] = \u2217\u02d9a\u03a6(x, \u03b8, \u03b8\u2217), i Q \u2212 591 (95.9) where Q\u2217\u02d9a are appropriate di\ufb00erential operators. To \ufb01gure out Qa and what they should be, we \ufb01rst introduce the anticommuting derivatives \u2202a \u2261 \u02d9a, which obey \u2202a\u03b8c = \u03b4a \u2202/\u2202\u03b8a and \u2202\u2217\u02d9a \u2261 \u02d9c. Note, \u2202/\u2202\u03b8\u2217 however, that complex conjugation should reverse the order of a product of Grassmann variables, in order to maintain consistency with hermitian \u02d9c, which conjugation. Then we have \u03b4a implies c = (\u2202a\u03b8c)\u2217 = \u03b8\u2217 c and \u2202\u2217\u02d9a\u03b8\u2217 \u02d9c(\u2202a)\u2217 = \u02d9c = \u03b4 \u02d9a (\u2202a)\u2217\u03b8\u2217 \u2212 (\u2202a)\u2217 = \u2202\u2217\u02d9a. (95.10) \u2212 Thus, our \ufb01rst guess for the di\ufb00erential operators in eqs. (95.8) and (95.9) Qa = \u2202a and is \u2202\u2217\u02d9a. However, this choice is inconsistent with 2\u03c3\u00b5 = Qa, Q\u2020\u02d9a} { Q\u2217\u02d9a = a \u02d9aP\u00b5. \u2212 \u2212 An alternative that avoids this pitfall is Qa = +\u2202a + i\u03c3\u00b5 \u2202\u2217\u02d9a \u2212 \u2217\u02d9a = \u2212 Q a \u02d9c\u03b8\u2217 i\u03b8c\u03c3\u00b5 \u02d9c\u2202\u00b5, c \u02d9a\u2202\u00b5. These obey the anticommutation relations = Qb} {Qa, {Qa, \u2217\u02d9a} Q {Q = \u2212 = 0, \u2217\u02d9a, \u2217\u02d9b } Q 2i\u03c3\u00b5 a \u02d9a\u2202\u00b5. It is straightforward to check that eqs. (95.5\u201395.12) are now mutually compatible. In particular, the Jacobi identity [\u03a6, Q], Q\u2020 { } + [\u03a6, Q\u2020], Q { } \u2212 [\u03a6, Q, Q\u2020 { ] = 0 } (95.15) is satis\ufb01ed. Next we introduce the supercovariant derivatives Da = +\u2202a \u2212 \u2217\u02d9a = i\u03c3\u00b5 a \u02d9c\u03b8\u2217 \u2202\u2217\u02d9a + i\u03b8c\u03c3\u00b5 D \u2212 \u02d9c\u2202\u00b5, c \u02d9a\u2202\u00b5. (95.16) (95.17) These obey = = 0, {Da, Db} {Da, = (95.20) \u2217\u02d9b } \u2217\u02d9b } Q D\u2217\u02d9a\u03a6 = 0 Because of eq. (95.20), we could impose the condition on a super\ufb01eld, and this condition would be preserved by the supersymmetry transformations of eqs. (95.8) and (95.9). A super\ufb01eld that obeys = 0. Da\u03a6 = 0 or \u2217\u02d9a, \u2217\u02d9b } D {D = 2i\u03c3\u00b5 a \u02d9a\u2202\u00b5, \u2217\u02d9a, = \u2217\u02d9a} \u2217\u02d9a, {D {Da, {Da, (95.19) (95.18) Qb} Qb} {D Q = D \u2217\u02d9a\u03a6(x, \u03b8, \u03b8\u2217) = 0 D (95.21) (95.11) (95.12) (95.13) (95.14) 95: Supersymmetry 592 is a left-handed ch", "iral super\ufb01eld. Its hermitian conjugate \u03a6\u2020(x, \u03b8, \u03b8\u2217) obeys Da\u03a6\u2020(x, \u03b8, \u03b8\u2217) = 0, and is a right-handed chiral super\ufb01eld. We can solve eq. (95.21) by introducing and noting that y\u00b5 = x\u00b5 \u2212 i\u03b8c\u03c3\u00b5 c \u02d9c\u03b8\u2217 \u02d9c, (95.22) (95.23) \u2217\u02d9a\u03b8a = 0 \u2217\u02d9ay\u00b5 = 0. (95.24) D\u2217\u02d9ay\u00b5 = 0, remember that there is a minus sign from (When verifying pulling the \u2202\u2217\u02d9a through \u03b8c.) Thus, any super\ufb01eld \u03a6(y, \u03b8) that is a function of y and \u03b8 only is a left-handed chiral super\ufb01eld. and D D We can expand \u03a6(y, \u03b8) in powers of \u03b8; because \u03b8 is an anticommuting variable with a two-valued index, we have \u03b8a\u03b8b\u03b8c = 0, and so the expansion terminates after the quadratic term. We thus have \u03a6(y, \u03b8) = A(y) + \u221a2\u03b8\u03c8(y) + \u03b8\u03b8F (y), (95.25) where A(y) and F (y) are complex scalar \ufb01elds, \u03c8a(y) is a left-handed Weyl \ufb01eld, and we have used our standard index-suppression conventions: \u03b8\u03c8 = \u03b8a\u03c8a and \u03b8\u03b8 = \u03b8a\u03b8a. The factor of root-two is conventional. We can now substitute in eq. (95.23), and continue to expand in powers of \u03b8 and \u03b8\u2217. Making use of the spinor identities \u03b8a\u03b8b = + 1 \u03b8\u2217\u02d9a\u03b8\u2217\u02d9b = \u2212 2 \u03b8\u03b8\u03b5ab, 1 2 \u03b8\u2217\u03b8\u2217\u03b5 \u02d9a\u02d9b, \u03b8\u2217 (95.26) 1 \u02d9a\u03b8\u2217 \u03b8a\u03b8b = \u2212 \u02d9b = + 1 2 \u03b8\u03b8\u03b5ab, 2 \u03b8\u2217\u03b8\u2217\u03b5 \u02d9a\u02d9b, \u02d9a, along with the Fierz identity (95.27) where \u03b8\u03b8 = \u03b8a\u03b8a and \u03b8\u2217\u03b8\u2217 = (\u03b8\u03b8)\u2217 = \u03b8\u2217\u02d9a\u03b8\u2217 (\u03b8\u03c3\u00b5\u03b8\u2217)(\u03b8\u03c3\u03bd\u03b8\u2217) = 1 2 \u03b8\u03b8\u03b8\u2217\u03b8\u2217g\u00b5\u03bd, \u2212 we \ufb01nd \u03a6(x, \u03b8, \u03b8\u2217) = A(x) + \u221a2\u03b8\u03c8(x) + \u03b8\u03b8F (x) 1 \u221a2 \u2212 i\u03b8\u03b8\u03b8\u2217\u00af\u03c3\u00b5\u2202\u00b5\u03c8(x) + 1 4 \u03b8\u03b8\u03b8\u2217\u03b8\u2217\u22022A(x). i(\u03b8\u03c3\u00b5\u03b8\u2217)\u2202\u00b5A(x) \u2212 (95.28) (95.29) Let us investigate the properties of a left-handed chiral super\ufb01eld under It is a supersymmetry transformation, given by eqs. (95.8) and (95.9). easiest to use the y and \u03b8 coordinates, since b, Qa\u03b8b = \u03b4a \u2217\u02d9a\u03b8b = 0, Q Qay\u00b5 = 0, \u2217\u02d9ay\u00b5 = \u2212 Q 2i\u03b8c\u03c3\u00b5 c \u02d9a. (95.30) 95: Supersymmetry We thus have Qa\u03a6(y, \u03b8) = \u2202a\u03a6(y, \u03b8) = \u221a2\u03c8a(y) + 2\u03b8aF (y), \u2217\u02d9a\u03a6(y, \u03b8) = Q = \u2212 \u2212 2i\u03b8c\u03c3\u00b5 2i\u03b8c\u03c3\u00b5 c \u02d9a\u2202\u00b5\u03a6(y, \u03b8) c \u02d9a\u2202\u00b5A(y) + i\u221a2\u03b8\u03b8\u2202\u00b5\u03c8c", "(y)\u03c3\u00b5 c \u02d9a, 593 (95.31) (95.32) where \u2202\u00b5 is with respect to y; we used eq. (95.26) to get the last line. We can now \ufb01nd the supersymmetry transformations of the component \ufb01elds A, \u03c8, and F by matching powers of \u03b8 on each side of eqs. (95.8) and (95.9). Remembering that the Q and Q\u2020 operators anticommute with \u03b8 and \u03b8\u2217, we get i\u221a2\u03c8a, i\u221a2\u03b5acF, \u2212 [A, Qa] = \u03c8c, Qa} \u2212 { [F, Qa] = 0, = [A, Q\u2020\u02d9a] = 0, \u03c8c, Q\u2020\u02d9a} = { \u2212 [F, Q\u2020\u02d9a] = \u221a2\u2202\u00b5\u03c8c\u03c3\u00b5 \u221a2\u03c3\u00b5 c \u02d9a, c \u02d9a\u2202\u00b5A, (95.33) (95.34) (95.35) where all component \ufb01elds have spacetime argument y. However, y is arbitrary, and so we are free to replace it with x. Eq. (95.35) is the most important: it tells us that the supersymmetry d4x F (x) transformation of the F \ufb01eld is a total derivative. Therefore, is invariant under a supersymmetry transformation, and hence could be a term in the action of a supersymmetric theory. R The product of two left-handed chiral super\ufb01elds is another left-handed chiral super\ufb01eld; this is obvious from eq. (95.25), and the fact that the \u03b8 expansion always terminates with the quadratic term. For two chiral super\ufb01elds \u03a61(y, \u03b8) and \u03a62(y, \u03b8), we have \u03a61\u03a62 = A1A2 + \u221a2\u03b8(A1\u03c82 + A2\u03c81) + \u03b8\u03b8(A1F2 + A2F1 \u2212 More generally, given a set of left-handed chiral super\ufb01elds \u03a6i, we can consider a function of them W (\u03a6); this function is itself a left-handed chiral super\ufb01eld. Its F term (the coe\ufb03cient of \u03b8\u03b8) is \u03c81\u03c82). (95.36) = F \u2202W (A) \u2202Ai Fi \u2212 1 2 \u22022W (A) \u2202Ai\u2202Aj \u03c8i\u03c8j, (95.37) W (\u03a6) where repeated indices are summed. The spacetime integral of this term (like the spacetime integral of any F term) is invariant under supersymmetry, and hence could be a term in the action of a supersymmetric theory. In this case, the function W (\u03a6) is called the superpotential. 95: Supersymmetry 594 We still need kinetic terms. To get them, we \ufb01rst investigate the prop- erties of a vector super\ufb01eld. A vector super\ufb01eld V (x, \u03b8, \u03b8\u2217) is hermitian, [V (x, \u03b8, \u03b8\u2217)]\u2020 = V (x, \u03b8, \u03b8\u2217), (95.38) but is not subject to any other constraint. Its component expansion is V (x, \u03b8, \u03b8\u2217) = C(x) + \u03b8\u03c7(x) + \u03b8\u2217\u03c7\u2020(x) + \u03b8\u03b8M (x) + \u03b8\u2217\u03b8\u2217M \u2020(x) + \u03b8\u03c3\u00b5\u03b8\u2217v\u00b5(x) + \u03b8\u03b8\u03b8\u2217\u03bb\u2020(x) + \u03b8\u2217\u03b8\u2217\u03b8\u03bb(x) + 1 2 \u03b8\u03b8\u03b8\u2217\u03b8\u2217D(x), (95.39) where C and D are real scalar \ufb01elds, M is a complex scalar \ufb01eld, \u03c7 and \u03bb are left-handed Weyl \ufb01elds, and v\u00b5 is a real vector \ufb01eld. Following the analysis that led to eq. (95.35), we \ufb01nd [D, Qa] = \u03c3\u00b5 a ", "\u02d9c\u2202\u00b5\u03bb\u2020 \u02d9c, \u2212 [D, Q\u2020\u02d9a] = +\u2202\u00b5\u03bbc\u03c3\u00b5 c \u02d9a. (95.40) We see that the supersymmetry transformation of the D component of a d4x D(x) is invariant vector super\ufb01eld is a total derivative. Therefore, under a supersymmetry transformation, and hence could be a term in the action of a supersymmetric theory. R Consider the product of a left-handed chiral super\ufb01eld \u03a6(x, \u03b8, \u03b8\u2217), as given by eq. (95.29), and its hermitian conjugate \u03a6\u2020(x, \u03b8, \u03b8\u2217) = A(x) + \u221a2\u03b8\u2217\u03c8\u2020(x) + \u03b8\u2217\u03b8\u2217F \u2020(x) + i(\u03b8\u03c3\u00b5\u03b8\u2217)\u2202\u00b5A\u2020(x) + 1 \u221a2 i\u03b8\u2217\u03b8\u2217\u2202\u00b5\u03c8\u2020(x)\u00af\u03c3\u00b5\u03b8 + 1 4 \u03b8\u03b8\u03b8\u2217\u03b8\u2217\u22022A\u2020(x). (95.41) The product \u03a6\u2020\u03a6 is obviously hermitian, and so is a vector super\ufb01eld. After considerable use of eqs. (95.26\u201395.28), we \ufb01nd that the D term (the coe\ufb03cient of \u03b8\u03b8\u03b8\u2217\u03b8\u2217) of this vector super\ufb01eld is \u03a6\u2020\u03a6 = D 1 2 \u2202\u00b5A\u2020\u2202\u00b5A + 1 2 i\u03c8\u2020 \u00af\u03c3\u00b5\u2202\u00b5\u03c8 4 A\u22022A\u2020 + 1 2 i\u2202\u00b5\u03c8\u2020 \u00af\u03c3\u00b5\u03c8 1 \u2212 + 1 4 A\u2020\u22022A \u2212 + F \u2020F. (95.42) The spacetime integral of this term (like the spacetime integral of any D term) is invariant under supersymmetry, and hence could be a term in the action of a supersymmetric theory. After some integrations by parts, and dropping total divergences, we \ufb01nd \u03a6\u2020\u03a6 = D \u2212 \u2202\u00b5A\u2020\u2202\u00b5A + i\u03c8\u2020 \u00af\u03c3\u00b5\u2202\u00b5\u03c8 + F \u2020F. (95.43) We see that we have standard kinetic terms for the complex scalar \ufb01eld A and the left-handed Weyl \ufb01eld \u03c8. We also have a term with no derivatives 95: Supersymmetry 595 for the complex scalar \ufb01eld F. The F \ufb01eld is therefore called an auxiliary \ufb01eld. If we consider a set of left-handed chiral super\ufb01elds \u03a6i, we get a hermi- tian, supersymmetric action if we take as the lagrangian = \u03a6\u2020i \u03a6i L + D W (\u03a6) F + h.c., (95.44) where the index in the \ufb01rst term is summed. Since Fi appears only quadratically and without derivatives, we can easily perform the path integral over it. The result is equivalent to solving the classical equation of motion for Fi, \u2202 L \u2202Fi = F \u2020i + \u2202W (A) \u2202Ai = 0, (95.45) and substituting the solution back into the lagrangian; the result is = L \u2212 \u2202\u00b5A\u2020i \u2202\u00b5Ai + i\u03c8\u2020i \u00af\u03c3\u00b5\u2202\u00b5\u03c8i 2 \u2202W (A) \u2202Ai 1 2 \" \u22022W (A) \u2202Ai\u2202Aj \u2212 where the indices are summed in each term. \u2212 \u03c8i\u03c8j + h.c. #, (95.46) As an example, let us consider a single left-handed chiral super\ufb01eld, with superpotential W (A) = 1 2 mA2 + 1 6 gA3. (95.47) This is the Wess-Zumino model. The scalar potential is V (A) = 2 \u2202W/\u2202A | | = m2A\u2020A + 1 2 gm(A\u2020A2 + A\u2020 2A) + 1 4 g2(A\u2020A)2. (95.48) We see that the scalar has mass m. The last term", " in eq. (95.46) becomes Lmass+Yuk = \u2212 1 2m\u03c8\u03c8 \u2212 1 2 gA\u03c8\u03c8 + h.c.. (95.49) We see that the fermion also has mass m, and a Yukawa interaction with the scalar. The Yukawa couping is related (by supersymmetry) to the cubic and quartic self-interactions of the scalar. Next we would like to introduce gauge \ufb01elds. Recall that the vector super\ufb01eld V (x, \u03b8, \u03b8\u2217) has among its components a real vector \ufb01eld v\u00b5(x) that could be identi\ufb01ed as an abelian gauge \ufb01eld. (Later we will add an adjoint index to the super\ufb01eld in order to get a nonabelian gauge \ufb01eld.) We need to generalize the notion of a gauge transformation to super\ufb01elds. We begin by noting that if \u039e is a left-handed chiral super\ufb01eld, then \u039e) is a vector super\ufb01eld. We then de\ufb01ne a supergauge transformation i(\u039e\u2020 \u2212 V \u2192 V + i(\u039e\u2020 \u039e). \u2212 (95.50) 95: Supersymmetry 596 We will attempt to construct actions that are invariant under eq. (95.50). Following the pattern of eqs. (95.29) and (95.41), we write \u039e(x, \u03b8, \u03b8\u2217) = B(x) + \u03b8\u03be(x) + \u03b8\u03b8G(x) i(\u03b8\u03c3\u00b5\u03b8\u2217)\u2202\u00b5B(x) +..., \u2212 (95.51) \u039e\u2020(x, \u03b8, \u03b8\u2217) = B\u2020(x) + \u03b8\u2217\u03be\u2020(x) + \u03b8\u2217\u03b8\u2217G\u2020(x) +i(\u03b8\u03c3\u00b5\u03b8\u2217)\u2202\u00b5B\u2020(x) +.... (95.52) If we set B = 1 2 (b + ia), where a and b are real scalar \ufb01elds, we \ufb01nd i(\u039e\u2020 \u2212 \u039e) = a \u2212 i\u03b8\u03be + i\u03b8\u2217\u03be\u2020 \u2212 i\u03b8\u03b8G + i\u03b8\u2217\u03b8\u2217G\u2020 \u2212 (\u03b8\u03c3\u00b5\u03b8\u2217)\u2202\u00b5b +.... (95.53) From eq. (95.39), we see that the supergauge transformation of eq. (95.50) results in C \u03c7 C + a, \u03c7 i\u03be, \u2212 \u2192 \u2192 iG, \u2202\u00b5b. M \u2192 v\u00b5 \u2192 M \u2212 v\u00b5 \u2212 The last of these is the usual abelian gauge transformation. The \ufb01rst three allow us to gauge away the C, \u03c7, and M components of a vector super\ufb01eld. That is, we can make a supergauge transformation with a = i\u03c7 iM ; in this gauge, known as Wess-Zumino gauge, the vector and G = super\ufb01eld becomes (95.54) C, \u03be = \u2212 \u2212 \u2212 V = (\u03b8\u03c3\u00b5\u03b8\u2217)v\u00b5 + \u03b8\u03b8\u03b8\u2217\u03bb\u2020 + \u03b8\u2217\u03b8\u2217\u03b8\u03bb + 1 2 \u03b8\u03b8\u03b8\u2217\u03b8\u2217D. (95.55) Note that we still have the freedom to make the supergauge transformation of eq. (95.50) with B(x) = 1 2 b(x), and that this still implements the ordinary abelian gauge transformation of eq. (95.54). Now consider a left-handed chiral super\ufb01eld \u03a6 that has charge +1 under a U(1) gauge group. We take the kinetic term for \u03a6 to be where g is the gauge coupling. The vector super\ufb01eld \u03a6\u2020e\u2212 invariant under the supergauge transformation Lkin = \u03a6\u2020e\u2212 2gV \u03a6, (95.56) D \u03a6 \u03a6\u2020 V \u2192 \u2192 \u2192 e\u2212 2ig\u039e \u03a6, \u03a6\u2020e+2ig\u039e\u2020, V + i(\u039e\u2020 \u039e). \u2212 2gV \u03a6 is clearly (95.57) (95.58) (95.59", ") 95: Supersymmetry 597 Let us evaluate eq. (95.56) in Wess-Zumino gauge, where we have 1 2 \u03b8\u03b8\u03b8\u2217\u03b8\u2217v\u00b5v\u00b5, V 2 = \u2212 V 3 = 0. The exponential factor in eq. (95.56) becomes e\u2212 2gV = 1 2g(\u03b8\u03c3\u00b5\u03b8\u2217)v\u00b5 \u2212 \u2212 \u03b8\u03b8\u03b8\u2217\u03b8\u2217(gD + g2v\u00b5v\u00b5). 2g\u03b8\u03b8\u03b8\u2217\u03bb\u2020 2g\u03b8\u2217\u03b8\u2217\u03b8\u03bb \u2212 \u2212 The relevant terms in \u03a6\u2020\u03a6 are \u03a6\u2020\u03a6 = A\u2020A + \u221a2\u03b8\u2217\u03c8\u2020A + \u221a2\u03b8\u03c8A\u2020 + (\u03b8\u03c3\u00b5\u03b8\u2217)(\u03c8\u2020 \u00af\u03c3\u00b5\u03c8 +... + \u03b8\u03b8\u03b8\u2217\u03b8\u2217(\u03a6\u2020\u03a6)D, \u2212 iA\u2020\u2202\u00b5A + iA\u2202\u00b5A\u2020) (95.60) (95.61) (95.62) (95.63) where we used 2(\u03b8\u2217\u03c8\u2020)(\u03b8\u03c8) = (\u03b8\u03c3\u00b5\u03b8\u2217)(\u03c8\u2020 \u00af\u03c3\u00b5\u03c8) to get the \ufb01rst term in the second line, and (\u03a6\u2020\u03a6)D is given by eq. (95.42). Combining eqs. (95.62) and (95.63), taking the D term, and performing the same integrations by parts that led to eq. (95.43), we \ufb01nd \u03a6\u2020e\u2212 2gV \u03a6 = D (D\u00b5A)\u2020D\u00b5A + i\u03c8\u2020 \u00af\u03c3\u00b5D\u00b5\u03c8 + F \u2020F \u2212 + \u221a2g\u03c8\u2020\u03bb\u2020A + \u221a2gA\u2020\u03bb\u03c8 gA\u2020DA, \u2212 (95.64) where D\u00b5 = \u2202\u00b5 \u2212 of charge +1. igv\u00b5 is the usual gauge covariant derivative acting on \ufb01eld We still need a kinetic term for the vector super\ufb01eld. To get it, we \ufb01rst introduce a super\ufb01eld that carries a left-handed spinor index, \u02d9a 1 4 D \u2217 \u2217\u02d9aD Wa \u2261 DaV. Since the two components of D\u2217\u02d9aD\u2217\u02d9b D\u2217\u02d9c = 0. D\u2217 anticommute, we have Thus Wa obeys D\u2217\u02d9aWa = 0, and is therefore a left-handed chiral super\ufb01eld. Furthermore, Wa is invariant under the supergauge transformation of eq. (95.50). To see this, we \ufb01rst note that Da annihilates \u039e\u2020 (because (95.65) \u02d9a D\u2217\u02d9aD\u2217 Da does). Then, we use eq. (95.19) to write \u2217\u02d9aDa + 2i\u03c3\u00b5 a \u02d9a\u2202\u00b5) Da = Da also annihilates \u039e (because D\u2217 \u02d9a D\u2217\u02d9aD\u2217 ( D \u2217\u02d9aD \u2212 D \u02d9a \u2217 D Thus, invariant under eq. (95.50). \u02d9a does). Therefore, Wa is \u02d9a. \u2217 (95.66) Since Wa is a left-handed chiral super\ufb01eld, it has an expansion in the form of eq. (95.25). To \ufb01nd the component \ufb01elds of Wa, we set x = y + i\u03b8\u03c3\u00b5\u03b8\u2217 in eq. (95.55), and expand in \u03b8 and \u03b8\u2217. The result is V = (\u03b8\u03c3\u00b5\u03b8\u2217)v\u00b5 + \u03b8\u03b8\u03b8\u2217\u03bb\u2020 + \u03b8\u2217\u03b8\u2217\u03b8\u03bb + 1 2 \u03b8\u03b8\u03b8\u2217\u03b8\u2217(D i\u2202\u00b5v\u00b5), \u2212 (95.67) 95: Supersymmetry 598 where all component \ufb01elds have spacetime argument y. From eq. (95.24), \u2202\u2217\u02d9a when it acts on a function of y, \u03b8, and \u03b8\u2217. We also we see", " that have D\u2217\u02d9a = \u2212 Da = ( Da\u03b8c)\u2202c + ( Da\u03b8\u2217 \u02d9c)\u2202\u2217\u02d9c + ( Day\u00b5)\u2202\u00b5 = \u2202a + 0 2i\u03c3\u00b5 a \u02d9a\u03b8\u2217 \u02d9a\u2202\u00b5, (95.68) \u2212 where \u2202\u00b5 is with respect to y. Using eq. (95.68), we \ufb01nd DaV = \u03b8\u2217\u03b8\u2217 \u03bba + \u03b8a(D h +.... i\u2202 v) \u00b7 \u2212 \u2212 i(\u03c3\u00b5 \u00af\u03c3\u03bd\u03b8)a\u2202\u00b5v\u03bd + i\u03b8\u03b8(\u03c3\u00b5\u2202\u00b5\u03bb\u2020)a i (95.69) When we act on DaV with of \u03b8\u2217\u03b8\u2217 survives. Since \u2202\u2217\u02d9a\u2202\u2217 \u02d9a = \u2202\u2217\u02d9a\u2202\u2217 D\u2217\u02d9aD\u2217 \u02d9a(\u03b8\u2217\u03b8\u2217) = 4, we \ufb01nd \u02d9a to get Wa, only the coe\ufb03cient Wa = \u03bba + \u03b8a(D i\u2202 v) \u00b7 \u2212 \u2212 i(\u03c3\u00b5 \u00af\u03c3\u03bd \u03b8)a\u2202\u00b5v\u03bd + i\u03b8\u03b8(\u03c3\u00b5\u2202\u00b5\u03bb\u2020)a. (95.70) We can simplify eq. (95.70) by using the identity (\u03c3\u00b5 \u00af\u03c3\u03bd)a b = g\u00b5\u03bd \u03b4a c \u2212 \u2212 2i(S\u00b5\u03bd L )a b. (95.71) Remembering that S\u00b5\u03bd strength L is antisymmetric on \u00b5 \u2194 \u03bd, and de\ufb01ning the \ufb01eld F\u00b5\u03bd \u2261 \u2202\u00b5v\u03bd \u2212 \u2202\u03bd v\u00b5, we get Wa = \u03bba + \u03b8aD (S\u00b5\u03bd L )a \u2212 c\u03b8cF\u00b5\u03bd + i\u03b8\u03b8\u03c3\u00b5 a \u02d9a\u2202\u00b5\u03bb\u2020 \u02d9a. (95.72) (95.73) We see that Wa involves the vector \ufb01eld v\u00b5 only through its gauge-invariant \ufb01eld strength F\u00b5\u03bd. Since Wa is supergauge invariant, this is to be expected. Next, consider the F term of W aWa. This term is Lorentz invariant, and its spacetime integral (like the spacetime integral of any F term) is invariant under supersymmetry, and hence could be a term in the action of a supersymmetric theory. Working out the components, we \ufb01nd W aWa = 2i\u03bba\u03c3\u00b5 \u02d9a a \u02d9a\u2202\u00b5\u03bb\u2020 1 2 Tr(S\u00b5\u03bd L S\u03c1\u03c3 L )F\u00b5\u03bd F\u03c1\u03c3 + D2, (95.74) \u2212 F L ) = (S\u00b5\u03bd L S\u03c1\u03c3 where Tr(S\u00b5\u03bd c(S\u03c1\u03c3 form, we used the fact that (S\u00b5\u03bd the identity L )a L )c L )ac is symmetric on a a. To get the spin matrices into this c. Now we use \u2194 Tr(S\u00b5\u03bd L S\u03c1\u03c3 L ) = 1 2 (g\u00b5\u03c1g\u03bd\u03c3 g\u00b5\u03c3g\u03bd\u03c1) 1 2 i\u03b5\u00b5\u03bd\u03c1\u03c3 \u2212 \u2212 (95.75) to get W aWa F = 2i\u03bba\u03c3\u00b5 \u02d9a a \u02d9a\u2202\u00b5\u03bb\u2020 1 2 F \u00b5\u03bdF\u00b5\u03bd \u2212 \u2212 1 2 i \u02dcF \u00b5\u03bdF\u00b5\u03bd + D2, (95.76) 95: Supersymmetry 599 where \u02dcF \u00b5\u03bd = 1 vector super\ufb01eld as 2 \u03b5\u00b5\u03bd\u03c1\u03c3F\u03c1\u03c3. We can now identify the kinetic term for the Lkin = 1 4 W aWa F + h.c. = i\u03bb\u2020\u00af\u03c3\u00b5\u2202\u00b5\u03bb 1 4 F \u00b5\u03bdF\u00b5\u03bd + 1 2 D2. \u2212 (95.77) We integrated by parts and dropped total divergences to get the second line. We see that we have the standard kinetic terms for the gauge \ufb01eld v\u00b5 and the gaugino \ufb01eld \u03bb, while D is an auxiliary \ufb01eld. All of this generalizes in a straightforward", " way to the nonabelian case. We de\ufb01ne a matrix-valued vector super\ufb01eld V = V aT a R, and a matrixvalued chiral super\ufb01eld \u039e = \u039eaT a R, where a is an adjoint group index. A chiral super\ufb01eld \u03a6 in the representation R still transforms according to eqs. (95.57) and (95.58), but for the vector \ufb01eld we have 2gV e\u2212 \u2192 2ig\u039e\u2020 e\u2212 e\u2212 2gV e+2ig\u039e ; this reduces to eq. (95.59) in the abelian case. The \ufb01eld-strength super\ufb01eld is now \u2217\u02d9aD Under a supergauge transformation, \u2212 Wa = 1 8g D \u02d9ae+2gV \u2217 2gV. Dae\u2212 Wa \u2192 e\u2212 2ig\u039e Wa e+2ig\u039e. (95.78) (95.79) (95.80) Eq. (95.73) still holds, but the derivative that acts on \u03bb\u2020 is now the gauge covariant derivative for the adjoint representation, and F \u00b5\u03bd now includes the usual nonabelian commutator term. These changes also apply to eq. (95.77), where we must also trace over the group indices and (for proper normalization) divide by the index T (R). Reference Notes Introductions to supersymmetry can be found in Martin, Siegel, Weinberg III, and Wess & Bagger. Problems 95.1) Use eq. (95.6) to show that the hamiltonian is positive semide\ufb01nite, and that a state with zero energy must be annihilated by all the supercharges. 95: Supersymmetry 95.2) Supersymmetry is spontaneously broken if the ground state annihilated by all the supercharges. 600 is not 0 i | = 0. 0 i 6 | a) Use the \ufb01rst of eqs. (95.34) to show that supersymmetry is spontaneously broken if F 0 | h. Use the result to show that supersymmetry is \u03bba, Qb} { b) Compute spontaneously broken if 0 i 6 | 95.3) Consider a supersymmetric theory with three chiral super\ufb01elds A, B, v2), where m and and C, and a superpotential W = mBC + \u03baA(C 2 v are parameters with dimensions of mass, and \u03ba is a dimensionless coupling constant. This is the O\u2019Raifeartaigh model. D 0 | h = 0. \u2212 a) Show that one or more F components is nonzero at the minimum of the potential, and hence that supersymmetry is spontaneously broken. b) Show that the potential is minimized along a line in \ufb01eld space, and \ufb01nd the masses of the particles at an arbitrary point on this line. You should \ufb01nd that there is a massless Goldstone fermion or goldstino that is related by supersymmetry to the linear combination of F \ufb01elds that gets a nonzero vacuum expectation value. 95.4) Supersymmetric Quantum Electrodynamics. Consider a supersymmetric U(1) gauge theory with chiral super\ufb01elds \u03a6 and \u03a6 with charges 1, respectively. Here the bar over the \u03a6 in the \ufb01eld \u03a6 is part +1 and of the name of the \ufb01eld, and does not denote any sort of conjugation. We include a gauge invariant superpotential W = m\u03a6\u03a6. \u2212 a) Work out the lagrangian in terms of the component \ufb01elds. b) Eliminate the auxilliary \ufb01elds F, F, and D. 95.5) In a supersymmetric gauge theory with a U(1) factor, we can add a Fayet-Illiopoulos term LFI = e\u03beD to the lagrangian, where D is the auxilliary \ufb01eld for the U(1) gauge \ufb01eld, and \u03be is a parameter with dimensions of mass-squared. a) Explain why adding this term preserves supersymmetry. Explain why the corresponding gauge \ufb01eld cannot be nonabelian. b) Add this term to the SQED lagrangian that you found in problem 94.4, and eliminate the auxilliary \ufb01elds. c) Minimize the resulting potential. Show that supersymmetry is spontaneously broken if \u03be is in a certain range. 95.6)", " R symmetry. Given a supersymmetric gauge theory (abelian or nonabelian), consider a global U(1) transformation that changes the 95: Supersymmetry 601 phase of the gaugino \ufb01elds, \u03bba \u2192 have R charge +1. e\u2212 i\u03b1\u03bba; we say that the gauginos a) If R symmetry is to be a good symmetry of the lagrangian, what relation must hold between the R charges of the scalar and fermion components of a chiral super\ufb01eld that couples to the gauge \ufb01elds? b) In additon, what conditions must be placed on the superpotential? c) Identify the R symmetry, if any, of supersymmetric quantum electrodynamics. 96: The Minimal Supersymmetric Standard Model 602 96 The Minimal Supersymmetric Standard Model Prerequisite: 89, 95 Having seen how to construct a general supersymmetric gauge theory in section 95, we can now write down a supersymmetric version of the Standard Model. \u00d7 To do so, we introduce vector super\ufb01elds for the gauge group SU(3) \u00d7 U(1), which include the usual gauge bosons along with their ferSU(2) I, and \u00afD\u03b1 mionic partners, the gauginos; chiral super\ufb01elds LiI, \u00afEI, Q\u03b1iI, \u00afU \u03b1 I that include three generations of quarks and leptons along with their scalar partners, the squarks and sleptons; and chiral super\ufb01elds Hi and \u00afHi in the 2 ), which include two copies of the representations (1, 2, usual Higgs \ufb01eld along with their fermionic partners, the higgsinos. We give all these \ufb01elds supersymmetric, gauge invariant kinetic terms, and a superpotential 2 ) and (1, 2, + 1 \u2212 1 W = \u2212 yIJ \u03b5ijHiLjI \u00afEJ \u2212 y\u2032IJHiQ\u03b1jI \u00afD\u03b1 J \u2212 y\u2032\u2032IJ \u00afH iQ\u03b1iI \u00afU \u03b1 J \u2212 \u00b5\u03b5ij \u00afHiHj, (96.1) where \u00b5 is a mass parameter. This superpotential generates the usual Yukawa couplings, among others, and gives a positive mass-squared to both Higgs \ufb01elds. We forbid terms of the form \u00afH iLiI (which are allowed by the gauge symmetry) by invoking a discrete symmetry, R parity. Under R parity, all Standard Model \ufb01elds (including both Higgs scalars) are taken to be even, and all superpartner \ufb01elds (gauginos, higgsinos, squarks, sleptons) are taken to be odd. Supersymmetry is clearly not an exact symmetry of the real world, and so if present must be spontaneously broken. Correct phenomenology requires supersymmetry breaking to be triggered by \ufb01elds other than those listed above. There are many possibilities for the dynamics of the \ufb01elds in this hidden sector, and an exploration of them is beyond our scope. However, we can parameterize the e\ufb00ects of the hidden sector via a spurion \ufb01eld. This is a constant chiral super\ufb01eld of the form S = m2 S \u03b8\u03b8, (96.2) where mS is the supersymmetry breaking scale. We couple S to the quark, lepton, and Higgs super\ufb01elds via D terms of the form Lspur, D = m\u2212 2 M S\u2020S C (H)H \u2020H + C ( \u00afH) \u00afH \u2020 \u00afH IJ L\u2020ILJ + C ( \u00afE) C (L) + + C ( \u00afU) \u00afE\u2020I \u00afUJ + C ( \u00afD) IJ h X \u00afU \u2020I IJ IJ IJ \u00afEJ + C (Q) IJ Q\u2020IQJ \u00afD\u2020I \u00afDJ. (96.3) D i 96: The Minimal Supersymmetric Standard Model 603 Here we have suppressed all but generation indices; each C (\u03a6) is a dimensionless hermitian matrix in generation space. The parameter mM is the messenger scale. Eq. (96.3) gives masses of order m2 S/mM to the scalars. IJ We also couple S to the chiral gauge super\ufb01elds W (i) a via F terms of the form Lspur, gauge = m\u2212 1 M S Xi=1,2,3 C (i)W (i)a", "W (i) a + h.c., (96.4) F where the sum is over the three gauge-group factors; a is a spinor index. Eq. (96.4) gives masses of order m2 S/mM to the gauginos. Finally, we couple S to the chiral super\ufb01elds via F terms of the form Lspur, F = m\u2212 1 M S 28 CAWA + h.c., (96.5) XA=1 where WA is one of the 28 gauge-invariant terms in the superpotential. In particular, eq. (96.5) includes a mass term of the form \u00afHH + h.c. for the Higgs scalars. F Eqs. (96.1) and (96.3\u201396.5) specify the Minimal Supersymmetric Standard Model, or MSSM for short. Obviously, it has a complicated phenomenology that is beyond our scope to explore in detail. One point worth noting is that R parity implies that the lightest superpartner, or LSP, is absolutely stable. Reference Notes Supersymmetric versions of the Standard Model are discussed in Martin, Ramond II, and Weinberg III. Problems 96.1) Explain why we need two Higgs doublets. 96.2) a) Write the mass terms for the Higgs scalars as LHiggs mass = \u2212 m2 1H \u2020H m2 2 \u00afH \u2020 \u00afH \u2212 \u2212 m2 3\u03b5ij( \u00afHiHj + h.c.), (96.6) and compute the quartic terms in H and \u00afH that arise from eliminating the auxilliary SU(2) and U(1) D \ufb01elds. b) Find the conditions on the mass parameters in eq. (96.6) in order for the potential to be bounded below. c) Find the conditions on the mass parameters in eq. (96.6) in order to have spontaneous breaking of the SU(2) U(1) symmetry. \u00d7 96: The Minimal Supersymmetric Standard Model 604 d) Show that there are \ufb01ve physical Higgs particles, two charged and three neutral. e) Let tan \u03b2 \u2261 \u00afv/v be the ratio of the two Higgs VEVs. Show that tan \u03b2 + cot \u03b2 = (m2 1 + m2 2)/m2 3. (96.7) 97: Grand Uni\ufb01cation 605 97 Grand Unification Prerequisite: 89 The Standard Model is based on the gauge group SU(3) \u00d7 with left-handed Weyl \ufb01elds in three copies of the representation (1, 2, \u2295 (\u00af3, 1, 2 (1, 1, +1) 3 ), and a complex scalar \ufb01eld 3 ) \u2212 1 2 ). The lagrangian includes all terms of mass in the representation (1, 2, dimension four or less that are allowed by the gauge symmetries and Lorentz invariance. (3, 2, + 1 6 ) U(1), 1 2 ) (\u00af3, 1, + 1 SU(2) \u2295 \u2212 \u00d7 \u2212 \u2295 \u2295 To complete the speci\ufb01cation of the Standard Model, we need twenty real numbers: the three gauge couplings; the three diagonal entries of each of the three diagonalized Yukawa coupling matrices (for the up quarks, the down quarks, and the charged leptons); the four angles in the CKM mixing matrix for the quarks; the vacuum angles for the SU(3) and SU(2) gauge groups; the scalar quartic coupling; and the scalar mass-squared. The goal of grand uni\ufb01cation is to construct a more compact model with fewer parameters by supposing that the Standard Model is the result of the spontaneous breaking of a larger gauge symmetry. The simplest model along these lines is the Georgi\u2013Glashow SU(5) model. Its starting point is the grand uni\ufb01ed gauge group SU(5). We include a scalar \ufb01eld \u03a6 = \u03a6aT a in the adjoint or 24 representation, and assume that the scalar potential for this \ufb01eld results in a vacuum expectation value 2 )V. As we saw in (VEV) of the form \u2212 section 84, this VEV spontaneously breaks the gauge symmetry down to U(1). The generator of the unbroken U(1) subgroup is SU(3) \u00d7 1 2 ), where c2 = 3/5. It will prove convenient T 24 = c diag( 3, to write = diag SU(2 24 = 3 5 Y, (97.1) q and express the U", "(1) charge as the value of Y rather than the value of T 24. We also note that the SU(5) breaking scale V must be considerably larger 250 GeV in order to suppress the than the SU(2) observable e\ufb00ects of the extra gauge \ufb01elds. U(1) breaking scale v \u00d7 \u223c Under the SU(3) SU(2) fundamental representations of SU(5) transform as \u00d7 \u00d7 U(1) subgroup, the fundamental and anti- 5 5 \u2192 \u2192 (3, 1, 1 3 ) \u2212 (\u00af3, 1, + 1 3 ) \u2295 \u2295 (1, 2, + 1 (1, 2, \u2212 2 ). 1 2 ). Next we use eq. (97.2) to \ufb01nd that the product 5 5 transforms as 5 5 \u2297 \u2192 (6, 1, \u2212 (\u00af3, 1, 2 3 )S \u2295 \u2212 2 3 )A \u2295 (3, 2, + 1 6 )S \u2295 (3, 2, + 1 \u2295 \u2297 (\u00af3, 1, 6 )A \u2295 2 3 )S (1, 3, +1)A, \u2212 (97.2) (97.3) (97.4) 97: Grand Uni\ufb01cation 606 where the subscripts indicate the symmetric and antisymmetric parts of the product. In terms of SU(5), we have 5 = 15 S \u2295 Comparing eqs. (97.4) and (97.5), we \ufb01nd \u2297 5 10A. 10 (\u00af3, 1, 2 3 ) \u2212 \u2295 \u2192 (3, 2, + 1 6 ) \u2295 (1, 3, +1). (97.5) (97.6) From eqs. (97.2) and (97.6), we see that one generation of quark and lepton \ufb01elds \ufb01ts exactly into the representation 5 10 of SU(5). We therefore de\ufb01ne a left-handed Weyl \ufb01eld \u03c8i in the 5 representation \u03c7ji in the 10 representation. of SU(5), and a left-handed Weyl \ufb01eld \u03c7ij = The gauge covariant derivatives of these \ufb01elds are \u2212 \u2295 (D\u00b5\u03c8)i = \u2202\u00b5\u03c8i ig5Aa \u2212 = \u2202\u00b5\u03c8i + ig5Aa \u00b5(T a 5 )i \u00b5(T a)j j\u03c8j i\u03c8j, (D\u00b5\u03c7)ij = \u2202\u00b5\u03c7ij \u2212 = \u2202\u00b5\u03c7ij \u2212 ig5Aa ig5Aa \u00b5(T a 10)ij \u00b5[(T a)i kl\u03c7kl k\u03c7kj + (T a)j (97.7) l\u03c7il], (97.8) where g5 is the SU(5) gauge coupling and T a is the generator matrix in the fundamental representation. The kinetic terms for these \ufb01elds are Lkin = i\u03c8\u2020i \u00af\u03c3\u00b5(D\u00b5\u03c8)i + 1 2 i\u03c7\u2020 ij \u00af\u03c3\u00b5(D\u00b5\u03c7)ij, (97.9) where the implicit sum over i and j is unrestricted; this necessitates the prefactor of one-half in the second term to avoid double counting. The interaction terms with the gauge \ufb01elds then work out to be \u2212 g5 \u00b5)i j \u00af\u03c3\u00b5\u03c8j + \u03c7\u2020 \u03c8\u2020i (AT h Lint = \u00b5T a is the matrix-valued gauge \ufb01eld, and AT where A\u00b5 = Aa \u00b5 is its transpose. Note that we have written the factors in matrix-multiplication order (with a trace for the second term). (97.10) k \u00af\u03c3\u00b5\u03c7kj ji(A\u00b5)i i, We can identify the components of \u03c8i and \u03c7ij as \u03c8i = ( \u00afd r 0 \u00afug \u2212 \u03c7ij = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed \u00afd b \u00afug 0 \u00afd g \u2212 \u00afub \u00afur 0 \u00afub \u2212 ur \u2212 dr \u2212 \u2212 \u2212 ur ub \u2212 db \u2212 ug 0 dg \u2212 \u00afe e ur ub ug \u03bd ), \u2212 (97.11) dr db \uf8f6 dg \u00af97.12) 97: Grand Uni\ufb01cation 607 where r,", " b, and g stand for the three colors (red, blue, and green). We can also write the gauge \ufb01elds as Gr 1 3 cB r r \u2212 Gb \uf8eb Gb b b Gr Gr g g Gb g 1 3 cB b \u2212 Gg 1 X 1 r \u221a2 1 X 1 b \u221a2 1 X 1 g \u221a2 2 W 32 1 \u221a2 1 \u221a2 1 W + \u221a2 2 W 3+ 1 1 r 1 Gg Gg 2 cB cB \u2212 g X \u2020 1 AaT a = 1 \u221a2 1 \u221a2 1 \u221a2 1 \u221a2 1 \u221a2 1 \u221a cB \uf8f7 \uf8f7 \uf8f8 (97.13) where the Lorentz index has been omitted. Here B is the hypercharge iW 2 are the SU(2) gauge \ufb01elds. gauge \ufb01eld, and W 3 and \u221a2 W \u00b1 = W 1 g = 0. The The gluon \ufb01elds Gi X i \u03b1 \ufb01elds correspond to the broken generators of SU(5), and hence become massive; here i is an SU(2) index and \u03b1 is an SU(3) index. As we saw in section 84, the X i U(1), and their mass is MX = 5 6\u221a2 \u03b1 \ufb01elds transform as (3, 2, g5V. j are subject to the constraint Gr 5 6 ) under SU(3) r + Gb b + Gg SU(2) 1 \u221a2 \u2212 \u00d7 \u00b1 \u2212 \u00d7 If we substitute eqs. (97.11\u201397.13) into eq. (97.10), we \ufb01nd the usual U(1) gauge \ufb01elds with the quarks and SU(2) interactions of the SU(3) leptons, but with \u00d7 \u00d7 5 3 g1 = g2 = g3 = g5. (97.14) q These relations among the gauge couplings hold in the MS renormalization scheme; later we will discuss a modi\ufb01ed scheme that is more appropriate at energies well below MX. We also \ufb01nd the couplings of the X \ufb01eld to quarks and leptons; these work out to be LX,int = \u2212 1 \u221a2 g5 \u03b1 1\u00b5 ( \u00afd \u2020\u03b1 \u00af\u03c3\u00b5e X \u2020 h \u03b1 2\u00b5 ( + X \u2020 \u00afe\u2020 \u00af\u03c3\u00b5d\u03b1 + u\u2020 \u00afd \u2020\u03b1 \u00af\u03c3\u00b5\u03bd + \u00afe\u2020 \u00af\u03c3\u00b5u\u03b1 + d\u2020 \u2212 \u2212 \u03b2 \u00af\u03c3\u00b5 \u00afu\u03b3\u03b5\u03b1\u03b2\u03b3) \u03b2 \u00af\u03c3\u00b5 \u00afu\u03b3\u03b5\u03b1\u03b2\u03b3) i + h.c. \u03b5ij \u00afe\u2020\u00af\u03c3\u00b5qj\u03b1 + q\u2020 \u03b2i\u00af\u03c3\u00b5 \u00afu\u03b3\u03b5\u03b1\u03b2\u03b3) + h.c. = \u2212 \u2261 \u2212 1 \u221a2 1 \u221a2 \u03b1 g5X \u2020 i\u00b5 (\u03b5ij \u00afd \u2020\u03b1\u00af\u03c3\u00b5\u2113j \u2212 \u03b1 i\u00b5 J i\u00b5 \u03b1 + h.c., g5X \u2020 (97.15) where the last line de\ufb01nes the current J \u00b5 that couples to the X\u00b5 \ufb01eld. (To include more than one generation, we add a generation index I to each quark and lepton \ufb01eld, and sum over it.) The most interesting feature of eq. (97.15) is that the \ufb01rst two terms in the current have baryon number B = + 1 2 3 and L = 0. Thus exchange of an X boson can violate baryon and lepton 3 and lepton number L = +1, while the third term has B = \u2212 97: Grand Uni\ufb01cation 608 number conservation, leading to phenomena such as proton decay. Proton e+\u03c00 is decay has not been observed; the limit on the rate 1/\u03c4 for p p/8\u03c0M 4 \u03c4 > 1033 yr. A rough estimate of 1/\u03c4 from eq. (97.15) is g4 X. Taking g5 \u223c \u00d7 1 2 ). The smallest complete representation of SU(5) that includes this piece is the 5. Call the corresponding \ufb01eld H i; we can identify its components as We still need a scalar \ufb01eld in the representation (1, 2, 0.6, we \ufb01nd that we must have MX > 3 \u2192 5m5 1015 GeV. g2 \u223c \u2212 H i = ( \u03c6r", " \u03c6b \u03c6g \u03d5\u2212 \u2212 \u03d50 ). (97.16) The possible gauge-invariant Yukawa couplings with the \u03c8i and \u03c7ij \ufb01elds are LYuk = yH i\u03c8j\u03c7ij \u2212 \u2212 1 8 y\u2032\u2032\u03b5ijklmH \u2020i \u03c7jk\u03c7lm + h.c. ; (97.17) with three generations, y and y\u2032\u2032 become matrices in generation space. We can write out LYuk using eqs. (97.11), (97.12), and (97.16); the result is LYuk = \u2212 \u2212 y\u03b5ij\u03d5i\u2113j \u00afe y\u03b5\u03b1\u03b2\u03b3\u03c6\u03b1 \u00afd\u03b2 \u00afu\u03b3 \u2212 y\u03b5ij\u03d5iq\u03b1j \u00afd\u03b1 y\u2032\u2032\u03d5\u2020 \u2212 y\u03b5ij\u03c6\u03b1q\u03b1i\u2113j \u2212 \u2212 iq\u03b1i \u00afu\u03b1 y\u2032\u2032\u03c6\u2020\u03b1 \u00afu\u03b1\u00afe + h.c.. (97.18) The terms on the \ufb01rst line are those of the Standard Model, except that the down-quark Yukawa coupling matrix (called y\u2032 in section 89) is the same as the charged-lepton Yukawa coupling matrix (called y in section 88). Since the quark and lepton masses are directly proportional to the Yukawa couplings, eq. (97.18) predicts mb = m\u03c4, ms = m\u00b5, md = me. (97.19) These relations hold in the MS renormalization scheme; later we will discuss a modi\ufb01ed scheme that is more appropriate at energies well below MX. The terms on the second line of eq. (97.18) are the couplings of the colored scalar \ufb01eld \u03c6 to the quarks and leptons; we see that these couplings, like those of the X\u00b5 \ufb01eld, violate baryon and lepton number conservation. Since \ufb01rstgeneration Yukawa couplings are smaller than gauge couplings by a factor of 105, the limit on M\u03c6 from proton decay is roughly M\u03c6 > 1010 GeV. To compute M\u03c6, we need the complete scalar potential. For simplic\u03a6. Then the most general ity, we assume a Z2 symmetry under \u03a6 renormalizable potential is \u2194 \u2212 V (\u03a6, H) = 1 2 m2 \u2212 + m2 \u03a6 Tr \u03a62 + 1 HH \u2020H + 1 \u03a6, m2 4 \u03bb1Tr \u03a64 + 1 4 \u03bb2(Tr \u03a62)2 2 \u03ba2H \u2020\u03a62H. 4 \u03ba1(H \u2020H)2 We take all the parameters (m2 H, \u03bb1, \u03bb2, \u03ba1, \u03ba2) to be positive. We found in problem 84.1 that in this case the \ufb01rst line of eq. (97.20) is mini2, + 1 \u03a6/(7\u03bb1 + 30\u03bb2). mized by \u03a6 = diag( 2 )V, with V 2 = 36m2 3, + 1 (97.20 \u2212 \u2212 \u2212 97: Grand Uni\ufb01cation 609 1 1 M 2 (1, 2, H \u2212 (97.21) (\u00af3, 1, + 1 \u03c6 = m2 \u223c \u2212 m2 \u03d5 = m2 From the \ufb01rst and third terms on the second line of eq. (97.20), we \ufb01nd that 1 3 ) scalar \ufb01elds are 2 ) and \u03c6 the masses-squared of the \u03d5 \u223c 8 \u03ba2V 2, 18 \u03ba2V 2. H \u2212 (100 GeV)2 and M 2 \u03c6 > +(1010 GeV)2. This requires We want m2 8 \u03ba2V 2 to at least sixteen signi\ufb01cant digits (but not m2 exactly). There is no obvious reason for the parameters in eq. (97.20) to satisfy this odd relation; we would more naturally expect m2 \u03c6 to g2 5 and have the same order of magnitude (and furthermore to have \u03ba2 \u223c MX > 1015 GeV). This is the \ufb01ne tuning problem of grand hence M\u03c6 \u223c uni\ufb01ed theories. More generally, we can ask why the breaking scale of the grand uni\ufb01ed group is so much larger than the breaking scale of the electroweak subgroup; this is gauge hierarchy problem. H to be equal to 1 \u03d5 and M 2 \u03d5 \u223c \u2212 (97", ".22) Since the X\u00b5 and \u03c6 \ufb01elds are so heavy, we can integrate them out, generating e\ufb00ective interactions among the quarks and leptons.1 For the X\u00b5 \ufb01eld, we get the leading term (in a double expansion in powers of g5 and inverse powers of MX) by ignoring the kinetic energy and other interactions of the X\u00b5 \ufb01eld, solving the equations of motion for X\u00b5 that follow from LX,mass + LX,int, where LX,int is given by eq. (97.15) and \u03b1 i\u00b5 X i\u00b5 M 2 \u03b1, XX \u2020 LX,mass = \u2212 (97.23) LX,int. This is and \ufb01nally substituting the solutions back into equivalent to evaluating tree-level Feynman diagrams with a single X exchanged. The result is LX,mass + LX,e\ufb00 = 1 2M 2 X \u03b1 i\u00b5 J i\u00b5 \u03b1. J \u2020 (97.24) Keeping only \ufb01elds from the \ufb01rst generation, we \ufb01nd that the baryon- and LX,e\ufb00 are lepton-number violating terms in \u00afe\u2020\u00af\u03c3\u00b5qi\u03b1)\u00afu\u2020\u03b2 \u00af\u03c3\u00b5qj\u03b3 + h.c. \u2206B =1 X,e\ufb00 = | | L = g2 5 2M 2 X g2 5 M 2 X \u2212 \u03b5ij\u03b5\u03b1\u03b2\u03b3( \u00afd \u2020\u03b1 \u00af\u03c3\u00b5\u2113i \u2212 \u03b5ij\u03b5\u03b1\u03b2\u03b3 (\u2113iqj\u03b3)( \u00afd \u2020\u03b1 \u00afu\u2020\u03b2) + (\u00afe\u2020 \u00afu\u2020\u03b3)(qi\u03b1qj\u03b2) i h + h.c., (97.25) where the second line follows from a Fierz identity and a relabeling of the color indices in the second term. We can treat the \u03c6 \ufb01eld similarly; 1We should also integrate out the heavy components of \u03a6, which transform as (8, 1, 0)\u2295 (1, 3, 0) \u2295 (1, 1, 0) under SU(3) \u00d7 SU(2) \u00d7 U(1), but these do not couple directly to quarks and leptons. 97: Grand Uni\ufb01cation 610 see problem 97.2. We will compute the decay rate for p eq. (97.25) in problem 97.4. \u2192 e+\u03c00 from Once the heavy \ufb01elds have been integrated out, we can apply the MS renormalization scheme to the theory with the remaining light \ufb01elds. This is not the same as MS for the original theory, because now only light \ufb01elds circulate in loops. (Loops of heavy \ufb01elds contribute to corrections to the e\ufb00ective interactions.) With the usual \ufb01elds of the Standard Model, we \ufb01nd from our results in sections 66 and 73 that the one-loop beta functions for the three gauge couplings are given by \u00b5 d d\u00b5 gi = with bi 16\u03c02 g3 i + O(g5 i ), b3 = \u2212 b2 = b1 = + 20 \u2212 22 11 + 97.26) (97.27) (97.28) (97.29) where n = 3 in the number of generations; the + 1 6 contributions to b2 (We and b1 are from the \u03d5 \ufb01eld. These formulae apply for \u00b5 < MX. are assuming that the heavy scalars do not have masses much less than MX.) For \u00b5 MX, we must restore the heavy \ufb01elds, and then eq. (97.14) applies. If we now neglect the higher-loop corrections, integrate eq. (97.26) for each coupling, impose eq. (97.14) at \u00b5 = MX, and set g2 = e/ sin \u03b8W and g1 = e/ cos \u03b8W, we \ufb01nd for \u00b5 < MX that \u2265 1 \u03b13(\u00b5) sin2 \u03b8W(\u00b5) \u03b1(\u00b5) cos2 \u03b8W(\u00b5) \u03b1(\u00b5) = = = 1 \u03b15(MX) 1 \u03b15(MX) 5/3 \u03b15(MX) + + + b3 2\u03c0 b2 2\u03c0 b1 2\u03c0 ln(MX/\u00b5), (97.30) ln(MX/\u00b5), (97.31) ln(MX/\u00b5). (97.32) The quantities on the left-hand sides are measured at \u00b5", " = MZ to be \u03b13(MZ) = 0.1187 0.0020. \u00b1 1/\u03b1(MZ) = 127.91 \u00b1 sin2 \u03b8W(MZ) = 0.23120 0.02, 0.00015. \u00b1 (97.33) (97.34) (97.35) 97: Grand Uni\ufb01cation 611 We can use eqs. (97.30\u201397.32) to solve for 1/\u03b15(MX) and ln(MX/MZ) in terms of the known parameters \u03b1(MZ) and \u03b13(MZ); the result is 1 \u03b15(MX) = ln(MX/MZ) = 1 b1+b2\u2212 2\u03c0 b1+b2\u2212 b3 \u2212 \u03b1(MZ) + 8 3 b3 1 8 3 b3 \u03b1(MZ) \u2212 b1+b2 \u03b13(MZ) 8/3 \u03b13(MZ),. (97.36) (97.37) Plugging in eqs. (97.27\u201397.29) and eqs. (97.33\u201397.34), we \ufb01nd 1/\u03b15(MX) = 1014 GeV. 41.5 and MX = 7 This value of MX is about an order of magnitude below the lower limit imposed by proton decay. 1014 GeV; two-loop corrections lower MX to 4 \u00d7 \u00d7 We can also use eqs. (97.30\u201397.32) to express sin2 \u03b8W(MZ) in terms of \u03b1(MZ) and \u03b13(MZ); the result is sin2 \u03b8W(MZ) = 1 b1+b2\u2212 b2\u2212 8 3 b3 b3 + (b1 5 3 b2) \u03b1(MZ) \u03b13(MZ). (97.38) \u2212 This is a prediction of the SU(5) model that we can test. Plugging in eqs. (97.27\u201397.29) and eqs. (97.33\u201397.34), we \ufb01nd sin2 \u03b8W(MZ) = 0.207. 0.001. Comparing with eq. (97.35), Two-loop corrections raise this to 0.210 we see that the SU(5) prediction is too low by about 10%. \u00b1 The situation improves considerably if we consider the Minimal SuIn this case the persymmetric Standard Model, discussed in section 96. beta-function coe\ufb03cients become b3 = \u2212 9 + 2n, 6 + 2n + 1, b2 = b1 = + 10 \u2212 3 n + 1. (97.39) (97.40) (97.41) Now we \ufb01nd sin2 \u03b8W(MZ) = 0.231, with two-loop corrections raising it to 0.234; there are, however, numerous sources of uncertainty related to the 1016 GeV; masses of the supersymmetric particles. We also \ufb01nd MX = 2 this result (which is not changed signi\ufb01cantly by two-loop corrections) is high enough to avoid too-rapid proton decay. \u00d7 Next, let us consider the predicted equality of the down quark and charged lepton masses, eq. (97.19). These relations are subject to renormalization; see problem 97.5. However, the one-loop corrections from gaugeboson exchange cancel in the predicted ratios me m\u00b5 = md ms, m\u00b5 m\u03c4 = ms mb. (97.42) 97: Grand Uni\ufb01cation 612 Alas, these predictions are not satis\ufb01ed, in the \ufb01rst case by an order of magnitude. Resolving this problem requires either a more complicated set of Higgs \ufb01elds, and/or including higher-dimension, nonrenormalizable terms in the lagrangian that are suppressed by inverse powers of the some 1019 GeV. mass scale larger than MX, such as the Planck mass MP = 1.2 To get neutrino masses in the SU(5) model (see section 91), we must add left-handed Weyl \ufb01elds \u00af\u03bdI that are singlets of SU(5), and couple them to I \u00af\u03bdJ. the neutrinos via \u02dcyIJH \u2020i \u03c8i \u00d7 \u00af5 A more elegant scheme starts with SO(10) as the grand uni\ufb01ed gauge group. SO(10) has a complex sixteen dimensional spinor representation that transforms as 1 10 under the SU(5) subgroup; thus, each generation of fermions (including \u00af\u03bd) \ufb01ts into a single", " Weyl \ufb01eld in this 16 representation. A scalar \ufb01eld in the 10 representation is needed for the Standard Model Higgs \ufb01eld, and additional scalars in higher dimensional representations (such as the 45 dimesional adjoint representation and the 16) are needed to break SO(10) down to SU(3) SU(2) U(1). \u2295 \u2295 LYuk = \u2212 A great variety of grand uni\ufb01ed models can be constructed, with and without supersymmetry. Which, if any, are relevant to the natural world is a question yet to be answered. \u00d7 \u00d7 Problems 97.1) Is the gauge symmetry of the SU(5) model anomalous? If it is, modify the model to turn it into a consistent quantum \ufb01eld theory. Prerequisite: section 75. Hint: see problem 70.4. 97.2) Compute the \u2206B = from \u03c6 exchange. \u00b1 1 terms in the e\ufb00ective lagrangian that arise 97.3) Let us write eq. (97.25) as ZC1C1O1 \u2212 Le\ufb00 = \u03b5ij\u03b5\u03b1\u03b2\u03b3(\u2113iqj\u03b3)( \u00afd \u2020\u03b1 \u00afu\u2020\u03b2) and \u2212 ZC2C2O2 + h.c., (97.43) O1 \u2261 O2 \u2261 \u03b5ij\u03b5\u03b1\u03b2\u03b3(\u00afe\u2020 \u00afu\u2020\u03b3)(qi\u03b1qj\u03b2), where ZC1 and ZC2 are renormalizing factors, and C1 and C2 are coe\ufb03cients that depend on the MS renormalization scale \u00b5. At \u00b5 = MX, we have C1(MX) = C2(MX) = 4\u03c0\u03b15(MX)/M 2 X. a) Working in Lorenz gauge, and using the results of problems 88.7 and 89.5, show that the one-loop contribution to ZC1 from gaugeboson exchange is given by Zm in spinor electrodynamics in Lorenz gauge, with 1)(+1)e2 ( \u2212 \u2192 0 + \u03b5\u03b1\u2032\u03b2\u2032\u03b3(T a h 3 )\u03b1\u2032 \u03b1(T a 3 )\u03b2\u2032 \u03b2/\u03b5\u03b1\u03b2\u03b3 g2 3 i 97: Grand Uni\ufb01cation + + Evaluate these coe\ufb03cients. \u03b5i\u2032j\u2032 h ( \u2212 h (T a i(T a 2 )i\u2032 2 )(+ 1 2 )j\u2032 6 ) + (+ 1 j/\u03b5ij + 0 i g2 1. 3 )( g2 2 1 \u2212 2 3 ) i 613 (97.44) b) Similarly, compute the one-loop contribution to ZC2 from gaugeboson exchange. c) Compute the corresponding anomalous dimensions \u03b31 and \u03b32 of C1 and C2. d) Compute the numerical values of C1(\u00b5) and C2(\u00b5) at \u00b5 = 2 GeV. For simplicity, take the top quark mass equal to MZ, and all other quark masses less than 2 GeV. Ignore electromagnetic renormalization below MZ. 97.4) Consider the proton decay mode with the most stringent experimental e+\u03c00. The terms in eq. (97.43) relevant for this mode are bound, p \u2192 (97.45) Le\ufb00 = C1\u03b5\u03b1\u03b2\u03b3(eu\u03b3)( \u00afd \u2020\u03b1 \u00afu\u2020\u03b2) + 2C2\u03b5\u03b1\u03b2\u03b3(\u00afe\u2020 \u00afu\u2020\u03b3)(d\u03b1u\u03b2) + h.c.. SU(2)R global symmetry of QCD that Note that, under the SU(2)L \u00d7 we discussed in section 83, the operator u( \u00afd \u2020 \u00afu\u2020) transforms as the \ufb01rst component of a (2, 1) representation, while the operator \u00afu\u2020(du) is related by parity, and transforms as the \ufb01rst component of a (1, 2) representation. At low energies, we can replace these operators (up to an overall constant factor) with hadron \ufb01elds with the same properties SU(2)R \u00d7 under Lorentz and SU(2)L \u00d7 a) Show that PL(u u = exp[ i\u03c0aT a/f\u03c0], where \u03c0a is the triplet of pion \ufb01elds, and the Dirac \ufb01eld for the proton-neutron doublet. )1 transform appropriately. Here Ni is U(1)V transformations. )1 and PR(u\u2020N N b) Show that the low-energy version of eq. (97.45) is then Le\ufffd", "\ufffd\ufffd = C1A CPL(u )1 + 2C2A CPR(u\u2020 )1 + h.c., (97.46) E N E C is the charge conjugate of the Dirac \ufb01eld for the electron where E C is the Dirac \ufb01eld for the positron), and A is a (in other words, constant with dimensions of mass-cubed. Lattice calculations have yielded a value of A = 0.0090 GeV3 for \u00b5 = 2 GeV. N E c) Write out the terms in eq. (97.46) that contain the proton \ufb01eld and either zero or one \u03c00 \ufb01elds. e+\u03c00. Note that there are two d) Compute the amplitude for p \u2192 contributing Feynman diagrams: one where eq. (97.46) supplies the 97: Grand Uni\ufb01cation 614 proton-positron-pion vertex, and one where the proton emits a pion via the interaction in eq. (83.30), and then converts to a positron via the no-pion terms in eq. (97.46). Neglect the positron mass. Hint: your result should be proportional to 1 + gA. e+\u03c00. Use the e) Compute the spin-averaged decay rate for p values of C1 and C2 for \u00b5 = 2 GeV that you computed in problem 97.3. How does your answer compare with the naive estimate we 1 + 4C 2 made earlier? Hint: your result should be proportional to C 2 2. \u2192 97.5) Consider the Yukawa couplings for the the down quark and charged lepton of one generation, LYuk = \u2212 Zyy\u03b5ij\u03d5i\u2113j \u00afe \u2212 Zy\u2032y\u2032\u03b5ij\u03d5iq\u03b1j \u00afd\u03b1, (97.47) where we have included renormalizing factors. a) Consider one-loop contributions from gauge-boson exchange to Zy\u2032/Zy. Show that the only contributions of this type that do not cancel in the ratio are those where the gauge boson connects the two fermion lines. b) Show that, in Lorenz gauge, these contributions to Zy\u2032 and Zy are given by Zm in spinor electrodynamics in Lorenz gauge, with a replacement analogous to eq. (97.44) that you should specify. c) Let r y\u2032/y, and compute the anomalous dimension of r. \u2261 d) Take r(MX) = 1, and evaluate r(MZ). For simplicity, take the top quark mass equal to MZ. e) Below MZ, treat the top quark as heavy, and neglect the small electromagnetic contribution to the anomalous dimension of r. Compute r(mb), where mb = mb(mb) = 4.3 GeV is the bottom quark mass parameter. Use your results to predict the tau lepton mass. How does your prediction compare with its observed value, m\u03c4 = 1.8 GeV? Bibliography \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 M. V. Berry and K. E. Mount, Semiclassical approximations in wave mechanics, Rep. Prog. Phys 35, 315 (1972), available online at http://www.phy.bris.ac.uk/people/berry mv/publications.html. Lowell S. Brown, Quantum Field Theory (Cambridge 1994). Sidney Coleman, Aspects of Symmetry (Cambridge 1985). Ta-Pie Cheng and Ling-Fong Li, Gauge Theory of Elementary Particles (Oxford 1988). John Collins, Renormalization (Cambridge 1984). Howard Georgi, Weak Interactions and Modern Particle Theory (Benjamin/Cummings 1984). Claude Itzykson and Jean-Bernard Zuber, Quantum Field Theory (McGraw\u2013Hill 1980). Stephen P. Martin, A Supersymmetry Primer, available online at http://arxiv.org/abs/hep-ph/9709356. MILC Collaboration, Phys. Rev. D 70, 114501 (2004), available online at http://arxiv.org/abs/hep-lat/0407028. T. Muta, Foundations of Quantum Chromodynamics (World Scienti\ufb01c 1998). Abraham Pais, Inward Bound (Oxford 1986). Michael E. Peskin and Daniel V. Schroeder, An Introduction to Quantum Field Theory (Westview 1995). Chris Quigg, Gauge Theories of the Strong, Weak and Electromagnetic Interactions (Westview 1983). Pierre Ramond (I), Field Theory: A Modern Primer (Addison-Wesely", " 1990). Pierre Ramond (II), Journeys Beyond the Standard Model (Perseus 1999). Warren Siegel, Fields, available online at http://arxiv.org/abs/hep-th/9912205. 615 Jan Smit, Introduction to Quantum Fields on a Lattice (Cambridge 2002). George Sterman, An Introduction to Quantum Field Theory (Cambridge 1993). Steven Weinberg, The Quantum Theory of Fields, Volumes I, II, and III (Cambridge 1995). Julius Wess and Jonathan Bagger, Supersymmetry and Supergravity (Princeton 1992). Anthony Zee, Quantum Field Theory in a Nutshell (Princeton 2003). \u2022 \u2022 \u2022 \u2022 \u2022 Errata for this book are maintained at http://www.physics.ucsb.edu/ mark/qft.html. \u223c Current experimental results on elementary particles are available from the Particle Data Group at http://pdg.lbl.gov. Essentially all papers on elementary particle theory written after 1992 can be found in the hep-th (high energy physics, theory), hep-ph (high energy physics, phenomonology), and hep-lat (high energy physics, lattice) archives at http://arXiv.org. 616 B) if Cij = Aij + Bij for all i and j. Clearly, for this to work, A and B must have the same dimension. \u2022 A matrix C is the product of matrices A and B (C = AB) if Cij = k AikBkj (1.2) for all i and j. Clearly, for this to work, if A is m \u00d7 n then B must have dimension n \u00d7 p for some integer p, and the product C will have dimension m \u00d7 p. \u2022 The identity matrix In is an n \u00d7 n matrix defined by (In)ij = \u03b4ij, where \u03b4ij is the Kronecker delta (\u03b4ij = 1 if i = j and 0 otherwise). \u2022 The inverse matrix of an n \u00d7 n matrix A is denoted A\u22121, and satisfies AA\u22121 = A\u22121A = In. The inverse matrix is not guaranteed to exist, and a matrix for which the inverse exists is said to be invertible or nonsingular. \u2022 The transpose matrix of a matrix A is given by (AT)ij = Aji. Clearly if A is m \u00d7 n then AT is n \u00d7 m. 1.4 Eigenvalues and Eigenvectors 19 Finally, we will define the determinant of a matrix, something which maps a square, real-valued matrix to a real number. Let A be an n \u00d7 n matrix. Then the determinant is det(A) := j1j2...jn j1j2...jn A1j1A2j2... Anjn, (1.3) where the permutation symbol j1j2...jn is +1 if (j1, j2,..., jn) is an even permutation of (1, 2,..., n) (i.e., can be obtained from (1, 2,..., n) by exchanging adjacent pairs of numbers an even number of times), it is \u22121 if (j1, j2,..., jn) is an odd permutation of (1, 2,..., n), and is 0 otherwise (i.e., if any number is repeated). While this serves as a formal definition of the determinant, it is unwieldy except for small matrices. However, we will be most concerned with 2 \u00d7 2 matrices in optics, where the determinant is given by A B C D det = AD \u2212 BC. (1.4) The determinant has the important property that it is nonzero if and only if the matrix is nonsingular. 1.4 Eigenvalues and Eigenvectors Let A be an n \u00d7 n matrix. We want to consider cases where A \u00b7 x = \u03bbx, (1.5) where x is an n-dimensional vector and \u03bb \u2208 R. If there are \u03bb and x that satisfy this relation then \u03bb is said to be an eigenvalue of A with corresponding eigenvector x. Why should we consider eigenproblems? Eigenvalues typically represent physically important values of physical quantities, and eigenvectors typically represent physically important elements of vector spaces (typically providing a physically significant basis for a vector space). As a simple example, consider a system of coupled oscillators (mechanical or otherwise). Each uncoupled oscillator would satisfy an equation of the form \u00a8x = \u2212\u03c92x. When coupled together, the system of oscillators would more generally satisfy a matrix equation of the form Upon making the ansatz x(t) = x(0) e", "\u2212i\u03c9t, we obtain an eigenvalue equation, \u00a8x = A \u00b7 x. A \u00b7 x = \u2212\u03c92x. (1.6) (1.7) Thus, the eigenvalues of A represent the distinct frequencies of oscillation for the coupled system, while the eigenvectors represent how different oscillators move together to make each distinct \u2018\u2018mode\u2019\u2019 of oscillation corresponding to each frequency. How do we find the eigenvalues and eigenvectors? First, note that the eigenvalue condition above implies that (A \u2212 \u03bbIn) \u00b7 x = 0, so that A \u2212 \u03bbIn is a singular matrix. Thus, we have the condition that det(A \u2212 \u03bbIn) = 0, (1.8) which yields the characteristic polynomial in \u03bb. polynomial. For a 2 \u00d7 2 matrix A, the characteristic polynomial is simple (it is handy to know this): The eigenvalues are the roots of the characteristic \u03bb2 \u2212 Tr(A)\u03bb + det(A) = 0. (1.9) Here, Tr(A) is the trace of the matrix, defined as the sum over the diagonal elements. The eigenvector corresponding to an eigenvalue \u03bb can then be found by solving the homogenous linear system (A\u2212\u03bbIn)\u00b7x = 0. If the eigenvectors are linearly independent (i.e., it is not possible to write any one of them as a linear combination of the others), then they form a \u2018\u2018nice basis\u2019\u2019 for the vector space in the sense that in this basis, the linear transformation represented by A is now represented by a diagonal matrix. Mathematically, let P be a matrix such that the columns are eigenvectors of A. Then P\u22121AP is a diagonal matrix. In fact the diagonal elements are the eigenvalues. If the eigenvectors are furthermore mutually orthogonal (as is the case for real, symmetric matrices with distinct eigenvalues) and normalized, then we have PT = P\u22121 (i.e., P is an orthogonal matrix), and thus the diagonal matrix is PTAP. 20 1.5 Exercises Problem 1.1 Chapter 1. Review of Linear Algebra (a) The trace of a square matrix A is defined by Tr [A] := j Ajj. Let A and B be n \u00d7 n matrices. Prove that the trace of the product is order-invariant, Tr [AB] = Tr [BA]. (1.10) (1.11) (b) Prove that the trace of the product of n \u00d7 n matrices A(1), A(2),..., A(N ) is invariant under cyclic permutation of the product. Problem 1.2 Let M and N be vector spaces of dimension M and N, respectively, and let \u03c6 : M \u2212\u2192 N be a linear transformation. Show that \u03c6 can be represented by an N \u00d7 M matrix; that is, show that there is a matrix A such that A \u00b7 x = \u03c6(x) for every vector x \u2208 M. Problem 1.3 Let A be an n \u00d7 n matrix. Show that det(A) = (\u22121)ndet(\u2212A). Problem 1.4 Show that if A is a 2 \u00d7 2 matrix, the characteristic polynomial is given by \u03bb2 \u2212 Tr(A)\u03bb + det(A) = 0, and thus that the eigenvalues are given by \u03bb1,2 = Tr(A) 2 \u00b1 Tr(A) 2 2 \u2212 det(A). Further, show explicitly that if det(A) = 1, then \u03bb2 = 1/\u03bb1. Problem 1.5 (1.12) (1.13) Let A be a diagonalizable n \u00d7 n matrix (i.e., there exists a matrix P such that A = PDP\u22121, where D is a diagonal matrix with the eigenvalues of A along the diagonal). Show that You may use the property det(AB) = det(A) det(B) of the determinant for n \u00d7 n matrices A and B. log[det(A)] = Tr[log(A)]. (1.14) Problem 1.6 (a) Is the trace of a matrix (viewing the trace operation as a function) a linear transformation? If yes, prove it; if no, give a counterexample. (b) Repeat (a), but for the determinant of a matrix. Chapter 2 Ray Optics 2.1 Introduction Ray optics, or geometrical optics, is the simplest theory of optics. The \u2018\u2018mathematical\u2019\u2019 definition is simply that the ray is a path. The more", " intuitive \u2018\u2018definition\u2019\u2019 is that a ray represents the center of a thin, slowly diverging beam of light. While this theory gets a lot of things right, it also misses many phenomena. But even when we get to wave optics, geometrical optics will still provide a lot of basic intuition that we need to understand the more complex behavior of optical waves. For now, we will ignore any wave effects, and any width or spreading of the ray (though by examining many rays, it is possible to model the imaging behavior of an optical system). 2.2 Ray Optics and Fermat\u2019s Principle Ray optics boils down to a single statement, which we will get to very shortly. But first, we will start by noting that the fundamental assumption in ray optics is that light travels in the form of rays. Again, as mentioned above, a single ray could represent a beam of light, but typically you would use many rays to model light propagation (e.g., to model the performance of an imaging system). The optical rays propagate in optical media. To keep things simple, we will assume that the media are lossless, and thus we can characterize them completely by their index of refraction, which we will denote by n. Usually, n \u2265 1, with n = 1 corresponding to vacuum. And while many media are uniform, meaning that the refractive index is uniform throughout the medium, many media also have refractive indices n(r) that vary spatially. The only effect that we require of the refractive index is that it changes the speed of light. The speed of light in a medium of refractive index n is simply c0/n, where c0 is the vacuum speed of light (defined to be exactly 2.997 924 58 \u00d7 108 m/s). Now to the fundamental principle of ray optics, called Fermat\u2019s Principle. Consider a path r(s) inside an optical medium between points A and B, parameterized by the variable s, so that A corresponds to r(0), and B corresponds to r(d), where both s and d have the dimensions of length. Then the optical path length for this path is the length of the path, but weighted by the local refractive index. Mathematically, we can define the optical path length functional as [r] := d 0 n(r) ds. (2.1) (optical path length) ABro(s) 22 Chapter 2. Ray Optics This quantity is proportional to the time light takes to traverse the path, \u2206t = /c0, and is just the ordinary length of the path in the case n = 1. Then Fermat\u2019s Principle states that optical rays traverse paths that satisfy \u03b4 = 0. (2.2) (Fermat\u2019s Principle) The \u2018\u2018\u03b4\u2019\u2019 here is like a derivative, but for functions, and \u2018\u2018\u03b4\u2019\u2019 here can be read as \u2018\u2018the variation of \u2019\u2019 or the \u2018\u2018the first variation of.\u2019\u2019 What the statement \u03b4 = 0 means is that nearby paths have the same path length. Formally, this means that [r + \u03b4r] \u2212 [r] = 0, (2.3) at least to first order in, where \u03b4r is an arbitrary path satisfying \u03b4r(0) = \u03b4r(d) = 0. The proper mathematical framework for all this is the calculus of variations, but we won\u2019t really go much more into this here. However, it is for this reason that Fermat\u2019s Principle is called a variational principle. It turns out that variational principles are extremely important in many areas of physics. Variational principles are conceptually odd, however. Normally you think of light starting someplace, and then asking, where does it go? When applying Fermat\u2019s Principle, the idea is to stipulate the endpoints, or where the light begins and ends up, and then ask, what does it do in the meantime? Often, as happens in standard calculus, the condition \u03b4 = 0 yields a minimum for. Hence Fermat\u2019s Principle is often referred to as a \u2018\u2018principle of least time.\u2019\u2019 However, this isn\u2019t necessarily the case. Often, the stationary condition turns out to be an inflection point or more commonly a saddle point, where is a minimum along one direction but a maximum along another. Obviously \u03b4 = 0 can never yield a true global maximum. Given an stationary path, there is always a nearby path that is slightly longer. 2.3 Fermat\u2019s Principle: Examples We will now consider some applications of Fermat\u2019s Principle. 1. Homogeneous Medium. In a homogeneous medium, the refractive index n is constant. Thus, the", " minimum optical path length occurs for the path with the minimum length, which is a straight line. Then the optical path length is simply = n B A ds = nd, (2.4) where d is the regular length of the path. This quantity is thus minimized for a straight line connecting A and B. We know experimentally that light travels in straight lines, and this is one of the motivations for Fermat\u2019s Principle. More formally, recall that we must have \u03b4 = 0, which is satisfied for a path that minimizes. But to see that the variation vanishes for the straight line, consider that any small perturbation \u00b1\u03b4r to the path r(s) will change the path length by the same amount for either sign, just due to the symmetry of the straight line. Thus, \u03b4 = 0, at least to lowest order in. Note that for any path that isn\u2019t straight, we couldn\u2019t come to the same conclustion: for almost any perturbation to the path, one of the perturbations \u00b1\u03b4r will tend to increase the path length, while the other will decrease it. 2. Plane Mirror. Again, we will start by fixing the endpoints of the ray, in the diagram below we label these by A and A. ABro(s)dl = 0ABro(s) 2.3 Fermat\u2019s Principle: Examples 23 We will assume a homogeneous medium to the left of the mirror with refractive index n, and we have already shown the ray to be straight while in this medium. We can guess from symmetry that AP A is the minimum-length path (of the paths that bounce off the mirror). But let\u2019s prove it. Consider a nearby point P. The length of AP A is = a2 + (b \u2212 \u03b5)2 + a2 + (b + \u03b5)2. n Differentiating with respect to the perturbation \u03b5, \u2202 \u2202\u03b5 n = \u03b5 \u2212 b a2 + (b \u2212 \u03b5)2 + \u03b5 + b a2 + (b + \u03b5)2. (2.5) (2.6) This quantity vanishes for \u03b5 = 0, and so the point P represents the extremal path. Thus, we conclude that (2.7) (Law of Reflection) thus arriving at the Reflection Law for optical rays. 3. Refractive Interface. \u03b8i = \u03b8r, At a planar interface between two optical media of different refractive index, the ray splits. We have to assume this as an experimental fact at this point, since we really need full electromagnetism to show this from first principles. The reflected ray behaves according to the law of reflection that we just derived. The other, refracted ray is a little different. Assume that the refracted ray begins at point A and ends at point C (i.e., we take both A and C to be fixed. We will assume it crosses the A (-a, b)Aoo' (-a, -b)Poooo' (0, e)Po (0, 0)qiqrn1n2Ao (0, 0)Bo (a, b)Co (c, d)qrq1q2 24 Chapter 2. Ray Optics refractive interface at point B, and now we will compute where the point B must be according to Fermat\u2019s Principle, \u03b4(ABC) = 0. The path length is = n1 a2 + b2 + n2 (c \u2212 a)2 + (d \u2212 b)2. (2.8) Differentiating with respect to the moveable coordinate b of B, we arrive at the extremal condition \u2202 \u2202b = \u221a n1b a2 + b2 \u2212 n2(d \u2212 b) (c \u2212 a)2 + (d \u2212 b)2 = 0. Using the angles marked in the diagram, we can rewrite this condition as or which is simply Snell\u2019s Law. Note that n1 sin \u03b81 \u2212 n2 sin \u03b82 = 0, n1 sin \u03b81 = n2 sin \u03b82, sin \u03b82 = n1 n2 sin \u03b81 \u2264 1. If n1 > n2 then there is a critical angle \u03b8c given by n1 n2 sin \u03b8c = 1, (2.9) (2.10) (2.11) (Snell\u2019s Law) (2.12) (2.13) such that if \u03b81 > \u03b8c, then there is no possible transmitted ray. All the light is instead reflected, and this phenomenon is called total internal reflection. 4. Spherical Mirror. All rays from the center to the outer edge and back have the same (minimum) optical path", " length (i.e., the minimum for reflected rays). Thus, a spherical mirror focuses rays from an object at the center point back onto itself. 5. Elliptical Mirror. We can define an ellipse as the set of all points {P : AP A = d} for some constant distance d. The points A and A are the foci of the ellipse. Fermat\u2019s Principle tells us immediately that since for any point P on the ellipse, we know that AP A is constant (and a minimal length for reflecting paths), and thus rays starting at the point A will end at the point A. Thus an elliptical mirror images an object at A to the other focus A. A\u00a2AP 2.4 Paraxial Rays 6. Parabolic Mirror. 25 We can define a parabola as the set of all points {P : AP = P A where P A \u22a5 directrix}. Adding P A to AP = P A, we find AP + P A = AA, which is constant. So AP A is constant for all rays. Thus, a parabolic mirror collimates all rays starting at A, which is the focus of the parabola. Reversing the sense of propagation, we can also conclude that all incoming parallel rays orthogonal to the directrix are concentrated to the focus A. 2.4 Paraxial Rays Now we will set up a formalism for keeping track of optical rays more precisely. We will represent a ray by a vector, which keeps track of the position and direction of the ray with respect to the optical axis\u2014the reference axis for optical propagation. The displacement y from the optical axis and the direction \u03b8 are sufficient to completely specify the ray at a particular longitudinal position z. In the paraxial approximation that we will get to below, we will see that the angle \u03b8 is approximately equivalent to the slope y of the ray, so we will write vectors interchangeably with the angle and slope of the ray: y \u03b8 \u2248 y y. (2.14) We want to be able to compute the change in the ray vector for any optical system. In this sense, we will model an optical system as a transformation of ray vectors. In the most general case, we can write or in vector form, y2 = f1(y1, y 1); 2 = f2(y1, y y 1), y2 y 2 = f y1 y 1, (2.15) (2.16) where the vector function f models the optical system. APA\u00a2A\u00a2\u00a2directrixoptical axiszyyo(z)qo(z) \u00e0 yooo'(z)OAoptical systemy1q1 \u00e0 y1'y2q2 \u00e0 y2' 26 y 1: Assume that y\u03b1 and y \u03b1 are small. Then we can Taylor-expand the function to lowest order in y1 and Chapter 2. Ray Optics y2 = f1(y1, y 1) = 2 = f2(y1, y y 1) = \u2202f1 \u2202y1 \u2202f2 \u2202y1 y1=y y1=y y1 + 1=0 y1 + 1=0 \u2202f1 \u2202y 1 \u2202f2 \u2202y 1 y1=y y1=y 1 + higher-order terms in y1, y y 1 1=0 1 + higher-order terms in y1, y y 1. 1=0 (2.17) Note that we are assuming f1(0, 0) = f2(0, 0) = 0, as we have not written down the zeroth-order terms in the expansion. This assumes that any ray following the optical axis on the input side will continue to do so on the output side, and therefore optics like spherical lenses, interfaces, and so on are centered on the optical axis. It also excludes certain optics that deflect all rays, such as tilted mirrors and prisms. However, by defining the optical axis carefully, we can still accommodate these optics (see the example of the plane mirror on p. 28; tilted mirrors and prisms can be \u2018\u2018unwrapped\u2019\u2019 in the same way). Now we can rewrite the above expansion in matrix form: y2 y 2 = \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 \u2202f1 \u2202y1 \u2202f2 \u2202y1 \u2202f1 \u2202y 1 \u2202f2 \u2202y 1 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb y1=y y1 y 1 1=0 + higher-order terms in y1, y 1 (2.18) In the paraxial approximation, we will ignore the quadratic terms and model the optical system using only linear transformations. This approximation is valid for", " small y\u03b1 and y \u03b1 (or equivalently, \u03b8\u03b1, so that \u03b1, which justifies our interchangeable use of \u03b8 and y). The matrix of derivatives is \u03b8\u03b1 \u2248 sin \u03b8\u03b1 \u2248 tan \u03b8\u03b1 = y the transfer matrix that represents the optical system in the paraxial approximation. Note that for this approximation to be valid, the ray must always stay close to the optical axis. While the choice of the optical axis is in principle arbitrary, it is best chosen such that the paraxial approximation is good for optical rays of interest. The higher order corrections that we are neglected are treated in aberration theory, which we will not treat here. 2.5 Matrix Optics Recall from before that the most general linear transformation of a two-dimensional vector is a 2 \u00d7 2 matrix. We have written a matrix above as an expansion of a more general transformation, but for the general paraxial case we will use the notation y2 y 2 = A B C D y1 y 1. (2.19) The matrix representing the optical system is referred to as an \u2018\u2018ABCD matrix,\u2019\u2019 \u2018\u2018ray matrix,\u2019\u2019 or \u2018\u2018raytransfer matrix.\u2019\u2019 We will now derive the fundamental matrices. (notation for ABCD-matrix propagation of rays) 1. Free-Space Propagation. M = 1 d 1 0 OAdy1y1'y2y2' 2.5 Matrix Optics 27 One of the simplest cases is propagation in free space over a distance d. The ray travels in a straight line, so the angle does not change By comparison to the matrix equation y the slope is y 1, the position changes according to 2 = y y 1. 2 = Cy1 + Dy 1, we can conclude that C = 0 and D = 1. Since (2.20) y2 = y1 + y 1d. (2.21) By comparison to y2 = Ay1 + By matrix is simply 1, we can conclude that A = 1 and B = d. Thus, the free-space M = 1 d 1 0. (2.22) (free-space propagation matrix) Note that this is matrix still valid for propagation over a distance d within a refractive medium, independent of n. The rays still change height according to their angle; the \u2018\u2018compression\u2019\u2019 effect on the optical length happens because the angles themselves change when the rays enter and leave the medium, say, for a planar window of glass. 2. Thin Lens Because the lens is thin, the ray does not propagate over any distance \u2206z. The ray is continuous, so y2 = y1. Thus, A = 1 and B = 0. The ray is deflected, however, such that it satisfies the thin lens law: 1 di Here do is the object distance, di is the image distance, and f is the focal length, the single parameter that completely characterizes a thin lens. The sign convention for the focal length is that f > 0 for a convex lens, and f < 0 for a concave lens, as shown here (assuming a larger index for the lens than the surroundings, as for glass lenses in air): (2.23) 1 do 1 f = +. The line drawings shown to the right in the above figure are common schematic representations of convex and concave lenses in diagrams. To arrive at the rest of the ray-matrix elements, we can dodiOAy1,2y1oo'y2oo'orconvex lens (f > 0):orconcave lens (f < 0): 28 Chapter 2. Ray Optics take the object and image distances to be where the ray crosses the axis before and after the lens, respectively. Thus, we can write the initial slope as y 1 = y1 do, and similarly we can write y 2 = \u2212 y2 di = \u2212 y1 di = \u2212y1 1 f \u2212 1 do = \u2212 y1 f + y 1, (2.24) (2.25) where we used y2 = y1 and the thin lens law to eliminate y2. Thus C = \u22121/f and D = 1, and we can write the ray matrix for a thin lens as 2.26) (thin-lens ABCD matrix) 3. Plane Mirror. M = 1 0 0 1 Again, this is a thin optic, so y2 = y1. The reflection law says that \u03b82 = \u03b81. Thus the ray matrix is simply the identity matrix: M = 1 0 0 1. (2.27) (planar mirror ABCD matrix) So if the ray matrix is the identity, what is the effect of a planar mirror? Really it is just to reverse the direction of propagation. If we adopt an \u2018\u2018un", "folded\u2019\u2019 convention, where the (z) optical axis always points in the general direction of the ray travel, then the mirror is really equivalent to nothing, as shown schematically in the above sketch. 4. Spherical Mirror. OAq2 = q1q1OAy2oo' = y1oo'y1oo'OAconcave (shown): R < 0 -q\u2122q\u00a1y\u00a1 = y\u2122j(-R)M=\uf8ee\uf8f0102R1\uf8f9\uf8fbconvex: R > 0 2.5 Matrix Optics 29 For the spherical mirror, we use the sign convention that R < 0 for a concave mirror (as shown here) and R > 0 for a convex mirror. Thus we will use (\u2212R) > 0 in the figure. We will also mark the angle shown as (\u2212\u03b82) because the ray, as it is drawn, points downward (compare to the plane mirror sketch). Again, y2 = y1 in the paraxial approximation (i.e., the mirror is a thin optic). We can also write the angle with the radius line as \u03d5 = y1 \u2212R. The Law of Reflection implies that the angles on either side of the radius line are equal: We can rewrite this as Thus, we can write the ray matrix as \u03d5 \u2212 \u03b81 = \u2212\u03b82 \u2212 \u03d5. \u03b82 = \u03b81 \u2212 2\u03d5 = \u03b81 + 2y1 R. (2.28) (2.29) (2.302.31) (spherical-mirror ABCD matrix) Comparing this matrix to the thin-lens matrix, we see that in the paraxial approximation, a spherical mirror is equivalent to a thin lens with a focal length f = \u2212 R 2, (2.32) with, of course, the reversal of the optical axis. 5. Planar Refractive Interface. A planar refractive interface is thin, so the beam height doesn\u2019t change, but the angles change according to Snell\u2019s Law (within the paraxial approximation). M = 1 0 0 n1/n2 We leave the derivation of this ABCD matrix as an exercise (Problem 2.7). 6. Spherical Refractive Interface. A spherical refractive interface is a bit more complicated, in the way that the angle changes through the interface, but Snell\u2019s Law still of course governs the behavior. (Within the paraxial approximation, this interface is a thin optic, so the height again doesn\u2019t change.) \uf8ee \uf8f0 M = 1 (n2 \u2212 n1) n2R \u2212 \uf8f9 \uf8fb 0 n1 n2 convex (shown): R > 0 concave: R < 0 We also leave the derivation of this ABCD matrix as an exercise (Problem 2.7). OAn1n2ROAn1n2 30 Chapter 2. Ray Optics 2.6 Composite Systems Now we can consider more general optical systems, or composite optical systems made up of the more basic optical elements. When propagating the ray through the composite system, we can start on the first component by applying the first matrix: y1 y 1 = M1 y0 y 0. We can repeat this for the second optical element: y2 y 2 = M2 y1 y 1 = M2M1 y0 y 0. Iterating this procedure, we can arrive at the transformation for the entire system: yn y n = MnMn\u22121 \u00b7 \u00b7 \u00b7 M2M1 y0 y 0 =: Mcomposite y0 y 0. (2.33) (2.34) (2.35) So, the ray matrix of a composite systems is simply the product of the individual ray matrices. Note the right-to-left ordering of the product. M1 acts first on the input ray, so it must be the rightmost in the product. Mcomposite = MnMn\u22121 \u00b7 \u00b7 \u00b7 M2M1 (ABCD matrix for composite optical system) (2.36) 2.6.1 Example: Thin Lens We can regard a thin lens as a composition of two cascaded refractive interfaces. Since the lens is thin, we assume that there is no distance between the interfaces. The composite matrix is thus \uf8ee \uf8f0 M = 1 (n1 \u2212 n2) n1R2 \u2212 0 n2 n1 \uf8f9 \uf8ee \uf8fb \uf8f0 1 (n2 \u2212 n1) n2R1 \u2212 0 n1 n2 \uf8f9 \uf8ee \uf8fb = \uf8f0 (n2 \u2212 n1) n1 1 1 R2 \u2212 1 R1 \uf8f9 \uf8fb. 0 1 If we take n1 =", " 1, this simplifies to \uf8ee \uf8f0 M = 1 1 R2 \u2212 1 R1 \uf8f9 \uf8fb. 0 1 (n2 \u2212 1) (2.37) (2.38) OAMcomposite = MnoMno-o1 \u2026\u2026\u2026 M2oM1M1M2M3Mny0y0'y1y1'y2y2'y3y3'yno-o1yno-o1'ynyn'n1 = 1n2 > 1R1R2 2.7 Resonator Stability 31 If we compare this to the standard thin-lens matrix, we can equate the C matrix entries and write 1 f = \u2212(n2 \u2212 1) 1 R2 \u2212, 1 R1 (2.39) (Lensmaker\u2019s formula) which is known as the Lensmaker\u2019s formula. The sign conventions work out as follows. For a convex lens, we have R1 \u2265 0 and R2 \u2264 0, which means that 1 f = (n2 \u2212 1) 1 |R2| + 1 |R1| =\u21d2 f > 0, (convex lens) (2.40) and thus a positive focal length. For a concave lens, we have R1 \u2264 0 and R2 \u2265 0, which means that 1 f = \u2212(n2 \u2212 1) 1 |R2| + 1 |R1| =\u21d2 f < 0, (convex lens) (2.41) and thus a negative focal length. 2.7 Resonator Stability We want to consider resonators, or optical systems that trap light rays. Such things are very important for the operation of lasers, where light often needs to pass through a gain medium many times, or for interferometry, as we\u2019ll get to later. As a basic example, let\u2019s look at a resonator composed of two spherical mirrors separated by a distance d: It is easier to analyze this if we \u2018\u2018unwrap\u2019\u2019 the system into an equivalent waveguide of lenses as follows: This is just the \u2018\u2018unit cell\u2019\u2019 of the waveguide, which repeats over and over again for each round-trip of the ray in the cavity. We have exploited the equivalence of spherical mirrors and thin lenses here. As we have drawn it, i.e., for 2 concave mirrors, f1,2 = |R1,2/2|. The matrix for one round trip (or the waveguide unit cell) is the product of two free-space propagation matrices and two thin-lens (spherical mirror) matrices f1 \u2212 \uf8f9 \uf8fb f2 \u2212 \uf8f9 \uf8fb 0 1 1 d 1 0. (2.42) Note the proper order of multiplication, which follows the path of the ray through the cavity/waveguide: first (the rightmost matrix) is left-to-right propagation of distance d, then reflection off the right mirror, R\u2122R\u00a1df\u2122 = -R\u2122/2f\u00a1 = -R\u00a1/2dd 32 Chapter 2. Ray Optics right-to-left propagation, and reflection off the left mirror (the leftmost matrix in the product). Multiplying this all out, we get d 2 \u2212 d f2 \uf8f9 d f1f2 1 \u2212 d f1 1 \u2212 \uf8fa \uf8fa \uf8fb \u2212 d f1 (2.43) d f2 \uf8ee M = \uf8ef \uf8ef \uf8f0 1 \u2212 \u2212 1 f1 \u2212 1 f2 for the cavity round-trip matrix. 2.7.1 Stability Condition d f2 + The question we want to ask now is, does the resonator confine the ray? In other words, is the cavity stable? Consider the ray after n round trips in the cavity: yn y n = A B C D n y0 y 0 (2.44) To answer this question, we can diagonalize the matrix. Recall that the characteristic polynomial for a 2 \u00d7 2 matrix is \u03bb2 \u2212 Tr(M)\u03bb + det(M) = 0. For a ray matrix, there is a general result that states that (see Problem 2.9) det(M) = n1 n2, (2.45) (2.46) where n1 is the refractive index at the input of the optical system and n2 is the refractive index at the output. For the matrix describing a single pass through a resonator, the input and output are exactly the same place, thus n1 = n2 and det(M) = 1. Thus, the characteristic polynomial becomes \u03bb2 \u2212 Tr(M)\u03bb + 1 = 0. The eigenvalues of M", " are the roots of this polynomial, given by \u03bb\u00b1 = \u03b2 \u00b1 \u03b22 \u2212 1, where \u03b2 := Tr(M)/2 = (A + D)/2. (2.47) (2.48) So how does this help? Remember that we can decompose an arbitrary vector into eigenvectors. In particular, for the initial condition vector we can write y+ y + y0 y 0 = \u03b1+ + \u03b1\u2212 y\u2212 y \u2212 for some constants \u03b1\u00b1, and the vectors [y\u00b1 y \u00b1]T are the eigenvectors corresponding to \u03bb\u00b1: y\u00b1 y \u00b1 y\u00b1 y \u00b1 = \u03bb\u00b1. M Then after one round trip in the cavity, y0 y 0 M = \u03b1+\u03bb+ y+ y + + \u03b1\u2212\u03bb\u2212 y\u2212 y \u2212, and after n passes, yn y n = Mn y0 y 0 = \u03b1+\u03bbn + y+ y + + \u03b1\u2212\u03bbn \u2212 y\u2212 y \u2212. Let\u2019s simplify things a bit by only considering the positions yn. Then we can write yn = (\u03b1+y+)\u03bbn + + (\u03b1\u2212y\u2212)\u03bbn \u2212 =: \u03b3+\u03bbn + + \u03b3\u2212\u03bbn \u2212, (2.49) (2.50) (2.51) (2.52) (2.53) 2.7 Resonator Stability 33 where we are introducing the new constants \u03b3\u00b1 := \u03b1\u00b1y\u00b1 for notational convenience. There are two possibilities that we have to consider: either |\u03b2| \u2264 1 or |\u03b2| > 1. Let\u2019s consider the |\u03b2| > 1 case first. Then clearly \u03bb\u00b1 are real, since the argument of the radical is positive: \u03b22 \u2212 1 > 0. We can also see that |\u03bb+| > 1 and |\u03bb\u2212| < 1 (in fact, \u03bb+\u03bb\u2212 = 1, since det(M) = \u03bb+\u03bb\u2212 = 1). Now let\u2019s reexamine the solution: yn = \u03b3+\u03bbn + + \u03b3\u2212\u03bbn \u2212. (2.54) The first term grows exponentially with n, while the second damps exponentially away. Thus, for generic initial conditions (i.e., \u03b3+ = 0), the solution grows exponentially as yn \u223c \u03bbn +. This is the unstable case, since the solution runs away to infinity. Now let\u2019s consider the other case, |\u03b2| \u2264 1. Then \u03b22 \u2212 1 < 0, so we can write \u03bb\u00b1 = \u03b2 \u00b1 i 1 \u2212 \u03b22, and clearly now the eigenvalues are complex. Note also that both eigenvalues have unit modulus: |\u03bb\u00b1|2 = \u03bb\u00b1\u03bb\u2217 \u00b1 = \u03b2 \u00b1 i 1 \u2212 \u03b22 \u03b2 \u2213 i 1 \u2212 \u03b22 = \u03b22 + (1 \u2212 \u03b22) = 1. (2.55) (2.56) So |\u03bb\u00b1| = 1, and thus it follows that |\u03bbn so this is the stable case. Let\u2019s see this more explicitly: define \u03c6 := cos\u22121 \u03b2 \u00b1| = 1. We can already see that yn will stay bounded as n increases, \u03b2 = cos \u03c6, 1 \u2212 \u03b22 = sin \u03c6. Then \u03bb\u00b1 = exp(\u00b1i\u03c6), and \u03bbn \u00b1 = exp(\u00b1in\u03c6). Thus we can write the solution as yn = \u03b3+ein\u03c6 + \u03b3\u2212e\u2212in\u03c6 = ymax sin(n\u03c6 + \u03c60) (2.57) (2.58) for some constants ymax and \u03c60, which can be obtained from \u03b3\u00b1 noting that yn must be real-valued. In other words, each pass through the cavity simply increments the phase of a harmonic oscillation by some fixed amount \u03c6. As a side note, in both cases we can determine the constants of the motion from the initial condition. We can do this by finding the eigenvectors and decomposing the initial ray vector, or, for example, we can equivalently use y0 and y1: y0 = \u03b3+ + \u03b3\u2212, y1 = \u03b3+\u03bb+ + \u03b3\u2212\u03bb\u2212. Solving these two equations leads to \u03b3\u00b1 = y1 \u2212 y0\u03bb\u2213 \u03bb\u00b1 \u2212 \u03bb\u2213 (2.59) (2.60) as a compact formula for the coefficients. Thus the stability condition for the ray to remain bounded in the long term is simply |\u03b2| \u2264 1. We can also write or |Tr(M)| \u2264 2 |A + D| \u2264 2 (2.61) (resonator stability condition) (2.62) (resonator stability condition)", " for the stability condition explicitly in terms of the matrix elements. 2.7.2 Periodic Motion A condition more restrictive than the stability condition is the periodic ray condition, which states that the ray repeats itself exactly after s passes through the cavity, where s is some integer: ym+s = ym for all m. (2.63) 34 Chapter 2. Ray Optics If s is the smallest integer for which this is true, then s is called the period of the ray. Note that in the paraxial approximation, the existence of a single periodic ray implies that all rays for the optical system are periodic (except in trivial cases), because the matrix Ms collapses to the identity. More generally, though, in nonlinear systems both periodic and nonperiodic rays are possible in the same system, just depending on the initial condition. We can explore this a bit further mathematically. Obviously for a ray to be periodic the resonator must be stable. We can thus rewrite the periodic-ray condition (2.63) as \u03b3+ein\u03c6+is\u03c6 + \u03b3\u2212e\u2212in\u03c6\u2212is\u03c6 = \u03b3+ein\u03c6 + \u03b3\u2212e\u2212in\u03c6. This holds for arbitrary coefficients if the phases differ by exact multiples of 2\u03c0, i.e., (2.64) (2.65) (2.66) (periodic-ray condition) for some integer q. Thus, s\u03c6 = 2\u03c0q \u03b2 = cos 2\u03c0q s for integers q and s for periodic motion to occur. 2.7.3 Resonator Stability: Standard Form For the two-mirror resonator that we started out with, we can write out the stability condition more explicitly. Starting with |(A + D)/2| \u2264 1, we can write, (2.67) which, after inserting the matrix elements from Eq. (2.43) and a bit of algebra, we can rewrite the stability condition as 0 \u2264 1 \u2212 d 2f1 1 \u2212 d 2f2 \u2264 1. (2.68) (resonator stability condition) It is conventional to define the stability parameters g1,2 := 1 \u2212 d 2f1,2 = 1 +, d R1,2 (2.69) (cavity stability parameters) where the rightmost expression applies to the original two-mirror resonator rather than the equivalent lens waveguide. In terms of these parameters, the stability condition is particularly simple: 0 \u2264 g1g2 \u2264 1. (2.70) (resonator stability condition) Then we can sketch the stability diagram according to this inequality. 2.7 Resonator Stability 35 The shaded regions correspond to unstable resonators, the light regions are stable. There are four special cases marked in the diagram that we should consider. All four cases are on the borders of the stability regions, so these are all marginally stable cases. Case 1. g1 = g2 = 1 =\u21d2 R1,2 = \u221e (Planar resonator) Case 2. g1,2 = 0 =\u21d2 R1,2 = \u2212d (Confocal resonator) Case 3. g1,2 = \u22121 =\u21d2 R1,2 = \u2212d/2 (Spherical resonator/symmetrical concentric) Case 4. g1 = 0, g2 = 1 =\u21d2 R1 = \u2212d, R2 = \u221e (Confocal\u2013planar resonator, or half of a spherical resonator) The planar resonator is special in the following sense. We can compute \u03b2 = 2g1g2 \u2212 1 = 1. (2.71) Recalling that we defined the phase angle \u03c6 = cos\u22121 \u03b2, we find \u03c6 = 0 for this case. From Eq. (2.58), we see that yn = y0 for all n. Hence every ray is periodic with period 1 for a planar resonator. [We can also see this directly by noting that the cavity round-trip matrix (2.43) reduces to M = I2 for f1,2 = \u221e.] Actually, unstablestable111-1g\u00a1g\u2122-1423RRRR 36 Chapter 2. Ray Optics this is an artifact of the eigenvalue formalism; in this marginally stable (\u2018\u2018parabolic\u2019\u2019) case, the solutions can increase linearly rather than exponentially: yn = y0 + ny 0. The confocal resonator is one of the most important resonators, as we will see when we discuss optical spectrum analyzers. For the confocal resonator, d = \u2212R, so \u03b2 = 2g1g2 \u2212 1 = \u22121 and \u03c6 = cos\u22121(\u22121) = \u03c0. From Eq. (2.", "58), we see that yn = (\u22121)yn\u22121 = (\u22121)ny0. (2.72) Thus, every ray is periodic with period 2, so each ray repeats itself after 2 passes through the cavity. [Again, we can also see this directly by noting that the cavity round-trip matrix (2.43) reduces to M = \u2212I2 for f1,2 = d/2.] On a single pass, the position reverses itself (this happens for the angle as well), leading to \u2018\u2018figure-eight\u2019\u2019 orbits in the cavity. 2.8 Nonparaxial Ray Tracing with Interfaces Tracing paraxial rays lends itself to analytic calculations and even simple numerical calculations, using the matrix formalism. Tracing nonparaxial rays is much more difficult in general, and analytic calculations typically depend on the existence of fairly simple solutions, symmetries, and a clever setup. Ray tracing for fairly general interfaces can be done readily on a computer however. You can spend hundreds to thousands of dollars on commercial ray-tracing software. Or, armed with Snell\u2019s Law, the Law of Reflection, and a brain, you can get pretty far on your own for free. We will set up the basic formalism for ray tracing here, and go through a simple example to show the general idea, with code in an open-source, mathematical programming language (Octave). First, let\u2019s go through the basic idea for rays confined to a plane (valid for cylindrical optics, or axially symmetric optics). This will give us some intuition for the problem, and then we\u2019ll redo everything later in a more general form that extends to fully three-dimensional rays. The setup is as follows. The optical axis lies along the z-direction, and a surface (reflective or refractive interface) is defined by the relation f (y, z) = 0. (2.73) (representation of interface) The incident angle is \u03b81, and we wish to find the refracted or reflected angle \u03b82, given that the ray encounters the interface at the point (y, z). To apply Snell\u2019s Law or the Law of Reflection, we will need to track the direction normal to the surface, which we represent via the angle \u03c6 (the angle between the optical axis and the surface-normal direction). To represent this angle, note that \u03c6 is also the angle between the y-axis and the tangent to the surface at the intersection point. n1n2fo(y,oz)o=o0q1qiqrqtq2fyz 2.8 Nonparaxial Ray Tracing with Interfaces 37 And given that f (y, z) = 0 can be solved for the alternate function representation z(y) of the surface, we then have \u03c6 = tan\u22121 dz dy. (2.74) (normal angle of interface) Then for the case of refraction, we can relate the input and output angles via Snell\u2019s Law and reading off the various angles as: Similarly, in the case of reflection, we have \u03b8i = \u03b81 + \u03c6 \u03b8t = sin\u22121 n2 n1 \u03b82 = \u03b8t \u2212 \u03c6. sin \u03b8i \u03b8i = \u03b81 + \u03c6 \u03b8r = \u03b8i \u03b82 = \u03b8r + \u03c6, (2.75) (refraction) (2.76) (reflection) where \u03b82 here (not explicitly shown in this case) is still the angle of the outgoing ray with respect to the optical axis. In the reflection case, of course, the z-component of the reflected ray should have the opposite sign with respect to the incident ray. Given that we can find the point of intersection of the incident ray with the interface, then we can trace the outgoing refracted and reflected rays, and then repeat for subsequent interfaces encountered by the continuing ray. We will develop this procedure more explicitly in the more general case below. 2.8.1 Refraction and Reflection Laws: Coordinate-Free Form The angle-based formalism that we developed above is perfectly adequate for rays confined to a plane. However, for more general ray propagation, the generalization in terms of angles becomes more cumbersome, because the direction of the ray is represented by two angles, and applying Snell\u2019s Law involves a reorientation with respect to two other angles specifying the surface normal at the intersection point. Therefore, in the more general case, we will take a slightly different approach, and consider refraction at an interface, where we adapt the coordinate system to the surface. The incident ray, whose direction is represented by the unit vector \u02c6r1, is incident on the surface as shown, and is either reflected or refracted into the", " final direction, which we will represent by the unit vector \u02c6rr or \u02c6rt, respectively. z(y)q1ffyz 38 Chapter 2. Ray Optics In the diagram, we show a planar interface, but this is really just a zoomed version of a general, possibly curved surface, such that it appears flat, and \u02c6n is the unit normal vector to the surface at the point of intersection of the incident ray with the surface. In adapting the coordinate system we are not losing generality, because the idea is to eliminate all explicit references to coordinates, and only express the output direction \u02c6rr or \u02c6rt in terms of the incident direction \u02c6r1 and the normal vector \u02c6n. Note that there are two sensible orientations for \u02c6n, corresponding to \u00b1\u02c6z in this coordinate system, and we should be careful about this. Starting with the reflected and refracted angles \u02c6rr = sin \u03b81 \u02c6y \u2212 cos \u03b81 \u02c6z \u02c6rt = sin \u03b82 \u02c6y + cos \u03b82 \u02c6z, we can eliminate \u02c6y and \u02c6z in favor of \u02c6r1 and \u02c6n to write (Problem 2.23) \u02c6rr = \u02c6r1 \u2212 2 sgn(\u02c6n \u00b7 \u02c6r1) cos \u03b81 \u02c6n \u02c6r1 + sgn(\u02c6n \u00b7 \u02c6r1) cos \u03b82 \u2212 \u02c6rt = n1 n2 cos \u03b81 \u02c6n. n1 n2 (2.77) (2.78) Then eliminating the remaining angles, and thus any explicit references to the coordinate system, we can obtain the coordinate-independent forms (Problem 2.24) \u02c6rr = \u02c6ri \u2212 2(\u02c6n \u00b7 \u02c6ri)\u02c6n \uf8ee \u02c6rt = n1 n2 \u02c6r1 + \uf8f0sgn(\u02c6n \u00b7 \u02c6r1) 1 \u2212 2 n1 n2 |\u02c6n \u00d7 \u02c6r1|2 \u2212 \uf8f9 (\u02c6n \u00b7 \u02c6r1) \uf8fb \u02c6n. n1 n2 (2.79) for the reflected and refracted rays. Again, we have been careful to keep our results invariant under the replacement \u02c6n \u2212\u2192 \u2212\u02c6n, so that either possible normal vector of the surface is suitable for our purposes. (reflected and refracted directions, coordinate-independent) At this point it is worth expanding on a couple of details. In the formula for the refracted ray in Eqs. (2.79), note that the cross product |\u02c6n \u00d7 \u02c6r1| can take any value between 0 and 1. Thus, if n1 > n2, it is possible for the argument of the square root to become negative, and in this case \u02c6rt becomes imaginary. This means that there is in fact no transmitted ray; the entire ray is reflected, and this phenomenon is called total internal reflection, a topic to which we will return in the context of waves (Section 9.3). Similarly, we have mentioned that at an arbitrary interface the possibilities of both reflection and refraction of a ray. The fraction reflected vs. refracted is best discussed in the context of wave optics (Chapters 9 and 10, and is complicated somewhat by the possibility of wave interference from reflections from different interfaces. Furthermore, when tracing all possible reflected and refracted rays, the number of rays proliferates exponentially with the number of interfaces involved. Typically, a ray-tracing analysis focuses on one set of rays to keep the problem tractable (e.g., tracing the refracted rays through a compound-lens system, or the reflection from a single optical surface), and therefore requires some thought and care in the setup of the analysis. n1n2q1r^1q1r^rq2r^tn^yz 2.8 Nonparaxial Ray Tracing with Interfaces 39 2.8.2 Ray Tracing: A Recipe Now with the formulae (2.79) in hand to compute the direction of the ray after interacting with the interface, the only remaining tasks are to find the intersection of the incoming ray with the surface, and then to determine the normal vector. Since we define the interface in terms of the vanishing of a function, f (r) = 0, (2.80) (representation of interface) finding the intersection of a ray with the surface amounts to a root-finding problem. If we assume homogeneous media between interfaces, then the rays are straight lines, and this reduces our task to a one-dimensional root-finding problem,", " which is a fairly simple one. More explicitly, we can take the ray, described by its location r, to be parameterized by the location along the optical (z) axis (at least, between encounters with interfaces, where ray may reverse its direction with respect to the optical axis), so that we may write the ray as r(z). Then we are finding the root of f (z) = f [r(z)], or solving f (z) = f x0 + rix riz (z \u2212 z0), y0 + (z \u2212 z0), z = 0, riy riz (2.81) where r0 = (x0, y0, z0) is some reference location for the ray (say the \u2018\u2018source point\u2019\u2019 of the ray), and \u02c6ri = (rix, riy, riz) is the direction unit vector of the incident ray (the same as the incident vector in the previous section). Here we have written the x and y coordinates of the ray in terms of the initial point and the tangents rix/riz and riy/riz of the ray-propagation angles. In general, finding the intersection with a surface requires searching the function for various z to first bracket the interface. Mathematically, two points z1 and z2 bracket the interface [i.e., bracket the root of f (z)] if the function changes sign between these two points: (interface-intersection condition) f (z1)f (z2) < 0. (2.82) (root-bracketing condition) This can often be done through knowledge of the geometry, but in more general-purpose software, a range of z must be searched systematically in an attempt to bracket the intersection, if one exists (the ray can, of course, sail past the edge of an optic). Once bracketed, there are several simple algorithms to find the root. The bisection algorithm for example, says to try the midpoint \u00afz = (z1 + z2)/2, and then use the same condition to see whether (z1, \u00afz) or (\u00afz, z2) bracket the root. The process can be iterated to find an arbitrarily narrow interval that brackets the root. Once we determine the intersection of the ray with a surface, we need to find the direction of the outgoing ray. The unit-normal vector, specifying the orientation of the surface at the point of incidence, is simply \u02c6n = \u2207f |\u2207f |, (2.83) (normal vector to surface f = 0) evaluated at the point of incidence. Intuitively, this is because the surface is defined by the constant value f = 0, and so the maximum change in f occurs perpendicular to the surface. Then the outgoing-ray direction is set by Eqs. (2.79); recall that the sign of \u02c6n (which could vary because f (r) is not unique\u2014\u2212f (r) is as good as f (r) for specifying the surface) doesn\u2019t affect the results. So to summarize the procedure: 1. Begin with the ray at some initial reference point r0, traveling in the direction \u02c6ri, incident on some interface described by f (r) = 0. 2. Find the point of intersection of the ray with the interface, by bracketing and then finding the root of Eq. (2.81). 3. Determine the normal vector \u02c6n of the surface at this point, using Eq. (2.83). 40 Chapter 2. Ray Optics 4. Use Eqs. (2.79) to find the orientation of the reflected and/or refracted ray. 5. Lather, rinse, repeat: set the reference point r0 to the intersection point, use the refracted or reflected direction as the new orientation vector, and repeat the whole process for the next interface (or for the outside boundary of the plot, if this was the last interface). You may need to take care if the direction along the optical axis became reversed by the prior interface. 2.8.3 Example: Parabolic Interface As an example, let\u2019s consider a parabolic interface, where the parabola is specified by We can write this in our standard notation f (r) = 0 for root-finding as f (y, z) = z \u2212 zc \u2212 ay2 = 0. z \u2212 zc = ay2. (2.84) (2.85) Here zc is an offset along the optical axis, and a controls the curvature of the ray. In the case where a > 0 (such that the parabola opens to the right), and the refractive index to the right is greater than to the left of the interface, the interface will act to focus incoming parallel rays. 2.8.3.1 Ref", "raction The diagram below shows the refracted rays for the case a = 1/2, n1 = 1 to the left of the interface, and n2 = 2 to the right. 2.8 Nonparaxial Ray Tracing with Interfaces 41 The focusing action of the interface is clear, as well as the severe aberrations for rays far from the optical axis. Note that we have also marked the location of the effective focal point with a green dot. This is the common point on the optical axis through which horizontal rays very close to the optical axis will pass. The effective focal length is given by (Problem 2.25) feff = n2 2a(n2 \u2212 n1) (2.86) with respect to the location of the parabola along the optical axis; thus the absolute z location of the focal point is zc + feff. A program in the Octave1 mathematical scripting language that traces rays refracting through this surface to generate this disgram is given below. Note that we are using the recipe and the coordinateindependent form of the refraction formula. We only worry about one refraction, and we don\u2019t bother with rays that reverse direction, because this doesn\u2019t happen here. For finding the incident point of rays on the parabola, we define the surfacefunc function, which implements Eq. (2.81). We then use the fzero built-in function to find the root of surfacefunc, and thus the intersection point. Since fzero expects a one-dimensional function, we must pass other parameters to the function via global variables. We also compute the normal vector in gradfunc, by explicitly programming the gradient (\u2202f /\u2202y)\u02c6y + (\u2202f /\u2202z)\u02c6z. In a more general-purpose program, we could just estimate the gradient by computing f (r) at several different points, but the explicit form is pedagogically clearer here. This script is a good basis for ray tracing in other simple systems (e.g., Problem 2.27). The script is available at http://atomoptics.uoregon.edu/~dsteck/teaching/octave/parabolic_surface_refract.m and listed below. defined by f(y,z) = z-zc - a*y^2 = 0. % parabolic_surface_refract.m % % Trace incoming parallel rays for parabolic-surface \"lens\" % % % Represent rays by a position y at z, and also a direction unit vector % % % Use coordinate-free form of Snell's Law in terms of surface normal vector. rhat = [ry; rz]; z is the direction along the optical axis. global gY gRhat gZ0 gA gZc a = 0.5; zc = -0.5; dy = 0.06; ymax = 6*a; zmin = -6*a; zmax = 6*a; n1 = 1; n2 = 2; % parabolic parameter % shift of parabola center % increment of initial y position of rays % max y ray to trace % launch z % max z % index to left of parabola % index to right of parabola hence the global statement. % In the following function we need to pass extra parameters, % % % % % gY -> y at z0 gRhat -> ray direction (unit) vector, in form [ry; rz] gZ0 -> reference coordinate z0 gA -> copy of A 1Octave is free, open-source, mostly compatible with MATLAB, and runs on most platforms: http://octave.org. 42 Chapter 2. Ray Optics gZc -> copy of zc % % Note: global variables are labeled with a 'g' to avoid confusion gA = a; gZc = zc; % function for finding intersections with the parabolic surface function fout = surfacefunc(z) % z is the position along the optical axis global gY gRhat gZ0 gA gZc % extrapolate ray straight to z y = y0 + (z-z0)*ry/rz; % ray_y = gY + (z-gZ0)*gRhat(1)/gRhat(2); % function to define surface via f(y,z) = (z-zc) - a*y^2 = 0 fout = (z-gZc) - gA*ray_y^2; end %function % function for finding the normal vector to the surface % function fout_vec = gradfunc(y,z) returns gradient of f = z-zc - a*y^2 global gA fout_vec = [... -2*gA*y; 1", "... % partial f/partial y... % partial f/partial z ]; fout_vec = fout_vec/sqrt(dot(fout_vec,fout_vec)); % make unit vector end %function % plot parabola, using parametric form dt = 0.01; t=(zmin:dt:zmax)'; plot(a*t.^2+zc, t,'b-'); % plot as blue line % set axes using z dimensions, and square aspect ratio axis([zmin, zmax, zmin, zmax], \"square\"); xlabel('z (mm)'); ylabel('y (mm)'); title(sprintf('ray trace of parabolic interface, n = %.2f',n2)); hold on; % subsequent plots will be overlayed %%% loop over rays nrays = 2*floor(ymax/dy); for ray = 0:nrays, y0 = (ray-nrays/2) *dy; yarr = [y0]; zarr = [zmin]; % array of z positions % array of y positions to plot % find first intersection with the parabolic surface % look for intersection (root) between zmin and 10*zmax rhat = [0; 1]; % direction of initial ray, horizontal = zhat gY = y0; gRhat = rhat; gZ0 = zmin; 2.8 Nonparaxial Ray Tracing with Interfaces 43 z = fzero('surfacefunc', [zmin; 10*zmax]); y = gY; % add point to plot yarr = [yarr; y]; zarr = [zarr; z]; % compute refracted ray direction, update direction vector rhat nhat = gradfunc(y,z); % surface normal vector nhat3 = [0; nhat]; rhat3 = [0; rhat]; crossthing = (n1/n2)*cross(nhat3,rhat3); % (n1/n2)*(n x r) rhat = (n1/n2)*rhat... % add x dimension for cross product + ( sign(dot(nhat,rhat))*sqrt(1-dot(crossthing,crossthing))... - (n1/n2)*dot(nhat,rhat) )*nhat; % extrapolate to the end of the optical axis z0 = z; z = zmax; yarr = [yarr; y + (z-z0)*rhat(1)/rhat(2)]; zarr = [zarr; z]; % plot ray plot(zarr, yarr, 'r-'); % plot as red line end %for feff = n2/(2*a*(n2-n1)); % effective, paraxial focal length % mark effective focal length, account for offset of surface plot([feff+zc], [0], 'g.'); hold off; 2.8.3.2 Reflection With relatively minor modifications to the script, we can use it to trace the reflected rays from the parabola. 44 Chapter 2. Ray Optics We list only the changed section below, which is the last part of the loop over rays where we find the direction of the outgoing ray. We simply change from the refraction to the reflection formula. Also, we must now handle the possibility of a backwards-propagating ray, so we insert a condition based on the sign of the z-component of \u02c6rr, and propagate the final ray to either zmin or zmax for backward- or forward-propagating rays, respectively. % compute reflected ray direction, update direction vector rhat nhat = gradfunc(y,z); % surface normal vector rhat = rhat - 2*dot(nhat,rhat)*nhat; % extrapolate to the end of the optical axis % if ( rhat(2) > 0 ), % forward case handle backwards rays based on sign of z component of rhat z0 = z; z = zmax; yarr = [yarr; y + (z-z0)*rhat(1)/rhat(2)]; zarr = [zarr; z]; else, % backward case z0 = z; z = zmin; yarr = [yarr; y + (z-z0)*rhat(1)/rhat(2)]; zarr = [zarr; z]; end %if The full script to produce the above plot is available at http://atomoptics.uoregon.edu/~dsteck/teaching/octave/parabolic_surface_reflect.m 2.9 Exercises 2.9 Exercises Problem 2.1 45", " A fish is 1 m beneath the surface of a pool of water. How deep does it appear to be, from the point of view of an observer above the pool? The refractive index of water is n = 1.33. Problem 2.2 A corner-cube reflector is an arrangement of 3 planar mirrors to form, appropriately enough, the corner of a cube. The reflecting surfaces face the interior of the cube. Commercial corner cubes often use internal reflections to form the mirror surfaces, as in this photograph of the back side of a mounted corner-cube prism. Show that any ray entering the corner cube is reflected such that the exiting ray is parallel (but opposite) to the incident ray. You may assume that the incident ray\u2019s direction is such that it reflects from all 3 surfaces. Problem 2.3 Suppose that a ray in air (n = 1) is incident on a planar window of uniform thickness T and refractive index n. If the angle of incidence is \u03b8, show that the transmitted beam is parallel to the incident beam but has a horizontal displacement given by \u03b4 = T sin \u03b8 1 \u2212, cos \u03b8 n cos \u03b8 (2.87) where n sin \u03b8 = sin \u03b8. Problem 2.4 An anamorphic prism pair is used to expand or shrink a beam in one dimension without deflecting its angle, as shown here. aaq 46 Chapter 2. Ray Optics The pair consists of two idential prisms with wedge angle \u03b1 and refractive index n. The beam enters each prism at normal incidence to the front surface. As a model of the beam, it is useful to consider parallel rays as shown. (a) Write down an expression that relates the deflection angle \u03b8 after the first prism to \u03b1 and n. (b) By how much is the beam reduced after the first prism? Write your answer in terms of n and \u03b1 only. (c) By how much is the beam reduced after the second prism? Problem 2.5 A thin lens is submerged in water (n = 1.33). If the focal length is f = 1 m in air, what is the focal length in water? Assume the lens is made of fused silica (n = 1.46). Problem 2.6 Consider2 a reflection off the center P of a parabolic mirror (x = ky2) as shown. By symmetry of the reflection at the center, we can fix the endpoints of the reflected ray to have coordinates (a, b) and (a, \u2212b), with a > 0. (a) Consider a point P a small distance (with vertical component \u03b5) from P. Show that the optical path length of AP A is an extremum when P coincides with P. (b) Show that AP A is a minimum when k < kc and a maximum when k > kc, where kc := a 2(a2 + b2). (2.88) (c) The locus of all points B such that ABA = AP A is clearly an ellipse with foci A and A. Show that the equation describing this ellipse is given by (x \u2212 a)2 \u03b32 \u2212 b2 + y2 \u03b32 = 1, (2.89) where 2\u03b3 := ABA. Argue that this result is consistent with the result of part (b). Problem 2.7 (a) Derive the ABCD matrix for a refractive spherical boundary: 2Adapted from H. A. Buchdahl, Introduction to Hamiltonian Optics (Dover, 1993). A (a, b)xyx = ky\u2122PA\u00a2 (a, -b)P \u00a2 2.9 Exercises 47 \uf8ee \uf8f0 M = 1 (n2 \u2212 n1) n2R \u2212 \uf8f9 \uf8fb 0 n1 n2 convex (shown): R > 0 concave: R < 0 The convention is that R > 0 for a convex surface (as shown here) and R < 0 for a concave surface. Note that in the paraxial approximation, the height of the ray does not change across the boundary. (b) Derive the ABCD matrix for a refractive planar boundary: M = 1 0 0 n1/n2 (c) Find the determinants of the ABCD matrices in parts (a) and (b). Problem 2.8 (a) Derive the ray-transfer matrix for free-space propagation, followed by a thin lens, followed by more free-space propagation, as shown in the Figure. (b) Show that applying the thin-lens law, 1 d1 + 1 d2 = 1 f, (2.90) all rays originating from a single point y1 in the input plane reach the output plane at the single point y2, independent of the input angle y (c) Show that if d", "2 = f, all parallel incident rays are focused to a single point in the output plane. 1. Compute the magnification y2/y1. Problem 2.9 Let M be the ray matrix for an arbitrary optical system where the input and output refractive indices are n1 and n2, respectively. Prove that det(M) = n1 n2 (2.91) (as you saw in Problem 6(c)), using the following outline, which exploits the formal equivalence of ray optics to classical Hamiltonian mechanics. Recall the action principle (Fermat\u2019s principle) for ray optics: \u03b4 n(x, y, z) ds = 0, (2.92) ROAn1n2OAn1n2d\u2122fd\u00a1 48 Chapter 2. Ray Optics where ds2 = dx2 + dy2 + dz2 and n(x, y, z) is a refractive-index profile that models the optical system. Compare to the action principle for classical mechanics. Take the coordinate z to be the \u2018\u2018time\u2019\u2019 variable and the coordinate y to be the position coordinate. Let\u2019s consider the two-dimensional case, so x is an ignorable coordinate, and note that z is also ignorable in the sense of being completely determined by x, y, and s. Then for the optical case, write down the Lagrangian. Show that the conjugate momentum p for the position y is n dy/ds, and then write down the Hamiltonian. Now consider the following transformation relating the canonical coordinates before and after the optical system, y2 p2 = M y1 p1, (2.93) which of course is valid in the paraxial approximation (where it is also true that s \u2248 z). Because y and p are canonical variables and M represents \u2018\u2018time evolution\u2019\u2019 of a Hamiltonian system, M represents a canonical transformation and in particular M is a symplectic matrix, which implies that det(M) = 1. This is essentially the content of Liouville\u2019s theorem. Using this result, transform to the standard (noncanonical) variables y and y, and compute the determinant of M. A very brief review of variational principles in classical mechanics may help. Recall that the action functional is given by the integral S[L] := t2 t1 L(q, \u02d9q; t) dt, (2.94) where the Lagrangian L is typically of the form L = T ( \u02d9q) \u2212 V (q) in particle mechanics. The variational principle (Hamilton\u2019s principle) is \u03b4S[L] = 0, which for our purposes implies the Euler-Lagrange equation d dt \u2202L \u2202 \u02d9q\u03b1 \u2212 \u2202L \u2202q\u03b1 = 0 (2.95) under the condition that the endpoints of the variation are fixed (\u03b4q(t1) = \u03b4q(t2) = 0). The Hamiltonian is given by a Legendre transformation of the Lagrangian via H(q, p; t) := \u02d9q\u03b1p\u03b1 \u2212 L(q, \u02d9q; t), (2.96) where the conjugate momenta are defined by p\u03b1 := \u2202L/\u2202 \u02d9q\u03b1. Problem 2.10 (a) Suppose that two thin lenses of focal length f1 and f2 are placed in contact. Show that the combination acts as a thin lens with a focal length given by 1 f = 1 f1 + 1 f2. (2.97) (b) The optical power of a lens is defined as 1/f, where f is the focal length of the lens. Typically the lens power is measured in diopters, defined as 1/f where f is measured in meters (i.e., a lens with a 100 mm focal length has a power of 10 diopters). Based on your answer for part (a), why is the optical power a natural way to characterize a thin lens? Problem 2.11 Show that the effective focal length feff of two lenses having focal lengths f1 and f2, separated by a distance d, is given by 1 feff 1 f1 1 f2 d f1f2 = + \u2212. (2.98) 2.9 Exercises 49 Note that this system is no longer a thin lens, so for this to work out you should show that the effect of the two-lens optical system is equivalent to that of a single lens of focal length feff, with free-space propagation of distances d1 and d2 before and after the single lens, respectively, where 1 d1 = 1 d \u2212 1 f1 + f2 f1d, 1 d2 = 1 d \u2212 1 f2 + f", "1 f2d. (2.99) This problem is a simple model for one realization of a \u2018\u2018zoom\u2019\u2019 or variable focus lens, e.g., for still photography. (A \u2018\u2018true zoom\u2019\u2019 lens would also shift the location of the pair to maintain focus as the focal length changes, whereas a simpler variable focus lens merely changes the separation to achieve different focal lengths, possibly requiring a refocusing adjustment.) Problem 2.12 Because of dispersion, the index of refraction varies slightly with the wavelength of light, and thus the focal length of a thin lens varies slightly with optical wavelength. This effect is called chromatic aberration. A common technique to correct for this aberration is to cement two lenses together of different materials to form an achromatic doublet or achromat. For this problem, assume a simple linear model of the refractive-index variation: n(\u03bb) \u2248 n(\u03bb0) + dn d\u03bb (\u03bb \u2212 \u03bb0), \u03bb=\u03bb0 (2.100) where \u03bb0 is some wavelength in the center of the region of interest. The dispersion of an optical glass is often characterized by its refractive indices at three special wavelengths, the Fraunhofer C, D, and F lines, given by \u03bbC = 656.3 nm, \u03bbD = 587.6 nm, and \u03bbF = 486.1 nm, named after Fraunhofer\u2019s catalog of the dark features in the solar spectrum. In terms of the three indices, we can define the Abb\u00e9 v-constant by vD := nD \u2212 1 nF \u2212 nC, in terms of which we can write the refractive index as n(\u03bb) \u2248 1 + (nD \u2212 1) 1 \u2212 \u03b4\u03bb vd(\u03bbC \u2212 \u03bbF), (2.101) (2.102) where \u03b4\u03bb := \u03bb \u2212 \u03bbD. Design a thin, achromatic doublet with the following materials: BK7 borosilicate crown glass (nF = 1.52238, nD = 1.51673, nC = 1.51432) and F2 flint glass (nF = 1.63208, nD = 1.61989, nC = 1.61503). Use BK7 for the first section in the shape of a biconvex lens, and F2 for the second section in the shape of a plano-concave lens, as shown. Obviously, the radii of curvature at the cemented interface should match. Make the thin-lens approximation and choose the two radii of curvature to achieve a lens with f = 100 mm over the visible spectrum. Problem 2.13 A refracting telescope is an optical system consisting of two thin lenses with a fixed space in between. Light first enters the objective lens of focal length fo, propagates over a distance L, and goes through the ocular lens (eyepiece lens) of focal length fe. The length of the telescope satisfies L = fo + fe. F2BK7R\u00a1R\u2122R\u00a3 = \u2022 50 Chapter 2. Ray Optics (a) Construct the ABCD matrix for propagation through a telescope (from left to right in this diagram). (b) Show that a telescope produces angular magnification. That is, incoming rays with angle \u03b81 from the optical axis exit the system at angle \u2212(fo/fe)\u03b81, independent of the initial ray position y1. (c) Show that a telescope can also act as a beam reducer (or expander): i.e., show that a bundle of rays of diameter d traveling parallel to the optical axis has diameter (fe/fo)d when exiting the eyepiece. (d) Sketch a Keplerian telescope, where both focal lengths are positive. Draw in the parallel rays corresponding to part (c). Also sketch a ray that is initially not parallel to the optical axis; use the thin lens law to justify the minus sign in the angular magnification of part (b). (e) Clearly, a telescope with positive angular magnification (i.e., a telescope that produces an upright image) has exactly one lens with negative focal length so that (\u2212fo/fe) > 0. Such a refracting telescope is known as a Galilean telescope. Which of the two lenses can have negative focal length if the telescope produces a magnified (not reduced) image? Problem 2.14 A retroreflector is any optic that reflects an incident ray, such that the exiting ray is parallel (but opposite) to the incoming ray. One version of a \u2018\u2018cat\u2019s eye\u2019\u2019 retroreflector uses a thin lens of focal length f and a mirror as shown. Set up the ray matrix for this optical system", " to prove that it is, indeed, a retroreflector. Problem 2.15 (a) A ball lens is a sphere of optical material (e.g., glass or sapphire) that is used as a lens, especially for coupling light into or out of an optical fiber. Within the paraxial approximation, derive an expression for the location where a beam of collimated incoming light (aimed at the sphere\u2019s center) will come to a focus, assuming a sphere of diameter D and refractive index n, surrounded by vacuum. (Your answer should be in the form of a distance from the center of the ball lens, thus giving its effective focal length.) (b) What is the condition for incoming rays parallel to the optical axis to be imaged onto the back surface of the sphere? Argue that under these conditions, the sphere acts as a retroreflector (incoming rays are reflected back along the same direction of incidence). This is the traditional realization of a \u2018\u2018cat\u2019s eye\u2019\u2019 retroreflector, and is the reason, for example, that some painted lines on roads and road signs look so bright at night under headlights (small silica spheres embedded in the paint reflect headlight back to your eyes). Problem 2.16 Consider a two-mirror resonator, as shown here. One mirror is concave (R < 0) and the other is flat. Lfofeobjective lensocular lensf 2.9 Exercises 51 (a) What is the round-trip ray matrix for this resonator? Derive the matrix for a ray starting just to the right of the curved mirror, traveling to the right. (b) For what range of d is the cavity stable? (c) Derive the ray matrix for two round trips for the special case R = \u22122d. Sketch an example ray that illustrates your answer. Problem 2.17 Suppose you have two convex, spherical mirrors in a gas laser resonator, with unknown and possibly different radii of curvature R1 and R2. Suppose also that you can vary the length d of the cavity, and that after much labor you find that the laser operates in the ranges d < 50 cm and 100 cm < d < 150 cm. What are the numerical values of R1 and R2? Keep in mind that gas lasers have relatively low gain per pass, and thus proper laser operation requires that the light makes many round trips inside the resonator before leaking out. Problem 2.18 (a) Consider a cavity consisting of two planar mirrors and identical thin lenses of focal length f, regularly spaced as shown. For a given set of lenses, what is the range of d for which the cavity is stable? (b) Write down the eigenvalues of the round-trip ray matrix for this cavity. Problem 2.19 Consider a symmetric cavity consisting of two planar mirrors separated by 1 m. Suppose that a thin lens of focal length f is placed inside the cavity against one of the mirrors. For what range of f is the cavity stable? Problem 2.20 Consider the ring cavity shown below, where the optical axis forms an equilateral triangle, with sides of length d, and a single lens of focal length f centered in the bottom leg of the cavity. dRR\u2122R\u00a1dfd/2dfffffffd/2dddddd 52 Chapter 2. Ray Optics (a) For what range of f is the cavity stable? (b) Discuss how moving the lens to the left or right influences the stability of the cavity. Problem 2.21 Suppose an optical resonator is represented by the round-trip matrix M, which satisfies 2.103) (a) What are the two eigenvalues of M? (b) Is the resonator stable? Explain. (c) If your answer to (b) was \u2018\u2018yes,\u2019\u2019 are there any rays that diverge after many round trips? If your answer to (b) was \u2018\u2018no,\u2019\u2019 are there any rays that remain stable? In either case, give an example and give a physical interpretation of the rays. Problem 2.22 In the matrix formalism of optics, we have stuck to modeling only one dimension (y), transverse to the optical axis (z). Suppose we want to model both transverse directions x and y, by an analogous matrix formalism, with a four-vector of the form \uf8f9 \uf8ee \uf8ef \uf8ef \uf8f0 x x y y \uf8fa \uf8fa \uf8fb (2.104) for the state of the vector at some point along the optical axis. Write down the ray-transfer matrix for this four-vector for free-space propagation over a distance d. Problem 2.23 Using the setup in this ray-tracing diagram, dfn1n2q1r^1q1r^rq2r", "^tn^yz 2.9 Exercises 53 show that the unit vectors for the reflected and refracted rays may be written in the form of Eqs. (2.78), \u02c6rt = n1 n2 Problem 2.24 \u02c6rr = \u02c6r1 \u2212 2 sgn(\u02c6n \u00b7 \u02c6r1) cos \u03b81 \u02c6n \u02c6r1 + sgn(\u02c6n \u00b7 \u02c6r1) cos \u03b82 \u2212 cos \u03b81 \u02c6n. n1 n2 (2.105) Starting with the results and setup of Problem 2.23, derive the coordinate-independent forms for the normal vectors of the refracted and reflected rays from Eqs. (2.79), \u02c6rr = \u02c6ri \u2212 2(\u02c6n \u00b7 \u02c6ri)\u02c6n \uf8ee \u02c6rt = n1 n2 \u02c6r1 + \uf8f0sgn(\u02c6n \u00b7 \u02c6r1) 1 \u2212 2 n1 n2 |\u02c6n \u00d7 \u02c6r1|2 \u2212 \uf8f9 (\u02c6n \u00b7 \u02c6r1) \uf8fb \u02c6n. n1 n2 Problem 2.25 For a parabolic refracting surface of the form (2.84) z \u2212 zc = ay2, (2.106) (2.107) with index n1 to the left of the surface and n2 to the right show that the effective focal length can be written feff = n2 2a(n2 \u2212 n1). (2.108) Do this by considering incident, paraxial rays from the left, parallel to the optical axis, and where they focus by crossing the optical axis. In doing this, you will establish Eq. (2.86). Problem 2.26 Suppose you have an optical system, and you want to find its effective focal length, which we define as the point where rays parallel and close to the optical axis cross the axis. One way to do this with ray matrices is to propagate a parallel ray through the optical system (represented by M), and then over a free-space distance d: 1 d 1 0 (2.109) M = yout y out yin y in Then we simply find the d such that yout = 0. We can regard this as a root-finding problem, yout(d) = 0, and we can use a root-finder on the computer to do solve this. Note that in the paraxial approximation, this works for any input ray that is parallel to the optical axis, so we can choose yin to be any (nonzero) number. Note also that this is a rather trivial root-finding problem: given the output ray from the optical system M, it is not a difficult trigonometry exercise to figure out the crossing point. However, the point of this exercise is to get used to using the root finder and multiplying together ray matrices in Octave, as a prelude for the following problem on computational ray-tracing. The goal of this exercise is for you to write an Octave script to carry out the above calculation for a ball lens of diameter D = 2 mm and refractive index n = 1.77 (appropriate as an average index of sapphire around 500 nm). Reference your calculated focal length to the center of the ball, and compare your numerical answer to your analytic expression in Problem 2.15. To get you started, study the example script below, which uses the same method to find the focal length of a thin lens of focal length f (which, as you might imagine, should turn out to be f ). Then modify it to solve the ball-lens case. (To run it: start in the same directory as the script file, start Octave, and type the name of the script into the Octave shell.) You can download this script at 54 Chapter 2. Ray Optics http://atomoptics.uoregon.edu/~dsteck/teaching/octave/thin_lens_matrix_root.m and the code listing follows. % thin_lens_matrix_root % % Trace rays through thin lens using ray matrices, and then find the % focus via root-finder. global gF % parameters f = 50; % focal length of thin lens in whatever units you like (let's say mm) % we need to pass a parameters to our function via global variables; % gF = f; prepend global names with a 'g' and keep them separate for clarity inputs: d -> distance of propagation after the lens % function to return ray height at the end of the imaging system % % % % % function yout = rayfunc(d) outputs: yout -> height of ray after propagation through lens and distance", " d % get and rename global variables global gF f = gF; thin_lens_M = [1, 0; -1/f, 1]; exterior_free_space_M = [1, d; 0, 1]; % note order of multiplication! % composite_M =... use '...' to continue an expression on the next line exterior_free_space_M *... thin_lens_M; % now trace an initially parallel ray, height = 1 yin_vec = [1; 0]; yout_vec = composite_M * yin_vec; % return the height of the final ray yout = yout_vec(1); end %function rayfunc % find distance where ray crosses the axis, using 'fzero' root finder i.e., we want to find the distance d such that rayfunc(d) = 0 % Note that we need to supply an initial guess, say 10f here, just % 2.9 Exercises 55 % to make it not so trivial. d_focus = fzero('rayfunc', 10*f); % print result printf('numerically computed focal length = %f mm\\n', d_focus); % compare to what you know it should be printf('analytically computed focal length = %f mm\\n', f); Problem 2.27 Study the code for refractive ray-tracing for a parabolic interface in the text (Section 2.8.3). Modify this code to trace rays through a ball lens (spherical dielectric in air) of radius 2 mm, and make plots for refractive indices n = 1.33 (water), n = 1.77 (an average index of sapphire around 500 nm), and n = 2.42 (diamond, around 589 nm) of the lens. As an example of what you should get, the n = 1.77 case is shown here. Some notes: \u2022 You will need to handle refraction at two interfaces. \u2022 As it turns out, you will not need to worry about total internal reflection. 56 Chapter 2. Ray Optics \u2022 However, as you can see from the example, you will need to handle rays that are refracted into the backwards direction, so study the modifications for reflective ray tracing in the parabola example. Make sure to choose your incident rays to show off some of these backward rays. \u2022 Be sure to mark the paraxial focal point of the lens, using the results of Problem 2.15. Chapter 3 Fourier Analysis Fourier analysis, the study of representing functions as sums of harmonic functions (sines and cosines), is one of the most important concepts in essentially every area of physics. 3.1 Periodic Functions: Fourier Series Before getting into defining the Fourier transform, it is helpful to motivate it by first considering the simpler Fourier series for a periodic function (or equivalently, a function defined on a bounded domain). Suppose f (t) is a periodic function with period T, so that it satisfies for all t. The frequency corresponding to the period is given by f (t) = f (t + T ), \u03c9 = 2\u03c0\u03bd = 2\u03c0 T, (3.1) (3.2) where \u03bd is the frequency and \u03c9 is the angular frequency. The distinction between these different frequency conventions can cause confusion when specific numbers are involved. For clarity, \u03bd can be reported in Hz, and \u03c9 can be reported in rad/s or as \u03c9/2\u03c0 (in Hz). The basic point is that we can think of the harmonic functions (sines and cosines) as being fundamental building blocks for functions. So let\u2019s try to build up f (t) out of harmonic functions. Since we know that f (t) is periodic, we will only use those harmonic functions that are also periodic with period T : f (t) = a0 + 2 \u221e n=1 an cos(n\u03c9t) + 2 \u221e n=1 bn sin(n\u03c9t). (3.3) (Fourier series) This expansion uses only expansion coefficients for positive n, but we can make things a bit easier to generalize if we define negative ones too: a\u2212n := an, b\u2212n := \u2212bn (3.4) for all nonnegative n (in particular, b0 = 0 by our definition). Then for any integer n (positive and negative), we can define the complex Fourier coefficient by cn := an + ibn. This definition allows us to write things a bit more compactly. Let\u2019s rewrite the Fourier series: f (t) = a0 + b0 + \u221e n=1 ein\u03c9t + e\u2212in\u03c9t \u2212 i an \u221e n=1 ein\u03c9t \u2212 e\u2212in\u03c9t. bn (3.5) (", "3.6) 58 Chapter 3. Fourier Analysis In each term of the form exp(\u2212in\u03c9t) we can make the transformation n \u2212\u2192 \u2212n to simplify the sums: f (t) = \u221e n=\u2212\u221e ane\u2212in\u03c9t + i \u221e n=\u2212\u221e bne\u2212in\u03c9t. (3.7) Using the complex coefficients, this simplifies greatly: f (t) = \u221e n=\u2212\u221e cne\u2212in\u03c9t. (3.8) (complex Fourier series) Now we can see that this Fourier series is a sum over complex harmonic functions with frequency \u03c9n = n\u03c9. The convention is that the harmonic function of the form exp(\u2212i\u03c9t) corresponds to a \u2018\u2018positive\u2019\u2019 frequency \u03c9, whereas the conjugate function exp(i\u03c9t) = exp[\u2212i(\u2212\u03c9)t] corresponds to a \u2018\u2018negative\u2019\u2019 frequency \u2212\u03c9. It might sound strange to talk about positive and negative frequencies, but for a real function f (t) the positive frequency \u2018\u2018contributes as much as\u2019\u2019 its negative counterpart in the sense that the coefficients must obey the constraint cn = c\u2217 \u2212n. However, the series (3.8) is more general than the original series (3.3) in that the second series can also represent complex-valued functions. Of course, given the cn coefficients, we can obtain the coefficients in the original Fourier series (3.3): an = bn = 1 2 1 2i (cn + c\u2217 n) = (cn + c\u2212n) (cn \u2212 c\u2217 n) = (cn \u2212 c\u2212n). 1 2 1 2i (3.9) But how do we obtain the cn coefficients in the first place? First, observe that for some integer n, we can evaluate the integral (real and complex coefficients related) T 0 e\u2212in\u03c9tein\u03c9t dt = T e\u2212i(n\u2212n)\u03c9t dt 0 1 \u03c9 2\u03c0 0 1 \u2212i\u03c9 e\u2212i(n\u2212n)x dx 2\u03c0 e\u2212i(n\u2212n)x n \u2212 n 0 = = (3.10) = 0 if n = n, where we defined x := \u03c9t. We need to be more careful with the n = n case, though, since there is a removable singularity here. If we define s = n \u2212 n, then e\u2212i2\u03c0s \u2212 1 s lim s\u21920 = lim s\u21920 [1 \u2212 i2\u03c0s + O(s2)] \u2212 1 s = \u22122\u03c0i, (3.11) so that the integral (3.10) takes the value 2\u03c0/\u03c9 = T if n = n. We can rewrite this relation in the more meaningful form 1 T T 0 e\u2212in\u03c9t\u2217 e\u2212in\u03c9t dt = \u03b4nn, (3.12) where \u03b4nn is the Kronecker delta (\u03b4nn = 1 if n = n and 0 otherwise). This is the orthogonality relation for the harmonic functions. The harmonic functions are basis vectors in a vector space of functions, and the orthogonality relation is a special case of the inner product defined by the same integral: (orthonormality of harmonic functions) f1, f2 := 1 T T 0 f \u2217 1 (t)f2(t) dt. (3.13) (inner product of periodic functions) 3.1 Periodic Functions: Fourier Series 59 So we are still doing linear algebra, but the infinite-dimensional, continuous version instead of the finite, discrete version with matrices that we reviewed in the beginning. Now consider the inner product of a basis vector with f (t): e\u2212in\u03c9t, f = 1 T T 0 e\u2212in\u03c9t\u2217 f (t) dt = \u221e n=\u2212\u221e cn T T 0 e\u2212in\u03c9t\u2217 e\u2212in\u03c9t dt = \u221e n=\u2212\u221e cn\u03b4nn = cn. (3.14) Thus, we can use the orthonormal properties of basis functions to project out the coefficients of exp(\u2212in\u03c9t): cn = 1 T T 0 ein\u03c9tf (t) dt. (3.15) (complex Fourier coefficient) Of course, as we noted above, we can also now calculate the an and bn coefficients in terms of the cn. It can be shown for \u2018\u2018reasonable\u2019\u2019 functions that N n=\u2212N cne\u2212in\u03c9t \u2212\u2192 f (t) as N \u2212\u2192 \u221e (3.16) at each point t except possibly", " on a set of zero measure, and that the set of coefficients cn that accomplish this is uniquely defined. 3.1.1 Example: Rectified Sine Wave As a simple example, let\u2019s compute the Fourier series for the function |sin \u03c9t|. Note that because of the \u2018\u2018rectification,\u2019\u2019 this function has a period of \u03c0/\u03c9, so the effective frequency of this function is 2\u03c9. Thus, letting T = \u03c0/\u03c9, cn = = = = = = Thus, we can write the series as |sin \u03c9t|ein(2\u03c9)t dt T 0 T 1 2i 0 T 1 T 1 T 1 2iT 1 2\u03c0i 1 2\u03c0i ei\u03c9t \u2212 e\u2212i\u03c9t ein(2\u03c9)t dt ei(2n+1)\u03c9t \u2212 ei(2n\u22121)\u03c9t dt 0 \u03c0 ei(2n+1)x \u2212 ei(2n\u22121)x dx 0 ei(2n+1)\u03c0 \u2212 1 i(2n + 1) \u2212 ei(2n\u22121)\u03c0 \u2212 1 i(2n \u2212 1) 1 \u03c0(2n + 1) \u2212 1 \u03c0(2n \u2212 1) = 2 \u03c0(1 \u2212 4n2). f (t) = \u221e n=\u2212\u221e 2 \u03c0(1 \u2212 4n2) e\u2212i2n\u03c9t. (3.17) (3.18) As reflected in the initial setup of this problem, the rectified sine wave contains only the even harmonics of the original pure harmonic wave, and none of the initial frequency. This is a useful feature to keep in mind, for example, when designing a device to double the frequency of an input signal. Note that cn = c\u2212n = c\u2217 n in this example because f (t) is real and even, whereas in the more general case of a real function we would only expect the less restrictive case cn = c\u2217 \u2212n. As in the plot here, the Fourier series converges fairly rapidly as a function of the number N of frequencies included, even already for N = 5 terms (beyond the dc component). 60 Chapter 3. Fourier Analysis For more terms, such as N = 20, the sum is visually indistinguishable from the original function. Other functions, particularly those with discontinuities, have series that do not converge so well as this example. 3.2 Aperiodic Functions: Fourier Transform We can also use the same harmonic functions e\u2212i\u03c9t to build up aperiodic functions. Recall that if T is the period, then we were using a discrete (countable) set of harmonic functions with a frequency spacing given by \u2206\u03c9 = 2\u03c0/T. An aperiodic function corresponds to T \u2212\u2192 \u221e, so we must use a representation where \u2206\u03c9 \u2212\u2192 0. Thus, we need a continuous spectrum to represent an aperiodic function, since there is much more \u2018\u2018information\u2019\u2019 in the function than in the periodic case. So let\u2019s define the Fourier transform as a generalization of the Fourier series, but with a slightly different normalization: f (t) = \u221e n=\u2212\u221e cne\u2212in\u03c9t \u2212\u2192 f (t) = 1 2\u03c0 \u221e \u2212\u221e \u02dcf (\u03c9) e\u2212i\u03c9t d\u03c9. (3.19) Thus, \u02dcf (\u03c9)/2\u03c0 is the amplitude of the frequency component e\u2212i\u03c9t (the normalization coefficient depends on how we define the density of the continuum of basis functions, as we will see shortly). The other usual nomenclature is that \u02dcf (\u03c9) is the Fourier transform of f (t). The above mathematical operation is the inverse Fourier transform, since we are finding f (t) from its Fourier transform. (generalization to inverse Fourier transform) In the same way as before, we can do a projection to find the amplitudes (Fourier transform). The analogue of the Fourier-series projection is cn = 1 T T 0 ein\u03c9tf (t) dt \u2212\u2192 \u02dcf (\u03c9) = \u221e \u2212\u221e f (t) ei\u03c9t dt. (3.20) The pair of equations (3.19) and (3.20) is one of the most important tools in physics, so they deserve to be written again: (generalization to Fourier transform) \u221e f (t) = 1 2\u03c0 \u02dcf (\u03c9) e\u2212i\u03c9t d\u03c9, \u02dcf (\u03c9) = f (t) ei\u03c9t d", "t. \u2212\u221e \u2212\u221e \u221e (3.21) The functions f (t) and \u02dcf (\u03c9) are said to be a Fourier transform pair, and their relationship is of fundamental importance in understanding linear systems in physics. (Fourier and inverse Fourier transforms) |osin wt|No=o1No=o3No=o5wt/2p-110foo(t)01 3.2 Aperiodic Functions: Fourier Transform 61 Again, if f (t) is a real function, then the Fourier transform satisfies \u02dcf (\u03c9) = \u02dcf \u2217(\u2212\u03c9), in which case f (t) = 2Re 1 2\u03c0 \u221e \u2212\u221e \u02dcf (\u03c9) e\u2212i\u03c9t d\u03c9, (3.22) so that the positive and negative frequencies contribute equally in amplitude to form a real-valued function. Note that there is an important alternate convention, where instead of \u03c9 we can use the standard frequency \u03bd = \u03c9/2\u03c0. This changes the density of the basis functions in a way that makes the normalizations more symmetric: \u221e \u221e f (t) = \u00aff (\u03bd) e\u2212i2\u03c0\u03bdt d\u03bd, \u00aff (\u03bd) = f (t) ei2\u03c0\u03bdt dt. \u2212\u221e \u2212\u221e (3.23) Here, we have defined an alternate Fourier transform by \u00aff (\u03bd) := \u02dcf (\u03c9/2\u03c0). It is a bit easier to remember this form to work out where the (1/2\u03c0) goes in the \u03c9 form of the transform equations. (Fourier and inverse Fourier transforms, \u03bd-t convention) We haven\u2019t been too concerned with rigor here, but it is interesting to ask, when is it possible to have a Fourier transform? As with the Fourier series, the f (t) must be a \u2018\u2018reasonable\u2019\u2019 function for \u02dcf (\u03c9) to exist. If f (t) is defined on the real line, then one possible set of sufficient conditions is as follows: 1. \u221e \u2212\u221e f (t) dt exists. 2. f has only a finite number of discontinuities and a finite number of maxima and minima in any finite interval. 3. f has no infinite discontinuities. Naturally, functions useful in physics, e.g., to model wave phenomena, tend to be reasonable in precisely this sense. Finally, let\u2019s reemphasize the connection to linear algebra: the Fourier transform is a linear transformation between vector spaces of functions. If we denote the Fourier transform by the symbol \u2018\u2018F,\u2019\u2019 we can write the above definitions in the compact form \u02dcf (\u03c9) = F [f (t)], f (t) = F \u22121[ \u02dcf (\u03c9)]. Then again, linearity of the Fourier transform means that F [\u03b1f (t) + \u03b2g(t)] = \u03b1 \u02dcf (\u03c9) + \u03b2\u02dcg(\u03c9) iff F [f (t)] = \u02dcf (\u03c9), F [g(t)] = \u02dcg(\u03c9) for all \u03b1, \u03b2 \u2208 R. Again, this is like a matrix transformation, but in the infinite, continuous limit. 3.2.1 Example: Fourier Transform of a Gaussian Pulse One of the most useful Fourier transforms that can be easily calculated is of the Gaussian pulse, Using the Fourier transform definition (3.20), f (t) = Ae\u2212\u03b1t2. \u02dcf (\u03c9) = \u221e \u2212\u221e Ae\u2212\u03b1t2+i\u03c9t dt. (3.24) (3.25) (3.26) (3.27) To evaluate the integral, we need to perform a mathematical trick, completing the square in the exponent. That is, let\u2019s rewrite the argument in the exponent in the form a(t \u2212 b)2 + c = at2 \u2212 2abt + c \u2212 ab2. Equating powers of t, we find the new coefficients: t2: a = \u03b1 t1: \u22122ab = i\u03c9 =\u21d2 b = \u2212i\u03c9 2\u03b1 t0: c \u2212 ab2 = 0 =\u21d2 c = \u2212 \u03c92 4\u03b1. (3.28) 62 Chapter 3. Fourier Analysis Substituting the new form of the exponent and letting t \u2212\u2192 t + b, \u02dcf (\u03c9) = \u221e \u2212\u221e Ae\u2212at2+c dt = Aec \u03c0 a = A \u03c0 \u03b1 exp \u2212, \u03c92 4\u03b1 (3.29) where we evaluated the integral by comparison to the standard normalized form", " of the Gaussian (you should memorize this, by the way), \u221e \u2212\u221e \u221a 1 2\u03c0\u03c3 exp \u2212 x2 2\u03c32 dx = 1, (3.30) (Gaussian normalization) by taking the standard deviation \u03c3 to be 1/ Gaussian is a Gaussian. \u221a 2\u03b1. Hence we see that the Fourier transform of a Notice that while the standard deviation of the original Gaussian is \u03c3t = 1 \u221a 2\u03b1, the standard deviation of the Fourier transform is \u221a 2\u03b1. \u03c3\u03c9 = (3.31) (3.32) Since the standard deviation is a measure of the width of a function (more precisely the square root of the variance of the function), we can see that as a increases, the width of the original pulse decreases, but the width of the transform increases. This reflects a general property of Fourier transforms, that \u03b4t \u223c 1 \u03b4\u03c9, (3.33) (uncertainty relation) where \u03b4t is the \u2018\u2018width\u2019\u2019 of f (t) and \u03b4\u03c9 is the width of \u02dcf (\u03c9) (for root-mean-square widths and Gaussian functions, there is a strict equality here). This is precisely the same uncertainty principle that is a fundamental principle in quantum mechanics. 3.3 The Fourier Transform in Optics We will spend essentially the rest of the course using the Fourier transform to understand wave optics. But first let\u2019s review a few of the physically important Fourier transforms that you should know, and briefly discuss examples of situations where they come up. The major use of the Fourier transform, as we will see, is in Fourier optics. The central effect behind Fourier optics is that within the paraxial approximation, a thin lens acts as a Fourier-transform computer. That is, given a scalar electric field E(x) one focal length before the lens, the field after the lens is related to F [E(x)]. But Fourier transforms show up everywhere in optics as well as in the rest of physics. Some of the most important and useful ones are summarized here. You should memorize these. 1. The Fourier transform of a Gaussian is a Gaussian. F e\u2212t2/2 = e\u2212\u03c92/2 (3.34) (Fourier transform of Gaussian) Eo(x)Fo[Eo(x)]ff 3.4 Delta Function 63 Recall that the spherical mirrors of resonators act as lenses. We will see that Gaussian beams are electromagnetic field modes of a spherical-mirror resonator. A simple way to see this is that the modes must be Fourier transform of itself in order to resonate (repeat itself) in the cavity. The Gaussian function is the most localized function that has this property. 2. The Fourier transform of an exponential is a Lorentzian. F e\u2212|t| = 2 1 + \u03c92 (3.35) (Fourier transform of exponential) Atoms decay exponentially due to spontaneous emission. If Ne is the number of atoms in the excited state, the decay law is Ne(t) = Ne(0) exp(\u2212\u0393t). It turns out that the Fourier transform of the time dependence of the emission gives the radiation spectrum for spontaneous emission. For atoms obeying the exponential decay law, the emission spectrum is Lorentzian. 3. The Fourier transform of a square pulse is a sinc function. The rectangular function is defined by F [rect(t)] = sinc(\u03c9/2) := sin(\u03c9/2)/(\u03c9/2) (Fourier transform of square pulse) (3.36) rect(t) := \uf8f1 \uf8f2 \uf8f3 1 1/2 0 if if if |t| < 1/2 |t| = 1/2 |t| > 1/2. (3.37) The far-field diffraction pattern of a uniformly illuminated slit is a sinc function for this reason, a fact that is related to the Fourier-transform property of a lens as mentioned above. 4. The Fourier transform of a constant is a delta function. F 1 2\u03c0 = \u03b4(\u03c9) (3.38) (Fourier transform of constant) We haven\u2019t yet defined the delta function but we will do so shortly. For now it suffices to think of the delta function \u03b4(t) as a unit-area pulse in the limit as \u03b4t tends to zero. In other words, this relation is the extreme limit of the uncertainty principle. 5. The Fourier transform of a delta function is a constant. F [\u03b4(t)] = 1 (3.39) (Fourier transform of delta function) This is the opposite extreme of the uncertainty principle", ": an arbitrarily short pulse has an arbitrarily broad spectrum. 3.4 Delta Function As we just mentioned, a delta function is an idealized limit of a very short pulse. But this is a sloppy definition. To do this in a mathematically well-defined way, we need to consider a sequence of functions hn(t) that have the following properties: 1. The hn are \u2018\u2018reasonable\u2019\u2019 (e.g., simply peaked around t = 0). 2. The width of hn converges to zero as n \u2212\u2192 \u221e. 3. The hn are normalized: \u221e \u2212\u221e hn(t) dt = 1 for all n. 64 Chapter 3. Fourier Analysis For example, we can use Gaussian functions. In normalized form, we can write and if we choose \u03c3n = 1/ \u221a 2\u03c0n, the functions become hn(t) = \u221a 1 2\u03c0\u03c3n exp \u2212, t2 2\u03c32 n Then consider the integral hn(t) = n exp \u2212\u03c0n2t2. \u221e \u2212\u221e hn(t)f (t) dt, (3.40) (3.41) (3.42) for an arbitrary test function f (t), which we again assume to be \u2018\u2018reasonable.\u2019\u2019 We will define the delta function \u03b4(t) such that \u221e \u2212\u221e \u03b4(t)f (t) dt := lim n\u2212\u2192\u221e \u221e \u2212\u221e hn(t)f (t) dt, (3.43) (definition of delta function) or in inner-product notation, \u03b4, f := lim \u03b4, hn n\u2212\u2192\u221e (3.44) (definition of delta function) Notice that the order of the integral sign and the limit are important; i.e., the integral and the limit don\u2019t commute. This is because limn\u2212\u2192\u221e hn(t) does not exist. Thus, the delta function only makes sense as part of the argument of an integral. But as physicists, we often write \u03b4(t) = lim n\u2212\u2192\u221e hn(t) (sloppy!) (3.45) (\u2018\u2018definition\u2019\u2019 of delta function) as a shorthand for the above (more careful) definition. Similarly, note that limn\u2212\u2192\u221e hn(t) = 0 for all t = 0, so it is useful to think of \u03b4(t) as a \u2018\u2018unit-area function that is zero everywhere but \u221e at t = 0,\u2019\u2019 as long as you are careful about it. Now we will review and derive a few properties of the delta function. The list that follows is by no means exhaustive. 1. \u03b4(t) is normalized: \u221e \u2212\u221e \u03b4(t) dt = 1 (3.46) (normalization of delta function) Proof. Since the functions hn(t) are normalized, 2. Projection property of \u03b4(t): \u221e \u2212\u221e \u03b4(t) dt = lim n\u2212\u2192\u221e \u221e \u2212\u221e hn(t) dt = 1. (3.47) \u221e \u2212\u221e \u03b4(t)f (t) dt = f (0). (projection property of delta function) (3.48) Proof. Assuming f has a Taylor expansion about t = 0, f (t) = f (0) + f (0)t + 1 2 f (0)t2 + \u00b7 \u00b7 \u00b7 (3.49) Noting that \u221e \u2212\u221e hn(t)tm dt = n \u221e \u2212\u221e tm exp(\u2212\u03c0n2t2) dt = 1 + (\u22121)m 2 1 + m 2 \u0393 \u03c0\u2212(m+1)/2 n\u2212m (3.50) 3.4 Delta Function for the Gaussian hn(t)\u2019s defined above, we can write f (0) 2\u03c0n2 + O As n \u2212\u2192 \u221e, all the higher order terms vanish, leaving just f (0). hn(t)f (t) dt = f (0) + \u2212\u221e \u221e 65 (3.51). 1 n4 3. Shifted projection property of \u03b4(t): \u221e \u2212\u221e \u03b4(t \u2212 a)f (t) dt = f (a). (3.52) This can be proved by letting t \u2212\u2192 t + a in the integral, and then using Eq. (3.48) to evaluate the result. (shifted projection property of delta function) 4. Fourier transform of \u03b4(", "t): F [\u03b4(t \u2212 a)] = ei\u03c9a. (3.53) (Fourier transform of delta function) The proof of this follows trivially from the previous property and the definition (3.20) of the Fourier transform. 5. Integral representation of \u03b4(t): \u03b4(t \u2212 a) = 1 2\u03c0 \u221e \u2212\u221e e\u2212i\u03c9(t\u2212a) d\u03c9. (3.54) This important representation of the delta function follows from the inversion of the previous property, F \u22121[ei\u03c9a] = \u03b4(t \u2212 a), (integral representation of delta function) (3.55) and the application of the inverse transform integral (3.19). This final property above, the integral representation (3.54) of \u03b4(t) is useful, for example, for showing the orthogonality of the harmonic basis functions. As in the periodic case, we can define two suitable functions f (t) and g(t) to be orthogonal if \u221e \u2212\u221e f \u2217(t) g(t) dt = 0, (3.56) (orthogonality of functions) which is just f, g = 0 in inner-product notation. For harmonic functions, exp(\u2212i\u03c9t) and exp(\u2212i\u03c9t) are orthogonal because \u221e \u2212\u221e exp(i\u03c9t) exp(\u2212i\u03c9t) dt = 2\u03c0\u03b4(\u03c9 \u2212 \u03c9), (3.57) which is zero if \u03c9 = \u03c9. Also, since \u03b4(t) is normalized, we see that in the form exp(\u2212i\u03c9t)/2\u03c0 [or alternately, exp(i2\u03c0\u03bdt)], the harmonic functions can form an orthonormal set. Subject to the conditions on hn(t) above, the delta function (and the properties we have derived) are independent of the particular choice of hn(t). However, sometimes, some ambiguous integrals of delta functions sometimes crop up, such as (orthogonality of harmonic functions) \u221e dt \u03b4(t) =?, (3.58) which has no well-defined value. Depending on the functions hn(t) used to define \u03b4(t), this function can take on essentially any value (though more reasonably, one could expect it to be between zero and unity). This is one case where the more rigorous definition helps in physics; the physical model leading to the delta function defines the choice of hn(t), and thus the value of this integral. 0 66 3.5 Exercises Problem 3.1 Chapter 3. Fourier Analysis Compute the Fourier transform \u02dcf (\u03c9) of the rectangular-pulse function f (t) = 1, 0 |t| < a/2 elsewhere. (3.59) Show from your answer that as the pulse width a decreases, the width of \u02dcf (\u03c9) increases. Problem 3.2 Suppose that f (t) is a real, odd [f (\u2212t) = \u2212f (t)], periodic function; what constraints are satisfied by the Fourier coefficients cn for this function? Problem 3.3 If cn is the nth Fourier coefficient for the periodic function f (t), write down an expression for the nth Fourier coefficient for f (t + \u03b1) in terms of cn. Problem 3.4 (a) Compute the Fourier series for a square wave of unit period, given by where sgn(x) is the sign of x, i.e., f (t) = sgn[sin(2\u03c0t)] sgn(x) := \uf8f1 \uf8f2 \uf8f3 1 0 \u22121 if x > 0 if x = 0 if x < 0. (b) Make plots of the truncated Fourier series, fN (t) = N n=\u2212N cne\u2212in\u03c9t, (3.60) (3.61) (3.62) along with the exact function for N = 1, 3, 5, 21, and 101 to check how good the finite approximations are. You should notice a ringing behavior near the discontinuities, where the approximant fN (t) overshoots the actual function f (t). This is called the Gibbs phenomenon, and persists at an essentially constant (nonzero) value even for arbitrarily large N. From your plots, estimate the magnitude of the Gibbs overshoot. (c) We said that the functions fN (t) converge to f (t). But the Gibbs phenomenon says that for any N, there is always some t such that fN (t) is at least some minumum distance away from f (t). Explain how these two statements are consistent. Problem", " 3.5 One application of Fourier series is in synthesizing waveforms or images with relatively little information. For example, the JPEG image compression algorithm achieves much of its compression by removing harmonic components, and (analog) electronic organs can use relatively few harmonics (on the order of 10) to make a passable imitation of a musical instrument.1 A common feature of truncated Fourier series is that relatively few terms are needed to make a fairly good approximant, but for a high accuracy approximant, many terms may be necessary. Mathematically, the initial convergence of the Fourier series is rapid, but the convergence rate becomes very slow 1A nice discussion of this example appears in T. W. K\u00f6rner, Fourier Analysis (Cambridge, 1988). 3.5 Exercises 67 by the time many terms have accumulated. (This is even the case for continuous functions, where the Gibbs phenomenon is absent.) As a simple model, suppose you want to synthesize a triangle wave, f (t) := 1 \u2212 4|t|/T, f (t + T ) = f (t \u2212 T ) |t| < T /2 elsewhere, using as few harmonics as possible. (a) Compute the Fourier series for the triangle wave as defined above. (b) Consider the partial sum, fN (t) = N n=\u2212N cne\u2212in\u03c9t, and let\u2019s define the fractional error of the partial sum to be normalized root-mean-square error: \u03b5N := T |fN (t) \u2212 f (t)|2 dt 0 T 0 |f (t)|2 dt. (3.63) (3.64) (3.65) How large should N be in the partial sum to obtain an accuracy of 10%? 1%? 0.1%? 0.01%? (A computer may be of great help in solving this problem.) Problem 3.6 Let f (t) be a function with Fourier transform \u02dcf (\u03c9). Show that the Fourier transform of f (t) cos(\u03b1t) is [ \u02dcf (\u03c9 + \u03b1) + \u02dcf (\u03c9 \u2212 \u03b1)]/2. Problem 3.7 Prove the following form of the Poisson sum rule: \u221e \u03b4(t \u2212 n) = \u221e n=\u2212\u221e n=\u2212\u221e cos(2\u03c0nt). (3.66) Problem 3.8 Let \u02dcf (\u03c9) be the Fourier transform of f (t). Show that the Fourier transform of g(t) := f (t) cos2(\u03b1t) is \u02dcg(\u03c9) = \u02dcf (\u03c9) + 1 2 1 4 \u02dcf (\u03c9 + 2\u03b1) + \u02dcf (\u03c9 \u2212 2\u03b1). (3.67) Problem 3.9 Show that if f (t) and \u02dcf (\u03c9) form a Fourier transform pair, then so do f (\u03b1t) and \u02dcf (\u03c9/\u03b1)/|\u03b1| (for \u03b1 = 0). Problem 3.10 (a) Compute the Fourier transform of f (t) = UH(t) e\u2212at, where UH(t) is the Heaviside step function [i.e., UH(t > 0) = 1 and UH(t < 0) = 0], and a > 0. (b) Use your result from (a) to compute the Fourier transform of g(t) = UH(t) te\u2212at for a > 0. (Try to do this without any integration other than what you have already done.) Chapter 4 Review of Electromagnetic Theory 4.1 Maxwell Equations in Vacuum Electromagnetism is the fundamental theory that underlies all of wave optics, so before proceeding with this more general theory of optics, we will stop and review the basics. Maxwell\u2019s equations form the basis of electromagnetism. We will write Maxwell\u2019s equations in free space as \u2207 \u00b7 E = 0 \u2207 \u00b7 H = 0 (Gauss\u2019 law in free space) (no magnetic monopoles) \u2207 \u00d7 E = \u2212\u00b50 \u2202H \u2202t \u2207 \u00d7 H = 0 \u2202E \u2202t (Faraday\u2019s law) (Amp\u00e9re\u2019s law with Maxwell\u2019s correction in free space) where E(r, t) is the electric field, H(r, t) is the magnetic field, (Maxwell\u2019s equations) (4.1) 0 8.85 \u00d7 10\u221212 F m (4.2) is the electric permittivity of the vacuum, and \u00b50 := 4\u03c0 \u00d7 10\u22127 N A2 (4.3) (magnetic permeability of the vacuum", ") is the magnetic permeability of the vacuum. Also recall the notation \u2207 := \u02c6x \u2202 \u2202x + \u02c6y \u2202 \u2202y + \u02c6z \u2202 \u2202z (4.4) for the vector derivative operator that we used in writing down the Maxwell equations. Actually, we won\u2019t use Maxwell\u2019s equations very often in the form we wrote. Rather, we will use them to derive the wave equation, the fundamental equation of optics. To do this, we first apply (\u2207\u00d7) to Faraday\u2019s law: Now we will use the vector identity \u2207 \u00d7 (\u2207 \u00d7 E) = \u2212\u00b50 \u2202 \u2202t (\u2207 \u00d7 H). \u2207 \u00d7 (\u2207 \u00d7 A) = \u2207(\u2207 \u00b7 A) \u2212 \u22072A, (4.5) (4.6) 70 Chapter 4. Review of Electromagnetic Theory which holds for any vector A to evaluate the left-hand side of this equation, and we will use Amp\u00e9re\u2019s law to evaluate the right-hand side: \u2207(\u2207 \u00b7 E) \u2212 \u22072E = \u2212\u00b50 \u2202 \u2202t. 0 \u2202E \u2202t (4.7) The first term vanishes, as we can see from Gauss\u2019 law. Then cleaning things up a bit, we arrive at the standard form of the wave equation: Here, \u22072E \u2212 1 c 2 0 \u22022E \u2202t2 = 0. c0 = \u221a 1 0\u00b50 (4.8) (wave equation) (4.9) (speed of light in vacuum) is the speed of light in vacuum. This is actually now a defined quantity, given by Since \u00b50 and c0 are both exactly defined quantities, the exact value of 0 is defined by (definition of speed of light in vacuum) (4.10) c0 := 2.997 924 58 \u00d7 108 m/s. 0 := 1 \u00b50c 2 0. (4.11) (electric permittivity of the vacuum) Finally, we note that the magnetic field H satisfies the same wave equation as the electric field. 4.2 Intensity One of the most important features of electromagnetic waves (time-dependent solutions of the Maxwell or wave equations) is that they transport energy. The flow of energy is described by the Poynting vector, defined by S := E \u00d7 H. (4.12) (Poynting vector) The Poynting vector points in the direction of energy flow; note that it is orthogonal to both the electric and magnetic field vectors. Energy transport and the Poynting vector are explored further in Problem 4.4. More commonly, energy transport is quantified by the intensity of the electromagnetic wave. The intensity is defined as simply the magnitude of the time-averaged Poynting vector, I := |S|, (4.13) (intensity) where the average is taken over time scales long compared to an optical cycle (a few fs) but short compared to other time scales of interest (the angle brackets here denote the time average). Physically, the intensity is a measure of what a realistic optical detector would register is response to the field. Detectors can register time variation of the field energy in up into the MHz or GHz range, but not optical frequencies (which are in the range of hundreds of THz). 4.3 Maxwell Equations in Media 71 4.3 Maxwell Equations in Media Maxwell\u2019s equations take on a slightly different form for electromagnetic fields in a mediumB \u2202t \u2202D \u2202t. (4.14) Here, D is the electric flux density or electric displacement, and B is the magnetic flux density. These equations are still valid only in a source-free region (i.e., free of charges and currents). (Maxwell\u2019s equations in magnetodielectric media) The relation between the E and D fields is determined by the electric properties of the medium: D = 0E + P. (4.15) (electric flux density in terms of polarization) Here, P is the polarization density, defined to be the electric dipole moment per unit volume. Usually the polarization is induced by the electric field, though some materials can be fabricated with permanent polarization (electrets). Similarly the relation between the H and B fields is related to the magnetic properties of the medium: B = \u00b50H + M. (4.16) Here, M is the magnetization density, defined as the magnetic dipole moment per unit volume. Sometimes the magnetization is induced by an external magnetic field, but permanent magnets are relatively commonplace. (magnetic flux density in terms of magnetization) In free space, P = M = 0, so D = 0E and B = \u00b50H,", " in which case we recover the free-space Maxwell equations (4.1) from Eqs. (4.14). In optics, we usually deal with dielectric materials. Thus, from now on we will assume M = 0, so that B = \u00b50H. 4.4 Simple Dielectric Media We will now discuss dielectric media in more detail, but here we will only treat the simplest possible such media. For our purposes here it will be useful to think of the electric field E as something externally applied to the medium, while the polarization P is the response of the medium to the field. Thus, we can think of the polarization as a function of the electric field, as well as space and time: P(r, t) = P(E(r, t); r, t). The simple dielectric medium has a number of constraints on its properties, which greatly simplify things. 1. Linearity: |P| \u221d |E|, so that the superposition principle still holds in the medium. dielectricmediumEP 72 Chapter 4. Review of Electromagnetic Theory 2. Nondispersivity: the speed of light does not depend on the frequency of the electromagnetic waves. In the time domain, this means that there is no hysteresis, since it amounts to saying that P(t) is a function of only E(t), never E(t) for t < t. This is always an idealization, but sufficient for many purposes. 3. Homogeneity: P(E) is independent the position r inside the medium. 4. Isotropy: P(E) is independent of the orientation of the medium with respect to the field, so there is no \u2018\u2018preferred direction\u2019\u2019 in the medium. The only way for this to work is for P E. 5. Spatial nondispersivity (locality): P(r) is a function of only E(r), never E(r) for r = r. To satisfy all these conditions, the simple dielectric has a polarization that can be written as a simple proportionality: (4.17) (polarization in terms of susceptibility) The constant \u03c7 is the electric susceptibility, which competely characterizes the simple medium. Thus, we can rewrite Eq. (4.15) as P = 0\u03c7E. where D = E, (4.18) (electric flux density in simple dielectric) := 0(1 + \u03c7) (4.19) (electric permittivity of dielectric) is the electric permittivity of the medium, and the ratio 0 = 1 + \u03c7 (4.20) (dielectric constant) is the dielectric constant. If we rederive the wave equation using the Maxwell equations for a dielectric, we find that the equation has the same form as before, \u22072E \u2212 1 c 2 \u22022E \u2202t2 = 0, (4.21) (wave equation for dielectric medium) (H also satisfies this equation) except that c0 is replaced by c := c0 n, (4.22) (speed of light in dielectric medium) which gives the speed of light in the dielectric medium in terms of the index of refraction, defined as the square root of the dielectric constant: n := 0. (4.23) (index of refraction) Thus we see how the dielectric properties of the medium relate to the refractive index that we used in ray optics. Finally, note that in a simple medium (as well as in vacuum), each component of E and H separately satisfies the wave equation. Thus, we can write a scalar wave equation, \u22072E \u2212 1 c 2 \u22022E \u2202t2 = 0, (4.24) (scalar wave equation) where E now stands for any component of E or H. Note that this equation only holds in cases like simple media where the different vector components don\u2019t get mixed. For example, you have to be more careful when analyzing a refractive interface, which in general mixes different field components except in a special basis. However, even when the scalar wave equation is incorrect, it can be a useful approximation, and is certainly easier to handle than the full vector case. 4.5 Monochromatic Waves and Complex Notation 73 4.5 Monochromatic Waves and Complex Notation In general, any function of the form E(x, t) = E(x \u00b1 ct) is a solution to the scalar wave equation in a homogeneous medium. These solutions represent functions that do not change shape but propagate rightward (x \u2212 ct) or leftward (x + ct) with speed c. There are many possible functions of this form, and to handle more general media, we can study monoch", "romatic waves, which have a harmonic time dependence with a single angular frequency \u03c9: E(r, t) = E(r) cos(\u03c9t + \u03c6), H(r, t) = H(r) cos(\u03c9t + \u03c6). (4.25) Here, \u03c6 and \u03c6 are constant phase offsets. As in the Fourier analysis case, we can instead use complex harmonic functions and define the complex monochromatic fields E(\u00b1) by breaking up the real field into its positive- and negative-frequency parts: (monochromatic fields) E(r, t) = E(r) e\u2212i\u03c6 2 e\u2212i\u03c9t + E(r) ei\u03c6 2 ei\u03c9t =: E(+)(r)e\u2212i\u03c9t + E(\u2212)(r)ei\u03c9t. (4.26) Recall that we are defining E(\u00b1) to go with e\u2213i\u03c9t since by convention e\u2212i\u03c9t corresponds to the positive frequency \u03c9 and ei\u03c9t = e\u2212i(\u2212\u03c9)t corresponds to the negative frequency (\u2212\u03c9). Of course, we can do the same for all of the other fields H, D, B, P. The physical field is just the sum of the positive- and negativefrequency parts. But notice that these parts are complex conjugates, as is required to get a real (physical) field. Thus, we can always write the physical field as E(+) with its conjugate: E(+)(r)e\u2212i\u03c9t E(r, t) = E(+)(r) e\u2212i\u03c9t + c.c. = 2Re (complex monochromatic fields) (4.27). Since all of the time dependence in E(+) is of the form e\u2212i\u03c9t, keeping track of only E(+) will greatly simplify electromagnetism calculations. In particular, the Maxwell equations take on a simpler form for complex monochromatic fields. Because the time dependence is of the simple form e\u2212i\u03c9t, and because we can always make the formal identification \u2202 \u2202t e\u2212i\u03c9t = \u2212i\u03c9e\u2212i\u03c9t, for a monochromatic field. Then the Maxwell equations become \u2202 \u2202t \u2261 \u2212i\u03c9 \u2207 \u00b7 D(+) = 0 \u2207 \u00b7 B(+) = 0 \u2207 \u00d7 E(+) = i\u03c9B(+) \u2207 \u00d7 H(+) = \u2212i\u03c9D(+), (4.28) (4.29) with D(+) = 0E(+) + P(+) and B(+) = \u00b50H(+). (Maxwell\u2019s equations, monochromatic field) (4.30) Note that all of this complex notation stuff works out because the wave and Maxwell equations are all linear. That is, the real and imaginary parts (equivalently, positive- and negative-frequency parts) satisfy these equations separately, so they are equivalent to the regular Maxwell equations. The imaginary part is just an extra piece that helps to simplify calculations, which doesn\u2019t do any harm so long as you get rid of it at the end. In fact, in writing down the Maxwell equations for monochromatic fields, we are implicitly taking a Fourier transform of the general field. Again, this follows from the linearity of the field equations along with the harmonic time dependence of the harmonic fields. We can write the solutions to the monochromatic 74 Chapter 4. Review of Electromagnetic Theory Maxwell equations as E(+)(r, \u03c9) (and so on for H(+), D(+), B(+), etc.), and thereby obtain the general solution as an inverse Fourier transform, E(r, t) = 1 2\u03c0 \u221e \u2212\u221e E(+)(r, \u03c9)e\u2212i\u03c9t d\u03c9, (4.31) where we can identify E(\u2212)(r, \u03c9) = E(+)(r, \u2212\u03c9). In other words, the monochromatic Maxwell equation establishes the forms of the harmonic basis functions. We can then build up arbitrary solutions out of these basis functions. (arbitrary field in terms of complex monochromatic components) 4.5.1 Alternate Complex Notation Another common convention for complex notation involves working with a single complex field (that we will write as E ) with time dependence e\u2212i\u03c9t, such that the real part is the physical field: E(r, t) = Re EEE (r)e\u2212i\u03c9t. (4.32) This is the same idea as before, but rather than thinking of a complex field as always accompanied by its conjugate, the idea here is to carry around an extra imaginary field EEE (r) sin(\u03c9", "t + \u03c6), where \u03c6 is the phase of EEE. This extra, imaginary field is discarded at the end of the calculation by taking the real part of the answer. This alternate notation is common in textbooks on optics and electrodynamics, and the E(\u00b1) notation is more common in quantum optics. The E(\u00b1) notation has some advantages, including a better emphasis on the physical field E (since E(\u00b1) is part of E), and it is a bit less prone to causing confusion when computing nonlinear quantities of the field, such as the intensity. We will stick to the E(\u00b1) notation here. 4.6 Intensity in Complex Notation Quantities that are not linear in the fields are a bit more subtle. Still, for monochromatic waves, the Poynting vector and intensity are relatively simple: S = E \u00d7 H E(+)e\u2212i\u03c9t + E(\u2212)ei\u03c9t = = E(+) \u00d7 H(\u2212) + E(\u2212) \u00d7 H(+) + E(+) \u00d7 H(+)e\u2212i2\u03c9t + E(\u2212) \u00d7 H(\u2212)ei2\u03c9t. \u00d7 H(+)e\u2212i\u03c9t + H(\u2212)ei\u03c9t (4.33) For intensity calculations, we time-average the 2\u03c9 terms, so S = E(+) \u00d7 H(\u2212) + E(\u2212) \u00d7 H(+) = E(+) \u00d7 H(\u2212) + c.c. = 2Re E(+) \u00d7 H(\u2212). (4.34) The optical intensity then is simply the magnitude of the Poynting vector, I = E(+) \u00d7 H(\u2212) + c.c. = 2Re E(+) \u00d7 H(\u2212), (intensity of monochromatic waves) (4.35) but now in terms of the complex field components. 4.6.1 Complex Notation for Simple Dielectric Media Since D = E and B = \u00b50H for a simple dielectric medium, we can rewrite the Maxwell equations for a monochromatic field in a simple dielectric as \u2207 \u00b7 E(+) = 0 \u2207 \u00b7 H(+) = 0 \u2207 \u00d7 E(+) = i\u03c9\u00b50H(+) \u2207 \u00d7 H(+) = \u2212i\u03c9E(+). (Maxwell\u2019s equations, simple dielectric) (4.36) 4.7 Plane Waves 75 This is essentially the same set as the original Maxwell equations, but with \u2202/\u2202t \u2212\u2192 \u2212i\u03c9. Thus, the same replacement applies in the vector wave equation (4.21), and we obtain the vector Helmholtz equation \u22072 + k2 E = 0, (4.37) (vector Helmholtz equation) \u221a where k = \u03c9/c = \u03c9 \u00b50 = nk0 is the wave number in the medium, and k0 = \u03c9/c0 is the wave number in vacuum. In a simple dielectric, each component of each field again satisfies this equation separately, and so each satisfies the scalar Helmholtz equation \u22072 + k2 E = 0, (scalar Helmholtz equation) (4.38) (scalar Helmholtz equation) This is the time-independent version of the scalar wave equation (4.24), where we must keep in mind that a solution to the Helmholtz equation has an implied time dependence of e\u2212i\u03c9t. 4.7 Plane Waves The plane wave is one of the simplest and most important solutions of the scalar wave equation. We can define the plane wave as the monochromatic field E(+)(r) = E(+) 0 eik\u00b7r. (4.39) Recall that this expression has an implied harmonic time dependence: E(+)(r, t) = E(+) 0 ei(k\u00b7r\u2212\u03c9t). (4.40) (scalar plane wave) Here, E(+) the inner product k \u00b7 r is simply kxx + kyy + kzz. 0 is a complex constant scalar and k is the wave vector. In terms of the components (kx, ky, kz), The plane wave satisfies the scalar Helmholtz equation. We can see this by observing that \u02c6x \u2202 \u2202x E(+) = \u02c6xikxE(+), (4.41) with similar relations holding for y and z. As for the time dependence of the monochromatic field, we can make the formal identification \u2202/\u2202x \u2261 ikx for the plane wave. Combining all three components, Iterating the differentiation once, \u2207E(+) = ikE(+). \u22072E(+) = \u2212k \u00b7", " kE(+) = \u2212k2E(+). (4.42) (4.43) Thus, we recover the Helmholtz equation, where we can identify the wave number we defined before with the magnitude of the wave vector k. E(+) 0, Now let\u2019s examine the phase of the plane wave. If we define \u03c6 to be the phase of the complex amplitude 0 =: |E(+) E(+) 0 |ei\u03c6, (4.44) then we can write (4.45) 0 Since ei\u03c6 = ei(\u03c6+2\u03c0), the wave repeats itself every 2\u03c0 in phase. Thus, the wave fronts (surfaces of constant phase) are given by E(+)(r) = |E(+) |ei(k\u00b7r+\u03c6). k \u00b7 r = kxx + kyy + kzz = 2\u03c0q + \u03c6 (4.46) (plane-wave fronts) 76 Chapter 4. Review of Electromagnetic Theory for integer q. These surfaces of constant k \u00b7 r are planes perpendicular to k, separated by the wavelength \u03bb := 2\u03c0 k (4.47) (wavelength) in space (i.e., \u03bb is the distance along k corresponding to a phase shift of 2\u03c0). Let\u2019s simplify a bit by orienting the z-axis along k, so that k = k\u02c6z. Then the plane wave has the form E(+)(r) = E(+) 0 eikz. (4.48) The corresponding real wave is E(r, t) = E(+)(r) e\u2212i\u03c9t + c.c. = |E(+) 0 | ei(kz\u2212\u03c9t+\u03c6) + c.c. = 2|E(+) 0 | cos(kz \u2212 \u03c9t + \u03c6) (4.49) In standard form, the real plane wave is E(r, t) = E0 cos(kz \u2212 \u03c9t + \u03c6), (4.50) (real plane wave) where E0 := 2|E(+) 0 |. Note that we can also write the plane wave in the form where again c = \u03c9/k is the speed of light. As time evolves, the points of stationary phase\u2014constant (z \u2212 ct)\u2014move with phase velocity c. E(r, t) = E0 cos[k(z \u2212 ct) + \u03c6] (4.51) In a medium, as compared to the vacuum, the spatial quantities are multiplied by the refractive index, whereas the time quantities are unchanged: c = c0/n \u03bb = \u03bb0/n k = nk0 \u03c9 = \u03c90. (4.52) Here, as before, the subscripted quantities refer to the vacuum. klwave frontslzFixed t:T = 2p/wtFixed z: 4.8 Vector Plane Waves 77 4.8 Vector Plane Waves In the general case, the electric field is a vector, and sometimes it is insufficient to treat the electromagnetic field as a scalar. Thus we should treat the polarization or orientation of the field explicitly\u2014along with the direction of motion that we have already treated\u2014and write E(r, t) = E(+)(r)e\u2212i\u03c9t + c.c. = E (+) 0 ei(k\u00b7r\u2212\u03c9t) + c.c., (4.53) and since each component of this vector field separately satisfies the (homogeneous-medium) scalar wave equation (4.24), this field satisfies the vector wave equation (4.8). We can also guess that this form holds for the magnetic field, since each of its components satisfies the scalar wave equation: H(r, t) = H(+)(r)e\u2212i\u03c9t + c.c. = H (+) 0 ei(k\u00b7r\u2212\u03c9t) + c.c., Using the Maxwell equations from Eqs. (4.36), \u2207 \u00d7 H(+) = \u2212i\u03c9E(+) \u2207 \u00d7 E(+) = i\u03c9\u00b50H(+), we can use the identification \u2207 \u2261 ik for the plane wave to write k \u00d7 H(+) = \u2212\u03c9E(+) k \u00d7 E(+) = \u03c9\u00b50H(+). (4.54) (4.55) (4.56) By inspecting this pair of equations, we see that k, H0, and E0 are all mutually orthogonal (i.e., they determine a basis in three dimensions), and they must be arranged as shown below. Since E0 and H0 are orthogonal to k, plane waves are often called Transverse Electromagnetic (TEM) waves. Taking the magnitudes of", " the above two Maxwell equations, we find H0 = H0 = \u03c9 k k \u03c9\u00b50 E0 E0, (4.57) where again E0 := 2|E(+) 0 | and similarly H0 := 2|H (+) 0 |. For these two equations to be consistent, we require which leads to \u03c9 k = k \u03c9\u00b50, \u221a k = \u03c9 \u00b50 = \u03c9 c = n\u03c9 c0 = nk0, (4.58) (4.59) which is the same as the condition for the wave to satisfy the Helmholtz equation. EkH 78 Chapter 4. Review of Electromagnetic Theory 4.8.1 Wave Impedance Now if we consider the ratio of the electric and magnetic fields, E0 H0 = \u03c9\u00b50 k = \u00b50c0 n = 1 n \u00b50 0, (4.60) we can define this constant ratio of E0 to H0 \u03b7 := 1 n \u00b50 0, (4.61) (impedance of dielectric medium) which has the dimensions of impedance. Thus \u03b7 is the wave impedance of the medium. We can also write \u03b7 = \u03b70 n, (4.62) (wave impedance modified by refractive index) where \u03b70 := \u00b50 0 = \u00b50c0 = 1 0c0 \u2248 377 \u2126 (4.63) (impedance of vacuum) is the wave impedance of the vacuum. Returning to the Poynting vector, S = E(+) \u00d7 H(\u2212) + c.c. = 0 H (\u2212) E(+) 0 + c.c. \u02c6k = |2 |E(+) 0 \u03b7 + c.c. \u02c6k = E 2 0 4\u03b7 + c.c. \u02c6k = E 2 0 2\u03b7 \u02c6k, (4.64) where \u02c6k := k/k. Thus, for a plane wave, we can write the intensity in terms of the real amplitude E0 as Alternately, we can also write the intensity I = E 2 0 2\u03b7. (4.65) (intensity in terms of real field amplitude) I = 2|E(+)| 2 \u03b7 (4.66) (intensity in terms of complex amplitude) in terms of the complex field |E(+)|. As an example, a typical light bulb uses 100 W of electric power. Assuming a light conversion efficiency of \u223c 2.5% and a diameter of \u223c 5 cm, the intensity at the surface is the optical power per unit area: The electric field is thus at the surface. I = P A = 2.5 W 4\u03c0(2.5 cm)2 = 0.03 W cm2. E0 = 2\u03b70I = 5 V cm (4.67) (4.68) A laser, as we will see later, has much higher intensity and electric field even for a much more modest power. A low-power He-Ne laser might typically have P \u223c 2 mW of output power and a beam diameter of d \u223c 0.8 mm. Making the simplistic estimate I = P /d2, we find an intensity of 0.3 W/cm2 and an electric field of 15 V/cm, values larger than the light bulb despite a much smaller power rating. 4.9 Exercises 4.9 Exercises Problem 4.1 79 (a) Prove the vector identity \u2207 \u00d7 (\u2207 \u00d7 A) = \u2207(\u2207 \u00b7 A) \u2212 \u22072A (4.69) for an arbitrary vector A. (Hint: try doing this first for only one Cartesian component of this equation and then argue that the other components follow in the same way.) (b) Use this identity and Maxwell\u2019s equations satisfies the wave equation to show that the magnetic field H(r, t) in free space 1 c2 0 which is the same equation satisfied by the electric field E(r, t). Here, c0 is the speed of light in vacuum. \u22022 \u2202t2 H = 0, \u22072H \u2212 (4.70) Problem 4.2 For a plane-wave solution of the wave equation, E(r, t) = E0 cos(k \u00b7 r \u2212 \u03c9t + \u03c60), (4.71) we can write the constant E0 as \u02c6eE0, where the unit vector \u02c6e is the polarization and E0 is a scalar constant. Show that the wave cannot be polarized along the k direction. Prove this directly; do not make any assumption about the form of the magnetic field. Problem 4.3 Recall the wave equation for the electric field in a simple, linear dielectric, Consider the particular solution", ", \u22072E \u2212 1 c2 \u22022 \u2202t2 E = 0. E(r, t) = \u02c6zE0 cos(kx \u2212 \u03c9t + \u03c60), (4.72) (4.73) where E0 and \u03c60 are constants and \u02c6z is the unit vector along the z-direction. (a) Show explicitly that this solution satisfies the wave equation provided \u03c9/k = c. (b) Show explicitly that the solution is invariant under transformations of the form t \u2212\u2192 t + T, where T = 2\u03c0/\u03c9, and thus show that the solution is periodic in time with period T (i.e., the time required for the wave to repeat itself at fixed position is T ). (c) Show that the wave is spatially periodic with period \u03bb, where \u03bb = 2\u03c0/k. (d) Using a similar argument, show that \u03c9/k is the phase velocity (the velocity at which the wave peaks travel) of the wave. (e) Show that the corresponding solution for the magnetic field is H(r, t) = \u2212\u02c6y E0 \u00b50c cos(kx \u2212 \u03c9t + \u03c60). (4.74) (f) Write down the complex form E(+) of the wave in terms of the above real wave. Be explicit about the dependence on all the parameters in the real wave. Problem 4.4 The energy density (energy stored per unit volume) of an electromagnetic field in a simple, linear dielectric is w = E \u00b7 E + \u00b50H \u00b7 H 1 2 (4.75) 80 Chapter 4. Review of Electromagnetic Theory for real field E and H. Consider a plane wave in complex notation of the form E(+)(r, t) = \u02c6zE(+) 0 ei(kx\u2212\u03c9t), with E(+) (a) Show that the time-averaged energy density is given by a complex constant. 0 w = 1 4 E 2 0 + \u00b50H 2 0, (4.76) (4.77) where E0 = 2|E(+) not the definition of E0.) 0 |. What is the significance of the extra factor of 1/2 in this expression? (Hint: it\u2019s (b) What is the real wave corresponding to the complex wave given above? Be explicit about the parameter dependence. (c) Calculate w, the Poynting vector S, and the intensity I for the real plane wave of part (b). (d) Show, for the real plane wave of part (b), that S = cw\u02c6x. With the interpretation that the Poynting vector represents an energy flux density, argue that this means that the electromagnetic energy of a plane wave propagates with velocity c. Problem 4.5 A laser with \u03bb = 200\u03c0 nm points along the \u2212x-direction, and its electric field is polarized in the z-direction. Assume the laser output is a plane wave in free space with intensity 1 W/cm2. Write down expressions for the electric and magnetic fields for the laser beam, giving numerical values for any parameters you use. Don\u2019t worry too much about arithmetic; just reduce your numerical answers to reasonably simple fractions, powers, etc. Problem 4.6 Write down an expression for both the electric and magnetic field for a monochromatic plane wave in vacuum with the following properties: the wave vector is (\u2212k, k, 0)/ 2, the electric field amplitude is E0, and the polarization is along the positive z-direction. Your expressions should only involve these parameters and fundamental constants, and you should explain each aspect of what you write down. (Write down the physical fields, not the complex versions.) \u221a Problem 4.7 Two possible definitions of the intensity in terms of the Poynting vector are the magnitude of the time-averaged Poynting vector, and the time-average of the magnitude of the Poynting vector, I := |S|, I := |S| = S. (4.78) (4.79) The first definition is the one we use here, while the second definition is common in textbooks on optics and electromagnetism. The point of this problem is to examine whether the order of the two operations matters, and if so which ordering is more sensible. Consider the composite field (+) E(+)(r, t) = E 1 (r, t) + E (+) 2 (r, t), (4.80) where the first field is an optical plane wave propagating in the z-direction and polarized in the xdirection, (+) 1 (r, t) = \u02c6xE(+) 10 ei(kz\u2212\u03c9t), E (4.81) 4.", "9 Exercises while the second is a dc field, polarized in the z-direction: (+) 2 (r, t) = \u02c6zE(+) 20. E 81 (4.82) To make algebra easier, take the special case z = 0, with E(+) 20 both being real numbers. (+) (r, t) that is consistent with E 1 (r, t). 10 and E(+) (+) (a) Write down the magnetic field H 1 (+) (b) We will take the magnetic field corresponding to the second electric field to be zero, H 2 Explain why this is consistent with Maxwell\u2019s equations. (r, t) = 0. (c) Compute the Poynting vector for the combined field. (d) Compute the time average of the Poynting vector, and then the intensity according to the first definition. (e) Compute the magnitude of the Poynting vector. The time average will then be difficult to compute; to make it easier, expand your result to lowest nontrivial order in \u03b1 := E(+) 20 E(+) 10. (4.83) Then compute the time average, and hence the intensity according to the second definition. (f) Give a physical argument for why one definition of the intensity is better than the other. (Consider what intensity is intended to describe, and remember that an alternative is not preferable merely because it is easier to calculate!) Problem 4.8 In the Coulomb-gauge formulation of electromagnetism, there is a field A (the vector potential) that satisfies \u2207 \u00b7 A = 0, and defines the electromagnetic fields via E = \u2212\u2202A/\u2202t and B = \u00b50H = \u2207 \u00d7 A. Use these relations and Maxwell\u2019s equations to show that A satisfies the same wave equation as E (and B and H). For simplicity, stick to vacuum. Problem 4.9 (a) The electromagnetic fields can be generally defined in terms of the scalar potential \u03c6 and vector potential A as E = \u2212\u2207\u03c6 \u2212 \u2202tA, B = \u2207 \u00d7 A. (4.84) Use Maxwell\u2019s equations to derive a set of two coupled wave equations for the potentials in a dielectric medium described by permittivity (r). (b) Then show that in the generalized Lorenz gauge, specified by the gauge condition \u2207 \u00b7 A + 2 0 \u00b50\u2202t\u03c6 = 0, (4.85) the two wave equations decouple. Chapter 5 Interference 5.1 Superposition of Two Plane Waves Interference occurs when multiple waves are added together, and it is one of the most obvious effects not accounted for by geometrical optics. First, we will consider the simplest case of adding two scalar, monochromatic waves E(+) (r). Both have the same frequency, and thus an implicit time dependence of 2 e\u2212i\u03c9t. The superposition of the two waves is just the sum (r) and E(+) 1 Clearly the combined field E(+)(r) is also monochromatic. Recall from Eq. (4.66)that the intensity of each component wave is given by E(+)(r) = E(+) 1 (r) + E(+) 2 (r). To simplify our notation a bit, we can introduce the fields U1,2(r), defined by I1,2 = 2|E(+) 1,2 | 2 \u03b7. (5.1) (5.2) U1,2 := 2 \u03b7 E(+) 1,2 (r), (5.3) (scaled electric fields) so that the intensity is simply the square of the U field without any extra coefficients: I1,2 = |U1,2|2. (5.4) (intensity in terms of scaled electric fields) In this notation, the intensity of the superposition is simply I = |U |2 = |U1 + U2|2 = |U1|2 + |U2|2 + U \u2217 1 U2 + U1U \u2217 2. Writing out explicitly the phases of U1,2, U1,2 = I1,2 ei\u03c61,2, (5.5) (5.6) we find that I = I1 + I2 + = I1 + I2 + 2 I1I2 ei(\u03c62\u2212\u03c61) + c.c. I1I2 cos(\u03c62 \u2212 \u03c61). (5.7) Notice that all the phase dependence is in the factor cos(\u03c62 \u2212 \u03c61), which ranges from \u22121 to 1. Putting these extreme values into this expression, we see that the intensity ranges from ( (combined intensity of two waves)", " I2)2 to ( I2)2. I1 \u2212 I1 + \u221a \u221a \u221a \u221a 84 Chapter 5. Interference A useful and simple special case is where the intensities are equal: In this case the intensity becomes I1 = I2 =: I0. I = 2I0 [1 + cos(\u03c62 \u2212 \u03c61)]. (5.8) (5.9) If \u03c62 = \u03c61, the intensity attains its maximum value I = 4I0, and this situation is fully constructive interference. If \u03c62 \u2212 \u03c61 = \u03c0, the intensity attains its minimum value I = 0, and this situation is fully destructive interference. Comparing to the intensity range above, we see that the case of equal intensities is the case of maximum intensity variation. The intensity \u00afI averaged over all phases is 2I0, as we might expect on average for adding two beams of intensity I0 from energy conservation. Of course, adding two waves should always conserve energy regardless of the phase, and we will see how this works out when we consider interferometers below, in particular in how waves are combined using beam splitters. 5.2 Mach-Zehnder Interferometer An interferometer is a device that uses the interference effect to do something useful. One of the simplest interferometers is the Mach-Zehnder interferometer, shown here. A beam of intensity I0 is split into two components by a beam splitter (a beam splitter is simply a device that deflects part of the intensity of an optical beam, while allowing the rest to pass). For simplicity we will assume a 50% lossless beam splitter, so that half the intensity goes into either arm of the interferometer. The two components are then recombined on the second beam splitter and the intensities I+ and I\u2212 can be monitored at their respective output ports. The wave in path 1, just before the second beam splitter, is U1 = I0 2 ei\u03c61, where \u03c61 = k1d1 = 2\u03c0n1d1 \u03bb0, (5.10) where d1 is the length of path 1, n1 is the refractive index of path 1, and \u03bb0 is the optical wavelength (in vacuum). Similarly, for the wave in path 2 just before the second beam splitter U2 = I0 2 ei\u03c62, where \u03c62 = k2d2 = 2\u03c0n2d2 \u03bb0. (5.11) f\u2122 - f\u00a12p3pp0-p-2pImaxI\u00baI = 2I\u00baI = 4I\u00bainput beamI\u00f2I\u00baI\u00ba/2I\u00ba/2I\u2013output portsmirrormirrorbeam splitterbeam splitterpath 2path 1 5.3 Stokes Relations 85 Then the intensities of the two output ports are given by squaring the sums of the intensities. At the first port, I+ = U1\u221a 2 + U2\u221a 2 2 = I0 2 [1 + cos(\u03c62 \u2212 \u03c61)]. At the second port, we also add the fields, but with an extra minus sign: I\u2212 = U1\u221a 2 \u2212 U2\u221a 2 2 = I0 2 [1 \u2212 cos(\u03c62 \u2212 \u03c61)]. (5.12) (5.13) Why the extra minus sign in I\u2212? Well, first of all, we can see that energy conservation works in this case, since the total output intensity is I+ + I\u2212 = I0, which is the same as the input intensity. This wouldn\u2019t have worked out if we omitted the extra minus sign. But we can do a better job of justifying it by considering an important time-reversal argument. 5.3 Stokes Relations The Stokes relations follow elegantly from the time-reversal invariance of electromagnetism. We can see this invariance directly from the electromagnetic wave equation: \u22072E \u2212 1 c 2 0 \u22022E \u2202t2 = 0. (5.14) Since time appears only as a second derivative, the equation is invariant under the replacement t \u2212\u2192 \u2212t. Consider a beam of light at an interface between two media (e.g., a dielectric interface). From geometrical optics, we know that part of the beam will transmit (refract) and part will reflect, although we don\u2019t know how much of the light will go in each direction. So we\u2019ll just define coefficients r and t, such that if U0 is the amplitude of the incident field, then the amplitude of the reflected field is rU0, and the amplitude of the transmitted field is tU0. In standard terminology", ", r is the field reflection coefficient and t is the field transmission coefficient. if two waves of amplitude rU0 and Time-reversal invariance states that this should also work in reverse: tU0 are incident from opposite sides of the same interface, they should combine to form a single beam that travels away from the interface. Of course, this happens by constructive interference. In the other opposite location where you might expect to see a beam going away from the interface, there really is none due to complete destructive interference. Note that in principle this is possible and it works for the purposes of this derivation. In practice, it is often very difficult to implement ideal time-reversed situations such as this. Now let\u2019s consider all the reflections and transmissions from the two incident beams that form the two outgoing beams in the time-reversed case. On the top side, the outgoing beam comprises a reflection of the rU0 and a transmission from the tU0 beam. So we can write this field as rrU0 + ttU0. The primed coefficients r and t denote the reflection and transmission coefficients for a ray incident from below, which are possibly different from r and t. Similarly, the outgoing beam on the bottom comprises a transmission of the rU0 beam and a reflection of the tU0 beam, so we can write this field as trU0 + rtU0. U\u00barU\u00batU\u00baU\u00barU\u00batU\u00ba 86 Chapter 5. Interference But the last two diagrams must be consistent with each other, so for the top beam we require that and the bottom beam must vanish, so r2U0 + ttU0 = U0, trU0 + rtU0 = 0. Simplifying these two equations a bit, we arrive at the Stokes relations (5.15) (5.16) r = \u2212r tt = 1 \u2212 r2 (5.17) (Stokes relations) in standard form. Now we can use the Stokes relations to analyze what happens at a physical beam splitter. A physical beam splitter is typically something like a slab of glass, where the splitting occurs at one interface and the no splitting occurs at the other (often this is accomplished with special dielectric coatings). If the reflection coefficient is r for the beam that impinges on the beam splitter from the outside, the beam in the other arm of the interferometer impinges on the same interface from within the beam splitter. Thus, for the other arm, the reflection coefficient must be \u2212r. We can\u2019t work out exactly what happens to the transmission coefficients from the Stokes relations, but it works out that they are equal. Since r2 = 0.5 for the 50% splitters (the split ratio refers to intensity), we have that tt = 1 \u2212 r2 = 0.5, so there is no sign change between t and t. 5.4 Mach-Zehnder Interferometer: Applications Since the Mach-Zehnder interferometer has two spatially separated but otherwise equivalent paths, it is typically used to measure an optical phase shift (path-length change) due to an object in one arm. The other arm acts as a \u2018\u2018phase reference\u2019\u2019 for the phase-shifted beam. For example, you could use this technique to measure the index of refraction of gas in a cell by measuring the phase shift as a function of the gas pressure. Or you can visualize a phase object such as a flame or a gas flow, where the shape of the optical fringes reflects the spatial index profile of the object. rrU\u00ba + t\u00a2tU\u00batrU\u00ba + r\u00a2tU\u00barU\u00batU\u00baU\u00baU\u00barU\u00bar\u00a2U\u00baoo=oo-rU\u00batU\u00bat\u00a2U\u00ba 5.5 Michelson Interferometer 87 5.5 Michelson Interferometer The Mach-Zehnder interferometer can be \u2018\u2018folded\u2019\u2019 so that the two beam splitters coincide and can be replaced by a single splitter. This new configuration is the Michelson interferometer. Note that in this diagram, the input port is also one of the output ports, but often in this configuration you would only monitor the other output port. Just as in the Mach-Zehnder case, the Michelson interferometer measures the path-length difference between the two arms. Assuming the same refractive index for both arms and a 50/50 beam splitter, the output intensity is Iout = I0 2 1 + cos 2\u03c0n(d2 \u2212 d1) \u03bb. (5.18) This interferometer is very useful for measuring lengths. For example, one arm can act as a reference distance, while the output changes depending on how far the other \u2018\u2018probe", "\u2019\u2019 mirror moves. Also, if two different laser beams are coupled into the same interferometer, measuring the fringes of both lasers as a mirror is moved gives a measure of the relative wavelength\u2014useful as an optical \u2018\u2018wave meter\u2019\u2019 or \u2018\u2018\u03bb-meter\u2019\u2019 for measuring the wavelength of an unknown laser in terms of a known reference laser. (output intensity of Michelson interferometer) One example is the interferometric gravimeter, which is a Michelson interferometer where one of the mirrors is actually a corner-cube reflector (a prism that retroreflects the incident beam in the same direction independent of small changes in its orientation). The corner cube acts as a freely falling mass, and by monitoring the output it is possible to get a real-time position measurement of the cube\u2019s position. This method is still the state-of-the-art method for sensitive measurements of g. Another important application of the Michelson interferometer was its original version and namesake in the Michelson-Morley experiment. The idea here was to search for an \u2018\u2018aether,\u2019\u2019 the fundamental medium through which light propagates, which would define a \u2018\u2018rest frame\u2019\u2019 for light. The interferometer would show a fringe shift if it moved through the aether (because it was stationary on Earth) depending on its orientation with respect to its velocity. No effect was observed, the conclusion being that there is no preferred rest frame for light propagation. input beamd\u00a1d\u2122output portmirrormirrorbeamsplitterd\u00a1d\u2122detectorfixed mirrorfree-falling corner cube 88 Chapter 5. Interference Yet another example is the LIGO (Laser Interferometric Gravitational Wave Observatory) experiment, which uses some of the largest and most impressive Michelson interferometers built thus far. The arms of the LIGO interferometers are 4 km long, with many technical refinements to detect displacements of mirrors due to gravitational waves. The strain sensitivities achieved so far are better than 10\u221221, corresponding to detecting the displacement of a macroscopic mirror to a precision of much less than a nuclear diameter. 5.6 Sagnac Interferometer The Sagnac interferometer is another variation on the Mach-Zehnder configuration, where the second beam splitter is replaced by a mirror. The mirror causes the beams to continue around, so really there are not two distinct \u2018\u2018arms\u2019\u2019 of this interferometer. Rather, two beams traverse the same ring but in opposite directions. Since there is no path-length difference, there is no sense of distance measurement in this system. However, if the interferometer rotates, the symmetry between the two beams is broken due to the Doppler effect, giving an effective phase shift between the two beams. Thus, the Sagnac interferometer is a very effective rotation sensor. One common instrument that uses this effect is the ring-laser gyroscope, which uses this interferometer as the cavity for a laser for sensitive detection of rotation. 5.7 Interference of Two Tilted Plane Waves A simple but useful effect to understand is the interference pattern between two plane waves with slightly different angles. Consider a plane wave propagating in the z-direction, U1 = I0 eikz, and a second wave of equal intensity propagating at an angle \u03b8 with respect to the first, U2 = I0 ei[k(cos \u03b8)z+k(sin \u03b8)x]. (5.19) (5.20) z = 0xzI\u00baI(x)U\u00a1U\u2122qI\u00ba 5.8 Multiple-Wave Interference 89 For a screen at z = 0, the phase difference between the two waves is just (k sin \u03b8)x, so from Eq. (5.9), the intensity on the screen is Iscreen = 2I0 {1 + cos[(k sin \u03b8)x]}. (interference pattern of two tilted plane waves) (5.21) Thus, the interference pattern varies sinusoidally with spatial period 2\u03c0 k sin \u03b8 = \u03bb sin \u03b8. (5.22) This is useful as a simple model of misalignment in a two-beam interferometer. Obviously this result is a useful \u2018\u2018interferometer\u2019\u2019 in its own right: this gives you an accurate method for measuring angles or printing accurate periodic patterns lithographically (e.g., for holographically fabricating diffraction gratings). 5.8 Multiple-Wave Interference A useful generalization of the two-beam interference theme is the interference due to N waves: In general, this is a very complicated problem, but we can easily treat the special case of equal", " amplitudes and constant phase differences: U = U1 + U2 + \u00b7 \u00b7 \u00b7 + UN. (5.23) Un = I0 ei(n\u22121)\u03c6, n = 1, 2,..., N. The algebra simplifies if we define h := ei\u03c6, so that Un = I0 hn\u22121. Then U = I0 1 + h + h2 + \u00b7 \u00b7 \u00b7 + hN \u22121 = I0 1 \u2212 hN 1 \u2212 h I0 = 1 \u2212 eiN \u03c6 1 \u2212 ei\u03c6. The corresponding intensity is the squared magnitude of this expression: I = I0 e\u2212iN \u03c6/2 \u2212 eiN \u03c6/2 e\u2212i\u03c6/2 \u2212 ei\u03c6/2 2. (5.24) (5.25) (5.26) (5.27) Here, we multiplied through by a factor of e\u2212iN \u03c6/2/e\u2212i\u03c6/2 inside the magnitude, which doesn\u2019t change anything. Finally, we have (5.28) This corresponds to an interference pattern with tall, narrow peaks of height N 2I0 with N \u2212 2 sub-lobes in between. I = I0 sin2 (N \u03c6/2) sin2(\u03c6/2) (interference of N waves, equal phase difference). f2p4p6p0ImaxI = NI\u00baNo-o2minor lobes}I = N I\u00ba\u21222\u03c0N 90 Chapter 5. Interference The corresponding average intensity is \u00afI = 1 2\u03c0 2\u03c0 0 I d\u03c6 = N I0, (5.29) again as we expect from energy conservation. This problem is a useful model for diffraction through N thin, equally spaced slits, or a diffraction grating where N rulings are illuminated. However, these problems are better dealt with in the more general and powerful formalism of Fourier optics (Chapter 12). 5.9 Exercises 5.9 Exercises Problem 5.1 91 The Young double slit experiment is another example of a simple interferometer. Two slits in a planar wall are illuminated by the same light source. Interference fringes appear in the light on a distant, planar screen (parallel to the first wall), because the path lengths of the two beams to the screen vary slightly with the propagation direction of the waves. (a) Sketch the setup and show that the angular distance between successive dark bands (areas of destructive interference) is \u2206\u03b8 = \u03bb/a, where \u03bb is the wavelength of the light and a is the separation between slits. Because the screen is distant, you can assume the two waves propagating to any particular point on the screen are approximately parallel. (b) Suppose that the two slits are 0.5 mm apart and diffract monochromatic light onto a screen 10 m away. You measure the fringe spacing (distance between dark bands) on the screen to be 1.266 cm. What is the wavelength of the light? (c) What type of laser could plausibly be the source of the light in part (b)? Problem 5.2 Consider a Michelson interferometer illuminated by laser light. Assume that the incident light is well described by a plane wave. Suppose that one of the retroreflecting mirrors has a small misalignment of angle \u03b4\u03b8. Describe the interference pattern on a screen at the output of the interferometer, relating aspects of the interference pattern to \u03b4\u03b8 as appropriate. Also describe how the pattern changes as either mirror moves. Problem 5.3 Consider a Mach-Zehnder interferometer with a gas cell of length 10 cm in one arm as shown. The input light is \u03bb = 200\u03c0 nm. The cell is initially evacuated, and all of the interferometer light exits output port 1. As the cell is pressurized, you observe that port 1 goes dark then becomes bright again a total of 10 complete cycles. What is the refractive index of the gas at the final pressure? Problem 5.4 Consider a Michelson interferometer, operating at wavelength \u03bb, containing an electro-optic modulator (EOM). For the purposes of this problem, think of the EOM as a slab of glass of thickness d, whose refractive index n can be changed slightly via an electrical bias signal. gas cellport 1 92 Chapter 5. Interference Suppose the output of the interferometer on the detector is at the maximum of a bright fringe (at refractive index n0 of the EOM). By how much does the refractive index change so that the output goes to minimum intensity? (Compute the smallest change that does this.) Compute a numerical value for the refractive-index change for d = 10 cm and \u03bb", " = 1 \u00b5m. Incidentally, this is one method for making fast optical switches (as used in telecommunications, but these are usually Mach-Zehnder interferometers implemented with optical waveguides). Problem 5.5 Consider a Michelson interferometer, operating at wavelength \u03bb. A slab of glass of thickness d and refractive index n is inserted in one arm, parallel to the closest mirror as shown. Derive an expression for the number of fringes the output goes through as the slab is rotated (clockwise in the diagram) through an angle \u03b8. Problem 5.6 Consider a Michelson interferometer, with mirrors at distances d1 and d2 from the 50/50 beam splitter. Suppose a Gaussian beam enters the interferometer (assume the beam and mirrors are all ideally aligned). Consider the intensity pattern on a screen at the output, where s is the distance from the beam splitter to the screen. Show that there will be a \u2018\u2018bull\u2019s-eye\u2019\u2019 intensity pattern of concentric interference fringes on the screen, provided d1 = d2. Compare the fringe spacing near the center vs. near the edge of the pattern. Hint: derive the expression for the intensity pattern due to two interfering waves of arbitrary intensity profiles I1,2 and phase \u03c61,2, and then replace with Gaussian-beam expressions as appropriate. Try to focus only on the important stuff. input beamdetectorEOMinput beamdetectorglass slab Chapter 6 Gaussian Beams 6.1 Paraxial Wave Equation Just as paraxial rays greatly simplified ray optics by allowing us to use linear propagation equations, paraxial waves are much simpler to treat than waves in the general case. A paraxial wave is one where the curves normal to the wave fronts are paraxial rays. Paraxial waves propagate (more or less) along the optical axis, so they should be fairly close to a plane wave. Thus, we can factor out the plane-wave part of the paraxial wave. For a scalar electric field propagating along the z-direction, E(+)(r) = \u03c8(r)eikz. (6.1) Here, \u03c8 is the envelope and eikz is the carrier wave. For a paraxial wave, the envelope \u03c8 should be smooth on the scale of \u03bb. Plugging this into the wave (Helmholtz) equation, (\u22072 + k2)E(+) = 0, we find \u22072 T + i2k \u2202 \u2202z + \u22022 \u2202z2 \u03c8 = 0, where \u22072 T := \u22022 \u2202x2 + \u22022 \u2202y2 is the transverse Laplacian. If \u03c8 is smooth (i.e., varies slowly) on the scale of \u03bb, then which we can rewrite as We can similarly write \u2202\u03c8 \u2202z \u03c8 \u03bb, \u2202\u03c8 \u2202z k\u03c8. \u22022\u03c8 \u2202z2 k \u2202\u03c8 \u2202z, (6.2) (6.3) (6.4) (6.5) (6.6) zOA 94 Chapter 6. Gaussian Beams and thus we can neglect the \u22022/\u2202z2 term in Eq. (6.2). This is called the slowly varying envelope approximation, and the resulting wave equation is the paraxial wave equation: \u22072 T + i2k \u2202 \u2202z \u03c8 = 0. (6.7) (paraxial wave equation) Remember that in treating the field in terms of \u03c8, we have already factored out the plane-wave part eikz of the phase. It is interesting to note that this is exactly the same as the two-dimensional Schr\u00f6dinger equation in quantum mechanics for the wave function \u03c8(x, y, t): \u2212 \u00afh2 2m \u22072 T \u03c8 + V (x, y)\u03c8 = i\u00afh \u2202\u03c8 \u2202t. (Schr\u00f6dinger wave equation) (6.8) We can see this equivalence by rewriting this equation as \u22072 T \u2212 2m \u00afh2 V (x, y) + i2m \u00afh \u2202 \u2202t \u03c8 = 0, (6.9) and then making the formal identifications z \u2212\u2192 t and k \u2212\u2192 m/\u00afh. The potential V (x, y) turns out to be related to the refractive index profile, though in a somewhat subtle way\u2014the obvious correspondence is to the Schr\u00f6dinger equation in free space. 6.2 Gaussian Beams The simplest waves are the plane wave, which is of the form cos(kx \u2212 \u03c9t),", " and the spherical wave, which is of the form cos(kR \u2212 \u03c9t)/R. These are useful models, so why would we want to study something more complicated like a Gaussian beam? The Gaussian beam is the simplest model of a directed beam that satisfies Maxwell\u2019s equations, at least in the paraxial approximation. It also turns out that the outputs of sphericalmirror resonators and lasers are often Gaussian beams. The Gaussian beam is just like the Gaussian wave packet in quantum mechanics, being just complex enough to be interesting but simple enough to be tractable. Let\u2019s start by just writing it down (we will take it for granted that this satisfies the paraxial wave equation, at least until we study Fourier optics): E(+)(r) = E(+) 0 w0 w(z) exp \u2212 r2 w2(z) ikz \u2212 i tan\u22121 exp z z0 exp ik. r2 2R(z) (6.10) This is the Gaussian or TEM0,0 (for \u2018\u2018transverse electromagnetic\u2019\u2019, as for the plane wave; the subscript comes from the hierarchy of Hermite\u2013Gaussian beams, which we will come to later) beam. In this expression, we have used polar coordinates (r, z) with r = x2 + y2; E(+) is as usual an overall field-amplitude constant; z0 is a constant called the Rayleigh length (or \u2018\u2018Rayleigh range\u2019\u2019); (Gaussian beam) 0 w0 := \u03bbz0 \u03c0 (6.11) (beam waist parameter) is the beam waist parameter or beam radius, (that\u2019s a \u2018\u2018double-u,\u2019\u2019 not an \u2018\u2018omega,\u2019\u2019 by the way), where the inverted relation z0 = \u03c0w 2 0 \u03bb (6.12) (Rayleigh length in terms of w0) for the Rayleigh length in terms of the beam waist parameter is also useful; w(z) := w0 1 + 2 z z0 (6.13) (beam waist along optical axis) 6.2 Gaussian Beams 95 is the beam waist as a function of z; and R(z) := z 1 + 2 z0 z (6.14) (wave-front radius of curvature) is the radius of curvature of the wave fronts, as we will see. This pretty much looks like a big mess, right? It\u2019s not nearly as bad as it looks, we just have to break it down and analyze each piece to see what is going on. First, let\u2019s try rewriting the Gaussian beam solution in a way that\u2019s partitioned into three smaller factors: E(+)(r) = E(+) 0 \u2212 exp w0 w(z) ikz \u2212 i tan\u22121 r2 w2(z) z z0 \u00d7 exp (amplitude factor) (longitudinal phase factor) \u00d7 exp ik. r2 2R(z) (radial phase factor) (Gaussian beam, organized form) (6.15) We will now analyze each factor separately. 6.2.1 Amplitude Factor The first factor in Eq. (6.15) is the amplitude factor, which completely describes the intensity profile of the Gaussian beam. In this sense, this is the most useful factor for understanding the behavior of the Gaussian beam. There are several important features to understand here: Intensity Profile. The intensity is just the squared modulus of Eq. (6.15), but the other two factors make no contribution: I(r, z) = I0 2 w0 w(z) exp \u2212. 2r2 w2(z) (6.16) Here, we have defined I0 := 2|E(+) |2/\u03b7 as the maximum intensity of the Gaussian beam (which occurs at the center r = 0). Clearly, the intensity falls off in the radial direction like a Gaussian function, hence the name \u2018\u2018Gaussian beam.\u2019\u2019 Also, note that w(z) is one measure of the width of the Gaussian, and it reduces to the minimum value w0 at z = 0. There are differing conventions for quoting the width of the beam (and even for what w0 represents), so when using w0 or w(z) to represent the size of a Gaussian beam, for clarity it is good to call it the \u2018\u20181/e2 radius\u2019\u2019 of the beam. (intensity profile of Gaussian beam) 0 Total Power. The total power of the Gaussian beam is given by integrating the intensity over a transverse plane: P = \u221e 0 I(r,", " z) 2\u03c0r dr = I0 2 \u03c0w 2 0. (6.17) (total power of Gaussian beam) This result holds independent of z. Thus we can rewrite the intensity profile as I(r, z) = 2P \u03c0w2(z) exp \u2212. 2r2 w2(z) (6.18) It is useful to consider some similar integrals over a disc of finite radius, as a measure of how \u2018\u2018large\u2019\u2019 is a Gaussian beam. For example, a circle of radius R = w(z) contains 86% of the optical power of a Gaussian beam. A circle of radius R = 1.5w(z) contains 99% of the power of a Gaussian beam. (intensity profile in terms of total power) 96 Chapter 6. Gaussian Beams Beam Divergence. Note that the beam waist w(z) traces out a hyperbolic curve in z. Near the focus at z = 0, the waist achieves its minimum value w0. At large distances from the focus, the hyperbola approaches its asymptotes, given by (w0/z0)z. The Rayleigh length z0 marks the crossover between these two regimes; thus, w(z) \u223c w0 for |z| z0 w(z) \u223c w0 z0 z for |z| z0 (6.19) for the regions near to and far away from the focus, respectively. From this we see that in the far field, the beam propagates in the form of a cone of half angle \u03b80, where In the paraxial regime, we have tan \u03b80 := w0 z0 = \u03bb \u03c0w0. (6.20) \u03b80 \u2248 w0 z0 = \u03bb \u03c0w0. (6.21) (far-field divergence half-angle) Thus, a tightly focused beam (small w0) diverges more quickly, whereas a larger beam (large w0) diverges less. This is a manifestation of the uncertainty principle in optics. Also, a larger wavelength causes more far-field divergence: generally speaking, diffraction effects are more important for longer wavelengths. As for the intensity, in the far field we have I(r = 0, z) \u2248 I0 2, z0 z (6.22) so the Gaussian beam is consistent with the inverse-square intensity law in the far field (as is also the case for the spherical wave). Depth of Focus. As we just mentioned, the best focus occurs at z = 0. At a distance of one Rayleigh length z0 from the focus, the beam waist increases by 41% compared to the focal value: w(\u00b1z0) = \u221a 2 w0. (6.23) Again, z0 marks the transition from focused behavior to far-field divergence. So we can say that the beam is roughly focused between \u00b1z0, so the \u2018\u2018depth of focus\u2019\u2019 for a Gaussian beam is 2z0, where again 2z0 = 2\u03c0w 2 0 \u03bb. (6.24) (depth of focus) From this relation we can see explicitly that tightly focused beams have a smaller depth of focus. This is related to a general principle in imaging optics, which is that larger apertures produce smaller depths of focus (as occurs in photography). w\u00baz\u00ba-z\u00baw(z)-w(z)-q\u00bazq\u00bazq\u00ba 6.2 Gaussian Beams 97 6.2.2 Longitudinal Phase Factor Recall that the longitudinal phase factor has the form ikz \u2212 i tan\u22121 exp. z z0 (6.25) (longitudinal phase factor) The first term in the phase is simply the phase of a plane wave ikz propagating in the same direction and with the same optical frequency as the Gaussian beam. The second term is called the Gouy phase shift1 and represents a small departure from planarity. The longitudinal phase is dominated by the ikz term, but don\u2019t go underestimating the importance of the Gouy term. It represents a phase retardation compared to the plane wave (in that the total phase here changes more slowly than in the plane wave). The tan\u22121 form implies a monotonically decreasing phase shift (relative to the plane-wave part), amounting to a total of \u2212\u03c0 change in phase over all z. Gouy effects are generic to focusing-beam-type solutions to the wave equation. As we will see, the Gouy phase is important in computing the resonant frequencies of optical resonators. Also, it implies that the phase velocity of a Gaussian beam is slightly larger than c, since the spacing between wave fronts is slightly larger than \u03bb. 6.2.3 Radial Phase Factor Recall that the", " radial phase factor is of the form where exp ik, r2 2R(z) R(z) = z 1 + 2. z0 z (6.26) (radial phase factor) (6.27) (radius of curvature) This factor gives the dependence of the phase on r, whereas the longitudinal phase factor simple gave the on-axis phase behavior. Thus the entire phase factor is ikz \u2212 i tan\u22121 exp z z0 + ik. r2 2R(z) (6.28) (total Gaussian phase) To understand this, we need to compare this phase to the phase of a spherical wave, E(+)(r) = E(+) 0 R eikR = E(+) 0 R \u221a eik r2+z2, where R = |r|. In the paraxial case, r z, so E(+)(r) = = E(+) 0 R E(+) 0 R E(+) 0 R E(+) 0 R exp ik|z| 1 + r2 z2 exp ik|z| 1 + r2 2z2 r2 2|z| r2 2R. exp ik|z| + ik exp ik|z| + ik (6.29) (6.30) 1C. R. Gouy, \u2018\u2018Sur une propri\u00e9t\u00e9 nouvelle des ondes lumineuses,\u2019\u2019 Comptes Rendus Hebdomadaires des S\u00e9ances de l\u2019Acad\u00e9mie des Sciences (Paris) 110, 1251 (1890); C. R. Gouy, \u2018\u2018Sur la Propagation Anomale Des Ondes,\u2019\u2019 Annales de Chimie et de Physique, 6e s\u00e9rie 24, 145 (1891). 98 Chapter 6. Gaussian Beams Comparing this phase to Eq. (6.28), we see that in the paraxial approximation, the spherical wave and Gaussian beam have the same form except for the absolute value of z, which accounts for the fact that the spherical wave propagates outward from R = 0; the Gouy phase, which represents only a small correction (and in some sense \u2018\u2018patches\u2019\u2019 together the wave fronts on either side of the focus, due to the reversal of half of the wave fronts as compared to the spherical wave); and the fact that the radius of curvature changes with z for the Gaussian beam. Thus, R(z) in the Gaussian beam represents the radius of curvature of the wave fronts. Examining the radius of curvature, we see that near and far away from the focus, R(z) \u223c z 2 0 z for |z| z0 R(z) \u223c z for |z| z0 (6.31) Again, z \u2248 R in the paraxial approximation, so the Gaussian beam approaches a spherical wave for large |z|. We can also see where R(z) is minimum (i.e., the curvature is maximum) by differentiating R: dR dz = d dz z2. (6.32) Thus, dR/dz = 0 when z = \u00b1z0, so maximum wave-front curvature occurs at the Rayleigh length. Closer to the focus, the wave fronts become flat (as we could guess from symmetry considerations), and farther out, the wave-front curvature decreases as for a spherical wave. The minimum radius of curvature is R(\u00b1z0) = 2z0. In a cavity, the boundary conditions imposed by the cavity mirrors require that the curvature of the spherical mirrors and the curvature of the wave fronts match. This allows the wave to map back on itself. This is one reason why Gaussian beams can exist in resonators. But in particular, recall that for a confocal cavity, the length d of the cavity is the radius of curvature of both mirrors R1,2. A Gaussian beam with 2z0 = |R1,2| has wave fronts that exactly match the mirror curvatures at z = \u00b1z0. Hence, z0 is also referred to as the confocal parameter. 6.3 Specification of Gaussian Beams Despite the complex form of the Gaussian beam, relatively little information is needed to completely specify it. For example, if we know where z = 0 is located, the value of w0, and the optical wavelength \u03bb all the other parameters are uniquely fixed. Alternately, it is sufficient to know w0 and R(z) at some distance z, or it is sufficient to know w(z) and R(z) at some distance z. From a Fourier-transform point of view, this is because a Gaussian wave packet is uniquely determined by its center and", " width. By construction of the Gaussian beam, the center is zero. The divergence is set by the width of the Fourier transform of the wave packet, which we know is fixed via the uncertainty principle. Again, the tradeoff between tighter focus and larger divergence angle is precisely the manifestation of the uncertainty principle in the Gaussian beam. z\u00ba2z\u00bazR(z)\u2122~oz\u00bao/z~oz \u2202Ex \u2202x + \u2202Ey \u2202y + \u2202Ez \u2202z = 0. \u2207T \u00b7 E + \u2202Ez/\u2202z = 0. ET 2w0 + kEz = 0, 6.4 Vector Gaussian Beams 99 6.4 Vector Gaussian Beams We have thus far only treated Gaussian beams as solutions to the scalar wave equation. It turns out that for the full vector electromagnetic field case, the only difference is in the overall amplitude. That is, supposing we have a Gaussian beam propagating along the z-direction and polarized in the x-direction, we can obtain the vector field from the scalar field by the simple replacement 0 \u2212\u2192 E(+) E(+) 0 \u2212\u02c6x + \u02c6z. x z \u2212 iz0 (modification to obtain vector Gaussian beam, linear polarization) (6.33) Note that the polarization is mostly along the x-direction in the region where the paraxial approximation holds, since |x| |z + iz0|. Thus, the field is mostly transverse. Optical beams are nearly transverse (TEM), but the transverse property is not completely necessary (as it is for the plane wave), as long as \u2207 \u00b7 E = 0. Writing out this condition, The first two terms represent a transverse divergence, so that Putting in approximate numbers, x + E 2 where ET is the transverse field they must be on the same order, and thus E 2 y, and 2w0 is the beam diameter. Thus, for these terms to cancel, Ez ET \u223c 1 2kw0 = \u03bb 4\u03c0w0. (6.37) So as long as the beam intensity falls off on a length scale large compared to \u03bb, which happens for a Gaussian beam if w0 \u03bb, the field is mostly transverse (i.e., Ez is small). In the paraxial approximation, the field is transverse to good approximation, justifying the label of TEM0,0 for the Gaussian beam. 6.5 ABCD Law The ABCD Law is a quick method for propagating Gaussian beams. We\u2019ll begin by just stating it, and then we\u2019ll justify it later. It is convenient to define a complex q parameter by q(z) := z \u2212 iz0. (6.38) (q parameter for Gaussian beams) It is worth reiterating that the longitudinal coordinate z is measured with respect to the beam focus. Note that q determines the Rayleigh length and the location of the beam focus, and therefore determines the entire geometry of the Gaussian beam. Alternately, we can compute the inverse of q, 1 z \u2212 iz0 Recalling that w0 = \u03bbz0/\u03c0, w(z) = w0 more useful form 1 q = = z + iz0 z2 + z 2 0 = z z2 + z 2 0 + i z0 z2 + z 2 0. (6.39) 1 + (z/z0)2, and R(z) = z[1 + (z0/z)2], we can write 1/q in the 1 q(z) = 1 R(z) + i\u03bb \u03c0w2(z). (6.40) (inverse of q parameter) (6.34) (6.35) (6.36) 100 Chapter 6. Gaussian Beams Again, q determines both the spot size and the radius of curvature at a given location, so q completely determines the Gaussian beam geometry. We can see that this simplifies things greatly by rewriting the Gaussian beam in a compact form in terms of q: E(+)(r) = E(+) 0 q0 q(z) exp ikr2 2q(z) (Gaussian beam in terms of q parameter) exp(ikz). (6.41) Here, we have used the notation q0 := q(0) = \u2212iz0. The Gouy phase ends up being just the phase angle of 1/q, while q also represents the width and curvature of the beam. Here is the other nice thing about q: it transforms through an optical system according to a simple law related to the geometrical optics ABCD matrix for the system. In terms of the matrix elements, the ABCD law reads q2 = Aq1 + B Cq1 +", " D, (6.42) (ABCD Law) where q1 and q2 are the q parameters before and after the optical system, respectively. or in alternate form, 1 q2 = C + D/q1 A + B/q1. (6.43) (ABCD Law) Let\u2019s check this for a couple of simple examples, and then use them to argue that the ABCD law is generally valid. 6.5.1 Free-Space Propagation Recall that the matrix for propagation over distance d in free space is Thus, applying the ABCD law gives M = 1 d 1 0. q2 = Aq1 + B Cq1 + D = q1 + d. Writing out the real and imaginary parts explicitly, z2 \u2212 iz02 = (z1 + d) \u2212 iz01. (6.44) (6.45) (6.46) The imaginary part of this equation states that the Rayleigh length is unchanged, z02 = z01, and thus the shape of the beam is unaffected. The real part states that z2 = z1 + d, that is, the position along the beam is shifted by d, as we expect for a propagation of the same distance. Thus, the ABCD law gives a sensible result. Note that the argument carries through for a dielectric medium of refractive index n, by repeating the argument with the replacement d \u2212\u2192 d/n. However, this point is a bit subtle: propagation over a distance d through a medium of refractive index n is described by the same matrix as in free space, independent of n. However, propagation through a refractive slab of thickness d (surrounded by vacuum) is given by the composite matrix for two planar, refractive interfaces separated by propagation over distance d/n = 1 d/n 1 0 (6.47) Again, this matrix works for the Gaussian beam, since the propagation distance is effectively reduced by n, due to the refraction at the boundaries modifying the divergence angle. 6.5 ABCD Law 6.5.2 Thin Optic 101 Recall that for a thin optic, the position of the ray does not change (y2 = y1). Thus, the matrix for a thin optic has the general form M = 1 0 C D. (6.48) This condition is also equivalent to the condition that the Gaussian beam waist does not change across the optic (w02 = w01), so that power density on each side of the optic is consistent. In ray optics, the matrix affects the transmitted angle, 2 = Cy1 + Dy y 1. (6.49) In the paraxial approximation, the Gaussian beam is equivalent to a spherical wave of radius R1 or R2 before or after the optic, respectively. For the (paraxial) rays normal to the wave fronts, we can write Putting this into Eq. (6.49), we find y 1,2 \u2248 y1,2 R1,2. 1 R2 = C + D R1. Using 1/R = 1/q \u2212 i\u03bb/\u03c0w2 and D = det(M) = n1/n2 = \u03bb2/\u03bb1, we find 1 q2 = C + D q1 =\u21d2 q2 = q1 Cq1 + D, so we recover the ABCD law in this restricted case. 6.5.3 Cascaded Optics (6.50) (6.51) (6.52) As in ray optics, the ABCD law works for cascaded (composite) systems. It is possible to verify directly by substitution that if then q3 = A2q2 + B2 C2q2 + D2 and q2 = A1q1 + B1 C1q1 + D1, q3 = Aq1 + B Cq1 + D provided that the new matrix elements are obtained by matrix multiplication in the proper order: A B C D = A2 B2 C2 D2 A1 B1 C1 D1. Thus, we can handle Gaussian-beam propagation through any composite system via the composite ray matrix. 6.5.4 Factorization of a General Matrix We showed that the ABCD law works in the case of propagation in free space (and thus also a homogenous medium), as well as for a thin optic. We now know that the ABCD law works for any cascade of propagations and thin optics. Even for continuous media, we can always in principle break any medium up into a cascade of infinitesimal propagations and weak, thin optics. In principle, the ABCD law should then work for any optical system. We can argue this more precisely without so much hand-waving, however. Consider a general ray matrix, M = A B C D. (6.56) (6.53) (6", ".54) (6.55) 102 Chapter 6. Gaussian Beams We can write down an explicit factorization of this matrix in terms of propagation and thin-optic matrices. Consider a system comprising (1) a propagation in free space over distance d1; (2) a thin lens of focal length f ; (3) a planar refractive interface to a homogenous medium of index n; and finally (4) a propagation over distance d2/n in the medium. The ray matrix is M = 1 d2 1 0 1 0 0 1/n 1 \u22121/f 1 d1 1 0 0 1. Then these two systems are equivalent (M = M) provided that n = 1 det(M) f = \u2212 det(M) C d1 = = D \u2212 det(M) C d2 = A \u2212 1 C (6.57) (6.58) We have already verified that the ABCD law works for any of the component matrices, thus by the cascading rule it works for the matrix M. 6.5.5 \u201cDeeper Meaning\u201d of the ABCD Law It seems amazing that a solution to the wave equation can be propagated using classical rules. But really what this is saying is that, in some sense, wave phenomena are absent in the paraxial approximation. As we will see, this transformation rule applies in more general situations as long as the paraxial approximation holds. But let\u2019s focus on the Gaussian beam for concreteness. The Gaussian beam stays Gaussian through any optical system as long as the paraxial approximation is valid. As soon as the paraxial approximation breaks down (i.e., nonlinear terms become important), the beam will become non-Gaussian. For example, one would not expect a beam in an unstable resonator to remain Gaussian. However, within the paraxial approximation, a bundle of geometrical rays with a Gaussian density profile (in both y and y, in a way that satisfies the uncertainty principle) follows the Gaussian beam exactly. Showing this mathematically really requires the Wigner transform, so we won\u2019t get into this here. But the important point is that Gaussian beam optics is really just ray optics; Gaussian beams do not show any manifestly wave-like behavior, at least within the paraxial approximation. All the diffraction-type effects can be mimicked by an appropriate ray ensemble. Again, since paraxial wave propagation is equivalent to quantum mechanics, this is equivalent to the relation between quantum and classical mechanics for linear systems. For linear systems (i.e., free-space propagation and the harmonic oscillator), quantum and classical mechanics turn out to be equivalent. That is, you can attribute any quantum effects in a linear system to the initial condition, not to the time evolution. Nonlinear systems dynamically generate quantum effects, even for a classical initial state. 6.5.6 Example: Focusing of a Gaussian Beam by a Thin Lens As a more useful example of the ABCD law in action, let\u2019s consider the focusing of a Gaussian beam by a lens. We will consider the special case where the lens is placed at the waist of the beam to be focused, so that the initial radius of curvature R1 = \u221e. The ray matrix is \uf8ee \uf8f0 6.59) 6.5 ABCD Law and the initial beam parameter is, starting from Eq. (6.40), 1 q1 = 1 R1 + i\u03bb \u03c0w 2 1 = i\u03bb \u03c0w 2 01 = i z01, 103 (6.60) where we recall the initial Rayleigh length is z01 = \u03c0w 2 parameter simply in terms of the Rayleigh length at the focus of the beam. Applying the ABCD law, 01/\u03bb. Remember that we can only represent the q We should then compare this relation to the general form (6.40) for 1/q: 1 q2 = C + D/q1 A + B/q1 = \u2212 1 f + i \u03bb \u03c0w 2 01. 1 q2 = 1 R2 + i\u03bb \u03c0w 2 2. (6.61) (6.62) Equating the real and imaginary parts, we find that the radius of curvature at the output is R2 = \u2212f, corresponding to a focusing beam if f > 0. Also, w2 = w01, which means that the beam size is unchanged\u2014 again, as we expect for any thin optic. 6.5.7 Example: Minimum Spot Size by Lens Focusing We can do a similar analysis to compute the minimum spot size obtained by a focusing lens for a Gaussian beam. The setup is the same as for the last example, but we will also consider a free-space propagation over a distance d after the lens. Thus, the system matrix", " is Applying the ABCD law, 1 q2 = C + D/q1 A + B/q1 = \u22121 + if /z01 f \u2212 d + idf /z01. Separating the real and imaginary parts, 1 q2 = \u2212(f \u2212 d) + df 2/z 2 01 (f \u2212 d)2 + (f d/z01)2 + i (f \u2212 d)(f /z01) + df /z01 (f \u2212 d)2 + (f d/z01)2. (6.63) (6.64) (6.65) From Eq. (6.40), we can equate the real and imaginary parts of this expression with 1/R2 and 1/z02, respectively (recall that we can only make the latter identification under the assumption that we are at the focus). Thus, R2 = (f \u2212 d)2 + (f d/z01)2 \u2212(f \u2212 d) + df 2/z 2 01. (6.66) But plane 2 is at the beam waist, so R2 = \u221e and thus the denominator of this expression must vanish: \u2212(f \u2212 d) + df 2/z 2 01 = 0. (6.67) 2w\u00ba\u00a1dbeamwaistbeamwaist2w\u00ba\u2122 104 Chapter 6. Gaussian Beams Solving this for the distance, we find d = f 1 + f 2/z 2 01. (6.68) (focus distance for Gaussian beam) Note that the focus is not located a distance f away as in classical optics\u2014the focus occurs at a slightly smaller distance. We only recover the geometrical result d = f in the limit of a large input beam, z01 \u2212\u2192 \u221e (i.e., the plane-wave limit). Roughly speaking, a beam with a finite z01 (or w01) is still \u2018\u2018like\u2019\u2019 a converging beam at the focus. This isn\u2019t really a large effect. For some sample numbers, let\u2019s consider f = 0.5 m, \u03bb = 2\u03c0 \u00d7 10\u22127 m (close to a He-Ne laser line), and w01 = 1 cm. In this case, z01 = \u03c0w 2 01/\u03bb = 500 m. Since z01 f, the effect is very small (only a 1 ppm reduction in the focal distance). Now we can come to the main point, the minimum spot size. The imaginary part of Eq. (6.65) gives z02 = \u03c0w 2 02 \u03bb = (f \u2212 d)2 + (f d/z01)2 (f \u2212 d)(f /z01) + f d/z01. (6.69) We will consider the limit z01 f, which simplifies the algebra here, and is the most important case in practical optical systems when trying to obtain a small spot size (i.e., using a lens to focus a large, collimated beam, where the large spot size implies a correspondingly large Rayleigh length). Then d \u2248 f and Thus, z02 \u2248 f d z01 \u2248 f 2 z01. (6.70) w02 \u2248 \u03bbf \u03c0w01 \u2248 3 \u03c0 \u03bb(f /#), (6.71) (ideal minimum spot size) where f /# (the \u2018\u2018f -number\u2019\u2019) is given by (f /#) := f D, (6.72) (definition of f /#) with D denoting the diameter of the lens. In writing down this expression, we are assuming a diameter of 2 \u00b7 1.5 \u00b7 w01, which passes 99% of the total incident beam power. This calculation is for an ideal lens, where we can ignore aberrations. A simple approximate correction to account for aberrations2 is to multiply this result by 4/3: w02 \u2248 4 \u03c0 \u03bb(f /#). (6.73) (more realistic minimum spot size) However, this is a heuristic formula based on experience with typical lenses, and not justifiable on fundamental grounds. (In particular, spherical aberrations should lead to larger corrections for larger (f /#) and thus smaller spot sizes.) 6.6 Hermite\u2013Gaussian Beams If we don\u2019t require axial symmetry, we can obtain more general solutions to the paraxial wave equation. The Hermite\u2013Gaussian modes form one possible set of solutions l,m (r) = E(+) E(+) 0 w0 w(z) 1 \u221a 2 x w(z) Hl Hm \u221a 2 y w(z) 2l+ml!m! ikz \u2212 i(1 + l +", " m) tan\u22121 exp r2 w2(z) \u00d7 exp \u2212 z z0 exp ik. r2 2R(z) (TEMl,m mode) (6.74) 2As recommended by the Melles-Griot catalog. 6.6 Hermite\u2013Gaussian Beams 105 These are also referred to as TEMl,m modes. Here, a mode refers to field profiles that, when confined in resonators (as we will see) have profiles that are time-invariant. (So far, we are only considering propagation in free space, in which case the analogous property is that the intensity profiles of these modes are invariant along the beam except for an overall scaling, which is described by w(z) here.) To facilitate comparison with the Gaussian beam, we can write the general Hermite\u2013Gaussian mode in terms of the simple Gaussian beam as l,m (r) = E(+) E(+) Gauss(r) 1 2l+ml!m! \u221a 2 x w(z) Hl Hm \u221a 2 y w(z) \u2212i(l + m) tan\u22121 exp, z z0 (6.75) where E(+) Gauss(r) is the Gaussian beam from Eq. (6.10). The differences compared to the Gaussian beam are: (1) the presence of the Hermite polynomials Hn(z); (2) the normalization factor just before the Hermite polynomials, chosen so that the total beam power for a given field E(+) is the same for all l and m; and (3) the factor of (1 + l + m) in the Gouy phase. (TEMl,m mode in terms of Gaussian beam) 0 The Hermite polynomial Hn(z) is a polynomial of degree n. We can write them out explicitly as H0(z) = 1 H1(z) = 2z H2(z) = 2(2z2 \u2212 1) Hn(z) = (\u22121)nez2 dn dzn e\u2212z2 (explicit formula) (6.76) Hn+1(z) = 2zHn(z) \u2212 2nHn\u22121(z). (recurrence relation) For our purposes here, it is most important to know that Hn(z) is an nth degree polynomial with n zeros on the real axis. Note that since H0(z) = 1, we see that for l = m = 0 we recover the Gaussian beam, as is consistent with our earlier notation. Hence, the Gaussian beam is also called the TEM0,0 beam. In general, the intensity pattern of the TEMl,m mode has l dark bands across the x-direction and m dark bands across the y-direction. Alternately, the intensity pattern has a grid of l + 1 bright spots in the x-direction and m + 1 in the y-direction. Note that the Hermite polynomials diverge for large |x| or |y|. However, the Gaussian factor always cuts off the intensity far away from the optical axis, so the intensity is still concentrated (relatively) close to the optical axis. Like the Gaussian beam, the Hermite\u2013Gaussian beams have intensity patterns that stay the same as they propagate in space, except for the overall scale. They have the same far-field divergence characteristics as for the Gaussian beam (except for the different spatial profile, which again is constant up to stretching over the entire extent of the beam). The higher-order modes are wider for the same value of w0. However, all the same transformation rules apply to the Hermite\u2013Gaussian beams, including the ABCD law. Again, there is an important connection with quantum mechanics. The Hermite\u2013Gaussian beams are equivalent to the eigenfunctions of the quantum harmonic oscillator. Thus, the Hermite\u2013Gaussian modes form a complete set\u2014that is, any beam can be represented as a linear combination of Hermite\u2013Gaussian modes, and its spatial evolution is a manifestation of the interferences between the component modes. 106 Chapter 6. Gaussian Beams TEM0,0 TEM0,1 TEM0,2 TEM0,3 TEM1,0 TEM2,0 TEM1,1 TEM2,4 TEM5,6 6.6.1 Doughnut and Laguerre\u2013Gaussian Modes Before finishing, it is worth mentioning one more beam, the equal superposition of TEM0,1 and TEM1,0 beams, with a relative phase of \u00b1i. This beam has an axially symmetric intensity pattern, where the intensity vanishes at the origin. This beam is called the", " doughnut mode or TEM\u2217 0,\u00b11 modes. This mode is important in describing the output of lasers as the dominant component for slightly distorted (but still 6.6 Hermite\u2013Gaussian Beams 107 axially symmetric) Gaussian beams. TEM\u2217 0,\u00b11 = TEM0,1 \u00b1 iTEM1,0 The doughnut mode is a particular example of another class of beams, the Laguerre\u2013Gaussian beams, where the general form is given by 3 l,m (r) = E(+) E(+) Gauss(r) 2l! \u03c0(l + |m|)! \u221a 2 r w(z) |m| L|m| l \u2212i(2l + |m|) tan\u22121 exp z z0 e\u2212im\u03c6, (6.77) where l \u2265 0 and m \u2208 Z. These beams are radially symmetric in intensity, so that the only azimuthal dependence (on the azimuthal angle \u03c6) is in the phase. These are the analogous solutions to the paraxial wave equation to the Hermite\u2013Gaussian beams, but when separated in cylindrical rather than Cartesian coordinates. l,m mode in terms of Gaussian beam) The polynomials L|m| l (z) here are the associated Laguerre polynomials 2r2 w2(z) (TEM\u2217 Lm 0 (z) = 1 Lm Lm 2 (z) = 1 (z) = \u2212z + m + 1 z2 2 z\u2212mez n! here defined for m \u2265 0. Again, L|m| (z) is a polynomial of degree l, and thus has l zeros. Here, these are manifested in increasing numbers of dark circles or dark spots at the center (for odd l) for higher-order beams. \u2212 (m + 2)z + dn dzn (m + 2)(m + 1) 2 (explicit formula), e\u2212zzn+m n (z) = (6.78) Lm Note that the azimuthal phase in Eq. (6.77) varies as eim\u03c6, which gives the two options m = \u00b11 for two doughnut modes TEM\u2217 0,\u00b11. These correspond to the phase \u2018\u2018winding\u2019\u2019 in different directions around the doughnut. More generally, the phase winding in these doughnut modes and the Laguerre\u2013Gaussian beams represents angular momentum carried by the electromagnetic field (with larger |m| values representing larger angular momenta). l 3For an operator-based derivation of both the Laguerre\u2013Gaussian and Hermite\u2013Gaussian beams, see Francesco Pampaloni and J\u00f6rg Enderlein, \u2018\u2018Gaussian, Hermite-Gaussian, and Laguerre-Gaussian beams: A primer,\u2019\u2019 (2004) (arXiv: physics/0410021). 108 6.7 Exercises Problem 6.1 Chapter 6. Gaussian Beams A common technique in the laboratory for measuring the beam waist parameter of a Gaussian beam is illustrated in the diagram. A Gaussian beam is incident on an optical power meter, which registers the total power of the incident beam. A knife edge can be translated in the transverse direction to block part of the beam (i.e., if the position of the knife edge is xknife, then the parts of the beam with x < xknife is blocked from reaching the power meter). The \u2018\u201810-90\u2019\u2019 rule is to measure the knife edge position x10% where the power meter reads 10% of the total beam power, and then the position x90% where the power meter reads 90% of the total beam power. Then the beam radius w(z) at the knife-edge location z along the beam is given by w(z) = \u03b1|x10% \u2212 x90%|, (6.79) where \u03b1 is some constant factor. Calculate the numerical value of \u03b1. Problem 6.2 Suppose you have a 500 mW, TEM0,0 (i.e., Gaussian) beam from a continuous-wave Ti:sapphire laser (\u03bb = 850 nm). The measured beam radius at the output coupler (mirror) is w = 0.7 mm, and the beam radius in the laser rod is w = 330 \u00b5m. (a) What is the optical distance between the output coupler and the laser rod? Assume that the length of the rod is negligible (it is actually around 20 mm) and that the focus of the beam occurs in the center of the rod. (b) How far will this beam propagate (past the output coupler) before the", " spot size is 1 cm? (c) What is the radius of curvature of the phase front at the distance you calculated in (b)? (d) What is the amplitude of the electric field at the center of the focus? Assume a refractive index of n = 1.75 for sapphire. Problem 6.3 Recall that the Gaussian (TEM0,0) beam is a solution to the wave equation in the paraxial approximation. Also recall that the Gaussian beam is specified completely in terms of relatively few parameters. (a) What are the parameters? (b) State (general) conditions on the parameters that guarantee the paraxial approximation is valid. Problem 6.4 Show that within the paraxial approximation, a thin lens converts a plane wave into a spherical wave. Use the following procedure. (a) Use Fermat\u2019s principle to show that a thin lens of focal length f introduces an optical delay with an equivalent path length given by lens(y) = d \u2212, (6.80) y2 2f where d is an arbitrary length constant and y is as usual the normal distance from the optical axis. xzpower meterknife edge 6.7 Exercises (b) A spherical wave centered on the optical axis propagating inward has the form E(+)(R) = E(+) 0 R exp(\u2212ikR), 109 (6.81) where R2 = y2 + (z \u2212 z0)2 is the radial distance from the center of the wave at (y, z) = (0, z0). Note that we are suppressing the x-dependence for simplicity, but it isn\u2019t difficult to extend the argument to the full three-dimensional case. Take z0 = f, make the paraxial approximation (expanding to lowest order in y) and show that the field at z = 0 can be written E(+)(R) = E(+) 0 f exp(\u2212ikf ) exp \u2212ik. y2 2f (6.82) (c) What is the phase shift corresponding to lens(y)? Write down a plane wave that matches the spherical wave provided the lens is placed at z = 0. Problem 6.5 A Gaussian beam passes through a thin lens. Show that R2 = R1f f \u2212 R1, (6.83) where f is the focal length of the lens, R1 is the radius of curvature of the input beam at the lens, and R2 is the radius of curvature of the output beam at the lens. (Use the ABCD Law.) Problem 6.6 A Gaussian beam with w0 = 1 cm is focused by a thin lens of focal length f = 2 cm. The lens is placed at the focus of the original Gaussian beam. Assume an optical wavelength of \u03bb = 1.0 \u00b5m. (a) At what distance from the lens does the new focus occur? (b) What is the spot size at the new focus? Give numbers for both an ideal and a realistic lens. (c) What is the far-field expansion angle? Problem 6.7 Verify that the compact form for the Gaussian beam in terms of the complex beam parameter, Eq. (6.41), E(+)(r) = E(+) 0 q0 q(z) exp ikr2 2q(z) exp(ikz). is equivalent to the standard form for the Gaussian beam, Eq. (6.10), E(+)(r) = E(+) 0 w0 w(z) exp \u2212 r2 w2(z) ikz \u2212 i tan\u22121 exp z z0 exp ik. r2 2R(z) Problem 6.8 (a) What is the Hermite\u2013Gaussian beam pictured below? (6.84) (6.85) 110 Chapter 6. Gaussian Beams (b) Suppose the horizontal distance between the vertical dark bands is 100 \u00b5m at the focus. What is w0? Problem 6.9 (a) For the TEM0,0, TEM0,1, and TEM1,1 modes, show that the fraction of the total power contained in a disc of radius \u03c1 is P0,0(\u03c1) P0,0(\u03c1 \u2192 \u221e) for the TEM0,0 mode, = 1 \u2212 exp \u22122 2 \u03c1 w0 P0,1(\u03c1) P0,1(\u03c1 \u2192 \u221e) = 1 \u2212 exp \u22122 \u03c1 w0 2 1 + 2 2 \u03c1 w0 for the TEM0,1 mode, and P1,1(\u03c1) P1,1(\u03c1 \u2192 \u221e) = 1 \u2212 exp \u22122 2 \u03c1 w0 1 + 2 2 \u03c1 w0 + 2 \u03c1 w0 4 for the TEM1,1", " mode. Note: you will need to evaluate an integral of the form a 0 x2n+1e\u2212bx2 dx. This is easy for n = 0, since the integral is just a Gaussian: a 0 xe\u2212bx2 dx = 1 2b exp(\u2212bx2) a 0 = 1 2b 1 \u2212 exp(\u2212ba2). Other cases may be evaluated recursively, noting that a 0 x2n+1e\u2212bx2 dx = \u2212 \u2202 \u2202b a 0 x2n\u22121e\u2212bx2 dx. For example, (6.86) (6.87) (6.88) (6.89) (6.90) (6.91) a 0 x3e\u2212bx2 dx = \u2212 \u2202 \u2202b a 0 xe\u2212bx2 dx = \u2212 \u2202 \u2202b 1 2b 1 \u2212 exp(\u2212ba2) = 1 2b2 1 \u2212 exp(\u2212ba2)(1 + ba2). (6.92) (b) For each mode, what radius contains 99% of the total beam power? 6.7 Exercises Problem 6.10 111 A laser of wavelength \u03bb emits a Gaussian beam, where the focus of the Gaussian beam occurs at the front of the laser. The laser beam hits a screen, which is a distance d from the front of the laser. (a) What beam-waist parameter w0 produces the smallest spot on the screen, and what is the corresponding 1/e2 beam radius on the screen? (b) What is the 1/e2 beam radius on the screen for \u03bb = 200\u03c0 nm and d = 10 cm? Problem 6.11 Here we will consider a simple model for the pinhole camera. The idea is to compute the optimal size of the pinhole to get the best image resolution: the image on the screen of a distant point source must be at least the size of the aperture, but also is blurred by diffraction if the aperture is too small. The optimal diameter can be chosen as a compromise between the two effects; Lord Rayleigh4 found an \u03bbd for a uniform circular aperture. optimal pinhole diameter of 1.9 \u221a Consider an aperture illuminated by a plane wave. The light passing through the aperture is viewed on a screen a distance d away. (a) If the incident light is treated in geometrical optics, what is the size of the aperture that gives the smallest spot on the screen? Explain. (Hint: what do plane waves correspond to in geometrical optics?) (b) Now answer the same question in wave optics. Assume that the aperture has a Gaussian transmission coeffcient for the electric field of t = exp \u2212, r2 a2 (6.93) (i.e., Eout = tEin), so that a is the \u2018\u2018radius\u2019\u2019 of the aperture. Now what is the size of the aperture that gives the smallest spot on the screen? Problem 6.12 (a) What is the Hermite\u2013Gaussian beam pictured below? 4\u2018\u2018Pinhole camera,\u2019\u2019 Wikipedia entry (http://en.wikipedia.org/wiki/Pinhole_camera) as of 25 February 2006. dscreenapertureplane wave 112 Chapter 6. Gaussian Beams (b) Suppose this beam has the same fractional power contained in a circular disc of radius a as a Gaussian beam. Compare (qualitatively) the far-field divergence of the two beams. Problem 6.13 Consider two monochromatic, Gaussian beams: a \u2018\u2018red\u2019\u2019 beam of frequency \u03c9, and a \u2018\u2018blue\u2019\u2019 beam of frequency 2\u03c9. Both have identical transverse intensity profiles at their respective foci. Describe completely how the geometries (both near- and far-field) of the two fields differ. Problem 6.14 A monochromatic Gaussian beam (with beam waist parameter w0) is focused onto a thin, nonlinear optical element, whose effect on the incident field is Eout = \u03b1(Ein)2, (6.94) where \u03b1 is some constant. Show that if the optical frequency of the input beam is \u03c9, the output field has a component with optical frequency 2\u03c9. Describe completely how the geometry (both near- and far-field) of the 2\u03c9 output field differs from the input field. Hint: The nonlinear optical element only acts at z = 0, so consider what the field looks like there. Problem 6.15 The set of all Hermite\u2013Gaussian modes with the same beam parameter w0 form a complete set. Denoting the TEMl,m mode by E(+) l,m (r; w0) (6.", "95) to emphasize the dependence on w0, this means that an arbitrary electric field profile E(+)(r) at z = 0 can be written as a superposition of all the modes at z = 0: E(+)(x, y) = \u221e l,m=0 cl,mE(+) l,m (x, y; w0). (6.96) (a) Noting that the Hermite\u2013Gaussian modes are all orthogonal, explain how you would compute the coefficients cl,m. (Set up any necessary equations but don\u2019t refer to the explicit form of E(+) l,m (r; w0) or evaluate any complicated integrals.) (b) Once you know these coefficients, you can compute the field at any location along the z-axis by E(+)(x, y, z) = \u221e l,m=0 cl,mE(+) l,m (x, y, z; w0). (6.97) Find the flaw in the following argument: each mode E(+) l,m (r; w0) has the same far-field divergence angle (\u03b80 = tan\u22121(w0/z0)); thus, every field, no matter what its profile (or size) at z = 0, has a far-field divergence angle of \u03b80. (For examply, you can apply this argument to a Gaussian of size 15w0, but this clearly has a different far-field divergence angle than \u03b80 for the constituent beams.) Hint: how is the far-field divergence angle for a Gaussian beam defined? How would modifying the definition give a different answer? Problem 6.16 Suppose that an optical cavity is described by the round-trip ray matrix M(zi) = A B C D, (6.98) 6.7 Exercises 113 which is computed for a ray starting at location zi. Assume that a Gaussian beam is the characteristic mode of the cavity, and let q = q(zi) denote the complex beam parameter for the beam at z = zi. (a) Show that q = A \u2212 D 2C \u2212 i 1 C 2 \u2212 A + D 2C 2. (b) Show that the radius of curvature at z = zi is and the beam radius at z = zi is R(zi) = \u2212 2B A \u2212 D w(zi) = \u03c0 1 \u2212 \u03bb|B| A + D 2. 2 (6.99) (6.100) (6.101) Hint 1: what should happen to the q parameter after one round trip if it is to be the resonant mode of a cavity? Hint 2: recall that det(M) = 1 for a round-trip matrix. Problem 6.17 Consider a Gaussian beam propagating along the +z direction in vacuum, with beam waist w0 and vacuum wavelength \u03bb0. Now compare this to the case where the beam shines into a thick dielectric slab of refractive index n > 1, extending to z = +\u221e, and with the (planar) vacuum\u2013dielectric interface occurring a distance d in front of the original focus (i.e., at z = \u2212d). The beam is perpendicular to the interface, and the new focus occurs inside the dielectric slab. (a) Use the ABCD Law to derive an expression for the location of the new focus relative to the interface. (b) What is the beam waist in the dielectric case, in terms of the beam waist w0 in the vacuum case? Problem 6.18 The Hermite polynomials may be defined as in Eq. (6.76) by the explicit formula (Rodrigues formula) but they can also be defined via the generating function Hn(x) = (\u22121)n ex2 dn dxn e\u2212x2, g(x, t) := e2xt\u2212t2 = \u221e n=0 Hn(x) tn n!. That is, the Hermite polynomials are the coefficients in the Taylor expansion of g in t. (a) Show that differentiating the generating function with respect to t leads to the recurrence Hn+1(x) = 2xHn(x) \u2212 2nHn\u22121(x). (b) Show that differentiating the generating function with respect to x leads to the recurrence H n(x) = 2nHn\u22121(x). (c) Show that the Rodrigues formula leads to the same recurrence as in (a). (d) Show that the Rodrigues formula leads to the same recurrence as in (b). (e) Thus argue that the two definitions for the Hermite polynomials are equivalent. (6.102) (6.", "103) (6.104) (6.105) Chapter 7 Fabry\u2013Perot Cavities Now let\u2019s return to the subject of resonators, which we treated earlier in Section 2.7 in the context of ray optics. Now we will perform a better analysis, which includes the wave nature of the light. Interference is a central feature of resonators with waves, and thus resonators are also interferometers. Because of this, the conventional two-mirror resonator is referred to as a Fabry\u2013Perot interferometer, Fabry\u2013Perot cavity, Fabry\u2013Perot etalon, or an optical spectrum analyzer. 7.1 Resonance Condition We will start with the simplest case of a planar cavity of length d, where the mirrors are perfect reflectors. In order for a plane wave to exist in the cavity, it must return to exactly the same phase after one round trip through the cavity. In this case, it will constructively interfere with itself. Otherwise, the phase will precess on each successive round trip, and eventually lead to destructive interference. Thus, the round-trip accumulated phase must be some integer multiple of 2\u03c0: 2kd = 2\u03c0q (q = 0, 1, 2,...). (7.1) (cavity resonance condition) Thus, the allowed stationary waves, or modes, of the cavity are plane waves that satisfy kq = \u03c0q d. (7.2) (cavity allowed wave numbers) We can also rewrite this resonance condition in terms of related quantities: \u03bbq = 2d q ; \u03bdq = cq 2d ; \u03c9q = \u03c0cq d. The corresponding spectrum is a set of regularly spaced, arbitrarily sharp peaks. (cavity allowed wavelengths/frequencies) (7.3) dn\u0153n\u0153o\u00f2o\u00a1n\u0153o\u2013o\u00a1nIfsr = c/2d 116 Chapter 7. Fabry\u2013Perot Cavities The frequency spacing is a special quantity called the free spectral range, defined as FSR := \u03bdq+1 \u2212 \u03bdq = c 2d = c0 2nd, (7.4) (free spectral range) where n is the refractive index of the medium filling the cavity. An alternative to this constructive-interference approach to the resonance condition is to view the reflectors as imposing boundary conditions on the wave. The wave inside the cavity is a standing-wave mode, which is a superposition of left- and right-going waves: For ideal mirrors, the amplitudes of the two waves are equal, so that E(+)(r) = E(+) 01 eikz + E(+) 02 e\u2212ikz. |E(+) 01 | = |E(+) 02 | =: E(+) 0. Thus, we can write E(+)(r) = E(+) 0 = E(+) 0 = E(+) 0 eikz+i\u03c61 + e\u2212ikz\u2212i\u03c62 ei(\u03c61\u2212\u03c62)/2 ei\u2206\u03c6/2 cos(kz + \u00af\u03c6), eikz+i(\u03c61+\u03c62)/2 + e\u2212ikz\u2212i(\u03c61+\u03c62)/2 (7.5) (7.6) (7.7) where \u03c61 and \u2212\u03c62 are the phases of the first and second fields, respectively, \u2206\u03c6 := \u03c61 \u2212 \u03c62, and \u00af\u03c6 := (\u03c61 + \u03c62)/2. The wave must vanish at the mirrors, which we can view as perfect conductors. If the left-hand mirror is at z = 0, E(+)(z = 0) = 0 =\u21d2 \u00af\u03c6 = \u2212\u03c0/2, so that the cosine changes to a sine: E(+)(r) = E(+) 0 ei\u2206\u03c6/2 sin(kz). Then the right-hand mirror is at z = d: E(+)(z = d) = 0 =\u21d2 sin(kd) = 0 =\u21d2 kd = q\u03c0. (7.8) (7.9) (7.10) This is the same resonance condition as before. The term resonance here suggests that when the cavity is exposed to light with one of the allowed frequencies, the cavity will respond, but not otherwise. We will discuss this more below, after generalizing what we have done so far to \u2018\u2018leaky\u2019\u2019 cavities. 7.2 Broadening of the Resonances: Cavity Damping Real cavities involve some loss of the light on each round trip. The losses damp the amplitude of the cavity. Just as in the", " case of a driven mechanical oscillator, the damping leads to broadening of the resonances. Consider the circulating field inside a cavity. In particular, we will track the amplitude of the field on its nth round trip in the cavity. rEo(+)n 7.2 Broadening of the Resonances: Cavity Damping Suppose that the amplitude of the wave is reduced by a factor of r on each round trip: |E(+) n+1| = |rE(+) n |. 117 (7.11) n Here, E(+) is the electric field amplitude of the plane wave on the nth round trip through the cavity. The loss due to r could be some transmission of the mirrors (as indicated in the diagram here), losses around the edges of finite-size mirrors, scattering from gases or objects inside the cavity, and so on. On one round trip, the phase of the plane wave changes by \u2206\u03c6 = 2kd =, (7.12) 4\u03c0d \u03bb so that n+1 = rei2kdE(+) n. We may assume that r is real, as any nonzero phase angle can be absorbed into the phase factor as an additional effective length. The total wave is (7.13) E(+) E(+) = E(+) 0 + E(+) 1 + E(+) 2 + \u00b7 \u00b7 \u00b7 = E(+) 0 1 + rei2kd + (rei2kd)2 + \u00b7 \u00b7 \u00b7 = E(+) 0 1 \u2212 rei2kd, (7.14) where we have used \u221e n=0 xn = 1/(1 \u2212 x). The total intensity is I = I0 |1 \u2212 rei2kd|2 = = = = = I0 |1 \u2212 r cos(2kd) \u2212 i sin(2kd)|2 I0 [1 \u2212 r cos(2kd)]2 + r2 sin2(2kd) I0 1 \u2212 2r cos(2kd) + r2 cos2(2kd) + r2 sin2(2kd) I0 1 + r2 \u2212 2r cos(2kd) I0 (1 \u2212 r)2 + 4r sin2(kd), (7.15) where I0 = 2|E(+) 0 |2/\u03b7. 7.2.1 Standard Form In standard form, we can write the resonator intensity as where the maximum intensity is and the finesse, defined by I = Imax 1 + 2 2F \u03c0 sin2(kd), (7.16) (cavity circulating intensity) Imax := I0 (1 \u2212 r)2, (7.17) (maximum circulating intensity) F := \u221a \u03c0 r 1 \u2212 r, (7.18) (cavity finesse) 118 Chapter 7. Fabry\u2013Perot Cavities is an important parameter that characterizes the lossy nature of the cavity. The only dependence of the intensity on the cavity length d is in the sin function, so the intensity is periodic in d with period \u03bb/2. In terms of the free spectral range c/2d, we can write the intensity as I = Imax 1 + 2 2F \u03c0 sin2 \u03c0\u03bd FSR, (7.19) (cavity circulating intensity) so that in frequency the intensity is also periodic, where the period is the free spectral range. 7.2.2 Maximum and Minimum Intensity The intensity is maximized when the denominator is smallest. This occurs when the sinusoidal term vanishes, which gives the condition sin2 \u03c0\u03bd FSR = 0 =\u21d2 \u03bd = q(FSR). (7.20) This is just the resonance condition for the lossless cavity. At these points, the cavity attains the maximum intensity Imax. The minimum intensity occurs when the denominator is maximum. Thus the sinusoidal term becomes unity, and Imin = 1 + Imax 2F \u03c0 2. (7.21) The minimum intensity only goes to zero in the limit of large finesse. 7.2.3 Width of the Resonances As a way to characterize the width of the resonances, we will now find the frequencies such that the intensity falls to Imax/2. In this case, the width condition is sin2 \u03c0\u03bd (7.22) 2 =, \u03c0 2F FSR which is satisfied by the frequencies \u03bd = \u00b1 FSR \u03c0 sin\u22121 \u03c0 2F, (7.23) up to an added multiple of the free spectral range (again, because of the periodicity of the sin function). The width of the resonances is really only sensible concept in the limit of large finesse, when the resonances are well resolved. In this limit", ", we can use the small-angle approximation for the sin function and write the half-maximum intensity frequencies as \u03bd \u2248 \u00b1 FSR 2F. (7.24) In this approximation, we can thus write the full width of the resonance at half maximum as \u03b4\u03bdFWHM = FSR F. (7.25) Notice that in the limit of arbitrarily large finesse, \u03b4\u03bdFWHM \u2212\u2192 0 while Imax diverges\u2014the resonances become \u03b4 functions, perfectly defined resonances as in the lossless cavity. (cavity resonance full width at half maximum) We can summarize the characteristics of the Fabry-Perot cavity in the diagram, recalling that two parameters (FSR and F ) are necessary to completely specify the cavity behavior; all other parameters are determined by these two. 7.2 Broadening of the Resonances: Cavity Damping 119 The next diagram compares the cavity intensities for several different finesses, showing the transition from broad resonances to sharply defined peaks. 7.2.4 Survival Probability Losses in a cavity can come from a number of sources. The mirrors can lead to intensity loss due to partial transmission, losses of beams around the mirror edges, absorption of the light in the reflective coatings, and scatter due to surface roughness. A medium inside the cavity can also absorb or scatter the light. To generalize the treatment of losses, we can define the survival probability Ps, which is the fraction of the intensity that stays in the cavity after one round trip. If the two mirrors have reflectances (intensity reflection coefficients) R1,2, for example, then the survival probability is just Ps = R1R2. (7.26) Additional factors can model other losses due, for example, to other objects in the cavity. In terms of the survival probability, we can write F = \u03c0P 1/4 s \u221a Ps 1 \u2212 (7.27) (finesse in terms of survival probability) as a more convenient expression for the finesse and Imax := I0 \u221a, Ps)2 (1 \u2212 (maximum cavity intensity in terms of survival probability) (7.28) for the maximum intensity. 7.2.5 Photon Lifetime The name \u2018\u2018survival probability\u2019\u2019 more appropriately describes the fate of a photon after one round trip through the cavity. The probability that the photon is lost is simply 1 \u2212 Ps. Thus, a photon lasts for n = 1 1 \u2212 Ps (7.29) round trips on average. We can see this as follows. The probability for a photon to exit the cavity on the nth round trip is the product of the probability P n\u22121 to survive for n \u2212 1 round trips, multiplied by the probability (1 \u2212 Ps) for the photon to not survive the nth round trip: s P (n) = P n\u22121 s (1 \u2212 Ps). (7.30) nqo(fsr)(qo-o1)(fsr)(qo+o1)(fsr)IminImaxI oo/2maxIfsr/FIF = 2F = 10F = 50nqo(fsr)(qo-o1)(fsr)(qo+o1)(fsr) 120 Chapter 7. Fabry\u2013Perot Cavities The average number of round trips survived is then the mean of this distribution function, noting that the photon can only fail to survive after one round trip: n = = \u221e n=1 \u221e n=0 nP (n) nP (n) = (1 \u2212 Ps) = (1 \u2212 Ps) = (1 \u2212 Ps) = (1 \u2212 Ps) \u221e n=0 \u2202 \u2202Ps \u2202 \u2202Ps nP n\u22121 s \u221e P n s n=0 1 1 \u2212 Ps 1 (1 \u2212 Ps)2 (7.31) 1 1 \u2212 Ps Here, we have again used the geometric sum \u221e derivative to transform the sum into a more standard one. =. n=0 xn = 1/(1 \u2212 x), and note the handy trick of using the Since the time for one round trip is \u03c4rt = 2d/c, we can more generally write Then the \u2018\u2018lifetime\u2019\u2019 of a photon inside the cavity is the product of the round-trip time and the average number of round trips that the photon will survive: \u03c4rt = 1 FSR. (7.32) (round-trip time). (7.33) (cavity photon lifetime) \u03c4p = 1 (FSR)(1 \u2212 Ps) \u221a F \u2248 \u03c0 \u221a 1 \u2212 Ps For a \u2018\u2018good\u2019\u2019 resonator, Ps is close to unity, so lowest order in (1\u2212 Thus, Ps), which amounts to setting P 1/4 \u221a", " Ps is also close to unity. In this limit we can expand to \u221a Ps). s \u2248 1 and (1\u2212Ps) = (1\u2212 Ps) \u2248 2(1\u2212 Ps)(1+ \u221a \u221a (7.34) (7.35) and so \u03c4p \u2248 1 2(FSR)(1 \u2212 \u221a Ps) = F 2\u03c0(FSR) = 1 2\u03c0\u03b4\u03bdFWHM. Thus we arrive at an \u2018\u2018uncertainty relation\u2019\u2019 \u03c4p\u03b4\u03bdFWHM = 1 2\u03c0 (7.36) (cavity uncertainty relation) for optical cavities between frequency and time. 7.2.6 Q Factor The Q factor represents the \u2018\u2018quality\u2019\u2019 of any oscillator. The general definition is Q := 2\u03c0 \u00b7 stored energy energy loss per oscillation cycle. (7.37) (Q factor for general oscillator) 7.2 Broadening of the Resonances: Cavity Damping 121 For a \u2018\u2018good\u2019\u2019 oscillator\u2014that is, one that does not dissipate energy very rapidly\u2014the Q factor is large because the denominator in the definition is small. A cavity is an oscillator, but at optical frequencies, which are of the order of a few hundred THz. Thus, we can already see that optical Q factors are very large. In fact, the finesse plays the same role as Q for optical cavities, but generally produces more manageable numbers. For an optical cavity, the fraction of energy lost per round trip is 1 \u2212 Ps. Following the same logic as in the last section, the rate at which energy leaves the cavity is 1/\u03c4p, relative to the amount of energy in the cavity. This may not be completely obvious, so let\u2019s quickly show this. In terms of the survival probability, the probability to survive n round trips in the cavity is s = (Ps)t/\u03c4rt = e(t/\u03c4rt) log Ps \u2248 e\u2212(t/\u03c4rt)(1\u2212Ps) = e\u2212t/\u03c4p, P n (7.38) where we used log(1 + x) \u2248 x for small x = 1 \u2212 Ps, and \u03c4p = \u03c4rt/(1 \u2212 Ps), which follows from Eqs. (7.32) and (7.33). Thus, the energy loss is exponential in time (with the assumption of a good cavity), with time constant \u03c4p, and hence loss rate 1/\u03c4p. Continuing with Q, the energy loss per oscillation cycle is just the optical period multiplied by this rate, or 1/\u03bdq\u03c4p. Putting this into the above definition, we see that Q = 2\u03c0\u03bdq\u03c4p. (7.39) (Q for optical cavity) We can also write this in terms of the cavity line width, using the uncertainty relation in Eq. (7.36): Q = \u03bdq \u03b4\u03bdFWHM. (7.40) (Q for optical cavity) We can compare this to the form of the finesse from Eq. (7.25), and see that Q and F are the same except that a different (smaller) oscillator frequency is used for the finesse. Since \u03bdq = q(FSR), we see that Q and F differ by a factor of the resonance order q: F = FSR \u03b4\u03bdFWHM, (7.41) Q = qF. (7.42) (Q related to finesse) The order q is typically a very large integer for optical cavities, since the cavity length often far exceeds the optical wavelength. Thus, the Q factor for optical cavities is typically huge compared to the finesse. That\u2019s why the finesse is a preferred quantity in the optical regime\u2014the numbers are much easier to deal with.1 7.2.7 Example: Finesse and Q Consider a planar cavity where both mirrors reflect 99% of the incident intensity. Then Ps = 0.992, and so F \u2248 300. For a cavity length d = 15 cm, the round-trip time is \u03c4rt = 2d/c = 1 ns. (A related handy number to remember for the speed of light is c0 \u2248 1 ft/ns.) Then the free spectral range is FSR = 1/\u03c4rt \u2248 1 GHz. The photon lifetime is \u03c4p = F /2\u03c0(FSR) \u2248 50 ns. This leads to a cavity line width of \u03b4\u03bdFWHM = 1/2\u03c0\u03c4p \u2248 3 MHz. A typical optical frequency is 300 THz, so that Q = \u03bdq/\u03b4\u03bdFWHM = 108, a really large number compared to the finesse. 1Nevertheless, the Q factor is a useful measure of the stability", " of the best oscillators, which now operate at optical frequencies. In 2010 the record for an observed Q = 4.2 \u00d7 1014 was set using a clock based on an Al+ ion operating at 1.121 PHz, with a FWHM width of 2.7 Hz: C. W. Chou, D. B. Hume, T. Rosenband, and D. J. Wineland, \u2018\u2018Optical Clocks and Relativity,\u2019\u2019 Science 329, 1630 (2010) (doi: 10.1126/science.1192720). 122 Chapter 7. Fabry\u2013Perot Cavities The corresponding wavelength is \u03bb = c0/\u03bdq = 1 \u00b5m, which is in the near-infrared (approximately the wavelength of a Nd:YAG laser). The order is thus q = \u03bdq/(FSR) = 3 \u00d7 105. Recall that q = 2d/\u03bb, and it is because typical cavities are long compared to the optical wavelength that the order q is often very large. Again, one reason that Q is huge is because the optical frequency is so high. In the microwave, cavity Q factors are more reasonable. For example, in the microwave \u03bdq \u223c 1 GHz, and so \u03b4\u03bdFWHM \u223c 1 MHz would yield a Q of 103. But note that some microwave devices can have extremely narrow line widths, so microwave Q factors can be large as well. The best \u2018\u2018supercavities\u2019\u2019 in the optical have finesses in the range of 105 to 106 at present.2 7.3 Cavity Transmission If you think about it in a certain way, a Fabry\u2013Perot cavity is an amazing device. Imagine a mirror that is nearly perfectly reflecting. A laser beam impinging on the front mirror will mostly bounce off, only transmitting a small fraction of the intensity. But by putting a second mirror behind that mirror, just the right distance away, all the laser power can instead transmit through the mirror. Makes your brain tingle just thinking about it, doesn\u2019t it? Such is the power of interference. Let\u2019s see how this works out. We now need to be a bit more precise about the intensities and electric fields in and around the cavity. We will refer to the left- and right-hand mirrors as mirrors 1 and 2, respectively. The total circulating intensity I within the cavity is the intensity propagating to the right of and away from mirror 1. We will also use the conventions in the diagram for the field reflection and transmission coefficients (with primed coefficients for the reverses of the directions indicated). The circulating intensity is related to the intensity of the initial wave before the resonant buildup by Eq. (7.19): I = Imax 1 + 2 2F \u03c0 sin2 \u03c0\u03bd FSR, (7.43) where again Imax = I0/(1 \u2212 Ps = r 2 \u221a 1 r 2 2. The output (transmitted) intensity is related to the intracavity intensity by Ps)2. In terms of the field reflection coefficients from the inside of the cavity, Iout = |t2|2I, while the initial intensity I0 is similarly related to the input intensity: I0 = |t 1|2Iin. Thus we can define the cavity intensity transmission coefficient Tcav by (7.44) (7.45) Tcav := Iout Iin = |t 1t2|2 I I0 = Tcav,max 1 + 2 2F \u03c0 sin2 \u03c0\u03bd FSR, 2A finesse of 1.9 \u00d7 106 was reported by G. Rempe, R. J. Thompson, H. J. Kimble, and R. Lalezari, \u2018\u2018Measurement of ultralow losses in an optical interferometer,\u2019\u2019 Optics Letters 17, 363 (1992). (cavity intensity transmission coefficient) (7.46) mirror 1mirror 2IE (+)E (+)t1oE (+)t2oE (+)r1oE (+)r2oE (+) 7.3 Cavity Transmission 123 where Tcav,max := 1t2|2 |t (1 \u2212 r1r2)2 (7.47) (maximum cavity transmission) is the maximum fraction of the input intensity that the cavity can transmit. So here\u2019s the statement that we started with: if the two mirrors are identical, then r1 = r2 =: r and t1 = t2 =: t. Then the maximum cavity transmission becomes Tcav,max = |tt|2 (1 \u2212 r2)2. (7.48) But one of the Stokes relations, Eq. (5.17), states that tt = 1 \u2212 r2, so for two identical mirrors", ", (7.49) Thus, when the two mirrors are identical, a resonant Fabry-Perot cavity transmits all the input light, independent of the reflectivity of the mirrors, at least in steady state after the circulating field builds up. (maximum cavity transmission, identical mirrors) Tcav,max = 1. 7.3.1 Reflected Intensity So what happened to the input light that supposedly reflects off of the input mirror? It must work out that it disappears via destructive interference, and we can now show this explicitly. Part of the intracavity wave transmits through the front mirror as well. From the way we defined I, the part transmitted through the front mirror is also reduced by the reflection coefficient of the second mirror: Iout (front) = |t1|2|r2|2I = |r2|2|1 \u2212 r2 |t 1|2 1|2 I. (7.50) Again, we used the Stokes relation tt = 1 \u2212 r2 to arrive at the last expression. In the case of identical mirrors, then, and on resonance, I = Imax = I0/(1 \u2212 r2)2, so Iout (front) = |r|2|1 \u2212 r2|2 |t|2 I, Iout (front) = r2 |t|2 I0 = r2Iin. (7.51) (7.52) This is exactly the same intensity as the part of the input beam that reflects off the first mirror. But in terms of the field, there is a \u03c0 phase shift of the reflected beam compared to the transmitted beam, due to the other Stokes relation r = \u2212r. Thus, these two beams destructively interfere in steady state. That\u2019s how all the power can transmit through the resonant cavity and allow for energy conservation in steady state. 7.3.2 Intracavity Buildup Notice that these effects require that on resonance, the cavity circulating intensity must be much larger than the input or output intensity in the regime of large finesse. This is most easily seen from Eq. (7.35), which we can rewrite as \u03c4p \u03c4rt F 2\u03c0 \u2248. (7.53) Thus, we can conveniently identify F /2\u03c0 (actually, the correct factor will be F /\u03c0, as we discuss below) as a \u2018\u2018buildup factor,\u2019\u2019 since light entering a resonant cavity makes this many round trips inside the cavity before leaving it. We can also see this directly from the intensities, since the output intensity for a symmetric cavity is related to the circulating intensity by Iout = |t|2I = |t|2Imax. From energy conservation, the reflected and 124 Chapter 7. Fabry\u2013Perot Cavities transmitted intensities of a mirror must add to the original intensity, so |t|2 = 1 \u2212 R, where R is the intensity \u221a reflection coefficient for the mirror, and 1 \u2212 R = 1 \u2212 Ps for a symmetric cavity. Hence, so that Imax = Iout \u221a 1 \u2212, Ps (7.54) Imax \u2248 F \u03c0 Iout = F \u03c0 Iin (7.55) (symmetric cavity buildup) in the limit of large finesse. Note that this differs from the photon-lifetime estimate by a factor of 2, because the loss for a resonant, symmetric cavity only occurs due to the output mirror (recall that the output through the input mirror destructively interferes with the reflected input light). Also we used that Iin = Iout for a symmetric cavity, so the cavity circulating intensity is a factor of F /\u03c0 larger than the input in the good-cavity limit. For an asymmetric cavity, we can combine Eqs. (7.28) and (7.45) to obtain Imax Iin = 1|2 |t \u221a (1 \u2212 Ps)2, (7.56) and again, in the limit where Eq. (7.34) applies, Imax Iin \u2248 1|2F 2 |t \u03c02, (7.57) (asymmetric cavity buildup) so the buildup factor is slightly more complicated in this case. For a symmetric cavity, again 1|2 = 1 \u2212 R = 1 \u2212 |t Ps \u2248 \u03c0 F, (7.58) so that we recover the symmetric-cavity expression (7.55). Note that the way we defined the circulating intensity I, this is only the intensity of the rightwardpropagating field in the cavity. But the field inside a linear cavity, as pictured at the beginning of this chapter, is a superposition of left- and right-going waves and thus a standing wave. That is, a field probe placed in the cavity would see more", " than I. The intensity I does, however, properly represent the circulating field in a ring cavity, where the intensity only circulates in one direction. For the linear cavity, the left-going wave is reduced in amplitude because of reflection from the righthand (output) mirror. In the good-cavity limit, this reduction is small, and the total field is an interference pattern of left- and right- going waves, each of intensity I\u2014a standing-wave pattern. The total intracavity intensity Icav for a linear cavity is thus Icav(z) = \u221a Ieikz \u2212 \u221a 2 Ie\u2212ikz (circulating intensity inside linear, planar cavity) = 2I[1 \u2212 cos(2kz)] = 4I sin2(kz). (7.59) Thus, the cavity intensity is a sinusoidal pattern with maximum intensity 4I \u2248 2F /\u03c0 and period \u03bb/2. Recall that for consistency with the boundaries that the field must vanish at z = 0 and d, so we chose the proper phase of the left-going field. 7.4 Optical Spectrum Analyzer In the good-cavity regime, a Fabry\u2013Perot cavity transmits light only at well-defined frequencies. Thus the Fabry\u2013Perot cavity is extremely useful in the laboratory as an optical spectrum analyzer\u2014a device that can measure the spectrum of a input light source. The width \u03b4\u03bdFWHM is the resolution of the analyzer, since it cannot distinguish frequencies within this narrow range. We now also see the reason for the terminology free spectral range for the spacing between cavity resonances, since this is the longest range in frequency that the cavity can measure without repetition or \u2018\u2018wrapping\u2019\u2019 of the spectrum. 7.4 Optical Spectrum Analyzer 125 Very often, a Fabry\u2013Perot spectrum analyzer can be scanned or tuned by changing the cavity length d. This can be accomplished by mounting one of the mirrors on a piezoelectric stack. If the length cavity is changed by an amount \u03b4d, the corresponding change in the resonance frequencies is given by \u03b4\u03bdq = qc 2(d + \u03b4d) \u2212 qc 2d \u2248 \u2212 qc\u03b4d 2d2 = \u2212\u03bdq \u03b4d d, (7.60) where we have expanded to lowest order in \u03b4d. Note that by a similar argument, the free spectral range changes by the much smaller amount [\u2212(FSR)\u03b4d/d], so the change in free spectral range is a second-order effect (a factor of \u223c 105 smaller for typical optical cavities). In general, it is very useful to note that small changes in any cavity parameter translates into linear changes in other parameters, \u03b4\u03bbq \u03bb = \u03b4d d \u2248 \u2212 \u03b4\u03bdq \u03bdq = \u2212 \u03b4kq kq, (7.61) (shifts in cavity parameters) at least to lowest order in the perturbation. These relations follow from the fact that these quantities are all proportional or inversely proportional. Most useful spectrum-analyzer-type problems can be solved with similar relations. This is what a Fabry\u2013Perot optical spectrum analyzer looks like in the laboratory. This is actually a confocal cavity, which we\u2019ll get to later, so the parameters work out a little differently than for the planar cavity. The cavity length is d = 5 cm giving FSR = 1.5 GHz. The mirrors are coated for 99.2% reflectivity, so that F = 200. Here is the same cavity disassembled. On the left you can see one of the mirrors mounted on a threaded barrel. The barrel is used to adjust the cavity length to match precisely the mirror curvatures. The other mirror is attached to the right-hand side of the main body on a tubular piezo stack, which gives a total displacement of \u223c 1 \u00b5m for a 0\u201315 V control signal. The mirror cover on the right contains a photodiode to monitor the transmitted light. 126 Chapter 7. Fabry\u2013Perot Cavities 7.5 Spherical-Mirror Cavities: Gaussian Modes In the more general case that we analyzed in ray optics, resonators are composed of spherical mirrors. Plane waves are no longer compatible with the mirrors. Instead, we must turn to Gaussian beams. As in the planar case, the waves must vanish at the mirror boundary. Because Gaussian beams have spherical wave fronts, they are a natural choice for cavity modes that match the spherical boundary conditions. Consider the spherical-mirror cavity shown here. We will try to \u2018\u2018fit\u2019\u2019 a Gaussian beam to it. As usual the Gaussian beam focus occurs at z = 0, and the mirrors of curvature radii R1 and R2 are placed at z", "1 and z2, respectively, and they are separated by a total distance d. Thus, the length condition is z2 \u2212 z1 = d. (7.62) (As drawn in the diagram, z2 > 0 but z1 < 0.) Matching the wavefront curvature to the curvature of mirror 1, R(z1) = R1 =\u21d2 R1 = z1 + z 2 0 z1. At the second mirror, R(z2) = \u2212R2 =\u21d2 \u2212R2 = z2 + z 2 0 z2. (7.63) (7.64) Note the difference in minus sign in the two cases, because the ray-optics convention for the mirror radius of curvature differs from the Gaussian-beam convention for the wave-front curvature R(z): for concave mirrors R\u00a1R\u2122zz\u00a1zo=o0z\u2122dfocus 7.5 Spherical-Mirror Cavities: Gaussian Modes 127 as in the diagram, for example, R1,2 < 0, but the matching wave-front curvature R(z) \u2277 0 for z \u2277 0. These two relations are also consistent even for convex-concave cavities, since the focus turns out to be outside the cavity. These three equations have the solution z1 = z2 = z 2 0 = \u2212d(R2 + d) R1 + R2 + 2d d(R1 + d) R1 + R2 + 2d \u2212d(R1 + d)(R2 + d)(R1 + R2 + d) (R1 + R2 + 2d)2. (7.65) We have determined the location of z = 0 relative to the mirrors along with z0, so we have completely determined the Gaussian beam geometry. 7.5.1 Physical Modes For a physical (bounded) mode, the Rayleigh length z0 must be real. If z0 is imaginary, w2(z) becomes negative at large z, and thus the Gaussian amplitude factor exp[\u2212r2/w2(z)] diverges for large r. Thus, the physical mode condition is z 2 0 \u2265 0. (7.66) Inserting the expression from Eqs. (7.65) and performing much algebra, we find that this condition becomes 0 \u2264 g1g2 \u2264 1, g1,2 = 1 + d R1,2, (7.67) which is just the same as the stability condition from ray optics. 7.5.2 Symmetric Cavities In the special case of a symmetric resonator, \u2212R1 = \u2212R2 =: R > 0 (i.e., both mirrors are concave), the solutions (7.65) simplify greatly. By symmetry we can guess that the focus occurs at the center of the cavity. Also, z1 = z2 = z0 = \u2212d 2 d 2 d 2 2R d \u2212 1. The corresponding beam waist at the focus is w 2 0 = \u03bbd 2\u03c0 2R d \u2212 1, and the beam waist at the mirrors is w 2 1,2 = w2(z1,2) = \u03bbd/\u03c0 2 \u2212. d R d R Note that for z0 \u2208 R, we must have R > d/2, which is the stability condition of a symmetric cavity. (7.68) (7.69) (7.70) 128 Chapter 7. Fabry\u2013Perot Cavities 7.5.3 Special Cavities Recall the special resonator casesfrom Section 2.7.3. Let\u2019s look at the symmetric ones again in the context of Gaussian-beam modes. Case 1. g1 = g2 = 1 =\u21d2 R1,2 = \u221e (Planar resonator) In this case, z0 \u2212\u2192 \u221e, w0 \u2212\u2192 \u221e, and w1,2 \u2212\u2192 \u221e. This is the plane-wave limit of the Gaussian beam, so we recover our analysis of the planar cavity. Case 2. g1,2 = \u22121 =\u21d2 R1,2 = \u2212d/2 (Spherical resonator/symmetrical concentric) Here z0 \u2212\u2192 0, w0 \u2212\u2192 0, and w1,2 \u2212\u2192 \u221e. This is the spherical-wave limit of the Gaussian-beam mode, which is what we expect in the limit of a spherical boundary. Case 3. g1,2 = 0 =\u21d2 R1,2 = \u2212d (Confocal resonator) For the confocal resonator, z0 = d/2, w0 = \u03bbd/2\u03c0, and w1,2 = 2w0. These relations imply that each mirror is a distance", " z0 from the focus. Alternately, the focal length of both mirrors is f = \u2212R/2 = d/2 = z0, so the cavity length is 2z0. Again, this is the reason that z0 is also called the confocal parameter. \u221a 7.5.4 Resonance Frequencies Recall from Eq. (6.28) that the total phase of a Gaussian beam is \u03c6(r, z) = kz \u2212 tan\u22121 z z0 + k r2 2R(z). (7.71) We already matched the shape of the phase fronts to the mirrors, so it suffices to consider the on-axis phase \u03c6(0, z). Thus, we will only consider the plane-wave and Gouy phases: \u03c6(0, z) = kz \u2212 tan\u22121. z z0 The round-trip phase is \u2206\u03c6rt = 2 [\u03c6(0, z2) \u2212 \u03c6(0, z1)] = 2k(z2 \u2212 z1) \u2212 tan\u22121 z2 z0 \u2212 tan\u22121. z1 z0 (7.72) (7.73) For constructive interference, this phase change must be \u2206\u03c6rt = 2\u03c0q for some integer q. Solving for k and using z2 \u2212 z1 = d, kq = q + \u03c0 d 1 d tan\u22121 z2 z0 \u2212 tan\u22121 z1 z0. (7.74) RRR 7.6 Spherical-Mirror Cavities: Hermite\u2013Gaussian Modes Rewriting this in terms of frequency using k = 2\u03c0\u03bd/c, \u03bdq = (FSR) q + 1 \u03c0 tan\u22121 z2 z0 \u2212 tan\u22121. z1 z0 129 (7.75) Using Eqs. (7.65), we can write down a mess of stuff that is difficult to simplify because of the signs of the factors under the radical signs. Through a rather long sequence of algebraic steps, the contribution of the Gouy phase reduces to a startlingly simple expression, \u03bdq = (FSR) q + cos\u22121 \u221a g1g2, 1 \u03c0 (7.76) where again g1,2 = 1 + d/R1,2. The Gouy-phase term (with the cos\u22121) is again a geometry-dependent term that is independent of q. Since q is very large at optical frequencies, this terms leads to an overall shift of the spectrum that is small compared to the optical frequency \u03bdq. However, the frequency shift is not necessarily small on the scale of the free spectral range. The shift vanishes for in the planar and spherical-concentric cavities, but for the confocal limit, the Gouy frequency shift is exactly half of the free spectral range. 7.5.5 Algebraic Digression How does one go from Eq. (7.75) to Eq. (7.76)? A (very) brief outline of the algebra is as follows. First, write the beam-parameter solutions of Eqs. (7.65) in terms of the stability parameters, with the result = z1 d z2 d z 2 0 d2 = = \u2212g2(1 \u2212 g1) g1 + g2 \u2212 2g1g2 g1(1 \u2212 g2) g1 + g2 \u2212 2g1g2 g1g2(1 \u2212 g1g2) (g1 + g2 \u2212 2g1g2)2. (7.77) In terms of these parameters, the signs are easier to handle since g1g2 > 0 for a stable cavity. Then it is relatively easy to show that tan\u22121 cos z2 z0 \u2212 tan\u22121 = z1 z0 after a bit of algebraic manipulation. z 2 0 + z1z2 1 ) (z 2 0 + z 2 (z 2 0 + z 2 2 ) \u221a = g1g2 (7.78) 7.6 Spherical-Mirror Cavities: Hermite\u2013Gaussian Modes The same idea applies to the higher-order TEMl,m modes of the cavities, since the wave fronts off all the higher-order modes are essentially the same as for the Gaussian. The main difference is that the Gouy term in the on-axis phase becomes \u03c6l,m(0, z) = kz \u2212 (1 + l + m) tan\u22121 z z0 (7.79) for the TEMl,m mode, as we can see by examination of Eq. (6.74). Repeating the above argument shows that only the Gouy term is modified, so that \u03bdl,m,q = (FSR) q + g1g", "2, (7.80) (1 + l + m) cos\u22121\u00b1 \u221a 1 \u03c0 g1g2 is positive if g1, g2 > 0, while the sign is negative if g1, g2 < 0.3 Now the Gouy term where the sign of represents a frequency shift that depends on the cavity geometry and the order of the Hermite\u2013Gaussian \u221a 3Anthony E. Siegman, Lasers (University Science Books, 1986) (ISBN: )0935702113 Section 19.3, p. 761. 130 Chapter 7. Fabry\u2013Perot Cavities mode. Thus, the cavity modes must be labeled by three indices as shown. The terminology is that modes corresponding to different q values are the longitudinal modes, which are similar in structure to the plane-wave resonances. The modes corresponding to different values of l and m are the different transverse modes, since they have different transverse intensity patterns. Note that there are a number of degeneracies among the modes due to the axial symmetry of the resonator. 7.6.1 Confocal Cavity Why was the laboratory cavity in the photographs (see p. 125) a confocal cavity? It would seem that a cavity with spherical mirrors would have a more complicated mode structure than the planar cavity and thus more difficult to use as an optical spectrum analyzer. As it turns out, though, the mode structure is simple for this case as well. For a symmetric confocal cavity, d = \u2212R1 = \u2212R2, so g1 = g2 = 0. Then g1g2 = 0, and since cos\u22121(0) = \u03c0/2, the mode frequencies become \u03bdl,m,q (confocal) = (FSR) q + (1 + l + m). 1 2 We can write this in terms of an effective index s as where s = 2q + 1 + l + m is a positive integer, and \u03bds (confocal) = (FSRconfocal)s, FSRconfocal := FSR 2 = c 4d = c0 4nd, (7.81) (7.82) (7.83) is an effective free-spectral range for the confocal cavity. Thus, we recover the spectrum for a planar cavity, but with a length of 2d. Recall the typical trajectory of the confocal cavity that makes 4 passes per round trip. That\u2019s why the cavity is effectively twice as long as for the planar cavity; the off-axis modes travel twice as far before repeating themselves. And that\u2019s why laboratory spectrum analyzers are often confocal cavities: the modes are degenerate, in the sense that they all fall into two series of frequency resonances, according to whether s is even or odd (for example, \u03bd0,0,q = \u03bd1,1,q\u22121, but there is no equivalent frequency of the form \u03bd1,0,q for any q). Notice, however, that this argument only works if several transverse modes are excited by the input source. If only the TEM0,0 mode is present, then the free spectral range is back to the old value. That\u2019s because longitudinal modes are still spaced by FSR, not FSRconfocal. The effective confocal free-spectral range only works if modes operate in the ring-trajectory manner shown. Thus, in the laboratory, the input beam to a confocal resonator is usually slightly misaligned to ensure that the \u2018\u2018in-between\u2019\u2019 modes are populated as well, so the relevant free spectral range is FSRconfocal. A confocal cavity operated in this manner is said to be operated in \u2018\u2018ring mode.\u2019\u2019 Notice also that the finesse effectively changes, since the line width is unaffected but the free spectral range is different. Thus, F confocal = FSRconfocal \u03b4\u03bdFWHM = F 2 (7.84) for a confocal cavity in ring mode. 7.7 Exercises 7.7 Exercises Problem 7.1 131 Consider a cavity constructed from two planar mirrors separated by 1 cm. (a) What is the free spectral range if the cavity is completely empty? (b) What is the free spectral range if the cavity is otherwise empty but a 1 mm glass plate (n = 1.5) is placed in the cavity? Ignore any reflections from the surfaces of the glass plate. Problem 7.2 Consider a ring resonator consisting of three planar mirrors, arranged at the vertices of an equilateral triangle of side length d as shown in the diagram. (a) What is the free spectral range of the cavity? (b) Suppose that the top mirror is moved vertically as shown in the diagram. If the cavity is resonant with", " light of wavelength \u03bb, by how much should the mirror move to go through one complete interference fringe? Problem 7.3 One method for boosting the intensity of continuous-wave (cw) light is to use a cavity dumper, which is a ring cavity with an optical switch as shown. The cavity dumper operates as follows: the input light is turned on, and the circulating light builds up in the cavity to a large intensity. Then the optical switch is activated, deflecting the circulating intensity out of the cavity. (a) What are the round-trip time and free spectral range? Assume the beam path is a rectangle of dimensions 12 \u00d7 20 cm. Also model the optical switch as a block of glass (n = 1.5) of length 2 cm. (b) What are the survival probability, finesse, and photon lifetime of the cavity? Assume an intensity reflection coefficient of 99.8% for all the mirrors and an intensity loss of 0.2% per pass due to the switch. dopticalswitch 132 Chapter 7. Fabry\u2013Perot Cavities (c) Suppose the input power is 1 W and that the circulating power is allowed to build up to its steady-state value. What are the duration and power of the output pulse when the switch is activated? Assume the switch couples all the light out of the cavity. (d) What is the maximum repetition rate of the cavity dumper if a 50 W output pulse is desired? Problem 7.4 Consider a symmetric confocal optical spectrum analyzer operating at a nominal wavelength of 780 nm. The nominal cavity length is 5 cm, which can be changed precisely (on a sub-\u03bb scale) using a piezo stack attached to one of the mirrors. The cavity transmission is monitored as the cavity length d is scanned. (a) Here is a measured transmission spectrum, and this is the same set of data but with the horizontal axis zoomed by a factor of 20: Compute the finesse and resolution \u03b4\u03bdFWHM of the cavity, keeping in mind that this cavity is confocal. (Note: in the first scan, the distances between the peaks are slightly different due to the nonlinearity of the piezo stack displacement with voltage; it is best to average the two distances to get an accurate value.) Dd (arb. units)cavity transmission (arb. units)Dd (arb. units, 20\u00a5 zoom)cavity transmission (arb. units) 7.7 Exercises 133 (b) The cavity length is changed by a relatively large amount (compared to \u03bb, say by a threaded barrel adjuster, such that the cavity is slightly longer than required to be confocal. The measured spectrum now looks more complicated: The same data zoomed in by a horizontal factor of 10 shows a structure of many peaks for each longitudinal mode: Note that the horizontal units here do not correspond to the plots in (a). From the data here, estimate the new cavity length by using the cavity line splittings to calculate the new stability parameter. Hint 1: The cavity no longer satisfies the confocal condition, so don\u2019t treat is as a confocal cavity. Note that the spectrum almost repeats after some distance and then really repeats after that. Hint 2: You should calculate the new cavity length by understanding the above spectrum. However, as a sanity check, this new spectrum was taken when the threaded barrel that sets the overall cavity length was rotated by about 1 1 4 turns. The barrel was machined with a nominal pitch of 40 threads/inch. Problem 7.5 One limitation to the finesse of a Fabry\u2013Perot cavity is the surface roughness of the mirrors. As a simple estimate of this effect, consider a planar resonator of length d0 with perfectly reflective mirrors. Dd (arb. units)cavity transmission (arb. units)Dd (arb. units, 10\u00a5 zoom)cavity transmission (arb. units) 134 Chapter 7. Fabry\u2013Perot Cavities (a) Recall the normalized form of the normal (Gaussian) distribution with zero mean, f (x; \u03c3) = \u221a 1 2\u03c0\u03c3 exp \u2212. x2 2\u03c32 (7.85) Here, \u03c3 is the standard deviation or root-mean-square (rms) deviation. Compute the full width at half maximum \u03b4xFWHM of this distribution in terms of \u03c3. (b) Model the effective cavity length as d = d0+\u03b4d1+\u03b4d2, where the \u03b4d1,2 are the (small) deviations due to mirror roughness. Assume that these are uncorrelated random variables over the mirror surfaces. Also assume that they are normally distributed with \u03b4d1,2 = 0 and \u03b4d 2 1,2 = \u03b52 (i.e., the rms roughness is", " \u03b5), where the angle brackets denote an average over the mirror surfaces. Show that the effective cavity finesse is F = \u221a 8 \u03bb log 2 \u03b5 (7.86) due to the surface roughness. (c) A typical rms surface roughness for a laser-grade mirror substrate is 3 \u00c5, while a typical roughness for a superpolished mirror substrate is 0.5 \u00c5. Assuming that the reflective coating of the mirror has the same roughness as the substrate, compute the roughness-limited finesse for each case at \u03bb = 780 nm. Problem 7.6 Show that if \u03bdq = (FSR) q + 1 \u03c0 tan\u22121 z2 z0 \u2212 tan\u22121. z1 z0 (7.87) is the resonance frequency for a Gaussian mode of a stable, spherical-mirror cavity, then the following expression is equivalent: \u03bdq = (FSR) q + g1g2. (7.88) cos\u22121 \u221a 1 \u03c0 Problem 7.7 Consider a planar cavity of nominal length d = 1 cm. The cavity is resonant with light of \u03bb = 1 \u00b5m. (a) Compute the ratio Q/F for this cavity at this wavelength. (b) Suppose that you change the wavelength by \u2206\u03bb before you find the next cavity resonance. What is \u2206\u03bb? (c) Suppose that the mirrors have very high reflectivity, so that the cavity is \u2018\u2018good.\u2019\u2019 State whether the following quantities would be larger, smaller, or the same as for a \u2018\u2018bad\u2019\u2019 cavity of the same geometry: 1. survival probability 2. Q factor 3. finesse 4. round-trip time 5. free spectral range 6. resonance width (\u03b4\u03bdFWHM) 7. photon lifetime Problem 7.8 A laser of nominal wavelength \u03bb = 1 \u00b5m emits at two slightly different wavelengths, where the difference is \u2206\u03bb \u03bb. The transmission of the laser light through a Fabry\u2013Perot spectrum analyzer of nominal length d = 5 cm is observed as the analyzer\u2019s length is scanned. 7.7 Exercises 135 (a) The transmission of the analyzer is observed to be periodic in the analyzer length with period \u2206d. What is \u2206d? (b) Suppose that the analyzer is set so that one wavelength component of the laser is maximally transmitted. You observe that when the analyzer\u2019s length is changed by \u2206d/10, the other laser wavelength is maximally transmitted. What is \u2206\u03bb? (c) The laser itself consists of a Fabry\u2013Perot cavity filled with gas (n \u2248 1). Suppose that the two output wavelengths correspond to two adjacent resonances of the laser cavity. What is the length of the laser cavity, assuming it is longer than the analyzer? Problem 7.9 A laser of nominal wavelength \u03bb = 200\u03c0 nm is on resonance with a Fabry\u2013Perot spectrum analyzer of nominal length d = 10 cm. Suppose that the two mirrors are identical, with intensity reflection coefficients R = 99%. (a) By how much must you increase d so that the laser excites the next cavity resonance? (b) By how much must the laser wavelength increase to go from one resonance to the next? (c) By how much must the laser frequency change to go from one resonance to the next? (d) Suppose again that the laser is tuned to the analyzer\u2019s resonance, where the analyzer is initially evacuated. Now suppose the analyzer is filled with a gas of refractive index n, and over the course of the filling, the laser ends up again on resonance, but two resonances away from the starting point. What is n? (e) Going back to the initial setup, what fraction of the intensity transmits through the analyzer (on resonance)? By how much must the analyzer length change so that the transmitted intensity drops to half this value? Problem 7.10 We showed [Eq. (7.29)] that the mean number of round trips that a photon survives in a resonator is Derive an expression for the variance \u2206n2 := (n \u2212n)2 = n2 \u2212n2 Hint: try working with n(n + 1).. n =. (7.89) 1 1 \u2212 Ps Problem 7.11 All parts of this problem refer to the resonator below, consisting of three flat mirrors, and a beam path that forms an equilateral triangle, where each side has length d. Each leg of the path has a thin lens of focal length f, centered between the two mirrors; each leg of the resonator is identical to the others. Assume the optics are surrounded by vacuum. d/2d/2f 136 Chapter 7. Fabry\u2013Perot", " Cavities (a) For what range of f is the resonator stable? (b) For any resonant mode of wavelength \u03bb, by how much must d change to go from one resonance to the next? (c) Consider the TEM0,0 mode of the resonator. Where do the points of minimum beam size (e.g., diameter) occur? Explain briefly. (d) Compute the beam-waist parameter w0 at the locations in (c). (e) Suppose each mirror has a reflectance of R, and ignore any absorption losses in the mirrors, and also ignore any losses due to the lenses. Derive an expression for the steady-state circulating power inside the cavity, assuming the cavity is on resonance, and given an input power Pin. (Write your answer in terms of R and Pin only.) (f) Describe quantitatively (in the sense of defining a function you can plot; a plot is also okay if it is labeled well) the time dependence of the output-beam power at the lower-right mirror, given that an input beam of power Pin at the lower-left mirror is turned on suddenly, such that the initial turn-on transient hits the input mirror at t = 0. Assume the cavity is on resonance, and do not make any assumptions about whether the cavity is \u2018\u2018good\u2019\u2019 or \u2018\u2018bad.\u2019\u2019 (g) Suppose the resonator is flooded with a fluid of index nbath. Give a new expression for the stability condition for the resonator, assuming the lenses have a refractive index of nlens > nbath. Chapter 8 Polarization 8.1 Vector Plane Waves Recall that a plane wave in complex notation has the form where the vector amplitude has only transverse components: E(+)(r) = E (+) 0 eikz, (+) 0 = E(+) 0x \u02c6x + E(+) 0y \u02c6y. E The corresponding physical field, with the explicit time dependence, is E(r, t) = E(+)(r, t) + c.c. = E (+) 0 ei(kz\u2212\u03c9t) + c.c. (8.1) (8.2) (8.3) The polarization of the wave is related to the direction of E so we need to work out some details before defining particular polarization states.. However, this direction is time dependent, (+) 0 8.2 Polarization Ellipse We will now examine the \u2018\u2018trajectory\u2019\u2019 of the electric-field vector. Writing out the explicit phases of the components, 0x = |E(+) E(+) 0y = |E(+) E(+) 0x |ei\u03c6x 0y |ei\u03c6y, we can write the real field as where E(r, t) = Ex(r, t)\u02c6x + Ey(r, t)\u02c6y, Ex = E0x cos(kz \u2212 \u03c9t + \u03c6x) Ey = E0y cos(kz \u2212 \u03c9t + \u03c6y). (8.4) (8.5) (8.6) Recall that for consistency among the real and complex notations, E0x = 2|E(+) 0y |. These equations are the parametric equations for a general ellipse. With a bit of algebra it is possible to expand the cosines using the sum-angle formula and then eliminate the explicit time dependence to obtain the equation for an ellipse, 0x | and E0y = 2|E(+) 2 Ex E0x + Ey E0y 2 \u2212 2 ExEy E0xE0y cos \u03c6 = sin2 \u03c6, (8.7) (polarization ellipse) 138 Chapter 8. Polarization where \u03c6 := \u03c6x \u2212 \u03c6y is the relative phase between the components. This form isn\u2019t completely transparent, but it turns out that this is the relation for a ellipse rotated by an angle \u03b1, where tan 2\u03b1 = 2E0xE0y cos \u03c6 0x \u2212 E 2 0y E 2. (8.8) (rotation angle of polarization ellipse) In the special case \u03b1 = 0, we recover the usual form for the unrotated ellipse, 2 Ex E0x + Ey E0y 2 = 1, (8.9) of width 2E0x and height 2E0y. 8.2.1 Simple Cases We will now consider a few of the simplest polarizations, corresponding to special values of \u03c6, Ex, and Ey. 1. In-phase components (\u03c6 = 0). In this case, Eq. (8.7) reduces to Ex E0x ExEy E", "0xE0y Ey E0y \u2212 2 2 2 + which is equivalent to Ex E0x = Ey E0y. = 0, (8.10) (8.11) Thus, the electric-field vector remains on a straight line in the Ex-Ey plane (say, at z = 0) at an angle \u03b1 from the Ex-axis, where tan \u03b1 = E0y E0x (8.12) Thus, this is the case of linear polarization. In the full three-dimensional space, the electric-field vector traces out a sinusoidal waveform, always staying within a single plane: E\u2248aE\u00a5 8.2 Polarization Ellipse 139 In general, the polarization is linear if \u03c6 = 0 or \u03c0, or if either of E0x or E0y are zero (in which case \u03c6 may as well be zero). 2. Equal-magnitude components in quadrature (\u03c6 = \u03c0/2, E0x = E0y =: E0). In this case, Eq. (8.7) for the polarization ellipse reduces to 8.13) the equation for a circle of radius E0. We can work out the direction of the time-dependent motion of the vector as follows: At z = 0, the x-component has the form cos(\u03c0/2 \u2212 \u03c9t), while the x-component has the form cos(\u2212\u03c9t); at t = 0, (x, y) = (0, 1), which changes to (1, 0) at t = T /4, where T = 2\u03c0/\u03c9 is the optical period. Thus, we see that the electric-field vector traces out a circle in the Ex-Ey plane in the clockwise direction, viewed such that the propagation direction is out of the page (i.e., along the +z-direction): By convention, this polarization is called right-circular polarization (RCP). In three-dimensional space, the electric-field vector traces out a right-handed corkscrew, hence the name (curl your fingers in the direction of the electric-field trajectory, and your thumb points along the direction of propagation. Note that in this picture, the wave propagates to the right. 3. Equal-magnitude components in opposite quadrature (\u03c6 = \u2212\u03c0/2, E0x = E0y =: E0). This case is the same as for RCP light, but the electric-field vector travels counterclockwise: E\u2248zoo=ook^^E\u00a5E\u2248zoo=ook^^E\u00a5 140 Chapter 8. Polarization Thus, this is left-circular polarization (LCP). In three-dimensional space, the electric-field vector traces out a left-handed corkscrew. One important point to note from this analysis: the only difference between linear, RCP, and LCP light is the relative phase between the two component, at least in the proper coordinate system. In the general case, as we noted before, the electric-field vector traces out some ellipse in the Ex-Ey plane, representing an elliptical polarization somewhere between linear and circular. 8.3 Polarization States: Jones Vectors To make the notation more explicit, we can write out the electric-field vector E component vector: (+) 0 as an explicit two- (+) 0 = E \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 E(+) 0x E(+) 0y \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb. (8.14) (Jones vector) A Jones vector is simply this explicit two-component form. Thus, a Jones vector models the polarization state of an electromagnetic wave. Often to represent different polarizations the Jones vectors are written in normalized form, such that E(+) and so that the overall field is written as a separate factor\u2014for polarization purposes, we only care about the magnitudes and phases of the two components relative to each other. E(+) (8.15) = 1, + 0x 0y 2 2 Thus, the some of the basic Jones vectors are as follows: 1. linear polarization (along the x-direction): y x 1 0 8.3 Polarization States: Jones Vectors 141 2. linear polarization (angle \u03b1): cos \u03b1 sin \u03b1 3. right-circular polarization (RCP): 1 \u2212i 1\u221a 2 4. left-circular polarization (LCP): 1 i 1\u221a 2 y y y \u03b1 x x x Notice that the 90\u25e6 phase shift for the circular polarizations are represented by the complex phase difference between the two components. Also, all of these vectors can have an extra overall phase, which doesn\u2019t amount to anything\u2014it\u2019s like a translation in time but it doesn\ufffd", "\ufffdt affect the geometry of the way. Not unless, that is, there is interference, in which case the phase is relative to some other phase. 8.3.1 Vector Properties Being vectors, Jones vectors inherit all the usual properties we associate with general vectors. We will just reiterate two of the more useful ones here. Two Jones vectors are orthogonal if their inner product vanishes: \u2217 (+) 01 E \u00b7 E (+) 02 = E (\u2212) 01 \u00b7 E (+) 02 = 0. (8.16) (orthogonality condition) Note that since these are complex-valued vectors, the inner product involves complex conjugation of the first vector before the usual dot product. Thus, by this definition, linear-x and linear-y polarizations are orthogonal, and LCP and RCP are also orthogonal. Jones vectors also have a magnitude given by E(+) 0x 2 + 2 E(+) 0y. (8.17) As we mentioned above, a Jones vector is normalized if its magnitude is unity. 142 Chapter 8. Polarization 8.4 Polarization Devices: Jones Matrices A polarization device is some optical element or system that transforms the polarization state of an optical field. Since we are representing the state of the field by a Jones vector, the most general linear transformation of the polarization state is a 2 \u00d7 2 matrix. Thus, we can model any linear polarization device by a 2 \u00d7 2 matrix. Such a matrix is called a Jones matrix. Mathematically, (+) 02 = TE (+) 01, E (8.18) (Jones-matrix transformation) where T is the Jones matrix modeling the system, E (+) 01 is the input field, and E (+) 02 is the output field. 8.4.1 Linear Polarizer A linear polarizer (usually referred to simply as a \u2018\u2018polarizer\u2019\u2019) is a device that selectively annihilates the electric field along one direction. The corresponding Jones matrix for a polarizer that blocks the y-component of the field is which induces the transformation \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 E(+) 0x E(+) 0y \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb \u2212\u2192 (+) 0x 0 (8.19) (linear polarizer along x) (8.20) Thus, independent of the initial state of the field, the final state is linear-x. Thus, this is a polarizer with its transmission axis along the x-direction. Obviously, the Jones matrix for a polarizer that transmits along the y-axis is T = 0 0 0 1. (8.21) (linear polarizer along y) 8.4.2 Wave Retarder A wave retarder is a device that adds a phase shift to one component of the field. The Jones matrix for a wave retarder that retards the phase of the y-direction by \u2206\u03c6 is T = 1 0. 0 ei\u2206\u03c6 (8.22) (wave retarder along y) This is a retardation of phase in time, since our phase convention here is exp(\u2212i\u03c9t) for the time dependence. For this retarder, the x-axis is the \u2018\u2018fast axis,\u2019\u2019 and the y-axis is the \u2018\u2018slow axis.\u2019\u2019 Let\u2019s look at two special cases more closely. \u2206\u03c6 = \u03c0/2 \u2212\u2192 \u2018\u2018quarter-wave plate\u2019\u2019 (\u03bb/4-plate) \u2206\u03c6 = \u03c0 \u2212\u2192 \u2018\u2018half-wave plate\u2019\u2019 (\u03bb/2-plate) (8.23) (special wave retarders) These two cases are especially important in the optics laboratory. The quarter-wave plate induces the transformation 1 1 \u2212\u2192 1 i, (8.24) 8.4 Polarization Devices: Jones Matrices 143 and thus converts linear polarization (with the right orientation) to LCP (or RCP). This is one common laboratory method for producing circular polarization. The half-wave plate induces the transformation 1 1 \u2212\u2192 1 \u22121, (8.25) and thus rotates linearly polarized light by 90\u25e6, at least if it has the right initial orientation (of 45\u25e6 from the slow axis). More generally, the half-wave plate flips the polarization about the fast (x) axis. Now we can visualize how a wave retarder changes the polarization state of light. The wave propagation is to the right (along +z); the x-direction is vertical, and the y-direction is horizontal. The initial polarization state is linear at 45\u25e6, with Jones vector 1 \u221a 2 1 1. (8.26) The wave is shown as the dark wave, and the x- and y- components are", " shown in a lighter color. As the y-component is retarded, the polarization changes to LCP (\u03bb/4 retardation) to linear with a 90\u25e6 rotation (\u03bb/2 retardation), to RCP (3\u03bb/4 retardation), and finally back to the original state (\u03bb retardation). 0 retardation (\u2206\u03c6 = 0) \u03bb/8 retardation (\u2206\u03c6 = \u03c0/4) \u03bb/4 retardation (\u2206\u03c6 = \u03c0/2) 3\u03bb/8 retardation (\u2206\u03c6 = 3\u03c0/4) \u03bb/2 retardation (\u2206\u03c6 = \u03c0) 5\u03bb/8 retardation (\u2206\u03c6 = 5\u03c0/4) 3\u03bb/4 retardation (\u2206\u03c6 = 3\u03c0/2) 7\u03bb/8 retardation (\u2206\u03c6 = 7\u03c0/4) \u03bb retardation (\u2206\u03c6 = 2\u03c0) 144 Chapter 8. Polarization The same evolution of the polarization is shown here as an animation in the electronic version of this document. 8.4.3 Polarization Rotator The Jones matrix for a polarization rotator with rotation angle \u03b8 is T(\u03b8) = cos \u03b8 \u2212 sin \u03b8 cos \u03b8 sin \u03b8. (8.27) (polarization rotator) cos \u03b81 sin \u03b81 \u2212\u2192 cos \u03b82 sin \u03b82 (8.28) This matrix induces the transformation provided \u03b82 = \u03b81 + \u03b8. 8.4.4 Cascaded Systems Of course, the power of this matrix formalism is that this works for cascaded systems via matrix multiplication, just like in ray optics (although there are now new matrices). Just as in ray optics, the order of matrix multiplication is critical: the first polarization device that the field encounters is the rightmost matrix in the product. 8.5 Coordinate Transformations Consider a coordinate change from (x, y) coordinates to a new set of coordinates (x, y) that are rotated by an angle \u03b8 with respect to the original coordinates. 8.5 Coordinate Transformations 145 (+) If E 0 given by is the electric-field vector in the original coordinates, then the vector E (+) 0 in the new coordinates is (+) 0 (+) = R(\u03b8)E 0. E (8.29) (effect of coordinate-system rotation) Here, R(\u03b8) is the rotation matrix, defined by R(\u03b8) := cos \u03b8 \u2212 sin \u03b8. sin \u03b8 cos \u03b8 (8.30) (rotation matrix) Notice that the rotation of the coordinates by an angle \u03b8 is the same as a rotation of the vector by an angle (+) vector is closer to the x-axis (\u2212\u03b8), or R(\u03b8) = T(\u2212\u03b8). This is evident from the drawing, since the E 0 vector is to the x-axis. Mathematically, you can see this by comparing the rotation matrix, than the E Eq. (8.30), to the Jones matrix for the polarization rotator, Eq. (8.28)\u2014that is, the Jones matrix for a polarization rotator of angle \u03b8 is R(\u2212\u03b8). (+) 0 What about rotations of polarization devices? That is, given a Jones matrix T, what is the Jones matrix T\u03b8 for the same device rotated by an angle \u03b8? The intuitive way to handle this is to consider an (+) equivalent picture where the field goes through the unrotated device. To compensate, the input field E 0 should be rotated by an angle (\u2212\u03b8), so that the input to the rotated device is R(\u03b8)E. After passing through the device, we should restore things to the original coordinates by undoing the rotation, rotating the output vector by an angle \u03b8. Thus, we have argued that (+) 0 T\u03b8E (+) 0 = R(\u2212\u03b8)TR(\u03b8)E (+) 0. (8.31) This should be true for any input polarization, so T\u03b8 = R(\u2212\u03b8)TR(\u03b8) (8.32) (Jones matrix for rotated optical device) is the transformation law for rotating the polarization device by \u03b8. We can also derive this transformation more formally. The inner product of two vectors \u2217 (+) 02 E \u00b7 E (+) 01 (8.33) is a scalar, and thus is independent of coordinate rotations. Thus, we can consider the scalar quantity (+) 02 E \u2217 \u00b7 TE (+) 01. Rotating both the device and the vectors by an angle \u03b8, we have R(\u2212\u03b8)E \u2217 \u00b7 (+) 02 T\u03b8R(\u2212\u03b8)E. (+) 01 (8.34) (", "8.35) These combined rotations are equivalent to a coordinate rotation by angle (\u2212\u03b8), so the two scalars are the same. Using the fact that the dot product is a matrix product with a transposition, a\u00b7b = aTb, the transpose property (AB)T = BTAT, and the fact that [R(\u03b8)]T = [R(\u03b8)]\u22121 = R(\u2212\u03b8), we can rewrite the expression (8.35) as \u2217 (+) 02 E \u00b7 R(\u03b8)T\u03b8R(\u2212\u03b8)E (+) 01. (8.36) xx\u00a2qyE\u00ba\u00f9ooo\u00f4\u00fb, E\u00ba\u00f9ooo\u00f4\u00e6\u00fby\u00a2 146 Chapter 8. Polarization For consistency of this expression with (8.34), we must have which we can rewrite as which is the same as Eq. (8.32). T = R(\u03b8)T\u03b8R(\u2212\u03b8), T\u03b8 = R(\u2212\u03b8)TR(\u03b8), (8.37) (8.38) As with the polarization vector, the effect of rotating the coordinates is the opposite: T = R(\u03b8)TR(\u2212\u03b8). (Jones matrix after coordinate rotation by \u03b8) (8.39) Keeping the signs straight is important, but it can be a bit tricky. 8.6 Normal Modes Polarization devices are represented by matrices, which obviously have eigenvectors and eigenvalues. Recall that the eigenvectors (eigenpolarizations) are defined by the relation TE (+) 0 = \u03bbE (+) 0, (8.40) (normal-mode condition) where \u03bb (the eigenvalue) is a constant. There are at most two such vectors for a 2 \u00d7 2 matrix that satisfy this (+) 0,\u03bb2, for the eigenvectors corresponding to \u03bb1 and \u03bb2, respectively. equation, and we can call them E These eigenvectors are called the normal modes of a system. (+) 0,\u03bb1 and E The normal modes are easy to propagate through the system, since the effect of the Jones matrix is the same as multiplication by a constant factor. A vector can be decomposed into a superposition of the eigenvectors, and then easily propagated through the device: TE (+) 0 = \u03bb1\u03b11E (+) 0,\u03bb1 + \u03bb2\u03b12E (+) 0,\u03bb2. (8.41) (propagation via normal modes) Note that in general, the eigenvalues could have a zero value (e.g., for a polarizer). Polarization devices can be lossy, so there is no general expression for the determinant of a Jones matrix, in contrast to the ray matrix case where the Hamiltonian structure ensures a nonvanishing determinant. 8.7 Polarization Materials 8.7.1 Birefringence Now that we have covered the mathematical formalism of polarization, we will discuss some of the materials used to make polarization devices. One important class of optical materials are birefringent materials, where the material is anisotropic: the wave sees a different index of refraction for different polarizations. Materials could be birefringent, for example, due to anisotropic crystal structure or mechanical stress. Another example of material birefringence is the electro-optic effect, where certain crystals become birefringent when placed in a dc, uniform electric field. Examples of electro-optic crystals are lithium tantalate (LTA), potassium dihydrogen phosphate (KDP), potassium dideuterium phosphate (KD*P), and ammonium dihydrogen phosphate (ADP). Consider a wave incident on a birefringent plate as shown. 8.7 Polarization Materials 147 Suppose that ny > nx. Then in the notation above, the y-direction is the \u2018\u2018slow axis,\u2019\u2019 and the x-direction is the \u2018\u2018fast axis.\u2019\u2019 Then the optical path length for x-polarization is and the optical path length for the y-polarization is kxd = 2\u03c0nxd \u03bb0, kyd = 2\u03c0nyd \u03bb0. (8.42) (8.43) Thus, the relative phase shift between the two components is 2\u03c0(ny \u2212 nx)d/\u03bb0. So for example, we can make a \u03bb/4-plate by choosing (ny \u2212 nx)d = \u03bb0 4. This gives a \u03c0/2 phase retarder along the y-direction, with Jones matrix 1 0 0 ei\u03c0/2 = 1 0, 0 i as we saw above. (8.44) (\u03bb/4-", "plate) (8.45) On the practical side, making a wave plate based on the criterion (8.44) can lead to impractically thin\u2014and thus fragile\u2014wave plates (which would only be a few optical wavelengths thick). More common is a multiple-order wave plate, which adds a large multiple of 2\u03c0 in phase retardation. Thus, the criterion becomes something like (ny \u2212 nx)d = m + 1 4 \u03bb0, (8.46) (multiple-order \u03bb/4-plate) for a \u03bb/4-plate, where m is an integer that makes the resulting optic \u223cmm thick. The wave plate represented by (8.44) is a zero-order wave plate, since it corresponds to m = 0. Multiple-order wave plates are relatively economical, but they suffer from a problem: wave plates are designed to achieve a certain retardation at a certain wavelength, but give different retardations at other wavelengths (since d \u221d \u03bb0). The refractive indices also vary slightly with wavelength, which can further compound the problem. While such errors may be tolerable for a zero-order wave plate, any such errors are magnified by a factor of order m for a multiple-order wave plate. Zero-order wave plates are similarly much less sensitive to temperature variations, which cause the thickness of the plate to change. In practice, zero-order wave plates are constructed by combining two multiple-order wave plates of order with retardations of (m + 1/4)\u03bb0 and m\u03bb0 (in the case of a \u03bb/4-plate), with fast axes oriented at 90\u25e6 with respect to each other. Thus, all the excess retardation cancels, giving the equivalent of a thin \u03bb/4-plate. 8.7.1.1 Multiple Refraction Birefringent materials also have more complicated behavior for rays at non-normal incidence. The basic effect is multiple refraction, where different polarizations are refracted differently. Consider a wave with mixed polarization incident from free space into a birefringent medium. n\u2248zdn\u00a5E\u2248E\u00a5 148 Chapter 8. Polarization Let\u2019s define a bit of notation that we will return to when introducing the Fresnel relations. The incident ray lies in the x-y plane, which we will call the plane of incidence. The polarization that lies inside the plane of incidence is P-polarization (for \u2018\u2018parallel\u2019\u2019 to the plane of incidence), whereas the orthogonal polarization, normal to the plane of incidence, is S-polarization (for \u2018\u2018senkrecht,\u2019\u2019 the German word for \u2018\u2018perpendicular\u2019\u2019). The S-polarized component gives rise to the ordinary wave, which sees only the index nz in the material. Thus, Snell\u2019s law gives sin \u03b8i = nz sin \u03b8o. (8.47) (ordinary-wave refraction) The P-polarizated component gives rise to the extraordinary wave, which is more complicated, because it sees an angle-dependent index: n(\u03b8e) = ny cos \u03b8e + nx sin \u03b8e. (refractive index for extraordinary wave) (8.48) Thus, Snell\u2019s law cannot be solved analytically for \u03b8e: sin \u03b8i = n(\u03b8e) sin \u03b8e. (8.49) (extraordinary-wave refraction) First, this relation leads to a different refraction angle for the extraordinary wave, compared to the ordinary wave. In some cases, this is a useful effect. For example, the separation of polarizations is useful in constructing polarizers (polarizing beam splitters). But multiple refraction is even a bit more subtle. It turns out that for the extraordinary waves, the geometrical rays are not in general orthogonal to the wave fronts (hence the name \u2018\u2018extraordinary\u2019\u2019). For this reason, it is even possible to get separation of the polarizations for normal incidence. Multiple refraction due to a birefringent calcite crystal is shown here in this photograph. n\u00a1o=o1(air)(birefringent)extraordinarymixedpolarizationwaveordinarywaveiqoqeqn\u2248,oon\u00a5,oon\u03a9xyz 8.7 Polarization Materials 149 8.7.2 Optical Activity Whereas a birefringent material has different refractive indices for different linear optical polarizations, an optically active material has different indices for the two circular polarizations. In mathematical notation, we can denote the two indices by n+ and n\u2212 for RCP and LCP light, respectively. Optical activity is characteristic of media with helical", " molecules. The helicity of the medium breaks the symmetry between the two polarizations. Examples of such media include quartz and sugar in the solid phase, or sugar solution and turpentine in the liquid phase. Note that given a material with a particular helicity, the same material with the opposite helicity exists in principle as well\u2014it would just be the mirror image of the orginal material. For example, quartz can have n+ > n\u2212 or n+ < n\u2212, depending on the crystal structure. Glucose and fructose are two different sugars that give rise to opposite senses of optical activity. One other important manifestation of optical activity is the Faraday effect. Some crystals, such as YIG (yttrium-indium-garnet), exhibit optical activity when they are placed in a uniform magnetic field. Optical activity causes optical rotation of input linear polarization. To see this, note that linearly polarized light (say, linear-x) is a linear combination of RCP and LCP: E (+) 0,in = i 1 2 After passing through an optically active medium of thickness d, the Jones vector becomes 1 i 1 \u2212i (+) 0,out = ei\u03c6+ + ei\u03c6\u2212, E 1 2 1 2 where \u03c6\u00b1 = k\u00b1d = 2\u03c0n\u00b1d/\u03bb0. We can rewrite this in the form cos \u03c6 sin \u03c6 0,out = ei\u03c60 (+) E, (8.50) (8.51) (8.52) where \u03c6 = (\u03c6\u2212 \u2212 \u03c6+)/2 and \u03c60 = (\u03c6\u2212 + \u03c6+)/2. Thus, linear polarization is rotated by the angle \u03c6. It is more useful to specify a rotation per unit length, d\u03c6 dz = \u03c0(n\u2212 \u2212 n+) \u03bb0. (8.53) (polarization rotation per unit length) 150 Chapter 8. Polarization For the Faraday effect, it is conventional to specify the rotation per unit length as where the constant V is the Verdet constant. The rotation is typically proportional to the magnetic field in regimes of interest for laboratory polarization rotators. d\u03c6 dz = V B, (8.54) (Faraday effect) 8.8 Exercises 8.8 Exercises Problem 8.1 151 Write down the Jones matrix for an ideal polarizer that transmits x-polarized light. A realistic polarizer along the x-direction blocks some fraction \u03b1 (of the electric field) of the x polarization, and transmits some fraction \u03b2 of the y polarization. (Both \u03b1 and \u03b2 should be small for a decent polarizer.) Write down the Jones matrix for this realistic polarizer. Problem 8.2 (a) Consider a linear polarizer and a wave linearly polarized at an angle \u03b8 with respect to the polarizer\u2019s transmission axis. Show that the intensity of the wave is reduced by cos2 \u03b8 after passing through the polarizer (this is called the Law of Malus). (b) Consider a system of N cascaded polarizers. The polarizers have their transmission axes at angles \u03c0/2N, 2\u03c0/2N, 3\u03c0/2N,..., \u03c0/2 from the x-axis, in the order that an input wave sees them. That is, the last polarizer is oriented along the y-direction. Suppose that an input wave is polarized in the x-direction. Compute the intensity transmission coefficient for the system. Show that the transmission coefficient approaches unity as N \u2212\u2192 \u221e. This is a simple realization of the quantum Zeno effect, where each polarizer acts as a \u2018\u2018measurement\u2019\u2019 of the polarization state\u2014the polarization is \u2018\u2018dragged\u2019\u2019 by the measurements as long as they are sufficiently frequent. Problem 8.3 One important optical polarization system is the optical isolator, which prevents laser light from being reflected back into the laser (which could cause instability or even damage). An optical isolator typically consists of a polarizer (say, along x), followed by a 45\u25e6 polarization rotator, followed by a polarizer at 45\u25e6. Note that if T is the Jones matrix for an optic, the Jones matrix for the optic rotated by an angle \u03b8 is T\u03b8 = R(\u2212\u03b8) T R(\u03b8), where R(\u03b8) is the rotation matrix. (a) Derive the Jones matrices for forward and backward propagation through the isolator. Assume ideal polarizers and note that the rotator produces the same rotation independent of the light\u2019s direction. Show that x-polarized light passes forward through the system unattenuated, but that any light traveling backwards through the isolator will be completely extinguished. (Realistic isolators attenuate the reverse beam by about", " 40 dB.) (b) A cheaper isolator is a polarizer (say, along x) followed by a quarter-wave retarder oriented at 45\u25e6. Show that if a laser passes through this isolator, the isolator will block any light returning from a direct mirror reflection, but will not block arbitrary return polarizations. Problem 8.4 A beam of light can be described by its position and direction as a geometrical ray as well as by its polarization Jones vector. Suppose the position/direction vector and polarization vector transform respectively according to y2 \u03b82 = M y1 \u03b81 = A B C D y1 \u03b81 ; Ax,2 Ay,2 = T Ax,1 Ay,1 = A B C D Ax,1 Ay,1. (8.55) Write down an expression for the matrix that models both aspects of the optical system; that is, the matrix that gives the transformation \uf8ee \uf8ef \uf8ef \uf8f0 y2 \u03b82 Ax,2 Ay,2 \uf8f9 \uf8fa \uf8fa \uf8fb = \uf8ee \uf8ef \uf8ef \uf8f0 \uf8f9 \uf8fa \uf8fa \uf8fb. y1 \u03b81 Ax,1 Ay,1 (8.56) 152 Problem 8.5 Chapter 8. Polarization One method of Q-switching a laser is to include a polarizer and a switchable retarder (Pockels cell) oriented at 45\u25e6 in the cavity. If we ignore losses other than due to the polarization rotation, the cavity finesse is unchanged. To \u2018\u2018Q-spoil\u2019\u2019 the cavity, we can switch on some phase retardation to introduce extra loss. Calculate the cavity finesse for the ring cavity of Problem 1 without the gain medium, for retardations of \u2206\u03c6 = 0, \u2206\u03c6 = \u03c0/4, and \u2206\u03c6 = \u03c0/2. (Obviously you should not make any low-loss approximations here.) Problem 8.6 Suppose you have a laser beam parallel to the optical table, and the light polarization is perpendicular to the table. You need light that is polarized parallel to the table. How can you do this with only two mirrors? (And without turning the laser on its side!) Problem 8.7 Consider a planar cavity. Suppose that an ideal (linear) polarizer and a polarization rotator (which rotates the polarization by \u2206\u03b8) are placed inside the cavity. Assuming the mirrors are perfect reflectors, what is the finesse of the cavity? Chapter 9 Fresnel Relations 9.1 Optical Waves at a Dielectric Interface One situation where the vector nature of light is crucial is in the reflection and refraction at a dielectric interface. In Chapter 2, we considered reflection and refraction in ray optics as a consequence of Fermat\u2019s principle. But that only fixed the reflection and refraction angles. We need wave optics to establish the amplitude of the reflected and refracted components. Consider the setup in the diagram. We are assuming a priori that there is an incident, reflected, and transmitted wave. We will assume that the waves are monochromatic, so that the three fields have the form E E E (+) i = E (+) r = E (+) t = E (+) (+) 0i eiki\u00b7r 0r eikr\u00b7r 0t eikt\u00b7r, (+) (9.1) where the subscripts \u2018\u2018i,\u2019\u2019 \u2018\u2018r,\u2019\u2019 and \u2018\u2018t\u2019\u2019 refer to the incident, reflected, and transmitted waves, respectively. Note that we are making some pretty specific assumptions about the field, since we are anticipating the result. We can make whatever assumptions we want, so long as the solution turns out to be self-consistent\u2014 the self-consistency justifies the assumptions. The thing to worry about is the generality of the solution, but we will return to this later. These three waves are not independent: the electromagnetic boundary conditions at the dielectric interface impose constraints among the three waves. The electromagnetic boundary conditions are 1. The tangential components (to the dielectric interface) of E and H are continuous across the boundary. n1n2x = 0zyxqiqrqtE0i(+)H0i(+)E0r(+)H0r(+)E0t(+)H0t(+) 154 Chapter 9. Fresnel Relations 2. The normal components of D and B are continuous across the boundary. These boundary conditions follow directly from the Maxwell equations, most straightforwardly in integral form. It turns out that the first boundary condition is sufficient to completely constrain the fields. Let\u2019s now apply the boundary conditions. First, we\u2019ll use the fact that E", "\u00b7 \u02c6z (the tangential component) is continuous across the boundary: (E (+) 0i \u00b7 \u02c6z)eikiy sin \u03b8i + (E (+) 0r \u00b7 \u02c6z)eikry sin \u03b8r = (E (+) 0t \u00b7 \u02c6z)eikty sin \u03b8t. (9.2) The left-hand side is the sum of the incident and reflected waves, while the right-hand side is the transmitted wave. We have defined our coordinate system such that the interface is located at x = 0. For this to be true for all y, the exponents must all be identically equal: the phases of the waves must agree everywhere along the boundary. For the phases of the incident and reflected waves to be equal, we require ki sin \u03b8i = kr sin \u03b8r. (9.3) But ki = kr, since both waves see the same refractive index, and k = nk0, where k0 is the vacuum wave number. Thus, we recover the law of reflection: \u03b8i = \u03b8r. (9.4) (Law of Reflection) Equating the incident and transmitted phases, This simplifies to give Snell\u2019s Law. ki sin \u03b8i = kt sin \u03b8t. n1 sin \u03b8i = n2 sin \u03b8t. (9.5) (9.6) (Snell\u2019s Law) Thus, we have recovered everything that we derived from ray optics. The geometrical properties of reflection and refraction are purely a consequence of the wave geometry. Now to consider the amplitudes. The continuity of the electric field becomes E (+) 0i + E (+) 0r = E (+) 0t, because the electric fields lie entirely along the z-direction. We can rewrite this relation as 0i + E(+) E(+) E(+) 0t 0r = 1, (9.7) (9.8) which will be more convenient for later manipulations. The continuity of H is only slightly more complicated. The tangential component is the component in the y-direction, so the continuity condition becomes H (+) 0i \u00b7 \u02c6y + H (+) 0r \u00b7 \u02c6y = H (+) 0t \u00b7 \u02c6y, (9.9) where we have already used the equality of the phases. Using the fact that the magnetic and electric fields are related by H0 = E0/\u03b7 = nE0/\u03b70, where \u03b7 is the wave impedance and \u03b70 is the vacuum wave impedance, this boundary condition becomes \u2212 n1E(+) 0i \u03b70 cos \u03b8i + n1E(+) 0r \u03b70 cos \u03b8r = \u2212 n2E(+) 0t \u03b70 cos \u03b8t. Again, we can rewrite this as 0i \u2212 E(+) E(+) E(+) 0t 0r = n2 cos \u03b8t n1 cos \u03b8i. (9.10) (9.11) 9.1 Optical Waves at a Dielectric Interface Adding Eqs. (9.8) and (9.11), 2 E(+) 0i E(+) 0t = 1 + n2 cos \u03b8t n1 cos \u03b8i = n1 cos \u03b8i + n2 cos \u03b8t n1 cos \u03b8i. 155 (9.12) Now we can define the field transmission coefficient tS as the ratio of the transmitted to incident field: We can rewrite Eq. (9.8) as tS := E(+) 0t E(+) 0i = 2n1 cos \u03b8i n1 cos \u03b8i + n2 cos \u03b8t. E(+) 0r E(+) 0i = E(+) 0t E(+) 0i \u2212 1, (9.13) (9.14) so if we define the field reflection coefficient rS as the ratio of the reflected to incident field amplitude, rS := E(+) 0r E(+) 0i = n1 cos \u03b8i \u2212 n2 cos \u03b8t n1 cos \u03b8i + n2 cos \u03b8t. (9.15) Note that we have assumed a particular input polarization for the light. The notation is that the polarization is transverse electric (TE), since the electric field is tangential to the interface. A more common notation in optics is S-polarization (\u2018\u2018S\u2019\u2019 is for \u2018\u2018senkrecht,\u2019\u2019 which means \u2018\u2018perpendicular\u2019\u2019 in German), because the electric-field vector is perpendicular to the plane of incidence (the x-y plane in our coordinates here). Thus, Eqs.", " (9.15) and (9.13), rewritten here, rS = tS = n1 cos \u03b8i \u2212 n2 cos \u03b8t n1 cos \u03b8i + n2 cos \u03b8t 2n1 cos \u03b8i n1 cos \u03b8i + n2 cos \u03b8t. (9.16) (Fresnel relations, S-polarization) are the Fresnel relations for S-polarization. We can repeat this derivation for the other polarization, parallel to the plane of incidence, called transverse magnetic (TM) or P-polarization. The setup is shown here in the following diagram. For the vector convention shown, the Fresnel relations turn out to be rP = tP = n1 cos \u03b8t \u2212 n2 cos \u03b8i n1 cos \u03b8t + n2 cos \u03b8i 2n1 cos \u03b8i n1 cos \u03b8t + n2 cos \u03b8i. (9.17) (Fresnel relations, P-polarization) n1n2x = 0zyxqiqrqtE0i(+)H0i(+)H0r(+)E0r(+)E0t(+)H0t(+) 156 Chapter 9. Fresnel Relations These are the Fresnel relations for P-polarization. Here is a graphical representation of incident, reflected, and transmitted fields at an air-glass interface that matches our setup here (air on the left, glass on the right). 9.1.1 Phase Changes and the Brewster Angle The phase factors of all the waves must be equal, so the relative phases are in some sense fixed. The reflection and transmission coefficients are also real, so there is no phase shift of the components, except possibly for an overall minus sign. Thus, there is the possibility of a \u03c0 phase shift of the wave after it encounters the boundary. Transmission. By examination of Eqs. (9.16) and (9.17) it is evident that both tS > 0 and tP > 0. Thus, for either polarization, the transmitted wave is in phase with the incident wave. Reflection: S-polarization. The reflection coefficient is more subtle. For S-polarization, the sign of the reflection coefficient depends on the sign of (n1 cos \u03b8i \u2212 n2 cos \u03b8t). If n1 > n2, then from Snell\u2019s law, we know that \u03b8t > \u03b8i, and so cos \u03b8t < cos \u03b8i. Thus, rS > 0, and the reflected wave is in phase with the incident wave. One the other hand, If n1 < n2, then \u03b8t < \u03b8i, and so cos \u03b8t > cos \u03b8i. In this case, rS < 0, and the reflected wave has a \u03c0 phase shift. Thus, the reflected wave only picks up a \u03c0 phase shift when the light is incident on a more dense medium (e.g., from air onto glass). Reflection: P-polarization. This case is even a bit more complex. From Eqs. (9.17), the sign of rP depends on the sign of n1 cos \u03b8t \u2212n2 cos \u03b8i. Suppose that this quantity is positive. Since n1/n2 = sin \u03b8t/ sin \u03b8i, the condition is equivalent to n1 cos \u03b8t \u2212 n2 cos \u03b8i > 0 sin \u03b8t cos \u03b8t \u2212 sin \u03b8i cos \u03b8i > 0. This, in turn, is equivalent to the condition sin(\u03b8t \u2212 \u03b8i) cos(\u03b8t + \u03b8i) > 0, (9.18) (9.19) (9.20) which follows from the general trigonometric identity cos(A + B) sin(A \u2212 B) = sin A cos A \u2212 sin B cos B. Eq. (9.20) is true if the two factors are either both positive or both negative. Let\u2019s summarize these two cases as follows: 1. n2 < n1: both factors are positive when \u03b8t > \u03b8i and \u03b8t + \u03b8i < \u03c0/2 2. n2 > n1: both factors are negative when \u03b8t < \u03b8i and \u03b8t + \u03b8i > \u03c0/2 9.2 Reflectance and Transmittance 157 There is a \u03c0 phase change on reflection if neither set of conditions holds. The borderline case occurs when \u03b8i + \u03b8t = \u03c0/2, which means that the reflected ray is perpendicular to the refracted ray. Also, this is the case where the numerator in the expression for rP in Eq. (9.17) van", "ishes, so the borderline case corresponds to rP = 0, so that the reflected wave vanishes. From Snell\u2019s law, the this borderline case corresponds to an incidence angle \u03b8i,B determined by n1 sin \u03b8i,B = n2 sin \u03b8t = n2 sin \u2212 \u03b8i,B \u03c0 2 = n2 cos \u03b8i,B. (9.21) Thus, the incidence angle for the borderline case is \u03b8i,B = tan\u22121 n2 n1, (9.22) (Brewster\u2019s angle) which is called Brewster\u2019s angle. Note that the reflection coefficient only vanishes for P-polarization. There is no analogous behavior for S-polarization. Thus, Brewster\u2019s angle is also called the polarizing angle because light incident at this angle is polarized on reflection from a dielectric interface\u2014only S-polarized light reflects, while part of the S-polarization and all of the P-polarization transmit. 9.2 Reflectance and Transmittance Now we must characterize the energy transport due to the fields at the dielectric interface. The energy fluxes at the surface due to the the incident, reflected, and transmitted waves are given in terms of the respective Poynting vectors, Si \u00b7 \u02c6x, Sr \u00b7 \u02c6x, and St \u00b7 \u02c6x. Note that since these are plane waves, the incident and reflected waves overlap. Thus, it seems that we should consider interference cross terms when calculating the Poynting vectors. However, we can treat them separately by realizing that plane waves are an idealization: physical beams have finite width, so we can go far away from the interface until the incident and reflected beams no longer overlap. To consider the energy balance, we note that (see Problem 4.4) the Poynting vector can be written in terms of the electromagnetic energy density as S = cw\u02c6k. (9.23) Energy balance requires that the energy flux towards the interface is the same as the energy flux away: Si \u00b7 \u02c6x = St \u00b7 \u02c6x + Sr \u00b7 (\u2212\u02c6x). Writing this out in terms of the energy densities, c0 n2 wi cos \u03b8i = c0 n1 wt cos \u03b8t \u2212 c0 n1 wr cos \u03b8i. We can now define the reflectance as the ratio of reflected to incident energy fluxes: R := \u2212Sr \u00b7 \u02c6x Si \u00b7 \u02c6x = c0 n1 c0 n1 wr cos \u03b8i wi cos \u03b8i = wr wi. (9.24) (9.25) (9.26) (reflectance) Similarly, the transmittance as the ratio of transmitted to incident energy fluxes: T = St \u00b7 \u02c6x Si \u00b7 \u02c6x = n1wt cos \u03b8t n2wi cos \u03b8i. (9.27) (transmittance) Now we can use the explicit forms for the energy densities in terms of the fields: wi = 2n 2 1 0|E(+) i |2; wr = 2n 2 1 0|E(+) r |2; wt = 2n 2 2 0|E(+) t |2. (9.28) 158 Chapter 9. Fresnel Relations Thus, the reflectance is the relative intensity, but the transmittance is not, in general, the obvious expression in terms of the field transmission coefficient (T = |t|2). Rather, R = |r|2, (9.29) (reflectance in terms of reflection coefficient) T = n2 n1 cos \u03b8t cos \u03b8i |t|2. In terms of the reflectance and transmittance, Eq. (9.25) becomes (transmittance in terms of transmission coefficient) (9.30) R + T = 1. (9.31) (conservation of energy) Note that the expression (9.30) may seem counterintuitively complicated. In practice, we usually care about reflection at a dielectric interface that is at one end of, say, a slab of glass (such as a glass window used as a beam splitter). In this case, the incident and transmitted energies are measured outside the dielectric medium. In particular, they are measured in the same surrounding medium (say, air), in which case the transmittance reduces to |t|2, the relative transmitted intensity. Thus, for reflection from the surface of a glass window, the reflected and transmitted intensities (detected outside the window) must add up to the incident intensity. 9.3 Internal Reflection Recall that Snell\u2019s law reads As in the geometrical case, if n1 > n2 there is", " a critical incident angle given by n1 sin \u03b8i = n2 sin \u03b8t. (9.32) sin \u03b8i,c = n2 n1, (9.33) (critical angle) which comes from the condition \u03b8t = \u03c0/2, the maximum possible refraction angle. Beyond this incident angle, the transmitted angle becomes a complex number. The simple interpretation is that there is no transmitted plane wave, but we will explore this phenomenon in more detail. The transmitted wave has the form for either polarization. Supposing that \u03b8i > \u03b8i,c, we have E (+) t = E (+) 0t exp {ikt [(cos \u03b8t) x + (sin \u03b8t) y]} sin \u03b8t = n1 n2 sin \u03b8i > 1, so that cos \u03b8t = \u00b1 1 \u2212 sin2 \u03b8t = \u00b1i 2 n1 n2 sin2 \u03b8i \u2212 1, (9.34) (9.35) (9.36) where the final form shows explicitly that the cosine is an imaginary number. Thus, the transmitted wave becomes E (+) t = E (+) 0t exp ikt \uf8f0 \uf8ed\u00b1i \uf8f1 \uf8f2 \uf8f3 \uf8ee \uf8eb 2 n1 n2 \uf8f6 \uf8f9 \uf8fc \uf8fd \uf8fe sin2 \u03b8i \u2212 1 \uf8f8 x + (sin \u03b8t) y \uf8fb. (9.37) 9.3 Internal Reflection 159 For a bounded solution, we must choose the +i branch of the square root in Eq. (9.36). Then we see that the transmitted wave, \uf8ee E (+) t = E (+) 0t exp \uf8f0\u2212kt 2 n1 n2 \uf8f9 sin2 \u03b8i \u2212 1 x \uf8fb exp ikt n1 n2 (sin \u03b8i) y, (9.38) (evanescent wave) which is an exponentially damped harmonic wave: the left exponential factor represents exponential damping along the x-direction, while the right exponential factor has the form of a traveling plane wave along the y-direction. This damped field is called the evanescent field. From the damping factor in Eq. (9.38), the wave decays as where the skin depth \u03b4, given by E (+) t \u223c e\u2212x/\u03b4, \uf8ee \u03b4 = \uf8f0kt 2 n1 n2 \uf8f9 \u22121 sin2 \u03b8i \u2212 1 \uf8fb, (9.39) (9.40) (skin depth) is the characteristic distance over which the evanescent wave damps away. Since the (second) propagatingwave factor in Eq. (9.38) only involves y, the wave fronts of the evanescent wave propagate along the interface, as shown. This is one way to see that the evanescent wave carries no energy away from the interface. The other, which we will see shortly, is that R = 1. Thus, this situation (\u03b8 > \u03b8i,c) is called total internal reflection. Here is a visualization of internal reflection at a glass-air interface. The incoming wave is incident at an angle just slightly past the critical angle, so that the skin depth is relatively large. You can clearly see the propagation of the wave fronts parallel to the interface. n\u00a1iqexponentialwave frontstransversedampingikrkn\u2122oo<oon\u00a1xyz 160 Chapter 9. Fresnel Relations Actually, what I said above about the evanescent wave isn\u2019t quite true. It can carry away some energy, provided that the evanescent wave is interrupted. For example, for internal reflection at a glass-air interface, the wave is damped in the air. But if another glass interface is placed very close to the first, so that the air is present only in a thin (\u223c\u03bb) gap, the damped wave will begin to propagate again in the second glass region. This phenomenon is called frustrated total internal reflection and is the analog of quantum tunneling. This effect can be used, for example, to make a variable-ratio beam splitter by adjusting the size of a small air gap between two prisms1. 9.3.1 Phase Shifts The reflection coefficients from Eqs. (9.16) and (9.17) are rS = rP = n1 cos \u03b8i \u2212 n2 cos \u03b8t n1 cos \u03b8i + n2 cos \u03b8t n1 cos \u03b8t \u2212 n2 cos \u03b8i n1 cos \u03b8t + n2 cos \u03b8i. From Eq. (9.36), we have (upon choosing the positive branch of the square root", ") cos \u03b8t = i n1 n2 2 sin2 \u03b8i \u2212 1, so that we can eliminate \u03b8t in the reflection coefficients: rS = cos \u03b8i \u2212 i cos \u03b8i + i sin2 \u03b8i \u2212 (n2/n1)2 sin2 \u03b8i \u2212 (n2/n1)2 rP = \u2212 (n2/n1)2 cos \u03b8i \u2212 i sin2 \u03b8i \u2212 (n2/n1)2 (n2/n1)2 cos \u03b8i + i sin2 \u03b8i \u2212 (n2/n1)2. Notice that both of these reflection coefficients are of the form \u00b1 \u03b1 \u2212 i\u03b2 \u03b1 + i\u03b2, (9.41) (9.42) (9.43) (9.44) 1D. Bertani, M. Cetica, and R. Polloni, \u2018\u2018A simple variable-ratio beam splitter for holography,\u2019\u2019 Journal of Physics E: Scientific Instruments 16, 602 (1983). 9.4 Air-Glass Interface: Sample Numbers 161 so that the reflectance for both polarizations is unity: RS = RP = 1. (9.45) (total internal reflection) However, there is a phase shift on reflection, since the reflection coefficients themselves are not unity in general. The phase angle in each case is exp(\u22122\u03c6), (up to a minus sign) where tan \u03c6 = \u03b2/\u03b1, so we can write rS = exp \uf8ee \uf8f0\u2212i2 tan\u22121 \uf8eb \uf8ed sin2 \u03b8i \u2212 (n2/n1)2 cos \u03b8i \uf8f6 \uf8f9 \uf8f8 \uf8fb \uf8ee \uf8f0i\u03c0 \u2212 i2 tan\u22121 \uf8eb \uf8ed rP = exp sin2 \u03b8i \u2212 (n2/n1)2 (n2/n1)2 cos \u03b8i \uf8f6 \uf8f9 \uf8f8 \uf8fb (9.46) for the explicit phase shifts of the two polarizations. Apart from an overall minus sign (\u03c0 phase shift), there is a refractive index ratio in the expression for rP that is absent in the expression for rS. Hence, if both polarization components are present, there is a relative phase shift between them. (phase shifts for internal reflection) This phase shift can be very useful. In Section 8.7.1, we discussed how a birefringent material can make a wave retarder (wave plate). But the phase shifts due to internal reflection obviously also act as wave retarders. One optical element, the Fresnel rhomb, uses two internal reflections to produce a \u03c0/2 (\u03bb/4) phase shift. For glass with n = 1.52, the correct incidence angle turns out to be \u03b8 = 55.5\u25e6. The rhomb has the sometimes inconvenient property of translating the beam, but unlike typical wave plates, the retardation of a rhomb is relatively insensitive to the optical wavelength used. 9.4 Air-Glass Interface: Sample Numbers To get a feel for the theory presented here, let\u2019s work out specific numbers for an interface between air (n1 = 1) and glass (n2 = 1.52). At normal incidence, cos \u03b8i = cos \u03b8t = 1, so and r = rS = rP = n1 \u2212 n2 n1 + n2 = \u22120.21 R = |r|2 = 4.3%. (9.47) (9.48) Thus, a glass window attenuates light due to reflection near normal incidence by about twice this amount, around 8%. At 45\u25e6 incidence, the reflectances for the two polarizations are quite different. First, the refraction angle is \u03b8t = sin\u22121 n1 n2 sin \u03b8i = 27.7\u25e6. (9.49) q 162 Chapter 9. Fresnel Relations Thus, the reflectances are RS = |rS|2 = RP = |rP|2 = n1 cos \u03b8i \u2212 n2 cos \u03b8t n1 cos \u03b8i + n2 cos \u03b8t n1 cos \u03b8t \u2212 n2 cos \u03b8i n1 cos \u03b8t + n2 cos \u03b8i 2 2 = 9.7% = 0.94%, (9.50) or about 10% and 1%, respectively. (These numbers are useful to remember in the laboratory for making simple 10% beam splitters.) The reflectance for P-polarization is much smaller because this incidence angle is fairly close to the Brewster angle of \u03b8i,B = tan\u22121 n2 n1 = 56.7\u25e6. (9.51) The reflectances", " for the air-glass interface for both polarizations are shown here as a function of the incident angle. On the other hand, for a glass-air interface, the reflectance curves look the same except that they are horizontally compressed. The reflectances are unity beyond the critical incident angle \u03b8i,c = sin\u22121 nair nglass = 41.1\u25e6, (9.52) qi (degrees)09056.745re\ufb02ectance100.043S polarizationP polarizationqi (degrees)09033.345re\ufb02ectance100.043S polarizationP polarization 9.5 Reflection at a Dielectric-Conductor Interface 163 as is evident from the plot. 9.5 Reflection at a Dielectric-Conductor Interface One useful generalization of reflection at a dielectric interface is to treat the case where one of the materials is a conductor. This covers the important case of metallic reflectors, which are some of the most common in everyday experience as well as the laboratory. We will not deal explicitly with the case of reflection at the interface of two conductors, as this is more subtle. But to deal with reflection from a single conductor, we will need to treat wave propagation in a conductive medium. 9.5.1 Propagation in a Conducting Medium In Chapter 4, we considered only Maxwell\u2019s equations in dielectric media without sources. With field sources, the Maxwell equations become \u2207 \u00b7 B \u2202t \u2202D \u2202t \u2207 \u00d7 H = + J, (9.53) (Maxwell equations coupled to sources) where \u03c1 is the charge density and J is the current density. For an ohmic material, the current density is proportional to the local electric field, J = \u03c3E, (9.54) (Ohm\u2019s Law) where \u03c3 is the conductivity. We can think of the conductivity as a real constant, but in reality at optical frequencies it is best modeled by a complex number, due to the lag of the response of the conduction electrons to the electric field (see Section 17.2.2 for a model of the conductivity). Charge and current are related by the continuity equation \u2202\u03c1 \u2202t = \u2212\u2207 \u00b7 J, (9.55) (continuity equation) which is essentially equivalent to the statement that charge is conserved. If the current density at some point is increasing, then there must be a convergence of current density to transport charge to that point. We can use the relation D = E and Eq. (9.54) to write \u2202\u03c1 \u2202t = \u2212\u03c3(\u2207 \u00b7 E) = \u2212 \u03c3 \u03c1. (9.56) This means that \u03c1 \u2212\u2192 0 exponentially with a time constant /\u03c3. For a good conductor (large \u03c3), the damping time is very short, so to good approximation we can set \u03c1 = 0. On longer time scales of optical interest, the charge density relaxes to nothing, and this is why we can justify ignoring it. This is a simple example of an adiabatic approximation. For our purposes, this means that the first Maxwell equation in Eq. (9.53) becomes \u2207 \u00b7 D = 0, (9.57) (first Maxwell equation for conductor) so that the only source is the current density, which we can write in terms of E. 164 Chapter 9. Fresnel Relations We can now repeat the derivation of the wave equation from Chapter 4, with the result \u22072E \u2212 \u00b50 \u22022E \u2202t2 \u2212 \u00b50\u03c3 \u2202E \u2202t = 0, (9.58) (wave equation for conductor) with the same result for H. Note that we are ignoring magnetic properties of the conductor by assuming \u00b5 = \u00b50. We can again try the plane-wave ansatz Substitution shows that the plane wave is a solution provided that E(+)(r, t) = E (+) 0 ei(kz\u2212\u03c9t). k2 = \u00b50\u03c92 + i\u00b50\u03c3\u03c9 = \u03c92\u00b50 + i. \u03c3 \u03c9 (9.59) (9.60) Note that the frequency dependence in this equation is deceptively simple, as in general \u03c3 and are themselves functions of frequency. If we define a complex permittivity \u02dc = + i \u03c3 \u03c9, (9.61) (complex permittivity) then the dispersion relation (9.60) takes the same form as for an ordinary dielectric, k2 = \u03c92\u00b50\u02dc, (9.62) (dispersion relation for conductor) with the replacement \u2212\u2192 \u02dc. Similarly, since the refractive index for a dielectric is the square root of the dielectric constant, n2 = 0, (", "9.63) we can define the complex refractive index \u02dcn as the square root of the complex dielectric constant: \u02dcn2 = \u02dc 0. (9.64) (complex refractive index) We can also write the complex refractive index itself in terms of real and imaginary parts as \u02dcn =: n(1 + i\u03ba), (9.65) (complex refractive index) where \u03ba is the attenuation index or extinction coefficient. This is the central point of electromagnetic wave propagation in conductors: everything is the same as for an ordinary dielectric, so long as you use a complex refractive index. Let\u2019s compute explicitly the real and imaginary parts of the refractive index. First, computing \u02dcn2 from both Eqs. (9.64) and (9.65), \u02dcn2 = n2(1 \u2212 \u03ba2) + i2n2\u03ba = 0 + i \u03c3 \u03c90. Matching the real and imaginary parts, we have the two equations n2(1 \u2212 \u03ba2) = 0, 2n2\u03ba = \u03c3 \u03c90. These equations have the solutions (9.66) (9.67) n2 = (Re[\u02dcn])2 = n2\u03ba2 = (Im[\u02dcn]) 20 20 + 1 \u2212 1. (real/imaginary parts of complex refractive index) (9.68) 9.5 Reflection at a Dielectric-Conductor Interface Similarly, we can compute the wave vector in terms of the refractive index as where k = \u03c9\u00b50\u02dc = \u03c9 \u221a \u00b500 \u02dcn = k+ + ik\u2212, k 2 \u00b1 = \u03c92\u00b50 2 1 + 2 \u03c3 \u03c9 \u00b1 1. 165 (9.69) Thus, we can rewrite the plane-wave solution as (real/imaginary parts of wave number) (9.70) E(+)(r) = E (+) 0 eikz = E (+) 0 e\u2212k\u2212zeik+z. (9.71) (damped plane wave) As we saw for internal reflection, the plane wave breaks up into propagating and damping factors\u2014the imaginary part of the refractive index leads to damping of the wave, while the real part represents the propagating aspect of the wave. We can again define a skin depth \u03b4 for the damped wave by e\u2212k\u2212z = e\u2212z/\u03b4, (9.72) so that \u03b4 = 1 k\u2212 = 1 \u00b500n\u03ba \u221a \u03c9 = \u03bb0 2\u03c0n\u03ba = \u03bb0 2\u03c0Im[\u02dcn]. (9.73) (skin depth in conductor) As before, this is the length scale over which the wave damps exponentially away. Typically, for a good conductor, n\u03ba is large, so that \u03b4 \u03bb0. This is called the skin effect: for good conductors, an electromagnetic wave penetrates the surface by only a small fraction of the vacuum wavelength. 9.5.2 Inductive Heating Since the transmitted wave gets damped, something must be taking up that dissipated energy. Of course, this goes into the metal as heat. While being undesirable from the point of view of making mirrors for highpower lasers, this effect can be useful in the machine shop or foundry for heat-treating or melting metals. An induction heater drives a high-voltage, radio-frequency signal through a coil that surrounds the conductive object to be heated. The electromagnetic wave generated by the coil couples into the conductive object, and generates heat as it is absorbed. Of course, the equivalent picture is that the radiation induces currents in the object, causing Joule heating\u2014from our analysis above, we saw that it is precisely these induced currents that damp the electromagnetic wave after it penetrates the surface. The following photographic sequence shows a quarter being heated and then melted by an inductive heater. Roughly speaking, you can think of the coil as being a solenoid, which ideally produces a uniform, oscillating magnetic field along the solenoid axis. The corresponding electric field is roughly a cylindrical wave collapsing onto the quarter, with the wave polarization being in the plane of the quarter. The wavelength for the 200 kHz radiation is quite large, around 1.5 km. However, the copper inside the quarter is an extremely good conductor; assuming a conductivity of 60 \u00d7 106 (m \u00b7 \u2126)\u22121, the skin depth turns out to be only 0.15 mm. Thus, due to the skin effect, the radiation is absorbed in a thin shell on the outside of the quarter. You can see this in the photo series, because the outside of the quarter is visibly much hotter than the middle, since 166 Chapter 9. Fresnel Relations the middle is only being heated by thermal", " conduction. Here is a movie of the heating process, where you can see the heat flowing in from the outside of the quarter to the inside. 9.5 Reflection at a Dielectric-Conductor Interface 167 9.5.3 Fresnel Relations It is worth reiterating: with the conductor present, everything is the same as in the dielectric case, except that the refractive index is now complex. In particular, the Fresnel reflection coefficients still hold with the complex refractive index: rS = rP = n1 cos \u03b8i \u2212 \u02dcn2 cos \u03b8t n1 cos \u03b8i + \u02dcn2 cos \u03b8t n1 cos \u03b8t \u2212 \u02dcn2 cos \u03b8i n1 cos \u03b8t + \u02dcn2 cos \u03b8i. (9.74) (Fresnel relations with complex index) Here, n1 is the refractive index of the dielectric (from which the wave is incident), and \u02dcn2 is the complex refractive index of the conductor. Note that this is even a bit more complicated than before, though, since from Snell\u2019s law, \u03b8t is a complex number too! We won\u2019t go into much detail regarding the wave in the medium beyond what we\u2019ve already done, since we are usually only interested in this wave if it is then transmitted back into free space where it can be detected. This situation is better treated with the matrix formalism for multilayer films that we will get to shortly. Let\u2019s plug in some numbers for the simple example case of a polished silver mirror at normal incidence. The reflectance is R = n1 \u2212 \u02dcn2 n1 + \u02dcn2 2. (9.75) For silver at 780 nm,2 \u02dcn = 0.27 + 4.47i, which give R = 95.0%. Thus silver is a good reflector, but some of the incident energy is lost in the metal. This is because the wave penetrates a short distance into the metal. Putting in numbers, we find that the skin depth is \u03b4 = \u03bb0 2\u03c0n\u03ba = \u03bb0 28.1 = 27.8 nm (9.76) at 780 nm. Here is a visualization of wave reflection from a dielectric-conductor interface. The specific indices used are vacuum (n1 = 1) on the left and slightly conductive glass (\u02dcn2 = 1.52 + 0.152i) on the right. Note that the wave damping is similar to the case of internal reflection, but now the transmitted wave propagates away from the interface rather than along it. 2J. H. Weaver and H. P. R. Frederikse, \u2018\u2018Optical Properties of Selected Elements,\u2019\u2019 in CRC Handbook of Chemistry and Physics, 82nd ed., David R. Lide, Ed. (CRC Press, Boca Raton, 2001), p. 12-133. 168 Chapter 9. Fresnel Relations 9.6 Exercises 9.6 Exercises Problem 9.1 We derived the Fresnel relations for S-polarized light giving the reflection coefficient and transmission coefficient rS = n1 cos \u03b8i \u2212 n2 cos \u03b8t n1 cos \u03b8i + n2 cos \u03b8t tS = 2n1 cos \u03b8i n1 cos \u03b8i + n2 cos \u03b8t 169 (9.77) (9.78) of the electric field at a dielectric interface. Use the setup from the text to derive the Fresnel relations for the reflection coefficient, rP = n1 cos \u03b8t \u2212 n2 cos \u03b8i n1 cos \u03b8t + n2 cos \u03b8i and transmission coefficient tP = 2n1 cos \u03b8i n1 cos \u03b8t + n2 cos \u03b8i for P-polarized light. Problem 9.2 (9.79) (9.80) When we worked out the Fresnel relations for the reflection and coefficients at a planar, dielectric interface, we ignored two boundary conditions, namely that the components of D and B normal to the interface are continuous across the boundary. Write out both of these boundary conditions in terms of the Fresnel coefficients r and t for both polarizations. Problem 9.3 One way of making a cheap but good polarizer (except for wavefront distortion) is to use a stack of microscope slides. Assuming an index of refraction n = 1.52 and that the slides are all oriented at Brewster\u2019s angle with respect to a randomly polarized laser beam, how many slides does it take to attenuate one polarization by a factor of 10\u22124 in intensity compared to the other? Keep in mind that both sides of each slide are at Brewster\u2019s angle. Ignore multiple reflections in", " your calculation. Problem 9.4 Diode lasers in the near infrared (say, around 780 nm) are made from GaAlAs crystals. For low-power diodes (in the 5-15 mW range), the crystal ends are cleaved to form flat surfaces that act as the two flat reflectors of the laser cavity. Model a typical diode laser as a crystal (n = 3.5) of length 300 \u00b5m, surrounded by air (n = 1). (a) For this resonator, calculate the round-trip time, the free spectral range, the finesse, the photon lifetime, and the frequency width (FWHM) of the modes. (Note that for higher power diodes, the output facet is typically antireflection coated to couple out more of the light in the cavity.) (b) Assuming that the diode laser frequency is that of a single longitudinal mode, calculate the change in laser frequency per degree Celsius for a temperature change. Assume a nominal operating (vacuum) wavelength of 780 nm and a coefficient of thermal expansion (\u2206L/L) for GaAlAs of 7 \u00d7 10\u22126/\u25e6C. (In practice this tuning rate is only valid for small temperature changes; for larger changes, the change in the band gap is more important, causing much larger average tuning rates as the laser \u2018\u2018hops\u2019\u2019 between longitudinal modes.) Problem 9.5 Plot the intensity reflection coefficients as a function of incident angle for light incident from air (n = 1) onto crown glass (n = 1.52) for both S- and P-polarized light. Repeat for light incident from within the glass. 170 Problem 9.6 Chapter 9. Fresnel Relations For total internal reflection from a glass\u2013air interface (n1 = 1.52 for glass, n2 = 1 for air), plot the skin depth normalized to the vacuum wavelength (\u03b4/\u03bb0) as a function of incident angle. Problem 9.7 Derive the wave equation for the magnetic field H in a homogeneous medium of permittivity and conductivity \u03c3. Problem 9.8 To describe the propagation of electromagnetic waves in a conductor, we derived a wave equation for E, coupled to the current density induced by the field in the conductor. An alternate approach is to consider the wave equation to be coupled to the polarization P of the medium. As the first step in this method, recall that the electric displacement, electric field, and polarization density are related by and ignoring magnetic effects, we can assume D = 0E + P, B = \u00b50H. Then use Maxwell\u2019s equations in a source-free, dielectric medium = \u2212\u2202tB \u2207 \u00d7 H = \u2202tD, (9.81) (9.82) (9.83) to derive a wave equation driven by P. You may assume the polarization field to be transverse, i.e., \u2207 \u00b7 P = 0. Problem 9.9 Compute the reflectance and skin depth for light at normal incidence from vacuum at wavelength 780 nm for polished surfaces of gold (\u02dcn = 0.08 + 4.60i) and copper (\u02dcn = 0.24 + 4.80i). Problem 9.10 (a) Show that in the limit of low frequency, the skin depth for a good (nonmagnetic) conductor can be written \u03b4 \u2248 2 \u00b50\u03c9\u03c3. (9.84) (b) Evaluate the skin depth at 200 kHz for titanium (\u03c3 = 2 \u00d7 106 (m \u00b7 \u2126)\u22121). Recall that \u00b50 = 4\u03c0 \u00d7 10\u22127 N/A2. (c) Suppose you have a disc of titanium the size and shape of a quarter. The disc is in the center of a solenoidal coil, which is driven at 200 kHz, and it lies in the plane orthogonal to the coil axis. Assume that the solenoid length is much larger than all other length scales in this setup. What is the (qualitative) distribution of power absorption in the disc? Explain. Problem 9.11 It is conventional to represent the mode functions of a compact region in space by normalized mode 9.6 Exercises 171 functions fk,\u03b6(r), where k is the wave vector (which takes on discrete values), and \u03b6 is an index marking the two possible polarizations. These mode functions are normalized in the sense that V d3r |fk,\u03b6(r)|2 = 1, (9.85) where V is the volume over which the mode exists (i.e., the cavity size). More generally, the orthonormality relation is V d3r fk,\u03b6(r) \u00b7 f\u2217 k,\u03b6(r) = \u03b43 k,k\u03b4\u03b6,\u03b6. (9.86) However", ", for mode functions defined over all space, the integration volume is unbounded and k takes on any value in R3, and this orthonormality relation generalizes to d3r fk,\u03b6(r) \u00b7 f\u2217 k,\u03b6(r) = (2\u03c0)3\u03b43(k \u2212 k) \u03b4\u03b6,\u03b6, assuming f (x) = 1 2\u03c0 \u221e \u2212\u221e \u02dcf (k) eikx dk, \u02dcf (k) = \u221e \u2212\u221e f (x) e\u2212ikx dx, as the Fourier-transform convention in each dimension. (9.87) (9.88) In the presence of a vacuum\u2013dielectric interface, the mode function for TE polarization can be written fk,TE = \u02c6\u03b5k,TE eik\u00b7r\u0398(z) + rTEeikr\u00b7r\u0398(z) + tTEeikt\u00b7r\u0398(\u2212z), (9.89) where the dielectric (of permittivity, and index n) occupies the half-space z \u2264 0, rTE and tTE are the Fresnel reflection and transmission coefficients for TE (S) polarization, and \u0398(z) is the Heaviside step function. Show that this mode function satisfies the orthonormality relation d3r (r) 0 k,TE(r) \u00b7 fk,TE(r) = (2\u03c0)3\u03b43(k \u2212 k), f\u2217 (9.90) which reduces to the free-space relation as (r) \u2212\u2192 0. For simplicity, suppose that fk,TE(r) and fk,TE(r) are both incident from the vacuum side. Note: there is a bit of a trick to getting the delta function to come out right on the transmitting side. You might want to refer to the article that first proved this for help: C. K. Carniglia and L. Mandel, \u2018\u2018Quantization of Evanescent Electromagnetic Waves,\u2019\u2019 Physical Review D 3, 280 (1971) (doi: 10.1103/PhysRevD.3.280). Chapter 10 Thin Films Now with reflections from single interfaces under our belt, we can tackle multiple interfaces in the form of thin films. The most obvious everyday examples of thin optical films are the swirling colors in films of oil on puddles of water, or similar colors in bubbles of soap or sea foam, as shown in the photo below. However, optical thin films have widespread applications in modifying the reflection properties of surfaces, including antireflection coatings (e.g., of eyeglasses and camera lenses), high-reflection coatings for laser optics, wavelength filters for spectroscopy and color separation in projectors and photography, polarizers, and so on. We will start out with a simple treatment of a single-layer film in the style of a Fabry\u2013Perot cavity, and then rederive the result in a much more powerful matrix formalism. We will be implicitly assuming dielectric thin films, but of course we can also handle conductive films simply by using complex refractive indices. 174 Chapter 10. Thin Films 10.1 Reflection-Summation Model We will now consider the multiple reflections due to an input wave E considering only a single input polarization (S or P), but the analysis works for either one separately. (+) 0i at a single thin film. We are implicitly The convention is that ra, ta, rb, and tb are the reflection and transmission coeffiencients at interfaces a and b for left-to-right incidence. For right-to-left incidence, the coefficients are r b. The interfaces a and b divide the media with refractive indices ni (for incident medium), n1 (the coating), and ns (the substrate or transmission medium); we will use the same subscripts to denote waves inside these media (e.g., k1 is the wave vector in medium of index n1). b, and t a, r a, t Then the total reflected field is the sum of all the individual reflected waves. After the first term, a pattern emerges because each successive reflection is the same as the previous reflection but with one more round trip inside the thin film. E (+) 0r = raE (+) 0i + t arbtaE (+) 0i ei\u03c6 + t arbr arbtaE (+) 0i ei2\u03c6 + \u00b7 \u00b7 \u00b7. (10.1) Here, \u03c6 is the propagation phase accumulated between successive reflections as measured along the same wave front. This is a somewhat subtle point that requires a careful geometric construction. As shown in the diagram, the correct form for \u03c6 follows from the following argument. The phase difference between successive reflections", " is the phase of the ray after one round trip inside the thin film as it exits abnin1nsE0i(+)raE0i(+)t'arbtaE0i(+)taE0i(+)rbtaE0i(+)r'arbtaE0i(+)tbtaE0i(+)tbr'arbtaE0i(+)dhiqiqf\u00a1f\u00ba1q 10.1 Reflection-Summation Model 175 (call this phase \u03c61), but we need to subtract the propagation of the directly reflected ray to the same perpendicular position along the ray (call this second phase \u03c60). That\u2019s how displaced plane waves interfere: the relative phase of the parts that align along the same wave front (normal to the ray) are what deterimine the interference. Thus, where and Thus, \u03c6 = \u03c61 \u2212 \u03c60, \u03c61 = 2k1 \u00b7 r1 = 2k1 sin \u03b81 h + 2k1 cos \u03b81 d, \u03c60 = k0 \u00b7 r0 = 2ki sin \u03b8i h. The first term vanishes by Snell\u2019s law, so we are left with \u03c6 = \u03c61 \u2212 \u03c60 = 2h(k1 sin \u03b81 \u2212 ki sin \u03b8i) + 2k1d cos \u03b81. (10.2) (10.3) (10.4) (10.5) Now we can write out the explicit sum for the reflected wave by completing the pattern from Eq. (10.1): \u03c6 = 2k1d cos \u03b81. (10.6) (round-trip phase in thin film) E (+) 0r = E (+) 0i ra + t arbtaei\u03c6 rbr aei\u03c6n. \u221e n=0 We can evaluate this using the formula \u221e n=0 xn = 1 1 \u2212 x, which is valid provided |x| < 1, with the result rfilm := E E (+) 0r (+) 0i = ra + arbtaei\u03c6 t aei\u03c6. 1 \u2212 rbr (10.7) (10.8) (10.9) Using the Stokes relations t the result ata = 1 \u2212 r2 a and ra = \u2212r a, we can eliminate the transmission coefficients, with rfilm = ra + (1 \u2212 r2 a)rbei\u03c6 1 + rarbei\u03c6. (10.10) Rewriting all this as a single fraction, we finally come to rfilm = ra + rbei\u03c6 1 + rarbei\u03c6, (10.11) (thin film reflection coefficient) which is the main result of this section. 10.1.1 Example: Single Glass Plate as a Fabry\u2013Perot Etalon To get a better feeling for the meaning behind Eq. (10.11), let\u2019s look at the special case of ni = ns. For example, this could be a glass plate surrounded by air. In this case, ra = \u2212rb, so that rfilm = ra(1 \u2212 ei\u03c6) aei\u03c6, 1 \u2212 r2 (10.12) and the reflectance becomes Rfilm = |rfilm|2 = 2r2 1 + r4 a(1 \u2212 cos \u03c6) a \u2212 2r2 a cos \u03c6, (10.13) (etalon reflectance) 176 Chapter 10. Thin Films assuming that ra is real, as for an air\u2013glass interface. Notice that Rfilm drops to zero when cos \u03c6 = 1. This happens when \u03c6 = 2\u03c0q for some integer q, or k1 cos \u03b81 = \u03c0q d. Comparing this to the resonance condition for a planar-mirror Fabry-Perot cavity [Eq. (7.2)], kq = \u03c0q d, (10.14) (10.15) we see that we have just rederived the Fabry\u2013Perot resonance condition, but now generalized for non-normal incidence. As we mentioned in Chapter 7, a glass window can act as a planar Fabry\u2013Perot etalon. Even though the window thickness is nominally fixed (neglecting thermal expansion and mechanical stress), we can tune the etalon by changing its orientation with respect to the incident wave. The transmission of the plate is then Tfilm = 1 \u2212 Rfilm = = = 1 + r4 (1 \u2212 r2 a)2 (1 \u2212 r2 a \u2212 2r2 a cos \u03c6 a)2 a sin2(\u03c6/2) (1 \u2212 r2 a)2 + 4r2 1 2, sin2(\u03c6/2) 1 + 2F \ufffd", "\ufffd (10.16) where F = \u03c0ra 1 \u2212 r2 a (10.17) (etalon finesse) is the finesse of the etalon. The transmission angle is, by two applications of Snell\u2019s law, just the same as the incident angle. We can think of an arbitrary thin film as an etalon, with finesse F = \u221a rarb \u03c0 1 \u2212 rarb. (10.18) (finesse of single film) Of course, the finesse depends implicitly on the incident angle via the angle-dependence of ra. The transmitted angle is just given in terms of the incident angle by ni sin \u03b8i = ns sin \u03b8t. 10.2 Thin Films: Matrix Formalism Now we will rederive the results from the last section by matching boundary conditions, as in Chapter 9. We will assume S-polarization, and later modify the results to handle P-polarization. The setup is shown in the diagram. 10.2 Thin Films: Matrix Formalism 177 We are assuming fields as follows: Incident Region: E Thin-Film Region 1: E (+) (+) 0i eiki\u00b7r + E 0t1 eikt1\u00b7r + E (+) 0t eikt\u00b7(r\u2212d\u02c6x). (+) 0r eikr\u00b7r (+) 0r1 eikr1\u00b7r Substrate/Transmission Region: E (10.19) This is the same setup as for reflection at a single interface, except for the addition of left- and rightpropagating fields inside the film. Notice also the phase offset in the transmitted field, so that the transmitted field has zero phase at boundary b. As before, we apply boundary conditions. The transverse component of the electric field (E \u00b7 \u02c6z) must be continuous across each boundary: 0r = E(+) 0i + E(+) E(+) 0t1 ei\u03b4 + E(+) E(+) 0t1 + E(+) 0r e\u2212i\u03b4 = E(+) 0r1 =: E(+) 0t =: E(+) a b (interface a) (interface b). (10.20) Here, \u03b4 := (kt1 \u00b7 \u02c6x)d = 2\u03c0 \u03bb0 n1d cos \u03b8t1 (10.21) (film propagation phase) is the phase change traveling directly across the interface. Note that \u03b4 = \u03c6/2, where we justified \u03c6 before in Eq. (10.6). In this case, we can think of matching phases everywhere, but considering the phases at constant y to ensure they are comparable, hence the \u2018\u2018straight across\u2019\u2019 phase. Note that a tilted wave in the film region has an effective wavelength of \u03bb0/ cos \u03b8t1, along a horizontal slice of the wave. Similarly, the transverse component of the magnetic field (H \u00b7 \u02c6y) must be continuous across the bound- ary: H (+) 0i 0r cos \u03b8i = H (+) cos \u03b8i \u2212 H (+) H (+) 0t1 cos \u03b8t1ei\u03b4 \u2212 H (+) 0t1 cos \u03b8t1 \u2212 H (+) 0r1 cos \u03b8t1e\u2212i\u03b4 = H (+) 0r1 cos \u03b8t1 =: H (+) 0t cos \u03b8t =: H (+) a b (interface a) (interface b). (10.22) We can rewrite the magnetic fields as \u03b1i(E(+) 0i \u2212 E(+) 0r ) = \u03b11(E(+) 0t1 \u2212 E(+) 0r e\u2212i\u03b4) = \u03b1sE(+) 0r1 ) = H (+) 0t = H (+) 0t1 ei\u03b4 \u2212 E(+) \u03b11(E(+) a b, (10.23) a(x = 0)b(x = d)nin1nszyxqiqt1qtE0i(+)H0i(+)E0r(+)H0r(+)E0t(+)H0t(+)E0t1(+)H0t1(+)E0r1(+)H0r1(+) 178 Chapter 10. Thin Films where the media are characterized by the \u03b1 constants. The constants for the incident, film, and substrate media are \u03b1i := \u03b11 := \u03b1s := ni \u03b70 n1 \u03b70 ns \u03b70 cos \u03b8i cos \u03b8t1 cos \u03b8t, (10.24) (geometry coefficients) where as usual, \u03b70 is the vacuum wave impedance. Our goal is now to find a relation between the fields at interfaces a and b.", " Adding and subtracting the expressions for E(+) and H (+) b b, E(+) 0t1 = E(+) 0r1 = \u03b11E(+) \u03b11E(+) b b + H (+) 2\u03b11 b \u2212 H (+) 2\u03b11 b e\u2212i\u03b4 ei\u03b4. Using the definitions of E(+) a and H (+) a in terms of the intra-film fields, E(+) a = E(+) 0t1 + E(+) 0r1 = cos \u03b4 E(+) b \u2212 H (+) a = \u03b11(E(+) 0t1 \u2212 E(+) 0r1 ) = \u2212i\u03b11 sin \u03b4 E(+) H (+) b i sin \u03b4 \u03b11 b + cos \u03b4 H (+) b (10.25) (10.26). We can conveniently rewrite this in matrix form: E(+) a H (+) a = \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 cos \u03b4 \u2212i sin \u03b4 \u03b11 \u2212i\u03b11 sin \u03b4 cos \u03b4 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb E(+) b H (+) b =: F. E(+) b H (+) b The matrix F is the thin-film transfer matrix, and has the general form (thin-film transfer matrix) (10.27) F = A B C D. (10.28) The matrix \u2018\u2018transfers\u2019\u2019 the fields from interface a to interface b. As usual for matrix formalisms, this works for multiple films via matrix multiplication: Here, where the phase shifts are E(+) 0 H (+) 0 Fj = \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 = F1F2F3 \u00b7 \u00b7 \u00b7 FN. E(+) N H (+) N (composition of thin-film matrices) (10.29) cos \u03b4j \u2212i sin \u03b4j \u03b1j \u2212i\u03b1j sin \u03b4j cos \u03b4j \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb, (10.30) (thin-film matrices) \u03b4j = 2\u03c0 \u03bb0 njdj cos \u03b8tj (10.31) (thin-film phase shifts) 10.2 Thin Films: Matrix Formalism 179 (dj is the thickness of the jth layer), the geometry coefficients are and the transmission angles are given by \u03b1j = nj \u03b70 cos \u03b8tj, (10.32) (geometry coefficients) \u03b8tj = sin\u22121 ni nj sin \u03b8i. (10.33) (transmission angles) Now we can solve for the reflected and transmitted fields. Using the definitions 0r 0i + E(+) E(+) a = E(+) b = E(+) E(+) 0t H (+) E(+) 0i \u2212 E(+) a = \u03b1i b = \u03b1sE(+) H (+) 0t 0r (10.34) in the matrix equation (10.27), we find 0i + E(+) E(+) 0r 0i \u2212 E(+) \u03b1i(E(+) 0r ) A B C D =. E(+) 0t \u03b1sE(+) 0t This is equivalent to the system of equations 1 + r = At + B\u03b1st \u03b1i(1 \u2212 r) = Ct + D\u03b1st, (10.35) (10.36) Ea(+)Ha(+)Eb(+)Hb(+)nin1n2n3nNns\u2026 180 Chapter 10. Thin Films where as usual r := E0r/E0i and t := E0t/E0i. Solving these equations, we find the film reflection and transmission coefficients rfilm = tfilm = \u03b1iA + \u03b1i\u03b1sB \u2212 C \u2212 \u03b1sD \u03b1iA + \u03b1i\u03b1sB + C + \u03b1sD 2\u03b1i \u03b1iA + \u03b1i\u03b1sB + C + \u03b1sD in terms of the transfer-matrix elements. The above analysis assumed S-polarization. It turns out that for P-polarization, the coefficients in Eqs. (10.37) are still valid, with a slight reinterpretation. The only modification is that in these relations and in the transfer matrix in Eq. (10.27), we must calculate the geometric coefficients \u03b1j differently. For the two polarizations, we have (film reflection/transmission coefficients) (10.37) \u03b1j = \u03b1j = nj \u03b70 cos \u03b8tj (S-polarization/TE polarization) nj \u03b70 cos \u03b8tj (P-polarization/TM polarization).", " (10.38) (The same modifications apply to the incident and substrate coefficients \u03b1i and \u03b1s.) With these changes, the expression for the reflection coefficient is still valid, though the expression for the transmission coefficient is more complicated. For most purposes, it is sufficient to note that the film reflection coefficient (recipe for both polarizations) rfilm = \u03b1iA + \u03b1i\u03b1sB \u2212 C \u2212 \u03b1sD \u03b1iA + \u03b1i\u03b1sB + C + \u03b1sD (10.39) has the same form for both polarizations, and the transmittance can be computed from Tfilm = 1 \u2212 Rfilm = 1 \u2212 |rfilm|2. (film reflection coefficient, both polarizations) Even though the formulae get a bit complicated, the general idea is simple: compute the \u03b1j, compute the transfer matrix for the whole film, and then compute the reflection coefficient, and reflectance and transmittance if desired. 10.3 Optical Coating Design Now we\u2019ll consider a few of the simplest optical coating designs in common use. 10.3.1 Single-Layer Antireflection Coating Consider a single-layer film at normal incidence. The transfer matrix is just the single-layer matrix from Eq. (10.30) \uf8f9 \uf8ee F = \uf8ef \uf8ef \uf8ef \uf8f0 cos \u03b4 \u2212i sin \u03b4 \u03b11 \u2212i\u03b11 sin \u03b4 cos \u03b4. \uf8fa \uf8fa \uf8fa \uf8fb (10.40) Putting the matrix elements into the expression from Eqs. (10.37) for the film reflection coefficient, we find r = \u03b1i cos \u03b4 \u2212 i \u03b1i cos \u03b4 \u2212 i \u03b1i\u03b1s \u03b11 \u03b1i\u03b1s \u03b11 sin \u03b4 + i\u03b11 sin \u03b4 \u2212 \u03b1s cos \u03b4 sin \u03b4 \u2212 i\u03b11 sin \u03b4 + \u03b1s cos \u03b4 = n1(ni \u2212 ns) cos \u03b4 \u2212 i(nins \u2212 n 2 n1(ni + ns) cos \u03b4 \u2212 i(nins + n 2 1 ) sin \u03b4 1 ) sin \u03b4, (10.41) 10.3 Optical Coating Design where for normal incidence, \u03b1i = ni \u03b70 ; \u03b11 = n1 \u03b70 ; \u03b1s = ns \u03b70, 181 (10.42) and ni is the refractive index of the incidence material (say, air), n1 is the index of the coating and ns is the index of the substrate (say, glass). We can then calculate the reflectance (intensity reflection coefficient) of the film as R = |r|2 = n 2 1 (ni \u2212 ns)2 cos2 \u03b4 + (nins \u2212 n 2 1 (ni + ns)2 cos2 \u03b4 + (nins + n 2 n 2 1 )2 sin2 \u03b4 1 )2 sin2 \u03b4 (reflectance, single-layer film). (10.43) This is still a bit messy, but we\u2019ll now make it simpler. An important but simple case is the quarter-wave film (still at normal incidence). The thickness is d = \u03bb 4 = \u03bb0 4n1, (10.44) where it is important to note that the layer is a quarter wavelength for the wavelength inside the medium. This thickness leads to a phase shift of \u03b4 = 2\u03c0 \u03bb0 n1d cos \u03b8t1 = \u03c0 2, (10.45) which means that cos \u03b4 = 0 and sin \u03b4 = 1. This leads to a reflectance of R = nins \u2212 n 2 1 nins + n 2 1 2. Notice that the reflectance vanishes if the film index is chosen such that (reflectance, single quarter-wave film) (10.46) n1 = \u221a nins. (10.47) (single-layer antireflection condition) As an aside, let\u2019s quickly compare this to the Fabry\u2013Perot etalon discussion from the first part of this chapter. The reflectance vanishes even though there is only \u03bb/2 of propagation over one full round trip. Normally, a Fabry\u2013Perot cavity requires an integer multiple of \u03bb of propagation distance per round trip, but now we have to be careful about phase changes on reflection. For example, if ni < n1, then the antireflection condition (10.47) implies that n1 < ns (and n1 > ns in the case that ni > n1). Thus, exactly one of the reflections inside the film will produce a \u03c0 phase shift, playing the role of an extra \u03bb/2 of propagation distance. This is different from the etalon that we considered early on in this chapter, which was for the case ni = ns, which requires a thickness", " of \u03bb/2 for R to vanish. Let\u2019s now consider the example of a real, single-layer antireflection (AR) coating. For an air\u2013glass interface, we have ni = 1 and ns = 1.52 for optical crown glass. Ideally, n1 = 1.23 for an AR coating (where R vanishes). The closest material suitable for making an optical coating (with good optical properties and durability) is MgF2, which has n = 1.38 (for visible wavelengths). Putting this into the formula (10.46), we find R = 1.3%, which is much better than the uncoated reflectance of R = 4.3% but not really that close to zero. The cheapest optical coatings in the visible are single-layer MgF2. 10.3.2 Two-Layer Antireflection Coating The basic problem with single-layer antireflection coatings is that there is a limited choice of suitable coating materials, and so it is difficult to match the optimum index condition very precisely. One solution to this problem is to add a second layer, in the hope that an extra free parameter will make the choice of materials more flexible. 182 Chapter 10. Thin Films Let\u2019s assume a stack of two quarter-wave films at normal incidence. The matrix for a single quarter- wave layer is \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 F1 = cos \u03b4 \u2212i sin \u03b4 \u03b11 \u2212i\u03b11 sin \u03b4 cos \u03b11 \u2212i\u03b11 0 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb. The transfer matrix for the two-layer stack is the product of two of these matrices: F = F1F2 = \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 0 \u2212 i \u03b11 \u2212i\u03b11 \u03b12 \u2212i\u03b12 0 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb = \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 \u2212 \u03b12 \u03b11 0 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb. 0 \u2212 \u03b11 \u03b12 (10.48) (10.49) Again, putting the matrix elements into the expression from Eqs. (10.37) for the film reflection coefficient, we find r = \u2212 \u2212 The reflectance is thus \u03b1i\u03b12 \u03b11 \u03b1i\u03b12 \u03b11 + \u2212 \u03b11\u03b1s \u03b12 \u03b11\u03b1s \u03b12 = n 2 2 ni \u2212 n 2 2 ni + n 2 n 2 1 ns 1 ns. (10.50) R = n 2 2 ni \u2212 n 2 2 ni + n 2 n 2 1 ns 1 ns 2. (10.51) (two-layer reflectance) The reflectance vanishes if n 2 2 ni = n 2 1 ns, which gives the AR condition n2 n1 = ns ni, (10.52) (two-layer antireflection condition) allowing us to pick the ratio of two refractive indices rather than a single absolute index. As an example, again for crown glass (ni = 1 and ns = 1.52), the ideal AR ratio is n2/n1 = 1.23. We can choose ZrO2 (n2 = 2.1) and CeF3 (n1 = 1.65), which gives n2/n1 = 1.27, a value that is pretty close to the ideal value. Working out the reflectance, we find R = 0.1% for this coating (about as good as you can expect given fabrication tolerances). This kind of coating is called a \u2018\u2018vee\u2019\u2019 coating due to the shape of the reflectance plot as a function of wavelength: there is a relatively narrow range (say, 10 nm or so) of good antireflection for a coating in the visible. In the example plotted here, the design wavelength is \u03bb0 = 532 nm, which is marked in the plot, and the uncoated air\u2013glass reflectance is also marked in the plot. l0350750532500600400700re\ufb02ectance0.1700.043 10.3 Optical Coating Design 183 It is possible to make a more broadband AR coating by adding a third layer, but we won\u2019t go into that here. 10.3.3 High Reflector: Quarter-Wave Stack ns/ni, From the analysis of the two-layer AR coating, we note that if the ratio n2/n1 is very different from then the reflectance is relatively large. Thus, the idea behind making a high reflectance (HR) coating is to essentially reverse the order of the layers in the antireflection coating: n1 should have a larger index than n2 for good reflection. As an", " example, let\u2019s take the same materials as for the AR coating [ZrO2 (n2 = 2.1) and CeF3 (n1 = 1.65)], but reverse the order of the layers. In this case, the reflectance turns out to be R = 18%. This is a much higher reflectance than the AR coating, although it is not as high as, say, a good metallic coating. Of course, it is possible to get higher reflectances by choosing other materials with a greater index contrast. But these improvements will be marginal. How can we get a reflectance approaching unity? The general idea is to stack many of these double layers to form a quarter-wave stack. Consider N double layers. The transfer matrix for one of the double layers at normal incidence is \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 \u2212 n2 n1 \u2212 n1 n2 Thus, the transfer matrix for the whole stack is FN = \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 \u2212 N n2 n1 0 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb. 0 n1 n2 N \u2212 The reflection coefficient is r = ni \u2212 ni \u2212 n2 n1 n1 n2 N N \u2212 ns \u2212 \u2212 ns \u2212 n1 n2 n2 n1 N N = ni ns ni ns \u2212 \u2212 n2 n1 n2 n1 2N 2N \u2212 1. + 1 Thus, the reflectance is R = \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 ni ns ni ns \u2212 \u2212 n2 n1 n2 n1 2N 2N 10.53) (10.54) (10.55) For the same materials as above, ni/ns = 1/1.52 and n2/n1 = 1/1.27, so for (reflectance of dielectric stack, normal incidence) (10.56) N = 2 :R = 36% N = 3 :R = 53% N = 4 :R = 68% N = 8 :R = 94% N = 12 :R = 99%. (10.57) Thus, we see that as the number of double layers increases, the reflectance rapidly (exponentially) converges to unity. We could use fewer layers with a better choice of materials (higher index contrast), although we should note that the intermediate reflectances shown here are also useful as beam splitters. The periodic structure of these quarter-wave stacks gives a simple example of a photonic crystal or Bragg reflector. 184 Chapter 10. Thin Films 10.4 Gradient-Index Layers So far we have only considered propagation and reflection due to layers of homogeneous media, and interfaces between them. But what if we have a layer where the refractive index varies continuously? We will again consider a planar geometry with a refractive index that varies only in the x-direction. Specifically, we will consider the case of a homogeneous \u2018\u2018incident\u2019\u2019 medium of refractive index ni occupying the region x < 0, a \u2018\u2018substrate\u2019\u2019 medium of index ns occupying x > d, and an arbitrarily and continuously varying index profile n(x) in the region 0 \u2264 x \u2264 d. Note that the index profile need not be continuous at x = 0 or x = d, and n(x) itself need not be continuous, though it should at least be piecewise continuous, and the motivation for this calculation comes from the continuous case. The main motivation for such gradient-index (GRIN) layers is as an alternative to thin interference films to produce antireflection coatings for optical surfaces. The idea stems from the observation that abrupt changes in the refractive index cause reflections\u2014the Fresnel reflection coefficient at normal incidence is (n1 \u2212 n2)/(n1 + n2), so larger index steps cause larger reflections. One possible strategy is to try to avoid the abrupt index change at an optical surface by introducing a layer that smoothly interpolates the refractive index from air to glass. This could in principle be accomplished by many layers of gradually increasing index, but in practice the materials aren\u2019t available to do this well. More practically, it is possible to shape a surface with subwavelength structures, such as the pattern shown here. At first glance, this surface doesn\u2019t seem too promising as an antireflection surface, as it is basically a diffraction grating. But if the period of the grating is sufficiently small (smaller than the optical wavelength), the light does not \u2018\u2018see\u2019\u2019 the surface as a diffraction grating (this occurs when the angle of the first diffracted orders is so large as to be nonsensical, so that no orders except for the zeroth order propagate). In this case, we can coarse-", "grain the permittivity, or equivalently, the square of the refractive index, at each position x normal to the surface plane, by averaging the density of the material in the other directions at fixed x. (Note that the susceptibility and thus the permittivity give the direct polarization response of the medium to the incident field, and this is why the permittivity or squared index, not the index, should be coarsegrained.) If the index in the coarse-grained sense then changes sufficiently slowly in space, we might then expect reflections to be suppressed, in a way that should be robust to angle and wavelength changes, as compared to interference-based dielectric coatings. Such coatings are already available in commercial SLR camera lenses.1 To work out the reflection coefficient for the GRIN layer, we will apply the thin-film matrix formalism by dividing up the GRIN layer into N sublayers of width \u03b4x = d N. (10.58) If N is large, then the sublayers are thin, and the refractive index is approximately constant over each layer, so that the matrix formalism applies. 1For example, see http://www.canon.com/news/2008/sep17e.html, where the subwavelength structures seem to be cone- shaped. xnxo=o0xo=odno(x)nins 10.4 Gradient-Index Layers 185 We take advantage of the fact that \u03b4x is small to simplify the matrices, and we will also explicitly take the continuum limit N \u2212\u2192 \u221e to develop a modified formalism to handle the continuous layer. As in the matrix-formalism development, we will assume S-polarization, and we will indicate the changes necessary to handle P-polarization at the end. 10.4.1 Finite Stack of Very Thin Films In a stack of thin films, the transfer matrix of the jth film is given by \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 Fj = cos \u03b4j \u2212i sin \u03b4j \u03b1j \u2212i\u03b1j sin \u03b4j cos \u03b4j \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb. (10.59) Assuming the layer is very thin, of thickness \u03b4x, then the phase shift is similarly small, \u03b4j = 2\u03c0 \u03bb0 nj(\u03b4x) cos \u03b8tj = k0nj(\u03b4x) cos \u03b8tj, (10.60) and we can expand the transfer matrix to lowest order in \u03b4j: Then with \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 Fj \u2248 1 \u2212i\u03b4j \u03b1j \u2212i\u03b1j\u03b4j 1 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb. \u03b1j = nj \u03b70 cos \u03b8tj, (10.61) (10.62) Ea(+)Ha(+)Eb(+)Hb(+)nin1n2n3nNns\u2026 186 Chapter 10. Thin Films we can write this matrix directly in terms of refractive indices and angles as \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 Fj \u2248 1 \u2212i\u03b70k0 \u03b4x \u2212i k0 \u03b70 j cos2 \u03b8tj \u03b4x n 2 1 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb. But using Snell\u2019s law, we can rewrite the angle factor as j cos2 \u03b8tj = n 2 n 2 j \u2212 n 2 j sin2 \u03b8tj = n 2 j \u2212 n 2 i sin2 \u03b8i, and thus we have \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 Fj \u2248 \u2212i k0 \u03b70 1 \u2212i\u03b70k0 \u03b4x n 2 j \u2212 n 2 i sin2 \u03b8i \u03b4x 1 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb (10.63) (10.64) (10.65) as the transfer matrix in the limit of very thin layers. 10.4.2 Continuous Medium Rather than multiply the N matrices together as in the normal matrix formalism, we will take the continuum limit first, and then handle the propagation of the fields by converting the matrix relation into a system of coupled differential equations. Recall from Eq. (10.27) that the matrix Fj \u2018\u2018transfers\u2019\u2019 the parallel components of E(+) and H(+) from the output edge to the input edge of the jth layer. If we put this into continuous notation, where \u03b4x \u2212\u2192 dx, nj \u2212\u2192 n(x), and we regard the", " jth layer of the medium as occupying the region x \u2212 dx to x, we have E(x \u2212 dx) H(x \u2212 dx) = \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 1 \u2212i\u03b70k0 dx \u2212i k0 \u03b70 n2(x) \u2212 n 2 i sin2 \u03b8i dx 1 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb E(x) H(x), (10.66) where again E and H refer only to the components of the fields parallel to the film surface (i.e., Ez and Hy in our coordinate system, for S-polarization), and note that we are dropping the (+) superscripts to simplify the notation. Expanding the left-hand side to first order in dx, \uf8ee \uf8ef \uf8f0 E(x) \u2212 H(x) \u2212 dE dx dH dx dx dx \uf8f9 \uf8fa \uf8fb = \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 1 \u2212i\u03b70k0 dx \u2212i k0 \u03b70 n2(x) \u2212 n 2 i sin2 \u03b8i dx (x) H(x) and canceling the common E(x) and H(x) terms, we have the linear differential system d dx E(x) H(x) = ik0 \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 0 1 \u03b70 n2(x) \u2212 n 2 i sin2 \u03b8i \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb \u03b70 0 E(x) H(x). If we condense our notation by defining (10.67) (10.68) n\u03b8(x) := n2(x) \u2212 n 2 i sin2 \u03b8i, (10.69) (angle-corrected refractive index) 10.4 Gradient-Index Layers the linear system becomes d dx E(x) H(x) = ik0 \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 0 n 2 \u03b8 (x) \u03b70 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb \u03b70 0 E(x) H(x), or 187 (10.70) ik0n 2 \u03b8 (x) \u03b70 if we write them out as separate equations. These equations allow us to propagate the fields from x = d to x = 0, the procedure that takes the place of multiplying together the N matrices. E(x). = (10.71) = ik0\u03b70H(x) dE dx dH dx 10.4.3 Reflection Coefficient To relate the fields to the reflection coefficient, we can then use the same boundary conditions at the interfaces as in the matrix formalism. Rewriting the boundary conditions (10.34) for the transverse components of the fields at the incident (a), at x = 0, and substrate (b), at x = d, boundaries of the continuous film, 0r 0i + E(+) E(+) a = E(+) b = E(+) E(+) 0t H (+) E(+) 0i \u2212 E(+) a = \u03b1i b = \u03b1sE(+) H (+). 0r 0t (10.72) The coefficients \u03b1i and \u03b1s are defined in Eqs. (10.24). But instead of working with the individual fields, we will work with the ratios of the (parallel components of the) fields, which we will call effective impedances (effective because they only refer to parallel components) \u02dc\u03b7(x) := E(+)(x) H (+)(x). At the incident boundary, Eqs. (10.72) give E(+) a H (+) a \u02dc\u03b7(0) = = \u03b1i 0i + E(+) E(+) 0i \u2212 E(+) E(+) 0r 0r = 1 + r \u03b1i (1 \u2212 r) and at the transmitting boundary, \u02dc\u03b7(d) = E(+) b H (+) b = 1 \u03b1s In the region 0 < x < d, we can deduce the dependence of \u02dc\u03b7 by differentiating the definition (10.73): d\u02dc\u03b7 dx = 1 H (+) dE(+) dx \u2212 E(+)(x) [H (+)(x)]2 dH (+) dx. (10.73) (10.74) (10.75) (10.76) Then using Eqs. (10.71) to eliminate the field derivatives, we find that the effective impedance obeys the differential equation2 d\u02dc\u03b7 dx = ik", "0\u03b70 \u2212 ik0n 2 \u03b8 (x) \u03b70 \u02dc\u03b72(x). (10.77) (propagation equation for fields) 2We have basically concocted an alternate derivation of the tapered-transmission-line formalism in Daniel H. Raguin and G. Michael Morris, \u2018\u2018Analysis of antireflection-structured surfaces with continuous one-dimensional surface profiles,\u2019\u2019 Applied Optics, 32, 2582 (1993) (doi: 10.1364/AO.32.002582); see Eq. (26) in particular. 188 Chapter 10. Thin Films The procedure for calculating the reflection coefficient is as follows: first, use Eq. (10.75) as the initial condition for \u02dc\u03b7(d), then use this differential equation to integrate backwards to \u02dc\u03b7(0), and finally, use Eq. (10.74) in the form r = \u03b1i \u02dc\u03b7(0) \u2212 1 \u03b1i \u02dc\u03b7(0) + 1 (10.78) (GRIN-layer reflection coefficient) for the reflection coefficient. 10.4.4 Solution of Ricatti Equations The differential equation is nonlinear in \u02dc\u03b7, which makes it difficult to solve in general. However, it is of a special form, being first-order and only quadratic. An ODE of this form is called a Ricatti equation dy dx = a(x)y2 + b(x)y + c(x), (10.79) (Ricatti equation) which can be converted to a linear (but second-order) ODE by a change of variables. To do this, define z(x) such that Differentiating this expression, we have y = \u2212 z(x) a(x)z(x). y = z a2z2 (az + az) \u2212 z az = az a2z + z2 az2 \u2212 z az. But combining Eqs. (10.79) and (10.80), we have Equating these last two expressions for y, we have y = ay2 + by + c = z2 az2 \u2212 bz az + c z \u2212 + b a a z + acz = 0 (10.80) (10.81) (10.82) (10.83) after multiplying through by az. Assuming that we can solve this equation (which still may be tricky due to non-constant coefficients), the solution to the original equation (10.79) then follows from the transformation (10.80). 10.4.5 Example: Single, Homogeneous Film As a simple check and illustration of the technique, we can work out a solution that we have already solved: the case where n(x) = n1 is a constant over the full range 0 < x < d. From Eq. (10.69), we have \u03b8 (x sin2 \u03b8i = n 2 \u03b8, which is now constant, greatly simplifying the solution. Then we can take the coefficients a = \u2212 ik0n 2 \u03b8 \u03b70, b = 0, c = ik0\u03b70, and thus we should solve the equation z + acz = 0, (10.84) (10.85) (10.86) with constant coefficients. We can certainly do this, because a and c are constants; we can parameterize the general solution to this equation as z = \u02dc\u03b2 cos \u221a ac x + \u03b2 sin \u221a ac x, (10.87) 10.4 Gradient-Index Layers 189 where \u03b2 and \u02dc\u03b2 are undetermined coefficients. With the change of variables (10.80), the solution to the propagation equation (10.77) is \u02dc\u03b7(x) = \u2212 c a \u221a \u2212 sin \u221a cos \u221a ac x + \u03b2 cos \u221a ac x + \u03b2 sin ac x ac x. With the boundary condition (10.75) \u02dc\u03b7(d) = 1 \u03b1s, we can equate this with the solution \u02dc\u03b7 at x = d to fix the coefficient \u221a \u221a \u03b2 = \u03b1s sin \u03b1s cos ac d \u2212 (a/c) cos \u221a ac d + (a/c) sin \u221a ac d ac d. Then we require the solution at the incident boundary, \u02dc\u03b7(0) = \u2212 c a \u03b2, and the S-polarized reflection coefficient (10.78) is r = \u03b1i \u02dc\u03b7(0) \u2212 1 \u03b1i \u02dc\u03b7(0) + 1 = (c/a) \u03b2 + 1 (c/a) \u03b2 \u2212 1, \u03b1i \u03b1i or with Eq. (10.90), r = \u03b1i\u03b1s", " \u03b1i\u03b1s (c/a) sin (c/a) sin \u221a \u221a ac d \u2212 \u03b1i cos ac d + \u03b1i cos \u221a \u221a ac d + \u03b1s cos ac d \u2212 \u03b1s cos \u221a \u221a ac d + (a/c) sin ac d + (a/c) sin \u221a \u221a ac d ac d. (10.88) (10.89) (10.90) (10.91) (10.92) (10.93) To compare this with our previous results in the matrix formalism in Section 10.3.1, we can write argument of the trigonometric functions as 1 \u2212 n 2 n 2 c/a or \u221a ac d = k0n\u03b8d = k0d i sin2 \u03b8i = k0d 1 \u2212 n 2 n 2 1 sin2 \u03b8t1 = n1k0d cos \u03b8t1 = \u03b4, (10.94) and the other factor we need is is negative, and there is a phase ambiguity: do we choose back to the source of this factor, recall from Eq. (10.88) that positive, there is no phase ambiguity here, and we may proceed: a/c. Note that we have to be careful about this, because c/a a/c to have a phase of +\u03c0/2? To go ac/a; since ac is c/a was a shorthand for c/a or \u221a so that c a = \u221a ca a = k0n\u03b8 \u2212ik0n 2 \u03b8 /\u03b70 = i\u03b70 n\u03b8 = i\u03b70 n1 cos \u03b8t1 = i \u03b11, (10.95) r = \u2212i \u2212i \u03b1i\u03b1s \u03b11 \u03b1i\u03b1s \u03b11 sin \u03b4 + \u03b1i cos \u03b4 \u2212 \u03b1s cos \u03b4 + i\u03b11 sin \u03b4 sin \u03b4 + \u03b1i cos \u03b4 + \u03b1s cos \u03b4 \u2212 i\u03b11 sin \u03b4. (10.96) This is the same expression we wrote out before for the reflection coefficient in the matrix formalism, Eq. (10.41). 10.4.6 Example: Linear Permittivity Gradient As a less-simple example, we will now work out the reflection coefficient due to a linear ramp in the permittivity, which interpolates linearly from the incident vacuum (i = 0), to the substrate, of susceptibility \u03c7 [s = 0(1 + \u03c7)]. This transition happens over a distance d. 190 Chapter 10. Thin Films This is a coarse-grained model of a subwavelength diffraction grating with triangular ridges, as illustrated here, where the substrate medium (e.g., glass) has susceptibility \u03c7. Note that this is only a quantitative model of this particular grating for S-polarization and the polarization vector is parallel to the grooves; otherwise the effective index profile has a different functional form,3 because gratings are highly anisotropic, and the coarse-grained surface acts as if highly birefringent. In terms of refractive indices, then, we have ni = 1 n(x) = 1 + \u03c7 d ns = 1 + \u03c7. x From Eq. (10.69), we have n 2 \u03b8 (x) = 1 + \u03c7 d x \u2212 sin2 \u03b8i = cos2 \u03b8i + \u03c7 d x, and the Ricatti coefficients in Eq. (10.79) become a = \u2212 ik0n 2 \u03b8 (x) \u03b70, b = 0, c = ik0\u03b70, and thus we should solve the equation z \u2212 a a z + acz = 0, which becomes z \u2212 \u03c7/d cos2 \u03b8i + (\u03c7/d)x z + k 2 0 cos2 \u03b8i + x \u03c7 d z = 0 (10.97) (10.98) (10.99) (10.100) (10.101) after putting in the explicit forms of the a and c coefficients. To solve this equation, we will compare it to a known differential equation. The Airy equation is and two independent solutions of the Airy equation are the Airy functions4 y \u2212 xy = 0, Ai(x) := Bi(x) := 1 \u03c0 1 \u03c0 \u221e dt cos 0 \u221e dt 0 t3 3 exp \u2212 + xt t3 3 + xt + sin + xt t3 3 (10.102) (10.103) 3Daniel H. Raguin and G. Michael Morris, op. cit., Eqs. (5) and (6). 4Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions (", "Dover, 1965), Section 10.4. Available online at http://people.math.sfu.ca/~cbm/aands/page_446.htm. The properties of the Airy function that we will use, including asymptotic forms, can be found there. xexo=o0xo=odeo(x)o=oe0o(1o+ocx/d)eio=oe0eso=oe0o(1o+oc) 10.4 Gradient-Index Layers 191 of the first (Ai) and second (Bi) kind. These functions are plotted below, where we will be interested mainly in the oscillatory part at negative x. Taking the Airy equation and dividing through by x, 1 x y \u2212 y = 0, and then differentiation the equation, we find Changing variables to z = y, we obtain the equation 1 x y \u2212 1 x2 y \u2212 y = 0. z(x) \u2212 1 x z(x) \u2212 xz(x) = 0, (10.104) (10.105) (10.106) to which the solutions are the derivatives Ai(x) and Bi(x). Now changing the independent variable x \u2212\u2192 \u2212(k 2 0 \u03c7/d)1/3x, we obtain the equation to which the solutions are Ai[\u2212(k 2 Eq. (10.101) by letting 0 \u03c7/d)1/3x]. Finally, we can change this into z + z \u2212 1 x 0 \u03c7/d)1/3x] and Bi[\u2212(k 2 k 2 0 \u03c7 d xz = 0, x \u2212\u2192 x + d \u03c7 cos2 \u03b8i, and thus we may write the general solution to Eq. (10.101) as z(x) = \u02dc\u03b2 Ai(\u2212\u02dck\u02dcx) + \u03b2Bi(\u2212\u02dck\u02dcx), where \u03b2 and \u02dc\u03b2 are undetermined constants, and we are using the shorthand, shifted coordinate and scaled wave number \u02dcx := dn 2 \u03b8 (x) \u03c7 = x + d \u03c7 cos2 \u03b8i \u02dck := k 2 0 \u03c7 d 1/3. (10.107) (10.108) (10.109) (10.110) (10.111) x-10100-55Aio(x), Bio(x)1-10-0.50.5Aio(x)Bio(x) 192 Chapter 10. Thin Films Then using Eq. (10.80), the solution to the Ricatti equation (10.77) in this case is \u02dc\u03b7(x) = \u2212 \u03b70d ik0\u03c7\u02dcx \u02dck Ai(\u2212\u02dck\u02dcx) + \u03b2Bi(\u2212\u02dck\u02dcx) Ai(\u2212\u02dck\u02dcx) + \u03b2Bi(\u2212\u02dck\u02dcx) Ai(\u2212\u02dck\u02dcx) + \u03b2Bi(\u2212\u02dck\u02dcx) Ai(\u2212\u02dck\u02dcx) + \u03b2Bi(\u2212\u02dck\u02dcx) Ai(\u2212\u02dck\u02dcx) + \u03b2Bi(\u2212\u02dck\u02dcx) Ai(\u2212\u02dck\u02dcx) + \u03b2Bi(\u2212\u02dck\u02dcx), = k0\u03b70 i\u02dck3 \u02dcx \u02dck2 \u02dcx = \u2212 ik0\u03b70 \u02dck where the Airy equation (10.102) implies the second derivatives Ai(x) = xAi(x), Bi(x) = xBi(x). The next step is to fix the undetermined constant \u03b2 using the boundary condition (10.75) Letting and we find Then defining \u02dc\u03b7(d) = 1 \u03b1s. \u02dc\u03b1 := \u02dck k0\u03b70 \u02dcd := \u02dcx|x=d = d 1 + cos2 \u03b8i \u03c7, \u03b2 = \u2212 \u03b1sAi(\u2212\u02dck \u02dcd) \u2212 i\u02dc\u03b1Ai(\u2212\u02dck \u02dcd) \u03b1sBi(\u2212\u02dck \u02dcd) \u2212 i\u02dc\u03b1Bi(\u2212\u02dck \u02dcd). \u02dcd0 := \u02dcx|x=0 = d cos2 \u03b8i \u03c7, the effective impedance at the incident boundary is \u02dc\u03b7(0) = 1 i\u02dc\u03b1 Ai(\u2212\u02dck \u02dcd0) + \u03b2Bi(\u2212\u02dck \u02dcd0) Ai(\u2212\u02dck \u02dcd0) + \u03b2Bi(\u2212\u02dck \u02dcd0). Finally, the reflection coefficient (10.78) becomes r = \u03b1iAi(\u2212\u02dck \u02dcd0) + \u03b1i\u03b2Bi(\u2212\u02dck \u02dcd0) \u2212 i\u02dc\u03b1", "Ai(\u2212\u02dck \u02dcd0) \u2212 i\u02dc\u03b1\u03b2Bi(\u2212\u02dck \u02dcd0) \u03b1iAi(\u2212\u02dck \u02dcd0) + \u03b1i\u03b2Bi(\u2212\u02dck \u02dcd0) + i\u02dc\u03b1Ai(\u2212\u02dck \u02dcd0) + i\u02dc\u03b1\u03b2Bi(\u2212\u02dck \u02dcd0) = \u03b1iAi(\u2212\u02dck \u02dcd0) \u2212 i\u02dc\u03b1Ai(\u2212\u02dck \u02dcd0) + \u03b2 \u03b1iAi(\u2212\u02dck \u02dcd0) + i\u02dc\u03b1Ai(\u2212\u02dck \u02dcd0) + \u03b2 \u03b1iBi(\u2212\u02dck \u02dcd0) \u2212 i\u02dc\u03b1Bi(\u2212\u02dck \u02dcd0) \u03b1iBi(\u2212\u02dck \u02dcd0) + i\u02dc\u03b1Bi(\u2212\u02dck \u02dcd0). (10.112) (10.113) (10.114) (10.115) (10.116) (10.117) (10.118) (10.119) (10.120) Then inserting Eq. (10.117), we finally find, condensing our notation yet further, the reflection coefficient 10.4 Gradient-Index Layers 193 for S-polarization rS = A\u2212 A+ 0 B\u2212 \u2212 B\u2212 0 B\u2212 \u2212 B+ 0 A\u2212 0 A\u2212 A\u00b1 B\u00b1 0 := \u03b1iAi(\u2212\u02dck \u02dcd0) \u00b1 i\u02dc\u03b1Ai(\u2212\u02dck \u02dcd0), 0 := \u03b1iBi(\u2212\u02dck \u02dcd0) \u00b1 i\u02dc\u03b1Bi(\u2212\u02dck \u02dcd0), cos2 \u03b8i \u02dcd0 := \u02dcd := d \u03c7 1/3 1 +, \u02dck := k 2 0 \u03c7 d \u03b1i := cos \u03b8i \u03b70 = 1 \u03b70 \u02dck k0\u03b70, \u02dc\u03b1 := \u02dcd0\u03c7 d, A\u2212 := \u03b1sAi(\u2212\u02dck \u02dcd) \u2212 i\u02dc\u03b1Ai(\u2212\u02dck \u02dcd) B\u2212 := \u03b1sBi(\u2212\u02dck \u02dcd) \u2212 i\u02dc\u03b1Bi(\u2212\u02dck \u02dcd) d cos2 \u03b8i \u03c7 \u221a 1 + \u03c7 \u03b70 cos \u03b8t = 1 \u03b70 \u03c7 + cos2 \u03b8i = \u02dcd\u03c7 d, 1 \u03b70 \u03b1s := where we used \u03b2 = \u2212A\u2212/B\u2212 in our condensed notation. (reflection coefficient, S-polarization, linear GRIN film) (10.121) 10.4.7 Asymptotic Forms 10.4.7.1 Sharp-Boundary Limit As a simple check on our calculation, we can compute the reflection coefficient in the limit d = 0, corresponding to a hard boundary between vacuum and the dielectric, without any smooth index gradient. First, analyzing the numerator of the reflection coefficient (10.121), note that terms proportional to \u03b1i\u03b1s and \u02dc\u03b12 will cancel, leaving A\u2212 0 B\u2212 \u2212 B\u2212 0 A\u2212 = \u2212i\u03b1i \u02dc\u03b1Ai(0) Bi(0) + i\u03b1s \u02dc\u03b1Ai(0) Bi(0) \u2212 i\u03b1s \u02dc\u03b1Ai(0) Bi(0) + i\u03b1i \u02dc\u03b1Ai(0) Bi(0) = i Ai(0) Bi(0) \u2212 Ai(0) Bi(0)\u03b1i \u02dc\u03b1 + i Ai(0) Bi(0) \u2212 Ai(0) Bi(0)\u03b1s \u02dc\u03b1 (10.122) = a0(\u03b1i \u2212 \u03b1s), where we have defined a0 := i\u02dc\u03b1 Ai(0) Bi(0) \u2212 Ai(0) Bi(0) = \u2212i\u02dc\u03b1 Ai(0) Bi(0) \u2212 \u03be0Ai(0) Bi(0) = \u2212 \u221a 2i\u02dc\u03b1 3\u0393(1/3)\u0393(2/3), (10.123) and the last two equalities here follow from the expressions Ai(0) = 1 32/3\u0393(2/3), Ai(0) = \u2212 1 31/3\u0393(1/3), Bi(0) = 1 31/6\u0393(2/3), Bi(0) = 31/6 \u0393(1/3). (10.124) The exact form of a0 is unimportant, as it will cancel here anyway. The denominator of the reflection coefficient (10.121) is the same, except A\u2212 0, which changes the signs of", " all terms involving \u03b1s. Thus, the reflection coefficient is 0 are replaced by A+ 0 and B\u2212 0 and B+ rS = \u03b1i \u2212 \u03b1s \u03b1i + \u03b1s = cos \u03b8i \u2212 cos \u03b8i + 1 + \u03c7 cos \u03b8t 1 + \u03c7 cos \u03b8t, (10.125) \u221a \u221a which is the correct Fresnel coefficient for S polarization incident from vacuum onto a dielectric. 10.4.7.2 Small-Reflection Limit In the limit of large d, which smooths the transition between the vacuum and dielectric, we expect a small reflection coefficient. In this limit, we can use the asymptotic forms for real x for the Airy functions Ai(\u2212x) \u223c \u221a 1 \u03c0x1/4 sin 2x3/2 3 +, \u03c0 4 Bi(\u2212x) \u223c \u221a 1 \u03c0x1/4 cos 2x3/2 3 +, \u03c0 4 (10.126) 194 and the derivatives Chapter 10. Thin Films Ai(\u2212x) \u223c \u2212 x1/4 \u221a \u03c0 2x3/2 3 +, \u03c0 4 cos Bi(\u2212x) \u223c x1/4 \u221a \u03c0 2x3/2 3 +. \u03c0 4 sin (10.127) Then examining the components of the refection coefficient (10.121), we note that both terms in each 0, A\u2212, and B\u2212 scale as d\u22121/6, if we note that \u02dck \u02dcd and \u02dck \u02dcd0 scale as d2/3, and \u02dc\u03b1 scales as coefficient A\u00b1 d\u22121/3. 0, B\u00b1 In the reflection coefficient (10.121), rS = A\u2212 A+ 0 B\u2212 \u2212 B\u2212 0 B\u2212 \u2212 B+ 0 A\u2212 0 A\u2212, (10.128) we will begin by evaluating the first term in the numerator: A\u2212 \u03b1iAi(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd) + \u03b1sAi(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd) 0 B\u2212 = \u03b1i\u03b1sAi(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd) \u2212 i\u02dc\u03b1 \u2212 \u02dc\u03b12Ai(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd). (10.129) Comparing the first and last terms here in the limit of large d, we can see from the asymptotic forms (10.126) and (10.127) that the dependence on trigonometric functions will be the same. To compare them further, we can see that \u03b1i\u03b1sAi(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd) \u02dc\u03b12Ai(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd) = \u03b1i\u03b1s \u02dcd0 \u02dc\u03b12\u02dck = \u02dcd \u03c7 0 \u02dc\u03b12\u02dckd \u03b7 2 = k 2 0 \u03c7 \u02dck3d = 1, (10.130) where we used Eqs. (10.121) to reduce the ratio. Thus, the first and last terms in Eq. (10.129) cancel in the large-d limit, and we have A\u2212 0 B\u2212 = \u2212i\u02dc\u03b1 \u03b1iAi(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd) + \u03b1sAi(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd). (10.131) Proceeding to the second term in the numerator of Eq. (10.128), we see that this is the same as the first term (10.131) if we interchange d and d0, and then \u03b1i with \u03b1s. Then from the last two equations in (10.121), we have and writing out the numerator of (10.121), \u03b1s \u03b1i = \u02dcd \u02dcd0, A\u2212 0 B\u2212 \u2212 B\u2212 0 A\u2212 = \u2212i\u03b1i \u02dc\u03b1 Ai(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd) + Ai(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd) \u02dcd \u02dcd0 \u2212 i\u03b1i \u02dc\u03b1 \u02dcd \u02dcd0 Ai(\u2212\u02dck \u02dcd)Bi(\u2212\u02dck \u02dcd0) + Ai(\u2212\u02dck \u02dcd)Bi(\u2212\u02dck \u02dcd0) (10.132) (10.133) we can put in the asymptotic forms (10.126) and (10.", "127) to see that this vanishes. In the denominator, we have different signs, and so the first cancellation of the first and last terms in Eq. (10.129) does not occur, because the sign of the \u02dc\u03b12 term changes (as does the sign of the \u02dc\u03b1\u03b1s term), and so we have A+ 0 B\u2212 = 2\u03b1i\u03b1sAi(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd) \u2212 i\u02dc\u03b1 \u03b1iAi(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd) \u2212 \u03b1sAi(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd). (10.134) Again, the other term in the denominator has the same form, but with \u02dcd0 and \u02dcd interchanged, and with \u03b1i \u2212\u2192 \u2212\u03b1s and \u03b1s \u2212\u2192 \u2212\u03b1i: A\u2212 0 B+ = 2\u03b1i\u03b1sAi(\u2212\u02dck \u02dcd)Bi(\u2212\u02dck \u02dcd0) + i\u02dc\u03b1 \u03b1sAi(\u2212\u02dck \u02dcd)Bi(\u2212\u02dck \u02dcd0) \u2212 \u03b1iAi(\u2212\u02dck \u02dcd)Bi(\u2212\u02dck \u02dcd0). (10.135) 10.4 Gradient-Index Layers 195 From the asymptotic forms (10.126) and (10.127), along similar lines to Eq. (10.133), we see that the imaginary terms are equal in magnitude but opposite in sign, so that the denominator becomes A+ 0 B\u2212 \u2212 A\u2212 0 B+ = 2\u03b1i\u03b1s Ai(\u2212\u02dck \u02dcd0)Bi(\u2212\u02dck \u02dcd) + Ai(\u2212\u02dck \u02dcd)Bi(\u2212\u02dck \u02dcd0) + 2i\u02dc\u03b1 \u03b1sAi(\u2212\u02dck \u02dcd)Bi(\u2212\u02dck \u02dcd0) \u2212 \u03b1iAi(\u2212\u02dck \u02dcd)Bi(\u2212\u02dck \u02dcd0) \u02dcd 0 \u2212 \u02dcd 3/2 \u02dcd0 \u02dck3/2 \u02dcd 3/2 2i\u02dc\u03b1\u03b1i \u03c0 2 3 \u2212 1/4 = \u03c0 2\u03b1i\u03b1s 1/4 sin \u02dck2 \u02dcd0 \u02dcd 2 3 cos \u02dck3/2 \u02dcd 3/2 0 \u2212 \u02dcd 3/2, (10.136) which is nonvanishing. Since we have a vanishing numerator and a nonvanishing denominator, we have established that rS \u2212\u2192 0 as d \u2212\u2192 \u221e. Note that the denominator scales as d\u22121/3 for large d. The terms of the same order ended up canceling in the numerator. The next-order terms in the the asymptotic forms (10.126) and (10.127) are smaller by factors that scale as x\u22123/2 for large x. Since x here refers to \u02dck \u02dcd and \u02dck \u02dcd0, both of which scale as d2/3 the next-order part of the numerator should scale as 1/d compared to the denominator. Thus, for large d, the reflection coefficient scales as 1/d for large d, or the reflectance scales as R \u223c 1/d2. To illustrate all this, below is plotted the reflectance at normal incidence for a vacuum\u2013glass interface (assuming n = 1.52 for glass), connected by a linear permittivity gradient. Here, we can see that the reflectance starts at 4.3% at d = 0, which is what we expect for a vacuum\u2013glass interface. We also see oscillations in the reflectance, and in the large-d regime, the envelope of the oscillations shows the expected 1/d2 scaling. To gain a bit more insight into the oscillations, note that asymptotically, the Airy functions conspire to form trigonometric functions as in Eq. (10.136), with argument \u02dck3/2 \u02dcd 3/2 \u2212 \u02dcd 3//2 d3/2 3/2 1 + 1 \u03c7 \u2212 d \u03c7 3/2 = 2k0d 3\u03c7 (1 + \u03c7)3/2 \u2212 1, (10.137) where to keep things simple we are considering only normal incidence. This has the form of the optical path length of the GRIN film, d 0 dx n(x) = 1 + dx d 0 \u03c7 d x = 2d 3\u03c7 (1 + \u03c7)3/2 \u2212 1, (10.138) multiplied by k0, and so the", " oscillations here are similar to what we expect in for a thin, homogeneous film, provided we interpret the result in terms of the more general path length. GRIN film thickness d/l0051234reflectance10-110-510-210-310-4 196 10.4.8 P-Polarization Chapter 10. Thin Films So far we have calculated the reflection coefficient in the case of S-polarization. But recall that we can adapt the calculation to P-polarization by moving the angle cosines from the numerators to the denominators of the \u03b1 coefficients. Beginning with Eq. (10.61), we use along with \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 Fj \u2248 1 \u2212i\u03b4j \u03b1j \u2212i\u03b1j\u03b4j 1 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb. \u03b1j = nj \u03b70 cos \u03b8tj \u03b4j = k0nj(\u03b4x) cos \u03b8tj to write the alternate transfer matrix \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 Fj \u2248 1 \u2212i\u03b70k0 cos2 \u03b8tj \u03b4x \u2212i k0 \u03b70 n 2 j \u03b4x i\u03b70k0 n 2 i sin2 \u03b8i j \u2212 n 2 n 2 j \u03b4x \u2212i k0 \u03b70 n 2 j \u03b4x 1 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb. Passing over to the continuous limit, the differential system for the fields (10.70) becomes d dx E(x) H(x) = ik0 \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 0 n2(x) \u03b70 \u03b70n 2 \u03b8 (x) n2(x) 0 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb E(x) H(x), (10.139) (10.140) (10.141) (10.142) (10.143) where the definition (10.69) of n\u03b8(x) still holds. Again E and H refer only to the components of the fields parallel to the film surface (now Ey and Hz in our coordinate system, for P-polarization). Written as two coupled equations, we have dE dx = ik0\u03b70 n 2 \u03b8 (x) n2(x) H(x) ik0n2(x) \u03b70 Then proceeding with the effective impedance \u02dc\u03b7(x) defined in Eq. (10.73), we can eliminate the derivatives in Eq. (10.76) to find in place of Eq. (10.77) dH dx E(x). = (10.144) d\u02dc\u03b7 dx = ik0\u03b70 n 2 \u03b8 (x) n2(x) \u2212 ik0n2(x) \u03b70 \u02dc\u03b72. (10.145) That is, part of the space dependence n 2 \u03b8 (x)/n2(x) of the variable index moves from the nonlinear term to the constant term. The procedure is then the same as before for calculating the reflection coefficient: first, use Eq. (10.75) (propagation equation, P-polarization) E(+) b H (+) b as the initial condition for \u02dc\u03b7(d) (with suitably redefined \u03b1s), then use this differential equation to integrate backwards to \u02dc\u03b7(0), and finally, use Eq. (10.74), (10.146) \u02dc\u03b7(d) = 1 \u03b1s = r = \u03b1i \u02dc\u03b7(0) \u2212 1 \u03b1i \u02dc\u03b7(0) + 1, (10.147) 10.4 Gradient-Index Layers 197 for the reflection coefficient, with the modified \u03b1i. 10.4.8.1 Example: Linear Permittivity Gradient Now we return to the example of the linear permittivity gradient from Section 10.4.6, and work out the reflection coefficient for P-polarization. The Ricatti coefficients in Eq. (10.79) in this case become a = \u2212 ik0n2(x) \u03b70, b = 0, c = ik0\u03b70 n 2 \u03b8 (x) n2(x), where Eq. (10.98) still holds for n\u03b8(x) in this case. We should then solve the equation z \u2212 a a z + acz = 0, which becomes z \u2212 \u03c7/d 1 + (\u03c7/d)x z + k 2 0 cos2 \ufffd", "\ufffdi + x \u03c7 d z = 0, (10.148) (10.149) (10.150) which is the same as Eq. (10.101), except that the cosine in the coefficient of z is no longer there. Unfortunately, the variable-transformation tricks we used in the S-polarization case do not carry over here, and this equation does not have an obvious solution\u2014except, that is, at normal incidence, when this case is equivalent to the S-polarization case. Nevertheless, it is possible to solve Eq. (10.145) numerically to obtain the reflection coefficient. Doing this for the vacuum\u2013glass (n = 1.52) interface for normal incidence, and 45\u25e6 incidence for both polarizations leads to the plot below. In the d = 0 limit, S-polarization reflects more than does P, because 45\u25e6 is not far from Brewster\u2019s angle. This behavior persists for all d, with similar oscillations in each case (but with longer period in the angledincidence cases). Note, however, that while something analogous to the polarizing effect at Brewster\u2019s angle, the suppression is not as complete as for a simple interface, as shown in the plot below, which shows both polarizations for three film thicknesses (with the d = 0 case corresponding to the usual vacuum\u2013glass interface). GRIN film thickness d/l0051234reflectance10-110-610-210-310-410-5normal incidence45\u00c4, S45\u00c4, P 198 Chapter 10. Thin Films However, notice that the reflectance is suppressed over a wide range of angles by the GRIN film. angle of incidence0\u00c490\u00c460\u00c430\u00c4reflectance10010-710-110-210-310-410-510-6S, do=o0P, do=o0S, do=ol0P, do=ol0S, do=o10l0P, do=o10l0 10.5 Exercises 10.5 Exercises Problem 10.1 199 Consider a planar glass window of refractive index n and thickness d, used as an etalon for light of wavelength \u03bb incident at angle \u03b8 with respect to the window surface. Derive an expression for the changes in resonance wavelength \u03b4\u03bb and frequency \u03b4\u03bd, given a small change \u03b4\u03b8i in the incident angle. (That is, linearize the solution in \u03b4\u03b8i.) Problem 10.2 Consider a variable beam splitter consisting of two 45\u25e6-45\u25e6-90\u25e6 prisms5 of refractive index n, separated by an air (n = 1) gap \u2206 and operated as shown with light of (vacuum) wavelength \u03bb0. Assume that n is large enough that frustrated internal reflection occurs at the gap at the incident angle of 45\u25e6. (a) Write down an expression for the intensity reflection coefficient due to the air gap (ignore reflections from all other prism surfaces). Everything you write down should be defined in terms of parameters given in this problem, and make sure to treat both possible polarizations. (b) Obtain the limits of your expression for small and large \u2206, and comment on whether or not they make sense. (c) Plot the reflectance of the beamsplitter as a function of \u2206/\u03bb0, assuming n = 1.52 for glass. Also include in your plot for comparison the simple estimate 1 \u2212 e\u22122\u2206/\u03b4, where \u03b4 is the skin depth for internal reflection, which is based on the normalized intensity of the evanescent wave where it enters the second prism (hence, this fraction of the intensity transmits through the second prism). Problem 10.3 Light of wavelength \u03bb0 is incident from air (n = 1) onto a single dielectric thin film (of index nf, and thickness \u03bb/4, where \u03bb is the wavelength inside the film), which covers a glass substrate (index ng). (a) Write down an expression for the film reflectance, assuming the light is at normal incidence, using the results of the reflection-summation formalism. (b) Derive the value of nf that makes a perfect anti-reflection coating. Problem 10.4 Repeat the derivation of the thin-film matrix formalism for P-polarization, showing that the transfermatrix formalism is the same as for the S-polarization case that we already derived, provided the \u03b1j 5for a realization of this beam splitter, see D. Bertani, M. Cetica, and R. Polloni, \u2018\u2018A simple variable-ratio beam splitter for holography,\u2019\u2019 Journal of Physics E: Scientific Instruments 16, 6", "02 (1983) (doi: 10.1088/0022-3735/16/7/007). For application to Q-switched laser cavities, see Ian N. Court and Frederick K. von Willisen, \u2018\u2018Frustrated Total Internal Reflection and Application of Its Principle to Laser Cavity Design,\u2019\u2019 Applied Optics 3, 719 (1964) (doi: dx.doi.org/10.1364/AO.3.000719). vary DDno=o1nfngE0i(+) 200 Chapter 10. Thin Films coefficients are suitably redefined. Compute the film reflection and transmission coefficients for this case. Problem 10.5 Discuss the advantage of a two-layer antireflection coating over a single-layer coating. Assume that you are only concerned with suppressing reflection at one incident wavelength and normal incidence, and distinguish practical vs. in principle performance of the coating in your answer. (Two or three sentences should suffice here.) Problem 10.6 Plot the intensity reflection coefficients as a function of angle for light incident from air onto crown glass with a single-layer antireflection coating of MgF2. (Make plots for both polarizations.) Assume n = 1.38 for MgF2, and a coating thickness of \u03bb/4 (where \u03bb is the wavelength inside the coating!), where the design wavelength is \u03bb = 550 nm (in vacuum). How thick is the coating in nm? Problem 10.7 Plot the intensity reflection coefficients as a function of \u03bb0 for light incident from air onto crown glass with a double-layer antireflection coating. The thin-film stack consists of a \u03bb/4 layer of ZrO2 (n = 2.1) directly on top of the crown glass, followed by a \u03bb/4 layer of CeF3 (n = 1.65). Assume a design wavelength of 550 nm (in vacuum) and extend the plot over the visible spectrum (400-700 nm). Consider only the case of normal incidence. How thick are the layers in nm? Problem 10.8 Consider a \u2018\u2018slab\u2019\u2019 (film) of glass of index n surrounded on both sides by vacuum. Use the transfermatrix method to derive the reflection coefficient for the film, assuming it is very thin (d \u03bb0, \u03bb), Taylor expanding to linear order in the film thickness d (at every stage, to keep it simple), and at normal incidence. Chapter 11 Fourier Analysis II: Convolution 11.1 Motion Blurring in Photography To motivate the idea behind a convolution, let\u2019s consider the common situation of photographing a moving object. As an example, we\u2019ll choose a white box on a black background as the object, as shown below, to the left. As far as the camera is concerned, this is only the image that you will record in the limit of a very short shutter time. For longer shutter times, the image will be blurred, as shown below to the right. Let\u2019s represent this more mathematically, and consider the one-dimensional intensity function I(x), representing the horizontal cross section through the middle of the figure. Again, this object is moving, say to the right. During the shutter exposure, we must sum over the image in each location it occupies while it is moving along. We can represent this as the exposure-summed intensity Iexp(x) = d/2 \u2212d/2 I(x \u2212 x) dx. (11.1) xoo'Io(xoo') 202 Chapter 11. Fourier Analysis II: Convolution Here, we assume that the object moves a distance d = vtexp at speed v during the exposure time texp. In the integral, we add the intensity function centered at each x within this range, and for concreteness, we take the range of the image\u2019s motion to be from \u2212d/2 to d/2. The result is a blurred object, which would look something like the plot below, if the blurring effect is not too heavy. Suppose that we introduce the box function of width d: h(x) = 1, 0, |x| \u2264 d/2, |x| > d/2. Then we can rewrite the exposure integral as Iexp(x) = \u221e I(x \u2212 x) h(x) dx. \u2212\u221e (11.2) (11.3) Here, h(x) is unity to represent the locations of the object where the shutter is open, and it is zero when the shutter is closed. It need not be the case that h(x) is only unity or zero; for example, h(x) could ramp smoothly from zero to one as the shutter smoothly opens, or h(x) could be more complicated if the object\u2019", "s velocity is nonuniform, since it spends different amounts of time in different locations. Considering this integral further, let\u2019s change variables by letting x \u2212\u2192 x + x and then x \u2212\u2192 \u2212x: Iexp(x) = = \u221e I(\u2212x) h(x + x) dx \u2212\u221e \u221e I(x) h(x \u2212 x) dx. \u2212\u221e (11.4) What this is saying is that the blurring effect is equivalent when we have a stationary object, but a moving camera, represented by a moving function h(x \u2212 x) in the integral (that is, we have just exchanged the dependences of I and h). This integral is precisely what we mean by a convolution. 11.2 Convolution More abstractly, the convolution is a mathematical operation that maps a pair of functions to a function. Roughly speaking, the convolution is an operation that \u2018\u2018smears\u2019\u2019 one function with another. Formally, we can define the convolution f \u2217 g of two functions f and g as the function given by the integral (f \u2217 g)(x) := \u221e \u2212\u221e f (x) g(x \u2212 x) dx. (11.5) (convolution of functions) The interpretation of this integral is as follows: as a function of x, we displace (scan the location) of g(x). It is most intuitive to consider the (common) case of a localized function g(x), which we will refer to as the convolution kernel. Then the value of the convolution at location x is the \u2018\u2018amount\u2019\u2019 of f (x) that \u2018\u2018passes through\u2019\u2019 g(x) when g is displaced by an amount x. Notice that this is proportional to the projection of f (x) onto g(x \u2212 x). As an example, consider the perfectly localized convolution kernel g(x) = \u03b4(x). The convolution with a function f (x) is (f \u2217 \u03b4)(x) = \u221e \u2212\u221e f (x) \u03b4(x \u2212 x) dx = f (x). (11.6) xoo'Iexpo(xoo') 11.2 Convolution 203 The effect of convolution with a delta function is simply to do nothing; there is no \u2018\u2018smearing\u2019\u2019 due to a perfectly localized function. Typically, we will use centered kernels; the effect of a displaced kernel is simply to displace the convolution by the same amount. For example, if g(x) = \u03b4(x \u2212 x0), then (f \u2217 g)(x) = \u221e \u2212\u221e f (x) \u03b4(x \u2212 x0 \u2212 x) dx = f (x \u2212 x0), and thus, convolution with a displaced delta function is just a displacement of the function. 11.2.1 Example: Convolution of Box Functions As a slightly more complicated example, consider the convolution of box functions, both given by f (x) = g(x) = 1, 0, |x| \u2264 1/2, |x| > 1/2. (11.7) (11.8) (11.9) The convolution consists of displacing g(x) by x, multiplying the functions together, and integrating. For this simple case (box functions of unit height), the convolution (product) just turns out to be the area where the two functions overlap. When the displacement is large, |x| > 1, the boxes don\u2019t overlap at all, so the convolution is zero. Otherwise, the overlap area varies linearly with the displacement, so the convolution is a triangle function, peaking at unity. (Note that f \u2217 f in the figure is the same as f \u2217 g for this special case of f = g.) We see now the \u2018\u2018smoothing\u2019\u2019 or \u2018\u2018blurring\u2019\u2019 effect of the convolution. The original functions were discontinuous, but the convolution is continuous. The convolution is also wider than the original functions. As we will see, continued convolutions will make the distribution look Gaussian. 11.2.2 Example: Photographic Blurring Let\u2019s return to the original photographic example. A photograph is a two-dimensional image, but the above convolution (and convolution theorem) generalize readily to two dimensions (just do a convolution in both directions, or equivalently, do a two-dimensional Fourier transform, multiply, and invert the transform). Here, we will start off with the original photo on the left.1 1RIP Charliepoo. xoo'xoo'o=o01/2-1/2xfo(x", "oo')go(xo-oxoo')(fo*og)o(x)xxo=o01/2-1/21-1fo(x)(fo*ofo)o(x) 204 Chapter 11. Fourier Analysis II: Convolution \u2217 = In the middle, we have the convolution kernel. To be precise, this is a 2048 \u00d7 2048 image, which is all zero except for a unit-height region in the center, 16 pixels high (this is intended to be as small as possible, but is this large so that the bright region shows up in this figure) and 256 pixels wide. Think of this as essentially a delta function in the vertical direction, and a square pulse in the horizontal direction. There is essentially no blurring in the vertical direction in the convolved image, but significant blurring (on the scale of the width of the white line) in the image. This models motion of the camera during an exposure along the white line; more complicated trajectories of the camera would correspond to a more complicated white squiggle in the kernel image. Note the dark borders in the convolution image occur because the image is assumed to be zero (black) outside its borders. 11.3 Convolution Theorem The convolution theorem gives an easy way to evaluate the convolution integral in Eq. (11.5), both in an intuitive and a computational sense. The convolution theorem states that the Fourier transform of the convolution is the product of the Fourier transforms of the individual functions: F [f \u2217 g] = F [f ]F [g]. (11.10) (convolution theorem) Before working through the proof, we need to review the Fourier transform, and in particular the alternate convention for Fourier transforms between position and spatial frequency (x-k) vs. time and frequency (t-\u03c9). 11.3.1 Spatial Fourier Transforms Previously, in Chapter 3, we\u2019ve been dealing with time-frequency Fourier transforms. However, we will soon be dealing with Fourier transforms between space and spatial frequency. So before proceeding with the convolution theorem, we\u2019ll have a short digression to introduce the different convention. Recall that the formulae for the Fourier transform and the inverse transform are f (t) = 1 2\u03c0 \u221e \u2212\u221e \u02dcf (\u03c9) e\u2212i\u03c9t d\u03c9, \u02dcf (\u03c9) = \u221e \u2212\u221e f (t) ei\u03c9t dt. (11.11) The basis function was exp(\u2212i\u03c9t), and this determines the form of the transform integrals. Recall that the right-going plane wave has the form exp[i(kx \u2212 \u03c9t)], so the basis harmonic function for the spatial part is exp(ikx). So to get the spatial convention from the temporal convention, we must make the replacements (Fourier transform\u2013\u03c9-t convention) t \u2212\u2192 x, \u03c9 \u2212\u2192 k, i \u2212\u2192 \u2212i. (11.12) Thus, the transform equations for spatial functions are f (x) = 1 2\u03c0 \u221e \u2212\u221e \u02dcf (k) eikx dk, \u02dcf (k) = \u221e \u2212\u221e f (x) e\u2212ikx dx, where again \u02dcf (k) = F [f (x)] = F [f ](k). (Fourier transform\u2013k-x convention) (11.13) 11.3 Convolution Theorem 205 11.3.2 Proof To prove the convolution theorem, we\u2019ll just compute the explicit form of F [f \u2217 g]. This will be very much a physicist\u2019s proof, not a mathematician\u2019s proof, in that we\u2019ll just assume the functions are nice enough that all the integrals simply exist. F [f \u2217 g] = F \u221e dx f (x) g(x \u2212 x) \u2212\u221e \u221e dx \u221e dx f (x) g(x \u2212 x) e\u2212ikx = = = \u2212\u221e \u221e \u2212\u221e \u221e \u2212\u221e \u221e dx f (x) e\u2212ikx dx dx f (x) e\u2212ikx \u221e \u2212\u221e g(x \u2212 x) e\u2212ik(x\u2212x) dx g(x \u2212 x) e\u2212ik(x\u2212x). Letting x \u2212\u2192 x + x in the second integral, \u2212\u221e \u2212\u221e F [f \u2217 g] = \u221e \u2212\u221e f (x) e\u2212ikx dx \u221e \u2212\u221e g(x) e\u2212ikx dx = F [f ]F [g]. (11.14) (11.15) Thus, to convolve two functions, just follow this recipe: Fourier transform both functions, multiply them", " together, then compute the inverse Fourier transform. That is, f \u2217 g = F \u22121 {F [f ]F [g]}, (11.16) (convolution-theorem recipe) if we write this procedure out mathematically. 11.3.3 Example: Convolution of Two Gaussians Since it\u2019s easy to compute the Fourier transform of Gaussian distributions, let\u2019s use the convolution theorem to convolve two Gaussians. Let\u2019s write the two functions as f (x) = Ae\u2212x2/\u03b12, g(x) = Be\u2212x2/\u03b22. We computed the Fourier transform of a Gaussian in Chapter 3, so we can write \u221a \u221a F [f ](k) = \u02dcf (k) = A\u03b1 \u03c0e\u2212\u03b12k2/4, F [g](k) = \u02dcg(k) = B\u03b2 \u03c0e\u2212\u03b22k2/4. Then the product of the Fourier transforms is (F [f ]F [g]) (k) = AB\u03b1\u03b2\u03c0e\u2212(\u03b12+\u03b22)k2/4. (11.17) (11.18) (11.19) Finally, we invert the Fourier transform to obtain the convolution. To do this, we can adapt the formulae \u03c0/\u03b12 + \u03b22. Then we obtain for f (x) and \u02dcf (k) by first letting \u03b1 \u2212\u2192 \u03b12 + \u03b22 and then A \u2212\u2192 AB\u03b1\u03b2 the same Fourier transform above, so the inversion yields AB\u03b1\u03b2\u03c0e\u2212(\u03b12+\u03b22)k2/4 (f \u2217 g)(x) = F \u22121 \u221a \u221a = \u2212. (11.20) exp AB\u03b1\u03b2 \u03c0 \u03b12 + \u03b22 x2 \u03b12 + \u03b22 Recall that the standard (normalized) form of the Gaussian is \u221a 1 2\u03c0 \u03c3 exp \u2212 (x \u2212 \u00b5)2 2\u03c32, (11.21) 206 Chapter 11. Fourier Analysis II: Convolution \u221a where the \u00b5 is the mean and \u03c3 is the standard deviation (\u03c32 is the variance), which we\u2019ll come back to shortly. The standard deviation is one measure of the width of a Gaussian function. Note that f (x) has standard (\u03b12 + \u03b22)/2. deviation \u03b1/ Thus, the convolution of Gaussians is still Gaussian, but the blurring effect of the convolution makes the convolved Gaussian wider than the original functions. Indeed, the convolved width is the width of the originals summed in quadrature. 2, and (f \u2217 g)(x) has standard deviation 2, g(x) has standard deviation \u03b2/ \u221a 11.4 Application: Error Analysis One important application of the convolution is in probability theory, which carries over to analysis of let X1 and statistical error of experimental data in physics. The mathematical problem is as follows: X2 be independent random variables with probability density functions f1(x) and f2(x), respectively. That is, the probability that X1,2 is between x and x + dx is f1,2(x) dx. These could represent the errors in measurements of quantities x1 and x2, respectively, so that the measurement results are x1 + X1 and x2 + X2. Then we can ask, what is the probability density of X1 + X2, which corresponds to the error of the combined (derived) measurement of x1 + x + 2? To answer this, we can note that X1 + X2 = x for any pair of values of X1 and X2 that happen to add up to x. But then we must sum over all such pairs. The probability that both X1 and X2 will both have particular probabilities is the product of the individual probabilities since the variables are independent. Thus, expressing what we said in equation form, Prob(X1 + X2 between x and x + dx) = Prob(X1 between x and x + dx) \u00d7 Prob(X2 between x and x + dx|x = x + x). x,x (11.22) We can translate this statement in terms of the probability densities and implement the constraint as a delta function. Letting f+(x) denote the probability density of X1 + X2, f+(x) dx = \u221e dx \u221e dx f1(x) f2(x) \u03b4(x + x \u2212 x) dx. (11.23) \u2212\u221e \u2212\u221e The factor of dx here tempers the infinite height of the delta function, so that the delta function registers unity when the constraint is met (it is useful to think of the delta function as", " being zero everywhere, but 1/dx at the origin, so that multiplying by dx and summing over all x gives unity). Evaluating the x integral, we see that the probability density of the sum is the convolution of the individual densities: \u221e f+(x) dx = dx f1(x) f2(x \u2212 x) dx = (f1 \u2217 f2)(x) dx. \u2212\u221e Note that this result is general in that it doesn\u2019t assume any particular form for f1(x) or f2(x). In the special case where both distributions are Gaussian, from Eqs. (11.20) and (11.21) we can see (probability density of sum) (11.24) that the variances are related by Var[X1 + X2] = Var[X1] + Var[X2]. The variance here is defined by Var[X] := (X \u2212 X) for a random variable X. The angle brackets denote the expectation value, (11.25) (variance of sum) (11.26) (variance) h(X) := \u221e \u2212\u221e h(x) f (x) dx, (11.27) (expectation value) 11.5 Application: Central Limit Theorem 207 where h(x) is an arbitrary function and f (x) is the probability density function of X. For the Gaussian probability density in the form of Eq. (11.21), the variance turns out to be \u03c32. Recalling that the standard deviation is the square root of the variance, \u03c3X := Var[X], (11.28) (standard deviation) we can see that for Gaussians, when you add the random variables, the standard deviation adds in quadrature: \u03c3X1+X2 = \u03c3 2 X1 + \u03c3 2 X2. (11.29) (error propagation for sums) This is the law of error propagation for addition in error analysis: if two independent measurements have uncertainties \u03c31 and \u03c32, then the uncertainty of the sum is 2. Actually, in terms of variances, these results turn out to hold also for non-Gaussian variables, so long as the variances are well-defined. 1 + \u03c3 2 \u03c3 2 11.5 Application: Central Limit Theorem Often, we make many independent measurements of some quantity; however, we might not know the underlying distribution, in which case it\u2019s difficult to make use of a result like Eq. (11.24). Fortunately, there is a very important result that often makes it unnecessary to know the underlying distribution. This is also the reason that the Gaussian probability distribution is so important. Let X1,..., XN be independent, identically distributed random variables. Let f (x) be the probability density function of each of the Xj. Defining the sum by SN := N j=1 Xj, (11.30) (sum of random variables) we will now ask, what is the probability density fSN (x) of SN? Evidently, we can iterate Eq. (11.24) to obtain fSN (x) = (f \u2217 f \u2217 \u00b7 \u00b7 \u00b7 \u2217 f )(x), (11.31) (probability density of SN ) where the result is the successive convolution of N copies of f (for N \u2212 1 total convolution operations). However, it turns out that this distribution becomes simple for large enough N. The central limit theorem states that, provided that the mean and variance of the Xj exist, with the mean \u00b5 = Xj and variance \u03c32 = Var[Xj], the distribution fSN (x) becomes asymptotically Gaussian for large N with SN = N \u00b5, Var[SN ] \u2212\u2192 N \u03c32. (11.32) (central limit theorem) (The mean is an exact result, whereas the distribution and variance are valid for large N.) This is a rough statement, since \u2018\u2018becomes asymptotically Gaussian\u2019\u2019 is an imprecise statement. So let\u2019s clean this up a bit. The central limit theorem states alternately that the probability density function fZN (x) of the cen- tered, scaled statistic ZN := SN \u2212 N \u00b5 \u221a \u03c3 N (11.33) converges to the \u2018\u2018standard normal\u2019\u2019 (Gaussian) distribution fZN (x) \u2212\u2192 1 \u221a 2\u03c0 e\u2212x2/2, (11.34) (asymptotic scaled distribution) which is the special Gaussian with mean 0 and unit variance. 208 Chapter 11. Fourier Analysis II: Convolution Let\u2019s prove this now2. To", " evaluate the convolutions in Eq. (11.31), we need to employ the convolution theorem. Taking the Fourier transform of f (x), \u221e \u02dcf (k) = f (x) e\u2212ikx dx \u2212\u221e \u221e \u221e f (x) \u2212\u221e (\u2212ik)j j! j=0 \u221e j=0 = = X j (\u2212ikx)j j! dx (11.35) = 1 \u2212 ik\u00b5 \u2212 k2(\u03c32 + \u00b52) 2 + O(k3). Here, we Taylor-expanded e\u2212ikx and then used the fact that the terms of the expansion were proportional to expectation values X j, especially the first two, which follow from Xj = \u00b5, Var[Xj] = \u03c32 = X 2 j \u2212 \u00b52. (11.36) In particular, note that in probability theory the characteristic function of a probability density, given by \u02dcf (\u2212k) = eikX, (11.37) is an important tool for the manipulation of probabilities. This is more cumbersome than necessary, so let\u2019s recompute the expansion in Eq. (11.35) for the centered, scaled variable Xj \u2212 \u00b5 \u221a N \u03c3 with corresponding probability density fZ(x). The centering effectively zeroes the mean, and the rescaling changes the factor in front of the variance, with the result (11.38) Zj =, Since \u02dcfZ(k) = 1 \u2212 k2 2N + O k \u221a N 3. ZN = N j=1 Zj, (11.39) (11.40) the convolution theorem says that to calculate the transform of the N -fold convolution, we just compute \u02dcfZ(k) to the N th power: \u02dcfZN (k) = \u02dcfZ(k) N = 1 \u2212 k2 2N + O k \u221a N 3N. (11.41) As N becomes large, we can neglect the higher order terms beyond the first, and then use the formula to see that for large N, the transform becomes lim n\u2192\u221e 1 + n x n = ex \u02dcfZN (k) = exp \u2212. k2 2 (11.42) (11.43) But now the inverse Fourier transform of exp(\u2212k2/2) is exp(\u2212x2/2)/ normal distribution as N \u2212\u2192 \u221e. \u221a 2\u03c0, so fZN converges to a standard 2This is the physicist\u2019s proof; the rigorous version is in T. W. K\u00f6rner, Fourier Analysis (Cambridge, 1988), starting on p. 349. 11.5 Application: Central Limit Theorem 209 11.5.1 Central Limit Theorem Application: Random Walk The central limit theorem has one (of many) of its important applications in statistical mechanics. Suppose a random walker takes a random step of size X with probability density f (x) between periodic intervals of duration \u2206t. Let\u2019s assume that X = 0, Var[X] = \u03c32. (11.44) After N steps (N large), what has the walker done? The central limit theorem says that the probability density of the accumulated displacement SN = N j=1 Xj is Gaussian with zero mean and variance N \u03c32. That is, the width (standard deviation) is \u03c3 probability distribution thus becomes asymptotically Gaussian with a time-dependent width of (11.45) \u221a N. The \u03c3(t) = \u03c3 t \u2206t. (11.46) (asymptotic distribution width) This random-walk behavior is characteristic of a diffusion process, which is a transport process by which the distribution grows as t1/2, \u2206x \u223c D t1/2, (11.47) (diffusion law) where for the random walker the diffusion coefficient is D = \u03c3/ \u2206t. Note that within certain restrictions (we\u2019ll explore this more in the homework), the final distribution is Gaussian, independent of the one-step distribution. Some one-step distributions, such as the Cauchy distribution, lead to anomalous diffusion where the diffusion coefficient diverges. \u221a 11.5.2 Central Limit Theorem Application: Standard Deviation of the Mean Returning again to error analysis, suppose we make independent measurements X1,..., XN of some quantity in the laboratory. The sample mean is We can rewrite this as \u00b5N := 1 N N j=1 Xj. \u00b5N = SN N = \u00b5 + \u03c3ZN\u221a N, (11.48) (11.49) where the first term represents the true mean, and the second is the experimental error", " (statistical fluctuation in the sample mean). Applying the central limit theorem, ZN is approximately standard normal for large N, so \u00b5N is Gaussian with mean \u00b5 and standard deviation \u03c3mean = \u03c3 \u221a N, (11.50) (standard deviation of the sample mean) where \u03c3 is the standard deviation of a single measurement. Thus, the standard deviation of the mean (also called the standard error) is \u03c3/ N. This is why, by making many measurements, it is possible to increase the accuracy of a measured quantity. (However, the convergence is fairly slow with N, so for a high-accuracy measurement, N must be very large, or \u03c3 must be small.) \u221a 210 Chapter 11. Fourier Analysis II: Convolution 11.6 Application: Impulse Response and Green Functions Let\u2019s consider a simple mechanical system, such as a damped harmonic oscillator subject to a forcing function f (t): \u00a8x + \u03b3 \u02d9x + \u03c9 2 0 x = f (t). (11.51) (forced, damped harmonic oscillator) The three terms on the left-hand side represent, respectively, the acceleration, damping, and harmonic restoring force. Rather than solving this directly for an arbitrary forcing function, let\u2019s assume a particular form for the forcing function: an impulse (delta function) at time t: \u00a8x + \u03b3 \u02d9x + \u03c9 2 0 x = \u03b4(t \u2212 t). (11.52) The solution to this is fairly straightforward. There is no force before the impulse (t < t), so the oscillator is at rest, (11.53) assuming the boundary conditions x(\u2212\u221e) = 0 and \u02d9x(\u2212\u221e) = 0. At t = t, the impulse \u2018\u2018kicks\u2019\u2019 the oscillator. We can see this by solving Eq. (11.52) and integrating over a short interval around t = t. x(t < t) = 0, t+\u03b5 \u00a8x dt = t+\u03b5 t\u2212\u03b5 t\u2212\u03b5 \u03b4(t \u2212 t) \u2212 \u03b3 \u02d9x \u2212 \u03c9 2 0 x dt. (11.54) In the limit as \u03b5 \u2212\u2192 0, the last two terms in the right-hand integral become negligible (being proportional to \u03b5). Thus, as \u03b5 \u2212\u2192 0, we find \u2206 \u02d9x = 1 (11.55) across the impulse. This makes intuitive sense: an impulsive force changes the velocity, but there isn\u2019t yet time for any position change to take place. Since the velocity just before t is zero, the velocity just after t is 1. Thus, the solution for t > t is the usual damped harmonic oscillator solution with the initial conditions x(t) = 0 and \u02d9x(t) = 1, and so we can write (without deriving the solution) where x(t > t) = 1 \u03c9\u03b3 e\u2212(\u03b3/2)(t\u2212t) sin[\u03c9\u03b3(t \u2212 t)], \u03c9\u03b3 := \u03c9 2 0 \u2212 (\u03b3/2)2 (11.56) (11.57) is the oscillation frequency in the presence of damping. Let\u2019s rewrite this as the impulse-response solution, the solution x(t) to the equation in response to a unit impulsive force at t = t: g(t, t) = \uf8f1 \uf8f4\uf8f2 \uf8f4\uf8f3 0, 1 \u03c9\u03b3 t < t e\u2212(\u03b3/2)(t\u2212t) sin[\u03c9\u03b3(t \u2212 t)], t \u2265 t. Then g(t, t) satisfies Eq. (11.52), where we write out the derivative operators separately: (Green function for damped harmonic oscillator) (11.58) \u22022 \u2202t2 + \u03b3 + \u03c9 2 0 \u2202 \u2202t g(t, t) = \u03b4(t \u2212 t). (11.59) (Green-function condition) We note also the property of the delta function f (t) = \u221e \u2212\u221e dt f (t) \u03b4(t \u2212 t), (11.60) 11.6 Application: Impulse Response and Green Functions so we can multiply through Eq. (11.59) by f (t) and integrate over t, with the result \u22022 \u2202t2 + \u03b3 \u2202 \u2202t + \u03c9 2 0 \u221e dt f (t) g(t, t) = f (t). \u2212\u221e This is equivalent to the original equation of motion (11.51), \u22022 \u2202t2 + \u03b3 + \ufffd", "\ufffd 2 0 \u2202 \u2202t x(t) = f (t). 211 (11.61) (11.62) provided we identify x(t) = \u221e \u2212\u221e f (t) g(t, t) dt. (11.63) (Green-function solution) Thus, given the impulse-response function g(t, t), we can write down the solution for any forcing function f (t) in terms of an integral. In fact, noting that in this particular case that the time dependence is such that g(t, t) = g(t \u2212 t), we can write x(t) = \u221e \u2212\u221e f (t) g(t \u2212 t) dt = (f \u2217 g)(t). (11.64) so that the general solution is the convolution of f (t) with g(t) := g(t, 0). This form holds whenever the only explicit time dependence in the system is in the forcing function. This is an example of a convolution method for solving linear system. This works because of the superposition principle and because any function can be represented as a superposition of delta functions, which is essentially the content of Eq. (11.60). This convolution or impulse-response method is an example of a more general method called the Green-function solution (g(t, t) is the Green function for the damped harmonic oscillator that is initially undisturbed). All this works more generally, as the Green-function equation (11.59) has the general form (Green-function solution as convolution) where L(t) is a linear operator. g(t, t) = g(t \u2212 t): Of course then we can just shift t to write L(t)g(t, t) = \u03b4(t \u2212 t), (11.65) If L(t) is time-independent, as it is for the harmonic oscillator, then Lg(t \u2212 t) = \u03b4(t \u2212 t). Lg(t) = \u03b4(t). (11.66) (11.67) Then we can write the formal solution for the Green function as g(t) = L\u22121\u03b4(t). (11.68) (formal solution for general Green function) All of the integration above is essentially just working out the specific expression for L\u22121, which is of course different for every differential equation. 11.6.1 Frequency Domain Just for fun, let\u2019s look at the Fourier transform of the Green function. A quick trick for doing so is to take the Fourier transform of Eq. (11.59). This is easy, since it is essentially just assuming that the time dependence is of the form e\u2212i\u03c9t. Thus, to take the Fourier transform, we just make the following replacements: \u2202 \u2202t \u2212\u2192 \u2212i\u03c9, g(t, t) \u2212\u2192 \u02dcg(\u03c9, t), \u03b4(t \u2212 t) \u2212\u2192 \u221e \u2212\u221e \u03b4(t \u2212 t) ei\u03c9t dt = ei\u03c9t. (11.69) The resulting equation is (\u2212\u03c92 \u2212 i\u03b3\u03c9 + \u03c9 2 0 )\u02dcg = ei\u03c9t. (11.70) 212 Chapter 11. Fourier Analysis II: Convolution Solving for the Fourier transform \u02dcg, \u02dcg(\u03c9, t) = ei\u03c9t \u03c9 2 0 \u2212 \u03c92 \u2212 i\u03b3\u03c9, (11.71) which we see is a Lorentzian function in frequency. To employ the convolution theorem, we really just wanted the Fourier transform for the impulse at t = 0 (the t is only there to facilitate the convolution), so \u02dcg(\u03c9) = \u02dcg(\u03c9, t = 0) = 1 \u03c9 2 0 \u2212 \u03c92 \u2212 i\u03b3\u03c9 (transfer function of harmonic oscillator). We can also assume we know the Fourier transform of the driving function, \u02dcf (\u03c9) = F [f (t)], so the convolution theorem gives an alternate way to write the general solution: x(t) = F \u22121 \u02dcf (\u03c9) \u02dcg(\u03c9) = 1 2\u03c0 \u221e d\u03c9 \u2212\u221e \u02dcf (\u03c9) e\u2212i\u03c9t \u03c9 2 0 \u2212 \u03c92 \u2212 i\u03b3\u03c9. (11.72) (11.73) (11.74) In this case, we see that in the frequency domain, the kernel g(t) acts as a bandpass filter (centered on \u03c90, with a width controlled by \u03b3) that passes some of the frequencies of the driving function f (t) to produce the output x(t). Thus \u02dcg(\u03c9", ") is the transfer function for the damped harmonic oscillator. The same idea holds in electronics, and what we have done here is formally equivalent to a signal f (t) going through a band-pass filter (R-L-C resonant circuit). (transfer-function solution) In the language of the linear operator L as in Eq. (11.68), we have computed the Fourier transform of this equation in the sense \u02dcg(\u03c9) = 1 F [L], (11.75) where the Fourier transform of the operator L is given by the same replacements in Eqs. (11.69). The Fourier transform of L thus defines transfer function, or the filtering action of the system or oscillator. (formal Green-function solution, frequency domain) 11.7 Application: Spectral Transmission Finally, we can get to an application in optics. Recall from Chapter 7 that a Fabry\u2013Perot cavity can act as a spectrum analyzer, so that we can view the frequency content of an arbitrary input wave. We only treated the Fabry\u2013Perot cavity in the case of monochromatic light, however, so let\u2019s explore this in more detail. Suppose we have a Fabry\u2013Perot cavity with transmission function Tcav = 1 + 2F \u03c0 Tcav,max 2 sin2 \u03c0\u03bd FSR. (11.76) This depends only on a single frequency, so it isn\u2019t so apparent how scanning the cavity length and an input spectrum would fit in. So let\u2019s massage things a little. As a spectrum analyzer, we want to analyze the transmission with respect to small length changes, so we can write \u03c0\u03bd FSR = \u03c0(\u03bdq + \u03b4\u03bd) FSR = \u03c0q + \u03c0\u03b4\u03bd FSR = = 2\u03c0d \u03bb 2\u03c0d0 \u03bb \u03c0\u03b4\u03bd + FSR 2\u03c0\u03b4d \u03bb + (11.77) + \u03c0\u03b4\u03bd FSR. 11.7 Application: Spectral Transmission 213 Here, d0 is the nominal cavity length, \u03b4d is the (small) length change while scanning the cavity, and \u03b4\u03bd is a small frequency deviation from the nominal resonance. Defining \u03c60 := 2\u03c0d0/\u03bb as the overall offset phase due to the nominal cavity length (essentially a constant), we can write where \u03c0\u03bd FSR = \u03c0(\u03b4\u03bd \u2212 \u03b4\u03bdd) FSR + \u03c60, \u03b4\u03bdd := \u2212 FSR \u03bb/2 \u03b4d (11.78) (11.79) is the change in the resonance frequency due to the cavity length change. Thus, we can write the functional dependence of Tcav = Tcav(\u03b4\u03bd \u2212 \u03b4\u03bdd) on both the input frequency and cavity length changes. If the input spectrum is the \u2018\u2018line-shape function\u2019\u2019 g(\u03bd), we will assume that the line shape is sharply peaked about some center frequency \u03bd0. Thus, we can define a centered line-shape function gc(\u03b4\u03bd) := g(\u03bd0 + \u03b4\u03bd), (11.80) where \u03b4\u03bd = \u03bd \u2212 \u03bd0. Then we can write the total transmission through the cavity as the amplitude gc(\u03b4\u03bd) at frequency \u03b4\u03bd, weighted by the cavity transmission coefficient at the same frequency, integrated over all frequencies: Tcav,total(\u03b4d) \u2248 \u221e \u2212\u221e Tcav(\u03b4\u03bd \u2212 \u03b4\u03bdd) gc(\u03b4\u03bd) d(\u03b4\u03bd). (11.81) (spectrum-analyzer output) The approximation here is the extension of the integration to all frequencies, which is valid if g(\u03bd) is sharply peaked. Thus, the cavity transmission is the convolution of the input spectrum with the cavity transmission spectrum. Since the latter is periodic in frequency (due to the free spectral range), the total transmission is still periodic, but the individual transmission peaks are broadened by the input line shape due to the convolution. As a spectrum analyzer, the cavity line shape (for a given finesse) represents a resolution limit, since any input spectrum on transmission through the cavity will have an apparent line width of at least \u03b4\u03bdFWHM due to the convolution. 214 Chapter 11. Fourier Analysis II: Convolution 11.8 Exercises Problem 11.1 Show that the order of the convolution does not matter: (f \u2217 g)(x) = (g \u2217 f )(x). (11.82) Problem 11.2 Show that the area of a convolution is the product of the areas. \u221e dx (f \u2217 g)(x) = \u221e dx f (x", ") \u221e dx g(x). \u2212\u221e \u2212\u221e \u2212\u221e (11.83) (In particular, note that convolution with a normalized kernel does not change the area of a function.) Problem 11.3 Show that convolution within the integral of a product can be associated by flipping the convolution kernel. That is, show that \u221e dx (f \u2217 g)(x) h(x) = \u221e dx f (x) (g\u2212 \u2217 h)(x), \u2212\u221e \u2212\u221e (11.84) where g\u2212(x) := g(\u2212x). (Note that in the important case of an even convolution kernel, g\u2212(x) = g(x).) Problem 11.4 Prove the convolution derivative formula (f \u2217 g)(x) = (f \u2217 g)(x) = (f \u2217 g)(x). (11.85) Problem 11.5 Let f (x) denote the unit box function, f (x) = 1, 0 |x| \u2264 1/2 elsewhere (11.86) (a) Let f \u2217N (x) denote the convolution of f (x) with itself N \u2212 1 times. Write down the asymptotic form for f \u2217N (x) in the limit of large N. (b) Make plots of f (x), (f \u2217 f )(x), (f \u2217 f \u2217 f )(x), and (f \u2217 f \u2217 f \u2217 f )(x). Include the corresponding asymptotic distribution on each plot. Comment on your results. Problem 11.6 The \u2018\u2018Gaussian blur\u2019\u2019 in image-processing programs is just a convolution with a Gaussian kernel. Other types of image blurring can be thought of as convolutions with other kernels (e.g., the trajectory of motion for camera shake, or the response function of a lens system for an unfocused camera). As a model of this, compute the convolution of a single Fourier component with a (normalized) Gaussian blurring kernel f (x) = cos(kx) g(x) = \u221a 1 2\u03c0\u03c3 e\u2212x2/2\u03c32. (11.87) (11.88) Interpret your result in the limits of large and small spatial frequency k, and large and small blurring width \u03c3. 11.8 Exercises Problem 11.7 215 Consider independent, identically distributed random variables X1,..., XN with Cauchy (Lorentzian) probability density functions 1 \u03c0(1 + x2). (11.89) (a) Show that the Fourier transform is given by f (x) = (11.90) it\u2019s much easier to compute the inverse Fourier transform of \u02dcf (k) and use the fact that the \u02dcf (k) = e\u2212|k|. Hint: Fourier transform is invertible for reasonable functions. (b) Show that the probability density of the mean \u00b5N := 1 N N j=1 Xj (11.91) is also f (x). (c) A naive application of the central limit theorem leads to the conclusion that \u00b5N has a Gaussian probability distribution for large N, but in (b) you have shown that this is not the case. Explain why there is no contradiction. Problem 11.8 The harmonically forced oscillator has the following equation of motion: Use the Green function method in the time domain to show that the motion of the driven, damped oscillator (in steady state) is \u00a8x + \u03b3 \u02d9x + \u03c9 2 0 x = f0 sin(\u03c9t). (11.92) where x(t) = f0 sin(\u03c9t + \u03b4) 0 \u2212 \u03c92)2 + \u03c92\u03b32 (\u03c9 2, tan \u03b4 = \u03c9\u03b3 0 \u2212 \u03c92. \u03c9 2 (11.93) (11.94) You may find the following integral formula useful: \u221e 0 e\u2212ax sin(bx) sin(cx + d) dx = b[2ac cos d + (a2 + b2 \u2212 c2) sin d] (a2 + b2)2 + 2(a2 \u2212 b2)c2 + c4 (a, b, c, d real, a > 0). (11.95) Problem 11.9 Work out the Green function g(t, t) for the differential equation for arbitrary \u2018\u2018forcing\u2019\u2019 function f (t). Write the solution x(t) as a convolution involving f (t) and g(t, t) = g(t \u2212 t). Assume the boundary condition g(t \u2212\u2192", " \u2212\u221e, t) = 0. \u2202x \u2202t = f (t), (11.96) Problem 11.10 Work out the Green function g(t, t) for the differential equation \u2202nx \u2202tn = f (t), for positive integer n and arbitrary \u2018\u2018forcing\u2019\u2019 function f (t), given the boundary condition that g(t < t) = 0 and any derivatives up to the (n \u2212 2)th of g(t, t) vanish at t = t. Write the solution x(t) as a convolution involving f (t) and g(t, t) = g(t \u2212 t). (11.97) 216 Chapter 11. Fourier Analysis II: Convolution Problem 11.11 Derive the Green function for the ODE \u02d9x + \u03b3x = f (t), (11.98) and write down the general solution x(t) in terms of the forcing function f (t). Problem 11.12 In this problem you will calculate the Green function for the Poisson equation \u22072\u03c6(r) = \u2212 \u03c1(r) 0 (11.99) for the electric potential \u03c6 in terms of the source charge density \u03c1(r). We can do this by using the three-dimensional Fourier-transform, f (r) = 1 (2\u03c0)3 d3k \u02dcf (k) eik\u00b7r, \u02dcf (k) = d3r f (r) e\u2212ik\u00b7r. (11.100) The Green function satisfies \u22072G(r, r) = \u2212\u03b43(r \u2212 r) = \u2212\u03b4(x \u2212 x) \u03b4(y \u2212 y) \u03b4(z \u2212 z). (11.101) (a) Start by finding the Fourier transform \u02dcG(k, r) of G(r, r) by computing the Fourier transform of Eq. (11.101), and solving the resulting equation. (b) Then find G(r, r) by inverting the Fourier transform. Evaluate the k integral in spherical coordinates. You may find the following integral formula useful: \u221e 0 dx sin(ax) ax = \u03c0 2|a| (a \u2208 R). (11.102) (c) Finally, use your expression for G(r, r) to write down the Green-function solution (i.e., an integral solution) to Poisson\u2019s equation for an arbitrary charge density \u03c1(r). Comment on your result, and the physical significance of the Green function. Problem 11.13 In this problem we will work with the Hermite\u2013Gaussian function Hn(x) e\u2212x2 = (\u22121)n dn dxn e\u2212x2, (11.103) consisting of the product of a Hermite polynomial and a Gaussian factor. The expression on the right-hand side follows from the explicit formula for the Hermite polynomials [see Eq. (6.102) in Problem 6.18]: Hn(x) = (\u22121)n ex2 dn dxn e\u2212x2. (11.104) (a) Show that the Fourier transform is given by Hn(x) e\u2212x2 F (k) = \u221a \u03c0(\u2212ik)ne\u2212k2/4. (b) Derive the convolution formula Hm(x) e\u2212x2 \u2217 Hn(x) e\u2212x2 = \u221a \u03c0 2(m+n+1)/2 Hm+n x \u221a 2 e\u2212x2/2 (11.105) (11.106) 11.8 Exercises 217 for these Hermite\u2013Gauss functions. (c) Derive the convolution formula e\u2212x2/\u03b12 \u2217 Hn x \u03b2 e\u2212x2/\u03b22 = \u221a \u03c0 \u03b1\u03b2n+1 (\u03b12 + \u03b22)(n+1)/2 Hn x \u03b12 + \u03b22 e\u2212x2/(\u03b12+\u03b22). (11.107) That is, a convolution of the Hermite\u2013Gauss function with a Gaussian leads to a stretched version of the original Hermite\u2013Gauss function. Problem 11.14 Another Hermite\u2013Gauss function can be written Hn(x) e\u2212x2/2 = (\u22121)nex2/2 dn dxn e\u2212x2, (11.108) differing from Eq. (11.108) in the factor of 2 in the exponential. This is a better representation 2 appears in the of the profile of a Hermite\u2013Gaussian beam [see Eq. (6.74), where the factor of Hermite-polynomial argument but", " not in the exponential], compared to the Hermite\u2013Gauss function from Problem 11.13. However, this function is somewhat more difficult to work with. \u221a (a) Recalling the generating function for the Hermite polynomials, Eq. (6.103), g(x, t) := e2xt\u2212t2 = \u221e n=0 Hn(x) tn n!, show that and then show F g(x, t) e\u2212x2/2 (k) = \u221a 2\u03c0 g(k, \u2212it) e\u2212k2/2, F Hn(x) e\u2212x2/2 (k) = \u221a 2\u03c0 (\u2212i)nHn(k) e\u2212k2/2. (11.109) (11.110) (11.111) That is, up to a constant factor, the Fourier transform of a Hermite\u2013Gauss function has the same form as the original. (b) What does the result from (a) say about Fraunhofer diffraction of a Hermite\u2013Gauss beam, and how is this (qualitatively) consistent with the Hermite\u2013Gaussian solution to the paraxial wave equation? (c) Write down an integral expression for the convolution of two Hermite\u2013Gauss functions of different orders. Chapter 12 Fourier Optics 12.1 Fourier Transforms in Multiple Dimensions Before introducing Fourier optics, we must extend the spatial convention for Fourier transforms that we introduced before. As opposed to time, space takes up three dimensions, so we need a Fourier transform in multiple dimensions. In Cartesian coordinates, this turns out to be simple: just take a one-dimensional Fourier transform along each direction. For example, in two dimensions, the inverse transform is f (x, y) = 1 (2\u03c0)2 \u221e dkx \u221e dky \u02dcf (kx, ky) ei(kxx+kyy), \u2212\u221e \u2212\u221e (inverse Fourier transform, 2D) (12.1) and the Fourier transform is \u02dcf (kx, ky) = \u221e dx \u221e dy f (x, y) e\u2212i(kxx+kyy), \u2212\u221e \u2212\u221e (12.2) (Fourier transform, 2D) where again \u02dcf (kx, ky) = F [f (x, y)]. Fourier transforms can be extended to arbitrarily many dimensions, but two is sufficient for our purposes here. 12.2 Wave Propagation in Homogeneous Media 12.2.1 Fingerprints of Propagation The electromagnetic wave equation gives us a formal method for propagating an optical wave forward in space. Essentially, we can do this because we know how plane waves propagate, and any field can be regarded as a superposition of plane waves. That\u2019s where the \u2018\u2018Fourier\u2019\u2019 comes into Fourier optics: we use a Fourier transform to break up an arbitrary field into plane-wave components. Let\u2019s see how this works. Recall the scalar plane wave (we won\u2019t do any vector theory, which is much more complicated) solution to the wave equation: E(+)(r) = E(+) 0 eik\u00b7r = E(+) 0 ei(kxx+kyy+kzz), (12.3) (scalar plane wave) where the wave vector k = (kx, ky, kz) indicates the propagation direction of the wave. Let\u2019s let the optical axis lie along the z-axis. Then the wave vector k has a direction determined by its angles with respect to the z-axis, \u03b8x = sin\u22121, \u03b8y = sin\u22121 kx k, ky k where kx and ky are the transverse spatial frequencies. (propagation angles of plane wave) (12.4) 220 Chapter 12. Fourier Optics That is, in the z = 0 plane, the electric field has harmonic spatial dependence in the x- and y-directions: E(+)(x, y, z = 0) = E(+) 0 eikxxeikyy. (12.5) This is the central point of Fourier optics: harmonic phase variation in the (x, y)-plane (say, at z = 0) corresponds to a plane wave in a particular direction determined by Eqs. (12.4). That is, the wavelength of the transverse harmonic variation determines the direction of propagation\u2014the wavelength of the wave is fixed (since we are assuming monochromatic light), and so only the proper angle can match the wave to the phase variation at z = 0. In this sense, the transverse harmonic variation is a \u2018\ufffd", "\ufffdfingerprint\u2019\u2019 of propagation in a certain direction. Often, we will only want to look at these in the paraxial approximation, where these relations are simpler. In the paraxial regime, we can write \u03b8x \u2248 kx k, \u03b8y \u2248 ky k, (12.6) (paraxial propagation angles) since kx,y k. 12.2.2 Decomposition A general field profile E(+)(x, y) at z = 0 can thus be written as a superposition of plane waves via the Fourier transform: E(+)(x, y) = dkx dky \u02dcE(+)(kx, ky) ei(kxx+kyy). 1 (2\u03c0)2 \u221e \u221e \u2212\u221e \u2212\u221e We know how individual plane waves travel with z, ei(kxx+kyy) \u2212\u2192 ei(kxx+kyy)eikzz, where, since k2 =, kz = k2 \u2212 k 2 x \u2212 k 2 y. (12.7) (12.8) (12.9) So to propagate the full field forward in z, we just multiply each plane wave in the superposition by eikzz: \u221e \u221e E(+)(x, y, z) = dkx dky \u02dcE(+)(kx, ky) ei(kxx+kyy)eikzz. (12.10) 1 (2\u03c0)2 \u2212\u221e \u2212\u221e It is important to keep in mind the functional dependence kz = kz(kx, ky) here. xzykqxqyEoo(+)o(x,ozo=o0)Eoo(+)o(x,oz) 12.2 Wave Propagation in Homogeneous Media 12.2.3 Reverse Waves There is also a subtle point here in writing down Eq. (12.9), since kz = \u2212 k2 \u2212 k 2 x \u2212 k 2 y. 221 (12.11) is also a solution consistent with k2 = k 2 z. These solutions correspond to k vectors with a component along the \u2212z-direction, and thus correspond to backward-propagating solutions. Thus, in writing down Eq. (12.9), we are building in the assumption that we have an entirely forward-propagating wave (i.e., all plane-wave components are propagating in the +z-direction). Without this assumption, the timeindependent wave equation + k2 E(+)(r) = 0 is not well-posed as an initial-value problem by specifying E(+)(x, y, z = 0) and propagating to other z. This is because the wave equation is second-order in z, and thus requires more information [e.g., specifying both E(+)(x, y, z = 0) and \u2202zE(+)(x, y, z = 0), or our forward-propagating assumption]. Of course, the paraxial wave equation (6.7) (12.12) (12.13) i2k\u2202z + \u22072 \u03c8 = 0 T is only first-order in z, and thus only requires the initial condition \u03c8(x, y, z = 0) to be well-posed. Recall that in deriving the wave equation, we explicitly assumed forward propagation, by assuming that all planewave components were close to a central \u2018\u2018carrier\u2019\u2019 wave. Recall that the paraxial wave equation is formally equivalent to the Schr\u00f6dinger equation in quantum mechanics, where the initial-value problem is well-posed by specifying only \u03c8(r, t = 0) and evolving to arbitrary time. The quantum-mechanical analog of our problem here is the Klein\u2013Gordon equation, which for plane waves is basically the wave equation. In the Klein\u2013Gordon equation, the analogues to backward-propagating waves are negative-energy solutions, which can be interpreted as antiparticles (particles traveling backwards in time). In optics, backward waves are formed from forward waves by gradients in the refractive index (such as refractive interfaces, which cause reflections); plane waves are eigenstates of homogeneous media, so only an inhomogeneity can scatter waves (In quantum mechanics, these inhomogeneities correspond to timebetween different plane-wave states. dependent potentials). 12.2.4 Fourier-Transform Recipe The theoretical development thus leads us to a fairly simple (in principle) recipe for propagating the field forward in space, given the form of the field in some plane. We can see that the information in the entire three-dimensional (monochromatic) field is thus encoded in any two-dimensional slice, a fact", " which motivates the field of holography. But for now, we wish to use this recipe as a calculational tool for understanding diffraction, so to summarize: 1. Start with an initial field E(+)(x, y) at z = 0. 2. Compute the Fourier transform of the initial field: \u02dcE(+)(kx, ky) = \u221e dx \u221e dy E(+)(x, y) e\u2212i(kxx+kyy). \u2212\u221e \u2212\u221e (decomposition of initial field into plane waves) (12.14) 3. Multiply the transform E(+)(kx, ky) by the free-space transfer function eikzz: \u02dcE(+)(kx, ky) \u2212\u2192 \u02dcE(+)(kx, ky) eikzz. (12.15) (k-space propagation) Again, we need to apply this to the Fourier transform rather than the initial function because kz = kz(kx, ky). In the paraxial approximation, we can alternatively use the paraxial form (12.18) below for the transfer function. 222 Chapter 12. Fourier Optics 4. Finally, invert the Fourier transform to find the propagated field: E(+)(x, y, z) = 1 (2\u03c0)2 \u221e dkx \u221e \u02dcE(+)(kx, ky) eikzz dky ei(kxx+kyy). \u2212\u221e \u2212\u221e (field propagated to arbitrary z) (12.16) This recipe works because the superposition principle holds in linear optics and because it is easy to propagate plane waves. 12.2.5 Paraxial Propagation The integral in Eq. (12.16) is in general difficult to evaluate because of the functional form of kz in the exponent. However, in the paraxial approximation, the integral greatly simplifies. Since, kx,y k, we can write k2 x + k 2 k 2 y 2k = k \u2212. (12.17) kz = k2 \u2212 k 2 x \u2212 k 2 y \u2248 k Thus, the free-space transfer function becomes eikzz \u2248 eikze\u2212i(k 2 x +k 2 y )z/2k, (12.18) where the first factor is an overall phase factor corresponding to plane-wave propagation along the optical axis, and the second factor generates the evolution of the spatial profile. (free-space transfer function, paraxial approximation) Note that in the paraxial limit, because kx and ky are assumed to be small compared to k, we can still keep the limits of integration in Eq. (12.16) to \u00b1\u221e because we assume that \u02dcE(+)(kx, ky) has negligible \u2018\u2018stuff\u2019\u2019 at large kx,y. 12.2.5.1 Solution of the Paraxial Wave Equation We have shown that the transfer function (12.18) produces the proper solution to the wave equation, once we have made the paraxial approximation. Of course, then, this should represent the solution to the paraxial wave equation (6.7) \u22072 T + i2k \u2202 \u2202z \u03c8 = 0,, where the transverse Laplacian is and the field envelope \u03c8 is defined by If we write the wave equation as we have a differential equation of the form for which the formal solution is \u22072 T := \u22022 \u2202x2 + \u22022 \u2202y2, E(+)(r) = \u03c8(r)eikz. \u2202 \u2202z \u03c8 = i 2k \u22072 T \u03c8, \u2202 \u2202z \u03c8 = A\u03c8, \u03c8(z) = eAz\u03c8(0). (12.19) (12.20) (12.21) (12.22) (12.23) (12.24) Here, \u2018\u2018formal\u2019\u2019 means that we have to be careful that the exponential of the operator A exists (it turns out that this is an okay assumption); if A is a scalar, then there is no problem. Since A is an operator, the 12.2 Wave Propagation in Homogeneous Media 223 operator exponential is defined by the Taylor expansion (powers of operators are obviously well-defined). Thus, the solution for the paraxial wave equation is \u03c8(z) = ei\u22072 T z/2k\u03c8(0) = ei(\u2202 2 x +\u2202 2 y )z/2k\u03c8(0). (12.25) Recall that when operating on a", " plane wave with spatial dependence eik\u00b7r, the derivatives are equivalent to k-vector components: \u2202 \u2202x \u2202 \u2202y \u2212\u2192 ikx, \u2212\u2192 iky. (12.26) Thus, for plane waves (where the vector k is well-defined), we can replace the derivatives according to these replacements, so that the solution is \u03c8(z) = e\u2212i(k 2 x +k 2 y )z/2k\u03c8(0). (12.27) When kx and ky are not well defined, they can be thought of as shorthands for the derivatives. Then changing back to the regular field, E(+)(z) = eikze\u2212i(k 2 x +k 2 y )z/2kE(+)(0). (12.28) Thus, we recover the paraxial transfer function (12.18). Again for kx and ky to make sense as numbers, the electric field should be decomposed into its plane-wave components by a Fourier transform. 12.2.6 Nonparaxial Propagation and the Diffraction Limit If the paraxial approximation isn\u2019t valid, we have to live with the square-root form of kz(kx, ky), and the general propagation problem isn\u2019t easy (although it is still straightforward on a computer). However, we can still get some insight by examining the propagation equation (12.16) a bit more. Recall that through the Fourier transform, E(+)(x, y) is composed of different spatial frequencies. In particular, broad features in E(+)(x, y) are represented by small spatial frequencies, whereas fine details are represented by large spatial frequencies. Very fine details on spatial scales smaller than \u03bb are represented by spatial frequencies kx,y > k, which may seem strange in view of the form of Eq. (12.16). In fact, these high frequencies correspond to evanescent waves and don\u2019t propagate, as we can see from the form of the propagation factor: eikzz = ei k2\u2212k 2 x \u2212k 2 y z = e\u2212 x +k 2 k 2 y \u2212k2 z (if k 2 x + k 2 y > k2). (12.29) Here we have observed that k2 \u2212 k 2 y < 0, and thus factored a minus sign out of the square root, producing a factor of i outside the (now real) square root. This produces either an exponentially damping solution, which is physically sensible (note that the backward version of this wave would be exponentially growing in z, which is only physically sensible if it propagates along \u2212z). Thus, in this form, the wave damps exponentially with a length scale x + k 2 so that for the evanescent-wave solutions. eikzz = e\u2212z/\u03b4(kx,ky) \u03b4(kx, ky) = k 2 x + k 2 y \u2212 k2\u22121/2, (12.30) (12.31) This effect gives rise to a diffraction limit in imaging: propagation over long distances wipes out fine (sub-\u03bb) details in the initial image E(+)(x, y), because the components that carry these details rapidly decay away. Note that this is still true in the ideal limit where all kx,y up to k can propagate, corresponding to propagation up to 180\u25e6 from the optical axis; typical imaging systems aperture the light to a much smaller solid angle, wiping out yet more of the fine detail. In any case, this is why it is often said that it is impossible to optically resolve objects that are smaller than the optical wavelength \u03bb. 224 Chapter 12. Fourier Optics 12.3 Fraunhofer Diffraction 12.3.1 Far-field Propagation Fraunhofer diffraction refers to what happens to an optical pattern after propagation over a very large distance z in the paraxial approximation. Often, the \u2018\u2018diffraction pattern\u2019\u2019 refers to the far-field pattern of an aperture (say, a slit or a pinhole) illuminated by a uniform, monochromatic field, such as an expanded laser beam (plane wave) at normal incidence. We will speak more generally here, however, and refer to the diffraction pattern of an arbitrary initial field E(+)(x, y) in the z = 0 plane. It turns out that the Fraunhofer diffraction pattern is simple to calculate: with the proper scaling factors, it\u2019s just the Fourier transform of the initial pattern. We\u2019ll go through a heuristic derivation of this result here, and then do the mathematical version in the next section. Consider the diffraction setup shown here. The initial field E(+)(x, y) breaks", " up into its Fourier components. As we argued before in Section 12.2.1, each Fourier component corresponds to a wave propagating at a particular angle. In the paraxial approximation, we can write this out as kx,y k Thus, we can write the Fourier transform of E(+)(x, y) as \u03b8x,y \u2248. \u02dcE(+)(kx, ky) = \u02dcE(+) (k\u03b8x, k\u03b8y). (12.32) (12.33) Noting that in the paraxial approximation, \u03b8x \u2248 x/z and \u03b8y \u2248 y/z, we can write E(+)(x, y, z) = k 2\u03c0z \u02dcE(+) kx z, ky z, where \u02dcE(+)(kx, ky) = F E(+)(x, y), (12.34) where we have neglected an overall phase. The factor k/2\u03c0z is less obvious here, but is what is required to make the total power of the diffracted wave equal to the power of the initial wave (in particular, the 1/z dependence enforces the inverse-square law). We will establish it more firmly in the next section. For now it is sufficient to realize that Eq. (12.34) gives the Fraunhofer diffraction pattern in terms of the Fourier transform, and in terms of angle the relative field components in terms of the angles \u03b8x,y are given by \u02dcE(+) (k\u03b8x, k\u03b8y). While we are being fairly heuristic here, we will clean this up and be more formal when we redo this via Fresnel diffraction. (Fraunhofer diffraction pattern) The Fraunhofer diffraction pattern is clearly an approximation to the \u2018\u2018exact\u2019\u2019 pattern obtained by the propagation recipe. So when is this approximation valid? There are two conditions to be satisfied for the Fraunhofer approximation to hold. The first is that the size of the diffraction pattern must be much larger than the size of the initial aperture pattern E(+)(x, y). Eoo(+)o(x,oy)distant screenzoo=oo0zyxzqx 12.3 Fraunhofer Diffraction 225 We can view each component at angle \u03b8 as a \u2018\u2018beam\u2019\u2019 with a size determined by the size of E(+)(x, y), rather than as a plane wave. Then each beam at each angle must separate spatially from the others by the time the field hits the screen; then we can treat each Fourier component as a separate \u2018\u2018spot\u2019\u2019 on the screen. In other words, it must be that E(+)(x, y) looks like a point source to an observer on the screen. Letting \u03b4xsource denote the size (\u2018\u2018radius\u2019\u2019) of E(+)(x, y) in the x-direction (the same argument will hold for the y-direction), we can recall from the uncertainty relation (presented in temporal form in Eq. (7.36), but equally applicable to space)that the spatially confined field implies a spread in spatial frequency of \u03b4kx \u223c 1 \u03b4xsource. Using Eq. (12.32), the uncertainty relation implies a maximum angle of max(\u03b8x) \u223c \u03bb \u03b4xsource. (12.35) (12.36) An angle \u03b8x corresponds to a separation over distance z of z\u03b8x, so for the size of the diffraction pattern to be much larger than the initial pattern, we require \u03bbz/\u03b4xsource \u03b4xsource, or \u03b4xsource \u221a \u03bbz. (12.37) (Fraunhofer small-source condition) We can write this condition equivalently as z (\u03b4xsource)2 \u03bb, (12.38) (Fraunhofer far-field condition) and so this is the far-field requirement for the approximation to be valid. One thing that we have neglected comes from the propagation phase factor eikzz \u2212\u2192 eikz exp \u2212i x + k 2 k 2 y 2k z (12.39) that accompanies free-space propagation. We can drop the overall phase exp(ikz), but the other phase factor represents some phase variation across the screen. At the screen, this phase factor becomes exp \u2212ik x2 + y2 2z, (12.40) since we make the identification kx \u2212\u2192 kx/z and ky \u2212\u2192 ky/z. This phase factor is in general not negligible, even in the paraxial regime. For", " example, to make a paraxial argument, we could consider only a region of \u2018\u2018radius\u2019\u2019 \u03b4xscreen at the screen, where \u221a \u03b4xscreen \u03bbz. (12.41) This is essentially a small-angle restriction, more restrictive than the usual paraxial restriction, and because we require the diffraction pattern to be much larger than the source (which obeys the same condition), dx 226 Chapter 12. Fourier Optics this restriction would amount to looking only at the (kx, ky) = (0, 0) region of the diffraction pattern. To understand this phase factor, note that we considered each Fourier component propagating away from the source at different angles. For all of them to propagate over the same distance r, we would have to consider a spherical screen of radius r, a distance r from the source. Thus, we can correctly write E(+)(x, y, r) = k 2\u03c0r \u02dcE(+) kx r, ky r, where \u02dcE(+)(kx, ky) = F E(+)(x, y), r = x2 + y2 + z2 (12.42) for the diffraction pattern at constant r, whereas the expression (12.34) should include the phase factor (12.40). The phase factor here is the spherical (parabolic in the paraxial approximation) phase correction for continuing this field onto a flat screen. When computing the screen intensity, this phase factor vanishes anyway, but strictly speaking the far-field pattern is only the Fourier transform without this factor on a spherical screen. (Fraunhofer diffraction pattern) 12.3.2 Thin Lens as a Fourier Transform Computer The whole Fourier-transform argument above also works for a thin lens. Recall that for a thin lens, parallel rays propagating at an angle \u03b8 with respect to the optical axis get mapped to a transverse position \u03b8f in the focal plane of the lens. With this fact in mind, let\u2019s consider the setup shown here, where a field profile is followed by a thin lens placed one focal length f away, and a screen is another focal length past the lens. We can now make a few observations: 1. The lens images E(+)(x, y) to z = \u221e. Eoo(+)o(x,oy)distant screenzoo=oo0ryxzqxqqffEoo(+)o(x,oy)Fo[Eoo(+)o(x,oy)]ff 12.3 Fraunhofer Diffraction 227 2. Plane waves at different angles \u03b8x,y get mapped to different points \u03b8x,yf in the focal plane (on the screen). This is the wave-optics version of the parallel-ray argument above. 3. The lens maps the image at z = \u2212\u221e to the focal plane. 4. Because an image at infinity emits parallel rays, we can identify the images at z = \u00b1\u221e as being equivalent. The field at z = \u2212\u221e is the Fraunhofer diffraction pattern of E(+)(x, y) because the image at z = \u221e in the absence of the lens is the Fraunhofer pattern of the original image. 5. The lens then maps the field at \u2212\u221e to the focal plane. Thus, we can conclude that the image at the focal plane of the lens (i.e., the screen) is the Fourier transform of E(+)(x, y), but where f is now the propagation length scale: E(+) focalplane(x, y) = k 2\u03c0f \u02dcE(+) kx f, ky f. (12.43) (Fourier-transform via lens) Thus, at least within the paraxial approximation, thin lenses act as analog Fourier transform computers. Note that the phase factor (12.40) is negligible here, because in the paraxial limit, both the source and image fields are narrowly confined near the optical axis. In the Fraunhofer-diffraction case, the image was large due to the far-field requirement, but the lens takes care of compressing the image down to a small size here. 12.3.3 Example: Diffraction from a Double Slit As an easy example, let\u2019s compute the Fraunhofer diffraction pattern due to two narrow slits separated by distance a. The slits are uniformly illuminated by a monochromatic plane wave at normal incidence to the mask. Assuming the slit widths are much smaller than a, we can model the slits as \u03b4-functions: E(+)(x) = A [\u03b4(x \u2212 a/2) + \u03b4(x + a/2)]. (12.44", ") The Fourier transform is given by \u02dcE(+)(kx) = A \u221e \u2212\u221e e\u2212ikxx [\u03b4(x \u2212 a/2) + \u03b4(x + a/2)] dx = A eikxa/2 + e\u2212ikxa/2 = 2A cos kxa 2. (12.45) Thus, the far-field diffraction pattern at distance z is E(+)(x, z) = k 2\u03c0z \u02dcE(+) kx z = 2k \u03c0z A cos kax 2z = 2A \u221a \u03bbz cos \u03c0ax \u03bbz. (12.46) k/2\u03c0z as the normalization factor, rather than k/2\u03c0z, because this is a oneNote that we are using dimensional transform (we associate half of the full factor with each dimension\u2014note that the units come out correctly this way). In terms of angle, the diffraction pattern is (up to an overall factor) (double-slit diffraction pattern) \u02dcE(+) (k\u03b8x) = 2A cos ka\u03b8x 2 = 2A cos \u03c0a\u03b8x \u03bb. (12.47) This function vanishes when the argument is \u00b1\u03c0/2, \u00b13\u03c0/2, and so on, which corresponds to angles \u03b8x of \u00b1\u03bb/2a, \u00b13\u03bb/2a, and so on. Thus, the angular separation between fringes is \u2206\u03b8 = \u03bb/a. This result is consistent with the result of a simple relative-path-length calculation (Problem 5.1). (double-slit angular diffraction pattern) 228 Chapter 12. Fourier Optics 12.3.4 Example: Diffraction from a Sinusoidal Intensity-Mask Grating Another simple example is the diffraction of a plane wave by a sinusoidal intensity mask (or intensity grating). If the intensity mask has a transmission function tmask = cos2 gx, and is illuminated by a normally incident plane wave E(+), then the field is 0 E(+)(x) = tmaskE(+) 0 = E(+) 0 cos2 gx = E(+) 0 2 + E(+) 0 4 cos(2gx) = E(+) 0 2 + E(+) 0 4 ei2gx + e\u2212i2gx. (12.48) Using the Fourier-transform relation F ei\u03b1x = \u221e \u2212\u221e e\u2212ix(kx\u2212\u03b1) dx = 2\u03c0\u03b4(kx \u2212 \u03b1), (12.49) we can easily compute the Fourier transform of Eq. (12.48): F E(+)(x) = \u02dcE(+)(kx) = \u03c0E(+) 0 \u03b4(kx) + \u03c0E(+) 0 2 [\u03b4(kx \u2212 2g) + \u03b4(kx + 2g)]. (12.50) Thus, the angular diffraction pattern is \u02dcE(+)(k\u03b8x) = \u03c0E(+) 0 \u03b4(k\u03b8x) + \u03c0E(+) 0 2 [\u03b4(k\u03b8x \u2212 2g) + \u03b4(k\u03b8x + 2g)]. (12.51) Using \u03b4(ax) = \u03b4(x)/|a|, this becomes \u02dcE(+)(k\u03b8x) = \u03c0E(+) 0 k \u03b4(\u03b8x) + \u03c0E(+) 0 2k [\u03b4(\u03b8x \u2212 2g/k) + \u03b4(\u03b8x + 2g/k)]. (12.52) Thus, there are diffraction peaks at angles \u03b8x = 0 and \u00b12g/k = \u00b1g\u03bb/\u03c0. There are only 3 peaks, so this grating is special because it\u2019s sinusoidal: there are no higher-order diffraction peaks from this grating (we\u2019ll say more about this in a bit). It\u2019s important to see the Fourier-transform behavior at work here again\u2014as the period of the grating gets longer, g decreases, and so the diffraction pattern narrows, as you\u2019d expect from the uncertainty principle. (angular diffraction pattern of sinusoidal grating) Briefly, the corresponding pattern on a distant screen is E(+)(x, z) = k 2\u03c0z \u02dcE(+)(kx/z) = E(+) 0 2 \u03b4(x) + E(+) 0 4 [\u03b4(x \u2212 2gz/k) + \u03b4(x + 2gz/k)] (12.53) when the screen is at distance z. 12.", "3.5 Example: Diffraction from an Arbitrary Grating For a general periodic transmission mask of period \u03c0/g, with a normally incident plane wave, we can repeat the above analysis by decomposing the function into its Fourier coefficients: E(+)(x) = \u221e n=\u2212\u221e E(+) 0 cnei2gnx, (12.54) where the cn are the Fourier coefficients of the transmission function. Then the Fourier transform is \u02dcE(+)(kx) = 2\u03c0E(+) 0 \u221e n=\u2212\u221e cn\u03b4(kx \u2212 2gn), (12.55) 12.4 Fresnel Diffraction 229 so the angular diffraction pattern is \u02dcE(+)(k\u03b8x) = 2\u03c0E(+) 0 k \u221e n=\u2212\u221e cn\u03b4(\u03b8x \u2212 2gn/k). (12.56) Thus, for an arbitrary pattern, there are many diffraction peaks, with the diffraction peak of order n located at (angular diffraction pattern of arbitrary grating) \u03b8n = 2gn k, (12.57) (angle of nth diffraction order) and this peak has amplitude cn (or relative intensity |cn|2). 12.4 Fresnel Diffraction 12.4.1 Convolution Revisited If we stay in the paraxial approximation, but we don\u2019t require that the observation screen is in the far field, then we have the problem of Fresnel diffraction. Actually, Fresnel diffraction is also used to refer to the case where the wavefronts at z = 0 are not planar, but this is already accounted for in the Fraunhofer formalism that we presented in the last section, as it only modifies the particular form of the initial function E(+)(x, y). Thus, we will be concerned with the propagation of the optical wave to any distance, not just the far field. Actually we were already doing this when we introduced the Fourier recipe for wave propagation from Section 12.2.4. Recalling this procedure, we start with a field E(+)(x, y) at z = 0. Again, this is completely general, so this could be an arbitrary wave E(+) in (x, y) incident on a phase or amplitude mask (aperture) with transmission taperture, so that E(+)(x, y) = tapertureE(+) in (x, y). Then the frequency-domain solution (Eq. (12.16) is E(+)(x, y, z) = F \u22121 F E(+)(x, y) eikzz. In the paraxial approximation, but not necessarily in the far field, this becomes y )z/2k E(+)(x, y, z) = F \u22121 eikze\u2212i(k 2 E(+)(x, y) x +k 2 F Now we\u2019ll use the identity F \u22121 e\u2212i(k 2 x +k 2 y )z/2k = k 2\u03c0iz eik(x2+y2)/2z, (12.58) (12.59) (12.60) (12.61). which follows from the Gaussian Fourier-transform formula (see Section 3.2.1), but with a little care. Since the exponent is purely imaginary, the transform integral doesn\u2019t actually converge. But letting z \u2212\u2192 z \u2212 i\u03b2, where \u03b2 > 0, we can do the resulting convergent integral and then let \u03b2 \u2212\u2192 0 afterwards (this is saying that arbitrarily high spatial frequencies, corresponding to arbitrary fine detail, don\u2019t physically contribute). So, using Eq. (12.61) in Eq. (12.60), we can write E(+)(x, y, z) = F \u22121 F E(+)(x, y) F keikz 2\u03c0iz eik(x2+y2)/2z. (12.62) By the convolution theorem, E(+)(x, y, z) = E(+)(x, y) \u2217 g(x, y; z) \u221e \u221e = dx dy E(+)(x, y) g(x \u2212 x, y \u2212 y; z), \u2212\u221e \u2212\u221e (paraxial propagation as convolution) (12.63) 230 where Chapter 12. Fourier Optics g(x, y; z) := keikz 2\u03c0iz eik(x2+y2)/2z = eikz i\u03bbz eik(x2+y2)/2z. (12.64) This is the convolution or space-domain solution to the paraxial-propagation or", " Fresnel-diffraction problem. (Green function for paraxial propagation) 12.4.2 Paraxial Impulse Response So what exactly is this convolution kernel g(x, y)? Suppose we have an optical source that is a unit point, E(+)(x, y, z = 0) \u2212\u2192 \u03b4(x, y), and let\u2019s propagate it forward. The Fourier transform is easy: F [\u03b4(x, y)] = 1. (12.65) Then multiplying by the propagation factor exp(ikzz) = exp(ikz) exp[\u2212i(k 2 transform, the propagation induces the transformation x + k 2 y )z/2k] and inverting the \u03b4(x, y) \u2212\u2192 keikz 2\u03c0iz eik(x2+y2)/2z = g(x, y; z). (12.66) Thus, g(x, y) is the impulse-response function or Green function for the diffraction problem, and Eq. (12.63) is the Green function solution to the diffraction problem (in the paraxial approximation). (paraxial propagation of point source) 12.4.3 Far-Field (Fraunhofer) Limit In the far field, we should be able to recover the Fraunhofer diffraction pattern and thus rigorously derive the results of Section 12.3. Let\u2019s start by writing out the Green function solution (12.63) explicitly: E(+)(x, y, z) = \u221e dx \u221e dy E(+)(x, y) \u2212\u221e \u2212\u221e keikz 2\u03c0iz eik[(x\u2212x)2+(y\u2212y)2]/2z. (12.67) We can assume the Fraunhofer limit \u03b4xsource exp[ik(x2 + y2)/2z] is approximately unity (i.e., neglecting x2 compared to x2), so Eq. (12.67) becomes \u03bbz (far field/small source), so that the phase factor \u221a E(+)(x, y, z) = keikz 2\u03c0iz eik(x2+y2)/2z \u221e dx \u221e dy E(+)(x, y) exp \u2212\u221e \u2212\u221e \u2212i kx z x exp \u2212i. y ky z (12.68) Now defining \u03bax := kx/z and \u03bay := ky/z, we can write E(+)(x, y, z) = keikz 2\u03c0iz eik(x2+y2)/2z \u221e dx \u221e dy E(+)(x, y) ei(\u03baxx+\u03bayy). \u2212\u221e \u2212\u221e (Fraunhofer diffraction pattern) (12.69) This is just the Fourier transform, so E(+)(x, y, z) = keikz 2\u03c0iz eik(x2+y2)/2z \u02dcE(+)(\u03bax, \u03bay) = keikz 2\u03c0iz eik(x2+y2)/2z \u02dcE(+) kx z, ky z. (12.70) Thus, we recover the Fourier transform form for the Fraunhofer diffraction pattern, including the correct scaling factor and the phase factors that we merely indicated in our heuristic treatment. This includes the phase factor exp[ik(x2 + y2)/2z] for projecting a spherical wave front onto the flat screen at constant z. We can see this directly by noting that in the paraxial approximation, (Fraunhofer diffraction pattern) z = r2 \u2212 x2 \u2212 y2 \u2248 r \u2212 x2 + y2 2z, (12.71) 12.4 Fresnel Diffraction 231 so that upon making this replacement in the exp(ikz) factor, the diffraction pattern becomes E(+)(x, y, r) = keikr 2\u03c0ir \u221e dx \u2212\u221e \u221e dy E(+)(x, y) ei(\u03baxx+\u03bayy), (12.72) where we have also replaced z \u2248 r in the denominator. In this form, the extra phase factor is gone when the diffraction pattern is viewed on a spherical screen at constant r. \u2212\u221e (Fraunhofer diffraction pattern, constant r) 12.4.4 Example: Fresnel Diffraction from a Slit As one of the simplest examples, let\u2019s consider the Fresnel diffraction pattern from a single slit of width a. In this case, Eq. (12.63) becomes E(+)(x, z) = E(+) 0 k 2\u03c0iz eikz", " a/2 \u2212a/2 eik(x\u2212x)2/2z dx, (12.73) where we assume the y-integration has already been completed, and we\u2019ve also suppressed the y-dependence. Again, in suppressing the y-dependence, note that we are also suppressing half of the normalization factor k/2\u03c0iz from the 2D Gaussian transform (12.61)\u2014the y profile of the beam will simply remain a plane wave, not diffracting. Letting x \u2212\u2192 x + x, E(+)(x, z) = E(+) 0 k 2\u03c0iz eikz \u2212x+a/2 \u2212x\u2212a/2 eikx2/2z dx. Now letting x \u2212\u2192 \u03c0z/k x, E(+)(x, z) = E(+) 0 \u2212i 2 eikz k/\u03c0z (\u2212x+a/2) k/\u03c0z (\u2212x\u2212a/2) ei\u03c0x2/2 dx. Now defining the Fresnel integrals (12.74) (12.75) C(x) := x 0 \u03c0t2 2 cos dt S(x) := x 0 \u03c0t2 2 dt, sin (12.76) (Fresnel integrals) we can write the diffraction pattern in standard form as E(+)(x, z) = E(+) 0 \u2212i 2 eikz C x + k \u03c0z + iS a 2 \u2212 C x \u2212 k \u03c0z x + k \u03c0z a 2 \u2212 iS a 2 k \u03c0z x \u2212 a 2, (12.77) where we have used the fact that C(x) and S(x) are odd functions, because they are both integrals from 0 to x of an even integrand. The Fresnel integrals are easily evaluated on a computer these days. (Fresnel diffraction by a slit) The Fresnel integrals are plotted here. Note their oscillatory nature, which ultimately gives rise to the fringes in the diffraction patterns, and that they asymptotically settle to \u00b11/2. 232 Chapter 12. Fourier Optics The Fresnel integrals are famously plotted as the \u2018\u2018Cornu spiral,\u2019\u2019 which is a parametric plot of S(t) vs. C(t). The Fresnel integrals can be evaluated graphically using the Cornu spiral, but their evaluation much easier on a computer. Some diffraction patterns for the single slit are shown here. The slit edges are marked by vertical lines. Notice the gradual transition from something rectangular to the familiar far-field sinc pattern as the distance increases. z = 0.0025(a2/2\u03bb) z = 0.025(a2/2\u03bb) x-10100-11Co(x)0.8-0.80-0.50.5x-10100-1.411.41So(x)0.8-0.80-0.50.5Co(t)-110So(t)1-10(k/pz)1/2ox-15150normalized intensity01(k/pz)1/2ox-660normalized intensity01 12.5 Spatial Filters 233 z = 0.125(a2/2\u03bb) z = 0.25(a2/2\u03bb) z = 0.5(a2/2\u03bb) z = 5(a2/2\u03bb) 12.5 Spatial Filters The fact that a thin lens computes the Fourier transform of an optical field profile opens up a lot of intriguing possibilities\u2014namely, that of optical information processing. The basic setup for doing this is a system of two lenses. For simplicity we\u2019ll assume that the two lenses have the same focal length f. Then the propagation is over a distance 4f as shown here. For this reason this is called the 4-f imaging system for spatially filtering an image. The Fourier transform of the original field profile appears in the plane midway between the two lenses. The second lens undoes the Fourier transform and reconstructs the original image on the output plane. (Technically, the second lens performs a second Fourier transform, which is equivalent to the inverse transform with x \u2212\u2192 \u2212x and y \u2212\u2192 \u2212y, so the final image is inverted. We\u2019ll ignore this in the discussions here.) (k/pz)1/2ox-440normalized intensity01(k/pz)1/2ox-440normalized intensity01(k/pz)1/2ox-660normalized intensity01(k/pz)1/2ox-15150normalized intensity0.120Eoo(+)o(x,oy)inF", "o[Eoo(+)o(x,oy)]inEoo(+)o(x,oy)outffff 234 Chapter 12. Fourier Optics The point of this setup is that we can put in filters to modify the Fourier transform of the image, in either the amplitude or phase. In doing so, we can perform many useful enhancements or modifications to the image. The aperture in the Fourier plane defines the transfer function for the system. in (x, y), then in the Fourier-transform plane, the field is given in Recall that in the original wave is E(+) terms of the Fourier transform \u02dcE(+) in (kx, ky) by E(+) focalplane(x, y) = k 2\u03c0f \u02dcE(+) in kx f, ky f. (12.78) The spatial locations in the Fourier plane correspond to different spatial frequencies and thus to details in the images of different length scales. The low frequencies (and thus broad image features) are near the center of the transform plane, with the dc component being right in the center. The fine details are encoded in the high frequencies away from the center. This observation motivates some of the simplest spatial filters. 1. Low-pass filter. A circular aperture in the Fourier plane will block the high spatial frequencies and pass the low ones. The effect is to remove the fine details, blurring the image. If the filter is an aperture of diameter a, the cutoff frequency is given by the condition kr = kr/f, giving a cutoff frequency of kcutoff = ka 2f. That is, structures smaller than the length scale of \u03bbcutoff = 2\u03c0 kcutoff = 4\u03c0f ka = 2\u03bbf a (12.79) (12.80) are removed from the image. From the convolution theorem, as we\u2019ll come back to later, the output image is the convolution of the original image with the Fourier transform of the aperture. The aperture\u2019s Fourier transform is the Airy disk, and so the image is \u2018\u2018smeared\u2019\u2019 by the Airy disk, causing the blurring. 2. High-pass filter. A circular, absorbing spot in the Fourier does the opposite, blocking the low frequencies and passing the high. For a spot of diameter a, the cutoff is the same as for the low-pass filter. Thus, the image filter keeps only those structures smaller than 2\u03bbf /a. This filter is also called an edge detector because sharp edges generate high spatial frequencies and thus make it through the filter. 3. Directional filter. The high- and low-pass filters can also work only in one direction, with the obvious modification of the aperture. For example, a vertical slit preferentially passes vertical structures while blocking horizontal structures. Examples of these filters are shown here in simulations. The original photo (of Oregon moss on trees) has structures on many different scales. The low-pass filter wipes out the fine details of the moss, leaving k\u00a5broadfinedetails (y)finedetails (y)finedetails (x)finedetails (x)featuresk\u2248 12.5 Spatial Filters 235 only the branches. The high-pass filter does just the opposite: the structure of the branches is mostly wiped out, while the moss detail is preserved. The directional mask in the final photo is oriented at 45\u25e6; notice that only the branches perpendicular to the slit survive the filter, while the parallel branches are mostly erased. That\u2019s because branches parallel to the slit represent high spatial frequencies perpendicular to the slit, which get wiped out. original photo (intensity) low-pass filter high-pass filter directional mask 12.5.1 Spatial Filtering of a Gaussian Beam One important application of spatial filtering in real optical setups is the cleaning of a Gaussian beam. When propagating through real optical elements, Gaussian beams get contaminated with spatial intensity noise due to dust, imperfections, diffraction at the edges of objects, etalon fringes (when, say, passing through an uncoated window), aberrations, and so on. For example, an acousto-optic modulator acting as a switch tends to distort the beam profile, and so it is often desirable to restore it. The idea is to use a low-pass filter, here in the form of a pinhole. The Gaussian beam is focused through the pinhole, and under the right conditions a nearly ideal Gaussian profile appears on the other side. 236 Chapter 12. Fourier Optics Let\u2019s simplify things from the basic 4-f setup. We\u2019ll use a lens of focal length f to focus the beam through the pinhole, and let\u2019s assume that the waist of the unfocused beam occurs at the lens. The pinhole is much", " smaller than the unfocused beam, so the pinhole is a distance f from the pinhole. After the pinhole, a second lens of focal length fout recollimates the beam. In the absence of the pinhole, this is simply a telescope beam expander, and so the diameter of the beam is magnified by a factor fout/f. The most important part of the setup is the pinhole, so let\u2019s focus on that. Let the incident beam have radius w0. Then we can use the results in Section 6.5 from the ABCD Law, which gives the beam radius at the pinhole: w0,pinhole = \u03bbf \u03c0w0. (12.81) This assumes, of course, a clean Gaussian beam. The commonly recommended pinhole diameter is one that passes 99% of the incident power of a clean beam, which is Dpinhole = 2 \u00b7 1.5 \u00b7 w0,pinhole. (12.82) In this case, nearly all of the Gaussian beam passes through the pinhole. But noise, especially high-frequency noise, has its power distributed away from the pinhole in the Fourier-transform plane. So the pinhole blocks the power associated with spatial intensity noise on the beam. Thus, the recollimated beam is a very clean Gaussian to good approximation. There are two caveats here. One is that a small amount of intensity noise is left, corresponding to the same length scales as the beam diameter. This is usually not noticeable, and corresponds to shifts or slight distortion of the profile. In cases where there is significant low-frequency noise (such as the highly distorted output of a single-mode diode laser), it may be necessary to use a smaller pinhole and lose much more power. The other caveat is that the output isn\u2019t quite Gaussian. From the convolution theorem, the output is the convolution of a Gaussian and the Airy disk, so there are generally faint rings around a spatially filtered beam. These can be clipped off if necessary with an aperture after the recollimating lens. All this information is best visualized in the simulated example shown below. The first plot shows the intensity profile of a clean Gaussian beam. The second is the same profile, but with intensity noise superposed. Now the task is to clean this up as much as possible. Focusing the beam produces a very narrow Gaussian profile, shown in the next plot. The Gaussian itself is rather clean, but what is only apparent with a large vertical magnification (fourth plot) is that the baseline is contaminated with broadly distributed noise. This is what we expect from the Fourier transform argument above. We clip this noise away with pinhole (fifth and sixth plots, edges of pinhole marked by vertical lines). The very tips of the Gaussian profile are clipped as well, but essentially all of the noise is gone. The beam on the output is a nice-looking Gaussian (seventh plot), but notice that the amplitude is reduced compared to the first plot: power is lost in the cleaning process, since we are essentially projecting the beam into the Gaussian mode. The faint Airy rings are not visible on this plot, but they are on a logarithmic plot (last plot). The side lobes in this plot do not occur on the original beam. These are visible to the human eye, which itself has a logarithmic response to intensity. 2w\u00bafpinholerecollimatedbeaminput (dirty)beamoutf 12.5 Spatial Filters 237 original Gaussian beam beam contaminated with noise focused beam before pinhole focused beam (vertical zoom) after pinhole (vertical zoom) after pinhole (horizontal zoom) cleaned beam cleaned beam (log plot) xI(x)xI(x)xI(x)xI(x)xI(x)xI(x)xI(x)xlog[I(x)] 238 Chapter 12. Fourier Optics 12.5.2 Visualization of Phase Objects The \u2018\u2018obvious\u2019\u2019 spatial filters in the beginning of this section (low-pass, high-pass, directional) are intuitive examples of what you can do with optical spatial filters. But these days, it seems far easier just to digitize the image and do the same processing on a computer. So why do optical information processing? Sometimes it is convenient or useful to do so optically, as in cleaning a Gaussian beam. But sometimes it is impossible to digitize or even record optical information before optical processing. An important class of examples is in imaging phase objects, or objects that only influence the phase of probe light. We can model a phase object by a transmission coefficient of the form If this object is illuminated by a plane wave E(+) 0 (in the z = 0 plane), the resulting field becomes t(x, y) \u223c ei\u03c6", "(x,y). The intensity is 0 \u2212\u2192 E(+) E(+) 0 t(x, y) \u223c E(+) 0 ei\u03c6(x,y). I(x, y) \u223c 2|E(+) 0 \u03b7 |2, (12.83) (12.84) (12.85) which is a constant and thus has no information about the phase \u03c6(x, y). The point is that the transmission function has unit amplitude everywhere, and so doesn\u2019t modify the intensity of the transmitted light. So we can\u2019t just record the light on film and post-process the information. We could get at this information via interferometry or holographic techniques. Instead, we will look at how spatial filters can reveal this information. 12.5.2.1 Zernike Phase-Contrast Imaging One widely used phase-imaging method was developed by Frits Zernike, who won the 1953 Nobel Prize in Physics for this work. This method is commonly used, for example in biological microscopy to view transparent objects such as bacteria (that would otherwise require staining). Let\u2019s consider a phase object with a small phase shift of the form t(x, y) = ei\u03c6(x,y) \u2248 1 + i\u03c6(x, y) + O\u03c62. As we saw before, when uniformly illuminated, the intensity transmitted is I \u223c |1 + i\u03c6(x, y)|2 = 1 + O\u03c62, (12.86) (12.87) so that there is no intensity change (we are discarding the higher order terms in \u03c6 to be consistent with the initial expansion). Essentially, this is because the linear cross-terms cancel when squaring the field. The idea behind the Zernike method starts with the observation that the unity part of Eq. (12.86) is the dc component of the signal, while \u03c6(x, y) represents spatial structure, and thus has a wideband spectrum. The vanishing cross-terms are thus a product of the dc component and the rest of the spectrum. The idea is, what if we modify one of these to prevent the cancellation? Specifically, let\u2019s try a \u03c0/2 phase shift of the dc component: ei\u03c0/2 + i\u03c6(x, y) i [1 + \u03c6(x, y)] Now the linear term stays, so the intensity reflects the phase information. Actually, since the intensity change is linear in the phase (for small phase shifts), this method gives an image that is simple to interpret: other than an overall background, the intensity is a direct image of the phase. The other imaging methods we will discuss here produce more complicated functions of the phase. = 1 + 2\u03c6(x, y) + O\u03c62. (12.88) I \u223c = 2 2 Since the intensity increases with increasing phase shift, this method is called positive phase contrast imaging. We can also try a 3\u03c0/2 phase shift of the dc component, with the result I \u223c ei3\u03c0/2 + i\u03c6(x, y) 2 = (\u2212i) [1 \u2212 \u03c6(x, y)] 2 = 1 \u2212 2\u03c6(x, y) + O\u03c62. (12.89) 12.5 Spatial Filters 239 Now the intensity decreases with increasing phase, so this version is called negative phase contrast imaging. The requisite phase shift is accomplished here by a filter in the Fourier plane of the 4-f system that could be, for example, a glass plate with a transparent dielectric thin-film dot at the center that causes a relative phase shift of \u00b1\u03c0/2 of the small portion at the center of the image. 12.5.2.2 Central Dark-Ground Method An alternative to the Zernike method is the central dark-ground method. The idea is very similar, but instead of a mask with a phase-shifting dot, the mask has an opaque dot at the center to block the dc component. The details are left as an exercise (see Problem 12.12), but we can summarize the results. The advantage is that the background (dc component) is suppressed, giving a dark background. The disadvantage is that the intensity varies nonlinearly with phase (as \u03c62 for small \u03c6). 12.5.2.3 Schlieren Method The last method we will discuss here is important in the visualization of fluid flows, but requires considerably more mathematical sophistication to handle than the last two cases. The upshot of all this is that the schlieren method shows gradients in the phase of the object in a particular direction. The name comes from the German word schlieren, which means \u2018\u2018streaks.\u2019\u2019 One application of this method was to test glass lenses for impurities introduced", " during the fabrication process that would cause inhomogeneities in the refractive index. These would show up as streaks in the schlieren test, and hence the name. The setup is the same 4-f filtering setup. But rather than some sort of spot in the Fourier plane to modify the dc component, the schlieren method uses a knife edge in the Fourier plane to block half of the frequency spectrum. So we can represent the transmission function of the Fourier-plane mask as the Heaviside step function: tmask(x, y) = UH(x) := 1, x > 0 0, x < 0. (12.90) Since the y-direction is trivial to handle, we\u2019ll only do a one-dimensional analysis. Following the field through the 4-f setup, 1. Input field: E(+) in (x) 2. Transform plane: \u02dcE(+) in (kx) 3. After knife edge: \u02dcE(+) \u02dcE(+) in (kx) UH(kx) = in (kx) UH(kx) 4. Image plane: F E(+) in \u2217 \u02dcUH The last expression follows from the convolution theorem, and \u02dcUH(x) is the Fourier transform of the Heaviside function, which we must now evaluate. Note that for simplicity, we aren\u2019t being particularly careful with distinguishing Fourier transforms vs. inverse Fourier transforms, which means we are dropping factors of 2\u03c0 (though this will turn out not to matter) and ignoring some image flips (which do matter\u2014the final image should be flipped compared to what we will derive). (x) So let\u2019s now compute F [UH(x)]. Unfortunately, this is a little tricky, so we\u2019ll spend some time working this transform out. It turns out that the answer is F [UH(x)] \u2261 \u02dcUH(kx) = \u03c0\u03b4(k) + 1 ik. (12.91) Rather than derive this, we\u2019ll simply verify that the inverse transform recovers the Heaviside function. The inverse transform is (Fourier transform of step function) F \u22121 \u02dcUH(kx) = 1 2\u03c0 \u221e \u2212\u221e \u03c0\u03b4(k)eikx dk + 1 2\u03c0 \u221e \u2013 \u2212\u221e 1 ik eikx dk. (12.92) 240 Chapter 12. Fourier Optics Note the special notation in the integral in the last term. This is because this integral is singular due to the 1/k dependence. To handle the singularity, we take the Cauchy principal value by splitting the integral about the singularity, and then approaching the singularity symmetrically: \u221e \u2013 \u2212\u221e \u00b7 \u00b7 \u00b7 \u2261 lim \u03b4\u21920 \u2212\u03b4 \u221e \u00b7 \u00b7 \u00b7. \u00b7 \u00b7 \u00b7 + \u2212\u221e \u03b4 (12.93) (Cauchy principal value) In this way, the singular part cancels; this works because the 1/k changes sign on either side of k = 0. Thus, even though \u221e \u2013 \u2212\u221e dx x = 0, \u221e \u2212\u221e dx x (12.94) (12.95) diverges. This amounts to excluding the symmetric, shaded region in the graph below for the integrand 1/x, and then taking the limit as the width of the \u2018\u2018excised\u2019\u2019 region vanishes. Of course, if the singularity occurs elsewhere, the splitting is understood to be there. Now back to evaluating Eq. (12.92). F \u22121 \u02dcUH(kx) = 1 2 + 1 2\u03c0i lim \u03b4\u21920 \u221e \u03b4 eikx k dk + \u2212\u03b4 \u2212\u221e eikx k dk. (12.96) Flipping the integration limits in the last term, F \u22121 \u02dcUH(kx) = 1 2 + 1 \u03c0 \u221e 0 sin(kx) k dk = 1 2 + 1 2 sgn(x) = UH(x). (12.97) Here we used the result \u221e 0 sin(kx) k dk = \u00b1 \u03c0 2, (12.98) where the result is \u03c0/2 for x > 0 and \u2212\u03c0/2 for x < 0 (proving this requires a contour integral, which we won\u2019t get into here). Also, sgn(x) is the signum function, which returns the sign of the input (i.e., it is +1 for x > 0, \u22121 for x < 0, and 0 for x = 0). So now we\u2019ve verified that we have the correct Fourier transform of", " UH(x). But that\u2019s not quite what we wanted. What we really need is the inverse Fourier transform of UH(kx), to get \u02dcUH(x). So to handle the overall scaling and sign conventions, we just need to divide our result by 2\u03c0 and let i \u2212\u2192 \u2212i to obtain \u02dcUH(x) = 1 2 \u03b4(x) + i 2\u03c0x. (12.99) Now, as we concluded above in tracing the field through the filtering system, the output field is the convolution of the input field with \u02dcUH(x): E(+) out = E(+) in \u2217 \u02dcUH (x). (12.100) x1/xd-d 12.5 Spatial Filters 241 Recalling the form of the convolution integral from Eq. (11.5), we can thus write the output field for the schlieren setup as E(+) out = E(+) in (x) 2 + i 2\u03c0 \u221e \u2013 \u2212\u221e E(+) in (x) x \u2212 x dx. (output of schlieren imaging system) If E(+) in (x) is produced by a uniformly illuminated phase object, then for small phase shifts in (x) = E(+) E(+) 0 ei\u03c6(x) \u2248 E(+) 0 (1 + i\u03c6(x)). Then the output intensity is Iout Iin = 1 2 + i\u03c6(x) 2 \u2212 1 2\u03c0 \u221e \u2013 \u2212\u221e \u03c6(x) x \u2212 x dx + O \u03c62 2, where we have already used \u221e \u2013 \u2212\u221e dx x \u2212 x = 0. Multiplying out the intensity expression and keeping only first-order terms in \u03c6, the result is (12.101) (12.102) (12.103) (12.104) Iout Iin = 1 4 \u2212 1 2\u03c0 \u221e \u2013 \u2212\u221e \u03c6(x) x \u2212 x dx. Thus, we now see how the phase information is reflected in the intensity. The integral that appears in Eqs. (12.101) and (12.105) is of special significance, and it is called the Hilbert transform. For a function f (x), the Hilbert transform \u02c6f of f is given by (intensity output of schlieren imaging system) (12.105) \u02c6f (x) := 1 \u03c0 \u221e \u2013 \u2212\u221e f (x) x \u2212 x dx. (12.106) (Hilbert transform) As for the Fourier transform, we can also use an operator notation \u02c6f \u2261 H [f ] (12.107) (Hilbert-transform notation) to simplify some algebra. So, we see from Eq. (12.101) that the output field of the schlieren system is (up to an offset and scaling factor) the Hilbert transform of the input field, and from Eq. (12.105) we see that the output intensity is the Hilbert transform of the input phase, in the limit of small phase shifts. So we can write Eq. (12.105) more compactly as Iout Iin = 1 4 + \u02c6\u03c6(x) 2. (12.108) (schlieren intensity as Hilbert transform) But what does this mean intuitively? Well, the Hilbert transform is the convolution of a function f (x) with 1/x (up to a minus sign). But 1/x diverges at x = 0, and right where it diverges it suddenly changes sign. When you multiply f (x) by 1/x and integrate, you thus expect that it will take on a large value only when something upsets the cancellation at x = 0. That is, the value will be large when f (x) has a large slope at x = 0, so that the two sides of 1/x are unequally weighted. So the Hilbert transform is \u2018\u2018something like\u2019\u2019 a derivative. The complication is that 1/x falls off so slowly that the Hilbert transform picks up long-range information about the function, while the derivative is defined by information arbitrarily close a single point. In this sense, there isn\u2019t even really a limit (like \u2018\u2018slowly varying functions\u2019\u2019) for arbitrary functions in which the Hilbert transform closely approximates the derivative. But as far as visualizing the schlieren image is concerned, the derivative is a reasonable intuitive guide. 242 Chapter 12. Fourier Optics Let\u2019s go through a sloppy argument1 that \u2018\u2018shows\u2019\u2019 that the schlieren image is the derivative of the phase. We\u2019ll start off with the property that the Hilbert transform is its own inverse, up", " to a minus sign: H \u22121 = \u2212H, (12.109) (involution property of Hilbert transform) or H 2 = \u22121. This statement can be proved by using the fact that the Hilbert transform is linear, decomposing a test function into harmonic components (via the Fourier transform), and computing the Hilbert transform of the harmonic functions (see Problem 12.16). In any case, we can use Eq. (12.109) to write \u03c6(x) = \u2212 1 \u03c0 \u221e \u2013 \u2212\u221e \u02c6\u03c6(x) x \u2212 x dx. (12.110) Differentiating, we find 1 \u03c0 = \u2212 d\u03c6 dx \u221e \u2013 \u2212\u221e \u02c6\u03c6(x) (x \u2212 x)2 dx = \u2212 Here is the first problem: the integral with 1/x2 doesn\u2019t converge for generic functions \u02c6\u03c6, even using the Cauchy principal value. But let\u2019s go with it, since it gives a good intuitive result. Now something even sloppier: if \u02c6\u03c6 varies slowly, then 1/x2 is something like a \u03b4-function. Again, the integral of 1/x2 is divergent, while the integral of the \u03b4-function is not, and 1/x2 has no length scale, so it isn\u2019t clear what we mean by slowly varying. But, if we make the replacement, then the convolution is with the \u03b4-function and so, dropping constant factors, (12.111) \u02c6\u03c6(x) \u2217 1 x2. 1 \u03c0 d\u03c6 dx \u223c \u02c6\u03c6(x). (12.112) Thus, the Hilbert transform in Eq. (12.105) varies as the derivative of \u03c6(x). So, the conclusion is that the schlieren image is approximately an intensity background plus the derivative of the phase. But if we\u2019re concerned about actually being correct, it\u2019s better to stick with the Hilbert transform. 12.5.2.4 Numerical Examples Now we\u2019ll look at simulated images of a phase object to get a better idea of how the different phasevisualization methods compare. The original image is a plot of the phase shift, where the phase shifts are concentrated in randomly placed Gaussian blobs. The first set of images is in the small-phase regime (the largest phase shift is 0.2\u03c0), so all our linearized conclusions above hold. The other three images show intensity patterns at the output of the 4-f imaging system (neglecting inversion and magnification). The Zernike image almost perfectly reproduces the original phase, but has an intensity background. The darkground image has a dark background, but the Gaussian spots are somewhat distorted. The derivative nature of the schlieren image is apparent in the last image, giving a \u2018\u2018three-dimensional\u2019\u2019 sort of look. The second set of images is in the large phase-shift regime, where the largest phase shift is 3\u03c0. The images are visually more interesting but harder to interpret. Similar comments to the first set of images hold here, except that the \u2018\u2018wrapping\u2019\u2019 of the large phase shifts is apparent in the image. 1from John B. DeVelis and George O. Reynolds, Theory and Applications of Holography (Addison-Wesley, 1967). 12.5 Spatial Filters 243 phase plot (dark: 0, bright: 0.2\u03c0) Zernike phase-contrast image dark-ground image schlieren image 244 Chapter 12. Fourier Optics phase plot (dark: 0, bright: 3\u03c0) Zernike phase-contrast image dark-ground image schlieren image 12.6 Holography Recall from our earlier discussion of Fourier-optics propagation that an arbitrary monochromatic field E(+)(x, y, z) is completely determined by the field in a single plane, say E(+)(x, y, z = 0). This is because we can simply apply the propagation factor eikzz to E(+)(x, y, z = 0) (in frequency space) to obtain the field at any location z. We can exploit this idea that the three-dimensional field can be represented in two dimensions. A hologram (at least ideally) is a thin optical element with a complex transmission coefficient t(x, y) \u221d E(+)(x, y, 0). (12.113) 12.6 Holography 245 When uniformly illuminated by a monochromatic field, the holographic mask reconstructs E(+)(x, y, 0), up to an overall constant. Thus the hologram reconstructs the entire three-dimensional field E(+)(x, y,", " z), at least in the z > 0 region past the holographic mask. 12.6.1 Example: Single-Frequency Hologram Suppose that the hologram has a transmission of t(x, y) = ei(k sin \u03b8)x, (12.114) so that it modulates the signal with a single spatial frequency k sin \u03b8 in the x-direction. Illuminating this with an on-axis plane wave E(+)(x, y) = E(+), the transmitted (reconstructed) field is 0 Then propagating this field forward to arbitrary z, we find t(x, y)E(+)(x, y) = E(+) 0 ei(k sin \u03b8)x. t(x, y)E(+)(x, y, z) = E(+) 0 ei(k sin \u03b8)xei(k cos \u03b8)z, which is a plane wave with wave vector k = k (\u02c6x sin \u03b8 + \u02c6z cos \u03b8). (12.115) (12.116) (12.117) The original direction was k = k\u02c6z, so this hologram deflects a plane wave by an angle \u03b8. In other words, this hologram acts like a prism. In fact, a phase delay of exp[i(k sin \u03b8)x] = exp(ikd), where d = x sin \u03b8, literally is a prism. 12.6.2 Film Holograms But now the question is, how can we make a hologram? Reconstructing the phase information in the last example requires a glass wedge of varying thickness (i.e., a prism). But photographic detectors such as film and ccd cameras respond to intensity, not to the phase of the light. A hologram needs both intensity and phase information, so we need to employ some tricks. The key point in holography is this: we can use interference to make a hologram, mixing E(+) (the \u2018\u2018object field\u2019\u2019 that we want to record) with a reference plane wave E(+) records the intensity of the superposition: ref (x, y) = E(+) obj (x, y) ref,0. The photograph Irecord = 2|E(+) obj + E(+) ref |2 \u03b7. (12.118) Expanding out the product, Irecord = Iobj + Iref + 2 \u03b7 obj E(\u2212) E(+) ref + 2 \u03b7 obj E(+) E(\u2212) ref. (12.119) (recorded film signal) Alternately, we may write out the conjugate terms to make the interference more obvious, Irecord = Iobf + Iref + 2IobjIref cos[\u03c6obj(x, y)], (12.120) (recorded film signal) where \u03c6obj is the object phase, given by obj = |E(+) E(+) obj |ei\u03c6obj(x,y). (12.121) The hologram is the developed film that is exposed to this superposition. Both the phase and amplitude information are encoded in the intensity and hence the photograph (like in an interferometer). But now getting the information back out is a bit subtle because of the encoding. 246 Chapter 12. Fourier Optics How do we recover the information recorded on the film? The idea is to illuminate the holographic mask with the same reference light E(+) \u221a Iref. Then ref \u221d ref \u221d E(+) E(+) = tE(+) ref Iref + E(+) ref Iobj + IrefE(+) obj + E(\u2212) obj E(+) ref 2. 2 \u03b7 (12.122) Dividing through by E(+) ref, \u221d Iref + Iobj(x, y) + 2E(\u2212) ref \u03b7 E(+) obj (x, y) + 2E(+) ref \u03b7 E(\u2212) obj (x, y). E(+) E(+) ref (12.123) The first term is just a constant offset, and the second is the intensity profile of the object. The original field is recovered in the last two terms. The third term is proportional to the original field, which is what we\u2019re trying to get. The fourth term is the conjugate wave, the complex conjugate of the original wave. This term is a consequence of the fact that the film hologram is fundamentally a real-valued transmission mask. (reconstructed hologram field) 12.6.3 Hologram of a Plane Wave and Off-Axis Holography A simple example, similar to the example of Section 12.6.1, is the hologram of a single plane wave, Then the reconstructed", " wave is obj \u223c Iobjeik(sin \u03b8)x. E(+) E(+) obj\u221a Iref \u223c Iref + Iobj + IrefIobjei(k sin \u03b8)x + IrefIobje\u2212i(k sin \u03b8)x. (12.124) (12.125) Again, the third term is the reconstructed plane wave, while the fourth term is the extra conjugate wave. The conjugate wave turns out to be a plane wave at the same angle but reflected about the z-axis. filmreferencewaveEoo(+)refobjectwaveEoo(+)objreconstructfilmdevelopedreferencewaveEoo(+)refreconstructedEoo(+)obj(but get Eoo(-)obj too)filmEoo(+)refEoo(+)objreconstructfilmdevelopedEoo(+)refEoo(+)objEoo(-)objIrefoo+ooIobj 12.6 Holography 247 Once again, the reconstruction is not perfect because the film is fundamentally a real-valued transmitter, hence the conjugate wave. This hologram turns out to be just the sinusoidal diffraction grating from Section 12.3.4. However, this example motivates a method to get at the reconstructed object field along, called off-axis holography. The idea is to ensure that E(+) is at a sufficiently large angle (larger than its obj angular spread). Then the forward wave Iref + Iobj and conjugate wave E(\u2212) obj are all spatially separated from the reconstructed wave E(+) obj. 12.6.4 Setup: Off-Axis Reflection Hologram Then the optical setup for an off-axis reflection hologram is as shown here. A laser beam (monochromatic light source) is split, with half illuminating the object to form the object wave, and half going directly to the screen to form the reference wave. Notice that both beams must be expanded appropriately to cover the object and the film. Also, while recording, the relative phase of the two waves must be held very constant, so it is important to minimize the vibrations of the optics. Also, generally holographic film takes the form of glass plates for structural rigidity during exposure. Holographic plates must also have high resolution and are consequently rather expensive. To reconstruct the hologram, the idea is to illuminate the hologram with the same reference wave (although the exact details of the reference wave aren\u2019t too important). Then the wave after the hologram forms a virtual image of the original object in the same location behind the film. Another interesting thing to note is that it is possible to make synthetic holograms of objects that never really existed, for example on a computer. The appropriate pattern can be computed and printed out on a transparency, although usually the pattern will still need to be photographically reduced to achieve the required spatial resolution. (film)hologramreferencelaserbeamsplitterobjectwavewave(film)hologramlaservirtualobserverimage 248 12.7 Exercises Problem 12.1 Chapter 12. Fourier Optics Consider the Gaussian field profile E(+)(x, y, z = 0) = E(+) 0 ]. Use the Fouriertransform formalism for propagating the field to derive the form of E(+)(x, y, z) (in the paraxial approximation), and thus show that your results are consistent with the Gaussian beam solution to the paraxial wave equation. exp[\u2212(x2 + y2)/w 2 0 Problem 12.2 We derived a Fourier-transform recipe for propagating a field E(+)(x, y, z = 0) to arbitrary z under the assumption of a forward-propagating wave. Write down the analogous recipe for propagating the field to arbitrary z under the assumption of a backward-propagating wave. Explain. Problem 12.3 Consider a plane wave incident on a glass\u2013air interface (from the glass side), in total internal reflection. Use the language of Fourier optics and the diffraction limit to explain this phenomenon, and rederive expressions for the critical angle and skin depth from your discussion. (Note: what is the spectrum in kx for a steeply angled plane wave? Since the spectrum should be continuous across the interface, which acts as a thin optic, what is the difference in kx before vs. after the interface?) Problem 12.4 (a) Compute the far-field (Fraunhofer) diffraction pattern on a screen at location z due to a rectangular aperture of width b at z = 0. Assume the aperture is uniformly illuminated by a plane wave at normal incidence to the aperture. You may restrict your calculation to one dimension. (b) Write down the far-field diffraction pattern if the aperture is instead a pair of narrow slits separated by", " a distance a. (c) Use the convolution theorem to write down the far-field diffraction of an aperture consisting of two parallel slits of width b with centers separated by distance a. Problem 12.5 (a) Show that the far-field (Fraunhofer) diffraction pattern due to a uniformly illuminated (by a normally incident plane wave), circular aperture of radius a is E(+) 0 2\u03c0a2 \u03bbz J1(2\u03c0ar/\u03bbz) 2\u03c0ar/\u03bbz = E(+) 0 a r J1 2\u03c0ar \u03bbz, (12.126) 0 where E(+) is the amplitude of the illuminating wave, r is the radial distance from the center of the screen, and J1(x) is an ordinary Bessel function. [The first expression here emphasizes the connection to the \u2018\u2018jinc\u2019\u2019 function J1(r)/r.] The Bessel functions are defined by the integral Jn(x) = 1 \u03c0 \u03c0 0 cos(n\u03c6 \u2212 x sin \u03c6) d\u03c6, (12.127) and it may help you to know that they satisfy the recurrence relation nJn(x) + xJ n(x) = xJn\u22121(x), (12.128) which is most useful here for n = 1. (b) Make an intensity plot of this diffraction pattern. This is a well-known pattern called the Airy disk. 12.7 Exercises Problem 12.6 249 (a) Compute the far-field (Fraunhofer) diffraction pattern due to an infinite, regularly spaced array of narrow slits, where the distance between adjacent slits is a. (This is a simple model of a diffraction grating.) Assume the slits are uniformly illuminated. (b) Compute the far-field intensity pattern due to N regularly spaced slits with spacing a. Problem 12.7 Suppose that when illuminated by a plane wave at normal incidence, an aperture diffracts into an intensity pattern I(x, z) on a screen a long distance z away. We\u2019ll ignore the y-direction for simplicity. What is the effect on the intensity pattern if the incident wave makes a small angle \u03b8 with respect to the x-axis? The answer should be reasonably obvious but you should prove your answer mathematically. Problem 12.8 In diffraction theory, Babinet\u2019s principle is useful for calculating the diffraction pattern of an aperture if you know its complement. More specifically, suppose you illuminate an aperture with a normally incident plane wave in (x, y) = E(+) E(+) 0. (12.129) We will represent the aperture by the transmission function t(x, y), so that to find the diffraction pattern, we propagate the initial field to find the diffraction field E(+)(x, y, z). Let us denote the complement aperture by \u00aft(x, y), such that E(+)(x, y) = t(x, y)E(+) in (x, y) (12.130) t(x, y) + \u00aft(x, y) = 1, (12.131) and the diffraction field for the complement aperture by \u00afE(+)(x, y, z). The idea behind Babinet\u2019s principle is that if we add the fields for the aperture and complement, we just obtain the incident plane wave: E(+)(x, y) + \u00afE(+)(x, y) = t(x, y)E(+) in (x, y) + \u00aft(x, y)E(+) in (x, y) = E(+) in (x, y) = E(+) 0. (12.132) Note that a plane wave propagates to arbitrary z just as a plane wave (it does not diffract into a delta function in the far field\u2014there is no way to satisfy the far-field condition for a plane wave!). Thus, the propagation of the plane wave should be 0 \u2212\u2192 E(+) E(+) 0 eikz. (12.133) But the superposition principle (linearity of the wave equation) says that this should be the same as the sum of the diffracted fields of the two apertures: E(+)(x, y, z) + \u00afE(+)(x, y, z) = E(+) 0 eikz. Solving for the diffraction pattern of the complement aperture, \u00afE(+)(x, y, z) = E(+) 0 eikz \u2212 E(+)(x, y, z). (12.134) (12.135) This says that the diffraction pattern of a particular aperture is equivalent to the shadow of the complement aperture in the incident beam.", " There is, unfortunately, one complication in applying Babinet\u2019s principle, and it relates to the phase factor (\u2212i) that comes up in the coefficient k/2\u03c0iz in the free-space transfer function (particularly in k/2\u03c0iz, so the far-field limit). (Recall that in a one-dimensional diffraction problem, this factor is 250 Chapter 12. Fourier Optics \u221a \u2212i.) The same phase should be present in the plane wave for consistency. the corresponding phase is In the far-field limit, this is an artifact of noncommuting limits. For example, suppose we consider diffraction a Gaussian aperture, which you know should produce a Gaussian beam with longitudinal phase factor (12.136) including the Gouy phase factor. If we take the far-field limit z \u2212\u2192 \u221e, then tan\u22121(z/z0) \u2212\u2192 \u03c0/2, and the longitudinal phase becomes eikze\u2212i tan\u22121(z/z0), eikze\u2212i\u03c0/2 = \u2212ieikz. (12.137) Then if we take the plane-wave limit w0 \u2212\u2192 \u221e (which implies z0 \u2212\u2192 \u221e), the phase factor is the same. On the other, hand, if we take the plane-wave limit first, the Gouy phase reduces to tan\u22121(z/z0) \u2212\u2192 0, so the longitudinal phase reduces to eikz, as we expect for a plane wave. Thus, the extra (\u2212i) is essentially a generic Gouy phase, which results from assuming a confined field inside the aperture, and taking the far-field limit. The point is that instead of using the simple plane wave E(+) eikz in Eq. (12.135), you should use the limit of the diffraction pattern as the aperture disappears, so that the diffraction pattern and the reference plane wave have appropriately matching phases. Normally you don\u2019t have to worry about this, if you just want the diffraction pattern, because this is an overall phase. But now, we\u2019re concerned with interference involving the diffraction pattern, so it\u2019s important. 0 Of course, you can avoid all this by just considering the intensity of the diffraction pattern, and you should do this for this problem. The same argument above carries through for the intensities of the diffraction pattern, with the result \u00afI (+)(x, y, z) = I0 \u2212 I(x, y, z), (12.138) where I and \u00afI are the regular and complementary diffraction patterns, respectively, and I0 is the intensity of the incident plane wave. Use Babinet\u2019s principle to compute the Fraunhofer diffraction pattern for a plane wave around an opaque strip (e.g., wire) of material of width b. (Be careful to get the phase right when you set up the diffraction for the complement aperture.) Verify that your answer makes sense in any appropriate limits. Problem 12.9 Consider a Gaussian beam, with the focus occurring at z = 0, and propagating along the z-axis. Suppose a circular aperture is introduced in the z = 0 plane, centered with respect to the Gaussian beam\u2019s axis. Describe qualitatively the far-field (large z) profile of the apertured beam, paying special attention to the limits of large and small aperture diameter. Describe the mathematical form of the far-field profile for arbitrary aperture size. Note: you should mention convolutions in your answer. Problem 12.10 (a) Write down an expression for the Fresnel diffraction pattern for a knife edge illuminated by a plane wave at normal incidence. Write your answer in terms of the Fresnel integrals C(x) and S(x). (b) Make a few representative plots showing the diffraction pattern in the near, intermediate, and far fields. (Better yet, try to do all this in one plot!) Problem 12.11 Suppose you have a cruddy Gaussian beam of radius w0 = 1 mm and wavelength \u03bb = 780 nm that you want to clean up and expand. (a) Using a lens of focal length f = 25.4 mm to focus the beam, what is the best pinhole diameter according to the \u2018\u2018rule of thumb\u2019\u2019 to spatially filter the beam? You may assume that the lens is at the focus of the original beam. 12.7 Exercises 251 (b) If you want to expand the beam to a final radius of w0 = 10 mm, what focal length lens should you use to recollimate the beam after the pinhole? Problem 12.12 Following the treatment for the Zernike phase-contrast imaging system, work out the output intensity of a central dark-ground imaging system, which is a 4-f system with a filter consisting of an", " absorbing dot that blocks the dc component in the Fourier plane. Assume that the system input is from a phase object, of the form exp[i\u03c6(x)]. Problem 12.13 Go through the derivation in the notes of the output intensity of a schlieren imaging system, where the input is a phase object. Do this in detail, filling in the missing steps where appropriate. Also go through the argument that the Hilbert transform is \u2018\u2018something like\u2019\u2019 a derivative, so the schlieren image is something like the derivative of the phase shift. Problem 12.14 Compute the Hilbert transforms of the following functions: sin(x), cos(x), and exp(ix). From your answer, explain the following statement: for narrowband signals, the Hilbert transform is proportional to the derivative, without any silly approximations. Hint: use the convolution theorem to evaluate the transform integral. Problem 12.15 Compute the Hilbert transform of the unit box function rect(x): rect(x) := 1 0 if if |x| < 1/2 |x| > 1/2. Problem 12.16 Recall that the Hilbert transform is defined by H [f (x)] \u2261 \u02c6f (x) := 1 \u03c0 \u221e \u2013 \u2212\u221e f (x) x \u2212 x dx. (12.139) (12.140) Show that the Hilbert transform is its own inverse, provided you throw in an extra minus sign; that is, show that H \u22121 = \u2212H. (12.141) One method to prove this is as follows. Recall from Problem 12.14 that H [e\u00b1ix] = \u00b1ie\u00b1ix. From this you can show that H [eikx] = sgn(k)ieikx. Then compute what the function H [f (x)] looks like in frequency space, and finally do the same for H 2[f (x)]. Problem 12.17 Consider a normalized Gaussian of the form G(x) = 1 \u221a \u03c0 e\u2212x2. Show that the Hilbert transform \u02c6G(x) may be written where is Dawson\u2019s integral, plotted below. \u02c6G(x) = \u2212 2 \u03c0 F (x), F (x) := e\u2212x2 x dx ex2 0 (12.142) (12.143) (12.144) 252 Chapter 12. Fourier Optics Use the following outline to do this. (a) Start by assuming the relation between F (x) and \u02c6G(x) holds; we will transform this into the definition for F (x) using reversible steps. Write out the Hilbert transform: F (x) = \u2212 \u03c0 2 \u02c6G(x) = \u2212 1 2 \u221e \u2013 \u2212\u221e dx G(x \u221e \u2013 \u2212\u221e dx e\u2212x2 x \u2212 x. (12.145) Then change variables by letting x \u2212\u2192 x + x, and use the result to show that F (x) satisfies the differential equation F (x) = 1 \u2212 2xF (x). (b) Now the idea is to integrate this differential equation. Show that the differential equation e\u2212x2 d dx ex2 F (x) = 1, and integrate both sides from 0 to x, to obtain the definition of is equivalent to the one in part (a). (c) Now multiply through by e\u2212x2 F (x). (d) Plot \u02c6G(x) and G(x), and comment on the statement that the Hilbert transform is \u2018\u2018something like a derivative.\u2019\u2019 (e) To make the distinction more formal, obtain a large-|x| asymptotic form for \u02c6G(x) [and thus for F (x)], and show that it is a much slower decay than the tails of G(x). Problem 12.18 Consider the 4-f imaging system shown here, and suppose that there is nothing between the two lenses of focal length f. (12.146) (12.147) x-550Foo(x)0.6-0.60Eoo(+)o(x,oy)inFo[Eoo(+)o(x,oy)]inEoo(+)o(x,oy)outffff 12.7 Exercises Recall that the field in the focal plane is given by E(+) focalplane(x, y) = k 2\u03c0f \u02dcE(+) in kx f, ky f. 253 (12.148) Show, by computing the Fourier transform (not the inverse transform) of the focal-plane field, that the field in the output plane is out (x, y) = E(+) E(+) in (\u2212x, \u2212y). (12.149) Strictly speaking, this is", " only true in the paraxial approximation. This is the image inversion of the 4-f system that we ignored. Problem 12.19 (a) What is the diffraction limit that occurs in optical-wave propagation? Describe the effect mathematically, using the Fourier-transform propagation recipe. (b) Consider the 4-f imaging system from Problem 3, and suppose again that there is nothing between the two lenses of focal length f. Suppose also that the lenses are of diameter D. What is the new diffraction limit for this situation? For simplicity, you may assume that the image object represented by E(+) in (x, y) is much smaller than D, and you can also ignore the effect of the right-hand lens. Hint: Start by sketching some rays from the object through the system, to see how the image forms in the output plane. Then recall the argument for why the Fourier transform of the input image appears in the focal plane of the first lens. How is the field in the transform plane affected by the finite lens diameter? Problem 12.20 The Hilbert transform, being a convolution with 1/x, is \u2018\u2018something like\u2019\u2019 a derivative. But a true derivative, written as a convolution, involves a delta function. Prove (by induction) that f (n)(x) = (\u22121)n(f \u2217 \u03b4(n))(x), (12.150) where f (n)(x) is the nth derivative of f (x), and \u03b4(n)(x) is the nth derivative of \u03b4(x). Problem 12.21 Consider the integral \u221e x dx. \u2212\u221e (12.151) (a) Explain why this integral is problematic. (b) Give a definition of the Cauchy principal value of this integral (suitably generalized from the 1/x example from class), \u221e \u2013 \u2212\u221e x dx. (12.152) What is the value of this integral? (c) Give an alternate definition of this integral (let\u2019s call it the \u2018\u2018Schmauchy principal value\u2019\u2019) such that the value of the integral is 1. (Thus, you are giving another reason why the original integral is not well-defined without being careful.) Problem 12.22 (a) Write down a formal definition for the Cauchy principal value integral in the special case of the integrand \u221e \u2013 \u2212\u221e dx (x + a)(x \u2212 a). (12.153) 254 (b) Show that Chapter 12. Fourier Optics \u221e \u2013 \u2212\u221e dx (x + a)(x \u2212 a) = 0 (a \u2208 R, a = 0). (12.154) Hint: write out the partial-fraction decomposition of the integrand; that is, set the integrand equal to c+/(x + a) + c\u2212/(x \u2212 a) for undetermined constants c\u00b1. Problem 12.23 Compute the Hilbert transform of f (x) = 1/x. Hint: play with the transform integral to try to guess the answer, up to a constant factor. This is most of the way there. Then verify your answer using the inverse transform. You may also find the following formula useful (Problem 12.22): \u221e \u2013 \u2212\u221e dx (x + a)(x \u2212 a) = 0 (a \u2208 R, a = 0). (12.155) Problem 12.24 Consider a hologram of an object, recorded on a holographic film plate with light of wavelength \u03bb. Compare the reconstructed image when using a reference wave of wavelength \u03bb vs. a reference wavelength of some other wavelength \u03bb. For concreteness, take \u03bb < \u03bb, and be quantitative in your answer. (Hint: think of the holographic plate as basically a complex diffraction grating, and use diffraction theory to analyze this problem.) Chapter 13 Acousto-Optic Diffraction One important application of diffraction occurs when you can make a dynamical diffraction grating. The acousto-optic effect is one way to accomplish a dynamical grating. The idea is that a piezoelectric transducer driven by a radiofrequency (rf) wave (typically in the 10 MHz to 1 GHz range) causes an acoustic wave to propagate through an optical crystal. The sound wave leads to local compressions in the crystal and thus local density variations. This in turn leads to a spatial variation in the refractive index due to the acoustic strain. Generally, the other end of the crystal is cut at an angle to avoid a backreflected acoustic wave and thus a standing wave. For a monochromatic acoustic traveling wave, the resulting refractive index variation is n(x) = n0 + \u03b4n cos(krfx \u2212 \u03c9rft),", " (13.1) where n0 is the natural refractive index of the crystal and \u03b4n is the modulation amplitude induced by the acoustic strain. Typically, \u03b4n n0. As implied by the name, this device is useful for modulating the amplitude of a wave (since the \u2018\u2018amplitude\u2019\u2019 of the diffraction grating can be controlled electronically), and it turns out that this device can also change the frequency of a laser beam. A commercial acousto-optic modulator is shown below. This is an 80 MHz, TeO2 modulator designed for 780 nm light. The crystal is the chunk of transparent material on the left-hand side of the modulator. Notice the red piezo transducer on the right-hand side of the crystal, and notice the angle-cut of the left-hand side of the crystal to suppress the reflected acoustic wave. The crystal is surrounded by aluminum to help it dissipate heat from the acoustic wave. The rf power enters via the coaxial cable on the right, and the coils in the modulator form an impedance-matching circuit to maximize the coupling efficiency into the piezo. qinput lightrf drivesound wavediffracted beamspiezo~xz 256 Chapter 13. Acousto-Optic Diffraction This photo shows the same modulator passing a He-Ne laser beam. In the far-field, the beam is diffracted into several spots. This modulator is designed to maximize the power in a single diffraction order (operating in the Bragg regime, as we will discuss below), but in this regime and alignment multiple diffracted orders are visible. 13.1 Raman\u2013Nath Regime There are distinct regimes of acousto-optic diffraction, essentially depending on the optical path length of the crystal. We will first treat the simpler Raman\u2013Nath regime, where the acoustic wave is confined to a small width \u2206z. What this means is that we can treat the effect of the modulator as simply leaving a phase imprint on the beam, as if it were an aperture. That is, we don\u2019t have to worry about the fact that the beam 13.1 Raman\u2013Nath Regime 257 diffracts a bit from the front of the crystal before getting to the back of the crystal. Thus, this problem is essentially that of computing the Fraunhofer diffraction pattern due to a sinusoidal phase grating, although we will be a bit more careful and go beyond the paraxial approximation. The phase imprint due to the acoustic wave is just eik\u00b7\u2206r, where \u2206r is the path length through the acoustic wave and the dependence on the refractive index is implicit in the longitudinal spatial frequency kz. We can write this out explicitly: exp(ik \u00b7 \u2206r) \u2248 exp ik\u2206z n0 cos \u03b8 [n0 + \u03b4n cos(krfx \u2212 \u03c9rft)], (13.2) where \u03b8 is the incident angle (with respect to the z-axis), and the factor of cos \u03b8 accounts for the fact that the acoustic beam \u2018\u2018looks\u2019\u2019 wider to the optical wave when it propagates through at angle \u03b8 from the normal (in the past, we ignored this in the paraxial approximation). Again, this is an approximation assuming that \u2206z is small, so that the diffraction grating acts as a thin optic. It is also important to note the convention here: we are doing all the diffraction analysis inside the crystal, so k and \u03bb refer to the spatial frequency and wavelength inside the crystal, so that they are changed by a factor of n0 and 1/n0 from their respective vacuum values. Also, \u03b8 refers to the incident angle inside the crystal. Generally it is most useful to analyze the incident and diffracted beams outside the crystal, which requires a simple application of Snell\u2019s law. We can now write Eq. (13.2) in two parts as ik\u2206z cos \u03b8 exp(ik \u00b7 \u2206r) \u2248 exp exp ik\u2206z \u03b4n n0 cos \u03b8 cos(krfx \u2212 \u03c9rft). (13.3) The first factor is just an overall phase, which we can discard. We can define K := k\u2206z \u03b4n n0 cos \u03b8 (13.4) (phase-shift amplitude) as the amplitude of the phase imprint, so that exp(ik \u00b7 \u2206r) \u2248 exp [iK cos(krfx \u2212 \u03c9rft)]. Now we need to break this phase imprint into simpler components. One important identity is the generating function for the ordinary Bessel functions: exp x 2 t \u2212 1 t = \u221e j=\u2212\u221e Jj(x)tj. (", "13.5) (13.6) This function generates the Bessel functions Jn(x) in the sense that Jn(x) is the coefficient of tn in the power-series (Laurent) expansion. Letting t \u2212\u2192 ieix and x \u2212\u2192 K, we obtain a useful identity: eiK cos x = \u221e j=\u2212\u221e ijJj(K)eijx. (13.7) (modulation identity) Essentially, we just computed the Fourier series of eiK cos x. If the input is the plane wave E(+) in exp[i(k\u00b7r\u2212\u03c9t)], then after the phase imprint, we can use Eqs. (13.5) and (13.7) to write the effect of the diffraction as in exp[\u2212(k \u00b7 r \u2212 \u03c9t)] \u2212\u2192 E(+) E(+) in exp[i(k \u00b7 r \u2212 \u03c9t)] \u221e j=\u2212\u221e ijJj(K) exp[i(krfx \u2212 \u03c9rft)]. (13.8) Again, we have technically just computed the Fraunhofer pattern, taking into account a different incident angle, but we will improve on this in a moment. (Raman\u2013Nath diffraction pattern) 258 Chapter 13. Acousto-Optic Diffraction 13.1.1 Diffraction Amplitudes: Bessel Functions Each component in the sum corresponds to a different diffracted wave. The jth wave, or the jth order diffracted wave, has an amplitude of Jj(K), ignoring an overall phase. Each wave has a different wave vector and thus propagates in a different direction. We will analyze this fact more carefully below, but for now we will simply note that this means that it makes sense to compute the intensities of each diffracted wave separately. We are treating the waves as plane waves, but really they are beams with finite diameter, and we will generally be interested in the intensities of the waves in the far field when they are spatially separated. The intensity of the jth diffracted wave is thus given by Ij Iin = |2 |E(+) j |E(+) in |2 = J 2 j (K). (13.9) To get a feeling for what this means, we need to explore the Bessel functions a bit more. For our purposes here, we can define the Bessel function of order \u03bd via the integral J\u03bd(x) = \u221a (x/2)\u03bd \u03c0\u0393(\u03bd + 1/2) \u03c0 0 cos(x cos \u03b8) sin2\u03bd \u03b8 d\u03b8, (13.10) for \u03bd > \u22121/2, and where \u0393(\u03bd) is the Gamma function (which satisfies \u0393(n + 1) = n! for integer arguments). In Eq. (13.9), we only need the Bessel functions of integer order n, where the integral simplifies to \u03c0 Jn(x) = cos(x sin \u03b8 \u2212 n\u03b8) d\u03b8. (13.11) 1 \u03c0 0 The Bessel functions satisfy the recurrence relations J\u03bd+1(x) = 2\u03bd x J\u03bd(x) \u2212 J\u03bd\u22121(x); \u03bd x J\u03bd(x) = J\u03bd\u22121(x) \u2212 J \u03bd(x). (13.12) For integer orders, it is handy to know that Jn(x) = (\u22121)nJ\u2212n(x), and for real x, we can also write Jn(\u2212x) = (\u22121)nJn(x). For small arguments, the Bessel functions have the form Jn(x) \u2248 xn 2nn! (13.13) for n \u2265 0. In particular, J0(0) = 1 and Jn=0(0) = 0. As x increases from zero, the higher-order Bessel functions \u2018\u2018turn on later\u2019\u2019 compared to their low-order counterparts. The Bessel functions oscillate, and as x becomes large, they decay as x\u22121/2. The first few ordinary Bessel functions are plotted here. x02051015Jn(x)1-0.500.5no=o0no=o1no=o2no=o3no=o4 13.1 Raman\u2013Nath Regime For fixed argument x and large order \u03bd, the Bessel functions are approximately J\u03bd(x) \u2248 \u221a 1 2\u03c0\u03bd ex 2\u03bd \u03bd. 259 (13.14) In particular, notice that for large order \u03bd, the amplitude decays exponentially with \u03bd. We can conclude that as a function of n, Jn(x) has possibly", " large amplitude (oscillating with n) until n exceeds x. Then, the amplitude of Jn(x) drops quickly off to zero. Here is a plot of Jax(a) for several different values of a. Whenever x exceeds 1, the value of the Bessel function drops rapidly off, as we expect. The point of all this Bessel stuff is this: Practically, for Raman-Nath diffraction, the maximum diffraction order that contains substantial power is K. So we expect to see about 2K + 1 spots in the diffraction pattern of the device. 13.1.2 Frequency Shifts The jth order diffracted wave has a time dependence of the form exp(\u2212i\u03c9jt), where the frequency is \u03c9j = \u03c9 + j\u03c9rf. (13.15) (diffracted frequencies) Thus, each diffracted wave also experiences a frequency shift. There are two intuitive interpretations of this shift. The first is the view that the diffraction is a nonlinear scattering process of photons from a stream of phonons. A phonon is the sound-wave analog of a photon, or in other words a quantum of the acoustic field. Diffraction into the +|j| order corresponds to the absorption of j phonons by the photon, hence raising the photon energy. Diffraction into the \u2212|j| order, on the other hand, corresponds to the stimulated emission of j phonons by the photon, hence lowering the photon energy. Thus, we can view the frequency shift as a conservation of energy principle, where part of the energy is carried by the phonon. The other interpretation is purely classical: The diffraction grating is moving, and thus the deflected (diffracted) beams are frequency-shifted due to the Doppler effect. 13.1.3 Momentum Conservation Now we need to examine the spatial dependence of the diffracted waves. From Eq. (13.8), the wave vector of the jth diffracted wave is kj = k + jkrf, (13.16) (momentum conservation) x021Jax(a)0.5-0.50ao=o5ao=o10ao=o20ao=o100ao=o500 260 Chapter 13. Acousto-Optic Diffraction where krf = krf \u02c6x. Since the momentum of a photon is proportional to its wave vector, we can regard this relation as conservation of momentum. This is a vector relation, though so it is a bit more complicated than conservation of energy. Note that in this diagram, jkrf points opposite krf, so this corresponds to some negative diffraction order. Optical and acoustic frequencies are very different; optical frequencies are around 1014 Hz, while acoustic (rf/microwave) frequencies are in practical limited to the few GHz range, and are much more commonly around 100 MHz. Thus, \u03c9rf \u03c9, (13.17) so while relative frequency shifts are observable in very sensitive setups, for our purposes here we can write for absolute frequencies \u03c9j \u2248 \u03c9, (13.18) which implies that the spatial frequency of the diffracted wave is equal to that of the incident wave to very good approximation: kj \u2248 k. (13.19) Thus, the momentum-vector triangle must be isosceles, so that k and kj are symmetrical about the z-axis. From the diagram, we can see that or sin \u2206\u03b8j 2 = krf 2k, \u2206\u03b8j = 2 sin\u22121 krf 2k = 2 sin\u22121 \u03bb 2\u03bbrf, (13.20) (13.21) (diffraction angles) where \u03bbrf = 2\u03c0/krf is the wavelength of the rf acoustic wave. Now, it\u2019s important to realize that from the argument we just presented, the statement krf = krf \u02c6x (13.22) can\u2019t be quite true, even though the acoustic wave propagates in the x-direction. If we assume that kj \u2248 k to conserve energy, the only way to satisfy momentum conservation (i.e., to get the triangle to close) is to have the isosceles triangle shown above. But if krf = krf \u02c6x, then for any given input angle \u03b8, there can be at most one diffraction order that closes the momentum triangle. In particular, this can only be true for the jth order if so that \u03b8j = \u03b8 + \u2206\u03b8j = \u03b8 = \u2212 \u2206\u03b8j 2, \u2206\u03b8j 2. (13.23) (13.24) qDq\u2206k\u2206jokrfkxzqDq\u2206k\u2206jokrfkxz 13.2 Bragg Regime 261", " This seems to imply that there is only one diffracted beam. However, in the Raman\u2013Nath regime, many diffraction orders are observed. How is this possible? Well, we started off with the assumption that the acoustic wave was confined to a very narrow beam. The acoustic wave is subject to diffraction effects, so the narrow spatial profile implies that it has a large spread in the direction of krf, just as it would if it were an optical wave\u2014diffraction patterns from smaller apertures are larger in the far field for precisely this reason, and this is another manifestation of the uncertainty relation \u03b4x \u03b4kx \u223c 1 that we saw in the discussion of Fraunhofer diffraction. Since there are many different available orientations of krf available, conservation of energy and momentum \u2018\u2018pick out\u2019\u2019 the right phonon that closes the triangle. It is the thinness of the acoustic beam that makes the proper phonons available for many diffraction orders. The length of krf is the same; it\u2019s only the orientation that changes, so for \u2018\u2018nonresonant\u2019\u2019 diffraction orders, the momentum triangle has the same shape, but the entire thing is rotated appropriately. 13.2 Bragg Regime The Bragg regime of acousto-optic diffraction is the opposite extreme from Raman\u2013Nath diffraction. Instead of a narrowly confined acoustic wave, in the Bragg regime, we assume that the acoustic wave is very broad but relatively weak. In this case, the acoustic wave is approximately a plane wave, so that krf has a well-defined direction. The arguments of the last section hold here, so we can already conclude that diffraction can only occur into a single order. This turns out to be a useful fact, since nearly all of the incident wave can be diffracted into a single order. For this reason most acousto-optic modulators in laboratory use are Bragg-regime devices, since the diffracted beam is the intensity-modulated beam. It is thus desirable to have as much modulation range as possible as well as to not waste too much of the incident beam. We will thus consider only first-order diffraction. Simplifying the setup from the Raman\u2013Nath discus- sion above, we see that the incident angle must satisfy the Bragg condition, \u03b8 = \u03b8B := \u2206\u03b81 2 = sin\u22121 \u03bb 2\u03bbrf (13.25) (Bragg condition) Again, diffraction in this configuration corresponds to absorbing a single phonon. Diffraction to other orders in this geometry is suppressed. Dq\u2206k\u2206jokrfkxz~xzxzk\u00a1k\u00bakrfkqbqbqb 262 Chapter 13. Acousto-Optic Diffraction The diffracted wave makes an angle 2\u03b8B with respect to the incident wave. Notice that the Bragg condition is equivalent to the following condition: the variations in the optical wave in the direction of the acoustic wave must have twice the wavelength of the acoustic wave. There is an interpretation to this, which is as follows: each of the \u2018\u2018bumps\u2019\u2019 in the refractive-index profile reflects a little bit of the light in the opposite direction. If the small reflections from successive bumps interfere constructively, then a lot of the light can be reflected in the opposite direction. The condition for this is that one round-trip between two bumps is 2\u03c0 in phase\u2014one round trip in the direction of the acoustic wave, that is. Thus, the array of bumps acts as a sort of generalized Fabry\u2013Perot cavity, whereby if the transverse optical wavelength matches the grating wavelength, efficient diffraction occurs. Of course this argument works for larger angles, when the transverse optical wavelength is smaller (e.g., when the transverse optical wavelength exactly matches the acoustic wavelength): these larger angles correspond to higher diffraction orders. Recall that forward propagation over a distance z is equivalent to multiplication by exp(ikzz). We will want to be a little more general than this, so let\u2019s say that when you propagate the wave forward in the z-direction from z to z + dz, the field picks up some phase change d\u03c6: Expanding the left-hand side, we can write E(+)(z + dz) = ei d\u03c6E(+)(z). E(+)(z + dz) = E(+)(z) + \u2202E(+) \u2202z dz, (13.26) (13.27) since for infinitesimals, dz2 = 0. Expanding the exponential in the right-hand side, we can similarly write since d\u03c62 = 0. Thus, we can write Eq. (13.26) in", " differential form as ei d\u03c6E(+)(z) = E(+)(z)[1 + i d\u03c6], \u2202E(+) \u2202z = i \u2202\u03c6 \u2202z E(+). (13.28) (13.29) In principle, this evolution equation is the same as the wave equation, but this equation is first-order: it implicitly considers only forward-propagating waves (kz > 0), whereas the full wave equation handles propagation in any direction. This is fine as long as there isn\u2019t anything like a dielectric interface that would generate a reverse wave. Now let\u2019s calculate the phase change d\u03c6 as the wave propagates in the acousto-optic medium. Note that we will want to keep track of the phase evolution along each of the diffracted waves, so we need to compute the phase evolution along the wave vectors k0 and k1. As z \u2212\u2192 z + dz, the path length along k0,1 is dz/ cos \u03b80,1, where \u03b80,1 are the angles made by k0,1 from the z-axis. From our discussion above, the Bragg condition is |\u03b80| = |\u03b81| = \u03b8B for efficient diffraction, but we will consider this more general setup to handle angular misalignment. The two angles will still be subject to the constraint |\u03b80| + |\u03b81| = 2\u03b8B. Recalling that k = k0 \u2248 k1, we can write the phase change as d\u03c60,1 = n n0 k dr0,1 = n n0 k dz cos \u03b80,1 = k dz n0 cos \u03b80,1 [n0 + \u03b4n cos(krfx \u2212 \u03c9rft)], (13.30) where dr0,1 is the path length along k0,1 corresponding to a displacement dz along z. Then \u2202\u03c60,1 \u2202z = k cos \u03b80,1 + k \u03b4n 2n0 cos \u03b80,1 ei(krfx\u2212\u03c9rft) + e\u2212i(krfx\u2212\u03c9rft). (13.31) xzk\u00a1k\u00bakrfkq\u00baq\u00a1q\u00ba 13.2 Bragg Regime 263 The first term here represents free-space evolution in the absence of the acoustic wave, the second term represents coupling of diffraction order j \u2212\u2192 j + 1, and the third term represents coupling of diffraction order j \u2212\u2192 j \u2212 1. Let\u2019s simplify this by defining the phase-rotation rate \u03baj := k cos \u03b8j = k 1 \u2212 (kj,x/k)2 \u2248 k + k 2 j,x 2k, (13.32) (phase-precession rate) where the last expression holds in the paraxial approximation. We will also define the Rabi frequency \u2126 := k \u03b4n n0 cos \u03b8j. (13.33) (Rabi frequency) In principle the Rabi frequency is different for each wave, so we should keep track of two Rabi frequencies \u21260,1. However, we will assume that |\u03b80| is very close to |\u03b81|. We will see that even small differences in the two angles have a substantial impact on the diffraction via the \u03baj, but since \u03b4n/n0 is small, any difference in the Rabi frequencies is a much smaller effect and thus ignorable. With these definitions, the phase shift becomes \u2202\u03c60,1 \u2202z = \u03ba0,1 + \u2126 2 ei(krfx\u2212\u03c9rft) + e\u2212i(krfx\u2212\u03c9rft). (13.34) Now let\u2019s consider the evolution of the zeroth and first diffraction orders. We\u2019ll ignore the others, since they don\u2019t satisfy the Bragg condition and thus they shouldn\u2019t be excited. We\u2019ll denote the zeroth-order wave by E(+)(k0, z) and the first by E(+)(k1, z), where the wave vectors are given by kj = k0 + jkrf = k0 + jkrf \u02c6x. The total field is the sum of the two waves, and with the explicit time dependence we can write E(+)(z, t) = E(+)(k0, z) e\u2212i\u03c90t + E(+)(k1, z) e\u2212i\u03c91t. (13.35) Note that we\u2019ve already included the exp(ikj \u00b7 r) in the z-dependence", " of E(+)(kj, z), so we haven\u2019t written it out explicitly here. Then recalling that the phase evolution \u03c6 depends on the wave vector, the evolution equation (13.29) becomes E(+)(k0, z) e\u2212i\u03c90t + E(+)(k1, z) e\u2212i\u03c91t = \u2202 \u2202z i\u03ba0E(+)(k0, z) e\u2212i\u03c90t + i\u03ba1E(+)(k1, z) e\u2212i\u03c91t + + i\u2126 2 i\u2126 2 E(+)(k0, z) eikrfxe\u2212i\u03c91t + E(+)(k1, z) eikrfxe\u2212i\u03c92t + i\u2126 2 i\u2126 2 E(+)(k0, z) e\u2212ikrfxe\u2212i\u03c9\u22121t E(+)(k1, z) e\u2212ikrfxe\u2212i\u03c90t, (13.36) where \u03c9j = \u03c90 + j\u03c9rf. Picking out the frequency component of this equation at \u03c90 (e.g., by multiplying through by exp(i\u03c90t) and integrating over a time interval of length 2\u03c0/\u03c90) and discarding the common time dependence, we obtain the equation of motion \u2202 \u2202z E(+)(k0, z) = i\u03ba0E(+)(k0, z) + i\u2126 2 E(+)(k1, z) e\u2212ikrfx. Similarly, for the component at frequency \u03c91, \u2202 \u2202z E(+)(k1, z) = i\u03ba1E(+)(k1, z) + i\u2126 2 E(+)(k0, z) eikrfx. (13.37) (13.38) In these two equations, the factors of exp(\u00b1ikrfx) account for the momentum change on absorbing or emitting a phonon. Notice that E(+)(kj, z) \u223c exp(ikj \u00b7 r) and kj = k0 + jkrf \u02c6x, so all terms in Eq. (13.37) have the 264 Chapter 13. Acousto-Optic Diffraction common dependence of exp(ik0 \u00b7 r), and all terms in Eq. (13.38) have the common dependence of exp(ik1 \u00b7 r). As long as we keep this in mind, we don\u2019t have to write out the explicit phase dependence, since what we\u2019re really after is the relative amplitudes of the two waves. Thus, we\u2019ve made things much simpler, reducing the wave equation to a pair of ordinary differential equations. Defining the normalized electric fields, Eqs. (13.37) and (13.38) become cj(z) := E(+)(kj, z) E(+) in, \u2202zc0 = i\u03ba0c0 + i \u2202zc1 = i\u03ba1c1 + i \u2126 2 \u2126 2 c1 c0, (13.39) (13.40) where \u2202z \u2261 \u2202/\u2202z. We can solve these coupled equations by decoupling them. Differentiating the \u2202zc0 equation, Using the other equation for \u2202zc1, \u2202 2 z c0 = i\u03ba0\u2202zc0 + i \u2126 2 \u2202zc1. \u2202 2 z c0 = i\u03ba0\u2202zc0 + i \u2126 2 (i\u03ba1c1) + 2 i \u2126 2 c0, and using the \u2202zc0 equation to eliminate c1, we find \u2202 2 z c0 \u2212 i(\u03ba0 + \u03ba1)\u2202zc0 \u2212 \u03ba0\u03ba1 \u2212 \u21262 4 c0 = 0. Repeating this procedure yields the same equation of motion for c1: \u2202 2 z c1 \u2212 i(\u03ba0 + \u03ba1)\u2202zc1 \u2212 \u03ba0\u03ba1 \u2212 \u21262 4 c1 = 0. (13.41) (13.42) (13.43) (13.44) Let\u2019s concentrate on obtaining an explicit solution for c1(z). We can factor Eq. (13.44) and write it as \u2202z \u2212 i \u2126+ 2 \u2202z \u2212 i \u2126\u2212 2 c1 = 0 where the quadratic roots are given by where The general solution to Eq. (13.45) is \u2126\u00b1 = (\u03ba0 + \u03ba1) \u00b1 \u02dc\u2126, ", "\u02dc\u2126 := (\u03ba0 \u2212 \u03ba1)2 + \u21262. c1(z) = A+ei\u2126+z/2 + A\u2212ei\u2126\u2212z/2, (13.45) (13.46) (13.47) (13.48) where A\u00b1 are constants to be determined by the initial conditions. Note that given the form of \u2126\u00b1, we can equivalently write c1(z) = ei(\u03ba0+\u03ba1)/2 A1 cos + A2 sin, (13.49) \u02dc\u2126z 2 \u02dc\u2126z 2 where A1,2 are (other) constants to be determined by the initial conditions. 13.2 Bragg Regime 265 Initially, upon entering the sound field, all of the incident light is undiffracted, so c0(z = 0) = 1 and c1(z = 0) = 0. Thus, we can see that A1 = 0. Also, from the \u2202zc1 equation in Eq. (13.40) gives while differentiating Eq. (13.49) gives so that A2 = i\u2126/ \u02dc\u2126. Thus, the solution is \u2202zc1(z = 0) = i \u2126 2, \u2202zc1(z = 0) = \u02dc\u2126A2 2, (13.50) (13.51) c1(z) = iei(\u03ba0+\u03ba1)/2 \u2126 \u02dc\u2126 sin \u02dc\u2126z 2. (13.52) (Bragg-diffraction amplitude) Since we\u2019re only interested in relative amplitudes, we can obtain |c1| from the conservation constraint |c0|2 + |c1|2 = 1. The intensity of the diffracted beam is given by I1 Iin = |c1(z)|2 = 2 \u2126 \u02dc\u2126 sin2 \u02dc\u2126z 2, (13.53) (Bragg-diffraction intensity) where of course I1 + I0 = Iin. We can also write this in the form I1 Iin = |c1(z)|2 = \u2126 \u02dc\u2126 2 1 2 \u2212 1 2 cos \u02dc\u2126z. (13.54) (Bragg-diffraction intensity) In this form we can see that the intensity oscillates sinusoidally with frequency \u02dc\u2126. Now we are in a position to make a few observations about acousto-optic Bragg diffraction 1. Rabi flopping. The intensity oscillates between the incident and diffracted wave at rate \u02dc\u2126. This behavior (oscillation of energy between two coupled oscillators) occurs often in physics, but in particular this is equivalent to a two-level atom (the two optical modes) driven by a classical laser field (the sound wave, in this case). With this analogy, the intensity oscillations are Rabi oscillations, and \u02dc\u2126 is the generalized Rabi frequency. 2. Resonance condition. The (peak-to-peak) amplitude of the intensity oscillations from Eq. (13.53) is \u21262 \u02dc\u21262 = \u21262 \u21262 + (\u03ba0 \u2212 \u03ba1)2. (13.55) Since this amplitude \u2264 1, there are incomplete Rabi oscillations unless the resonance condition \u03ba0 = \u03ba1 is satisfied. Putting in the definitions of the \u03baj, this condition is equivalent to |k0,x| = |k1,x|. But \u2206kx = krf for first-order diffraction, so the resonance condition occurs when \u2212k0,x = k1,x = krf/2. Expanding this further, of course, just leads to the Bragg condition \u03b80 = \u03b8B = sin\u22121(\u03bb/2\u03bbrf). (13.56) (13.57) (13.58) 266 Chapter 13. Acousto-Optic Diffraction 3. Resonant Rabi flopping. On resonance (\u03ba0 = \u03ba1), the generalized Rabi frequency \u02dc\u2126 reduces to the ordinary Rabi frequency \u2126. Then the diffraction efficiency becomes I1 Iin = sin2 \u2126z 2 = sin2 k \u03b4n 2n0 cos \u03b8B z. (13.59) Since the index modulation amplitude \u03b4n is related to the rf drive power, we can see that for best diffraction efficiency there is an optimum rf drive power for a given crystal length. 4. Mirror image. We did the analysis for the +1 diffraction", " order, but the analysis works just as well for diffraction into the \u22121 order, the differences being the orientation and that the diffracted beam is downshifted in frequency instead of upshifted like the +1 order. 5. Resonant selection. Only one order can be on resonance (i.e., satisfy the Bragg condition) at a time. Again krf is well-defined, so we need a particular input angle \u03b80. Diffraction to the jth order requires that |k0,x| = |kj,x|, which in turn requires that the zeroth and jth order diffracted waves to subtend equal angles about the normal to krf. 6. Higher-order scattering. We only considered first-order diffraction, but higher-order Bragg diffraction is also possible by absorbing or emitting multiple phonons. The intermediate waves are not resonantly coupled, so this process proceeds exponentially more slowly (in terms of interaction length for fixed \u03b4n) than first-order diffraction. Thus, practical devices usually stick to first-order diffraction. 13.2.1 Efficiency We can manipulate the form of the (optimally aligned) diffraction efficiency in the Bragg regime from Eq. (13.59): I1 Iin = sin2 k \u03b4n 2n0 cos \u03b8B z = sin2 \u03c0 \u03b4n z n0\u03bb cos \u03b8B = sin2 \u03c0z 2 cos \u03b8B \u221a \u03bb \u221a 2 \u03b4n n0. (13.60) Now we can translate the index modulation \u03b4n into properties of the crystal and acoustic wave. We can write M Iacoustic := 2(\u03b4n)2 n 2 0, (13.61) (figure of merit) where Iacoustic is the intensity of the acoustic wave and M is the figure of merit that characterizes the acousto-optic crystal. Thus, we can write the efficiency as I1 Iin = sin2 \u03c0z 2 cos \u03b8B \u221a \u03bb M Iacoustic. (13.62) (Bragg-diffraction efficiency) Noting that the dimensions of the acoustic wave are the interaction length z and the crystal height h, we can write the acoustic intensity as Iacoustic = Pacoustic zh, (13.63) (acoustic intenstiy) where Pacoustic is the total power in the acoustic wave. qbqbqbk\u2013\u00a1k\u00bak 13.3 Borderline 267 13.2.2 Example: TeO2 Modulator (Bragg Regime) To get a feeling for the numbers, we\u2019ll look at a commercial TeO2 (\u2018\u2018tellurium dioxide\u2019\u2019) modulator designed to work with a 150 MHz rf drive at 780 nm (vacuum) optical wavelength. For TeO2, the figure of merit is M = 34.5 \u00d7 10\u221215 s3/kg, the acoustic-wave velocity is vrf = 4.26 km/s, and the refractive index is n0 = 2.35. The crystal dimensions are z = 15 mm thickness and h = 1 mm height. We will use the paraxial approximation throughout. The Bragg angle is \u03b8B \u2248 \u03bb\u03c9rf 4\u03c0vrf = 14 mrad = 0.79\u25e6. (13.64) The deflection angle of the diffracted beam is 2\u03b8B = 1.6\u25e6. Recall that all geometrical quantities refer to their values inside the crystal, so after exiting the crystal, the deflection angle is 3.7\u25e6. Maximum diffraction efficiency occurs when Since cos \u03b8B \u2248 1, we can use Eq. (13.62) to write an expression for the optimal acoustic intensity, I1 Iin = sin2(\u03c0/2). Iacoustic, optimum = \u03bb2 2z2M, or equivalently for the optimal power Pacoustic, optimum = \u03bb2h 2zM = 0.1 W. (13.65) (13.66) (13.67) The actual power applied to the piezo for the best diffraction efficiency is actually 0.8 W, which is of the same order of magnitude, but implies that much power is wasted in the piezo transducer in converting electrical to acoustic energy. We can also use Eq. (13.60) to see that at maximum diffraction efficiency, the index modulation amplitude is given by \u03b4n n0 \u2248 \u03bb 2z = 1.1 \u00d7 10\u22125. (13.68) This is a very small index change, but its effect builds up coherently over the optical path length of the modulator. 13.3 Borderline We have discussed the operation of acousto-optic modulators in the Raman\u2013", "Nath and Bragg regimes. But how do we distinguish which regime applies to a particular device? Recall that what distinguishes the two regimes is whether or not the direction of krf is well-defined, which says whether or not diffraction to many orders is allowed. One approach is to examine the Klein\u2013Cook parameter1 Q := (angular spread scale of diffracted beams) (angular spread of rf phonons), which we can evaluate as Q = \u2248 = (2\u03b8B) (1/krfz) (krf/k) 1/krfz k2 rfz k, (13.69) (13.70) 1W. Klein and B. Cook, \u201dUnified approach to Ultrasonic Light Diffraction,\u201d IEEE Transaction on Sonics and Ultrasonics, SU-14, 123 (1967). 268 Chapter 13. Acousto-Optic Diffraction where we used the paraxial form for \u03b8B, and finally if again vrf is the phase velocity of the acoustic wave in the medium, Q = \u03bb\u03c92 rfz 2\u03c0v2 rf. (13.71) (Klein\u2013Cook parameter) Then we can distinguish the two regimes as follows: 1. Raman\u2013Nath (large angular spread of phonons): Q 1 2. Bragg (small angular spread of phonons): Q 1 The borderline is usually taken to be from 1 to 2\u03c0, which is some intermediate regime where neither treatment is really appropriate. However, Q = 1 is a practical dividing line between the two regimes. For the TeO2 modulator in the previous example, Q = 39, so this device is clearly in the Bragg regime. 13.4 Exercises 13.4 Exercises Problem 13.1 269 A phase-modulated optical wave has the form E(+)(x, t) = E(+) 0 ei[kx\u2212\u03c9t+\u03b4\u03c6 sin(\u03c9modt)], (13.72) where \u03c9mod is the modulation frequency. Such a wave could result, for example, by running the wave through an electro-optic crystal, whose refractive index is modulated by an applied ac signal with frequency \u03c9mod. (a) For a wave with time dependence exp[\u2212i\u03c6(t)], we can define the instantaneous frequency as \u03c9inst := d\u03c6 dt. (13.73) Compute the instantaneous frequency of the phase-modulated wave and thus show that the frequency oscillates about \u03c9. That is, phase modulation is in some sense the same as frequency modulation. (b) Write the phase-modulated wave as a sum of plane waves, with the general form \u221e j=\u2212\u221e cjei(kx\u2212\u03c9j t). (13.74) Hint: start by writing down a Bessel series for the function exp(iK sin x). (c) From your answer to (b), argue that the intensity spectrum (as viewed through a Fabry\u2013Perot spectrum analyzer) consists of a series of peaks with relative intensity J 2 j (\u03b4\u03c6). You may assume the response of the Fabry\u2013Perot analyzer is slow compared to the modulation frequency. This phasemodulation technique is commonly used in the laboratory to shift the frequency of a laser or to generate multiple laser frequencies. Problem 13.2 Suppose you observe that for a particular acousto-optic modulator in the Bragg regime, 780 nm light (wavelength in air) diffracts efficiently into the first order, with an angle of 2\u25e6 between the undiffracted and first-order beams. The most efficient diffraction is observed for 0.8 W of input rf power. (a) If 850 nm light is used instead, what is the new diffraction angle? (b) What is the rf power required for most efficient diffraction at the new wavelength? (c) The coupled-wave theory of Bragg diffraction predicts that the best diffraction efficiency is 100%. However, in practice the best efficiencies are more like 90%, so obviously we made some assumptions that aren\u2019t quite true. Give an example of something we didn\u2019t take into account that could explain the lower practical efficiencies of acousto-optic modulators. Problem 13.3 In the Bragg regime, the optimum diffraction efficiency occurs when the incident angle is \u03b8B from the normal to the acoustic-wave propagation direction. (a) Show that for small \u03b8B, an angular misalignment of \u03b4\u03b8 = \u03b4n 2n0\u03b8B (13.75) causes the diffraction efficiency to drop by a factor of 2. You can use the following general outline: First, note that the best diffraction efficiency is given by \u21262/ \u02dc\ufffd", "\ufffd2, and this is the part of the efficiency that is sensitive to angle. This drops to 1/2 when (\u03ba0 \u2212 \u03ba1)2 = \u21262. Now draw out the diffraction geometry 270 Chapter 13. Acousto-Optic Diffraction for the misaligned case, noting that the total diffracted angle is still 2\u03b8B, so that |k1,x/k| \u2248 |\u03b8B \u2212 \u03b4\u03b8| and |k0,x/k| \u2248 |\u03b8B + \u03b4\u03b8|. Use these to eliminate \u03ba0,1 and obtain the result above. (b) For the 150 MHz TeO2 modulator that we discussed as an example in class, what is \u03b4\u03b8? Problem 13.4 Consider an acousto-optic modulator operating in the Raman-Nath regime at \u03bb0 = 780 nm (vacuum) wavelength. Assume the modulator is made from TeO2. (a) If the modulator is 80 MHz, what is the maximum thickness of the crystal in the direction of beam propagation? Assume the beam is normal to the acoustic wave. (b) What is the maximum achievable relative intensity in any diffracted order (other than the zeroth)? This is why Bragg modulators are potentially much more efficient at diffracting light into a single order. Assuming z = 0.1 mm, what would the index modulation \u03b4n/n0 need to be to achieve this maximum intensity? (c) What would the index modulation \u03b4n/n0 need to be to completely deplete the input beam (zeroth order)? For parts (b) and (c) you will need to consult a table of Bessel functions or use some mathematical software. Problem 13.5 Consider a Gaussian beam focused to a waist w0 in a Bragg-mode acousto-optic modulator. What is the turn-on time of the diffracted beam, assuming that the piezo transducer is suddenly switched on? Assume that the only contribution to the rise time is due to the finite propagation velocity of the acoustic wave front, and ignore any distortion of the wave front due to dispersive propagation in the crystal. For concreteness, define the rise time as the time it takes the diffracted beam to go from 10% to 90% of its maximum intensity. Problem 13.6 A glass acousto-optic modulator (n0 \u2248 1.5) diffracts He\u2013Ne laser light (vacuum \u03bb \u2248 200\u03c0 nm). Suppose that you observe that the angle between the zeroth and first diffracted order is 3\u25e6 (obviously, you are measuring this angle outside the modulator). However, some diabolical fiend has arranged it such that you cannot tell if any other diffracted orders are present. In order to save the world, you must figure out whether or not the modulator diffracts substantial power into other orders. How did you get yourself into this situation, anyway? (a) What is the Bragg angle \u03b8B? (b) Under what conditions would you expect there to be many diffracted beams? Give a rough range of optical path lengths of the crystal that would lead to many diffracted beams. (You need not compute any particular parameter, just go through the reasoning for what distinguishes the Raman\u2013Nath and Bragg regimes.) (c) Suppose that you observe that I0 Iin \u2248 J 2 0 (17), (13.76) where I0 is the the intensity of the zeroth diffraction order and Iin is the input intensity. Estimate the number of diffracted beams you would expect. Assume that the modulator operates in the regime where there are many diffracted peaks. Problem 13.7 Describe how you can use an acousto-optic modulator as a spectrum analyzer for an rf signal. Also 13.4 Exercises 271 describe how set up the analyzer to maximize the frequency resolution. You may assume that any diffraction angles are small, and that the acoustic-wave velocity is independent of frequency. Chapter 14 Coherence When considering interference of monochromatic waves in Chapter 5, we made the comment that assuming a monochromatic wave with time dependence of the form exp(\u2212i\u03c9t) is implicitly computing a Fourier transform. Now that we have the proper mathematical machinery in place, we will explore this idea in more detail. Let\u2019s now consider an interference experiment that involves a range of frequencies. Recalling the interference of two monochromatic plane waves, we can write the superposition of the two waves as sum(r, t) = E(+) E(+) 10 ei(kz\u2212\u03c9t) + E(+) 20 ei(kz\u2212\u03c9t)", "ei\u03c6. (14.1) Here, \u03c6 is a relative phase difference between the two waves. Recall that we are just writing the positivefrequency components, so that the physical fields also must include the negative-frequency parts. The intensity of the superposition is Isum = |Esum \u00d7 Hsum| = (Esum)2. 1 \u03b7 Writing this out explicitly in terms of the component fields, Isum = 1 \u03b7 10 ei(kz\u2212\u03c9t) + E(+) E(+) 20 ei(kz\u2212\u03c9t)ei\u03c6 + E(\u2212) 10 e\u2212i(kz\u2212\u03c9t) + E(\u2212) 20 e\u2212i(kz\u2212\u03c9t)e\u2212i\u03c62 The optical terms of the form exp(\u00b1i2\u03c9t) vanish in the time average, so we obtain (14.2). (14.3) Isum = 2 \u03b7 10 E(+) E(\u2212) 10 + 2 \u03b7 2 \u03b7 10 E(+) E(\u2212) = I1 + I2 + 20 E(+) E(\u2212) 20 + 20 ei\u03c6 + c.c.. 2 \u03b7 20 ei\u03c6 + 10 E(+) E(\u2212) 2 \u03b7 20 E(+) E(\u2212) 10 e\u2212i\u03c6 (14.4) Again, the interference is in the terms with the relative phase \u03c6. Suppose that the phase difference represents a difference in optical path length, in the form of a time delay \u03c4. Then \u03c6 = \u2212\u03c9\u03c4, and so Isum = I1 + I2 + 10 E(+) E(\u2212) 20 e\u2212i\u03c9\u03c4 + c.c.. 2 \u03b7 (14.5) Now let\u2019s handle the case of multiple frequencies. To simplify things, we\u2019ll assume that the two waves have equal amplitude and come from a common source. Then the intensity density at frequency \u03c9 is Isum(\u03c9) = 2I(\u03c9) + |E(+) 0 (\u03c9)|2e\u2212i\u03c9\u03c4 + c.c.. 2 \u03b7 (14.6) 274 Chapter 14. Coherence Note that the notation here is a little funny; the frequency-dependent quantities I(\u03c9) and E(+)(\u03c9) don\u2019t have the respective dimensions of intensity and electric field; rather, I(\u03c9) d\u03c9 and |E(+)(\u03c9)|2 d\u03c9 are the intensity and (squared) electric field, respectively, in the frequency interval between \u03c9 and \u03c9 + d\u03c9. Now we can sum over all frequencies to find the total intensity: Itotal = Isum(\u03c9) d\u03c9 \u221e 0 \u221e 0 = 2 I(\u03c9) d\u03c9 + 2 \u03b7 \u221e 0 |E(+) 0 (\u03c9)|2e\u2212i\u03c9\u03c4 d\u03c9 + c.c.. (14.7) Note that the frequency integral ranges only over positive frequencies; we\u2019ve already accounted for the negative frequencies by including the complex conjugate terms. Thus the intensity spectrum Isum(\u03c9) is a one-sided spectrum, which is common when working with intensities and powers. We can now recognize the second integral in the last expression as a \u2018\u2018one-sided\u2019\u2019 Fourier transform of |E(+) (\u03c9)|2, where we recall the normalization convention for \u03c9-t Fourier transforms: 0 f (t) = 1 2\u03c0 \u221e \u02dcf (\u03c9) e\u2212i\u03c9t d\u03c9, \u02dcf (\u03c9) = \u2212\u221e \u221e f (t) ei\u03c9t dt. \u2212\u221e (14.8) (\u03c9-t Fourier transform) But what is this, when we don\u2019t know the general form of the electric field? 14.1 Wiener\u2013Khinchin Theorem Recall the convolution theorem for functions f and g (see Section 11.3): F [f \u2217 g] = F [f ]F [g]. Writing out the convolution integral explicitly, \u221e (f \u2217 g)(t) = f (t) g(t \u2212 t) dt = F \u22121 [F [f ]F [g]]. \u2212\u221e If we make the particular choice g(t) = f \u2217(\u2212t), then (14.9) (14.10) F [g(t)] = F [f \u2217(\u2212t)] = \u221e f \u2217(\u2212t) ei\u03c9t dt = \u221e f \u2217(t) e\u2212i\u03c9t dt = (F [f (t)])\u2217. (14.11) \u2212\u221e \u2212\u221e Thus, Eq. (14.10) becomes \u221e f (", "t) f \u2217(t \u2212 t) dt = F \u22121 F [f ] 2. (14.12) \u2212\u221e Inverting the transform and letting t \u2212\u2192 t + t, we obtain the Wiener\u2013Khinchin theorem: \u221e F f \u2217(t) f (t + t) dt = |F [f ]|2. \u2212\u221e (14.13) (Wiener\u2013Khinchin theorem) The function on the left-hand side, \u221e f \u2217(t) f (t + \u03c4 ) dt, \u2212\u221e (14.14) (autocorrelation function) is the autocorrelation function of f (t). Essentially, it compares f to itself but shifted by an amount \u03c4 by computing an overlap integral. We can understand the right-hand side by noting that F [f ] is the spectrum of f, and so |F [f ]|2 is the energy spectral density of f. Essentially, the energy spectrum is the square of the usual spectrum, with the phase information removed. This is consistent with the notion of energy going 14.2 Optical Wiener\u2013Khinchin Theorem 275 as the square of a signal amplitude. Thus, the Wiener\u2013Khinchin theorem states that the Fourier transform of the autocorrelation function gives the energy spectrum. There is one subtle point to these definitions: for some signals, such as steady optical signals, the correlation integral diverges: \u221e f \u2217(t) f (t + \u03c4 ) dt \u2212\u2192 \u221e. \u2212\u221e (14.15) In this case, we should consider a time average instead of the normal integral. For an averaging time of T, f \u2217(t) f (t + \u03c4 )T := 1 T T /2 \u2212T /2 f \u2217(t) f (t + \u03c4 ) dt. (14.16) For bounded signals, this integral is guaranteed to converge. To be physically sensible, T should be a suitably long observation time (e.g., long enough to resolve the frequency spectrum). For such signals, we can write the Wiener\u2013Khinchin theorem as Here we have defined a finite-time Fourier transform by F [f \u2217(t) f (t + \u03c4 )T ] = F \u2217[f ] FT [f ]. FT [f ] := 1 T T /2 \u2212T /2 f (t) ei\u03c9t dt. (14.17) (14.18) Defining this seems a bit funny, but it avoids problems with singular spectra. Now the Wiener\u2013Khinchin theorem says that the Fourier transform of the (time-averaged) correlation function is the power spectral density, or the energy spectral density per unit time. For a stationary process, the correlation function is independent of t (generally for a sufficiently long averaging time T ). Then we can extend the averaging time T \u2212\u2192 \u221e. Denoting this long-time averaging limit as f \u2217(t) f (t + \u03c4 ) = f \u2217(t) f (t + \u03c4 )T \u2192\u221e, (14.19) we can thus write the Wiener\u2013Khinchin theorem as F [f \u2217(t) f (t + \u03c4 )] = lim T \u2192\u221e 1 T T /2 \u2212T /2 2 f (t) ei\u03c9t dt. (14.20) Again, the right-hand side is the power spectral density, and in this form it is more clear that this is the energy density per unit time. (Wiener\u2013Khinchin theorem, time-average form) 14.2 Optical Wiener\u2013Khinchin Theorem In terms of stationary optical fields, the Wiener\u2013Khinchin theorem (14.20) becomes \u221e I(\u03c9) e\u2212i\u03c9\u03c4 d\u03c9 = 0 2 \u03b7 E(\u2212)(t)E(+)(t + \u03c4 ). (14.21) This is because the intensity density I(\u03c9) = |S| is the time-averaged power spectral density of the optical field. Note that from the inverse Fourier transform, there is conventionally a factor of 1/2\u03c0, but it is missing here because it is already implicitly included in I(\u03c9). We can see this from the boundary condition at \u03c4 = 0, which gives the total intensity: (optical Wiener\u2013Khinchin theorem) \u221e I(\u03c9) d\u03c9 = 0 2 \u03b7 E(\u2212)(t)E(+)(t) = |E(+)(t)|2. 2 \u03b7 (14.22) 276 Chapter 14. Coherence In optics, as in statistics and other fields, it is conventional to define a normalized correlation function g(1)(\u03c4 ) := E(\u2212)(t)E(+)(", "t + \u03c4 ) E(\u2212)(t)E(+)(t), so that (degree of first-order temporal coherence) (14.23) 2 \u03b7 E(\u2212)(t)E(+)(t + \u03c4 ) = \u221e I(\u03c9) d\u03c9 g(1)(\u03c4 ). 0 (14.24) The normalization is such that g(1)(\u03c4 = 0) = 1. (Also, the \u2018\u2018(1)\u2019\u2019 superscript is a notation that indicates this is the first in a hierarchy of increasingly complex correlation functions that we won\u2019t consider here.) That is, a correlation value of unity indicates perfect correlation. Note that we could just as well have written the correlation function as E(+)(t + \u03c4 )E(\u2212)(t), but it turns out that the order of E(+) and E(\u2212) matters in quantum mechanics, so we\u2019ll be careful to stick to the form in Eq. (14.23). Notice also that from the definition of g(1)(\u03c4 ), the correlation function is subject to the constraint g(1)(\u2212\u03c4 ) = [g(1)(\u03c4 )]\u2217. (14.25) (time symmetry of correlation function) In optics, g(1)(\u03c4 ) is called the degree of first-order temporal coherence. The light is said to be coherent if |g(1)(\u03c4 )|2 = 1, incoherent if |g(1)(\u03c4 )|2 = 0, and partially coherent otherwise (for \u03c4 = 0). Returning to the interference result of Eq. (14.7), we find Itotal = 2 Isum(\u03c9) d\u03c9 + (\u03c9)|2e\u2212i\u03c9\u03c4 d\u03c9 + c.c. \u221e 0 \u221e = I(\u03c9) d\u03c9 2 + \u221e |E(\u2212) 0 2 \u03b7 g(1)(\u03c4 ) + c.c. 0, (14.26) and thus 0 Itotal = 2 \u221e I(\u03c9) d\u03c9 1 + Re g(1)(\u03c4 ). 0 (14.27) It is worth keeping in mind that the form (14.13) is still relevant for pulsed fields, in which case the result here is modified so that the coefficient in front is the integrated power rather than intensity. (interferometer signal in terms of g(1)(\u03c4 )) 14.2.1 Application: FTIR Spectroscopy and the Michelson Interferometer One example of where the optical Wiener\u2013Khinchin theorem is very useful is in Fourier-transform infrared (FTIR) spectroscopy. The idea in FTIR spectroscopy is to use a Michelson interferometer to monitor a stationary input signal E(+)(t). The interferometer splits and interferes the beam with itself, and the path-length difference \u03c4 varies with the displacement of one of the mirrors. The photodetector at the output measures the time average of the product of the fields, up to an overall constant, and thus measures g(1)(\u03c4 ), just as in Eq. (14.27). The idea is then to digitize the output of the Eo(+)o(t)detector\u00b7Eo(-)o(t)ooEo(+)o(to+ot)\u201a + c.c.vary t 14.2 Optical Wiener\u2013Khinchin Theorem 277 photodetector as a function of the mirror displacement, effectively recording Re[g(1)(\u03c4 )]. Computing the Fourier transform of the signal on the computer gives the spectrum I(\u03c9). (Note that I(\u03c9) is real, and so the imaginary parts of g(1)(\u03c4 ) don\u2019t contribute to the spectrum.) This technique is common in molecular spectroscopy, because it turns out that obtaining a spectrum this way gives better rejection of thermal detector noise compared to, say, a recording the same spectrum with the same detector on a diffractiongrating spectrometer. 14.2.2 Example: Monochromatic Light As a simple example, let\u2019s compute the correlation function for monochromatic light and verify that the Wiener\u2013Khinchin theorem makes sense. Let\u2019s take a monochromatic wave of the form The correlation function is E(+)(t) = E(+) 0 e\u2212i\u03c90t. E(\u2212)(t)E(+)(t + \u03c4 ) = 2 E(+) 0 e\u2212i\u03c90\u03c4. In normalized form, this function becomes g(1)(\u03c4 ) = e\u2212i\u03c90\u03c4. (14.28) (14.29) (14.30) Thus, a wave with harmonic time dependence leads to a harmonic correlation function. This is true independent of the phase of the input field; that is, the", " correlation function does not reflect any extra phase in E(+) 0. The power spectrum is easy to calculate via the Wiener\u2013Khinchin theorem: F E(\u2212)(t)E(+)(t + \u03c4 ) = E(+) 0 2 F e\u2212i\u03c90\u03c4 = 2 E(+) 0 \u03b4(\u03c9 \u2212 \u03c90). (14.31) Again, the harmonic time dependence produces a single frequency in the power spectrum. Of course, with our convention of positive and negative frequencies, there would be a matching \u03b4(\u03c9 + \u03c90) component, but this is already included in the one-sided spectrum I(\u03c9). Let\u2019s now compute the power spectrum directly, using Eq. (14.17). The normal spectrum is F E(+)(t) = E(+) 0 \u03b4(\u03c9 \u2212 \u03c90). (14.32) The finite-time transform is E(+)(t) = FT 1 T T /2 \u2212T /2 E(+) 0 ei(\u03c9\u2212\u03c90)t dt = E(+) 0 sinc [(\u03c9 \u2212 \u03c90)T /2]. (14.33) Note that the value of the sinc function is 1 at \u03c9 = \u03c90. Thus, the spectrum is F \u2217 E(+)(t) E(+)(t) FT = E(\u2212) 0 \u03b4(\u03c9 \u2212 \u03c90)E(+) 0 sinc [(\u03c9 \u2212 \u03c90)T /2] = E(+) 0 2 \u03b4(\u03c9 \u2212 \u03c90). (14.34) This result is consistent with Eq. (14.31). Note that without being careful with finite-time transforms, we would run into something bad involving the square of a \u03b4-function. 14.2.3 Normalized One- and Two-Sided Spectra Now we will be a bit more precise about the nature of the spectrum, to avoid confusion with different possible conventions for the power spectrum. First, it\u2019s convenient to define a normalized spectral density (lineshape function) s(\u03c9) := I(\u03c9) \u221e I(\u03c9) d\u03c9. 0 (14.35) (normalized spectral density) 278 Chapter 14. Coherence Note that this spectrum extends only over positive frequencies, as the intensity spectrum I(\u03c9) corresponded to physical frequency components. Examining the inverse Fourier transform, we combine Eqs. (14.21) and (14.24) to obtain g(1)(\u03c4 ) = \u221e s(\u03c9) e\u2212i\u03c9\u03c4 d\u03c9. 0 (14.36) However, the usual inverse Fourier transform has an integral extending over both positive and negative frequencies. We can obtain something of this form by considering the real part of the correlation function, (first-order coherence in terms of normalized, one-sided spectral density) Re g(1)(\u03c4 ) = = g(1)(\u03c4 ) + g(1)(\u03c4 )\u2217 2 \u221e = s(\u03c9) e\u2212i\u03c9\u03c4 d\u03c9 + 1 2 0 g(1)(\u03c4 ) + g(1)(\u2212\u03c4 ) 2 \u221e s(\u03c9) ei\u03c9\u03c4 d\u03c9, 1 2 0 (14.37) so that if we define a two-sided spectrum s\u2194(\u03c9) (for all \u03c9 \u2208 R) via s\u2194(\u03c9) := \uf8f1 \uf8f2 \uf8f3 s(\u03c9)/2 s(\u03c9) s(\u2212\u03c9)/2 (\u03c9 > 0) (\u03c9 = 0) (\u03c9 < 0), which satisfies we obtain (two-sided spectrum in terms of one-sided spectrum) (14.38) s\u2194(\u2212\u03c9) = s\u2194(\u03c9), (14.39) (symmetry of two-sided spectrum) g(1)(\u03c4 ) = Re \u221e s\u2194(\u03c9) e\u2212i\u03c9\u03c4 d\u03c9. \u2212\u221e (14.40) We have lost the imaginary part, but only the real part can contribute to a physical result: g(1)(\u03c4 ) must always be accompanied by its conjugate, as we saw in the interference experiment. (first-order coherence in terms of two-sided spectrum) Inverting this last relation, we have s\u2194(\u03c9) = = = \u221e Re \u2212\u221e \u221e ei\u03c9\u03c4 d\u03c4 g(1)(\u03c4 ) g(1)(\u03c4 ) cos \u03c9\u03c4 d\u03c4 1 2\u03c0 1 2\u03c0 1 2\u03c0 (two-sided spectral density in terms of first-order coherence) g(1)(\u03c4 ) cos \u03c9\u03c4 d\u03c4 + c.c. (\u03c9 \u2208 R). \u2212\u221e \u221e 0 (14.41) The one-sided spectrum can then be written in terms of the double", "-sided spectrum as s(\u03c9) = \uf8f1 \uf8f2 \uf8f3 s\u2194(\u03c9) + s\u2194(\u2212\u03c9) = 2s\u2194(\u03c9) s\u2194(\u03c9) 0 (\u03c9 > 0) (\u03c9 = 0) (\u03c9 < 0), (14.42) so that the one-sided spectrum simply concentrates all the power at positive and negative frequencies on the positive side. Of course, this cannot be done for any spectrum, but the power spectral density\u2014the Fourier transform of the coherence function\u2014has a special structure due to the symmetry of the correlation (two-sided spectrum in terms of one-sided spectrum) 14.3 Visibility function. Thus, we can write the one-sided spectrum as s(\u03c9) = = = \u221e Re \u2212\u221e \u221e ei\u03c9\u03c4 d\u03c4 g(1)(\u03c4 ) g(1)(\u03c4 ) cos \u03c9\u03c4 d\u03c4 1 \u03c0 1 \u03c0 1 \u03c0 (one-sided spectral density in terms of first-order coherence) g(1)(\u03c4 ) cos \u03c9\u03c4 d\u03c4 + c.c. (\u03c9 \u2265 0) \u2212\u221e \u221e 0 279 (14.43) in terms of the coherence function. Combining Eqs. (14.40) and (14.42), we then find g(1)(\u03c4 ) = Re \u221e s(\u03c9) cos \u03c9\u03c4 d\u03c9 0 for the reverse transformation. Finally, note that it can be more convenient to associate spectra separately with g(1)(\u03c4 ) and its con- (first-order coherence in terms of one-sided spectrum) (14.44) jugate, by defining s\u00b1(\u03c9) := 1 4\u03c0 \u221e g(1)(\u03c4 ) e\u00b1i\u03c9\u03c4 d\u03c4 = \u2212\u221e 1 4\u03c0 \u221e g(1)(\u03c4 ) e\u00b1i\u03c9\u03c4 d\u03c4 + c.c. (\u03c9 \u2208 R), 0 (component spectral densities in terms of first-order coherence) (14.45) so that Then the relations s+(\u03c9) = s\u2212(\u2212\u03c9). (14.46) (symmetry of component spectra) s\u2194(\u03c9) = s+(\u03c9) + s\u2212(\u03c9) = s+(\u03c9) + s+(\u2212\u03c9) s(\u03c9) = 2[s+(\u03c9) + s\u2212(\u03c9)] = 2[s+(\u03c9) + s+(\u2212\u03c9)] (\u03c9 \u2208 R) (\u03c9 \u2265 0) recover the total spectra (14.41) and (14.43), respectively. (component spectral densities in terms of first-order coherence) (14.47) 14.3 Visibility The example in Section 14.2.2 of the correlation function for a monochromatic wave is special in the sense that the correlation function does not decay with \u03c4. This is important, because the correlation function is the magnitude of the interference terms. To quantify this better, we can define the fringe visibility as V := Imax \u2212 Imin Imax + Imin, (14.48) (fringe visibility) where Imax and Imin are the maximum and minimum intensities achieved for phase variations on the order of several \u03c0. For example, complete interference results in intensity variation from 0 to some maximum, and so V = 1. For no interference, the intensity does not vary with phase and so V = 0. Partial coherence is represented by intermediate values of the visibility. Writing out the explicit phase of the correlation function, and so Eq. (14.27) becomes g(1)(\u03c4 ) = |g(1)(\u03c4 )|ei\u03c6(\u03c4 ), Itotal = 2 \u221e I(\u03c9) d\u03c9 1 + |g(1)(\u03c4 )| cos \u03c6(\u03c4 ). 0 (14.49) (14.50) 280 Chapter 14. Coherence The cosine varies from \u22121 to 1, so the visibility is just the magnitude of the correlation function: For monochromatic light, V = 1. Actually, this is only true if the amplitudes of the input waves are equal. For two monochromatic waves of unequal amplitude, the visibility becomes V = |g(1)(\u03c4 )|. (14.51) (visibility in terms of coherence) V = \u221a 2 I1I2 I1 + I2. (14.52) (monochromatic, unbalanced interference) The important thing to note is that for interference of monochromatic light, the visibility is independent of \u03c4. 14.4 Coherence Time, Coherence Length, and Uncertainty Measures As we have just seen, a peculiarity of monochromatic light is that the coherence does not decay with \u03c4. In the generic case, where light is composed of a range of", " frequencies, the visibility drops as the phase difference increases, since g(1)(\u03c4 ) \u2212\u2192 0 as \u03c4 \u2212\u2192 \u221e. Intuitively, this is because as light waves with different wavelengths propagate, the monochromatic components tend to dephase. Mathematically, we can express this as an uncertainty relationship. Recall from Section 14.2.3 that the coherence Re[g(1)(\u03c4 )] and the normalized (twosided) spectral density (lineshape function) form a Fourier-transform pair. Actually, 2\u03c0s\u2194(\u03c9) is the second half of the pair, due to the form of the transforms (14.40) and (14.41): recall that in the \u03c9-t convention, there is normally a factor of 1/2\u03c0 in front of the inverse-transform integral, as in Eqs. (14.8). Thus, if we define the root-mean-square (rms) widths of these two functions, regarding Re[g(1)(\u03c4 )]2 and [s\u2194(\u03c9)]2 as (unnormalized) probability distributions, we can write down an uncertainty relation, as in quantum mechanics: \u03b4\u03c9rms \u03b4\u03c4rms \u2265 1 2. (14.53) (rms uncertainty relation) For many distributions, we can say that the equality is more or less satisfied to \u03b4\u03c9rms \u03b4\u03c4rms \u223c 1 2, (14.54) and thus see that the \u2018\u2018widths\u2019\u2019 of g(1)(\u03c4 ) and s\u2194(\u03c9) are inversely related. The problem is that for some useful functions in optics, these uncertainties can diverge (e.g., for Lorentzian functions) or vary by orders of magnitude. The uncertainty inequality is always satisfied, of course, but as a practical relation for the temporal and frequency widths, this is less useful. Thus we will adopt other uncertainty conventions in time and frequency1 We can define the coherence time as the power-equivalent width of the correlation function: \u03b4\u03c4 := \u221e Re \u2212\u221e g(1)(\u03c4 ) 2 d\u03c4. (14.55) The idea here is to first note that g(1)(\u03c4 = 0) = 1. Thus the width \u03b4\u03c4 is the width of a box of unit height, with the same area as |Re[g(1)(\u03c4 )]|2. That is, \u03b4\u03c4 is the width of a unit-height box signal with the same power as |Re[g(1)(\u03c4 )]|2. (coherence time, power-equivalent width) 1as in Max Born and Emil Wolf, Principles of Optics, 7th (expanded) ed. (Cambridge, 1999). 14.4 Coherence Time, Coherence Length, and Uncertainty Measures 281 We\u2019ll take a slightly different convention for the width of the frequency spectrum. Let\u2019s define the average value of the one-sided normalized spectrum as \u221e s2(\u03c9) d\u03c9 = 2 \u00afs = \u221e 0 s 2 \u2194(\u03c9) d\u03c9. \u2212\u221e (14.56) That is, we\u2019re regarding s(\u03c9) as a probability distribution (because it has the appropriate normalization), and then we\u2019re calculating the expectation value of s(\u03c9) with respect to itself. Then we\u2019ll define the effective frequency width by the reciprocal of the average value: \u03b4\u03c9 := (\u00afs)\u22121 = \u221e s2(\u03c9) d\u03c9 \u22121. 0 (14.57) (spectral width) Thus, \u03b4\u03c9 is the width of the box function of unit area and height \u00afs. Note that we have constructed the diagram as if s(\u03c9) is two-sided, but the argument carries through for the one-sided case as well, being careful to keep track of factors of 2. The definitions for \u03b4\u03c4 and \u03b4\u03c9 look like inverses, but they\u2019re really quite different because of the different normalizations of the two functions. The big advantage of these definitions is that they are related in a simple way. We can write \u03b4\u03c4 \u03b4\u03c9 = \u221e Re \u2212\u221e \u221e g(1)(\u03c4 ) 2 s 2 \u2194(\u03c9) d\u03c9 \u2212\u221e 2 d\u03c4. (14.58) We can use Parseval\u2019s theorem to evaluate this ratio, which states that the signal power is equivalently measured in either the time or frequency basis: \u221e |f (t)|2 dt = \u2212\u221e 1 2\u03c0 \u221e \u2212\u221e 2 \u02dcf (\u03c9) d\u03c9. (14.59) (Parseval\u2019s theorem) Noting again that Re[g(1)(\u03c4 )] and 2\u03c0s\u2194", "(\u03c9) form a Fourier-transform pair, and recalling the factor of two for using the one-sided spectrum, we can use this to write \u03b4\u03c4 \u03b4\u03c9 = \u03c0. (14.60) (uncertainty relation) tReo[go(1)o(t)]2dt1same areawso(w)dws-unit area 282 Chapter 14. Coherence This \u2018\u2018uncertainty relation\u2019\u2019 is a strict equality, valid for any functions as long as the measures exist and are finite. Neat, eh? The point of all this is, for time (optical path length) delays larger than the coherence time \u03b4\u03c4, the fringe visibility is mostly gone. This coherence time corresponds to a physical path length difference c := c \u03b4\u03c4, (14.61) (coherence length) which is called the coherence length. Here are some examples. For a He-Ne laser, the laser line width is \u03b4\u03bd = \u03b4\u03c9/2\u03c0 \u223c 1 GHz. This corresponds to a coherence time of \u03b4\u03c4 \u223c 1 ns, or a coherence length c \u223c 30 cm. On the other hand for a light bulb that spans the visible wavelength range of 400-700 nm, the line width is \u03b4\u03bd \u223c \u03bd \u03b4\u03bb \u03bb = c \u03b4\u03bb \u03bb2 = 300 THz. (14.62) This gives a coherence time \u03b4\u03c4 \u223c 3 fs and a coherence length c \u223c 1 \u00b5m. So in fact it is possible to see interference of white light in a Michelson, but it\u2019s very difficult because the path lengths must be matched to \u00b5m accuracy. On the other hand, it\u2019s much easier to observe interference or record a hologram with light from a He-Ne laser, because it remains coherent on the scale of about a foot. 14.5 Interference Between Two Partially Coherent Sources In general, we can now look at the interference pattern between two partially coherent sources, represented by the two fields E(+) (t). The second field has an adjustable time delay of \u03c4. Then the intensity 1 of the superposition of these waves is (t) and E(+) (+) E(+) 1 (t) + E(+) 2 (t + \u03c4 ) (t) + 2 \u03b7 E(+) 2 = I1 + I2 + 2 I1I2 Re g(1) 12 (\u03c4 ), (t + \u03c4 ) + 2 \u03b7 E(\u2212) 1 (t)E(+) 2 (t + \u03c4 ) + c.c. (14.63) where g(1) 12 (\u03c4 ) is the normalized cross-correlation function: g(1) 12 (\u03c4 ) := E(\u2212) 1 E(\u2212) (t)E(+) 2 (t)E(+) 2 1 (t + \u03c4 ). (t) (14.64) (normalized cross-correlation) Again, the visibility is V = \u221a 2 I1I2 I1 + I2 g(1). 12 (\u03c4 ) (14.65) (visibility for two different fields) This tells us that very different beams, resulting in little correlation, don\u2019t interfere very well. 14.6 Exercises 14.6 Exercises Problem 14.1 Prove Parseval\u2019s Theorem: if f (x) and \u02dcf (k) form a Fourier transform pair, then \u221e \u2212\u221e |f (x)|2 dx = 1 2\u03c0 \u221e \u2212\u221e | \u02dcf (k)|2 dk 283 (14.66) Problem 14.2 Go through the derivation in the notes and fill in the details of the \u2018\u2018uncertainty relation\u2019\u2019 \u03b4\u03c4 \u03b4\u03c9 = \u03c0, where the coherence time \u03b4\u03c4 is the power-equivalent width of g(1)(\u03c4 ) and \u03b4\u03c9 is the effective spectral width. Problem 14.3 Find the coherence times \u03b4\u03c4 according to the power-equivalent width definition for the following coherence functions: (a) Lorentzian: g(1)(\u03c4 ) = 1/[1 + (\u03c4 /\u2206\u03c4 )2], where \u2206\u03c4 is the full width at half maximum. (b) Gaussian: g(1)(\u03c4 ) = exp[\u2212\u03c4 2/(\u2206\u03c4 )2], where \u2206\u03c4 is the 1/e time. (c) Exponential: g(1)(\u03c4 ) = exp[\u2212|\u03c4 /\u2206\u03c4 |], where \u2206\u03c4 is the 1/e time. Problem 14.4 Consider a Michelson interferometer, where the input is bichromatic with air wavelengths \u03bb1 and \u03bb2. Ass", "ume that the splitting is small, |\u2206\u03bb| \u03bb, where \u2206\u03bb := \u03bb2 \u2212 \u03bb1 and \u03bb := (\u03bb1 + \u03bb2)/2. Show that the visibility of the output fringes varies periodically as one mirror is displaced, with period \u2206d = \u03bb2 2\u2206\u03bb (14.67) in terms of the mirror displacement. Chapter 15 Laser Physics 15.1 Overview Now we\u2019ll explore in a fair amount of detail the operation of a laser, which is very important as a source of intense and/or highly coherent light. The name \u2018\u2018laser\u2019\u2019 is an acronym for \u2018\u2018light amplification by the stimulated emission of radiation,\u2019\u2019 emphasizing that the laser is an active, amplifying device. There are three key components to any laser: 1. The pump is the energy source that is converted to laser light. 2. The gain medium performs the actual conversion of energy to coherent light, by amplifying light passing through it. 3. The optical resonator, which we have already studied in detail, recycles the light through the gain medium to achieve better amplification. Also, since an optical resonator is a frequency-selective filter, it narrows the frequency spectrum of the output and thus increases the coherence of the laser light. Put together to form a laser, these are the ingredients for an active optical oscillator,1 in contrast to the passive optical elements that we have studied so far. Laser light has a number of potentially desireable properties. Among its key features are: 1. Monochromaticity. A laser tends to have a relatively narrow frequency spectrum, either in the absolute sense or in the sense of being limited by the time-frequency uncertainty principle (i.e., for \u2018\u2018transform-limited\u2019\u2019 laser pulses). 2. Directionality. The output of a laser has small beam divergence, typically diffraction-limited or nearly so. The output of a laser is commonly a Gaussian beam. 3. High power. Not all lasers have high output powers, but many of the most intense man-made optical sources are lasers. Now let\u2019s examine the main components of a laser in more detail. 15.1.1 Laser Pumps The laser pump is the energy reservoir for the laser. Often, the energy comes from electricity, although sometimes indirectly. Examples of pumps that directly involve electricity are gas discharges and injection currents in semiconductor lasers. Less direct are optical pumps, in the form of flashlamps or even another laser. A more bizarre example of a pump is a nuclear device, for the \u2018\u2018pop-up x-ray laser\u2019\u2019 defense system. 1in fact, to emphasize the importance of the oscillator nature of the laser, one could use the name \u2018\u2018light oscillation by the stimulated emission of radiation,\u2019\u2019 but this would be a loser, and not nearly as cool. (I stole this joke from Leno Pedrotti, from way back when I learned laser physics from him at the University of Dayton.) 286 15.1.2 Gain Media Chapter 15. Laser Physics The gain medium converts the pump energy into the desired light. Invariably the gain medium involves quantum systems, where the transition energies determine the output wavelength. 15.1.2.1 Gas-Phase Atoms One important class of gain media is atoms in the gas phase. Examples include the He\u2013Ne laser and the argon-ion laser. These gain media typically lase in the visible to ultraviolet range, and they are pumped by electrical discharge. Commonly these lasers operate in CW (continuous-wave) mode, although pulsed gas lasers are possible (e.g., nitrogen and copper-vapor lasers). The efficiencies of gas lasers tend to be appalling, in the neighborhood of 0.05% for an Ar+ laser. Output powers vary from 1 mW (He\u2013Ne) to 20 W (Ar+). 15.1.2.2 Atoms Embedded in Transparent Solids Atoms can also be suspended in (doped into) transparent solids. Important examples are Nd:YAG, Nd:glass, Ti:sapphire (sapphire is crystalline Al2O3), ruby (Cr:sapphire), Er:fiber, and Yb:fiber. Here, the notation is \u2018\u2018dop-ant:medium.\u2019\u2019 These lasers operate mostly in the near-infrared (in the range of 0.7-1.6 \u00b5m), and can operate as either CW or pulsed lasers. The efficiencies of these solid-state lasers are much higher than gas lasers, in the neighborhood of 1%. The efficiency varies widely depending on the details of the pump; for example, a laser-pumped Ti:sapphire laser can have an efficiency approaching", " 10%. Output powers are in the range of 100 mW (CW lasers) to PW peak pulse powers for some truly huge lasers (P = peta = 1015). Here is a closeup view of part of a Ti:sapphire ring-cavity laser. The blue-green pump light (from an Ar+ laser) is clearly visible as it enters through the input mirror (with a special coating to transmit blue-green light but reflect the 850 nm laser light). The input lens aligns and matches the pump-beam profile to the laser cavity mode. The Ti:sapphire crystal is Brewster-cut to suppress reflection loss at the surfaces and glows bright red due to amplified spontaneous emission. The other components (birefringent filter, Brewster etalons, ICA) all help force the laser to oscillate in a single cavity mode. The output mirror is to the left of the area shown. For overall scale, the holes in the table surface are spaced at 1\u2019\u2019 intervals. BirefringentfilterTi:sapphirecrystalInputmirrorInputlensBrewstertuning platesICA 15.1 Overview 15.1.2.3 Molecules 287 Molecular lasers come in two main flavors. A gas-phase molecular laser is CO2, and this is a particularly important laser in industrial cutting and welding. Molecules can also be contained in liquid solvent, as in the manifold organic laser dyes. The wavelengths of these lasers range from the infrared (10.6 \u00b5m for CO2) to the ultraviolet/visible/near-infrared, the wide range covered by the various available laser dyes. The efficiencies of molecular lasers are also relatively high, in the range of 1-10%, depending on the details of the pump. Output powers range from around 100 mW (the low end of dye lasers) to 10 kW for CO2 lasers. One of the most amusing (and nearly edible) lasers falls in this category: the gelatin laser, where the molecules included sodium fluorescein (marginally edible) and rhodamine 6G (much less edible). The \u2018\u2018normal slight tremor\u2019\u2019 when holding the gain medium by hand was observed to enhance the operation of the medium.2 15.1.2.4 Semiconductor Lasers Diode lasers have become enormously important in recent years. Materials include GaAs, GaAlAs, and InGaN/GaN/AlGaN. Most diodes lase in the red- to near-infrared, although InGaN/GaN/AlGaN lases in the 400-420 nm range. Diode lasers are pumped via injection current flowing across the p-n junction. Efficiencies of diode lasers are as high as it gets (up to 65%), and output powers are in the range of 1 mW to 50 W (for diode arrays). A single-mode, low-power laser diode is shown in the photo below. This diode would normally have a cylindical, protective metal \u2018\u2018can\u2019\u2019 surrounding the front end of the diode, with a window for the laser output, which has been removed here. For overall scale, the outer, circular flange of the diode is 9 mm in diameter. The laser cavity itself is a tiny dot in this photo; the active wave-guiding region has transverse dimensions in the range of about 1\u20136 \u00b5m, though the transverse extent of the chunk of semiconductor is larger than this. The length of the cavity is of the order of 100 \u00b5m. The front of the laser emits out of the page, down, and to the left of the photo. The back of the cavity is also transparent and emits some light, which striked a monitor photodiode (photodetector). Electrical connections come in through three pins in the back. The leftmost pin (as viewed in the photo) protrudes out next to the semiconductor, and a thin wire is visible connecting the pin to the top of the laser cavity. The bottom of the laser cavity is connected to case ground. 2T. W. H\u00e4nsch, M. Pernier, A. L. Schawlow, \u2018\u2018Laser Action of Dyes in Gelatin,\u2019\u2019 IEEE Journal of Quantum Electronics QE-7, 45 (1971). LasercavityPhotodiodeElectricalconnections 288 Chapter 15. Laser Physics The rightmost pin is connected to the photodiode by another thin wire, visible to the right of the photodiode. The photodiode shares a common ground connection with the laser. 15.1.3 Optical Resonator The laser cavity or resonator is often a spherical-mirror resonator. Typically, the laser resonator is stable for CW operation, but sometimes unstable resonators are used in pulsed-laser", " designs. 15.1.4 A Simple Model of Laser Oscillation: Threshold Behavior We can get a great deal of insight about laser operation just by considering a very simple model.3 Consider a laser, composed of a two-mirror resonator, with mirror reflectances R1,2. The laser contains a thin gain medium, with single-pass intensity gain G. \u221a The effect of the mirrors after one round trip on the intracavity intensity is Iafter = R1,2Ibefore. The effect of the gain medium after a single pass is Iafter = \u221a GIbefore. (15.1) (15.2) Note that G is the round-trip gain, where G > 1 for net gain and G < 1 for net loss (e.g., due to scattering loss). The reflectances R1,2 < 1, representing loss. Thus, the intensity after a round trip is Iafter = GR1R2Ibefore. (15.3) If \u03c4rt = 2d/c is the cavity round-trip time (ignoring any contribution due to the thin gain medium), we can write I(t + \u03c4rt) = GR1R2I(t). (15.4) Let\u2019s assume that the intensity varies slowly on the time scale of \u03c4rt. Then we can expand to lowest order in \u03c4rt: I(t) + dI(t) dt \u03c4rt + O(\u03c4 2 rt) = GR1R2I(t). (15.5) Rearranging gives the differential equation dI(t) dt = (GR1R2 \u2212 1) \u03c4rt I(t). (15.6) (simple laser rate equation) Note that this is a very simple model: we are ignoring the fact that the gain G should depend on the pump, and we are assuming that the laser light is resonant with the cavity. But with the assumption of constant G, we can write the explicit solution to Eq. (15.6) as I(t) = I0 exp (GR1R2 \u2212 1) \u03c4rt t. (15.7) (laser evolution) 3In this and the next subsection we are following the excellent treatment in Frank L. Pedrotti, Leno M. Pedrotti, and Leno S. Pedrotti, Introduction to Optics, 3rd ed. (Benjamin Cummings, 2006). dR\u00a1thin gain mediumR\u2122\u221aG 15.1 Overview 289 Note that there is a threshold behavior represented in this solution, depending on the value of the gain: G > G < 1 R1R2 1 R1R2 =\u21d2I(t) \u2212\u2192 \u221e =\u21d2I(t) \u2212\u2192 0. (15.8) (laser threshold behavior) That is, the gain must be larger than the threshold gain of 1/R1R2 (i.e., the gain must exceed the loss) in order for the intensity to start growing. But an exponentially growing (divergent) solution is unphysical! We\u2019ve arrived at this funny solution because the linear model with constant G is too simple. 15.1.5 A Less-Simple Model of Laser Oscillation: Steady-State Oscillation A better (but still quite simplified) model for the gain is G(I) = 1 +, g0 1 + I Isat (15.9) (gain-saturation model) where g0/g is the small-signal gain coefficient (with g the length of the thin gain medium), and Isat is the saturation intensity. For small intensities (I Isat), G(I) \u2248 1 + g0 =: G0, (15.10) and so the small-signal gain G0 is a constant, and is also the maximum possible value of G. For large intensities (I Isat), the gain becomes G(I) \u2248 1 +, (15.11) g0Isat I and thus the gain decreases as I increases, with G \u2212\u2192 1 as I \u2212\u2192 \u221e. That is, there is no gain in the limit of large input intensity. This nonlinear effect is called gain saturation, and is a universal feature of physical amplifiers, which have finite energy reservoirs. The saturation intensity Isat marks the crossover between these two regimes. If we write the gain as where the integrated gain coefficient g(I) is related to the small-signal gain coefficient by G(I) = 1 + g(I), g(I) =. g0 1 + I Isat (15.12) (15.13) Then the saturation intensity is defined such that when I = Isat, the gain drops to half its small-signal value: g(Isat) = g0 2. (15.14) Now let\ufffd", "\ufffds look at the solution to Eq. (15.6) when we include gain saturation, assuming we start with some small, positive intensity. I(t)texponential growth (I < oooI )sat<gain saturation kicks insettles to steady state 290 Chapter 15. Laser Physics Assuming we are above threshold, we only see the exponential rise in intensity while I Isat. Once I becomes comparable to Isat, gain saturation begins to kick in, tempering the growth of the intensity. The gain continues to decrease due to gain saturation until it just balances the losses, and the intensity settles into a steady state. We can calculate the steady-state intensity Iss by setting dI/dt = 0. This happens when G(Iss)R1R2 = 1. (15.15) Putting in the gain from Eq. (15.9) and solving for Iss, Iss = (g0 + 1)R1R2 \u2212 1 1 \u2212 R1R2 Isat. (15.16) (steady-state intensity) Again, this is only true if the small-signal gain G0 is above threshold (G0R1R2 > 1). If the small-signal gain is below threshold (G0R1R2 < 1), then any initial intensity decays away, and so Iss = 0. Roughly speaking, Isat is a fixed property of a given gain medium, and g0 is proportional to the pump power (at least for small g0). Noting that the steady-state laser output is Iout = (1 \u2212 R2)Iss, (15.17) where the second mirror is the output coupler, we can see the more physical threshold behavior of a laser. Above the threshold value g0,th of the small-signal integrated gain coefficient, the steady-state output intensity rises linearly with slope R1R2 1 \u2212 R1R2 (1 \u2212 R2). (15.18) For a given pump, and thus gain g0, there is an optimum output coupler reflectance R2 that maximizes the output power, because this slope and the threshold gain are coupled together. However, we will defer this calculation until we have treated the laser in more detail. 15.2 Light\u2013Atom Interactions 15.2.1 Quantization Here we are going to adopt a fairly simplistic view of how the electromagnetic field interacts with atoms. In particular, we will be using the language of photons, but we will stick to a strictly semiclassical treatment, not really treating the atoms or the field quantum mechanically. But we will start with the observation that the energy in an electromagnetic field is quantized. This means that a monochromatic field of frequency \u03c9 (typically restricted to some \u2018\u2018quantization volume\u2019\u2019 such as an optical cavity) has possible energies given by En = n + 1 2 \u00afh\u03c9, (15.19) (allowed field energies) I oo(t)g\u00baI oo=oo0ss,othoutthreshold (g\u00baoooooooooo)(~ pump power)(steadystate)linear increase after threshold 15.2 Light\u2013Atom Interactions 291 where n is a nonnegative integer, representing the number of photons in the field. This may be familiar as the energy-level structure of the quantum harmonic oscillator. The photon number is always defined with respect to a particular mode (fixing the direction, polarization, and frequency characteristics). The energies for atoms and molecules are also quantized, although the exact energy-level structure depends on the specific atom or molecule. If we denote the quantized energies by En, then the differences in energy levels correspond to frequencies via \u2206Emn := En \u2212 Em = \u00afh\u03c9mn. (15.20) (transition energies) The idea is that atoms with an energy difference \u2206Emn prefer to interact with resonant fields of frequency \u03c9mn. In this case, the energy of a single photon matches the atomic energy difference, and energy is conserved. There are different types of transitions, generally corresponding to different types of radiation. Electronic transitions in atoms are the most energetic of the type we will consider, and they correspond to visible optical frequencies. Vibrational transitions in a molecule correspond to different amplitudes and types of motion internal to the molecule, and generally correspond to radiation in the infrared. Rotational transitions in molecules have yet lower energy, and they correspond to microwave radiation (which enables the maser, the microwave predecessor to the laser). 15.2.2 Fundamental Light\u2013Atom Interactions There are three fundamental interactions between light and atoms. In all cases we will consider only a twolevel atom with ground-state energy E1 and excited-state energy E2. We will also assume resonant light, \u03c9 = (E2 \u2212 E1)/\u00afh. 1. Absorption (stimulated). In the absorption process,", " a photon is destroyed and the atom is promoted to the excited state. More generally, if there are n photons to start with in some resonant mode, then there are n \u2212 1 photons after the absorption process. 2. Stimulated Emission. This process involves the atom initially being in the excited state, in the presence of n photons in some resonant mode. After the stimulated-emission event, the atom is demoted to the ground state and the field is left with n + 1 photons. In some sense, this process is the opposite of stimulated absorption, although absorption ending with 0 photons is possible while stimulated emission beginning with 0 photons is not. 3. Spontaneous Emission. This process is much like stimulated emission, but when the atom is demoted, a photon is created in some mode that is initially unpopulated. Thus, a photon can go into a wide range of possible modes by spontaneous emission. It is possible to view spontaneous emission as stimulated emission due to quantum vacuum fluctuations in addition to classical radiation reaction (the bound electron radiates because it is accelerating). E1E2photonE1E2(no photon)E1E2n photonsE1E2no+o1 photonsE1E2(no photons)E1E21 photon 292 Chapter 15. Laser Physics Generally, we can associate stimulated absorption and emission with the laser mode, while spontaneous emission is additionally associated with all other modes. Absorption and stimulated emission are the most important once the laser gets going\u2014absorption attenuates the laser mode, while stimulated emission amplifies the laser mode. 15.2.3 Einstein Rate Equations New let\u2019s consider an ensemble of two-level atoms interacting with light. Let N1,2 denote the number density of atoms with energy E1,2. Then the Einstein rate equation for the excited state is dN2 dt = \u2212A21N2 \u2212 B21\u03c1(\u03c9)N2 + B12\u03c1(\u03c9)N1. (15.21) Here, \u03c1(\u03c9) is the energy density of the electromagnetic field (the energy density in the frequency interval \u03c9 to \u03c9 + d\u03c9). The first term corresponds to spontaneous emission, and we can see that it reduces the excited-state population, even in the absence of any field. The second and third terms are proportional to \u03c1(\u03c9), and correspond to stimulated emission and absorption, respectively, as we can see from their overall signs. By convention, the constant A21 is called the Einstein A coefficient, while B21 and B12 are called the Einstein B coefficients. (Einstein rate equation, excited state) To be consistent, N1 + N2 must add up to some constant, assuming that we really have two-level atoms and that the atoms stay in place (something that even works fairly well for gas lasers as long as we modify A21 appropriately). Thus, dN2/dt = \u2212dN1/dt, and so it is easy to write down the rate equation dN1 dt = A21N2 + B21\u03c1(\u03c9)N2 \u2212 B12\u03c1(\u03c9)N1. for the ground-state population N1. It is most convenient to look at the steady-state behavior to see if we can have laser operation. Steady state occurs when dN2/dt = 0, whence it follows from Eq. (15.21) that (Einstein rate equation, ground state) (15.22) N2 N1 = B12\u03c1(\u03c9) A21 + B21\u03c1(\u03c9). (15.23) If the energy levels are not degenerate, it turns out that B12 = B21, as we will see in the next section. That is, stimulated emission and absorption are exactly symmetric from the rate-equation point of view. Then we can rewrite the steady-state solution as N2 N1 = 1 A21 B21\u03c1(\u03c9). + 1 (15.24) (steady-state solution, two-level atom) We can see from this that N2 < N1 in steady state. Thus, there is no steady-state population inversion in a two-level system. The spontaneous emission breaks the symmetry of the stimulated emission and absorption processes and guarantees that there will be always more atoms in the ground state than in the excited state. In the limit of large intensity, \u03c1(\u03c9) \u2212\u2192 \u221e, the best we can do is to equalize the population. Thus, we can\u2019t use a two-level system for a laser gain medium. We need a population inversion for stimulated emission (amplification) to win out over absorption (attenuation). 15.2.4 Relations Between the Einstein Coefficients Now we briefly outline Einstein\u2019s derivation of the relation between the A and B coefficients.4 If the energy levels are degenerate, we can define the degeneracy factors g", "1,2 as the number of ways of having energy E1,2. 4A. Einstein, \u2018\u2018Zur Quantentheorie der Strahlung,\u2019\u2019 Physikalische Zeitschrift 18, 121 (1917). For an English translation, see A. Einstein, \u2018\u2018On the Quantum Theory of Radiation,\u2019\u2019 in The World of the Atom, H. A. Boorse and L. Motz, Eds. (Basic Books, 1966), vol. 2, p. 888. Einstein\u2019s main aim in this paper was in fact to derive the Planck blackbody distribution by solving the rate equations. 15.2 Light\u2013Atom Interactions 293 For example g1,2 = 2J1,2 + 1 for atomic angular-momentum states. Then the steady-state population ratio from Eq. (15.22) can be written also via Boltzmann statistics as N2 N1 = g2 g1 e\u2212\u00afh\u03c9/kBT = B12\u03c1(\u03c9) A21 + B21\u03c1(\u03c9). Solving for \u03c1(\u03c9), B12g1 B21g2 This is equivalent to the Planck blackbody distribution5 \u03c1(\u03c9) = A21 B21 1 e\u00afh\u03c9/kBT \u2212 1. \u03c1(\u03c9) = 8\u03c0h \u03bb3 1 e\u00afh\u03c9/kBT \u2212 1 (15.25) (15.26) (15.27) if we make the identifications and g2B21 = g1B12. (15.28) (relation between B coefficients) 8\u03c0h \u03bb3. Recall that \u03bb here is the wavelength within the gain medium. A21 B21 = (15.29) (relation between A and B coefficients) 15.2.5 Line Shape and Spectral Distributions So far, we\u2019ve considered only monochromatic light and two-level atoms with sharply defined energy levels. Now it\u2019s time to improve our model of the two-level atom and its interaction with light. We will first introduce a line-shape function s(\u03c9) to model the fact that the energy levels have some width. The line shape is defined such that s(\u03c9) d\u03c9 is the probability that a spontaneously emitted photon will have frequency between \u03c9 and \u03c9 + d\u03c9. We can also interpret this as the relative probability of stimulated emission or absorption of a photon with frequency between \u03c9 and \u03c9 + d\u03c9. Since s(\u03c9) represents a probability density, it is appropriately normalized: \u221e 0 s(\u03c9) d\u03c9 = 1. (15.30) Note that as in our discussion of coherence in Chapter 14, we are using a \u2018\u2018one-sided spectrum\u2019\u2019 that ranges only over positive frequencies. In terms of the \u2018\u2018two-sided spectrum\u2019\u2019 s\u2194(\u03c9) with both positive and negative frequencies, the one-sided spectrum satisfies s(\u03c9) := s\u2194(\u03c9) + s\u2194(\u2212\u03c9) for \u03c9 \u2265 0 and s(\u03c9) = 0 for \u03c9 < 0. When we apply the line shape and sum over all frequencies, the rate equation becomes dN2 dt = \u2212A21N2 \u2212 B21N2 \u221e 0 \u03c1(\u03c9)s(\u03c9) d\u03c9 + B12N1 \u221e 0 \u03c1(\u03c9)s(\u03c9) d\u03c9. (15.31) Qualitatively, we can picture the line shape function as a relatively sharply peaked distribution centered around the resonant optical frequency \u03c90. (rate equation including spectra) 5P. W. Milonni and M.-L. Shih, \u2018\u2018Zero-point energy in early quantum theory,\u2019\u2019 American Journal of Physics 59, 684 (1991). dw/2p ~ 10oo -10 o Hzw\u00ba/2p ~ 10oo -10 o Hz7141512ws(w) 294 Chapter 15. Laser Physics Often, s(\u03c9) turns out to be a Lorentzian, a Gaussian, or a convolution of the two (a Voigt profile). The line-shape function models transition width due to spontaneous emission, collisions, Doppler shifts in gas lasers, and local crystal structure effects on the dopant atoms. Note that in the absence of radiation, the rate equation is dN2/dt = \u2212A21N2, which has an exponentially damping solution. The Fourier transform of an exponential is a Lorentzian, so the line shape for spontaneous emission (the \u2018\u2018natural line shape\u2019\u2019) is Lorentzian, with a half-width at half maximum of A21. Collisions are often modeled by a spontaneouse", "mission-like term, and thus also lead to Lorentzian line shapes. Doppler shifts lead to Gaussian line shapes If multiple, independent broadening because the Maxwell\u2013Boltzmann velocity distribution is Gaussian. effects contribute, their combined effect can be modeled by the convolution of the individual line shapes. Now we will consider two limiting cases for the light spectrum. Both are important in understanding laser operation, but the second is the more useful case for understanding laser gain. 15.2.5.1 Broadband Light Light is broadband (relative to the transition) if \u03c1(\u03c9) is much broader than s(\u03c9). Then we can evaluate the integral in Eq. (15.31) by noting that \u03c1(\u03c90) varies slowly over the width of s(\u03c9), so that we can pull it out of the integral: \u221e \u221e \u03c1(\u03c9)s(\u03c9) d\u03c9 \u2248 \u03c1(\u03c90) s(\u03c9) d\u03c9 = \u03c1(\u03c90). (15.32) Thus, we recover the previous rate equations, corresponding to Eq. (15.21), with sharp energy levels. 0 0 15.2.5.2 Nearly Monochromatic Light For nearly monochromatic light, the field spectrum is narrow, so s(\u03c9) is much broader than \u03c1(\u03c9). Thus, we can evaluate the integral with the same slowly varying approximation as for the broadband case: \u221e 0 \u03c1(\u03c9)s(\u03c9) d\u03c9 \u2248 s(\u03c9field) \u221e 0 \u03c1(\u03c9) d\u03c9. (15.33) The integral on the right-hand side is the total field energy density, summed over all frequencies. Let\u2019s denote this by \u03c1total. Then the rate equation becomes dN2 dt = \u2212A21N2 \u2212 B21N2s(\u03c9field)\u03c1total + B12N1s(\u03c9)\u03c1total. (15.34) The energy density is related to the intensity by \u03c1total = Itotal/c (see Problem 4.4), so dN2 dt = \u2212A21N2 \u2212 \u03c3(\u03c9field) Itotal \u00afh\u03c9field N2 \u2212 N1. g2 g1 (rate equation for monochromatic light) (15.35) Here, we have defined the laser cross-section \u03c3(\u03c9) = A21 \u03bb2 4 s(\u03c9). (15.36) (laser cross-section) The cross-section has the dimensions of area, and is defined such that \u03c3(\u03c9)I(\u03c9) is the power absorbed by a single atom when irradiated by intensity I(\u03c9) (in the weak-excitation limit). Note that for a Lorentzian line shape s(\u03c9), the resonant cross section \u03c3(\u03c90) is given by s(\u03c9) = \u2206\u03c9 2\u03c0 [(\u03c90 \u2212 \u03c9)2 + (\u2206\u03c9/2)2], (15.37) (15.38) \u03c3(\u03c90) = A21 \u2206\u03c9 \u03bb2 2\u03c0. 15.3 Light Amplification 295 For homogenous broadening, \u2206\u03c9 is the natural line width given by \u2206\u03c9 = A21, so that the natural cross section is \u03c3(\u03c90) =. (15.39) \u03bb 2 0 2\u03c0 This answer is consistent with a fully quantum-mechanical calculation. In fact, this answer assumes an average over all possible atomic orientations, since the blackbody distribution of Eq. (15.27) assumes isotropic radiation. For atomic dipole moments aligned with the field polarization, the resonant cross section is since the coupling that would normally be \u2018\u2018distributed\u2019\u2019 among three orthogonal directions is concentrated into one. \u03c3(\u03c90) = 3\u03bb 2 0 2\u03c0, (15.40) 15.3 Light Amplification In writing down the atomic rate equations, we have developed a quantitative model for the influence of light on atoms. To complete our picture of light-matter interactions to model a laser, we have to quantify the influence of atoms on light\u2014the other aspect of the fundamental radiation processes of absorption, stimulated emission, and spontaneous emission. This will allow us to quantify the gain of the laser gain medium in terms of atomic (or molecular) parameters. To do so, we will consider the amplification (or absorption) of a monochromatic input intensity component I(\u03c9) by a thin gain medium of thickness dz. Note that we will assume the input light to be in a single mode; however, the gain medium can in principle radiate into many different modes, and we will need to physically distinguish input photons from amplification photons or spontaneously emitted photons.6 15.3.1 Gain Coefficient Now we can compute", " the change in intensity through the gain medium and then through the filtering and detection system above. We will treat the contribution of each of the three fundamental processes separately. In each case, we will consider an infinitesimal intensity change of the form dI(\u03c9) = photon-interaction rate atom energy photon atoms cross-sectional area line-shape weight (15.41) due to the gain medium of thickness dz. We have simply converted the rate at which photons are absorbed or emitted by a single atom to an intensity, essentially by changing units: we multiplied by the energy per photon to give the rate at which energy is absorbed or emitted per atom; then we multiplied by the atoms per unit area of the gain medium to give an intensity; then finally, we multiplied by the spectral line shape to account for the frequency dependence (resonance behavior) of the atom, producing an intensity density as we expect for I(\u03c9). 15.3.1.1 Stimulated Emission The intensity change for stimulated emission, according to the general template of Eq. (15.41), becomes dI(\u03c9) = B21 I(\u03c9) c + \u00afh\u03c9 N2 dz s(\u03c9). (stimulated emission) (15.42) Let\u2019s go through these factors again in a bit more detail. From the atomic rate equation, stimulated-emission events occur at the rate of B21\u03c1(\u03c9) = B21I(\u03c9)/c per atom, and every emission event adds the photon energy \u00afh\u03c9 to the atom. This must be weighted by the relative emission probability s(\u03c9) at the incident optical 6Here we are following the construction, and most of the notation, of Joseph T. Verdeyen, Laser Electronics, 3rd ed. (Prentice Hall, 1995). 296 Chapter 15. Laser Physics frequency. We then multiply this rate per atom by the number of excited-state atoms per unit area, N2 dz. Due to the nature of stimulate emission, the emitted photons are in the same mode as the incoming photons, and so we count every stimulated-emission event as contributing to the amplification of the input mode. 15.3.1.2 Absorption Absorption leads to a similar expression for the intensity change, dI(\u03c9) = B12 I(\u03c9) c \u2212 \u00afh\u03c9 N1 dz s(\u03c9), (absorption) (15.43) with the only two changes being the sign of the photon energy (and hence the overall sign), since absorption attenuates the wave, and the number of atoms per unit area is now N1 dz, since only the ground-state atoms are relevant here. 15.3.1.3 Spontaneous Emission For spontaneous emission, we have a more complicated expression, because spontaneous emission can occur into any mode, but we want to count only those in the input mode: dI(\u03c9) = A21 + \u00afh\u03c9 N2 dz s(\u03c9) d\u2126 4\u03c0 \u2206\u03c9 \u2206\u03c9gain 1 2. (spontaneous emission) (15.44) The emission rate is similar to the stimulated-emission version, but the overall rate per atom is now given in terms of the Einstein A coefficient, and of course the energy density of the input mode does not enter here. The other differences lie in the fraction of spontaneously emitted photons that are emitted into the input mode (i.e., the laser-resonator mode if we are considering amplification within the laser). Only a small fraction of the photons have the right direction to match the input mode, which we assume subtends the solid angle d\u2126, so we include the factor d\u2126/4\u03c0. The fraction of photons with the proper frequency in the width \u2206\u03c9 of the input mode is \u2206\u03c9/\u2206\u03c9gain, where \u2206\u03c9gain is the width of s(\u03c9). Finally, we include a factor of 1/2, since spontaneous emission can occur into either of 2 possible polarizations, and only one will match the input mode. Generally, the angle and frequency factors are the most restrictive, and hence spontaneous emission accounts for only a small fraction of the detected intensity. Thus, we can ignore it for the purposes of computing gain once the laser is going, but it is still important to get the laser going, since it always provides a source of power, even for an initially empty cavity. 15.3.1.4 Combined Effects Now we can combine the contributions of Eqs. (15.42) and (15.43), while neglecting spontaneous emission. We can use the relationsof Eqs. (15.28) and (15.28) between the Einstein coefficients and the definition of the laser cross section, Eq. (15.36), to write dI(\u03c9) dz = \u00afh\u03c9B21 c N2 \u2212 I(", "\u03c9) g2 g1 N1 = \u03c3(\u03c9)I(\u03c9) N2 \u2212 N1, g2 g1 (15.45) which becomes dI(\u03c9) dz = \u03b3(\u03c9)I(\u03c9), (15.46) (intensity change by medium) if we define the gain coefficient (gain per unit length) \u03b3(\u03c9) := \u03c3(\u03c9) N2 \u2212 N1. g2 g1 (15.47) (gain coefficient) Thus, we see that in terms of the atomic parameters, the gain is related only to the cross section and the population inversion. How does gain saturation fit into this? Without any optical intensity, the population 15.3 Light Amplification 297 relaxes to some steady state depending on other pumping fields and decay mechanisms. In the limit of small intensity, the light does not change the populations, and so the gain coefficient keeps the value of the small-signal gain coefficient \u03b30(\u03c9). Strong light, however, depletes the population inversion, since each added photon reduces the inversion. Thus, N2 \u2212 (g2/g1)N1 \u2212\u2192 0 for strong intensities, and the gain coefficient reduces to zero. This is the fundamental root of the gain saturation that we explored earlier. 15.3.2 Threshold Behavior and Single-Mode Operation Again, for small signals, the gain coefficient \u03b3(\u03c9) \u2248 \u03b30(\u03c9), and is thus constant. The solution to Eq. (15.46) is therefore just an exponential, I(\u03c9, z) = I0(\u03c9)e\u03b3(\u03c9)z, (15.48) and so we can write the total (small-signal) round-trip power gain as G0 = exp[2\u03b30(\u03c9)g], (15.49) (small-signal, round-trip gain) where g is the length of the gain medium. The factor of two here arises because we are assuming a linear laser cavity, so that a round trip entails two passes through the medium. Recall that the threshold condition is that the gain matches the loss. For oscillation in a two-mirror linear cavity, we should have G0R1R2 \u2265 1. We can rewrite this as exp[2\u03b30(\u03c9)g]R1R2 \u2265 1, (15.50) or solving for \u03b30, \u03b30(\u03c9) \u2265 1 2g log 1 R1R2 =: a, (15.51) where a is the effective \u2018\u2018loss per unit length\u2019\u2019 per round trip, distributed along the length of the gain medium. In terms of this coefficient, the threshold condition is \u03b30(\u03c9) \u2265 a for oscillation to occur. Depicting this graphically, we can gain great insight into the frequency dependence of laser operation. (threshold condition for gain coefficient) The small-signal gain profile \u03b30(\u03c9) reflects the line shape s(\u03c9), and so with a sufficient pump rate, there will be a range of frequencies that will be above threshold. Any cavity modes that fall within this range are candidates for laser operation. However, for many lasers, the laser operates in only a single cavity mode, even when there are several candidate modes. This occurs if s(\u03c9) is the same for all the atoms in the medium (where the width of s(\u03c9) is said to be due to \u2018\u2018homogenous broadening\u2019\u2019), and all the atoms interact with the cavity modes in the same way. Under these conditions, we get single-mode operation: only the mode with the largest gain will oscillate. To see how this works, consider this graphical example. w\u00ba w g\u00ba(w) arange of possible laser operation (loss/unit length) (gain/unit length) 298 Chapter 15. Laser Physics Three modes are above threshold and can potentially oscillate. Initially, they all start to oscillate. As gain the gain profile \u03b3(\u03c9) is reduced in magnitude. In equilibrium, the gain must balance saturation kicks in, the loss, so the equilibrium condition is that \u03b3(\u03c9) = a. Thus, the gain profile will continue to decrease in magnitude until the equilibrium condition is satisfied at the laser frequency. The other modes with smaller gain end up below threshold (\u03b3 < a), and are thus extinguished. This argument does not hold for certain lasers such as gas lasers (e.g., He\u2013Ne), where \u2018\u2018Doppler broadening\u2019\u2019 is a major component to the width of s(\u03c9). Doppler broadening arises because atoms with different velocities experience different Doppler shifts of the laser light, and thus effectively have different resonant frequencies. Thus, different atomic velocity classes can contribute to different modes, and so the gain profile is", " not necessarily reduced uniformly by gain saturation. In fact, it is possible for many modes to oscillate simultaneously. profile is reduced only locally around each of the laser frequencies. In multi-mode operation, the gain 15.4 Pumping Schemes Now that we have established how the laser gain coefficient is related to the atomic inversion, we need to develop some details of specific gain media to compute the inversion. Let\u2019s simplify things by assuming equal degeneracies, g1 = g2. (This is equivalent to setting N 1 = (g2/g1)N1, so there isn\u2019t any loss of generality in doing this.) Recall that for a two-level atom, without a pump, has the rate equation [from Eq. (15.21) with B21 = 0] dN2 dt = \u2212A21N2 = \u2212N2/\u03c421, (15.52) where \u03c421 = 1/A21 is the lifetime of the excited state. This is so named because this rate equation has the exponential solution N2(t) = N2(0) e\u2212A21t = N2(0) e\u2212t/\u03c421, (15.53) and thus \u03c421 is the time scale for exponential decay from the excited level. w g\u00ba(w) a3 modes could oscillate { w g(w) aw g(w) a 15.4 Pumping Schemes When we include the pump, we can write the rate equation (15.21) as dN2 dt = \u2212A21N2 \u2212 \u03c3(\u03c9)I \u00afh\u03c9 (N2 \u2212 N1), 299 (15.54) where \u03c9 is the pump frequency, and I is the total optical intensity (i.e., I(\u03c9) integrated over all frequencies), if we assume a nearly monochromatic pump. The basic point that we made earlier is that because the pump enters the rate equation with a term of the form (N2 \u2212 N1), the pump tries to equalize N1 and N2. Since spontaneous decay is asymmetric, there is no possibility for population inversion in the two-level atom. Without inversion, there is no gain, and thus no laser oscillation. 15.4.1 Three-Level Laser The simplest change that we can make to achieve a population inversion is to add a third level. The ruby laser is one example of a three-level laser. The level scheme is shown here. The new level (with highest energy) decays quickly, while the laser (2 \u2212\u2192 1) transition decays slowly. That is, we will assume A21 R13, A32. Also, for a monochromatic pump (e.g., the pump is another laser), we canuse Eq. (15.35) to write R13 = \u03c3(\u03c9)I \u00afh\u03c9, (15.55) (pumping rate, monochromatic pump) where again I is the total intensity, while if the pump is broadband (e.g., a flash lamp), we canuse Eq. (15.32) to write R13 = I(\u03c90) \u00afh\u03c90 \u221e 0 \u03c3(\u03c9) d\u03c9 = A21 \u03bb2I(\u03c90) 4\u00afh\u03c90, where \u03c90 = (E3 \u2212 E1)/\u00afh is the pump transition frequency. Then we can write the rate equations as (pumping rate, broadband pump) (15.56) dN3 dt dN2 dt dN1 dt = \u2212R13(N3 \u2212 N1) \u2212 A32N3 = A32N3 \u2212 A21N2 (15.57) (rate equations, three-level atom) = A21N2 + R13(N3 \u2212 N1), where of course one of the equations is redundant since N1 + N2 + N3 must be a constant of the motion. Now we will make the adiabatic approximation to eliminate energy level 3. The idea is that the A32 term in the dN3/dt equation is a damping term (i.e., like a friction term) for N3. In the limit of large A32 (A32 R13), the adiabatic approximation says that N3 damps to equilibrium very quickly, and so we can assume that it is always in quasiequilibrium. Thus, the equilibrium condition for N3 gives dN3 dt \u2248 0 =\u21d2 N3 \u2248 R13 A32 + R13 N1 \u2248 R13 A32 N1. (15.58) 12B12A21123pumpR13A21o(slow)A32o(fast)lasertransition} 300 Chapter 15. Laser Physics The population N3 is said to ad", "iabatically follow the N1 population, and it is now a redundant quantity since it is completely determined by N1. Furthermore, R13/A32 1, so N3 N1, so level 3 is essentially unpopulated. Then we can use N3 \u2212 N1 \u2248 \u2212N1 and write effective rate equations (15.57) for the remaining two levels as dN2 dt dN1 dt = R13N1 \u2212 A21N2 = A21N2 \u2212 R13N1. These are the rate equations for a two-level atom, but with an asymmetric pump. (rate equations, adiabatic approximation) (15.59) The presence of the third level breaks the symmetry of the pump, and now we can get a population inversion if the pump rate exceeds the spontaneous decay rate. In steady state, dN2/dt = dN1/dt = 0, which gives the equilibrium population ratio N2 N1 = R13 A21. (15.60) Again, we see that we have a population inversion if the ratio exceeds unity and thus if the pump rate exceeds the spontaneous emission rate from level 2, R13 > A21. To compute the inversion, note that Similarly, N2 \u2212 N1 = N2 N1 \u2212 1 N1 = R13 A21 \u2212 1 N1. N2 + N1 = N = R13 A21 + 1 N1, (15.61) (15.62) where N is the total atomic number density, and we are neglecting the population N3. Using Eq. (15.62) to eliminate N1 in the right-hand side of Eq. (15.61), we find an inversion of N2 \u2212 N1 = R13 \u2212 A21 R13 + A21 N = R \u2212 1 R + 1 N, where R := R13/A21. Again, we can see that a pump rate R > 1 corresponds to a population inversion. (population inversion, three-level atom, adiabatic approximation) (15.63) 15.4.2 Four-Level Laser Another important level scheme is the four-level laser, where we add yet another extra level. The Nd:YAG laser is one example of a four-level laser. Four-level media are particularly important since, as we will see, they are more efficient in general than three-level media. The level scheme is shown here. The new level 0 is now the ground state, with the laser transition (2 \u2212\u2192 1) sandwiched between the two auxiliary levels. 12R13A211230pumpR03A21o(slow)A32o(fast)A10o(fast)A20(slow)lasertransition} 15.5 Gain Coefficient 301 Other decay paths besides the ones we have included, but we assume them to be slow, in which case they do not affect our basic results. The rate equations are dN3 dt dN2 dt dN1 dt dN0 dt = \u2212R03(N3 \u2212 N0) \u2212 A32N3 = A32N3 \u2212 (A21 + A20)N2 = A21N2 \u2212 A10N1 = A20N2 + A10N1 + R03(N3 \u2212 N0). Again, we will assume that all decay rates except those from level 2 are fast: A32, A10 R03, A21, A20. Now both levels 1 and 3 have fast damping terms, so we can adiabatically eliminate both of them. (rate equations, four-level atom) (15.64) Setting dN3/dt \u2248 0 leads to while setting dN1/dt \u2248 0 leads to N3 \u2248 N1 \u2248 R03 A32 A21 A10 N0 N0, N2 N2. (15.65) (15.66) Again, we can neglect the small populations in levels 1 and 3, and in particular N3 \u2212 N0 \u2248 \u2212N0. Thus, the effective rate equations for the remaining two levels are = R03N0 \u2212 (A21 + A20)N2 dN2 dt dN0 dt (rate equations, four-level atom, adiabatic approximation) = (A21 + A20)N2 \u2212 R03N0. (15.67) Again, this corresponds to a two-level atom with an asymmetric pump. However, the difference is that the ground level 0 of this effective atom is not involved in the laser transition. The ground level of the laser transition is level 1, which is almost completely depleted, and this is why it is easier to develop an inversion in the four-level laser than in the three-level version. In steady state, N2 N", "0 \u2248 R03 A21 + A20. (15.68) Using N0 + N2 \u2248 N and a procedure similar to the three-level calculation, we can eliminate N0 to get an expression for N2. Since N1 \u2248 0, this is the population inversion: N2 \u2212 N1 \u2248 N2 \u2248 R 1 + R N, (15.69) where R := R03/(A21 + A20). Note that there is an inversion for any pump R > 0, whereas we saw that in the three-level laser, the inversion was of the form N (R \u2212 1)/(R + 1), so that a certain minimum pump was required. (population inversion, four-level atom, adiabatic approximation) Again, the whole point of this is: auxiliary levels break the symmetry of the pump and allow for population inversion. Other auxiliary levels might make the pumping process easier. 15.5 Gain Coefficient With the specifics of the atomic media in hand, we can proceed to explore the gain coefficient in more detail. The complication is that now we will have to consider the interaction of the gain medium with two optical 302 Chapter 15. Laser Physics fields: the pump field and the laser field. For simplicity, we will only do this for the four-level laser, but similar results hold for the three-level laser. Going back to the four-level laser scheme, we will consider it to be pumped by a monochromatic laser at a rate R2 = \u03c3pumpIpump \u00afh\u03c9pump, (15.70) and we can regard this to be a unidirectional pump since we assume level 3 will decay quickly to level 2. Recall from Eq. (15.67) that the effective rate equation is dN2 dt = R2N0 \u2212 A2N2 \u2212 \u03c3I \u00afh\u03c9 (N2 \u2212 N1), (15.71) where A2 := A21 + A20. The last term accounts for the interaction of the atom with the laser field, so that \u03c3, I, and \u03c9 all refer to the laser field. Again, for the four-level laser we can neglect N1 compared to N2, with the result dN2 dt \u03c3I \u00afh\u03c9 \u2248 R2N0 \u2212 A2N2 \u2212 N2. (15.72) Of course, the stimulated absorption/emission terms for the pump and laser light have the same form, except that they interact with different levels, and they have different effective signs after the adiabatic elimination of levels 1 and 3. Steady state occurs when dN2/dt = 0, which gives the steady-state population N2 =. R2N0 \u03c3I \u00afh\u03c9 A2 + (15.73) Following the algebraic steps of the last section, when we have an equation of the form N2 = (stuff)N0 with N0 + N2 \u2248 N, the solution is (N2/N ) = (stuff)/(1 + stuff). Thus, R2 A2 + \u03c3I \u00afh\u03c9 R2 1 + A2 + \u03c3I \u00afh\u03c9 N2 N = where and \u03c42 = A \u22121 2. = A2 + R2 \u03c3I \u00afh\u03c9 R2\u03c42 = + R2 1 + R2\u03c42 +, I Isat Isat = \u00afh\u03c9 \u03c3\u03c42, (15.74) (15.75) Thus, the gain coefficient for the four-level laser is \u03b3(\u03c9) = \u03c3(\u03c9)[N2 \u2212 N1] = \u03c3(\u03c9)R2\u03c42N I Isat 1 + R2\u03c42 +. From this expression, we can see that there are two types of saturation in the gain medium. (gain coefficient, four-level laser) (15.76) 1230effectivepumpR2A21(slow)fast decayfast decayh-wlasertransition} 15.5 Gain Coefficient 303 1. Gain saturation. Consider the limit of a weak pump (R2\u03c42 1). Then the gain coefficient (15.74) becomes \u03b3(\u03c9) = \u03c3(\u03c9)R2\u03c42N 1 + I Isat. (15.77) In the small-signal regime (I Isat), the gain coefficient takes on the small-signal value so that \u03b30(\u03c9) = \u03c3(\u03c9)R2\u03c42N, (small-signal gain coefficient, weak pump) (15.78) \u03b3(\u03c9) =. \u03b30(\u03c9) I Isat 1 + (15.79) (gain coefficient, weak pump) Again, the saturation intensity is defined such that at the saturation intensity, the gain coefficient drops to half its small", "-signal value: \u03b3(I = Isat) =. (15.80) \u03b30 2 In the limit of large intensity, \u03b3(I \u2212\u2192 \u221e) = 0. This is precisely the gain-saturation effect that we discussed before in the context of the simple laser model of Section 15.1.5. 2. Pump saturation. In the limit of a very strong pump, R2\u03c42 \u2212\u2192 \u221e, the gain coefficient saturates at a value \u03b3(\u03c9) \u2212\u2192 \u03c3(\u03c9)N. (15.81) (strong-pump gain coefficient) This limit has a nice physical interpretation: there is only a finite number N of atoms, and each can only add an energy \u00afh\u03c9 of energy as soon as it is stimulated by an incoming laser photon. Thus, the gain can also be limited by the finite number of atoms in the gain medium, so that eventually the gain coefficient does not continue to increase with an increasing pump intensity. 15.5.1 Gain in a Medium of Finite Length Now let\u2019s revisit the gain medium, and compute the gain of the medium in terms of the gain coefficient. The medium has length g, and the input and output intensities are I1 and I2, respectively. The gain coefficient is defined such that dI(\u03c9) dz = \u03b3(\u03c9)I(\u03c9), so we can write this explicitly in terms of the small-signal gain coefficient as 1 I(\u03c9) dI(\u03c9) dz =. \u03b30(\u03c9) I Isat 1 + (15.82) (15.83) Recall that the small-signal gain coefficient \u03b30(\u03c9) is defined by Eq. (15.78) in the weak-pump regime, and the saturation intensity Isat is given by Eq. (15.75). Since the right-hand side of this differential equation involves the intensity, the solution is more complicated than a simple growing exponential. lgI\u00a1I\u2122 304 Chapter 15. Laser Physics To find the solution, we solve for dz: 1 I(\u03c9) + 1 Isat dI(\u03c9) = \u03b30(\u03c9) dz. Integrating over the length of the gain medium, and thus I2 1 I1 I(\u03c9) + 1 Isat dI(\u03c9) = g 0 \u03b30(\u03c9) dz, log I2 I1 + I2 \u2212 I1 Isat = \u03b30(\u03c9)g. Defining the single-pass gain as we can rewrite Eq. (15.86) as G := I2 I1, (15.84) (15.85) (15.86) (15.87) log G + I1 Isat (G \u2212 1) = \u03b30(\u03c9)g. (15.88) (single-pass gain-factor equation) This is the central result of this section. We cannot get a general expression for G in terms of \u03b30, but rather we can get a transcendental equation that relates them. In the small-signal regime (I1, I2 Isat), we can ignore the (G \u2212 1) term in Eq. (15.88). In this case, we recover the small-signal result: G0 = exp[\u03b30(\u03c9)g]. (15.89) (small-signal gain factor) In the high-intensity limit (I1 Isat), we can ignore the log G term in Eq. (15.88), with the result I2 \u2212 I1 = [\u03b30(\u03c9)Isat]g. (15.90) (high-intensity gain) The interpretation of this relation is that the gain medium adds an amount \u03b30Isatg to the input intensity. Writing out the gain coefficient and the saturation intensity, \u03b30Isatg = [\u03c3(\u03c9)R2\u03c42N ] g = (N g)(\u00afh\u03c9)R2. (15.91) \u00afh\u03c9 \u03c3(\u03c9)\u03c42 Thus, the gain medium adds an intensity given by the number of atoms per unit of transverse area, multiplied by the photon energy, multiplied by the pumping rate per atom. We can conclude from this that we can only extract energy from the medium at the rate that energy is pumped into it. That is, an intense input signal extract all possible energy from the gain medium. 15.6 Laser Output: CW Now we\u2019ll examine in more detail the output characteristics of a continuous-wave (CW) laser, and in particular look at its efficiency and how to optimize it. Let\u2019s consider a ring-cavity laser as shown, so that our above analysis of the single-pass gain remains valid (in a linear cavity, the two counterpropagating waves saturate each", " other, making things more complicated). gainoutputR\u00a1R\u2122R\u00a3R\u00a2I\u2122I\u00a1medium 15.6 Laser Output: CW 305 In addition, we will assume that there is an \u2018\u2018optical diode\u2019\u2019 in the cavity to force the laser to operate in only one direction (typically, an optical diode consists of a Faraday rotator and a birefringent plate, which leaves the polarization unaffected in only one direction). If all the losses are due to partial mirror transmission, then the survival probability is Ps = R1R2R3R4. (15.92) This expression can easily be generalized to include other cavity losses (e.g., due to reflections at the gain medium surfaces). In steady state, the gain equals the loss, and so Thus, the gain equation (15.88) becomes GPs = 1. Solving for I2, we find log 1 Ps + I2 Isat (1 \u2212 Ps) = \u03b30(\u03c9)g. I2 = Isat[\u03b30g \u2212 log(1/Ps)] 1 \u2212 Ps. (15.93) (15.94) (15.95) Ideally, there is no loss in the output coupler (mirror), so that T1 = 1 \u2212 R1. Then the output intensity is simply I2T1. 15.6.1 Optimum Output How can we optimize the output of the laser? In other words, given a certain gain and loss in the laser, what reflectance should we choose for the output coupler to maximize the output power? Let\u2019s work in the low-loss approximation where (1 \u2212 Ps) is small. Then log 1 Ps = \u2212 log[1 \u2212 (1 \u2212 Ps)] \u2248 1 \u2212 Ps. (15.96) Also, 1 \u2212 Ps = 1 \u2212 (1 \u2212 T1)R2R3R4 = (1 \u2212 R2R3R4) + T1R2R3R4 = Pl + T1(1 \u2212 Pl) \u2248 L + T1, (15.97) where we have defined the loss probability Pl := 1 \u2212 R2R3R4, and we have made the low-loss approximation 1 \u2212 Pl \u2248 1. Then from Eq. (15.95), Iout = T1Isat[\u03b30g \u2212 log(Pl + T1)] Pl + T1 = T1Isat \u03b30g Pl + T1 \u2212 1. This is an optimum when \u2202Iout/\u2202T1 = 0, which gives an optimum transmission T1(opt) = \u03b30gPl \u2212 Pl. The corresponding optimal output intensity is Iout(opt) = Isat \u03b30g \u2212 Pl 2. (15.98) (15.99) (15.100) (15.101) Sometimes, in real lasers it is necessary to choose an output coupler that has lower reflectance than this theoretical minimum. This can enhance the stability of the laser at the expense of some output power. 306 Chapter 15. Laser Physics 15.6.2 Quantum Efficiency In the ideal case, there is no loss (Pl = 0). Then the optimal output intensity is Iout(opt) = \u03b30gIsat = \u00afh\u03c9R2gN. If A is the area of the gain medium, then the corresponding output power is Pout(opt) = AIout(opt) = \u00afh\u03c9R2N (gA), (15.102) (15.103) where gA is the volume of the gain medium, and thus N gA is the total number of atoms in the gain medium. Thus the output power is Pout = \u00afh\u03c9 \u00b7 (total pump rate). In other words, the output power consists of one output photon energy per input photon. This sounds like perfect efficiency, but it isn\u2019t quite: the pump photon is more energetic than the laser photon because they correspond to different energy transitions. Suppose that \u03b7q is the laser efficiency in this ideal case. Then we can write \u03b7q as the ratio of the output power to the input power: \u03b7q = \u00afh\u03c9 \u00b7 (total pump rate) \u2206Epump \u00b7 (total pump rate) = \u00afh\u03c9 \u2206Epump. (15.104) Here, \u2206Epump is the energy of the pumping transition. This ideal efficiency is called the quantum efficiency of the laser, and represents the fundamental limit to efficiency for any given laser gain medium. Other limits to efficiency include cavity losses (extraction efficiency) and the fact that not all pump photons lead to stimulated emission (pumping efficiency). For example, a Nd:YAG laser has a pump wavelength of \u03bbpump = 810 nm when pumped from a diode laser (these lasers can also be flashlamp-pumped in the blue", " with correspondingly less efficiency). The laser wavelength is \u03bblaser = 1064 nm, giving a relatively high quantum efficiency of \u03b7q \u223c 0.8. A He\u2013Ne laser, on the other hand, operates by electric discharge to a state of \u223c20 eV in He. The energy is transferred via collision to a state of \u223c20 eV in Ne. The laser is on a Ne transition, with a 632.8 nm wavelength corresponding to \u223c 2 eV. The quantum efficiency in this case is much lower, around \u03b7q \u223c 0.1. This is one reason why gas-discharge lasers tend to be less efficient than optically pumped lasers: the discharge leads to very high excitation, and most of this excitation energy is not extracted by the laser field. 15.7 Laser Output: Pulsed Many lasers don\u2019t work in CW mode, but instead work in pulsed mode. There are many reasons why pulsed lasers are preferred in some applications. For example, pulsed lasers are capable of much higher peak powers than CW lasers (with state-of-the-art lasers reaching the petawatt range). With high powers or certain gain media, it might not be possible to sustain the pump or to cool the laser system efficiently enough. Also, when the laser is used to probe or visualize a fast process, short pulses may be needed to strobe and effectively freeze the dynamics during the probe. 15.7.1 Laser Spiking In the treatment of CW lasers above, we focused on the steady-state behavior of lasers. To treat pulses, though we must treat the laser dynamics. One kind of pulsed laser output occurs in the initial transient behavior of a CW laser. From the simple laser model of Section 15.1.5 (see the plot on p. 289), we might expect a CW laser to turn on smoothly, especially for a weak or slowly activated pump where the laser intensity adiabatically follows the inversion until steady state is reached. However, especially with high-gain lasers such as flashlamp-pumped Nd:YAG lasers, there is another possibility, called laser spiking, where the dynamics of the inversion N2 \u2212 N1 and the intensity I conspire to produce intensity oscillations. The laser spikes are also called relaxation oscillations, although this term often refers to the low-amplitude oscillations after the gain spikes settle down to near equilibrium (or when the laser is only slightly perturbed). 15.7 Laser Output: Pulsed 307 As in the simulation shown here, the initial transient can look like a periodic sequence of short intensity pulses. This process occurs due to the following process. 1. A fast pump rapidly excites the gain medium far above the threshold inversion. This happens on a fast time scale, before the cavity intensity can respond. 2. The intensity, as a result, builds up rapidly, quickly depleting the gain medium as the laser oscillates. The cavity intensity becomes so large that the gain medium drops below threshold. 3. We are more or less at the initial condition again, so the process repeats. This process causes the intensity oscillations or laser spikes. Since both the laser output and spontaneous emission act as damping (friction-like) processes, the oscillations damp away and the laser eventually relaxes to steady state. Below is the measured intensity from a diode-laser pumped Yb:fiber laser operating at 1085 nm that shows laser spiking. The pump diodes are turned on quickly, and the resulting laser spikes are clearly visible before the laser power settles down to the steady-state value of about 10 W. During the first spike, the laser jumps to well above the steady-state power. I(t) t N\u2122o-oN\u00a1Ntht0500t (\u00b5s)300output power (W) 308 15.7.2 Q-Switching Chapter 15. Laser Physics The idea behind Q-switching is to avoid the laser spiking by compressing all the relaxation oscillations in to a single, big pulse. For example, for a flashlamp-pumped laser, the cavity is made lossy (i.e., the cavity Q is \u2018\u2018switched low\u2019\u2019) during the flashlamp pulse to allow the inversion to build up way beyond the high-Q threshold value. Near the end of the flashlamp pulse, the cavity is restored to its former high-Q condition, so that the laser finds itself way above threshold. In this case, the intensity is extracted in one short, enormous pulse, as shown in the simulation below. For a high-gain laser (such as Nd:YAG with an unstable resonator), the output pulse can be on the order of the cavity round-trip time (10 ns pulses are typical for a 1 m linear Nd:YAG laser). Several techniques are available for switching the", " cavity Q. One common method is to use a Pockels cell (electro-optic crystal) in the cavity, which is essentially a nonlinear crystal that becomes birefringent when high voltage is applied across it. When the voltage is applied, the polarization is rotated going through the cell, and a polarizer in the cavity strongly attenuates the mode. (Actually it turns out that a polarizer isn\u2019t necessary, the birefringence is sufficient to spoil the Q.) Another method is to use a rotating cavity mirror, so that the cavity Q is only high at the moment when the mirror reaches the right angle. Finally, an \u2018\u2018automatic\u2019\u2019 Q-switch can be made from a saturable absorber. This is a medium that absorbs light, but due to the same saturation effect as in gain medium, high intensities pass through the medium nearly unaffected (this saturation effect is also called \u2018\u2018bleaching\u2019\u2019). The saturable absorber will spoil the Q until the gain reaches the point where an amplified spontaneous wave exceeds the saturation threshold, at which point the absorber bleaches and the cavity Q is switched high. I(t) I(t)t tN\u2122o-oN\u00a1Nthtflashlampt 15.7 Laser Output: Pulsed 15.7.3 Cavity Dumper 309 Another simple source of pulse laser light is the cavity dumper (see Problem 7.3). Actually, this is more of a device to convert CW light into pulses, since it is not itself a laser. The idea is to take a ring cavity and inject CW laser light into one port. The intensity in the cavity builds up to \u223c(F /2\u03c0)Iin in steady state, so a high cavity finesse is important in attaining large pulse intensities. Once the light has built up, the idea is to suddenly turn on an optical switch (such as an acousto-optic modulator) inside the cavity to couple out the high-intensity light. For efficient out-coupling, the pulse duration is equal to the round-trip time of the cavity. 15.7.4 Mode Locking Mode locking is another method for converting a CW laser into a pulsed laser. This is also the method that provides some of the shortest possible pulses. The idea is that a gain medium with inhomogeneous broadening can support simultaneous lasing on many modes, as we discussed in Section 15.3.2. This is because different modes interact with different atoms, and thus do not compete for pump energy. Suppose that the gain medium can support modes in a range of order \u2206\u03c9. The modes are spaced by the usual cavity spacing \u03b4\u03c9 = 2\u03c0/\u03c4rt. Usually, the modes will have no particular phase relationship. The laser output is thus the interference of all the beams, but with random phases, giving a noisy output that fluctuates about the mean intensity \u00afI. But what if it is possible to give them all the same relative phase? Recall the Poisson sum rule (Problem 3.7), \u221e n=\u2212\u221e cos(2\u03c0nt) = \u221e n=\u2212\u221e \u03b4(t \u2212 n). (15.105) We can interpret the left-hand side as the sum of many waves of equal amplitude, frequency \u03c9n = 2\u03c0n, and zero phase offset. The right-hand side is a periodic series of arbitrarily short pulses. In terms of the laser, opticalswitchw Dwdwoo=oo2p(fsr)oo=oo2p/trtttrt2trt3trt0 I I(CW) 310 Chapter 15. Laser Physics if we can lock the relative phases of many modes, we can expect the output to be a series of short, intense pulses. Let\u2019s consider a slightly less simple model of mode locking. Recall from Section 5.8 that the interference of N equal-amplitude waves of the form exp[i(n \u2212 1)\u03c6] is I = I0 sin2(N \u03c6/2) sin2(\u03c6/2), (15.106) where I0 is the intensity of a single wave. Interpreting \u03c6 \u2212\u2192 (\u03b4\u03c9)t, where \u03b4\u03c9 is much smaller than the central optical frequency \u03c90, we have a sum of the form which leads to an intensity N n=1 I0ei\u03c90tei(n\u22121)(delta\u03c9)t 2, I = I0 sin2[N (\u03b4\u03c9)t/2] sin2[(\u03b4\u03c9)t/2]. (15.107) (15.108) This is a sequence of short pulses, with duration \u223c1/\u2206\u03c9, period \u03c4rt, and height N \u00afI. The output for 10 equalampl", "itude, phase-locked modes is shown here, compared to the random-phase version shown as a dashed line. If there are many modes available, the pulses can be extremely intense and short. For example, a Ti:sapphire laser is popular for mode-locking because of its very wide emission range, spanning 600-1050 nm. The full width at half maximum of the emission range is more like \u2206\u03bb \u223c 200 nm (from 700-900 nm), which gives \u2206\u03c9/2\u03c0 \u223c 9.5 THz, so the pulse width is \u03b4t \u223c 1/\u2206\u03c9 \u223c 1.7 fs. Real lasers have much longer pulse lengths due to a number of effects, but fs-scale pulses are now routine through this method. There several ways to mode-lock a laser. One way is to \u2018\u2018encourage\u2019\u2019 only the large intensities of the mode-locked operation and \u2018\u2018discourage\u2019\u2019 the random-phase CW operation. For example, just as in the Q-switched laser, a saturable absorber can select the mode-locked behavior, since the pulses will bleach the absorber, while the CW light will be blocked. A similar effect is achieved by the Kerr lens, which takes advantage of the nonlinear optical behavior of the gain medium. At high intensities, the refractive index of the gain medium increases with the applied intensity. Thus, a high-intensity Gaussian beam effectively creates a lens for itself in the gain medium. An appropriately placed aperture (or effective aperture) can block the cavity light except when the Kerr effect focuses the mode through it, thus selecting again the highintensity pulses. Lasers can also be mode-locked actively. For example, an electro-optic phase modulator in the cavity can put sidebands on each mode, separated by the modulation frequency (see Problem 13.1). If the modulation frequency matches the free spectral range of the laser cavity (or possible some multiple of it), the sidebands lead to coupling of the modes. The modes then act as coupled nonlinear oscillators, and tend to synchronize (you can also think of the sidebands seeding the other modes with light of the proper phase). A similar effect can be achieved by modulating the amplitude of the intracavity light, e.g., by periodically driving an acousto-optic modulator, where the modulation frequency matches the cavity free-spectral range. ttrt2trt3trt0 I I(CW)~NItrtN~1Dw 15.8 Exercises 15.8 Exercises Problem 15.1 311 Consider a two-mirror cavity with a gain medium. The mirrors have intensity reflection coefficients R1 and R2. The gain medium is thin and has a round-trip intensity gain coefficient G. (All this is in the text, the point is to go through it and fill in the details.) (a) Write down the cavity intensity at time t + \u03c4rt in terms of the cavity intensity at time t, where \u03c4rt is the round-trip time of the cavity. (b) Expand the above expression to first order in \u03c4rt, and hence obtain a differential equation for I(t). What are the conditions under which this approximation is justified? (c) Apply the result of part (b) to a cavity without a gain medium, to derive an expression for the cavity photon lifetime \u03c4p, defined as the time required for the cavity intensity to decay to 1/e times its initial value. Problem 15.2 This is a prelude to problems below that require numerically solving differential equations (i.e., laser rate equations). Get thee to a computer running Mathematica, and run the commands in the following notebook: http://atomoptics.uoregon.edu/~dsteck/teaching/mathematica/ode_example.nb This notebook is also posted in pdf format here: http://atomoptics.uoregon.edu/~dsteck/teaching/mathematica/ode_example.pdf Run the supplied commands to solve two simple systems of ordinary differential equations. Give the analytic solutions to these equations, and comment on whether the numerical solutions are sensible. Make sure to understand the syntax, so you can apply the same commands to plotting the solutions to the laser rate equations in later problems. Problem 15.3 (a) Obtain an analytic solution to the laser cavity intensity equation, where the gain G saturates according to dI dt = (G(I) R1R2 \u2212 1) \u03c4rt I, G(I) = 1 +, g0 1 + I Isat (15.109) (15.110) and where g0 is a constant. Treat the case of a perfect cavity, R1R2 = 1 (for this part only). You will not be able to obtain an explicit solution, I", "(t), but you can get an implicit solution t(I). (Hint: we did something very similar to this in the text.) (b) Find the steady-state intensity for a laser modeled by the above equations. (c) Plot the solution I(t) to the above equations. For the purposes of the plot, take g0 = 2, I0 = 1 \u00b5W/cm2, Isat = 1 W/cm2, R1R2 = 0.99, and plot the solution as a function of t/\u03c4rt. Solve the differential equation numerically to make the plot. Problem 15.4 Starting with the gain relation for a medium of length g, log G + I1 Isat (G \u2212 1) = \u03b30(\u03c9)g, (15.111) 312 Chapter 15. Laser Physics show that in the limit of a thin gain medium, the gain coefficient takes the form that we considered in the simple laser model of Eq. (15.9): G(I) = 1 +. g0 1 + I Isat (15.112) What is g0 in terms of the gain parameters? Problem 15.5 Examine the validity of the adiabatic approximation in the three-level laser model as follows. (a) Write down the rate equations for the three-level atom, both with and without the adiabatic elimination of the third level. Then modify these in two ways. First, eliminate N1 by letting N be the total number density of the atoms, and then use conservation of atom number. Write the equations in terms of the normalized populations nj := Nj/N. Second, measure time in units of 1/A21. In other words, write the equations in terms of the scaled time parameter t := A21t. (b) Assume a pumping rate of R13 = 2A21. Make plots of n2(t) for different values of A32, corresponding to where the adiabatic approximation is bad, marginal, and good (say, A32/R13 = 1, 10, and 100). Include the plot of the solution of the adiabatic rate equations for comparison. In all cases start with the atoms entirely in the ground state. Problem 15.6 Suppose you build your own He-Ne laser. You use a symmetric cavity (geometrically speaking) by sealing two concave mirrors with the same radius of curvature R = 100 cm onto the ends of a glass tube of length L. You then fill the tube with the appropriate mixture of helium and neon, install proper electrodes, hook up the power supply, and so on. Assume that the laser operates on the red line at 632.8 nm, and that the refractive index of the helium/neon mixture is n = 1. Also assume that the output mirror has a reflectance of Rout = 99%, while the other (back-end) mirror is perfectly reflecting. (a) Gas lasers have relatively small single-pass gain, and so the light must pass through the gain medium many times before exiting the cavity to sustain oscillation. For what range of length L might you expect a working laser? (b) Suppose you settle on a length L = 75 cm. Calculate the beam waist parameter w0, the spot size w at the mirrors, and the far-field divergence angle of your laser, assuming it lases. (c) Calculate the free spectral range of the resonator. (d) Calculate the finesse and Q factor of the resonator, and the order q of the lasing mode. (e) Calculate the frequency width of the resonator modes. This is a very crude estimate of the frequency width of the laser output, as it assumes that the gain medium plays negligible role in determining the frequency spectrum. This is only a good estimate if the laser operates close to threshold. (f) The He-Ne gain medium can support laser operation within a frequency range of about 1.3 GHz due to Doppler broadening at room temperature. Do you expect your laser to operate in a single longitudinal mode? Assume that the laser operates in a single transverse mode (TEM0,0). (g) Suppose that the laser can oscillate in either the TEM0,0 or the TEM1,0 mode. For the same longitudinal mode, calculate the difference in the lasing frequencies of these two modes. (h) Suppose you change the temperature of the laser resonator by 1 \u25e6C, say by cooling it off with air current from a fan. By how much does the frequency of the output light change? Assume that you constructed your laser using a glass tube made out of BK7 borosilicate crown glass, and that the laser is not too much hotter than room temperature. (Part of this exercise is to locate a reputable source for optical material properties, such as the thermal expansion", " coefficient of a popular optical glass.) (i) Calculate the minimum round-trip intensity gain G required for your laser to oscillate (lase). 15.8 Exercises 313 (j) Suppose that the output power of your laser is 5 mW. Calculate the optical power of the light circulating in the resonator (assuming that all the power is either reflected from or transmitted through the output mirror) and the peak intensity inside the resonator. (k) You notice that your laser is behaving strangely. Upon closer inspection, you find that you managed to trap a tiny bug inside the cavity while you were filling it. You observe that the bug is resting near the center of the output mirror, and as it moves around slightly the laser jumps between the Gaussian and donut modes. Estimate the diameter of the bug, using these assumptions and observations: (1) the bug is a perfectly absorbing disc, and is otherwise impervious to the laser radiation; (2) in addition to the transmission through the output mirror, the only sources of loss are absorption by the bug and loss around the edges of the 2 mm diameter mirrors; (3) the bug is much smaller than the beam diameter; (4) the TEM0,0 and donut modes compete so that only the mode with the least loss lases at any given time; and (5) the gains of the two modes in the medium are roughly the same. Problem 15.7 In this problem you will examine numerically the dynamics of a CW Nd:YAG laser. Assume that the laser is well modeled as a four-level laser producing output at a wavelength of 1.064 \u00b5m, and that the refractive index of YAG is n = 1.83 at the laser wavelength (be careful to note that 1.064 \u00b5m is the wavelength in vacuum!). (a) What does \u2018\u2018CW Nd:YAG\u2019\u2019 mean? (b) Assume the following parameters for the gain medium:7 the lasing transition has a lifetime \u03c421 = 1.2 ms and a linewidth \u2206\u03c9/2\u03c0 = 120 GHz. Assume that you can model the lineshape function as a Lorentzian, s(\u03c9) = \u2206\u03c9 2\u03c0 [(\u03c90 \u2212 \u03c9)2 + (\u2206\u03c9/2)2], (15.113) where \u03c90 is the resonance frequency. Use this form to write down an expression for s(\u03c90). What is the Einstein A coefficient for the transition? What is the corresponding resonant laser cross section \u03c3(\u03c90)? The saturation intensity Isat on resonance? (Check your answers: Isat = 262 W/cm2.) (c) Assume a cavity with a perfectly reflecting back mirror and a 95% reflecting output mirror (5% transmission), with all other losses negligible. Also assume a 10 cm long gain medium. What is the threshold population inversion density N2 \u2212 N1 required for the laser to oscillate? (Assume g2 = g1 for this laser.) (d) Assuming an ideal 4-level laser, we can use the rate equation dN2 dt = RpumpN0 \u2212 A21N2 (15.114) to model the level population densities, with N3 \u2248 0, N1 \u2248 0, and N \u2248 N2 + N0. (Right now we are still excluding the effect of laser light.) Find an expression for the steady-state inversion (N2 \u2212 N1)/N. (e) At what rate Rpump must you pump the medium to achieve an inversion density of 5 times the threshold value you found in (c)? Assume a dopant density N = 1019 cm\u22123. (Check your answers: Rpump = 1.8 s\u22121.) (f) Now including the laser light, the rate equation is dN2 dt = RpumpN0 \u2212 A21N2 \u2212 2\u03c3(\u03c90)I \u00afh\u03c90 N2, (15.115) where I is the laser intensity, and the extra factor of 2 in the last term accounts for the fact that the laser light counterpropagates through the medium. We will assume the cavity intensity equation dI dt = (R1R2 \u2212 1) \u03c4rt I + 2g\u03c3(\u03c90)N2 \u03c4rt I + \u03b7SEg\u00afh\u03c90A21N2 \u03c4rt, (15.116) 7Bahaa E. A. Saleh and Malvin Carl Teich, Fundamentals of Photonics (Wiley, 1991) p. 478 314 Chapter 15. Laser Physics where the second term now accounts for the gain (saturation is built into the rate equation for N2), and the third term is a small contribution to the intensity due to spontaneous emission. Again, the second term here", " has a factor of 2 to account for the counterpropagating mode in the linear cavity. These equations are valid when the laser intensity changes slowly. Here we have defined \u03b7SE as the fraction of spontaneously emitted photons that end up in the cavity mode (with proper direction, frequency, polarization, etc.; note that we are not properly treating effects due to amplified spontaneous emission). Assume a cavity length of 1 m (make sure to account for the laser rod in calculating the round-trip time \u03c4rt!). Solve these simultaneous equations and make plots of the laser intensity and the population inversion N2/N. Assume also a small initial intensity (due to spontaneous emission) to get things started, and generate solutions over a few photon lifetimes. For solving ODEs, numerical integrators don\u2019t like really big or really small numbers, nor units for that matter. Therefore you should switch to dimensionless variables. Solve this set of differential equations, which is equivalent to the set of unscaled equations above: dn2 dt = R dI dt = \u2212I + pump(1 \u2212 n2) \u2212 n2 nth I + \u03b7SE \u03c4p \u03c421 n2. nth n2 \u2212 \u03c4p \u03c421 2I n2 (15.117) Here, t := t/\u03c4p, I := I/Isat, n2 := N2/N, R pump := Rpump\u03c4p, n0,2 = N0,2/N, \u03c4p := \u03c4rt/(1 \u2212 R1R2) is the photon lifetime, and nth := (1 \u2212 R1R2)/(2g\u03c3(\u03c90)N ) is the threshold inversion density. (Use scaled variables in the plots as well.) Use initial conditions n2(0) = 0 and I(0) = 0, and assume \u03b7SE = 10\u221212. Use the parameters you calculated or were given above to set up the parameters for these scaled equations. How does changing the ratio \u03c4p/\u03c421 affect the laser dynamics? Perform sanity checks on your results, e.g., by calculating expressions for steady-state values to compare with your data. Problem 15.8 This is a follow-up to Problem 15.7 that models the CW Nd:YAG laser. Here you will examine numerically the dynamics of a pulsed Nd:YAG laser. Use these assumptions, which are different from the last problem: 1. The laser uses a ring cavity, comprising four mirrors at each of the corners of the square beam path. Each leg of the beam path is 50 cm long, and the gain medium sits in one leg of the cavity. Appropriate measures have been taken to ensure the laser oscillates in only one direction. 2. The output coupler has an intensity reflection coefficient of R1 = 80%. The other three mirrors have reflectances of R2 = R3 = R4 = 99.5%. Any light not reflected is transmitted through the mirror. 3. The laser will be pumped at a rate Rpump that produces an inversion density N2 of 10 times the threshold inversion Nth (in the absence of laser oscillation). Everything else should be the same as last time: 1. The laser is well modeled as a four-level laser producing output at a wavelength of 1.064 \u00b5m. Ignore degeneracy of the laser transition levels. 2. The refractive index of YAG is n = 1.83 at the laser wavelength, so the laser wavelength inside the gain medium is 580 nm. The gain medium has length g = 10 cm. 3. The lasing transition has a lifetime \u03c421 = 1.2 ms and a line width \u2206\u03c9/2\u03c0 = 120 GHz (with a Lorentzian line-shape function s(\u03c9)). These parameters produce a resonant laser cross section 2 \u03c3(\u03c90) = 5.9 \u00d7 10\u221219 cm2 and a saturation intensity Isat = 262 W/cm. 4. The Nd dopant density is N = 1019 cm\u22123. 15.8 Exercises 315 (a) Recalculate the threshold inversion density N2 required for this laser to oscillate. You must generalize the formula you used last week slightly; do not use the formula given below for Nth. What is the corresponding pump rate that produces an inversion of 10 times the threshold value (if the laser is not oscillating)? (Check your answer: Rpump = 35 s\u22121.) (b) The output mirror we chose is better for creating short pulses, since it quickly couples energy out the resonator. What reflectance of the output coupler would maximize the output intensity for CW operation? What is the corresponding output intensity? (c) We will model the laser with the pair of equations dN2 dt dI dt = RpumpN0 \u2212 A21N2 \u2212", " = (Ps \u2212 1) \u03c4 I + g\u03c3(\u03c90)N2 \u03c4 \u03c3(\u03c90)I \u00afh\u03c90 I + N2, \u03b7SEg\u00afh\u03c90A21N2 \u03c4. (15.118) where Ps := R1R2R3R4 is the round-trip survival probability, \u03c4rt is the cavity round-trip time, I is the intensity circulating in the cavity, and \u03b7SE is the fraction of spontaneously emitted photons that end up in the cavity mode (with proper direction, frequency, polarization, etc.; note that we are not properly treating effects due to amplified spontaneous emission). These are the same equations as before, but with spontaneous emission into the cavity mode explicitly accounted for. These equations are really only valid for a high-finesse resonator, so they will be marginally accurate but good enough for our purposes. Using the scaled and otherwise defined variables t := t/\u03c4p, I := I/Isat, Nth := (1 \u2212 Ps)/g\u03c3(\u03c90), nth := Nth/N, n2 := N2/N, R pump := Rpump\u03c4p, \u03c4p := \u03c4rt/(1 \u2212 Ps), Isat := \u00afh\u03c90/\u03c3(\u03c90)\u03c421, and A21 = \u03c4 \u22121 21, show that the above rate equations are equivalent to the following equations: dn2 dt = R dI dt = \u2212I + pump(1 \u2212 n2) \u2212 n2 nth I + \u03b7SE \u03c4p \u03c421 n2. nth n2 \u2212 \u03c4p \u03c421 I n2 (15.119) pump \u2212\u2192 R (d) Assume that the gain medium is pumped by a flashlamp modeled by a square pulse of duration \u03c4pulse = 1 ms (i.e., define a pulse function fpulse(t) which is unity between t = 0 and \u03c4pulse, which can be written in terms of the Heaviside step function as fpulse(t) = UH(t) \u2212 UH(t \u2212 \u03c4pulse/\u03c4p), and then replace R pumpfpulse(t) in the above equations). Plot the solution to the scaled rate equations for these conditions, with the initial conditions n2(0) = I (0) = 0 (assume \u03b7SE = 10\u221212). (e) Now plot the laser output for a Q-switched pulse. We will model Q-switching the cavity by allowing nth to be effectively time dependent. Let fQ(t) have the value 12 during the flashlamp pulse (to model a Q-spoiled cavity) and then have the value unity immediately afterwards; then make the replacement nth \u2212\u2192 nthfQ(t) in the above equations (in addition to the modification you made in (d)). Compare your results to what you saw for the \u2018\u2018long-pulse mode\u2019\u2019 of (d). What is the duration of the Q-switched pulse (in ns)? Problem 15.9 Recall the simple laser model where the gain G saturates according to dI dt = (GR1R2 \u2212 1) \u03c4 I Isat (15.120) (15.121) 316 Chapter 15. Laser Physics Under the conditions where this equation is valid, show that this model is equivalent to the model in Problem 15.7 dI dt = (R1R2 \u2212 1) \u03c4 I + 2g\u03c3(\u03bd0)N2 \u03c4 I, (15.122) which is specific to the four-level laser (note that we have dropped the spontaneous emission term). Do this in the following steps: (a) Argue that both G and R1R2 must be close to 1. Then use this to argue that the first model can be rewritten (approximately) as dI dt = (R1R2 \u2212 1) \u03c4 I + (G \u2212 1) \u03c4 I. (15.123) (b) Derive an expression for the round-trip gain G in the regime where G is close to 1, and hence recover the equation for the second model above. (c) Complete the argument by using the results from the four-level laser to write an expression for g. Be careful to make sure your results are valid for a round trip, not just a single pass through the gain medium. Problem 15.10 For a ring laser modeled by the gain equations (as in Problem 15.8) \u03c4p \u03c421 n2 \u2212 dn2 dt = R dI dt = \u2212I + pump(1 \u2212 n2) \u2212 n2 nth I + \u03b7SE \u03c4p \u03c421 n2, nth I n2 (15.124) where Ps is the round-trip survival probability, \u03c4rt is the cavity round-trip time, I is", " the intensity circulating in the cavity, \u03c4p := \u03c4rt/(1 \u2212 Ps), t := t/\u03c4p, I := I/Isat, Nth := (1 \u2212 Ps)/g\u03c3(\u03c90), nth := pump := Rpump\u03c4p, Isat := \u00afh\u03c90/\u03c3(\u03c90)\u03c421, and A21 = \u03c4 \u22121 Nth/N, n2 := N2/N, R 21, you will compute the frequency and damping rates of the relaxation oscillations for small disturbances about the equilibrium (these are the same oscillations as in the laser spikes when they have nearly damped away). Show that the intensity oscillation frequency and damping rate are given by \u03c9\u03b3 and \u03b3/2, respectively, where \u03c9 = \u03c9\u03b3 = \u03c92 \u2212 (\u03b3/2)2 r \u2212 1 \u03c4p\u03c421 r \u03c421(1 \u2212 nth) R pump R r := \u03b3 =, pump,th (15.125) pump,th is the scaled threshold pumping rate (the value of R and R You can do this via the following outline. (a) Neglect spontaneous emission and show that in steady state, n2 = nth and I R pump,th? pump above which the laser oscillates). ss = r \u2212 1. What is (b) Ignore spontaneous emission and linearize the above pair of rate equations about the equilibrium solution by letting n2 = nth + \u03b4n2 and I = I ss + \u03b4I, putting these into the rate equations, and discarding terms beyond first order in \u03b4n2 and \u03b4I. You should now have a pair of rate equations for the small disturbances \u03b4n2 and \u03b4I. (c) Recall that the damped harmonic oscillator is governed by the equation m\u00a8x + m\u03b3 \u02d9x + m\u03c92x = 0. (15.126) 15.8 Exercises 317 Recall that in the underdamped regime, the solution with p(0) = 0 is x(t) = x(0)e\u2212(\u03b3/2)t cos(\u03c9\u03b3t), (15.127) where \u03c9\u03b3 is defined above. Show that the harmonic oscillator equation can be rewritten as the pair of first-order equations \u02d9x = p m \u02d9p = \u2212m\u03c92x \u2212 \u03b3p, (15.128) where p = m \u02d9x. Now draw a formal analogy between the laser gain equations and the damped harmonic oscillator, and write down expressions for the oscillation frequency and damping rate. Remember that the above rate equations measure time in units of \u03c4p, so you must undo this to get the above answers. (d) To get a sense of scale, put in some numbers to see a typical oscillation frequency. Assume a low-loss ring cavity (nth 1, \u03c4p \u223c 10\u03c4rt) just above threshold (r = 1.1) with a \u223c1 m cavity path length (\u03c4rt \u2248 3 ns), and a Nd:YAG gain medium (\u03c421 = 1.2 ms). Roughly how many oscillations do you expect to see before they damp away? Problem 15.11 (This is a simple review question to help you tie together some of the concepts.) Consider a laser whose gain medium is described by a frequency-dependent gain coefficient \u03b3(\u03c9). Your answers to the questions below should be qualitative and need not involve calculations or equations. (a) What is the condition for the laser to oscillate? (b) Suppose that many cavity modes satisfy the condition in (a), but in steady state, the laser operates only in a single mode. What is the underlying effect that causes single-mode operation? Explain how this effect leads to single-mode operation. (c) A different gain medium in the same cavity causes the laser to oscillate in many modes simultaneously. Explain how the medium is different from the medium in part (b). (d) What is mode-locking? Give an example of how to achieve mode locking in a multimode laser. Problem 15.12 Due to temperature fluctuations, the length and therefore the frequency of a helium-neon (He-Ne) laser drifts slightly. Suppose you observe that a particular He-Ne laser oscillates in only one (longitudinal) mode at some temperatures, but two modes at others. Estimate the length of the laser cavity, assuming an operating wavelength \u03bb = 632.8 nm, a gain linewidth \u2206\u03bd = 1.5 GHz, and TEM0,0 transverse-mode operation. Problem 15.13 Consider the three-level laser scheme shown here. (The first three parts of this problem are a review of the chapter material.) (a) Write down the rate equations for this laser. (b) What are the conditions on R", "13, A32, and A21 that ensure the best laser operation? 123pumpR13A21oA32olasertransition} 318 Chapter 15. Laser Physics (c) In the regime you described in part (b), carry out the adiabatic elimination of the appropriate state to obtain a set of rate equations for an effective two-level atom. (d) Suppose we add an extra decay path as shown below, described by the (possibly large) coefficient A31. Perform a similar analysis (with the same adiabatic elimination), and write down an expression for the small-signal gain coefficient \u03b30 in terms of the coefficient for the gain medium in parts (a)-(c), assuming everything else is the same. That is, what is the effect of A31 on the gain coefficient? Problem 15.14 Consider the inverted three-level laser scheme shown here. (a) Write down the rate equations for this laser. (b) What are the conditions on R02, A21, and A10 that ensure the best laser operation? (c) In the regime you described in part (b), carry out the adiabatic elimination of the appropriate state to obtain a set of rate equations for an effective two-level atom. (d) The usual three-level laser that we covered in Section 15.4.1 has the lower two levels defining the laser transition. Argue qualitatively that in the best possible cases (optimal pumping and decay rates), for the same pumping rate, laser cross section, and atomic number density, the small-signal gain coefficient for the usual scheme is twice the small-signal gain coefficient for the inverted scheme shown here. Recall that the gain coefficient is given by \u03b3(\u03c9) = \u03c3(\u03c9)[Ne \u2212 Ng], where \u2018\u2018e\u2019\u2019 and \u2018\u2018g\u2019\u2019 refer to the excited and ground levels of the laser transition, respectively. Hint: you don\u2019t need to do any calculations, just reason out what the steady-state populations would be in the optimal cases. (e) Give a qualitative argument for why the saturation intensity for the inverted three-level scheme will be twice that of the saturation intensity for the usual three-level scheme. Assume the laser cross sections are the same in both cases. 123pumpR13A21oA32oA31olasertransition}012pumpR02A10oA21olasertransition} Chapter 16 Dispersion and Wave Propagation Now that we have discussed the generation of laser light and the propagation of laser light through a gain medium, we will take a deeper look at propagation in media. In particular, we will look at the effects of dispersion\u2014dependence of dielectric material properties on the optical frequency\u2014on light propagation. 16.1 Causality and the Kramers\u2013Kronig Relations Recall from Section 4.3 that the polarization density of a dielectric is the material response to the electric field, P = 0\u03c7E, (16.1) where \u03c7 is the susceptibility. For dispersive media, the susceptibility is frequency-dependent (\u03c7 = \u03c7(\u03c9)), \u02dcP(\u03c9) = 0\u03c7(\u03c9) \u02dcE(\u03c9), (16.2) where \u02dcE(\u03c9) and \u02dcP(\u03c9) is the magnitude of the component exp(\u2212i\u03c9t). We can also think of \u03c7(\u03c9) as the (complex-valued) frequency-space transfer function for the medium. It turns out that just by knowing that \u03c7(\u03c9) is a response function corresponding to a physical system, and thus describing a causal response, we can put substantial constraints on its form. By the convolution theorem, the frequency-space multiplication in Eq. (16.2) is equivalent to a convo- lution in the time domain of the electric field with the inverse fourier transform g\u03c7(t) of \u03c7(\u03c9): P(t) = 0 \u221e \u2212\u221e g\u03c7(t)E(t \u2212 t) dt. (16.3) Since P(t) is the material response to E(t), causality requires that P(t) can\u2019t depend on E(t) for future times t > t. Thus, from the convolution, we can see that causality implies g\u03c7(t) = 0 for t < 0. Equivalently, we can write causality as the mathematical statement where UH(t) is the Heaviside step function. Recalling from Section 12.5.2.3 that g\u03c7(t) = g\u03c7(t)UH(t), F [UH(t)] \u2261 \u02dcUH(\u03c9) = \u03c0\u03b4(\u03c9) + i \u03c9, (16", ".4) (16.5) we can compute the Fourier transform of Eq. (16.4), using the convolution theorem again, but this time in 320 Chapter 16. Dispersion and Wave Propagation the opposite direction than usual: \u03c7(\u03c9) = F [g\u03c7(t)] = F [g\u03c7(t)UH(t)] = F F \u22121[\u03c7(\u03c9)]F \u22121[ \u02dcUH(\u03c9)] = = = = 1 2\u03c0 1 2\u03c0 \u03c7(\u03c9) 2 \u03c7(\u03c9) 2 \u2212\u221e + \u2212 i 2\u03c0 i 2 \u03c7(\u03c9) \u2217 \u02dcUH(\u03c9) \u221e \u03c0\u03b4(\u03c9 \u2212 \u03c9) + \u03c7(\u03c9) \u221e \u2013 \u2212\u221e \u03c7(\u03c9) \u03c9 \u2212 \u03c9 d\u03c9 H [\u03c7(\u03c9)]. Here again, H denotes the Hilbert transform, defined by H [f (x)] := 1 \u03c0 \u221e \u2013 \u2212\u221e f (x) x \u2212 x dx. i \u03c9 \u2212 \u03c9 d\u03c9 (16.6) (16.7) The factor of 1/2\u03c0 in applying the convolution theorem in the above derivation comes again from applying the convolution theorem in the reverse sense: the usual convolution theorem reads F [f (t) \u2217 g(t)] = F [f (t)]F [g(t)], but in the opposite direction we have F \u22121[ \u02dcf (\u03c9) \u2217 \u02dcg(\u03c9)] = 2\u03c0F \u22121[ \u02dcf (\u03c9)]F \u22121[\u02dcg(\u03c9)], due to the factor of 2\u03c0 that appears in the inverse Fourier transform equation (3.21). Thus, from Eq. (16.6) we have that Writing out the real and imaginary parts of this equation, \u03c7(\u03c9) = \u2212iH [\u03c7(\u03c9)]. Re [\u03c7(\u03c9)] = H {Im [\u03c7(\u03c9)]} Im [\u03c7(\u03c9)] = \u2212H {Re [\u03c7(\u03c9)]}. (16.8) (16.9) Writing out the Hilbert transform explicitly, we arrive at the standard form of the Kramers\u2013Kronig relations: Re[\u03c7(\u03c9)] = 1 \u03c0 Im[\u03c7(\u03c9)] = \u2212 d\u03c9 \u221e \u2013 \u2212\u221e \u221e 1 \u2013 \u03c0 \u2212\u221e Im[\u03c7(\u03c9)] \u03c9 \u2212 \u03c9 Re[\u03c7(\u03c9)] \u03c9 \u2212 \u03c9 d\u03c9. (16.10) Notice that this is a strong constraint on the form of \u03c7(\u03c9): the real part of \u03c7 determines the imaginary part via the Hilbert transform and vice versa. The Kramers\u2013Kronig relations are often written in terms of the complex permittivity (dielectric constant) \u02dc(\u03c9) = 0[1 + \u03c7(\u03c9)], where we recall from Section 4.3 that the electric displacement is given in terms of the permittivity by \u02dcD(\u03c9) = \u02dc(\u03c9) \u02dcE(\u03c9). The Kramers\u2013Kronig relations for the permittivity then read Re[\u02dc(\u03c9)] = 0 + Im[\u02dc(\u03c9)] = \u2212 1 \u03c0 d\u03c9 \u221e \u2013 \u2212\u221e 1 \u03c0 \u221e \u2013 \u2212\u221e Im[\u02dc(\u03c9)] \u03c9 \u2212 \u03c9 Re[\u02dc(\u03c9)] \u2212 0 \u03c9 \u2212 \u03c9 d\u03c9. (16.11) Again, the real part of \u02dc determines the imaginary part via the Hilbert transform and vice versa. 16.1 Causality and the Kramers\u2013Kronig Relations 321 16.1.0.1 DC Component But one should then ask, why do the Kramers\u2013Kronig relations for the susceptibility [Eqs. (16.10)] and the permittivity [Eqs. (16.11)] have different forms, where 0 needs to be subtracted from Re[\u02dc(\u03c9)]? After all, from \u02dcD(\u03c9) = \u02dc(\u03c9) \u02dcE(\u03c9), \u02dc looks like an input-output response function just like \u03c7. The difference lies in the dc component of (\u03c9), which is absent in \u03c7. This is because at very large frequencies, the material electrons fail to respond to the electric field, and thus lim \u03c9\u2192\u00b1\u221e \u03c7(\u03c9) = 0; lim \u03c9\u2192\u00b1\u221e \u02dc(\u03c9) = 0. (16.12) If we repeat the derivation of Eqs. (16.6), we find that if \u03c7 had an asymptotic dc component of \u03c7\u221e := \u03c7(\u03c9 \u2192 \u221e", "), the response function g\u03c7(t) picks up the additional term \u03c7\u221e\u03b4(t). It works out if UH(t) passes \u2018\u2018half\u2019\u2019 of \u03b4(t), so that \u03c7(\u03c9) \u2212 \u03c7\u221e = \u2212iH [\u03c7(\u03c9) \u2212 \u03c7\u221e]. (16.13) Thus, any causal response function with its asymptotic value subtracted away satisfies the Kramers\u2013Kronig relations. Thus, Eqs. (16.11) follow from the fact that [\u02dc(\u03c9) \u2212 0] is the appropriate response function for the Kramers\u2013Kronig relations. 16.1.1 Refractive Index Notice that the Kramers\u2013Kronig relations are not specific to electromagnetism. Rather, they apply to any frequency-space transfer function that corresponds to a physical, causal response. For example, they could apply to the response function of an electronic amplifier or feedback loop. For our purposes, it most relevant to note that the Kramers\u2013Kronig relations also apply to the refractive index. Recall the propagation of a plane wave in a medium with a complex refractive index \u02dcn, which carries over into the dispersive case: E(+)(z) = E(+) 0 ei\u02dcnk0z \u2212\u2192 \u02dcE(+)(\u03c9; z) = \u02dcE(+) 0 (\u03c9)ei\u02dcn(\u03c9)k0z. The corresponding infinitesimal evolution is \u02dcE(+)(\u03c9; z + dz) = \u02dcE(+)(\u03c9; z)[1 + i\u02dcn(\u03c9)k0dz], which is equivalent to the differential equation d \u02dcE(+)(\u03c9; z) dz = ik0 \u02dcn(\u03c9) \u02dcE(+)(\u03c9; z). (16.14) (16.15) (16.16) This is another input-output relation, with input \u02dcE(+), output d \u02dcE(+)/dz, and transfer function proportional to \u02dcn(\u03c9). Thus, the refractive index (\u02dcn(\u03c9) \u2212 1, after subtracting the asymptotic dc part as in the discussion of Section 16.1.0.1, since \u02dcn tends to unity at high frequencies) obeys the Kramers\u2013Kronig relations or in inefficient form, \u02dcn(\u03c9) \u2212 1 = \u2212iH [\u02dcn(\u03c9) \u2212 1], Re[\u02dcn(\u03c9)] = 1 + Im[\u02dcn(\u03c9)] = \u2212 1 \u03c0 d\u03c9 \u221e \u2013 \u2212\u221e 1 \u03c0 \u221e \u2013 \u2212\u221e Im[\u02dcn(\u03c9)] \u03c9 \u2212 \u03c9 Re[\u02dcn(\u03c9)] \u2212 1 \u03c9 \u2212 \u03c9 d\u03c9. (16.17) (16.18) in(w)out(w)response(w) 322 Chapter 16. Dispersion and Wave Propagation Recall that the complex refractive index corresponds to both phase delay and absorption (or gain). This is evident from Eq. (16.16), where we can write the solution as \u02dcE(+)(\u03c9; z) = \u02dcE(+) 0 (\u03c9)eiRe[\u02dcn(\u03c9)]k0ze\u2212Im[\u02dcn(\u03c9)]k0z. (16.19) The first exponential factor is a propagating plane wave with wave number Re[\u02dcn(\u03c9)]k0. The second exponential factor represents exponential decay (or gain if Im[\u02dcn(\u03c9)] < 0). Since Re[\u02dcn(\u03c9)] controls the phase shift of propagating light and Im[\u02dcn(\u03c9)] controls the absorption (gain), we see why the Kramers\u2013Kronig relations are so significant. Since the Hilbert transformation is \u2018\u2018something like\u2019\u2019 a derivative (see Section 12.5.2.3), derivatives in the real part of \u02dcn(\u03c9) induce large imaginary parts, and vice versa. Thus, for example, variation with frequency of the (real) refractive index implies regions of absorption. Similarly, variation in the absorption coefficient implies regions of strong (real) refractive index. 16.1.1.1 Example: Lorentzian Absorption One particularly important absorption line shape is the Lorentzian [see Eq. (15.37)], which we can write as Im[\u02dcn(\u03c9)] = a(\u03c9) 2k0 = (a0/2k0) (\u2206\u03c9/2)2 (\u03c90 \u2212 \u03c9)2 + (\u2206\u03c9/2)2. (16.20) The convention here is that a(\u03c9) is the intensity absorption", " coefficient, so that I(z) = I0 exp[\u2212a(\u03c9)z], and a0 = a(\u03c90). Note that a0 > 0 corresponds to absorption, while a0 < 0 corresponds to gain. Now we need to evaluate the integral Re[\u02dcn(\u03c9)] \u2212 1 = 1 \u03c0 \u221e \u2013 \u2212\u221e Im[\u02dcn(\u03c9)] \u03c9 \u2212 \u03c9 d\u03c9 (16.21) to find the phase behavior. Let\u2019s first do this with a simple, approximate method. Assume that a(\u03c9) is sharply peaked around \u03c90 (\u2206\u03c9 \u03c90). Then we can approximate the Lorentzian by a \u03b4-function, by considering the normalized form of the line shape: In this case, so that \u2206\u03c9 2\u03c0[(\u03c90 \u2212 \u03c9)2 + (\u2206\u03c9/2)2] \u2248 \u03b4(\u03c9 \u2212 \u03c90). Im[\u02dcn(\u03c9)] \u2248 a0 2k0 \u03c0\u2206\u03c9 2 \u03b4(\u03c9 \u2212 \u03c90), Re[\u02dcn(\u03c9)] \u2212 1 \u2248 a0 4k0 \u221e \u2206\u03c9 \u2013 \u2212\u221e \u03b4(\u03c9 \u2212 \u03c90) \u03c9 \u2212 \u03c9 d\u03c9 = a0 4k0 \u2206\u03c9 (\u03c90 \u2212 \u03c9). (16.22) (16.23) (16.24) Notice that the refractive index scales as (\u03c90 \u2212 \u03c9) for large |\u03c90 \u2212 \u03c9|, while the absorption coefficient scales as (\u03c90 \u2212 \u03c9)2. This turns out to be an important result in atom optics. It isn\u2019t too difficult to compute the Hilbert transform without approximations. Let\u2019s compute the Hilbert transform of the scaled Lorentzian 1/(1 + x2). We can do this using the convolution theorem, the fact that the Fourier transform of 1/(1 + x2) is \u03c0 exp(\u2212|k|) (see Problem 11.7), and the fact that the Fourier transform of 1/x is i sgn(k) (see Problem 12.16): H 1 1 + x2 = \u2212F \u22121[i\u03c0 sgn(k) exp(\u2212|k|)] \u221e 0 \u221e \u03c0 exp(\u2212k)eikx dk + c.c. e\u2212(1\u2212ix)k dk + c.c. (16.25) \u2212i 2\u03c0 \u2212i 2 = = = 0 i 2(1 \u2212 ix) x 1 + x2. = \u2212 + c.c. 16.2 Pulse Propagation and Group Velocity Then with x = 2(\u03c9 \u2212 \u03c90)/\u2206\u03c9, we can write We will derive this formula again when we examine models of atomic media. Re[\u02dcn(\u03c9)] = 1 + 2(\u03c90 \u2212 \u03c9) \u2206\u03c9 Im[\u02dcn(\u03c9)]. 323 (16.26) 16.2 Pulse Propagation and Group Velocity 16.2.1 Phase Velocity Now we will consider the velocity of light propagation in a dispersive medium. We will start out by considering the monochromatic case, and then we will generalize to the case of pulses with a spectrum of finite width. Recall the plane-wave solution to the electromagnetic wave equation: E(+)(x, t) = E(+) 0 ei(kx\u2212\u03c9t). (16.27) The phase of the wave is thus \u03c6 = kx \u2212 \u03c9t. We can define the phase velocity as the velocity of the points of constant phase. To derive an expression for the phase velocity, consider the infinitesimal transformation x \u2212\u2192 x + dx and t \u2212\u2192 t + dt, while \u03c6 is held constant. Then the phase is the same before and after the transformation, so that k dx = \u03c9 dt. This defines the phase velocity vph as \u03c6 = k(x + dx) \u2212 \u03c9(t + dt) = kx \u2212 \u03c9t, vph := dx dt = \u03c9 k. Recall that for a plane wave to be a solution of the wave equation, vph = \u03c9 k = c = c0 n, (16.28) (16.29) (16.30) where n is the refractive index, and c0 = 1/ 0\u00b50 is what we defined to be the speed of light in vacuum in Section 4.1. In general, the refractive index depends on the optical wavelength and thus on the wave number, n = n(k), for a dispersive dielectric medium. Thus, the phase velocity vph = vph(k) is likewise wavelength-dependent. By the Kramers\u2013Kronig relations,", " we know that the dispersive refractive index must also be complex, so when we write n here, we are referring only to the real part of \u02dcn. \u221a 16.2.2 Group Velocity But what happens with pulses, which contain a continuum of frequency components? In a dispersive medium, each component sees a different phase velocity: E(+)(x, t) = 1 2\u03c0 \u221e 0 \u02dcE(+)(k)ei(kx\u2212\u03c9t) dk. (16.31) Im[n~(w)]Re[n~(w)]o-o1w (16.33) (16.34) 324 Chapter 16. Dispersion and Wave Propagation Here we account for dispersion by regarding \u03c9 = \u03c9(k) and use k as the independent variable, which we can do via \u03c9(k) = kvph(k) = c0k/\u02dcn(k). This works for small deviations of the refractive index from unity, such that k(\u03c9) is an invertible function, but the ideas here are generally applicable. For \u03c9 = \u03c90, where \u03c90 is a constant, all the components have the same phase velocity, so all the time dependence in Eq. (16.31) is of the form x \u2212 t \u03c90 k = (x \u2212 vpht). (16.32) Thus, E(+)(x, t) = E(+)(x \u2212 vpht), so the pulse propagates at the phase velocity vph = \u03c90/k. Now suppose we have a dispersive medium, but let\u2019s assume that the dispersion relation is linear. More generally, we can expand an arbitrary dispersion relation to first order about a center wave number kc. In either case, \u03c9(k) = \u03c90 + (k \u2212 kc) = \u03c90 + vg(k \u2212 kc), d\u03c9 dk kc where \u03c90 = \u03c9(kc), and we have defined the group velocity vg := d\u03c9 dk kc, whose significance will become apparent presently. Now we can write the pulse function (16.31) as \u221e 0 \u221e 0 E(+)(x, t) = = = 1 2\u03c0 1 2\u03c0 1 2\u03c0 \u02dcE(+)(k)ei[kx\u2212\u03c9(k)t] dk \u02dcE(+)(k)ei(kx\u2212\u03c90t)e\u2212ivg(k\u2212kc)t dk (16.35) eik0vgte\u2212i\u03c90t \u221e 0 \u02dcE(+)(k)eik(x\u2212vgt) dk Now the phases outside the integral are uninteresting overall phase factors, and all the spacetime dependence is of the form E(+)(x, t) = E(+)(x \u2212 vgt). (16.36) Thus, the pulse propagates at the group velocity vg = d\u03c9/dk. The expansion of Eq. (16.33) is valid if the pulse spectrum is narrow enough for the higher-order terms to be neglected. Note that there are multiple relevant propagation velocities here: while the phase fronts of the individual waves still propagate at the appropriate phase velocities, the overall envelope of the pulse propagates at the group velocity. Recall that the usual refractive index, which controls the phase velocity, can be written n = c0 vph, so we can define an analogous group index for the group velocity by ng := c0 vg = c0 dk d\u03c9 = n + \u03c9 dn d\u03c9, since k = n\u03c9/c0. Also, the vacuum wavelength is \u03bb0 = 2\u03c0 k0 = 2\u03c0c0 \u03c9, (16.37) (16.38) (16.39) vgvph 16.2 Pulse Propagation and Group Velocity where k0 = k/n is the vacuum wave number. Then d\u03bb0 d\u03c9 = \u2212 2\u03c0c0 \u03c92 = \u2212 \u03bb0 \u03c9, and so we can write the group index (16.38) as ng = n + \u03c9 dn d\u03c9 = n + \u03c9 dn d\u03bb0 d\u03bb0 d\u03c9 = n \u2212 \u03bb0 dn d\u03bb0 \u2248 n \u2212 \u03bb dn d\u03bb, 325 (16.40) (16.41) where the last equality follows from assuming that n(\u03bb) is a slowly varying function. This form for the group index is convenient, since the normal refractive index n is commonly specified in terms of wavelength. There are, broadly speaking three cases to consider with respect to the group index. 1. No dispersion. (dn/d\u03bb0 = 0) This", " is the nondispersive case that we already discussed, when the group and refractive indices are equivalent: ng = n. 2. Normal dispersion. (dn/d\u03bb0 < 0) Here the pulse envelope travels more slowly than the phase fronts; this happens when the index decreases with wavelength, and implies that ng > n. This case is called \u2018\u2018normal\u2019\u2019 because this is the observed behavior when white light is dispersed by a glass prism: blue wavelengths see a larger refractive index than red wavelengths, and are thus deflected more by the prism. 3. Anomalous dispersion. (dn/d\u03bb0 > 0) Here the pulse envelope travels more quickly than the phase fronts; this happens when the index increases with wavelength, and implies that ng < n. For the Lorentzian line-shape function (Section 16.1.1.1), normal dispersion occurs everywhere except near the line center, where the slope of the index curve changes sign. Here, there is a region of width \u2206\u03c9 (usually very narrow) of anomalous dispersion. In the anomalous case, it is even possible to have vg > c0. This \u2018\u2018superluminal\u2019\u2019 propagation seems to violate (relativistic) causality, since it seems possible to send a signal pulse faster than the speed of light. However, a more careful analysis of the pulse shape shows that the signal can\u2019t be sent at a higher velocity than c0. 16.2.3 Pulse Spreading Higher order terms in the dispersion relation lead to pulse spreading (second order) or distortion (third and higher orders). For example, up to second order, where \u03c9(k) = \u03c90 + vg(k \u2212 kc) + \u03c92(k \u2212 kc)2, \u03c92 := 1 2 d2\u03c9 dk2 kc. (16.42) (16.43) Then the pulse function (16.31) becomes E(+)(x, t) = 1 2\u03c0 eik0vgte\u2212i\u03c90t \u221e 0 \u02dcE(k)(+)eik(x\u2212vgt)ei\u03c92(k\u2212kc)2t dk (16.44) Im[n~(w)]normaldispersionnormaldispersionRe[n~(w)]o-o1wanomalousdispersion 326 Chapter 16. Dispersion and Wave Propagation For a Gaussian pulse, E(+)(x, t = 0) = E(+) 0 e\u2212x2/4(\u2206x)2 eik0x, so that the Fourier transform is approximately \u02dcE(k) \u2248 E(+) 0 2 \u221a \u03c0(\u2206x)e\u2212(\u2206x)2(k\u2212kc)2. Thus, the propagation to time t leads to a larger effective width (\u2206x)2 \u2212\u2192 (\u2206x)2 + i\u03c92t, (16.45) (16.46) (16.47) which disperses the pulse in the same way as the complex q parameter of the Gaussian beam of Section 6.5 (p. 99). To see this, we can define a q parameter for this problem by where q0 = 2(\u2206x)2 and \u03c4 = 2\u03c92t, so that the propagation replacement becomes q := \u03c4 \u2212 iq0, (\u2206x)2 \u2212\u2192 (\u2206x)2 + i\u03c92t = iq 2. (16.48) (16.49) Now we can take the inverse Fourier transform, thus evaluating the integral of Eq. (16.44). The Fourier transform after propagation through time t is \u02dcE(+)(k, t) = E(+) 0 2 \u221a \u03c0(\u2206x)eiq(k\u2212kc)2 e\u2212ikvgt, and so up to an overall phase, the inverse transform is E(+)(x, t) = E(+) 0 \u2206x q exp ik(x \u2212 vgt)2 2q eikcx. This has the same form as the Gaussian beam in terms of the q parameter [Eq. (6.41)], E(+)(r) = E(+) 0 q0 q(z) exp ikr2 2q(z) exp(ikz), where q = z \u2212 iz0 = z \u2212 i\u03c0w 2 0 /\u03bb. Since the Gaussian beam satisfies the expansion law [Eq. (6.13)], w(z) = w0 1 + 2, z z0 (16.50) (16.51) (16.52) (16.53) the analogue in pulse propagation in a dispersive medium", " (by comparing the two forms of the q parameter) is that the pulse width \u2206x increases with a similar functional form with the time parameter \u03c4. The other aspect of the complex q parameter is that it controls the radius of curvature of the Gaussian beam [Eq. (6.14)], R(z) = z 1 + 2. z0 z (16.54) The Gaussian beam acquires curvature because it is a superposition of plane waves with slightly different directions in their wave vector k, and as the beam propagates, these begin to separate. For propagation in the +z-direction, the plane waves with wave vectors with positive x-components will move towards one edge of the beam profile, while the ones with negative x-components move towards the other. The analogue in the pulse propagation is that the Gaussian pulse is a superposition of plane waves of different wavelength, and the red/blue wavelength components begin to move to the front/back of the pulse (or the other way around, depending on the sign of \u03c92). Here, \u03c92 is related to the group velocity dispersion, given by dvg/d\u03c9, which controls this dispersion effect. 16.3 Slow and Fast Light 327 16.3 Slow and Fast Light A recent \u2018\u2018hot topic\u2019\u2019 in the quantum optics community has been the slow and fast propagation of light in atomic media.1 The general question is, how is it possible to make vg very small or very large? Answering this question is fundamentally interesting, and can also be technologically interesting: for example, a \u2018\u2018slow\u2019\u2019 medium can act as a compact delay line or even memory storage for optical information. Let\u2019s consider a medium with Lorentzian absorption (appropriate for stationary gas-phase atoms). From Section 16.1.1.1, the absorptive part of the refractive index is thus given by while the phase index is Im[\u02dcn(\u03c9)] = a(\u03c9) 2k0 = (a0/2k0) (\u2206\u03c9/2)2 (\u03c90 \u2212 \u03c9)2 + (\u2206\u03c9/2)2, n = Re[\u02dcn(\u03c9)] = 1 + 2(\u03c90 \u2212 \u03c9) \u2206\u03c9 Im[\u02dcn(\u03c9)]. (16.55) (16.56) The maximum value of |n \u2212 1| occurs for (\u03c9 \u2212 \u03c90) = \u00b1\u2206\u03c9/2, giving a maximum deviation of n from unity of \u03b4nmax = a0 4k0. (16.57) To evaluate the group index [Eq. (16.38)], we need to consider the derivative dn/d\u03c9. The maximum value of the derivative occurs at (\u03c9 \u2212 \u03c90) = \u00b1 3\u2206\u03c9/2, giving \u221a dn d\u03c9 = max a0 8k0\u2206\u03c9 = \u03b4nmax 2\u2206\u03c9. Thus, we can write the maximum group index as ng,max = n + \u03c9 dn d\u03c9 max = 1 + 1 + \u03c9 2\u2206\u03c9 \u03b4nmax. (16.58) (16.59) Let\u2019s put in some typical numbers to see the magnitude of the effect. For alkali vapors, the refractive index is not too large (n \u223c 1.1), the resonant frequency is in the optical (\u03c90/2\u03c0 \u223c 400 THz), but the line width is quite narrow (\u2206\u03c9/2\u03c0 \u223c 6 MHz). Putting in these numbers gives ng,max \u223c 3 \u00d7 106, (16.60) an absolutely enormous index (so that pulses should propagate at \u223c 100 m/s. Again, this is valid for stationary atoms. For a vapor at room temperature, the line width is more like \u2206\u03c9/2\u03c0 \u223c 1 GHz due to Doppler broadening (in fact the line-shape function should be Gaussian rather than Lorentzian, but we\u2019ll ignore this here). The corresponding group index is smaller, but still quite large: ng,max \u223c 2 \u00d7 104. (16.61) However, it is practically impossible to observe these effects in ordinary vapors, because a large group index is accompanied by strong absorption: a \u223c (\u03b4n)4k0 = 4\u03c9(\u03b4n) c0 \u223c 3 \u00d7 104 cm\u22121. (16.62) Thus, we need to be more crafty when setting up a medium with an extreme group index. 1For a good review, see Robert W. Boyd and Daniel J. Gauthier, \u2018\u2018\u2018Slow\u2019 and \u2018Fast\u2019 Light,\u2019\u2019 in Progress in Optics, vol. 43, E. Wolf, ed. (Elsevier, Amsterdam, 2002),", " p. 497. We roughly follow some of their arguments here. 328 Chapter 16. Dispersion and Wave Propagation 16.3.1 Quantum Coherence: Slow Light The solution to the problem of acheiving a high group index in a transparent medium is to use a quantum interference effect called electromagnetically induced transparency (EIT). Roughly speaking, this effect comes about in a medium of three-level atoms in the \u2018\u2018\u039b\u2019\u2019 configuration, where two fields (the pump and probe) couple the two ground states to the excited state. Due to quantum interference, the pump saturates the probe transition, but only if the two detunings match precisely: \u03c9probe \u2212 \u03c9eg2 = \u03c9pump \u2212 \u03c9eg1. (16.63) This creates a narrow \u2018\u2018dip\u2019\u2019 in the usual Lorentzian absorption spectrum for the probe transition. For small intensities, the width of the dip can be small, on the order of the decay rate between the ground states (which can be as low as Hz to kHz). The physical significance of the dip is that the medium is transparent to the probe field, when it is properly tuned. The dip also induces a steep slope in the refractive index in the same frequency range. Since dn/d\u03c9 is large, the group index is correspondingly large, even though the absorption is small. A pulse propagating through the medium is thus delayed compared to pulse propagation without the medium. Again, this is a quantum-mechanical effect, so we can\u2019t describe this via the rate-equation approach that we used to model lasers. Rather, we need a fully quantum-mechanical description of the atom, which we will not get into here. probepump|g\u00a1\u00d2|g\u2122\u00d2|e\u00d2Im[n~(w)]Re[n~(w)]o-o1w 16.3 Slow and Fast Light 329 A number of slow-light experiments have achieved impressively small group velocities, down to 8 m/s for the slowest achieved velocities.2 16.3.2 Fast Light The alternative to slow light is \u2018\u2018fast\u2019\u2019 light. One approach that was realized experimentally by Wang et al.3 is to use two pump beams to induce two gain peaks in the refractive-index profile. These appear as negative peaks in the absorption curve. This arrangement flips the slope of the phase index, so that the sign of dn/d\u03c9 becomes negative. The magnitude of the derivative can be large enough that the group index can become negative, as we can see from Eq. (16.38). In the experiment of Wang et al., the group index was ng = \u2212310, so that the group velocity was vg = \u2212c0/310. Thus, a pulse entering the medium must propagate backwards through it. Given the continuity conditions at the boundaries, this happens if a \u2018\u2018pulse pair\u2019\u2019 is created at the far boundary, where the backwards-propagating pulse meets the incoming pulse and annihilates it. The outgoing pulse is thus advanced with respect to the \u2018\u2018normal\u2019\u2019 propagation of the incoming pulse, and the propagation appears to be \u2018\u2018faster than c0.\u2019\u2019 2D. Budker, D. F. Kimball, S. M. Rochester, and V. V. Yashchuk, \u2018\u2018Nonlinear Magneto-optics and Reduced Group Velocity of Light in Atomic Vapor with Slow Ground State Relaxation,\u2019\u2019 Physical Review Letters 83, 1767 (1999). 3L. J. Wang, A. Kuzmich and A. Dogariu, \u2018\u2018Gain-assisted superluminal light propagation,\u2019\u2019 Nature 406, 277 (2000). noooooo>oonntngIm[n~(w)]Re[n~(w)]o-o1w 330 Chapter 16. Dispersion and Wave Propagation Again, there can be no violation of causality, and in fact this behavior is completely consistent with the Kramers\u2013Kronig relations that enforce causality. noooooo<oo0ntng 16.4 Exercises 16.4 Exercises Problem 16.1 331 Consider an optical medium, where you observe that at a particular wavelength (say, 500 nm), the gas has a refractive index n = 1.1, but exactly zero absorption. From this information, what can you conclude about the existence of absorption of the gas at other wavelengths? Explain. Problem 16.2 Show that the Kramers\u2013Kronig relations for the complex permittivity Re[\u02dc(\u03c9)] = 0 + Im[\u02dc(\u03c9)] = \u2212 1 \u03c0 can be written in", " the equivalent form Re[\u02dc(\u03c9)] = 0 + Im[\u02dc(\u03c9)] = \u2212 2\u03c9 \u03c0 d\u03c9 \u221e \u2013 \u2212\u221e 1 \u03c0 \u221e \u2013 \u2212\u221e Im[\u02dc(\u03c9)] \u03c9 \u2212 \u03c9 Re[\u02dc(\u03c9)] \u2212 0 \u03c9 \u2212 \u03c9 d\u03c9 \u221e 2 \u2013 \u03c0 0 \u221e \u2013 0 \u03c9Im[\u02dc(\u03c9)] \u03c92 \u2212 \u03c92 d\u03c9 d\u03c9. Re[\u02dc(\u03c9)] \u2212 0 \u03c92 \u2212 \u03c92 (16.64) (16.65) To do this, you will need to argue that the time-domain polarization response g\u03c7(t) applies to real fields and polarizations, and thus \u02dc(\u2212\u03c9) = \u02dc\u2217(\u03c9). Problem 16.3 Show that for a rarefied medium (|\u03c7| 1), the refractive index n \u2261 Re[\u02dcn] and intensity absorption coefficient a can be written in terms of the susceptibility as n \u2248 1 + Re[\u03c7] 2 a \u2248 k0 Im[\u03c7]. (16.66) Problem 16.4 Show that the pulse spread due to propagation in a fiber composed of a dispersive material can be written4 \u2206tpulse \u2248 (\u2206\u03bb0) d, (16.67) \u03bb0 c0 d2n d\u03bb 2 0 where \u2206\u03bb0 is the wavelength range of the pulse, and d is the propagation distance. Use the following outline. (a) Show that the group velocity may be written vg \u2248 c0 n 1 +. \u03bb0 n dn d\u03bb0 (16.68) (b) Then use the fact that the pulse propagation time is tpulse = d/vg and \u2206vg \u2248 (dvg/d\u03bb0)\u2206\u03bb0 to obtain the above expression, in the regime where (dn/d\u03bb0)2 |d2n/d\u03bb 2 0 |. 4See, for example, Peter W. Milonni and Joseph H. Eberly, Lasers (Wiley, 1988). 332 Chapter 16. Dispersion and Wave Propagation (c) Compute the pulse spread per unit length per unit of wavelength spread for a single-mode, fusedsilica fiber where the material makes the dominant contribution to dispersion. Calculate the dispersion at the popular diode-laser wavelengths of 850 nm, 1310 nm, and 1550 nm. Give your answers in units of ps/(nm \u00b7 km). To get the refractive-index derivatives, we can use the Sellmeier formula for fused silica5 n2(\u03bb0) \u2212 1 = 0.68374049400 \u03bb 2 0 \u03bb 2 0 \u2212 0.00460352869 + 0.42032361300 \u03bb 2 0 \u03bb 2 0 \u2212 0.01339688560 + 0.58502748000 \u03bb 2 0 \u03bb 2 0 \u2212 64.49327320000, (16.69) where \u03bb0 is in \u00b5m. This gives d2n/d\u03bb 2 1310 nm, and d2n/d\u03bb 2 0 = \u22120.00472 \u00b5m\u22122 at 1550 nm. 0 = 0.0296 \u00b5m\u22122 at 850 nm, d2n/d\u03bb 2 0 = \u22120.00112 \u00b5m\u22122 at Problem 16.5 Consider a bichromatic optical wave of frequencies \u03c91 and \u03c92 propagating through a dispersive medium. (a) Show that the optical phase of the wave propagates at velocity \u03c9/k, while the envelope of the interference pattern propagates at velocity \u2206\u03c9/\u2206k, where \u03c9 := (\u03c91 + \u03c92)/2, \u2206\u03c9 := \u03c92 \u2212 \u03c91, k := (k1 + k2)/2, \u2206k := k2 \u2212 k1, \u03c91 = c0k1/n1, \u03c92 = c0k2/n2, and n1 and n2 are the refractive indices at frequencies \u03c91 and \u03c92, respectively. Assume that |\u2206\u03c9| \u03c9 and that the two frequency components have equal amplitudes. (b) What are the conditions for the envelope to propagate backwards? 5\u2018\u2018HPFS Fused Silica KrF Grade,\u2019\u2019 Corning Incorporated data sheet, 2003. Chapter 17 Classical Light\u2013Atom Interactions We will now model the interaction between light and atoms, using a classical model of the atom. This will allow us to treat a variety of phenomena from the refractive index of atomic vapors to the conductivity of metals to laser cooling and trapping of atoms. We will model the atom as a classical harmonic oscillator, an electron bound", " to the nucleus by a harmonic force (linear spring): m\u00a8x + m\u03c9 2 0 x = 0. (17.1) Here, x represents the average position of the electron, since quantum-mechanically, the electron is not localized, and \u03c90 is the resonant frequency of the harmonic potential. The above equation is also in centerof-mass coordinates, so that we can ignore the motion of the nucleus. Thus, m is the reduced mass of the electron, given by m = memn me + mn, where me is the electron mass, and mn is the nuclear mass. Generally me mn, so m \u2248 me 1 \u2212, me mn (17.2) (17.3) and generally, it is a good approximation to use m \u2248 me. Why use a classical calculation, when an atom is a manifestly quantum-mechanical object? It turns out that the classical calculation gets many phenomena correct, and these results can be justified by a quantum calculation. Essentially, the classical calculation is good for small excitations, when the harmonic potential, the lowest-order approximation to an arbitrary potential, is an accurate model. In particular, the classical model does not predict any saturation effects, and as we will see, it requires a bit of patching to make it quantitatively correct, even in the limit of small intensity. 17.1 Polarizability We will now consider the interaction of the atom with a monochromatic field of the form E(+)(t) = \u02c6\u03b5E(+) 0 e\u2212i\u03c9t, (17.4) where \u02c6\u03b5 is the unit polarization vector. In writing down this expression, we are making the dipole approximation: we are assuming that the size of the atom is much smaller than the optical wavelength, so that the electron only sees the field at the nuclear position. Thus, we need not consider the spatial dependence or propagation direction of the field. The force on the electron due to the field is F(+) = \u2212eE(+), (17.5) 334 Chapter 17. Classical Light\u2013Atom Interactions where e is the fundamental charge, the magnitude of the electron charge (so that the electron charge is \u2212e). Then the equation of motion for the electron becomes m\u00a8x + m\u03c9 2 0 x = \u2212\u02c6\u03b5eE(+) 0 e\u2212i\u03c9t. (17.6) We need only worry about the electron motion in the direction of the electric field; we will ignore any motion except that induced by the field, as we will justify when considering the damped version of the harmonic oscillator. We now see if we can assume that the solution has the same time dependence as the field: With this solution, Eq. (17.6) becomes x(+)(t) = \u02c6\u03b5x(+) 0 e\u2212i\u03c9t. \u2212m\u03c92x(+) 0 + m\u03c9 2 0 x(+) 0 = \u2212eE(+) 0 which we can solve for x(+) 0 to obtain the solution x(+) 0 = eE(+) 0 /m \u03c92 \u2212 \u03c9 2 0. e\u2212i\u03c9t, (17.7) (17.8) (17.9) Again, we are breaking the electron displacement into its positive and negative components x(t) = x(+)(t) + x(\u2212)(t). The dipole moment of the atom is d(+) = \u2212ex(+), (17.10) where x = \u02c6\u03b5x. Since the dipole moment is induced by the field (the electron displacement is zero in equilibrium), we can define the polarizability \u03b1 to describe how easily the field induces the dipole moment by d(+) = \u03b1(\u03c9)E(+). From Eqs. (17.9) and (17.10), we can write the polarizability as \u03b1(\u03c9) = e2/m 0 \u2212 \u03c92. \u03c9 2 (17.11) (17.12) The polarizability completely characterizes the response of the atom to the applied field. Of course, this is the frequency-space response function, which we have obtained via an implicit Fourier transform of the applied field. 17.1.1 Connection to Dielectric Media Recall that the polarization density P (from Section 4.3) is the dipole moment per unit volume. Thus, for an atomic vapor of number density N, P(+) = N d(+) = N \u03b1(\u03c9)E(+) = \u02c6\u03b5 N e2/m \u03c92 \u2212 \u03c9 2 0 E(+) 0 e\u2212i\u03c9t. (17.13) This expression is valid for a rarefied medium, where the interactions between the atoms are negligible. In dense media, correlations between dipoles cause deviations from these results. We can thus write the susceptibility for", " the vapor as, (17.14) in view of the defining relation P = 0\u03c7E. \u03c7(\u03c9) = N e2/m0 \u03c92 \u2212 \u03c9 2 0 17.2 Damping: Lorentz Model 335 17.1.2 Conducting Media: Plasma Model In conducting media, the electrons are not tightly bound to the nucleus. We can treat this by letting \u03c90 \u2212\u2192 0, modeling the conductor as a plasma. Then the susceptibility becomes where \u03c7(\u03c9) = \u2212 N e2/m0 \u03c9 2 = \u2212 2, \u03c9p \u03c9 \u03c9p := N e2 m0 (17.15) (17.16) is the plasma frequency. The complex refractive index (see Section 9.5) is given in terms of the susceptibility by \u02dcn(\u03c9) = 1 + \u03c7(\u03c9) = 1 \u2212 2. \u03c9p \u03c9 (17.17) The behavior is quite different in the limiting cases of low and high frequency. 1. Low frequency (\u03c9 < \u03c9p). In this case, \u02dcn is purely imaginary, so the reflection coefficient (Section 9.5) is of the form r = n1 \u2212 in2 n1 + in2, (17.18) where n1 and n2 are real, so that |r| = 1. Thus, we find perfect reflection from a metal surface for frequencies below the plasma frequency. 2. High frequency (\u03c9 \u03c9p). Here \u02dcn \u2212\u2192 1, so that the metal becomes transparent at high frequencies. For typical metals, the plasma frequency occurs in the ultraviolet, so that metals are transparent to ultraviolet and shorter wavelengths. Again, this model is valid for a rarefied plasma, so it is only qualitatively correct for metals, where density-induced correlations are more important. 17.2 Damping: Lorentz Model A better model of the atom is a damped harmonic oscillator. This improved model is known as the Lorentz model of the atom, and the equation of motion is m\u00a8x + m\u03b3 \u02d9x + m\u03c9 2 0 x = \u2212\u02c6\u03b5eE(+) 0 e\u2212i\u03c9t. (17.19) The damping (\u2018\u2018friction\u2019\u2019) term models radiation reaction due to the charge acceleration (the classical analogue of spontaneous emission) and collisions with other atoms. A quantum-mechanical calculation shows that for an isolated atom, the damping rate is the same as the Einstein A coefficient (spontaneous emission rate): \u03b3 = A21. Again, we assume a solution of the form x(+)(t) = \u02c6\u03b5x(+) 0 e\u2212i\u03c9t. Following the method above, the solution is x(+) 0 = eE(+) \u03c92 \u2212 \u03c9 2 0 /m 0 + i\u03b3\u03c9. (17.20) Now the displacement is complex, reflecting a phase lag of the displacement behind the field, with phase angle. (17.21) The phase lag approaches zero for \u03c9 \u03c90 and \u03c0 for \u03c9 \u03c90 (\u03b4 = \u03c0/2 exactly on resonance). Then for this case, the polarizability becomes \u03b4 = tan\u22121 \u03b3\u03c9 \u03c9 2 0 \u2212 \u03c92 \u03b1(\u03c9) = e2/m \u03c9 2 0 \u2212 \u03c92 \u2212 i\u03b3\u03c9. (17.22) 336 Chapter 17. Classical Light\u2013Atom Interactions The susceptibility likewise becomes \u03c7(\u03c9) = N e2/m0 \u03c9 2 0 \u2212 \u03c92 \u2212 i\u03b3\u03c9. (17.23) It is worth reiterating here that \u03b1 and \u03c7 are complex quantities defined for the positive-rotating fields via d(+) = \u03b1(\u03c9)E(+) and P(+) = 0\u03c7E(+), and therefore must be treated appropriately. If \u03c7 is small (as for a dilute vapor), \u02dcn(\u03c9) = 1 + \u03c7(\u03c9) \u2248 1 + \u03c7(\u03c9) 2 = 1 + N e2 2m0 (\u03c9 2 0 \u2212 \u03c92) 0 \u2212 \u03c92)2 + \u03b32\u03c92 + i (\u03c9 2 N e2 2m0 \u03b3\u03c9 0 \u2212 \u03c92)2 + \u03b32\u03c92. (\u03c9 2 (17.24) Then we can read off the phase index as the real part, n(\u03c9) \u2248 1 + N e2 2m0 (\u03c9 2 0 \u2212 \u03c92) 0 \u2212 \u03c92)2 + \u03b32\u03c92, (\u03c9 2 while the absorption coefficient becomes a(\u03c9) = 2k0Im[\u02dcn(\u03c9)] \u2248 N e2\u03c92 m0c0 \u03b3 0 \u2212 \u03c92)2", " + \u03b32\u03c92. (\u03c9 2 (17.25) (17.26) The region where significant absorption occurs for small detunings of the field from the atomic resonance, |\u03c9 \u2212 \u03c90| \u03c90. Then 0 \u2212 \u03c92 = (\u03c90 \u2212 \u03c9)(\u03c90 + \u03c9) \u2248 2\u03c9(\u03c90 \u2212 \u03c9). \u03c9 2 (17.27) This is effectively equivalent to the rotating-wave approximation in quantum optics. With this approximation, the phase index and absorption become n(\u03c9) \u2248 1 + N e2 2m0 (\u03c90 \u2212 \u03c9)/2\u03c9 (\u03c90 \u2212 \u03c9)2 + (\u03b3/2)2 a(\u03c9) \u2248 N e2 m0c0\u03b3 (\u03b3/2)2 (\u03c90 \u2212 \u03c9)2 + (\u03b3/2)2. (17.28) Hence, we recover the Lorentzian absorption profile from Sections 15.2.5 and 16.1.1.1 (hence the name) with full width at half maximum \u2206\u03c9 = \u03b3 and resonant absorption coefficient a0 = N e2/m0c0\u03b3. Also, we see that in the same regime, n \u2212 1 = 2(\u03c90 \u2212 \u03c9) \u03b3 N e2 2m0\u03b3\u03c9 (\u03b3/2)2 (\u03c90 \u2212 \u03c9)2 + (\u03b3/2)2 = 2(\u03c90 \u2212 \u03c9) \u03b3 Im[\u02dcn(\u03c9)], (17.29) as required by the Kramers\u2013Kronig relations [Eq. (16.26)]. In general, we can have atoms with multiple electrons that we need to sum over. Then the polarizability and susceptibility become \u03b1(\u03c9) = \u03c7(\u03c9) = j j e2 m \u03c9 2 N e2 m0 f0j 0j \u2212 \u03c92 \u2212 i\u03b3j\u03c9 f0j 0j \u2212 \u03c92 \u2212 i\u03b3j\u03c9. \u03c9 2 (17.30) Here, f0j is the absorption oscillator strength, which acts as a weighting factor for each electron. The quantum-mechanical (and correct) interpretation of these expressions is that each term in the sum represents a transition from the ground level 0 to excited level j. The oscillator strength can only be obtained from a quantum calculation, and is necessary to make the classical calculation quantitatively correct. Because of the quantitative importance of this factor, we will explore it in more detail. 17.2 Damping: Lorentz Model 337 17.2.1 Oscillator Strength Since the absorption coefficient scales as the susceptibility and thus the oscillator strength (for a dilute gas), the oscillator strength also scales with the cross section [see Eq. (15.36) for the laser cross section that we defined earlier]. On resonance, the cross section for absorption is defined by a(\u03c90) = \u03c3(\u03c90)N = \u03c30N. Thus, using Eq. (17.26), we can write the classical absorption cross section as \u03c3classical(\u03c90) = e2\u03c92 m0c0 \u03b3 0 \u2212 \u03c92)2 + \u03b32\u03c92 (\u03c9 2 \u03c9=\u03c90 = e2 m0c0\u03b3. (17.31) (17.32) This cross section is not quantitatively correct, as the correct cross section from Eq. (15.39) for the transition to level j is \u03c30j = \u03c3j(\u03c90) = \u03bb 2 0j 2\u03c0 = \u03c0c 2 0 \u03c9 2 0j. (17.33) Note that this cross section assumes an orientational average (and thus a factor of 1/3) that is generally appropriate for our purposes. We can then define the absorption oscillator strength to be the \u2018\u2018fudge factor\u2019\u2019 to fix the classical cross section: We can also write the cross section as f0j := \u03c30j \u03c30, classical = 0 \u03b3j 2\u03c00mc 3 e2\u03c9 2 0j \u03c30j = f0j e2 m0c0\u03b3j,. (17.34) (17.35) which will be useful later. More commonly, the absorption oscillator strength is defined to include the degeneracy of the level structure,1 f0j = 0 \u03b3j 2\u03c00mc 3 e2\u03c9 2 0j gj g0, (17.36) with a separate expression defined for the emission oscillator strength f0j (which just flips the degeneracy ratio). Also, in the limit of large frequency, the susceptibility of Eqs. (17.30) becomes \u03c7(\u03c9) \u2212\u2192 \u2212 N e2 m0\u03c92 f", "0j. j (17.37) In this limit, the induced electron displacements are small, and thus the damping and harmonic-potential forces are not important. We thus expect to recover the behavior of the free-electron plasma of Eq. (17.15) in the high-frequency limit (i.e., the conductor without damping): \u03c7(\u03c9) = \u2212 2 \u03c9p \u03c9 = \u2212 N e2 m0\u03c92. (17.38) Comparing these two expressions, we find the Thomas\u2013Reiche\u2013Kuhn sum rule for the oscillator strength: j f0j = 1. (17.39) Since f0j > 0, the sum rule tells us that f0j < 1. The interpretation is that the classical cross section represents the maximum possible cross section, which turns out to be distributed over all the possible transitions from the ground level. Note that transitions to unbound (ionized) states are also included in this sum, making it difficult to verify this with atomic transition data.2 1Alan Corney, Atomic and Laser Spectroscopy (Oxford, 1987). 2See Peter W. Milonni and Joseph H. Eberly, Lasers (Wiley, 1988), p. 239. 338 Chapter 17. Classical Light\u2013Atom Interactions 17.2.2 Conductor with Damping: Drude Model We can again use the Lorentz model to derive an improved model of conductors or plasmas. This model, the Drude model, accounts for damping, and accounts for electron collisions in the metal or plasma. We again take the Lorentz-model results and let \u03c90 \u2212\u2192 0 so that the electrons are not bound. In this limit, Eqs. (17.30) give The complex refractive index becomes \u03c7(\u03c9) = \u2212N e2/m0 \u03c92 + i\u03b3\u03c9 = \u2212\u03c9 2 p \u03c92 + i\u03b3\u03c9. In the dc limit (\u03c9 \u2212\u2192 0), \u02dcn(\u03c9) = 1 + \u03c7(\u03c9) = 1 \u2212 \u03c9 2 p \u03c92 + i\u03b3\u03c9. \u02dcn(\u03c9) \u2212\u2192 1 + i \u03c9 2 p \u03b3\u03c9 \u2248 i \u03c9 2 p \u03b3\u03c9. Thus, the refractive index is not purely imaginary at any nonzero frequency, even though the imaginary part is much larger than the real part. This leads to a reflectance that is slightly less than unity, which is a distinct improvement over the plasma-model result of perfect reflectance. In the high-frequency limit, the damping part is not important, and the Drude model reduces to the undamped plasma model. The imaginary (damping) part of the conductivity here becomes significant at infrared and higher frequencies, gradually rolling off the material reflectance below the plasma frequency. Another useful quantity for a conductor is the induced current density, which we can write as Since \u02d9x = \u2212i\u03c9x, we can use the solution for x(+) 0 from Eq. (17.20) to write J = \u2212N e \u02d9x. J = N e2/m \u03b3 \u2212 i\u03c9 E (17.43) (17.44) in the unbound limit. Recalling that the conductivity \u03c3 is defined by J = \u03c3E, we can write the Drude-model conductivity as where \u03c3 = N e2/m \u03b3 \u2212 i\u03c9 = \u03c30 1 \u2212 i\u03c9/\u03b3, \u03c30 := N e2 m\u03b3 is the dc conductivity. We thus have a model for the frequency dependence of the metal conductivity that we ignored in Section 9.5. It is worth reiterating that this is only a simple model and valid for rarefied plasmas. 17.3 Atom Optics: Mechanical Effects of Light on Atoms Now we will have a brief look at the field of atom optics, or optics with matter (de Broglie) waves. We will only be looking here at how to trap and cool atoms with laser light using the classical Lorentz model of the atom, so in a sense we will be doing \u2018\u2018geometrical atom optics.\u2019\u2019 Broadly speaking, there are two types of mechanical forces that light can have on atoms. The first, the dipole force, is related to the potential energy of the induced dipole in the electric field, and is thus related to the real part of \u03b1(\u03c9) [see Eq. (17.53)]. The second is radiation pressure due to absorption and rescattering of the incident light, which is thus related to the imaginary part of \u03b1(\u03c9) [see Eq. (17.89)]. (17.40) (17.41) (17.42) (17.45) (17.46) 17.3", " Atom Optics: Mechanical Effects of Light on Atoms 339 17.3.1 Dipole Force The dipole moment of the atom is induced by the external field, so we can write the potential energy of the induced dipole as Vdipole = \u2212 = \u2212. (17.47) d \u00b7 E 2 d E 2 The extra factor of 1/2 compared to the usual dipole energy is because the dipole is induced, and thus Vdipole = \u2212 E 0 (d) dE = \u2212 E 0 \u03b1E dE = \u2212 1 2 \u03b1E2. (17.48) Since we found the solution for the positive-frequency component of the field, we should write out the potential in terms of the same components: Vdipole = \u2212 1 2 d(+) + d(\u2212) E(+) + E(\u2212). \u00b7 Noting that we can see that the terms of the form d(\u00b1) \u223c e\u2213i\u03c9t, E(\u00b1) \u223c e\u2213i\u03c9t, d(\u00b1) \u00b7 E(\u00b1) \u223c e\u2213i2\u03c9t (17.49) (17.50) (17.51) rotate at twice the optical frequency, which is too fast for the atoms to respond mechanically. So we will drop these terms in the time average (the same average that leads to the intensity). The terms of the form d(\u00b1) \u00b7 E(\u2213) \u223c 1 (17.52) are dc, so we can keep these. Thus, Vdipole = \u2212 = \u2212 1 2 1 2 1 2 d(\u2212) \u00b7 E(+) 1 2 \u00b7 E(\u2212) \u2212 \u03b1(\u03c9)E(\u2212) d(+) \u00b7 E(\u2212) \u2212 \u03b1(\u03c9)E(+) E(+) Re[\u03b1(\u03c9)]I(r). 2 = \u2212Re[\u03b1(\u03c9)] = \u2212 \u03b70 2 \u00b7 E(+) (17.53) Here, \u03b70 is the vacuum wave impedance from Eq. (4.63), and we are regarding the electric-field envelope E(+)(r) to be a slowly varying function of position. Putting in the explicit form for the polarizability, we can write the dipole potential as Vdipole = \u2212e2 2m0c0 0 \u2212 \u03c92 \u03c9 2 0 \u2212 \u03c92)2 + \u03b32\u03c92 I(r). (\u03c9 2 (17.54) Thus, the atom sees a spatial potential proportional to I(r) and to (n \u2212 1). This potential-shift effect (also known as the ac Stark shift) is the atomic counterpart of the phase shift (due to n \u2212 1) of a beam propagating through a vapor. Both effects follow from the coupling of the field to the atom. The corresponding force is given by the potential gradient Fdipole = \u2212\u2207Vdipole \u221d \u2207I(r). (17.55) Thus, the dipole force responds to intensity gradients. If the dipole is viewed as two slightly separated, opposite charges, there is only a net force if the two charges see a different electric field, which is only possible in the ideal dipole limit if the field has a gradient. 340 Chapter 17. Classical Light\u2013Atom Interactions The sign of the dipole potential is set solely by the detuning of the field from the atomic resonance. Defining the detuning \u2206 := \u03c9 \u2212 \u03c90, we can write the dipole potential as Vdipole = e2 2m0c0 (\u03c90 + \u03c9)\u2206 [(\u03c90 + \u03c9)\u2206]2 + \u03b32\u03c92 I(r). (17.56) Everything in this expression is positive except for the factor of \u2206 in the numerator. Thus, for positive \u2206 (\u03c9 > \u03c90, or blue detuning), Vdipole > 0, while for negative \u2206 (\u03c9 < \u03c90, or red detuning), Vdipole < 0. That is, a bright spot in space (e.g., due to a tightly focused Gaussian beam) will repel an atom for blue detunings, forming a potential barrier, while for red detunings, the spot attracts atoms and forms a potential well. The sign dependence of Vdipole makes sense in terms of the phase lag (17.21). Recall that for small frequencies (\u2206 < 0), the phase lag of the dipole behind the field is smaller than \u03c0/2, while for large frequencies (\u2206 > 0), the phase lag is between \u03c0/2 and \u03c0. Since Vdipole \u221d \u2212d \u00b7 E, the phase lag is important because \u2277 0 then d and E are mostly aligned or mostly opposed for \u2206 < 0 and \u2206 > 0", ", respectively. Thus, Vdipole for \u2206 \u2277 0. 17.3.1.1 Dipole Potential: Standard Form By writing the dipole potential in a more standard form, we can see that it matches the result of a quantum calculation, at least in the limit of low intensity. To do this, we first need to patch the classical result of Eq. (17.54) as before by including the oscillator strength and summing over all transitions: Vdipole = \u2212 e2f0j 2m0c0 j 0j \u2212 \u03c92 \u03c9 2 0j \u2212 \u03c92)2 + \u03b3 2 (\u03c9 2 j \u03c92 I(r). (17.57) Now to put this in more standard form, we need to define the saturation intensity for the atom. We defined the saturation intensity before in Eq. (15.75) for a four-level laser, but we need to redo this for the two-level atom. Recall that a nondegenerate, two-level atom has the steady state solution when driven on resonance from Eq. (15.23) N2 N1 = B12\u03c1(\u03c90) A21 + B21\u03c1(\u03c90). Using N1 + N2 = N and B12 = B21, we can solve this to find N2 N = B21\u03c1(\u03c9) A21 + 2B21\u03c1(\u03c9) = \u03c30I/\u00afh\u03c90 A21 + 2\u03c30I/\u00afh\u03c90 = I/2Isat 1 + I/Isat, where \u03c30 is the resonant cross section and we have defined the saturation intensity as Isat := \u00afh\u03c90\u03b3 2\u03c30, (17.58) (17.59) (17.60) and we have identified \u03b3 = A21. Note that this definition differs by a factor of 2 from the four-level laser definition (15.75), since here the ground state is not naturally depleted. For the maximum possible resonant cross section of \u03c30 = 3\u03bb2/2\u03c0 (where there is no average over the dipole orientation), the saturation intensity is Isat = 1.10 mW/cm2 for 133Cs on the D2 transition (852 nm), while for 87Rb on the same transition (780 nm), the saturation intensity is Isat = 1.67 mW/cm2. We can also write the saturation intensity in terms of the oscillator strength by using Eq. (17.35), with the result Isat,j = \u00afh\u03c90jm0c0\u03b3 2 j 2e2f0j. (17.61) Even though the above numerical values are often quoted for the saturation intensity, this is actually a context-dependent quantity (as we can see from the difference of its value in the two- and four-level cases). A safe but cumbersome approach is to use the quantum-mechanical formalism for angular momentum to directly calculate the cross section and thus saturation intensity.3 3Daniel A. Steck, \u2018\u2018Cesium D Line Data,\u2019\u2019 2003. Available online at http://steck.us/alkalidata. 17.3 Atom Optics: Mechanical Effects of Light on Atoms Using Eq. (17.61), we can write the dipole potential (17.57) as Vdipole = \u2212 \u00afh\u03c90j\u03b3 2 j 4 j 0j \u2212 \u03c92 \u03c9 2 0j \u2212 \u03c92)2 + \u03b32 (\u03c9 2 j \u03c92 341 (17.62) I(r) Isat,j. This is the general expression for any frequency, so long as the intensity is small. To simplify this, we can look at the functional form far away from all resonances (|\u03c90j \u2212 \u03c9| \u03b3j for all j) so that Vdipole = \u00afh\u03c90j\u03b3 2 j 4 j 1 \u03c92 \u2212 \u03c9 2 0j I(r) Isat,j = j \u00afh\u03b3 2 j 8 1 \u03c9 \u2212 \u03c90j \u2212 1 \u03c9 + \u03c90j I(r) Isat,j. (17.63) The first term in the parentheses is the inverse of the detuning, and represents the Stark shift due to the atomic resonances. The second term can be interpreted as the weak, additional Stark shift due to resonances at the corresponding negative frequencies. This secondary shift is always negative (like a red detuning), and corresponds to the Bloch\u2013Siegert shift. Note that this expression also recovers the dc Stark shift (or equivalently, the dc polarizability up to some universal factor) when \u03c9 = 0, when both terms contribute equal, negative energy shifts. If one resonance is dominant (that", " is, the laser is tuned far away from resonance, but much closer to one than all the others), then we can make the rotating-wave approximation and neglect the second term in the parentheses of Eq. (17.63) to obtain Vdipole = \u00afh\u03b3 2 j 8\u2206 I(r) Isat, (17.64) where again \u2206 = \u03c9 \u2212 \u03c90 is the detuning from resonance. Note that for a far-detuned, linearly polarized laser creating this potential, it turns out that \u03c30 = \u03bb 2 0 /2\u03c0 is the appropriate resonant cross section, so the above saturation intensity values should be multiplied by 3 before being used in this formula. Typically, a focused, red-detuned, Gaussian laser beam is used to make a dipole trap or far-off resonance trap (FORT)4 for atoms via the dipole force.5 Below is an example image of about 105 87Rb atoms confined in a dipole trap formed by a 10 W, 1090 nm Gaussian laser beam (far below the 780 and 794 nm main resonances) focused to a 31 \u00b5m beam waist (1/e2 radius), implying a Rayleigh length (depth of focus along the beam direction) of 2.8 mm. The dipole trap clearly runs from left to right with a slight downward angle; the dimensions of the image are 270 \u00d7 29 CCD pixels (6.59 \u00d7 0.71 mm). This is an absorption image, where the shadow cast by the atoms in a brief, resonant laser probe is imaged and recorded on a CCD camera. (The image greyscale is inverted so the atoms appear bright rather than dark.) But now the important question to address is under what conditions the trap is stable, since as the atom scatters photons, it heats up until it boils out of the trap. So we will need to take a closer look at the radiation of the Lorentz atom. 17.3.2 Radiation Pressure Now we will examine the forces due to absorption and reemission of the incident light. We will begin by examining the radiation process in detail. 4Steven L. Rolston, Christoph Gerz, Kristian Helmerson, P. S. Jessen, Paul D. Lett, William D. Phillips, R. J. Spreeuw, and C. I. Westbrook, \u2018\u2018Trapping atoms with optical potentials,\u2019\u2019 Proceedings of SPIE 1726, 205 (1992) (doi: 10.1117/12.130392). 5The first observation of atoms trapped in a dipole-force potential was Steven Chu, J. E. Bjorkholm, A. Ashkin, and A. Cable, \u2018\u2018Experimental Observation of Optically Trapped Atoms,\u2019\u2019 Physical Review Letters 57, 314 (1986) (doi: 10.1103/PhysRevLett.57.314). 342 Chapter 17. Classical Light\u2013Atom Interactions 17.3.2.1 Dipole Radiation The electric and magnetic fields for an oscillating dipole are6 E(+)(r, t) = 1 4\u03c00 H(+)(r, t) = c0 4\u03c0 (\u02c6\u03b5 \u00d7 \u02c6r) [3(\u02c6\u03b5 \u00b7 \u02c6r)\u02c6r \u2212 \u02c6\u03b5] \u02d9d(+)(tr) c0r2 + + d(+)(tr) r3 \u00a8d(+)(tr) c 2 0 r \u02d9d(+)(tr) c0r2 + 1 4\u03c00, [(\u02c6\u03b5 \u00b7 \u02c6r)\u02c6r \u2212 \u02c6\u03b5] \u00a8d(+)(tr) c 2 0 r (17.65) where tr = t \u2212 r/c0 is the retarded time, and \u02c6\u03b5 is the polarization unit vector of the applied field (and thus the dipole orientation vector). Only the 1/r terms actually transport energy to infinity (i.e., they correspond to radiation), so we can drop the rest to obtain E(+)(r, t) \u2248 H(+)(r, t) \u2248 1 4\u03c00c 2 0 1 4\u03c0c0 [(\u02c6\u03b5 \u00b7 \u02c6r)\u02c6r \u2212 \u02c6\u03b5] \u00a8d(+)(tr) r (\u02c6\u03b5 \u00d7 \u02c6r) \u00a8d(+)(tr) r. (17.66) The energy transport is governed by the Poynting vector, which we can write as S = E(+) \u00d7 H(\u2212) + c.c. = = 1 16\u03c020c 3 0 \u02c6r 16\u03c020c 3 0 | \u00a8d(+)|2 r2 | \u00a8d(+)|", "2 r2 [(\u02c6\u03b5 \u00b7 \u02c6r)\u02c6r \u2212 \u02c6\u03b5] \u00d7 (\u02c6\u03b5\u2217 \u00d7 \u02c6r) + c.c. (17.67) 1 \u2212 |\u02c6r \u00b7 \u02c6\u03b5|2 + c.c., [(\u02c6\u03b5 \u00b7 \u02c6r)\u02c6r \u2212 \u02c6\u03b5] \u00d7 (\u02c6\u03b5\u2217 \u00d7 \u02c6r) = 1 \u2212 |\u02c6r \u00b7 \u02c6\u03b5|2 \u02c6r (17.68) where we have used for the angular dependence. There are two main possibilities for the polarization vector: the incident light can be linearly or circularly polarized. 1. Linear polarization (\u02c6\u03b5 = \u02c6z): 1 \u2212 |\u02c6r \u00b7 \u02c6\u03b5|2 = sin2 \u03b8. This is the usual \u2018\u2018doughnut-shaped\u2019\u2019 radiation pattern for an oscillating dipole. 2. Circular polarization (\u02c6\u03b5 = \u02c6\u03b5\u00b1 := \u2213(\u02c6x \u00b1 i\u02c6y)/ shaped\u2019\u2019 radiation pattern for a rotating dipole. \u221a 2): 1 \u2212 |\u02c6r \u00b7 \u02c6\u03b5|2 = (1 + cos2 \u03b8)/2. This is a \u2018\u2018peanut- Here, \u03b8 is the angle from the z-axis, while \u03c6 is the angle around the azimuth. Note that any arbitrary polarization can be represented as a superposition of these three basis vectors. The (intensity/power) radiation patterns for the linear and circular dipole cases are shown here. 6See John David Jackson, Classical Electrodynamics, 3rd ed. (Wiley, 1999), p. 411 or Peter W. Milonni and Joseph H. Eberly, Lasers (Wiley, 1988), p. 44. 17.3 Atom Optics: Mechanical Effects of Light on Atoms 343 The three-dimensional distributions are generated by sweeping these patterns around the z-axis. The corresponding electric fields for the dipole radiation are polarized. From Eq. (17.66), we can see that the polarization vector is proportional to (\u02c6\u03b5\u00b7 \u02c6r)\u02c6r \u2212 \u02c6\u03b5. For linear polarization (\u02c6\u03b5 = \u02c6z), this factor turns out to be sin \u03b8 \u02c6\u03b8, while for circular polarization (\u02c6\u03b5 = \u02c6\u03b5\u00b1 = \u2213(\u02c6x \u00b1 i\u02c6y)/ 2), the polarization vector is proportional to (cos \u03b8 \u02c6\u03b8 \u2213 i \u02c6\u03c6)e\u2213i\u03c6/ 2. \u221a \u221a Now let\u2019s define the angular-distribution function via f\u02c6\u03b5(\u03b8, \u03c6) := 1 \u2212 |\u02c6r \u00b7 \u02c6\u03b5|2. 3 8\u03c0 For linear and circular polarization, this takes the form sin2(\u03b8) f\u02c6z(\u03b8, \u03c6) = f\u00b1(\u03b8, \u03c6) = 3 8\u03c0 3 16\u03c0 1 + cos2(\u03b8). (17.69) (17.70) This function has the nice property that it is normalized, and thus represents a probability distribution for photon emission in quantum mechanics: f\u02c6\u03b5(\u03b8, \u03c6) d\u2126 = 1. Here, d\u2126 = sin \u03b8 d\u03b8 d\u03c6 is the usual solid-angle element. Now we can write the Poynting vector in terms of the angular-distribution function as The power radiated per unit solid angle is then S = \u02c6r 3\u03c00c 3 0 | \u00a8d(+)|2 r2 f\u02c6\u03b5(\u03b8, \u03c6). and the total radiated power is dPrad d\u2126 = r2S \u00b7 \u02c6r = | \u00a8d(+)|2 3\u03c00c 3 0 f\u02c6\u03b5(\u03b8, \u03c6), Prad = d\u2126 dPrad d\u2126 = | \u00a8d(+)|2 3\u03c00c 3 0 = e2|\u00a8x(+)|2 3\u03c00c 3 0. Of course, the incident intensity is contained implicitly in the electron acceleration \u00a8x. (17.71) (17.72) (17.73) (17.74) circular dipolelineardipolexz 344 Chapter 17. Classical Light\u2013Atom Interactions 17.3.2.2 Damping Coefficient Now we can connect the radiated power to the damping term in the Lorentz model,7 Eq. (17.6). Note that the radiated power in Eq. (17.74) is the time-", "averaged power, since we used the complex representation. In terms of the real displacement, we can make the replacement where the angle brackets denote the time average. Then the average work done by radiation reaction must balance the energy emitted into the field: |\u00a8x(+)|2 \u2212\u2192 \u00a8x2 2, (17.75) x x0 Frr \u00b7 dx = t t0 Frr \u02d9x(t) dt = \u2212 = \u2212 e2 6\u03c00c 3 0 e2 6\u03c00c 3 0 t t0 \u02d9x\u00a8x (\u00a8x)2 dt t t0 \u2212 t t0 ___ x dt \u02d9x. (17.76) Here Frr refers to the radiation-reaction force. If we pick t\u2212t0 to be an integer multiple of the optical period, the boundary term vanishes (it is also negligible for large t \u2212 t0). Then the radiation-reaction force is Going back to the complex representation and noting that the displacement is a harmonic function, Frr = e2 6\u03c00c 3 0 ___ x. If we define F(+) rr = e2 6\u03c00c 3 0 ___ x (+) \u2248 \u2212 e2\u03c9 2 0 6\u03c00c 3 0 \u02d9x(+). \u03b3 = e2\u03c9 2 0 6\u03c0m0c 3 0, then we recover the damping term in the harmonic oscillator: F(+) rr = \u2212m\u03b3 \u02d9z(+). (17.77) (17.78) (17.79) (17.80) This is the classical result for the spontaneous emission rate, which isn\u2019t quite correct. Again, we can patch this with the substitution e2/m \u2212\u2192 (e2/m)f0j, with the result \u03b3j = e2\u03c9 2 0jf0j 6\u03c0m0c 3 0. (17.81) This is consistent with Eq. (17.34) if we take \u03c30j = 3\u03bb 2 0 /2\u03c0 (i.e., no orientational average for the dipole). Again, there are some subtleties here regarding the cross sections and orientational averages that are better handled by angular-momentum algebra. 17.3.2.3 Photon Scattering Rate Now we can compute the rate of photon scattering as a way to get to the rate of momentum transfer. We can write the total radiated power from Eq. (17.74) in terms of the polarizability as Prad = \u03c94|\u03b1(\u03c9)|2 0 c 4 6\u03c0 2 0 I, (17.82) where we used d(+) = \u03b1(\u03c9)E(+) 0 e\u2212i\u03c9t. 7This argument follows Alan Corney, Atomic and Laser Spectroscopy (Oxford, 1987), p. 230. 17.3 Atom Optics: Mechanical Effects of Light on Atoms 345 As a brief aside, though, we can write down the scattering cross section from the total radiated power, given the defining relation Prad = \u03c3I: \u03c3Rayleigh = \u03c94|\u03b1(\u03c9)|2 0 c 4 6\u03c0 2 0. (17.83) The overall scaling is as \u03c94 (neglecting the small modification due to the polarizability). This is the usual explanation for why the sky is blue and sunsets are red: blue wavelengths are preferentially scattered by the atmosphere, while red wavelengths are preferentially transmitted. We can continue by writing out explicitly the polarizability in Eq. (17.82), using Eq. (17.30): Prad = e2\u03c94 6\u03c0m2 2 0 c 2 0 j f0j \u03c9 2 0j \u2212 \u03c92 \u2212 i\u03b3j\u03c9 2 I. Using Eq. (17.81) to eliminate the oscillator strengths, Prad = 6\u03c0c 2 0 j \u03c92 \u03c9 2 0j \u03b3j \u03c9 2 0j \u2212 \u03c92 \u2212 i\u03b3j\u03c9 = \u00afh 2 \u03c92 \u03c90j \u221a j \u03b33/2 j \u03c9 2 0j \u2212 \u03c92 \u2212 i\u03b3j\u03c9 2 I I Isat,j 2, where we used \u03c30j = 3\u03bb 2 0 /2\u03c0 to write the saturation intensity as Isat,j = \u00afh\u03c90j\u03b3j 2\u03c30j = \u00afh\u03c9 3 0j\u03b3j 4\u03c0c 2 0. The photon scattering rate Rsc is the radiated power divided by the photon energy \u00afh\u03c9: Rsc = Prad \u00afh\u03c9 = j \u03c93/2 2\u03c90j \u03b33/2 j \u03c9 2 0j \u2212 \u03c92 \u2212 i\u03b3j\u03c9 I Isat,j 2. (", "17.84) (17.85) (17.86) (17.87) Again, this expression simplifies greatly for certain detunings. Near one resonance, we can ignore the contribution of the others: Rsc \u2248 = \u03c93 2\u03c90 \u03c93 2\u03c90 \u03b33 0 \u2212 \u03c92 \u2212 i\u03b3\u03c9|2 \u03b33 0 \u2212 \u03c92)2 + \u03b32\u03c92 I Isat I Isat |\u03c9 2 (\u03c9 2 Using Eq. (17.30) restricted to a single resonance, we find Rsc = \u03b70 \u00afh \u03c92 \u03c9 2 0 Im[\u03b1]I(r), (17.88) (17.89) which shows the connection of the scattering rate (and hence the radiation pressure force below) to the absorptive part of the polarizability. Far away from the dominant resonance (|\u2206| \u03b3), but still close enough for the resonance to still dominate, we find that Rsc \u2248 \u03b33 8\u22062 I Isat = \u03b3 \u00afh\u2206 Vdipole, (17.90) where in the last formula we have used Eq. (17.64) for the dipole potential. This result is of prime importance in the design of an optical dipole trap. The photon scattering rate represents heating of the atoms, as we 346 Chapter 17. Classical Light\u2013Atom Interactions will discuss, because of the random nature of photon emission. But the scattering rate and dipole potential scale as I \u22062 ; Vdipole \u221d so that for a given desired potential depth, the scattering (heating) rate can be made very small by making the detuning \u2206 large, and compensate by increasing the intensity. Thus, dipole traps with small heating rates and hence long lifetimes (up to minutes for dipole traps created by CO2 laser light) can be created in this way. Note that for a linearly polarized, far-detuned dipole trap where these scalings are valid, the same saturation intensities are to be used to calculate the dipole force and scattering rate, as discussed above. (17.91) Rsc \u221d I \u2206, 17.3.2.4 Scattering Force Each photon carries a momentum of \u00afhk0. Thus, the photon scattering rate implies a rate of momentum transfer and thus a radiation pressure force of Frad = \u00afhk0Rsc \u02c6k, (17.92) where \u02c6k is the unit vector along the direction of the wave vector (propagation direction) k. Note that even though we have invoked the concept of the photon here, which will be convenient when discussing the heating rate, everything here is really classical, since the classical field momentum is related to the absorbed beam power: F = dp/dt = Pabs/c0 = \u03c3I/c0. To get a sense of scale of the momenta involved we can compute the recoil velocity vr, defined as the velocity corresponding to one photon recoil momentum \u00afhk0: vr = \u00afhk0 m. (17.93) For 133Cs at 852 nm, vr = 3.5 mm/s, and for 87Rb at 780 nm, vr = 5.9 mm/s, so the recoil velocity is orders of magnitude smaller than typical room-temperature velocities. Close to a single resonance, the scattering rate from Eq. (17.88) is Rsc = (\u03b3/2)3 \u22062 + (\u03b3/2)2 I Isat, so that the radiation pressure force becomes Frad = \u00afhk0(\u03b3/2)3 \u22062 + (\u03b3/2)2 I Isat \u02c6k. (17.94) (17.95) Again, depending on the polarization and exactly how close the detuning is (i.e., whether or not the hyperfine structure is resolved), the appropriate value of the saturation intensity might be very different, so some caution is necesary in applying these formulae. 17.3.3 Laser Cooling: Optical Molasses Now let\u2019s explore how we can use the radiation-pressure force to cool atoms. The simplest setup we can consider is an atom moving with velocity v, exposed to identical but counterpropagating laser fields along the velocity direction. The radiation-pressure force on the atom due to the two fields from Eq. (17.95) is Frad = \u00afhk0(\u03b3/2)3 1 1 + (\u03b3/2)2 \u2212 \u2206 2 1 2 + (\u03b3/2)2 \u2206 2 I Isat, (17.96) laser 1laser 2v 17.3 Atom Optics: Mechanical Effects of Light on Atoms 347 where \u22061,2 are the effective det", "unings of the two lasers. The detunings of the two lasers are the same in the laboratory frame, but the idea behind Doppler cooling is to tune the lasers below the atomic resonance, so that the beam that opposes the atomic velocity is Doppler-shifted into resonance, thus tending to stop the atom. With the pictured setup, the frequency of laser 1 is Doppler shifted (red shifted) by \u2212k0v, while the frequency of laser 2 is Doppler shifted (blue shifted) by +k0v. Since the detunings are given by \u22061,2 = \u03c91,2 \u2212 \u03c90, we can write \u22061 = \u2206 \u2212 k0v \u22062 = \u2206 + k0v, where \u2206 = \u03c9 \u2212 \u03c90 is the detuning in the laboratory frame. Then the force is Frad = \u00afhk0(\u03b3/2)3 1 (\u2206 \u2212 k0v)2 + (\u03b3/2)2 \u2212 1 (\u2206 + k0v)2 + (\u03b3/2)2 I Isat. (17.97) (17.98) Regarded as a function of velocity, this expression is the difference of two Lorentzians, each displaced by |\u2206|/k0 from zero velocity. This force is plotted for \u2206 = \u2212\u03b3/2, and the two offset Lorentzians are shown as dashed lines. For small velocity [v max(|\u2206|, \u03b3)/k0], we can expand to lowest order in v to obtain the viscous damping (\u2018\u2018friction\u2019\u2019) force: Frad = \u00afhk 2 0 \u03b33 2 \u2206 [\u22062 + (\u03b3/2)2]2 I Isat v. (17.99) Because this force is damping for \u2206 < 0, typically leading to heavily overdamped motion for trapped atoms, this light configuration is called optical molasses.8 The maximum damping rate occurs for \u2206 = \u2212\u03b3/2 3, although it turns out that the optimal detuning is actually something else for reasons we will soon discuss. The velocity capture range is the range in velocity for which the force is appreciable. Thus, the capture \u221a range is on the order of \u00b1 \u223c \u00b1 |\u2206| k \u03b3 2k assuming \u2206 \u223c \u2212\u03b3/2. For both 133Cs (\u03b3 = 32.8 \u00d7 106 s\u22121, \u03bb0 = 852 nm) and 87Rb (\u03b3 = 38.1 \u00d7 106 s\u22121, \u03bb0 = 780 nm), the capture range is about \u00b12 m/s. Thus, only fairly slowly moving atoms can be cooled at all with this method. Traditionally to load atomic traps, atoms were slowed by other methods from hot atomic beams to below the capture velocity and then trapped. However, it is possible to load a trap from room-temperature vapor with this method by capturing only the small fraction of the atoms with small enough velocity. (17.100) \u03b3\u03bb 4\u03c0 = \u00b1, 8Steven Chu, J. E. Bjorkholm, A. Ashkin, and A. Cable, \u2018\u2018Experimental Observation of Optically Trapped Atoms,\u2019\u2019 Physical Review Letters 57, 314 (1986). F\u00b1|D|/k\u00baradv 348 Chapter 17. Classical Light\u2013Atom Interactions 17.3.3.1 Doppler Cooling Limit For laser cooling in three dimensions, it is sufficient to simply combine three of the above one-dimensional setups, one along each axis.9 Then we can write the force vector for small velocities as, Frad = \u00afhk 2 0 \u03b33 2 \u2206 [\u22062 + (\u03b3/2)2]2 I Isat v. (17.101) where I is still the intensity of a single beam. Our treatment so far makes it appear as though the atomic velocity may be damped completely away. However, we have only considered the average cooling force. There are also fluctuations of the cooling force that lead to a temperature limit. We will now derive this temperature limit, the Doppler limit, for the cooling mechanism presented here. Let\u2019s look at the variance of the velocity distribution: v2 = 2 d dt v \u00b7 dv dt = 2 ma v \u00b7 dp dt = 2 ma v \u00b7 Frad. (17.102) Here, ma is the atomic mass, and the angle brackets denote an ensemble average. With the small-velocity expression (17.99) for the average cooling force, this equation of motion becomes d dt v2 = 0 \u03b33 \u00afhk 2 ma \u2206 [\u22062 + (\u03b3/2)2]2 I Isat v2. (17", ".103) Again, according to this differential equation, the velocity damps to zero for \u2206 < 0. Now we will include the force fluctuations heuristically, since the fluctuations are quantum\u2013mechanical in origin (although there is a more general connection between damping and fluctuations known as the fluctuation\u2013dissipation theorem). In the course of scattering a photon from one of the laser beams, there is a photon absorption and a photon emission. Each absorption leads to a momentum \u2018\u2018kick\u2019\u2019 of magnitude \u00afhk0, and the direction is random but along one of the six beams. The emission is also in a random direction (not in a dipole-radiation pattern if we assume all polarizations to be equally present), leading to a second kick of magnitude \u00afhk0 in a random direction. Thus, a scattering event is effectively equivalent to two steps in a random walk in velocity space, where the step size is \u00afhk0/ma. These scattering events happen at the scattering rate Rsc = (\u03b3/2)3 \u22062 + (\u03b3/2)2 6I Isat, (17.104) since there are six beams present. Recall from Section 11.5.1 that for a random walk, each step increases v2 by a fixed amount, given by the variance after one step starting from the origin. The three-dimensional probability distribution for a single scattering event is confined to a shell of radius \u00afhk0/ma in velocity space 9Graphics by Windell Oskay. 17.3 Atom Optics: Mechanical Effects of Light on Atoms 349 for either the absorption or emission event. The probability distribution is also inversion symmetric in either case. Thus if the atom is initially at rest, then after one step in the random walk, we can write v2 initial = 0 \u2212\u2192 v2 final = 2, \u00afhk0 ma so that v2 increases at the rate 2Rsc 2. \u00afhk0 ma Including the heating rate in Eq. (17.103), we find d dt v2 = 0 \u03b33 \u00afhk 2 ma \u2206 [\u22062 + (\u03b3/2)2]2 I Isat v2 + 3\u03b33 2 1 \u22062 + (\u03b3/2)2 I Isat 2. \u00afhk0 ma In steady state, we can set the right-hand side to zero, with the result v2 = 3\u00afh\u03b3 4ma 1 + (2\u2206/\u03b3)2 (\u22122\u2206/\u03b3). (17.105) (17.106) (17.107) (17.108) This is an expression for the equilibrium kinetic energy, which we can convert to a temperature via where kB is the Boltzmann constant. This gives 1 2 mav2 = 3 2 kBT, kBT = \u00afh\u03b3 4 1 + (2\u2206/\u03b3)2 (\u22122\u2206/\u03b3). The temperature is minimized for the detuning \u2206 = \u2212\u03b3/2, giving the Doppler temperature TD: kBTD = \u00afh\u03b3 2. (17.109) (17.110) (17.111) This temperature is the best expected for Doppler cooling. For 133Cs at 852 nm, TD = 125 \u00b5K, and for 87Rb at 780 nm, TD = 146 \u00b5K. These temperatures are extremely low. We can compare these temperatures to the recoil temperature Tr, which is the temperature corresponding to atoms with an average momentum of one photon recoil \u00afhk0 (i.e., a one-dimensional rms momentum of one photon recoil): kBTr = (\u00afhk0)2 ma. (17.112) For 133Cs, Tr = 198 nK, and for 87Rb, Tr = 362 nK, so the Doppler limit is TD = 631 Tr for 133Cs and TD = 403 Tr for 87Rb. Since the (one-dimensional) rms velocity is vrms = TD Tr, \u00afhk0 ma (17.113) which is 8.8 cm/s for 133Cs and 12 cm/s for 87Rb. These velocites are about three orders of magnitude slower than room-temperature rms velocities. It turns out that for alkali vapors, typical laser-cooled samples exhibit temperatures well below the Doppler limit. Such \u2018\u2018sub-Doppler\u2019\u2019 cooling is due to the degenerate level structure of alkali atoms.10 For example, 133Cs can be laser cooled with the same general setup described above to about 2.5 \u00b5K.11 10Sub-Doppler temperature were observed in some of the first laser cooling experiments. The first reported observation is P. D. Lett, W. D. Phillips, S. L. Rolston, C", ". E. Tanner, R. N. Watts, and C. I. Westbrook, \u2018\u2018Optical molasses,\u2019\u2019 Journal of the Optical Society of America B 6, 2084 (1989). A classic treatment of sub-Doppler cooling mechanisms is J. Dalibard and C. Cohen-Tannoudji, \u2018\u2018Laser cooling below the Doppler limit by polarization gradients: simple theoretical models,\u2019\u2019 Journal of the Optical Society of America B 6, 2023 (1989). 11C. Salomon, J. Dalibard, W. D. Phillips, A. Clairon, and S. Guellati, \u2018\u2018Laser Cooling of Cesium Atoms below 3 \u00b5K,\u2019\u2019 Europhysics Letters 12, 683 (1990). 350 Chapter 17. Classical Light\u2013Atom Interactions 17.3.3.2 Magneto-Optical Trap Optical molasses tends to stop atoms, making them \u2018\u2018stuck,\u2019\u2019 but it does not confine atoms to a particular place. A slight modification to the three-dimensional optical molasses is to impose the magnetic field due to two opposed current loops in the \u2018\u2018anti-Helmholtz\u2019\u2019 configuration. This arrangment is called the magnetooptical trap (MOT).12 The magnetic field vanishes at the center point of the trap, thus defining a point for atoms to accumulate. We will not go into the operation of the trap in detail, but essentially the idea is very similar to laser cooling. The additional complication is that the laser beams must all be correctly (circularly) polarized to address magnetic substates in the degenerate excited level. The magnetic field gives a position-dependent \u2018\u2018Zeeman\u2019\u2019 shift of the transition frequency, such that if the atom is away from the center of the trap, the appropriate beam comes into resonance and pushes it towards the trap center.13 12Graphics by Windell Oskay. 13Jean Dalibard proposed the idea for the magneto-optical trap, and the MOT was first demonstrated by E. L. Raab, M. Prentiss, Alex Cable, Steven Chu, and D. E. Pritchard, \u2018\u2018Trapping of Neutral Sodium Atoms with Radiation Pressure,\u2019\u2019 Physical Review Letters 59, 2631 (1987). 17.4 Exercises 17.4 Exercises Problem 17.1 351 What are the two types of optical forces on atoms? How are they related to the atomic polarizability? Index ABCD law, 99\u2013104 cascading rule, 101 deeper meaning, 102 ABCD matrix, 26, 100 factorization of general, 101\u2013102 Q factor, 120 Q switch, 152 Q-switching, 307\u2013308 f -number, 104 Abb\u00e9 v-constant, 49 aberration theory, 26 absorption, 291, 296 absorption image, 341 absorption oscillator strength, 336\u2013337 acousto-optic diffraction, 255\u2013271 Bragg regime, 261\u2013270 Klein\u2013Cook parameter, 267 Raman\u2013Nath regime, 256\u2013261, 268, 270\u2013271 rise time, 270\u2013271 TeO2 modulator, 266\u2013268 acousto-optic modulator as spectrum analyzer, 270\u2013271 action functional, 48 adiabatic approximation, 312 air-glass interface cool movie, 159 Airy disk, 248 amplified spontaneous emission, 286 anamorphic prism pair, 45 antireflection (AR) coating, 199 single layer, 180\u2013181, 200 two layer, 181\u2013183, 200 approximations silly, 251 associated Laguerre polynomials, 107 atom optics, 338\u2013351 dipole force, 338\u2013341 Doppler temperature, 347\u2013349 magneto-optical trap (MOT), 349\u2013350 optical molasses, 346\u2013349 radiation pressure, 341\u2013349 attenuation index, 164 autocorrelation function, 274 Babinet\u2019s principle, 249\u2013250 ball lens, 50 beam radius, 94 beam splitter, 84, 86 beam waist parameter, 94 Bessel function, 248 Bessel functions, 257 generating function for, 257 big mess, 95 birefringence, 146\u2013149 BK7 borosilicate crown glass, 49 bleaching, 308 Boltzmann statistics, 293 boundary conditions, 153\u2013154 Bragg reflector, 183 Brewster\u2019s angle, 157 bug in laser, 313 calculus of variations, 22 cat convolution of, 203\u2013204 cat\u2019s eye, 50 Cauchy principal value, 240, 253\u2013254 Cauchy probability distribution, 215 cavity dumper, 131, 308\u2013309 CeF3, 182, 183 central dark-ground imaging, 239, 251 central limit theorem, 207\u2013209 characteristic function, 208 characteristic", " polynomial, 19 coherence, 273\u2013283 length, 280\u2013282 time, 280\u2013282 coherence time, 283 complex dielectric constant, 164 complex notation, 72\u201375 complex refractive index, 164 conductivity, 163 Drude model, 338 conductors Drude model, 337\u2013338 Fresnel relations for, 167\u2013168 plasma model, 334\u2013335 353 354 Index wave propagation in, 163\u2013168 confocal parameter, 98 confocal resonator, 35 convolution, 201\u2013216 definition, 202 of box functions, 203\u2013204 of two Gaussians, 205\u2013206 with delta function, 202 convolution theorem, 204\u2013205, 248 corner-cube reflector, 45 cross section absorption, 337 laser, 294 natural, on resonance, 294, 295 Dawson\u2019s integral, 251\u2013252 degree of coherence first-order temporal, 276 delta function, 63\u201365 derivative of, 253 determinant, 19 dielectric constant, 72 complex, 164 dielectric medium simple, 71\u201372 diffraction acousto-optic, 255\u2013271 Babinet\u2019s principle, 249\u2013250 Fraunhofer, 223\u2013231, 248\u2013250 Fresnel, 229\u2013233, 250 diffraction grating, 227\u2013229, 248\u2013249 diffraction limit, 223 diffusion coefficient, 209 diffusion process, 209 diode laser, 169 diopter, 48 dipole force, 338\u2013341 directional filter, 234 dispersion, 49, 319\u2013332 anomalous, 325 fast light, 326\u2013327, 329\u2013330 material, for optical fiber, 331\u2013332 normal, 325 slow light, 326\u2013329 Doppler broadening, 294, 298, 312 Doppler temperature, 347\u2013349 Drude model, 337\u2013338 eigenvalues, 19 eigenvectors, 19 Einstein A and B coefficients, 292\u2013293 Einstein rate equations, 292\u2013295 EIT, 328 electric displacement, 71, 320 electric flux density, 71 electric permittivity, 72 electro-optic effect, 146 electromagnetic field energy density, 80 electromagnetically induced transparency, 328 energy density, 80 energy spectral density, 274 error analysis, 206\u2013207 Euler-Lagrange equation, 48 evanescent field, 159 extinction coefficient, 164 extraordinary wave, 148 F2 flint glass, 49 Fabry\u2013Perot cavity, 115\u2013136, 173, 212\u2013213 Q factor, 120 confocal, 130, 132\u2013133 convolution and, 212\u2013213 finesse, 117 free spectral range, 116 Gaussian modes, 126\u2013129 glass plate etalon, 175\u2013176 Hermite\u2013Gaussian modes, 129\u2013130 line width, 118 photon lifetime, 119 physical mode condition, 127 resonance condition, 115 resonators in ray optics, 31\u201336 roughness limit to finesse, 133\u2013134 survival probability, 119 Fabry\u2013Perot etalon, 115 Fabry\u2013Perot interferometer, 115 far-off resonance trap (FORT), 341 Faraday effect, 149, 150 Verdet constant, 150 Fermat\u2019s Principle, 21\u201325 Fermat\u2019s principle, 47 fiend, diabolical, 270 film reflection, 173\u2013180 matrix formalism, 176\u2013180 reflection-summation model, 173\u2013176 finesse, 117, 176 fish underwater, 45 fluctuation\u2013dissipation theorem, 348 Fourier optics, 219\u2013247 central point, 220 decomposition into plane waves, 220\u2013221 diffraction limit, 223 holography, 244\u2013247 nonparaxial regime, 223 Index 355 paraxial regime, 222\u2013223 recipe, 221\u2013222 spatial filters, 233\u2013244 Fourier series, 57\u201360 accuracy of, 67 of rectified sine wave, 59\u201360 Fourier transform, 60\u201363, 73 computed by lens, 62 multiple dimensions, 219 of Gaussian, 61\u201362 spatial convention, 204 thin lens as analog computer, 226\u2013227 Fraunhofer diffraction, 223\u2013231, 248\u2013250 Airy disk, 248 circular aperture, 248 diffraction grating, 227\u2013229, 248\u2013249 double slit, 227, 248 rectangular aperture, 248 thin lens, 226\u2013227 validity conditions, 224\u2013226 Fraunhofer lines, 49 free spectral range, 116 Fresnel diffraction, 229\u2013233, 250 knife edge, 250 single slit, 231\u2013233 Fresnel relations, 153\u2013169 Fresnel rhomb, 161 frustrated total internal reflection, 160 FTIR spectroscopy, 276\u2013277 function, 17 functions orthogonal, 65 gain coefficient, 295\u2013298 gain saturation, 289\u2013290, 298 Gaussian beam, 235\u2013237, 248, 250\u2013251 knife-edge measurement (10-90 rule), 108 Gaussian beams, 94\u2013104 ABCD law, 99\u2013104 focusing by thin lens, 102\u2013103 minimum spot size for focusing lens, 103\u2013104 parameter specification of, 98 vector, 98\u201399", " geometrical optics, 21\u201336 Gibbs phenomenon, 66 Gouy phase, 96\u201397, 100 grating, diffraction, 227\u2013229, 248\u2013249 gravimeter, interferometric, 87 Green function, 209\u2013212, 215\u2013216 group index, 324\u2013325 group velocity, 323\u2013325 group velocity dispersion, 326 half-wave plate, 142, 143 Hamilton\u2019s principle, 48 Hamiltonian mechanics, 47 harmonic oscillator, 333\u2013338 damped, 209\u2013212, 215, 335\u2013338 Heaviside step function, 67, 239\u2013240 Fourier transform, 239\u2013240 Helmholtz equation, 75 Hermite polynomials, 105, 113 generating function, 113 Hermite\u2013Gauss function, 216\u2013217 Hermite\u2013Gaussian beam, 104\u2013107 high reflection (HR) coating, 183 high-pass filter, 234 Hilbert transform, 241, 251\u2013252, 254, 320 holography, 244\u2013247, 254 off-axis, 246\u2013247 homogenous medium, 22 impedance, 77\u201378 of vacuum, 78 impulse-response function, 209\u2013212 index of refraction, 72 induction heater, 165 intensity, 70 interface, air-conductor cool movie, 167 interface, air-glass cool movie, 156 interference between partially coherent sources, 282 coherence, 273\u2013283 constructive, 84 destructive, 84 multiple waves, 89\u201390 two beams, 83\u201384 two tilted plane waves, 88\u201389 visibility, 279\u2013280 interferometer Mach-Zehnder, 84\u201386, 91 Michelson, 86\u201388, 91, 276\u2013277 internal reflection, 158\u2013161 evanescent field, 159 skin depth, 159 jinc, 248 Jones matrix, 141\u2013146 cascaded system, 144 linear polarizer, 142 normal modes, 146 polarization rotator, 144 realistic polarizer, 151 rotation of, 144\u2013146 356 wave retarder, 142\u2013144 Jones vector, 140\u2013141 normalization, 141 orthogonality, 141 Kerr lens, 310 Klein\u2013Cook parameter, 267 Kramers\u2013Kronig relations, 319\u2013323, 331 Lagrangian, 48 Laguerre\u2013Gaussian beam, 106\u2013107 laser Q switch, 152 diode, 169 He\u2013Ne, 306 Nd:YAG, 306, 313\u2013315 Ti:sapphire, 310 laser spiking, 306\u2013307 lasers, 285\u2013318 Q-switching, 307\u2013308 cavity dumper, 308\u2013309 continuous-wave, 304\u2013306 four-level, 300\u2013301 gain, 288, 303\u2013304 gain coefficient, 295\u2013298, 301\u2013304 gain media, 285\u2013288 gain saturation, 289\u2013290, 298, 302\u2013303 gain, small-signal, 289 key components of, 285 laser spiking, 306\u2013307 light\u2013atom interactions, 290\u2013295 mode locking, 309\u2013310 multi-mode operation, 298 optimum output, 305 optimum output coupler, 290 properties of, 285 pulsed, 306\u2013310 pump saturation, 303 pumping schemes, 298\u2013301 pumps, 285 resonator, 288 single-mode operation, 297\u2013298 steady-state behavior, 290 three-level, 299\u2013300, 312 threshold behavior, 288\u2013290, 297 Law of Malus, 151 Legendre transformation, 48 Lensmaker\u2019s formula, 31 light\u2013atom interactions, 290\u2013295 LIGO, 87 line shape, 293 natural, 294 linear algebra, 17\u201319 Index linear transformation, 18, 20, 26 Liouville\u2019s theorem, 48 Lord Rayleigh, 111 Lorentz model, 335\u2013338 damping coefficient, 343\u2013344 Lorentzian absorption, 322\u2013323, 327, 336 line shape, 294 loss probability, 305 low-pass filter, 234 Mach-Zehnder interferometer, 84\u201386, 91 magnetic flux density, 71 magnetization density, 71 magneto-optical trap (MOT), 349\u2013350 Malus, Law of, 151 matrix characteristic polynomial, 19\u201320 definition, 17 determinant, 19\u201320 eigenvalues, 19\u201320 eigenvectors, 19 identity, 18 inverse, 18 linear transformation as, 20 product, 18 sum, 18 trace, 20 transpose, 18 Maxwell equations, 79 in dielectric media, 70\u201374 in vacuum, 69\u201370 melting coin cool movie, 166 MgF2, 181 Michelson interferometer, 86\u201388, 91, 276\u2013277 Michelson-Morley experiment, 87 mirror elliptical, 24 parabolic, 25, 46 plane, 22 spherical, 24 mode, 105 mode locking, 309\u2013310 monochromatic waves, 72\u201374 multiple refraction, 147 natural line width, 295 optical activity, 149\u2013150 optical axis, 25 optical isolator, 151 optical molasses, 346\u2013349 Index 357 optical path length, 21 optical power (of lens), 48 optical spectrum analyzer, 115, 124\u2013", "126 confocal, 130 ordinary wave, 148 orthogonal functions, 65 oscillator strength, 336\u2013337 paraxial approximation, 25, 26 paraxial wave equation, 93\u201394 Parseval\u2019s Theorem, 283 Parseval\u2019s theorem, 281 permeability, 69 permittivity, 69, 320 permutation symbol, 19 phase velocity, 323 photon lifetime, 311 photon scattering rate, 344\u2013346 photonic crystal, 183 pinhole camera, 111 planar resonator, 35 Planck blackbody distribution, 293 plane wave, 79 plane waves, 75\u201378 plasma frequency, 335 plasma model, 334\u2013335 Pockels cell, 152 Poisson equation Green function, 216 Poisson sum rule, 67, 309 polarizability, 334 polarization, 77, 79, 137\u2013152 birefringence, 146\u2013149 circular, 139\u2013140 cool movie, 144 electro-optic effect, 146 ellipse, 137\u2013138 elliptical, 140 Faraday effect, 149, 150 Jones matrix, 141\u2013146 Jones vector, 140\u2013141 linear, 138\u2013139 optical activity, 149\u2013150 S and P, 155 polarization (of dielectric), 72 polarization density, 71, 319, 334 polarizer (linear), 142 polarizing angle, 157 power spectral density, 275 one- vs. two-sided, 277\u2013279 power-equivalent width, 280, 283 Poynting vector, 70, 80 pulse spread in optical fiber, 331\u2013332 quantum efficiency, 306 quantum harmonic oscillator, 105 quantum Zeno effect, 151 quarter-wave plate, 142 quarter-wave stack, 183 radiation pressure, 341\u2013349 radiation reaction, 343\u2013344 radiation, dipole, 341\u2013344 angular distribution, 343 random walk, 208\u2013209, 348 rate equations Einstein\u2019s, 292\u2013295 ray matrix, 26 ray optics, 21\u201336 ray tracing, 36\u201344 ray-transfer matrix, 26 Rayleigh length, 94 recoil temperature, 349 recoil velocity, 346 reflectance, 157 refractive index complex, 164, 321\u2013323 fused silica, 332 refractive interface, 23\u201324 resonator confocal, 36 confocal\u2013planar, 35 planar, 35 spherical, 35 resonators, 31\u201336 equivalent waveguide, 31 periodic rays, 33\u201334 stability condition, 32\u201336 Rodrigues formula, 113 rotation matrix, 145 sample mean, 209 saturable absorber, 308, 310 saturation intensity, 289 for four-level laser, 302 for two-level atom, 340 scattering rate, photon, 344\u2013346 schlieren imaging, 239\u2013242, 251 Schr\u00f6dinger equation, 94 Sellmeier formula for fused silica, 332 set, 17 cartesian product, 17 signum function, 240 skin depth, 159, 165, 170 Index wave plate, 142 multiple-order, 147 zero-order, 147 Wiener\u2013Khinchin theorem, 274\u2013279 optical, 275\u2013279 Wigner transform, 102 Young double slit, 91, 227 Zernike phase-contrast imaging, 238\u2013239 negative, 239 positive, 238 ZrO2, 182, 183 358 skin effect, 165 small-signal gain coefficient, 289 Snell\u2019s Law, 23\u201324, 29 spatial filter, 233\u2013244 4-f imaging system, 233, 236, 239, 242 4-f setup, 252\u2013254 central dark-ground imaging, 239, 251 cleaning Gaussian beam, 235\u2013237, 250\u2013251 directional, 234 high-pass, 234 low-pass, 234 phase-object visualization, 237\u2013244 schlieren imaging, 239\u2013242, 251 Zernike phase-contrast imaging, 238\u2013239 speed of light, 21, 70 spherical wave, 108 spontaneous emission, 291, 296 standard deviation, 206, 207, 209 of mean, 209 step function, 239\u2013240 Fourier transform, 239\u2013240 stimulated emission, 291, 295\u2013296 Stokes relations, 85\u201386 applied to beam splitter, 86 survival probability, 305 susceptibility, 319, 334 telescope, 49 TEM (transverse electromagnetic) wave, 99 TEMl,m mode, 105 thin-film reflection, 173\u2013180 matrix formalism, 176\u2013180 reflection-summation model, 173\u2013176 thin-lens law, 47 Thomas\u2013Reiche\u2013Kuhn sum rule, 337 total internal reflection, 38, 158\u2013161, 248 frustrated, 160 transfer matrix, 26 transmittance, 157 uncertainty principle, 96, 98 uncertainty relation, 280, 282, 283 variational principle, 22 vector, 18 vector space, 17\u201318 Verdet constant, 150 visibility, 279\u2013280 Voigt profile, 294 wave equation, 70, 79, 163\u2013164, 170 paraxial, 93\u201394 wave fronts, 75 wave meter, 87in Figure 1.12b is obtained from the distortion of either the face-centered orthorhombic lattice or the body-centered", " orthorhombic lattice. For the two monoclinic systems, the lattice constants are a \u2260 b \u2260 c and the angles are \u03b1 = \u03b3 = 90\u00b0, \u03b2 \u2260 90\u00b0: The triclinic system, shown in Figure 1.12c, is obtained by pulling the top of a monoclinic solid sideways relative to the bottom. Thus, all the faces become diamonds, and the only remaining symmetry is inversion symmetry. For the triclinic system, the lattice constants are a \u2260 b \u2260 c and the angles are \u03b1, \u03b2, \u03b3 \u2260 90\u00b0. 1.3 LATTICE PLANES AND MILLER INDICES A lattice plane is defined to be a plane that has at least three noncollinear Bravais lattice points. In fact, such a plane would contain an infinite number of two-dimensional Bravais lattice points. A three-dimensional Bravais lattice is represented as a family of parallel equally spaced lattice planes. Such lattice planes can be constructed in a lattice in a variety of ways. Figures 1.13a and b represent two ways of representing the same simple cubic Bravais lattice as a family of lattice planes. The Miller indices are used to label a crystal plane. They are obtained by using the following procedure. All the Bravais lattice points lie on a chosen assembly of equally spaced parallel lattice planes (whether cubic or otherwise). Certain planes of the assembly (sometimes all of them) will always intersect the coordinate axes at the lattice points. Therefore, every plane in this chosen set of parallel planes would intercept the coordinate axes that bear a definite rational ratio to one another. To define the Miller indices, one adopts the following procedure. One 12 CHAPTER 1 Basic Properties of Crystals (a) (b) FIGURE 1.13 Lattice planes in a simple cubic lattice. The shaded regions in (a) and (b) are two of the infinite ways in which a family of lattice planes can be represented in the same cubic lattice. of the lattice points is chosen as the origin, and the coordinate axes are chosen such that they are not coplanar. In some cases, such as the cubic lattices, the coordinate axes are chosen as orthogonal and parallel to the sides of the cubic unit cell. This property of the Bravais lattice was used to define the Miller indices by crystallographers and was later used for the proper definition of reciprocal lattice vectors. The Miller indices are obtained by using the following prescription: 1. Determine the intercepts on the axes a1, a2, a3 in the units of lattice constants. Here, a1, a2, a3 need not be primitive vectors. For example, determine a1 c, where a, b, c are the lattice constants of the three-dimensional lattice (note that a = b = c for a cubic lattice). These intercepts have rational ratios although they are not, in general, integers. b, a3 a, a2 2. Take the reciprocal of each number, i.e., a, c a3 a1 3. Then reduce these numbers to the three smallest integers h, k, l that have the same ratio. The, b a2 : integers h, k, l are called the Miller indices. \u00de \u00f0 hkl denote a single crystal plane or a set of parallel planes. 4. The parentheses 5. If a plane cuts an axis on the negative side of the origin, the negative index is indicated by. If a plane is parallel to a particular axis, because placing a minus sign above the index the intercept with the axis occurs at infinity, the corresponding Miller index is zero (reciprocal of infinity). Thus, the cube faces of a cubic crystal are (100), (010), (001), (100), (010), and (001). hkl \u00de \u00f0 6. Planes equivalent by symmetry are denoted by curly brackets f of a cubic crystal described previously can also be described as hkl. Thus, the set of cube faces 7. A direction in a lattice is denoted by square brackets hkl $ \u00bd is always perpendicular to a plane having the same directions. However, this is not true for other crystals. Figure 1.14 shows the lattice planes and their Miller indices in a simple cubic lattice.. In cubic crystals, the direction hkl $ hkl \u00de \u00f0 \u00bd g 100 : g f 1.4 Bravais Lattices and Crystal Structures 13 a1 a3 a2 a3 a1 a2 a3 a1 a2 (010) FIGURE 1.14 (110) (111) Lattice planes and their Miller indices in a simple cubic lattice. 1.4 BRAVAIS LATTICES AND CRYSTAL STRUCTURES 1.4.1 Crystal Structure A crystal structure is", " obtained when an identical basis of atoms is attached to each lattice point. We will give a simple example of two atoms A and B (of different types) to illustrate a crystal structure. Figure 1.15 shows an example of a two-dimensional crystal structure with two types of atoms A and B. Figure 1.15a shows a two-dimensional rectangular lattice. Figure 1.15b shows a basis of two different types of atoms, A and B. It has to be emphasized that the basis is the definition of the same physical unit of atoms or ions (ranging from 1 for some elements to nearly 100 for some complex proteins) that is located symmetrically at each point of the Bravais lattice. Thus, a basis can be translated through all the vectors of a Bravais lattice, and another identical basis with the same location around another lattice point would be reached. The crystal structure is defined (Figure 1.15c) as a lattice with a basis. In fact, this symmetry of the crystals makes it possible to study their physical properties. We note that the lattice points in the rectangular lattice can be symmetrically shifted in the crystal structure as long as the basis of atoms is grouped symmetrically around each lattice point. 1.4.2 Lattice with a Basis We also note that there is a dichotomy in the usage of the words \u201clattice with a basis.\u201d In addition to describing a crystal structure, in which case the basis is the description of identical atoms or ions based symmetrically around each lattice point, this expression is also used to describe a Bravais lattice as a lattice with a basis. As examples of the most commonly used lattices, the bcc Bravais lattice can be described as a simple cubic lattice (the unit cell) with a two-point basis 0, a 2 \u00f0 ^x + ^y + ^z, \u00de and the fcc lattice can be described as a four-point basis 0, a 2 \u00f0 ^x + ^y a 2 \u00f0 ^y + ^z, \u00de, \u00de a 2 \u00f0 ^z + ^x : \u00de (1.6) (1.7) 14 CHAPTER 1 Basic Properties of Crystals (a) A B (b) (c) FIGURE 1.15 (a) A two-dimensional rectangular lattice; (b) basis of two atoms, A and B; (c) Crystal structure (two dimensions): lattice + basis. 1.4.3 Packing Fraction The packing fraction of a crystal structure is defined as the fraction of space occupied by the atoms or ions considered as contacting hard spheres. For example, a bcc crystal has a packing fraction of 0.68, and a fcc crystal has a packing fraction of 0.74. 1.5 CRYSTAL DEFECTS AND SURFACE EFFECTS 1.5.1 Crystal Defects It has to be noted at this point that no crystal, either grown naturally or in the laboratory, is perfect. There are a certain number of impurities, varying in range depending on the laboratory conditions during crystal growth, that are invariably present. The average number of impurities in reasonably pure crystal is 1 in 10,000. There are also lattice defects in the crystal in the sense that an atom is found not at the designated lattice point, but either it is missing or found at an interstitial. This is called a lattice defect, and such defects are also in the range of 1 in 10,000. The crystal impurities and lattice defects, together known as crystal defects, do play a significant role in certain physical properties of the crystal. 1.5.2 Surface Effects There is another significant difference between a lattice and a crystal structure even if there are only single atoms or ions (called monatomic lattice) at each lattice point in the crystal structure. As defined earlier, a lattice is a mathematical concept that extends to infinity in all dimensions, whereas every 1.6 Some Simple Crystal Structures 15 crystal has finite dimensions. In general, a crystal has a huge number of atoms (around 1023). This difference between the infinite lattice and the finite lattice structure is not significant inside the crystal, and as we will discuss later, one gets around the finite size of the crystal by using periodic boundary conditions. Therefore, the concept of a Bravais lattice can be extended to a crystal. However, this concept breaks down when one approaches the surface of the crystal because the periodicity of the lattice, which is the cornerstone of most of the basic theoretical concepts of solid state physics, is no more valid. Therefore, the physical properties at or near the surface of a crystal are very different. The study of these surface properties, constrained by the nonperiodicity in a third dimension perpendicular to the surface, requires application of different techniques and is by itself a vast and fascinating field.", " 1.6 SOME SIMPLE CRYSTAL STRUCTURES 1.6.1 Sodium Chloride Structure The sodium chloride structure is shown in Figure 1.16. There are equal numbers of sodium and chlorine ions placed at alternate points of a simple cubic lattice. Each ion has six of the other kind of ions as its nearest neighbor. Thus, the coordination number is 6. The crystal structure can be described as a Bravais fcc lattice with a basis. The basis consists of one sodium ion at 0 and one chlorine ion at the center of the cubic unit cell; i.e., at ^x + ^y + ^z The compounds that have the sodium chloride structure include AgBr, AgCl, AgF, BaO, BaS, BaSe, BaTe, CaO, CaS, CaSe, CaTe, CsF, KBr, KCl, KF, KI, LiBr, LiCl, LiF, LiH, LiI, MgO, MgS, MgSe, MnO, NaBr, NaCl, NaF, NaI, RbBr, PbS, RbCl, RbF, RbI, SrO, SrS, SrSe, SrTe, and UO. a/2 \u00f0 \u00de\u00f0. \u00de 1.6.2 Cesium Chloride Structure The cesium chloride structure is shown in Figure 1.17. There is one molecule per unit cubic cell with the ions in the body-centered positions; i.e., Cs+: 000 and Cl\u2212: 1 2. Thus, the cesium chloride 1 2 1 2 FIGURE 1.16 The sodium chloride structure. FIGURE 1.17 CsCl structure. 16 CHAPTER 1 Basic Properties of Crystals structure can be described as a simple cubic lattice with a basis. The cesium ion is at the origin 0, and the chlorine ion is at : Because each ion is at the center of a cube of ions of the other type of ions, the coordination number is 8. ^x + ^y + ^z a/2 \u00de\u00f0 \u00f0 \u00de The compounds that have the cesium chloride structure include AlNi, BeCu, CsBr, CsCl, CsI, LiHg, NH4Cl, RbCl, TlBr, TlCl, and TlI. \u00f0 a/4 ^x + ^y + ^z 1.6.3 Diamond Structure The diamond structure is shown in Figure 1.18. The diamond cubic structure (Ref. 2a) consists of two interpenetrating fcc Bravais lattices displaced from each other by one-quarter of a body diagonal. Alternately, it can be considered as a fcc lattice with basis at 0 and at : The diamond structure is a result of covalent bonding. The covalent bond between two atoms is a very strong bond between two electrons, one from each atom, with directional properties. The spins of the two electrons are antiparallel, and the electrons forming the bond tend to be localized in the region between the two atoms. Because the C, Ge, and Si atoms each lack four electrons to form filled shells, these elements can have attractive interaction due to charge overlap, a type of interaction not found between atoms with filled shells because of the Pauli exclusion principle and the consequent repulsive interaction. There are eight atoms in a unit cube, and each atom has 4 nearest neighbors and 12 next nearest neighbors in the diamond lattice. Hence, the diamond lattice, which is not a Bravais lattice, is relatively empty. The 4 nearest neighbors of each atom form the vertices of a regular tetrahedron, and the covalent bonding between the neighboring atoms in the diamond structure is also known as tetrahedral bonding. The maximum proportion of the volume that is available to be filled by hard spheres is 0.34. The elements that crystallize in the diamond structure are C (diamond), Si, Ge, and \u03b1-Sn (grey). The coordination number of the diamond lattice is 4, and the packing fraction is 0.34. \u00de\u00f0 \u00de FIGURE 1.18 The diamond cubic structure. 1.6 Some Simple Crystal Structures 17 1.6.4 Zincblende Structure The zincblende (or cubic zinc sulfide) structure, shown in Figure 1.19, is obtained from the diamond structure (Figure 1.18) when Zn ions are placed on one fcc lattice and S ions are placed on the other fcc lattice. Thus, each ion has four of the opposite type as nearest neighbors. Some crystals with zincblende structure are AgI, AlAs, AlP, AlSb, BeS, BeSe, BeTe, CdS, CdTe, CuBr, CuCl, CuF, Cu", "I, GaAs, GaP, GaSb, HgS, HgSe, HgTe, InAs, InP, InSb, MnS(red), MnSe, SiC, ZnS, ZnSe, and ZnTe. 1.6.5 Hexagonal Close-Packed (hcp) Structure The hexagonal close-packed (hcp) structure is not a Bravais lattice, but a large number of elements crystallize in this form. If we assume atoms to be hard spheres, close-packed planes with hexagonal symmetry (see Figure 1.20) can be formed. If one starts with the atom in position A, there are two other kinds of spaces between the atoms, B or C. If the second plane is placed on the B positions, a third nesting layer can be placed either over site A or site C. The stacking ABABAB\u2026 yields the hcp structure while the stacking ABCABC\u2026 gives the fcc structure (see Problems 1.8 and 1.11). Both types of stacking have the same density of packing, and the packing fraction is 0.74. The unit cell of the hcp structure is the hexagonal primitive cell, and the basis contains two = 1:633. Figures 1.21 and elements of the same type. The ideal c/a ratio for hcp structures is 1.22 represent two alternate ways of representing the hcp structure. 8 3 In Figure 1.21, the unit cell of the hcp structure is shown. The unit cell of the hcp structure is the hexagonal primitive cell, and the basis contains two atoms. One atom of the basis is at the origin (000),, which means at r = 2/3a + 1/3b + 1/2c. The c/a ratio of hexagonal and the other atom is at p \ufb03\ufb03 2 3 1 3 \u00f0 1 2\u00de A C B FIGURE 1.19 Zincblende structure. FIGURE 1.20 The three sites A, B, and C in a close-packed plane of atomic spheres (the [111] plane of a fcc structure or the basal plane of a hcp structure). 18 CHAPTER 1 Basic Properties of Crystals \u2192 c \u2192 b \u2192 a FIGURE 1.22 FIGURE 1.21 The hexagonal close-packed (hcp) structure. The basic unit of hcp structure as a trigonal cell with two identical atoms as the basis at (000) and. (Note the differences in origin between Figures 1.21 and 1.22.) 1 2\u00de 1 3 \u00f0 1 3 close-packing of spheres is 1.633. However, several crystals, of which the c/a ratio is quite different, such as zinc with a c/a ratio of 1.85, are also called hcp structures. The lattice constants of elements with hcp structure are shown in Table 1.3. The coordination number is the same for both fcc and hcp structures. Therefore, at appropriate temperatures, many metals transform easily between fcc and hcp structures. This is called martinestic transformation. However, the alternate way to describe the basic unit of hcp structure is a trigonal cell containing two atoms, which is shown in Figure 1.22. The basic unit of the unit cell of the hcp structure described in Figure 1.22 is the trigonal cell. We note that the origins and the lattice vectors are containing two identical atoms at (000) and chosen differently in Figures 1.21 and 1.22. The choice of any one of the two alternate methods of representing the same hcp crystal structure depends on convenience. 1 2\u00de \u00f0 1 3 1 3 Table 1.3 Lattice Constants of Elements with hcp Structure Element Beryllium Cadmium Cerium Chromium Cobalt Dysprosium Erbium Gadolinium Hafnium Helium (2 K and 26 atm) a(A\u2218) 2.29 2.98 3.65 2.72 2.51 3.59 3.55 3.56 3.20 3.57 c(A\u2218) 3.58 5.62 5.96 4.43 4.07 5.65 5.59 5.80 5.06 5.83 (Continued ) 1.7 Bragg Diffraction 19 Table 1.3 Lattice Constants of Elements with hcp Structure\u2014cont\u2019d Element Holmium Hydrogen (molecule) Lanthanum Lutetium Magnesium Neodymium Nickel Osmium Praseodymium Rhenium Ruthenium Scandium Terbium Titanium Thallium Thulium Yttrium Zinc Zirconium a(A\u2218) 3.58 3.75 3.75 3.50 3.21 3.66 2.65 2.", "74 3.67 2.76 2.70 3.31 3.60 2.95 3.46 3.54 3.65 2.66 3.23 c(A\u2218) 5.62 6.49 6.07 5.55 5.21 5.90 4.33 4.32 5.92 4.46 4.28 5.27 5.69 4.69 5.53 5.55 5.73 4.95 5.15 Source: R. W. G. Wyckoff, Crystal Structures, vol. 1 (J. Wiley, 1963). 1.7 BRAGG DIFFRACTION W. L. Bragg considered a crystal as made up of a set of parallel lattice planes of ions spaced equal distances d apart, as shown in Figure 1.23. If the incident waves are reflected specularly, the reflected rays (also known as diffracted rays) would interfere constructively if the path difference is 2d sin \u03b8 = n\u03bb, (1.8) where \u03b8 is the angle of incidence and n is an integer, also known as the order of the corresponding reflection. This is the famous Bragg law of X-ray diffraction, (Ref. 2a) although the same law is valid for other types of waves (such as electron waves) as long as there is specular reflection from the ions in the crystal lattice. The Bragg law considers the periodicity of the lattice but does not include the basis of ions or atoms at each lattice point. In fact, it is the composition of the basis that determines the intensity of diffraction for various orders of n. In addition, it is pertinent to note that the crystal planes can be arranged in an infinite number of ways. Even for the same incident ray, both the direction and intensity of the reflected (note that we are using diffraction and reflection interchangeably) rays would depend on the orientation of the crystal planes. One immediate consequence of the Bragg law (Eq. 1.8) is that Bragg reflection can occur only for wavelengths \u03bb \u2264 2d: Originally, Bragg diffraction was used by crystallographers to study the 20 CHAPTER 1 Basic Properties of Crystals \u03b8 d sin \u03b8 \u03b8 \u03b8 d sin \u03b8 \u03b8 d FIGURE 1.23 Bragg reflection from a family of lattice planes. crystal structure of solids. Later, after the discovery of diffraction of electrons from a periodic crystal by Davisson and Germer, Bragg diffraction became important in understanding the theory of crystalline solids. 1.8 LAUE METHOD Von Laue considered the X-ray diffraction from a crystal by considering it as composed of identical atoms or ions (basis) placed at the lattice sites R of a Bravais lattice. Each ion or atom at these sites would reradiate the incident radiation in all directions (Laue did not make any assumptions about specular reflections). However, sharp peaks would be observed only at appropriate wavelengths and directions when the scattered radiations from the ions or atoms (basis) placed at sites R would interfere constructively. We consider two ions, separated by the vec- tor d (see Figure 1.24). If ^n is the direction of the incident radiation of wavelength \u03bb, the incident wave vector k = 2\u03c0 ^n/\u03bb. If the path difference between the radiation scattered by each of the two ions is m\u03bb, where m is an integer, the scattered radiation (assuming elastic scattering) would be observed in the direction ^n\u2032 with the same wavelength \u03bb and wave vector k\u2032 = 2\u03c0^n\u2032/\u03bb. This leads to constructive interference, the condition for which is (Figure 1.24) d. ^n \u2212 ^n\u2032 \u00de \u00f0 = m\u03bb: (1.9) \u2227 n \u2227 n \u2019 \u2227\u2192 d cos \u03b8 = d \u22c5n k k \u03b8 d \u03b8 \u2032 \u2227 d cos \u03b8 \u2032=\u2212 d \u22c5n \u2032 \u2192 k \u2032 k \u2032 FIGURE 1.24 Path difference of X-rays scattered by two points separated by d. 1.9 Reciprocal Lattice 21 Multiplying Eq. (1.9) by 2\u03c0/\u03bb and using the definitions of k and k\u2032, we can write Eq. (1.9) in the alternate form d. k \u2212 k\u2032 = 2\u03c0m: (1.10) \u00de If we consider all the ions at each site of the Bravais lattice, the condition of constructive interference of all the scattered radiations is obtained from a generalization of Eq. (1.10), which can be written in the form \u00f0 where R is a direct lattice vector of the Bravais lattice. If we write R. k \u2212 k\u2032 \u00de \u00f0 = 2\u03c0m, K = k \u2212", " k\u2032 = k\u2032 \u2212 k, (1.11) (1.12) where K is the set of wave vectors that satisfy the condition of constructive interference of all the scattered radiations (Eq. 1.11), i.e., K. R = 2\u03c0m: (1.13) In Section 1.9, we will study the properties of the reciprocal lattice constructed from the set of wave vectors K by first constructing them using the basic properties of the direct lattice vectors R. 1.9 RECIPROCAL LATTICE 1.9.1 Definition If R is a set of points that constitute a Bravais lattice as defined in Eq. (1.1), R = n1a1 + n2a2 + n3a3, there would exist a particular set of wave vectors K that would yield plane waves with the periodicity of the lattice. The analytic definition of the set of vectors K would be (1.14) for any vector r and all direct lattice vectors R in the Bravais lattice. Eq. (1.14) can be written in the alternate form eiK. \u00f0 r+R \u00de = eiK.r, eiK.R = 1: (1.15) The set of wave vectors K that satisfy Eq. (1.14) for any direct lattice vector R that generates a Bravais lattice is defined as the reciprocal lattice of that Bravais lattice. It is obvious that each Bravais lattice has its own reciprocal lattice. It can be shown that the reciprocal lattice can be generated from the three primitive vectors b1, b2, and b3, where b1, b2, b3 are defined by b1 = 2\u03c0 b2 = 2\u03c0 b3 = 2\u03c0 a2 \u00d7 a3 a2 \u00d7 a3 \u00f0 a3 \u00d7 a1 a2 \u00d7 a3 \u00f0 a1 \u00d7 a2 a2 \u00d7 a3 \u00f0 a1. a1. a1.,, : \u00de \u00de \u00de (1.16) 22 CHAPTER 1 Basic Properties of Crystals It is easy to show from Eq. (1.16) that (Problem 1.13) ai. bj = 2\u03c0\u03b4ij, (1.17) where \u03b4ij is the Kronecker delta function, defined by \u03b4ij = 0 for i \u2260 j and \u03b4ij = 1 for i = j. We also note from Eqs. (1.16) and (1.17) that the vectors b1, b2, and b3, are in the reciprocal space and are not in the same plane because the primitive lattice vectors a1, a2, and a3 are not coplanar. Therefore, any vector k in the reciprocal space can be written as k = k1b1 + k2b2 + k3b3: From Eqs. (1.1), (1.16), and (1.18), we obtain k. R = 2\u03c0 \u00f0 For Eq. (1.15) to hold true for any K, k1n1 + k2n2 + k3n3 : \u00de K. R = 2\u03c0m, (1.18) (1.19) (1.20) where m is an integer. Because n1, n2, and n3 are integers, and we obtain from Eqs. (1.19) and (1.20) that m = k1n1 + k2n2 + k3n3, (1.21) whenever k = K. Eq. (1.21) must hold good for any choice of the integers ni. It follows that k1, k2, and k3 are also integers whenever k = K. Thus, we can write m in the alternate form m = m1 + m2 + m3, (1.22) where each mi is an integer. If we write m1 = hn1, m2 = kn2, and m3 = ln 3, where h, k, l are integers, the reciprocal lattice vectors can be expressed as K = hb1 + kb2 + lb3, (1.23) Later, we will identify h, k, l as the Miller indices introduced earlier. As mentioned earlier, any Bravais lattice has its own reciprocal lattice. It has sometimes been compared to each person having his or her own shadow. 1.9.2 Properties of the Reciprocal Lattice We can enumerate some of the properties of the reciprocal lattice from the previous definitions. a. Each vector of the reciprocal lattice is normal to a set of lattice planes of the direct lattice. Proof: We have obtained from Eq. (1.20), K. R = 2\u03c0m", ", where m is an integer. Therefore, the projection of vector R on the direction of K has the length d, j However, there is an infinite number of points in the direct lattice with the same property. To show this, let us consider a lattice point R\u2032 represented by the integers j d = : (1.24) 2\u03c0m K n\u20321 = n1 \u2212 pl; n\u20322 = n2 \u2212 pl; n\u20323 = n1 + p h + k, \u00de \u00f0 (1.25) 1.9 Reciprocal Lattice 23 where p is an arbitrary integer; n1, n2, and n3 are the set of integers that represent a lattice point R, and h, k, and l are the integers that define a reciprocal lattice vector. It can be easily shown that K. R\u2032 = K. R = 2\u03c0m: (1.26) b. is inversely proportional to the spacing of the lattice planes normal to K, if the components Therefore, R\u2032 has the same projection on K as R and is on the plane normal to K, at a distance d from the origin. Thus, there is an infinite number of lattice points on this plane if it has at least one lattice point. K j of K have no common factor. Proof: We have shown in Eq. (1.24) that d = 2\u03c0m K j factor, then a lattice vector R\u2032\u2032 with components : If (h, k, l), the components of K have no common can always be found such that j j \u00f0 K. R\u2032\u2032 = 2\u03c0 n\u2032\u20321, n\u2032\u20322, n\u2032\u20323\u00de m + 1 : \u00f0 \u00de Here, Thus, the lattice plane containing R\u2032\u2032 is at a distance m + 1 = hn\u2032\u20321 + kn\u2032\u20322 + ln\u2032\u20323: 2\u03c0 d\u2032\u2032 = m + 1 K \u00de j \u00f0 j (1.27) (1.28) (1.29) j j K from the origin. Comparing Eqs. (1.24) and (1.29), we note that the lattice plane containing R\u2032\u2032 is spaced 2\u03c0/ from the lattice plane containing R. The simplest way of identifying the planes of a direct lattice is by their normals, which are the vectors of the reciprocal lattice. The planes that are most densely populated with lattice sites are usually the most prominent planes in a direct lattice. These are also the most widely separated because the density of direct lattice sites is constant. Therefore, the most prominent planes are those with the smallest reciprocal lattice vectors. c. The Miller indices h, k, l, which identify the direct lattice planes, are also the integers that identify the reciprocal lattice vectors normal to those planes. Proof: Consider a lattice plane with normal K such that K. R = 2\u03c0m (Eq. 1.20) is satisfied by all lattice points identified by R in that plane. If a lattice point has n2 = n3 = 0, from Eq. (1.21), we obtain The intercept of this plane along the a1 axis has the length n1 = m/h: d1 = n1a1 = a1: m h \" # (1.30) (1.31) 24 CHAPTER 1 Basic Properties of Crystals One can similarly obtain and d2 = n2a2 = a2, m k \" # d3 = n3a3 = m l \" # a3: (1.32) (1.33) Thus, the intercepts of this plane among the axes, measured in the units of the corresponding basis vectors, are inversely proportional to h, k, l. These integers are precisely the definition of the Miller indices (after removal of common factors) of the plane. d. The volume of a unit cell of the reciprocal lattice is inversely proportional to the volume of the unit cell of a direct lattice. Proof: The primitive vectors of the reciprocal lattice are b1, b2, and b3. The volume of a unit cell of the reciprocal lattice is b1. \u00f0 b2 \u00d7 b3 = \u00de 2\u03c0 \u00f0 a2 \u00d7 a3 a1. \u00de \u00f0 a2 \u00d7 a3. b2 \u00d7 b3 \u00f0 \u00de \u00de, (1.34) which can be simplified as By using the vector identity \u00f0 b1. \u00f0 b2 \u00d7 b3 = \u00de \u22122\u03c0b2. a1. \u00d7 b3 \u00f0 \u00f0 a2 \u00d7 a3 \u00de a2 \u00d7 a3 a2 \u00d7 a3 \u00de \u00de \u00d7 A = \u2212B \u00f0 \u00d7 b3 = \u2212a2 A. C \u00de b3. a", "3 \u00f0 + a3 \u00de b3. a2 \u00de \u00f0 \u00f0, we obtain From Eqs. (1.17) and (1.36), we obtain From Eqs. (1.17), (1.35), and (1.37), we obtain a2 \u00d7 a3 \u00f0 \u00de \u00d7 b3 = \u22122\u03c0a2: 2\u03c0 \u00de a2 \u00d7 a3 \u00de Because the volume of the primitive cell in the direct lattice is b2 \u00d7 b3 \u00de b1. a1. \u00f0 \u00f0 = \u00f0 : 3 we obtain from Eqs. (1.32) and (1.33), v = a1. a2 \u00d7 a3 \u00de, \u00f0 b1. \u00f0 b2 \u00d7 b3 = \u00f0 2\u03c0 v \u00de 3 \u00de : : (1.35) : \u00de (1.36) (1.37) (1.38) (1.39) (1.40) 1.9 Reciprocal Lattice 25 e. The reciprocal lattice of a simple cubic Bravais lattice with side a is a simple cubic lattice with cubic primitive cell of side 2\u03c0/a: Proof: For a simple cubic (sc) lattice, the primitive lattice vectors are a1 = a^x, a2 = a^y, a3 = a^z: From Eqs. (1.11) and (1.35), we obtain the reciprocal lattice vectors of the sc lattice, b1 = 2\u03c0 a ^x, b2 = 2\u03c0 a ^y, b3 = 2\u03c0 a ^z: (1.41) (1.42) f. The direct lattice is the reciprocal of its own reciprocal lattice. Proof: This can be easily shown by inspection of Eq. (1.40) or (1.42). g. The reciprocal lattice of a bcc Bravais lattice with conventional unit cell of side a is a fcc lattice with conventional unit cell of side 4\u03c0/a. Proof: We have seen in Eq. (1.2) that for a bcc lattice, the symmetric set of primitive vectors is a1 = a 2 \u00f0 ^y + ^z \u2212 ^x ; a2 = \u00de a 2 \u00f0 ^z + ^x \u2212 ^y ; a3 = \u00de a 2 \u00f0 ^x + ^y \u2212 ^z : \u00de From Eqs. (1.11), the reciprocal lattice vectors are b1 = 4\u03c0 a 1 2 \u00f0 ^y + ^z \u00de ; b2 = 4\u03c0 a 1 2 \u00f0 ^z + ^x ; b3 = \u00de 4\u03c0 a 1 2 \u00f0 ^x + ^y : \u00de (1.43) This has the form of the fcc primitive vectors (Eq. 1.3), provided the side of the cubic cell is taken to be 4\u03c0/a. h. One can similarly show (Problem 1.14) that the reciprocal lattice of the fcc Bravais lattice with conventional unit cell of side a is a bcc lattice with conventional unit cell of side 4\u03c0/a. i. The unit cell of the reciprocal lattice need not be a parallelepiped. 1.9.3 Alternative Formulation of the Laue Condition We have derived the Laue condition (Eq. 1.12) for constructive interference of the incident radiation reradiated by the ions or atoms in all directions as K = k\u2032 \u2212 k, where K is a reciprocal lattice vector. Because the incident and scattered radiations have the same wavelengths \u03bb = \u03bb\u2032 for elastic scattering, it follows that From Eqs. (1.12) and (1.44), we obtain = k j j k\u2032 j j = k: k\u2032 = k \u2212 K j j = k: (1.44) (1.45) 26 CHAPTER 1 Basic Properties of Crystals By squaring Eq. (1.45), we obtain k2 = k2 + K2 \u2212 2k. K, which can be written in the alternate form k. ^K = 1/2K: (1.46) (1.47) Eq. (1.47) implies that the incident wave vector k would satisfy the Laue condition only if the tip of k is on a plane that is a perpendicular bisector of the line joining the origin to K. This is shown in Figure 1.25, and such planes in k space are known as Bragg planes. In Figure 1.26, the Laue condition is shown in an alternate way such that its equivalence to Bragg refection can be demonstrated. In fact, it can be easily shown that the Bragg and Laue formulations of X-ray diffraction from a crystal", " are equivalent. We write K = nK0, where n is an integer and K0 is the shortest reciprocal lattice vector parallel to K. Further, Eq. (1.24) can be written in the alternate form We can easily show from Figure 1.26, From Eqs. (1.48) and (1.49), we obtain K = 2\u03c0n d : K = 2k sin \u03b8: k sin \u03b8 = n\u03c0 d : (1.48) (1.49) (1.50 \u2192\u2192 k \u2032 FIGURE 1.25 The Laue condition. A typical Bragg plane is shown in the diagram. 1.10 Brillouin Zones 27 \u2192 \u2192 \u2192 K = k \u2032 \u2212 k \u2192 \u2212k \u2192 k \u2032 \u2192 k FIGURE 1.26 Because K = k\u2032\u2212k and = k j j k\u2032 j j, k and k\u2032 have the same angle \u03b8 with the plane perpendicular to K: Using the expression k = 2\u03c0/\u03bb, we can write Eq. (1.50) as 2d sin \u03b8 = n\u03bb, (1.51) which is precisely Eq. (1.8), the Bragg condition. We have also shown that the order n of the Bragg reflection is n = K / j j j K0 : j 1.10 BRILLOUIN ZONES 1.10.1 Definition The Brillouin zone is a very important concept in solid state physics; it plays a major role in the theoretical understanding of the elementary ideas of electronic energy bands. The first Brillouin zone is defined as the Wigner\u2013Seitz primitive cell of the reciprocal lattice. Thus, it is the set of points in the reciprocal space that is closer to K = 0 than to any other reciprocal lattice point. We have shown in Figure 1.21 that the Bragg planes bisect the lines joining 0 (the origin) to the reciprocal lattice points. Thus, we can also define the first Brillouin zone as the set of points that can be reached from 0 without crossing any Bragg planes. Here, the points common to the surface of two or more zones have not been considered. The second Brillouin zone is the set of points that can be reached from the first Brillouin zone by crossing only one plane. Similarly, the nth Brillouin zone can be defined as the set of points that can be reached by crossing n \u2013 1 Bragg planes. We will first describe Brillouin zones of one- and two-dimensional (square) lattices to explain the fundamental methods of obtaining the Brillouin zones before describing the Brillouin zones of some important lattices. 28 CHAPTER 1 Basic Properties of Crystals 1.10.2 One-Dimensional Lattice Consider a one-dimensional (linear) lattice of lattice constant a, which is taken in the ^x direction. By definition, the direct lattice vector is R = n1 a ^x, where n1 is an integer. The reciprocal lattice vectors for this linear lattice are K = 2\u03c0/a h^x, \u00de \u00f0 (1.52) where h is an integer. Thus, the reciprocal lattice is also a linear lattice of side b = 2\u03c0/a. As described earlier, the Bragg planes (points in one dimension also known as zone boundaries) bisect the lines joining a reciprocal lattice point with its neighbors. The first three Brillouin zones of the linear lattice are shown in Figure 1.27. K0 j As we can see from Eq. (1.52), the shortest reciprocal lattice vector = 2\u03c0/a. If the origin 0 is chosen at the center, the first Bragg plane (zone boundary) is at \u2013\u03c0/a and \u03c0/a. Similarly, the second Bragg plane is at \u20132\u03c0/a and 2\u03c0/a, and the third Bragg plane is at \u20133\u03c0/a and 3\u03c0/a. Thus, the first Brilloun zone (shown by horizontal lines) extends from \u2013\u03c0/a to \u03c0/a, the second Brillouin zone (shown by lines \\\\\\\\\\\\\\) is between \u20132\u03c0/a and \u2013\u03c0/a as well as between \u03c0/a and 2\u03c0/a. Similarly, the third Brillouin zone (shown by lines ////////) is between \u20133\u03c0/a and \u20132\u03c0/a as well as between 2\u03c0/a and 3\u03c0/a. These are consistent with the definition of the Brillouin zones; i.e., a point in the first zone does not cross any Bragg plane (point in one dimension), a point in the second zone crosses only one Bragg plane (zone boundary), and a point in the third Brillouin zone crosses two Bragg planes (zone boundaries).", " j 1.10.3 Two-Dimensional Square Lattice The direct lattice vectors of a two-dimensional square Bravais lattice are R = n1a^x + n2a^y, (1.53) kz \u22123\u03c0 /a \u22122\u03c0 /a \u2212\u03c0 /a O \u03c0 /a 2\u03c0 /a 3\u03c0 /a 2 3 1 2 3 FIGURE 1.27 The Brillouin zones of the linear lattice. 1.10 Brillouin Zones 29 FIGURE 1.28 The Brillouin zones for a two-dimensional square lattice of side 4\u03c0/a [[4\u03c0/\u03b1]]. where n1 and n2 are integers and a is the lattice constant. The reciprocal lattice vectors are K = 2\u03c0 a \u00f0 h^x + k^y, \u00de (1.54) where h and k are integers. Thus, the reciprocal lattice is also a square lattice of side b = 2\u03c0/a. The Brillouin zones are constructed according to the method outlined earlier. In Figure 1.28, all the Bragg lines (in two dimensions) that are in a square of side 4\u03c0/a centered on the origin are shown. The Bragg lines divide the square into regions belonging to different zones. 1.10.4 bcc Lattice We have shown in Eq. (1.43) that the primitive translational vectors of the reciprocal lattice of a bcc lattice are given by b1 = 2\u03c0/a \u00f0 \u00de\u00f0 ^y + ^z \u00de ; b2 = 2\u03c0/a \u00f0 ^z + ^x \u00de\u00f0 \u00de ; b3 = 2\u03c0/a ^x + ^y : \u00de \u00de\u00f0 \u00f0 The reciprocal lattice vectors are K = hb1 + kb2 + lb3 = \u00f0 2\u03c0/a k + l \u00de \u00de\u00bd\u00f0 ^x + \u00f0 h + l ^y + \u00de h + k \u00f0 ^z \u00de $. (1.55) 30 CHAPTER 1 Basic Properties of Crystals kz P N kx \u0393 ky H FIGURE 1.29 The first Brillouin zone for the bcc lattice. From Eq. (1.55), we note that the shortest nonzero K\u2032s are the 12 vectors 2\u03c0/a \u00f0 \u00de\u00f0 \u00b1^x \u00b1 ^y ; \u00de 2\u03c0/a \u00f0 \u00de\u00f0 \u00b1^y \u00b1 ^z ; \u00de 2\u03c0/a \u00f0 \u00de\u00f0 \u00b1^z \u00b1 ^x : \u00de (1.56) The first Brillouin zone is the primitive cell formed from the planes normal to the 12 vectors (the Bragg planes) of Eq. (1.56). Thus, the 12 vectors from the origin to the center of each face of the first Brillouin zone are \u03c0/a \u00f0 \u00b1^x \u00b1 ^y \u00de\u00f0 ; \u00de \u03c0/a \u00f0 \u00de\u00f0 \u00b1^y \u00b1 ^z ; \u00de \u03c0/a \u00f0 \u00de\u00f0 \u00b1^z \u00b1 ^x : \u00de (1.57) The first Brillouin zone for the bcc lattice is shown in Figure 1.29. This regular 12-faced solid is a rhombic dodecahedron. The important symmetry points are marked by conventional symbols. By convention, \u0393 is denoted as the center of a zone. 1.10.5 fcc Lattice The primitive translation vectors b1, b2, b3 of the reciprocal lattice of the fcc lattice are (Problem 1.14) b1 = b2 = b3 = 2\u03c0/a 2\u03c0/a 2\u03c0/a \u00f0 \u00f0 \u00f0 ^y + ^z \u2212 ^x ^z + ^x \u2212 ^y ^x + ^y \u2212 ^z \u00de\u00f0 \u00de\u00f0 \u00de\u00f0,,. \u00de \u00de \u00de (1.58) Thus, the reciprocal lattice vectors of the fcc lattice are K = hb1 + kb2 + lb3 = \u00f0 2\u03c0/a \u2212h + k + l \u00de \u00de\u00bd\u00f0 ^y + ^1.59) 1.11 Diffraction by a Crystal Lattice with a Basis 31 kz W U K \u0393 L X kxkx ky FIGURE 1.30 The first Brillouin zone for the fcc lattice. While the shortest nonzero K\u2032s are the eight vectors the corners of the octahedron formed by the eight planes normal to these vectors at their midpoints are truncated by the planes that are the perpendicular bisectors of the six reciprocal lattice vectors 2\u03c0/a \u00f0 \u00b1^x \u00b1 ^y \u00b1 ^z \u00de\u00f0, \u00de (", "1.60) \u00de The first Brillouin zone of the fcc lattice is the truncated octahedron shown in Figure 1.30. The \u00de \u00f0 \u00f0 \u00b1 4\u03c0/a ^x; \u00b1 4\u03c0/a ^y; \u00b1 \u00f0 4\u03c0/a ^z: \u00de (1.61) conventional symbols of the important symmetry points are also shown in the figure. 1.11 DIFFRACTION BY A CRYSTAL LATTICE WITH A BASIS The analysis of diffraction of the incident radiation by a crystal lattice with a basis and, consequently, the intensity of radiation in a given Bragg peak is more complicated than a monoatomic crystal because we have to consider the scattering of the radiation from each ion or atom of the basis at each lattice point. A crystal lattice with a basis can have two or more ions or atoms, either of the same type or of different types. In general, each lattice point in a primitive cell would have a basis of atoms or ions associated with it, and the crystal structure consists of a repetitive unit of this basis. 1.11.1 Theory We consider n scatterers at positions r1, r2, :::, rn, in each primitive cell with the lattice point considered as the origin of rj. It is important to note that in an ideal crystal, each lattice point has an identical basis 32 CHAPTER 1 Basic Properties of Crystals around it so that the scatterers are located at the same relative position in each cell. We have already noted that a Bragg peak is associated with a change in the wave vector K = k\u2032 \u2212 k and the phase difference between two rays scattered from ri and rj would be K.. The amplitudes of the rays scattered at r1, :::, rn would be in the ratios eiK.r1, :::, eiK.rn. Thus, the amplitude of the scattered ray from the primitive cell, which is the sum of the scattered rays from the n identical scatterers of the primitive cell, will be equal to ri \u2212 rj \u00f0 \u00de The amplitude of the scattered rays (Eq. 1.62) can be expressed as S = \u2211 R \u2211 j fje\u2212i \u00f0 R+rj.\u0394k: \u00de e\u2212iR.\u0394kSK, S = \u2211 R where SK is the structure factor obtained by the Bragg condition \u0394k = K, n SK = \u2211 j = 1 fj eiK.rj : (1.62) (1.63) (1.64) Here, fj is the atomic scattering factor or form factor that is a measure of the scattering power of the jth atom or ion in the basis. 1.11.2 Geometrical Structure Factor We will first consider the basis as consisting of n identical atoms or ions around each lattice point. f. Then we can This is equivalent to stating that each atomic form factor fj has the same value fj write Eq. (1.64) in the form % n SK = f \u2211 j = 1 eiK.rj f SK: % (1.65) SK is known as the geometrical structure factor. Because SK is proportional to the amplitude, is proportional to the intensity of the Bragg peak and indicates the extent to which the waves scattered from the ions or atoms in the basis within the primitive cell interfere to reduce it. SK j j 2 Next, we will discuss the structure factor of a few important lattices. 1.11.3 Application to bcc Lattice If we consider the bcc lattice as a simple cubic lattice of side a with identical atoms or ions at r1 = 0 and at r2 =, the structure factor (Eq. 1.65) can be expressed as ^x + ^y + ^z a/2 \u00f0 \u00de\u00f0 \u00de Further, for a simple cubic lattice, SK = \u00f0 1 + eiK. \u00bd 1/2a \u00f0 ^x+^y+^z \u00de$ : \u00de K = 2\u03c0 a \u00f0 h^x + k^y + l^z : \u00de (1.66) (1.67) 1.11 Diffraction by a Crystal Lattice with a Basis 33 From Eqs. (1.66) and (1.67), we obtain SK = 1 + ei\u03c0 \u00f0 \u22121 \u00f0 = 1 + h+k+l \u00de\u00de h+k+l \u00de \u00f0 = $ 0, 2, h + k + l odd even h + k + l : % (1.68) (1.69) From Eq. (1.69), we note that whenever h + k + l is odd, there is no Bragg", " diffraction (reflection). Because in this derivation, the bcc lattice has been considered as a simple cubic lattice with a basis, this implies that the odd reciprocal lattice vectors h + k + l of the reciprocal simple cubic lattice do not exist. Therefore, the actual reciprocal lattice is a fcc lattice, which could have been obtained directly if we had considered the bcc lattice as the primitive lattice. 1.11.4 Application to fcc Lattice If we consider a fcc lattice as a simple cubic lattice of side a and basis at r1 = 0, r2 = a 2 \u00f0 ^y + ^z \u00de, r3 = a 2 \u00f0 ^z + ^x, r4 = \u00de a 2 \u00f0 ^x + ^y, \u00de the structure factor (Eq. 1.65) can be expressed as SK = 1 + eiK.a/2 \u00f0 ^y+^z \u00de + eiK.a/2 ^z+^x \u00f0 \u00de + eiK.a/2 \u00f0 ^x+^y \u00de: From Eqs. (1.67) and (1.70), we obtain Eq. (1.71) can be written in the alternate form SK = 1 + ei\u03c0 \u00f0 k+l \u00de + ei\u03c0 \u00f0 h+l \u00de + ei\u03c0 \u00f0 h+k \u00de: (1.70) (1.71) SK = 1 + \u22121 \u00f0 k+l + \u22121 h+l + \u22121 h+k: \u00de \u00f0 From Eq. (1.72), we note that SK = 4, if each value of the integers h, k, l is either even or odd. However, SK = 0, if either only one of the three integers h, k, l is even and the other two are odd, or one of the three integers is odd and the other two are even. Thus, there can be no Bragg reflection if the indices h, k, l are partly odd and partly even. In contrast, there is Bragg reflection if each of the indices h, k, l is either even or odd. This is true for a reciprocal bcc lattice. Hence, the reciprocal lattice of a fcc lattice is a bcc lattice, which we stated earlier, and has been assigned as a problem. \u00de \u00de \u00f0 (1.72) 1.11.5 The Atomic Scattering Factor or Form Factor In Eq. (1.64), we defined fj as the atomic scattering factor or form factor, which was a component of the geometrical structure factor. However, in subsequent discussions, we considered that all the atoms or ions in a basis are identical. Thus, each fj = f and f are factored out of the summation. We wrote the geometrical structure factor as SK = f SK, where SK is the structure factor. 34 CHAPTER 1 Basic Properties of Crystals However, if the atoms or the ions in a basis are not identical, the atomic form factor fj at the site dj in the basis is K fj \u00f0 \u00de = \u2212 1 e Z dr eiK.r\u03c1j\u00f0 r : \u00de (1.73) Here, \u03c1j\u00f0 Eqs. (1.64) and (1.73) that the geometrical structure factor would no longer vanish. is the electronic charge density of the ion of type j placed at r = 0. It is evident from r \u00de PROBLEMS 1.1. Show that the volume of the primitive Bravais cell of a bcc lattice is a3/2, where a is the side of the unit cube. 1.2. Show that the volume of the primitive Bravais cell of a fcc lattice is a3/4, where a is the side of the unit cube. 1.3. A plane in a lattice with primitive vectors a1, a2, and a3 has intercepts at 3a1, 2a2, and \u20132a3. Calculate the Miller indices of the plane. Label the direction perpendicular to this plane. 1.4. Draw a sketch of the 102 \u00f0 \u00de plane in a simple cubic lattice. 1.5. Prove that in a cubic crystal, a direction [hkl] is perpendicular to the plane (hkl) having the same indices. 1.6. Show that the actual volume occupied by the spheres in the simple cubic structure (assuming that they are contacting hard spheres) is 52.4% of the total volume. 1.7. Show that the actual volume occupied by the spheres in the bcc structure (packing fraction: assuming that they are contacting hard spheres) is 0.68. 1.8. Show that", " the actual volume occupied by the spheres in the fcc structure (packing fraction: assuming that they are contacting hard spheres) is 0.74. 1.9. Show that the angle between any two of the lines (bonds) joining a site of the diamond lattice : to its four nearest neighbors is cos\u22121 \u22121/3 \u00f0 \u00de 1.10. Show that the ideal c/a ratio of the hexagonal close-packed structure is 8/3p. 1.11. Show that the packing fraction of the hexagonal close-packed structure is 0.74. \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 1.12. Sodium transforms from bcc to hcp at about 23 K, which is also known as the \u201cmartenistic\u201d transformation. The lattice constant in the cubic phase is a = 4.23 A\u2218. Determine the lattice constant a of the hexagonal phase. Assume that the c/a ratio in the hexagonal phase is indistinguishable. 1.13. Show that if ai are the three direct lattice primitive vectors and bj are the three primitive vectors of the reciprocal lattice (as defined in Eq. 1.16), then where \u03b4ij is the Kronecker delta function. ai. bj = 2\u03c0\u03b4ij, References 35 1.14. Show that the primitive translation vectors b1, b2, b3 of the reciprocal lattice of the fcc lattice, and prove that the ^z + ^x \u2212 ^y are b1 = 2\u03c0/a reciprocal lattice of a fcc lattice of side a is a bcc lattice of side 4\u03c0/a. ^x + ^y \u2212 ^z ^y + ^z \u2212 ^x, b3 =, b2 = 2\u03c0/a 2\u03c0/a \u00de\u00f0 \u00de\u00f0 \u00de\u00f0 \u00f0 \u00de \u00f0 \u00de \u00f0 \u00de 1.15. Show that for a monatomic diamond lattice, the structure factor is SK = 1 + ei \u00bd 2\u03c0 \u00f0 h+k+l \u00de$ 2, h + k + l 1\u00b1i, = ( is twice an even number, is odd, is twice an odd number: Interpret these conditions geometrically. : 9 = ; 1.16. a. Show that the reciprocal lattice of the sodium chloride structure is bcc and a reciprocal lattice vector can be written as K = 4\u03c0 a \u00f0 n1 ^x + n2^y + n3^z, \u00de where a is the side of the cube and all the coefficients ni of a of a reciprocal lattice vector are integers or integer+1/2. b. If the atomic form factors of the two types of ions are f1 and f2, show that the geometrical structure factor SK = f1 + f2 if each ni in a set is an integer and SK = f1 \u2212 f2 if each ni in a set is an integer+1/2. References 1. Aschroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 2a. Bragg WL. Structure of some crystals as indicated by their diffraction of x-rays. Proc Roy Soc. London; 1913; A89. 2b. Bragg WH and Bragg WL. Structure of the diamond. Proc Roy Soc. London; 1913;A89:277. 3. Buerger MJ. Crystal structure analysis. New York: John Wiley & Sons; 1960. 4. Kittel C. Introduction to solid state physics. New York: John Wiley & Sons; 1976. 5. Marder MC. Condensed matter physics. New York: John Wiley & Sons; 2000. 6. Myers HP. Introduction to solid state physics. London: Taylor & Francis; 1990. 7. Wyckoff RWG. Crystal structures, vol. 1. New York: John Wiley & Sons; 1963. Phonons and Lattice Vibrations CHAPTER 2 CHAPTER OUTLINE 2.1 Lattice Dynamics........................................................................... 37 2.1.1 Theory..........................................................................", ".... 37 2.1.2 Normal Modes of a One-Dimensional Monoatomic Lattice............................... 41 2.1.3 Normal Modes of a One-Dimensional Chain with a Basis................................ 44 2.2 Lattice Specific Heat....................................................................... 48 2.2.1 Theory.............................................................................. 48 2.2.2 The Debye Model of Specific Heat..................................................... 49 2.2.3 The Einstein Model of Specific Heat................................................... 52 2.3 Second Quantization....................................................................... 53 2.3.1 Occupation Number Representation................................................... 53 2.3.2 Creation and Annihilation Operators................................................... 54 2.3.3 Field Operators and the Hamiltonian................................................... 58 2.4 Quantization of Lattice Waves............................................................... 61 2.4.1 Formulation......................................................................... 61 2.4.2 Quantization of Lattice Waves......................................................... 65 Problems..................................................................................... 66 References..........................................", ".......................................... 68 2.1 LATTICE DYNAMICS 2.1.1 Theory In a crystalline solid, at finite temperatures, the ions or the atoms are not stationary but vibrate around an equilibrium position. In 1907, Einstein proposed a theory of the heat capacity of a solid based on Planck\u2019s quantum hypothesis. He assumed that each atom of the solid vibrates around its equilibrium position with a frequency \u03bdE, known as the Einstein frequency. Each atom vibrates like a simple harmonic oscillator that is in the potential well of the force field of its neighbors. The atoms have the same frequency \u03bdE and vibrate independent of the other atoms. Thus, a mole of solid with N atoms is assumed to have 3N independent harmonic oscillators. The excitation spectrum of the crystalline solid is composed of levels that are spaced at a distance \u0127\u03bdE from each other. Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00002-5 \u00a9 2012 by Elsevier Inc. All rights reserved. 37 38 CHAPTER 2 Phonons and Lattice Vibrations The basic assumption of the Einstein model was that the atoms vibrate independent of each other, which could be justified only if the temperature is very high. However, at normal temperatures this assumption breaks down, because if two or more atoms move in unison, the restoring forces between them, which tend to restore each of them to their equilibrium position, would be reduced. In such a scenario, the required energy to excite a quantum would be reduced. In fact, the correlation between the motion of the adjacent atoms would play a significant role in solving the problem of lattice vibrations. We consider the ground state of the lattice with a basis as the state at zero temperature, where each ion or atom of mass Mn is located at the equilibrium position dn, which is a vector connecting the local origin of the cell with a basis to the atom or ion of mass Mn. We assume that if the lattice has only one atom per unit cell, the atom is located at the lattice point and dn = 0: At a finite temperature, the displacement of the nth atom or the ion in the ith unit cell (the local origin of which is located at the direct lattice vector Ri, and we have assumed that the origin of the lattice is one of the lattice points) from its equilibrium position at dn is the vector uni. The definition of the vector uni is schematically shown in Figure 2.1. We note that uni can also be written as uni = r \u2212 Ri \u2212 dn, (2.1) where r is the instantaneous position of the atom or ion of mass Mn located in the unit cell of which the local origin is at Ri: The kinetic energy of the crystalline lattice can be written as T = \u2211 ni 1 2 Mn _uni j 2: j (2.2) The potential energy of the crystal depends on the structure of the cell as well as the interatomic forces. However, we assume that at any given instant, the function V describes the potential energy of the crystal in terms of the instantaneous positions of all atoms, i.e., in terms of their actual displacements from the equilibrium positions. In addition, it is assumed that the perfect lattice is a configuration of stable equilibrium. The study of phonons is based on the assumption that the deviation uni of the ions or atoms from their equilibrium position Ri + dn is so small that one can make a Taylor expansion in powers of the variables uni: Thus, we can write uni \u00de \u00f0 V = V0 + \u2211 ni\u03b1 \u2202V \u2202u\u03b1 ni! \"0 u\u03b1 ni + 1 2 \u2211 nn\u2032 \u2211 ii\u2032 \u2211 \u03b1\u03b2 \" \u22022V ni\u2202u\u03b2 n\u2032i\u2032 \u2202u\u03b1 #0 niu\u03b2 u\u03b1 n\u2032i\u2032 + \u2026, (2.3) O FIGURE 2.1 Definition of uni for a lattice with a basis. \u2192 r \u2192 uni Mn \u2192 d n \u2192 Ri 2.1 Lattice Dynamics 39 where u\u03b1 basis vectors of three-dimensional space. ni are the Cartesian components of uni and \u03b1 and \u03b2 are Cartesian indices running over the The first term V0 is essentially the cohesive energy of the crystal, but for the present purpose it is a constant term that can be neglected without affecting the calculation. Because the lowest energy state is a minimum as a function of the locations of the ions or atoms, the linear term in Eq. (2.3) must vanish near equilibrium. The first important term in Eq. (2", ".3) is the quadratic term in Eq. (2.3). This is known as the harmonic approximation. If one considers the higher-order terms for study of certain properties of the crystal, those are known anharmonic terms. However, in the present discussion, we shall restrict ourselves to the harmonic approximation. We follow the Lagrangian procedure in classical mechanics to solve Eqs. (2.2) and (2.3). If we define the Lagrangian function (Symon, 1971, p. 366) the Lagrangian equations are L = T \u2212 V, d dt \u2202L \u2202 _q k# \u2212 $ \u2202L \u2202qk = 0, k = 1, \u2026, 3N: From Eqs. (2.2), (2.3), and (2.5), we obtain (Problem 2.1) Mn \u20acu\u03b1 ni = \u2212 \u2211 n\u2032i\u2032\u03b2 \u22022V ni\u2202u\u03b2 n\u2032i\u2032 \u2202u\u03b1 \" #0 u\u03b2 n\u2032i\u2032 : We define a Cartesian tensor \u03a6 such that its components are obtained by the relation \u03a6 \u03b1\u03b2 ni\u02b9;n\u2032i\u2032\u2261 \u22022V ni\u2202u\u03b2 n\u2032i\u2032 \u2202u\u03b1 \" : #0 From Eqs. (2.6) and (2.7), we obtain the vector equation Mn \u20acuni = \u2212\u2211 n\u2032i\u2032 \u03a6ni;n\u2032i\u2032. un\u2032i\u2032: (2.4) (2.5) (2.6) (2.7) (2.8) Eq. (2.8) can be interpreted as the force acting on the nth atom in the ith cell (the cell of which the local origin is at a distance Ri from the origin of the lattice, where Ri is a direct lattice vector) due to the displacement un\u2032i\u2032 of the atom on the n\u2032th site of the i\u2032th cell. In fact, much of the theory of phonons can be developed without considering how to calculate \u03a6ni;n\u2032i\u2032. However, \u03a6ni;n\u2032i\u2032 cannot depend on the absolute position of Ri and Ri\u2032 in the crystalline lattice. Thus, the tensor \u03a6 has to be a function of their relative position Ri and Ri\u2032: If we write the Cartesian tensor can be expressed in the alternate form Ri\u2032\u2212Ri = Rl, \u03a6ni,n\u2032i\u2032 = \u03a6nn\u2032\u00f0 Rl : \u00de (2.9) (2.10) 40 CHAPTER 2 Phonons and Lattice Vibrations From Eqs. (2.8) and (2.10), we obtain Mn\u20acuni = \u2212\u2211 n\u2032l Rl \u03a6nn\u2032\u00f0 \u00de. un\u2032,i+l: (2.11) Here, un,i+l denotes the instantaneous displacement of the nth atom in the i + lth cell. We note that because the summation over the direct lattice vector Rl spans the entire lattice, Eq. (2.11) is translationally invariant. For example, if we change the label Rl to Rl\u2032, we obtain Mn\u20acuni = \u2212\u2211 n\u2032l\u2032 \u03a6nn\u2032\u00f0 Rl\u2032\u00de. un\u2032,i+l\u2032, (2.12) which yields the same set of equations as Eq. (2.11). We also note another property of the tensor \u03a6 is that the energy of the crystal cannot change if all the ions (or atoms) are displaced by a single vector, i.e., \u2211 l\u2032 \u03a6nn\u2032\u00f0 Rl\u2032\u00de = 0: (2.13) Because Eq. (2.11) must satisfy Bloch\u2019s theorem, if the set of functions of the time that describes for each value of Ri has been found, then according to the Bloch condition, there t the value of uni \u00de would be a wave vector q such that \u00f0 uni t \u00f0 \u00de = eiq. Riun,0 t, \u00de \u00f0 (2.14) t \u00de \u00f0 is the displacement of the nth atom in the cell that has been chosen as the origin for the where un,0 direct lattice vectors Ri: It is obvious from Eq. (2.14) that the atom or ion located at every site dn (measured from the local origin of the unit cell, as shown in Figure 2.1) moves with the same amplitude and direction. However, the phase would vary for each cell. From Eqs. (2.11) and (2.14), we obtain Mn\u20acun,0eiq. Ri = \u2212\u2211", " n\u2032l Rl \u03a6nn\u2032\u00f0 \u00de. un\u2032,0eiq. Rleiq. Ri: (2.15) In Eq. (2.15), we cancel the eiq. Ri from both sides, and because the origin 0 is arbitrary, we consider a solution with a definite value of q by writing and obtain un,0 = Un,q, Mn \u20acUn,q = \u2212\u2211 n\u2032! \u2211 l \u03a6nn\u2032\u00f0 Rl \u00de eiq. Rl. \u20acUn\u2032,q: \" We define the Fourier transform of the force tensor \u03a6 as q \u03a6nn\u2032\u00f0 \u2261 \u2211 l \u00de \u03a6nn\u2032\u00f0 Rl \u00de eiq. Rl, (2.16) (2.17) (2.18) 2.1 Lattice Dynamics 41 and rewrite Eq. (2.17) as Mn \u20acUn,q = \u2212\u2211 n\u2032 \u03a6nn\u2032\u00f0 q \u00de. Un\u2032,q, (2.19) Eq. (2.19) is a set of 3m equations (assuming that there are m atoms per unit cell and there are three component equations\u2014due to the three Cartesian components of the vectors\u2014for each of the m values of n) in contrast to Eq. (2.8), which had a set of 3mN equations. This enormous simplification was possible because of the translational invariance. Using the classical theory of vibrations, we obtain from Eqs. (2.19) and (2.20), U\u03b1 n,q\u00f0 t \u00de = U\u03b1 n,qei\u03c9t, \u03a6 \u2211 n\u2032\u03b2\u00bd \u03b1\u03b2 nn\u2032\u00f0 q \u00de \u2212 \u03c92Mn\u03b4nn\u2032\u03b4\u03b1\u03b2 U\u03b1 n\u2032,q = 0: $ (2.20) (2.21) Eq. (2.21) is an eigenvalue equation with 3m solutions that are solved by finding the roots of the equation in \u03c92 when the determinant of the matrix [ ] is equal to zero. In a sense, we are solving 3m normal modes of vibration of m atoms in a unit cell that are assumed to interact via the force tensor, which is different for each value of q, is a sum of interactions of : This force tensor, \u03a6nn\u2032\u00f0 \u03a6nn\u2032\u00f0 all n-type atoms (those atoms located at site n in each unit cell) with all the atoms on site n\u2032, and includes the effect of their relative phases. q \u00de q \u00de 2.1.2 Normal Modes of a One-Dimensional Monoatomic Lattice We consider a set of ions of mass M (we have been using ions and atoms interchangeably for the lattice vibrations) located at the lattice points separated by a distance a. The one-dimensional Bravais lattice vectors are Ri = nia: In this one-dimensional chain of lattice, one atom (or ion) of the same type is located at the lattice point. If the lattice constant is a, we define L = Na such that periodic boundary conditions are applied to the linear chain. This boundary condition requires that the atoms located at s and at s + N vibrate with the same amplitude and phase. The linear chain is shown in Figure 2.2. L = Na N cells sa FIGURE 2.2 (s + N)a A linear chain of lattice points with periodic boundary conditions. The linear boundary condition is best illustrated if we construct an endless circular chain of lattice points. This endless circular chain is illustrated in Figure 2.3. We note that although this illustration is possible for a onedimensional lattice, such periodic boundary conditions have to be imagined for a threedimensional lattice. 42 CHAPTER 2 Phonons and Lattice Vibrations s s + N a s s + 1 us FIGURE 2.4 s + p us +p FIGURE 2.3 An endless circular chain of lattice points. A linear chain of identical atoms, each located at a lattice point. Figure 2.4 shows a linear chain of atoms of identical point mass m in a linear lattice of lattice constant a. The atoms are located at the lattice points. The atom (or ion) of which the equilibrium position is sa is displaced from equilibrium by an amount us, as shown in Figure 2.4. For simplicity, we consider only nearest-neighbor interactions between the atoms (ions). The potential energy V in Eq. (2.3) is of the form V harm = 1 2 u K\u2211 s \u00bd \u00f0 sa \u00de \u2212 u \u00f0\u00bd s + 1 2, a $ \u00de$ (2.22) where K is the interaction energy of two ions (popularly known as the spring constant)", " \u2212\u03a6i,i+1. \u03a6ii is determined by the condition set in Eq. (2.13). The equations of motion are obtained from Eq. (2.22) as M\u20acu sa \u00de \u00f0 = \u2212 \u2202V harm sa \u2202u \u00f0 \u00de = \u2212K sa u \u00bd\u00f0 \u00f0 \u00de \u2212 2u s \u2212 1 \u00f0\u00bd \u2212 u a $ \u00de \u00f0\u00bd\u00f0 s + 1 : a $ \u00de$ The solution of Eq. (2.23) is of the type Using the periodic boundary condition (Figure 2.3), sa, t u \u00f0 \u00de = Aei \u00f0 qsa\u2212\u03c9t \u00de: we obtain the expression for q, eiqNa = 1, q = s N 2\u03c0 a, (2.23) (2.24) (2.25) (2.26) 2.1 Lattice Dynamics 43 where s is an integer and a is the lattice constant of the one-dimensional lattice. The q values lie between \u2212 \u03c0 a : Substituting Eq. (2.24) in Eq. (2.23), we obtain a and \u03c0 which leads to the solution M\u03c92 = 2K 1\u2212 cos qa \u00f0 \u00de = 4K sin2 qa/ sin K M \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 qa 2, \u2019 & (2.27) (2.28) where we have taken only the positive root of Eq. (2.27) because \u03c9 is an even function of q. We plotted \u03c9 as a function of q in Figure 2.5. We note that there is a maximum vibrational frequency \u03c9 = 2 and the behavior is periodic with period 2\u03c0/a. In fact, all possible vibrations are given by K M \ufb03\ufb03\ufb03\ufb03 values of q in the range 2.29) which is the Brillouin zone (we will discuss the Brillouin zone in detail in Chapter 4 by using the nearly free electron model in a periodic lattice potential) for a linear lattice. One can also count the number of modes in the following way. From Eq. (2.26), we obtain an expression for the density of modes in one-dimensional q space as where L is the length of the sample. We also note that when qa \u226a 1, Na 2\u03c0 = L 2\u03c0, d\u03c9 dq = \u03c9 q = K M \ufb03\ufb03\ufb03\ufb03\ufb03 r a = constant: \u03c9 \u03c9 max (2.30) (2.31) 2\u03c0 /a 0 \u22122\u03c0 /a 2\u03c0 /a FIGURE 2.5 The vibrational frequencies of a linear chain of identical point masses. 44 CHAPTER 2 Phonons and Lattice Vibrations Thus, the group velocity is equal to the phase velocity, and this proportionality of frequency to the wave number is also the property of the ordinary elastic waves in a continuum. This is also the velocity of sound waves and is known as the acoustic mode. However, at large values of q, the velocity of the wave is not constant. In fact, when q = \u03c0/a = 2\u03c0/\u03bb, i.e., when the wavelength \u03bb = 2a, \u03c9q bends over to a horizontal tangent (see Figure 2.5). This shows the property of dispersion. 2.1.3 Normal Modes of a One-Dimensional Chain with a Basis We consider a one-dimensional Bravais lattice of lattice constant a with two ions or atoms of masses M1 and M2 per unit cell. This is shown in Figure 2.6. The basic assumption is that each ion interacts only with the nearest neighbors, which are at a distance a/2 from each other, and M1 > M2: Thus, the lattice constant of the linear chain is a. If \u039a is the force constant, from Eq. (2.21) we obtain V harm = K 2 \u2211 s \u00bd u1 \u00f0 sa \u00de \u2212 u2 sa \u00f0 \u00de$ 2 + K 2 u2 \u2211 s \u00bd sa s + 1 \u2212 u1 \u00bd $ a \u00de \u00f0 2, $ (2.32) where u1 sa \u00de \u00f0 ment of the ion that oscillates around sa + d: The equations of motion are is the displacement of the ion that oscillates about the site sa and u2 sa \u00de \u00f0 is the displace- M1 \u20acu1 sa \u00de \u00f0 = \u2212 M2 \u20acu2 sa \u00de \u00f0 = \u2212 \u2202V harm sa \u2202u1 \u00de \u00f0 \u2202V harm sa \u2202u2 \ufffd", "\ufffd \u00f0 = \u2212K 2u1 sa \u00de \u00f0 \u00bd \u2212 u2 sa \u00de \u00f0 \u2212 u2 s \u2212 1 \u00f0\u00bd, a $ \u00de$ = \u2212K 2u2 sa \u00de \u00f0 \u00bd \u2212 u1 sa \u00de \u00f0 \u2212 u1 s + 1 \u00f0\u00bd : a $ \u00de$ The solutions of Eq. (2.33) are of the type and sa, t u1 \u00f0 \u00de = \u22081ei \u00f0 qsa\u2212\u03c9t \u00de sa, t u2 \u00f0 \u00de = \u22082ei \u00f0 qsa\u2212\u03c9t \u00de: Substituting Eq. (2.34) in Eq. (2.33), we obtain \u22082 \u2212 2\u22081 + \u22082e\u2212iqa \u00f0 \u22081eiqa \u2212 2\u22082 + \u22081 \u00de ei \u00f0 ei \u00f0 qsa\u2212\u03c9t \u00de qsa\u2212\u03c9t \u00de: \u00de \u2212\u03c92M1\u22081ei \u00f0 qsa\u2212\u03c9t \u00de = K \u2212\u03c92M2\u22082 ei \u00f0 qsa\u2212\u03c9t \u00de = K and a M1 M2 FIGURE 2.6 \u00f0 Diatomic linear chain of masses M1 and M2. ( ( ( ( We cancel the ei \u00f0 and solve the determinantal equation qsa\u2212\u03c9t \u00de term from both sides 2K \u2212 M1\u03c92 1 + eiqa \u2212K \u00f0 \u00de = 0: (2.36) 1 + e\u2212iqa \u2212K 2K \u2212 M2\u03c92 \u00f0 \u00de ( ( ( ( (2.33) (2.34) (2.35) 2.1 Lattice Dynamics 45 There are two roots of the solution of Eq. (2.36), which yields (Problem 2.4) \u03c92 \u00b1 = K 1 M1 + 1 M2 # \u00b1 K $ v u u t 2 4 sin2 qa/2 1 M1 # 1 M2 $ + \u2212 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 # \" \u00f0 M1M2 \u00de : (2.37) The two solutions of Eq. (2.37) are the two branches of the phonon dispersion relation. For small q, the two roots of Eq. (2.37) are and \u03c9\u2212 = K M1 + M2 2 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u00de \u00f0 r qa \u03c9+ = 2K M1 + M2 \u00de \u00f0 M1M2 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 : s (2.38) (2.39) We also note from Eq. (2.37) that if qa = \u00b1\u03c0 (the Brillouin zone boundary), the expressions for \u03c9\u00b1 reduce to Thus, we obtain and \u03c92 \u00b1 = K #! 1 M1 + 1 M2 \u00b1 K \"! 1 M2 \u2212 1 M1 : \" $ (2.40) \u03c9 + = 2K M2r \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 (2.41) (2.42) \u03c9\u2212 = 2K : M1r \ufb03\ufb03\ufb03\ufb03\ufb03\ufffd", "\ufffd At the Brillouin zone boundary, q = \u00b1 \u03c0 a : We also note that because M1 > M2, \u03c9 + > \u03c9\u2212. Another interesting point to note is that from Eq. (2.34), u1 and u2 are periodic with q = \u00b12\u03c0/a. Therefore, the dispersion relation repeats itself for each Brillouin zone. From the previous discussions, we obtain the following results. The vibrational frequency of a diatomic linear chain of mass M1 and M2 is shown in Figure 2.7. The first branch, \u03c9\u2212, which tends to become zero at q = 0, is known as the acoustic mode. \u03c9+ \u03c9\u2212 Optical Acoustic \u2212\u03c0 /a FIGURE 2.7 O q \u03c0 /a Optical and acoustic phonon branches of a diatomic linear chain. 46 CHAPTER 2 Phonons and Lattice Vibrations It is in fact the analogue of a long-wavelength vibration of the linear chain, conceived as an elastic continuum. The normal value of the velocity of sound is reproduced in this case just as in the case of a monoatomic lattice. In the acoustic mode, the atoms vibrate in unison, whereas in the optical mode, the atoms in the unit cell vibrate out of phase. Essentially, the two sublattices of the two types of atoms move rigidly in opposition to one another. This can also be stated in the alternate form that the diatomic molecules vibrate as if they are independent of the neighbors. If these crystals are ionic crystals, the two types of atoms are of opposite electric charge. This yields an oscillating dipole moment that is optically active. We also note that if the masses of, there would be no gap at the zone the two atoms in the linear chain were equal (M1 = M2 \u00de boundaries. The configuration of atoms in the acoustic and optical modes in the diatomic linear chain is shown in Figure 2.8. In the acoustic mode, atoms within a unit cell move in concert, whereas in the optical mode they vibrate against each other. In the previous example, we discussed only a linear chain of atoms. Thus, we obtained dispersion relation for longitudinal acoustic (LA) and longitudinal optical (LO) phonons. However, if we consider two atoms per primitive cell in a three-dimensional lattice, as in the NaCl or diamond structure, for each polarization mode in a direction of propagation, there are two branches in the dispersion relation \u03c9 versus q. In addition to LA and LO phonons, of which the dispersion relations are essentially the same as shown in Figure 2.7, one obtains transverse acoustical (TA) and transverse optical (TO) phonons. Figure 2.9 shows the TA and TO modes for a diatomic linear lattice at the same wavelength. The optical modes of oscillation of individual ions in both transverse and longitudinal modes are shown in Figure 2.10. e m T i Acoustic Optical Space FIGURE 2.8 Configuration of atoms in acoustic and optical modes in the diatomic linear chain. 2.1 Lattice Dynamics 47 + + \u2212 \u2212 + + \u2212 \u2212 + \u2212 Optical mode + \u2212 + \u2212 \u2212 + Acoustical mode + q + q FIGURE 2.9 Transverse optical and acoustical waves for the same wavelength in a diatomic linear lattice. q q Basis: Anion Cation (a) (b) FIGURE 2.10 The direction of displacement of individual ions in (a) transverse optical (TO) and (b) longitudinal optical (LO) modes of oscillation. 48 CHAPTER 2 Phonons and Lattice Vibrations 2.2 LATTICE SPECIFIC HEAT 2.2.1 Theory The thermal excitation of the lattice contributes to the specific heat of solids. To calculate the lattice specific heat, one has to use quantum mechanical operators instead of the classical coordinates uni and their momenta. In the earlier formulation of the theory of lattice vibrations, we obtained a set of equations of motion that are those of an assembly of independent simple harmonic oscillators. However, in a quantum mechanical treatment, the excitations are of the Bose\u2013Einstein type. We introduced the term phonons to describe the normal modes of the crystal in the same manner the term photons was introduced to describe the quantum theory of electromagnetic field. In that theory, \u0127\u03c9, where \u03c9 is the angular frea normal mode of the radiation field in a cavity is given by quency of the mode. The photons are the quanta of radiation field that describes classical light. Similarly, the phonons are the quanta of the ions\u2019 displacement field that describes classical sound. The nomenclature of normal modes and phonons are equivalent although the latter is much more convenient. n + 1 2\u00de \u00f0 If we want to specify", " the energy levels of an N-ion harmonic crystal that can be regarded as 3N independent oscillators, the contribution to the energy \u03b5qs of a particular normal mode with angular frequency \u03c9s can have the discrete set of values q \u00f0 \u00de \u03b5qs = nqs + \u0127\u03c9s q, \u00de \u00f0 (2.43) 1 2 \u2019 & where there are nqs phonons of type s with wave vector q present in the crystal. We note that an equivalent classical description would be that the normal mode of branch s with wave vector q is in its nqsth excited state. We note that the total energy E is the sum of the energies of the individual normal modes, E = \u2211 qs nqs + & 1 2 \u0127\u03c9s \u2019 q : \u00de \u00f0 (2.44) When we consider an assembly of independent simple harmonic oscillators, the excitations must be bosons. From the theory of Bose\u2013Einstein statistics, nqs, the mean number of bosons with energy in thermal equilibrium at temperature T, when the chemical potential \u03bc is taken to be zero \u0127\u03c9s (\u03bc = 0 because the total number of phonons in thermal equilibrium is determined by the temperature and hence is not an independent variable), is given by q \u00de \u00f0 where nqs = 1 q e\u03b2\u0127\u03c9s \u00f0, \u00de\u22121 \u03b2 = 1/kBT: The quanta in the qth mode will contribute an energy \u03b5qs = nqs + & 1 2 \u0127\u03c9s \u2019 q, \u00de \u00f0 (2.45) (2.46) (2.47) 2.2 Lattice Specific Heat 49 which includes the zero-point energy. From Eqs. (2.45) and (2.47), the average total energy of the system (neglecting the zero-point energy) is \u03b5 = \u2211 qs \u0127\u03c9s e\u03b2\u0127\u03c9s q \u00f0 \u00de q \u00de\u22121 \u00f0 : (2.48) Here, the summation is over all polarizations s and different modes q. Thus, the specific heat is given by CV = 1 V \u2202\u03b5 \u2202T = 1 V \u2202 \u2202T \u2211 qs \u0127\u03c9s e\u03b2\u0127\u03c9s q \u00de \u00f0 q \u00de\u22121 \u00f0, (2.49) where V is the volume of the crystal. We will now evaluate the specific heat in the high-temperature and intermediate-temperature. We will evaluate low-temperature limits in Problem 2.5. 1. In the high-temperature limit, kBT/\u0127 is large compared with all the phonon frequencies. Thus, if we write x = \u0127\u03c9 kBT \u226a 1, 1 ex\u22121 \u2248 1 x! 1\u2212 x 2 + x2 12 + \u2026 : \" From Eqs. (2.49) and (2.50), we obtain CV \u2243 3N V, (2.50) (2.51) which is the classical law of Dulong and Petit. The higher-order terms in Eq. (2.50) yield the quantum corrections to the classical Dulong and Petit law. 2. In intermediate-temperature limits, both the Debye and Einstein models of specific heat are obtained by making different approximations. We note that in a large crystal, we can replace the sum in Eq. (2.49) by an integration over the wave vectors q that satisfy the Born\u2013von Karman boundary conditions. Thus, Eq. (2.49) can be expressed as CV = \u2202 \u2202T \u2211 s Z dq 3 2\u03c0 \u0127\u03c9s e\u03b2\u0127\u03c9s \u00f0 q \u00f0 \u00de q \u00de \u22121, (2.52) \u00f0 where the integral is over the first Brillouin zone. We can retrieve both the Debye and the Einstein models of specific heat of solids from Eq. (2.52), although they were originally derived under very different assumptions. \u00de 2.2.2 The Debye Model of Specific Heat The Debye model of specific heat can be obtained by making two basic assumptions. First, all branches of the vibration spectrum are replaced with three branches of the acoustic mode such that the linear dispersion = sq, \u03c9 q \u00de \u00f0 (2.53) 50 CHAPTER 2 Phonons and Lattice Vibrations where s is the velocity of sound. Second, the integral over the first Brillouin zone is replaced by an integral over a sphere of the same volume and radius qD in reciprocal space. It is assumed that the sphere contains exactly N allowed wave vectors just as in the case of the first Brillouin zone. Thus, the radius of the Debye sphere, qD, is such", " that and hence N = V 8\u03c03 4 3 \u03c0q3 D, qD = 6\u03c02 vc# 1/3, $ where vc is the volume of the Wigner\u2013Seitz cell, which can be written in the alternate form where n is the density of ions. From Eqs. (2.53) through (2.55), we obtain N = 3 qD 6\u03c02, We define a Debye frequency and a Debye temperature If we write Eq. (2.57) can be written as CV = \u2202 \u2202T 3\u0127c 2\u03c02 qD Z 0 q2dq e\u03b2\u0127sq\u22121 : \u03c9D = qDs \u0398D = \u0127\u03c9D/kB: x = \u0127sq/kBT, CV = 9NkB 3 T \u0398D# $ \u0398D/T Z 0 x4exdx 2: x \u2212 1 \u00de \u00f0 (2.54) (2.55) (2.56) (2.57) (2.58) (2.59) (2.60) (2.61) where N is the number of cells in the crystal. At low temperatures, the upper limit of the integral \u0398D/T tends to infinity, in which case the integral \u221e Z 0 x4exdx x \u2212 1 \u00de \u00f0 2 = 4\u03c04 5 : (2.62) 2.2 Lattice Specific Heat 51 1.0 0.8 B k N 3 / v C 0.6 0.4 0.2 0.0 0.0 0.5 1.0 Temperature T/\u0398D 1.5 2.0 FIGURE 2.11 Specific heat in the Debye approximation. Thus, we obtain from Eqs. (2.61) and (2.62), CV \u2248 12\u03c04 5 NkB 3 T \u0398D# $, (2.63) which is the Debye\u2019s T 3\u2014 law of lattice specific heats: The Debye law of specific heat is shown schematically in Figure 2.11. The Debye\u2019s theory of specific heat works very well for solids. However, the Debye temperature, \u0398D, is often interpolated from the observed specific heat, thereby allowing it to depend on the temperature T. In fact, from Eqs. (2.58) and (2.59), \u0398D is related to the velocity of sound, s, by the relation, \u0398D = \u0127qDs kB, (2.64) and therefore should be calculated directly for a solid. In any case, \u0398D for most solids is listed in tables. It is interesting to note that at temperatures well above \u0398D, the integrand form in Eq. (2.61) can be replaced by its form for small x and one obtains Dulong and Petit\u2019s law. Therefore, the Debye temperature, \u0398D, is a measure of the temperature separating the low-temperature region where quantum statistics is used from the high-temperature region where classical mechanics can be used. However, Debye\u2019s theory of specific heat has its limitations. In most cases, the sharp cutoff frequency \u03c9D = kB\u0398D is not justified. In general, there is a spread with several peaks that correspond \u0127 to the modes of different polarization, which have different velocities. In addition, there is a peak at high frequencies due to the strong dispersion near the zone boundary. The contrast between the Debye spectrum and an actual lattice spectrum is shown in Figure 2.12. 52 CHAPTER 2 Phonons and Lattice Vibrations ) ( \u03c9 D ) ( \u03c9 D \u03c9 2 (a) \u03c9D \u03c9 O \u03c9 (b) FIGURE 2.12 (a) Debye spectrum; (b) actual lattice spectrum. 2.2.3 The Einstein Model of Specific Heat As we noted earlier, in a lattice with a basis, there are both acoustic modes and optical modes. The three acoustic modes can be easily treated by the Debye model. The optical modes can be treated by the Einstein model in which each mode has the same frequency, \u03c9E, which is independent of q. Therefore, from Eq. (2.45), we obtain that each optical branch will contribute to the thermal energy density in the Einstein approximation, \u03b5 = N\u0127\u03c9E e\u03b2\u0127\u03c9E \u22121, (2.65) and if there are p such branches, Coptical V = 1 V \u2202E \u2202T We can also define an Einstein temperature \u0398E by \u0398E = \u0127\u03c9E kB : (2.67) It should be noted that Einstein was the first to derive the theory of specific heat", " of solids by using quantum statistics instead of classical statistics. From Eq. (2.66), we note that when T \u226b \u0398E, \u03b2\u0127\u03c9E \u2261 \u0398E T is very small and we can write Coptical V = pNkB, (2.68) so that each optical mode contributes kB/V, as required by Dulong and Petit\u2019s law. However, when T \u226a \u0398E, \u03b2\u0127\u03c9E is very large so that the = pNkB \u00f0 \u03b2\u0127\u03c9E \u00de e\u03b2\u0127\u03c9E \u22121 2 e\u03b2\u0127\u03c9E 2 \u00de : (2.66) Debye Einstein 0.1 0.2 0.3 0.4 0.5 T/q 0.6 0.7 0.8 0.9 1.0 FIGURE 2.13 Comparison of Debye\u2019s and Einstein\u2019s specific heat as a function of T \u0398 : 2.3 Second Quantization 53 contribution to CV drops exponentially. Thus, it is very difficult to excite the optical modes at low temperatures. A comparison of Debye\u2019s and Einstein\u2019s expressions for specific heat as a function of T \u0398 shown in Figure 2.13. is 2.3 SECOND QUANTIZATION 2.3.1 Occupation Number Representation The behavior of systems with a very large number of particles is studied in statistical physics. Suppose we have a system of N noninteracting particles that are found in states with wave functions \u03c61, \u03c62, \u2026, \u03c6N which form a complete orthonormal set. For noninteracting particles, the wave function \u03c6i corresponds to one of the plane wave states They obey the orthonormality condition, \u03c6k = eik. r: as well as dr \u03c6%i \u00f0 r \u03c6j\u00f0 \u00de r \u00de = \u03b4i,j, Z \u2211 i \u03c6%i \u00f0 r \u03c6j\u00f0 r\u2032 \u00de \u00de = \u03b4 r \u2212 r\u2032 \u00f0 \u00de : (2.69) (2.70) (2.71) The system can evidently be described by specifying the number of particles in states \u03c61, \u03c62, \u2026, \u03c6N: The complete wave function can be specified by using a new representation. We represent the basis function in the many-body wave function as n1, n2, n3, \u2026, nk \u2026 >, j with nk as the number of particles in state \u03c6k: The total number of particles \u2211ni = N: (2.72) (2.73) For Fermions (electrons, protons, neutrons, and He3 atoms), which obey Fermi\u2013Dirac statistics, the restriction is that ni = 0, or 1: (2.74) For bosons (photons, phonons, and He4 atoms), which obey Bose\u2013Einstein statistics, there is no restriction for ni. For photons and phonons, \u2211 i ni = N T, \u00de \u00f0 (2.75) 54 CHAPTER 2 Phonons and Lattice Vibrations where N(T) depends on the temperature T. Further, N = 0 at T = 0 and N \u2260 0 at T \u2260 0: (2.76) 2.3.2 Creation and Annihilation Operators (a) Bosons We will first consider the creation and annihilation operators for bosons. We define the operators ak and a\u2020 k by and n1, n2,\u2026, nk, \u2026 > = nkp ak j n1, n2, \u2026, nk \u2212 1, \u2026 > j \ufb03\ufb03\ufb03\ufb03\ufb03 a\u2020 k j n1. n2, \u2026, nk, \u2026 > = nk + 1 j n1, n2, \u2026, nk + 1, \u2026 >: Here, ak is the annihilation operator because it decreases p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 and a\u2020 k is the creation operator because it increases nk! nk \u2212 1, The vacuum state is defined as nk! nk + 1: 0, 0, \u2026,0, \u2026 >, j (2.77) (2.78) (2.79) (2.80) (2.81) whereas the other occupation numbers remain unchanged. From Eqs. (2.77) and (2.78), we obtain a\u2020 kak j n1, n2, \u2026, nk, \u2026 > = n1, n2, \u2026, nk \u2212 1, \u2026 > nkp a\u2020", " k j j \ufb03\ufb03\ufb03\ufb03\ufb03 = nk n1, n2, \u2026, nk, \u2026 >: (2.82) From Eq. (2.82), we obtain that a\u2020 kak \u21d2 nk is the number operator in state k. Similarly, we obtain aka\u2020 k j n1, n2, \u2026, nk, \u2026 > = = p nk + 1 nk + 1 \u00dej \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u00f0 n1, n2, \u2026, nk + 1, \u2026 > ak n1, n2, \u2026, nk, \u2026 >: j (2.83) 2.3 Second Quantization 55 Subtracting Eq. (2.82) from Eq. (2.83), we obtain \u00f0 Eq. (2.84) implies that aka\u2020 k \u2212 a\u2020 kak n1, n2, \u2026, nk, \u2026 > = n1, n2, \u2026, nk, \u2026 >: j \u00dej aka\u2020 k \u2212 a\u2020 kak = 1, which can be written in the alternate form where A, B $\u2212 is the commutator \u00bd If k \u2260 k\u2032, ak, a\u2020 k$\u2212 = 1, \u00bd A, B \u00bd $\u2212 = AB \u2212 BA: a\u2020 k\u2032ak j n1, n2, \u2026, nk\u2032, \u2026, nk, \u2026 > = = nkp a\u2020 k\u2032 j p nk\u2032 + 1 nkp \ufb03\ufb03\ufb03\ufb03\ufb03 n1, n2, \u2026, nk\u2032, \u2026, nk\u22121, \u2026 > n1, n2, \u2026, nk\u2032+1, \u2026, nk\u22121 \u2026 >: j (2.84) (2.85) (2.86) (2.87) (2.88) n1, n2, \u2026, nk\u2032, \u2026, nk, \u2026 > = ak \ufb03\ufb03\ufb03\ufb03\ufb03 aka\u2020 k\u2032 j \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 p nk\u2032 + 1 p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 j nk\u2032 + 1 = nkp n1, n2, \u2026, nk\u2032+1, \u2026, nk \u2026 > n1. n2, \u2026, nk\u2032+1, \u2026, nk \u2026 >: j (2.89) Subtracting Eq. (2.89) from Eq. (2.88), we obtain \ufb03\ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 k\u2032ak \u2212 aka\u2020 a\u2020 k\u2032\u00dej \u00f0 n1, n2, \u2026, nk\u2032+1, \u2026, nk, \u2026 > = 0: (2.90) From Eqs. (2.86) and (2.90), we obtain the commutation relation between the creation and annihilation operators of bosons, ak, a\u2020 k\u2032$\u2212 = \u03b4k,k\u2032, \u00bd where \u03b4k,k\u2032 is the Kronecker delta function, i.e., \u03b4k,k\u2032 = 1 if k = k\u2032, and \u03b4k,k\u2032 = 0 if k \u2260 k\u2032: Similarly, we can obtain the other commutation relations for bosons, and ak, ak\u2032$\u2212 = 0, \u00bd k, a\u2020 a\u2020 \u00bd k\u2032$\u2212 = 0: (2.91) (2.92) (2.93) 56 CHAPTER 2 Phonons and Lattice Vibrations Eqs. (2.91) and (2.93) are the three commutation relations for bosons. The total number of bosons like He4 atoms, <nk> = N, \u2211 k (2.94) where <nk> is the thermally average occupation number representation in state k at temperature T or, equivalently, the number of particles in state k at temperature T. According to the Bose\u2013Einstein distribution function, where <nk> = 1 \u03b5k", "\u2212\u03bc, \u00de \u2212 1 e\u03b2 \u00f0 \u03b5k = \u01272k2/2m, (2.95) (2.96) and \u03bc = chemical potential. For He4 atoms, \u03bc \u2260 0 at high temperatures. However, for photons and phonons, \u03b5k = sk and \u03bc \u2261 0: (2.97) In Eq. (2.97), and s = c = speed of light for photons, s = speed of sound for phonons: (b) Fermions The creation and destruction operators for Fermions obey very different commutation relations because the total number of particles in a level ni can be 0 or 1. The basis function in the many-body representation is with the constraint, ni = 0, or ni = 1: The Fermion operators ai and a\u2020 i are defined as n1, n2, \u2026, ni, \u2026 >, j n1, n2, \u2026, ni, \u2026 > = nip ai j \u22121 \u00f0 m \u00de j n1, n2, \u2026, ni \u2212 1, \u2026 >, \ufb03\ufb03\ufb03\ufb03 m = \u2211 j<i nj where and (2.98) (2.99) (2.100) a\u2020 i j n1, n2, \u2026, ni, \u2026 > = 1 \u2212 ni \u22121 \u00f0 m \u00de j p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 n1, n2, \u2026, ni + 1, \u2026 >: (2.101) 2.3 Second Quantization 57 Here, the function \u22121 \u00de for Fermions. From Eqs. (2.99) and (2.101), we obtain \u00f0 m comes from the requirement that the wave function is antisymmetric a\u2020 i ai j n1, n2, \u2026, ni, \u2026 > = = nip nip \ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03 \u22121 \u00f0 p ma\u2020 i j \u00de 2 \u2212 ni \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 Similarly, n1, n2, \u2026, ni \u2212 1, \u2026 > \u22121 n1, n2, \u2026, ni \u2026 >: 2m \u00de j aia\u2020 i j n1, n2, \u2026, ni \u2026 > = = p p mai \u22121 \u00f0 \u00de p 1 + ni j \u00f0 n1, n2, \u2026, ni + 1, \u2026 > \u22121 n1, n2, \u2026, ni, \u2026 >: 2m \u00de j 1 \u2212 ni 1 \u2212 ni \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 If ni = 0, we obtain from Eqs. (2.102) and (2.103), \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 i ai + aia\u2020 a\u2020 i \u00dej \u00f0 n1, n2, \u2026, ni, \u2026 > = n1, n2, \u2026, ni, \u2026 >: j Similarly, if ni = 1, we obtain from Eqs. (2.102) and (2.103), i ai + aia\u2020 a\u2020 i \u00dej \u00f0 n1, n2, \u2026, ni, \u2026 > = n1, n2, \u2026, ni, \u2026 >: j From Eqs. (2.104) and (2.105), we obtain which can be written in the alternate form i ai + aia\u2020 a\u2020 i = 1, ai, a\u2020 \u00bd i $+ = 1: In general, we can show that and ai, a j j $+ = 0, i \u2260 j, ai, aj $+ = \u00bd i, a\u2020 a\u2020 j $+ = 0: \u00bd \u00bd From Eqs. (2.107) and (2.108), the commutation relations for Fermions can be written as and We also obtain from Eq. (2.102), ai, a\u2020 j $+ = \u03b4ij, \u00bd ai, aj \u00bd $+ = \u00bd i, a\u2020 a\u2020 j $+ = 0: a\u2020 i ai =", " nip 2 \u2212 ni = ni, p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03 (2.102) (2.103) (2.104) (2.105) (2.106) (2.107) (2.108) (2.109) (2.110) 58 CHAPTER 2 Phonons and Lattice Vibrations because if and if ni = 0, ni = 0 ni = 1, ni = 1: (2.111) Thus, a\u2020 i ai = ni = ni is the number operator for state i. From the Fermi\u2013Dirac distribution function, we obtain the expression for the average number of particles in state k at temperature T, <a\u2020 kak> = <nk> = 1 \u03b5k\u2212\u03bc : \u00de + 1 e\u03b2 \u00f0 2.3.3 Field Operators and the Hamiltonian Here, we introduce the operators of a field of particles: \u03c6i\u00f0 = \u2211 i \u03c8 \u03be \u00de \u00f0 \u03be ai \u00de and \u03c8 \u2020 \u03be \u00f0 \u00de = \u2211 i \u03c6i% \u03be a\u2020 i : \u00de \u00f0 Here, ai and a\u2020 ai and a\u2020 i are operators for bosons, i are the second quantization operators, and \u03c6i\u00f0 (2.112) (2.113) (2.114) is the particle in the state i. If \u03be \u00de \u03c8 \u03be \u03c8 \u2020 \u00f0 \u03be\u2032 \u00de \u00de \u00f0 \u2212 \u03c8 \u2020 \u03be,j \u03be \u03c6i\u00f0 \u03c6%j \u00f0 \u00de \u03be\u2032 aia\u2020 \u00de\u00bd j \u2212 a\u2020 j ai : $ From Eqs. (2.91) and (2.115), we obtain \u03c8 \u03be \u00f0 \u00de \u03c8 \u2020 \u03be\u2032 \u00f0 \u00de \u2212 \u03c8 \u2020 \u03c8 \u03be \u03c6i\u00f0 \u03c6%i \u00f0 \u00de \u03be \u00de : (2.115) (2.116) We note that because \u03c6i\u00f0 \u03be \u00de \u21d2 \u03c61\u00f0 \u03be \u00de, \u03c62\u00f0 \u2211 i \u03be \u03be, \u03c63\u00f0, \u2026, forms a complete orthonormal set, it satisfies \u00de \u00de \u03be \u03c6%i \u00f0 \u03be \u03c6i\u00f0 \u03be \u2212 \u03be\u2032 = \u03b4 : \u00de \u00de \u00de \u00f0 (2.117) From Eqs. (2.116) and (2.117), we obtain By using a similar procedure, we can easily derive $\u2212 = \u03b4 \u03be\u2032 \u00f0 \u03be \u2212 \u03be\u2032 \u00de \u00de$\u2212 = i are Fermion operators, we can derive, by using a similar procedure, \u00de$\u2212 = 0: \u03be\u2032 \u03c8 \u00bd, \u03c8 \u03be If ai and a\u2032 \u00de$+ = \u03b4 \u00f0 \u03be \u2212 \u03be\u2032 \u00de, (2.118) (2.119) (2.120) 2.3 Second Quantization 59 and, \u03c8 \u03c8 \u03be \u00f0 \u00de \u00bd \u03be\u2032 \u00de$+ = \u2032 \u00de$+ = 0: (2.121) We will now consider the Hamiltonian that describes a system of interacting particles (for example, the electrons in a solid): H = \u2211 i 1 2m pi 2 + \u2211 i V ri ri \u2212 rj : \u00de (2.122) Here, pj = \u2212i\u0127\u2207j = \u2212i\u0127 \u2202 \u2202rj 2 tial, and V \u00f0 denote pi ri \u2212 rj \u00de 2/2m and V \u00f0 \u00de\u00f0 \u00de is the momentum operator for the jth particle, V is the periodic poten- ri \u00f0 \u00de is the interaction energy between the ith particle and the jth particle. We will ri as single-particle operators and V \u00f0 as two-particle operators. ri \u2212 rj 2 \u00de\u00f0 \u00de The Hamiltonian in Eq. (2.122) can be expressed in terms of the second quantization operators H = \u2211 \u03b1 Z \u2212 \u01272 2m! \u03c8 \u2020 r \u03b1\u00f0 \u22072\u03c8\u03b1\u00f0 \u00de r \u00de + V r \u03c8 \u2020 \u03b1\u00f0 \u00de \u00f0 r \u03c8 \u03b1\u00f0 \u00de r \u00de dr \" + 1 2 \u2211 \u03b1\u03b2 Z Z \u03c8 \u2020 \u03b1\u00f0 r \u00de \u03c8 \u2020 \u03b2\u00f0 r\u2032 \u00de V", " \u00f0 2 \u00de r\u2212r\u2032 \u03c8 \u03b2\u00f0 r\u2032 \u00de \u03c8 \u03b1\u00f0 \u00de r \u00de \u00f0 dr dr\u2032: (2.123) In Eq. (2.123), we have written \u03c8 \u03be \u00f0 \u00de = \u03c8 \u03b1\u00f0 r \u00de, where \u03b1 is the spin index and r is the coordinate. The expression for the Hamiltonian given by Eq. (2.123) is equivalent to the corresponding expression in Eq. (2.122). We also note that the total number of particles N = <N> = \u2211 k nk = \u2211 k e\u03b2 1 \u03b5k\u2212\u03bc \u00f0 \u00de + 1 : The chemical potential \u03bc is determined by the condition that when T = 0, \u03bc = \u03b5F = the Fermi energy: (2.124) (2.125) We consider a system of free particles moving in space. The single-particle wave function can be written as This leads to the free-particle Schrodinger equation, \ufb03\ufb03\ufb03\ufb03 r \u03c6k\u00f0 \u00de = eik. r: 1 Vp \u2212\u01272 2m \u22072\u03c6k\u00f0 r \u00de = \u03b5k\u03c6k\u00f0 r \u00de = \u01272k2 2m r \u03c6k\u00f0 \u00de : (2.126) (2.127) 60 CHAPTER 2 Phonons and Lattice Vibrations Here, k is the momentum of the particle, and its kinetic energy is \u01272k2 have different states. Thus, the field operators become 2m : For different k values, we and = \u03c8 r \u00de \u00f0 1 Vp \u2211 k ak eik. r \ufb03\ufb03\ufb03\ufb03 r \u03c8 \u2020 \u00f0 \u00de = 1 Vp \u2211 a\u2020 k e\u2212ik. r: (2.128) (2.129) In Eqs. (2.128) and (2.129), the spin indices are neglected for the present. Substituting Eqs. (2.128) \ufb03\ufb03\ufb03\ufb03 and (2.129) in Eq. (2.123), and letting V = 1, we obtain H = \u2211 k \u01272k2 2m a\u2020 kak + \u2211 kq V q \u00de \u00f0 a\u2020 k+qak + 1 2 \u2211 kpq V \u00f0 2 \u00de q \u00de \u00f0 a\u2020 k\u2212q a\u2020 p+qap aq: (2.130) We note that if V q \u00de \u00f0 2 = V \u00f0 q \u00de\u00f0 \u00de = 0, the Hamiltonian is diagonalized and Eq. (2.130) reduces to H = \u2211 k \u03b5k a\u2020 k ak = \u2211 k \u03b5knk: Eq. (2.131) reduces to the previous result a\u2020 kak = nk: If we define V(q) as the Fourier transform of V(r), eiq. r dr, (2.133) and the interaction V(q) a\u2020 sented by Figure 2.14. k+q ak can be repre2.131) (2.132) 2 V \u00f0 Similarly, we define the Fourier transform of \u00de\u00f0 as eiq. r dr: For example, if 2 V \u00f0 = r \u00de \u00f0 \u00de e2 r, q 2 V \u00f0 4\u03c0e2 q2 : a\u2020 k+q a\u2020 The interaction V \u00f0 represented by Figure 2.15. q \u00de\u00f0 \u00de \u00f0 = \u00de \u00de 2 (2.134) (2.135) (2.136) FIGURE 2.14 a\u2020 k+qak\u2212q FIGURE 2.15 2 V \u00f0 q \u00de\u00f0 \u00de a\u2020 k+qa\u2020 p\u2212qapak. p\u2212q ap ak can be 2.4 Quantization of Lattice Waves 61 2.4 QUANTIZATION OF LATTICE WAVES 2.4.1 Formulation We consider a cubic lattice of lattice constant a. We assume that at each lattice site, we have atoms with mass M vibrating around this site. The Hamiltonian that describes this system of atoms is p2 i 2M di \u2212 dj, \u00de (2.137) \u00f0 di \u2212 dj where V is the interaction potential energy between atoms at the sites di and dj: Here, di = Ri + ui and dj = Rj + uj, where ui and uj are the displacements of the atoms from the lattice sites located at Ri and Rj: Figure 2.16 shows these displacements, for convenience, on", " the surface of a cubic lattice. \u00de For small ui and uj ui and uj \u226a a, we have \u00de \u00f0 di \u2212 dj \u2211 & Ri \u2212 Rj ui. + \u2211 i \u2260 j \u00de \u22022V \u2202Ri\u2202Rj. uj: (2.138) The first term on the right side of Eq. (2.138) is a constant term and can be ignored. If we define the tensor $G \u2261 \u22022V \u2202Ri\u2202Rj, Eq. (2.138) can be written in the alternate form di \u2212 dj \u2211 ui. $G. uj: (2.139) (2.140) \u2192 ui \u2192 di \u2192 Ri \u2192 dj \u2192 uj \u2192 Rj 0 FIGURE 2.16 Definitions of di and ui in a cubic lattice of lattice constant a. a a 62 CHAPTER 2 Phonons and Lattice Vibrations Thus, the Hamiltonian can be written as H = \u2211 i P2 i 2M 1 2 + \u2211 i \u2260 j ui. $G. uj = T + V: In Eq. (2.141), From Lagrange\u2019s equation of motion, Pi = M _u i : L = T \u2013 V and we obtain from Eq. (2.141), We make a Fourier transformation, d dt \u2202L \u2202 _ul \u2212 \u2202L \u2202ul = 0, M\u20acul = \u2212\u2211 l\u2032 $Gll\u2032. _ul\u2032: Uq eiq. l, ul = \u2211 q and obtain from Eqs. (2.145) and (2.146), M\u2211 q \u20acUl eiq. l = \u2212\u2211 l\u2032 $Gll\u2032. \u2211 q Uqeiq. l\u2032, which can be written in the alternate form Eq. (2.148) implies If we write Eq. (2.149) can be written as M \u20acUq eiq. l + \u2211 l\u2032 \u2211 q h $Gll\u2032. Uqeiq. l\u2032 = 0: i M \u20acUq + \u2211 l\u2032 $Gll\u2032. Uqeiq. l\u2032\u2212l \u00f0 \u00de = 0: = Uqei\u03c9qt, Uq t \u00f0 \u00de M\u03c92 q ^\u2208q = \u2211 l\u2032 $Gll\u2032. ^\u2208q eiq. \u00f0 l\u2212l\u2032 \u00de: (2.141) (2.142) (2.143) (2.144) (2.145) (2.146) (2.147) (2.148) (2.149) (2.150) (2.151) 2.4 Quantization of Lattice Waves 63 Here, ^\u2208q = Uq Uq j j = unit vector along Uq = polarization vector: ^\u2208q could be along \u03bb = x, y, z directions. Thus, ^\u2208q\u03bb \u21d2 \u03c9q\u03bb: Thus, Eq. (2.151) can be written in the alternate form M\u03c92 q\u03bb ^\u2208q\u03bb = \u2211 l\u2032 $Gll\u2032. ^\u2208q\u03bbeiq. \u00f0 l\u2212l\u2032 \u00de = $G. ^\u2208q\u03bb: q \u00de \u00f0 From Eqs. (2.141) and (2.154), we can write the Hamiltonian in the alternate form We can express where We also write Here, H = \u2211 l P2 l 2m + 1 2 \u2211 l \u2260 l\u2032 ul. $Gll\u2032. ul\u2032: Pl = M N \ufb03\ufb03\ufb03\ufb03\ufb03 r \u2211 k\u03bb ^\u2208k\u03bb Pk\u03bb e\u2212ik. l, ^\u2208kx \u22a5 ^\u2208ky \u22a5 ^\u2208kz: Ukeik. l: ul = \u2211 k Uk = 1 MNp ^\u2208k\u03bb Qk\u03bb, where N is the number of lattice sites (or atoms). From Eqs. (2.157) and (2.158), we obtain \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 From Eq. (2.156), we obtain \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 ul = 1 MNp \u2211 k ^\u2208k\u03bb Qk\u03bb eik. l: P2 l 2M \u2211 l = 1 2N \u2211 k\u03bb,k\u2032\u03bb\u2032, ^\u2208k\u2032\u03bb\u2032", ". ^\u2208k\u03bb Pk\u2032\u03bb\u2032 Pk\u03bb \u2211 l k+k\u2032 e\u2212 \u00f0. l: \u00de Using the relation e\u2212i \u00f0 k+k\u2032. l = N \u03b4k,\u2212k\u2032, \u00de \u2211 l (2.152) (2.153) (2.154) (2.155) (2.156) (2.157) (2.158) (2.159) (2.160) (2.161) 64 CHAPTER 2 Phonons and Lattice Vibrations we can write Eq. (2.160) in the alternate form P2 l 2M \u2211 l = 1 2 \u2211 k,\u03bb,\u03bb\u2032 ^\u2208\u2212k\u03bb\u2032. ^\u2208k\u03bb P\u2212k\u03bb\u2032 Pk\u03bb: Because Further, and we can substitute \u03c9k\u03bb = \u03c9\u2212k\u03bb, ^\u2208k\u03bb\u2032 = ^\u2208\u2212k\u03bb\u2032: ^\u2208k\u03bb\u2032. ^\u2208k\u03bb = \u03b4\u03bb\u03bb\u2032, P\u2212k\u03bb = P\u2020 k\u03bb: Substituting Eq. (2.164) in Eq. (2.162), we obtain P2 l 2M \u2211 l = 1 2 \u2211 k\u03bb P\u2020 k\u03bb Pk\u03bb: We now consider the term (2.162) (2.163) (2.164) (2.165) (2.166) 1 2 \u2211 ll\u2032 ul. $Gll\u2032. ul\u2032 = 1 2MN \u2211 k\u03bb,\u03ba\u2032\u03bb\u2032,ll\u2032 ^\u2208k\u03bb Qk\u03bb. $Gll\u2032. ^\u2208k\u2032\u03bb\u2032Qk\u2032\u03bb\u2032eik. l+ik\u2032. l\u2032: (2.167) We can write eik. l+ik\u2032. l\u2032 = eik\u2032. l\u2032\u2212l \u00de ei \u00f0 \u00f0 k+k\u2032. l: \u00de Eq. (2.154) can be written in the alternate form (by changing q to k\u2019, and \u03bb to \u03bb\u2032) Further, M\u03c92 k\u2032\u03bb\u2032 ^\u2208k\u2032\u03bb\u2032 = \u2211 l\u2032 Gll\u2032. ^\u2208k\u2032\u03bb\u2032 eik\u2032. \u00f0 l\u2032\u2212l \u00de: k+k\u2032 ei \u00f0. l = N\u03b4k,\u2212k\u2032: \u00de \u2211 l From Eqs. (2.167) through (2.170), we obtain 1 2 \u2211 ll\u2032 ul. $Gll\u2032. ul\u2032 = = 1 2 1 2 ^\u2208k\u03bb. ^\u2208k\u2032\u03bb\u2032Qk\u03bbQk\u2032\u03bb\u2032\u03b4k,\u2212k\u2032\u03c92 k\u2032\u03bb\u2032, \u2211 k\u03bb,k\u2032\u03bb\u2032 ^\u2208k\u03bb. ^\u2208\u2212k\u03bb\u2032Qk\u03bbQ\u2212k\u03bb\u2032\u03c92 \u2212k\u03bb\u2032: \u2211 k,\u03bb,\u03bb\u2032 (2.168) (2.169) (2.170) (2.171) (2.172) 2.4 Quantization of Lattice Waves 65 We use the identities in Eq. (2.172), and obtain ^\u2208\u2212k\u03bb\u2032 = ^\u2208k\u03bb\u2032 ^\u2208k\u03bb. ^\u2208k\u03bb\u2032 = \u03b4\u03bb\u03bb\u2032 \u03c9\u2212k\u03bb = \u03c9k\u03bb, 1 2 \u2211 ll\u2032 ul. $Gll\u2032. ul\u2032 = 1 2 \u2211 k\u03bb \u03c92 k\u03bb Qk\u03bbQ\u2212k\u03bb = 1 2 \u2211 k\u03bb k\u03bb Q\u2020 \u03c92 k\u03bbQk\u03bb: From Eqs. (2.155), (2.166), and (2.174), we obtain H = 1 2 \u2211 k\u03bb P\u2020 k\u03bb Pk\u03bb + 1 2 \u2211 k\u03bb k\u03bb Q\u2020 \u03c92 k\u03bb Qk\u03bb: In Eq. (2.175), we have expressed the Hamiltonian in normal coordinates Pk\u03bb and Qk\u03bb: (2.173) (2.174) (2.175) 2.4.2 Quantization of Lattice Waves The quantization of lattice waves starts from the fundamental ideas of quantum mechanics. If we define A, B $ \u00bd = AB \u2212 BA, Pl = l, Py Px l, Pz l \u00de, \u00f0 ul = l, uy ux l, uz l \u00de, \u00f0 and we have the commutation relations l, u\u03b2 P\u03b1 l\u2032$ \u00bd = \u0127 i \u03b4ll\u2032\u03b4\u03b1\u03b2, \u03b1, \u03b2 = x, y, z l, u\u03b2 u\u03b1 l\u2032$ \u00bd Introducing the creation", " and annihilation operators, a\u2020 l, P\u03b2 P\u03b1 l\u2032$ \u00bd = = 0: k\u03bb and ak\u03bb, and substituting and Pk\u03bb = \u0127\u03c9k\u03bb 2 # 1 2 $ a\u2020 k\u03bb \u2212 a\u2212k\u03bb i \u00f0 \u00de Qk\u03bb = \u0127 2\u03c9k\u03bb # 1 2 \u00f0 $ ak\u03bb + a\u2020 \u2212k\u03bb\u00de, (2.176) (2.177) (2.178) (2.179) (2.180) (2.181) (2.182) 66 CHAPTER 2 Phonons and Lattice Vibrations and using the commutation relations from Problem 2.9, we obtain and ak\u03bb, a\u2020 \u00bd k\u2032\u03bb\u2032$ = \u03b4kk\u2032\u03b4\u03bb\u03bb\u2032: ak\u03bb, ak\u2032\u03bb\u2032$ \u00bd = k\u03bb, a\u2020 a\u2020 \u00bd k\u2032\u03bb\u2032$ = 0: (2.183) (2.184) These operators satisfy the commutation relations for bosons. Substituting Eqs. (2.181) and (2.182) in Eq. (2.175), we obtain (Problem 2.10) H = \u2211 k\u03bb \u0127\u03c9k\u03bb a\u2020 & k\u03bb ak\u03bb + 1 2 \u2019 : (2.185) We denote a\u2020 \u03c9k\u03bb is the vibrational frequency in the state k and polarization \u03bb. The term 1 zero-point vibrational energy. The other definitions are k\u03bbak\u03bb = nk\u03bb, as the occupation number operator for phonons with energy \u0127\u03c9k\u03bb: Here, \u0127\u03c9k\u03bb in H is the 2 \u2211 k\u03bb nk> = the state with nk phonons, nk> = nk> = nk>, j nkp nk\u22121>, p nk> = j 0> = 0, no phonons to destroy: nk + 1>, j nk + 1 \ufb03\ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 j j a\u2020 kak ak j a\u2020 k j ak j (2.186) PROBLEMS 2.1. Derive Eq. (2.6) from Eqs. (2.2), (2.3), and (2.5). 2.2. Derive Eq. (2.21) from Eq. (2.20). 2.3. Derive Eq. (2.28) from Eq. (2.27). 2.4. Derive Eq. (2.37) from Eq. (2.36). 2.5. The general expression for specific heat of solids is given in Eq. (2.52): Cv = \u2202 \u2202T \u2211 s Z dq 3 2\u03c0 \u0127\u03c9s e\u03b2\u0127\u03c9s q \u00f0 \u00de q \u00de\u22121 \u00f0, (1) \u00f0 where the integration is over the first Brillouin zone. Show that at very low temperatures, where the optical modes can be neglected, the three acoustic branches can be written as \u03c9 = \u03c9s cs integration over the first Brillouin zone can be replaced by an integration over all space, = are the long-wavelength phase velocities of the acoustic mode), and the k-space q (cs ^q ^q q \u00de \u00f0 \u00de \u00de \u00de \u00f0 \u00f0 Cv = \u2202 \u2202T \u2211 s Z dq 2\u03c0 \u00f0 3 \u00de \u0127cs e\u03b2\u0127cs ^q q \u00f0 \u00de q\u22121 ^q \u00f0 \u00de : (2) Problems 67 In spherical coordinates, dq = q2dq d\u03a9: By writing z \u2261 \u03b2\u0127cs written as \u00f0 ^q \u00de q, show that Eq. (2) can be Cv = \u2202 \u2202T 4 \u00de 3 kBT \u00f0 \u0127s \u00f0 \u00de 3 2\u03c02 \u221e Z 0 z3dz ez \u2212 1, where 1 s3 is the average of the third power of the long-wavelength phase velocities, Using the result of the integral, 1 s3 = 1 3 \u2211 s d\u03a9 4\u03c0 3: 1 ^q \u00f0 \u00de cs \u221e Z 0 x3dx ex\u22121 = \u03c04 15, show from Eqs. (3) and (5) that at very low temperatures, Cv = 2\u03c02 5 kB 1 \u03b2\u0127c 3 : $ # 2.6. The lattice specific heat is of the form Eq. (2.", "52) or Eq. (1) in Problem 2.5, Cv = \u2202 \u2202T \u2211 s Z dq 2\u03c0 \u00f0 \u00de 3 F \u03c9s : q \u00f0 \u00de\u00de \u00f0 Show that Eq. (1) can be written in the alternate form Cv = \u2202 \u2202 d\u03c9, \u03c9 where D \u00f0 levels. Hence, D \u00f0 divided by its volume. Show that \u03c9 \u00de is the density of normal modes per unit volume, i.e., the phonon density of d\u03c9 is the total number of modes between \u03c9 and \u03c9 + d\u03c9 in the crystal \u00de It can also be shown that D \u00f0 \u03c9 \u00de = \u2211 s Z dq 2\u03c0 \u00f0 \u00de \u03c9 \u2212 \u03c9s 3 \u03b4 \u00f0 : q \u00f0 \u00de\u00de \u03c9 D \u00f0 \u00de = \u2211 s Z d\u03a9 2\u03c0 \u00f0 3 \u00de 1 \u2207\u03c9s \u00f0 j q \u00dej, where the integral is over the surface of the first Brillouin zone on which \u03c9s \u2261 \u03c9: q \u00f0 \u00de (3) (4) (5) (6) (1) (2) (3) (4) 68 CHAPTER 2 Phonons and Lattice Vibrations 2.7. In the Debye approximation, all the wave vectors of the normal modes lie within a sphere of radius qD, and in all three branches of the spectrum the linear dispersion relation \u03c9 = cq: Show by using Eq. (3) of Problem 2.6 that the density of normal modes, D, is given by \u03c9 \u00f0 \u00de D \u00f0 \u03c9 \u00de = 3 2\u03c02 ) = 0, \u03c92 s3, \u03c9 = \u03c9D = qDcs \u00f0 \u03c9 > \u03c9D: q, \u00de * 2.8. Derive Eq. (2.145) from Eqs. (2.141), (2.143), and (2.144). 2.9. Using the expressions and Pl = M N \ufb03\ufb03\ufb03\ufb03\ufb03 \u2211 k\u03bb r ^\u2208k\u03bb Pk\u03bb e\u2212k. l Ql = 1 MNp \u2211 k\u03bb ^\u2208k\u03bb Qk\u03bb eik. l, show that \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 and Pk\u03bb, Qk\u2032\u03bb\u2032$ \u00bd = \u0127 i \u03b4kk\u2032 \u03b4\u03bb\u03bb\u2032, Pk\u03bb, Pk\u2032\u03bb\u2032$ \u00bd = Qk\u03bb, Qk\u2032\u03bb\u2032$ \u00bd \u2261 0: 2.10. Substituting Eqs. (2.181) and (2.182) in Eq. (2.175), show that H = \u2211 k\u03bb k\u03bb ak\u03bb + \u0127\u03c9k\u03bb a\u2020 & 1 2 : \u2019 (1) (2) (3) (4) (1) References 1. Ashcroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 2. Born M, Huang K. The dynamical theory of crystal lattices. London: Oxford University Press; 1954. 3. Callaway J. Quantum theory of the solid state. New York: Academic Press; 1976. 4. Carruthers P. Theory of thermal conductivity of solids at low temperatures. Rev Mod Phys 1961;33:92. 5. Kittel C. Quantum theory of solids. New York: John Wiley & Sons; 1963. 6. Kittel C. Introduction to solid state physics. New York: John Wiley & Sons; 1976. 7. Madelung O. Introduction to solid state theory. New York: Springer-Verlag; 1978. 8. Mahan GD. Many-particle physics. New York: Plenum Press; 1986. 9. Maradudin A. Lattice dynamical aspects of the resonance absorption of gamma rays by nuclei bound in a crystal. Rev Mod Phys 1964;36:417. References 69 10. Maradudin A, Montroll EW, Weiss GH, Ipatova IP. Theory of lattice dynamics in the harmonic approxi- mation, solid state physics, suppl. 3. New York: Academic Press; 1971. 11. Marder MP. Condensed matter physics. New York: John Wiley & Sons; 2000. 12. Messiah A. Quantum mechanics, vol. 1. Amsterdam: Elsevier; 1961. 13. Peierls RE. Quantum theory of solids. London: Oxford University Press; 1955. 14. Schewber S. An introduction to relativistic quantum field theory. New York: Harper; 1961.", " 15. Symon KR. Mechanics. Reading, MA: Addison-Wesley; 1971. 16. Wannier G. Elements of solid state theory. London: Cambridge University Press; 1959. 17. Ziman JM. Principles of the theory of solids. Cambridge: Cambridge University Press; 1972. Free Electron Model CHAPTER 3 CHAPTER OUTLINE 3.1 The Classical (Drude) Model of a Metal....................................................... 71 3.2 Sommerfeld Model......................................................................... 73 3.2.1 Introduction......................................................................... 73 3.2.2 Fermi Distribution Function........................................................... 74 3.2.3 Free Electron Fermi Gas.............................................................. 77 3.2.4 Ground-State Energy of the Electron Gas............................................... 79 3.2.5 Density of Electron States............................................................ 81 3.3 Fermi Energy and the Chemical Potential..................................................... 82 3.4 Specific Heat of the Electron Gas............................................................ 84 3.5 DC Electrical Conductivity................................................................... 86 3.6 The Hall Effect............................................................................ 87 3.7 Failures of the Free Electron Model.......................................................... 89 Problems...........................................................................", ".......... 90 References.................................................................................... 93 3.1 THE CLASSICAL (DRUDE) MODEL OF A METAL Around 1900, Drude developed the theory of electrical and thermal conductivity of metals by considering a metal as a classical gas of electrons, the properties of which were governed by Maxwell\u2013 Boltzmann statistics. According to his model, the electrons were wandering around in a metal with a background of immobile but heavy positive charges. A sketch of his model is shown in Figure 3.1. Drude had no idea as to the nature or origin of these positive charges, although he recognized that in order to be electrically neutral, a metal would have to have an equal number of positive and negative charges. Later, these heavy positive immobile charges were identified as the ions (constituted of the nucleus and the surrounding core electrons) of the neutrally charged atom stripped of its valence electron(s). The conduction electrons in a metal were the valence electrons stripped from their parent atoms because of the strong attractive interaction between the valence electron(s) of an atom and the positively charged ions surrounding it. In what follows, we will treat these positive charges as the ions described previously, although this concept was introduced by Bohr in his model of the atom long after Drude formulated a theory of metals. This concept of a sea of valence electrons moving around in a background of static positively charged ions and holding them together like a glue was in a sense an extension of the model of a hydrogen molecule in which the Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00003-7 \u00a9 2012 by Elsevier Inc. All rights reserved. 71 72 CHAPTER 3 Free Electron Model + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + FIGURE 3.1 Drude model of scattering of electrons in a metal. Isolated Na atom Two Na atoms Isolated Na atom FIGURE 3.2a A system of two sodium atoms. y g r e n E Separation FIGURE 3.2b two valence electrons were stripped from their parent atoms and moving around the positively charged protons, thereby forming a strong bond. An example of formation of bands in a sodium atom is shown Figure 3.2. Figure 3.2a shows a system of two sodium atoms. The two 3s electrons occupy a lower energy level than the isolated atoms. Figure 3.2b shows the energy levels of the outer 3s electrons in a pair of sodium atoms as a function of the separation between the two atoms. This figure shows how the energy level widens with smaller distance. The energy levels of two 3s electrons of sodium atoms as function of separation. When there are N sodium atoms, where N is a very large number, these energy levels form bands. In Figure 3.2c, we show the 1s, 2s, 2p, and 3s energy bands of sodium. However, the bands become wider for electrons that are less tightly bound to the parent nucleus. We note the significant difference between the 1s and the 3s bands. Drude introduced the concept of electron density n = N/V, where V is the volume of the metal and N is the number of atoms per mole (the Avogadro\u2019s number). He also introduced the electron density rs, which is defined as the radius of a sphere of which the volume is equal to the available volume for each conduction electron. According to his model, 1 n = 4\u03c0r3 s 3 = V N, (3.1) 3.2 Sommerfeld Model 73 from which it follows that 3s rs = 1/3 : (3.2) 3 4\u03c0n! \" 2s 1s 2p y g r e n E a0 Separation Drude applied the kinetic theory of gas to these conduction electrons even though their electron densities are approximately a thousand times greater than those of a classical gas at normal pressure and temperature. He introduced several interesting approximations (in his kinetic theory model) to develop theories of both electrical and thermal conductivity of metals. First, the electron gas was assumed to be free and independent in the sense that there was no electron\u2013electron or electron\u2013ion interactions. An electron moved in a straight line (as per Newton\u2019s law) in the absence of an external electric field. Second, when an external field is applied, the electron continues", " to move in a straight line between collisions with the ions (the collisions with other electrons were neglected in his theory). The velocity of an electron would change immediately after a collision, which was assumed to be an instantaneous event. Third, the time between two collisions was the same value \u03c4 (often known as the relaxation time) for each electron. Fourth, immediately after each collision, an electron\u2019s velocity was not related to the velocity with which it was traveling before the collision. The new speed of the electron depended on the temperature at the time of collision, and the direction of the new velocity was randomly directed. In a sense, the electron achieved thermal equilibrium with its surroundings. The formation of 1s, 2s, 2p, and 3s energy bands in sodium. Note that 1s is in the lowest energy state, whereas 3s is in the highest energy state. FIGURE 3.2c The two major achievements of Drude\u2019s model were the derivation of electrical and thermal conductivity (see problems) in a very simplistic way. However, it had several drawbacks. Some of these were solved by Sommerfeld, who recognized the fact the free electrons in a metal were not classical but quantum gas, for which the application of Fermi\u2013Dirac statistics (instead of Maxwell\u2013 Boltzmann statistics) was appropriate. 3.2 SOMMERFELD MODEL 3.2.1 Introduction The major failure of Drude\u2019s theory was the prediction that the specific heat of a metal was Cv = 3/2NkB, where N is the number of electrons in a metal and kB is the Boltzmann constant. In practice, the specific heat of a metal was more than 100 times smaller than this value at room temperature. This anomaly was explained by Sommerfeld only after the introduction of quantum mechanics. The Pauli exclusion principle requires that no two electrons can have identical quantum numbers, which implies that a quantum state can be occupied by at most one electron. Sommerfeld recognized the significance of the Pauli exclusion principle, and as a consequence, the Fermi\u2013Dirac statistics (instead 74 CHAPTER 3 Free Electron Model of the classical Maxwell\u2013Boltzmann statistics) were applicable to a system of identical electron gas (or other fermions) in thermal equilibrium. Sommerfeld essentially adapted Drude\u2019s model of the metal as a free electron gas moving in a background of static positively charged ions but modified it by using the Fermi\u2013Dirac distribution function for the electron velocity. In the process, he was able to explain the (apparently) anomalous low specific heat of the electron gas as well as other thermal properties. 3.2.2 Fermi Distribution Function (a) Grand Canonical Ensemble The Fermi distribution function is easily derived from the concept of a Grand Canonical Ensemble, which is an ensemble of \u2115 identical systems (labeled as 1, 2, \u2026, \u2115) mutually sharing a total number of particles \u2115N and a total amount of energy \u2115E. If nr, s denotes the number of systems that have, at any time t, the number \u2115r of particles and the amount of energy \u0395s(r, s = 0, 1, 2, \u2026), nr,s = \u2115, \u2211 r,s nr,s Nr = \u2115N, \u2211 r,s nr,s Es = \u2115E: \u2211 r,s (3.3) (3.4) (3.5) We consider this ensemble of \u2115 identical systems that are characterized by a Hamiltonian operator ^\u0397. At any time t, the physical states of the various systems in the ensemble are characterized by the denote the normalized wave function characterizing the physical wave functions \u03c8 state in which the kth state of the system of the ensemble happens to be at the time. We can write. Let \u03c8 k r, t \u00f0 r, t \u00de \u00de \u00f0 = i!h _\u03c8 k \u00f0 We introduce a complete set of orthonormal functions \u03d5n\u00f0 \u00de r, t \u00f0 ak n\u00f0 ^\u0397\u03c8 k = \u2211 n \u03c8 k r, t \u00de \u00de \u00f0 t r \u00de \u03d5n\u00f0 r, t : \u00de \u2261 \u03d5n\u00f0 r : \u00de r1, r2, \u2026, rN \u00de and express (3.6) (3.7) We can write t = ak n\u00f0 \u00de where d\u03c4 is the volume element of the coordinate space of the system. Thus, ak are the probability t n\u00f0 \u00de 2 represents ak, and amplitudes for the various systems of the ensemble to be in the states \u03d5n\u00f0 n\u00f0 the probability that a measurement of time t finds the kth system of the ensemble to be in", " the particular state \u03d5n\u00f0. Thus, we obtain (3.8) \u03d5n# \u03c8 k d\u03c4, r, t \u00dej ak n\u00f0 \u00dej 2 = 1 for all k: (3.9) Eq. (3.9) can also be derived from Eq. (3.7) in a straightforward manner. From Eq. (3.7), we obtain \u03c8 k r, t \u00f0 2 d\u03c4 = \u2211 n,m \u00dej j Z = \u2211 n,m ak #n \u00f0 ak #n \u00f0 t \u00de t \u00de ak t m\u00f0 ak m\u00f0 t \u03d5n# r \u00de \u00f0 \u00de Z \u03b4n,m = \u2211 \u00de r \u03d5m\u00f0 ak n\u00f0 n j d\u03c4 t \u00dej 2 = 1: (3.10) \u00de 3.2 Sommerfeld Model 75 (b) Density Operator The density operator ^\u03c1 t \u00f0 \u00de is defined by the matrix elements t \u03c1mn\u00f0 \u00de = 1 \u2115 \u2115 \u2211 k =1 ak t m\u00f0 ak #n \u00f0 \u00de t \u00de : (3.11) The diagonal element \u03c1nn\u00f0 2, represents the probability of a system, chosen at random from the ensemble at time t, and is found to be in the state \u03d5n. From Eqs. (3.10) and (3.11), we obtain, which is the ensemble average of the probability \u00de ak t n\u00f0 \u00dej t j # $ One can show that if the system is in a state of equilibrium,8 \u03c1nn = 1: \u2211 n and tr where < ^G> is the expectation value of a physical quantity G. i!h _^\u03c1 = ^H, ^\u03c1 \u00bd < ^G> = tr = 0, % ^\u03c1 ^G \u00de, \u00f0 = 1, ^\u03c1 \u00f0 \u00de In the Grand Canonical Ensemble, the density operator ^\u03c1 operates on a Hilbert space in an indefinite number of particles. Therefore, the density operator commutes not only with the Hamiltonian operator ^H, but also with a number operator ^n of which the eigenvalues are 0, 1, 2. We can write8 ^\u03c1 = 1 \u03bc, V, T \u00f0 \u03bc, V, T Z \u00de is defined as e\u2212\u03b2 ^H\u2212\u03bc^n \u00f0 \u00de, where the Grand partition function Z \u00f0 \u03bc, V, T Z \u00f0 = \u2211 r, s \u00de The chemical potential \u03bc is defined as \u00de e\u2212\u03b2 Er\u2212\u03bcNs \u00f0 \u00de = tr ^H\u2212\u03bc^n e\u2212\u03b2 \u00f0 : \u00de n o where F is the Helmholtz free energy defined by the relation \u03bc = \u2202F \u2202N! \", T,V,N Here, E is the energy, S is the entropy, T is the absolute temperature, and V is the volume. For a free electron gas, F = E \u2212 TS: and Therefore, ^H = \u2211 k k \u03b5 \u00f0 \u00de c\u2020 k\u03c3ck\u03c3 Ns = 0 or 1: Er = Ns \u03b5 k \u00de = 0 or \u03b5 \u00f0 \u00f0 k : \u00de (3.12) (3.13) (3.14) (3.15) (3.16) (3.17) (3.18) (3.19) (3.20) (3.21) (3.22) 76 CHAPTER 3 Free Electron Model From Eqs. (3.17) and (3.20) through (3.22), we obtain (omitting spin [\u03c3]), and The average number in state k is Zk = 1 + e\u2212\u03b2 \u03b5 \u00f0 \u00f0 k \u2212\u03bc \u00de \u00de ^\u03c1k = 1 Zk e\u2212\u03b2 \u03b5 k \u2212\u03bc \u00de \u00de \u00f0 \u00f0 c\u2020 kck : (3.23) (3.24) nk = tr nk ^\u03c1kg f \u03b5 = e\u2212\u03b2 \u00f0 \u00f0 1 + e\u2212\u03b2 k \u00de \u03b5 \u00f0 = \u2211 nk < nk nk ^\u03c1k j j nk > = <1 1> ^\u03c1k j j \u2212\u03bc k \u00de \u2212\u03bc = 1 \u22123.25) e\u03b2 \u03b5 k \u00f0 \u03b5, T \u00de + 1 is the Fermi distribution, which is a function of energy and temperature. The Fermi Here, f k \u00de \u00f0 gives the probability that a state at energy \u03b5i is occupied in an ideal distribution function f electron gas in thermal equilibrium. We also note that the Fermi distribution is an eigenvalue of the statistical operator ^\u03c1k In", " practice, because the electrons in an electron gas are indistinguishable and are constantly moving around, the probability that a state is occupied by an electron is the average number of electrons in the state and \u03b5i f \u2211 i \u00f0 \u00de = N, (3.26) where N is the total number of electrons in the free electron gas (note that we do not include the core electrons of the atoms that are bound to the parent nuclei and are not free to move around). In Eq. (3.26), \u03b5i are the energy levels. In semiconductor physics, \u03bc is also called the Fermi level. The Fermi energy \u03b5F is defined as the energy of the topmost filled level of the electron states at T = 0. All the electron states having energy greater than the Fermi energy are empty (there are no electrons in these states) at T = 0. From Eq. (3.26), we note that the alternate definition of the Fermi energy at T = 0 is We also note from Eq. (3.23) that f \u03b5i \u00f0 \u00de = 1, \u03b5i \u2264 \u03b5F, = 0, \u03b5i > \u03b5F: lim 0 T! f \u03b5i \u00f0 \u00de = 1, \u03b5i \u2264 \u03bc, = 0, \u03b5i > \u03bc: (3.27) (3.28) Comparing Eqs. (3.27) and (3.28), we note that at T = 0, \u03b5F = \u03bc: Thus, in the limit T 0, when \u03b5i = \u03b5F = \u03bc, the Fermi\u2013Dirac distribution function changes abruptly from 1 to 0. At higher tempera= 1/2 when \u03b5i = \u03bc. The distribution of the Fermi distribution function and its tures (T > 0), f derivative at T = 0 and at finite temperature is shown in Figure 3.3.! \u03b5i \u00de \u00f0 From Figure 3.3, we note that the Fermi distribution function at a large temperature T is significantly different from that at zero temperature. When the metal is heated from T = 0, electrons are transferred from the region \u03b5/\u03bc < 1 to the region \u03b5/\u03bc > 1: The rest of the electrons deep inside the 3.2 Sommerfeld Model 77 Fermi level are not affected when the electron gas is heated. However, as the temperature increases, there is a significant change in the Fermi distribution function. We will discuss this in Section 3.3. On the other hand, each electron in a classical electron gas would gain energy when the system is heated. This immediately explains the anomalous results for the specific heat of the electron gas. As we will see, this property of the fermions is very significant in the derivation of other physical quantities of metals. \u2212\u2202 f \u00b0/\u2202\u03b5 T = 0 f \u00b0(\u03b5) \u03b5 \u03b5F kT FIGURE 3.3 3.2.3 Free Electron Fermi Gas By definition, a free electron gas consists of N independent electrons confined in a volume V. There are positively charged static ions inside the metal to keep it electrically neutral. For simplicity, we will consider this volume to be a cube of side L. An electron is completely free to move inside this cube. There are no other interactions, either with other electrons or with the positively charged background (the lattice). Because there is no potential energy term, the Schrodinger equation for an electron in state k can be written as Fermi distribution function and its derivative at T = 0 and at a finite temperature. \u2212!h2 2m \u22072\u03c8 k\u00f0 r \u00de = \u03b5 k r \u03c8 k\u00f0 \u00de \u00de : \u00f0 (3.29) r Here, \u03c8 k\u00f0 ized solution of Eq. (3.29) is easily obtained, by requiring \u03c8 k\u00f0 cube (the particle in a box problem in quantum mechanics), as is the wave function and \u03b5(k) is the energy of the electron in the state k. The normalto vanish at the surface of the r \u00de \u00de r \u03c8 k\u00f0 = 8/V 1/2 sin \u00de n1\u03c0x/L sin n2\u03c0y/L sin n3\u03c0z/L, (3.30) \u00f0 \u00de \u00f0 \u00f0 \u00de where L is the side of the cubic metal of volume V(V = L3) and n1, n2, and n3 are positive integers. For simplicity, we have considered the metal as a cube, although we could have chosen a parallelepiped, which would yield the same results but the derivation would have been more complicated. However, we note that the solutions obtained in Eq. (3.30) are", " standing-wave solutions and the 2 of the electron in k state varies with its position in real space, thereby probability density yielding an unrealistic picture for free electrons. To obtain more realistic solutions for free electron gas, we introduced the Born\u2013von Karman (or periodic) boundary conditions. These conditions are \u03c8 k\u00f0 \u00dej r \u00de \u00f0 \u00de j x + L, y, z \u00f0 \u00de x, y + L, z \u00de \u00f0 x, y, z + L \u00de \u00f0 These periodic boundary conditions were imposed by imagining that each face of the cube is joined to the face opposite it. When an electron meets the surface of the metal, instead of being reflected x, y, z \u00f0 x, y, z \u00f0 x, y3.31) \u03c8 \u03c8 \u03c8 78 CHAPTER 3 Free Electron Model by the surface, it emerges at an equivalent point on the opposite surface. This is obviously an improbable situation in a three-dimensional solid, but the periodic boundary conditions yield free electron wave functions for the free electron gas. It may be noted that these approximations hold good as long as one is not considering the physical properties of the electrons on or close to the surface, in which case they have to be modified. It may be further noted that it is not necessary to consider the metal as a cube of side L; the results would still hold good for a parallelepiped. The general solutions of Eq. (3.29) are the well-known plane waves r \u03c8 k\u00f0 \u00de = eik. r: 1 Vp (3.32) In general, k is a continuous variable in the reciprocal space. However, if we invoke the periodic boundary conditions as stated in Eq. (3.31), we obtain only discrete values of k, i.e., \ufb03\ufb03\ufb03\ufb03 Eq. (3.33) implies that kx, ky, and kz are discrete variables, eikxL = eikyL = eikzL = 1: kx = 2\u03c0n1 L, ky = 2\u03c0n2 L, kz = 2\u03c0n3 L : (3.33) (3.34) Here, n1, n2, and n3 are integers (zero, positive, or negative). Thus, the components of the wave vector k are discrete in the k-space and, along with the spin components ms, are the quantum numbers of the problem. We also obtain from Eqs. (3.29) and (3.32), =!h2k2 2m : k \u03b5 \u00f0 \u00de The plane wave is also an eigenfunction of the linear momentum p, r p\u03c8 k\u00f0 \u00de = \u2212i!h\u2207 1 Vp eik. r =!hk\u03c8 k\u00f0 r : \u00de (3.35) (3.36) Thus, the electron velocity v in the state k is \ufb03\ufb03\ufb03\ufb03 given by ky FIGURE 3.4 Points in a two-dimensional k-space. kx 2\u03c0 L!hk=m: v \u00bc (3.37) We can now use Eq. (3.34) to build the k-space. The k-space is a three-dimensional space with an extremely large number (N) of allowed discrete k points. Figure 3.4 illustrates the points in a two-dimensional k-space that are separated by 2\u03c0 L in both kx and ky directions (as per Eq. 3.34). As seen in Figure 3.4, the discrete k values are separated by 2\u03c0/L in each dimension. In a three-dimensional k-space, the volume per each k-point is (2\u03c0/L)3 = 8\u03c03/V, where the volume of the metal V = L3. Therefore, a k-space of volume \u03a9 will contain \u03a9V/8\u03c03 allowed values of k. 3.2 Sommerfeld Model 79 kz Thus, the density of states, \u03c1, in the k-space (number of allowed k-values per unit volume of k-space) is \u03c1 = V/8\u03c03: (3.38) kx ky kF FIGURE 3.5 Schematic diagram of the Fermi sphere. We note that each k state can contain exactly two electrons (spin up or spin down) at zero temperature as per the Fermi\u2013Dirac distribution function in Eq. (3.25). Thus, when we build the filled k-space at zero temperature in three dimensions, we are essentially filling up these discrete k states starting with the lowest energy, which is zero. Because the k states are densely packed and the total number of electrons is N, which is a huge", " number (~1023), the constant energy levels of the electrons can be considered as the surface of a sphere (as per Eq. 3.33). Thus, all the N electrons will fill a sphere of volume \u03a9 = 4\u03c0/3 k3 F, which is known as the Fermi sphere. kF, the radius of the sphere, is known as the Fermi wave vector. The surface of the Fermi sphere that separates the filled k states from the empty k states (at zero temperature) is known as the Fermi surface. The energy of an electron on the Fermi surface is called the Fermi energy \u03b5F =!h2k2 2m, and the momentum pF =!hkF is known as the Fermi momentum. A schematic diagram of the Fermi sphere is shown in Figure 3.5. F The volume of the Fermi sphere is \u03a9 = 4\u03c0k3 F/3 and, as shown earlier, a k-space of volume \u03a9 will contain \u03a9V/8\u03c03 allowed values of k. Hence, the allowed values of k in the Fermi sphere are \u03a9V 8\u03c03 = V 8\u03c03 4\u03c0k3 F 3 = k3 FV 6\u03c02 : (3.39) & Because there are two electrons per each k state and the total number of electrons is N, we obtain from Eq. (3.39) \u2019 & \u2019 N 2 = k3 FV 6\u03c02 : The electron density n \u2261 N/V is easily obtained from Eq. (3.40), n = k3 F 3\u03c02 : (3.40) (3.41) 3.2.4 Ground-State Energy of the Electron Gas To make a summation over allowed values of k, we use the following procedure. If F(k) is a function of k states, we can write, using Eq. (3.38), \u2211 k F k \u00de \u00f0 = V 8\u03c03 \u2211 k F k \u00f0 \u00de \u0394k: (3.42) 80 CHAPTER 3 Free Electron Model In the limit V \u2192 \u221e, \u0394k k F integral dk. Thus, we can write Eq. (3.42) in the alternate form 0 and the summation over k on the right side, \u2211 k! \u00f0 \u00de R lim 8\u03c03 Z F k \u00f0 dk: \u00de F k \u00f0 \u0394k approaches the \u00de (3.43) In general, the approximation V \u2192 \u221e is fairly good because the volume of a metal is infinitely large compared to the volume of a primitive cell. The ground-state energy of N free electrons in a metal is calculated by adding the energies of the k states inside the Fermi sphere and multiplying by 2 (because each k state has two electrons of opposite spin), From Eqs. (3.35) and (3.44), From Eqs. (3.43) and (3.45), E = 2\u2264 kF!h2k2 2m : E = V 4\u03c03 Zk\u2264 kF dk!h2k2 2m : Because dk = 4\u03c0k2dk, Eq. (3.46) can be easily integrated, and we obtain E = V!h2k5 F 10\u03c02m : (3.44) (3.45) (3.46) (3.47) Here, E is the total energy of the electron gas. To find the energy per electron, E/N, we obtain from Eqs. (3.40) and (3.47), which can be written in the alternate form E N = 3!h2k2 F 10m, E N = 3 5 \u03b5F: (3.48) (3.49) The average energy of an electron at zero temperature, derived in Eq. (3.49), differs significantly from the energy of a classical ideal gas, which is zero at T = 0. This significant difference is due to the fact that the electrons obey the Pauli exclusion principle, and therefore, the Fermi\u2013Dirac distribution function has to be applied to the electron gas instead of the Maxwell\u2013Boltzmann distribution function, which is applicable for a classical ideal gas. We now define the Fermi temperature using the relation \u03b5F = kBTF: (3.50) 3.2 Sommerfeld Model 81 From Eqs. (3.49) and (3.50), we obtain TF = 5E 3NkB : (3.51) The Fermi temperature TF is of the order of 104 K for simple metals (see Problem 3.7), an incredibly large value for the temperature of the electrons on the Fermi surface. We note that for", " a classical ideal gas, the temperature T = 0 irrespective of the location of the gas molecule. The wavelength of an electron at the Fermi surface is given by \u03bbF = 2\u03c0 kF : (3.52) 3.2.5 Density of Electron States Because the k states are discrete variables, normally one makes a summation over all the k states. In Eq. (3.43), we approximated the summation of a function F(k) over all the k states as an integration over the k states. An alternate way to consider the summation over all the k states is to consider the same summation as integration over the energy levels. Such integrations over the energy levels are done by using the concept of the density of electron states. As an example, we consider the summation \u2211 k F \u03b5 k \u00f0 \u00f0 \u00de\u00de, where F is a function of the k \u03b5 \u00f0 \u00f0 \u00de\u00de energy of the electrons in the filled k states. We can write Eq. (3.43) in the alternate form 1 V \u2211 k F \u03b5 \u00f0 k \u00f0 \u00de\u00de = 1 4\u03c03 Z dkF k \u03b5 \u00f0 \u00f0 \u00de\u00de : (3.53) We note that because each k state has two electrons with the same energy, we have multiplied the right side of Eq. (3.43) by a factor of 2. Because \u03b5 =!h2k2/2m, we can write k \u00f0 dk = 4\u03c0k2dk: \u00de From Eqs. (3.53) and (3.54), we obtain Here, 1 4\u03c03 Z dkF k \u03b5 \u00f0 \u00f0 \u00de\u00de = \u221e Z 0 k2dk \u03c02 F k \u03b5 \u00f0 \u00f0 \u00de\u00de = \u2212\u221e Z \u221e d\u03b5g \u03c02!h2 2m\u03b5, h2 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 r = 0, \u03b5 > 0 \u03b5 \u2264 0: (3.54) (3.55) (3.56) 82 CHAPTER 3 Free Electron Model EF Energy FIGURE 3.6 Density of states versus energy.0 0. EF Energy (a) EF Energy (b) EF Energy (c) FIGURE 3.8 The density of occupied states (c) obtained from the Fermi\u2013Dirac distribution (a) and the density of states (b). Energy FIGURE 3.7 Comparison of density of states for a typical metal (solid line) with the predicted value (dashed curve). Thus, the density of electron states g(\u03b5)d\u03b5 is defined as the number of one-electron levels per unit volume of the metal in the energy range \u03b5 and \u03b5 + d\u03b5. From Eqs. (3.41) and (3.56), we can rewrite g(\u03b5) in the alternate form = g \u03b5 \u00de \u00f0 3n 2 & 1/2 \u03b5 \u03b53 F \u2019, \u03b5 > 0 = 0, \u03b5 \u2264 0: (3.57) \u03b5 The number of single electron states N is equal to the area under the density of states curve g up to the Fermi energy. The plot of \u03b5 g versus \u03b5 is shown in Figure 3.6. \u00de However, the density of states for a typical metal, which is plotted in Figure 3.7, is different from the curve shown in Figure 3.6. \u00de \u00f0 \u00f0 3.3 FERMI ENERGY AND THE CHEMICAL POTENTIAL We will derive an expression between the chemical potential \u03bc and the Fermi energy \u03b5F at a reasonably low temperature (we note that \u03bc = \u03b5F at T = 0). The electron density n = N/V can be written in the form n = 1 4\u03c03 Z dk f \u03b5 \u00f0 \u00f0 k \u00de\u00de : (3.58) 3.3 Fermi Energy and the Chemical Potential 83 From Eqs. (3.55) and (3.58), we obtain \u221e n = g \u03b5 d\u03b5: (3.59) \u03b5 f \u00f0 \u00de \u00f0 \u00de Z \u2212\u221e Using the Sommerfeld expansion (Problem 3.8) and retaining terms of the order T 2, we obtain n = \u221e Z \u2212\u221e d\u03b5 + g \u03b5 \u00de \u00f0 \u03c02 6 \u00f0 kBT 2g\u2032 \u00de \u03b5F : \u00de \u00f0 From Eqs. (3.57) and (3.60), we obtain \u03bc d\u03b5", " + g \u03b5 \u00de \u00f0 \u03c02 6 \u00f0 kBT 2g\u2032 \u00f0 \u00de \u03b5F : \u00de n = Z 0 Eq. (3.61) can be written in the alternate form \u03bc \u03b5F n = d\u03b5 + g \u03b5 \u00de \u00f0 Z \u03b5F Z 0 d\u03b5 + g \u03b5 \u00f0 \u00de \u03c02 6 \u00f0 kBT 2 \u03b5F g\u2032 \u00f0 \u00de : \u00de We note that at zero temperature, the electron density n can also be expressed as n = \u03b5F Z 0 g \u03b5 \u00f0 d\u03b5: \u00de (3.60) (3.61) (3.62) (3.63) In addition, at low temperatures, g tion of g in this range can be approximated as \u2248 g \u03b5F \u03b5 \u00de \u00f0 \u00de \u00f0 \u03b5 \u00f0 \u00de in the small energy range of \u03b5F to \u03bc and the integra- \u03bc Z \u03b5F d\u03b5 \u2248 g \u03b5 \u00de \u00f0 \u00f0 \u03bc \u2212 \u03b5F \u03b5F g \u00f0 \u00de : \u00de From Eqs. (3.62) through (3.64), we obtain \u03bc \u2212 \u03b5F \u00f0 \u00de g \u00f0 \u03b5F \u00de + \u03c02 6 \u00f0 kBT 2g\u2032 \u00de \u03b5F \u00f0 \u00de = 0: From Eq. (3.65), the chemical potential \u03bc can be expressed as \u03b5F\u00de \u00f0 \u03b5F\u00de \u00f0 \u03bc = \u03b5F \u2212 \u03c02 6 \u00f0 2 g\u2032 g kBT \u00de : From Eqs. (3.64) and (3.66), we obtain \u03bc = \u03b5F 1\u2212 \" \u03c02 12 2 kBT \u03b5F& : # \u2019 (3.64) (3.65) (3.66) (3.67) Eq. (3.67) expresses the chemical potential \u03bc in terms of the Fermi energy \u03b5F (up to the order of T 2). 84 CHAPTER 3 Free Electron Model f (\u03b5) 1.0 0.8 0.6 0.4 0.2 0.0 f (\u03b5): kBT =.25 \u00b5 f (\u03b5): kBT =.005 \u00b5 f (\u03b5): kBT =.025 \u00b5 exp [\u2013\u03b2(\u03b5 \u2013 \u00b5)]: kBT =.25 \u00b5 \u00b5 \u03b5 FIGURE 3.9 Variation of the Fermi\u2013Dirac distribution function with temperature. The variation of the Fermi\u2013Dirac distribution function for various values of kBT is shown in Figure 3.9. The Fermi function differs slightly from a step function around the chemical potential \u03bc. It differs significantly from 0 or 1 in the region in which the width is kBT: For \u03b5 \u03bc, it is hard to \u2019 distinguish the Fermi function from the classical Boltzmann distribution function e\u2212\u03b2\u03b5: 3.4 SPECIFIC HEAT OF THE ELECTRON GAS is the At a finite temperature T, the k states will be filled by 2 f Fermi\u2013Dirac distribution function, which determines the probability that a k state is occupied by an electron. The factor 2 is multiplied because each k state can have two electrons of opposite spin as prescribed by the Pauli exclusion principle. Thus, the total internal energy of the electrons at temperature T is given by electrons, where f k \u00f0 \u03b5 \u00f0 \u00de\u00de \u00de\u00de k \u03b5 \u00f0 \u00f0 The electronic specific heat of the metal is given by Eel T \u00f0 \u00de = 2 \u00de\u00de :! By converting the summation over k states to integration and following the procedure outlined in Eq. (3.53), we can rewrite Eq. (3.68) as Cel = 1 V \u2202Eel \u2202T : N \" From Eqs. (3.55) and (3.70), we obtain Eel V = 1 4\u03c03 dk \u221e Eel V = \u03b5g Z \u2212\u221e d\u03b5: \u03b5 \u00de \u00f0 (3.68) (3.69) (3.70) (3.71) 3.4 Specific Heat of the Electron Gas 85 By using the Sommerfeld expansion (Problem 3.8), we obtain from Eq. (3.71) Eel V = \u03bc Z \u2212\u221e \u03b5g \u03b5 d\u03b5 + \u03c02 6 \u00f0 \u00de \u00f0 kBT \u00de 2 d \u03b5g \u03b5 \u00de\u00de j \u03b5 = \u03bc + higher order terms in T: \u00f0 (3.72) d\u03b5\u00f0 We note from Eq. (3.57) that the density of one-electron levels g tion,", " we rewrite Eq. (3.72) as \u00f0 \u03b5 \u00de = 0 for \u03b5 \u2264 0. Using this condi- Eel V = \u03b5F Z 0 \u03b5g \u03b5 d\u03b5 + \u00de \u00f0 \u03bc Z \u03b5F d\u03b5 + \u03b5g \u03b5 \u00de \u00f0 \u03c02 6 \u00f0 kBT 2 \u00de g \u03bc \u00de \u00f0 + \u03bcg\u2032 \u00f0 \u00bd \u03bc \u00de% + higher order terms in T: (3.73) From Eqs. (3.67), which specifies the relation between \u03bc and \u03b5F, retaining the second-order terms in T, we obtain \u03bc Z \u03b5F \u03b5g \u03b5 \u00de \u00f0 d\u03b5 \u2248 \u03b5F \u03bc \u2212 \u03b5F \u03b5F g \u00f0 \u00de : \u00de \u00f0 (3.74) Substituting Eq. (3.74) in Eq. (3.73) and consistently retaining terms only up to the second order in T g, we obtain and \u03bc \u2248 \u03b5F in Eq: \u2248 g \u03b5F \u03bc \u00f0 \u00de \u00f0 \u00de 3:73 \u00f0 \u00de Eel V = \u03b5F Z 0 d\u03b5 + \u03b5g \u03b5 \u00de \u00f0 \u03c02 6 \u00f0 kBT 2g \u00de \u03b5F \u00f0 \u00de + \u03b5F \u03bc \u2212 \u03b5F \u03b5F g \u00f0 \u00de \u00de + \u00f0 ( \u03c02 6 \u00f0 kBT 2g\u2032 \u00de \u03b5F \u00f0 \u00de : ) (3.75) From Eqs. (3.65) and (3.75), Eel V = \u03b5F Z 0 d\u03b5 + \u03b5g \u03b5 \u00de \u00f0 \u03c02 6 \u00f0 kBT 2g \u00de \u00f0 \u03b5F : \u00de (3.76) From Eqs. (3.69) and (3.76), noting that the first term on the right side of Eq. (3.76) is independent of temperature for constant N, we obtain an expression for the electronic specific heat for a free electron gas in the metal, From Eqs. (3.57) and (3.77), Cel = \u03c02 3 kB 2Tg : \u03b5F \u00f0 \u00de Cel = BT n\u03c02k2 2\u03b5F : (3.77) (3.78) We note the result for the electronic specific heat obtained by using Fermi\u2013Dirac statistics is significantly different and approximately 100 times smaller than Drude\u2019s result for a classical electron gas (Cv = 3/2 nkB) obtained by using Maxwell\u2013Boltzmann statistics. The major difference 86 CHAPTER 3 Free Electron Model is that while each electron in a classical gas gains an energy 3/2 kBT when heated to a temperature T, only the electrons in a range kBT near the Fermi level are excited to the higher energy states, whereas the rest of the electrons inside the Fermi surface are unaffected. 3.5 DC ELECTRICAL CONDUCTIVITY We have shown in Eq. (3.36) that p =!hk: Using Newton\u2019s second law, we can see the force F on the electron with momentum p is In an external electric field E, the force on each electron of charge \u2013e is F = dp dt =!h dk dt : From Eqs. (3.80) and (3.81), we obtain F = \u2212eE:!h dk dt = \u2212eE: (3.79) (3.80) (3.81) (3.82) If the electric field E is applied to the metal at time t = 0, after a time \u03c4, the center of the Fermi sphere (within which all the electrons are located at T = 0), which was at the origin of the k space, would shift by \u0394k = \u2212 eE\u03c4 m : (3.83) The displacement of the Fermi sphere due to an applied electric field is shown in Figure 3.10. Because the change of momentum of each electron is \u0394p =!h\u0394k = m\u0394v, (3.84) and the electric current density of n electrons per unit volume is j = \u2212ne\u0394v, (3.85) \u2206k Ex we obtain from Eqs. (3.83) through (3.85) FIGURE 3.10 Displacement of the Fermi sphere in an applied electric field. j = ne2\u03c4 m E: (3.86) The electrical conductivity \u03c3 is defined by (note that conductivity is a tensor, but we are treating it here as a scalar quantity) j = \u03c3E: (3", ".87) 3.6 The Hall Effect 87 From Eqs. (3.86) and (3.87), we obtain \u03c3 = ne2\u03c4 m : (3.88) Eq. (3.88) for the electrical conductivity \u03c3, derived by considering the displacement of the Fermi sphere in an electric field, is identical to the expression derived by using Drude\u2019s theory of metals (Problem 3.1), in which electrons were treated as a classical electron gas. However, one significant difference is that although the relaxation time defined by Drude was the average collision time between the electrons and the static positively charged ions while in the present model, the collisions of the electrons with impurities, lattice imperfections, and phonons can be included. The two results appear identical because of the approximation of a relaxation time \u03c4 in both models. The following are the major drawbacks of the theory of DC electrical conductivity. First, it may be noted that in this derivation, the average relaxation time \u03c4 was not properly defined, although it was implicitly borrowed from the concept of the relaxation time defined by Drude. Second, because the origin of the collision time was left unanswered, the temperature dependence of the DC conductivity in metals could be explained only by introducing a temperature dependence in the relaxation time \u03c4. Third, the current density j is not parallel to the electric field E in some metals, and the DC conductivity depends on the orientation of the metal with respect to E. 3.6 THE HALL EFFECT Here, we consider a conductor in the shape of a rod that has a rectangular cross-section, which is placed under a magnetic field B in the z direction. There is a longitudinal electric field Ex. The electric and magnetic fields are so adjusted that the current cannot flow out of the rod in the y direction jy = 0. This configuration is shown in Figure 3.11. The equation of motion of the displacement \u03b4k, a Fermi sphere of particles acted on a force F, \u00f0 \u00de is given by F =!h! d dt + 1 \u03c4 \" B \u03c5x \u03c5y Ex jy jx EH FIGURE 3.11 The geometry of the Hall effect. \u03b4k, (3.89) where \u03c4 is the relaxation time. The Lorentz force on carrier of charge q (which can be positive or negative) in an electric field E and a magnetic field B is 3.90) 1 c! \" Here, we note that q = \u2212e for electrons, but the Hall effect is equally applicable for determining the sign of other carriers in a solid. If!h \u03b4k = m\u03b4V, Eq. (3.90) can be written as + 1 \u03c4 m d dt! \u03b4V = q \" E + \u00f0 1 c \u03b4V \u00d7 B : \u00de (3.91) 88 CHAPTER 3 Free Electron Model Here, \u03b4V is the average of V over the Fermi sphere. In the steady state, the acceleration d\u03b4V We can write Eq. (3.91) in the steady state (writing V for \u03b4V for convenience), dt = 0: V = q\u03c4 3.92) Because the static electric field E lies in the xy plane and B is along the z axis (Figure 3.7), we can write the components of Eq. (3.92) as and Here, Vx = Vy = q\u03c4 m q\u03c4 m Ex \u2212 \u03c9c\u03c4Vy, Ey + \u03c9c\u03c4Vx: \u03c9c \u2261 \u2212qB mc (3.93) (3.94) (3.95) is the cyclotron frequency, usually defined for free electrons (in which case q = \u2212e). By solving Eqs. (3.93) and (3.94), we obtain Vx = Vy = and The conductivity is q\u03c4/m \u03c9c\u03c4 \u00f0 1 + Ex\u2212\u03c9c\u03c4Ey, \u00de 2 \u00f0 \u00de q\u03c4/m \u03c9c\u03c4 \u00f0 1 + Ey + \u03c9c\u03c4Ex : \u00de 2 \u00f0 \u00de and the components of charge current density from Eqs. (3.96) through (3.98), \u03c30 = nq2\u03c4 m, jx = nqVx = jy = nqVy = \u03c30 \u03c9c\u03c4 \u00f0 \u03c30 \u03c9c\u03c4 + Ex \u2212 \u03c9c\u03c4Ey Ey + \u03c9c\u03c4Ex, \u00de, \u00de and because the z-component of the current is not affected by the magnetic field, jz = nqVz = \u03c30Ez: (3.96) (3.97) (3.98) (3.99) (3.100) (3.101) In the Hall effect", ", the external fields are E = Ex and Bz = B: There is a drift of the charges in the y direction due to the magnetic field B. However, no current can flow in this direction because of lack 3.7 Failures of the Free Electron Model 89 of circuit continuity. Instead, the charges pile up on the surface of the sample, thereby setting up an electric field Ey: This field, which nullifies the Lorentz force, is known as the \u201cHall field.\u201d The Hall field is obtained by setting jy = nqvy = 0: From Eqs. (3.100) and (3.102), we obtain \u03c30 1 + \u03c9c \u03c9c\u03c4Ex + Ey = 0, \u00de 2\u03c42 \u00f0 which yields Ey = EH = \u2212\u03c9c\u03c4Ex = qB m \u03c4Ex: From Eqs. (3.99) and (3.104), we obtain jx = \u03c30 1 + \u03c9c Ex + \u03c9c 2\u03c42 \u00f0 2\u03c42Ex \u00de = \u03c30Ex: The Hall coefficient RH is defined as The Hall resistance is defined as RH = EH jxB = q\u03c4 \u03c30m = 1 nq : \u211cH = Vy Ix : (3.102) (3.103) (3.104) (3.105) (3.106) (3.107) The sign of the Hall coefficient is determined by the charge carrier q, and for metals, these are the electrons of which the charge is \u2212e: Later, we will see the impact of the Hall effect in determining the sign of the charge carriers in semiconductors as well as study the importance of the quantum Hall effect. 3.7 FAILURES OF THE FREE ELECTRON MODEL The failures of the free electron model can be briefly summarized as follows: a. The temperature and the directional dependence of the DC electrical conductivity could not be adequately explained by the Sommerfeld theory. b. The free electron model explains the contribution of the linear term in T to the specific heat of metals, but the magnitude of this contribution is inadequate for many metals in the sense that it is either too large or too small. In addition, there is a cubic term (T3) that contributes to the specific heat of metals but cannot be explained by the free electron model. c. The major failure of the free electron model is that it does not explain the significant difference between metals, insulators, and semiconductors. To be specific, it fails to explain why some elements crystallize as good conductors (metals) of which the conductivity decreases with the increase of temperature, some others crystallize as insulators that are very poor conductors, and the rest 90 CHAPTER 3 Free Electron Model crystallize as semiconductors of which the conductivity is very poor at low temperatures but increases with increase of temperature unlike that of metals. In fact, it has been commented that the difference ohm-cm of electrical resistivity at room temperature between that of good conductors and of good insulators (1014\u22121022 ) ohm-cm is one of the widest range of any physical property ever found in nature. In contrast, the electrical resistivity of semiconductors at room temperature is in the range ohm-cm but is strongly dependent on temperature. 10\u221210 \u221210\u22126 10\u22122\u2212109 \u00de \u00f0 \u00f0 \u00de To understand this major classification of solids, we have to include the effect of the positively charged stationary ions that are arranged in a periodic array. As we will show, the periodic array of ions leads to a periodic potential that results in the band theory of solids. The band theory of solids explains the difference between the electrical properties of metals, semiconductors, and insulators. In addition, it distorts the spherical Fermi surface to a more complex shape, and as we have noted, the Fermi surface determines many physical properties of the solid. There are other significant deficiencies in the free electron model. First, it introduces the relaxation time of the electrons in an unrealistic manner in the sense that the role of the background positive ions is confined only to scatter an electron moving in an external field. Second, it ignores the electron\u2013electron interactions, which has significant contribution to some physical properties. Third, the positively charged ions are not stationary but oscillate around their equilibrium position. Since these ions are positively charged, they affect the motion of the negatively charged electrons (the electron\u2013phonon interaction). Fourth, the free electron model completely ignores the effects of lattice defects and lattice impurities. Fifth, the free electron model ignores the surface effects because of the periodic boundary condition invoked as a necessary condition of the model. PROBLEMS 3.1. Show that in Drude\u2019s model, the average electron velocity in an external electric field E is m and", " the current density j = \u2212nev = ne2\u03c4 vavg = \u2212 eE\u03c4 per unit volume. Because j = \u03c3E, where \u03c3 is the DC electrical conductivity of the metal, show that \u03c3 = ne2\u03c4 m. E: Here, n is the number of electrons! \" m 3.2. The basic assumption of the Drude model is that the thermal current in a metal is carried by the conduction electrons. If we define the thermal current density jq = \u2212\u03ba \u2207T, where \u03ba is the thermal conductivity, it can be easily shown that \u03ba = 1 specific heat. By applying classical ideal gas laws to the electron gas, cv = 3 2 kBT (where kB is the Boltzmann constant), show that 3 3v2\u03c4cv, where cv is the electronic 2 nkB, and 1 2mv2 = \u03ba \u03c3 = 3 2 T: 2 kB e! \" This relation between the electrical and thermal conductivities of a metal is known as the Wiedemann\u2013Franz law. 3.3. Show that the Fermi\u2013Dirac distribution function reduces to the Maxwell\u2013Bolzmann = e\u2212 distribution function f \u00f0 condition to make this classical approximation is e\u2212\u03bc/kBT \u03b5 \u00de \u00f0 \u00de /kBT for very high temperatures when \u03b5 \u03b5\u2212\u03bc \u03bc: An alternate ( 1. \u2019 Problems 91 3.4. Derive an expression for rs (defined in Eq. 3.2) by using the classical approximation outlined in Problem 3.3. 3.5. Show that the wave function of a free electron confined in a cubic box of side L is! 3.6. Show from Eqs. (3.2) and (3.40) that kF = \u00f0 r \u03c8 k\u00f0 \u00de = 1/2 sin 8 L3 & \u2019 n2\u03c0y L sin \"! n3\u03c0z L : \" n1\u03c0x L sin \" 1/3! = 1:92 rs. 9\u03c0/4 rs \u00de 3.7. Show that the Fermi temperature TF (defined in Eq. 3.50) can be written in the alternate form where a0 is the Bohr radius. TF = 58:2 rs/a0 2 \u00d7 104 K, \u00de \u00f0 3.8. If a function F chemical potential \u03bc, by making a Taylor series expansion of the form, does not vary rapidly in the energy range of the order of kBT about the \u03b5 \u00f0 \u00de Define a function so that dn d\u03b5n F \u03b5 \u00f0 \u00dej \u03b5 = \u03bc \u00f0 \u03b5 \u2212 \u03bc n \u2212\u221e F \u03b5\u2032 \u00de \u00f0 d\u03b5\u2032 = F \u03b5 \u00de \u00f0 d\u03d5 \u03b5 \u00f0 d\u03b5 \u00de : Integrating by parts, show that \u221e F Z \u2212\u221e \u03b5 \u03b5 f \u00de \u00f0 \u00f0 \u00de \u221e d\u03b5 = \u03d5 Z \u2212\u221e \u2202f \u2202\u03b5 \u03b5 \u00de \u00f0 \u2212 & d\u03b5, \u2019 (1) (2) (3) (4) where f \u03b5 \u00de \u00f0 is the Fermi function. Expand By using the identity \u00de \u03b5 \u2212 \u03bc n! \u00f0 * n \u00de + * dn\u03d5 \u03b5 \u00f0 d\u03b5n \u00de : +\u03b5 = \u03bc in a Taylor series about \u03b5 = \u03bc and show that (5) (6) \u221e \u00f0 Z \u2212\u221e \u2212\u2202f /\u2202\u03b5 \u00de d\u03b5 = 1, 92 CHAPTER 3 Free Electron Model and the fact that \u2202f /\u2202\u03b5 is an even function of \u03b5 \u2212 \u03bc (only terms with even n in Eq. 5 contribute), show that \u221e Z \u2212\u221e d\u03b5 \u2212\u221e Z \u221e \u00f0 \u221e d\u03b5 + \u2211 \u00de n = 1 Z \u2212\u221e 2n \u03b5 \u2212 \u03bc 2n \u00f0 \u00de! \u00de \u2202f \u2202\u03b5 \u2212 & d\u03b5 \u2019 d2n\u22121 d\u03b52n\u22121 F \u03b5 \u00dej \u03b5 = \u03bc: \u00f0 Show that the integration can also be expressed as \u221e F Z \u2212\u221e \u03bc \u03b5 \u03b5 f \u00de \u00f0 \u00f0 \u00de d\u03b5 = F Z \u2212\u221e \u03b5 \u00f0 where \u221e d\u03b5 + \u2211 \u00de n = 1 \u00f0 kBT 2nan \u00de d2n\u22121 d\u03b52n\u22121 F \u03b5 \u00dej \u03b5 = \u03bc, \u00f0 an = \u221e Z \u2212\u221e \u00f0 x2n 2n \u00de! & \u2212 d dx \u00f0 1 ex + 1 dx: \u2019 \u00de This is known as", " the Sommerfeld expansion. Here, an are dimensionless constants of the order of unity and are related to the Riemann zeta function, \u03be, as n \u00f0 \u00de The Riemann zeta function, \u03be 1 n\u22121 \u00f0 \u03be \u2019 \u00f0 \u00de 2n : \u00de an = 2\u2212 &, is defined as 22 2n + 1 3n + 1 4n + ::: (7) (8) (9) (10) (11) 3.9. From Eqs. (3.99) through (3.101), show that the current density can be written in the matrix form jx jy jz 0 = 1 \u03c30 1 + \u03c92 c\u03c42 \u00f0 0 \u00de 1 \u2212\u03c9c\u03c4 0 \u03c9c\u03c4 0 1 0 1 + \u03c9c 0 Ex Ey Ez : 1 1 0 2\u03c42 (1) The matrix on the right side of Eq. (1) gives the nine components of the magnetoconductivity tensor. Hence, show that @ A @ A @ A \u03c3xx = \u03c3yy = \u03c30 1 + \u03c9c 2\u03c42, and \u03c3xy = \u2212\u03c3yx = \u03c30\u03c9c\u03c4 1 + \u03c9c 2\u03c42 : (2) (3) Thus, the magnitude of the diagonal components \u03c3xx and \u03c3yy decrease monotonically as the magnetic field B is increased, whereas the off-diagonal components \u03c3xy and \u03c3yx at first increase and then decrease as the magnetic field is increased. References 93 References 1. Ashcroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 2. Harrison WA. Solid state theory. New York: McGraw-Hill; 1969. 3. Kittel C. Introduction to solid state physics. New York: John Wiley & Sons; 1967. 4. Landau LD, Lifshitz EM. Statistical physics, part 1. Oxford: Pergamon Press; 1980. 5. Madelung O. Introduction to solid state theory. New York: Springer-Verlag; 1978. 6. Marder MP. Condensed matter physics. New York: John Wiley & Sons; 2000. 7. Myers HP. Introductory solid state physics. London: Taylor & Francis; 1990. 8. Pathria RK. Statistical mechanics. Oxford: Butterworth-Heinemann; 2000. 9. Reif F. Fundamentals of statistical and thermal physics. New York: McGraw-Hill; 1965. 10. Seitz F. The modern theory of solids. New York: McGraw-Hill; 1940. 11. Ziman JM. Electrons and phonons. Oxford: Oxford University Press; 1960. 12. Ziman JM. Principles of the theory of solids. Cambridge: Cambridge University Press; 1972. Nearly Free Electron Model CHAPTER 4 CHAPTER OUTLINE 4.1 Electrons in a Weak Periodic Potential...................................................... 96 4.1.1 Introduction........................................................................ 96 4.1.2 Plane Wave Solutions............................................................... 97 4.2 Bloch Functions and Bloch Theorem......................................................... 99 4.3 Reduced, Repeated, and Extended Zone Schemes............................................ 99 4.3.1 Reduced Zone Scheme........................................................", "..... 100 4.3.2 Repeated Zone Scheme.............................................................. 100 4.3.3 Extended Zone Scheme............................................................ 101 4.4 Band Index............................................................................. 101 4.5 Effective Hamiltonian.................................................................... 102 4.6 Proof of Bloch\u2019s Theorem from Translational Symmetry....................................... 103 4.7 Approximate Solution Near a Zone Boundary................................................ 105 4.8 Different Zone Schemes.................................................................. 109 4.8.1 Reduced Zone Scheme............................................................. 109 4.8.2 Extended Zone Scheme............................................................ 110 4.8.3 Repeated Zone Scheme............................................................ 111 4.9 Elementary Band Theory of Solids......................................................... 111 4.9.1 Introduction....................................................................... 111 4.9.2 Energy Bands in One Dimension.................................................... 112 4.9.3 Number of States in a Band........................................................ 112 4.10 Metals, Insulators, and Semiconductors.......................", "............................. 112 4.11 Brillouin Zones.......................................................................... 117 4.12 Fermi Surface........................................................................... 119 4.12.1 Fermi Surface (in Two Dimensions)................................................. 119 4.12.2 Fermi Surface (in Three Dimensions)................................................ 121 4.12.3 Harrison\u2019s Method of Construction of the Fermi Surface.............................. 121 Problems.................................................................................... 124 References.................................................................................. 130 Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00004-9 \u00a9 2012 by Elsevier Inc. All rights reserved. 95 96 CHAPTER 4 Nearly Free Electron Model 4.1 ELECTRONS IN A WEAK PERIODIC POTENTIAL 4.1.1 Introduction In the nearly free electron approximation, it is assumed that there are no electron\u2013electron or electron\u2013phonon interactions. This means that a valence electron, stripped from its parent atom due to the attractive interaction of the neighboring positively charged ions, does not interact either with other electrons or with the vibrating motion of the ions at a finite temperature. However, unlike the free electron approximation, the electron is subjected to a weak periodic potential due to the background of symmetric array of positively charged ions in the crystal lattice. We will first show that this potential is periodic with the periodicity of a lattice vector. As an example, we consider a two-dimensional rectangular lattice, as shown in Figure 4.1. If OO\u2032 = R!i and OE = r!, the potential energy at the electron E (charge \u2013e) due to the positively charged ions of the crystal lattice (one ion of charge ze is assumed to be located at each lattice site) is given by V r!\u00de \u00f0 = \u2211 i \u2212ze2 r!\u2212 R!i j : j (4.1) Here, we have considered the fundamental principle of electrostatics that in a spherical charge distribution, the potential at a point outside the sphere is the same as that of the potential due to the net charge considered to be located at the center of the sphere. If r! is translated by a lattice vector Rj!, Eq. (4.1) can be written in the alternate form V r! + R!j \u00de \u00f0 = \u2211 i \u2212ze2 r! \u2212 Ri! + R!j : j j One can write R!i \u2212 R!j = Rl!, in which case Eq. (4.2) can be written in the alternate form V r! + R!j \u00de \u00f0 = \u2211 l \u2212ze2 r! \u2212 R!l : j j \u2212e E O\u2032 r", "\u2192 \u2192 Ri ze ze ze O (4.2) (4.3) FIGURE 4.1 Two-dimensional rectangular lattice with O as the global origin and O\u2032 as the local origin of the unit cell within which the electron of charge \u2013e is located at E. The charge of each ion is ze. 4.1 Electrons in a Weak Periodic Potential 97 Because the summation over both R!i and R!l spans the entire lattice vectors, we obtain from Eqs. (4.1) and (4.3) V r!\u00de is a periodic potential with the periodicity of a direct Eq. (4.4) clearly demonstrates that V lattice vector. It may be noted that this proof was based on the simple assumption that an ion of charge ze is located at each lattice point. However, the proof can be generalized to an identical cluster of ions (a basis), located symmetrically around each lattice point. r! + R!j \u00f0 r!\u00de (4.4) = V \u00f0 \u00de \u00f0 : 4.1.2 Plane Wave Solutions For simplicity, we consider a linear lattice of lattice constant a. Later, we will generalize our results to a three-dimensional lattice. From Eq. (4.4), the periodic potential in a one-dimensional lattice can be written as where n is an integer. If we express the periodic potential V V x \u00de \u00f0 = \u2211 q\u2032 V q\u2032 \u00f0 \u00de \u00f0 eiq\u2032x, as a Fourier series x \u00de we obtain x + na V \u00f0 = \u2211 q\u2032 V q\u2032 \u00de \u00f0 \u00de x+na eiq\u2032 \u00f0 \u00de: From Eqs. (4.5) through (4.7), it is easy to show that \u2211 q\u2032 V q\u2032 \u00de \u00f0 eiq\u2032x = \u2211 q\u2032 V q\u2032 \u00de \u00f0 x+na eiq\u2032 \u00f0 \u00de: Eq. (4.8) has to be valid for each value of the integer n. This is possible only if eiq\u2032na = 1, (4.5) (4.6) (4.7) (4.8) (4.9) for all values of n and q\u2032. This condition is satisfied only when q\u2032 = 2\u03c0m/a, where m is any integer. This is precisely the definition of a reciprocal lattice vector K in one dimension and hence q\u2032 = K: in Eq. (4.3) can therefore be expressed as The periodic potential eiKx: The Schrodinger equation of the electron in a one-dimensional lattice is easily obtained, where E is the energy eigenvalue and \u03c8 Karman boundary conditions imply that \u2212 \u01272 2m! \u22022 \u2202x2 + \u2211 x K \u00f0 \u00de V K \u00f0 \u00de eiKx \u03c8 \" x \u00de \u00f0 = E\u03c8, x \u00f0 \u00de is the wave function of the electron. The Born\u2013von \u03c8 \u00f0 x + Na = 4.12) (4.10) (4.11) 98 CHAPTER 4 Nearly Free Electron Model can also be expanded in terms of the plane waves, which are a complete set of functions eiqx: From Eqs. (4.11) and (4.13), we obtain eiqx + \u2211 K \u03b50 qa \u2211 q q \u00de \u00f0 where \u2211 +K ei \u00f0 \u00de x = E\u2211 q a q \u00de \u00f0 eiqx, \u03b50 q = \u01272q2 2m : (4.13) (4.14) (4.15) We assume that the one-dimensional crystal has a length L. Multiplying Eq. (4.14) by e\u2212iq\u2032x and integrating over x from 0 to L, we obtain L \u03b50 qa q \u00f0 \u2211 q \u00de 0 Z q\u2212q\u2032 ei \u00f0 \u00de xdx + \u2211 ei q\u2212q\u2032+K \u00f0 xdx = E ei \u00f0 q\u2212q\u2032 xdx: \u00de (4.16) The Born\u2013von Karman boundary conditions for a linear lattice lead to the conditions for the \u00de. These conditions imply that both q and q\u2032 must \u00de and eiq\u2032x = eiq\u2032 \u00f0 x+L x+L \u00f0 plane waves that eiqx = eiq satisfy where n and m are integers. The integration q = 2\u03c0n L and q\u2032 = 2\u03c0m L, I = L Z 0 q\u2212q\u2032 ei \u00f0 xdx = \u00de ei \u00f0 i q\u2212q\u2032 L \u2212 1", " \u00de q \u2212 q\u2032 \u00de \u00f0 = \u00bd n\u2212m \u00de \u2212 1 n \u2212 m ei2\u03c0 \u00f0 2\u03c0i \u00f0 \u00de L $ = L\u03b4n,m = L\u03b4q,q\u2032, where \u03b4q,q\u2032 is the Kronecker delta function. Similarly, one can show that I\u2032 = L Z 0 ei q\u2212q\u2032+K \u00f0 xdx = L\u03b4q,q\u2032\u2212K: \u00de Substituting the results of Eqs. (4.18) and (4.19) in (4.16), we obtain \u03b5\u20320 q a q\u2032 \u2212 K q\u2032 q = Ea \u00f0 \u00de, \u00de which can be written in the alternate form, by substituting q for q\u2032, \u03b50 : Eq. (4.21) can be expressed as It is obvious from Eq. (4.22) that \u2212 \u03b50 q \u00de : is small unless E \u2248 \u03b50 q. q \u00de \u00f0 (4.17) (4.18) (4.19) (4.20) (4.21) (4.22) 4.3 Reduced, Repeated, and Extended Zone Schemes 99 We can also subtract an arbitrary reciprocal lattice vector K\u2032 from q in Eq. (4.21) and rewrite it as \u03b50 q \u2212 K \u2212 E a q \u2212 K\u2032 \u2212 K = 0: (4.23) \u00de \u00f0 Eq. (4.22) connects a \u00de, i.e., with the Fourier coefficients for which the wave vector differs from q by a reciprocal lattice vector K. This leads to a very important and useful result about the form of the eigenfunctions \u03c8. These wave functions of an electron in a periodic potential are called the Bloch functions. \u00de with every Fourier coefficient The equivalent proof correlating a q and a q \u2212 K assigned as a homework problem (see Problem 4.4). \u00f0 \u00de \u00f0 for a three-dimensional crystal lattice is \u00de 4.2 BLOCH FUNCTIONS AND BLOCH THEOREM In Eq. (4.12), we considered an arbitrary wave vector that appears in the summation over q and denoted it as k. We note from Eq. (4.21) that instead of the continuous Fourier coefficients a, \u00de ; i.e., the allowed K\u2032s in the wave function are of x only those of the form a the form k \u2212 K. Thus, we can write enter into \u03c8 k\u00f0 Eq. (4.24) can be written in the alternate form \u03c8 k\u00f0 x \u00de k \u2212 K a \u00f0 \u00de k\u2212K ei \u00f0 x: \u00de If we introduce we obtain x \u03c8 k\u00f0 \u2212iKx \u00de eikx: # uk \u2212iKx, \u00de We note from Eq. (4.26) that if m is an integer, x \u03c8 k\u00f0 = eikxuk : x \u00f0 \u00de because x + ma uk \u2212iK \u00de \u00f0 x+ma \u00de = uk x \u00de \u00f0 e\u2212iKma = e\u22122im\u03c0 = 1: (4.24) (4.25) (4.26) (4.27) (4.28) (4.29) x is referred to as a Bloch function and uk \u03c8 k\u00f0 \u00f0 tion because it has the periodicity of the lattice. \u00de is known as the periodic part of the Bloch func- x \u00de 4.3 REDUCED, REPEATED, AND EXTENDED ZONE SCHEMES We will now discuss the three types of zone schemes (reduced, repeated, and extended) used to describe electrons in a crystal lattice. For simplicity, we will first discuss these schemes for free electrons in a one-dimensional lattice to introduce the concept of the band index before we extend our discussion to electrons in a three-dimensional lattice as well as to electrons in a periodic potential. 100 CHAPTER 4 Nearly Free Electron Model Reduced zone k \u22122\u03c0/a \u2212\u03c0/a k\u2032 O g g k \u03c0/a 2\u03c0/a k 4\u03c0/a FIGURE 4.2 All k points reduce to k\u2032 in the one-dimensional reciprocal lattice. 4.3.1 Reduced Zone Scheme If we consider a one-dimensional lattice, the reciprocal lattice vectors K can also be relabeled as gn where n is an integer (positive or negative), gn = n 2\u03c0 a : (4.30) If we restrict k to the first Brillouin zone, i.e., if we assign a state k, any wave number in the", " set k = k\u2032 + 2\u03c0 a n, (4.31) k is only defined modulo (2\u03c0/a). Thus, all the k points in Figure 4.2 are equivalent. One can therefore consider k\u2032 as the representative of all these k values, with restricted to the first Brillouin zone. Thus, it is always possible to choose the value of k such that k4.32) \u03b5k C 0 k A \u03c0 a A\u2032 \u2212 \u03c0 a FIGURE 4.3 Reduced zone scheme for energy-wave number relations of free electrons in a one-dimensional lattice. This way of restricting the wave numbers to the first Brillouin zone is known as the reduced zone scheme. In Figure 4.3, the reduced zone scheme is shown by drawing the energy-wave number relation for free electrons, \u03b50 k \u00de \u00f0 We will now describe the energy-wave number relations \u03b50 = \u01272k2/2m for the repeated and = \u01272k2/2m. k \u00f0 \u00de extended zone schemes. 4.3.2 Repeated Zone Scheme It is often convenient to repeat the first Brillouin zone and the other zones reduced to the first Brillouin zone through all of k space. Thus, in the repeated zone scheme, \u03b50 k+K in one k = \u03b50 4.4 Band Index 101 \u03b5k \u22123\u03c0 /a \u2212\u03c0 /a 0 k \u03c0 /a 3\u03c0 /a FIGURE 4.4 Energy-wave number relations for a one-dimensional lattice in the repeated zone scheme. \u03b5k \u22123\u03c0 /a \u2212\u03c0 /a 0 k \u03c0 /a 3\u03c0 /a FIGURE 4.5 Energy-wave number relations in the extended zone scheme. k = \u03b50 k+K in three dimensions. All values of \u03b50 dimension and \u03b50 k\u2212K, etc. are plotted against k, k + K, k \u2212 K, etc. in one dimension. The repeated zone scheme is useful in describing several physical properties of the solid, specifically the connectivity of electron orbits in a magnetic field. The repeated zone scheme for free electrons in a one-dimensional lattice is shown schematically in Figure 4.4. k+K, \u03b50 k, \u03b50 4.3.3 Extended Zone Scheme In the extended zone scheme, the k values extend throughout all reciprocal space, and the energy k = \u01272k2/2m, are plotted against the wave number k. Thus, for free electrons, one eigenvalues, \u03b50 obtains a parabola. The extended zone scheme for free electrons in a one-dimensional lattice is shown schematically in Figure 4.5. 4.4 BAND INDEX A large number of eigenfunctions and eigenvalues correspond to the same wave vector k (wave numbers k in one dimension) in the reduced zone scheme. To distinguish these eigenfunctions and eigenvalues in the reduced zone scheme, we introduce an additional index n (called band index). The band index has a much greater significance when one considers the eigenfunctions and eigenvalues by including the periodic potential in the Hamiltonian. The periodic potential opens up an energy gap at the zone boundaries, and the band index plays a much greater role in the classification of solids as metals, insulators, and semiconductors. The necessity of using a band index also follows as a natural consequence when we discuss the effective Hamiltonian. 102 CHAPTER 4 Nearly Free Electron Model 4.5 EFFECTIVE HAMILTONIAN The necessity of a band index n will be evident by constructing an effective Hamiltonian. The Schrodinger equation can be written as \u01272 2m = Ekeik.ruk eik.ruk \u22072 + V (4.33) r H\u03c8 k\u00f0 \" We can rewrite Eq. (4.33) in the form eik.r \u01272 2m \u00bd $ Canceling eik.r from both sides, we obtain \u2212\u22072 \u2212 2ik. \u2207! + k2 uk r \u00de \u00f0 $ + V uk r \u00f0 \u00de \u00f0 r \u00de % = eik.rEk uk r \u00de \u00f0 r : \u00de \u00f0 \u01272 2m \u00bd $ \u2212\u22072 \u2212 2ik. \u2207! + k2 + V r \u00de \u00f0 $ % uk r \u00de \u00f0 = Ek uk r \u00de \u00f0 r : \u00de \u00f0 Eq. (4.35) can be expressed as where the effective Hamiltonian Heff uk r \u00de \u00f0 = Ek uk r \u00de \u00f0 r, \u00de \u00f0 Heff = \u01272 2m \u00bd \u2212\u22072 \u2212 2ik. \u2207 + k2 + V", " r : \u00de \u00f0 $ (4.34) (4.35) (4.36) (4.37) The boundary conditions are that whenever r lies on one boundary of the unit cell and r + R is another boundary of the unit cell, then r + R uk \u00f0 = uk r \u00de \u00f0 \u00de ^n r. \u2207!uk r = \u2212 ^n r + R \u00de is a unit normal to the cell boundary at r. \u00f0 \u00f0 \u00de \u00f0. \u2207!uk \u00de r + R, \u00de \u00f0 (4.38) (4.39) and where ^n r \u00f0 \u00de Thus, Eq. (4.36) can be considered as a Hermitian eigenvalue problem that is restricted to a single primitive cell of the crystal. Because the eigenvalue problem is in a fixed finite volume, there will be an infinite family of solutions with discretely spaced eigenvalues. These are labeled with the band index n. The importance of the band index will become apparent when we consider the effect of the periodic potential by using perturbation theory. If we include the band index n, Eq. (4.27) can now be rewritten as x \u03c8 nk\u00f0 It is easy to generalize Eq. (4.27) to a three-dimensional crystal lattice (see Problem 4.4), \u03c8 k\u00f0 = eik.ruk x r r \u00de \u00de \u00f0 \u00de \u00f0 \u00de :, = eikxunk where For a three-dimensional lattice, in analogy with Eq. (4.31), we can write r + R uk \u00f0 = uk r : \u00de \u00f0 \u00de k = k\u2032 + K, (4.40) (4.41) (4.42) (4.43) 4.6 Proof of Bloch\u2019s Theorem from Translational Symmetry 103 where k\u2032 is restricted to the first Brillouin zone. From Eq. (4.41) and (4.43), we obtain \u00de \u00de r k\u2032+K \u00f0 r + Rl \u03c8 k\u00f0.Rl\u03c8 k\u00f0 = ei = eiK.Rl eik\u2032.Rl \u03c8 k\u00f0 = eik\u2032.Rl\u03c8 k\u00f0 satisfies the Bloch\u2019s theorem with the wave vector k\u2032. Thus, every state has a large number of possible wave vectors, differing from each other by the reciprocal lattice vectors K. If we choose the value of K such that k\u2032 lies in the first Brillouin zone (which is the reduced zone scheme, in which we relabel k\u2032 as k), there will be a large number of eigenfunctions and eigenvalues corresponding to the same wave vector k. From Eq. (4.44), it is evident that \u03c8 k\u00f0 (4.44) \u00de r r r \u00de \u00de \u00de : If we introduce the band index n, which follows as a consequence of restricting the wave vector k to the first Brillouin zone, Eq. (4.44) can be rewritten as = eik.runk r \u03c8 nk\u00f0 \u00de r \u00f0 \u00de and r + R \u00de r Eq. (4.45) is known as the Bloch theorem; \u03c8 nk\u00f0 unk \u00f0 is known as the periodic part of the Bloch function. The Bloch theorem can also be proved by using the translational symmetry of the crystal lattice. is known as the Bloch function, and unk r \u00de \u00de \u00f0 r : \u00de \u00f0 = unk 4.6 PROOF OF BLOCH\u2019S THEOREM FROM TRANSLATIONAL SYMMETRY We will now prove Bloch\u2019s theorem by using the translational symmetry of the crystal lattice. Through use of a three-dimensional equivalence of Problem 4.1, it can be easily shown that the translation operator ^T Ri \u00f0 \u00de is defined by ^T Ri f r \u00de \u00f0 \u00de \u00f0 = f r + Ri \u00f0 : \u00de The Hamiltonian of the electron in the periodic potential can be written as From Eqs. (4.4), (4.47), and (4.48), we obtain ^H r \u00f0 \u00de = \u2212 \u01272 2m \u22072 + V r : \u00de \u00f0 Because f r \u00de \u00f0 ^T Ri ^H r f r = ^H r + Ri f r + Ri \u00f0 \u00de \u00f0 \u00f0 is any arbitrary function of r, we obtain \u00de \u00de \u00de \u00f0 \u00f0 \u00de = ^H r f r + Ri \u00f0 \u00de \u00de \u00f0 = ^H", " ^T r \u00de \u00f0 \u00f0 Ri f r : \u00de \u00f0 \u00de ^T Ri ^H r \u00de \u00f0 \u00de \u00f0 = ^H ^T r \u00de \u00f0 \u00f0 Ri : \u00de It is also easy to show that from which we have ^T Ri ^T \u00de \u00f0 Rj r f \u00de \u00f0 \u00de \u00f0 = f r + Ri + Rj \u00f0 \u00de = ^T Rj \u00f0 ^T \u00de \u00f0 Ri f r, \u00de \u00f0 \u00de ^T Ri Rj ^T \u00de \u00f0 \u00de \u00f0 = ^T Rj Ri ^T \u00de \u00f0 \u00de \u00f0 = ^T \u00f0 Ri + Rj : \u00de (4.45) (4.46) (4.47) (4.48) (4.49) (4.50) (4.51) (4.52) (4.53) (4.54) 104 CHAPTER 4 Nearly Free Electron Model From Eqs. (4.49) and (4.52), we note that the Hamiltonian ^H and the translation operators ^T Ri \u00de (corresponding to each Bravais lattice vector Ri) form a mutually commuting set of operators. Therefore, according to an important theorem in quantum mechanics,5 these operators will have a complete set of common eigenfunctions. \u00f0 If \u03c8 r \u00f0 \u00de is one of the eigenfunctions of the Hamiltonian with eigenvalue \u03b5, ^H\u03c8 = \u03b5\u03c8 r r, \u00f0 \u00de \u00f0 \u00de it follows from the previous theorem that \u03c8 \u00de \u00f0 are the eigenvalues of the translation operators ^T \u03c8 \u00de = C = \u03c8 Ri Ri r r \u00de \u00de \u00f0 \u00f0 \u00f0 \u00f0 ^T r + Ri : \u00de. It also follows from Eq. (4.52) Ri \u00f0 \u00de Here, C Ri \u00f0 and Eq. (4.53) that \u00de C \u00de \u00de Because Ri and Rj are Bravais lattice vectors, they can be expressed as Ri + Rj Ri \u00f0 = C = C Rj Rj Ri C C \u00f0 \u00de \u00de \u00f0 \u00de \u00f0 \u00f0 : and Ri = n1a1 + n2a2 + n3a3 Rj = m1a1 + m2a2 + m3a3, (4.55) (4.56) where a1, a2, and a3 are the three primitive vectors of the Bravais lattice and n1, n2, n3, m1, m2, and m3 are appropriate integers corresponding to the lattice vectors Ri and Rj: From Eqs. (4.55) and (4.56), it is obvious that C must be an exponential of the form (4.57) \u00de where pi, which could be a complex number, has to be determined. From Eqs. (4.56) and (4.57), we obtain ai C \u00f0 = epi, ai \u00f0 \u00de From Eqs. (4.57) and (4.58), we obtain Ri C \u00f0 \u00de = C a1 \u00f0 \u00de n1 C a2 \u00f0 \u00de n2C n3: a3 \u00f0 \u00de = en1p1+n2p2+n3p3 : C Ri \u00f0 \u00de (4.58) (4.59) We now restate the Born\u2013von Karman boundary conditions (originally stated for a cubic crystal in Eq. 3.11) for the wave functions of the electrons in a more general form (instead of restricting these conditions to a cubic crystal), \u00de \u00f0 where M1, M2, and M3 are the number of primitive vectors in the directions a1, a2, and a3, respectively. Obviously, the total number of primitive cells in the crystal \u00de \u00de \u00f0 \u03c8 r + Miai = eMipi\u03c8 r = \u03c8 i = 1, 2, 3,, r \u00f0 (4.60) From Eq. (4.61), we obtain N = M1M2M3: eM1p1 = eM2p2 = eM3p3 = 1: Eq. (4.62) yields the necessary condition that pi = 2\u03c0imi Mi. i = 1, 2, 3, (4.61) (4.62) (4.63) 4.7 Approximate Solution Near a Zone Boundary 105 where, m1, m2, and m3 are a set of integers. If we define the Bloch wave vectors as 3 k = \u2211 i", " = 1 pi 2\u03c0i bi, 0 \u2264 mi < Mi, (4.64) where bi, the primitive vectors of the reciprocal lattice, were defined in Eq. (1.16). According to Eq. (4.64), the total number of k states in a Brillouin zone is M1M2M3, which is also equal to the number of lattice points in the lattice. It is possible to have k outside the first Brillouin zone by allowing the integers to be greater than Mi: However, the resulting k would differ from the k within the first Brillouin zone by a reciprocal lattice vector K. Because eik.Ri = ei.Ri, the resulting eigenfunction would be the same according to Eq. (4.45) as well as from Eq. (4.67) (which we will prove), and because k is a label for this eigenvalue, two such k values are physically identical. Thus, the number of physically distinct values of the Bloch wave vector k equals the number of lattice sites of the original Bravais lattice. k+K \u00de \u00f0 We obtain from Eqs. (4.60), (4.63), and (4.64), eiMik.ai = 1 and = eik.Ri: We obtain from Eqs. (4.52) and (4.66), the Bloch theorem, Ri C \u00f0 \u00de (4.65) (4.66) \u03c8 r + Ri = C Ri \u03c8 r = eik.Ri \u03c8 r : (4.67) \u00de If we identify the eigenfunctions \u03c8 with a band index n and wave vector k (we will show the importance and the necessity of the band index), Eq. (4.67) can be written in the more general form nk\u00f0 The Bloch theorem stated in Eq. (4.68) can also be rewritten in the alternate form \u03c8 nk\u00f0 r \u00f0 is known as the periodic part of the Bloch function. In fact, from Eqs. (4.68) and = eik.runk r + Ri (4.68) (4.69) r r r \u00de \u00de \u00de \u00de :, = eik.Ri \u03c8 nk\u00f0 where unk (4.69), we obtain \u00f0 \u00de from which the term periodic part is self-evident. r + Ri unk \u00f0 = unk r, \u00de \u00f0 \u00de 4.7 APPROXIMATE SOLUTION NEAR A ZONE BOUNDARY To understand the occurrence of band gaps, we consider the results of Problem 4.4 (Eq. 10), where \u03b50 k\u2212K \u2212 \u03b5 \u00f0 Ck\u2212K + \u2211 K\u2032 \u00de VK\u2032\u2212KCk\u2212K\u2032 = 0, \u03b50 k\u2212K \u2261 \u01272 2 k \u2212 K \u00f0 2m \u00de (4.70) (4.71) (4.72) 106 CHAPTER 4 Nearly Free Electron Model is the free electron energy eigenvalue for an electron of wave vector k \u2212 K. We will also denote \u03b50 k as the free electron eigenvalue for wave vector k. We can rewrite Eq. (4.71) as \u03b5 \u2212 \u03b50 k\u2212K\u00de \u00f0 Ck\u2212K = \u2211 K\u2032 \u2260 K VK\u2032\u2212KCk\u2212K\u2032 (4.73) because we have assumed that V0 = 0: Eq. (4.73) includes the terms K = 0 and K\u2032 = 0. If we use nondegenerate perturbation theory and assume that k\u2212K \u2212 \u03b50 \u03b50 k\u2212K\u2032 \u226b VK\u2032\u2212K, for all K\u2032 \u2260 K and fixed k, Eq. (4.73) can be rewritten as & & & Ck\u2212K = \u2211 K\u2032\u2260 K & & & VK\u2032\u2212KCk\u2212K\u2032 \u03b5 \u2212 \u03b50 k\u2212K (4.74) (4.75) : For another coefficient Ck\u2212K1 corresponding to the reciprocal lattice vector K1 (where K1 satisfies the condition of Eq. 4.74), Ck\u2212K1 = VK\u2212K1Ck\u2212K \u03b5 \u2212 \u03b50 k\u2212K1 + \u2211 K\u2032\u2260 K \u2260 K1 VK\u2032\u2212K1Ck\u2212K\u2032 \u03b5 \u2212 \u03b50 k\u2212K\u2032 : (4.76) In deriving Eq. (4.76), we have made the basic assumption that the free electron eigenvalue \u03b50 is not nearly degenerate to any other \u03b50 Eq. (4.78) in", " second order and higher terms in V would not be valid. k\u2212K k\u2212K\u2032 in the set. Otherwise, the expansion of the energy in From Eqs. (4.73) and (4.76), we obtain \u03b5 \u2212 \u03b50 k\u2212K\u00de \u00f0 VK\u2032\u2212KVK\u2212K\u2032 Ck\u2212K = \u2211 K\u2032\u2260 K k\u2212K\u2032 + higher-order terms in V: \u03b5 \u2212 \u03b50 Ck\u2212K + \u2211 K\u2033\u2260 K\u2032\u2260 K \u00f0 VK\u2032\u2212KVK\u2212K\u2032VK\u2212K\u2033 \u03b5 \u2212 \u03b50 \u03b5 \u2212 \u03b50 k\u2212K\u2032\u00de\u00f0 k\u2212K\u2033\u00de Ck\u2212K (4.77) 2 or higher-order Because the perturbed energy \u03b5 differs from the free electron energy \u03b50 j terms (in the specific case of energy values that are neither degenerate nor nearly degenerate), we retain the terms up to the second order in V, use the relation V\u2212K = V %K in Eq. (4.77), substitute \u03b5 by \u03b50 k\u2212K in the denominator of the first term on the right, and obtain the expression for \u03b5: k\u2212K by V j \u03b5 = \u03b50 k\u2212K + \u2211 K\u2032\u2260 K 2 VK\u2032\u2212K j k\u2212K \u2212 \u03b50 \u03b50 j k\u2212K\u2032 + O V 3 : \u00de \u00f0 (4.78) Eq. (4.78) is valid as long as nondegenerate perturbation theory can be applied to the problem, i.e., as long as \u03b50 k\u2212K\u2032 (or sufficiently close in values so that the perturbation theory breaks down). The simplest example is when k lies near a zone boundary, in which case the second-order perturbation theory breaks down. k\u2212K \u2260 \u03b50 If k lies near a zone boundary (for simplicity, we assume that it lies near the boundary bisecting the vector K), the electron undergoes a Bragg reflection by the lattice, similar to the situation as if it would have been an external electron beam. In such a case, we will use degenerate perturbation theory, consider only Ck and Ck\u2212K, and neglect the other coefficients. 4.7 Approximate Solution Near a Zone Boundary 107 Eq. (4.73) can be rewritten as \u03b50 k \u2212 \u03b5 \u00f0 V\u2212KCk + Ck + VKCk\u2212K = 0 \u00de \u03b50 Ck\u2212K = 0: k\u2212K \u2212 \u03b5 \u00f0 \u00de From Eqs. (4.77) and (4.79), we obtain \u03b50 k \u2212 \u03b5 \u00f0 V %KCk + Ck + VKCk\u2212K = 0 \u00de \u03b50 Ck\u2212K = 0: k\u2212K \u2212 \u03b5 \u00f0 \u00de (4.79) (4.80) k \u2248 \u03b50 We have \u03b50. k j j It is evident from Figure 4.6(a) that this is possible only when k lies on the Bragg plane that bisects the line joining the origin of k space to the reciprocal lattice point K. \u226b V, when K \u2260 K, 0. This is possible only when k \u2212 \u03b50 \u03b50 j k\u2212K and k \u2212 K j k\u2212Kj = j Eq. (4.80) can be solved from the determinant O 1 2 : 2 O \u03b50 k \u2212 \u03b5 VK \u03b50 k\u2212K \u2212 \u03b5 V %K = 0: (4.81) & & & & & & & & & & The solutions of the quadratic equation are \u03b5\u00b1 = k \u00de \u00f0 1 2\u00f0 k +\u03b50 \u03b50 k\u2212K\u00de k \u2212\u03b50 2 k\u2212K 2 + VK j j \u00b1 \u03b50 ( \u2019 \u00f0 \u00de : k # and \u03b5 \u2212 k and \u03b50 Thus, k \u2212 K.r with energy \u03b50 ) (4.82) the free electron states eik.r and ei k\u2212K are combined into two other states \u03c8 + and \u03c8 \u2212 with energy \u03b5 + k \u00f0 \u00de It is easy to analyze Eq. (4.82) for points lying on the Bragg plane because |k| = |k \u2212 K| and \u03b50 k\u2212K. This implies that k must lie on the Brillouin zone boundary (see Figure 4.6a). Further, at all points on the Bragg plane, one, whereas the other energy level is raised by VK j : Thus, when k is energy level is lowered by j on a", " single Bragg plane, we can write k = \u03b50 j VK j \u00de \u00f0 (4.83) \u03b5\u00b1 k = \u03b50 k \u00b1 : VK j j \u00f0 \u00de Hence, there is an energy gap of 2 when k = 1 2 K. This is shown in Figure 4.7. This is known as the band gap because the energy levels are split into two bands. When k is closer to the origin (far away from the Bragg plane), the energy levels are practically the same as the free electron energy levels. Vk j j K k \u2212 K 1 K2 k 1 K2 k (a) k \u2212 1 K2 (b) FIGURE 4.6 (a) k lies in the Bragg plane determined by K if |k| = |k \u2212 K|; (b) k \u2212 1 if k lies in the Bragg plane. 2 K is parallel to the Bragg plane 108 CHAPTER 4 Nearly Free Electron Model In addition, when \u03b50 k = \u03b50 k\u2212K, we obtain from Eq. (4.82), \u2202\u03b5 \u2202k = \u01272 m K : (4.84) k \u2212 1 2 \u2019 # Eq. (4.84) implies that when k is on the Bragg plane, the gradient of \u03b5 is parallel to the Bragg plane (see Figure 4.6b). Therefore, the constant-energy surfaces at the Bragg plane are perpendicular to the plane because the gradient is perpendicular to the surfaces on which a function is constant. j j It is easy to plot the energy bands from Eq. (4.82) if k is parallel to K (see Figure 4.7). When k =\u00bd K, the two bands are separated by a band gap 2. Vk It is much easier to consider the energy bands in one dimension. In one dimension, if we consider k at the zone boundary at \u00bdK, we note that (k \u2212 K)2 = k2 and \u03b50 k\u2212K: Thus, in Eqs. (4.20) and (4.23), we retain only the terms involving a, k \u2212 K k \u00f0 \u00f0 and we obtain and write E = \u03b5 k = \u03b50 and a k \u00f0 \u00de \u00de \u00de \u03b5 0 2 Vk k 1 K2 FIGURE 4.7 Energy bands when k is parallel to K. \u03b50 k \u2212 \u03b5 \u00f0 V\u2212Ka k k \u00f0 + a k + VKa k \u2212 K = 0, \u00de \u00f0 \u00de\u00de \u03b50 k\u2212K \u2212 \u03b5 \u00f0 a \u00de\u00de \u00de k \u2212 K \u00f0 k \u00f0 = 0: \u00f0 \u00de, Eqs. (4.85) and (4.86) can be solved by the determinant equation \u00de \u00f0 Because VK = V\u2212K = VK % which can be rewritten as Thus, the two roots are & & & & \u2212 \u03b5 2 k \u03b5 \u00f0 \u00de \u03b50 k \u2212 \u03b5 V %K \u00f0 k \u00de VK \u03b50 k\u2212K \u2212 \u03b5 k = 0, \u00f0 \u00de & & & & k\u2212K\u03b50 k \u2212 k \u00f0 \u00de\u00f0 k\u2212K + \u03b50 \u03b50 k\u00de + \u03b50 2 = 0: VK j j \u03b5\u00b1 = k \u00f0 \u00de 1 2 \u00f0 k\u2212K + \u03b50 \u03b50 k\u00de When k = K/2 (at the zone boundary), k\u2212K \u2212 \u03b50 \u03b50 k\u00de 2 + 2 VK j j \u00b1 1 4 \u00f0 h 1 2: i Substituting Eq. (4.90) in Eqs. (4.85) and (4.86), we obtain \u03b5\u00b1 \u00f0 K/2 \u00de = \u03b50 K/2 \u00b1 : VK j j for the two roots marked \u00b1. \u00b1VK K/2 a \u00f0 j \u00de j = \u2212VKa \u2212K/2, \u00de \u00f0 (4.85) (4.86) (4.87) (4.88) (4.89) (4.90) (4.91) 4.8 Different Zone Schemes 109 The corresponding eigenstates \u03c8 \u00b1 When k = K/2, assuming that VK is negative, a a K/2 \u00de for the positive root. Thus, we obtain K/2 \u00f0 \u2212K/2 = \u2212 are obtained from Eqs. (4.24), (4.85), and (4.86). for the negative root and = a \u2212K/2 \u03c8 \u2013 r \u00f0 \u00de = a K/2 \u00f0 ei/2Kr + e\u2212i/2Kr \u00de\u00bd $ and :", " $ Using the normalization conditions for the eigenstates, we obtain ei/2Kr \u2212 e\u2212i/2Kr \u00de\u00bd \u03c8 + K/2 = a r \u00f0 \u00f0 \u00de and \u03c8 \u2013 r \u00f0 \u00de = 1 2p \u00bd ei/2Kr + e\u2212i/2Kr = 2p cos $ 1 2 Kr, \ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03 \u03c8 + r \u00f0 \u00de = 1 2p \u00bd ei/2Kr \u2212 e\u2212i/2Kr 2p i sin 1 2 = $ Kr: (4.92) (4.93) (4.94) (4.95) When k is near the zone boundary, we can define a wave vector \u03b4, which measures the difference of k from the zone boundary by \ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03 From Eqs. (4.89) and (4.96), we obtain \u03b4 = K/2 \u2212 k: \u03b5\u00b1 k \u00f0 \u00de = \u2245 \u01272/2m \u00f0 \u01272/2m \u00de \u00f0 1 4 \u00de 1 \u2019 4 \u2019 K2 + \u03b42 K2 + \u03b42 4\u03b50 K/2\u00f0 \u00bd \u01272\u03b42/2m + VK 2 1 2 $ j 2 \u00de \u03b50 K/2/ j VK j j \u00de\u00f0 VK j j\u00bd 1 + 2 \u00f0 \u01272\u03b42/2m \u00b1 \u00b1 # # From Eqs. (4.90) and (4.97), we obtain \u03b5\u00b1 k = \u03b5\u00b1 K/2 + \u01272\u03b42/2m 1 \u00b1 2 \u00de 0, or far from a zone boundary (in the extended zone scheme), \u00de\u00bd \u00f0 \u00de \u00f0 \u00f0 \u00f0 j $ \u03b50 K/2/ VK j 2 : When k & (4.96) (4.97) (4.98) 1 2: \u00de$ \u03b50 k & which is a free electron parabola. We will represent these results in the reduced, extended, and repeated zone schemes described earlier for free electrons. \u01272k2/2m, (4.99) \u00de & k \u00f0 \u03b5 4.8 DIFFERENT ZONE SCHEMES 4.8.1 Reduced Zone Scheme In the reduced zone scheme, the wave vector k always lies within the first Brillouin zone. If a wave vector k\u2032 lies outside the first Brillouin zone, one can always find a lattice vector K\u2032 such that k = k\u2032 \u2212 K\u2032 lies within the first Brillouin zone. We show in Figure 4.8 the energy bands of a linear lattice (with a periodic potential) in the reduced zone scheme. 110 CHAPTER 4 Nearly Free Electron Model \u03c8 + \u03b5 (k) \u00de \u00f0 k However, 4.8.2 Extended Zone Scheme In the extended zone scheme, the energy \u03b5 is plotted against the wave vector k. We had seen that for free electrons, the curve is the free eleck tron parabola because \u03b5 = \u01272k2 2m : in the presence of a periodic potential, as we have seen, the parabola must meet the zone boundary normally, and an energy gap of 2|VK| develops between the lower and the upper band. This gap increases as K increases. The energy bands in the extended zone scheme of a linear lattice with periodic potential are shown in Figure 4.9. \u00f0 \u00de \u03c8 \u2212 O \u2192 k 1 K2 Figure 4.10 shows the energy bands for a linear lattice with periodic potential and succinctly demonstrates the development of the forbidden part of the zone (or more commonly referred to as band gap) that increases as the energy of the band increases. 1 K\u2212 2 FIGURE 4.8 The energy bands of a linear lattice in the reduced zone scheme. E Forbidden \u03b5 (k) Forbidden 2 Vk K k FIGURE 4.10 FIGURE 4.9 The energy bands of a linear lattice in the extended zone scheme. The energy gaps in a linear lattice. The width in energy increases as the energy of the band increases. In pure materials, there are no eigenstates for electrons with energy lying within these energy gaps. 0 \u03c0 /a 2\u03c0 /a 3\u03c0 /a k 4.9 Elementary Band Theory of Solids 111 E 3 2 1 1 2 3 \u2212(k1 + K) k1 + K E x t e n d e d \u2212(k1 \u2212 K) k1 \u2212 K Z o n e \u22123\u03c0 /a \u22122\u03c0 /a k1 Scheme 0 \u2212\u03c0 /a Reduced zone", " \u03c0 /a 2\u03c0 /a 3\u03c0 /a FIGURE 4.11 The repeated zone scheme for a linear lattice and its comparison with the reduced and extended zone schemes shown in Figures 4.9 and 4.10. k + K 4.8.3 Repeated Zone Scheme. In fact, When the first Brillouin zone is periodically repeated through all k space, \u03b5 \u03b5 : This type of construction of energy bands is known as \u00f0 the repeated zone scheme. This scheme is particularly useful in demonstrating the electron orbits in a magnetic field. The repeated zone scheme and its connection with the reduced and extended zone schemes is shown in Figure 4.11. is the same energy band as.9 ELEMENTARY BAND THEORY OF SOLIDS 4.9.1 Introduction We will now discuss the elementary band theory of solids using the one-dimensional lattice and analyze how crystalline solids, of which the basic components are negatively charged electrons and positively charged ions, have a wide diversity in physical properties. For example, some solids are metals that are good conductors, some others are metals but poor conductors, some crystallize as insulators, and the rest are crystals that are semiconductors. Each of these types of solids has widely divergent properties. For example, whereas some metals such as the alkali metals are very good conductors, some others such as the alkaline-earth metals are comparatively poor conductors. Another striking feature is that the ratio of the resistivity of metals (good conductors) and insulators is of the order of 10\u221220 at room temperature. As has been often remarked, this is one of the widest 112 CHAPTER 4 Nearly Free Electron Model divergences in the physical properties occurring in nature. Another example is the difference between the temperature dependence of the resistivity of metals and semiconductors. The resistivity of metals, which is small at absolute zero, increases with increase of temperature. In contrast, the resistivity of pure semiconductors, which are insulators (if the material is pure) at absolute zero, decreases with increase of temperature. We will try to explain this wide variety of exotic properties by using a simple one-dimensional band theory. In subsequent chapters, we will discuss the various techniques used in the energy band theory of solids in three dimensions and discuss the properties of various types of solids in a more rigorous manner. 4.9.2 Energy Bands in One Dimension We have discussed how a periodic potential breaks the free electron energy curve, which is a parabola, into discrete segments of interval \u03c0/a: Thus, there is a forbidden region for eigenstates of electrons in pure materials that is known as the energy gap. This energy gap increases with the increase in energy; i.e., the higher the energy of the band, the larger its width in energy. This significant fact, which is essentially the elementary band theory of solids, allows us to understand many of the characteristic features of solids. The basic idea of the formation of these energy bands was shown in Figure 4.10. 4.9.3 Number of States in a Band Here, we consider a one-dimensional crystal constructed of primitive cells of lattice constant a. The length of the crystal is L = Na, where N is the number of primitive cells. As we have noted, in one dimension, the allowed values of the electron wave vector k in the first Brillouin zone are given by k = 0; \u00b1 2\u03c0 L ; \u00b1 4\u03c0 L ; \u2026, N\u03c0 L : (4.100) We note that because N\u03c0/L \u2261 \u03c0/a, the point, defined as \u2212N\u03c0/L \u2261 \u2212\u03c0/a, is connected by a reciprocal lattice vector K with \u03c0/a, and hence cannot be counted as an independent point. The total number of points given in Eq. (4.100) is N. This result is also carried over to three dimensions; i.e., each primitive cell contributes one independent value of k to each energy band. If one considers the spin of the electron, each energy band will have 2N independent states. 4.10 METALS, INSULATORS, AND SEMICONDUCTORS We can now discuss the reason crystalline solids have to be grouped into four extremely dissimilar varieties: metals (good conductors), semimetals (poor conductors), insulators, and semiconductors. First, by using the elementary band theory, we will discuss some of the general features that are responsible for distinguishing solids into these four categories. Later, we will discuss some specific examples in each category to illustrate the characteristic features. If there is a single atom of valence one in each primitive cell, the first band (the bands are stacked above each other with [increasing] energy gaps between them, as shown in Figure 4.11) will be half-filled with electrons, and the solid will be a metal", " (good conductor) because there are enough empty states available for electrons to be excited whenever an electric field E is applied. In 4.10 Metals, Insulators, and Semiconductors 113 fact, when an electric field E is applied, the force on the electron of charge \u2013e is F = \u2212eE. The force is also the rate of change of momentum, F = \u2212eE = \u0127 dk dt : (4.101) Because the alkali metals and noble metals have one valence electron per primitive cell, they are good conductors. As an example, we consider sodium. Each Na atom has the atomic configuration 1s22s22p63s1. Thus, there is one valence electron in the 3s state in each separated Na atom, while the 3s state could accommodate two valence electrons. When N such atoms are bound in a solid, the 3s energy band has N electrons, and therefore, it is only half-filled (a band can accommodate 2N electrons). Thus, sodium is a good conductor because there are a large number of energy levels available just above the filled ones and the valence electrons can be easily raised to a higher energy state by an electric field, as shown in Eq. (4.101). In fact, as a rule of thumb, all monovalent solids are good conductors. According to the same rule of thumb, all divalent solids like the alkaline-earth metals that have two valence electrons per primitive cell should be insulators. However, this is not true if we consider a three-dimensional band picture. For example, we consider magnesium, of which the atomic configuration is 1s22s22p63s2. The 3s energy band is full, and as per the one-dimensional band picture, magnesium should be an insulator. However, this is not true because in a three-dimensional band picture, there is an overlap between the 3s and 3p bands, which is shown in Figure 4.12. In fact, the same overlap was also there for sodium in a three-dimensional band picture, but we did not have to take that fact into consideration because the 3s band was only half full. Because of this overlap between 3s and 3p bands, magnesium, like all alkaline-earth metals, is a metal, but some y g r e n E 0 y \u03b5F g r e n E \u03b5F k (a) \u03c0 a 0 \u03c0 a k (b) FIGURE 4.12 (a) The overlap of the 3s and 3p energy bands in three dimensions. (b) In Al, the 3s band is full, but the 3p band is not full. The upper energy level is the Fermi energy \u03b5F. 114 CHAPTER 4 Nearly Free Electron Model divalent solids such as Sr and Ba are poor conductors because the overlap is small and a relatively small number of electrons are excited when an electric field is applied. We note in Figure 4.12a that the three components of the wave vector k are in different directions in the 3s and 3p bands. Thus, the lowest energy levels of the 3p band are lower in energy than the highest energy levels of the 3s band, a fact that would be impossible in a one-dimensional band picture. Because the electrons tend to occupy the lowest energy states, some electrons have spilled over to the 3p band. We also note that the highest energy level in both bands is the Fermi level \u03b5F, which is in conformity with the definition of the Fermi level. All trivalent solids such as Al, Ga, In, and Tl are good metals because there are three valence electrons per primitive cell, and hence, they can fill one and a half bands. For example, the atomic configuration of aluminum is 1s22s22p63s23p1. The valence electrons are in the 3p band that is half empty and can be easily excited to higher energy states by an electric field. Therefore, aluminum is a very good conductor. This is schematically shown in Figure 4.12b. A crystalline solid becomes an insulator if it has only completely filled bands, provided the energy gap between the last-filled band (known as the valence band) and the next allowed empty band (known as the conduction band) is very large. The condition that the energy gap between the valence and conduction bands must be very large is due to the fact that the electrons can be thermally excited to the conduction band at room temperature if the energy gap is small. In addition, when an electric field is applied, the electrons are excited due to the external force (Eq. 4.101). The energy gap has to be large enough to prevent the excitation of the electrons from the valence band to the conduction band at reasonably large electric fields. The ionic crystals", " are good examples of insulators. The energy bands of an ionic crystal such as NaCl are from the Na+ (1s22s22p6) and Cl\u2212 ions. Because both of these ions have a closed-shell structure, all the occupied bands of NaCl are full and the energy gap between the highest occupied band (valence band) and the next empty (conduction) band is very large. Another example of an insulator that is a tetravalent solid is diamond, but the energy gap is very large for it to become a semiconductor. 1s22s22p63s23p6 \u00f0 \u00de In an insulator, the energy gap is large enough to prevent the valence electrons to be excited to the conduction band. However, if the applied electric field is greater than a critical value (known as the critical field), such that the valence electrons gain energy that is equal to or greater than the energy gap, they can be excited to the conduction band. In such cases, the insulator behaves like a good conductor as long as the applied electric field is greater than the critical field. Thus, insulators, which have large energy gaps, are used as breaking devices in high-voltage transmission. When the external electric field is greater than the critical field required to cross the energy gap, there is good transmission of electric current because the insulator behaves as a good conductor, but when the applied electric field drops below the critical value, the insulator stops the flow of current. The transmission of electricity is restored by appropriate repairs such that the external electric field is again larger than the critical field. In case of intrinsic semiconductors (pure semiconductors are called intrinsic in order to distinguish them from impurity [or doped] semiconductors), the energy gap between the valence band and the conduction band is sufficiently low (0.7 eV for Ge and 1.09 eV for Si). Although an intrinsic semiconductor behaves as an insulator at absolute zero temperature, some valence electrons are thermally excited to the conduction band, leaving behind an equal number of unoccupied states (holes) in the valence band. We will later show that these holes act like positive charges. In an applied electric field, both the (few) electrons in the conduction band and the holes in the valence band are excited and move in opposite directions, thereby conducting electricity. However, because 4.10 Metals, Insulators, and Semiconductors 115 the number of conducting electrons and holes is much smaller than that compared to metals, the resistivity of semiconductors is very large compared to that of metals. When the temperature is increased, more valence electrons are excited into the conduction band, leaving behind more (positively charged) holes in the valence band. Thus, the resistivity of semiconductors decreases with increase of temperature because of the increase in the number of carriers. In contrast, the resistivity of metals increases with the increase of temperature because the electrons are scattered by the lattice ions (phonons) and lattice impurities due to thermal vibrations. \u00de \u00f0 and Ge 1s22s22p63s23p63d104s24p2 We will now discuss the typical case of the most commonly used semiconductors such as Si 1s22s22p63s23p2. We note that Si has two 3s and two 3p \u00f0 electrons, and Ge has two 4s and two 4p electrons. Normally, we would expect Si and Ge to be conductors because each one of them has four unfilled p states. However, the 3s and 3p levels (for Si) and the 4s and 4p levels (for Ge) mix when they form covalent bonds. The energy of the electron levels corresponding to the four space-symmetric wave functions, one for the 2s levels and three for the 2p levels, is lowered. The energy of the other four levels, one 2s and three 2p, is raised. Thus, the valence band has four levels per atom that are filled, whereas the conduction band is empty. \u00de An interesting example is Sn, which is also a tetravalent solid. It has two phases: in one phase it is metallic, whereas in another phase it is a semiconductor. The shape of the Brillouin zone changes when the crystal structure is changed, and hence, it becomes possible to have large energy gaps to hold all the electrons. On the other hand, Pb, which is a tetravalent solid is a metal because of the band structure such that the electrons in the conduction band can be excited to higher energy states by an electric field. To summarize, the elements in Group IV of the periodic table have a wide range of properties. C in the form of a diamond is an insulator, Si and Ge are semiconductors, Sn can either be", " a metal or a semiconductor, whereas Pb is a metal. The pentavalent solids such as As, Sb, and Bi have 5 electrons per atom. However, their crystal structure is such that there are 2 atoms per unit cell. Thus, there are 10 electrons per unit cell. These 10 electrons would normally fill 5 bands. However, due to the effect of the band structure, the fifth band is not quite full because there is a little overlap (schematically very similar to Figure 4.12a) with the sixth band. Therefore, even at zero temperature, a few electrons in both the fifth and the sixth bands are always available to be excited (to carry the current) when an external electric field is applied. These are poor conductors and are known as semimetals. The iron group of the transition metals (Cr, Mn, Fe, Co, Ni) and the groups that are higher in the periodic table have incomplete d-shells. For example, only 6 out of the 10 states in the 3d-shell of Fe atom are filled, while two more electrons fill the outer 4s state. The d-orbitals in a solid overlap to form a \u201cd-band\u201d that can be treated by a tight-binding or LCAO method (see Chapter 5). The two electrons that form an s-band (in some transition metals there are electrons in both s and p states, and the corresponding band is known as the s-p-band) hybridize with the narrow d-band that is capable of accommodating up to 10 electrons per atom. This hybridization between d-bands and s-bands is shown in Figure 4.13. These bands are called resonance bands, and the hybridization is important in understanding the magnetic phenomena. Because neither of the bands is full, these solids are metallic and the conduction is mainly metallic. The four possible band structures for a solid are shown in Figure 4.14. Thus, it is possible to explain the occurrence of metals (good conductors), semimetals (poor conductors), insulators, and (intrinsic) semiconductors from a simple one-dimensional picture of band theory. We will later discuss in detail the characteristic properties of each of these solids. 116 CHAPTER 4 Nearly Free Electron Model We have summarized, by using the one-dimensional band theory of solids, the classification of solids into metals (good conductors), semimetals (poor conductors), insulators, and semiconductors. The rule of thumb is that each Brillouin zone has room for two electrons per primitive cell of a sample. If we consider a linear lattice that has one monovalent atom per primitive cell, the Brillouin zone is half filled. The electrons near the Fermi surface (the surface that separates the highest filledenergy states from the empty states) can be accelerated by an applied electric field, and because there are many empty states available, the metal is a good conductor. If there is one divalent atom per primitive cell, the first zone should be normally filled with electrons. However, in a threedimensional band picture, there is usually overlap between the top of the electron states in the first zone and the bottom of the empty electron states in the second zone. The energy gap (in different k directions) disappears. This leads the electrons to spill over from the top of the first zone to the bottom of the second zone, and the Fermi surface is in both zones. Such metals are not very good metals because of the small number of electrons that are excited in an external electric field. If the atom in each lattice point is trivalent, the first Brillouin zone is completely filled, but the second zone is half full. s-band d-bands \u03b5 \u03b5 O Z.B. O Z.B. k (a) FIGURE 4.13 k (b) (a) d-bands crossing s-bands; (b) s-d hybridization. The Fermi surface is in the second zone, and because a large number of electron states above the Fermi surface are empty, the solid is a metal and a good conductor. If there is a quadrivalent atom per primitive cell, the solid is either an insulator or a semiconductor depending on the magnitude of the energy gap. If there are two quadrivalent atoms per primitive cell (examples Forbidden Allowed, empty Allowed, occupied Key Conductor (a) Insulator (b) Conductor (c) Semiconductor (d) FIGURE 4.14 Four possible band structures for a solid: (a) conductor because the band is partially full, (b) insulator because of large energy gap between the filled and the empty bands, (c) semimetals because the allowed bands overlap, and (d) semiconductor because of the very small energy gap between the filled and the empty bands. 4.11 Brillouin Z", "ones 117 are diamond, silicon, and germanium), there are eight valence electrons per primitive cell. Because the bands do not overlap, diamond is an insulator (because of the large energy gap), and both silicon and germanium are intrinsic semiconductors because of the small energy gap. In both cases, there is no Fermi surface in the usual sense, but for semiconductors, the Fermi level is usually located at the center of the energy gap. The electrons in metals in the highest occupied states have immediate access to the empty states, and the surface that separates these states is called the Fermi surface. However, the highest occupied electron states in insulators and semiconductors (at zero temperature) are separated from each other by energy gaps. Thus, the Fermi surface plays a vital role in determining the properties of metals. The Fermi surface of free electrons is a sphere in three dimensions. However, the Fermi surface is much more complex in a metal because of the periodic potential. To be able to understand the increasing complexity of the Fermi surface in such solids, we need to first understand the properties of two- and three-dimensional lattices and the Brillouin zones. In the nearly free electron approximation, a constant-energy surface is perpendicular to a Bragg plane when they intersect. 4.11 BRILLOUIN ZONES We will first discuss the Brillouin zones for a two-dimensional square lattice, which we discussed in Chapter 1. The Bragg planes bisect the line joining the origin to points of the reciprocal lattice. The first Brillouin zone is defined as the set of points reached from the origin without crossing any Bragg plane (except that the points lying on the Bragg planes are common to two or more zones). The second Brillouin zone is the set of points that can be reached from the first zone by crossing only one Bragg plane. One can make a generalization of this definition and define the nth Brillouin zone as the set of points that can be reached from the origin by crossing no fewer than n \u2013 1 Bragg planes. The first four zones of the two-dimensional square Bravais lattice are shown in Figure 4.15. In general, a Brillouin zone can be constructed by using the rule that an incoming wave scatters strongly off a lattice with reciprocal lattice vector K, only when k. K =\u00bd K2: (4.102) The set of points that satisfy Eq. (4.102) is a plane that is perpendicular to the vector connecting the origin to K and lying midway between 0 and K. When many such planes are constructed using all possible K values, the origin would be enclosed within a solid region. This is the first Brillouin zone because all points inside are closer to the origin than any reciprocal lattice vector. An example of this construction is the Brillouin zone of a two-dimensional centered rectangular lattice shown in Figure 4.16. As explained earlier, the nth Brillouin zone is constituted of the set of points in reciprocal space that is closer to the n \u2013 1 reciprocal points 1 2 4 3 4 FIGURE 4.15 Brillouin zones for a two-dimensional square Bravais lattice. The first three zones are contained entirely in the square. 118 CHAPTER 4 Nearly Free Electron Model 1 2 4 3 4 FIGURE 4.16 The zone boundaries for a two-dimensional centered rectangular lattice are obtained by drawing perpendicular bisectors between the origin and the nearby reciprocal points. than it is to the origin. The construction of the first three Brillouin zones for a rectangular centered lattice is obviously more complicated than a square lattice. Such construction for the first three Brillouin zones, shaded in different ways, is shown in Figure 4.17. The first zone is the set of points closer to the origin than any other reciprocal lattice point. The second zone is constituted of the set of points that one reaches by crossing only one zone boundary. The third zone is the set of points that one reaches by crossing a minimum of two zone boundaries FIGURE 4.17 The first three Brillouin zones of a two-dimensional centered rectangular lattice. kz X R \u2206 \u039b S T \u0393 M\u03a3 Z ky kx The construction of Brillouin zones for a three-dimensional lattice gets more complicated. For example, the first Brillouin zone of a simple cubic lattice is simple cubic, but the first Brillouin zones of a bcc and a fcc lattice are much more complicated. The first Brillouin zone of a simple cubic lattice with the symmetry points is shown in Figure 4.18. (The symmetry points are explained in Appendix A.) First Brillouin zone of the simple cubic lattice. FIGURE 4.18 In Figure 4.18, the point \u0393 is at the center of the", " zone. R is at the corner of the cube that is connected to the other corners so that all eight corners are a single point. \u0393 and R have the same representation, the cubic group. X is at the intersection of the kz axis with the lower face of the cube. M is at the intersection of the kxky plane with the vertical edges (there are three equivalent points to M). M and X have the same symmetry elements 4/mmm. T is equivalent to the three points on the other vertical edges. The points T and \u0394 have the same point group, 4mm. The point \u039b has point group 3m. The points \u03a3 and S are holomorphic to 2mm. The point Z has two mirror planes and a two-fold axis. kz H \u2206 kx G \u0393 F P D N \u039b \u03a3 ky kx 4.12 Fermi Surface 119 X kz \u2206 L \u039b \u03a3 K W \u0393 U ky FIGURE 4.19 The first Brillouin zone of the bcc lattice. The first Brillouin zone of the body-centered cubic lattice, which is a rhombic dodecahedron, is shown in Figure 4.19 along with the symmetry points and the axes. The symmetry operations of \u0393, \u0394, \u039b, \u03a3 are the same as the similar points in the simple cubic lattice shown in Figure 4.19. H has the full cubic symmetry like \u0393. The Brillouin zone of a face-centered cubic lattice is a truncated octahedron that is shown in Figure 4.20. Here, \u0393 is at the center of the zone, L is at the center of each hexagonal face, X is at the center of each square face, and W is at each corner formed from one square and two hexagons. FIGURE 4.20 The Brillouin zone of the fcc lattice showing the symmetry points FIGURE 4.21 The Brillouin zone of the hcp structure. The Brillouin zone of the hexagonal close-packed structure is shown with the symmetry points in Figure 4.21. 4.12 FERMI SURFACE 4.12.1 Fermi Surface (in Two Dimensions) For free electrons, the Fermi surface is a circle in two dimensions and a sphere in three dimensions. The two-dimensional Fermi circles corresponding to one, two, and three electrons per atom for a square lattice are shown in Figure 4.22. 120 CHAPTER 4 Nearly Free Electron Model c1 c2 c3 FIGURE 4.22 FIGURE 4.23 Two-dimensional Fermi circles corresponding to one, two, and three electrons per atom in the Brillouin zones of a square lattice (without distortion at the zone boundaries). The weak periodic potential causes a distortion of the Fermi circle of a two-dimensional electron gas as it approaches the zone boundary. For example, in Figure 4.22, the free electron circle c1 is entirely within the first Brillouin zone. However, the free electron circles c2 and c3 intersect the zone boundaries. Because there is an energy gap between the electron states in the zone boundaries, the Fermi circle is distorted as it approaches the zone boundaries. In addition, the energy curves must be normal to the zone boundaries and develop \u201cnecks.\u201d The distortion of the Fermi circle by the weak periodic potential is shown in Figure 4.23. The distortion of the Fermi circle of a two-dimensional electron gas due to a periodic potential. ky kx FIGURE 4.24 Distortion of the free electron Fermi circles as energy contours approach the zone boundaries. However, the distortion of the Fermi circle does not suddenly develop only when the circle approaches the Brillouin zone boundary. The energy contours develop bumps that increase as they approach the zone boundary. The free electron Fermi circles are distorted much before they approach the zone boundary and develop \u201cnecks\u201d at the zone boundary because of the energy gap. The distortion of these energy contours is shown in Figure 4.24. We note from Figure 4.24 that there is a decrease in the constant energy contour when the Fermi circle has contact with the Brillouin zone. 4.12 Fermi Surface 121 4.12.2 Fermi Surface (in Three Dimensions) The occupied states of the free electron gas lie within a sphere. The radius of this sphere is the Fermi radius, and the surface is the Fermi surface. In Figure 4.25a, it is shown that when VK = 0, the free electron Fermi sphere meets the zone boundary at a distance \u00bdK from the origin O, but there is no distortion of the Fermi surface. In Figure 4.25b, VK \u2260 0 and there is distortion of the Ferm", "i sphere at the zone boundary. The Fermi surface intersects the plane in two circles. It can be shown (Problem 4.6) that the radii r1 and r2 of these circles are related by the equation 2 \u2212 r2 2 r1 \u00f0 = \u00de 4m \u01272 j VK : j (4.103) In Figure 4.26, we show a free electron Fermi surface completely enclosing the first Brillouin zone of a two-dimensional centered rectangular lattice. We note that the shape of the Fermi surface is modified near the zone boundaries. Figure 4.27 shows the portion of the Fermi surface in the second Brillouin zone that is mapped back into the first zone so that the energy surface is continuous. This is essentially achieved by using the reduced zone scheme. The portion of the Fermi surface is mapped back to the first Brillouin zone by appropriate translations through reciprocal lattice vectors so that the energy surface is contiguous, as shown in Figure 4.27. However, this method of mapping back the Fermi surface to the first Brillouin zone by any single reciprocal lattice vector becomes increasingly complicated even when there are electron states in the third Brillouin zone. In that case, it is not possible to map the contiguous portions of the third Brillouin zone into the first Brillouin zone by a single reciprocal lattice vector. In such cases, Harrison\u2019s method of construction of the Fermi surface becomes very useful. 4.12.3 Harrison\u2019s Method of Construction of the Fermi Surface When the band structure of a solid gets more complicated and the number of valence electrons per atom is large, it becomes very difficult to draw the Fermi surface of a metal. As we have noted, the shape and contours of the Fermi surface are important in determining the physical properties of a metal. To make this task simpler, Harrison2 proposed a method of constructing the Fermi surface of O K1 2 r1 r2 O K1 2 FIGURE 4.25 (a) (b) (a) Free electron sphere cutting Bragg plane when VK = 0; (b) free electron sphere cutting Bragg plane when VK \u2260 0. 122 CHAPTER 4 Nearly Free Electron Model FIGURE 4.26 FIGURE 4.27 Free electron Fermi surface completely enclosing the Brillouin zone of a two-dimensional centered rectangular lattice. The portion of the Fermi surface in the second Brillouin zone mapped back to the first zone. a metal of valence Z by using the periodic zone scheme. According to his method, if the perturbing potential is very small, the energy surfaces must be spheres. The radius of the sphere, which contains \u00bdZ times the volume of a zone (there are two electrons per each k state because of spin), is drawn with the center at the origin. The same sphere is drawn about each point of the reciprocal lattice, and one obtains a pattern that has the periodicity of the repeated zone scheme. Harrison\u2019s construction of the free electron Fermi surface is shown in Figure 4.28. From Figure 4.28, one can choose various parts that are continuously fitted together such that the surfaces are repeated in each zone. These different figures are either a branch of the Fermi surface or a part of the Fermi surface in the second and the third zone. The first zone is completely filled and therefore does not have a Fermi surface. The different parts of the Fermi surface for the second and the third zones are shown in Figures 4.29a and b. Free electron Fermi surface constructed by using Harrison\u2019s method. FIGURE 4.28 We note from Figures 4.29a and b that these surfaces have cusps where the parts join because they are drawn for spherical Fermi surfaces (free electrons). However, in the nearly free electron model, the Fourier components of the potential would round off the corners, and one would obtain smooth 4.12 Fermi Surface 123 (a) (b) FIGURE 4.29 (a) Fermi surface of the second zone in the reduced zone obtained from Harrison\u2019s construction (Figure 4.28). The orbit is a hole orbit. (b) Fermi surfaces in the third Brillouin zone. The orbit in the top-right corner (rosettes) is in the reduced zone scheme. FIGURE 4.30 FIGURE 4.31 One corner of the third zone of Figure 4.29b due to the effect of the periodic potential. The Harrison construction of the Fermi surfaces for a two-dimensional centered rectangular lattice. geometrical objects. This rounding off of the corners and the fact that the line of constant energy intersects the zone boundary at normal incidence are shown for a corner of the third zone in Figure 4.30.", " A more visual construction of Harrison\u2019s method (see Figure 4.31) shows how the surface in the nth Brillouin zone looks when it is mapped into the first Brillouin zone (i.e., the reduced zone scheme). 124 CHAPTER 4 Nearly Free Electron Model First zone\u2014full Second zone\u2014holes \u0393 (a) U K W X W W L \u0393 Third zone\u2014electrons Fourth zone\u2014electrons FIGURE 4.32 Free electron Fermi surface of aluminum in the reduced zone scheme obtained by Harrison. Reproduced from Harrison2 with the permission of the American Physical Society. Figure 4.31 shows how the surface of the nth Brillouin zone looks when it is mapped back into the first Brillouin zone. The Fermi sphere in the second Brillouin zone is identified by all points in the first Brillouin zone that are inside two or more spheres. The Fermi sphere in the third Brillouin zone is identified by all points in the first Brillouin zone that are inside three or more spheres. One can extend this method of obtaining Fermi surfaces in three dimensions by using the operations of constructive solid geometry. The free electron Fermi surface of aluminum in the reduced zone scheme, as obtained by Harrison, is shown in Figure 4.32. PROBLEMS 4.1. It can be shown in quantum mechanics that the momentum operator is the generator of infinitesimal translations \u03b5 (equivalent to an analogous relation in classical mechanics) ^T \u03b5 \u00de \u00f0 = 1 \u2212 i\u03b5 \u0127 ^p: (1) Problems 125 (2) (3) Using \u03b5 = a/N in Eq. (1), one obtains a/N ^T \u00f0 \u00de = 1 \u2212 ia \u0127N ^p: By using the formula show that the operator ^T obtained by a \u00de \u00f0 e\u2212ax = lim! N 1 \u2212 ax N \u00de \u221e \u00f0 N, corresponding to a finite translation a (in one dimension) can be ^T a \u00de \u00f0 = lim N! ^T \u221e \u00bd \u00f0 a/N \u00de$ N = e\u2212ia^p/\u0127: (4) \u00f0 \u00de fKeiK.r 4.2. In general, any function f (r) can be expanded in terms of the plane waves that form a complete set of functions. However, if a function f Bravais lattice, then it is easy to show that r + R \u00de for all r and all k in the = f r \u00f0 = \u2211 K because only eiK.r has the periodicity of the lattice. Show that the Fourier coefficients fK are given by r \u00f0 \u00de f (1) fK = 1 v ZC dr e\u2212iK.rf r, \u00de \u00f0 where v is the volume of the primitive cell C. To prove Eq. (2), first show that dreiK. \u00f0 r+l \u00de = dreiK.r = dreiK.r, ZC ZC\u2032 ZC (2) (3) where C\u2032 is the translated cell when C is translated through a vector l. From Eq. (3), one obtains eiK.l \u2212 1 \u00f0 \u00de ZC dr eiK.r = 0, from which it follows that (because eiK.l \u2260 1 \u00de dr eiK.r = 0, which is needed to prove Eq. (2) from Eq. (1). ZC (4) (5) 4.3. By using the Born\u2013von Karman boundary conditions, one obtains the periodicity for a crystal lattice (which can be considered as a very large Bravais lattice with the volume of the primitive cell V, the volume of the crystal), It has been shown that a vector of the reciprocal to this lattice has the form = f f r \u00de \u00f0 \u00f0 r + Miai \u00de, i = 1, 2, 3: 3 k = \u2211 i =1 mi Mi bi: (1) (2) 126 CHAPTER 4 Nearly Free Electron Model In a manner similar to Problem 4.2, show that if then and f r \u00f0 \u00de = \u2211 k fkeik.r, dr eik.r = 0, ZV fk = 1 V Z dre\u2212ik.rf r : \u00de \u00f0 4.4. It was shown in Eq. (4.4) that the lattice potential V r has the periodicity of the lattice Therefore, using the results from Problem 4.2, one obtains eiK.r From Eq. (2) of Problem 4.2, one obtains V K \u00f0 \u00de = 1 v Z", "C dr e\u2212iK.rV. r \u00de \u00f0 Assume that V(0) = 0. Show that because V symmetry, r \u00de \u00f0 is real and if the crystal has inversion Because the wave function \u03c8 r \u00f0 \u00de V K = V \u2212K = V K %: \u00f0 \u00de \u00f0 can be expanded in the set of plane waves Cqeiq.r, (3) (4) (5) (1) (2) (3) (4) (5) 3 where the q\u2032s are wave vectors of the form q = \u2211 i =1 can be written as \u01272 2m \u22072 + V \u01272 2m \u2212 \" $!! mi Mi bi, show that the Schrodinger equation q2 \u2212 E Cq + \u2211 K\u2032 \" VK\u2032Cq\u2212K\u2032 eiq.r = 0: (6) % The coefficient of each term in Eq. (6) must vanish (because the plane waves are orthogonal), \u01272 2m q2 \u2212 E Cq + \u2211 K\u2032 VK\u2032Cq\u2212K\u2032 = 0: \" If q = k \u2212 K, where k lies in the first Brillouin zone and changing the variables to K\u2032 K\u2032 \u2212 K, show that Eq. (7) can be written as! \u01272 2m Ck\u2212K \" + \u2211 K\u2032 VK\u2032\u2212KCk\u2212K\u2032 = 0:! (7) (8) Problems 127 Eq. (8) shows that for a fixed k in the first Brillouin zone, only wave vectors that differ from k by a reciprocal lattice vector are coupled. Rewriting \u03c8 and (8), show that and E as \u03b5, from Eqs. (5) as \u03c8 K\u00f0 r r \u00f0 \u00de \u00de = eik.r r \u03c8 k\u00f0 \u00de \u2211 G \u00f0 Ck\u2212Ke\u2212iK,r, \u00de \u01272 2m \u00f0! k \u2212 K 2 \u2212 \u03b5 \u00de Ck\u2212K + \u2211 K\u2032 \" VK\u2032\u2212KCk\u2212K\u2032 = 0: and Define uk r \u00de \u00f0 = \u2211 K Ck\u2212Ke\u2212iK.r, and hence show from Eqs. (9) and (11) that r \u03c8 k\u00f0 \u00de = eik.ruk : r \u00f0 \u00de From Eq. (11), show that \u00f0 is the Bloch function, and uk uk r = uk r + R \u00de is the periodic part of the Bloch function (in : \u00de r \u00f0 \u00f0 \u00de Here, \u03c8 k\u00f0 three dimensions). r \u00de (9) (10) (11) (12) (13) 4.5. In the Kronig\u2013Penney model, an electron in a one-dimensional lattice is in the presence of a potential V x \u00f0 \u00de = \u2211 n V0\u03d5 \u00f0 x \u2212 nb \u03c6 \u00f0 \u00de nb + c \u2212 x, \u00de (1) where b > c, n is zero or a positive integer, b + c = L and \u03c6 is the Heaviside unit-step function The one-dimensional periodic potential can be represented as = \u03c6 x \u00de \u00f0 $ 0, 1, x < 0: x > 0: : % (2) \u00de as shown in Figure P4.1. The potential energy as a function of distance is given in Figure P4.1 Because there is symmetry under the displacement by L, it can be shown that the eigenfunction is 3) \u03c8 x \u00de \u00f0 = eikxu x, \u00de \u00f0 (4) where u(x + L) = u(x) and k is arbitrary, and as we will show, it is the propagation constant. In addition, we consider the one-dimensional periodic potential visualized as a ring of circumference NL, such that \u03c8 x + NL \u00de \u00f0 = \u03c8 x : \u00de \u00f0 (5) 128 CHAPTER 4 Nearly Free Electron Model c First cell Second cell b L V0 0 0 E x FIGURE P4.1 The Kronig\u2013Penny model of the square potential lattice. From Eqs. (4) and (5), we obtain from which, we obtain eik \u00f0 x+NL \u00de = eikx, k = 2\u03c0n NL, where n = 0, \u00b11, \u00b12, :::: Here, k is called the propagation constant of the state. We represent the wave function as \u03c8 \u00f0 where x + L x \u03c8 \u00f0 \u00de \u00de, = eikL\u03c8 = eikxu x \u00f0 x", " \u00f0 \u00de, \u00de \u00de The advantage of Eq. (8) is that if we know \u03c8 can be calculated for any other cell for any one cell of the periodic lattice, it (6) (7) (8) (9) To solve this problem, one can use the one-dimensional equivalence of Eq. (4.35) (except that k is now called the propagation constant of the state k = 2\u03c0n NL ), d2u dx2 + 2ik du dx + 2m \u01272 E \u2212 V ( \u2212 x \u00de \u00f0 \u01272k2 2m = 0: u x \u00de \u00f0 ) From Figure P4.1 and Eq. (9), we have the periodicity condition u Introducing the notations and du x + L \u00f0 \u00de /dx = du /dx: x \u00f0 \u00de k1 = 2mE \u01272 \u2019 # 1 2 and k2 = 2m \u01272 \u00f0 V0 \u2212 E h 1 2, i \u00de the solution of Eq. (10) for the square lattice can be written as x, x + Be\u2212i = A ei k1+k k1\u2212k \u00f0 \u00de \u00de \u00f0 u1 x \u00f0 \u00de for the region of the well (from x = 0 to x = b), and = Ce\u00f0 k2\u2212ik x + De\u2212 \u00f0 \u00de k2+ik x, \u00de u2 x \u00de \u00f0 (10) (11) (12) (13) (14) Problems 129 for the region of the hill (x = b to x = L). Using the periodicity conditions mentioned previously, show that and A + B = e\u2212ikL \u00bd Cek2L + De\u2212k2L, $ A \u2212 B ik1 \u00f0 \u00de = k2e\u2212ikL \u00bd Cek2L \u2212 De\u2212k2L : $ Show from the continuity conditions that Aeik1b + Be\u2212ik1b = Cek2b + De\u2212k2b, and (15) (16) (17) $ Eqs. (15) through (18) have nontrivial solutions only if the determinant of the matrix of the coefficients vanishes. Show that \u00bd $ ik1 Aeik1b \u2212 Be\u2212ik1b = k2 Cek2b \u2212 De\u2212k2b \u00bd : (18) cos k1b cosh k2c \u2212 It can also be shown that cos k1b cos k2c \u2212 k2 1 \u2212 k2 2 2k1k2 k2 1 + k2 2 2k1k2 sin k1b sinh k2c = cos kL, E < V0: (19) sin k1b sin k2c = cos kL, E > V0: (20) Eqs. (19) and (20) can be solved numerically, and the results are shown in Figure P4.2. The remarkable feature is that the right sides of the eigenvalue equations are bound between \u20131 and +1. Thus, only those values of E that make the left side of these equations also lie in the same interval and all other values are excluded. This is the origin of the band structure in solids. +1 0 \u22121 1 2 3 E/V0 FIGURE P4.2 The left sides of Eqs. (19) and (20) are plotted as functions of E that join smoothly at E = V0. The heavy lines display the allowed range of energy values. (1) (2) (3) 130 CHAPTER 4 Nearly Free Electron Model 4.6. If we write Eq. (4.82) can be rewritten as k =\u00bd K + q, = \u03b50 K/2 + \u03b5 k \u00f0 \u00de \u01272q2 2m \u00b1 4\u03b50! K/2 \u01272q2 k2m + 2 VK j j \" 1 2, is the parallel component of q. We can also write \u03b5F = \u03b50 K/2 \u2212 + \u03b4: VK j j where q k Show that: a. When \u03b4 < 0, no Fermi surface intersects the Bragg plane. b. When 0 < \u03b4 < 2|VK|, the Fermi surface intersects the Bragg plane in a circle of radius, r = : 2m\u03b4 \u01272 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 r c. When \u03b4 > |2VK|, the Fermi surface cuts the Bragg plane in two circles (because it lies in both bands) of radii r1 and r2 and r1 2 \u2212 r2 2 =", " 4m VK j \u01272 j : (4) References 1. Ashcroft NW, Mermin ND. Solid state physics. Toronto: Thomson Learning; 1976. 2. Harrison WA. Band structure of aluminum. Phys Rev 1960;118:1182. 3. Harrison WA. Solid state theory. New York: McGraw-Hill; 1969. 4. Kittel C. Introduction to solid state physics. New York: John Wiley & Sons; 1976. 5. Liboff RI. Introductory quantum mechanics. Reading, MA: Addison-Wesley; 1980. 6. Madelung O. Introduction to solid state theory. New York: Springer-Verlag; 1978. 7. Marder MP. Condensed matter physics. New York: John Wiley & Sons; 2000. 8. Myers HP. Introductory solid state physics. London: Taylor & Francis; 1990. 9. Seitz F. The modern theory of solids. New York: McGraw-Hill; 1940. 10. Wannier GH. Elements of solid state theory. Cambridge: Cambridge University Press; 1954. 11. Wilson AH. The theory of metals. Cambridge: Cambridge University Press; 1958. 12. Ziman JM. Principles of the theory of solids. Cambridge: Cambridge University Press; 1972. Band-Structure Calculations CHAPTER 5 CHAPTER OUTLINE 5.1 Introduction............................................................................. 131 5.2 Tight-Binding Approximation.............................................................. 131 5.3 LCAO Method............................................................................ 135 5.4 Wannier Functions....................................................................... 140 5.5 Cellular Method......................................................................... 142 5.6 Orthogonalized Plane-Wave (OPW) Method.................................................. 145 5.7 Pseudopotentials........................................................................ 147 5.8 Muffin-Tin Potential...................................................................... 149 5.9 Augmented Plane-Wave (APW) Method..................................................... 150 5.10 Green\u2019s Function (KKR) Method.......................................", "..................... 152 5.11 Model Pseudopotentials.................................................................. 156 5.12 Empirical Pseudopotentials............................................................... 157 5.13 First-Principles Pseudopotentials.......................................................... 158 Problems.................................................................................... 160 References.................................................................................. 163 5.1 INTRODUCTION In the preceding chapter, the nearly free electron approximation was described in detail. In this approximation, the valence electrons are considered to be nearly free electrons moving in a background of positive charges that are arranged in a regular array such that the potential is periodic. The positive charges are actually the cores of the atoms in which the valence electrons have been stripped from the parent atoms. These cores either are located at the lattice points or a group of cores is located symmetrically around each lattice point. The cores are constituted from atomic orbitals that are essentially localized. In general, these atomic orbitals do not overlap. However, in some cases, the localized atomic orbitals overlap to an extent such that they form Bloch functions in the crystalline solid. 5.2 TIGHT-BINDING APPROXIMATION It is useful to formulate a linear combination of these atomic orbitals such that they would be Bloch functions as required for wave functions in a crystalline solid. Thus, the tight-binding approximation, formulated for the core electrons, complements the nearly free electron approximation formulated for Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00005-0 \u00a9 2012 by Elsevier Inc. All rights reserved. 131 132 CHAPTER 5 Band-Structure Calculations the valence electrons. The tight-binding method is very useful in calculating the energy bands of partially filled d-shells of transition metal atoms as well as for describing the electronic structure of insulators. \u00de In its simplest form, the tight-binding method can be expressed as follows. We assume that is an atomic orbital for a free atom located at a lattice point at the origin; i.e., it is the ground r \u03d5a\u00f0 of an isolated atom such that the energy is \u03b5a. As state of an electron moving in the potential va \u00de a first approximation, we consider a monatomic lattice with one atom located at each lattice point. We assume that the free atom is in an s state since the bands obtained from the other (p, d, \u2026) states are much more complicated because the atomic levels are degenerate. We further assume that the influence of one atom on another is small. If the bound levels of the atomic Hamiltonian Ha are localized, the Schrodinger equation can be written as = \u03b5a\u03d5a\u00f0 Ha\u03d5a\u00f0 r (5.1) r \u00f0 r \u00f0 \u00de \u00de \u00de, where Ha = \u2212!h2 2m \u22072 + va: (5.2) \u00de r The range of \u03d5a\u00f0 is very small when r exceeds the distance of a lattice constant. However, in the crystalline solid, the Hamiltonian H would differ from the atomic Hamiltonian because of the as the (small) correction corrections to the atomic potential in the crystal lattice. If we write \u0394", "V to the atomic potential in the lattice (the difference between the periodic potential in the crystal lattice and the potential of an isolated atom), the crystal Hamiltonian H can be written as r \u00de \u00f0 H = Ha + \u0394V r : \u00de \u00f0 (5.3) The periodic potential V r \u00de \u00f0 is the sum of the atomic potential va r \u00de \u00f0 r \u00de \u00f0 are drawn r In Figure 5.1, r\u03d5a\u00f0 \u00de along a linear chain of the atomic sites that are located at the lattice points. and \u0394V = va V r r \u00de \u00f0 \u00f0 The Schrodinger equation can be written as \u2212!h2 2m! \u22072 + V r \u00de \u00f0 \" r \u03c8 k\u00f0 \u00de = Ek\u03c8 k\u00f0 r \u00de : (5.5) r We note that the eigenfunctions \u03c8 k\u00f0 have to be Bloch functions. As a zero-order approximation, we assume that \u0394V = 0 in the region \u00de \u2260 0: In that case, we can construct where \u03d5a\u00f0 the Bloch functions as a linear combination of the atomic orbitals r r \u00f0 \u00de \u00de + \u0394V r : \u00de \u00f0 \u00de FIGURE 5.1 and the correction term \u0394V r, \u00f0 \u00de (5.4) r\u03c6a \u2206V(r) r r \u03c8 k\u00f0 \u00de = \u2211 i eik.Ri \u03d5a\u00f0 r \u2212 Ri, \u00de (5.6) Here, r\u03d5a(r) and \u0394V (r) are plotted against r in a linear chain of atomic sites. 5.2 Tight-Binding Approximation 133 where the Ri are the complete set of Bravais lattice vectors in the crystal lattice. We will now prove that \u03c8 k\u00f0 are Bloch functions. From Eq. (5.6), r \u00de r + Rj \u03c8 k\u00f0 \u00de r + Rj \u2212 Ri Ri\u2212Rj eik.Ri\u03d5a\u00f0 eik. \u00f0 eik.Rl\u03d5a\u00f0 = \u2211 i = eik.Rj \u2211 i = eik.Rj \u2211 l = eik.Rj \u03c8 k\u00f0 r, \u00de \u00de\u03d5a\u00f0 r \u2212 Rl \u00de \u00de r + Rj \u2212 Ri \u00de (5.7) which are the criteria for the Bloch functions. The expectation value of the energy can be written as = Z k \u03b5 \u00f0 \u00de!h2 2m \u22072 + V r \u00f0 \u03c8 #k \u2212 # \u03c8 #k\u00f0 r \u00de \u03c8 k\u00f0 r \u00de Z dr r \u00de : \u00de \u03c8 k\u00f0 $ dr Assuming that the overlap between the neighboring cells is small, We obtain from Eqs. (5.4) and (5.8), \u03c8 #k\u00f0 r r \u03c8 k\u00f0 dr \u2248 1, j eik. Ri\u2212Rj \u00f0 \u00de r \u2212 Rj \u03d5#a\u00f0 \u00de # Z \u2212!h2 2m \u22072 + va + \u0394V r \u00de \u00f0 \u03d5a\u00f0 r \u00f0 \u00de $ r \u2212 Ri dr: \u00de From Eqs. (5.1) and (5.10), we can write \u03b5 k in the alternate form \u00f0 \u00de e\u2212ik.Rl k \u03b5 \u00f0 \u00de = \u03b5a + N\u2211 l r \u2212 Rl \u03d5a\u00f0 \u0394V r \u03d5a\u00f0 \u00de \u00f0 r \u00de dr: \u00de Z (5.8) (5.9) (5.10) (5.11) Here, N is the number of atoms in the monatomic crystal lattice. Neglecting all integrals except those at the atomic sites and between the nearest-neighbor atomic sites and defining Rn as the nearest-neighbor lattice vectors of a primitive lattice in which the origin is located, we can rewrite Eq. (5.11) in the alternate form where and k \u03b5 \u00f0 \u00de \u2248 \u03b5a \u2212 \u03b2 \u2212 \u03b3\u2211 l e\u2212ik.Rl, \u03b2 = \u2212N Z r \u03d5#a\u00f0 \u00de \u0394V r \u03d5a\u00f0 \u00de \u00f0 r \u00de dr, \u03b3 = \u2212N Z r \u2212 Rn \u03d5#a\u00f0 \u0394V r \u03d5a\u00f0 \u00de \u00f0 r \u00de dr: \u00de (5.12) (5.13) (5.14) 134 CHAPTER 5 Band-Structure Calculations As an example, we consider a simple cubic lattice. The nearest-neighbor positions of a simple cubic lattice", " of lattice constant a are Rn = : Thus, 0, \u00b1a, 0 \u00b1a, 0, 0 \u00de \u00f0 Eq. (5.12) becomes 0, 0, \u00b1a, and \u00de \u00de \u00f0 \u00f0, = \u03b5a \u2212 \u03b2 \u2212 2\u03b3 k \u03b5 \u00f0 \u00de \u00f0 coskxa + coskya+ coskza : \u00de (5.15) $ The energies are located in a band of width 12 \u03b3: If ka 1, Eq. (5.15) yields \u03b5 \u00f0 k \u00de\ufb03 \u03b5a\u2212\u03b2\u22126\u03b3+\u03b3k2a2: (5.16) In this derivation, we have considered only one state (the s state) of the free atom and obtained one band. We note that usually for s states, \u03b2 =0: We also note that Eq. (5.15) is periodic in k and therefore only those values of k lying in the first Brillouin zone will define independent wave functions. The number of states in the first Brillouin zone that corresponds to a nondegenerate atomic level is equal to 2N. In Figure 5.2, we schematically show \u03b5 plotted as a function of k along the cube axis by using the tighthas a minimum at k binding method. Here, \u03b5 \u00f0 = 0 along the cube direction and a maximum at k= k \u00de k \u00de \u00f0 \u03b5 (k) \u03b5a 2r \u2212\u03c0 /a FIGURE 5.2 O k 100 \u03c0 /a Here, \u03b5(k) is plotted as a function of k along the cube axis. \u03b5 Bands in solid \u03b5c \u03b5b N-fold N-fold \u03b5a N-fold Levels of free atoms \u00de, which is the zone boundary. \u03c0/a,0,0 \u00f0 Thus, for every state of an electron in the free atom, there will be a band of energies in the crystal. The complexities of the problem increase for higher atomic levels. One can generalize by stating that when N identical atoms are kept far apart, and each atom has several different atomic levels (orbitals), there will be N-fold degenerate states for a single electron. When these atoms are brought closer, the atomic orbitals overlap, and a band that has N states is formed. This is shown in Figure 5.3. FIGURE 5.3 O a Formation of bands from atomic levels as the atoms come closer. Thus, one can refer to the 3s-band, 3p-band, 3d-band, and so on. It is too simplistic to consider these higher atomic states in the simple tight-binding method discussed previously. However, in Figure 5.3, we have schematically shown the formation of bands in solids by using the tight-binding method. Figure 5.4 shows the broadening of the bands that ultimately overlap. It is important to note at this stage that when these bands, formed from different atomic orbitals, broaden and overlap, the tight-binding method has to be modified and one uses the linear combination of atomic orbitals (LCAO method). We will discuss specific cases of such overlaps when we consider the p- and d-bands as well as the s\u2011d \u201cresonance\u201d bands. 5.3 LCAO Method 135 V(r) Energy levels r (Spacing)\u2212a) (b) N-fold degenerate levels FIGURE 5.4 (a) Nondegenerate atomic levels; (b) energy levels of N such atoms in a periodic array plotted as a function of mean interatomic spacing. 5.3 LCAO METHOD In the linear combination of atomic orbitals (LCAO) method (Ref. 18), the Bloch wave function of the crystalline solid is expanded in terms that are linear combinations of atomic orbitals. Thus, Eq. (5.1) is rewritten as where r \u03c8 k\u00f0 \u00de = \u2211 i eik.Ri \u03d5c\u00f0 r \u2212 Ri, \u00de r \u03d5c\u00f0 \u00de = \u2211 j r \u03b2j\u03d5j\u00f0 \u00de, (5.17) (5.18) \u03b2j are arbitrary constants to be determined, and \u03d5j\u00f0 atom located at the origin. In general, \u03c8 k\u00f0 minimizing the ground-state energy. r \u00de r is one of the set of localized atomic orbitals on the is used as a trial wave function and \u03b2j are obtained by \u00de The crystal Schrodinger equation can be rewritten as \u03c8 k\u00f0 H\u03c8 k\u00f0 = \u03b5 k r \u00de \u00de \u00f0 r, \u00de where H = Hat + \u0394V r : \u00de \u00f0 (5.19) (5.20) Multiplying Eq", ". (5.19) by one of the atomic wave functions, \u03d5a# obtain r, and integrating over all r, we \u00de \u00f0 \u03d5a# r \u00f0 Hat + \u0394V \u00de\u00bd r r \u03c8 k\u00f0 \u00de \u00de\u2019 \u00f0 dr = \u03b5 k \u00de \u00f0 Z \u03d5a# r \u03c8 k\u00f0 \u00de \u00f0 r \u00de dr, Z (5.21) 136 CHAPTER 5 Band-Structure Calculations which can be rewritten in the alternate form \u03b5 k \u00f0 \u00de \u00f0 \u2212 \u03b5a \u00de Z \u03d5a# r \u03c8 k\u00f0 \u00de \u00f0 r \u00de dr = Z \u03d5a# \u0394V r \u00de \u00f0 r \u03c8 k\u00f0 \u00de \u00f0 r \u00de dr, where we have used Eq. (5.1) to obtain \u03b5a: Because the atomic orbitals are orthonormal, i.e., we obtain from Eqs. (5.17), (5.18), and (5.23), \u03d5a# r \u03d5a\u2032\u00f0 \u00de \u00f0 r \u00de dr = \u03b4aa\u2032, Z \u03d5a# r \u03c8 k\u00f0 \u00de \u00f0 r \u00de dr = \u03b2a + \u2211 j \u03b2j \u2211 i\u22600 Z Z \u03d5a# r \u03d5j\u00f0 \u00de \u00f0 r \u2212 Ri \u00de eik.Ri dr :! From Eqs. (5.21), (5.23), and (5.24), we obtain k \u03b5 \u00f0 \u00de % \u2212 \u03b5a \u03b2a = & \u03b5a \u2212 \u03b5 k \u00f0 \u2211 \u00de j \u00f0 \u03b2j \u2211 \u03d5a# r \u00de \u00f0 r \u2212 Ri \u03d5j\u00f0 \u00de eik.Ridr! r \u03d5j\u00f0 \u00de \u00f0 r \u00de dr \" i\u22600 Z \u0394V r \u00de \u03b2j + \u2211 j! Z \u03d5a# \u00f0 (5.22) (5.23) (5.24) (5.25) + \u2211 j \u03b2j \u2211 i\u22600 Z \u03d5a# \u0394V r \u00de \u00f0 r \u03d5j\u00f0 \u00de \u00f0 r \u2212 Ri \u00de eik.Ridr :! We note that each of the three terms on the right side of the equation is small (for different reasons) unless the atomic levels are degenerate. The first term contains the overlap integrals \u03d5a# r \u03d5j\u00f0 \u00de \u00f0 \u00de r \u2212 Ri dr where Ri \u2260 0. The atomic wave functions are centered on different lattice Z sites, and therefore, the overlap is very small compared to unity because the atomic orbitals are well localized. The second term on the right side of the equation is small because at large distances, where \u0394V (which is the difference between the periodic and atomic potentials) is significant, the atomic wave functions are small. The third term on the right side is small for the same reason as the first term, because they also contain atomic wave functions centered at different sites. r \u00f0 \u00de \u00de \u00f0 k for the three bands, and \u03b2 If the atomic levels are nondegenerate, i.e., for an s-level, Eq. (5.25) is essentially the same as Eq. (5.13). For bands arising from atomic p-levels, which are triple-degenerate, Eq. (5.25) would give a set of three homogeneous equations. One has to solve a 3 \u00d7 3 secular problem. The, the appropriate linear combination of eigenvalues would give \u03b5 the atomic orbitals at the various k values in the Brillouin zone, would be obtained from the solutions. Similarly, the d-levels are five-fold degenerate and yield five homogeneous equations. One has to solve a 5 \u00d7 5 problem to obtain the eigenvalues as well as the values of \u03b2(k). In transition metals, the s and d electrons overlap in energy. In addition, because the s electrons have a large degree of plane wave character, they have essentially a uniform spatial distribution. Therefore, there is a spatial overlap between the s and the d orbitals. There is an intra-atomic s-d resonance that hybridizes the s- and d-bands due to this overlap in both real space and energy. Therefore, for transition metals, one has to solve a 6 \u00d7 6 secular problem that includes both d- and s-levels. k \u00de \u00f0 5.3 LCAO Method 137 L \u0393 X W K FIGURE 5.5 \u0393\u03a7 direction of the first Brillouin zone of the fcc structure. E \u0393 X FIGURE 5.6 The five d sub-bands of a transition metal", " atom in the \u0393\u03a7 direction of a fcc structure. The five d sub-bands of a transition metal in the \u0393\u03a7 direction in the fcc structure that is shown in Figure 5.5 are shown in Figure 5.6. The progression of the 3d-band through the Fermi energy for elements in and near the long series of transition metals is shown in Figure 5.7. The bands that are primarily of sp character are shown by the heavier lines. The tight-binding approximation is more complicated in crystalline solids in which the Bravais lattice is not monatomic. For example, the hexagonal close-packed metals are simple hexagonal with a two-point basis. There are two ways of solving this problem. One procedure is to consider the two-point basis as a molecule, of which the wave functions are known, and treat the problem in the same manner as we have done for a monatomic lattice, except that in this case, molecular wave functions are used instead of atomic wave functions. For s-levels, for which the nearest-neighbor overlaps are small, the overlap would be small in the molecule. Thus, an atomic s-level will give rise to two nearly degenerate molecular levels and hence will yield two tight-binding bands. The alternate 138 CHAPTER 5 Band-Structure Calculations Ar fcc Ti hcp V bcc Cr bcc Fe bcc EF Co fcc Ni fcc Cu fcc Zn hcp Key EF sp-band d-bands FIGURE 5.7 The progression of the 3d-band through the Fermi energy for elements in and near the first long series of transition metals. \u0393 X 5.3 LCAO Method 139 method is to make linear combinations of atomic levels centered at the Bravais lattice points and at the basis points. The LCAO wave function would be of the form r \u03c8 k\u00f0 \u00de = \u2211 i a1\u03d5 r \u2212 Ri + a2\u03d5 \u00f0 \u00de r \u2212 Ri + l \u00de \u00f0 % where l is the vector connecting the two basis atoms in the Bravais lattice. eik.Ri, & (5.26) In these methods of tight-binding calculations, we have considered only the spin-independent linear combinations of atomic orbitals, thereby implicitly neglecting spin-orbit coupling, which is important in calculating the atomic levels in the heavier elements. We can include spin-orbit coupling by considering the interaction of the electron spin on the orbital of its parent atom located at the origin as well as by including the interaction between the spin of that electron and the electric. In this method, we must use linear combinations of the field of all the other ions in \u0394V \u00de spin-dependent atomic wave functions. r \u00f0 The LCAO method is not very satisfactory for the quantitative calculation of Bloch functions in solids. It is very difficult to calculate the three-center integrals and nonorthogonal basis functions. In addition, representing the valence electron states in a metal or a semiconductor by an expression such as Eq. (5.17) is basically wrong. We consider the interstitial region, the creation of which is illustrated in Figures 5.8 and 5.9. Figure 5.8 shows the bound atomic orbitals of two free atoms. In Figure 5.9, these two bound atomic orbitals are superposed, thereby creating an interstitial region with a small constant potential. \u03b5 = 0 \u03b5 r + \u03c5a FIGURE 5.8 Potential wells centered on the nuclei of two free atoms. \u03b5 r \u03c8 \u2032 FIGURE 5.9 The Bloch functions in a crystalline lattice. \u03b5 r (i) \u03c6a \u03c5a \u03c8k 140 CHAPTER 5 Band-Structure Calculations The individual atomic orbitals, used in Eq. (5.17), do not exist in the interstitial regions between the main potential wells centered on the atomic nuclei because they would lie above the energy of the barriers between the atomic spheres. It is also inappropriate to represent a Bloch function, which is a consequence of a periodic potential V(r) in the crystalline solid, by a linear combination of the atomic orbitals \u03d5a\u00f0 Finally, the set of functions used in the LCAO method is incomplete because it does not include the scattered-wave eigenstates in the Schrodinger equation in the continuum, above the energy zero of : They are useful in the construction of the Bloch functions inside the atomic core but inappropriva ate in the interstitial region, where the use of a linear combination of plane waves is more appropriate., which arise due to an atomic potential va and rapidly tend to zero as r\u2192\u221e. r r r \u00de \u00f0 \u00f0 \u00de \u00de The main disadvantage of the tight-binding as well as the LCAO methods is that the", " independent electron approximation is one of the bases of these approximations. This approximation works very well for both insulators and the low-lying bands in metals. In such cases, the tight-binding bands are very low in energy and completely filled. However, the independent electron approximation in the formulation of the tight-binding method fails when there are partially filled bands that are obtained from localized atomic orbitals with small overlap integrals. In particular, when we consider the properties of the narrow tightbinding bands that are obtained from partially filled atomic shells, the failure of the independent electron approximation becomes more apparent and leads to anomalous results. When one uses the tight-binding method for the d- and f -shells in metals, the inclusion of the electron\u2013electron interaction becomes very important. In particular, when there is a magnetic structure, this failure is even more apparent. One of the important reasons for going beyond the tight-binding approximation is that when there is a second electron at a given atomic site, the strong repulsion between the first electron and the second electron at the same site cannot be treated by the independent electron approximation. An interesting consequence of the tight-binding approximation is the hypothetical question as to what would happen if the distance between the atoms is continuously increased so that there is a slow but steady transition from the metallic to the atomic state. The overlap integrals would become smaller with the increase of the lattice constant, and eventually all bands, including the partially filled conduction band(s), become very narrow tight-binding bands. The decrease of electrical conductivity of a metal would continuously drop with the decrease of the overlap integrals as the lattice constant increases and eventually would become zero. Thus, the metal would become an insulator. This is known as a Mott transition. In practice, if electron\u2013electron interactions are included, the conductivity would abruptly become zero at a Mott transition. The Mott transition has indeed been observed in certain transition metal oxides. They are nor- mally insulators but suddenly become good conductors above a certain temperature. 5.4 WANNIER FUNCTIONS One of the main problems of the tight-binding formulation is caused by the difficulties of orthogonalization since a linear combination of the atomic orbitals is used. This orthogonalization problem can be circumvented by defining a set of orthonormal wave functions that can be constructed from Bloch functions and are localized on the atomic sites. These functions are known as Wannier functions. The Wannier function (Ref. 19) centered at the lattice site Ri is defined as r \u2212 Ri wn \u00f0 = \u00de 1 Np \u2211 k e \u2212ik.Ri\u03c8 nk\u00f0 r, \u00de \ufb03\ufb03\ufb03\ufb03 (5.27) 5.4 Wannier Functions 141 where N is the number of lattice sites as well as the number of k states in the first Brillouin zone. We will now show that the Wannier functions form an orthonormal set: dr w#m\u00f0 r \u2212 Rj r \u2212 Ri wn \u00de \u00f0 = \u00de dr\u2211 k \u2211 k\u2032 Z Z 1 N e\u2212ik.Ri + ik\u2032.Rj\u03c8 #mk\u2032\u00f0 r \u03c8 nk\u00f0 \u00de r \u00de : Because the Bloch functions are orthonormal (Problem 5.3), i.e., Eq. (5.28) can be rewritten in the alternate form dr \u03c8 #mk\u2032\u00f0 r \u03c8 nk\u00f0 \u00de r dr = \u03b4m,n\u03b4k,k\u2032, \u00de Z dr w#m\u00f0 r \u2212 Rj wn \u00de \u00f0 r \u2212 Ri = \u00de 1 N \u2211 k \u2211 k\u2032 e\u2212ik.Ri+ik\u2032.Rj\u03b4m,n\u03b4k, k\u2032: Z Because it can be easily shown that (Problem 5.4) e\u2212ik. \u00f0 Ri\u2212Rj \u00de = N\u03b4i, j, \u2211 k we obtain from Eqs. (5.30) and (5.31) the orthonormal conditions of the Wannier functions, dr w#m\u00f0 r \u2212 Rj wn \u00de \u00f0 r \u2212 Ri \u00de = \u03b4m,n\u03b4i, j: Z (5.28) (5.29) (5.30) (5.31) (5.32) If the Wannier functions are known, one can obtain the Bloch functions from Eq. (5.27) by multiplying both sides by eik\u2032.Ri and summing over all direct lattice vectors Ri,.Ri\u03c8 nk\u00f0 eik\u2032.Riwn r \u2212 Ri k\u2212k\u2032 \u00de (5.33) e\u2212i = r \u00f0", " \u00de \u00de : \u00f0 1 Np \u2211 k \u2211 Ri \u2211 Ri It can be easily shown that (Problem 5.5) \ufb03\ufb03\ufb03\ufb03 e\u2212 \u00f0 k\u2212k\u2032 \u00de.Ri = N\u03b4k,k\u2032: \u2211 Ri From Eqs. (5.33) and (5.34), we obtain r \u03c8 nk\u00f0 \u00de = 1 Np \u2211 Ri eik.Riwn r \u2212 Ri \u00f0 \u00de : (5.34) (5.35) It may be noted that any Bloch function is determined only within an overall phase factor. So the Wannier functions defined in Eq. (5.27) can also be defined as \ufb03\ufb03\ufb03\ufb03 wn r \u2212 Ri \u00f0 \u00de = 1 Np \u2211 k e\u2212ik.Ri + i\u03c6 k \u00f0 r \u00de\u03c8 nk\u00f0 \u00de, (5.36) k \u00de where \u03c6 is an arbitrary real function. The Wannier functions can be optimized by making them drop off as fast as possible when r \u2212 Ri starts increasing. This can be achieved by manipulating the arbitrary function \u03c6 : The Wannier functions on adjacent sites can be schematically shown as in Figure 5.10. \u00f0 \ufb03\ufb03\ufb03\ufb03 k \u00de \u00f0 142 CHAPTER 5 Band-Structure Calculations FIGURE 5.10 The Wannier functions on adjacent sites. 5.5 CELLULAR METHOD The cellular method was originally proposed by Wigner and Seitz (Ref. 20) to calculate the band structure of sodium. Later, with the availability of powerful computers that could solve differential equations, the cellular method was considerably improved and generalized to calculate the band structure of metals. The essential principle behind the cellular method is that because of the Bloch condition r + Ri \u03c8 k\u00f0 = eik.Ri \u03c8 k\u00f0 r, \u00de (5.37) \u00de it is sufficient to solve the Schrodinger equation in Eq. (5.5) within a single primitive cell C. The wave function in other primitive cells can be obtained from its values in C by using Eq. (5.37). must be continand \u2207\u03c8 k\u00f0 However, both \u03c8 k\u00f0 uous as r crosses the cell boundary. In the original Wigner\u2013Seitz formulation, the Wigner\u2013Seitz cell of a fcc or a bcc lattice is a polyhedron (see Figure 5.11) that could be approximated as a sphere of equal volume of which the radius is rs. We will denote this sphere as the Wigner\u2013Seitz sphere. r r \u00de \u00de In addition, they considered the lowest state energy of sodium metal so that k = 0. Thus, the boundary conditions reduce to, r r + Ri = \u03c8 k\u00f0 \u03c8 k\u00f0 is periodic from cell to cell, r and because \u03c8 k\u00f0 it should have a horizontal tangent at the surface of the Wigner\u2013Seitz sphere (5.38) \u00de \u00de \u00de rs FIGURE 5.11 \u2202\u03c8 k \u2202r # $r=rs = 0: (5.39) Wigner\u2013Seitz cell of sodium metal (bcc structure) approximated as a sphere. 5.5 Cellular Method 143 \u03c8a V(r) rs Conduction band \u03b5 (0) FIGURE 5.12 The ground-state energy \u03b5(0) of sodium metal in the Wigner\u2013Seitz method. For a potential that is symmetric inside the sphere, the energy \u03b5 Figure 5.12. One can also write, by using the Bloch condition, can be defined as shown in 0 \u00de \u00f0 r \u03c8 k\u00f0 \u00de = eik.ruk r, \u00de \u00f0 where uk r \u00de \u00f0 is the periodic part of the Bloch function. From Eqs. (5.5) and (5.40), \u2212!h2 2m \u00f0 \u22072 + 2ik. \u2207 \u2212 k2 uk r \u00de \u00f0 \u00de + V uk r \u00de \u00f0 r \u00f0 \u00de = \u03b5 uk k \u00de \u00f0 r, \u00de \u00f0 (5.40) (5.41) where uk obtained by using this method. However, the structure of the solid is neglected in this procedure. has a horizontal tangent at r = rs. The shape of the band can be approximately r \u00de \u00f0 There has been considerable improvement in the techniques of the cellular method for calculation of the band structure since Wigner and Seitz\u2019s original papers in 1933 and 1934 (Ref. 21", "). The must be continuous as r crosses the r starting point is the boundary conditions that \u03c8 k\u00f0 \u00de boundary of the primitive cell C (usually located at the origin, i.e., R = 0) to the neighboring cell and \u2207\u03c8 k\u00f0 r \u00de 144 CHAPTER 5 Band-Structure Calculations located at the lattice point R. These boundary conditions, which introduce the wave vector k in the solution as well as retain the discrete set of energies \u03b5n = e \u2212ik.R\u03c8 k\u00f0 = \u03b5, can be restated as \u00de r + R \u03c8 k\u00f0 (5.42) k r \u00de \u00de \u00f0 and \u2207\u03c8 k\u00f0 r \u00de = e \u2212ik.R\u2207\u03c8 k\u00f0 r + R \u00de (5.43) for pairs of points on the surface separated by R. If the point r is located on the surface of the primitive cell C, the normals ^n to surface of the cell at r and r + R are oppositely directed. Therefore, the continuity condition (Eq. 5.43) can be rewritten as (Problem 5.7) ^n r \u00de \u00f0. \u2207\u03c8 k\u00f0 r \u00de = \u2212 e \u2212 ik.R^n r + R \u00f0. \u2207\u03c8 k\u00f0 \u00de r + R : \u00de (5.44) In the cellular method, the periodic potential V(r) in the Wigner\u2013Seitz cell C (see Figure 5.13) is replaced by a potential V\u2032 \u00f0 that has spherical symmetry (see Figure 5.14) about the origin. Because we have made the approximation that the potential V\u2032 \u00f0 is spherically symmetric inside the cell C, a complete set of solutions to the Schrodinger equation (Eq. 5.5) is of the form (see Goswami7 or any book on quantum mechanics) r r \u00de \u00de (5.45) (5.46) where Ylm \u03b8, \u03d5 \u00de \u00f0 \u00f0 are the spherical harmonics and R\u03b5l \u00de \u00de r r \u03c8 \u03b5lm\u00f0 = R\u03b5l r Ylm, \u00de \u03b8, \u03d5 \u00f0 satisfies the differential equation \u00f0 \u00de 1 r2 d dr \u00f0 r2 d dr\u00de + 2m!h2 # \u03b5 \u2212 V\u2032 \u00f0!!h2l l + 1 \u00f0 2mr2 \u00de \u2212 r \u00de R\u03b5l r \u00f0 \u00de = 0: \" $ FIGURE 5.13 The actual potential V (r) inside a primitive cell. FIGURE 5.14 The approximate potential V\u2032(r) with spherical symmetry. 5.6 Orthogonalized Plane-Wave (OPW) Method 145 Here, we have written \u03b5 instead of n (which is appropriate for an atom) for the quantum number related to energy. Any linear combination of Eq. (5.45) would also be a solution of Eq. (5.5). Thus, we obtain r, \u03b5 \u03c8 \u00f0 = \u2211 lm \u00de BlmR\u03b5l r \u00f0 Ylm \u00de \u00f0 \u03b8, \u03d5, \u00de (5.47) where the arbitrary coefficients Blm are determined from the boundary conditions in Eqs. (5.42) and (5.44). In actual computations, a finite number of Blm are chosen, and an equal number of points on the surface of the cell are chosen where the boundary conditions in Eqs. (5.42) and (5.44) are imposed. One obtains a set of k-dependent linear homogeneous equations for Blm. The energies are those values of \u03b5 for which the determinant of the homogeneous equations for Blm k \u03b5n \u00de vanishes. \u00f0 \u00de 5.6 ORTHOGONALIZED PLANE-WAVE (OPW) METHOD The nearly free electron method does not account for the rapid oscillatory behavior in the core region of the atom. On the other hand, the tight-binding method, which matches reasonably well to inside the atomic core, does not account for the scattered-wave eigenstates of the Schrodinger r \u03c8 k\u00f0. The Bloch states in equation in the continuum above the energy zero of the atomic potential va the interstitial region must behave like a combination of free electron plane wave states. This anomaly of finding a complete set of wave functions that can account for the entire region in the crystal was resolved by the orthogonalized plane-wave (OPW) method first introduced by Herring. This method is based on the basic fact that the eigenfunctions of the crystalline Hamiltonian must be orthogonal at every point in the crystal. r \u00de \u00f0", " In the OPW method, one starts with the construction of a complete set of Bloch functions for the core states. Here, the core states are defined as the atomic orbitals of the ions. A tight-binding combination of the core states, which are Bloch functions, as constructed in Eq. (5.17), can be written as r \u03d5ck\u00f0 \u00de = \u2211 Ri eik.Ri\u03d5c\u00f0 r \u2212 Ri, \u00de (5.48) \u00de r is one of the core orbital c. Here, the core wave functions are localized around the where \u03d5c\u00f0 is a linear combination of degenerate onek atomic sites and are assumed to be known. In fact, \u03d5c\u00f0 electron states. Thus, Eq. (5.48) is one of the solutions of the Schrodinger equation of the crystal. As we have noted in the discussion of the tight-binding method, this would k with energy \u03b5c \u00de form a fully occupied narrow band. If we denote \u03c7k\u00f0 as a wave function for one of the higher states, it should be orthogonal to the must have the properties of a core states defined in Eq. (5.48). In addition, the higher states \u03c7k\u00f0 free electron wave function (plane wave)eik.r in the interstitial region. Thus, we can define the OPW as r r \u00de \u00de \u00de \u00f0 r \u03c7k\u00f0 \u00de = eik.r \u2212 \u2211 c bc\u03d5ck\u00f0 r \u00de, (5.49) 146 CHAPTER 5 Band-Structure Calculations where the sum is over all the core levels. The coefficients bc are determined from the orthogonalization condition From Eqs. (5.49) and (5.50), we obtain dr \u03d5#ck\u00f0 r \u03c7k\u00f0 \u00de r \u00de = 0: Z dr \u03d5#ck\u00f0 r eik.r: \u00de bc = Z (5.50) (5.51) In deriving Eq. (5.51), we have made the basic assumption that the core states are orthogonal, i.e., dr \u03d5#ck\u00f0 r \u03d5c\u2032k\u00f0 \u00de r \u00de = \u03b4c,c\u2032: Z (5.52) \u00de The OPW state \u03c7k\u00f0 defined in Eq. (5.49) is appropriate for the interstitial region because the core r behaves like the plane wave r states are localized, and consequently, each \u03d5ck is small. Thus, \u03c7k\u00f0 eik.r in the interstitial region. However, because \u03c7k\u00f0 is orthogonal to the core state within the core, it is appropriately a higher atomic orbital than the occupied atomic states. For example, if the will behave like a 3s or 3p orbital with an extra nodal core states are 2s and 2p orbitals, \u03c7k\u00f0 surface. r r \u00de \u00de \u00de A schematic representation of an OPW is shown in Figure 5.15. The eigenfunctions of the Schrodinger equation in the crystalline lattice can be expressed as a linear combination of OPWs that would form the basis states. The wave function can be written as r \u03c8 k\u00f0 \u00de = \u2211 K Ck\u2212K \u03c7k\u2212K\u00f0 r \u00de : (5.53) The coefficients Ck\u2212K are obtained by using the variational principle, to minimize the expectation value of the energy. The variational principle is briefly explained here. If we define the energy functional!h2 2m j! \u2207\u03d5 r \u00f0 \u00dej 2 + V r \u00f0 \u00dej \u03d5 r \u00f0 \u00dej 2 dr \", = Z \u03d5 \u03b5 \u00bd \u2019 it follows from Eq. (11) of Problem 5.8 that which is the variational principle. r \u03d5 j \u00f0 2dr \u00dej Z \u03b5 \u03c8 k\u2019 \u00bd = \u03b5k, (5.54) (5.55) 5.7 Pseudopotentials 147 FIGURE 5.15 Plane waves, core functions, and orthogonalized plane waves. 5.7 PSEUDOPOTENTIALS In 1959, Phillips and Kleinman14 developed the pseudopotential method that is extensively used to calculate the band structure of metals. This method has been refined through the past 50 years. We will essentially discuss the original theory proposed by Phillips and Kleinman. They showed that advantage, which are smooth parts of the can be taken of the crystal symmetry to construct wave functions \u03d5k\u00f0 contains an additional term of simr symmetrized Bloch functions. The wave equation satisfied by \u03d5k\u00f0 \u00de ple character that corresponds to the usual OPW terms and has a simple physical interpretation as", " an effective repulsive potential. The sum of the crystal potential and this repulsive potential is known as the pseudopotential. The cancellation between the attractive periodic potential and the repulsive potential in the core region is responsible for the rapid convergence of the OPW calculations for s states. r \u00de, as expressed in Eq. (5.53) The pseudopotential method starts with the assumption that \u03c8 k\u00f0 and with appropriate coefficients Ck\u2212K obtained by the use of the variational method, is the exact crystal wave function. We construct a linear combination of the plane waves, r \u00de r \u03c6k\u00f0 \u00de = \u2211 K Ck\u2212Kei \u00f0 k\u2212K.r, \u00de (5.56) 148 CHAPTER 5 Band-Structure Calculations which have the same coefficients Ck\u2212K. Because \u03c8 k\u00f0 described in Eq. (5.48), we obtain r \u00de must be orthogonal to the core states \u03d5ck\u00f0 r \u00de r \u03c8 k\u00f0 \u00de = \u03c6k\u00f0 r \u00de \u2212 \u2211 c r < \u03d5ck\u00f0 \u00dej r \u03c6k\u00f0 \u00de > \u03d5ck\u00f0 r \u00de : The Schrodinger equation in the crystal is From Eqs. (5.57) and (5.58), we obtain H\u03c8 k\u00f0 r \u00de = \u03b5k\u03c8 k\u00f0 r \u00de : (5.57) (5.58) r H\u03c6k\u00f0 \u00de \u2212 \u2211 c < \u03d5ck\u00f0 r \u00dej r \u03c6k\u00f0 \u00de > H\u03d5ck\u00f0 r \u00de = \u03b5k\u03c6k\u00f0 r \u00de \u2212 \u03b5k\u2211 c < \u03d5ck\u00f0 r \u00dej r \u03c6k\u00f0 \u00de > \u03d5ck\u00f0 r \u00de : (5.59) Using the relation in Eq. (5.59) and rearranging the terms, we obtain H\u03d5ck\u00f0 r \u00de = \u03b5ck r \u03d5ck\u00f0 \u00de \u00f0 r \u00de, r H\u03c6k\u00f0 \u00de + \u2211 c \u00f0 \u03b5k \u2212 \u03b5ck < \u03d5ck\u00f0 r \u00dej r \u03c6k\u00f0 \u00de > \u03d5ck\u00f0 r \u00de = \u03b5k\u03c6k\u00f0 r \u00de : \u00de Eq. (5.61) can be expressed as \u00de where VR is the repulsive part of the potential, \u00f0 H + VR \u03c6k\u00f0 r \u00de r = \u03b5k\u03c6k\u00f0, \u00de VR\u03c6k\u00f0 r \u00de ( \u03b5k \u2212 \u03b5ck \u2211 c \u00f0 r < \u03d5ck\u00f0 \u00dej r \u03c6k\u00f0 \u00de > \u03d5ck\u00f0 r \u00de : \u00de (5.60) (5.61) (5.62) (5.63) Thus, Eq. (5.62) is an effective Schrodinger equation satisfied by \u03c6k\u00f0 the Bloch function. Because r \u00de, which is the smooth part of H + VR = \u2212!h2 2m \u22072 + V + VR, r \u00de \u00f0 the pseudopotential is defined as From Eqs. (5.62), (5.64), and (5.65), we obtain Vps = V + VR: r \u00de \u00f0 = \u2212! The \u201csmoothed\u201d wave function \u03c6k\u00f0 Hps\u03c6k\u00f0 r \u00de \u00de!h2 2m \u22072 + Vps r \u03c6k\u00f0 \" \u00de = \u03b5k\u03c6k\u00f0 r \u00de : is often referred to as the pseudo wave function that satisfies the effective Schrodinger equation, which has a pseudo-Hamiltonian Hps in which the potential r (5.64) (5.65) (5.66) 5.8 Muffin-Tin Potential 149 is relatively small because the pseudopotential Vps is the sum of the lattice potential and the repulsive potential VR: However, we note from Eq. (5.63) that VR is a nonlocal operator. Thus, we define such that VR r, r\u2032 = \u2211 c \u00f0 \u00de \u00f0 \u03b5k \u2212 \u03b5ck \u03d5ck\u00f0 \u00de r \u03d5#ck\u00f0 \u00de r\u2032 \u00de VR\u03c6k\u00f0 r \u00de = Z VR r, r\u2032 \u00de \u03c6k\u00f0 r\u2032 \u00de \u00f0 dr\u2032: (5.67) (5.68) Thus, VR is different when it operates on different functions with different angular momentum. This implies that the effect of the pseudopotential Vps on a wave function is not just to multiply it by a function of r. Further, because Vps depends on the energy \u03b5", "k that is supposed to be the quantity being calculated, one cannot apply some of the basic theorems of quantum mechanics. For example, one cannot apply the theorem that the eigenfunctions belonging to different eigenvalues of Hps are orthogonal. The pseudopotential formulation has another problem in the sense that there is no unique method to construct the pseudopotential. One can easily show (Problem 5.9) that the valence eigenvalues of the Hamiltonian H + VR are the same for any operator of the form VR\u03c6k\u00f0 r \u00de = \u2211 c < \u03b8ck r \u00f0 \u00dej r \u03c6k\u00f0 \u00de > \u03d5ck\u00f0 r \u00de, (5.69) where \u03b8ck r \u00de choose the functions \u03b8ck pseudopotential is small. \u00f0 \u00f0 \u00de are arbitrary functions. This is also an advantage because one can appropriately such that there is good cancellation between V(r) and VR such that the r As an example, consider From Eqs. (5.65), (5.69), and (5.70), we obtain \u03b8ck r \u00de \u00f0 = \u2212 V r \u00f0 \u03d5ck\u00f0 \u00de r \u00de : Vps\u03c6k\u00f0 r \u00de = V r \u03c6k\u00f0 \u00de \u00f0 r \u00de \u2212 \u2211 c < \u03d5ck j V r \u00f0 \u00dej r \u03c6k\u00f0 \u00de > \u03d5ck\u00f0 r \u00de : (5.70) (5.71) Thus, one can subtract from V(r) any sum of the core functions. This cancellation is at the core of the pseudopotential theory because it explains why the valence electrons in metals and semiconductors appear not to interact strongly with the ions of the crystal lattice. The success of the nearly free electron model is a consequence of the cancellation principle implicitly included in the pseudopotential concept. There has been significant improvement in the application of the pseudopotential theory in the band calculations in metals, but here we have essentially discussed the original theory developed by Phillips and Kleinman. 5.8 MUFFIN-TIN POTENTIAL The muffin-tin potential is a very convenient description of the metal in the sense that it represents an isolated ion within a sphere of radius ri about each lattice point and constant (zero) everywhere else. Therefore, the periodic potential in the metals considered as a muffin-tin potential is shown in Figure 5.16. 150 CHAPTER 5 Band-Structure Calculations \u00f0 \u00de r The muffin-tin potential U is considered to be a constant (zero) in the interstitial region and spherically symmetrical within radius ri about each ion (the core or the atomic region). We assume that there is one ion located at the center of each Wigner\u2013Seitz cell and the spheres do not overlap; i.e., the radius ri of the sphere is smaller than the Wigner\u2013Seitz radius rs: Thus, the periodic potential of the lattice can be defined as V r \u00de \u00f0 = \u2211 i U \u00f0j r \u2212 Ri : j\u00de (5.72) FIGURE 5.16 Muffin-tin potential. The Schrodinger equation can be solved exactly within each sphere because of the spherical symmetry as well as in the interstitial region where the potential is zero. These solutions are matched on the surface of each sphere, and the Schrodinger equation for the crystal is obtained. r The basis states \u03d5k\u03b5\u00f0 1. \u03d5k\u03b5 = eik:r in the interstitial region (outside the muffin hole). We note that because \u03d5k\u03b5 is obtained by matching the solutions at the boundaries of the sphere, there is no precondition that \u03b5 =!h2k2 2m. are defined as follows: \u00de 2. In the spherical region about Ri, \u03d5k\u03b5 satisfies the atomic Schrodinger equation \u2212!h2 2m \u22072\u03d5k\u03b5\u00f0 r \u00de + U \u00f0j r \u2212 Ri r \u03d5k\u03b5\u00f0 \u00de = \u03b5\u03d5k\u03b5\u00f0 r \u00de, j\u00de j r \u2212 Ri < ri: j (5.73) 3. \u03d5k\u03b5 is continuous at the boundary of the sphere. There are two elegant methods to solve the eigenvalue problem of this type of Schrodinger equation: the augmented plane-wave (APW) method and the Green\u2019s function (KKR) method proposed independently by Korringa, and Kohn and Rostoker (Ref. 11). We will first describe the APW method. 5.9 AUGMENTED PLANE-WAVE (APW) METHOD The augmented plane-wave (APW) method was proposed by Slater15. The potential U is a spherically symm", "etric function within the sphere (muffin-tin hole), and therefore, we write it as of an electron in a spherically symmetric Ylm U(r). We express the solutions \u03c8 \u03b5\u00f0 potential (note that the energy quantum number is written as \u03b5 instead of n and the spherical harmo. From atomic physics, we know that the radial part of the ^r are expressed as Yl.m nics Yl, m \u00de\u00de \u00f0 Schrodinger equation for an atom satisfies \u03b8, \u03d5 \u00f0 r \u2212 Ri = R\u03b5l ^r \u00de \u00f0 r \u00f0 j\u00de \u00f0j r \u00de \u00de \u00de \u2212!h2 2mr2 \u2202 \u2202r r2 \u2202 \u2202r R\u03b5l h2 l + 1 \u00de 2mr2 $ R\u03b5l r \u00f0 \u00de = \u03b5R\u03b5l r : \u00de \u00f0 (5.74) 5.9 Augmented Plane-Wave (APW) Method 151 Ignoring the solution of Eq. (5.74) that diverges at the origin, one can rewrite Eq. (5.73) as \u221e \u03d5k\u03b5 = \u2211 l=0 l \u2211 m=\u2212l Blm Ylm k \u00f0 \u00de R\u03b5l ^r \u00de \u00f0 : r \u00f0 \u00de (5.75) k The coefficients Blm are arbitrary, which are obtained from the boundary condition that the wave function is continuous across the boundary of each sphere (muffin hole). The wave function in the interstitial is a plane wave eik.r. It is well known that the plane wave can be expanded in terms of the spherical harmonics, \u00de \u00f0 eik.r = 4\u03c0 \u2211 \u221e l=0 l \u2211 m=\u2212l kr iljl\u00f0 ^k Y #lm\u00f0 \u00de \u00de Ylm, ^r \u00de \u00f0 (5.76) kr is the spherical Bessel function. At the boundary of each sphere, r = Ri, where Ri is where jl\u00f0 the radius of each sphere. Matching Eqs. (5.75) and (5.76) at the surface of each sphere where r = Ri, we obtain \u00de Blm k \u00f0 \u00de = 4\u03c0 iljl\u00f0 kRi R\u03b5l ^k Y #lm\u00f0 \u00de \u00de Ri \u00de \u00f0 : From Eqs. (5.75) and (5.77), we obtain r \u03d5k\u03b5\u00f0 \u00de \u221e = 4\u03c0 \u2211 l=0 l \u2211 m=\u2212 l iljl\u00f0 kRi R\u03b5l ^k Y #lm\u00f0 \u00de \u00de Ri \u00de \u00f0 Ylm R\u03b5l ^r \u00de \u00f0 \u00f0 r \u00de : (5.77) (5.78) \u00de r We have a function \u03d5k\u03b5\u00f0 for each value of k and \u03b5. These APW functions have a discontinuity in slope at the boundary of the muffin tin. The boundary conditions at the edge of the primitive cell can be matched (Problem 5.7) by noting that the APW functions are plane waves at the edge of the primitive cell. The plane waves obey the Bloch condition. Therefore, one can write the solution of the Schrodinger equation of the crystal by making a linear combination of the APWs, all of the same energy \u03b5: r \u03c8 k\u00f0 \u00de = \u2211 K bk + K\u03d5k + K, \u03b5 k \u00f0 \u00de\u00f0 r, \u00de (5.79) where the coefficients bk + K are to be determined. The best way is to use the variational principle (Problem 5.8), which is briefly described here. We define an energy functional,!h2 2m j! \u2207\u03c8 r \u00f0 \u00dej 2 + V r \u00f0 \u00dej \u03c8 r \u00f0 \u00dej 2 dr \" : (5.80) = Z E \u03c8 \u00bd \u2019 In the variational technique (Problem 5.8), E respect to a differentiable function \u03c8 Eqs. (5.79) and (5.80), one can show that the condition that E when Eq. (5.80) is stationary with that satisfies the Bloch condition with wave vector k. From is stationary leads to \u03c8 k\u2019 \u00bd 2dr \u03c8 r \u00f0 \u00dej j Z \u03c8 \u00bd 152 CHAPTER 5 Band-Structure Calculations \u2202E/\u2202bk+K = 0, which yields a set of homogeneous equations in bk + K. When the determinant of the coefficients of these equations is equal to zero, one obtains an equation of which the roots determine \u03b5 k", " : \u00de \u00f0 5.10 GREEN\u2019S FUNCTION (KKR) METHOD Korringa (1947) and Kohn and Rostoker11 independently proposed a Green\u2019s function method to calculate the band structure of metals. The KKR method essentially uses a Green\u2019s function method to solve the Schrodinger equation of a crystalline solid with a periodic potential. The KKR method starts with the objective to find the \u201cpropagating\u201d solutions of the Schrodinger equation in the lattice \u2212!h2 2m # + V \u2212 E r \u00de \u00f0 r \u03c8 k\u00f0 \u00de = 0: $ (5.81) Here, V(r) is the periodic potential, and the boundary conditions in the central polyhedron (Wigner\u2013 Seitz cell) that surrounds the origin are r1 \u03c8 k\u00f0 \u00de = eik.R\u03c8 k\u00f0 r \u00de (5.82) and /\u2202n1 = \u2212e\u2212ik.R\u2202\u03c8 k\u00f0 The conjugate points r and r1 are defined (see Figure 5.17) as the points on the surface of the polyhedron separated by the lattice translation vector R. \u2202\u03c8 k\u00f0 (5.83) /\u2202n: r1 r \u00de \u00de The Green\u2019s function is defined by!h2 2m! \u22072 + \u03b5 G\u03b5 \" r \u2212 r\u2032 \u00de \u00f0 = \u03b4 \u00f0 r \u2212 r\u2032 : \u00de (5.84) For conjugate boundary points r and r1, \u2192 r1 r1 \u2212 r\u2032 G\u03b5 \u00f0 \u00de = eik.RG\u03b5 r \u2212 r\u2032 \u00de \u00f0 (5.85) and \u2202G\u03b5 r1 \u2212 r\u2032 \u00de \u00f0 /\u2202n1 = \u2212eik.R\u2202G\u03b5 r \u2212 r\u2032 \u00f0 \u00de /\u2202n: (5.86) It can be shown that G\u03b5 in the alternate form r \u2212 r\u2032 \u00de \u00f0 can be rewritten r \u2212 r\u2032 G\u03b5 \u00f0 \u00de = \u2212 2m!h2 r\u2212r\u2032 j j r \u2212 r\u2032 ei\u03ba 4\u03c0 j, j (5.87) R\u2192 r\u2192 FIGURE 5.17 Conjugate boundary points r and r1. where and 5.10 Green\u2019s Function (KKR) Method 153 \u03ba = 2m\u03b5!h2, \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 r \u03b5 > 0, = i s \u2212 \u03b5 2m, \u00f0!h2 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u00de \u03b5 < 0: (5.88) (5.89) An alternate way of expressing G is obtained by multiplying Eq. (5.81) by G#\u00f0 plex conjugate of Eq. (5.84) by \u03c8 k\u00f0 hedron (Problem 5.10) to obtain and the com, subtracting and integrating over r in the interior of the poly- r \u2212 r\u2032 r \u00de \u00de r \u03c8 k\u00f0 \u00de = Z dr\u2032G\u03b5 k \u00f0 \u00de\u00f0 r \u2212 r\u2032 \u00de V r\u2032 \u00de r\u2032 \u03c8 k\u00f0 \u00f0 \u00de : Consider the muffin-tin potential ^V r \u00de \u00f0 \u2212 V0, r \u2264 ri = V r \u00f0 \u00de = 0, r > ri, (5.90) (5.91) Here, ri is the radius of the inscribed sphere, and V0 is the average value of the constant potential V in the space between the inscribed sphere and the boundary of the polyhedron. Thus, we can write the periodic potential as r \u00f0 \u00de From Eqs. (5.87), (5.90), and (5.92), we obtain V r \u00f0 \u00de = \u2211 i ^V \u00f0 r \u2212 Ri : \u00de i Z We move the origin such that r\u2033 = r\u2032 \u2212 Ri, and Eq. (5.93) can be rewritten as r \u03c8 k\u00f0 \u00de = \u2211 G\u03b5 k \u00f0 \u00de\u00f0 r \u2212 r\u2032 \u00de ^V \u00f0 r\u2032 \u2212 Ri \u03c8 k\u00f0 r\u2032 \u00de \u00de dr\u2032: r \u03c8 k\u00f0 \u00de = \u2211 G\u03b5 k \u00f0 \u00de\u00f0 r \u2212 r\u2033 \u2212 Ri ^V \u00de r\u2033 \ufffd", "\ufffd k\u00f0 \u00de \u00f0 r\u2033 + Ri : \u00de Using the Bloch theorem \u03c8 k\u00f0 r\u2033 + R \u00de and replacing r\u2033 with r\u2032 in Eq. (5.94), we obtain r \u03c8 k\u00f0 \u00de = r \u2212 r\u2032 \u00de k \u00de\u00f0 ^V r\u2032 \u00de \u03c8 k\u00f0 r\u2032 \u00de, \u00f0 (5.95) i Z = eik.R\u03c8 k\u00f0 dr\u2032 G0 k,\u03b5 \u00f0 r\u2033 \u00de Zr\u2032<ri\u2212\u03b5 where the integration is done over a single cell, and the structural Green function or Greenian is defined as G0 k,\u03b5 \u00f0 r \u2212 r\u2032 \u00de = \u2211 i k \u00de\u00f0 G\u03b5 k \u00f0 \u00de\u00f0 r \u2212 r\u2032 \u2212 Ri eik.Ri : \u00de (5.96) (5.92) (5.93) (5.94) 154 CHAPTER 5 Band-Structure Calculations r r \u2032\u2032 r \u2032 l r r \u2032\u2032 r \u2032\u2032 r \u2032\u2032 l (a) G(r \u2013 r \u2032) propagating from r \u2032 to r (b) G0 (r \u2013 r \u2032) combines waves from all points in the lattice. k, \u03b5 (k) FIGURE 5.18 (a) The ordinary Green function; (b) the structural Green function. From Eqs. (5.87) and (5.96), G0 k,\u03b5 \u00f0 k \u00de\u00f0 r \u2212 r\u2032 \u00de can also be rewritten in the alternate form G0 k,\u03b5 \u00f0 r \u2212 r\u2032 \u00de k \u00de\u00f0 = \u2212 2m!h2 1 4\u03c0 \u2211 i j ei\u03ba r\u2212r\u2032 \u2212 Ri r \u2212 r\u2032 \u2212 Ri j eik.Ri: j j The ordinary and structural Green functions are shown in Figure 5.18. From Eq. (5.96), it can be easily shown that!h2 2m! \u2207\u20322 + \u03b5 G0 k,\u03b5 \" r \u2212 r\u2032 = \u03b4 \u00f0 \u00de r \u2212 r\u2032, \u00de k \u00de\u00f0 \u00f0 r, r\u2032< ri, (5.97) (5.98) where ri is the radius of the spherical region in muffin-tin potential. We can also write by using the mathematical identity G0 k,\u03b5 \u2207\u20322\u03c8 k\u00f0 r\u2032 \u00de k \u00f0 \u00de = \u2207\u2032. G0 k,\u03b5 \u00f0 k \u00f0 \u00de r\u2032 \u2207\u2032\u03c8 k\u00f0 \u00de \u2212 \u2207\u2032\u03c8 k\u00f0 r\u2032 G0 k,\u03b5 \u00de k \u00de\u00de \u00f0 + \u03c8 k\u00f0 r\u2032 \u00de \u2207\u20322Gk,\u03b5 : k \u00f0 \u00de (5.99) From Eqs. (5.95), (5.98), and (5.99), we can show that dr\u2032 \u2207\u2032. G0 k,\u03b5 \u00bd k \u00f0 \u00de\u00f0 r \u2212 r\u2032 \u2207\u2032\u03c8 k\u00f0 \u00de r\u2032 \u00de \u2212 \u03c8 k\u00f0 r\u2032 \u00de \u2207\u2032G0 k,\u03b5 \u00f0 k \u00de\u00f0 r \u2212 r\u2032 \u00de\u2019 = 0: (5.100) Zr\u2032\u2212\u03b5 5.10 Green\u2019s Function (KKR) Method 155 Using the Gauss theorem to transform the volume integral to a surface integral over the sphere of radius ri \u2212 \u03b5, we obtain ds\u2032 G0 k,\u03b5 h k \u00f0 \u00de\u00f0 Zr\u2032\u2212\u03b5 ri\u03b8\u03d5, ri\u03b8\u2032\u03d5\u2032 \u2202 \u2202r\u2032 \u00de \u03c8 k\u00f0 r\u2032\u03b8\u2032\u03d5\u2032 \u00dej r\u2032= ri \u2212 \u03c8 k\u00f0 ri\u03b8\u2032\u03d5\u2032 \u2202 \u2202r\u2032 \u00de Gk,\u03b5 k \u00de\u00f0 \u00f0 ri\u03b8\u03d5, r\u2032\u03b8\u2032\u03d5\u2032 \u00dej r\u2032= ri It can be easily shown11 that for r < r\u2032< ril ClmRl r Ylm \u00de \u00f0 \u00f0 \u03b8, \u03d5, \u00de = 0: i (5.101) (5.102) and Go k,\u03b5 Here, r \u2212 r\u2032 \u00de k \u00f0 \u00de\u00f0 = \u2211 l,m \u2211 l\u2032,m\u2032\u00bd Alm,l\u2032m\u2032jl \u03bar jl\u2032\u00f0 \u03bar\u2032 \u00de \u00de \u00f0 + \u03ba\u03b4ll\u2032\u03b4mm\u2032jl \u03bar nl \u00de \u00f0 \u00f0 \u03bar\u2032 Ylm \u00de\u2019 \u03b8,", " \u03d5 Y #l\u2032m\u2032\u00f0 \u00de \u00f0 \u03b8\u2032, \u03d5\u2032 : \u00de (5.103) = = jl nl x \u00de \u00f0 x \u00de \u00f0 1/2Jl + 1/2 \u03c0/2x x \u00f0 \u00de 1/2J \u2212 l \u2212 1/2 \u03c0/2x \u00de \u00f0 \u00f0 \u00de x \u00f0, \u00de (5.104) x are the Bessel functions, and Alm;l\u2032m\u2032 are functions of k and \u03b5, which are characteristics for the J\u03bd \u00f0 \u00de lattice under consideration. Substituting Eqs. (5.102) and (5.103) in Eq. (5.101), multiplying by = 1, and Y #lm\u00f0 finally letting \u2208\u21920, we obtain, integrating over the sphere r = ri \u2212 \u2208, using the normalization condition Rl ri \u00f0 \u03b8, \u03d5 \u00de \u00de \u2211 l\u2032,m\u2032 jl Alm;l\u2032m\u2032\u00f0 \u00bd jl\u2032Ll\u2032 \u2212 jl\u2032\u2032 \u00de + \u03ba\u03b4ll\u2032\u03b4mm\u2032\u00f0 nl\u2032Ll\u2032 \u2212 nl\u2032\u2032 \u00de\u2019 Cl\u2032m\u2032 = 0, (5.105) where Ll = r dRl \u00f0 dr \u00de /Rl r \u00f0 \u00dej r = ri \u00de j r = ri (5.106) djl \u03bar \u00f0 dr dnl \u03bar jl\u2032 = nl\u2032 = \u00f0 dr \u00de : j r = ri Before equating the determinant of Eq. (5.105) to zero, we divide each row by jl and each col- umn by jl\u2032Ll\u2032 \u2212 jl\u2032\u2032 \u00de \u00f0 and obtain the secular equation Det Alm;l\u2032m\u2032 + \u03ba\u03b4ll\u2032\u03b4mm\u2032 \u00f0 \u00f0 nlLl \u2212 nl\u2032 \u00de jlLl \u2212 jl\u2032 \u00de ( ( ( ( = 0: ( ( ( ( (5.107) 156 CHAPTER 5 Band-Structure Calculations Eq. (5.107) is used by first tabulating the structure constants Alm,l\u2032m\u2032 as functions of \u03b5 and k for each type of lattice. The logarithmic derivatives Ll for the first few l are obtained as functions of energy. The convenient way of solving Eq. (5.107) is to fix \u03b5 (and hence \u03ba) to find those k\u2019s (k enters through the A\u2019s) that make Eq. (5.107) vanish. It may be noted that the same expression (Eq. 5.107) can be also obtained by using a variational method. One may note a good deal of similarity between the Green function (KKR) and the APW method. In the APW method, the expansion is in terms of spherical harmonics. Then a secular determinant is solved for contributions from different lattice vectors. In contrast, in the KKR method, the summation is over lattice vectors (in practice, the summation is over reciprocal lattice vectors by first making a Fourier transformation of G0 ) and then have a secular determinant in the contributions k,\u03b5 \u00de from different spherical harmonics. Anderson (Ref. 1) has proposed an approximate first-principle method for calculating the band structure of closely-packed structures. In his atomic sphere model (ASM), the atomic polyhedra are replaced by spheres of the same volume and the potential is spherically symmetric within each sphere. The ASM model has no interstitial region. r \u2212 r\u2032 \u00de\u00f0 k \u00f0 5.11 MODEL PSEUDOPOTENTIALS It is obvious from the preceding discussions that the APW and KKR methods do not depend on the atomic potentials, but only on the gradient of Rl at the surface of the atomic sphere. It can be shown from the partial wave theory of scattering that for r, \u03b5 a spherical potential, the radial solution Rl \u00de can be matched to a free electron wave of the same energy \u03b5 =!h2\u03ba2/2m and angular momen\u03b5 tum l, through a phase shift \u03b7l\u00f0 \u00de \u2212 tan \u03b7l\u00f0 \u2212 tan \u03b7l\u00f0 Ri, \u03b5 \u00f0 Ri, \u03b5 \u00f0 defined by. nl\u2032. nl\u2032 Rl\u2032 Rl \u03bar \u03bar \u03bar \u03bar jl\u2032 jl \u03c8 | Ll ( = \u03c8 \u00f0 \u03c6, r \u00de j r = Ri \u00de (5.108) where Ri for a muffin-tin potential is defined in E", "q. (5.73) and the spherical Bessel functions and their derivatives were defined earlier. In the model pseudopotential approach, each atomic potential va is replaced by a weak potential wa, which has the same scattering amplitude as conduction electrons. The energy \u03b5 of the crystal would be \u00de identical to this hypothetical material. Figure 5.19 shows how the model potential w, with wave function \u03d5 inside the atom, replaces the effect of the true potential V with the true wave function \u03c8: The analytical pseudopotentials of Phillips and Kleinman are nonlocal, energy dependent, and arbitrary. Thus, one can obtain an infinitely k \u00f0 \u03b5 \u03c9 \u03c5 r FIGURE 5.19 The true potential V with wave function \u03c8 is replaced by model potential w with wave function \u03d5 inside the atom. 5.12 Empirical Pseudopotentials 157 Rl \u03b5MTZ \u03c5MT j0 (\u03ba Rl) R0(r) A0 Rl A1 Rl A2 Rl l = 0 l = 1 l = 2 FIGURE 5.20 Model pseudopotential for KKR matrix elements. \u00de r, \u03b5, that can be used to obtain the true radial functions Rl large number of localized potentials, wl\u00f0 would not work for a different outside the atomic core. However, the same model potential wl\u00f0 \u03b5 or l. Thus, the various angular momentum components in a wave function have to be separated by introduction of appropriate operators. However, it is more convenient to introduce the functional for computational convenience. The appropriate choice form of each model pseudopotential wl for a muffin-tin lattice is a delta-function singularity for each value of l, to reproduce the scattering phase shift \u03b7l\u00f0 The model pseudopotential for the KKR matrix elements is shown in Figure 5.20. It can be shown, by the use of the analytical properties of spherical Bessel functions and plane at the surface of the sphere. r, \u03b5 r, \u03b5 \u03b5 \u00f0 \u00de \u00de \u00de waves, that the pseudopotential matrix elements used in the KKR method can be written as jl\u00f0j k \u2212 K 2l + 1 Ri \u00de j, V KKR ps K, K\u2032 \u00de \u00f0 = \u2212 4\u03c0N \u03ba \u2211 l \u00f0 Pl cos \u03b8KK\u2032\u00de \u00f0 tan \u03b7l\u2032 \u00de k \u2212 K\u2032 2 Ri j jl\u00f0 jl\u00f0j \u00de \u03baRi \u00de\u00de\u2019 \u00bd where cot \u03b7l\u2032 cot \u03b7l \u2212 nl\u00f0 \u03baRi /jl\u00f0 \u00de \u03baRi : \u00de ( (5.109) (5.110) 5.12 EMPIRICAL PSEUDOPOTENTIALS The pseudopotential theory of Phillips and Kleinman established the existence of weak potentials due to cancellations. The empirical pseudopotential concept is based on the assumption that one can choose weak potentials to match the important features of experimental results. An example of the empirical pseudopotential is the empty-core potential. The Ashcroft pseudopotential U(r) is obtained from the approximation that and U r \u00f0 \u00de = 0, r < Rc, = \u2212U0e\u2212r/d/r, r > Rc, U r \u00f0 \u00de (5.111) (5.112) 158 CHAPTER 5 Band-Structure Calculations where Rc is a cutoff radius, and d is the exponential decay length. The Ashcroft (Ref. 2) emptycore pseudopotential is shown in Figure 5.21. The three parameters in the empty-core pseudopotential are U0, Rc, and d. These parameters are adjusted to fit the measurement of magnetic or optical experimental results. In general, these types of potentials yield satisfactory results for alkali metals or aluminum. 0 ) r ( U \u2212U0 0 Rc Distance r FIGURE 5.21 The Ashcroft empty-core pseudopotential. 5.13 FIRST-PRINCIPLES PSEUDOPOTENTIALS The starting point for a convenient method of obtaining first-principles pseudopotentials (probably a misnomer) is to start with the Kohn\u2013Sham equations12 obtained by using the local density approximation (LDA), \u2212!h2 2m \u22072\u03c8 i\u00f0 r \u00de + dr\u2032 + V # r \u00f0 \u00de e2n r\u2032 \u00de \u00f0 r \u2212 r\u2032 j + \u03b4\u03b5xc \u00f0 \u03b4n n \u00de r \u03c8 i\u00f0 $ \u00de r = \u03b5l\u03c8 i\u00f0 \u00de j is one of N single-electron wave functions, n(r) is the density defined by Z, (5.113) where \u03c8 i\u00f0 1", " j 2, r \u03c8 i\u00f0 \u00dej (5.114) and \u03b5xc is the exchange-correlation energy of the uniform electron gas, which means that it can be chosen freely to ensure that properties of the uniform electron gas are obtained correctly. In the first-principle pseudopotential method, one chooses an atom and makes the approximation that n(r) = \u2212 Ze2, and all the is spherically symmetric about the nucleus. Thus, the atomic potential is V r j is the radial wave function and solutions are of the form \u03c8 Ylm is a spherical harmonic. With these approximations, Eq. (5.113) can be rewritten as, where Rnl = Rnl \u03b8, \u03c6 \u03b8, \u03c6 Ylm h2 2m # 1 r \u22022 \u2202r2 r \u2212 l \u00f0 l + 1 r2 \u00de Rnl + r \u00f0 \u00de $ # Z e2n r \u00de \u00f0 r \u2212 r\u2032 j j dr\u2032 \u2212 Ze2 r + \u03b4\u03b5xc \u03b4n \u2212 \u03b5ni r \u00f0 \u00de Rnl $ = 0: (5.115) Eq. (5.115) is solved for all the electrons of the atom, and the energies of these states are \u03b5ni: The outermost states that are in partially filled shells are singled out for special treatment because they contribute to the bonding between atoms and solids. The radial wave functions and the pseudo radial wave functions of such a solid are schematically shown in Figure 5.22. The radial wave functions for a solid with partially filled shells can be obtained from, Z, \u03b5xc, and \u03b5ni. The corresponding pseudo by picking a point beyond the r Eq. (5.115) by using the appropriate values for n radial function Rps nl \u00f0 rightmost node and drawing a smooth curve into the origin. This pseudo radial function Rps nl \u00f0 is obtained from each radial function Rnl.13 First-Principles Pseudopotentials 159 R ps 4d R4d Rps 5s ) r ( l n R 0.0 R5s R5p R ps 5p 0.0 r FIGURE 5.22 Schematic diagram of the radial wave functions Rnl \u00f0 r and 5p states of a solid with partially filled shells. and pseudo radial wave functions Rps r nl \u00f0 \u00de \u00de for 4d, 5s, should join on to the actual radial function Rnl with two continuous derivatives, should be with\u00de out any nodes, but vanish as rl at the origin. In addition, the pseudo wave functions built from these pseudo radial functions have to be normalized. We can now replace the Coulomb potential \u2212 ze2 r in Eq. (5.115) with the pseudopotential r \u00f0 V ps l \u00f0 r \u00de =!h2 2m 1 r\u2032Rps nl \" \u22022r\u2032Rps nl \u2202r2 \u2212 l \u00f0 l + 1 r2 \u00de \u2212 # e2nps r\u2032 \u00f0 r \u2212 r\u2032 \u00de j \"Z j dr\u2032 + \u03b4\u03b5xc \u03b4nps \u2212 \u03b5nl, # (5.116) such that when one solves the Kohn\u2013Sham equations, the radial functions Rps are obtained instead of R. However, there is a different V ps for each angular momentum state. To be able to resolve l \u00f0 this problem that the pseudopotential is nonlocal, we write an arbitrary wave function as a sum of its angular momentum components, r \u00de \u03c8 r \u00de \u00f0 = \u2211 lm r \u03c8 lm\u00f0 \u00de Ylm \u00f0 \u03b8, \u03d5 : \u00de We multiply Eq. (5.117) by Y #l\u2032m\u2032\u00f0 \u03b8, \u03d5 \u00de, and integrating over \u03b8 and \u03d5, we obtain d\u03b8d\u03d5 Y #l\u2032m\u2032\u00f0 \u03b8, \u03d5 \u03c8 r \u00de \u00f0 \u00de = \u2211 lm r \u03c8 lm\u00f0 \u00de Z Z d\u03b8d\u03d5 Y #l\u2032m\u2032\u00f0 \u03b8, \u03d5 Ylm \u03b8, \u03d5 \u00f0 : \u00de \u00de Using the orthonormality condition of the spherical harmonics, d\u03b8d\u03d5Y #l\u2032m\u2032\u00f0 \u03b8, \u03d5 Yl,m \u00de \u00f0 \u03b8, \u03d5 \u00de = \u03b4l,l\u2032\u03b4m.m\u2032, Z (5.117) (5.118) (5.119) 160 CHAPTER 5 Band-Structure Calculations we can rewrite Eq. (5.118) as r \u03c8 lm\u00f0 \u00de = Z d\u03b8d\u03d5 Y #l", "m\u00f0 \u03b8, \u03d5 \u03c8 r : \u00de \u00f0 \u00de (5.120) The pseudopotential V ps l \u00f0 forming the Hamiltonian. r \u00de defined in Eq. (5.116) is multiplied by \u03c8 lm\u00f0 r \u00de defined in Eq. (5.120) in PROBLEMS 5.1. The 12 nearest neighbors of the origin of a fcc crystal are R = a 2 \u00f0 \u00b11, \u00b11, 0 a 2 \u00f0, \u00de \u00b11, 0, \u00b11 a 2 \u00f0, \u00de 0, \u00b11, \u00b11 : \u00de Show that for an s-level, of which the wave function depends only on the magnitude r, \u03b5 k \u00de \u00f0 = \u03b5s \u2212 \u03b2 \u2212 4\u03b3 cos \u00f0 1 2 kxa cos 1 2 kya + cos 1 2 kya cos 1 2 kza + cos 1 2 kza cos 1 2 kxa, \u00de where \u03b3 = \u2212 dr \u03d5# Z x, y, z \u0394U x, y, y \u2212 1 2 a, z : \u00de Show that in the limit of small ka, Eq. (2) can be rewritten as = \u03b5s \u2212 \u03b2 \u2212 12\u03b3 + \u03b3k2a2: k \u03b5 \u00f0 \u00de (1) (2) (3) (4) 5.2. Show that the tight-binding expression for energies of an s-band in a fcc crystal can be written as where 0 \u2264 \u03bc \u2264 1: \u03b5 = \u03b5s \u2212 \u03b2 \u2212 4\u03b3 1 + 2 cos \u03bc\u03c0 \u00f0 \u00de along \u0393X, 5.3. Show that the Bloch functions of different bands and different k are orthonormal, i.e., \u03c8 #mk\u2032\u00f0 r \u03c8 nk\u00f0 \u00de r \u00de dr = \u03b4n,m\u03b4k,k\u2032: Z 5.4. Show that \u2211 k e \u2212 ik. \u00f0 Ri \u2212 Rj \u00de = N\u03b4i, j, where the summation is over all values of k in the first Brillouin zone and Ri and Rj are direct lattice vectors. 5.5. Show that \u2211 Ri vectors Ri. e\u2212i \u00f0 k\u2212k\u2032.Ri = N\u03b4k,k\u2032, where the summation is over all the direct lattice \u00de 5.6. Show that for a simple cubic lattice of side a, the Wannier function at the origin is Problems 161 wn r \u00de \u00f0 = Np unk sin \u00f0 r \u00f0 \u00de \u03c0x/a sin \u03c0y/a \u00f0 \u03c0y/a \u03c0z/a sin \u00f0 \u00de \u03c0z/a \u00de \u00de\u00f0 \u00de\u00f0 \u00de, \u00de \u03c0x/a \u00f0 \ufb03\ufb03\ufb03\ufb03 is the periodic part of the Bloch function. where unk r \u00de \u00f0 5.7. The periodic part of the Bloch function has the property uk r \u00de \u00f0 = uk \u00f0 r + R, \u00de where R is a Bravais lattice vector. If r lies on the boundary of the unit cell and r + R is another boundary point of the cell, then ^n r \u00de \u00f0. \u2207uk r \u00de \u00f0 = \u2212 ^n \u00f0 r + R \u00de. \u2207uk r + R, \u00de \u00f0 where ^n r \u00de \u00f0 show that is normal to the cell boundary. Because = eik.ruk r \u03c8 k\u00f0 \u00de r, \u00de \u00f0 5.8. Define the functional eik.R^n r \u00de \u00f0. \u2207\u03c8 k\u00f0 r \u00de = \u2212 ^n r + R \u00f0 \u00de. \u2207\u03c8 k\u00f0 h2 2m j! \u2207\u03d5 r \u00f0 \u00dej 2 + V r \u00f0 \u00dej \u03d5 r \u00f0 \u00dej 2 dr \", 2dr \u03d5 r \u00f0 j \u00dej Z where Here, \u03c8 k\u00f0 r \u00de is the wave function in the Schrodinger equation, \u03d5 r \u00de \u00f0 = \u03c8 k\u00f0 r \u00de + \u03b4\u03d5 r : \u00de \u00f0 \u03c8 k\u00f0 V(r) is the periodic potential in the lattice, and both \u03d5 condition. From Eqs. (1) and (2), show that + V \u2212 \u22072\u03c8 k\u00f0 r r \u00de \u00f0 \u00de r, r = \u03b5k\u03c8 k\u00f0 \u00de and \u03c8 k\u00f0 r \u00de \u00de \u00f0!h2 2m r satisfy the Bloch \u00de Def", "ine the functional \u03d5 \u03b5 \u00bd \u2019 = \u03b5 \u03c8 k\u2019 \u00bd + O \u00f0 \u03b4\u03d5 \u00de 2: \u03b8, \u03c6 f \u00bd \u2019 = dr!h2 2m! Z \u2207\u03b8#. \u2207\u03c6 + V r \u00de \u00f0 \u03b8#\u03c6 : \" (1) (1) (2) (3) (4) (1) (2) (3) (4) (5) 162 CHAPTER 5 Band-Structure Calculations Using the Dirac notation, show that = \u03d5 \u03b5 \u00bd \u2019 \u03d5 Using the integration-by-parts formulas for any functions \u03c61\u00f0 periodicity of the Bravais lattice, r \u00de and \u03c62\u00f0 r \u00de that have the and show that and dr \u03c61\u2207\u03c62 = \u2212 dr \u03c62\u2207\u03c62 Zc dr \u03c61\u22072\u03c62 = dr \u03c62\u22072\u03c61, Zc Zc Zc f \u03c6, \u03c8 k\u2019 \u00bd = \u03b5k< \u03c6 \u03c8 k > j \u03c8 k, \u03c6 f \u00bd = \u03b5k< \u03c8 k j \u2019 \u03c6 >, where \u03c6 also satisfies the Bloch condition. Hence, show that f \u03d5, \u03d5 \u00bd \u2019 = \u03b5k < \u03b4\u03d5 > + < \u03b4\u03d5 < \u03b4\u03d5 > + < \u03b4\u03d5 \u03c8 k > + o \u03b4\u03d5 \u00f0 2, \u00de g \u03c8 k > + o \u03b4\u03d5 \u00f0 2, \u00de j j and = \u03b5 \u03d5 \u00bd \u2019 f \u00bd < \u03d5 \u03d5, \u03d5 \u2019 \u03d5 > j = \u03b5k + O \u00f0 \u03b4\u03d5 \u00de 2: (6) (7) (8) (9) (10) (11) (12) (13) 5.9. Show that the valence eigenvalues of the Hamiltonian H + VR are the same for any operator of the form VR\u03c6k\u00f0 r \u00de = \u2211 c < \u03b8ck r \u00f0 \u00dej r \u03c6k\u00f0 \u00de > \u03d5ck\u00f0 r \u00de, where \u03b8ck are completely arbitrary functions. 5.10. The Schrodinger equation for a crystalline solid with a periodic potential V is r \u00de \u00f0 \u2212!h2 2m! \u2207, \u03c8 r \u00f0 \u00de \" (1) (1) and the Green\u2019s function is defined as References 163 \u2212!h2 2m \u22072 \u2212 \u03b5 G \" r \u2212 r\u2032 = \u2212 \u03b4 r \u2212 r\u2032 Multiply Eq. (1) by G#\u00f0 integrate over r, use the Hermitian property of the Green\u2019s function and the complex conjugate of Eq. (2) by \u03c8 \u00de and show that G\u03b5 r \u2212 r\u2032 \u00f0 \u00de = G#\u03b5\u00f0 r\u2032 \u2212 r, \u00de r \u03c8 k\u00f0 \u00de = Z dr\u2032G\u03b5 k \u00f0 \u00de\u00f0 r \u2212 r\u2032 V r\u2032 \u03c8 k\u00f0 \u00de r\u2032 \u00de : \u00f0 \u00de r, subtract and \u00de \u00f0 (2) (3) (4) References 1. Anderson OK. Simple approach to the band structure problem. Solid State Commun 1973;13:133. 2. Ashcroft NW. Electron-ion pseudopotentials in metals. Phys Lett 1966;23:48. 3. Ashcroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 4. Bullett DW. Solid State Phys: Adv Res Appl 1982;35:129. 5. Callaway JC. Quantum theory of the solid state. New York: Academic Press; 1976. 6. Car R, Parrinello M. United approach for molecular dynamics and density functional theory. Phys Rev Lett 1985;55:2471. 7. Goswami A. Quantum mechanics. Dubuque: Wm. C. Brown; 1997. 8. Harrison WA. Pseudopotentials in the theory of metals. New York: Benjamin; 1966. 9. Heine V. Solid State Phys: Adv Res Appl 1970;24:1. 10. Heine V, Wearie D. Solid State Phys: Adv Res Appl 1970;24:250. 11. Kohn W, Rostoker N. Phys Rev 1954;94:1111. 12. Kohn W, Sham LJ. Self-consistent equations including exchange and correlation effects. Phys Rev 1965;140:A1133. 13. Kohn W, Rostoker N. Solution of the Schrodinger equation in periodic lattice with an application to metallic lithium. Phys Rev 1954", ";94:1111. 14. Korringa J. On the calculation of the energy of a Bloch wave in a metal. Physica 1947;13:392. 15. Marder MP. Condensed Matter Physics. New York: Brooks/Cole; 2000. 16. Phillips JC, Kleinman L. New Method for calculating wave functions in crystals and molecules. Phys Rev 1959;116:287. 17. Slater JC. Wave funcions in a periodic potential. Phys Rev 1937;51:846. 18. Slater JC, Koster GF. Wave functions in a periodic potential. Phys Rev 1956;94:1498. 19. Wannier GH. Elements of solid state theory. Cambridge: Cambridge University Press; 1959. 20. Wigner E, Seitz F. On the constitution of metallic sodium. Phys Rev 1933;43:804. 21. Wigner E, Seitz F. Solid State Phys: Adv Res Appl 1955;1:97. 22. Ziman JM. Principles of the theory of solids. Cambridge: Cambridge University Press; 1972. Static and Transport Properties of Solids CHAPTER 6 CHAPTER OUTLINE 6.1 Band Picture............................................................................ 166 6.2 Bond Picture............................................................................ 167 6.3 Diamond Structure....................................................................... 168 6.4 Si and Ge............................................................................... 168 6.5 Zinc-Blende Semiconductors.............................................................. 170 6.6 Ionic Solids............................................................................. 172 6.7 Molecular Crystals....................................................................... 174 6.7.1 Molecular Solids.................................................................... 174 6.7.2 Hydrogen-Bonded Structures......................................................... 174 6.8 Cohesion of Solids................................................................", "....... 174 6.8.1 Molecular Crystals: Noble Gases..................................................... 174 6.8.2 Ionic Crystals....................................................................... 176 6.8.3 Covalent Crystals................................................................... 177 6.8.4 Cohesion in Metals.................................................................. 178 6.9 The Semiclassical Model................................................................. 179 6.10 Liouville\u2019s Theorem...................................................................... 182 6.11 Boltzmann Equation...................................................................... 183 6.12 Relaxation Time Approximation............................................................ 184 6.13 Electrical Conductivity................................................................... 186 6.14 Thermal Conductivity..................................................................... 187 6.15 Weak Scattering Theory of Conductivity.................................................... 188 6.15.1 Relaxation Time and Scattering Probability.......................................... 188 6.15.2 The Collision Term................................................................. 188 6.15.3 Impurity Scattering......................................................", ".......... 189 6.16 Resistivity Due to Scattering by Phonons................................................... 192 Problems.................................................................................... 194 References.................................................................................. 196 Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00006-2 \u00a9 2012 by Elsevier Inc. All rights reserved. 165 166 CHAPTER 6 Static and Transport Properties of Solids 6.1 BAND PICTURE In Chapter 4, we discussed how the different types of solids can be classified as metals, semimetals, insulators, and semiconductors according to the band picture. In Chapter 5, we considered some of the methods used for calculation of the energy bands of solids. To summarize the band picture of solids, one constructs Brillouin zones, which are the Wigner\u2013Seitz cells of the reciprocal lattice. Each Brillouin zone can have N k states (where N is the number of primitive cells in the crystal) and hence can accommodate 2N electrons (because each k state can contain two electrons of opposite spin). The electrons in the lattice move in a periodic potential in a background of the positively charged ions (because the valence electrons have been stripped off from their parent atoms) that opens up an energy gap 2 j at each Brillouin zone boundary. In pure materials, there are no eigenstates for electrons with energies lying within these energy gaps. This can be restated as \u201cthe electron states are forbidden in this energy gap.\u201d Normally, these energy gaps appear in the free electron parabola at the Brillouin zone boundaries because of the periodic potential. This is called the extended zone scheme because it is extended over the entire reciprocal space. However, any k state in the extended zone scheme can be written as VK j k = k1 \u00b1 K, (6.1) where K is a reciprocal lattice vector, and k1 is a vector in the first Brillouin zone. Therefore, we can map the different segments of the extended zone into the first zone and use a band index n to identify them. These are called bands because each segment is an independent region of energy levels. For example, the state kn i in the nth zone can be written as kin = kn i \u2212 K, (6.2) where kin is the corresponding state in the reduced zone. In the reduced zone scheme, these electron energy bands are stacked above each other in the first Brillouin zone. The energy gap between the bands increases as one proceeds from the lowest zone to the higher zones. This is shown in Figure 6.1. It is easy to classify the solids in the band picture (as we did in Chapter 4) by counting the number of electrons per unit cell. Thus, all monovalent and trivalent solids are good metals because they are either half-filled or one-and-a-half-filled bands. A divalent solid, which normally would be an insulator, is actually a poor conductor because of the overlap of the bands in three dimensions. Solids with five electrons per atom such as As, Sb, and Bi have two atoms per unit cell that would fill five bands, but due to a small overlap of the fifth and sixth band, there are a few electrons in both bands that can be excited to higher energy states, and hence these are poor conductors. These are known as semimetals. Solids with four electrons per atom range from insulators (diamond) to semiconductors (Ge, Si). Sn is either a metal or a semiconductor, depending on the phase in which it crystallizes, whereas Pb is a pure metal. \u03b5 2|V3k| 2|V2k| 2|Vk| k 0 K1 2 FIGURE 6.1 The bands stacked above each other in the reduced zone scheme. 6.2 Bond Picture 167 6.2 B", "OND PICTURE To be able to discuss the properties of diamond as well as that of semiconductors such as Ge and Si, which crystallize in the \u201cdiamond structure,\u201d we need to discuss the bond picture of solids. The simplest bond picture is that of a covalent bond shown in Figure 6.2, which is formed by a hydrogen molecule. There is a repulsive force between the two nuclei, but the attractive force between the two electrons and the nuclei lowers the energy of the system. This is the simplest example of a covalent bond. The simplest example of a tetrahedral bond is that of a methane molecule, shown in Figure 6.3. Diamond, Si, and Ge crystallize in the diamond structure, which is essentially a fcc lattice with a basis of two atoms at (0, 0, 0) and (\u00bc, \u00bc, \u00bc) per unit cell. It can be visualized as having an atom at each lattice point of a fcc lattice and another atom at each point of a second fcc lattice of which the corner is at one-fourth of the distance along the main diagonal of the cube. The valence electrons in the ground state of the free atoms have the configuration 2s22p2, 3s23p2, and 4s24p2 for diamond, Si, and Ge, respectively. In the crystal, the ground state is formed from the configuration nsnp3, where n = 2, 3, and 4 for diamond, Si, and Ge, respectively. The valence electrons form directed sp3 tetrahedral bonding orbitals (from the orthonormal s and p orbitals) of the form (assuming no overlap) H + H + e\u2212 e\u2212 1 \u03d51 = 2 \u00f0 \u03d52 = 1 2 \u00f0 1 2 \u00f0 1 2 \u00f0 \u03d53 = \u03d54 = s + px + py + pz s \u2212 px + py + pz s + px \u2212 py \u2212 pz s \u2212 px \u2212 py \u2212 pz FIGURE 6.2 Covalent bond of a hydrogen molecule6.3) There is a tetrahedral bond between the atom that forms the center of the tetrahedron and the neighboring four atoms. In fact, there is only one way in which these tetrahedra can be arranged so that each atom forms bonds with four others. In the diamond structure, each atom is at the center of a tetrahedron with the nearest-neighbor atoms at the vertices. The diamond structure is shown in Figure 6.4. The positions of the atoms in the unit cell of the diamond structure projected on a cube face are shown in Figure 6.5. The points 0 and \u00bd are on the fcc lattice, and those at \u00bc and \u00be are on a similar lattice displaced among the body diagonal by one-fourth of its length. FIGURE 6.3 A simple example of tetrahedral bonding (methane molecule). 168 CHAPTER 6 Static and Transport Properties of Solids FIGURE 6.4 The diamond structure. Each atom forms the center of a tetrahedron. FIGURE 6.5 Positions of the atoms in the unit cell of the diamond structure projected on a cube face. \u00de \u00de \u00f0 2s22p2 6.3 DIAMOND STRUCTURE Diamond obviously crystallizes in the diamond structure. The orbital electrons of carbon atom have : When they form the diamond crystal, the valence electrons are in the state the structure : They form the directional sp3 orbitals as shown in Eq. (6.3). The tetrahedral bonding in 2s2p3 \u00f0 diamond is due to the bonding between a carbon atom and its four nearest carbon atoms. However, diamond is an insulator because of the large energy gap between the filled valence band and the empty conduction band. At room temperature, the valence band is completely filled with electrons, whereas the conduction band is empty because the energy gap is too large for electrons to be excited to the conduction band. When an external electric field is applied, the electrons cannot be excited to the higher energy states that lie in the energy gap. 6.4 Si AND Ge Si and Ge crystallize in the diamond structure. The Bravais lattice of the diamond structure has a basis of two atoms, each of which has eight sp electron states, but only four of them are occupied by electrons. Thus, the Brillouin zones have to accommodate 16 electron states, but only eight electrons (from the two atoms) fill them. Consequently, the band structure has eight sub-bands. Four of these are completely filled, and the other four are completely empty at 0\u00b0 K: 6.4 Si and Ge 169 Si and Ge are typical examples of semiconductors because the energy gap between the valence band and the con", "duction band is sufficiently small. At absolute zero, these are insulators, but at room temperature, a few electrons are excited from the valence band to the conduction band. The conductivity is small but increases with increase of temperature. The property of these intrinsic semiconductors can be easily understood from the bond picture that will help us understand the property of impurity or doped semiconductors as well as that of the p-n junctions and their importance in electronics. A simple example can illustrate the formation of the energy gap in these solids by considering the fact that Si and Ge have covalent bonding. The sp states in a free atom form an eight-fold-degenerate level. In an Si dimer, the electron states would interact and form bonding and antibonding levels, as shown in Figure 6.6. \u00de \u00f0 \u00de 3s3p3 3s23p2 The tetrahedral bonding of Si is shown in has four valence Figure 6.7. An Si atom \u00f0 electrons. In the crystalline form, a Silicon atom, which has the valence configuration forms four sp3 hybrids and results in the tetrahedral bonding with the four nearest Si atoms. Therefore, each bond has two electrons shared by two Si atoms. In principle, the wave functions of these electrons would extend throughout the crystalline solid. However, the electron cloud, which is essentially a pair of electrons of opposite spin, is mostly located in the bond between the neighboring Si ions, as shown schematically in Figure 6.7. This is also confirmed by band calculations. \u00f0 \u00de Ge also crystallizes in the diamond structure. In the Ge atom, the valence electrons have the 4s24p2 configuration, whereas in the crystalline \u00f0 4s4p3 form, the valence electrons have the \u00de configuration. These form the sp3 hybrids, as explained in Eq. (6.3). Each of these four orbital wave functions is directed toward the vertex of a tetrahedron, and hence, the electronic structure of Ge is almost identical to that of Si. The band structures of Si and Ge, calculated by Chelikowsky and Cohen3 using the empirical nonlocal pseudopotential method, are shown in Figures 6.8 and 6.9. Antibonding Bonding FIGURE 6.6 Bonding and antibonding levels in a silicon dimer. Si Si Si Si Si Si Si Si Si We note that the energy gaps in both Ge and Si are indirect. The highest occupied state lies at \u0393. However, the lowest lying unoccupied state is near the \u03a7 point, and thus the gap is \u201cindirect.\u201d FIGURE 6.7 Electronic structure (schematic) of Si. 170 CHAPTER 6 Static and Transport Properties of Solids L3 L1 Si L3\u2019 L1 L2\u2019 6 4 2 0 \u22122 \u22124 \u22126 \u22128 \u221210 \u221212 \u03932\u2019 \u039315 \u039325\u2019 \u03932\u2019 \u039315 \u039325\u2019 X4 X1 X1 \u03931 \u03931 L A \u0393 \u2206 X U, K \u03a3 \u0393 FIGURE 6.8 Band structure of Si calculated by Chelikowsky and Cohen by using nonlocal pseudopotential. The dashed lines are results obtained by using local pseudopotential. Reproduced from Chelikowsky and Cohen3 with the permission of the American Physical Society2 \u22124 \u22126 \u22128 \u221210 \u221212 \u03938 \u03936 \u03937 \u03938 \u03937 L4,5 L6 Ge \u03938 \u03934 \u03937 \u03938 \u03937 X5 X5 X5 L6 L4,5 L6 L6 L6 \u03936 \u03936 L A \u0393 \u2206 X U, K \u03a3 \u0393 FIGURE 6.9 Band structure of Ge calculated by Chelikowsky and Cohen by using nonlocal pseudopotential. Reproduced from Chelikowsky and Cohen3 with the permission of the American Physical Society. 6.5 ZINC-BLENDE SEMICONDUCTORS The III\u2013V zinc-blende semiconductors such as GaAs and InSb have the same basis core of closed shells as Si and Ge. However, Ga and In have three outer electrons, whereas As and Sb have five outer electrons. In addition, Ga or In occupies all the A sites in the diamond structure, whereas As or Sb occupies all the B sites (see Figure 6.4). The bond picture of GaAs is shown in Figure 6.10. The bond structure of InSb is similar to that of GaAs. We note that the bond structures of GaAs and InSb are different from that of Si or Ge in the sense that because Ga (or In) has three outer electrons and As (or Sb) has five outer electrons, there is a shrinking of", " the charge clouds toward As (or Sb), and away from Ga (or In). These compounds are also semiconductors like Si and Ge, but with a smaller energy gap. Their band structures, as calculated by Chelikowsky and Cohen using a nonlocal pseudopotential method, are shown in Figures 6.11 and 6.12. ZnS, which is a II\u2013VI compound, has the zinc-blende structure (in fact, the name is derived from it). It is also a semiconductor like the III\u2013V compounds. In ZnS, Zn has two electrons, and S has six electrons in the outermost shells. However, the S6+ ions attract the eight electrons surrounding it and try to pull them further in through a process known as electron affinity. The S6+ ions try to form a closed shell, and the Zn2+ ions barely hold on to the two outermost electrons. In the process, the two electrons in each bond are much closer to S than to Zn because the bond electrons are greatly polarized by the residual charge on the ions. The zinc sulphide bond is shown schematically in Figure 6.13. 6.5 Zinc-Blende Semiconductors 171 ) 2 \u22124 \u22126 \u22128 \u221210 \u221212 \u03938 \u03937 \u03936 \u03938 \u03937 L4,5 L6 GaAs \u03938 \u03937 \u03936 \u03938 \u03937 L6 L4,5 L6 L6 L6 X7 X7 X6 X6 X6 X6 \u03936 L A \u0393 \u03936 \u03a3 \u0393 \u2206 X U, K \u2192 Wave vector k FIGURE 6.11 Band structure of GaAs calculated by Chelikowsky and Cohen. Reproduced from Chelikowsky and Cohen3 with the permission of the American Physical Society. Ga As Ga As Ga As Ga As Ga FIGURE 6.10 Bond structure of GaAs. 6 4 2 0 L4,5 L6 \u03938 \u03937 X7 InSb X6 L6 L4,5 \u03936 \u03938 \u03938 \u03937 \u03938 \u03936 \u03937 L6 \u22122 \u03937 y g r e n E \u22124 \u22126 \u22128 \u221210 \u221212 L6 L6 X7 X6 X6 X6 Zn S Zn S Zn S \u03936 \u03936 L A \u0393 \u2206 X U, K \u03a3 \u0393 Wave vector k Zn S Zn FIGURE 6.12 Band structure of InSb calculated by Chelikowsky and Cohen. Reproduced from Chelikowsky and Cohen3 with the permission FIGURE 6.13 of the American Physical Society. Schematic picture of zinc sulphide bond. 172 CHAPTER 6 Static and Transport Properties of Solids 6.6 IONIC SOLIDS In CuCl, the copper atom has one electron in its outermost shell, whereas the chlorine atom has seven electrons. In the crystalline form, the Cu atom loses its electron, which is captured by the chlorine atom. CuCl crystallizes in an alternating lattice. The attraction is between the Cu+ and Cl\u2212 ions. In fact, there are two alternate crystal structures in which the ionic crystals crystallize: the NaCl structure and the CsCl structure. The important factor in the crystal structure in an ionic crystal is to have as many negative ions as possible around a positive ion. A schematic diagram of sodium chloride is shown in Figure 6.14. The sodium chloride crystal structures are shown in Figure 6.15. In the sodium chloride struc- ture, the space lattice is fcc, and the basis has one Na+ ion at (000) and one Cl\u2212 ion at (\u00bd\u00bd\u00bd). The cesium chloride crystal structure is shown in Figure 6.16. The space lattice is simple cubic, and the basis has one Cs+ ion at (000) and one Cl\u2212 ion at (\u00bd\u00bd\u00bd). We have discussed the progression of the band structure from perfectly covalent silicon in which the four electrons are distributed around the Si4+ ion cores in tetrahedral bonding, to covalent gallium arsenide, where there is more charge of the electron cloud surrounding the As5+ core than Ga3+: Consequently, GaAs has a small ionic character. The zinc sulphide crystal is weakly covalent because Zn2+ has very few valence electrons while S6+ is almost full; therefore, the electron cloud is mostly around S6+: ZnS is also partly ionic in character. Sodium chloride is perfectly ionic because the Na + ion has lost its valence electron to the chlorine atom that becomes the Cl\u2212 ion. The attraction between the ions of opposite sign results in the ionic bond. A schematic diagram of the progression from perfectly covalent to perfectly ion", "ic crystals is shown in Figure 6.17. FIGURE 6.14 FIGURE 6.15 A schematic diagram of sodium chloride. Sodium chloride crystal structure. 6.6 Ionic Solids 173 FIGURE 6.16 Cesium chloride structure. Si 4+ Si 4+ Ga3+ As5+ Si 4+ Si 4+ As5+ Ga3+ Zn2+ S 6+ Na+ Cl7+ S 6+ Zn2+ Cl7+ Na+ FIGURE 6.17 A schematic diagram of progression from perfectly covalent to perfectly ionic crystals. 174 CHAPTER 6 Static and Transport Properties of Solids 6.7 MOLECULAR CRYSTALS 6.7.1 Molecular Solids The column VIII elements are good examples of molecular solids. The solid noble gases (except helium) crystallize in monatomic fcc Bravais lattices. Each atom has a stable closed-shell type, which gets deformed in the solid. However, the solid is held together by the van der Waals or fluctuating dipole forces. These forces are very weak and can be qualitatively calculated by considering two atoms (A and B) at a distance r. At any instant, if atom A has an instantaneous dipole moment pA (of which the time-averaged value vanishes), then the electric field at atom B due to atom A is proportional to pA=r3: The dipole moment at atom B due to this electric field is given by pB # \u03b1pA r3, (6.4) where \u03b1 is the polarizability of the atom. The interaction energy between two dipoles of moment pA and pB is given by pApB r3 \u2248 \u2212\u03b1\u03b2 where \u03b2 is the polarizability of atom B and A = \u03b1\u03b2 pA atoms due to the van der Waals force is very small. Vi = \u2212\u03b2 2 A r6 = \u2212 pA r6 2: Thus, the lowering of energy between two (6.5), 6.7.2 Hydrogen-Bonded Structures In the hydrogen-bonded structures, the forces between the molecular groups are through a shared proton. The chemical properties of the molecules dominate the properties of these structures. 6.8 COHESION OF SOLIDS The cohesive energy is the difference between the energy of a solid and the energy of a gas of widely separated atoms from which the solid is eventually constituted. Essentially, the determination of the cohesive energy answers two separate questions. First, it explains how a large group of atoms form a crystalline solid when they come together. Second, it explains which crystal structure leads to the lowest energy. As we have seen, crystals can be divided into five categories: molecular, ionic, covalent, metallic, and hydrogen bonded. We will discuss the cohesive energy of each type of solids. 6.8.1 Molecular Crystals: Noble Gases The van der Waals interaction between two molecules, which are well separated at a distance r, is derived (using a classical theory) in Eq. (6.5). However, the term that is proportional to r\u22126 (Eq. 6.5) represents the interaction energy between the molecules at large distances that do not have a permanent dipole moment. We note that the induced dipole moment in van der Waals interaction is such that it is always attractive and hence lowers the energy of the system. It may be noted that quantum or thermal fluctuations always induce small dipole moments in atoms that normally do not have any dipole moments. However, when the atoms come close together such that their separation 6.8 Cohesion of Solids 175 is of the same order as that of the atomic radii, there is an additional repulsive force between them. This additional force is proportional to r\u221212: The resulting potential, which is the sum of the attractive force arising out of the van der Waals term (Eq. 6.5) and the repulsive force, can be written as \u03d5 r \u00f0 \u00de = \u2212 A r6 + B r12, (6.6) 1/6! \" where A and B are positive constants. For convenience, we introduce the parameters \u03c3 = B A and \u2208 = A2 4B, so that Eq. (6.6) can be written as \u03d5 \u03c3 r # $ This is known as the Lennard\u2013Jones 6-12 potential. The Lennard\u2013Jones parameters \u2208 and \u03c3 are appropriately chosen to reproduce the thermodynamic properties of gaseous neon, argon, krypton, and xenon at low temperatures. The Leonard\u2013Jones potential \u03d5/4\u2208 has been plotted against r/a in Figure 6.18. \u03c3 r # $ = 4\u2208 (6.7) : & r \u00f0 \u2212 % \u00de 12 6 The total potential energy U of a crystal of N atoms separated by", " R (using periodic boundary conditions so that each atom has identical surroundings) is given by (from Eq. 6.7) U = 1 2 N \u2211 R\u22600 \u03d5 R \u00de \u00f0 = 2N\u2208 \u2211 R\u22600 12 6.8) where the sum is over all nonzero vectors for the bcc lattice (He) or the fcc lattice (all other noble gases). Here, the factor \u00bd is multiplied because the interaction energy of each atom has been counted twice. The potential energy per atom is given by u = U/N = 2\u2208 \u2211 R\u22600 12 \u2212 \u03c3 R # $ 6 \u03c3 R # $ : & % If the nearest-neighbor distance is r, we can rewrite Eq. (6.9) as (6.9) u = 2\u2208 A12 % \u2212A6 12 \u03c3 r # $ 6 \u03c3 r # $, & where (6.10) \u03c6 /4\u03b5 0.6 0.5 0.4 0.3 0.2 0.1 : (6.11) 0.0 1.2 1.4 1.6 1.8 2.0 r/\u03c3 An = \u2211 R\u22600 n r R # $ It can be shown that A6 = 14.45 for a fcc structure and 12.25 for a bcc structure while A12 = 12.13 for a fcc structure and 9.11 for a bcc structure. The nearest-neighbor equilibrium spacing r0 is easily obtained from Eq. (6.10) from the expression, \u22120.1 \u22120.2 FIGURE 6.18 \u2202u=dr = 0 r = r0: j (6.12) The Leonard\u2013Jones 6-12 potential. 176 CHAPTER 6 Static and Transport Properties of Solids From Eqs. (6.10) and (6.12), we obtain, 2A12 A6\u2019 The cohesive energy per ion pair at equilibrium, u0 \u2261 \u03b5/N, is obtained from Eqs. (6.10) and (6.13), 2 A6 2A12 \u2248 \u22128:6 \u2208, u0 = \u2212\u2208 r0 = \u03c3 ( : (6.13) (6.14) 1/6 for a fcc lattice. The Bulk modulus B is given by B = V \u22022\u03b5 \u2202V 2 : For a fcc lattice, Therefore, at equilibrium, V N = r3 2p = v, \u2202 \u2202v = 2p 3r3 \ufb03\ufb03\ufb03 : \u2202 \u2202r \ufb03\ufb03\ufb03 \u22022u \u2202r2 j r = r0 = B0 = 2p 9r0 \ufb03\ufb03\ufb03 4\u2208 \u03c33 A12 5/2 A6 A12\u2019 ( : (6.15) (6.16) (6.17) 6.8.2 Ionic Crystals The ionic crystals crystallize in four common lattice structures: the sodium chloride, the cesium chloride, the zinc-blende, and the wurtzite structures. The largest term in the interaction energy is the Coulomb interaction because the particles in the ionic crystals are electrically charged ions. The dipole interaction, considered for molecular crystals, which is inversely proportional to the sixth power of the interionic distance, is easily neglected. However, the short-range core\u2012core repulsion due to the Pauli principle prevents the crystal from collapsing. We assume that in the first unit cell, the negative ion is at the origin, and the positive ion is at a distance d (d is a translation vector through a/2 of a cubic side). The total cohesive energy per ion pair is \u00f0 As an example, we consider the sodium chloride structure that is a fcc Bravais of negative ions at site Ri and a second Bravais lattice of positive cations displaced by d from the first lattice. The total potential energy of a single anion or cation is given by \u00de \u00de \u00de \u00f0 = ucoulomb d + ucore d : u d \u00f0 (6.18) u = \u2212 e2 d 1 + \u2211 i \u2260 0 ( d Ri + d j j \u2212 d Ri \u2019 : ) ( The total potential energy for N/2 ions pair (or N ions) is given by U = \u2212 Ne2 2d 1 + \u2211 i \u2260 0 ( d Ri + d j j \u2212 d Ri : ) ( \u2019 (6.19) (6.20) 6.8 Cohesion of Solids 177 The energy per ion pair is obtained by dividing Eq. (6.20) by N/2 ions", ", and we obtain ucoul = \u2212 d \u00f0 \u00de e2 d 1 + \u2211 i \u2260 0 ( d Ri + d j j \u2212 d Ri : ) ( \u2019 (6.21) The evaluation of the summation is complicated and depends on the lattice structure in addition to containing diverging terms. It can be shown, by performing a Ewald summation (Problem 6.4), that ucoul = \u2212 d \u00f0 \u00de \u03b1e2 d, (6.22) where \u03b1 is known as the Madelung constant, which depends on the crystal structure. The repulsive term ucore, which is essential in preventing a collapse, can only be estimated approximately by a phenomenological term, d \u00f0 \u00de ucore = d \u00f0 \u00de A 12d : The total energy per pair of ions is obtained from Eqs. (6.22) and (6.23), The equilibrium distance d0 is obtained from the relation d u \u00f0 \u00de = \u2212\u03b1 e2 d + A d12 : From Eqs. (6.24) and (6.25), we obtain \u2202u d \u00f0 \u2202d \u00de = 0: \u03b1 e2 d2 0 = 12, A d3 0 which leads to d0 = 1/11 : 12A \u03b1e2 \u2019 ( (6.23) (6.24) (6.25) (6.26) (6.27) The phenomenological constant A is obtained by matching the minimum of the potential to the experimental results. 6.8.3 Covalent Crystals The theory of cohesion in the covalent crystals involves much more complex calculation than that of ionic or molecular crystals. The main reason is that in the crystalline form, the valence electrons in covalent crystals are distributed in a very different way than they are in a group of isolated atoms or ions. Previously, for noble gases and ionic crystals, we obtained reasonable results by calculating the potential energy of a large number of deformed ions or atoms that were arranged in the relevant crystal structure. To calculate the cohesive energy of covalent crystals, we have to calculate their band structure by following techniques discussed in Chapter 5. In fact, the theory of chemical bonding in molecules is used to describe the cohesion in covalent insulators. 178 CHAPTER 6 Static and Transport Properties of Solids We can make a qualitative discussion by considering a specific case. For example, we place a set of carbon atoms in the sites of a diamond structure, where the lattice constant is large enough so that there is no interaction between the ions, and consequently, the cohesive energy is zero. As the lattice constant is gradually reduced, the atomic orbitals would start overlapping, and the interaction energy between the atoms placed at the designated lattice sites would correspondingly decrease. If the outermost shell of the isolated atom were filled, this overlap would lead to core\u2012core repulsion due to the Pauli exclusion principle, and the energy would increase. However, because the outermost shells of the carbon atoms are partially filled, the electrons are flexible enough to rearrange themselves whenever there is overlap between neighboring wave functions. The primary reason for this flexibility is due to the fact that in the same atomic shell, there are empty energy levels. Thus, when the lattice constant of the (hypothetical) diamond structure is reduced to an appropriate value (the actual lattice constant of diamond), the cohesion energy will be sufficient to form the crystal. The overlap of the outermost shells leads to lowering of the energy. The new levels formed by the electrons are no longer localized around a single atom or ion. The bond picture of covalent solids in general, and the tetrahedral bonding in diamond structure in particular, partially explains the cohesion in such crystals. However, as noted earlier, the theory of cohesion in covalent crystals requires more complex calculations. 6.8.4 Cohesion in Metals In general, the theory of the cohesion of metals also requires band theory calculations. However, the explanation of cohesion for the alkali metals is comparatively easier because one can consider an alkali metal to consist of a sea of electrons moving around in a background of localized positively charged atomic cores that are located at the lattice sites. There are three types of contributions to the cohesive energy of alkali metals: the interaction between the sea of electrons and the background of positive ions, the average kinetic energy of the electrons, and the exchange energy. The interaction between the ion cores and the sea of electrons can be expressed as e2n dr n \u2211 i\u2260j e2 2 e2 r \u2212 Ri j dr1dr2 \u03b5el = \u2212 1 Ri \u2212 Rj j where n(r) is the electron density. We note that the first term in Eq. (6.28) is the sum of the attractive Coulomb interaction between the electrons and the ions; the second term is the", " sum of the repulsive interaction between the positively charged ions of charge e at the lattice sites Ri and Rj (the factor 1/2 prevents double counting); and the third term is the direct electron\u2012electron interaction term that is a function of the electron densities n(r1) and n(r2). If n(r) is taken as a constant, n = N=V: It can be easily shown by adapting a technique used similar to that for ionic crystals in Problem 6.4, (6.28) n r1 \u00f0 \u00de \u00f0 r1 \u2212 r2 j r2 Z \u00de j j, where \u03b1 is the Madelung constant and rs (defined in Chapter 3) is the free space around each electron in an electron gas, \u03b5el N = \u2212 1 2 \u03b1 e2 rs, (6.29) rs = 3V 4\u03c0N # 1/3 : $ (6.30) 6.9 The Semiclassical Model 179 We also showed in Chapter 3 that the average kinetic energy of the electrons is given by \u03b5k N = 3 5 \u03b5F = 2 3\u01272kF 10m = 3\u01272 10m 2/3 1 rs 2 : 9\u03c0 4 # $ (6.31) The exchange term arises due to the fact that electrons at r1 and r2 flip places while interacting with each other. Because the wave function is antisymmetric, a negative sign is introduced due to this interaction. The exchange integral can be written as \u03b5ex = \u2212 e2dr1dr2 r1 \u2212 r2 j j Z \u2211 n<m\u00bd \u03d5%n\u00f0 r1 \u03d5%m\u00f0 \u00de r2 r2 \u03d5n\u00f0 \u00de \u03d5m\u00f0 \u00de r1 \u00de, \u03b4\u03c7i\u03c7j& (6.32) where \u03d5\u2032s are the one-particle wave functions and \u03c7\u2032s are the spin index. In the jellium model, the exchange energy can be shown to be \u03b5ex N = \u2212 3e2 4\u03c0 1/3 1 rs 9\u03c0 4 # $ : (6.33) Adding Eqs. (6.29), (6.31), and (6.33); using the value \u03b1 = 1:792 for fcc, bcc, and hcp lattices; expressing the results in terms of Bohr radius a0 = 0:529 A\u00b0; and measuring the energy in units of electron volts per atom, we obtain \u03b5 N = \u221224:35 rs/a0 \u00de \u00f0 \" + 30:1 2 \u2212 rs/a0 \u00de \u00f0 \u00f0 12:5 rs/a0 # \u00de eV/atom: (6.34) The minimum of \u03b5 N occurs when rs a0 \u2248 1:6: However, for alkali metals, the observed values of rs a0 are between 2 and 6. The significant difference between the theoretical value and the experimental results is due to the neglect of correlation energy in this derivation. The treatment of correlation energy is beyond the scope of the present discussions because they are treated by using quantum field theory. Qualitatively, one can summarize the effect of correlation by stating that the correlations tend to keep the electrons apart, thereby further decreasing the energy of the system derived in Eq. (6.34). We will now formulate the semiclassical model of electron dynamics and then derive the Boltzmann equations, which are essential for understanding the study of solids. 6.9 THE SEMICLASSICAL MODEL The semiclassical model predicts the position r and the wave vector k of each electron evolved in the presence of external electric field E and magnetic field B in the absence of collisions. The basic assumptions of the semiclassical model are as follows: 1. The band index n is a constant of motion, and there are no \u201cinterband transitions.\u201d 2. The position of an electron in a crystal with inversion symmetry evolves according to _r = vn = k \u00de \u00f0 1 \u0127 \u2202\u03b5nk \u2202k : (6.35) 180 CHAPTER 6 Static and Transport Properties of Solids 3. The electron wave vector obeys the equation of motion \u0127 _k = \u2212eE r, t \u00f0 \u2212 e c _r \u00d7 B r, t, \u00de where E are electric and magnetic fields that may vary spatially. Eqs. (6.35) and \u00f0 (6.36) can be derived from the Lagrangian formulation in classical mechanics as follows (see Symon10). and B r, t r, t \u00de \u00f0 \u00de \u00f0 \u00de (6.36) \u00f0 If the forces acting on a dynamical system depend on the velocities, we can define a function, called the velocity-dependent potential (", "which includes the ordinary q1, q2, :::, qf ; _q1, _q2, :::, _qf t U potential energy V), such that the generalized force Qk associated with the coordinate qk is given by \u2202U \u2202qk k = 1, :::, f : (6.37) Qk = d dt \u2212 \u00de, \u2202U \u2202 _qk The Lagrangian function is defined as and the equations of motion can be written as L = T \u2212 U, d dt \u2202L \u2202 _qk \u2212 \u2202L \u2202qk = 0: (6.38) (6.39) If we consider a system described in terms of a fixed system of coordinates, the kinetic energy T is a homogeneous quadratic function of the generalized velocities _q1, _q2, :::, _qn, we obtain from Euler\u2019s theorem, f \u2211 k = 1 _qk \u2202T \u2202 _qk = 2T: (6.40) If V, the potential energy, is a function of the coordinates q1, q2, :::, qf, the total energy from Eqs. (6.40) and (6.41) is The Hamiltonian function is defined as We can also write E = T + V: f H = \u2211 k = 1 pk _qk \u2212 L: f dL = \u2211 k = 1 f = \u2211 \u2202L \u2202 _q k d _qk + \u2202L \u2202qk \u2019 pkd _qk + _pkdqk k = 1\u00f0 + \u2202L \u2202t dt dt: dqk ( + \u2202L \u2202t \u00de From Eqs. (6.42) and (6.43), we obtain f dH = \u2211 k = 1\u00f0 _qkdpk \u2212 _pkdqk \u2202L \u2202t \u2212 \u00de dt: (6.41) (6.42) (6.43) (6.44) 6.9 The Semiclassical Model 181 Thus, we obtain _qk = \u2202H \u2202pk and _pk = \u2212 \u2202H \u2202qk, k = 1, \u2026, f, Eq. (6.42) can be written in the generalized form that whenever there is a Lagrangian L also possible to derive a Hamiltonian, using the formulae _QlPl \u2212 L H = \u2211 l and Pl = \u2202L \u2202 _Ql : (6.45) Q, _Q \u00f0, it is \u00de (6.46) (6.47) Here, the Q\u2019s are the three components of rc and the three components of kc (the subscript c is for classical values that we will drop in the future). In an electromagnetic field, the electromagnetic force on an electron of charge \u2212e can be written as F = \u2212eE \u2212 e c V \u00d7 B: (6.48) The Hamiltonian can also be written by replacing the Hamiltonian in a pure perfect crystal lattice by an, where n is the band index. Thus, the electron is treated as equivalent Hamiltonian operator \u03b5n, the modified kinetic energy. If the free because the effect of the lattice potential is included in \u03b5n, the equivalent Hamiltonian operator can be written in the alterpotential of the external field is U nate form, \u2212i\u2207 \u00de \u2212i\u2207 \u00de r \u00f0 \u00f0 \u00de \u00f0 H r, p \u00de = \u03b5n \u2212i\u2207 + U r : \u00f0 According to the quantum mechanical formulation, the Hamiltonian operator is obtained by replacing the classical momentum variable p by \u2212i\u0127\u2207 in the classical Hamiltonian function. Thus, by reversing the steps, we obtain the equivalent classical Hamiltonian from Eq. (6.49) by replacing \u2212i\u2207 by p/\u0127, \u00de \u00f0 \u00f0 \u00de (6.49) H r, p = \u03b5n p/\u0127 + U r : (6.50) \u00de Here, the basic assumption is that the energy functions \u03b5n are known from band structure calculations, and thus, there is no need to include the explicit form of the periodic potential of the ions. k \u00f0 \u00f0 \u00de \u00de \u00de \u00f0 \u00f0 The equations of motion can be written in the canonical Hamiltonian form (Eq. 6.45), and _r = \u2202H \u2202p, _p = \u2212 \u2202H \u2202r : From Eqs. (6.50) and (6.51), we obtain _r \u2261 vn k \u00de \u00f0 = \u2202H", " \u2202p = \u2202 \u2202p f p/\u0127 \u03b5n \u00f0 \u00de + U = r \u00f0 \u00deg \u2202\u03b5n p/\u0127 \u00f0 \u2202p \u00de = 1 \u0127 \u2202\u03b5n k \u00f0 \u2202k \u00de : (6.51) (6.52) (6.53) 182 CHAPTER 6 Static and Transport Properties of Solids From Eqs. (6.48), (6.50), and (6.52), we obtain p_ = \u2212 \u2202H \u2202r = \u2212 \u2202U \u2202r = F = \u2212e E r, t + 1 c \u00de vn k \u00f0 \u00de \u00f0 \u00f0 \u00de \u00d7 B r, t : (6.54) h An alternate way to approach the problem is by defining the vector and scalar potentials A and \u03d5 as i and B = \u2207 \u00d7 A E = \u2212\u2207\u03d5 \u2212 1 c \u2202A \u2202t : From Eqs. (6.48), (6.55), and (6.56), we can write F as F = e\u2207\u03d5 + e c \u2202A \u22026.55) (6.56) (6.57) One can show from Eqs. (6.41) and (6.42) that the Hamiltonian for the electrons in the nth band can be written as and H r, p \u00de \u00f0 = \u03b5n p + e c r $ \u2212 e\u03d5 r, t \u00f0 \u00de where the variables \u0127k are defined as \u0127 _k = \u2212eE \u2212 ev c \u00d7 B, \u0127k = p + e c r, t A \u00f0 : \u00de (6.58) (6.59) (6.60) The canonical crystal momentum p (which is the canonical momentum in the Hamiltonian formulation) can be written as p = \u0127k \u2212 e c r, t A \u00f0 : \u00de (6.61) 6.10 LIOUVILLE\u2019S THEOREM We will now briefly discuss Liouville\u2019s theorem before proceeding further. The configuration and the motion of a system is specified by the coordinates and momenta, q1, q2, :::, qf : p1, p2, :::, pf : The 2f -dimensional space is called the phase space of the mechanical system. The velocity of the phase points is given by the Hamilton equations in Eq. (6.45). The possible state of the mechanical system is represented by each phase point that is occupied by a \u201cparticle\u201d that moves according to the equation of motion (Eq. 6.45). These particles trace out paths that represent the history of the mechanical system. The solutions of the equations of motion are uniquely determined because the positions and momenta are given, and hence, there is only one possible path through each phase point. Liouville\u2019s theorem states that the phase \u201cparticles\u201d move as an incompressible fluid. The phase volume occupied by a set of \u201cparticles\u201d is a constant. The proof of Liouville\u2019s theorem can be found in any standard book of classical mechanics (see Symon10). 6.11 Boltzmann Equation 183 6.11 BOLTZMANN EQUATION The Boltzmann equation can be obtained by first considering the continuity equation. Here, we consider the motion of particles illustrated in Figure 6.19. We consider a large number of a conserved : In a collection of identical particles at position x, moving with a velocity v time \u0394t, the total number of particles moving into a small region dx minus the number of particles that are moving out is equal to and density g x x \u00f0 \u00f0 \u00de \u00de In the limit \u0394t 0,! \u0394g \u0394t \u2212 v \u00f0 x + dx g \u00f0 \u00de x + dx \u0394t: \u00de \u2202g \u2202t = \u2212 \u2202 \u2202x g x v \u00f0 \u00de \u00f0 x, t : \u00de Eq. (6.63) can be generalized for many variables as \u2202g \u2202t = \u2212\u2211 i \u2202 \u2202xi vi g x!\u00de \u00f0 \u00f0 x!, t : \u00de (6.62) (6.63) (6.64) We define grk Thus, grk t t \u00f0 \u00de \u00f0 \u00de as the occupation number of electrons at time t with position r and wave vector k. is the probability that a state is occupied at time t. From Eq. (6.64), we obtain \u2202grk \u2202t = \u2212 \u2202 \u2202r. r_ grk \u2212. _kgrk: \u2202 \u2202k (6.65", ") In an electromagnetic field, we have derived _r = \u2202\u03b5 1 \u2202k \u210f \u210f _k = \u2212eE \u2212 = v, k : \u00de (6.66) E does not depend on k, and v does not depend on r. Further, \u2202 \u2202k \u00d7 v = 0 because v = 1 \u0127 : \u2202\u03b5 \u2202k (6.67) From Eqs. (6.65) through (6.67), we obtain \u2202g \u2202t = \u2212r_. \u2202 \u2202r g \u2212 _k. \u2202 \u2202k g: (6.68) From Eqs. (6.66) and (6.68), we obtain \u2202g \u2202t + v. \u2202 \u2202r g + 1 \u0127 F. \u2202 \u2202k g = 0: (6.69) However, there are momentum transfers caused by thermal fluctuations and impurities, which are not included in Eq. (6.69). To include these, g(x) g(x + dx) \u03c5(x) \u03c5(x + dx) x x + dx FIGURE 6.19 Density of particles at x and x + dx. 184 CHAPTER 6 Static and Transport Properties of Solids Boltzmann added a collision term to Eq. (6.69), and the resulting equation, known as the Boltzmann equation, can be written as \u2202g \u2202t + v. \u2202 \u2202r g + 1 \u0127 F. \u2202 \u2202k g = \u2202g \u2202t \u2019 (coll : (6.70) The terms on the left side of the equation are known as the drift terms, and the term on the right side is known as the collision term. 6.12 RELAXATION TIME APPROXIMATION In thermal equilibrium, the distribution function reduces to the Fermi function frk = 1 e\u03b2r\u00f0 \u03b5k\u2212\u03bcr\u00de + 1, (6.71) where T(r) is the temperature at r and \u03bcr is the constant that produces the correct density n collision term \u2202g \u2202t equilibrium. The simplest approximation for this to happen is the relaxation time approximation, # $. The in Eq. (6.70) must be such that the distribution g must relax toward thermal coll r \u00de \u00f0 \u2202g \u2202t \u2019 (coll = \u2212 1 \u03c4\u03b5 \u00f0 grk \u2212 frk, \u00de (6.72) where \u03c4\u03b5 is defined as the relaxation time. In the simplest solutions, \u03c4\u03b5 is considered as a constant \u03c4., If we approach the same problem in an alternate method, we can argue that because g = g r, k, t we can write dg dt = \u2202g \u2202t + _r. \u2202g \u2202r + _k. \u2202g \u2202k : From Eqs. (6.70), (6.72), and (6.73), we obtain dg dt Integrating Eq. (6.74), we obtain = \u2212 g \u2212 f \u03c4\u03b5 : r, k, t g \u00f0 = \u00de t Z \u2212\u221e dt\u2032f, k r t\u2032 \u00de \u00f0 \u00f0 \u00f0 t\u2032 \u00de\u00de e\u2212 /\u03c4\u03b5 \u00de t\u2212t\u2032 \u00f0 \u03c4\u03b5 : \u00f0 \u00de (6.73) (6.74) (6.75) Here, r r(t) = r and k(t) = k. Integrating Eq. (6.75) by parts, we obtain and k \u00f0 t\u2032 t\u2032 \u00f0 \u00de \u00de, which are solutions of the semiclassical equations, evolve in time such that r, k, t g \u00f0 = f r, k \u00f0 \u00de \u2212 \u00de t Z \u2212\u221e dt\u2032 e\u2212 \u00f0 t\u2212t\u2032 /\u03c4\u03b5 d dt\u2032 \u00de f r t\u2032 \u00de \u00f0 \u00f0, k : t\u2032 \u00f0 \u00de\u00de Because f is the Fermi function, which is evaluated at local temperature and chemical potential, d dt\u2032 f r t\u2032 \u00f0 \u00de \u00f0, k t\u2032 \u00f0 \u00de\u00de = _r # t\u2032 \u00de \u00f0. \u2202 \u2202r + _k t\u2032 \u00f0 \u00de. \u2202 \u2202k f r t\u2032 \u00de, k \u00f0 \u00f0 \u00f0 t\u2032 \u00de\u00de : $ (6.76) (6.77) 6.12 Relaxation Time Approximation 185 From Eq. (6.71), we obtain \u2202f \u2202r = \u2202f \u2202\u03b5 \u2202f \u2202k = \u2212 % \u2202f \u2202\u03b5", " \u2202\u03bc \u2202r \u2212 \u00f0 \u03b5\u2212\u03bc T \u00de \u2202T \u2202r, & \u2202\u03b5 \u2202k = \u0127vk \u2202f \u2202\u03b5, and In addition, from Eq. (6.66), we have \u2202f \u2202\u03b5 = \u2212 \u2202f \u2202\u03bc : _k. \u2202f \u2202k = \u2212evk. E: Substituting Eqs. (6.77) through (6.81) in Eq. (6.76), we obtain r, k, t g \u00f0 = f r, k \u2212 \u00de \u00f0 \u00de t Z \u2212\u221e dt\u2032 e\u2212 \u00f0 t\u2212t\u2032 \u00de /\u03c4\u03b5 vk. eE + % \u2202\u03bc \u2202r + \u03b5k\u2212\u03bc T \u2202T \u2202r \u2202f t\u2032 \u00f0 \u00de \u2202\u03bc : & (6.78) (6.79) (6.80) (6.81) (6.82) Neglecting the time dependence of all terms except the exponential factor, the integral can be easily done, and we obtain r, k g \u00f0 \u00de = f r, k \u00de \u00f0 \u2212 \u03c4\u03b5vk. eE + \u2207!\u03bc + h \u03b5k \u2212 \u03bc T \u2207!T \u2202f \u2202\u03bc : i The current density can be written as which can be rewritten in the alternate form Z J = \u22122e evkg k \u00f0 dk, \u00de J = \u22122 eVkf Z dk \u2212 2 k \u00de \u00f0 Z e2\u03c4eVk Vk. E \u00f0 \u2202f \u2202\u03b5 \u00de \u2212 e T \u00f0 K1. \u2212\u2207T, \u00de\u00de \u00f0 where Kn = 1 4\u03c02 \u03c4e \u0127 ZZ VkVk \u03b5 \u2212 \u03bc \u00de \u00f0 n \u2212 \u00f0 \u2202f \u2202\u03b5n\u00de dS vk d\u03b5. \u2212\u2207T, \u00de \u00f0 (6.83) (6.84) (6.85) (6.86) dk = 1 8\u03c02 d\u03b5 for integration over a constant energy surface (Problem 6.7). and we have used Eq. (6.86) indicates that even if there were no electric field, the thermal gradient, \u2207T, would give rise to an electric current. This is known as the thermo-electric effect, which we will discuss later. We note that Eq. (6.85) can be rewritten in the alternate form RR RR dS \u0127vk J = \u22122 eVk f Z k \u00f0 dk + e2K0. E \u2212 \u00de K1. e T \u2212\u2207T : \u00de \u00f0 (6.87) 186 CHAPTER 6 Static and Transport Properties of Solids 6.13 ELECTRICAL CONDUCTIVITY In an electric field E, which is independent of time, there is no variation of g, f, \u03bc, and T with position. We can rewrite Eq. (6.83) in the simpler form To calculate the electrical conductivity, we need to first calculate the current density \u2212 \u03c4\u03b5vk. \u2202f \u2202\u03bc & : eE \u00bd From Eqs. (6.88) and (6.89), we obtain J = \u22122e evkg Z dk: k \u00de \u00f0 where we have used the identity, Z J = \u22122 evkf dk \u2212 2 k \u00de \u00f0 Z e2\u03c4\u03b5vk. vk. E \u00f0 \u2202f \u2202\u03b5, \u00de \u2202f \u2202\u03bc = \u2212 \u2202f \u2202\u03b5 : In Eq. (6.90), because fk is the equilibrium Fermi distribution function, From Eqs. (6.90) and (6.92), we obtain \u2212 evkfkdk \u2261 0: Z J = 1 4\u03c03 ZZ e2\u03c4\u03b5vk vk. E \u00de \u00f0 \u2202f \u2202\u03b5 ( \u2212 \u2019 dS \u0127vk d\u03b5, (6.88) (6.89) (6.90) (6.91) (6.92) (6.93) where we have changed an integral over a volume of k space into integrations over surfaces of constant energy (Problem 6.7). In a metal, \u2212 \u2202f \u2202\u03b5 \u2019 ( \u03b5 \u2212 \u03b5F = \u03b4 \u00f0 : \u00de We obtain from Eqs. (6.93) and (6.94), J = 1 4\u03c03 e2\u03c4\u03b5 \u0127 Z vkvkdSF vk. E: The equation for electric conductivity tensor can be expressed as", " J = \u03c3. E: From Eqs. (6.95) and (6.96), we obtain \u03c3 = 1 4\u03c03 e2\u03c4\u03b5 \u0127 vkvkdSF vk : Z (6.94) (6.95) (6.96) (6.97) 6.14 Thermal Conductivity 187 In crystals with cubic symmetry, the conductivity tensor is a scalar and if E and J are both in the same x direction, From Eqs. (6.97) and (6.98), we obtain vkvk. E \u00dex = Evx 2 = \u00f0 v2E: 1 3 where \u039b is the mean free path \u03c3 = e2 12\u0127\u03c03 Z \u039b dSF, \u039b = \u03c4v: (6.98) (6.99) (6.100) 6.14 THERMAL CONDUCTIVITY The thermal conductivity experiment is usually done by keeping the specimen in an open circuit. This implies that the specimen has an electric field E, but there is no electric current; i.e., J = 0. From Eqs. (6.87) and (6.92), we have an electric field in the wire The thermal conductivity can be calculated from the total flux of heat (per unit volume) E = \u2212 1 e K\u22121 0 K1 1 T. \u2207T: From Eqs. (6.83), (6.86), and (6.102), we obtain U = 2 fk k \u03b5 \u00f0 f \u00de \u2212\u03bc g vkdk: Z From Eqs. (6.101) and (6.103), we obtain U = \u2212eK1. E + K2. 1 T \u2212\u2207T \u00f0 \u00de : U = 1 T \u00f0 K1K\u22121 0 K1. \u2207T\u2212K2. \u2207T, \u00de which can be rewritten in the alternate form U = 1 T \u00f0 K2\u2212K1K\u22121 0 K1. \u00f0 \u00de \u2212\u2207T : \u00de The thermal conductivity \u03ba is defined as U = \u03ba. \u2212\u2207T : \u00de \u00f0 If we neglect the second term of Eq. (6.105) for metals, we obtain from Eqs. (6.105) and (6.106) Using the expansion (Problem 6.8) \u03ba \u2248 1 T K2: d\u03b5 F \u03b5 \u00de \u00f0 Z \u2212\u2202f0 \u2202\u03b5 \u2019 d\u03b5 = F ( + \u03bc \u00f0 \u00de 1 6 \u03c02 \u00f0 kBT 2 \u22022F \u03b5 \u00f0 \u2202\u03b52 % \u00de \u00de &\u03b5 = \u03bc + :::, (6.101) (6.102) (6.103) (6.104) (6.105) (6.106) (6.107) (6.108) 188 CHAPTER 6 Static and Transport Properties of Solids and \u03c3 = e2K0 [from Eq. (6.87) under isothermal conditions], we can write \u03ba = \u03c02 3 k2 e2 T\u03c3: (6.109) This is also known as the Wiedemann\u2013Franz law. 6.15 WEAK SCATTERING THEORY OF CONDUCTIVITY 6.15.1 Relaxation Time and Scattering Probability We will now consider a more realistic description of the collisions by assuming that there is a scattering probability Wk,k\u2032dk\u2032dt 2\u03c0 \u00f0, 3 \u00de which is the probability in an infinitesimal time interval dt that an electron with wave vector k is scattered into any one of the group of levels with the same spin in an infinitesimal k-space volume element dk\u2032 about k\u2032: These levels are not forbidden by the exclusion principle in the sense that is the fraction of levels in the volume element they are not occupied. Because, by definition, g k\u2032, and hence, the total probdk\u2032 about k\u2032 that are occupied, the fraction that is unoccupied is 1\u2212g \u00de ), is given by (from ability per unit time for a collision (which is precisely the definition of Eq. 6.110) k6.110) 1 k \u00f0 \u00de \u03c4 = Z dk\u2032 2\u03c0 \u00de \u00f0 3 Wk,k\u2032\u00bd 1 \u2212 g k\u2032 : \u00de& \u00f0 (6.111) We note that \u03c4 the distribution function g k \u00de \u00f0 k \u00f0 : \u00de is not the usual relaxation time defined in Eq. (6.72) but explicitly depends on 6.15.2 The Collision Term The number of electrons scattered out of the volume element dk about k in time interval dt is given", " by dnout = \u2212 k dg \u00f0 dt \u00de \u2019 (out 3 dt: dk 2\u03c0 \u00f0 \u00de (6.112) The total number of electrons per unit volume in dk about k that suffer a collision in time interval dt is also obtained from Eq. (6.111) as From Eqs. (6.111) through (6.113), we obtain dnout = dt k \u00f0 \u00de \u03c4 g k \u00de \u00f0 dk 2\u03c0 3 : \u00de \u00f0 k dg \u00f0 dt \u2019 \u00de (out = \u2212 g k \u00de \u00f0 Z dk\u2032 2\u03c0 \u00de \u00f0 3 Wk,k\u2032\u00bd 1\u2212g : k\u2032 \u00de& \u00f0 (6.113) (6.114) 6.15 Weak Scattering Theory of Conductivity 189 The electrons not only scatter out of the level k, but are also scattered into it from other levels. By making a similar argument as before (see Problem 6.8), we obtain k dg \u00f0 dt \u00de \u2019 (in = g0 Z dk\u2032 2\u03c0 \u00f0 3 Wk\u2032,kg \u00de k\u2032 : \u00de \u00f0 If we define (coll we obtain from Eqs. (6.114) through (6.116) \u2019 k dg \u00f0 dt \u00de = k dg \u00f0 dt \u2019 \u00de + (in \u2019 k dg \u00f0 dt \u00de, (out (6.115) (6.116) k dg \u00f0 dt \u2019 \u00de = (coll Z dk\u2032 2\u03c0 g k\u2032 3 f \u00f0 \u00de\u00bd \u00f0 In the relaxation-time approximation, Eq. (6.104) can be rewritten as \u00de 1 \u2212 g k Wk\u2032,k \u2212 g \u00f0 \u00de& k 1 \u2212 g k\u2032 \u00de& Wk,k\u2032g : \u00f0 (6.117) \u00de\u00bd \u00f0 k dg \u00f0 dt \u2019 \u00de = (coll % g0 6.118) 6.15.3 Impurity Scattering In any real specimen, at very low temperatures, the main source of collisions would be impurities because the scattering of electrons by the thermal vibrations of the ions and the electron\u2012electron scattering become increasingly weak as the temperature is lowered. This impurity-electron scattering will be elastic as long as the energy gap between the impurity ground state and the excited state is large compared to kBT: We consider the case of elastic scattering by fixed substitutional impurities that are located at random lattice sites R in the crystal. The basic assumptions made in this calculation are as follows: 1. There is a relatively small number of impurities in the crystal, such that one can consider the electrons to be interacting with one impurity at a time. 2. There are very few excited impurity ions that lose energy to electrons in collisions. 3. There are very few empty electronic levels that are low enough in energy to receive an electron after it has lost enough energy to excite an impurity ion. 4. The scattering potential U, which describes the interaction between an electron and a single impurity, is spherically symmetrical and weak. 5. The energies of the occupied electronic states \u03b5k are isotropic and depend only on the magnitude of k. 6. The probability of scattering between two levels k and k\u2032 vanishes unless k = k\u2032 and depends on the common value of their energies and on the angle between k and k\u2032: The rate at which an electron at k makes a transition to k\u2032 is given by Fermi\u2019s golden rule (see, for example, Goswami4), Wk,k\u2032 = 2\u03c0 \u2032 Ut <k j \u00de\u00dej r \u00f0 \u00dej \u00f0 k\u2032> j 2, (6.119) 190 CHAPTER 6 Static and Transport Properties of Solids where Here, U write \u00f0 r\u2212R \u00de Ut 6.120) are spherically symmetric potentials centered at sites R. In simple cases, one can where ni is the number of impurities per unit volume. In more rigorous derivations, one can conas the sum of pseudopotentials of impurity atoms added to a perfect crystal or of atoms sider Ut dislocated from their regular positions in the crystal. Here, r \u00f0 \u00de Ut r \u00f0 \u00de = niU \u00f0 \u00de r, (6.121) and the Bloch functions \u03c8 nk\u00f0 r \u00de are normalized, <k U j k\u2032> = j Z dr \u03c8 %nk\u00f0 r \u00de U r \u03c8 nk\u00f0 \u00de \u00f0 r \u00de, dr r \u03c8 nk\u00f0 j \u00dej 2 = vcell", ": Zcell (6.122) (6.123) We note from Eq. (6.119) that Wk,k\u2032 = 0 unless \u03b5. In addition, Wk,k\u2032 is independent of the electron distribution function in the independent electron approximation. Further, because is spherically symmetric, Wk,k\u2032 depends only on the angle between k and k\u2032, and therefore, U \u00de we obtain k This symmetry is also known as \u201cdetailed balancing.\u201d From Eqs. (6.117) and (6.124), we obtain Wk,k\u2032 = Wk\u2032,k: (6.124) (6.125) (6.126) \u2202g k \u00f0 \u2202t \u00de \u2019 = \u2212 (coll Z dk\u2032 2\u03c0 \u00de \u00f0 g 3 Wk,k\u2032 \u00bd \u00f0 k \u00de \u2212 g k\u2032 : \u00de& \u00f0 We further assume that the solution of the Boltzmann equation g has the form gk = fk + a. k, \u03b5 \u00f0 \u00de dk\u2032 2\u03c0 where a is a vector that depends on k only through its magnitude \u03b5 brium distribution function (Fermi function). We define the relaxation time through k \u00f0 \u00de and fk is the local equili- 1 \u03c4\u03b5 \u00bd g k \u00de \u00f0 \u2212 f k \u00de& \u00f0 = 3 Wk,k\u2032\u00bd g k \u00de \u00f0 \u2212 g k\u2032 : \u00de& \u00f0 (6.127) Because Wk,k\u2032 vanishes unless \u03b5 (6.127), \u00f0 k \u00de = \u03b5 \u00f0 k\u2032 \u00de and we obtain from Eqs. (6.126) and \u03b5\u2032 \u00f0 \u00de \u00de From Eq. (6.128) and Eq. (3) of Problem 6.8, we obtain \u00f0 Z 1 \u03c4\u03b5 dk\u2032 2\u03c0 3 Wk,k\u2032\u00f0 k \u2212 k\u2032 \u00de : 1 \u03c4\u03b5 = Z dk\u2032 2\u03c0 \u00de \u00f0 3 Wk,k\u2032\u00f0 1 \u2212 ^k. ^k\u2032 : \u00de (6.128) (6.129) 6.15 Weak Scattering Theory of Conductivity 191 When k and k\u2032 are on the Fermi surface, k = k\u2032 = kF: If we define we obtain or q = k \u2212 k\u2032, q2 = 2k2 F\u00f0 1 \u2212 ^k. ^k\u2032 \u00de 1 \u2212 ^k. ^k\u2032 \u00f0 = 2 \u00de q 2kF\u2019 2 : ( We can now write from Eqs. (6.119) and (6.130), Eq. (6.133) can be rewritten in the alternate form Wk,k\u2032 = dr V eiq.r \u2211 U R \u00f0 Z 2\u03c0 * * * * * Wk, k\u2032 = 2\u03c0 \u0127 1 V 2 We define the static structure factor \u2211 * R * * * * S q \u00f0 \u00de and the Fourier transform of the scattering potential U(r) \u00de* * * * *, eiq.R Z dreiq. eiq.R = 1 Ns * * * * dr eiq. \u03b5F \u2212 \u03b5 : k\u2032 \u00de\u00de \u00f0 2 \u03b4 \u00f0 \u03b5F \u2212 \u03b5 : k\u2032 \u00f0 \u00de\u00de (6.130) (6.131) (6.132) (6.133) (6.134) (6.135) (6.136) Z where Ns is the number of scatterers. From Eqs. (6.130) and (6.134) through (6.136), we obtain Wk,k\u2032 = 2\u03c0Ns \u0127V 2 S \u00f0 q \u00dej U q \u00f0 2 \u03b4 \u00dej \u03b5F \u2212 \u03b5 \u00f0 \u00f0j k \u2212 q j\u00de\u00de : Substituting Eqs. (6.132) and (6.137) in Eq. (6.129), we obtain 1 \u03c4\u03b5 = 2\u03c0Ns \u0127V dk\u2032 2\u03c0 \u00f0 \u00de 3S \u00f0 Z q U \u00dej q 2 \u03b4 \u00dej \u00f0 \u00f0 \u03b5F \u2212 \u03b5 k \u2212 q 2 \u00dej \u00de\u00de \u00f0j q 2kF\u2019 2 : ( (6.137) (6.138) Eq. (6.138) can be rewritten in the alternate form, 1 \u03c4\u03b5 = Ns \u0127V 1 4\u03c0 q4 k2 F Z dq S q \u00dej U q \u00f0 \u00dej \u00f0 1 2 Z \u22121 cos \u03b8 d \u00f0 \u00de \ufffd", "\ufffd \u03b5F \u2212 \u03b5 # k2 F + q2 \u2212 2kFq cos \u03b8 # \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q $$ : (6.139) We also use the results of the angular integration, 1 Z \u22121 d \u00f0 cos \u03b8 \u03b4 \u00de \u03b5F \u2212 \u03b5 # k2 F + q2 \u2212 2kFq cos \u03b8 # \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q $$ = 2kF \u2212 q \u2202\u03b5/\u2202kF \u03b8 \u00f0 q \u00f0 \u00de \u00de, (6.140) 192 CHAPTER 6 Static and Transport Properties of Solids where \u03b8 x \u2212 a \u00de \u00f0 is the step function, x \u2212 a \u03b8 \u00f0 = 1, = 0, \u00de for x > a for x < a: The step function is related to the Dirac delta function, From Eqs. (6.139) through (6.142), we obtain (Problem 6.10) x \u2212 a \u03b4 \u00f0 \u00de = d\u03b8 x \u2212 a /dx: \u00de \u00f0 1 \u03c4\u03b5 = 1 4\u03c0\u01272k2 FvF Ns V 2kF Z 0 dq q3S \u00f0 q U \u00dej 2: q \u00f0 \u00dej The resistivity (in the relaxation time approximation) is given by \u03c1 = m ne2\u03c4\u03b5 = 3\u03c0Ns Fk4 4e2\u0127v2 f V 2kF Z 0 dq q3 S \u00f0 q U \u00dej 2: q \u00f0 \u00dej (6.141) (6.142) (6.143) (6.144) 6.16 RESISTIVITY DUE TO SCATTERING BY PHONONS We consider only one atom per unit cell that is located at the lattice site (its equilibrium position) at zero temperature. At a finite temperature, the deviation of the atom or ion from equilibrium position is (from Chapter 2) The static structure factor is defined in Eq. (1.64). We obtain ul = r \u2212 Rl \u2212 d: eiq. \u00f0 Rl+ul \u00de = \u2211 l eiq.Rl \u00bd \u2211 l 1 + iq. ul + ::: : & (6.145) (6.146) The structure factor of the unperturbed lattice, \u2211 l Eqs. (2.158) and (2.182), we have the Fourier transform of eiq.Rl, has no contribution to the resistivity. From where ^ul = 1 MNp ^Q ^\u2208 \u2211 k k!\u03bb k!\u03bb eik.Rl, \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 ^Q k!\u03bb = 1 2 3 5 ^a \u2019 k!\u03bb + ^a\u2020 \u2212k!\u03bb : ( \u0127 2\u03c9 k!\u03bb 2 4 We can rewrite from Eqs. (6.147) and (6.148), ^uk\u03bbeik.Rl + ^u\u2020 k\u03bbe\u2212ik.Rl \u2211 k\u03bb\u00bd, & ul = 1 Np \ufb03\ufb03\ufb03\ufb03 (6.147) (6.148) (6.149) 6.16 Resistivity Due to Scattering by Phonons 193 where ^uk\u03bb = \u0127 2M\u03c9 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 k!\u03bb s \u2208!k\u03bb^ak\u03bb:", " From Eqs. (6.146) and (6.149), we obtain eiq. \u00f0 Rl+ul \u00de = \u2211 l 1 Np \u2211 lk i \u00bd ^uk. qei \u00f0 k+q.Rl + ^u\u2020 k \u00de. qei \u00f0 q\u2212k \u00de.Rl, & where only the longitudinal mode (q||k) is retained. Because \ufb03\ufb03\ufb03\ufb03 ei \u2211 l q+k \u00f0.Rl \u00de \u00de = N\u03b4K,q+k, Eq. (6.151) can be rewritten in the alternate form eiq. \u00f0 Rl+ul \u00de = \u2211 l Np \u2211 Kk i ^uk. q\u03b4K,q+k + ^u\u2020 k \u00bd. q\u03b4K,q\u2212k : & (6.150) (6.151) (6.152) (6.153) In the N-process (normal scattering), K = 0 and q = k; i.e., the wave vectors of electrons and phonons are equal. In the Umklapp process (U-process), K \u2260 0. In the following discussion, we will consider the N-process. \ufb03\ufb03\ufb03\ufb03 From Eqs. (6.135) and (6.153), we obtain = S q \u00de \u00f0 \u2211 l 1 N \u2248< ^u\u2020 q < eiq. \u00f0 j 2>,. q j Rl+ul 2> \u00de j j because \u03b40,q\u2212k \u2260 0 can be nonzero only when both k and q are in the first Brillouin zone. From Eqs. (6.137) and (6.141), we obtain = S q \u00de \u00f0 \u0127 2M\u03c9q j \u2208!: q! j 2<^aq^a\u2020 q + ^a\u2020 q^aq>, (6.154) (6.155) where \u2208! is the polarization vector of the longitudinal mode. S(q) can be rewritten in the alternate form, = S q \u00de \u00f0 \u0127q2 2M\u03c9q \u00f0 2nq + 1 : \u00de From Eq. (6.144) and (6.156), we obtain \u03c1 = 3\u03c0 e2\u0127v2 F 2kF 1 4k4 F Z 0 Ns V # $ dq q3 \u0127q2 2M\u03c9q \u00f0 2nq + 1 U \u00dej q \u00f0 \u00dej 2: It can be shown after some algebra that \u03c1 can be written as \u03c1 = 3\u03c0 e2\u0127v2 F Ns V # $ 1 4k4 F \u0127 2Mc kBT \u0127c # $ Z 0 dz z4 ez + 1 ez \u2212 1 2\u0398/T 5 zkFT \u0398 # U * * * * 2 : $* * * * (6.156) (6.157) (6.158) 194 CHAPTER 6 Static and Transport Properties of Solids In Eq. (6.158), \u0398 = \u0127kFc kFT, \u0398 is the Debye temperature, c is the velocity of the longitudinal kB wave, and M is the mass of the ion. At large temperatures, \u03c1 is linear in T, while at very low temperatures, \u03c1 ; z = q\u0398 T 5: # PROBLEMS 6.1. Derive Eq. (6.10) from Eq. (6.9). 6.2. Two noble gas atoms of charge Ze, each surrounded by Z electrons, are at 0 and R, where R is so large that there is no interaction between the electronic charges around the two nuclei. The Hamiltonian of the system can be written as where the interaction Hamiltonian HI is given by H = H1 + H2 + HI, HI = e2 Z2 r\u00f0 \u00dej \u2212 R j + Z 2 r\u00f0 \u00dej 0 @ 1 A z + \u2211 i, j = 1 1 1 r\u00f0 \u00dei \u2212 r\u00f0 2 \u00dej j : 3 5 (1) (2) a. Show that the effect of the first-order perturbation is exponentially small because the ground state of the two atoms is spherically symmetric and the charge distributions hardly overlap because they are far apart. b. Show that in the second-order perturbation theory, all contributions to the relevant integrals are small unless j 1 r\u00f0 \u00dej \u226a R and j \u226a R: Thus, one can expand Eq. (2) as 2 \u00dej \u2212 R r\u00f0 j 3 1 r\u00f0 \u00dej \u00f0", " j. R \u00de\u00f0 R2 2 r\u00f0 \u00dei \u2212 R HI \u2248 \u2212 e2 R3 \u2211 i, j 2 4 \u00de \u2212 r\u00f0 1 \u00dej. 2 r\u00f0 \u00dei \u2212 R \u00f0 : 3 \u00de (3) 5 c. Hence, show that \u0394E = <0 HI j 0> + \u2211 k j the leading term in the second-order perturbation theory, <0 k> 2 j HI j E0 \u2212 Ek j j, varies as \u2212 1 R6 : This is essentially the quantum theory of the van der Waals force. 6.3. Derive Eq. (6.17) for the Bulk modulus at equilibrium. 6.4. We have derived in Eq. (6.21) ucoul = \u2212 d \u00f0 \u00de e2 d 1 + \u2211 i \u2260 0 ( d Ri + d \u2212 d Ri, ) ( j \u2019 j Rewrite the expression as e2 d f S(d) can be written as (because Ri is a vector) ucoul = d \u00de \u00f0 dS 0 \u00de \u00f0 \u2212 dS \u2212 \u2212 Ri j, j (1) (2) (3) Problems 195 which can be expressed in the integral form = S d \u00de \u00f0 \u221e Z 0 2 dr \u03c0p \u2211 i \u2260 0 e\u2212r2 j d\u2212Ri 2 : j Eq. (4) can also be rewritten as \ufb03\ufb03\ufb03 = S d \u00de \u00f0 \u221e Z 0 2 dr \u03c0p \ufb03\ufb03\ufb03 Z dk \u03c03p r3 \u2211 Ri \u2260 0 e\u2212k2/r2 + 2ik. \u00f0 d\u2212Ri \u00de: \ufb03\ufb03\ufb03\ufb03\ufb03 Using the relation, \u2211 K is the Dirac delta function and \u03a9 is the volume of the unit cell, show that \u2211 i \u00f0 \u00de \u00f0 \u00de \u03b4 q\u2212K, 2\u03c0 eiq.Ri = 3 1 \u03a9 where \u03b4 \u00de Eq. (5) can be rewritten as q\u2212 dr \u03c0p Z dk \u03c03p r3 \" ( 3 2\u03c0 \u00de \u03a9 \u00f0 \u2211 K \u03b4 \u00f0 2k\u2212K \u00de \u22121 # ) e\u2212k2/r2+2ik.d: \ufb03\ufb03\ufb03\ufb03\ufb03 Show that Eq. (7) can be simplified as \ufb03\ufb03\ufb03 = S d \u00de \u00f0 \u221e Z 0 2 dr \u03c0p r3 % \u03c03 \u03c03p 1 \u03a9 \u2211 K e\u2212K2/4r2 + iK.d\u2212ed2r2, & \ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03 4\u03c0 \u03a9K2 eiK.d\u2212 1 d : = \u2211 K In deriving Eqs. (4) and (9), we have used the identity (4) (5) (6) (7) (8) (9) \u221e = 1 d 2 dr \u03c0p e\u2212d2r2 : (10) Z 0 Eq. (9) diverges for the K = 0 term. S(0) also diverges. However, S(d) \u2013 S(0) does not diverge, which can be shown by rewriting S(d) from Eqs. (4) and (8) as \ufb03\ufb03\ufb03 = d S \u00f0 \u00de G Z 0 2 dr \u03c0p \u03c03 \u03c03p 1 \u03a9 \u2211 K\u22600 r3 \" e\u2212K2/4r2 + iK.d\u2212ed2r2 \u221e Z G + # 2 dr \u03c0p \u2211 Ri\u22600 e\u2212r2 \u00f0 d\u2212Ri 2 \u00de, (11) where G is of the order of a reciprocal lattice vector. Eq. (11) can be rewritten by using the procedure followed earlier as \ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03 d S \u00f0 \u00de = \u2211 K \u2260 0 4\u03c0 \u03a9K2 e\u2212K2/4G2 + iK.d\u2212 G Z 0 2 dr \u03c0p e\u2212d2r2 + Z \u221e G 2 dr \u03c0p \u2211 Ri \u2260 0 e\u2212r2 \u00f0 d\u2212Ri 2 \u00de : (12) Each of the terms in Eq. (12", ") converges. Similarly, one can show that S(0) also converges. Thus, dS converges, although the individual summations in Eq. (9) do not 0 \ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03 \u2212dS d \u00de \u00f0 \u00f0 \u00de 196 CHAPTER 6 Static and Transport Properties of Solids converge. Hence, we define where \u03b1 is the Madelung constant. 6.5. Derive Eq. (6.35) from Eq. (6.34). dS d \u00de \u00f0 \u2212 dS 0 \u00de \u00f0 + 1 = \u03b1, (13) 6.6. Derive Eq. (6.82) from Eqs. (6.72) through (6.81). 6.7. Derive Eq. (6.93) from Eq. (6.92). 6.8. According to the Sommerfeld expansion rule, \u221e Z \u2212\u221e d\u03b5 \u221e \u221e d\u03b5 + \u2211 \u00de n = 1 H \u03b5 \u00f0 kbT an \u00f0 2n d2n\u22121 d\u03be2n\u22121 H \u00de \u03b5 \u00dej \u03b5 = \u03bc: \u00f0 2n = 22n\u22121 \u03c02n Here, \u03be 2n \u00de function, and Bn are the Bernoulli numbers. If! Bn, f \u03b5 \u00de \u00f0 \u00de \u00f0 \u00f0 is the Fermi distribution function, \u03be is the Riemann zeta show (by using the values \u03be dF \u03b5 \u00f0 d\u03b5 \u00de, H \u03b5 = \u00de \u00f0 6 and B1 = 1 6\u00de \u2212\u2202f \u2202\u03b5 = F + \u03bc \u00de \u00f0 ( = \u03c02 2 \u00de \u00f0 \u221e Z \u2212\u221e d\u03b5 F \u03b5 \u00f0 \u00de \u2019 that \u03c02 6 \u00f0 kBT 2 \u22022F \u03b5 \u00f0 \u2202\u03b52 % \u00de \u00de : &\u03b5 = \u03bc 6.9. Derive Eq. (6.115) by making the same arguments made for Eq. (6.114). 6.10. Write the vector k\u2032 in terms of its components + k\u2032\u22a5 = k\u2032 = k\u2032 k ^k. k\u2032 \u00de \u00f0 ^k + k\u2032\u22a5: (1) (2) (3) (1) Show that because the collision probability Wk,k\u2032 depends only on the angle between k and k\u2032 (for elastic scattering), Hence, show that because for k \u2260 k\u2032, Wk,k\u2032 = 0, Z dk\u2032 Wk,k\u2032k\u2032\u22a5 = 0: 6.11. Derive Eq. (6.143) from (6.139) through (6.142). dk\u2032 Wk,k\u2032 = k Z Z dk\u2032 Wk,k\u2032\u00f0 ^k. ^k\u2032 : \u00de (2) (3) References 1. Aschroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 2. Bernandes N. Theory of Solid Ne, A, Kr and Xe at 0 K. Phys Rev 1958;112:1534. References 197 3. Chelikowsky JR, Cohen ML. Nonlocal pseudopotential calculation for the electronic structure of eleven diamond and zinc-blende structures. Phys Rev B 1976;14:556. 4. Goswami A. Quantum mechnics. Boston: Wm. C. Brown; 1997. 5. Hirshfelder JO, Curtiss CF, Bird BB. Molecular theory of gases and liquids. New York: John Wiley & Sons; 1954. 6. Kittel C. Introduction to solid state physics. New York: John Wiley & Sons; 1967. 7. Lennard Jones JE. Proc R Soc (London) 1924;A106:463. 8. Madelung O. Introduction to solid state theory. New York: Springer-Verlag; 1978. 9. Marder MP. Condensed matter physics. New York: John Wiley & Sons; 2000. 10. Symon KR. Mechanics. Reading, MA: Addison-Wesley; 1971. 11. Ziman JM. Principles of theory of solids. Cambridge: Cambridge University Press; 1972. Electron\u2013Electron Interaction CHAPTER 7 CHAPTER OUTLINE 7.1 Introduction....................................................................", "......... 199 7.2 Hartree Approximation.................................................................... 200 7.3 Hartree\u2013Fock Approximation.............................................................. 203 7.3.1 General Formulation............................................................... 203 7.3.2 Hartree\u2013Fock Theory for Jellium.................................................... 204 7.4 Effect of Screening...................................................................... 207 7.4.1 General Formulation............................................................... 207 7.4.2 Thomas\u2013Fermi Approximation...................................................... 208 7.4.3 Lindhard Theory of Screening....................................................... 209 7.5 Friedel Sum Rule and Oscillations......................................................... 214 7.6 Frequency and Wave-Number-Dependent Dielectric Constant.................................. 217 7.7 Mott Transition.......................................................................... 222 7.8 Density Functional Theory................................................................ 223 7.8.1 General Formulation............................................................... 223 7.8.2 Local Density Approximation....................................................... 224 7.9 Fermi Liquid Theory..................", ".................................................... 225 7.9.1 Quasiparticles..................................................................... 225 7.9.2 Energy Functional................................................................. 227 7.9.3 Fermi Liquid Parameters........................................................... 230 7.10 Green\u2019s Function Method................................................................. 232 7.10.1 General Formulation............................................................... 232 7.10.2 Finite-Temperature Green\u2019s Function Formalism for Interacting Bloch Electrons........ 233 7.10.3 Exchange Self-Energy in the Band Model............................................ 234 Problems.................................................................................... 235 References.................................................................................. 241 7.1 INTRODUCTION In the nearly free electron model, we assumed that the motion of the electrons is independent of each other. The independent electron was assumed to obey the one-electron Schrodinger equation, and the lattice potential was considered to be due to the motion of the electron in a background of static positive charges in the lattice. In this approximation, we neglect the electron\u2013electron interaction as well as the motion of the ions at a finite temperature. Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00007-4 \u00a9 2012 by Elsevier Inc. All rights reserved. 199 200 CHAPTER 7 Electron\u2013Electron Interaction However, in a rigorous solution of the underlying quantum mechanical problem, the Hamiltonian of the solid should include the motion of the interacting N electrons as well as the motion of the nuclei of which the mass is much heavier than that of the electrons. Because the nuclei are much more massive than the electrons and the interaction between the conduction electrons and the lattice waves is a complex problem, as a first step in solving the Hamiltonian, we will apply the Born\u2013Oppenheimer approximation. According to this approximation, the electronic problem is first solved by assuming that the nuclei are static, classical potentials. The motion of the nuclei gradually increases around their equilibrium positions in the lattice when the temperature increases. As we saw in Chapter 2, the quanta of these lattice vibrations are called phonons. In the Born\u2013Oppenheimer approximation, it is assumed that the cloud of negative charge of the electrons follow the nuclei in", " their motion. Because this motion is followed by charge redistribution, the energies involved in moving the nuclei also depend on the electron energy. This interaction between the conduction electrons and lattice waves is a dynamical problem that can be treated by perturbation theory. In the adiabatic approximation of Born and Oppenheimer, the eigenfunction of the Hamiltonian can be written as a product of the N-electron wave function satisfies a Schrodinger equation for the wave functions \u03a8 is the wave function of the N electrons in a static lattice, frozen of the ions and \u03a8 with the lth ion at the point Rl of the Bravais lattice. Here, the instantaneous positions of the N electrons are ri(i = 1, 2, \u2026, N) and the corresponding spins are si. We will consider the ionic motion in a later, which is the total section, where we will show that we would need to add an additional term \u03b5e energy of the electrons as a function of the ions, in the Schrodinger equation for the ions. and \u03a6 r1s1, r2s2, \u2026, rNsN r1s1, r2s2, \u2026, rNsN \u00f0, where If we include the electron\u2013electron interactions, the Schrodinger equation for the N-electron wave function can be written as N H\u03a8 = \u2211 i = 1 \u01272 2m \u2212! \u22072 i \u03a8 \u2212 ze2\u2211 l 1 ri \u2212 Rl j \u03a8 + j \" 1 2 \u2211 i\u2260 j e2 ri \u2212 rj j \u03a8: j (7.1) It is impossible to solve Eq. (7.1), even with the fastest computer available to physicists, because the total number of electrons is of the order of 1023. The only possible way to solve Eq. (7.1) is by making drastic approximations. We will start with the simplest approximation, known as the Hartree approximation. 7.2 HARTREE APPROXIMATION There are two alternate methods to derive the Hartree approximation. We will first consider the simplest method. We proceed with the basic idea of \u201cderiving\u201d from Eq. (7.1) the one-electron Schrodinger equation for the electrons in a lattice potential of the form \u2212 \u01272 2m \u22072\u03c8 = \u03b5\u03c8 r, \u00de \u00f0 (7.2) by obviously making suitable (and drastic) approximations in Eq. (7.1). The potential due to the static ions in the lattice is Uion r \u00f0 \u00de = \u2212ze2\u2211 l 1 r \u2212 Rl j : j (7.3) 7.2 Hartree Approximation 201 can be approximated as an electron moving in a field The effective electron\u2013electron potential Uee produced by the sum of all other electrons. If the electrons are considered as a smooth distribution of negative charge of charge density \u2212en, where n(r) is the number density of electrons, r r \u00f0 \u00de \u03c8 j\u00f0 the potential energy of the electron in this field would be = \u2211 j r \u00de \u00f0 j r 2, \u00dej \u00f0 \u00de n (7.4) (7.5) Uee r \u00de \u00f0 = dr\u2032 Z e2n r\u2032 \u00de \u00f0 r \u2212 r\u2032 j j : The one-electron Schrodinger equation is obtained by adding U = Uion + Uee, \u2212 \u01272 2m \u22072\u03c8 i\u00f0 r \u00de \u2212 Ze2\u2211 l 1 r \u2212 Rl j r \u03c8 i\u00f0 \u00de + e2\u2211 \" j Z j dr\u2032 \u03c8 j\u00f0 r\u2032 j 2 \u00dej 1 r \u2212 r\u2032 j j # r \u03c8 i\u00f0 \u00de = \u03b5i\u03c8 i\u00f0 r \u00de : (7.6) Eq. (7.6) is known as the Hartree equation. The Hartree equation can also be derived from Eq. (7.1) by following a variational method as follows. We write the N-electron wave function as \u00f0 \u03a8 r1s1, r2s2, \u2026, rNsN \u03c8 2\u00f0 are a set of orthonormal one-electron wave functions. We have to consider \u03a8 as a trial where \u03c8 i\u00f0 wave function for a variational calculation and find the equation that is satisfied by the single-particle functions \u03c8 i\u00f0 so that the trial function (Eq. 7.7) minimizes \u2026 \u03c8 N\u00f0 = \u03c8 1\u00f0 rNsN (7.7) r1s1 r2s2 ri,si risi \u00de \u00de \ufffd", "\ufffd \u00de \u00de \u00de, Here, \u03a8 is normalized because the set \u03c8 i\u00f0 Eq. (7.1) as <H> = \u2211 s1 \u2211 s2 \u2026 \u2211 sN Z risi \u00de dr1dr2 \u2026 drN\u03a8#H\u03a8: (7.8) is orthonormal. If we rewrite the Hamiltonian in H = \u2211 i Hi + 1 2 Vij, \u2211 i\u2260 j (7.9) we note that Hi operates only on the coordinate of the ith electron and Vij operates on the two-body coordinates of both i and j. From Eqs. (7.7) through (7.9), we obtain <H> = \u2211 i Z \u03c8 #i Hi\u03c8 idr + 1 2 \u2211 i\u2260 j Z Z dri drj \u03c8 #i \u03c8 #j Vij\u03c8 i\u03c8 j: (7.10) If we minimize <H> with respect to variation of \u03c8 #i, we obtain \u03b4<H> = \u2211 \u03b4\u03c8 #i # where the variations \u03b4\u03c8 #i satisfy the equation i Z Hi + \u2211 j\u2260i Z \u03c8 #j Vij \u03c8 j drj \u03c8 i dri = 0, $ \u03b4\u03c8 #i \u03c8 i dri = 0, Z (7.11) (7.12) 202 CHAPTER 7 Electron\u2013Electron Interaction because of their normalization conditions. Eq. (7.12) acts as a constraint on the variation of <H>. We follow the method of Lagrangian multipliers by multiplying Eq. (7.12) by a multiplier \u2208i and subtract the sum from Eq. (7.11). We obtain \u2211 i Z \u03b4\u03c8 #i # Hi + \u2211 j\u2260i Z \u03c8 #j Vij \u03c8j drj \u2212 \u2208i \u03c8idri = 0: (7.13) $ Because the variations \u03b4\u03c8 #i are independent, the coefficient of each \u03b4\u03c8 #i must vanish; i.e., we must have Hi + \u2211! j\u2260i Z \u03c8 #j Vij \u03c8 j drj \u03c8 i = \u2208i \u03c8 i, \" (7.14) which, along with Eqs. (7.1) and (7.9), leads to the Hartree equation, \u2212 \u01272 2m \u22072\u03c8 i\u00f0 ri \u00de \u2212 Ze2\u2211 l 1 ri \u2212 Rl j j ri \u03c8 i\u00f0 \u00de + e2\u2211 \" j\u2260i Z 2 drj rj \u03c8 j\u00f0 j \u00dej 1 ri \u2212 rj j # j ri \u03c8 i\u00f0 \u00de = \u03b5i\u03c8 i\u00f0 ri \u00de : (7.15) Comparing Eqs. (7.6) and (7.15), we note that the result of the formal derivation of the Hartree equation (Eq. 7.15) is almost identical to that derived earlier (Eq. 7.6) by using a qualitative argument except that each electron interacts not with the full charge density of the system (Eq. 7.6), but with the charge density minus the density of the electron itself. However, in general, the set of equations in Eq. (7.6) is known as the Hartree equations and is easier to solve. In what follows, we will assume that Eq. (7.6) is the Hartree equation even though it fails to represent the way in which a particular configuration of the N \u2013 1 electrons affects the electron under consideration. The nonlinear Hartree equations (Eq. 7.6) for one-electron wave functions and energies are usually solved by first guessing a form for Uee (the term in the square bracket in Eq. 7.6) and the one-particle Schrodinger equation is solved. The value of Uee is again computed from the, and the Schrodinger equation is again solved using these values. This new wave functions \u03c8 i\u00f0 iteration procedure is continued until there is no significant change in the potential. In fact, for this reason, the Hartree approximation is also known as the \u201cself-consistent field approximation.\u201d r \u00de The major failure of the Hartree approximation is that in the formal derivation (by using the variational principle), we started with Eq. (7.7) in which the basic approximation is that the full N-electron wave function \u03a8 is a product of the one-electron levels. However, this simple assumption is not compatible with the Pauli principle according to which the N-electron wave function is antisymmetric; that is, the sign", " of \u03a8 changes when two of its arguments are interchanged, i.e., \u03a8 \u00f0 r1s1, \u2026, risi, \u2026, rlsl, \u2026, rNsN = \u2212\u03a8 \u00f0 \u00de r1s1, \u2026, rlsl, \u2026, risi, \u2026, rNsN : \u00de (7.16) Eq. (7.7) does not satisfy the constraints imposed by Eq. (7.16) unless \u03a8 = 0 identically. In addition to the noncompliance of the Pauli principle, the other failure of the Hartree approximation is that it does not include the well-known effects such as \u201cexchange,\u201d \u201ccorrelation,\u201d and \u201cscreening.\u201d In what follows, we will discuss \u201cexchange\u201d by deriving the Hartree\u2013Fock equations and \u201cscreening\u201d by discussing \u201cThomas\u2013Fermi theory\u201d and \u201cLindhard theory,\u201d but we will ignore \u201ccorrelation\u201d because it is much harder to discuss and is in the realm of \u201cmany-body problems,\u201d which requires the use of \u201cfield theory,\u201d a subject beyond the scope of this book. 7.3 Hartree\u2013Fock Approximation 203 7.3 HARTREE\u2013FOCK APPROXIMATION 7.3.1 General Formulation Fock and Slater showed that the simplest way to ensure that the N-electron wave function obeys the Pauli principle is to construct a Slater determinant of orthonormal one-electron wave functions that is antisymmetric: \u03a8 r1s1, r2s2, \u2026, rNsN \u00f0 \u00de = 1 N!p \ufb03\ufb03\ufb03\ufb03\ufb03 \u03c81\u00f0 \u00de ::::::::::::::::::: \u03c81\u00f0 \u00de \u03c8N\u00f0 \u00de :::::::::::::::::: \u03c8N\u00f0 \u00de r2s2... r2s2 \u03c81\u00f0 r1s1... r1s1 \u03c8N\u00f0 & & & & & & & & & & \u00de rNsN... rNsN (7.17 The Slater determinant can be rewritten in the alternate form, r1s1, r2s2, \u2026, rNsN \u03a8 \u00f0 = \u00de 1 N!p \u2211 n \u00f0 \u22121 \u00de n \u03c8 n1\u00f0 r1s1 \u00de \u03c8 n2\u00f0 r2s2 \u00de \u2026 \u03c8 nN \u00f0 rNsN \u00de, (7.18) where the sum is over all permutations n of 1\u2026 N. Eq. (7.18) can be rewritten in the alternate form = 1 N!p r1s1 \u2026 rNsN n \u220f j \u03c8 nj\u00f0 \u2211 n \u00f0 \u22121 \u00de rjsj : \u00de \u03a8 \ufb03\ufb03\ufb03\ufb03\ufb03 \u00f0 \u00de (7.19) Here, if the Hamiltonian does not involve the spin explicitly, \u03c8 i\u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03 rjsj \u00de has the simple form sj rjsj \u03c8 i\u00f0 = \u03d5i\u00f0 = \u03b4sj,1 for \u201cspin-up\u201d functions and \u03c7i\u00f0 sj where \u03c7i\u00f0 the wave function It can be shown, by using the Slater determinant in Eq. (7.1) and a variational technique, very similar to that used for the \u03a8 Hartree approximation but involving much more tedious algebra, that the Hartree\u2013Fock equation can be rewritten as (Problem 7.1) \u00de = \u03b4sj,\u22121 for \u201cspin-down\u201d functions. r1s1, r2s2, \u2026, rNsN (Eq. 7.19) (7.20) \u03c7i\u00f0 for rj sj \u00de \u00de \u00f0 \u00de \u00de \u00de, \u2212\u01272 2m \u22072\u03d5i\u00f0 r \u00de \u2212 ze2\u2211 l 1 r \u2212 Rl j r \u03d5i\u00f0 \u00de + e2\u2211 \" j Z j dr\u2032 \u03d5j\u00f0 r\u2032 j \u00dej 2 dr\u2032 \u2212\u2211 j Z e2 r \u2212 r\u2032 j j \u03d5j# r\u2032 r\u2032 \u03d5i\u00f0 \u00de \u00f0 1 r \u2212 r\u2032 r \u03d5i\u00f0 \u00de # j \u03d5j\u00f0 \u00de r \u00de j \u03b4\u03c7i, \u03c7j = \u03b5i\u03d5i\u00f0 r : \u00de Eq. (7.21) can also be", " expressed as \u2212 \u01272 2m \u22072\u03d5i\u00f0 r \u00de + Uion r \u03d5i\u00f0 \u00de \u00f0 r \u00de + Uel r \u03d5i\u00f0 \u00de \u00f0 r \u00de + Uex r \u00f0 \u00de \u03d5i\u00f0 r \u00de r = \u03b5i\u03d5i\u00f0 : \u00de (7.21) (7.22) The Hartree\u2013Fock equation derived in Eq. (7.21) differs from the Hartree equation derived in Eq. (7.15) in the sense that there is an additional term (the last term) on the left side of the equation known as the exchange term. It can be interpreted by stating that particles 1 and 2 flip places during the course of interaction and the negative sign in the exchange integral is due to the antisymmetry 204 CHAPTER 7 Electron\u2013Electron Interaction of the wave function. The exchange term is in fact an integral operator of the type Uex r \u03d5 r \u00de \u00f0 \u00de \u00f0 = dr\u2032. Thus, the Hartree\u2013Fock equations are a complicated set of nonlinear equations U r, r\u2032 \u00f0 \u03d5 r\u2032 \u00f0 \u00de \u00de Z that can only be solved numerically. 7.3.2 Hartree\u2013Fock Theory for Jellium The only case in which the Hartree\u2013Fock equations can be solved exactly is the jellium model where the electrons in a metal are considered to be a set of free electrons of which the solutions are the familiar plane waves. In the jellium model, the positively charged ions are represented by a uniform distribution of positive charge with the same density as the electronic charge. If there are N electrons in a volume V, the wave function of the free electrons is the plane waves The kinetic energy term is r \u03d5i\u00f0 \u00de =.r eiki Vp : \ufb03\ufb03\ufb03\ufb03 \u2212 \u01272 2m \u22072 1 Vp!.r eiki = \" 1 Vp 2 \u01272ki 2m eiki.r = 2 \u01272ki 2m \u03d5i\u00f0 r : \u00de (7.23) (7.24) \ufb03\ufb03\ufb03\ufb03 The potential due to the interaction of the ions with the electrons in Eq. (7.22) in the jellium model can be expressed from Eq. (7.21) as \ufb03\ufb03\ufb03\ufb03 Uion = \u2212 r \u00de \u00f0 N V dr\u2032 Z j e2 r \u2212 r\u2032 : j (7.25) The potential due to the Coulomb interaction in Eq. (7.22) in the jellium model can be expressed from Eq. (7.21) as For plane waves, From Eqs. (7.26) and (7.27), Uel = r \u00de \u00f0 N dr\u2032 \u2211 j = 1 e2 r \u03d5j\u00f0 \u00dej j r \u2212 r \u03d5j\u00f0 \u00dej. N V Uel r \u00de \u00f0 = N V dr\u2032 Z e2 r \u2212 r\u2032 j. j From Eqs. (7.25) and (7.28), we obtain Uion + Uel = 0: (7.26) (7.27) (7.28) (7.29) 7.3 Hartree\u2013Fock Approximation 205 We note that this type of cancellation is true only for the jellium model. We will now discuss the exchange term that arises due to the antisymmetry of the wave function and distinguishes the Hartree approximation from the Hartree\u2013Fock approximation. The exchange term can be written as Uex\u03d5i\u00f0 r \u00de N = \u2212 \u2211 j =1 #Z dr\u2032 e2 r \u2212 r\u2032 j j \u03d5#j \u00f0 r\u2032 \u00de r\u2032 \u03d5i\u00f0 \u00de r \u03d5j\u00f0 \u00de \u03b4\u03c7i, \u03c7j: $ From Eqs. (7.23) and (7.30), we obtain Uex\u03d5i\u00f0 r \u00de = \u2212 N e2 V 3/2 \u2211 j = 1 #Z ei \u00f0 dr\u2032 ki\u2212kj.r\u2032eikj \u00de r \u2212 r\u2032 j j.r $ \u03b4\u03c7i, \u03c7j : Changing the variable of integration r\u2032\u2032 = r\u2032 \u2212 r, we can rewrite Eq. (7.31) as Uex\u03d5i\u00f0 r \u00de = \u2212 Using the relation N e2 V 3/2 \u2211 j = 1#Z dr\u2032\u2032.r\u2032\u2032 \u00de ei \u00f0 ki\u2212kj r\u2032\u2032 $ eiki.r \u03b4\u03c7i,", " \u03c7j: and the Fourier transformation Eq. (7.32) can be rewritten as r \u03d5i\u00f0 \u00de = eik.ri, 1 Vp dr\u2032\u2032.r\u2032\u2032 \u00de ei \u00f0 ki\u2212kj r\u2032\u2032 Z \ufb03\ufb03\ufb03\ufb03 = 4\u03c0 ki \u2212 kj 2, j j Uex\u03d5i\u00f0 r \u00de = \u2212e2\u03d5i\u00f0 r \u00de N \u2211 j = 1 1 V 4\u03c0 ki \u2212 kj j 2 \u03b4\u03c7i, \u03c7j: j (7.30) (7.31) (7.32) (7.33) (7.34) (7.35) Before we proceed further, it is appropriate to mention that there is a divergence when ki \u2192 kj. This divergence arises due to the unphysical assumption in the Hartree\u2013Fock approximation, which decays very slowly at long treats only two electrons at a time, that the Coulomb interaction distances. In practice, the effective interaction between two electrons falls off much more rapidly due to the effect of screening, i.e., where the rest of the electrons play a role in reducing this interaction. We will treat the effect of screening in a subsequent section. e2 r \u2212 r\u2032 j j The sum over kj can be converted to an integration over k space as outlined in Chapter 3: \u2211 k F k \u00de \u00f0 = V 8\u03c03 Z dk F k. \u00de \u00f0 (7.36) 206 CHAPTER 7 Electron\u2013Electron Interaction The density of states is usually multiplied by a factor of 2 because each k state can have two elecreduces the density of states by trons of opposite spin. However, the Kronecker delta function \u03b4\u03c7i, \u03c7 a factor of 1 j 2. Thus, Eq. (7.35) can be rewritten as kF Uex\u03d5i\u00f0 r \u00de = \u2212e2\u03d5i\u00f0 r \u00de dk 8\u03c03 \u00f0 Z 4\u03c0 k2 + k2 i \u2212 2kki cos \u03b8 : \u00de (7.37) Because dk = k2dk sin \u03b8 d\u03b8 d\u03d5, integrating Eq. (7.37) over d\u03d5 yields a factor of 2\u03c0. Then integrating over d d\u03b8 and subsequently over dk (Problem 7.2), we can easily show that cos \u03b8 \u00f0 \u00de Uex\u03d5i\u00f0 r \u00de = \u2212e2\u03d5i\u00f0 r \u00de 1 2\u03c0ki \u00f0 # F \u2212 k2 k2 i \u00de ln kF + ki kF\u2212ki! \" + 2kikF : $ We define a Lindhard dielectric function as and express Eq. (7.38) as \u2212 x2 4x \u00de ln 1 + x 1 \u2212 x & & & & & & Uex\u03d5i\u00f0 r \u00de = \u2212 2e2 \u03c0 kFF ki kF! \" \u03d5i\u00f0 r : \u00de From Eqs. (7.22), (7.24), (7.29), and (7.40), we obtain \u03b5i k \u00de \u00f0 = \u01272k2 i 2m \u2212 2e2 \u03c0 kFF ki : kF! \" (7.38) (7.39) (7.40) (7.41) We note that the group velocity of the electrons at the Fermi surface, 1, is infinite, an \u0127 unphysical result due to the fact that screening has been neglected in deriving the Hartree\u2013Fock results. \u2202\u03b5 \u2202ki \u2019 ( j ki = kF The total energy of the N-electron system is obtained by summing over \u03b5i, multiplying the first term in Eq. (7.41) by 2 (because each k has two spin levels), and then dividing the second term by 2 (to avoid counting each electron pair twice while summing the interaction energy of an electron with all electrons). We obtain k \u00f0 \u00de E = 2 \u2211 k\u2264kF \u01272k2 2m \u2212 e2kF \u03c0 k2 F \u2212 k2 2kkF ln 1 + \u2211 k\u2264kF # kF + k kF \u2212 k : $ (7.42) & & & & & & & & The first term was evaluated in Chapter 1, and the second term is evaluated in Problem 7.3. We obtain E = N \u03b5F \u2212 3 5 # 3 4 e2kF \u03c0 : $ (7.43) 7.4 Effect of Screening 207 Eq. (7.43) can also be rewritten as (Problem 7.4) E N = = 3", " 5 \u00f0 e2 2a0 # 2:21 rs/a0 \" \u00f0 \u00de kFa0 2 \u2212 \u00de 3 2\u03c0 \u00f0 kFa0 \u00de $ 2 \u2212 0:916 rs/a0 \u00f0 Ry, # \u00de (7.44) where a0 is the Bohr radius and rs is the radius of the average free space (sphere) for each electron (Chapter 1). It may be noted that the second term in Eq. (7.44) is comparable to the first term for simple metals, and hence, the electron\u2013electron interaction term cannot be ignored in the theory of metals. However, as we indicated earlier, there is a logarithmic singularity in the expression for energy when k = kF. The singularity arises due to the fact that we have ignored the presence of other electrons when considering the effect of electron\u2013electron interaction in the Hartree\u2013Fock approximation. One has to consider the effect of screening to eliminate this divergence. There are two theories to include the effect of screening. We will first discuss the general effect of screening and then discuss both the Thomas\u2013Fermi and Lindhard theories of screening. 7.4 EFFECT OF SCREENING 7.4.1 General Formulation It is shown in Problem 7.5 that the Fourier transform of the external potential due to an external positive charge (for example, substituting a Zn ion for a Cu ion), \u03d5e, is related to the Fourier \u00f0 transform of the total potential (external plus induced charge), \u03d5, by the relation where \u2208 is the wave-vector-dependent dielectric constant. This type of relation is normally used in \u00f0 dielectric materials where the wave vector dependence is not considered because the fields are uniform. r \u00f0 is linearly related to the total potential \u03d5 The induced charge density \u03c1i, \u00de is the potential due to the extra positively charged r \u00f0 is the potential due to the cloud of screening electrons provided \u03d5 particle introduced in the metal, and \u03d5i induced by it. In such a case, one can write Fourier transforms as is sufficiently weak. Here, \u03d5e = \u03d5e + \u03d5i 7.45) q \u00de \u00f0 \u03d5e \u2208 q \u00f0 q \u00f0 : \u00de q \u00f0, the dielectric constant. The Poisson\u2019s equa- (7.46) We will now derive a relation between \u03c7 \u00f0 tions for the particle\u2019s charge density \u03c1e r \u00de \u00f0 \u2212\u22072\u03d5e the Fourier transform of which is \u03c1i q \u03d5 = \u03c7 q \u00de \u00f0 and \u2208 \u00de \u00f0 q q \u00de \u00f0 can be written as \u00de r \u00de \u00f0 = 4\u03c0\u03c1e r, \u00de \u00f0 q2\u03d5e q \u00de \u00f0 = 4\u03c0\u03c1e q \u00f0 : \u00de (7.47) (7.48) 208 CHAPTER 7 Electron\u2013Electron Interaction If the total charge density \u03c1 r \u00de \u00f0 = \u03c1e r \u00de \u00f0 the Fourier transform of which is + \u03c1i r, the Poisson\u2019s equation is \u00de \u00f0 \u2212\u22072\u03d5 r \u00de \u00f0 = 4\u03c0\u03c1 r, \u00de \u00f0 \u00f0 From Eqs. (7.46), (7.48), and (7.50), we obtain q2\u03d5 q = 4\u03c0\u03c1! \u03d5e q \u00f0 \u00de 4\u03c0 q2 \u03c7 \u00f0 1 \u2212 : q \u00de \" Comparing Eqs. (7.45) and (7.51), we derive the relation between \u2208 q \u00f0 \u00de and \u2212 4\u03c0 q2 \u03c7 q \u00f0 : \u00de (7.49) (7.50) (7.51) (7.52) 7.4.2 Thomas\u2013Fermi Approximation Thomas and Fermi argued that if the total local potential \u03d5 electron, \u03b5 cal approximation, we obtain k \u00f0 \u00de is slowly varying, the energy of an, will be modified from the free electron value. Treating this modification in a classi- r \u00f0 \u00de The induced charge density is = \u01272k2 2m k \u03b5 \u00f0 \u00de \u2212 e\u03d5 : r \u00f0 \u00de (7.53) (7.54), and en0 is the charge (7.55) (7.56) where \u2212en density of the positive background \u00de is the charge density when there is a local potential. We can rewrite Eq. (7.54) in the alternate form \u03c1i r = \u2212e n r \u2212 n0, % \u03c1i where \u00f0 r \u00f0", " \u00de = \u2212e \u00de n0 \u00f0 \u00de \u03bc + e\u03d5 r \u00f0 \u00de\u00de \u2212 n0, \u03bc \u00f0 \u00de% \u00f0 \u00bd n0 \u03bc \u00de \u00f0 = 1 4\u03c03 Z 1 \u01272k2/2m dk \u03b2 e\u00bd \u00f0\u00f0. \u2212\u03bc \u00de \u00de% + 1 Because \u03d5 terms cancel, and Eq. (7.55) can be rewritten in the alternate form is small, we make a Taylor expansion of the first term in Eq. (7.55). The leading r \u00de \u00f0 \u03c1i r \u00de \u00f0 \u2248 \u2212e2 \u2202n0 \u2202\u03bc \u03d5 : r \u00f0 \u00de (7.57) 7.4 Effect of Screening 209 We make a Fourier transformation of \u03c1i r and \u03d5 r, and from Eqs. (7.46) and (7.57), we obtain \u00f0 \u00de \u03c7TF\u00f0 q \u00de \u00f0 \u00de = \u2212e2 \u2202n0 \u2202\u03bc : From Eqs. (7.52) and (7.58), we obtain \u2208TF q \u00de \u00f0 = 1 + 4\u03c0e2 q2 \u2202n0 \u2202\u03bc, (7.58) (7.59) where \u2208TF ing factor \u00f0 q \u00de is the dielectric constant in the Thomas\u2013Fermi approximation. If we define a screen- Eq. (7.60) can be rewritten as TF = 4\u03c0e2 \u2202n0 \u03bb2 \u2202\u03bc, \u2208TF q \u00f0 \u00de = 1 + \u03bb2 TF q2 : (7.60) (7.61) As an example, if there is a point charge Q at r, it can be shown that (Problem 7.6) = \u03d5e\u00f0 r This is known as the screened Coulomb potential because there is an exponential damping factor \u03bbTF. \u03d5 r \u00de \u00de \u00f0 (7.62) e\u2212\u03bbTFr: 7.4.3 Lindhard Theory of Screening In a quantum mechanical treatment of Eq. (7.53), the one-electron Schrodinger equation can be written as \u2212 \u01272 2m \u22072\u03c8 k\u00f0 r \u00de \u2212 e\u03d5 r \u03c8 k\u00f0 \u00de \u00f0 r \u00de = \u03b5k\u03c8 k\u00f0 r \u00de, (7.63) where \u2212e\u03d5 function is a plane wave that we denote by | k >. Using first-order perturbation theory, we obtain is the potential energy. In the absence of the potential, the one-electron wave r \u00f0 \u00de r \u03c8 k\u00f0 \u00de = k > + \u2211 q j bk+q j k + q >, where and bk + q = < k + q \u03b50 k \u00f0 \u00de k > \u2212 e\u03d5 j \u2212 \u03b50 r \u00dej \u00f0 k + q \u00f0 \u00de k > = eik.r: j (7.64) (7.65) (7.66) 210 CHAPTER 7 Electron\u2013Electron Interaction The charge density is \u03c1 r \u00f0 \u00de = \u2212e\u2211 k f 0 k \u00f0 \u00dej r \u03c8 k\u00f0 \u00dej 2 = \u03c10 r \u00de \u00f0 + \u03c1i r, \u00de \u00f0 where f 0 k \u00f0 \u00de is the equilibrium Fermi distribution. We can write the charge density as \u03c1 r \u00de \u00f0 = \u2212e\u2211 k \u00bd f 0 k \u00f0 \u00dej r \u03c8 k\u00f0 \u00dej 2 % = \u2212e\u2212ik.r + \u2211 b#k+qe\u2212i \u00f0 q k+q.r \u00de eik.r + \u2211 q bk+qei k+q \u00f0 \u00de.r. # * *) Eq. (7.69) can be approximated as \u03c1 r \u00de \u00f0 \u2248 \u2212e\u2211 k f0 k \u00f0 \u00de 1 + \u2211 q # bk+qeiq.r + \u2211 q b#k+qe\u2212iq.r $ or \u2248 \u03c10 e2\u03d5 q \u00f0 ) k + q f 0 \u00de\u00bd eiq.r \u00de%, * (7.67) (7.68) (7.69) (7.70) (7.71) where we have written k for k \u2212 q, k + q for k, and eiq.r for e\u2212iq.r as the labels for the second term. Because \u03c1 r \u00f0 \u00de = \u03c10 r \u00de \u00f0 + \u03c1i \ufffd", "\ufffdi r \u00de \u00f0 = \u2211 q \u03c1i q \u00de \u00f0 r, \u00f0 \u00de eiq.r, from Eqs. (7.71) through (7.73), we obtain \u03c1i q \u00de \u00f0 = e2 From Eqs. (7.46) and (7.74), we obtain \u2212 \u03c7L\u00f0 \u00de = e27.72) (7.73) (7.74) (7.75) where \u03c7L\u00f0 obtain the expression for the dielectric constant for static screening in the Lindhard theory: is the expression derived from Lindhard theory. From Eqs. (7.52) and (7.75), we q \u00de \u2208L q \u00f0 \u00de = 1 \u2212 4\u03c0e2 q2 \u2211 7.76) 7.4 Effect of Screening 211 We first consider the dielectric constant when q \u2192 0: We can write and q. \u2207k\u03b5 q. \u2202f 0 k \u00f0 \u2202\u03b5 \u00de \u00de \u2207k\u03b5 k : \u00de \u00f0 From Eqs. (7.76) through (7.78), we obtain \u2208L q \u00f0 \u00de \u2248 1 \u2212 4\u03c0e2 q2 \u2211 k \u2202f 0 k \u00f0 \u2202\u03b5, \u00de which can be rewritten in the alternate form, \u2208L q \u00de \u00f0 \u2248 1 + 4\u03c0e2 q2 \u2212 Z! \u2202f 0 k \u00f0 \u2202\u03b5 d\u03b5, \u03b5 \u00f0 \u00de \u00de g \" where g \u03b5 \u00de \u00f0 is the density of states. Because, at very low temperatures, \u2202f 0 k \u00f0 \u2202\u03b5 \u00de \" \u2212! \u03b5 \u2212 \u03b5F \u2248 \u03b4 \u00f0, \u00de Eq. (7.80) can be rewritten in the alternate form \u2208L q \u00de \u00f0 \u2248 1 + 4\u03c02 q2 g \u00f0 \u03b5F : \u00de (7.77) (7.78) (7.79) (7.80) (7.81) (7.82) From Eq. (7.82), we obtain \u2208 \u2192 \u221e as q \u2192 0: Thus, an external field of long wavelength is almost entirely screened due to the flow of electrons. We can show that the same results are obtained through a simpler method by using the Thomas\u2013Fermi approximation. We can also write and \u01272 \u00f0 k. q/m \u00de \u2211 k F k \u00de \u00f0 = dk 4\u03c03 F : k \u00de \u00f0 Z We note that Eq. (7.84) is different from Eq. (7.34) because we have considered V = 1 and multiplied the density of states by a factor of 2 to be able to accommodate two electrons of opposite spin in each k state. We obtain from Eqs. (7.71) through (7.73) and (7.83) through (7.84), \u03c1i q \u00de \u00f0 = \u2212e2 Z dk 4\u03c03 f0 \u00f0 k \u00de \u01272 k + q \u2212 f0 \u00f0 k. q/7.85) (7.83) (7.84) 212 CHAPTER 7 Electron\u2013Electron Interaction Comparing with the definition of \u03c1i, in Eq. (7.46), we obtain \u00de q \u03c7L\u00f0 \u00de = \u2212e2 Z dk 4\u03c03 f0 k \u00de \u00f0 \u01272 k + q /m \u2212 f0 \u00f0 k. q \u00de \u00f0 \u00de : We note that the equilibrium Fermi function for free electrons is f0 k \u00de \u00f0 = 1 \u01272k2/2m\u2212\u03bc, \u00de + 1 e\u03b2 \u00f0 f0 k \u00de \u00f0 \u2212 f0 \u00f0 k + q \u2248 \u00de \u01272 m k. q \u2202 \u2202\u03bc f0 k + O \u00f0 \u00de \u00f0 q2 : \u00de From Eqs. (7.86) and (7.88), we obtain q \u03c7L\u00f0 \u00de \u2248 \u2212e2 \u2202n0 \u2202\u03bc = \u03c7TF\u00f0 q, \u00de which is identical to the Thomas\u2013Fermi result derived in Eq. (7.58). (7.86) (7.87) (7.88) (7.89) However, the general result for \u03c7L\u00f0 (7.86) at T = 0. It can be shown that (Problem 7.7) q \u00de in Lindhard theory can be obtained by integrating Eq. q \u03c7L\u00f0 \u00de = \u2212e2 mkF \u01272\u03c02! \" #", " 1 2 + 1 \u2212 x2 4x where x = q 2kF. We note that when T << TF, ln n0 \u2202\u03bc = g = \u03b5F \u00f0 \u00de mkF \u01272\u03c02 : From Eqs. (7.90) through (7.91), we obtain (7.90) (7.91) = \u03c7TF\u00f0 We have shown that the function in the square bracket, which is 1 at x = 0 and is equal to the function F(x) appearing in the Hartree\u2013Fock energy, is the Lindhard correction to the Thomas\u2013Fermi result. Thus, the dielectric constant \u2208L = 1 \u2212 4\u03c0\u03c7L/q2 is not analytic at q = 2kF. In fact, the static dielectric constant in the Lindhard theory can be written as (Problem 7.8) (7.92) \u03c7L\u00f0 L q \u00de \u00f0 = 1 + 4\u03c0e2 q2 mkF \u01272\u03c02 1 2 # + 4k2 F \u2212 q2 8kFq ln 2kF + q 2kF \u2212 q : $ & & & & & & & & It can also be shown that at large distances and at T = 0, the screened potential \u03d5 charge is of the form \u03d5 r \u00f0 \u00de& 1 r3 cos 2kFr: (7.93) of a point r \u00de \u00f0 (7.94) 7.4 Effect of Screening 213 If we write where \u03bbL is the screening parameter, we obtain \u2208L q \u00de \u00f0 = 1 + \u03bb2 L q2, (7.95) L = 4\u03c0e2 mkF \u03bb2 \u01272\u03c02 1 2 # + 4k2 F \u2212 q2 8kFq ln 2kF + q 2kF \u2212 q & & & L as a function of q is shown in Figure 7.1. & & $ & & & The variation of \u03bb2 : (7.96) We note from Eqs. (7.60) and (7.96) that as q \u2192 0, \u03bbL \u2192 \u03bbTF. The effective screening length 1/\u03bbL increases as q increases, and it becomes increasingly difficult to screen out the potentials of short wavelength. When q = qc = 2kF, there is a logarithmic singularity in the expression for \u03bb2 k + q in Eq. (7.76). If we consider the values of k, where either L. This singularity arises due to the term f 0 occurring in the summation over k in the expression for k + q > is \u2208L empty or vice versa, these lie in two regions covering the surface of the Fermi sphere. Figure 7.2 shows these regions of the Fermi sphere. k > is occupied and \u2212 \u03bb2 0 2kF q FIGURE 7.1 Variation of Lindhard screening parameter with q. kF q (a) FIGURE 7.2 q (b) q (c) Contribution to the dielectric constant from different regions of the Fermi sphere. 214 CHAPTER 7 Electron\u2013Electron Interaction These regions gradually expand as q increases. Thus, the sum over k increases until a value of q = qc = 2kF is reached and the whole Fermi sphere contributes to the sum. Because there are no additional terms in the sum over k, the functional form of the sum changes. However, each term remains a continuous function of q. The singularity is not serious because the contribution from the last few points on the sphere before reaching q = qc = 2kF is small. However, at q = 2kF, \u2202\u2208L/\u2202q \u2192 \u221e. If we consider the effect of an extra positive point charge in the jellium model, the \u201cexternal field\u201d is and the Fourier transform in a three-dimensional box is = e2/r, V r \u00de \u00f0 The \u201cscreened\u201d potential is and = 4\u03c0e2/q2 /\u2208\u2212iq.rdq: \u00de q \u00f0 From Eqs. (7.93) and (7.98) through (7.100), we obtain U r \u00de \u00f0 = 4\u03c0e2 q2 + Z ) 4me2kF \u01272\u03c0 1 2 # + 4k2 F \u2212 q2 8kFq ln (7.97) (7.98) (7.99) (7.100) \u22121 e\u2212iq.rdq: (7.101) 2kF + q 2kF \u2212 q & & & & & * $ & & & Due to the singularity at q = 2k", "F, there will be a special contribution to U that will contain oscillations of wave number 2kF. These oscillations are known as Friedel oscillations or Ruderman\u2013 Kittel oscillations depending on the context. We will now derive the Friedel oscillations using a much simpler approach. r \u00f0 \u00de 7.5 FRIEDEL SUM RULE AND OSCILLATIONS Friedel derived an equivalent formula for the oscillations by considering the effect of a spherically (impurity) placed in the electron gas. First, he derived a sum rule for the symmetric potential U valence difference between the impurity and the solvent metal. Then he derived an expression for the oscillating charge density associated with the singularity in Eq. (7.101). It is well known in quantum mechanics14 that when an incident plane wave (for convenience, we consider it moving in the z direction) is scattered by a spherically symmetric potential (see Figure 7.3), the wave function at a distance far from the scattering target can be expressed as r \u00de \u00f0 r \u2192 \u221e \u00f0 \u00de r, \u03b8 \u03c8 k\u00f0 \u00de = eikz + eikr : f \u03b8 \u00f0 \u00de r (7.102) 7.5 Friedel Sum Rule and Oscillations 215 FIGURE 7.3 Plane wave is incident in the z direction on spherically symmetric potential at the center. The scattered outgoing wave is spherical. It has been shown in Liboff\u2014Eqs. (14.11) through (14.13)\u2014that at large distances, r, \u03b8 \u03c8 k\u00f0 \u00de \u221e = \u2211 l = 0 il \u00f0 2l + 1 ei\u03b4l 1 kr \u00de kr \u2212 sin \u00f0 l\u03c0 2 + \u03b4l Pl \u00f0 \u00de cos \u03b8 : \u00de Here, Pl cos \u03b8 \u00de \u00f0 is a Legendre polynomial, and \u03b4l is a phase shift for the lth partial wave r, \u03b8 \u03c8 k,l\u00f0 \u00de = Ak,l 1 r sin kr \u2212 \u00f0 l\u03c0 2 + \u03b4l Pl \u00f0 \u00de cos \u03b8 : \u00de (7.103) (7.104) Friedel used the boundary condition that \u03c8 k = 0 at r = \u211c, as if the spherically symmetric potential was placed at the center of a sphere of radius \u211c. Thus, the values of k that are appropriate are k\u211c \u2212 l\u03c0 2 + \u03b4l = m\u03c0, (7.105) where m is an integer. We note that the phase shift \u03b4l would have been zero if the sphere had been empty. In that case, the \u201callowed\u201d values of k would be m + l 2 \u211c km =! \u03c0 \" : (7.106) There are many values of l for each value of m, and there are (2l + 1 values of \u03c8 for each l. However, the total number of states with \u03b5 \u2264 \u03b5F, i.e., states of which the wave vector is less than kF, \u00de 216 CHAPTER 7 Electron\u2013Electron Interaction \u03b4i(k)/ k \u03c0 / k\u2032 FIGURE 7.4 New \u201callowed\u201d values between k and k\u2032. would be the same as that for the cubic lattice considered in Chapter 1. When the impurity with the spherically symmetric potential is at the center of the sphere, \u03b4l \u2260 0 and the \u201callowed\u201d values of k do not become equal to km. They will be shifted by \u03b4l/\u211c. In addition, \u03b4l varies with k. This is shown schematically in Figure 7.4. Thus, between k and k\u2032, the set will have k = 0 \u00f0 /\u03c0 new \u201callowed\u201d values. Because \u00de = 0, counting a factor of 2 for spin states of an electron and the fact that for each l states, the total number of extra electrons needed to fill up to the Fermi wave \u2212 \u03b4l k\u2032 \u03b4l \u00de% k \u00f0 \u00f0 \u00bd \u03b4l \u00de there are (2l + 1 vector kF is \u00de \u03be = 2 \u03c0 \u2211 l \u00f0 2l + 1 kF \u03b4l \u00de \u00f0 : \u00de (7.107) Here, \u03be is the difference between the valence of the spherically symmetric impurity and the metal, which is the number of electrons necessary to neutralize the charge of the impurity. Eq. (7.107) is the Friedel sum rule, which was derived with two basic assumptions that the charge of a spherically symmetric impurity must be neutralized by extra electrons within", " a finite distance, and at large distances, kF is the same as in the crystal without any impurity. At large distances from the spherical potential at the center of the sphere \u211c, the change in the electron charge density associated with the phase-shifted waves in Eq. (7.104) is given by \u03b4\u03c1 = \u2212e\u03b4n, where \u03b4n is the change in electron density, \u03b4\u03c1 = \u2212e\u2211 l \u00f0 2l + 1 kF \u00de Z 0 h \u03b4l \u03c8 k,l\u00f0 & & & 2\u211c \u03c0 dk: 2 \u2212 \u03c8 k,l\u00f0 & & & \u00de & & & 0 \u00de 2 i & & & Here, we have used the expression from Eq. (7.106), km \u2212 km\u22121 = \u03c0 \u211c, (7.108) (7.109) and there are two spin states for each k. The wave function in Eq. (7.104) has been normalized in a sphere of radius \u211c in which the terms going to zero more rapidly at large values of r have been neglected, \u03c8 k,l\u00f0 j r, \u03b4l \u00dej Z 2 r2 dr sin \u03b8 d\u03b8 d\u03d5 = A2 l \u211c \u03c0 2\u03c0 0 Z Z 0 Z 0 dr sin2 kr \u2212 \u00f0 l\u03c0 2 + \u03b4l P2 l \u00f0 \u00de cos \u03b8 \u00de sin \u03b8 d\u03b8 d\u03d5: (7.110) 7.6 Frequency and Wave-Number-Dependent Dielectric Constant 217 x P2 l \u00f0 \u00de dx = 2 2l + 1 and sin2\u03c6 = 1 2 \u00f0 1 \u2212 cos 2\u03c6, neglecting the oscillatory terms, we obtain \u00de Because 1 Z \u22121 2l + 1 2\u03c0\u211c 1/2 1/2 : \u00de \u00de Al \u2248 \u00f0 \u00f0 From Eqs. (7.104) and (7.111), we obtain r, \u03b4l \u03c8 k,l\u00f0 \u00de 2l + 1 2\u03c0\u211c 1/2 \u00de 1/2 \u00de \u2248 \u00f0 \u00f0 1 r sin kr \u2212 \u00f0 l\u03c0 2 + \u03b4l : \u00de From Eqs. (7.108) and (7.112), we obtain \u03b4\u03c1 = \u2212 e \u03c02 \u2211 l \u00f0 2l + 1 kF \u00de Z 0 h sin2 kr \u2212 \u00f0 l\u03c0 2 + \u03b4l \u00de \u2212 sin2 kr \u2212 \u00f0 l\u03c0 2 \u00de 1 r2 dk: i Using the formula we obtain from Eq. (7.113) sin2 \u03b1 \u00de \u00f0 \u2212 sin2 \u03b2 \u00f0 \u00de = sin \u03b1 + \u03b2 sin \u00f0 \u00de \u03b1 \u2212 \u03b2, \u00de \u00f0 \u03b4\u03c1 = \u2212 e 2\u03c02 \u2211 l \u00f0 2l + 1 kf \u00de Z 0 2kr \u2212 l\u03c0 + \u03b4l sin \u00f0 sin \u03b4l \u00de 1 r3 d \u00f0 2kr : \u00de By integrating Eq. (7.115), we obtain \u03b4\u03c1 \u2248 \u2212e 2\u03c02 \u2211 l \u00f0 2l + 1 \u22121 \u00de \u00de\u00f0 l sin \u03b4l cos \u00f0 2kFr + \u03b4l r3 \u00de : (7.111) (7.112) (7.113) (7.114) (7.115) (7.116) Here, the terms going to zero more rapidly at r (the limit of the integral at k = 0) have been dropped. Eq. (7.116) gives the oscillating charge density that arises due to the singularity obtained earlier in Eq. (7.101). The oscillating charge is not a negligible effect because it varies as 1/r3. Further, the electrons are driven away by the spherically symmetric impurity in some regions because \u03b4\u03c1 becomes negative. 7.6 FREQUENCY AND WAVE-NUMBER-DEPENDENT DIELECTRIC CONSTANT We consider a homogeneous electron gas that has a number density n = N/v, where v is the volume is applied, fluctuations in the electron gas are r, t of the crystal. When an external perturbation \u03d5e\u00f0 induced such that \u00de n = n0 + \u03b4n: (7.117) 218 CHAPTER 7 Electron\u2013Electron Interaction, which essentially describes the screening The fluctuations give rise to an internal potential \u03d5i\u00f0 effect of the electron gas on an electron at r. These fluctuations are obtained by the Poisson equation r, t \u00de \u00de The total potential acting on the electron is \u22072\ufffd", "\ufffdi\u00f0 r, t = \u22124\u03c0e2\u03b4n r, t \u00f0 \u00de : r, t \u03d5 \u00f0 = \u03d5i\u00f0 \u00de r, t + \u03d5e\u00f0 \u00de r, t : \u00de (7.118) (7.119) The statistical mean values of any physical quantity G are determined through the statistical operator < ^G> = tr (Eq. 3.14) where ^\u03c1k is defined in Eq. (3.24) as ^\u03c1 ^G \u00f0 \u00de ^\u03c1k = Zk \u22121e\u2212\u03b2 k \u03b5 \u00f0 \u00f0 \u00de \u2212\u03bc \u00de, where Zk is the grand partition function defined in Eq. (3.23) as ^Zk = 1 + e\u2212\u03b2 \u00f0 \u03b5 \u00f0 k \u00de \u2212\u03bc \u00de, where the spin has been omitted in both Eqs. (7.120) and (7.121). It can be easily shown that the equilibrium state of the electron gas (index 0) is given by ^H0 j k > = \u03b5 k >, k \u00dej \u00f0 ^\u03c10 j k > = f0 k >, k \u00f0 \u00dej and k where f0 \u00de perturbation, \u00f0 and We have is the Fermi distribution function. In the presence of the time-dependent external ^H = ^H0 + ^\u03d5 r, t \u00f0 \u00de ^\u03c1 = ^\u03c10 + \u03b4^\u03c1 r, t : \u00de \u00f0 (7.124) (7.125) i\u0127\u03b4 _^\u03c1 j k > = ^H, ^\u03c1 k > = ^H0, \u03b4 ^\u03c1 k > + k >: (7.126) j j i k > in Eq. (7.126). We note that ^H0 commutes with ^\u03c10. i h j ^\u03d5, ^\u03c10 h h i ^\u03d5, \u03b4 ^\u03c1 \u00bd %j Here, we have neglected the term We can rewrite Eq. (7.126) as i\u0127 < k\u2032 \u03b4 _^\u03c1 j j k > = k\u2032 \u00de% \u03b4 ^\u03c1 j j k > \u2212 k\u2032 f0 \u00f0 \u00bd \u00de \u2212 f0 k \u00f0 \u00de% < k\u2032 k >: ^\u03d5 j j The qth Fourier component of \u03d5 is given by t \u03d5q\u00f0 \u00de = 1 v Z e\u2212iq.r \u03d5 dr, r, t \u00f0 \u00de (7.127) (7.128) (7.120) (7.121) (7.122) (7.123) 7.6 Frequency and Wave-Number-Dependent Dielectric Constant 219 where We can express q = k\u2032 \u2212 k: r, t \u03d5 \u00f0 \u00de = ZZ q, \u03c9 \u03d5 \u00f0 \u00de eiq.rei\u03c9te\u03b1t dq d\u03c9, (7.129) (7.130) where the term e\u03b1t has been multiplied so that the oscillation of frequency \u03c9 and wave vector k\u2032 \u2212 k grows slowly with a time constant \u03b1. In the final result, \u03b1 \u2192 0: Because \u03b4\u03c1 has the same q time dependence as \u03d5, from Eqs. (7.127) and (7.128), we obtain for each Fourier component (Problem 7.10) r \u2212 i\u0127 \u00f0 i\u03c9 + \u03b1 \u00de\u03c1 j j f0 \u00f0 \u00bd k + q \u2212 f0 k t \u03d5q\u00f0 \u00de% \u00de, \u00f0 \u00de where \u03d5e\u00f0 From Eqs. (7.118), (7.128), and (7.132), we obtain = \u03d5eq\u00f0 + \u03d5iq\u00f0 \u03d5q\u00f0 = \u00de \u00de \u00de t t t \u00bd q, \u03c9 \u00de q, \u03c9 + \u03d5i\u00f0 \u00de% i\u03c9t+\u03b1t e\u00f0 \u00de: q2\u03d5iq\u00f0 t \u00de = 4\u03c0e2 \u03b4nq: It can be easily shown that the particle concentration \u03b4n is given by We can also express from Eqs. (7.121) and (7.125) r0, t \u03b4n \u00f0 = \u2211 q \u00de r0, t \u00f0 \u00de eiq.r0 \u03b4nq. r0, t \u03b4n \u00f0 \u00de = tr \u03b4 \u00bd \u00f0 r \u2212 r0 \u03b4\u03c1 % \u00de < k \u03b4 \u00f0 j = \u2211 kk\u2032 1 v = \u2211 kk\u2032 r \u2212 r0 > \u00de.r0 < k\u2032 k\u2032 > < k\u2032 k > \u03b4\u03c1 j j j k > \ufffd", "\ufffd\u03c1 j j k\u2212k\u2032 ei \u00f0 \u00de From Eqs. (7.133) through (7.135), we obtain (taking v = 1) = 1 v \u2211 q eiq.r0 \u2211 k < k + q k >: \u03b4\u03c1 j j t \u03d5iq\u00f0 \u00de = 4\u03c0e2 q2 \u2211 k < k + q k >: \u03b4\u03c1 j j From Eqs. (7.131), (7.132), and (7.136), we obtain q, \u03c9 \u03d5i\u00f0 \u00de = \u03d5 \u00de q, \u03c9 \u00f0 4\u03c0e2 q2 \u2211 k q, \u03c9 \u2212 \u03d5e\u00f0 f0 \u2212 f0 k \u00de + \u0127\u03c9 \u2212 i\u0127\u03b1 \u00f0 \u00de q, \u03c9 \u03d5 \u00f0 : \u00de = (7.131) (7.132) (7.133) (7.134) (7.135) (7.136) (7.137) 220 CHAPTER 7 Electron\u2013Electron Interaction We obtain from Eq. (7.137), q, \u03c9 \u03d5e\u00f0 \u00de = \u03d5 \u00f0 q, \u03c9 1 \u2212 \u00de # 4\u03c0e2 q2 \u2211 k \u03b5 \u00f0 If we define the dielectric constant as f0 \u2212 f0 k \u00de + \u0127\u03c9 \u2212 i\u0127\u03b1 $ \u00de : (7.138) \u2208 \u00f0 q, \u03c9 = \u00de q, \u03c9 q, \u03c9 \u03d5e\u00f0 \u03d5 \u00f0 \u00de \u00de, (7.139) we obtain from Eqs. (7.138) and (7.139) q, \u03c9 \u2208 \u00f0 \u00de = 1 \u2212 4\u03c0e2 q2 \u2211 + \u0127\u03c9 \u2212 i\u0127\u03b1 k \u00de \u00f0 \u00f0 : (7.140) Eq. (7.140) is the Lindhard equation for the dielectric constant of the electron gas. We note that for, Eq. (7.140) becomes identical to Eq. (7.76) derived by the case of static screening (\u03c9 = 0, \u03b1 = 0 \u00de using time-independent perturbation theory. We can write Thus, we obtain where and Lim 1 z \u2212 i\u03b1 = P 1 z! \" + i\u03c0\u03b4 : z \u00de \u00f0 \u2208 \u00f0 q, \u03c9 = \u22081 \u00f0 \u00de q, \u03c9 \u00de + i\u22082 q, \u03c9, \u00de \u00f0 \u22081 q, \u03c9 \u00de \u00f0 = 1 \u2212 4\u03c0e2 q2 + \u0127\u03c9 # \u22082 q, \u03c9 \u00de \u00f0 = 4\u03c02e2 q2 \u2211 % \u00f0 \u2212 \u03b5 k \u00de \u00f0 \u2212 \u0127\u03c9 : \u00de (7.141) (7.142) (7.143) (7.144) is related to the absorption constant of the electron gas and pair exciThe imaginary part \u22082 tations are involved in this absorption. The conservation of energy indicated by the \u03b4\u2212 function indicates that in Eq. (7.143), the denominator always has q, \u03c9 \u00de \u00f0 \u0127\u03c9 > % \u00f0 \u00de (7.145) for each term. We divide the summation over k into two parts and introduce new summation indices k + q and \u2212k into the first and second part, respectively. We bring the two parts together and neglect all energy differences between states k + q and k. 7.6 Frequency and Wave-Number-Dependent Dielectric Constant 221 We obtain q, \u03c9 \u22081 \u00f0 \u00de = 1 \u2212 4\u03c0e2 q2 \u2211 + \u0127\u03c9 \u2212 \u2211 \u2212k \u2212k \u03b5 \u00f0 \u00de f 0 \u2212 \u03b5 \u2212k \u00f0 \u00de \u2212k + q \u00f0 \u00de, \u2212 \u0127\u03c9 * = 1 + 4\u03c0e2 q2 \u2211 k 2f 0 k \u00f0 \u00def \u03b5 \u2212 \u03b5 k \u00f0 \u00de \u01272\u03c92 \u00f0 k \u2212 q \u00deg : (7.146) (7.147) Here, we have rearranged the sum over k and neglected the energy differences in the denominator because \u0127\u03c9 \u226b in the plasma oscillation region. in powers of q, the first term vanishes. We obtain \u03b5 \u00bd Expanding % \u00de and \u2248 \u2212 \u01272q2 2m f 0 k \u00de \u00f0 \u2211 k = n: Substituting Eqs. (7.147) and (7.148) in Eq. (7.149), we obtain \u22081 \u03c9 \u00de \u00f0 = 1 \u2212", " 4\u03c0ne2 m\u03c92 = 1 \u2212 2 \u03c9p \u03c92, (7.148) (7.149) (7.150) where \u03c9p is the plasma frequency. Eq. (7.150) shows that when \u03c9 \u2192 \u03c9p, \u22081\u2192 0: However, from Eq. (7.138), q, \u03c9 \u03d5 \u00f0 = \u00de q, \u03c9 q, \u03c9 \u03d5e\u00f0 \u2208 \u00f0 \u00de \u00de \u2192\u221e: (7.151) An infinitesimal external field \u03d5e\u00f0 gives rise to an extremely large effective field. Thus, the system is self-exciting. \u03c9p is a natural mode of oscillation of the electron gas. It is called a plasma mode. q, \u03c9 \u00de We note that Eq. (7.140) is also known as Lindhard dielectric susceptibility. Thus, if \u03d5e\u00f0 is the actual external potential, the Fourier transformation is given by r, t \u00de r, t \u03d5e\u00f0 \u00de = ZZ q, \u03c9 \u03d5e\u00f0 \u00de eiq.rei\u03c9tdq d\u03c9: The effective potential U r, t \u00f0 \u00de seen by the electrons is given by r, t U \u00f0 \u00de = \u03d5e\u00f0 \u2208L \u00f0 q, \u03c9 \u00de q, \u03c9 \u00de ZZ eiq.rei\u03c9tdq d\u03c9: (7.152) (7.153) 222 CHAPTER 7 Electron\u2013Electron Interaction 7.7 MOTT TRANSITION The Mott transition is an example of a metal-insulator transition that has been actually observed in transition metal oxides. This involves discussion of the Hubbard model, which we will consider later. However, we will consider the basic argument for metal-insulator transition first proposed by Mott. He argued that if one has a collection of hydrogen or other monovalent atoms, it is unlikely that they will form a conductor even if they are far apart from each other. Individual hydrogen atoms are insulators because the electron is bound to the parent atom. If, through some process, an electron is taken out from one of the array of hydrogen (or any other group of monovalent) atoms, it will leave behind a positively charged ion. However, the electron will still be attracted to the positively charged ion and might even form a bound state. Thus, it would not be able to carry current. The solid would still be an insulator. There has to be a certain criterion when such types of insulators would become conductors (metals). When many electrons are excited, the electron gas would screen the electron\u2012ion interaction. The potential is of the form (because of screening of the ion) V r \u00f0 \u00de = \u2212 e2 r e\u2212\u03bbr: (7.154) If the density of ionized electrons is n, we have derived in Eq. (7.96) an expression for the screening parameter \u03bbL, which can be written for small q values as Because kF = 3\u03c02n 1/3, \u00de \u00f0 \u03bb2 \u2248 4e2mkF \u01272\u03c0 : \u03bb2 \u2248 1 3 : 4me2n \u01272 The radius of the ground state of a hydrogen atom is given by ao = \u01272 me2 : (7.155) (7.156) (7.157) There cannot be any bound state if 1/\u03bb < a0. The criterion that the ionized atom, of which the potential is expressed in Eq. (7.154), is capable of recapturing the electron that has been removed from it is From Eqs. (7.156) through (7.158), this condition can be expressed as n\u22121/3 > 4a0: \u03bb < a\u22121 0 : (7.158) (7.159) Thus, if the average spacing of the atoms is greater than 4 atomic units, the system would be an insulator. As the atoms come closer, the transition from insulator to metal would be quite sharp. This is known as a Mott transition. 7.8 Density Functional Theory 223 7.8 DENSITY FUNCTIONAL THEORY 7.8.1 General Formulation Hohenberg and Kohn hypothesized that the electron density n r has all the information about the many-electron wave function. In this case\u03a8 N \u2211 l =1 j \u03b4 \u00f0 r \u2212 rl \u00dej \u03a8 >: of a many-electron system at point is defined as r \u00de \u00f0 (7.160) r1, r2, \u2026, rN Here, \u03a8 Hartree and Hartree\u2013Fock approximations. Eq. (7.160) can be", " rewritten in the alternate form is the many-electron wave function defined earlier both in the context of the \u00f0 \u00de n r \u00de \u00f0 = N dr1dr2 \u2026 drN \u03a8# Z r1, r2, \u2026, rN r \u2212 r1 \u03b4 \u00f0 \u00de \u03a8 \u00f0 \u00de r1, r2, \u2026, rN : \u00de \u00f0 (7.161) \u00f0 \u00de According to Hohenberg and Kohn, the electronic density n r determines the external potentials U and the number of electrons that are inside these potentials. If the kinetic energy of the electrons is T, the ground-state energy is \u03b5, the electrons that obey Schrodinger\u2019s equation interact via the Coulomb potentials Uee, and the potential due to the ions is U, one can write an expression, for the functional for the ground-state energy is found, the ground-state energy density n % % \u00bd % n \u00bd + Uee : n \u00bd If the functional \u03b5 n \u00bd % is a constraint that the total number of electrons is (7.162) minimizes it. However, there r \u00de \u00f0 (7.163) (7.164) (7.165) Because U n % \u00bd depends only on the density, one can also write N = Z dr n : r \u00de \u00f0 If we write = U n % \u00bd Z dr n U r \u00de \u00f0 r : \u00de \u00f0 FHK n \u00bd % = T n \u00bd % + Uee, n \u00bd % it is a universal function for all systems of N particles. We obtain from Eqs. (7.162) and (7.165), We can also define a functional F density n of F = FHK n \u00bd, which is the minimum over all wave functions producing % (7.166 % \u2019 min\u03a8\u2192n < \u03a8 T + Uee j \u03a8 >: j (7.167) In theory, after the universal function F particle system through the potential U r \u00de \u00f0 n is found, any set of nuclei can be added to the many, which minimizes \u00bd. Then one has to find the function n r % \u00f0 \u00de 224 CHAPTER 7 Electron\u2013Electron Interaction can be obtained by making appropriate it to solve the Schrodinger equation. In practice, F approximations such that the results would agree with the experimental results, and hence, the form of F varies with the nature of the experiment. n n % \u00bd \u00bd % 7.8.2 Local Density Approximation Kohn and Sham introduced the local density approximation (LDA), in which they proposed to instead express the electron density as a function of a set of N single-electron wave functions \u03c8 i\u00f0 of a function of all material properties1 j 2. r \u03c8 i\u00f0 \u00dej The kinetic energy term of the energy functional is written as T n % \u00bd = \u2211 i \u01272 2m j 2. \u2207\u03c8 ij (7.168) (7.169) The rest of the quantities (including exchange) are obtained from the results of a homogeneous electron gas. We derived (the second term in Eq. 7.43) that and we derived in Chapter 1 that from which we obtain \u03b5ex = \u2212 3 4 N e2kF \u03c0, n = k3 F 3\u03c02, \u03b5ex n \u00bd % = \u2212 3e2 4 1/3 3 \u03c0 \u2019 ( Z n4/3 dr: r \u00de \u00f0 (7.170) (7.171) (7.172) From Eqs. (7.169) and (7.172) and using standard expressions for Uee n % \u00bd and U n \u00bd %, we obtain N = \u2211 i =1 \u01272 2m j n \u03b5 \u00bd % 2 + \u2207\u03c8 ij drn dr dr\u2032 e2n n r \u00f0 \u00f0 \u00de r \u2212 r\u2032 r\u2032 \u00de \u2212 j 3e2 4 1/3 3 \u03c0 \u2019 ( Z drn4/3 r \u00f0 : \u00de (7.173) j If we vary Eq. (7.173) with respect to \u03c8 #i (because we know the density as a functional of the wave function from Eq. 7.168), we obtain \u2212\u01272 2m \u22072\u03c8 i\u00f0 dr\u2032 e2n r\u2032 \u00de \u00f0 r \u2212 r\u2032 j j \u2212 e2 3 \u03c0 \u2019 n r \u00f0 1/3 \u00de ( * r \u03c8 i\u00f0 \u00de = \u03b5i\u03c8 i\u00f0 r \u00de : (7.174) 7.9 Fermi Liquid Theory 225 In general, if one wants to add the correlation terms (any corrections to the exchange", " terms because of the many-body nature of the N-electron system), one can rewrite Eq. (7.174) in the alternate form Here, \u03b5xc n \u00f0 \u00de \u2212\u01272 2m \u22072\u03c8 i\u00f0 dr\u2032 e2n r\u2032 \u00de \u00f0 r \u2212 r\u2032 j j + n \u00de \u2202\u03b5xc \u00f0 \u2202n * r \u03c8 i\u00f0 \u00de = \u03b5i\u03c8 i\u00f0 r \u00de : (7.175) is the exchange-correlation energy of the electron gas. The approximation of the form (Eq. 7.175) is known as the local density approximation (LDA). A variety of band calculations are based on the density functional theory in general and the LDA in particular. The success of these band calculations primarily depends on the use of appropriate exchange-correlation functions. In fact, the correlation functions are a consequence of many-body theory and can be calculated only by using field-theoretical methods. In band calculations, one can only make intelligent guesses of these functions depending on the available experimental results. However, a variety of first principles calculations of the exchange functionals are available to calculate the total energies of atoms. It has been shown that the recent calculations using these functionals yield more accurate results for the energy of atoms than the results obtained by using the Hartree\u2013Fock approximation. In fact, Kohn was awarded the Nobel prize in chemistry for his pioneering work on density functional theory. 7.9 FERMI LIQUID THEORY 7.9.1 Quasiparticles The Fermi liquid theory was originally developed by Landau to explain the properties of liquid 3He. However, it has also been applied to the theory of electron\u2013electron interactions in metals. In particular, the Fermi liquid theory explains the success of the independent electron approximation even though the electron\u2013electron interactions are significant. The Fermi liquid theory also indicates how the effect of electron\u2013electron interactions can be qualitatively taken into account in the study of the properties of metals in general and transport properties in particular. In Landau\u2019s argument, if one starts with a set of noninteracting electrons and gradually turns on the interactions between electrons, there would be two types of effects. The energy of the one-electron level would be modified, and it can be treated by the Hartree\u2013Fock and other types of approximations described earlier in this chapter. However, in the Hartree\u2013Fock approximation, the one-electron levels are stationary in spite of the interacting system. In contrast, when the interactions between the electrons are turned on, the electrons would be scattered in and out of the single-electron levels, which are no longer stationary. If the scattering is sufficiently low, one can introduce a relaxation time and try to solve the problem by using a relaxation-time mechanism used to treat transport theories. However, the electron\u2013electron relaxation time is usually much larger than those used in other transport theories. The basic idea of the Landau Fermi Liquid theory is to consider the excitations of the strongly interacting system instead of concentrating on the nature of the ground state. The scattering rate of the fermions is considerably reduced due to the Pauli exclusion principle. In addition, Landau termed the elementary excitations, which act like particles, as quasiparticles. These quasiparticles interact with each other, but not as strongly as the particles from which they are constructed. 226 CHAPTER 7 Electron\u2013Electron Interaction \u03b5F \u2192 \u03b51, k (a) \u03b5F \u03b54 \u03b53 \u03b52 (b) FIGURE 7.5 (a) An excited state \u03b51 > \u03b5F of wave vector k: (b) \u03b52 < \u03b5F, \u03b53 > \u03b5F, \u03b54 > \u03b5F: To highlight Landau\u2019s theory, we consider an N-electron state at T = 0 filling a Fermi sphere of radius kF and energy \u03b5F. In addition, there is a single excited electron (quasiparticle) in a level \u03b51 > \u03b5F. To be scattered, this electron can interact with an electron of energy \u03b52, but because only electron levels of energy less than \u03b5F are occupied, \u03b52 < \u03b5F. Both of these electrons can scatter only into energy states \u03b53 and \u03b54, which are unoccupied; i.e., these states are above the Fermi sphere. Thus, \u03b53 > \u03b5F and \u03b54 > \u03b5F. These are schematically shown in Figure 7.5. The law of conservation of energy requires that \u03b51 + \u03b52 = \u03b53 + \u03b54: (7.176) Because \u03b51 > \u03b5F at T =", " 0, some phase space is available for scattering, and \u03b52, \u03b53, and \u03b54 can vary near the Fermi surface. After within a shell of which the thickness is on the order of \u03b52 and \u03b53 are fixed within the shell, there cannot be any new choice for the value of \u03b54 because of Eq. (7.176). Thus, at T \u2192 0, the scattering rate is given by \u03b51 \u2212 \u03b5F \u03b51 \u2212 \u03b5F j j j j 1 \u03c4 j T= 0 \u2248 a1 \u00f0 \u03b51 \u2212 \u03b5F 2: \u00de (7.177) If we consider the scattering at a temperature T, there would be partially occupied levels of width 2. The scatterkBT around \u03b5F. Thus, there will be an additional scattering rate proportional to ing rate near the Fermi surface at temperature T is given by kBT \u00de \u00f0 1 \u03c4 \u2248 a1 \u00f0 \u03b51 \u2212 \u03b5F 2 + a2 kBT \u00f0 2: \u00de \u00de (7.178) These derivations are the same as that for an independent N-electron system. However, we will discuss Landau\u2019s formulation of quasiparticles and his arguments that lead to a one-to-one correspondence with the fermions in the N-electron system. We will first derive an approximate formula for the \u03b5F, the first term in Eq. (7.178) temperature-dependent part of the relaxation time in Eq. (7.178). If \u03b51 can be neglected compared to the second term; \u03c4 will essentially be proportional to 1/T 2. We further assume that the dependence of \u03c4 on the electron\u2013electron interaction is through the Fourier transform of the interaction potential, which can be estimated by the Thomas\u2013Fermi screened potential defined in Eq. (7.61). Because \u2202n0 \u2202\u03bc >1, according to Eq. (7.61), the Thomas\u2013Fermi screened potential is everywhere less than 4\u03c0e2 \u03bb2 TF. Thus, neglecting the first term in Eq. (7.178), we assume that & \u2019 ( 1 \u03c4 \u221d 4\u03c0e2 \u03bb2 TF!\u00f0 kBT 2: \u00de (7.179) 7.9 Fermi Liquid Theory 227 2: (7.180) From Eqs. (7.60), (7.91), and (7.179), we obtain 1 \u03c4 \u221d \u01272\u03c02 mkF! 2 kBT \u00f0 \u00de \" The proportionality constant \u2018a\u2019 in Eq. (7.180) can be written as a dimensionless number by multiplying Eq. (7.180) by m3/\u01277 such that 1 \u03c4 has the dimensions of inverse time, 1 \u03c4 = 2 \u00de : a \u00f0 kBT \u0127\u03b5F (7.181) It has been estimated that the range of the dimensionless number \u2018a\u2019 is between 1 and 100. The empirical formula for scattering in the presence of electron\u2013electron interaction, using the Thomas\u2013Fermi theory for screening, was derived essentially by turning on electron\u2013electron interaction on an N-particle independent electron system. Landau validated these arguments by using the concept of quasiparticles, according to which the low-lying states of the strongly interacting N-electron system evolve in a continuous way such that there is a one-to-one correspondence with the noninteracting electron system. In a noninteracting electron system, if n electrons of wave vectors k1, \u2026, kn above kF have been excited from the states k\u20321, \u2026, k\u2032n below kF, the energy of the excited state is E = Eg + \u03b5 k1 + \u2026 + \u03b5 kn \u2212 \u03b5 \u2212 \u2026 \u2212\u03b5 k\u2032n, (7.182) \u00f0 \u00f0 \u00f0 = \u01272k2 where Eg is the ground-state energy, and for free electrons, \u03b5 i /2m. The quasiparticles are the states of the interacting system where n quasiparticles have been excited out of levels with wave vectors k\u20321, \u2026, k\u2032n below kF and the same number of excited quasiparticles with wave vectors k1, \u2026, kn are present above kF. In this case, we have the relation \u00f0 \u00f0 \u00de \u00de \u00de \u00de k\u20321\u00de ki E = Eg + \u03b5 k1 \u00f0 \u00de + \u2026 + \u03b5 kn \u2212\u03b5 k\u20321\u00de \u00f0 \u00de \u00f0 \u2212 \u2026 \u2212\u03b5, k\u2032n \u00f0 \u00de (7.183) \u00de \u00de are very different from the free electron energies. In fact,", " the quasiparticle relation ki but \u03b5 \u00f0 versus ki is very difficult to determine because particles with the same spin have an interaction \u03b5 ki \u00f0 that differs from the interaction between particles with a different spin. However, there is indeed a one-to-one correspondence between the free electrons and quasiparticles. Landau showed, by using Green\u2019s function methods, that for all orders of perturbation theory, every interacting Fermi system is normal in the sense that the quasiparticle representation is valid. Landau\u2019s argument is very complex, and we will not discuss his theory any further here. The summary of Landau\u2019s model of quasiparticles is that there exists a set of wave functions that are in one-to-one correspondence with low-lying excited states of the noninteracting Fermi gas. They behave under translations like noninteracting particles with index k. Similarly, states that correspond to low-energy holes of the noninteracting electron gas can be constructed. Although these states are not true eigenfunctions, they decay very slowly near the Fermi surface. 7.9.2 Energy Functional Landau proposed a phenomenological description of a quantum state with a large number of quasiparticles. Let nk\u03c3 describe the occupation number of state k. In the ground state, nk\u03c3 = 1 below the 228 CHAPTER 7 Electron\u2013Electron Interaction Fermi surface and nk\u03c3 = 0 above it. The difference between the occupation of the state k\u03c3 and its occupation in the ground state is given by \u03b4nk\u03c3. Suppose that the energy of the quantum state can be expanded in terms of the occupation number \u03b4nk\u03c3 by \u03b5 \u03b4n \u00bd % = \u03b50 + \u2211 k\u03c3 0 \u00dek \u03b4nk\u03c3 + \u03b5\u00f0 1 2 \u2211 k\u03c3,k\u2032\u03c3\u2032 \u03b4nk\u03c3f k\u03c3,k\u2032\u03c3\u2032 \u00de \u00f0 \u03b4nk\u2032\u03c3\u2032 + \u2026, (7.184) where \u03b4nk\u03c3 is the change in the occupation number of the quasiparticle mode of wave vector k and spin \u03c3 in the ground state, and \u03b4nk\u2032\u03c3\u2032 corresponds to the higher energy level. The function \u03b5\u00f0 is the energy of the noninteracting particles. For a metal, it is the energy of the Bloch electrons. The f-function, which depends on the mutual spins of the two quasiparticles, describes interactions between the quasiparticles. The exchange force favors parallel alignment of spin because the Pauli exclusion principle keeps two electrons with the same spin away from the same point in space. Thus, the correlation effect is enhanced while the mutual electrostatic potential energy is reduced. However, in general, it is difficult to compute the f-function. 0 \u00dek At zero temperature, the energy needed to add one quasiparticle \u03b4nk\u03c3 above the Fermi surface is \u03b5k\u03c3 \u2019 0 \u00dek\u03c3 + \u2211 \u03b5\u00f0 k\u2032\u03c3\u2032 f k\u03c3, k\u2032\u03c3\u2032 \u00de \u00f0 \u03b4nk\u2032\u03c3\u2032: (7.185) Because the chemical potential \u03bc increases when a quasiparticle is added, the system will be filled with quasiparticles until the cost of adding them rises above \u03bc. Similarly, the quasiparticle states will empty out until \u03bc = \u03b5k\u03c3. At T = 0, \u03bc = \u03b5F and the occupation numbers nk\u03c3 are \u00f0 The Fermi wave vector is defined by \u03b5kF = \u03b5F for an isotropic system. Because there is a one-toone correspondence between the quasiparticles and the free particles, \u00de (7.186) 0 n\u00f0 \u00dek\u03c3 \u2019 \u03b8 \u03b5F \u2212 \u03b5k\u03c3 : n = N V = 1 V fk\u03c3 = \u2211 k\u03c3 k3 F 3\u03c02. It can also be shown after some algebra that \u03b4nk\u03c3 = nk\u03c3 \u2212 n\u00f0 0 \u00dek\u03c3 = 1 \u03b5k\u03c3\u2212\u03bc e\u03b2 \u00f0 \u00de + 1 \u2212 n\u00f0 0 \u00dek\u03c3 : (7.187) (7.188) Eq. (7.188) indicates the probability that the quasiparticle states are occupied is given by an implicit expression for the Fermi function with the value of energy given in Eq. (7.185). The effective mass of the quasiparticles can be obtained from the relation vF \u2019 1 \u0127 \u2202\u03b5k \u2202k j kF \u0127kF m#, \u2019 (7.189) & & & & & & & & where \u03b5k was defined in Eq. (7.185), from which it", " becomes obvious that the effective mass m# is different from the free electron mass. The effective mass can be computed by calculating the particle current of quasiparticles in two different ways and then by equating them. 7.9 Fermi Liquid Theory 229 Because the quasiparticle states are eigenfunctions of the momentum, the total flow of particles JN is given by JN = \u2211 k\u03c3 \u0127k\u03c3 m nk\u03c3, which can be rewritten in the alternate form JN = \u2211 k\u03c3 \u0127k\u03c3 m \u03b4nk\u03c3 (7.190) (7.191) because n\u00f0 change in energy when their momenta change by a small amount. It can be shown that 0 \u00dek\u03c3 is spherically symmetric. The particle current can also be calculated by calculating the where JN = \u2211 k\u03c3 vk\u03c3nk\u03c3, vk\u03c3 = 1 \u0127 \u2202\u03b5k\u03c3 \u2202k : From Eqs. (7.185), (7.192), and (7.193), we obtain JN = 1 \u0127 \u2211 k\u03c3 0 \u2202\u03b5\u00f0 \u00dek\u03c3 \u2202k nk\u03c3 + 1 \u0127 nk\u03c3 \u2211 kk\u2032\u03c3\u03c3\u2032 \u2202 \u2202k \u00bd f k\u03c3, k\u2032\u03c3\u2032 \u00de \u03b4nk\u2032\u03c3\u2032% : \u00f0 We can write Eq. (7.194) by using Eq. (7.188) in the alternate form JN = 0 \u2202\u03b5\u00f0 \u00dek\u03c3 \u2202k 1 \u0127 \u2211 k\u03c3 \u03b4nk\u03c3 + 1 \u0127 \u2211 kk\u2032\u03c3\u03c3\u2032\u00bd \u03b4nk\u03c3 + n0 \u2202 \u2202k \u00bd f \u00f0 k\u03c3, k\u2032\u03c3\u2032 \u03b4nk\u2032\u03c3\u2032% \u00de : k\u03c3% Eq. (7.195) can be rewritten with the help of Eq. (7.185), JN = \u2211 k\u03c3 vk\u03c3\u03b4nk\u03c3 + 1 \u0127 \u2211 kk\u2032\u03c3\u03c3\u2032 n0 k\u03c3 \u2202 \u2202k \u00bd f k\u03c3, k\u2032\u03c3\u2032 \u00de \u03b4nk\u2032\u03c3\u2032% : \u00f0 Integrating the second term by parts, we obtain JN = \u2211 k\u03c3 vk\u03c3\u03b4nk\u03c3 \u2212 1 \u0127 \u2211 kk\u2032\u03c3\u03c3\u2032 0 \u2202n\u00f0 \u00dek\u03c3 \u2202k f \u00f0 k\u03c3, k\u2032\u03c3\u2032 \u03b4nk\u2032\u03c3\u2032, \u00de which can be rewritten in the alternate form by interchanging k\u03c3 and k\u2032\u03c3\u2032, JN = \u2211 k\u03c3 vk\u03c3\u03b4nk\u03c3 + \u2211 kk\u2032\u03c3\u03c3\u2032 f k\u03c3, k\u2032\u03c3\u2032 \u00de \u00f0 vk\u2032\u03c3\u2032\u03b4 \u00f0 0 \u00dek\u2032\u03c3\u2032 \u2212 \u03b5F \u03b5\u00f0 \u03b4nk\u03c3: \u00de (7.192) (7.193) (7.194) (7.195) (7.196) (7.197) (7.198) 230 CHAPTER 7 Electron\u2013Electron Interaction Comparing Eqs. (7.190) and (7.198) for the case when one \u03b4nk\u03c3 is nonzero, we obtain \u0127k m = vk + \u2211 k\u2032\u03c3\u2032 f \u00f0 k\u03c3, k\u2032\u03c3\u2032 vk\u2032\u03c3\u2032\u03b4 \u00f0 \u00de 0 \u00dek\u2032\u03c3\u2032 \u2212 \u03b5F \u03b5\u00f0 : \u00de Using the definition of the effective mass from Eq. (7.189), we obtain \u0127k m = \u0127k m# + \u2211 k\u2032\u03c3\u2032 f k\u03c3, k\u2032\u03c3\u2032 \u00de \u00f0 \u0127k\u2032 m# \u03b4 \u00f0 0 \u00dek\u2032\u03c3\u2032 \u2212 \u03b5F \u03b5\u00f0 : \u00de (7.199) (7.200) We take dot products on both sides with k and assume that the Fermi surface is spherical. In addition, because the Fermi liquid theory is valid near the Fermi surface, the magnitudes of both k and k\u2032 are approximately kF. Thus, we can rewrite Eq. (7.200) in the alternate form m# m \u2248 1 + \u2211 k\u2032\u03c3\u2032 f \u00f0 k\u03c3, k\u2032\u03c3\u2032 k.k\u2032 k2 F \u00de \u03b4 \u00f0 0 \u00dek\u2032\u03c3\u2032 \u2212 \u03b5F \u03b5\u00f0 : \u00de (7.201) The summation over k\u2032 can be converted as an integration, and we can write m m# = 1 + V Z dk\u2032k\u20322g \u03b4 k\u2032 \u00de \u00f0 \u00f0 0 \u00dek\u2032\u03c3\u2032 \u2212", " \u03b5F \u03b5\u00f0 \u03b8 2\u03c0 \u00de 0Z Z 0 d\u03b8 sin \u03b8 d\u03d5 f k\u03c3, k\u2032\u03c3\u2032 \u00de \u00f0 cos \u03b8, (7.202) where \u03b8 is the angle between ^k and ^k\u2032. Because the density of states g angular integral, dividing g by 4\u03c0, we obtain \u03b5 normally includes the \u03b5 \u00de \u00f0 \u00f0 \u00de k\u20322\u03b4 g k\u2032 \u00de \u00f0 Z \u03b50 k\u2032 \u2212 \u03b5F \u00f0 \u00de dk\u2032 = 1 4\u03c0 Z g 0 \u03b5\u00f0 \u00dek\u2032 \u00de \u00f0 \u03b4 0 \u00dek\u2032 \u2212 \u03b5F \u03b5\u00f0 \u00de \u00f0 d\u03b5 = g \u03b5F \u00f0 4\u03c0 \u00de : (7.203) Further, because f \u00f0 obtain from Eqs. (7.202) and (7.203) k\u03c3, k\u2032\u03c3\u2032 \u00de depends on the angle between k and k\u2032 near the Fermi surface, we m# m = 1 + 1 2 Vg \u03b5F \u00f0 \u00de 1 Z \u22121 cos \u03b8 d \u00f0 f k\u03c3,k\u2032\u03c3\u2032 \u00de \u00f0 \u00de cos \u03b8. (7.204) We note that the effective mass depends on the weighted average of the interactions over the Fermi surface, but the latter is not known for strongly correlated systems. We will now derive an expression for the effective mass from Eq. (7.204) and the Fermi liquid parameters. 7.9.3 Fermi Liquid Parameters If we assume that the particles with the same spin have the possibility of a different interaction compared to the particles with different spin, we define f f k \u00f0 \" k \u00f0, k\u2032 \"\u00de, k\u2032 #\u2032 #\u00de, k\u2032 \"\u00de = f s = f s kk\u2032 + f a kk\u2032, kk\u2032 \u2212 f a kk\u2032, (7.205) (7.206) 7.9 Fermi Liquid Theory 231 where the symmetric and antisymmetric functions are represented by s and a. These formulae may be inverted, and \u221e f s kk\u2032 = \u2211 l = 0 f s l Pl \u00f0 cos \u03b8 \u00de \u221e f a kk\u2032 = \u2211 l = 0 f a l Pl \u00f0 cos \u03b8, \u00de (7.207) (7.208) where Pl cos \u03b8 \u00f0 \u00de is the Legendre polynomial. We can invert Eqs. (7.207) and (7.208) to write and f s l = 2l + 1 2 f a l = 2l + 1 2 1 Z \u22121 1 Z \u22121 cos \u03b8 d \u00f0 cos \u03b8 Pl \u00de \u00f0 f s kk\u2032 \u00de cos \u03b8 d \u00f0 Pl \u00de \u00f0 cos \u03b8 f a kk\u2032: \u00de We define the dimensionless Fermi liquid parameters and Fs l \u2019 Vg \u03b5F \u00f0 f s l \u00de Fa l \u2019 \u03b5F Vg \u00f0 \u00de f a l : (7.209) (7.210) (7.211) (7.212) \u03b5F is the density of energy states at the Fermi surface. Fs Here, Vg l are known as Fermi liquid parameters. For example, it can be shown that the effective mass defined in Eq. (7.204) and the Fermi liquid parameters defined in Eqs. (7.205) and (7.206) have the form l and Fa \u00de \u00f0 m# m = 1 + 1 2 Vg \u03b5F \u00f0 1 \u00de Z \u22121 cos \u03b8 d \u00f0 P1 cos \u03b8 \u00f0 \u00de \u00de #, k\u2032 f k \u00f0 \" \"\u00de, k\u2032 k \" \u00f0 #\u00de + f 2 : $ From Eqs. (7.209), (7.211), and (7.212), we obtain m# m = 1 + 1 3 Fs 1: (7.213) (7.214) The effective mass equation is widely used in Fermi liquid theory, especially in highly correlated systems. 232 CHAPTER 7 Electron\u2013Electron Interaction 7.10 GREEN\u2019S FUNCTION METHOD 7.10.1 General Formulation The Green\u2019s function method has been widely used in solving many-body problems that go beyond the electron\u2013electron interactions. It starts with the idea that amplitude for finding a particle at site R > at time t, when it was at site 0 > at time 0, is given by j j < R t ^G \u00f0 j \u00dej 0 > =", " < R e\u2212i ^Ht/\u0127 j 0 >: j The Fourier transformation of ^G is given by t \u00f0 \u00de ^G \u03be \u00f0 \u00de = 1 i\u0127 \u221e Z 0 dt ei\u03bet/\u0127 ^G t \u00f0 : \u00de (7.215) (7.216) Eq. (7.216) converges if \u03be has a positive imaginary part, and hence, \u03be varies in a complex plane. In fact, the physical significance of the Green\u2019s functions depends on the complex part of the energy. From Eqs. (7.125) and (7.126), we obtain If the Hamiltonian is perturbed, we can write \u03be ^G \u00f0 \u00de = \u00f0 \u03be \u2212 ^H \u22121: \u00de If we define we obtain from Eqs. (7.217) and (7.219) ^H = ^H0 + ^H1: ^G0 = \u03be \u2212 ^H0 \u22121, \u00de \u00f0 ^G = \u03be \u2212 ^H0 \u2212 ^H1 \u00f0 \u22121 \u00de \u03b5 \u2212 ^H0 = \u00f0\u00f0 1 \u2212 \u00de\u00f0 \u00f0 \u03be \u2212 ^H0 \u22121 ^H1 \u00de \u22121 \u00de\u00de 1 \u2212 ^G0 ^H1 = \u00f0 \u22121 ^G0 \u00de = ^G0 + ^G0 ^H1 ^G0 + ^G0 ^H1 ^G0 ^H1 ^G0 + \u2026 = ^G0 + ^G0 ^H1 ^G = ^G ^H1 ^G0 ^G0 + ^G0 ^T ^G0. \u2019 Here, the ^T matrix is the operator that satisfies the equality in Eq. (7.224). (7.217) (7.218) (7.219) (7.220) (7.221) (7.222) (7.223) (7.224) 7.10 Green\u2019s Function Method 233 7.10.2 Finite-Temperature Green\u2019s Function Formalism for Interacting Bloch Electrons In the finite-temperature Green\u2019s function formalism for an interacting system of Bloch electrons in the presence of a periodic potential V(r), the one-particle propagator G satisfies the equation \u03bel \u2212 H G \u00de r, r\u2032, \u03bel\u00de \u00f0 + \u00f0 Z dr\u2032\u2032\u03a3 r, r\u2032, \u03bel\u00de \u00f0 G r\u2032\u2032, r\u2032, \u03bel\u00de \u00f0 = \u03b4 \u00f0 r \u2212 r\u2032, \u00de where \u03a3 is the exact proper self-energy operator, \u03bel is the complex energy, \u03bc is the chemical potential, l = 0, \u00b11, \u00b12, \u2026, and ^H is the one-particle Hamiltonian, \u03bel = \u00f0 2l + 1 \u00de i\u03c0/\u03b2 + \u03bc, Both G and \u03a3 have the symmetry and ^H = p2 2m +, r\u2032 + R, \u03bel\u00de = G r, r\u2032, \u03bel\u00de \u00f0 \u03a3 r + R, r\u2032 + R, \u03bel\u00de \u00f0 = \u03a3 r, r\u2032, \u03bel\u00de : \u00f0 (7.225) (7.226) (7.227) (7.228) (7.229) We can write the equation of motion in the Bloch representation, i.e., in terms of the basis functions, r \u03d5nk\u03c1\u00f0 \u00de = eik.rUnk\u03c1 r, \u00de \u00f0 (7.230) where Unk\u03c1 vector, and \u03c1 is the spin index. Using the Bloch representation, we can rewrite Eq. (7.225) as is a periodic two-component function, n is the band index, k is the reduced wave r \u00de \u00f0 p2 2m \u00f0 \u00f0 \u00de r r r\u2212r\u2032\u2032 \u2212 V \u00de \" \u03bel \u2212 eik\u2032\u2032. \u00d7 G \u00f0 \u2211 n\u2032\u2032,\u03c1\u2032\u2032,k\u2032,k\u2032\u2032 Z dr dr\u2032dr\u2032\u2032e\u2212ik.rU#nk\u03c1\u00f0! eik\u2032.r\u2032 r\u2032 Un\u2032k\u2032\u03c1\u2032\u00f0 r\u2032\u2032, r\u2032, \u03bel\u00de drdr\u2032dr\u2032\u2032dr\u2032\u2032\u2032e\u2212ik.rU#nk\u03c1\u00f0 Un\u2032k\u2032\u03c1\u2032\u00f0 r\u2032\u2032\u2032, r\u2032, \u03bel\u00de Eq. (7.231) can be rewritten in the alternate form (Problem 7.11) r, r\u2032\u2032, \u03bel\u00de \ufffd", "\ufffd eik\u2032.r\u2032 = \u03b4nn\u2032\u03b4\u03c1\u03c1\u2032: \u00de n\u2032\u2032,\u03c1\u2032\u2032,k\u2032,k\u2032\u2032 Z \u00d7 G \u00f0 + \u2211 eik\u2032\u2032. r\u2032\u2032\u2212r\u2032\u2032\u2032 \u03a3 \u00de r\u2032 r \u00de \u00f0 \u00deUn\u2032\u2032k\u2032\u2032\u03c1\u2032\u2032\u00f0 r U#n\u2032\u2032k\u2032\u2032\u03c1\u2032\u2032\u00f0 \u00de r\u2032\u2032 \u00de (7.231) r\u2032\u2032 \u00de U#n\u2032\u2032k\u2032\u2032\u03c1\u2032\u2032\u00f0 r\u2032\u2032\u2032 \u00de \u00deUn\u2032\u2032k\u2032\u2032\u03c1\u2032\u2032\u00f0 \u2211 n\u2032\u2032,\u03c1\u2032\u2032 \u00bd \u03bel \u2212 H k\u2032, \u03bel\u00de%nk\u03c1,n\u2032\u2032k\u03c1\u2032\u2032Gn\u2032\u2032k\u03c1\u2032\u2032,n\u2032k\u03c1\u2032\u00f0 k\u2032, \u03bel\u00dej k\u2032 = k = \u03b4nn\u2032\u03b4\u03c1\u03c1\u2032, \u00f0 (7.232) 234 CHAPTER 7 Electron\u2013Electron Interaction H k\u2032, \u03bel\u00de \u00f0 = 1 2m \u00f0 p + \u0127k\u2032, \u03bel\u00de, \u00f0 drdr\u2032 U#n k\u03c1 r \u00de \u00f0 e\u2212ik\u2032. \u00f0 r\u2212r\u2032 \u00de\u03a3 r, r\u2032, \u03bel\u00de \u00f0 Un\u2032\u2032k\u03c1\u2032\u2032\u00f0 r\u2032 \u00de, (7.233) (7.234) where and \u03a3nk\u03c1,n\u2032\u2032k\u03c1\u2032\u2032\u00f0 k\u2032, \u03bel\u00de = Gn\u2032\u2032k\u03c1\u2032\u2032,n\u2032k\u03c1\u2032\u00f0 k\u2032, \u03bel\u00de = Z Z drdr\u2032 U#n\u2032\u2032k\u03c1\u2032\u2032\u00f0 r G \u00f0 \u00de e\u2212ik\u2032. r, r\u2032, \u03bel\u00de r\u2212r\u2032 \u00f0 \u00deUn\u2032k\u03c1\u2032\u00f0 r\u2032 \u00de : (7.235) Because the Unk\u03c1\u2019s form a complete set of periodic functions, Eq. (7.232) can be rewritten in the alternate form Eq. (2.236) can also be rewritten as \u03bel \u2212 H k, \u03bel\u00de% \u00f0 G \u00f0 k, \u03bel\u00de \u00bd = I: G \u00f0 k, \u03bel\u00de = 1 \u03bel \u2212 H \u00f0 : k, \u03bel\u00de 7.10.3 Exchange Self-Energy in the Band Model The exchange contribution to the self-energy is nonlocal in r space, \u03a3 r, r\u2032, \u03bel\u00de \u00f0 = \u2212 1 \u03b2 \u2211 \u03bel veff r, r\u2032 \u00de G \u00f0 r, r\u2032, \u03bel \u2212 \u03bel\u2032\u00de, \u00f0 (7.236) (7.237) (7.238) where a simple static screening approximation is made in obtaining veff approximation, the self-energy is independent of \u03bel, and one can write r, r\u2032 \u00f0 \u00de from v r, r\u2032 \u00f0 \u00de. In this r, r\u2032 \u03a3 \u00f0 \u00de = \u2212 1 \u03b2 \u2211 \u03bel veff r, r\u2032 \u00de G \u00f0 r, r\u2032, \u03bel\u00de : \u00f0 \u03a3 and G can be expanded in terms of Bloch states as follows: and \u03a3 r, r\u2032 \u00de \u00f0 = \u2211 n,m,k,\u03c1,\u03c1\u2032 \u03a3n\u03c1,m\u03c1\u2032\u00f0 k \u03c8 nk\u03c1\u00f0 \u00de r r\u2032 \u03c8 #mk\u03c1\u2032\u00f0 \u00de \u00de G \u00f0 r, r\u2032 \u00de = \u2211 n,m,k,\u03c1,\u03c1\u2032 k Gn\u03c1,m\u03c1\u2032\u00f0 \u03c8 nk\u03c1\u00f0 \u00de r \u03c8 #mk\u03c1\u2032\u00f0 r\u2032 \u00de : \u00de (7.239) (7.240) (7.241) Substituting Eqs. (7.240) and (7.241) in Eq. (7.239), we obtain Problems 235 k \u03c8 nk\u03c1\u00f0 r \u00de \u03c8 #mk\u03c1\u2032\u00f0 r\u2032 \u00de \u00de \u2211 n,m,\u03c1,\u03c1\u2032 \u03a3n\u03c1,m\u03c1\u2032\u00f0 1 \u03b2 = \u2212 \u2211 p,q,k\u2032,\u03c1,\u03c1\u2032 veff r, r\u2032 Gp\u03c1,q\u03c1\u2032\u00f0 \u00de k\u2032 \u03c8 pk\u2032\u03c1\u00f0 \u00de r \u03c8 qk\u2032\u03c1\u2032\u00f0 r\u2032 \u00de : \u00de \u00f0 (7.242) \u2211 \u03bel If the effective electron\u2013electron interaction is spin independent, then \u03c1 = \u03c1", ", \u03c1\u2032 = \u03c1\u2032, and we have \u03a3n\u03c1,m\u03c1\u2032\u00f0 k \u00de = \u2212 1 \u03b2 \u2211 k\u2032, \u03bel,p,q < nm veff j \u00f0 k, k\u2032 \u00dej pq > \u03c1\u03c1\u2032Gp\u03c1,q\u03c1\u2032\u00f0 k, \u03bel\u00de, (7.243) where <nm veff j \u00f0 k, k\u2032 \u00dej pq>\u03c1\u03c1\u2032 = Z r \u03c8 #nk\u03c1\u00f0 \u00de \u03c8 mk\u03c1\u2032\u00f0 r\u2032 \u00de veff r, r\u2032 \u00de \u03c8 pk\u2032\u03c1\u00f0 \u00f0 r r\u2032 \u03c8 #qk\u2032\u03c1\u2032\u00f0 \u00de \u00de dr dr\u2032: (7.244) Equation (7.243) is the expression for exchange self-energy in the band model. The problem of exchange self-energy in the band model, which includes the effect of a magnetic field where the self-energy has been expanded in different orders in the magnetic field, is discussed in more detail in Chapter 12 (Section 12.4.5). PROBLEMS 7.1. In Eq. (7.1), we obtained N H\u03a8 = \u2211 i = 1 \u01272 2me \u2212! \u22072 i \u03a8 \u2212 ze2\u2211 l 1 ri \u2212 Rl e2 ri \u2212 rj j \u03a8: j The Hamiltonian can be written as H = \u2211 i Hi + 1 2 \u2211 i\u2260j Vij, where Hi operates on the coordinate of the ith electron, and Vij operates on the two-body coordinates of both i and j. In Eq. (7.19), we showed that r1s1 \u2026 rN \u03a8 \u00f0 = \u00de 1 N!p n \u2211 n \u00f0 \u22121 \u00de \u220f k \u03c8 nk\u00f0 rksk : \u00de Show that the expectation value of the one-electron terms in the Hamiltonian <Hi> is \ufb03\ufb03\ufb03\ufb03\ufb03 dNr 1 N! \u2211 nn\u2032 \u00f0 \u22121 \u00de n+n\u2032 \u2211 s1:::sN Z \u220f k # rksk \u03c8 #nk \u00f0 \u2211 i! \u00de $ \u2212\u01272\u22072 i 2m ze2 ri \u2212 Rl \u2212 \u2211 l j \u220f k\u2032 \u03c8 n\u2032k\u2032\u00f0 : rk\u2032sk\u2032\u00de $ \"# j Because the \u03c8\u2019s are orthonormal, only n = n\u2032 terms survive the summation and (1) (2) (3) (4) integration \u2211 s1:::sN Z dNr. The sum over n results in a factor (N \u2013 1)! for all indices other than i. 236 CHAPTER 7 Electron\u2013Electron Interaction Hence, show that Eq. (4) can be rewritten as \u2211 i \u2211 si Z dri 1 N! \u2211 n risi \u03c8 #ni \u00f0 \u2212\u01272\u22072 i 2m \u00de! ze2 ri \u2212 Rl \u2212 \u2211 l j \" j risi \u03c8 ni\u00f0 : \u00de The sum over n results in a factor (N \u2013 1)! for all indices other than i. However, si ranges over all values that can be written as a sum over i\u2032, and one can drop the dummy index i when integrating over risi. Show that Eq. (5) can be rewritten as \u2211 i \u2211 s Z dr 1 N \u2211 i\u2032 rs \u03c8 #i\u2032\u00f0 \u00de! \u2212\u01272\u22072 2m ze2 r \u2212 Rl \u2212 \u2211 l j \u03c8 i\u2032\u00f0 \" j rs : \u00de (5) (6) risi, the sum over the spin index can be eliminated. The sum over Because \u03c8 i\u00f0 \u00de i yields a factor of N, and the sum over i\u2032 can be rewritten as a sum over i. Show that Eq. (6) can be expressed as = \u03d5i\u00f0 \u03c7i\u00f0 ri si \u00de \u00de N \u2211 i = 1 Z dr \u03d5#i \u00f0 r \u00de \u2212\u01272\u22072 2m! \u2212 \u2211 l j \u2212ze2 r \u2212 Rl r \u03d5i\u00f0 \u00de : \" j The expectation value of the Coulomb interaction term in the Hamiltonian is obtained from Eqs. (2) and (3) as \u2211 s1 ::: sN Z dNr \u2211 n,n\u2032 1 N! 1 2 \u2211 i \u2260 j n+n\u2032 e2 \u22121 \u00de \u00f0 ri \u2212 rj j \u220f k,k\u2032 \u03c8 #n", "k\u00f0 rksk \u03c8 n\u2032k\u2032\u00f0 rk\u2032sk\u2032\u00de : \u00de j (7) (8) Show that Eq. (8) can be rewritten in the alternate form \u2211 s1\u2026sN Z dNr \u2211 n,n\u2032 1 N! 1 2 \u2211 i \u2260 j e2 n+n\u2032 \u22121 \u00f0 \u00de ri \u2212 rj j \u03c8 #ni\u00f0 j h ri \u03c8 #nj\u00f0 rj\u00de \u03c8 n\u2032i\u00f0 ri \u00de rj \u03c8 n\u2032j\u00f0 \u00de \u220f k,k\u2032\u2260i,j \u00de rk \u03c8 #nk\u00f0 \u00de \u03c8 n\u2032k\u2032\u00f0 rk\u2032\u00de (9), i where \u03c8 ni\u00f0 (which leaves two permutations n\u2032 for given n), show that Eq. (9) can be rewritten as etc: for brevity. Integrating over all the terms except ri and rj \u03c8 ni\u00f0 ri, si \u00de \u2019 ri \u00de 1 2 \u2211 i \u2260 j \u2211 sisj Z dri drj\u2211 n 1 N! e2 ri \u2212 rj j j h ri \u03c8 ni\u00f0 j \u00dej 2 2 rj \u03c8 nj\u00f0 j \u00dej \u2212 \u03c8 #ni\u00f0 ri \u00de rj \u03c8 #nj\u00f0 \u00de rj \u03c8 ni\u00f0 \u00de ri \u03c8 nj\u00f0 \u00de (10) : i Show that because the integrations can be performed over the dummy variables 1 and 2, the sum over i \u2260 j yields a factor of N(N \u2013 1). Hence, show that Eq. (10) can be rewritten as \u2211 s1s2 Z dr1dr2 N \u2212 2 2 \u00f0! \u00de e2 r1 \u2212 r2 j \u2211 n j h 2 \u03c8 n1\u00f0 j r1 \u00dej \u03c8 n2\u00f0 j r2 \u00dej 2 \u2212 \u03c8 #n1\u00f0 r1 \u00de r2 \u03c8 #n2\u00f0 \u00de \u03c8 n1\u00f0 r2 \u00de \u03c8 n2\u00f0 r1 (11) : \u00de i Summing over the permutations over n, show that Eq. (11) can be rewritten as Problems 237 1 2 \u2211 s1s2 Z e2dr1dr2 r1 \u2212 r2 j \u2211 i \u2260 j h r1 \u03c8 i\u00f0 j 2 \u00dej r2 \u03c8 j\u00f0 j 2 \u00dej r1 \u2212 \u03c8 #i \u00f0 \u03c8 #j \u00f0 \u00de r2 \u03c8 i\u00f0 \u00de r2 r1 \u03c8 j\u00f0 \u00de \u00de : i j Using Eq. (7.20), show that Eq. (12) can be rewritten as 1 2 Z e2dr1dr2 r1 \u2212 r2 j \u2211 i \u2260 j h r1 \u03d5i\u00f0 j 2 \u00dej j r2 \u03d5j\u00f0 j 2 \u00dej \u2212 \u03d5#i \u00f0 r1 r2 \u03d5#j \u00f0 \u03d5i\u00f0 \u00de \u00de r2 \u03d5j\u00f0 \u00de r1 \u03b4\u03c7i, \u03c7j \u00de : i j From Eqs. (7) and (13), we obtain (12) (13) ze2 r \u2212 Rl \u2212 \u2211 l \u03d5i\u00f0 j j r1 2 j \u00dej r \u00de \u03d5i\u00f0 \" j r2 \u03d5j\u00f0 \u00dej 2 N <H> = \u2211 i = 1Z 1 + 2 dr\u03d5#i \u00f0 r \u00de! e2dr1dr2 r1 \u2212 r2 j ZZ j \u03d5#i H1\u03d5idr + = \u2211 i Z \u2212\u01272\u22072 2m \u2211 ZZ \u2211 i \u2260 j ZZ \u2212 1 2 \u2212 \u03d5#i \u00f0 r1 r2 \u03d5#j \u00f0 \u00de \u03d5i\u00f0 \u00de r2 \u03d5j\u00f0 \u00de r1 \u03b4\u03c7i, \u03c7j \u00de i dr1dr2\u03d5#i \u00f0 r1 \u00de \u03d5#j \u00f0 r2 r1 H12\u03d5i\u00f0 \u03d5j\u00f0 \u00de \u00de r2 \u00de dr1dr2\u03d5#i \u00f0 \u03d5#j \u00f0 r1 \u00de r2 H12\u03d5i\u00f0 \u00de r2 \u03d5j\u00f0 \u00de r1 \u00de \u03b4\u03c7i, \u03c7j, where and H1 \u2019 \u2212\u01272\u22072 2m! \u2212 \u2211 l j ze2 r \u2212 Rl \" j H12 e2 r", "1 \u2212 r2. j \u2019 j Following a variational technique similar to that adopted in Eqs. (7.11) through (7.15), derive the Hartree\u2013Fock equation in Eq. (7.21). 7.2. We derived Uex\u03d5i\u00f0 r \u00de = \u2212e2\u03d5i\u00f0 r kF \u00de Z kF dk 8\u03c03 \u03c0 = \u2212e2\u03d5i\u00f0 r \u00de 0Z Z 0 \u03c0 \u00f0 4\u03c0 k2 + k2 i \u2212 2kki cos \u03b8 \u00f0 \u00de k2dk sin \u03b8d\u03b8 k2 i + k2 \u2212 2kki cos \u03b8 : \u00de First, integrate over d\u03b8 and then integrate over dk to obtain Uex\u03d5i\u00f0 r \u00de = \u2212e2\u03d5i\u00f0 r \u00de 1 2\u03c0ki \u00f0 # F \u2212 k2 k2 i \u00de ln kF + ki kF \u2212 ki! \" + 2kikF : $ (1) (2) 238 CHAPTER 7 Electron\u2013Electron Interaction 7.3. Show, by integration, that the total energy of N electrons E = 2 \u2211 k\u2264kF \u01272k2 2m \u2212 e2kF \u03c0 k2 F \u2212 k2 2kkF ln 1 + \u2211 k\u2264kF # = N 3 5 # \u03b5F \u2212 3 4 e2kF \u03c0 : $ kF + k kF \u2212 k & & & & $ & & & & The first term was evaluated in Chapter 1. 7.4. We derived in Eq. (7.43), # Show that Eq. (1) can also be rewritten as E = N 3 5 \u03b5F \u2212 3 4 e2kF \u03c0 : $ E N = = e2 2a0 3 5 \u00f0 kFa0 2 \u2212 \u00de 3 2\u03c0 \u00f0 kFa0 h 2:21 rs/a0 \u00de \" \u00f0 2 \u2212 0:916 rs/a0 \u00f0 Ry: # \u00de \u00de i 1 Ry = e2/2a0 = 13:6 eV : \u00de \u00f0 7.5. Assume that a positively charged particle is placed at position r (an example is the substitution of a Zn ion for a copper ion in a metal lattice), which creates a surplus of negative charge, thereby screening its field. The electrostatic potential can be written as \u03d5 r = \u03d5e r + \u03d5i r : \u00de is potential due to the positively charged particle, and \u03d5i Here, \u03d5e induced charge density because of the presence of the positively charged particle. The Poisson\u2019s equation can be written as arises due to the where the total charge density \u03c1 r \u00de \u00f0 \u2212\u22072\u03d5 r = 4\u03c0\u03c1 \u00f0 \u00f0 can be expressed as \u00de r, \u00de \u03c1 r \u00de \u00f0 = \u03c1e r \u00de \u00f0 + \u03c1i r : \u00de \u00f0 are the external and induced charge densities, respectively. Here, \u03c1e and \u03c1i r \u00f0 \u00de r \u00f0 \u00de Show that in a spatially uniform electron gas, \u03d5e r \u00f0 the difference between their position, in an equation of the form and \u03d5 r \u00f0 \u00de are linearly related through \u00de \u03d5e = r \u00de \u00f0 Z dr\u2032 \u2208 \u00f0 r \u2212 r\u2032 \u00de \u03d5 : r\u2032 \u00de \u00f0 (4) (1) (2) (1) (2) (1) (2) (3) By making appropriate Fourier transformations, and show that \u2208 q \u00f0 \u00de = \u03d5e q \u00f0 \u00de = Z Z dr e\u2212iq.r \u2208 r, \u00de \u00f0 dr e\u2212iq.r\u03d5e r, \u00de \u00f0 \u03d5 r\u2032 \u00f0 \u00de = dq Z \u00f0 2\u03c0 \u00de 3 eiq.r\u2032\u03d5 q, \u00de \u00f0 \u03d5e Eq. (8) can be rewritten in the alternate form = \u03d5 q \u00de \u00f0 \u03d5e \u2208 q \u00f0 q \u00f0 \u00de \u00de, Problems 239 (5) (6) (7) (8) (9) where \u2208 q \u00de \u00f0 is the dielectric constant of the metal. 7.6. If the external potential in a metal (in the jellium model) is due to a positive point charge Q located at r, show that \u03d5e = q \u00de \u00f0 4\u03c0Q q2 Hence, show from Eq. (7.61) that = \u03d5 q \u00f0 \u00de Using Fourier transformation, show that", " \u03d5e \u2208 q \u00f0 q \u00f0 \u00de \u00de = 4\u03c0Q q2 + \u03bb2 : = \u03d5 r \u00de \u00f0 dq Z \u00f0 2\u03c0 \u00de eiq.\u2212\u03bbr: 7.7. Show that at T = 0, the integral = \u2212e2 \u03c7 q \u00de \u00f0 Z dk 4\u03c03 f0 \u00f0 k \u00de \u01272 can be performed explicitly to give k + q /m \u2212 f0 e2 mkF \u01272\u03c02! \" # 1 2 + 1 \u2212 x2 4x ln 1) (2) (3) (1) (2) 240 CHAPTER 7 Electron\u2013Electron Interaction where x = q 2kF : 7.8. The expression for the dielectric constant for static screening in the Lindhard theory, derived in Eq. (7.76), is given by \u2208L q \u00f0 \u00de = 1 \u2212 4\u03c0e2 q2 \u2211 1) By adjusting the summation over f 0 an integration by using polar coordinates, show that (Eq. 7.93) to and converting the summation over k to 7.9. If we define and \u2208L q \u00de \u00f0 = 1 + 4\u03c0e2 q2 mkF \u01272\u03c02 1 2 # + 4k2 F \u2212 q2 8kFq ln 2kF + q 2kF \u2212 kk\u2032 = \u2211 l = 0 f s l Pl \u00f0 cos \u03b8 \u00de \u221e f a kk\u2032 = \u2211 l = 0 f a l Pl \u00f0 cos \u03b8, \u00de by using the properties of the Legendre polynomial, show that and l = 2l + 1 f s 2 f a l = 2l + 1 2 1 Z \u22121 1 Z \u22121 cos \u03b8 d \u00f0 cos \u03b8 Pl \u00de \u00f0 f s kk\u2032 \u00de cos \u03b8 d \u00f0 Pl \u00de \u00f0 cos \u03b8 f a kk\u2032: \u00de 7.10. Derive from Eq. (7.127) i\u0127 < k\u2032 \u03b4 _\u03c1 j j k > = k\u2032 \u03b5 \u00bd \u00f0 \u00de \u2212 \u03b5 k \u00de% \u00f0 < k\u2032 j \u03b4\u03c1 j k > \u2212 f0 k\u2032 \u00de \u00f0 \u00bd \u2212 f0 k \u00f0 < k\u2032 \u00de% k >, \u03d5 j j and the fact that \u03b4\u03c1 has the same time dependence as \u03d5 \u00f0 r \u2212 i\u0127 \u00f0 i\u03c9 + \u03b1 \u00de% < k + q k > = \u03b4\u03c1 j j \u00bd \u00de f0 in Eq. (7.130), show that k + q \u2212 f0 k, t \u03d5q\u00f0 \u00de \u00de% \u00f0 \u00de \u00f0 (2) (1) (2) (3) (4) (1) (2) References 241 where t \u03d5q\u00f0 \u00de = \u03d5eq\u00f0 t \u00de + \u03d5iq\u00f0 t \u00de = \u03d5e\u00f0 \u00bd q, \u03c9 + \u03d5i\u00f0 \u00de q, \u03c9 e\u00f0 \u00de% i\u03c9t+\u03b1t \u00de: 7.11. We derived in Eq. (7.231) p2 2m \u03bel \u2212 r\u2032 \u00de + \u2211 \u2211 n\u2032\u2032,\u03c1\u2032\u2032,k\u2032,k\u2032\u2032 Z drdr\u2032dr\u2032\u2032e\u2212ik.rU#nk\u03c1\u00f0! eik\u2032.r\u2032 Un\u2032k\u2032\u03c1\u2032\u00f0 r\u2032\u2032, r\u2032, \u03bel\u00de drdr\u2032dr\u2032\u2032dr\u2032\u2032\u2032e\u2212ik.rU#nk\u03c1\u00f0 Un\u2032k\u2032\u03c1\u2032\u00f0 r\u2032\u2032\u2032, r\u2032, \u03bel\u00de Show that Eq. (1) can be rewritten in the alternate form k\u2032, \u03bel\u00de%nk\u03c1,n\u2032\u2032k\u03c1\u2032\u2032Gn\u2032\u2032k\u03c1\u2032\u2032,n\u2032k\u03c1\u2032\u00f0 r, r\u2032\u2032, \u03bel\u00de \u00f0 eik\u2032.r\u2032 = \u03b4nn\u2032\u03b4\u03c1\u03c1\u2032: n\u2032\u2032,\u03c1\u2032\u2032,k\u2032,k\u2032\u2032 Z \u03bel \u2212 H r\u2032\u2032,\u03c1\u2032\u2032 \u00bd eik\u2032\u2032. \u00f0 r\u2212r\u2032\u2032 \" \u00deUn\u2032\u2032k\u2032\u2032\u03c1\u2032\u2032\u00f0 r U#n\u2032\u2032k\u2032\u2032\u03c1\u2032\u2032\u00f0 \u00de r\u2032\u2032 \u00de eik\u2032\u2032. \u00f0 r\u2032\u2032\u2212r\u2032\u2032\u2032 \u00deUn\u2032\u2032k\u2032\u2032\u03c1\u2032\u2032\u00f0 r\u2032\u2032 U#n\u2032\u2032k\u2032\u2032\u03c1\u2032\u2032\u00f0 \u00de r\u2032\u2032\u2032 \u00de k", "\u2032, \u03bel\u00dej k\u2032= k = \u03b4nn\u2032\u03b4\u03c1\u03c1\u2032, where and H k\u2032, \u03bel\u00de \u00f0 = 1 2m \u00f0 p + \u0127k\u2032, \u03bel\u00de \u00f0, \u03a3nk\u03c1,n\u2032\u2032k\u03c1\u2032\u2032\u00f0 k\u2032, \u03bel\u00de = Z drdr\u2032 U#n k\u03c1 r \u00de \u00f0 e\u2212ik\u2032. \u00f0 r\u2212r\u2032 \u00de\u03a3 r, r\u2032, \u03bel\u00de \u00f0 Un\u2032\u2032k\u03c1\u2032\u2032\u00f0 r\u2032 \u00de, Gn\u2032\u2032k\u03c1\u2032\u2032,n\u2032k\u03c1\u2032\u00f0 drdr\u2032 U#n\u2032\u2032k\u03c1\u2032\u2032\u00f0 Because the Unk\u03c1\u2019s form a complete set of periodic functions, show that Eq. (5) can be rewritten in the alternate form \u00deUn\u2032k\u03c1\u2032\u00f0 r, r\u2032, \u03bel\u00de k\u2032, \u03bel\u00de G \u00f0 : \u00de r\u2032 = Z r \u00de \u00f0 e\u2212ik\u2032. r\u2212r\u2032 \u03bel \u2212 H \u00bd k, \u03bel\u00de% \u00f0 G k, \u03bel\u00de \u00f0 = I: References 1. Abrikosov AA. Introduction to the theory of normal metals. New York: Academic Press; 1972. 2. Ashcroft NA, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 3. Born M, Oppenheimer JR. On the quantum theory of metals. Ann Phys 1927;84:457. 4. Callaway J. Quantum theory of the solid state. New York: Academic Press; 1976. 5. Dirac PAM. Quantum mechanics of many-electron systems. Proc R Soc (Lond) 1929;A123:714. (3) (1) (2) (3) (4) (5) (6) 242 CHAPTER 7 Electron\u2013Electron Interaction 6. Ferrell RA. Rigorous validity criterion for testing approximations to the electron gas correlation energy. Phys Rev Lett 1958;1:444. 7. Fock V. A method for the solution of many-body problem in quantum mechanics. Z Physik 1930;61:126. 8. Hartree DR. The wave mechanics of an atom with a non-Coulomb central field. Proc Camb Phil Soc 1928;24:89. 9. Hohenberg PC, Kohn W. Inhomogeneous electron gas. Phys Rev 1964;80:864. 10. Kittel C. Quantum theory of solids. New York: John Wiley & Sons; 1987. 11. Kohn W, Sham LJ. Self-consistent equations including exchange and correlation effects. Phys Rev 1965;140:A1113. 12. Landau LP, Lifshitz EM. Quantum mechanics, non-relativistic theory. Reading, MA: Addison-Wesley; 1965. 13. Landau LD. The theory of a Fermi liquid. Sov Phys JETP 1957;3:920; Oscillations in a Fermi liquid. 1957;5:101. 14. Liboff RL. Quantum mechanics. Reading, MA: Addison-Wesley; 1980. 15. Lieb EH. Thomas-Fermi and related theories of atoms and molecules. Rev Mod Phys 1981;53:603. 16. Lindhard J. Kgl Danske Videnskab. Selskab Mat.-Fys. Medd. 1954;28(8). 17. Marder MP. Condensed matter physics. New York: Brooks/Cole; 2000. 18. Noziers P, Blandin A. Kondo effect in real metals. J Phys (Paris) 1980;41:193. 19. Slater JC. A simplification of the Hartree-Fock method. Phys Rev 1951;81:385. 20. Varma CM, Nussinov Z, Sarloos W. Singular or non-Fermi liquids. Phys Rep 2002;361:267. 21. Wigner E. On the interaction of electrons in metals. Phys Rev 1934;46:1002. 22. Wilson AH. The theory of metals. Cambridge: Cambridge University Press; 1960. 23. Ziman JM. Principles of the theory of solids. Cambridge: Cambridge University Press; 1972. Dynamics of Bloch Electrons CHAPTER 8 CHAPTER OUTLINE 8.1 Semiclassical Model..................................................................... 243 8", ".2 Velocity Operator........................................................................ 244 8.3 k \u00b7 p Perturbation Theory.................................................................. 245 8.4 Quasiclassical Dynamics................................................................. 246 8.5 Effective Mass.......................................................................... 247 8.6 Bloch Electrons in External Fields......................................................... 248 8.6.1 Time Evolution of Bloch Electrons in an Electric Field.................................. 250 8.6.2 Alternate Derivation for Bloch Functions in an External Electric and Magnetic Field...... 252 8.6.3 Motion in an Applied DC Field....................................................... 253 8.7 Bloch Oscillations....................................................................... 254 8.8 Holes.................................................................................. 255 8.9 Zener Breakdown (Approximate Method).................................................... 258 8.10 Rigorous Calculation of Zener Tunneling................................................... 261 8.11 Electron\u2013Phonon Interaction.............................................................. 264 Problems.................................................................................... 271 References..........................................", "........................................ 274 8.1 SEMICLASSICAL MODEL We determined in Chapter 4 (Eq. 4.45) that the eigenfunction of the Hamiltonian of an electron that can be moving in a crystalline solid with a periodic potential V expressed as is a Bloch function \u03c8 nk\u00f0 r r \u00de \u00de \u00f0 r \u03c8 nk\u00f0 \u00de = eik.runk\u00f0 r, \u00de (8.1) \u00de where unk\u00f0 is the periodic part of the Bloch function. Here, n is a band index, k is a vector in the first r Brillouin zone in the restricted zone scheme but extends to infinity in the periodic zone scheme, and has the unique property that it remains unchanged when translated by any direct lattice vector Ri r unk\u00f0 (Eq. 4.46), i.e., \u00de Thus, the Schrodinger equation unk\u00f0 r + Ri\u00de r = unk\u00f0 \u00de : H\u03c8 nk\u00f0 r \u00de = \u03b5n\u00f0 k r \u03c8 nk\u00f0 \u00de \u00de Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00008-6 \u00a9 2012 by Elsevier Inc. All rights reserved. (8.2) (8.3) 243 244 CHAPTER 8 Dynamics of Bloch Electrons can be rewritten in the alternate form (Eq. 4.36) where Hkunk\u00f0 r \u00de = \u03b5n\u00f0 k r unk\u00f0 \u00de \u00de, Hk = \u01272 2m \u00f0! \u2212i8.4) (8.5) In this chapter, we have written Heff as Hk for convenience. Here, the boundary conditions for unk\u00f0 are specified in Eq. (8.2). In the periodic zone scheme, which will be used in the derivations that r follow, the eigenfunctions and eigenvalues are periodic functions of k in the reciprocal lattice, i.e., \u00de and r \u03c8 nk\u00f0 \u00de = \u03c8 n, k + K\u00f0 r \u00de k \u03b5n\u00f0 \u00de = \u03b5n\u00f0 k + K : \u00de (8.6) (8.7) The set of electronic levels \u03b5n\u00f0, for each n, is known as the energy band because the band index n is a \u00de vary continuously as k varies because k is constant of motion. In addition, the energy levels \u03b5n\u00f0 essentially a parameter in Eq. (8.5). Further, the size of the crystal is not a factor in Eq. (8.4), and as we noted earlier, the Born\u2013von Karman boundary condition (Eq. 3.31), k k \u00de \u03c8 k\u00f0 \u03c8 k\u00f0 \u03c8 k\u00f0 x + L, y, z x, y + L, z x, y, z + L = \u03c8 k\u00f0 = \u03c8 k\u00f0 = \u03c8 k\u00f0 x, y, z \u00de x, y, z \u00de x, y, z \u00de, \u00de \u00de \u00de (8.8) specifies that k is continuous when L \u221e (Eq. 3.14). We will use the property that k is a continuous parameter in what follows. These same boundary conditions are also valid for the Bloch functions \u03c8 nk\u00f0! r \u00de : 8.2 VELOCITY OPERATOR For a crystal with inversion symmetry, the velocity operator can be rewritten in the alternate form v = _r = \u2212 i \u0127 \u00bd r, H $ v = \u2212 i\u0127 m \u2207: If we define the velocity operator using the effective Hamiltonian in Eq. (8.5), vk = \u2212 i \u0127 \u00bd r, Hk$, (8.9) (8.10) (8.11) 8.3 k \u00b7 p Perturbation Theory 245 we obtain vk = \u0127 m \u00f0 \u2212i\u2207 + k : \u00de (8.12) We will use this relation for the velocity operator to obtain an expression for the velocity of the Bloch electrons. 8.3 k \u00b7 p PERTURBATION THEORY We assume that k is increased by an infinitesimal amount \u0394k. When k rewritten in the form! k + \u0394k, Eq. (8.5) can be Hk+\u0394k = \u01272 2m \u00f0! \u2212i\u2207 + k + \u0394k \u00de 2 + V r, \" \u00de \u00f0 which can be rewritten in the alternate form Hk+\u0394k = H", "k + \u01272 m \u00f0 \u2212i\u2207 + k. \u0394k + \u01272 2m \u00f0 \u00de \u0394k 2: \u00de We use nondegenerate second-order perturbation theory and write Hk+\u0394k = Hk + H\u2032k: The perturbation term H\u2032k is obtained from Eqs. (8.14) and (8.15), H\u2032k = \u01272 m \u00f0 \u2212i\u2207 + k \u00de. \u0394k + \u01272 2m \u00f0 \u0394k 2: \u00de If we express the energy eigenvalue in different orders of \u03b5 as \u03b5n\u00f0 k + \u0394k \u00de k = \u03b5n\u00f0 \u00de 1 + \u03b5\u00f0 \u00den \u00f0 k \u00de 2 + \u03b5\u00f0 \u00den \u00f0 k \u00de + \u2026, we obtain from second-order perturbation theory, k \u03b5n\u00f0 \u00de = <unkj Hkj H unk> = <\u03c8 nkj \u03c8 nk> j and where the term \u01272 2m \u00f0 \u2207k \u00de \u00de = <unkj 1 k \u03b5\u00f0 \u00den \u00f0 2, which is second order in \u2207k, has been included in \u03b5\u00f0 unk> = <unkj unk>, j \u2212i\u2207 + k H\u2032kj. \u0394k \u00de \u01272 m \u00f0 2 \u00den \u00f0 2 k \u03b5\u00f0 \u00den \u00f0 \u00de = \u01272 2m \u00f0 \u2207k \u00de 2 + \u2211 n\u2032\u2260n <unkj j \u01272 m \u0394k. \u2212i\u2207 + k \u00f0 \u03b5nk \u2212 \u03b5n\u2032k un\u2032k> \u00dej 2 j : (8.13) (8.14) (8.15) (8.16) (8.17) (8.18) (8.19) (8.20) k, \u00de 246 CHAPTER 8 Dynamics of Bloch Electrons If we also expand \u03b5n\u00f0 k + \u0394k \u00de as \u03b5n\u00f0 k + \u0394k \u00de k = \u03b5n\u00f0 \u00de + \u2211 i k \u2202\u03b5n\u00f0 \u2202ki \u00de \u0394ki + 1 2 \u2211 ij \u22022\u03b5n\u00f0 k \u2202ki\u2202kj \u00de \u0394ki\u0394kj + \u2026, (8.21) we obtain from Eqs. (8.12), (8.17), (8.19), and (8.21), vkj which can be rewritten in the alternate form (Problem 8.2) = \u0127<unkj \u00de unk>, k \u2202\u03b5n\u00f0 \u2202k k \u2202\u03b5n\u00f0 \u2202k \u00de = \u0127<\u03c8 nkj v j \u03c8 nk> = \u0127vnk: Thus, the mean velocity of a Bloch electron is given by k \u2202\u03b5n\u00f0 \u2202k vnk = 1 \u0127 \u00de : (8.22) (8.23) (8.24) It is interesting to note that one can derive a similar expression for the velocity of electrons in a periodic potential and an external field. We will show by using quasiclassical dynamics that vk = \u0127k m = 1 \u0127 \u2202\u03b5 k \u00f0 \u00de \u2202k : (8.25) It is also interesting to note that the group velocity of a wave packet moving freely in space is given by the same expression as Eq. (8.25). 8.4 QUASICLASSICAL DYNAMICS The classical Hamilton equations are expressed as11 _qk = \u2202H \u2202pk, _pk = \u2212 \u2202H \u2202qk, (8.26) where H is the classical Hamiltonian function, and qk and pk are the generalized coordinates and momenta, respectively. In the quantum mechanical formulation, the Hamiltonian operator is obtained by replacing the classical momentum p by \u2212i\u0127\u2207. In an external electrostatic field, the Hamiltonian operator can be written as ^H = ^H0 + U r = ^\u03b5 \u00f0 \u2212i\u2207 \u00de + U r, \u00f0 where ^H0 is the Hamiltonian operator in the perfect lattice, U external field, and ^\u03b5 \u00f0 replacing \u2212i\u2207 by p/\u0127, the classical Hamiltonian function is given by \u2212i\u2207 \u00de is the perturbing potential of the r \u00f0 is the equivalent Hamiltonian operator. If we reverse these steps by \u00de \u00de \u00de \u00f0 (8.27) r, p H \u00f0 \u00de = \u03b5 p/8.28) 8.5 Effective Mass 247 From Eqs. (8", ".26) and (8.28), we obtain where _r = v = \u2202H \u2202p = \u2202\u03b5 p/\u0127 \u00f0 \u2202p \u00de = 1 \u0127 \u2202\u03b5 k \u00f0 \u00de \u2202k, p = \u0127k: (8.29) (8.30) It may be noted that \u0127k is the crystal momentum and not the actual momentum of an electron, as was obtained in the Sommerfeld model for free electrons. This is a consequence of the fact that in the Bloch formulation, the effect of the periodic potential has been already included. The second Hamiltonian equation in Eq. (8.26) can be written as _p = \u0127 _k = \u2212 \u2202H \u2202r = \u2212\u2207U r : \u00de \u00f0 (8.31) Because U F acting on the Bloch electron. According to Newton\u2019s law, r \u00de \u00f0 is the potential energy of the electron in an external field, \u0127 _k is essentially the force where m is the mass of the electron. F = m dv dt = \u0127 _k, (8.32) 8.5 EFFECTIVE MASS It is often convenient to define an effective mass to describe the motion of Bloch electrons in an external field. Later, we will use the same concept to describe the motion of electrons and holes in a semiconductor. If we assume that the external field (electrostatic) acting on the electrons is weak and the change in k is slow, we can write d dt vnk = \u2211 ij ^\u2208i \u2202vi nk \u2202kj \u2202kj \u2202t, (8.33) where ^\u2208i is one of the three unit vectors ^x, ^y, and ^z. We can also introduce the effective mass tensor M\u22121 (it is actually an inverse effective mass tensor) of the Bloch electrons (it is important to n \u00f0 note that the effect of the periodic potential of the lattice has already been included in the Bloch formulation) in analogy with Eq. (8.32) as k \u00de From Eqs. (8.24), (8.33), and (8.34), we obtain d dt vnk = \u0127M\u22121 n \u00f0 k \u00de. _k: M\u22121 n \u00f0 \u00bd k \u00de$ij = 1 \u01272 \u22022\u03b5n \u2202ki\u2202kj : (8.34) (8.35) 248 CHAPTER 8 Dynamics of Bloch Electrons We can derive an expression for the effective mass tensor from the results of the perturbation theory. From Eqs. (8.20) and (8.21), we obtain <unkj j \u01272 m \u0394k. k \u03b5n\u00f0 \u2212i\u2207 + k \u00f0 \u2212 \u03b5n\u2032\u00f0 k \u00de \u00de un\u2032k> 2 j \u00dej (8.36) 2 \u03b5\u00f0 \u00den \u00f0 k \u00de = \u01272 2m \u00f0 2 + \u2211 n\u2032\u2260n \u2207k \u00de \u22022\u03b5n\u00f0 k \u2202ki\u2202kj \u00de \u0394ki\u0394kj: = 1 2 \u2211 ij From Eqs. (8.35) and (8.36), we obtain (Problem 8.4) M\u22121 k n \u00f0 \u00bd \u00de$ij = 1 m \u03b4ij + \u01272 m2 \u2211 n\u2032\u2260n <nk j \u2212i\u2207ij n\u2032k >< n\u2032k \u2212i\u2207jj j k \u2212 \u03b5n\u2032\u00f0 k \u03b5n\u00f0 \u00de \u00de nk> + c:c: : (8.37) Here, the summation is over all band indices n\u2032 except n: The inverse mass tensor plays an important role in the formulation of dynamics of Bloch electrons. As we will see, the inverse effective mass tensor can be either positive or negative. Because the concept of a negative effective mass associated with the negatively charged Bloch electrons is contrary to our physical understanding, we will also introduce the concept of \u201choles,\u201d which are positively charged particles associated with positive effective mass. The holes play a very important role in semiconductors. In addition, the mystery of a positive charge in the Hall coefficient (in the free electron model, RH = \u2212nec), in the measurement of the Hall effect in certain metals such as aluminum in high magnetic fields, is also explained by the concept of holes. 8.6 BLOCH ELECTRONS IN EXTERNAL FIELDS In an external electric field E, the force acting on a free electron is given by The effect of a magnetic field B is included by assuming that the Lorentz force equation is valid", ", i.e., \u0127 _k = \u2212eE: (8.38) \u0127 _k = \u22128.39) The proof of an equivalent result for Bloch electrons is much more difficult for a magnetic field. We will first derive an equivalent result for Bloch electrons in a static electric field and then generalize the results by using a more rigorous method. In a static electric field E, the general expression reduces to rA r, t \u00f0 \u2202t \u00de \u2212\u2207\u03d5 r, t \u00f0 \u00de E = \u2212\u2207\u03d5 r : \u00de \u00f0 (8.40) (8.41) 8.6 Bloch Electrons in External Fields 249 Here, A(r, t) is the vector potential, and \u03d5 Bloch electrons, \u00f0 r \u00de is the scalar potential. We showed in Eq. (8.24) that for _r = vn\u00f0 k \u00de = 1 \u0127 k \u2202\u03b5n\u00f0 \u2202t \u00de : The wave packet moves in such a way that the energy \u03b5 = \u03b5n\u00f0 k t \u00f0 \u00de\u00de \u2212 e\u03d5 r t \u00f0 \u00f0 \u00de\u00de = constant: (8.42) (8.43) Because the energy \u03b5 of the wave packet is constant, we obtain, by taking the time derivative of the energy, \u2202\u03b5 \u2202t = k \u2202\u03b5n\u00f0 \u2202t. _k \u2212 e\u2207\u03d5. _r = 0: \u00de From Eqs. (8.42) and (8.44), we obtain Vn\u00f0 \u00de From Eq. (8.45), we obtain the desired result k \u0127 _k \u2212 e\u2207\u03d5. \u00bd = 0: $ \u0127 _k = e\u2207\u03d5 = \u2212eE: (8.44) (8.45) (8.46) However, there are problems associated in this derivation because the periodic boundary conditions that were implicitly assumed in this derivation (by assuming that the eigenfunctions of the Hamiltonian are the Bloch functions) are no longer valid. The electrostatic potential U(r), due to the = \u2212E. r: Therefore, when a metal is uniform electric field E, grows linearly in space because U placed in an electric field, the surface charges build up and cancel the interior field. Thus, the basic assumption that the Bloch formulation is still valid is questionable, and the periodicity associated with our formulation of the problem in a Brillouin zone with a band index n and crystal momentum k is invalid. We will derive Eq. (8.46) by using a second method where the periodicity of the potential is not lost. However, we will first use the previous derivation to justify (in a hand-waving way) the effect of a magnetic field. r \u00f0 \u00de is k It may be noted that the results of Eq. (8.45) are valid if any term perpendicular to Vn\u00f0 added in the square bracket. For example, anticipating that the Lorentz equations are valid, we can rewrite Eq. (8.45) as \u00de k Vn\u00f0 \u00de. \u00bd \u0127 _k \u2212 e\u2207\u03d5 + e c k Vn\u00f0 \u00de \u00d7 B $ = 0: (8.47) Because e\u2207\u03d5 = \u2212eE (from Eq. 8.46), Eq. (8.47) leads to the desired semiclassical expression for the Bloch electron in an electric field E and magnetic field B, \u0127 _k = \u2212eE \u2212 e c k Vn\u00f0 \u00de \u00d7 B: (8.48) We note that Eq. (8.48) is in no way a rigorous derivation of the motion of the Bloch electrons in an electric and magnetic field. 250 CHAPTER 8 Dynamics of Bloch Electrons 8.6.1 Time Evolution of Bloch Electrons in an Electric Field If we use the expression for the electric field E through a time-dependent vector potential A that increases linearly with time and set the scalar potential \u03d5 to zero, i.e., r \u00f0 \u00de A = \u2212cEt, \u03d5 = 0, r \u00de \u00f0 (8.49) we verify that E = \u2212 1 c \u2202A \u2202t \u2212 \u2207\u03d5 = E: (8.50) Such a vector potential A, which allows the electric field E to coexist with the periodic boundary conditions (in a ring-shaped crystal) is shown in Figure 8.1. The use of a time-dependent vector potential instead of a scalar potential allows us to use the periodic boundary conditions. We can generalize this method to a Bloch state under the influence of a time-dependent electric field, E", "(t), turned on at t = 0.6 The time-dependent Schrodinger equation for an electron in a Bloch state under the influence of an electric field, E(t), turned on at t = 0, is given by R E FIGURE 8.1 The vector potential A = \u2212cEt can be established in a ring-shaped crystal by changing the magnetic flux at a uniform rate Bz = 2\u03c0RcEt \u03b4 \u00f0 solenoid inside the ring. through an infinite r \u00de where H\u03c8 = r/c \u00f0 2m c t Z 0 E t\u2032 \u00f0 \u00de dt\u2032, \u03c8 = i\u0127 \u2202\u03c8 \u2202t, 3 \u00de 7 5 (8.51) (8.52) V(r) is the lattice periodic potential, and the time-dependent electric field is included as a vector potential, as described earlier. We can obtain \u03c8 by using eigenfunction expansion r, t of which the elements are from \u03c8 r, 0 \u00f0 \u00de \u00de \u00f0 A e/c \u00f0 2m = \u03b5i\u00f0 \u00de \u00de r, t \u03d5\u2032i \u00f0 : \u00de r, t \u03d5\u2032i \u00f0 3 7 5 (8.53) 8.6 Bloch Electrons in External Fields 251 For each t, the Hamiltonian is invariant under a crystal lattice translation because A is independent of r for a homogeneous electric field, and V is periodic. If we substitute r \u00f0 \u00de in Eq. (8.53), we obtain r, t \u03d5\u2032 i\u00f0 \u00de = e\u2212ieA.r/\u0127c\u03d5i\u00f0 r, t, \u00de p2 2m! + V r \u00f0 \u00de \" \u03d5i = \u03b5i\u03d5i: (8.54) (8.55) The solutions of Eq. (8.55) are the Bloch functions and energy band functions of the unperturbed crystal, and From Eqs. (8.54) and (8.56), we obtain \u03d5i = \u03d5nk\u00f0 r \u00de \u03b5i = \u03b5n\u00f0 k : \u00de r, t \u03d5\u2032 i\u00f0 \u00de = e\u2212ieA.r/\u0127c\u03d5nk\u00f0 r : \u00de (8.56) (8.57) (8.58) Because H is invariant under a lattice translation, the allowed values of k are defined using periodic boundary conditions on the \u03d5\u2032 i. We obtain k \u2212 eA \u0127c = \u2211 i ni Ni Ki, (8.59) i = 1, 2, 3 where Ki\u00f0 are the primitive reciprocal-lattice translation vectors, Ni are the number of cells in the i direction, and \u2212Ni/2 < ni \u2264 Ni/2. For the periodic boundary condition to be satisfied, k must be a function of t and \u00de \u0127 _k = _A = e c e c \u00f0 \u2212cE t \u00f0 \u00de\u00de = \u2212eE : t \u00f0 \u00de (8.60) t \u00de From Eq. (8.59) and the time dependence of A, it follows that the Brillouin zone describing the \u00de allowed k values is time dependent. The periodic boundary conditions that lead to Brillouin zone \u00f0 boundaries move in time. Thus, the wave vector does not undergo an Umklapp process (or U-process) back to the other side of the Brillouin zone, but continues with continuous values of k in the time-dependent zone. Having considered the general case of a time-dependent electric field, if we restrict to a time-independent electric field, t \u00f0 \u00f0 \u00de t and \u0127 _k = \u2212eE A = \u2212cEt: (8.61) (8.62) 252 CHAPTER 8 Dynamics of Bloch Electrons Thus, the k vectors are time dependent and should be written as k also obtain the Bloch equation \u00f0 t \u00de : From Eqs. (8.55) and (8.56), we \u03d5nk r \u00de t \u00f0 \u00de\u00f0 = eik t \u00f0 \u00de.runk \u00f0 r : \u00de t \u00de\u00f0 (8.63) The functions defined in (8.53) and (8.58) are also known as Houston functions. It may be noted that the Houston functions are not exact solutions of the Schrodinger equation because the eigenvalues are time dependent. The amplitude of Houston states with nearby k becomes nonzero and grows, a behavior typical of wave packets. In addition, the electron can jump from one band to another, which is known as Zener tunneling and which we will discuss later. We note that if we use", " the Houston functions instead of the Bloch functions, we obtain expressions for the effective Hamiltonians by a using a similar technique followed earlier (Problem 8.5). The effective Hamiltonian for Unk\u00f0 r, E, t is \u00de 1 2m \u00f0 Heff = \u2212i\u0127\u2207 + \u0127k \u2212 eEt 2 + V \u00de r : \u00de \u00f0 (8.64) One important aspect to note is that the electric field E is incorporated through k Eq. (8.61). \u00f0 t \u00de as described in When the electron jumps from one band to another because of a strong electric field, it is called electric breakdown or the Zener breakdown. It is easier for the electron to jump to a neighboring band in a magnetic field. This is known as magnetic breakdown. We will discuss these breakdowns later. We will describe the Zener breakdown after discussing an alternate (but essentially flawed) derivation of the motion of Bloch electrons in an electric and magnetic field. The controversy in this derivation is that one cannot use Bloch functions for Hamiltonians that are not periodic. 8.6.2 Alternate Derivation for Bloch Functions in an External Electric and Magnetic Field The Hamiltonian in an electric and magnetic field can be written as H = 1 2m 2 p + eA c $ # k0, r + V + eE. r: r \u00de \u00f0 In a small interval dt, the Bloch function \u03c8 = \u03c8 n\u00f0 = e\u2212i/\u0127Hdt \u03c8 n\u00f0 \u03c8 n\u00f0 k, r, dt \u00de k0, r \u00de at t = 0 will change by \u00de \u2248 1 \u2212 # Hdt i \u0127 k0, r : \u00de \u03c8 n\u00f0 $ If we operate with the translation operator, with properties described in Eq. (4.47), we obtain which can be written as ^T Ri\u00de \u00f0 \u03c8 = ^T Ri\u00de \u00f0 1 \u2212 Hdt i \u0127 # k0, r, \u00de \u03c8 n\u00f0 $ ^T Ri\u00de \u00f0 \u03c8 = 1\u2212 # Hdt i \u0127 $ ^T Ri\u00de \u00f0 \u03c8 n \u2212 i \u0127 dt T Ri\u00de \u00f0 \u00bd, H \u03c8 n: $ (8.65) (8.66) (8.67) (8.68) 8.6 Bloch Electrons in External Fields 253 It can be shown that (Problem 8.6) where ^T \u00bd Ri\u00de \u00f0, H $ = \u03be! = % e c \u03be!. Ri + e2B2 2m R2 ix ^T &, Ri\u00de \u00f0 v \u00d7 B + eE + _A: e c We neglect the last term in Eq. (8.69) in a weak magnetic field approximation and obtain ^T Ri\u00de!. Ri\u00de dt ^T i \u03c8 n = Ri\u00de \u00f0 \u2026 \u00f0 \u00de eik0.Ri \u03c8 n = eik.Ri\u03c8: Here, or k = k0\u2212 \u03be! \u0127 dt \u0127 _k = \u2212 \u03be!: (8.69) (8.70) (8.71) (8.72) (8.73) If we neglect the last term in Eq. (8.70), we obtain \u0127 _k = \u22128.74) Eq. (8.74) is the standard Lorentz force equation. 8.6.3 Motion in an Applied DC Field Eq. (8.61) can be rewritten in the alternate form eEt \u0127, (8.75) from which we obtain vn\u00f0 (8.76) = vn k eEt \u0127 \u00de\u00de is periodic in We note from Eq. (8.42) that vn\u00f0 the reciprocal lattice vector K and when E K, Eq. (8.76) is oscillatory. This oscillatory behavior in one dimension is shown in Figure 8.2. The velocity, which is linear in k near the band minimum, reaches a maximum and then decreases to zero at the Brillouin zone boundary. It is surprising \u03c0 /a 0 k \u03c0 /a FIGURE 8.2 v k \u00f0 \u00de and \u03b5 k \u00f0 \u00de versus k in one dimension. 254 CHAPTER 8 Dynamics of Bloch Electrons to note that in the region in which the velocity decreases with increasing k, the acceleration of the electron is opposite to the external applied field. k \u00f0 We first note the periodic behavior of the \u03b5 versus k curve in Figure 8.2. The Bloch electrons \u00de have positive effective masses at k = 0 and negative effective masses at k = \u00b1\u03c0/a: The oscillatory behavior of the", " velocity of the Bloch electron with k (and with time that is related to k through Eq. 8.69) is a consequence of the effect of the periodic potential that exerts an additional force. Hypothetically, the external DC electric field could induce alternating current, but collisions with phonons and impurities present in a crystal do not allow any such possibility. In fact, in the absence of damping, a perfect crystal with a periodic potential would not have any resistivity at zero temperature. 8.7 BLOCH OSCILLATIONS The Bloch oscillations result from the fact that \u03b5nk is a periodic function of k. For example, we consider the energy of a band in one dimension using the tight-binding method. From Eq. (5.13), we obtain the expression for energy of a one-dimensional crystal of lattice constant a (in the x direction) as where the constants were defined in Eq. (5.13). When there is an external electric field E parallel to the linear chain of the lattice (in the x direction), we have from Eq. (8.61) k \u03b5 \u00f0 \u00de = \u03b5a \u2212 \u03b2 \u2212 2\u03b3 cos ka, (8.77) and From Eqs. (8.42), (8.77), and (8.79), we have \u0127 _k = \u2212eE k = \u2212eEt/\u0127: (8.78) (8.79) dr dt = _r = 1 \u0127 \u2202\u03b5 \u2202t = \u2212 2a\u03b3 \u0127 sin aeEt \u0127 (8.80) : $ # Integrating with respect to t, we obtain r = 2\u03b3 eE cos aeEt \u0127 # $ : (8.81) Eq. (8.81) implies that when the external electric field is very large, the Bloch electrons would oscillate around a mean position. Instead of being a good conductor, a metal would become an insulator. However, it can be easily shown that a small amount of damping can destroy this oscillation. The existence of such Bloch oscillations was experimentally observed by Dahan et al (Ref. 3). Dahan et al. prepared ultracold cesium 0.0.5 \u22122 \u22121 0 Fta( k) 1 2 FIGURE 8.3 Bloch oscillations of cesium atoms in an optical periodic potential driven by a constant external force observed experimentally. The negative values of Fta were measured by changing the sign of F. Reproduced from Dahan et al.3 with the permission of the American Physical Society. 8.8 Holes 255 t atoms in the ground-state energy band of the potential induced by an optical standing wave. The periodic potential results from the light shift of the ground state of cesium atoms illuminated by a laser standing wave. The constant external force was mimicked by introducing a tunable frebetween two counter propagating laser waves creating the optical potenquency difference \u03b4v \u03bb/2. tial. The reference frame in which the optical potential is stationary moves with velocity \u03b4v, a constant inertial force F = \u2212ma = \u2212m\u03bb d /2 t For a linear variation of time in \u03b4v is exerted on the atoms in this frame. They observed Bloch oscillations of the atoms driven by this constant inertial force shown in Figure 8.3. The recoil velocity of the cesium atom was 0.35 cm/s, and the acceleration was \u00b10:85 m/s2: Figure 8.3 shows the results for potential depth U0 = 4:4 ER: t \u00de \u00f0 dt \u03b4v \u00f0 \u00f0 \u00de \u00de \u00f0 \u00de t 8.8 HOLES We can write the expression for the contribution of all the electrons in a given band to the current density as j = \u2212e Zoccupied dk 4\u03c03 v k : \u00de \u00f0 Because the current in a completely filled band is zero, we can also write j = e Zunoccupied dk 4\u03c03 v k : \u00de \u00f0 If only one electron is missing from the occupied band, the current is j = ev : k \u00de \u00f0 (8.82) (8.83) (8.84) The absence of an electron in state k from an otherwise filled band is known as a \u201chole\u201d in state k. There are a couple of important physical properties to note about a hole: 1. The hole is a positive charge, as evidenced by Eq. (8.84). 2. The velocity of the \u201cwave packet\u201d of the hole will be the same as the velocity of the electrons in either side of the hole, i.e., vnk = 1 \u0127 k \u2202\u03b5n\u00f0 \u2202k \u00de : (8", ".85) There are several ways in which a hole can be formed in a filled band. An electron can absorb a photon and can be excited to the empty band above the filled band. Similarly, an electron can be thermally excited at room temperature to the empty band above the filled band provided the energy gap is small, as is the case for intrinsic semiconductors like Si or Ge. In fact, \u201choles\u201d play a significant role in semiconductors, which is the topic of Chapter 9. The formation of a hole by the absorption of a photon by an electron in state k to an empty band is shown in Figure 8.4. 256 CHAPTER 8 Dynamics of Bloch Electrons \u03b5 Empty band Because the unoccupied levels normally lie near the top of the band and the band energy has its maximum value at km, we can k \u03b5n\u00f0 \u00de expand \u03b5n\u00f0 about km assuming that k is sufficiently close to km: We obtain (assuming that km is a point of high symmetry) k \u00de Electron k \u03b5n\u00f0 \u00de \u2248 \u03b5n\u00f0 km\u00de + a k \u2212 km\u00de \u00f0 \u2212 b k \u2212 km\u00de \u00f0 2 + \u2026 (8.86) However, the linear term on the right side will is a maximum and the vanish because \u03b5n\u00f0 coefficient of b will be negative. From Eqs. (8.85) and (8.86), we obtain km\u00de \u2192 k Hole and k vn\u00f0 \u00de = \u2212 bk 2 \u0127 d dt k vn\u00f0 \u00de = a = \u2212 b _k: 2 \u0127 (8.87) (8.88) Filled band FIGURE 8.4 An electron of vector k absorbs a photon and is excited to the empty band, leaving behind a positively charged \u201chole\u201d of the same vector k in the filled band. Comparing Eq. (8.88) with Eq. (8.34) for the effective mass tensor, we obtain \u0127M\u22121 n \u00f0 k \u00de. _k = \u2212 2 \u0127 b _k = a: (8.89) Thus, the inverse mass tensor is negative if k is near a band maximum. Because a negative mass is physically unacceptable, a hole can be considered as a positive charge with a positive effective mass. Thus, the general expression of the equation of motion can be written as From Eqs. (8.74) and (8.90), we obtain k Mn\u00f0 \u00de. a = \u0127 _k: Mn\u00f0 k \u00de. a = \u2213e E + \u00f0 1 c k vn\u00f0 \u00de \u00d7 B : \u00de (8.90) (8.91) It is easier to understand the motion of the electrons and holes if we consider only the electric field E. In that case, Eq. (8.91) can be rewritten as k Mn\u00f0 \u00de. a = \u2213eE: (8.92) 8.8 Holes 257 As shown in Figure 8.5, the symbol \u2218 indicates the position of the hole in k space. However, the behavior of the hole, which is governed by its k vector, is marked \u2297: In an electric field, the electron., respectively, always move rigidly like and hole states in the same band, which are marked beads on a string. However, dk/dt for these two particles, which have opposite charges, are of opposite signs. For the holes, dk/dt is always positive, and the points \u2297 move in the same direction as the moves opposite applied electric field. Because the electrons have negative charge, the electron state to the direction of the electric field. and. \u2218 \u00f0 \u00de \u00de \u00f0 \u00f0 \u00de The positive and negative effective mass (in units of mass of cesium atom) as a function of the potential depth, when k = 0 and k = K, respectively (in a simulated periodic potential), were demonstrated in the experiment described previously. The experimental results are shown in Figure 8.6. It is often convenient to describe the physical behavior in terms of the hole states, especially when a Brillouin zone is nearly full. An example is shown in Figure 8.7, which shows a plane 1 \u22122 0 FIGURE 8.6 k = Q k =K 2 U0[ER] 4 6 Effective masses m% for k = 0 and k = K (in units of cesium atomic mass) versus potential depth U0: \u03c0 /a \u2192 k \u2212\u03c0 /a \u2192 \u2212k FIGURE 8.5. \u00de \u00f0 and hole The electrons states always move together under the action of an electric field E, as shown by arrows. The behavior of the holes is governed by its k vector and is marked \u2297. \u2218 \u00de \u00f0 258 CHAPTER", " 8 Dynamics of Bloch Electrons + + + + FIGURE 8.7 FIGURE 8.8 Occupied states in the first zone shown by a plane section perpendicular to (001). The circular areas of hole states at the zone corners (in two dimensions). They are spheres in three dimensions. section perpendicular to (001) in a cubic zone structure where the occupied states are only in the first zone. We note that the empty states of the nearly filled zone are in the corners of the zone. These empty states form spheres of hole states in the repeated zone scheme in the three-dimensional k space, a two-dimensional equivalent of which is shown in Figure 8.8. It is much easier to deal with these spheres of positive charges (holes) than the electrons in the Brillouin zone. The Fermi surface of divalent metals has small sections of holes in the first Brillouin zone as a consequence of electrons spilling over into the second Brillouin zone. The electrons in the second zone are ellipses of occupied states (ellipsoids in three dimensions), as shown in Figure 8.9. FIGURE 8.9 The electrons of the second zone in the Fermi surface of a divalent metal in the repeated zone scheme. The major portion of the first zone is filled with electrons, but these states are replaced by the hole states. 8.9 ZENER BREAKDOWN (APPROXIMATE METHOD) We will consider the Zener breakdown (Ref. 12) in an electric field in two alternate ways. First, we will consider a simple and traditional method that is very controversial because the fundamental postulate on which the Bloch functions and the concept of the Brillouin zones have been built\u2014 i.e., the potential is periodic and the crystal has symmetry in the sense that one can consider the opposite ends to be equivalent\u2014is lost in an external electric field. In fact, as we have seen, the wave vector k becomes time dependent, and the Brillouin zone boundary keeps moving with time. Nevertheless, we will first discuss the simple case neglecting these objections. Later, we will discuss the more rigorous theory of Zener breakdown (Ref. 12). 8.9 Zener Breakdown (Approximate Method) 259 When we consider the motion of a Bloch electron (in one dimension) in an external electric field E along the x-axis, there are two ways in which we can represent the motion of the electron. The motion along the k space is shown in Figure 8.10 in the repeated zone scheme. These are the familiar Bloch oscillations in which the electron travels from 0 to A and then to B and C, and so on. The zone boundaries A and C are equivalent, and one can describe the oscillatory motion of the electron by stating that the electron has jumped from C to A. However, the path of the electron in the external electric field in real space (in one dimension) shown in Figure 8.11 appears very different. The electron slows down as it moves from 0 to A, where it has a Bragg reflection. It cannot go forward because there is an energy gap at A, and the electron is forbidden to move in that region. The electron reverses direction until it reaches B and then accelerates again until it reaches the zone boundary at C. \u03b5 (k) E A\u2032 A k B 0 FIGURE 8.10 Electron trajectory in k space. z B 0 FIGURE 8.11 In addition, in a strong electric field E, the bands are tilted as shown in Figure 8.12. An electron moving from P to Q will be reflected back into the band, or it can move from Q to R by crossing the energy gap \u03b5g if the electric field is strong enough to satisfy the condition \u03b5g = eEd, where d = QR. Thus, we obtain C d \u2248 \u03b5g eE : (8.93) It may be noted that QR is a forbidden region, and the electron has to tunnel through this region. This tunneling problem was originally solved by Zener by using WKB approximation. However, this is a semiclassical derivation with a lot of controversies because the periodicity of the Bloch functions is lost in an electric field. We will first outline a brief derivation of tunneling \u2192 eE \u03b5 y g r e n E P FIGURE 8.12 R Q x C A x Path of electron in real space in an electric field E along the x axis. The energy bands in the \u03b5 \u2212 x diagram are tilted in an applied electric field. 260 CHAPTER 8 Dynamics of Bloch Electrons using the WKB approximation and then derive the effect of electric field on the Bloch functions for derivation of a more rigorous formula. The tunneling of an electron in a square potential barrier is shown in Figure 8.13. This is solved by the WKB approximation.4 The approximate solution of the Schrodin", "- ger equation in the WKB approximation is V 0 d x E FIGURE 8.13 Square potential barrier. Penetration of a barrier by a wave function \u03c8(x) when V > d: = Ae \u03c8 x \u00de \u00f0 i/\u0127 x R dx p 2m \u00f0 \u03b5\u2212V \u00de : \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 (8.94) If V \u2212 \u03b5 = u, where u is positive, an electron wave function that starts on the left side of Figure 8.13 and travels distance d through the gap will have a transmission coefficient7 T = e\u22122d p 2mu/\u01272 : \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 (8.95) However, in the case of a Bloch electron moving on an electric field, if \u03b5v and \u03b5c are the energies of the valence and conduction bands, respectively, while \u03b5 \u2212 \u03b5v is negative in the energy gap, \u03b5c \u2212 \u03b5 = 0 = \u03b5v \u2212 \u03b5 at the zone boundaries, and once the electron tunnels to the conduction band, \u03b5v \u2212 \u03b5 is positive. If we include these specific conditions for tunneling through the bands, a rough estimate of the Zener tunneling problem can be made by rewriting the WKB approximation, and the probability for an electron to tunnel from one band to another is of the form d 2i/\u0127 dx T = e 0 R p p \u00de\u00f0 2m \u03b5c\u2212\u03b5 \u00f0 \u03b5v\u2212\u03b5 \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 : (8.96) Eq. (8.96) can be rewritten in the alternate form (in a very rough approximation) From Eqs. (8.87) and (8.97), we obtain the expression for tunneling probability as \u22122d T \u2248 e 2m\u03b5g \u01272 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 : q \u2212 T \u2248 e 2\u03b5g eE 2m\u03b5g \u01272 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 : q (8.97) (8.98) The probability of the electron to tunnel through the barrier obtained in Eq. (8.98) is a very rough estimate of the tunneling problem. 8.10 Rigorous Calculation of Zener Tunneling 261 8.10 RIGOROUS CALCULATION OF ZENER TUNNELING We expanded \u03c8 r, t \u00f0 \u00de \u2014in Eqs. (8.53) and (8.58)\u2014in terms of the Houston states ai\u00f0 = \u2211 i \u03d5\u2032i\u00f0 r, t r, t \u03c8 \u00de \u00de \u00f0 \u00de t, which can be rewritten in the alternate form from Eq. (8.58) r, t \u03c8 \u00f0 = \u2211 n\u2032k\u2032 t an\u2032k\u2032\u00f0 \u00de \u00de e\u2212ieA.r/\u0127c\u03d5n\u2032k\u2032 r, \u00de t \u00f0 \u00de\u00f0 H\u03c8 r, t = \u2211 n\u2032k\u2032 \u00de \u00f0 t an\u2032k\u2032\u00f0 \u03b5n\u2032\u00f0 \u00de k\u2032 t \u00f0 \u00de\u00de ei \u00f0 k\u2032\u2212eA/\u0127c \u00de.rUn\u2032k\u2032 r : \u00de t \u00f0 \u00de\u00f0 (8.99) (8.100) (8.101) Here, we replaced ai\u00f0 Houston functions as t \u00de by an\u2032k\u2032\u00f0 t \u00de and replaced the summation over i by n\u2032k\u2032 and expressed the We also obtain \u03d5nk r \u00de t \u00f0 \u00de\u00f0 = eik.rUnk r : \u00de t \u00f0 \u00de\u00f0 i\u0127 \u2202 \u2202t r, t \u03c8 \u00f0 = i\u0127\u2211 n\u2032k", "\u2032 \u00de t \u2202an\u2032k\u2032\u00f0 \u2202t ei k\u2032\u2212 \u00f0 \u00de eA \u0127c\u00de.rUn\u2032k\u2032 t \u00f0 \u00de\u00f0 r \u00de + i\u0127\u2211 n\u2032k\u2032 t an\u2032k\u2032\u00f0 \u00de ei k\u2032\u2212 \u00f0 eA \u0127c\u00de.r \u2202 \u2202t r Un\u2032k\u2032\u00f0 \u00de : Here, we used the condition that k\u2032\u2212 eA \u0127c is time independent from Eq. (8.59). Further, \u2202 \u2202t Unk r \u00de t \u00f0 \u00de\u00f0 = \u2202 \u2202kx Unk dkx dt = \u2212 eE \u0127 \u2202 \u2202kx Unk: (8.102) (8.103) (8.104) Substituting Eqs. (8.101) through (8.103) in Eq. (8.51), multiplying both sides by e\u2212i \u00f0 using E = ^iE the Bloch functions (Problem 8.9), we obtain, \u00de, integrating over the volume of the crystal, and using the orthonormal conditions of k\u2212 r \u00de \u00f0 t eA \u0127c\u00de.rU%nk\u00f0 where k, t \u03b5n\u00f0 ank \u00de t \u00f0 \u00de = i\u0127 \u2202ank \u2202t t \u00f0 \u00de \u2212 eE t \u2211 n\u2032 \u00de \u00f0 k Ann\u2032\u00f0 \u00f0 t \u00de\u00de an\u2032k, t \u00f0 \u00de k Ann\u2032\u00f0 \u00f0 t \u00de\u00de & \u2212i Z U%nk t \u00f0 \u00de \u2202 \u2202kx Un\u2032k t \u00f0 \u00de d\u03c4: (8.105) (8.106) From Eq. (8.105), it follows that the coefficients ank \u00f0 same k. Therefore, for a given k, we can write t \u00de are coupled only to an\u2032k t \u00f0 \u00de, i.e., coupled to the i \u0127 \u2212 e t 0 R \u03b5n t\u2032 k \u00f0 dt\u2032 \u00de \u00f0 : ank t \u00f0 \u00de = \u03b1n\u00f0 t \u00de (8.107) 262 CHAPTER 8 Dynamics of Bloch Electrons Substituting Eq. (8.107) in Eq. (8.105), we obtain t _\u03b1 n\u00f0 \u00de = t ieE \u00f0 \u0127 \u00de \u2211 n\u2032 t \u03b1n\u2032\u00f0 k Ann\u2032\u00f0 t \u00f0 \u00de \u00de\u00de exp t Z 0 i \u0127 2 4 k \u03b5n\u00f0 \u00bd \u00f0 t\u2032 \u00de\u00de k \u2212 \u03b5n\u2032\u00f0 \u00f0 t\u2032 \u00de\u00de$ dt\u2032 : 3 (8.108) 5 This general expression can be used for derivation of Wannier\u2013Stark ladders for time-dependent E as well as for Zener tunneling between the valence and conduction bands. At t = 0, before the field E is turned on, k Bloch wave in band n with ; then at t = 0, t = 0 \u00f0 \u00de \u00f0, the electron state can be described by a \u00de \u03b1n\u2032 = \u03b4nn\u2032: (8.109) After a time T = \u2212 \u0127K time so that \u03b1n\u2032 \u226a 1, n\u2032 \u2260 n, we can substitute Eq. (8.109) in Eq. (8.108) to obtain eE, where T is the period of one Bloch oscillation, which is a sufficiently short \u03b1n\u2032\u00f0 t \u00de = t Z 0 t\u2032 \u00de ieE \u00f0 \u0127 k An\u2032n\u00f0 \u00f0 t\u2032 \u00de\u00de exp i \u0127 2 t\u2032 Z 0 k \u03b5n\u2032\u00f0 \u00bd \u00f0 t\u2033 \u00de\u00de k \u2212 \u03b5n\u00f0 \u00f0 t\u2033 \u00de\u00de$ dt\u2033 dt\u2032: 3 (8.110) 4 We consider the case in which E is constant and is in the direction of the reciprocal-lattice vector K (which is along the x axis) and use \u0127 _k = \u2212eE to change the variables 5 dt = \u2212 \u0127 eE dk: (8.111) From Eqs. (8.110) and (8.111), it can be shown that the transmission probability per period T, Pnn\u2032 = T \u03b1n\u2032\u00f0 j \u00dej 2 = ( ( ( ( ( ( K/2 Z \u2212K/2 kx Z 0 k An\u2032n\u00f0 \u00de exp \u2212 i eE 2 4 k\u2032 x, k\u22a5\u00de \u03b5n\u2032\u00f0 \u00bd \u2212 \u03b5n\u00f0 k\u2032 x", ", k\u22a5\u00de$ dk\u2032 x dkx 3 5 2 : (8.112) ( ( ( ( ( ( Any calculation of Pnn\u2032 requires actual band calculations for the particular metal or the semiconduc, and evaluation of the integrals requires the use of a computer. However, in tor to calculate An\u2032n\u00f0 what follows, we will make several drastic approximations to derive an expression for the Zener tunneling between a valence band and a conduction band. In the process, we will explain why the Zener tunneling between the two bands is much easier for semiconductors than for metals. k \u00de To be able to consider Zener tunneling, we consider two parabolic bands that are of the form and \u03b51k = \u03b5v \u2212 \u01272k2 2m%v \u03b52k = \u03b5c + \u01272k2 2m%c : (8.113) (8.114) 8.10 Rigorous Calculation of Zener Tunneling 263 Thus, \u03b52k \u2212 \u03b51k = \u03b5g + \u01272k2 2m% where \u03b5g = \u03b5c \u2212 \u03b5v, and m\u2217 is the reduced mass defined by, 1 m% = 1 m%v + 1 m%c : With these conditions, Eq. (8.105) can be rewritten as from which we obtain \u03b51k t \u00f0 \u00de a1k t \u00f0 \u00de = i\u0127 \u2202a1k \u2202t t \u00f0 \u00de, a1k t \u00f0 \u00de = exp \u2212 2 i \u0127 t Z 0 dt\u2032\u03b51k\u00f0 t\u2032 : 3 \u00de (8.115) (8.116) (8.117) (8.118) If the time interval is short such that \u03b1n\u2032 \u226a 1, n\u2032 \u2260 n, we can substitute Eq. (8.118) in Eq. (8.108) to obtain an expression for the rate of tunneling from band 1 to band 2 after time t as 4 5 t \u03b12\u00f0 \u00de = t Z 0 ieE \u0127 k A21\u00f0 \u00f0 t\u2032 exp \u00de\u00de i \u0127 2 t\u2032 Z 0 \u03b52k \u00bd t\u2033 \u00f0 \u00de \u2212 \u03b51k dt\u2033 t\u2033 \u00de$ \u00f0 dt\u2032: 3 (8.119) 4 For a linear lattice, it can be shown that apart from the oscillatory terms in k, iA21 \u2248 a, where a is the lattice constant. Further, if \u03c4 is the time needed for k to move by a reciprocal lattice vector K = 2\u03c0/a, 5 \u03c4 \u03b12\u00f0 \u00de \u2248 a 2\u03c0/a Z 0 dk exp k Z 0 \u2212i eE! dk\u2032 \u00f0 \u03b52k\u2032 \u2212 \u03b51k\u2032\u00de : \" (8.120) Here, we used the relation \u0127 _k = \u2212eE to change the variables dt = \u2212 \u0127 (8.120), we obtain eE dk. From Eqs. (8.115) and \u03c4 \u03b12\u00f0 \u00de \u2248 a 2\u03c0/a Z 0 dk exp \u2212i eE 2 k Z 0 dk\u2032 \u03b5g + \u01272k\u20322 % 2m% & : 3 (8.121) The integral is impossible to perform exactly, but an approximate result achieved by using the method of steepest descent yields the expression for the rate of tunneling: 4 5 \u03c4 \u03b12\u00f0 \u00de \u2248 a exp \u2212 \" 2\u03b53/2 g 3eE 2m% \u01272 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 : # r (8.122) 264 CHAPTER 8 Dynamics of Bloch Electrons 2, is much smaller in metals than in The probability of tunneling, which is given by semiconductors because both the band gaps and the effective masses are much smaller for semiconductors. \u03b12\u00f0 \u00dej \u03c4 j 8.11 ELECTRON\u2013PHONON INTERACTION The electron\u2013phonon interaction process is basically the absorption (annihilation) or emission of a k \u00b1 q, \u03c3 >: Here, phonon \u03c3 is the spin index and \u03bb = x, y, z directions. These two processes are shown as (a) and (b) in Figure 8.14 (the spin index \u03c3 and \u03bb are not shown). with a simultaneous change of the electron states from k, \u03c3 > to qb) (d) k k + q (f) k \u2212 q q k (a) k \u2212 q k q k k (c) k \u2212 q q (e)", " q q k q k k\u2032 k + q k\u2032 \u2212 q q (g) FIGURE 8.14 Graphs for various types of electron\u2013phonon interaction. The Umklapp process is not shown here. 8.11 Electron\u2013Phonon Interaction 265 Process (a) is called phonon emission, and process (b) is called phonon absorption. In processes (c) and (d), the time axis runs from left to right. The electrons are assumed to run backward in time, and holes run forward in time. Process (c) describes the recombination of an electron-hole pair where a phonon is emitted. Process (d) describes the generation of an electron-hole pair by a phonon. One can describe processes (a) through (d) by using first-order perturbation theory. The conservation laws for the sum of the wave vectors (momentum) and energy are obeyed. Processes (e) through (g) are due to contributions of perturbation calculations of higher order. Process (e) describes the emission and reabsorption of a virtual phonon. Process (f) describes the emission and reabsorption of a virtual electron-hole pair. Process (g) describes the electron\u2013electron interaction of a virtual phonon. In these graphs, the intermediate states are not stationary states of the system. The laws of conservation of energy do not apply to these intermediate states because these processes occur in a very short time. The uncertainty relation between energy and time holds for these virtual states. The electron\u2013electron interaction by exchange of virtual phonons will be neglected in the following discussions. The Hamiltonian of a crystal can be written as Hc = Hel + Hion + Hel\u2212ion: (8.123) Here, the contribution due to exchange is neglected. Using the Born\u2013Oppenheimer (adiabatic) approximation discussed in Chapter 7, we can separate the movement of the electrons in a stationary lattice and the movement of the ions in a uniform space of electrons. The Hamiltonian Hion is considered separately and is used to study lattice dynamics, which we considered in Chapter 2. We write Hel\u2212ion = H0 el\u2212ion + Hel\u2212ph, (8.124) where H0 model. el\u2212ion describes the electron interaction with the periodic potential in the one-electron band We express He = Hel + H0 el\u2212ion: (8.125) Thus, we can write the Schrodinger equation for the electrons\u2014from Eqs. (8.123) through (8.125)\u2014as We also note that we can write H\u03c8 = He + Hel\u2212ph\u00de \u00f0 \u03c8 = E\u03c8: Hel\u2212ion = \u2211 j,i,\u03b1 V rj \u2212 Ri\u03b1\u00f0 t \u00f0 \u00de\u00de, (8.126) (8.127) (see Figure 2.1 in Chapter 2). Here, where rj is the position of an electron, and Ri\u03b1\u00f0 the position vector of the ion is the sum of the equilibrium position of the \u03b1th ion in the ith Wigner\u2013 Seitz cell, and u\u03b1i\u00f0 = Ri + d\u03b1 + u\u03b1i\u00f0 is the instantaneous deviation of the \u03b1th ion from the equilibrium position. \u00de \u00de \u00de t t t 266 CHAPTER 8 Dynamics of Bloch Electrons We assume that the interaction potential depends only on the electron\u2013ion separation (Nordheim\u2019s rigid ion model). We can expand the potential V\u03b1\u00f0 rj \u2212 Ri\u03b1 \u2212 u\u03b1i\u00de \u2248 V\u03b1\u00f0 rj \u2212 Ri\u03b1\u00de \u2212 \u2211 \u03b1ij u\u03b1i. \u2207V\u03b1\u00f0 rj \u2212 Ri\u03b1\u00de : From Eqs. (8.22), (8.127), and (8.128), we obtain Hel\u2212ph = \u2212\u2211 \u03b1ij u\u03b1i. \u2207V\u03b1\u00f0 rj \u2212 Ri\u03b1\u00de : (8.128) (8.129) In Eq. (2.159), we also derived (note that we have changed the notation from k to q because we will use k for electrons and generalized to a lattice with a basis, and from n to \u03b1) u\u03b1i = 1 M\u03b1Np \u2211 q\u03bb q ^\u2208\u03b1\u03bb\u00f0 \u00de ^Qq\u03bbeiq.Ri, where the expression for the normal coordinate Qq\u03bb was derived in Eq. (2.182) as \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 % From Eqs. (8.129) through (8.131), we obtain ^Qq\u03bb = \u0127 2\u03c9q\u03bb 1 2 \u00f0 & ^aq\u03bb + ^a\u2020 \u2212q\u03bb\u00de : Hel\u2212ph = \u2212", "\u2211 \u03b1ij 1 M\u03b1Np \u2211 q\u03bb ^Qq\u03bbeiq.Ri ^\u2208\u03b1\u03bb\u00f0 q \u00de. \u2207V\u03b1\u00f0 rj \u2212 Ri\u03b1\u00de : (8.130) (8.131) (8.132) We will now convert quantum mechanical equations for He from the r-representation into the occupation number representation. We assume that the Hamiltonian is a sum of one-electron operators \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 ^H = \u2211 i ^H ri, si\u00de \u00f0 : We recall that, for fermions, we derived in Chapter 2 (Eq. 2.110) that ^c\u2020 k^ck = nk (8.133) (8.134) (In this chapter, we use ^c\u2020 from the phonon operators.) k and ^ck for creation and annihilation operators for electrons to distinguish Similarly, from Eqs. (2.99) through (2.101) and (2.109), we derived ck, c\u2020 k$ + = \u03b4k, k\u2032, \u00bd n1, n2, \u2026, nk \u2026, > = 1\u2212 nk c\u2020 kj \u22121 \u00f0 \u00de m j n1, n2, \u2026, nk + 1, \u2026, > and p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 (8.135) (8.136) n1, n2, \u2026, nk \u2026, > = nkp ckj \u22121 \u00f0 \u00de m j n1, n2, \u2026, nk \u2212 1, \u2026, >, (8.137) \ufb03\ufb03\ufb03\ufb03\ufb03 8.11 Electron\u2013Phonon Interaction 267 where m = \u2211 j<k ni: (8.138) We also showed in Chapter 7 (Eq. 7.18) that the wave function is a Slater determinant, which can be written in the form \u03a8 r1s1, r2s2, \u2026, rNsN\u00de \u00f0 = 1 N!p \u22121 \u2211 n \u00f0 n\u03c8 n1\u00f0 \u00de r1s1\u00de \u03c8 n2\u00f0 r2s2\u00de \u2026 \u03c8 nN\u00f0, rNsN\u00de (8.139) where the sum n is over all permutations n of 1 \u2026 N. We can rewrite Eq. (8.139) in the alternate form \ufb03\ufb03\ufb03\ufb03\ufb03 \u03a8 = 1 N!p \u2211 P P \u00f0 \u22121 P\u03c8 \u03b1\u00f0 \u00de r1s1\u00de \u03c8 \u03b2\u00f0 r2s2\u00de \u2026 \u03c8 \u03c9\u00f0 rNsN\u00de, (8.140) \ufb03\ufb03\ufb03\ufb03\ufb03 where the sum P is over all permutations of the indices \u03b1, \u03b2, \u2026, etc. From Eq. (8.140), we can write H\u03a8 r1s1, r2s2, \u2026, rNsN\u00de \u00f0 = 1 N!p H\u2211 P P \u00f0 \u22121 P\u03c8 \u03b1\u00f0 \u00de r1s1\u00de \u03c8 \u03b2\u00f0 r2s2\u00de \u2026 \u03c8 \u03c9\u00f0 rNsN\u00de : (8.141) \ufb03\ufb03\ufb03\ufb03\ufb03 From Eqs. (8.133) and (8.141), we obtain H\u03a8 = 1 N!p \u2211 i \u2211 P P \u00f0 \u22121 P\u03c8 \u03b1\u00f0 \u00de r1s1\u00de \u03c8 \u03b2\u00f0 r2s2\u00de \u2026 H risi\u00de \u03c8 \u03bb\u00f0 risi\u00de \u00f0 \u2026 \u03c8 \u03c9\u00f0 rNsN\u00de : (8.142) If the Hamiltonian does not involve the spin explicitly (Eq. 7.20), \ufb03\ufb03\ufb03\ufb03\ufb03 \u03c8 %\u03bb\u00f0 risi\u00de H ri\u00de \u03c8 \u03bb\u2032\u00f0 ris\u2032 i\u00de \u00f0 = \u03d5%\u03bb\u00f0 ri\u00de H ri\u00de \u00f0 \u03d5\u03bb\u2032\u00f0 Z Z \u03b4si, s\u2032 i ri\u00de because We can write the Hamiltonian as a sum of operators on single particles, \u03c8 \u03bb\u00f0 rjsj\u00de = \u03d5\u03bb\u00f0 rj\u00de \u03c7\u03bb\u00f0 sj\u00de : H = \u2211 i H", " risi\u00de & \u00f0 \u2211 i Hi: (8.143) (8.144) (8.145) Because \u03c8 \u03bb forms a complete set, we can rewrite Eq. (8.145) in a more convenient way by writing ^H = \u2211 i\u03bb\u03bb\u2032j \u03c8 \u03bb\u2032\u00f0 i \u00de i >< \u03c8 \u03bb\u2032\u00f0 \u00dej ^Hij i \u03c8 \u03bb\u00f0 \u00de > < \u03c8 \u03bb\u00f0 i \u00dej j : The one-particle equivalent of Eq. (8.146) is risi\u00de risi\u00de \u03c8 \u03bb\u00f0 H \u00f0 = \u2211 \u03bb\u2032 \u03c8 \u03bb\u2032\u00f0 risi\u00de < \u03bb\u2032 Hij j \u03bb >, (8.146) (8.147) 268 CHAPTER 8 Dynamics of Bloch Electrons which has i-independent matrix elements. Here, the spin index s is dropped from \u03bb because the Hamiltonian H does not explicitly contain spin terms, whereas the summation over \u03bb\u2032 explicitly contains spin summation. We define the operator \u00de The operator ^A\u03bb\u2032\u03bb searches for each electron (one at a time) in state \u03c8 \u03bb and moves it to \u03c8 \u03bb\u2032. From ^A\u03bb\u2032\u03bb = \u2211 i j i \u03c8 \u03bb\u2032\u00f0 >< \u03c8 \u03bb\u00f0 i : \u00dej (8.148) Eqs. (8.146) and (8.148), we obtain ^H = \u2211 \u03bb\u03bb\u2032 ^A\u03bb\u2032\u03bb < \u03c8 \u03bb\u2032\u00f0 1 ^H1j \u03c8 \u03bb\u00f0 1 \u00de \u00dej >, (8.149) where the label 1 is used instead of i because the matrix elements of a one-particle operator do not depend on which particle is involved. We first consider a wave function \u03a8 and wish to evaluate (8.150) (8.151) N \u2211 i = 1 < \u03a8aj \u03c8 \u03bb\u2032\u00f0 i \u00de i >< \u03c8 \u03bb\u00f0 \u00dej \u03a8b >: We consider one term \u00de 1 1 \u03c8 \u03bb\u2032\u00f0 >< \u03c8 \u03bb\u00f0 < \u03a8aj \u03a8b >, \u03c8 \u03bb is unoccupied and \u03c8 \u03bb\u2032 is occupied while in j \u03a8b >: \u00dej \u03a8a >, \u03c8 \u03bb This term is nonzero only if in is unoccupied and \u03c8 \u03bb\u2032 is occupied, and otherwise \u03a8a and \u03a8b are identical: Thus, one has to permute m\u03bb\u2032. Similarly, \u03c8 \u03bb\u2032 past all the states in the ordering to obtain \u22121 1 \u03c8 \u03bb\u2032\u00f0 \u00de \u00f0 j m\u03bb to permute \u03c8 \u03bb past all the states below it to obtain >. Here, \u22121 one obtains a factor >, and one obtains a factor of 1 \u00de j \u03c8 \u03bb\u00f0 j \u00de \u00f0 \u00de Thus, we obtain m\u03bb = \u2211 i<\u03bb ni and m\u03bb\u2032 = \u2211 i<\u03bb\u2032 ni: N \u2211 i = 1 < \u03a8aj \u03c8 \u03bb\u2032\u00f0 i \u00de >< \u03c8 \u03bb\u00f0 i \u00dej \u03a8b > = \u22121 \u00de \u00f0 m\u03bb+m\u03bb\u2032, (8.152) (8.153) if it is not zero. Thus, in the occupation number representation, ^A\u03bb\u2032\u03bbj n1, n2, n\u03bb\u2032\u2026 n\u03bb\u2032\u2026 > = \u22121 m\u03bb\u2032 \u00de \u00f0 \u00f0 \u22121 \u00de m\u03bb n1, n2, \u2026, n\u03bb\u2032+1 \u2026 n\u03bb\u22121 \u2026 >: (8.154) j n\u03bb 1 \u2212 n\u03bb\u2032\u00de \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 p From Eqs. (8.149) and (8.154), we obtain (Problem 8.11) ^H n1, n2, \u2026, n1, n2, \u2026, n\u03bb\u2032, \u2026, n\u03bb, \u2026 > j = \u2211 \u03bb, \u03bb\u2032 p n\u03bb\u00f0 1 \u2212 n\u03bb\u2032\u00de \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u22121 \u00de m\u03bb+m\u2032 \u03bb\u2032", "< \u03bb\u2032 j \u03bb > H1j j n1, n2, \u2026, n\u03bb\u2032+1, \u2026, n\u03bb\u22121, \u2026 > if \u03bb \u2260 \u03bb\u2032, (8.155) = \u2211 \u03bb n\u03bb < \u03bb \u03bb > H1j j j n1, n2, \u2026, n\u03bb\u2026 >, if \u03bb = \u03bb\u2032: (8.156) 8.11 Electron\u2013Phonon Interaction 269 Here, the terms \u2211 either zero or 1. 1 \u2212 n\u03bb\u00de n\u03bb\u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 From Eqs. (8.152) through (8.154), the operator p and n\u03bb occur due to the property of the fermions that n\u03bb can be ^A\u03bb\u2032\u03bb = ^c\u2020 \u03bb\u2032 ^c\u03bb, (8.157) where the creation and annihilation operators for fermions were defined in Eqs. (8.136) and (8.137). From Eqs. (8.149), (8.153), and (8.155) through (8.157), we obtain by replacing \u03bb and \u03bb\u2032 by ks and k\u2032s, H n1, n2, \u2026, nk \u2026 > = \u2211 kk\u2032s j < k\u2032s r H1\u00f0 j \u00dej ks > c\u2020 k\u2032scksj n1, n2, \u2026, nk\u2032 \u2026 nk \u2026 >: (8.158) The Hamiltonian H in the occupation number representation becomes where H = \u2211 k,k\u2032,s < k\u2032s r H1\u00f0 j \u00dej ks > c\u2020 k\u2032scks, <k\u2032s r H1\u00f0 j ks > = \u00dej Z \u03c8 %k\u2032s\u00f0 r H1\u00f0 \u00de r r \u03c8 ks\u00f0 \u00de \u00de dr: (8.159) (8.160) Here, we dropped the index i in ri: We note that because spin is conserved in k k\u2032 transition, we have to carry out only one spin summation. The matrix elements on the right side of Eq. (8.160) are formed from the Bloch functions.! To calculate Hel\u2212ph from Eqs. (8.132) and (8.160), we expand V\u03b1\u00f0 V\u03b1\u00f0 r \u2212 Ri\u00de \u00deV\u03b1\u03ba! ei \u03ba!. r\u2212Ri \u00f0, = \u2211 \u03ba! r \u2212 Ri\u03b1\u00de in a Fourier series from which we obtain < k\u2032s \u2207V\u03b1j j ks > = \u2211 \u03ba! e\u2212i \u03ba!. R!iV\u03b1\u03ba! i \u03ba! < k\u2032s ei \u03ba!. r! j ks >: j From Eqs. (8.132) and (8.162), the sum over i contains the factor q!\u2212 \u03ba!\u00de ei \u00f0 \u2211 i. R!i = N \u2211 K!l \u03b4 : \u03ba!, q!+K!l (8.161) (8.162) (8.163) Thus, in the sum over \u03ba! in Eq. (8.162), only the term \u03ba! = q! + K!l survives. If we write the Bloch functions (note that we have also used bold letters as vectors) in Eq. (8.162), the & integrand I contains the term eik.runk\u00f0 k> r \u00de j I = dr ei k+q+Kl\u2212k\u2032 \u00f0 Z \u00de.ru%n\u2032k\u2032\u00f0 r unk\u00f0 \u00de r \u00de : (8.164) 270 CHAPTER 8 Dynamics of Bloch Electrons Because u%n\u2032k\u2032\u00f0 r r unk\u00f0 \u00de \u00de is a periodic function in R, I = Z dr u%n\u2032k\u2032\u00f0 r unk\u00f0 \u00de r \u00de \u03b4k\u2032, k+q+Kl: (8.165) From Eqs. (8.132), (8.159), and (8.162) through (8.165), and letting n\u2032 = n (because the electron remains in the same band n = n\u2032 when making the transition k to k\u2032 except for the U-process to be discussed later), we obtain Hel\u2212ph = \u2212 \u2211 kKlq\u03b1\u03bbs i N M\u03b1r \ufb03\ufb03\ufb03", "\ufb03\ufb03\ufb03 u%n, q+Kl+k\u00f0 V\u03b1, q+Kl \u00f0 r unk\u00f0 r \u00de. ^\u2208\u03b1\u03bb \u0127 q + Kl\u00de 2\u03c9q\u03bb & % ^c\u2020 ^a\u2020 k+q+Kl,sck,s: \u2212q\u03bb + ^aq\u03bb\u00de dr \u00de \u00f0 \u00d7 Z (8.166) We note that if k and q are added vectorially, the resultant vector can lie outside the first Brillouin zone (where k, k\u2032, q, and q\u2032 lie in the reduced zone scheme in the repeated zone scheme, so that k\u2032 = k + q + Kl: The sum over Kl in Eq. (8.164) is thus reduced to one term. If Kl = 0, the transition is called a normal process (N-process) while if Kl \u2260 0, the transition is called the Umklapp process (U-process). (Umklapp is a German word for \u201cflopping over.\u201d) \u00de To simplify further, we assume that we restrict ourselves to Bravais lattices so that there is only one atom in the Wigner\u2013Seitz cell. Thus, we can omit the index \u03b1 and \u03bb and count the different acoustic branches since the optical branches do not exist. We also restrict the derivation to the N-process such that Kl = 0. We also assume that phonons are either longitudinal or transverse so that ^\u2208\u03bb is either parallel or perpendicular to q. With these assumptions, Eq. (8.166) can be written as Hel\u2212ph = \u2212 \u2211 kq\u03bbs i \u0127N 2M\u03c9q\u03bb \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 Vqq. ^\u2208\u03bb \u00d7 dr u%n, k+q\u00f0 r unk\u00f0 \u00de r \u00de \u00f0 ^c\u2020 ^a\u2020 k+q, sck, s: \u2212q\u03bb + ^aq\u03bb\u00de (8.167) Z Because ^\u2208\u03bb \u22a5 q for transverse phonons, ^\u2208\u03bb coupled to the electrons. Eq. (8.167) can be rewritten in the alternate form r. q = 0. Thus, only longitudinal acoustic phonons are Hel\u2212ph = \u2211 kqs Bkq\u00f0 ^c\u2020 ^a\u2020 \u2212q + ^aq\u00de k+q, s^ck,s: (8.168) We can calculate the probability of transition of an electron from state | k > into state Because the spin is unchanged in this transition, it can be ignored. j k + q>. From Fermi\u2019s \u201cgolden rule\u201d (originally derived by Dirac from perturbation theory), the transition probability P i \u00f0! = f \u00de 2\u03c0 \u0127 j < f Hel\u2212phj j i > 2 \u03b4 \u00f0 j \u03b5f \u2212 \u03b5i\u00de : (8.169) Here, the initial states are nk and nk+q of the electron state and nq and n\u2212q of the phonon states involved in the transition i > and the final states j f > are characterized by the occupation numbers j nk+q, nk; nq, n\u2212q>: j (8.170) Problems 271 If we consider the transition involving the absorption of a phonon, we apply the operator ^c\u2020 to Eq. (8.170). It can be shown that (Problem 8.12) k+q^ckaq < nk+q + 1, nk\u22121; nq\u22121 ^c\u2020 k+q^ck^aqj j nk+q, nk; nq> = nknq 1 \u2212 nk+q\u00de \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 : q (8.171) The matrix elements vanish for all cases except when nk = 1; nk+q = 0. The energy relations for the absorption of a phonon are Similarly, the transitions involving emission of phonons can be written as (Problem 8.13) \u03b5f \u2212 \u03b5i = \u2212 \u0127\u03c9q: < nk+q + 1, nk\u22121; n\u2212q + 1 k+q", "^ck^a\u2020 ^c\u2020 j \u2212qj nk+q, nk; n\u2212q > = n\u2212q + 1 nk\u00f0 1\u2212nk+q\u00de \u00de \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03, q and From Eqs. (8.168) through (8.174), we obtain \u03b5f \u2212 \u03b5i = + \u0127\u03c9q: (8.172) (8.173) (8.174) k P \u00f0! k + q \u00de = 2\u03c0 \u0127 j + 2nk\u00f0 Bkqj 1 \u2212 nk+q\u00def k + q \u03b5 \u03b4 n\u2212q + 1 \u00f0 \u00de \u00f0 \u00f0 nq\u03b4 \u03b5 k \u00f0 \u00de \u2212 \u0127\u03c9q\u00de (8.175) \u2212 \u03b5 k + q \u00de + \u0127\u03c9q\u00deg : \u00f0 \u00f0 k \u00f0 \u00de \u2212 \u03b5 \u00de In this case, nk = 1, nk+q = 0, and the matrix element vanishes in all other cases. However, the factor nk is retained because if and when we consider a large number of states instead of the 1 \u2212 nk+q\u00de \u00f0 transition probability from one occupied state to one empty state, the Fermi and Bose distributions have to be used for electrons and phonons if the system is in equilibrium. PROBLEMS 8.1. Derive Eqs. (8.19) and (8.20) from Eq. (8.16). 8.2. We can write Show that Hence, show that \u2212i\u2207 + k \u00de \u00f0 unk\u00f0 r \u00de = \u2212i\u2207 + k \u00de \u00f0 e\u2212ik.r\u03c8 nk\u00f0 r : \u00de \u2212i\u2207 + k \u00de \u00f0 e\u2212k.r\u03c8 nk\u00f0 r \u00de = e\u2212k.r \u2212i\u2207 \u00de \u03c8 nk\u00f0 \u00f0 r \u00de : < unkj\u00f0 \u2212i\u2207 + k unk > = < \u03c8 nkj \u2212i\u2207 j \u00dej \u03c8 nk >: (1) (2) (3) 272 CHAPTER 8 Dynamics of Bloch Electrons 8.3. Show that the inverse effective mass tensor can be written as M\u22121 k n \u00f0 \u00bd \u00de$ij = 1 \u01272 \u22022\u03b5n \u2202ki\u2202kj : 8.4. Using Eqs. (8.35) and (8.36), show that M\u22121 n \u00f0 \u00bd k \u00de$ij = 1 m \u03b4ij + \u01272 m2 \u2211 n\u2032\u2260n < nk j \u2212i\u2207ij n\u2032k >< n\u2032k j \u03b5n\u00f0 \u2212 \u03b5n\u2032\u00f0 k \u2212i\u2207jj k \u00de \u00de nk > + c:c: : 8.5. Show that if in Eq. (8.53), we can write r, t \u03d5\u2032 i\u00f0 \u00de = eik.runk\u00f0 r, E, t, \u00de the effective Hamiltonian for unk\u00f0 1 2m \u00f0 r, E, t is \u00de \u2212i\u0127\u2207 + \u0127k \u2212 eEt 2 + V \u00de r : \u00de \u00f0 8.6. Show that ^T \u00bd Ri\u00de \u00f0, H $ = \u00f0 \u03be!. Ri + e2B2 2m ^T R2 ix\u00de : Ri\u00de \u00f0 8.7. The eigenfunction \u03b5n\u00f0 k \u00de is periodic in k space and can be expanded as k \u03b5n\u00f0 \u00de Cnmeik.Rm : = \u2211 m If k is replaced by \u2212i\u2207, show that \u03b5n\u00f0 \u2212i\u2207 \u00de \u03c8 n\u00f0 k, r = \u2211 m \u00de CnmeiRm.\u2207\u03c8 n\u00f0 k, r k = \u03b5n\u00f0 \u00de \u00de k, r \u03c8 n\u00f0 : \u00de If we write the time-dependent Schrodinger equation as \u2212 \u01272 2m! \u22072 + V \u2212 e\ufffd", "\ufffd r \u00de \u00f0 \u03c8 = i\u0127 _\u03c8, \" (1) (1) (1) (2) (1) (1) (2) (3) and represent the electron as a wave packet constructed from all the Bloch states of all bands, show that \u03c8 = \u2211 nk An\u00f0 k, t \u03c8 n\u00f0 \u00de k, r, \u00de \u2212i\u2207 En\u00f0 \u00bd \u00de \u2212 e\u03d5 $ \u03c8 = i\u0127 _\u03c8 : (4) (5) Problems 273 8.8. If we substitute \u03d5\u2032 i\u00f0 r, t \u00de = eieA.r/\u0127c\u03d5i\u00f0 r, t \u00de in p + A e/c \u00f0 2m \u2032 i\u00f0 3 \u00de \u00f0 r, t t = \u03b5i\u00f0 \u03d5\u2032 i\u00f0 \u00de \u00de r, t, \u00de $ 7 5 show that Eq. (1) reduces to p2 2m! + V r \u00f0 \u00de \" \u03d5i = \u03b5i\u03d5i: 8.9. Derive the expression 8.10. Use the expression k, t \u03b5n\u00f0 ank t \u00f0 \u00de \u00de = i\u0127 \u2202ank \u00f0 \u2202t t \u00de \u2212 eE t \u00f0 \u2211 n\u2032 \u00de k Ann\u2032\u00f0 \u00f0 t an\u2032k \u00de\u00de : t \u00f0 \u00de t _\u03b1 n\u00f0 \u00de = t ieE \u00f0 \u0127 \u00de \u2211 n\u2032 t \u03b1n\u2032\u00f0 Ann\u2032\u00f0 \u00de k \u00f0 t \u00de\u00de exp i \u0127 2 t Z 0 k \u03b5n\u00f0 \u00bd \u00f0 t\u2032 \u00de\u00de k \u2212 \u03b5n\u2032\u00f0 \u00f0 t\u2032 \u00de\u00de$ dt\u2032, 3 4 5 (1) (2) (1) (1) which is a general expression used for derivation of Wannier\u2013Stark ladders for time-dependent E as well as for Zener tunneling between the valence and conduction bands. At t = 0, before the field E is turned on, k, the electron state can be described by a Bloch wave in band n with, then at t = 0 \u03b1n\u2032 = \u03b4nn\u2032: (2) After a time T = \u2212 \u0127K eE, where T is the period of one Bloch oscillation, which is a sufficiently short time so that \u03b1n\u2032 \u226a 1, n\u2032 \u2260 n, show that we can substitute Eq. (2) in Eq. (1) to obtain t \u03b1n\u2032\u00f0 \u00de = t Z 0 t\u2032 ieE \u00f0 \u0127 \u00de k An\u2032n\u00f0 \u00f0 t\u2032 \u00de\u00de exp i \u0127 2 t\u2032 Z 0 k \u03b5n\u2032\u00f0 \u00bd \u00f0 t\u2033 \u00de\u00de k \u2212 \u03b5n\u00f0 \u00f0 t\u2033 \u00de\u00de$ dt\u2033 dt\u2032: 3 (3) 4 5 8.11. Using Eqs. (8.149) and (8.154), show that n1, n2, \u2026, n1, n2, \u2026, n\u03bb\u2032, \u2026, n\u03bb, \u2026 > ^H j = \u2211 \u03bb, \u03bb\u2032 n\u03bb\u00f0 1 \u2212 n\u03bb\u2032\u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 p \u22121 \u00de \u00f0 m\u03bb+m\u2032 \u03bb\u2032< \u03bb\u2032 j \u03bb > H1j j n1, n2, \u2026, n\u03bb\u2032+1, \u2026 n\u03bb\u22121, \u2026 > if \u03bb \u2260 \u03bb\u2032, = \u2211 \u03bb n\u03bb < \u03bb \u03bb > H1j n1, n2, \u2026, n\u03bb\u2026 >, if \u03bb = \u03bb\u2032: j j (1) (2) 274 CHAPTER 8 Dynamics of Bloch Electrons 8.12. By using the expressions derived in Eqs. (2.75) and (2.76) for bosons, a\u2020 qj n1, n2, \u2026, nq, \u2026 > = nq + 1 n1, n2, \u2026, nq + 1, \u2026 >, j p n1, n2, \u2026, nq, \u2026 > = \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 nqp n1, n2, \u2026, nq \u2212 1", ", \u2026 >, aqj j \ufb03\ufb03\ufb03\ufb03\ufb03 and from Eqs. (8.133) through (8.136) for fermions, show that < nk+q + 1, nk \u2212 1; nq \u2212 1 > ^c\u2020 k+q^ckaqj j nk+q, nk; nq > = nknq 1 \u2212 nk+q\u00de \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 : q 8.13. Using the properties of the creation and annihilation operators for bosons and fermions outlined in Problem 8.12, show that < nk+q + 1, nk \u2212 1; n\u2212q + 1 k+q^ck^a\u2020 ^c\u2020 j \u2212qj nk+q, nk; n\u2212q > = n\u2212q + 1 nk\u00f0 1 \u2212 nk+q\u00de \u00de \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 : q (1) (2) (3) (1) References 1. Aschcroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 2. Cllaway J. Quantum theory of the solid state. New York: Academic Press; 1976. 3. Dahan BM, Peik E, Reichel J, Castin Y, Solomon C. Bloch oscillations of aroms in an optical potential. Phys Rev Lett 1996;76:4508. 4. Goswami A. Quantum mechanics. Dubuque: William C. Brown; 1997. 5. Kittel C. Quantum theory of solids. New York: John Wiley & Sons; 1987. 6. Kreiger JB, Iafrate GJ. Time evolution of Bloch electrons in homogeneous electric field. Phys Rev B 1986;33:5494. 7. Liboff R. Quantum mechanics. Reading, MA: Addison-Wesley; 1980. 8. Madelung O. Introduction to solid state theory. New York: Springer-Verlag; 1978. 9. Marder MP. Condensed matter physics. New York: John Wiley & Sons; 2000. 10. Shoenberg D. Magnetic oscillations in metals. Cambridge: Cambridge University Press; 1984. 11. Symon KR. Mechanics. Reading, MA: Addison-Wesley; 1971. 12. Zener C. Non-adiabatic crossing of energy levels. Proc R Soc Lond 1932;A137:696. 13. Ziman JM. Principles of the theory of solids. Cambridge: Cambridge University Press; 1972. Semiconductors CHAPTER 9 CHAPTER OUTLINE 9.1 Introduction............................................................................. 275 9.2 Electrons and Holes...................................................................... 278 9.3 Electron and Hole Densities in Equilibrium.................................................. 279 9.4 Intrinsic Semiconductors.........................", "........................................ 283 9.5 Extrinsic Semiconductors................................................................. 284 9.6 Doped Semiconductors................................................................... 285 9.7 Statistics of Impurity Levels in Thermal Equilibrium.......................................... 288 9.7.1 Donor Levels....................................................................... 288 9.7.2 Acceptor Levels..................................................................... 288 9.7.3 Doped Semiconductors.............................................................. 289 9.8 Diluted Magnetic Semiconductors......................................................... 290 9.8.1 Introduction........................................................................ 290 9.8.2 Magnetization in Zero External Magnetic Field in a DMS............................... 291 9.8.3 Electron Paramagnetic Resonance Shift.............................................. 291 9.8.4 k!. \u03c0! Model....................................................................... 295 9.9 Zinc Oxide.............................................................................. 296 9.10 Amorphous Semiconductors............................................................... 296 9.10.1 Introduction.....................................", ".................................. 296 9.10.2 Linear Combination of Hybrids Model for Tetrahedral Semiconductors................. 297 Problems.................................................................................... 300 References.................................................................................. 303 9.1 INTRODUCTION In Chapter 4 (section 4.9), we discussed that by using elementary band theory, crystalline solids can be divided into three major categories: metals, insulators, and homogeneous semiconductors. The metals are good conductors (with the exception of the divalent metals) because either the conduction band is half-filled (monovalent or trivalent metals), or there is significant overlap between the valence and conduction band (divalent metals). The crystalline solids with four valence electrons per unit cell can either be an insulator or a homogeneous semiconductor, depending on the energy gap between the valence band and the conduction band. The energy gap Eg is defined as the energy between the bottom of the lowest-filled band(s) and the top of the highest-filled bands(s). In the case of both insulators and semiconductors, the lowest unoccupied band is known as the conduction band, and the highest occupied Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00009-8 \u00a9 2012 by Elsevier Inc. All rights reserved. 275 276 CHAPTER 9 Semiconductors E Conduction band band is known as the valence band. A schematic diagram of the valence and conduction bands, which are the energy bands in the reduced zone scheme, is shown in Figure 9.1. Eg FIGURE 9.1 Valence band Schematic diagram of the valence and conduction bands. At T = 0, the valence band is full, and the conduction band is empty for both insulators and semiconductors. Thus, the conductivity of both types of solids is zero because no carriers would be available in either of the bands to be excited by an external electric field unless the electric field (DC) is sufficiently large to cause Zener tunneling (Chapter 8) or the frequency \u03c9 of the AC electric field is such that \u0127\u03c9 > Eg. At T \u2260 0, a few electrons would be thermally excited to the conduction band, leaving behind a few positively charged holes in the valence band (as we will show, the probability of such transition is e\u2212Eg/2KBT ). If the energy gap is large so that very few electrons are thermally excited from the valence band to the conduction band at the room temperature, a negligible number of carriers in either of the bands would be available to conduct electricity, and essentially no current would be generated. This type of solid, which carries no current in an electric field, is known as an insulator. However, if the energy gap is small enough, a significant number of electrons are thermally excited at room temperature to the conduction band(s), leaving an equal number of positively charged \u201choles\u201d at the top of the valence band(s). Thus, there are both \u201cpositively\u201d charged carriers in the valence band and \u201cnegatively\u201d charged carriers in the conduction band to conduct electricity and generate a perceptible current when an external field is applied. These solids are known as homogeneous semiconductors. The homogeneous semiconductors are also known as intrinsic semiconductors to distinguish them from impurity (doped) semiconductors. Thus, the distinction between the semiconductors and insulators depends essentially on the magnitude of the energy gap, and as a rule of the thumb, solids with Eg < 2 eV are semiconductors, whereas solids with Eg > 2 eV are insulators. However, most semiconductors have a much smaller energy gap. There are two types of homogeneous semiconductors: the semiconducting elements and the semiconducting", " compounds. The most popular and widely used semiconducting elements are Si and Ge, both of which belong to column IV of the periodic table and crystallize in the diamond structure. The Bravais lattice of the diamond structure has a basis of two atoms, each of which has 8 sp electron states, but only 4 of these are occupied by electrons. The Brillouin zones have to accommodate 16 electron states, but only 8 electrons (from the two atoms) fill them. Therefore, the band structure has 8 subbands, 4 of which are completely filled and the other 4 are completely empty at 0\u00b0 K. The energy gap of Si is 1.11 eV and that of Ge is 0.74 eV. These are also known as indirect semiconductors because the bottom of the conduction band does not lie directly above the top of the valence band, and thus, they have an indirect energy gap. The absorption of a photon creates an electron-hole pair, in which both the energy and momentum have to be conserved, but the momentum of a photon is negligible compared to that of the electrons. Thus, neither Si nor Ge is a good material for most optical applications because the two bands are not directly above each other. However, Si and Ge are extensively used in electronics because they can be easily doped with impurities. The other elemental semiconductors 9.1 Introduction 277 Table 9.1 Comparison of Calculated Bandwidth with Photoemission Data for the Homopolar Materials (Energy in eV) Quasiparticle Theory Expt. Diamond Si Ge 23.0 12.0 12.8 Reproduced from Louie8 with the permission of Elsevier. 24.2 \u00b1 1 12.5 \u00b1 0.6 12.9 \u00b1 0.2 from column IV of the periodic table are gray tin, which has a very small energy gap (0.1 eV), red phosphorus, boron (1.5 eV), selenium, and tellurium (0.35 eV), which are solids with complex crystal structures. These semiconductors are neither used in optical applications nor in electronics. An example of the compound semiconductors of column IV of the periodic table is SiC, which has an indirect energy gap of 2.2 eV. An example of the calculated and experimental bandwidth for the homopolar materials Si, Ge (both semiconductors), and diamond (insulator) is shown in Table 9.1. The III\u2013V semiconductors are crystals composed from columns III and V of the periodic table and have zincblende structure with predominantly covalent bonding. In the III\u2013V zincblende semiconductors such as GaAs and InSb, Ga and In have three outer electrons, whereas As and Sb have five outer electrons. Ga or In occupies all the A sites in the diamond structure, whereas As or Sb occupies all the B sites (see Figure 6.4 in Chapter 6). The most popular of these is GaAs, which has a direct energy gap of 1.43 eV and thus facilitates the absorption of photons creating electron-hole pairs. GaAs is therefore widely used in optical applications. InSb has a direct energy gap of 0.18 eV. The other III\u2013V semiconductors are GaN, GaSb, InP, and InAs, which have direct energy gaps of 3.44 eV, 0.7 eV, 1.34 eV, and 0.36 eV, respectively. The essential features of the band structure of Ge are shown in Figure 9.2. In Ge, which ) V e ( y g r e n E 10 8 6 4 2 0 \u22122 \u22124 \u22126 \u22128 L4,5 L6 L4,5 L6 L6 \u221210 L6 \u221212 \u221214 L Ge X5 X5 X5 \u03936 \u03937 \u03938 \u03938 \u03937 Theory Experiment Typical error \u03936 \u039b \u0393 \u2206 X \u2192 Wave vector k FIGURE 9.2 Calculated quasiparticle energies of Ge versus direct (\u00b0) and inverse (\u2662) photoemission data. Reproduced from Louie8 with permission of Elsevier. 278 CHAPTER 9 Semiconductors has an indirect energy gap, the top of the valence band is at \u0393, but the bottom of the conduction band is at L. In contrast, GaAs (not shown in the figure), which has a direct energy gap, both the top of the valence band and the bottom of the conduction band are at \u0393: 9.2 ELECTRONS AND HOLES In an intrinsic semiconductor at room temperature, the carriers are the electrons excited to the bottom of the conduction band from the top of the valence band. Because the filled valence band has no net charge (there are as many electrons as positively charged ions), the absence of an electron creates a net positive charge", " in the band. In addition, we consider the effective inverse mass tensor of an electron introduced in Eq. (8.35): M\u22121 k n \u00f0 \u00bd \u00de$ij = 1 \u01272 \u22022\u03b5n \u2202ki\u2202kj : We also derived an expression for the mass tensor in Eq. (8.37) by using perturbation theory, M\u22121 n \u00f0 \u00bd k \u00de$ij = 1 m \u03b4ij + \u01272 m2 \u2211 n\u2032\u2260n < nk j \u2212i\u2207ij n\u2032k > < n\u2032k j \u2212 \u03b5n\u2032\u00f0 k \u03b5n\u00f0 \u2212i\u2207jj k \u00de \u00de nk > + c:c: : (9.1) (9.2) At the top of the valence band, the inverse effective mass tensor can either be positive or negative. Because the concept of a negative mass is contrary to our physical understanding, the empty state at the top of the valence band can be considered as a positively charged \u201chole\u201d with a positive effective mass. This concept of the positively charged holes is very important in formulating a theory for the semiconductors. We introduce the general definition of the effective mass tensor for both electrons and holes, M\u22121 n \u00f0 \u00bd k \u00de$ij = \u00b1 1 \u01272 \u22022\u03b5n \u2202ki\u2202kj, (9.3) where the positive sign is for electrons and the negative sign is for holes. Because the bottom of the conduction band is at \u03b5c and the top of the valence band is at \u03b5v, we can express the energy of the electrons and holes as k k \u03b5e\u00f0 \u00f0 \u00de\u00de \u03b5h\u00f0 \u00f0 \u00de\u00de and k \u03b5e\u00f0 \u00de = \u03b5c + \u01272 2 \u2211 ij ki\u00f0 M\u22121 e \u00deijkj k \u03b5h\u00f0 \u00de = \u03b5v \u2212 \u01272 2 \u2211 ij ki\u00f0 M\u22121 h \u00deijkj: (9.4) (9.5) However, there are two types of effective hole masses (light holes and heavy holes) for Si, Ge, and GaAs. At the valence band maximum, there are two degenerate bands at \u0393, while the third band (there are three degenerate bands in the absence of spin) is lowered due to spin-orbit interaction. The band 9.3 Electron and Hole Densities in Equilibrium 279 with the low curvature results in holes with large effective mass (heavy holes), whereas the band with the high curvature has holes with small effective mass (light holes), which is evident from Eq. (9.3). One can write the effective mass tensor in terms of a set of orthogonal principal axes, = \u03b5c + \u01272 2 e \u00deiik2 M\u22121 \u03b5e\u00f0 \u2211 i \u00f0 k \u00de i and k \u03b5h\u00f0 \u00de = \u03b5v \u2212 \u01272 2 \u2211 i \u00f0 h \u00deiik2 M\u22121 i : We can redefine the electron and hole masses by rewriting Eqs. (9.6) and (9.7) as and k \u03b5e\u00f0 \u00de = \u03b5c + \u01272 2 3 \u2211 i = 1 1 me i k2 i k \u03b5h\u00f0 \u00de = \u03b5v \u2212 \u01272 2 3 \u2211 i = 1 1 mh i k2 i : (9.6) (9.7) (9.8) (9.9) 9.3 ELECTRON AND HOLE DENSITIES IN EQUILIBRIUM To determine the number of carriers in each band in a semiconductor, we will modify the expression between the electron density and the density of states derived in Eq. (3.58), which was obtained for free electrons, n = \u221e \u2212\u221e Z g \u03b5 f \u00de \u00f0 \u00f0 \u03b5 \u00de d\u03b5: Here, f \u03b5 \u00de \u00f0 is the Fermi distribution function, which for electrons is = \u03b5 fe\u00f0 \u00de \u03b5\u2212\u03bc e\u00f0 \u00de 1 /kBT + 1 : (9.10) (9.11) In the case of free electrons, the chemical potential \u03bc at T = 0 is equal to the Fermi energy \u03b5F, which is defined as the energy at the boundary between the filled and the empty states. In the case of semiconductors, the filled and empty states are separated by an energy gap Eg: Thus, we can argue that the chemical potential \u03bc, known as the Fermi level in the case of semiconductors, lies somewhere between the energy gap. Later, we will show that for", " an intrinsic semiconductor, the Fermi level lies exactly at the middle of the gap at T = 0. In addition, we are considering the density of electrons in the conduction band for which Eq. (9.10) is modified as T nc\u00f0 \u00de = \u221e \u03b5c Z gc\u00f0 \u03b5 fe\u00f0 \u00de \u03b5 \u00de d\u03b5, (9.12) 280 CHAPTER 9 Semiconductors \u03b5 is the density of states of the electrons in the conduction band. We can write a similar where gc\u00f0 expression for the holes in the valence band except that the distribution function for a hole can be written as \u00de \u03b5 fh\u00f0 \u00de = 1 \u2212 1 /kBT + 1 \u00de = \u03b5\u2212\u03bc e\u00f0 1 /kBT + 1 \u00de : \u03bc\u2212\u03b5 e\u00f0 (9.13) In addition, the density of states for holes lies below the valence band edge. Thus, the density of holes in the valence band can be written as T pv\u00f0 \u00de = \u03b5v Z \u2212\u221e gv\u00f0 \u03b5 fh\u00f0 \u00de \u03b5 \u00de d\u03b5, (9.14) \u00de \u03b5 where gv\u00f0 is the density of states of the holes in the valence band. For semiconductors, the conduction bands are nearly empty, and the valence bands are nearly full. We assume that the band shapes are nearly parabolic as in the case of free electrons. Therefore, we use the expression for the density of states for a free electron gas derived in Eq. (3.56) and suitably modify it for semiconductors by substituting effective masses m%n and m%p for the free electron mass m, and the fact that the energy of the electrons is \u03b5 \u2265 \u03b5c and the energy of the holes is \u2264 \u03b5v (we note that the energy of the holes is negative), \u03b5 j j \u03b5 gc\u00f0 \u00de = 3 n \u03b5\u2032 2m% p\u03c02\u01273 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u03b7c, \u03b5\u2032 > \u03b5c where and where = 0, \u03b5\u2032 \u2264 \u03b5c, \u03b5 \u2212 \u03b5c, \u03b5\u2032 & \u03b5 gv\u00f0 \u00de = 3 \u03b5\u2033 2m% p j j q \u03c02\u01273 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03, = 0, > \u03b5v \u2264 \u03b5v, \u03b5\u2033 j j \u03b5\u2033 j j \u03b5\u2033 j&j j \u03b5 \u2212 \u03b5vj : (9.15) (9.16) (9.17) (9.18) Here, \u03b7c is the number of symmetrically equivalent minima in the conduction band (six for Si and eight for Ge). m%n and m%p are the effective mass of electrons and holes that are obtained from the relation and m%n = \u00f0 1/2 me 1me 2me 3\u00de m% 3/2 p = m%pl\u00de \u00f0 3/2 + 3/2, m%ph\u00de \u00f0 (9.19) (9.20) 9.3 Electron and Hole Densities in Equilibrium 281 where and m%pl = \u00f0 mlh 1 mlh 2 mlh 3 \u00de 1/2 m%ph = \u00f0 mhh 1 mhh 2 mhh 3 \u00de 1/2: (9.21) (9.22) Here, m%pl and m%ph are the effective masses of the light and heavy holes defined in Eqs. (9.15) and (9.17), respectively. We note that in Eq. (9.17), the energy of the holes is zero at the valence band and negative downwards. For most semiconductors, the following approximations can be easily made: and \u03b5c \u2212 \u03bc \u226b kBT \u03bc \u2212 \u03b5v \u226b kBT: (9.23) (9.24) The semiconductors for which the approximations (9.23) and (9.24) are valid are known as nondegenerate semiconductors, whereas those for which these approximations are not valid are known, respecas degenerate semiconductors and one has to use Eqs. (9.11) and (9.13) for fe\u00f0 \u00de tively. For nondegenerate semiconductors, we can rewrite Eqs.", " (9.11) and (9.13) as and fh\u00f0 \u03b5 \u03b5 \u00de and \u03b5 fe\u00f0 \u00de \u03b5\u2212\u03bc \u2248 e\u2212 \u00f0 /kBT \u2248 e\u00bd \u00de \u2212\u03b5c\u2212 \u03b5\u2032\u2212\u03bc \u00f0 \u00de$ /kBT \u2248 e\u2212\u03b5c/kBT fe\u00f0 \u03b5\u2032 \u00de \u03b5 fh\u00f0 \u00de \u03b5\u2212\u03bc \u2248 e\u00f0 /kBT \u2248 e\u00bd \u00de \u03b5v+ \u00f0j \u03b5\u2212\u03b5v \u2212\u03bc j \u00de$ /kBT \u2248 e\u03b5v/kBT fh\u00f0j : \u03b5\u2033 j\u00de The density of carriers (in the conduction and valence bands) is obtained from the relations, and \u03b5\u2032 gc\u00f0 \u03b5\u2032 fe\u00f0 \u00de d\u03b5\u2032 \u00de T nc\u00f0 \u00de = \u221e 0 Z \u221e T pv\u00f0 \u00de = 0 Z gv\u00f0j \u03b5\u2033 j\u00de fh\u00f0j \u03b5\u2033 j\u00de d : \u03b5\u2033 j j From Eqs. (9.15), (9.25), and (9.27), the expression for nc\u00f0 is obtained as T \u00de T nc\u00f0 \u00de = \u03b7c 3 2m% n p\u03c02\u01273 e\u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u03bc\u2212\u03b5c /kBT \u00de \u03b5\u20321/2e\u2212\u03b5\u2032/kBT d\u03b5\u2032: Z 0 \u221e From Eqs. (9.17), (9.26), and (9.28), the expression for pv\u00f0 T \u00de is obtained as T pv\u00f0 \u00de = 3 2m% p q\u03c02\u01273 e\u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u03b5v\u2212\u03bc /kBT \u00de \u221e Z 0 1/2e\u2212 \u03b5\u2033 j j /kBT d \u03b5\u2033 j j : \u03b5\u2033 j j (9.25) (9.26) (9.27) (9.28) (9.29) (9.30) 282 CHAPTER 9 Semiconductors It can be easily shown that \u221e Z 0 \u03b51/2 e\u2212\u03b5/kBT d\u03b5 = 1 2 \u00f0 kBT \u00de 3/2\u03c01/2: From Eqs. (9.29) and (9.31), we obtain T nc\u00f0 \u00de = \u2135c\u00f0 T \u00de \u03bc\u2212\u03b5c e\u00f0 /kBT \u00de T \u2135c\u00f0 \u00de = 2\u03b7c m%nkBT 2\u03c0\u01272 \" 3/2 : # can be expressed numerically as where T \u2135c\u00f0 \u00de m%n m \" # Similarly, from Eqs. (9.30) and (9.31), we obtain = 2:51\u03b7c T \u2135c\u00f0 \u00de 3/2 T 300 K 3/2 % $ \u00d7 1019cm\u22123: where T pv\u00f0 \u00de = \u2118v\u00f0 T \u00de e\u00f0 \u03b5v\u2212\u03bc /kBT, \u00de T \u2118v\u00f0 \u00de = 2 m%pkBT 2\u03c0\u01272 3/2 : # \" T \u2118v\u00f0 \u00de can be expressed numerically as T \u2118v\u00f0 \u00de = 2:51 3/2 m%p m \" # $ From Eqs. (9.32) and (9.35), we obtain 3/2 1019cm\u22123: T 300 K % nc\u00f0 where Eg is the energy gap between the conduction and valence band, = \u2135c\u00f0 T pv\u00f0 \u00de \u2118v\u00f0 \u00de T T T \u00de \u00de e\u2212Eg/kBT, Eg = \u03b5c \u2212 \u03b5v: (9.31) (9.32) (9.33) (9.34) (9.35) (9.36) (9.37) (9.38) (9.39) Eq. (9.38) is known as the law of mass action. It states that at a given temperature, one can obtain the density of one type of carrier if one knows the density of the other type of carrier, the effective masses of both carriers (including that of light and heavy holes), the number of symmetrically equivalent minima in the conduction band, and the energy gap of the semiconductor. We note that Eqs. (9.32) and (9.35)", " are valid for both intrinsic (pure) and extrinsic (semiconductors with natural or doped impurities) semiconductors. We will first consider the case of intrinsic semiconductors. 9.4 Intrinsic Semiconductors 283 9.4 INTRINSIC SEMICONDUCTORS In an intrinsic semiconductor, because the number of electrons nc\u00f0 T equal to the number of holes pv\u00f0 \u00de in each band from Eq. (9.38), as in the conduction band is in the valence band, we can express ni, the number of carriers T \u00de T pv\u00f0 nc\u00f0 \u00de \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 From Eqs. (9.34), (9.37), and (9.40), we obtain ni\u00f0 v\u00f0 \u2135c\u00f0 \u00de \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 e\u2212Eg/2kBT : T ni\u00f0 \u00de = 2:51\u03b71/2 c 3/4 m%nm%p m2 T 300 K 3/2 e\u2212Eg/2kBT 1019 cm\u22121: # The chemical potential of an intrinsic semiconductor is obtained from the equality relation \" $ % and from Eqs. (9.32) and (9.35), ni\u00f0 T \u00de T = nc\u00f0 \u00de T = pv\u00f0 \u00de, T \u2135c\u00f0 e\u00f0 \u00de \u03bci\u2212\u03b5c \u00de /kBT = \u2118v\u00f0 T \u00de e\u00f0 \u03b5v\u2212\u03bci\u00de /kBT (9.40) (9.41) (9.42) (9.43) or \u03bc = \u03bci = \u03b5v + 1 2 Eg + 1 2 kBT ln\u2061 T \u2118v\u00f0 T \u2135c\u00f0 \u00de \u00de : (9.44) From Eqs. (9.33), (9.36), and (9.44), we obtain \u03bci = \u03b5v + 1 2 Eg + 3 4 kBT ln\u2061 m%p m%n\" # \u2212 1 2 kBT ln\u2061\u03b7c: (9.45) From Eq. (9.45), at T = 0, \u03bci has a simple form, \u03bci = \u03b5v + 1 2 Eg, (9.46) which is precisely at the center of the energy gap. Even at reasonable temperatures, because m%p \u2248 m%n for most semiconductors, \u03bci is close to the center of the energy gap. This is shown in Figure 9.3. From Eq. (9.32) and in analogy with the free electron model, we can write the conductivity \u03c3i of an intrinsic semiconductor as \u03b5c \u2212 \u00b5 \u00b5 \u2212 \u03b5v Eg >> kBT \u03b5 \u03b5c \u00b5 1 \u03b5v + Eg 2 \u03b5v FIGURE 9.3 \u03c3i = nie2\u03c4e m%n + pie2\u03c4h m%p, (9.47) The chemical potential for an intrinsic semiconductor lies within the energy gap at T \u2260 0. 284 CHAPTER 9 Semiconductors where \u03c4e and \u03c4h are the relaxation times for electrons and holes. Eq. (9.47) can be rewritten in the alternate and more familiar form \u03c3i = nie\u03bce + pie\u03bch, (9.48) where \u03bce and \u03bch are the mobility of the electrons and holes, which is the velocity (always defined as positive) of the carriers in unit electric field. 9.5 EXTRINSIC SEMICONDUCTORS If there are impurities present in a semiconductor, which contribute a significant number of electrons to the conduction band or holes to the valence band, the semiconductor is known as an extrinsic semiconductor. This can be obtained either by doping (which is the base of modern electronics) or by contamination. In either case, From Eqs. (9.38) and (9.40), we have \u0394n T \u00f0 \u00de T = nc\u00f0 \u00de T \u2212 pv\u00f0 \u00de \u2260 0: which can be rewritten in the alternate form (dropping the T in the", " bracket), T nc\u00f0 T pv\u00f0 \u00de \u00de T = ni\u00f0 \u00de 2, or pv + \u0394n \u00f0 pv = n2 i, \u00de v + \u0394npv \u2212 n2 p2 i = 0: Solving the quadratic equation, we have pv = \u2212 1 2 \u0394n + Following a similar procedure, we obtain 2 + 4n2 i \u0394n \u00de 1 2 \u00f0 h 1 2: i nc = \u0394n + 1 2 1 2 \u0394n 2 + 4n2 i \u00f0 \u00de 1 2: i h We dropped the negative sign before the square root in Eqs. (9.53) and (9.54) because both pv and nc are positive. We also obtain from Eqs. (9.32), (9.35), and (9.42), nc = ni e\u00f0 \u03bc\u2212\u03bci\u00de /kBT and pv = pi e\u2212 \u03bc\u2212\u03bci\u00de \u00f0 \u03bc\u2212\u03bci\u00de /kBT = ni e\u2212 \u00f0 /kBT : (9.55) (9.56) (9.49) (9.50) (9.51) (9.52) (9.53) (9.54) 9.6 Doped Semiconductors 285 From Eqs. (9.49), (9.55), and (9.56), we have nc \u2212 pv = ni\u00bd e\u00f0 \u03bc\u2212\u03bci\u00de \u03bc\u2212\u03bci\u00de /kBT \u2212 e\u2212 \u00f0 /kBT = 2ni sinh\u2061 \u03bc \u2212 \u03bci\u00de \u00bd\u00f0 /kBT : $ $ From Eqs. (9.49) and (9.57), we obtain \u0394n ni = 2 sinh\u2061 \u03bc \u2212 \u03bci\u00de \u00bd\u00f0 /kBT, $ which can be expressed in the alternate form \u03bc = \u03bci + kBT sinh\u2061\u22121 \u0394n 2ni& : \u2019 (9.57) (9.58) (9.59) Because \u03bci is at the center of the energy gap, \u0394n must exceed ni by many orders of magnitude before the chemical potential \u03bc violates the condition of \u201cnondegeneracy\u201d stated earlier (Eqs. 9.23 and 9.24). The exception is in a region of \u201cextreme extrinsic behavior.\u201d We also note from Eqs. (9.53) and (9.54), if \u0394n j j \u226b ni, and if \u0394n is positive, in which case, nc \u226b pv and the semiconductor is called n-type. If \u0394n is negative, nc \u2248 \u0394n and pv \u2248 nc 2, ni \u0394n $ % pv \u2248 \u0394n j j and nc \u2248 pv ni \u0394n j \" 2, # j (9.60) (9.61) in which case, pv \u226b nc and the semiconductor is called p-type. These types of semiconductors are known as doped semiconductors and are the base on which modern electronics is built. 9.6 DOPED SEMICONDUCTORS Doped impurities, which contribute additional electrons to the conduction band such that the semiconductor becomes n-type, are known as donors, whereas those that contribute holes to the valence band to make the semiconductor p-type are known as acceptors. The simplest example is that when in a group IV semiconductor such as a crystal of pure Si, an Si atom is replaced with a P atom, and there is an extra electron (because phosphorous has five valence electrons) that does not participate in the covalent bonds of Si. A deliberate substitution of quite a large number of P in Si is called doping, and because the excess electrons that are originally bound to their parent phosphorus atoms eventually end up in the conduction band at room temperature (because, as we will show, this binding energy is small), the semiconductor is called the n-type. A schematic diagram of this type of doping is shown in Figure 9.4. The problem of the substitutional impurity can be simplified by ignoring the structural difference between silicon and phosphorus ion cores. In addition, as shown in Figure 9.5, the extra electron, which is bound to the parent phosphorus atom, can essentially be considered as a particle of charge \u2212e and mass m% moving in the presence of an attractive center of charge e in a medium of dielectric constant \u2208. This problem is equivalent to that of the ground state of a hydrogen atom with two modifications: the dielectric constant \u22080 of a vacuum (in a hydrogen atom) is replaced by the dielectric constant \u2208 of the semiconductor, and the free electron mass (me) is replaced by the effective", " mass m%e: 286 CHAPTER 9 Semiconductors Si Si Si \u2212 Si P Si Si Si Si Si Si Si FIGURE 9.4 Phosphorus atom in a pure silicon crystal (donor impurity). \u2212e \u22124e \u221210e +15e (a) FIGURE 9.5 \u2212e +e (b) (a) The electrons in a phosphorus impurity. (b) The 10 inner electrons and the 4 valence electrons, which participate in the covalent bonding, screen the nucleus, which has an effective attractive charge e. The binding energy of the donor electron is given by m%ee4 4\u03c0 \u2208 \u0127 \u03b5 = \u2212 2 \u00f0 and the radius of the first Bohr orbit is given by r = m m%e \u2208 a0, \u00de 2, (9.62) (9.63) where a0 is the first Bohr radius and m is the free electron mass. Because the dielectric constant \u2208 is large 100 A\u00b0. This justifies the use of a semiclassical model, and the binding energy is obtained from the expression and the effective mass m%e is small, r \u00de 0:1 m \u00f0\u2019 \u00f0\u2019 20 \u2019 \u00de \u03b5 \u2248 m%e m 1 \u22082 \u00d7 13:6 eV: (9.64) \u2019 Using the same arguments, one can easily show that the binding energy of the donor electron \u22120:01 5 eV. The bound impurity level is formed relative to the energy of the conduction band, is, is and hence, the binding energy, which is much smaller than the energy gap ( \u00de measured relative to these levels. At room temperature, the donor impurity is ionized, and the electron jumps to the conduction band. When a large number of such extra electrons are introduced in the conduction band by doping Si with P (or other elements from group V), the crystal is known as an n-type semiconductor. Eg \u2248 1:14 eV for Si \u00f0 It is easy to make the same argument for acceptor impurities by substituting an Si atom with an element from group III in a silicon crystal. An example of substituting an Si atom with an Al atom in a silicon crystal is shown in Figure 9.6. As shown in Figure 9.6, there is a deficiency of one electron in the formation of a covalent bond that requires four electrons while aluminum has only three valence electrons. Thus, a hole is created 9.6 Doped Semiconductors 287 that is initially bound to the parent aluminum atom. This bound state is known as an acceptor state with a role reversal of a hydrogen-like atom in the sense that there is a net charge of \u2018\u2212e\u2019 and mass m at the center and hole with charge \u2018e\u2019 and effective mass m%h orbiting around it. One can easily show that the radius of the first Bohr orbit is given by Si Si Si Si Al Si Si Si Si Si Si Si FIGURE 9.6 An example of acceptor impurity: aluminum atom in a silicon crystal. rh = m m%h \u2208 a0: (9.65) The binding energy of the ground state is nearly the same for both electrons and holes, i.e., on the order of \u22120:015 eV: However, the binding energy of the donor levels is measured relative to the conduction band, and the binding energy of the acceptor levels is measured relative to the valence band (the zero of the valence band is at the top of the band, and the energy of the holes is measured positive downwards). Thus, both the donor and the acceptor levels are formed in the energy gap, as shown in Figure 9.7. However, at room temperature, the thermal energy is sufficient for both the electrons and holes to fall into the conduction and valence bands, respectively. Donor levels E 0 E 0 Acceptor levels (a) (b) FIGURE 9.7 Donor and acceptor levels for (a) n-type and (b) p-type semiconductors. 288 CHAPTER 9 Semiconductors 9.7 STATISTICS OF IMPURITY LEVELS IN THERMAL EQUILIBRIUM 9.7.1 Donor Levels For simplicity, we consider a semiconductor with donor states that have a binding energy \u03b5d located just below the bottom of the conduction band. Ignoring electron\u2013electron interaction, the three possibilities for the donor level are either the level could be empty, or the donor could trap one electron of either spin (up or down). The donor level cannot bind two electrons of opposite spin at the same time. Using the Fermi statistics (in a grand canonical ensemble), we obtain fd, the mean number of electrons in the donor level, which can be written in the form", " fd = nie\u2212\u03b2 \u03b5i\u2212\u03bcNi \u00f0 \u00de \u2211 i e\u2212\u03b2 \u00f0 \u03b5i\u2212\u03bcNi \u00de \u2211 i, fd = 0 + 2e\u2212\u03b2 1 + 2e\u2212\u03b2 \u03b5d \u2212\u03bc \u00f0 \u03b5d \u2212\u03bc \u00f0 \u00de \u00de : A more appropriate way to write Eq. (9.67) is fd = 1 + 1 e\u03b2 1 2 : \u03b5d\u2212\u03bc \u00f0 \u00de (9.66) (9.67) (9.68) If the density of donors per unit volume is Nd, we can express nd, the number density of electrons bound to the donor sites, as nd = Ndfd\u00de \u00f0 nd = 1 + Nd 1 e\u03b2 2 : \u03b5d \u2212\u03bc \u00f0 \u00de (9.69) \u2212e 9.7.2 Acceptor Levels The statistics of the holes can be simplified by considering the holes as the \u201cabsence\u201d of electrons. An acceptor level, which is placed at an energy \u03b5a above the valence band, can be either singly or doubly occupied but cannot be empty. An acceptor impurity is essentially a fixed, negatively charged attractive center superimposed on an unaltered host atom, which can weakly bind one hole. The binding energy of the hole is \u03b5a \u2212 \u03b5v, and a second electron moves into the acceptor level when the hole is \u201cionized.\u201d There is little probability that two holes would be localized in the presence of the acceptor impurity because the Coulomb repulsion between two holes would be very large. This scenario corresponds to no electrons in the acceptor level. In addition, we must account for the fact that the valence maximum is four-fold degenerate (including spin degeneracy). Thus, we obtain an expression for fd\u2032, the occupation probability of electrons in the acceptor level, \u00de \u00f0 fd\u2032 = 4e\u03b2\u03bc + 2e\u2212\u03b2 4e\u03b2\u03bc + e\u2212\u03b2 \u03b5a\u22122\u03bc \u00f0 \u03b5a\u22122\u03bc e\u03b2 \u03bc\u2212\u03b5a \u00f0 \u00de e\u03b2 \u03bc\u2212\u03b5a \u00f0 \u00de : (9.70) 9.7 Statistics of Impurity Levels in Thermal Equilibrium 289 The average number of holes in the acceptor level is the difference between the maximum number of electrons the level can hold (two) and the mean number of electrons in the level, which can be rewritten in the alternate form (for the mean number of holes) fa = 2 \u2212 fd e\u03b2 \u03bc\u2212\u03b5a \u00f0 \u00de e\u03b2 \u03bc\u2212\u03b5a \u00f0 \u00de = 1 + 1 e\u03b2 \u00f0 1 4 : \u03bc\u2212\u03b5a \u00de (9.71) If the semiconductor is doped with Na acceptor impurities per unit volume, pa\u00f0 ber of holes in the acceptor levels is pa = Nafa\u00de, the num- pa = Na e\u03b2 1 + 1 4 : \u03bc\u2212\u03b5a \u00f0 \u00de (9.72) 9.7.3 Doped Semiconductors We assume that a semiconductor is doped with Nd donor impurities and Na acceptor impurities (per unit volume) and assume that Nd \u2265 Na. Thus, Na of the Nd electrons, which are supplied by the donor atoms, will drop from the donor levels into the acceptor levels. The eventual scenario is that in the ground state (T = 0), both the valence band and the acceptor levels are filled. In addition, Nd \u2212 Na of the donor levels are filled, but the conduction bands are empty. However, at finite temperature, the number of empty (electron) levels is pa + pv in the valence band and the acceptor levels. So, we can write Nd \u2212 Na = nc + nd \u2212 pv \u2212 pa, (9.73) where nc and nd are the number of electrons in the conduction band and donor levels. To simplify the formulation, we assume (known as the conditions of nondegeneracy) that and \u03b5d \u2212 \u03bc \u226b kBT \u03bc \u2212 \u03b5a \u226b kBT: (9.74) (9.75) From Eqs. (9.69), (9.70), (9.74), and (9.75), we conclude that the impurities are essentially ionized and hence nd \u226a Nd and pa \u226a Na: Therefore, Eq. (9.74) can be approximated as From Eqs. (9.53), (9.54), and (9.76), we obtain Nd \u2212 Na \u2248 nc \u2212 pv \u2248 \u0394n: and nc = 1 2 \u00bd\u00f0 Nd \u2212 Na\u00de$ + 1 2", " \u00bd\u00f0 Nd \u2212 Na\u00de 2 + 4n2 i $ 1 2 pv = 1 2 \u00bd\u00f0 Na \u2212 Nd\u00de$ + 1 2 \u00bd\u00f0 Nd \u2212 Na\u00de 2 + 4n2 i $ 1 2: (9.76) (9.77) (9.78) 290 CHAPTER 9 Semiconductors In addition, Eq. (9.59) can be rewritten in the alternate form \u03bc = \u03bci + kBT sinh\u2061\u22121 Nd \u2212 Naj \u00f0j must exceed ni by many orders of magnitude before the conditions of nondegeneThus, racy outlined in Eqs. (9.74) and (9.75) are violated. From Eqs. (9.69) and (9.72), we obtain nd \u2248 Nd and pa \u2248 Na, thereby ensuring that most of the impurities are fully ionized. From Eq. (9.77), we obtain in the extrinsic regime Nd \u2212 Naj j /2ni\u00de, when Nd \u226b Na, (9.79) ni \u226a : \u00f0 Nd \u2212 Naj\u00de j and nc \u2248 Nd \u2212 Na pv \u2248 n2 i Nd \u2212 Na : (9.80) (9.81) Eqs. (9.80) and (9.81) state that the number of donors is essentially equal to the number of mobile electrons while the number of holes in the valence band is very small. In the other extrinsic regime, when Na \u226b Nd, we obtain from Eq. (9.78), and pv \u2248 Na \u2212 Nd nc \u2248 n2 i Na \u2212 Nd : (9.82) (9.83) Eqs. (9.82) and (9.83) state that the number of acceptors is essentially equal to the number of mobile holes in the valence band while the number of electrons in the conduction band is very small. 9.8 DILUTED MAGNETIC SEMICONDUCTORS 9.8.1 Introduction The diluted magnetic semiconductors (DMSs), which are ternary or quaternary alloys, in which a part of the nonmagnetic cations of the host material has been substituted by magnetic ions, have attracted considerable attention in recent years. In theory, any semiconductor with a fraction of its constituent ions replaced by ions bearing a net magnetic moment is considered as a DMS. However, in practice, the majority of DMS involve Mn2+ ions embedded in various AIIBVI or AIVBVI hosts because Mn2+ can be incorporated in the host without affecting the crystallographic quality of the resulting material. In addition, Mn2+ possesses a relatively large magnetic moment characteristic of a half-filled d-shell (S = 5/2). Further, because Mn2+ is electrically neutral in AIIBVI or AIVBVI hosts, it does not have accepting or donating centers. The Mn containing IV\u2013VI DMSs can be divided into two groups, according to their carrier concentrations. The magnetic behavior of the first group (charge\u2013carrier concentration, or p-type, range 1017\u22121019cm\u22123 can be ascribed to antiferromagnetic interactions of the superexchange type between the Mn ions. This group has a spin-glass phase at low temperatures. The magnetic behavior of the second group (charge-carrier concentration, or p, of 1021cm\u22123) has a ferromagnetic ordering, \u00de 9.8 Diluted Magnetic Semiconductors 291 induced by RKKY interactions (an indirect exchange mechanism in which the interaction between the magnetic ions is mediated by the itinerant charge carriers, or p) at low temperatures. The magnetic properties of Sn1\u2212xMnxTe and Pb0:28\u2212xSn0:72MnxTe (which can be considered as partly like semimetals and partly like semiconductors) are very diverse. The ferromagnetic, spinglass (SG) and reentrant-spin-glass phases have been observed, but the transition to a paramagnetic phase has not been observed at T > 1:5 K: The occurrence of these magnetic phases depends not only on the Mn concentration but also on the concentration of free carriers. The significance of carrier-mediated magnetism in a DMS is mainly due to the search for materials suitable information processing and storage (in for the light of developments in the rapidly emerging area of spintronics, which will be discussed in Chapter 11). Some experimental results of the threedimensional (T, x, p) magnetic phase diagram for Sn1\u2212xMnxTe are shown in Figure 9.8. 15 10 ) K ( T 5 0 20 C P F SG F arrier c (1 o n 02", " SG F 10 c e 0 c m ntratio \u2212 3 ) n p SG 5 0 0.05 0.10 Concentration x FIGURE 9.8 (T, x, p) phase diagram for Sn1\u2212xMnxTe. F = ferromagnetic phase; P = paramagnetic phase; SG = spin-glass phase. Reproduced from Vennix et al.,18 with the permission of the In subsequent experiments, the magnetic phase of the samples included magnetization, AC susceptibility, specific heat, and neutron diffraction. In Figure 9.9, the magnetic phase is displayed as a function of both the Mn concentration and the carrier concentration. The carrier concentration (the theoretical calculation of which is described later) has a significant effect on the magnetic phase of the material. At a carrier concentration of pc = 3 \u00d7 1020 cm\u22123 (T > 1.5 K), there is an abrupt change from the, there is a gradual transiparamagnetic to ferromagnetic phase. At high carrier concentrations tion to an intermediate reentrant-spin-glass phase and eventually to the spin-glass state. The location of these transitions is shifted to higher carrier concentrations if the Mn concentration is increased. American Physical Society. p \u226b pc\u00de \u00f0 9.8.2 Magnetization in Zero External Magnetic Field in a DMS Another interesting property of some DMSs is that magnetization in a zero external magnetic field can be induced by optically excited carriers in Hg1\u2212xMnxTe and Cd1\u2212xMnxTe: The observed magnetization is due to the orientation of the Mn ions caused by the spin dynamics of the magnetic response on a picosecond time scale and is the result of spin-polarized electrons. This can be calculated by the spin-EPR shift, of which the theory is briefly described. 9.8.3 Electron Paramagnetic Resonance Shift The electron paramagnetic resonance (EPR) shift is a measure of the internal field created at the magnetic ion site in magnetic materials or materials with magnetic impurities by the partial 292 CHAPTER 9 Semiconductors PM FM 10\u2212 10\u22122 Pc RSG SG 1020 2 FIGURE 9.9 5 1021 Carrier concentration (cm\u22123) 2 5 1022 Magnetic-phase diagram of Pb 0.28\u2212xSn0.72MnxTe (squares) and Sn1\u2212xMnxTe (circles). PM = paramagnetic, FM = ferromagnetic, RSG = reentrant spin glass, SG = spin glass. Reproduced from Eggenkamp et al.,5 with the permission of the American Physical Society. \u03a9 = 1 \u03b2 Tr ln \u00f0 \u2212 G\u03bel \u00de : e r!\u00de polarization of electrons and/or carriers in an applied magnetic field. It can be shown that the contribution to the spin-EPR shift at the Mn2 + ion on the jth site of a DMS is given by the expression P\u03bcv j \u00de \u00f0 ( ( where \u03a9 is the thermodynamic potential and is given by ( P\u03bcv j = \u2212 \u22022 \u03a9 \u2202B\u03bc\u2202Mv j, 0, M! j!! 0 B!! (9.84) (9.85) G is related to the one-particle Green\u2019s function Here, M!j is the local moment at the jth site, and, spin-orbit interaction, applied magnetic (Eq. 7.219) in the presence of a periodic potential V field B!, and local magnetic moment. Tr involves summation of over both imaginary frequencies and e one-particle states, forming a complete orthonormal set. The one-particle Green\u2019s function G satisfies the equation \u00f0 r!, r\u2032!, B!, M!j, \u03bel\u00de \u00f0 where the complex energy and in a symmetric gauge, \u03bel \u2212 H G \u00f0 r!, r\u2032!, B!, M!j, \u03bel\u00de \u00de = \u03b4 r!\u2212 r\u2032! \u00de \u00f0, \u00f0 \u03bel = \u00f0 i\u03c0 2l + 1 \u03b2 \u00de + \u03bc, l = 0, \u00b11, \u00b12, \u2026, G \u00f0 r!, r!, B!, M!j, \u03bel\u00de = ei h!. r! \u00d7 r\u2032! G r!, r\u2032!, B!, M!j, \u03bel\u00de, \u00f0 e (9.86) (9.87) (9.88) 9.8 Diluted Magnetic Semiconductors 293 where e B! 2\u0127c G is the Green\u2019s function that satisfies the lattice translational symmetry, and ei h!. r! \u00d7 r\u2032! is the Here, Peierls phase factor. The phase factor has the effect of translating the", " origin of the vector potential. (9.89) h! = : e Recently, Trellakis16 used a singular gauge transformation based on a lattice of magnetic flux lines, an equivalent quantum system with a periodic vector potential. However, his theory is beyond the scope of this book. The one-particle Hamiltonian is ^H = 1 2m \u00f0 p! + e c A! \u00de 2 + V r!\u00de \u00f0 + \u0127 4m2c2 \u03c3!. \u2207!V \u00d7 p! + \u00f0 e c + A! \u00de 1 2 g0\u03bc0 B!. \u03c3! + ^HI, where ^HI = 1 2gJ\u03bc0 \u2211 j! M!j. \u03c3!\u2111 r! \u2212 R!j\u00de : \u00f0 (9.90) (9.91) is the strength of the exchange interaction r! \u2212 R!j\u00de Here, A! is the magnetic vector potential, \u2111 \u00f0 between the conduction electrons and/or the carriers and the local moment at the jth site, and the other symbols have their usual meaning. One can write the equation of motion in a representation, where k! is the reduced defined by the periodic part u k!\u03c1\u00f0 wave vector and \u03c1 is the spin index. of the Bloch function \u03c8 k!\u03c1\u00f0 r!\u00de r!\u00de It has been shown (Misra et al.,11 p. 1903) that in this representation, Eqs. (9.86), (9.88), (9.90), and (9.91) can be rewritten as where H \u03ba!\u00de \u00f0 \u03bel \u2212 H G \u03ba!\u00de$ \u00f0 k!, \u03bel\u00de \u00f0 \u00bd = I, e = H0\u00f0 k! + H\u2032 \u00de \u00f0 2 + V + \u01272, \u03ba!\u00de H0\u00f0 k! \u00de = p! + \u0127 k! \u00de 4m2c2 \u03c3!: \u2207!V \u00d7, p! + \u0127 k! \u00de \u00f0 1 2m \u00f0 = \u2212i \u0127 m H\u2032 \u00f0 \u03ba!\u00de h\u03b1\u03b2\u03c0\u03b1\u2207\u03b2 k + 1 2 g0\u03bc0\u03c3\u03bcH\u03bc + 1 2\u03bc0 1 gj \u2211 j Mv j \u03c3v\u2111, \u03ba! = k! + i h! \u00d7 \u2207!k, \u03c0! = p! + \u0127 4mc2 \u03c3! \u00d7 \u2207!V, (9.92) (9.93) (9.94) (9.95) (9.96) (9.97) h\u03b1\u03b2 = \u2208\u03b1\u03b2\u03bch\u03bc, where \u2208\u03b1\u03b2\u03bc is the antisymmetric tensor of the third rank, and we follow Einstein summation convention. It may be noted that \u03ba! (Eq. 9.96) is the Misra\u2013Roth operator. Eq. (9.85) can be further simplified by writing the frequency summation as \u03a9 = \u2212 1 2\u03c0i tr Ic \u03d5 \u03be \u03be G \u00f0 \u00de \u00de \u00f0 d\u03be, e (9.98) 294 CHAPTER 9 Semiconductors where \u2212\u03b2 ln\u2061 \u00bd \u00f0 \u03bc\u2212\u03be \u00de $ (9.99) and the contour c encircles the imaginary axis in a counterclockwise direction. Eq. (9.92) can be k!, \u03be solved by using a perturbation expansion of G, G \u00f0 k!, \u03be = k!, \u03be + k!, \u03be H\u2032 \u00de \u00de Here, the terms only up to second order are retained because the EPR shift independent of the applied field and the local moment is calculated. In Eq. (9.100), satisfies the equation e \u00de \u00de \u00de \u00de k!, \u03be H\u2032 k!, \u03be H\u2032 \u00de k!, \u03be : (9.100) G0\u00f0 e G0\u00f0 e G0\u00f0 e \u00f0 \u00de + k!, \u03be G0\u00f0 e e G0\u00f0 e G0\u00f0 e G0\u00f0 e k!, \u03be \u00de and is diagonal in the basis u k!\u03c1\u00f0 r!\u00de \u2207\u03b1 G0\u00f0 k e we obtain \u00bd k! k!, \u03be \u03be \u2212 H0\u00f0 G0\u00f0 : Using the identity in Eq. (9.100), e = I \u00de$ \u00de k!, \u03be = \u0127 m \u00de k!, \u03be \u03c0\u03b1 \u00de G0\u00f0 e k!, \u03be, \u00de G0\u00f0 e \u00de G \u00f0 k!, \u03be = \u2211 j j B\u03bc 1", " Mv 2gj G0\u03c3\u03bc \u00f0 i \u2212 m e Here, G0\u00f0 Expressing the contributions of the two terms in Eq. (9.103) to \u03a9 as e G0\u03c3v\u2111 G0\u03c3\u03bc G0\u03c3v\u2111 G0 + G0\u00de G0\u03c3v\u2111 G0\u03c3v\u2111 G0\u03c0\u03b2 G0\u03c0\u03b1 G0 + \u03b5\u03b1\u03b2\u03bc\u00f0 e e e e e e e G0 is the compact form of \" e e e \u03a9 = \u03a91 + \u03a92, G0\u03c0\u03b1 G0\u03c0\u03b2 G0\u00de e e e. Eq. (9.98) is evaluated by using Eqs. (9.99) and (9.103). k!, \u03be \u00de e e (9.101) (9.102) : # (9.103) we obtain and where and \u03a91 = \u2211 j H\u03bcMv j \u2211 n, k!,\u03c1,\u03c1\u2032 \u03c3v\u2111 1 2gj & \u03c3\u03bc n\u03c1\u2032:n\u03c1f \u2032 \u2019n\u03c1,n\u03c1\u2032 \" En k!\u03c1 # \u03a92 = \u2211 j H\u03bcMv j \u2211 n,m,k!,\u03c1,\u03c1\u2032,\u03c1\u2033m \u2260 n i m \u03b5\u03b1\u03b2\u03bc \u03c3v\u2111 1 2gj & n\u03c1\u2032,m\u03c1\u2033\u03c0\u03b2 \u03c0\u03b1 m\u03c1\u2033,n\u03c1 \u2019n\u03c1,n\u03c1\u2032 Emn En k!\u03c1 f \u2032 \", # Emn & Em k! \u2212 En k! H0un k!\u03c1\u00f0 r!\u00de = En k! un k!\u03c1\u00f0 r!\u00de : (9.104) (9.105) (9.106) (9.107) (9.108) 9.8 Diluted Magnetic Semiconductors 295 The matrix elements are of the type An\u03c1,n\u03c1\u2032 & Z u% n k!\u03c1\u00f0 r!\u00de Aun k!\u03c1\u00f0 r!\u00de d r!, (9.109) En k!\u00de is the first derivative of the Fermi function. The expressions for \u03a91 and \u03a92 derived in and f \u2032 \u00f0 Eqs. (9.105) and (9.106) are substituted in Eq. (9.84), and the complex spin-orbit terms are neglected. After considerable algebra, we obtain the expression for the spin contribution to the EPR shift at the jth site, Pv\u03bc js = \u2212 1 2 \u2211 n k!\u03c1\u03c1\u2032 \" 1 2gj \u03c3v\u2111 g\u03bc nn\u00f0 k! \u00de \u03c3\u03bc n\u03c1\u2032,n\u03c1 f \u2032 \" En k!, # #n\u03c1,n\u03c1\u2032 where the effective g factor g\u03bc nn\u00f0 is defined as k! \u00de g\u03bc nn\u00f0 n\u03c1\u2032,n\u03c1 = g0\u03c3\u03bc \u03c3\u03bc k! \u00de n\u03c1\u2032,n\u03c1 + 2i m \u2208\u03b1\u03b2\u03bc \u2211 m \u2260 n,\u03c1\u2033 n\u03c1\u2032,m\u03c1\u2033\u03c0\u03b2 \u03c0\u03b1 Emn m\u03c1\u2033,n\u03c1 : (9.110) (9.111) Eq. (9.110) can be used to calculate the EPR shift of any diluted magnetic semiconductor of which the band structure is known. A brief discussion of the calculation of the EPR shift from Eq. (9.110) by using the k!. \u03c0! model is explained in the next section. \u00f0 k!. p!\u00de \u03c7j\u00f0 9.8.4 k!. \u03c0! Model The k!. p! model of Luttinger and Kohn9 (p. 869) was modified by Tripathi et al.17 (p. 3091) to include the effect of magnetic fields. Their method of calculation is known as the k!. \u03c0! model. model, a complete orthonormal set of functions, In the Luttinger\u2013Kohn k!, r!\u00de = ei \u00f0 k!\u2212 k!0 \u00de. r!\u03c8 j\u00f0 k!0, r!\u00de, (9.112) is defined, where k!0 is a reference point in the Brillouin zone at which the energy bands and wave functions \u03c8 j\u00f0 have been determined. The unknown wave function is expanded as k!0, r!\u00de \u03c7j\u00f0 In the k!. \u03c0! model, the effective equation of motion is k!, r!\u00de = \u2211 j Anj\u00f0 \u03c8 n\u00f0 k! \u00de k!, r!\u00de : \u2211 j & \"", " Ej\u00f0 k!0\u00de \u2212 En\u00f0 k! \u00de + \u01272 k2 \u2212 k2 0\u00de 2m \u00f0 \u03b4jl + \u0127 m \u00f0 # k!\u2212 k!0\u00de. \u03c0!lj Anj\u00f0 k! \u00de = 0, \u2019 where \u03c0!lj = \u00f0 3 2\u03c0 \u00de \u03a9c Zcell d3r u%l \u00f0 k!0, r!\u00de \u03c0! uj\u00f0 k!0, r!\u00de : (9.113) (9.114) (9.115) 296 CHAPTER 9 Semiconductors There is one equation for each value of the band index l, and the condition for this infinite set of simultaneous, linear, and homogeneous equations to have a nontrivial solution is that the determinant \u03b4jl, with of the coefficients should vanish. A general element of the determinant has the form Hjl \u2212 E Hjl = Ej\u00f0 \" k!0\u00de One can use Eq. (9.116) to determine the spin contribution to the EPR shift of any diluted magnetic semiconductor. An example of using this method to determine the EPR shift of Pb1\u2212xMnxTe as a function of carrier concentration can be found in Das et al.4 k! \u2212 k!0\u00de (9.116) Ps\u00de \u00f0 k2 \u2212 k2 0\u00de. \u03c0!lj: \u03b4jl + \u0127 m \u00f0 # + \u01272 2m \u00f0 k! \u00de \u00f0 9.9 ZINC OXIDE The semiconductor ZnO has attracted widespread attention in recent years for its optoelectronic properties and, more recently, for the possibility of finding room-temperature ferromagnetism when doped with magnetic and nonmagnetic impurities. The electronic structure of ZnO has been studied for the past 50 years. However, in spite of extensive energy band calculations, there still exists a controversy with regard to the valence band ordering in ZnO. The wurtzite ZnO conduction band is mainly constructed from the s-like state having \u03937 symmetry, whereas the valence band is p-like, which is split into three bands due to the crystal field and spin-orbit interactions. A schematic picture of the band diagram is given in Figure 9.10. By treating the wurtzite energy levels as a perturbation over those of the zincblende, a formula has been derived for the valence band mixing, the extent of which is controlled by the relative magnitudes of the spin-orbit and crystal field splittings. No satisfactory theory has yet been proposed for the possibility of finding room-temperature ferromagnetism in ZnO, when doped with magnetic and nonmagnetic impurities. Some of the fascinating theoretical models include the k \u00b7 p + U formalism, where U is the many-body Hubbard term. 9.10 AMORPHOUS SEMICONDUCTORS 9.10.1 Introduction In recent years, amorphous semiconductors have attracted considerable attention. The amorphous structures can be obtained by rapid cooling from the melt or by evaporation onto a cooled substrate. They can also be obtained by sputtering the components on to a cooled substrate. The evaporation or sputtering of Si or Ge onto substrates held below 300\u00b0 C produces amorphous films, which are stable up to 425\u00b0 C: In the amorphous (a) state, the sp3 covalent bonds are strong in Si and Ge. However, although these bonds exist between nearest neighbors, the \u03937c Eg \u03939v A \u03937v B \u03937v C FIGURE 9.10 Schematic picture of valence band ordering for the wurtzite ZnO. 9.10 Amorphous Semiconductors 297 bonding is not perfect. There is four-fold coordination over small regions of the solid, but the tetrahedral symmetry is lost over extended regions. The bond angles are distorted from the ideal value, and many atoms can have only three neighbors. One can consider the system as a random network of imperfectly bonded atoms in which even small voids can exist. \u00f0 \u03c7e\u00de Recent experiments of amorphous semiconductors exhibit many interesting properties. Tetrahedral and quasitetrahedral materials exhibit an unusual enhancement of diamagnetic susceptibility, \u03c7 (450% for Si, 270% for Ge, and 150% for CdGeAs2) in the amorphous (a) phase relative to the. In contrast, in chalcogenides, crystalline (c) phase but no change in dielectric susceptibility there is very little change in \u03c7 in either phase but appreciable reduction of \u03c7e in the (", "a) phase. These unusual properties are related to the nature of the chemical bonding and the presence or absence of long-range order. The chemical bond approach to the study of electronic properties of solids is much simpler than the band theory; emphasizes the bond aspect of the crystal structure; and is valuable in studying chemical trends such as covalency, polarity, and metallicity. This approach has a certain degree of flexibility and is well suited to the study of amorphous as well as periodic ones. In the chemical bond approach, each atom in the amorphous system has four nearest neighbors, but the bonds are distorted in bond lengths and directions. Thus, one starts with a crystalline phase and considers the covalent bonds that give rise to bonding and antibonding orbitals, leading to valence and conduction bands. Then distortion in both bond lengths and angles is introduced, and the random network model of Polk13 (p. 365) is used to derive appropriate expressions for physical properties of amorphous structures. The electron energy levels of amorphous systems can be derived from a modified bond orbital (LCH) theory typical of the crystalline phase. Even if they are randomly arranged, the covalent bonds give rise to some system of bonding and antibonding levels, which leads to valence and conduction bands. However, due to distortion of the bonds, the bands become broader. We first consider a linear combination of hybrids (LCH) model for tetrahedral semiconductors. 9.10.2 Linear Combination of Hybrids Model for Tetrahedral Semiconductors We describe a linear combination of hybrids (LCH) model for a chemical bond approach to tetrahedral semiconductors (Sahu and Misra,14 p. 6795). In this model, the basis set for the valence bands is a linear combination of sp3 hybrids forming a bond, in which their relative phase factors, which have been neglected in the usual chemical bond models, adapted from the bond orbital models for molecules (Sukhatme and Wolff,15 p. 1369; Chadi et al.,2 p. 1372), have been included properly. A basis set for the conduction bands, which are orthogonal to the valence band functions, has also been constructed. It can be shown that the basic assumption of the earlier chemical bond models\u2014 i.e., that the localized functions have the character of chemical bonds\u2014is equivalent to ignoring the relative Bloch phase factor eik.dj (where dj is a bond length) between the hybrids forming a bond. is a lattice vector, these relative phase factors play an important However, because dj \u2212 dj\u2032\u00f0 role in solids, unlike the case of molecules where it can be neglected. j \u2260 j\u2032 \u00de We will now discuss the LCH model in detail. In the zincblende structure, each atom is surrounded tetrahedrally by four identical atoms, which may be of the second type. The primitive cell contains two basic atoms at site i, with four sp3 hybrids h1 pointing from atom I to the nearest r \u2212 Ri\u00de j \u00f0 neighbors (atom II) along the directions j( j = 1, \u2026, 4) and four other sp3 hybrids h2 r \u2212 Ri \u2212 dj\u00de j \u00f0 pointing from these nearest neighbors to atom I. Here, Ri is a lattice vector for site i and locates 298 CHAPTER 9 Semiconductors atoms of type I (one of the atomic sites I is chosen as the origin) and dj is a nearest-neighbor vector joining atom I with atom II. The hybrids can be expressed as (9.117) (9.118) h1 j \u00f0 r \u2212 Ri\u00de = 1 2 \u00bd s1 + 3p \u00f0 j px1 + \u03bey \u03bex j py1 + \u03bez j pz1\u00de$ and h2 j \u00f0 \ufb03\ufb03\ufb03 = 1 2 \u00bd s2 \u2212 3p j px2 + \u03bey \u03bex \u00f0 j py2 + \u03bez, j pz2\u00de$ r \u2212 Ri \u2212 dj\u00de 1, 1, 1 \u00de 3p \u00f0 1, 1, 1 ; \u03be!2 = 1 where \u03be!1 = 1 are the atomic orbi3p \u00f0 tals at sites I and II, respectively. The Bloch-type tight-binding sums for valence-band basis functions \ufb03\ufb03 \ufb03\ufb03 are constructed by taking a linear combination of the hybrids forming a bond r \u2212 Ri \u2212 dj\u00de r \u2212 Ri\u00de ; and \u03be!4 = 1 + \u03bbh2 j \u00f0 ; \u03be!3 = 1 f v k j \u00f0 (9.119) 1, 1, 1 h1 j \u00f0 eik.Ri \u03c7v j", " \u00f0 eik.dj = \u2211 i 3p \u00f0 3p \u00f0 r, k \u00de \u00de \u00de \u00de \u00de \ufb03\ufb03 \ufb03\ufb03 \u00bd $, \ufb03\ufb03\ufb03 1, 1, 1 where 1 + \u03bb2 + 2\u03bbS cos k. dj\u00de$ \u00de Here, S is the overlap integral, and \u03bb2/ is the probability of the electron being around atom II for III\u2013V semiconductors and is related to Coulson\u2019s ionicity (Coulson et al.,3 p. 357; Nucho et al.,12 p. 1843) by the expression \u00f0 \u00bd 1 + \u03bb2 \u00f0 (9.120) f v j \u00f0 \u22121/2: = N k \u00de fc = \u00f0 1/2 1 \u2212 S 1 \u2212 \u03bb2 \u00f0 1 + \u03bb2 + 2\u03bbS \u00de \u00f0 \u00de \u00de : (9.121) We note that the LCH model discussed here is different from the usual bond-orbital model12 in the sense that a relative phase factor eik.dj between the two hybrids forming a bond has been included to properly account for the origin. The basis functions for the conduction band \u03c7c j \u00f0 are obtained by constructing functions r, k \u00de orthogonal to \u03c7v j \u00f0 r, k \u03c7c j \u00f0 r eik.Ri \u00de \u03bb + Seik.dj \u00bd\u00f0 h1 j \u00f0 \u00de r \u2212 Ri\u00de \u2212 \u00f0 \u03bbS + eik.dj h2 j \u00f0 \u00de r \u2212 Ri \u2212 dj\u00de$, (9.122) where \u2212 S2 \u03bb + Se\u2212ik.dj 1 + \u03bb2 + 2\u03bbS cos \u2061k. dj\u00de\u00f0 \u03bb + Seik.dj \u00f0 The Bloch eigenfunctions for the valence and conduction bands are \u00de\u00f0 and r, k \u03c8 n\u00f0 = \u2211 j \u03b1v k jn\u00f0 \u03c7v j \u00f0 \u00de \u00de r, k \u00de r, k \u03c8 m\u00f0 = \u2211 j \u03b1c jm\u00f0 k \u03c7c j \u00f0 \u00de \u00de r, k, \u00de 1/2 : # \u00de (9.123) (9.124) (9.125) where and 9.10 Amorphous Semiconductors 299 k \u03b1v jn\u00f0 \u03b1v\u2020 nj\u2032\u00f0 \u00de k \u00de \u2211 n = \u03b4jj\u2032 k \u03b1c jm\u00f0 \u03b1c\u2020 mj\u2032\u00f0 \u00de k \u00de \u2211 m = \u03b4jj\u2032: (9.126) (9.127) The conduction and valence bands for the tetrahedral semiconductors as well as the expression for the energy gap can be obtained by using these Bloch functions in the same way they were derived earlier in Chapter 4. The physical properties of model amorphous semiconductors (a-phase), such as magnetic susceptibility and dielectric constant, are obtained by introducing disorder in the bond angles and the bond lengths. The nearest-neighbor coordinations of the a-phase are not supposed to change appreciably from the c-phase. However, the bands become broader due to distortion of the bonds. In addition, the disorder causes a spread of the energy levels into the energy gap region. These are known as tail states, which arise due to the distorted bonds. These tail states are nonconducting and localized. A schematic picture of these tail states is shown in Figure 9.11. E In addition, an amorphous tetrahedral semiconductor has imperfectly coordinated atoms, and thereby, the bonds are uncompensated. These are known as \u201cdangling\u201d bonds, each of which has limited degeneracy. A dangling bond has an empty state and an electron associated with it. Because these bonds are uncompensated, both the electron and the empty state are localized at 0\u00b0 K. Thus, the immobile states are separate but overlapping, and the distributions are obtained within the energy gap. In fact, the concentration of the dangling bonds is on the order of 1025 m\u22123. Due to this unusually high concentration of the dangling bonds, each of which produces a localized electron and a localized empty is at the center of the \u03bc state, the Fermi level \u00de \u00f0 energy gap. Thus, the amorphous semiconductors essentially become insensitive to doping. The \u201cband\u201d picture of an amorphous tetrahedral semiconductor is shown in Figure 9.12. Eg FIGURE 9.11 The nonconducting tail states in a distorted tetrahedral semiconductor arise from the distorted bonds. 300 CHAPTER 9 Semiconductors E \u00b5 FIGURE", " 9.12 The band picture of amorphous tetrahedral semiconductors including tail states. PROBLEMS 9.1. The effective mass of electrons and holes in a semiconductor can be expressed as \u00de$ij = \u00b1 where the positive sign is for electrons and the negative sign is for holes. Because the bottom of the conduction band is at \u03b5c and the top of the valence band is at \u03b5v, show that the energy of the electrons can be written as and holes k k \u00bd, (1) M\u22121 n \u00f0 1 \u01272 \u22022\u03b5n \u2202ki\u2202kj \u03b5e\u00f0 \u00f0 \u00de\u00de \u03b5h\u00f0 k \u00de \u00f0 \u03b5e\u00f0 k \u00de\u00de = \u03b5c + \u01272 2 \u2211 ij ki\u00f0 M\u22121 e \u00deijkj and k \u03b5h\u00f0 \u00de = \u03b5v \u2212 \u01272 2 \u2211 ij ki\u00f0 M\u22121 h \u00deijkj: 9.2. Show that the density of states in the conduction and valence bands can be expressed as \u03b5 gc\u00f0 \u00de = 3 n \u03b5\u2032 2m% p\u03c02\u01273 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u03b7c, \u03b5 > \u03b5c = 0, \u03b5 \u2264 \u03b5c, (2) (3) (1) where and where \u03b5\u2032 & \u03b5 \u2212 \u03b5c \u03b5 gv\u00f0 \u00de = 3/2 \u03b5\u2033 2m% p j j q \u03c02\u01273 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03, = 0, > \u03b5v \u2264 \u03b5v, \u03b5\u2033 \u03b5\u2033 j j j j \u03b5\u2033 j j&j \u03b5 \u2212 \u03b5vj : Problems 301 (2) (3) (4) Here, \u03b7c is the number of symmetrically equivalent minima in the conduction band (six for Si and eight for Ge). m%n and m%p are the density of states\u2019 effective mass of electrons and holes that are obtained from the relation m%n = \u00f0 me 1me 2me 3\u00de 1/2 and where and m% 3/2 p = m%pl\u00de \u00f0 3/2 + 3/2, m%ph\u00de \u00f0 m%pl = \u00f0 mlh 1 mlh 2 mlh 3 \u00de 1/2 m%ph = \u00f0 mhh 1 mhh 2 mhh 3 \u00de 1/2: (5) (6) (7) (8) Here, m%pl and m%ph are the effective masses of the light and heavy holes. 9.3. From Eq. (9.32) and in analogy with the free electron model, one can write the conductivity \u03c3i of an intrinsic semiconductor as \u03c3i = nie2\u03c4e m%n + pie2\u03c4h m%p, (1) where \u03c4e and \u03c4h are the relaxation times for electrons and holes. Show that Eq. (9.48) can be written in the alternate and more familiar form \u03c3i = nie\u03bce + pie\u03bch, where \u03bce and \u03bch are the mobility of the electrons and holes, which is the velocity (always defined as positive) of the carriers in a unit electric field. 9.4. We have derived an expression for the valence band T pv\u00f0 \u00de = 3 2m% p q\u03c02\u01273 e\u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u03b5v\u2212\u03bc /kBT \u00de \u221e Z 0 1/2e\u2212 j \u03b5\u2033 /kBT d j \u03b5\u2033 j j : \u03b5\u2033 j j (2) (1) 302 CHAPTER 9 Semiconductors It can be easily shown that \u221e Z 0 \u03b51/2e\u2212\u03b5/kBT d\u03b5 = 1 2 \u00f0 kBT \u00de 3/2\u03c01/2: From Eqs. (1) and (2), show that T pv\u00f0 \u00de = \u2118v\u00f0 T \u00de e\u00f0 \u03b5v\u2212\u03bc /kBT, \u00de where T \u2118v\u00f0 \u00de \u2118v\u00f0 can be expressed numerically as = 2 T \u00de m%pkBT 2\u03c0", "\u01272 3/2 : # \" T \u2118v\u00f0 \u00de = 2:51 3/2 m%p m \" # $ T 300 K 3/2 1019 cm\u22123: % (2) (3) (4) (5) 9.5. Substitute the expressions for \u03a91 and \u03a92 derived in Eqs. (9.105) and (9.106) in Eq. (9.84) and neglect the complex spin-orbit terms. Show that the expression for the spin contribution to the EPR shift at the jth site is Pv\u03bc js = \u2212 1 2 \u2211 n k!\u03c1\u03c1\u2032 1 2gj \" \u03c3v\u2111 #n\u03c1,n\u03c1\u2032 g\u03bc nn\u00f0 k! \u00de \u03c3\u03bc n\u03c1\u2032,n\u03c1 f \u2032 \u00f0, En k!\u00de where the effective g factor g\u03bc is defined as k! nn\u00f0 \u00de n\u03c1\u2032,n\u03c1 = g0\u03c3\u03bc \u03c3\u03bc n\u03c1\u2032,n\u03c1 + g\u03bc nn\u00f0 k! \u00de 2i m \u2208\u03b1\u03b2\u03bc \u2211 m \u2260 n,\u03c1\u2033 n\u03c1\u2032,m\u03c1\u2033\u03c0\u03b2 \u03c0\u03b1 Emn m\u03c1\u2033,n\u03c1 : 9.6. If the Bloch-type tight-binding sums for valence-band basis functions are constructed by taking a linear combination of the hybrids forming a bond \u03c7v j \u00f0 r eik.Ri \u00de h1 j \u00f0 \u00bd r \u2212 Ri\u00de + \u03bbh2 j \u00f0 r \u2212 Ri \u2212 dj\u00de eik.dj, $ where + \u03bb2 + 2\u03bbS cos \u2061k. dj\u00de$ r, k \u00de show that the basis functions for the conduction band \u03c7c j \u00f0 functions orthogonal to \u03c7v r + Seik.dj r \u2212 Ri\u00de : \u00de eik.Ri = \u2211 i h1 j \u00f0 \u03c7c j \u00f0 r, k \u2212 \u00bd\u00f0 \u00de \u00de \u00de \u22121/2, are obtained by constructing \u03bbS + eik.dj h2 j \u00f0 \u00de, r \u2212 Ri \u2212 dj\u00de$ \u00f0 where k f c j \u00f0 \u00de = 1 \u2212 S2 N \" \u00f0 \u00de\u00f0 \u03bb + Se\u2212ik.dj 1 + \u03bb2 + 2\u03bbS cos \u2061k. dj\u00de\u00f0 \u03bb + Seik.dj 1/2 : # \u00de (1) (2) (1) (2) (3) (4) References 303 9.7. Show that the Bloch eigenfunctions for the valence and conduction bands \u03c8 n\u00f0 \u03b1v jn\u00f0 = \u2211 j \u03c7v j \u00f0 r, k r, k k \u00de \u00de \u00de and r, k \u03c8 m\u00f0 = \u2211 j k \u03b1c jm\u00f0 \u00de \u00de r, k \u03c7c j \u00f0 \u00de are orthonormal. Here, \u03b1\u2019s are elements of (4 \u00d7 4 \u00de \u03b1v\u2020 k nj\u2032\u00f0 \u03b1v jn\u00f0 \u2211 n k \u00de = \u03b4jj\u2032 \u00de unitary matrices and k \u03b1c jm\u00f0 \u03b1c\u2020 mj\u2032\u00f0 \u00de k \u00de \u2211 m = \u03b4jj\u2032: (1) (2) (3) (4) References 1. Ashcroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 2. Chadi DJ, White RM, Harrison WA. Theory of the magnetic susceptibility of tetrahedral semiconductors. Phys Rev Lett 1975;35:1372. 3. Coulson CA, Redei LB, Stocker D. The electronic properties of tetrahedral intermetallic compounds I. charge distribution. Proc R Soc 1962;270:357. 4. Das RK, Tripathi GS, Misra PK. Theory of the spin EPR shift: application to Pb1-xMnxTe. Phys Rev B 2005;72:035216. 5. Eggenkamp PJT, Swegten HJM, Story T, Litvinov VI, Swuste CHW, de Jonge WJM. Calculations of the ferromagnet-to-spin-glass transition in diluted magnetic systems with RKKY interaction. Phys Rev 1995; 51:15250. 6. Furdyna JK, Kossut J, eds. Diluted magnetic semiconductors. Boston: Academic Press; 1988. 7. Kittel C. Introduction to solid state physics. New York: John Wiley & Sons; 1976. 8. Louie SG. Quasiparticle excitations and photoemission. Stud Surf Sci Catal 1992", ";74:32. 9. Luttinger JM, Kohn W. Motion of electrons and holes in perturbed periodic fields. Phys Rev 1969;97:869. 10. Marder MP. Condensed matter physics. New York: John Wiley & Sons; 2000. 11. Misra SK, Misra PK, Mahanti SD. Many-body theory of magnetic susceptibility of electrons in solids. Phys Rev B 1982;26:1903. 12. Nucho RN, Ramos JG, Wolff PA. Chemical-bond approach to the magnetic susceptibility of semiconduc- tors. Phys Rev B 1978;17:1843. 13. Polk DE. Structural model for amorphous silicon and germanium. J Non-Cryst Solids 1971;5:365. 14. Sahu T, Misra PK. Magnetic susceptibility of tetrahedrally coordinated solids. Phys Rev B 1982;26:6795. 15. Sukhatme VP, Wolff PA. Chemical-bond approach to the magnetic susceptibility of tetrahedral semicon- ductors. Phys Rev Lett 1975;35:1369. 16. Trellakis A. Nonperturbative solution for Bloch electrons in constant magnetic fields. Phys Rev Lett 2003;91:056405. 17. Tripathi GS, Das LK, Misra PK, Mahanti SD. Theory of spin-orbit and many-body effects of the Knight shift. Phys Rev B 1982;25:3091. 18. Vennix CWHM, Frikkee E, Eggenkamp PJT, Swagten HJM, Kopinga K, de Jonge WJM. Neutrondiffraction study of the carrier-concentration-induced ferromagnet-to-spin-glass transition in the diluted magnetic semiconductor Sn1-xMnxTe. Phys Rev B 1993;48:3770. Electronics CHAPTER 10 CHAPTER OUTLINE 10.1 Introduction............................................................................. 305 10.2 p-n Junction............................................................................ 306 10.2.1 Introduction....................................................................... 306 10.2.2 p-n Junction in Equilibrium........................................................ 307 10.3 Rectification by a p-n Junction............................................................ 311 10.3.1 Equilibrium Case.................................................................. 311 10.3.2 Nonequilibrium Case (V \u2260 0)....................................................... 313 10.4 Transistors........................................................................", "..... 318 10.4.1 Bipolar Transistors................................................................. 318 10.4.2 Field-Effect Transistor............................................................. 319 10.4.3 Single-Electron Transistor.......................................................... 321 10.5 Integrated Circuits....................................................................... 325 10.6 Optoelectronic Devices................................................................... 325 10.7 Graphene............................................................................... 329 10.8 Graphene-Based Electronics.............................................................. 332 Problems.................................................................................... 333 References.................................................................................. 336 10.1 INTRODUCTION The importance of semiconductors is due to their use in modern electronics. The original semiconductor device was the Schottky diode, constructed from a metal-semiconductor interface, which was used instead of a thermoionic diode for rectification. The Schottky diode was constructed by placing a metal whisker against an n-doped semiconductor crystal (the p-doped semiconductor-metal contact does not have any rectifying properties). The work function of the semiconductor is less than that of the metal. Because of the higher chemical potential, the electrons rush from the semiconductor to the metal, and the voltage of the metal is lowered until further motion of charges is prevented by electrostatic forces. The increase in voltage of the semiconductor compensates for the \u2212eV difference in the chemical potential \u03bc, which is equal to the Fermi energy \u03b5F of the metal. This also adds an electrostatic potential \u2212eV to the conduction and valence band levels, which are bent by the potentials formed across the junction. This scenario is shown in Figure 10.1. \u00de\u00de x x \u00de \u00f0 \u00f0 \u00f0 Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00010-4 \u00a9 2012 by Elsevier Inc. All rights reserved. 305 306 CHAPTER 10 Electronics Metal n-doped Semiconductor \u03b5c \u2212 eV(x) \u00b5 = \u03b5F \ufffd", "\ufffd\u03c5 \u2212 eV(x) FIGURE 10.1 The charges move from the n-doped semiconductor to the metal, and both the valence and conduction bands are increased by \u2212eV : The electrons flow to the metal when the voltage of the metal is raised \u00de relative to the semiconductor by eVA: x \u00f0 \u03c6b Metal \u03c6b \u2212 (\u03b5c \u2212 \u03b5F) \u2212 eVA \u03b5c \u2212 eV(x) \u00b5 = \u03b5F \u03b5\u03c5 \u2212 eV(x) FIGURE 10.2 There is no flow of current when VA is lowered. The height of the barrier is \u03d5b : The effect of an external voltage VA (which can be positive or negative) can be understood in a simple manner. When VA is positive, the electrons flow from the semiconductor into the metal because the barrier is lowered. However, when VA is negative, there is no change in the barrier, and hence, the current does not flow. This rectifying effect of the Schottky diode is shown in Figure 10.2. In practice, it is difficult to construct an ideal Schottky diode because semiconductor surfaces acquire oxide layers after cleaving, and it is virtually impossible to produce atomic flat surfaces. The age of modern electronics was started with the creation of the p-n junction in a semiconductor. 10.2 p-n JUNCTION 10.2.1 Introduction A simple p-n junction is fabricated by taking a single intrinsic semiconductor such as Si in which donor impurities are introduced in one region and acceptor impurities are introduced into another. It is easier to visualize the p-n junction as equivalent to two Si crystals doped with donor (n-type) and acceptor (p-type) impurities joined together with polished surfaces. In practice, this is not a good idea because the surface effects would interfere with the properties of the semiconductor. These semiconductors have different chemical potentials, \u03bc, respectively. Initially, most of the conduction electrons are at the n-type, and the holes are at the p-type semiconductor. This scenario is shown in Figure 10.3a. However, when the two materials are joined together, some of the conduction electrons will diffuse to the p-type material while some of the holes will diffuse to the n-type material. When an electron and \u03bc n p \u00f0 \u00de \u00de \u00f0 10.2 p-n Junction 307 \u00b5(p) p \u2212 + + \u2212 Eg n (a) \u00b5(n) Conduction band Valence band + \u2212 + \u2212 \u2212 + \u2212 + p \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 + + + + + + Depletion layer (b Conduction band \u00b5 Valence band e\u03c6 Eg n p Depletion layer (c) FIGURE 10.3 (a) The electrons (solids) and holes (open circles) in p- and n-type semiconductors. (b) When the p- and n-type regions are in contact, there are practically no carriers in the depletion region. The acceptor and donor ions are represented by \u2212 and + and electrons and holes are represented by circles with \u2296 and \u2295 signs. (c) The distribution of charge at a p-n junction produces a contact potential \u03d5 x \u00f0 across the junction. \u00de diffuses to the p-type layer, it will recombine with a hole due to the large concentration of holes in that layer. A similar situation will arise when a hole diffuses to the n-type layer. As more and more electrons and holes diffuse to the p- and n-type layers, the number of carriers in the junction will be greatly reduced. Thus, the layer on either side of the junction has much lower concentration of carriers than the rest of the semiconductor. This is known as the depletion layer, which is shown in Figure 10.3b and usually has an approximate length of 102 A\u00b0 : We note that the donor and acceptor ions are much too heavy to diffuse like the electrons and holes. Thus, the p-n junction is formed in the semiconductor with a common chemical potential \u03bc: This is shown in Figure 10.3c. 10.2.2 p-n Junction in Equilibrium We assume that the impurity concentration varies along the x-axis only in a small region around x = 0. If the \u201cabrupt junction,\u201d which is defined as the region about x = 0 where the impurity 308 CHAPTER 10 Electronics concentrations change, is narrow compared with the \u201cdepletion layer\u201d in which the carrier densities are not uniform, the donor impurities dominate at positive x, and the acceptor impurities dominate at negative x (see Figure 10.4). The depletion layer, which extends from xp to xn, is shown in Figure 10.5. The abrupt junction is at x", " = 0. The region to the left of xp is the p-doped mobile holes, and the region to right of xn is the n-doped mobile electrons (note that xp is negative). In the beginning, when the p-n junction is formed, the carrier concentration would be such that there would be charge neutrality everywhere in the crystal. However, the concentration of the electrons in the n-side would be very high and that of the holes in the p-side would be very low. Thus, the electrons would diffuse to the p-side, and the holes would diffuse to the n-side. The charge transfer will eventually build up an electric field that will prevent further diffusion. In fact, the effect of the field cancels the effect of the diffusion. To derive expressions for the carrier densities Impurity density n-type + + + + + + + + + + + + + + Nd x Na p-type \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 x = 0 FIGURE 10.4 \u201cAbrupt\u201d p-n junction in which the impurityconcentration change is narrow compared with the depletion layer shown in Figure 10.5. \u2212 + \u2212 + + \u2212 + \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 + + + + + + \u2212 + \u2212 + \u2212 xp xn \u2212 + + + \u2212\u2212 + + \u03c6(\u221e)\u2212 \u03c6(\u2212\u221e) FIGURE 10.5 The depletion layer has length between 102 A\u00b0 and 104 A\u00b0 extending from xp to xn:, one can use Eqs. (9.32) and at a position x at temperature T in the presence of a potential \u03d5 (9.35) subject to the semiclassical condition that each one-electron energy level is shifted by \u2212e\u03d5 x \u00f0 \u00de x \u00f0 : Thus, one obtains \u00de and \u03b5c \u2192 \u03b5c \u2212 e\u03d5 x \u00de \u00f0 \u03b5v \u2192 \u03b5v \u2212 e\u03d5 x : \u00de \u00f0 From Eqs. (9.32) and (10.1), we obtain and from Eqs. (9.35) and (10.2), we obtain x nc\u00f0 \u00de = \u2135c\u00f0 T \u00de e\u2212\u03b2 \u03b5c\u2212e\u03d5 \u00f0 \u00f0 \u2212\u03bc x \u00de \u00de, pv\u00f0 1/kBT: If we define the electrochemical potential \u03bce\u00f0 = \u2118v\u00f0 \u00de, T x \u00de \u00de \u03b5v\u2212e\u03d5 \u00f0 \u00f0 \u2212\u03bc x \u00de e\u03b2 as x \u00de where \u03b2 # x \u03bce\u00f0 \u00de = \u03bc + e\u03d5 x, \u00de \u00f0 (10.1) (10.2) (10.3) (10.4) (10.5) 10.2 p-n Junction 309 Eqs. (10.3) and (10.4) can be rewritten in the alternate form, x nc\u00f0 \u00de = \u2135c\u00f0 T \u00de e\u2212\u03b2 x \u03b5c\u2212\u03bce\u00f0 \u00f0 \u00de\u00de (10.6) and x pv\u00f0 T = \u2118v\u00f0 e\u03b2 \u03b5v\u2212\u03bce\u00f0 \u00f0 x \u00de\u00de: \u00de The total charge density must vanish in both the limits x \u2192 \u2212\u221e and x \u2192 \u221e, which requires that \u00de Nd = nc\u00f0 \u221e \u00de and Na = pv\u00f0 \u2212\u221e \u00de : From Eqs. (10.3) through (10.5), one can also derive (Problem 10.1) \u221e \u03bce\u00f0 \u2212 \u03bce\u00f0 \u00de \u2212\u221e \u00de = e\u0394\u03d5 = Eg + 1 \u03b2 ln NdNa T \u2118v\u00f0 \u2135c\u00f0! \u00de T, \" \u00de where \u0394\u03d5 \u221e \u03d5 \u00f0 \u00de # \u2212 \u03d5 \u2212\u221e : \u00de \u00f0 (10.8) (10.9) (10.7) x The effect of the internal potential \u03d5 in Figure 10.6, in which the electrochemical potential \u03bce\u00f0 x carrier densities at a point x are obtained by using \u03bce\u00f0 x \u03b5a\u00f0 Here, \u03b5d\u00f0 = \u03b5d \u2212 e\u03d5 = \u03b5a \u2212 e\u03d5 on the electron and hole densities of a p-n junction is shown is plotted along the p-n junction. The as the equivalent chemical potential. The alternate method of representing the carrier densities at any point x is shown in Figure 10.7. and \u03bc is the constant chemical potential. To calculate \u03d5 x equation in one dimension,, one uses the Poisson\u2019s \u00de \u00f0 \u03b5d \u03b5a \u03b5v FIGURE 10.", "6 \u03b5c \u00b5e(x) \u22022\u03d5 \u2202x2 = \u2212 x 4\u03c0\u03c1 \u00f0 \u2208 \u00de, (10.10) where the charge density \u03c1 x is obtained from \u00f0 x x \u03c1 = Nd\u00f0 \u00f0 \u2212Na\u00f0 and \u2208 is the dielectric constant. It can be easily shown that (Problem 10.2) + pv\u00f0 \u2212nc\u00f0 (10.11, \u03b5c and \u03b5v are conduction and valence bands; \u03b5d and \u03b5a are donor and acceptor levels. \u03b5d(x) \u03b5a(x) \u03b5v(x) FIGURE 10.7 Carrier densities at a point x obtained by using a fixed chemical potential \u03bc: and \u03b5c(x) \u00b5 x nc\u00f0 \u00de x pv\u00f0 \u00de = Nde\u2212e\u03b2 \u03d5 \u221e \u00f0 \u2212\u03d5 \u00de x \u00de% \u00f0 \u00bd (10.12) = Naee\u03b2 \u03d5 \u2212\u221e \u00f0 x \u2212\u03d5 \u00de%: \u00f0 \u00de \u00bd (10.13) It can also be shown that (Problem 10.3; except at the boundaries of the depletion layer), outside the depletion layer, = 0, \u03c1 x \u00de \u00f0 (10.14) 310 CHAPTER 10 Electronics and inside the depletion layer, The solution for \u03d5 x \u00de \u00f0 Nd\u00f0 is obtained in Problem 10.4 \u2212 Na\u00f0 \u00de% : \u00de (10.15) \u03d5 x = 8 \u00f0 \u00de >>>>< >>>>: and \u03d5\u2032 In addition, both \u03d5 \u00f0 \u00dejx = \u2208, where \u2208 \u2192 0, we obtain, \u2212\u221e \u00de \u2212\u221e \u00de \u2212, \u221e \u221e \u00de 2\u03c0eNa \u2208 2\u03c0eNd # \u2208 x \u2212 xp\u00de \u00f0 $ 2, x \u2212 xn\u00de \u00f0 $ + # x < xp 2, 0 > x > xp 0 < x < xn x > xn 9 >>>>= >>>>; \u00de must be continuous at x = 0. From the condition \u03d5 : (10.16 2\u03c0eNd \u2208 x2 n = \u03d5 \u2212\u221e + \u00de \u00f0 2\u03c0eNa \u2208 x2 p, which can be rewritten in the alternate form From the condition \u03d5\u2032 \u00f0 x \u00dejx = \u2212\u2208 = \u03d5\u2032 x \u00f0 \u00f0 2\u03c0e \u0394\u03d5 = p + Ndx2 Nax2 n\u00de \u2208 \u00dejx = \u2208, we obtain : From Eqs. (10.18) and (10.19), we obtain expressions for the lengths xn and xp, Ndxn = \u2212Naxp: 0 xn,p\u00f0 \u00de = \u00b1 \" \u00b11 Na/Nd\u00de \u2208 \u00f0 2\u03c0e \u00f0 \u0394\u03d5 Na + Nd\u00de 1/2, # x \u00dejx = \u2212\u2208 = \u00f0 (10.17) (10.18) (10.19) (10.20) \u00de 0 where xn,p\u00f0 denotes the values of xn and xp when the external potential V = 0: Because Na and Nd are of the order of 1018 cm\u22123 and e\u0394\u03d5 is of the order of 0.1 eV, the depletion layer is of the order of 102 to 104 A\u00b0 : It is important to note that because the depletion layer has no mobile charges, the resistance of this region is considerably higher than that of the doped regions. It is equivalent to a series circuit in which a high resistance is sandwiched between two low resistances. When there is no external potential (V = 0), the p-n junction is known as the unbiased junction. The potential \u03d5 for the unbiased junction is plotted versus the position x in Figure 10.8. x \u00f0 \u00de \u03c6(x) \u03c6(\u221e) = \u03c6 (0) + xp xn p n Nd x2 n 2\u03c0e \u2208 x \u03c6(\u2212\u221e) = \u03c6 (0) \u2212 2\u03c0e \u2208 Na x2 \u03c6 FIGURE 10.8 The potential for the unbiased junction \u03d5 x \u00f0 \u00de against position x. The depletion layer is from x = xp to x = xn \u00f0 note that xp is negative \u00de : plotted 10.3 Rectification by a p-n Junction 311 10.3 RECTIFICATION BY A p-n Junction 10.3.1 Equilibrium Case When the external potential V = 0, the change in potential from the p-side to the n-side of the depletion layer is given by \u0394\u03d5, the expression for which is obtained in E", "q. (10.8). When the external potential V \u2260 0, the potential drop (or gain) \u0394\u03d5\u2032 will occur mostly across the high-resistance region, which is the depletion layer. Thus, \u0394\u03d5\u2032 is given by \u0394\u03d5\u2032 = \u0394\u03d5 \u2212 V: (10.21) The size of the layer, which extends from xp to xn, would accordingly change because in Eq. (10.20), \u0394\u03d5 would be replaced by \u0394\u03d5\u2032 in the expressions for xn and xp: From Eqs. (10.20) and (10.21), we obtain V xn,p\u00f0 \u00de = \u00b1 \u00b11 Na/Nd\u00de \u2208 \u00f0 2\u03c0e \u00f0 \u0394\u03d5\u2032 Na + Nd\u00de \" 1/2 : # (10.22) Eq. (10.22) can be rewritten in the alternate form, by using the expressions for xn,p\u00f0 Eqs. (10.20) and (10.21), 0 \u00de and \u0394\u03d5\u2032 from V xn,p\u00f0 \u00de = xn,p\u00f0 0 V \u0394\u03d5 1/2 : \" 1 \u2212 \u00de! (10.23) This scenario is shown in Figures 10.9 and 10.10. We will first discuss the scenario when there is no external potential in the p-n junction. Due to the thermal excitation of electrons in the valence band on the n-side of the depletion layer, holes that move at random are generated. These holes are the \u201cminority carriers\u201d in the n-side compared to the electrons that are the \u201cmajority carriers,\u201d but they play an important role. The holes are swept away from the n-side to the p-side as soon as they wander into the depletion layer because there is a strong electric field \u0394\u03d5 across the layer. The current due to the motion of the holes from : In addition, the holes the n-side to the p-side is known as the hole generation current, jhg\u00f0 V = 0 0 \u00f0 \u00de \u00de \u03c6(x) xp xn n \u2212 p + x \u03c6(x) p \u2212 n + xp xn x FIGURE 10.9 FIGURE 10.10 The potential \u03d5 x \u00de \u00f0 bias). \u0394\u03d5\u2032 = \u03d5 \u221e \u00de \u00f0 versus x when V > 0 (forward \u2212 \u03d5 \u2212\u221e \u00de \u00f0 \u2212 V : The potential \u03d5 x \u00de \u00f0 bias). \u0394\u03d5\u2032 = \u03d5 \u221e \u00de \u00f0 versus x when V < 0 (reverse \u2212 \u03d5 \u2212\u221e \u00de \u00f0 + V : 312 CHAPTER 10 Electronics that are thermally generated and are the majority carriers in the p-side flow to the n-side of the junction. The motion of these holes, which are positively charged, is opposed by the electric field of the depletion layer, but those holes of which the thermal energy is greater than the potential barrier would reach the n-side and would recombine with the electrons. The current due to the motion of the holes from the p-side to the n-side of the depletion layer is known as the hole recombination : Because there is no external potential, the sum of the two currents is zero. Thus, current, jhr\u00f0 \u00de we obtain 0 0 jhg\u00f0 + jhr\u00f0 If we write jhg = e\u2135hg and jhr = e\u2135hr, where \u2135hg and \u2135hr are the number density of holes generated in 0 the n-side and the p-side of the depletion layer, respectively, it follows that at V = 0, \u2135hg\u00f0 \u00de and Eq. (10.24) is satisfied because the positively charged holes flow in opposite directions. We also note that \u2135hr is approximately the same as Na: Therefore, when an external potential V is applied across the p-n junction, = \u2135hr\u00f0 (10.24) = 0: 0 0 \u00de \u00de \u00de From Eqs. (10.21) and (10.25), we obtain \u2135hr \u221d e\u2212e\u03b2\u03d5\u2032: (10.25) = \u2135hr\u00f0 In contrast, the external potential V would have no effect on the number of holes crossing the V \u2135hr\u00f0 (10.26) 0 \u00de \u00de e\u03b2eV : depletion layer. Thus, we obtain The net number of holes moving across the depletion layer is given by V \u2135hg\u00f0 \u00de = \u2135hg\u00f0 0 \u00de = \u2135hg: (10.27) V = \u2135hr\u00f0 The expression for the hole current density, jh =", " e\u2135h\u00f0 \u2135h\u00f0 V \u00de \u00de V : V + \u2135hg\u00f0, is obtained from Eqs. (10.26) through (10.28), (10.28) \u00de \u00de jh = jhr \u2212 jhg = e\u2135hg\u00f0 e\u03b2eV \u2212 1 : \u00de (10.29a) We can make a similar argument for the motion of the electrons that are the majority carriers in the n-side and the minority carriers in the p-side. However, the electrons are negatively charged particles, and therefore, only those electrons that have enough thermal energy to cross the potential barrier can move from the n-side to the p-side. The current produced by the motion of the negatively charged elec. In contrast, the electrons that reach the edge of trons is known as the electron generation current jeg\u00f0 the p-side of the p-n junction will be swept across the depletion layer and will eventually recombine with the holes in the n-side. This is known as the electron recombination current, jer\u00f0 jeg\u00f0 + jer\u00f0 If we write jeg = \u2212e\u2135eg and jer = \u2212e\u2135er, where \u2135eg and \u2135er are the number current density of electrons and generated in the p-side and the n-side, respectively, it follows that at V = 0, \u2135eg\u00f0 : We can write = \u2135er\u00f0 (10.29b) = 0 FIGURE 10.11 The j (V ) characteristic of the p-n junction. 10.3 Rectification by a p-n Junction 313 Eq. (10.25) is satisfied because the electrons flow in opposite directions. We also note that \u2135er is approximately the same as Nd: Therefore, we can express \u00de 0 V e\u03b2eV : \u00de \u2135er\u00f0 = \u2135er\u00f0 However, the external potential does not make any difference in the motion of the electrons generated in the p-side and moving across the depletion layer to the n-side. Thus, we obtain (10.30) (10.31) V \u2135eg\u00f0 = \u2135eg\u00f0 0 = \u2135eg: \u00de \u00de The net number of electrons moving across the depletion layer is given by + \u2135eg\u00f0 (10.32) V \u00de \u00de : \u00de From Eqs. (10.30) through (10.32), we obtain V \u2135e\u00f0 V = \u2135er\u00f0 = jer\u00f0 The total electrical current density is obtained from Eqs. (10.29) and (10.33), = e\u2135eg\u00f0 \u2212 jeg\u00f0 je\u00f0 V V V \u00de \u00de \u00de \u00de : e\u03b2eV \u2212 1 V j \u00f0 \u00de V = jh\u00f0 \u00de V + je\u00f0 \u00de = e \u2135hg + \u2135eg\u00de\u00f0 \u00f0 e\u03b2eV \u2212 1 : \u00de (10.33) (10.34) Eq. (10.34) is the characteristic property of rectifiers and is at the heart of modern transistors. A schematic diagram of the j characteristic for the p-n junction is shown in Figure 10.11. V \u00f0 \u00de 10.3.2 Nonequilibrium Case (V \u2260 0) and then introduced the In the preceding discussion, we considered the equilibrium case external potential V in an ad hoc manner to explain the elementary theory of rectification by a p-n V \u2260 0 junction. We will now consider the nonequilibrium problem : The nonequilibrium problem is the direct generalization of the equilibrium problem : When V \u2260 0, one has to solve the \u00de Boltzmann equation for semiconductors. In the relaxation-time approximation, the Boltzmann equation for semiconductors is different from that of metals because of the presence of electrons and holes. In the relaxation-time approximation, the Boltzmann equation for electrons can be written as (in analogy with Eqs. 6.73 and 6.74g \u2202t = \u2212_r. \u2202g \u2202r \u2212 _k. \u2202g \u2202k \u2212 g \u2212 f \u03c4n, (10.35) where g is the density of particles at position r. In the presence of weak applied fields, g is replaced, which is the occupation number of electrons at position r, which have a wave vector k at t by grk\u00f0 time t. It can be easily shown from Eq. (6.83) that at constant temperature \u00de grk \u2248 frk \u2212 e\u03c4n \u2202f \u2202\u03bc vk. E: (10.36) 314 CHAPTER 10 Electronics The electron density nc at position r in the", " conduction band can be written as while the equilibrium density of electrons in the conduction band can be written as nc = 1 4\u03c03 Z grkdk, n0 c = 1 4\u03c03 Z f dk: (10.37) (10.38) Integrating Eq. (10.35) over dk, we obtain from Eqs. (10.35) through (10.38) (see Problem 10.6) \u2202nc \u2202t = \u2212 \u2202 \u2202r. <_r> + n0 c \u2212 nc \u03c4n : (10.39) Here, n0 over the Brillouin zone, c is the equilibrium density of the electrons in the conduction band, and <_r> is vk averaged <_r> = 1 4\u03c03nc Z dk grkvk = 1 4\u03c03nc Z dk f \u2212 \u03c4nvk. eE! % \u2202f \u2202\u03bc + \u2202f \u2202r vk: \" & Because fdk vanishes by symmetry, \u2202f /\u2202\u03bc = \u03b2f and f \u2248 g, Eq. (10.40) can be rewritten as R <_r> \u2248 1 4\u03c03nc Z dk \u2212\u03c4nvk. eE\u03b2g +! % \u2202g \u2202r & \" vk: We define the mobility \u03bcn and the diffusion constant Dn as and \u03bcn = e\u03b2 3 <\u03c4nv2 k> Dn = 1 3 <\u03c4nv2 k> = \u03bcn e\u03b2 : From Eqs. (10.41) through (10.43), we obtain <_r> = \u2212\u03bcnE \u2212 Dn nc \u2202nc \u2202r : The electron current is given by Similarly, the hole current can be obtained as je = \u2212enc<_r> = enc\u03bcnE + eDn \u2207!nc: jh = epv\u03bcpE \u2212 eDp \u2207!pv: From Eq. (10.39) and its analogous equations for holes, Eqs. (10.45) and (10.46), we obtain \u2202nc \u2202t = 1 e \u2207!. je + n0 c \u2212 nc \u03c4n (10.40) (10.41) (10.42) (10.43) (10.44) (10.45) (10.46) (10.47) 10.3 Rectification by a p-n Junction 315 and \u2202pv \u2202t = \u2212 1 e \u2207!. jh + p0 v \u2212 pv \u03c4p : (10.48) Because we have assumed that the carriers in the doped semiconductor primarily move in the x direction, Eqs. (10.45) and (10.46) can be rewritten as and je = enc\u03bcnE + eDn jh = epv\u03bcpE \u2212 eDp dnc dx dpv dx : Similarly, Eqs. (10.47) and (10.48) can be rewritten as \u2202nc \u2202t = 1 e \u2202je \u2202x + n0 c \u2212 nc \u03c4n \u2202pv \u2202t = \u2212 1 e \u2202jh \u2202x + p0 v \u2212 pv \u03c4p : and We note that i = n0 n2 cpv = p0 vnc: (10.49) (10.50) (10.51) (10.52) (10.53) In the steady state, \u2202nc \u2202t = \u2202pv \u2202t = 0: Thus, in the steady state, we can rewrite Eqs. (10.51) and (10.52) as and 1 e \u2202je \u2202x + n0 c \u2212 nc \u03c4n = 0 \u2212 1 e \u2202jh \u2202x + p0 v \u2212 pv \u03c4p = 0: (10.54) (10.55) In the regions, where E \u2192 0, the majority carrier density is constant. In these regions, from Eqs. (10.49) and (10.50), we obtain and je\u00f0 xp\u00de \u2248 eDn dnc dx jx = xp jh\u00f0 xn\u00de \u2248 \u2212eDp dpv dx jx = xn: (10.56) (10.57) 316 CHAPTER 10 Electronics Eqs. (10.56) and (10.57) essentially state that in the steady state, the minority carrier drift current can be ignored compared to the minority carrier diffusion current. From Eqs. (10.54) through (10.57), we obtain and Dn d2nc dx2 + n0 c \u2212 nc \u03c4n = 0 Dp d2pv", " dx2 + p0 v \u2212 pv \u03c4p = 0: It can be easily shown (Problem 10.7) that in the n-side of the depletion layer, for x \u2265 xn, x pv\u00f0 \u00de \u2212 pv\u00f0 \u221e \u00de = pv\u00f0 xn\u00de \u2212 pv\u00f0 \u00bd \u221e \u00de% x\u2212xn e\u2212 \u00f0 /dp, \u00de and in the p-side of the depletion layer, for x \u2264 xp, (10.58) (10.59) (10.60) xp\u00de nc\u00f0 where the diffusion lengths for holes in the n-doped region and electrons in the p-doped region are defined as \u2212 nc\u00f0 \u2212 nc\u00f0 nc\u00f0 x (10.61) \u2212\u221e \u2212\u221e e\u00f0 = \u00de% \u00de \u00de \u00bd \u00de x\u2212xp /dn, dp = Dp\u03c4p\u00de \u00f0 1/2 and dn = 1/2: Dn\u03c4n\u00de \u00f0 (10.62) In fact, we can define two diffusion regions, which extend over a distance of the order of the diffusion length in either side of the depletion layer, sandwiched between the depletion layer and the homogeneous regions. The diffusion regions do not exist when V = 0: Thus, we have three regions when V \u2260 0: The difference between the three regions is that in the depletion layer, the electric field, the space charge, and the carrier density gradients are large (Eq. 10.23), but there is no current in the equilibrium case because the drift and the diffusion currents are equal and opposite for both electrons and holes. When V \u2260 0, there is a net current in the depletion layer due to the difference between the drift and diffusion currents of each carrier type, which are large. In the diffusion regions, the drift current of the majority carrier density is quite large, whereas that of the minority carrier is small. However, the diffusion current due to both the carriers is appreciable. In contrast, the only current in the homogeneous regions is the majority carrier drift current. These regions are shown in Figure 10.12. V = 0 \u00f0 \u00de We will now calculate the total current j flowing in the p-n junction for a given value of V: We make the assumption that the passage of carriers across the depletion layer is so swift that the generation and recombination of electrons and holes within the layer are negligible. In the steady state, the total current j = je + jh of electrons and holes will be constant. Thus, je and jh can be evaluated separately at any arbitrary point in the depletion layer. The electron current je can easily be calculated at the boundary between the depletion layer and the diffusion region on the p-side, and the hole current jh can easily be calculated at the boundary between the two regions on the n-side xp\u00de \u00f0 10.3 Rectification by a p-n Junction 317 Diffusion region Diffusion region Homogeneous region Depletion layer Homogeneous region nc(\u2212\u221e) = 2 ni Na FIGURE 10.12 xp + dp xp O xn xn + dn x pv(\u221e) = 2 ni Nd The depletion layer, the two diffusion and homogeneous regions when V \u2260 0: in Figure 10.12. From Eq. (10.60) and Problem 10.4, we obtain the solution of the diffusion xn\u00de \u00f0 equations and x pv\u00f0 \u00de = x nc\u00f0 \u00de = n2 i Nd n2 i Na + pv\u00f0 \u00bd xn\u00de \u2212 n2 i Nd% x\u2212xn e\u2212 \u00f0 /dp, \u00de x \u2265 xn (10.63) + nc\u00f0! xp\u00de \u2212 n2 i Nd \" x\u2212xp e\u00f0 /dn, \u00de x \u2264 xp: (10.64) Note that both x and xp are negative in Eq. (10.64). From Eqs. (10.56), (10.57), (10.63), and (10.64), we obtain expressions for the minority carrier currents at the edges of the depletion layer (x = xn and x = xp), and From Problem 10.8, and je\u00f0 xp\u00de = eDn dn \u2212 xp\u00de nc\u00f0! n2 i Na \" jh\u00f0 xn\u00de = eDp dp \u2212 xn\u00de pv\u00f0! n2 i Nd : \" \u2248 Ndee\u03b2 \u00bd V x \u00f0 \u2212V \u00de xn \u00f0 \u00de% x nc\u00f0 \u00de \u2248 Nae\u2212e\u03b2 \u00bd V \u2212V x \u00de xp \u00f0 \u00f0 \u00de%: x pv\u00f0 \u00de (10.65) (10.66) (10", ".67) (10.68) 318 CHAPTER 10 Electronics From Eqs. (10.67), (10.68), and Problem 10.9, and From Eqs. (10.65) and (10.69), nc\u00f0 xp\u00de = Ndee\u03b2 V\u2212\u0394\u03d5 \u00f0 \u00de = pv\u00f0 xn\u00de = Naee\u03b2 V\u2212\u0394\u03d5 \u00f0 \u00de = n2 i Na ee\u03b2V n2 i Nd ee\u03b2V : je\u00f0 xp\u00de = en2 i Dn dnNa \u00f0 ee\u03b2V \u2212 1, \u00de and from Eqs. (10.66) and (10.70), The total current, j = je\u00f0 xp\u00de + jh\u00f0 Dp dpNd \u00f0 = en2 i xn\u00de jh\u00f0, is the sum of Eqs. (10.71) and (10.72), ee\u03b2V \u2212 1 \u00de : xn\u00de j = en2 i Dn dnNa + Dp dpNd \u2019 ee\u03b2V \u2212 1 : \u00de \u00f0 ( (10.69) (10.70) (10.71) (10.72) (10.73) Eq. (10.73) is the ideal diode, or Shockley, equation. We note that because Na and Nd appear in the denominator, the heavily doped side acts as a short circuit, whereas the current flow is in the lightly doped side. 10.4 TRANSISTORS 10.4.1 Bipolar Transistors In the bipolar transistor, three alternately doped layers of semiconductor (n-p-n or p-n-p) are sandwiched. We will discuss the n-p-n bipolar transistor in the following section. The middle layer (p) is the base that is very narrow, and the other two layers (n) are known as the emitter and the collector. Energy band diagrams of an n-p-n bipolar transistor with no applied voltage and with an applied voltage, which makes the emitter negative with respect to the collector, are shown in Figure 10.13. In Figure 10.13b, a voltage is applied such that the emitter is negative with respect to the collector. In such a scenario, the emitterbase junction is forward biased, whereas the Base p (a) p (b) Collector n Reverse bias n \u00b5 + Emitter n Forward bias \u2212 n FIGURE 10.13 (a) An n-p-n bipolar transistor with no applied voltage. (b) The same transistor with the emitter negative with respect to the collector. 10.4 Transistors 319 base-collector junction is reverse biased. In a p-n junction, each time an electron recombines with a hole in the base region, an acceptor ion is left behind that accumulates as more electrons flow through the junction and ultimately inhibits the flow of electrons. However, in a transistor, due to the contact with the base region, a current of holes flows into the base. The base remains electrically neutral because the number of holes drawn into the base compensates those that are lost due to recombination. The way in which a bipolar transistor acts as an amplifier is shown in Figure 10.14. When a small input of hole current iB is supplied to the base of the transistor, the base is positively charged, due to which the potential barrier between the base and the emitter is lowered. This results in an increase in the flow of electrons from the emitter to the base. Even if only a small percentage of the conduction electrons (from the emitter) recombine with the holes in the base while most of them flow into the collector, iE \u226b iB, so that base remains neutral. Thus, the flow of current due to the electrons in the collector, iC, is nearly equal to iE, and the current gain in the amplifier, \u03b1, which is defined as the ratio of change in the output current \u0394iC to the change in the input current \u0394iB, is given by FIGURE 10.14 iC iE iB p n n \u03b1 = \u0394iC \u0394iB : (10.74) Amplification of current in a bipolar transistor. Thus, a signal can be easily amplified by the bipolar transistor because a small change in the base current produces a very large change in the collector current. Source Gate Metal Oxide Drain n n p 10.4.2 Field-Effect Transistor The most common field-effect transistor (FET) is the metal-oxide-semiconductor FET, or the MOSFET. The structure of an n-type MOSFET is shown in Figure 10.15. In a MOSFET, a metal contact above the gate region is separated from the p-type semic", "onductor (Si) substrate by a thin oxide layer (usually SiO2), which is an insulator. When a positive voltage is applied to the metal contact above the gate region, the holes in the p-type semiconductor are repelled from the surface, whereas the electrons, which are the minority carriers, are attracted to the surface. This scenario is shown in Figure 10.16. FIGURE 10.15 The structure of an n-type MOSFET. + + + + + + + + \u2212 \u2212 \u2212 + + + + \u2212 \u2212 \u2212 \u2212 \u2212 FIGURE 10.16 Electrons are attracted to the surface when a positive voltage is applied to the gate. 320 CHAPTER 10 Electronics As the voltage is increased, more and more electrons, which are the minority carriers in the p-type semiconductor, are attracted to the surface. When the voltage increases beyond a the number of electrons threshold value, becomes greater than the number of holes near the surface even though it is a p-type semiconductor. This is known as the inversion layer, and the semiconductor behaves as if it is an n-type. This scenario is shown in Figure 10.17. \u2212+ \u2212 \u2212 \u2212 \u2212 \u2212 + \u2212 + + + \u2212 + + \u2212 \u2212 \u2212 FIGURE 10.17 Inversion layer near the surface, when the gate voltage is greater than the threshold voltage. Figure 10.17 explains how the MOSFET acts as a switch. The inversion layer allows the conduction electrons to flow from the source to the drain. The device is on when the current flows and off when the flow of current is stopped by decreasing the gate voltage below the threshold voltage. In such a situation, there is no inversion layer. An explanation of how the inversion layer is obtained is shown in Figure 10.18. The Fermi energy is close to the valence band edge in the p-type semiconductor when it is well below the interface with the oxide layer. However, the electron energies are lowered near the surface of the semiconductor because there is a positive charge on the other side of the oxide layer, and both the conduction and valence bands are bent. If the Fermi energy is closer to the conduction band than to the valence band in this region, as shown in Figure 10.18, the n-type behavior is restricted to this region. Because the semiconductor displays n-type behavior, the probability of inversion depends on the degree of band bending. The switching time of a transistor is defined as the minimum time period over which a transistor can be switched from off to on and again to the off position. The most important factor in designing integrated circuits is that a transistor must be in the on state until an electron (or hole) moves from the source to the drain. If it does not reach the drain before the transistor reaches the off position, no current will reach the drain, and the electronic device would not be switched on for any length of time. The MOSFET is eminently suitable to reduce the switching time in a Metal Oxide p-type semiconductor Inversion layer CB EF VB y g r e n E EF eVG FIGURE 10.18 The band picture of the inversion layer in the MOSFET. 10.4 Transistors 321 transistor. It may be noted that there are p-type MOSFETs, but because the mobility of the holes is much less than the mobility of the conduction electrons, the n-type MOSFETs can switch much faster than the p-type. 10.4.3 Single-Electron Transistor As we have noted, a MOSFET is essentially a parallel-plate capacitor, one plate of which is a metal and the other of which is silicon, a semiconductor. SiO2 is the insulator of the capacitor. Kastner et al.7 designed a single-electron transistor (SET) that had two metal gates. A schematic diagram of their device is shown in Figure 10.19. The electron gas, which is formed in the p-type Si by the positively biased upper gate, which is continuous as in the conventional MOSFET, is confined by the lower gate, which has a narrow gap of about 70 nm width. The narrow channel is approximately 20 nm wide and 1\u201310 \u03bcm long. The surface electrons are isolated from the bulk by the p-n junctions. The electron-rich region is called an inversion layer. When the voltage in the upper gate is positive with respect to the semiconductor while the bottom gate is neutral or negative, electrons are added to the semiconductor only under the gap in the bottom gate. Figure 10.19b shows how the electron gas is confined to move in one direction. If Vg is the gate voltage, it can be easily shown that (10.75) e\u0394 N/L = C/L \u00de \u0394Vg, \u00f0 where C is the capac", "itance, L is the length, e is the electronic charge, and N is the number of electrons added. It was observed that when conductance g was plotted against the electron density (N/L), oscillations of the conductance were periodic. This is shown in Figure 10.20. \u00de \u00f0 From Eq. (10.75) and Figure 10.20, it was observed that the transistor\u2019s conductance oscillated as a function of the number of electrons per unit length. In fact, there is a special length L0, such Inversion region Lower gate Upper gate 70 nm CVD oxide Lower gate Thermal oxide Si (p) (a) 70 nm (b) FIGURE 10.19 Schematic diagram of the (a) cross-section and (b) top view of the silicon transistor with a continuous upper gate and a gap in the lower gate. Reproduced from Kastner et al.7 with the permission of the American Physical Society. 322 CHAPTER 10 Electronics 2 \u00b5m Long T < 100 mK \u223c ).40 8.45 8.50 8.55 VG (volts) FIGURE 10.20 Conductance C versus gate voltage Vg in singletransistor Si devices. Reproduced from Kastner et al.7 with the permission of the American Physical Society. Charges Potential: FIGURE 10.21 Narrow channel through which the electrons move. The metal gate with the gap is shown by the shaded area. The random distribution of charges near the interface is shown by black diamonds. The bottom sketch shows the electrostatic potential resulting from such impurities versus position along the channel. that the voltage difference for one period is that necessary to add some fixed number of electrons to some fixed length, e\u0394N = L0\u00f0 (10.76) \u0394Vg: C/L \u00de In a separate experiment, it was also observed that the period was not correlated to the channel length. The sensitivity of the conductance fluctuations to thermal cycling lead to the conclusion that the period was determined by the distribution of charges at the Si-SiO2 interfaces. These charges are always present even at densities of approximately 1010 per cm2 for the best interfaces. It was calculated that for the transistor that had 2 \u03bcm \u00d7 1 \u03bcm in size, there were approximately two charges adjacent to it. In fact, each sample had random distribution of charges that changed each time the sample was warmed to the room temperature. Kastner et al.7 guessed that the charges create potential barriers along the length of the transistor, as shown schematically in Figure 10.21. They postulated that each period of conductance oscillations corresponded to the addition of one electron to the distance between the charges. The proof that the period of oscillations did not correspond to the addition of two electrons due to spin degeneracy or four electrons due to the energy-band structure of Si depended on the measurement of the length L0 defined in Eq. (10.76). The measurement of L0 was facilitated by the discovery of small transistors in GaAs by Meirav et al.13 Their device is shown in Figure 10.22. Reproduced from Kastner et al.7 with the permission of the Using molecular-beam epitaxy (MBE), a layer of AlGaAs, which has larger band gap than GaAs, is grown on a heavily doped GaAs : Then a layer of pure GaAs is crystal \u00de grown where the electrons accumulate. The density of the electrons is controlled by the positive voltage applied to the n + substrate. A metal gate, which is negatively biased so that the electrons are repelled from it, is deposited on the top by electron-beam lithography. The negative bias on the top gates creates a potential barrier for the electrons moving down the narrow channel. American Physical Society. n + \u00f0 The advantage of the GaAs-AlGaAs interface is that the density of charges near the semiconductor and insulator is smaller than the Si-SiO2 interface. Figure 10.23 shows the conductance as a function of gate voltage Vg for two devices with different length. 10.4 Transistors 323 Double-Barrier Channel Metal Gate IDES GaAs (140 nm) AIGaAs (100 nm) n+ substrate Top View: 450 nm 100 nm 1 \u00b5m 3 \u00b5m FIGURE 10.22 The double-barrier channel shows a one-dimensional electron gas (1DES) or narrow two-dimensional gas forms at the top of the GaAs-AlGaAs interface. The density is controlled by the substrate voltage Vg : The top view shows the top metal gate structure, which has a narrow channel with two constrictions. Reproduced from Kastner et al.7 with the permission of the American Physical Society. 15 10 Sample 2 0 5 Sample 3 15 10 5 15 10 10 Gate voltage (mV from threshold) 20 0 5 15 20 FIGURE 10.23", " Conductance as a function of Vg for L0 = 0:8 \u03bcm (sample 2) and L0 = 0:6 \u03bcm (sample 3). Reproduced from Kastner et al.7 with the permission of the American Physical Society. Each period corresponds to the addition of the same number of electrons to the regions between constrictions. The voltage to add one electron is larger for a shorter segment because the capacitance is smaller. However, it is necessary to know the absolute capacitance of a segment to be able to estimate whether the number of electrons added per period is one or two. The charge density was obtained by solving the Poisson equation and estimated by a method similar to the Thomas\u2013Fermi approximation. The 324 CHAPTER 10 Electronics capacitance was calculated at each gate voltage by integrating the charge contained in the region between the two constrictions. The capacitance was compared with the measured period. It was conclusively proved that the period of oscillations corresponds to the addition of one electron per oscillation. Thus, the one-electron transistor turns on and off again each time an electron is added. We note from Figure 10.23 that the conductance of a small region of electron gas, separated by tunnel junctions from its leads, oscillates with density, which is explained by the simple Coulomb blockade model. A schematic picture of the tunnel barrier with a metal particle is shown in Figure 10.24. lead Metal particle lead e2/C EF FIGURE 10.24 A Coulomb blockade system. Electrons tunnel from one lead onto a small metal particle and then to the other lead. Reproduced from Kastner et al.7 with the permission of the Due to the Coulomb interaction between the electrons in the metal particle, the electrons cannot tunnel from one plate of the capacitor to the other plate through the metal particle. For current to flow, an electron or hole of charge q has to be added to the particle. This costs an energy q2/ 2C, where C is the capacitance between the particle and the rest of the system. Thus, there is an energy gap in the single-particle density of states. If we consider the tunneling of an electron or a hole, the gap width is e2/C: This is shown in Figure 10.24. American Physical Society. \u00b1e \u00de \u00f0 Because the potential difference between the gate and the electron gas is Vg, the isolated region of the transistor has an electrostatic energy \u03b5 = q2 2C \u2212qVg, (10.77) where \u2212qVg is the attractive interaction between the positively charged gate and the charge in the isolated region, and q2/ 2C is the repulsive term between two charges in the isolated region. We note that Eq. (10.77) can be written in the alternate form \u03b5 = q \u2212 q0\u00de \u00f0 2/2C, (10.78) where q0 = CVg and a constant q2 0/ 2C has been added to \u03b5: Any value of q0 can be added to minimize the energy, but because the charge is quantized, only discrete values of energy are possible for a given qo: Thus, either q0 = ne or q0 = e, where n is an integer. These two cases are n + 1/2 illustrated in Figure 10.25. We note that when q0 = e are degenerate. The energy gap in the tunneling density of states disappears because the charge fluctuates between the two values even at zero temperature. Because the conductance is thermally activated at all values of e, it has sharp peaks at low temperatures. The the gate voltage except those for q0 = \u00de change in voltage to alter q0 from (n + 1/2)e to (n + 3/2)e is \u0394Vg = e/C, which is the period in Vg: The activation energy at the minimum is e2/2C: e, the states where q = ne and q = \u00de n + 1/2 \u00f0 n + 1/2 n + 1 \u00f0 \u00de \u00f0 \u00de \u00f0 To summarize, the number of electrons on the isolated segment is quantized because the time for tunneling onto and off the segment is long. The Coulomb interactions and the quantization of 10.6 Optoelectronic Devices 325 q0 = ne q0 = (n + 1 2 )e E E e2/2C n \u2212 2 n \u2212 1 n q(e(e) FIGURE 10.25 Energy versus charge for q0 = ne and q0 = of energy are shown by solid circles. n + 1/2 \u00f0 e: Because the charge is quantized, the allowed values \u00de Reproduced from Kastner et al.7 with the permission of the American Physical Society. charge suppress all charge fluctuations at zero temperature for all values of q0 not equal", " to e and at that value the charge fluctuates only by one electron. This device is called a n + 1/2 \u00f0 Coulomb island because the Coulomb interaction suppresses the charge fluctuations. \u00de 10.5 INTEGRATED CIRCUITS An integrated circuit consists of capacitors, resistors, transistors, and other metallic connections required for a complete electrical circuit. The most popular electrical circuits are the MOSFET circuits because the switching time can be easily reduced. In addition, the transistors switch faster if the devices are made smaller. However, it is important to keep in check the power dissipated by a transistor so that the amount of heat produced in an integrated circuit can be kept under control. Due to the improvement of the technology in building integrated circuits, primarily due to the decrease in the individual devices as well as in the increase in the area of the circuit, there has been a rapid growth in the number of transistors on an integrated circuit since the first such circuit was fabricated in 1961 with only four transistors. At present, a typical integrated circuit has about 80 million transistors. The single most important criterion is to keep in check the enormous heat produced by such circuits. 10.6 OPTOELECTRONIC DEVICES Semiconductors can be utilized for a variety of optical properties. The simplest way to consider the role of a photon is that if it has energy \u2208 = h\u03bd greater than the band gap, it can excite an electron from the valence band to the conduction band, leaving behind a hole in the valence band. It may be noted that both the laws of conservation of momentum and energy have to be obeyed by the photons and electrons involved in the collision process where the energy is absorbed from the photon by the electron. This is usually possible only in direct band-gap semiconductors such as GaAs where the minimum of the conduction band is directly above the maximum of the valence band. This process, shown in Figure 10.26a, is known as photoconductivity because a beam of light with appropriate frequency can produce a large number of electrons and holes, thereby significantly 326 CHAPTER 10 Electronics h\u03bd (a) Conduction band Eg Valence band h\u03bd (b) FIGURE 10.26 (a) A photon excites an electron from the valence band to the conduction band. (b) An electron in the conduction band recombines with a hole and produces a photon. increasing the conductivity. In the reverse process, shown in Figure 10.26b, an electron in the conduction band recombines with a hole in the valence band and produces a photon of energy \u2208 = h\u03bd: This is the basis of both the light-emitting diode (LED) and the semiconductor laser. As emphasized earlier, only semiconductors of which the minimum of the conduction band lies directly above the conduction band can be used for such a process. A light-emitting diode constitutes a forward-biased p-n junction made from a direct band-gap material. The band gap of the semiconductor determines the wavelength of the emitted light. One can obtain light across most of the visible spectrum by using GaAsP if one varies the phosphorus content of the alloy semiconductor. A photodiode, shown in Figure 10.27, is used to detect light. One uses a reverse-biased p-n junction. In this junction, when a valence electron in the depletion region on the p-type region absorbs a photon of the appropriate frequency, it is excited to the conduction band, where it is the minority carrier. The strong electric field sweeps it across the depletion region to the n-type region, and hence, the electron contributes to the drift current. The magnitude of this current is proportional to the intensity of light. This process can be used to construct a solar cell. A laser (light amplification by stimulation of radiation) that produces coherent light is different from an LED in the sense that although the production of photons in an LED by the recombination of electrons and holes is a spontaneous process, in a laser the recombination of one-electron-hole pair triggers similar events in other electrons and holes, simultaneously producing a large number of photons of the same frequency. This is known as stimulated emission. The original diode laser, shown in Figure 10.28, had more electrons at the bottom of the conduction band than at the top of the valence band, which is obtained in a heavily doped p-n junction. CB \u2212 \u2212 Light in VB \u2212 h\u03bd p + CB VB n FIGURE 10.27 A reverse-biased p-n junction used as a photodetector. In Figure 10.28, the population inversion exists only near the middle of the depletion region even though the recombination events take place in a wide region. Consequently, most of these events have spontaneous emission rather than stimulated emission as required in lasers. The efficiency of the", " lasing process is low. Because a major proportion of the electrons and holes are lost due to spontaneous emission, to replenish their loss, a large current must flow through the junction in a continuous basis. This large current produces so CB VB \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212\u2212 \u2212 \u2212 \u2212 \u2212 + + + + + + + + + + + + + + + + + + + + \u2212 \u2212 \u2212 \u2212 \u2212 Light out CB VB p n FIGURE 10.28 Schematic diagram of a diode laser. 10.6 Optoelectronic Devices 327 much heat that the diode laser can be used for only short bursts or pulses. The efficiency of the diode laser is very low. However, the efficiency of the laser is considerably improved by using two heterostructures instead of a p-n junction. A schematic diagram of the doubleheterostructure laser is shown in Figure 10.29. In a double-heterostructure laser, the conduction electrons and holes are confined in the GaAs layer. Due to the fact that both the conduction electrons and holes collect in the same region, the proportion of stimulated emission is much larger than a diode laser. In view of the above, this type of laser does not get heated very quickly, and therefore, it can be used for a longer period. Semiconducting lasers are very widely used in communication systems using optical fibers. For best results, the wavelength is matched to the performance of the fiber. An InGaAsP laser, which is widely used in optical communication systems, is shown in Figure 10.30. Recent studies on semiconductor lasers has shown that a semiconductor laser generates a numberphased squeezed state rather than a squeezed state, primarily due to the fact that a semiconductor laser is pumped by a shot-noise-free electric current. When the squeezed state becomes phase coherent with an independent local laser oscillator, the squeezed light can be detected by optical homodyne detectors and used for various interferometric measurements. Thus, a phase-coherent squeezed state can be generated by injection locking the squeezed slave laser with an external master laser. A setup for generating a squeezed vacuum state by destructively interfering with an amplitudesqueezed state from an injection-locked slave laser with strong coherent light from a master laser is shown in Figure 10.31. A constant-current-driven semiconductor laser, which is denoted as a slave laser, is injection locked by an external master laser. Thus, the two signals, which are phase coherent, are combined at a high transmission mirror where the coherent excitation of the squeezed AIGaAs GaAs AIGaAs \u2212 \u2212 \u2212 \u2212 + + + Light out + p n n contact n InP substrate n InP InGaAsP active layer p InP SiO2 isolation layer Stripe contact FIGURE 10.29 FIGURE 10.30 Schematic diagram of a double-heterostructure laser. Schematic diagram of an InGaAsP laser. 328 CHAPTER 10 Electronics output signal from the injection-locked slave laser is canceled by the destructive interference with the master laser signal. The squeezed output signal of the slave laser is not degraded because the noise of the master laser signal is attenuated by the mirror. The actual experimental setup is shown in Figure 10.32. The master laser is an AlGaAs singlemode high-power semiconductor laser, and the slave laser is a single-mode low-power GaAs Master laser Injection signal Slave laser FR a2 Half-wave plate Squeezed vacuum Reflected master laser output Master laser output PBS HTM Signal output (a) a1 Squeezed vacuum state (b) Slave laser output FIGURE 10.31 Squeezed-vacuum-state generation by mixing an amplitude-squeezed state with a coherent state. FR, Faraday rotator; PBS, polarization beam splitter, HTM, high-transmission mirror. Reproduced from Yamamoto et al.6 with the permission of Elsevier. Master laser LD Isolator x 2 HWP PBS HWP PBS HWP PBS HWP PZT PBS NPBS HWP PBS HTM Cryostat 77\u00b0K HWP HWP LD Rotator PBS HWP Slave laser PBS CRT Fabry-Perot interferometer Optical spectrum analyzer Spectrum analyzer + \u2212 Differential amplifier Delay line \u03c4 \u223c\u2212100 nsec FIGURE 10.32 The experimental setup for the squeezed-vacuum-state generation by a semiconductor laser system; HWP, half-wave plate; PBS, polarization beam splitter; NPBS, nonpolarization bean splitter; HTM, high-transmission mirror; PZT, pizeo translator. Reproduced from Yamamoto et al.6 with the permission of Elsevier. transverse-junction strip semiconductor laser with antireflection coating and high-reflection coating on the rear facet. \u223c90% \u00f0 \u00de 10.", "7 Graphene 329 \u223c10% \u00f0 \u00de on the front facet 10.7 GRAPHENE Graphene is a flat monolayer of carbon atoms that is tightly packed into a two-dimensional honeycomb lattice. Graphene can be wrapped up into zero-dimensional fullerenes or rolled into onedimensional nanotubes. It can also be stacked into three-dimensional graphite. Figure 10.33 shows the different ways in which graphene can be wrapped or stacked. FIGURE 10.33 Graphene is a 2D carbon material that can be wrapped into 0D buckyballs, rolled into 1D nanotubes, or stacked into 3D graphites. Reproduced from Geim5 with the permission of MacMillan Publishers Ltd. Copyright 2011. 330 CHAPTER 10 Electronics Graphene and its bilayer are both zero-gap semiconductors (which can also be considered as zero-overlap semimetals) with one type of electron and one type of hole. The conduction and valence bands start overlapping for three or more layers. Thus, single-, double-, and few- (3- to 9-) layer graphene can be considered as three types of 2D crystals. In these crystals, the charge carriers can travel thousands of interatomic distances. Single- and few-layer graphenes have been grown epitaxially by chemical vapor deposition of hydrocarbons on metal substrates and thermal decomposition of SiC. Few-layer graphene obtained on SiC shows high-mobility charge carriers. Although epitaxial growth of graphene is the viable route for experimental applications, current experiments mostly use samples obtained by micromechanical cleavage of bulk graphite, which provides high-quality graphene crystallites up to 100 \u03bcm in size. Graphene becomes visible in an optical microscope if placed on an Si wafer with an appropriate thickness of SiO2: This is due to feeble interference-like contrast with an empty wafer. These high-quality graphene crystallites are shown in Figure 10.34. There is a pronounced ambipolar electric field effect in graphene that is shown in Figure 10.35. 0 9 A\u02da 13 A\u02da 1 \u00b5m 1 \u00b5m (a) (b) Crystal faces 10 \u00b5m (c) FIGURE 10.34 (a) Graphene visualized by atomic force microscopy. The folded region exhibiting a relative height of \u2248 4 A\u00b0 indicates that it is a single crystal. (b) Transmission electron microscopy image of a graphene sheet freely suspended on a micrometer-sized metallic scaffold. (c) Scanning electron microscopy of relatively large graphene crystal, which shows that most of the crystal\u2019s faces are zigzag and armchair edges. Reproduced from Geim5 with the permission of MacMillan Publishers Ltd. Copyright 2011 Ef E ky kx Ef 0 \u221260 \u221230 0 Vg(v) 30 60 FIGURE 10.35 Ambipolar electric field effect in single-layer graphene. Reproduced from Geim5 with the permission of MacMillan Publishers. Copyright 2011. 10.7 Graphene 331 The insets in Figure 10.35 show the conical low-energy spectrum E(k), which shows the change in the Fermi energy EF with gate voltage Vg: Positive Vg induces electrons and negative Vg induces holes. The concentration n = \u03b1Vg, where \u03b1 = 7:2 \u00d7 1010 cm\u22122V\u22121 for field-effect devices with a 300 nm SiO2 layer used as a dielectric. The rapid decrease in resistivity \u03c1 on adding charge carriers indicates their high mobility and does not change significantly when the temperature is increased to 300\u00ba K. The charge carriers in graphene can be tuned continuously between electrons and holes in concentrations n \u223c1013 cm\u22122 and mobilities \u03bc \u2265 15,000 cm2 V\u22121 s\u22121. A unique property of graphene is that the interaction of the electrons with the periodic potential of the honeycomb lattice gives rise to quasiparticles that are described at low energies E by a (2 + 1)\u2013dimensional Dirac equation with an effective speed of light vF = 106 m\u22121s\u22121: These quasiparticles are essentially electrons that have lost their rest mass me and are called massless Dirac fermions. The conical sections of the energy spectrum for < 1 eV are a consequence of the fact that graphene is a zero-gap semiconductor. The low-E quasiparticles within each valley can be described by the Dirac-like Hamiltonian, which is a direct consequence of graphene\u2019s crystal symmetry. A detailed analysis of the physical properties of graphene, including the origin of Dirac fermions, is given in Chapter 18 (on novel materials). The Hamiltonian H can be written as (for details, see Chapter 18): E j j H = \u0127", "vF \u2019 0 kx + iky kx \u2212 iky 0 ( = \u0127vF \u03c3!. k: (10.79) Here, k is the quasiparticle momentum, \u03c3! is the 2D Pauli matrix (but pseudospin rather than real spin), and the Fermi velocity vF, which is independent of k, plays the role of the speed of light. The honeycomb lattice of graphene is made up of two equivalent carbon sublattices, A and B. The cosine-like energy bands associated with the sublattices give rise to the conical sections of the energy spectrum for |E| < 1 eV (Figure 10.33) because they intersect at zero E near the edges of the Brillouin zone. In addition to the unique feature of the band structure E = \u0127vFk, the electronic states at zero E, where bands intersect, are a mixture of states of different sublattices. One has to use two-component wave functions (spinors) by requiring an index to identify sublattices A and B, to account for their relative contributions in the makeup of the quasiparticles. Thus, in Eq. (10.79), \u03c3! is referred as the pseudospin. The real spin of the electrons is described by additional terms in the Hamiltonian. However, the pseudospin effects, which are inversely proportional to the speed of light c (note that vF plays the role of c) in quantum electrodynamics (QED), dominate the effects due to real spin because c/vF \u2248 300: 332 CHAPTER 10 Electronics In QED, chirality (which is positive or negative for electrons or holes, respectively) is defined as the projection of \u03c3! on the direction of motion k. In graphene, the intricate connection of chirality is a consequence of the fact that k electrons and \u2212k hole states originate from the same carbon sublattices. Both chirality and pseudospin are conserved in graphene. \u00f0 \u03bc \u00de 10.8 GRAPHENE-BASED ELECTRONICS of the charge carriers of graphene does not decrease even in the highest-fieldThe high mobility induced concentrations and remains unchanged by chemical doping. Thus, there can be ballistic transport on a submicrometer scale at room temperature. The switching time is reduced due to the large value of vF and low-resistance contacts without a Schottky barrier. The on\u2013off ratios, which, do not create any proare comparatively low for graphene because of poor conductivity blem for high-frequency applications. In fact, it has been shown that transistors can be operational at THz frequencies. \u2248100 \u00de \u00f0 The fact that graphene remains metallic even at the neutrality point creates a problem for mainstream logic applications. However, it has been shown that significant semiconductor gaps, \u0394E, can be induced in bilayer graphene, which can be used in tunable infrared lasers and detectors. \u0394E can also be induced in single-layer graphene by spatial confinement or lateral-superlattice potential. If graphene is epitaxially grown on top of crystals with matching lattices such as BN or SiC, such superlattice effects are likely to occur. \u22480:3 eV \u00f0 \u00de It can be shown that the confinement gap for graphene is \u0394E eV \u00f0 \u00de \u2248 \u03b1\u0127vF/d \u2248 1/d nm, \u00de \u00f0 (10.80) where the coefficient \u03b1 \u2248 0:5 for Dirac fermions. For room-temperature operations, d \u2248 10 nm, which is achievable with the rapidly advancing Si-based technology. However, hitherto, no technique has been found to make anisotropic etching of graphene to make devices with crystallographicdefined faces in order to avoid irregular edges. Electronic states associated with short irregular edges in short channels usually induce a significant sample-dependent conductance, whereas those associated in long channels lead to additional scattering. Graphene can also be used as a conductive sheet where the different nanometer-size structures can be carved to make a single-electron transistor circuit. Graphene nanostructures are stable down to nanometer sizes that allow the exploration of a region between SET and molecular electronics. Figure 10.36 shows an SET made from graphene by using electron-beam lithography and dry etching. Figure 10.36b shows that for a minimum feature size of \u224810 nm, the combined Coulomb and confinement gap reaches >3 kBT: This would allow a SET-like circuitry operational at room temperature, and resistive barriers can be used to induce a Coulomb blockade. There are two impediments for growth of graphene electronics. High-quality wafers suitable for industrial applications are yet to be developed despite the recent progress in epitaxial growth", " of graphene. In addition, individual features in graphene devices have to be controlled to provide reasonably accurate reproducibility in their properties. Problems 333 0.3\u00b0.05 0.1 Vg(V) (a) ) S n ( \u03c3 80 60 40 20 0 300\u00b0K 100 nm \u221210 10 0 Vg(V) (b) FIGURE 10.36 (a) Coulomb blockade in large quantum dots (diameter \u22480.25 \u03bcm) at low temperatures. Narrow constrictions in graphene serve as quantum barriers. (b) Here, 10 nm\u2013scale graphene structures are stable under ambient conditions and survive thermal cycling to liquid-helium temperature. The inset shows a scanning electron micrograph of two graphene dots of \u2248 40 nm in diameter with narrower (<10 nm) constrictions. Reproduced from Geim5 with the permission of MacMillan Publishers Ltd. Copyright 2011. PROBLEMS 10.1. In Eqs. (10.3) and (10.4), we derived nc\u00f0 x and = \u2135c\u00f0 T \u00de \u00de e\u2212\u03b2 x \u03b5c\u2212e\u03d5 \u00de \u00f0 \u00f0 \u2212\u03bc \u00de Because Nd = nc\u00f0 \u221e \u00de x \u03b5v\u2212e\u03d5 \u00f0 \u2212\u03bc \u00de \u00de: \u00f0 pv\u00f0 x \u00de \u2212\u221e T e\u03b2 \u00de = \u2118v\u00f0, show that \u00de \u2212\u221e \u00de \u00f0 = Eg + and Na = pv\u00f0 \u2212 e\u03d5 \u221e \u00de e\u03d5 \u00f0 1 \u03b2 ln NdNa T \u2118v\u00f0 \u2135c\u00f0! \u00de T : \" \u00de 10.2. From the preceding expressions, show that and = Nde\u2212e\u03b2 \u03d5 \u00bd \u221e \u00de x \u2212\u03d5 \u00f0 \u00f0 \u00de% = Naee\u03b2 \u03d5 \u00bd \u2212\u221e \u00f0 \u2212\u03d5 \u00de x \u00de%: \u00f0 x nc\u00f0 \u00de x pv\u00f0 \u00de 10.3. Within the boundaries of the depletion layer \u00f0 Outside the depletion layer, nc = Nd on the n-side and pv = Na on the p-side: Neglecting the boundary effects, show that, outside the depletion layer, x = xp and x = xn\u00de, nc \u226a Nd and pv \u226a Na: = 0, \u03c1 x \u00de \u00f0 (1) (2) (3) (1) (2) (1) 334 CHAPTER 10 Electronics and, inside the depletion layer, Nd\u00f0 10.4. In the \u201cabrupt junctions,\u201d one can approximate x \u00f0 = e \u03c1 \u00de \u00bd x \u2212 Na\u00f0 x \u00de% : \u00de and x Nd\u00f0 \u00de = % Nd, 0, x > 0 x < 0 x Na\u00f0 \u00de = % Na, 0, x < 0 x > 0 & : & Show that the one-dimensional Poisson\u2019s equation (Eq. 10.10) can be written as \u22022\u03d5 \u2202x2 = By integrating Eq. (3), show that 8 >>>>>< >>>>>: 0, \u2212 4\u03c0eNa \u2208 4\u03c0eNd \u2208, 0, x < xp, 0 > x > xp 0 < x < xn x > xn : 9 >>>>>= >>>>>; \u03d5 \u03d5 \u2212\u221e \u00de \u00f0 \u2212\u221e \u2212, # 2\u03c0eNa \u2208 # 2\u03c0eNd \u2208 x \u2212 xp\u00de \u00f0 $ 2, x \u2212 xn\u00de \u00f0 $ x < xp 2, 0 > x > xp 0 < x < xn x > xn 9 : >>>>>= >>>>>; = \u03d5 x \u00de \u00f0 8 >>>>>< >>>>>: 10.5. From Eq. (10.6) and (10.7), prove the law of mass action, pv\u00f0 Since, for an intrinsic semiconductor, nc\u00f0 nc\u00f0 nc\u00f0 x,T \u221e \u00de \u00de x,T \u00de x,T T e\u2212\u03b2Eg : T = \u2135c\u00f0 = pv\u00f0 \u00de \u2118v\u00f0 \u00de x,T \u00de \u00de = ni\u00f0 \u00de = Nd (the n-side of the depletion layer), using Eq. (1), show that T, and from Eq. (10.12), \u221e pv\u00f0 \u00de \u2248 n2 T i \u00f0 Nd \u00de, where Similarly, because pv\u00f0 \u2212\u221e \u00de e\u2212\u03b2Eg/2: = Na in the p-side of the depletion layer, show that = ni\u00f0 ni\u00f0 T 0 \u00de", " \u00de nc\u00f0 \u2212\u221e \u00de \u2248 n2 T i \u00f0 Na \u00de : (2) (1) (2) (3) (4) (1) (2) (3) (4) Problems 335 10.6. It has been shown that \u2202g \u2202t = \u2212_r: \u2202g \u2202r \u2212 _k. \u2202g \u2202k \u2212 g \u2212 f \u03c4n, (1) where g is the density of particles at position r. In the presence of weak-applied fields, g is, which is the occupation number of electrons at position r, which have a t replaced by grk\u00f0 wave vector k at time t. It can be easily shown from Eq. (6.83) that at constant temperature \u00de grk \u2248 frk \u2212 e\u03c4n \u2202f \u2202\u03bc vk. E: The electron density nc at position r in the conduction band can be written as while the equilibrium density of electrons in the conduction band can be written as nc = 1 4\u03c03 Z grkdk, n0 c = 1 4\u03c03 fdk: Z Integrating Eq. (1) over dk, show from Eqs. (1) through (4), \u2202nc \u2202t = \u2212 \u2202 \u2202r. <_r> + n0 c \u2212 nc \u03c4n : Here, n0 averaged over the Brillouin zone, c is the equilibrium density of the electrons in the conduction band and <_r> is vk <_r> = 1 4\u03c03nc Z dk grkvk = 1 4\u03c03nc Z dk f \u2212 \u03c4nvk. eE! % \u2202f \u2202\u03bc + \u2202f \u2202r vk: & \" 10.7. It has been shown that in the steady state, for the minority carriers, and Dn d2nc dx2 + n0 c \u2212 nc \u03c4n = 0 Dp d2pv dx2 + p0 v \u2212 pv \u03c4p = 0: Show that in the n-side of the depletion layer, for x \u2265 xn, pv\u00f0 \u2212 pv\u00f0 xn\u00de and in the p-side of the depletion layer, for x \u2264 xp, \u2212 pv\u00f0 pv\u00f0 \u221e = x \u00de \u00de \u00bd \u221e e\u2212 \u00f0 x\u2212xn /dp, \u00de \u00de% x nc\u00f0 \u00de \u2212 nc\u00f0 \u2212\u221e \u00de = nc\u00f0 xp\u00de \u00bd \u2212 nc\u00f0 \u2212\u221e \u00de% x\u2212xp e\u00f0 /dn, \u00de (2) (3) (4) (5) (6) (1) (2) (3) (4) (5) (1) (2) (3) (4) 336 CHAPTER 10 Electronics where the diffusion lengths for holes and electrons are defined as dp = Dp\u03c4p\u00de \u00f0 1/2 and dn = 1/2: Dn\u03c4n\u00de \u00f0 10.8. From Eqs. (10.49) and (10.50), we have je = enc\u03bcnE + eDn jh = epv\u03bcpE \u2212 eDp dnc dx dpv dx : = Nde\u03b2 \u00bd V \u2212V x \u00de \u00f0 \u00f0 xn x nc\u00f0 \u00de x je eNdDn Z xn \u00de% 1 + 2 4 dx\u2032e\u2212\u03b2 \u00bd V x\u2032 \u00f0 \u2212V \u00de xn \u00f0 \u00de% 3 5 and Show that and x pv\u00f0 \u00de = Nae\u2212\u03b2 \u00bd V x \u00f0 \u00de \u2212V xp \u00f0 \u00de% 1 \u2212 2 dx\u2032e\u03b2 V \u00bd x\u2032 \u00f0 \u2212V \u00de xp \u00f0 \u00de% : 3 x jh eNaDp Z xp 6 4 7 5 Hence, show that the second term in the square bracket can be neglected compared to the first term in both Eqs. (3) and (4). 10.9. We derived in Chapter 9 that We also derived in Eq. (10.8), n2 i = \u2135c\u00f0 T T \u2118v\u00f0 \u00de e\u2212\u03b2Eg: \u00de e\u0394\u03d5 = Eg + 1 \u03b2 ln NdNa T \u2118v\u00f0 \u00de T : \" \u00de \u2135c\u00f0! Show from Eqs. (1) and (2) that i = NdNae\u2212e\u03b2\u0394\u03d5: n2 (1) (2) (3) References 1. Ashcroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 2. Bardeen J. Surface states and rectification at a", " metal semi-conductor contact. Phys Rev 1947;71:717. 3. Bardeen J, BrattinWH. The transistor, a semi-conductor triode. Phys Rev 1948;74:230. References 337 4. Cohen ML, Chelikosky JR. Electronic structure and optical properties of semiconductors. Berlin: Springer- Verlag; 1989. 5. Geim AK, Novoselov KS. The rise of graphene. Nat Mater 2007;6:183. 6. Yamamoto Y, Inoue S, Bjork G, Heitmann H, Matinaga F. In: Kapon E, editor. Semiconductor lasers, vol. 1. Amsterdam: Elsevier; 1999:361. 7. Kastner MA. The single-electron transistor. Rev Mod Phys 1992;64:849. 8. Kimmerling LC, Kolenbrander KD, Michel J, Palm J. Light emission from silicon. Solid State Phys Adv Res Appl 1996;50:333. 9. Kittel C. Introduction to solid state physics. New York: John Wiley & Sons; 1976. 10. Klingston CF. Semiconductor optics. Berlin: Springer-Verlag; 1995. 11. Maiman TH. Stimulated optical radiation in ruby. Nature 1960;147:493. 12. Marder MP. Condensed matter physics. New York: John Wiley & Sons; 2000. 13. Meirav U, Kastner MA, Heiblum M, Wind SJ. Phys Rev B 1989;40:5871. Spintronics CHAPTER 11 CHAPTER OUTLINE 11.1 Introduction............................................................................. 339 11.2 Magnetoresistance...................................................................... 340 11.3 Giant Magnetoresistance................................................................. 340 11.3.1 Metallic Multilayers................................................................ 340 11.4 Mott\u2019s Theory of Spin-Dependent Scattering of Electrons...................................... 342 11.5 Camley\u2013Barnas Model.................................................................... 345 11.6 CPP-GMR............................................................................... 348 11.6.1 Introduction....................................................................... 348 11.6.2 Theory of CPP-GMR of Multilayered Nanowires....................................... 350 11.7 MTJ, TMR, and MRAM.....", "............................................................... 352 11.8 Spin Transfer Torques and Magnetic Switching.............................................. 356 11.9 Spintronics with Semiconductors.......................................................... 357 11.9.1 Introduction....................................................................... 357 11.9.2 Theory of an FM-T-N Junction...................................................... 358 11.9.3 Injection Coefficient............................................................... 361 Problems.................................................................................... 364 References.................................................................................. 367 11.1 INTRODUCTION As we noted in Chapter 10 on electronics, charges are manipulated by electric fields, but spins are generally ignored. Magnetic recording and other techniques use the spin only through the magnetization of a ferromagnet. When giant magnetoresistance (GMR) of the magnetic multilayers was discovered in 1988, an efficient control of the electrons was achieved through the orientation of their magnetization by acting on their spin. The application of the GMR to the read heads of the hard discs contributed significantly to the quick rise in the density of stored information. This led to the extension of hard disc technology and consumer electronics. A large number of phenomena related to the control and manipulation of spin currents were developed through a new area of physics/materials science called spintronics. The rapidly growing area of research in spintronics includes such phenomena as spin transfer, molecular spintronics, spintronics with semiconductors, and single-electron spintronics. Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00011-6 \u00a9 2012 by Elsevier Inc. All rights reserved. 339 340 CHAPTER 11 Spintronics 11.2 MAGNETORESISTANCE In Chapter 3, we discussed the Hall effect by considering a conductor in the shape of a rod that has a rectangular cross-section, which is placed under a magnetic field B in the z direction. There is a longitudinal electric field Ex: The electric and magnetic fields are so adjusted that the current cannot : The charges pile up on the surface of the sample, flow out of the rod in the y direction thereby setting up an electric field Ey: This field, which nullifies the Lorentz force (Eq. 3.90), is known as the Hall field. The Hall coefficient RH, which defines the size of the carrier, is defined as RH = Ey=Bjx: The component of the resistivity tensor \u03c1, \u03c1yx, is the Hall resistance defined as \u03c1yx = Ey=jx, whereas the diagonal component \u03c1xx is the magnetoresistance, defined as", " jy = 0 \u00de \u00f0 \u03c1xx = Ex jx, and the magnetoresistance ratio is defined as \u0394\u03c1 \u03c10 = \u03c1 \u2212 \u03c10 \u03c10, (11.1) (11.2) where \u03c1 and \u03c10 are the resistivities (along a given direction) in the presence and absence of a magnetic field, respectively. The dependence of the electrical resistance of the material on an applied magnetic field, usually perpendicular to the direction of the current, represents this effect. 11.3 GIANT MAGNETORESISTANCE 11.3.1 Metallic Multilayers A typical multilayer unit is shown in Figure 11.1. In a magnetic multilayer system, one of the metals (A) is magnetic, whereas the other (B) is nonmagnetic, which is referred to as the spacer layer. When the thickness of the spacer layer is varied, there are oscillations in the magnetic coupling between the magnetic layers. The thickness of these thin films can vary from a few tenths of a nanometer to tens of nanometers. The magnetization directions of the ferromagnetic layers are coupled to each other through an exchange interaction. The sign of this coupling oscillates as the thickness of the spacer layer is varied. The best multilayer samples have around 30 periods of oscillations. Some metallic multilayers exhibit drastic changes in magnetoresistance and, hence, the name giant magnetoresistance (GMR). The drastic change of magnetoresistance in metallic multilayers has resulted in a new definition of the magnetoresistance ratio, MR, as MR = RAP \u2212 RP RP \u00d7 100, (11.3) where RP and RAp are the resistances of the parallel and antiparallel magnetic configurations, respectively. In the mid-1980s, it became possible to develop molecular-beam epitaxy (MBE) and other techniques to fabricate multilayers composed of very thin individual layers. The explanation of the Capping layer Magnetic layer (A) Spacer layer (B) A B A B A Substrate H = 0 FIGURE 11.1 H > HC A multilayer structure and the changes in magnetization directions with an applied magnetic field H. The parallel magnetizations lead to a large magnetoresistance compared to that of an antiparallel structure. Reproduced from Fernando 6 with the permission of Elsevier. 11.3 Giant Magnetoresistance 341 M M M M FIGURE 11.2 Spin-dependent electron scattering and redistribution of scattering events upon anti-alignment of magnetization. Reproduced from Grunberg11 with the permission of the American Physical Society. GMR effect includes spin-dependent electron scattering and redistribution of scattering events due to anti-alignment of magnetizations. This scenario is shown in Figure 11.2. The microscopic explanation for the GMR effect is that the scattering rates of the electrons depends on the orientation (parallel or antiparallel) of the electron spins with respect to the local magnetizations. Figure 11.2 shows two structures, one with parallel and the other with antiparallel alignment with the magnetization. The electrons with spin parallel to the local magnetization in their random walk are not scattered in the ideal situation. In Figure 11.2, only one passage from the left to the right is shown as a sample of the electron motion. A short circuit is caused by the electrons that are not scattered. However, when the electron with spin-up enters the layer where the magnetization has been turned around, its spin is opposite to the local magnetization. The bottom diagram in Figure 11.2 shows the increase in resistivity due to the removal of the short circuit. In practice, although both types of electrons are scattered, the resistivity due to the antiparallel magnetization is more than that due to parallel magnetization. In Figure 11.3, the arrows represent the majority spin direction in the magnetic layers. In the ferelectrons are weakly scattered everywhere, which romagnetic (F) configuration, the spin+ \u00de gives a short circuit effect and hence a small resistivity. In the antiferromagnetic (AF) configuration, each spin direction is scattered at every second magnetic layer, and the resistivity is higher because there is no short circuit effect. sz = 1/2 \u00f0 The short circuit effect for the ferromagnetic (F) configuration is visually explained in Figure 11.4. experience little or In the parallel (P) configuration, majority electrons in the (+) channel sz = 1/2 \u00f0 \u00de 342 CHAPTER 11 Spintronics M NM M M NM M M M NM \u2212 + AF FIGURE 11.3 \u2212 + F Schematic picture of the electron trajectories. The electron trajectories between two scatterings are represented by straight lines and the scatterings by abrupt change in direction. The + and \u2013 are for electron spins sz = 1/2 and sz = \u22121/2, respectively", ". Reproduced from Fert et al.10 with the permission of Elsevier. (+) Channel (\u2212) Channel FIGURE 11.4 Schematic diagram of the short circuit effect. Reproduced from Fernando6 with the permission of Elsevier. no resistance, and hence, a short circuit effect occurs. In the antiparallel (AP) configuration (not channels will experience a significant resistance shown), electrons in the (+) and (\u2013) when going through the slab with opposite magnetization with no short circuit effects. sz = \u22121/2 \u00f0 \u00de 11.4 MOTT\u2019S THEORY OF SPIN-DEPENDENT SCATTERING OF ELECTRONS A simple model of a scattering of electrons, which was given by Mott15 will be presented here. It was modified later by more complex models. We consider a ferromagnetic d-band metal with a magnetization M(T) at temperature T. The schematic density of states representing s-, p-, and d-bands of a transition metal is shown in Figure 11.5. \u00f0 T 1 \u2212 \u03b2 =M0, a fraction \u00de If M0 is the saturation magnetization at T = 0, =2 of the and \u03b2 = M \u00de unoccupied d states will have their spins parallel to M, and a fraction =2 will have their \u00de E E spins antiparallel. If we define \u03c11\u00f0 \u00de as the respective density of states, then they will have a parabolic form, and \u03c12\u00f0 1 + \u03b2 \u00f0 \u00de \u00f0 E \u03c11\u00f0 \u00de = a E1 \u2212 E \u00de \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 p (11.4) and E d (E) \u03c1 \u2193 (E) \u03c1\u2193 E = EF d \u03c12\u00f0 E \u00de = a E2 \u2212 E \u00de \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03, (11.5) s & p p where E1 and E2 are the highest energies the two spins can have. However, above the Curie temperature, where the system becomes paramagnetic FIGURE 11.5 Density of states plotted against E for s, p, and d states of a transition metal. 11.4 Mott\u2019s Theory of Spin-Dependent Scattering of Electrons 343 and the number of parallel spins equals the number of antiparallel spins, E1 = E2 = E0: It can be shown that the corresponding relaxation times are related to the density of states as and 1 \u03c41 E \u221d \u03c11\u00f0 \u00de + \u03b4 1 \u03c42 E \u221d \u03c12\u00f0 \u00de + \u03b4, (11.6) (11.7) where \u03b4 is a contribution from spin-independent scattering. From Eqs. (11.4) through (11.7), it can be shown that (Problem 11.1) 1 \u03c41 = A \u00f0 T/MI\u03b82 \u00de \" and 1 \u03c42 T/MI\u03b82 = A \u00f0 \u00de \" E1 \u2212 E \u00de \u00f0 E0 \u2212 \u03b60\u00de p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 p E2 \u2212 E \u00de \u00f0 E0 \u2212 \u03b60\u00de p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 + \u03b1 # + \u03b1, # (11.8) (11.9) p where \u03b1 \u2248 1/4, \u03b8 is the Debye temperature, and A is a constant. If \u03b60 represents the energy at T", " = 0 and \u03b2 = 0, and \u03b60\u2032 represents the energy at the highest occupied state (Fermi energy) at T = 0 when the states are split, because the density of states is proportional to n1/3 0, we can write (Problem 11.2) and E1 \u2212 \u03b60\u2032 E0 \u2212 \u03b60 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 s = 1 \u2212 \u03b2 \u00f0 1/3 \u00de E2 \u2212 \u03b60\u2032 E0 \u2212 \u03b60 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 s 1 + \u03b2 = \u00f0 1/3: \u00de (11.10) (11.11) One can write the expression for conductivity by using a Drude-type formula, but including the energy dependence of the relaxation time, \u03c3 = \u2212 Ne2/m \u03c41 + \u03c42\u00de 2 \u00f0 \u2202f \u2202E dE: Z (11.12) The resistivity \u03c1 Eq. (11.12) to be nonzero only when E = \u03b60\u2032: We obtain, after some algebra (Problem 11.3), = 1/\u03c3 is obtained by approximating the partial derivative of the integrand in \u03b2, T \u00f0 \u00de \u03b2, T \u03c1 \u00f0 = A \u00f0 \u00de T/m\u03b82 1 1//3 + \u03b1 1 + \u03b2 \u00de \u00f0 \u22121 : # (11.13) 344 CHAPTER 11 Spintronics Because \u03b1 \u2243 0:25, the ratio of the resistances between the saturated magnetization case paramagnetic case is given by and the \u00de \u00de If we define the resistances for up and down channels as R\u2191 and R\u2193, then the resistances RP and RAP \u00f0 \u00f0 \u2243 0:34: (11.14) \u03c1 \u03c1 \u03b2 = 1, T \u03b2 = 0, T for the parallel and antiparallel alignments of spins relative to a given magnetic layer are given by and RP = \u00f0 1/R\u2191 + 1/R\u2193 \u22121 \u00de RAP = \u00f0 R\u2191 + R\u2193 /4, \u00de (11.15) (11.16) when the mean path of the electrons is much higher than the repeat length of the multilayer. From Eqs. (11.3), (11.15), and (11.16), we obtain where MR = \u00f0 2 1 \u2212 \u03b7 \u00de 4\u03b7 \u00d7 100%, \u03b7 = R\u2193 R\u2191 : (11.17) (11.18) Because \u03b7 is a positive parameter, it follows from Eq. (11.17) that RAP > RP: This is a very simplified explanation of the higher resistance encountered in the antiparallel case. When an applied field changes an alignment from antiferromagnetic (AF) to ferromagnetic (F) alignment, the difference in resistivity is the largest. The AF alignment is usually provided by interlayer exchange or by coercivities of successive magnetic layers, by pinning the magnetization using an antiferromagnetic material in direct contact, known as exchange biasing. If GMR is obtained by exchange biasing, it is called a spin-valve system. The first discovery of GMR was done by Baibich et al.1 on Fe/Cr magnetic multilayers, in which it is possible to switch the relative orientation in adjacent magnetic layers from antiparallel to parallel by applying a magnetic field. The resistivity is strongly enhanced in the antiparallel magnetic configuration of two adjacent layers A and B because the electrons in each channel are slowed down at every second magnetic layer. There is no such enhancement in layers A and B in the parallel magnetic configuration because the electrons can go easily through all the magnetic layers, and the short circuit through this channel leads to a small resistance. This opens up the possibility of switching between high and low resistivity states by changing the relative orientation of the magnetizations of A and B layers from parallel to antiparallel. Similar GMR effects were discovered by Binash et al.2 in Fe/Cr/Fe trilayers. These effects are shown later in Figure 11.9. Camley and Barnas3 presented a theoretical description of the GMR effects in Fe/Cr/Fe trilayers by calculating the resistivity using Boltzmann transport equations, with spin-dependent scattering at the interface. 11.5 Camley\u2013Barnas Model 345 11.5 CAMLEY\u2013BARNAS MODEL The Camley\u2013Barnas model3 includes the idea that there is an antifer", "romagnetic coupling between Fe films due to the intervening Cr films. At the zero field, the resulting magnetic moments of neighboring Fe films are antiparallel to each other. However, in a strong external magnetic field, all the magnetic moments of the Fe films can be forced to lie in the same direction. Earlier experimental results suggested that the resistance is the largest when the magnetic moments in neighboring Fe films are antiparallel and smallest when they are parallel. Further, multilayer structures with thin Fe films have a much larger magnetoresistance than a single sandwich structure of Fe/Cr/Fe. In addition, the magnetoresistance is increased by a factor of 2 or 3 when the temperature is changed from room temperature to that of liquid He. Camley and Barnas interpreted these results to conclude that a spin-dependent scattering is responsible for the observed effects and the relative lengths of the mean-free path are more important than the thickness of the various films. To develop a theory for the multilayer, Camley and Barnas first considered a single sandwich structure of Fe/Cr/Fe, as shown in Figure 11.6, and computed the conductivity by using the Boltzmann equation. In Figure 11.6, the dashed line in the center of the Cr films is the position at which the change in axis of quantization for the electron spin is calculated. In each region, the Boltzmann equation reduces to a differential equation that depends on the coordinate z only, \u2202g \u2202z + g \u03c4vz = eE mvz \u2202f0 \u2202vx, (11.19) where f0 is the equilibrium distribution function, \u03c4 is the relaxation time, E is the external field in the x direction, and g is the correction to the distribution function due to scattering. Fe A Cr B C Fe D E y x z Z FIGURE 11.6 The geometry of the Fe/Cr/Fe sandwich structure. 346 CHAPTER 11 Spintronics In the region A, the contributions to g for regions for spin-up or spin-down electrons moving to the right (positive vz) or left (negative vz) are g = gA+\u2191\u00f0 vz, z + gA+\u2193\u00f0 \u00de vz, z \u00de and g = gA\u2212\u2191\u00f0 From Eqs. (11.19) through (11.21), it can be easily shown that (Problem 11.4) + gA\u2212\u2193\u00f0 vz, z vz, z \u00de \u00de : and gA\u00b1\u2191\u00f0 vz, z \u00de = eE\u03c4 m \u2202f0 \u2202vx 1 + A\u00b1\u2191 exp \u2213z vzj\" \u03c4 % # j $ gA\u00b1\u2193\u00f0 vz, z \u00de = eE\u03c4 m \u2202f0 \u2202vx 1 + A\u00b1\u2193 exp \u2213z : vzj\" \u03c4 % # j $ (11.20) (11.21) (11.22) (11.23) The coefficients A\u00b1\u2191 and similar coefficients for the regions B, C, and D are determined through \u00f0 the boundary conditions. The distribution function g for an electron leaving the surface (at the outer surfaces of the sandwich) is equal to the distribution function g for an electron of the same spin striking the surface multiplied by a specular scattering event R0: Thus, we obtain \u2193 \u00de and gA+\u2191 = R0gA\u22121 at z = \u2212b gD\u2212\u2191 = R0gD+\u2191 at z = +b: (11.24) (11.25) Similar equations are obtained for down-spins. Camley and Barnas3 assumed a model system of two equivalent simple metals that have the same Fermi energies, mean-free path values, and so on. They neglected the angular dependence of scattering and assumed that there is only transmission or diffusive scattering at the Fe/Cr interfaces. They further assumed that the scattering at the outer boundaries is purely diffusive (the reflection coefficients are zero). If the transmission coefficients are T\u2191 for up-spins and T\u2193 for down-spins, for up-spins at z = \u2212a, and gA\u2212\u2191 = T\u2191 gB\u2212\u2191 \u2193 \u00f0 \u00de \u2193 \u00f0 \u00de \u2193 \u00f0 \u00de gB+\u2191 \u00f0 \u2193 \u00de = T\u2191 gA+\u2191 \u00f0 \u2193 \u00de : \u2193 \u00f0 \u00de (11.26) (11.27) and gC\u2212\u2191 A similar set of equations holds for the z = +a interface. Similar boundary conditions are obtained for, except that one has to account for the fact that the magnetic moments in the two gB\u2212\u2191 \u00f0 films are in different directions by an angle \u03b8, which is the angle between the magnetization vectors in the two Fe films. We", " define the transmission coefficients, \u2193 \u00f0 \u2193 \u00de \u00de T\u2191\u2191 = T\u2193\u2193 = cos\u20612 \u03b8/2 \u00de \u00f0 (11.28) 11.5 Camley\u2013Barnas Model 347 and T\u2191\u2193 = T\u2193\u2191 = sin\u20612 \u03b8/2, \u00de \u00f0 (11.29) where T\u2191\u2191 is the probability of an electron of spin-up (in layer A, with respect to the magnetization) at z = \u22120 to continue as a spin-up (in layer D, with respect to magnetization) at z = 0, and the other symbols are defined similarly. We obtain (11.30) (11.31) (11.32) (11.33) (11.34) and gB\u2212\u2191 \u2193 \u00f0 \u00de = gC\u2212\u2191 cos\u20612 \u03b8/2 \u00de \u00f0 \u2193 \u00f0 \u00de + gC\u2212\u2193 sin\u20612 \u03b8/2 \u00de \u00f0 \u2191 \u00f0 \u00de \u00de The current density at different fields are obtained by using the expression \u00de \u00f0 \u00f0 \u00de \u00de \u00de gC+\u2191 \u2193 \u00f0 = gB+\u2191 \u2193 \u00f0 cos\u20612 \u03b8/2 + gB+\u2193 sin\u20612 \u03b8/2 : \u2191 \u00f0 The current of the whole structure is obtained by integrating J \u00f0 Camley and Barnas defined the diffusive scattering parameter, z \u00de over the coordinate z: = J z \u00de \u00f0 Z vx, z vxg \u00f0 \u00de d3v: D\u2191 = 1 \u2212 T\u2191, and the asymmetry in up-spin and down-spin scattering, N = D\u2191/D\u2193: Their results are plotted in Figure 11.7. 0.08 0.06 0.04 0.02 Fe Cr Fe E DCBA x y z a b \u2212b \u2212a 0.2 0.4 0.6 D1 0.8 N = 12.5 N = 2 N = 1.5 N = 1 1.0 FIGURE 11.7 The maximum normalized change in resistance as a function of D1 and N: The inset shows the geometry shown in Figure 11.6. Reproduced from Camley and Barnas3 with the permission of the American Physical Society. 348 CHAPTER 11 Spintronics 1.6 1.2 0.8 0. ( Theory Experiment \u22121.8 \u22121.2 \u22120.6 0 B [10\u22121 T ] 0.6 1.2 1.8 FIGURE 11.8 The percentage change of resistance as a function of an applied field results are from Binash et al.2 from Figure 11.9b. N = 6, D1 = 0:48 \u00f0 : The experimental \u00de Reproduced from Camley and Barnas3 with the permission of the American Physical Society. The calculations shown in Figure 11.7 were made to show how the magnetoresistance effect depends on D\u2191 and N for a (120 \u00c5 Fe)/(10 \u00c5 Cr)/(120 \u00c5 Fe) sandwich structure with a mean-free path of \u03bb = 180 \u00c5. The theoretical value of the percentage change in resistance as a function of an applied field for a (120 \u00c5 Fe)/(10 \u00c5 Cr)/(120 \u00c5 Fe) sandwich structure is shown in Figure 11.8. In the preceding calculations, the mean-free path is \u03bb = 180 \u00c5, and the angle \u03b8 between the magnetizations in the two Fe films was calculated by minimizing the sum of the exchange, anisotropy, and Zeeman energies for the Fe/Cr/Fe sandwich. The first observation of giant magnetoresistance was discovered by Baibich et al.1 and is shown in Figure 11.9a. It may be noted that if we define MR = (RAP \u2212 RP\u00de /RP \u00d7 100, MR = 85% for the (Fe 3 nm/Cr 0.9 nm) multilayer. The experimental results of Binash et al.2 for Fe/Cr/Fe trilayers is shown in Figure 11.9b. A schematic diagram of the electrons in parallel (low resistance) and antiparallel (high resistance) spin configurations is shown in Figure 11.9c. As we noted earlier, GMR can be obtained in the current in plane (CIP) and current perpendicular to plane (CPP) geometry. At present, the CIP configuration is used for most sensor applications, and in most experiments the current flows in the CIP geometry. However, Pratt et al.17 showed that for the Ag/Co multilayers, CPP-MR is several times larger than the CIP-GMR. There has been considerable interest in CPP-MR, which is more", " likely to be used in future sensor applications. 11.6 CPP-GMR 11.6.1 Introduction The first experiment of CPP-GMR was done by sandwiching a magnetic multilayer between superconducting electrodes. This restricted the use of such multilayers only at very low temperatures. However, Fert and Piraux9 as well as other groups have fabricated magnetic nanowires by electrodepositing into 11.6 CPP-GMR 349 R/R(H = 0) 1 (Fe 30 A\u00b0 /Cr 18 A\u00b0 )30 HS 0.8 0.7 0.6 0.5 80% (Fe 30 A\u00b0 /Cr 12 A\u00b0)35 HS (Fe 30 A\u00b0 /Cr 9 A\u00b0)60 HS 20 30 40 \u221240 \u221230 \u221220 \u221210 0 10 Magnetic field (kG) (a) 1.5 1.0 0.5 0 ) % ( 1 1 R R \u2206 / MR = 1.5% Fe 250 A\u00b0 F1 M F2 F1 M F2 \u22121500 \u22121000 \u2212500 500 1000 1500 0 B0 (10\u22126 T) (b) (c) FIGURE 11.9 (a) GMR in Fe/Cr(001) multilayers. MR = 85% for the (Fe 3 nm/Cr 0.9 nm) multilayer. (b) GMR in Fe/Cr/Fe trilayers. (c) Schematic of the mechanism of GMR. Reproduced from Fert 7 with the permission of American Physical Society. 350 CHAPTER 11 Spintronics pores of membranes. These nanowires usually have a diameter in the range 30\u2013500 nm for a length of 10 \u03bcm: The multilayered nanowires can be composed of a stack of layers of different metals with thicknesses in the nanometer range. The general technique for fabrication of magnetic multilayered nanowires in nanoporous polymer membranes (Figure 11.10b) consists of a pulse-plating method in which two metals are deposited from a single solution by switching between the deposition potential of two constituents. These types of multilayers include Co/Cu, NiFe/Cu, Ni/Cu, and Fe/Cu. Other types of multilayered nanowires such as Ni/NiO/Co heterostructures and Co/Pb multilayers have been grown by electrodeposition. We will now summarize the theory of the perpendicular magnetoresistance in magnetic multilayers (CPP-MR) derived by Valet and Fert22 (the V\u2013F model) and in magnetic nanowires by Fert and Piraux.9 11.6.2 Theory of CPP-GMR of Multilayered Nanowires The V\u2013F theory uses the Boltzmann equation to calculate the transport properties of magnetic multilayers for currents perpendicular to the layer. Their model takes into account both volume and interface scattering and includes a spin-lattice relaxation term \u03c4sf, which describes the relaxation of spin accumulation by spin-flip scattering. The data on the perpendicular magnetoresistance can be used to separate the volume and interface scattering. The notations generally used in the calculation of the magnetoresistance of a multilayer are as follows. The spin \u2191 (majority) and spin \u2193 (minority) resistivities of the ferromagnetic metal are defined by where \u03b2 is the bulk-scattering spin asymmetry coefficient. The (equal) spin resistivities of the nonmagnetic metal are defined by \u03c1F \u2193 \u2191 \u00f0 \u00de = 2\u03c1#F\u00bd 1 \u2212 \u00f0 +\u03b2, \u00de% (11.35) \u03c1N \u2193 \u2191 \u00f0 \u00de = 2\u03c1#N, (11.36) and the interface resistances per unit area, r\u2191 and r\u2193, are defined by r\u2191 \u00f0 \u2193 = 2r#b\u00bd 1 \u2212 + \u03b3, \u00de % \u00f0 \u00de where \u03b3 is the interface asymmetry coefficient, tN and tf are defined as the thickness, and lN sf as the spin diffusion length (SDL) of the nonmagnetic and ferromagnetic layers, respectively. In the long SDL limit, where tN \u226a lN sf, Valet and Fert22 showed that the resistances RP (parallel) and RAP (antiparallel) of a unit area of the multilayer (which is also valid for nanowires for a true antiparallel configuration and for a state with zero net magnetization in a volume of the cube of the SDL) are sf and tF \u226a lF sf and lF (11.37) and where N is the number of periods. RAP = N \u03c1#FtF + \u03c1#NtN + 2r#b\u00de \u00f0 RP", " = RAP \u2212 f \u03b2\u03c1#FtF + 2\u03b3r#bg RAP 2 N2, (11.38) (11.39) 11.6 CPP-GMR 351 CIP-GMR T = 4.2\u00b0 K T = 300\u00b0 K Polycarbonate 40 nm AF ) % ( o i t a r R M 80 70 60 50 40 30 20 10 0 0 70 60 50 40 30 20 10 ) % ( / P R R \u2206 80 60 40 20 0 \u22128 \u22124 4 0 H (kG) 8 F AF F AF 6 nm 10 20 30 40 50 60 70 Cu thickness (A\u00b0) (a) 60 nm 10 ) % ( Cu layer Co layer Cu film (b) CPP-GMR 400 nm 0 \u221210 \u22125 5 10 0 H(kG) (c) FIGURE 11.10 0 100 200 300 400 500 Co thickness (nm) (d) (a) Variation of the GMR ratio of Co/Cu multilayers as a thickness of Cu in the CIP geometry. (b) Schematic of an array of multilayered nanowires in nanoporous track-etched polymer membranes. (c) The variation of /Cu the CPP-GMR \u00de (dotted lines) multilayered nanowires as a function of H. The inset shows the same plot at 4.2\u00ba K for the Ni80Fe20/Cu sample. (d) Variation of the CPP-GMR ratio of Co/Cu multilayered nanowires as a thickness of the Co layers. at 77\u00ba K for Ni80Fe20\u00f0 (solid lines) and Co(10 nm)/Cu(5 nm) \u0394R/RP \u00f0 4 nm \u00f0 12 nm \u00de \u00de Reproduced from Fert7 with the permission of the American Physical Society. From Eqs. (11.38) and (11.39), it can be shown that (Problem 11.5) RAP \u2212 RP RAP \" \u22121/2 # = \u03c1#FtF + 2r#b \u03b2\u03c1#FtF + 2\u03b3r#b + \u03c1#NtN \u03b2\u03c1#FtF + 2\u03b3r#b : (11.40) The magnetoresistance properties of multilayered nanowires helps considerably in the understanding of the CPP-GMR and the determination of the spin diffusion length. In addition, in current-induced 352 CHAPTER 11 Spintronics switching (or reversal) of magnetization, nanowires can be used. They are ideal for high-injection density currents in order to probe the change in spin configuration of multilayers. In Figure 11.10a, the large and oscillatory GMR effects in Co/Cu, which were discovered simultaneously by Mosca et al.13 and Parkin et al.16, are shown. In fact, these effects became the archetypical GMR system. In 1991, Dieny et al.5 observed the GMR in spin valves, i.e., trilayered structures in which the magnetization of one magnetic layer is pinned by coupling with the antiferromagnetic layer while the magnetization by the second layer is free. A small magnetic field can reverse the magnetization of the free layer; this concept is now used in most applications in spin valves. Figure 11.10b shows a schematic picture of an array of multilayered nanowires in nanoporous tracketched polymer membranes and the technique of fabrication, the theory of which was described earlier. Due to the spin accumulation effects that occur in the CPP geometry, the length scale of the spin transport becomes the long spin diffusion length. In fact, the CPP-GMR has demonstrated the spin accumulation effects that determine the propagation of a spin-polarized current through magnetic and nonmagnetic materials. The CPP-GMR plays an important role in all recent developments of spintronics. As one can see in Figures 11.10c and d, in the CPP, the GMR is not only higher than in CIP, but also subsists in multilayers with relatively thick layers, up to the micron range. The CPP-GMR has demonstrated the spin accumulation effects that govern the propagation of a spin-polarized current through a series of magnetic and nonmagnetic materials. Thus, it plays an important role in the development and future use of spintronics. The key mechanism driving a spin-polarized current at a large distance from the interface is the diffusion current induced by the accumulation of spins at the magnetic\u2013nonmagnetic interface. For example, spin-polarized currents can be transported in long carbon nanotubes because the SDL is quite large beyond the micron range. When an electron flux crosses the interface between a ferromagnetic and a nonmagnetic", " material, far from the magnetic side, the current is large in one of the spin channels. However, when the flux is far from the interface on the other side, the current is equally distributed in both channels. This scenario is shown in Figure 11.11. Figure 11.11a shows the spin-up and spin-down currents far from an interface between ferromagnetic and nonmagnetic conductors outside the spin accumulation zone. Figure 11.11b shows the splitting of the chemical potentials EF\u2191 and EF\u2193 at the interface. When the current travels through the spin-accumulation zone, it is polarized due to the inversion of the spin accumulation and opposite spin flips. The spin flips control the gradual depolarization of the current due to the left and the right. Figure 11.11c shows the variation of the current spin polarization on both sides of metal/metal, where there is a balance between the spin flips on both sides, and the metal/nonmagnetic semiconductor (without spin-dependent interface resistance), where the spin flips dominate in the left side. Thus, the current is nearly depolarized when it enters the semiconductor. One can introduce a tunnel junction, which results in a discontinuity of the spin accumulation at the interface, thereby increasing the depolarization from the metallic to the semiconductor side. However, due to the large tunnel resistances, it is difficult to efficiently transform the spin information into an electrical signal. 11.7 MTJ, TMR, AND MRAM There have been significant advances in the research on the tunneling magnetoresistance (TMR) of magnetic tunnel junctions (MTJ). The MTJs are tunnel junctions with ferromagnetic electrodes of which the resistances are different for parallel and antiparallel configurations. In addition, the two 11.7 MTJ, TMR, and MRAM 353 (a) Ferromagnetic Nonmagnetic Zone of spin accumulation LF sf LN sf (b) E Spin accumulation \u2206EF = EFup \u2212 EFdown Z EFup EFdown (c) Metal/metal Current spin polarization = ( jup \u2212 jdown)/( jup + jdown) Metal/semiconductor Z FIGURE 11.11 Schematic representation of the spin accumulation at an interface between a ferromagnetic and a nonmagnetic layer. Reproduced from Fert7 with the permission of the American Physical Society. 354 CHAPTER 11 Spintronics metallic layers that are called electrodes are separated by an insulating layer thin enough to allow some tunneling current. Julliere13 did one of the early experiments on MTJ by using the Fe-Ge-Co system. The conductance G(V) measurements, made for Fe and Co when the average magnetizations were parallel and antiparallel, showed a difference related to the spin polarizations of the tunneling conduction electrons. Julliere used a simple (stochastic) model for tunneling electrons and denoted the fractions of electrons, of which the magnetic moments are parallel to the magnetizations in Fe and Co, as a and a\u2032: The conductance of the Fe-Ge-Co junction when the magnetizations in Fe and Co are parallel can be expressed as21 and antiparallel Gp\u00de \u00f0 \u00f0 aa\u2032 + GAP\u00de \u00f0 Gp \u221d \u00bd 1 \u2212 a 1 \u2212 a\u2032 \u00de\u00f0 \u00de% and Assuming that the spin is conserved, Gap \u221d a 1 \u2212 a\u2032 \u00de + a\u2032 \u00f0 \u00f0 \u00bd 1 \u2212 a : \u00de% where the conduction electron spin polarization of the two ferromagnetic metals are13 TMR = GP \u2212 Gap Gap = 2PP\u2032 1 \u2212 PP\u2032, \u00de \u00f0 P = 2a \u2212 1 and P\u2032 = 2a\u2032 \u2212 1 : (11.41) (11.42) (11.43) (11.44) The original value measured by Julliere was 14%, mainly due to interface roughness. Recent experiments indicate that in MTJs consisting of ferromagnetic amorphous CoFeB and MgO, the TMR can be as large as 500%. In fact, it appears that MgO is crucial for achieving large TMR values. The polarization of tunneling electrons depends on the barrier height of the insulator in addition to the polarization of the ferromagnets. Slonczewski19 made the simple assumption that the ferromagnets had parabolic but spin-split bands that are separated by an insulating barrier. The schematic picture is shown in Figure 11.12. A schematic picture of the one- and two-band parabolic models is shown in Figure 11.13. U U0 1 Z 2 Z \u2032 Z 3 \u03b8 h = hA Y X EF h = 0 h = hB X \u2032 Y \u2032 0 d \u03be FIGURE 11.12 Schematic picture of an insulator between two different types of", " ferromagnets. Reproduced from Slonczewski19 with the permission of the American Physical Society. 11.7 MTJ, TMR, and MRAM 355 E EF = EF2 h0 EF = EF1 \u2013 0 \u2212h0 0 When the barrier has low energy and thin size, the wave function of the two ferromagnets has to be matched at the interfaces. Slonczewski19 showed that P\u2032, the polarization of the tunneling electrons, depends on the polarization of the ferromagnets as well as on the barrier height, P\u2032 = k\u2191 \u2212 k\u2193 k\u2191 + k\u2193 $ % \u03ba2 \u2212 k\u2191k\u2193 \u03ba2 + k\u2191k\u2193 $, % (11.45) \u03c1 where k\u2191 and k\u2193 are the Fermi wave vectors for the up- and down-spin bands, and \u03c1 FIGURE 11.13 \u03c1\u2191\u00de \u00f0 and spin-down Density of spin-up electrons. The positions of the Fermi energy for oneand two-band band ferromagnet are shown schematically. models of a EF 1\u00de \u00f0 EF 2\u00de \u00f0 \u03c1\u2193\u00de \u00f0 Reproduced from Slonczewski19 with the permission of the American Physical Society., p \u0127\u03ba = 2m \u00f0 (11.46) Vb \u2212 EF% \u00bd \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 where Vb is the barrier height, and EF is the Fermi level of the ferromagnet. This barrier-dependent factor can vary from \u22121 to 1. It may be noted that for parabolic, free electron bands, the polarization associated with the ferromagnets is P = k\u2191 \u2212 k\u2193 k\u2191 + k\u2193 \u03c1\u2191 \u2212 \u03c1\u2193 \u03c1\u2191 + \u03c1\u2193 : (11.47) % Thus, Eqs. (11.46) and (11.47) change Julliere\u2019s result obtained in Eq. (11.44), especially when the tunnel barrier is high. The signs of the polarization can even change in certain cases. $ $ % The magnetic random access memory (MRAM) is built from the concept of MTJs, as shown in Figure 11.14. \u201cBit\u201d lines \u201cCross-point\u201d architecture \u201cWord\u201d lines FIGURE 11.14 MRAM 1,200 1,000 800 600 400 200 ) % ( 300 \u20181\u2019 \u20180\u2019 (a) \u2212200 \u2212100 100 0 External field (0e) 200 300 (b) Principles of the magnetic random access memory (MRAM) in the basic \u201ccross-point\u201d structure. Reproduced from Fert7 with the permission of the American Physical Society. 356 CHAPTER 11 Spintronics In Figure 11.14a, the binary information \u201c0\u201d and \u201c1\u201d is recorded on the two opposite orientations of the magnetization of the MTJs, which are connected to two crossing points of two perpendicular arrays of parallel conducting lines. The current pulses are sent through one line of each array for \u201cwriting,\u201d and only at the crossing point, the resulting magnetic field is high enough to orient the magnetization of the free layer. The resistance between the two lines between the addressed cell is measured for \u201creading.\u201d Figure 11.14b shows the TMR = /Rmin for the stack annealed at 475\u00ba C after growth. 2:1 nm Co25Fe75\u00de80B20\u00f0 \u00f0 The measurements were done at room temperature (closed circles) and low temperatures (open circles). In the first MRAMs, the memory cells are MTJs with an alumina barrier. The \u201cword\u201d and \u201cbit\u201d lines generate magnetic fields that switch the magnetic configuration. The future of the technology of computers is based on the ST-RAM, which is based on MgO tunnel junctions and switching by spin transfer. Co25Fe75\u00de80B20\u00f0 Rmax\u2061\u2212 Rmin\u00de \u00f0 4:3 nm /MgO 4 nm \u00f0 \u00f0 \u00de \u00de \u00de / 11.8 SPIN TRANSFER TORQUES AND MAGNETIC SWITCHING A spin current, $Q, which consists of moving spins, can be written as where P is the polarization (scalar). The spin and current densities can be written as $Q = \u0127/2 \u00f0 \ufffd", "\ufffd P^s \u2297 j, r s!\u00f0 \u00de = \u2211 i\u03c3\u03c3\u2032 r \u03c8 #i\u03c3\u00f0 \u00de ^S\u03c3,\u03c3\u2032\u03c8 #i\u03c3\u2032\u00f0 r \u00de (11.48) (11.49) and \u03c8 #i\u03c3\u00f0 where ^S\u03c3,\u03c3\u2032 and ^v are spin and velocity operators. The continuity equation, which expresses the conservation of number of electrons, is given by = \u2211 i\u03c3\u03c3\u2032 (11.50) $Q \u00f0, \u00de\u00de ^S\u03c3,\u03c3\u2032 \u2297 ^v\u03c8 #i\u03c3\u00f0 Re r r r \u00de \u00f0 \u00de \u2202n \u2202t = \u2212\u2207. j, (11.51) where n is the number density, and j is the current density. It can be shown that there is a similar equation for the spin density, s, which consists of an extra term arising due to the noncommutivity of spin density with magnetocrystalline anisotropy, \u2202s \u2202t = \u2212\u2207. $Q + next, (11.52) where \u2207. $Q = \u2202kQik: Here, next is the external torque density that rotates the spins. Eq. (11.52) is equivalent to the continuity equation stated in Eq. (11.51) with additional terms next: In general, there are two contributions to the spin current. If the direction of magnetization is nonuniform in a ferromagnet, the left and right spin currents do not cancel, and the gradient of this current gives rise to a spin torque nex = \u2212\u2207. $Qex: There is also a second contribution to the spin current at the interfaces due to the imbalance in the population of the spin states at or near the Fermi energy. It can be 11.9 Spintronics with Semiconductors 357 Cobalt/Copper/Cobalt e\u2212 e\u2212 S F1 Transverse component S F2 FIGURE 11.15 Illustration of the spin-transfer concept. The spin-transfer torque described previously acts on F2: Reproduced from Fert7 with the permission of the American Physical Society. shown that the spin current along the direction of the magnetization is conserved. Further, the reflected and transmitted components do not (generally) have any transverse components. Therefore, the main mechanism for transfer of angular momentum is the exchange interaction experienced by the electrons in the ferromagnet, which exerts a torque on the spin current. The scattering of spin at a ferromagnetic interface is a growing area of research, and we have considered only an elementary discussion of this topic. To summarize, in spin-transfer phenomena, the magnetic moments of a ferromagnet are manipulated by the transfer of spin-angular momentum from a spin-polarized current without the application of a magnetic field. The transfer of a transverse spin current to a \u201cfree\u201d magnetic layer F2 is due to the torque acting on the magnetic moment, as briefly described previously. The spin torque can induce irreversible switching of the magnetic moment or can generate precession of the moment in the microwave frequency range in the presence of a magnetic field in a second regime. This scenario is schematically shown in Figure 11.15. The first magnetic layer, F, prepares a spinpolarized current with an obliquely oriented spin polarization with respect to the magnetization axis of a second layer, F2: When this current goes through F2, the exchange interaction aligns its spin polarization along the magnetization axis. Because the exchange interaction is spin conserving, the transverse spin polarization lost by the current is transferred to the total spin of F2: This can lead to a magnetic switching of the F2 layer or to magnetic oscillations in the microwave frequency range. 11.9 SPINTRONICS WITH SEMICONDUCTORS 11.9.1 Introduction Spintronics with semiconductors has the tremendous advantage of combining the potential of the magnetic materials such as the control of current by spin manipulation, nonvolatility, etc., with the potential of the semiconductors (control of current by gate, coupling with optics, etc.). Datta and Das4 358 CHAPTER 11 Spintronics proposed the concept of a spin-effect transistor (spin FETs) based on spin transport in semiconductor lateral channels between spin-polarized sources and drains with control of the spin transmission by a field-effective gate. The spin precession can be controlled by spin-orbit coupling. The three ingredients for a spin transistor are (1) long relaxation time of a semiconductor, (2) gate voltage control of the spin-orbit coupling, and (3) high spin injection coefficients. Optical experiments have established that electron spins of semiconductors have long relaxation time. Modulation of the spin-orbit splitting at the Ferm", "i level by gate voltage has been reported for both electrons and holes for a variety of semiconductors. However, the spin injection from a ferromagnetic (FM) source into a semiconductor is a very difficult task and yields a maximum of 1% spin polarization. The conductivity mismatch between an FM metal emitter and a semiconductor is the primary reason for the difficulty in spin injection. It has been shown that if we define the spin injection coefficient in a diffusive regime, \u03b3, as where \u03c3N and \u03c3F are the conductivities of a normal (N) and FM contacts, then \u03b3 \u221d \u03c3N/\u03c3F, if N is a paramagnetic metal, and \u03c3N/\u03c3F \u226b 1 \u03c3N/\u03c3F \u226a 1 (11.53) (11.54) (11.55) if N is a semiconductor. This explains why the spin injection from an FM source into a paramagnetic metal is very efficient, whereas the same from an FM source to a semiconductor is practically impossible. Rashba18 showed that tunnel contacts (T) can significantly increase spin injection and solve the problem of the mismatch between the conductivities of a ferromagnetic metal (FM) and a semiconductor (N) microstructure. The tunnel resistance, rc, should be larger than the competing \u201ceffective resistances\u201d making the total contact resistance, where LF and LN are spin diffusion lengths in the FM and N conductors and w is the width of the N conductor. rc \u2265 LF/\u03c3F, min\u2061 LN, w g f /\u03c3N, (11.56) 11.9.2 Theory of an FM-T-N Junction and assumed that the T contact, at Rashba considered a semi-infinite FM x = 0, has different spin conductivities, \u03a3\u2191 and \u03a3\u2193, for up- and down-spins, and there is no spin relaxation in it. The currents j\u2191,\u2193\u00f0 carried by up- and down-spins can be written as and is the space derivative of the electrochemical potentials \u03be\u2191,\u2193\u00f0 \u00f0 \u00de \u00de x j\u2191,\u2193\u00f0 = \u03c3\u2191,\u2193 \u03be\u2191,\u2193\u2032 x, (11.57), which are related to in the FM region, x where \u03be\u2191,\u2193\u2032 \u00f0 the nonequilibrium parts n\u2191,\u2193\u00f0 \u00de x \u00de x of the electron concentrations and \u03c6F\u00f0 \u03be\u2191,\u2193\u00f0 eD\u2191,\u2193/\u03c3\u2191,\u2193\u00de \u2212 \u03c6F\u00f0 n\u2191,\u219311.58) 11.9 Spintronics with Semiconductors 359 Here, D\u2191,\u2193 are diffusion coefficients, and \u03c3\u2191,\u2193 are conductivities of up- and down-spin electrons. To maintain the charge neutrality under the spin injection conditions, The continuity equation is x n\u2191\u00f0 \u00de x + n\u2193\u00f0 \u00de = 0: j\u2191\u2032 x \u00de \u00f0 = en\u2191\u00f0 x \u00de /\u03c4F s, where \u03c4F s is the spin-relaxation time. Because the charge is conserved, J = j\u2191\u00f0 = constant: + j\u2193\u00f0 x x \u00de \u00de Introducing the notations and x \u03beF\u00f0 \u00de x = \u03be\u2191\u00f0 \u00de \u2212 \u03be\u2193\u00f0 x \u00de x jF\u00f0 \u00de x = j\u2191\u00f0 \u00de x \u2212 j\u2193\u00f0 \u00de, one can show (Problem 11.7) that the diffusion equation can be written as where and DF\u03beF\u2033 x \u00de \u00f0 = \u03beF\u00f0 x \u00de /\u03c4F s, DF = \u03c3\u2193D\u2191 + \u03c3\u2191D\u2193\u00de \u00f0 /\u03c3F \u03c3F = \u03c3\u2191 + \u03c3\u2193: From Eqs. (11.57) through (11.66), it can be shown that (Problem 11.8) At T = 0, the Einstein relations are \u03c6F\u2032 x \u00de \u00f0 = D\u2191 \u2212 D\u2193\u00de \u00bd\u00f0 /DF%\u00f0 \u03c3\u2191\u03c3\u2193/\u03c32 F\u00de \u03beF\u2032 x \u00f0 \u00de \u2212 J/\u03c3F: e2D\u2191,\u2193 = \u03c3\u2191,\u2193/\u03c1\u2191,\u2193, (11.59) (11.60) (11.61) (11.62) (11.63) (11.64) (11.65) (11.66) (11.67) (11.68) where \u03c1\u2191,\u2193 are the densities of states at the Fermi level. One can show (Problem 11.9) from Eqs. (11.65) through (11", ".68) that and e2DF = \u03c3\u2191\u03c3\u2193/\u03c3F\u00de\u00f0 \u03c1F/\u03c1\u2191\u03c1\u2193\u00de \u00f0 \u03c1\u2193\u03c3\u2191 \u2212 \u03c1\u2191\u03c3\u2193\u00de \u00f0 /\u03c1F\u03c3F = \u0394\u03c3/\u03c3F\u00de \u00bd\u00f0 \u2212 \u0394\u03c1/\u03c1F\u00de% /2: \u00f0 It can be easily shown (Problem 11.10) from Eqs. (11.67), (11.69), and (11.70) that \u03c6F\u2032 x \u00de \u00f0 = \u0394\u03c3/\u03c3F\u00de \u00bd\u00f0 \u2212 \u0394\u03c1/\u03c1F\u00de% \u00f0 \u03beF\u2032 \u00f0 x \u00de /2 \u2212 J/\u03c3F: (11.69) (11.70) (11.71) 360 CHAPTER 11 Spintronics From Eqs. (11.57) and (11.71), it is easy to show that x = 2 jF\u00f0 \u03be\u2032F\u00f0 Eqs. (11.64), (11.65), (11.71), and (11.72) are a complete system of bulk equations for the F region. One can also show (Problem 11.11) that + \u03be\u2193\u00f0 J: \u0394\u03c3/\u03c3F\u00de \u03c3\u2191\u03c3\u2193/\u03c3F\u00de \u0394\u03c1/\u03c1F\u00de x 2\u03c6F\u00f0 (11.72) (11.73) \u03beF\u00f0 \u03be\u2191\u00f0 = \u2212 + + \u00de and \u03c1\u2191\u03c1\u2193/\u03c1F\u00de \u00f0 The equations for the N region can be obtained from the equations for the F region by substituting the following in Eqs. (11.64), (11.67), and (11.72): x \u03beF\u00f0 (11.74) n\u2191\u00f0 = x \u00de \u00de : \u03c3\u2191 = \u03c3\u2193 = \u03c3N/2, \u0394\u03c1 = \u0394\u03c3 = 0, DN = D\u2191 = D\u2193: DN\u03beN\u00f0 x = \u03beN\u00f0 x \u00de \u00de = \u2212J/\u03c3N, x \u03c6\u2032N\u00f0 \u00de /\u03c4N s, (11.75) (11.76) (11.77) (11.78) and We obtain and x jN\u00f0 = \u03c3N\u03be\u2032N\u00f0 In Eqs. (11.77) through (11.79), the symbol N in the prefix has been used for the N region, instead of the prefix F, which was used in the F region in the previous equations. Because there is no spin relaxation at the interface x = 0, the boundary conditions are j\u2191\u00f0 is continuous at x = 0 and hence jF\u00f0 = jN\u00f0 (11.79) /2 : Substituting these in Eqs. (11.72) and (11.79), we obtain \u03c3\u2191\u03c3\u2193/\u03c3F\u00de \u0394\u03c3/\u03c3F\u00de are related to the conductivities of the T contact, \u03c3N\u03be\u2032N\u00f0 \u03be\u2032F\u00f0 \u2212 4 = 2 J: 0 0 \u00f0 \u00f0 \u00de \u00de The currents j\u2191,\u2193\u00f0 0 \u00de 0 j\u2191,\u2193\u00f0 \u00de = \u03a3\u2191,\u2193\u00f0 \u03beN \u2191,\u2193 \u2212 \u03be\u2191,\u2193\u00de : Using Eq. (11.62) and its equivalent for \u03beN, we can rewrite Eq. (11.81) as \u2212 \u03beF\u00f0 rcJ + 2rc j \u0394\u03a3/\u03a3 = \u22122 \u03beN\u00f0 0 \u00f0 0 0 \u00f0 \u00de \u00de \u00de \u00de, where \u0394\u03a3 = \u03a3\u2191 \u2212 \u03a3\u2193, \u03a3 = \u03a3\u2191 + \u03a3\u2193 and rc = \u03a3/4\u03a3\u2191\u03a3\u2193: (11.80) (11.81) (11.82) (11.83) 11.9 Spintronics with Semiconductors 361 From Eq. (11.73), its equivalent for \u03c6N and Eq. (11.83), we obtain \u03beF\u00f0 Eq. (11.84) implies that even when rc = 0, because \u0394\u03c1 \u2260 0, there is a finite potential drop at the interface, \u2212 \u03c6N\u00f0 0 \u03c6F\u00f0 (11.84) rcj \u00de\u00de : = rcJ \u2212 \u0394\u03a3 \u03a3 + \u0394\u03c1 2\u03c1F \u03c6F\u00f0 \u00bd \u2212 \u03c6N\u00f0 \u00de \u00de% \u221d J: 11.9.3 Injection Coefficient The solutions of Eqs. (11.64) and (11.78) are and where and = Ae\u2212LF x \u03beF\u00f0 x \u00de = Be\u2212", "LN x, x \u03beN\u00f0 \u00de LF = 1/2 DF\u03c4F s \u00de \u00f0 LN = DF\u03c4N s \u00de 1/2: \u00f0 Here, LF and LN are known as the diffusion lengths. Thus, we obtain and 0 \u03be\u2032N\u00f0 \u00de = \u2212\u03beN\u00f0 0 \u00de /LN = 2\u03b3J/\u03c3N 0 \u03be\u2032F\u00f0 \u00de = \u03beF\u00f0 0 \u00de /LF: The injection coefficient is defined as If we eliminate \u03beF\u00f0 0 \u00de Here, and /J: \u00de \u00f0 from Eqs. (11.80) and (11.82), we obtain (Problem 11.14) \u03b3 = j 0 \u03b3 = rF\u00f0 \u0394\u03c3/\u03c3F\u00de \u00bd + rc\u00f0 \u0394\u03a3/\u03a3 rFN: \u00de% & rFN = rF + rN + rc, rF = LF\u03c3F/4\u03c3\u2191\u03c3\u2193, rN = LN/\u03c3N: (11.85) (11.86) (11.87) (11.88) (11.89) (11.90) (11.91) (11.92) (11.93) (11.94) (11.95) 362 CHAPTER 11 Spintronics Eq. (11.93) shows that rc, rF, and rN are connected in series. We note that if rF = rN, \u03b3 \u223c 1 if and only if rc \u2265 rN, a criterion that can be satisfied for narrow tunnel junctions of the atomic scale. If rc \u226b rN, rF, rFN \u223c rc from Eq. (11.93). Using this approximation, we obtain (from Eq. 11.92) the injection coefficient \u03b3 \u2248 \u0394\u03a3/\u03a3: The contact completely determines \u03b3 in this regime. Thus, the spin injection coefficient is controlled by the element of an FM-T-N junction having the largest effective resistance. There are alternate possibilities of spintronics with semiconductors based on the use of ferromagnetic semiconductors such as Ga1\u2212xMnxAs, where x \u226a 1: There is a good possibility of controlling the ferromagnetic properties with a gate voltage as well as having large TMR effects. An alternate approach is the spin accumulation effect due to spin-orbit coupling or anomalous scattering mechanisms. When a spin-unpolarized current flows in a metal, the spin-orbit interaction produces asymmetric scattering of the conduction electrons so that spin-up electrons have a larger probability to be scattered to the right compared to spin-down electrons, and spin-down electrons would tend to scatter to the left more than spin-up electrons. This results in a spin current that is generated in a direction transverse to the direction of the flow of current. If a spin-polarized current is present in a semiconductor, a Hall-like effect can be induced by the spin-orbit coupling without an external field. Zhang23 derived an expression for the spin Hall effect (SHE) using a semiclassical Boltzmann equation and extending it to the case where the spin diffusion effect is finite. He showed that when the Ga1-xMnxAs (30 nm) GaAs (10 ) AIAs (15\u201319 ) GaAs (5\u201310 nm) AIAs (15\u201319 ) GaAs (10 ) Ga1-xMnxAs (300 nm) 40 30 20 10 0 ) % ( R M T \u2212300 \u2212200 \u2212100 0 Magnetic field (Oe) 100 200 300 (b) 50 40 30 20 10 ) % ( R M T 1.45 nm 1.7 nm 1.95 nm (a) 0 1E\u22123 0.1 0.01 ARAP (\u2126cm2) (c) FIGURE 11.16 Spintronics of semiconductors illustrated by experimental results. Reproduced from Fert7 with the permission of the American Physical Society. 11.9 Spintronics with Semiconductors 363 formulation is applied to certain metals and semiconductors, the magnitude of the spin Hall voltage is much larger than that of magnetic multilayers. Because SHE is also found in nonmagnetic metals, further research in this area is very active. Spintronics with semiconductors is shown in Figure 11.16. The structure, shown in Figure 11.16a, is composed of a GaAs layer that is separated from the GaMnAs source and drain by tunnel barriers of AlAs. Figure 11.16b shows the MR curve at 4.2\u00ba K, which shows a difference in resistance of 40% between the parallel and antiparallel magnetic configuration between the source and the drain. Figure 11.16c shows the MR ratio as a function", " of the resistance of the tunnel barriers. It may be noted that spintronics is a very rapidly growing area of research. Recently, large GMR- and TMR-like effects were predicted in carbon-based molecules. In fact, due to small spin-orbit coupling, carbon molecules have a long spin lifetime. Recent experiments on carbon nanotubes between a ferromagnetic source and drain made of the metallic manganite La2/3Sr1/3MnO3 (LSMO) have been encouraging (see Figure 11.17). La2/3Sr1/3MnO3 (LSMO) La2/3Sr1/3MnO3 (LSMO) CNT (a) 5\u00b0 K 25 mV MR = 72% \u2212150 \u221275 75 150 0 B(mT) (b) MR = 60% 60 50 40 30 20 10 ) % ( R M 0 \u2013150 \u2013100 \u201350 0 50 100 150 B(mT) (c) FIGURE 11.17 (a) Sketch of spintronics with LSMO molecules. Magnetoresistance experimental results of Heuso et al.12 at 4.2\u00baK on carbon nanotubes between electrodes made of LSMO are shown in (b) and (c). A contrast of 72% and 60% is obtained for the parallel (high field) and antiparallel (peaks) magnetic configurations of the source and drain. Reproduced from Fert7 with the permission of the American Physical Society. 364 CHAPTER 11 Spintronics The results indicate that the resistances of the parallel and antiparallel configurations are between 60% and 70%, which is larger than that obtained for semiconductor channels. In addition, their high Fermi velocity, which is responsible for short dwell time, is an advantage over semiconductors. The research is very active in the general area of spintronics and, in particular, on graphene-based devices. In view of the above, any review of the general area of spintronics would very quickly become outdated. Therefore, this chapter may be considered as an introduction to the topic. PROBLEMS 11.1. By using Mott\u2019s two-current model, in which the spin transport in a metal is perceived as being due to two independent spin channels, and by using Eqs. (11.4) through (11.7), show that the energy-dependent relaxation times can be expressed as 1 \u03c41 = A \u00f0 T/MI\u03b82 \u00de \" and 1 \u03c42 = A \u00f0 T/MI\u03b82 \u00de \" E1 \u2212 E \u00de \u00f0 E0 \u2212 \u03c20\u00de p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 p E2 \u2212 E \u00de \u00f0 E0 \u2212 \u03c20\u00de p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u00f0 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 p where \u03b1 \u2248 1/4 and \u03b8 is the Debye temperature. 11.2. Show that and E1 \u2212 \u03b6\u20320 E0 \u2212 \u03b60 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 r 1 \u2212 \u03b2 = \u00f0 1/3 \u00de E2 \u2212 \u03b6\u20320 E0 \u2212 \u03b60 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 r 1 + \u03b2 = \u00f0 1/31) (2) (1) (2) where \u03b60\u2032 represents the energy of the highest occupied state (Fermi energy) at T = 0 when the states are split and \u03b60 when \u03b2 = 0: 11.3. By using Eqs. (11.8) through (11.12), derive the equation for resistivity (", "Eq. (11.13), \u03b2, T \u03c1 \u00f0 \u00de = A \u00f0 T/m\u03b82 1 1/1 1 1/1) Problems 365 11.4. From Eqs. (11.19) through (11.21), show that and gA\u00b1\u2191\u00f0 vz, z \u00de = eE\u03c4 m \u2202f0 \u2202vx 1 + A\u00b1\u2191exp\u2061 \u2213z vzj\" \u03c4 j $ # % gA\u00b1\u2193\u00f0 vz, z \u00de = eE\u03c4 m \u2202f0 \u2202vx 1 + A\u00b1\u2193exp\u2061 \u2213z vzj\" \u03c4 j $ : # % 11.5. We obtained in Eqs. (11.38) and (11.39) and Show that RAP = N \u03c1#FtF + \u03c1#NtN + 2r#b\u00de \u00f0 RP = RAP \u2212 f \u03b2\u03c1#FtF + 2\u03b3r#bg RAP 2 N2 : RAP \u2212 RP RAP \" \u22121/2 # = \u03c1#FtF + 2r#b \u03b2\u03c1#FtF + 2\u03b3r#b + \u03c1#NtN \u03b2\u03c1#FtF + 2\u03b3r#b : 11.6. Using a simple model, we have shown in Eqs. (11.41) and (11.42) that and Gp \u221d aa\u2032 + \u00bd 1 \u2212 a \u00f0 \u00de\u00f0 1 \u2212 a\u2032 \u00de% \u00de Assuming that the spin is conserved, show that \u00f0 \u00bd Gap \u221d a 1 \u2212 a\u2032 + a\u2032 \u00f0 1 \u2212 a : \u00de% TMR = GP \u2212 Gap Gap = 2PP\u2032 1 \u2212 PP\u2032 \u00f0 \u00de, where the conduction electron spin polarization of the two ferromagnetic metals is P = 2a \u2212 1 and P\u2032 = 2a\u2032 \u2212 1 : 11.7. Introducing the notations and x \u03beF\u00f0 \u00de x = \u03be\u2191\u00f0 \u00de x \u2212 \u03be\u2193\u00f0 \u00de jF\u00f0 x \u00de x = j\u2191\u00f0 \u00de x \u2212 j\u2193\u00f0 \u00de, (1) (2) (1) (2) (3) (1) (2) (3) (4) (1) (2) 366 CHAPTER 11 Spintronics show that the diffusion equation can be rewritten as where and DF\u03be\u2033F\u00f0 x \u00de = \u03beF\u00f0 x \u00de /\u03c4F s, DF = \u03c3\u2193D\u2191 + \u03c3\u2191D\u2193\u00de \u00f0 /\u03c3F \u03c3F = \u03c3\u2191 + \u03c3\u2193: 11.8. From Eqs. (11.57) through (11.66), show that x \u03c6\u2032F\u00f0 \u00de = D\u2191 \u2212 D\u2193\u00de \u00bd\u00f0 /DF%\u00f0 \u03c3\u2191\u03c3\u2193/\u03c32 F\u00de \u03be\u2032F\u00f0 x \u00de \u2212 J/\u03c3F: 11.9. By using Eqs. (11.65) through (11.68), show that and where and e2DF = \u03c3\u2191\u03c3\u2193/\u03c3F\u00de\u00f0 \u03c1F/\u03c1\u2191\u03c1\u2193\u00de \u00f0 \u03c1\u2193\u03c3\u2191 \u2212 \u03c1\u2191\u03c3\u2193\u00de \u00f0 /\u03c1F\u03c3F = \u0394\u03c3/\u03c3F\u00de \u00bd\u00f0 \u2212 \u0394\u03c1/\u03c1F\u00de% \u00f0 /2, \u0394\u03c3 = \u03c3\u2191 \u2212 \u03c3\u2193, \u0394\u03c1 = \u03c1\u2191 \u2212 \u03c1\u2193, \u03c1F = \u03c1\u2191 + \u03c1\u2193: 11.10. Show from Eqs. (11.67), (11.69), and (11.70) that x \u03c6\u2032F\u00f0 \u00de = \u0394\u03c3/\u03c3F\u00de \u00bd\u00f0 \u2212 \u0394\u03c1/\u03c1F\u00de% \u03be\u2032F\u00f0 \u00f0 x \u00de /2 \u2212 J/\u03c3F: 11.11. From Eqs. (11.64), (11.65), (11.71), and (11.72), show that and x \u03be\u2191\u00f0 \u00de x + \u03be\u2193\u00f0 \u00de = \u2212 x 2\u03d5F\u00f0 \u00bd \u00de + \u0394\u03c1/\u03c1F\u00de \u03beF\u00f0 \u00f0 x \u00de% x n\u2191\u00f0 \u00de = \u03c1\u2191\u03c1\u2193/\u03c1F\u00de x \u03beF\u00f0 \u00de : \u00f0 11.12. Using Eq. (11.62) and its equivalent for \u03beN, show that Eq. (11.81) can be rewritten as 0 \u03beN\u00f0 \u00de 0 \u2212 \u03beF\u00f0 \u00de = \u22122 \u0394\u03a3", "/\u03a3 \u00de \u00f0 rcJ + 2rc j 0, \u00de \u00f0 (3) (4) (5) (1) (1) (2) (3) (4) (5) (1) (1) (2) (1) References 367 where \u0394\u03a3 = \u03a3\u2191 \u2212 \u03a3\u2193, \u03a3 = \u03a3\u2191 + \u03a3\u2193 and rc = \u03a3/4\u03a3\u2191\u03a3\u2193: Here, rc is the effective contact resistance. 11.13. From Eq. (11.73), its equivalent for \u03c6N, and Eq. 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Diamagnetism and Paramagnetism CHAPTER 12 CHAPTER OUTLINE 12.1 Introduction............................................................................. 370 12.2 Atomic (or Ionic) Magnetic Susceptibilities................................................. 371 12.2.1 General Formulation............................................................... 371 12.2.2 Larmor Diamagnetism.............................................................. 372 12.2.3 Hund\u2019s Rules..................................................................... 373 12.2.4 Van Vleck Paramagnetism.......................................................... 374 12.2.5 Land\u00e9 g Factor.................................................................... 375 12.2.6 Curie\u2019s Law....................................................................... 377 12.3 Magnetic Susceptibility of Free Electrons in Metals.......................................... 378 12.3.1 General Formulation............................................................... 378 12.3.2 Landau Diamagnetism and Pauli Paramagnetism.............", "........................ 380 12.3.3 De Haas\u2013van Alphen Effect........................................................ 383 12.4 Many-Body Theory of Magnetic Susceptibility of Bloch Electrons in Solids...................... 388 12.4.1 Introduction....................................................................... 388 12.4.2 Equation of Motion in the Bloch Representation..................................... 388 12.4.3 Thermodynamic Potential.......................................................... 390 12.4.4 General Formula for \u03c7.............................................................. 390 12.4.5 Exchange Self-Energy in the Band Model............................................ 393 12.4.6 Exchange Enhancement of \u03c7s....................................................... 394 12.4.7 Exchange and Correlation Effects on \u03c7o.............................................. 395 12.4.8 Exchange and Correlation Effects on \u03c7so............................................. 396 12.5 Quantum Hall Effect...................................................................... 396 12.5.1 Introduction....................................................................... 396 12.5.2 Two-Dimensional Electron Gas...................................................... 396 12.5.3 Quantum Transport of a Two-Dimensional Electron Gas in a Strong Magnetic Field................................................................. 397 12.5.4 Quantum Hall Effect from Gauge Invariance......................................... 400", " 12.6 Fractional Quantum Hall Effect............................................................ 400 Problems.................................................................................... 401 References.................................................................................. 407 Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00012-8 \u00a9 2012 by Elsevier Inc. All rights reserved. 369 370 CHAPTER 12 Diamagnetism and Paramagnetism 12.1 INTRODUCTION The magnetization density of a quantum-mechanical system is defined as Mv = \u2212 1 V \u2202\u03a9 \u2202Bv, (12.1) where \u03a9 is the thermodynamic potential, V is the volume of the solid, and Bv is the vth component of the magnetic induction B, the microscopic field perceived by the nuclei or the electrons due to an external magnetic field H. The magnetic susceptibility is defined as \u03c7\u03bcv = lim\u2061 0! B \u2202Mv \u2202B\u03bc = \u2212 1 V \u22022 \u03a9 \u2202B\u03bc\u2202Bv : lim\u2061 0 B! (12.2) It may be noted that the magnetic susceptibility is a tensor, whereas the magnetization is a vector. If a system has positive magnetic susceptibility, it is known as paramagnetic, and if it has negative magnetic susceptibility, it is known as diamagnetic. In a linear medium, B = \u03bcH, where \u03bc is known \u2202Mv as the magnetic permeability. Because an alternate definition of susceptibility is \u03c7\u03bcv = lim\u2061 \u2202H\u03bc, 0 a linear system is paramagnetic if \u03bc > 1 and diamagnetic if \u03bc < 1:! H There exists a lot of confusion in the literature as to when to use H and when to use B. H is the external applied magnetic field usually produced by external currents. Therefore, experiments that control external currents control H more directly than B, and hence, it is appropriate to consider H as the experimentally applied field. However, when this external magnetic field is applied to the system through external currents, the microscopic field perceived by the nuclei or the electrons is the magnetic induction B. Thus, the microscopic Hamiltonian should be written in terms of B, whereas the experimental results should be expressed in terms of H., and spin-orbit interactions In this chapter, we will first calculate by using approximate methods, the atomic susceptibility, the susceptibility of insulators with filled shells that leads to Larmor diamagnetism, and the susceptibility of a collection of magnetic ions with partially filled shells that leads to paramagnetism. Then we will calculate the magnetic susceptibility of free electrons in metals that leads to Pauli paramagnetism and Landau diamagnetism, as well as the de Haas\u2013van Alphen effect. Finally, we will outline the many-body theory of magnetic susceptibility of Bloch electrons in a magnetic field. We, spin will express the total magnetic susceptibility as a sum of the contributions of the orbital, and discuss the effects of exchange and correlation on each \u03c7s\u00de \u00f0 of these terms. Here, the spin susceptibility \u03c7s includes the effect of spin-orbit interaction on the spin, whereas \u03c7so is the contribution to magnetic susceptibility from the effect of spin-orbit coupling on the orbital motion of Bloch electrons. In the usual derivations of many-body theories of magnetic susceptibility, attention is focused either on the orbital part or on the spin part of the Hamiltonian, and the effects of spin-orbit coupling are accounted for in \u03c7o through the modifications of the Bloch functions and in \u03c7s by replacing the free electron g factor by the effective g factor. In this process, \u03c7so, which is of the same order as \u03c7s for solids with large effective g factors, has been neglected. \u03c7so\u00de \u03c7o\u00de \u00f0 \u00f0 12.2 Atomic (or Ionic) Magnetic Susceptibilities", " 371 12.2 ATOMIC (OR IONIC) MAGNETIC SUSCEPTIBILITIES 12.2.1 General Formulation It can be easily shown that the classical Hamiltonian for an electron in a magnetic field B is given by H = p + 1 2m! r eA \u00f0 c \u00de 2, \" (12.3) is the vector potential obtained from the relation where p is the classical momentum, and A \u00f0 B = \u2207 \u00d7 A and \u2207. A = 0: In quantum mechanics, the vector p is replaced by the operator \u2212i\u0127\u2207: Thus, the Hamiltonian for a free electron with spin operator ^s in a magnetic field B is given by r \u00de ^H = 1 2m! \u2212i\u0127\u2207 + r eA \u00f0 c \u00de 2 \" + g0\u03bcBB. ^s = ^p + 1 2m! r eA \u00f0 c 2 \u00de \" + g0\u03bcBB. ^s, (12.4) where ^p and ^s are the momentum and spin operators. The Hamiltonian of an atom (or ion) in a uniform magnetic field B (in the z direction) can be written as ^H = 1 2m \u2211 i ri\u00de eA \u00f0 c 2 \" ^pi +! + g0\u03bcBB\u2211 i ^si z, (12.5) where ri, pi, and ^si are the position, momentum, and spin operator of each electron in the atom (or ion), and A \u00f0 is the vector potential such that ri\u00de B = \u2207 \u00d7 A and \u2207. A = 0: (12.6) Here, g0 \u2248 2:0023 and \u03bcB = e\u0127/2mc: The first term in the Hamiltonian in Eq. (12.5), in which the sum is taken over all the electrons in the atom, is due to ^T, the kinetic energy operator of the electrons, which includes the interaction of the orbital magnetic momentum with the magnetic induction. This in the momentum operator pi of interaction is accounted for by including the vector potential A \u00f0 the electron at ri: The second term in Eq. (12.5) is due to the interaction of B with the electron spin and is a consequence of Dirac\u2019s relativistic theory. We will assume that ri\u00de A = \u2212 1 2 r \u00d7 B, (12.7) which satisfies both conditions in Eq. (12.6). Thus, the kinetic energy operator, ^T, in Eq. (12.5) can be written as ^T = 1 2m \u2211 i ^p2 i + \u03bcB ^L. B + e2 8mc2 B2\u2211 i \u00f0 i + y2 x2 i \u00de, where ^L is the total electronic orbital angular momentum operator, ^L = 1 \u0127 \u2211 i ri \u00d7 ^pi: (12.8) (12.9) 372 CHAPTER 12 Diamagnetism and Paramagnetism From Eqs. (12.5), (12.8), and (12.9), we obtain (Problem 12.1) ^H = 1 2m \u2211 i ^p2 i + \u03bcB\u00f0 ^L + 2^S \u00de. B + e2 8mc2 B2\u2211 i \u00f0 i + y2 x2 i \u00de, where we have approximated g0 \u2248 2, and ^S = \u2211 i ^si: (12.10) (12.11) We can rewrite Eq. (12.10) in the alternate form ^H = ^H0 + \u0394 ^H, where ^H0 is the Hamiltonian in the absence of the magnetic field, and \u0394 ^H is the perturbation due to the magnetic field, (12.12) \u0394 ^H = \u03bcB\u00f0 ^L + 2^S. B + \u00de e2B2 8mc2 \u2211 i \u00f0 i + y2 x2 i \u00de : (12.13) If the states by using second-order perturbation theory, n> are the orbital states of the electrons in the absence of the magnetic field, we obtain, j where and En = E0 n + \u0394En, E0 n = <n n> H0j j \u0394En = <n \u0394 ^H j n> + \u2211 j n\u2032\u2260 n j <n \u0394 ^H n\u2032 j j En\u2212En\u2032 j 2> : From Eqs. (12.13) and (12.16), retaining terms through those quadratic in B, we obtain \u0394En = \u03bcBB. <n ^L + 2^S j n> + \u2211 j n\u2032\u2260n <n j \u03bcBB. j ^L", " + 2^S \u00f0 En\u2212En\u2032 n\u2032> 2 j \u00dej + e2B2 8mc2 <n \u2211 i \u00f0 j i + y2 x2 n> : i \u00dej (12.14) (12.15) (12.16) (12.17) The magnetic susceptibility of atoms, ions, and molecules is obtained from Eq. (12.17). Usually, the first term on the right side of the equation is the dominant term unless it vanishes. As we will see, the first term will be zero for either closed shells or shells with one electron short of being half filled (J = 0). However, for magnetic ions, the first term is much larger than the second and third terms, which can be neglected. The collection of ions becomes paramagnetic in such cases. 12.2.2 Larmor Diamagnetism If a solid has atoms or ions where all the electronic shells are filled, each atom has zero orbital and spin angular momentum in its ground state. Therefore, 0> = ^L ^J j j 0> = ^S 0> = 0: j (12.18) 12.2 Atomic (or Ionic) Magnetic Susceptibilities 373 In the ground state, the third term in Eq. (12.17) is the only term that contributes to the magnetic susceptibility, \u0394E0 = e2B2 8mc2 <0 \u2211 i \u00f0 j i + y2 x2 i \u00dej 0> = e2B2 12mc2 <0 \u2211 i j r2 i j 0>, (12.19) which follows from the spherical symmetry of the atom or ion. Assuming that the free energy is equal to the ground-state energy (which is true if and only if J = 0), the magnetic susceptibility of the solid of N such ions in a volume V (in a semiclassical approximation) is given by \u03c7 \u2248 \u2212 \u22022 \u0394E0 \u2202B2 = \u2212 N V <0 \u2211 i j 0> : r2 i j (12.20) Eq. (12.20) is known as Larmor diamagnetic susceptibility or sometimes as Langevin susceptibility. It is valid only for solids composed of atoms or ions of filled shells. We will now discuss Hund\u2019s rules, which are needed to discuss atoms or ions with partially filled shells. 12.2.3 Hund\u2019s Rules Hund\u2019s rules, which are valid for atoms or ions with partially filled shells, were obtained from the analysis of atomic spectra as well as by rigorous theoretical calculations. They are valid for incomplete shells of which the one-electron levels are characterized by orbital angular momentum l: The shell would have 2 \u00f0 one-electron levels (including spin). If n is the number of electrons in the shell, 2l + 1 \u00de The degeneracy of these levels is lifted by electron\u2013electron interaction as well as by the spin-orbit interaction. 0 <n <2 2l + 1 : \u00de \u00f0 (12.21) We will first discuss Russsell\u2013Saunders coupling, which states that the Hamiltonian of the atom or ion (with partially filled shells) commutes with the total angular momentum ^J = ^L + ^S and with the total electronic orbital and spin angular momentum ^L and ^S (provided the spin-orbit coupling is not too large such that it can be considered as a small perturbation). Thus, the partially filled shells can be indexed by the quantum numbers J, Jz, L, Lz, S, and Sz: This indexing is based on the fact that the, and ^Sz are J eigenvalues of the operators ^J, and Sz:, ^Lz, ^S, ^Jz, ^L, Lz, S, Jz Hund\u2019s rules are as follows: 1. In an incomplete shell, the electrons that lie lowest in energy have the largest total spin S, which is consistent with the exclusion principle. The largest value S can have is equal to the largest magnitude that Sz can have. When n \u2264 2l + 1, each electron can have parallel spin without multiple occupation of any one-electron level in the shell, provided each electron has a different value of lz: Thus, when n \u2264 2l + 1, S = 1 2 n: The exclusion principle requires that when n > 2l + 1, the spin of each additional electron has spin opposite to the first 2l + 1 electrons. Therefore, S is reduced by half a unit from its maximum vale of l + 1 2 for each electron after the first 2l + 1 electrons. 374 CHAPTER 12 Diamagnetism and Paramagnetism 2. Once S has been determined, as per Hund\u2019s first rule, the total angular momentum L of the electrons of", " the lowest-lying states has the largest value. For example, the first electron in = l, which is its maximum value. The second electron is in the level the shell is in the level lzj = 0: The spin l \u2212 1 j of the electrons that fill the second half of the shell are opposite to those in the first half, and using the same arguments, L = 0: = l \u2212 1: When the electrons are half filled + \u2026 lzj \u00de$ \u00de \u00f0 \u00f0 j 2L + 1 3. The values of L and S or the states of lowest energy are obtained from the first two rules. possible states that are degenerate. The degeneracy of these However, there are : The total angular momentum states is lifted by spin-orbit coupling, which is of the form \u03bb J can take on all integral values between and L + S: Spin-orbit coupling favors minimum j J if \u03bb is positive (for shells that are less than half filled) and maximum J if \u03bb is negative (for shells that are more than half filled). Thus, in the ground state, 2S + 1 L \u2212 S ^L. ^S \u00de\u00f0, j, n \u2264 n \u2265 \u00f0 2l + 1 \u00f0 2l + 1 \u00de, \u00de. (12.22) We note that when the shell is half full, L = 0 and there is no jump in J because J = S: We further note that Hund\u2019s rules apply to partially filled d and f shells but not to partially filled p shells, which contain valence electrons and broaden into bands in the solid. 12.2.4 Van Vleck Paramagnetism We consider the susceptibility of insulators containing ions with a partially filled shell. First, we consider the case where J = 0 (shells that are one electron short of being half filled). The first term in Eq. (12.17) still vanishes, as in the case of a filled shell. However, the second and third terms contribute to the shift in the ground-state energy, and we obtain \u0394E0 = \u2212\u2211 n j <0 j \u03bcBB. ^L + 2^S \u00f0 En \u2212 E0 n> \u00dej 2 j + e2B2 8mc2 \u2211 i <0 j\u00f0 i + y2 x2 i \u00dej 0>: (12.23) Because J = 0, assuming as before that the free energy is equal to the ground-state energy, the magnetic susceptibility is given by 2E0 \u2202B2 \u22122\u03bc2 \" B\u2211 n j <0 j n> ^Lz + 2^Szj En \u2212 E0 2 j + e2 4mc2 <0 \u2211 i \u00f0 j i + y2 x2 0> i \u00dej (12.24) : # The first term in Eq. (12.24), which is positive, is known as Van Vleck paramagnetism, and the second term is the Larmor diamagnetism derived for ions of filled shells. Thus, the magnetic susceptibility of ions with a shell one electron short of being half filled is obtained by the sum of the two terms. The Van Vleck paramagnetism also exists in molecules that have a more complex structure than single atoms or ions. It may be noted that the basic assumption made in these derivations is that the ground state is occupied with appreciable probability in thermal equilibrium, and hence, the free energy is equal to 12.2 Atomic (or Ionic) Magnetic Susceptibilities 375 the ground-state energy. However, if the next state is close to the J = 0 ground state, the free energy is not just the ground-state energy, and the derivation of the formula for magnetic susceptibility is much more complicated. 12.2.5 Land\u00e9 g Factor If J \u2260 0 for the shell, the first term for the shift in energy in Eq. (12.17) becomes the dominant term compared to the other two terms, which yield the Larmor diamagnetism and Van Vleck paramagnetism. If we ignore the last two terms in Eq. (12.17), we have to consider the matrix elements\u2019 dominant term j Here, we have assumed that B = B^z: The ground state -fold degenerate in the absence of the magnetic field. The matrix elements can be evaluated by diagonalizing and evaluating the matrix elements of the -dimensional square matrix 2J + 1 \u00de \u00f0 j \u0394En \u2248 \u03bcBB <n n> : ^Lz + 2^Szj n> is (12.25) 2J + 1 \u00f0 \u00de where <JLSJzj ^Lz + 2^Szj JLSJ\u2032z >, Jz,J\u2032z = \u2212J, \u2026, J: (12.26) (", "12.27) To evaluate these matrix elements, we use the Wigner\u2013Eckart theorem according to which the matrix elements of any vector operator ^A in the -dimensional space of eigenstates of J2 and Jz with a given value of J are proportional to the matrix elements of ^J itself: 2J + 1 \u00f0 \u00de <JLSJzj The proportionality constant g ^A j JLSJ\u2032z > = g JLS <JLSJzj \u00de ^J JLSJ\u2032z > : j \u00f0 (12.28) depends on ^A but does not depend on the values of Jz and J\u2032z : JLS \u00de \u00f0 Applying the Wigner\u2013Eckart theorem to magnetism, <JLSJzj ^Lz + 2^Szj JLSJ\u2032z > = g JLS <JLSJzj \u00de \u00f0 JLSJ\u2032z > = g ^Jzj JLS \u00f0 \u00de Jz\u03b4JzJ\u2032z : (12.29) Here, g JLS \u00de ground state Eqs. (12.25) and (12.29), we obtain \u00f0 j is known as the Land\u00e9 g factor. The matrix is diagonal in the states of definite Jz, and the n>, which is (2J + 1)-fold degenerate, is split into states with definite values of Jz: From The energies are uniformly split by g logy with Eq. (12.28), \u00f0 JLS \u00de \u03bcBB: The Land\u00e9 g factor can be evaluated by writing, in ana- \u0394EJz \u2248 g JLS \u00de \u00f0 \u03bcBBJz \u03b4JzJ\u2032z : (12.30) <JLSJzj We can also rewrite Eq. (12.31) in the alternate form (Problem 12.2) <JLSJzj\u00f0 JLSJ\u2032z > = g JLS \u00dej \u00f0 \u00de ^L + 2^S JLSJ\u2032z > : ^J j <JLSJzj ^L + 2^S j J\u2032L\u2032S\u2032J\u2032z > = g JLS <JLSJzj \u00de ^J j \u00f0 J\u2032L\u2032S\u2032J\u2032z >, (12.31) (12.32) 376 CHAPTER 12 Diamagnetism and Paramagnetism because both matrix elements vanish unless J = J\u2032, L = L\u2032, and S = S\u2032: By using the completeness relation, \u2211 J\u2033L\u2033S\u2033J\u2033 j J\u2033L\u2033S\u2033J\u2033z > <J\u2033L\u2033S\u2033J\u2033z j = 1, we can rewrite Eq. (12.32) in the alternate form \u2211 J\u2033L\u2033S\u2033J\u2033z = g <JLSJzj\u00f0 JLS \u2211 J\u2033L\u2033S\u2033J\u2033z \u00de \u00f0 ^L + 2^S \u00dej <JLSJzj J\u2033L\u2033S\u2033J\u2033z >. <J\u2033L\u2033S\u2033J\u2033z j ^J J\u2032L\u2032S\u2032J\u2032z > j J\u2033L\u2033S\u2033J\u2033z >. <J\u2033L\u2033S\u2033J\u2033z j ^J ^J j j J\u2032L\u2032S\u2032J\u2032z >. (12.33) (12.34) Because the sum over Eq. (12.34) is taken over a complete set, from Eqs. (12.31), (12.32), and (12.34), we obtain (Problem 12.3) j We have, from the relation (Problem 12.4), \u00de <JLSJzj\u00f0 ^L + 2^S. ^J JLSJ\u2032z > = g JLS ^J <JLSJzj \u00de \u00f0 2 JLSJ\u2032z > : j ^J = ^L + ^S, ^L + 2^S \u00f0 \u00de. \u00f0 ^L + ^S \u00de 2 = ^L + 2^S 2 + ^L. ^J + 2^S. ^J = 1 2 \u00bd 2 3^J \u2212 ^L 2 2 + ^S : $ From Eqs. (12.35) through (12.37), we obtain 3J \u00f0 = \u00bd J + 1 JLS g \u00f0 \u00de \u00de \u2212L \u00f0 2J $ : (12.35) (12.36) (12.37) (12.38) We note that for the 2J + 1 -dimensional set of states that make up the degenerate atomic ground state in the zero field, Eq. (12.32) can often be rewritten without the state vectors as", " \u00f0 \u00de ^L + 2^S = g JLS J, \u00de \u00f0 (12.39) as long as the matrix elements are diagonal in J, L, and S: If the splitting between the zero-field atomic ground-state multiplet and the first excited multiplet is large compared with kBT, then the states in the ground-state multiplet will contribute significantly to the free energy. In this case, we can rewrite of the field B with a magthe first term of Eq. (12.32) by using Eq. (12.39), as the interaction netic moment, 2J + 1 \u2212\u03bc!. B \u00f0 \u00de \u00f0 \u00de \u03bc! = \u2212g JLS \u03bcBJ: \u00de \u00f0 (12.40) The magnetic susceptibility has to be obtained from the free energy because the free energy cannot be equated with the ground-state energy in this case. When B 2J + 1 \u00de lowest-lying states would be small compared with kBT: As we will see, the magnetic susceptibility of a collection of magnetic ions is paramagnetic and leads to Curie\u2019s law. 0, the splitting of the! \u00f0 12.2 Atomic (or Ionic) Magnetic Susceptibilities 377 12.2.6 Curie\u2019s Law To calculate the Helmholtz free energy F, we assume that only the lowest-lying spin multiplet contributes to the statistical mechanical sums. Thus, only the lowest 2J + 1 states are thermally excited with appreciable probability. The Helmholtz free energy F, for a single ion in magnetic field B, is given by J e\u2212\u03b2F = \u2211 Jz = \u2212J e\u2212\u03b2g JLS \u00de \u00f0 J \u03bcBBJz = \u2211 Jz = \u2212J e\u2212\u03b2\u03b3BJz, where JLS \u00f0 By summing over the geometric series, we obtain \u03b3 = g \u00de \u03bcB: The magnetization of N ions in a volume V is defined as e\u2212\u03b2F = e\u03b2\u03b3B 2\u00de \u2212 e\u2212\u03b2\u03b3B J+1 J+1 2\u00de \u00f0 \u00f0 e\u03b2\u03b3B/2 \u2212 e\u2212\u03b2\u03b3B/2 : M = \u2212 N V \u2202F \u2202B : From Eqs. (12.43) and (12.44), we obtain (Problem 12.5) M = N V \u03b3JBJ\u00f0 \u03b2\u03b3JB, \u00de where the Brillouin function BJ\u00f0 = \u00f0 x x BJ\u00f0 \u00de \u00de 2J + 1 2J is defined as If \u03b2\u03b3B \u226a 1, x \u226a 1, and \u00de coth\u2061 2J + 1 2J 1 2J x \u2212 $ # coth\u2061 1 2J # x : $ coth x3, \u00de \u00f0 from Eqs. (12.46) and (12.47), we obtain x BJ\u00f0 \u00de \u2248 J + 1 3J x \u2248 J + 1 3 \u03b2\u03b3B \u2248 J + 1 3 g \u00f0 JLS \u00de \u03b2\u03bcBB: From Eqs. (12.45) and (12.48), we obtain The magnetic susceptibility is M = N V \u03b2g2\u03bc2 B 3 J + 1 BJ \u00f0 : \u00de \u03c7 = \u2202M \u2202B = N V 2 g\u03bcB\u00de 3 \u00f0 J J + 1 \u00f0 kBT \u00de : (12.41) (12.42) (12.43) (12.44) (12.45) (12.46) (12.47) (12.48) (12.49) (12.50) 378 CHAPTER 12 Diamagnetism and Paramagnetism This is known as Curie\u2019s law and characterizes paramagnetic systems with \u201cpermanent moments\u201d of which the alignment is opposed by thermal disorder but favored by the magnetic field. We note that Curie\u2019s law, as derived here, is valid as long as the system of magnetic ions is considered as noninteracting. We can rewrite Curie\u2019s law in the alternate form \u03c7 = 1 3 N V \u03bc2 Bp2 kBT, where the \u201ceffective Bohr magneton number\u201d p is given by p = g JLS \u00f0 \u00de p J J + 1 \u00f0 \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 (12.51) (12.52) : 12.3 MAGNETIC SUSCEPTIBILITY OF FREE ELECTRONS IN METALS 12.3.1 General Formulation The diamagnetic and paramagnetic susceptibilities of free", " electrons in a metal were calculated by Landau4 and Pauli, respectively. The diamagnetic susceptibility arises out of the orbital motion of the electrons, and the paramagnetic susceptibility arises out of the realignment of spins in a magnetic field. It can be shown that Pauli paramagnetism is three times the value of Landau diamagnetism with the signs reversed. In addition, there is an oscillatory contribution to the magnetic susceptibility, which is known as the de Haas\u2013van Alphen effect5. The de Haas\u2013van Alphen effect is an effective tool to measure the contours of the Fermi surface of a metal. The spin-orbit interaction is not included in the following derivation for magnetic susceptibility of free electrons. We will first calculate the magnetic susceptibility of a free electron gas confined in a box. For a free electron gas, one can write the magnetization density M as where F is the free energy M = \u2212 1 V \u2202F \u2202B, F = N\u03bc\u2212 1 \u03b2 \u2211 k ln \u2061 Zk, (12.53) (12.54) \u03b2 = 1/kBT, N is the total number of electrons, \u03bc is the chemical potential, and the partition function is The Schrodinger equation for free electrons in a magnetic field B (neglecting spin) is given by Zk = 1 + e\u2212\u03b2 E \u00f0 k \u00f0 \u2212\u03bc \u00de \u00de: H\u03c8 = r \u00de \u00f0 1 2m \u00f0 \u2212i\u0127\u2207 + 2\u03c8 eA c \u00de r \u00de \u00f0 = E\u03c8 r : \u00de \u00f0 (12.55) (12.56) 12.3 Magnetic Susceptibility of Free Electrons in Metals 379 We assume that the magnetic field is along the z direction, B = B^z, in which case A = Bx^y (from B = \u2207 \u00d7 A). This is also known as the Landau gauge. In the Landau gauge, we can write H = 1 2m \u00bd x + p2 p2 z + \u00f0 py + m\u03c9cx 2, $ \u00de \u03c9c = eB/2mc: where the cyclotron frequency If we make the substitution py = \u0127ky, pz = \u0127kz, and x = \u2212 \u0127 m\u03c9c ky + q = x0 + q, Eq. (12.57) can be rewritten in the alternate form H = 1 2m \u00f0 x + m2\u03c92 p2 cq2 + \u01272 2m \u00de k2 z : (12.57) (12.58) (12.59) (12.60) We consider the electrons to be confined in a rectangular parallelepiped of sides Lx, Ly, and Lz with the Born\u2013von Karman periodic boundary conditions. The solution of Eq. (12.56), which is a plane wave eik.r when A = 0, is modified (by using Eq. 12.60) as x \u00f0 Substituting Eqs. (12.60) and (12.61) in (12.56), we obtain r \u00f0 \u00de\u03d5 \u03c8 \u00de = ei kyy+kzz \u00f0 : \u00de \u2212 \u01272 2m x \u03d5\u2033 \u00f0 \u00de + m\u03c92 c 2 \u00f0 x \u2212 x0\u00de 2\u03d5 x \u00de \u00f0 = E\u2212 \u01272k2 z 2m \u03d5! x, \u00de \u00f0 (12.61) (12.62 Wave number kz kz FIGURE 12.1 Sketch of the one-dimensional energy sub-bands in the free electron model. where Eq. (12.62) is the Schrodinger equation of the one-dimensional oscillator centered at x0: The eigenvalues are Ev = \u01272k2 z 2m + v + # 1 2 $ \u0127\u03c9c, v = 0,1, 2, \u2026: (12.63) Thus, the energy of the electron is the sum of the kinetic energy of its undisturbed motion along the z direction, and the quantized energy of the oscillatory motion in the plane orthogonal to the field direction. This part of the energy is the contribution due to the orbital motion. The energy of the one-dimensional sub-bands is plotted as a function of the wave number kz in Figure 12.1.The wave function in Eq. (12.61) depends on ky and kz. directly and on kz and v indirectly through \u03d5 x \u00f0 \u00de 380 CHAPTER 12 Diamagnetism and Paramagnetism Because one can choose any ky for a particular value of kz and v, the state ky is degenerate. Further, x0 lies in the", " range \u2212Lx/2 < x0 < Lx/2: (12.64) If we neglect spin, the states occur on the ky axis in the intervals of 2\u03c0/Ly: One can write E d ) E ( g go(E)dE g(E)dE \u2212 1 2\u0127 m\u03c9cLy < ky < 1 2\u0127 m\u03c9cLx: (12.65) FIGURE 12.2 5 10 15 20 E \u00f0 \u00de\u00f0 Ly/2\u03c0 m\u03c9cLx/\u0127 different \u00de in the range dkz, the Thus, ky can have values. Furthermore, z-component of k will have multiplying by a factor of 2 for spin and dividing by the volume of the cube (Lx Ly Lz\u00de dkz = Lz/2\u03c0 dkz: g \u00de \u00f0 dkz different values. The density of states is obtained by Sketch of the density of states for B = 0 and B \u2260 0:, (12.66) v, kz\u00de \u00f0 m\u03c9c \u0127 2 2\u03c0 \u00f0 2 \u00de From Eqs. (12.63) and (12.66), we obtain (Problem 12.6) 2m \u01272! \" The total density of states is obtained by summation over all bands that lie \u201cbelow\u201d the energy E, E\u2212 v + # 2 2\u03c0 \u00f0 (12.67) dE = \u0127\u03c9c dE: E h\u03c9c 2 \u22121/2 3/2 g E \u00f0 \u00de v\u2032 dE = \u2211 v = 0 g \u00f0 dE: E, v \u00de (12.68) It can be easily shown (Problem 12.7) that Eq. (12.68) has the same density of states as the zero-field case. The average density of states is unaffected by a magnetic field. The states are redistributed due to the magnetic field B, which pulls together a large number of states into a single level (see Figure 12.2). \u03bcBB, depending on its direction, where the Bohr magneton \u03bcB = e\u0127/2mc and the g factor for free electrons is g = 2. If we include the contribution of spin, the energy E is modified as The contribution due to spin is \u00b1 \u00f0 g/2 \u00de E\u00b1 = \u01272k2 z 2m + \u00f0 2v + 1 \u03bcBB \u00b1 \u00de g 2 \u03bcBB = E \u00b1 \u03bcBB: (12.69) 12.3.2 Landau Diamagnetism and Pauli Paramagnetism From Eqs. (12.53) through (12.55) and (12.69), we obtain the expression for magnetization per unit volume (Problem 12.8), M = \u2212 d dB n\u03bc ln \u2061 \u00bd e\u03b2 \u03bc\u2212E \u00f0 \u00de + 1 dE + \u2212 $ 1 \u03b2 \u221e Z 0 g E\u2212\u00de \u00f0 ln \u2061 \u00bd e\u03b2 \u03bc\u2212E \u00f0 \u00de + 1 dE\u2212 $ (12.70) : 9 = ; 12.3 Magnetic Susceptibility of Free Electrons in Metals 381 In Eq. (12.70), the summation over the energy states is replaced by an integration because the energy levels are very close together. Here, n is the electron concentration, and \u03bc is the chemical potential. We have n = \u221e Z 0 dE = dE, where f E \u00f0 \u00de is the Fermi function, \u00de The density of states in k space is given by \u00f0 f E = 1 + e\u03b2 \u00bd E \u00f0 k \u00de \u00f0 \u2212\u03bc \u00de \u22121: $ E g \u00f0 \u00de dE = 1 2\u03c02 3/2 2m \u01272! \" E1/2 dE: From Eqs. (12.71) through (12.73), we obtain n = 2 mkBT 2\u03c0\u01272! 3/2 \" 2 \u03c0p F, \u03b2\u03bc \u00de \u00f0 where the Fermi integral F is given by x \u00f0 \u00de = F x \u00de \u00f0 \ufb03\ufb03\ufb03 \u221e Z 0 y1/2 1 + e\u00f0 y\u2212x \u00de dy. It can be shown (Problem 12.9) that and F x \u00de \u00f0 \u2248 \u03c0p 2 \ufb03\ufb03\ufb03 ex, for x < 0 F x \u00f0 \u00de \u2248 x3/2, 2 3 for x > 0: (12.71) (12.72) (12.73) (12.74) (12.75) (12.76) Using a similar expansion for Eq. (12.70", ") and for low temperatures, in which the first term is retained, one can show that (Problem 12.10) M = 3n\u03bc2 BB 2EF 2 1\u2212 6 6 6 4 1 3 + \u03c0kBT \u03bcBB EF \u03bcBB! 1/1 \u00de vp cos \u2061 \u03c0v \u00de \u00f0 \ufb03\ufb03\ufb03 sin\u2061 \u03c0 4 \u2212 \u03c0vEF \u03bcBB \u03c02vkBT \u03bcBB! sinh\u2061 7 7 7 5 \" 3 : (12.77) Here, EF is the Fermi energy (value of \u03bc at T = 0). The magnetic susceptibility is obtained from the expression \u03c7 = \u2202M \u2202B : (12.78) 382 CHAPTER 12 Diamagnetism and Paramagnetism For most cases, M is linear in B for attainable field strengths, and the definition reduces to \u03c7 \u2248 M B : (12.79) From Eqs. (12.77) and (12.79), we obtain (retaining only the first term in the series expansion), where \u03c7P is the Pauli spin paramagnetism, \u03c7 = \u03c7P + \u03c7L + \u03c7dH\u2212vA, \u03c7L is the Landau diamagnetism, \u03c7P = 3n\u03bc2 B 2EF, \u03c7L = \u2212 n\u03bc2 B 2EF, and \u03c7dH\u2212vA is the de Haas\u2013van Alphen effect, which is oscillatory with a period additional diamagnetic term given by (12.80) (12.81) (12.82) 2\u03bcB EF : This is an \u03c7dH\u2212vA \u2248 3n\u03c0 2\u03b2 \u03bcB EFB! \" 1/2 sin! sinh 0 B B @ \u2212 \u03c0EF \u03bcBB \u03c02 \u03bcB\u03b2B 12.83) We note from Eq. (12.83) that because the sin term is oscillatory, sin \u2061 \u03c0 4 \u2212 \u03c0EF \u03bcBB = sin \u03c0 4! \u2212 \u03c0EF \u03bcBB \"! + 2n\u03c0, \" \u03c7dH\u2212vA is oscillatory whenever or \u03c0EF \u03bcBB = \u03c0EF \u03bcBB + 2n\u03c0 1 B = 1 B + 2n\u03bcB EF : (12.84) (12.85) (12.86) \u03c7dH\u2212vA is periodic in 1/B with a temperature-independent period 2\u03bcB/EF: However, these oscillations can be observed only at low temperatures and high magnetic fields. It will be shown that the condition for such oscillations is \u0127\u03c9c = 2\u03bcBB \u226b 1 \u03b2 : (12.87) Otherwise, the distribution of electrons in the region of EF is widely spread, and the oscillations are spread out. 12.3 Magnetic Susceptibility of Free Electrons in Metals 383 EF 0 (a) 0 Densities of states g g, (b) 0 (c) FIGURE 12.3 Sketch of densities of states g\u2193 and g\u2191 as a function of energy. The origin of each of the previous three contributions to the magnetic susceptibility of a free electron gas can be explained in simple manner. The origin of the Pauli spin paramagnetism is the additional contribution \u00b1\u03bcBB to the kinetic energy of the electrons due to the magnetic field, depending on their spin directions. This is shown in Figure 12.3. In Figure 12.3a, the occupied states for both spins are the same in the absence of a magnetic field. The highest occupied states for each spin have the energy equal to the Fermi energy EF: In Figure 12.3b, the states of opposite spins are shifted in a magnetic field B: However, the \u201coccupied\u201d states above EF with spin \u201cup\u201d flow to the \u201cunoccupied\u201d states with spin \u201cdown\u201d until the states are filled. The imbalance of the density of states in a magnetic field between the two spin states contributes to the positive Pauli paramagnetism. The Landau diamagnetic term represents the orbital quantization of the electrons in a magnetic field. It is one-third the contribution of the Pauli paramagnetism and is of the opposite (negative) sign. The theory of diamagnetic susceptibility of metals was derived by Misra and Roth (Ref. 12). The theory of magnetic susceptibility of Bloch electrons was derived by Misra and Kleinman (Ref. 13). 12.3.3 De Haas\u2013van Alphen Effect The de Haas\u2013van Alphen effect5, which has been used extensively to investigate the Fermi surfaces of metals, can be explained by first assuming that the Fermi level is approximately constant as B is varied. The validity of this assumption can be as seen in Figure", " 12.4. We note G the energy increases. Here, G are the number of states below the energy E in the presence and absence of the magnetic field. Because the number of states below the Fermi level EF is half of the number of electrons at T = 0 (each state can have two electrons), EF is approximately constant as B is varied. E E \u223c G0\u00f0 \u00f0 \u00de and G0\u00f0.2 1.1 1.0 1.9 0.8 0.7 E 5 10 15 20 Energy E FIGURE 12.4 The ratio of the number of states G(E ) below E in a magnetic field to the number G0(E ) in the absence of a magnetic field. 384 CHAPTER 12 Diamagnetism and Paramagnetism If we consider the case at T = 0, all states are filled up to the Fermi level EF, and all states above it are empty. If we consider a plane slab in k space of thickness \u03b4kz at kz, the number of allowed values of kz in the range \u03b4kz is \u03b4kz: The total degeneracy of the state v (neglecting spin) per unit volume in the slice \u03b4kz is (from Eq. 12.66) Lz/2\u03c0 \u00de \u00f0 m\u03c9c 4\u03c02\u0127 \u03b4kz = \u03beB: (12.88) Thus, the degeneracy parameter (apart from spin), which is defined as the degeneracy per unit magnetic field per unit volume, is \u03be = e\u03b4kz 4\u03c02\u0127c : (12.89) We consider the de Haas\u2013van Alphen effect for a free electron gas at absolute zero. Figure 12.5 shows the spectrum of the Landau levels when the energy E is plotted versus B. At T = 0, all levels in the slice \u03b4kz will be filled (Figure 12.5) for which v + \u00f0 1 2\u00de \u0127\u03c9c + \u01272k2 z 2m \u2264 EF, E 10 9 8 7 6 5 4 3 which can be rewritten in the alternate form, v + 1 2\u00de \u00f0 \u0127\u03c9c \u2264 \u03b5\u2032F, (12.91) where \u03b5\u2032F = EF \u2212 \u01272k2 z 2m : (12.92) At T = 0, if v\u2032 is the highest filled level, n, the number of electron states in the slice of thickness \u03b4kz is (from Eq. 12.89 and the fact that v = 0 is a filled level) n = v\u2032 + 1 \u03beB: \u00de \u00f0 (12.93) When B is increased, n increases linearly with B until v\u2032 coincides with \u03b5\u2032F: As B is further increased, all the electrons in v\u2032 will have energy greater than \u03b5\u2032F, and hence, they will empty out into the orbits in other slices with different values of kz and \u03b5\u2032F: This oscillatory evacuation occurs when v\u2032 + 1 2\u00de \u00f0 = \u03b5\u2032F \u0127\u03c9c = mc\u03b5\u2032F e\u0127 1 B : (12.94) (12.90) \u03b5\u2032F 2 1 \u03bd = 0 B FIGURE 12.5 Sketch of the spectrum of Landau levels versus B. 12.3 Magnetic Susceptibility of Free Electrons in Metals 385 Hence, the population \u03b4n is approximately a periodic function of 1/B, with period e\u0127 mc\u03b5\u2032F of the electrons in the slice \u03b4kz in a magnetic field B such that the population n0 is : The energy v\u2032 E0 = \u0127\u03beB\u03c9c\u2211 v + 1 2 # $ + n0 \u01272k2 z 2m = 1 2 \u0127\u03beB\u03c9c\u00f0 v\u2032 + 1 \u00de 2 + n0 \u01272k2 z 2m : From Eqs. (12.89) and (12.95), we obtain E0 = 1 2 \u0127\u03c9c \u03beB n2 0 + n0 \u01272k2 z 2m, (12.95) (12.96) where n0 = v\u2032 + 1 \u00f0 v\u2032 for the highest filled state, \u00de \u03beB from Eq. (12.89) for this value of B: For a nearby B that has the same value of E = 1 2 \u0127\u03c9c \u03beB n2 + n \u01272k2 z 2m + n0 \u2212 n \u00f0 \u00de \u03b5F, (12.97) n0 \u2212 n where \u03b5F is the change in energy arising out of the transfer of n \u2212 n0 electrons at the Fermi level while the first two terms are the energy of the electrons in the slice", " \u03b4kz: From Eqs. (12.92), (12.96), and (12.97), we obtain \u00f0 \u00de where \u03b4E = E \u2212 E0 = \u03bc \u03be \u00f0 n2 \u2212 n2 0\u00de + \u00f0 n0 \u2212 n \u03b5\u2032F, \u00de \u03bc = e\u0127 2mc : Further, from Eq. (12.94), \u00f0 From Eqs. (12.98) and (12.100), we obtain \u03b5\u2032F \u2248 n0/\u03beB \u00de \u0127\u03c9c = 2\u03bcn0/\u03be: \u03b4E = \u03bc \u03be \u00f0! \" 2: n \u2212 n0\u00de From Eq. (12.101), \u03b4M, the magnetization of the slice at T = 0, \u03b4M = \u2212 \u2202E \u2202B = \u2212 2\u03bc \u03be \u00f0 n \u2212 n0\u00de dn dB : From Eqs. (12.93), (12.94), and (12.100), we have From Eqs. (12.102) and (12.103), we obtain dn dB \u2248 \u03b5\u2032F \u03be 2\u03bcB : \"! \u03b4M \u2248 \u2212 \u03b5\u2032F B \u00f0 n \u2212 n0\u00de : (12.98) (12.99) (12.100) (12.101) (12.102) (12.103) (12.104) 386 CHAPTER 12 Diamagnetism and Paramagnetism As B is increased, n \u2212 n0 oscillates with extrema \u00b1 1 2 \u03beB because the population of the level v\u2032 varies between \u03beB and 0. Thus, the magnetization varies between \u2213 1 2 \u03be\u03b5\u2032F: This oscillation of the magnetization as a periodic function of B is known as the de Haas\u2013van Alphen effect, a more rigorous analysis of which was presented earlier in this section. The de Haas\u2013van Alphen oscillations are sensitive probes of the geometrical property of the Fermi surface. The areas of the extremals of the electron orbits can be determined from the period of oscillation. Because the de Haas\u2013van Alphen effect is an important tool and is widely used to experimentally measure the contours of the Fermi surface of crystalline solids, we will present the sequence of events that occur when a magnetic field is applied to a free electron gas in a rectangular parallelepiped, the Fermi surface of which is a sphere prior to the application of a magnetic field. Figure 12.6 shows a plane section in the k space. The electron states are uniformly distributed in plane. The magnetic field B is directed normally into the plane section. The k states are kx, ky\u00de the \u00f0 subjected to a Lorenz force, and all the k states rotate with the cyclotron frequency about an axis through the origin and parallel to the field direction. The magnetic field causes a redistribution of the k states that lie on rings that correspond to the energies Ev = v + 1 2\u00de \u00f0 \u0127\u03c9c = \u01272k2 v 2m : (12.105) These rings are shown in Figure 12.7. Because the magnetic field B is in the z direction, kz is not affected by it. Thus, in the k space, the representative points lie on cylinders (Landau cylinders) of which the cross-sections are the Landau rings. At T = 0, the k states are within the Fermi sphere of radius kF: Thus, the Landau cylinders, shown in Figure 12.8, are either partially occupied or fully empty. ky 0 1 2 3 kx \u03bd \u03bd \u00d7 B k F FIGURE 12.6 A magnetic field B is directed normally into the plane section of k space. The Landau levels have a concentric circular form plane but have cylindrical surfaces kx, ky \u00de in the \u00f0 in the k space. FIGURE 12.7 12.3 Magnetic Susceptibility of Free Electrons in Metals 387 FIGURE 12.8 \u03bd \u03bd \u2032 \u03bd \u2033 Landau cylinders. As the magnetic field B is increased, the cylinders expand until they become empty when they cross the Fermi sphere. FIGURE 12.9 The de Haas\u2013van Alphen oscillations for a metal with complex structure with decreasing magnetic field. 1 B Figure 12.9 shows the de Haas\u2013van Alphen oscillations for metals with complicated crystal structures as B is decreased. Usually, the Fermi surface of such metals has a thin neck and a thick belly. The large-scale oscillations, of which the amplitude decreases with decreasing B, are due to the extremal orbits around the thin neck, whereas the barely resolved small-scale oscillations are due to the", " extremal orbits of the thick belly. 388 CHAPTER 12 Diamagnetism and Paramagnetism 12.4 MANY-BODY THEORY OF MAGNETIC SUSCEPTIBILITY OF BLOCH ELECTRONS IN SOLIDS 12.4.1 Introduction of Bloch elecMisra et al. (Ref. 14) derived an expression for the total magnetic susceptibility trons in solids (including spin-orbit interaction) that includes both many-body and interband effects.14 They used a finite-temperature Green\u2019s function formalism to express the thermodynamic potential \u03a9 for an interacting electron system in the presence of a periodic potential V (r), spin-orbit interaction, and external magnetic field B in terms of the exact one-particle propagator G: They showed that the total magnetic susceptibility of a nonferromagnetic solid can be expressed as the sum of three terms, T, V, \u03bc, B \u03c7 \u00f0 \u00de \u00f0 \u00de \u03c7 = \u03c7o + \u03c7s + \u03c7so, (12.106) where \u03c7o is the orbital contribution, \u03c7s is the spin contribution, and \u03c7so is the contribution of the spin-orbit coupling on the orbital motion of the Bloch electrons. An important aspect of their derivation is the analysis of exchange and correlation effects on each of these terms that have been explicitly calculated. Misra et al. (Ref. 14) have shown that if they make a simple approximation for the self-energy, their expression for \u03c7o is reduced to the earlier results. If they make drastic assumptions while solving the matrix integral equations for the field-dependent part of the self-energy, their expression for \u03c7s is equivalent to the earlier results for the exchange-enhanced \u03c7s but with the g factor replaced by the effective g factor, a result that has been intuitively used but not yet rigorously derived. An important aspect of their derivation is the analysis of exchange and correlation effects on \u03c7so that are more subtle and cannot be included in an intuitive way. 12.4.2 Equation of Motion in the Bloch Representation The exact one-particle propagator G satisfies the equation \u03bel \u2212 ^H \u00f0 G r, r\u2032, \u03bel\u00de \u00f0 \u00de + Z dr\u2033^\u03a3 r, r\u2033, \u03bel\u00de \u00f0 G r\u2033, r\u2032, \u03bel\u00de \u00f0 = \u03b4 \u00f0 r \u2212 r\u2032, \u00de (12.107) where ^\u03a3 is the exact proper self-energy operator, \u03bel is the complex energy, \u03bel = \u00f0 \u03c0i 2l + 1 \u03b2 \u00de + \u03bc, l = 0, \u00b11, \u00b12, :::, (12.108) and ^H is the Hamiltonian of the Bloch electron in a magnetic field, + \u01272 + \u01272. \u2207!V \u00d7 p + ^H = + V p + r 2 1 2m # eA c $ 4m2c2 \u03c3! eA c $ # In the absence of the magnetic field, both G and ^\u03a3 have the symmetry 8m2c2 \u22072V + 1 2 g\u03bcBB. \u03c3!: (12.109) \u00f0 \u00de G r + R, r\u2032 + R, \u03bel\u00de \u00f0 = G r, r\u2032, \u03bel\u00de \u00f0 (12.110) 12.4 Many-Body Theory of Magnetic Susceptibility of Bloch Electrons in Solids 389 and ^\u03a3 r + R, r\u2032 + R, \u03bel\u00de \u00f0 = ^\u03a3 r, r\u2032, \u03bel\u00de : \u00f0 (12.111) The vector potential in the Hamiltonian destroys this symmetry. It can be shown that in a, both ^G and ^\u03a3 can be written as the product of a Peierls phase A = 1 symmetric gauge factor and a part that has the preceding symmetry, 2 B \u00d7 r \u00f0 \u00de and where G r, r\u2032, \u03bel, h \u00f0 \u00de = eih.r\u00d7r\u2032 G r, r\u2032, \u03bel, h \u00de \u00f0 e ^\u03a3 r, r\u2032, \u03bel, h \u00de \u00f0 = eih.r\u00d7r\u2032 \u03a3 r, r\u2032, \u03bel, h \u00de, \u00f0 e h = eB 2\u0127c : (12.112) (12.113) (12.114) Substituting Eqs. (12.112) and (12.113) in Eq. (12.108), commuting the differential operator through the Peierls phase factor, and then multiplying the left side by e\u2212ih.r \u00d7 r\u2032", ", we obtain (Problem 12.11) \u03bel \u2212 \u00f0 \u2212 1 2 1 2m\u00bd g\u03bcBB. \u03c3!\u00de p + \u0127h \u00d7 r\u2212r\u2032 \u00f0 2 \u00de$ \u2212 G r, r\u2032, \u03bel, h \u00de \u00f0 \u2212 Z \u0127 4m2c2 \u03c3! dr\u2033eih. \u00f0. \u2207!V \u00d7 p + \u0127h \u00d7 \u00bd \u00f0 r \u2212 r\u2032 \u00de$ \u2212V \u2212 r \u00de \u00f0 \u01272 8m2c2 \u22072V (12.115) r\u2032\u00d7r+r\u00d7r\u2033+r\u2033\u00d7r\u2032 \u00de \u03a3 r, r\u2033, \u03bel, h \u00de \u00f0 G r\u2033, r\u2032, \u03bel, h \u00de \u00f0 = \u03b4 \u00f0 r\u2212r\u2032. \u00de One can write the equation of motion in the Bloch representation, i.e., in terms of the basis functions, e e e r \u03c8 nk\u03c1\u00f0 \u00de = eik.rUnk\u03c1\u00f0 r, \u00de (12.116) is a periodic two-component function, n is the band index, k is the reduced wave vector, r where Unk\u03c1\u00f0 and \u03c1 is the spin index. Using the Bloch representation, one can show that Eq. (12.115) can be rewritten as (Problem 12.12) \u00de where \u03bel \u2212 ^H \u00bd G \u03ba!, \u03bel\u00de$ \u00f0 k, \u03bel\u00de \u00f0 = I, e ^H \u03ba!, \u03bel\u00de \u00f0 = 1 2m\u00f0 p+\u0127 \u03ba!\u00de 2 +V + \u0127 4m2c2 \u03c3! r \u00de \u00f0. \u2207!V \u00d7 p+\u0127 \u03ba!\u00de \u00f0 + \u01272 8m2c2 \u22072V + 1 2 g\u03bcBB. \u03c3!+\u03a3 and \u03ba! = k + ih \u00d7\u2207k: (12.117) \u00f0 \u03ba!, \u03bel\u00de (12.118) (12.119) 390 CHAPTER 12 Diamagnetism and Paramagnetism 12.4.3 Thermodynamic Potential The grand partition function of a system is defined as f where \u03bc is the chemical potential, and ^N is the operator giving the number of particles. If we write ZG in the form g ZG = Tr e\u2212\u03b2 ^H\u2212\u03bc ^N \u00f0 \u00de, (12.120) ZG = e\u2212\u03b2\u03a9 T,V,\u03bc,B \u00f0 \u00de, (12.121) all the thermodynamic properties may be derived from \u03a9 namic potential. It can be easily shown that the mean energy E is given by T, V, \u03bc, B \u00f0 \u00de, which is called the thermody- \u03a9 = E \u2212 \u03bcN \u2212 TS, (12.122) where N is the mean number of particles, and S is the entropy. Because S last term in Eq. (12.122) is neglected.! 0 when T 0, the! It can be shown that the thermodynamic potential for an interacting system is given by 1 \u03b2 1 \u03b2 \u00bd \u2212 \u03a9 = Tr ln G\u03bel\u00de e e are the one-particle Green\u2019s function and the proper self-energy, Here, respectively. \u03bel stands for the imaginary frequencies, \u03a9qp is the contribution from quasiparticles, and \u03a9corr describes the corrections from electron correlations. The functional G\u03bel\u00de$ % e G\u03bel % e \u03a3\u03bel % e \u03a9qp + \u03a9corr: G\u03bel + \u03d5 (12.123) G\u03bel\u00de G \u00f0 \u03bel\u00de \u2212Tr and \u03a3\u03bel + \u03a3 e e e \u00f0 \u00f0 \u00f0 \u03d5 \u00f0 G\u03bel \u00de e = lim\u2061 1 \u03bb! Tr\u2211 n \u03bbn 2n n \u00de \u03a3\u00f0 \u00f0 G\u03bel \u00de e e e G\u03bel, (12.124) \u00f0 where Tr involves summation over both the imaginary frequencies and one-particle states, and n is the nth-order self-energy part, where only the interaction parameter \u03bb occurring explicitly G\u03bel \u00de \u03a3\u00f0 \u00de is defined through the decomposition of in Eq. (12.123) is used to determine the order. In fact, \u03d5 n : Differinto skeleton diagrams. There are 2n G\u03bel \u00de G\u03bel lines for the nth-order diagrams in \u03d5 e \u03a3", "\u00f0 \u00de G\u03bel has the effect of \u201copening\u201d any of the 2n lines of the nth-order entiating \u03d5 diagram, and each will give the same contribution when Tr is taken. e e with respect to G\u03bel\u00de e G\u03bel\u00de e G\u03bel\u00de e e \u00f0 \u00f0 \u00f0 \u00f0 12.4.4 General Formula for \u03c7 The magnetic susceptibility is calculated from the expression \u22022 \u03a9 \u2202B\u03bc\u2202Bv : From Eqs. (12.121) through (12.123), it can be shown that14 \u03c7\u03bcv = \u2212 lim\u2061 0 B! 1 V \u03c7\u03bcv = 1 V\u03b2 \u2212 \u22022 \u2202B\u03bc\u2202Bv Tr ln \u00f0 \u2212 \" \u22022 \u03a3\u03bel \u2202B\u03bc\u2202Bv e + Tr G\u03bel\u00de e G\u03bel + Tr e \u2202 \u03a3\u03bel \u2202B\u03bc e G\u03bel \u2202 \u2202Bv e : #B 0! (12.125) (12.126) 12.4 Many-Body Theory of Magnetic Susceptibility of Bloch Electrons in Solids 391 We can expand \u03a3 \u03ba!, B, \u03bel\u00de \u00f0 = \u03a3 k, B, \u03bel\u00de \u00f0 \u2212 ih\u03b1\u03b2 \u2202 \u03a3 \u00f0 k, B, \u03bel\u00de \u2202k\u03b1 \u2207\u03b2 k\u2212 1 2 h\u03b1\u03b2h\u03b3\u03b4 \u03a3 \u22022 \u2202k\u03b1\u2202k\u03b3 \u2207\u03b2 k\u2207\u03b4 k + \u2026 and e e e e \u03a3 k, B, \u03bel\u00de \u00f0 = \u03a30 k, \u03bel\u00de \u00f0 + B\u03bc\u03a31,\u03bc k, \u03bel\u00de \u00f0 + B\u03bcBv\u03a32,\u03bcv k, \u03bel\u00de \u00f0 + \u2026, where e h\u03b1\u03b2 = \u2208\u03b1\u03b2\u03b3 h\u03b3, (12.127) (12.128) (12.129) \u2208\u03b1\u03b2\u03b3 is the antisymmetric tensor of the third rank, and we follow the Einstein summation convention. From Eqs. (12.118), (12.119), (12.127), and (12.128) (Problem 12.13), we obtain ^H \u03ba!, \u03bel\u00de \u00f0 = ^H0\u00f0 k, \u03bel\u00de + ^H\u2032 \u00f0, k, \u03bel\u00de (12.130) where ^H0\u00f0 k, \u03bel\u00de = 1 2m \u00f0 p + \u0127k \u00de 2 + V r \u00de \u00f0 + \u03a30 k, \u03bel\u00de \u00f0 + \u01272 8mc2 \u22072V + \u0127 4m2c2 \u03c3!. \u2207!V \u00d7 p + \u0127k \u00de \u00f0 (12.131) and ^H\u2032 \u00f0 k, \u03bel\u00de = \u2212ih\u03b1\u03b2\u220f \u03b1 \u2207\u03b2 \u2212 1 2 h\u03b1\u03b2h\u03b3\u03b4 k + \u01272 m & g\u03bcBB\u03bc\u03c3\u03bc + B\u03bc\u03a31,\u03bc 0 1 2 \u03b4\u03b1\u03b3 + \u22022 \u2207\u03b2 k\u2207\u03b4 \u03a3 \u2202k\u03b1\u2202k\u03b3 k, \u03bel\u00de \u00f0 k + B\u03bcBv\u03a32,\u03bcv \u2019 \u2212ih\u03b1\u03b2B\u03bc \u2202\u03a3 1,\u03bc \u2202k\u03b1 \u2207\u03b2 k, k, \u03bel\u00de \u00f0 (12.132) where the terms up to the second order in the magnetic field are retained, and \u220f operator,!=\u0127 is the velocity (! = \u0127 \u220f m \u00f0 p! + \u0127k \u00de + \u01272 4m2c2 \u03c3! \u00d7 \u2207!V + \u2207k\u03a30 k, \u03bel\u00de : \u00f0 (12.133) We can make a perturbation expansion ( k, \u03bel\u00de = G0\u00f0 k, \u03bel\u00de + G0\u00f0 k, \u03bel\u00de H\u2032G0\u00f0 k, \u03bel\u00de + G0\u00f0 k, \u03bel\u00de H\u2032G0\u00f0 k, \u03bel\u00de H\u2032G0\u00f0 k, \u03bel\u00de + :::,\u22ef (12.134) G \u00f0 where e G0\u00f0 k, \u03bel\u00de = 1 \u03bel \u2212 H0\u00f0, k, \u03bel\u00de (12.135) and only terms up to the second order in the magnetic", " field are retained. It can be shown that (Problem 12.14) \u2207\u03b1 kG0\u00f0 k, \u03bel\u00de = G0\u00f0 k, \u03bel\u00de \u220f \u03b1G0\u00f0 k, \u03bel\u00de (12.136) 392 CHAPTER 12 Diamagnetism and Paramagnetism and where \u2207\u03b1 k\u2207\u03b3 kG0\u00f0 k, \u03bel\u00de = G0 \u01272 m & \u03b4\u03b1\u03b3 + \u03a7\u03b1\u03b3 G0 + G0\u220f \u2019 \u03b1G0\u220f \u03b3G0 + G0\u220f \u03b3G0\u220f \u03b1G0, (12.137) \u03a7\u03b1\u03b3 = \u2207\u03b1 k \u01272 4m2c2 \u00f0 & \u03c3! \u00d7 \u2207!V \u03b3 + \u2207\u03b3 \u00de 0 k \u2211 k, \u03bel\u00de \u00f0 : \u2019 (12.138) After considerable algebra (for details, see Misra et al.14), the general expression for the total magnetic susceptibility of nonferromagnetic solids (including exchange and correlation effects) is obtained as where \u03c7\u03bcv 0 = \u2211 k \u00f0 + 1 + \u03b4\u03bcv\u00de( e2\u2208\u03b1\u03b2\u03bc\u2208\u03b3\u03b4v 4\u01272c2 \" e2\u2208\u03b1\u03b2\u03bc\u2208\u03b3\u03b4v 48\u01272c2 \u2207\u03b1 \u220f 2\u01272 m \u2212 \u00f0 \u03b3 m\u03c1\u2032,q\u03c1\u2033\u220f E ln \u2061EqnEmn \u03b3 m\u03c1\u2032,q\u03c1\u2033\u220f \u03b2 n\u03c1,m\u03c1\u2032\u220f E2 mnEqn \u03b1\u03b3 q\u03c1\u2033,n\u03c1 \u03b4 m\u03c1\u2032,q\u03c1\u2033\u03a7 + + + \u220f \u03b1 n\u03c1,n\u03c1\u220f \u220f \u03b2 n\u03c1,m\u03c1\u2032\u220f EmnEqn \u03c7\u03bcv = \u03c7\u03bcv 0 + \u03c7\u03bcv s + \u03c7\u03bcv so, k\u2207\u03b3 kEn \u2207\u03b2 k\u2207\u03b4 kEn f \u2032 En\u00de \u00f0 \u03b1 n\u03c1,m\u03c1\u2032\u220f E2 mn \u03b3 m\u03c1\u2032,n\u03c1 \u03b4\u03b2\u03b4 + 2 \u220f \u03b1 n\u03c1,m\u03c1\u2032\u220f \u03b2 n\u03c1\u2033,q\u03c1\u2033\u2032\u220f \u03b3 m\u03c1\u2032,n\u03c1\u2033\u220f E2 mnEqn \u03b2 q\u03c1\u2033,n\u03c1 \u03b4 m\u03c1\u2032,q\u03c1\u2033\u220f \u03b3 n\u03c1,m\u03c1\u2032\u220f EmnE2 qn \u03b4 q\u03c1\u2033,n\u03c1 \u03b4 q\u03c1\u2033,n\u03c1 \u220f \u2212 \u03b2 n\u03c1,m\u03c1\u2032\u03a7 \u03b1\u03b3 m\u03c1\u2032,q\u03c1\u2033\u220f EmnEqn \u03b4 q\u03c1\u2033,n\u03c1 \u03b2 m\u03c1\u2032,q\u03c1\u2033\u220f \u03a7 \u03b1\u03b3 n\u03c1,m\u03c1\u2032\u220f EmnEqn \u03a7 \u2212 \u03b1\u03b3 n\u03c1,n\u03c1\u220f \u03b2 n\u03c1,m\u03c1\u2032\u220f E2 mn \u03b4 m\u03c1\u2032,n\u03c1 # \u03b1 n\u03c1,m\u03c1\u2032\u220f \u220f \u22122 \u03b2 q\u03c1\u2033,l\u03c1\u2033\u2032\u220f \u03b4 l\u03c1\u2033\u2032,n\u03c1 \u2212 \u220f \u03b1 n\u03c1,n\u03c1\u220f \u03b4 q\u03c1\u2033\u2032,n\u03c1 (12.139) (12.140) \u220f + ie 4\u0127c \u2208\u03b1\u03b2v \" where \u03b2 n\u03c1,m\u03c1\u2032Y \u03b1\u03bc Emn m\u03c1\u2032,n\u03c1 Y \u03b1\u03bc n\u03c1,m\u03c1\u2032\u220f Emn \u2212 \u03b2 m\u03c1\u2032,n\u03c1 f En\u00de \u00f0 # ), and repeated indices means summation over band and spin. Similarly, one can show14 that the effective Pauli spin susceptibility, including the exchange and correlation effects is, Emn % Em \u2212 En, (12.141) \u03c7\u03bcv s = \u2212 1 8 \u00f0 1 + \u03b4\u03bcv\u00de \u03bc2 B \u2211 n, k,\u03c1,\u03c1\u2032 k gv nn\u00f0 \u00de n\u03c1,n\u03c1\u2032 g\u03bc \u03c3v! nn\u00f0 k \u00de \u03c3\u03bc n\u03c1\u2032,n\u03c1 + 2 \u03bcB 1,\u03bc n\u03c1\u2032,n\u03c1 \u03a3, En\u00de f \u2032 \u00f0 \" where the effective g matrix is defined as k gv nn\u00f0 \u00de \u03c3v n\u03c1,n\u03c1\u2032 = ie \u03bcB\u0127c \u2208\u03b1\u03b2v \u2211 m,\u03c1\u2033 \u03b2 m\u03c1\u2033,n\u03c1\u2032 \u220f \u03b1 n\u03c1,m\u03c1\u2033\u220f Emn + g\u03c3v n\u03c1,n\u03c1\u2032: (12.142) (12.", "&nk\u03c1\u00f0 r \u03c8 mk\u03c1\u2032\u00f0 \u00de r\u2032 veff \u00f0 r, r\u2032 \u00de \u03c8 pk\u2032\u03c1\u00f0 \u00de r \u03c8 &qk\u2032\u03c1\u2032\u00f0 \u00de r\u2032 drdr\u2032: \u00de (12.153) Eq. (12.153) is the exchange self-energy in the band model. One can obtain \u03a3 G. We make the further approximation (defined in Eq. 12.128), by expanding 0, \u03a3 1, \u03a3 2, and so on \u00dej From Eqs. (12.152) and (12.154), j <nn k, k\u2032 veff \u00f0 pq> \u03c1\u03c1\u2032 \u2248 <nn e veff \u00f0 k, k\u2032 j pp> \u03b4pq = vnp\u00f0 j k, k\u2032 \u00de \u03b4pq: (12.154) \u03a3n\u03c1,n\u03c1\u2032\u2032, \u03bel,\u03c1 vnp\u00f0 k, k\u2032. k\u2032, \u03bel\u00de (12.155) Gp\u03c1,p\u03c1\u2032\u00f0 \u00de e Substituting only the value of the first-order terms of B occurring in G from Eq. (12.134) on the right side of Eq. (12.155), and neglecting the terms proportional to f, we obtain (Problem 12.16) \u03a3 1,\u03bc n\u03c1,n\u03c1\u2032\u00f0 k \u00de \u2243 \u2212\u2211 mk\u2032 vnm\u00f0 k, k\u2032 \u03a3 1,\u03bc m\u03c1,m\u03c1\u2032\u00f0 k\u2032 \u00de f \u2032 m\u00f0 \u00de k\u2032 \u2212 \u00de 1 2 \u03bcB \u2211 mk\u2032 vnm\u00f0 k, k\u2032 g\u03bc mm\u00f0 \u00de e k\u2032 \u00de \u03c3\u03bc m\u03c1,m\u03c1\u2032f \u2032 m\u00f0 k\u2032 : \u00de (12.156) Similarly, to calculate \u03a3 k, we assume 1 n\u03c1,m\u03c1\u2032\u00f0 \u00de <nm \u00dej From Eqs. (12.152) and (12.157), we obtain veff \u00f0 j k, k\u2032 pq> \u03c1\u03c1\u2032 = vnm\u00f0 k, k\u2032 \u00de \u03b4np\u03b4mq: k \u03a3n\u03c1,m\u03c1\u2032\u00f0 e \u00de = \u2212 1 \u03b2 \u2211 k\u2032, \u03bel vnm\u00f0 k, k\u2032 \u00de. k\u2032, \u03bel\u00de Gn\u03c1,m\u03c1\u2032\u00f0 e (12.157) (12.158) 12.4.6 Exchange Enhancement of \u03c7s We will first discuss how \u03c7\u03bc\u03bc form s gets exchange enhanced. One can rewrite Eq. (12.142) in the alternate s = \u03c7\u03bc\u03bc \u03c7\u03bc\u03bc 0,s + \u03c7\u03bc\u03bc 1.s, (12.159) 12.4 Many-Body Theory of Magnetic Susceptibility of Bloch Electrons in Solids 395 where \u03c7\u03bc\u03bc 0,s = \u2212 1 4 \u03bc2 B \u2211 n,k,\u03c1,\u03c1\u2032 nn\u03c3\u03bc g\u03bc n\u03c1,n\u03c1\u2032g\u03bc nn\u03c3\u03bc n\u03c1\u2032,n\u03c1 f \u2032 En\u00de \u00f0 is the effective Pauli spin susceptibility for noninteracting Bloch electrons, and \u03c7\u03bc\u03bc 1,s = \u2212 1 2 \u03bcB \u2211 n,k,\u03c1,\u03c1\u2032 g\u03bc nn\u03a3 1,\u03bc n\u03c1,n\u03c1\u2032\u03c3\u03bc n\u03c1\u2032,n\u03c1 f \u2032 \u00f0 En\u00de (12.160) (12.161) is the contribution due to exchange and correlation. If we consider the individual band enhancement 1,\u03bc n\u03c1,n\u03c1\u2032 in Eq. (12.156), make an average and neglect interband interactions in the expression for \u03a3 exchange enhancement ansatz, and assume vnm \u2243 vnn\u03b4nm, which is equivalent to the assumption that \u03a3 1,\u03bc is independent of k, we obtain 1,\u03bc n\u03c1,n\u03c1\u2032 = \u03a3 1 2 \u03b1n 1\u2212\u03b1n \u03bcBg\u03bc nn\u03c3\u03bc n\u03c1,n\u03c1\u2032, where \u03b1n = \u2212 \u2211 k\u2032,m vnm\u00f0 k, k\u2032 f \u2032 \u00f0 Em\u00f0 \u00de k\u2032 \u00de\u00de : From Eqs. (12.159) through (12.162), we obtain (12.162) (12.163) \u03c7\u03bc\u03bc s = \u2211 n \u03c7\u03bc 0s,n, 1\u2212\u03b1n\u00de where \u03c7\u03bc\u03bc 0s.n is the contribution to effective Pauli susceptibility for each band. Eq. (12.164) is known as the Stoner enhancement, which was obtained by making drastic assumptions while solving 1,\u03bc n\u03c1,n\u03c1\u2032", ": However, the neglect of coupling of interband terms, i.e., couthe matrix integral equations for \u03a3 1,\u03bc n\u03c1,n\u03c1\u2032, might be too drastic for systems such as Be, Cd, and so on. It can be easily pling between \u03a3 shown14 that even in a simple two-band model, the exchange enhancement of \u03c7s is quite different from the simple form obtained from Eq. (12.164). (12.164) \u00f0 12.4.7 Exchange and Correlation Effects on \u03c7o The exchange and correlation effects on \u03c7o are very complicated. We consider only the first term of Eq. (12.140), 0 \u2248 \u03c7qp \u03c7\u03bcv LP = \u2211 k \u00f0 1 + \u03b4\u03bcv\u00de e2\u2208\u03b1\u03b2\u03bc\u2208\u03b3\u03b4v 48 \u01272c2 \u2207\u03b1 k\u2207\u03b3 kEn \u2207\u03b2 k\u2207\u03b4 kEn f \u2032, En\u00de \u00f0 (12.165) which is the familiar Landau\u2013Peierls susceptibility for quasiparticles, because the energy in the Landau\u2013Peierls term is the quasiparticle energy. This is the well-known Sampson\u2013Seitz prescription that had been stated without proof. If we include the effects of electron\u2013electron interaction through an effective mass and ignore the band effects, we obtain the well-known results for \u03c7o,21 \u03c7qp LP\u00de \u00f0 \u03c7qp LP = \u03c7LP 1 + A1/3, (12.166) 396 CHAPTER 12 Diamagnetism and Paramagnetism where A1 is the Fermi-liquid parameter. The second through fifth terms in Eq. (12.140) are corrections to the Landau\u2013Peierls term, which are zero for free electrons, but are of the same order as \u03c7LP for band electrons even in the absence of electron\u2013electron interactions.12 Therefore, while one considers many-body effects on \u03c7o, it is wrong to consider only \u03c7qp LP as was done in earlier calculations. 12.4.8 Exchange and Correlation Effects on \u03c7so The effect of electron\u2013electron interactions is different on the various terms in \u03c7so in Eq. (12.144). A discussion of these effects is beyond the scope of this book except to note that even in the absence of exchange and correlation effects, the contributions of \u03c7so are of the same order as \u03c7s for some metals and semiconductors. 12.5 QUANTUM HALL EFFECT 12.5.1 Introduction von Klitzing et al.20 first observed the integer quantum Hall effect (QHE) in a two-dimensional electron gas formed by an inversion layer at an Si/SiO2 interface (discussed in Chapter 11). We note that a two-dimensional electron gas can be formed at the semiconductor surface if the electrons are fixed close to the surface by an external electric field in either Silicon MOSFETS or GaAs-AlxGa1\u2013xAs heterostructures (Chapter 11). In fact, a two-dimensional electron gas is essential for the observation of the quantum Hall effect. In addition to the quantum phenomena connected with the confinement of electrons within a two-dimensional layer, the Landau quantization of the electron motion in a strong magnetic field is necessary for the interpretation of the quantum Hall effect. We will discuss the integer quantum Hall effect in detail and briefly mention the fractional quantum Hall effect, discovered soon after, because a detailed discussion involving many-body theory is beyond the scope of this book. 12.5.2 Two-Dimensional Electron Gas The energy of mobile electrons in semiconductors can be written as E = \u01272 Positive charge (ionized impurities) Insulator AlxGa1\u2212xAs Semiconductor (p \u2212 GaAs) FIGURE 12.10 Two-dimensional electron gas formed close to the semiconductor surface of GaAs-AlxGa1\u2013xAs heterostructures by an external electric field along the z direction. x + k2 k2 y + k2 : z \u00de (12.167) 2m& \u00f0 When the energy for the motion in the z direction is fixed by using a triangular potential with an infinite barrier at the surface (z = 0) and with a constant electric field Fs for z \u2265 0 (z is positive downward), one obtains a quasi-two-dimensional electron gas (as shown in Figure 12.10). The electrons are confined close to the surface due the electrostatic field Fs normal to the interface originating from the positive charges, which causes a drop in the electron potential toward the surface. 12.5 Quantum Hall Effect 397 Source Gate Drain Drain Source Gate SiO2 n+ n-inversion", " layer (2DEG) (a) FIGURE 12.11 \u03c1 \u2212 Si 2DEG (b) GaAs (undoped) Si-doped 20 nm 140 nm Al0.3Ga0.7As Undoped GaAs (undoped) 5 nm 2DEG 1 \u00b5m GaAs (semi-insulating) (c) Typical geometries and cross-sections of devices used in the quantum Hall experiments. From left to right: (a) Long silicon device (Hall geometry) with potential probes (for Rx measurements) and Hall probes (for RH measurements): typical length: 0.5 mm. (b) Circular MOS device for \u03c3xy measurements. (c) Crosssection and top view of a GaAs\u2212Al0:3Ga0:7As heterostructure with Hall geometry. Reproduced from von Klitzing 19 with the permission of Elsevier. The energy of the carriers is grouped into electric sub-bands corresponding to the quantized levels for the motion in the z direction provided the potential well is small compared to the de Broglie wavelength of the electrons. At very low temperatures (T < 4\u00ba K), if the carrier densities of the two-dimensional electron gas are small such that only the lowest electric sub-band E0 is occupied with electrons (the electric quantum limit that occurs when the Fermi energy EF is small compared with the sub-band separation E1 \u2013 E0), the energy spectrum can be written as where k jj is a wave vector within the plane of the electron gas. E = E0 + \u01272k2 \u2225 jj 2m&, (12.168) The experimental arrangement for QHE measurements is shown in Figure 12.11. For measurement of current, heavily doped n + contacts are used as current contacts and potential probes at the semiconductor surface. The resistivity component \u03c1xx to \u03c3xx = \u03c3yy, \u03c3xy = \u2212\u03c3yx. Hence, \u03c1xx = \u03c3xx/ : This means that the condition \u03c3xx = 0 (fully occupied Landau levels) leads to \u03c1xx = 0: Thus, a correct value for the quantized Hall resistance is RH = h/e2n, which is expected only under the condition \u03c1xx = 0: is directly proportional xx + \u03c32 \u03c32 yy\u00de \u00f0 12.5.3 Quantum Transport of a Two-Dimensional Electron Gas in a Strong Magnetic Field When a strong magnetic field B is applied such that Bz is normal to the interface, the two-dimensional electrons move in cyclotron orbits parallel to the surface. The energy levels can be expressed as En = E0 + n + # 1 2 $ \u0127\u03c9c + gs\u03bcBB, (12.169) 398 CHAPTER 12 Diamagnetism and Paramagnetism, and s is the spin quantum number, s = \u00b1 1 where \u03c9c is the cyclotron frequency 2 : Laughlin (Ref. 8) derived an expression of a 2DEG in a strong magnetic field by considering the isotropic effective-mass Hamiltonian \u03c9c = \u2212eB/m& \u00f0 \u00de where the y coordinate is related to the vector potential (in the Landau gauge), ^H = 1 2m& p + e c 2 A \u2212eE0y, h i A = By^x: From Eqs. (12.170) and (12.171), the wave functions are given by y \u2212 y0\u00de where \u03d5n is the solution of the harmonic-oscillator equations \u03c8 k,n = eikx\u03d5n\u00f0, and The energy of the state is p2 y + e c \u00f0 B \u00de 2y2 1 2m& h i \u03d5n = n + \u00f0 1 2\u00de \u0127\u03c9c\u03d5n y0 = 1 \u03c9c \u0127k m& h \u2212 cE0 B : i En,k = n + \u00f0 1 2\u00de \u0127\u03c9c\u2212eE0y0 + 1 2 m& cE0B \u00f0 2: \u00de (12.170) (12.171) (12.172) (12.173) (12.174) (12.175) The y0 are changed by a vector potential increment \u0394A^x only through the location of their centers, y0! It is obvious from Eqs. (12.175) and (12.176) that the energy changes linearly with \u0394A: y0 \u2212 \u0394A/B: (12.176) From Eq. (12.174), one can obtain the degeneracy factor for each Landau level that is given by the number of center coordinates y0 (note that y \u2212 y0 is a good quantum number) in the sample. For a two-dimensional electron", " gas confined in a device of dimensions Lx, Ly, \u0394y0 = 1 \u03c9c \u0127\u0394k m&c = \u2212\u0127 eB \u0394k = \u2212\u0127 eB 2\u03c0 Lx = \u2212h eBLx : The degeneracy factor is given by N0 = Ly \u0394y0 = \u2212LxLyeB h, (12.177) (12.178) (note that e is negative) which is the same as the number of the flux quanta in the sample. The degeneracy factor per unit area (from Eq. 12.178) is N = N0 LxLy = \u2212eB h : (12.179) 12.5 Quantum Hall Effect 399 Thus, the degeneracy factor for each Landau level is independent of the effective mass and other semiconductor parameters. It can be shown that the Hall voltage UH of a two 2DEG with a surface carrier density ns (ns = nN, when n energy levels are fully occupied) is UH = \u2212 B nse I, where I is the current in the sample. From Eqs. (12.179) and (12.180), the Hall resistance RH = UH I = \u2212 B nse = \u2212 B nNe = h ne2, where n = 1, 2, 3, \u2026. Thus, whenever n = \u2212 nsh eB (12.180) (12.181) (12.182) is an integer (by adjusting the magnetic field B and the density of states ns), the Hall resistance is quantized. When the condition outlined in Eq. (12.182) is satisfied, there is no current flow in the direction of the electric field, and hence, the conductivity \u03c3xx = 0: The electrons move like free particles perpendicular to the electric field. The quantized plateaus in the Hall resistance are shown in Figure 12.12. h/2e2 RH h/3e2 B = 18.9 T T = 1.5\u00b0 K h/4e2 RX h/6e2 n =1 h/8e2 12 10 8 6 4 2 0 0 10 20 vg /v 30 FIGURE 12.12 Gate voltage dependence of the Hall resistance RH and resistivity Rx at B = 18.9 T for a long silicon MOS device at B = 18.9 T and T = 1.5\u00b0 K. Reproduced from von Klitzing 19 with the permission of Elsevier. 400 CHAPTER 12 Diamagnetism and Paramagnetism 12.5.4 Quantum Hall Effect from Gauge Invariance Laughlin (Refs. 7-9) considered a two-dimensional metallic loop (see Figure 12.13) pierced by a magnetic field B normal to its surface. A voltage UH is applied between the two edges of the ring. When \u03c3xx = 0, the energy is conserved and Faraday\u2019s law of induction can be written as I = c \u2202E \u2202\u03d5, (12.183) I UH B where E is the total energy of the system, and \u03d5 is the magnetic flux threading the loop. If \u03d5! \u03d50 = \u2212hc/e is a flux quan\u03d5 + \u0394\u03d5, where \u0394\u03d5 tum, the wave function enclosing the flux changes by a factor 2\u03c0, which implies k k + 2\u03c0 %! \u00f0 \u00de The change in energy \u0394E when the states are transported from one edge to the other is /L, where L is the circumference of the ring. FIGURE 12.13 Quantized Hall resistance for a model two-dimensional metallic loop. \u0394E = \u2212neUH, (12.184) where n corresponds to the filled Landau levels. From Eqs. (12.183) and (12.184), we obtain an expression for the dissipationless Hall current and the Hall voltage, I = c \u2202E \u2202\u03d5 = c \u0394E \u0394\u03d5 = \u2212 cneUH \u03d50 = ne2UH h : (12.185) The quantized Hall resistance is obtained from the expression RH = UH I = h ne2 : (12.186) 1 3 2 5 12.6 FRACTIONAL QUANTUM HALL EFFECT There have been numerous papers on the fractional quantum Hall effect. It is generally interpreted on the elementary excitations of quasiparticles with a charge e/3, e/5, e/7, and so on. A typical experimental result is shown in Figure 12.14. 2 3 1 4 3 \u03bd = 2 \u03c1xy = Vy J 0 50 100 B(kG) 150 200 FIGURE 12.14 Schematic diagram of fractional quantum Hall effect in GaAs/Ga1\u2212x AlxAs heterostructure. Problems 401 PROBLEMS 12", ".1. The Hamiltonian of an atom (or ion) in a uniform magnetic field B (in the z direction) can be written as If H = 1 2m \u2211 i pi +! ri\u00de eA \u00f0 c 2 \" + g0\u03bcBB\u2211 i ^si z: show that Eq. (1) can be rewritten in the alternate form A ri\u00de \u00f0 = \u2212 1 2 r \u00d7 B, ^H = 1 2m \u2211 i ^p2 i + \u03bcB\u00f0 ^L + g0^S. B + \u00de e2 8mc2 B2\u2211 i \u00f0 i + y2 x2 i \u00de : 12.2. We have derived <JLSJzj\u00f0 ^L + 2^S \u00dej JLSJ\u2032z > = g JLS <JLSJzj \u00de ^J j \u00f0 JLSJ\u2032z >: Show that Eq. (1) can be rewritten in the alternate form <JLSJzj ^L + 2^S J\u2032L\u2032S\u2032J\u2032z > = g j \u00f0 JLS <JLSJzj \u00de ^J j J\u2032L\u2032S\u2032J\u2032z >: 12.3. We have proved that \u2211 J\u2033L\u2033S\u2033J\u2033z = g <JLSJzj\u00f0 JLS \u2211 J\u2033L\u2033S\u2033J\u2033z \u00de \u00f0 ^L + 2^S J\u2033L\u2033S\u2033J\u2033z >. <J\u2033L\u2033S\u2033J\u2033z j \u00dej <JLSJzj J\u2033L\u2033S\u2033J\u2033z >. <J\u2033L\u2033S\u2033J\u2033z j ^J j ^J J\u2032L\u2032S\u2032J\u2032z > j J\u2032L\u2032S\u2032J\u2032z >: ^J j (1) (2) (3) (1) (2) (1) Because the sum over Eq. (1) is taken over a complete set, using Eqs. (12.31) and (12.32), show that Eq. (1) can be rewritten in the alternate form <JLSJzj\u00f0 ^L + 2^S. ^J \u00de JLSJ\u2032z > = g j \u00f0 JLS <JLSJzj \u00de 2 ^J j JLSJ\u2032z >: 12.4. Using the standard relation for the total angular momentum of an electron in an atom, show that ^J = ^L + ^S, ^L + 2^S ^L + ^S. \u00f0 \u00de \u00de \u00f0 2 = ^L + 2^S 2 + ^L. ^J + 2^S. ^J = 1 2 \u00bd 2 3^J \u2212 ^L 2 2 + ^S : $ (2) (1) (2) 402 CHAPTER 12 Diamagnetism and Paramagnetism 12.5. The Helmholtz free energy is obtained from e\u2212\u03b2F = e 1 2 $ \u03b2\u03b3B J + \u2212\u03b2\u03b3B J + \u2212 e # e\u03b2\u03b3B/2 \u2212 e\u2212\u03b2\u03b3B/2 # 1 2 $ : The magnetization of N ions in a volume V is defined as M = \u2212 N V \u2202F \u2202B : From Eqs. (1) and (2), show that where the Brillouin function BJ\u00f0 x x BJ\u00f0 \u00de = \u00f0 12.6. From the equations N V M = \u03b3JBJ\u00f0 is defined by \u03b2\u03b3JB \u00de, \u00de 2J + 1 2J \u00de coth\u2061 # 2J + 1 2J 1 2J x \u2212 $ coth\u2061 1 2J x : $ # and show that E, v g \u00f0 \u00de 12.7. In the limiting case, B! Ev = \u01272k2 z 2m 1 2 + v + # \u0127\u03c9c, v = 0, 1, 2, \u2026 $ g v, kz\u00de \u00f0 dkz = 2 2\u03c0 \u00f0 2 \u00de m\u03c9c \u0127 dkz, dE = 2 2\u03c0 2 \u0127\u03c9c 2 3/2 2m \u01272! \" 1 2 E\u2212 v + # # \u22121/2 dE: \u0127\u03c9c $ $ \u00f0 \u00de 0, the summation v\u2032 dE = \u2211 0 \u00de g \u00f0 E g \u00f0 E, v dE \u00de (1) (2) (3) (4) (1) (2) (3) (1) can be replaced by an integration because the sub-bands of different quantum numbers move very close to each other. By carrying out this integration (using Eq. 3 from Problem 12.6), and", " substituting B = 0, show that g E dE = \u00de \u00f0 1 2\u03c02 3/2 2m \u01272! \" E1/2dE, (2) which is the density of states of free electrons. Problems 403 12.8. Show from Eqs. (12.53) through (12.55) and (12.69) that the expression for magnetization per unit volume can be written as M = \u2212 n\u03bc\u2212 d dB ln e\u03b2 \u00bd \u00f0 \u03bc\u2212E \u00de + 1 dE + \u2212 $ 1 \u03b2 \u221e Z 0 g E\u2212\u00de \u00f0 ln \u00bd e\u03b2 \u03bc\u2212E \u00f0 \u00de + 1 dE\u2212 $ 12.9. The Fermi integral F : is given by x \u00f0 \u00de Show that and = F x \u00f0 \u00de \u221e Z 0 y1/2 1 + e\u00f0 y\u2212x \u00de dy. F x \u00f0 \u00de \u2248 \u03c0p 2 \ufb03\ufb03\ufb03 ex, for x < 0 F x \u00de \u00f0 \u2248 x3/2, 2 3 for x > 0: : 9 = ; (1) (1) (2) 12.10. The magnetization per unit volume is given by (Problem 12.8) M = \u2212 n\u03bc\u2212 d dB ln \u00bd e\u03b2 \u03bc\u2212E \u00f0 \u00de + 1 dE + \u2212 $ 1 \u03b2 \u221e Z 0 g E\u2212\u00de \u00f0 ln \u00bd e\u03b2 \u03bc\u2212E \u00f0 \u00de + 1 (1) dE\u2212 $ : 9 = By retaining the first term in the expansion similar to the one outlined in the text, show that at very low temperatures : ; M = 3n\u03bc2 BB 2EF 2 1\u2212 1 3 + \u03c0kBT \u03bcBB EF \u03bcBB! 1/1 \u00de vp cos \u03c0v \u00de \u00f0 \ufb03\ufb03\ufb03 where EF is the Fermi energy (the value of \u03bc at T = 0. \u00de 6 6 6 4 sin \u03c0 4! sinh \u2212 \u03c0vEF \u03bcBB \u03c02vkBT \u03bcBB \" 3, (2) 7 7 7 5 12.11. Substituting Eqs. (12.112) and (12.113) in Eq. (12.108), commuting the differential operator through the Peierls phase factor, and then multiplying the left side of the equation by e\u2212ih.r\u00d7r\u2032, show that \u03bel\u2212 \u00f0 1 p + \u0127h \u00d7 2m \u00bd 1 g\u03bcBB. \u03c3!\u00de 2. r \u2212 r$ \u2212 r \u2212 r\u2032 \u00f0 r, r\u2032, \u03bel, h \u00f0 \u00de Z \u0127 4m2c2 \u03c3! \u2212 dr\u2033eih.. \u2207! V \u00d7 p + \u0127h \u00d7 \u00bd r\u2032\u00d7r+r\u00d7r\u2033+r\u2033\u00d7r\u2032 \u00f0 r \u2212 r\u2032 \u2212V \u00f0 \u00de$ r, r\u2033, \u03bel \u2212 \u01272 8m2c2 \u22072V r\u2033, r\u2032, \u03bel, h \u00de \u00f0 \u00de e (1) 404 CHAPTER 12 Diamagnetism and Paramagnetism 12.12. Using the Bloch representation, \u03c8 nk\u03c1\u00f0 can be rewritten as r \u00de = eik. rUnk\u03c1\u00f0 r \u00de, show that Eq. (1) in Problem 12.11 \u03bel\u2212 1 2m \u00bd p + \u0127h \u00d7 r \u2212 r\u2032 \u00de$ \u00f0 2 \u2212 \u0127 4m2c2 \u03c3!. \u2207!V \u00d7 p + \u0127h \u00d7 \u00bd r \u2212 r\u2032 \u00f0 \u00de$ \u2211 n\u2033,\u03c1\u2033, k\u2032,k\u2033 Z drdr\u2032dr\u2033e\u2212ik.rU&nk\u03c1\u00f0 r \u00de 0 \u2212V r \u00f0 \u00de B B @ r\u2033 U&n\u2033k\u2033\u03c1\u2033\u00f0 G \u00f0 \u00de \u2212 \u01272 8m2c2 \u22072V\u2212 r\u2033, r\u2032, \u03bel, h 1 2 Un\u2032k\u2032\u03c1\u2032\u00f0 \u00de g\u03bcBB. \u03c3! eik\u2032.r\u2032 r\u2032 \u00de \u00d7 e\u2212k\u2033. \u00f0 r \u2212 r\u2033 r \u00deUn\u2033k\u2033\u03c1\u2033\u00f0 \u00de + \u2211 n\u2033,\u03c1\u2033, k\u2032,k\u2033 Z drdr\u2032dr\u2033dr\u2032\u2032\u2032e\u2212ik. rU&nk\u03c1\u00f0 e eih. r \u00de r\u2032 \u00d7 r + r \u00d7 r\u2033 + r\u2033 \u00d7 r\u2032 \u00f0 \u00de \u03a3 r, r\u2033 \u00de \u00f0 eik\u2033. \u00f0 r\u2033", "\u2212r\u2032\u2032\u2032 \u00de Un\u2033k\u2033\u03c1\u2033\u00f0 r\u2033 \u00de U&n\u2033k\u2033\u03c1\u2033\u00f0 r\u2032\u2032\u2032 \u00de \u00d7 G r\u2032\u2032\u2032, r\u2032, \u03bel, h \u00f0 \u00de Un\u2032k\u2032\u03c1\u2032\u00f0 r\u2032 \u00de e eik\u2032.r\u2032 = \u03b4nn\u2032\u03b4\u03c1\u03c1\u2032. 1 C C A (1) e By introducing a change of variables R1 = r\u2033 \u2212 r\u2032, and R2 = 1 2 \u00f0 R1 = r \u2212 r\u2033, R2 = 1 r\u2032 + r\u2032\u2032\u2032 \u00de using partial integration of the type, R3 = r\u2032\u2032\u2032 \u2212 r\u2032, and R4 = 1 r + r\u2033 \u00de 2 \u00f0 2 \u00f0 r\u2032 + r\u2033 \u00de in the first term, in the second term, and by r \u2212 r\u2032 \u2211 k\u2033 \u00f0 eik\u2033. \u00de r\u2212r\u2032 \u00f0 \u00deeik\u2033. r\u2032\u2212r\u2033 \u00f0 \u00deUn\u2033k\u2033\u03c1\u2033\u00f0 r eik\u2033. r\u2212r\u2032 \u00f0 \u00dei\u2207k\u2033eik\u2033. r\u2032\u2212r\u2033 \u00f0 = \u2211 k\u2033 r\u2033 U&n\u2033k\u2033\u03c1\u2033\u00f0 \u00de \u00de U&n\u2033k\u2033\u03c1\u2033\u00f0 r \u00deUn\u2033k\u2033\u03c1\u2033\u00f0 \u00de, r\u2033 \u00de show that Eq. (1) can be rewritten in the form \u03bel\u2212 ^H \u2211 n\u2033,\u03c1\u2033 \u00bd \u03ba!\u2032, \u03bel\u00de$nk\u03c1,n\u2033k\u03c1\u2033 \u00f0 k\u2032, \u03bel\u00dejk\u2032 = k = \u03b4nn\u2032\u03b4\u03c1\u03c1\u2032, Gn\u2033k\u03c1\u2033,n\u2032k\u2032\u03c1\u2032\u00f0 e = 1 p + \u0127 \u03ba!\u00de 2m \u00f0 \u03ba! = k + ih \u00d7 \u2207k: 2 + V + \u0127 4m2c2 \u03c3! r \u00f0 \u00de. \u2207!V \u00d7 p + \u0127 \u03ba!\u00de \u00f0 + \u01272 8m2c2 \u22072V + 1 2 g\u03bcBB. \u03c3! + \u03a3 e where ^H \u03ba!,\u03bel\u00de \u00f0 \u03ba!\u2032,\u03bel\u00de = Z \u03a3nk\u03c1,n\u2033k\u03c1\u2033\u00f0 e and drdr\u2032U&n\u2033k\u03c1\u2033\u00f0 r \u00de e\u2212i \u03ba!\u2032. \u00f0 r\u2212r\u2032 \u00de \u03a3 r, r\u2032, \u03bel\u00de Un\u2033k\u03c1\u2033\u00f0 r\u2032 \u00de, \u00f0 e Gn\u2033k\u03c1\u2033,n\u2032k\u03c1\u2032\u00f0 e k\u2032,\u03bel\u00de = Z drdr\u2032U&n\u2033k\u03c1\u2033\u00f0 r \u00de G r, r\u2032, \u03bel\u00de \u00f0 e\u2212ik\u2032. r\u2212r\u2032 \u00f0 \u00deUn\u2032k\u03c1\u2032\u00f0 r\u2032 \u00de : e (2) (3) (4) \u00f0, \u03ba!, \u03bel\u00de (5) 6 \u00de \u00f0 (7) Problems 405 Hence, show that because the Unk\u03c1\u2032s form a complete set of functions, Eq. (3) can be rewritten in the alternate form \u03bel \u2212 ^H \u00bd \u03ba!, \u03bel\u00de$ G \u00f0 k, \u03bel\u00de \u00f0 = I: 12.13. From Eqs. (12.118), (12.119), (12.127), and (12.128), show that e where ^H \u03ba!, \u03bel\u00de \u00f0 = ^H0\u00f0 k, \u03bel\u00de + ^H\u2032, k, \u03bel\u00de \u00f0 (8) (1) ^H0\u00f0 k, \u03bel\u00de = 1 2m \u00f0 p + \u0127k 2 + V \u00de r \u00f0 \u00de + \u03a30 k, \u03bel\u00de \u00f0 + \u01272 8mc2 \u22072V + \u0127 4m2c2 \u03c3!. \u2207!V \u00d7 p + \u0127k \u00de, \u00f0 (2) and ^H\u2032 k, \u03bel\u00de \u00f0 = \u2212ih\u03b1\u03b2\u220f \u03b1 \u2207 \u03b2 k + g\u03bcBB\u03bc\u03c3\u03bc + B\u03bc\u03a31,\u03bc 1 2 \u03b4\u03b1\u03b3 + \u22022 \u2212 1 2 h\u03b1\u03b2h\u03b3\u03b4 \u01272 m & 0 \u03a3 \u2202k\u03b1\u2202k\u03b3 \u2019 \u2207\u03b2 k\u2207\u03b4 \u2212 ih\u03b1\u03b2B\u03bc \u2202\u03a3 1,\u03bc \u2202k\u03b1 \u2207\u03b2 k \u00f0", " k, \u03bel\u00de k + B\u03bcBv\u03a32,\u03bcv, k, \u03bel\u00de \u00f0 (3) where the terms up to the second order in the magnetic field are retained and \u220f velocity operator,!=\u0127 is the (! = \u0127 \u220f m \u00f0 p! + \u0127k \u00de + \u01272 4m2c2 \u03c3! \u00d7 \u2207!V + \u2207k\u03a30 k, \u03bel\u00de : \u00f0 ( 12.14. The temperature Green\u2019s function operator is defined as 1 \u03bel \u2212 H0\u00f0 is defined in Eq. (12.131). Show that k, \u03bel\u00de G0\u00f0 = k, \u03bel\u00de, where H0\u00f0 k, \u03bel\u00de and where \u2207\u03b1 k\u2207\u03b3 kG0\u00f0 \u2207\u03b1 kG0\u00f0 k, \u03bel\u00de = G0\u00f0 k, \u03bel\u00de \u220f \u03b1G0\u00f0 k, \u03bel\u00de k, \u03bel\u00de = G0 \u01272 m & \u03b4\u03b1\u03b3 + \u03a7\u03b1\u03b3 G0 + G0\u220f \u2019 \u03b1G0\u220f \u03b3G0 + G0\u220f \u03b3G0\u220f \u03b1G0, \u03a7\u03b1\u03b3 = \u2207\u03b1 k \u01272 4m2c2 \u00f0 & \u03c3! \u00d7 \u2207!V \u00de \u03b3 + \u2207\u03b3 k \u2211 0 k, \u03bel\u00de \u00f0 : \u2019 (4) (1) (2) (3) (4) 406 CHAPTER 12 Diamagnetism and Paramagnetism 12.15. We have derived k \u2211 n,m,\u03c1,\u03c1\u2032 \u03a3n\u03c1,m\u03c1\u2032\u00f0 1 = \u2212 e \u03b2 r\u2032 r \u03c8 nk\u03c1\u00f0 \u00de veff \u00f0 \u03c8 &mk\u03c1\u2032\u00f0 r, r\u2032 \u00de \u00de \u00de Gp\u03c1,q\u03c1\u2032\u00f0 e If the effective electron\u2013electron interaction is spin independent, then \u03c1 = \u03c1, \u03c1\u2032 = \u03c1\u2032: Show that \u2211 \u03bel,p,q,k\u2032,\u03c1,\u03c1\u2032 \u03c8 &qk\u2032\u03c1\u2032\u00f0 k\u2032, \u03bel\u00de \u03c8 pk\u2032\u03c1\u00f0 r\u2032 \u00de r \u00de. (1) \u03a3n\u03c1,m\u03c1\u2032\u00f0 k \u00de = \u2212 1 \u03b2 \u2211 k\u2032,\u03bel,p,q <nm j veff \u00f0 k, k\u2032 pq > \u03c1\u03c1\u2032 \u00dej k\u2032, \u03bel\u00de, Gp\u03c1,q\u03c1\u2032\u00f0 e (2) where <nm j veff \u00f0 k, k\u2032 pq > \u03c1\u03c1\u2032 = \u00dej Z \u03c8 &nk\u03c1\u00f0 r \u03c8 mk\u03c1\u2032\u00f0 \u00de r\u2032 \u00de veff \u00f0 r, r\u2032 \u00de \u03c8 pk\u2032\u03c1\u00f0 r \u03c8 &qk\u2032\u03c1\u2032\u00f0 \u00de r\u2032 \u00de drdr\u2032: (3) 12.16. From Eqs. (12.152) and (12.154), we have obtained \u03a3n\u03c1,n\u03c1\u2032\u2032,\u03bel,\u03c1 vnp\u00f0 k, k\u2032 \u00de. k\u2032, \u03bel\u00de (1) Gp\u03c1,p\u03c1\u2032\u00f0 e Substituting only the value of the first-order terms of B occurring in the right side of Eq. (1), and neglecting the terms proportional to f, show that G from Eq. (12.134) on \u03a3 1,\u03bc n\u03c1,n\u03c1\u2032\u00f0 k \u00de \u2243 \u2212 \u2211 mk\u2032 vnm\u00f0 k, k\u2032 \u00de \u03a3 1,\u03bc m\u03c1,m\u03c1\u2032\u00f0 k\u2032 \u00de f \u2032 m\u00f0 k\u2032 \u00de \u2212 1 2 \u03bc0 \u2211 mk\u2032 e \u03c3\u03bc m\u03c1,m\u03c1\u2032f \u2032 m\u00f0 k\u2032 vnm\u00f0 \u00de k : \u00de 12.17. The isotropic effective-mass Hamiltonian (Eq. 12.170) is given by where the y coordinate is related to the vector potential (in the Landau gauge), ^H = 1 2m& p + e c 2 A \u2212eE0y, h i A = By^x: From Eqs. (1) and (2), show that the wave functions are given by where \u03d5n is the solution of the harmonic-oscillator equations \u03c8 k,n = eikx\u03d5n\u00f0 y \u2212 y0\u00de, and p2 y + e c \u00f0 B 2y2 \u00de 1 2m& h", " i \u03d5n = n + \u00f0 1 2\u00de \u0127\u03c9c\u03d5n y0 = \u0127k m& \u2212 cE0 B 1 \u03c9c h : i (2) (1) (2) (3) (4) (5) References 407 References 1. Aschroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 2. Harrison WA. Solid state theory. New York: McGraw-Hill; 1969. 3. Kittel C. Introduction to solid state physics. New York: John Wiley & Sons; 1976. 4. Landau LD. Diamagnetism of metals. Z Physik 1930;64:629. 5. Landau LD. On the de Haas-van Alphen effect. Proc Roy Soc (London) 1939;A170:363. 6. Landau LD, Lifshitz LM. Statistical physics part 1. Oxford: Pergamon Press; 1980. 7. Laughlin RB. Quantized Hall conductivity in two dimensions. Phys Rev B 1981;23:5632. 8. Laughlin RB. Impurities and edges in the quantum Hall effect Surf. Sci 1982;22:113. 9. Laughlin RB. Anomalous quantum Hall effect: An incompressible quantum fluid with fractionally charged excitations. Phys Rev Lett 1983;50:1395. 10. Madelung O. Introduction to solid state theory. New York: Springer-Verlag; 1978. 11. Marder MP. Condensed matter physics. New York: John Wiley & Sons; 2000. 12. Misra PK, Roth LM. Theory of diamagnetic susceptibility of metals. Phys Rev 1969;177:1089. 13. Misra PK, Kleinman L. Theory of magnetic susceptibility of Bloch electrons. Phys Rev B 1972;5:4581. 14. Misra SK, Misra PK, Mahanti SD. Many-body theory of magnetic susceptibility of electrons in solids. Phys Rev B 1982;26:1903. 15. Myers HP. Introduction to solid state physics. London: Taylor & Francis; 1990. 16. Prang RE, Girvin SM. The quantum hall effect. New York: Springer-Verlag; 1987. 17. ter Haar D, editor. Collected papers of L.D. Landau. New York: Gordon and Breach; 1965. 18. van Vleck JH. Theory of electric and magnetic susceptibilities. Oxford: Oxford University Press; 1932. 19. von Klitzing K. The quantized Hall effect. Rev Mod Phys 1986;58:519. 20. von Klitzing K, Dorda G, Pepper M. New method for high-accuracy determination of the fine structure constant based on quantized Hall resistance. Phys Rev Lett 1980;45:494. 21. White RM. Quantum theory of magnetism. New York: McGraw-Hill; 1970. 22. Ziger HJ, Pratt GW. Magnetic interactions in solids. Oxford: Oxford University Press; 1970. 23. Ziman JM. Principles of the theory of solids. Cambridge: Cambridge University Press; 1972. Magnetic Ordering CHAPTER 13 CHAPTER OUTLINE 13.1 Introduction............................................................................. 410 13.2 Magnetic Dipole Moments................................................................ 411 13.3 Models for Ferromagnetism and Antiferromagnetism.......................................... 412 13.3.1 Introduction....................................................................... 412 13.3.2 Heitler\u2013London Approximation................................................", "...... 412 13.3.3 Spin Hamiltonian.................................................................. 414 13.3.4 Heisenberg Model................................................................. 416 13.3.5 Direct, Indirect, and Superexchange................................................ 416 13.3.6 Spin Waves in Ferromagnets: Magnons.............................................. 417 13.3.7 Schwinger Representation.......................................................... 417 13.3.8 Application to the Heisenberg Hamiltonian.......................................... 418 13.3.9 Spin Waves in Antiferromagnets.................................................... 421 13.4 Ferromagnetism in Solids................................................................. 422 13.4.1 Ferromagnetism Near the Curie Temperature......................................... 422 13.4.2 Comparison of Spin-Wave Theory with the Weiss Field Model.......................... 424 13.4.3 Ferromagnetic Domains............................................................ 425 13.4.4 Hysteresis......................................................................... 426 13.4.5 Ising Model....................................................................... 427 13.5 Ferromagnetism in Transition Metals....................................................... 427 13.5.1 Introduction.................................................................", "...... 427 13.5.2 Stoner Model...................................................................... 428 13.5.3 Ferromagnetism in Fe, Co, and Ni from Stoner\u2019s Model and Kohn\u2013Sham Equations..... 430 13.5.4 Free Electron Gas Model........................................................... 431 13.5.5 Hubbard Model.................................................................... 433 13.6 Magnetization of Interacting Bloch Electrons................................................ 434 13.6.1 Introduction....................................................................... 434 13.6.2 Theory of Magnetization............................................................ 434 13.6.3 The Quasiparticle Contribution to Magnetization...................................... 435 13.6.4 Contribution of Correlations to Magnetization........................................ 436 13.6.5 Single-Particle Spectrum and the Criteria for Ferromagnetic Ground State.............. 437 13.7 The Kondo Effect........................................................................ 439 13.8 Anderson Model......................................................................... 439 Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00013-X \u00a9 2012 by Elsevier Inc. All rights reserved. 409 410 CHAPTER 13 Magnetic Ordering 13.9 The Magnetic Phase Transition............................................................ 440 13.9.1 Introduction....................................................................... 440 13.9.2 The Order Parameter.................................", ".............................. 441 13.9.3 Landau Theory of Second-Order Phase Transitions................................... 441 Problems.................................................................................... 443 References.................................................................................. 448 13.1 INTRODUCTION The phenomena of magnetic ordering such as ferromagnetism, antiferromagnetism, and ferrimagnetism are very complex. Ferromagnets have been known to exist for thousands of years in the shape of loadstones. However, ferromagnetism in transition metals, which is one of the most important as well as complex phenomena in physics, remains one of the major unsolved problems in solid state physics and is not well understood compared to most other physical properties. We will discuss some of the important models used in the theory of magnetic ordering used in solids before discussing some specific cases, including that of ferromagnetism in transition metals. There are three different types of \u201cmagnetic ordering,\u201d known as ferromagnetism, antiferromagnetism, and ferrimagnetism. The ferromagnetic solid has a nonvanishing magnetic moment even in the absence of an external magnetic field. This is known as \u201cspontaneous magnetization,\u201d which is a result of parallel orientation of the individual magnetic moments that must be due to interactions between these moments. In a ferromagnet, because all the individual moments are aligned in the same direction, there is a net total moment even in the absence of a magnetic field. As the temperature is increased, these orientations become gradually disordered, and at a critical temperature known as the Curie temperature, the spontaneous magnetization vanishes. This situation is far more complicated than the simple model used at T = 0. In the preceding discussion, we tacitly assumed that the electrons of an ion in a lattice are tightly bound so that each ion has a net magnetic moment. In some other types of solids, the ions of the nearest neighbors have antiparallel spin. The ground state consists of two sublattices of identical ions having opposite spin directions. This is known as an antiferromagnet, in which the two sublattices have mutually compensating magnetic moments. Thus, the net magnetic moment of an antiferromagnet is zero. In ferrimagnets, there are usually two types of basis atoms and therefore two sublattices because the ions in the individual sublattices are different. The individual ions in each sublattice will possess a magnetic moment, and each sublattice will have a net magnetic moment in the ground state. The total moment in the ground state will be the vector sum of the moments of the sublattices. For sublattices with opposite magnetic moments, the net magnetic moment will be the difference between the two moments. Such types of solids are known as ferrimagnets. These differences are illustrated in Figure 13.1. Ferromagnetic, antiferromagnetic, and ferrimagnetic states. Ferromagnet Antiferromagnet FIGURE 13.1 Ferrimagnet 13.2 Magnetic Dipole Moments 411 13.2 MAGNETIC DIPOLE MOMENTS The simplest problem involves magnetic dipole moments and how nearby magnetic dipoles interact with each other even though such interaction does not lead to ferromagnetism. The magnetic dipole moment m of a current distribution j can be defined as m = dr Z 1 2c r \u00d7 j r : \u00de \u00f0 (13.1) We note that m will be independent of the origin provided dr j r \u00de \u00f0 = 0, which implies that the Z current distribution is over a closed loop and vanishes except at the origin. The Lorentz force on a current distribution is F = 1 c Z dr Because j r \u00de \u00f0 vanishes except at the origin, we can expand B in a Taylor series and write", " r \u00de \u00f0 F = 1 c Z r dr. \u2207! B 0 \u00de \u00f0 \u00de + ::: : $ (13.2) (13.3) Using vector identities, we can easily show from Eqs. (13.1) through (13.3) that (Problem 13.1) F = \u00f0 m \u00d7 \u2207! \u00de \u00d7 B = \u2207! m. B, \u00de \u00f0 where B B 0 \u00de \u00f0 % : Thus, the potential energy U of a dipole in an external magnetic field is U = \u2212m. B: (13.4) (13.5) When two dipoles are close to each other, they interact with each other\u2019s magnetic fields. The induction B produced by a magnetic dipole of moment m1 at a distance r, where a second dipole of moment m2 is located, is given by B = \u2207! m1. \u2207! h 1 r i = 3^r m1. ^r r3 \u00f0 \u00de \u2212 m1 : (13.6) From Eqs. (13.5) and (13.6), we obtain the expression for the direct dipolar interaction energy of two magnetic dipoles separated by r, U = 1 r3 \u00bd m1. m2 \u2212 3 m1. ^r m2. ^r : \u00de$ \u00de\u00f0 \u00f0 (13.7) One can calculate the energy scale for the dipole interaction from Eq. (13.7), which is of the order of 10\u22124 eV (equivalent to 1\u00b0 K) for solids in which the distance between magnetic moments is of the order of 2 \u00c5, while the electrostatic energy difference between two atomic states is on the order of 0.1\u20131.0 eV. Thus, dipolar interactions, which are important in explaining the phenomenon of ferromagnetic domains, are not the source of magnetic interaction responsible for ferromagnetism. 412 CHAPTER 13 Magnetic Ordering 13.3 MODELS FOR FERROMAGNETISM AND ANTIFERROMAGNETISM 13.3.1 Introduction Various models describe ferromagnetism and antiferromagnetism. The direct exchange model assumes that the electrons of a lattice ion, which contribute to the magnetic moment, are tightly bound so that the ions can be assumed to be isolated. However, the nearest neighbors are sufficiently close enough for a significant exchange interaction between them. The spins of nearest neighbors in the Bravais lattice of a ferromagnet (where electrons are tightly bound in the lattice ion) are aligned parallel in the ground state by the exchange interaction. In the absence of such interactions, the moments would be thermally disordered due to random orientations, and there would be no magnetic moment. In the superexchange model, an exchange between magnetic ions occurs over large distances in an insulator, in which a paramagnetic ion between them facilitates the interaction. An example is MnO, in which two metallic ions (Mn) with unfilled d-shells are linked by an oxygen atom that has two p-electrons with spins in opposite directions. Each d-electron will interact with one of the two p-electrons, and because the two p-electrons are linked by the Pauli principle, there is an effective interaction between the two d-electrons that is known as the superexchange. In the indirect exchange (RKKY) model (Refs. 10, 23, 31), the localized spins of a lattice ion interact with the conduction electrons of a metal. Essentially, the electrons mediate between the interaction of the lattice ions. This ion\u2013ion interaction via conduction electrons plays a major role in rare-earth metals, which have a variety of ordered magnetism. In the transition metals, which form the most significant group of ferromagnetic metals, the spins of itinerant electrons give rise to ferromagnetism. The d- and s-bands are only partly filled, and hence, there is a superposition of 3d and 4s bands. We will discuss the occurrence of ferromagnetism in transition metals as a special category. 13.3.2 Heitler\u2013London Approximation The Heitler\u2013London (Ref. 6) approximation is designed to describe the interaction between two spins arising from Coulomb forces between the two electrons of two adjacent atoms, as shown in Figure 13.2. It was originally designed to explain the bonding of a hydrogen molecule and is also an example of the valence bond method. We further note that the Heitler\u2013 London theory does not yield good results for the binding energy of the hydrogen molecule. The reason is that when two protons are close, the system looks like the helium atom with Z = 2, whereas in the Heitler\u2013London theory, the electrons are in", " s states for Z = 1. However, it features the essential ingredients of the exchange \u2192 R1 \u2192 R2 FIGURE 13.2 The overlapping wave functions of electrons (of opposite spin) of two adjacent hydrogen atoms. The protons are located at R!1 and R!2. The arrows are the directions of the spins of the electrons. 13.3 Models for Ferromagnetism and Antiferromagnetism 413 interaction used in the theory of magnetism in insulators, and hence, we will discuss the theory in some detail. When the two hydrogen atoms (i = 1 and i = 2) are infinitely apart, the Schrodinger equation for each atom can be written as \u01272\u22072 i 2m e2 ri \u2212 Rij \u2212 \u2212 \u03d5i\u00f0 \" which is the spatial part of the wave function. The wave function of the two atoms is just the product of the individual atomic wave functions. When the two atoms come much closer, there will be considerable overlap, and a molecule is formed. The individual atomic wave functions will not be orthogonal. In fact, = \u03b50\u03d5i\u00f0 (13.8) ri\u00de ri\u00de! j, \u03d52\u00f0 \u00de where I is the overlap integral, which can be positive or negative. The Hamiltonian can be written as dr = I, \u03d5&1\u00f0 (13.9) Z r r \u00de \u01272\u22072 2 2m \u2212\u01272\u22072 1 2m j + \u2212 \u2212 \u2212 \u2212 \u2212 ^H = e2 r2 \u2212 R2j e2 r1 \u2212 R1j j e2 r1 \u2212 R2j j e2 r2 \u2212 R1j Here, the electrostatic interactions between the two electrons and the two protons, with all permutations, have been included in the Hamiltonian of the molecule. It may be noted that the wave function of the molecule is no longer the product of the individual atomic functions. The spin component of the wave function has to be included such that the total wave function will be antisymmetric when the particle numbers are exchanged. The spin operators commute with the spatial part of the and ^Sz Hamiltonian. The spin eigenfunctions are the eigenfunctions of the commuting operators ^S and form either one singlet or three triplet states. The spin singlet state of the molecule is given by e2 R1 \u2212 R2j e2 r1 \u2212 r2j (13.10) + : j j j 2 \u03c7s = 1 2p \u00bd \u03c7\u2191\u00f0 1 \u03c7\u2193\u00f0 \u00de 2 \u00de \u2212 \u03c7\u2193\u00f0 1 \u03c7\u2191\u00f0 \u00de 2, \u00de$ which is antisymmetric, and the three-spin triplet states are \ufb03\ufb03\ufb03 (13.11) (13.12) 2 1 \u03c7t = \u03c7\u2191\u00f0 1 = 2p \u00bd = \u03c7\u2193\u00f0 1 \ufb03\ufb03\ufb03 \u03c7\u2191\u00f0 \u00de \u03c7\u2191\u00f0 1 \u03c7\u2193\u00f0 \u00de \u00de 2, \u00de 2 \u03c7\u2193\u00f0 ; \u00de + \u03c7\u2193\u00f0 2 \u03c7\u2191\u00f0 \u00de 1, \u00de$ \u00de which are all symmetric. Here, the symbols in the parentheses denote the spinors for the electrons at r1 and r2. Thus, the normalized spatial part of the wave functions of the singlet and triplet spin states are \u00f0 where the s, t symbols as well as the + \u03c8 s,t\u00f0 r1, r2\u00de = 2 \u00bd \u2212 \u00de$ r1\u00de \u03d51\u00f0 \u00b1 \u03d52\u00f0 signs are for the singlet (triplet) spin states., r2\u00de$ \u03d52\u00f0 \u03d51\u00f0 r1\u00de r2\u00de \u00bd \u00de From Eqs. (13.8) through (13.10) and Eq. (13.13), we obtain (Problem 13.3) \u00f0 1 \u00b1 I2 \u2212 1 2 (13.13) \u03b5s,t = < ^H >s,t = 2\u03b50 + e2 R12 + e2 Z dr1dr2 \u03c8 &s,t\u00f0 r1, r2\u00de 1 r1 \u2212 r2j! j \u2212 j 1 r1 \u2212 R2 \u2212 1 r2 \u2212 R1j \" j \u03c8 s,t\u00f0 r1, r2\u00de, (13.14) 414 CHAPTER 13 Magnetic Ordering where 1 R12 % 1 R1 \u2212 R2j j : We define the Coulomb and exchange integrals as Vc\u00f0 R12\u00de = e2 Z dr1dr2j \u03d51\u00f0 2 r1\u00dej \u03d52\u00f0 j r2\ufffd", "\ufffdj 2 1 r1 \u2212 r2j j! \u2212 1 r1 \u2212 R1j j \u2212 1 r2 \u2212 R2j j \" and (13.15) (13.16) = e2 Vex\u00f0 R12\u00de 1 \u03d5&2\u00f0 r2 \u2212 R1j j From Eqs. (13.14) through (13.17), it can be shown that (Problem 13.4) 1 r1 \u2212 R2j j 1 r1 \u2212 r2j dr1dr2\u03d5&1\u00f0 r2\u00de! r1\u00de \u2212 \u2212 Z j \u03b5s = 2\u03b50 + e2 R Vc\u00f0 + R12\u00de \u00f0 + Vex\u00f0 1 + I2 \u00de R12\u00de and \u03b5t = 2\u03b50 + Vc\u00f0 + e2 R R12\u00de \u2212 Vex\u00f0 1 \u2212 I2 \u00f0 \u00de R12\u00de : From Eqs. (13.18) and (13.19), we obtain \u03d52\u00f0 r1\u00de \u03d51\u00f0 r2\u00de : \" (13.17) (13.18) (13.19) I2Ve \u2212 Vs\u00de 1 \u2212 I4 Heitler and London (Ref. 6) found that J is negative, which implies that \u03b5t > \u03b5s: Because the singlet state is of lower energy, the spins of the two atoms are in opposite directions, which is an example of two-atom antiferromagnetism. \u03b5t \u2212 \u03b5s = (13.20) = \u2212J: 2 \u00f0 13.3.3 Spin Hamiltonian Dirac and Heisenberg argued that the original Hamiltonian in Eq. (13.10) acts only on the spatial degrees of freedom and yields two eigenvalues, \u03b5s and \u03b5t (one singlet and three degenerate triplet states, depending on whether the spins of the two electrons are parallel or antiparallel). They showed that the same results can be obtained by considering a Hamiltonian that involves only the spin degrees of freedom and that is more useful in the study of magnetism. The actual wave function of the Heitler\u2013London problem (Ref. 6) is the product of the spin (Eqs. 13.11 or 13.12) and spatial part of the wave function (Eq. 13.13), i.e., and \u03a8s = \u03c8s \u03c7s \u03a8t = \u03c8t \u03c7t: (13.21) (13.22) It is much more convenient to represent the Hamiltonian in a spin representation as long as the states derived in Eqs. (13.18) and (13.19) by using the energy in the singlet and triplet \u03b5s\u00de \u00f0 \u03b5t\u00de \u00f0 13.3 Models for Ferromagnetism and Antiferromagnetism 415 spatial Hamiltonian of the two-electron system are the same in the new representation. This is possible because the coupling in the spin Hamiltonian depends only on the relative orientation of the two spins but not on their directions on R1 \u2212 R2: If there are dipolar interactions or spin-orbit coupling, which breaks the rotational symmetry of the Hamiltonian in the Heitler\u2013London model, one must add additional terms to the spin Hamiltonian. Thus, the spin Hamiltonian can be written as where \u03bb1 and \u03bb2 are such that and spin ^H = \u03bb1 + \u03bb2^S1. ^S2, spin ^H \u03a8s = \u03b5s\u03a8s spin ^H \u03a8t = \u03b5t\u03a8t: We also note that for a two-electron system, where ^S1. ^S2 = 1 2 \u00bd + ^S 1 ^S\u2212 2 + ^S + 2 + ^S z 1 ^S z 2, ^S\u2212 1 $ \u00b1 ^S = ^Sx \u00b1 i^Sy: (13.23) (13.24) (13.25) (13.26) (13.27) It can be easily shown from Eqs. (13.11), (13.12), and (13.24) through (13.26) that (Problem 13.5) and ^S1. ^S2\u03c7s = \u2212 3 4 \u03c7s ^S1. ^S2\u03c7t = 1 4 \u03c7t: From Eqs. (13.24) through (13.29), we obtain spin ^H = 1 4 \u00f0 \u03b5s + 3\u03b5t\u00de \u2212 \u03b5s \u2212 \u03b5t\u00de \u00f0 ^S1. ^S2, such that and spin ^H \u03a8s = \u03b5s\u03a8s spin ^H \u03a8t = \ufffd", "\ufffdt\u03a8t for each of the three triplet states. Comparing Eqs. (13.23) and (13.30), we obtain and \u03bb1 = 1 4 \u00f0 \u03b5s + 3\u03b5t\u00de \u03bb2 = \u03b5s \u2212 \u03b5t\u00de : \u00f0 (13.28) (13.29) (13.30) (13.31) (13.32) (13.33) (13.34) 416 CHAPTER 13 Magnetic Ordering From Eqs. (13.18) through (13.20), (13.23), (13.33), and (13.34), we obtain spin ^H = 2\u03b50 + e2 R + 1 4! Vc\u00f0 R12\u00de + Vex\u00f0 1 + I2 R12\u00de + 3 Vc\u00f0 R12 \u2212 Vex\u00f0 1 \u2212 I2 R12\u00de \u2212 J^S1. ^S2: \" By shifting the zero of energy, we can rewrite Eq. (13.35) in the convenient form spin ^H = \u2212J^S1. ^S2: (13.35) (13.36) Because J = \u03b5s \u2212 \u03b5t (from Eq. 13.20), a positive value of J means that \u03b5t < \u03b5s. This implies that the two spins are aligned in the same direction (ferromagnetism). Similarly, a negative value of J implies that the spins are aligned in the opposite direction. This alternate change in the direction of the spins is known as antiferromagnetism. 13.3.4 Heisenberg Model Heisenberg (Ref. 5) extended Eq. (13.36) for the general case of a large collection of magnetic ions placed in a lattice. He postulated that the Hamiltonian can be written as spin ^H = \u2212\u2211\u2032 ij Jij^Si. ^Sj, (13.37) where the exchange integral Jij is a function of the positions of the lattice sites Ri and Rj (note that. We further note that the exchange integral Jij, which is a function of Ri \u2212 Rj, is large only Ri \u2260 Rj\u00de for one- or two-lattice spacing. It is not possible to \u201cderive\u201d the Heisenberg Hamiltonian, but there have been many attempts to \u201cderive\u201d it or at least make reasonable attempts to justify it. Nevertheless, it remains the starting point for the theory of ferromagnetism and antiferromagnetism. 13.3.5 Direct, Indirect, and Superexchange The Heisenberg model is the result of the direct overlap of wave functions of two magnetic ions. This is due to the direct exchange interaction between localized spins of nearest neighbors, and the basic assumption is that the electrons (of a lattice ion) are tightly bound such that the ion can be considered as isolated, but the nearest neighbors are close enough for exchange interaction to occur. However, there are other mechanisms for exchange. In indirect exchange, the localized spins of the lattice ions interact with the conduction electrons of a metal. Thus, the information on the spin over a given ion is passed on by an electron to another ion. Hence, the interaction between the two ions is mediated by conduction electrons. This indirect ion\u2013ion interaction is known as the RKKY interaction (Refs. 10, 20, 23) and plays a major role in the rare-earth metals (Tm and Gd). In the superexchange mechanism in an insulator, the exchange between magnetic ions often occurs over large distances. A paramagnetic ion (an ion with closed electronic shells) between the two magnetic ions facilitates the interaction. For example, in MnO, two metallic ions with unfilled d-shells are linked by an oxygen atom. Each d-electron interacts with one of the two p-electrons of the spin-saturated outermost electron pair of the oxygen atom. There is an effective interaction between the two d-electrons, known as superexchange, because the two spins of the two p-electrons 13.3 Models for Ferromagnetism and Antiferromagnetism 417 are linked by the Pauli principle. These three types of interactions are shown schematically in Figure 13.3. 13.3.6 Spin Waves in Ferromagnets: Magnons For ferromagnets, the ground state of the Heisenberg model is such that all the Jij are positive and all the spins point in the same direction. It can be easily shown from Eq. (13.35) that if all the spins point in the same direction (for instance, z), the ground-state energy is <\u2191\u2191\u2191\u2191.... ^H \u2191\u2191\u2191\u2191..", ".. :::> = \u2212\u2211\u2032 <ii\u2032> j j : Jii\u2032 4 (13.38) (a) (b) The ferromagnetic state of the ground state is degenerate. Even though the spins have to point in the same direction, there is no preference for a particular direction like z direction. The lowenergy excitons can be constructed by slowly twisting the local spin orientation while propagating through the crystal. These propagations are known as spin waves. (a) Direct exchange, (b) superexchange, and (c) indirect exchange. FIGURE 13.3 (c) 13.3.7 Schwinger Representation We note that in an excited state, the spin of one or more ions at a lattice site is reversed from \u2191 to \u2193. Each time such a spin reversal occurs, the net spin change is 1. Thus, the appropriate operators to represent such changes are Bose operators. Schwinger (Ref. 24) proposed a formal theory of spin waves by using a representation in which the spins are represented by Bose operators even though the Heisenberg Hamiltonian is a product of Fermion operators. In this representation, the Bose operators are defined as \u03b1 ^S = 1 2 \u2211 ij ^a\u2020 i \u03c3\u03b1 ij^aj, (13.39) where \u03c3\u03b1 are the Pauli spin matrices (\u03b1 = x, y, z (Problem 13.6) \u00de. It is easy to show from Eq. (13.39) that x ^S = y ^S = z ^^a2 + ^a\u2020 ^a\u2020 2^a1 \u2212 ^a\u2020 ^a\u2020, and, 2^a1\u00de 1^a2\u00de 2^a2\u00de : 1^a1 \u2212 ^a\u2020 ^a\u2020 (13.40) 418 CHAPTER 13 Magnetic Ordering It can be easily shown from Eq. (13.40) that If one introduces the operators y x, ^S ^S \u00bd $ = i^S z : \u00b1 ^S = ^S x \u00b1 i^S y, it can be shown from Eqs. (13.40) and (13.42) that and + ^S = ^a\u2020 1^a2 ^S\u2212 = ^a\u2020 2^a1: Because the total spin is 2S, one can write 1^a1 + ^a\u2020 ^a\u2020 2^a2 = 2S: (13.41) (13.42) (13.43) (13.44) (13.45) Through use of the Holstein\u2013Primakoff transformation (Ref. 7) (which is an unusual transformation to be used for operators, but seems to be valid), Eq. (13.45) becomes From Eqs. (13.40) and (13.46), we obtain ^a2 = 2S \u2212 ^a\u2020 1^a1 : \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q (13.46) ^S\u2212 = z ^S = One can easily show that (Problem 13.7) + ^S = ^a\u2020 1, 2S \u2212 ^a\u2020 1^a1 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q 2S \u2212 ^a\u2020 ^a1 1^a1 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q ^a\u2020 1^a1 \u2212 S \u00f0 \u00de :, and (13.47) +, ^S\u2212 ^S \u00bd $ = 2^S z : (13.48) 13.3.8 Application to the Heisenberg Hamiltonian If we assume that the direct exchange interaction between the nearest neighbors is dominant, for simple lattices, Eq. (13.37) can be rewritten as ^H = \u2212J\u2211 i,\u03b4 \u2032^Si. ^Si+\u03b4, (13.49) \u03b4 = 1, 2, :::, z \u00de where Rj = Ri + R\u03b4, R\u03b4 \u00f0 is a vector to the nearest neighbor of the ith ion and Ji,i+\u03b4 = J for all \u03b4. If J is positive in E", "q. (13.49), it leads to ferromagnetism. The ion spins are assumed to be so aligned that in the ground state their z-components have the maximum values S. If S>n represents the spin of the nth ion in state s, the ground state can be written as j \u03a60 = \u03a0nj S>n: (13.50) 13.3 Models for Ferromagnetism and Antiferromagnetism 419 Eq. (13.49) can be rewritten in the alternate form ^H = \u2212J\u2211 i, j where j = i + \u03b4: Thus, we obtain \u2032 ^Siz^Sjz + h ^S + i j + ^S\u2212 ^S\u2212 i ^H\u03a60 = \u03b50\u03a60 = \u2212S2J \u2211 i,i+\u03b4 \u00f0 1 \u00de \u03a60 = \u2212JS2zN\u03a60, (13.51) (13.52) where N is the total number of sites, z is the coordination number of each site, and the application of S + to a function with maximum spin leads to zero. Here, N is the total number of ions. Thus, we obtain the expression for the energy of the ground state for a ferromagnet, \u03b50 = \u2212JNzS2: (13.53) However, if we consider the state \u03a6m in which the mth spin is reduced by 1, the new state can be written as From Eqs. (13.51) and (13.54), we can write \u03a6m = ^S\u2212 m\u03a0nj S >n: ^H\u03a6m = \u2212J\u2211 ij \u2032 ^S h z i ^S z j ^S\u2212 m + 1 2 \u00f0 ^S + i ^S\u2212 j m + ^S\u2212 ^S\u2212 i ^S + j ^S\u2212 m\u00de \u03a60: i (13.54) (13.55) It can be easily shown by using the commutation relation of the spin operators (Problem 13.8) that ^H\u03a6m = E0\u03a6m + 2JS\u2211 R\u03b4 \u00f0 \u03a6m \u2212 \u03a6m+\u03b4\u00de : (13.56) Thus, \u03a6m is not an eigenstate of the spin Hamiltonian. In fact, the deviation of spin at one site spreads into the neighboring sites, creating spin waves. To solve this, we take S to be large and expand in powers of 1/S. For example, we obtain from Eqs. (13.47) and (13.51) ^H = \u2212J\u2211 i, j \u00f0! 1 ^a\u2020 j 2 + S \u2212 ^a\u2020 i ^ai\u00de\u00f0 2S \u2212 ^a\u2020 j ^aj \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q S \u2212 ^a\u2020 + j ^aj\u00de 2S \u2212 ^a\u2020 i ^ai \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 1 2 ^a\u2020 i ^ai q, \" q 2S \u2212 ^a\u2020 i ^ai \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 ^aj 2S \u2212 ^a\u2020 j ^aj \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q (13.57) where j = i + \u03b4, \u03b4 being the nearest neighbors. For large values of S, we can write Sp bi\u00de, is very small, and bi is determined by minimizing the Hamiltonian. The constants where \ufb03\ufb03\ufb03 bi are determined by minimizing the Hamiltonian while the series in 1/S is obtained by expanding the remainder. Thus, Eq. (13.57) can be rewritten as Sp bi\u00de \ufb03\ufb03\ufb03 Sp bi + ^ai \u2212 \u00f0 (13.58) ^ai = ^ai \u2212 \ufb03\ufb03\ufb03 \u00f0 ^H = \u2212J\u2211", " ij S2 1 2 \u00f0! b&i bj + bjb&i \u00de 2 bij 2 \u2212 j \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 2 bjj 2 \u2212 j \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q Here, it is tacitly assumed that i and j are nearest pairs. To obtain the ground-state energy, we minimize the Hamiltonian with respect to all values of bi and bj: The only possible way in which the q \" + 1 \u2212 \u00f0 2 bij\u00de j \u00f0 1 \u2212 2 bjj\u00de j : (13.59) 420 CHAPTER 13 Magnetic Ordering ground-state energy \u03b50 obtained from Eq. (13.59) will be the same as obtained in Eq. (13.53) is to assume bi = bj = b (Problem 13.9). In fact, if the spins rotate in any direction as long as they all point together, the ground-state energy is independent of b, which can therefore be chosen as b = 0. The operator ^a is treated as small (Eq. 13.58) if we continue with the expansion of the Hamiltonian. From Eqs. (13.53) and (13.59), we can write ^H (up to the first order in S), treating ^a as small, b = 0, and multiplying the second term by 2 because <ij> is a sum over nearest-neighbor pairs and each pair appears twice (Problem 13.10), ^H = \u2212JNzS2 \u2212 2JS \u2211 <ij>\u00f0 We make a Fourier transformation i ^aj + ^a\u2020 ^a\u2020 j ^ai \u2212 ^a\u2020 i ^ai \u2212 ^a\u2020 j ^aj\u00de : (13.60) ^ai = 1 Np \u2211 k ^bke\u2212ik.Ri, (13.61) where k (for cyclic boundary conditions) is limited to the N values inside a Brillouin zone in the k \ufb03\ufb03\ufb03\ufb03 space. From Eqs. (13.60) and (13.61), we obtain where and ^H = \u2212JNzS2 + \u2211 k \u0127\u03c9k^nk, ^nk = ^b \u2020 k ^bk \u0127\u03c9k = 2JS\u2211 l \u00f0 1 \u2212 cos\u2061 k. Rl\u00de : (13.62) (13.63) (13.64) Eq. (13.64) gives the spin-wave dispersion relation. Here, Rl are the nearest neighbors. The first term in Eq. (13.64) is the energy of the ground state, the second term is the energy contained in the magnons, and nk = ^b ^bk is the magnon particle number. The magnon energy can be expressed as \u2020 k where \u03b2 = 1/kBT. In the isotropic case, \u0127\u03c9k \u221d k2 = \u03b3k2 (from Eq. 13.65), and we obtain E = \u2211 k \u0127\u03c9k e\u03b2\u0127\u03c9k \u2212 1, E = V 4\u03c03 kmax\u2061 Z 0 4\u03c0\u03b3k4dk e\u03b3k2/kBT \u2212 1 : (13.65) (13.66) At very low temperatures, when only a few magnons are excited, the upper limit of the integration can be considered as infinity. Thus, Eq. (13.66) can be rewritten as E = \u03b3V \u03c02 kBT \u03b3 \u221e 5/2 x4dx ex2 \u2212 1 = \u03b1T 5/2, (13.67) Z 0 where \u03b1 is a constant. Thus, the specific heat is proportional to T 3/2, which agrees with the experimental results. % $ 13.3 Models for Ferromagnetism and Antiferromagnetism 421 13.3.9 Spin Waves in Antiferromagnets When J is negative in Eq. (13.49), the neighboring spins are aligned in opposite directions. If all the ions are of the same type, this leads to antiferromagnetism. In antiferromagnets, the neighboring spins are antiparallel at zero temperature; this is known as a N\u00e9el state (Ref. 21). The ground state of an antiferromagnet is schematically illustrated in Figure 13.4 FIGURE 13.", "4 The spins are divided into two interpenetrating sublattices A and B in the N\u00e9el state. The spin states can be described as two interpenetrating sublattices, A and B. Each spin state in A has spins in the B sublattice as the nearest neighbors. One way of solving the complicated problem is as follows. The spin Hamiltonian in the antiferromagnetic state can be obtained from that of the ferromagnetic state by rotating all the spin operators on the B sublattice by 180\u00b0 about the x-axis, while keeping the spin operators intact in the A sublattice. Thus, x \u2212z: If all the Ri vectors are in the A sublattice and the Rj vectors are in the B sublattice, the spin operators defined in Eqs. (13.40) and (13.42) for ferromagnets can be rewritten for antiferromagnets as \u2212y, and z x, y!!! and ^S z j! \u2212^S z j ^S \u00b1 j! ^S\u2213 j : (13.68) (13.69) The Hamiltonian in Eq. (13.51) for ferromagnets can be rewritten for antiferromagnets as (Problem 13.11) ^H = 2 J j \u2211 j <ij> \u2212^Siz^Sjz + h ^S + i ^S 1 2 \u00f0 + j + ^S\u2212 i ^S\u2212 j \u00de, i (13.70) where the factor of 2 is multiplied because each neighboring pair <ij> appears twice in the summation over i and j. Using a procedure adapted to obtain Eq. (13.57) with the modifications for antiferromagnets (Eqs. 13.68 through 13.70), we obtain (Problem 13.12) S \u2212 ^a\u2020 i ^ai\u00de\u00f0 S \u2212 ^a\u2020 j ^aj\u00de + 1 2 ^a\u2020 i \u2212 \u00f0 ^a\u2020 j 2S \u2212 ^a\u2020 i ^ai \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q 2S \u2212 ^a\u2020 j ^aj \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q (13.71) ^H = 2 j + J \u2211 j <ij> 1 2 q! 2S \u2212 ^a\u2020 i ^ai \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 ^ai 2S \u2212 ^a\u2020 j ^aj \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 ^aj. \" q We follow a 1/S expansion method similar to the procedure outlined in Eq. (13.57) and from Eqs. (13.70) and (13.71), we obtain (Problem 13.13) ^H = \u2212 J NzS2 + 2 j j J j S \u2211 <i, j>\u00bd j i ^ai + ^a\u2020 ^a\u2020 j ^aj + ^a\u2020 i ^a\u2020 j + ^ai^aj$, (13.72) where N is the number of lattice sites, and z is the number of nearest neighbors of each site. We define the operators ^ai = 1 Np \u2211 k eik.Ri ^bk \ufb03\ufb03\ufb03\ufb03 (13.73) 422 CHAPTER 13 Magnetic Ordering and ^a\u2020 i = 1 Np \u2211 k e\u2212ik.Ri^b \u2020 k: Substituting Eqs. (13.73) and (13.74) in Eq. (13.72), we obtain (Problem 13.14) \ufb03\ufb03\ufb03\ufb03 ^H = \u2212 J j j NzS2 + S J j j \u2211 k,l\u00bd j 2^b \u2020 k ^bk + ^b \u2020 k ^b \u2020 \u2212k + ^bk^b\u2212k\u00de \u00f0 cos \u2061 \u00f0 k", ". Rl\u00de$, (13.74) (13.75) where Rl are nearest-neighbor vectors. To diagonalize the Hamiltonian, we introduce two new operators through the transformation, ^bk = ^c\u2020 \u2212k + sinh\u2061 \u03b2k\u00de \u00f0 \u00f0 cosh\u2061\u03b2k\u00de ^ck, (13.76) where \u03b2 is real. Substituting Eq. (13.76), we can show that the Hamiltonian in Eq. (13.75) is diagonalized provided (Problem 13.14) tanh 2\u03b2k = \u2212 1 z \u2211 l cos\u2061 k. Rl\u00de \u00f0 : Substituting Eqs. (13.76) and (13.77) in Eq. (13.75), we obtain ^H = \u2212N J j zS b \u2020 k ^bk + 1 2 zS\u2211 k & 1 \u2212 tanh\u20612 2\u03b2k \u2019 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 p : Thus, the ground-state energy is given by (13.77) (13.78) \u03b50 = \u2212N J j zS \u00f0 j S + 1 + J j zS\u2211 k j \u00de r 1 z2 \u00f0 cos\u2061 k. Rl\u00de 1 \u2212 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u2211 l 2, (13.79) and the energy of a magnon of wave number k is given by \u03b5k = 2 J j zS j r 1 z2 \u00f0 cos\u2061 k. Rl\u00de 1 \u2212 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u2211 l 2 : (13.80) 13.4 FERROMAGNETISM IN SOLIDS 13.4.1 Ferromagnetism Near the Curie Temperature The spontaneous magnetization of a ferromagnet vanishes above the Curie temperature. This phenomenon can be explained by the exchange interaction by using the molecular field approximation. The Hamiltonian for the exchange interaction in the presence of an external field is given by ^H = \u2212\u2211\u2032 ijJij^Si. ^Sj \u2212 g\u03bcBB. \u2211 N i=1 ^Si: (13.81) 13.4 Ferromagnetism in Solids 423 In the mean-field approximation, of which the validity is in general questionable, one of the operators appearing in the exchange integral is replaced by the mean value, so we can rewrite Eq. (13.81) as N ^H = \u2212\u2211 i=1\u00f0 N g\u03bcBB + \u2211 j=1, j\u2260i Jij <^Sj> \u00de. ^Si, which we can rewrite in the alternate form N ^H = \u2212\u2211 i=1 g\u03bcB\u00f0 B + BM\u00de : (13.82) (13.83) Here, BM is equivalent to the internal field originally introduced by Weiss to explain ferromagnetism. Assuming nearest-neighbor interaction between the spins, we obtain from Eqs. (13.82) and (13.83) BM = zJ g\u03bcB <^Sj>: If we further assume that <^Sj> is in the same direction as the magnetization M, and M = Ng\u03bcB <^Sj> BM = \u03bbM, (13.84) (13.85) (13", ".86) where \u03bb is the constant originally introduced by Weiss, known as the Weiss constant. From Eqs. (13.84) through (13.86), the relation between the Weiss constant \u03bb and the exchange integral J is given by \u03bb = zJ Ng2\u03bc2 B : (13.87) We obtain an expression for the spontaneous magnetic moment M for ferromagnetic solids by following a procedure similar to that used earlier for the derivation of M for paramag- B = 0 \u00f0 netic solids in Eq. (12.45), \u00de where the Brillouin function Bs\u00f0 x \u00de g\u03bcBS\u03bbM kBT M = NgS\u03bcBBs, % was defined in Eq. (12.46), $ (13.88) : (13.89) x BS\u00f0 \u00de = \u00f0 2S + 1 2S \u00de coth\u2061 x \u2212 coth\u2061 2S + 1 2S 1 2S x 2S & We note that for ferromagnetic solids, J is replaced with S in Eqs. (12.45) and (12.46). In addition, the external field B = 0. We can easily check the accuracy of Eq. (13.89) by noting that when T = 0, coth x = 1 and BS = 1. From Eqs. (13.88) and (13.89), we obtain the expression for saturation magnetization, \u2019 M = Ng\u03bcBS: (13.90) 424 CHAPTER 13 Magnetic Ordering When the temperature increases, the spins become randomly oriented, and the spontaneous magnetization gradually decreases until it disappears. The critical temperature TC at which the spontaneous magnetization disappears is obtained from the following approximation for the Brillouin function BS\u00f0 0, which occurs when M. When x!! 0, x \u00de Substituting Eq. (13.91) in (13.88), we obtain T x BS\u00f0 2S + 1 \u2212 1 x3 45 : 4 4 \u00de \u00de (13.91) 2S \u00f0 TC\u00de, \u00f0 M \u2248 Ng2\u03bc2 BS! S + 1 \u03bbM/3kBTC, \u00f0 from which we obtain the expression for the critical temperature, known as the Curie temperature, \u00de (13.92) TC = Ng2\u03bc2 BS \u00f0 3kB S + 1 \u03bb \u00de : (13.93) In the paramagnetic phase, the temperature dependence of magnetization can be described by the Curie law, provided the internal field is included along with the external field, or Further, we can write M = C T \u00f0 B + \u03bbM \u00de M = C T \u2212 \u03bbC B: M = \u03c7B = C T \u2212 TC B, from which we obtain the expression for the paramagnetic susceptibility \u03c7, \u03c7 = C T \u2212 TC, (13.94) (13.95) (13.96) (13.97) where TC is the Curie temperature. From Eqs. (13.95) and (13.96), the Curie constant to be inserted in Eq. (13.97) is given by C = TC \u03bb, (13.98) where an expression for TC was obtained in Eq. (13.93). 13.4.2 Comparison of Spin-Wave Theory with the Weiss Field Model Both the spin-wave theory and the Weiss field model use the concept of direct exchange interaction between localized spins of nearest neighbors. The basic assumption in both models is that the electrons of a lattice ion contributing to the magnetic moment are tightly bound for the ions to be isolated, but the 13.4 Ferromagnetism in Solids 425 nearest neighbors are sufficiently close for a significant interaction to arise. However, the results obtained by using the two models differ significantly at low and high temperatures. For example, from Eqs. (13.88) and (13.93), it can be easily shown that (Problem 13.15\u22123TC / S+1 \u00f0 T, \u00de (13.99) while the experimental results agree with the T 3/2 law obtained by using spin-wave theory. 1.0 0.8 0.6 0.4 0.2 \u00b5 N M / 0 0.2 0.4 0.6 0.8 1.0 T/TC This result suggests that for temperatures near the ground state (T = 0), the spin-wave theory can be derived by using the method of elementary excitations rather than using other approximations. At higher temperatures, it is more appropriate to use semiclassical methods such as the Weiss field model, which can be justified by the general exchange-interaction concept. However, many aspects of the behavior of a ferromagnet near the Curie temperature can be understood by using the concept of", " magnons. Even at temperatures above the Curie temperature, paramagnons can be used to understand the properties in the paramagnetic region. The temperature dependence of the spontaneous magnetization, obtained from Eq. (13.88), is shown in Figure 13.5. Saturation magnetization as a function of temperature. FIGURE 13.5 13.4.3 Ferromagnetic Domains It is a common experience to note that a ferromagnetic material is not necessarily magnetized when the temperature is lower than the Curie temperature. However, it is strongly attracted by magnetic fields and can be \u201cmagnetized\u201d by stroking it with a \u201cpermanent magnet.\u201d The basic question is how the atomic magnetic dipoles are aligned below the Curie temperature and yet produce zero magnetization. The key to explain this phenomenon is the fact that we have only considered the Heisenberg Hamiltonian while deriving an expression for the magnetization and neglected the magnetic dipolar interaction between the spins introduced in Eq. (13.7). The main reason for omitting the latter is that the dipolar coupling between nearest neighbors is much smaller than the exchange coupling. However, the exchange interaction is very short-ranged and decreases exponentially with spin separation in a ferromagnetic insulator. In contrast, the dipolar interaction falls off as the inverse cube of the separation. The magnetic configuration of a macroscopic sample depends on both interactions, especially when a large number of spins is involved where the dipolar energies that have been hitherto neglected become quite significant. It can be easily shown that the dipolar energy can be substantially reduced by dividing the ferromagnetic solid into uniformly magnetized domains of much smaller size and of which the magnetization vectors point in different directions. The concept of ferromagnetic domains was introduced by Weiss. Figure 13.6 shows a schematic diagram of the domains in a ferromagnet (Ref. 7). The atomic dipoles are aligned in the same direction within a domain but have no common link with the neighboring domains. 426 CHAPTER 13 Magnetic Ordering FIGURE 13.6 Ferromagnetic domains. The dimensions are 10 \u03bcm to 100 \u03bcm. FIGURE 13.7 Schematic diagram of the change in orientation of magnetic dipoles in neighboring domains. Figure 13.7 shows schematically a change in orientation of magnetic dipoles, in which each dipole is slightly misaligned with the neighboring dipole. The change in the orientation of the dipoles from the direction of one domain to that of the neighboring domain takes place over a distance of a few hundred atomic spaces. This narrow region between the adjacent domains is called a Bloch wall. 13.4.4 Hysteresis Figure 13.8 shows the magnetization curve that describes the process by which a ferromagnetic material can be converted from a nonmagnetic state to a ferromagnetic state with the application of a reasonably small magnetic field. This curve is known as the hysteresis curve (Ref. 27). The magnetization curve is plotted as B versus H, where H is the external magnetic field and the magnetic induction, B = H + 4\u03c0M. The external field H is applied to an initially unmagnetized sample until the magnetization reaches the saturation value. B = H + 4\u03c0 M \u2212Hc H FIGURE 13.8 The hysteresis curve. ^H = \u2212J\u2211 i, j 13.5 Ferromagnetism in Transition Metals 427 After the sample has become fully magnetized, the field is subsequently reduced, and the magnetization decreases to a constant value when the external field is zero. To return the sample to its original state (i.e., to demagnetize it), a magnetic field has to be applied in the opposite direction until B = 0 when H = \u2212Hc. Usually, this is considered as the definition of the coercive force. 13.4.5 Ising Model The Ising model is a very simplified version of the Heisenberg model. In this model, the Hamiltonian is written as \u2032^Siz^Sjz \u2212 g\u03bcBB\u2211 ^Siz: (13.100) N i=1 Thus, the terms S + and S\u2212 are essentially dropped from the Heisenberg model, and the magnetic field is taken in the z direction. Because all ^Siz commute, the Hamiltonian ^H is diagonal in the representation in which each ^Siz is diagonal. Hence, all the eigenfunctions and the eigenvalues of the Hamiltonian are known. Thus, the Ising model is very convenient in describing the statistical theory of phase transitions\u2014for example, in describing the system when the ferromagnetic state is changed to a paramagnetic state at the Curie temperature. However, in spite of these simplifications, only by using a twodimensional Ising model for simple lattices (i.e., square, triangular, honeycomb) can one calculate the exact free", " energy in zero magnetic field and the spontaneous magnetization. We will not discuss this model in more detail. 13.5 FERROMAGNETISM IN TRANSITION METALS 13.5.1 Introduction The problem of the origin of ferromagnetism in some transition metals, of which the common feature, remains one of the major is that they have narrow unfilled d-bands (3d) as well as filled s-bands \u00de unsolved problems in solid state physics. There have been many theoretical explanations, but no theory can satisfactorily explain why Fe, Co, and Ni are ferromagnetic metals while a large number of transition elements with narrow unfilled d-bands are not ferromagnetic; i.e., they do not have spontaneous magnetization. For example, the configurations of these elements are s2 \u00f0 3d74s2 $ These three elements (Fe, Co, and Ni) are ferromagnetic metals, whereas Mn and Pd become ferromagnetic under certain conditions. 3d84s2 $ 3d64s2, Ni Co Ar Ar Ar Fe \u00f0\u00bd \u00f0\u00bd \u00f0\u00bd \u00de \u00de \u00de $ :, There have been many attempts to solve this problem of band ferromagnetism (Ref. 4). It is indeed easy to understand the origin of an atomic magnetic moment arising out of an intra-atomic 428 CHAPTER 13 Magnetic Ordering exchange but difficult to explain the cooperative interaction that couples the moments on different atoms. Such cooperative phenomena require a band model of ferromagnetism. However, one of the essential features of the band model is to delocalize the moment from the atom. Thus, the Heisenberg model is inappropriate for use in ferromagnetism in transition metals. We will discuss the existing theories, with the cautionary note that not one of them is adequate in explaining band ferromagnetism. The magnetic moments of transition metals can indeed be calculated with reliable accuracy by using a combination of band calculations and density functional theory. However, this approach does not provide a specific model for band ferromagnetism. 13.5.2 Stoner Model Stoner (Ref. 26) considered a collective electron model in which there is an interaction term between the pairs of electrons of opposite spin. Thus, Hi = U N \u2211 k,k\u2032 nk\u2191nk\u2032\u2193, (13.101) \u00de \u2193 \u00f0 is the occupation number of the states >. Each pair of electrons of opposite spin where nk\u2191 contributes a positive \u201cexchange\u201d energy U/N. It should be made clear at this point that the nature of this positive \u201cexchange\u201d energy between opposite spins in the same d-shell was neither explained by Stoner nor has it been explained by any other group since then. In a later section, we will present an alternate interpretation for U. The contributions from electrons of the same spin in the d-shell are included in the definition of zero of energy and therefore not counted in Eq. (13.101). If we define the total number of electrons n\u03c3 per atom of spin \u03c3, the energy of an electron of \u2191 spin is k\u2191 \u2193 \u00de \u00f0 j \u2193 \u00f0 \u00de The number of electrons in each state is given by two Fermi\u2013Dirac distributions with the same chemical potential \u03bc. Thus, we have \u03b5k\u2191 \u00f0 \u2193 \u00de = \u03b5 k \u00de \u00f0 \u2213 \u03bcBB + Un\u2193 : \u2191 \u00f0 \u00de (13.102 \u03b5k\u2191 \u2193 \u00f0 \u00de\u00de d\u03b5, (13.103) n\u2191 \u2193 \u00f0 \u00de = Z \u03b5k\u2191 \u00f0 where f 0 is the density of states per spin. The chemical potential is the same for the two electron distributions. To satisfy this, the following conditions have to be met: is the Fermi\u2013Dirac distribution function, and D \u2193 \u00de\u00de \u03b5 \u00f0 \u00de \u00f0 and n <\u03bc> = \u03bcB \u221e Z 0 f 0 \u03b5k\u2191\u00de \u00f0 f \u2212 f 0 \u03b5k\u2193\u00deg \u00f0 D \u00f0 \u03b5 \u00de d\u03b5 n = \u221e Z 0 f 0 \u03b5k\u2191\u00de \u00f0 f + f 0 \u03b5k\u2193\u00deg \u00f0 D \u00f0 \u03b5 \u00de d\u03b5: The magnetic moment is given by M = \u03bcB\u00f0 n\u2191 \u2212 n\u2193\u00de : (13.104) (13.105) (13.106) 13.5 Ferromagnetism in Transition Metals 429 From Eqs. (13.102), (13.103), and (13.106), we obtain M = \u03bcB \u221e f Z 0 f 0 \u03b5 \u2212 \u03bcBB + Un\u2193\u00de \u00f0 \u2212 f 0 \u03b5 + \u03bc", "BB + Un\u2191\u00deg \u00f0 d\u03b5, (13.107) which can be written in the alternate form in the limit of B 0, T 0,!! M \u2248 \u00bd MU + 2\u03bc2 BB $ \u221e Z 0 $ \u2212 \u2202f 0 \u2202\u03b5 d\u03b5 D \u03b5 \u00f0 \u00de : % At T! 0, we have, for any function F \u03b5, \u00de \u00f0 \u221e \u2202f 0 \u2202\u03b5 \u2212 Z 0 d\u03b5 = F F \u03b5 \u00de \u00f0 \u03b5F\u00de : \u00f0 From Eqs. (13.108) and (13.109), we obtain Eq. (13.110) can be rewritten in the alternate form M \u2248 \u00bd\u00f0 MU + 2\u03bc2 B B \u00de D \u03b5F\u00de$ : \u00f0 The magnetic susceptibility is easily obtained from Eq. (13.111), M \u2248 2\u03bc2 \u03b5F\u00de BBD \u00f0 \u03b5F\u00de 1 \u2212 UD \u00f0 : (13.108) (13.109) (13.110) (13.111) When the \u201cexchange field\u201d is sufficiently large so that \u03c7 = M B = 2\u03bc2 \u03b5F\u00de BD \u00f0 \u03b5F\u00de 1 \u2212 UD \u00f0 : (13.112) \u03b5F\u00de Eq. (13.112) is unstable. In the Stoner model, this leads to a transition to ferromagnetism. In addition, from Eq. (13.106), to have a permanent magnetic moment, (13.113) UD \u00f0 > 1, n\u2191 \u226b n\u2193: (13.114) However, Stoner\u2019s model does not explain this large difference between n\u2191 and n\u2193: In addition, Stoner\u2019s model neither explains the origin of the positive \u201cexchange interaction\u201d U nor does it explain why only Fe, Co, and Ni are ferromagnetic metals among such a large number of elements in the transition group with narrow d-bands. The typical density of states of iron and the other transition metals in the same group with 3d narrow bands and 4s wide bands is shown in Figure 13.9. Because the Fermi energy lies inside the 3d band, the density of states at the Fermi energy is very high 3d 4s E EF FIGURE 13.9 Schematic diagram of density of states of 3d and 4s bands in iron and other transition metals. 430 CHAPTER 13 Magnetic Ordering 13.5.3 Ferromagnetism in Fe, Co, and Ni from Stoner\u2019s Model and Kohn\u2013Sham Equations In the 3d electrons in bulk transition metals, there is a significant overlap between neighboring atoms. The 4s electrons form broad free-electron-like bands of width 20\u201330 eV, whereas the 3d electron bandwidths are typically 5\u201310 eV. Due to hybridization and crystal environment, one electron is transferred from s- to d-bands. Thus, there are 7, 8, and 9 electrons per atom, respectively, in the d-bands in Fe, Co, and Ni. The predicted spin moments from Hund\u2019s rule are \u03bcspin\u00f0 = 3 \u03bcB, = \u03bcB. However, the actual values are noninteger and smaller because = 2 \u03bcB, and \u03bcspin\u00f0 \u03bcspin\u00f0 = 2:12 \u03bcB, \u03bcspin\u00f0 of partial delocalization; i.e., \u03bcspin\u00f0 Due to the large density of states, there are many unoccupied states just above the Fermi energy, which allows the promotion of electrons from minority spin to majority spin states at a modest energy cost. We have obtained the Stoner criterion for ferromagnetic susceptibility, = 1:58 \u03bcB, and \u03bcspin\u00f0 = 0:56 \u03bcB. Co Co Fe Fe Ni Ni \u00de \u00de \u00de \u00de \u00de \u00de \u03b5F\u00de The parameter U can be evaluated by perturbation theory based on non-spin-polarized solutions of the Kohn\u2013Sham equations of density functional theory described in detail in Section 7.8. (13.115) > 1: UD \u00f0 Janak9 showed that for infinitesimal Stoner splitting, one can write where \u03b3 r \u00de \u00f0 is essentially a normalized local density of states at the Fermi energy, U = dr \u03b32 Z r \u00f0 K \u00dej \u03b5F \u2212 \u03b5i\u00dej \u03c8 i\u00f0 \u03b5F\u00de D \u00f0 2 \u00dej, (13.116) (13.117) and \u03b5i and \u03c8 i\u00f0 is a kernel giving the exchange-correlation enhancement of the field due to magnetization defined by are the self-consistent energies and wave", " functions of the Kohn\u2013Sham equations. K(r) r \u00de \u03b42Exc\u00bd \u03b4m r \u03b4m \u00de \u00f0 nm $ r\u2032 \u00f0 ( )m = 0 \u00de = 2K r \u03b4 r \u2212 r\u2032 \u00de, \u00f0 \u00de \u00f0 (13.118) where Exc\u00bd density approximation of Kohn and Sham. Here, n and m are defined in the usual manner, is the exchange-correlation functional, which is defined in Eq. (7.165) in the local n; m $ r n\u03c1\u00f0 \u00de = \u2211 i \u03b8 \u03b5F \u2212 \u03b5i\u03c1\u00dej \u00f0 \u03c8 i\u03c1j 2, n r \u00f0 \u00de r = n\u2191\u00f0 \u00de r + n\u2193\u00f0 \u00de, and \u2212 n\u2193\u00f0 Janak (Ref. 9) calculated the values of both U and D = n\u2191\u00f0 m r r \u00de \u00de \u00f0 as functions of atomic number Z. The exchange-correlation-enhanced spin susceptibilities of 32 elements from Li through In were calculated using the spin-polarized exchange-correlation functional of von Barth and Hedin. The methods of Janak\u2019s calculation are summarized in his paper, and the results are tabulated in Table 1 of the same \u00f0 r : \u00de \u03b5F\u00de (13.119) (13.120) (13.121) 13.5 Ferromagnetism in Transition Metals 431 \u00f0 \u03b5f \u00de \u03b5F\u00de \u03b5F\u00de versus Z, where D are for both paper [note that his results for D \u00f0 as a spins]. He has also plotted U and D \u03b5F\u00de \u00f0 function of atomic number Z in Figure 13.1 of his paper. In Figure 13.10, we have plotted is the density 1 \u2212 UD \u00f0 of states per spin. We note that according to the data, both Fe (Z = 26) and Ni (Z = 28) are ferromagnetic according to the Stoner criterion = 0:97 for Co (Z = 27). In con\u03b5F\u00de while UD \u00f0 trast, calculations by Gunnarson (Ref. 4) indicate is in the range 1.6\u20131.8 for Co. \u03b5F\u00de that UD \u00f0 To summarize, the ferromagnetism of Fe, Co, and Ni can be explained by using densityfunctional theory through a combination of band theory and Stoner criterion, but we still lack a fundamental theory of ferromagnetism in transition metals.50 1.00 0.50 0.00 \u22120.50 \u22121.00 \u22121.50 0 10 20 30 40 50 60 FIGURE 13.10 Z Using data from Figure 1 of Janak (Ref. 9), 1 \u2212 UD Blackman JA, p.231 of Ref. 19.). is plotted against Z (by Binns C and \u03b5F \u00de \u00f0 Reproduced from Misra19 with permission of Elsevier. 13.5.4 Free Electron Gas Model It is well known that the free electron gas model does not explain the ferromagnetism in metals. However, one can derive conditions for the onset of ferromagnetism using the free electron gas model without using spin-dependent interactions. The free electron gas model provides a simple theory that is grossly inadequate but nevertheless provides a glimpse into the much harder and yet unsolved problem. It deals with the itinerant aspect of exchange through the use of the Hartree\u2013 Fock approximation of the uniform electron gas. We derived in Eq. (7.44) an expression for the ground-state energy of the free electrons, E = N e2 2a0 h 3 5 \u00f0 kFa0\u00de 2 \u2212 3 2\u03c0 \u00f0 kFa0\u00de, i (13.122) where a0 is the Bohr radius, the first term is the total kinetic energy, and the second term is the exchange energy that is between the electrons of the same spin. Eq. (13.122) was derived with the assumption that each occupied one-electron state was occupied with two electrons of opposite spin. However, if it so happens that there is a spin imbalance, then each one-electron state has k less than some k\u2191 with spin-up electrons and similarly k < k\u2193 for some spin-down electrons. Because the exchange interaction is between electrons of the same spin, we obtain from Eq. (13.122) an equation for each spin population, The total energy E \u2191,\u2193 \u00f0 \u00de = N \u2191,\u2193 \u00f0 \u00de e2 2a0 3 5 \u00f0 h k \u2191,\u2193 \u00f0 \u00de 2 a0\u00de \u2212 3 2\u03c0 \u00f0 k \u2191,\u2193 \u00f0 \u00de a0\ufffd", "\ufffd : i E = E\u2191 + E\u2193, (13.123) (13.124) 432 CHAPTER 13 Magnetic Ordering and the total number of electrons N = N\u2191 + N\u2193 = V \u2191 + k3 k3 6\u03c02 \u2193 = V! k3 F 3\u03c02 : If N\u2191 > N\u2193, the ground state will have a nonvanishing magnetic density which leads to a ferromagnetic electron gas. In the other extreme, if M = \u2212g\u03bcB N\u2191\u2212N\u2193 V, N\u2193 = N, N\u2191 = 0, E = E\u2193, k\u2193 = 21/3kF \u00f0 from Eq: 13:125 : \u00de From Eqs. (13.124), (13.125), and (13.127), we obtain (13.125) (13.126) (13.127) 22/3 3 5 2 3 2\u03c0 21/3 E\u2193 = N kFa0\u00de Comparing Eqs. (13.122) with (13.128), we note that the energy of the fully magnetized state is lower than the energy of the unmagnetized state when the exchange energy dominates the kinetic energy. It can be shown from Eq. (13.128) that when E = E\u2193, transition to a fully magnetized state occurs, in which case kFa0\u00de (13.128) \u2212 h i \u00f0 \u00f0 : kFa0 = 5 2\u03c0 1 21/3 + 1 : (13.129) We earlier defined the electron density rs, rs = 1/3 = 3V 4\u03c0N 1:92 kF : (13.130) \u2019 From Eqs. (13.129) and (13.130), the condition for transition from a nonmagnetic to a ferromagnetic state is & rs a0 = 2\u03c0 5 \u00f0 21/3 + 1 1/3 9\u03c0 4 & \u2019 \u00de = 5:45: (13.131) The only metal with such a low conduction electron density is cesium. There are some metallic compounds that also satisfy the criteria that rs/a0 > 5.45. However, none of these are ferromagnetic metals, even though their band structures are reasonably described by the free electron model. Thus, one has to know the specific features of the band structure and the itinerant exchange interactions to be able to account for magnetic ordering. The starting point is the fact that the d-electrons form a narrow band. Next, we will briefly discuss a few models for the theory of ferromagnetic metals. 13.5 Ferromagnetism in Transition Metals 433 13.5.5 Hubbard Model Hubbard (Ref. 8) proposed a simple model that yields both bandlike and localized behavior in suitable limits. He proposed an extension of the tight-binding model by adding a term that includes an energy penalty U for any atomic site occupied by more than one electron. Hubbard\u2019s main argument was that because the off-diagonal elements of the Coulomb interaction were much smaller than the diagonal elements, the Hamiltonian can be approximated as ^H = \u2211 <ij>,\u03c1 \u2212t \u00bd i\u03c1^aj\u03c1 + ^a\u2020 ^a\u2020 j\u03c1^ai\u03c1$ + U\u2211 i i\u2191^ai\u2191^a\u2020 ^a\u2020 i\u2193^ai\u2193, (13.132) where the sum <ij> is taken over nearest-neighbor pairs. Here, the off-diagonal states have nonvanishing matrix elements t, known as the hopping parameter, between those pairs of states that differ only by a single electron that has moved from an ion to one of its neighbors without change of spin. This set of terms leads to a tight-binding model of one-electron Bloch levels. Even this oversimplified Hamiltonian can be solved only by using mean-field theory. Eq. (13.132) can be rewritten as If we write ^H = \u2211 <ij>,\u03c1 \u2212t \u00bd i\u03c1^aj\u03c1 + ^a\u2020 ^a\u2020 j\u03c1^ai\u03c1$ + U\u2211 i ^ni\u2191^ni\u2193: ^ni\u03c1 = n\u03c1 + ^ni\u03c1 \u2212 n\u03c1\u00de, \u00f0 and consider the second term in Eq. (13.134) as small, Eq. (13.133) can be rewritten as ^H = \u2211 <ij>,\u03c1 \u2212t i\u03c1^aj\u03c1 + ^a\u2020 ^a\u2020 \u00bd j\u03c1^ai\u03c1$ + U\u2211 i \u00f0 ^ni\u2191n\u2193 + n\u2191^ni\u2193 \u2212 n\u2191n\u2193\u00de : Making a Fourier transformation of the type ^ai\u03c1 = 1 Np \u2211 k eik.Ri ^bk\u03c1, from Eqs. (13", ".135) and (13.136), we obtain (Problem 13.17) \ufb03\ufb03\ufb03\ufb03 ^H = \u2211 k,Rl,\u03c1 \u2212t^b \u2020 k\u03c1 ^bk\u03c1 cos\u2061 Rl. k + U\u2211 k \u00f0 ^nk\u2191n\u2193 + n\u2191^nk\u2193 \u2212 n\u2191n\u2193\u00de, (13.133) (13.134) (13.135) (13.136) (13.137) where Rl are nearest-neighbor vectors. The Hamiltonian in Eq. (13.137) becomes diagonal in k space. Specific calculations for a ferromagnetic solid (Ref. 30) can be made depending on the lattice structure and the dimension. In addition to the difficulties in making specific calculations in the Hubbard model, its major drawback is that it was designed for solutions that lead to ferromagnetism in metals. We will present an alternate model for ferromagnetism in the transition metals starting with a theory of magnetization of interacting Bloch electrons. 434 CHAPTER 13 Magnetic Ordering 13.6 MAGNETIZATION OF INTERACTING BLOCH ELECTRONS 13.6.1 Introduction Recently, Tripathi and Misra29 derived a theory of magnetization of interacting electrons in the presence of a periodic potential, spin-orbit interaction, and an applied magnetic field in the paramagnetic limits. Starting from a thermodynamic potential, which includes both the quasiparticle and correlation contributions, they showed that the modifications brought about by the electron\u2013electron interactions for the magnetization in the quasiparticle approximation are precisely canceled by the contributions due to electron correlations. The magnetization is expressed as a product of the spin density and effective g factor, due mainly to the spin-orbit interaction. Tripathi and Misra (Ref. 29) also showed the importance of the self-energy corrections in the single-particle energy spectrum. By considering a variant of the Hubbard Hamiltonian in the momentum space, their theory can predict whether or not the ground state of the interacting electron system is magnetic. Tripathi and Misra\u2019s model for ferromagnetism appears as a variant of the Stoner model but from a very different perspective. 13.6.2 Theory of Magnetization The magnetization of an interacting electron system is given by Mv = \u2212 \u2202\u03a9 \u2202Bv, where \u03a9 is the thermodynamic potential, and in Eq. (12.123), Misra et al. (Ref. 18) showed where \u03a9 = \u03a9qp + \u03a9corr, \u03a9qp = 1 \u03b2 Tr ln \u2212 \u00f0 = \u2212 1 2i\u03c0 Tr G\u03bel\u00de e G F \u03be \u00f0 \u00de \u03be \u00de \u00f0 d\u03be, IC e (13.138) (13.139) (13.140) ln! and the contour encircles the imaginary axis in a counterclockwise direction and Tr involves summation over one-particle states.18 Misra et al.18 derived (Eq. 12.117) (13.141\u2212\u03b2 \u03be\u2212\u03bc 1 \u03b2 where \u03bel \u2212 ^H \u00bd \u03ba!, \u03bel\u00de$ \u00f0 ^G \u00f0 k, \u03bel\u00de = I, \u03ba! = k + ih \u00d7 \u2207k, (from Eq. 12.119) and h = eB 2\u0127c (from Eq. 12.114). Misra et al. (Ref. 18) also derived (Eqs. 12.130 through 12.132) ^H \u03ba!, \u03bel\u00de \u00f0 = ^H0\u00f0 k, \u03bel\u00de + ^H\u2032 \u00f0, k, \u03bel\u00de (13.142a) (13.142b) (13.143) 13.6 Magnetization of Interacting Bloch Electrons 435 where ^H0\u00f0 k, \u03bel\u00de = 1 2m \u00f0 p + \u0127k 2 + V \u00de r \u00de \u00f0 + \u03a30 k, \u03bel\u00de \u00f0 + \u01272 8mc2 \u22072V + \u0127 4m2c2 \u03c3!. \u2207!V \u00d7 p + \u0127k \u00de, \u00f0 (13.144) and retaining terms up to the first order in the magnetic field in Eq. (12.132), k + 1 2 In Eq. (13.145), h\u03b1\u03b2 = \u2208\u03b1\u03b2\u03bc h\u03bc, where \u2208\u03b1\u03b2\u03bc h\u03bc, where \u2208\u03b1\u03b2\u03bc is the antisymmetric tensor of the third rank and we follow Einstein summation convention. g\u03bcBB\u03bc\u03c3\u03bc + B\u03bc\ufffd", "\ufffd1,\u03bc = \u2212ih\u03b1\u03b2\u03a0\u03b1\u2207\u03b2 k, \u03bel\u00de k, \u03bel\u00de (13.145) H\u2032 \u00f0 \u00f0 : Here, \u03a0! was defined in Eq. (12.133), \u03a0! = \u0127 m \u00f0 p + \u0127k \u00de + \u01272 4m2c2 \u03c3! \u00d7 \u2207!V + \u2207k\u03a30 : k, \u03bel\u00de \u00f0 In addition, from Eqs. (12.134) through (12.136), Misra et al. (Ref. 18) derived = G0\u00f0 ^H\u2032G0\u00f0 + G0\u00f0 k, \u03bel\u00de k, \u03bel\u00de k, \u03bel\u00de k, \u03bel\u00de + ::: G \u00f0 and where (from Eq. 12.135) e \u2207\u03b1 kG0\u00f0 k, \u03bel\u00de = G0\u00f0 k, \u03bel\u00de \u03a0\u03b1G0\u00f0, k, \u03bel\u00de From Eqs. (13.145) through (13.148), it can be shown that (Problem 13.18) G0\u00f0 k, \u03bel\u00de = 1 \u03bel \u2212 ^H0\u00f0 : k, \u03bel\u00de where = G0 \u2212 G0 ih\u03b1\u03b2\u03a0\u03b1\u2207\u03b2 h k \u2212 1 2 g\u03bcBBvFv G0, i G \u00f0 k, \u03bel\u00de e Fv = \u03c3v + 1,v \u03a3 2 g\u03bcB (13.146) (13.147) (13.148a) (13.148b) (13.149) (13.150) was defined in Eq. (12.147) and is the renormalized spin vertex in the presence of the electron\u2013 electron interaction. e 13.6.3 The Quasiparticle Contribution to Magnetization From Eqs. (12.123) and (13.149), one can obtain the quasiparticle contribution to the thermodynamic potential, after evaluating the trace and contour integration (Problem 13.19) as \u03a9qp = 1 2 \u03bcBBv \u2211 nk\u03c1m\u03c1\u2032,m \u2260 n \" gFv n\u03c1,n\u03c1\u2032 + ie \u0127c\u03bcB \u2208\u03b1\u03b2v \u03a0 \u03b2 \u03b1 n\u03c1,m\u03c1\u2032\u03a0 m\u03c1\u2032,n\u03c1 Emn f Enk\u03c1\u00de \u00f0, # where f Enk\u03c1\u00de \u00f0 is the Fermi distribution function for an electron in band n and spin \u03c1, \u03a0\u03b1 n\u03c1,m\u03c1\u2032 = d3ru&nk\u03c1\u03a0\u03b1umk\u03c1\u2032 Zcell (13.151) (13.152) 436 CHAPTER 13 Magnetic Ordering and Using Emn = Em\u00f0 k \u00de \u2212 En\u00f0 k \u00de : Mv qp = \u2212 \u2202\u03a9qp \u2202Bv, and Eqs. (13.150) and (13.151), Tripathi and Misra (Ref. 29) obtain where Mv qp = \u2212 1 2 \u03bcB \u2211 nk\u03c1\u03c1\u2032 gv k nn\u00f0! \u03c3v n\u03c1,n\u03c1\u2032 + \u00de \u03a3 1,v n\u03c1,n\u03c1\u2032 2 \u03bcB Enk\u03c1\u00de \u00f0, f \" e k gv nn\u00f0 \u00de n\u03c1,n\u03c1\u2032 = g\u03c3v \u03c3v n\u03c1,n\u03c1\u2032 + ie \u0127c\u03bcB \u2208\u03b1\u03b2v \u2211 m \u2260 n,\u03c1\u2032 \u03a0 \u03b1 n\u03c1,m\u03c1\u2032\u03a0 Emn \u03b2 mv\u2032,n\u03c1, (13.153) (13.154) (13.155) (13.156) and is the effective g factor in the presence of spin-orbit interaction. In the absence of spin-orbit interactions, the second term in Eq. (13.156) vanishes, and the effective g factor reduces to the free electron g factor. In the absence of many-body interactions, the second term in the square brackets of Eq. (13.155) vanishes. However, Tripathi and Misra (Ref. 29) showed that the inclusion of Mcorr would precisely cancel this term, and the final expression for magnetization would be free from any explicit many-body corrections\u2014a surprising result that we will discuss in detail at the end of this section. 13.6.4 Contribution of Correlations to Magnetization In Eq. (12.123), Misra et al. (Ref. 18) derived The contribution of correlation to magnetization is (Ref. 29) e e \u03a9corr = 1 \u03b2! \u2212Tr", " \u03a3\u03bel G\u03bel + \u03d5 : \" (13.157) G\u03bel\u00de \u00f0 e \u2202\u03a9corr \u2202Bv = Simplifying Eq. (13.158) with the use of Eqs. (12.127), (12.128), (13.155), and the Luttinger\u2013Ward (Ref. 15) prescription for frequency summation, we obtain corr = \u2212 (13.158) G\u03bel: Mv Tr e 1 \u03b2 \u2202 \u03a3\u03bel \u2202Bv e 1 \u03b2 1 \u03be \u2212 ^H0\u00de where \u03930 encircles the real axis in a clockwise direction, and f obtained (Problem 13.20) after evaluating the tr over a complete set of single particle states, 1 \u03bel \u2212 ^H0\u00de 1 2i\u03c0 m = \u2211 \u03bel \u03be \u00de d\u03be, m f Tr Z\u03930 \u03be \u00f0 \u00f0 \u00f0 \u00f0 \u00de is the Fermi distribution function (13.159) Mv corr = \u2211 nk\u03c1 1,v n\u03c1,n\u03c1\u2032 f \u03a3. Enk\u03c1\u00de \u00f0 e (13.160) 13.6 Magnetization of Interacting Bloch Electrons 437 From Eqs. (13.155) and (13.160), Mv = \u2212 1 2 \u03bcB \u2211 nk\u03c1\u03c1\u2032 gv k nn\u00f0 \u00de \u03c3v n\u03c1,n\u03c1\u2032f Enk\u03c1\u00de \u00f0 : (13.161) Thus, the renormalization of the spin vertex by the electron\u2013electron interactions in the quasiparticle contribution is precisely canceled by Mv corr. It may appear surprising that although both the spin contribution to the susceptibility (Ref. 18) (see Chapter 12) and spin Knight shift (Ref. 28) are exchange enhanced by electron\u2013electron interactions, the self-energy corrections do not explicitly appear in the expression for magnetization. The reason is that the magnetic susceptibility and Knight shift arise from second-order effects in the magnetic field, where both the spin vertices are renormalized, and the renormalization of only one of the vertices is canceled due to the electron correlations. The renormalization of the other spin vertex, in a Hartree\u2013Fock analysis, contributes to the exchange enhancement of the spin contributions to the susceptibility (Chapter 12) and Knight shift. In contrast, the single-particle energies appearing in Eqs. (13.144) and (13.145) depend on electron\u2013electron interaction and the magnetic field through the self-energy. In the next section, we will show the consequences of these self-energy corrections on the single-particle spectrum in predicting whether or not the ground state of the interacting system is magnetic. Assuming the effective magnetic field to be in the z direction and considering a single band, Tripathi and Misra (Ref. 29) obtained a tractable expression by averaging over the g matrix and denoting it as the effective g factor geff. They showed that Eq. (13.161) can be expressed as M = \u2212 1 2 \u03bcBgeff \u2211 k\u03c1\u03c1\u2032 \u03c3\u03c1\u03c1\u2032f : Ek\u03c1\u00de \u00f0 Here, \u03c1 and \u03c1\u2032 take both \u2191 and \u2193 states. Eq. (13.162), therefore, can be rewritten as where M = \u2212 1 2 \u03bcBgeff \u00f0 n\u2191 \u2212 n\u2193\u00de, n\u2191 \u2193 \u00f0 \u00de f = \u2211 k \u00f0 Ek\u2191 : \u2193 \u00f0 \u00de\u00de (13.162) (13.163) (13.164) 13.6.5 Single-Particle Spectrum and the Criteria for Ferromagnetic Ground State The single-particle energies Ek\u03c1 appearing in Eq. (13.154) are the eigenvalues of the field-independent Hamiltonian described in Eq. (13.144). If the spin-orbit interaction is neglected, \u03c1 becomes a pure spin state, and Ek\u03c1 can be obtained by the following procedure. The many-body Hamiltonian can be written as ^H0 = \u2211 k\u03c1 \u03b5kc\u2020 k\u03c1ck\u03c1 + 1 2 \u2211 qkk\u2032\u03c1\u03c1\u2032 V q \u00de \u00f0 k+q,\u03c1c\u2020 c\u2020 k\u2032\u2212q,\u03c1\u2032ck\u2032,\u03c1\u2032ck,\u03c1, (13.165) where c\u2020 spin \u03c1, and V \u00f0 q \u00de k\u03c1 and ck\u03c1 are the creation and annihilation operators for an electron with wave vector k and is the Fourier transformation of the Coulomb interaction. 438 CHAPTER 13 Magnetic Ordering Ek\u03c1 can be obtained by using the mean-field approximation, Ek\u03c1 = 1", " N From Eqs. (13.165) and (13.166), we obtain ck\u03c1, ^H0$, c\u2020 k\u03c1 : o+ \u00bd n * Ek\u03c1 = \u03b5k \u2212 1 N \u2211 q V q f \u03b5k\u2212q,\u03c1\u00de \u03b5k\u03c1\u00de \u00f0 + f \u03b5k,\u2212\u03c1\u00de$ \u00f0 : (13.166) (13.167) Because V is assumed to be a constant U in the approximation of the averaged Coulomb interaction ansatz, we can rewrite Eq. (13.167) as where Ek\u03c1 = \u03b5k + 1 N U\u2211 k f \u03b5k,\u2212\u03c1\u00de % \u00f0 \u03b5k + Un\u2212\u03c1, n\u2212\u03c1 = 1 N \u2211 k f \u03b5k,\u2212\u03c1\u00de : \u00f0 (13.168) (13.169) Eq. (13.168) is the Hartree\u2013Fock representation of the single-particle spectrum. If we write n = n\u2191 + n\u2193 and m = n\u2191 \u2212 n\u2193 where n is the number of electrons/atoms and m is the average magnetization per atom in a Bohr magneton, the total energy per atom is E = 1 N \u2211 k\u03c1 \u03b5k f Ek\u03c1\u00de \u00f0 + Un\u2191n\u2193: The nonmagnetic solution in the ground state is obtained from the criteria f Ek\u03c1\u00de \u00f0 = 1 if Ek\u03c1 < EF = 0 if Ek\u03c1 > EF: (13.170) (13.171) of elecThe Fermi energy EF is determined from the condition that one has the right number trons per atom. To study the stability of this state, let us transfer some electrons in an energy range \u03b4E below EF from the down-spin states to the up-spin states. The change in the kinetic energy is n \u00f0 \u00de where D \u00f0 action energy is EF\u00de is the density of states per atom per spin at the Fermi energy. The change in the inter- \u0394T = D \u00f0 EF\u00de\u00f0 \u03b4E2, \u00de (13.172) \u0394Eint\u2061 = U n 2 + D\u03b4E n 2 \u2212 D\u03b4E \u2212 1 4 Un2 \u2212UD2 EF\u00de\u00f0 \u00f0 \u03b4E2 : \u00de % (13.173) From Eqs. (13.172) and (13.173), the change in the total energy is & \u2019 & \u2019 \u0394E = D \u00f0 EF\u00de\u00f0 \u00de \u00bd The nonmagnetic state would be stable if UD \u03b5F\u00de < 1 and the condition for the ferromagnetic > 1. This appears similar to the Stoner condition for ferromagnetic stability stability is UD except that here U is well defined (Eq. 13.168) instead of an arbitrary positive attractive interaction between electrons of opposite spin in an atom. It can be shown that the preceding conditions could (13.174) 1 \u2212 UD EF\u00de$ \u03b5F\u00de \u03b4E \u00f0 \u00f0 \u00f0 : 2 13.8 Anderson Model 439 also be obtained if one considers the second-order terms in the field in the thermodynamic potential and calculates the magnetic susceptibility. 13.7 THE KONDO EFFECT In 1936, de Haas and van den Berg observed in experiments with gold (probably containing a small amount of iron impurities) that electrical resistivity dropped as the normal temperature decreased until about 8\u00ba K, and below that it increased. This observation was in contrast to the theory of electrical resistivity of normal metals (nonsuperconductors) where the resistance is supposed to fall with temperature as the atomic vibrations decrease and, below about 10\u00ba K, remain constant. This low-temperature residual resistivity is due to scattering by defects in the material. Kondo (Ref. 14) used second-order perturbation theory to calculate the scattering of the conduction electrons of a noble metal by magnetic defects, using a Heisenberg model to describe the interaction between conduction electron spins and the spin moments and the defects. Kondo observed that spin-flip processes can occur in the second order, which results in a resistance that increases logarithmically when the temperature is lowered. These results are valid above a temperature TK (known as the Kondo temperature), but the perturbation theory breaks down for T < TK. In 1963, Nagaoka20 published a self-consistent treatment that yielded solutions both above and below the Kondo temperature and reproduced the resistance minimum. 13.8 ANDERSON MODEL In 1961, Anderson (Ref. 1) proposed a model for the Hamiltonian of a single magnetic impurity to a Fermi sea of electrons, H = \u2211 k\u03c3 \u03b5kc\u2020 k\u03c3ck\u03c3 +", " \u03b5d\u2211 \u03c3 d\u2020 \u03c3d\u03c3 + \u2211 k\u03c3 \u00f0 Vdkd\u2020 \u03c3ck\u03c3 + V &dkc\u2020 k\u03c3d\u03c3\u00de + Ud\u2020 \u2191d\u2191d\u2020 \u2193d\u2193: (13.175) The first term is the kinetic energy of the electrons, where c\u2020 k\u03c3 and ck\u03c3 are the creation and annihilation operators for electrons with spin \u03c3 in state k. The magnetic state is represented by a single state with energy \u03b5d and electron operators d\u2020 \u03c3 and d\u03c3. The third term represents the s \u2212 d hybridization of the d-level with the conduction band. The fourth term represents the cost, in Coulomb energy, of double occupancy of the d-level. U is assumed to be large and \u03b5 < \u03b5F, where \u03b5F is the Fermi energy. In the absence of the hybridization term, we have a single occupied (magnetic) d state. The Anderson model does not attempt to describe the details of the atomic 3d orbitals, but it captures the essential physics of the Kondo problem. The Anderson Hamiltonian leads to the spin exchange process. The electron can tunnel from the localized impurity state to an unoccupied state just above the Fermi energy, and another electron from the Fermi energy sea replaces it. If these two electrons have opposite spin, it is known as a spin-flip process. An alternate spin-flip process involves double occupancy of the impurity state, with a cost of Coulomb energy. When many such processes are taken together, a new many-body process known as the Kondo resonance is generated. The Kondo resonance, which has a spin-singlet state, has a width of kBTK and is pinned at the Fermi level, as schematically shown in Figure 13.11. 440 CHAPTER 13 Magnetic Ordering \u03b5\u2191 \u03b5F \u03b5d kBTK \u0393 FIGURE 13.11 Schematic diagram of Kondo resonance of width kBTK. The impurity state is broadened to width \u0393 by hybridization with the conduction electrons. D(\u03b5)\u2192 The impurity state at \u03b5d is shifted and broadened through hybridization with electron states of the Fermi sea. This is known as a d-resonance. The net effect is an alignment of the spins of the conduction electrons near the impurity atom so that the local moment is screened. It has been shown by scaling laws and parameters of the Anderson model that kBTK = 1 2 \u00f0 \u0393U 1 2 exp\u2061 \u00bd \u00de \u03c0\u03b5d\u00f0 \u03b5d + U /\u0393U : $ \u00de (13.176) 13.9 THE MAGNETIC PHASE TRANSITION 13.9.1 Introduction There are several types of phase transitions, but the most thoroughly investigated type is the paramagnetic to ferromagnetic phase transition. When the temperature is lowered through a critical temperature Tc of the system, a long-range order appears that was absent above Tc. The order, which usually has an infinite derivative at Tc, rapidly increases below Tc. In the absence of external fields, the ordering is not completely determined, and the transition is characterized by divergence of certain quantities like the magnetic susceptibility and the specific heat. 13.9 The Magnetic Phase Transition 441 13.9.2 The Order Parameter In the case of magnetization, the order parameter \u03c3 order in the phase transition, is defined as r \u00de \u00f0, which describes the sudden appearance of where Mz is the magnetization that defines the z-axis, and M0 is the maximum possible value of magnetization, \u03c3 r \u00de \u00f0 = Mz/M0, (13.177) M0 = Ng\u03b2S: (13.178) The order parameter, which is zero above but not below Tc, can continuously approach zero as Tc from below if the transition is not of the first order. However, it is not completely deterT mined below Tc in the absence of external fields.! 13.9.3 Landau Theory of Second-Order Phase Transitions According to Landau theory (Ref.15), the free energy is defined as a function of the order parameter is the free energy density. The entropy S is given by G = G \u00f0 T, \u03c3 \u00de = dr, g r \u00de \u00f0 Z where g r \u00de \u00f0 Landau introduced a quantity h r \u00f0 \u00de S = \u2212 \u2202G \u2202T : & \u03c3 \u2019, which is related to the external field H0\u00f0 r \u00de \u00f0 = M0H h r : \u00de can be expanded as a power series in \u03c3, \u00f0 by r \u00de Landau assumed that g r g r T = g0\u00f0 \u00f0 \u00de \u2212 h", " \u00f0 Here, g is minimized (which determines the most probable value of \u03c3) by requiring that \u00dej \u00de$ \u00dej \u00de$ \u00de\u00bd \u00de\u00bd 2 + ::::: \u2207\u03c3 r \u00f0 from which we obtain Eq. (13.184) yields g r dr = 0, \u00de \u00f0 \u03b4 Z d\u03c3 \u2212h r \u00de \u00f0 \u00bd Z + 2a\u03c3 + 4b\u03c33 \u2212 2c \u22072\u03c3 r \u00f0 dr = 0: \u00de$ \u00f0 If \u03c3 and h are assumed to be independent of r, Eq. (13.185) yields \u00de$ \u00de \u00f0 \u00de \u00f0 \u00bd \u22122c\u22072\u03c3 r + 2a + 4b\u03c32 \u03c3 r = h r \u00f0 r : \u00de If h = 0, the two possible solutions are 2a + 4b\u03c32 \u00bd \u03c3 = h: $ \u03c3 = 0 (13.179) (13.180) (13.181) (13.182) (13.183) (13.184) (13.185) (13.186) (13.187) 442 CHAPTER 13 Magnetic Ordering and \u03c3 = \u00b1 \u00f0 \u2212a/2b 1/2: \u00de (13.188) From Eqs. (13.182) and (13.183), if a > 0, Eq. (13.187) minimizes the free energy, whereas if a < 0, Eq. (13.188) minimizes the free energy. The free energy should describe a system with nonzero magnetization in the absence of external fields for T < Tc. This suggests the assumption where \u03b1 is a constant. From Eqs. (13.188) and (13.189), we obtain T a \u00f0 \u00de = \u03b1 T \u2212 Tc\u00de, \u00f0 (13.189) 1/2, \u00de Eq. (13.190) is consistent with the molecular field theory. To obtain the reduced susceptibility, we consider h \u2260 0. The reduced susceptibility is defined as T < Tc\u00de (13.190) Tc \u2212 T \u2248 T \u03c3 \u00f0 \u00f0 \u00de \u00f0 : \u03c7r = \u2202\u03c3/\u2202h \u00deT : \u00f0 (13.191) From Eq. (13.186) and (13.191), we obtain 2a\u03c7r + 12\u03c32\u03c7r = 1: Because \u03c32 is small and, hence, can be neglected for T > Tc, Eq. (13.192) can be rewritten in the alternate form with the help of Eq. (13.189), (13.192) \u00f0 Eq. (13.193) is the Curie\u2013Weiss law. Below Tc, from Eqs. (13.190) and (13.192), we obtain \u00f0 \u03c7r = 1/2\u03b1 T \u2212 Tc\u00de, T > Tc\u00de : (13.193) At a constant external field, the specific heat is obtained from the relation \u03c7r = \u22121/4a = \u22121/4\u03b1 Tc \u2212 T, \u00de \u00f0 T < Tc\u00de : \u00f0 CM = \u2212T \u22022G \u2202T 2 : $ h = 0 %h. Because \u03c3 = 0 for T > Tc, First, we consider the case of zero external field \u00f0 C = C0 = \u2212Td2/dT 2 \u00de r g0\u00f0 dr, T > Tc\u00de : \u00f0 \u00de Z For a uniform system below Tc, it can be shown that (Problem 13.22) Thus, from Eqs. (13.195) and (13.197), the specific heat is G = dr Z g0 \u2212 \u00bd \u00f0 a2/4b : \u00de$ (13.194) (13.195) (13.196) (13.197) C = C0 + T T \u2212 Tc\u00de Comparing Eqs. (13.196) and (13.198), we note that there is a finite contribution from magnetization to the specific heat just below Tc that results in a finite discontinuity to C at Tc: T < Tc\u00de : \u00f0 (13.198) dr, $ db/dT \u2212 \u00bd\u00f0 Z \u00de \u00f0 $ \u00bd \u03b12/2b \u03b12 2/4b2 Problems 443 PROBLEMS 13.1. Using vector identities, show from Eqs. (13.1) through (13.3) that F = \u00f0 m \u00d7 \u2207! \u00de \u00d7 B = \u2207! m. B, \u00de \u00f0 where B B 0 : \u00de \u00f0 % 13.", "2. Show that the induction B produced by a magnetic dipole of moment m1 at a distance r, where a second dipole of moment m2 is located, is given by 13.3. From Eqs. (13.8) through (13.10) and Eq. (13.13), show that B = \u2207! m1. \u2207! h 1 r i = 3^r m1. ^r r3 \u00f0 \u00de \u2212 m1 : \u03b5s,t = < ^H>s,t = 2\u03b50 + e2 R12 + e2 dr1dr2 \u03c8 &s,t\u00f0 r1, r2\u00de Z 1 r1 \u2212 r2j! j \u2212 1 r1 \u2212 R2j j \u2212 1 r2 \u2212 R1j j \" \u03c8 s,t\u00f0 r1, r2\u00de, (1) where 1 R12 % 1 R1\u2212R2j j : 13.4. From Eqs. (13.14) through (13.17), show that \u03b5s = 2\u03b50 + e2 R + and Vc\u00f0 R12\u00de \u00f0 + Vex\u00f0 1 + I2 \u00de R12\u00de 13.5. For a two-electron system, we can express \u03b5t = 2\u03b50 + Vc\u00f0 + e2 R R12\u00de \u00f0 \u2212 Vex\u00f0 1 \u2212 I2 \u00de R12\u00de : ^S1. ^S2 = 1 2 \u00bd ^S + 1 ^S\u2212 2 + ^S + 2 + ^S z 1 ^S z 2, ^S\u2212 1 $ where Show that and \u00b1 ^S = ^Sx \u00b1 i^Sy: ^S1. ^S2\u03c7s = \u2212 3 4 \u03c7s ^S1. ^S2\u03c7t = 1 4 \u03c7t, where \u03c7s and \u03c7t are the spin singlet and triplet states. (2) (1) (2) (1) (2) (3) (4) 444 CHAPTER 13 Magnetic Ordering 13.6. In the Schwinger representation, the Bose operators are defined as \u03b1 ^S = 1 2 \u2211 ii\u2032 ^a\u2020 i \u03c3\u03b1 ii\u2032^ai\u2032, where \u03c3\u03b1 are the Pauli spin matrices \u00f0 x ^S y ^S z ^S \u03b1 = x, y, z. Show from Eq. (1) that \u00de 1^a2 + ^a\u2020 ^a^a1\u00de 1^a2\u00de 2^a2\u00de : 2^a1 \u2212 ^a\u2020 ^a\u2020 1^a1 \u2212 ^a\u2020 ^a\u2020 13.7. By using the Holstein\u2013Primakoff transformation, we have obtained + p ^S ^S\u2212 = z ^S = 2S \u2212 ^a1^a1 ^a1 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 = ^a\u2020 1 p 2S \u2212 ^a1^a1 ^a\u2020 1^a1 \u2212 S : \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u00f0 \u00de Show that 13.8. In Eq. (13.54), we defined +, ^S\u2212 ^S \u00bd $ = 2^S z : \u03a6m = ^S\u2212 m\u03a0nj S>n: From Eqs. (13.51) and (13.54), we can write ^H\u03a6m = \u2212J\u2211 ij \u2032 ^S h z i ^S z j ^S\u2212 m + 1 2 \u00f0 ^S + i ^S\u2212 j m + ^S\u2212 ^S\u2212 i ^S + j \u03a60: ^S\u2212 m\u00de i Show by using the commutation relation of the spin operators that ^H\u03a6m = E0\u03a6m + 2JS\u2211 Rl \u00f0 \u03a6m \u2212 \u03a6m+l\u00de : (1) (2) (1) (2) (1) (2) (3) 13.9. Show that the only possible way in which the ground-state energy \u03b50 obtained from Eq. (13.59) will be the same as obtained in Eq. (13.53) is to assume bi = bj = b. 13.10. From Eq. (13", ".53) and (13.59), show that we can write ^H (up to the first order in S), treating ^a as small, b = 0, and multiplying the second term by 2 because <ij> is a sum over nearestneighbor pairs (each pair appears twice), ^H = \u2212JNzS2 \u2212 2JS \u2211 <ij>\u00f0 i ^aj + ^a\u2020 ^a\u2020 j ^ai \u2212 ^a\u2020 i ^ai \u2212 ^a\u2020 j aj\u00de : (1) Problems 445 13.11. If all the Ri vectors are in the A sublattice and the Rj vectors are in the B sublattice, the spin operators defined in Eqs. (13.40) and (13.42) for ferromagnets can be rewritten for antiferromagnets as and ^S z j! \u2212^S z j ^S \u00b1 j! ^S\u2213 j : Show that the Hamiltonian in Eq. (13.51) for ferromagnets can be rewritten for antiferromagnets as ^H = 2 J j j \u2211 <ij> h \u2212^Siz^Sjz + ^S + i ^S 1 2 \u00f0 + j + ^S\u2212 i ^S\u2212 j \u00de : i (1) (2) (3) 13.12. Using a procedure adapted to obtain Eq. (13.57) with the modifications for antiferromagnets (Eqs. 13.68 through 13.70), show that ^H = 2 j J \u2211 j <ij> 1 2 \u2212 S \u2212 ^a\u2020 \u00f0! 2S \u2212 ^a\u2020 i ^ai \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 i ^ai\u00de\u00f0 S \u2212 ^a\u2020 j ^aj\u00de + 1 2 ^a\u2020 i ^a\u2020 j 2S \u2212 ^a\u2020 i ^ai \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q 2S \u2212 ^a\u2020 j ^aj \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q (1) 2S \u2212 ^a\u2020 j ^aj \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 13.13. By following a 1/S expansion method similar to the procedure outlined in Eqs. (13.59) and q q ^ai ^aj + \" : Eq. (13.70), derive from Eq. (1) of Problem 13.12 13.14. We derived ^H = \u2212 J j j NzS2 + 2 J S \u2211 <i,j>\u00bd j j i ^ai + ^a\u2020 ^a\u2020 j ^aj + ^a\u2020 i ^a\u2020 j + ^ai^aj$ : ^H = \u2212 J NzS2 + j j S J j j \u2211 k,l\u00bd j 2^b \u2020 k ^bk + ^b \u2020 k ^b \u2020 \u2212k + ^bk^b\u2212k\u00de \u00f0 cos\u2061 k. Rl\u00de$, \u00f0 (1) (1) where Rl are nearest-neighbor vectors. To diagonalize the Hamiltonian, we introduced two new operators through the transformation, sinh\u2061\u03b2k\u00de where \u03b2 is real. Show that by substituting Eq. (2), the Hamiltonian in Eq. (1) is diagonalized provided cosh\u2061\u03b2k\u00de ^ck, \u00f0 \u00f0 ^c\u2020 \u2212k + ^bk = tanh\u20612\u03b2k = \u2212 1 z \u2211 l cos\u2061 k. Rl\u00de : \u00f0 (2) (3) 13.15. Derive from Eqs. (13.88) and (13.91) that the ratio of saturation magnetization M(T) at temperature T and M(0) at zero temperature is \u22123TC/ T : S+1 \u00de \u00f0 (1) 446 CHAPTER 13 Magnetic Ordering 13.16.", " The chemical potential is the same for the two electron distributions with up- and down-spins. Show that, to satisfy this, the following conditions have to be met: and n<\u03bc> = \u03bcB \u221e Z 0 1 2f f 0 \u03b5k\u2191\u00de \u00f0 \u2212 f 0 \u03b5k\u2193\u00deg \u00f0 D \u00f0 \u03b5 \u00de d\u03b5 n = \u221e Z 0 1 2f f 0 \u03b5k\u2191\u00de \u00f0 + f 0 \u03b5k\u2193\u00deg \u00f0 D \u03b5 \u00f0 d\u03b5: \u00de 13.17. We derived in Eq. (13.135) ^H = \u2211 <ij>,\u03c1 \u2212t \u00bd i\u03c1^aj\u03c1 + ^a\u2020 ^a\u2020 j\u03c1^ai\u03c1$ + U\u2211 i \u00f0 ^ni\u2191n\u2193 + n\u2191^ni\u2193 \u2212 n\u2191n\u2193\u00de : Making a Fourier transformation of the type show that ^ai\u03c1 = 1 Np \u2211 k eik.Ri^bk\u03c1, \ufb03\ufb03\ufb03\ufb03 ^H = \u2211 k,Rl,\u03c1 \u2212t^b \u2020 k\u03c1 ^bk\u03c1 cos\u2061 k. Rl + U\u2211 k \u00f0 ^nk\u2191n\u2193 + n\u2191^nk\u2193 \u2212 n\u2191n\u2193\u00de : 13.18. We derived in Eq. (13.145) H\u2032 k, \u03bel\u00de \u00f0 = \u2212ih\u03b1\u03b2\u03a0\u03b1\u2207\u03b2 k + 1 2 g\u03bcBB\u03bc\u03c3\u03bc + B\u03bc\u03a31,\u03bc : k, \u03bel\u00de \u00f0 Here, \u03a0! was defined in Eq. (12.133), \u03a0! = \u0127 m \u00f0 p + \u0127k \u00de + \u01272 4m2c2 \u03c3! \u00d7 \u2207!V + \u2207k\u03a30 : k, \u03bel\u00de \u00f0 In addition, from Eqs. (13.147) and (13.148), we have k, \u03bel\u00de k, \u03bel\u00de = G0\u00f0 + G0\u00f0 k, \u03bel\u00de G \u00f0 ^H\u2032G0\u00f0 + ::: k, \u03bel\u00de and e \u2207\u03b1 kG0\u00f0 k, \u03bel\u00de = G0\u00f0 k, \u03bel\u00de \u03a0\u03b1G0\u00f0 k, \u03bel\u00de : From Eqs. (1) through (4), show that where G k, \u03bel\u00de \u00f0 e = G0 \u2212 G0 ih\u03b1\u03b2\u03a0\u03b1\u2207\u03b2 h k \u2212 1 2 g0\u03bcBBvFv G0, i Fv = \u03c3v + 2 g\u03bcB 1,v \u03a3 e (1) (2) (1) (2) (3) (1) (2) (3) (4) (5) (6) Problems 447 was defined in Eq. (12.147) and is the renormalized spin vertex of the electron\u2013electron interaction. 13.19. From Eqs. (12.123) and (13.149), the quasiparticle contribution to the thermodynamic potential is obtained. After evaluating the trace and contour integration, show that \u03a9qp = 1 2 \u03bcBBv \u2211 nk\u03c1m\u03c1\u2032,m \u2260 n \" gFv n\u03c1,n\u03c1\u2032 + ie \u0127c\u03bcB \u2208\u03b1\u03b2v \u03b2 m\u03c1\u2032,n\u03c1 \u03a0 \u03b1 n\u03c1,m\u03c1\u2032\u03a0 Emn f, Enk\u03c1\u00de \u00f0 # where f Enk\u03c1\u00de \u00f0 is the Fermi distribution function for an electron in band n and spin \u03c1, and \u03a0\u03b1 n\u03c1,m\u03c1\u2032 = d3ru&nk\u03c1\u03a0\u03b1umk\u03c1\u2032 Zcell Emn = Em\u00f0 k \u00de k \u2212 En\u00f0 \u00de : 13.20. The contribution of correlation to magnetization is \u2202\u03a9corr \u2202Bv = Simplifying Eq. (1) with the use of Eqs. (12.127), (12.128), (13.145), (13.147), and the Luttinger\u2013Ward (Ref. 16) prescription for frequency summation, corr = \u2212 G\u03bel : Mv Tr e 1 \u03b2 \u2202 \u03a3\u03bel \u2202Bv e 1 \u03b2 \u2211 \u03bel 1 \u03bel \u2212 ^H0\u00de \u00f0 m = 1 2i\u03c0 Tr 1 \u03be \u2212 ^H0\u00de Z\u03930 \u00f0 m f \u03be \u00f0 \u00de d\u03be,", " (1) (2) (3) (1) (2) where \u03930 encircles the real axis in a clockwise direction, and f function, show that \u03be \u00f0 \u00de is the Fermi distribution e 13.21. The many-body Hamiltonian can be written as Mv corr = \u2211 nk\u03c1 1,v n\u03c1,n\u03c1\u2032 f \u03a3 Enk\u03c1\u00de : \u00f0 ^H0 = \u2211 k\u03c1 \u03b5kc\u2020 k\u03c1ck\u03c1 + 1 2 \u2211 qkk\u2032\u03c1\u03c1\u2032 V q \u00de \u00f0 k+q,\u03c1c\u2020 c\u2020 k\u2032\u2212q,\u03c1\u2032ck\u2032,\u03c1\u2032ck,\u03c1, where c\u2020 q vector k and spin \u03c1, and V \u00de Ek\u03c1 can be obtained by using the mean-field approximation, k\u03c1 and ck\u03c1 are the creation and annihilation operators for an electron with wave is the Fourier transformation of the Coulomb interaction. \u00f0 Ek\u03c1 = 1 N ck\u03c1, ^H0$ \u00bd, c\u2020 k\u03c1 : o+ * n (3) (1) (2) 448 CHAPTER 13 Magnetic Ordering From Eqs. (1) and (2), show that Ek\u03c1 = \u03b5k \u2212 1 N \u2211 q V q f \u03b5k\u2212q,\u03c1\u00de \u03b5k\u03c1\u00de \u00f0 \u00f0 \u2212 f \u03b5k,\u2212\u03c1\u00de$ : \u00f0 13.22. Show that for a uniform system below Tc, Hence, show by using Eq. (13.195) that the specific heat is G = Z dr g0 \u2212 \u00bd a2/4b \u00f0 \u00de$ : C = C0 + T \u00bd\u00f0 Z \u03b12/2b \u2212 \u03b12 \u00bd T \u2212 Tc\u00de \u00f0 \u00de 2/4b2 db/dT $ $ dr: T < Tc\u00de \u00f0 (3) (1) (2) References 1. Anderson PW. Localized magnetic states in metals. Phys Rev 1961;124:41. 2. Ashcroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 3. Callaway J. Quantum theory of the solid state. New York: Academic Press; 1976. 4. Gunnarson O. Band model for magneticism of transition metals in the spin-density-funcional formalism. J Phys F, Met Phys 1976;6:587. 5. Heisenberg W. On the theory of ferromagnetism. Z Physik 1926;49:619. 6. Heitler W, London F. Interaction of neutral atoms and homopolar binding in quantum mechanics. Z Physik 1927;44:455. 7. Holstein T, Primakoff H. Field dependence of the intrinsic domain magnetization of a ferromagnet. Phys Rev 1940;58:1098. 8. Hubbard J. Electron correlations in narrow energy bands. Proc R Soc Lond 1963;A277:237. 9. Janak JF. Uniform susceptibilities of metallic elements. Phys Rev B 1976;16:255. 10. Kasuya T. A theory of metallic ferro- and antiferromagnetism on Zener\u2019s model. Prog Theor Phys (Kyoto) 1956;16:45. 11. Kittel C. Introduction to solid state physics. New York: John Wiley & Sons; 1976. 12. Kittel C. Quantum theory of solids. New York: John Wiley & Sons; 1987. 13. Kittel C. Indirect Exchange Interaction in Metals. Solid State Phys 1968;22:1. 14. Kondo J. Resistance minimum in dilute magnetic alloys. Prog Theor Phys 1964;32:37. 15. Landau LD. On the theory of phase transition. Phhysikalliische Zeitschraft der Sowjetunion 1937;11:26 and 545. 16. Luttinger JM, Ward JC. Ground-state energy of a many fermion system II. Phys. Rev. 1960;118:1417. 17. Marder MP. Condensed matter physics. New York: John Wiley & Sons; 2000. 18. Misra SK, Misra PK, Mahanti SD. Many-body theory of magnetic susceptibility of electrons in solids. Phys Rev B 1982;26:1903. 19. Misra PK, series editor, Blackman J, editor. Metallic nanoparticles. Amsterdam: Elsevier; 2008. 20. Nagaoka Y. Self-consistent treatment of Kondo effect in dilute alloys. Phys Rev 1965;138:A1112. 21. Neel L. Magnetic properties of ferrites: Ferromagnetism and sntiferromagnetism. Annales de Physique", " 1948;3:137. 22. Rado GT, Suhl H, editors. Magnetism. New York: Academic Press; 1963. 23. Ruderman MA, Kittel C. Indirect exchange coupling of nuclear magnetic moments by conduction electrons. Phys Rev 1954;96:99. 24. Schwinger J. Quantum theory of angular momentum. New York: Academic Press; 1965. References 449 25. Stoner EC. Magnetism in matter. London: Methuen; 1934. 26. Stoner EC. Collective electron ferromagnetism II. Energy and specific heat. Proc R Soc Lond A 1939;169:339. 27. Stoner EC, Wohlfarth EP. A Mechanism of Magnetic Hysteresis in Heterogeneous Alloys. Phil Trans Lond A 1948;240:599. 28. Tripathi GS, Das LK, Misra PK, Mahanti SD. Theory of spin-orbit and many-body effects on the Knight shift. Phys. Rev. B 1982;25:3091. 29. Tripathi GS, Misra PK. Many-body theory of magnetization of Bloch electrons in te presence pf spin-orbit interaction. J Magn Mag Mater 2010;322:88. 30. Vosko SH, Pedrew V. Theory of the spin susceptibility of an inhomogeneous electron gas via the density functional formalism. Can J Phys 1975;53:1385. 31. Yosida K. Many-body theory of magnetization of Bloch electrons in the presence of spin-orbit interaction. Phys Rev 1957;106:893. 32. Ziman JM. Principles of the theory of solids. Cambridge: Cambridge University Press; 1972. Superconductivity CHAPTER 14 CHAPTER OUTLINE 14.1 Properties of Superconductors............................................................ 452 14.1.1 Introduction....................................................................... 452 14.1.2 Type I and Type II Superconductors................................................. 453 14.1.3 Second-Order Phase Transition..................................................... 454 14.1.4 Isotope Effect..................................................................... 454 14.1.5 Phase Diagram.................................................................... 454 14.2 Meissner\u2013Ochsenfeld Effect............................................................... 455 14.3 The London Equation..................................................................... 455 14.4 Ginzburg\u2013Landau Theory.........................................................", "........ 456 14.4.1 Order Parameter................................................................... 456 14.4.2 Boundary Conditions............................................................... 457 14.4.3 Coherence Length................................................................. 457 14.4.4 London Penetration Depth.......................................................... 458 14.5 Flux Quantization........................................................................ 459 14.6 Josephson Effect........................................................................ 460 14.6.1 Two Superconductors Separated by an Oxide Layer................................... 460 14.6.2 AC and DC Josephson Effects...................................................... 462 14.7 Microscopic Theory of Superconductivity................................................... 462 14.7.1 Introduction....................................................................... 462 14.7.2 Quasi-Electrons................................................................... 463 14.7.3 Cooper Pairs...................................................................... 464 14.7.4 BCS Theory....................................................................... 466 14.7.5 Ground State of the Superconducting Electron Gas................................... 466 14.7.6 Exc", "ited States at T = 0............................................................. 469 14.7.7 Excited States at T \u2260 0............................................................. 470 14.8 Strong-Coupling Theory.................................................................. 472 14.8.1 Introduction....................................................................... 472 14.8.2 Upper Limit of the Critical Temperature, Tc.......................................... 472 14.9 High-Temperature Superconductors........................................................ 473 14.9.1 Introduction....................................................................... 473 14.9.2 Properties of Novel Superconductors (Cuprates)...................................... 474 Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00014-1 \u00a9 2012 by Elsevier Inc. All rights reserved. 451 452 CHAPTER 14 Superconductivity 14.9.3 Brief Review of s-, p-, and d-wave Pairing........................................... 474 14.9.4 Experimental Confirmation of d-wave Pairing......................................... 476 14.9.5 Search for a Theoretical Mechanism of High Tc Superconductors...................... 481 Problems.................................................................................... 481 References.................................................................................. 485 14.1 PROPERTIES OF SUPERCONDUCTORS 14.1.1 Introduction The main source of electrical resistance in a metal is the electron\u2013phonon interaction caused by the scattering of the electrons due to vibrations of the ions. This is the lattice excitations in bulk metals,, where d is the amplitude of the vibrations d/L \u226a 1 which corresponds", " to small ionic vibrations and L is the lattice period. This is described as acoustic quanta (phonons) with energies, \u00de \u00f0 \u03b5i ph = \u0127\u03a9i q, (14.1) \u00de = 2\u03c0/\u03bb, where \u03bb is the phonon\u2019s wavelength. where q = \u0127k is the momentum of the phonon, and Here, i corresponds to the various phonon branches (longitudinal, transverse, and optical). The electron\u2013lattice interaction, i.e., the energy exchange between the electrons and lattice, is due to the radiation and adsorption of phonons and is known as the electron\u2013phonon interaction. As the temperature is lowered, the amplitude of the ions becomes smaller, and the electrical resistance is reduced. \u00f0 k j j But the resistivity of a normal metal does not become zero at zero temperature because there are other sources of electrical resistance due to the presence of impurities and imperfections in the crystal structure. As the temperature is reduced, there is a residual resistivity \u03c10 at absolute zero that is approximately 1% of the resistivity of a pure sample. The temperature around which the resistivity is constant is approximately 1\u00b0 K. The resistivity of a metal can be expressed as \u03c1 \u03c10 0 FIGURE 14.1 Variation of resistivity of a normal metal. = \u03c10 + AT 5, T \u03c1 \u00f0 \u00de (14.2) where \u03c10 is the residual resistivity, and AT 5 is the term arising from electron\u2013phonon scattering. The variation of resistivity with temperature for a nonsuperconducting metal is shown in Figure 14.1. The phenomenon of superconductivity was discovered by Kammerlingh Onnes in 1911.15 He measured the resistivity of platinum and found that the resistivity followed a curve similar to the curve shown in Figure 14.1. However, when he performed a similar experiment with a sample of mercury, he found that the resistance of the sample dropped sharply to a value of zero at 4.2\u00ba K. The metals of which the resistivity becomes zero at a particular temperature are known as superconductors. The temperature at T 14.1 Properties of Superconductors 453 or below which a metal becomes a superconductor is called the critical temperature, Tc. Thus, there are two types of metals: some that become superconductors at or below Tc and others that continue to have resistivity even at 0\u00ba K. The variation of resistivity with temperature for a superconducting metal is shown in Figure 14.2. \u03c1 14.1.2 Type I and Type II Superconductors Tc T FIGURE 14.2 The experimental results displayed in the previous section were for metals when there was no external magnetic field. However, when an external magnetic field H is applied, the superconducting state is destroyed above a critical external magnetic field Hc: This type of a superconductor is called a type I superconductor. From experiment results, one can obtain a relationship between the critical magnetic field HT, which destroys superconductivity at a given temperature T, Variation of resistivity with temperature of a superconducting metal. 2 HT = Hc 1 \u2212 T \" Tc! \" : # (14.3) This experimental result is plotted in Figure 14.3. One can distinguish type I and type II superconductors in the following manner. In type I superconductors, the diamagnetism grows linearly with the magnetic field until the critical value is reached. This is shown in Figure 14.4. \u00b50M H Hc Normal Superconducting Tc T FIGURE 14.3 Variation of external magnetic field H against temperature T in a type I superconductor. FIGURE 14.4 Type I superconductor. Bc B0 454 CHAPTER 14 Superconductivity In type II superconductors, the diamagnetization is linear with the external field up to a value Bc1, but the magnetization decreases when the flux begins to penetrate the metal. This is shown in Figure 14.5. \u00b50M 14.1.3 Second-Order Phase Transition The transition from the normal (N) state to the superconducting (S) state is known as a second-order phase transition. In the first-order phase transitions, the heat capacity is continuous, whereas in second-order phase transitions, the heat capacity is discontinuous. Figure 14.6 schematically shows the variation of heat capacity from a normal state to a superconducting state. 14.1.4 Isotope Effect It was discovered in 1950 that the different isotopes of the same element possess different transition temperatures to enter the superconducting state, TcM\u03b1 i = constant, (14.4) where Mi is the isotopic mass of the same element. Usually, \u03b1 \u2248", " 1/2. This was a strong indication that the interaction of the electrons with ions in the lattice was important for superconductivity. 14.1.5 Phase Diagram Superconductivity is destroyed by application of a large magnetic field. In addition, superconductivity is destroyed if the current exceeds a \u201ccritical current.\u201d This is known as the Silsbee effect. To summarize, superconductivity depends on three factors: (a) the temperature T, (b) the external magnetic field H, and (c) the current density j. This type of dependence on three different experimental parameters is shown as a three-dimensional curve in Figure 14.7. Bc1 Bc2 FIGURE 14.5 Type II superconductor. V 120 ) 80 40 S N 0 1 2 3 T(K) 4 5 6 FIGURE 14.6 Second-order phase transition. J Jc Normal Superconducting H Hc Tc T FIGURE 14.7 Critical surface separating the superconducting and normal states. 14.2 MEISSNER\u2013OCHSENFELD EFFECT Meissner and Ochsenfeld observed that an external magnetic field (as long as it is not too large) cannot penetrate inside a superconductor.22 Thus, a superconductor behaves as a perfect diamagnet. If a normal metal is cooled below the critical temperature in a magnetic field, the magnetic flux is expelled abruptly. The transition to the superconducting state in a magnetic field is accompanied by the surface currents necessary to cancel the magnetic field inside the specimen. This scenario is shown in Figure 14.8. 14.3 The London Equation 455 \u2192 B \u2192 J FIGURE 14.8 Flux lines cannot pass through the current loop in a superconductor. \u00f0 T 14.3 THE LONDON EQUATION London and London explained the Meissner\u2013Ochsenfeld effect by adopting the two-fluid model of Gorter and Casimir.18 The basic assumption of their model is that there are two types of electrons, (density of superconducting electrons) and n (density of total number of conducting electrons), ns in a metal in the superconducting state. At temperatures T < Tc, only a fraction ns /n of the total Tc \u00b7 n \u2212 ns cannot carry number of conduction electrons can carry a supercurrent. ns an electric current without dissipation. In a small electric field, the normal electrons that flow parallel to the superconducting electrons are inert and therefore ignored while discussing the motion of electrons. In the presence of an electric field E, \u2192 0 as _vs = \u2212eE, (14.5) where vs is the mean velocity of superconducting electrons. Because the current density j = \u2212evsns, Eq. (14.5) can be rewritten in the alternate form From Eq. (14.6) and Faraday\u2019s law of induction, dj dt = nse2 m E: we obtain \u2207! \u00d7 E = \u2212 1 c \u2202B \u2202t, \u2202 \u2202t! \u2207! \u00d7 j + nse2 mc B = 0: \" (14.6) (14.7) (14.8) It is easy to show that Maxwell\u2019s equation (neglecting the displacement current D as well as replacing H with B because j is the mean microscopic current) can be written as (Problem 14.1) \u2207! \u00d7 B = 4\u03c0 c j: (14.9) 456 CHAPTER 14 Superconductivity Because Eqs. (14.8) and (14.9) are consistent with any static magnetic field and the Meissner\u2013Ochsenfeld (Ref. 22) effect requires that the magnetic field be expelled in the superconducting state, London and London (Ref. 18) postulated that \u2207! \u00d7 j +! nse2 mc B = 0: \" Using the vector identity, we obtain from Eqs. (14.9) through (14.11), \u2207! \u00d7 \u2207! \u00d7 V \u00de \u00f0 = \u2207! \u00f0 \u2207!. V \u00de \u2212 \u22072V, and \u22072B = 4\u03c0nse2 mc2 B \u22072j = 4\u03c0nse2 mc2 j: (14.10) (14.11) (14.12) (14.13) Eqs. (14.12) and (14.13) predict that the magnetic fields and currents in a superconductor can exist within a layer of thickness \u03bbL, known as the London penetration depth, \u03bbL = mc2 4\u03c0nse2! 1 2 : \" One can rewrite \u03bbL in the alternate form \u03bbL = 41:9 3/2 rs a0! \" 1/2 n ns! \" A\u02da : (14.14", ") (14.15) n \u2248 ns the surface currents that screen out the applied magnetic field occur Thus, when T << Tc, within a layer of 102 \u2212 103 \u00c5. The magnetic field drops continuously to zero within this layer. It has indeed been observed experimentally that the field penetration is incomplete in superconducting films that are thinner than \u03bbL: \u00f0 \u00de 14.4 GINZBURG\u2013LANDAU THEORY 14.4.1 Order Parameter Landau and Ginzburg17 described the superconductivity state through a position-dependent order para\u00de, which describes the macroscopic properties of a superfluid condensate.17 The meter \u03a8 r \u00dej \u00de is a wave function that vanishes at Tc. Landau = superfluid density ns and Ginzburg (Ref. 17) guessed that to be able to study the magnetic and thermodynamic properties of superconductors, the total free energy Fs with respect to its value in the normal state Fn should be 2, suggesting that \u03a8 \u00dej ei\u03d5 Fs = Fn + dr 8\u03c0 B2 + 1 2m# \u0127 i # h # # \u2207! + e14.16) 14.4 Ginzburg\u2013Landau Theory 457 Here, the first two terms are from Landau\u2019s general theory of second-order phase transition borrowed from their theory for liquid 4He. According to their phenomenological model, one assumes that the comfor N electrons is known. If an extra electron is added, plete ground-state wave function \u03c8 N\u00f0 one knows \u03c8 N+1\u00f0 r1\u2026:::rN+1. We define r1\u2026\u2026rN \u00de \u00de = \u03a8 r \u00de \u00f0 Z dNr\u03c8 #N\u00f0 r1\u2026\u2026rN \u03c8 N+1\u00f0 \u00de r1\u2026:::rN, r : \u00de (14.17) Far away from r, \u03a8N and \u03a8N+1 coincide, but in its neighborhood \u03a8N+1 accommodates one extra particle. The last two terms are due to the fact that the superconducting electron wave function might interact with the vector potential like a single macroscopic particle of effective charge e# (it was later found that e# = 2e, consistent with Cooper pairs (Ref. 7)). Minimizing Eq. (14.16) with respect to A, we obtain where = j r \u00f0 \u00de \u2212e#\u0127 2im# \u2207! \u00d7 B = 4\u03c0 c j, \u03a8# \u2207!\u03a8 \u2212 \u03a8 \u2207!\u03a8# h i (14.18) (14.19) 2 \u2212 e# m#c A\u03a8#\u03a8: Minimizing Eq. (14.16) with respect to \u03a8#, by first integrating such that all spatial derivatives act on \u03a8 and then taking functional derivatives with respect to \u03a8#, we obtain (Problem 14.3 2m# $ \u0127 i & \u2207! + e# c A 2 % \u2019 \u03a8 = 0: (14.20) Eqs. (14.19) and (14.20) are known as the Landau\u2013Ginzburg (Ref. 17) equations. 14.4.2 Boundary Conditions No current can flow out of the boundary when a superconductor is in contact with a vacuum. Thus, one can write e# c The additional boundary condition imposed by Landau and Ginzburg17 is that \u03a8 = 0 was not considered as an acceptable condition, because in that case, it would be impossible to obtain solutions for thin superconducting films. ^n. \u0127 i & (14.21) \u2207! + = 0: A \u2019 14.4.3 Coherence Length The coherence length \u03be specifies the scale for variation of the order parameter \u03a8. The coherence length is defined by \u03be = \u01272 2m#j $ \u03b1 1 2, % j (14.22) 458 CHAPTER 14 Superconductivity where \u03b1 is defined in Eq. (14.20). If the external magnetic field vanishes, Eq. (14.20) can be rewritten as 2 = \u03a82 0 = \u2212 \u03b1 \u03b2 \u03a8 j j or \u03a8 j j 2 = 0: (14.23) Because \u03b2 is positive (otherwise, F is minimized when \u03a8 uniform superconducting state. The free energy per unit volume is given by Eq. (14.16), \u221e), \u03b1 < 0, which corresponds to a! It can be easily shown that Fs \u2212 FN = \u2212 \u03b12 2\u03b2 : \u0394F = FN \u2212 Fs = H2 c 8\u03c0 : From Eqs. (14.24) and (14.25), we obtain H2", " c = 4\u03c0\u03b12 \u03b2 : We can scale the order parameter for superconductors, \u03c8 = \u03a8 \u03a80 : From Eqs. (14.20), (14.22), and (14.27), we obtain for A = 0, 2 = 0: \u2212\u03be2\u22072\u03c8 \u2212 \u03c8 + \u03c8 \u03c8 j j (14.24) (14.25) (14.26) (14.27) (14.28) Thus, the coherence length \u03be is the characteristic scale on which \u03c8 varies (in the absence of a magnetic field). 14.4.4 London Penetration Depth In Eq. (14.19), in the presence of a weak magnetic field, \u03a8 \u2248 \u03a80 for x < 0: Thus, we obtain from Eqs. (14.18) and (14.19), and j = c 4\u03c0 2 \u2207! \u00d7 B = \u2212 e# m#c \u03a82 0A \u2207 \u00d7 \u2207 \u00d7 B = \u2212 4\u03c0 c 2 e# m#c 0B = \u2212 B \u03a82 \u03bb2 L : (14.29) (14.30) Because m# = 2m, e# = 2e, and \u03a82 Eq. (14.14). The parameter 0 = \u2212 \u03b1 \u03b2, \u03bbL is the London penetration depth defined in is the only parameter in Landau\u2013Ginzburg theory. \u03ba = \u03bbL/\u03be = m#c e\u0127 \u03b2 2\u03c0 \ufb03\ufb03\ufb03\ufb03\ufb03 r (14.31) 14.5 Flux Quantization 459 It can be easily shown that the surface energy is positive if \u03ba < 1/ 2p. The magnetic flux can easily penetrate a superconductor if the surface energy is negative and can \ufb03\ufb03\ufb03 form type II superconductors, which are interlocking regions of normal and superconducting metal. Type I superconductors are those for which \u03ba < 1/ 2p and the flux cannot penetrate. 2p and negative if \u03ba > 1/ \ufb03\ufb03\ufb03 Consider a magnetic field H > Hc so that \u03a8 = 0 because superconductivity is destroyed. When the field is gradually lowered, considering only the first-order terms in \u03a8 (because j = 0), \ufb03\ufb03\ufb03 1 2m# \u03b1 + $ & \u0127 i \u2207! + e# c A % \u2019 2 \u03a8 = 0: (14.32) Eq. (14.32) is an eigenvalue equation of a charge (e# = 2e) in a constant magnetic field, the lowest energy is \u0127\u03c9c/2, and because Hc2 is the largest magnetic field that permits a solution, From Eqs. (14.32) and (14.33), or \u03c9c = e#Hc2 m#c : \u2212\u03b1\u03a8 = \u0127e#Hc2 2m#c \u03a8 = \u03b1 j j \u0127eHc2 m#c : From Eqs. (14.26), (14.31), and (14.35), we obtain Hc2 Hc = 2p \u03ba: \ufb03\ufb03\ufb03 14.5 FLUX QUANTIZATION In Eq. (14.18), if we substitute where \u03d5 r \u00de \u00f0 is real, we obtain from Eqs. (14.18) and (14.37) = \u03a80ei\u03d5 \u00f0 r \u00de, \u03a8 r \u00de \u00f0 j = \u2212 \u03a82 0 m# 2 e# c! A + e#\u0127 \u2207!\u03d5, \" which can be rewritten in the alternate form \u2207!\u03d5 = \u2212 1 \u0127 m# e#\u03a82 0 j + e# c A :! (14.33) (14.34) (14.35) (14.36) (14.37) (14.38) (14.39) If we consider a superconductor in the shape of a ring (Figure 14.9), which shows a path encir- cling the aperture but lying well within the interior, j. dl = 0: I (14.40) 460 CHAPTER 14 Superconductivity From Eqs. (14.23) and (14.24), we obtain \u2207!\u03d5. dl = \u2212 I From Stoke\u2019s theorem, I e#A \u0127c. dl: (14.41) A. dl = \u2207! \u00d7 A. dS = B. dS = \u03a6, (14.42) FIGURE 14.9 A path encircling the aperture ring", " of superconducting material. I I I where \u03a6 is the flux enclosed by the ring. We note that because the magnetic field cannot penetrate a superconducting material, the enclosed flux is independent of the choice of the path. In addition, the order parameter \u03a8 is single-valued, and hence, its phase changes by 2\u03c0n (n is an integer) when the ring is encircled. Thus, we have From Eqs. (14.41) through (14.43), we obtain \u2207!\u03d5. dl = 2\u03c0n: I Defining a fluxoid or a flux quantum, \u2212e# \u0127c \u03a6 = 2\u03c0n: \u03a60 $ hc 2e = 2:0679 G cm2, we obtain from Eqs. (14.43) and (14.44) (14.43) (14.44) (14.45) = 2ne e# The integrated magnetic flux that penetrates a hole through a superconductor is quantized in units of \u03a60. Thus, e# = 2e, confirming the fact that the microscopic theory of superconductivity 2e/e#\u00de \u00f0 involves pairing of electrons. Flux quantization was experimentally observed independently by Deaver and Fairbank8 and Doll and Nabauer.12 (14.46) \u03a60: \u03a6 j j 14.6 JOSEPHSON EFFECT 14.6.1 Two Superconductors Separated by an Oxide Layer In 1962, Josephson (Refs. 13, 14) predicted that if two superconductors were separated by a small strip of nonsuperconducting material (shown in Figure 14.10), the wave function would oscillate because it would interfere with itself. The two superconductors would interact through their small residues while decaying through the barrier. Due to tunneling, the change in energy \u0394\u03b5 is given by \u0394\u03b5 = dr f r \u00f0 \u00de & Z r \u03a8#1\u00f0 \u00de \u03a82 r \u00de \u00f0 + \u03a81 r \u00f0 \u03a8#2\u00f0 \u00de r : \u00de \u2019 (14.47) 14.6 Josephson Effect 461 At some reference point in the bulk, if the macroscopic wave functions are \u03a81 and \u03a82, \u0394\u03b5 = \u2208 \u03a8#1\u03a82 + \u03a81\u03a8#2\u00de \u00f0 where \u2208 is a very small quantity. The Schrodinger equations for \u03a81 and \u03a82 can be written as (14.48), i\u0127 \u2202\u03a81 \u2202t = \u03b51\u03a81 + \u2208 \u03a82 (14.49) S1 S2 Oxide layer FIGURE 14.10 The oxide layer between two superconductors S1 and S2. and i\u0127 \u2202\u03a82 \u2202t = \u2208\u03a81 + \u03b52\u03a82: The wave functions are taken of the form \u03a8j = njp ei\u03d5j, where nj Eq. (14.51) in Eqs. (14.49) and (14.50), we obtain j = 1 or 2 are the superconducting electron densities n = \ufb03\ufb03\ufb03\ufb03 \u00de \u00f0 (14.50) (14.51) n1n2p and n1 \u2248 n2: Substituting \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 i\u0127 _n1 n1p! \ufb03\ufb03\ufb03\ufb03\ufb03 + i n1p _\u03d51 ei\u03d51 = \" \ufb03\ufb03\ufb03\ufb03\ufb03 \u03b51 n1p ei\u03d51 + \u2208 n2p ei\u03d52 \u00f0 \u00de \ufb03\ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03 and _n2 n2p i\u0127! + i n2p _\u03d52 ei\u03d52 = \u03b52 n2p ei\u03d52 + \u2208 n1p ei\u03d51 : \u00de \u00f0 \" \ufb03\ufb03\ufb03\ufb03\ufb03 It can be easily shown from Eqs. (14.52) and (14.53) that \ufb03\ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03 and _n1 = 2 \u2208n \u0127 \u03d52 sin \u2061 \u00f0 \u2212 \ufffd", "\ufffd1\u00de = \u2212 _n2 = j e# _\u03d52 \u2212 _\u03d51 = 1 \u0127 \u00f0 \u03b51 \u2212 \u03b52 = \u00de e# \u0127 \u00f0 V2 \u2212 V1 : \u00de (14.52) (14.53) (14.54) (14.55) function phase gauge-invariant, we have to add the line integral e# \u0127c We note that the phase of a wave function changes when there is a magnetic field. To make a wave dl. A, taken from some arbitrary reference point, to the phase. Incorporating this and taking into account the fact that e# = 2e, we obtain 0 R j = j0 sin \u2061 \u03d52 0 \u2212 \u03d51 + @ 2e \u0127c 2 Z 1 dl. A 1 A (14.56) 462 CHAPTER 14 Superconductivity and \u2212 1 \u0127 \u00f0 \u03b52 \u2212 \u03b51 \u00de = 2eV \u0127 = \u2202 \u2202t \u03d52 \u2212 \u03d51 + 2e \u0127c 0 @ 2 dl. A Z 1 (14.57) 1 A Here, the energy difference \u03b52 \u2212 \u03b51 is the difference in energies of electron pairs, and V is the difference of the voltage between the two superconductors. 14.6.2 AC and DC Josephson Effects If we place two superconductors at different volthen \u03b51 \u2212 \u03b52 \u2260 0. Because tages in contact, \u2212 \u03d51 drifts in time, n1 and n2 oscillate about \u03d52 their mean values at 484 MHz/\u03bcV. This is known as the AC Josephson effect. However, if \u03b51 \u2212 \u03b52 = 0, \u03d51 : \u2192 A m n i J 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 \u22124 \u22123 \u22122 \u22121 1 0 H in Gauss \u2192 2 3 4 FIGURE 14.11 Maximum zero-voltage current in an Sn-SnO-Sn junction at T = 1.9\u00ba K (sketch of results of R.C. Jaklevic (Ref. 23)). \u2212 \u03d52 is independent of time. Because \u03d51 \u2260 \u03d52, there is a steady current flow according to Eq. (14.56). This phenomenon is known as the DC Josephson effect. If \u2212 \u03d51 adjusts to a nonzero value. The maxithe electrons are injected into a Josephson junction, \u03d52 \u2212 \u03d51 is \u03c0/2: Thus, there is a current flow in the absence of any voltage difference mum value of \u03d52 (see Figure 14.11). This is a unique property of the superconductors and is a significant consequence of the Josephson effect. 14.7 MICROSCOPIC THEORY OF SUPERCONDUCTIVITY 14.7.1 Introduction In Figure 8.14g, the graph shows an electron emitting a virtual phonon that is absorbed by a second electron. The lattice is deformed (polarized) in the vicinity of the first electron. A second electron near this polarized cloud experiences a force of repulsion or attraction that is independent of the Coulomb interaction between the two electrons. This is termed an effective electron\u2013electron interaction via virtual phonons and is responsible for superconductivity under certain conditions. In the graph of Figure 8.14g, there are two possible intermediate states: electron k emits a phonon \u2212q, which is absorbed by electron k\u2032; or electron k\u2032 emits a phonon q, which is absorbed by electron k. In Eq. (8.165), we derived an expression for electron\u2013phonon coupling for longitudinal acoustic (LA) phonons, ^Hel\u2212ph = \u2211 kqs Mq \u00f0 ^a\u2020 \u2212q + ^aq ^c\u2020 k+q,s^ck,s: \u00de (14.58) We note that the direction of the spin is not changed by this interaction. 14.7 Microscopic Theory of Superconductivity 463 14.7.2 Quasi-Electrons We now consider a system of electrons and phonons and write a Hamiltonian of the form ^H = \u2211 k,s E k ^c\u2020 ks^cks + \u2211 \u00de q \u00f0 \u0127\u03c9q^a\u2020 q^aq + \u2211 kq Mq \u00f0 ^a\u2020 \u2212q + ^aq ^c\u2020 k+q,s^ck,s = ^H0 + ^H1, \u00de (14.59) where ^H1 is the electron\u2013phonon interaction term. We use a canonical transformation, ^HS = e\u2212^S ^He^S = ^H", "0 + ^H1 + \u00f0 ^H0, ^S \u00bd &\u00de + 1 2 \u00bd\u00f0 ^H1 + ^H0, ^S +, ^S & & \u00bd ^H1, ^14.60) If we choose ^S such that ^H0, ^S (14.61) the electron\u2013phonon interaction ^H1 is eliminated apart from a higher-order term. We choose ^S of the form ^H1 + = 0, & \u00bd ^S = \u2211 kqs\u2032 Mq \u00f0 \u03b1^a\u2020 \u2212q + \u03b2^aq ^c\u2020 k+q,s\u2032^ck,s\u2032: \u00de From Eqs. (14.61) and (14.62), it can be easily shown that (Problem 14.5) \u03b1\u2212 \u2212 \u0127\u03c9q and \u03b2\u2212 + \u0127\u03c9q: \u00de (14.62) (14.63) (14.64) From Eqs. (14.60) and (14.61), the next-order interaction term is 1 2 \u00bd that contain operator products of the form ^H1, ^S, which is a sum of terms & k\u2032+q\u2032,s\u2032^ck\u2032,s\u2032^c\u2020 One can show (Problem 14.6) that because q\u2032 = \u2212q (from momentum conservation), only one out of all the possible combinations does not contain any phonon operators, \u00b1q^a\u00b1q\u2032^c\u2020 ^a\u2020 k+q,s^ck,s: (14.65) when k+q,s^c\u2020 ^c\u2020 k\u2032\u2212q,s\u2032^ck\u2032,s\u2032^ck,s, k\u2032 \u2260 k, k + q: (14.66) (14.67) From. Eqs. (14.60), (14.62), and (14.66), the explicit form for this interaction is (Problem 14.7) ^Heff = 1 2 \u2211 k,k\u2032,q,s,s\u2032 j Mq 2 j \u00f0 \u03b1 \u2212 \u03b2 k+q,s^c\u2020 ^c\u2020 \u00de k\u2032\u2212q,s\u2032^ck\u2032,s\u2032^ck,s: (14.68) Substituting the values of \u03b1 and \u03b2 from Eqs. (14.63) and (14.64) in Eq. (14.68), we obtain (Problem 14.8) < f ^Heff j i> = j 1 2 < f \u2211 k,k\u2032,q,s,s\u2032 Vkq^c\u2020 j k+q,s^c\u2020 k\u2212q,s\u2032^ck\u2032s\u2032^ck,s i>, j (14.69) 464 CHAPTER 14 Superconductivity where Vkq = 2 Mq j \u2212 E 2\u0127\u03c9q 2 \u2212 k j E k + q \u0127\u03c9q 2, \u00de (14.70) \u00de& f > are the initial and final states, respectively. Vkq is essentially the Fourier coefficient < \u0127\u03c9q, Vkq is negative and the interaction is and of the effective interaction. When attractive, whereas the interaction is repulsive when Vkq is positive. i> and j k + q \u2212 E \u00dej Omitting all the terms in the transformed Hamiltonian that contains phonon creation or annihila- tion operators, we obtain from Eqs. (14.59) and (14.70), ^HS = \u2211 ks E k ^c\u2020 k,s^ck,s + \u2211 \u00de k,k\u2032,q,s,s\u2032 j \u00f0 Mq 2 j k + q E \u00bd \u00f0 \u00de \u0127\u03c9q k \u2212 E \u00f0 \u00de& 2 ^c\u2020 k+q,s^c\u2020 k\u2032\u2212q,s\u2032^ck\u2032,s\u2032^ck,s: (14.71) 2 \u2212 \u0127\u03c9q \u00f0 \u00de 14.7.3 Cooper Pairs We consider a state containing N noninteracting electron gas, which fills the Fermi sphere in k\u2212 space. The ground state G > is the filled Fermi sphere. We introduce two electrons k1 and k2 into this system and take as the interaction between these electrons the positive part of Vkq (Eq. 14.70), which is repulsive. Thus, the interaction involving phonon exchange will take place only when \u2264 \u0127\u03c9q. The wave function of the electron pair can be written as \u2212 \u00dej \u03c812 = \u2211 k1k2s1s2 As1s2\u00f0 k1, k2 ^c\u2020 k1s1 \u00de ^c\u2020 k2s2 j G >: (14.72) The summation over k", "1 and k2 is carried out subject to the condition that K = k1 + k2 = constant in order to form a state with definite momentum. This scenario is shown in Figure 14.12. If we consider two electrons just above the Fermi sphere, an interaction will occur only when \u2264 EF + \u0127\u03c9q. The regions in k space that are summed in Eq. (14.72) are shown in the shaded k1 E areas because K = k1 + k2. These regions are at a maximum when K = 0. \u00f0 \u00de Assuming that K = 0, and the electron spins are antiparallel, we can rewrite Eq. (14.72) as \u03c8 12 = \u2211 k b k \u00de \u00f0 ^c\u2020 k,sc\u2212k,\u2212s G>: j (14.73) We will explain later why only antiparallel spins are included in Eq. (14.73). We make a further approximation in Eq. (14.70) by considering Vkq to be a constant in the range of attractive interaction (Vkq = \u2212U), and Vkq = 0 otherwise. In fact, from Eq. (14.71), we obtain the condition EF EF + h\u03c9 q k1 kF k2 K E j U \u2260 0, when k + q \u00de The Debye frequency \u03c9D is the maximum value of \u03c9q in the Debye approximation. \u2264 \u0127\u03c9q: (14.74) \u00dej k \u00f0 \u00f0 \u2212 E FIGURE 14.12 Here, k vectors of two interacting electrons (K = k1 + k2) that lie within a shell of thickness \u0127\u03c9q above EF. 14.7 Microscopic Theory of Superconductivity 465 Substituting Eq. (14.74) in Eq. (14.71), we obtain k+q,s^c\u2020 ^c\u2020 k,s^ck,s \u2212 U ^c\u2020 k 2 \u00de ^H = \u2211 k,s \u2211 k,q,s E \u00f0 \u2212k\u2212q,\u2212s^c\u2212k,\u2212s^ck,s: (14.75) Here, we note that the factor \u00bd in the second term in Eq. (14.75) is due to the summation over only one spin index. From Eqs. (14.73) and (14.75), we obtain an expression for the energy E, E = <\u03c8 ^H j \u03c8> = 2\u2211 k j E k \u00f0 \u00dej b k \u00f0 2 \u2212 U \u2211 \u00dej k,q b where b k \u00de \u00f0 is obtained by varying E subject to the condition: \u00dej By varying E in Eq. (14.76), we obtain \u2202 \u2202b#k\u2032 h E \u2212 \u03b2\u2211 k\u2032 j from which we have 2 b \u00f0 k\u2033 \u00dej i = 2E b k\u2032 \u00de \u00f0 \u00f0 k\u2032 \u00de \u2212 U\u2211 q b \u00f0 k\u2032 \u2212 q \u00de \u2212 \u03b2b k\u2032 \u00f0 \u00de = 0, k 2E \u2211 k\u2032 b k\u2032 : \u00de \u00f0 (14.76) (14.77) (14.78) (14.79) Here, \u03b2 is Lagrange\u2019s parameter and U \u2260 0 only for the energies EF to EF + \u0127\u03c9q. One can show (Problem 14.9) that if we take the complex conjugate of Eq. (14.79), multiply by b, and sum over k, we obtain an equation that agrees with Eq. (14.76) provided \u03b2 = E. k \u00f0 \u00de Similarly, the sum over b k\u2032 \u00f0 \u00de is over a small number of k\u2032, such that we can write \u2211 k\u2032 k\u2032 b \u00f0 \u00de = B: From Eqs. (14.79) and (14.80), we obtain \u2211 k b k \u00de \u00f0 or = B = \u2211 k E \u00f0 \u00de UB k \u00de \u00f0 \u2212 \u03b2 2E = 1: \u2211 k E 2E U k \u00f0 \u2212 \u03b2 \u00de We note that if we had considered parallel spins in Eq. (14.73), the constant B would have been zero because of the antisymmetry of the spatial part of the wave function. If we take the complex conjugate of Eq. (14.79), multiply by b, and sum over k, we obtain an equation that agrees with Eq. (14.76), provided \u03b2 = E (Problem 14.9). Because U \u2260 0 only in the range EF and EF + \u0127\u03c9q, we can rewrite Eq. (14.82) in", " the alternate form k \u00de \u00f0 \u00de \u00f0 (14.80) (14.81) (14.82) EF+\u0127\u03c9q U Z EF D dx x \u00de \u00f0 2x \u2212 E = 1, (14.83) 466 CHAPTER 14 Superconductivity where D tion and obtain (Problem 14.10) x \u00f0 \u00de is the density of states. We approximate D x \u00de \u00f0 by D \u00f0 EF \u00de in the narrow range of integra- E = 2EF \u2212 2\u0127\u03c9q exp 1 \u2212 exp \u22122/D EF \u00f0 \u00bd \u22122/D EF U \u00de U & \u2248 2EF \u2212 2\u0127\u03c9q exp \u22122/D EF U : (14.84) \u00bd \u00f0 \u00de & \u00bd \u00f0 \u00de & The expression on the extreme right in Eq. (14.84) is valid for weak interactions (small U). We note that we have made the approximation Vkq = \u2212U in the range of attractive interaction and zero elsewhere. Thus, the energy of the electron pair is less than 2EF in the absence of the interaction. It can be shown that all other solutions lead to energies greater than 2EF: Hence, the lowest energy state of the electron pair is a bound state known as a Cooper pair (Ref. 7) that is formed from a pair of electrons with opposite spin and opposite wave vector. The formation of a bound state by the additional electron pair means that when two electrons from states directly below the Fermi surface are excited into states above it, the energy is lowered. Thus, the filled Fermi sphere is unstable, and one can gain energy by combining electrons into Cooper pairs. 14.7.4 BCS Theory The theory of superconductivity was proposed by Bardeen, Cooper, and Schrieffer on the basic assumption that electron\u2013phonon interaction provides a means for creating Cooper pairs.2 A ground state of a superconductor consists of a condensate, which is a single quantum state available only to interacting Cooper pairs. The condensate is the product of the pair wave functions, and its center of mass is stationary in the absence of electric and magnetic fields. If one wants to remove a pair of electrons from the condensate, an energy 2\u0394 (the binding energy of the Cooper pair) is required to \u201cbreak\u201d a pair. The BCS model Hamiltonian (Ref. 2) can be written as ^HBCS = \u2211 k,s Ek^c\u2020 ks^cks + \u2211 kk\u2032 Ukk\u2032^c\u2020 k\u2191^c\u2020 \u2212k\u2193^c\u2212k\u2032\u2193^ck\u2032\u2191: (14.85) Bardeen et al.2 used the grand canonical ensemble and showed that the wave function for coherent states can be written as \u03a6> = \u220f k \u00bd j 1 + gk^c\u2020 k\u2191^c\u2020 \u2212k\u2193&j \u0398> = ^\u03a6 j \u0398>, (14.86) which creates all possible pairs of 2N particles with various weights through gk: They used a variational procedure by using the constraint that the average particle number is maintained at N. However, we will use the Bogoliubov\u2013Valatin transformation and obtain the same result as the BCS theory.5 14.7.5 Ground State of the Superconducting Electron Gas We can write the Hamiltonian in Eq. (14.75) in a simpler form, H = \u2211 k E k \u00f0 k^ck + ^c\u2020 ^c\u2020 \u2212k^c\u2212k \u00de\u00f0 \u2212 U \u2211 kk\u2032 \u00de k\u2032^c\u2020 ^c\u2020 \u2212k\u2032^c\u2212k^ck, (14.87) 14.7 Microscopic Theory of Superconductivity 467 where a positive spin is associated with the index k, and a negative spin is associated with the index \u2212k. Eq. (14.87) is rearranged by introducing the following creation and annihilation operators: ^\u03b1k = uk^ck \u2212 vk^c\u2020 \u2212k ^\u03b1\u2212k = uk^c\u2212k + vk^c\u2020 k ^\u03b1\u2020 k = uk^c\u2020 \u2212 vk^c\u2212k k \u2212k = ukc\u2020 \u03b1\u2020 \u2212k + vk^ck: (14.88) k + v2 k = 1, uk = u\u2212k, vk = \u2212v\u2212k: In addiIn the preceding transformations, we use the conditions u2 tion, uk and vk are nonzero outside and inside the Fermi sphere, respectively. These conditions guarantee that the commutation relations for the ^c\u2212 operators also apply to the ^\u03b1\u2212 operators (Problem 14.11). We avoid the confusion created by the situation that", " a hole in the state k has a momentum \u2212\u210fk while an electron has a momentum \u210fk by defining quasi-particles which have momentum \u210fk both outside and inside the Fermi sphere. Since a quasi-particle of momentum \u210fk is created by the operator c\u2020 k when it is outside the Fermi sphere and by c\u2212k when it is inside the Fermi sphere, we have in Eq. (14.88), k > kF k < kF: We also introduce the operator ^H = ^H \u2212 EF ^N and the energy \u03b5 the ^c\u2212 to the ^\u03b1\u2212 operators (Bogoliubov\u2013Valatin transformation) leads to (Problem 14.12) uk = 1, vk = 0 for uk = 0, vk = 1 for k \u00de = E k \u00f0 \u00f0 \u00de \u2212 EF: The transition from (14.88a) \u03b5 k ^H = \u2211 \u00f0 \u00de\u00bd k \u2212 U \u2211 u2 k 2v2 k + \u00f0 ukvkuk\u2032vk\u2032\u00f0 kk\u2032f\u00bd u2 \u2212 v2 uk\u2032vk\u2032\u00f0 k\u00de k k^\u03b1\u2020 k^\u03b1\u2212k^\u03b1k \u2212 v2 u2 + + \u00f0 \u00f0 \u2212 v2 k^\u03b1k + ^\u03b1\u2020 ^\u03b1\u2020 k\u00de\u00f0 1 \u2212 ^\u03b1\u2020 \u2212k\u2032^\u03b1\u2212k\u2032 \u2212k^\u03b1\u2212k \u2212 ^\u03b1\u2020 \u00de k\u2032^\u03b1k\u2032\u00de\u00f0 k^\u03b1\u2020 \u03b1\u2020 + 2ukvk \u00f0 \u2212k^\u03b1\u2212k \u2212 ^\u03b1\u2020 1 \u2212 ^\u03b1\u2020 k^\u03b1k \u2212k + ^\u03b1\u2212k^\u03b1k \u00de& 1 \u2212 ^\u03b1\u2020 k\u03b1\u2020 \u2212k\u2032^\u03b1\u2212k\u2032 k\u2032^\u03b1\u2020 u2 \u2212k\u00de\u00f0 \u2212 ^\u03b1\u2020 k\u2032^\u03b1k\u2032\u00de\u00f0 k\u2032\u03b1\u2020 \u2212 v2 \u2212k\u2032 ^\u03b1\u2212k^\u03b1k + ^\u03b1\u2020 k\u2032^\u03b1\u2212k\u2032^\u03b1k\u2032\u00deg : k^\u03b1\u2020 \u2212k\u00de \u00de& 14:89 \u00de \u00f0 The last term of the first line in Eq. (14.89) vanishes from the condition that uk and vk are zero inside and outside the Fermi sphere (Eq. 14.88 (a)). Normally, the first term (energy of the filled Fermi sphere) and the second term (energy of the quasiparticles defined by \u03b1\u2020 k\u03b1k) remain, but in the ground state, the second term vanishes. As we will show, the terms containing the products of \u03b1\u2020 \u2212k and \u03b1\u2212k\u03b1k can be eliminated by choosing different values of uk and vk: The last term in Eq. (14.89) contributes very little to the final result and hence can be neglected. Thus, in the ground state, Eq. (14.89) can be rewritten as k\u03b1\u2020 HG = 2\u2211 k \u03b5 \u00f0 k \u00de v2 k \u2212 U \u2211 k,k\u2032 ukvkuk\u2032vk\u2032 + \u2211 k 2ukvk\u03b5 h k \u00f0 \u00de \u2212 u2 k \u00f0 \u2212 v2 k\u00de U\u2211 k\u2032 uk\u2032vk\u2032 \u00f0 i For the square bracket in Eq. (14.90) to vanish, we first introduce a constant, \u0394 = U\u2211 k ukvk, k^\u03b1\u2020 ^\u03b1\u2020 \u2212k + ^\u03b1\u2212k^\u03b1k : \u00de (14.90) (14.91) (14.92) (14.93) (14.94) (14.95) (14.96) 468 CHAPTER 14 Superconductivity and then set the condition, In addition, 2ukvk\u03b5 k \u00de \u00f0 = \u0394 u2 k \u00f0 \u2212 v2 k\u00de : k + v2 u2 k = 1: From Eqs. (14.92) and (14.93), it is easy to show that \u03bek = \u03b5 k, \u00f0 \u00f0 k \u00de + \u03942 \u03b52 \u00de q \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 1 u2 k = 1 + \u03bek\u00de 2 \u00f0 1 1 \u2212 \u03bek\u00de : 2 \u00f0 v2 k = From Eqs. (14.91), (14.94), and (", "14.95), one can show (Problem 14.12) \u0394 = U\u2211 k ukvk = U 2 \u2211 k \u0394 : + \u03942 \u03b52 \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 Because \u0394 \u2260 0 (no vanishing interaction), U \u2260 0 only in the range over one spin direction, the density of states is D k to an integration, we can write Eq. (14.96) in the alternate form k \u00f0 q \u03b5 \u00f0 \u2264 \u0127\u03c9q. Since the sum is /2. Thus, while converting the summation over \u00de \u00dej k \u03b5 \u00f0 j U 4 \u0127\u03c9q Z \u2212\u0127\u03c9q We obtain from Eq. (14.97) \u2248 UD EF \u00f0 4 \u00de D d\u03b5 \u03b5 \u00de \u00f0 \u03b52 + \u03942 p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u0127\u03c9q p Z \u2212\u0127\u03c9q d\u03b5 \u03b52 + \u03942 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 = 1: (14.97) \u00bd This is precisely the binding energy of the Cooper pair. Because the energy of the filled Fermi sphere is \u00de \u00f0 & \u0394 = 2\u0127\u03c9q exp \u22122/D EF U : (14.98) H0 = 2 \u2211 k<kF \u03b5 k, \u00de \u00f0 (14.99) from Eqs. (14.90) and (14.99), we obtain the difference between the Hamiltonians with and without interactions (note that the coefficient of the square bracket in Eq. 14.90 is equal to zero), which, in the absence of any operators, is the energy difference between the ground state of the interacting and noninteracting electron gas: E = HG \u2212 H0 = 2\u2211 k v2 k k \u03b5 \u00f0 \u00de \u2212 2 \u2211 k<kF \u03b5 k \u00f0 \u00de \u2212 U \u2211 kk\u2032 ukvkuk\u2032vk\u2032: (14.100) 14.7 Microscopic Theory of Superconductivity 469 Substituting the values of uk, vk, and \u03bek from Eqs. (14.94) and (14.95) in Eq. (14.100), we obtain E = \u2211 \u03942 \u03b52 + \u03942 When we convert the sum over k to an integration (with only one spin direction), \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u03b5 j j \u03b52 + \u03942 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u03b5 \u03b52 + \u03942 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u03b5 1 \u2212! + \u2211 k>kF \u2212 \u2211 k k<kF 14.101) E = D EF \u00f0 \u0127\u03c9q \u03b5 \u2212 1 2! \u00de Z 0 2\u03b52 + \u03942 p \u03b52 + \u03942 \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \" d\u03b5: (14.102) After integrating Eq. (14.102) (see Problem 14.13), we obtain for weak interactions, 1 + \u0394 \u0127\u03c9q! \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \" E is known as the condensate energy of the ground state.", " 2 1 \u2212 s2 \u0127\u03c9q E = 4 \u00de \u00f0 \u00de EF D \u00f0 2 2 \u2248 \u2212 D \u03942 : EF \u00f0 4 \u00de 3 5 \u0394 \u226a \u0127\u03c9q, \u00de \u00f0 (14.103) 14.7.6 Excited States at T = 0 From Eqs. (14.89) and (14.90), we can write the Hamiltonian of the excited states, He, as He = HG + \u2211 k \u03b5 k \u00f0 \u00de\u00f0 u2 k \u2212 v2 k\u00de + U\u2211 k\u2032 2ukvkuk\u2032vk\u2032 k^\u03b1k + ^\u03b1\u2020 ^\u03b1\u2020 \u2212k^\u03b1\u2212k h Eq. (14.104) can be expressed by using Eq. (14.91), i& He = HG + \u2211 k \u03b5 k \u00f0 \u00de\u00f0 u2 k \u2212 v2 k\u00de + 2\u0394ukvk k^\u03b1k + ^\u03b1\u2020 ^\u03b1\u2020 \u2212k^\u03b1\u2212k i& Eq. (14.105) can be rewritten with the help of Eq. (14.92), h He = HG + \u2211 k 2ukvk \u0394 \u03b52 k \u00de \u00f0 + \u03942 \u00f0 \u2019 & nk\u2191 + n\u2212k\u2193\u00de : From Eqs. (14.96) and (14.106), we obtain The energy of a quasiparticle is given by E \u2212 E0 = \u2211 k + \u03942 k \u00de \u00f0 \u03b52 h 1/2 nk14.104) (14.105) (14.106) (14.107) = \u03b5 k \u00f0 \u00de k + \u03942 \u03b52 \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u00f0 q is measured from the Fermi surface: : (14.108) Note that here the energy \u03b5 k \u00de \u00f0 \u00f0 Thus, the ground and the first excited states are separated by an energy gap \u0394. This is shown in Figure 14.13. \u00de \u00f0 \u03b5 k = E \u2212 EF: k \u00de (14.109) 470 CHAPTER 14 Superconductivity 2 \u2206 + 2 \u03b5 = \u03b5\u2212 \u2206 \u22124\u2206 \u22123\u2206 \u22122\u2206 \u2212\u2206 0 \u2206 2\u2206 3\u2206 4\u2206 FIGURE 14.13 Energy \u03b5 of the quasiparticles plotted as a k \u00de \u00f0 function of \u03b5 k \u00f0 \u2212 EF : = E k \u00f0 \u00de \u00de Because in a scattering process, a quasiparticle is created in pairs, corresponding to the electron-hole pair of the noninteracting electron gas, the threshold energy of an excitation is 2\u0394. This leads to the conclusion that the Cooper pairs must be broken up during the scattering process. Thus, both Cooper pairs and single quasiparticles are present in the excited state, and while individual particles are scattered, Cooper pairs lead to a current that flows without resistance. We now are back to a two-fluid model. 14.7.7 Excited States at T \u2260 0 The states k\u2191 and k\u2193 are fermions, and at temperatures T \u2260 0, replacing the particle numbers nk by their statistical average, we obtain by taking the Fermi distribution as the occupational probability, <nk> fk = 1 k \u00de + 1, e\u03b2\u03b5 \u00f0 is always positive (from Eq. 14.108), except that \u03b5 where the energy of the quasiparticles \u03b5 note that \u03b5 \u00de at T \u2260 0. Because nk \u2260 0 for T \u2260 0, the energy gap \u0394 is now temperature dependent, \u0394 the bracket in Eq. (14.90) has to be replaced by the condition uk\u2032vk\u2032\u00f0 $ replaces the usual energy difference E \u2212 \u03bc. We further \u2212 \u03bc k \u00f0 \u00de. Thus, 1 \u2212 2fk\u2032\u00de is defined as \u03b5 = E T \u00f0 \u2212 v2 k\u00de (14.111) 2ukvk\u03b5 = 0: k \u00de k \u00de k \u00de k \u00de u2 \u2211 k\u2032 (14.110) Similarly, \u0394 \u00f0 T \u00de is defined as T \u0394 \u00f0 \u00de = U\u2211 k\u2032 uk\u2032vk\u2032\u00f0 1 \u2212 2fk\u2032\u00de = U 2 \u2211 k\u2032 \u0394 T \u00f0 \u00de + \u03942 \u03b52 \u00de", " \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 k\u2032 \u00de T \u00f0 \u00f0 q 1 \u2212 2fk\u2032\u00de \u00f0 : (14.112) Because the critical temperature Tc is the highest one for which Eq. (14.112) has a solution for from both \u2260 0, and the sum over k\u2032 is over one spin state, we can eliminate \u0394 which \u0394 sides, and by using a procedure similar to Eq. (14.97), we obtain T \u00f0 T \u00de \u00f0 \u00de EF UD \u00f0 4 \u00de \u0127\u03c9q Z \u2212\u0127\u03c9q d\u03b5 \u03b52 + \u03942 \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 T \u00f0 q 1 \u2212 2f 8 >< T \u03b52 + \u03942 \u00de q kBT \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u00f0 0 B @ = 1: (14.113) 9 1 >= C A >; >: Eq. (14.113) can be rewritten in the alternate form \u0127\u03c9q Z \u2212\u0127\u03c9q q d\u03b5 \u03b52 + \u03942 \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 T \u00f0 \u0127\u03c9q \u2212 Z \u2212\u0127\u03c9q q d\u03b5 \u03b52 + \u03942 \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 T \u00f0 2f 0 B @ T \u03b52 + \u03942 \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q kBT \u00f0 = 4 EF UD \u00f0 : \u00de 1 C A (14.114) 14.7 Microscopic Theory of Superconductivity 471 \u226a \u0127\u03c9q, it can be shown from Eqs. (14.98) and (14.114) that (Problem 14.14) Because \u0394 0 \u00f0 \u00de ln 2 \u221e Z 0 dx x2 + kBT 0 \u0394 \u00f0 \u00de $ \u22121! % = F x2 + 1 p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 T 0 \u0394 \u00f0 \u0394 \u00f0! kBT 14.115) \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 T Because Eq. (14.115) is a function of two parameters \u0394, Tc is calculated from \u00f0 \u00de. It can be shown by numerical integration that 0 the condition that \u0394 \u00de \u00f0 \u00f0 kBTc \u2248 0:57\u0394 \u00f0 = 0 is linearly dependent on \u0394 and kBT/\u0394 (14.116) /\u0394 \u00f0 Tc 0 0 0 \u00de \u00f0 \u00de \u00de \u00de : Eq. (14.116) can be rewritten in the alternate form 2\u0394 0 \u00de \u00f0 kBTc \u2248 3:53: (14.117) Approximating \u03c9q by \u03c9D (the Debye frequency), replacing the Debye temperature \u03b8D =", " \u0127\u03c9D/kB, from Eqs. (14.98) and (14.117), we obtain Tc \u2248 1:13 \u03b8D exp\u2061 \u2212 [Be aware that in the notations of BCS theory, N U = V.] \u00f0 EF $ \u00de 2 EF \u00de EF D \u00f0 = D \u00f0 : U % (14.118) /2 is the density of states per spin and \u00de 1.0 The BCS theory of superconductivity (Ref. 2) can be briefly summarized as follows. When a uniform electric field is applied to a superconductor, a current is generated because all the pairs in the condensate experience the same force and move in the same direction. When a voltage difference V exists across the two ends of the superconductor, an energy 2 eV would be gained by a Cooper pair and different parts of the condensate would have different energy, and eventually, the condensate would be lost. Thus, the condensate must move and carry current without any potential difference; i.e., it must exist as a phaselocked entity. The condensate cannot receive any energy less than 2\u0394, the energy required T \u00f0 to break a single pair of the condensate. However, when the temperature is above zero, some pairs are broken due to thermal excitation. Because the density of Cooper pairs is large at absolute zero, the initial reduction of numbers has little effect on 2\u0394, but as T approaches a critical temperature T = Tc, 2\u0394 = 0 and the \u00f0 condensate ceases to exist. The Cooper pairs are broken into individual electrons, and superconductivity is destroyed. A diagram that displays the reduced energy gap \u0394 as a function \u00f0 of the reduced temperature T/Tc is shown in Figure 14.14. Reduced energy gap \u2206(T )/\u2206(0) as a function of reduced temperature T/Tc. FIGURE 14.14 ) 0 ( \u2206 / ) T ( \u2206 /\u0394 \u00f0 T/Tc \u2192 Tc 1.0 0.4 0.4 0.8 0. 472 CHAPTER 14 Superconductivity It can also be shown that the condensation energy per unit volume is given by W = \u22121/4 D \u00f0 EF \u00de \u03942: (14.119) If we distinguish the density of states in a normal conductor as Dn state as Ds k = E \u2212 EF (Eq. 14.108), because no states are lost,, where \u03b5 E \u00f0 d\u03b5 k k \u03b5 \u00de dE and in its superconducting \u00f0 \u00f0 \u00de\u00de \u00f0 \u00de \u00f0 \u00de from which we obtain Dn E \u00f0 \u00de dE = Ds k \u03b5 \u00f0 \u00f0 \u00de\u00de d\u03b5 k, \u00de \u00f0 Ds \u03b5 \u00de \u00f0 = Dn dE d\u03b5 : E \u00f0 \u00de Eq. (14.121) can be rewritten in the alternate form by using Eqs. (14.108) and (14.109), Ds \u03b5 \u00de \u00f0 = Dn E \u00f0 \u00de = 0, \u03b5 \u03b52 \u2212 \u03942 \u00f0 1/2, \u00de > \u0394, < 0: \u03b5 \u03b5 j j j j (14.120) (14.121) (14.122) 14.8 STRONG-COUPLING THEORY 14.8.1 Introduction The BCS theory is based on a weak coupling approximation, i.e., the electron\u2013phonon coupling constant \u03bb \u226a 1 at T = 0. Here, \u03bb reflects the strength of the electron\u2013lattice interaction. However, for many superconductors, \u03bb \u2265 1: For example, \u03bb = 1:4 for lead, \u03bb = 1:6 for mercury, and \u03bb = 2:1 for Pb0:65Bi0:35. Thus, the more universal approach of the strong-coupling theory was developed after the formulation of the BCS theory. The strong-coupling theory is based on the Green\u2019s function method of the many-body theory, and we will introduce only the main results of this theory as derived by McMillan21 and modified by Dynes (Ref. 11). 14.8.2 Upper Limit of the Critical Temperature, Tc The electron\u2013phonon coupling constant can be written as \u03bb = 2 \u03b12 Z F \u03a9 \u00f0 \u00de \u03a9 \u00de \u00f0 \u22121d\u03a9, \u03a9 (14.123) where \u03a9 is the phonon frequency (we note that \u0127\u03a9D = kB\u03b8D), F and \u03b12 acteristic phonon frequency is the phonon density of states, is", " a measure of the phonon-frequency-dependent electron\u2013phonon interaction. The char- \u03a9 is defined as \u03a92>1/2, where the average is determined by < \u03b12 \u00de \u03a9 \u00f0 \u00de F \u00f0 \u03a9 \u00de d\u03a9: (14.124) (14.125) 14.9 High-Temperature Superconductors 473 McMillan (Ref. 21) introduced a convenient expression for the coupling constant, \u03bb, where \u03bd is the bulk density of states, \u03bb = \u03bd<I2>/M\u03a9 2, \u03bd = m#pF/2\u03c02, (14.126) (14.127) and <I2> contains the average value of the electron\u2013phonon matrix element I. The expression in the strong-coupling limit, for the critical temperature Tc, obtained by McMillan21 (later modified by Dynes (Ref. 11)) is given by and \u03bc# can be expressed as the Coulombic pseudopotential \u03bc# = Vc 1 + Vc ln \u03b50/ \u00f0 \u03a9 \u00bd Tc = \u03b8D 1:2 exp \u2212 $ 1:04 \u03bb \u2212 \u03bc#\u00f0 1 + \u03bb \u00f0 1 + 0:62\u03bb \u00de, % \u00de \u03b50 \u2248 EF \u22121: \u00de& \u00f0, \u00de (14.128) (14.129) When we use a rough estimate of upper limit, each of these values is \u03b8D \u2248 400\u00b0 K, \u03bb \u2248 0:8, and \u03bc# \u2248 0:1: In Eq. (14.128), we obtain the upper limit for Tc \u2248 30\u00b0 K in the strong coupling limit. In fact, until 1986, Nb3Ge was the material that had the highest critical temperature, Tc = 23\u00b0 K: e 14.9 HIGH-TEMPERATURE SUPERCONDUCTORS 14.9.1 Introduction A new class of superconducting materials, the high Tc copper oxides (often called cuprates), was discovered by Bednroz and Muller in 1986.3 They found that La1:85Ba0:15CuO4 became superconducting at a critical temperature of Tc \u2248 30\u00b0 K. The atomic structure of this remarkable superconductor is shown in Figure 14.15. La1:85Ba0:5CuO4 is very different from traditional superconductors in the sense that it is not a conventional metal but brittle ceramic, which is an antiferromagnetic insulator, carefully doped so as to produce metallic and superconducting phases. The discovery of La1:85Ba0:15CuO4 set out a frenzy to discover more high Tc superconductors. compound the The La1:82Sr0:18CuO4 moved the transition temperature close to Tc \u2248 40\u00b0 K: The first high-temperature superconductor of which the critical temperature exceeded the liquid nitrogen temperature is synthesis of similar La, Ba Cu O FIGURE 14.15 The atomic structure of La1.85Ba0.15CuO4. 474 CHAPTER 14 Superconductivity Ba Y Ba Copper chains Apical oxygen Copper planes Copper planes Copper Oxygen FIGURE 14.16 Structure of the YBCO compound. YBa2Cu3O7\u2212x (x = 0.1) with Tc \u2248 93\u00b0 K, which was discovered by Wu et al.29 in February 1987. In fact, it was soon found that YBa2Cu3O7\u2212x is superconducting in the orthogonal structure although it is not superconducting in 0 \u2264 x \u2264 0:6 \u00de \u00f0 the tetragonal structure. This is the most \u00de \u00f0 studied high Tc compound. The orthogonal YBCO structure is shown in Figure 14.16. The highest observed value of Tc is 150\u00ba K for the HgBa2Ca2Cu3O8+x compound under pressure. x > 0:6 14.9.2 Properties of Novel Superconductors (Cuprates) All cuprates have a layered structure. The main structural unit typical for the whole family is the Cu-O plane (see Figure 14.16), where the pairing originates and the charge reservoir is located. The YBCO compound contains the Cu-O chains, and the change in the oxygen content in the chain layers leads to charge transfer between these two subsystems. The charge transfer occurs through the apical oxygen ion located between the chains and the planes, as shown in Figure 14.16. American Physical Society. Reproduced from Kresin and Wolf 16 with the permission of the Whereas the undoped parent compounds of the cuprates are insulators, the novel superconductors are doped materials. Doping leads to", " conductivity and, for larger concentration, to superconductivity. The doping is provided either by Sr substitution in La2\u2212xSrxCuO4). changing the oxygen content or by chemical substitution (La The dopants create electrons, whereas the holes are produced by doping, which removes electrons. For example, some cuprates like YBCO contain carriers that are holes, whereas other cuprates like T max\u2061 is obtained for a Nd-Ce-Cu-O contain carriers that are electrons. The maximum value of Tc c characteristic value nm of the carrier concentration. Tc < T max\u2061 for both n < nm (the underdoped region) and n > nm (the overdoped region).! $ c 14.9.3 Brief Review of s-, p-, and d-wave Pairing Next, we will review the three possible choices for pairing\u2014the s-, p-, and d-wave pairings\u2014and discuss the differences between these types of pairings. The Landau\u2013Ginzburg (Ref. 17) order parameter \u03a8 (defined in Eq. 14.17) becomes nonzero only in the presence of superconductivity. Under a gauge transformation, r \u00de \u00f0 A! A + \u2207!\u03d5, \u03a8e\u22122ie\u03d5/\u0127c: \u03a8! (14.130) (14.131) 14.9 High-Temperature Superconductors 475 In fact, Gorkov showed that the pair potential \u0394r (used in Bogoliubov theory of superconductivity) is proportional to the order parameter \u03a8 r \u00f0 and also transforms as \u0394re\u22122ie\u03d5/\u0127c: \u00de \u0394r! (14.132) \u00f0 \u00de r r, r\u2032 depended only on the center-of-mass coordinate r = The implicit assumption in the BCS theory of superconductivity was that the effective potential U was isotropic and \u03a8 /2. This is known as s-wave superconductivity. However, in general, the order parameter is a function of, and one can define R = r \u2212 r\u2032. When \u03a8 is independent of the direction of R, the super\u03a8 \u00f0 conductor is called s-wave. When \u03a8 decays rapidly as a function of R, \u03a8 has the symmetry of. R (proportional to cos \u03b8, as in the case of superfluid 3He); this is called p-wave. When \u03a8 x! 2, which is proportional to depends on the direction of R, the symmetry is \u00de cos \u20612\u03b8 and is known as d-wave. r + r! y! \u00de \u00de \u00f0 \u00f0 One can also gain insight into the nature of the pair-condensate state based on symmetry considerations. For example, the parity of a superconductor with inversion symmetry can be specified using the Pauli exclusion principle. Because the crystal structures of bulk superconductors are all characterized by a center of inversion, they can be classified by the parity of the pair state. The spin-triplet state (total spin S = 1) has a superconducting order parameter (gap function) with odd parity, while the spin-singlet pair state (S = 0) corresponds to an orbital pair wave function \u03c8 (k) \u03b1 \u0394(k) with even parity, i.e., \u0394. Because spin-orbit interaction is relatively small in cuprate superconductors, the spin-singlet and -triplet states are well defined. It can be shown from group-theoretical considerations that the gap function \u0394 for each pair state can be expanded as a function of kx, ky, and kz: Some examples follow: \u2212 \u0394s k \u00de \u00f0 = \u03940 s + \u03941 s \u00f0 cos kx + cos \u2061ky + \u03942 s cos\u2061kz +, \u2019 \u2019 \u2019 \u00de \u0394dx2\u2212y2 \u00f0 k \u00de = \u03940 dx2 \u2212y2 \u00f0 cos\u2061kx \u2212 cos\u2061ky +, \u2019 \u2019 \u2019 \u00de \u0394dxy\u00f0 k \u00de = \u03940 dxy\u00f0 sin\u2061kx sin\u2061ky + : \u2019 \u2019 \u2019 \u00de (14.133) (14.134) (14.135) Thus, the order parameters for both the possible d-wave pair states have node lines. All cuprate superconductors are characterized by a relatively high ratio of c-axis to a-axis lattice constants. For example, the c/a ratio is 3.0 for YBCO, 5.7 for BI-2212, and 7.6 for Tl-2212. These ratios of c/a translate into a flattened Brillouin zone possessing the basic symmetric properties of the unit cell of a square/", "rectangular lattice. Recent studies of interplane dc and ac intrinsic Josephson effects has shown that high Tc superconductors such as Bi-2212 act as stacks of two-dimensional superconducting CuO2-based layers coupled by Josephson interactions (Ref. 13). It has also been shown that the vortex state can be understood in terms of stacks of two-dimensional pancake vortices. The cores of these 2D vortices, localized in the CuO2 layers, are connected by Josephson vortices with cores confined in the nonsuperconducting charge-reservoir layers (see Figure 14.16). Thus, the pairing symmetry should reflect the underlying CuO2 square/rectangular lattices. It is more convenient to consider our study of pairing symmetry in a square lattice. A k-space representation of allowed symmetry basis functions for the C4v symmetry is shown in Figure 14.17. 476 CHAPTER 14 Superconductivity Group-theoretic notation Order parameter basis function Wave function name Schematic representation of \u2206(k) in B.Z. A1g A2g B1g constant xy(x2-y2) x2-y2 s-wave g dx2-y2 B2g xy dxy ky kx FIGURE 14.17 A k-space representation of allowed symmetry basis functions for the C4v symmetry. Reproduced from Tsuei and Kirtley 27 with the permission of the American Physical Society. A schematic presentation in k space is shown where black and white represent opposite signs of the order parameter. 14.9.4 Experimental Confirmation of d-wave Pairing Angle-Resolved Photoemission A number of both non-phase-sensitive and phasesensitive experimental techniques confirm that pairing in cuprates is highly anisotropic with a line of nodes in the superconducting gap. We will first discuss the angle-resolved photoemission spectroscopy (ARPES), which has the advantage of directly investigating the momentum space of the gap. We will discuss Bi \u2212 2212 Bi2Sr2CaCu2O8+x, of which the complex structure consists of a superlattice of orthorhombic units. However, the basic orthorhombic has subunit of Bi2Sr2CaCu2O8+x essentially tetragonal symmetry with lattice parameters a \u2248 b. This pseudotetragonal subunit is closely approximated by a body-centered tetragonal structure of which the primitive cell is shown in Figure 14.18. The key elements of this structure are the presence of two Cu-O sheets similar to those in the 40\u00b0 and 90\u00b0 K materials, as well as a double layer of edge-sharing Bi-O octahedrals, Bi \u2212 2212 Ca Sr Bi Cu O FIGURE 14.18 Primitive cell for body-centered tetragonal Bi2Sr2CaCu2O8+x. O(1), O(2), and O(3) denote oxygens in the Cu, Bi, and Sr planes, respectively. The a and the b axes at 45\u00b0 to x and y are shown. Reproduced from Stavola et al.25 with the permission of the American Physical Society. 14.9 High-Temperature Superconductors 477 which fulfill a structural role that is analogous to the Cu-O chains in YBa2Cu3O7: The main band features at the Fermi level include a pair of half-filled two-dimensional Cu-O 3d\u22122p bands similar to those found in other Cu-O planar superconductors, as well as slightly filled 6p bands, which provide additional carriers in the Bi-O planes. \u00f0 The ARPES study shows that the gap in Bi\u22122212 Bi2Sr2CaCu2O8+x (Ref. 10) is largest \u00de along the \u0393 \u2212 M direction (parallel to a or b) and smallest among \u0393 \u2212 Y (the diagonal line between them), as expected for a dx2\u2212y2 superconductor. Figure 14.19 shows the inferred value of the energy gap as a function of the angle in the Fermi surface (solid circles), compared to the prediction of a simple d-wave model (solid line). There is remarkable agreement between experimental and theoretical results. However, ARPES is not phasesensitive and cannot distinguish between d-wave and highly anisotropic s-wave pairing. M\u2212 1 \u0393 Y E M\u2212 15 1 20 40 60 FS angle (deg) 80 ) V e m ( | \u2206 | 15 40 30 20 10 0 0 FIGURE 14.19 Energy gap in Bi \u2212 2212, measured with ARPES as a function of an angle in the Fermi surface: solid curve, with fits to the data using a d-wave order parameter.", " Inset indicates the data points in the Brillouin zone. Reproduced from Ding et al.10 with permission of the American Nuclear Magnetic Resonance Nuclear-magnetic-resonance (NMR) measurements can probe the electronic properties of individual atomic sites on the CuO2 sheets of the high-temperature superconductors. There is no Hebel-Slichter peak (found in normal superconductors due to density of states in the gap edge) and an increase in the nuclear relaxation rate T \u22121 1 near Tc, for both Cu and O in-plane sites. These properties can be explained by using a dx2\u2212y2 model with Coulomb correlations that yield (a) a weaker quasiparticle density-of-states singularity at the gap edge compared with an s-wave BCS gap, (b) the vanishing of for a dx2\u2212y2 gap, and (c) inelastic\u2013scatterthe coherence factor for quasiparticle scattering for q \u223c ing suppression of the peak, which is similar for both d-waves and s-waves. There has been excellent T \u22121 T \u22121 1 \u00dec and the trans1 \u00deab/ agreement between the experimental results for both the anisotropy ratio \u00f0 \u00f0 verse nuclear relaxation rate T \u22121 G for 63Cu when a d-wave model is used, but absolutely no agreement when an s-wave model is used for the theoretical calculations. Physical Society. 2 \u00f0 \u03c0, \u03c0 \u00de \u00de \u00f0 Josephson Tunneling As we discussed earlier, Josephson first pointed out that Cooper pairs can flow through a thin insulating barrier between two superconductors.13 A schematic representation of a Josephson tunnel junction between a pure dx2\u2212y2 superconductor on the left and a superconductor with some admixture of s in a predominantly dx2\u2212y2 state on the right is shown in Figure 14.20. The gap states align with the crystal-line axes, which are rotated by angles \u03b8L and \u03b8R with respect to the junction normals nL and nR. 478 CHAPTER 14 Superconductivity \u03b8L nR \u03b8R nL FIGURE 14.20 Schematic diagram of a Josephson junction showing the tunnel barrier sandwiched between two junction electrodes, with order parameters \u0394i \u00f0 ei\u03d5i : ki \u00de \u0394i j Reproduced from Tsuei and Kirtley with the permission of the American Physical Society.27 = j The supercurrent Is, proportional to the tunneling rate of Cooper pairs through the barrier, was given by Josephson as where \u03b3 is the gauge-invariant phase difference at the junction, Is = Ic sin \u2061\u03b3, \u03b3 = \u03d5L \u2212 \u03d5R + R 2\u03c0 \u03a60Z L A. dl, (14.136) (14.137) where A is the vector potential, and dl is the element of line integration from the left electrode (L) to the right electrode (R) across the barrier. It was shown by Cohen et al. (Ref. 6) that the supercurrent Is at zero temperature is given by \u2212 \u03b3R\u00de where Tk,l is the time-reversal-symmetry-invariant tunneling matrix element, Is = \u2211 k,l j EL \u00bd Tk, l sin \u03b3L \u00f0 \u00f0 \u00de \u00de \u00de& k I \u00f0 \u00de \u00f0 \u00f0 j 2 \u0394L EL k k \u0394R ER I \u00de \u00f0 I \u00de \u00f0 1 + ER, (14.138) \u00f0 \u00de \u00de k = Ei k \u00f0 + \u03942 \u22082 k k i \u00f0 i \u00f0 \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 and \u2208 is the one-electron energy. Eq. (14.138) is used to determine the parity of superconductors. It was shown by Pals et al. (Ref. 24). that Is vanishes up to the second order in Tk,l, in tunnel junctions between spin-singlet (even parity) and -triplet (odd parity) superconductors. However, because pair tunneling exists for Josephson junctions made of a cuprate superconductor and low-Tc conventional superconductor such as Nb or Pb, this confirms that the superconducting state in cuprates, just as in low-Tc conventional", " superconductors, is that of even-parity spin-singlet pairing. (14.139) q \u00de, \u03c0\u2212Rings The sign changes in the pair critical current Ic are arbitrary for a particular junction because an arbitrary phase can always be added to either side of the junction. However, the signs of the critical currents in a closed ring of superconductors that is interrupted by Josephson weak links can be assigned self-consistently. One can determine if a particular geometry is frustrated by counting these sign changes. A frustrated geometry has a local maximum in its free energy with zero \u0393j \u00f0 \u00de 14.9 High-Temperature Superconductors 479 circulating supercurrent in the absence of an external magnetic field. A negative pair-tunneling critical current Ic can be considered as a phase shift of \u03c0 at the junction interface, i.e., Is = \u2212 Ic j j sin\u2061\u03b3 = sin Ic j j \u00f0 \u03b3 + \u03c0 : \u00de (14.140) A superconducting ring with an odd number of \u03c0 shifts is frustrated. In the Ginzburg\u2013Landau formalism (Ref. 17), the order parameter is small near Tc, and hence, the free energy of the Josephson junction can be expanded as a power series of the order parameter. of the Further, the gap function can be expressed as a linear combination of the basis functions irreducible representation \u00f0, which corresponds to the highest Tc of the order parameter, \u03c7j \u03bc\u00de lj = \u2211 \u03bc=1 where lj is the dimensionality of \u0393j, and the expansion coefficient \u03b7\u03bc is invariant under all symmetry operations of the normal-state group G. \u03b7\u03bc\u03c7j \u03bc\u00f0 (14.141) \u0394 \u00f0 k k \u00de \u00de, The free energy per unit area of Josephson coupling between two superconducting electrodes with order parameters \u03c8 L and \u03c8 R (Figure 14.20) can be written as Fj = W ds Z \u03c8 L\u03c8 #R exp h i h \u00f0 2\u03c0/\u03a60 R \u00de Z L A. dl + c:c:, i i (14.142) where W is a measure of the Josephson coupling strength, and the integral is over the junction interface. By minimizing the total free energy with respect to \u03c8 L and \u03c8 R (Problem 14.15), we can obtain an expression for the Josephson current density Js, flowing perpendicular to the junction interface from superconductor L to R, \u00de n n \u03b7Rj \u03c7R\u00f0 sin \u2061\u03b3 = Jc sin\u2061\u03b3: Js = tL,R\u03c7L\u00f0 \u03b7Ljj \u00dej k Here, Jc is the critical current density; \u03c7L,R, the basis function, is related to the gap function \u0394 \u00de \u00f0, where n is the unit vector normal to the junction \u03c7L,R\u00f0 through Eq. (14.141); \u0394L,R \u00de ei\u03d5L. R ; and \u03b3 is defined in Eq. (14.137). tL,R is a constant characteristic of \u03b7L,Rj = interface, \u03b7L,R\u00f0 \u00de j the junction. The basis functions \u03c7 of a Josephson junction electrode with tetragonal symmetry (point group C4v) are listed in Table 14.1. Here, nx, ny are the projections of the unit vector n onto the crystallographic axes x and y, respectively. = \u03b7L,R\u00f0 n (14.143 It can be shown from Eq. (14.143), for Josephson junctions between two d-wave superconductors, \u03c7 n \u00de \u00f0 = n2 x \u2212 n2 y, and of which the interface is clean and smooth, Js = As cos\u2061 \u00f0 2\u03b8L \u00de cos\u2061 2\u03b8R \u00f0 \u00de sin\u2061\u03b3, (14.144) Table 14.1 Basis Functions \u03c7 Symmetry n \u00de \u00f0 Irreducible Representation Basis function \u03c7 n \u00f0 \u00de A1g s 1 for a Josephson Junction Electrode with Tetragonal Crystal A2g g nxny n2 x \u00f0 \u2212 n2 y \u00de B1g dx2\u2212y2 n2 x \u2212 n2 y B2g dxy nxny 480 CHAPTER 14 Superconductivity where \u03b8L and \u03b8R are the angles of the crystallographic axes with respect to the interface, and As is a constant characteristic of the junction. However, in real Josephson junctions made with cuprates, the electron wave vector normal to the junction interface can be significantly distorted by interface roughness, oxygen deficiency, strain, and so", " on. In this limit, Js = As cos\u20612 \u03b8L + \u03b8R sin \u2061\u03b3: \u00de \u00f0 Flux Quantization of a Superconducting Ring The flux quantization of a superconducting ring can be written as \u03a6a + IsL + \u03a60 2\u03c0 \u2211 ij \u03b3ij = n\u03a60, (14.145) (14.146) where \u03a6a is due to the flux of an external field, Is is the supercurrent circulating in the ring Is = Iij c \u00f0 \u03b8i, \u03b8j \u00de sin\u2061\u03b3ij, (14.147) \u03b3ij is defined in Eq. (14.137), and \u03a60 is a flux quantum. It can be shown easily (Problem 14.16) that the ground state of a superconducting ring containing an odd number of sign changes (\u03c0 ring) has a spontaneous magnetization of a half-magnetic flux quantum, IsL \u2248 \u03a60, when the external field is zero. Because Is = 0 in the ground state for an even number of \u03c0 shifts, the magnetic-flux state has an even number of quantization. To summarize, 1/2 \u00f0 \u00de and where N is an integer. \u03a6 = n\u03a60 for N even 0 ring, \u00de \u00f0 1 2 \u03a6 = n + & \u03a60 for N odd \u2019 \u03c0 ring \u00f0 \u00de, (14.148) (14.149) Tricrystal Magnetometry The multiple-junction ring consists of deliberately oriented cuprate crystals for defining the direction of the pair wave function. The presence or absence of the half-integer flux-quantum effect in such samples as a sample configuration differentiates between various pairing symmetries. In the first tricrystal experiment of Tsuei et al. (Ref. 24), an epitaxial YBCO film (1200 \u00c5 thick) was deposited using laser ablation on a tricrystal (100) SrTiO3 substrate. In addition to the three-junction ring located at the tricrystal meeting point, two two-junction rings and one ring with no junction were also made as controls. The design was such that IcL \u226b \u03a60 for observing the half-integer flux quantization. The magnetic flux threading through the superconducting cuprate rings in the tricrystal magnetometry experiments were directly measured by a high-resolution SQUID microscope. A series of tricrystal experiments with various geometrical configurations confirmed that only the dx2\u2212y2 frustrated configuration showed the half-integer flux-quantum effect. d-Wave Pairing Symmetry The evidence from both phase-sensitive symmetry as well as non-phase-sensitive symmetry techniques has conclusively proved that the cuprates have d-wave pairing symmetry. The identification of Problems 481 d-wave symmetry is based on group theory and the macroscopic quantum coherence phenomena of pair tunneling and flux quantization. However, it does not necessarily specify a mechanism for high-temperature superconductivity. 14.9.5 Search for a Theoretical Mechanism of High Tc Superconductors The identification of d-wave pairing symmetry does not necessarily specify a mechanism for hightemperature superconductors. The solution of this problem should include pairing symmetry, pairing interactions (mediated by phonons, spin fluctuations, or some other bosons) in the presence of strong correlations, and taking into consideration the anomalous normal state and charge segregation and the stripe phase. The observation of a pseudogap in the normal state, with a d-wave-like k dependence of many underdoped cuprate semiconductors, has generated renewed interest for a suitable theory. Recently, Kresin and Wolf proposed some experiments that will unambiguously resolve the issue.16 Anderson noted, \u201cThe consensus is that there is absolutely no consensus on the theory of highTc superconductivity.\u201d Bardeen et al.2 proposed a microscopic theory of superconductivity in 1957, which was unanimously accepted 46 years after Kammerlingh Onnes discovered superconductivity in Hg in 1911.15 We do not know how long we have to wait for a satisfactory theory for high Tc superconductivity, which was discovered by Bednroz and Muller in 1986.3 PROBLEMS 14.1. Show that Maxwell\u2019s equation (neglecting the displacement current D as well as replacing H with B because j is the mean microscopic current) can be written as \u2207! \u00d7 B = 4\u03c0 c j: 14.2. Minimizing Eq. (14.16) with respect to A, show that \u2207! \u00d7 B = 4\u03c0 c j, where = r j \u00f0 \u00de e#\u0127 2im# \u00bd \u03a8#", " \u2207!\u03a8 \u2212 \u03a8 \u2207!\u03a8# 2 \u2212 e# m#c & A\u03a8#\u03a8: 14.3. Minimizing Eq. (14.16) with respect to \u03a8#, by first integrating such that all spatial derivatives act on \u03a8 and then taking functional derivatives with respect to \u03a8#, show that \u03b1 + \u03b2 2 + \u03a8 j j 1 2m# $ \u0127 i & \u2207! + e# c A 2 % \u2019 \u03a8 = 0: 14.4. Substituting Eq. (14.54) in Eqs. (14.52) and (14.53), we obtained i\u0127 _n1 n1p! \ufb03\ufb03\ufb03\ufb03 + i n1p _\u03d51 ei\u03d51 = \" \ufb03\ufb03\ufb03\ufb03 \u03b51 n1p ei\u03d51 + \u2208 n2p ei\u03d52 \u00f0 \u00de \ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03 (1) (1) (2) (1) (1) 482 CHAPTER 14 Superconductivity and i\u0127 _n2 n2p! Show from Eqs. (1) and (2) that + i n2p _\u03d52 ei\u03d52 = \" \u03b52 \u00f0 n2p ei\u03d52 + \u2208 n1p ei\u03d51 \u00de : \ufb03\ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03\ufb03\ufb03 _n1 = 2 \u2208n \u0127 sin \u03d52 \u00f0 \u2212 \u03d51\u00de = \u2212 _n2 = j e# and where n = n1n2p : 14.5. If we choose ^S such that \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 _\u03d52 \u2212 _\u03d51 = 1 \u0127 \u00f0 \u03b51 \u2212 \u03b52 = \u00de e# \u0127 \u00f0 V2 \u2212 V1 \u00de, ^H0, ^S \u00bd in which case, the electron\u2013phonon interaction ^H1 is eliminated apart from a higher-order term. If we choose ^S of the form ^H1 + = 0, & ^S = \u2211 kqs\u2032 Mq \u00f0 \u03b1^a\u2020 \u2212q + \u03b2^aq ^c\u2020 k+q,s\u2032^ck,s\u2032: \u00de From Eqs. (1) and (2), show that and \u03b1\u2212 \u2212 \u0127\u03c9q \u03b2\u2212 + \u0127\u03c9q: (2) (3) (4) (1) (2) (3) (4) 14.6. From Eqs. (14.60) and (14.61), the next-order interaction term is 1 2 \u00bd terms that contain operator products of the form ^H1, ^S &, which is a sum of \u00b1q^a\u00b1q\u2032^c\u2020 ^a\u2020 k\u2032+q\u2032,s\u2032^ck\u2032,s\u2032^c\u2020 k+q,s^ck,s: Show that because q\u2032 = \u2212q (from momentum conservation), out of all the combinations, only one does not contain any phonon operators, when k+q,sc\u2020 c\u2020 k\u2032\u2212q,s\u2032^ck\u2032,s\u2032^ck,s, k\u2032 \u2260 k, k + q: 14.7. Show from Eqs. (14.60), (14.62), and (14.66) that the proper form of Eq. (2) is Heff = 1 2 \u2211 k,k\u2032,q,s,s\u2032 j Mq 2 j \u00f0 \u03b1 \u2212 \u03b2 k+q,s^c\u2020 ^c\u2020 \u00de k\u2032\u2212q,s\u2032^ck\u2032,s\u2032^ck,s: (1) (2) (3) (1) 14.8. Substituting the values of \u03b1 and \u03b2 from Eqs. (14.63) and (14.64) in Eq. (1) of Problem 14.7, Problems 483 show that where < f Heff j i> = j 1 2 \u2211 k,k\u2032,q,s,s\u2032 < f j Vkq^c\u2020 k+q,s^c\u2020 k\u2212q,s\u2032^ck\u2032s\u2032^ck,s Vkq = 2 E k +", " q \u00de \u00f0 \u00bd Mq j \u2212 E 2\u0127\u03c9q j 2 \u2212 k \u00f0 \u00de& 2 : \u0127\u03c9q \u00f0 \u00de i>, j (1) (2) 14.9. Show that if we take the complex conjugate of Eq. (14.79), multiply by b k, we obtain an equation that agrees with Eq. (14.76), provided \u03b2 = E., and sum over k \u00de \u00f0 14.10. In Eq. (14.83), we derived EF+\u0127\u03c9q U Z EF D dx x \u00de \u00f0 2x \u2212 E = 1, is the density of states. Show that if we approximate D x \u00de \u00f0 by D EF \u00f0 \u00de in the \u00de x where D \u00f0 narrow range of integration, we obtain \u22122/D \u00bd \u22122/D 2\u0127\u03c9q exp 1 \u2212 exp E = 2EF \u2212 EF \u00f0 EF U \u00de U \u00de & \u00f0 \u00bd & \u2248 2EF \u2212 2\u0127\u03c9q exp \u22122/D \u00bd EF U : & \u00de \u00f0 14.11. We introduced the ^\u03b1\u2212 operators in Eq. (14.88): ^\u03b1k = uk^ck \u2212 vk^c\u2020 \u2212k ^\u03b1\u2212k = uk^c\u2212k + vk^c\u2020 k k = uk^c\u2020 ^\u03b1\u2020 \u2212 vk^c\u2212k k \u2212k = ukc\u2020 \u03b1\u2020 \u2212k + vk^ck: k + v2 In addition, we use the conditions u2 k = 1, uk = u\u2212k, and vk = v\u2212k: Show that these conditions guarantee that the commutation relations for the c-operators also apply to the \u03b1-operators. 14.12. From Eqs. (14.91), (14.94), and (14.95), show that 14.13. In Eq. (14.102), we obtained + \u03942 \u03b52 \u00de \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \u00f0 E = D EF \u00f0 \u00de \u0127\u03c9q \u03b5 \u2212 1 2! Z 0 d\u03b5: 2\u03b52 + \u03942 \u03b52 + \u03942 p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \" (1) (2) (1) (1) (1) 484 CHAPTER 14 Superconductivity After integrating Eq. (1), show that E = D EF \u00f0 2 \u00de \u00f0 \u0127\u03c9q 2 1 \u2212 \u00de s2 2 1 + \u0394 \u0127\u03c9q! \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 \" \u2248 \u2212 D \u00f0 3 EF 4 \u03942 \u00de : 4 5 14.14. Because \u0394 \u00f0 0 \u00de \u226a \u0127\u03c9q, from Eqs. (14.98) and (14.114), show that ln 2 \u221e Z 0 dx x2 + 1 p f \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 T 0 \u0394 \u00f0 \u0394 \u00f0 \u00de \u00de kBT 0 \u0394 \u00f0 \u00de $ \u2212 kBT 0 \u0394 \u00f0 \u00de : \"! x2 + 1 p \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 (2) (1) 14.15. The free energy per unit area of Josephson coupling between two superconducting electrodes with order parameters \u03c8 L and \u03c8 R (Figure 14.20) can be written as Fj = W Z ds \u03c8L\u03c8#R exp\u2061 \u00bd \u00bd i 2\u03c0/\u03a60 \u00f0 R \u00de Z L A. dl & + c:c:, & (1) where W is a measure of the Josephson coupling strength, and the integral is over the junction interface. By minimizing the total free energy with respect to \u03c8L and \u03c8R", ", show that the expression for the Josephson current density Js, flowing perpendicular to the junction interface from superconductor L to R, is n Js = tL,R\u03c7L\u00f0 \u03b7Ljj \u00dej Here, Jc is the critical current density; \u03c7L,R, the basis function, is related to the gap function, where n is the unit vector normal to n \u0394 \u00f0 \u00de \u00f0 \u03b7L,Rj the junction interface, \u03b7L,R\u00f0 j constant characteristic of the junction. = \u03b7L,R\u00f0 ei\u03d5L.R ; and \u03b3 is defined in Eq. (14.137). tL,R is a through Eq. (14.141); \u0394L,R = sin \u2061\u03b3 = Jc sin \u2061\u03b3: \u03c7L,R\u00f0 n \u03c7R\u00f0 \u03b7Rj 2) 14.16. The flux quantization of a superconducting ring can be written as \u03a6a + IsL + \u03a60 2\u03c0 \u2211 ij \u03b3ij = n\u03a60, (1) where \u03a6a is due to the flux of an external field, Is is the supercurrent circulating in the ring Is = Iij c \u00f0 \u03b8i, \u03b8j \u00de sin \u2061\u03b3ij, (2) where \u03b3ij is defined in Eq. (14.137), and \u03a60 is a flux quantum. Show that the ground state of a superconducting ring containing an odd number of sign changes (\u03c0 ring) has a spontaneous magnetization of a half-magnetic flux quantum, IsL \u2248 external field is zero. Because Is = 0 in the ground state for an even number of \u03c0 shifts, the magnetic-flux state has an even number of quantization. To summarize, \u03a60, when the 1/2 \u00f0 \u00de \u03a6 = n\u03a60 for N even (0 ring), (3) References 485 and where N is an integer. 1 2 \u03a6 = n + & \u03a60 for N odd \u2019 \u03c0 ring, \u00de \u00f0 (4) References 1. Aschroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 2. Bardeen J, Cooper LN, Schrieffer JR. Theory of Superconductivity. Phys Rev 1957;108:1175. 3. Bednroz G, Muller KA. Possible high Tc superconductivity in the Ba-Ca-Cu-O system. Z Physik 1986; B64:189. 4. Bednroz G, Muller KA. Perovskite-type oxides-The new approach to high-Tc superconductivity. Rev Mod Phys 1988;60:585. 5. Bogoliubov NN. A new method in the theory of superconductivity. Soviet Phys JETP 1958;7:41. 6. Cohen MH, Falicov LM, and Phillips JC. Superconducting Tunneling. 7. Cooper LN. Bound Electron Pairs in a Degenerate Fermi Gas. Phys Rev 1986;104:1189. 8. Deaver BS, Fairbank MS. Experimental Evidence for Quantized Flux in Superconducting Cylinders. Phys Rev Lett 1961;7:43. 9. de Gennes P-G. Superconductivity of metals and alloys. Reading, MA: Addison-Wesley; 1992. 10. Ding H, Norman MR, Campuzino JC, Randeria M, Bellmar AF, Yokoya T, et al. Angle-resolved photoemission spectroscopy study of the superconducting gap anisotropy in Bi2Sr2CaCu2O8+x. Phys Rev B 1996;54:R9678. 11. Dynes R. McMillan\u2019s equation and the Tc of superconductors. Solid State Commun. 1972;10:615. 12. Doll R, Nabauer M. Experimental Proof of Magnetic Flux Quantization in a Superconducting Ring. Phys Rev Lett 1961;7:51. 13. Josephson BD. Possible new effects in superconducting tunneling. Phys Lett 1962;1:251. 14. Josephson BD. The discovery of tunneling supercurrents. Rev Mod Phys 1974;46:251. 15. Kammerlingh Onnes H. The resistance of platinum at helium temperature. Comm Phys Lab Univ Leiden 1911;119b:19. 16. Kresin VJ, Wolf SA. Electron-lattice interaction and its impact in high-Tc superconductivity. Rev Mod Phys 2009;81:481. 17. Landau LD, Ginzburg VL. On the theory of superconductivity", ". JETP (USSR) 1950;20:1064. 18. London F, London H. The Electromagnetic Equations of the Superconductor. Proc. Roy. Soc. London 1935;A149:71. 19. Madelung O. Introduction to solid state physics. New York: Springer-Verlag; 1978. 20. Marder MP. Condensed matter physics. New York: John Wiley & Sons; 2000. 21. McMillan W. Transition Temperature of Strong-Coupled Superconductors. Phys Rev 1968;167:331. 22. Meissner W, Ochsenfeld R. A new effect in penetration of superconductors. Die Naturwissenschaften 1933;21:787. 23. Mercereau. Superconductivity vol. 1, editor. Parks RD. New York: Marcel Dekker; 1961, p. 393. 24. Pals JA, Haeringen W, van Maaren MH. Phys. Rev. B 1977;15:2592. 25. Stavola M, Krol DM, Schneemeyer LF, Sunshine SA, Fleming RN, Waszczak JV, et al. Raman scattering from single crystals of the 84-K superconductor Bi2.2La0.8Sr2Cu2O8+\u0394. Phys Rev B 1988;41:R5110. 26. Tinkham M. Introduction to superconductivity. New York: McGraw-Hill; 1996. 27. Tsuei CS, Kirtley JR. Pairing Symmetry in Cuprate Superconductors. Rev Mod Phys 2000;72:969. 486 CHAPTER 14 Superconductivity 28. Tsuei CC, Kirtley JR, Chi CC, Yu-Jahnes LS, Gupta A, Shaw T, Sun J-Z, Ketchen MB. Pairing Symmetry and Flux Quantization in a Tricrystal Superconducting Ring of YBa2Cu3O7-\u0394. Phys. Rev. Lett. 1994;73:593. 29. Wu MK, Ashburn JR, Torng CJ, Hor PH, Meng RL, Gao L, et al. Superconductivity at 93 K in a new mixed phase Y-Ba-Cu-O compound system. Phys Rev Lett 1987;58:908. Heavy Fermions CHAPTER 15 CHAPTER OUTLINE 15.1 Introduction............................................................................. 488 15.2 Kondo-Lattice, Mixed-Valence, and Heavy Fermions.......................................... 490 15.2.1 Periodic Anderson and Kondo-Lattice Models........................................ 490 15.2.2 Mixed-Valence Compounds......................................................... 492 15.2.3 Slave Boson Method............................................................... 493 15.2.4 Cluster Calculations............................................................... 494 15.3 Mean-Field Theories..................................................................... 498 15.3.1 The Local Impurity Self-Consistent App", "roximation.................................... 498 15.3.2 Application of LISA to Periodic Anderson Model..................................... 499 15.3.3 RKKY Interaction.................................................................. 500 15.3.4 Extended Dynamical Mean-Field Theory............................................. 501 15.4 Fermi-Liquid Models..................................................................... 502 15.4.1 Heavy Fermi Liquids............................................................... 502 15.4.2 Fractionalized Fermi Liquids....................................................... 505 15.5 Metamagnetism in Heavy Fermions........................................................ 506 15.6 Ce- and U-Based Superconducting Compounds.............................................. 508 15.6.1 Ce-Based Compounds.............................................................. 508 15.6.2 U-Based Superconducting Compounds.............................................. 509 15.7 Other Heavy-Fermion Superconductors..................................................... 513 15.7.1 PrOs4Sb12........................................................................ 513 15.7.2 PuCoGa5.......................................................................... 513 15.7.3 PuRhGa5..........................................................", "............... 515 15.7.4 Comparison between Cu and Pu Containing High-Tc Superconductors.................. 516 15.8 Theories of Heavy-Fermion Superconductivity............................................... 516 15.9 Kondo Insulators........................................................................ 516 15.9.1 Brief Review...................................................................... 516 15.9.2 Theory of Kondo Insulators......................................................... 517 Problems.................................................................................... 519 References.................................................................................. 524 Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00015-3 \u00a9 2012 by Elsevier Inc. All rights reserved. 487 488 CHAPTER 15 Heavy Fermions 15.1 INTRODUCTION Heavy fermions, which are also sometimes referred to as heavy electrons, are a loosely defined collection of intermetallic compounds containing lanthanide (mostly Ce, Yb) or actinide (mostly U, Np) elements. They also include other compounds such as quasi two-dimensional CeCoIn5 and \u201cSkutterdites\u201d such as PrOs4Sb12. The common feature of the heavy fermions is that they have large effective mass m! (50\u20131000 times greater than the mass of a free electron) below a coherence temperature T!. The effective mass is estimated through the electronic specific heat. In general, for very low temperatures, the specific heat C of a metal can be expressed as where C/T = \u03b3 + \u03b2T 2, \u03b3 = VmkFk2 Bm!/3\u01272: (15.1) (15.2) Here, Vm is the molar volume, kF is the Fermi vector, m! is the effective mass of the electron, T is the absolute temperature, \u03b3 is the electronic contribution, and \u03b2 is the contribution of the phonons to the specific heat. There is an additional spin-fluctuation term \u03b4T 3 ln \u2061T in the specific heat of UPt3 and UAl2. For normal metals such as copper or aluminum, \u03b3 is of the order 1 mJ/mol K2 at low temperatures. A generally accepted definition of heavy fermions is those systems that have \u03b3 > 400 mJ/f atom mol K2 below the coherence temperature T!. \u03b3 is generally normalized to a mole of f atoms so that there can be a comparison between systems with different structure. Some of the other properties of heavy fermions include (a) an enhanced Pauli spin susceptibility indicating a large effective mass; (b) a Wilson ratio of approximately one; (c) a huge T2 term in the electrical resistivity; and (d) highly temperature-dependent de Haas\u2013van Alphen oscillation amplitudes at very low temperatures. The Wilson ratio (Ref. 35)", " R is defined as ) s t i n u R = \u03c02k2 g2\u03bc2 BJ 15.3) \u00f0 \u00f0 \u00f0 \u00de \u00f0 \u00de 0 0 and \u03b3 \u00de Here, \u03c7 are the magnetic susceptibility and specific heat at zero temperature, J is the total angular momentum, gJ is the Land\u00e9 g factor, and the other symbols have their usual meanings. CeAl3, which earlier had been considered a mixed-valence compound, was the first heavy-fermion system discovered by Andres et al. in 1975. They found that below 0.2\u00b0 K, \u03b3 = 1620 mJ mole/K2 and the coefficient of the T2 term in \u03c1 = AT2, A = 35 \u03bc\u03a9 cm/K2. The intense interest in heavy-fermion systems started with the discovery of superconductivity in CeCu2Si2 by Steglich et al.29 in 1979. Their results are shown in Figures 15.1 and 15.2(Tesla) 0 0.1 (\u03c1) Tc 1 \u22122 0.1 0.5 T(K) 1 (\u03c7) Tc 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 Temperature (K) FIGURE 15.1 Resistivity (main part) and low-field ac susceptibility (inset) of CeCu2Si2 as a function of temperature. Arrows give transition temperatures Tc ( \u03c7) = 0.54 \u00b1 0.03\u00b0 K. Reproduced from F. Steglich et al.29 with permission of the (\u03c1) = 0.60 \u00b1 0.03\u00b0 K and Tc American Physical Society. 2..0 0.4 T(K) 0.6 CeCu2Si2 ) c) Tc 0 0.3 FIGURE 15.2 LaCu2Si2 1 3 10 T (K) Molar specific heat of CeCu2Si2 as a function of temperature on a logarithmic scale. Inset shows the specific heat jumps of two other samples. Reproduced from F. Steglich et al.29 with the permission of the American Physical Society. 15.1 Introduction 489 CeCu2Si2 UBe13 1000 J m 2 K l o m C T 800 600 400 200 0 0 20 40 80 100 120 60 T 2(K2) FIGURE 15.3 Specific heat of nonsuperconducting single crystals (\u25cf) and a piece of superconducting crystal (\u25a0) of UBe13. Reproduced from Stewart30 with the permission of the American Physical Society. The main part of Figure 15.2 shows, in a logarithmic scale, the molar specific heat of CeCu2Si2 at B = 0 as a function of temperature. The inset in Figure 15.2 shows in a C/T plot the specific heat-jumps of two other CeCu2Si2 samples that do not look very profound. The specific heat-jumps below the coherence temperature T*, which is characteristic of heavyfermion systems, are elegantly displayed when one plots C/T versus T2. Stewart30 plotted C/T versus T2 of nonsuperconducting single crystals of CeCu2Si2 and a piece of a superconducting single crystal of UBe13. These results are reproduced in Figure 15.3, in which the line through UBe13 serves as a guide to the eye. Since 1974, approximately 50 heavy-fermion compounds have been discovered, but there is no uniformity in their properties. For example, UBe13 is a superconductor in the ground state with non-Fermi-liquid properties in the normal state, whereas UPt3 orders antiferromagnetically below the N\u00e9el temperature (TN), exhibits a heavy Fermi-liquid state well below TN, and has unconventional superconductivity with a multicomponent superconducting parameter. CeAl2 and U2Zn17 are antiferromagnets with weak moments at very low temperatures, and CeNiSn and Ce3Bi4Pt3 are 490 CHAPTER 15 Heavy Fermions narrow-gap semiconductors with quasiparticles having large effective masses. Some heavy-fermion superconductors such as CeCoIn5 are quasi two-dimensional. The only common feature is the large effective mass below the coherence temperature and the fact that all these are highly correlated electron systems. In addition, some heavy fermions such as CeRu2Si2 exhibit metamagnetism, which has a wide variety of technological applications. There are many factors that lead to the conclusion that the large effective mass of heavy fermions below the coherence temperature is not due to band-structure renormalization. For example, the magnitude of", " the nuclear relaxation rate of UBe13 and the ultrasonic attenuation in UPt3 in the normal state are the same as ordinary metals. The thermal conductivity measurements in CeCu2Si2, UBe13, and UPt3 yield results similar to ordinary metals. There have been several powerful techniques applied to discuss the theory of these strongly correlated systems. However, the theory of these systems lags behind the experiment. In this chapter, we will discuss the properties of the wide variety of these correlated systems without going into the details of the complex theories. 15.2 KONDO-LATTICE, MIXED-VALENCE, AND HEAVY FERMIONS 15.2.1 Periodic Anderson and Kondo-Lattice Models It has been noted that the majority of the rare-earth and actinide compounds have local moments and can be classified as systems in the magnetic regime. The f orbitals have no charge fluctuation in this region and have integral valence. Therefore, they can be considered to be in a Mott insulating stage. Weak residual spin polarization of the conduction electrons, Rudderman\u2013Kittel\u2013 Kasuya\u2013Yosida (RKKY) interactions between the local moments (Refs. 12, 21, 36), magnetic transition at low temperatures, and spin-wave excitations occur. The spin waves scatter the conduction electrons at low temperatures. To correlate and to study their dependence on the various relevant parameters, the simplest Hamiltonian is the orbitally nondegenerate periodic Anderson model (Ref. 1). The periodic Anderson model for a system consisting of a set of N sites is denoted by sites i, j. On each site, there are two orthogonal nondegenerate orbitals that will be referred to as C and f. The Hamiltonian is assumed to have the form H = t \u2211 i\u2260j,\u03c3 ^C \u2020 i\u03c3 ^Cj\u03c3 + V \u2211 i\u2260j,\u03c3\u00f0 ^C \u2020 i\u03c3 ^f j\u03c3 + ^f \u2020 j\u03c3 ^Ci\u03c3\u00de + \u03b5f \u2211 i,\u03c3 \u2020 ^f i,\u03c3 ^f i\u03c3 + U\u2211 i \u2020 ^f i\u2191 ^f i\u2191 \u2020 ^f i\u2193 ^f i\u2193: (15.4) Here, t (which can be positive or negative) is the transfer (hopping) integral of the extended orthoi\u03c3 and ^Cj\u03c3 are gonal orbitals between sites i and j (restricted to nearest neighbors in our model). ^C the creation and annihilation operators for these extended orbitals at sites i and j with spin \u03c3. There is one extended orbital per site per spin with a mean energy that is the origin of the energy scale. \u2020 ^f i\u03c3 and ^f i\u03c3 are the creation and annihilation operators for the localized f orbitals (i denotes the site) with energy \u03b5f : V is a positive hybridization parameter between the localized and the band orbitals in neighboring sites. The third term represents the single-particle energy of the isolated f orbitals. The fourth term is an interaction of the Hubbard type between electrons of the f orbitals on the same site. U is the Coulomb repulsion between two electrons of opposite spin in the f orbital and \u2020 15.2 Kondo-Lattice, Mixed-Valence, and Heavy Fermions 491 describes a short-range interaction between them. U is positive, whereas t and \u03b5f can have either sign. When we consider the f orbital only on a single site, the model (Eq. 15.4) is reduced to a single-impurity Anderson model (Ref. 1), as discussed in Section 13.9, except that the localized magnetic moment for rare-earth metals is due to s \u2212 f mixing instead of s \u2212 d mixing as originally visualized in the single-impurity Anderson model. The Hamiltonian (Eq. 15.4) can be augmented by additional terms such as second-neighbor hopping or Coulomb repulsion between extended orbitals and f electrons, or between electrons on different sites. Because the orbital degeneracy is neglected, there is no Hund\u2019s rule coupling between the f orbitals in this model. However important such terms are in applications to real systems, they are ignored here in the belief that they would contribute nothing really essential to the qualitative physics. When each f orbital is occupied by a single electron (either up-spin or down-spin), the system is described as the Kondo regime. The empty sites and doubly occupied sites become virtual states. The low-energy physics of the periodic Anderson model (Eq. 15.4) can be described by an effective model where the f -electron degrees of freedom are represented by localized spins. Schrieffer and Wolff (Ref. 24) (Problem 15.", "1) used a second-order perturbation with respect to V to obtain an effective Hamiltonian H = t \u2211 j\u2260i,\u03c3\u00f0 C\u2020 i\u03c3Cj\u03c3 + H:C: + J\u2211 i \u00de Si. Sc i, where Si = 1 2 \u2211 \u03c3,\u03c3\u2032 \u03c4\u03c3,\u03c3\u2032f \u2020 i\u03c3 fi\u03c3\u2032, i = 1 Sc 2 \u2211 \u03c3,\u03c3\u2032 \u03c4\u03c3,\u03c3\u2032C\u2020 i\u03c3Ci\u03c3\u2032, (15.5) (15.6) (15.7) \u00f0 \u00de and \u03c4 are the Pauli spin matrices. Thus, Sc i are the spin-density operators of the conduction electrons, and Si are the localized spins. J is the exchange interaction, which is antiferromagnetic J > 0 and inversely proportional to U. Under symmetric conditions, J = 8V 2/U. Thus, the rare-earth compounds that have either localized four f -electrons (Ce, Yb) or five f electrons (U, Np) can be considered as a Kondo-lattice (Ref. 9), where at each lattice site a local moment interacts via an exchange coupling J with the spin of any conduction electron sitting at the site. The Hamiltonian in Eq. (15.5) is also known as the Kondo-lattice model (Ref. 9). The exchange coupling is the source of interesting many-body effects in the Kondo-lattice model. The complexity of solving the Kondo-lattice model arises due to the complex correlation effect involving both the localized spin and the itinerant electron degrees of freedom. In fact, a conduction electron undergoes a spin-flip process with a localized spin if the spin is antiparallel. The conduction electron leaves a trace of its spin exchange processes with the localized spins while moving around the lattice. The direction of the localized spins is determined by the history of the electrons that passed through this site. Thus, the conduction electrons are no longer independent. There are similar correlation effects in the periodic Anderson model due to the dynamic aspects of the localized electrons. Because these systems are highly correlated, most of the theoretical models developed during the past 30 years are approximate treatments of the complex problem. 492 CHAPTER 15 Heavy Fermions 15.2.2 Mixed-Valence Compounds The properties of rare-earth metals and their compounds as well as the actinide compounds have been the subject of a great deal of interest for the past 40 years. A subclass of these rare-earth and the actinide compounds is known as mixed-valence compounds, which are poor metals but have a fluctuating valence. Clear indication that two ionic valence states are present in these compounds is provided by X-ray photoelectron spectra in which the two valence states are seen side by side. They are also evident by both photoemission measurements as well as by isomer-shift measurements. In these compounds, near the Fermi energy, the s and d electrons as well as the much heavier f electrons are present. A simple explanation is that because in the ground state, both f n and configurations are present, their energies must be very close. The difference of energy is on the order of the hopping line width. The extra electron is assumed to go into an extended state, so its energy is equal to the Fermi energy. The extra available f orbital can be described as a localized state, with energy \u2208f nearly equal to the Fermi energy EF, and that can accept one electron but not two. f n\u22121 + conduction electron \u00f0 \u00de The mixed-valence compounds generally form with rare-earth elements only at the beginning, the middle, and the end of the rare-earth series. The reason that the beginning and the end of the rare-earth series are favored is that a closed shell screens the nuclear charge very effectively. Hence, the 4f electron in Ce and the 4f hole in Yb are loosely bound and not far off from the 5d configuration. The middle of the rare-earth series is favored because of the importance of Hund\u2019s rule coupling. The final occupied f-level, even for Sm, is not far below the d-level. A typical example of a mixed-valence compound is SmS, which is a semiconductor at normal pressure. Sm has the electronic structure [Xe] 4f 55d06s2, and S has the electronic structure [Ne] 3s23p4. In compounds, the d-level broadens into a band and hybridizes with the 6s band, but the f-levels are essentially unaffected. In a schematic electronic structure, one can visualize a localized f 6-level in the gap between the 5d \u2212 6s band and the s \u2212 p bands. Under pressure,", " the lower of the crystal field split d-bands broadens and moves down in energy relative to the f-level and ultimately crosses it. When the f \u2013 d gap goes to zero, a metal insulator transition occurs. The electronic structure and the density of states for the metallic state are shown schematically in Figure 15.5. The f-levels hybridize with the d level on a neighboring atom because they cannot hybridize with the d-level of the same atom (the f 6 configuration has a total J = 0). The \u201cbandwidth\u201d of the hybridized f-band is very narrow so that in the density of states, over the smooth s \u2013 d background, there is a sharppeak attributable to the f -like atomic character in a tight-binding representation. The wave functions near this peak are linear combinations of f -like and d-like wave functions and can be written as 5d \u2212 6s band f 6 s \u2013 p bands FIGURE 15.4 Schematic structure of SmS in the semiconducting phase. (15.8) r \u03c8 k\u00f0 = ak\u03d5d\u00f0 r + bk\u03d5f \u00f0 r, \u00de \u00de \u00de where the proportion ak to bk varies rapidly near the peak. The f 6-level, which is nondegenerate due to correlation energy, accommodates only one electron per atom. Because this peak is derived from the f 6-level, the integrated density 15.2 Kondo-Lattice, Mixed-Valence, and Heavy Fermions 493 f n\u22122 f n\u22121 f n f n+1 E EF s \u2212 d band in it is also f -like character of states of one-electron per atom. Hence, at T = 0, the Fermi level is pinned to lie in the f peak. This characteristic is in a sense the definition of mixed-valence compounds. The wave function in Eq. (15.8) represents the linear combination of the atomic orbital states, which is partly f 5d and partly f 6. The d electron is not affected by the local atomic exchange and correlations because it is relatively represents a linear combination r free. Thus, \u03c8 k\u00f0 of 3+ and 2+ valence states on the rare-earth ion, and the compound is known as a mixed-valence system. The average valence can be defined as \u00de FIGURE 15.5 Electronic energy levels for mixed-valence materials. A wide sd band overlaps one of the configurations of the multiplet splitting of f electrons. Reproduced from Varma32 with the permission of the American Physical Society. Vav = \u2211 k j 2= akj \u2211 k j j bkj 2, (15.9) where the sum is over the occupied part of the occupied band. However, the instantaneous valence Vinst will be different from Vav because the f electrons have a nonzero bandwidth. 15.2.3 Slave Boson Method The slave boson method was developed for the U = \u221e periodic Anderson model that had the constraint nf \u2264 1 at each site. The essential feature of this method is that the localized electron operators are written as a composition of a fermion ^f and a boson ^b, where we may consider the boson as an f vacancy. Every site is occupied either by an ^f fermion or a ^b boson. The localized electron operators are written as a composition of a boson ^b and a fermion ^f : One defines The operator equality, i\u03c3 = ^f f \u2020 \u2020 i\u03c3 ^bi and fi\u03c3 = ^b \u2020 i ^f i\u03c3: \u2020 ^f i\u03c3 ^fi\u03c3 + ^b \u2020 i ^bi = 1, \u2211 \u03c3 (15.10) (15.11) satisfies the preceding condition. The Anderson lattice Hamiltonian can be written as H = \u2211 k,\u03c3 \u03b5kc\u2020 k\u03c3ck\u03c3 + \u03b5f \u2211 i,\u03c3 \u2020 ^f i\u03c3 ^fi\u03c3 + V\u2211 i,\u03c3 \u00f0 c\u2020 i\u03c3 ^b \u2020 i ^fi\u03c3 + ^f \u2020 i\u03c3 ^bici\u03c3\u00de \u03bbi + \u2211 i \u2211 \u03c3! \u2020 ^f i\u03c3 fi\u03c3 + ^b \u2020 i bi \u2212 1 : (15.12) \" Here, \u03bbi is a Lagrangian multiplier for the site i and is needed to impose the local constraints. The properties of the Hamiltonian (Eq. 15.12) are usually discussed in a mean-field approximation. It is assumed that the bosons have Bose condensations, <^bi> = b0, and the Lagrange multiplier \u03bbi = \u03bb0 for all sites. Thus, the constraint is obeyed only on the average over the whole system. 494", " CHAPTER 15 Heavy Fermions 15.2.4 Cluster Calculations To correlate the mixed-valence, Kondo, and heavy-fermion behavior and to study how they correspond to different regimes of one fundamental phenomenon (at least in Ce systems), Misra et al.17 considered the application of the periodic Anderson model to finite clusters with periodic boundary conditions. Although the phrase \u201cperiodic Anderson model\u201d is somewhat inappropriate when applied to a small system, they used it in reference to a four-atom cluster in which each site has a localized orbital and an extended orbital with appropriate Coulomb repulsion, hybridization, and transfer matrix elements. In a later paper, they extended the number of electrons to eight particles, but the results were similar. The value of small cluster calculations is that exact solutions of the Hamiltonian are obtained. However, it has to be recognized that in some respects, small clusters are not representative of bulk materials. For example, at sufficiently low temperatures the specific heat of a cluster model will vanish exponentially, and the magnetic susceptibility will either be infinity or zero. The large number of states obtained even for a small cluster suggests that statistical mechanics may give results that fairly represent a large system over a reasonable range of temperatures. Misra et al. (Ref. 17) applied the periodic Anderson model (Eq. 15.4) to four-site tetrahedral clusters of equal length with periodic boundary conditions, thereby including the band structure effects. For example, their model Hamiltonian for a tetrahedron is identical to that of an fcc lattice if the Brillouin zone sampling is restricted to four reciprocal-lattice points, the zone center \u0393, and the three square-face-center points X. They studied the region of crossover between the magnetic, Kondo, and mixed-valence regimes by varying the different parameters U/|t |, V/|t|, and Ef /|t|, and their results for the tetrahedron, which reflect the properties of cerium alloys, are presented in Figure 15.6. One can distinguish the three regimes by considering n electrons per site. Consider the non-f electrons to constitute an electron reservoir. In Figure 15.6, Ed is the Fermi level when there are n non-f electrons per site, and Ed\u2032, is the Fermi level when there are n \u2212 1 electrons per site (if there is no interaction with the f electrons). EF is the chemical potential when the f electrons are in contact with the electron reservoir. Let Ef be the energy boundary such that at T = 0, the ion will be in the f 0 state if EF < Ef and in the f 1 state if EF > Ef. Ef + U is another ionization boundary separating the f 1 state from the f 2 state. The crossover from one regime to another depends sensitively on the various parameters U,V, and Ef as well as on the geometry (band structure). E Ef + U Ef f2 f1 f 0 Ed EF Ed\u2032 FIGURE 15.6 A schematic diagram illustrating the dependence of the three different regimes on the position of the f-level. The Hamiltonian (Eq. 15.4) is conveniently considered on a basis of states diagonal in occupation numbers; Misra et al. (Ref. 17) calculated the many-body eigenstates and eigenvalues. Because spin is a good quantum number, the states can be classified as spin singlets, triplets, and quintets. For n = 4, there are 784 singlet, 896 triplet, and 140 quintet states. For n = 8 (Ref. 3), there are 12,870 states available for eight particles. These rather large numbers of states should tend to make the results somewhat representative of large systems except at extremely low temperatures (lower 15.2 Kondo-Lattice, Mixed-Valence, and Heavy Fermions 495 than the separation between the ground state and the first excited state). The significant results are projected through the diagram in Figure 15.7. Misra et al. (Ref. 17) constructed a computer program to diagonalize the Hamiltonian within subspaces of fixed values of Sz: They calculated the f-state occupation (nf ), temperature dependence of specific heat (Cv), and the magnetic susceptibility ( \u03c7f ) of the f electrons (by using a canonical ensemble) for a large number of parameters. In Figure 15.8, a typical example is presented by plotting Cv /T against T for Ef ranging from \u22125:0 to \u22124:0 (nf varies from 0.9943 to 0.9788). We notice that for Ef = \u22125:0, Cv/T increases very rapidly at very low temperatures (which mimics the onset of heavy-fermion behavior) but gradually decreases as Ef is increased until the 1.0 0.", "9 0.8 0.7 0.6 f N 0.5 0.4 0.3 0.2 0.1 0 \u22128 0.1 0.5 1.5 2.0 2.0 1.5 0.5 0.1 \u22126 \u22124 \u22122 0 2 4 6 Ef / t FIGURE 15.7 The f occupation number per site, nf, in the four-electron ground state in terms of Ef and various hybridization energies V for t = \u22121, U = 50 for a tetrahedron. Reproduced from Misra et al.17 with the permission of the American Physical Society. 496 CHAPTER 15 Heavy Fermions 3..0 1.K) FIGURE 15.8 Cv/T/|t| versus T/|t| for various Ef for E negative, U = 50, and V = 0.1 for a tetrahedron. Curve (a), Ef = \u22125.0; curve (b), Ef = \u22124.8; curve (c), Ef = \u22124.6; curve (d), Ef = \u22124.4; curve (e), Ef = \u22124.2; curve (f), Ef = \u22124.0. (All parameters in units of |t |.) Reproduced from Misra et al.17 with the permission of the American Physical Society. heavy-fermion feature has practically disappeared when Ef = \u22124:2. To explain the unusual increase in Cv/T, Misra et al. (Ref. 17) plotted the energy-level diagram (Figure 15.9) of the first few manybody states for each of these Ef as well as for Ef = \u22123:0. In Figure 15.9, for Ef = \u22125:0, the ground state is a singlet, but the next two higher-energy states are a triplet and a quintet, which are nearly degenerate with the ground state. The low-temperature rise in Cv is determined by these three levels. As Ef increases, the separation between the lowest three levels increases, and the rise in Cv/T correspondingly decreases. Thus, the heavy-fermion behavior is obtained when the many-body ground state is a singlet but nearly degenerate to two other magnetically ordered states. The same pattern is repeated for a tetrahedron for t =1, except that in some cases the ground state is a magnetically ordered triplet state. In such cases, the ground state of the heavy-fermion system would be magnetically ordered. In Figure 15.10, Misra et al. plotted kB\u03c7f T/ g\u03bc2 versus T/|t| to compare their results with the benchmark results for the single-impurity Anderson model. They defined a \u201cfrozen-impurity\u201d, \u03c7f T \u2248 0:125 \u03c7f T = 0 regime \u00de. In addition, they defined an \u201cintermediate regime\u201d for and a local moment regime which 0 < \u03c7f T < 0:125, but \u03c7f T essentially remains a constant in this regime., a free orbital regime \u00f0 \u03c7f T \u2248 0:25 \u00de, a valence-fluctuation regime \u03c7f T \u2248 0:167 B\u00de\u00f0$ \u03c7f T \u00de \u00f0 \u00f0 \u00de \u00de \u00f0 \u00f0 We note from Figure 15.10 that when Ef = \u22125:0 (nf = 0.994), there is a transition from the frozen-impurity to the local-moment regime. For Ef = \u22123:0 (nf = 0.725), there is a transition from 0.01 0.009 0.008 0.007 0.006 t / i E 0.005 0.004 0.003 0.002 0.001 15.2 Kondo-Lattice, Mixed-Valence, and Heavy Fermions 497 2.0 0.24.16 0.08 1.0 b c d e 0 \u221210 0 2.0 3.0 1.0 In T t FIGURE 15.10 g \u03bc2 b \u00de \u00f0 kB\u03c7f T / versus ln T/|t | for t = \u22121, U = 50, and V = 0.1 for various Ef for a tetrahedron. Curve (a), Ef = \u22125.0; curve (b), Ef = \u22123.0; curve (c), Ef = \u22121.0; curve (d), Ef = \u22120.5; curve (e), Ef = 0.5. (All parameters in units of |t |.) Reproduced from Misra et al.17 with the permission of", " the American Physical Society. 0\u22125.0 \u22124.8 \u22124.6 \u22124.4 \u22124.2 \u22124.0 \u22123.0 0 Ef/t FIGURE 15.9 Energy-level diagram of the first few many-body states for various Ef for t negative, U = 50, V = 0.1 for a tetrahedron. (All parameters in units of |t |.) Reproduced from Misra et al.17 with the permission of the American Physical Society. the frozen-impurity to the valence-fluctuation regime. When Ef is further increased, the transition is from the frozen-impurity to the free orbital regime. The high-temperature results are in excellent agreement with the single-impurity \u201cbenchmark\u201d results. When we compare the specific heat curves with the \u03c7T curves for the same parameters, the specific heat maxima generally occur below the temperature at which \u03c7T reaches its high-temperature value (i.e., the crossover temperature from enhanced Pauli- to Curie-like susceptibility). The main reason is that at low temperatures where Cv is a maximum, the many-body states with magnetic moments are still just becoming thermally populated. The same broad features have also been observed experimentally. It was generally believed that as Ef is increased from far below Ed (Figure 15.6), there would be a transition from the magnetic to the Kondo-lattice regime. However, Misra et al. (Ref. 17) found that for some choice of parameters, the system undergoes a transition from a Kondo-lattice to a magnetic regime as Ef is increased. Subsequently, it reenters a Kondo-lattice regime for higher values of Ef. This unusual feature of reentry to the Kondo-lattice regime is very sensitive to the hybridization parameter and occurs only for low V/|t| values, which are the most important parameters in determining nf as well as the thermodynamic properties. 498 CHAPTER 15 Heavy Fermions 15.3 MEAN-FIELD THEORIES 15.3.1 The Local Impurity Self-Consistent Approximation17 The dynamical mean-field theory is a very powerful tool for studying the strongly correlated system. In this approach, a lattice model is replaced by a single-site quantum impurity problem embedded in an effective medium determined self-consistently. This leads to an intuitive picture of the local dynamics of a quantum many-body problem. Because the impurity problem has been extensively studied, the self-consistency condition incorporates the translation invariance and the coherence effects of the lattice. This approach is now popularly known as the local impurity selfconsistent approximation (LISA). The LISA freezes spatial fluctuations but includes local quantum fluctuations and is therefore characterized as a dynamical mean-field theory. The on-site quantum problem is still a many-body problem that can be addressed by using a variety of techniques. The dynamical mean-field theory becomes exact in the limit of large spatial dimensions d \u2192 \u221e or in the limit of large lattice coordination. In the mean-field theory, a lattice problem with many degrees of freedom is approximated by a single-site effective problem. The dynamics at a given site are the interaction of the degrees of freedom at this site with an external bath created by the degrees of freedom on the other sites. A simple example is an application to the Hubbard model in which the Hamiltonian is H = \u2212 \u2211 <ij>,\u03c3 tij\u00f0 i\u03c3Cj\u03c3 + C\u2020 C\u2020 j\u03c3Ci\u03c3\u00de + U\u2211ni\u2191ni\u2193: An imaginary-time action (the local effective action) for the fermionic degrees of freedom at site o is \u00f0 (15.13) Co\u03c3, C\u2020 o\u03c3\u00de \u03b2 \u03b2 Seff = \u2212 d\u03c4 Z 0 Z 0 d\u03c4\u2032\u2211 \u03c3 C\u2020 o\u03c3\u00f0 \u03c4 g\u22121 0 \u00f0 \u00de \u03c4 \u2212 \u03c4\u2032 \u00de Co\u03c3\u00f0 \u03c4\u2032 \u00de + U \u03b2 Z 0 \u03c4 d\u03c4n\u03c3\u2191\u00f0 n\u03c3\u2193\u00f0 \u00de \u03c4 \u00de : (15.14) \u00de \u03c4 \u2212 \u03c4\u2032, the generalized \u201cWeiss function,\u201d is the effective amplitude for a fermion to be Here, g0\u00f0 created on the isolated site at time \u03c4 (coming from the \u201cexternal bath\u201d) and destroyed at time \u03c4\u2032 (going back to the bath). Because g0 is a function of time, it accounts for local quantum fluctuations. It can be shown that (Problem 15.2) G \u00f0, the on-site interacting Green\u2019s function, is calculated from \u22121 = i\u03c9n + \u03bc + G \u00f0 i\u03c9n\u00de i\u03c9n\u00de g0\u00f0", " \u2212 R \u22121 \u00bd i\u03c9n\u00de \u22121 G i\u03c9n\u00de \u00f0 \u03c4 \u2212 \u03c4\u2032 G \u00f0 \u00de = \u2212< TC C\u2020 \u03c4 \u00f0 \u00de \u03c4\u2032 \u00de \u00f0 > Seff G i\u03c9n\u00de \u00f0 = \u03b2 Z 0 d\u03c4G \u03c4 \u00f0 \u00de eiwn\u03c4, \u03c9n $ \u03c0 2n + 1 \u03b2 \u00de \u00f0 : : & (15.15) (15.16) (15.17) Here, R(G) is the reciprocal function of the Hilbert transform of the density of states corresponding to the lattice. As an example, in the Hubbard model \u2212 \u03b5k\u00de, \u00f0 \u03b5k = \u2211 ij tijei k!. # Ri!\u2212 Rj! : $ (15.18) 15.3 Mean-Field Theories 499 The Hilbert transform D \u00f0 \u03be \u00de and its reciprocal function R are defined by \u03be D \u00f0 \u00de $ \u221e d\u03b5 Z \u2212\u221e & = \u03be: (15.19) Eqs. (15.14) through (15.16) are the basic equations of the LISA method. However, the major difficulty lies in the solution of Seff : It can be shown that solving these equations yields the local quantities, and all the k!-dependent correlation functions of the original lattice Hubbard model can be obtained. j j \u2193\u232a, \u2191\u232a, j Co\u03c3, C\u2020 It may be noted that the LISA approach freezes spatial fluctuations but retains local quantum fluctuations. Each site undergoes transition between the four possible quantum states \u2191, \u2193\u232a by exchanging electrons with the rest of the lattice or \u201cthe external bath.\u201d As an 0\u232a, j as an impurity orbital. The bath can be described as a \u201cconexample, one can consider duction band\u201d described by the operators, and the Hamiltonian is the well-known singleimpurity Anderson Hamiltonian 2l a\u2020 Eq. (15.20) is quadratic in a\u2020 e Eq. (15.14), provided l\u03c3, al\u03c3, and integrating these gives rise to Seff of the form given in a\u2020 l\u03c3Co\u03c3 + C\u2020 Vl\u00f0 o\u03c3Co\u03c3 + Un\u03c3\u2191n\u03c3\u2193: HAM = \u2211 l\u03c3 l\u03c3al\u03c3 + \u2211 l\u03c3 al\u03c3, a\u2020 l\u03c3\u00de \u2212 \u03bc\u2211 \u03c3 o\u03c3al\u03c3\u00de (15.20) o\u03c3\u00de C\u2020 \u00f0 \u00f0 g\u22121 0 \u00f0 i\u03c9n\u00de \u221e AM = i\u03c9n + \u03bc \u2212 d\u03c9 Z \u2212\u221e \u03c9 \u0394 \u00de \u00f0 i\u03c9n \u2212 \u03c9 (15.21) and \u03c9 \u0394 \u03b5i\u00de If the parameters Vl, \u03b5l are chosen to obtain g0, the solution of the mean-field equations, HAM \u03b5l\u2019s are effective parameters and not \u03b5k, the becomes the Hamiltonian representation of Seff. Here, single-particle energy. In addition, \u0394(\u03c9), the conduction bath density of states, is obtained when the self-consistent problem is solved. = \u2211 l\u03c3 (15.22 Thus, by using the LISA approach, one obtains the Anderson impurity embedded in a selfconsistent medium from the Hubbard model. The dynamical mean-field equations are solved such that the proper g0 is obtained. When this g0 is inserted into the Anderson model, the resulting Green\u2019s function should obey the self-consistency condition in Eq. (15.15). The mapping onto impurity models, which have been studied by a variety of analytical and numerical techniques, is used to study the strongly correlated lattice models in large dimensions. However, it is important to solve Seff by using reliable methods. e 15.3.2 Application of LISA to Periodic Anderson Model We will now briefly describe the application of the LISA method to heavy-fermion systems and the Kondo insulators (Ref. 20). This is done by using the periodic Anderson model (PAM). This model describes a band of conduction electrons that hybridize with localized f \u2013 electrons at each lattice site. The PAM Hamiltonian can be written as i\u03c3Ci\u03c3\u00de f \u2020 i\u03c3fi\u03c3 + U\u2211 i \u00f0 k\u03c3Ck\u03c3 + V\u2211 i\u03c3 \u00f0 i\u03c3fi\u03c3 + f \u2020 C\u2020 + \u2208f \u2211 i\u03c3 H = \u2211 k\u03c3 nfi\u2191 \u2212 1/2 nfi\u2193 \u2212 1/2 (15.23) \u2208kC\u2020 \u00de\u00f0 \u00de, 500 CHAPTER 15 Heavy Fermions where the terms were defined", " in Section 15.2. In the d \u2192 \u221e limit, the local interaction gives rise to k-independent self-energy, and the various Green\u2019s functions are obtained in the form, \u22121 = i\u03c9n \u2212 \u2208k \u2212 i\u03c9n, k \u00de \u22121 = i\u03c9n \u2212 \u2208f \u2212 \u03a3f \u00f0 i\u03c9n, k \u00de \u22121 = i\u03c9n \u2212 \u2208k\u00de\u00f0 Gc\u00f0 Gf \u00f0 Gcf \u00f0 where \u03a3f \u00f0 is the self-energy of the f electrons, and \u03bc, the chemical potential, is absorbed in the definitions of \u2208k and \u2208f : It can be shown by reducing to a self-consistent single-site model that the effective action is V 2 i\u03c9n\u00de i\u03c9n \u2212 \u2208f \u2212 \u2211f \u00f0 V 2 i\u03c9n\u00de, \u2212 i\u03c9n \u2212 \u2208k i\u03c9n \u2212 \u2208f \u2212 \u03a3f \u00f0 i\u03c9n, k \u00de 1 V f\u00bd\u00f0 i\u03c9n\u00de& i\u03c9n\u00de (15.24) \u2212 V 2 g, Seff = \u2212 \u03b2 d\u03c4 0 Z 0 Z \u03b2 d\u03c4\u2032\u2211 \u03c3 \u03c4 f \u2020 \u03c3 \u00f0 g\u22121 0 \u00f0 \u00de \u03c4 \u2212 \u03c4\u2032 \u00de f\u03c3\u00f0 \u03c4\u2032 \u00de + U \u03b2 0 Z The f self-energy is obtained from d\u03c4 \u03c4 nf \u2191\u00f0 \u00bd \u00de \u2212 1/2 \u03c4 nf \u2193\u00f0 \u00de &\u00bd \u2212 1/2 : & (15.25), Gf $ Because the self-consistency condition requires that the Green\u2019s function of the impurity problem must be equal to the local f Green\u2019s function of the lattice model, we obtain (15.26) \u03a3f = g0 \u2212 G\u22121 f \u2212<Tff \u2020> Seff : \u221e Gf \u00f0 i\u03c9n\u00de = d\u2208 D \u2208 \u00f0 \u00de \u2212 V 2 i\u03c9n\u00de i\u03c9n \u2212 \u2208f \u2212 \u2211f \u00f0 is the density of states (noninteracting) of the conduction electrons. i\u03c9n \u2212 \u2208 \u00f0 % \u2212\u221e Z \u00de : (15.27) The temperature dependence of the electronic transport of the heavy-fermion systems can be calculated by using a self-consistent second-order perturbation theory in terms of the Coulomb repulsions U. Here, D \u2208 \u00de \u00f0 15.3.3 RKKY Interaction There are two competing interactions in the heavy-fermion system: the indirect exchange between the moments mediated by the RKKY interaction (Refs. 12, 21, 36) and the Kondo exchange between the conduction electrons and the moments. The conducting electrons and the moments retain their identities and interact weakly. The RKKY interaction is described in the following section. Ruderman and Kittel21 considered the problem of nuclear-spin ordering in a metal and used second-order perturbation theory to derive an expression for the indirect nuclear spin\u2013spin interaction (Problem 15.5), HRKKY = \u2212 9\u03c0 8 n2 c J2 \u03b5F \u2211 <ij> Si. Sj r3 ij 2kF cos\u2061 2kFrij\u00de \u00f0 sin\u2061 \u00f0 \u2212 2kFrij\u00de rij, \u2019 & where kF is the Fermi wave vector, and nc is the density of conduction electrons. The spin\u2013spin interaction is long ranged and changes its sign depending on the distance between the pair of spins. Kasuya discussed the magnetic properties of rare-earth metals based on Eq. (15.28).12 Yosida36 (15.28) 15.3 Mean-Field Theories 501 showed that the oscillatory behavior originates from the Friedel oscillation of the spin polarization of conduction electrons induced by a localized spin. Therefore, Eq. (15.28) is known as the RKKY interaction. For rare-earth metals, the Fourier transform of the RKKY interaction is given by \u03c7(q), the susceptibility of the conduction electrons for wave number q. The ground state is usually ferromagnetic if \u03c7(q) is maximum at q = 0. If the maximum of \u03c7(q) occurs at q = Q, the antiferromagnetic wave vector, the ground state becomes antiferromagnetic. The ground state may have a spiral spin ordering if \u03c7(q) becomes maximum at a general wave vector. The Kondo effect is suppressed whenever there is any type of magnetic ordering. The low-energy physics of the Kondo effect is given by the Kondo temperature TK = \u03b5Fe\u22121/", "J\u03c1 \u03b5F \u00f0 \u00de: (15.29) However, the characteristic energy of the RKKY interaction is given by J2/\u03b5F: This energy dominates over the Kondo temperature in the weak-coupling regime. In the strong-coupling regime, the local moments are quenched because of the formation of local singlets. The Kondo effect or the effect of singlet formation is not considered for the derivation of the RKKY interaction. The relation between RKKY interaction and the Kondo effect depends on the conduction electron density, dimensionality, and the exchange coupling. As an example, we consider two localized spins, S1 and S2. The direct exchange coupling between the two spins can be expressed as H = JRKKYS1. S2, (15.30) where JRKKY, the intersite coupling constant, is arbitrary. For J > 0, the Kondo coupling is antiferromagnetic, and the ground state is a singlet. When J > JRKKY, each of the two localized spins forms a singlet with conduction electrons, and hence, the interaction between the singlets is weak. When JRKKY\u00bb J, the two localized spins form a singlet by themselves, and J is no longer important. There is a difference among theorists as to whether the change between the two regimes is smooth or sharp. 15.3.4 Extended Dynamical Mean-Field Theory16 The extended dynamical mean-field theory (EDMFT), which is an extension of DMFT, is particularly suitable to solve problems such as the competition between the exchange interaction and kinetic energy. In the EDMFT, the local quantum fluctuations are treated on the same level as the intersite quantum fluctuations. This is achieved by reducing the correlated lattice problem to a novel effective impurity problem corresponding to an Anderson impurity model with additional self-consistent bosonic baths. These bosonic baths reflect the influence of the rest of the lattice on the impurity site. As an example, they represent the fluctuating magnetic fields induced by the intersite spin-exchange interactions in the magnetic case. The intersite quantum fluctuations are included through self-consistency. Smith and Si28 applied the EDMFT method to the two-band Kondo-lattice model H = \u2211 <ij>,\u03c3 tijC\u2020 i\u03c3Cj\u03c3 + \u2211 i JK Si!. sci! \u2212 \u2211 <ij> Jij Si!. Sj!, (15.31). \u03c7\u22121 s,0\u00f0, \u00de \u00de \u03c4 \u2212 \u03c4\u2032 \u03c4\u2032 S! \u00f0 i (15.32) s,0 are determined by 502 CHAPTER 15 Heavy Fermions where Si! is the impurity spin at site i, and s!ci is the spin of conduction electrons at site i, tij is the hopping integral, and Jij is the spin-exchange interaction. In the large D (dimension) limit, with Dp, they derived an expression for the impurity action t0 = tij Dp and J0 = Jij c\u2212 \u00de \u00f0 \ufb03\ufb03\ufb03\ufb03 SMF = Stop + \u03b2 \ufb03\ufb03\ufb03\ufb03 d\u03c4JK S!i. s!c \u2212 \u03b2 d\u03c4 \u03b2 d\u03c4\u2020 \u03c3\u00f0 G\u22121 0 \u00f0 \u00de \u03c4 \u2212 \u03c4\u2032 \u00de C\u03c3\u00f0 \u03c4\u2032 \u00de + S! \u00f0 \u03c4 \u00de where Stop is the Berry phase of the impurity spin. The Weiss fields G\u22121 the self-consistency equations, 0 and \u03c7\u22121 G\u22121 0 \u00f0 i\u03c9n\u00de = i\u03c9n + \u03bc \u2212 \u2211 ij ti0t0j\u00bd Gij\u00f0 i\u03c9n\u00de \u2212 Gi0\u00f0 i\u03c9n\u00de G0j\u00f0 Gloc\u00f0 i\u03c9n\u00de % i\u03c9n\u00de& and \u03c7\u22121 s,0 = \u2211 ij Ji0J0j\u00f0 \u03c7s,ij \u2212 \u03c7s,i0 \u03c7s,0j=\u03c7s,loc\u00de : (15.33) (15.34) Here, \u03c7s is the spin susceptibility. Smith and Si (Ref. 28) also showed that the effective action can be written in terms of the impurity problem, Himp = \u2211 k\u03c3 Ek\u03b7\u2020 k\u03c3\u03b7k\u03c3 + \u2211 q \u2020 wq \u03d5! q. \u03d5!q \u2212 \u03bc\u2211 \u03c3 C\u2020 \u03c3C\u03c3 C\u2020 \u03c3\u03b7k\u03c3 + H:C: + t\u2211 k\u03c3 \u00f0 q where Ek, t, wq, and g are determined from the Weiss fields G\u22121 = G\u22121 0 \u00f0 i\u03c9n", " + \u03bc \u2212 t2\u2211 k iwn \u2212 Ek\u00de + JK S!. s!c + g\u2211 1 \u00de 0 and \u03c7\u22121 i\u03c9n\u00de S!. \u2020 \u03d5!q + \u03d5! \u2212q\u00de, \u00f0 s,0 specified by \u00f0 % and g2\u2211 q wq 2 i\u03bdn\u00de \u00bd\u00f0 \u2212 w2 q& = \u03c7\u22121 s,0 \u00f0 : i\u03bdn\u00de % (15.35) (15.36) (15.37) 15.4 FERMI-LIQUID MODELS 15.4.1 Heavy Fermi Liquids A number of universal features are associated with the coherent Fermi-liquid state in heavy-fermion systems. They can be summarized as follows: a. The dimensionless Wilson ratio (Ref. 35) \u03c7 0 \u00f0 R = is close to the value of unity. \u03bc2 B g2 JJ 0 J + 1 \u00f0 \u00de \u03c02k2 B \u00de \u03b3 % \u00f0 \u00de % function of the form T 3 ln T. c. The resistivity \u03c1 is proportional to T 2: d. The low-T susceptibility \u03c7T also appears to vary as T 2. b. The specific heat CV has a rapid downturn with increasing temperature, which has been fit to a 15.4 Fermi-Liquid Models 503 In addition, the evidence of universal behavior of heavy fermions is the observation of Kadowaki and Woods that \u03c1/T 2 \u2261 A is the same multiple of \u03b32 for essentially all materials.11 A large number of the heavy-fermion systems become heavy Fermi liquids at low temperatures in the sense that Landau\u2019s Fermi-liquid theory is still adequate to describe the physics, provided the large effective mass is included. In the Fermi-liquid theory, the specific-heat enhancement Cv/Cv0 is related, at low temperatures, to the quasiparticle density of states at the Fermi surface. This is equivalent to an average Fermi velocity or to that of an average mass. In view of the above, the effective mass is defined as Cv Cv0 = m! m : (15.38) Here, it is important to comment on the physical interpretation of the quasiparticles. The f electrons are supposed to be hopping from site to site. There are a large number of f electrons, and the Luttinger theorem (described later) requires the Fermi surface to contain the total number of states and not a volume containing the mobile holes in the f-band. The f electrons have very large mass due to the weak effective hybridization. Thus, the quasiparticles are essentially f electrons, and the quasiparticle bands are f -bands that have moved up to the Fermi energy and have been narrowed by correlation. The heavy Fermi liquid arises due to the Kondo screening of the localized moments at each lattice site. In a sense, the localized moments \u201cdissolve\u201d into the Fermi sea. We will now summarize the concepts of the various heavy Fermi-liquid models by following the elegant but brief review of Senthil et al.25 The Kondo-lattice model can be written as HK = \u2211 k \u03b5kC\u2020 k\u03b1Ck\u03b1 + JK 2 \u2211 r Sr. C\u2020 r\u03b1\u03c3\u03b1\u03b1\u2032Cr\u03b1\u2032: (15.39) Here, nc is the density of conduction electrons with dispersion \u03b5k, C\u2020 k\u03b1 and Ck\u03b1 are the creation and annihilation operators of conduction states, k is the momentum, and \u03b1 = \u2191, \u2193 is a spin index. The conduction electrons interact with f electron spins Sr via the antiferromagnetic Kondo exchange coupling constant JK. Here, r is a lattice position, and \u03c3 are the Pauli spin matrices. In the heavy-fermion liquid models, the charge of the fr\u03b1 electrons is fully localized on the rare-earth sites. These electrons occupy a flat dispersionless band, as shown in Figure 15.11a. Because this band is half filled, it is placed at the Fermi level. The Cr\u03b1 electrons occupy their own conduction band. The Kondo exchange turns on a small hybridization between these two bands. The hybridization can be represented by a bosonic operator C\u2020 (15.40) br \u223c \u2211 \u03b1 r\u03b1 fr\u03b1: Because <br> is nonzero, renormalized bands are formed (Figure 15.11b) due to the mixture of the two bands. Because the f-band was initially dispersionless, the renormalized bands do not overlap. One now applies the Fermi-surface sum", " rule by Luttinger (Ref. 13), also known as the Luttinger theorem (Ref. 14). According to Luttinger, the volume enclosed by the Fermi surface is entirely determined by only the electron density. The volume is independent of the type and strength of an interaction, if the system remains a Fermi liquid and no phase transition occurs. Using this theorem leads to the conclusion that the occupied states are entirely within the lower band, and a single Fermi surface is obtained within wave vector kF. The volume within kF is obtained by the total density of f and c electrons. 504 CHAPTER 15 Heavy Fermions c f k (a) kF k (b) FIGURE 15.11 A completely flat f electron \u201cband\u201d\u2014the dashed line in (a)\u2014mixes with the conduction electrons to obtain the renormalized bands in (b). The single Fermi surface at kF in the Fermi-liquid state contains states of which the wave number equals the sum of the c and f electrons. Reproduced from Senthil, Sachdev, and Vojta25 with the permission of Elsevier. Hot Fermi surface Cold Fermi surface Fermi liquid Spinon Fermi surface Small electron Fermi surface Fractionalized Fermi liquid FIGURE 15.12 Fermi-surface evolution from FL to FL*: The FL phase has two Fermi-surface sheets (the cold c and the hot f sheets) close to the transition. The f sheet becomes the spinon Fermi surface while the c sheet is the small conduction-electron Fermi surface on the FL* side. Reproduced from Senthil, Vojta, and Sachdev27 with the permission of the American Physical Society. The Fermi surface is in a region (Figure 15.11b) where the electrons primarily have an f character, and the band is flat. According to this model, this accounts for the large effective mass of the fermionic quasiparticles. Because the charge fluctuations are quenched at the f electron sites, every rare-earth site has a constraint f \u2020 r\u03b1 fr\u03b1 = 1, \u2211 \u03b1 (15.41) 15.4 Fermi-Liquid Models 505 which is obeyed at each rare-earth site. This implies that the theory is invariant under the spacetime-dependent U(1) gauge transformation f \u2020 r\u03b1! \u03c4 fr\u03b1ei\u03d5r \u00f0 \u00de, (15.42) where \u03c4 is imaginary time. 15.4.2 Fractionalized Fermi Liquids Senthil et al.26,27 showed the existence of nonmagnetic translation-invariant small-Fermi-surface states, originally with a focus on two-dimensional Kondo lattices. These states are obtained when a due to intermoment interactions. local-moment system settles into a fractionalized spin liquid A weak Kondo coupling to conduction electrons leaves a sharp (but small) Fermi surface of quasiparticles (FL) of which the volume counts the conduction density, but the structure of the spin liquid is undisturbed. These states have fractionalized excitations that coexist with conventional Fermi-liquid-like quasiparticles. FL! \u00f0 \u00de In this paper, Senthil et al. considered a three-dimensional lattice by using U states. They spin-liquid state with fermionic spinons that form a Fermi focused on a three-dimensional U spin-liquid state is stable to a weak Kondo coupling to conduction electrons. The surface. The U 1 FL! state consists of a spinon Fermi surface coexisting with a separate Fermi surface of U conduction electrons. Senthil et al. used a mean-field theory to describe a U FL! state and its \u00de transition to a heavy FL. They considered a three-dimensional Kondo\u2013Heisenberg model on a cubic latticekC\u2020 k\u03b1Ck\u03b1 + JK 2 \u2211 r S!r. C\u2020 r\u03b1 \u03c3!\u03b1\u03b1\u2032Cr\u03b1\u2032 + JH \u2211 <rr\u2032> S!r. S!r\u2032: (15.43) Here, Ck\u03b1 is the conduction electron destruction operator, S!r are the spin-1/2 local moments, and summation over repeated spin indices \u03b1 is implied. In a fermionic \u201cslave-particle\u201d representation of the local moments S!r = 1/2 f \u2020 r\u03b1 \u03c3!\u03b1\u03b1\u2032 fr\u03b1\u2032, (15.44) where fr\u03b1 is a spinful fermion destruction operator at site r. The decoupling of the Kondo and the Heisenberg exchange is made using two auxiliary fields by a", " saddle-point approximation, and the mean-field Hamiltonian is Hmf = \u2211 k \u2208k C\u2020 k\u03b1Ck\u03b1 \u2212 \u03c70 \u2211 <rr\u2032>\u00f0 f \u2020 r\u03b1 fr\u2032\u03b1 + H:C: + \u03bcf \u2211 r \u00de f \u2020 r\u03b1 fr\u03b1 \u2212 b0\u2211 k \u00f0 C\u2020 k\u03b1 fk\u03b1 + H:C: : \u00de (15.45) Here, b0 and \u03c70 are assumed to be real, and additional constants to H are dropped. The mean-field parameters b0, \u03c70, and \u03bcf are obtained from (Problem 15.3) 1 = < f \u2020 r\u03b1 fr\u03b1>, b0 = JK 2 <C\u2020 r\u03b1 fr\u03b1>, % 2 < f \u2020 \u03c70 = JH r\u03b1 fr\u2032\u03b1>, % (15.46) (15.47) (15.48) 506 CHAPTER 15 Heavy Fermions where r and r\u2032 are nearest neighbors. At zero temperature, in the Fermi-liquid (FL) phase, \u03c70, b0, and \u03bc0 are nonzero. In the FL* phase, b0 = \u03bc0 = 0 but \u03c70 \u2260 0: In this state, the conduction electrons are decoupled from the local moments and form a small Fermi surface. The local-moment system is described as a spin fluid with a Fermi surface of neutral spinons. The mean-field is diagonalized by the transformation (Problem 15.4), and Ck\u03b1 = uk\u03b3k\u03b1 + + vk\u03b3k\u03b1\u2212 fk\u03b1 = vk\u03b3k\u03b1 + \u2212 uk\u03b3k\u03b1\u2212: The Hamiltonian can be written in terms of the new fermionic operators \u03b3k\u03b1\u00b1, Hmf = \u2211 k\u03b1 where Ek + \u03b3\u2020 k\u03b1 + \u03b3k\u03b1 + + Ek\u2212\u03b3\u2020 k\u03b1\u2212\u03b3k\u03b1\u2212, Here, \u2208kf = \u03bcf \u2212 \u03c70\u2211a = 1,2,3 cos\u2061 \u00f0 ka\u00de Ek\u00b1 = \u2208k + \u2208kf 2 \u00b1 2 \u2208k \u2212 \u2208kf 2 + b2 0: \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03! r : The uk, vk are determined by \" uk = \u2212 b0vk Ek + \u2212 \u2208k, k + v2 u2 k = 1: (15.49) (15.50) (15.51) (15.52) (15.53) For the FL* phase, b0 = \u03bc0 = 0 but \u03c70 \u2260 0: The conduction-electron dispersion \u2208k determines the electron Fermi surface and is small. The spinon Fermi surface encloses one spinon per site and has volume half that of the Brillouin zone. Senthil et al. assumed that the conduction\u2012electron filling is less than half, and the electron Fermi surface does not intersect the spinon Fermi surface. In the FL phase near the transition (small b0), there are two bands corresponding to Ek\u00b1: one derives from the c electrons with f character (c-band), whereas the other derives from the f particles with weak c character ( f-band). As shown in Figure 15.12, for small b0, the Fermi surface consists of two sheets because both bands intersect the Fermi energy. The total volume is large because it includes both local moments and conduction electrons. When b0 decreases to zero, the transition moves to FL*, the c-Fermi surface expands in size to match onto the small Fermi surface of FL*, and the f-Fermi surface shrinks to match onto the spinon Fermi surface of FL*. 15.5 METAMAGNETISM IN HEAVY FERMIONS The name metamagnetism was originally introduced for antiferromagnetic (AF) materials where, at low temperatures, for a critical value of the magnetic field (H), the spin flips, which gives rise to a first-order phase transition. This was extended to paramagnetic (Pa) systems where field reentrant ferromagnetism (F) would appear in itinerant magnetism. Eventually, it was used to describe a crossover", " inside a persistent paramagnetic state between low-field Pa phase and an enhanced paramagnetic polarized (PP) phase. 15.5 Metamagnetism in Heavy Fermions 507 In heavy-fermion systems, the f electrons are located near the border between itinerant and local moment behavior, as shown in the phase diagram in Figure 15.13. Doniach considered a onedimensional analog of a system of conduction electrons exchange-coupled to a localized spin in each cell of a lattice.6 He suggested that a second-order transition from an antiferromagnetic to a Kondo spin-compensated ground state would occur as the exchange coupling constant J increased to a critical value Jc. For J near to, and slightly smaller than Jc, there would exist antiferromagnets with very weak, \u201cnearly quenched\u201d moments, even though the f electrons are in a state with a welldefined local nonzero spin state. The existence of this transition can be understood by comparing the binding energy of the Kondo singlet with that of an RKKY antiferromagnetic state WK \u223c N 0 \u00de \u00f0 \u22121e\u22121/N 0 \u00f0 J, \u00de WAF \u223c CJ2N 0, \u00de \u00f0 (15.54) (15.55) where N(0) is the density of conduction electron state, and C is a dimensionless constant that depends on the band structure. As shown in Figure 15.13, for JN(0) less than a critical value, the RKKY state dominates, whereas above this, the Kondo singlet binding dominates. The RKKY binding again takes over at large J, but the weak coupling formula (Eq. 15.54) breaks down in this regime. The heavy fermions that exhibit metamagnetism are CeRu2Si2, Sr3Ru2O7, CeCu6\u2212xAux, UPt3, UPd2Al3, URu2Si2, CePd2Si2, YbRh2Si2, and CeIr3Si2: There have been many theories proposed for metamagnetism of heavy fermions, but no satis- factory model is yet available. A review of the various theoretical models was made by Misra.16 e\u22121/J J 2/2.5 10\u22121 10\u22122 0 0.5 1 1.5 J FIGURE 15.13 Comparison of AF with Kondo energies. Reproduced from Doniach6 with the permission of Elsevier. 508 CHAPTER 15 Heavy Fermions 15.6 CE- AND U-BASED SUPERCONDUCTING COMPOUNDS16 15.6.1 Ce-Based Compounds Since 1979, approximately 25 unconventional superconductors have been discovered in heavyfermion systems. Although most of these systems are Ce- and U-based compounds, a few others are quasi-two-dimensional in nature and filled skutterdites. The multiphase diagrams in UPt3 and U (Be1\u2212xThx)13 indicate unusual superconductivity with multicomponents. In fact, UPt3 is the first oddparity superconductor to be discovered in heavy-fermion systems. UPd2Al3 and UNi2Al3 are unconventional superconductors coexisting with the AF phase and are considered to have even- and odd-parity pairing states, respectively. There is coexistence of hidden-order and unconventional superconductivity in URu2Si2. UPt3, URu2Si2, UNi2Al3, and UPd2Al3 have the following common features: (a) they order antiferromagnetically below TN, ranging from 5 to 17\u00b0 K; and (b) they exhibit, well below TN and coexisting with AF order, a heavy Landau Fermi-Liquid (LFL) state that becomes unstable against a superconducting transition at Tc (ranging between 0.5 and 2\u00b0 K). Recently, a variety of heavy-fermion Ce-based superconductors were discovered due to progress in experiments under pressure. They include CeCu2Ge2, CePd2Si2, CeRh2Si2, CeNi2Ge2, and CeIn3. These materials, which have the same ThCr2Si2-type crystal structure as CeCu2Si2 (except CeIn3), are AF metals at ambient pressure, whereas under high pressures, the AF phases abruptly disappear accompanied by SC transitions. The family of CeTIn5 (T = Co, Rh, and Ir), which has a HoCoGa5-type crystal structure (Ref. 16), has attracted a great deal of attention because they possess a relatively high transition temperature (Tc) such as Tc = 2.3\u00b0 K for", " CeCoIn5, which is the highest among Ce- and U-based heavy-fermion superconductors. It has been proposed that valence fluctuations are responsible for the superconductivity in some Ce-based compounds. This is due to the fact that in metallic cerium, the phase diagram shows a first-order valence discontinuity line. This line separates the \u03b3-Ce with a 4f shell occupation nf = 1.0 from the \u03b1-Ce with nf \u2248 0:9. The valence transition is isostructural, and the line has a critical end point in the vicinity of pcr = 2 GPa and Tcr = 600\u00b0 K. In cases in which pcr is positive, either Tcr is very high or Tcr is negative, and only a crossover regime is accessible even at T = 0. The exceptions are CeCu2Si2 and CeCu2Ge2, for which Tcr is likely positive although small. In such a situation, the associated low-energy valence fluctuations can mediate superconductivity. Holmes et al.10 proposed that the superconducting phase diagram for CeCu2(Ge,Si)2, shown in Figure 15.14, exhibits a maximum in the transition temperature in close vicinity to a valencechanging critical point. Miyake (Ref. 16) has argued that superconductivity may develop around the region where the critical end point is suppressed to zero, to become a quantum critical point. The heavy-fermion superconductor CeCoIn5 is a quasi-two-dimensional (2D) system, and the de Haas\u2013van Alphen effect data indicate a quasi-2D Fermi surface. These properties have led to the possibility of a Fulde\u2013Ferrell\u2013Larkin\u2013Ovchinnikov (FFLO) (Refs. 7, 13) superconducting state in CeCoIn5. These states result from the competition between superconducting condensate energy and the magnetic Zeeman energy that lowers the total energy of the electrons in the normal state. This competition is strong when the superconductivity is of a spin-singlet nature. In this case, the superconducting Cooper pairs form with opposite spins, and the electrons cannot lower the total energy of the system by preferentially aligning their spins along the magnetic field. This effect, called Pauli limiting, leads to suppression of superconductivity in the magnetic field. The characteristic 15.6 Ce- and U-Based Superconducting Compounds 509 T TN Ce3+ Ce4+ Tc Tc SC AFM SC P FIGURE 15.14 Schematic phase diagram for CeCu2(Ge,Si)2 illustrating a possible valence fluctuation critical point beneath the superconducting dome at high pressures. Reproduced from P. Coleman, (Ref. 5), with the permission of Elsevier. Pauli field HP determines the upper limit of the superconducting upper critical field Hc2. When Pauli limiting is the dominant mechanism for suppression of superconductivity, a new inhomogeneous superconducting FFLO state would appear at high fields between the normal and the mixed, or the vortex, state below the critical temperature TFFLO with planes of normal electrons that can take advantage of Pauli susceptibility. In the FFLO state, pair breaking due to the Pauli paramagnetic effect is reduced by the formation of a new pairing state (k\u2191, \u2013k + q\u2193), with |q| ~ 2\u03bcBH/\u0127vF (vF is the Fermi velocity) between the Zeeman split parts of the Fermi surface. One of the intriguing features is the T and H phase dependence of the phase boundary between the FFLO and non-FFLO FFLO (H \ufe8d\ufe8d ab) exhibits an unusually large shift to higher fields at higher superconducting state. H \ufe8d\ufe8d temperatures. The results of Bianchi et al.2 are shown in Figure 15.15. 15.6.2 U-Based Superconducting Compounds The first two U-based heavy-fermion superconductors, UBe13 (Tc = 0.9\u00b0 K) and UPt3 (Tc = 0.54\u00b0 K), were discovered in 1983 by Ott et al. (Ref. 19) and in 1984 by Stewart et al.,31 respectively. It was evident within a few years that UPt3 had three superconducting phases, which created great impetus for further study of this unusual heavy-fermion superconductor. UBe13 was the first actinide-based heavy-fermion compound that was found to be a bulk superconductor below approximately 0.9\u00b0 K. The cubic UBe13 is also one of the most fascinating HF superconductors because superconductivity develops out of a", " highly unusual normal state characterized by a large and strongly T-dependent resistivity. In addition, upon substituting a small amount of Th for U in U1\u2212xThxBe13, a nonmonotonic evolution of Tc and a second-phase transition of Tc2 below Tc1, the superconducting one, is observed in a critical concentration range of x. It was also shown that the superconducting state is formed by heavy-mass quasiparticles. This was demonstrated by plotting Cp/T versus T (at low temperatures), which is shown in Figure 15.16. The anomaly at Tc is compatible with the large \u03b3 parameter in the normal state at this temperature. The temperature dependence of the specific heat of UBe13 well below Tc was the first indication of the unconventional superconductivity. Figure 15.17 shows the nonexponential but power-law-type 510 CHAPTER 15 Heavy Fermions 0.2 0.4 0.6 0.8 T(K) 1. 11.4 11.2 11.0 H (T) 5.0 2.9 1.9 1.2 0.8 0 10.8 10.6 10.4 10.2 10.0 1.2 9.8 (a) ) T ( H 11.4 11.2 11.0 10.8 10.6 10.4 10.2 10.0 9.8 TFFLO Tc 0.0 0.2 0.4 0.8 1.0 1.2 0.6 T (K) (b) FIGURE 15.15 (a) Electronic specific heat of CeCoIn5 divided by temperature with H \ufe8d\ufe8d [110] collected with the temperature decay method, as a function of field and temperature. (b) Contour plot of the data in (a) in the H-T plane. Gray lines indicate the superconducting phase transition Tc and the FFLO-mixed state TFFLO anomaly. The color scale is the same in (a) and (b). Reproduced from Bianchi et al.2 with the permission of the American Physical Society. 15.6 Ce- and U-Based Superconducting Compounds 511 UBe13 specific heat 1 2 3 4 5 6 7 T (K.5 2.0 1.5 1 0.5 0 0 FIGURE 15.16 Electronic specific heat of UBe13 below 7\u00b0 K. The solid line represents the BCS approximation of the anomaly at and below Tc. Reproduced from Ott18 with the permission of Elsevier. 101 100 10\u22121 10\u2212 UBe13 specific heat ABM(w.c.) 10\u22123 0 3 5 BCS 7 Tc /T ABM(e.c.) 9 11 13 FIGURE 15.17 Normalized electronic specific heat of UBe13 below Tc, plotted versus Tc /T. The solid and broken lines represent calculations assuming point nodes in the gap. Reproduced from Ott18 with the permission of Elsevier. 512 CHAPTER 15 Heavy Fermions decrease of Cp\u00f0 excitation spectrum. T \u00de that was interpreted as being the consequence of nodes in the gap of the electronic It was also found that when small amounts of U atoms in UBe13 were replaced with other elements, there was a substantial reduction of the critical temperature. Tc is also first substantially reduced with the alloys U1\u2212xThxBe13 as x is increased. However, when x > 0.018, Tc increases again until it passes over a willow maximum at x = 0.033 and gradually decreases with a reduced slope when x is further increased. Further, in the range 0.019 < x < 0.05, a second transition at Tc2 below Tc was discovered by measuring the specific heat of these alloys at very low temperatures. Measurements of \u03c1(T) and \u03c7(T) confirmed that the phase at temperatures below the second anomaly of Cp(T) was superconducting. The phase diagram of superconductivity of U1\u2212xThxBe13, from these observations as well as from thermodynamic arguments, is shown in Figure 15.18. One can identify three different superconducting phases: F, L, and U. ) K ( c T 1.0 0.8 0.6 0.4 0.2 0 0 U1\u2212xThxBe13 U L F 2 4 %Th 6 FIGURE 15.18 An x, T phase diagram for superconducting U1\u2212xThxBe13 as derived from the measurements of the specific heat. The letters F, L, and U denote three superconducting phases. Reproduced from Ott18 with the permission of Elsevier. 15.7 Other Heavy-Fermion Superconductors 513 60 50 40 30 20 10 ) K ( T Superconductivity has also been", " discovered in UPt3, and its alloys, URu2Si2; and in UPd2Al3 and its alloys, UNi2Al3, UGe2, URhGe, and UIr. The discovery of superconductivity in UGe2 in single crystals of UGe2 under pressure below Pc \u223c 16 kbar was very surprising. The sensational part of this discovery is that the pressure P \u223c 12 kbar, where the superconducting temperature TS = 0.75\u00b0 K is strongest, the Curie temperature TC \u223c 35\u00b0 K is two orders of magnitude higher than TS; superconductivity occurs in a very highly polarized state (\u03bc(T \u2192 0\u00b0 K) \u223c \u03bcB). The superconductivity in UGe2 disappears above a pressure Pc \u2248 16 kbar that coincides with the pressure at which the ferromagnetism is suppressed. The pressure-temperature phase diagram of UGe2 is shown in Figure 15.19. UGe2 Tc Ferromagnetism 10 \u00d7 Ts SC 10 P (kbar) 15 20 0 0 5 FIGURE 15.19 The pressure-temperature phase diagram of UGe2. Reproduced from Demuer et al.4 with permission of Elsevier. 15.7 OTHER HEAVY-FERMION SUPERCONDUCTORS 15.7.1 PrOs4Sb12 The filled skutterdite PrOs4Sb12 becomes superconducting at Tc = 1:85\u00b0 K. It appears to involve heavy-fermion quasiparticles with effective mass m! \u223c 50 me. There is speculation that the quadrupolar fluctuations play a role in the heavy-fermion superconductivity of PrOs4Sb12. The ground state of Pr3+ ions in the cubic CEF appears to be the \u03933 nonmagnetic doublet. Therefore, the heavy-fermion behavior possibly involves the interaction of the Pr3+ \u03933 quadrupole moments and the charges of the conduction electrons. In such a case, the quadrupolar fluctuations would play a role in the heavy-fermion superconductivity of PrOs4Sb12. The variation of C at low temperature and the magnetic phase diagram inferred from C, the resistivity and magnetization, show that there was a doublet ground state. The two distinct superconducting anomalies in C provide evidence of two superconducting critical temperatures at TC1 = 1.75\u00b0 K and TC2 = 1.85\u00b0 K. This could arise from a weak lifting from of the ground-state degeneracy, which supports the theory of quadrupolar pairing; i.e., superconductivity in PrOs4Sb12 is neither of electron\u2013 phonon nor of magnetically mediated origin. The H-T superconducting phase diagram of PrOs4Sb12 determined by specific heat measure- ments is shown in Figure 15.20. 15.7.2 PuCoGa5 The discovery of superconductivity in the transuranium compound PuCoGa5 with Tc \u2248 18.5\u00b0 K, which is by far the highest critical temperature for any heavy-fermion superconductor, has attracted considerable attention. PuCoGa5 crystallizes in the HoCoGa5 structure, the same type as the CeMIn5 materials. The H-T phase diagram of PuCoGa5, inferred from the heat capacity data as a function of temperature in a magnetic field applied along the three orthogonal directions, is shown in Figure 15.21. 514 CHAPTER 15 Heavy Fermions Hc2 H\u2032 3.5 3 2.5 2 1.5 1 0..5 1 T (K) 1.5 FIGURE 15.20 H-T superconducting phase diagram of PrOs4Sb12. The field dependences of Tc1 and Tc2 are identical. The dashed-dotted line is the same fit with the same parameters as the other lines but without paramagnetic limitation. Reproduced from Measson et al.15 with the permission of Elsevier. Slope =\u2212 9.9 T/ / 450 400 350 300 9 T 17 15 10 18 T(K) 19 PuCoGa5 H || a H || b H || c 0 17.0 17.5 18.0 18.5 T (K) FIGURE 15.21 H-T diagram of PuCoGa5 inferred from single-crystal heat capacity measurements with magnetic field applied along three orthogonal directions. The inset shows representative heat capacity data from which Tc(H) was inferred. Reproduced from J.L. Sarrao et al. (Ref. 23), with the permission of Elsevier. 15.7 Other Heavy-Fermion Superconductors 515 15.7.3 PuRhGa5 The discovery of superconductivity in PuRhGa", "5 with Tc \u2248 9\u00b0 K was reported by Wastin et al. (Ref. 34). PuRhGa5 crystallizes in the tetragonal HoCoGa5 structure with the lattice parameters a = 4.2354 \u00c5 and c = 6.7939 \u00c5. This structure has a two-dimensional feature, where alternating PuGa3 and RhGa2 layers are stacked along the c-axis. There are two crystallographically inequivalent Ga sites in this structure, which are denoted Ga(1) (the 1c site) and Ga(2) (the 4i site), respectively. The Ga(1) site is surrounded by four Pu atoms in the c plane, whereas the Ga(2) site is surrounded by two Pu and two Rh atoms in the a plane. The high-pressure measurements on PuRhGa5 are shown in Figure 15.22, in which the electrical resistance is plotted against temperature for pressures up to 18.7 GPa. This figure displays a metallic shape in the normal state, but an NFL behavior (\u03c1(T) \u223c T1.3) develops up to 50\u201360\u00b0 K. The inset of Figure 15.22 shows the plot of Tc of both PuRhGa5 and PuCoGa5 against pressure. The variation of Tc as a function of pressure (Figure 15.22) suggests that the pairing mechanism is differently affected by pressure for the two materials. The layered crystal structure associated with the quasi-2D Fermi surface calculated for these materials suggests that anisotropic properties might be the cause for this difference. 125 100 75 50 25 ) \u2126 m R ( 0 0 PuCoGa5 PuRhGa5 24 20 16 12 ) K ( c T 0 5 10 P(GPa) 15 8 20 PuRhGa5 1.6 Gpa 2.8 Gpa 9.4 Gpa 18.7 Gpa 100 200 300 T (K) FIGURE 15.22 Evolution of the electrical resistance of PuRhGa5 crystal up to 18.7 GPa. The inset shows Tc behavior of PuCoGa5 as a function of the applied pressure. Reproduced from Griveau et al.8 with the permission of Elsevier. 516 CHAPTER 15 Heavy Fermions The (T-P) phase diagram of PuRhGa5 and PuCoGa5 can be compared to that of CeCoIn5. In these isostructural compounds, the superconducting transition temperature increases with increasing pressure and reaches a maximum before decreasing at higher pressure. The NFL behavior is also maintained over a large range of pressure. 15.7.4 Comparison between Cu and Pu Containing High-Tc Superconductors Recently, Wachter33 compared the Cu \u201chigh Tc superconductors\u201d with equivalent measurements on \u201chigh Tc\u201d PuCoGa5 and PuRhGa5. He observed the following common features. First, in all materials, spin pseudogaps were observed, which necessitates at least antiferromagnetic short-range order, i.e., in clusters. Second, all Cu and Pu superconductors are of mixed valence, as photoemission data have shown. The majority ions (Cu or Pu) are magnetic, and the minority ions are nonmagnetic and act as spin holes. Only short-range correlations remain because these spin holes have a concentration of 10% and hence dilute the antiferromagnetic order. According to Wachter, two dimensionality is not essential and n- or p-type conductivity is not important. 15.8 THEORIES OF HEAVY-FERMION SUPERCONDUCTIVITY The superfluid 3He, the physical properties of which were extensively studied prior to the discovery of heavy-fermion superconductivity, exhibited gap anisotropy and nodal structures like some heavy-fermion compounds. After the discovery of heavy-fermion superconductors, it was natural to compare them with superfluid 3He to be able to understand the former. However, there are many differences between the two systems. For example, the presence of a crystal field and the fact that charged particles are paired in heavy fermions instead of pairing of the neutral atoms in 3He are important. In addition, the strong correlation effects and the spin-orbit interaction in heavy-fermion systems are major factors to be considered. In heavy-fermion compounds, the f-shell electrons are strongly correlated. These f electrons determine the properties of the quasiparticles at the Fermi level, which gives rise to a large effective mass. It is generally believed that superconductivity is mainly by the heavy quasiparticles. These quasiparticles with f characters would have difficulty forming ordinary s-wave Cooper pairs, characteristic of the BCS theory of superconductivity, due to the strong Coulomb repulsion. To avoid a large overlap of", " the wave functions of the paired particles, the system would rather choose an anisotropic channel, such as a p-wave spin triplet (as is done in superfluid 3He) or a d-wave spin singlet state to form pairs. We cannot review here in detail the theory of superconductivity of each heavy-fermion compound. In addition, heavy-fermion systems are one of the areas in physics where the experimentalists are well ahead of the theorists and superconductivity in various heavy-fermion compounds has a different origin. 15.9 KONDO INSULATORS 15.9.1 Brief Review The strongly correlated f-electron materials called Kondo insulators have recently attracted much attention because of their unusual physical properties. At high temperatures, they behave like metals with a local magnetic moment, whereas at low temperatures, they behave as paramagnetic insulators 15.9 Kondo Insulators 517 with a small energy gap at the Fermi level. It appears that a gap in the conduction band opens at the Fermi energy as the temperature is reduced. Despite intensive theoretical and experimental studies, the mechanism of gap formation is still unclear, and there is considerable controversy on how to describe the physics of Kondo insulators. We will concentrate on the 4f and 5f compounds, i.e., those f-element compounds that are in a certain sense \u201cvalence\u201d compounds. The general properties of these materials are characterized by a small gap. The f elements that are present in these compounds have unstable valence, with the valence corresponding to the nonmagnetic f state of the element satisfying the valence requirements of the other elements in the material. The Kondo insulator can be viewed as a limiting case of the correlated electron lattice: exactly one half-filled band interacting with one occupied f-level. This can also be viewed as the limiting case of the Kondo lattice with one conduction electron to screen one moment at each site. However, there has been no clear definition of Kondo insulators, and this situation stems from the confusion over how to understand various types of Kondo insulators consistently. Following is a variety of Kondo insulators, some of which are semiconductors that become Kondo insulators with application of pressure: CeNiSn, Ce3Bi4Pt3, CeRhAs, CeRhSb, CeNiSn, CeRu4Sn6, URu2Sn, CeFe4P12, CeRu4P12, CeOs4Sb12, UFe4P12, TmSe, URu2Sn, YbB12, SmB6, and SmS. A detailed review of the experimental properties of each one of these Kondo insulators is available in Misra.16 15.9.2 Theory of Kondo Insulators The Anderson Lattice Model The Anderson lattice model provides a basic description of the electronic properties of the heavyfermion materials. The solution of the model at half-filling is expected to exhibit an indirect gap in the density of states. The chemical potential lies directly in the gap making the system semiconducting. Thus, if there are four states per atom\u2014two states per atom in the upper hybridized band and two states per atom in the lower hybridized band\u2014then at half-filling, two electrons per atom completely fill the doubly degenerate lower hybridized band and the noninteracting system is semiconducting. According to Luttinger\u2019s theorem, if the interactions are turned on adiabatically so that perturbation theory converges, the ground state of the interacting system will remain insulating. The Hamiltonian can be written as H = Hf + Hd + Hfd, (15.56) where Hf is the Hamiltonian of the lattice of localized f electrons, Hd is the Hamiltonian of the conduction electron states, and Hfd is the hybridization Hamiltonian, Hf = \u2211 i,\u03b1 Ef f \u2020 i,\u03b1 fi,\u03b1 + \u2211 i,\u03b1,\u03b2 Uff 2 i,\u03b1 f \u2020 f \u2020 i,\u03b2 fi,\u03b2 fi,\u03b1, Hd = \u2211 k,\u03b1 \u03b5d\u00f0 k \u00de d\u2020 k,\u03b1dk,\u03b1, and Hfd = N\u22121/2 s \u2211 i,k,\u03b1\u00bd V k \u00de exp\u2061 \u00f0 \u00f0 \u2212ik. Ri\u00de f \u2020 i,\u03b1dk,\u03b1 + V! exp\u2061 k \u00de \u00f0 ik. Ri\u00de \u00f0 d\u2020 k,\u03b1 fi,\u03b1& : (15.57) (15.58) (15.59) 518 CHAPTER 15 Heavy Fermions k \u00de is the dispersion relation for the d-bands; the operators d\u2020 Here, Ef is the binding energy of a single", " f electron to a lattice site, and Uff is the Coulomb repulsion between a pair of f electrons located on the same lattice site. Due to the spin and orbital degrees of freedom, the total degeneracy of each f orbital is 14. This degeneracy can be lifted by spin-orbit coupling and crystal field splitting. We will consider the degeneracy of the lowest f multiplet to be N = 2. The operators f \u2020 i\u03b1( fi\u03b1) create (destroy) an f electron at site i with a combined spin-orbit label \u03b1. The summation is over all lattice sites and all degeneracy labels. create and annihilate an \u03b5d\u00f0 electron in the \u03b1th d sub-band state labeled by the Bloch wave vector k. The hybridization between the f states and the states of the d-band is governed by Hfd. The first term represents a process in which a conduction electron in the Bloch state k hops into the f orbital located at site i. However, \u03b1 is conserved in the process. The Hermitian conjugate term describes an electron in the f orbital at site i tunneling into the conduction band state labeled by the Bloch state k. The summation runs over the total number of lattice sites Ns and over the k values of the first Brillouin zone. dk\u03b1\u00de k\u03b1\u00f0 Riseborough\u2019s Theory 0) is exactly soluble and Riseborough (Ref. 20) showed that the noninteracting Hamiltonian (Uff! the electronic states fall into two quasiparticle bands of mixed f and conduction band character. He showed that in this limit, the binding energy of the f-levels falls within the width of the unhybridized conduction band, which has a width of 2 W = 12t in the tight-binding approximation. The indirect gap is between the zone boundary of the upper branch and the k = 0 state of the upper branch. The direct gap occurs for k values halfway along the body diagonal and has a magnitude of 2V. A sketch of the hybridized bands is shown in Figure 15.23. Each band can contain a maximum of 2N electrons. 6 4 2 0 \u22122 \u22124 t / E E+(k) E\u2212(k) \u22126 \u22121 \u22120.5 FIGURE 15.23 0 k/G/2 0.5 1 Sketch of the hybridized band structure, for k vectors along the body diagonal of the first Brillouin zone. Reproduced from Riseborough20 with the permission of Taylor & Francis Ltd. Thus, the noninteracting system is a semiconductor. If the interactions are turned on adiabatically, Luttinger\u2019s theorem implies that the ground state of the system will be an insulator. In the mean-field approximation, one can use the slave boson technique described earlier. This \u221e limit of the Anderson approach to the Uff! lattice model projects the states of double f occupancy. The f electron operators are replaced by a product of an f quasiparticle operator and a slave boson field, Ef \u2020 i,\u03b1bi, \u2020 i,\u03b1, f f \u2020 i,\u03b1 = f fi,\u03b1 = b\u2020 i e (15.60) e where bi and b\u2020 i are the annihilation and creation operators for the site i, and the f quasiparticle f i,\u03b1: These operators satisfy operators are f the constraints e f i,\u03b1 + b\u2020 e i bi = Qi = 1: \u2020 i,\u03b1 and (15.61) \u2020 i,\u03b1 f \u2211 \u03b1 e e The slave boson field satisfies the equation of motion i\u0127 \u2202 \u2202t \u00f0 = \u03bbib\u2020 b\u2020 i \u00de 1 N1/2 s \u2211 k,\u03b1 i + V \u00f0 Problems 519 k \u00de exp\u2061 \u00bd ik. Ri& \u2020 i,\u03b1dk,\u03b1: f (15.62) The lowest-order approximation, the terms of zeroth order in the boson fluctuation operators bi, is retained. If b0 is finite, this corresponds to a time-independent macroscopic equation of the k = 0 state that is equivalent to assuming that the boson field has undergone Bose\u2013Einstein condensation. In this approximation, Eq. (15.62) can be written as (Problem 15.6) \u03bbib!0 = \u22121 i,\u03b1dk,\u03b1i N1/2 s e Here, b0 and \u03bbi can be determined self-consistently from Eqs. (15.62) and (15.63). The hybridization matrix element is renormalized through ik. Ri&h (15.63) exp\u2061 \u2211 k, and the f -level energy is renormalized through V \u00f0 = b0V", " k, \u00de \u00f0 e Ef = Ef + \u03bb: (15.64) (15.65) This moves the quasiparticle component of the f structure from the incoherent bare f -level component of the density of states to a position near the chemical potential. It can be shown that the quasiparticle dispersion relations are obtained as (Problem 15.7) e Ef \u2212 \u03b5d\u00f0 In this formulation, the amplitude of the slave boson condensate bi is temperature dependent and e vanishes at a critical temperature, Tc, for the semiconductor system. It can be shown that Ef + \u03b5d\u00f0 e k E\u00b1\u00f0 (15.66) = \u00b1 \u00dej \u00de& \u00f0\u00bd /2 1 2 \u00bd Ef \u2212 \u03bc N\u0394& 2/W, W represents approximately half the width of the conduction band, and the where \u0394 = j direct gap has a magnitude of 2V. It is interesting to note that this temperature dependence is related to the Kondo temperature, in which the effects of both band edges are taken into account. kBTc = 1:14W exp\u2061 (15.67), \u2019 V j It may be noted that the slave boson mean-field theory is exact only when the degeneracy of the f-level approaches infinity. In addition, it is valid only when the lower band is fully occupied, which is true only for some Kondo insulators but not others, such as SmB6. The effect of the magnetic field gives rise to a Zeeman splitting of the quasiparticle bands, reducing the hybridization gap. It has been shown by using the periodic Anderson lattice (in the limit of infinite spatial dimen\u221e) that the semiconductor-to-metal transition associated with the high field closing the sions, d gap may be of the first order.! PROBLEMS 15.1. The Schrieffer-Wolff transformation24 can be easily used to relate the Anderson model of a localized magnetic moment to that of Kondo. The two models can be shown to be equivalent in small s \u2212 f mixing. The Anderson Hamiltonian for a single localized orbital f is ^H = \u2211 k\u03c3 \u2208k nk\u03c3 + \u2211 \u03c3 \u2208f nf \u03c3 + Unf \u2191nf \u2193 + \u2211 k\u03c3 \u00bd \u2020 Vkf ^C k\u03c3 ^Cf \u03c3 + V!kf ^C \u2020 f \u03c3 ^Ck\u03c3& = ^H0 + ^H1, (1) 520 CHAPTER 15 Heavy Fermions where \u2208k and \u2208f are the one-electron energies of the conduction and localized orbitals, measured relative to the Fermi energy, and ^H0 is the sum of the first three terms in Eq. (1). The model can be characterized by two dimensionless ratios where and r\u00b1 $ \u0393\u00b1, \u2208\u00b1j j % \u2208f = \u2208f + U, \u03b1 = +, \u03b1 = \u2212, = \u2208f, (2) (3) 2 AVE: Vkf j \u0393\u03b1 = \u03c0N \u2208\u03b1\u00de \u2208\u03b1\u00dej \u00f0 is the density of band states in the perfect crystal at energy \u2208\u03b1, and the matrix N \u00f0 elements are averaged over k states of this energy. If \u2208 + > 0 and \u2208\u2212 < 0, then for Vkf! the ground state is given by the filled Fermi sea and a single electron occupying the f orbital. A localized moment occurs even at zero temperature because the states with f -electron spin \u2191 and \u2193 are degenerate. For r\u03b1 \u226a 1, these two spin states are mixed by electrons hopping on and off the f orbital due to V. Because arbitrarily small energy denominators \u2208k \u2212 \u2208k\u2032 \u2243 0 occur in fourth and higher orders of V, V cannot be treated directly by perturbation theory. However, the interactions that dominate the dynamics of the system for r\u03b1 \u226a 1 can be isolated. Show that one can perform a canonical transformation, 0, (4) by requiring that Vkf is eliminated to the first order, where eSHe\u2212S, H $ Show from Eqs. (4) and (6) that H0, S & \u00bd = H1: S = \u2211 k\u03c3\u03b1 Vkf \u2208k \u2212 \u2208\u03b1 f,\u2212\u03c3C\u2020 n\u03b1 k\u03c3Cf \u03c3 \u2212 H:C:, where the projection operators n\u03b1 f,\u2212\u03c3 are defined by n\u03b1 f,\u2212\u03c3 = nf,\u2212\u03c3, \u03b1 = +, = 1 \u2212 nf,\u2212\u03c3, \u03b1 = \u2212: Show also that in the limit r\u03b1 \u226a 1, H \u2248 H0 + H", "2, where H2 = 1 2 \u00bd S, H1& \u2248 Hex = \u2212\u2211 kk\u2032 Jk\u2032k\u00f0 \u03a8. \u2020 k\u2032S\u03a8k\u00de \u03a8 \u00f0 \u2020, f S\u03a8f \u00de where 2S = \u03c4 are the Pauli spin matrices, \u03a8k and \u03a8f are the field operators \u03a8k = ck\u2191 ck\u2193) *, \u03a8d = cf \u2191, cf \u2193) * (5) (6) (7) (8) (9) (10) (11) Problems 521 and Jk\u2032k = Vk\u2032f Vfkf\u00f0 \u2208k \u2212 \u2208 + \u00de For k and k\u2032 \u2243 kF, Jkk\u2032 is given by \u22121 + \u22121 \u2208k\u2032 \u2212 \u2208 + \u00de \u00f0 \u2212 \u00f0 \u22121 \u2208k\u2032 \u2212 \u2208\u2212\u00de \u2212 \u2208k\u2032 \u2212 \u2208\u2212\u00de \u00f0 \u22121 : g (12) VkF f j This coupling is antiferromagnetic. If there is an f electron at every site of the lattice, from Eq. (10), the f-electron degrees of freedom are represented by localized spins. JkF kF $ J0 = 2 \u2208f \u00f0 \u00de j : (13) 2 U \u2208f + U 15.2. Show that i\u03c9n\u00de, the on-site interacting Green\u2019s function, is calculated from the effective action i\u03c9n\u00de \u00f0 i\u03c9n\u00de \u2212 R \u00bd g0\u00f0 \u22121 = i\u03c9n + \u03bc + G G \u22121 \u22121 \u00f0 : & (1) Here, G \u00f0 Seff defined in Eq. (15.14), i\u03c9n\u00de \u03c4 \u2212 \u03c4\u2032 G \u00f0 \u00de = \u2212 < TC C\u2020 \u03c4 \u00f0 \u00de \u03c4\u2032 \u00f0 \u00de > Seff G \u00f0 i\u03c9n\u00de = \u03b2 Z 0 \u03c4 d\u03c4G \u00f0 \u00de ei\u03c9n\u03c4, \u03c9n $ \u03c0 2n + 1 \u03b2 \u00de \u00f0 : G Here, R \u00f0 corresponding to the lattice. The noninteracting density of states is is the reciprocal function of the Hilbert transform of the density of states \u00de where \u2208 k\u00de, tijeik. Ri\u2212Rj \u00f0 \u00de: \u2208k $ \u2211 ij The Hilbert transform D \u03be \u00f0 \u00de and its reciprocal function R are defined by = D \u03be \u00de \u00f0 \u221e d\u2208 \u2212\u221e Z D \u2208 \u00de \u00f0 \u03be \u2212 \u2208 and D \u00f0 In principle, G can be computed as a functional of g0, using the impurity action Seff. Thus, Eqs. (15.14), (1), and (2) form a complete system of functional equations for the on-site Green\u2019s function G and the Weiss function g0: = \u03be: \u00de& R \u03be \u00bd 15.3. The Hamiltonian of a three-dimensional Kondo\u2013Heisenberg model on a cubic lattice is JK 2 r\u03b1 \u03c3!\u03b1\u03b1\u2032Cr\u03b1\u2032 + JH \u2211 <rr\u2032> H = \u2211 k S!r. S!r\u2032: k\u03b1Ck\u03b1 + S!r. C\u2020 \u2208k C\u2020 \u2211 r (2) (3) (4) (5) (6) (7) (1) 522 CHAPTER 15 Heavy Fermions Here, Ck\u03b1 is the conduction electron destruction operator, S!r are the spin-1/2 local moments, and summation over repeated spin indices \u03b1 is implied. In a fermionic \u201cslave-particle\u201d representation of the local moments S!r = 1/2f \u2020 r\u03b1 \u03c3!\u03b1\u03b1\u2032fr\u03b1\u2032, (2) and fr\u03b1 is a spinful fermion destruction operator at site r. The decoupling of the Kondo and the Heisenberg exchange is made using two auxiliary fields by a saddle-point approximation, and the mean-field Hamiltonian is Hmf = \u2211 k \u2208k C\u2020 k\u03b1Ck\u03b1 \u2212 \u03c70 \u2211 <rr\u2032>\u00f0 f \u2020 r\u03b1 fr\u2032\u03b1 + H:c: \u00de + \u03bcf \u2211 r f \u2020 r\u03b1 fr\u03b1 \u2212 b0\u2211 k \u00f0 C\u2020 k\u03b1 fk\u03b1 + H:C: \u00de : (3) Here, b0 and \u03c70 are assumed to be real and additional constants to H are dropped. Show that the mean-field parameters b0, \u03c70, and \u03bcf are obtained from 1 = < f \u2020", " r\u03b1 fr\u03b1>, b0 = JK/2 <C\u2020 r\u03b1 fr\u03b1>, (4) (5) \u03c70 = JH/2 < f \u2020 where r and r\u2032 are nearest neighbors. At zero temperature, in the Fermi-liquid (FL) phase, \u03c70, b0, and \u03bc0 are nonzero. In the FL* phase, b0 = \u03bc0 = 0, but \u03c70 \u2260 0: In this state, the conduction electrons are decoupled from the local moments and form a small Fermi surface. The local-moment system is described as a spin fluid with a Fermi surface of neutral spinons. r\u03b1 fr\u2032\u03b1>, (6) 15.4. In Problem 15.3, the mean-field is diagonalized by the transformation (Senthil et al.26), and Ck\u03b1 = uk\u03b3k\u03b1 + + vk\u03b3k\u03b1\u2212 fk\u03b1 = vk\u03b3k\u03b1 + \u2212 uk\u03b3k\u03b1\u2212: Show that the Hamiltonian (Eq. 3 in Problem 15.3) can be written in terms of the new fermionic operators \u03b3k\u03b1\u00b1, Hmf = \u2211 k\u03b1 where Ek + \u03b3\u2020 k\u03b1 + \u03b3k\u03b1 + + Ek\u2212\u03b3\u2020 k\u03b1\u2212\u03b3k\u03b1\u2212, Here, \u2208kf = \u03bcf \u2212 \u03c70\u2211a = 1,2,3 cos\u2061 \u00f0 Ek\u00b1 = \u2208k + \u2208kf 2 \u2208k \u2212 \u2208kf 2 \u00b1 + b2 0: \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03! : The uk, vk are determined by r \" 2 ka\u00de uk = \u2212 b0vk Ek + \u2212 \u2208k, k + v2 u2 k = 1: (1) (2) (3) (4) (5) 15.5. The metals of the rare-earth (lanthanide) group have very small 4f n magnetic cores immersed in a sea of conduction electrons from the 6s\u20136p bands. The magnetic properties of these metals can be understood in detail in terms of an indirect exchange interaction between the magnetic Problems 523 cores via the conduction electrons. If the spins of the local magnetic moments at r = ri are Si and at r = rj are Sj, the second-order interaction between the two spins is given by Here, x H\u2033 \u00f0 \u00de = \u2211 \u2032 kk\u2032ss\u2032 < ks H j j k\u2032s\u2032 >< k\u2032s\u2032 \u03b5k \u2212 \u03b5k\u2032 H j j ks > : H = \u2211 kk\u2032ss\u2032 Z & d3x\u03d5!k\u2032s\u2032\u00f0 x A \u00f0 x \u2212 ri\u00de \u00de S. Si\u03d5ks\u00f0 x \u00de c\u2020 k\u2032s\u2032cks, \u2019 (1) (2) is the interaction (which is proportional to the delta function) between the spin where A \u00f0 of the electron S and the spin Si of the local moment at site ri: Here, \u03d5ks are the Bloch functions \u03d5ks = \u03d5kj x \u2212 ri\u00de s >, and S operates on the spin part of \u03d5ks: Show that Eq. (2) can be rewritten as k\u2032\u2193ck\u2191 + S\u2212 k\u2032\u2191ck\u2191 \u2212 c\u2020 c\u2020 H = 1 i c\u2020 S\u2020 k\u2032, k i c\u2020.Ri J k\u2212k\u2032 ei \u00f0, \u00de k\u2032\u2191ck\u2193 + Sz i \u00f0 k\u2032\u2193ck\u2193& \u00f0 \u00de\u00bd 2 \u2211 k,k\u2032 where If J k, k\u2032 \u00de \u00f0 = Z A \u00f0 From Eqs. (1), (2), and (6), show that J \u00f0 x = J\u03b4 x, \u00de \u00f0 \u00de k\u2032, k = J: \u00de d3x \u03d5!k\u2032\u00f0 x A \u00f0 \u00de x \u03d5k\u00f0 \u00de x : \u00de x H\u2033 \u00f0 \u00de S. Si\u00de\u00f0 = \u2211 s \u00f0 S. Sj\u00de mJ2\u0127\u22122 kF \u221e 2\u03c0 \u00f0 \u22126P \u00de Z o d3", "k d3k\u2032 Z kF \u00de :x k\u2212k\u2032 \u00f0 e\u2212i k2 \u2212 k\u20322 + cc: The sum over electron spin states is done with the help of the standard relation between Pauli operators, \u03c3. Si\u00de\u00f0 \u03c3. Sj\u00de \u00f0 = Si. Sj + i\u03c3. Si \u00d7 Sj: Because the trace of any component of \u03c3 vanishes, S. Sj\u00de S. Si\u00de\u00f0 \u2211 s \u00f0 = 1 2 Si. Sj: From Eqs. (7) and (9), by performing the integrations, show that x H\u2033 \u00f0 \u00de = \u00f0 4J2mk4 F 3\u01272r4 ij \u00bd 2\u03c0 \u00de 2kFrij cos 2kFrij\u00de \u2212 sin \u00f0 \u00f0 2kFrij\u00de& Si. Sj: The density of conduction electrons, nc = k3 F 3\u03c02 : (8) (9) (10) (11) (3) (4) (5) (6) (7) 524 CHAPTER 15 Heavy Fermions From Eqs. (10) and (11), we obtain the RKKY interaction, HRKKY = \u2212 9\u03c0 8 n2 c J2 \u03b5F \u2211 <ij> Si. Sj r3 ij 2kF cos 2kFrij\u00de \u00f0 \u2212 & 2kFrij\u00de sin \u2061 \u00f0 rij, \u2019 (12) where a factor of \u00bd has been multiplied to avoid double counting of i and j. The spin\u2013spin interaction is long ranged and changes its sign depending on the distance between the pair of spins. 15.6. The slave boson field satisfies the equation of motion i\u0127 \u2202 \u2202t \u00f0 = \u03bbib\u2020 b\u2020 i \u00de k \u00f0 1 N1/2 s \u2211 k,\u03b1 i + V \u00de exp\u2061 \u00bd ik. Ri& \u2020 i,\u03b1dk,\u03b1: f (1) The lowest-order approximation, the terms of zeroth order in the boson fluctuation operators bi, is retained. If b0 is finite, this corresponds to a time-independent macroscopic equation of the k = 0 state, which is equivalent to assuming that the boson field has undergone Bose\u2013 Einstein condensation. In this approximation, show that Eq. (1) can be rewritten as e 15.7. Show that in the slave boson mean-field theory, the quasiparticle dispersion relations are \u03bbib!0 = \u22121 N1/2 s \u2211 k,\u03b1 V k \u00de exp\u2061 \u00bd \u00f0 ik. Ri&h \u2020 : f i,\u03b1dk,\u03b1i e obtained as k E\u00b1\u00f0 \u00de = 1 2 \u00bd Ef + \u03b5d\u00f0 e \u00b1 k \u00de k Ef \u2212 \u03b5d\u00f0 \u00f0\u00bd e 2 + 4 \u00de& k V j \u00f0 2 1/2 \u00dej \u00de : & e (2) (1) References 1. Anderson PW. Localized Magnetic States in Metals. Phys Rev B 1961;124:41. 2. Bianchi A, Movshovich R, Caban C, Pagluso PG, Sarrao JL. A possible Fulde-Ferrel-Larkin-Ovchinnikov superconducting state in Ce Co In5. Phys Rev Lett 2003;91:187004. 3. Callaway J, Chen DP, Kanhere DG, Misra PK. Cluster Simulation of the Lattice Anderson Model. Phys Rev B 1988;38:2583. 4. Demuer A, Sheikin I, Braithwaite D, Fak B, Huxley A, Raymond S, Flouquet J. J. Magn. Matter 2001;17:226. 5. Coleman P. Physica B 2006;378\u2013380:1160. 6. Doniach S. The Kondo lattice and weak antiferromagnetism. Physica B 1977;91:231. 7. Fulde P, Ferrell RA. Superconductivity in Strong Spin-Exchange Field. Phys Rev A 1964;135:550. 8. Griveau J-C, Boulet P, Collineau E, Wastin F, Rebizant J. Pressure effect on PuMGd5(M=Co, Rh, Zr). Physica B 2005;359\u2013361:1093. 9. Hewson AC. The Kondo problem in heavy fermions. Cambridge: Cambridge University Press; 1993. 10. Holmes AT, Jaccard D, Miyake K. Signatures of Valence Fluctuations in CeCu2Si2 under high pressure", ". Phys Rev B 2004;69:024508. 11. Kadowaki K, Woods SB. A universal relationship of the resistivity and specific heat of heavy-Fermion compounds. Solid State Commun 1986;58:507. References 525 12. Kasuya T. A Theory of Metallic Ferro- and Antiferromagnetism in Zener\u2019s model. Prog Theor Phys (Kyoto) 1956;16:45. 13. Larkin AI, Ovchinnikov YN. Inhomogeneous state of superconductor. Sov Phys JETP 1965;20:762. 14. Luttinger JW. Fermi Surface and Some Simple Equilibrium Properties of a system of Interacting Fermions. Phys Rev 1960;119:1153. 15. Measson M-A, Brison JP, Seyfarth G, Braithwaite D, Lapertat G, Salce B, et al. Superconductivity of the filled skutterdite-PrOs4Sb12. Physica 2005;359\u2013361:827. 16. Misra PK. Heavy-fermion systems. Amsterdam: Elsevier; 2008. 17. Misra PK, Kanhere DG, Callaway J. Periodic Anderson Model for four-site clusters. Phys Rev B 1987;35:5013. 18. Ott HR. Heavy-electrons and non-Fermi liquids, the early times. Physica B 2006;378\u2013380:1. 19. Ott HR, Rudigier H, Delsing P, Fisk Z. UBe3:An unconventional Actinide Superconductor. Phys Rev Lett 1983;50:1595. 20. Riseborough PS. Heavy fermion semiconductors. Adv Phys 2000;49:257. 21. Ruderman MA, Kittel C. Indirect Exchange Coupling of Nuclear Magnetic Moments by Conduction Electrons. Phys Rev 1954;96:99. 22. Sarrao JL, Morales LA, Thompson JD, Scole BL, Stewart GR, Wastin F, et al. Plutonium-based Supercon- ductivity with a transition temperature above 18K. Nature 2002;420:297. 23. Sarrao JL, Brauer ED, Morales LA, Thompson JD. Structural tuning and anisotropy in PuCoGa5. Physica B 2005;359\u2013361:1144. 24. Schrieffer JR, Wolff PA. Rekation between the Anderson and Kondo Hamiltonian. Phys Rev 1966;149:491. 25. Senthil T, Sachdev S, Vojta M. Quantum phase transition out of the heavy Fermi liquid. Physica 2005;359\u2013361:9. 26. Senthil T, Sachdev S, Vojta M. Fractionalized Fermi liquids. Phys Rev Lett 2003;90:216403. 27. Senthil T, Vojta M, Sachdev S. Weak magnetism and non-Fermi liquids near heavy-fermion critical points. Phys Rev B 2004;69:035111. 28. Smith JL, Si Q. Spatial correlations in dynamical mean-field theory. Phys Rev B 2000;61:5184. 29. Steglich F, Aarts J, Bredl CD, Lieke W, Meschede D, Franz W, et al. Superconductivity in the Presence of Strong Pauli Paramagnetism: CeCu2Si2. Phys Rev Lett 1979;43:1892. 30. Stewart GR. Heavy-fermion systems. Possibility of Coexistence of Bulk Supercon. Rev Mod Phys 1984;56:755. 31. Stewart GR, Rudigier H, Delsing P, Fisk Z. Possibility of Coexistence of Bulk Supercon. Phys Rev Lett 1984;52:679. 32. Varma CM. Mixed-Valence Compounds. Rev Mod Phys 1976;48:219. 33. Wachter P. Similarities between Cu and Pu containing \u201chigh-Tc Superconductor.\u201d Physica C 2007;453:1. 34. Wasrtin F, Boule P, Rebizant J, Collineau E, Lander GH. Advances in the preparation and characterization of transuranium systems. J. Phys. Condens. Matter 2003;15:S2279. 35. Wilson KG. The renormalization group: Critical phenomena and the Kondo problem. Rev Mod Phys 1975;47:773. 36. Yosida K. Magnetic Properties of Cu-Mn Alloys. Phys Rev 1957;106:893. Metallic Nanoclusters CHAPTER 16 CHAPTER OUTLINE 16.1 Introduction..................", "........................................................... 528 16.1.1 Nanoscience and Nanoclusters..................................................... 528 16.1.2 Liquid Drop Model................................................................. 528 16.1.3 Size and Surface/Volume Ratio..................................................... 528 16.1.4 Geometric and Electronic Shell Structures........................................... 530 16.2 Electronic Shell Structure................................................................ 531 16.2.1 Spherical Jellium Model (Phenomenological )........................................ 531 16.2.2 Self-Consistent Spherical Jellium Model............................................. 532 16.2.3 Ellipsoidal Shell Model............................................................. 535 16.2.4 Nonalkali Clusters................................................................. 535 16.2.5 Large Clusters..................................................................... 535 16.3 Geometric Shell Structure................................................................ 537 16.3.1 Close-Packing..................................................................... 537 16.3.2 Wulff Construction................................................................. 537 16.3.3 Polyhedra......................................", "................................... 538 16.3.4 Filling between Complete Shells.................................................... 540 16.4 Cluster Growth on Surfaces............................................................... 540 16.4.1 Monte Carlo Simulations........................................................... 540 16.4.2 Mean-Field Rate Equations......................................................... 541 16.5 Structure of Isolated Clusters............................................................. 542 16.5.1 Theoretical Models................................................................ 542 16.5.2 Structure of Some Isolated Clusters................................................. 546 16.6 Magnetism in Clusters................................................................... 547 16.6.1 Magnetism in Isolated Clusters..................................................... 547 16.6.2 Experimental Techniques for Studying Cluster Magnetism............................. 549 16.6.3 Magnetism in Embedded Clusters................................................... 553 16.6.4 Graphite Surfaces................................................................. 555 16.6.5 Study of Clusters by Scanning Tunneling Microscope................................. 555 16.6.6 Clusters Embedded in a Matrix..................................................... 557 16.7 Superconducting State of Nanoclusters.............................................", "........ 558 16.7.1 Qualitative Analysis................................................................ 558 16.7.2 Thermodynamic Green\u2019s Function Formalism for Nanoclusters......................... 559 Problems.................................................................................... 562 References.................................................................................. 565 Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00016-5 \u00a9 2012 by Elsevier Inc. All rights reserved. 527 528 CHAPTER 16 Metallic Nanoclusters 16.1 INTRODUCTION 16.1.1 Nanoscience and Nanoclusters Nano objects have a size that is intermediate between atoms or molecules and bulk matter. The recent invention of a variety of tools for studying systems at the atomic level, coupled with the development of techniques for producing nanoclusters, has led to the use of nanoscience as a new field of study. The scanning probe microscopes (the first of which was the scanning tunneling microscope, or STM), make it possible to \u201csee\u201d individual atoms and molecules on surfaces of materials as well as to move them on the nanoscale. New sources to produce clusters in the gas phase were developed in the 1960s and 1970s, but it was in 1980s that Knight et al.16 first produced clusters of alkali metals with approximately 100 atoms and systematically studied their properties. Nanoclusters can now be formed from most elements of the periodic table. They can be classified as metallic, semiconductor, ionic, rare gas, or molecular, according to their constituents. Clusters are classified as homogeneous if they contain a single type of atom or heterogeneous if they comprise more than one constituent. They may be neutral or charged (anions or cations). There have been two main approaches in creating nanostructures: top-down and bottom-up. In the top-down method, starting from a large piece of material, a nanostructure is formed by removing material from it through etching or machining by using an electron beam or focused ion beam lithography. In the bottom-up approach, nanoparticles or molecules are produced by chemical synthesis followed by ordered structures by physical or chemical interactions between the units. In this chapter, we will concentrate on the study of metallic nanoclusters. 16.1.2 Liquid Drop Model The simplest description of a metallic nanocluster is the liquid drop model (LDM). The cluster is represented as a sphere of radius R, which is related to the number of atoms N through the Wigner\u2013Seitz radius rs, R = N1/3rs: (16.1) Here, the Wigner\u2013Seitz radius rs, originally defined as the radius of the volume occupied by each valence electron, is equivalent to the volume occupied by each atom in a monovalent nanometal. The internal structure of the cluster is ignored in this model. This model is equivalent to the free electron theory of solids. However, the solid box has macroscopic dimensions with a continuum of energy levels, whereas the cluster box is a nanoscale entity and the energy levels are discrete. The first few energy levels for noninteracting electrons in a spherical box and the number of electrons required for complete filling of the shells are shown in Figure 16.1. 16.1.3 Size and Surface/Volume Ratio The fraction of atoms that are on the surface of a cluster distinguishes the difference of the properties of the cluster from the bulk. One simple way to analyze this is to cut a regularly shaped object 16.1 Introduction 529 58 40 34 20 18 8 2 FIGURE 16.2 FIGURE 16.1 First few energy levels for noninteracting electrons in a spherical box. A 561-atom cut from a cubocta", "hedral cluster cut from a bulk fcc crystal. Reproduced from Misra 28 with the permission of Elsevier. from a fcc lattice and count the surface atoms for different size clusters. Such clusters have eight triangular and six square faces, as shown in Figure 16.2. The fraction of atoms that are on the surfaces of the clusters is shown in Figure 16.3. Clusters represent a state of matter that is intermediate between atoms and the solid or liquid state, with properties that depend on the size, shape, and material of the particle. The arbitrary property per atom, x, can be expressed as.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 10 102 103 104 105 106 Size of cluster, N N x \u00f0 \u00de = a + bN\u22121/3, (16.2) FIGURE 16.3 where the first term is the \u201cbulk\u201d contribution, and the second term is the \u201csurface\u201d contribution. Fraction of atoms on cuboctohedral clusters of N atoms. Reproduced from Misra 28 with the permission of Elsevier. 530 CHAPTER 16 Metallic Nanoclusters 16.1.4 Geometric and Electronic Shell Structures Knight et al.16 produced clusters through the supersonic expansion of a metal/carrier gas mixture. The alkali metal is first vaporized and then seeded in the inert carrier gas (typically argon). The supersonic expansion of the mixture results in adiabatic cooling. The cluster is then ionized and passed through a mass spectrometer. Knight et al.16 observed marked peaks in the mass spectrum of the clusters indicating high stability at particular sizes. Their experiment results are reproduced in Figure 16.4.6 1.2 0.8 0.a) 2p 0.4 0 1f 1g 1p 2s 1d \u22120.4 8 40 20 Number of sodium atoms per cluster, N 58 34 FIGURE 16.4 (b) 92 1h 92 3s 2d 70 (a) The mass spectrum of sodium clusters with N = 4\u201375 and 75\u2013100 (inset). (b) The calculated change in the electronic energy difference, \u0394 orbitals. versus N. The peak labeling corresponds to closed-shell Reproduced from Knight et al.16 with the permission of the American Physical Society. 16.2 Electronic Shell Structure 531 The marked peaks or steps at special numbers in the sodium-cluster mass spectra are called \u201cmagic numbers.\u201d These magic numbers were associated with complete filling of electronic shells, a concept of the electronic shell model borrowed from nuclear physics. From Figure 16.4, the magic number peaks are seen at cluster sizes N = 8, 20, 40, 58, and 92. 16.2 ELECTRONIC SHELL STRUCTURE 16.2.1 Spherical Jellium Model (Phenomenological ) In the spherical jellium model, the ionic cores are replaced by a uniform positive charge background of radius R. The electrons are treated as independent particles moving in a parametrized phenomenological potential. The basic parameter of the model is the Wigner\u2013Seitz radius rs defined in Eq. (16.1). Further, the wave function for a spherically symmetric potential can be written as = Rnl\u00f0 There are three empirical potentials used to describe the nanoclusters, all borrowed from basic ideas of nuclear physics. The simplest model is the harmonic oscillator potential. \u03c8 nlm\u00f0 r Ylm\u00f0 (16.3) \u03b8, \u03d5 r \u00de \u00de \u00de : Harmonic Oscillator Potential The harmonic oscillator potential is described by The energy of the harmonic oscillator potential is given by = V r \u00f0 \u00de 1 2 m\u03c90r2: E\u03bd = + \u03bd 3 2! \u0127\u03c90: \" (16.4) (16.5) The quantum numbers \u00f0 with the same value of quantum number \u00f0 2\u03bd + l \u2212 2 \u00f0 \u00de \u00de n, l \u00de 2n + l can be used for any spherically symmetric potential. Thus, all orbitals are degenerate and the energies are written in terms of the single. The potential energy due to the background charge is r2 3 # 3e2N 8\u03c0\u03b50R3 e2Z e2N 4\u03c0\u03b50r \u2212 R2 if r < R, $ if r > R: (16.6) = V r \u00f0 \u00de 8 >>>>>< >>>>>: So the harmonic oscillator potential mimics the potential felt by the electrons inside the cluster, provided the electron\u2013electron interaction is ignored. Spherical Square-Well Potential The spherical square-well potential is described by V r \u00f0 \u00de = C for r < R, = \u221e otherwise: (16", ".7) 532 CHAPTER 16 Metallic Nanoclusters Here, C is a constant. The radial wave function Rnl\u00f0, where terms of the spherical Bessel function jl\u00f0 \u03banlR \u00de for the square-well potential is written in r \u00de 1/2 : j & (16.8) \u03banl = 2m E j \u01272 % = 0 and, for each l, the first The energy levels are determined by the boundary conditions jl\u00f0 zero of jl is given the quantum number n = 1, the second n = 2, and so on. The order of the energy levels (which are borrowed from nuclear physics and are different from atomic physics) are 1s, 1p, 1d, 2s, 1f, 2p, 1g, 2d, and so on. If two solutions have the same number of radial nodes, the one with higher l has higher energy. In this notation, the principal quantum number in atomic physics is equal to n + l. The interior of the jellium cluster will be electrically neutral if one includes the exchange-correlation contribution to the electrostatic potential of the electrons, and the effective potential will be nearly constant. The square-well potential essentially represents this phenomenon. \u03banlR \u00de Woods\u2013Saxon Potential The Woods\u2013Saxon potential R yields a better phenomenological representation of a potential that is flat in the middle of the cluster and rounded at the edges. This potential is described by = U r \u00f0 \u00de exp\u2061 \u2212U0 r \u2212 R \u00bd\u00f0 : + 1 $ (16.9) /\u03b5 \u00de U0 is the sum of the Fermi energy and the work function of the bulk metal. R is determined by Eq. (16.4) with rs, the Wigner-Seitz radius of the bulk. The parameter \u03b5 is taken to match the variation in the potential at the surface. This potential is flat in the middle of the cluster but rounded at the edges. The three potentials are shown in Figure 16.5. In Figure 16.5, the degeneracy of the states of the Woods\u2013Saxon potential is similar to that of the square-well potential, but the ordering of the energy levels is different. Knight et al.16 used the Woods\u2013Saxon potential (Ref. 35) in the analysis of the mass spectra shown in Figure 16.3. The ordering of the energy levels is shown, and the cumulative totals of electrons are indicated above the energy levels. At the bottom of the figures are sketches of the potentials as a function of r. 16.2.2 Self-Consistent Spherical Jellium Model For an interacting electron gas, the density functional theory (DFT) based on the Hohenberg\u2013Kohn (HK) theorem (discussed in detail in Section 7.8.1) is used. The basic assumption of DFT is that the total energy of the system is a functional of the electron density n. The HK theorem states that the exact ground-state energy of a correlated electron system is a functional of the density, and the minimum is the ground-state density. One can apply the HK theorem to clusters. The density nI\u00f0 of the smeared-out positive charge of the ions can be expressed as r \u00de r \u00de \u00f0 nI\u00f0 r \u00de = n0\u03b8 r \u2212 R \u00f0 \u00de, (16.10) where R is the cluster radius given in Eq. (16.1), \u03b8 is a step function, and n0 is the constant bulk density of the metal n0 = 4\u03c0rs 3 \u22121 : \"! (16.11) 16.2 Electronic Shell Structure 533 \u03bd 6 5 4 3 2 1 0 168 112 70 40 20 8 2 n,l 2g 1l 3p 2f 1h 3s 2d 1g 2p 1f 2s 1d 1p 1s 156 138 112 106 92 70 68 58 40 34 20 18 8 2 156 138 132 106 92 90 68 58 40 34 20 18 8 2 (a) (b) (c) FIGURE 16.5 Comparison of the (a) harmonic, (b) Woods\u2013Saxon, and (c) square-well potentials. Reproduced from Misra 28 with the permission of Elsevier. The total energy can be written as = Ees\u00bd is the kinetic energy of a system of independent particles of density n, + Eex\u00bd, $ E n n n \u00bd $ $ where Ek\u00bd n $ n Ek\u00bd $ = \u2211 i \u01272 2m j \u2207\u03c8 ij 2: where Ees\u00bd n $ is the electrostatic energy, n Ees\u00bd $ = e2 2 ZZ dr\u2032 dr \u00bd n r \u00de \u00f0 r r\u2032 n \u2212 nI\u00f0 \u00de \u00f0 \u00de$\u00bd r \u2212 r", "\u2032 j j \u2212 nI\u00f0 r\u2032 \u00de$ : (16.12) (16.13) (16.14) 534 CHAPTER 16 Metallic Nanoclusters Following the procedure outlined in Section 7.8.2 (Ref. 17) (subject to the modification for a metallic cluster), we obtain (Problem 16.1) where and \u2212 \u01272 2m \u22072\u03c8 i\u00f0 r \u00de + VKS\u00f0 r \u03c8 i\u00f0 \u00de r \u00de = \u03b5i\u03c8 i\u00f0 r \u00de, r VKS\u00f0 \u00de r = VH\u00f0 \u00de + Vxc\u00f0 r \u00de r VH\u00f0 \u00de = 2e2 dr\u2032 Z r\u2032 \u00de : n \u00f0 r\u2032 \u2212 n \u00f0 \u00de r \u2212 r\u2032 j j (16.15) (16.16) (16.17) The local density approximation assumes that n(r) is slowly varying. The exchange-correlation part of the total energy can be written as and n Exc\u00bd $ = Z drn r \u00f0 \u00de \u03b5xc\u00f0 n r, \u00de\u00de \u00f0 r Vxc\u00f0 \u00de = n \u00de \u2202\u03b5xc\u00f0 \u2202n + \u03b5xc\u00f0 n \u00de : (16.18) (16.19) Here, \u03b5xc\u00f0 is the exchange and correlation energy of a uniform electron gas of density n. A number of functional forms are used for \u03b5xc: The form used most often is given by Ekardt,7 n \u00de \u03b5xc\u00f0 n r \u00f0 \u00de\u00de = \u2212 0:916 rs\u00f0 r \u00de \u2212 0:0666G x r : \u00de\u00de \u00f0 \u00f0 The first term is the exchange part, where 3/4\u03c0n \u00bd and the second term is the correlation part, where rs\u00f0 = r \u00de r 1/3, \u00de$ \u00f0 and = G x \u00de \u00f0 1 + x3 \u00f0 \u00de ln 1 + h 1 x i \u2212 x2 + x 2 \u2212 1 3 (16.20) (16.21) (16.22) = rs\u00f0 The wave functions and hence the energy are obtained self-consistently from the Kohn\u2013Sham equations (Ref. 17). The Kohn\u2013Sham potentials are fairly flat in the center of the cluster and analogous to the Woods\u2013Saxon potential discussed earlier. (16.23) /11:4: r r x \u00f0 \u00de \u00de Due to the absence of the ionic structure, the simple jellium model, which works reasonably well for alkali metals, does not work out for higher-density materials such as aluminum. Recently, 16.2 Electronic Shell Structure 535 in the \u201cstabilized jellium model,\u201d the ionic pseudopotentials have been introduced through a modified exchange-correlation potential. 16.2.3 Ellipsoidal Shell Model The use of spheres is a justified approximation only for closed-shell structures. However, for openshell structures, the ellipsoidal shell model (borrowed from nuclear physics) is used. The model is based on the harmonic oscillator Hamiltonian H = p2 2m + 1 2 m\u03c92 0 x2 + y2 + \u03a92 z z2 \u00de \u03a92 \u22a5\u00f0 h i \u2212 U\u0127\u03c90\u00bd l2 \u2212 < l2>n$ : (16.24) It is essentially a spheroid that has two equal axes (x and y) and one unequal axis (z). Constant volume is maintained by imposing the condition \u03a9 2 \u22a5\u03a9z = 1, and the distortion is expressed by a parameter \u03b4, \u03b4 = 2 \u03a9\u22a5 \u2212 \u03a9z\u00de \u00f0 \u03a9\u22a5 + \u03a9z : (16.25) The last term in Eq. (16.24) is an empirical addition that splits the states of different angular momenta and gives the same ordering as the Woods\u2013Saxon potential. The potential is elongated in the x and y directions, and the cluster is elongated in the z-axis (prolate distortion), but when \u03b4 is negative, there is oblate distortion as well as expansion in the x-y plane. In Eq. (16.24), < l2>n = n + 3 (Problem 16.2). 1 2\u00de n \u00f0 \u00de \u00f0 One can obtain the wave function self-consistently from the Kohn\u2013Sham equation outlined, where \u039b is the earlier. The wave functions are characterized by the quantum numbers component of l along z, and n, n1, n2, n3 are defined in Problem 16.3. Shell filling occurs with a prolate distortion until half-filling and then reverts to oblate", ". Closed-shell clusters are spherical, but there are subshell closings that are ellipsoidal. n, n3, \u039b \u00de \u00f0 16.2.4 Nonalkali Clusters The noble metals (Cu, Ag, and Au) lie at the end of the 3d, 4d, and 5d periods, respectively, with a filled d-shell of 10 electrons and a single valence electron. In bulk, the d-band falls well below the Fermi level, and the valence electrons are expected to behave similarly to that observed in alkali metal clusters. In the experiments of Katakuse et al.,15 the clusters were created as positively charged ions, so the number of electrons in a cluster was N \u2013 1, and the magic numbers corresponding to electronic shell charges had fair agreement with experimental results. The electronic shell model also applies to divalent metals where 2N corresponds to the shell-filling numbers shown in Figure 16.4, leading to magic numbers at N = 4, 9, 10, 17, 20, 29, 34, 35, 46, 53, 56, \u2026. This is also experimentally confirmed by another experiment by Katakuse et al.15 16.2.5 Large Clusters The mass spectra of sodium clusters have been recorded up to sizes of about 25,000 atoms, in which the magic numbers are evident. However, the electronic energy levels of large clusters tend 536 CHAPTER 16 Metallic Nanoclusters to bunch together in groups so that there is a series of approximately degenerate levels (shells). From experimental results, it is apparent that these shells fill on a scale that is proportional to N1/3. One can gain useful insight by considering the electronic shells that occur for potentials for which exact degeneracy occurs and for which there is an analytic solution. We consider the harmonic oscillator potential and the energy-level scheme shown in Figure 16.5, all states with denotes the number of atoms in a cluster with complete levels = \u03bd in Figure 16.5). If the states are labeled as n, l \u00de \u00f0 and define a shell index K K = 2n + l are degenerate. If N filling up to and including the shell K K \u00f0 \u00f0 \u00de K = 0, 1, 2,:::, \u00de\u00f0 \u00de\u00f0 K3 3 : (16.26) The shell index associated with the magic numbers is plotted against N1/3 in Figure 16.6. If all states are degenerate 1/r potential. The number of states for a complete filling up to shell K is, K = n + l are degenerate. (The notation nat in atomic \u00de \u00f0 physics is nat = K3: (16.27) Thus, the general filling scale is K \u221d N1/3: Figure 16.6 shows the shell index K plotted against N1/3. Figure 16.6 distinctly shows that there is a break in behavior at about N 1500. For N < 1500, the electronic shell model yields satisfactory results. However, for N > 1500, one has to use a different model. The model used is known as the geometric shell model 20 18 16 14 12 10 8 6 4 2 0 Experiment Pseudo q.n. Shells of electrons s m ells o f a t o h S Experiment Icosahedra 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 N1/3 FIGURE 16.6 Shell index K plotted against N1/3 for NaN clusters. Reproduced from Martin 25 with the permission of Elsevier. 16.3 Geometric Shell Structure 537 16.3 GEOMETRIC SHELL STRUCTURE 16.3.1 Close-Packing We have noted that for sodium clusters, for N > 1500, a new criterion for stability sets is based on the close-packing of atoms (the fcc or hcp structure in bulk) suitably modified by the presence of a surface. This is known as the geometric shell model. In fact, for other metallic clusters, the geometric shell model is the starting point for studies of stability and associated magic numbers. The bonding is dominated by the short-ranged d \u2212 d interaction in metals such as nickel, and the configuration tends to maximize the number of nearest neighbors. The transition metals with unfilled d\u2212 shells have directional dependence in their bonding. An example is the bcc and hcp structure in bulk iron and cobalt atoms. 16.3.2 Wulff Construction Wulff proposed that at equilibrium, the polyhedron is such that the perpendicular distance from the center of the particle to a face of the polyhedron is proportional to the surface energy of the face.36 When one uses this procedure for a fcc metal, a polyhedron comprising eight (111) and six (100) faces, shown in Figure 16.7, is obtained. If one defines the perpendicular distance of the two faces as", " p111 and p100, it can be shown that (Problem 16.4) the lengths of the sides of the faces are \u03b2 and 6p p 111, as shown in Figure 16.7. The scale factor \u03b2 is given by 1 \u2212 2\u03b2 in units of \u00f0 \u00de \ufb03\ufb03\ufb03 \u03b2 = 1 \u2212 p100= 3p p111: (16.28) If \u03b3100 and \u03b3111 are the surface energies of the two faces, we have from Eq. (16.28) \ufb03\ufb03\ufb03 \u03b2 = 1\u2212 \u03b3100 3p \u03b3111 : (16.29) \ufb03\ufb03\ufb03 \u00de \u00f0 \u03d5 Assuming pairwise interaction between nearest neighbors, the specific surface energy \u03b3 = nd\u03d5/A, where nd is the number of neighbors a surface atom is deficient, and A is the surface area per atom. An atom on the (100) surface has a deficiency of four neighbors compared with an atom in the bulk. Therefore, (100) \u03b2 \u03b2 \u03b3100 = 4\u03d5/d2, (16.30) where d is the nearest-neighbor distance. An atom on the (111) surface has nine neighbors and has a deficiency of three compared with an atom in the bulk. We thus have \u03b3111 = 2 3p \u03d5/d2: (16.31) \ufb03\ufb03\ufb03 (111) < 1 \u2212 2 \u03b2 > FIGURE 16.7 Polyhedral structure based on Wulff construction. Reproduced from Marks (Ref. 24) with the permission of Elsevier. 538 CHAPTER 16 Metallic Nanoclusters From Eqs. (16.29) through (16.31), we obtain \u03b2 = 1 3 : (16.32) Thus, \u03b2 = 1 \u2212 2\u03b2 and, hence, the lengths of the sides of the hexagonal and square faces are equal. This is shown in Figure 16.8 for a 586-atom polyhedron. It may be noted that the Wulff criterion actually applies for large sizes when vertex and edge effects can be neglected. 16.3.3 Polyhedra fcc A basic shape that can be constructed from a symmetric cluster is the octahedron (see Figure 16.9) with eight triangular faces that are close-packed (111) planes with low surface energy. In Figure 16.9, we defined a shell number K as the number of atoms along the edge of a face. A cluster with an even value of K is built around an elementary octahedron of six atoms while a cluster with an odd value of K has a single atom at its center. The total number of atoms N in a cluster containing K octahedral shells is N = 1 3 \u00f0 2K3 + K : \u00de (16.33) FIGURE 16.8 A 586-atom truncated octahedron with square and hexagonal faces. Reproduced from Misra 28 with the permission of Elsevier. A cuboctahedron, which is an octahedron truncated by a cube, was shown earlier in Figure 16.2. It has eight (111) and six square (100) faces. The polyhedron has a central atom and can be considered as built of successive shells covering interior shells. The Kth shell atoms. It can be shown contains that the total number of atoms in a cluster with K shells (Problem 16.5) is 10K2 + 2 \u00f0 \u00de N = 1 3 \u00f0 10K3 \u2212 15K2 + 11K \u2212 3 : \u00de (16.34) From Eq. (16.34), the magic numbers associated with geometric shell filling are 1, 13, 55, 147, 309, 561, \u2026. FIGURE 16.9 A 489-atom octahedron. Reproduced from Misra 28 with the permission of Elsevier. 16.3 Geometric Shell Structure 539 Mackay Icosahedron It can be shown that the total surface energy can be minimized if all the faces of the polyhedron are (111) planes and the surface area can be kept at a minimum. The icosohedral structure shown in Figure 16.10a has 20 triangular faces and has this feature. The Mackay icosahedron is made up of 20 distorted tetrahedra. The distortion is shown in Figure 16.10b. The vertex O is at the center. If the three edges meeting at the center (OA, OB, OC) are of unit length, the three sides of the equilateral triangle ABC are extended to 1.05146. The angles subtended at O are 63\u00b026\u2032. The icosohedra can be constructed by arranging 12 neighbors of a central atom at the corners of an icosohedron. One can build larger clusters by", " covering the 13-atom core with a second layer of 42 atoms, a third of 92 atoms, and so on. The total number of atoms is identical to that of the cuboctohedron with triangular faces, as in Eq. (16.34). The reason for assignment of magic numbers to geometrical shell structures for large sodium clusters is that experimental results support (16.34) with K = 10\u201319, thereby confirming an icosohedral or cuboctahedral configuration. In addition, in the K against N1/3 plots of Figure 16.6, there is a good fit to Eq. (16.36) in the large K limit (for 0:21 K3 fits for small N > 1500 % clusters. One can show that the later equation can be obtained by using the Woods\u2013Saxon potential. the values from Eq., whereas N \u00de bcc The 369-atom rhombic dodecahedron shown in Figure 16.11 is one of the main cluster shapes based on the bcc lattice. (a) A B O C (b) FIGURE 16.10 (a) An icosahedron of 561 atoms; (b) the Mackay icosahedron with vertex O at the center. Reproduced from Misra 28 with the permission of Elsevier. There are 12 identical faces of which the 2p, and diagonals have lengths with the ratio 1: this shape is similar to the cuboctahedron. The number of atoms, N, for a cluster of K shells is \ufb03\ufb03\ufb03 Thus, the cluster sizes at successive shell fillings are 1, 15, 65, 369, \u2026. N = 4K3 \u2212 6K2 + 4K \u2212 1: (16.35) 540 CHAPTER 16 Metallic Nanoclusters FIGURE 16.11 The 360-atom rhombic dodecahedron. Reproduced from Misra 28 with the permission of Elsevier. 16.3.4 Filling between Complete Shells There are two different models for filling of sub-shells in between the completion of the main shells of the Mackay (Ref. 21) icosohedra. In one model, one starts from a complete icosohedron and adds atoms that preferentially sit on the three-fold hollow formed by the three atoms on the (111) face, thereby maximizing their coordinate number. The atoms rearrange themselves to the Mackay packing arrangement when the shell is half filled. This model has been applied to rare-earth clusters. In the alternate model, also known as the umbrella model, the atoms are added to icosohedral sites on the surface of the core icosohedron. The covering of a complete face achieves enhanced stability. 16.4 CLUSTER GROWTH ON SURFACES 16.4.1 Monte Carlo Simulations The easiest way to understand the key ideas of cluster growth on surfaces is through Monte Carlo simulation of cluster (or island) nucleation and growth during vapor deposition. Usually, through the use of a comprehensive set of simulations, an empty lattice is made to represent a substrate. The monomers are deposited at random into this lattice at a constant deposition rate. They can diffuse by random nearest-neighbor hops, and when they collide, they nucleate new point islands. The immobile islands \u201cgrow\u201d whenever new monomers diffuse onto them. They remain as single points on the lattice while their recorded size increments by the number of adsorbed monomers. The 16.4 Cluster Growth on Surfaces 541 density of islands nucleated in the simulation depends on the ratio, R, of the monomer deposition rate to the monolayer deposition rate. R dictates the Monte Carlo simulation procedure, by changing the probabilities of the next simulation step being a deposition event or a monomer diffusion event. Typically, R \u223c 105 \u2212 1010 in experimental growth conditions. The Monte Carlo simulations throw up many questions relevant to experiment, such as (1) how does the island density depend on growth rate and temperature, (2) how does the critical island size i change with these conditions, (3) how does the island size distribution (ISD) depend on this critical island size, and (4) why it displays scale invariance? 16.4.2 Mean-Field Rate Equations The following assumptions are made in deriving the mean-field rate equations: a. Only monomers are mobile. b. Islands of size s \u2265 2 grow by capturing the diffusing monomers. c. Monomers are produced on the substrate by random deposition from vapor or by release from an existing island. d. The direct impingement from vapor onto existing monomers and islands can be neglected in the early stages of nucleation and growth. The time evolution of monomer density N1 and density Ns of islands of size s is and dN1 dt = F \u2212 2D\u03c31N", "2 1 \u2212 DN1 \u2211 s>2 \u03c3sNs + 2 N2 \u03c42 Ns \u03c4s + \u2211 s\u22653 dNs dt = DN1\u00f0 \u03c3s\u22121Ns\u22121 \u2212 \u03c3sNs\u00de + Ns+1 \u03c4s+1 \u2212 Ns \u03c4s : (16.36) (16.37) Here, F is the monomer deposition rate (monolayer per second), D is the monomer diffusion rate, \u03c3s is the so-called \u201ccapture number\u201d of monomers by an island of size s, and 1/\u03c4s is the dissociation rate (rate of monomer release) of an island of size s. The preceding equations can be numerically integrated up to some finite time and some maximum possible island size at that time, provided one knows \u03c3s, the capture numbers, and 1/\u03c4s, the dissociation rate of an island of size s. \u03c3s can be obtained by considering the monomer density s/\u03c0p at the origin and by using cylindrical n1\u00f0 symmetry. The monomer density is zero at the adsorbing island edge and rises to the global average N1 a long way from the island. The diffusive flux into the island is given by in the vicinity of a circular island of radius rs = \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 r, t \u00de DN1\u03c3s = D2\u03c0rs \u2202n1 \u2202r jrs, (16.38) from which the capture number \u03c3s can be calculated. The monomer density field can be obtained from the diffusion equation by first assuming that the monomers can evaporate from the substrate at 542 CHAPTER 16 Metallic Nanoclusters the rate 1/\u03c4a and that N1 \u2192 F\u03c4a, a long way from the island. The later assumption implies that the density of islands is sufficiently low. The diffusion equation can be written as r, t \u2202n1\u00f0 \u2202t \u00de = D\u22072n1\u00f0 r, t \u2212 \u00de r, t n1\u00f0 \u03c4a \u00de + F: (16.39) In the steady state t > \u03c4a, it can be shown from Eqs. (16.38) and (16.39) that the capture number \u03c3s can be written as (Problem 16.6) \u03c3s = 2\u03c0Xs K1\u00f0 K0\u00f0 Xs\u00de Xs\u00de, where K0 and K1 are the Bessel functions and Xs = rs D\u03c4ap : (16.40) (16.41) If we assume that the islands around the central one (being considered) are randomly distributed and each one has the same size <s> and the same capture number \u03c3x, \u03c4s in Eq. (16.40) can be replaced by \u03c4 such that \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 1 \u03c4 = 1 \u03c4a + D\u03c3xN, (16.42) \u00de 1/\u03c4a = 0 Ns is the density of stable islands. The evaporation of monomers can be switched off where N = \u2211 s>i in Eq. (16.42) to model the \u201ccomplete condensation\u201d regime of the film growth. \u00f0 In an improved method using the results from kinetic Monte Carlo simulations, it has been shown (Bales and Chrazan1) that a self-consistent solution for all \u03c3s can be used to reproduce the evolution of the monomer and total island density. In their work, Eq. (16.42) for the average capture number is replaced by 1 \u03c4 = 2D\u03c31N1 + \u2211 s\u22652 D\u03c3sNs + F\u03ba1, (16.43) which allows for monomer trappings by other monomers and direct hits from vapor deposition on existing monomers as well as capture by islands. 16.5 STRUCTURE OF ISOLATED CLUSTERS 16.5.1 Theoretical Models Hartree\u2013Fock Methods The earliest method employed was the Hartree\u2013Fock method (discussed in Chapter 4), but because it uses a single-determinant wave function, it does not describe electrons with different spins. The neglect of correlations produces energy values that are larger than the actual ones. There have been various methods used to include the effect of correlations in post-Hartree\u2013Fock theories. 16.5 Structure of Isolated Clusters 543 In the configuration interaction (CI) methods, one or more occupied orbitals in the Hartree\u2013Fock determinant are replaced by virtual orbitals where each replacement is equivalent to an electron excitation. In the full CI method, the wave function is formed as a", " linear combination of the Hartree\u2013Fock determinant and all possible substitute determinants. However, it is necessary to truncate the expansion at a fairly low level of substitution (single and double excitations) except for the smallest systems. There are refinements that use Rayleigh\u2013Schrodinger perturbation theory and add the higher excitations as noniterative corrections. Usually, these methods are limited to small clusters of about 20 atoms. Density Functional Theory The local density approximation (LDA) in density functional theory (DFT) can be used by first writing the exchange-correlation energy as a sum of exchange and correlation part Exc = Ex + Ec: (16.44) r If \u03c1\u03b1\u00f0 local exchange functional is given by the standard Dirac\u2013Slater form are the electron densities of the two spin components, and \u03c1\u03b2\u00f0 r \u00de \u00de ELDA x = \u2212 3 2 1/3 3 4\u03c0! \" Z dr \u03b1 + \u03c14/3 \u03c14/3 \u03b2 $, \u00bd \u03b1 = \u2191 and \u03b2 = \u2193, the \u00de \u00f0 \u00de \u00f0 (16.45) and the Kohn\u2013Sham (Ref. 17) wave functions are atom-centered. The local spin-density correlation energy functional (Vosko et al.34) can be written as ELDA c \u03c1\u03b1, \u03c1\u03b2$ \u00bd = dr\u03c1\u03b5c\u00f0 \u03c1\u03b1, \u03c1\u03b2\u00de, (16.46) Z where \u03b5c is a complicated function of \u03c1 and several constants. However, the correlation energy is overestimated by 100% in this approximation. A more widely used exchange-energy functional based on the generalized-gradient approxima- tion (GGA) of Becke (Ref. 2) is given by EGGA x = ELDA x \u2212\u03b2 \u2211 \u03c3 = \u03b1,\u03b2 Z dr \u03c14/3 \u03c3 x2 \u03c3 1 + 6\u03b2x\u03c3sinh\u2061\u22121, x\u03c3\u00de \u00f0 where x\u03c3 is the dimensionless ratio x\u03c3 = \u03c1\u22124/3 \u03c3 \u2207\u03c1\u03c3j j : In general, the GGA exchange-correlation energy functionals can be expressed as EGGA xc \u03c1\u2191, \u03c1\u2193$ \u00bd = Z dr f \u03c1\u2191, \u03c1\u2193, \u2207\u03c1\u2191, \u2207\u03c1\u2193\u00de : \u00f0 (16.47) (16.48) (16.49) An interesting technique to go beyond GGA is to use hybrid functionals that are formulated as a mixture of Hartree\u2013Fock and DFT exchange coupled with DFT correlation. However, a number of authors have suggested that hybrid methods should be avoided for metal clusters because these techniques do not yield reliable results. 544 CHAPTER 16 Metallic Nanoclusters x, y, z Tight-Binding Methods The tight-binding model is used mainly in the study of transition metal clusters. The Hamiltonian is expressed in terms of matrix elements in an orthogonal basis set composed from the valence atomic orbitals. The intersite matrix elements are s, p determined by the Slater\u2013Koster hopping integrals (Slater and Koster,31) ss\u03c3, sp\u03c3, sd\u03c3, pp\u03c3, pd\u03c0, dd\u03b4, which decay exponentially with distance, but the three-center integrals are ignored. In the model proposed by Mehl and Papaconstantopoulos,27 the total energy is obtained by summing up the occupied energy levels, xy, yz, zx, x2 \u2212 y2, 3z2 \u2212 r2, d \u00de \u00f0 \u00f0 \u00de \u03b5i = ai + b\u03bb\u03c12/3 i + c\u03bb\u03c14/3 i + d\u03bb\u03c12 i, where \u03c1i = \u2211 j\u2260i exp\u2061 \u2212\u03b3 % # rij r0\u22121 : & $ (16.50) (16.51) The three on-site terms depend on the orbital angular momentum \u03bb = s, p, d. In Eq. (16.51), r0 is the interaction term in the bulk, and a cutoff is included in the distance in the sum. The repulsive energy is incorporated by varying it as a function of the total density of each function. The Hamiltonian for the model is written as H = \u2211 i\u03bb\u03c3 \u03b5i\u03bbc\u2020 i\u03bb\u03c3ci\u03bb\u03c3 + \u2211 i\u2260j \u03bb\u03bc\u03c3 ij c\u2020 \u03b2\u03bb\u03bc i\u03bb\u03c3cj\u03bc\u03c3: (16.52) In Eq. (16.52), \u03bb and \u03bc are orbital labels, \u03c3 is the spin label, \u03b2\u03bb\u03bc ij are the hopping integrals between sites i and j, and the other symbols have their usual meanings. The different parameters are obtained by fitting to", " bulk properties. There are a large number of variations of the tight-binding model. Specifically, spin-polarized systems, a Hubbard-like term, Hint\u2061 = \u2212J\u03bb\u2211 i\u03bb\u03c3 \u03c3c\u2020 i\u03bb\u03c3ci\u03bb\u03c3, is added to Eq. (16.52). In Eq. (16.53), J\u03bb = 1 2 \u2211 \u03bc J\u03bb\u03bc\u00f0 ni\u03bc \u2191 \u2212ni\u03bc\u2193\u00de, (16.53) (16.54) where J\u03bb\u03bc is an exchange integral, and ni\u03bc\u03c3 is the component of electron density at site i associated with orbitals \u03bc and spin \u03c3. A variety of additional terms are added to this basic model to obtain the computational results that can relate better to the cluster properties. Semi-Empirical Potentials To study large clusters, several semi-empirical potentials have been developed that contain manybody contributions and can be used with Monte Carlo or molecular dynamics simulations. The various models are known as embedded atom model (EAM), effective medium theory (EMT), glue 16.5 Structure of Isolated Clusters 545 model, and second-moment methods. All these methods start with a common functional form for the total energy\u2260i V rij\u00de \u00f0 + F, \u03c1i\u00de # \u00f0 (16.55) \u00f0 V rij\u00de 2 \u2211 j\u2260i where 1 is a repulsive pair potential between atoms separated by a distance rij. In the EAM (Daw and Baskes5; Foiles et al.10), each term of a metal is considered as an impurity embedded in \u03c1i\u00de is the energy required to embed atom i in an elec- a host provided by the rest of the electrons. F tron gas of density \u03c1i, where \u00f0 \u03c1i = \u2211 j\u2260i \u03c1 : rij\u00de \u00f0 (16.56) One has to use empirical fits to bulk properties to be able to generate the embedding functions and pair interactions. In the second-moment approximations (Finnis and Sinclair,9; Sutton,33) the total energy is E = \u03b5\u2211 i \u00bd \u2211 j\u2260i V rij\u00de \u00f0 \u2212 c \u03c1ip : $ (16.57) One can see the similarity as well as the contrast between Eqs. (16.55) and (16.57). In Eq. (16.57), \u03b5 is a parameter that has dimensions of energy, and c is a dimensionless parameter. \u03c1i is expressed as a sum of pair potentials \ufb03\ufb03\ufb03\ufb03 \u03c1i = \u2211 j\u2260i \u03d5 : rij\u00de \u00f0 (16.58) \u03c1i can be described as the bond energy. We consider just one orbital site and local density of states E di\u00f0 at site i with a center of gravity \u03b5i. The bond energy is given by \u00de EF Ebond = 2\u2211 i Z \u2212\u221e E \u2212 \u03b5i\u00de E di\u00f0 \u00f0 \u00de dE: The second moment of the local density of states can be written as 2 \u00dei = \u03bc\u00f0 \u221e Z \u2212\u221e E \u2212 \u03b5i\u00de \u00f0 E 2di\u00f0 \u00de dE: (16.59) (16.60) From Eqs. (16.59) and (16.60), we obtain an approximation in which moments no higher than the second are used (Problem 16.7), i \u00debond = A E\u00f0 2 \u03bc\u00f0, \u00dei \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 q (16.61) 546 CHAPTER 16 Metallic Nanoclusters where A is a constant. It can be shown that the local density of states can be written as where H is the Hamiltonian for the system, and (16.62), we obtain an expression for the second moment, j E di\u00f0 >, j \u00f0 i > is the orbital on-site i. From Eqs. (16.60) and (16.62) \u221e 2 \u00dei = \u03bc\u00f0 Z \u2212\u221e = < i E \u2212 \u03b5i\u00de \u00f0 \u03b5i \u2212 H j\u00f0 \u00dej i > dE We can write the Hamiltonian H as H = \u2211 i i > < i \u03b5ij + \u2211 i\u2260j \u03b2ijj j i > < j, j (16.63) (16.64) where \u03b5i is the energy at site i, and \u03b2ij is the hopping integral from site i to j. From Eqs. (16.63) and (16.64), we obtain 2 \u00dei", " = \u2211 \u03bc\u00f0 j\u2260i \u03b22 ij: (16.65) The second-moment approximation is very effective for metals that tend to form close-packed structures. However, one needs the fourth moment, which controls the stability of the bcc structure, and the fifth and sixth moments if hcp and fcc packing are to be differentiated. There are several other semi-empirical potentials that we will not discuss in this chapter. There are several alternate models to choose the potential in Eqs. (16.57) and (16.58). Once a potential is chosen, there are many global optimization techniques that can be used to find the lowest energy of a cluster. The standard Monte Carlo or molecular dynamics procedure is used after incorporating some algorithms. The computation involved is very massive, but fairly good global minima have been obtained for clusters of more than 100 atoms. 16.5.2 Structure of Some Isolated Clusters A schematic configuration of a number of commonly occurring small clusters is shown in Figure 16.12. There is usually some departure from the symmetric configurations shown in the figure. The linear and planar clusters usually show some departure from one or two dimensionality, and triangular configuration in 3b is isosceles instead of equilateral. The three-dimensional clusters occur at various sizes. One extreme is nickel, which has a tetrahedral structure at N = 4, whereas the other extreme is gold clusters, which are planar until N > 10: The alkali metals are between these two extremes. An interesting aspect is that anions and cations have different structures than neutral clusters. A large number of experimental research is focused on probing clusters of special sizes to verify whether magic number clusters, such as icosahedral or cuboctahedral (13, 55, 147), truncated octahedral (38), or Marks decahedral (75, 146, 192), are more stable than clusters of other sizes as suggested by the geometrical shell model. 16.6 Magnetism in Clusters 547 3a 5a 3b 4a 4b 4c 4d 5b 5c 6a 6b 6c 7 8a 8b 12 13 19 20 FIGURE 16.12 Schematic representation of some small clusters. The 3D figures are (4d) tetrahedron, (5b) trigonal bipyramid, (5c) square pyramid, (6b) pentagonal pyramid, (6c) octahedron, (7) pentagonal bipyramid, (8a) bicapped octahedron, (8b) bisphenoid, (12) incomplete icosahedron, (13) icosahedron, (19) double icosahedron, and (20) large tetrahedron. Reproduced from Misra 28 with the permission of Elsevier. 16.6 MAGNETISM IN CLUSTERS 16.6.1 Magnetism in Isolated Clusters Superparamagnetism and Blocking Temperature There has been a rapidly growing interest in the nanostructures formed from magnetic clusters because of their enormous potential in the development of high-performance magnetic materials and devices. There has been considerable development of technologies in the manufacture of quantum dots, monolayers, self-organized islands, quantum wires, and deposited nanoclusters. However, magnetism in small clusters, which is in the mesoscopic regime, is not well understood in either the atomic or bulk states. It has been shown (Skomski,30) that nanoclusters are single-domain particles of which the size is well below the critical radius RSD, above which it is favorable to form domain walls. For a sphere, RSD = AKp 36 \u03bc0M2 s \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03, (16.66) where Ms is the spontaneous magnetization, K is an anisotropy constant, and A is related to the exchange stiffness. At low temperatures (well below the equivalent of the Curie temperature), the nanoparticles can be considered as giant magnetic moments of ferromagnetically coupled spins. This phenomenon is 548 CHAPTER 16 Metallic Nanoclusters known as superparamagnetism. In an external field, the reorientation of this giant moment takes place coherently and the individual moments remain aligned with each other. The different magnetic alignments are separated by an anisotropy boundary KV, where V is the volume of the nanoparticle and K is the anisotropic constant. At high temperatures, because kBT \u226b KV, the anisotropic boundary becomes unimportant, but there is a competition between the external field, which tends to orient the particles, and the thermal fluctuations of the magnetic moments, which tend to magnetically disorder the array. The volume V of the transition metal clusters is V \u2264 10\u221225m3, and at room temperature,", " kBT \u226b KV. Because all magnetization directions have equal probability, the magnetization M of a cluster along the direction of an applied field H can be obtained from the classical Langevin function L, where \u03bcc is the giant magnetic moment of each cluster and \u2112 M = \u03bcc \u2112 \u03bccH kBT #, $ (16.67) \u2212 1/x. x \u00f0 \u00de = coth\u2061 x \u00f0 \u00de An assembly of nanoclusters, each of which has several hundred atomic spins, can be easily saturated, unlike an assembly of atoms. At temperatures less than 50\u00b0 K, kBT \u223c KV, and the magnetization in a given field deviates from superparamagnetism. At very low temperatures, the moment in each particle becomes static. The blocking TB is the temperature at which half the cluster moments have relaxed during the time of a measurement. Because a narrow temperature region around TB separates the frozen moments from superparamagnetic behavior, these can be used in magnetic recording technology provided one can obtain deposited small clusters with TB greater than the room temperature. In fact, the deposition of clusters or their embedding in a matrix is essential for practical applications. Cluster Magnetism The magnetism in small clusters is usually significantly different from bulk magnetism of the same atoms because a much larger percentage of atoms in a cluster lies at the surface. For example, 162 of the atoms in a 309-atom cuboctahedral cluster lie at the surface, and these atoms have a reduced coordination and are in a lower symmetry environment. In the second-moment approximation discussed earlier, the bandwidth W is proportional to z1/2, where z is the coordination number. For simplicity, if we consider a rectangular band for the d electrons, the density of states scales as 1/W (Problem 16.8). The reduced coordination of the surface atoms results in an enhanced density of states D We have derived in Eq. (13.111),. \u03b5F\u00de \u00f0 M % 2\u03bc2 \u03b5F\u00de BBD \u00f0 \u03b5F\u00de 1 \u2212 UD \u00f0 : (16.68) Thus, due to the Stoner criterion for ferromagnetic instability, UD \u03b5F\u00de ferromagnetic instability in clusters that was absent in the bulk material. \u00f0 > 1, there is possibility of The reduced zi, the coordination number of the surface atoms, will result in an increase of the local moments \u03bci on the atom i on the surface compared with the bulk value \u03bcbulk. 16.6 Magnetism in Clusters 549 A crude approximation is \u03bci = zbulk zi# 1/2 $ \u03bcbulk: (16.69) However, a reduced coordination in the second-moment approximation causes a contraction in the interatomic distances, which reduces the magnetic moment. A simple expression proposed for the average moment \u03bcN of a cluster of N atoms (Jensen and Bennemann,14) is given by \u03bcN = \u03bcbulk + \u03bcsurf \u2212 \u03bcbulk\u00de \u00f0 N\u22121/3: (16.70) The experiments indicate a more complex size dependence but confirm the trend of the decrease in moment toward the bulk value with increasing size. 16.6.2 Experimental Techniques for Studying Cluster Magnetism Chemical Probe Methods The chemical probe method is usually used to investigate the geometric structure of clusters of Fe, Co, and Ni (and other transition metals) up to the sizes of 200 atoms. Atoms and small molecules react with transition metal clusters in ways that are analogous to the physisorption and chemisorption processes that occur on metal surfaces. The geometric structure of the cluster can be inferred from the fact that the reactivity of the cluster is dependent on the environment of the atoms. The main probe molecules that have been used are NH3, N2, H2O, and H2 or D2. The determination of the number of binding sites of a cluster for a particular molecule yields important clues as to possible structures. Because the d-orbitals have significant spatial extent and are involved in the transition metal bonding, the chemical properties are sensitive to both the number and the configuration of the d electrons. The experimental arrangement is as follows. The clusters are produced by a pulsed laser vaporization of a metal target in a flow tube upstream of a flow-tube reactor (FTR) using helium gas. To ensure that the cluster growth is finished and clusters have cooled to ambient temperature before they enter the FTR, one uses a narrow flow tube and low helium pressure to ensure a rapid decrease in metal atom density. The reagent gas is introduced when the clusters enter the FTR. The reaction between the cluster, MN, and the reagent molecule, A, produces an internally excited complex, MN + A \u21c4 MNA&: (16.71) The excited complex requires a collision with a third body (a carrier gas atom, C) to stabilize", " it, MNA& + C \u2192 MNA + C: (16.72) The clusters and reaction products expand out of the nozzle and form into a molecular beam. The clusters are then pulsed laser ionized and mass analyzed in a time-of-flight mass spectrometer. There are different binding rules for the various reagent molecules. Surface studies indicate that the NH3 molecule binds to the surface through the donation of the N-localized lone pair to a metal atom. However, in the case of clusters, the binding preferentially occurs on the low-coordination metal atoms, which means that saturation with ammonia counts the number of vertex atoms on a 550 CHAPTER 16 Metallic Nanoclusters cluster. Clusters of cobalt and nickel yield more straightforward clues than iron about their structure. The reason is iron and other elements in the middle of the transition metal series have fewer filled d-orbitals, and the binding of the atoms in the cluster is much more directional. It is interesting to note that, although cobalt and nickel are structurally similar, there are significant differences for small clusters. This is shown in Figure 16.13. n Same structures: Nin.Con Triangle 3 Different structures: Nin n Con 4 Tetrahedron, square 12 Icosahedron minus one atom fcc/hcp minus one atom 5 Trigonal bipyramid 6 Octahedron 7 Capped octahedron Transition region: n Nin Con 8 Bisdisphenoid Bicapped octahedron 13 Icosahedron fcc/hcp 14 Bicapped hexagonal antiprism Twinned tetrahedron 15 Bicapped icosahedron Centered hexagonal antiprism 9 Tricapped trigonal prism Tricapped trigonal prism or tricapped octahedron 16 Tricapped icosahedron C2v 10 Tricapped pentagonal bipyramid FIGURE 16.13 19 Double icosahedron hcp Comparison of structures of 3\u201319 atom nickel and cobalt structures. Reproduced from Riley 29 with the permission of Elsevier. 16.6 Magnetism in Clusters 551 Gradient-Field Deflection (Stern\u2013Gerlach) The gradient-field deflection method is based on the classic Stern-Gerlach experiment that first detected the electron spin. A schematic diagram of the experiment for measuring the magnetic moment of free clusters is shown in Figure 16.14. A collimated cluster beam that is generated by a pulsed laser evaporation source with a variable temperature nozzle is guided into a magnetic field gradient dB/dz, which deflects a particle of magnetic moment M vertically by d = M dB dz L2 \u00f0 2D/L + 1 2mv2 x \u00de : (16.73) Here, m is the mass of the cluster, and vx is its velocity when it enters the magnet. D is the distance from the end of the magnet of length L to the detector. vx is measured by a controlled delay between the evaporation laser and the ionizing laser. A mechanical chopper is used in front of the source to define the pulse start time. Figure 16.15 shows the low-temperature total moments of FeN, CoN, and NiN (in \u03bcB/atom) as a function of cluster size. The right scale in Figure 16.15 indicates the spin imbalance obtained from the equation Mspin Mtot = 2 g, (16.74) where g is the gyromagnetic ratio. One can see in Figure 16.15 that clusters with different sizes show locked-moment behavior. The moment of the Ni cluster drops rapidly to the bulk value at N = 160 and then increases slowly 350. The moments of Fe and Co clusters fall less until it reaches the bulk value again at N % rapidly. They reach the bulk value between N 400\u2212500. % Ion detector Extractor grids Ionising laser Evaporation laser Chopper Gas pulse Sample rod FIGURE 16.14 Schematic of the experiment for measuring the magnetic moment of free clusters. Reproduced from Misra 28 with the permission of Elsevier. 552 CHAPTER 16 Metallic Nanoclusters 1.2 1.0 0.8 0.6 0.4 2.6 2.4 2.2 2.0 1.8 1.6 1...0 2.6 2.2 1.8 NiN at T = 78 K Bulk 0 100 200 300 400 500 600 700 (a) CoN at T = 78 K Bulk 0 100 200 300 400 500 600 700 (b) FeN at T = 120 K Bulk 0 100 200 300 400 500 600 700 Cluster size N (c) 1.0 0.8 0.6 0.4 2.2 2.0 1.8 1.6 1.4 3.4 3.0 2.6 2.2 1. FIGURE 16.15 Low-temperature total moments (in \u03bcB/", "atom spin imbalance defined in Eq. (16.74). \u00de of (a) NiN, (b) CoN, and (c) FeN. The right scale indicates the Reproduced from Isabella et al.13 with the permission of Elsevier. 16.6 Magnetism in Clusters 553 The magnetic behavior of clusters primarily depends on their structure. Most of the theoretical work has been done using density functional theory and geometry optimization; however, the details of the theoretical calculations are beyond the scope of this book. 16.6.3 Magnetism in Embedded Clusters Switching and Blocking Temperature The interest in magnetic nanoparticles has recently increased considerably because of the device applications such as in ultra-high-density recording as well as in spintronics. For device applications, the nanoclusters are either deposited on a surface or embedded in another material. The embedding material can modify the properties of the magnetic clusters to a great extent. Switching (from magnetism to superparamagnetism) is a major factor in miniaturization of devices. When the energy barrier due to anisotropy, KV, becomes very small, the probability of switching due to the thermal energy becomes significant, which creates problems in such devices. The probability, P, of not switching in time t is given by the Boltzmann statistics and the Arrhenius relation \u2212 t \u03c4, P = e 1 \u03c4 = f0 e \u2212 KV kBT,! \" (16.75) (16.76) where f0 is the attempt frequency. The blocking temperature, which marks the onset of instability during an observation time \u03c4, is obtained from Eq. (16.76) as For a spherical particle of diameter D, V = 1/6 \u03c0D3, and from Eq. (16.77), we obtain TB = KV kB ln f0\u03c4 \u00de \u00f0 : TB = \u03c0KD3 6kB ln f0\u03c4 : \u00f0 10 years, which yields ln 0:2 MJ/m3, and Eq. (16.77) yields for a nanocluster with D = 14 nm, T For data storage, the time constant \u03c4 40. For a typical uniaxial 520\u00b0 K anisotropy K for a 10-year blocking temperature, which is well above the operating temperature of a memory device. However, for a nanocluster for which D = 7 nm, T 65\u00b0 K, which is far lower than the room temperature. f0\u03c4 \u00f0 \u00de % % % % % \u00de One option to increase the blocking temperature is to increase the anisotropy K by depositing ferromagnetic particles on a platinum surface (which has large spin-orbit interaction) or the formation of alloys such as CoPt or FePt. An alternate approach to increase the effective anisotropy is through relying on the exchange bias phenomenon. This is based on the enhanced stability of the ferromagnetic component due to the exchange coupling at the interface of a ferromagnetic and antiferromagnetic material as well as the high anisotropy in the antiferromagnet. (16.77) (16.78) 554 CHAPTER 16 Metallic Nanoclusters Small Clusters and 2D Nanostructures In this section, we first study the introduction of small 3d metal clusters on surfaces of a noble metal (Cu(001)). The results of the calculations are based on density functional theory and the KKR method in which a Green\u2019s function technique is used to treat the perturbation of a surface into the ideal crystal and then placing an adatom or cluster on the surface. Figure 16.16 shows the spin magnetic moment of 3d metal clusters and monolayer on Cu(001). One can note that the moments on the Co and Fe atoms plotted in Figure 16.16 are relatively insensitive to the island size. The deposition of atoms of the 3d elements onto a ferromagnetic surface shows that the magnetic moments of the adatoms may align parallel or antiparallel to the direction of magnetization of the substrate. There is further complexity with 2d clusters or nanolayers if there is a tendency for antiferromagnetic ordering within the nanostructure itself. The spin magnetic moments obtained from using KKR Green\u2019s function method with exchange and correlation effects on monolayers on bcc Fe(001) are shown in Figure 16.17. Similar results have been obtained for fcc Ni(001) substrate except for Mn. The total energy calculations for fcc Ni(001) substrate on Mn indicate ferromagnetic coupling for all positions. Further, the enhancement of the impurity moments is larger with Ni(001) than with Fe(001) because the hybridization is weaker due to the fact that the Ni wave function is less extended. The dimers on nextnearest-neighbor sites on top of Fe", "(001) couple to the surface with the same configuration as the adatoms except for Mn. The moments of the dimer atoms are parallel to each other. However, for the Mn dimer, the two states are energetically degenerate, one with the moments on both atoms ferromagnetically coupled to the substrate, whereas the other has the moments of the dimer atoms antiparallel. Monomer Dimer Trimer Tetramer Monolayer Cu(001 Sc Ti V Cr Mn Fe Co Ni FIGURE 16.16 Spin magnetic moment of 3d metal clusters and monolayer on Cu(001). Reproduced from Stepanyuk et al.32 with the permission of Elsevier1 \u22122 \u22123 \u22124 Ferromagnetic 3d/Fe(100) V Cr Mn Fe Co Ni Antiferromagnetic FIGURE 16.17 16.6 Magnetism in Clusters 555 3.5 3 2.5 2 1.5 1 0. Spin Orbital Theory 0 0 2 6 4 Cluster size 8 10 FIGURE 16.18 Experimental results from Lau et al.15 of spin and orbital moments (in \u03bcB/atom from Martinez et al.26 for spin moments for Fe clusters deposited on Ni/Cu (001) clusters. and theoretical results \u00de Reproduced from Misra 28 with the permission of Elsevier. Magnetic moments of the monolayer on the surface: p(1 x 1) ferromagnetic coupling (solid squares/solid line); layered antiferromagnetically (open squares); c(2 x 2) ferromagnetic (diamonds/ dashed line); 3d monolayers on Ag(001) for comparison. Reproduced from Handschuh and Blugel 11 with the permission It is interesting to note that the spin moments of Fe clusters deposited on the Ni(001) surface are enhanced compared with bulk bcc Fe. The results of theoretical calculation, using a modified embedded atom model, show that the spin moments of N = 2\u20139 Fe clusters on Ni(001), which adopt a planar geometry on top of the Ni(001) surface, have spin moments that vary smoothly from 3.15 \u03bcB to 2.85 \u03bcB for N = 2 to N = 9 (Martinez et al.26). In contrast, the experimental results (Lau et al.20) indicate oscillations. These contradictory results are shown in Figure 16.18. of Elsevier. 16.6.4 Graphite Surfaces The spin magnetic moments of the 3d transition metal adatoms and dimers on a graphite surface from the calculations of Duffy and Blackman6 and of 3d monolayers by Kruger et al. (Ref. 19) are shown in Figure 16.19. 16.6.5 Study of Clusters by Scanning Tunneling Microscope The ability to image single impurity atoms on a surface by scanning tunneling microscope (STM) has become very convenient to study the properties of clusters on the surface of a metal. The images of magnetic atoms on the surface of a metal and the Kondo resonance can be obtained from the current versus voltage characteristics. STM experiments measure the bias voltage dependence of 556 CHAPTER 16 Metallic Nanoclusters ) Adatom Dimer Monolayer Sc Ti V Cr Mn Fe Co Ni FIGURE 16.19 Composite showing the results of calculations of spin moments of 3d transition metal adatoms, dimers (Duffy and Blackman,6 and monolayers (Kruger et al.19) on graphite. The Mn monolayer has antiferromagnetic and ferrimagnetic ground states that are degenerate. Reproduced from Misra 28 with the permission of Elsevier. the differential conductance dI/dV. The electrons tunnel between states at \u03b5F on the tip and states at \u03b5F + eV on the surface, so the structure in the measured differential conductance reflects the surface density of states at that energy. Because the STM probes a localize state immersed in a Fermi sea, the spectroscopy is that of a discrete auto-ionized state. We discuss the case of noninteracting impurity resonances that have no Coulomb interactions. This is the situation when the s-orbital of an adatom hybridizes with a metal surface. Such a system can be described by a resonant-level model, where the electron is allowed to hop between the discrete atomic orbital and the continuum of the electronic band states. Thus, the Kondo resonance can be analyzed as a type of Fano resonance (Fano,8). Fano showed that the Hamiltonian for a discrete state and a continuum, ignoring the spin, can be written as ^H0 = \u2211\u03b5k^c\u2020 k^ck + \u03b5a^a\u2020^a + \u2211 k \u00f0 Vak^a\u2020^ck + V &ak^c\u2020 k^a, \u00de (16.79) where", " \u03b5a, ^a\u2020, and ^a are, respectively, the energy, creation, and annihilation operators of an electron residing in the discrete atomic state; and Vak is the hybridization matrix element connecting the atomic state to the kth band state. The advanced atomic-state Green\u2019s function can be written as (Madhavan et al.23) 1 \u03b5 Re\u2061\u03a30\u00f0 where the real and imaginary parts of the self-energy are \u03b5a + \u03b5 \u2212 aa\u00f0 G0 = \u03b5 \u00de \u00bd \u00bd, + i Im\u2061 \u03a30\u00f0 \u03b5 \u00de \u00de$$ and Re\u2061\u03a30\u00f0 \u03b5 \u00de = \u2211 k j Vakj 1 \u03b5 \u2212 \u03b5k 2P # $ Im\u2061\u03a30\u00f0 \u03b5 \u00de = \u03c0\u2211 k j Vakj 2\u03b4 \u03b5 \u2212 \u03b5k\u00de : \u00f0 (16.80) (16.81) (16.82) Here, P denotes the Cauchy principal value. In the context of STM, the energy \u03b5 = eV. In a tunneling experiment, a second electrode (the tip) is added to the system, and electrons tunnel between the electrodes. The tip is modeled by a single state with energy \u03b5t, and ^t removes an electron from that state. ^M is a transfer Hamiltonian term (treated as a perturbation) that induces electrons to tunnel from one electrode to the other and can be expressed as ^M = Mat^a\u2020^t + H:c: \u00f0 \u00de + \u2211 k \u00f0 Mkt^c\u2020 k^t + H:c: : \u00de (16.83) 16.6 Magnetism in Clusters 557 Mat and Mkt are the tunnel matrix elements that connect the STM tip to the discrete atomic state and continuum states, respectively. From Eqs. (16.79) and (16.83), we obtain ^H = ^Ho + ^\u03b5t^t\u2020^t + ^M: It can be shown that the differential conductance can be written as (Problem 16.9) dI dV = 2\u03c0e2 \u0127 2 \u03b5\u2032 + q 1 + \u03b5\u20322 \u03c1tip\u2211 \u00de k j \u00f0 2\u03b4 Mktj eV \u2212 \u03b5k\u00de \u00f0 + constant, where q = A/B, A = Mat + \u2211 k MktVakP B = \u03c0\u2211 k MktVak\u03b4 1 eV \u2212 \u03b5k, $ # eV \u2212 \u03b5k\u00de \u00f0, \u03b5\u2032 = eV \u2212 \u03b5a \u2212 Re\u2061\u03a3 eV Im\u2061\u03a3 \u00f0 \u00de eV \u00f0 \u00de, and \u03c1tip is the tip density of states. \u03c1tip is usually treated as a constant to reflect the broadening of the tip state by contact with the remainder of the tip electrode. Here, the quantity A in Eq. (16.87) is an amplitude for tunneling to the discrete state modified by hybridization with the continuum (the second term in Eq. 16.87). The quantity B in Eq. (16.88) is an amplitude for tunneling into a set of continuum states contained in the range of energies on the order of the width of the resonance. If q \u226b 1, tunneling via the resonance dominates, but the continuum states if q \u226a 1, dominate the tunneling. The experimental results for Co atoms on the Au(111) substrate are reproduced in Figure 16.20. This figure shows how the dI/dV spectra vary as the STM tip is moved with a clear feature of the intermediate q type appearing when the tip is in the vicinity of the Co atom, indicating Kondo resonance.5 2.0 1.5 1.0 0.5 0.0 \u2212100 \u221250 0 50 100 Sample voltage (mV) 16.6.6 Clusters Embedded in a Matrix The magnetic properties of particles are modified when they are embedded in a matrix of another material. The magnetic moment of a FIGURE 16.20 dI/dV spectra at various lateral distances away from the center of a single Co atom on Au(111) taken with the STM tip. Reproduced from Madhavan et al.22 with the permission of AAAS. (16.84) (16.85) (16.86) (16.87) (16.88) (16.89) 12A\u00b0 9A\u00b0 5A\u00b0 3A\u00b0 2A\u00b0 0A\u00b0 \u22123A\u00b0 \u22124A\u00b0 \u22127A\u00b0 \u221210A\u00b0 \u221212A\u00b0 558 CHAPTER 16 Metallic Nanoclusters free particle is enhanced over the bulk value because the", " coordination of the atoms at the surface of the particle is reduced. In contrast, the embedded particle has the coordination of its surface atoms restored. The probability of the magnetic moment reverting to its bulk value depends on the electronic band structures of the particle and the host matrix. Further, if the particle and the matrix have the same bulk crystal structure with similar lattice spacings, there is the possibility of embedding with fairly good epitaxy. In contrast, if the two materials have different structures, the particle structure could either remain robust or adapt to that of the host matrix. The overall picture for the full concentration range can be visualized by considering the magnetic phase diagram for films of deposited 3 nm diameter Fe nanoparticles embedded in Ag matrices, as shown in Figure 16.21. Superparamagnetic 100 Interacting superparamagnetic (ISP) Correlated superspin glass (CSSG) ) K ( T 10 Collective blocking Single-particle blocking 75 50 25 Fe volume fraction (%) 100 1 0 FIGURE 16.21 Magnetic phase diagram for films of deposited 3 nm diameter Fe nanoparticles embedded in Ag matrices as a function of volume fraction and temperature. Reproduced from Binns et al.3 with the permission of the Ideal superparamagnetism occurs at the lowest concentrations above a blocking temperature. When the volume filling fraction (VFF) increases, the magnetic behavior is determined by the dipolar interactions and the aggregation. A correlated superspin glass (CSSG) state occurs at higher concentrations. Blocking occurs at low temperatures, and the particles are aligned to their anisotropy axes. There is a significant difference between singleparticle blocking and collective blocking. The single-particle blocking occurs in isolated clusters because the intraparticle anisotropy is the only characteristic to stabilize the magnetic moment. Strong interparticle interactions lead to collective blocking, which occurs at higher temperatures. Collective blocking occurs due to the fact that a particle, which would be thermally activated if it was isolated, would have its magnetization stabilized by a neighbor that is slightly larger or more anisotropic. Institute of Physics. 16.7 SUPERCONDUCTING STATE OF NANOCLUSTERS 16.7.1 Qualitative Analysis In Chapter 14, expressions for the critical temperature Tc in both the weak coupling (BCS) and the strong coupling (McMillan) theories were derived. The expression for the critical temperature Tc in the weak-coupling limit (Eq. 14.116) is Tc % 1:13 \u03b8D exp \u2212 % 2 \u03b5F\u00de \u00f0, U & D (16.90) 16.7 Superconducting State of Nanoclusters 559 and in the strong coupling limit (Eq. 14.127), it is 1:04 \u03bb \u2212 \u03bc&\u00f0 In Eq. (16.91), \u03bb is the coupling constant (Eq. 14.124), exp \u2212 % Tc % \u03b8D 1:2 1 + \u03bb \u00f0 1 + 0:62\u03bb \u00de \u03bb = \u03bd < I 2> /M\u03a9 2, and \u03bd is the bulk density of states (Eq. 14.125), \u03bd = m&pF/2\u03c02: : & \u00de (16.91) (16.92) (16.93) It is obvious from Eqs. (16.90) and (16.91) that Tc, the critical temperature of a superconductor, increases if the density of states increases at the Fermi level. In fact, it has been observed that the Tc of Al films is nearly double the value for bulk samples. Granular Al has Tc \u223c 3\u00b0 K, which is almost three times the critical temperature of Al. These increases were explained by size quantization and corresponding increase in the effective density of states in films and isolated granules. \u223c2:1\u00b0 K \u00de \u00f0 The metallic nanoclusters contain delocalized electrons of which the states organize into shells, similar to those in atoms or nuclei. In some clusters, the shells are completely filled all the way up to the highest occupied shell. These are known as the \u201cmagic numbers,\u201d N = Nm = 20, 40, 58, 92, 138, 168, :::, and the clusters are spherical. These magic clusters have similarity with atoms in the sense that their electronic states are labeled by radial quantum number n and orbital momentum l. The Cooper pairs are formed by electrons with opposite projections of orbital momentum, similar to the pairing in atomic nuclei. The degeneracy of the shell, which is 2, is large if l is large, and the effective density of states corre\u00f0 spondingly increases. Further, the energy spacing \u0394E between neighboring shells varies. Some of them are separated by only a small \u0394E. 2l + 1 \u00de The combination of high degeneracy and a small energy spacing \u0394E between the", " highest occupied shell (HOS) and the lowest unoccupied shell (LUS) results in a large increase in the strength of the superconducting pairing interaction in the clusters. One can understand this result qualitatively from the fact that if the HOS is highly degenerate, the shell has a large number of electrons, which is equivalent to having a sharp peak in the density of states at the Fermi level. It is obvious from Eqs. (16.90) and (16.91) that the critical temperature of a superconductor increases if the density of states increases at the Fermi level. 16.7.2 Thermodynamic Green\u2019s Function Formalism for Nanoclusters In this section, we will derive an expression for the critical temperature for the nanoclusters. We will show how it is possible that the critical temperature for some of these nanoclusters can be of the order of room temperature. We follow a method outlined by Kresin and Wolf.18 However, we will not discuss in detail the strong coupling theory based on Green\u2019s function method of the many-body theory. In the standard thermodynamic Green\u2019s function formalism of Abrikosov et al. (1963), the equa, can be written as tion for the pairing order parameter, \u0394 \u00f0 \u03c9 \u00de \u03bbT\u2211 \u03c9n\u2032 % \u0394 Z \u03c9n\u00de \u00f0 \u03c9n \u2212 \u03c9n\u2032; \u03a9 D \u00f0 F + \u00de, \u03c9n\u2032\u00de \u00f0 e (16.94) 560 CHAPTER 16 Metallic Nanoclusters where Z is the renormalization function that describes the \u201cdressing\u201d of electrons moving through the lattice, \u03bb is the electron\u2013phonon coupling constant, \u03bb = 2 Z \u03b12 \u03a9 \u00f0 F \u00de \u00f0 \u03a9 \u00de \u03a9\u22121d\u03a9, (16.95) \u00f0 \u00de \u03a9 is the phonon density of states, \u03a9 is the phonon frequency, and \u03b12 F phonon-frequency-dependent electron\u2013phonon interaction. Here, D is the phonon Green\u2019s function, \u03c9n \u2212 \u03c9n\u2032\u00de \u00bd\u00f0 is a measure of the D = \u03a92 2 + \u03a92 (16.96) \u22121, \u03a9 \u00de \u00f0 $ where and \u03c9n = 2n + 1 \u03c0T \u00de \u00f0 \u03a9 = < \u03a92>1=2: The average < f \u03a9 > is determined by d\u03a9 f \u03b12 \u22121, and the pairing Green\u2019s function, introduced by Gorkov, is given by F + = \u0394 = \u03c9n\u00de \u00f0 \u00bd n + \u03be2 + \u03942 \u03c92 \u03c9n\u00de$ : \u00f0 (16.97) (16.98) (16.99) (16.100) In Eq. (16.100), \u03be is the electron energy relative to the chemical potential. The metallic clusters contain delocalized electrons of which the states organize into shells. For some clusters that are spherical, the shells are filled up to the highest occupied shell, i.e., those with the magic numbers N = Nm = 20, 40, 58, 92, 138, 168, :::: The electronic states in the magic clusters are labeled by radial quantum number n and orbital momentum l. The Cooper pairs are formed by electrons with opposite projections of orbital momentum, and if the orbital momentum is large, the shell has a degeneracy of 2 2l + 1 For nanoclusters, of volume V, Eqs. (16.92), (16.93), (16.94), (16.96), and (16.100) are modified as \u00f0 : \u00de where and \u0394 \u03c9n\u00de \u00f0 Z = \u03b7 T 2V \u2211 \u03c9n\u2032 \u2211 s D \u00f0 \u03c9n\u2212\u03c9n\u2032, \u03a9 F + s \u00f0, \u03c9n\u2032\u00de \u00de e \u03b7 = < I2> /M 2, \u03a9 \u03c9n\u2212\u03c9n\u2032, \u03a9 D \u00f0 2 = \u03a9 e \u03c9n\u2212\u03c9n\u2032\u00de \u00bd\u00f0 2 2 + \u03a9 \u22121, $ \u00de F + s \u00f0 \u03c9n\u2032\u00de e = \u0394 \u00f0 e \u03c9n\u2032\u00de\u00bd n\u2032 + \u03be2 \u03c92 s + \u03942 e \u03c9n\u2032\u00de$ \u00f0 \u22121, \u03bes = Es \u2212 \u03bc: (16.101) (16.102) (16.103) (16.104) (16.105) 16.7 Superconducting State of Nanoclusters 561 \u03c9n\u00de are the vibrational propagator and the pairing function, \u0394 \ufffd", "\ufffd is the order Here D and F\u00fes \u00f0 parameter, Z is the renormalization function, Es is the energy of the sth electronic state, and \u03bc is the chemical potential. The summation is over all the discrete electronic states. There are two aspects in which these equations differ from the corresponding equations for the bulk. First, they contain a summation over discrete energy levels ES instead of integration over a continuous energy spectrum, as in a bulk superconductor. Second, because the number of electrons is fixed, the position of the chemical potential \u03bc is determined by N and T and hence different from the value of Fermi level EF: \u03c9n\u00de Kresin and Wolf (Ref. 18) also showed that because \u201cmagic clusters\u201d have a spherical shape, Gj, where Gj is the one can replace the summation over states by summation over shells \u03a3s\u2192\u2211 j shell degeneracy \u00f0 and lj is the orbital momentum. They showed that for such clusters, Eqs. (16.101) through (16.104) can be written in the form \u00de Gj = 2 2lj + 1, (16.106) \u0394 \u03c9n\u00de \u00f0 Z = \u03bb 2EF 3N Gj \u2211 \u03c9n\u2032 \u2211 j 2 \u03a9 + 2 \u03a9 \u03c9n \u2212 \u03c9n\u2032\u00de \u00f0 e 2 2 \u03c9n\u2032\u00de \u0394 \u00f0 n\u2032 + \u03be2 \u03c92 l j Tc: (16.107) In Eq. (16.107), the bulk coupling constant \u03bb (Eq. 16.92) and the Fermi energy EF are used because the characteristic vibrational frequency is close to the bulk value due to the fact that the pairing is mediated mostly by the short-wave part of the vibrational spectrum. e However, if the shell is incomplete, there is a Jahn\u2013Teller deformation in the cluster. Because the shape becomes ellipsoidal, the s states are classified by their projection of the orbital momentum \u2265 1. In the weak coupling case, \u03b7/V \u226a 1, m j j \u03a9. In Eqs. (16.101), (16.103), and (16.104), one should substitute Z = 1 and which leads to \u03c0Tc \u226a D = 1. This leads to the BCS (weak coupling) theory. \u2264 l, and each level contains up to four electrons for m j j e It may be noted that Eqs. (16.101), (16.103), and (16.104) are different from the strong-coupling theory of bulk superconductors in two aspects. First, they contain a summation over discrete energy levels ES, whereas one integrates over a continuous energy spectrum (over \u03be) in bulk superconductors. Second, because the clusters have a finite Fermi system, the number of electrons N is fixed. Hence, the position of the chemical potential \u03bc is different from the Fermi level EF and is determined by the value of N and T: The value of the critical temperature Tc in a nanocluster depends on the number of valence elec\u03a9: One can obtain a trons N; the energy spacing \u0394E = EL \u2212 EH; and on the values of \u03bbb, EF, and high value of Tc (as high as 100\u00b0 K) by using realistic values of these parameters. The pairing in an isolated cluster can be observed from the strong temperature dependence of the excitation spectrum. When T \u223c 0\u00b0 K, the excitation energy is strongly modified by the gap parameter and significantly exceeds the gap parameter when T > Tc: For example, the minimum absorption energy for Gd83 clusters at T > Tc corresponds to \u0127\u03c9 6 meV, whereas for T \u226a Tc, its value is much larger: \u0127\u03c9 34 meV: Recently, Cao et al.4 measured the heat capacity of an isolated cluster. They observed a jump in 200\u00b0 K: The value of Tc as well as heat capacity for selected Al clusters (i.e., for Al\u2212 the amplitude of the jump and its width are in agreement with this theory. 35 ions) at T % % % e 562 CHAPTER 16 Metallic Nanoclusters PROBLEMS 16.1. We have shown that for a metallic cluster, the total energy can be written as n + Ees\u00bd is the kinetic energy of a system of independent particles of density n, n + Exc\u00bd = Ek\u00bd n \u00bd E n $ $ $ $, where Ek\u00bd n $ n Ek\u00bd $ = \u2211 i \u01272 2m j 2: \u2207\u03c8 ij Ees\u00bd n $ is the electrostatic energy, n Ees\u00bd $ = e2 2 ZZ n dr\u2032 dr \u00bd \u00f0 r", " \u00de \u2212 nI\u00f0 \u2212 nI\u00f0 r\u2032 \u00de$, r\u2032 n r \u00de \u00f0 \u00de$\u00bd r \u2212 r\u2032 j j (1) (2) (3) n and Exc\u00bd Section 7.8.2 (subject to the modification for a metallic cluster), show that is the exchange-correlation term of the energy. Following the procedure outlined in $ \u2212 \u01272 2m \u22072\u03c8 i\u00f0 r \u00de + VKS\u00f0 r \u03c8 i\u00f0 \u00de r \u00de = \u03b5i\u03c8 i\u00f0 r, \u00de VKS\u00f0 r \u00de = VH\u00f0 r r VH\u00f0 \u00de = 2e2 dr\u2032 Z \u00de, r + Vxc\u00f0 \u00de r\u2032 \u2212 nI\u00f0 r\u2032 n \u00de \u00f0 r \u2212 r\u2032 j j \u00de, Vxc\u00f0 is the exchange-correlation energy of the electron gas. \u03b5xc\u00f0 = n r \u00de \u00de \u00de n \u2202\u03b5xc\u00f0 \u2202n where and n Here, \u03b5xc\u00f0 \u00de 16.2. Show that in Eq. (16.24), < l2> 16.3. The ellipsoidal model is based on the harmonic oscillator Hamiltonian H = p2 2m + 1 2 m\u03c92 0\u00bd \u03a92 \u22a5\u00f0 x2 + y2 + \u03a92 z z2 \u2212 U\u0127\u03c90\u00bd l2 \u2212 < l2>n$ : $ \u00de Ignoring the anharmonic terms in Eq. (1), show that E n1, n2, n3\u00de \u00f0 = \u0127\u03c90 \u03a9\u22a5\u00f0 h n1 + n2 + 1 \u00de + \u03a9z n3 +! 1 2 : \" i (4) (5) (6) (7) (1) (1) (2) Problems 563 By using the definition of the distortion parameter defined in the text, \u03b4 = 2 \u03a9\u22a5 \u2212 \u03a9z\u00de \u00f0 \u03a9\u22a5 + \u03a9z, show that Eq. (2) can be written as E n1, n2, n3\u00de \u00f0 = \u0127\u03c90 \u2212 3n3\u00de + \u03b42 18 \u00f0 n + 3n3 + 3, & \u00de where n = n1 + n2 + n3: 16.4. If one defines the perpendicular distance of the two faces as p111 and p100, show that the lengths of the sides of the faces of the Wulff polyhedron (Figure 16.7) are \u03b2 and 1 \u2212 2\u03b2 units of 6p p111. The scale factor \u03b2 is given by \u03b2 = 1 \u2212 p100= 3p p111. \u00f0 (3) (4) (5) in \u00de 16.5. The polyhedron has a central atom and can be considered as built of successive shells \ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03 covering interior shells. The Kth shell contains of atoms in a cluster with K shells is \u00f0 10K 2 + 2 \u00de atoms. Show that the total number N = 1 3 \u00f0 10K3 \u2212 15K2 + 11K \u2212 3 : \u00de 16.6. The diffusive flux into the island is given by DN1\u03c3s = D2\u03c0rs \u2202n1 \u2202r jrs, (1) (1) from which the capture number \u03c3s can be calculated. The monomer density field can be obtained from the diffusion equation by first assuming that the monomers can evaporate from the substrate at the rate 1/\u03c4a and that N1 \u2192 F\u03c4a, a long way from the island. The latter assumption implies that the density of islands is sufficiently low. The diffusion equation can be written as r, t \u2202n1\u00f0 \u2202t \u00de = D\u22072n1\u00f0 r, t \u2212 \u00de r, t n1\u00f0 \u03c4a \u00de + F: In the steady state t > \u03c4a, which can be shown from Eqs. (1) and (2), show that the capture number \u03c3s can be written as where K0 and K1 are the Bessel functions and \u03c3s = 2\u03c0Xs K1\u00f0 K0\u00f0 Xs\u00de Xs\u00de, Xs = rs D\u03c4ap : \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 (2) (3) (4) 564 CHAPTER 16 Metallic Nanoclusters 16.7. We considered the simple case of one orbital site and local density of states di\u00f0 with center of gravity di\u00f0 E, \u00de EF Ebond = 2\u2211 \u00f0 i Z \u2212\u221e E", " \u2212 \u03b5i\u00de di\u00f0 E \u00de dE: The second moment of the local density of states can be written as 2 \u03bc\u00f0 \u00dei = \u221e Z \u2212\u221e E \u2212 \u03b5i\u00de \u00f0 E 2di\u00f0 \u00de dE: at site i E \u00de (1) (2) E by a Gaussian centered at \u03b5i and at width We represent di\u00f0 condition of charge neutrality to confirm that other factors arising out of Gaussian integration of Eq. (1) are site-independent, show that, in an approximation in which moments no higher i \u00debond is given by than the second are used, the bond energy E\u00f0. By invoking the q \u00de 2 \u03bc\u00f0 \u00dei \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 where A is a constant. i \u00debond = A E\u00f0 q, 2 \u03bc\u00f0 \u00dei \ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 16.8. In the second-moment approximation discussed earlier, show that the bandwidth W is proportional to z1/2, where z is the coordination number. For simplicity, if we consider a rectangular band for the d electrons, show that the density of states scales is 1/W. 16.9. Show that the low-temperature STM differential conductivity can be written as dI dV = 2\u03c0e2 \u0127 \u00f0 where 2 \u03b5\u2032 + q 1 + \u03b5\u20322 \u03c1tip\u2211 \u00de k j Mktj 2\u03b4 eV \u2212 \u03b5k\u00de \u00f0 + constant, q = A/B, A = Mat + \u2211 k MktVakP # 1 eV \u2212 \u03b5k, $ B = \u03c0\u2211 k MktVak\u03b4, eV \u2212 \u03b5k\u00de \u00f0 \u03b5\u2032 = eV \u2212 \u03b5a \u2212 Re\u2061\u03a3 eV Im\u2061\u03a3 \u00f0 \u00de eV \u00f0 \u00de, (3) (1) (2) (3) (4) (5) and \u03c1tip is the tip density of states. \u03c1tip is usually treated as a constant to reflect the broadening of the tip state by contact with the remainder of the tip electrode. (Hint: Most of the derivation is outlined in the appendix of the research paper by Madhavan et al.23). References 565 References 1. Bales GL, Chrazan DC. Dynamics of irreversible growth during submonolayer epitaxy. Phys Rev B 1994;50:6057. 2. Becke AD. 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J Non-Cryst Solids 1996;205\u2013207:781. 30. Skomski R. Nanomagnetics. J Phys Condens Matter 2003;15:R 841. 31. Slater JC, Koster GF. Simplified LCAO Method for the Periodic Potential Problem. Phys Rev 1954;94:1498. 32. Stepanyuk VS, Hergert A, Rennert P, Wildberger K, Zeller R, Dedelichs PA. Transition metal magnetic nanostructures on metal surfaces. Surface Sci 1997;377\u2013379:495. 33. Sutton AP. Electronic Structure of Materials. Oxford: Oxford University Press; 1993. 34. Vosko SJ, Wilk L, Nusair M. Accurate spin-dependent electron liquid correlation energies for local spin density calculations: a critical analysis. Can J Phys 1980;58:1200. 35. Woods BD, Saxon DS. Diffuse Surface Optical Mode for Nucleon-Nucleon Scattering. Phys Rev 1954;95:577. 36. Wulff G. Z Kirst 1901;34:449. Complex Structures CHAPTER 17 CHAPTER OUTLINE 17.1 Liquids................................................................................. 568 17.1.1 Introduction.", "...................................................................... 568 17.1.2 Phase Diagram.................................................................... 568 17.1.3 Van Hove Pair Correlation Function................................................. 569 17.1.4 Correlation Function for Liquids.................................................... 570 17.2 Superfluid 4He.......................................................................... 570 17.2.1 Introduction....................................................................... 570 17.2.2 Phase Transition in 4He............................................................ 570 17.2.3 Two-Fluid Model for Liquid 4He.................................................... 571 17.2.4 Theory of Superfluidity in Liquid 4He............................................... 571 17.3 Liquid 3He.............................................................................. 573 17.3.1 Introduction....................................................................... 573 17.3.2 Possibility of Superfluidity in Liquid 3He............................................ 574 17.3.3 Fermi Liquid Theory............................................................... 574 17.3.4 Experimental Results of Superfluidity in Liquid 3He.................................. 575 17.3.5 Theoretical Model for the A and A1 Phases.............", "............................. 575 17.3.6 Theoretical Model for the B Phase.................................................. 577 17.4 Liquid Crystals.......................................................................... 578 17.4.1 Introduction....................................................................... 578 17.4.2 Three Classes of Liquid Crystals.................................................... 578 17.4.3 The Order Parameter............................................................... 580 17.4.4 Curvature Strains.................................................................. 581 17.4.5 Optical Properties of Cholesteric Liquid Crystals...................................... 581 17.5 Quasicrystals........................................................................... 583 17.5.1 Introduction....................................................................... 583 17.5.2 Penrose Tiles...................................................................... 583 17.5.3 Discovery of Quasicrystals.......................................................... 584 17.5.4 Quasiperiodic Lattice.............................................................. 584 17.5.5 Phonon and Phason Degrees of Freedom............................................ 586 17.5.6 Dislocation in the Penrose Lattice..........................", "......................... 589 17.5.7 Icosahedral Quasicrystals........................................................... 589 Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00017-7 \u00a9 2012 by Elsevier Inc. All rights reserved. 567 568 CHAPTER 17 Complex Structures 17.6 Amorphous Solids....................................................................... 590 17.6.1 Introduction....................................................................... 590 17.6.2 Energy Bands in One-Dimensional Aperiodic Potentials............................... 591 17.6.3 Density of States.................................................................. 593 17.6.4 Amorphous Semiconductors........................................................ 593 Problems.................................................................................... 594 References.................................................................................. 597 17.1 LIQUIDS 17.1.1 Introduction It is well known that matter exists in three different (solid, liquid, and gaseous) phases. Solids have an ordered arrangement of atoms or molecules as evidenced by rigid and sharp Bragg reflections in a diffraction experiment. Liquids and gases are fluids and will flow even under a small shear stress. In diffraction experiments, they yield only diffuse rings, showing that there is no long arrangement of molecules. In addition, there are glasses and amorphous solids that blur the distinction between solids and fluids. The atoms or molecules in glasses are arranged at random, whereas those in amorphous solids have short-range order. Van der Waals first pointed out the continuity of liquid and gaseous states. At low temperatures below a critical temperature, two fluid phases can coexist in equilibrium: the dense phase is called liquid, and the less-dense phase is called gas. By heating above the critical temperature, compressing, and cooling, one can pass continuously from low-temperature gas to low-temperature liquid. The difference between liquid and gas is essentially a difference in density. For roughly spherical molecules (rare gases), there is disorder in only translational motion. There is the possibility of rotational disorder in molecules that are far from spherical. In plastic crystals, both kinds of disorder occur, whereas in \u201cliquid crystals,\u201d there can be translational order but rotational disorder 250 200 150 100 Melting solid Freezing liquid Liquid-vapour coexistence Tt 0.5 1.0 1.5 2.0 Density (gm cm\u22123)", " 17.1.2 Phase Diagram The study of the physics of liquids is mainly to understand why particular phases are stable in particular ranges of density, as shown in Figure 17.1. One has to relate the stability, structure, and dynamical properties of fluid phases to FIGURE 17.1 Phase diagram for the 6-12 fluid (solid lines) and for argon (dashed lines and circles). Here, \u2218 \u2208/k = 119\u00b0 K, \u03c3 = 3:405 A. Reproduced from Barker and Henderson5 with the permission of the American Physical Society. 17.1 Liquids 569 the size and shape of the molecules, atoms, or ions, and the nature of forces between them. Because the interactions that determine the bulk properties of matter are basically electrostatic in character (apart from small relativistic and retardation effects), and arise from the Coulomb interactions between nuclei and electrons, one way to predict the properties would be to solve the many-body Schrodinger equation describing the motion of nuclei and electrons. However, we will restrict the discussion to the concept of order parameters, a function that is present when order is present (solids) and vanishes when the desired order is absent (liquids). 17.1.3 Van Hove Pair Correlation Function The best way to distinguish between liquids and solids is to first introduce the van Hove pair correlation function p defined as r1, r2, t \u00f0 \u00de r1, r2\u2260j r1 \u2212 Xi\u00f0 \u00f0 0 \u03b4 r2 \u2212 Xj\u00f0 t \u00f0 \u00de\u00de \u00de\u00de : \" (17.1) \u00f0 r1, r2, t Here, p \u00de is to be found at position r2 at time t1 + t. The brackets mean a thermal average, and Xi\u00f0 t location of the particles. Here, Xi\u00f0 is the probability that if a particle is found at position r1 at time t1, some other particle tracks the is the Heisenberg operator defined for all i and all t by the equation \u00de \u00de t Xi\u00f0 where H is the Hamiltonian of the system. t \u00de = eiHt/\u0127 Xie\u2212iHt/\u0127, (17.2) The static structure factor, the dimensionless measure of scattering in a scattering experiment that measures the thermal average (because such experiments last much longer than the time scale of atomic motions), is defined as = q S \u00f0 \u00de 1 N \u2211 i, j h eiq. Xi\u2212Xj \u00f0 \u00de : i Eq. (17.3) can be rewritten in the alternate form \u2211 i,j Z From Eqs. (17.1) and (17.4), we obtain \u00f0 \u00de S q = 1 N dr1dr2 eiq. r1\u2212r2 \u00f0 \u00de \u03b4 \u00f0 r1 \u2212 Xi\u00de \u03b4 \u00f0 r2 \u2212 Xj\u00dei : h We define = 1 + q S \u00f0 \u00de 1 N Z dr1 dr2 p r1, r2, 0 \u00de \u00f0 eiq. r1\u2212r2 \u00f0 \u00de: = q p \u00f0 \u00de 1 V ZZ dr dr\u2032 p \u00f0 r + r\u2032, r, 0 eiq.r\u2032, \u00de where V is the volume of the system. From Eqs. (17.5) and (17.6), we obtain = 17.3) (17.4) (17.5) (17.6) (17.7) 570 CHAPTER 17 Complex Structures 17.1.4 Correlation Function for Liquids In general, liquids are homogeneous and isotropic. Therefore, one can assume that the pair correla. If the density of the liquid is tion function p \u03c1 = N/V, a dimensionless correlation function g r1 \u2212 r2j j can be defined as will depend on the distance r = r1, r2\u00de \u03c12 : \u00f0 \u00de One obtains from Eqs. (17.5) and (17.8) Eq. (17.9) can be rewritten in the alternate form S q \u00de \u00f0 = 1 + \u03c1 Z dr g r \u00f0 \u00de eiq.r dr eiq.r + \u03c1 Z dr g r \u00f0 \u00de \u00bd \u2212 1 eiq.r: $ (17.8) (17.9) (17.10) The second term in Eq. (17.10) is a delta function, which is zero except when directly along the scattering beam. Thus, it can be dropped while analyzing the experimental results of scattering, and one obtains from Eq. (17.10 dr r g \u00bd \u00f0 \u00de \u2212 1 $ eiq.r: (17", ".11) 17.2 SUPERFLUID 4He 17.2.1 Introduction Kammerlingh Onnes discovered in 1908 that liquid helium never solidified under its own vapor pressure. The interaction between the helium atoms is very weak because helium is an inert gas. The liquid phase is very weakly bound, and the normal boiling point is very low (4.2\u00b0 K). The large-amplitude quantum mechanical zero-point vibrations due to the small atomic masses and the weak interactions do not permit the liquid to freeze into the crystalline state. Liquid 4He solidifies only when a pressure of at least 25 atmospheres is applied. Therefore, it is possible to study liquid 4He all the way down to the neighborhood of absolute zero. 17.2.2 Phase Transition in 4He At 2.17\u00b0 K, a remarkable phase transition was discovered in liquid 4He under saturated vapor pressure. When the liquid was cooled through this temperature, all boiling ceased, and the liquid became perfectly quiescent. This effect occurs because liquid helium becomes an enormously good heat conductor. The thermal inhomogeneities that give rise to bubble nucleation are absent. The specific heat versus temperature curve of liquid 4He was shaped like a Greek letter \u03bb, characteristic of a 17.2 Superfluid 4He 571 second-order phase transition. This temperature is called the lambda point, and the experimental results are shown in Figure 17.2. The temperature at the lambda point is usually denoted by T\u03bb: Below this temperature, liquid 4He has remarkable flow as well as the \u201csuperheat\u201d transport properties. If a small test tube containing the liquid were raised above the surrounding helium bath, a mobile film of the liquid would be transported up the inner walls. Eventually, it would drip back into the bath, and the test tube would be emptied. Kapitza (Ref. 10) showed that liquid 4He could flow through the tiniest pores and cracks. Allen and Jones (Ref. 10) found that if a glass tube packed tightly with a powder was partially immersed in a 4He bath and then heated, a fountain of helium rising high above the level of the surrounding helium bath was produced.0 2.5 2.0 1.5 1.0 0.5 0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 T (K) FIGURE 17.2 Schematic diagram of the specific heat of liquid helium versus temperature. 17.2.3 Two-Fluid Model for Liquid 4He Landau (Ref. 10) and Tisza independently developed a two-fluid model to describe these phenomena. According to the two-fluid model, below T\u03bb, liquid 4He can be thought of as two interpenetrating fluids known as the normal and superfluid components. The superfluid component does not carry entropy and is involved in superflow through pores and cracks. In addition, it does not interact with the walls of a vessel containing the fluid in a dissipative fashion. In contrast, the normal component transports heat and exhibits viscosity, which allows transfer of energy between the liquid and the walls. The normal fluid density decreased with decreasing temperature, whereas the superfluid density increased, becoming dominant at the lowest temperature. The superfluid component replaces the normal fluid, which carries heat away from the heat source. The flow of the superfluid component toward a source of heat manifests in the fountain effect mentioned previously. The normal fluid consists of a gas of quantized thermal excitations that include the phonons (longitudinal sound waves) and rotons (short-wavelength compact excitations). It was predicted that heat transport would obey a wave equation that describes the compression and rarefactions in the proton/roton \u201cgas,\u201d which is known as second sound. 17.2.4 Theory of Superfluidity in Liquid 4He London (Ref. 15) noted that as the temperature of liquid 4He is reduced through the transition temperature, the occupancy of the one-particle ground state becomes macroscopic and can be thought of as a Bose\u2013Einstein (BE) condensate, which is the superfluid component of the two-fluid picture, although strong interactions between the atoms in the liquid modify this picture. According to London, the superfluid atoms are governed by a wave-function-like entity called the order parameter 572 CHAPTER 17 Complex Structures that was originally introduced by Ginzburg and Landau to explain the phenomenon of superconductivity (described in detail in Section 14.4.1). The order parameter \u03a8 for a superfluid 4He is given by \u03a8 = \u03a80ei\u03d5, (17.12) where \u03a80 is roughly thought of as the square root of the density of the superfluid component, and \u03d5 is a phase factor. The fact that the macroscopic order parameter is also", " described by a definite phase is known as broken gauge symmetry. It has been shown that the superfluid velocity is proportional to the gradient of the phase. The macroscopic order parameter picture describes how the helium atoms march in \u201clock step\u201d during superfluid flow. The existence of quantized vortices in superfluid 4He is also a consequence of this model. In fact, this phenomena is also seen in superfluid liquid 3He (to be discussed in the next section) and in superconductivity in solids, where a quantized current vortex must enclose a quantum of flux (discussed in Chapter 14). The fundamental assumption that underlies the modern theory of superfluidity in a Bose system such as liquid 4He is that the superfluid phase is characterized by a generalized Bose\u2013Einstein Condensation (BEC). We assume that at any given time t, it is possible to find a complete orthonormal basis (which may itself depend on time) of single-particle states such that one and only one of these states is occupied by a finite fraction of all the particles, while the number of particles in any other single-particle state is of order 1 or less. The corresponding single-particle wave function is called a condensate wave function, and the N0 particles occupying it, the condensate. The \u03c70\u00f0 T = 0 condensate fraction N0/N \u223c 0:1, where N is the total number of particles in the system. The macroscopic occupation occurs only in a single-particle state because according to the Hartree\u2013Fock approximation, the macroscopic occupation of more than one state is always energetically unfavorable provided the effective low-energy interaction is repulsive, as is the case for 4He. r, t \u00de Because the BEC occurs in the sense defined earlier, at any given time there exists one and only that is macroscopically occupied, and the conceptual basis for super- r, t one single-particle state \u03c70\u00f0 fluidity is quite simple. We can write \u00de and define the superfluid velocity vs\u00f0 from which we obtain \u03c70\u00f0 r, t by \u00de vs\u00f0 r, t = \u03c70\u00f0 j \u00de r, t r,t ei\u03d5 \u00f0 \u00dej \u00de r, t \u00de & \u0127 m \u2207\u03d5 r, t, \u00de \u00f0 \u2207 \u00d7 vs = 0: (17.13) (17.14) (17.15) Thus, the superfluid flow is irrotational. In addition, because no \u201cignorance\u201d is associated with the single state \u03c70, the entropy must be carried entirely by the \u201cnormal\u201d component, i.e., the particles occupying single-particle states other than \u03c70: These two observations provide the basis for Landau\u2019s phenomenological two-fluid hydrodynamics. However, the superfluid density \u03c1s, which occurs in the latter, is given by \u03c1s! is everywhere nonzero, the application of the Stokes theorem to the curl of Eq. (17.14) leads to the conclusion that the integral of vs around any closed curve is zero (Problem 17.1). However, we consider a line or a region infinite in one dimension on which N/V while N0! In a region where 0! \u00de \u03c70j j 0:1 N. T \u00f0 r, t \u03c70\u00f0 j \u00dej 17.3 Liquid 3He 573 vanishes. This would happen if the liquid is physically excluded from this region, or while atoms are present in this region, the single-particle state into which BEC has taken place happens to have a nodal line there. When we integrate Eq. (17.14) around a circuit that encloses the one-dimensional region, the fact that the phase of the wave function \u03c70 must be single-valued modulo 2\u03c0 leads to the Onsager\u2013Feynman quantization condition (Problem 17.2) vs. dl = nh/m: I (17.16) In a simply connected region of space, Eq. (17.16) can be satisfied by a \u201cvortex,\u201d which is a pattern of flow in which vs \u223c 1/r, where r is the perpendicular distance from the \u201ccore.\u201d Because vanishes at r = 0, vs is not defined, and hence, the singularity that appears at the core is physically irrelevant. \u03c70j j Thus, in a superfluid system, the circulation is quantized according to Eq. (17.16). In practice, the values n = \u00b11 are of interest because vortices with higher values of n are unstable against decay into these. However, vortices can be metastable for", " astronomical times even under equilibrium conditions. is not a directly observable quantity, whereas the is observable. Landau (Refs. 10-13) showed in his phenomenological theory We note that the superfluid velocity vs\u00f0 mass current J that in stable or metastable equilibrium, this quantity is given by r, t r, t \u00f0 \u00de \u00de J r, t \u00f0 \u00de = \u03c1svs\u00f0 r, t + \u03c1nvn\u00f0 \u00de r, t, \u00de (17.17) \u00de \u00de r, t r, t should be zero in the frame of reference in which the walls of the vessel are at rest. \u03c1 \u2212 \u03c1s are functions of only the temperature. is assumed to behave like the velocity of a normal (nonsuperfluid) liquid. In where the \u201csuperfluid\u201d and \u201cnormal\u201d densities \u03c1s and \u03c1n & The normal velocity vn\u00f0 equilibrium, vn\u00f0 In the 70 years since Landau\u2019s original proposal, although there has been almost universal belief that the key to superfluidity in liquid 4He is the onset of BEC at the lambda temperature, it has proved very difficult to verify the latter phenomenon directly. The main evidence comes from the high-energy neutron scattering and from the spectrum of atoms evaporated from the surface of the liquid. Although both are consistent with the existence of a condensate fraction of approximately 10%, neither can be said to establish it beyond all possible doubt. 17.3 LIQUID 3He 17.3.1 Introduction 3He is a rare isotope of helium that occurs in both solid and liquid states. Because the nuclear moments are very small, solid 3He undergoes nuclear magnetic ordering at a temperature of about 1 mK. Therefore, in the range of temperatures above 0.01\u00b0 K, the nuclear spins of the 3He atoms comprised of the solid are almost fully disordered. Hence, for spin 1/2 nuclei, the entropy Ssolid = R ln 2 per mole (Problem 17.3). In contrast, liquid 3He obeys Fermi\u2013Dirac statistics. Well below the Fermi, which is less than 1\u00b0 K, both the specific heat and the entropy will be degeneracy temperature linear functions of the temperature, Sliquid = \u03b3T: The liquid\u2013solid phase equilibrium is determined by the Clausius\u2013Clapeyron equation, according to which the melting curve is given by TF\u00de \u00f0 dP dT = Sliquid \u2212 Ssolid Vliquid \u2212 Vsolid = Latent Heat Vliquid \u2212 Vsolid\u00de \u00f0 : T (17.18) 574 CHAPTER 17 Complex Structures For 3He because Vliquid >Vsolid, the denominator is always positive. In the Fermi degenerate region (at the lowest temperatures), Sliquid < Ssolid and hence the slope of the melting curve becomes negative; i.e., it takes heat to freeze liquid 3He. 17.3.2 Possibility of Superfluidity in Liquid 3He In 1972, Osheroff, Richardson, and Lee (Ref. 20) and Osheroff, Gully, Richardson, and Lee21 discovered that liquid 3He possesses three anomalous phases below 5 mK, each of which appears to display most of the properties expected of a superfluid. Because liquid 3He obeys Fermi rather than Bose statistics, the mechanism of superfluidity cannot be simple BEC as in liquid 4He. It is believed that just as in metallic superconductors, the fermions pair up to form \u201cCooper pairs,\u201d which are sort of giant diatomic quasimolecules of which the characteristic \u201cradius\u201d is much larger than the typical interatomic distance. These molecules, which are composed of two fermions, effectively obey Bose statistics and hence can undergo BEC. We will discuss superfluidity in liquid 3He in the next section. The strong short-range repulsion of the quasiparticles in liquid 3He prevents pairing through so that the members of a pair do not rotate l = 0 Cooper pairs that have zero angular momentum \u00f0 around one another. However, over the years, a number of higher orbital angular momentum pairing states for a hypothetical superfluid state of liquid 3He were proposed. It is interesting to note that both p-wave (l = 1) and d-wave (l = 2) states of relative orbital angular momentum were proposed. However, the proposals for p-wave pairing by Anderson and Morel (Ref. 1) and Bailan and Werthamer (Ref. 4) were later identified as the actual superfluid phases of liquid 3He. The basic characteristics of the hypothetical superfluid 3He were that (1) there would be an intrinsic pairing mechanism not mediated by an ionic lattice (phonons), and (2) the", " resulting Cooper pairs would have internal degrees of freedom. These properties would distinguish superfluid 3He from superfluid 4He and superconducting electrons. \u00de 17.3.3 Fermi Liquid Theory Liquid 3He is composed of neutral atoms with nuclear spin angular momentum of \u0127/2 and a nuclear magnetic moment. The 3He atom has an odd number of elementary particles, so it obeys Fermi\u2013Dirac statistics and the Pauli exclusion principle. Because the atoms in the liquid interact strongly, Landau developed the Fermi liquid theory (Section 7.9) to explain the properties of liquid 3He. The basic idea of the Fermi liquid theory is to consider the excitations of the strongly interacting system instead of concentrating on the nature of the ground state. The scattering rate of the fermions is considerably reduced due to the Pauli exclusion principle. Landau termed the excitations, which act like particles, as quasiparticles. The various properties of normal liquid 3He qualitatively resembled the properties of ideal Fermi gas, but the numerical factors were obtained from the Fermi-liquid parameters (Section 7.9.3). In a Fermi liquid at low temperatures, the thermally excited quasiparticles occur in a narrow band near the Fermi surface with energy width on the order of kBT: Only the quasiparticles in this narrow band participate in scattering or in thermal excitations. The width of the band shrinks as T is lowered, and fewer quasiparticles can participate in such events. Consequently, the specific heat C and the entropy S depend linearly on the temperature (C = \u03b1T), and the mean free path is proportional to T \u22122: The thermal conductivity has a 1/T dependence, and the viscosity has a 1/T 2 dependence. At the lowest temperatures in a Fermi 17.3 Liquid 3He 575 liquid, the collisions are absent, and ordinary sound, which is produced as a result of propagation of waves of compression and rarefaction brought about by collisions of the molecules, dies away. This new mode of sound propagation is known as the zero sound and arises at the lowest temperatures due to self-consistent rearrangements of quasiparticles under the influence of Fermi-liquid interactions. 2:7 mK and TB % 17.3.4 Experimental Results of Superfluidity in Liquid 3He The classical experiment to find superfluidity in 3He was performed by Osheroff, Richardson, and Lee (Ref. 20) who used a Pomerchunk cell to observe two phase transitions (denoted as A and B) at 2:1 mK, respectively. They originally misinterpreted the results temperatures TA % as a second-order magnetic phase transition in solid 3He. However, subsequent nuclear magnetic resonance experiments made at the suggestion of Gully (Osheroff, Gully, Richardson, and Lee21) showed that the A and B phases were both superfluid phases of liquid 3He. In addition, it was observed that the A phase split into two phases (A and A1) in a magnetic field, whereas at about 0.6 Tesla, the B phase no longer existed. The early specific-heat measurements of liquid 3He near the superfluid transition were done by Webb et al.26 and are shown in Figure 17.3. The shape is characteristic of a BCS pairing transition. The phase diagram of liquid 3He in a magnetic field was investigated by Paulson et al.22 at pressures below melting pressures; they studied the static magnetization of the liquid via SQUID interferometry. The A phase narrowed and finally vanished at a point called the polycritical point (PCP) at about 22 bar. In a larger magnetic field, the B phase is suppressed in favor of the A phase even at the lowest pressure, and the PCP disappears, as shown in Figure 17.4. Liquid 3He-A 33.41 bar 0.030 0.035 0.025 A schematic P-T-H diagram showing the general topology of the superfluid phases, A, A1, and B of liquid 3He, is shown in Figure 17.5. The A1 phase occurs between the surfaces labeled A1 and A2: The A phase occurs at temperatures below the boundary labeled A2: The boundary between phases A and B is labeled B: The surface labeled S corresponds to the melting curve. 17.3.5 Theoretical Model for the A and A1 Phases The 3He A phase corresponds to the p-wave equal spin pairing state first considered by Anderson and Morel.1 It is a p-wave pairing state with total L = 1 and S = 1. It is an orbital R n / C 0.020 0.015 0.010 0.005 0 2.3 2.5 2.7 2.9 3.1 3.3 T-mK FIGURE 17.", "3 Specific-heat measurements of liquid 3He near the superfluid transition. Reproduced from Webb et al.26 with the permission of the American Physical Society. TB \u2212 A 49 Gauss 3He-A P S 576 CHAPTER 17 Complex Structures 32 30 28 26 24 22 20 18 16 14 12 10 r a b - P TA \u2212 B 378 Gauss 3He-B PCP TC NORMAL FERMI LIQUID 8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 T-mK FIGURE 17.4 Experimental data for the phase diagram in a magnetic field. Reproduced from Paulson et al.22 with the permission of the American Physical Society. B A2 A1 T H FIGURE 17.5 A schematic P-T-H diagram showing the superfluid phases, A, A1, and B of liquid 3He. Reproduced from Lee14 with the permission of the American Physical Society. m state along some direction ^l and spin m = 0 state along direction ^d, where ^d is the direction of zero spin projection. The Anderson\u2013Morel order parameter (Ref. 1) can be expressed as \u03c8 AM = orbital part \u00d7 spin part : (17.19) \u00f0 \u00f0 \u00de \u00de Here, the orbital part is in momentum space, and the spherical harmonic Y11 \u223c ei\u03c6 sin \u03b8 defines a polar axis corresponding to the direction of the pair orbital angular momentum. Thus, the Anderson\u2013 Morel order parameter (Ref. 1) can be defined as \u00de, % \u2193\u2191 + \u2191\u2193 \u03c8 AM \u223c ei\u03c6 sin \u03b8 1 2p \u00f0 $ where the spherical harmonic Y11 \u223c ei\u03c6 sin \u03b8 defines a polar axis ^l corresponding to the direction of the pair orbital angular momentum. Because in the spin-triplet pair-wave function, the spin part appears along the ^d-axis, only the component occurs. The three-dimensional repre\u00de sentation of the Anderson\u2013Morel order parameter is shown in Figure 17.6a. The vector ^l defines the axes of the order parameter. The amplitude is zero along this axis, which corresponds to sin \u03b8 dependence where \u03b8 is the polar angle with respect to ^l: The vector d! has the same direction for all points on the Fermi surface. The shaded region in Figure 17.17b shows the anisotropic energy gap and the two nodes along ^l. \u2193\u2191 + \u2191\u2193 \u00f0 (17.20) \ufb03\ufb03\ufb03 ^ d ^ l \u2206(T ) ^ l (a) (b) FIGURE 17.6 (a) A three-dimensional representation of the Anderson\u2013Morel order parameter; (b) the anisotropic energy gap is indicated by the shaded region. Reproduced from Lee14 with the permission of the American Physical Society. 17.3 Liquid 3He 577 It can be shown by using a classical argument that the dipolar interaction combined with spontaneously broken symmetry would favor the state for which ^l\u2225^d (Problem 17.4). The order parameter has nodes at \u03b8 = 0 and \u03b8 = \u03c0. Because the behavior of the BCS energy gap follows the order parameter, the gap nodes also appear at \u03b8 = 0 and \u03b8 = \u03c0, which is shown in Figure 17.6b. The three-dimensional picture is obtained by a revolution about the ^l-axis. The direction of ^l, which is perpendicular to the walls of the container, is also sensitive to flow and to the magnetic field. The spin state in Eq. (17.20) can be rotated in spin space, which results in the equal spin pairing version of the Anderson\u2013Morel order parameter, (17.21) \u03c8 AM \u223c ei\u03c6 sin \u03b8 \u2191\u2191\u232a + ei\u03a6 j \u2193\u2193\u232a, j \u00bd\u00f0j where \u03a6 is a phase factor. The A1 phase has the orbital properties described by the Anderson\u2013Morel state but has only \u2191\u2191\u232a spin pairs. \u00de$ The concept of spin fluctuation effect was introduced earlier by Layzer and Fay, who noted that because the nuclear magnetic susceptibility of liquid 3He was much higher than that of an ideal Fermi gas of comparable density, there was some tendency for the liquid to be ferromagnetic. When a 3He quasiparticle passed through the liquid, it would polarize spins of neighboring quasiparticles parallel to its own spin because of this ferromagnetic tendency. Anderson and Brinkman2 showed that the spin fluctuation feedback effect could indeed lead to a stable Anderson\u2013Morel phase in a zero magnetic field. After this paper, the Anderson-Morel phase was known as the Anderson\u2013", "Brinkman\u2013Morel (ABM) phase. \u2227 n axis ^ d \u2206(T) 17.3.6 Theoretical Model for the B Phase The simplest Balian\u2013Werthamer (BW) state (Ref. 4) is the 3P0 state, represented by \u03c8 BW \u223c Y1,\u22121j \u2191\u2191\u232a + Y10j \u2191\u2193 + \u2193\u2191\u232a + Y11j \u2193\u2193\u232a: (17.22) (a) (b) FIGURE 17.7 (a) The order parameter for superfluid 3He B; (b) the isotropic energy gap of the BW order parameter. Reproduced from Lee14 with the permission of the American Physical Society. 578 CHAPTER 17 Complex Structures = a^k, where a Because J = 0 for the 3P0 state, the BW state is specified in terms of the vector ^d is a constant. The order parameter for superfluid 3He B showing ^d vectors (thick lines) rotated by 104\u00b0 (Problem 17.4) about a vector along the radial directions (thin lines) for all points on the Fermi sphere is shown in Figure 17.7a. Figure 17.7b shows the isotropic energy gap of the BW order parameter. k \u00de \u00f0 17.4 LIQUID CRYSTALS 17.4.1 Introduction Liquid crystal is a state of matter intermediate between that of an isotropic liquid and a crystalline solid. Liquid crystals have many of the properties of a liquid, such as high fluidity, formation and coalescence of droplets, and inability to support shear. They are also similar to crystals in the sense that they exhibit anisotropy in electric, magnetic, and optical properties. There are two broad types of liquid crystals. Liquid crystals that are obtained by melting a crystalline solid are called thermotropic where temperature and (secondarily) the pressure are the controllable parameters. Liquid crystalline behavior is also found in certain colloidal solutions and certain polymers. This type of liquid crystal is called lyotropic, for which concentration and (secondarily) the temperature are the controllable parameters. Liquid crystals are found among certain organic compounds that may be of a variety of chemical types. However, the molecules forming liquid crystal phases have certain structural features that can be summarized as follows: a. The molecules are elongated and have flat segments, as in benzene rings. b. The long axis of the molecule is defined by a fairly rigid backbone containing double bonds. c. The molecule should have strong dipoles and easily polarizable groups. d. The extremities of the molecules are not very important. 17.4.2 Three Classes of Liquid Crystals Para-azoxyanisole (PAA) and 2-p-methoxybenzylidene n-butylaniline (MBBA), the two liquid crystals that have been extensively studied, are shown in Figure 17.8. CH3O CH3\u2013O FIGURE 17.8 N N \u2193 O N C H O\u2013CH3 PAA C4H9 MBBA Molecular structure of para-azoxyanisole (PAA) and 2-p-methoxybenzylidene n-butylaniline (MBBA). Reproduced from Stephen and Straley 25 with the permission of American Physical Society. Liquid crystals are divided into three main classes: (a) nematic, (b) cholesteric, and (c) smectic phases. They are shown in Figure 17.9. (a) Nematics: In the nematic phase, the long planar molecules are symbolized by ellipses. The long axes of the molecules align along a preferred direction, which indicates that there is a long-range orientational order. The locally preferred direction usually varies throughout the medium in the strained nematic. One defines a vector field n(r), known as the director, which gives its local orientation. Its magnitude is taken 17.4 Liquid Crystals 579 as unity. The director field can be aligned by electric and magnetic fields and by surfaces that have been properly prepared. Some structural perturbations appear as threads on optical examinations, which indicate that there is no idealized equilibrium configuration. There is short-range order in nematics as in ordinary liquids, but no long-range order in the centers of mass of the molecules. There is no preferential arrangement of the two ends of the molecules if they differ, but they can rotate about their long axes. Thus, the sign of the director has no physical significance, and optically, a nematic behaves as a uniaxial material with a center of symmetry, which is confirmed by the absence of ferroelectric phenomena. It has been suggested on the basis of X-ray and optical data that there exists another type of nematic phase, known as the", " cybotactic phase. In this phase, the molecules are arranged in groups such that the centers of mass of the molecules in each group lie in a plane. (b) Cholesterics: The cholesteric phase differs from the nematic phase in the sense that the director n varies in direction in the medium in a regular way. If a nematic initially aligned along the y-axis is twisted about the x-axis, a cholesteric configuration would be obtained. The director and the Fresnel ellipsoid rotate as one proceeds along the twist axis. The long axes of the molecules tend to align along a single preferred direction in any plane perpendicular to the twist axis. However, in a series of equidistant parallel planes, the preferred direction rotates as shown in Figure 17.9b. The distance measured along the twist axis over which the director rotates through a full circle is known as the pitch (P) of the cholesteric. Because n and \u2013n are indistinguishable, the periodicity length of the cholesteric is P/2. The pitch of cholesterics, which is sensitive to temperature flow, chemical composition, and applied magnetic or electric fields, is comparable with visible light because it is of the order of several thousand angstroms. (a) 1 2 P (b) (c) FIGURE 17.9 The arrangement of molecules in the (a) nematic, (b) cholersteric, and (c) smectic A phase. Reproduced from Stephen and Straley 25 with the permission of the American Physical Society. 580 CHAPTER 17 Complex Structures The characteristic colors of the cholesterics in reflection (through Bragg reflection by the periodic structure) and their very large optical rotatory power are due to the spiral arrangement. (c) Smectics: The molecules in the smectics are arranged in layers, and in addition to orientational layering, they exhibit orientational ordering. There are a number of different classes of smectics, which are briefly listed here: Smectic A: In this phase, the molecules are aligned perpendicular to the layers, but there is no long-range crystalline order within a layer (see Figure 17.9c). The layers can slide freely over one another. Smectic B: In this phase, there is a hexagonal crystalline order within the layers. The layers can slip on each other but cannot rotate on each other. Smectic C: In this phase, the preferred axis is not perpendicular to the layers so that the phase has biaxial symmetry. Smectic D: Optically, the D phase appears to have a cubic structure, and the X-ray patterns are consistent with a cubic packing. Smectic E: The X-ray patterns obtained from the smectic E phase show the presence of a layered structure and a high degree of order arrangement within the planes. 17.4.3 The Order Parameter If we assume that the molecules of a nematic or cholesteric liquid crystal are rigid and rodlike in i shape, then we can describe the orientation of the ith molecule by introducing a unit vector v!\u00f0 i along its axis. v!\u00f0 \u00de is different from the director n, which gives the average preferred direction of the molecules. It is not possible to introduce a vector order parameter for liquid crystals, which possess a i \u00de vanishes. Hence, the order parameter can be center of symmetry due to which the average of v!\u00f0 expressed only as a second-rank tensor \u00de r S\u03b1\u03b2\u00f0 \u00de = 1 N \u2211 i i i \u00de\u03b1 v\u00f0 v\u00f0 \u00de\u03b2 \u2212 1 3 \u03b4\u03b1\u03b2, (17.23) & where the sum is over all the N molecules in a macroscopic volume located at r, and the v\u03b1 are the components of v! referred to by a set of laboratory-fixed axes. S\u03b1\u03b2 is a symmetric traceless tensor of rank two and has five independent components. In the isotropic case, where the molecules have random orientation, S\u03b1\u03b2 = 0 (Problem 17.5). \u2019 To express the order parameter for nonlinear rigid molecules, one can introduce a Cartesian coordinate system x\u2032y\u2032z\u2032 fixed in the molecules. In the case of a uniaxial liquid crystal, the order parameter tensor is defined by r S\u03b1\u2032\u03b2\u2032\u00f0 \u00de = < cos \u03b8\u03b1\u2032 cos \u03b8\u03b2\u2032 \u2212 1 3 \u03b4\u03b1\u2032\u03b2\u2032>, (17.24) where cos \u03b8\u03b1\u2032 is the angle between the \u03b1\u2032 molecular axis, and the preferred direction or the optic axis. The angle brackets indicate an average over the molecules in a small but macroscopic volume. It can be shown that Eq. (17.24) is equivalent to Eq. (17.23) in the case of linear molecules or molecules with a well-defined long", " axis about which they rotate rapidly (Problem 17.6). 17.4 Liquid Crystals 581 In real liquid crystals, different parts of the molecules might have to be described by different S\u03b1\u03b2 tensors. It is preferable to define the order parameter through a macroscopic property such as the anisotropy in the diamagnetic susceptibility, Q\u03b1\u03b2 = \u03c7\u03b1\u03b2 \u2212 1 3 \u03b4\u03b1\u03b2\u03c7\u03b3\u03b3, (17.25) where \u03c7\u03b1\u03b2 is the magnetic susceptibility tensor per unit volume. By convention, repeated indices are to be summed over. Here, Q\u03b1\u03b2 is a symmetric traceless tensor of rank two and has five independent components. The diamagnetic susceptibility is approximately the sum of the susceptibilities of individual 0 0 0 \u00de3, and \u00de2, and \u03c7\u00f0 \u00de1, \u03c7\u00f0 molecules. We choose the principal susceptibilities of a rigid molecule to be \u03c7\u00f0 choose the fixed axes x\u2032, y\u2032, and z\u2032 of the molecule to coincide with the principal axes of the susceptibility. It can be easily shown for a uniaxial liquid crystal (Problem 17.7), Qxx = Qyy = 2N Sy\u2032y\u2032 + Sz\u2032z\u2032\u00de \u00bd\u00f0 0 \u00de1 + \u03c7\u00f0 Sz\u2032z\u2032 + Sx\u2032x\u2032\u00de \u00f0 0 \u00de2 + \u03c7\u00f0 Sx\u2032x\u2032 + Sy\u2032y\u2032\u00de \u00f0 0 \u03c7\u00f0 \u00de3 $ and Qzz = N 0 \u00de1 + Sy\u2032y\u2032\u03c7\u00f0 Sx\u2032x\u2032\u03c7\u00f0 \u00f0 0 0 \u00de2 + Sz\u2032z\u2032\u03c7\u00f0, \u00de3 \u00de where N is the number of molecules per unit volume. Further, it can be easily shown that Sx\u2032x\u2032 + Sy\u2032y\u2032 + Sz\u2032z\u2032 = 0: (17.26) (17.27) (17.28) Hence, there are only two independent parameters on the right side of Eqs. (17.26) and (17.27). 17.4.4 Curvature Strains There is a preferred axis along which the molecules orient themselves in the microscopic region of a liquid crystal. The direction of this axis varies from place to place in equilibrium, and it can also be forced to vary by external forces. The relative orientations away from the equilibrium position are known as curvature strain, and the restoring forces are known as curvature stresses. The free energy density is a quadratic function of the curvature strains. The theory of the curvature elastic energy is based on the symmetry properties of the liquid crystal. \u00f0 r The three distinct curvature strains of a liquid crystal are shown in Figure 17.10 by assuming that ^n is a unit vector giving the direction of the preferred orientation at the point r and varies slowly from point to point in molecules with permanent dipole moments. At r, we introduce a coordinate system with z parallel to n. The curvature has six components, resulting in three curvature strains: splay, twist, and bend. \u00de 17.4.5 Optical Properties of Cholesteric Liquid Crystals The unusual optical properties of cholesteric liquid crystals include the color effects seen in reflection under white light. These effects are due to the interactions of the light with the twisted arrangement of the molecules as well as the spatial variation of the dielectric constant. The variation of the dielectric constant through the medium is small, and we will assume that the normal \u00f0 form of the waves propagating through the medium is approximately that of ordinary circularly \u0394n/n \u223c 0:03 \u00de 582 CHAPTER 17 Complex Structures (a) (b) (c) FIGURE 17.10 The three curvature strains of liquid crystals: (a) splay: s1 = \u2202nx /\u2202x, s2 = \u2202ny /\u2202y ; (b) twist: t1 = \u2212\u2202ny /\u2202x, t2 = \u2202nx /\u2202y ; (c) bend: b1 = \u2202nx /\u2202z; b2 = \u2202ny /\u2202z: Reproduced from Stephen and Straley 25 with the permission of the American Physical Society. \u03b5 c\u03c9/ 0 1.1 Right polarized \u03c9+ 0.9 Left polarized Q/t0 \u03c9\u2212 0.9 FIGURE 17.11 Dispersion of light in a right-hand twisting cholesteric. Reproduced from Stephen and Straley 25 with the permission of the American Physical Society. polarized waves. These waves would be strongly affected only if the half-wavelength nearly matches the periodicity length (or its projection on the wave vector) and if the sense of its rotation is the same as the twisted structure. Otherwise, the effect on the dispersion relation of the wave would", " be the same as in other periodic structures. Because n and \u2013n are indistinguishable, the periodicity is one-half the pitch, and according to the Bragg formula, a \u201cband gap\u201d would appear at a matching wavelength \u03bb0 = p sin \u03b8, (17.29) where p is the periodicity length. The periodic perturbation determines the size of the band gap. The form of the dispersion relation is shown in Figure 17.11. The right-hand circularly polarized wave is strongly affected when its half-wavelength nearly matches the periodicity length. The left-hand circularly polarized wave of the same wavelength is unaffected. The extrema of the band gaps are obtained from the fact that light of wavelength \u03bb0 can travel at two distinct speeds with its electric vector aligned with the principal axes of the dielectric constant. Thus, the extrema of the band gap are \u03c9 + = 2\u03c0c/n\u2212\u03bb0, \u03c9\u2212 = 2\u03c0c/n + \u03bb0: (17.30) Here, n + and n\u2212 are the refraction indices. If light of frequencies between \u03c9 + and \u03c9\u2212 and of the appropriate circular polarization to match the twist is directed at the liquid crystal, it is 17.5 Quasicrystals 583 totally reflected, and the polarization of the reflected light matches the sense of the twist. Thus, a right-hand cholesteric reflects right-polarized light, whereas left-polarized light suffers only weak reflection. Because the reflected band is narrow, a pure color is reflected; \u00de this color depends on the pitch as well as the angle of incidence. \u0394\u03c9/\u03c9 \u223c \u0394n/n \u223c 0:03 \u00f0 Liquid crystals are extensively used in fiber-optic devices applied in telecommunication circuits. The optics of liquid-crystal devices (LCD) has evolved extensively in the past decade. 17.5 QUASICRYSTALS 17.5.1 Introduction It had been generally accepted among solid state physicists that crystals can have translational periodicity as well as one-, two-, three-, four-, and six-fold rotational symmetries. However, five-fold rotational symmetry could not exist in equilibrium-condensed phases. An icosahedron, which is the most locally densely packed arrangement, had been observed in liquids and amorphous solids. It was thought that an icosahedral rotational symmetry contradicts the translational periodicity and is unlikely to be found. 17.5.2 Penrose Tiles Penrose introduced the concept of two-dimensional tiles in 1974 (the concept was published in 1977)23; he proposed that it is possible to cover (tile) any flat two-dimensional space with only two different tile shapes (known as fat and skinny rhombus) in an infinite number of aperiodic ways. The tiles, which are rhombi, must be placed such that the matching arrows are always adjacent. The Penrose tiles are shown in Figure 17.12. The smaller angle of the fat rhombus is 2\u03c0/5, and the smaller angle of the skinny rhombus is \u03c0/5. If the length of each side is 1, the long diagonal of rhombus (dashed line) has length the fat /2, and the short diagonal of the \u03c4 = skinny rhombus (dashed line) has length 1/\u03c4: 5p + 1 \u00de \u00f0 \ufb03\ufb03\ufb03 The Penrose lattice is obtained by tiling a plane using a collection of the two Penrose tiles. The Penrose lattice, shown in Figure 17.13, shows the local regions of five-fold symmetry. FIGURE 17.12 Penrose tiles. FIGURE 17.13 The Penrose lattice. 584 CHAPTER 17 Complex Structures 31.72 63.43 FIGURE 17.14 Diffraction pattern of five-fold symmetry (area proportional to the intensity) and stereogram of the icosahedral group. Reproduced from Misra18 with the permission of Elsevier. 17.5.3 Discovery of Quasicrystals In 1984, Shectman et al. (Ref. 24) discovered sharp diffraction patterns of icosahedral symmetry in Al-Mn alloy rapidly quenched from the melt. The icosahedral phase in rapidly solidified Al-Mn alloy is resistant to crystallization up to 350\u00b0 C and is metastable. The results of Shectman et al. are shown schematically in Figure 17.14. In view of the above, Levine and Steinhardt generalized the notion of crystal to include a quasiperiodic translational order and named it \u201cquasicrystal.\u201d A quasicrystal has long-range quasiperiodic order as well as long-range crystallographically forbidden orientation symmetry. The classes of quas", "icrystal include Al-based transition metal alloys (TM alloys: AlMn, AlMnSi, AlCuFe, and AlPdMn); alloys having similar composition to that of the Frank\u2013Kasper phase with tetrahedrally close-packed structure (FK alloys: Al-Cu-Li, Zn-Mg-RE; RE = rare earth); and stable binary alloys (Cd alloys: Cd5.7Yb and Cd17Ca3). One can observe a decagonal phase with a diffraction pattern of 10-fold rotational symmetry in AlMn, AlFe, AlCuCo, and AlCoNi. There is a one-dimensional translational periodicity along the 10-fold rotational axis in these alloys. CrNiSi and VNiSi are two-dimensional octagonal quasicrystals. The two-dimensional dodecagonal quasicrystals are TaTe and a polymer alloy. 17.5.4 Quasiperiodic Lattice The various methods used to mathematically generate the structures of quasicrystals include the inflation\u2013deflation operation, utilization of matching rules, grid method, strip projection method, cut projection method, or generalized dual method. We will discuss the strip projection method and note that the other methods are closely related. In the strip projection method, the lattice points in the hypercubic lattice in the n-dimensional space E\u2033 are projected on vertices in the d-dimensional quasiperiodic lattice. E\u2033 is decomposed into two subspaces, 17.5 Quasicrystals 585 E\u2033 = E\u2225 + E\u22a5, (17.31) where the subspace E\u2225 is the d-dimensional physical space, and E\u22a5 is orthogonal to E\u2225. E\u22a5 is called the \u201cperpendicular space\u201d (perp-space). The projection of a unit \u201chypercube\u201d onto E\u22a5 is known as the \u201cwindow\u201d W. For any x0 in E\u2225, we write W0 for W + x0. When a lattice point in E\u2033 is located inside W0, its projection onto E\u2225 is selected as a vertex in the physical space E\u2225. If the \u201cslope\u201d of the hyperplane E\u2225 is rational to the hypercubic lattice in E\u2033, the projected lattice is periodic. However, if the \u201cslope\u201d of the hyperplane E\u2225 is irrational, the projected lattice becomes quasiperiodic. The hypercubic lattice, the directions of the two orthogonal subspaces E\u2225 and E\u22a5, and W must be invariant under operations of the noncrystalline group of the quasicrystal. In Figure 17.15, we show a one-dimensional quasiperiodic lattice obtained by projection from a two-dimensional square lattice. The Fibonacci lattice is generated when the E\u2225-axis is generated at \u03b8 = arctan 1/\u03c4 to the x-axis. Here, the golden mean \u03c4 = 5p + 1 /2. In Figure 17.16, we show a Fibonacci lattice, obtained in this manner, that can also be described : The twoas a one-dimensional section \u00de dimensional periodic structure consists of a periodic arrangement of a line segment extending in the direction of E\u22a5. The line segment is called an atomic surface. A point sequence is obtained on the E\u2225 section comprised of two spacings L and S. The irrational shape indicates the lack of periodicity in the arrangement of L and S. \ufb03\ufb03\ufb03 of a two-dimensional periodic function The distribution function of lattice points \u03c1 x of a two-dimensional square lattice is given by x \u03c1 \u00f0 \u00de = \u03c1 \u00f0 x\u2225, x\u22a5 = \u2211 jl \u03b4 \u00f0 \u00de \u00d7 \u03b4 \u00f0 \u00de x\u22a5 + j sin \u03b8 \u2212 l cos \u03b8, \u00de (17.32) \u00f0 \u00de x\u2225 \u2212 j cos \u03b8 \u2212 l sin \u03b8 and the distribution function on the projected lattice on E\u2225 is (Problem 17.8) x\u2225 \u03c10\u00f0 \u00de = \u221e Z \u2212\u221e dx\u22a5\u03c1 \u00f0 x\u2225, x\u22a5 W x\u22a5 : \u00de \u00f0 \u00de (17.33) EII E EII E W FIGURE 17.15 One-dimensional quasiperiodic lattice (crossmarks on the E\u2225-axis) generated by projection. FIGURE 17.16 A Fibonacci latt", "ice. Reproduced from Misra18 with the permission of Elsevier. Reproduced from Misra18 with the permission of Elsevier. 586 CHAPTER 17 Complex Structures Here, W x\u22a5 \u00f0 \u00de = ( 1, if 2 < x\u22a5 j j \u00f0 cos \u03b8 + sin \u03b8 \u00de = 0, otherwise: ( (17.34) The diffraction pattern of the one-dimensional Fibonacci lattice is expressed by a structure factor \u221e = f k \u00f0 \u00de % Here, and dx\u2225e\u2212ikx\u2225 x\u2225 \u00de \u03c10\u00f0 sin k\u22a5\u03c9 k\u22a5\u03c9 Z \u2212\u221e \u221e \u2211 m,n =\u2212\u221e k \u2212 2\u03c0 \u03b4 \u00f0 \u00f0 m sin \u03b8 + n cos \u03b8 : \u00de\u00de \u03c9 = 1 2 \u00f0 cos \u03b8 + sin \u03b8 \u00de k\u22a5 = 2\u03c0 \u00f0 m cos \u03b8 + n cos \u03b8 : \u00de (17.35) (17.36) (17.37) This procedure of projection and calculation of structure factor is generalized to higher-dimensional cases. For example, the pentagonal quasicrystal is obtained by a projection of the hypercubic lattice to a two-dimensional space and the icosahedral quasicrystal by that of the six-dimensional one to a three-dimensional space. The three-dimensional icosahedral quasicrystals are constructed by two rhombohedral units: prolate and oblate. 17.5.5 Phonon and Phason Degrees of Freedom The diffraction intensity pattern I of a solid is, in general, given by q \u00f0 \u00de where q I \u00f0 \u00de &j S q 2, \u00dej \u22122\u03c0iq.rdr, (17.38) (17.39) q is the wave vector, and \u03c1 characteristics of the function I(q) observed experimentally are as follows: is the atomic-density function in real space. For a quasicrystal, the r \u00f0 \u00de a. It consists of \u03b4\u2212 functions. b. The number of basis vectors necessary for indexing the positions of the \u03b4\u2212 functions exceeds the number of dimensions. c. It shows a rotational symmetry forbidden in the conventional crystallography. E Ell (a) E\u22a5 O ull (c) Ell 17.5 Quasicrystals 587 E\u22a5 U O (b) E\u22a5 u\u22a5 O (d) Ell Ell FIGURE 17.17 (a) A Fibonacci lattice, the structure resulting from a displacement of (b) U, (c) u, and (d) w. Reproduced from Misra18 with the permission of Elsevier. As an example, we consider the Fibonacci lattice in Figure 17.17a. We consider the translation of the two-dimensional periodic structure by a vector U with respect to the origin of the physical space E\u2225 (Figure 17.17b). It can be shown that the two structures on E\u2225 before and after the displacement U can be overlapped out to large finite distances by a finite translation in E\u2225. These two structures are said to belong to the same local isomorphic class (LI class). Because these structures are physically indistinguishable, they give the same diffraction intensity function I(q) and have the same energy. The vector U can be written as U = u + w, (17.40) where u represents the degrees of freedom of d-dimensional translation in physical space that crystals also possess, and w represents degrees of freedom characteristic of a quasiperiodic system. Figure 17.17c shows u results in a translation of the Fibonacci lattice in E\u2225, and Figure 17.17d shows how w generates a rearrangement of L and S. N \u2212 d \u00f0 \u00de Here, u and w are called phonon and phason displacements. When these displacements vary spatially, their gradients yield a strain. The gradient of u yields the conventional elastic strain (phonon strain), whereas the gradient of w yields the phason strain. A phonon-strained Fibonacci lattice is shown in Figure 17.18. A uniform phonon strain is introduced by a compression deformation of the two-dimensional structure. 588 CHAPTER 17 Complex Structures O O FIGURE 17.18 FIGURE 17.19 Phonon-strained Fibonacci lattice. Phason-strained Fibonacci lattice. Reproduced from Misra18 with the permission of Elsevier. Reproduced from Misra18 with the permission of Elsevier. FIGURE 17.20 Examples of phason flips in the two-dimensional Penrose lattice. Reproduced from", " Misra18 with the permission of Elsevier. A uniform phason strain is introduced by a shear deformation of the Fibonacci lattice, as shown in Figure 17.19. A phason displacement results in a local rearrangement of points (atoms) such as LS\u2194SL, which is known as the phason flip. Examples of the phason flip in two-dimensional Penrose lattice, which is known as a two-dimensional decagonal quasicrystal, are shown in Figure 17.20. To decompose properly the total N degrees of freedom into d phonon and phason degrees of freedom, i.e., to embed properly a given d-dimensional quasiperiodic structure into an N-dimensional hypercrystal, one needs to know the point group symmetry of the system. N \u2212 d \u00f0 \u00de 17.5 Quasicrystals 589 FIGURE 17.21 A perfect edge dislocation in the Penrose lattice. The shaded tiles are destroyed to produce intratile phason defects after gliding the dislocation to the left. Reproduced from Misra18 with the permission of Elsevier. 17.5.6 Dislocation in the Penrose Lattice The perfect edge dislocation introduced in the two-dimensional Penrose lattice (or Penrose tiling) is shown in Figure 17.21. The phason strain field cannot be easily recognized around the dislocation. However, the tiling pattern changes when the dislocation position is translated to the left by a distance represented by the arrow. 17.5.7 Icosahedral Quasicrystals An icosahedron, shown in Figure 17.22, is a regular polyhedron with 20 identical equilateral triangular sides. The main characteristics of a quasicrystal structure are that it is a combination of a quasiperiodic lattice and a cluster decorating it. Three types of icosahedral quasilattices, P-, F-, and I-types, are known theoretically, out of which two types, P and F, have been observed experimentally. These two types are distinguished by reflection conditions. The reflection vector of g of a icosahedral quasicrystal can be written as g = 1 a6D 6 \u2211 1 miei//, (17.41) where the six ei// vectors with length 1/2 are parallel to the lines connecting to the center and vertices of an icosahedron, and a6D is the lattice parameter of the six-dimensional hypercubic lattice in the framework of the section. There are no restrictions for the indices in the case of P-type reflections, whereas either all odd or all even indices appear in the F-type. The reflection condition in the F-type exhibits a \u03c4-scaling rule (\u03c4: golden mean), whereas the P-type has a \u03c43-scaling rule. However, in real space, a P-type quasilattice can be decomposed into two F-type sublattices. Thus, the F-type quasicrystal can be interpreted as an ordered phase, in which two kinds of atomic clusters with different atomic configurations are arranged regularly. The icosahedron. FIGURE 17.22 In the case of Al-transition metal quasicrystals, the Mackay-type cluster is considered as a basic structural unit. The 54 atoms form the triple shells as presented in Figure 17.23a. The first and the 590 CHAPTER 17 Complex Structures 2.45 \u00c5 4.70 \u00c5 (a) 4.83 \u00c5 2.48 \u00c5 4.50 \u00c5 4.93 \u00c5 6.78 \u00c5 (b) FIGURE 17.23 Icosahedral atomic clusters. (a) Mackay type in Al-Mn-Si approximant and (b) Bergman type in Zn-Mg-Al approximant. The radius in each shell is inserted. Reproduced from Misra18 with the permission of Elsevier. second shell, which are composed of Al, are an icosahedron and an icosidodecahedron, respectively. The third shell is a larger icosahedron of the transition metal. The five ternary alloys, which are of the stable Mackay type, are Al-Cu-(Fe, Ru, Os) and Al-Pd-(Mn, Re). The Zn-based quasicrystals are the Bergman clusters shown in Figure 17.23b. They include 104 atoms. This cluster has four concentric shells: an icosahedron, a dodecahedron, a larger icosahedron, and a truncated icosahedron (soccer ball). The 10 ternary alloys, which are of the Bergman type, are Al-Li-Cu", ", Zn-Mg-Ga, Ti-Zr-Ni, Mg-Al-Pd, and Zn-Mn-(Y, Dy, Gd, Ho, Tb, and Er). 17.6 AMORPHOUS SOLIDS 17.6.1 Introduction Amorphous solids have attracted a great deal of attention in recent years. There are an infinite number of ways in which the geometrical arrangement of atoms can be visualized. Sometimes, amorphous solids have been visualized as frozen-liquid structures, but there are some significant differences between the two types of states. For example, there are distinct differences between the radial distribution of functions for a true liquid and an amorphous solid. Another interesting difference in their properties is that whereas liquid Ge and Si are metals, amorphous Si and Ge 17.6 Amorphous Solids 591 are semiconductors. The bonds are broken in liquid Ge and Si, where presumably the sp3 configuration in the solid is changed to an s2p2 configuration in the liquid, which makes it metallic. However, in the amorphous state, the sp3 bonds are still present between nearest neighbors. These bonds are imperfect because the amorphous films are produced at relatively low temperature. Due to this bonding, amorphous Ge and Si are semiconductors like their crystalline counterpart, but there are significant differences arising out of the fact that (1) the bonding is imperfect because there is no true tetrahedral symmetry in spite of four-fold coordination over small regions; (2) different regions are not linked because there is no long-range order; and (3) many atoms could have only three nearest neighbors due to which the bond angles are severely distorted from the ideal value. Amorphous solids can be obtained in a variety of ways. In some cases, they are obtained in the frozen-liquid amorphous state either by evaporation into a cooled substrate or by extremely rapid cooling from the melt. The more popular method is by a process known as \u201csputtering\u201d the components onto a cooled substrate. In this process, the atoms in a solid are knocked out by energetic ions of inert gas such as argon. The inert gas is at a reduced pressure. It is ionized by an electrical discharge. A substrate is placed above the solid, which is in contact with an electrode, and the atoms are discharged by the impingement of the energetic argon ionized by the electric discharge. The atoms condense above the substrate and form a thin film. The sputtering process has the advantage that a thin layer of an amorphous film is deposited on the substrate. The disadvantage of this method is that both argon and oxygen are invariably present as impurities. 17.6.2 Energy Bands in One-Dimensional Aperiodic Potentials To consider the localized states, we consider the energy bands in one-dimensional aperiodic potentials.8 We consider a finite segment of a line a < x < b, where b \u2212 a = L: The potential energy of the segment is periodic but can be derived from a periodic potential by a disordering process. The two linearly independent real solutions of the Schrodinger equation are and b Z a b \u03c8 2 1 dx = 1, where \u03c81\u00f0 a \u00de = \u03c81\u00f0 b \u00de, \u03c81\u00f0 x \u03c82\u00f0 \u00de x \u00de dx = 0 dx = 1: (17.42) (17.43) Economou and Cohen (Ref. 8) postulated that the solutions of the aperiodic case that most resemble the Bloch functions of the periodic case are those combinations of \u03c6 that satisfy an extremal condition in relation to the average value of momentum <^p>. \u03c6 is so chosen that the real part of <^p> is as extremum. One can write \u03c6 as \u03c6 = \u03c81 + \u03b1 + i\u03b2 \u03c8 2= \u00bd \u00de \u00f0 1 + \u03b12 + \u03b22 1 2, $ (17.44) 592 CHAPTER 17 Complex Structures where x and y are real. The average value of ^p (Problem 17.9) is <^p> = \u2212i\u0127 d\u03c6/dx \u03c6\u2019 \u00f0 \u00de dx = \u0127 Z \u03b2 \u03c012 \u2212 \u03c021\u00de \u00f0 \u2212 i \u00bd where \u03c011 + \u03b12 + \u03b22 \u00de 1 + \u03b12 + \u03b22 \u00f0 \u03c022 + \u03b1 \u03c012 + \u03c021\u00de$, \u00f0 (17.45) \u03c0ij = b Z a \u03c8 id\u03c8 j, i, j = 1, 2 : The extremum requirement on Re <^p> gives \u03b1 = 0, \u03b2 = 1 or \u03b1 = 0, \u03b2 = \u22121: (17.46) (17.47) To avoid obtaining", " the nonzero imaginary part of the expectation value of the extremum value of <^p> from the preceding solutions, Economou and Cohen imposed periodic boundary conditions, = \u03c82\u00f0 so that the energy eigenvalues originating from Eq. (17.48) are exactly the same as the first, third, fifth, and so on, obtained by setting V = \u221e outside (a, b. Thus, there are half the number of values \u00de in this problem as in the infinite-barrier problem. (17.48) \u03c82\u00f0 b a \u00de \u00de, From Eqs. (17.44) and (17.47), there are two functions that make <^p> an extremum (Problem 17.10), and \u03c6 = \u03c81 + i\u03c82\u00de \u00f0 2p / \ufb03\ufb03\ufb03 (17.49) \u03c6\u2019 = \u03c81\u2212 i\u03c82\u00de \u00f0 2p : / (17.50) The corresponding values of <^p> are \ufb03\ufb03\ufb03 log (density of states) log (|pext|) Periodic 1 2 \u00f0 \u00de x x \u0127 (17.51), \u03c6\u2019n\u00f0 pext = \u00b1 = \u00b1\u0127\u03c012: \u03c012 \u2212 \u03c021\u00de One can easily show that the set of functions \u03c6n\u00f0, n = 1, 2, 3, \u2026 is complete and ortho\u00de normal and makes each momentum expectation value extremal. This set is the closest one can come to a set of Bloch functions in the aperiodic case. A schematic representation of these results is shown in Figure 17.24. Here, Ec is the energy at which the states change from localized to extended. EB is the energy at which the density of states vanishes, the bound of the spectrum. They coincide in the periodic case and move in the opposite direction as the aperiodicity is increased. Aperiodic EC EC = EB EB E FIGURE 17.24 Density of states (solid curve) and distribution of the extended states (dashed curve) for an energy band of an ordered and disordered lattice. Reproduced from Economou and Cohen (Ref. 8) with the permission of the American Physical Society. 17.6 Amorphous Solids 593 17.6.3 Density of States Concepts such as crystal momentum, effective mass, or band structure function E(k) cannot be used for localized one-electron states. However, the concept that is used and still valid is the density of states, g, which can be defined as E \u00f0 \u00de = E g \u00f0 \u00de 1 Vg \u2211 i \u03b4 E \u2212 Ei\u00de \u00f0 : (17.52) In delocalized states of disordered solids, bands can exist in which tails with localized states are attached to their edges. The region without states between the tails of two adjacent bands is called a gap, or if the tails overlap, the region is known as a pseudogap. 17.6.4 Amorphous Semiconductors We first assume that each Ge or Si atom in the amorphous state has four nearest neighbors as in the crystalline state. However, the covalent bonds in the crystalline phase are distorted in the amorphous phase. The band edges of the conduction and valence band contain localized states. They are separated from the extended states by the mobility edges. The abrupt band edges are shown in Figure 17.25. The electrons occupying the tail states cannot take part in the conduction because they are loca- lized due to the disorder in the potential. Normally, each Si or Ge atom, which has four valence electrons, shares one electron in the covalent bond with its neighbor. However, in the amorphous state, the three-fold-coordinated atoms produce a dangling bond because one bond remains uncompensated. A dangling bond essentially means an electron and an empty state. The electron in the dangling bond is localized at 0\u00b0 K because there are no adjacent sites available for it to move in the amorphous state. The concentration of these dangling bonds is very high (on the order of 1025m\u22123, and they control the position of the Fermi level. \u00de EG However, at finite temperatures, the process of electrical conduction occurs in three different regimes: the propagating regime, the jumping regime, and the hopping regime. The propagating regime, which is commonplace in the crystalline solid state theory, is dominant in disordered systems only at very high temperatures. At intermediate temperatures, electronic conduction in such systems takes place by diffusion or Brownian motion. This type of conduction, encountered near the mobility edges of amorphous solids, is known as jumping conduction. At low temperatures, the hopping regime", " is prevalent. In this regime, the electrons can move only through phonon-assisted hopping. At relatively higher temperatures in the hopping regime of conduction, the largest tunnel contribution arises from jumps to unoccupied levels of nearest-neighbor centers. At lower temperatures, the number and Localized states at the edges of the valence and conduction bands in amorphous semiconductors. FIGURE 17.25 O E \u00b5 594 CHAPTER 17 Complex Structures EC\u2032 E \u00b5 O EC\u2032 E \u00b5 O FIGURE 17.26 The overlap of the tails of the localized states in amorphous tetrahedral semiconductors. FIGURE 17.27 Impurity band of local defects in the gap or the pseudogap. energy of phonons available for absorption decrease so that tunneling is restricted to seek centers that are not nearest neighbors but lie energetically closer. This type of conduction is known as variable-range hopping. The overlap of tails of the localized states in amorphous tetrahedral semiconductors is shown in Figure 17.26. The regime of variable-range hopping is subdivided into two ranges. At relatively higher temperatures of this regime, variable-range hopping transport is done by tunneling conduction in a band tail with somewhat large activation energy. At the lower temperature range, hopping transport by tunneling conduction near the Fermi level is favored. The local defects, such as dangling bonds, can cause impurity bands to appear in the gap or pseudogap. This is shown in Figure 17.27. The extended states above the mobility gap are important in deciding the optical properties. However, at low temperatures, the transport properties of amorphous semiconductors are mainly determined by the states of the local defects rather than the localized states of the tail states. PROBLEMS 17.1. Show that in a region where is everywhere nonzero, the application of the Stokes theorem to the curl of Eq. (17.14) leads to the conclusion that the integral of vs around any closed curve is zero. \u03c70j j 17.2. Show that when we integrate Eq. (17.14) around a circuit that encloses the one-dimensional region, the fact that the phase of the wave function \u03c70 must be single-valued modulo 2\u03c0 leads to the Onsager\u2013Feynman quantization condition vs. dl = nh/m: I (1) Problems 595 17.3. Show that because, in the range of temperatures above 0.01\u00b0 K, the nuclear spins of the 3He atoms comprising the solid are almost fully disordered, the entropy Ssolid = R ln 2 (per mole). 17.4. The dipolar interaction is calculated by taking a quantum mechanical average of the dipolar Hamiltonian over the order parameter (pair-wave function). It can be shown that the dipolar free energies are given by (1) (2) \u0394FD = 8 < : where T 3 gD\u00f0 \u2212 5 4 T gD\u00f0 5 \u00de 2 1 \u2212 d!. l! \u00de \u00de\u00bd cos \u03b8 + 2 cos2 \u03b8 +, A phase, 3 4 $, B phase, o n 1 \u2212 gD % T Tc 10\u22123ergs/cm3: ) ( Show that to minimize the free energy, dipolar energy is minimized for \u03b8 = cos\u22121 l! and d! must be parallel in the A phase, and the \u00f0 17.5. For a nematic or cholesteric liquid crystal where the molecules are rigid and rodlike in = 104\u00b0 in the B phase: \u2212 1 4\u00de shape, the order parameter can be expressed as a second-rank tensor r S\u03b1\u03b2\u00f0 \u00de = 1 N \u2211 i i i \u00de\u03b1 v\u00f0 v\u00f0 \u00de\u03b2 \u2212 1 3 \u03b4\u03b1\u03b2, (1) & where the sum is over all the N molecules in a macroscopic volume located at r, and the v\u03b1 are the components of v! referred to by a set of laboratory-fixed axes. Show that in the isotropic case, where the molecules have random orientation, S\u03b1\u03b2 = 0. \u2019 17.6. In the order parameter for nonlinear rigid molecules, one can introduce a Cartesian coordinate system x\u2032y\u2032z\u2032 fixed in the molecules. In the case of a uniaxial liquid crystal, the order parameter tensor is defined by r S\u03b1\u2032\u03b2\u2032\u00f0 \u00de = < cos \u03b8\u03b1\u2032 cos \u03b8\u03b2\u2032 \u2212 1 3 \u03b4\u03b1\u2032\u03b2\u2032>, (1) where cos \u03b8\u03b1\u2032 is the angle between the \u03b1\u2032 molecular axis, and the preferred direction or the optic axis. The angle brackets indicate an average over the molecules in a small but macroscopic volume. Show that Eq. (1) in Problem (17.6) is equivalent to", " Eq. (1) in Problem (17.5) in the case of linear molecules or molecules with a well-defined long axis about which they rotate rapidly. 17.7. Show that for a uniaxial liquid crystal, Qxx = Qyy = 2N Sy\u2032y\u2032 + Sz\u2032z\u2032\u00de \u00bd\u00f0 0 \u00de1 + \u03c7\u00f0 Sz\u2032z\u2032 + Sx\u2032x\u2032\u00de \u00f0 0 \u00de2 + \u03c7\u00f0 Sx\u2032x\u2032 + Sy\u2032y\u2032\u00de \u00f0 0 \u03c7\u00f0 \u00de3 $ and Qzz = N \u00f0 0 \u00de1 + Sy\u2032y\u2032\u03c7\u00f0 Sx\u2032x\u2032\u03c7\u00f0 0 0 \u00de2 + Sz\u2032z\u2032\u03c7\u00f0 \u00de3 \u00de, (1) (2) where N is the number of molecules per unit volume where the symbols are defined in the text. 596 CHAPTER 17 Complex Structures 17.8. In a quasicrystal, the distribution function of lattice points \u03c1 lattice is given by x \u00f0 \u00de of a two-dimensional square \u03c1 x \u00de \u00f0 = \u03c1 \u00f0 x\u2225, x\u22a5 = \u2211 jl \u03b4 \u00f0 \u00de x\u2225 \u2212 j cos \u03b8 \u2212 l sin \u03b8 \u00d7 \u03b4 x\u22a5 + j sin \u03b8 \u2212 l cos \u03b8 \u00f0 \u00de : \u00de Show that the distribution function on the projected lattice on E\u2225 is Here, x\u2225 \u03c10\u00f0 \u00de = \u221e Z \u2212\u221e dx\u22a5\u03c1 x\u2225, x\u22a5 W x\u22a5 \u00f0 : \u00de \u00de \u00f0 W x\u22a5 \u00f0 \u00de = 1, ( if 2 j x\u22a5 j < \u00f0 cos \u03b8 + sin \u03b8 \u00de = 0, otherwise: ( 17.9. Show that if one writes \u03c6 as \u03c6 = \u03c81 + \u03b1 + i\u03b2 \u03c82= \u00bd \u00de \u00f0 1 + \u03b12 + \u03b22 1 2, $ where x and y are real, the average value of the momentum ^p is (1) (2) (3) (1) <^p> = \u2212i\u0127 \u03c6\u2019 Z d\u03c6/dx \u00f0 \u00de dx = \u0127 \u03b2 \u03c012 \u2212 \u03c021\u00de \u2212 i \u00bd \u00f0 where \u03c011 + \u03b12 + \u03b22 \u00de 1 + \u03b12 + \u03b22 \u00f0 \u03c022 + \u03b1 \u03c012 + \u03c021\u00de$, \u00f0 (2) b \u03c0ij = \u03c8id\u03c8j, i, j = 1, 2 : Z a (3) 17.10. Show from Eqs. (17.44) and (17.47) that the two functions that make <^p> an extremum are and \u03c6 = \u03c81 + i\u03c82\u00de \u00f0 2p / \ufb03\ufb03\ufb03 \u03c6\u2019 = \u03c81\u2212 i\u03c82\u00de \u00f0 2p : / \ufb03\ufb03\ufb03 (1) (2) References 597 References 1. Anderson PW, Morel P. Generalized Bardeen-Cooper-Schrieffer States and the proposed low-temperature Phase of Liquid He3. Phys Rev 1961;123:1911. 2. Anderson PW, Brinkman WF. Anisotropic Superfluidity in He3: A Possible Interpretation of its Stability as a Spin-Fluctuation Effect. Phys Rev Lett 1973;30:1108. 3. Ashcroft NW, Mermin ND. Solid state physics. New York: Brooks/Cole; 1976. 4. Balian R, Werthamer NR. Phys. Rev. 1963;131:1553. Superconductivity with Pairs in a Relative p Wave. 5. Barker JA, Henderson DH. What is \u201cliquid\u201d? Understanding the states of matter. Rev Mod Phys 1976;48:587. 6. Chandrasekhar S. Liquid Crystals. Cambridge: Cambridge University Press; 1992. 7. de Gennes P-G, Prost J. The physics of liquid crystals. Oxford: Clarendon Press; 1993. 8. Economou EN, Cohen MH. Energy Bands in One-Dimensional Aperiodic Potential. Phys Rev Lett 1970;24:218. 9. Kramer P. Non-periodic Central space filling with icosahedral symmetry using copies of seven elementary cells. Acta Crystallogr 1982;A38:257. 10. Landau LD. J Phys. USSR 1941;5:71. 11. Landau LD. JETP 1957;3:920. 12. Landau LD. JET", "P 1957;5:101. 13. Landau LD, Lifshitz EM. Statistical physics, part 1. Oxford: Pergamon Press; 1980. 14. Lee DM. The Extraordinary phases of liquid He3. Rev Mod Phys 1997;69:645. 15. London F. On the Bose-Einstein Condensation. Phys Rev 1938;54:947. 16. Mackay AL. Cryatallography and the Penrose patterns. Physica 1982;114A:609. 17. Marder MP. Condensed matter physics. New York: John Wiley & Sons; 2000. 18. Misra PK, Fujiwara T, Ishii Y. Quasicrystals. Amsterdam: Elsevier; 2008. 19. Mott NF, Davis EA. Electronic properties of noncrystalline materials. Oxford: Clarendon Press; 1979. 20. Osheroff DD, Richardson RC, Lee DM. Evidence for a New Phase of Solid He3. Phys Rev Lett 1972;28:885. 21. Osheroff DD, Gully WJ, Richardson RC, Lee DM. New magnetic Phenomena in Liquid He3 below 300 K. Phys Rev Lett 1972;29:920. 22. Paulson DN, Kojima H, Wheatley JC. Preferred Effect of a Magnetic Field in the Phase Diagram of Super- fluid 3He. Phys Rev Lett 1974;32:1098. 23. Penrose R. Bull Inst Math Appl 1974;10:266. 24. Shechtman D, Blech L, Gratias D, Cahn JW. Metallic Phase with Long Range Orientational Order and no Translational Symmetry. Phys Rev Lett 1984;53:1951. 25. Stephen MA, Straley JP. Physics of Liquid Crystals. Rev Mod Phys 1974;46:617. 26. Webb RA, Greytak JT, Johnson RT, Wheatley. Observation of a second-order Phase Transition and its Associated P-T phase Diagram in liquid He3. Phys Rev Lett 1973;30:210. Novel Materials CHAPTER 18 CHAPTER OUTLINE 18.1 Graphene............................................................................... 600 18.1.1 Introduction....................................................................... 600 18.1.2 Graphene Lattice.................................................................. 601 18.1.3 Tight-Binding Approximation....................................................... 602 18.1.4 Dirac Fermions.................................................................... 606 18.1.5 Comprehensive View of Graphene................................................... 608 18.2 Fullerenes.............................................................................. 608 18.2.1 Introduction...........................", "............................................ 608 18.2.2 Discovery of C60................................................................... 609 18.3 Fullerenes and Tubules................................................................... 613 18.3.1 Introduction....................................................................... 613 18.3.2 Carbon Nanotubeles............................................................... 614 18.3.3 Three Types of Carbon Nanotubes.................................................. 614 18.3.4 Symmetry Properties of Carbon Nanotubes........................................... 616 18.3.5 Band Structure of a Fullerene Nanotube............................................. 617 18.4 Polymers............................................................................... 617 18.4.1 Introduction....................................................................... 617 18.4.2 Saturated and Conjugated Polymers................................................. 618 18.4.3 Transparent Metallic Polymers...................................................... 621 18.4.4 Electronic Polymers................................................................ 621 18.5 Solitons in Conducting Polymers.......................................................... 622 18.5.1 Introduction...................................", ".................................... 622 18.5.2 Electronic Structure............................................................... 623 18.5.3 Tight-Binding Model............................................................... 623 18.5.4 Soliton Excitations................................................................. 624 18.5.5 Solitons, Polarons, and Polaron Excitations.......................................... 626 18.6.6 Polarons and Bipolarons........................................................... 626 18.6 Photoinduced Electron Transfer........................................................... 627 Problems.................................................................................... 627 References.................................................................................. 630 Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00018-9 \u00a9 2012 by Elsevier Inc. All rights reserved. 599 600 CHAPTER 18 Novel Materials 18.1 GRAPHENE 18.1.1 Introduction Carbon has four perfect crystalline forms: graphite, diamond, \u201cBuckminsterfullerene\u201d and a fullerene nanotube. In addition, graphene is a one-atom-thick allotrope of carbon, which is a honeycomb lattice of carbon atoms. Graphene also has two-dimensional Dirac-like excitations. We discussed the properties of graphene as well as its possible applications in electronics in Sections 10.7 and 10.8. In the following sections, we will discuss graphene as a building block for all novel materials of carbon as well as derive the theory of Dirac fermions discussed in Section 10.7. One can view graphite as a stack of graphene layers, and carbon nanotubes can be considered as rolled cylinders of graphene. \u201cBuckminsterfullerene\u201d can be viewed as molecules obtained by introducing pentagons on the hexagonal lattice of wrapped graphene. These are shown in Figure 18.1. Diamond is not shown in the diagram because it is primarily used in making jewelry due to its beauty and elegance, and it does not have any major applications in materials science, presumably because of its cost. In addition, each atom in diamond is surrounded in all three directions in space by a full coordination. Because all directions are taken up, it would be nearly impossible for an atom in a diamond lattice to have any bonding with any other atom in the outside 3D space. C60\u00de \u00f0 Graphene is a two-dimensional (2D) allotrope of carbon that can be imagined to be", " benzene rings stripped out from the hydrogen atoms. Fullerenes are molecules where carbon atoms are arranged spherically and are zero-dimensional (0D) objects that have discrete energy states. Fullerenes can be thought of as wrapped-up graphene because they are obtained from graphene with the introduction of pentagons, which create positive curvature defects. Carbon nanotubes, which have only hexagons and can be thought of as one-dimensional (1D) objects, are obtained by rolling graphene along a definite direction and reconnecting the carbon bonds. Graphite, which is a threedimensional (3D) allotrope of carbon, is made out of stacks of graphene layers that are weakly coupled by van der Waals forces. Two-dimensional materials like graphene were presumed not to exist until 2004, when it was obtained in liquid suspension.2 Graphene could also be obtained on top of noncrystalline substrates3\u20125 and was eventually spotted in optical microscopes due to the subtle optical effects made by it on top of an SiO2 substrate. Graphene exhibits high crystal quality, in which charge carriers can travel thousands of interatomic distances without scattering. FIGURE 18.1 Clockwise: graphene (2D), graphite (3D), \u201cBuckminsterfullerene\u201d (0D), and carbon nanotubes (1D). Reproduced from Castro Neto et al.1 with the permission of the American Physical Society. The Coulomb interactions are considerably enhanced in small geometries such as graphene quantum dots that lead to Coulomb blockade effects. The transport properties of graphene lead to a variety of applications, which range from single molecule detection to spin injection. Because 18.1 Graphene 601 graphene has unusual structural and electronic flexibility, it can be tailored: deposition of metal atoms or molecules on top; incorporation of boron and/or nitrogen in its structure; and using different substrates that modify the electronic structure. The control of graphene properties can be extended in new directions that would allow for the creation of graphene-based systems with magnetic and superconducting properties. 18.1.2 Graphene Lattice Carbon has four valence electrons, three of which form tight bonds with neighboring atoms in the plane. Their wave functions are of the form 1 3p 2s + \u03c8e\u00f0 \u00de \u00de # wave function for carbon, and \u03c8 e\u00f0 \" \ufb03\ufb03\ufb03 2p \u03c8e\u00f0 \ufb03\ufb03\ufb03 \u03c4i2p, i = 1, 2, 3, \u00de \u00f0 (18.1) \u00de \u00de 2s is the 2s \u00f0 wave functions of which where \u03c8 e\u00f0 the axes are in the directions \u03c4i joining the graphite atom to its three neighbors in the plane. The fourth electron is in the 2pz state. Its nodal plane is the lattice plane and its axis of symmetry perpendicular to it. Because the three electrons forming coplanar bonds do not play any part in the conductivity, graphene can be considered to have one conduction electron in the 2pz state. are the \u03c4i2p 2p \u00f0 \u00de \u00de The unit cell of the hexagonal layer, designated as PQRS in Figure 18.2, contains two carbon atoms A and B. The distance AB \u2248 a = 1:42 A\u00b0 : The fundamental lattice displacements are a1 = AA\u2032 3p \u00d7 1:42 A\u00b0 = 2:46 A\u00b0 : The reciprocal lattice vectors and a2 = AA\u2033, and their magnitude is a1 = have magnitude 8\u03c0/3a and are in the directions AB and AS, respectively. Hence, the first Brillouin zone is a hexagon (see Figure 18.3) of which the sides are at a distance 4\u03c0/3a from its center. The density of electron states in k space is 2A, where A is the area of the crystal. The zone has exactly one electron per atom. Therefore, the first Brillouin zone of graphene has 2N electron states, and the second Brillouin zone is empty. As we discussed in Sections 10.7 and 10.8, it becomes a semiconductor at finite temperatures. \ufb03\ufb03\ufb03 P A Q \u2192 \u03b4 3 \u2192 a1 B A\u2032 \u2192 \u03b4 1 \u2192 \u03b4 2 R S \u2192 a2 A\u2032\u2032 FIGURE 18.2 Honeycomb lattice structure of graphene, made out of two interpenetrating triangular lattices. The lattice unit vectors a!1 and a!2 and the nearest-neighbor vectors \u03b4!1, \u03b4!2, and \u03b4!3. AB = a. ky \u2192 b 1 \u0393 K M K\u2032 \u2192 b 2 kx N FIGURE 18.3 First Brillouin zone of the honeycomb lattice and", " the Dirac points K and K \u2032 at the corners. 602 CHAPTER 18 Novel Materials The lattice vectors of graphene can be written as (Figure 18.2) a1 = 3a 2 1 3p ^x + ^y ; a2 = \" # 3a 2 ^x \u2212 1 3p \" # ^y, (18.2) where the carbon\u2013carbon distance is a \u2248 1:42 A\u00b0. The reciprocal lattice vectors (shown in Figure 18.3) are given by (Problem 18.1) \ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03 b1 = 2\u03c0 3a ^x + 3p ^y ; b2 = ^x \u2212 3p ^y : (18.3) 2\u03c0 3a \" The positions of the two Dirac points K and K\u2032, located at the corners of the Brillouin zone (of which the significance is to be explained later), are (Problem 18.2) \" # \ufb03\ufb03\ufb03 # \ufb03\ufb03\ufb03 K = 2\u03c0 3a ^x + $ 1 3p ; K\u2032 = ^y % 2\u03c0 3a ^x \u2212 $ 1 3p ^y : % The three nearest-neighbor vectors in real space are given by \ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03 \u03b4!1 = a 2 ^x + 3p ^y ; \u03b4!2 = a 2 ^x \u2212 3p ^y ; \u03b4!3 = \u2212a^x: # The six second-nearest neighbors are located at \u03b4!\u20321 = \u00b1a1; \u03b4!\u20322 = \u00b1a2, and \u03b43\u2032 = \u00b1 \u00f0 # \" \" \ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03 (18.4) (18.5). a2 \u2212 a1\u00de 18.1.3 Tight-Binding Approximation Wallace (Ref. 19) developed a \u201ctight-binding\u201d method for the band theory of graphite. Because the compared with the hexagonal spacing of spacing of the lattice planes of graphite is large 1:42 A\u00b0 the layer, he neglected, as a first approximation, the interactions of the planes and \u00de assumed that conduction takes place in the layers. This is precisely graphene, which at that time was merely a concept. We note that some the notations have different values in that paper, presumably because it was published in 1947, but these have been modernized in the present derivation. 3:37 A\u00b0 \u00de \u00f0 \u00f0 If \u03c7 r is the normalized orbital 2pz wave function for an isolated atom, the wave function in the tight-binding approximation has the form \u00f0 \u00de where and \u03c8 = \u03d51 + \u03bb\u03d52, \u03d51 = 1 Np \u2211 A eik.rA \u03c7 r \u2212 rA\u00de \u00f0 \ufb03\ufb03\ufb03\ufb03 1 Np \u2211 B \u03d52 = eik.rB \u03c7 r \u2212 rB\u00de : \u00f0 (18.6) (18.7) (18.8) Here, the first sum is taken over A and all the lattice points generated from it by primitive lattice translations, and the second sum is similarly over the points generated from B (Figure 18.2). Neglecting the overlap integrals, \ufb03\ufb03\ufb03\ufb03 \u03c7 r \u2212 rA\u00de \u00f0 \u03c7 r \u2212 rB\u00de \u00f0 dr = 0, Z (18.9) 18.1 Graphene 603 H\u03c8 = E\u03c8, H11 + \u03bbH12 = E H12 + \u03bbH22 = \u03bbE: (18.10) (18.11) (18.12) and substituting in Eq. (18.6), we obtain (Problem 18.3) and Here, and H11 = H22 = \u03d51 # H\u03d52d\u03bd; H12 = H21# = \u03d51 # H\u03d52d\u03bd (18.13) Z Z Z \u03d51 # \u03d51d\u03bd = \u03d52 # \u03d52d\u03bd = 1: Z Eliminating \u03bb from Eqs. (18.11) and (18.12), we obtain the secular equation From Eq. (18.15), it is easy to show that H11 \u2212 E H21 H12 H22 \u2212 E & & & = 0: & & & E = 1 2 ( H11 + H22 \u00b1 \u00f0 \" H11 \u2212 H22\u00de 2 + 4 2 H12j j 1 2 : ) # Because H11 = H", "22, Eq. (18.16) can be rewritten in the alternate form E\u00b1 = H11 \u00b1 H12j : j (18.14) (18.15) (18.16) (18.17) The positive sign in Eq. (18.17) will apply to the outside of the hexagonal zone and the negative sign to the inside. The discontinuity of energy across the zone boundary is From Eqs. (18.7), (18.13), and (18.17), we obtain \u0394E = 2 : H12j j H11 = 1 N \u2211 A, A\u2032 e\u2212ik. \u00f0 rA\u2212rA\u2032\u00de Z \u03c7 # r \u2212 rA\u00de H\u03c7 r \u2212 rA\u2032\u00de \u00f0 \u00f0 d\u03bd: Keeping only the nearest-neighbor integrals among the atoms A and writing and E0 = Z \u03c7 # H\u03c7 r \u00de \u00f0 r \u00de \u00f0 d\u03bd \u03b30\u2032 = \u2212 Z \u03c7 # r \u2212 \u03c1!\u2032 H\u03c7 r \u00de \u00f0 \u00de \u00f0 d\u03bd, (18.18) (18.19) (18.20) (18.21) 604 CHAPTER 18 Novel Materials where \u03c1!\u2032 = a1\u00f0 (Problem 18.4) say \u00de is a vector joining the nearest neighbor among atoms A, we can show that H11 = E0 \u2212 2\u03b30\u2032 cos \u2019 + 2cos 3p kya \" \ufb03\ufb03\ufb03 # 3 2 \" kxa cos # 3p 2 \" \ufb03\ufb03\ufb03 kya : ( # Writing where H0 is the Hamiltonian of an isolated carbon atom, and using H \u2212 H0 = V \u2212 U < 0, H = H0 + H \u2212 H0\u00de, \u00f0 (18.22) (18.23) (18.24) where U is the potential field of an isolated atom and V is the periodic potential of the lattice because H0\u03c7 = E\u03c7, (18.25) (E is the energy of an electron in the 2pz state in carbon), from Eqs. (18.20), (18.21), (18.24), and (18.25), we obtain and E0 = E \u2212 Z \u03c7 # r \u00f0 \u00de\u00f0 U \u2212 V r \u03c7 \u00de \u00f0 \u00de d\u03bd \u03b30\u2032 = Z \u03c7 # \u00f0 r \u2212 \u03c1!\u2032 \u00de\u00f0 U \u2212 V \u03c7 r \u00f0 \u00de \u00de d\u03bd > 0: (18.26) (18.27) Similarly, we obtain the expression for H12, H12 = 1 N \u2211 A,B e\u2212ik. rA\u2212rB \u00f0 \u00de \u03c7 # r \u2212 rA\u00de \u00f0 H\u03c7 r \u2212 rB\u00de \u00f0 : (18.28) Z Considering only the nearest-neighbor interactions in the lattice (between atoms of type A and type B and vice versa), we write (in analogy with Eq. 18.27) where \u03b30 = Z \u03c7 # r \u2212 \u03c1!\u00de\u00f0 \u00f0 U \u2212 V \u03c7 r \u00de \u00f0 \u00de d\u03bd > 0, \u03c1! = AB: It can be shown that (Problem 18.5) H12 = \u2212\u03b30 e\u2212ikxa + 2 cos h 3p 2 \" \ufb03\ufb03\ufb03 kya # i e 3 2 \" kxa # i and H12j j 2 = \u03b32 0 1 + 4 cos2 \u2019 3p 2 \" \ufb03\ufb03\ufb03 kya + 4 cos # 3 2 \" kxa cos # 3p 2 \" \ufb03\ufb03\ufb03 kya : ( # (18.29) (18.30) (18.31) (18.32) 18.1 Graphene 605 From Eqs. (18.22) and (18.32), we can write E\u00b1\u00f0 k \u00de = H11 \u00b1 H12 = E0 \u2212 \u03b30\u2032 f k \u00b1 \u03b30\u00bd \u00de \u00f0 3 + f k \u00f0 1/2, \u00de% (18.33a) where f k \u00de \u00f0 = 2 cos \u2019 \u00f0 3p kya \u00de + 2 cos 3 2 \ufb03\ufb03\ufb03 \" # 3p 2 \" \ufb03\ufb03\ufb03 ( # kxa cos kya : (18.33b) The energies at the various points in the Brillouin zone can be written as \u0393: E = E0 \u2212 3\u03b30 \u2212 6\u03b30\u2032, N: E = E0 +", " 3\u03b30 \u2212 6\u03b30\u2032, K: E = E0 + 3\u03b30\u2032, (18.34a) M inside \u00f0 outside \u00f0 : E = E0 \u2212 \u03b30 + 2\u03b30\u2032, \u00de : E = E0 + \u03b30 + 2\u03b30\u2032: \u00de M Across the boundary at any point over a side of the zone (Figure 18.3), there is a discontinuity of energy of amount 3p 2 \" \ufb03\ufb03\ufb03 which is a maximum at the center and decreases to zero at the corners. The degeneracy at K and similar points (called Dirac points) and the zero-energy gap at these points are consequences of the symmetry of the lattice and are independent of any approximation. 2\u03b30 2 cos \u2019 (18.34b) \u2212 1 kya # (, The energy contours are given by E = E0 \u2212 3\u03b30 \u2212 6\u03b30\u2032 + 3 4 \u00f0 \u03b30 + 6\u03b30\u2032 \u00de\u00f0 x + k2 k2 y \u00de a2: The curves of constant energy are shown in Figure 18.4. It may be noted that near the corners K or K\u2032 (Dirac points), (18.35) (18.36) a \u2212 j 9 4 \u03b30\u2032 k \u2212 K j j 2a2: K K\u2032 E \u2212 EKj \u03b30j The surfaces of constant energy are circular. If one neglects \u03b30\u2032 relative to \u03b3, Eq. (18.36) can be rewritten near the corners of the zone, = 3\u03b30\u2032 \u00b1 j 3 2 k \u2212 K E \u2212 EKj j \u2248 \u2248 3 2 3 2 \u03b30a k \u2212 K j j + O \u00bd\u00f0 q K2 \u00de% ) \u03b30a q j j + O q \u00bd\u00f0 K2, \u00de% (18.37) which can be rewritten in the alternate form ) q E\u00b1\u00f0 \u00de \u2248 vFj q j + O q \u00bd\u00f0 K, 2 \u00de % (18.38) ) FIGURE 18.4 Curves of constant energy. Reproduced from Wallace19 with the permission of the American Physical Society. 606 CHAPTER 18 Novel Materials where q is the momentum (in units \u0127 = 1) measured relatively to the Dirac points, and vF is the Fermi velocity, The numerical value of vF \u2248 1 \u00d7 106 m/s. From Eqs. (18.36) and (18.39), we obtain vF = 3\u03b30a/2: E\u00b1\u00f0 q \u00de q = 3\u03b30\u2032 \u00b1 vFj j + O \u00bd\u00f0 q K : 2 \u00de % (18.39) (18.40) Conduction in the graphene layer will take place through the electrons excited into the upper band and through the equal number of positive holes created in the lower band, as shown in Figure 18.5. over the whole range in which the Fermi E \u2212 EKj is different from its value at absolute zero. One can write \u03be = EK and express For moderate temperatures, N is even in \u2208 = E \u00de \u00f0 j ) distribution \u2208/kBT + 1 : \u00de (18.41) \u00f0 ) It is interesting to note that the original tight-binding method, used by Wallace19 in 1947 as a first approximation for the calculation of band structure of a single layer of graphite, is now being widely used to study the energy bands of graphene. The tight-binding Hamiltonian for electrons in graphene can be written in the second-quantiza- tion form1 (in units such that \u0127 = 1), ^H = \u2212t \u2211 <i,j>,\u03c3\u00f0 ^a\u2020 i,\u03c3 ^bj,\u03c3 + H:c: \u00de \u2212t\u2032 \u2211 <<i,j>>,\u03c3\u00f0 ^a\u2020 i,\u03c3^aj,\u03c3 + ^b \u2020 i,\u03c3 ^bj,\u03c3 + H:C:, \u00de (18.42) i,\u03c3\u00f0 where ^a\u2020 ^ai,\u03c3\u00de lattice A, and ^b Here, t are the creation and annihilation operators with spin \u03c3 \u2020 bj,\u03c3\u00de j,\u03c3\u00f0 \u2248 2:7 eV \u00de \u00f0 \u03c3 = \u2191, \u2193 \u00de are the corresponding operators on site Rj on sublattice B. is the nearest-neighbor hopping energy (between A and B), and t\u2032 = \u22120:2 t is the next nearest-neighbor hopping energy (between two A\u2019s or two B\u2019s). We note that t = \u03b30 and t\u2032 = \u03b30\u2032 in Wallace\u2019s theory. on site Ri on sub- \u00f0", " The electronic dispersion in the honeycomb lattice is shown in Figure 18.6, for finite values of t = 2:7 eV and t\u2032 = \u22120:2 t: We also note the most striking difference between the results of Eq. (18.39) = \u01272q2/2m, where m is the electron mass. In Eq. (18.39), the Fermi and the usual case in which \u2208 2E/mp velocity vF does not depend on the energy and momentum while in the usual case, v = \u0127k/m =, and hence the velocity changes substantially with energy. We also note that the presence of the second\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03\ufb03 order terms (arising due to t\u2032 in Eq. 18.40) shifts in energy the position of the Dirac point and breaks the electron-hole symmetry. q \u00de N(E) \u00f0 EK E FIGURE 18.5 The form of the electronic energy states, N near EK. E, \u00de \u00f0 18.1.4 Dirac Fermions Graphene\u2019s charge carriers have a particularly unique nature. Its charge carriers mimic relativistic particles and are described starting with the Dirac equation rather than the Schrodinger equation. The interaction of the electrons with the graphene\u2019s honeycomb lattice gives rise to new 18.1 Graphene 607 4 2 0 Ek \u22122 \u22124 \u22122 0 kx 4 2 ky 0 \u22122 2 \u22124 4 FIGURE 18.6 Left: Energy spectrum (in units of t) for t = 2:7 eV and t \u2032 = \u22120:2 t : Right: Close-up of the energy bands near to one of the Dirac points. Reproduced from Castro Neto et al.1 with the permission of the American Physical Society. quasiparticles, which, at low energies E, are accurately described by the (2 + 1)-dimensional Dirac equation with an effective speed of light vF \u2248 106 m\u22121s\u22121. These quasiparticles are called massless Dirac fermions. They can be viewed as electrons that have lost their rest mass me or as neutrinos that have acquired the electron charge e. The reason the quasiparticles are known as Dirac fermions is as follows. The Dirac equation for an electron in a periodic potential V can be written as where ^H\u03d5i = \u00f0 c \u03b1!. p! + \u03b2mec2 + IV \u03d5i = \u03b5i\u03d5i, \u00de \u03b1! = \u03c3! 0 \u03c318.43) (18.44) \u03c3! is the Pauli spin matrix vector, E is a 2 \u00d7 2 unit matrix, p! is the momentum operator, me is the rest mass of the electron, and \u03d5i is a four-component Bloch function with an energy \u03b5i: The suffix i signifies a set of the wave vector, band index, and spin direction and is limited to positive energy states. Graphene is a zero-gap semiconductor, in which low-E quasiparticles within each valley can be described by the Dirac-like Hamiltonian ^H = \u0127vF $ 0 kx + iky kx \u2212 iky 0 % = \u0127vF \u03c3!. k!: (18.45) Eq. (18.43) can be approximated by Eq. (18.45) when the k-independent Fermi velocity vF plays the role of the velocity of light c, p! = \u0127k!, and because the electrons are fermions, they are called Dirac fermions. The honeycomb lattice is made up of two equivalent carbon sublattices A and B, and the cosine-like energy bands associated with the sublattices intersect at zero E near the edges of the Brillouin zone, giving rise to conical sections of the energy spectrum. The electronic states at the intersection of the bands are composed of states belonging to the different sublattices, and their relative 608 CHAPTER 18 Novel Materials A B \u2192 a2 \u2192 a1 B1 \u2192 a2 B2 \u2192 a1 A1 B2 A2 \u2227 x A1 \u2227 y B1 (a) (b) (c) FIGURE 18.7 B3 \u2227 y B1 A2 B2 \u2227 x A1 A1 (d) ky kx \u2192 b2 \u2192 b1 K\u2032 T\u2032 M \u2211 K T \u0393 (e) A top view of (a) unit cell of monolayer graphene, showing the inequivalent atoms A and B and unit vectors a1 and", " a2; (b) real-space bilayer graphene in which the light-/dark-gray dots and black circles/black dots represent the carbon atoms in the upper and lower layers; (c) the unit cell and the ^x and ^y unit vectors of bilayer graphene; (d) the same as (c) for trilayer graphene; and (e) the reciprocal space unit cell showing the first Brillouin zone with its high symmetry points. Reproduced from Malard et al.12 with the permission of Elsevier. contributions are taken into account by using two-component wave functions (spinors). The index to indicate sublattices A and B is known as pseudospin \u03c3! because it is similar to the spin index (up and down) in quantum electrodynamics (QED). The real spin of the electrons must be described by additional terms in the Hamiltonian. Because the QED-specific phenomena are often inversely proportional to c and because c/vF \u2248 300, the pseudospin effects usually dominate over those due to the real spin. One can introduce the concept of chirality, which is formally a projection of \u03c3! on the direction of motion k!, and is positive (negative) for electrons (holes). Chirality in graphene signifies the fact that k! electrons and \u2212 k! hole states originate from the same sublattice. The concepts of chirality and pseudospin are important because they are conserved quantities. 18.1.5 Comprehensive View of Graphene A comprehensive view of the unit cell of monolayer, bilayer, and trilayer graphene and the first Brillouin zone with its high symmetry points are shown in Figure 18.7. 18.2 FULLERENES 18.2.1 Introduction If one forms a vapor of carbon atoms and lets them condense slowly while keeping the temperature high, as the intermediate species grow, there is a path where the bulk of all reactive kinetics follows that make spheroidal fullerenes. There are two types of fullerenes that are famous for different reasons. The \u201cBuckminsterfullerene\u201d is the most symmetric of all possible molecules. In addition, it is possible by adding a few percent of other atoms (nickel and cobalt) to trick the carbon into making tubes. The (10,10) fullerene nanotube is the most famous nanotube. The propensity for bonding that causes C60 to be the end point of 30\u221240% of all the reactive kinetics leads to the (10,10) C60\u00de \u00f0 18.2 Fullerenes 609 nanotube. The metal atoms (nickel and cobalt) prevent the addition of the seventh, eighth, and ninth pentagons, and ultimately, the growing tubelet can anneal to its most energetically favored form. The idea that C60 would form a stable molecule originated from Euler\u2019s rule stating that a solid figure with any even number n of 24 or more vertices could be constructed with 12 pentagons and /2 hexagons. The spheroidal carbon\u2013cage carbon molecules consisting only of pentagons n \u2212 20 \u00f0 and hexagons were given the generic name \u201cfullerenes.\u201d \u00de 18.2.2 Discovery of C60 The truncated icosahedron form of C60 is shown in Figure 18.8. It was discovered by Kroto et al.9 by using a supersonic laser-vaporization nozzle source, as shown in Figure 18.9a. C60 is chemically a very stable structure. Cluster \u201ccooking\u201d reactions in the \u201cintegrating cup\u201d were responsible for the C60 cluster\u2019s becoming over 50 times more intense than any other cluster in the nearby size range. The up-clustering reactions with small carbon chains and rings reacted away nearly all clusters except for C60, which survived because of its perfect symmetry. C60 does not have any dangling bonds because the valences of every carbon atom are satisfied. There is no specific point of chemical attack because every atom is equivalent by symmetry. While curving the intrinsically planar system of double bonds into a spherical shape, strain is introduced. However, this strain is uniformly and symmetrically distributed over the molecule. No other structure has this high degree of symmetry, and hence, the experimental observation that carbon-vapor condensation conditions could be found where the intensity of the mass spectrum peak of the C60 in the carbon cluster beam was many times the intensity of any of its near neighbors in mass is shown in Figure 18.9b. Truncated icosahedron C60, popularly known as \u201cBuckminsterfullerene.\u201d Reproduced from Curl 2 with the permission of the American FIGURE 18.8 Physical Society. Vaporization laser 10", " atm helium Integration cup Rotating graphite disk FIGURE 18.9a Schematic cross-sectional drawing of the supersonic laser-vaporization nozzle source used in the discovery of fullerenes. Reproduced from Smalley 16 with the permission of the American Physical Society. 610 CHAPTER 18 Novel Materials To confirm the existence of C60, Kroto and coworkers9 made two pioneering NMR experiments. The first experiment was done on a solution of C60 with benzene, which yielded a very strong resonance line at 128 ppm (for benzene) and a very tiny NMR trace in which C60 resonance was identified at 143 ppm. However, a second experiment (Ref. 18) in which C-NMR spectrum obtained from chromatographically purified samples of soluble material extracted from arc-processed graphite, yielded a spectrum of purified C60, in which a strong resonance was obtained at 143 ppm. This result is shown in Figure 18.10. 44 52 60 68 76 84 Carbon atoms per cluster FIGURE 18.9b Intensity of the mass spectrum peak of the C60 in the carbon cluster beam relative to its neighbors in mass. Reproduced from Curl 2 with the permission of the American Physical Society. (a) (b) (c) C60 C60 + C70 C60 C70 d c b a e ppm 150 140 130 FIGURE 18.10 C-NMR spectrum of (a) purified C60 (143 ppm); (b) mixed sample; (c) purified C70 (five lines). Reproduced from Kroto8 with the permission of the American Physical Society. 18.2 Fullerenes 611 The fullerenes have a wide variety of technological applications. An example of a fullerene-based n-channel FET is shown in Figure 18.11. A highly doped n-type silicon wafer takes the place of the gate metal, a \u223c30\u2212300 nm thick layer of SiO2 serves as the oxide, and the fullerene film serves as the semiconductor. When an appropriate positive gate voltage VG is applied, the drain current ID increases, which indicates that a conduction channel is formed near the fullerene-insulator interface. Another application of fullerene is in C60 photolithography. The sequence of steps (deposition, exposure, development, and pattern transfer) used in photolithography, in which C60 acts as a negative photoresist, is shown in Figure 18.12. One of the many important potential applications of fullerenes is the nature of the fullerenes and metallic and semiconductor substrates. Direct rectification between solid C60 and p-type crystalline Si has been shown in Nb/Co60/p-Si and Ti/Co60/p-Si heterojunctions, which are strongly rectifying. Because the potential barriers at the Nb-C60 and Ti-C60 interfaces are close to (VD \u2265 0) D VD ID N + ID VD D (drain) Au Cr S N + G VG SiO2 P-Si B (a) C60 SiO2 n+Si VG G (gate) (b) S (source) Au Cr FIGURE 18.11 (a) The terminal designations and blasting conditions for Si-based MOSFETs. G, D, B, and S, respectively, denote the ground, drain, base, and source. (b) The corresponding structure for the fullerene C60 device. Reproduced from Dresselhaus et al.3 with the permission of Elsevier. 612 CHAPTER 18 Novel Materials C60 Photolithography Si water C60 (a) C60 deposition by thermal sublimation Mask (b) UV exposure Organic solvents Thermal sublimation Photon irradiation (c) Negative developing (d) Postprocess etching (e) Removal of C60 in solvents FIGURE 18.12 Sequence of steps of C60 photolithography. Reproduced from Dresselhaus et al.3 with the permission of Elsevier. 18.3 Fullerenes and Tubules 613 Vbias Nb or Ti C60 p-Si Al ohmic contact FIGURE 18.13 Schematic cross-section of an Nb/C60/p-Si structure used as a heterojunction diode. Reproduced from Dresselhaus et al.3 with the permission of Elsevier. zero, it is the C60/p-Si interface that is responsible for the strong rectifying properties of the heterostructure. A schematic cross-section of the Nb/C60/p-Si interface is shown in Figure 18.13. 18.3 FULLERENES AND TUBULES 18.3.1 Introduction The fullerene nanotube (10,10) mentioned in the introduction of the previous section, with one end open, is shown in", " Fig. 18.14. The (10,10) tube is formed because the metal atoms frustrate the ability of the open edge to curve in and close. The addition of the seventh, eighth, and ninth pentagons is prevented, and by appropriate choice of temperature and reaction rate, the growing tubelet can anneal to its most energetically favored form. The closed end is a hemifullerene dome (one half of C240), whereas the other end is left open. These ends are directly amenable to the formation of excellent C-O, C-N, or C-C covalent bonds to attach any molecule, enzyme, membrane, or surface to the end of the tube. If two objects A and B are attached to the two ends, they will communicate with each other by metallic transport along the tube. Thus, the (10,10) tube is a metallic wave guide for electrons. FIGURE 18.14 Section of a (10,10) fullerene nanotube with one end open. 614 CHAPTER 18 Novel Materials 18.3.2 Carbon Nanotubeles It is convenient to specify a general carbon nanotubule in terms of the tubule diameter dt and the chiral angle \u03b8, which are shown in Figure 18.15 as the rectangle bounded by the chiral vector OA or Ch. The chiral vector Ch = na1 + ma2 (18.46) is defined on the honeycomb lattice by unit vectors a1 and a2. Ch connects two crystallographically equivalent sites O and A on a two-dimensional graphene sheet where a carbon atom is located at each vertex of the honeycomb structure. Figure 18.15 shows the chiral angle \u03b8 of the nanotube with and the unit vectors a1 and a2 of the hexagonal honeycomb respect to the zigzag direction lattice. The armchair tubule (Figure 18.16a) corresponds to \u03b8 = 30\u00b0 on this construction. An ensemble of chiral vectors can be specified by Eq. (18.46) in terms of pairs of integers, and this ensemble is shown in Figure 18.17. Each pair of integers defines a different set of rolling the graphene sheet to form a carbon nanotube. n, m \u00de \u00f0 \u03b8 = 0 n, m \u00de \u00f0 \u00de \u00f0 Along the zigzag axis \u03b8 = 0\u00b0: Also shown in the figure is the basic translation vector OB = T of the 1D tubule unit cell, and the rotation angle \u03c8 and the translation \u03c4, which constitute the basic uniquely determine the tubular diameter dt and \u03b8. symmetry operation R = \u03c4 \u00de The diagram is constructed for. The integers 4, 2 = n, m \u03c8 \u00f0 \u00de j n\u2032 A Ch x a1 a2 \u03b8 B T \u03c4 O \u03c8 FIGURE 18.15 The 2D graphene sheet is shown along with the vector that specifies the chiral nanotube. Reproduced from Dresselhouse et al.4 with the permission of Elsevier. 18.3.3 Three Types of Carbon Nanotubes When the two ends of the vector Ch are superimposed, the cylinder connecting the two hemispherical caps of Figure 18.16 is formed. The line AB\u2032 (in Figure 18.15) is joined to the parallel line OB, where the lines OB and AB\u2032 are perpendicular to the vectors Ch at each end. There are no distortions of the bond angles in the chiral tubule except the distortions caused by the cylindrical curvature of the tubule. Differences in the tubular diameter dt give rise to the differences in the various properties of carbon denote zigzag nanotubes. The vectors \u00de tubules, and the vectors denote armchair tubules. The larger the value of n, the larger the tubule diameter. The have high symmetry and exhibit a mirror symmetry plane normal to the tubular axis. The other vectors correspond to chiral nanotubes. Because both right- and left-handed chirality are n, 0 \u00f0 n, n n, m \u00de and n, 0 n, n \u00f0 \u00f0 \u00de \u00f0 \u00de \u00f0 \u00de 18.3 Fullerenes and Tubules 615 (a) (b) (c) FIGURE 18.16 Three types of nanotubes obtained by rolling a graphene sheet into a cylinder and capping each end of the cylinder with half of a fullerene molecule; a \u201cfullerene-derived tubule\u201d that is one atomic layer in thickness is formed: (a) \u03b8 = 30\u00b0 (an armchair tubule); (b) \u03b8 = 0\u00b0 (a) zigzag tubule; and (c) chiral tubule. Reproduced from Dresselhaus et al.4 with the permission of Elsevier. possible for chiral nanotubes,", " the chiral tubules are optically active to either right- or left-circularly polarized light propagating along the tubule axis. The tubular diameter dt is given by dt = Ch/\u03c0 = where aC\u2212C is the nearest-neighbor C\u2013C distance, Ch is the length of the chiral vector C!h, and the chiral angle \u03b8 is given by m2 + mn + n2 1/2/\u03c0, \u00de (18.47) 3p aC\u2212C\u00f0 \ufb03\ufb03\ufb03 The three types of carbon nanotubes are shown in Figure 18.16. \ufb03\ufb03\ufb03 \u03b8 = tan\u22121 3p m/ \u00f0 \u00bd m + 2n : \u00de% (18.48) \u00de \u00f0 n, m Figure 18.17 shows the number of distinct caps that can be formed theoretically from pentagons and hexagons, such that each cap fits continuously onto the cylinder of the tubule, specified by a pair. It shows that the hemispheres of C60 are the smallest caps that satisfy these given requirements, so that the smallest carbon nanotube is expected to be 7 A\u00b0, which is in agreement with the experiment. Figure 18.17 also shows that the number of possible caps increases rapidly is listed the number of distinct with increasing tubular diameter. Below each pair of integers. n, m caps that can be joined continuously to the cylindrical carbon tubule denoted by \u00de n, m will give rise to equivalent nanotubes. Therefore, one restricts consideration to the nanotubes arising from the 30\u00b0 wedge of the 3D Bravais lattice shown in Figure 18.17. Because the length-to-diameter ratio of Due to the point group symmetry of the honeycomb lattice, several values of n, m \u00f0 \u00f0 \u00f0 \u00de \u00de 616 CHAPTER 18 Novel Materials (0,0) (1,0) (2,0) (3,0) (4,0) (5,0) (6,0) (7,0) (8,0) zigzag (9,0) 1 (10,0) 3 (11,0) 17 (12,0) 24 (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (7,1) (8,1) (9,1) (10,1) 3 (2,2) (3,2) (4,2) (5,2) (6,2) (7,2) (8,2) (9,2) 10 (10,2) 19 (11,1) 20 (11,2) 56 a1 (3,3) (4,3) (5,3) (6,3) (7,3) (8,3) 7 (8,3) 18 (9,3) 48 y (4,4) (5,4) (6,4) (7,4) 5 (8,4) 17 (9,4) 43 (10,4) 92 a2 (5,5) 1 (6,5) 1 (7,5) 13 (8,5) 37 (9,5) 80 (6,6) 15 (6,7) 32 (7,7) 87 x Metal Semiconductor Armchair FIGURE 18.17 The 2D graphene sheet that specifies the chiral nanotube. Reproduced from Dresselhaus et al.4 with the permission of Elsevier. carbon nanotubes is >103 while the diameter is only \u223c10 A\u00b0, carbon nanotubes are an important system for studying one-dimensional physics. 18.3.4 Symmetry Properties of Carbon Nanotubes To study the properties of carbon nanotubes as 1D systems, we define the lattice vector T along the tubule axis normal to the chiral vector Ch defined in Eq. (18.46) and Figure 18.15. The vector T defines the unit cell of the 1D carbon nanotube. The length T of the translation vector T corresponds to the first lattice point of the 2D graphene sheet through which the vector T passes. Thus, we obtain from Figure 18.15 and these definitions with a length T = 2m + n a1 \u2212 \u00de \u00f0 \u00bd\u00f0 2n + m \u00de dR, a2% ) T = 3p Ch dR: The length Ch is defined in Eq. (18.47). Defining d as the highest common divisor of have \ufb03\ufb03\ufb03 ) d if n \u2212 m is not a multiple of 3d 3d if n \u2212 m is a multiple of 3d: dR = *", " (18.49) (18.50) n, m \u00f0 \u00de, we (18.51) The relation between the fundamental symmetry vector R = pa1 + qa2 of the 1D unit cell and the, the chiral vector Ch, and translation vector T two vectors that specify the carbon nanotube n, m \u00de are shown in Figure 18.18. \u00f0 T \u03c4 O R X Ch FIGURE 18.18 Relation between R, Ch, and T. Reproduced from Dresselhaus et al.3 with the permission of Elsevier. Band structure of the (10,10) tube ) V e ( E 9 6 3 0 \u22123 \u22126 \u22129 B1 B2 / A2 A1 E E B2 B1 \u2212\u03c0 \u22122\u03c0/3 \u2212\u03c0/3 0 ka FIGURE 18.19 18.4 Polymers 617 N d R The projection of R on the Ch and T axes yields \u03c8 and \u03c4. X in the figure is \u03c8 scaled by Ch/2\u03c0: 18.3.5 Band Structure of a Fullerene Nanotube The electronic band structure of a (10,10) fullerene nanotube was first calculated by Dresselhaus et al.3 by using tight-binding methods and by using zone folding from the band structure of an infinite 2D graphene sheet. Their results are shown in Figure 18.19. As one can see in Figure 18.19, the two bands that cross the Fermi energy at ka = \u22122\u03c0/3 have different symmetry and guarantee that the tube will be a metallic conductor. 18.4 POLYMERS 18.4.1 Introduction A long chain of molecules that has a backbone of carbon atoms is known as a polymer. The basic building block, which usually but not necessarily consists of one carbon atom and two hydrogen atoms, is known as a monomer. A polymer is formed by repeating the structure of the monomer over and over again. In fact, a single polymer can be constituted from several thousand monomers. Band structure of a (10,10) fullerene nanotube, calculated by Dresselhaus et al., using zone folding from the band structure of an infinite 2D graphene sheet. Reproduced from Dresselhaus et al. (Ref. 3) with the permission Materials composed from polymers, such as bone, wood, skin, and fibers, have been used by man since prehistoric times. However, polymer science was developed in the twentieth century by Hermann Staudinger, who developed the concept of macromolecules in the 1920s. Wallace Carothers showed the great industrial potential of synthetic polymers and invented nylon in 1935. Synthetic polymers are now used in large quantities in a variety of applications. In the 1950s, Ziegler and Natta discovered polymerization catalysts, which led to the development of the modern plastics industry. Some of the most popularly known polymers are rubber, plastic, and Teflon. They do not have any common property other than the fact that they are lightweight, flexible, resistant to corrosion, and easy to of Elsevier. 618 CHAPTER 18 Novel Materials mold or cut into any desired shape. Some polymers such as rubber are very elastic and can be deformed very easily. Other polymers that have similar elastic property to rubber are known as elastomers. Paul Flory created modern polymer science through both experimental and theoretical studies of \u201cmacromolecules.\u201d A polymer is essentially a giant molecule. There are a variety of ways in which giant molecules can be obtained. Other examples of such giant molecules are branched polymers, in which hydrogen atoms are replaced by any of the halogen elements that need one electron to have a filled subshell. When the hydrogen atoms are replaced by molecules, more complex polymers are obtained. A third alternative is that the carbon chain can be replaced by silicon atoms. In view of the fact that there is a huge variety of ways in which polymers can be obtained, and the fact that they can be easily and cheaply produced, there is a great deal of excitement in finding new ways of obtaining and using complex polymers. 18.4.2 Saturated and Conjugated Polymers In the saturated polymers studied by Staudinger, Flory, Ziegler, and Natta, all four valence electrons of carbon are used up in covalent bonds, and hence, they are insulators. Therefore, they are viewed as unsuitable for use as electronic materials. In contrast, in conjugated polymers, the chemical bonding leads to one unpaired electron (the \u03c0 electron) per carbon atom. The carbon orbitals are in the sp2pz configuration in the \u03c0 bonding. Because the orbitals of the successive carbon atoms overlap, the electrons are delocalized along the backbone of the polymer, which provides the path for charge mobility along the polymer chain. The molecular structure", " of some conjugated polymers is shown in Figure 18.20. The bond-alternated structure of polyacetylene is characteristic of conjugated polymers, which are typically semiconductors. The electronic structure of conducting polymers can be determined by the symmetry of the chain. These polymers can exhibit either metallic or semiconducting properties. These electrically conducting polymers are known as the \u201cfourth generation of polymeric materials.\u201d The electronic structure in conducting polymers is determined by the number and kinds of atoms within the repeat unit (chain symmetry). The classic example is trans- and cis-polyacetylene, (-CH)n, which is shown in Figure 18.21. In polyacetylene, if the carbon\u2013carbon bond lengths were equal, the chemical formula, (-CH)n with one unpaired electron per formula unit, would yield a metallic state. If the electron\u2013electron interactions were too strong, (-CH)n would be an antiferromagnetic Mott insulator, a possibility that has been eliminated through a variety of studies together with the fact that these are easily converted to a metallic state on doping. \u00f0& CH = CH The structure in polyacetylene is dimerized due to Peierls instability with two carbon atoms in \u00den. Thus, the \u03c0 band is divided into \u03c0 and \u03c0# bands, each of which can the repeat unit, hold two electrons per atom (spin-up and spin-down). The \u03c0 \u2212 \u03c0# energy gap Eg implies that there are no partially filled bands and polyacetylene is a semiconductor. However, the electrical conductivity of polyacetylene can be increased by more than a factor of 107 to a level approaching that of a metal by using a dopant. The electrical conductivity of trans-(CH), as a function of (AsF5) dopant concentration, is shown in Figure 18.22. However, until 1990, there were no known examples of stable metallic polymers. It was shown that polyaniline (PANI) could be rendered, conducting either through oxidation of the Polyacetylene (PA) Polythiophene (PT) Poly(3-alkyl) thiophene (P3AT) (R-methyl, butyl, etc.) Polypyrrole (PPy) Poly isothianaphthene (PITN) Polyethylene dioxythiophene (PEDOT Alkoxy-substituted poly para-phenylene vinylene (MEH-PPV) O O n 18.4 Polymers 619 Polyparaphenylene vinylene (PPV) Poly (2,5 dialkoxy) paraphenylene vinylene (e.g. MEH-PPV) OR1 n n R2O Polyparaphenylene (PPP) n R2 R1 Ladder-type polyparaphenylene (LPPP) R1 n R2 Polyparaphenylene sulphide (PPS) Polyheptadiyne (PHT) Poly (3-hexyl) thiophene (P3HT Polyaniline, PANI H | N N N FIGURE 18.20 Molecular structures of some conjugated polymers (bond-alternated structures). Reproduced from Heeger6 with the permission of the American Physical Society. 620 CHAPTER 18 Novel Materials ab) FIGURE 18.21 (a) All trans- and (b) all cis-polyacetylene. 1000 100 10 1 10\u22121 10\u22122 10\u22123 10\u22124 10\u2212 Trans[CH(AsF5)y] TRANS CIS 10\u22126 0.00 0.05 0.10 0.15 0.20 0.05 0.30 y FIGURE 18.22 CH Electrical conductivity of trans\u00f0 of (AsF5) dopant concentration. \u00de as a function Reproduced from Heeger6 with the permission of the American Physical Society. leucoemeraldine base or protonation of the emeraldine base. Eventually, protonic acids were used to both convert PANI to metallic form and render the resulting PANI complex soluble in common organic solvents. The processibility of PANI induced by the \u201csurfactant\u201d counterions has made possible the fabrication of conducting polymer blends with a variety of host polymers. The \u201csurfactant\u201d counterions led to the formation of a self-assembled network, due to which the PANI network remains connected and conducting even after the removal of the host polymer. This led to the fabrication of novel electrodes for use in electronic devices. The chemistry and physics of these polymers in the semiconducting state are also of great interest since their application to \u201cplastic electronic\u201d devices. The polymer diodes were fabricated in the 1980s and the light-emitting diodes (LEDs) in 1990. The other", " \u201cplastic\u201d optoelectronic devices include lasers, highsensitivity plastic photodiodes, photovoltaic cells, ultrafast image processors, thin-film transistors, and all-polymer integrated circuits. All these are fabricated semiconducting and metallic polymers that are thin-film devices in which the active layers are fabricated by casting the semiconducting and/or metallic polymers from solution. 18.4 Polymers 621 18.4.3 Transparent Metallic Polymers In conventional metals, the length of the interchain spacing and of the repeat unit are large compared to the interatomic distances. However, in metallic polymers, N, the number of electrons per unit volume is on the order of N \u223c 2 \u2212 5 \u00d7 1021 cm\u22123: The plasma frequency (the frequency below which the metals reflect light) is given by \u03c92 p = 4\u03c0Ne2/m#, where m# is the effective mass of the electrons. Thus, the plasma frequency of metallic polymers is on the order of 1 eV: Hence, they are semitransparent in the visible part of the spectrum but exhibit high reflectance in the infrared. Therefore, optical-quality thin films of metallic polymers are used as transparent electrodes. Transparent conducting films are used as antistatic coatings, as electrodes in liquid-crystal display cells or in polymer LEDs, or for fabricating electrochromic windows. (18.52) 18.4.4 Electronic Polymers Electronic polymers are those polymers of which the conductivity can be increased by several orders of magnitude, which can be obtained by doping. The increase of conductivity of electronic polymers by doping is shown in Figure 18.23. In Figure 18.23, trans-(CH)x and the emeraldine base form of polyamine are shown to give an example of the increase in electrical conductivity by doping. Electronic polymers are extensively used in light-emitting diodes. In addition, superconductivity has been discovered in regioregular Ag, Cu Fe Mg In, Sn Ge Si AgBr Glass Diamond Nylon Quartz 106 104 102 100 10\u22122 10\u22124 10\u22126 10\u22128 10\u221210 10\u221212 10\u221214 10\u221216 Doped trans-(CH)x [105 S/cm] Doped polyaniline [103 S/cm trans-(CH)x [10\u22125 S/cm] Polyaniline [10\u221210 S/cm] FIGURE 18.23 Increase of conductivity of electronic polymers by doping. Reproduced from MacDiarmid11 with the permission of the American Physical Society. 622 CHAPTER 18 Novel Materials poly (3-hexylthiophene) around a critical temperature of 2\u00ba K. The future application of this amazing discovery is still being speculated. 18.5 SOLITONS IN CONDUCTING POLYMERS 18.5.1 Introduction Trans-polyacetylene (trans-(CH)x) was the first highly conducting organic polymer. The simple molecular structure \u2212CH\u2212 units are repeated, which implies that each carbon contributes a single pz electron to the \u03c0 band, and as a result, the \u03c0 band would be half-filled. Thus, polyacetylene would be a one-dimensional metal. However, in 1955, Peierls showed that 1D metals are unstable with respect to a structural distortion that opens up an energy gap at the Fermi level; thereby, 1D metals end up as semiconductors. The periodicity of the Peierls distortion is \u039b = \u03c0/kF, where the Fermi wave vector kF = 2\u03c0/a for the half-filled band of trans-(CH)x. Hence, it converts trans-polyacetylene into trans\u00dex, which is essentially alternating single and double bonds, as shown in Figure 18.24a. \u2212HC = CH\u2212 However, a chain of monomers can be dimerized in two distinct patterns, both of which have \u00f0 the same energy (the degenerate A and B phases), as shown in Figure 18.24b. Thus, in addition to electron and hole excitations in a dimerized semiconductor, a domain wall separates regions of different bonding structures or different vacua, which would be a new type of excitation. The large width of the domain walls leads to a small effective mass for the excitations, on the order of electron mass instead of the ionic mass. The domain-wall excitation, which propagates freely, has been called a \u201csoliton.\u201d Because a moving soliton converts A\u2212 phase material into B\u2212 phase material (or vice versa), these objects can only be created or destroyed in pairs. The creation of a soliton is shown in Figure 18.24c. As we will see, because the midgap state is a solution to the Schrodinger equation", " in the presence of the structural kink, it can be occupied with zero, one, or two electrons. However, a charged soliton, which has either zero or two electrons in the gap state, carries a charge \u00b1e and has spin zero, rather than the spin 1 as for 2 an electron or a hole. These reversed spincharge relations are a fundamental feature of the soliton model of polyacetylene and are supported by experiment. The \u03c0 band of (CH)x is split into two subbands: a fully occupied \u03c0 band (valence band) and an empty \u03c0# band (the conduction band), each with a wide bandwidth and significant distortion. The resulting band structure results from the opening of the band gap that originates from the doubling of the unit cell. This is shown in Figure 18.25. \u223c5 eV \u00de \u00f0 (a) (b) (c) FIGURE 18.24 (a) Dimerized structure due to Peierls instability. (b) Degenerate A and B phases of dimerized structure. (c) Soliton in trans-polyacetylene. E 2t0 \u22060 0 \u2212\u22060 \u22122t0 k \u03c0 2a FIGURE 18.25 Band structure of polyacetylene. Energy opening at k = 2\u03c0/a due to Peierls distortion. 18.5 Solitons in Conducting Polymers 623 18.5.2 Electronic Structure The electronic structure of conjugated polymers (SSH) was described by Su et al.17 by using a tight-binding model in which the \u03c0 electrons are coupled to distortions in the polymer backbone by the electron\u2013phonon interaction. In the SSH model, photoexcitation across the \u03c0 \u2212 \u03c0# band gap creates the excitations of conducting polymers: solitons (in degenerate ground-state systems), polarons, and bipolarons. Direct photogeneration occurs due to the overlap between the uniform chain in the ground state and the distorted chain in the excited state. 18.5.3 Tight-Binding Model Each (CH) group (Figure 18.26) has six degrees of freedom for nuclear translation. Su et al.17 postulated that only the dimerization coordinate un, which specifies the displacement of the nth group along the molecular symmetry group, is important. The tight-binding Hamiltonian can be written as H = H\u03c0 + H\u03c0\u2212ph + Hph = \u2211 n,s\u00bd \u2212t0 + \u03b1 un+1 \u2212 un\u00de%\u00f0 \u00f0 c\u2020 n+1,scn.s + c\u2020 n,scn+1,s\u00de p2 n 2M + 1 2 + \u2211 n \u2019 K un+1 \u2212 un\u00de \u00f0 2, ( (18.53) where pn are the nuclear momenta, un are the displacements from equilibrium, M is the carbon mass, and K is an effective spring constant. The c\u2020 n,s, cn,s are the fermion creation and annihilaelectrons along the chain tion operators for site n. The first term describes the hopping of \u03c0 without spin flip. The second term describes the \u03c0\u2212 electron\u2013phonon interaction where the terms linear in un dominate higher-order terms for the weak-coupling systems. The last two terms are, respectively, a harmonic \u201cspring constant\u201d term, which represents the increase in potential energy that results from displacement from the uniform bond lengths in (CH)x, and a kinetic energy term, where M is the mass of the (CH) group and pn is the momentum conjugate to un: pz\u00de \u00f0 un un\u22121 un+1 x FIGURE 18.26 Dimerization coordinate un defined for trans-(CH)x. Reproduced from Heeger et al. (Ref. 7), with the permission of the American Physical Society. 624 CHAPTER 18 Novel Materials 18.5.4 Soliton Excitations The ground state of a one-dimensional metal is spontaneously distorted to form a charge-density wave <un> \u2260 0, as per Peierls theorem. The strongest instability occurs for a charge-density wave of wave number Q = kF = \u03c0/a, and hence, we consider the adiabatic ground-state energy E0 as a function of the mean amplitude of distortion u, where the un\u2019s are considered to be un! <un> = (18.54) nu: \u00de \u22121 \u00f0 For un given by Eq. (18.54), H\u03c0\u2212ph is invariant under spatial translation 2ma, m = \u00b11, \u00b12, \u2026, and and H can be diagonalized in k space in the reduced zone, \u2212\u03c0/2a < k < \u03c0/2a, for the valence \u00f0 conduction", " (+) bands. For a chain of monomers of ring geometry, we obtain (Problem 18.6) n2\u03b1u + 2NKu2: (18.55) ^H \u2212 u \u00de n+1,s^cn,s + ^c\u2020 ^c\u2020 t0 + \u22121 = \u2212\u2211 n,s\u00bd \u00f0 \u00de \u00f0 \u00de %\u00f0 n,s^cn+1,s\u00de For \u03b1 = 0, H u \u00f0 \u00de can be made diagonal by using the Bloch operators, ^cks = N \u22121/2 \u00f0 \u00de \u2211 n,s e\u2212ikna^cns (18.56) in the extended zone, where \u2212\u03c0/a < k \u2264 \u03c0/a. For \u03b1 \u2260 0, the big zone can be folded into the little zone, as shown in Figure 18.27. The valence- and conduction-band operators are defined as and ^cks\u2212 = N \u00f0 \u00de \u22121/2 \u2211 n,s e\u2212ikna ^cns ^cks + = \u2212i N \u00f0 \u00de \u22121/2 e\u2212ikna n^cns: \u22121 \u00de \u00f0 \u2211 n,s (18.57) (18.58) From Eqs. (18.55), (18.57), and (18.58), we obtain (Problem 18.8) ^H u \u00de \u00f0 = \u2211 ks h ks + ^cks\u2212 \u2212 ^c\u2020 ^c\u2020 \u03b5k\u00f0 ks\u2212^cks\u2212\u00de \u03b5k C V \u2212 \u03c0 a \u2212 \u03c0 2a \u03c0 2a \u03c0 a k FIGURE 18.27 The reduced and extended zone schemes. ks + ^cks\u2212 + ^c\u2020 ^c\u2020 + \u0394k\u00f0 ks\u2212^cks + \u00de + 2NKu2, (18.59) i where \u0394k, the energy gap parameter, is defined as (18.60) \u0394k = 4\u03b1u sin \u2061ka: In the reduced zone, the unperturbed band energy is defined as \u03b5k = 2t0 cos\u2061ka, \u2212\u03c0 2a / 2a, \u00de (18.61) where \u03b5k describes particles for (+) and holes for. We now convert H in Eq. (18.59) to a diag\u2212 \u00f0 onal form by making the transformation \u00de ^aks\u2212 = \u03b1k^cks\u2212 \u2212 \u03b2k^cks+ 18.5 Solitons in Conducting Polymers 625 and We have, in analogy with BCS theory of superconductivity, ^aks + = \u03b2k^cks\u2212 + \u03b1k^cks+: (18.62) \u03b2kj j so that the a\u2019s satisfy the Fermi anticommutation relations. Because H has no term that mixes the (18.63) \u03b1kj j 2 + 2 = 1, and + \u00f0 \u00de \u2212 \u00de \u00f0 operators, we obtain from Eq. (18.59) through (18.62) H = \u2211 k Ek\u00f0 nks + \u2212 nks\u2212\u00de + 2NKu2: The quasiparticle energy is given by (in analogy with BCS superconductivity theory) Ek = \u22082 k + \u0394k\u00de 1/2, \u00f0 \u03b1k = 1 + \u03b5k/Ek\u00de \u00bd\u00f0 /2 1/2, \u00de% \u03b2k = 1 \u2212 \u03b5k/Ek\u00de \u00bd\u00f0 /2 1/2 sgn\u2061\u0394k, \u00de% and (18.64) (18.65) (18.66) (18.67) \u03b1k\u03b2k = \u0394k/ 2Ek\u00de nu, the chain would behave as a conIf one were to clamp un at the ground-state mean-field value, ventional semiconductor with electron and hole excitations. However, because of the two-fold-degenerate, the system supports nonlinear excitations, which act as moving walls = E ground-state E separating the A phase.This is shown in Figure 18.28, which implies that \u2212u0\u00de \u00f0 the nonlinear excitations, solitons, will bef important. To determine the soliton, one introduces \u2212u0\u00de \u00f0 + u0\u00de \u00f0 and B phase (18.68) u0\u00de \u22121 \u00f0 \u00f0 \u00f0 \u00de : \u22121 \u00f0 so that in the A phase, \u03c6n = u0, and in the B phase, \u03c6n = \u2212u0: However, numerical calculations have shown that the form of \u03c6n, for the preferred width \u03be of the soliton that minimizes the total energy, is \u03c6n", " = nun, \u00de E \u2191 (18.69) \u00bd\u00f0 a/\u03be : \u00de% (18.70) \u03c6n = u0 tanh\u2061 n \u2212 n0\u00de It has been calculated that \u03be \u2248 7a, and the energy to create a soliton at rest is Es \u2248 0:42 eV. Thus, Es < 0:5 \u0394, where \u0394 is the single-particle gap. Thus, it is less costly to create a soliton than an electron or hole, and they are spontaneously generated by photoexcitation, by thermal generation, or by injection of electrons and/or holes. It \u2212u0 u0 u FIGURE 18.28 The total energy of the dimerized polyacetylene chain. 626 CHAPTER 18 Novel Materials can also be shown that for each widely spaced soliton or antisoliton (a reverse boundary from the B phase back to the A phase), there exists a normalized single-electron state in the gap center that can accommodate zero, one, or two electrons. 18.5.5 Solitons, Polarons, and Polaron Excitations It can be seen from Figure 18.29c that because the nonbonding or atomic state formed by the chain relaxation can be mapped to a specific atomic site, the resulting distribution of charge and spin can be understood. If the state is unoccupied (doubly occupied), the carbon atom at the boundary is left with a positive (negative) charge, but there are no unpaired spins. Therefore, the charged soliton is positively (negatively) charged but spinless. Single occupation of the soliton state neutralizes the electronic charge of the carbon nucleus, while introducing an unpaired spin onto the chain. The localized electronic state associated with the soliton is a nonbonding state at an energy that lies at the middle of the \u03c0 \u2212 \u03c0# gap, between the bonding and antibonding levels of the perfect chain. (a) Polaron s = 1/2, q = \u2212|e| (b) Negative soliton s = 0, q = \u2212|e| Positive solition s = 0, q = |e| (c) FIGURE 18.29 (a) Schematic picture of a polaron in polyparaphenylene (PPP); (b) band diagram of an electron polaron\u2014the lower gap state is single occupied and the upper gap state is empty for a hole polaron; (c) band diagrams for positive and negative solitons. Reproduced from Heeger 6 with the permission of the American Physical Society. 18.6.6 Polarons and Bipolarons A polaron can be thought of as a bound state of a charged soliton and a neutral soliton of which the midgap energy states hybridize to form bonding and antibonding levels. The neutral soliton contributes no charge and a single spin. The charged soliton carries a charge of \u00b1e and no spin. The resulting polaron is a fermion with the charge-spin relationship q = \u00b1e and s = 1=2. The positive polaron is a radical cation; the negative polaron is a negative anion. Both are quasiparticles consisting of a single electronic charge dressed with a local geometrical relaxation of the bond lengths. The schematic picture of polarons in PPP is shown in Figures 18.29a and b. A bipolaron is a bound state of two charged solitons of like charges or two polarons of which the neutral solitons annihilate each other with two corresponding midgap levels. The bipolarons are shown in Figure 18.30. Because each charged soliton has a single electronic charge and no spin, the bipolaron has charge \u00b12e and zero spin. Both positive and negative bipolarons are doubly charged bound states of two polarons bound together by enhanced geometrical relaxation of the bond lengths or the overlap of a common lattice distortion. Problems 627 (a) h\u03bd RO OR RO OR RO OR e\u2212 Bipolaron s = 0, q = \u22122|e| (b) RO OR FIGURE 18.30 Bipolarons in polymer with nondegenerate ground state. (a) Negative bipolaron in PPP; (b) band diagram for a negative bipolaron. For a positive bipolaron, both gap states are unoccupied. FIGURE 18.31 Photoinduced electron transfer from a conjugated semiconducting polymer to C60. Reproduced from Heeger 6 with the permission of the American Reproduced from Heeger 6 with the permission of the American Physical Society. Physical Society. 18.6 PHOTOINDUCED ELECTRON TRANSFER The electrons are promoted to the antibonding \u03c0# band of semiconducting polymers when they are photoexcited. The photoinduced nonlinear excitations (such as polarons) on the conjugated polymer backbone are quite stable. Thus, semiconducting poly", "mers are electron donors when they are photoexcited. On the other hand, Buckministerfullerene, C60, is an excellent electron acceptor capable of accepting up to six electrons. It forms charge-transfer salts with several types of donors. The photoinduced transfer process from a conjugated semiconducting polymer to C60 is shown schematically in Figure 18.31. A study of the dynamics of the photoinduced electron transfer from semiconducting polymers to C60 by Lanzani et al. (Ref. 10) demonstrated that the charge transfer occurs within 50 fs after photoexcitation. This charge transfer rate is more than 1000 times faster than any other process. The quantum efficiency for charge separation approaches one. In addition, the charge separation state is metastable. PROBLEMS 18.1. Show that the reciprocal lattice vectors of honeycomb lattice of graphene (Figure 18.3) are b1 = 2\u03c0 3a \u00f0 ^i + 3p ^j \u00de ; b2 = 2\u03c0 3a \u00f0 ^i \u2212 3p ^j \u00de : (1) \ufb03\ufb03\ufb03 18.2. Show that the positions of the two Dirac points K and K\u2032, located at the corners of the \ufb03\ufb03\ufb03 Brillouin zone (Figure 18.3), are K = 2\u03c0 3a ^i + $ 1 3p ; K\u2032 = 2\u03c0 ^j 3a % ^i \u2212 $ 1 3p ^j : % \ufb03\ufb03\ufb03 \ufb03\ufb03\ufb03 (1) 628 CHAPTER 18 Novel Materials 18.3. Show that by substituting in Eq. (18.6), we obtain and Here, and H\u03c8 = E\u03c8, H11 + \u03bbH12 = E H12 + \u03bbH22 = \u03bbE: (1) (2) (3) H11 = \u03d51 # H\u03d51 d\u03bd, H12 = H21# = \u03d51 # H\u03d51 d\u03bd; H22 = \u03d52 # H\u03d52 d\u03bd (4) Z Z Z \u03d51 # \u03d51d\u03bd = \u03d52 # \u03d52d\u03bd = 1: Z Z 18.4. It has been shown in Eq. (18.19) that H11 = 1 N \u2211 A, A\u2032 e\u2212ik. rA\u2212rA\u2032\u00de \u00f0 Z \u03c7 # r\u2212rA\u00de \u00f0 H\u03c7 r \u2212 rA\u2032\u00de \u00f0 d\u03bd: Keeping only the nearest-neighbor integrals among the atoms A, and writing and where \u03c1!\u2032 = a1\u00f0 E0 = Z \u03c7 # r H\u03c7 \u00de \u00f0 \u00f0 r \u00de d\u03bd \u03b30\u2032 = \u2212 Z \u03c7 # r \u2212 \u03c1!\u2032 H\u03c7 r \u00de \u00f0 \u00de \u00f0 d\u03bd, is a vector joining the nearest neighbor among atoms A, show that say \u00de H11 = E0 \u2212 2\u03b30\u2032 \u00bd cos\u2061 \u00f0 2\u03c0kya + 2 cos\u2061 \u00f0 \u00de \u03c0kxa 3p cos\u2061 \u00f0 \u00de \u03c0kya : \u00de% 18.5. It has been shown in Eq. (18.28) that H12 = 1 N \u2211 A, B e\u2212ik. rA\u2212rB \u00f0 \u00de \u03c7 # r \u2212 rA\u00de \u00f0 r \u2212 rB\u00de : \u00f0 Z Considering only the nearest-neighbor interactions in the lattice (between atoms of type A and type B and vice versa), we write (in analogy with Eq. 18.27) \ufb03\ufb03\ufb03 H\u03c7 (5) (1) (2) (3) (4) (1) \u03b30 = Z \u03c7 # r \u2212 \u03c1!\u00de\u00f0 \u00f0 U \u2212 V \u03c7 r \u00de \u00f0 \u00de dv > 0, (2) Problems 629 where Show that and H12j j \u03c1! = AB: H12 = \u2212\u03b30\u00bd e\u2212ikxa + 2 cos \u2061 3p kya \u00f0 \u00de e3ikxa/2 % \ufb03\ufb03\ufb03 2 = \u03b32 0 1 + 4 cos\u20612 \u2019 3p 2 \" \ufb03\ufb03\ufb03 kya + 4 cos # 3p 2 \" \ufb03\ufb03\ufb03 kya cos \u2061 # \" 3 2 kxa : ( # 18.6. The translation vector of a general chiral tubule can be written as T = 2m + n a1 \u2212 \u00de 2n + m \ufffd", "\ufffd \u00de \u00bd\u00f0 dR: a2% ) The length T of the translation vector T corresponds to the first lattice point of the graphene sheet. Show that T = 3p Ch/dR, (2) where the length Ch is defined in Eq. (18.47). Defining d as the highest common divisor of \ufb03\ufb03\ufb03 n, m \u00de, we have \u00f0 d if n \u2212 m is not a multiple of 3d 3d if n \u2212 m is a multiple of 3d: dR = * 18.7. The tight-binding Hamiltonian for the polymer trans-(CH)x is given by (Eq. 18.53) H = H\u03c0 + H\u03c0\u2212ph + Hph = \u2211 n,s\u00bd \u2212t0 + \u03b1 n+1,s^cn:s + ^c\u2020 ^c\u2020 un+1 \u2212 un\u00de%\u00f0 n,s^cn+1,s\u00de \u00f0 p2 n 2M + 1 2 + \u2211 n \u2019 K un+1 \u2212 un\u00de \u00f0 2 : ( Show that if one can write un! <un> = nu, \u22121 \u00de \u00f0 H u \u00de \u00f0 = \u2212\u2211 n,s\u00bd t0 + \u22121 n2\u03b1u \u00de \u00f0 %\u00f0 n+1,s^cn,s + ^c\u2020 ^c\u2020 n,s^cn+1,s\u00de + 2NKu2 for a chain of monomers in a ring geometry. 18.8. Show that for \u03b1 = 0, H Bloch operators, \u00f0 u \u00de in Eq. (3) of Problem 18.7 can be made diagonal by using the ^cks = N\u22121/2 \u00f0 \u2211 n,s \u00de e\u2212ikna^cns, (1) in the extended zone, where \u2212\u03c0/a < k \u2264 \u03c0/a. (3) (4) (5) (1) (3) (1) (2) (3) 630 CHAPTER 18 Novel Materials 18.9. Show that by using the operators and ^cks\u2212 = \u22121/2 N \u00f0 \u00de \u2211 n,s e\u2212ikna ^cns ^cks+ = \u2212i N \u00f0 \u00de \u22121/2 e\u2212ikna n^cns, \u22121 \u00de \u00f0 \u2211 n,s (1) (2) Eq. (3) in Problem 18.7 can be rewritten as H u \u00de \u00f0 = \u2211 ks \u2019 ks + ^cks\u2212 \u2212 ^c\u2020 ^c\u2020 \u03b5k\u00f0 ks\u2212^cks\u2212\u00de ks + ^cks\u2212 + ^c\u2020 ^c\u2020 + \u0394k\u00f0 ks\u2212^cks + \u00de ( + 2NKu2, (3) where \u0394k, the energy gap parameter, is defined as \u0394k = 4\u03b1u sin \u2061ka: 18.10. From Eqs. (18.59) through (18.62), show that H = \u2211 k Ek\u00f0 nks+ \u2212 nks\u2212\u00de + 2NKu2: (4) (1) References 1. Castro Neto AH, Guinea F, Peres NMR, Noboselov KS, Geim AK. The electronic properties of graphene. Rev Mod Phys 2009;81:109. 2. Curl RF. Dawn of the fullerenes: experiment and conjecture. Rev Mod Phys 1997;69:691. 3. Dresselhaus MS, Dresselhaus G, Ekland PC. Science of fullerenes and nanoparticles. Amsterdam: Elsevier Science; 1996. 4. Dresselhaus MS, Dresselhaus G, Saito R. Physics of Carbon Nanotubes. Carbon 1995;33:883. 5. Geim AK, Novoselov KS. The rise of Graphene. Nature materials 2007;6:183. 6. Heeger AJ. Semiconducting and metallic polymers: The fourth generation of polymeric materials. Rev Mod Phys 2001;73:681. 7. Heeger AJ, Kivelson S, Schrieffer JR, Su W-P. Solitons in conducting Polymers. Rev Mod Phys 1988;60:781. 8. Kroto HW. Symmetry, space, stars and C60. Rev Mod Phys 1997;69:703. 9. Kroto HW, Heath JR, O\u2019Brien SC, Curl RF, Smalley RE. C60: Buckminsterfullerene. Nature 1985; 318:162. 10. Lanzani G, Zenz C, Cerullo G, Graupner W, Leising G, Scherf U, De Silvestri S. Synth Met 2000; 111\u2012112:", "493. 11. MacDiarmid AG. \u201cSynthetic Metals.\u201d A novel role for organic polymers. Rev Mod Phys 2001;73:701. 12. Malard LM, Pimenta MA, Dresselhaus G, Dresselhaus MS. Raman Spectroscopy in Graphene. Phys Rep 2009;473:51. 13. Novoselov KS, Geim AK, Morozov SV, Jiang D, Zhang Y, Grigoneva IV, et al. Electric field effect in atomically thin carbon films. Science 2004;306:666. References 631 14. Novoselov KS, Geim AK, Morozov SV, Jiang D, Katsnelson MI, Grigoneva IV, et al. Two-dimensional gas of massless Dirac fermions in graphene. Nature 2005;438:197. 15. Novoselov KS, Jiang D, Schedin F, Booth TJ, Khotkevich VV, Morozov SV, et al. Two-dimensional atomic crystals. Proc Natl Acad Sci 2005;102:10451. 16. Smalley RE. Discovering the fullerenes. Rev Mod Phys 1997;69:723. 17. Su WP, Schrieffer JR, Heeger AJ. Solitons in Polyacetylene. Phys Rev Lett 1979;42:1698. 18. Taylor R, Hare JP, Abdul-Sada AK, Kroto J. Chem Soc Chemical Commu 1990;1423. 19. Wallace PR. The Band Theory of Graphite. Phys Rev 1947;71:622. Elements of Group Theory APPENDIX A CHAPTER OUTLINE A.1 Symmetry and Its Consequences............................................................ 633 A.1.1 Symmetry of Crystals................................................................ 633 A.1.2 Definition of a Group................................................................ 633 A.1.3 Symmetry Operations in Crystal Lattices.............................................. 634 A.2 Space Groups............................................................................ 634 A.2.1 Introduction........................................................................ 634 A.2.2 Space Group Operations............................................................. 634 A.3 Point Group Operations.................................................................... 636 A.3.1 Introduction....................................................................", ".... 636 A.3.2 Description of Point Groups.......................................................... 636 A.3.3 The Cubic Group Oh................................................................. 638 A.1 SYMMETRY AND ITS CONSEQUENCES A.1.1 Symmetry of Crystals Crystalline solids possess translational symmetry as well as symmetries involving rotations and reflections. The fundamental principle is that the quantum mechanical operators representing a symmetry operation of a crystal must commute with the Hamiltonian of the crystal. This facilitates the derivations of restrictions limiting possible Hamiltonians, the classification of eigenstates according to symmetry, as well as determining the selection rules for transitions between states. Often, one has to use group theory to consider these symmetry considerations. A.1.2 Definition of a Group A group has to satisfy the following three basic conditions: a. The group must contain the unit element, generally denoted by E. b. The product of any two elements in the group must be another element in the group. c. The inverse of every element in the group must belong to the group. If the members commute with each other, the group is Abelian. Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00019-0 \u00a9 2012 by Elsevier Inc. All rights reserved. 633 634 APPENDIX A Elements of Group Theory A.1.3 Symmetry Operations in Crystal Lattices For the purpose of considering the symmetry operations of crystals as well as their reciprocal lattices, we consider a set of symmetry operations (matrices) that form a group. We denote this group as. There are two types of such groups known as space groups and point groups. G f g A.2 SPACE GROUPS A.2.1 Introduction The elements of symmetry groups relevant to solids include translations as well as proper and improper rotations. These are also known as space groups. A Bravais lattice is generally characterized by the specification of all rigid operations that take the lattice into itself. This set of operations is known as the space group or symmetry group of the Bravais lattice. All translations through the lattice vectors of a Bravais lattice are included in the operations of a symmetry group. In addition, rotations, reflections, and inversions can also take the Bravais lattice into itself. For example, a cubic Bravais lattice can be taken into itself by a rotation through 90\u00b0 about a line of lattice points in a <100> direction, a rotation through 120\u00b0 about a line of lattice points in a <111> direction, and reflection of all points in a lattice plane. In contrast, a simple hexagof nal Bravais lattice can be taken into itself by a reflection in a lattice plane perpendicular to the c-axis, or rotation through 60\u00b0 about a line of points parallel to the c-axis. 100 g The symmetry operation of a Bravais lattice can be compounded out of a translation TR through a lattice vector R and a rigid operation leaving at least one point fixed. A simple example of two such consecutive operations (symmetry operation S and translation T\u2212R through \u2212R) is also known as a composite operation T\u2212RS. In this example, because S transports the origin of the lattice O to R while T\u2212R carries R back to the origin O, T\u2212RS also leaves at least one lattice point (O) fixed. In contrast, if the operation TR is performed after performing the operation T\u2212RS, the result is the operation S alone. In the process, TR has reversed the previous application of T\u2212R. Hence, S can be compounded out of operations that leave a point fixed (T\u2212RS), translations through Bravais lattice vectors (TR), or through successive applications of operations of both types. A.2.2 Space Group Operations \u03b1 denotes An operator of a space group contains a part that is denoted by the symbol a proper or improper rotation and t denotes a translation part. The coordinate transformation from x to x\u2032 can be expressed as, where \u03b1 g f t j e e x\u2032 = \u03b1x + t: (A.1) Here, \u03b1 can be represented by a 3 \u00d7 3 orthogonal matrix. One can multiply two such operators by e = \u03b1 \u03b2 t (A.2)", " t\u2032 j f gf j g \u03b2t + t\u2032 \u03b2\u03b1 j f : g e e e e e It is easily verified that the unit operator is is of the operator \u03b1 t f j g A.2 Space Groups 635. It can be shown from Eq. (A.2) that the inverse 1 = g f f \u03b1\u22121 j j \u03b1\u22121t \u2212 : g (A.3) In the block notation, we can write Eq. (A.1) in the alternate form, where the position vector x is e denoted by e e x = x1 x2 x3 0 @ 1 A, t = t1 t2 t3 0 @ 1 A Eq. (A.1) can be written as, and \u03b1 = e \u03b111 \u03b121 \u03b131 e e e 0 @ \u03b112 \u03b122 \u03b132 e e e \u03b113 \u03b123 \u03b133 e e e : 1 A 1 x1\u2032 x2\u2032 x3 t1 t2 t3 0 \u03b111 \u03b121 \u03b131 e e e 0 \u03b112 \u03b122 \u03b132 e e e 0 \u03b113 \u03b123 \u03b133 e e e 1 x1 x2 x3 @ or in the block notation 1 x\u2032! \" =! 1 t From Eqs. (A.2) and (A.6), we obtain 1 : x! \" \" 0 \u03b1 e and the inverse is 1 t\u2032 + \u03b2t = \"! e ; \" 0 \u03b1 \u03b2 e e \u22121 0 \u03b1 \"! 1 \u03b1\u22121t \u2212 0 \u03b1\u22121 : \" (A.4) (A.5) (A.6) (A.7) (A.8) f f define a few common terms used in group theory. has an inverse is an operator that represents a lattice translation through Ri, each e e, and because the sum Ri + Ri\u2032 = Rj, the lattice translations form a group. We will now \u03b5 Rig j \u2212Rig \u03b5 j Abelian: If the members commute with each other, the group is Abelian. The operators of the Rig e f \u03b5 j full space group, which contains both rotations and translations, are not necessarily Abelian. Invariant subgroup: If A is a member of a subgroup and X is a member of the full group and if B = XAX\u22121 is in the subgroup concerned for all A and all X, the subgroup is said to be invariant. This can be proved as follows. 636 APPENDIX A Elements of Group Theory Proof: Let Ri be any direct lattice vector and \u03b1 be the rotational part of any space group opera- tion, then \u03b1Ri is also a lattice vector. It is easy to show that t e \u03b1Rig j Thus, if is also a lattice translation and Eq. (A.9) proves \u03b1Rig \u03b5 e e j f that the subgroup is invariant. Thus, a space group is defined as a group of operators of the form is a lattice translation, then e Rigf j Rig (A.9) \u03b1 j gf gf = \u2212 \u22121 \u03b1\u22121t \u22121 = t g e \u03b1Ri + t \u03b1,t f g, which possesses an invariant subgroup of pure translations. e e A.3 POINT GROUP OPERATIONS A.3.1 Introduction It can be easily shown that a point group can contain rotations ples of these, but it forbids five-fold rotational symmetry. Proof: If R is a direct lattice vector, 3 R = \u2211 j =1 njaj; \u03b1 through 60\u00b0, 90\u00b0, and (or) multi- e (A.10) then integers. Thus, Because \u03b1R is also a direct lattice vector. Here, aj are the primitive translation vectors and the nj\u2032s are \u03b1R can be expressed in terms of aj with integer coefficients. e \u03b1 can be a rotation through an angle \u03b8 about some axis, in considering the operation of \u03b1 on R, one has to include the possibility that aj may not be orthogonal and assume that the aj are A. If the original vector is represented by n related to a Cartesian system by a nonsingular matrix (with components nj) and the rotated vector is denoted by n\u2032, we have e e e n A \u03b1 = n\u2032 e A; which can be written in the alternate form e e e \u22121 n A \u03b1 A = n\u2032: (A.11) (A.12) \u22121 \u22121 \u03b1 A A as well as the trace of The elements of must be integers because the elements of both n A \u03b1 A e e e and n\u2032 are arbitrary integers. Because the trace of a matrix is invariant under a similarity transformation, e e = 1 + 2 cos \u03b8 = integer: (A.13)", " e \u03b1 = tr tr A \u22121 e e e \u03b1 Eq. (A.13) implies that \u03b8 = 60\u00b0, 90\u00b0, or an integral multiple of these angles.3.2 Description of Point Groups There are 32 possible crystal point groups. There are two widely used but different notations for point groups: the Schoenflies system and the international system. We will first discuss the Schoenflies notation (see Table A.1). A.3 Point Group Operations 637 Table A.1 Shoenflies Symbol Operation Shoenflies Symbol Identity Rotation through 2\u03c0/n Inversion Improper rotation through 2\u03c0/n Reflection in a plane Reflection in a plane perpendicular to highest symmetry axis Reflection in a plane containing the highest symmetry axis Reflection in a plane containing the highest symmetry axis and bisecting the angle between two-fold axes perpendicular to symmetry axis E Cn i Sn \u03c3 \u03c3h \u03c3v \u03c3d Reproduced from Quantum Theory of the Solid State by J. Callaway (Academic Press, New York, 1976) with the permission of Elsevier. 1. Cn: These point groups contain only a single axis of symmetry, around which rotations through angles of 2\u03c0/n are permitted. As we have seen, n = 1, 2, 3, 4, 6: 2. Cnv: In addition to the operation Cn, these groups have a mirror plane that contains the axis of rotation, plus as many mirror planes as the existence of the n-fold axis requires. The reflection symmetry must be consistent with the rotational symmetry, and n-fold rotational symmetry about some axis demands the existence of n reflection planes at angles of \u03c0/n if there are any reflection planes containing the axis. The allowed values of n are 2, 3, 4, 6. 3. Cnh: These groups contain, in addition to the n-fold axis, a single mirror plane that is perpendicular to the axis. Thus, there is a \u201chorizontal\u201d reflection plane (operation \u03c3h), which is a reflection in a plane to the origin perpendicular to the axis of highest symmetry. The group also contains the inversion operation i if n is even. In this case, n = 1, 2, 3, 4, 6: 4. Sn: These groups have n-fold improper rotation (rotation combined with improper reflection in a plane perpendicular to the axis of rotation). Because the group S3 is identical with C3h, it is not counted. Thus, the distinct groups are S2 (simple inversion), S4, and S6: 5. Dn: In addition to the n-fold rotation axis, these groups contain a two-fold axes perpendicular to the highest symmetry (Cn) axis plus as many additional two-fold axes as are required by the existence of the n-fold axis. Thus, one can have n = 2, 3, 4, 6: 6. Dnh: These (the most symmetric of the groups) contain all the elements of Dn plus mirror planes containing the n-fold axes, which bisect the angles between the two-fold axes. Therefore, there are twice as many elements in Dnh as in Dn: 7. Dnd: These contain the elements of Dn plus reflections in a \u201cdiagonal\u201d plane (\u03c3d) containing the symmetry axis and bisecting the angle between the two-fold axis. The diagonal planes are special cases of vertical planes. There are two cases: D2d and D3d: The international notation for point groups and the three Schoenflies equivalents are given in Table A.2. 638 APPENDIX A Elements of Group Theory Table A.2 International Notation for Point Group Symbols Axis n-fold rotation axis Improper n-fold rotation axis Rotation axis with reflection plane perpendicular to it n-fold rotation axis with two-fold axis perpendicular to it n-fold rotation axis with reflection plane containing the axis Improper rotation axis with two-fold axis perpendicular to it Improper rotation axis with reflection planes containing the axis Rotation axis with a perpendicular reflection and a set of reflection planes containing it Symbol* n n n m n2 nm n2 nm mm or n n m m Schoenflies Equivalent n \u2194 Cn n22 \u2194 Dn nmm \u2194 Cnv * n is an integer 1, 2, 4, 6. Reproduced from Quantum Theory of the Solid State by J. Callaway (Academic Press, New York, 1976) with the permission of Elsevier. A.3.3 The Cubic Group Oh The operations of the full cubic group Oh, which is the point group of highest symmetry, may be described as follows: a. The identity E. b. Rotations by \u00b190\u00b0 about a coordinate (four-fold) axis. Class C4, six operations. c. Rotations by 180\u00b0 about", " the same axis. There are six such axes. Class C2, six operations. d. Rotations by \u00b1120\u00b0 about a three-fold axis (body diagonal of the cube). There are four such axes. Class C3, eight operations. e. The inversion with respect to the origin, class J, one operation. f. X Classes JC4, JC4 2, JC2, and JC3: Combinations of the preceding operations with the inversion. These four classes contain 23 operations. The cubic group Oh is described in Table A.3. The operations on a position vector r (with com- ponents x, y, z) can be specified as possible permutations of x, y, z with change of signs. The operations contained in the classes E, C4 subgroup Td are composed of E, C4 2, JC4, and C3. 2, C4, C2, and C3 form the subgroup O. The A class is a set of elements of a group that are conjugate to each other. For example, if X and A are members of a group, the element B = XAX \u20131 is conjugate to A. If two elements A and C are conjugate to a third member D, they are conjugate to each other. Table A.3 The Cubic Group Oh Class E 2 C4 C4 C2 C3 Operation z x y \u2212x \u2212y z y \u2212z z x x \u2212y \u2212z \u2212x \u2212y y \u2212x x \u2212z x z y z \u2212z y z \u2212y y \u2212x y x x \u2212z x z y z \u2212y \u2212x \u2212y \u2212x \u2212z \u2212z \u2212y \u2212x \u2212x \u2212z \u2212x \u2212y \u2212y \u2212z \u2212z \u2212x \u2212y \u2212z y \u2212z \u2212x z \u2212x x \u2212y Class J 2 JC4 JC4 JC2 JC3 Reference 639 Operation \u2212x \u2212y \u2212z y \u2212z x z \u2212x y x \u2212y z y \u2212x \u2212z \u2212y x \u2212z \u2212x z \u2212y y \u2212x \u2212z \u2212z \u2212y z \u2212y \u2212x \u2212y \u2212x y \u2212x \u2212z x \u2212z \u2212z \u2212x \u2212y \u2212y \u2212z \u2212x y x \u2212z y z \u2212x x \u2212y x y \u2212z z \u2212x \u2212y y z x z Reproduced from Quantum Theory of the Solid State by J. Callaway (Academic Press, New York, 1976) with the permission of Elsevier. Reference 1. Callaway J. Quantum theory of the solid state. New York: Academic Press; 1986. Mossbauer Effect APPENDIX B CHAPTER OUTLINE B.1 Introduction.............................................................................. 641 B.2 Recoilless Fraction....................................................................... 642 B.3 Average Transferred Energy................................................................ 643 Reference................................................................................... 644 B.1 INTRODUCTION We will first consider the case of free nucleus at rest of mass M. We assume that the nucleus is initially in an excited state of energy Ex above the ground state. When it emits a \u03b3 ray of energy E\u03b3 (and momentum p = E\u03b3/c, some recoil energy R is transferred to the nucleus. The recoil energy is given by \u00de R = p2/2M = E\u03b3 2/2Mc2: Because R is much smaller than Ex, the energy of the \u03b3 ray is Ex\u00bd Ey = Ex \u2212 R = Ex \u2212 2/2Mc2 \u00de # E\u03b3", " \u00f0 1 \u2212 (B.1) (B.2) Ex/2Mc2 \u00f0 : \u00de% # # 14 keV and R 0:002 eV. However, the natural linewidth \u0393 The typical values are 10 keV< Ex< 100 keV while 2 \u00d7 10\u22124 < R < 5 \u00d7 10\u22122 eV: For example, for 57Fe, 4:6 \u00d7 10\u22129 eV. An energy E\u03b3 + R Ex # must be supplied to a nucleus in its ground state to absorb a \u03b3 ray and make a transition to an excited state. However, because R \u226b \u0393, the emitted \u03b3 rays will not be reabsorbed with appreciable probability. If the radioactive nucleus is rigidly bound in a solid, the macroscopic mass of the solid, instead of the nuclear mass, appears in Eq. (B.1) and hence R = 0: This is an ideal case, because the binding of an atom to a solid is not rigid, and the lattice possesses vibrational degrees of freedom at finite temperature, which can be excited by a displaced nucleus. In general, a fraction of the emission of \u03b3 rays takes place with no perceptible recoil energy, and a width close to the natural width. This is known as the Mossbauer effect because Mossbauer (Ref. 1) had observed in 1958 that when radioactive nuclei, which emit low-energy \u03b3 rays, are bound in a lattice (solid), a considerable portion of the decay occurs without any transfer of energy to the lattice. Thus, the \u03b3 ray spectrum is extremely sharp and the resonant absorption of the \u03b3 rays by unexcited atoms can be easily observed. The Mossbauer effect is a significant tool for determining the hyperfine interactions, i.e., Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00020-7 \u00a9 2012 by Elsevier Inc. All rights reserved. 641 642 APPENDIX B Mossbauer Effect the interactions between a nucleus and the surrounding electrons. The hyperfine interactions involve the product of a nuclear quantity, such as the nuclear magnetic dipole moment and the electric quadrupole moment, as well as an atomic quantity, such as the electron density at the nucleus or the electric field gradient at the nuclear site. This larger fraction f (known as the recoilless fraction and originally observed by Mossbauer) will be calculated in Section B.2. A smaller fraction R, the average energy transferred to the lattice in the emission process, was calculated by Lipkin by using a sum rule. Thus, in a solid, one can find both a sharp \u03b3 ray line unshifted by frequency and a broad shifted background. The relative proportions are temperature dependent, and the width of the sharp line is independent of temperature. The natural width is usually dominant. B.2 RECOILLESS FRACTION The interaction of a single radioactive nucleus bound in a crystal with a radiation field can be written (using a semiclassical theory) as HI = \u2212 e/mp\u00de \u2211 i \u00f0 A xi\u00de \u00f0. pi, (B.3) where mp is the proton mass, xi and pi are the proton coordinate and momentum, A is the vector potential, and the sum includes all the protons in the nucleus. Here, the interaction with the magnetic moment has been neglected. xi\u00de \u00f0 We consider a monatomic lattice and introduce the following notations to simplify the problem. Consider the center of mass of the nucleus located at the position X\u03bd = R\u03bd + u\u03bd, and relative coordinates for the protons in a given nucleus xi = X\u03bd + ri: (B.4) (B.5) The vector potential A \u00f0 a wave vector. When the nucleus makes a transition i \u2192 f, the matrix element for the transition is is proportional to eik\u03bc.xi where the plane electromagnetic wave vector has xi\u00de where MT = < f nf \u00f0 q \u00degj.X\u03bd eik\u03bc q ni\u00f0 jf \u00de > g < f a j \u00f0 i>, k\u03bc\u00dej a k\u03bc\u00de \u00f0 = K\u2211 j eik\u03bc.rjA0. pj, (B.6) (B.7) K is a proportionality constant, and A0 is the amplitude of the radiation field. < f k\u03bc\u00de pendent of the lattice position because a involves all the nuclear physics of the problem. The matrix element for a transition between phonon states is such that the transition rate is i> is inde- k\u03bc\u00dej \u00f0 a \u00f0 j proportional to P nf, ni\u00de \u00f0 = j.X\u03bd eik\u03bc < nf j 2: ni > j j", " (B.8) B.3 Average Transferred Energy 643 The recoilless fraction f is the probability that a transition would occur between phonon states (ni \u2192 nf \u00de without a change in energy of the phonon distribution and can be written as f = \u2211 f < nij j.X\u03bd eik\u03bc nf > j 2\u03b4 j Ef \u2212 Ei\u00de \u00f0 : (B.9) of a phonon distribution at temHowever, Eq. (B.9) should be multiplied by the probability Pni\u00f0 perature T and summing over all states ni: In addition, it can be assumed that the equilibrium position of the radioactive atom is at the origin, i.e., X0 = u0, the displacement of this atom. Incorporating these, we rewrite Eq. (B.9) as T \u00de f = \u2211 if Pni\u00f0 T < nij \u00dej eik\u03bc.u0 nf > j 2 \u03b4 j Ef \u2212 Ei\u00de : \u00f0 (B.10) Because the natural width of the \u03b3 ray line is negligible compared to phonon energies, the transition can be considered to be sharp. If the final state in emission differs from the initial state by energy \u0127\u03c9, Eq. (B.10) can be generalized as f \u03c9 \u00f0 \u00de = \u2211 if Pni\u00f0 T < nij \u00dej eik\u03bc.u0 nf > j 2 \u03b4 j \u00f0 Ef \u2212 Ei \u2212 \u0127\u03c9 : \u00de (B.11) B.3 AVERAGE TRANSFERRED ENERGY The eigenstates of the Hamiltonian H contain a definite number of phonons. Because the kinetic energy operator for the radioactive nucleus, p2/2M, is the only portion of H that does not commute with eik\u03b3.u0, we obtain H, eik\u03bc.u0 \u00bd % = \u00bd p2 /2M, eik\u03bc.u0 = eik\u03bc.u0 % \u01272k\u03bc 2/2M \u0127/M + \u00f0 \u00de \u00de k\u03bc. p : % \u00bd\u00f0 (B.12) Similarly, H,eik\u03bc.u0 \u00bd %, e\u2212ik\u03bc.u0 = 2H \u2212 eik\u03bc.u0He\u2212ik\u03bc.u0 \u2212 e\u2212ik\u03bc.u0 Heik\u03bc.u0 = \u2212 \u01272k\u03bc 2/M: (B.13) From Eq. (B.13), we obtain! \" H, eik\u03bc.u0, e\u2212ik\u03bc.u0 % ni> = 2Ei \u2212 <nij j eik\u03bc.u0 He\u2212ik\u03bc.u0 ni> \u2212 <nij j e\u2212ik\u03bc.u0Heik\u03bc.u0 ni>: j (B.14) By inserting a complete set of final states,, Eq. (B.14) can be written in the alternate form \" <nij\u00bd! <nij\u00bd! H, eik\u03bc.u0, e\u2212ik\u03bc.u0 % j \" j nf ><nf j ni> = 2Ei \u2212 \u2211 <nij f e\u2212ik\u03bc.u0 + \u2211 f <nij = 2Ei \u2212 2\u2211 f Ef j <nij e\u2212ik\u03bc.u0 ni> j eik\u03bc.u0 H j Heik\u03bc.u0 nf ><nf j j nf ><nf j eik\u03bc.u0 nf > j 2: j j ni> j (B.15) (B.16) 644 APPENDIX B Mossbauer Effect We use the identity \u2211 nf j <nij eik\u03bc.u0 nf > j 2 = \u2211 j nf <nij eik\u03bc.u0 nf ><nf j j e\u2212ik\u03bc.u0 ni> = <nij j ni> = 1: From Eqs. (B.8), (B.13), (B.16), and (B.17), we obtain \u2211 f \u00f0 Ef \u2212 Ei\u00de P \u00f0 nf, ni\u00de = \u01272k\u03bc 2/2M: Thus, the average energy transferred to the phonon system is equal to the recoil energy which was the result originally derived by Lipkin. (B.17) (B.18) \u01272k\u03bc2 2M, The advantage of the Mossbauer effect is that one can measure the magnetic fields at the nuclear sites through the Zeeman effect. If a nucleus has spin I, the magnetic moment \u03bc!n = \u03b3n\u0127I (\u03b3n is the nuclear gyromagnetic ratio", ") would interact with an effective magnetic field Heff : The interaction is. Heff, which removes the (2I + 1) degeneracy of the spin orientation. This spin given by Hz = \u2212 \u03bc!n splitting is easily observable. Reference 1. Mossbauer R. Nuclear resonance fluroscence of gamma rays in Ir 191. Zeitschrift. fur Physik 1958;181:124. Introduction to Renormalization Group Approach APPENDIX C CHAPTER OUTLINE C.1 Critical Behavior.......................................................................... 645 C.2 Theory for Scaling........................................................................ 646 C.3 Renormalization Group Approach........................................................... 648 References.................................................................................. 649 C.1 CRITICAL BEHAVIOR We consider a ferromagnet in equilibrium at temperature T and under the action of a uniform magnetic field H. The reduced temperature variable is defined as T \u2212 Tc\u00de where Tc is the critical temperature, and we consider the properties as t \u2192 0 with H = 0. The initial susceptibility in this regime diverges as (C.1) /Tc, t = \u00f0 \u00de # where \u03b3 is known as the critical exponent. For ferromagnets like Fe and Ni, the critical exponent has values near 1.36, which implies that the mean-field, or \u201cclassical,\u201d prediction \u03b3 = 1 is incorrect. The zero-field specific heat can be written as T \u03c70\u00f0 C/t\u03b3, (C.2) or in the alternate form CH = 0\u00f0 T \u00de # A/t\u03b1, CH = 0\u00f0 T \u00de # A \u00f0 t\u2212\u03b1 \u2212 1 /\u03b1, \u00de (C.3) (C.4) where \u03b1 \u2243 \u2212 0:1 for Ni and other isotropic magnets. The scaling theory of critical behavior asserts that the singular part of the free energy F(T, H) varies asymptotically as where the gap exponent T, H f \u00f0 = \u2212 \u00f0 \u00de kBT \u22121Fsing\u00f0 \u00de T, H \u00de # t2\u2212\u03b1Y H/t\u0394, Physics of Condensed Matter. DOI: 10.1016/B978-0-12-384954-0.00021-9 \u00a9 2012 by Elsevier Inc. All rights reserved. (C.5) (C.6) 645 646 APPENDIX C Introduction to Renormalization Group Approach Y( y) is known as the scaling function and depends only on a single variable but is not explicitly given by the theory. From Eq. (C.4), we find that the spontaneous magnetization vanishes when t \u2192 0 as M0\u00f0 where \u03b2 is predicted by the exponent relation T \u03b2C.7) (C.8) \u03b2 0:36 for magnets. Further, the equation of state M = M T, H \u00f0 \u00de # can be written in the scaled form # where W(y) is a single-variable scaling function. M/t\u03b2 H/t\u0394 W \u00f0, \u00de The basic two-point correlation function is defined as G x, T \u00f0 \u00de = < S!0. S!x>, (C.9) (C.10) where S!x denotes a localized spin at site x. The variation of the scattering intensity in the critical region is given by At the critical point, ^G \u00f0 q, T = \u2211 x \u00de eiq.x G x, T \u00f0 \u00de : Gc\u00f0 x ^Gc\u00f0 q", " \u00de # \u00de # Dc/xd\u22122+\u03b7 ^Dc/q2\u2212\u03b7 as x \u2192 \u221e, as q \u2192 0: Here, the critical exponent \u03b7 lies in the range 0.03\u20130.1. In zero field, as t \u2192 0, the form predicted by scaling is \u00de # where \u03be is the correlation length. Eq. (C.13) can be written in the equivalent form x, T G \u00f0 x\u2212d+2\u2212\u03b7D x/\u03be \u00f0, \u00de \u03be\u223ct\u2212\u03bd, The correlation length exponent is given by q, T ^G \u00f0 \u00de # Ct\u2212\u03b3 ^D q2/t2\u03bd : \u00de \u00f0 The scaling function ^D z2 \u00f0 \u00de \u03bd = \u03b3/ \u00f0 represents the scattering \u201cline shape\u201d near Tc. 2 \u2212 \u03b7 \u00de : (C.11) (C.12) (C.13) (C.14) (C.15) C.2 THEORY FOR SCALING Any adequate theory for scaling should (a) show how to calculate the exponents \u03b1, \u03b3, and \u03b7; (b) pre, etc; and dict or justify scaling; (c) lead to explicit calculations of scaling functions Y \u00de, ^D \u00f0,W z2 y y \u00de \u00f0 \u00de \u00f0 C.2 Theory for Scaling 647 (d) describe the corrections of the asymptotic scaling laws and give a concrete estimate of their magnitude. The first step in building a theory is to describe a lattice structure of spacing a, generated by a set of nearest-neighbor vectors \u03b4!. In usual treatments of such problems, we consider the lattice with i = 1, 2, d, where d = 3 and is known as the spatial sites with coordinate vectors x = dimensionality. However, because d enters theoretical calculations in an essential way only through space and momentum (or wave number integrals), the definition of dimensionality can be extended to continuous values of d beyond 3. It turns out that the difference xi\u00de \u00f0 \u2208 = 4 \u2212 d (C.16) is a small but important parameter. The lattice, as constructed above with sites x, is populated with spins S!x. S!x has n components, i.e., Sx = with \u03bc = 1, 2, \u2026, n, which enter equally into inter\u00de actions. Some known examples are as follows: Sx \u00f0 \u03bc a. n = 3, Heisenberg spins: S = b. n = 2, XY or \u201cplanar\u201d spins: S = c. n = 1, uniaxial or Ising spins: S = Sz. Sx, Sy, Sz : \u00de Sx, Sy \u00f0 \u00f0 \u00de : One can show that for systems with short range, isotropic coupling is represented by an interac- tion Hamiltonian of the form s\u2032 xg = Hiso:exch: = \u2212 Hf 1 2 \u2211 x,x\u2032 J x \u2212 x\u2032 \u00de \u00f0 s!x. s!x\u2032: (C.17) The parameters n and d are the only ones that apparently determine the critical exponents. In Eq. (C.17), s! is the normalized spins or local variables \u00f0 It is often convenient to allow the spin length 1/2: \u00de% to vary continuously but introduce a spin weighting function in order to restrict the fluctuations in spin length. The partition function for a system of N spins is s! = S!/ (C.18) S! j S + 1 S j \u00bd where the reduced Hamiltonian, H, is defined as ZN\u00bdH% = TrNf exp H, g H = \u2212 H /kBT + : W \u00de \u00f0 (C.19) (C.20) is a spin weighting function, which is a sum of identical terms \u2212 w S! becomes very large and negative as j for each spin. W becomes large. The specification of H determines the W temperature, the external magnetic field, pressure, and so on, as well as all the interactions that are translationally invariant. The thermodynamics follow from the free energy per spin (the thermodynamic limit N \u2192 \u221e is essential if the critical behavior is to be investigated), s!, H \u00f0 /kBT = lim \u00de \u221e! N N\u22121 ln ZN\u00bd H : % (C.21) 648 APPENDIX C Introduction to Renormalization Group Approach C.3 RENORMALIZATION GROUP APPROACH The exact construction of a renormalization group is in general a very difficult task. The main ideas can be summarized as follows: a. The initial Hamiltonian H is ren", "ormalized to obtain a new Hamiltonian b. The renormalization group operator R reduces the spin variables (the degrees of freedom) from C.22) N to N\u2032 = N/bd, where b (b > 1) is known as the spatial rescaling factor and d is the dimensionality. c. Sometimes R is defined via a partial trace over (N \u2212 N\u2032) of the spin variables so that d. The partition function must be preserved by R, exp H \u2032 = TrN\u2212N\u2032f exp H : g (C.23) (C.24) % e. It is implicitly assumed that the renormalized Hamiltonian H \u2032 displays translational invariance. f. All spatial vectors entering into correlation functions are rescaled by the factor b according to g ZN\u2032\u00bd H \u2032 = ZNf H : (C.25) x \u21d2 x\u2032 = x/b, (C.26) so that the spatial density of degrees of freedom (of spins) is preserved. Thus, the momenta are rescaled by q \u21d2 q\u2032 = bq: (C.27) g. The basic spin fluctuation magnitude is preserved by rescaling the renormalized spin vectors as s!x \u21d2 s!\u2032x\u2032 = s!x/c, (C.28) where the spin rescaling factor c depends on H, i.e., c = c. This step is an essential feature of linear renormalization groups where the local variable is transformed into an equivalent renormalized variable to which it is linearly related. H \u00bd % The renormalization group operator depends on b and c. From Eqs. (C.20) and (C.25), we obtain that the free energy is transformed according to and from Eq. (C.28), we find that the basic spin\u2013spin correlation function transforms as (for a linear renormalization group) f H \u2032 % \u00bd = bd f H \u00bd, % (C.29) x; H G \u00bd % = c2G x/b; H \u2032 \u00bd : % (C.30) Eqs. (C.29) and (C.30) lead to scaling properties. The steps followed in completing the renormalization process are as follows: References 649 a. The transformation of the renormalization group operator R (defined in Eq. C.22) is iterated & by varying the parameters of the initial until one obtains a fixed-point Hamiltonian Hamiltonian, such that H R = & &: (C.31) % b. It can be shown from Eqs. (C.30) and (C.31) that one can obtain a functional equation with the \u00bdH H unique solution and x; G \u00bd & % H \u223c1/x2\u03c9 c& = c & % \u00bdH = b\u2212 \u03c9: Comparison of Eqs. (C.10), (C.12), and (C.33) shows that \u03b7 is determined by the equation c& = b\u2212 d\u22122+\u03b7 \u00f0 \u00de /2: (C.32) (C.33) (C.34) The preceding results are brief outlines only to introduce the reader to the fascinating field of the magnetic phenomena arising from different symmetry, spatial form, and magnitude of the Hamiltonians of magnetic materials. The field of renormalization group, which offers a systematic approach for studying the effect of such terms on critical behavior and distinguishing between them, is much more complex and requires much further reading and solving a host of existing problems before one can use them in practice. References 1. Fisher ME. The renormalization group in the theory of critical behavior. Rev Mod Phys 1974;46:597.2. 2. Wilson KG. Phys Rev B 1972;4:3174,3184. Index Page numbers in italics indicate figures and tables A Abelian, 633, 635 AC Josephson effect, 462 Acoustic mode in diatomic linear chain, 46, 46 Aluminum atom in silicon crystal, 287 Amorphous semiconductors, 296\u2013299 physical properties of, 299 Amorphous solids Band structure of fullerene nanotube, 617, 617 of trans-polyacetylene, 622, 623 Basic two-point correlation function, 646 bcc lattice, see Body-centered cubic lattice BCS theory microscopic theory of superconductivity, 466 strong coupling theory, 472 density of states, 593 energy bands in one-dimensional aperiodic potentials, 591\u2013592 introduction, 590\u2013594 semiconductors, 593\u2013594 Ge and Si, 590, 593 Bergman clusters, 590 Bipolar transistors, 318, 318\u2013319 amplification of current in, 319 Bipolarons, 626 Bloch electrons, 233\u2013234 Amorphous systems, electron energy", " levels of, 297 Anderson model, 439\u2013440 lattice model, 517\u2013518 Anderson\u2013Morel order parameter, 576\u2013577, 577 Angle-resolved photoemission spectroscopy (ARPES), 476, 476\u2013477 Antiferromagnetic (AF) alignment, 344 Antiferromagnetic (AF) materials, 506 Antiferromagnetic configuration, 341 Antiferromagnetic Mott insulator, 618 Antiferromagnetism, spin waves in, 421\u2013422 Antiparallel (AP) configuration, 342 APW method, see Augmented plane-wave method ARPES, see Angle-resolved photoemission spectroscopy Atomic magnetic susceptibility, 371\u2013378 Curie's law, 377\u2013378 formulation, 371\u2013372 Hund's rules, 373\u2013374 Land\u00e9 g factor, 375\u2013376 Larmor diamagnetism, 372\u2013373 Van Vleck paramagnetism, 374\u2013375 Atomic scattering factor, 33\u201334 Augmented plane-wave (APW) method, 150\u2013152, 156 Average transferred energy, 643\u2013644 B Balian\u2013Werthamer (BW) state, 577 Band gaps, extrema of, 582 Band index, 101 Band model, exchange self-energy in, 234\u2013235, 393\u2013394 Band picture, 166 of Ge, 170 of InSb, 171 of Si, 170 effective mass, 247\u2013248 electron\u2013phonon interaction, 264\u2013271 in external fields applied DC field, 253\u2013254 derivation for, 252\u2013253 Lorentz force equation, 248 time evolution of, 250\u2013252 holes electrons and, 257, 257 Fermi surface, 258, 258 negative effective mass, 257 positive effective mass, 256, 256\u2013257 k \u00b7 p perturbation theory, 245\u2013246 quasiclassical dynamics, 246\u2013247 semiclassical model, 243\u2013244 velocity operator, 244\u2013245 Zener breakdown, 258\u2013260 Zener tunneling, calculation of, 261\u2013264 Bloch functions, 99, 132, 591\u2013592 in crystalline lattice, 139 Bloch oscillations, 254, 254\u2013255 Bloch representation, equation of motion in, 388\u2013389 Bloch theorem, 99 effective Hamiltonian, 103 translational symmetry, 103\u2013105 Body-centered cubic (bcc) lattice, 5\u20137 application to, 32\u201333 Brillouin zones, 29\u201330, 30, 119, 119 lattice constants of, 6 polyhedra, 539, 540 primitive cell of, 5, 6 Wigner\u2013Seitz cell of, 6 Boltzmann equation, 183\u2013184 for semiconductors, 313 651 652 Index Bond picture, 167 of GaAs, 171 of InSb, 170 Born\u2013Oppenheimer approximation, 200 Born\u2013von Karman boundary conditions, 77 plane wave solutions, 97\u201398 Bose operators, 417 Bose\u2013Einstein (BE) condensation, 571\u2013572 Bosons, creation and annihilation operators, 54\u201358 Boundary conditions, Ginzburg\u2013Landau theory, 457 Bragg diffraction, 19\u201320 Bragg formula, 582 Bragg law, 19 Bragg planes, 26, 26, 28, 107, 107 Brillouin zones, 117 Bravais lattice, 2, 9\u201311, 137, 634 with basis, 13 and crystal structures, 13\u201314 of diamond, 276 in three dimensions, 4\u201311 in two dimensions, 2, 3, 4\u201311 Brillouin function, 423 Brillouin zones, 45, 134, 137, 276 bcc lattice, 29\u201330, 30, 119, 119 Bragg plane, 117 definition, 27 energies in, 605 fcc lattice, 30\u201331, 31, 119, 119 of graphene, 601 Harrison\u2019s method of construction, 121\u2013124, 122, 123, 124 hcp structure, 119, 119 of one-dimensional lattice, 28, 28 simple cubic lattice, 118, 118 two-dimensional centered rectangular lattice, 117, 118 two-dimensional square lattice, 28\u201329, 29, 117 Buckminsterfullerene (C60), 600 discovery of, 609\u2013613 C C-NMR spectrum of C60, 610, 610 C60 photolithography, 611, 612 Camley\u2013Barnas model, 345, 345\u2013348, 347, 348 Carbon nanotubes, 600, 614 symmetry properties of, 616\u2013617 types of, 614\u2013616 wave functions of, 601 Carrier-mediated magnetism, 291 Ce-based superconducting compounds, 508\u2013509 CeCu2Si2, molar specific heat of, 489 Cellular method, 142\u2013145 Cesium chloride structure, 15, 15\u201316, 173 Chemical probe methods, 549\u2013550, 550 Chirality in graphene, 608 Cholesteric liquid", " crystals, 579 optical properties of, 581\u2013583 CIP, see Current in plane Circuits, integrated, 325 cis polyacetylene, 620 Clusters calculations, 494\u2013497 embedded in matrix, 557\u2013558, 558 growth on surfaces mean-field rate equations, 541\u2013542 Monte Carlo simulations, 540\u2013541 isolated structure DFT, 543 Hartree\u2013Fock method, 542\u2013543 schematic representation of, 546, 547 semi-empirical potentials, 544\u2013546 tight-binding methods, 544 magnetism, 548\u2013549 in embedded clusters, 553\u2013555, 554\u2013555 experimental techniques for, 549\u2013553, 550\u2013552 graphite surfaces, 555, 556 in isolated clusters, 547\u2013549 STM, 555\u2013557, 557 Coherence length, Ginzburg\u2013Landau theory, 457\u2013458 Cohesion of solids, 174\u2013179 Condensate wave function, 572 Conducting polymers electronic structure of, 618 solitons in bipolarons, 626 electronic structure, 623 excitations, 624\u2013626 introduction, 622 polarons and polaron excitations, 626 tight-binding model, 623 Conduction bands, 275, 276 Conductivity electrical, 186\u2013187 thermal, 187\u2013188 weak scattering theory of, 188\u2013192 Conjugated polymers, 618\u2013620 electronic structure of, 623 Constant energy curves, 605, 605 Cooper pairs, 574 microscopic theory of superconductivity, 464, 464\u2013466 Coordination numbers, 5 Correlated super-spin glass (CSSG), 558 Correlation function for liquids, 570 Correlation length exponent, 646 Coulomb blockade system, 324 Index 653 Coulomb interactions, 204\u2013205, 324 graphenes, 600 Covalent crystals, 177\u2013178 CPP, see Current perpendicular to plane Crystal defects, 14 Crystal lattices, 2\u20133 basis, diffraction, 31\u201334 symmetry operations in, 634 Crystal structures, 15\u201318 Bravais lattice and, 13\u201314 packing fraction of, 14 Crystalline solids, categories of, 275 Crystals covalent, 177\u2013178 ionic, 176\u2013177 molecular, 174\u2013176 CSSG, see Correlated super-spin glass Cu superconductor, 516 Cubic group, 638, 639 Cuboctahedral cluster, 528\u2013529, 529 Curie temperature, 410, 422\u2013424 Curie\u2019s law, 377\u2013378, 422 Curie\u2013Weiss law, 441 Current in plane (CIP), 348 Current perpendicular to plane (CPP) geometry, 348 Current perpendicular to plane-GMR, 349, 348\u2013352 of multilayered nanowires, 350\u2013352, 351, 353 theory of, 350\u2013352 Curvature strains, liquid crystals, 581, 582 Cybotactic phase, 579 D 2D graphene sheet, 614, 616 2D nanostructures, 554\u2013555, 554, 555 d-wave pairing, 474\u2013476 experimental confirmation of ARPES, 476, 476\u2013477 flux quantization of superconducting ring, 480 high Tc superconductors, theoretical mechanism of, 481 Josephson tunneling, 477\u2013478, 478 NMR, 477 \u03c0\u2013rings, 478\u2013480, 478, 479 tricrystal magnetometry, 480 symmetry, 480\u2013481 Dangling bonds, 299 amorphous semiconductors, 593 DC electrical conductivity, 86\u201387 DC Josephson effect, 462, 462 de Haas\u2013van Alphen effect, 378, 380, 383\u2013387 Debye frequency, 50 Debye model of specific heat, 49\u201351, 51 Debye temperature, 49, 51 Degenerate semiconductors, 281 Degrees of freedom, phonon and phason, 586\u2013588 Delocalized states of disordered solids, 593 Density functional theory (DFT), 532 LDA, 224\u2013225 Schrodinger\u2019s equation, 223 structure of isolated clusters, 543 universal function, 223 Density of electron states amorphous solids, 592, 593 comparison of, 82, 82 concept of, 81 vs. energy, 81, 82 Density operator Fermi distribution function, 76 grand canonical ensemble, 75 DFT, see Density functional theory Diamagnetic susceptibility, 378, 581 Diamond, 600 Bravais lattice of, 168, 276 structure, 16, 16, 168, 168 Diluted magnetic semiconductors (DMS), 290\u2013296 Dimerization coordinate, 623 Dimerized polyacetylene chain, total energy, 625 Diode laser, 327 efficiency of, 327 Dirac delta function, 192 Dirac equation, 607 Dirac Fermions, graphene", ", 606\u2013608 Dirac points, 602 Direct exchange model, 416\u2013417, 417 DMS, see Diluted magnetic semiconductors Doped semiconductors, 285\u2013287 Doping in electronic polymers, 621, 621 Double-barrier channel, 323 Double-heterostructure laser, 327, 327 Drude model electrical and thermal conductivity, 73 electron density concept, 72 kinetic theory of gas, 73 3s electrons energy levels, 72, 72 scattering of electrons, 71, 72 sodium atoms, 72, 72 valence electrons, 71 Dulong and Petit law, 49, 51 E EAM, see Embedded atom model EDMFT, see Extended Dynamical Mean-Field Theory Effective Hamiltonian, 102\u2013103 Effective mass, 247\u2013248 tensor, 278 654 Index Effective medium theory (EMT), 544 Einstein frequency, 37 Einstein model of specific heat, 52\u201353 Einstein temperature, 52 Elastomers, 618 Electric breakdown, see Zener breakdown Electrical conductivity, 186\u2013187 and thermal conductivity, 73 of trans-(CH), 618, 620 Electron and holes, 278\u2013279 densities in equilibrium, 279\u2013282 in phosphorus impurity, 286 polarization of tunneling, 354 trajectories, 342 Electron paramagnetic resonance (EPR) shift, 291\u2013295 Electron\u2013electron interaction Born\u2013Oppenheimer approximation, 200 density functional theory LDA, 224\u2013225 Schrodinger\u2019s equation, 223 universal function, 223 Fermi liquid theory energy functional, 227\u2013230 Fermi liquid parameters, 230\u2013231 quasiparticles, 225\u2013227 frequency and wave-number-dependent dielectric constant Fermi distribution function, 218 Fourier component, 218 plasma mode, 221 time-independent perturbation theory, 220 Friedel sum rule and oscillations Legendre polynomial, 215 Plane wave, 215 wave function, 214, 216 Green\u2019s function method Bloch electrons, 233\u2013234 exchange self-energy, band model, 234\u2013235 Fourier transformation, 232 Hartree approximation Lindhard theory, 202 Pauli principle, 202 Schrodinger equation, 200 self-consistent field approximation, 202 Thomas\u2013Fermi theory, 202 Hartree\u2013Fock approximation jellium model, 204\u2013207 Pauli principle, 203 spin-up functions, 203 Mott transition, 222 Schrodinger equation, 200 screening effects Fourier transform, 207 Lindhard theory, 209\u2013214 Poisson\u2019s equations, 207 Thomas\u2013Fermi approximation, 208\u2013209 Electronic energy levels for mixed-valence materials, 493 Electronic energy states, graphene, 606 Electronic polymers, 621 Electronic shell structures, 530, 530\u2013531 ellipsoidal shell model, 535 large clusters, 533, 535\u2013536, 536 nonalkali clusters, 535 spherical jellium model harmonic oscillator potential, 531 self-consistent, 532\u2013535 spherical square-well potential, 531\u2013532 Woods\u2013Saxon potential, 532, 533 Electronic structure of conjugated polymers, 623 Electronics, graphene-based, 332 Electron\u2013phonon interaction, 264\u2013271 superconductors, 452 Electrons density concept, 72 in weak periodic potential plane wave solutions, 97\u201399 two-dimensional rectangular lattice, 96, 96 Elementary band theory of solids, 111\u2013112 Ellipsoidal shell model, 535 Embedded atom model (EAM), 544 Empirical pseudopotentials, 157\u2013158 EMT, see Effective medium theory Energy bands, 259, 259 in one-dimensional aperiodic potentials, 591\u2013592 Energy gap, 275\u2013276 EPR shift, see Electron paramagnetic resonance shift Equilibrium, p-n junction in, 308, 307\u2013310 Equivalent Hamiltonian operator, 181 Extended Dynamical Mean-Field Theory (EDMFT), 501\u2013502 Extended zone scheme, 101, 101, 110, 110 External electric fields, Bloch electrons in applied DC field, 253\u2013254 derivation for, 252\u2013253 Lorentz force equation, 248 time evolution of, 250\u2013252 Extrinsic semiconductors, 284\u2013285 F Face-centered cubic (fcc) lattice, 7\u20138 application to, 33 Brillouin zones, 31, 31, 119, 119 Index 655 lattice constants of, 8 polyhedra, 538, 538 primitive cell of, 7, 7 Wigner\u2013Seitz cell of, 7, 8 Faraday\u2019s law of induction, 400 fcc lattice, see Face-centered cubic lattice Fermi distribution, 606 function, 218, 435\u2013436 grand canonical ensemble, 74 Fermi energy, 79, 438, 617 and chemical potential, 82\u201384 Fermi function, 184 Fermi level, 383\u2013384 Fermi liquid", " theory, 503, 574\u2013575 energy functional effective mass, 228, 230 Fermi wave vector, 228 Pauli exclusion principle, 228 Fermi liquid parameters, 230\u2013231 quasiparticles Green\u2019s function, 227 Hartree\u2013Fock approximation, 225 Landau\u2019s argument, 225, 227 law of conservation of energy, 226 N-electron system, 226 Thomas\u2013Fermi screened potential, 226 Fermi liquids fractionalized, 505\u2013506 heavy, 502\u2013505 models, 502\u2013506 parameters, 230\u2013231 Fermi sphere, 79, 79 displacement of, 86, 87 Fermi surfaces, 503, 504 Harrison\u2019s method of construction, 121\u2013124, 122, 123, 124 of metal, 383 in three dimensions, 121, 121, 122 in two dimensions, 119\u2013120, 120 Fermi wave vector, 79, 228 Fermi\u2013Dirac distribution function density operator, 76 ground-state energy of electron gas, 80 specific heat of electron gas, 84\u201386 variation of, 84, 84 Fermi\u2013Dirac distributions, 428 Fermi\u2013Dirac statistics, 573\u2013574 Fermions creation and destruction operators for, 56 energy levels for, 60 Ferromagnetic domains, 425\u2013426 Ferromagnetic (F) configuration, 341 Ferromagnetic ground state, 437\u2013439 Ferromagnetism direct, indirect, and super exchange, 416\u2013417 Heisenberg Hamiltonian, application, 418\u2013420 Heisenberg model, 416 Heitler\u2013London approximation, 412\u2013414 magnons, 417 metallic ions, 412 Schwinger representation, 417\u2013418 in solids Curie temperature, 422\u2013424 ferromagnetic domains, 425\u2013426 hysteresis, 426\u2013427 Ising model, 427 spin Hamiltonian, 414\u2013416 in transition metals Fe, Co, and Ni, 430\u2013431 free electron gas model, 431\u2013432 Hubbard model, 433 magnetic moments of, 428 Stoner model, 428\u2013429 FET, see Field-effect transistor Fibonacci lattice, 585, 585, 587 diffraction pattern of, 586 Field-effect transistor (FET), 319\u2013321 First-order perturbation theory, 209 First-principles pseudopotentials, 158\u2013160 Five-fold symmetry, diffraction pattern of, 584 Flow-tube reactor (FTR), 549 Flux quantization, 459\u2013460, 460 of superconducting ring, 480 FM-T-N junction, theory of, 358\u2013361 Form factor, see Atomic scattering factor Fourier component, 218 Fourier transform, 207, 232 Fourth generation of polymeric materials, 618 Fractionalized Fermi liquids, 505\u2013506 Free-electron model failures of, 89\u201390 Fermi gas Born\u2013von Karman boundary conditions, 77 definition, 77 Fermi energy, 79 Fermi sphere, 79, 79 Fermi wave vector, 79 Schrodinger equation, 77 two-dimensional k-space, points, 78, 78 gas model, 431\u2013432 Frequency dependent dielectric constant Fermi distribution function, 218 Fourier component, 218 656 Index Frequency dependent dielectric constant (Cont.) plasma mode, 221 time-independent perturbation theory, 220 Friedel sum rule and oscillations Legendre polynomial, 215 plane wave, 215 wave function, 214, 216 FTR, see Flow-tube reactor Fulde\u2013Ferrell\u2013Larkin\u2013Ovchinnikov (FFLO) superconducting state, 508 Fullerene derived tubule, 615 Fullerenes carbon nanotubes, 614 symmetry properties of, 616\u2013617 types of, 614\u2013616 nanotube, 613, 613 band structure of, 617, 617 see also Buckminsterfullerene (C60) G GaAs-AlGaAs interface, 323 advantage of, 322 Gauss theorem, 155 Ge, 168\u2013169 band structure of, 168\u2013169, 277, 277 Geometric shell structure close-packing of atoms, 537 polyhedra bcc, 539, 540 fcc, 538, 538 Mackay icosahedron, 539, 539 Wulff construction, 537, 537, 538 Geometrical structure factor, 32 Giant magnetic resonance (GMR), 339, 340\u2013342 effect, 341, 352 metallic multilayers, 340\u2013342, 341\u2013342 Ginzburg\u2013Landau theory boundary conditions, 457 coherence length, 457\u2013458 London penetration depth, 458\u2013459 order parameter, 456\u2013457 GMR, see Giant magnetic resonance Gradient-field deflection method, 551, 551\u2013553, 552 Grand canonical ensemble, 74 density operator, 75\u201377 Fermi distribution function, 74 Graphene, 329, 329, 330 amb", "ipolar electric field effect in, 331 Dirac Fermions, 606\u2013608 introduction, 600\u2013601 lattice, 601\u2013602 tight-binding approximation, 602\u2013606 Graphene-based electronics, 332 Graphite, 600 surfaces, magnetism clusters, 555, 556 Green\u2019s function (KKR) method, 152\u2013156, 154, 554 Bloch electrons, 233\u2013234 exchange self-energy, band model, 234\u2013235 Fourier transformation, 232 Ground-state energy of electron gas, 79\u201381 Group, definition of, 633 Group theory point group operations, 636\u2013639 space groups, 634\u2013636 symmetry of crystals, 633\u2013634 H Hall effect, geometry of, 87, 87 Hall field, 89 Hamiltonian, renormalized group, 648\u2013649 Hamiltonian function, 180, 246 Harmonic approximation, 39 Harmonic oscillator potential, 531 Harmonic spring constant, 623 Harrison\u2019s method of construction Brillouin zones, 121\u2013124, 122, 123, 124 Fermi surface, 121\u2013124, 122, 123, 124 Hartree approximation Lindhard theory, 202 Pauli principle, 202 Schrodinger equation, 200 self-consistent field approximation, 202 Thomas\u2013Fermi theory, 202 Hartree\u2013Fock approximation, 572 exchange term, 203 jellium model Coulomb interaction, 204\u2013205 exchange term, 205 kinetic energy, 204 Kronecker delta function, 206 Lindhard dielectric function, 206 N-electron system, 206 Thomas\u2013Fermi theory, 207 Pauli principle, 203 spin-up functions, 203 Hartree\u2013Fock method, 542\u2013543 hcp structure, see Hexagonal close-packed structure Heat capacity of solid, Einstein theory, 37 Heavy Fermi liquids, 502\u2013505 Heavy-fermions, 488 properties of, 488 superconductivity theories, 516 Index 657 superconductors Ce-based compounds, 508\u2013509 PrOs4Sb12, 513, 514 PuCoGa5, 513, 514 PuRhGa5, 515, 515\u2013516 U-based compounds, 511, 509\u2013513 Heisenberg Hamiltonian, application of, 418\u2013420 Heisenberg model, 416 Heitler\u2013London model, 414 Helmholtz free energy, 377 Heterojunction diode, 613 Hexagonal close-packed (hcp) structure, 17\u201319, 18 Brillouin zones, 119, 119 lattice constants of elements, 19\u201318 unit cell of, 17 Hexagonal systems, 10, 10 Highest occupied shell (HOS), 559 Hilbert transform, 498 Hohenberg\u2013Kohn (HK) theorem, 532 Holstein\u2013Primakoff transformation, 418 Homogeneous semiconductor, see Intrinsic semiconductor Honeycomb lattice structure of graphene, 601, 607 electronic dispersion in, 606, 607 HOS, see Highest occupied shell Houston functions, 252 Hubbard model, 433, 498 Hund\u2019s rules, 373\u2013374 Hydrogen-bonded structures, 174 Hydrogen molecule, covalent bond, 167 Hyperfine interactions, 641 Hysteresis, 426\u2013427, 427 I Icosahedral atomic clusters, 590 Icosahedral group, stereogram of, 584 Icosahedral quasicrystals, 589\u2013590 Impurity band of local defects, 594 Impurity scattering, 189\u2013192 Indirect exchange model, 416\u2013417, 417 Indirect semiconductors, 276 InGaAsP laser, structure of, 327 Injection coefficient, 361\u2013364 Insulators, 112\u2013117, 276, 354 Integrated circuits, 325 Intrinsic semiconductor, 276, 278, 283\u2013284 chemical potential for, 283 types of, 276 Invariant subgroup, 635 Ionic crystals, 176\u2013177 Ionic magnetic susceptibility, see Atomic magnetic susceptibility Ionic solids, 172 Ising model, 427 Isotope effect, superconductors, 454 J Jellium model Coulomb interaction, 204\u2013205 exchange term, 205 kinetic energy, 204 Kronecker delta function, 206 Lindhard dielectric function, 206 N-electron system, 206 Thomas\u2013Fermi theory, 207 Josephson effect AC, 462 DC, 462, 462 two superconductors, oxide layer, 461, 460\u2013462 Josephson tunneling, 477\u2013478, 478!! model, 295\u2013296 K k!! k \u00b7 p perturbation theory, 245\u2013246 Kinetic theory of gas, 73 KKR, see Green\u2019s function method Kohn\u2013Sham equations, 158\u2013159, 534 Kondo effect, 439, 501 Kondo insulators, 516\u2013519 theory of, 517\u2013519 Kondo-", "lattice models, 490\u2013491, 501 Kronecker delta function, 206 L Lagrangian function, 180 Lambda point, 571 Landau diamagnetism, 380\u2013383 Landau Fermi-Liquid (LFL) state, 508 Landau gauge, 379 Landau levels, 384, 384, 386 Landau theory, second-order phase transitions, 441\u2013442 Landau\u2019s argument, 225, 227 Land\u00e9 g factor, 375\u2013376 Langevin susceptibility, see Larmor diamagnetic susceptibility Large clusters, 533, 535\u2013536, 536 Larmor diamagnetic susceptibility, 373 Larmor diamagnetism, 372\u2013373 Laser, 326 diode, 327, 327 double-heterostructure, 327, 327 InGaAsP, structure of, 327 master, 327\u2013328 semiconducting, 327 slave, 327\u2013328 658 Index Lattice planes, 11\u201312, 12, 13 specific heat Debye model of, 49\u201351 Einstein model of, 52\u201353 theory, 48\u201349 structure, graphene, 601\u2013602 translation, 636 Lattice constants, 4\u20135, 44, 61 of bcc lattices, 6 of elements with hcp structure, 18\u201319 of fcc lattices, 8 Lattice dynamics, 37\u201346 normal modes of one-dimensional chain with basis, 44\u201346 of one-dimensional monoatomic lattice, 41\u201344 theory, 37\u201341 Lattice points, 41, 41, 42 group, 3 Lattice waves, quantization of, 61\u201365 Laue condition, alternative formulation of, 25\u201327 Laue method, 20\u201321 Law of conservation of energy, 226 Law of mass action, 282 LCAO, see Linear combination of atomic orbitals method LCH model, see Linear combination of hybrids model LDA, see Local density approximation LDM, see Liquid drop model LED, see Light-emitting diode Legendre polynomial, 215 Leonard\u2013Jones potential, 175, 175 Light dispersion in cholesteric, 582 Light-emitting diode (LED), 326 Lindhard dielectric function, 206 Lindhard theory, 202 Fermi sphere, 213, 213 first-order perturbation theory, 209 Fourier transform, 214 Friedel oscillations, 214 Thomas\u2013Fermi approximation, 211 Linear combination of atomic orbitals (LCAO) method, 134\u2013140 Linear combination of hybrids (LCH) model for tetrahedral semiconductors, 297\u2013299, 299, 300 Liouville\u2019s theorem, 182 Liquid crystals, 568 classes of, 578\u2013580 curvature strains, 581, 582 optical properties of, 581\u2013583 order parameter, 580\u2013581 Liquid drop model (LDM), 528, 529 Liquid 3He Fermi liquid theory, 574\u2013575 liquid\u2013solid phase equilibrium, 573 superfluidity in, 574 experimental results of, 575 theoretical model A and A1 phases, 575\u2013577 B phase, 577\u2013578 Liquids correlation function, 570 phase diagram, 568\u2013569 specific heat vs. temperature, 571, 571 Van Hove pair correlation function, 569 Local density approximation (LDA), 158, 224\u2013225 Local impurity self-consistent approximation (LISA), 498\u2013499 application to periodic Anderson model, 499\u2013500 Localized states in amorphous semiconductors, 593, 594 overlap of tails, 594 London equation, 455\u2013456 London penetration depth, Ginzburg\u2013Landau theory, 458\u2013459 Longitudinal acoustic phonons, 270 Lorentz force equation, 253 Lowest unoccupied shell (LUS), 559 Luttinger theorem, 503, 518 Luttinger\u2013Kohn ( k!! Lyotropic liquid crystal, 578 p!) model, 295 M Mackay icosahedron, 539, 539 filling between complete shells, 540 Macroscopic order parameter, 572 Magnetic breakdown, 252 Magnetic dipole moments, 411 Magnetic ordering Anderson model, 439\u2013440 antiferromagnetism, spin waves in, 421\u2013422 ferrimagnetism, 410 ferromagnetism direct, indirect, and super exchange, 416\u2013417 Heisenberg Hamiltonian, application, 418\u2013420 Heisenberg model, 416 Heitler\u2013London approximation, 412\u2013414 magnons, 417 metallic ions, 412 Schwinger representation, 417\u2013418 in solids, 422\u2013427 spin Hamiltonian, 414\u2013416 in transition metals, 412, 427\u2013433 Kondo effect, 439 magnetic dipole moments, 411 spontaneous magnetization, 410 Index 659 Magnetic permeability, 370 Magnetic phase transition Landau theory, second-order phase transitions of, 441\u2013442 order parameter, 441 Magnetic random access memory (MRAM), 355 principles of, 355 Magnetic susceptibility of atoms", ", see Atomic magnetic susceptibility of Bloch electrons in solids, many-body theory, 388\u2013396 definition, 370 of free electrons in metals, 378\u2013387 de Haas\u2013van Alphen effect, 378, 382\u2013387 formulation, 378\u2013380 Landau diamagnetism, 380\u2013383 Pauli paramagnetism, 380\u2013383 general formula for, 390\u2013393 of nonferromagnetic solid, 388 orbital contribution to exchange and correlation effects, 395\u2013396 spin contribution to exchange enhancement, 394\u2013395 spin-orbit contribution to, 393 exchange and correlation effects on, 396 Magnetic switching, 356\u2013357 Magnetic tunnel junctions (MTJ), 352\u2013356 Magnetism in embedded clusters, 553\u2013555, 554, 555 2D nanostructures, 554, 554\u2013555, 555 small clusters, 554, 554\u2013555, 555 switching and blocking temperature, 553 experimental techniques for cluster chemical probe methods, 549\u2013550, 550 gradient-field deflection method, 551, 552, 551\u2013553 graphite surfaces, 555, 556 in isolated clusters, 547\u2013549 cluster magnetism, 548\u2013549 superparamagnetism and blocking temperature, 547\u2013548 Magnetization of Bloch electrons correlations contribution to, 436\u2013437 quasiparticle contribution to, 435\u2013436 theory of magnetization, 434\u2013435 in zero external magnetic field, 291 Magnetoresistance, 340 effect, 348 Mass spectrum peak of C60, 610 Massless Dirac fermions, 331, 607 Master laser, 327\u2013328 Matrix element, transition in phonon states, 642 Maxwell\u2013Boltzmann distribution function, 80 MBE, see Molecular-beam epitaxy Mean-field approximation, 423 Mean-field rate equations, 541\u2013542 Mean-field theories, 498\u2013502 Meissner\u2013Ochsenfeld effect, 455, 455 Metals, 112\u2013117, 275 cohesion in, 178\u2013179 Metamagnetism in heavy fermions, 506\u2013507 Miller indices, 11\u201312, 13 Mixed-valence compounds, 492\u2013493 Model pseudopotentials, 156\u2013157 for KKR matrix elements, 157 Molecular-beam epitaxy (MBE), 322 Molecular crystals, 174\u2013176 Molecular solids, 174 Molecular structures of conjugated polymers, 619 MBBA, 578 PAA, 578 Monoatomic lattice, 41\u201344 Monoclinic systems, 10\u201311, 11 Monomer, 617 Monte Carlo simulations, 540\u2013541 MOSFET, 320, 611 inversion layer in, 320, 320 n-type, structure of, 319, 319 Mossbauer effect advantage of, 644 average transferred energy, 643\u2013644 definition of, 641 recoilless fraction, 642\u2013643 Mott transition, 140, 222 Mott\u2019s theory of spin-dependent electron scattering, 342\u2013344 MRAM, see Magnetic random access memory MTJ, see Magnetic tunnel junctions Muffin-tin potential, 149\u2013150, 150 N n-doped semiconductor, 285\u2013286 charge movement to metal, 306 donor and acceptor levels for, 287 n-p-n bipolar transistor, 318, 318 n-type MOSFET, structure of, 319, 319 Nanoclusters nanoscience and, 528 superconducting state of qualitative analysis, 558\u2013559 thermodynamic Green\u2019s function, 559\u2013561 Negative effective mass, 257 Negative soliton, 626 Nematic liquid crystals, 578 molecule arrangement in, 579 Newton\u2019s law, 247 NMR, see Nuclear-magnetic-resonance Noble gases, 174\u2013176 660 Index Nonalkali clusters, 535 Nondegenerate semiconductors, 281 Nuclear-magnetic-resonance (NMR), 477 O One-dimensional aperiodic potentials, energy bands, 591\u2013592 One-dimensional electron gas (1DES), 323 One-dimensional quasiperiodic lattice, 585 Optical modes in diatomic linear chain, 46, 46 Optoelectronic devices, 325\u2013329 OPW method, see Orthogonalized plane-wave method Order parameter, 441 Ginzburg\u2013Landau theory, 456\u2013457 liquid crystals, 580\u2013581 for superfluid 3He, 577, 578 Orthogonalized plane-wave (OPW) method, 145\u2013146, 147 Orthorhombic systems, 9, 10 Phase transition, liquid 4He, 570\u2013571 Phason degrees of freedom, 586\u2013588 Phason flips in two-dimensional Penrose lattice, 588, 588 Phason-strained Fibonacci lattice, 588 Phonon-assisted hopping, 593 Phonon-strained Fibonacci lattice, 587, 5", "88 Phonons, 38\u201339, 48 absorption, 265 average energy transfer in, 643\u2013644 degrees of freedom, 586\u2013588 emission, 265 resistivity due to scattering by, 192\u2013194 Phosphorus atom in silicon crystal, 286 Photoconductivity, 325 Photodiode, 326 Photoexcitation, SSH model, 623 Photoinduced electron transfer, semiconducting polymer to P 3p energy bands, 113, 114 p-n junction, 306\u2013310, 307 in equilibrium, 307\u2013310, 308 rectification by, 311\u2013318 equilibrium case, 311\u2013313 nonequilibrium case, 313\u2013318 P-T-H diagram, 575, 576 p-type semiconductor, 285 donor and acceptor levels for, 287 p-wave pairing, 474\u2013476 PAM, see Periodic Anderson model Para-azoxyanisole (PAA), molecular structure, 578 Paramagnetism susceptibility, 378 Van Vleck, 374\u2013375 Pauli exclusion principle, 73, 228, 575 Pauli limiting, 508 Pauli paramagnetism, 380\u2013383 Peierls instability, 622, 622 Peierls theorem, 624 Penrose lattice, 583, 583 perfect edge dislocation in, 589, 589 Penrose tiles, 583 Pentagonal quasicrystal, 586 Periodic Anderson model (PAM), 490\u2013491 application of LISA to, 499\u2013500 Periodic zone scheme, 100\u2013101, 101, 111, 111, 243 Phase diagram of liquid 3He, 575 of liquids, 568, 568\u2013569 in magnetic field, 576 superconductors, 454, 454 C60, 627, 627 Photons, 48 absorption of, 276 Plane wave solutions Born\u2013von Karman boundary conditions, 97\u201398 Schrodinger equation, 97 Plasma frequency of metallic polymers, 621 Plasma mode, 221 Plastic crystals, 568 Point groups cubic group, 638, 639 description of, 636\u2013637 symbols, international notation for, 638 Poisson\u2019s equations, 207 Polarons, 626 excitations, 626 Polyacetylene structure, 618 Polyaniline (PANI) as conducting polymer, 618 Polycritical point (PCP), 575 Polyhedra bcc, 539, 540 fcc, 538, 538 Mackay icosahedron, 539, 539 Polymers, 617\u2013618 conducting polymers, solitons in bipolarons, 626 electronic structure, 623 excitations, 624\u2013626 introduction, 622 polarons and polaron excitations, 626 Polyparaphenylene (PPP) bipolaron in, 627 polaron in, 626 Positive effective mass, 257 Positive soliton, 626 Primitive cell, 2, 3, 3 of bcc lattice, 5, 6 of fcc lattice, 7, 7 of rectangular lattice, 3 of two-dimensional square Bravais lattice, 2 \u03c0\u2013rings, 478, 478\u2013480, 479 PrOs4Sb12, 513, 514 Pseudogap, 593 Pseudopotentials, 147\u2013149 empirical, 157\u2013158 first-principles, 158\u2013160 model, 156\u2013157 Pseudospin, 608 Pu superconductor, 516 PuCoGa5, 513, 514 PuRhGa5, 515, 515\u2013516 Q QED, see Quantum electrodynamics QHE, see Quantum Hall effect Quantization of lattice waves, 65\u201366 formulation, 61\u201365 Quantized Hall resistance, 397, 400, 400 Quantum electrodynamics (QED), 331, 608 chirality, 332 Quantum Hall effect (QHE) fractional, 400, 400 from gauge invariance, 400 in two-dimensional electron gas, 396\u2013397 in strong magnetic field, 397\u2013399 Index 661 R Reciprocal lattice definition, 21\u201322 properties of, 22\u201325 Recoil energy, 641 Recoilless fraction, 642\u2013643 Rectification in heterojunctions, 611 by p-n junction, 311\u2013318 Reduced zone scheme, 100, 100, 109\u2013110, 110 Relaxation time, 188 approximation, 184\u2013185 Renormalization group approach, 648\u2013649 Repeated zone scheme, 100\u2013101, 101, 111, 111 Rhombohedral systems, 10, 10 Riseborough\u2019s theory, 518\u2013519 Rudderman\u2013Kittel\u2013Kasuya\u2013Yosida (RKKY) interactions, 490, 500\u2013501 Russsell\u2013Saunders Coupling, 373 S 3s electrons energy levels, 72, 72 3s energy bands, 113, 114 s-wave pairing, 474\u2013476 Saturated polymers, 618\u2013", "620 Saturation magnetization, 425 sc lattice, see Simple cubic lattice Scaling theory, 646\u2013647 of critical behavior, 645 Scanning tunneling microscope (STM), 555\u2013557, 557 Scattering Quasi-electrons, microscopic theory of superconductivity, 463\u2013464 Quasiclassical dynamics, 246\u2013247 Quasicrystals of electrons, 71, 72 impurity, 189\u2013192 probability, 188 resistivity due to, 192\u2013194 discovery of, 584 icosahedral quasicrystals, 589\u2013590 Penrose tiles, 583 perfect edge dislocation in the Penrose lattice, 589, 589 phonon and phason degrees of freedom, 586\u2013588 quasiperiodic lattice, 584\u2013586 see also Liquid crystals Quasiparticles Green\u2019s function, 227 Hartree\u2013Fock approximation, 225 Landau\u2019s argument, 225, 227 law of conservation of energy, 226 N-electron system, 226 Thomas\u2013Fermi screened potential, 226 Quasiperiodic lattice, 584\u2013586 Schoenflies notation, point groups, 636, 637 Schottky diode, 305 rectifying effect of, 306 Schrieffer-Heeger (SSH) model, 623 Schrodinger equations, 77, 132, 135, 200, 223, 244 effective Hamiltonian, 102 Josephson effect, 461 plane wave solutions, 98 Schwinger representation, 417\u2013418 SDL, see Spin diffusion length Second-order perturbation theory, 245 Second-order phase transition, superconductors, 454, 454 Second quantization, 53\u201360 creation and annihilation operators, 54\u201358 field operators and Hamiltonian, 58\u201360 occupation number representation, 53\u201354 662 Index Semi-empirical potentials, 544\u2013546 Semiclassical model, 179\u2013182, 243\u2013244 Semiclassical theory, 642 Semiconductors, 112\u2013117 amorphous, 296\u2013299 physical properties of, 299 Boltzmann equation for, 313 degenerate, 281 DMS, 290\u2013296 doped, 285\u2013287 elements, 276 extrinsic, 284\u2013285 III-V zincblende, 277 III\u2013V semiconductors, 277 indirect, 276 intrinsic, see Intrinsic semiconductor n-type, 285\u2013286 donor and acceptor levels for, 287 p-type, 285 donor and acceptor levels for, 287 spintronics with, 357\u2013364 tetrahedral, LCH model, 297\u2013299, 299, 300 work function of, 305 zinc-blende, 170 Semimetals, 115 SET, see Single-electron transistor SHE, see Spin Hall effect Short circuit effect, 342, 342 Si, 168\u2013169 band structure of, 170 tetrahedral bonding, 167 Simple cubic (sc) lattice, 4, 13 Brillouin zones, 118, 118 Single-electron transistor (SET), 321\u2013325 Slater\u2013Koster hopping integrals, 544 Slave boson method, 493 Slave laser, 327\u2013328 Small clusters, 554, 554\u2013555, 555 Smetic liquid crystals, 579 molecule arrangement, 579 Sodium chloride, 172 structure, 15, 15 Sodium metal in Wigner\u2013Seitz method, 143 Solids cohesion of, 174\u2013179 ionic, 172 molecular, 174 Solitons in conducting polymers bipolarons, 626 electronic structure, 623 excitations, 624\u2013626 introduction, 622 polarons and polaron excitations, 626 tight-binding model, 623 Sommerfeld model density of electron states comparison of, 82, 82 concept of, 81 vs. energy, 81\u201382, 82 density operator, 75\u201377 Fermi distribution function, 74 free-electron Fermi gas Born\u2013von Karman boundary conditions, 77 definition, 77 Fermi energy, 79 Fermi sphere, 79, 79 Fermi wave vector, 79 Schrodinger equation, 77 two-dimensional k-space, points, 78, 78 ground-state energy of electron gas, 79\u201381 Pauli exclusion principle, 73 Space groups operations, 634\u2013636 see also Point groups Spatial dimensionality, 647 Specific heat Debye model of, 49\u201351, 51 Einstein model of, 52\u201353 of electron gas, 84\u201386 measurements of liquid 3He, 575 Spherical jellium model harmonic oscillator potential, 531 self-consistent, 532\u2013535 spherical square-well potential, 531\u2013532 Woods\u2013Saxon potential, 532, 533 Spherical square-well potential, 531\u2013532 Spin accumulation in ferromagnetic and nonmagnetic layer, 353 Spin-dependent electron scattering, 341 Mott\u2019s theory of, 342\u2013344 Spin diffusion length (SDL), nonmagnetic and", " ferromagnetic layers, 350, 352 Spin fluctuation effect, 577 Spin Hall effect (SHE), 362 Spin Hamiltonian, 414\u2013416 Spin-orbit interaction, 362 Spin-orbit splitting, modulation of, 358 Spin states, 421 Spin-transfer, 357 torques, 356\u2013357, 357 Spin transistor, 358 Spin-triplet pair-wave function, 576 Spin-up functions, 203 Spin-valve system, 344 Index 663 Spin weighting function, 647 Spintronics, 339 with semiconductors, 357\u2013364 Squeezed-vacuum-state generation, 328 Stabilized jellium model, 535 Step function, 192 Stern-Gerlach experiment, 551, 551\u2013553, 552 STM, see Scanning tunneling microscope Stoner model, 428\u2013429 Strong coupling theory, 472\u2013473 Super exchange model, 416\u2013417, 417 Superconducting compounds Ce-based, 508\u2013509 U-based, 509\u2013513, 511 Superconductivity, microscopic theory BCS theory, 466 Cooper pairs, 464, 464\u2013466 excited states at T = 0, 469\u2013470, 470 excited states at T \u2260 0, 470\u2013472, 471 quasi-electrons, 463\u2013464 superconducting electron gas, ground state, 466\u2013469 Superconductors high-temperature ARPES, 476, 476\u2013477 atomic structure of, 473, 473 d-wave pairing, 474\u2013476, 480\u2013481 flux quantization of superconducting ring, 480 Josephson tunneling, 477\u2013478, 478 NMR, 477 p-wave pairing, 474\u2013476 \u03c0\u2013rings, 478\u2013480, 478, 479 properties of, 474, 474 s-wave pairing, 474\u2013476 theoretical mechanism of, 481 tricrystal magnetometry, 480 properties of electron\u2013phonon interaction, 452 isotope effect, 454 phase diagram, 454, 454 resistivity, 452, 452\u2013453, 453 second-order phase transition, 454, 454 type I superconductors, 453, 453 type II superconductors, 454, 454 Superfluid component, 571 Superfluid 4He phase transition, 570\u2013571 theory of superfluidity, 571\u2013573 two-fluid model, 571 Superfluidity in liquid 3He, 574 Superparamagnetism, 547\u2013548 Surface effects, 14\u201315 Symmetry of crystals, 633, 634 T Temperature variable, reduced, 645 Tetragonal systems, 9, 9 Tetrahedral bonding, 167 of Si, 169 Tetrahedral semiconductors for LCH model, 297\u2013299, 299, 300 Theory of magnetization, 434\u2013435 Theory of superfluidity, liquid 4He, 571\u2013573 Thermal conductivity, 187\u2013188 Thermal equilibrium acceptor levels, 288\u2013289 donor levels, 288 impurity levels in, 288\u2013290 Thermodynamic Green\u2019s function, 559\u2013561 Thermodynamic potential, 390 Thermotropic liquid crystal, 578 Thomas\u2013Fermi approximation, 208\u2013209, 323 Tight-binding approximation, 131\u2013134 graphene, 602\u2013606 Tight-binding methods, 544 Tight-binding model, 623 Time-independent perturbation theory, 220 TMR, see Tunneling magnetoresistance trans-polyacetylene, 618, 620, 622 Transistors bipolar, 318, 318\u2013319 amplification of current in, 319 FET, 319\u2013321 SET, 321\u2013325 Transparent metallic polymers, 621 Transverse acoustical (TA) mode for diatomic linear lattice, 47 displacement direction of ions in, 47 Transverse optical (TO) mode for diatomic linear lattice, 47 displacement direction of ions in, 47 Transverse phonons, 270 Triclinic systems, 10\u201311, 11 Tricrystal magnetometry, 480 Trigonal systems, 10, 10 Truncated icosahedron, Buckminsterfullerene C60, 609, 609 Tunneling magnetoresistance (TMR), 352\u2013356 Two-fluid model, liquid 4He, 571 Type I superconductors, 453, 453 Type II superconductors, 454, 454 U U-based superconducting compounds, 509\u2013513, 511 Umklapp process, 250, 270 664 Index Unit cell, 3, 3 of cubic lattices, 4, 5 of hcp structure, 17 of monolayer graphene, 608 of rectangular lattice, 3 Universal function, 223 V Valence bands, 276, 276 Valence electrons, 71 van der Waals interaction, 174 Van Hove pair correlation function, 569 Van Vleck paramagnetism", ", 374\u2013375 Variable-range hopping, 594 Velocity operator, 244\u2013245 V\u2013F theory, 350 Volume filling fraction (VFF), 558 Vortex, 573 W Wannier functions, 140\u2013141, 142 Wave functions, 214, 216, 413 carbon, 601 Wave-number-dependent dielectric constant Fermi distribution function, 218 Fourier component, 218 plasma mode, 221 time-independent perturbation theory, 220 Weak scattering theory of conductivity, 188\u2013192 Weiss Field Model and Spin-wave theory, 424\u2013425 Wiedemann\u2013Franz law, 188 Wigner\u2013Eckart theorem, 375 Wigner\u2013Seitz cell, 3\u20134 of bcc Bravais lattice, 6 of fcc Bravais lattice, 7, 8 of reciprocal lattice, 27 of sodium metal, 142 of two-dimensional Bravais lattice, 3 Wilson ratio, 488 WKB approximation, 259\u2013260 Woods\u2013Saxon potential, 532, 533 Wulff construction, 537, 537, 538 Z Zeeman effect, 644 Zener breakdown, 252, 258\u2013260 Zener tunneling, calculation of, 261\u2013264 Zero-field specific heat, reduced, 645 Zero-gap semiconductor, graphene as, 607 Zinc-blende semiconductors, 170 structure, 17, 17 Zinc oxide (ZnO), 296 Zinc sulphide bond, 171 Zone boundary, 105\u2013109tellite\u2014 errors are not easy to correct later. Satellites, such as the Hubble Space Telescope shown here, have revolutionized scientific research, as well as communications. Think About This Physics research has led to many new technologies, such as satellite-based telescopes and communications. What are some other examples of tools developed from physics research in the last 50 years? physicspp.com 2 NASA 002-010 CH01-S1-845813 3/30/04 1:04 AM Page 3 Do all objects fall at the same rate? Question How does weight affect the rate at which an object falls? Procedure Analysis According to Aristotle, what should be the rate of fall of the single penny compared to the stack? What did you observe? Critical Thinking Explain which of the following properties might affect the rate of fall of an object: size, mass, weight, color, shape. The writings of the Greek philosopher Aristotle included works on physical science theories. These were a major influence in the late Middle Ages. Aristotle reasoned that weight is a factor governing the speed of fall of a dropped object, and that the rate of fall must increase in proportion to the weight of the object. 1. Tape four pennies together in a stack. 2. Place the stack of pennies on your hand and place a single penny beside them. 3. Observe Which is heaviest and pushes down on your hand the most? 4. Observe Drop the two at the same time and observe their motions. 1.1 Mathematics and Physics What do you think of when you see the word physics? Many people picture a chalkboard covered with formulas and mathematics: E mc2, I V/R, d 1 at2 v0t d0. Perhaps you picture scientists in 2 white lab coats, or well-known figures such as Marie Curie and Albert Einstein. Or, you might think of the many modern technologies created with physics, such as weather satellites, laptop computers, or lasers. What is Physics? Physics is a branch of science that involves the study of the physical world: energy, matter, and how they are related. Physicists investigate the motions of electrons and rockets, the energy in sound waves and electric circuits, the structure of the proton and of the universe. The goal of this course is to help you understand the physical world. People who study physics go on to many different careers. Some become scientists at universities and colleges, at industries, or in research institutes. Others go into related fields, such as astronomy, engineering, computer science, teaching, or medicine. Still others use the problem-solving skills of physics to work in business, finance, or other very different disciplines. Objectives \u2022 Demonstrate scientific methods. \u2022 Use the metric system. \u2022 Evaluate answers using dimensional analysis. \u2022 Perform arithmetic operations using scientific notation. Vocabulary physics dimensional analysis significant digits scientific method hypothesis scientific law scientific theory Section 1.1 Mathematics and Physics 3 Horizons Companies 002-010 CH01-S1-845813 3/30/04 2:08 PM Page 4 \u25a0 Figure 1-1 Physicists use mathematics to represent many different phenomena\u2014a trait sometimes spoofed in cartoons. \u00a91998 Bill Amend/Dist. by Universal Press Syndicate Mathematics in Physics Physics uses mathematics as a powerful language. As illustrated in Figure 1-1, this use of mathematics often is spoofed in cartoons. In physics, equations are important tools for modeling observations and for making predictions. Physicists rely on theories and experiments with numerical results to support", " their conclusions. For example, think back to the Launch Lab. You can predict that if you drop a penny, it will fall. But how fast? Different models of falling objects give different answers to how the speed of the object changes, or on what the speed depends, or which objects will fall. By measuring how an object falls, you can compare the experimental data with the results predicted by different models. This tests the models, allowing you to pick the best one, or to develop a new model. Electric Current The potential difference, or voltage, across a circuit equals the current multiplied by the resistance in the circuit. That is, V (volts) I (amperes) R (ohms). What is the resistance of a lightbulb that has a 0.75 amperes current when plugged into a 120-volt outlet? 1 Analyze the Problem \u2022 Rewrite the equation. \u2022 Substitute values. Known: I 0.75 amperes V 120 volts Unknown: R? 2 Solve for the Unknown Rewrite the equation so the unknown is alone on the left. Reflexive property of equality V IR IR V R V I vo ts l 20 1 Substitute 120 volts for V, 0.75 amperes for I. res e p am 5 0.7 160 ohms Resistance will be measured in ohms. Divide both sides by I. Math Handbook Isolating a Variable page 845 3 Evaluate the Answer \u2022 Are the units correct? 1 volt 1 ampere-ohm, so the answer in volts/ampere is in ohms, as expected. \u2022 Does the answer make sense? 120 is divided by a number a little less than 1, so the answer should be a little more than 120. 4 FOXTROT \u00a9 1998 Bill Amend. Reprinted with permission of UNIVERSAL PRESS SYNDICATE. All rights reserved. Chapter 1 A Physics Toolkit 002-010 CH01-S1-845813 3/30/04 1:06 AM Page 5 For each problem, give the rewritten equation you would use and the answer. 1. A lightbulb with a resistance of 50.0 ohms is used in a circuit with a 9.0-volt battery. What is the current through the bulb? 2. An object with uniform acceleration a, starting from rest, will reach a speed of v in time t according to the formula v at. What is the acceleration of a bicyclist who accelerates from rest to 7 m/s in 4 s? 3. How long will it take a scooter accelerating at 0.400 m/s2 to go from rest to a speed of 4.00 m/s? 4. The pressure on a surface is equal to the force divided by the area: P F/A. A 53-kg woman exerts a force (weight) of 520 Newtons. If the pressure exerted on the floor is 32,500 N/m2, what is the area of the soles of her shoes? \u25a0 Figure 1-2 What is a reasonable range of values for the speed of an automobile? Does it make sense? Sometimes you will work with unfamiliar units, as in Example Problem 1, and you will need to use estimation to check that your answer makes sense mathematically. At other times you can check that an answer matches your experience, as shown in Figure 1-2. When you work with falling objects, for example, check that the time you calculate an object will take to fall matches your experience\u2014a copper ball dropping 5 m in 0.002 s, or in 17 s, doesn\u2019t make sense. The Math Handbook in the back of this book contains many useful explanations and examples. Refer to it as needed. SI Units To communicate results, it is helpful to use units that everyone understands. The worldwide scientific community and most countries currently use an adaptation of the metric system to state measurements. The Syst\u00e8me International d\u2019Unit\u00e9s, or SI, uses seven base quantities, which are shown in Table 1-1. These base quantities were originally defined in terms of direct measurements. Other units, called derived units, are created by combining the base units in various ways. For example, energy is measured in joules, where 1 joule equals one kilogram-meter squared per second squared, or 1 J 1 kgm2/s2. Electric charge is measured in coulombs, where 1 C 1 As. Table 1-1 SI Base Units Base Quantity Base Unit Symbol Length Mass Time Temperature Amount of a substance Electric current Luminous intensity meter kilogram second kelvin mole ampere candela m kg s K mol A cd Section 1.1 Mathematics and Physics 5 Mark E. Gibson 002-010 CH01-S1-845813 3/30/04 1:07 AM Page 6 Scientific institutions have been created to define and regulate measures. The SI system is regulated by the International Bureau of Weights and Measures in S\u00e8", "vres, France. This bureau and the National Institute of Science and Technology (NIST) in Gaithersburg, Maryland keep the standards of length, time, and mass against which our metersticks, clocks, and balances are calibrated. Examples of two standards are shown in Figure 1-3. NIST works on many problems of measurement, including industrial and research applications. You probably learned in math class that it is much easier to convert meters to kilometers than feet to miles. The ease of switching between units is another feature of the metric system. To convert between SI units, multiply or divide by the appropriate power of 10. Prefixes are used to change SI units by powers of 10, as shown in Table 1-2. You often will encounter these prefixes in daily life, as in, for example, milligrams, nanoseconds, and gigabytes. Table 1-2 Prefixes Used with SI Units Prefix Symbol Multiplier Scientific Notation Example femto- pico- nano- micro- milli- centi- deci- kilo- mega- giga- tera.000000000000001 0.000000000001 0.000000001 0.000001 0.001 0.01 0.1 1000 1,000,000 1,000,000,000 1,000,000,000,000 1015 1012 109 106 103 102 101 103 106 109 1012 femtosecond (fs) picometer (pm) nanometer (nm) microgram (g) milliamps (mA) centimeter (cm) deciliter (dL) kilometer (km) megagram (Mg) gigameter (Gm) terahertz (THz) Dimensional Analysis You can use units to check your work. You often will need to use different versions of a formula, or use a string of formulas, to solve a physics problem. To check that you have set up a problem correctly, write out the equation or set of equations you plan to use. Before performing calculations, check that the answer will be in the expected units, as shown in step 3 of Example Problem 1. For example, if you are finding a speed and you see that your answer will be measured in s/m or m/s2, you know you have made an error in setting up the problem. This method of treating the units as algebraic quantities, which can be cancelled, is called dimensional analysis. Dimensional analysis also is used in choosing conversion factors. A conversion factor is a multiplier equal to 1. For example, because 1 kg 1000 g, you can construct the following conversion factors: k g 1 1 g 0 00 1 g 0 00 1 1 k g 1 \u25a0 Figure 1-3 The standards for the kilogram and meter are shown. The International Prototype Meter originally was measured as the distance between two marks on a platinum-iridium bar, but as methods of measuring time became more precise than those for measuring length, the meter came to be defined as the distance traveled by light in a vacuum in 1/299 792 458 s. Math Handbook Dimensional Analysis page 847 6 National Institute of Standards and Technology Chapter 1 A Physics Toolkit 002-010 CH01-S1-845813 3/30/04 1:07 AM Page 7 Choose a conversion factor that will make the units cancel, leaving the answer in the correct units. For example, to convert 1.34 kg of iron ore to grams, do as shown below. 1.34 kg1 g 1340 g 0 00 g k 1 You also might need to do a series of conversions. To convert 43 km/h to m/s, do the following: 43 1 m1 m 1 60 0 00 in n 12 m/s Use dimensional analysis to check your equation before multiplying. 5. How many megahertz is 750 kilohertz? 6. Convert 5021 centimeters to kilometers. 7. How many seconds are in a leap year? 8. Convert the speed 5.30 m/s to km/h. Significant Digits Suppose you use a meterstick to measure a pen, and you find that the end of the pen is just past 14.3 cm. This measurement has three valid digits: two you are sure of, and one you estimated. The valid digits in a measurement are called significant digits. The last digit given for any measurement is the uncertain digit. All nonzero digits in a measurement are significant. Are all zeros significant? No. For example, in the measurement 0.0860 m, the first two zeros serve only to locate the decimal point and are not significant. The last zero, however, is the estimated digit and is significant. The measurement 172,000 m could have 3, 4, 5, or 6 significant digits. This ambiguity is one reason to use scientific notation: it is clear that the measurement 1.7200105 m has five significant digits. Arithmetic with significant digits When you perform any arithmetic operation, it is important", " to remember that the result never can be more precise than the least-precise measurement. To add or subtract measurements, first perform the operation, then round off the result to correspond to the least-precise value involved. For example, 3.86 m 2.4 m 6.3 m because the least-precise measure is to one-tenth of a meter. To multiply or divide measurements, perform the calculation and then round to the same number of significant digits as the least-precise measurement. For example, 409.2 km/11.4 L 35.9 km/L, because the least-precise measure has three significant digits. Some calculators display several additional digits, as shown in Figure 1-4, while others round at different points. Be sure to record your answers with the correct number of digits. Note that significant digits are considered only when calculating with measurements. There is no uncertainty associated with counting (4 washers) or exact conversion factors (24 hours in 1 day). Math Handbook Significant Digits pages 833\u2014836 \u25a0 Figure 1-4 This answer to 3.9 7.2 should be rounded to two significant digits. Section 1.1 Mathematics and Physics 7 Horizons Companies 002-010 CH01-S1-845813 3/30/04 1:08 AM Page 8 Solve the following problems. 9. a. 6.201 cm 7.4 cm 0.68 cm 12.0 cm b. 1.6 km 1.62 m 1200 cm 10. a. 10.8 g 8.264 g b. 4.75 m 0.4168 m 11. a. 139 cm 2.3 cm b. 3.2145 km 4.23 km 12. a. 13.78 g 11.3 mL b. 18.21 g 4.4 cm3 Measuring Change Collect five identical washers and a spring that will stretch measurably when one washer is suspended from it. 1. Measure the length of the spring with zero, one, two, and three washers suspended from it. 2. Graph the length of the spring versus the mass. 3. Predict the length of the spring with four and five washers. 4. Test your prediction. Analyze and Conclude 5. Describe the shape of the graph. How did you use it to predict the two new lengths? Scientific Methods In physics class, you will make observations, do experiments, and create models or theories to try to explain your results or predict new answers, as shown in Figure 1-5. This is the essence of a scientific method. All scientists, including physicists, obtain data, make predictions, and create compelling explanations that quantitatively describe many different phenomena. The experiments and results must be reproducible; that is, other scientists must be able to recreate the experiment and obtain similar data. Written, oral, and mathematical communication skills are vital to every scientist. A scientist often works with an idea that can be worded as a hypothesis, which is an educated guess about how variables are related. How can the hypothesis be tested? Scientists conduct experiments, take measurements, and identify what variables are important and how they are related. For example, you might find that the speed of sound depends on the medium through which sound travels, but not on the loudness of the sound. You can then predict the speed of sound in a new medium and test your results. a b \u25a0 Figure 1-5 These students are conducting an experiment to determine how much power they produce climbing the stairs (a). They use their data to predict how long it would take an engine with the same power to lift a different load (b). 8 Laura Sifferlin Chapter 1 A Physics Toolkit 002-010 CH01-S1-845813 3/30/04 1:10 AM Page 9 \u25a0 Figure 1-6 In the mid-1960s, Arno Penzias and Robert Wilson were trying to eliminate the constant background noise in an antenna to be used for radio astronomy. They tested systems, duct-taped seams, and cleared out pigeon manure, but the noise persisted. This noise is now understood to be the cosmic microwave background radiation, and is experimental support for the Big Bang theory. \u25a0 Figure 1-7 Drawings of early telescope observations (a) showed channels on Mars; recent photos taken with improved telescopes do not (b). In this photo of Mars\u2019 surface from the Mars Global Surveyor spacecraft (c), these layered sedimentary rocks suggest that sedimentary deposits might have formed in standing water. Models, laws, and theories An idea, equation, structure, or system can model the phenomenon you are trying to explain. Scientific models are based on experimentation. Recall from chemistry class the different models of the atom that were in use over time\u2014new models were developed to explain new observations and measurements. If new data do not fit a model, both are re-examined. Figure 1-6 shows a historical example. If a very well-", "established model is questioned, physicists might first look at the new data: can anyone reproduce the results? Were there other variables at work? If the new data are born out by subsequent experiments, the theories have to change to reflect the new findings. For example, in the nineteenth century it was believed that linear markings on Mars showed channels, as shown in Figure 1-7a. As telescopes improved, scientists realized that there were no such markings, as shown in Figure 1-7b. In recent times, again with better instruments, scientists have found features that suggest Mars once had running and standing water on its surface, as shown in Figure 1-7c. Each new discovery has raised new questions and areas for exploration. A scientific law is a rule of nature that sums up related observations to describe a pattern in nature. For example, the law of conservation of charge states that in the various changes matter can undergo, the electric charge before and after stays the same. The law of reflection states that the angle of incidence for a light beam on a reflective surface equals the angle of reflection. Notice that the laws do not explain why these phenomena happen, they simply describe them. a b c Section 1.1 Mathematics and Physics 9 (t)Tom Pantages, (bc)NASA, (others)Photo Researchers 002-010 CH01-S1-845813 3/30/04 1:10 AM Page 10 \u25a0 Figure 1-8 Theories are changed and modified as new experiments provide insight and new observations are made. The theory of falling objects has undergone many revisions. Greek philosophers proposed that objects fall because they seek their natural places. The more massive the object, the faster it falls. Revision Galileo showed that the speed at which an object falls depends on the amount of time it falls, not on its mass. Revision Galileo\u2019s statement is true, but Newton revised the reason why objects fall. Newton proposed that objects fall because the object and Earth are attracted by a force. Newton also stated that there is a force of attraction between any two objects with mass. Revision Galileo\u2019s and Newton\u2019s statements still hold true. However, Einstein suggested that the force of attraction between two objects is due to mass causing the space around it to curve. A scientific theory is an explanation based on many observations supported by experimental results. Theories may serve as explanations for laws. A theory is the best available explanation of why things work as they do. For example, the theory of universal gravitation states that all the mass in the universe is attracted to other mass. Laws and theories may be revised or discarded over time, as shown in Figure 1-8. Notice that this use of the word theory is different from the common use, as in \u201cI have a theory about why it takes longer to get to school on Fridays.\u201d In scientific use, only a very well-supported explanation is called a theory. 1.1 Section Review 13. Math Why are concepts in physics described with formulas? 14. Magnetism The force of a magnetic field on a charged, moving particle is given by F Bqv, where F is the force in kgm/s2, q is the charge in As, and v is the speed in m/s. B is the strength of the magnetic field, measured in teslas, T. What is 1 tesla described in base units? 15. Magnetism A proton with charge 1.601019 As is moving at 2.4105 m/s through a magnetic field of 4.5 T. You want to find the force on the proton. a. Substitute the values into the equation you will use. Are the units correct? b. The values are written in scientific notation, m10n. Calculate the 10n part of the equation to estimate the size of the answer. c. Calculate your answer. Check it against your estimate from part b. d. Justify the number of significant digits in your answer. 16. Magnetism Rewrite F Bqv to find v in terms of F, q, and B. 17. Critical Thinking An accepted value for the acceleration due to gravity is 9.801 m/s2. In an experiment with pendulums, you calculate that the value is 9.4 m/s2. Should the accepted value be tossed out to accommodate your new finding? Explain. 10 Chapter 1 A Physics Toolkit physicspp.com/self_check_quiz 011-014 CH01-S2-845813 3/30/04 1:20 AM Page 11 Objectives \u2022 Distinguish between accuracy and precision. \u2022 Determine the precision of measured quantities. Vocabulary measurement precision accuracy 1.2 Measurement When you visit the doctor for a checkup, many measurements are taken: your height, weight, blood pressure, and heart rate. Even your vision is measured and assigned a number. Blood might be drawn so measurements can be made of lead or cholesterol levels. Measurements quantify", " our observations: a person\u2019s blood pressure isn\u2019t just \u201cpretty good,\u201d it\u2019s 110/60, the low end of the good range. What is a measurement? A measurement is a comparison between an unknown quantity and a standard. For example, if you measure the mass of a rolling cart used in an experiment, the unknown quantity is the mass of the cart and the standard is the gram, as defined by the balance or spring scale you use. In the Mini Lab in Section 1.1, the length of the spring was the unknown and the centimeter was the standard. Comparing Results As you learned in Section 1.1, scientists share their results. Before new data are fully accepted, other scientists examine the experiment, looking for possible sources of error, and try to reproduce the results. Results often are reported with an uncertainty. A new measurement that is within the margin of uncertainty confirms the old measurement. For example, archaeologists use radiocarbon dating to find the age of cave paintings, such as those from the Lascaux cave, in Figure 1-9, and the Chauvet cave. Radiocarbon dates are reported with an uncertainty. Three radiocarbon ages from a panel in the Chauvet cave are 30,940 610 years, 30,790 600 years, and 30,230 530 years. While none of the measurements exactly match, the uncertainties in all three overlap, and the measurements confirm each other. \u25a0 Figure 1-9 Drawings of animals from the Lascaux cave in France. By dating organic material in the cave, such as pigments and torch marks, scientists are able to suggest dates at which these cave paintings were made. Each date is reported with an uncertainty to show how precise the measurement is. Section 1.2 Measurement 11 PhotoEdit 011-014 CH01-S2-845813 3/30/04 1:21 AM Page 12 ) 15.0 14.5 14.0 13.7 Mini Lab Data 1 2 3 Student \u25a0 Figure 1-10 Three students took multiple measurements. Are the measurements in agreement? Is student 1\u2019s result reproducible? Suppose three students performed the Mini Lab from Section 1.1 several times, starting with springs of the same length. With two washers on the spring, student 1 made repeated measurements, which ranged from 14.4 cm to 14.8 cm. The average of student 1\u2019s measurements was 14.6 cm, as shown in Figure 1-10. This result was reported as (14.6 0.2) cm. Student 2 reported finding the spring\u2019s length to be (14.8 0.3) cm. Student 3 reported a length of (14.0 0.1) cm. Could you conclude that the three measurements are in agreement? Is student 1\u2019s result reproducible? The results of students 1 and 2 overlap; that is, they have the lengths 14.5 cm to 14.8 cm in common. However, there is no overlap and, therefore, no agreement, between their results and the result of student 3. Precision Versus Accuracy Both precision and accuracy are characteristics of measured values. How precise and accurate are the measurements of the three students? The degree of exactness of a measurement is called its precision. Student 3\u2019s measurements are the most precise, within 0.1 cm. The measurements of the other two students are less precise because they have a larger uncertainty. Precision depends on the instrument and technique used to make the measurement. Generally, the device that has the finest division on its scale produces the most precise measurement. The precision of a measurement is one-half the smallest division of the instrument. For example, the graduated cylinder in Figure 1-11a has divisions of 1 mL. You can measure an object to within 0.5 mL with this device. However, the smallest division on the beaker in Figure 1-11b is 50 mL. How precise were your measurements in the MiniLab? The significant digits in an answer show its precision. A measure of 67.100 g is precise to the nearest thousandth of a gram. Recall from Section 1.1 the rules for performing operations with measurements given to different levels of precision. If you add 1.2 mL of acid to a beaker containing 2.4102 mL of water, you cannot say you now have 2.412102 mL of fluid, because the volume of water was not measured to the nearest tenth of a milliliter, but to 100 times that. a b \u25a0 Figure 1-11 The graduated cylinder contains 41 0.5 mL (a). The flask contains 325 mL 25 mL (b). 12 Horizons Companies Chapter 1 A Physics Toolkit 011-014 CH01-S2-845813 3/30/04 1:21 AM Page 13 Accuracy describes how well the results of a measurement agree with the \u201creal\u201d value; that is, the accepted value as measured by competent experimenters. If", " the length of the spring that the three students measured had been 14.8 cm, then student 2 would have been most accurate and student 3 least accurate. How accurate do you think your measurements in the Mini Lab on page 8 were? What might have led someone to make inaccurate measurements? How could you check the accuracy of measurements? A common method for checking the accuracy of an instrument is called the two-point calibration. First, does the instrument read zero when it should? Second, does it give the correct reading when it is measuring an accepted standard, as shown in Figure 1-12? Regular checks for accuracy are performed on critical measuring instruments, such as the radiation output of the machines used to treat cancer. Techniques of Good Measurement To assure accuracy and precision, instruments also have to be used correctly. Measurements have to be made carefully if they are to be as precise as the instrument allows. One common source of error comes from the angle at which an instrument is read. Scales should be read with one\u2019s eye directly above the measure, as shown in Figure 1-13a. If the scale is read from an angle, as shown in Figure 1-13b, you will get a different, and less accurate, value. The difference in the readings is caused by parallax, which is the apparent shift in the position of an object when it is viewed from different angles. To experiment with parallax, place your pen on a ruler and read the scale with your eye directly over the tip, then read the scale with your head shifted far to one side. \u25a0 Figure 1-13 By positioning the scale head on (a), your results will be more accurate than if you read your measurements at an angle (b). How far did parallax shift the measurement in b? a b \u25a0 Figure 1-12 Accuracy is checked by measuring a known value. Distance to the Moon For over 25 years, scientists have been accurately measuring the distance to the Moon by shining lasers through telescopes. The laser beam reflects off reflectors placed on the surface of the Moon by Apollo astronauts. They have determined that the average distance between the centers of Earth and the Moon is 385,000 km, and it is known with an accuracy of better than one part in 10 billion. Using this laser technique, scientists also have discovered that the Moon is receding from Earth at about 3.8 cm/yr. Section 1.2 Measurement 13 Horizons Companies 011-014 CH01-S2-845813 5/28/04 3:19 PM Page 14 \u25a0 Figure 1-14 A series of expeditions succeeded in placing a GPS receiver on top of Mount Everest. This improved the accuracy of the altitude measurement: Everest\u2019s peak is 8850 m, not 8848 m, above sea level. The Global Positioning System, or GPS, offers an illustration of accuracy and precision in measurement. The GPS consists of 24 satellites with transmitters in orbit and numerous receivers on Earth. The satellites send signals with the time, measured by highly accurate atomic clocks. The receiver uses the information from at least four satellites to determine latitude, longitude, and elevation. (The clocks in the receivers are not as accurate as those on the satellites.) Receivers have different levels of precision. A device in an automobile might give your position to within a few meters. Devices used by geophysicists, as in Figure 1-14, can measure movements of millimeters in Earth\u2019s crust. The GPS was developed by the United States Department of Defense. It uses atomic clocks, developed to test Einstein\u2019s theories of relativity and gravity. The GPS eventually was made available for civilian use. GPS signals now are provided worldwide free of charge and are used in navigation on land, at sea, and in the air, for mapping and surveying, by telecommunications and satellite networks, and for scientific research into earthquakes and plate tectonics. 1.2 Section Review 18. Accuracy Some wooden rulers do not start with 0 at the edge, but have it set in a few millimeters. How could this improve the accuracy of the ruler? 22. Precision A box has a length of 18.1 cm and a width of 19.2 cm, and it is 20.3 cm tall. a. What is its volume? 19. Tools You find a micrometer (a tool used to measure objects to the nearest 0.01 mm) that has been badly bent. How would it compare to a new, highquality meterstick in terms of its precision? Its accuracy? 20. Parallax Does parallax affect the precision of a measurement that you make? Explain. 21. Error Your friend tells you that his height is 182 cm. In your own words, explain the range of heights implied by this statement. b. How precise is the measure of length? Of volume? c. How tall is a stack of 12 of these boxes? d. How precise is the measure of the height of one box? of 12 boxes? 23", ". Critical Thinking Your friend states in a report that the average time required to circle a 1.5-mi track was 65.414 s. This was measured by timing 7 laps using a clock with a precision of 0.1 s. How much confidence do you have in the results of the report? Explain. 14 Bill Crouse Chapter 1 A Physics Toolkit physicspp.com/self_check_quiz 015-019 CH01-S3-845813 3/30/04 1:26 AM Page 15 1.3 Graphing Data Awell-designed graph can convey information quickly and simply. Patterns that are not immediately evident in a list of numbers take shape when the data are graphed. In this section, you will develop graphing techniques that will enable you to display, analyze, and model data. Identifying Variables When you perform an experiment, it is important to change only one factor at a time. For example, Table 1-3 gives the length of a spring with different masses attached, as measured in the Mini Lab. Only the mass varies; if different masses were hung from different types of springs, you wouldn\u2019t know how much of the difference between two data pairs was due to the different masses and how much to the different springs. Table 1-3 Length of a Spring for Different Masses Mass Attached to Spring (g) Length of Spring (cm) Objectives \u2022 Graph the relationship between independent and dependent variables. \u2022 Interpret graphs. \u2022 Recognize common relationships in graphs. Vocabulary independent variable dependent variable line of best fit linear relationship quadratic relationship inverse relationship 0 5 10 15 20 25 30 35 13.7 14.1 14.5 14.9 15.3 15.7 16.0 16.4 A variable is any factor that might affect the behavior of an experimental setup. The independent variable is the factor that is changed or manipulated during the experiment. In this experiment, the mass was the independent variable. The dependent variable is the factor that depends on the independent variable. In this experiment, the amount that the spring stretched depended on the mass. An experiment might look at how radioactivity varies with time, how friction changes with weight, or how the strength of a magnetic field depends on the distance from a magnet. 15.5 16.5 16.0 m ) One way to analyze data is to make a line graph. This shows how the dependent variable changes with the independent variable. The data from Table 1-3 are graphed in black in Figure 1-15. The line in blue, drawn as close to all the data points as possible, is called a line of best fit. The line of best fit is a better model for predictions than any one point that helps determine the line. The problemsolving strategy on the next page gives detailed instructions for graphing data and sketching a line of best fit. \u25a0 Figure 1-15 The independent variable, mass, is on the horizontal axis. The graph shows that the length of the spring increases as the mass suspended from the spring increases. Length of a Spring for Different Masses c ( h t g n e L 15.0 14.5 14.0 13.5 0 5 10 15 20 25 30 35 Mass (g) Section 1.3 Graphing Data 15 015-019 CH01-S3-845813 5/28/04 10:56 AM Page 16 Plotting Line Graphs Use the following steps to plot line graphs from data tables. 1. Identify the independent and dependent variables in your data. The independent variable is plotted on the horizontal axis, the x-axis. The dependent variable is plotted on the vertical axis, the y-axis. 2. Determine the range of the independent variable to be plotted. 3. Decide whether the origin (0, 0) is a valid data point. 4. Spread the data out as much as possible. Let each division on the graph paper stand for a convenient unit. This usually means units that are multiples of 2, 5, or 10. 5. Number and label the horizontal axis. The label should include the units, such as Mass (grams). 6. Repeat steps 2\u20135 for the dependent variable. 7. Plot the data points on the graph. 8. Draw the best-fit straight line or smooth curve that passes through as many data points as possible. This is sometimes called eyeballing. Do not use a series of straight line segments that connect the dots. The line that looks like the best fit to you may not be exactly the same as someone else\u2019s. There is a formal procedure, which many graphing calculators use, called the least-squares technique, that produces a unique best-fit line, but that is beyond the scope of this textbook. 9. Give the graph a title that clearly tells what the graph represents. Linear Relationships Scatter plots of data may take many different shapes, suggesting different relationships. (The line of best fit may be called a", " curve of best fit for nonlinear graphs.) Three of the most common relationships will be shown in this section. You probably are familiar with them from math class. When the line of best fit is a straight line, as in Figure 1-15, the dependent variable varies linearly with the independent variable. There is a linear relationship between the two variables. The relationship can be written as an equation. Math Handbook Graphs of Relations pages 848\u2014852 \u25a0 Figure 1-16 To find an equation of the line of best fit for a linear relationship, find the slope and y-intercept. Length of a Spring for Different Masses rise ) m c ( h t g n e L 17.0 16.0 15.0 14.0 0 16 Chapter 1 A Physics Toolkit b 13.7 run Linear Relationship Between Two Variables y mx b 5 10 15 20 25 30 35 Mass (g) Find the y-intercept, b, and the slope, m, as illustrated in Figure 1-16. Use points on the line\u2014they may or may not be data points. 015-019 CH01-S3-845813 7/22/04 7:22 AM Page 17 The slope is the ratio of the vertical change to the horizontal change. To find the slope, select two points, A and B, far apart on the line. The vertical change, or rise, y, is the difference between the vertical values of A and B. The horizontal change, or run, x, is the difference between the horizontal values of A and B. y e is r Slope m u n r x The slope of a line is equal to the rise divided by the run, which also can be expressed as the change in y divided by the change in x. In Figure 1-16: m (16.0 cm 14.1 cm) (30 g 5 g) 0.08 cm/g If y gets smaller as x gets larger, then y/x is negative, and the line slopes downward. The y-intercept, b, is the point at which the line crosses the y-axis, and it is the y-value when the value of x is zero. In this example, b 13.7 cm. When b 0, or y mx, the quantity y is said to vary directly with x. Nonlinear Relationships Figure 1-17 shows the distance a brass ball falls versus time. Note that the graph is not a straight line, meaning the relationship is not linear. There are many types of nonlinear relationships in science. Two of the most common are the quadratic and inverse relationships. The graph in Figure 1-17 is a quadratic relationship, represented by the following equation. \u25a0 Figure 1-17 This graph indicates a quadratic, or parabolic, relationship. Quadratic Relationship Between Two Variables y ax2 bx c A quadratic relationship exists when one variable depends on the square of another. Math Handbook Quadratic Graphs page 852 Quadratic Equations page 846 A computer program or graphing calculator easily can find the values of the constants a, b, and c in this equation. In this case, the equation is d 5t2. See the Math Handbook in the back of the book for more on making and using line graphs. An object is suspended from spring 1, and the spring\u2019s elongation (the distance it stretches) is X1. Then the same object is removed from the first spring and suspended from a second spring. The elongation of spring 2 is X2. X2 is greater than X1. 1. On the same axes, sketch the graphs of the mass versus elongation for both springs. 2. Is the origin included in the graph? Why or why not? 3. Which slope is steeper? 4. At a given mass, X2 1.6 X1. If X2 5.3 cm, what is X1? Section 1.3 Graphing Data 17 015-019 CH01-S3-845813 3/30/04 1:28 AM Page 18 \u25a0 Figure 1-18 This graph shows the inverse relationship between resistance and current. As resistance increases, current decreases 30 25 20 15 10 5 0 Current v. Resistance at 120 V 10 20 30 40 Resistance (ohms) The graph in Figure 1-18 shows how the current in an electric circuit varies as the resistance is increased. This is an example of an inverse relationship, represented by the following equation. Inverse Relationship y a x A hyperbola results when one variable depends on the inverse of the other. The three relationships you have learned about are a sample of the simple relations you will most likely try to derive in this course. Many other mathematical models are used. Important examples include sinusoids, used to model cyclical phenomena, and exponential growth and decay, used to study radioactivity. Combinations of different mathematical models represent even more complex", " phenomena. 24. The mass values of specified volumes of pure gold nuggets are given in Table 1-4. a. Plot mass versus volume from the values given in the table and draw the curve that best fits all points. b. Describe the resulting curve. c. According to the graph, what type of relationship exists between the mass of the pure gold nuggets and their volume? d. What is the value of the slope of this graph? Include the proper units. e. Write the equation showing mass as a function of volume for gold. f. Write a word interpretation for the slope of the line. 18 Chapter 1 A Physics Toolkit Table 1-4 Mass of Pure Gold Nuggets Volume (cm3) Mass (g) 1.0 2.0 3.0 4.0 5.0 19.4 38.6 58.1 77.4 96.5 015-019 CH01-S3-845813 3/30/04 1:28 AM Page 19 \u25a0 Figure 1-19 Computer animators use mathematical models of the real world to create a convincing fictional world. They need to accurately portray how beings of different sizes move, how hair or clothing move with a character, and how light and shadows fall, among other physics topics. Predicting Values When scientists discover relations like the ones shown in the graphs in this section, they use them to make predictions. For example, the equation for the linear graph in Figure 1-16 is as follows: y (0.08 cm/g)x 13.7 cm Relations, either learned as formulas or developed from graphs, can be used to predict values you haven\u2019t measured directly. How far would the spring in Table 1-3 stretch with 49 g of mass? y (0.08 cm/g)(49 g) 13.7 cm 18 cm It is important to decide how far you can extrapolate from the data you have. For example, 49 kg is a value far outside the ones measured, and the spring might break rather than stretch that far. Physicists use models to accurately predict how systems will behave: what circumstances might lead to a solar flare, how changes to a circuit will change the performance of a device, or how electromagnetic fields will affect a medical instrument. People in all walks of life use models in many ways. One example is shown in Figure 1-19. With the tools you have learned in this chapter, you can answer questions and produce models for the physics questions you will encounter in the rest of this textbook. 1.3 Section Review 25. Make a Graph Graph the following data. Time is the independent variable. Time (s) 0 5 10 15 20 25 30 35 Speed (m/s) 12 10 8 6 4 2 2 2 26. Interpret a Graph What would be the meaning of a nonzero y-intercept to a graph of total mass versus volume? 27. Predict Use the relation illustrated in Figure 1-16 to determine the mass required to stretch the spring 15 cm. 28. Predict Use the relation in Figure 1-18 to predict the current when the resistance is 16 ohms. 29. Critical Thinking In your own words, explain the meaning of a shallower line, or a smaller slope than the one in Figure 1-16, in the graph of stretch versus total mass for a different spring. physicspp.com/self_check_quiz Section 1.3 Graphing Data 19 0017_3949.ps 020-021 CH01-LAB-845813 3/30/04 1:46 AM Page 20 Exploring Objects in Motion Physics is a science that is based upon experimental observations. Many of the basic principles used to describe and understand mechanical systems, such as objects in linear motion, can be applied later to describe more complex natural phenomena. How can you measure the speed of the vehicles in a video clip? QUESTION What types of measurements could be made to find the speed of a vehicle? Objectives Procedure \u25a0 Observe the motion of the vehicles seen in 1. Visit physicspp.com/internet_lab to view the the video. \u25a0 Describe the motion of the vehicles. \u25a0 Collect and organize data on the vehicle\u2019s motion. \u25a0 Calculate a vehicle\u2019s speed. Safety Precautions Materials Internet access is required. watch or other timer Chapter 1 lab video clip. 2. The video footage was taken in the midwestern United States at approximately noon. Along the right shoulder of the road are large, white, painted rectangles. These types of markings are used in many states for aerial observation of traffic. They are placed at 0.322-km (0.2-mi) intervals. 3. Observe What type of measurements might be taken? Prepare a data table, such as the one shown on the next page. Record your observations of the surroundings, other vehicles, and markings. On what color vehicle is the camera located, and what color is the pickup truck in the lane to the left? 4. Measure and", " Estimate View the video again and look for more details. Is the road smooth? In what direction are the vehicles heading? How long does it take each vehicle to travel two intervals marked by the white blocks? Record your observations and data. 20 020-021 CH01-LAB-845813 5/28/04 11:00 AM Page 21 Data Table Marker Distance (km) White Vehicle Time (s) Gray Pickup Time (s) Analyze Real-World Physics When the speedometer is observed by a front-seat passenger, the driver, and a passenger in the rear driver\u2019s-side seat, readings of 90 km/h, 100 km/h, and 110 km/h, respectively, are observed. Explain the differences. Design an Experiment Visit physicspp.com/ internet_lab to post your experiment for measuring speed in the classroom using remote-controlled cars. Include your list of materials, your procedure, and your predictions for the accuracy of your lab. If you actually perform your lab, post your data and results as well. To find out more about measurement, visit the Web site: physicspp.com 1. Summarize your qualitative observations. 2. Summarize your quantitative observations. 3. Make and Use Graphs Graph both sets of data on one pair of axes. 4. Estimate What is the speed of the vehicles in km/s and km/h? 5. Predict How far will each vehicle travel in 5 min? Conclude and Apply 1. Measure What is the precision of the distance and time measurements? 2. Measure What is the precision of your speed measurement? On what does it depend? 3. Use Variables, Constants, and Controls Describe the independent and the dependent variables in this experiment. 4. Compare and Contrast Which vehicle\u2019s graph has a steeper slope? What is the slope equal to? 5. Infer What would a horizontal line mean on the graph? A line with a steeper slope? Going Further Speed is distance traveled divided by the amount of time to travel that distance. Explain how you could design your own experiment to measure speed in the classroom using remote-controlled cars. What would you use for markers? How precisely could you measure distance and time? Would the angle at which you measured the cars passing the markers affect the results? How much? How could you improve your measurements? What units make sense for speed? How far into the future could you predict the cars\u2019 positions? If possible, carry out the experiment and summarize your results. 21 022 CH01-FEATURE-845813 3/30/04 2:34 AM Page 22 Computer History and Growth Computer History and Growth Each pixel of the animations or movies you watch, and each letter of the instant messages you send presents your computer with several hundred equations. Each equation must be solved in a few billionths of a second\u2014if it takes a bit longer, you might complain that your computer is slow. Early Computers The earliest computers could solve very complex arrays of equations, just as yours can, but it took them a lot longer to do so. There were several reasons for this. First, the mathematics of algorithms (problemsolving strategies) still was new. Computer scientists were only beginning to learn how to arrange a particular problem, such as the conversion of a picture into an easily-transmittable form, so that it could be solved by a machine. UNIVAC 1, an early computer, filled an entire room. Machine Size Second, the machines were physically large. Computers work by switching patterns of electric currents that represent binary numbers. A 16-bit machine works with binary numbers that are 16 bits long. If a 64-bit number must be dealt with, the machine must repeat the same operation four times. A 32-bit machine would have to repeat the operation only twice, thus making it that much faster. But a 32-bit machine is four times the size of a 16-bit machine; that is, it has four times as many wires and transistor switches, and even 8-bit machines were the size of the old UNIVAC shown above. Moreover, current travels along wires at speeds no greater than about two-thirds the speed of light. This is a long time if the computer wires are 15 m long and must move information in less than 109 s. 22 CORBIS Future Technology Memory Third, electronic memories were extremely expensive. You may know that a larger memory lets your computer work faster. When one byte of memory required eight circuit boards, 1024 bytes (or 1 K) of memory was enormous. Because memory was so precious, computer programs had to be written with great cleverness. Astronants got to the Moon with 64 K of memory in Apollo\u2019s on-board computers. Processor chips used in today\u2019s computers are tiny compared to the old computer systems. When Gordon Moore and others invented the integrated circuit around 1960, the size and cost of computer circuitry dropped drastically. Physically smaller, and", " thus faster, machines could be built and very large memories became possible. Today, the transistors on a chip are now smaller than bacteria. The cost and size of computers have dropped so much that your cell phone has far more computing power than most big office machines of the 1970s. Going Further 1. Research A compression protocol makes a computer file smaller and less prone to transmission errors. Look up the terms.jpg,.mp3,.mpeg, and.midi and see how they apply to the activities you do on your computer. 2. Calculate Using the example here, how long does it take for a binary number to travel 15 m? How many such operations could there be each second? 023-029 CH01-SG CA STP-845813 3/30/04 2:41 AM Page 23 1.1 Mathematics and Physics Vocabulary \u2022 physics (p. 3) \u2022 dimensional analysis (p. 6) \u2022 significant digits (p. 7) \u2022 scientific method (p. 8) \u2022 hypothesis (p. 8) \u2022 scientific law (p. 9) \u2022 scientific theory (p. 10) 1.2 Measurement Vocabulary \u2022 measurement (p. 11) \u2022 precision (p. 12) \u2022 accuracy (p. 13) 1.3 Graphing Data Vocabulary \u2022 independent variable (p. 15) \u2022 dependent variable (p. 15) \u2022 line of best fit (p. 15) \u2022 linear relationship (p. 16) \u2022 quadratic relationship (p. 17) \u2022 inverse relationship (p. 18) Key Concepts \u2022 Physics is the study of matter and energy and their relationships. \u2022 Dimensional analysis is used to check that an answer will be in the correct units. \u2022 The result of any mathematical operation with measurements never can be more precise than the least-precise measurement involved in the operation. \u2022 The scientific method is a systematic method of observing, experimenting, and analyzing to answer questions about the natural world. \u2022 Scientific ideas change in response to new data. \u2022 Scientific laws and theories are well-established descriptions and explanations of nature. Key Concepts \u2022 New scientific findings must be reproducible; that is, others must be able to measure and find the same results. \u2022 All measurements are subject to some uncertainty. \u2022 Precision is the degree of exactness with which a quantity is measured. Scientific notation shows how precise a measurement is. \u2022 Accuracy is the extent to which a measurement matches the true value. Key Concepts \u2022 Data are plotted in graphical form to show the relationship between two variables. \u2022 The line that best passes through or near graphed data is called the line of best fit. It is used to describe the data and to predict where new data would lie on the graph. \u2022 A graph in which data points lie on a straight line is a graph of a linear relationship. In the equation, m and b are constants. y mx b \u2022 The slope of a straight-line graph is the vertical change (rise) divided by the horizontal change (run) and often has a physical meaning. y e is r m u n r x \u2022 The graph of a quadratic relationship is a parabolic curve. It is represented by the equation below. The constants a, b, and c can be found with a computer or a graphing calculator; simpler ones can be found using algebra. y ax2 bx c \u2022 The graph of an inverse relationship between x and y is a hyperbolic curve. It is represented by the equation below, where a is a constant. y a x physicspp.com/vocabulary_puzzlemaker 23 023-029 CH01-SG CA STP-845813 3/30/04 2:41 AM Page 24 Concept Mapping 30. Complete the following concept map using the following terms: hypothesis, graph, mathematical model, dependent variable, measurement. experiment independent variable Mastering Concepts 31. Describe a scientific method. (1.1) 32. Why is mathematics important to science? (1.1) 33. What is the SI system? (1.1) 34. How are base units and derived units related? (1.1) 35. Suppose your lab partner recorded a measurement as 100 g. (1.1) a. Why is it difficult to tell the number of significant digits in this measurement? b. How can the number of significant digits in such a number be made clear? 36. Give the name for each of the following multiples of the meter. (1.1) 1 m a. 00 1 1 m b. 00 10 c. 1000 m 37. To convert 1.8 h to minutes, by what conversion factor should you multiply? (1.1) 38. Solve each problem. Give the correct number of significant digits in the answers. (1.1) a. 4.667104 g + 3.02105 g b. (1.70102 J) \u00f7 (5.922104 cm3) 39. What", " determines the precision of a measurement? (1.2) 40. How does the last digit differ from the other digits in a measurement? (1.2) 41. A car\u2019s odometer measures the distance from home to school as 3.9 km. Using string on a map, you find the distance to be 4.2 km. Which answer do you think is more accurate? What does accurate mean? (1.2) 42. How do you find the slope of a linear graph? (1.3) 43. For a driver, the time between seeing a stoplight and stepping on the brakes is called reaction time. The distance traveled during this time is the reaction distance. Reaction distance for a given driver and vehicle depends linearly on speed. (1.3) a. Would the graph of reaction distance versus speed have a positive or a negative slope? b. A driver who is distracted has a longer reaction time than a driver who is not. Would the graph of reaction distance versus speed for a distracted driver have a larger or smaller slope than for a normal driver? Explain. 44. During a laboratory experiment, the temperature of the gas in a balloon is varied and the volume of the balloon is measured. Which quantity is the independent variable? Which quantity is the dependent variable? (1.3) 45. What type of relationship is shown in Figure 1-20? Give the general equation for this type of relation. (1.3) y x \u25a0 Figure 1-20 46. Given the equation F mv2/R, what relationship exists between each of the following? (1.3) a. F and R b. F and m c. F and v 24 Chapter 1 A Physics Toolkit For more problems, go to Additional Problems, Appendix B. 023-029 CH01-SG CA STP-845813 5/28/04 11:23 AM Page 25 Applying Concepts 47. Figure 1-21 gives the height above the ground of a ball that is thrown upward from the roof of a building, for the first 1.5 s of its trajectory. What is the ball\u2019s height at t 0? Predict the ball\u2019s height at t 2 s and at t 5 s. Height of Ball v. Time ) m ( t h g e H i 25 20 15 10 5 0 1 2 3 4 \u25a0 Figure 1-21 Time (s) 48. Is a scientific method one set of clearly defined steps? Support your answer. 49. Explain the difference between a scientific theory and a scientific law. 50. Density The density of a substance is its mass per unit volume. a. Give a possible metric unit for density. b. Is the unit for density a base unit or a derived unit? 51. What metric unit would you use to measure each of the following? a. the width of your hand b. the thickness of a book cover c. the height of your classroom d. the distance from your home to your classroom 52. Size Make a chart of sizes of objects. Lengths should range from less than 1 mm to several kilometers. Samples might include the size of a cell, the distance light travels in 1 s, and the height of a room. 53. Time Make a chart of time intervals. Sample intervals might include the time between heartbeats, the time between presidential elections, the average lifetime of a human, and the age of the United States. Find as many very short and very long examples as you can. 54. Speed of Light Two students measure the speed of light. One obtains (3.001 0.001)108 m/s; the other obtains (2.999 0.006)108 m/s. a. Which is more precise? b. Which is more accurate? (You can find the speed of light in the back of this textbook.) 55. You measure the dimensions of a desk as 132 cm, 83 cm, and 76 cm. The sum of these measures is 291 cm, while the product is 8.3105 cm3. Explain how the significant digits were determined in each case. 56. Money Suppose you receive $5.00 at the beginning of a week and spend $1.00 each day for lunch. You prepare a graph of the amount you have left at the end of each day for one week. Would the slope of this graph be positive, zero, or negative? Why? 57. Data are plotted on a graph, and the value on the y-axis is the same for each value of the independent variable. What is the slope? Why? How does y depend on x? 58. Driving The graph of braking distance versus car speed is part of a parabola. Thus, the equation is written d av2 bv c. The distance, d, has units in meters, and velocity, v, has units in meters/second. How could you find the units of a, b, and c? What would they be? 59", ". How long is the leaf in Figure 1-22? Include the uncertainty in your measurement. \u25a0 Figure 1-22 60. The masses of two metal blocks are measured. Block A has a mass of 8.45 g and block B has a mass of 45.87 g. a. How many significant digits are expressed in these measurements? b. What is the total mass of block A plus block B? c. What is the number of significant digits for the total mass? d. Why is the number of significant digits different for the total mass and the individual masses? 61. History Aristotle said that the speed of a falling object varies inversely with the density of the medium through which it falls. a. According to Aristotle, would a rock fall faster in water (density 1000 kg/m3), or in air (density 1 kg/m3)? b. How fast would a rock fall in a vacuum? Based on this, why would Aristotle say that there could be no such thing as a vacuum? physicspp.com/chapter_test Chapter 1 Assessment 25 Laura Sifferlin 023-029 CH01-SG CA STP-845813 5/28/04 11:24 AM Page 26 62. Explain the difference between a hypothesis and a scientific theory. 63. Give an example of a scientific law. 73. Gravity The force due to gravity is F mg where g 9.80 m/s2. a. Find the force due to gravity on a 41.63-kg object. b. The force due to gravity on an object is 632 kgm/s2. 64. What reason might the ancient Greeks have had not What is its mass? to question the hypothesis that heavier objects fall faster than lighter objects? Hint: Did you ever question which falls faster? 65. Mars Explain what observations led to changes in scientists\u2019 ideas about the surface of Mars. 66. A graduated cylinder is marked every mL. How precise a measurement can you make with this instrument? Mastering Problems 1.1 Mathematics and Physics 67. Convert each of the following measurements to meters. a. 42.3 cm b. 6.2 pm c. 21 km d. 0.023 mm e. 214 m f. 57 nm 68. Add or subtract as indicated. a. 5.80109 s 3.20108 s b. 4.87106 m 1.93106 m c. 3.14105 kg 9.36105 kg d. 8.12107 g 6.20106 g 69. Rank the following mass measurements from least to greatest: 11.6 mg, 1021 g, 0.000006 kg, 0.31 mg. 70. State the number of significant digits in each of the following measurements. a. 0.00003 m b. 64.01 fm c. 80.001 m d. 0.720 g e. 2.40106 kg f. 6108 kg g. 4.071016 m 71. Add or subtract as indicated. a. 16.2 m 5.008 m 13.48 m b. 5.006 m 12.0077 m 8.0084 m c. 78.05 cm2 32.046 cm2 d. 15.07 kg 12.0 kg 72. Multiply or divide as indicated. a. (6.21018 m)(4.71010 m) b. (5.6107 m)/(2.81012 s) c. (8.1104 km)(1.6103 km) d. (6.5105 kg)/(3.4103 m3) 74. Dimensional Analysis Pressure is measured in pascals, where 1 Pa 1 kg/ms2. Will the following expression give a pressure in the correct units? (0.55 kg)(2.1 m/s) 9.8 m/s2 1.2 Measurement 75. A water tank has a mass of 3.64 kg when it is empty and a mass of 51.8 kg when it is filled to a certain level. What is the mass of the water in the tank? 76. The length of a room is 16.40 m, its width is 4.5 m, and its height is 3.26 m. What volume does the room enclose? 77. The sides of a quadrangular plot of land are 132.68 m, 48.3 m, 132.736 m, and 48.37 m. What is the perimeter of the plot? 78. How precise a measurement could you make with the scale shown in Figure 1-23? \u25a0 Figure 1-23 79. Give the measure shown on the meter in Figure 1-24 as precisely as you can. Include the uncertainty in your answer. 1 0 2 3 ACLASS A 4 5 A \u25a0 Figure 1-24 26 Horizons Companies Chapter 1 A Physics Toolkit For more problems, go to Additional Problems, Appendix B. 023-029 CH01-SG CA STP-", "845813 6/4/04 5:38 AM Page 27 80. Estimate the height of the nearest door frame in centimeters. Then measure it. How accurate was your estimate? How precise was your estimate? How precise was your measurement? Why are the two precisions different? 81. Base Units Give six examples of quantities you might measure in a physics lab. Include the units you would use. 82. Temperature The temperature drops from 24\u00b0C to 10\u00b0C in 12 hours. a. Find the average temperature change per hour. b. Predict the temperature in 2 more hours if the trend continues. c. Could you accurately predict the temperature in 24 hours? 1.3 Graphing Data 83. Figure 1-25 shows the masses of three substances for volumes between 0 and 60 cm3. a. What is the mass of 30 cm3 of each substance? b. If you had 100 g of each substance, what would be their volumes? c. In one or two sentences, describe the meaning of the slopes of the lines in this graph. d. What is the y-intercept of each line? What does it mean? Mass of Three Substances ) g ( s s a M 800 700 600 500 400 300 200 100 0 10 \u25a0 Figure 1-25 C B A 40 20 30 Volume (cm3) 50 84. During a class demonstration, a physics instructor placed a mass on a horizontal table that was nearly frictionless. The instructor then applied various horizontal forces to the mass and measured the distance it traveled in 5 seconds for each force applied. The results of the experiment are shown in Table 1-5. Table 1-5 Distance Traveled with Different Forces Force (N) Distance (cm) 5.0 10.0 15.0 20.0 25.0 30.0 24 49 75 99 120 145 a. Plot the values given in the table and draw the curve that best fits all points. b. Describe the resulting curve. c. Use the graph to write an equation relating the distance to the force. d. What is the constant in the equation? Find its units. e. Predict the distance traveled when a 22.0-N force is exerted on the object for 5 s. 85. The physics instructor from the previous problem changed the procedure. The mass was varied while the force was kept constant. Time and distance were measured, and the acceleration of each mass was calculated. The results of the experiment are shown in Table 1-6. Table 1-6 Acceleration of Different Masses Mass (kg) Acceleration (m/s2) 1.0 2.0 3.0 4.0 5.0 6.0 12.0 5.9 4.1 3.0 2.5 2.0 a. Plot the values given in the table and draw the curve that best fits all points. b. Describe the resulting curve. c. According to the graph, what is the relationship between mass and the acceleration produced by a constant force? d. Write the equation relating acceleration to mass given by the data in the graph. e. Find the units of the constant in the equation. f. Predict the acceleration of an 8.0-kg mass. physicspp.com/chapter_test Chapter 1 Assessment 27 023-029 CH01-SG CA STP-845813 5/28/04 11:24 AM Page 28 86. During an experiment, a student measured the mass of 10.0 cm3 of alcohol. The student then measured the mass of 20.0 cm3 of alcohol. In this way, the data in Table 1-7 were collected. Table 1-7 The Mass Values of Specific Volumes of Alcohol Volume (cm3) Mass (g) 10.0 20.0 30.0 40.0 50.0 7.9 15.8 23.7 31.6 39.6 a. Plot the values given in the table and draw the curve that best fits all the points. b. Describe the resulting curve. c. Use the graph to write an equation relating the volume to the mass of the alcohol. d. Find the units of the slope of the graph. What is the name given to this quantity? e. What is the mass of 32.5 cm3 of alcohol? Mixed Review 87. Arrange the following numbers from most precise to least precise 0.0034 m 45.6 m 1234 m 88. Figure 1-26 shows the toroidal (doughnut-shaped) interior of the now-dismantled Tokamak Fusion Test Reactor. Explain why a width of 80 m would be an unreasonable value for the width of the toroid. What would be a reasonable value? \u25a0 Figure 1-26 89. You are cracking a code and have discovered the following conversion factors: 1.23 longs 23.0 mediums, and 74.5 mediums 645 shorts. How many shorts are equal to one long? 90. You are given", " the following measurements of a rectangular bar: length 2.347 m, thickness 3.452 cm, height 2.31 mm, mass 1659 g. Determine the volume, in cubic meters, and density, in g/cm3, of the beam. Express your results in proper form. 91. A drop of water contains 1.71021 molecules. If the water evaporated at the rate of one million molecules per second, how many years would it take for the drop to completely evaporate? 92. A 17.6-gram sample of metal is placed in a graduated cylinder containing 10.0 cm3 of water. If the water level rises to 12.20 cm3, what is the density of the metal? Thinking Critically 93. Apply Concepts It has been said that fools can ask more questions than the wise can answer. In science, it is frequently the case that one wise person is needed to ask the right question rather than to answer it. Explain. 94. Apply Concepts Find the approximate mass of water in kilograms needed to fill a container that is 1.40 m long and 0.600 m wide to a depth of 34.0 cm. Report your result to one significant digit. (Use a reference source to find the density of water.) 95. Analyze and Conclude A container of gas with a pressure of 101 kPa has a volume of 324 cm3 and a mass of 4.00 g. If the pressure is increased to 404 kPa, what is the density of the gas? Pressure and volume are inversely proportional. 96. Design an Experiment How high can you throw a ball? What variables might affect the answer to this question? 97. Calculate If the Sun suddenly ceased to shine, how long would it take Earth to become dark? (You will have to look up the speed of light in a vacuum and the distance from the Sun to Earth.) How long would it take the surface of Jupiter to become dark? Writing in Physics 98. Research and describe a topic in the history of physics. Explain how ideas about the topic changed over time. Be sure to include the contributions of scientists and to evaluate the impact of their contributions on scientific thought and the world outside the laboratory. 99. Explain how improved precision in measuring time would have led to more accurate predictions about how an object falls. 28 Princeton Plasma Physics Laboratory Chapter 1 A Physics Toolkit For more problems, go to Additional Problems, Appendix B. 023-029 CH01-SG CA STP-845813 5/28/04 11:25 AM Page 29 Multiple Choice 1. Two laboratories use radiocarbon dating to measure the age of two wooden spear handles found in the same grave. Lab A finds an age of 2250 40 years for the first object; lab B finds an age of 2215 50 years for the second object. Which of the following is true? Lab A\u2019s reading is more accurate than lab B\u2019s. Lab A\u2019s reading is less accurate than lab B\u2019s. Lab A\u2019s reading is more precise than lab B\u2019s. Lab A\u2019s reading is less precise than lab B\u2019s. 2. Which of the following is equal to 86.2 cm? 8.62104 km 862 dm 8.62 m 0.862 mm 3. Jario has a problem to do involving time, distance, and velocity, but he has forgotten the formula. The question asks him for a measurement in seconds, and the numbers that are given have units of m/s and km. What could Jario do to get the answer in seconds? Multiply the km by the m/s, then multiply by 1000. Divide the km by the m/s, then multiply by 1000. Divide the km by the m/s, then divide by 1000. Multiply the km by the m/s, then divide by 1000. 4. What is the slope of the graph? 0.25 m/s2 0.4 m/s2 2.5 m/s2 4.0 m/s2 Stopping Distance ) 10 12 Time (s) 5. Which formula is equivalent to Dm Extended Answer 6. You want to calculate an acceleration, in units of m/s2, given a force, in N, and the mass, in g, on which the force acts. (1 N 1 kgm/s2) Rewrite the equation F ma so a is in terms of m and F. a. b. What conversion factor will you need to multiply by to convert grams to kilograms? c. A force of 2.7 N acts on a 350-g mass. Write the equation you will use, including the conversion factor, to find the acceleration. 7. Find an equation for a line of best fit for the data shown below. Distance v. Time ) 12 10 Time (s) Skip Around if You Can You may want to skip over difficult questions and come back to them later", ", after you\u2019ve answered the easier questions. This will guarantee more points toward your final score. In fact, other questions may help you answer the ones you skipped. Just be sure you fill in the correct ovals on your answer sheet. physicspp.com/standardized_test Chapter 1 Standardized Test Practice 29 030-033 CH02-S1-845813 3/19/04 1:32 PM Page 30 What You\u2019ll Learn \u2022 You will represent motion through the use of words, motion diagrams, and graphs. \u2022 You will use the terms position, distance, displacement, and time interval in a scientific manner to describe motion. Why It\u2019s Important Without ways to describe and analyze motion, travel by plane, train, or bus would be chaotic at best. Times and speeds determine the winners of races as well as transportation schedules. Running a Marathon As one runner passes another, the speed of the overtaking runner is greater than the speed of the other runner. Think About This How can you represent the motion of two runners? physicspp.com 30 AFP/Corbis 030-033 CH02-S1-845813 3/13/04 3:57 AM Page 31 Which car is faster? Question In a race between two toy cars, can you explain which car is faster? Procedure Analysis What data did you collect to show which car was moving faster? What other data could you collect to determine which car is faster? Critical Thinking Write an operational definition of average speed. 1. Obtain two toy cars, either friction cars or windup cars. Place the cars on your lab table or other surface recommended by your teacher. 2. Decide on a starting line for the race. 3. Release both cars from the same starting line at the same time. Note that if you are using windup cars, you will need to wind them up before you release them. Be sure to pull the cars back before release if they are friction cars. 4. Observe Watch the two cars closely as they move and determine which car is moving faster. 5. Repeat steps 1\u20133, but this time collect one type of data to support your conclusion about which car is faster. 2.1 Picturing Motion In the previous chapter, you learned about the scientific processes that will be useful in your study of physics. In this chapter, you will begin to use these tools to analyze motion. In subsequent chapters, you will apply them to all kinds of movement using sketches, diagrams, graphs, and equations. These concepts will help you to determine how fast and how far an object will move, whether the object is speeding up or slowing down, and whether it is standing still or moving at a constant speed. Perceiving motion is instinctive\u2014your eyes naturally pay more attention to moving objects than to stationary ones. Movement is all around you\u2014 from fast trains and speedy skiers to slow breezes and lazy clouds. Movements travel in many directions, such as the straight-line path of a bowling ball in a lane\u2019s gutter, the curved path of a tether ball, the spiral of a falling kite, and the swirls of water circling a drain. Objectives \u2022 Draw motion diagrams to describe motion. \u2022 Develop a particle model to represent a moving object. Vocabulary motion diagram particle model Section 2.1 Picturing Motion 31 Horizons Companies 030-033 CH02-S1-845813 3/13/04 3:58 AM Page 32 All Kinds of Motion What comes to your mind when you hear the word motion? A speeding automobile? A spinning ride at an amusement park? A baseball soaring over a fence for a home run? Or a child swinging back and forth in a regular rhythm? When an object is in motion, as shown in Figure 2-1, its position changes. Its position can change along the path of a straight line, a circle, an arc, or a back-and-forth vibration. Some of the types of motion described above appear to be more complicated than others. When beginning a new area of study, it is generally a good idea to begin with what appears to be the least complicated situation, learn as much as possible about it, and then gradually add more complexity to that simple model. In the case of motion, you will begin your study with movement along a straight line. Movement along a straight line Suppose that you are reading this textbook at home. At the beginning of Chapter 2, you glance over at your pet hamster and see that he is sitting in a corner of his cage. Sometime later, you look over again, and you see that he now is sitting by the food dish in the opposite corner of the cage. You can infer that he has moved from one place to another in the time in between your observations. Thus, a description of motion relates to place and time. You must be able to answer the questions of where and when an object is positioned to describe its motion. Next, you will look at", " some tools that are useful in determining when an object is at a particular place. Motion Diagrams Consider an example of straight-line motion: a runner is jogging along a straight path. One way of representing the motion of the runner is to create a series of images showing the positions of the runner at equal time intervals. This can be done by photographing the runner in motion to obtain a series of images. Suppose you point a camera in a direction perpendicular to the direction of motion, and hold it still while the motion is occurring. Then you take a series of photographs of the runner at equal time intervals. Figure 2-2 shows what a series of consecutive images for a runner might look like. Notice that the runner is in a different position in each image, but everything in the background remains in the same position. This indicates that, relative to the ground, only the runner is in motion. What is another way of representing the runner\u2019s motion? \u25a0 Figure 2-2 If you relate the position of the runner to the background in each image over equal time intervals, you will conclude that she is in motion. \u25a0 Figure 2-1 An object in motion changes its position as it moves. In this photo, the camera was focused on the rider, so the blurry background indicates that the rider\u2019s position has changed. 32 (t)Getty Images, (others)Hutchings Photography Chapter 2 Representing Motion 030-033 CH02-S1-845813 3/13/04 3:59 AM Page 33 Suppose that you stacked the images from Figure 2-2, one on top of the other. Figure 2-3 shows what such a stacked image might look like. You will see more than one image of the moving runner, but only a single image of the motionless objects in the background. A series of images showing the positions of a moving object at equal time intervals is called a motion diagram. The Particle Model Keeping track of the motion of the runner is easier if you disregard the movement of the arms and legs, and instead concentrate on a single point at the center of her body. In effect, you can disregard the fact that she has some size and imagine that she is a very small object located precisely at that central point. A particle model is a simplified version of a motion diagram in which the object in motion is replaced by a series of single points. To use the particle model, the size of the object must be much less than the distance it moves. The internal motions of the object, such as the waving of the runner\u2019s arms are ignored in the particle model. In the photographic motion diagram, you could identify one central point on the runner, such as a dot centered at her waistline, and take measurements of the position of the dot. The bottom part of Figure 2-3 shows the particle model for the runner\u2019s motion. You can see that applying the particle model produces a simplified version of the motion diagram. In the next section, you will learn how to create and use a motion diagram that shows how far an object moved and how much time it took to move that far. \u25a0 Figure 2-3 Stacking a series of images taken at regular time intervals and combining them into one image creates a motion diagram for the runner for one portion of her run. Reducing the runner\u2019s motion to a series of single points results in a particle model of her motion. 2.1 Section Review 1. Motion Diagram of a Runner Use the particle model to draw a motion diagram for a bike rider riding at a constant pace. 2. Motion Diagram of a Bird Use the particle model to draw a simplified motion diagram corresponding to the motion diagram in Figure 2-4 for a flying bird. What point on the bird did you choose to represent it? \u25a0 Figure 2-4 3. Motion Diagram of a Car Use the particle model to draw a simplified motion diagram corresponding to the motion diagram in Figure 2-5 for a car coming to a stop at a stop sign. What point on the car did you use to represent it? \u25a0 Figure 2-5 4. Critical Thinking Use the particle model to draw motion diagrams for two runners in a race, when the first runner crosses the finish line as the other runner is three-fourths of the way to the finish line. physicspp.com/self_check_quiz Section 2.1 Picturing Motion 33 Hutchings Photography 034-037 CH02-S2-845813 3/13/04 4:35 AM Page 34 2.2 Where and When? Objectives \u2022 Define coordinate systems for motion problems. \u2022 Recognize that the chosen coordinate system affects the sign of objects\u2019 positions. \u2022 Define displacement. \u2022 Determine a time interval. \u2022 Use a motion diagram to answer questions about an object\u2019s position or displacement. Vocabulary coordinate system origin position distance magnitude vectors scalars resultant time interval displacement Would it be possible to take measurements of distance and time from", " a motion diagram, such as the motion diagram of the runner? Before taking the photographs, you could place a meterstick or a measuring tape on the ground along the path of the runner. The measuring tape would tell you where the runner was in each image. A stopwatch or clock within the view of the camera could tell the time. But where should you place the end of the measuring tape? When should you start the stopwatch? Coordinate Systems When you decide where to place the measuring tape and when to start the stopwatch, you are defining a coordinate system, which tells you the location of the zero point of the variable you are studying and the direction in which the values of the variable increase. The origin is the point at which both variables have the value zero. In the example of the runner, the origin, represented by the zero end of the measuring tape, could be placed 6 m to the left of the tree. The motion is in a straight line; thus, your measuring tape should lie along that straight line. The straight line is an axis of the coordinate system. You probably would place the tape so that the meter scale increases to the right of the zero, but putting it in the opposite direction is equally correct. In Figure 2-6a, the origin of the coordinate system is on the left. You can indicate how far away the runner is from the origin at a particular time on the simplified motion diagram by drawing an arrow from the origin to the point representing the runner, as shown in Figure 2-6b. This arrow represents the runner\u2019s position, the separation between an object and the origin. The length of the arrow indicates how far the object is from the origin, or the object\u2019s distance from the origin. The arrow points from the origin to the location of the moving object at a particular time. \u25a0 Figure 2-6 In these motion diagrams, the origin is at the left (a), and the positive values of distance extend horizontally to the right. The two arrows, drawn from the origin to points representing the runner, locate his position at two different times (b). a b 0 5 10 15 meters 20 25 30 d 0 5 10 15 meters 20 25 30 d 34 Chapter 2 Representing Motion 034-037 CH02-S2-845813 5/28/04 11:38 AM Page 35 Is there such a thing as a negative position? Suppose you chose the coordinate system just described, placing the origin 4 m left of the tree with the d-axis extending in a positive direction to the right. A position 9 m to the left of the tree, 5 m left of the origin, would be a negative position, as shown in Figure 2-7. In the same way, you could discuss a time before the stopwatch was started. 5 0 5 10 15 meters 20 25 30 d \u25a0 Figure 2-7 The arrow drawn on this motion diagram indicates a negative position. Vectors and scalars Quantities that have both size, also called magnitude, and direction, are called vectors, and can be represented by arrows. Quantities that are just numbers without any direction, such as distance, time, or temperature, are called scalars. This textbook will use boldface letters to represent vector quantities and regular letters to represent scalars. You already know how to add scalars; for example, 0.6 0.2 0.8. How do you add vectors? Think about how you would solve the following problem. Your aunt asks you to get her some cold medicine at the store nearby. You walk 0.5 km east from your house to the store, buy the cold medicine, and then walk another 0.2 km east to your aunt\u2019s house. How far from the origin are you at the end of your trip? The answer, of course, is 0.5 km east 0.2 km east 0.7 km east. You also could solve this problem graphically, using the following method. Using a ruler, measure and draw each vector. The length of a vector should be proportional to the magnitude of the quantity being represented, so you must decide on a scale for your drawing. For example, you might let 1 cm on paper represent 0.1 km. The important thing is to choose a scale that produces a diagram of reasonable size with a vector that is about 5\u201310 cm long. The vectors representing the two segments that made up your trip to your aunt\u2019s house are shown in Figure 2-8, drawn to a scale of 1 cm, which represents 0.1 km. The vector that represents the total of these two, shown here with a dotted line, is 7 cm long. According to the established scale, you were 0.7 km from the origin at the end of your trip. The vector that represents the sum of the other two vectors is called the resultant. The resultant always points from the tail of the first vector to the tip of the last vector. Your house, Aunt s house Store 5 cm 2", " cm 7 cm \u25a0 Figure 2-8 Add two vectors by placing them tip to tail. The resultant points from the tail of the first vector to the tip of the last vector. Section 2.2 Where and When? 35 034-037 CH02-S2-845813 7/19/04 1:32 AM Page 36 \u25a0 Figure 2-9 You can see that it took the runner 4.0 s to run from the tree to the lamppost. The initial position of the runner is used as a reference point. The vector from position 1 to position 2 indicates both the direction and amount of displacement during this time interval. \u2022 Displacement vectors are shown in green. \u25a0 Figure 2-10 Start with two vectors, A and B (a). To subtract vector B from vector A, first reverse vector B, then add them together to obtain the resultant, R (b). A B Vectors A and B a b ti tf d 0 5 10 15 meters 20 25 30 d Time Intervals and Displacements When analyzing the runner\u2019s motion, you might want to know how long it took the runner to travel from the tree to the lamppost. This value is obtained by finding the difference in the stopwatch readings at each position. Assign the symbol ti to the time when the runner was at the tree and the symbol tf to the time when he was at the lamppost. The difference between two times is called a time interval. A common symbol for a time interval is t, where the Greek letter delta,, is used to represent a change in a quantity. The time interval is defined mathematically as follows. Time Interval t tf The time interval is equal to the final time minus the initial time. ti ti Although i and f are used to represent the initial and final times, they can be the initial and final times of any time interval you choose. In the example of the runner, the time it takes for him to go from the tree to the lamppost 5.0 s 1.0 s 4.0 s. How did the runner\u2019s position change is tf when he ran from the tree to the lamppost, as shown in Figure 2-9? The symbol d may be used to represent position. In common speech, a position refers to a place; but in physics, a position is a vector with its tail at the origin of a coordinate system and its tip at the place. Figure 2-9 shows d, an arrow drawn from the runner\u2019s position at the tree to his position at the lamppost. This vector represents his change in position, or displacement, during the time interval between ti and tf. The length of the arrow represents the distance the runner moved, while the direction the arrow points indicates the direction of the displacement. Displacement is mathematically defined as follows. Displacement d df Displacement is equal to the final position minus the initial position. di A B A (B) Resultant of A and (B) Again, the initial and final positions are the beginning and end of any interval you choose. Also, while position can be considered a vector, it is common practice when doing calculations to drop the boldface, and use signs and magnitudes. This is because position usually is measured from the origin, and direction typically is included with the position indication. 36 Chapter 2 Representing Motion 034-037 CH02-S2-845813 5/28/04 11:38 AM Page 37 a b df d di df d di \u25a0 Figure 2-11 The displacement of the runner during the 4.0-s time interval is found by subtracting df from di. In (a) the origin is at the left, and in (b) it is at the right. Regardless of your choice of coordinate system, d is the same. How do you subtract vectors? Reverse the subtracted vector and add. This is because A B A (B). Figure 2-10a shows two vectors, A, 4 cm long pointing east, and B, 1 cm long also pointing east. Figure 2-10b shows B, which is 1 cm long pointing west. Figure 2-10b shows the resultant of A and B. It is 3 cm long pointing east. To determine the length and direction of the displacement vector, d df di, draw di, which is di reversed. Then draw df and copy di with its tail at df\u2019s tip. Add df and di. In the example of the runner, his 25.0 m 5.0 m 20.0 m. He moved to the displacement is df right of the tree. To completely describe an object\u2019s displacement, you must indicate the distance it traveled and the direction it moved. Thus, displacement, a vector, is not identical to distance, a scalar; it is distance and direction. di What would happen if you chose a different coordinate system; that is, if you measured the", " position of the runner from another location? Look at Figure 2-9, and suppose you change the right side of the d-axis to be zero. While the vectors drawn to represent each position change, the length and direction of the displacement vector does not, as shown in Figures 2-11a and b. The displacement, d, in the time interval from 1.0 s to 5.0 s does not change. Because displacement is the same in any coordinate system, you frequently will use displacement when studying the motion of an object. The displacement vector is always drawn with its flat end, or tail, at the earlier position, and its point, or tip, at the later position. 2.2 Section Review 5. Displacement The particle model for a car traveling on an interstate highway is shown below. The starting point is shown. Here There Make a copy of the particle model, and draw a vector to represent the displacement of the car from the starting time to the end of the third time interval. 6. Displacement The particle model for a boy walking to school is shown below. Home School Make a copy of the particle model, and draw vectors to represent the displacement between each pair of dots. 7. Position Two students compared the position vectors they each had drawn on a motion diagram to show the position of a moving object at the same time. They found that their vectors did not point in the same direction. Explain. 8. Critical Thinking A car travels straight along the street from the grocery store to the post office. To represent its motion you use a coordinate system with its origin at the grocery store and the direction the car is moving in as the positive direction. Your friend uses a coordinate system with its origin at the post office and the opposite direction as the positive direction. Would the two of you agree on the car\u2019s position? Displacement? Distance? The time interval the trip took? Explain. physicspp.com/self_check_quiz Section 2.2 Where and When? 37 038-042 CH02-S3-845813 5/28/04 1:06 PM Page 38 2.3 Position-Time Graphs Objectives \u2022 Develop position-time graphs for moving objects. \u2022 Use a position-time graph to interpret an object\u2019s position or displacement. \u2022 Make motion diagrams, pictorial representations, and position-time graphs that are equivalent representations describing an object\u2019s motion. Vocabulary position-time graph instantaneous position Table 2-1 Position v. Time Time t (s) Position d (m) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 38 Chapter 2 Representing Motion When analyzing motion, particularly when it is more complex than the examples considered so far, it often is useful to represent the motion of an object in a variety of ways. As you have seen, a motion diagram contains useful information about an object\u2019s position at various times and can be helpful in determining the displacement of an object during time intervals. Graphs of the object\u2019s position and time also contain this information. Review Figure 2-9, the motion diagram for the runner with a location to the left of the tree chosen as the origin. From this motion diagram, you can organize the times and corresponding positions of the runner, as in Table 2-1. Using a Graph to Find Out Where and When The data from Table 2-1 can be presented by plotting the time data on a horizontal axis and the position data on a vertical axis, which is called a position-time graph. The graph of the runner\u2019s motion is shown in Figure 2-12. To draw this graph, first plot the runner\u2019s recorded positions. Then, draw a line that best fits the recorded points. Notice that this graph is not a picture of the path taken by the runner as he was moving\u2014the graphed line is sloped, but the path that he ran was flat. The line represents the most likely positions of the runner at the times between the recorded data points. (Recall from Chapter 1 that this line often is referred to as a best-fit line.) This means that even though there is no data point to tell you exactly when the runner was 30.0 m beyond his starting point or where he was at t 4.5 s, you can use the graph to estimate his position. The following example problem shows how. Note that before estimating the runner\u2019s position, the questions first are restated in the language of physics in terms of positions and times. \u25a0 Figure 2-12 A position-time graph for the runner can be created by plotting his known position at each of several times. After these points are plotted, the line that best fits them is drawn. The best-fit line indicates the runner\u2019s most likely positions at the times between the data", " points. Position v. Time ) 30.0 25.0 20.0 15.0 10.0 5.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 Time (s) 038-042 CH02-S3-845813 5/28/04 1:06 PM Page 39 When did the runner whose motion is described in Figure 2-12 reach 30.0 m beyond the starting point? Where was he after 4.5 s? 1 Analyze the Problem \u2022 Restate the questions. Question 1: At what time was the position of the object equal to 30.0 m? Question 2: What was the position of the object at 4.5 s? 2 Solve for the Unknown Question 1 Examine the graph to find the intersection of the best-fit line with a horizontal line at the 30.0-m mark. Next, find where a vertical line from that point crosses the time axis. The value of t there is 6.0 s 30.0 25.0 20.0 15.0 10.0 5.0 0.0 Question 2 Find the intersection of the graph with a vertical line at 4.5 s (halfway between 4.0 s and 5.0 s on this graph). Next, find where a horizontal line from that point crosses the position axis. The value of d is approximately 22.5 m. The two intersections are shown on the graph above. Position v. Time 1.0 2.0 3.0 4.0 5.0 6.0 Time (s) Math Handbook Interpolating and Extrapolating page 849 For problems 9\u201311, refer to Figure 2-13. 9. Describe the motion of the car shown by the graph. 10. Draw a motion diagram that corresponds to the graph. 11. Answer the following questions about the car\u2019s motion. Assume that the positive d-direction is east and the negative d-direction is west. a. When was the car 25.0 m east of the origin? b. Where was the car at 1.0 s? 12. Describe, in words, the motion of the two pedestrians shown by the lines in Figure 2-14. Assume that the positive direction is east on Broad Street and the origin is the intersection of Broad and High Streets. 13. Odina walked down the hall at school from the cafeteria to the band room, a distance of 100.0 m. A class of physics students recorded and graphed her position every 2.0 s, noting that she moved 2.6 m every 2.0 s. When was Odina in the following positions? a. 25.0 m from the cafeteria b. 25.0 m from the band room c. Create a graph showing Odina\u2019s motion. 150.0 100.0 50.0 0. 50.0 \u25a0 Figure 2-13 1.0 3.0 5.0 7.0 Time (s) Broad St. \u25a0 Figure 2-14 East ) High St. B West Time (s) Section 2.3 Position-Time Graphs 39 038-042 CH02-S3-845813 5/28/04 3:22 PM Page 40 a Table 2-1 b Position v. Time Position v. Time Time t (s) Position d (m) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 0.0 5.0 10.0 15.0 20.0 25.0 30. 30.0 25.0 20.0 15.0 10.0 5.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 Time (s) c Begin End \u25a0 Figure 2-15 The data table (a), position-time graph (b), and particle model (c) all represent the same moving object. How long did the runner spend at any location? Each position has been linked to a time, but how long did that time last? You could say \u201can instant,\u201d but how long is that? If an instant lasts for any finite amount of time, then the runner would have stayed at the same position during that time, and he would not have been moving. However, as he was moving, an instant is not a finite period of time. This means that an instant of time lasts zero seconds. The symbol d represents the instantaneous position of the runner\u2014the position at a particular instant. Equivalent representations As shown in Figure 2-15, you now have several different ways to describe motion: words, pictures (or pictorial representations), motion diagrams, data tables, and position-time graphs. All of these representations are equivalent. That is, they can all contain the same information about the runner\u2019s motion. However, depending on what you want to find out about an object\u2019s", " motion, some of the representations will be more useful than others. In the pages that follow, you will get some practice constructing these equivalent representations and learning which ones are the easiest to use in solving different kinds of problems. Considering the motion of multiple objects A position-time graph for two different runners in a race is shown in Example Problem 2. When and where does one runner pass the other? First, you need to restate this question in physics terms: At what time do the two objects have the same position? You can evaluate this question by identifying the point on the position-time graph at which the lines representing the two objects intersect. Niram, Oliver, and Phil all enjoy exercising and often go to a path along the river for this purpose. Niram bicycles at a very consistent 40.25 km/h, Oliver runs south at a constant speed of 16.0 km/h, and Phil walks south at a brisk 6.5 km/h. Niram starts biking north at noon from the waterfalls. Oliver and Phil both start at 11:30 A.M. at the canoe dock, 20.0 km north of the falls. 1. Draw position-time graphs for each person. 2. At what time will the three exercise enthusiasts be within the smallest distance interval? 3. What is the length of that distance interval? 40 Chapter 2 Representing Motion 038-042 CH02-S3-845813 5/28/04 1:07 PM Page 41 When and where does runner B pass runner A? 1 2 Analyze the Problem \u2022 Restate the question. At what time do A and B have the same position? Solve for the Unknown In the figure at right, examine the graph to find the intersection of the line representing the motion of A with the line representing the motion of B. These lines intersect at 45 s and at about 190 m. B passes A about 190 m beyond the origin, 45 s after A has passed the origin. Math Handbook Interpolating and Extrapolating page 849 ) 200 150 100 50 0 50 100 A B 15 25 35 45 55 Time (s) For problems 14\u201317, refer to the figure in Example Problem 2. 14. What event occurred at t 0.0 s? 15. Which runner was ahead at t 48.0 s? 16. When runner A was at 0.0 m, where was runner B? 17. How far apart were runners A and B at t 20.0 s? 18. Juanita goes for a walk. Sometime later, her friend Heather starts to walk after her. Their motions are represented by the position-time graphs in Figure 2-16. a. How long had Juanita been walking when Heather started her walk? b. Will Heather catch up to Juanita? How can you tell.0 5.0 4.0 3.0 2.0 1.0 0.0 \u25a0 Figure 2-16 Juanita e r a t h e H 0.5 1.0 1.5 2.0 Time (h) Section 2.3 Position-Time Graphs 41 038-042 CH02-S3-845813 5/28/04 1:07 PM Page 42 As you have seen, you can represent the motion of more than one object on a position-time graph. The intersection of two lines tells you when the two objects have the same position. Does this mean that they will collide? Not necessarily. For example, if the two objects are runners and if they are in different lanes, they will not collide. Later in this textbook, you will learn to represent motion in two dimensions. Is there anything else that you can learn from a position-time graph? Do you know what the slope of a line means? In the next section, you will use the slope of a line on a position-time graph to determine the velocity of an object. What about the area under a plotted line? In Chapter 3, you will draw other graphs and learn to interpret the areas under the plotted lines. In later chapters you will continue to refine your skills with creating and interpreting graphs. 2.3 Section Review 19. Position-Time Graph From the particle model in Figure 2-17 of a baby crawling across a kitchen floor, plot a position-time graph to represent his motion. The time interval between successive dots is 1 s. 0 20 40 60 80 100 120 140 160 Position (cm) \u25a0 Figure 2-17 20. Motion Diagram Create a particle model from the position-time graph of a hockey puck gliding across a frozen pond in Figure 2-18. 22. Distance Use the position-time graph of the hockey puck to determine how far it moved between 0.0 s and 5.0 s. 23. Time Interval Use the position-time graph for the hockey puck to determine how much time it took for the puck to go from 40 m beyond the origin to 80 m beyond the origin. 24. Critical Thinking Look at the particle", " model and position-time graph shown in Figure 2-19. Do they describe the same motion? How do you know? Do not confuse the position coordinate system in the partical model with the horizontal axis in the position-time graph. The time intervals in the partical model are 2 s. 140 120 100 80 60 40 20 ).0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 Time (s) \u25a0 Figure 2-18 For problems 21\u201323, refer to Figure 2-18. 21. Time Use the position-time graph of the hockey puck to determine when it was 10.0 m beyond the origin. 0 12 10 Position (m) 1 2 3 4 5 Time (s) \u25a0 Figure 2-19 42 Chapter 2 Representing Motion physicspp.com/self_check_quiz 043-047 CH02-S4-845813 5/28/04 1:12 PM Page 43 2.4 How Fast? You have learned how to use a motion diagram to show an object\u2019s movement. How can you measure how fast it is moving? With devices such as a meterstick and a stopwatch, you can measure position and time. Can this information be used to describe the rate of motion? Velocity Suppose you recorded two joggers on one motion diagram, as shown in Figure 2-20a. From one frame to the next you can see that the position of the jogger in red shorts changes more than that of the one wearing blue. In other words, for a fixed time interval, the displacement, d, is greater for the jogger in red because she is moving faster. She covers a larger distance than the jogger in blue does in the same amount of time. Now, suppose that each jogger travels 100.0 m. The time interval, t, would be smaller for the jogger in red than for the one in blue. Average velocity From the example of the joggers, you can see that both the displacement, d, and time interval, t, might be needed to create the quantity that tells how fast an object is moving. How might they be combined? Compare the lines representing the red and blue joggers in the position-time graphs in Figure 2-20b. The slope of the red jogger\u2019s line is steeper than the slope of the blue jogger\u2019s line. A steeper slope indicates a greater change in displacement during each time interval. Recall from Chapter 1 that to find the slope, you first choose two points on the line. Next, you subtract the vertical coordinate (d in this case) of the first point from the vertical coordinate of the second point to obtain the rise of the line. After that, you subtract the horizontal coordinate (t in this case) of the first point from the horizontal coordinate of the second point to obtain the run. Finally, you divide the rise by the run to obtain the slope. The slopes of the two lines shown in Figure 2-20b are found as follows: d d Red slope i f ti t f d d Blue slope i f ti t f m. 0 2 6 m.0 s 0. 1 s.0 3 2.0 m/s m. 0 2 3 m.0 s 0. 2 s.0 3 1.0 m/s a Objectives \u2022 Define velocity. \u2022 Differentiate between speed and velocity. \u2022 Create pictorial, physical, and mathematical models of motion problems. Vocabulary average velocity average speed instantaneous velocity \u25a0 Figure 2-20 The red jogger\u2019s displacement is greater than the displacement of the blue jogger in each time interval because the jogger in red is moving faster than the jogger in blue (a). The position-time graph represents the motion of the red and blue joggers. The points used to calculate the slope of each line are shown (b)..0 5.0 4.0 3.0 2.0 1.0 0.0 Red jogger.0 2.0 3.0 Section 2.4 How Fast? 43 Hutchings Photography 043-047 CH02-S4-845813 3/13/04 4:54 AM Page 44 Speed Records The world record for the men\u2019s 100-m dash is 9.78 s, established in 2002 by Tim Montgomery. The world record for the women\u2019s 100-m dash is 10.65 s, established in 1998 by Marion Jones. These sprinters often are referred to as the world\u2019s fastest man and woman. There are some important things to notice about this comparison. First, the slope of the faster runner is a greater number, so it is reasonable to assume that this number might be connected with the runner\u2019s speed. Second, look at the units of the slope, meters per second. In other words, the slope tells how many meters the runner moved in 1 s. These units are similar to miles", " per hour, which also measure speed. Looking at how the slope is calculated, you can see that slope is the change in position, divided by the time interval during which that change took place, or ti), or d/t. When d gets larger, the slope gets larger; (df when t gets larger, the slope gets smaller. This agrees with the interpretation above of the movements of the red and blue joggers. The slope of a position-time graph for an object is the object\u2019s average velocity and is represented by the ratio of the change of position to the time interval during which the change occurred. di) / (tf Average Velocity d f t f Average velocity is defined as the change in position, divided by the time during which the change occurred. d d v i ti t The symbol means that the left-hand side of the equation is defined by the right-hand side. It is a common misconception to say that the slope of a position-time graph gives the speed of the object. Consider the slope of the position-time graph shown in Figure 2-21. The slope of this position-time graph is 5.0 m/s. As you can see the slope indicates both the magnitude and direction. Recall that average velocity is a quantity that has both magnitude and direction. The slope of a position-time graph indicates the average velocity of the object and not its speed. Take another look at Figure 2-21. The slope of the graph is 5.0 m/s and thus, the object has a velocity of 5.0 m/s. The object starts out at a positive position and moves toward the origin. It moves in the negative direction at a rate of 5.0 m/s. Average speed The absolute value of the slope of a position-time graph tells you the average speed of the object; that is, how fast the object is moving. The sign of the slope tells you in what direction the object is moving. The combination of an object\u2019s average speed, v, and the direction in which it is moving is the average velocity, v. Thus, for the object represented in Figure 2-21, the average velocity is 5.0 m/s, or 5.0 m/s in the negative direction. Its average speed is 5.0 m/s. Remember that if an object moves in the negative direction, then its displacement is negative. This means that the object\u2019s velocity always will have the same sign as the object\u2019s displacement. 1 2 3 4 5 Time (s) As you consider other types of motion to analyze in future chapters, sometimes the velocity will be the important quantity to consider, while at other times, the speed will be the important quantity. Therefore, it is a good idea to make sure that you understand the differences between velocity and speed so that you will be sure to use the right one later. \u2022 Velocity vectors are red. \u2022 Displacement vectors are green. \u25a0 Figure 2-21 The object whose motion is represented here is moving in the negative direction at a rate of 5.0 m/s. 20 15 10 5 0 5 10 15 ) 44 Chapter 2 Representing Motion 043-047 CH02-S4-845813 6/4/04 5:41 AM Page 45 The graph at the right describes the motion of a student riding his skateboard along a smooth, pedestrian-free sidewalk. What is his average velocity? What is his average speed? 1 2 Analyze and Sketch the Problem \u2022 Identify the coordinate system of the graph. Solve for the Unknown Unknown: v? v? Find the average velocity using two points on the line. d v t Use magnitudes with signs indicating directions 12.0 9.0 6.0 3.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 Time (s) d d 1 2 t1 t 2 m Substitute d2 12 m 0..0 6 s 0. 8 4.0 s \uf8e7 1.5 m/s in the positive direction v 12.0 m, d1 6.0 m, t2 8.0 s, t1 4.0 s. Math Handbook Slope page 850 The average speed, v, is the absolute value of the average velocity, or 1.5 m/s. 3 Evaluate the Answer \u2022 Are the units correct? m/s are the units for both velocity and speed. \u2022 Do the signs make sense? The positive sign for the velocity agrees with the coordinate system. No direction is associated with speed. 25. The graph in Figure 2-22 describes the motion of a cruise ship during its voyage through calm waters. The positive d-direction is defined to be south. a. What is the ship\u2019s average speed? b. What is its average velocity? 26. Describe, in words, the motion of the cruise ship in", " the previous problem. Time (s 27. The graph in Figure 2-23 represents the motion \u25a0 Figure 2-22 of a bicycle. Determine the bicycle\u2019s average speed and average velocity, and describe its motion in words. 28. When Marilyn takes her pet dog for a walk, the dog walks at a very consistent pace of 0.55 m/s. Draw a motion diagram and position-time graph to represent Marilyn\u2019s dog walking the 19.8-m distance from in front of her house to the nearest fire hydrant 20 15 10 5 0 5 10 15 20 25 30 \u25a0 Figure 2-23 Time (min) Section 2.4 How Fast? 45 043-047 CH02-S4-845813 5/28/04 1:17 PM Page 46 Instantaneous Velocity Why is the quantity d/t called average velocity? Why isn\u2019t it called velocity? Think about how a motion diagram is constructed. A motion diagram shows the position of a moving object at the beginning and end of a time interval. It does not, however, indicate what happened within that time interval. During the time interval, the speed of the object could have remained the same, increased, or decreased. The object may have stopped or even changed direction. All that can be determined from the motion diagram is an average velocity, which is found by dividing the total displacement by the time interval in which it took place. The speed and direction of an object at a particular instant is called the instantaneous velocity. In this textbook, the term velocity will refer to instantaneous velocity, represented by the symbol v. Average Velocity on Motion Diagrams How can you show average velocity on a motion diagram? Although the average velocity is in the same direction as displacement, the two quantities are not measured using the same units. Nevertheless, they are proportional\u2014when displacement is greater during a given time interval, so is average velocity. A motion diagram is not a precise graph of average velocity, but you can indicate the direction and magnitude of the average velocity on it. Imagine two cars driving down the road at different speeds. A video camera records their motions at the rate of one frame every second. Imagine that each car has a paintbrush attached to it that automatically descends and paints a line on the ground for half a second every second. The faster car would paint a longer line on the ground. The vectors you draw on a motion diagram to represent the velocity are like the lines made by the paintbrushes on the ground below the cars. Red is used to indicate velocity vectors on motion diagrams. Figure 2-24 shows the particle models, complete with velocity vectors, for two cars: one moving to the right and the other moving more slowly to the left. Using equations Any time you graph a straight line, you can find an equation to describe it. There will be many cases for which it will be more efficient to use such an equation instead of a graph to solve problems. Take another look at the graph in Figure 2-21 on page 44 for the object moving with a constant velocity of 5.0 m/s. Recall from Chapter 1 that any straight line can be represented by the formula: y mx b where y is the quantity plotted on the vertical axis, m is the slope of the line, x is the quantity plotted on the horizontal axis, and b is the y-intercept of the line. For the graph in Figure 2-21, the quantity plotted on the vertical axis is position, and the variable used to represent position is d. The slope of the line is 5.0 m/s, which is the object\u2019s average velocity, v. The quantity plotted on the horizontal axis is time, t. The y-intercept is 20.0 m. What does this 20.0 m represent? By inspecting the graph and thinking about how the object moves, you can figure out that the object was at a position of 20.0 m when t 0.0 s. This is called the initial position of the object, and is designated di. Table 2-2 summarizes how the general variables in the straight-line formula are changed to the specific variables you have been using to describe motion. It also shows the numerical values for the two constants in this equation. \u25a0 Figure 2-24 Average velocity vectors have the same direction as their corresponding displacement vectors. Their lengths are different, but proportional, and they have different units because they are obtained by dividing the displacement by the time interval. Instantaneous Velocity Vectors 1. Attach a 1-m-long string to your hooked mass. 2. Hold the string in one hand with the mass suspended. 3. Carefully pull the mass to one side and release it. 4. Observe the motion, the speed, and direction of the mass for several swings. 5. Stop the mass from swinging. 6. Draw a diagram showing instantaneous velocity vectors at the following points: top of the swing, midpoint between top and bottom, bottom of the swing,", " midpoint between bottom and top, and back to the top of the swing. Analyze and Conclude 7. Where was the velocity greatest? 8. Where was the velocity least? 9. Explain how the average speed can be determined using your vector diagram. 46 Chapter 2 Representing Motion 043-047 CH02-S4-845813 3/13/04 4:56 AM Page 47 Table 2-2 Comparison of Straight Lines with Position-Time Graphs General Variable Specific Motion Variable Value in Figure 2-21 y m x b d v t di 5.0 m/s 20.0 m Based on the information shown in Table 2-2, the equation y mx b becomes d vt di, or, by inserting the values of the constants, d (5.0 m/s)t 20.0 m. This equation describes the motion that is represented in Figure 2-21. You can check this by plugging a value of t into the equation and seeing that you obtain the same value of d as when you read it directly from the graph. To conduct an extra check to be sure the equation makes sense, take a look at the units. You cannot set two items with different units equal to each other in an equation. In this equation, the left-hand side is a position, so its units are meters. The first term on the right-hand side multiplies meters per second times seconds, so its units are also meters. The last term is in meters, too, so the units on this equation are valid. Equation of Motion for Average Velocity d vt di An object\u2019s position is equal to the average velocity multiplied by time plus the initial position. This equation gives you another way to represent the motion of an object. Note that once a coordinate system is chosen, the direction of d is specified by positive and negative values, and the boldface notation can be dispensed with, as in \u201cd-axis.\u201d However, the boldface notation for velocity will be retained to avoid confusing it with speed. Your toolbox of representations now includes words, motion diagrams, pictures, data tables, position-time graphs, and an equation. You should be able to use any one of these representations to generate at least the characteristics of the others. You will get more practice with this in the rest of this chapter and also in Chapter 3 as you apply these representations to help you solve problems. 2.4 Section Review For problems 29\u201331, refer to Figure 2-25. 29. Average Speed Rank the position-time graphs according to the average speed of the object, from greatest average speed to least average speed. Specifically indicate any ties \u25a0 Figure 2-25 Time (s) C 30. Average Velocity Rank the graphs according to average velocity, from greatest average velocity to least average velocity. Specifically indicate any ties. 31. Initial Position Rank the graphs according to the object\u2019s initial position, from most positive position to most negative position. Specifically indicate any ties. Would your ranking be different if you had been asked to do the ranking according to initial distance from the origin? 32. Average Speed and Average Velocity Explain how average speed and average velocity are related to each other. 33. Critical Thinking In solving a physics problem, why is it important to create pictorial and physical models before trying to solve an equation? physicspp.com/self_check_quiz Section 2.4 How Fast? 47 048-049 CH02-LAB-845813 5/28/04 1:33 PM Page 48 Alternate CBL instructions can be found on the Web site. physicspp.com Creating Motion Diagrams In this activity you will construct motion diagrams for two toy cars. A motion diagram consists of a series of images showing the positions of a moving object at equal time intervals. Motion diagrams help describe the motion of an object. By looking at a motion diagram you can determine whether an object is speeding up, slowing down, or moving at constant speed. QUESTION How do the motion diagrams of a fast toy car and a slow toy car differ? Objectives Procedure \u25a0 Measure in SI the location of a moving object. \u25a0 Recognize spatial relationships of moving 1. Mark a starting line on the lab table or the surface recommended by your teacher. objects. \u25a0 Describe the motion of a fast and slow object. Safety Precautions Materials video camera two toy windup cars meterstick foam board 2. Place both toy cars at the starting line and release them at the same time. Be sure to wind them up before releasing them. 3. Observe both toy cars and determine which one is faster. 4. Place the slower toy car at the starting line. 5. Place a meterstick parallel to the path the toy car will take. 6. Have one of the members of your group get ready to operate the video camera. 7. Release the slower toy car from the starting line. Be sure to wind up the toy car before releasing it. 8. Use the video", " camera to record the slower toy car\u2019s motion parallel to the meterstick. 9. Set the recorder to play the tape frame by frame. Replay the video tape for 0.5 s, pressing the pause button every 0.1 s (3 frames). 10. Determine the toy car\u2019s position for each time interval by reading the meterstick on the video tape and record it in the data table. 11. Repeat steps 5\u201310 with the faster car. 12. Place a piece of foam board at an angle of approximately 30\u00b0 to form a ramp. 13. Place the meterstick on the ramp so that it is parallel to the path the toy car will take. 14. Place the slower toy car at the top of the ramp and repeat steps 6\u201310. 48 048-049 CH02-LAB-845813 3/13/04 5:02 AM Page 49 Data Table 1 Time (s) Position of the Slower Toy Car (cm) 0.0 0.1 0.2 0.3 0.4 0.5 Data Table 2 Data Table 3 Time (s) Position of the Faster Toy Car (cm) Time (s) Position of the Slower Toy Car on the Ramp (cm) 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 Analyze 1. Draw a motion diagram for the slower toy car using the data you collected. 2. Draw a motion diagram for the faster toy car using the data you collected. 3. Using the data you collected, draw a motion diagram for the slower toy car rolling down the ramp. Conclude and Apply How is the motion diagram of the faster toy car different from the motion diagram of the slower toy car? Going Further 1. Draw a motion diagram for a car moving at a constant speed. 2. What appears to be the relationship between the distances between points in the motion diagram of a car moving at a constant speed? 3. Draw a motion diagram for a car that starts moving fast and then begins to slow down. 4. What happens to the distance between points in the motion diagram in the previous question as the car slows down? 5. Draw a motion diagram for a car that starts moving slowly and then begins to speed up. 6. What happens to the distance between points in the motion diagram in the previous question as the car speeds up? Real-World Physics Suppose a car screeches to a halt to avoid an accident. If that car has antilock brakes that pump on and off automatically every fraction of a second, what might the tread marks on the road look like? Include a drawing along with your explanation of what the pattern of tread marks on the road might look like. To find out more about representing motion, visit the Web site: physicspp.com 49 050 CH02-FEATURE-845813 3/19/04 1:39 PM Page 50 aligned, the cesium atom is in one energy state. If they are oppositely aligned, the atom is in another energy state. A microwave with a particular frequency can strike a cesium atom and cause the outside spinning electron to switch its magnetic pole orientation and change the atom\u2019s energy state. As a result, the atom emits light. This occurs at cesium\u2019s natural frequency of 9,192,631,770 cycles/s. This principle was used to design the cesium clock. The cesium clock, NIST-F1, located at the NIST laboratories in Boulder, Colorado is among the most accurate clocks in the world. How Does the Cesium Clock Work? The cesium clock consists of cesium atoms and a quartz crystal oscillator, which produces microwaves. When the oscillator\u2019s microwave signal precisely equals cesium\u2019s natural frequency, a large number of cesium atoms will change their energy state. Cesium\u2019s natural frequency is equal to 9,192,631,770 microwave cycles. Thus, there are 9,192,631,770 cesium energy level changes in 1 s. Cesium clocks are so accurate that a modern cesium clock is off by less than 1 s in 20 million years. Going Further 1. Research What processes require the precise measurement of time? 2. Analyze and Conclude Why is the precise measurement of time essential to space navigation? Accurate Time What time is it really? You might use a clock to find out what time it is at any moment. A clock is a device that counts regularly recurring events in order to measure time. Suppose the clock in your classroom reads 9:00. Your watch, however, reads 8:55, and your friend\u2019s watch reads 9:02. So what time is it, really? Which clock or watch is accurate? Many automated processes are controlled by clocks. For example, an automated bell that", " signals the end of a class period is controlled by a clock. Thus, if you wanted to be on time for a class you would have to synchronize your watch to the one controlling the bell. Other processes, such as GPS navigation, space travel, internet synchronization, transportation, and communication, rely on clocks with extreme precision and accuracy. A reliable standard clock that can measure when exactly one second has elapsed is needed. The Standard Cesium Clock Atomic clocks, such as cesium clocks, address this need. Atomic clocks measure the number of times the atom used in the clock switches its energy state. Such oscillations in an atom\u2019s energy occur very quickly and regularly. The National Institute of Standards and Technology (NIST) currently uses the oscillations of the cesium atom to determine the standard 1-s interval. One second is defined as the duration of 9,192,631,770 oscillations of the cesium atom. The cesium atom has a single electron in its outermost energy level. This outer electron spins and behaves like a miniature magnet. The cesium nucleus also spins and acts like a miniature magnet. The nucleus and electron may spin in such a manner that their north magnetic poles are aligned. The nucleus and electron also may spin in a way that causes opposite poles to be aligned. If the poles are 50 National Institute of Standards and Technology Extreme Physics 051-055 CH02-SG CA STP-845813 3/13/04 5:12 AM Page 51 2.1 Picturing Motion Vocabulary \u2022 motion diagram (p. 33) \u2022 particle model (p. 33) Key Concepts \u2022 A motion diagram shows the position of an object at successive times. \u2022 In the particle model, the object in the motion diagram is replaced by a series of single points. 2.2 Where and When? Vocabulary \u2022 coordinate system (p. 34) \u2022 origin (p. 34) \u2022 position (p. 34) \u2022 distance (p. 34) \u2022 magnitude (p. 35) \u2022 vectors (p. 35) \u2022 scalars (p. 35) \u2022 resultant (p. 35) \u2022 time interval (p. 36) \u2022 displacement (p. 36) Key Concepts \u2022 You can define any coordinate system you wish in describing motion, but some are more useful than others. \u2022 A time interval is the difference between two times. t tf ti \u2022 A vector drawn from the origin of the coordinate system to the object indicates the object\u2019s position. \u2022 Change in position is displacement, which has both magnitude and direction. d df di \u2022 The length of the displacement vector represents how far the object was displaced, and the vector points in the direction of the displacement. 2.3 Position-Time Graphs Vocabulary \u2022 position-time graph (p. 38) \u2022 instantaneous position Key Concepts \u2022 Position-time graphs can be used to find the velocity and position of an object, as well as where and when two objects meet. (p. 40) \u2022 Any motion can be described using words, motion diagrams, data tables, and graphs. 2.4 How Fast? Vocabulary \u2022 average velocity (p. 44) \u2022 average speed (p. 44) \u2022 instantaneous velocity (p. 46) Key Concepts \u2022 The slope of an object\u2019s position-time graph is the average velocity of the object\u2019s motion. \u2206 d d v i \u2206 ti t d f t f \u2022 The average speed is the absolute value of the average velocity. \u2022 An object\u2019s velocity is how fast it is moving and in what direction it is moving. \u2022 An object\u2019s initial position, di, its constant average velocity, v, its position, d, and the time, t, since the object was at its initial position are related by a simple equation. d vt di physicspp.com/vocabulary_puzzlemaker 51 051-055 CH02-SG CA STP-845813 3/13/04 5:12 AM Page 52 Concept Mapping 34. Complete the concept map below using the following terms: words, equivalent representations, position-time graph. data table motion diagram Mastering Concepts 35. What is the purpose of drawing a motion diagram? (2.1) 36. Under what circumstances is it legitimate to treat an object as a point particle? (2.1) 37. The following quantities describe location or its change: position, distance, and displacement. Briefly describe the differences among them. (2.2) 38. How can you use a clock to find a time interval? (2.2) 39. In-line Skating How can you use the position-time graphs for two in-line skaters to determine if and when one in-line skater will pass the other one? (2.3) 40. Walking Versus Running A walker and a runner leave your front door at the same time. They move in the same direction at different constant velocities. Desc", "ribe the position-time graphs of each. (2.4) 41. What does the slope of a position-time graph measure? (2.4) 42. If you know the positions of a moving object at two points along its path, and you also know the time it took for the object to get from one point to the other, can you determine the particle\u2019s instantaneous velocity? Its average velocity? Explain. (2.4) Applying Concepts 43. Test the following combinations and explain why each does not have the properties needed to describe the concept of velocity: d t, d t, d t, t/d. 44. Football When can a football be considered a point particle? 45. When can a football player be treated as a point particle? 46. Figure 2-26 is a graph of two people running. a. Describe the position of runner A relative to runner B at the y-intercept. b. Which runner is faster? c. What occurs at point P and beyond Time (s) \u25a0 Figure 2-26 47. The position-time graph in Figure 2-27 shows the motion of four cows walking from the pasture back to the barn. Rank the cows according to their average velocity, from slowest to fastest. Elsie B essie M o o l i n d a olly Time (s) \u25a0 Figure 2-27 48. Figure 2-28 is a position-time graph for a rabbit running away from a dog. a. Describe how this graph would be different if the rabbit ran twice as fast. b. Describe how this graph would be different if the rabbit ran in the opposite direction Time (s) \u25a0 Figure 2-28 52 Chapter 2 Representing Motion For more problems, go to Additional Problems, Appendix B. 051-055 CH02-SG CA STP-845813 3/13/04 5:13 AM Page 53 Mastering Problems 2.4 How Fast? 49. A bike travels at a constant speed of 4.0 m/s for 5.0 s. How far does it go? 50. Astronomy Light from the Sun reaches Earth in 8.3 min. The speed of light is 3.00108 m/s. How far is Earth from the Sun? 51. A car is moving down a street at 55 km/h. A child suddenly runs into the street. If it takes the driver 0.75 s to react and apply the brakes, how many meters will the car have moved before it begins to slow down? 52. Nora jogs several times a week and always keeps track of how much time she runs each time she goes out. One day she forgets to take her stopwatch with her and wonders if there\u2019s a way she can still have some idea of her time. As she passes a particular bank, she remembers that it is 4.3 km from her house. She knows from her previous training that she has a consistent pace of 4.0 m/s. How long has Nora been jogging when she reaches the bank? 56. Figure 2-30 shows position-time graphs for Joszi and Heike paddling canoes in a local river. a. At what time(s) are Joszi and Heike in the same place? b. How much time does Joszi spend on the river before he passes Heike? c. Where on the river does it appear that there might be a swift current 18 16 14 12 10 8 6 4 2 0 Joszi Heike 0.5 1.0 1.5 2.0 2.5 Time (h) \u25a0 Figure 2-30 53. Driving You and a friend each drive 50.0 km. You travel at 90.0 km/h; your friend travels at 95.0 km/h. How long will your friend have to wait for you at the end of the trip? 57. Driving Both car A and car B leave school when a stopwatch reads zero. Car A travels at a constant 75 km/h, and car B travels at a constant 85 km/h. a. Draw a position-time graph showing the motion Mixed Review 54. Cycling A cyclist maintains a constant velocity of 5.0 m/s. At time t 0.0 s, the cyclist is 250 m from point A. a. Plot a position-time graph of the cyclist\u2019s location from point A at 10.0-s intervals for 60.0 s. b. What is the cyclist\u2019s position from point A at 60.0 s? c. What is the displacement from the starting position at 60.0 s? 55. Figure 2-29 is a particle model for a chicken casually walking across the road. Time intervals are every 0.1 s. Draw the corresponding positiontime graph and equation to describe the chicken\u2019s motion. This side The other side Time intervals are 0.1 s. \u25a0 Figure 2-29 of both cars. How far", " are the two cars from school when the stopwatch reads 2.0 h? Calculate the distances and show them on your graph. b. Both cars passed a gas station 120 km from the school. When did each car pass the gas station? Calculate the times and show them on your graph. 58. Draw a position-time graph for two cars traveling to the beach, which is 50 km from school. At noon, Car A leaves a store that is 10 km closer to the beach than the school is and moves at 40 km/h. Car B starts from school at 12:30 P.M. and moves at 100 km/h. When does each car get to the beach? 48.0 km 59. Two cars travel along a straight road. When a stopwatch reads t 0.00 h, car A is at dA moving at a constant 36.0 km/h. Later, when the watch reads t 0.50 h, car B is at dB 0.00 km moving at 48.0 km/h. Answer the following questions, first, graphically by creating a positiontime graph, and second, algebraically by writing equations for the positions dA and dB as a function of the stopwatch time, t. a. What will the watch read when car B passes car A? b. At what position will car B pass car A? c. When the cars pass, how long will it have been since car A was at the reference point? physicspp.com/chapter_test Chapter 2 Assessment 53 051-055 CH02-SG CA STP-845813 6/16/04 9:53 PM Page 54 60. Figure 2-31 shows the position-time graph depicting Jim\u2019s movement up and down the aisle at a store. The origin is at one end of the aisle. a. Write a story describing Jim\u2019s movements at the store that would correspond to the motion represented by the graph. b. When does Jim have a position of 6.0 m? c. How much time passes between when Jim enters the aisle and when he gets to a position of 12.0 m? What is Jim\u2019s average velocity between 37.0 s and 46.0 s? 14.0 12.0 10.0 8.0 6.0 4.0 2..00 10.0 20.0 30.0 40.0 50.0 60.0 \u25a0 Figure 2-31 Time (s) Thinking Critically 61. Apply Calculators Members of a physics class stood 25 m apart and used stopwatches to measure the time at which a car traveling on the highway passed each person. Their data are shown in Table 2-3. Table 2-3 Position v. Time Time Position (m) 0.0 1.3 2.7 3.6 5.1 5.9 7.0 8.6 10.3 0.0 25.0 50.0 75.0 100.0 125.0 150.0 175.0 200.0 Use a graphing calculator to fit a line to a positiontime graph of the data and to plot this line. Be sure to set the display range of the graph so that all the data fit on it. Find the slope of the line. What was the speed of the car? 62. Apply Concepts You plan a car trip for which you want to average 90 km/h. You cover the first half of the distance at an average speed of only 48 km/h. What must your average speed be in the second half of the trip to meet your goal? Is this reasonable? Note that the velocities are based on half the distance, not half the time. 63. Design an Experiment Every time a particular red motorcycle is driven past your friend\u2019s home, his father becomes angry because he thinks the motorcycle is going too fast for the posted 25 mph (40 km/h) speed limit. Describe a simple experiment you could do to determine whether or not the motorcycle is speeding the next time it is driven past your friend\u2019s house. 64. Interpret Graphs Is it possible for an object\u2019s position-time graph to be a horizontal line? A vertical line? If you answer yes to either situation, describe the associated motion in words. Writing in Physics 65. Physicists have determined that the speed of light is 3.00108 m/s. How did they arrive at this number? Read about some of the series of experiments that were done to determine light\u2019s speed. Describe how the experimental techniques improved to make the results of the experiments more accurate. 66. Some species of animals have good endurance, while others have the ability to move very quickly, but for only a short amount of time. Use reference sources to find two examples of each quality and describe how it is helpful to that animal. Cumulative Review 67. Convert each of the following time measurements to its equivalent in seconds. (Chapter 1", ") a. 58 ns b. 0.046 Gs c. 9270 ms d. 12.3 ks 68. State the number of significant digits in the following measurements. (Chapter 1) a. 3218 kg b. 60.080 kg c. 801 kg d. 0.000534 kg 69. Using a calculator, Chris obtained the following results. Rewrite the answer to each operation using the correct number of significant digits. (Chapter 1) a. 5.32 mm 2.1 mm 7.4200000 mm b. 13.597 m 3.65 m 49.62905 m2 c. 83.2 kg 12.804 kg 70.3960000 kg 54 Chapter 2 Representing Motion For more problems, go to Additional Problems, Appendix B. 051-055 CH02-SG CA STP-845813 7/13/04 1:50 PM Page 55 Multiple Choice 1. Which of the following statements would be true about the particle model motion diagram for an airplane taking off from an airport? The dots would form an evenly spaced pattern. The dots would be far apart at the beginning, but get closer together as the plane accelerated. The dots would be close together to start with, and get farther apart as the plane accelerated. The dots would be close together to start, get farther apart, and become close together again as the airplane leveled off at cruising speed. 2. Which of the following statements about drawing vectors is false? A vector diagram is needed to solve all physics problems properly. The length of the vector should be proportional to the data. Vectors can be added by measuring the length of each vector and then adding them together. Vectors can be added in straight lines or in triangle formations. Use this graph for problems 3\u20135. C IV D III B n o i t i s o P II I A Time 4. When is the person on the bicycle farthest away from the starting point? point A point B point C point D 5. Over what interval does the person on the bicycle travel the greatest distance? section I section II section III point IV 6. A squirrel descends an 8-m tree at a constant speed in 1.5 min. It remains still at the base of the tree for 2.3 min, and then walks toward an acorn on the ground for 0.7 min. A loud noise causes the squirrel to scamper back up the tree in 0.1 min to the exact position on the branch from which it started. Which of the following graphs would accurately represent the squirrel\u2019s vertical displacement from the base of the tree? Time (min Time (min Time (min) Time (min) Extended Answer 7. Find a rat\u2019s total displacement at the exit if it takes the following path in a maze: start, 1.0 m north, 0.3 m east, 0.8 m south, 0.4 m east, finish. 3. The graph shows the motion of a person on a bicycle. When does the person have the greatest velocity? section I section III point D point B Stock up on Supplies Bring all your test-taking tools: number two pencils, black and blue pens, erasers, correction fluid, a sharpener, a ruler, a calculator, and a protractor. physicspp.com/standardized_test Chapter 2 Standardized Test Practice 55 056-064 CH03-S1-845813 3/21/04 7:57 PM Page 56 What You\u2019ll Learn \u2022 You will develop descriptions of accelerated motion. \u2022 You will use graphs and equations to solve problems involving moving objects. \u2022 You will describe the motion of objects in free fall. Why It\u2019s Important Objects do not always move at constant velocities. Understanding accelerated motion will help you better decribe the motion of many objects. Acceleration Cars, planes, subways, elevators, and other common forms of transportation often begin their journeys by speeding up quickly, and end by stopping rapidly. Think About This The driver of a dragster on the starting line waits for the green light to signal the start of the race. At the signal, the driver will step on the gas pedal and try to speed up as quickly as possible. As the car speeds up, how will its position change? physicspp.com 56 Rob Tringali/SportsChrome 056-064 CH03-S1-845813 3/21/04 8:02 PM Page 57 Do all types of motion look the same when graphed? Question How does a graph showing constant speed compare to a graph of a vehicle speeding up? 7. Make and Use Graphs Make a graph of total distance versus interval number. Place data for both vehicles on the same plot and label each graph. Analysis Which vehicle moved with constant speed? Which one sped up? Explain how you determined this by looking at the timer tape. Critical Thinking Describe the shape of each graph. How does the shape", " of the graph relate to the type of motion observed? Procedure 1. Clamp a spark timer to the back edge of a lab table. 2. Cut a piece of timer tape approximately 50 cm in length, insert it into the timer, and tape it to vehicle 1. 3. Turn on the timer and release the vehicle. Label the tape with the vehicle number. 4. Raise one end of the lab table 8\u201310 cm by placing a couple of bricks under the back legs. CAUTION: Make sure the lab table remains stable. 5. Repeat steps 2\u20134 with vehicle 2, but hold the vehicle in place next to the timer and release it after the timer has been turned on. Catch the vehicle before it falls. 6. Construct and Organize Data Mark the first dark dot where the timer began as zero. Measure the distance to each dot from the zero dot for 10 intervals and record your data. 3.1 Acceleration Uniform motion is one of the simplest kinds of motion. You learned in Chapter 2 that an object in uniform motion moves along a straight line with an unchanging velocity. From your own experiences, you know, however, that few objects move in this manner all of the time. In this chapter, you will expand your knowledge of motion by considering a slightly more complicated type of motion. You will be presented with situations in which the velocity of an object changes, while the object\u2019s motion is still along a straight line. Examples of objects and situations you will encounter in this chapter include automobiles that are speeding up, drivers applying brakes, falling objects, and objects thrown straight upward. In Chapter 6, you will continue to add to your knowledge of motion by analyzing some common types of motion that are not confined to a straight line. These include motion along a circular path and the motion of thrown objects, such as baseballs. Objectives \u2022 Define acceleration. \u2022 Relate velocity and acceleration to the motion of an object. \u2022 Create velocity-time graphs. Vocabulary velocity-time graph acceleration average acceleration instantaneous acceleration Section 3.1 Acceleration 57 Horizons Companies 056-064 CH03-S1-845813 5/28/04 2:35 PM Page 58 A Steel Ball Race If two steel balls are released at the same instant, will the steel balls get closer or farther apart as they roll down a ramp? 1. Assemble an inclined ramp from a piece of U-channel or two metersticks taped together. 2. Measure 40 cm from the top of the ramp and place a mark there. Place another mark 80 cm from the top. 3. Predict whether the steel balls will get closer or farther apart as they roll down the ramp. 4. At the same time, release one steel ball from the top of the ramp and the other steel ball from the 40-cm mark. 5. Next, release one steel ball from the top of the ramp. As soon as it reaches the 40-cm mark, release the other steel ball from the top of the ramp. Analyze and Conclude 6. Explain your observations in terms of velocities. 7. Do the steel balls have the same velocity as they roll down the ramp? Explain. 8. Do they have the same acceleration? Explain. Changing Velocity You can feel a difference between uniform and nonuniform motion. Uniform motion feels smooth. You could close your eyes and it would feel as though you were not moving at all. In contrast, when you move along a curve or up and down a roller coaster, you feel pushed or pulled. Consider the motion diagrams shown in Figure 3-1. How would you describe the motion of the person in each case? In one diagram, the person is motionless. In another, she is moving at a constant speed. In a third, she is speeding up, and in a fourth, she is slowing down. How do you know which one is which? What information do the motion diagrams contain that could be used to make these distinctions? The most important thing to notice in these motion diagrams is the distance between successive positions. You learned in Chapter 2 that motionless objects in the background of motion diagrams do not change positions. Therefore, because there is only one image of the person in Figure 3-1a, you can conclude that she is not moving; she is at rest. Figure 3-1b is like the constant-velocity motion diagrams in Chapter 2. The distances between images are the same, so the jogger is moving at a constant speed. The distance between successive positions changes in the two remaining diagrams. If the change in position gets larger, the jogger is speeding up, as shown in Figure 3-1c. If the change in position gets smaller, as in Figure 3-1d, the jogger is slowing down. What does a particle-model motion diagram look like for an object with changing velocity? Figure 3-2 shows the particle-model motion diagrams below the motion diagrams of the jogger speeding up and slowing down. There are two major indicators of", " the change in velocity in this form of the motion diagram. The change in the spacing of the dots and the differences in the lengths of the velocity vectors indicate the changes in velocity. If an object speeds up, each subsequent velocity vector is longer. If the object slows down, each vector is shorter than the previous one. Both types of motion diagrams give an idea of how an object\u2019s velocity is changing. Velocity-Time Graphs Just as it was useful to graph a changing position versus time, it also is useful to plot an object\u2019s velocity versus time, which is called a velocitytime, or v-t graph. Table 3-1 on the next page shows the data for a car that starts at rest and speeds up along a straight stretch of road. a c \u25a0 Figure 3-1 By noting the distance the jogger moves in equal time intervals, you can determine that the jogger is standing still (a), moving at a constant speed (b), speeding up (c), and slowing down (d). 58 Chapter 3 Accelerated Motion b d 056-064 CH03-S1-845813 3/21/04 8:04 PM Page 59 The velocity-time graph obtained by plotting these data points is shown in Figure 3-3. The positive direction has been chosen to be the same as that of the motion of the car. Notice that this graph is a straight line, which means that the car was speeding up at a constant rate. The rate at which the car\u2019s velocity is changing can be found by calculating the slope of the velocity-time graph. The graph shows that the slope is (10.0 m/s)/(2.00 s), or 5.00 m/s2. This means that every second, the velocity of the car increased by 5.00 m/s. Consider a pair of data points that are separated by 1 s, such as 4.00 s and 5.00 s. At 4.00 s, the car was moving at a velocity of 20.0 m/s. At 5.00 s, the car was traveling at 25.0 m/s. Thus, the car\u2019s velocity increased by 5.00 m/s in 1.00 s. The rate at which an object\u2019s velocity changes is called the acceleration of the object. When the velocity of an object changes at a constant rate, it has a constant acceleration. Average and Instantaneous Acceleration The average acceleration of an object is the change in velocity during some measurable time interval divided by that time interval. Average acceleration is measured in m/s2. The change in velocity at an instant of time is called instantaneous acceleration. The instantaneous acceleration of an object can be found by drawing a tangent line on the velocity-time graph at the point of time in which you are interested. The slope of this line is equal to the instantaneous acceleration. Most of the situations considered in this textbook involve motion with acceleration in which the average and instantaneous accelerations are equal. \u25a0 Figure 3-2 The particle-model version of the motion diagram indicates the runner\u2019s changing velocity not only by the change in spacing of the position dots, but also by the change in length of the velocity vectors. \u25a0 Figure 3-3 The slope of a velocity-time graph is the acceleration of the object represented. Table 3-1 Velocity v. Time Time (s) Velocity (m/s) 0.00 1.00 2.00 3.00 4.00 5.00 0.00 5.00 10.0 15.0 20.0 25. 25.0 20.0 15.0 10.0 5.00 0.00 Velocity v. Time e 10.0 m/s s i m r 2.00 s u n r 5.00 m/s2 1.00 2.00 3.00 4.00 5.00 Time (s) Section 3.1 Acceleration 59 056-064 CH03-S1-845813 3/21/04 8:05 PM Page 60 \u25a0 Figure 3-4 Looking at two consecutive velocity vectors and finding the difference between them yields the average acceleration vector for that time interval. a vi b c vf vf vi vf vi \u2206v a Displaying Acceleration on a Motion Diagram \u2022 Acceleration vectors are violet. \u2022 Velocity vectors are red. \u2022 Displacement vectors are green. For a motion diagram to give a full picture of an object\u2019s movement, it also should contain information about acceleration. This can be done by including average acceleration vectors. These vectors will indicate how the velocity is changing. To determine the length and direction of an average acceleration vector, subtract two consecutive velocity vectors, as shown in Figures 3-4a and b. That is, v vf (vi). Then divide by the time interval, t. In Figures 3-4a and b, the time interval, t, is 1 s. This", " vec vi)/1 s, shown in violet in Figure 3-4c, is the average acceleration tor, (vf during that time interval. The velocities vi and vf refer to the velocities at the beginning and end of a chosen time interval. vf vi Velocity and Acceleration How would you describe the sprinter\u2019s velocity and acceleration as shown on the graph? 1 Analyze and Sketch the Problem \u2022 From the graph, note that the sprinter\u2019s velocity starts at zero, increases rapidly for the first few seconds, and then, after reaching about 10.0 m/s, remains almost constant. Known: v varies Unknown: a? 2 Solve for the Unknown Draw a tangent to the curve at t 1.0 s and t 5.0 s. Solve for acceleration at 1.0 s 12.0 6.0 0.00 The slope of the line at 1.0 s is equal to the acceleration at that time. e s i a r r u n 11.0 m/s 2.8 m/s 2.4 s 0.00 s 3.4 m/s2 5.00 Time (s) 10.0 Math Handbook Slope page 850 Solve for acceleration at 5.0 s: e s i a r r u n 10.3 m/s 10.0 m/s 10.0 s 0.00 s The slope of the line at 5.0 s is equal to the acceleration at that time. 0.030 m/s2 The acceleration is not constant because it changes from 3.4 m/s2 to 0.03 m/s2 at 5.0 s. The acceleration is in the direction chosen to be positive because both values are positive. 3 Evaluate the Answer \u2022 Are the units correct? Acceleration is measured in m/s2. 60 Chapter 3 Accelerated Motion 056-064 CH03-S1-845813 5/28/04 2:37 PM Page 61 1. A dog runs into a room and sees a cat at the other end of the room. The dog instantly stops running but slides along the wood floor until he stops, by slowing down with a constant acceleration. Sketch a motion diagram for this situation, and use the velocity vectors to find the acceleration vector. 2. Figure 3-5 is a v-t graph for Steven as he walks along the midway at the state fair. Sketch the corresponding motion diagram, complete with velocity vectors. Refer to the v-t graph of the toy train in Figure 3-6 to answer the following questions. a. When is the train\u2019s speed constant? b. During which time interval is the train\u2019s acceleration positive? c. When is the train\u2019s acceleration most negative? 4. Refer to Figure 3-6 to find the average acceleration of the train during the following time intervals. a. 0.0 s to 5.0 s b. 15.0 s to 20.0 s c. 0.0 s to 40.0 s 5. Plot a v-t graph representing the following motion. An elevator starts at rest from the ground floor of a three-story shopping mall. It accelerates upward for 2.0 s at a rate of 0.5 m/s2, continues up at a constant velocity of 1.0 m/s for 12.0 s, and then experiences a constant downward acceleration of 0.25 m/s2 for 4.0 s as it reaches the third floor. 43210 5 6 7 8 9 10 Time (s) \u25a0 Figure 3- 12.0 10.0 8.0 6.0 4.0 2.0 0.0 10.0 20.0 30.0 40.0 Time (s) \u25a0 Figure 3-6 Positive and Negative Acceleration Consider the four situations shown in Figure 3-7a. The first motion diagram shows an object moving in the positive direction and speeding up. The second motion diagram shows the object moving in the positive direction and slowing down. The third shows the object speeding up in the negative direction, and the fourth shows the object slowing down as it moves in the negative direction. Figure 3-7b shows the velocity vectors for the second time interval of each diagram, along with the corresponding acceleration vectors. Note \u2206t is equal to 1 s. a Begin End b v2 Begin End End v1 v a v a v2 v1 v2 v1 v a v2 v v1 a End Begin Begin \u25a0 Figure 3-7 These four motion diagrams represent the four different possible ways to move along a straight line with constant acceleration (a). When the velocity vectors of the motion diagram and acceleration vectors point in the same direction, an object\u2019s speed increases. When they point in opposite directions, the object slows down (b). Section 3.1 Acceleration 61 056-064 CH03-S1-845813 3/21/04", " 8:06 PM Page 62 East 0V l A E West \u25a0 Figure 3-8 Graphs A and E show motion with constant velocity in opposite directions. Graph B shows both positive velocity and positive acceleration. Graph C shows positive velocity and negative acceleration. Graph D shows motion with constant positive acceleration that slows down while velocity is negative and speeds up when velocity is positive. 62 Chapter 3 Accelerated Motion B D In the first and third situations when the object is speeding up, the velocity and acceleration vectors point in the same direction in each case, as shown in Figure 3-7b. In the other two situations in which the acceleration vector is in the opposite direction from the velocity vectors, the object is slowing down. In other words, when the object\u2019s acceleration is in the same direction as its velocity, the object\u2019s speed increases. When they are in opposite directions, the speed decreases. Both the direction of an object\u2019s velocity and its direction of acceleration are needed to determine whether it is speeding up or slowing down. An object has a positive acceleration when the acceleration vector points in the positive direction and a negative acceleration, when the acceleration vector points in the negative direction. The sign of acceleration does not indicate whether the object is speeding up or slowing down. Time (s) Determining Acceleration from a v-t Graph Velocity and acceleration information also is contained in velocity-time graphs. Graphs A, B, C, D, and E, shown in Figure 3-8, represent the motions of five different runners. Assume that the positive direction has been chosen to be east. The slopes of Graphs A and E are zero. Thus, the accelerations are zero. Both Graphs A and E show motion at a constant velocity\u2014Graph A to the east and Graph E to the west. Graph B shows motion with a positive velocity. The slope of this graph indicates a constant, positive acceleration. You also can infer from Graph B that the speed increased because it shows positive velocity and acceleration. Graph C has a negative slope. Graph C shows motion that begins with a positive velocity, slows down, and then stops. This means that the acceleration and velocity are in opposite directions. The point at which Graphs C and B cross shows that the runners\u2019 velocities are equal at that point. It does not, however, give any information about the runners\u2019 positions. Graph D indicates movement that starts out toward the west, slows down, and for an instant gets to zero velocity, and then moves east with increasing speed. The slope of Graph D is positive. Because the velocity and acceleration are in opposite directions, the speed decreases and equals zero at the time the graph crosses the axis. After that time, the velocity and acceleration are in the same direction and the speed increases. Calculating acceleration How can you describe acceleration mathematically? The following equation expresses average acceleration as the slope of the velocity-time graph. v v Average Acceleration a i ti t v f t f Average acceleration is equal to the change in velocity, divided by the time it takes to make that change. 056-064 CH03-S1-845813 5/28/04 2:39 PM Page 63 Suppose you run wind sprints back and forth across the gym. You first run at 4.0 m/s toward the wall. Then, 10.0 s later, you run at 4.0 m/s away from the wall. What is your average acceleration if the positive direction is toward the wall? v v a i ti t (4.0 m/s) (4.0 m/s) 10.0 s m.0 0.80 m/s2.s The negative sign indicates that the direction of acceleration is away from the wall. The velocity changes when the direction of motion changes, because velocity includes the direction of motion. A change in velocity results in acceleration. Thus, acceleration also is associated with a change in the direction of motion. Acceleration Describe the motion of a ball as it rolls up a slanted driveway. The ball starts at 2.50 m/s, slows down for 5.00 s, stops for an instant, and then rolls back down at an increasing speed. The positive direction is chosen to be up the driveway, and the origin is at the place where the motion begins. What is the sign of the ball\u2019s acceleration as it rolls up the driveway? What is the magnitude of the ball\u2019s acceleration as it rolls up the driveway? Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Draw the coordinate system based on the motion diagram. Known: vi vf 2.5 m/s 0.00 m/s at t 5.00 s Unknown: a? 1 2 Solve for the Unknown Find the magnitude of the acceleration from the slope of the graph. Solve for the change in velocity and the time taken to make that change. x x a Same point Begin End ).00 0", ".00 3.00 5.00 10.0 Time (s) v vf vi 0.00 m/s 2.50 m/s Substitute vf 2.50 m/s ti t tf 5.00 s 0.00 s 5.00 s Substitute tf 5.00 s, ti 0.00 s 0.00 m/s at tf 5.00 s, vi 2.50 m/s at ti 0.00 s Solve for the acceleration. v a t 2 / Substitute v 2.50 m/s, t 5.00 s Math Handbook Operations with Significant Digits pages 835\u2013836 0.500 m/s2 or 0.500 m/s2 down the driveway 3 Evaluate the Answer \u2022 Are the units correct? Acceleration is measured in m/s2. \u2022 Do the directions make sense? In the first 5.00 s, the direction of the acceleration is opposite to that of the velocity, and the ball slows down. Section 3.1 Acceleration 63 056-064 CH03-S1-845813 5/28/04 2:39 PM Page 64 6. A race car\u2019s velocity increases from 4.0 m/s to 36 m/s over a 4.0-s time interval. What is its average acceleration? 7. The race car in the previous problem slows from 36 m/s to 15 m/s over 3.0 s. What is its average acceleration? 8. A car is coasting backwards downhill at a speed of 3.0 m/s when the driver gets the engine started. After 2.5 s, the car is moving uphill at 4.5 m/s. If uphill is chosen as the positive direction, what is the car\u2019s average acceleration? 9. A bus is moving at 25 m/s when the driver steps on the brakes and brings the bus to a stop in 3.0 s. a. What is the average acceleration of the bus while braking? b. If the bus took twice as long to stop, how would the acceleration compare with what you found in part a? 10. Rohith has been jogging to the bus stop for 2.0 min at 3.5 m/s when he looks at his watch and sees that he has plenty of time before the bus arrives. Over the next 10.0 s, he slows his pace to a leisurely 0.75 m/s. What was his average acceleration during this 10.0 s? 11. If the rate of continental drift were to abruptly slow from 1.0 cm/y to 0.5 cm/y over the time interval of a year, what would be the average acceleration? There are several parallels between acceleration and velocity. Both are rates of change: acceleration is the time rate of change of velocity, and velocity is the time rate of change of position. Both acceleration and velocity have average and instantaneous forms. You will learn later in this chapter that the area under a velocity-time graph is equal to the object\u2019s displacement and that the area under an acceleration-time graph is equal to the object\u2019s velocity. 3.1 Section Review 12. Velocity-Time Graph What information can you obtain from a velocity-time graph? 13. Position-Time and Velocity-Time Graphs Two joggers run at a constant velocity of 7.5 m/s toward the east. At time t 0, one is 15 m east of the origin and the other is 15 m west. a. What would be the difference(s) in the position- time graphs of their motion? b. What would be the difference(s) in their velocity- time graphs? 14. Velocity Explain how you would use a velocitytime graph to find the time at which an object had a specified velocity. 15. Velocity-Time Graph Sketch a velocity-time graph for a car that goes east at 25 m/s for 100 s, then west at 25 m/s for another 100 s. 16. Average Velocity and Average Acceleration A canoeist paddles upstream at 2 m/s and then turns around and floats downstream at 4 m/s. The turnaround time is 8 s. a. What is the average velocity of the canoe? b. What is the average acceleration of the canoe? 17. Critical Thinking A police officer clocked a driver going 32 km/h over the speed limit just as the driver passed a slower car. Both drivers were issued speeding tickets. The judge agreed with the officer that both were guilty. The judgement was issued based on the assumption that the cars must have been going the same speed because they were observed next to each other. Are the judge and the police officer correct? Explain with a sketch, a motion diagram, and a position-time graph. 64 Chapter 3 Accelerated Motion physicspp.com/self_check_quiz 065-071 CH03-S2-845813 5/28/", "04 2:43 PM Page 65 3.2 Motion with Constant Acceleration You have learned that the definition of average velocity can be alge- braically rearranged to show the new position after a period of time, given the initial position and the average velocity. The definition of average acceleration can be manipulated similarly to show the new velocity after a period of time, given the initial velocity and the average acceleration. Velocity with Average Acceleration If you know an object\u2019s average acceleration during a time interval, you can use it to determine how much the velocity changed during that time. The definition of average acceleration, v a, can be rewritten as follows: t v at vf vi at The equation for final velocity with average acceleration can be written as follows. Objectives \u2022 Interpret position-time graphs for motion with constant acceleration. \u2022 Determine mathematical relationships among position, velocity, acceleration, and time. \u2022 Apply graphical and mathematical relationships to solve problems related to constant acceleration. Final Velocity with Average Acceleration vf The final velocity is equal to the initial velocity plus the product of the average acceleration and time interval. vi at In cases in which the acceleration is constant, the average acceleration, a, is the same as the instantaneous acceleration, a. This equation can be rearranged to find the time at which an object with constant acceleration has a given velocity. It also can be used to calculate the initial velocity of an object when both the velocity and the time at which it occurred are given. 18. A golf ball rolls up a hill toward a miniature-golf hole. Assume that the direction toward the hole is positive. a. If the golf ball starts with a speed of 2.0 m/s and slows at a constant rate of 0.50 m/s2, what is its velocity after 2.0 s? b. What is the golf ball\u2019s velocity if the constant acceleration continues for 6.0 s? c. Describe the motion of the golf ball in words and with a motion diagram. 19. A bus that is traveling at 30.0 km/h speeds up at a constant rate of 3.5 m/s2. What velocity does it reach 6.8 s later? 20. If a car accelerates from rest at a constant 5.5 m/s2, how long will it take for the car to reach a velocity of 28 m/s? 21. A car slows from 22 m/s to 3.0 m/s at a constant rate of 2.1 m/s2. How many seconds are required before the car is traveling at 3.0 m/s? Section 3.2 Motion with Constant Acceleration 65 065-071 CH03-S2-845813 3/21/04 8:11 PM Page 66 Table 3-2 Position-Time Data for a Car Time (s) Position (m) 0.00 1.00 2.00 3.00 4.00 5.00 0.00 2.50 10.0 22.5 40.0 62.5 \u25a0 Figure 3-9 The slope of a position-time graph of a car moving with a constant acceleration gets steeper as time goes on. \u25a0 Figure 3-10 The slopes of the p-t graph in Figure 3-9 are the values of the corresponding v-t graph (a). For any v-t graph, the displacement during a given time interval is the area under the graph (b). 66 Chapter 3 Accelerated Motion ) 70.0 60.0 50.0 40.0 30.0 20.0 10.0 0.00 Position-Time Graph for a Car.0 m 20.0 m 5.00 s 3.00 s m 60 20.0 m/s m 0. m 20 s. 00 3 10.0 m/s m.00 0 s.00 1 1.00 2.00 3.00 4.00 5.00 Time (s) Position with Constant Acceleration You have learned that an object experiencing constant acceleration changes its velocity at a constant rate. How does the position of an object with constant acceleration change? The position data at different time intervals for a car with constant acceleration are shown in Table 3-2. The data from Table 3-2 are graphed in Figure 3-9. The graph shows that the car\u2019s motion is not uniform: the displacements for equal time intervals on the graph get larger and larger. Notice that the slope of the line in Figure 3-9 gets steeper as time goes on. The slopes from the positiontime graph can be used to create a velocity-time graph. Note that the slopes shown in Figure 3-9 are the same as the velocities graphed in Figure 3-10a. A unique position-time graph cannot be created using a velocity-time graph because it does not contain any information about the object\u2019s position. However, the velocity-time graph does contain information about the object\u2019", "s displacement. Recall that for an object moving at a constant velocity, v v d/t, so d vt. On the graph in Figure 3-10b, v is the height of the plotted line above the t-axis, while t is the width of the shaded rectangle. The area of the rectangle, then, is vt, or d. Thus, the area under the v-t graph is equal to the object\u2019s displacement 25.0 20.0 15.0 10.0 5.00 0.00 10 20.0 m/s 15.0 m/s 4.00 s 3.00 s 5.00 m/s2 1.00 2.00 3.00 4.00 5.00 Time (s) v \u2206t 2 1 Time (s) 3 4 065-071 CH03-S2-845813 3/21/04 8:12 PM Page 67 Finding the Displacement from a v-t Graph The v-t graph below shows the motion of an airplane. Find the displacement of the airplane at t 1.0 s and at t 2.0 s. 1 Analyze and Sketch the Problem \u2022 The displacement is the area under the v-t graph. \u2022 The time intervals begin at t 0.0. Unknown: d? Known: v 75 m/s t 1.0 s t 2.0 s 2 Solve for the Unknown Solve for displacement during t 1.0 s. 82 80 78 76 74 72 ) vt (75 m/s)(1.0 s) 75 m Substitute v 75 m/s, t 1.0 s 0.0 Solve for displacement during t 2.0 s. \u2206t \u2206t 1.0 2.0 3.0 Time (s) d vt (75 m/s)(2.0 s) 150 m Substitute v 75 m/s, t 2.0 s Math Handbook Operations with Significant Digits pages 835\u2013836 3 Evaluate the Answer \u2022 Are the units correct? Displacement is measured in meters. \u2022 Do the signs make sense? The positive sign agrees with the graph. \u2022 Is the magnitude realistic? Moving a distance equal to about one football field is reasonable for an airplane. 22. Use Figure 3-11 to determine the velocity of an airplane that is speeding up at each of the following times. a. 1.0 s b. 2.0 s c. 2.5 s 23. Use dimensional analysis to convert an airplane\u2019s speed of 75 m/s to km/h. 24. A position-time graph for a pony running in a field is shown in Figure 3-12. Draw the corresponding velocity-time graph using the same time scale. 25. A car is driven at a constant velocity of 25 m/s for 10.0 min. The car runs out of gas and the driver walks in the same direction at 1.5 m/s for 20.0 min to the nearest gas station. The driver takes 2.0 min to fill a gasoline can, then walks back to the car at 1.2 m/s and eventually drives home at 25 m/s in the direction opposite that of the original trip. a. Draw a v-t graph using seconds as your time unit. Calculate the distance the driver walked to the gas station to find the time it took him to walk back to the car. b. Draw a position-time graph for the situation using the areas under the velocity-time graph 82 80 78 76 74 72 70 0.0 y \u25a0 Figure 3-11 1.0 2.0 3.0 Time (s) \u25a0 Figure 3-12 Time (s) x Section 3.2 Motion with Constant Acceleration 67 065-071 CH03-S2-845813 5/28/04 2:45 PM Page 68 vf vi ) 0V ti tf Time (s) \u25a0 Figure 3-13 The displacement of an object moving with constant acceleration can be found by computing the area under the v-t graph. Drag Racing A dragster driver tries to obtain maximum acceleration over a 402-m (quarter-mile) course. The fastest time on record for the 402-m course is 4.480 s. The highest final speed on record is 147.63 m/s (330.23 mph). 68 Chapter 3 Accelerated Motion The area under the v-t graph is equal to the object\u2019s displacement. Consider the v-t graph in Figure 3-13 for an object moving with constant acceleration that started with an initial velocity of vi. What is the object\u2019s displacement? The area under the graph can be calculated by dividing it into a rectangle and a triangle. The area of the rectangle can be found by drectangle t, and the area of the triangle can be found by vi 1 dtriangle vt. Because average acceleration, a, is", " equal to v/t, 2 v can be rewritten as at. Substituting v at into the equation for the (at)t, or 1 1 a(t)2. Solving for the total triangle\u2019s area yields dtriangle 2 2 area under the graph results in the following: d drectangle dtriangle vi(t) 1 a(t)2 2 When the initial or final position of the object is known, the equation can be written as follows: vi(t) 1 di a(t)2 or df 2 If the initial time is ti di 0, the equation then becomes the following. vi(t) 1 a(t)2 2 df Position with Average Acceleration df An object\u2019s position at a time after the initial time is equal to the sum of its initial position, the product of the initial velocity and the time, and half the product of the acceleration and the square of the time. di vitf 2 1 atf 2 An Alternative Expression Often, it is useful to relate position, velocity, and constant acceleration vi atf to solve for without including time. Rearrange the equation vf time: tf vi. vf a Rewriting df di vitf 2 by substituting tf yields the following: vi2 vi 1 vf a 2 This equation can be solved for the velocity, vf, at any time, tf. 1 atf 2 vi a vf di df a Velocity with Constant Acceleration vf The square of the final velocity equals the sum of the square of the initial velocity and twice the product of the acceleration and the displacement since the initial time. 2 vi 2 2a(df di) The three equations for motion with constant acceleration are summarized in Table 3-3. Note that in a multi-step problem, it is useful to add additional subscripts to identify which step is under consideration. Table 3-3 Equations of Motion for Uniform Acceleration vf Equation atf vi 1 atf 2 di) vitf 2 2a(df 2 di df 2 vi vf Variables tf, vf, a tf, df, a df, vf, a Initial Conditions vi di, vi di, vi 065-071 CH03-S2-845813 3/21/04 8:13 PM Page 69 Displacement An automobile starts at rest and speeds up at 3.5 m/s2 after the traffic light turns green. How far will it have gone when it is traveling at 25 m/s? Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Establish coordinate axes. \u2022 Draw a motion diagram. Unknown:? df Known: di vi vf a a 3.5 m/s2 0.00 m 0.00 m/s 25 m/s Begin Begin 1 2 vf Solve for the Unknown Solve for df. 2 vi di 2 2a(df di) 2 vf vi 2 a 0.00 m 89 m df 2 (25 m/s)2 (0.00 m/s)2 2(3.5 m/s2) End x End v a Math Handbook Order of Operations page 843 Substitute di 0.00 m, vf 25 m/s, vi 0.00 m/s 3 Evaluate the Answer \u2022 Are the units correct? Position is measured in meters. \u2022 Does the sign make sense? The positive sign agrees with both the pictorial and physical models. \u2022 Is the magnitude realistic? The displacement is almost the length of a football field. It seems large, but 25 m/s is fast (about 55 mph); therefore, the result is reasonable. 26. A skateboarder is moving at a constant velocity of 1.75 m/s when she starts up an incline that causes her to slow down with a constant acceleration of 0.20 m/s2. How much time passes from when she begins to slow down until she begins to move back down the incline? 27. A race car travels on a racetrack at 44 m/s and slows at a constant rate to a velocity of 22 m/s over 11 s. How far does it move during this time? 28. A car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achieve this speed? 29. A bike rider pedals with constant acceleration to reach a velocity of 7.5 m/s over a time of 4.5 s. During the period of acceleration, the bike\u2019s displacement is 19 m. What was the initial velocity of the bike? Section 3.2 Motion with Constant Acceleration 69 065-071 CH03-S2-845813 5/28/04 2:48 PM Page 70 Two-Part Motion You are driving a car, traveling at a constant velocity of 25 m/s, when you see a child suddenly run onto the road", ". It takes 0.45 s for you to react and apply the brakes. As a result, the car slows with a steady acceleration of 8.5 m/s2 and comes to a stop. What is the total distance that the car moves before it stops? 1 Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Choose a coordinate system with the motion of the car in the positive direction. \u2022 Draw the motion diagram and label v and a. Known: vreacting treacting a abraking vi, braking vf, braking 25 m/s 0.45 s 8.5 m/s2 25 m/s 0.00 m/s Unknown:? dreacting? dbraking dtotal? Begin Reacting Braking 0 Begin v a End x End 2 Solve for the Unknown Reacting: Solve for the distance the car travels at a constant speed. dreacting dreacting vreactingtreacting (25 m/s)(0.45 s) 11 m Substitute vreacting 25 m/s, treacting 0.45 s Braking: Solve for the distance the car moves while braking. vf, braking 2 vreacting 2 2abraking(dbraking) Solve for dbraking. dbraking 2 vreacting 2 vf, braking 2abraking (0.00 m/s)2 (25 m/s)2 2(8.5 m/s2) 37 m Math Handbook Isolating a Variable page 845 0.00 m/s, Substitute vf, braking vreacting 25 m/s, abraking 8.5 m/s2 The total distance traveled is the sum of the reaction distance and the braking distance. Solve for dtotal. dtotal dbraking dreacting 11 m 37 m 48 m Substitute dreacting 11 m, dbraking 37 m 3 Evaluate the Answer \u2022 Are the units correct? Distance is measured in meters. \u2022 Do the signs make sense? Both dreacting and dbraking are positive, as they should be. \u2022 Is the magnitude realistic? The braking distance is small because the magnitude of the acceleration is large. 70 Chapter 3 Accelerated Motion 065-071 CH03-S2-845813 3/21/04 8:14 PM Page 71 30. A man runs at a velocity of 4.5 m/s for 15.0 min. When going up an increasingly steep hill, he slows down at a constant rate of 0.05 m/s2 for 90.0 s and comes to a stop. How far did he run? 31. Sekazi is learning to ride a bike without training wheels. His father pushes him with a constant acceleration of 0.50 m/s2 for 6.0 s, and then Sekazi continues at 3.0 m/s for another 6.0 s before falling. What is Sekazi\u2019s displacement? Solve this problem by constructing a velocity-time graph for Sekazi\u2019s motion and computing the area underneath the graphed line. 32. You start your bicycle ride at the top of a hill. You coast down the hill at a constant acceleration of 2.00 m/s2. When you get to the bottom of the hill, you are moving at 18.0 m/s, and you pedal to maintain that speed. If you continue at this speed for 1.00 min, how far will you have gone from the time you left the hilltop? 33. Sunee is training for an upcoming 5.0-km race. She starts out her training run by moving at a constant pace of 4.3 m/s for 19 min. Then she accelerates at a constant rate until she crosses the finish line, 19.4 s later. What is her acceleration during the last portion of the training run? You have learned several different tools that you can apply when solving problems dealing with motion in one dimension: motion diagrams, graphs, and equations. As you gain more experience, it will become easier to decide which tools are most appropriate in solving a given problem. In the following section, you will practice using these tools to investigate the motion of falling objects. 3.2 Section Review 34. Acceleration A woman driving at a speed of 23 m/s sees a deer on the road ahead and applies the brakes when she is 210 m from the deer. If the deer does not move and the car stops right before it hits the deer, what is the acceleration provided by the car\u2019s brakes? 35. Displacement If you were given initial and final velocities and the constant acceleration of an object, and you were asked to find the displacement, what equation would you use? 36. Distance An in-line skater first accelerates from 0.0 m/s to 5.0 m/s in 4.5 s, then continues at this constant speed for another 4", ".5 s. What is the total distance traveled by the in-line skater? 37. Final Velocity A plane travels a distance of 5.0102 m while being accelerated uniformly from rest at the rate of 5.0 m/s2. What final velocity does it attain? 38. Final Velocity An airplane accelerated uniformly from rest at the rate of 5.0 m/s2 for 14 s. What final velocity did it attain? 39. Distance An airplane starts from rest and accelerates at a constant 3.00 m/s2 for 30.0 s before leaving the ground. a. How far did it move? b. How fast was the airplane going when it took off? 40. Graphs A sprinter walks up to the starting blocks at a constant speed and positions herself for the start of the race. She waits until she hears the starting pistol go off, and then accelerates rapidly until she attains a constant velocity. She maintains this velocity until she crosses the finish line, and then she slows down to a walk, taking more time to slow down than she did to speed up at the beginning of the race. Sketch a velocity-time and a position-time graph to represent her motion. Draw them one above the other on the same time scale. Indicate on your p-t graph where the starting blocks and finish line are. 41. Critical Thinking Describe how you could calculate the acceleration of an automobile. Specify the measuring instruments and the procedures that you would use. physicspp.com/self_check_quiz Section 3.2 Motion with Constant Acceleration 71 072-075 CH03-S3-845813 6/4/04 5:43 AM Page 72 3.3 Free Fall Objectives \u2022 Define acceleration due to gravity. \u2022 Solve problems involving objects in free fall. Vocabulary free fall acceleration due to gravity \u25a0 Figure 3-14 An egg accelerates at 9.80 m/s2 in free fall. If the upward direction is chosen as positive, then both the velocity and the acceleration of this egg in free fall are negative. 72 Richard Megna/Fundamental Photographs Chapter 3 Accelerated Motion Drop a sheet of paper. Crumple it, and then drop it again. Drop a rock or a pebble. How do the three motions compare with each other? Do heavier objects fall faster than lighter ones? A light, spread-out object, such as a smooth sheet of paper or a feather, does not fall in the same manner as something more compact, such as a pebble. Why? As an object falls, it bumps into particles in the air. For an object such as a feather, these little collisions have a greater effect than they do on pebbles or rocks. To understand the behavior of falling objects, first consider the simplest case: an object such as a rock, for which the air does not have an appreciable effect on its motion. The term used to describe the motion of such objects is free fall, which is the motion of a body when air resistance is negligible and the action can be considered due to gravity alone. Acceleration Due to Gravity About 400 years ago, Galileo Galilei recognized that to make progress in the study of the motion of falling objects, the effects of the substance through which the object falls have to be ignored. At that time, Galileo had no means of taking position or velocity data for falling objects, so he rolled balls down inclined planes. By \u201cdiluting\u201d gravity in this way, he could make careful measurements even with simple instruments. Galileo concluded that, neglecting the effect of the air, all objects in free fall had the same acceleration. It didn\u2019t matter what they were made of, how much they weighed, what height they were dropped from, or whether they were dropped or thrown. The acceleration of falling objects, given a special symbol, g, is equal to 9.80 m/s2. It is now known that there are small variations in g at different places on Earth, and that 9.80 m/s2 is the average value. The acceleration due to gravity is the acceleration of an object in free fall that results from the influence of Earth\u2019s gravity. Suppose you drop a rock. After 1 s, its velocity is 9.80 m/s downward, and 1 s after that, its velocity is 19.60 m/s downward. For each second that the rock is falling, its downward velocity increases by 9.80 m/s. Note that g is a positive number. When analyzing free fall, whether you treat the acceleration as positive or negative depends upon the coordinate system that you use. If your coordinate system defines upward to be the positive direction, then the acceleration due to gravity is equal to g; if you decide that downward is the positive direction, then the acceleration due to gravity is g. A strobe photo of a dropped egg is shown in Figure 3-14. The time interval between the images is 0", ".06 s. The displacement between each pair of images increases, so the speed is increasing. If the upward direction is chosen as positive, then the velocity is becoming more and more negative. Ball thrown upward Instead of a dropped egg, could this photo also illustrate a ball thrown upward? If upward is chosen to be the positive direction, then the ball leaves the hand with a positive velocity of, for example, 20.0 m/s. The acceleration is downward, so a is negative. That is, a g 9.80 m/s2. Because the velocity and acceleration are in opposite directions, the speed of the ball decreases, which is in agreement with the strobe photo. 072-075 CH03-S3-845813 3/21/04 8:18 PM Page 73 a 20..00 20.0 25.50 ).00 0 1 2 3 4 Time (s) 0.50 2.00 2.04 2.08 Time (s) 20.41 20.40 0 0 1 2 3 4 Time (s) 20.39 2.00 2.04 2.08 Time (s) After 1 s, the ball\u2019s velocity is reduced by 9.80 m/s, so it now is traveling at 10.2 m/s. After 2 s, the velocity is 0.4 m/s, and the ball still is moving upward. What happens during the next second? The ball\u2019s velocity is reduced by another 9.80 m/s, and is equal to 9.4 m/s. The ball now is moving downward. After 4 s, the velocity is 19.2 m/s, meaning that the ball is falling even faster. Figure 3-15a shows the velocity-time graph for the ball as it goes up and comes back down. At around 2 s, the velocity changes smoothly from positive to negative. Figure 3-15b shows a closer view of the v-t graph around that point. At an instant of time, near 2.04 s, the ball\u2019s velocity is zero. Look at the position-time graphs in Figure 3-15c and d, which show how the ball\u2019s height changes. How are the ball\u2019s position and velocity related? The ball reaches its maximum height at the instant of time when its velocity is zero. At 2.04 s, the ball reaches its maximum height and its velocity is zero. What is the ball\u2019s acceleration at that point? The slope of the line in the v-t graphs in Figure 3-15a and 3-15b is constant at 9.80 m/s2. Often, when people are asked about the acceleration of an object at the top of its flight, they do not take the time to fully analyze the situation, and respond that the acceleration at this point is zero. However, this is not the case. At the top of the flight, the ball\u2019s velocity is 0 m/s. What would happen if its acceleration were also zero? Then the ball\u2019s velocity would not be changing and would remain at 0 m/s. If this were the case, the ball would not gain any downward velocity and would simply hover in the air at the top of its flight. Because this is not the way objects tossed in the air behave on Earth, you know that the acceleration of an object at the top of its flight must not be zero. Further, because you know that the object will fall from that height, you know that the acceleration must be downward. \u25a0 Figure 3-15 In a coordinate system in which the upward direction is positive, the velocity of the thrown ball decreases until it becomes zero at 2.04 s. Then it increases in the negative direction as the ball falls (a, b). The p-t graphs show the height of the ball at corresponding time intervals (c, d). Section 3.3 Free Fall 73 072-075 CH03-S3-845813 3/21/04 8:18 PM Page 74 Free-fall rides Amusement parks use the concept of free fall to design rides that give the riders the sensation of free fall. These types of rides usually consist of three parts: the ride to the top, momentary suspension, and the plunge downward. Motors provide the force needed to move the cars to the top of the ride. When the cars are in free fall, the most massive rider and the least massive rider will have the same acceleration. Suppose the free-fall ride at an amusement park starts at rest and is in free fall for 1.5 s. What would be its velocity at the end of 1.5 s? Choose a coordinate system with a positive axis upward and the origin at the initial position of the car. Because the car starts at rest, vi would be equal to 0.00 m/s. To calculate the final velocity, use the equation for velocity with constant acceleration. vf atf vi", " 0.00 m/s (9.80 m/s2)(1.5 s) 15 m/s How far does the car fall? Use the equation for displacement when time and constant acceleration are known. vitf df 1 di atf 2 2 0.00 m (0.00 m/s)(1.5 s) 1 (9.80 m/s2)(1.5 s)2 2 11 m 42. A construction worker accidentally drops a brick from a high scaffold. a. What is the velocity of the brick after 4.0 s? b. How far does the brick fall during this time? 43. Suppose for the previous problem you choose your coordinate system so that the opposite direction is positive. a. What is the brick\u2019s velocity after 4.0 s? b. How far does the brick fall during this time? 44. A student drops a ball from a window 3.5 m above the sidewalk. How fast is it moving when it hits the sidewalk? 45. A tennis ball is thrown straight up with an initial speed of 22.5 m/s. It is caught at the same distance above the ground. a. How high does the ball rise? b. How long does the ball remain in the air? Hint: The time it takes the ball to rise equals the time it takes to fall. 46. You decide to flip a coin to determine whether to do your physics or English homework first. The coin is flipped straight up. a. If the coin reaches a high point of 0.25 m above where you released it, what was its initial speed? b. If you catch it at the same height as you released it, how much time did it spend in the air? 74 Chapter 3 Accelerated Motion 072-075 CH03-S3-845813 5/28/04 2:53 PM Page 75 You notice a water balloon fall past your classroom window. You estimate that it took the balloon about t seconds to fall the length of the window and that the window is about y meters high. Suppose the balloon started from rest. Approximately how high above the top of the window was it released? Your answer should be in terms of t, y, g, and numerical constants. Remember to define the positive direction when establishing your coordinate system. As motion problems increase in complexity, it becomes increasingly important to keep all the signs consistent. This means that any displacement, velocity, or acceleration that is in the same direction as the one chosen to be positive will be positive. Thus, any displacement, velocity, or acceleration that is in the direction opposite to the one chosen to be positive should be indicated with a negative sign. Sometimes it might be appropriate to choose upward as positive. At other times, it might be easier to choose downward as positive. You can choose either direction you want, as long as you stay consistent with that convention throughout the solution of that particular problem. Suppose you solve one of the practice problems on the preceding page again, choosing the direction opposite to the one you previously designated as the positive direction for the coordinate system. You should arrive at the same answer, provided that you assigned signs to each of the quantities that were consistent with the coordinate system. It is important to be consistent with the coordinate system to avoid getting the signs mixed up. 3.3 Section Review 47. Maximum Height and Flight Time Acceleration due to gravity on Mars is about one-third that on Earth. Suppose you throw a ball upward with the same velocity on Mars as on Earth. a. How would the ball\u2019s maximum height compare to that on Earth? b. How would its flight time compare? 48. Velocity and Acceleration Suppose you throw a ball straight up into the air. Describe the changes in the velocity of the ball. Describe the changes in the acceleration of the ball. 49. Final Velocity Your sister drops your house keys down to you from the second floor window. If you catch them 4.3 m from where your sister dropped them, what is the velocity of the keys when you catch them? 50. Initial Velocity A student trying out for the football team kicks the football straight up in the air. The ball hits him on the way back down. If it took 3.0 s from the time when the student punted the ball until he gets hit by the ball, what was the football\u2019s initial velocity? 51. Maximum Height When the student in the previous problem kicked the football, approximately how high did the football travel? 52. Critical Thinking When a ball is thrown vertically upward, it continues upward until it reaches a certain position, and then it falls downward. At that highest point, its velocity is instantaneously zero. Is the ball accelerating at the highest point? Devise an experiment to prove or disprove your answer. physicspp.com/self_check_quiz Section 3.3 Free Fall 75 076-077 CH03-LAB-845813 6/9/04 4:21 PM Page", " 76 Alternate CBL instructions can be found on the Web site. physicspp.com Acceleration Due to Gravity Small variations in the acceleration due to gravity, g, occur at different places on Earth. This is because g varies with distance from the center of Earth and is influenced by the subsurface geology. In addition, g varies with latitude due to Earth\u2019s rotation. di For motion with constant acceleration, the displacement is df 1 vitf a(tf 2 Dividing both sides of the equation by tf yields the following: df/tf The slope of a graph of df/tf versus tf, is equal to 1 a. The initial velocity, vi, is 2 determined by the y-intercept. In this activity, you will be using a spark timer to collect free-fall data and use it to determine the acceleration due to gravity, g. vi(tf ti) 1 2. atf 2 1 vi atf. 2 0, then the displacement is df ti)2. If di 0 and ti QUESTION How does the value of g vary from place to place? Objectives Procedure \u25a0 Measure free-fall data. \u25a0 Make and use graphs of velocity versus time. \u25a0 Compare and contrast values of g for different locations. Safety Precautions \u25a0 Keep clear of falling masses. Materials spark timer timer tape 1-kg mass C-clamp stack of newspapers masking tape 1. Attach the spark timer to the edge of the lab table with the C-clamp. 2. If the timer needs to be calibrated, follow your teacher\u2019s instructions or those provided with the timer. Determine the period of the timer and record it in your data table. 3. Place the stack of newspapers on the floor, directly below the timer so that the mass, when released, will not damage the floor. 4. Cut a piece of timer tape approximately 70 cm in length and slide it into the spark timer. 5. Attach the timer tape to the 1-kg mass with a small piece of masking tape. Hold the mass next to the spark timer, over the edge of the table so that it is above the newspaper stack. 6. Turn on the spark timer and release the mass. 7. Inspect the timer tape to make sure that there are dots marked on it and that there are no gaps in the dot sequence. If your timer tape is defective, repeat steps 4\u20136 with another piece of timer tape. 8. Have each member of your group perform the experiment and collect his or her own data. 9. Choose a dot near the beginning of the timer tape, a few centimeters from the point where the timer began to record dots, and label it 0. Label the dots after that 1, 2, 3, 4, 5, etc. until you get near the end where the mass is no longer in free fall. If the dots stop, or the distance between them begins to get smaller, the mass is no longer in free fall. 76 Horizons Companies 076-077 CH03-LAB-845813 6/15/04 11:46 PM Page 77 Data Table Time period (#/s) Interval Distance (cm) Time (s) Speed (cm/s) 1 2 3 4 5 6 7 8 10. Measure the total distance to each numbered dot from the zero dot, to the nearest millimeter and record it in your data table. Using the timer period, record the total time associated with each distance measurement and record it in your data table. Real-World Physics Why do designers of free-fall amusement-park rides design exit tracks that gradually curve toward the ground? Why is there a stretch of straight track? Analyze 1. Use Numbers Calculate the values for speed and record them in the data table. 2. Make and Use Graphs Draw a graph of speed versus time. Draw the best-fit straight line for your data. 3. Calculate the slope of the line. Convert your result to m/s2. Conclude and Apply 1. Recall that the slope is equal to 1 a. What is the 2 acceleration due to gravity? 2. Find the relative error for your experimental value of g by comparing it to the accepted value. Relative error Accepted value Experimental value Accepted value 100 3. What was the mass\u2019s velocity, vi, when you began measuring distance and time? Going Further What is the advantage of measuring several centimeters away from the beginning of the timer tape rather than from the very first dot? Communicate the average value of g to others. Go to physicspp.com/internet_lab and post the name of your school, city, state, elevation above sea level, and average value of g for your class. Obtain a map for your state and a map of the United States. Using the data posted on the Web site by other students, mark the values for g at the appropriate locations on the maps. Do you notice any variation in the acceleration due to gravity for different locations, regions and elev", "ations? To find out more about accelerated motion, visit the Web site: physicspp.com 77 078 CH03-FEATURE-845813 5/31/04 11:19 PM Page 78 Time Dilation at High Velocities Can time pass differently in two reference frames? How can one of a pair of twins age more than the other? Light Clock Consider the following thought experiment using a light clock. A light clock is a vertical tube with a mirror at each end. A short pulse of light is introduced at one end and allowed to bounce back and forth within the tube. Time is measured by counting the number of bounces made by the pulse of light. The clock will be accurate because the speed of a pulse of light is always c, which is 3108 m/s, regardless of the velocity of the light source or the observer. Suppose this light clock is placed in a very fast spacecraft. When the spacecraft goes at slow speeds, the light beam bounces vertically in the tube. If the spacecraft is moving fast, the light beam still bounces vertically\u2014at least as seen by the observer in the spacecraft. A stationary observer on Earth, however, sees the pulse of light move diagonally because of the movement of the spacecraft. Thus, to the stationary observer, the light beam moves a greater distance. Distance velocity time, so if the distance traveled by the light beam increases, the product (velocity time) also must increase. Because the speed of the light pulse, c, is the same for any observer, time must be increasing for the stationary observer. That is, the stationary observer sees the moving clock ticking slower than the same clock on Earth. Suppose the time per tick seen by the stationary observer on Earth is ts, the time seen by the observer on the spacecraft is to, the length of the light clock is cto, the velocity of the spacecraft is v, and the speed of light is c. For every tick, the spacecraft moves vts and the light pulse moves cto. This leads to the following equation: ts to c2 1 v 2 To the stationary observer, the closer v is to 78 Extreme Physics Observer inside the spacecraft Light clock MirrorMirror MirrorMirror t0 D D Observer on Earth c, the slower the clock ticks. To the observer on the spacecraft, however, the clock keeps perfect time. Time Dilation This phenomenon is called time dilation and it applies to every process associated with time aboard the spacecraft. For example, biological aging will proceed more slowly in the spacecraft than on Earth. So if the observer on the spacecraft is one of a pair of twins, he or she would age more slowly than the other twin on Earth. This is called the twin paradox. Time dilation has resulted in a lot of speculation about space travel. If spacecraft were able to travel at speeds close to the speed of light, trips to distant stars would take only a few years for the astronaut. Going Further 1. Calculate Find the time dilation ts/to for Earth\u2019s orbit about the Sun if vEarth 10,889 km/s. 2. Calculate Derive the equation for ts above. 3. Discuss How is time dilation similar to or different from time travel? 079-085 CH03-SG CA STP-845813 3/21/04 8:33 PM Page 79 3.1 Acceleration Vocabulary \u2022 velocity-time graph (p. 58) \u2022 acceleration (p.59) \u2022 average acceleration (p. 59) \u2022 instantaneous acceleration (p. 59) Key Concepts \u2022 A velocity-time graph can be used to find the velocity and acceleration of an object. \u2022 The average acceleration of an object is the slope of its velocity-time graph. v v a i ti t v f t f \u2022 Average acceleration vectors on a motion diagram indicate the size and direction of the average acceleration during a time interval. \u2022 When the acceleration and velocity are in the same direction, the object speeds up; when they are in opposite directions, the object slows down. \u2022 Velocity-time graphs and motion diagrams can be used to determine the sign of an object\u2019s acceleration. 3.2 Motion with Constant Acceleration Key Concepts \u2022 If an object\u2019s average acceleration during a time interval is known, the change in velocity during that time can be found. vf vi at \u2022 The area under an object\u2019s velocity-time graph is its displacement. \u2022 In motion with constant acceleration, there are relationships among the position, velocity, acceleration, and time. df di vitf 1 atf 2 2 \u2022 The velocity of an object with constant acceleration can be found using the following equation. vf 2 vi 2 2a(df di) 3.3 Free Fall Vocabulary \u2022 free fall (p. 72) \u2022 acceleration due to gravity (p. 72) Key Concepts \u2022 The acceleration due to gravity on Earth, g, is 9.80 m/s2 downward. The sign associated with g in equations depends upon the choice of the coordinate", " system. \u2022 Equations for motion with constant acceleration can be used to solve problems involving objects in free fall. physicspp.com/vocabulary_puzzlemaker 79 079-085 CH03-SG CA STP-845813 3/21/04 8:34 PM Page 80 Concept Mapping 53. Complete the following concept map using the following symbols or terms: d, velocity, m/s2, v, m, acceleration. Quantities of motion position 61. What quantity is represented by the area under a velocity-time graph? (3.2) 62. Write a summary of the equations for position, velocity, and time for an object experiencing motion with uniform acceleration. (3.2) 63. Explain why an aluminum ball and a steel ball of similar size and shape, dropped from the same height, reach the ground at the same time. (3.3) a 64. Give some examples of falling objects for which air resistance cannot be ignored. (3.3) m/s 65. Give some examples of falling objects for which air resistance can be ignored. (3.3) Mastering Concepts 54. How are velocity and acceleration related? (3.1) 55. Give an example of each of the following. (3.1) Applying Concepts 66. Does a car that is slowing down always have a negative acceleration? Explain. a. an object that is slowing down, but has a positive 67. Croquet A croquet ball, after being hit by a mallet, acceleration b. an object that is speeding up, but has a negative slows down and stops. Do the velocity and acceleration of the ball have the same signs? acceleration 56. Figure 3-16 shows the velocity-time graph for an automobile on a test track. Describe how the velocity changes with time. (3.1 30 25 20 15 10 5 0 5 10 15 20 25 30 35 Time (s) \u25a0 Figure 3-16 57. What does the slope of the tangent to the curve on a velocity-time graph measure? (3.1) 58. Can a car traveling on an interstate highway have a negative velocity and a positive acceleration at the same time? Explain. Can the car\u2019s velocity change signs while it is traveling with constant acceleration? Explain. (3.1) 59. Can the velocity of an object change when its acceleration is constant? If so, give an example. If not, explain. (3.1) 60. If an object\u2019s velocity-time graph is a straight line parallel to the t-axis, what can you conclude about the object\u2019s acceleration? (3.1) 68. If an object has zero acceleration, does it mean its velocity is zero? Give an example. 69. If an object has zero velocity at some instant, is its acceleration zero? Give an example. 70. If you were given a table of velocities of an object at various times, how would you find out whether the acceleration was constant? 71. The three notches in the graph in Figure 3-16 occur where the driver changed gears. Describe the changes in velocity and acceleration of the car while in first gear. Is the acceleration just before a gear change larger or smaller than the acceleration just after the change? Explain your answer. 72. Use the graph in Figure 3-16 and determine the time interval during which the acceleration is largest and the time interval during which the acceleration is smallest. 73. Explain how you would walk to produce each of the position-time graphs in Figure 3-17 Time Time \u25a0 Figure 3-17 80 Chapter 3 Accelerated Motion For more problems, go to Additional Problems, Appendix B. 079-085 CH03-SG CA STP-845813 3/21/04 8:34 PM Page 81 74. Draw a velocity-time graph for each of the graphs in Figure 3-18 Time Time Time \u25a0 Figure 3-18 75. An object shot straight up rises for 7.0 s before it reaches its maximum height. A second object falling from rest takes 7.0 s to reach the ground. Compare the displacements of the two objects during this time interval. 76. The Moon The value of g on the Moon is one-sixth of its value on Earth. a. Would a ball that is dropped by an astronaut hit the surface of the Moon with a greater, equal, or lesser speed than that of a ball dropped from the same height to Earth? b. Would it take the ball more, less, or equal time to fall? 77. Jupiter The planet Jupiter has about three times the gravitational acceleration of Earth. Suppose a ball is thrown vertically upward with the same initial velocity on Earth and on Jupiter. Neglect the effects of Jupiter\u2019s atmospheric resistance and assume that gravity is the only force on the ball. a. How does the maximum height reached by the ball on Jupiter compare to the maximum height reached on Earth? b. If the ball", " on Jupiter were thrown with an initial velocity that is three times greater, how would this affect your answer to part a? 78. Rock A is dropped from a cliff and rock B is thrown upward from the same position. a. When they reach the ground at the bottom of the cliff, which rock has a greater velocity? b. Which has a greater acceleration? c. Which arrives first? Mastering Problems 3.1 Acceleration 79. A car is driven for 2.0 h at 40.0 km/h, then for another 2.0 h at 60.0 km/h in the same direction. a. What is the car\u2019s average velocity? b. What is the car\u2019s average velocity if it is driven 1.0102 km at each of the two speeds? 80. Find the uniform acceleration that causes a car\u2019s velocity to change from 32 m/s to 96 m/s in an 8.0-s period. 81. A car with a velocity of 22 m/s is accelerated uniformly at the rate of 1.6 m/s2 for 6.8 s. What is its final velocity? 82. Refer to Figure 3-19 to find the acceleration of the moving object at each of the following times. a. during the first 5.0 s of travel b. between 5.0 s and 10.0 s c. between 10.0 s and 15.0 s d. between 20.0 s and 25.0 s 30.0 20.0 10..0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) \u25a0 Figure 3-19 83. Plot a velocity-time graph using the information in Table 3-4, and answer the following questions. a. During what time interval is the object speeding up? Slowing down? b. At what time does the object reverse direction? c. How does the average acceleration of the object in the interval between 0.0 s and 2.0 s differ from the average acceleration in the interval between 7.0 s and 12.0 s? Table 3-4 Velocity v. Time Time (s) Velocity (m/s) 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.0 11.0 12.0 4.00 8.00 12.0 14.0 16.0 16.0 14.0 12.0 8.00 4.00 0.00 4.00 8.00 physicspp.com/chapter_test Chapter 3 Assessment 81 079-085 CH03-SG CA STP-845813 5/31/04 11:23 PM Page 82 84. Determine the final velocity of a proton that has an initial velocity of 2.35105 m/s and then is accelerated uniformly in an electric field at the rate of 1.101012 m/s2 for 1.50107 s. 85. Sports Cars Marco is looking for a used sports car. He wants to buy the one with the greatest acceleration. Car A can go from 0 m/s to 17.9 m/s in 4.0 s; car B can accelerate from 0 m/s to 22.4 m/s in 3.5 s; and car C can go from 0 to 26.8 m/s in 6.0 s. Rank the three cars from greatest acceleration to least, specifically indicating any ties. 86. Supersonic Jet A supersonic jet flying at 145 m/s experiences uniform acceleration at the rate of 23.1 m/s2 for 20.0 s. a. What is its final velocity? b. The speed of sound in air is 331 m/s. What is the plane\u2019s speed in terms of the speed of sound? 3.2 Motion with Constant Acceleration 87. Refer to Figure 3-19 to find the distance traveled during the following time intervals. a. t 0.0 s and t 5.0 s b. t 5.0 s and t 10.0 s c. t 10.0 s and t 15.0 s d. t 0.0 s and t 25.0 s 88. A dragster starting from rest accelerates at 49 m/s2. How fast is it going when it has traveled 325 m? 89. A car moves at 12 m/s and coasts up a hill with a uniform acceleration of 1.6 m/s2. a. What is its displacement after 6.0 s? b. What is its displacement after 9.0 s? 90. Race Car A race car can be slowed with a constant acceleration of 11 m/s2. a. If the car is going 55 m/s, how many meters will it travel before it stops? b. How many meters will it take to stop a car going twice as fast? 91.", " A car is traveling 20.0 m/s when the driver sees a child standing on the road. She takes 0.80 s to react, then steps on the brakes and slows at 7.0 m/s2. How far does the car go before it stops? 95. Road Barrier The driver of a car going 90.0 km/h suddenly sees the lights of a barrier 40.0 m ahead. It takes the driver 0.75 s to apply the brakes, and the average acceleration during braking is 10.0 m/s2. a. Determine whether the car hits the barrier. b. What is the maximum speed at which the car could be moving and not hit the barrier 40.0 m ahead? Assume that the acceleration doesn\u2019t change. 3.3 Free Fall 96. A student drops a penny from the top of a tower and decides that she will establish a coordinate system in which the direction of the penny\u2019s motion is positive. What is the sign of the acceleration of the penny? 97. Suppose an astronaut drops a feather from 1.2 m above the surface of the Moon. If the acceleration due to gravity on the Moon is 1.62 m/s2 downward, how long does it take the feather to hit the Moon\u2019s surface? 98. A stone that starts at rest is in free fall for 8.0 s. a. Calculate the stone\u2019s velocity after 8.0 s. b. What is the stone\u2019s displacement during this time? 99. A bag is dropped from a hovering helicopter. The bag has fallen for 2.0 s. What is the bag\u2019s velocity? How far has the bag fallen? 100. You throw a ball downward from a window at a speed of 2.0 m/s. How fast will it be moving when it hits the sidewalk 2.5 m below? 101. If you throw the ball in the previous problem up instead of down, how fast will it be moving when it hits the sidewalk? 102. Beanbag You throw a beanbag in the air and catch it 2.2 s later. a. How high did it go? b. What was its initial velocity? Mixed Review 103. A spaceship far from any star or planet experiences uniform acceleration from 65.0 m/s to 162.0 m/s in 10.0 s. How far does it move? 92. Airplane Determine the displacement of a plane 104. Figure 3-20 is a strobe photo of a horizontally that experiences uniform acceleration from 66 m/s to 88 m/s in 12 s. 93. How far does a plane fly in 15 s while its velocity is changing from 145 m/s to 75 m/s at a uniform rate of acceleration? 94. Police Car A speeding car is traveling at a constant speed of 30.0 m/s when it passes a stopped police car. The police car accelerates at 7.0 m/s2. How fast will it be going when it catches up with the speeding car? moving ball. What information about the photo would you need and what measurements would you make to estimate the acceleration? \u25a0 Figure 3-20 82 Skip Peticolas/Fundamental Photographs Chapter 3 Accelerated Motion For more problems, go to Additional Problems, Appendix B. 079-085 CH03-SG CA STP-845813 6/1/04 1:24 AM Page 83 105. Bicycle A bicycle accelerates from 0.0 m/s to 4.0 m/s in 4.0 s. What distance does it travel? 106. A weather balloon is floating at a constant height above Earth when it releases a pack of instruments. a. If the pack hits the ground with a velocity of 73.5 m/s, how far did the pack fall? b. How long did it take for the pack to fall? 107. Baseball A baseball pitcher throws a fastball at a speed of 44 m/s. The acceleration occurs as the pitcher holds the ball in his hand and moves it through an almost straight-line distance of 3.5 m. Calculate the acceleration, assuming that it is constant and uniform. Compare this acceleration to the acceleration due to gravity. 108. The total distance a steel ball rolls down an incline at various times is given in Table 3-5. a. Draw a position-time graph of the motion of the ball. When setting up the axes, use five divisions for each 10 m of travel on the d-axis. Use five divisions for 1 s of time on the t-axis. b. Calculate the distance the ball has rolled at the end of 2.2 s. Table 3-5 Distance v. Time Time (s) Distance (m) 0.0 1.0 2.0 3.0 4.0 5.0 0.0 2.0 8.0 18.0 32.0 50.0 109. Engineers are developing new", " types of guns that might someday be used to launch satellites as if they were bullets. One such gun can give a small object a velocity of 3.5 km/s while moving it through a distance of only 2.0 cm. a. What acceleration does the gun give this object? b. Over what time interval does the acceleration take place? 110. Sleds Rocket-powered sleds are used to test the responses of humans to acceleration. Starting from rest, one sled can reach a speed of 444 m/s in 1.80 s and can be brought to a stop again in 2.15 s. a. Calculate the acceleration of the sled when starting, and compare it to the magnitude of the acceleration due to gravity, 9.80 m/s2. b. Find the acceleration of the sled as it is braking and compare it to the magnitude of the acceleration due to gravity. physicspp.com/chapter_test 111. The velocity of a car changes over an 8.0-s time period, as shown in Table 3-6. a. Plot the velocity-time graph of the motion. b. Determine the displacement of the car during the first 2.0 s. c. What displacement does the car have during the first 4.0 s? d. What is the displacement of the car during the entire 8.0 s? e. Find the slope of the line between t 0.0 s and t 4.0 s. What does this slope represent? f. Find the slope of the line between t 5.0 s and t 7.0 s. What does this slope indicate? Table 3-6 Velocity v. Time Time (s) Velocity (m/s) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 0.0 4.0 8.0 12.0 16.0 20.0 20.0 20.0 20.0 112. A truck is stopped at a stoplight. When the light turns green, the truck accelerates at 2.5 m/s2. At the same instant, a car passes the truck going 15 m/s. Where and when does the truck catch up with the car? 113. Safety Barriers Highway safety engineers build soft barriers, such as the one shown in Figure 3-21, so that cars hitting them will slow down at a safe rate. A person wearing a safety belt can withstand an acceleration of 3.0102 m/s2. How thick should barriers be to safely stop a car that hits a barrier at 110 km/h? \u25a0 Figure 3-21 Chapter 3 Assessment 83 Joel Bennett/Peter Arnold, Inc. 079-085 CH03-SG CA STP-845813 3/21/04 8:36 PM Page 84 114. Karate The position-time and velocity-time graphs of George\u2019s fist breaking a wooden board during karate practice are shown in Figure 3-22. a. Use the velocity-time graph to describe the motion of George\u2019s fist during the first 10 ms. b. Estimate the slope of the velocity-time graph to determine the acceleration of his fist when it suddenly stops. c. Express the acceleration as a multiple of the gravitational acceleration, g 9.80 m/s2. d. Determine the area under the velocity-time curve to find the displacement of the fist in the first 6 ms. Compare this with the positiontime graph 10.0 5.0 0.0 5.0 0.0 5.0 10.0 15.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (ms) \u25a0 Figure 3-22 115. Cargo A helicopter is rising at 5.0 m/s when a bag of its cargo is dropped. The bag falls for 2.0 s. a. What is the bag\u2019s velocity? b. How far has the bag fallen? c. How far below the helicopter is the bag? Thinking Critically 116. Apply CBLs Design a lab to measure the distance an accelerated object moves over time. Use equal time intervals so that you can plot velocity over time as well as distance. A pulley at the edge of a table with a mass attached is a good way to achieve uniform acceleration. Suggested materials include a motion detector, CBL, lab cart, string, pulley, C-clamp, and masses. Generate distancetime and velocity-time graphs using different masses on the pulley. How does the change in mass affect your graphs? 117. Analyze and Conclude Which has the greater acceleration: a car that increases its speed from 50 km/h to 60 km/h, or a bike that goes from 0 km/h to 10 km/h in the same time? Explain. 118. Analyze and Conclude An express train, traveling at 36.0 m/s,", " is accidentally sidetracked onto a local train track. The express engineer spots a local train exactly 1.00102 m ahead on the same track and traveling in the same direction. The local engineer is unaware of the situation. The express engineer jams on the brakes and slows the express train at a constant rate of 3.00 m/s2. If the speed of the local train is 11.0 m/s, will the express train be able to stop in time, or will there be a collision? To solve this problem, take the position of the express train when the engineer first sights the local train as a point of origin. Next, keeping in mind that the local train has exactly a 1.00102 m lead, calculate how far each train is from the origin at the end of the 12.0 s it would take the express train to stop (accelerate at 3.00 m/s2 from 36 m/s to 0 m/s). a. On the basis of your calculations, would you conclude that a collision will occur? b. The calculations that you made do not allow for the possibility that a collision might take place before the end of the 12 s required for the express train to come to a halt. To check this, take the position of the express train when the engineer first sights the local train as the point of origin and calculate the position of each train at the end of each second after the sighting. Make a table showing the distance of each train from the origin at the end of each second. Plot these positions on the same graph and draw two lines. Use your graph to check your answer to part a. Writing in Physics 119. Research and describe Galileo\u2019s contributions to physics. 120. Research the maximum acceleration a human body can withstand without blacking out. Discuss how this impacts the design of three common entertainment or transportation devices. Cumulative Review 121. Solve the following problems. Express your answers in scientific notation. (Chapter 1) a. 6.2104 m 5.7103 m b. 8.7108 km 3.4107 m c. (9.21105 cm)(1.83108 cm) d. (2.63106 m)/(4.08106 s) 122. The equation below describes the motion of an object. Create the corresponding position-time graph and motion diagram. Then write a physics problem that could be solved using that equation. Be creative. d (35.0 m/s) t 5.0 m (Chapter 2) 84 Chapter 3 Accelerated Motion For more problems, go to Additional Problems, Appendix B. 079-085 CH03-SG CA STP-845813 6/7/04 11:23 PM Page 85 Multiple Choice Use the following information to answer the first two questions. A ball rolls down a hill with a constant acceleration of 2.0 m/s2. The ball starts at rest and travels for 4.0 s before it stops. 1. How far did the ball travel before it stopped? 8.0 m 12 m 16 m 20 m 2. What was the ball\u2019s velocity just before it stopped? 2.0 m/s 8.0 m/s 12 m/s 16 m/s 3. A driver of a car enters a new 110-km/h speed zone on the highway. The driver begins to accelerate immediately and reaches 110 km/h after driving 500 m. If the original speed was 80 km/h, what was the driver\u2019s rate of acceleration? 0.44 m/s2 0.60 m/s2 8.4 m/s2 9.80 m/s2 4. A flowerpot falls off the balcony of a penthouse suite 85 m above the street. How long does it take to hit the ground? 4.2 s 8.3 s 8.7 s 17 s 5. A rock climber\u2019s shoe loosens a rock, and her climbing buddy at the bottom of the cliff notices that the rock takes 3.20 s to fall to the ground. How high up the cliff is the rock climber? 15.0 m 31.0 m 50.0 m 1.00102 m 6. A car traveling at 91.0 km/h approaches the turnoff for a restaurant 30.0 m ahead. If the driver slams on the brakes with an acceleration of 6.40 m/s2, what will be her stopping distance? 14.0 m 29.0 m 50.0 m 100.0 m 7. What is the correct formula manipulation to find 2 2ad? acceleration when using the equation vf (vf (vf 2 vi 2 vi 2)/d 2)/2d 2 vi vi)2/2d 2 vi 2)/2d (vf (vf 8. The graph shows the motion of a farmer\u2019s truck. What is the truck\u2019s total displacement? Assume that north is the", " positive direction. 150 m south 125 m north 300 m north 600 m south ) 25.0 15.0 5.00 0.00 5.00 15.0 25.0 5.00 15.0 25.0 35.0 45.0 Time (s) 9. How can the instantaneous acceleration of an object with varying acceleration be calculated? by calculating the slope of the tangent on a distance-time graph by calculating the area under the graph on a distance-time graph by calculating the area under the graph on a velocity-time graph by calculating the slope of the tangent on a velocity-time graph Extended Answer 10. Graph the following data, and then show calculations for acceleration and displacement after 12.0 s on the graph. Time (s) Velocity (m/s) 0.00 6.00 9.00 12.00 8.10 36.9 51.3 65.7 Tables If a test question involves a table, skim the table before reading the question. Read the title, column heads, and row heads. Then read the question and interpret the information in the table. physicspp.com/standardized_test Chapter 3 Standardized Test Practice 85 086-095 CH04-S1-845813 4/14/04 4:11 AM Page 86 What You\u2019ll Learn \u2022 You will use Newton\u2019s laws to solve problems. \u2022 You will determine the magnitude and direction of the net force that causes a change in an object\u2019s motion. \u2022 You will classify forces according to the agents that cause them. Why It\u2019s Important Forces act on you and everything around you at all times. Sports A soccer ball is headed by a player. Before play began, the ball was motionless. During play, the ball started, stopped, and changed directions many times. Think About This What causes a soccer ball, or any other object, to stop, start, or change direction? physicspp.com 86 Joe McBride/CORBIS 086-095 CH04-S1-845813 3/12/04 3:26 AM Page 87 Which force is stronger? Question What forces can act on an object that is suspended by a string? Procedure 1. Tie a piece of heavy cord around the middle of a book. Tie one piece of lightweight string to the center of the cord on the top of the book. Tie another piece to the bottom. 2. While someone holds the end of the top lightweight string so that the book is suspended in the air, pull very slowly, but firmly, on the end of the bottom lightweight string. Record your observations. CAUTION: Keep feet clear of falling objects. 3. Replace the broken string and repeat step 2, but this time pull very fast and very hard on the bottom string. Record your observations. Analysis Which string broke in step 2? Why? Which string broke in step 3? Why? Critical Thinking Draw a diagram of the experimental set-up. Use arrows to show the forces acting on the book. 4.1 Force and Motion Imagine that a train is speeding down a railroad track at 80 km/h when suddenly the engineer sees a truck stalled at a railroad crossing ahead. The engineer applies the brakes to try to stop the train before it crashes into the truck. Because the brakes cause an acceleration in the direction opposite the train\u2019s velocity, the train will slow down. Imagine that, in this case, the engineer is able to stop the train just before it crashes into the truck. But what if instead of moving at 80 km/h the train had been moving at 100 km/h? What would have to happen for the train to avoid hitting the truck? The acceleration provided by the train\u2019s brakes would have to be greater because the engineer still has the same distance in which to stop the train. Similarly, if the train was going 80 km/h but had been much closer to the truck when the engineer started to apply the brake, the acceleration also would need to be greater because the train would need to stop in less time. Objectives \u2022 Define force. \u2022 Apply Newton\u2019s second law to solve problems. \u2022 Explain the meaning of Newton\u2019s first law. Vocabulary force free-body diagram net force Newton\u2019s second law Newton\u2019s first law inertia equilibrium Section 4.1 Force and Motion 87 Horizons Companies 086-095 CH04-S1-845813 3/12/04 3:28 AM Page 88 System Ftable on book Fhand on book FEarth\u2019s mass on book \u25a0 Figure 4-1 The book is the system. The table, the hand, and Earth\u2019s mass (through gravity) all exert forces on the book. Force and Motion What happened to make the train slow down? A force is a push or pull exerted on an object. Forces can cause objects to speed up, slow down, or change direction as they move. When an engineer applies the brakes, the", " brakes exert a force on the wheels and cause the train to slow down. Based on the definitions of velocity and acceleration, this can be restated as follows: a force exerted on an object causes that object\u2019s velocity to change; that is, a force causes an acceleration. Consider a textbook resting on a table. How can you cause it to move? Two possibilites are that you can push on it or you can pull on it. The push or pull is a force that you exert on the textbook. If you push harder on an object, you have a greater effect on its motion. The direction in which the force is exerted also matters\u2014if you push the book to the right, the book moves in a different direction from the direction it would move if you pushed it to the left. The symbol F is a vector and represents the size and direction of a force, while F represents only the magnitude. When considering how a force affects motion, it is important to identify the object of interest. This object is called the system. Everything around the object that exerts forces on it is called the external world. In the case of the book in Figure 4-1, the book is the system. Your hand and gravity are parts of the external world that can interact with the book by pushing or pulling on it and potentially causing its motion to change. Contact Forces and Field Forces Again, think about the different ways in which you could move a textbook. You could touch it directly and push or pull it, or you could tie a string around it and pull on the string. These are examples of contact forces. A contact force exists when an object from the external world touches a system and thereby exerts a force on it. If you are holding this physics textbook right now, your hands are exerting a contact force on it. If you place the book on a table, you are no longer exerting a force on the book. The table, however, is exerting a force because the table and the book are in contact. There are other ways in which you could change the motion of the textbook. You could drop it, and as you learned in Chapter 3, it would accelerate as it falls to the ground. The gravitational force of Earth acting on the book causes this acceleration. This force affects the book whether or not Earth is actually touching it. This is an example of a field force. Field forces are exerted without contact. Can you think of other kinds of field forces? If you have ever experimented with magnets, you know that they exert forces without touching. You will investigate magnetism and other similar forces in more detail in future chapters. For now, the only field force that you need to consider is the gravitational force. Forces result from interactions; thus, each force has a specific and identifiable cause called the agent. You should be able to name the agent exerting each force, as well as the system upon which the force is exerted. For example, when you push your textbook, your hand (the agent) exerts a force on the textbook (the system). If there are not both an agent and a system, a force does not exist. What about the gravitational force? If you allow your textbook to fall, the agent is the mass of Earth exerting a field force on the book. 88 Chapter 4 Forces in One Dimension 086-095 CH04-S1-845813 5/31/04 11:28 PM Page 89 a y Contact with external world b y Frope on ball System Contact with external world Fhand on ball \u25a0 Figure 4-2 To make a physical model of the forces acting on an object, apply the particle model and draw an arrow to represent each force. Label each force, including its agent. System FEarth\u2019s mass on ball FEarth\u2019s mass on ball Free-body diagrams Just as pictorial models and motion diagrams are useful in solving problems about motion, similar representations will help you to analyze how forces affect motion. The first step in solving any problem is to create a pictorial model. For example, to represent the forces on a ball tied to a string or held in your hand, sketch the situations, as shown in Figures 4-2a and 4-2b. Circle the system and identify every place where the system touches the external world. It is at these places that contact forces are exerted. Identify the contact forces. Then identify any field forces on the system. This gives you the pictorial model. To make a physical representation of the forces acting on the ball in Figures 4-2a and 4-2b, apply the particle model and represent the object with a dot. Represent each force with a blue arrow that points in the direction that the force is applied. Try to make the length of each arrow proportional to the size of the force. Often, you will draw these diagrams before you know the magnitudes of all the forces. In such cases, make your best estimate. Always draw the force arrows pointing away from the particle, even", " when the force is a push. Make sure that you label each force. Use the symbol F with a subscript label to identify both the agent and the object on which the force is exerted. Finally, choose a direction to be positive and indicate this off to the side of your diagram. Usually, you select the positive direction to be in the direction of the greatest amount of force. This typically makes the problem easiest to solve by reducing the number of negative values in your calculations. This type of physical model, which represents the forces acting on a system, is called a free-body diagram. For each of the following situations, specify the system and draw a motion diagram and a free-body diagram. Label all forces with their agents, and indicate the direction of the acceleration and of the net force. Draw vectors of appropriate lengths. 1. A flowerpot falls freely from a windowsill. (Ignore any forces due to air resistance.) 2. A sky diver falls downward through the air at constant velocity. (The air exerts an upward force on the person.) 3. A cable pulls a crate at a constant speed across a horizontal surface. The surface provides a force that resists the crate\u2019s motion. 4. A rope lifts a bucket at a constant speed. (Ignore air resistance.) 5. A rope lowers a bucket at a constant speed. (Ignore air resistance.) Section 4.1 Force and Motion 89 086-095 CH04-S1-845813 3/12/04 3:29 AM Page 90 Force and Acceleration How does an object move when one or more forces are exerted on it? One way to find out is by doing experiments. As before, begin by considering a simple situation. Once you fully understand that situation, then you can add more complications to it. In this case, begin with one controlled force exerted horizontally on an object. The horizontal direction is a good place to start because gravity does not act horizontally. Also, to reduce complications resulting from the object rubbing against the surface, do the experiments on a very smooth surface, such as ice or a very well-polished table, and use an object with wheels that spin easily. In other words, you are trying to reduce the resistance to motion in the situation. To determine how force, acceleration, and velocity are related, you need to be able to exert a constant and controlled force on an object. How can you exert such a controlled force? A stretched rubber band exerts a pulling force; the farther you stretch it, the greater the force with which it pulls back. If you always stretch the rubber band the same amount, you always exert the same force. Figure 4-3a shows a rubber band, stretched a constant 1 cm, pulling a low-resistance cart. If you perform this experiment and determine the cart\u2019s velocity for some period of time, you could construct a graph like the one shown in Figure 4-3b. Does this graph look different from what you expected? What do you notice about the velocity? The constant increase in the velocity is a result of the constant acceleration the stretched rubber band gives the cart. How does this acceleration depend upon the force? To find out, repeat the experiment, this time with the rubber band stretched to a constant 2 cm, and then repeat it again with the rubber band stretched longer and longer each time. For each experiment, plot a velocity-time graph like the one in Figure 4-3b, calculate the acceleration, and then plot the accelerations and forces for all the trials to make a force-acceleration graph, as shown in Figure 4-4a. What is the relationship between the force and acceleration? It\u2019s a linear relationship where the greater the force is, the greater the resulting acceleration. As you did in Chapters 2 and 3, you can apply the straight-line equation y mx b to this graph.50 1.00 0.50 0 Cart Pulled by Stretched Rubber Band (1 cm) 0.50 1.00 1.50 2.00 2.50 3.00 3.50 Time (s) a v \u25a0 Figure 4-3 Because the rubber band is stretched a constant amount, it applies a constant force on the cart, which is designed to be low-friction (a). The cart\u2019s motion can be graphed and shown to be a linear relationship (b). a 90 Matt Meadows Chapter 4 Forces in One Dimension 086-095 CH04-S1-845813 3/12/04 3:29 AM Page 91 a Acceleration of Cart b Acceleration v. Force slo cart 2 carts 3 carts Force (N) \u25a0 Figure 4-4 The graph shows that as the force increases, so does the acceleration (a). You can see that the slope of the force-acceleration graph depends upon the number of carts (b). ).2 0.8 0.4 0.0 0 1 2 3 4 Application of force (in cm", " of stretch What is the physical meaning of this slope? Perhaps it describes something about the object that is accelerating. What happens if the object changes? Suppose that a second, identical cart is placed on top of the first, and then a third cart is added. The rubber band would be pulling two carts, and then three. A plot of the force versus acceleration for one, two, and three carts is shown in Figure 4-4b. The graph shows that if the same force is applied in each situation, the acceleration of two carts is 1 the accelera2 tion of one cart, and the acceleration of three carts is 1 the acceleration of 3 one cart. This means that as the number of carts is increased, a greater force is needed to produce the same acceleration. In this example, you would have to stretch the rubber band farther to get a greater amount of force. The slopes of the lines in Figure 4-4b depend upon the number of carts; that is, the slope depends on the total mass of the carts. If the slope, k in this case, is defined as the reciprocal of the mass (k 1/m), then a F/m, or F ma. What information is contained in the equation a F/m? It tells you that a force applied to an object causes that object to experience a change in motion\u2014the force causes the object to accelerate. It also tells you that for the same object, if you double the force, you will double the object\u2019s acceleration. Lastly, if you apply the same force to several different objects, the one with the most mass will have the smallest acceleration and the one with the least mass will have the greatest acceleration. What are the proper units for measuring force? Because F ma, one unit of force causes a 1-kg mass to accelerate at 1 m/s2, so one force unit has the dimensions 1 kgm/s2. The unit 1 kgm/s2 is called the newton, represented by N. One newton of force applied to a 1-kg object will cause it to have an acceleration of 1 m/s2. Do these units make sense? Think about a sky diver who is falling through the air. The properties affecting his motion are his mass and the acceleration due to the gravitational force, so these units do make sense. Table 4-1 shows the magnitude of some common forces. Force of a rocket motor Force of an accelerating car Description Force of gravity on a coin (nickel) Force of gravity on 1 lb (0.45 kg) of sugar Force of gravity on a 150-lb (70-kg) person Table 4-1 Common Forces F (N) 0.05 4.5 686 3000 5,000,000 Section 4.1 Force and Motion 91 086-095 CH04-S1-845813 5/31/04 11:30 PM Page 92 Combining Forces What happens if you and a friend each push a table and exert 100 N of force on it? When you and your friend push together, you give the table a greater acceleration than when you push against each other. In fact, when you push together, you give the table twice the acceleration that it would have if just one of you applied 100 N of force. When you push on the table in opposite directions with the same amount of force, as in Figure 4-5a, the table does not move at all. Figure 4-5b and c show free-body diagrams for these two situations. Figure 4-5d shows a third free-body diagram in which your friend pushes on the table twice as hard as you in the opposite direction. Below each free-body diagram is a vector representing the total result of the two forces. When the force vectors are in the same direction, they can be replaced by one vector with a length equal to their combined length. When the forces are in opposite directions, the resulting vector is the length of the difference between the two vectors. Another term for the vector sum of all the forces on an object is the net force. You also can analyze the situation mathematically. Assume that you are pushing the table in the positive direction with a 100 N force in the above cases. In the first case, your friend is pushing with a negative force of 100 N. Adding them together gives a total force of 0 N, which means there is no acceleration. In the second case, your friend\u2019s force is 100 N, so the total force is 200 N in the positive direction and the table accelerates in the positive direction. In the third case, your friend\u2019s force is 200 N, so the total force is 100 N and the table will accelerate in the negative direction. \u25a0 Figure 4-5 Pushing on the table with equal force in opposite directions (a) results in no net force on the table, as shown by the vector addition in the freebody diagram (b). However, there is a net force applied in (c) and", " (d), as shown by the free-body diagrams. a b x F1 F2 c 100 N F1 d 100 N F2 100 N F1 100 N F2 200 N F2 100 N F1 Fnet 0 N Fnet 200 N Fnet 100 N Equal forces Opposite directions Equal forces Same direction Unequal forces Opposite directions 92 Aaron Haupt Chapter 4 Forces in One Dimension 086-095 CH04-S1-845813 3/12/04 3:31 AM Page 93 Newton\u2019s Second Law You could conduct a series of experiments in which you and your friend vary the net force exerted on the table and measure the acceleration in each case. You would find that the acceleration of the table is proportional to the net force exerted on it and inversely proportional to its mass. In other words, if the net force of you and your friend acting on the table is 100 N, the table will experience the same acceleration as it would if only you were acting on it with a force of 100 N. Taking this into account, the mathematical relationship among force, mass, and acceleration can be rewritten in terms of the net force. The observation that the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object being accelerated is Newton\u2019s second law, which is represented by the following equation. F Newton\u2019s Second Law a net m The acceleration of an object is equal to the sum of the forces acting on the object, divided by the mass of the object. Notice that Newton\u2019s second law can be rearranged to the form F ma, which you learned about previously. If the table that you and your friend were pushing was 15.0 kg and the two of you each pushed with a force of 50.0 N in the same direction, what would be the acceleration of the table? To find out, calculate the net force, 50.0 N 50.0 N 100.0 N, and apply Newton\u2019s second law by dividing the net force of 100.0 N by the mass of the table, 15.0 kg, to get an acceleration of 6.67 m/s2. Here is a useful strategy for finding how the motion of an object depends on the forces exerted on it. First, identify all the forces acting on the object. Draw a free-body diagram showing the direction and relative strength of each force acting on the system. Then, add the forces to find the net force. Next, use Newton\u2019s second law to calculate the acceleration. Finally, if necessary, use kinematics to find the velocity or position of the object. When you learned about kinematics in Chapters 2 and 3, you studied the motion of objects without regard for the causes of motion. You now know that an unbalanced force, a net force, is the cause of a change in velocity (an acceleration). 6. Two horizontal forces, 225 N and 165 N, are exerted on a canoe. If these forces are applied in the same direction, find the net horizontal force on the canoe. 7. If the same two forces as in the previous problem are exerted on the canoe in opposite directions, what is the net horizontal force on the canoe? Be sure to indicate the direction of the net force. 8. Three confused sleigh dogs are trying to pull a sled across the Alaskan snow. Alutia pulls east with a force of 35 N, Seward also pulls east but with a force of 42 N, and big Kodiak pulls west with a force of 53 N. What is the net force on the sled? Section 4.1 Force and Motion 93 086-095 CH04-S1-845813 3/12/04 3:31 AM Page 94 Newton\u2019s First Law What is the motion of an object with no net force acting on it? A stationary object with no net force acting on it will stay at its position. Consider a moving object, such as a ball rolling on a surface. How long will the ball continue to roll? It will depend on the quality of the surface. If the ball is rolled on a thick carpet that offers much resistance, it will come to rest quickly. If it is rolled on a hard, smooth surface that offers little resistance, such as a bowling alley, the ball will roll for a long time with little change in velocity. Galileo did many experiments, and he concluded that in the ideal case of zero resistance, horizontal motion would never stop. Galileo was the first to recognize that the general principles of motion could be found by extrapolating experimental results to the ideal case, in which there is no resistance to slow down an object\u2019s motion. In the absence of a net force, the motion (or lack of motion) of both the moving ball and the stationary object continues as it was. Newton recognized this and generalized Galileo\u2019s results in a single statement. This statement, \u201can object that is at", " rest will remain at rest, and an object that is moving will continue to move in a straight line with constant speed, if and only if the net force acting on that object is zero,\u201d is called Newton\u2019s first law. Table 4-2 Some Types of Forces Force Symbol Definition Direction Friction Normal Spring Ff FN Fsp Tension FT Thrust Fthrust Weight Fg The contact force that acts to oppose sliding motion between surfaces Parallel to the surface and opposite the direction of sliding The contact force exerted by a surface on an object Perpendicular to and away from the surface A restoring force; that is, the push or pull a spring exerts on an object The pull exerted by a string, rope, or cable when attached to a body and pulled taut A general term for the forces that move objects such as rockets, planes, cars, and people A field force due to gravitational attraction between two objects, generally Earth and an object Opposite the displacement of the object at the end of the spring Away from the object and parallel to the string, rope, or cable at the point of attachment In the same direction as the acceleration of the object, barring any resistive forces Straight down toward the center of Earth 94 Chapter 4 Forces in One Dimension 086-095 CH04-S1-845813 5/31/04 11:31 PM Page 95 Shuttle Engine Thrust The Space Shuttle Main Engines (SSMEs) each are rated to provide 1.6 million N of thrust. Powered by the combustion of hydrogen and oxygen, the SSMEs are throttled anywhere from 65 percent to 109 percent of their rated thrust. Inertia Newton\u2019s first law is sometimes called the law of inertia. Is inertia a force? No. Inertia is the tendency of an object to resist change. If an object is at rest, it tends to remain at rest. If it is moving at a constant velocity, it tends to continue moving at that velocity. Forces are results of interactions between two objects; they are not properties of single objects, so inertia cannot be a force. Remember that because velocity includes both the speed and direction of motion, a net force is required to change either the speed or direction of an object\u2019s motion. Equilibrium According to Newton\u2019s first law, a net force is something that causes the velocity of an object to change. If the net force on an object is zero, then the object is in equilibrium. An object is in equilibrium if it is at rest or if it is moving at a constant velocity. Note that being at rest is simply a special case of the state of constant velocity, v 0. Newton\u2019s first law identifies a net force as something that disturbs a state of equilibrium. Thus, if there is no net force acting on the object, then the object does not experience a change in speed or direction and is in equilibrium. By understanding and applying Newton\u2019s first and second laws, you can often figure out something about the relative sizes of forces, even in situations in which you do not have numbers to work with. Before looking at an example of this, review Table 4-2, which lists some of the common types of forces. You will be dealing with many of these throughout your study of physics. When analyzing forces and motion, it is important to keep in mind that the world is dominated by resistance. Newton\u2019s ideal, resistance-free world is not easy to visualize. If you analyze a situation and find that the result is different from a similar experience that you have had, ask yourself if this is because of the presence of resistance. In addition, many terms used in physics have everyday meanings that are different from those understood in physics. When talking or writing about physics issues, be careful to use these terms in their precise, scientific way. 4.1 Section Review 9. Force Identify each of the following as either a, b, or c: weight, mass, inertia, the push of a hand, thrust, resistance, air resistance, spring force, and acceleration. a. a contact force b. a field force c. not a force 10. Inertia Can you feel the inertia of a pencil? Of a book? If you can, describe how. 11. Free-Body Diagram Draw a free-body diagram of a bag of sugar being lifted by your hand at a constant speed. Specifically identify the system. Label all forces with their agents and make the arrows the correct lengths. 12. Direction of Velocity If you push a book in the forward direction, does this mean its velocity has to be forward? 13. Free-Body Diagram Draw a free-body diagram of a water bucket being lifted by a rope at a decreasing speed. Specifically identify the system. Label all forces with their agents and make the arrows the correct lengths. 14. Critical Thinking A force of 1 N is the only force exerted on a block, and the acceleration of the block is measured. When the same force is the", " only force exerted on a second block, the acceleration is three times as large. What can you conclude about the masses of the two blocks? physicspp.com/self_check_quiz Section 4.1 Force and Motion 95 096-101 CH04-S2-845813 5/31/04 11:39 PM Page 96 4.2 Using Newton\u2019s Laws Objectives \u2022 Describe how the weight and the mass of an object are related. \u2022 Differentiate between actual weight and apparent weight. Vocabulary apparent weight weightlessness drag force terminal velocity \u25a0 Figure 4-6 The net force on the ball is the weight force, Fg. System v a Known a g m Unknown Fg Fg Fnet ma Fnet Fg and a g so Fg mg Newton\u2019s second law describes the connection between the cause of a change in an object\u2019s velocity and the resulting displacement. This law identifies the relationship between the net force exerted on an object and the object\u2019s acceleration. Using Newton\u2019s Second Law What is the weight force, Fg, exerted on an object of mass m? Newton\u2019s second law can help answer this question. Consider the pictorial and physical models in Figure 4-6, which show a free-falling ball in midair. With what objects is it interacting? Because it is touching nothing and air resistance can be neglected, the only force acting on it is Fg. You know from Chapter 3 that the ball\u2019s acceleration is g. Newton\u2019s second law then mg. Both the force and the acceleration are downward. The becomes Fg magnitude of an object\u2019s weight is equal to its mass times the acceleration it would have if it were falling freely. It is important to keep in mind that even when an object is not experiencing free-fall, the gravitational force of Earth is still acting on the object. This result is true on Earth, as well as on any other planet, although the magnitude of g will be different on other planets. Because the value of g is much less on the Moon than on Earth, astronauts who landed on the Moon weighed much less while on the Moon, even though their mass had not changed. Scales A bathroom scale contains springs. When you stand on the scale, the scale exerts an upward force on you because you are in contact with it. Because you are not accelerating, the net force acting on you must be zero. Therefore, the spring force, Fsp, pushing up on you must be the same magnitude as your weight, Fg, pulling down on you, as shown in the pictorial and physical models in Figure 4-7. The reading on the scale is determined by the amount of force the springs inside it exert on you. A spring scale, therefore, measures weight, not mass. If you were on a different planet, the compression of the spring would be different, and consequently, the scale\u2019s reading would be different. Remember that the proper unit for expressing mass is kilograms and because weight is a force, the proper unit used to express weight is the newton. a b y Fsp \u25a0 Figure 4-7 The upward force of the spring in the scale is equal to your weight when you step on the bathroom scale (a). The free-body diagram in (b) shows that the system is in equilibrium because the force of the spring is equal to your weight. 96 Aaron Haupt Chapter 4 Forces in One Dimension System Fg 096-101 CH04-S2-845813 3/12/04 3:40 AM Page 97 Fighting Over a Toy Anudja is holding a stuffed dog, with a mass of 0.30 kg, when Sarah decides that she wants it and tries to pull it away from Anudja. If Sarah pulls horizontally on the dog with a force of 10.0 N and Anudja pulls with a horizontal force of 11.0 N, what is the horizontal acceleration of the dog? 1 Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Identify the dog as the system and the direction in which Anudja pulls as positive. \u2022 Draw the free-body diagram. Label the forces. Known: m 0.30 kg FAnudja on dog FSarah on dog 11.0 N 10.0 N Unknown: a? 2 3 Solve for the Unknown FAnudja on dog Use Newton\u2019s second law. Fnet (FSarah on dog) F1 F2 F a net m (FSarah on dog) FAnudja on dog m Substitute Fnet FAnudja on dog (FSarah on dog) 11.0 N, FSarah on dog 10.0 N, m 0.30 kg 0.0 N Substitute FAnudja on dog 1 11.0 N g 0 k.3 0 3.3 m/s2 a 3.3 m/s2 toward Anudja Math Handbook Operations with", " Significant Digits pages 835\u2013836 Evaluate the Answer \u2022 Are the units correct? m/s2 is the correct unit for acceleration. \u2022 Does the sign make sense? The acceleration is in the positive direction, which is expected, because Anudja is pulling in the positive direction with a greater force than Sarah is pulling in the negative direction. \u2022 Is the magnitude realistic? It is a reasonable acceleration for a light, stuffed toy. 15. You place a watermelon on a spring scale at the supermarket. If the mass of the watermelon is 4.0 kg, what is the reading on the scale? 16. Kamaria is learning how to ice-skate. She wants her mother to pull her along so that she has an acceleration of 0.80 m/s2. If Kamaria\u2019s mass is 27.2 kg, with what force does her mother need to pull her? (Neglect any resistance between the ice and Kamaria\u2019s skates.) 17. Taru and Reiko simultaneously grab a 0.75-kg piece of rope and begin tugging on it in opposite directions. If Taru pulls with a force of 16.0 N and the rope accelerates away from her at 1.25 m/s2, with what force is Reiko pulling? 18. In Figure 4-8, the block has a mass of 1.2 kg and the sphere has a mass of 3.0 kg. What are the readings on the two scales? (Neglect the masses of the scales.) \u25a0 Figure 4-8 Section 4.2 Using Newton\u2019s Laws 97 096-101 CH04-S2-845813 3/12/04 3:41 AM Page 98 System y Fscale Fnet a v Fg \u25a0 Figure 4-9 If you stand on a scale in an elevator accelerating upward, the scale must exert an upward force greater than the downward force of your weight. Fscale Fg Apparent weight What is weight? Because the weight force is mg, Fg changes when g varies. On or near the defined as Fg surface of Earth, g is approximately constant, so an object\u2019s weight does not change appreciably as it moves around near Earth\u2019s surface. If a bathroom scale provides the only upward force on you, then it reads your weight. What would it read if you stood with one foot on the scale and one foot on the floor? What if a friend pushed down on your shoulders or up on your elbows? Then there would be other contact forces on you, and the scale would not read your weight. What happens if you stand on a scale in an elevator? As long as the elevator is in equilibrium, the scale will read your weight. What would the scale read if the elevator accelerates upward? Figure 4-9 shows the pictorial and physical models for this situation. You are the system, and upward is the positive direction. Because the acceleration of the system is upward, the upward force of the scale must be greater than the downward force of your weight. Therefore, the scale reading is greater than your weight. If you ride in an elevator like this, you would feel heavier because the floor would press harder on your feet. On the other hand, if the acceleration is downward, then you would feel lighter, and the scale would have a lower reading. The force exerted by the scale is called the apparent weight. An object\u2019s apparent weight is the force an object experiences as a result of all the forces acting on it, giving the object an acceleration. Imagine that the cable holding the elevator breaks. What would the scale read then? The scale and you would both accelerate with a g. According to this formula, the scale would read zero and your apparent weight would be zero. That is, you would be weightless. However, weightlessness does not mean that an object\u2019s weight is actually zero; rather, it means that there are no contact forces pushing up on the object, and the object\u2019s apparent weight is zero. Force and Motion When solving force and motion problems, use the following strategies. 1. Read the problem carefully, and sketch a pictorial model. 2. Circle the system and choose a coordinate system. 3. Determine which quantities are known and which are unknown. 4. Create a physical model by drawing a motion diagram showing the direction of the acceleration, and create a free-body diagram showing the net force. 5. Use Newton\u2019s laws to link acceleration and net force. 6. Rearrange the equation to solve for the unknown quantity. 7. Substitute known quantities with their units into the equation and solve. 8. Check your results to see if they are reasonable. 98 Chapter 4 Forces in One Dimension 096-101 CH04-S2-845813 3/12/04 3:41 AM Page 99 Real and Apparent Weight Your mass is 75.0 kg, and you are standing on a bathroom scale in an elevator", ". Starting from rest, the elevator accelerates upward at 2.00 m/s2 for 2.00 s and then continues at a constant speed. Is the scale reading during acceleration greater than, equal to, or less than the scale reading when the elevator is at rest? System y Fscale Fnet a v Fg Fscale Fg 1 Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Choose a coordinate system with the positive direction as upward. \u2022 Draw the motion diagram. Label v and a. \u2022 Draw the free-body diagram. The net force is in the same direction as the acceleration, so the upward force is greater than the downward force. Unknown:? Fscale Known: m 75.0 kg a 2.00 m/s2 t 2.00 s g 9.80 N 2 Solve for the Unknown Fnet Fnet ma Fscale (Fg) Solve for Fscale. Fnet Fscale Fg Fg is negative because it is in the negative direction defined by the coordinate system. Fscale Fg Elevator at rest: Fnet Fg mg (75.0 kg)(9.80 m/s2) 735 N The elevator is not accelerating. Thus, Fnet Substitute Fnet Substitute Fg Substitute m 75.0 kg, g 9.80 m/s2 0.00 N mg 0.00 N. Math Handbook Operations with Significant Digits pages 835\u2013836 Acceleration of the elevator: Fscale Fnet Fg ma mg m(a g) (75.0 kg)(2.00 m/s2 9.80 m/s2) 885 N mg Substitute Fnet Substitute m 75.0 kg, a 2.00 m/s2, g 9.80 m/s2 ma, Fg The scale reading when the elevator is accelerating (885 N) is larger than the scale reading at rest (735 N). 3 Evaluate the Answer \u2022 Are the units correct? kgm/s2 is the force unit, N. \u2022 Does the sign make sense? The positive sign agrees with the coordinate system. \u2022 Is the magnitude realistic? Fscale is larger than it would be at rest when Fscale would be 735 N, so the magnitude is reasonable. Section 4.2 Using Newton\u2019s Laws 99 096-101 CH04-S2-845813 5/31/04 11:40 PM Page 100 19. On Earth, a scale shows that you weigh 585 N. a. What is your mass? b. What would the scale read on the Moon (g 1.60 m/s2)? 20. Use the results from Example Problem 2 to answer questions about a scale in an elevator on Earth. What force would be exerted by the scale on a person in the following situations? a. The elevator moves at constant speed. b. It slows at 2.00 m/s2 while moving upward. c. It speeds up at 2.00 m/s2 while moving downward. d. It moves downward at constant speed. e. It slows to a stop at a constant magnitude of acceleration. Drag Force and Terminal Velocity It is true that the particles in the air around an object exert forces on it. Air actually exerts a huge force, but in most cases, it exerts a balanced force on all sides, and therefore it has no net effect. Can you think of any experiences that help to prove that air exerts a force? When you stick a suction cup on a smooth wall or table, you remove air from the \u201cinside\u201d of it. The suction cup is difficult to remove because of the net force of the air on the \u201coutside.\u201d So far, you have neglected the force of air on an object moving through the air. In actuality, when an object moves through any fluid, such as air or water, the fluid exerts a drag force on the moving object in the direction opposite to its motion. A drag force is the force exerted by a fluid on the object moving through the fluid. This force is dependent on the motion of the object, the properties of the object, and the properties of the fluid that the object is moving through. For example, as the speed of the object increases, so does the magnitude of the drag force. The size and shape of the object also affects the drag force. The drag force is also affected by the properties of the fluid, such as its viscosity and temperature. An air-track glider passes through a photoelectric gate at an initial speed of 0.25 m/s. As it passes through the gate, a constant force of 0.40 N is applied to the glider in the same direction as its motion. The glider has a mass of 0.50 kg. 1. What is the acceleration of the glider? 2. It takes the glider 1.3 s to pass through a second", " gate. What is the distance between the two gates? 3. The 0.40-N force is applied by means of a string attached to the glider. The other end of the string passes over a resistance-free pulley and is attached to a hanging mass, m. How big is m? 4. Derive an expression for the tension, T, in the string as a function of the mass, M, of the glider, the mass, m, of the hanging mass, and g. 100 Chapter 4 Forces in One Dimension 096-101 CH04-S2-845813 5/31/04 11:41 PM Page 101 If you drop a table-tennis ball, as in Figure 4-10, it has very little velocity at the start, and thus only a small drag force. The downward force of gravity is much stronger than the upward drag force, so there is a downward acceleration. As the ball\u2019s velocity increases, so does the drag force. Soon, the drag force equals the force of gravity. When this happens, there is no net force, and so there is no acceleration. The constant velocity that is reached when the drag force equals the force of gravity is called the terminal velocity. Fd Fg v Fd Fg v Fd Fg When light objects with large surface areas are falling, the drag force has a substantial effect on their motion, and they quickly reach terminal velocity. Heavier, more-compact objects are not affected as much by the drag force. For example, the terminal velocity of a table-tennis ball in air is 9 m/s, that of a basketball is 20 m/s, and that of a baseball is 42 m/s. Competitive skiers increase their terminal velocities by decreasing the drag force on them. They hold their bodies in an egg shape and wear smooth clothing and streamlined helmets. Sky divers can increase or decrease their terminal velocity by changing their body orientation and shape. A horizontal, spread-eagle shape produces the slowest terminal velocity, about 60 m/s. Because a parachute changes the shape of the sky diver when it opens, a sky diver becomes part of a very large object with a correspondingly large drag force and a terminal velocity of about 5 m/s. v Fd Fg v \u25a0 Figure 4-10 The drag force on an object increases as its velocity increases. When the drag force increases to the point that it equals the force of gravity, the object will no longer be accelerated. 4.2 Section Review 21. Lunar Gravity Compare the force holding a 10.0-kg rock on Earth and on the Moon. The acceleration due to gravity on the Moon is 1.62 m/s2. 22. Real and Apparent Weight You take a ride in a fast elevator to the top of a tall building and ride back down while standing on a bathroom scale. During which parts of the ride will your apparent and real weights be the same? During which parts will your apparent weight be less than your real weight? More than your real weight? Sketch freebody diagrams to support your answers. 23. Acceleration Tecle, with a mass of 65.0 kg, is standing by the boards at the side of an iceskating rink. He pushes off the boards with a force of 9.0 N. What is his resulting acceleration? 24. Motion of an Elevator You are riding in an elevator holding a spring scale with a 1-kg mass suspended from it. You look at the scale and see that it reads 9.3 N. What, if anything, can you conclude about the elevator\u2019s motion at that time? 25. Mass Marcos is playing tug-of-war with his cat using a stuffed toy. At one instant during the game, Marcos pulls on the toy with a force of 22 N, the cat pulls in the opposite direction with a force of 19.5 N, and the toy experiences an acceleration of 6.25 m/s2. What is the mass of the toy? 26. Acceleration A sky diver falls at a constant speed in the spread-eagle position. After he opens his parachute, is the sky diver accelerating? If so, in which direction? Explain your answer using Newton\u2019s laws. 27. Critical Thinking You have a job at a meat warehouse loading inventory onto trucks for shipment to grocery stores. Each truck has a weight limit of 10,000 N of cargo. You push each crate of meat along a low-resistance roller belt to a scale and weigh it before moving it onto the truck. However, right after you weigh a 1000-N crate, the scale breaks. Describe a way in which you could apply Newton\u2019s laws to figure out the approximate masses of the remaining crates. physicspp.com/self_check_quiz Section 4.2 Using Newton\u2019s Laws 101 102-107 CH04-S3-845813 5/31/04 11:47", " PM Page 102 4.3 Interaction Forces Objectives \u2022 Define Newton\u2019s third law. \u2022 Explain the tension in ropes and strings in terms of Newton\u2019s third law. \u2022 Define the normal force. \u2022 Determine the value of the normal force by applying Newton\u2019s second law. Vocabulary interaction pair Newton\u2019s third law tension normal force FA on B FB on A \u25a0 Figure 4-11 When you exert a force on your friend to push him forward, he exerts an equal and opposite force on you, which causes you to move backwards. You have learned that when an agent exerts a net force upon an object, the object undergoes acceleration. You know that this force can be either a field force or a contact force. But what causes the force? If you experiment with two magnets, you can feel each magnet pushing or pulling the other. Similarly, if you pull on a lever, you can feel the lever pulling back against you. Which is the agent and which is the object? Identifying Interaction Forces Imagine that you and a friend are each wearing in-line skates (with all the proper safety gear), and your friend is standing right in front of you, with his back to you. You push your friend so that he starts rolling forward. What happens to you? You move backwards. Why? Recall that a force is the result of an interaction between two objects. When you push your friend forward, you come into contact with him and exert a force that moves him forward. However, because he is also in contact with you, he also exerts a force on you, and this results in a change in your motion. Forces always come in pairs. Consider you (Student A) as one system and your friend (Student B) as another. What horizontal forces act on each of the two systems? Figure 4-11 shows the free-body diagram for the systems. Looking at this diagram, you can see that each system experiences a force exerted by the other. The two forces, FA on B and FB on A, are the forces of interaction between the two of you. Notice the symmetry in the subscripts: A on B and B on A. What do you notice about the directions of these forces? What do you expect to be true about their relative magnitudes? The forces FA on B and FB on A are an interaction pair. An interaction pair is two forces that are in opposite directions and have equal magnitude. Sometimes, this also is called an action-reaction pair of forces. This might suggest that one causes the other; however, this is not true. For example, the force of you pushing your friend doesn\u2019t cause your friend to exert a force on you. The two forces either exist together or not at all. They both result from the contact between the two of you. Newton\u2019s Third Law The force of you on your friend is equal in magnitude and opposite in direction to the force of your friend on you. This is summarized in Newton\u2019s third law, which states that all forces come in pairs. The two forces in a pair act on different objects and are equal in strength and opposite in direction. Newton\u2019s Third Law FA on B The force of A on B is equal in magnitude and opposite in direction of the force of B on A. FB on A Consider the situation of you holding a book in your hand. Draw one free-body diagram each for you and for the book. Are there any interaction 102 Chapter 4 Forces in One Dimension 102-107 CH04-S3-845813 6/7/04 5:49 AM Page 103 pairs? When identifying interaction pairs, keep in mind that they always will occur in two different free-body diagrams, and they always will have the symmetry of subscripts noted on the previous page. In this case, there is one interaction pair, Fbook on hand and Fhand on book. Notice also that each object has a weight. If the weight force is due to an interaction between the object and Earth\u2019s mass, then shouldn\u2019t each of these objects also exert a force on Earth? If this is the case, shouldn\u2019t Earth be accelerating? Consider a soccer ball sitting on a table. The table, in turn, is sitting on Earth, as shown in Figure 4-12. First, analyze the forces acting on the ball. The table exerts an upward force on the ball, and the mass of Earth exerts a downward gravitational force on the ball. Even though these forces are in the opposite direction on the same object, they are not an interaction pair. They are simply two forces acting on the same object, not the interaction between two objects. Consider the ball and the table together. In addition to the upward force exerted by the table on the ball, the ball exerts a downward force on the table. This is one pair of forces. The ball and Earth also have an interaction pair. Thus, the inter Ftable", " on ball and action pairs related to the soccer ball are Fball on table FEarth on ball. It is important to keep in mind that an interFball on Earth action pair must consist of two forces of equal magnitude pointing in opposite directions. These opposing forces must act on two different objects that can exert a force against each other. The acceleration caused by the force of an object interacting with Earth is usually a very small number. Under most circumstances, the number is so small that for problems involving falling or stationary objects, Earth can be treated as part of the external world rather than as a second system. Consider Example Problem 3 using the following problem-solving strategies. Interaction Pairs Use these strategies to solve problems in which there is an interaction between objects in two different systems. 1. Separate the system or systems from the external world. 2. Draw a pictorial model with coordinate systems for each system and a physical model that includes free-body diagrams for each system. 3. Connect interaction pairs by dashed lines. 4. To calculate your answer, use Newton\u2019s second law to relate the net force and acceleration for each system. 5. Use Newton\u2019s third law to equate the magnitudes of the interaction pairs and give the relative direction of each force. 6. Solve the problem and check the units, signs, and magnitudes for reasonableness. \u25a0 Figure 4-12 A soccer ball on a table on Earth is part of two interaction pairs\u2014the interaction between the ball and table and the interaction between the ball and Earth. (Not to scale) Tug-of-War Challenge In a tug-of-war, predict how the force you exert on your end of the rope compares to the force your opponent exerts if you pull and your opponent just holds the rope. 1. Predict how the forces compare if the rope moves in your direction. 2. Test your prediction. CAUTION: Do not suddenly let go of the rope. Analyze and Conclude 3. Compare the force on your end of the rope to the force on your opponent\u2019s end of the rope. What happened when you started to move your opponent\u2019s direction? Section 4.3 Interaction Forces 103 102-107 CH04-S3-845813 3/12/04 3:48 AM Page 104 Earth\u2019s Acceleration When a softball with a mass of 0.18 kg is dropped, its acceleration toward Earth is equal to g, the acceleration due to gravity. What is the force on Earth due to the ball, and what is Earth\u2019s resulting acceleration? Earth\u2019s mass is 6.01024 kg. 1 Analyze and Sketch the Problem \u2022 Draw free-body diagrams for the two systems: the ball and Earth. \u2022 Connect the interaction pair by a dashed line. Known: mball mEarth g 9.80 m/s2 0.18 kg 6.01024 kg Unknown: FEarth on ball? aEarth? 2 Solve for the Unknown Use Newton\u2019s second law to find the weight of the ball. Earth Softball y FEarth on ball Fball on Earth FEarth on ball mballa mball(g) (0.18 kg)(9.80 m/s2) 1.8 N Substitute a g Substitute m ball 0.18 kg, g 9.80 m/s2 Substitute FEarth on ball Fball on Earth Use Newton\u2019s third law to find Fball on Earth. FEarth on ball (1.8 N) 1.8 N Use Newton\u2019s second law to find aEarth. aEarth F n t e m th a E r 8 N 1. Substitute Fnet 6.0 24 kg 0 1 2.91025 m/s2 toward the softball 1.8 N Math Handbook Operations with Scientific Notation pages 842\u2013843 1.8 N, mEarth 6.01024 kg 3 Evaluate the Answer \u2022 Are the units correct? Dimensional analysis verifies force in N and acceleration in m/s2. \u2022 Do the signs make sense? Force and acceleration should be positive. \u2022 Is the magnitude realistic? Because of Earth\u2019s large mass, the acceleration should be small. 28. You lift a relatively light bowling ball with your hand, accelerating it upward. What are the forces on the ball? What forces does the ball exert? What objects are these forces exerted on? 29. A brick falls from a construction scaffold. Identify any forces acting on the brick. Also identify any forces that the brick exerts and the objects on which these forces are exerted. (Air resistance may be ignored.) 30. You toss a ball up in the air. Draw a free-body diagram for the ball while it is still moving upward. Identify any forces acting on the ball. Also identify any forces that the ball exerts and the objects on which these forces are exerted. \u25a0 Figure 4-13 31. A suitcase sits", " on a stationary airport luggage cart, as in Figure 4-13. Draw a free-body diagram for each object and specifically indicate any interaction pairs between the two. 104 Chapter 4 Forces in One Dimension 102-107 CH04-S3-845813 3/12/04 3:48 AM Page 105 Forces of Ropes and Strings Tension is simply a specific name for the force exerted by a string or rope. A simplification within this textbook is the assumption that all strings and ropes are massless. To understand tension in more detail, consider the situation in Figure 4-14, where a bucket hangs from a rope attached to the ceiling. The rope is about to break in the middle. If the rope breaks, the bucket will fall; thus, before it breaks, there must be forces holding the rope together. The force that the top part of the rope exerts on the bottom part is Ftop on bottom. Newton\u2019s third law states that this force must be part of an interaction pair. The other member of the pair is the force that the bottom part exerts on the top, Fbottom on top. These forces, equal in magnitude but opposite in direction, also are shown in Figure 4-14. Think about this situation in another way. Before the rope breaks, the bucket is in equilibrium. This means that the force of its weight downward must be equal in magnitude but opposite in direction to the tension in the rope upward. Similarly, if you look at the point in the rope just above the bucket, it also is in equilibrium. Therefore, the tension of the rope below it pulling down must be equal to the tension of the rope above it pulling up. You can move up the rope, considering any point in the rope, and see that the tension forces are pulling equally in both directions. Because the very bottom of the rope has a tension equal to the weight of the bucket, the tension everywhere in the rope is equal to the weight of the bucket. Thus, the tension in the rope is the weight of all objects below it. Because the rope is assumed to be massless, the tension everywhere in the rope is equal to the bucket\u2019s weight. Tension forces also are at work in a tug-of-war, like the one shown in Figure 4-15. If team A, on the left, is exerting a force of 500 N and the rope does not move, then team B, on the right, also must be pulling with a force of 500 N. What is the tension in the rope in this case? If each team pulls with 500 N of force, is the tension 1000 N? To decide, think of the rope as divided into two halves. The left-hand end is not moving, so the net 500 N. Similarly, force on it is zero. Thus, FA on rope 500 N. But the two tensions, Fright on left and FB on rope Fleft on right, are an interaction pair, so they are equal and opposite. Thus, the tension in the rope equals the force with which each team pulls, or 500 N. To verify this, you could cut the rope in half, tie the ends to a spring scale, and ask the two teams each to pull with 500 N of force. You would see that the scale reads 500 N. Fright on left Fleft on right FT (bottom on top) FT (top on bottom) Fg \u25a0 Figure 4-14 The tension in the rope is equal to the weight of all the objects hanging from it. FA 500 N FB 500 N Fnet 0 N \u25a0 Figure 4-15 In a tug-of-war, the teams exert equal and opposite forces on each other via the tension in the rope, as long as neither side moves. Section 4.3 Interaction Forces 105 Tim Fuller 102-107 CH04-S3-845813 3/12/04 3:49 AM Page 106 Lifting a Bucket A 50.0-kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3.0 m, it is moving at 3.0 m/s. If the acceleration is constant, is the rope in danger of breaking? 1 2 Analyze and Sketch the Problem \u2022 Draw the situation and identify the forces on the system. \u2022 Establish a coordinate system with the positive axis upward. \u2022 Draw a motion diagram including v and a. \u2022 Draw the free-body diagram, labeling the forces. y m 50.0 kg 0.0 m/s vi Known: vf d 3.0 m 3.0 m/s Unknown:? FT Solve for the Unknown Fnet is the sum of the positive force of the rope pulling up, FT, and the negative weight force, Fg, pulling down as defined by the coordinate system. Fnet FT (Fg) FT Fnet Fg ma mg m(a g) Substitute", " Fnet ma, Fg mg v FT (rope on bucket) a Fnet System Fg (Earth\u2019s mass on bucket) vi, vf, and d are known. Use this motion equation to solve for a. 2 2ad 2 vi vf 2 2 vi vf a 2 d 2 vf d 2 FT 2 m(a g) m( g) vf d 2 (50.0 kg)((3 570 N 2 m 0. 0 3. ( 9.80 m/s2) 2 s) / ) m Substitute vi 0.0 m/s2 2 v Substitute a f 2 d Math Handbook Isolating a Variable page 845 Substitute m 50.0 kg, vf 3.0 m/s, d 3.0 m, g 9.80 m/s2 The rope is in danger of breaking because the tension exceeds 525 N. 3 Evaluate the Answer \u2022 Are the units correct? Dimensional analysis verifies kgm/s2, which is N. \u2022 Does the sign make sense? The upward force should be positive. \u2022 Is the magnitude realistic? The magnitude is a little larger than 490 N, which is the mg (50.0 kg)(9.80 m/s2) 490 N weight of the bucket. Fg 32. You are helping to repair a roof by loading equipment into a bucket that workers hoist to the rooftop. If the rope is guaranteed not to break as long as the tension does not exceed 450 N and you fill the bucket until it has a mass of 42 kg, what is the greatest acceleration that the workers can give the bucket as they pull it to the roof? 33. Diego and Mika are trying to fix a tire on Diego\u2019s car, but they are having trouble getting the tire loose. When they pull together, Mika with a force of 23 N and Diego with a force of 31 N, they just barely get the tire to budge. What is the magnitude of the strength of the force between the tire and the wheel? 106 Chapter 4 Forces in One Dimension 102-107 CH04-S3-845813 3/12/04 3:50 AM Page 107 a b c FN FN 50.0 N FN 50.0 N mg mg mg The Normal Force Any time two objects are in contact, they each exert a force on each other. Think about a box sitting on a table. There is a downward force due to the gravitational attraction of Earth. There also is an upward force that the table exerts on the box. This force must exist, because the box is in equilibrium. The normal force is the perpendicular contact force exerted by a surface on another object. \u25a0 Figure 4-16 The normal force on an object is not always equal to its weight. In (a) the normal force is equal to the object\u2019s weight. In (b) the normal force is less than the object\u2019s weight. In (c) the normal force is greater than the object\u2019s weight. The normal force always is perpendicular to the plane of contact between two objects, but is it always equal to the weight of an object as in Figure 4-16a? What if you tied a string to the box and pulled up on it a little bit, but not enough to move the box, as shown in Figure 4-16b? When you Fg apply Newton\u2019s second law to the box, you see that FN Fstring on box. ma 0, which rearranges to FN Fstring on box Fg You can see that in this case, the normal force exerted by the table on the box is less than the box\u2019s weight, Fg. Similarly, if you pushed down on the box on the table as shown in Figure 4-16c, the normal force would be more than the box\u2019s weight. Finding the normal force will be important in the next chapter, when you begin dealing with resistance. 4.3 Section Review 34. Force Hold a book motionless in your hand in the air. Identify each force and its interaction pair on the book. 35. Force Lower the book from problem 34 at increasing speed. Do any of the forces or their interaction-pair partners change? Explain. 36. Tension A block hangs from the ceiling by a massless rope. A second block is attached to the first block and hangs below it on another piece of massless rope. If each of the two blocks has a mass of 5.0 kg, what is the tension in each rope? 37. Tension If the bottom block in problem 36 has a mass of 3.0 kg and the tension in the top rope is 63.0 N, calculate the tension in the bottom rope and the mass of the top block. 38. Normal Force Poloma hands a 13-kg box to 61-kg Stephanie, who stands on a platform. What is the normal force exerted by the platform on Stephanie? 39", ". Critical Thinking A curtain prevents two tug-ofwar teams from seeing each other. One team ties its end of the rope to a tree. If the other team pulls with a 500-N force, what is the tension? Explain. physicspp.com/self_check_quiz Section 4.3 Interaction Forces 107 108-109 CH04-LAB-845813 7/14/04 1:24 PM Page 108 Alternate CBL instructions can be found on the Web site. physicspp.com Forces in an Elevator Have you ever been in a fast-moving elevator? Was the ride comfortable? How about an amusement ride that quickly moves upward or one that free-falls? What forces are acting on you during your ride? In this experiment, you will investigate the forces that affect you during vertical motion when gravity is involved with a bathroom scale. Many bathroom scales measure weight in pounds mass (lbm) or pounds force (lbf) rather than newtons. In the experiment, you will need to convert weights measured on common household bathroom scales to SI units. QUESTION What one-dimensional forces act on an object that is moving in a vertical direction in relation to the ground? Objectives Safety Precautions \u25a0 Measure Examine forces that act on objects that move vertically. \u25a0 Compare and Contrast Differentiate between actual weight and apparent weight. \u25a0 Analyze and Conclude Share and compare data of the acceleration of elevators. \u25a0 Use caution when working around elevator doors. \u25a0 Do not interfere with normal elevator traffic. \u25a0 Watch that the mass on the spring scale does not fall and hit someone\u2019s feet or toes. Materials elevator bathroom scale spring scale mass Procedure 1. Securely attach a mass to the hook on a spring scale. Record the force of the mass in the data table. 2. Accelerate the mass upward, then move it upward at a constant velocity, and then slow the mass down. Record the greatest amount of force on the scale, the amount of force at constant velocity, and the lowest scale reading. 3. Get your teacher\u2019s permission and proceed to an elevator on the ground floor. Before entering the elevator, measure your weight on a bathroom scale. Record this weight in the data table. 108 Laura Sifferlin 108-109 CH04-LAB-845813 7/14/04 1:25 PM Page 109 Data Table Force (step 1) Highest Reading (step 2) Reading at Constant Velocity (step 2) Lowest Reading (step 2) Your Weight (step 3) Highest Reading (step 4) Reading at Constant Velocity (step 5) Lowest Reading (step 6) 4. Place the scale in the elevator. Step on the scale and record the mass at rest. Select the highest floor that the elevator goes up to. Once the elevator starts, during its upward acceleration, record the highest reading on the scale in the data table. 5. When the velocity of the elevator becomes constant, record the reading on the scale in the data table. 6. As the elevator starts to decelerate, watch for the lowest reading on the scale and record it in the data table. Analyze 1. Explain In step 2, why did the mass appear to gain weight when being accelerated upward? Provide a mathematical equation to summarize this concept. 2. Explain Why did the mass appear to lose weight when being decelerated at the end of its movement during step 3? Provide a mathematical equation to summarize this concept. 3. Measure in SI Most bathroom scales read in pounds mass (lbm). Convert your reading in step 4 in pounds mass to kilograms. (1 kg 2.21 lbm) (Note: skip this step if your scale measures in kilograms.) 4. Measure in SI Some bathroom scales read in pounds force (lbf). Convert all of the readings you made in steps 4\u20136 to newtons. (1 N 0.225 lbf) 5. Analyze Calculate the acceleration of the elevator at the beginning of your elevator trip using the equation Fscale ma mg. 6. Use Numbers What is the acceleration of the elevator at the end of your trip? Conclude and Apply How can you develop an experiment to find the acceleration of an amusement park ride that either drops rapidly or climbs rapidly? Going Further How can a bathroom scale measure both pounds mass (lbm) and pounds force (lbf) at the same time? Real-World Physics Forces on pilots in high-performance jet airplanes are measured in g\u2019s or g-force. What does it mean if a pilot is pulling 6 g\u2019s in a power climb? Communicate You can visit physicspp.com/ internet_lab to post the acceleration of your elevator and compare it to other elevators around the country, maybe even the world. Post a description of your elevator\u2019s ride so that a comparison of acceleration versus ride comfort can be evaluated. To find out more about forces and acceleration, visit the Web site: physicspp.com 109 110 CH04-FEATURE-", "845813 6/1/04 12:56 AM Page 110 Bathroom Scale The portable weighing scale was patented in 1896 by John H. Hunter. People used coin-operated scales, usually located in stores, to weigh themselves until the advent of the home bathroom scale in 1946. How does a bathroom scale work? 1 There are two long and two short levers that are attached to each other. Brackets in the lid of the scale sit on top of the levers to help evenly distribute your weight on the levers. 3 As the calibrating plate is pushed down by weight on the scale, the crank pivots. This, in turn, moves the rack and rotates the pinion. As a result, the dial on the scale rotates. Main spring Crank Dial Dial spring Calibrating plate Fg Pinion Rack Lever 4 When the spring force, Fsp, from the main spring being stretched is equal to Fg, the crank, rack, and pinion no longer move, and your weight is shown on the dial. 2 The long levers rest on top of a calibrating plate that has the main spring attached to it. When you step on the scale, your weight, Fg, is exerted on the levers, which, in turn, exert a force on the calibrating plate and cause the main spring to stretch. Thinking Critically 1. Hypothesize Most springs in bathroom scales cannot exert a force larger than 20 lbs (89 N). How is it possible that you don\u2019t break the scale every time you step on it? (Hint: Think about exerting a large force near the pivot of a see-saw.) 2. Solve If the largest reading on most scales is 240 lbs (1068 N) and the spring can exert a maximum of 20 lbs (89 N), what ratio does the lever use? 110 How It Works 111-117 CH04-SG CA STP-845813 6/1/04 1:04 AM Page 111 4.1 Force and Motion Vocabulary \u2022 force (p. 88) \u2022 free-body diagram (p. 89) \u2022 net force (p. 92) \u2022 Newton\u2019s second law (p. 93) \u2022 Newton\u2019s first law (p. 94) \u2022 inertia (p. 95) \u2022 equilibrium (p. 95) Key Concepts \u2022 An object that experiences a push or a pull has a force exerted on it. \u2022 Forces have both direction and magnitude. \u2022 Forces may be divided into contact and field forces. \u2022 In a free-body diagram, always draw the force vectors leading away from the object, even if the force is a push. \u2022 The forces acting upon an object can be added using vector addition to find the net force. \u2022 Newton\u2019s second law states that the acceleration of a system equals the net force acting on it, divided by its mass. Fn a et m \u2022 Newton\u2019s first law states that an object that is at rest will remain at rest, and an object that is moving will continue to move in a straight line with constant speed, if and only if the net force acting on that object is zero. \u2022 An object with no net force acting on it is in equilibrium. 4.2 Using Newton\u2019s Laws Vocabulary \u2022 apparent weight (p. 98) \u2022 weightlessness (p. 98) \u2022 drag force (p. 100) \u2022 terminal velocity (p. 101) Key Concepts \u2022 The weight of an object depends upon the acceleration due to gravity and the mass of the object. \u2022 An object\u2019s apparent weight is the force an object experiences as a result of the contact forces acting on it, giving the object an acceleration. \u2022 An object with no apparent weight experiences weightlessness. \u2022 The effect of drag on an object\u2019s motion is determined by the object\u2019s weight, size, and shape. \u2022 If a falling object reaches a velocity such that the drag force is equal to the object\u2019s weight, it maintains that velocity, called the terminal velocity. 4.3 Interaction Forces Vocabulary \u2022 interaction pair (p. 102) \u2022 Newton\u2019s third law (p. 102) \u2022 tension (p. 105) \u2022 normal force (p. 107) Key Concepts \u2022 All forces result from interactions between objects. \u2022 Newton\u2019s third law states that the two forces that make up an interaction pair of forces are equal in magnitude, but opposite in direction and act on different objects. FA on B FB on A \u2022 In an interaction pair, FA on B does not cause FB on A. The two forces either exist together or not at all. \u2022 Tension is the specific name for the force exerted by a rope or string. \u2022 The normal force is a support force resulting from the contact of two objects. It is always perpendicular to the plane of contact between the two objects. physicspp.com/vocabulary_puzzlemaker 111 111-117 CH04-SG CA STP-845813 6/1/", "04 1:04 AM Page 112 Concept Mapping 40. Complete the following concept map using the following term and symbols: normal, FT, Fg. force tension gravity FN Mastering Concepts 41. A physics book is motionless on the top of a table. If you give it a hard push with your hand, it slides across the table and slowly comes to a stop. Use Newton\u2019s laws to answer the following questions. (4.1) a. Why does the book remain motionless before the force of your hand is applied? b. Why does the book begin to move when your hand pushes hard enough on it? c. Under what conditions would the book remain in motion at a constant speed? 42. Cycling Why do you have to push harder on the pedals of a single-speed bicycle to start it moving than to keep it moving at a constant velocity? (4.1) 43. Suppose that the acceleration of an object is zero. Does this mean that there are no forces acting on it? Give an example supporting your answer. (4.2) 48. Ramon pushes on a bed that has been pushed against a wall, as in Figure 4-17. Draw a free-body diagram for the bed and identify all the forces acting on it. Make a separate list of all the forces that the bed applies to other objects. (4.3) \u25a0 Figure 4-17 49. Figure 4-18 shows a block in four different situations. Rank them according to the magnitude of the normal force between the block and the surface, greatest to least. Specifically indicate any ties. (4.3) \u25a0 Figure 4-18 44. Basketball When a basketball player dribbles a 50. Explain why the tension in a massless rope is ball, it falls to the floor and bounces up. Is a force required to make it bounce? Why? If a force is needed, what is the agent involved? (4.2) 45. Before a sky diver opens her parachute, she may be falling at a velocity higher than the terminal velocity that she will have after the parachute opens. (4.2) a. Describe what happens to her velocity as she opens the parachute. b. Describe the sky diver\u2019s velocity from when her parachute has been open for a time until she is about to land. 46. If your textbook is in equilibrium, what can you say about the forces acting on it? (4.2) 47. A rock is dropped from a bridge into a valley. Earth pulls on the rock and accelerates it downward. According to Newton\u2019s third law, the rock must also be pulling on Earth, yet Earth does not seem to accelerate. Explain. (4.3) constant throughout it. (4.3) 51. A bird sits on top of a statue of Einstein. Draw free-body diagrams for the bird and the statue. Specifically indicate any interaction pairs between the two diagrams. (4.3) 52. Baseball A slugger swings his bat and hits a baseball pitched to him. Draw free-body diagrams for the baseball and the bat at the moment of contact. Specifically indicate any interaction pairs between the two diagrams. (4.3) Applying Concepts 53. Whiplash If you are in a car that is struck from behind, you can receive a serious neck injury called whiplash. a. Using Newton\u2019s laws, explain what happens to cause such an injury. b. How does a headrest reduce whiplash? 112 Chapter 4 Forces in One Dimension For more problems, go to Additional Problems, Appendix B. 111-117 CH04-SG CA STP-845813 6/1/04 1:04 AM Page 113 54. Space Should astronauts choose pencils with hard or soft lead for making notes in space? Explain. 55. When you look at the label of the product in Figure 4-19 to get an idea of how much the box contains, does it tell you its mass, weight, or both? Would you need to make any changes to this label to make it correct for consumption on the Moon? \u25a0 Figure 4-19 56. From the top of a tall building, you drop two table- tennis balls, one filled with air and the other with water. Both experience air resistance as they fall. Which ball reaches terminal velocity first? Do both hit the ground at the same time? 57. It can be said that 1 kg equals 2.2 lb. What does this statement mean? What would be the proper way of making the comparison? 58. You toss a ball straight up into the air. a. Draw a free-body diagram for the ball at three points during its motion: on the way up, at the very top, and on the way down. Specifically identify the forces acting on the ball and their agents. 62. Breaking the Wishbone After Thanksgiving, Kevin and Gamal use the turkey\u2019s wishbone to make a wish. If Kevin pulls on it with a", " force 0.17 N larger than the force Gamal pulls with in the opposite direction, and the wishbone has a mass of 13 g, what is the wishbone\u2019s initial acceleration? 4.2 Using Newton\u2019s Laws 63. What is your weight in newtons? 64. Motorcycle Your new motorcycle weighs 2450 N. What is its mass in kilograms? 65. Three objects are dropped simultaneously from the top of a tall building: a shot put, an air-filled balloon, and a basketball. a. Rank the objects in the order in which they will reach terminal velocity, from first to last. b. Rank the objects according to the order in which they will reach the ground, from first to last. c. What is the relationship between your answers to parts a and b? 66. What is the weight in pounds of a 100.0-N wooden shipping case? 67. You place a 7.50-kg television on a spring scale. If the scale reads 78.4 N, what is the acceleration due to gravity at that location? 68. Drag Racing A 873-kg (1930-lb) dragster, starting from rest, attains a speed of 26.3 m/s (58.9 mph) in 0.59 s. a. Find the average acceleration of the dragster during this time interval. b. What is the magnitude of the average net force on the dragster during this time? c. Assume that the driver has a mass of 68 kg. What horizontal force does the seat exert on the driver? b. What is the velocity of the ball at the very top of 69. Assume that a scale is in an elevator on Earth. the motion? c. What is the acceleration of the ball at this same point? Mastering Problems 4.1 Force and Motion 59. What is the net force acting on a 1.0-kg ball in free-fall? 60. Skating Joyce and Efua are skating. Joyce pushes Efua, whose mass is 40.0-kg, with a force of 5.0 N. What is Efua\u2019s resulting acceleration? What force would the scale exert on a 53-kg person standing on it during the following situations? a. The elevator moves up at a constant speed. b. It slows at 2.0 m/s2 while moving upward. c. It speeds up at 2.0 m/s2 while moving downward. d. It moves downward at a constant speed. e. It slows to a stop while moving downward with a constant acceleration. 70. A grocery sack can withstand a maximum of 230 N before it rips. Will a bag holding 15 kg of groceries that is lifted from the checkout counter at an acceleration of 7.0 m/s2 hold? 61. A car of mass 2300 kg slows down at a rate of 71. A 0.50-kg guinea pig is lifted up from the ground. 3.0 m/s2 when approaching a stop sign. What is the magnitude of the net force causing it to slow down? What is the smallest force needed to lift it? Describe its resulting motion. physicspp.com/chapter_test Chapter 4 Assessment 113 Aaron Haupt 111-117 CH04-SG CA STP-845813 3/12/04 4:11 AM Page 114 72. Astronomy On the surface of Mercury, the gravitational acceleration is 0.38 times its value on Earth. a. What would a 6.0-kg mass weigh on Mercury? b. If the gravitational acceleration on the surface of Pluto is 0.08 times that of Mercury, what would a 7.0-kg mass weigh on Pluto? 73. A 65-kg diver jumps off of a 10.0-m tower. a. Find the diver\u2019s velocity when he hits the water. b. The diver comes to a stop 2.0 m below the surface. Find the net force exerted by the water. 74. Car Racing A race car has a mass of 710 kg. It starts from rest and travels 40.0 m in 3.0 s. The car is uniformly accelerated during the entire time. What net force is exerted on it? 4.3 Interaction Forces 75. A 6.0-kg block rests on top of a 7.0-kg block, which rests on a horizontal table. a. What is the force (magnitude and direction) exerted by the 7.0-kg block on the 6.0-kg block? b. What is the force (magnitude and direction) exerted by the 6.0-kg block on the 7.0-kg block? 76. Rain A raindrop, with mass 2.45 mg, falls to the ground. As it is falling, what magnitude of force does it exert on Earth? 77. A 90.0-kg man and a 55-kg man have a tug", "-of-war. The 90.0-kg man pulls on the rope such that the 55-kg man accelerates at 0.025 m/s2. What force does the rope exert on the 90.0-kg man? 78. Male lions and human sprinters can both accelerate at about 10.0 m/s2. If a typical lion weighs 170 kg and a typical sprinter weighs 75 kg, what is the difference in the force exerted on the ground during a race between these two species? 79. A 4500-kg helicopter accelerates upward at 2.0 m/s2. What lift force is exerted by the air on the propellers? 80. Three blocks are stacked on top of one another, as in Figure 4-20. The top block has a mass of 4.6 kg, the middle one has a mass of 1.2 kg, and the bottom one has a mass of 3.7 kg. Identify and calculate any normal forces between the objects. \u25a0 Figure 4-20 Mixed Review 81. The dragster in problem 68 completed a 402.3-m (0.2500-mi) run in 4.936 s. If the car had a constant acceleration, what was its acceleration and final velocity? 82. Jet A 2.75106-N catapult jet plane is ready for takeoff. If the jet\u2019s engines supply a constant thrust of 6.35106 N, how much runway will it need to reach its minimum takeoff speed of 285 km/h? 83. The dragster in problem 68 crossed the finish line going 126.6 m/s. Does the assumption of constant acceleration hold true? What other piece of evidence could you use to determine if the acceleration was constant? 84. Suppose a 65-kg boy and a 45-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface as in Figure 4-21. If the acceleration of the girl toward the boy is 3.0 m/s2, find the magnitude of the acceleration of the boy toward the girl.? 3.0 m/s2 \u25a0 Figure 4-21 85. Space Station Pratish weighs 588 N and is weightless in a space station. If she pushes off the wall with a vertical acceleration of 3.00 m/s2, determine the force exerted by the wall during her push off. 86. Baseball As a baseball is being caught, its speed goes from 30.0 m/s to 0.0 m/s in about 0.0050 s. The mass of the baseball is 0.145 kg. a. What is the baseball\u2019s acceleration? b. What are the magnitude and direction of the force acting on it? c. What are the magnitude and direction of the force acting on the player who caught it? 87. Air Hockey An air-hockey table works by pumping air through thousands of tiny holes in a table to support light pucks. This allows the pucks to move around on cushions of air with very little resistance. One of these pucks has a mass of 0.25 kg and is pushed along by a 12.0-N force for 9.0 s. a. What is the puck\u2019s acceleration? b. What is the puck\u2019s final velocity? 114 Chapter 4 Forces in One Dimension For more problems, go to Additional Problems, Appendix B. 111-117 CH04-SG CA STP-845813 3/12/04 4:12 AM Page 115 88. A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N. a. As the elevator moves up, the scale reading increases to 936 N. Find the acceleration of the elevator. b. As the elevator approaches the 74th floor, the scale reading drops to 782 N. What is the acceleration of the elevator? c. Using your results from parts a and b, explain which change in velocity, starting or stopping, takes the longer time. 89. Weather Balloon The instruments attached to a weather balloon in Figure 4-22 have a mass of 5.0 kg. The balloon is released and exerts an upward force of 98 N on the instruments. a. What is the acceleration of the balloon and instruments? b. After the balloon has accelerated for 10.0 s, the instruments are released. What is the velocity of the instruments at the moment of their release? c. What net force acts on the instruments after their release? d. When does the direction of the instruments\u2019 velocity first become downward? \u25a0 Figure 4-23 92. Two blocks, one of mass 5.0 kg and the other of mass 3.0 kg, are tied together with a massless rope as in Figure 4-24. This rope is strung over a massless, resistance-free pulley. The blocks are released from rest. Find the following. a. the", " tension in the rope b. the acceleration of the blocks Hint: you will need to solve two simultaneous equations. 98 N 5.0 kg \u25a0 Figure 4-22 90. When a horizontal force of 4.5 N acts on a block on a resistance-free surface, it produces an acceleration of 2.5 m/s2. Suppose a second 4.0-kg block is dropped onto the first. What is the magnitude of the acceleration of the combination if the same force continues to act? Assume that the second block does not slide on the first block. 91. Two blocks, masses 4.3 kg and 5.4 kg, are pushed across a frictionless surface by a horizontal force of 22.5 N, as shown in Figure 4-23. a. What is the acceleration of the blocks? b. What is the force of the 4.3-kg block on the 3.0 kg 5.0 kg \u25a0 Figure 4-24 Thinking Critically 93. Formulate Models A 2.0-kg mass, mA, and a 3.0-kg mass, mB, are connected to a lightweight cord that passes over a frictionless pulley. The pulley only changes the direction of the force exerted by the rope. The hanging masses are free to move. Choose coordinate systems for the two masses with the positive direction being up for mA and down for mB. a. Create a pictorial model. b. Create a physical model with motion and free- body diagrams. c. What is the acceleration of the smaller mass? 5.4-kg block? c. What is the force of the 5.4-kg block on the 4.3-kg block? 94. Use Models Suppose that the masses in problem 93 are now 1.00 kg and 4.00 kg. Find the acceleration of the larger mass. physicspp.com/chapter_test Chapter 4 Assessment 115 111-117 CH04-SG CA STP-845813 6/1/04 1:05 AM Page 116 95. Infer The force exerted on a 0.145-kg baseball by a bat changes from 0.0 N to 1.0104 N in 0.0010 s, then drops back to zero in the same amount of time. The baseball was going toward the bat at 25 m/s. a. Draw a graph of force versus time. What is the average force exerted on the ball by the bat? b. What is the acceleration of the ball? c. What is the final velocity of the ball, assuming that it reverses direction? 96. Observe and Infer Three blocks that are connected by massless strings are pulled along a frictionless surface by a horizontal force, as shown in Figure 4-25. a. What is the acceleration of each block? b. What are the tension forces in each of the strings? Hint: Draw a separate free-body diagram for each block. FT1 4.0 kg m2 2.0 kg m1 \u25a0 Figure 4-25 FT2 6.0 kg m3 F 36.0 N 97. Critique Using the Example Problems in this chapter as models, write a solution to the following problem. A block of mass 3.46 kg is suspended from two vertical ropes attached to the ceiling. What is the tension in each rope? 98. Think Critically Because of your physics knowledge, you are serving as a scientific consultant for a new science-fiction TV series about space exploration. In episode 3, the heroine, Misty Moonglow, has been asked to be the first person to ride in a new interplanetary transport for use in our solar system. She wants to be sure that the transport actually takes her to the planet she is supposed to be going to, so she needs to take a testing device along with her to measure the force of gravity when she arrives. The script writers don\u2019t want her to just drop an object, because it will be hard to depict different accelerations of falling objects on TV. They think they\u2019d like something involving a scale. It is your job to design a quick experiment Misty can conduct involving a scale to determine which planet in our solar system she has arrived on. Describe the experiment and include what the results would be for Pluto (g 0.30 m/s2 ), which is where she is supposed to go, and Mercury (g 3.70 m/s2), which is where she actually ends up. 99. Apply Concepts Develop a CBL lab, using a motion detector, that graphs the distance a freefalling object moves over equal intervals of time. Also graph velocity versus time. Compare and contrast your graphs. Using your velocity graph, determine the acceleration. Does it equal g? Writing in Physics 100. Research Newton\u2019s contributions to physics and write a one-page summary. Do you think his three laws of motion were his greatest accomplishments? Explain why or why not. 101. Review, analyze, and critique Newton\ufffd", "\ufffds first law. Can we prove this law? Explain. Be sure to consider the role of resistance. 102. Physicists classify all forces into four fundamental categories: gravitational, electromagnetic, strong nuclear, and weak nuclear. Investigate these four forces and describe the situations in which they are found. Cumulative Review 103. Cross-Country Skiing Your friend is training for a cross-country skiing race, and you and some other friends have agreed to provide him with food and water along his training route. It is a bitterly cold day, so none of you wants to wait outside longer than you have to. Taro, whose house is the stop before yours, calls you at 8:25 A.M. to tell you that the skier just passed his house and is planning to move at an average speed of 8.0 km/h. If it is 5.2 km from Taro\u2019s house to yours, when should you expect the skier to pass your house? (Chapter 2) 104. Figure 4-26 is a position-time graph of the motion of two cars on a road. (Chapter 3) a. At what time(s) does one car pass the other? b. Which car is moving faster at 7.0 s? c. At what time(s) do the cars have the same velocity? d. Over what time interval is car B speeding up all the time? e. Over what time interval is car B slowing down all the time 12 6 0 Position of Two Cars B A 2 4 6 8 Time (s) \u25a0 Figure 4-26 105. Refer to Figure 4-26 to find the instantaneous speed for the following: (Chapter 3) a. car B at 2.0 s b. car B at 9.0 s c. car A at 2.0 s 116 Chapter 4 Forces in One Dimension For more problems, go to Additional Problems, Appendix B. 111-117 CH04-SG CA STP-845813 6/1/04 1:06 AM Page 117 Multiple Choice 1. What is the acceleration of the car described by the graph below? 0.20 m/s2 0.40 m/s2 1.0 m/s2 2.5 m/s2 ) 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 6. A 45-kg child sits on a 3.2-kg tire swing. What is the tension in the rope that hangs from a tree branch? 310 N 4.4102 N 4.5102 N 4.7102 N 7. The tree branch in problem 6 sags and the child\u2019s feet rest on the ground. If the tension in the rope is reduced to 220 N, what is the value of the normal force being exerted on the child\u2019s feet? 2.2102 N 2.5102 N 4.3102 N 6.9102 N 0.0 2.0 4.0 6.0 Time (s) 8. According the graph below, what is the force being exerted on the 16-kg cart? 2. What distance will the car described by the above graph have traveled after 4.0 s? 13 m 40 m 80 m 90 m 3. If the car in the above graph maintains a constant acceleration, what will its velocity be after 10 s? 10 km/h 25 km/h 90 km/h 120 km/h 4. In a tug-of-war, 13 children, with an average mass of 30 kg each, pull westward on a rope with an average force of 150 N per child. Five parents, with an average mass of 60 kg each, pull eastward on the other end of the rope with an average force of 475 N per adult. Assuming that the whole mass accelerates together as a single entity, what is the acceleration of the system? 0.62 m/s2 E 2.8 m/s2 W 3.4 m/s2 E 6.3 m/s2 W 5. What is the weight of a 225-kg space probe on the Moon? The acceleration of gravity on the Moon is 1.62 m/s2. 139 N 364 N 1.35103 N 2.21103 N 16 N 32 N 4 N 8 N 8.0 6.0 4.0 2..0 1.0 2.0 3.0 4.0 Time (s) Extended Answer 9. Draw a free-body diagram of a dog sitting on a scale in an elevator. Using words and mathematical formulas, describe what happens to the apparent weight of the dog when: the elevator accelerates upward, the elevator travels at a constant speed downward, and the elevator falls freely downward. Maximize Your Score If possible, find out how your standardized test will be scored. In order to do your best, you need to know if there is a penalty for guessing, and if so,", " what the penalty is. If there is no random-guessing penalty at all, you should always fill in an answer, even if you have not read the question. physicspp.com/standardized_test Chapter 4 Standardized Test Practice 117 118-125 CH05-S1-845813 3/19/04 1:48 PM Page 118 What You\u2019ll Learn \u2022 You will represent vector quantities both graphically and algebraically. \u2022 You will use Newton\u2019s laws to analyze motion when friction is involved. \u2022 You will use Newton\u2019s laws and your knowledge of vectors to analyze motion in two dimensions. Why It\u2019s Important Most objects experience forces in more than one dimension. A car being towed, for example, experiences upward and forward forces from the tow truck and the downward force of gravity. Rock Climbing How do rock climbers keep from falling? This climber has more than one support point, and there are multiple forces acting on her in multiple directions. Think About This A rock climber approaches a portion of the rock face that forces her to hang with her back to the ground. How will she use her equipment to apply the laws of physics in her favor and overcome this obstacle? physicspp.com 118 CORBIS 118-125 CH05-S1-845813 3/13/04 6:21 AM Page 119 Can 2 N 2 N 2 N? Question Under what conditions can two different forces equal one other force? Procedure Analysis Does the sum of the forces measured by the two spring scales equal the weight of the hanging object? Is the sum greater than the weight? Less than the weight? Critical Thinking Draw an equilateral triangle, with one side vertical, on a sheet of paper. If the two sides of the triangle are 2.0 N, explain the size of the third side. How is it possible that 2 N 2 N 2 N? 1. Measure Use a spring scale to measure and record the weight of a 200-g object. 2. Obtain another spring scale, and attach one end of a 35-cm-long piece of string to the hooks on the bottom of each spring scale. 3. Tie one end of a 15-cm-long piece of string to the 200-g object. Loop the other end over the 35-cm-long piece of string and tie the end to the 200-g object. CAUTION: Avoid falling masses. 4. Hold the spring scales parallel to each other so that the string between them forms a 120\u00b0 angle. Move the string with the hanging object until both scales have the same reading. Record the readings on each scale. 5. Collect and Organize Data Slowly pull the string more and more horizontal while it is still supporting the 200-g object. Describe your observations. 5.1 Vectors How do rock climbers keep from falling in situations like the one shown on the preceding page? Notice that the climber has more than one support point and that there are multiple forces acting on her. She tightly grips crevices in the rock and has her feet planted on the rock face, so there are two contact forces acting on her. Gravity is pulling on her as well, so there are three total forces acting on the climber. One aspect of this situation that is different from the ones that you have studied in earlier chapters is that the forces exerted by the rock face on the climber are not horizontal or vertical forces. You know from previous chapters that you can pick your coordinate system and orient it in the way that is most useful to analyzing the situation. But what happens when the forces are not at right angles to each other? How can you set up a coordinate system and find for a net force when you are dealing with more than one dimension? Objectives \u2022 Evaluate the sum of two or more vectors in two dimensions graphically. \u2022 Determine the components of vectors. \u2022 Solve for the sum of two or more vectors algebraically by adding the components of the vectors. Vocabulary components vector resolution Section 5.1 Vectors 119 Horizons Companies 118-125 CH05-S1-845813 3/13/04 6:23 AM Page 120 40 N 40 N 80 N \u25a0 Figure 5-1 The sum of the two 40-N forces is shown by the resultant vector below them. \u25a0 Figure 5-2 Add vectors by placing them tip-to-tail and drawing the resultant from the tail of the first vector to the tip of the last vector. a b c Vectors Revisited Consider an example with force vectors. Recall the case in Chapter 4 in which you and a friend both pushed on a table together. Suppose that you each exerted 40 N of force to the right. Figure 5-1 represents these vectors in a free-body diagram with the resultant vector, the net force, shown below it. The net force vector is 80 N, which is what you probably expected. But how was this net force vector obtained? Vectors in Multiple Dimensions The process for adding", " vectors works even when the vectors do not point along the same straight line. If you are solving one of these two-dimensional problems graphically, you will need to use a protractor, both to draw the vectors at the correct angles and also to measure the direction and magnitude of the resultant vector. You can add vectors by placing them tip-to-tail and then drawing the resultant of the vector by connecting the tail of the first vector to the tip of the second vector, as shown in Figure 5-2. Figure 5-2a shows the two forces in the free-body diagram. In Figure 5-2b, one of the vectors has been moved so that its tail is at the same place as the tip of the other vector. Notice that its length and direction have not changed. Because the length and direction are the only important characteristics of the vector, the vector is unchanged by this movement. This is always true: if you move a vector so that its length and direction are unchanged, the vector is unchanged. Now, as in Figure 5-2c, you can draw the resultant vector pointing from the tail of the first vector to the tip of the last vector and measure it to obtain its magnitude. Use a protractor to measure the direction of the resultant vector. Sometimes you will need to use trigonometry to determine the length or direction of resultant vectors. Remember that the length of the hypotenuse of a right triangle can be found by using the Pythagorean theorem. If you were adding together two vectors at right angles, vector A pointing north and vector B pointing east, you could use the Pythagorean theorem to find the magnitude of the resultant, R. Pythagorean Theorem R2 A2 B2 If vector A is at a right angle to vector B, then the sum of the squares of the magnitudes is equal to the square of the magnitude of the resultant vector. If the two vectors to be added are at an angle other than 90\u00b0, then you can use the law of cosines or the law of sines. Law of Cosines R2 A2 B2 2AB cos The square of the magnitude of the resultant vector is equal to the sum of the magnitudes of the squares of the two vectors, minus two times the product of the magnitudes of the vectors, multiplied by the cosine of the angle between them. B A R Law of Sines b sin sin sin a The magnitude of the resultant, divided by the sine of the angle between two vectors, is equal to the magnitude of one of the vectors divided by the angle between that component vector and the resultant vector. 120 Chapter 5 Forces in Two Dimensions 118-125 CH05-S1-845813 6/1/04 2:02 AM Page 121 Finding the Magnitude of the Sum of Two Vectors Find the magnitude of the sum of a 15-km displacement and a 25-km displacement when the angle between them is 90\u00b0 and when the angle between them is 135\u00b0. 1 Analyze and Sketch the Problem \u2022 Sketch the two displacement vectors, A and B, and the angle between them. A 25 km B 15 km Known: 1 2 90\u00b0 135\u00b0 Unknown: R? 2 Solve for the Unknown When the angle is 90\u00b0, use the Pythagorean theorem to find the magnitude of the resultant vector. R2 A2 B2 R A2 B2 A B 2 R (25 km)2 (15 km)2 Substitute A 25 km, B 15 km 29 km Math Handbook Square and Cube Roots pages 839\u2013840 When the angle does not equal 90\u00b0, use the law of cosines to find the magnitude of the resultant vector. R2 A2 B2 2AB(cos 2) 2) R A2 B2 2AB(cos (25 km)2 (15 km)2 2(25 km)(15 km)(cos 135\u00b0) Substitute A 25 km, B 15 km, 37 km 2 135\u00b0 3 Evaluate the Answer \u2022 Are the units correct? Each answer is a length measured in kilometers. \u2022 Do the signs make sense? The sums are positive. \u2022 Are the magnitudes realistic? The magnitudes are in the same range as the two combined vectors, but longer. This is because each resultant is the side opposite an obtuse angle. The second answer is larger than the first, which agrees with the graphical representation. 1. A car is driven 125.0 km due west, then 65.0 km due south. What is the magnitude of its displacement? Solve this problem both graphically and mathematically, and check your answers against each other. 2. Two shoppers walk from the door of the mall to their car, which is 250.0 m down a lane of cars, and then turn 90\u00b0 to the right and walk an additional 60.0 m. What is the magnitude of the displacement of the shoppers\u2019 car from the mall door? Solve this problem both graphically and mathematically,", " and check your answers against each other. 3. A hiker walks 4.5 km in one direction, then makes a 45\u00b0 turn to the right and walks another 6.4 km. What is the magnitude of her displacement? 4. An ant is crawling on the sidewalk. At one moment, it is moving south a distance of 5.0 mm. It then turns southwest and crawls 4.0 mm. What is the magnitude of the ant\u2019s displacement? Section 5.1 Vectors 121 118-125 CH05-S1-845813 3/13/04 6:25 AM Page 122 a y Origin x b y A Ax Ay x \u25a0 Figure 5-3 A coordinate system has an origin and two perpendicular axes (a). The direction of a vector, A, is measured counterclockwise from the x-axis (b). \u25a0 Figure 5-4 The sign of a component depends upon which of the four quadrants the component is in. Second quadrant Ax 0 Ay 0 Ax 0 Ay 0 Third quadrant y First quadrant Ax 0 Ay 0 Ax 0 Ay 0 x Fourth quadrant Components of Vectors Choosing a coordinate system, such as the one in Figure 5-3a, is similar to laying a grid drawn on a sheet of transparent plastic on top of a vector problem. You have to choose where to put the center of the grid (the origin) and establish the directions in which the axes point. Notice that in the coordinate system shown in Figure 5-3a, the x-axis is drawn through the origin with an arrow pointing in the positive direction. The positive y-axis is located 90\u00b0 counterclockwise from the positive x-axis and crosses the x-axis at the origin. How do you choose the direction of the x-axis? There is never a single correct answer, but some choices make the problem easier to solve than others. When the motion you are describing is confined to the surface of Earth, it is often convenient to have the x-axis point east and the y-axis point north. When the motion involves an object moving through the air, the positive x-axis is often chosen to be horizontal and the positive y-axis vertical (upward). If the motion is on a hill, it\u2019s convenient to place the positive x-axis in the direction of the motion and the y-axis perpendicular to the x-axis. Component vectors Defining a coordinate system allows you to describe a vector in a different way. Vector A shown in Figure 5-3b, for example, could be described as going 5 units in the positive x-direction and 4 units in the positive y-direction. You can represent this information in the form of two vectors like the ones labeled Ax and Ay in the diagram. Notice that Ax is parallel to the x-axis, and Ay is parallel to the y-axis. Further, you can see that if you add Ax and Ay, the resultant is the original vector, A. A vector can be broken into its components, which are a vector parallel to the x-axis and another parallel to the y-axis. This can always be done and the following vector equation is always true. A Ax Ay This process of breaking a vector into its components is sometimes called vector resolution. Notice that the original vector is the hypotenuse of a right triangle. This means that the magnitude of the original vector will always be larger than the magnitudes of either component vector. Another reason for choosing a coordinate system is that the direction of any vector can be specified relative to those coordinates. The direction of a vector is defined as the angle that the vector makes with the x-axis, measured counterclockwise. In Figure 5-3b, the angle,, tells the direction of the vector, A. All algebraic calculations involve only the positive components of vectors, not the vectors themselves. In addition to measuring the lengths of the component vectors graphically, you can find the components by using trigonometry. The components are calculated using the equations below, where the angle,, is measured counterclockwise from the positive x-axis. A e d si t n e c ja d cos a x; therefore, Ax A p t e s u en o y h A e d si te i s o p p sin o y; therefore, Ay A t e s nu e o p y h When the angle that a vector makes with the x-axis is larger than 90\u00b0, the sign of one or more components is negative, as shown in Figure 5-4. A cos A sin 122 Chapter 5 Forces in Two Dimensions 118-125 CH05-S1-845813 3/13/04 6:25 AM Page 123 Algebraic Addition of Vectors y y B C By Cy You might be wondering why you need to resolve vectors into their components. The answer is that doing this often makes adding vectors together much easier mathematically. Two or more vectors (A, B,", " C, etc.) may be added by first resolving each vector into its x- and y-components. The x-components are added to form the x-component Cx. Similarly, the of the resultant: Rx y-components are added to form the y-component Cy. This process is illustrated graphically in By of the resultant: Ry Figure 5-5. Because Rx and Ry are at a right angle (90\u00b0), the magnitude of the resultant vector can be calculated using the Pythagorean theorem, R2 Rx 2. To find the angle or direction of the resultant, recall that the tangent of the angle that the vector makes with the x-axis is given by the following. 2 Ry Ax Ay Bx Ay Ax Cx Bx A R Ry x Rx x \u25a0 Figure 5-5 Rx is the sum of the x-components of A, B, and C. Ry is the sum of the y-components. The vector sum of Rx and Ry is the vector sum of A, B, and C. Angle of the Resultant Vector tan1 ( R y) R x The angle of the resultant vector is equal to the inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector. You can find the angle by using the tan1 key on your calculator. Note that when tan 0, most calculators give the angle between 0\u00b0 and 90\u00b0, and when tan 0, the angle is reported to be between 0\u00b0 and 90\u00b0. Vector Addition Use the following technique to solve problems for which you need to add or subtract vectors. 1. Choose a coordinate system. 2. Resolve the vectors into their x-components A cos, and their y-components A sin, where is the angle using Ax using Ay measured counterclockwise from the positive x-axis. 3. Add or subtract the component vectors in the x-direction. 4. Add or subtract the component vectors in the y-direction. to find the magnitude of the resultant vector. 5. Use the Pythagorean theorem, R Rx 6. Use tan1( y) to find the angle of the resultant 2 Ry 2, R R x vector. Math Review e id s te i s o p p sin o h e e s nu t o p y R y R e id s t n e c ja d cos a te te i s o p p tan ad Ry x Rx x Ry x Rx Section 5.1 Vectors 123 118-125 CH05-S1-845813 7/20/04 3:00 AM Page 124 Finding Your Way Home A GPS receiver indicates that your home is 15.0 km and 40.0\u00b0 north of west, but the only path through the woods leads directly north. If you follow the path 5.0 km before it opens into a field, how far, and in what direction, would you have to walk to reach your home? 1 Analyze and Sketch the Problem \u2022 Draw the resultant vector, R, from your original location to your home. \u2022 Draw A, the known vector, and draw B, the unknown vector. Known: A 5.0 km, due north R 15.0 km, 40.0\u00b0 north of west 140.0\u00b0 Unknown: B? 2 Solve for the Unknown Find the components of R. Rx Ry R cos (15.0 km) cos 140.0\u00b0 11.5 km R sin (15.0 km) sin 140.0\u00b0 9.64 km Substitute R 15.0 km, 140.0\u00b0 Substitute R 15.0 km, 140.0\u00b0 Because A is due north, Ax Use the components of R and A to find the components of B. 0.0 km and Ay 5.0 km. y A B R 40.0\u00b0 x 180.0\u00b0 40.0\u00b0 140.0\u00b0 Math Handbook Inverses of Sine, Cosine, and Tangent page 856 Bx By Ax Rx 11.5 km 0.0 km 11.5 km Ry Ay 9.64 km 5.0 km 4.6 km Substitute Rx The negative sign means that this component points west. 11.5 km, Ax 0.0 km 9.64 km, Ay Substitute Ry This component points north. 5.0 km Use the components of vector B to find the magnitude of vector B. B Bx 2 2 By (11.5 km)2 (4.6 km)2 12 km Substitute By 4.6 km, Bx 11.5 km Use the tangent to find the direction of vector B. B tan1 y B x m k.6 4 tan1 Substitute By 1.5 m k 1 22\u00b0 or 158\u00b0 4.6 km, Bx 11.5 km Tangent of an angle is negative in quadrants II and IV, Locate the tail of vector B at the origin of a coordinate system", " and draw the components Bx and By. The direction is in the third quadrant, at 158\u00b0, or 22\u00b0 north of west. Thus, B 12 km at 22\u00b0 north of west. so two answers are possible. 3 Evaluate the Answer \u2022 Are the units correct? Kilometers and degrees are correct. \u2022 Do the signs make sense? They agree with the diagram. \u2022 Is the magnitude realistic? The length of B should be longer than Rx because the angle between A and B is greater than 90\u00b0. 124 Chapter 5 Forces in Two Dimensions 118-125 CH05-S1-845813 7/20/04 3:27 AM Page 125 Solve problems 5\u201310 algebraically. You may also choose to solve some of them graphically to check your answers. 5. Sudhir walks 0.40 km in a direction 60.0\u00b0 west of north, then goes 0.50 km due west. What is his displacement? 6. Afua and Chrissy are going to sleep overnight in their tree house and are using some ropes to pull up a box containing their pillows and blankets, which have a total mass of 3.20 kg. The girls stand on different branches, as shown in Figure 5-6, and pull at the angles and with the forces indicated. Find the x- and y-components of the net force on the box. Hint: Draw a free-body diagram so that you do not leave out a force. 7. You first walk 8.0 km north from home, then walk east until your displacement from home is 10.0 km. How far east did you walk? 8. A child\u2019s swing is held up by two ropes tied to a tree branch that hangs 13.0\u00b0 from the vertical. If the tension in each rope is 2.28 N, what is the combined force (magnitude and direction) of the two ropes on the swing? 9. Could a vector ever be shorter than one of its components? Equal in length to one of its components? Explain. 10. In a coordinate system in which the x-axis is east, for what range of angles is the x-component positive? For what range is it negative? You will use these techniques to resolve vectors into their components throughout your study of physics. You will get more practice at it, particularly in the rest of this chapter and the next. Resolving vectors into components allows you to analyze complex systems of vectors without using graphical methods. 20.4 N 17.7 N 55.0\u00b0 120.0\u00b0 \u25a0 Figure 5-6 (Not to scale) 5.1 Section Review 11. Distance v. Displacement Is the distance that you walk equal to the magnitude of your displacement? Give an example that supports your conclusion. 12. Vector Difference Subtract vector K from vector L, shown in Figure 5-7. 5.0 4.0 M 37.0\u00b0 K \u25a0 Figure 5-7 L 6.0 13. Components Find the components of vector M, shown in Figure 5-7. 14. Vector Sum Find the sum of the three vectors shown in Figure 5-7. 15. Commutative Operations The order in which vectors are added does not matter. Mathematicians say that vector addition is commutative. Which ordinary arithmetic operations are commutative? Which are not? 16. Critical Thinking A box is moved through one displacement and then through a second displacement. The magnitudes of the two displacements are unequal. Could the displacements have directions such that the resultant displacement is zero? Suppose the box was moved through three displacements of unequal magnitude. Could the resultant displacement be zero? Support your conclusion with a diagram. physicspp.com/self_check_quiz Section 5.1 Vectors 125 126-130 CH05-S2-845813 6/1/04 6:31 AM Page 126 5.2 Friction Objectives \u2022 Define the friction force. \u2022 Distinguish between static and kinetic friction. Vocabulary kinetic friction static friction coefficient of kinetic friction coefficient of static friction \u25a0 Figure 5-8 There is a limit to the ability of the static friction force to match the applied force. Push your hand across your desktop and feel the force called friction opposing the motion. Push your book across the desk. When you stop pushing, the book will continue moving for a little while, then it will slow down and stop. The frictional force acting on the book gave it an acceleration in the direction opposite to the one in which it was moving. So far, you have neglected friction in solving problems, but friction is all around you. You need it to both start and stop a bicycle and a car. If you have ever walked on ice, you understand the importance of friction. Static and Kinetic Friction There are two types of friction. Both always oppose motion. When you pushed your book across the desk, it experienced a type of friction that acts on moving objects. This force is known", " as kinetic friction, and it is exerted on one surface by another when the two surfaces rub against each other because one or both of them are moving. To understand the other kind of friction, imagine trying to push a heavy couch across the floor. You give it a push, but it does not move. Because it does not move, Newton\u2019s laws tell you that there must be a second horizontal force acting on the couch, one that opposes your force and is equal in size. This force is static friction, which is the force exerted on one surface by another when there is no motion between the two surfaces. You might push harder and harder, as shown in Figures 5-8a and 5-8b, but if the couch still does not move, the force of friction must be getting larger. This is because the static friction force acts in response to other forces. Finally, when you push hard enough, as shown in Figure 5-8c, the couch will begin to move. Evidently, there is a limit to how large the static friction force can be. Once your force is greater than this maximum static friction, the couch begins moving and kinetic friction begins to act on it instead of static friction. A model for friction forces On what does a frictional force depend? The materials that the surfaces are made of play a role. For example, there is more friction between skis and concrete than there is between skis and snow. It may seem reasonable to think that the force of friction also might depend on either the surface area in contact or the speed of the motion, but experiments have shown that this is not true. The normal force between the two objects does matter, however. The harder one object is pushed against the other, the greater the force of friction that results. a b c 126 Chapter 5 Forces in Two Dimensions 126-130 CH05-S2-845813 3/13/04 6:32 AM Page 127 If you pull a block along a surface at a constant velocity, according to Newton\u2019s laws, the frictional force must be equal and opposite to the force with which you pull. You can pull a block of known mass along a table at a constant velocity and use a spring scale, as shown in Figure 5-9, to measure the force that you exert. You can then stack additional blocks on the block to increase the normal force and repeat the measurement. Plotting the data will yield a graph like the one in Figure 5-10. There is a direct proportion between the kinetic friction force and the normal force. The different lines correspond to dragging the block along different surfaces. Note that the line corresponding to the sandpaper surface has a steeper slope than the line for the highly polished table. You would expect it to be much harder to pull the block along sandpaper than along a polished table, so the slope must be related to the magnitude of the resulting frictional force. The slope of this line, designated k, is called the coefficient of kinetic friction between the two surfaces and relates the frictional force to the normal force, as shown below. \u25a0 Figure 5-9 The spring scale pulls the block with a constant force. Kinetic Friction Force Ff, kinetic The kinetic friction force is equal to the product of the coefficient of the kinetic friction and the normal force. kFN The maximum static friction force is related to the normal force in a similar way as the kinetic friction force. Remember that the static friction force acts in response to a force trying to cause a stationary object to start moving. If there is no such force acting on an object, the static friction force is zero. If there is a force trying to cause motion, the static friction force will increase up to a maximum value before it is overcome and motion starts. Static Friction Force Ff, static The static friction force is less than or equal to the product of the coefficient of the static friction and the normal force. sFN In the equation for the maximum static friction force, s is the coefficient of static friction between the two surfaces, and sFN is the maximum static friction force that must be overcome before motion can begin. In Figure 5-8c, the static friction force is balanced the instant before the couch begins to move. Kinetic Frictional Forces v. Normal Force Sandpaper Normal force \u25a0 Figure 5-10 There is a linear relationship between the frictional force and the normal force. Section 5.2 Friction 127 126-130 CH05-S2-845813 6/1/04 4:32 AM Page 128 Balanced Friction Forces You push a 25.0-kg wooden box across a wooden floor at a constant speed of 1.0 m/s. How much force do you exert on the box? 1 Analyze and Sketch the Problem \u2022 Identify the forces and establish a coordinate system. \u2022 Draw a motion diagram indicating constant v and a 0. \u2022 Draw the free-body diagram. Fp Ff Unknown:? Fp Known: m 25.0 kg v 1.", "0 m/s a 0.0 m/s2 k 0.20 (Table 5-1) Begin y v a 0 m/s2 x FN Fg Fp Ff Fnet 0 N 2 Solve for the Unknown The normal force is in the y-direction, and there is no acceleration. FN Fg mg (25.0 kg)(9.80m/s2) 245 N Substitute m 25.0 kg, g 9.80 m/s2 Math Handbook Operations with Significant Digits pages 835\u2013836 The pushing force is in the x-direction; v is constant, thus there is no acceleration. Fp kmg (0.20)(25.0 kg)(9.80 m/s2) 49 N Substitute k 0.20, m 25.0 kg, g 9.80 m/s2 3 Evaluate the Answer \u2022 Are the units correct? Performing dimensional analysis on the units verifies that force is measured in kgm/s2 or N. \u2022 Does the sign make sense? The positive sign agrees with the sketch. \u2022 Is the magnitude realistic? The force is reasonable for moving a 25.0-kg box. 17. A girl exerts a 36-N horizontal force as she pulls a 52-N sled across a cement sidewalk at constant speed. What is the coefficient of kinetic friction between the sidewalk and the metal sled runners? Ignore air resistance. 18. You need to move a 105-kg sofa to a different location in the room. It takes a force of 102 N to start it moving. What is the coefficient of static friction between the sofa and the carpet? 19. Mr. Ames is dragging a box full of books from his office to his car. The box and books together have a combined weight of 134 N. If the coefficient of static friction between the pavement and the box is 0.55, how hard must Mr. Ames push the box in order to start it moving? 20. Suppose that the sled in problem 17 is resting on packed snow. The coefficient of kinetic friction is now only 0.12. If a person weighing 650 N sits on the sled, what force is needed to pull the sled across the snow at constant speed? 21. Suppose that a particular machine in a factory has two steel pieces that must rub against each other at a constant speed. Before either piece of steel has been treated to reduce friction, the force necessary to get them to perform properly is 5.8 N. After the pieces have been treated with oil, what will be the required force? 128 Chapter 5 Forces in Two Dimensions 126-130 CH05-S2-845813 6/10/04 4:41 PM Page 129 Table 5-1 Typical Coefficients of Friction Surface Rubber on dry concrete Rubber on wet concrete Wood on wood Steel on steel (dry) Steel on steel (with oil) s 0.80 0.60 0.50 0.78 0.15 k 0.65 0.40 0.20 0.58 0.06 Note that the equations for the kinetic and maximum static friction forces involve only the magnitudes of the forces. The forces themselves, Ff and FN, are at right angles to each other. Table 5-1 shows coefficients of friction between various surfaces. Although all the listed coefficients are less than 1.0, this does not mean that they must always be less than 1.0. For example, coefficients as large as 5.0 are experienced in drag racing. Unbalanced Friction Forces If the force that you exert on the 25.0-kg box in Example Problem 3 is doubled, what is the resulting acceleration of the box? 1 Analyze and Sketch the Problem \u2022 Draw a motion diagram showing v and a. \u2022 Draw the free-body diagram with a doubled Fp. m 25.0 kg v 1.0 m/s Known: k Fp 0.20 2(49 N) 98 N Unknown: a? Begin v a 2 Solve for the Unknown The normal force is in the y-direction, and there is no acceleration. FN Fg mg Substitute Fg mg In the x-direction there is an acceleration. So the forces must be unequal. Fp Ff Fnet ma Fp Ff Ff Fp a m Substitute Fnet ma FN Fg y Ff Fnet Fp x Math Handbook Isolating a Variable page 845 Find Ff and substitute it into the expression for a. Ff kFN kmg kmg Fp a m 98 N (0.20)(25.0 kg)(9.80 m/s2) 25.0 kg Substitute FN mg Substitute Ff kmg Substitute Fp 98 N, m 25.0 kg, k 0.20, g 9.80 m/s2 2.0 m/s2 3 Evaluate the Answer \u2022 Are the units correct? a is measured in m/s2. \u2022 Does the sign make", " sense? In this coordinate system, the sign should be positive. \u2022 Is the magnitude realistic? If the force were cut in half, a would be zero. Section 5.2 Friction 129 126-130 CH05-S2-845813 6/1/04 4:34 AM Page 130 Causes of Friction All surfaces, even those that appear to be smooth, are rough at a microscopic level. If you look at a photograph of a graphite crystal magnified by a scanning tunneling microscope, the atomic level surface irregularities of the crystal are revealed. When two surfaces touch, the high points on each are in contact and temporarily bond. This is the origin of both static and kinetic friction. The details of this process are still unknown and are the subject of research in both physics and engineering. 22. A 1.4-kg block slides across a rough surface such that it slows down with an acceleration of 1.25 m/s2. What is the coefficient of kinetic friction between the block and the surface? 23. You help your mom move a 41-kg bookcase to a different place in the living room. If you push with a force of 65 N and the bookcase accelerates at 0.12 m/s2, what is the coefficient of kinetic friction between the bookcase and the carpet? 24. A shuffleboard disk is accelerated to a speed of 5.8 m/s and released. If the coefficient of kinetic friction between the disk and the concrete court is 0.31, how far does the disk go before it comes to a stop? The courts are 15.8 m long. 25. Consider the force pushing the box in Example Problem 4. How long would it take for the velocity of the box to double to 2.0 m/s? 26. Ke Min is driving along on a rainy night at 23 m/s when he sees a tree branch lying across the road and slams on the brakes when the branch is 60.0 m in front of him. If the coefficient of kinetic friction between the car\u2019s locked tires and the road is 0.41, will the car stop before hitting the branch? The car has a mass of 2400 kg. Here are a few important things to remember when dealing with frictional situations. First, friction always acts in a direction opposite to the motion (or in the case of static friction, intended motion). Second, the magnitude of the force of friction depends on the magnitude of the normal force between the two rubbing surfaces; it does not necessarily depend on the weight of either object. Finally, multiplying the coefficient of static friction and the normal force gives you the maximum static friction force. Keep these things in mind as you review this section. 5.2 Section Review 27. Friction In this section, you learned about static and kinetic friction. How are these two types of friction similar? What are the differences between static and kinetic friction? 28. Friction At a wedding reception, you notice a small boy who looks like his mass is about 25 kg running part way across the dance floor, then sliding on his knees until he stops. If the kinetic coefficient of friction between the boy\u2019s pants and the floor is 0.15, what is the frictional force acting on him as he slides? 30. Force The coefficient of static friction between a 40.0-kg picnic table and the ground below it is 0.43. What is the greatest horizontal force that could be exerted on the table while it remains stationary? 31. Acceleration Ryan is moving to a new apartment and puts a dresser in the back of his pickup truck. When the truck accelerates forward, what force accelerates the dresser? Under what circumstances could the dresser slide? In which direction? 29. Velocity Derek is playing cards with his friends, and it is his turn to deal. A card has a mass of 2.3 g, and it slides 0.35 m along the table before it stops. If the coefficient of kinetic friction between the card and the table is 0.24, what was the initial speed of the card as it left Derek\u2019s hand? 32. Critical Thinking You push a 13-kg table in the cafeteria with a horizontal force of 20 N, but it does not move. You then push it with a horizontal force of 25 N, and it accelerates at 0.26 m/s2. What, if anything, can you conclude about the coefficients of static and kinetic friction? 130 Chapter 5 Forces in Two Dimensions physicspp.com/self_check_quiz 131-135 CH05-S3-845813 3/13/04 6:42 AM Page 131 5.3 Force and Motion in Two Dimensions You have already worked with several situations dealing with forces in two dimensions. For example, when friction acts between two surfaces, you must take into account both the frictional force that is parallel to the surface and the normal force that is perpendicular to it. So far, you have considered only the motion along a", " level surface. Now you will use your skill in adding vectors to analyze situations in which the forces acting on an object are at angles other than 90\u00b0. Equilibrium Revisited Recall from Chapter 4 that when the net force on an object is zero, the object is in equilibrium. According to Newton\u2019s laws, the object will not accelerate because there is no net force acting on it; an object in equilibrium is motionless or moves with constant velocity. You have already analyzed several equilibrium situations in which two forces acted on an object. It is important to realize that equilibrium can occur no matter how many forces act on an object. As long as the resultant is zero, the net force is zero and the object is in equilibrium. Figure 5-11a shows three forces exerted on a point object. What is the net force acting on the object? Remember that vectors may be moved if you do not change their direction (angle) or length. Figure 5-11b shows the addition of the three forces, A, B, and C. Note that the three vectors form a closed triangle. There is no net force; thus, the sum is zero and the object is in equilibrium. Suppose that two forces are exerted on an object and the sum is not zero. How could you find a third force that, when added to the other two, would add up to zero, and therefore cause the object to be in equilibrium? To find this force, first find the sum of the two forces already being exerted on the object. This single force that produces the same effect as the two individual forces added together is called the resultant force. The force that you need to find is one with the same magnitude as the resultant force, but in the opposite direction. A force that puts an object in equilibrium is called the equilibrant. Figure 5-12 illustrates the procedure for finding this force for two vectors. Note that this general procedure works for any number of vectors. \u25a0 Figure 5-12 The equilibrant is the same magnitude as the resultant, but opposite in direction. A B B R A B A R Equilibrant Objectives \u2022 Determine the force that produces equilibrium when three forces act on an object. \u2022 Analyze the motion of an object on an inclined plane with and without friction. Vocabulary equilibrant A C B a b A B C \u25a0 Figure 5-11 An object is in equilibrium when all the forces on it add up to zero. Section 5.3 Force and Motion in Two Dimensions 131 131-135 CH05-S3-845813 3/13/04 6:42 AM Page 132 Find the equilibrant for the following forces. F1 F2 F3 4 61.0 N at 17.0\u00b0 north of east 38.0 N at 64.0\u00b0 north of east 54.0 N at 8.0\u00b0 west of north 93.0 N at 53.0\u00b0 west of north 65.0 N at 21.0\u00b0 south of west 102.0 N at 15.0\u00b0 west of south 26.0 N south 77.0 N at 22.0\u00b0 east of south 51.0 N at 33.0\u00b0 east of south 82.0 N at 5.0\u00b0 south of east F4 F5 F6 F7 F8 F9 F10 y 2 3 x 10 1 9 8 5 7 6 Motion Along an Inclined Plane You have applied Newton\u2019s laws to a variety of equilibrium situations, but only to motions that were either horizontal or vertical. How would you apply them in a situation like the one in Figure 5-13a, in which a skier glides down a slope? Start by identifying the forces acting on the object, the skier, as shown in Figure 5-13b and sketching a free-body diagram. The gravitational force on the skier is in the downward direction toward the center of Earth. There is a normal force perpendicular to the hill, and the frictional forces opposing the skier\u2019s motion are parallel to the hill. The resulting free-body diagram is shown in Figure 5-13c. You can see that, other than the force of friction, only one force acts horizontally or vertically, and you know from experience that the acceleration of the skier will be along the slope. How do you find the net force that causes the skier to accelerate? \u25a0 Figure 5-13 A skier slides down a slope (a). Identify the forces that are acting upon the skier (b) and draw a free-body diagram describing those forces (c). It is important to draw the direction of the normal and the friction forces correctly in order to properly analyze these types of situations. a y System Contact with outside world x b v Begin Fnet a End c y FN Ff x Fg 132 Chapter 5 Forces in Two Dimensions 131-135 CH05-S3-845813 3/13/04 6:43 AM Page 133 Components of Weight for", " an Object on an Incline A crate weighing 562 N is resting on a plane inclined 30.0\u00b0 above the horizontal. Find the components of the weight forces that are parallel and perpendicular to the plane. 1 Analyze and Sketch the Problem \u2022 Include a coordinate system with the positive x-axis pointing uphill. \u2022 Draw the free-body diagram showing Fg, the components Fgx and Fgy, and the angle. Unknown: Known:? 562 N Fgx Fg? 30.0\u00b0 Fgy 2 Solve for the Unknown Fgx and Fgy are negative because they point in directions opposite to the positive axes. y x Fgx Fgy y x Fg Fgx = Fg(sin ) Fgy = (562 N)(sin 30.0\u00b0) = 281 N Fg(cos ) (562 N)(cos 30.0\u00b0) 487 N Substitute Fg 562, 30.0\u00b0 Substitute Fg 562, 30.0\u00b0 Math Handbook Trigonometric Ratios page 855 3 Evaluate the Answer \u2022 Are the units correct? Force is measured in newtons. \u2022 Do the signs make sense? The components point in directions opposite to the positive axes. \u2022 Are the magnitudes realistic? The values are less than Fg. 33. An ant climbs at a steady speed up the side of its anthill, which is inclined 30.0\u00b0 from the vertical. Sketch a free-body diagram for the ant. 34. Scott and Becca are moving a folding table out of the sunlight. A cup of lemonade, with a mass of 0.44 kg, is on the table. Scott lifts his end of the table before Becca does, and as a result, the table makes an angle of 15.0\u00b0 with the horizontal. Find the components of the cup\u2019s weight that are parallel and perpendicular to the plane of the table. 35. Kohana, who has a mass of 50.0 kg, is at the dentist\u2019s office having her teeth cleaned, as shown in Figure 5-14. If the component of her weight perpendicular to the plane of the seat of the chair is 449 N, at what angle is the chair tilted? 36. Fernando, who has a mass of 43.0 kg, slides down the banister at his grandparents\u2019 house. If the banister makes an angle of 35.0\u00b0 with the horizontal, what is the normal force between Fernando and the banister? 37. A suitcase is on an inclined plane. At what angle, relative to the vertical, will the component of the suitcase\u2019s weight parallel to the plane be equal to half the perpendicular component of its weight? \u25a0 Figure 5-14 Section 5.3 Force and Motion in Two Dimensions 133 131-135 CH05-S3-845813 6/1/04 4:47 AM Page 134 Skiing Downhill A 62-kg person on skis is going down a hill sloped at 37\u00b0. The coefficient of kinetic friction between the skis and the snow is 0.15. How fast is the skier going 5.0 s after starting from rest? 1 Analyze and Sketch the Problem \u2022 Establish a coordinate system. \u2022 Draw a free-body diagram showing the skier\u2019s velocity and direction of acceleration. \u2022 Draw a motion diagram showing increasing v, and both a and Fnet in the x direction, like the one shown in Figure 5-13. y x Fnet v a y FN Ff x Fg Unknown: a?? vf Known: m 62 kg 37\u00b0 0.15 k 0.0 m/s vi t 5.0 s 2 3 Solve for the Unknown y-direction: Fnet, y may 0.0 N There is no acceleration in the y-direction, so ay 0.0 m/s2. Solve for FN. Fgy FN Fnet, y FN Fgy mg(cos ) x-direction: Solve for a. Fnet, x Fgx Ff Fgy is negative. It is in the negative direction as defined by the coordinate system. Substitute Fnet, y Substitute Fgy mg cos 0.0 N and rearrange Math Handbook Isolating a Variable page 845 Ff is negative because it is in the negative direction as defined by the coordinate system. max mg(sin ) mg(sin ) kFN kmg(cos ) ma, Fgx Substitute Fnet, x Substitute a ax because all the acceleration is in the x-direction; substitute FN mg cos mg sin, Ff kFN a g(sin kcos ) (9.80 m/s2)(sin 37\u00b0 (0.15)cos 37\u00b0) 4.7 m/s2 Substitute g 9.80 m/s2, 37\u00b0, k 0.15 Because vi, a, and t are all known,", " use the following. vf at vi 0.0 (4.7 m/s2)(5.0 s) 24 m/s Substitute vi 0.0 m/s, a 4.7 m/s2, t 5.0 s Evaluate the Answer \u2022 Are the units correct? Performing dimensional analysis on the units verifies that vf is \u2022 Do the signs make sense? Because vf and a are both in the x direction, the signs in m/s and a is in m/s2. do make sense. \u2022 Are the magnitudes realistic? The velocity is fast, over 80 km/h (50 mph), but 37\u00b0 is a steep incline, and the friction between the skis and the snow is not large. 134 Chapter 5 Forces in Two Dimensions 131-135 CH05-S3-845813 7/20/04 3:31 AM Page 135 38. Consider the crate on the incline in Example Problem 5. Calculate the magnitude of the acceleration. After 4.00 s, how fast will the crate be moving? 39. If the skier in Example Problem 6 were on a 31\u00b0 downhill slope, what would be the magnitude of the acceleration? 40. Stacie, who has a mass of 45 kg, starts down a slide that is inclined at an angle of 45\u00b0 with the horizontal. If the coefficient of kinetic friction between Stacie\u2019s shorts and the slide is 0.25, what is her acceleration? 41. After the skier on the 37\u00b0 hill in Example Problem 6 had been moving for 5.0 s, the friction of the snow suddenly increased and made the net force on the skier zero. What is the new coefficient of friction? The most important decision in problems involving motion along a slope is what coordinate system to use. Because an object\u2019s acceleration is usually parallel to the slope, one axis, usually the x-axis, should be in that direction. The y-axis is perpendicular to the x-axis and perpendicular to the surface of the slope. With this coordinate system, you now have two forces, the normal and frictional forces, in the directions of the coordinate axes; however, the weight is not. This means that when an object is placed on an inclined plane, the magnitude of the normal force between the object and the plane will usually not be equal to the object\u2019s weight. You will need to apply Newton\u2019s laws once in the x-direction and once in the y-direction. Because the weight does not point in either of these directions, you will need to break this vector into its x- and y-components before you can sum your forces in these two directions. Example Problem 5 and Example Problem 6 both showed this procedure. What\u2019s Your Angle? Prop a board up so that it forms an inclined plane at a 45\u00b0 angle. Hang a 500-g object from the spring scale. 1. Measure and record the weight of the object. Set the object on the bottom of the board and slowly pull it up the inclined plane at a constant speed. 2. Observe and record the reading on the spring scale. Analyze and Conclude 3. Calculate the component of weight for the 500-g object that is parallel to the inclined plane. 4. Compare the spring-scale reading along the inclined plane with the component of weight parallel to the inclined plane. 5.3 Section Review 42. Forces One way to get a car unstuck is to tie one end of a strong rope to the car and the other end to a tree, then push the rope at its midpoint at right angles to the rope. Draw a free-body diagram and explain why even a small force on the rope can exert a large force on the car. 43. Mass A large scoreboard is suspended from the ceiling of a sports arena by 10 strong cables. Six of the cables make an angle of 8.0\u00b0 with the vertical while the other four make an angle of 10.0\u00b0. If the tension in each cable is 1300.0 N, what is the scoreboard\u2019s mass? 44. Acceleration A 63-kg water skier is pulled up a 14.0\u00b0 incline by a rope parallel to the incline with a tension of 512 N. The coefficient of kinetic friction is 0.27. What are the magnitude and direction of the skier\u2019s acceleration? 45. Equilibrium You are hanging a painting using two lengths of wire. The wires will break if the force is too great. Should you hang the painting as shown in Figures 5-15a or 5-15b? Explain. a b \u25a0 Figure 5-15 46. Critical Thinking Can the coefficient of friction ever have a value such that a skier would be able to slide uphill at a constant velocity? Explain why or why not. Assume there are no other forces acting on the skier. physicspp.com/self_check_quiz Section", " 5.3 Force and Motion in Two Dimensions 135 136-137 CH05-LAB-845813 6/1/04 4:57 AM Page 136 Alternate CBL instructions can be found on the Web site. physicspp.com kFN, where The Coefficient of Friction Static and kinetic friction are forces that are a result of two surfaces in contact with each other. Static friction is the force that must be overcome to cause an object to begin moving, while kinetic friction occurs between two objects in motion relative to each other. The kinetic friction force, Ff, kinetic, is defined by Ff, kinetic k is the coefficient of kinetic friction and FN is the normal force acting on the object. The maximum static frictional force, Ff, max static, s is the coefficient of static friction and FN is is defined by Ff, static the normal force on the object. The maximum static frictional force that must be overcome before movement is able to begin is sFN. If you apply a constant force to pull an object along a horizontal surface at a constant speed, then the frictional force opposing the motion is equal and opposite to the applied force, Fp. Therefore, Fp when the object is on a horizontal surface and the applied force is horizontal. Ff. The normal force is equal and opposite to the object\u2019s weight sFN where QUESTION How can the coefficient of static and kinetic friction be determined for an object on a horizontal surface? Objectives \u25a0 Measure the normal and frictional forces acting on an object starting in motion and already in motion. \u25a0 Use numbers to calculate s and k. \u25a0 Compare and contrast values of s and k. \u25a0 Analyze the kinetic friction results. \u25a0 Estimate the angle where sliding will begin for an object on an inclined plane. Safety Precautions Materials pulley C-clamp masking tape wood surface Procedure string (1 m) spring scale, 0-5 N wood block 1. Check your spring scale to make sure that it reads zero when it is held vertically. If necessary, follow your teacher\u2019s instructions to zero it. 2. Attach the pulley to the edge of the table with a C-clamp. 3. Attach the string to the spring scale hook and the wood block. 4. Measure the weight of the block of wood, or other small object, and record the value as the normal force, FN, in Data Tables 1, 2, and 3. 5. Unhook the string from the spring scale and run it through the pulley. Then reattach it to the spring scale. 6. Move the wood block as far away from the pulley as the string permits, while having it remain on the wood surface. 7. With the spring scale oriented vertically so that a right angle is formed between the wood block, the pulley, and the spring scale, slowly pull up on the spring scale. Observe the force that is necessary to cause the wood block to begin sliding. Record this value for the static frictional force in Data Table 1. 136 Horizons Companies 136-137 CH05-LAB-845813 3/13/04 6:50 AM Page 137 Material Table Object material Surface material Data Table 1 Data Table 3 FN (N) Static Friction Force, Fs (N) FN(N) Fs (N) Ff (N) s k Trial 1 Trial 2 Trial 3 Average Data Table 2 Data Table 4 (Angle,, when sliding begins on an incline) FN (N) Kinetic Friction Force, Ff (N) * tan * Trial 1 Trial 2 Trial 3 Average 8. Repeat steps 6 and 7 for two additional trials. 9. Repeat steps 6 and 7. However, once the block begins sliding, pull just hard enough to keep it moving at a constant speed across the other horizontal surface. Record this force as the kinetic frictional force in Data Table 2. 10. Repeat step 9 for two additional trials. 11. Place the block on the end of the surface. Slowly raise one end of the surface to make an incline. Gently tap the block to cause it to move and overcome static friction. If the block stops, replace it at the top of the incline and repeat the procedure. Continue increasing the angle,, between the horizontal and the inclined surface, and tapping the block until it slides at a constant speed down the incline. Record the angle,, in Data Table 4. Analyze 1. Average the data for the static frictional force, Fs, max, from the three trials and record the result in the last column of Data Table 1 and in Data Table 3. 2. Average the data for the kinetic frictional force, Ff, from the three trials and record the result in the last column of Data Table 2 and in Data Table 3. Conclude and Apply 1. Compare and Contrast Examine your values k. Explain whether your results are for s and reasonable or not. 2. Use Models Draw a free-", "body diagram showing the forces acting on the block if it is placed on an incline of angle. Make certain that you include the force due to friction in your diagram. 3. From your diagram, assuming that the angle,, is where sliding begins, what does tan represent? 4. Compare your value for tan (experimental), s, and k. Going Further Repeat the experiment with additional surfaces that have different characteristics. Real-World Physics If you were downhill skiing and wished to determine the coefficient of kinetic friction between your skis and the slope, how could you do this? Be specific about how you could find a solution to this problem. 3. Use the data in Data Table 3 to calculate the coefficient of static friction, the value in Data Table 3. s, and record 4. Use the data in Data Table 3 to calculate the coefficient of kinetic friction, value in Data Table 3. k, and record the 5. Calculate tan for your value in Data Table 4. To find out more about friction, visit the Web site: physicspp.com 137 138 CH05-FEATURE-845813 3/19/04 1:55 PM Page 138 Roller Coasters Roller Coasters Why are roller coasters fun? A rollercoaster ride would be no fun at all if not for the forces acting on the coaster car and the rider. What forces do riders experience as they ride a roller coaster? The force of gravity acts on the rider and the coaster car in the downward direction. The seat of the car exerts a force on the rider in the opposite direction. When the coaster car makes a turn, the rider experiences a force in the opposite direction. Also, there are forces present due to the friction between the rider and the seat, the side of the car, and the safety bar. The Force Factor Designers of roller coasters take into account the magnitude of the forces exerted on the rider. They design the coaster in such a way that the forces thrill the rider without causing injury or too much discomfort. Designers measure the amount of force exerted on the rider by calculating the force factor. The force factor is equal to the force exerted by the seat on the rider divided by the weight of the rider. Suppose the rider weighs about 68 kg. When the roller coaster is at the bottom of a hill, the rider may experience a force factor of 2. That means that at the bottom of the hill, the rider will feel as though he or she weighs twice as much, or in this case 136 kg. Conversely, at the top of a hill the force factor may be 0.5 and the rider will feel as though he or she weighs half his or her normal weight. Thus, designers create excitement by designing portions that change the rider\u2019s apparent weight. The Thrill Factors Roller-coaster designers manipulate the way in which the body perceives the external world to create that \u201cthrilling\u201d sensation. For example, the roller coaster moves up the first hill very slowly, tricking the rider into thinking that the hill is higher than it is. The organs of the inner ear sense the position of the head both when it is still and when it is 138 courtesy of Six Flags Amusement Park Technology and Society moving. These organs help maintain balance by providing information to the brain. The brain then sends nerve impulses to the skeletal muscles to contract or relax to maintain balance. The constant change in position during a roller-coaster ride causes the organs of the inner ear to send conflicting messages to the brain. As a result, the skeletal muscles contract and relax throughout the ride. You know that you are moving at high speeds because your eyes see the surroundings move past at high speed. So, designers make use of the surrounding landscape along with twists, turns, tunnels, and loops to give the rider plenty of visual cues. These visual cues, along with the messages from the inner ear, can result in disorientation and in some cases, nausea. To enthusiasts the disorientation is part of the thrill. In order to attract visitors, amusement parks are constantly working on designing new rides that take the rider to new thrill levels. As roller-coaster technology improves, your most thrilling roller-coaster ride may be over the next hill. Going Further 1. Compare and Contrast Compare and contrast your experience as a rider in the front of a roller coaster versus the back of it. Explain your answer in terms of the forces acting on you. 2. Critical Thinking While older roller coasters rely on chain systems to pull the coaster up the first hill, newer ones depend on hydraulic systems to do the same job. Research each of these two systems. What do you think are the advantages and disadvantages of using each system? The thrill of a roller-coaster ride is produced by the forces acting on the rider and the rider\u2019s reaction to visual cues. 139-145 CH05-SG CA STP-845813 3/13/04 7:00 AM Page 139 5.1 Vectors Vocabulary", " \u2022 components (p. 122) \u2022 vector resolution (p. 122) Key Concepts \u2022 When two vectors are at right angles, you can use the Pythagorean theorem to determine the magnitude of the resultant vector. R2 A2 B2 \u2022 The law of cosines and law of sines can be used to find the magnitude of the resultant of any two vectors. R2 A2 B2 2AB cos B A R b sin sin sin a \u2022 The components of a vector are projections of the component vectors. A e d si t n e c ja d cos a x; therefore, Ax A o t e s u en p y h A e d si te i s o p p sin o y; therefore, Ay A t e s nu e o p y h ) tan1( R y R x A cos A sin \u2022 Vectors can be summed by separately adding the x- and y-components. Key Concepts \u2022 A frictional force acts when two surfaces touch. \u2022 The frictional force is proportional to the force pushing the surfaces together. \u2022 The kinetic friction force is equal to the coefficient of kinetic friction times the normal force. Ff, kinetic kFN \u2022 The static friction force is less than or equal to the coefficient of static friction times the normal force. Ff, static kFN 5.2 Friction Vocabulary \u2022 kinetic friction (p. 126) \u2022 static friction (p. 126) \u2022 coefficient of kinetic friction (p. 127) \u2022 coefficient of static friction (p. 127) 5.3 Force and Motion in Two Dimensions Vocabulary \u2022 equilibrant (p. 131) Key Concepts \u2022 The force that must be exerted on an object to cause it to be in equilibrium is called the equilibrant. \u2022 The equilibrant is found by finding the net force on an object, then applying a force with the same magnitude but opposite direction. \u2022 An object on an inclined plane has a component of the force of gravity in a direction parallel to the plane; the component can accelerate the object down the plane. physicspp.com/vocabulary_puzzlemaker 139 139-145 CH05-SG CA STP-845813 6/7/04 6:01 AM Page 140 Concept Mapping 47. Complete the concept map below with the terms sine, cosine, or tangent to indicate whether each function is positive or negative in each quadrant. Some circles could remain blank, and others can have more than one term. Quadrant I II III IV Mastering Concepts 48. How would you add two vectors graphically? (5.1) 49. Which of the following actions is permissible when you graphically add one vector to another: moving the vector, rotating the vector, or changing the vector\u2019s length? (5.1) 50. In your own words, write a clear definition of the resultant of two or more vectors. Do not explain how to find it; explain what it represents. (5.1) 51. How is the resultant displacement affected when two displacement vectors are added in a different order? (5.1) 52. Explain the method that you would use to subtract two vectors graphically. (5.1) 53. Explain the difference between A and A. (5.1) 54. The Pythagorean theorem usually is written c2 a2 b2. If this relationship is used in vector addition, what do a, b, and c represent? (5.1) 55. When using a coordinate system, how is the angle or direction of a vector determined with respect to the axes of the coordinate system? (5.1) 56. What is the meaning of a coefficient of friction that is greater than 1.0? How would you measure it? (5.2) 57. Cars Using the model of friction described in this textbook, would the friction between a tire and the road be increased by a wide rather than a narrow tire? Explain. (5.2) 58. Describe a coordinate system that would be suitable for dealing with a problem in which a ball is thrown up into the air. (5.3) 59. If a coordinate system is set up such that the positive x-axis points in a direction 30\u00b0 above the horizontal, what should be the angle between the x-axis and the y-axis? What should be the direction of the positive y-axis? (5.3) 60. Explain how you would set up a coordinate system for motion on a hill. (5.3) 61. If your textbook is in equilibrium, what can you say about the forces acting on it? (5.3) 62. Can an object that is in equilibrium be moving? Explain. (5.3) 63. What is the sum of three vectors that, when placed tip to tail, form a triangle? If these vectors represent forces on an object, what does this imply about the object? (5.3) 64. You are asked", " to analyze the motion of a book placed on a sloping table. (5.3) a. Describe the best coordinate system for analyzing the motion. b. How are the components of the weight of the book related to the angle of the table? 65. For a book on a sloping table, describe what happens to the component of the weight force parallel to the table and the force of friction on the book as you increase the angle that the table makes with the horizontal. (5.3) a. Which components of force(s) increase when the angle increases? b. Which components of force(s) decrease? Applying Concepts 66. A vector that is 1 cm long represents a displacement of 5 km. How many kilometers are represented by a 3-cm vector drawn to the same scale? 67. Mowing the Lawn If you are pushing a lawn mower across the grass, as shown in Figure 5-16, can you increase the horizontal component of the force that you exert on the mower without increasing the magnitude of the force? Explain. Fy F Fx \u25a0 Figure 5-16 140 Chapter 5 Forces in Two Dimensions For more problems, go to Additional Problems, Appendix B. 139-145 CH05-SG CA STP-845813 6/1/04 5:08 AM Page 141 68. A vector drawn 15 mm long represents a velocity of 30 m/s. How long should you draw a vector to represent a velocity of 20 m/s? 69. What is the largest possible displacement resulting from two displacements with magnitudes 3 m and 4 m? What is the smallest possible resultant? Draw sketches to demonstrate your answers. 70. How does the resultant displacement change as the angle between two vectors increases from 0\u00b0 to 180\u00b0? 71. A and B are two sides of a right triangle, where tan A/B. a. Which side of the triangle is longer if tan is greater than 1.0? b. Which side is longer if tan is less than 1.0? c. What does it mean if tan is equal to 1.0? 72. Traveling by Car A car has a velocity of 50 km/h in a direction 60\u00b0 north of east. A coordinate system with the positive x-axis pointing east and a positive y-axis pointing north is chosen. Which component of the velocity vector is larger, x or y? 73. Under what conditions can the Pythagorean theorem, rather than the law of cosines, be used to find the magnitude of a resultant vector? 74. A problem involves a car moving up a hill, so a coordinate system is chosen with the positive x-axis parallel to the surface of the hill. The problem also involves a stone that is dropped onto the car. Sketch the problem and show the components of the velocity vector of the stone. 75. Pulling a Cart According to legend, a horse learned Newton\u2019s laws. When the horse was told to pull a cart, it refused, saying that if it pulled the cart forward, according to Newton\u2019s third law, there would be an equal force backwards; thus, there would be balanced forces, and, according to Newton\u2019s second law, the cart would not accelerate. How would you reason with this horse? 76. Tennis When stretching a tennis net between two posts, it is relatively easy to pull one end of the net hard enough to remove most of the slack, but you need a winch to take the last bit of slack out of the net to make the top almost completely horizontal. Why is this true? 77. The weight of a book on an inclined plane can be resolved into two vector components, one along the plane, and the other perpendicular to it. a. At what angle are the components equal? b. At what angle is the parallel component equal to zero? 78. TV Towers The transmitting tower of a TV station is held upright by guy wires that extend from the top of the tower to the ground. The force along the guy wires can be resolved into two perpendicular components. Which one is larger? Mastering Problems 5.1 Vectors 79. Cars A car moves 65 km due east, then 45 km due west. What is its total displacement? 80. Find the horizontal and vertical components of the following vectors, as shown in Figure 5-17. a. E b. F c. A B(3.0) A(3.0) E(5.0) F(5.0) C(6.0) D(4.0) \u25a0 Figure 5-17 81. Graphically find the sum of the following pairs of vectors, whose lengths and directions are shown in Figure 5-17. a. D and A b. C and D c. C and A d. E and F 82. Graphically add the following sets of vectors, as shown in Figure 5-17. a. A, C, and D b. A, B, and E c. B", ", D, and F 83. You walk 30 m south and 30 m east. Find the magnitude and direction of the resultant displacement both graphically and algebraically. 84. Hiking A hiker\u2019s trip consists of three segments. Path A is 8.0 km long heading 60.0\u00b0 north of east. Path B is 7.0 km long in a direction due east. Path C is 4.0 km long heading 315\u00b0 counterclockwise from east. a. Graphically add the hiker\u2019s displacements in the order A, B, C. b. Graphically add the hiker\u2019s displacements in the order C, B, A. c. At what angle is the parallel component equal to c. What can you conclude about the resulting the weight? displacements? physicspp.com/chapter_test Chapter 5 Assessment 141 139-145 CH05-SG CA STP-845813 6/1/04 5:09 AM Page 142 85. What is the net force acting on the ring in Figure 5-18? y 500.0 N 400.0 N 40.0\u00b0 50.0\u00b0 x \u25a0 Figure 5-18 86. What is the net force acting on the ring in Figure 5-19? y 128 N 128 N 30.0\u00b0 64 N x \u25a0 Figure 5-19 87. A Ship at Sea A ship at sea is due into a port 500.0 km due south in two days. However, a severe storm comes in and blows it 100.0 km due east from its original position. How far is the ship from its destination? In what direction must it travel to reach its destination? 88. Space Exploration A descent vehicle landing on Mars has a vertical velocity toward the surface of Mars of 5.5 m/s. At the same time, it has a horizontal velocity of 3.5 m/s. a. At what speed does the vehicle move along its descent path? b. At what angle with the vertical is this path? 89. Navigation Alfredo leaves camp and, using a compass, walks 4 km E, then 6 km S, 3 km E, 5 km N, 10 km W, 8 km N, and, finally, 3 km S. At the end of three days, he is lost. By drawing a diagram, compute how far Alfredo is from camp and which direction he should take to get back to camp. 5.2 Friction 90. If you use a horizontal force of 30.0 N to slide a 12.0-kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor? 91. A 225-kg crate is pushed horizontally with a force of 710 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate. 92. A force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s2 along a horizontal surface. a. How large is the frictional force? b. What is the coefficient of friction? 93. Moving Appliances Your family just had a new refrigerator delivered. The delivery man has left and you realize that the refrigerator is not quite in the right position, so you plan to move it several centimeters. If the refrigerator has a mass of 180 kg, the coefficient of kinetic friction between the bottom of the refrigerator and the floor is 0.13, and the static coefficient of friction between these same surfaces is 0.21, how hard do you have to push horizontally to get the refrigerator to start moving? 94. Stopping at a Red Light You are driving a 2500.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an intersection, the traffic light turns red. You slam on the brakes. The car\u2019s wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 25.0 m. What is the coefficient of kinetic friction between your tires and the wet road? 5.3 Force and Motion in Two Dimensions 95. An object in equilibrium has three forces exerted on it. A 33.0-N force acts at 90.0\u00b0 from the x-axis and a 44.0-N force acts at 60.0\u00b0 from the x-axis. What are the magnitude and direction of the third force? 96. Five forces act on an object: (1) 60.0 N at 90.0\u00b0, (2) 40.0 N at 0.0\u00b0, (3) 80.0 N at 270.0\u00b0, (4) 40.0 N at 180.0\u00b0, and (5) 50.0 N at 60.0\u00b0. What are the magnitude and direction of a sixth force that would produce equilibrium? 97. Advertising Joe wishes to hang a sign weighing 7.50102 N so that cable A, attached to the", " store, makes a 30.0\u00b0 angle, as shown in Figure 5-20. Cable B is horizontal and attached to an adjoining building. What is the tension in cable B? 30.0\u00b0 A B \u25a0 Figure 5-20 142 Chapter 5 Forces in Two Dimensions For more problems, go to Additional Problems, Appendix B. 139-145 CH05-SG CA STP-845813 6/1/04 5:09 AM Page 143 98. A street lamp weighs 150 N. It is supported by two wires that form an angle of 120.0\u00b0 with each other. The tensions in the wires are equal. a. What is the tension in each wire supporting the street lamp? b. If the angle between the wires supporting the street lamp is reduced to 90.0\u00b0, what is the tension in each wire? 99. A 215-N box is placed on an inclined plane that makes a 35.0\u00b0 angle with the horizontal. Find the component of the weight force parallel to the plane\u2019s surface. 100. Emergency Room You are shadowing a nurse in the emergency room of a local hospital. An orderly wheels in a patient who has been in a very serious accident and has had severe bleeding. The nurse quickly explains to you that in a case like this, the patient\u2019s bed will be tilted with the head downward to make sure the brain gets enough blood. She tells you that, for most patients, the largest angle that the bed can be tilted without the patient beginning to slide off is 32.0\u00b0 from the horizontal. a. On what factor or factors does this angle of tilting depend? b. Find the coefficient of static friction between a typical patient and the bed\u2019s sheets. 101. Two blocks are connected by a string over a frictionless, massless pulley such that one is resting on an inclined plane and the other is hanging over the top edge of the plane, as shown in Figure 5-21. The hanging block has a mass of 16.0 kg, and the one on the plane has a mass of 8.0 kg. The coefficient of kinetic friction between the block and the inclined plane is 0.23. The blocks are released from rest. a. What is the acceleration of the blocks? b. What is the tension in the string connecting the blocks? 37.0\u00b0 \u25a0 Figure 5-21 102. In Figure 5-22, a block of mass M is pushed with a force, F, such that the smaller block of mass m does not slide down the front of it. There is no friction between the larger block and the surface below it, but the coefficient of static friction between the two blocks is for F in terms of M, m, s. Find an expression s, and g. m F M \u25a0 Figure 5-22 Mixed Review 103. The scale in Figure 5-23 is being pulled on by three ropes. What net force does the scale read? 27.0\u00b0 27.0\u00b0 75.0 N 75.0 N 150.0 N \u25a0 Figure 5-23 104. Sledding A sled with a mass of 50.0 kg is pulled along flat, snow-covered ground. The static friction coefficient is 0.30, and the kinetic friction coefficient is 0.10. a. What does the sled weigh? b. What force will be needed to start the sled moving? c. What force is needed to keep the sled moving at a constant velocity? d. Once moving, what total force must be applied to the sled to accelerate it at 3.0 m/s2? physicspp.com/chapter_test Chapter 5 Assessment 143 139-145 CH05-SG CA STP-845813 6/1/04 5:09 AM Page 144 105. Mythology Sisyphus was a character in Greek mythology who was doomed in Hades to push a boulder to the top of a steep mountain. When he reached the top, the boulder would slide back down the mountain and he would have to start all over again. Assume that Sisyphus slides the boulder up the mountain without being able to roll it, even though in most versions of the myth, he rolled it. a. If the coefficient of kinetic friction between the boulder and the mountainside is 0.40, the mass of the boulder is 20.0 kg, and the slope of the mountain is a constant 30.0\u00b0, what is the force that Sisyphus must exert on the boulder to move it up the mountain at a constant velocity? b. If Sisyphus pushes the boulder at a velocity of 0.25 m/s and it takes him 8.0 h to reach the top of the mountain, what is the mythical mountain\u2019s vertical height? 106. Landscaping A tree is being transported on a flatbed trailer by a landscaper, as shown in Figure 5-24. If the base of the tree slides on the trailer, the tree will fall over and be damaged", ". If the coefficient of static friction between the tree and the trailer is 0.50, what is the minimum stopping distance of the truck, traveling at 55 km/h, if it is to accelerate uniformly and not have the tree slide forward and fall on the trailer? \u25a0 Figure 5-24 Thinking Critically 107. Use Models Using the Example Problems in this chapter as models, write an example problem to solve the following problem. Include the following sections: Analyze and Sketch the Problem, Solve for the Unknown (with a complete strategy), and Evaluate the Answer. A driver of a 975-kg car traveling 25 m/s puts on the brakes. What is the shortest distance it will take for the car to stop? Assume that the road is concrete, the force of friction of the road on the tires is constant, and the tires do not slip. 108. Analyze and Conclude Margaret Mary, Doug, and Kako are at a local amusement park and see an attraction called the Giant Slide, which is simply a very long and high inclined plane. Visitors at the amusement park climb a long flight of steps to the top of the 27\u00b0 inclined plane and are given canvas sacks. They sit on the sacks and slide down the 70m-long plane. At the time when the three friends walk past the slide, a 135-kg man and a 20-kg boy are each at the top preparing to slide down. \u201cI wonder how much less time it will take the man to slide down than it will take the boy,\u201d says Margaret Mary. \u201cI think the boy will take less time,\u201d says Doug. \u201cYou\u2019re both wrong,\u201d says Kako. \u201cThey will reach the bottom at the same time.\u201d a. Perform the appropriate analysis to determine who is correct. b. If the man and the boy do not take the same amount of time to reach the bottom of the slide, calculate how many seconds of difference there will be between the two times. Writing in Physics 109. Investigate some of the techniques used in industry to reduce the friction between various parts of machines. Describe two or three of these techniques and explain the physics of how they work. 110. Olympics In recent years, many Olympic athletes, such as sprinters, swimmers, skiers, and speed skaters, have used modified equipment to reduce the effects of friction and air or water drag. Research a piece of equipment used by one of these types of athletes and the way it has changed over the years. Explain how physics has impacted these changes. Cumulative Review 111. Add or subtract as indicated and state the answer with the correct number of significant digits. (Chapter 1) a. 85.26 g 4.7 g b. 1.07 km 0.608 km c. 186.4 kg 57.83 kg d. 60.08 s 12.2 s 112. You ride your bike for 1.5 h at an average velocity of 10 km/h, then for 30 min at 15 km/h. What is your average velocity? (Chapter 3) 113. A 45-N force is exerted in the upward direction on a 2.0-kg briefcase. What is the acceleration of the briefcase? (Chapter 4) 144 Chapter 5 Forces in Two Dimensions For more problems, go to Additional Problems, Appendix B. 139-145 CH05-SG CA STP-845813 6/1/04 5:10 AM Page 145 Multiple Choice 1. Two tractors pull against a 1.00103-kg log. If the angle of the tractors\u2019 chains in relation to each other is 18.0\u00b0, and each tractor pulls with a force of 8102 N, what forces will they be able to exert? 250 N 1.52103 N 1.58103 N 1.60103 N 6. A string exerts a force of 18 N on a box at an angle of 34\u00b0 from the horizontal. What is the horizontal component of the force on the box? 10 N 15 N 21.7 N 32 N F 34\u00b0 1.00103 kg 9.0\u00b0 9.0\u00b0 2. An airplane pilot tries to fly directly east with a velocity of 800.0 km/h. If a wind comes from the southwest at 80.0 km/h, what is the relative velocity of the airplane to the surface of Earth? 804 km/h, 5.7\u00b0 N of E 858 km/h, 3.8\u00b0 N of E 859 km/h, 4.0\u00b0 N of E 880 km/h 45\u00b0 N of E 3. For a winter fair, some students decide to build 30.0-kg wooden pull-carts on sled skids. If two 90.0-kg passengers get in, how much force will the puller have to exert to move a pull-cart? The coefficient of maximum static friction between the cart and the snow is", " 0.15. 1.8102 N 3.1102 N 2.1103 N 1.4104 N 4. It takes a minimum force of 280 N to move a 50.0-kg crate. What is the coefficient of maximum static friction between the crate and the floor? 0.18 0.57 1.8 5.6 5. What is the y-component of a 95.3-N force that is exerted at 57.1\u00b0 to the horizontal? 51.8 N 80.0 N 114 N 175 N 7. Sukey is riding her bicycle on a path when she comes around a corner and sees that a fallen tree is blocking the way 42 m ahead. If the coefficient of friction between her bicycle\u2019s tires and the gravel path is 0.36, and she is traveling at 50.0 km/h, how much stopping distance will she require? Sukey and her bicycle, together, have a mass of 95 kg. 3.00 m 4.00 m 8.12 m 27.3 m Extended Answer 8. A man starts from a position 310 m north of his car and walks for 2.7 min in a westward direction at a constant velocity of 10 km/h. How far is he from his car when he stops? 9. Jeeves is tired of his 41.2-kg son sliding down the banister, so he decides to apply an extremely sticky paste that increases the coefficient of static friction to 0.72 to the top of the banister. What will be the magnitude of the static friction force on the boy if the banister is at an angle of 52.4\u00b0 from the horizontal? Calculators Are Only Machines If your test allows you to use a calculator, use it wisely. Figure out which numbers are relevant, and determine the best way to solve the problem before you start punching keys. physicspp.com/standardized_test Chapter 5 Standardized Test Practice 145 146-152 CH06-S1-845813 3/21/04 9:17 PM Page 146 What You\u2019ll Learn \u2022 You will use Newton\u2019s laws and your knowledge of vectors to analyze motion in two dimensions. \u2022 You will solve problems dealing with projectile and circular motion. \u2022 You will solve relative- velocity problems. Why It\u2019s Important Almost all types of transportation and amusement-park attractions contain at least one element of projectile or circular motion or are affected by relative velocities. Swinging Around Before this ride starts to move, the seats hang straight down from their supports. When the ride speeds up, the seats swing out at an angle. Think About This When the swings are moving around the circle at a constant speed, are they accelerating? physicspp.com 146 Gibson Stock Photography 146-152 CH06-S1-845813 3/21/04 9:19 PM Page 147 How can the motion of a projectile be described? Question Can you describe a projectile\u2019s motion in both the horizontal and the vertical directions? Critical Thinking Describe the motion of a horizontally launched projectile. Procedure 1. With a marked grid in the background, videotape a ball that is launched with an initial velocity only in the horizontal direction. 2. Make and Use Graphs On a sheet of graph paper, draw the location of the ball every 0.1 s (3 frames). 3. Draw two motion diagrams: one for the ball\u2019s horizontal motion and one for its vertical motion. Analysis How does the vertical motion change as time passes? Does it increase, decrease, or stay the same? How does the horizontal motion change as time passes? Does it increase, decrease, or stay the same? 6.1 Projectile Motion If you observed the movement of a golf ball being hit from a tee, a frog hopping, or a free throw being shot with a basketball, you would notice that all of these objects move through the air along similar paths, as do baseballs, arrows, and bullets. Each path is a curve that moves upward for a distance, and then, after a time, turns and moves downward for some distance. You may be familiar with this curve, called a parabola, from math class. An object shot through the air is called a projectile. A projectile can be a football, a bullet, or a drop of water. After a projectile is launched, what forces are exerted on the projectile? You can draw a free-body diagram of a launched projectile and identify all the forces that are acting on it. No matter what the object is, after a projectile has been given an initial thrust, if you ignore air resistance, it moves through the air only under the force of gravity. The force of gravity is what causes the object to curve downward in a parabolic flight path. Its path through space is called its trajectory. If you know the force of the initial thrust on a projectile, you can calculate its trajectory. Objectives \u2022 Recognize that the vertical and horizontal motions of a projectile are independent. \u2022", " Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, and then determine the range using the horizontal motion. \u2022 Explain how the trajectory of a projectile depends upon the frame of reference from which it is observed. Vocabulary projectile trajectory Section 6.1 Projectile Motion 147 Horizons Companies 146-152 CH06-S1-845813 6/1/04 5:16 AM Page 148 Independence of Motion in Two Dimensions Think about two softball players warming up for a game, tossing a ball back and forth. What does the path of the ball through the air look like? It looks like a parabola, as you just learned. Imagine that you are standing directly behind one of the players and you are watching the softball as it is being tossed. What would the motion of the ball look like? You would see it go up and back down, just like any object that is tossed straight up in the air. If you were watching the softball from a hot-air balloon high above the field, what motion would you see then? You would see the ball move from one player to the other at a constant speed, just like any object that is given an initial horizontal velocity, such as a hockey puck sliding across ice. The motion of projectiles is a combination of these two motions. Why do projectiles behave in this way? After a softball leaves a player\u2019s hand, what forces are exerted on the ball? If you ignore air resistance, there are no contact forces on the ball. There is only the field force of gravity in the downward direction. How does this affect the ball\u2019s motion? Gravity causes the ball to have a downward acceleration. Figure 6-1 shows the trajectories of two softballs. One was dropped and the other was given an initial horizontal velocity of 2.0 m/s. What is similar about the two paths? Look at their vertical positions. During each flash from the strobe light, the heights of the two softballs are the same. Because the change in vertical position is the same for both, their average vertical velocities during each interval are also the same. The increasingly large distance traveled vertically by the softballs, from one time interval to the next, shows that they are accelerated downward due to the force of gravity. Notice that the horizontal motion of the launched ball does not affect its vertical motion. A projectile launched horizontally has no initial vertical velocity. Therefore, its vertical motion is like that of an object dropped from rest. The downward velocity increases regularly because of the acceleration due to gravity. \u25a0 Figure 6-1 The ball on the right was given an initial horizontal velocity. The ball on the left was dropped at the same time from rest. Note that the vertical positions of the two objects are the same during each flash. Over the Edge Obtain two balls, one twice the mass of the other. 1. Predict which ball will hit the floor first when you roll them over the surface of a table and let them roll off the edge. 2. Predict which ball will hit the floor furthest from the table. 3. Explain your predictions. 4. Test your predictions. Analyze and Conclude 5. Does the mass of the ball affect its motion? Is mass a factor in any of the equations for projectile motion? 148 Chapter 6 Motion in Two Dimensions 146-152 CH06-S1-845813 3/21/04 9:20 PM Page 149 a Begin b Begin vx 0 ax ay vy y x Separate motion diagrams for the horizontal and vertical motions are shown in Figure 6-2a. The vertical-motion diagram represents the motion of the dropped ball. The horizontal-motion diagram shows the constant velocity in the x-direction of the launched ball. This constant velocity in the horizontal direction is exactly what should be expected because there is no horizontal force acting on the ball. In Figure 6-2b, the horizontal and vertical components are added to form the total velocity vector for the projectile. You can see how the combination of constant horizontal velocity and uniform vertical acceleration produces a trajectory that has a parabolic shape. \u25a0 Figure 6-2 A object\u2019s motion can be broken into its x- and y-components (a). When the horizontal and vertical components of the ball\u2019s velocity are combined (b), the resultant vectors are tangent to a parabola. Motion in Two Dimensions Projectile motion in two dimensions can be determined by breaking the problem into two connected one-dimensional problems. 1. Divide the projectile motion into a vertical motion problem and a horizontal motion problem. 2. The vertical motion of a projectile is exactly that of an object dropped or thrown straight up or straight down. A gravitational force acts on the object and accelerates it by an amount, g. Review Section 3.3 on free fall to refresh your problem-solving skills for vertical motion. 3. Analyzing the horizontal motion of a projectile is the same as solving a constant velocity problem. No horizontal force", " acts on a projectile when drag due to air resistance is neglected. Consequently, there are no forces acting in the horizontal direction and therefore, no horizontal acceleration; ax the same methods that you learned in Section 2.4. 0.0 m/s. To solve, use 4. Vertical motion and horizontal motion are connected through the variable of time. The time from the launch of the projectile to the time it hits the target is the same for both vertical motion and horizontal motion. Therefore, solving for time in one of the dimensions, vertical or horizontal, automatically gives you time for the other dimension. Section 6.1 Projectile Motion 149 146-152 CH06-S1-845813 3/21/04 9:21 PM Page 150 1. A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high. a. How long does it take the stone to reach the bottom of the cliff? b. How far from the base of the cliff does the stone hit the ground? c. What are the horizontal and vertical components of the stone\u2019s velocity just before it hits the ground? 2. Lucy and her friend are working at an assembly plant making wooden toy giraffes. At the end of the line, the giraffes go horizontally off the edge of the conveyor belt and fall into a box below. If the box is 0.6 m below the level of the conveyor belt and 0.4 m away from it, what must be the horizontal velocity of giraffes as they leave the conveyor belt? 3. You are visiting a friend from elementary school who now lives in a small town. One local amusement is the ice-cream parlor, where Stan, the short-order cook, slides his completed ice-cream sundaes down the counter at a constant speed of 2.0 m/s to the servers. (The counter is kept very well polished for this purpose.) If the servers catch the sundaes 7.0 cm from the edge of the counter, how far do they fall from the edge of the counter to the point at which the servers catch them? Projectiles Launched at an Angle When a projectile is launched at an angle, the initial velocity has a vertical component, as well as a horizontal component. If the object is launched upward, like a ball tossed straight up in the air, it rises with slowing speed, reaches the top of its path, and descends with increasing speed. Figure 6-3a shows the separate vertical- and horizontal-motion diagrams for the trajectory. In the coordinate system, the positive x-axis is horizontal and the positive y-axis is vertical. Note the symmetry. At each point in the vertical direction, the velocity of the object as it is moving upward has the same magnitude as when it is moving downward. The only difference is that the directions of the two velocities are opposite. Figure 6-3b defines two quantities associated with the trajectory. One is the maximum height, which is the height of the projectile when the vertical velocity is zero and the projectile has only its horizontal-velocity component. The other quantity depicted is the range, R, which is the horizontal distance that the projectile travels. Not shown is the flight time, which is how much time the projectile is in the air. For football punts, flight time often is called hang time. \u25a0 Figure 6-3 The vector sum of vx and vy at each position points in the direction of the flight. a ay vy vx 0 ax y x 150 Chapter 6 Motion in Two Dimensions b \u00b0 Begin ay Maximum height Range 146-152 CH06-S1-845813 7/20/04 3:32 AM Page 151 The Flight of a Ball A ball is launched at 4.5 m/s at 66\u00b0 above the horizontal. What are the maximum height and flight time of the ball? 1 Analyze and Sketch the Problem \u2022 Establish a coordinate system with the initial position of the ball at the origin. \u2022 Show the positions of the ball at the beginning, at the maximum height, and at the end of the flight. \u2022 Draw a motion diagram showing v, a, and Fnet. Known: 0.0 m 4.5 m/s yi vi i ay 66\u00b0 g Unknown:? ymax t? y vi \u00b0 ymax Begin x R v a 2 Solve for the Unknown Find the y-component of vi. vi(sin i) (4.5 m/s)(sin 66\u00b0) 4.1 m/s vyi Substitute vi vy0 v0 Fg Fnet vx0 4.5 m/s, i 66\u00b0 vy Find an expression for time. vyi ayt vyi gt vy vyi t g Substitute ay g Solve for t. ymax Solve for the maximum height. yi + vyit + 1 at2 2 yi +", " vyi( vyi 0.0 m + (4.1 m/s)(4.1 m 0.86 m vy) + 1 g 2 vyi (g)( vy)2 m/s) + 1 vy, a g vyi Substitute t g (9.80 m/s2)(4.. m/s)2 Solve for the time to return to the launching height. 0.0 m, Substitute yi vyi vy g 9.80 m/s2 4.1 m/s, 0.0 m/s at ymax, yf yi vyit 1 at2 2 0.0 m 0.0 m + vyit 1 gt2 2 g)(0.0 m) vyi 2 4(1 vyi g) 2(1 t 2 2 vy vyi i g vyi 2 g ) Substitute vyi.s 0.84 s Substitute yf 0.0 m, yi 0.0 m, a g Use the quadratic formula to solve for t. 0 is the time the ball left the launch, so use this solution. 4.1 m/s, g 9.80 m/s2 Math Handbook Quadratic Formula page 846 3 Evaluate the Answer \u2022 Are the units correct? Dimensional analysis verifies that the units are correct. \u2022 Do the signs make sense? All should be positive. \u2022 Are the magnitudes realistic? 0.84 s is fast, but an initial velocity of 4.5 m/s makes this time reasonable. Section 6.1 Projectile Motion 151 146-152 CH06-S1-845813 6/1/04 5:18 AM Page 152 4. A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0\u00b0 above the horizontal, as shown in Figure 6-4. Find each of the following. Assume that air resistance is negligible. a. the ball\u2019s hang time b. the ball\u2019s maximum height c. the ball\u2019s range 5. The player in problem 4 then kicks the ball with the same speed, but at 60.0\u00b0 from the horizontal. What is the ball\u2019s hang time, range, and maximum height? 6. A rock is thrown from a 50.0-m-high cliff with an initial velocity of 7.0 m/s at an angle of 53.0\u00b0 above the horizontal. Find the velocity vector for when it hits the ground below. 25 ) m ( y 0 Trajectory 60.0\u00b0 30.0\u00b0 x (m) 60 \u25a0 Figure 6-4 Trajectories Depend upon the Viewer Suppose you toss a ball up and catch it while riding in a bus. To you, the ball would seem to go straight up and straight down. But what would an observer on the sidewalk see? The observer would see the ball leave your hand, rise up, and return to your hand, but because the bus would be moving, your hand also would be moving. The bus, your hand, and the ball would all have the same horizontal velocity. Thus, the trajectory of the ball would be similar to that of the ball in Example Problem 1. Air resistance So far, air resistance has been ignored in the analysis of projectile motion. While the effects of air resistance are very small for some projectiles, for others, the effects are large and complex. For example, dimples on a golf ball reduce air resistance and maximize its range. In baseball, the spin of the ball creates forces that can deflect the ball. For now, just remember that the force due to air resistance does exist and it can be important. 6.1 Section Review 7. Projectile Motion Two baseballs are pitched horizontally from the same height, but at different speeds. The faster ball crosses home plate within the strike zone, but the slower ball is below the batter\u2019s knees. Why does the faster ball not fall as far as the slower one? 8. Free-Body Diagram An ice cube slides without friction across a table at a constant velocity. It slides off the table and lands on the floor. Draw free-body and motion diagrams of the ice cube at two points on the table and at two points in the air. 9. Projectile Motion A softball is tossed into the air at an angle of 50.0\u00b0 with the vertical at an initial velocity of 11.0 m/s. What is its maximum height? 10. Projectile Motion A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0\u00b0 below the horizontal. How far does the ball move horizontally before it hits the ground? 11. Critical Thinking Suppose that an object is thrown with the same initial velocity and direction on Earth and on the Moon, where g is one-sixth that on Earth. How will the following", " quantities change? a. vx b. the object\u2019s time of flight c. ymax d. R 152 Chapter 6 Motion in Two Dimensions physicspp.com/self_check_quiz 153-156 CH06-S2-845813 6/1/04 5:24 AM Page 153 6.2 Circular Motion Consider an object moving in a circle at a constant speed, such as a stone being whirled on the end of a string or a fixed horse on a merry-go-round. Are these objects accelerating? At first, you might think that they are not because their speeds do not change. However, remember that acceleration is the change in velocity, not just the change in speed. Because their direction is changing, the objects must be accelerating. Describing Circular Motion Uniform circular motion is the movement of an object or particle trajectory at a constant speed around a circle with a fixed radius. The positions of an object in uniform circular motion, relative to the center of the circle, are given by the position vectors r1 and r2, shown in Figure 6-5a. As the object moves around the circle, the length of each position vector does not change, but its direction does. To find the object\u2019s velocity, you need to find its displacement vector over a time interval. The change in position, or the object\u2019s displacement, is represented by r. Figure 6-5b shows two position vectors: r1 at the beginning of a time interval, and r2 at the end of the time interval. Remember that a position vector is a displacement vector with its tail at the origin. In the vector diagram, r1 and r2 are subtracted to give the resultant r, the displacement during the time interval. You know that a moving object\u2019s average velocity is d/t, so for an object in circular motion, v r/t. The velocity vector has the same direction as the displacement, but a different length. You can see in Figure 6-6a that the velocity is at right angles to the position vector, which is tangent to its circular path. As the velocity vector moves around the circle, its direction changes but its length remains the same. What is the direction of the object\u2019s acceleration? Figure 6-6a shows the velocity vectors v1 and v2 at the beginning and end of a time interval. The difference in the two vectors, v, is found by subtracting the vectors, as shown in Figure 6-6b. The average acceleration, a v/t, is in the same direction as v; that is, toward the center of the circle. Repeat this process for several other time intervals when the object is in different locations on the circle. As the object moves around the circle, the direction of the acceleration vector changes, but its length remains the same. Notice that the acceleration vector of an object in uniform circular motion always points in toward the center of the circle. For this reason, the acceleration of such an object is called center-seeking or centripetal acceleration. a v1 a v2 b r v1 v2 v Objectives \u2022 Explain why an object moving in a circle at a constant speed is accelerated. \u2022 Describe how centripetal acceleration depends upon the object\u2019s speed and the radius of the circle. \u2022 Identify the force that causes centripetal acceleration. Vocabulary uniform circular motion centripetal acceleration centripetal force a b v1 r1 r2 v2 r1 r r2 \u25a0 Figure 6-5 The displacement, r, of an object in circular motion, divided by the time interval in which the displacement occurs, is the object\u2019s average velocity during that time interval. \u25a0 Figure 6-6 The direction of the change in velocity is toward the center of the circle, and so the acceleration vector also points to the center of the circle. Section 6.2 Circular Motion 153 153-156 CH06-S2-845813 7/14/04 1:31 PM Page 154 Space Elevators Scientists are considering the use of space elevators as a low-cost transportation system to space. A cable would be anchored to a station at Earth\u2019s equator, and the cable would extend almost 35,800 km from Earth\u2019s surface. The cable would be attached to a counterweight and would stay extended due to centripetal force. Special magnetically powered vehicles would then travel along the cable. \u25a0 Figure 6-7 When the thrower lets go, the hammer initially moves in a straight line that is tangent to the point of release. Then it follows a trajectory like that of any object released into the air with an initial horizontal velocity. Centripetal Acceleration What is the magnitude of an object\u2019s centripetal acceleration? Compare the triangle made from the position vectors in Figure 6-5b with the triangle made by the velocity vectors in Figure 6-6b", ". The angle between r1 and r2 is the same as that between v1 and v2. Therefore, the two triangles formed by subtracting the two sets of vectors are similar triangles, and the ratios of the lengths of two corresponding sides are equal. Thus, r/r v/v. The equation is not changed if both sides are divided by t. v r t v t r However, v r/t and a v/t. 1 v r 1 t v t r Substituting v r/t in the left-hand side and a v/t in the right-hand side gives the following equation. a v v r Solve this equation for acceleration and give it the special symbol ac, for centripetal acceleration. 2 v Centripetal Acceleration ac r Centripetal acceleration always points to the center of the circle. Its magnitude is equal to the square of the speed, divided by the radius of motion. How can you measure the speed of an object moving in a circle? One way is to measure its period, T, the time needed for the object to make one complete revolution. During this time, the object travels a distance equal to the circumference of the circle, 2r. The object\u2019s speed, then, is represented by v 2r/T. If this expression is substituted for v in the equation for centripetal acceleration, the following equation is obtained. 2 r2 T r 2r 4 2 T ac Because the acceleration of an object moving in a circle is always in the direction of the net force acting on it, there must be a net force toward the center of the circle. This force can be provided by any number of agents. For Earth circling the Sun, the force is the Sun\u2019s gravitational force on Earth, as you\u2019ll learn in Chapter 7. When a hammer thrower swings the hammer, as in Figure 6-7, the force is the tension in the chain attached to the massive ball. When an object moves in a circle, the net force toward the center of the circle is called the centripetal force. To accurately analyze centripetal acceleration situations, you must identify the agent of the force that causes the acceleration. Then you can apply Newton\u2019s second law for the component in the direction of the acceleration in the following way. Newton\u2019s Second Law for Circular Motion Fnet The net centripetal force on an object moving in a circle is equal to the object\u2019s mass, times the centripetal acceleration. mac 154 Denis Charlet/AFP/Getty Images Chapter 6 Motion in Two Dimensions 153-156 CH06-S2-845813 3/21/04 9:27 PM Page 155 When solving problems, you have found it useful to choose a coordinate system with one axis in the direction of the acceleration. For circular motion, the direction of the acceleration is always toward the center of the circle. Rather than labeling this axis x or y, call it c, for centripetal acceleration. The other axis is in the direction of the velocity, tangent to the circle. It is labeled tang for tangential. You will apply Newton\u2019s second law in these directions, just as you did in the two-dimensional problems in Chapter 5. Remember that centripetal force is just another name for the net force in the centripetal direction. It is the sum of all the real forces, those for which you can identify agents that act along the centripetal axis. In the case of the hammer thrower in Figure 6-7, in what direction does the hammer fly when the chain is released? Once the contact force of the chain is gone, there is no force accelerating the hammer toward the center of the circle, so the hammer flies off in the direction of its velocity, which is tangent to the circle. Remember, if you cannot identify the agent of the force, then it does not exist. Uniform Circular Motion A 13-g rubber stopper is attached to a 0.93-m string. The stopper is swung in a horizontal circle, making one revolution in 1.18 s. Find the tension force exerted by the string on the stopper. 1 Analyze and Sketch the Problem \u2022 Draw a free-body diagram for the swinging stopper. \u2022 Include the radius and the direction of motion. \u2022 Establish a coordinate system labeled tang and c. The directions of a and FT are parallel to c. Unknown: Known: m 13 g? FT r 0.93 m T 1.18 s 2 Solve for the Unknown Find the centripetal acceleration. ac 2r 4 2 T 4 2( m) 3.9 0 )2 (1 8 s.1 26 m/s2 Substitute r 0.93 m, T 1.18 s Use Newton\u2019s second law to find the tension in the string. FT mac (0.013 kg)(", "26 m/s2) 0.34 N Substitute m 0.013 kg, ac 26 m/s2 r m v2 v1 a FT tang c Math Handbook Operations with Significant Digits pages 835\u2013836 3 Evaluate the Answer \u2022 Are the units correct? Dimensional analysis verifies that a is in m/s2 and F is in N. \u2022 Do the signs make sense? The signs should all be positive. \u2022 Are the magnitudes realistic? The force is almost three times the weight of the stopper, and the acceleration is almost three times that of gravity, which is reasonable for such a light object. Section 6.2 Circular Motion 155 153-156 CH06-S2-845813 3/21/04 9:27 PM Page 156 12. A runner moving at a speed of 8.8 m/s rounds a bend with a radius of 25 m. What is the centripetal acceleration of the runner, and what agent exerts force on the runner? 13. A car racing on a flat track travels at 22 m/s around a curve with a 56-m radius. Find the car\u2019s centripetal acceleration. What minimum coefficient of static friction between the tires and road is necessary for the car to round the curve without slipping? 14. An airplane traveling at 201 m/s makes a turn. What is the smallest radius of the circular path (in km) that the pilot can make and keep the centripetal acceleration under 5.0 m/s2? 15. A 45-kg merry-go-round worker stands on the ride\u2019s platform 6.3 m from the center. If her speed as she goes around the circle is 4.1 m/s, what is the force of friction necessary to keep her from falling off the platform? Path of car Path of passenger without car \u25a0 Figure 6-8 The passenger would move forward in a straight line if the car did not exert an inward force. A Nonexistent Force If a car makes a sharp left turn, a passenger on the right side might be thrown against the right door. Is there an outward force on the passenger? Consider a similar situation. If a car in which you are riding stops suddenly, you will be thrown forward into your safety belt. Is there a forward force on you? No, because according to Newton\u2019s first law, you will continue moving with the same velocity unless there is a net force acting on you. The safety belt applies the force that accelerates you to a stop. Figure 6-8 shows a car turning to the left as viewed from above. A passenger in the car would continue to move straight ahead if it were not for the force of the door acting in the direction of the acceleration; that is, toward the center of the circle. Thus, there is no outward force on the passenger. The socalled centrifugal, or outward force, is a fictitious, nonexistent force. Newton\u2019s laws are able to explain motion in both straight lines and circles. 6.2 Section Review 16. Uniform Circular Motion What is the direction of the force that acts on the clothes in the spin cycle of a washing machine? What exerts the force? 17. Free-Body Diagram You are sitting in the backseat of a car going around a curve to the right. Sketch motion and free-body diagrams to answer the following questions. a. What is the direction of your acceleration? b. What is the direction of the net force that is acting on you? c. What exerts this force? 18. Centripetal Force If a 40.0-g stone is whirled horizontally on the end of a 0.60-m string at a speed of 2.2 m/s, what is the tension in the string? 19. Centripetal Acceleration A newspaper article states that when turning a corner, a driver must be careful to balance the centripetal and centrifugal forces to keep from skidding. Write a letter to the editor that critiques this article. 20. Centripetal Force A bowling ball has a mass of 7.3 kg. If you move it around a circle with a radius of 0.75 m at a speed of 2.5 m/s, what force would you have to exert on it? 21. Critical Thinking Because of Earth\u2019s daily rotation, you always move with uniform circular motion. What is the agent that supplies the force that accelerates you? How does this motion affect your apparent weight? 156 Chapter 6 Motion in Two Dimensions physicspp.com/self_check_quiz 157-159 CH06-S3-845813 3/21/04 9:30 PM Page 157 6.3 Relative Velocity Suppose that you are in a school bus that is traveling at a velocity of 8 m/s in a positive direction. You walk with a velocity of 3 m/s toward the front of the bus. If a", " friend of yours is standing on the side of the road watching the bus with you on it go by, how fast would your friend say that you are moving? If the bus is traveling at 8 m/s, this means that the velocity of the bus is 8 m/s, as measured by your friend in a coordinate system fixed to the road. When you are standing still, your velocity relative to the road is also 8 m/s, but your velocity relative to the bus is zero. Walking at 3 m/s toward the front of the bus means that your velocity is measured relative to the bus. The problem can be rephrased as follows: Given the velocity of the bus relative to the road and your velocity relative to the bus, what is your velocity relative to the road? A vector representation of this problem is shown in Figure 6-9a. After studying it, you will find that your velocity relative to the street is 11 m/s, the sum of 8 m/s and 3 m/s. Suppose that you now walk at the same speed toward the rear of the bus. What would be your velocity relative to the road? Figure 6-9b shows that because the two velocities are in opposite directions, the resultant velocity is 5 m/s, the difference between 8 m/s and 3 m/s. You can see that when the velocities are along the same line, simple addition or subtraction can be used to determine the relative velocity. Take a closer look at how these results were obtained and see if you can find a mathematical rule to describe how velocities are combined in these relative-velocity situations. For the above situation, you can designate the velocity of the bus relative to the road as vb/r, your velocity relative to the bus as vy/b, and the velocity of you relative to the road as vy/r. To find the velocity of you relative to the road in both cases, you vectorially added the velocities of you relative to the bus and the bus relative to the road. vy/r. The more general Mathematically, this is represented as vy/b form of this equation is as follows. vb/r Relative Velocity va/c The relative velocity of object a to object c is the vector sum of object a\u2019s velocity relative to object b and object b\u2019s velocity relative to object c. vb/c va/b Phillipe whirls a stone of mass m on a rope in a perfect horizontal circle above his head such that the stone is at a height, h, above the ground. The circle has a radius of r, and the tension in the rope is T. Suddenly the rope breaks and the stone falls to the ground. The stone travels a horizontal distance, s, from the time the rope breaks until it impacts the ground. Find a mathematical expression for s in terms of T, r, m, and h. Does your expression change if Phillipe is walking 0.50 m/s relative to the ground? Objectives \u2022 Analyze situations in which the coordinate system is moving. \u2022 Solve relative-velocity problems. a vbus relative to street vyou relative to bus vyou relative to street b vbus relative to street vyou relative to bus vyou relative to street \u25a0 Figure 6-9 When a coordinate system is moving, two velocities are added if both motions are in the same direction and one is subtracted from the other if the motions are in opposite directions. Section 6.3 Relative Velocity 157 157-159 CH06-S3-845813 3/21/04 9:30 PM Page 158 vplane relative to air vair relative to ground vplane relative to ground \u25a0 Figure 6-10 The plane\u2019s velocity relative to the ground can be obtained by vector addition. This method for adding relative velocities also applies to motion in two dimensions. For example, airline pilots cannot expect to reach their destinations by simply aiming their planes along a compass direction. They must take into account the plane\u2019s speed relative to the air, which is given by their airspeed indicators, and their direction of flight relative to the air. They also must consider the velocity of the wind at the altitude they are flying relative to the ground. These two vectors must be combined, as shown in Figure 6-10, to obtain the velocity of the airplane relative to the ground. The resultant vector tells the pilot how fast and in what direction the plane must travel relative to the ground to reach its destination. A similar situation occurs for boats traveling on water with a flowing current. Relative Velocity of a Marble Ana and Sandra are riding on a ferry boat that is traveling east at a speed of 4.0 m/s. Sandra rolls a marble with a velocity of 0.75 m/s north, straight across the deck of the boat to Ana. What is the velocity of the marble relative to the water? 1 Analyze and", " Sketch the Problem \u2022 Establish a coordinate system. \u2022 Draw vectors to represent the velocities of the boat relative to the water and the marble relative to the boat. Known: vb/w vm/b 4.0 m/s 0.75 m/s Unknown: vm/w? 2 Solve for the Unknown Because the two velocities are at right angles, use the Pythagorean theorem. vm/w vm/w 2 vm/b 2 2 2 vm/b 2 vb/w vb/w (4.0 m/s)2 (0.75m/s)2 Substitute vb/w 4.1 m/s vm/b Find the angle of the marble\u2019s motion. w v / m v b/ b) tan1( s) tan1(0 / m 5.7 s / m.0 4 11\u00b0 north of east vm/w vb/w y x 4.0 m/s, vm/b 0.75 m/s Math Handbook Inverses of Sine, Cosine, and Tangent page 856 Substitute vb/w 4.0 m/s, vm/b 0.75 m/s The marble is traveling 4.1 m/s at 11\u00b0 north of east. 3 Evaluate the Answer \u2022 Are the units correct? Dimensional analysis verifies that the velocity is in m/s. \u2022 Do the signs make sense? The signs should all be positive. \u2022 Are the magnitudes realistic? The resulting velocity is of the same order of magnitude as the velocities given in the problem. 158 Chapter 6 Motion in Two Dimensions 157-159 CH06-S3-845813 3/21/04 9:31 PM Page 159 22. You are riding in a bus moving slowly through heavy traffic at 2.0 m/s. You hurry to the front of the bus at 4.0 m/s relative to the bus. What is your speed relative to the street? 23. Rafi is pulling a toy wagon through the neighborhood at a speed of 0.75 m/s. A caterpillar in the wagon is crawling toward the rear of the wagon at a rate of 2.0 cm/s. What is the caterpillar\u2019s velocity relative to the ground? 24. A boat is rowed directly upriver at a speed of 2.5 m/s relative to the water. Viewers on the shore see that the boat is moving at only 0.5 m/s relative to the shore. What is the speed of the river? Is it moving with or against the boat? 25. An airplane flies due north at 150 km/h relative to the air. There is a wind blowing at 75 km/h to the east relative to the ground. What is the plane\u2019s speed relative to the ground? Another example of combined relative velocities is the navigation of migrating neotropical songbirds. In addition to knowing in which direction to fly, a bird must account for its speed relative to the air and its direction relative to the ground. If a bird tries to fly over the Gulf of Mexico into too strong a headwind, it will run out of energy before it reaches the other shore and will perish. Similarly, the bird must account for crosswinds or it will not reach its destination. You can add relative velocities even if they are at arbitrary angles by using the graphical methods that you learned in Chapter 5. Remember that the key to properly analyzing a two-dimensional relative-velocity situation is drawing the proper triangle to represent the three velocities. Once you have this triangle, you simply apply your knowledge of vector addition from Chapter 5. If the situation contains two velocities that are perpendicular to each other, you can find the third by applying the Pythagorean theorem; however, if the situation has no right angles, you will need to use one or both of the laws of sines and cosines. 6.3 Section Review Biology Connection Biology Connection 26. Relative Velocity A fishing boat with a maximum speed of 3 m/s relative to the water is in a river that is flowing at 2 m/s. What is the maximum speed the boat can obtain relative to the shore? The minimum speed? Give the direction of the boat, relative to the river\u2019s current, for the maximum speed and the minimum speed relative to the shore. 27. Relative Velocity of a Boat A powerboat heads due northwest at 13 m/s relative to the water across a river that flows due north at 5.0 m/s. What is the velocity (both magnitude and direction) of the motorboat relative to the shore? 29. A Plane\u2019s Relative Velocity An airplane flies due north at 235 km/h relative to the air. There is a wind blowing at 65 km/h to the northeast relative to the ground. What are the plane\u2019s speed and direction relative to the ground?", " 30. Relative Velocity An airplane has a speed of 285 km/h relative to the air. There is a wind blowing at 95 km/h at 30.0\u00b0 north of east relative to Earth. In which direction should the plane head to land at an airport due north of its present location? What is the plane\u2019s speed relative to the ground? 28. Relative Velocity An airplane flies due south at 175 km/h relative to the air. There is a wind blowing at 85 km/h to the east relative to the ground. What are the plane\u2019s speed and direction relative to the ground? 31. Critical Thinking You are piloting a boat across a fast-moving river. You want to reach a pier directly opposite your starting point. Describe how you would navigate the boat in terms of the components of your velocity relative to the water. physicspp.com/self_check_quiz Section 6.3 Relative Velocity 159 160-161 CH06-LAB-845813 3/21/04 9:34 PM Page 160 On Target In this activity, you will analyze several factors that affect the motion of a projectile and use your understanding of these factors to predict the path of a projectile. Finally, you will design a projectile launcher and hit a target a known distance away. QUESTION What factors affect the path of a projectile? Objectives Procedure \u25a0 Formulate models and then summarize the factors that affect the motion of a projectile. \u25a0 Use models to predict where a projectile will land. Safety Precautions Possible Materials duct tape plastic ware rubber bands paper clips paper masking tape wood blocks nails hammer PVC tubing handsaw scissors coat hanger chicken wire wire cutter 1. Brainstorm and list as many factors as you are able to think of that may affect the path of a projectile. 2. Create a design for your projectile launcher and decide what object will be your projectile shot by your launcher. 3. Taking the design of your launcher into account, determine which two factors are most likely to have a significant effect on the flight path of your projectile. 4. Check the design of your launcher and discuss your two factors with your teacher and make any necessary changes to your setup before continuing. 5. Create a method for determining what effect these two factors will have on the path of your projectile. 6. Have your teacher approve your method before collecting data. 160 Horizons Companies 160-161 CH06-LAB-845813 6/1/04 5:46 AM Page 161 Data Table 1 Launch Angle (deg) Distance Projectile Travels (cm) Data Table 2 Distance Rubber Band Is Pulled Back (cm) Distance Projectile Travels (cm) Analyze Going Further 1. Make and Use Graphs Make graphs of your data to help you predict how to use your launcher to hit a target. 2. Analyze What are the relationships between each variable you have tested and the distance the projectile travels? Conclude and Apply 1. What were the main factors influencing the path of the projectile? 2. Predict the conditions necessary to hit a target provided by your teacher. 3. Explain If you have a perfect plan and still miss the target on your first try, is there a problem with the variability of laws of physics? Explain. 4. Launch your projectile at the target. If you miss, make the necessary adjustments and try again. 1. How might your data have varied if you did this experiment outside? Would there be any additional factors affecting the motion of your projectile? 2. How might the results of your experiment be different if the target was elevated above the height of the launcher? 3. How might your experiment differ if the launcher was elevated above the height of the target? Real-World Physics 1. When a kicker attempts a field goal, do you think it is possible for him to miss because he kicked it too high? Explain. 2. If you wanted to hit a baseball as far as possible, what would be the best angle to hit the ball? To find out more about projectile motion, visit the Web site: physicspp.com 161 162 CH06-FEATURE-845813 3/21/04 9:38 PM Page 162 Spinning Space Stations Spinning Space Stations There is a lot going on aboard the International Space Station (ISS). Scientists from different countries are conducting experiments and making observations. They have seen water drops form as floating spheres and have grown peas in space to test whether crops can be grown in weightlessness. One goal of the ISS is to examine the effects on the human body when living in space for prolonged periods of time. If negative health effects can be identified, perhaps they can be prevented. This could give humans the option of living in space for long periods of time. Harmful effects of weightlessness have been observed. On Earth, muscles have gravity to push and pull against. Muscles weaken from disuse if this resistance is removed. Bones can weaken for the same reason. Also, blood volume can decrease. On Earth, gravity pulls blood downward so it collects in the lower legs. In", " weightlessness, the blood can more easily collect in an astronaut\u2019s head. The brain senses the extra blood and sends a signal to make less of it. Long-term life in space is hindered by the practical challenges of weightlessness as well. Imagine how daily life would change. Everything must be strapped or bolted down. You would have to be strapped down to a bed to sleep in one. Your life would be difficult in a space station unless the space station could be modified to simulate gravity. How could this be done? The Rotating Space Station Have you ever been on a human centrifuge\u2014a type of amusement park ride that uses centripetal force? Everyone stands against the walls of a big cylinder. Then the cylinder begins to rotate faster and faster until the riders are pressed against the walls. Because of the centripetal acceleration, the riders are held there so that even when the floor drops down they are held securely against the walls of the whirling container. A space station could be designed that uses the effects of centripetal motion as a replacement for gravity. Imagine a space station in the form of a large ring. The space station and all the objects and occupants inside would float weightlessly inside. If the ring were made to spin, unattached objects would be held against the ring\u2019s outer edge because of the centripetal motion. If the space station spun 162 Future Technology This is an artist\u2019s rendition of a rotating space station. at the right rate and if it had the right diameter, the centripetal motion would cause the occupants to experience a force of the same magnitude as gravity. The down direction in the space station would be what an observer outside the station would see as radially outward, away from the ring\u2019s center. Centripetal acceleration is directly propor- tional to the distance from the center of a rotating object. A rotating space station could be built in the form of concentric rings, each ring experiencing a different gravity. The innermost rings would experience the smallest gravity, while outermost rings would experience the largest force. You could go from floating peacefully in a low-gravity ring to standing securely in the simulated Earth-gravity ring. Going Further 1. Research What factors must engineers take into account in order to make a rotating space station that can simulate Earth\u2019s gravity? 2. Apply You are an astronaut aboard a rotating space station. You feel pulled by gravity against the floor. Explain what is really going on in terms of Newton\u2019s laws and centripetal force. 3. Critical Thinking What benefits does a rotating space station offer its occupants? What are the negative features? 163-169 CH06-SG CA STP-845813 3/21/04 9:42 PM Page 163 6.1 Projectile Motion Vocabulary \u2022 projectile (p. 147) \u2022 trajectory (p. 147) Key Concepts \u2022 The vertical and horizontal motions of a projectile are independent. \u2022 The vertical motion component of a projectile experiences a constant acceleration. \u2022 When there is no air resistance, the horizontal motion component does not experience an acceleration and has constant velocity. \u2022 Projectile problems are solved by first using the vertical motion to relate height, time in the air, and initial vertical velocity. Then the distance traveled horizontally is found. \u2022 The range of a projectile depends upon the acceleration due to gravity and upon both components of the initial velocity. \u2022 The curved flight path that is followed by a projectile is called a parabola. 6.2 Circular Motion Vocabulary \u2022 uniform circular motion Key Concepts \u2022 An object moving in a circle at a constant speed accelerates toward the center (p. 153) of the circle, and therefore, it has centripetal acceleration. \u2022 centripetal acceleration \u2022 Centripetal acceleration depends directly on the square of the object\u2019s speed (p. 153) \u2022 centripetal force (p. 154) and inversely on the radius of the circle. 2 v r ac \u2022 The centripetal acceleration for an object traveling in a circle can also be expressed as a function of its period, T. 2r 4 2 T ac \u2022 A net force must be exerted toward the circle\u2019s center to cause centripetal acceleration. Fnet mac \u2022 The velocity vector of an object with a centripetal acceleration is always tangent to the circular path. 6.3 Relative Velocity Key Concepts \u2022 Vector addition can be used to solve problems involving relative velocities. vb/c va/c va/b \u2022 The key to properly analyzing a two-dimensional relative-velocity problem is drawing the proper triangle to represent all three velocity vectors. physicspp.com/vocabulary_puzzlemaker 163 163-169 CH06-SG CA STP-845813 6/1/04 5:53 AM Page 164 Concept Mapping 32. Use the following terms to complete the concept map below: constant speed, horizontal part of projectile motion,", " constant acceleration, relative-velocity motion, uniform circular motion. 37. To obtain uniform circular motion, how must the net force depend on the speed of the moving object? (6.2) 38. If you whirl a yo-yo about your head in a horizontal circle, in what direction must a force act on the yo-yo? What exerts the force? (6.2) Categories of Motion 39. Why is it that a car traveling in the opposite constant velocity vertical part of projectile motion Mastering Concepts 33. Consider the trajectory of the cannonball shown in Figure 6-11. (6.1) a. Where is the magnitude of the vertical-velocity component largest? b. Where is the magnitude of the horizontal- velocity component largest? c. Where is the vertical-velocity smallest? d. Where is the magnitude of the acceleration smallest? B A C D E \u25a0 Figure 6-11 34. A student is playing with a radio-controlled race car on the balcony of a sixth-floor apartment. An accidental turn sends the car through the railing and over the edge of the balcony. Does the time it takes the car to fall depend upon the speed it had when it left the balcony? (6.1) 35. An airplane pilot flying at constant velocity and altitude drops a heavy crate. Ignoring air resistance, where will the plane be relative to the crate when the crate hits the ground? Draw the path of the crate as seen by an observer on the ground. (6.1) 36. Can you go around a curve with the following accelerations? Explain. a. zero acceleration b. constant acceleration (6.2) direction as the car in which you are riding on the freeway often looks like it is moving faster than the speed limit? (6.3) Applying Concepts 40. Projectile Motion Analyze how horizontal motion can be uniform while vertical motion is accelerated. How will projectile motion be affected when drag due to air resistance is taken into consideration? 41. Baseball A batter hits a pop-up straight up over home plate at an initial velocity of 20 m/s. The ball is caught by the catcher at the same height that it was hit. At what velocity does the ball land in the catcher\u2019s mitt? Neglect air resistance. 42. Fastball In baseball, a fastball takes about 1 s to 2 reach the plate. Assuming that such a pitch is thrown horizontally, compare the distance the ball falls in the first 1 s with the distance it falls in the 4 second 1 s. 4 43. You throw a rock horizontally. In a second horizontal throw, you throw the rock harder and give it even more speed. a. How will the time it takes the rock to hit the ground be affected? Ignore air resistance. b. How will the increased speed affect the distance from where the rock left your hand to where the rock hits the ground? 44. Field Biology A zoologist standing on a cliff aims a tranquilizer gun at a monkey hanging from a distant tree branch. The barrel of the gun is horizontal. Just as the zoologist pulls the trigger, the monkey lets go and begins to fall. Will the dart hit the monkey? Ignore air resistance. 45. Football A quarterback throws a football at 24 m/s at a 45\u00b0 angle. If it takes the ball 3.0 s to reach the top of its path and the ball is caught at the same height at which it is thrown, how long is it in the air? Ignore air resistance. 46. Track and Field You are working on improving your performance in the long jump and believe that the information in this chapter can help. Does the height that you reach make any difference to your jump? What influences the length of your jump? 164 Chapter 6 Motion in Two Dimensions For more problems, go to Additional Problems, Appendix B. 163-169 CH06-SG CA STP-845813 3/21/04 9:43 PM Page 165 47. Imagine that you are sitting in a car tossing a ball straight up into the air. a. If the car is moving at a constant velocity, will the ball land in front of, behind, or in your hand? b. If the car rounds a curve at a constant speed, where will the ball land? 48. You swing one yo-yo around your head in a horizontal circle. Then you swing another yo-yo with twice the mass of the first one, but you don\u2019t change the length of the string or the period. How do the tensions in the strings differ? 49. Car Racing The curves on a race track are banked to make it easier for cars to go around the curves at high speeds. Draw a free-body diagram of a car on a banked curve. From the motion diagram, find the direction of the acceleration. a. What exerts the force in the direction of the acceleration? b. Can you have such a force without friction? 50. Driving on the Highway Explain", " why it is that when you pass a car going in the same direction as you on the freeway, it takes a longer time than when you pass a car going in the opposite direction. Mastering Problems 6.1 Projectile Motion 51. You accidentally throw your car keys horizontally at 8.0 m/s from a cliff 64-m high. How far from the base of the cliff should you look for the keys? 52. The toy car in Figure 6-12 runs off the edge of a table that is 1.225-m high. The car lands 0.400 m from the base of the table. a. How long did it take the car to fall? b. How fast was the car going on the table? v 1.225 m \u25a0 Figure 6-12 0.400 m 53. A dart player throws a dart horizontally at 12.4 m/s. The dart hits the board 0.32 m below the height from which it was thrown. How far away is the player from the board? 54. The two baseballs in Figure 6-13 were hit with the same speed, 25 m/s. Draw separate graphs of y versus t and x versus t for each ball. A B xA xB 60\u00b0 30\u00b0 yA yB \u25a0 Figure 6-13 55. Swimming You took a running leap off a high- diving platform. You were running at 2.8 m/s and hit the water 2.6 s later. How high was the platform, and how far from the edge of the platform did you hit the water? Ignore air resistance. 56. Archery An arrow is shot at 30.0\u00b0 above the horizontal. Its velocity is 49 m/s, and it hits the target. a. What is the maximum height the arrow will attain? b. The target is at the height from which the arrow was shot. How far away is it? 57. Hitting a Home Run A pitched ball is hit by a batter at a 45\u00b0 angle and just clears the outfield fence, 98 m away. If the fence is at the same height as the pitch, find the velocity of the ball when it left the bat. Ignore air resistance. 58. At-Sea Rescue An airplane traveling 1001 m above the ocean at 125 km/h is going to drop a box of supplies to shipwrecked victims below. a. How many seconds before the plane is directly overhead should the box be dropped? b. What is the horizontal distance between the plane and the victims when the box is dropped? 59. Diving Divers in Acapulco dive from a cliff that is 61 m high. If the rocks below the cliff extend outward for 23 m, what is the minimum horizontal velocity a diver must have to clear the rocks? 60. Jump Shot A basketball player is trying to make a half-court jump shot and releases the ball at the height of the basket. Assuming that the ball is launched at 51.0\u00b0, 14.0 m from the basket, what speed must the player give the ball? physicspp.com/chapter_test Chapter 6 Assessment 165 163-169 CH06-SG CA STP-845813 6/1/04 5:53 AM Page 166 6.2 Circular Motion 61. Car Racing A 615-kg racing car completes one lap in 14.3 s around a circular track with a radius of 50.0 m. The car moves at a constant speed. a. What is the acceleration of the car? b. What force must the track exert on the tires to produce this acceleration? 62. Hammer Throw An athlete whirls a 7.00-kg hammer 1.8 m from the axis of rotation in a horizontal circle, as shown in Figure 6-14. If the hammer makes one revolution in 1.0 s, what is the centripetal acceleration of the hammer? What is the tension in the chain? vtang 67. A 75-kg pilot flies a plane in a loop as shown in Figure 6-15. At the top of the loop, when the plane is completely upside-down for an instant, the pilot hangs freely in the seat and does not push against the seat belt. The airspeed indicator reads 120 m/s. What is the radius of the plane\u2019s loop? vtang 120 m/s \u25a0 Figure 6-14 63. A coin is placed on a vinyl stereo record that is making 331 revolutions per minute on a turntable. 3 a. In what direction is the acceleration of the coin? b. Find the magnitude of the acceleration when the coin is placed 5.0, 10.0, and 15.0 cm from the center of the record. c. What force accelerates the coin? d. At which of the three radii in part b would the coin be most likely to fly off the turntable? Why? 64. A rotating rod that is 15.3 cm long is spun with its axis through one", " end of the rod so that the other end of the rod has a speed of 2010 m/s (4500 mph). a. What is the centripetal acceleration of the end of the rod? b. If you were to attach a 1.0-g object to the end of the rod, what force would be needed to hold it on the rod? 65. Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a car can safely travel if the radius of the track is 80.0 m and the coefficient of friction is 0.40? 66. A carnival clown rides a motorcycle down a ramp and around a vertical loop. If the loop has a radius of 18 m, what is the slowest speed the rider can have at the top of the loop to avoid falling? Hint: At this slowest speed, the track exerts no force on the motorcycle at the top of the loop. \u25a0 Figure 6-15 6.3 Relative Velocity 68. Navigating an Airplane An airplane flies at 200.0 km/h relative to the air. What is the velocity of the plane relative to the ground if it flies during the following wind conditions? a. a 50.0-km/h tailwind b. a 50.0-km/h headwind 69. Odina and LaToya are sitting by a river and decide to have a race. Odina will run down the shore to a dock, 1.5 km away, then turn around and run back. LaToya will also race to the dock and back, but she will row a boat in the river, which has a current of 2.0 m/s. If Odina\u2019s running speed is equal to LaToya\u2019s rowing speed in still water, which is 4.0 m/s, who will win the race? Assume that they both turn instantaneously. 70. Crossing a River You row a boat, such as the one in Figure 6-16, perpendicular to the shore of a river that flows at 3.0 m/s. The velocity of your boat is 4.0 m/s relative to the water. a. What is the velocity of your boat relative to the shore? b. What is the component of your velocity parallel to the shore? Perpendicular to it? b v vw \u25a0 Figure 6-16 166 Chapter 6 Motion in Two Dimensions For more problems, go to Additional Problems, Appendix B. 163-169 CH06-SG CA STP-845813 6/1/04 5:54 AM Page 167 71. Studying the Weather A weather station releases a balloon to measure cloud conditions that rises at a constant 15 m/s relative to the air, but there is also a wind blowing at 6.5 m/s toward the west. What are the magnitude and direction of the velocity of the balloon? 72. Boating You are boating on a river that flows toward the east. Because of your knowledge of physics, you head your boat 53\u00b0 west of north and have a velocity of 6.0 m/s due north relative to the shore. a. What is the velocity of the current? b. What is the speed of your boat relative to the water? 73. Air Travel You are piloting a small plane, and you want to reach an airport 450 km due south in 3.0 h. A wind is blowing from the west at 50.0 km/h. What heading and airspeed should you choose to reach your destination in time? 78. A 1.13-kg ball is swung vertically from a 0.50-m cord in uniform circular motion at a speed of 2.4 m/s. What is the tension in the cord at the bottom of the ball\u2019s motion? 79. Banked Roads Curves on roads often are banked to help prevent cars from slipping off the road. If the posted speed limit for a particular curve of radius 36.0 m is 15.7 m/s (35 mph), at what angle should the road be banked so that cars will stay on a circular path even if there were no friction between the road and the tires? If the speed limit was increased to 20.1 m/s (45 mph), at what angle should the road be banked? 80. The 1.45-kg ball in Figure 6-17 is suspended from a 0.80-m string and swung in a horizontal circle at a constant speed such that the string makes an angle of 14.0\u00b0 with the vertical. a. What is the tension in the string? b. What is the speed of the ball? Mixed Review 74. Early skeptics of the idea of a rotating Earth said that the fast spin of Earth would throw people at the equator into space. The radius of Earth is about 6.38103 km. Show why this idea is wrong by calculating the following.", " a. the speed of a 97-kg person at the equator b. the force needed to accelerate the person in the circle c. the weight of the person d. the normal force of Earth on the person, that is, the person\u2019s apparent weight 75. Firing a Missile An airplane, moving at 375 m/s relative to the ground, fires a missile forward at a speed of 782 m/s relative to the plane. What is the speed of the missile relative to the ground? 76. Rocketry A rocket in outer space that is moving at a speed of 1.25 km/s relative to an observer fires its motor. Hot gases are expelled out the back at 2.75 km/s relative to the rocket. What is the speed of the gases relative to the observer? 77. Two dogs, initially separated by 500.0 m, are running towards each other, each moving with a constant speed of 2.5 m/s. A dragonfly, moving with a constant speed of 3.0 m/s, flies from the nose of one dog to the other, then turns around instantaneously and flies back to the other dog. It continues to fly back and forth until the dogs run into each other. What distance does the dragonfly fly during this time? 14.0\u00b0 \u25a0 Figure 6-17 81. A baseball is hit directly in line with an outfielder at an angle of 35.0\u00b0 above the horizontal with an initial velocity of 22.0 m/s. The outfielder starts running as soon as the ball is hit at a constant velocity of 2.5 m/s and barely catches the ball. Assuming that the ball is caught at the same height at which it was hit, what was the initial separation between the hitter and outfielder? Hint: There are two possible answers. 82. A Jewel Heist You are serving as a technical consultant for a locally produced cartoon. In one episode, two criminals, Shifty and Lefty, have stolen some jewels. Lefty has the jewels when the police start to chase him, and he runs to the top of a 60.0-m tall building in his attempt to escape. Meanwhile, Shifty runs to the convenient hot-air balloon 20.0 m from the base of the building and untethers it, so it begins to rise at a constant speed. Lefty tosses the bag of jewels horizontally with a speed of 7.3 m/s just as the balloon begins its ascent. What must the velocity of the balloon be for Shifty to easily catch the bag? physicspp.com/chapter_test Chapter 6 Assessment 167 163-169 CH06-SG CA STP-845813 3/21/04 9:45 PM Page 168 Thinking Critically 83. Apply Concepts Consider a roller-coaster loop like the one in Figure 6-18. Are the cars traveling through the loop in uniform circular motion? Explain. \u25a0 Figure 6-18 84. Use Numbers A 3-point jump shot is released 2.2 m above the ground and 6.02 m from the basket. The basket is 3.05 m above the floor. For launch angles of 30\u00b0 and 60\u00b0, find the speed the ball needs to be thrown to make the basket. 85. Analyze For which angle in problem 84 is it more important that the player get the speed right? To explore this question, vary the speed at each angle by 5 percent and find the change in the range of the attempted shot. 86. Apply Computers and Calculators A baseball player hits a belt-high (1.0 m) fastball down the left-field line. The ball is hit with an initial velocity of 42.0 m/s at 26\u00b0. The left-field wall is 96.0 m from home plate at the foul pole and is 14-m high. Write the equation for the height of the ball, y, as a function of its distance from home plate, x. Use a computer or graphing calculator to plot the path of the ball. Trace along the path to find how high above the ground the ball is when it is at the wall. a. Is the hit a home run? b. What is the minimum speed at which the ball could be hit and clear the wall? c. If the initial velocity of the ball is 42.0 m/s, for what range of angles will the ball go over the wall? 87. Analyze Albert Einstein showed that the rule you learned for the addition of velocities does not work for objects moving near the speed of light. For example, if a rocket moving at velocity vA releases a missile that has velocity vB relative to the rocket, then the velocity of the missile relative to an observer that is at rest is given by v (vA vB)/(1 vAvB/c2), where c is the speed of light, 3.00108 m/s. This formula gives the correct values for objects moving at slow", " speeds as well. Suppose a rocket moving at 11 km/s shoots a laser beam out in front of it. What speed would an unmoving observer find for the laser light? Suppose that a rocket moves at a speed c/2, half the speed of light, and shoots a missile forward at a speed of c/2 relative to the rocket. How fast would the missile be moving relative to a fixed observer? 88. Analyze and Conclude A ball on a light string moves in a vertical circle. Analyze and describe the motion of this system. Be sure to consider the effects of gravity and tension. Is this system in uniform circular motion? Explain your answer. Writing in Physics 89. Roller Coasters If you take a look at vertical loops on roller coasters, you will notice that most of them are not circular in shape. Research why this is so and explain the physics behind this decision by the coaster engineers. 90. Many amusement-park rides utilize centripetal acceleration to create thrills for the park\u2019s customers. Choose two rides other than roller coasters that involve circular motion and explain how the physics of circular motion creates the sensations for the riders. Cumulative Review 91. Multiply or divide, as indicated, using significant digits correctly. (Chapter 1) a. (5108 m)(4.2107 m) b. (1.67102 km)(8.5106 km) c. (2.6104 kg)/(9.4103 m3) d. (6.3101 m)/(3.8102 s) 92. Plot the data in Table 6-1 on a position-time graph. Find the average velocity in the time interval between 0.0 s and 5.0 s. (Chapter 3) Table 6-1 Position v. Time Clock Reading t (s) Position d (m) 0.0 1.0 2.0 3.0 4.0 5.0 30 30 35 45 60 70 93. Carlos and his older brother Ricardo are at the grocery store. Carlos, with mass 17.0 kg, likes to hang on the front of the cart while Ricardo pushes it, even though both boys know this is not safe. Ricardo pushes the cart, with mass 12.4 kg, with his brother hanging on it such that they accelerate at a rate of 0.20 m/s2. (Chapter 4) a. With what force is Ricardo pushing? b. What is the force the cart exerts on Carlos? 168 Bill Aron/PhotoEdit Chapter 6 Motion in Two Dimensions For more problems, go to Additional Problems, Appendix B. 163-169 CH06-SG CA STP-845813 7/20/04 3:34 AM Page 169 Multiple Choice 1. A 1.60-m-tall girl throws a football at an angle of 41.0\u00b0 from the horizontal and at an initial velocity of 9.40 m/s. How far away from the girl will it land? 4.55 m 5.90 m 8.90 m 10.5 m 2. A dragonfly is sitting on a merry-go-round 2.8 m from the center. If the tangential velocity of the ride is 0.89 m/s, what is the centripetal acceleration of the dragonfly? 0.11 m/s2 0.28 m/s2 0.32 m/s2 2.2 m/s2 3. The centripetal force on a 0.82-kg object on the end of a 2.0-m massless string being swung in a horizontal circle is 4.0 N. What is the tangential velocity of the object? 2.8 m/s2 3.1 m/s2 4.9 m/s2 9.8 m/s2 4. A 1000-kg car enters an 80.0-m-radius curve at 20.0 m/s. What centripetal force must be supplied by friction so the car does not skid? 5.0 N 2.5102 N 5.0103 N 1.0103 N 5. A jogger on a riverside path sees a rowing team coming toward him. If the jogger is moving at 10 km/h, and the boat is moving at 20 km/h, how quickly does the jogger approach the boat? 3 m/s 8 m/s 40 m/s 100 m/s 6. What is the maximum height obtained by a 125-g apple that is slung from a slingshot at an angle of 78\u00b0 from the horizontal with an initial velocity of 18 m/s? 0.70 m 16 m 32 m 33 m 7. An orange is dropped at the same time a bullet is shot from a gun. Which of the following is true? The acceleration due to gravity is greater for the orange because the orange is heavier. Gravity acts less on the bullet than on the orange because the", " bullet is moving so fast. The velocities will be the same. The two objects will hit the ground at the same time. Extended Answer 8. A colorfully feathered lead cannonball is shot horizontally out of a circus cannon 25 m/s from the high-wire platform on one side of a circus ring. If the high-wire platform is 52 m above the 80-m diameter ring, will the performers need to adjust their cannon (will the ball land inside the ring, or past it)? v 52 m 80 m (Not to scale) 9. A mythical warrior swings a 5.6-kg mace on the end of a magically massless 86-cm chain in a horizontal circle above his head. The mace makes one full revolution in 1.8 s. Find the tension in the magical chain. Practice Under Testlike Conditions Answer all of the questions in the time provided without referring to your book. Did you complete the test? Could you have made better use of your time? What topics do you need to review? physicspp.com/standardized_test Chapter 6 Standardized Test Practice 169 170-178 CH07 S1-845813 3/19/04 2:01 PM Page 170 What You\u2019ll Learn \u2022 You will learn the nature of gravitational force. \u2022 You will relate Kepler\u2019s laws of planetary motion to Newton\u2019s laws of motion. \u2022 You will describe the orbits of planets and satellites using the law of universal gravitation. Why It\u2019s Important Kepler\u2019s laws and the law of universal gravitation will help you understand the motion of planets and satellites. Comets Comet Hale-Bopp was discovered by Alan Hale and Thomas Bopp in 1995. The comet entered the inner solar system in 1997 and was visible from Joshua Tree National Park in California, providing spectacular views of its white dust tail and blue ion tail. Think About This Comets orbit the Sun just as planets and stars do. How can you describe the orbit of a comet such as Hale-Bopp? physicspp.com 170 Wally Pacholka/Astropics.com 170-178 CH07 S1-845813 3/5/04 11:19 AM Page 171 Can you model Mercury\u2019s motion? Question Do planets in our solar system have circular orbits or do they travel in some other path? Procedure 1. Use the data table to plot the orbit of Mercury using the scale 10 cm 1 AU. Note that one astronomical unit, AU, is Earth\u2019s distance from the Sun. 1 AU is equal to 1.5108 km. 2. Calculate the distance in cm for each distance measured in AU. 3. Mark the center of your paper and draw a horizontal zero line and a vertical zero line going through it. 4. Place your protractor on the horizontal line and center it on the center point. Measure the degrees and place a mark. 5. Place a ruler connecting the center and the angle measurement. Mark the distance in centimeters for the corresponding angle. You will need to place the protractor on the vertical zero line for certain angle measurements. 6. Once you have marked all the data points, draw a line connecting them. Analysis Describe the shape of Mercury\u2019s orbit. Draw a line going through the Sun that represents the longest axis of the orbit, called the major axis. Critical Thinking How does the orbit of Mercury compare to the orbit of comet Hale-Bopp, shown on page 170? Mercury\u2019s Orbit (\u00b0) d (AU) 4 61 122 172 209 239 266 295 330 350 0.35 0.31 0.32 0.38 0.43 0.46 0.47 0.44 0.40 0.37 7.1 Planetary Motion and Gravitation Since ancient times, the Sun, Moon, planets, and stars had been assumed to revolve around Earth. Nicholas Copernicus, a Polish astronomer, noticed that the best available observations of the movements of planets and stars did not fully agree with the Earth-centered model. The results of his many years of work were published in 1543, when Copernicus was on his deathbed. His book showed that the motion of planets is much more easily understood by assuming that Earth and other planets revolve around the Sun. Tycho Brahe was born a few years after the death of Copernicus. As a boy of 14 in Denmark, Brahe observed an eclipse of the Sun on August 21, 1560, and vowed to become an astronomer. Brahe studied astronomy as he traveled throughout Europe for five years. He did not use telescopes. Instead, he used huge instruments that he designed and built in his own shop on the Danish island of Hven. He spent the next 20 years carefully recording the exact positions of the planets and stars. Brahe concluded that the Sun and the Moon orbit Earth and that all other planets orbit the Sun. Objectives \u2022 Relate Kepler\u2019s laws to the law of universal gravitation. \u2022", " Calculate orbital speeds and periods. \u2022 Describe the importance of Cavendish\u2019s experiment. Vocabulary Kepler\u2019s first law Kepler\u2019s second law Kepler\u2019s third law gravitational force law of universal gravitation Section 7.1 Planetary Motion and Gravitation 171 170-178 CH07 S1-845813 3/5/04 12:05 PM Page 172 a b c \u25a0 Figure 7-1 Among the huge astronomical instruments that Tycho Brahe constructed to use on Hven (a) were an astrolabe (b) and a sextant (c). Kepler\u2019s Laws Johannes Kepler, a 29-year-old German, became one of Brahe\u2019s assistants when he moved to Prague. Brahe trained his assistants to use instruments, such as those shown in Figure 7-1. Upon his death in 1601, Kepler inherited 30 years\u2019 worth of Brahe\u2019s observations. He studied Brahe\u2019s data and was convinced that geometry and mathematics could be used to explain the number, distance, and motion of the planets. Kepler believed that the Sun exerted a force on the planets and placed the Sun at the center of the system. After several years of careful analysis of Brahe\u2019s data on Mars, Kepler discovered the laws that describe the motion of every planet and satellite. Kepler\u2019s first law states that the paths of the planets are ellipses, with the Sun at one focus. An ellipse has two foci, as shown in Figure 7-2. Like planets and stars, comets also orbit the Sun in elliptical orbits. Comets are divided into two groups\u2014long-period comets and short-period comets\u2014 based on orbital periods, each of which is the time it takes the comet to complete one revolution. Long-period comets have orbital periods longer than 200 years, and short-period comets have orbital periods shorter than 200 years. Comet Hale-Bopp, with a period of 2400 years, is an example of a long-period comet. Comet Halley, with a period of 76 years, is an example of a short-period comet. \u25a0 Figure 7-2 Planets orbit the Sun in elliptical orbits with the Sun at one focus. (Illustration not to scale) Foci Sun Planet Ellipse 172 Newberry Library/Stock Montage Chapter 7 Gravitation 170-178 CH07 S1-845813 3/5/04 11:22 AM Page 173 Kepler found that the planets move faster when they are closer to the Sun and slower when they are farther away from the Sun. Thus, Kepler\u2019s second law states that an imaginary line from the Sun to a planet sweeps out equal areas in equal time intervals, as illustrated in Figure 7-3. Planet Sun 1 12 11 10 9 8 7 6 5 4 \u25a0 Figure 7-3 A planet moves fastest when it is close to the Sun and slowest when it is farther from the Sun. Equal areas are swept out in equal amounts of time. (Illustration not to scale) from Kepler also found that there is a mathematical relationship between periods of planets and their mean distances away the Sun. Kepler\u2019s third law states that the square of the ratio of the periods of any two planets revolving about the Sun is equal to the cube of the ratio of their average distances from the Sun. Thus, if the periods of the planets are TA and TB, and their average distances from the Sun are rA and rB, Kepler\u2019s third law can be expressed as follows. 2 3 Kepler\u2019s Third Law The squared quantity of the period of object A divided by the period of object B, is equal to the cubed quantity of object A\u2019s average distance from the Sun, divided by object B\u2019s average distance from the Sun. Note that the first two laws apply to each planet, moon, and satellite individually. The third law, however, relates the motion of several objects about a single body. For example, it can be used to compare the planets\u2019 distances from the Sun, shown in Table 7-1, to their periods about the Sun. It also can be used to compare distances and periods of the Moon and artificial satellites orbiting Earth. Name Sun Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Average Radius (m) 6.96108 2.44106 6.05106 6.38106 3.40106 7.15107 6.03107 2.56107 2.48107 1.20106 Table 7-1 Planetary Data Mass (kg) 1.991030 3.301023 4.871024 5.981024 6.421023 1.901027 5.691026 8.681025 1.021026 1.251022 Mean Distance From Sun (m) \u2014 5.791010 1.081011 1.501011 2.281011 7.781011 1", ".431012 2.871012 4.501012 5.871012 Section 7.1 Planetary Motion and Gravitation 173 170-178 CH07 S1-845813 3/5/04 11:23 AM Page 174 Callisto\u2019s Distance from Jupiter Galileo measured the orbital sizes of Jupiter\u2019s moons using the diameter of Jupiter as a unit of measure. He found that Io, the closest moon to Jupiter, had a period of 1.8 days and was 4.2 units from the center of Jupiter. Callisto, the fourth moon from Jupiter, had a period of 16.7 days. Using the same units that Galileo used, predict Callisto\u2019s distance from Jupiter. 1 2 Analyze and Sketch the Problem \u2022 Sketch the orbits of Io and Callisto. \u2022 Label the radii. Known: TC T I r I 16.7 days 1.8 days 4.2 units Unknown Solve for the Unknown Solve Kepler\u2019s third law for rC. )3 ( )2 ( )2 3 ( 3 rI rC 3 rI )2 3 ( ay 3 (4.2 units)3 (1 3 6.4103 units3 19 units.7 6.8 1 d a d T C T rC I I )2 Substitute rI s y s Jupiter Io rI Callisto rC Math Handbook Isolating a Variable page 845 4.2 units, TC 16.7 days, TI 1.8 days 3 Evaluate the Answer \u2022 Are the units correct? rC should be in Galileo\u2019s units, like rI. \u2022 Is the magnitude realistic? The period is large, so the radius should be large. 1. If Ganymede, one of Jupiter\u2019s moons, has a period of 32 days, how many units are there in its orbital radius? Use the information given in Example Problem 1. 2. An asteroid revolves around the Sun with a mean orbital radius twice that of Earth\u2019s. Predict the period of the asteroid in Earth years. 3. From Table 7-1, on page 173, you can find that, on average, Mars is 1.52 times as far from the Sun as Earth is. Predict the time required for Mars to orbit the Sun in Earth days. 4. The Moon has a period of 27.3 days and a mean distance of 3.90105 km from the center of Earth. a. Use Kepler\u2019s laws to find the period of a satellite in orbit 6.70103 km from the center of Earth. b. How far above Earth\u2019s surface is this satellite? 5. Using the data in the previous problem for the period and radius of revolution of the Moon, predict what the mean distance from Earth's center would be for an artificial satellite that has a period of exactly 1.00 day. 174 Chapter 7 Gravitation 170-178 CH07 S1-845813 6/1/04 6:16 AM Page 175 Newton\u2019s Law of Universal Gravitation In 1666, 45 years after Kepler published his work, Newton began his studies of planetary motion. He found that the magnitude of the force, F, on a planet due to the Sun varies inversely with the square of the distance, r, between the centers of the planet and the Sun. That is, F is proportional to 1/r 2. The force, F, acts in the direction of the line connecting the centers of the two objects. It is quoted that the sight of a falling apple made Newton wonder if the force that caused the apple to fall might extend to the Moon, or even beyond. He found that both the apple's and Moon\u2019s accelerations agreed with the 1/r 2 relationship. According to his own third law, the force Earth exerts on the apple is exactly the same as the force the apple exerts on Earth. The force of attraction between two objects must be proportional to the objects\u2019 masses, and is known as the gravitational force. Newton was confident that the same force of attraction would act between any two objects, anywhere in the universe. He proposed his law of universal gravitation, which states that objects attract other objects with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. This can be represented by the following equation. m2 m1 Law of Universal Gravitation F G r2 The gravitational force is equal to the universal gravitational constant, times the mass of object 1, times the mass of object 2, divided by the distance between the centers of the objects, squared. According to Newton\u2019s equation, F is directly proportional to m1 and m2. Thus, if the mass of a planet near the Sun were doubled, the force of attraction would be doubled. Use the Connecting Math to Physics feature below to examine how changing one variable affects another. Figure 7-4 illustrates the inverse square law graph", "ically. Direct and Inverse Relationships Newton\u2019s law of universal gravitation has both direct and inverse relationships. F m1m2 1 F 2 r Change Result Change Result 2m1m2 3m1m2 2m1 3m2 1 m1m2 2 2F 3F 6F 1 F 2 2r 3r 4F 9F Force v. Distance Inverse Square Law ) N ( e c r o F 80 60 40 20 0 0.5 Distance (m) \u25a0 Figure 7-4 The change in gravitational force with distance follows the inverse square law. Section 7.1 Planetary Motion and Gravitation 175 170-178 CH07 S1-845813 6/1/04 6:16 AM Page 176 Sun r mp Planet \u25a0 Figure 7-5 A planet with mass mp and orbital radius r orbits the Sun with mass mS. (Illustration not to scale) Universal Gravitation and Kepler\u2019s Third Law Newton stated his law of universal gravitation in terms that applied to the motion of planets about the Sun. This agreed with Kepler\u2019s third law and confirmed that Newton\u2019s law fit the best observations of the day. ma, can be written as Fnet Consider a planet orbiting the Sun, as shown in Figure 7-5. Newton\u2019s mpac, where F is second law of motion, Fnet the gravitational force, mp is the mass of the planet, and ac is the centripetal acceleration of the planet. For simplicity, assume circular orbits. Recall from your study of uniform circular motion in Chapter 6 that, for a circular mpac may now be written as orbit, ac mp42r/T2. In this equation, T is the time required for the planet to Fnet make one complete revolution about the Sun. If you set the right side of this equation equal to the right side of the law of universal gravitation, you arrive at the following result: 42r/T2. This means that Fnet 2r 4 mp mp mS G 2 r2 T r3 T2 4 2 m G S r3 Thus, T 4 2 m G S The period of a planet orbiting the Sun can be expressed as follows. Period of a Planet Orbiting the Sun T 2 3 r m G S The period of a planet orbiting the Sun is equal to 2times the square root of the orbital radius cubed, divided by the product of the universal gravitational constant and the mass of the Sun. Squaring both sides makes it apparent that this equation is Kepler\u2019s third law of planetary motion: the square of the period is proportional to the cube of the distance that separates the masses. The factor 42/GmS depends on the mass of the Sun and the universal gravitational constant. Newton found that this derivation applied to elliptical orbits as well. Astronomers have detected three planets that orbit the star Upsilon Andromedae. Planet B has an average orbital radius of 0.059 AU and a period of 4.6170 days. Planet C has an average orbital radius of 0.829 AU and a period of 241.5 days. Planet D has an average orbital radius of 2.53 AU and a period of 1284 days. (Distances are given in astronomical units (AU)\u2014Earth\u2019s average distance from the Sun. The distance from Earth to the Sun is 1.00 AU.) 1. Do these planets obey Kepler\u2019s third law? 2. Find the mass of the star Upsilon Andromedae in units of the Sun\u2019s mass. B rB D rD C rC Upsilon Andromedae 176 Chapter 7 Gravitation 170-178 CH07 S1-845813 3/5/04 11:25 AM Page 177 Measuring the Universal Gravitational Constant How large is the constant, G? As you know, the force of gravitational attraction between two objects on Earth is relatively small. The slightest attraction, even between two massive bowling balls, is difficult to detect. In fact, it took 100 years from the time of Newton\u2019s work for scientists to develop an apparatus that was sensitive enough to measure the force of gravitational attraction. Cavendish\u2019s experiment In 1798, Englishman Henry Cavendish used equipment similar to the apparatus shown in Figure 7-6 to measure the gravitational force between two objects. The apparatus had a horizontal rod with two small lead spheres attached to each end. The rod was suspended at its midpoint by a thin wire so that it could rotate. Because the rod was suspended by a thin wire, the rod and spheres were very sensitive to horizontal forces. To measure G, Cavendish placed two large lead spheres in a fixed position, close to each of the two small spheres, as shown in Figure 7-7. The force of attraction between the large and the small spheres caused the rod to rotate. When the force required to twist the wire equaled the gravitational force between the spheres, the rod stopped rotating", ". By measuring the angle through which the rod turned, Cavendish was able to calculate the attractive force between the objects. The angle through which the rod turned is measured using the beam of light that is reflected from the mirror. He measured the masses of the spheres and the distance between their centers. Substituting these values for force, mass, and distance into Newton\u2019s law of universal gravitation, he found an experimental value for G: when m1 and m2 are measured in kilograms, r in meters, and F in newtons, then G 6.671011 Nm2/kg2. \u25a0 Figure 7-6 Modern Cavendish balances are used to measure the gravitational forces between two objects. Large lead sphere Small lead sphere Mirror Direction of rotation \u25a0 Figure 7-7 When the large lead spheres are placed near the small lead spheres, the gravitational attraction between the spheres causes the rod to rotate. The rotation is measured with the help of the reflected light ray. Section 7.1 Planetary Motion and Gravitation 177 courtesy of PASCO Scientific 170-178 CH07 S1-845813 6/3/04 10:27 AM Page 178 The importance of G Cavendish\u2019s experiment often is called \u201cweighing Earth,\u201d because his experiment helped determine Earth\u2019s mass. Once the value of G is known, not only the mass of Earth, but also the mass of the Sun can be determined. In addition, the gravitational force between any two objects can be calculated using Newton\u2019s law of universal gravitation. For example, the attractive gravitational force, Fg, between two bowling balls of mass 7.26 kg, with their centers separated by 0.30 m, can be calculated as follows: Fg (6.671011 Nm2/kg2)(7.26 kg)(7.26 kg) (0.30 m)2 3.9108 N You know that on Earth\u2019s surface, the weight of an object of mass m is a mg. If Earth\u2019s mass is repmeasure of Earth\u2019s gravitational attraction: Fg resented by mE and Earth\u2019s radius is represented by rE, the following is true: Fg m m m mg, and so g G G E E 2 2 rE rE This equation can be rearranged to solve for mE. 2 gr E G mE 6.38106 m, g 9.80 m/s2, and G 6.671011 Nm2/kg2, Using rE the following result is obtained for Earth\u2019s mass: mE (9.80 m/s2)(6.38106 m)2 6.671011 Nm/kg2 5.981024 kg When you compare the mass of Earth to that of a bowling ball, you can see why the gravitational attraction between everyday objects is not easily observed. Cavendish\u2019s experiment determined the value of G, confirmed Newton\u2019s prediction that a gravitational force exists between two objects, and helped calculate the mass of Earth. 7.1 Section Review 6. Neptune\u2019s Orbital Period Neptune orbits the Sun with an orbital radius of 4.4951012 m, which allows gases, such as methane, to condense and form an atmosphere, as shown in Figure 7-8. If the mass of the Sun is 1.991030 kg, calculate the period of Neptune\u2019s orbit. \u25a0 Figure 7-8 7. Gravity If Earth began to shrink, but its mass remained the same, what would happen to the value of g on Earth\u2019s surface? 8. Gravitational Force What is the gravitational force between two 15-kg packages that are 35 cm apart? What fraction is this of the weight of one package? 9. Universal Gravitational Constant Cavendish did his experiment using lead spheres. Suppose he had replaced the lead spheres with copper spheres of equal mass. Would his value of G be the same or different? Explain. 10. Laws or Theories? Kepler\u2019s three statements and Newton\u2019s equation for gravitational attraction are called \u201claws.\u201d Were they ever theories? Will they ever become theories? 11. Critical Thinking Picking up a rock requires less effort on the Moon than on Earth. a. How will the weaker gravitational force on the Moon\u2019s surface affect the path of the rock if it is thrown horizontally? b. If the thrower accidentally drops the rock on her toe, will it hurt more or less than it would on Earth? Explain. 178 Photodisc/Artbase Chapter 7 Gravitation physicspp.com/self_check_quiz 179-185 CH07 S2-845813 3/5/04 11:55 AM Page 179 Objectives \u2022 Solve orbital motion problems. \u2022 Relate weightlessness to objects in free fall. \u2022 Describe gravitational fields. \u2022 Compare views on gravitation. Vocabulary gravitational field inertial", " mass gravitational mass 7.2 Using the Law of Universal Gravitation The planet Uranus was discovered in 1781. By 1830, it was clear that the law of gravitation didn\u2019t correctly predict its orbit. Two astronomers proposed that Uranus was being attracted by the Sun and by an undiscovered planet. They calculated the orbit of such a planet in 1845, and, one year later, astronomers at the Berlin Observatory found the planet now called Neptune. How do planets, such as Neptune, orbit the Sun? Orbits of Planets and Satellites Newton used a drawing similar to the one shown in Figure 7-9 to illustrate a thought experiment on the motion of satellites. Imagine a cannon, perched high atop a mountain, firing a cannonball horizontally with a given horizontal speed. The cannonball is a projectile, and its motion has both vertical and horizontal components. Like all projectiles on Earth, it would follow a parabolic trajectory and fall back to the ground. If the cannonball\u2019s horizontal speed were increased, it would travel farther across the surface of Earth, and still fall back to the ground. If an extremely powerful cannon were used, however, the cannonball would travel all the way around Earth, and keep going. It would fall toward Earth at the same rate that Earth\u2019s surface curves away. In other words, the curvature of the projectile would continue to just match the curvature of Earth, so that the cannonball would never get any closer or farther away from Earth\u2019s curved surface. The cannonball would, therefore, be in orbit. Newton\u2019s thought experiment ignored air resistance. For the cannonball to be free of air resistance, the mountain on which the cannon is perched would have to be more than 150 km above Earth\u2019s surface. By way of comparison, the mountain would have to be much taller than the peak of Mount Everest, the world\u2019s tallest mountain, which is only 8.85 km in height. A cannonball launched from a mountain that is 150 km above Earth\u2019s surface would encounter little or no air resistance at an altitude of 150 km, because the mountain would be above most of the atmosphere. Thus, a cannonball or any object or satellite at or above this altitude could orbit Earth for a long time. v1 v3 v2 \u25a0 Figure 7-9 The horizontal speed v1 is not high enough and the cannonball falls to the ground. With a higher speed, v2, the cannonball travels farther. The cannonball travels all the way around Earth when the horizontal speed, v3, is high enough. Section 7.2 Using the Law of Universal Gravitation 179 179-185 CH07 S2-845813 3/5/04 11:58 AM Page 180 Geosynchronous Orbit The GOES-12 weather satellite orbits Earth once a day at an altitude of 35,785 km. The orbital speed of the satellite matches Earth\u2019s rate of rotation. Thus, to an observer on Earth, the satellite appears to remain above one spot on the equator. Satellite dishes on Earth can be directed to one point in the sky and not have to change position as the satellite orbits. A satellite in an orbit that is always the same height above Earth moves in uniform circular motion. Recall that its centripetal acceleration is given mac, can thus be rewritten as v2/r. Newton\u2019s second law, Fnet by ac mv2/r. If Earth\u2019s mass is mE, then this expression combined with Fnet Newton\u2019s law of universal gravitation produces the following equation: m m v2 m G E 2 r r Solving for the speed of a satellite in circular orbit about Earth, v, yields the following. E Speed of a Satellite Orbiting Earth v Gm r The speed of a satellite orbiting Earth is equal to the square root of the universal gravitational constant times the mass of Earth, divided by the radius of the orbit. A satellite\u2019s orbital period A satellite\u2019s orbit around Earth is similar to a planet\u2019s orbit about the Sun. Recall that the period of a planet orbiting the Sun is expressed by the following equation: T 2 3 r G m S Thus, the period for a satellite orbiting Earth is given by the following equation. Period of a Satellite Orbiting Earth T 2 The period for a satellite orbiting Earth is equal to 2times the square root of the radius of the orbit cubed, divided by the product of the universal gravitational constant and the mass of Earth. 3 r m G E The equations for the speed and period of a satellite can be used for any object in orbit about another. The mass of the central body will replace mE in the equations, and r will be the distance between the centers of the orbiting body and the central body. If the mass of the central body is much greater than the mass of the orbiting body, then r is equal to the distance between the centers of the orbiting body", " and the central body. Orbital speed, v, and period, T, are independent of the mass of the satellite. Are there any factors that limit the mass of a satellite? A satellite\u2019s mass Landsat 7, shown in Figure 7-10, is an artificial satellite that provides images of Earth\u2019s continental surfaces. Landsat images have been used to create maps, study land use, and monitor resources and global changes. The Landsat 7 system enables researchers to monitor small-scale processes, such as deforestation, on a global scale. Satellites, such as Landsat 7, are accelerated to the speeds necessary for them to achieve orbit by large rockets, such as shuttle-booster rockets. Because the acceleration of any mass must follow Newton\u2019s second law of motion, ma, more force is required to launch a more massive satellite into Fnet orbit. Thus, the mass of a satellite is limited by the capability of the rocket used to launch it. \u25a0 Figure 7-10 Landsat 7, a remote sensing satellite, has a mass of about 2200 kg and orbits Earth at an altitude of about 705 km. Geology Connection Geology Connection 180 Russ Underwood, Lockheed Martin Space Systems/NASA Chapter 7 Gravitation 179-185 CH07 S2-845813 7/20/04 4:53 AM Page 181 Orbital Speed and Period Assume that a satellite orbits Earth 225 km above its surface. Given that the mass of Earth is 5.971024 kg and the radius of Earth is 6.38106 m, what are the satellite\u2019s orbital speed and period? 1 Analyze and Sketch the Problem \u2022 Sketch the situation showing the height of the satellite\u2019s orbit. Known: h 2.25105 m 6.38106 m rE 5.971024 kg mE G 6.671011 Nm2/kg2 Unknown: v? T? Satellite rE h Earth r 2 Solve for the Unknown Determine the orbital radius by adding the height of the satellite\u2019s orbit to Earth\u2019s radius. r h rE 2.25105 m + 6.38106 m 6.61106 m Substitute h 2.25105 m, rE 6.38106 m Solve for the speed. Gm r v E (6.671011 Nm2/kg2)(5.971024 kg) 6.61106 m 7.76103 m/s Solve for the period. T 2 2 (6.61106 m)3 (6.671011 Nm2/kg2)(5.971024 kg) 5.35103 s 3 r m G E Substitute G 6.671011 Nm2/kg2, 5.971024 kg, r 6.61106 m mE Math Handbook Square and Cube Roots pages 839\u2013840 Substitute r 6.61106 m, G 6.671011 Nm2/kg2, mE 5.971024 kg This is approximately 89 min, or 1.5 h. 3 Evaluate the Answer \u2022 Are the units correct? The unit for speed is m/s and the unit for period is s. For the following problems, assume a circular orbit for all calculations. 12. Suppose that the satellite in Example Problem 2 is moved to an orbit that is 24 km larger in radius than its previous orbit. What would its speed be? Is this faster or slower than its previous speed? 13. Use Newton\u2019s thought experiment on the motion of satellites to solve the following. a. Calculate the speed that a satellite shot from a cannon must have to orbit Earth 150 km above its surface. b. How long, in seconds and minutes, would it take for the satellite to complete one orbit and return to the cannon? 14. Use the data for Mercury in Table 7-1 on page 173 to find the following. a. the speed of a satellite that is in orbit 260 km above Mercury's surface b. the period of the satellite Section 7.2 Using the Law of Universal Gravitation 181 179-185 CH07 S2-845813 6/3/04 10:31 AM Page 182 Acceleration Due To Gravity The acceleration of objects due to Earth\u2019s gravity can be found by using Newton\u2019s law of universal gravitation and his second law of motion. For a free-falling object, m, the following is true: m m m F G ma, so a G E E r2 2 r Because a g and r rE on Earth\u2019s surface, the following equation can be written: m g G E, thus, mE 2 rE m You found above that a G E for a free-falling object. Substituting the r2 above expression for mE yields the following: 2 gr E G 2 gr E G r2 E2 r r a G a g This shows that as you move farther from Earth\ufffd", "\ufffds center, that is, as r becomes larger, the acceleration due to gravity is reduced according to this inverse square relationship. What happens to your weight, mg, as you move farther and farther from Earth\u2019s center? Weight and weightlessness You probably have seen photos similar to the one in Figure 7-11 in which astronauts are on the space shuttle in an environment often called \u201czero-g\u201d or \u201cweightlessness.\u201d The shuttle orbits about 400 km above Earth\u2019s surface. At that distance, g 8.7 m/s2, only slightly less than on Earth\u2019s surface. Thus, Earth\u2019s gravitational force is certainly not zero in the shuttle. In fact, gravity causes the shuttle to orbit Earth. Why, then, do the astronauts appear to have no weight? Remember that you sense weight when something, such as the floor or your chair, exerts a contact force on you. But if you, your chair, and the floor all are accelerating toward Earth together, then no contact forces are exerted on you. Thus, your apparent weight is zero and you experience weightlessness. Similarly, the astronauts experience weightlessness as the shuttle and everything in it falls freely toward Earth. The Gravitational Field Recall from Chapter 6 that many common forces are contact forces. Friction is exerted where two objects touch, for example, when the floor and your chair or desk push on you. Gravity, however, is different. It acts on an apple falling from a tree and on the Moon in orbit. It even acts on you in midair as you jump up or skydive. In other words, gravity acts over a distance. It acts between objects that are not touching or that are not close together. Newton was puzzled by this concept. He wondered how the Sun could exert a force on planet Earth, which is hundreds of millions of kilometers away. Weightless Water This activity is best done outdoors. Use a pencil to poke two holes through a foam or paper cup: one on the bottom and the other on the side. Hold your fingers over the two holes to block them as your lab partner pours colored water into the cup until it is two-thirds full. 1. Predict what will happen as the cup is allowed to fall. 2. Test your prediction: drop the cup and watch closely. Analyze and Conclude 3. Describe your observations. 4. Explain your results. \u25a0 Figure 7-11 Astronaut Chiaki Mukai experiences weightlessness on board the space shuttle Columbia, as the shuttle and everything in it falls freely toward Earth. 182 NASA Chapter 7 Gravitation 179-185 CH07 S2-845813 6/3/04 10:32 AM Page 183 The answer to the puzzle arose from a study of magnetism. In the 19th century, Michael Faraday developed the concept of a field to explain how a magnet attracts objects. Later, the field concept was applied to gravity. Any object with mass is surrounded by a gravitational field in which another object experiences a force due to the interaction between its mass and the gravitational field, g, at its location. This is expressed by the following equation. M Gravitational Field g G 2 r The gravitational field is equal to the universal gravitational constant times the object\u2019s mass, divided by the square of the distance from the object\u2019s center. The direction is toward the mass\u2019s center. Suppose the gravitational field is created by the Sun. Then a planet of mass m has a force exerted on it that depends on its mass and the magnitude of the gravitational field at its location. That is, F mg, toward the Sun. The force is caused by the interaction of the planet\u2019s mass with the gravitational field at its location, not with the Sun millions of kilometers away. To find the gravitational field caused by more than one object you would calculate both gravitational fields and add them as vectors. The gravitational field can be measured by placing an object with a small mass, m, in the gravitational field and measuring the force, F, on it. The gravitational field can be calculated using g F/m. The gravitational field is measured in N/kg, which is also equal to m/s2. On Earth\u2019s surface, the strength of the gravitational field is 9.80 N/kg, and its direction is toward Earth\u2019s center. The field can be represented by a vector of length g pointing toward the center of the object producing the field. You can picture the gravitational field of Earth as a collection of vectors surrounding Earth and pointing toward it, as shown in Figure 7-12. The strength of the field varies inversely with the square of the distance from the center of Earth. The gravitational field depends on Earth\u2019s mass, but not on the mass of the object experiencing it. Two Kinds of Mass Recall that when the concept of mass was discussed in Chapter 4, it was defined as the slope of a graph of force versus acceleration. That is, mass is", " equal to the ratio of the net force exerted on an object to its acceleration. This kind of mass, related to the inertia of an object, is called inertial mass and is represented by the following equation. Inertial Mass minertial Inertial mass is equal to the net force exerted on the object divided by the acceleration of the object. Fn et a The inertial mass of an object is measured by exerting a force on the object and measuring the object\u2019s acceleration using an inertial balance, such as the one shown in Figure 7-13. The more inertial mass an object has, the less it is affected by any force\u2014the less acceleration it undergoes. Thus, the inertial mass of an object is a measure of the object\u2019s resistance to any type of force. \u25a0 Figure 7-12 Vectors representing Earth\u2019s gravitational field all point toward Earth\u2019s center. The field is weaker farther from Earth. \u25a0 Figure 7-13 An inertial balance allows you to calculate the inertial mass of an object from the period (T ) of the back-and-forth motion of the object. Calibration masses, such as the cylindrical ones shown here, are used to create a graph of T 2 versus the mass. The period of the unknown mass is then measured, and the inertial mass is determined from the calibration graph. Section 7.2 Using the Law of Universal Gravitation 183 courtesy of PASCO Scientific 179-185 CH07 S2-845813 3/5/04 12:00 PM Page 184 Newton\u2019s law of universal gravitation, F = Gm1m2/r2, also involves mass, but a different kind of mass. Mass as used in the law of universal gravitation determines the size of the gravitational force between two objects and is called gravitational mass. It can be measured using a simple balance, such as the one shown in Figure 7-14. If you measure the attractive force exerted on an object by another object of mass, m, at a distance, r, then you can define the gravitational mass in the following way. r2 F grav Gravitational Mass mgrav G m The gravitational mass of an object is equal to the distance between the objects squared, times the gravitational force, divided by the product of the universal gravitational constant, times the mass of the other object. How different are these two kinds of mass? Suppose you have a watermelon in the trunk of your car. If you accelerate the car forward, the watermelon will roll backwards, relative to the trunk. This is a result of its inertial mass\u2014its resistance to acceleration. Now, suppose your car climbs a steep hill at a constant speed. The watermelon will again roll backwards. But this time, it moves as a result of its gravitational mass. The watermelon is being attracted downward toward Earth. Newton made the claim that inertial mass and gravitational mass are equal in magnitude. This hypothesis is called the principle of equivalence. All experiments conducted so far have yielded data that support this principle. Albert Einstein also was intrigued by the principle of equivalence and made it a central point in his theory of gravity. Einstein\u2019s Theory of Gravity Newton\u2019s law of universal gravitation allows us to calculate the gravitational force that exists between two objects because of their masses. The concept of a gravitational field allows us to picture the way gravity acts on objects that are far away. Einstein proposed that gravity is not a force, but rather, an effect of space itself. According to Einstein, mass changes the space around it. Mass causes space to be curved, and other bodies are accelerated because of the way they follow this curved space. \u25a0 Figure 7-14 The platform balance shown here allows you to measure the force on a mass due to Earth\u2019s gravity. \u25a0 Figure 7-15 Matter causes space to curve just as an object on a rubber sheet curves the sheet around it. Moving objects near the mass follow the curvature of space. The red ball is moving clockwise around the center mass. 184 (t)Horizons Companies, (b)Ted Kinsman/Photo Researchers Chapter 7 Gravitation 179-185 CH07 S2-845813 6/3/04 10:33 AM Page 185 One way to picture how space is affected by mass is to compare space to a large, two-dimensional rubber sheet, as shown in Figure 7-15. The yellow ball on the sheet represents a massive object. It forms an indentation. A red ball rolling across the sheet simulates the motion of an object in space. If the red ball moves near the sagging region of the sheet, it will be accelerated. In the same way, Earth and the Sun are attracted to one another because of the way space is distorted by the two objects. Einstein\u2019s theory, called the general theory of relativity, makes many predictions about how massive objects affect one another. In every test conducted to date, Einstein\u2019", "s theory has been shown to give the correct results. Deflection of light Einstein\u2019s theory predicts the deflection or bending of light by massive objects. Light follows the curvature of space around the massive object and is deflected, as shown in Figure 7-16. In 1919, during an eclipse of the Sun, astronomers found that light from distant stars that passed near the Sun was deflected in agreement with Einstein\u2019s predictions. Another result of general relativity is the effect on light from very massive objects. If an object is massive and dense enough, the light leaving it will be totally bent back to the object. No light ever escapes the object. Objects such as these, called black holes, have been identified as a result of their effects on nearby stars. The radiation produced when matter is pulled into black holes has also been helpful in their detection. While Einstein\u2019s theory provides very accurate predictions of gravity\u2019s effects, it is still incomplete. It does not explain the origin of mass or how mass curves space. Physicists are working to understand the deeper meaning of gravity and the origin of mass itself. \u25a0 Figure 7-16 The light from a distant star bends due to the Sun\u2019s gravitational field, thereby changing the apparent position of the star. (Illustration not to scale) Reference star Star Actual position Apparent position Sun Moon Earth 7.2 Section Review 15. Gravitational Fields The Moon is 3.9105 km from Earth\u2019s center and 1.5108 km from the Sun\u2019s center. The masses of Earth and the Sun are 6.01024 kg and 2.01030 kg, respectively. a. Find the ratio of the gravitational fields due to Earth and the Sun at the center of the Moon. b. When the Moon is in its third quarter phase, as shown in Figure 7-17, its direction from Earth is at right angles to the Sun\u2019s direction. What is the net gravitational field due to the Sun and Earth at the center of the Moon? \u25a0 Figure 7-17 (Not to scale) Moon Sun Earth 16. Gravitational Field The mass of the Moon is 7.31022 kg and its radius is 1785 km. What is the strength of the gravitational field on the surface of the Moon? 17. A Satellite\u2019s Mass When the first artificial satellite was launched into orbit by the former Soviet Union in 1957, U.S. president Dwight D. Eisenhower asked his scientific advisors to calculate the mass of the satellite. Would they have been able to make this calculation? Explain. 18. Orbital Period and Speed Two satellites are in circular orbits about Earth. One is 150 km above the surface, the other 160 km. a. Which satellite has the larger orbital period? b. Which one has the greater speed? 19. Theories and Laws Why is Einstein\u2019s description of gravity called a \u201ctheory,\u201d while Newton\u2019s is a \u201claw?\u201d 20. Weightlessness Chairs in an orbiting spacecraft are weightless. If you were on board such a spacecraft and you were barefoot, would you stub your toe if you kicked a chair? Explain. 21. Critical Thinking It is easier to launch a satellite from Earth into an orbit that circles eastward than it is to launch one that circles westward. Explain. physicspp.com/self_check_quiz physicspp.com/self_check_quiz Section 7.2 Using the Law of Universal Gravitation 185 186-187 CH07-LAB-845813 3/19/04 2:04 PM Page 186 Modeling the Orbits of Planets and Satellites In this experiment, you will analyze a model that will show how Kepler\u2019s first and second laws of motion apply to orbits of objects in space. Kepler\u2019s first law states that orbits of planets are ellipses, with the Sun at one focus. Kepler\u2019s second law states that an imaginary line from the Sun to a planet sweeps out equal areas in equal time intervals. The shape of the elliptical orbit is defined by eccentricity, e, which is the ratio of the distance between the foci to the length of the major axis. When an object is at its farthest distance from the Sun along the major axis, it is at aphelion. When the object is at its closest distance from the Sun along the major axis, it is at perihelion. QUESTION What is the shape of the orbits of planets and satellites in the solar system? Objectives Procedure \u25a0 Formulate models to infer the shape of orbits 1. Place a piece of paper on a piece of cardboard of planets and satellites. \u25a0 Collect and organize data for aphelion distances and perihelion distances of objects as they orbit the Sun. \u25a0 Draw conclusions about Kepler\u2019s first and second laws of motion. Safety Precautions \u25a0 Pins are sharp and can puncture the skin. Materials piece of cardboard sheet of blank,", " white paper two push pins metric ruler sharp pencil or pen four small pieces of tape string (25 cm) using tape at the four corners. 2. Draw a line across the center of the paper, along the length of the paper. This line represents the major axis. 3. Mark the center of the line and label it C. 4. Use the string to tie a loop, which, when stretched, has a length of 10 cm. For each object listed in the data table, calculate the distance between the foci, d, using the following equation: cm).0 10 d 2e( 1 e 5. For the circle, place a pin at C. Put the loop of string over the pin and pull it tight with your pencil. Move the pencil in a circular fashion around the center, letting the string guide it. 6. For the next object, place one pin a distance of d/2 from C along the major axis. 7. Place a second pin a distance of d/2 on the opposite side of C. The two pins represent the foci. One focus is the location of the Sun. 8. Put the loop of string over both pins and pull it tight with your pencil. Move the pencil in a circular fashion, letting the string guide it. 9. Using the same paper, repeat steps 6-8 for each of the listed objects. 10. After all of the orbits are plotted, label each orbit with the name and eccentricity of the object plotted. 186 Horizons Companies 186-187 CH07-LAB-845813 6/3/04 10:35 AM Page 187 Data Table Object Circle Earth Pluto Comet Eccentricity (e) d (cm) Measured A Measured P Experimental e % Error 0 0.017 0.25 0.70 Analyze 1. Measure the aphelion distance, A, by measuring the distance between one focus and the farthest point in the orbit along the major axis. Record your data in the data table. 2. Measure the perihelion distance, P, by measuring the closest distance between one focus and the closest point in the orbit along the major axis. Record the data in the data table. 3. Calculate the experimental eccentricity for each of the objects and record your data in the data table. Use the following equation: P e A P A 4. Error Analysis Calculate the percent error for each object using the experimental eccentricities compared to the known eccentricities. Record your values in the data table. 5. Analyze Why is the shape of the orbit with e 0 a circle? 6. Compare How does Earth\u2019s orbit compare to a circle? 7. Observe Which of the orbits truly looks elliptical? Conclude and Apply 1. Does the orbit model you constructed obey Kepler\u2019s first law? Explain. 2. Kepler studied the orbit data of Mars (e 0.093) and concluded that planets move about the Sun in elliptical orbits. What would Kepler have concluded if he had been on Mars and studied Earth\u2019s orbit? 3. Where does a planet travel fastest: at aphelion or perihelion? Why? 4. Kepler\u2019s second law helps to determine the ratio between Pluto\u2019s velocity at aphelion and perihelion (vA/vP). To determine this ratio, first calculate the area swept out by Pluto\u2019s orbit. This area is approximately equal to the area of a triangle: Area 1 (distance to the Sun) 2 current velocity time. If the area that the orbit sweeps out in a fixed amount of time, such as 30 days, is the same at aphelion and perihelion, this relationship can be written P vPt 1 1 AvAt 2 2 What is the ratio vP /vA for Pluto? 5. Pluto\u2019s minimum orbital velocity is 3.7 km/s. What are the values for vP and vA? Going Further 1. You used rough approximations to look at Kepler\u2019s second law. Suggest an experiment to obtain precise results to confirm the second law. 2. Design an experiment to verify Kepler\u2019s third law. Real-World Physics Does a communications or weather satellite that is orbiting Earth follow Kepler\u2019s laws? Collect data to verify your answer. To find out more about gravitation, visit the Web site: physicspp.com 187 188 CH07-FEATURE-845813 3/5/04 12:20 PM Page 188 Black Holes What would happen if you were to travel to a black hole? Your body would be stretched, flattened, and eventually pulled apart. What is a black hole? What is known about black holes? A black hole is one of the possible final stages in the evolution of a star. When fusion reactions stop in the core of a star that is at least 20 times more massive than the Sun, the core collapses forever, compacting matter into an increasingly smaller volume. The infinitely", " small, but infinitely dense, object that remains is called a singularity. The force of gravity is so immense in the region around the singularity that nothing, not even light, can escape it. This region is called a black hole. Nothing Can Escape In 1917, German mathematician Karl Schwarzschild verified, mathematically, that black holes could exist. Schwarzschild used solutions to Einstein\u2019s theory of general relativity to describe the properties of black holes. He derived an expression for a radius, called the Schwarzschild radius, within which neither light nor matter escapes the force of gravity of the singularity. The Schwarzschild radius is represented by the following equation: Rs M 2G c2 mass can be determined by the gravitational field it generates. Mass is calculated by using a modified form of Kepler\u2019s third law of planetary motion. Studies using NASA\u2019s Rossi X-ray Timing Explorer have shown that black holes spin just as stars and planets do. A black hole spins because it retains the angular momentum of the star that formed it. Even though a black hole\u2019s electric charge has not been measured, scientists hypothesize that a black hole may become charged when an excess of one type of electric charge falls into it. Super-heated gases in a black hole emit X rays, which can be detected by X-ray telescopes, such as the space-based Chandra X-ray Observatory. Although not everything is known about black holes, there is direct and indirect evidence of their existence. Continued research and special missions will provide a better understanding of black holes. Hubble visible image of galaxy NGC 6240. Chandra X-ray image of two black holes (blue) in NGC 6240. In this equation, G is Newton\u2019s universal gravitational constant, M is the mass of the black hole, and c is the speed of light. The edge of the sphere defined by the Schwarzschild radius is called the event horizon. At the event horizon, the escape velocity equals the speed of light. Because nothing travels faster than the speed of light, objects that cross the event horizon can never escape. Indirect and Direct Evidence Black holes have three physical properties that can theoretically be measured\u2014mass, angular momentum, and electric charge. A black hole\u2019s 188 NASA Extreme Physics Going Further Solve The escape velocity of an object leaving the event horizon can be represented by the following equation: M v 2G Rs In this equation, G is Newton\u2019s universal gravitational constant, M is the mass of the black hole, and Rs is the radius of the black hole. Show that the escape velocity equals the speed of light. 189-195 CH07-SG CA STP-845813 3/5/04 12:24 PM Page 189 7.1 Planetary Motion and Gravitation Vocabulary \u2022 Kepler\u2019s first law (p. 172) \u2022 Kepler\u2019s second law (p. 173) \u2022 Kepler\u2019s third law (p. 173) \u2022 gravitational force (p. 175) \u2022 law of universal gravitation (p. 175) Key Concepts \u2022 Kepler\u2019s first law states that planets move in elliptical orbits, with the Sun at one focus. \u2022 Kepler\u2019s second law states that an imaginary line from the Sun to a planet sweeps out equal areas in equal times. \u2022 Kepler\u2019s third law states that the square of the ratio of the periods of any two planets is equal to the cube of the ratio of their distances from the Sun \u2022 Newton\u2019s law of universal gravitation states that the gravitational force between any two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The force is attractive and along a line connecting their centers. m2 m1 F G r2 \u2022 Newton\u2019s law of universal gravitation can be used to rewrite Kepler\u2019s third law to relate the radius and period of a planet to the mass of the Sun. r3 T2 4 2 m G S 7.2 Using the Law of Universal Gravitation Vocabulary \u2022 gravitational field (p. 183) \u2022 inertial mass (p. 183) \u2022 gravitational mass (p. 184) Key Concepts \u2022 The speed of an object in circular orbit is given by the following expression. E v Gm r \u2022 The period of a satellite in a circular orbit is given by the following expression. T 2 3 r m G E \u2022 All objects have gravitational fields surrounding them. m g G 2 r \u2022 Gravitational mass and inertial mass are two essentially different concepts. The gravitational and inertial masses of an object, however, are numerically equal. minertial Fn et a mgrav r2 F grav G m \u2022 Einstein\u2019s general theory of relativity describes gravitational attraction as a property of space itself. physicspp.com/vocabulary_puzzlemaker 189 189-195 CH07-SG CA STP-845813 3/5/04 1:39 PM Page 190 Concept Mapping 22. Create", " a concept map using these terms: planets, stars, Newton\u2019s law of universal gravitation, Kepler\u2019s first law, Kepler\u2019s second law, Kepler\u2019s third law, Einstein\u2019s general theory of relativity. Mastering Concepts 23. In 1609, Galileo looked through his telescope at Jupiter and saw four moons. The name of one of the moons that he saw is Io. Restate Kepler\u2019s first law for Io and Jupiter. (7.1) 24. Earth moves more slowly in its orbit during summer in the northern hemisphere than it does during winter. Is it closer to the Sun in summer or in winter? (7.1) 25. Is the area swept out per unit of time by Earth moving around the Sun equal to the area swept out per unit of time by Mars moving around the Sun? (7.1) 26. Why did Newton think that a force must act on the Moon? (7.1) 27. How did Cavendish demonstrate that a gravitational force of attraction exists between two small objects? (7.1) 28. What happens to the gravitational force between two masses when the distance between the masses is doubled? (7.1) 29. According to Newton\u2019s version of Kepler\u2019s third law, how would the ratio T 2/r 3 change if the mass of the Sun were doubled? (7.1) 30. How do you answer the question, \u201cWhat keeps a satellite up?\u201d (7.2) 31. A satellite is orbiting Earth. On which of the following does its speed depend? (7.2) a. mass of the satellite b. distance from Earth c. mass of Earth 32. What provides the force that causes the centripetal acceleration of a satellite in orbit? (7.2) 33. During space flight, astronauts often refer to forces as multiples of the force of gravity on Earth\u2019s surface. What does a force of 5g mean to an astronaut? (7.2) 34. Newton assumed that a gravitational force acts directly between Earth and the Moon. How does Einstein\u2019s view of the attractive force between the two bodies differ from Newton\u2019s view? (7.2) 35. Show that the dimensions of g in the equation g F/m are in m/s2. (7.2) 36. If Earth were twice as massive but remained the same size, what would happen to the value of g? (7.2) Applying Concepts 37. Golf Ball The force of gravity acting on an object near Earth\u2019s surface is proportional to the mass of the object. Figure 7-18 shows a tennis ball and golf ball in free fall. Why does a tennis ball not fall faster than a golf ball? \u25a0 Figure 7-18 38. What information do you need to find the mass of Jupiter using Newton\u2019s version of Kepler\u2019s third law? 39. The mass of Pluto was not known until a satellite of the planet was discovered. Why? 40. Decide whether each of the orbits shown in Figure 7-19 is a possible orbit for a planet. c Sun c Sun c Sun c Sun \u25a0 Figure 7-19 Barry Runk/Grant Heilman 190 Chapter 7 Gravitation For more problems, go to Additional Problems, Appendix B. 189-195 CH07-SG CA STP-845813 3/5/04 12:25 PM Page 191 41. The Moon and Earth are attracted to each other by gravitational force. Does the more-massive Earth attract the Moon with a greater force than the Moon attracts Earth? Explain. 42. What would happen to the value of G if Earth were twice as massive, but remained the same size? 48. If a mass in Earth\u2019s gravitational field is doubled, what will happen to the force exerted by the field upon the mass? 49. Weight Suppose that yesterday your body had a mass of 50.0 kg. This morning you stepped on a scale and found that you had gained weight. 43. Figure 7-20 shows a satellite orbiting Earth. a. What happened, if anything, to your mass? Examine the equation v E, relating the speed Gm r of an orbiting satellite and its distance from the center of Earth. Does a satellite with a large or small orbital radius have the greater velocity? Satellite r Earth \u25a0 Figure 7-20 (Not to scale) 44. Space Shuttle If a space shuttle goes into a higher orbit, what happens to the shuttle\u2019s period? 45. Mars has about one-ninth the mass of Earth. Figure 7-21 shows satellite M, which orbits Mars with the same orbital radius as satellite E, which orbits Earth. Which satellite has a smaller period? Mars rM M Earth rE E \u25a0 Figure 7-21 (Not to scale) b. What happened, if anything, to the ratio of your weight to your mass? 50. As", " an astronaut in an orbiting space shuttle, how would you go about \u201cdropping\u201d an object down to Earth? 51. Weather Satellites The weather pictures that you see every day on TV come from a spacecraft in a stationary position relative to the surface of Earth, 35,700 km above Earth\u2019s equator. Explain how it can stay in exactly the same position day after day. What would happen if it were closer? Farther out? Hint: Draw a pictorial model. Mastering Problems 7.1 Planetary Motion and Gravitation 52. Jupiter is 5.2 times farther from the Sun than Earth is. Find Jupiter\u2019s orbital period in Earth years. 53. Figure 7-22 shows a Cavendish apparatus like the one used to find G. It has a large lead sphere that is 5.9 kg in mass and a small one with a mass of 0.047 kg. Their centers are separated by 0.055 m. Find the force of attraction between them. 5.9 kg 0.055 m 0.047 kg 0.047 kg 0.055 m 5.9 kg \u25a0 Figure 7-22 46. Jupiter has about 300 times the mass of Earth and about ten times Earth\u2019s radius. Estimate the size of g on the surface of Jupiter. 47. A satellite is one Earth radius above the surface of Earth. How does the acceleration due to gravity at that location compare to acceleration at the surface of Earth? 54. Use Table 7-1 on p. 173 to compute the gravitational force that the Sun exerts on Jupiter. 55. Tom has a mass of 70.0 kg and Sally has a mass of 50.0 kg. Tom and Sally are standing 20.0 m apart on the dance floor. Sally looks up and sees Tom. She feels an attraction. If the attraction is gravitational, find its size. Assume that both Tom and Sally can be replaced by spherical masses. physicspp.com/chapter_test Chapter 7 Assessment 191 189-195 CH07-SG CA STP-845813 6/3/04 10:41 AM Page 192 56. Two balls have their centers 2.0 m apart, as shown in Figure 7-23. One ball has a mass of 8.0 kg. The other has a mass of 6.0 kg. What is the gravitational force between them? 67. Mimas, one of Saturn\u2019s moons, has an orbital radius of 1.87108 m and an orbital period of about 23.0 h. Use Newton\u2019s version of Kepler\u2019s third law to find Saturn\u2019s mass. 2.0 m 6.0 kg 8.0 kg \u25a0 Figure 7-23 57. Two bowling balls each have a mass of 6.8 kg. They are located next to each other with their centers 21.8 cm apart. What gravitational force do they exert on each other? 58. Assume that you have a mass of 50.0 kg. Earth has a mass of 5.971024 kg and a radius of 6.38106 m. a. What is the force of gravitational attraction between you and Earth? b. What is your weight? 59. The gravitational force between two electrons that are 1.00 m apart is 5.541071 N. Find the mass of an electron. 60. A 1.0-kg mass weighs 9.8 N on Earth\u2019s surface, and the radius of Earth is roughly 6.4106 m. a. Calculate the mass of Earth. b. Calculate the average density of Earth. 61. Uranus Uranus requires 84 years to circle the Sun. Find Uranus\u2019s orbital radius as a multiple of Earth\u2019s orbital radius. 62. Venus Venus has a period of revolution of 225 Earth days. Find the distance between the Sun and Venus as a multiple of Earth\u2019s orbital radius. 63. If a small planet, D, were located 8.0 times as far from the Sun as Earth is, how many years would it take the planet to orbit the Sun? 64. Two spheres are placed so that their centers are 2.6 m apart. The force between the two spheres is 2.751012 N. What is the mass of each sphere if one sphere is twice the mass of the other sphere? 65. The Moon is 3.9105 km from Earth\u2019s center and 1.5108 km from the Sun\u2019s center. If the masses of the Moon, Earth, and the Sun are 7.31022 kg, 6.01024 kg, and 2.01030 kg, respectively, find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon. 68. The Moon is 3.9108 m away from Earth and has a period of 27.33 days. Use Newton\u2019s version of Kepler\u2019s third law to find the mass of Earth. Compare this mass to the mass found in problem", " 60. 69. Halley\u2019s Comet Every 74 years, comet Halley is visible from Earth. Find the average distance of the comet from the Sun in astronomical units (AU). 70. Area is measured in m2, so the rate at which area is swept out by a planet or satellite is measured in m2/s. a. How quickly is an area swept out by Earth in its orbit about the Sun? b. How quickly is an area swept out by the Moon in its orbit about Earth? Use 3.9108 m as the average distance between Earth and the Moon, and 27.33 days as the period of the Moon. 7.2 Using the Law of Universal Gravitation 71. Satellite A geosynchronous satellite is one that appears to remain over one spot on Earth, as shown in Figure 7-24. Assume that a geosynchronous satellite has an orbital radius of 4.23107 m. a. Calculate its speed in orbit. b. Calculate its period. Satellite r Earth \u25a0 Figure 7-24 (Not to scale) 72. Asteroid The asteroid Ceres has a mass of 71020 kg and a radius of 500 km. a. What is g on the surface of Ceres? b. How much would a 90-kg astronaut weigh on Ceres? 73. Book A 1.25-kg book in space has a weight of 8.35 N. What is the value of the gravitational field at that location? 66. Toy Boat A force of 40.0 N is required to pull a 10.0-kg wooden toy boat at a constant velocity across a smooth glass surface on Earth. What force would be required to pull the same wooden toy boat across the same glass surface on the planet Jupiter? 74. The Moon\u2019s mass is 7.341022 kg, and it is 3.8108 m away from Earth. Earth\u2019s mass is 5.971024 kg. a. Calculate the gravitational force of attraction between Earth and the Moon. b. Find Earth\u2019s gravitational field at the Moon. 192 Chapter 7 Gravitation For more problems, go to Additional Problems, Appendix B. 189-195 CH07-SG CA STP-845813 6/3/04 10:42 AM Page 193 75. Two 1.00-kg masses have their centers 1.00 m apart. What is the force of attraction between them? 76. The radius of Earth is about 6.38103 km. A 7.20103-N spacecraft travels away from Earth. What is the weight of the spacecraft at the following distances from Earth\u2019s surface? a. 6.38103 km b. 1.28104 km 77. Rocket How high does a rocket have to go above Earth\u2019s surface before its weight is half of what it is on Earth? 78. Two satellites of equal mass are put into orbit 30.0 m apart. The gravitational force between them is 2.0107 N. a. What is the mass of each satellite? b. What is the initial acceleration given to each satellite by gravitational force? 79. Two large spheres are suspended close to each other. Their centers are 4.0 m apart, as shown in Figure 7-25. One sphere weighs 9.8102 N. The other sphere has a weight of 1.96102 N. What is the gravitational force between them? 4.0 m 9.8102 N 1.96102 N \u25a0 Figure 7-25 80. Suppose the centers of Earth and the Moon are 3.9108 m apart, and the gravitational force between them is about 1.91020 N. What is the approximate mass of the Moon? 81. On the surface of the Moon, a 91.0-kg physics teacher weighs only 145.6 N. What is the value of the Moon\u2019s gravitational field at its surface? 82. The mass of an electron is 9.11031 kg. The mass of a proton is 1.71027 kg. An electron and a proton are about 0.591010 m apart in a hydrogen atom. What gravitational force exists between the proton and the electron of a hydrogen atom? 83. Consider two spherical 8.0-kg objects that are 5.0 m apart. a. What is the gravitational force between the two objects? b. What is the gravitational force between them when they are 5.0101 m apart? 84. If you weigh 637 N on Earth\u2019s surface, how much would you weigh on the planet Mars? Mars has a mass of 6.421023 kg and a radius of 3.40106 m. 85. Using Newton\u2019s version of Kepler\u2019s third law and information from Table 7-1 on page 173, calculate the period of Earth\u2019s Moon if the orbital radius were twice the actual value of 3.9108 m. 86. Find the value of g, acceleration due to gravity, in the following situations.", " a. Earth\u2019s mass is triple its actual value, but its radius remains the same. b. Earth\u2019s radius is tripled, but its mass remains the same. c. Both the mass and radius of Earth are doubled. 87. Astronaut What would be the strength of Earth\u2019s gravitational field at a point where an 80.0-kg astronaut would experience a 25.0 percent reduction in weight? Mixed Review 88. Use the information for Earth in Table 7-1 on page 173 to calculate the mass of the Sun, using Newton\u2019s version of Kepler\u2019s third law. 89. Earth\u2019s gravitational field is 7.83 N/kg at the altitude of the space shuttle. At this altitude, what is the size of the force of attraction between a student with a mass of 45.0 kg and Earth? 90. Use the data from Table 7-1 on page 173 to find the speed and period of a satellite that orbits Mars 175 km above its surface. 91. Satellite A satellite is placed in orbit, as shown in Figure 7-26, with a radius that is half the radius of the Moon\u2019s orbit. Find the period of the satellite in units of the period of the Moon. Moon Earth rM rM1 2 Satellite \u25a0 Figure 7-26 92. Cannonball The Moon\u2019s mass is 7.31022 kg and its radius is 1785 km. If Newton\u2019s thought experiment of firing a cannonball from a high mountain were attempted on the Moon, how fast would the cannonball have to be fired? How long would it take the cannonball to return to the cannon? physicspp.com/chapter_test Chapter 7 Assessment 193 189-195 CH07-SG CA STP-845813 3/5/04 12:28 PM Page 194 93. The period of the Moon is one month. Answer the following questions assuming that the mass of Earth is doubled. a. What would the period of the Moon be? Express your results in months. b. Where would a satellite with an orbital period of one month be located? c. How would the length of a year on Earth be affected? 94. How fast would a planet of Earth\u2019s mass and size have to spin so that an object at the equator would be weightless? Give the period of rotation of the planet in minutes. 95. Car Races Suppose that a Martian base has been established and car races are being considered. A flat, circular race track has been built for the race. If a car can achieve speeds of up to 12 m/s, what is the smallest radius of a track for which the coefficient of friction is 0.50? 96. Apollo 11 On July 19, 1969, Apollo 11\u2019s revolution around the Moon was adjusted to an average orbit of 111 km. The radius of the Moon is 1785 km, and the mass of the Moon is 7.31022 kg. a. How many minutes did Apollo 11 take to orbit the Moon once? b. At what velocity did Apollo 11 orbit the Moon? Thinking Critically 97. Analyze and Conclude Some people say that the tides on Earth are caused by the pull of the Moon. Is this statement true? a. Determine the forces that the Moon and the Sun exert on a mass, m, of water on Earth. Your answer will be in terms of m with units of N. b. Which celestial body, the Sun or the Moon, has a greater pull on the waters of Earth? c. Determine the difference in force exerted by the Moon on the water at the near surface and the water at the far surface (on the opposite side) of Earth, as illustrated in Figure 7-27. Again, your answer will be in terms of m with units of N. Far tidal bulge \u25a0 Figure 7-27 (Not to scale) Near tidal bulge Earth Moon d. Determine the difference in force exerted by the Sun on water at the near surface and on water at the far surface (on the opposite side) of Earth. e. Which celestial body has a greater difference in pull from one side of Earth to the other? f. Why is the statement that the tides result from the pull of the Moon misleading? Make a correct statement to explain how the Moon causes tides on Earth. 98. Make and Use Graphs Use Newton\u2019s law of universal gravitation to find an equation where x is equal to an object\u2019s distance from Earth\u2019s center, and y is its acceleration due to gravity. Use a graphing calculator to graph this equation, using 6400\u20136600 km as the range for x and 9\u201310 m/s2 as the range for y. The equation should be of the form y c(1/x2). Trace along this graph and find y for the following locations. a. at sea level, 6400 km b. on top of Mt. Everest, 6410 km c", ". in a typical satellite orbit, 6500 km d. in a much higher orbit, 6600 km Writing in Physics 99. Research and describe the historical development of the measurement of the distance between the Sun and Earth. 100. Explore the discovery of planets around other stars. What methods did the astronomers use? What measurements did they take? How did they use Kepler\u2019s third law? Cumulative Review 101. Airplanes A jet airplane took off from Pittsburgh at 2:20 P.M. and landed in Washington, DC, at 3:15 P.M. on the same day. If the jet\u2019s average speed while in the air was 441.0 km/h, what is the distance between the cities? (Chapter 2) 102. Carolyn wants to know how much her brother Jared weighs. He agrees to stand on a scale for her, but only if they are riding in an elevator. If he steps on the scale while the elevator is accelerating upward at 1.75 m/s2 and the scale reads 716 N, what is Jared\u2019s usual weight on Earth? (Chapter 4) 103. Potato Bug A 1.0-g potato bug is walking around the outer rim of an upside-down flying disk. If the disk has a diameter of 17.2 cm and the bug moves at a rate of 0.63 cm/s, what is the centripetal force acting on the bug? What agent provides this force? (Chapter 6) 194 Chapter 7 Gravitation For more problems, go to Additional Problems, Appendix B. 189-195 CH07-SG CA STP-845813 3/5/04 12:29 PM Page 195 Multiple Choice TEXT TO COME 1. Two satellites are in orbit around a planet. One satellite has an orbital radius of 8.0106 m. The period of rotation for this satellite is 1.0106 s. The other satellite has an orbital radius of 2.0107 m. What is this satellite\u2019s period of rotation? 5.0105 s 2.5106 s 4.0106 s 1.3107 s 2. The illustration below shows a satellite in orbit around a small planet. The satellite\u2019s orbital radius is 6.7104 km and its speed is 2.0105 m/s. What is the mass of the planet around which the satellite orbits? (G 6.71011 Nm2/kg2) 2.51018 kg 4.01020 kg 2.51023 kg 4.01028 kg Satellite 6.7104 km Planet (Not to scale) 3. Two satellites are in orbit around the same planet. Satellite A has a mass of 1.5102 kg, and satellite B has a mass of 4.5103 kg. The mass of the planet is 6.61024 kg. Both satellites have the same orbital radius of 6.8106 m. What is the difference in the orbital periods of the satellites? no difference 1.5102 s 2.2102 s 3.0102 s 4. A moon revolves around a planet with a speed of 9.0103 m/s. The distance from the moon to the center of the planet is 5.4106 m. What is the orbital period of the moon? 1.2102 s 6.0102 s 1.2103 s 1.2109 s 5. A moon in orbit around a planet experiences a gravitational force not only from the planet, but also from the Sun. The illustration below shows a moon during a solar eclipse, when the planet, the moon, and the Sun are aligned. The moon has a mass of about 3.91021 kg. The mass of the planet is 2.41026 kg, and the mass of the Sun is 2.01030 kg. The distance from the moon to the center of the planet is 6.0108 m, and the distance from the moon to the Sun is 1.51011 m. What is the ratio of the gravitational force on the moon due to the planet, compared to its gravitational force due to the Sun during the solar eclipse? 5.0 7.5 Sun 0.5 2.5 6.0108 m Planet 1.51011 m (Not to scale) Extended Answer 6. Two satellites are in orbit around a planet. Satellite S1 takes 20 days to orbit the planet at a distance of 2105 km from the center of the planet. Satellite S2 takes 160 days to orbit the planet. What is the distance of Satellite S2 from the center of the planet? Plan Your Work and Work Your Plan Plan your workload so that you do a little work each day, rather than a lot of work all at once. The key to retaining information is repeated review and practice. You will retain more if you study one hour a night for five days in a row instead of cramming the night before a test. physicspp.com/standardized_test Chapter 7 Standardized Test Practice Chapter 7 Standardized", " Test Practice 195 195 196-200 CH08-S1-845813 3/19/04 2:07 PM Page 196 What You\u2019ll Learn \u2022 You will learn how to describe and measure rotational motion. \u2022 You will learn how torque changes rotational velocity. \u2022 You will explore factors that determine the stability of an object. \u2022 You will learn the nature of centrifugal and Coriolis \u201cforces.\u201d Why It\u2019s Important You encounter many rotating objects in everyday life, such as CDs, wheels, and amusement-park rides. Spin Rides Amusementpark rides that spin are designed to thrill riders using the physics of rotational motion. The thrill is produced by a \u201cforce\u201d that is present only when the ride spins. Think About This Why do people who ride amusement-park rides that spin in circles, such as this one, experience such strong physical reactions? physicspp.com physicspp.com 196 Paul L. Ruben 196-200 CH08-S1-845813 3/12/04 5:42 AM Page 197 How do different objects rotate as they roll? Question Do different objects of similar size and mass roll at the same rate on an incline? Critical Thinking Which of the objects\u2019 properties may have contributed to their behavior? List the properties that were similar and those that were different for each object. Procedure 1. You will need a meterstick, a piece of foam board, a ball, a solid can, and a hollow can. 2. Position the foam board on a 20\u00b0 incline. 3. Place the meterstick horizontally across the foam board, near the top of the incline, and hold it. 4. Place the ball, solid can, and hollow can against the meterstick. The solid can and hollow can should be placed sideways. 5. Simultaneously, release the three objects by lifting the meterstick. 6. As each object accelerates down the incline, due to gravity, observe the order in which each object reaches the bottom. 7. Repeat steps 2\u20135 two more times. Analysis List the objects in order from the greatest to the least acceleration. 8.1 Describing Rotational Motion You probably have observed a spinning object many times. How would you measure such an object\u2019s rotation? Find a circular object, such as a CD. Mark one point on the edge of the CD so that you can keep track of its position. Rotate the CD to the left (counterclockwise), and as you do so, watch the location of the mark. When the mark returns to its original position, the CD has made one complete revolution. How can you measure a fraction of one revolution? It can be measured in several 1 1 different ways. A grad is of of a revolution, whereas a degree is 60 3 00 4 a revolution. In mathematics and physics, yet another form of measurement is used to describe fractions of revolutions. In one revolution, a point on the edge travels a distance equal to 2 times the radius of the object. For this reason, the radian is defined as 1 of a revolution. In 2 other words, one complete revolution is equal to 2 radians. A radian is abbreviated \u201crad.\u201d Objectives \u2022 Describe angular displacement. \u2022 Calculate angular velocity. \u2022 Calculate angular acceleration. \u2022 Solve problems involving rotational motion. Vocabulary radian angular displacement angular velocity angular acceleration Section 8.1 Describing Rotational Motion 197 Horizons Companies 196-200 CH08-S1-845813 6/3/04 10:52 AM Page 198 \u25a0 Figure 8-1 The pie chart shows the radian measure of the most common angles, measured in the counterclockwise direction. Each angle is measured from 0 Angular Displacement The Greek letter theta,, is used to represent the angle of revolution. Figure 8-1 shows the angles in radians for several common fractions of a revolution. Note that counterclockwise rotation is designated as positive, while clockwise is negative. As an object rotates, the change in the angle is called angular displacement. As you know, Earth makes one complete revolution, or 2 rad, in 24 h. In 12 h, its rotation is through rad. Through what angle does Earth rotate in 6 h? Because 6 h is one-fourth of a day, Earth rotates through an angle of rad during that period. Earth\u2019s rotation as seen from the north pole is 2 positive. Is it positive or negative when viewed from the south pole? How far does a point on a rotating object move? You already found that a point on the edge of an object moves 2 times the radius in one revolution. In general, for rotation through an angle,, a point at a distance, r, from the center, as shown in Figure 8-2, moves a distance given by d r. If r is measured in meters, you might think that multiplying it by rad would result in d", " being measured in mrad. However, this is not the case. Radians indicate the ratio between d and r. Thus, d is measured in m. Angular Velocity How fast does a CD spin? How do you determine its speed of rotation? Recall from Chapter 2 that velocity is displacement divided by the time taken to make the displacement. Likewise, the angular velocity of an object is angular displacement divided by the time taken to make the displacement. Thus, the angular velocity of an object is given by the following equation, where angular velocity is represented by the Greek letter omega,. Angular Velocity of an Object t The angular velocity is equal to the angular displacement divided by the time required to make the rotation. \u25a0 Figure 8-2 The dashed line shows the path of the point on the CD as the CD rotates counterclockwise about its center. d r 198 Chapter 8 Rotational Motion 196-200 CH08-S1-845813 6/3/04 10:52 AM Page 199 Recall that if the velocity changes over a time interval, the average velocity is not equal to the instantaneous velocity at any given instant. Similarly, the angular velocity calculated in this way is actually the average angular velocity over a time interval, t. Instantaneous angular velocity is equal to the slope of a graph of angular position versus time. Angular velocity is measured in rad/s. For Earth, (2 rad)/ (24.0 h)(3600 s/h) 7.27105 rad/s. In the same way that counterclockwise rotation produces positive angular displacement, it also results in positive angular velocity. E Astronomy Connection Astronomy Connection If an object\u2019s angular velocity is, then the linear velocity of a point a distance, r, from the axis of rotation is given by v r. The speed at which an object on Earth\u2019s equator moves as a result of Earth\u2019s rotation is given by v r (6.38106 m)(7.27105 rad/s) 464 m/s. Earth is an example of a rotating, rigid body. Even though different points on Earth rotate different distances in each revolution, all points rotate through the same angle. All parts of a rigid body rotate at the same rate. The Sun, on the other hand, is not a rigid body. Different parts of the Sun rotate at different rates. Most objects that we will consider in this chapter are rigid bodies. Angular Acceleration What if angular velocity is changing? For example, if a car were accelerated from 0.0 m/s to 25 m/s in 15 s, then the angular velocity of the wheels also would change from 0.0 rad/s to 78 rad/s in the same 15 s. The wheels would undergo angular acceleration, which is defined as the change in angular velocity divided by the time required to make the change. Angular acceleration,, is represented by the following equation. Angular Acceleration of an Object t Angular acceleration is equal to the change in angular velocity divided by the time required to make that change. Angular acceleration is measured in rad/s2. If the change in angular velocity is positive, then the angular acceleration also is positive. Angular acceleration defined in this way is also the average angular acceleration over the time interval t. One way to find the instantaneous angular acceleration is to find the slope of a graph of angular velocity as a function of time. The linear acceleration of a point at a distance, r, from the axis of an object with angular acceleration,, is given by a r. Table 8-1 is a summary of linear and angular relationships. Table 8-1 Linear and Angular Measures Quantity Linear Angular Relationship Displacement Velocity Acceleration d (m) v (m/s) a (m/s2) (rad) (rad/s) (rad/s2) d r v r a r Section 8.1 Describing Rotational Motion 199 196-200 CH08-S1-845813 6/3/04 10:53 AM Page 200 1. What is the angular displacement of each of the following hands of a clock in 1 h? State your answer in three significant digits. a. the second hand b. the minute hand 2. c. the hour hand If a truck has a linear acceleration of 1.85 m/s2 and the wheels have an angular acceleration of 5.23 rad/s2, what is the diameter of the truck\u2019s wheels? 3. The truck in the previous problem is towing a trailer with wheels that have a diameter of 48 cm. a. How does the linear acceleration of the trailer compare with that of the truck? b. How do the angular accelerations of the wheels of the trailer and the wheels of the truck compare? 4. You want to replace the tires on your car with tires that have a larger diameter. After you change the tires, for trips at the same speed and over the same distance, how will the angular velocity and number of revolutions change? Angular frequency A rotating", " object can make many revolutions in a given amount of time. For instance, a spinning wheel can go through several complete revolutions in 1 min. Thus, the number of complete revolutions made by the object in 1 s is called angular frequency. Angular frequency is f /2. In the next section, you will explore the factors that cause the angular frequency to change. 8.1 Section Review 5. Angular Displacement A movie lasts 2 h. During that time, what is the angular displacement of each of the following? a. the hour hand b. the minute hand 6. Angular Velocity The Moon rotates once on its axis in 27.3 days. Its radius is 1.74106 m. a. What is the period of the Moon\u2019s rotation in seconds? b. What is the frequency of the Moon\u2019s rotation in rad/s? c. What is the linear speed of a rock on the Moon\u2019s equator due only to the Moon\u2019s rotation? d. Compare this speed with the speed of a person on Earth\u2019s equator due to Earth\u2019s rotation. 7. Angular Displacement The ball in a computer mouse is 2.0 cm in diameter. If you move the mouse 12 cm, what is the angular displacement of the ball? 8. Angular Displacement Do all parts of the minute hand on a watch have the same angular displacement? Do they move the same linear distance? Explain. 9. Angular Acceleration In the spin cycle of a clothes washer, the drum turns at 635 rev/min. If the lid of the washer is opened, the motor is turned off. If the drum requires 8.0 s to slow to a stop, what is the angular acceleration of the drum? 10. Critical Thinking A CD-ROM has a spiral track that starts 2.7 cm from the center of the disk and ends 5.5 cm from the center. The disk drive must turn the disk so that the linear velocity of the track is a constant 1.4 m/s. Find the following. a. the angular velocity of the disk (in rad/s and rev/min) for the start of the track b. the disk\u2019s angular velocity at the end of the track c. the disk\u2019s angular acceleration if the disk is played for 76 min 200 Chapter 8 Rotational Motion physicspp.com/self_check_quiz 201-210 CH08-S2-845813 6/3/04 10:57 AM Page 201 8.2 Rotational Dynamics How do you start the rotation of an object? That is, how do you change its angular velocity? Suppose you have a soup can that you want to spin. If you wrap a string around it and pull hard, you could make the can spin rapidly. Later in this chapter, you will learn why gravity, the force of Earth\u2019s mass on the can, acts on the center of the can. The force of the string, on the other hand, is exerted at the outer edge of the can, and at right angles to the line from the center of the can, to the point where the string leaves the can\u2019s surface. You have learned that a force changes the velocity of a point object. In the case of a soup can, a force that is exerted in a very specific way changes the angular velocity of an extended object, which is an object that has a definite shape and size. Consider how you open a door: you exert a force. How can you exert the force to open the door most easily? To get the most effect from the least force, you exert the force as far from the axis of rotation as possible, as shown in Figure 8-3. In this case, the axis of rotation is an imaginary vertical line through the hinges. The doorknob is near the outer edge of the door. You exert the force on the doorknob at right angles to the door, away from the hinges. Thus, the magnitude of the force, the distance from the axis to the point where the force is exerted, and the direction of the force determine the change in angular velocity. Lever arm For a given applied force, the change in angular velocity depends on the lever arm, which is the perpendicular distance from the axis of rotation to the point where the force is exerted. If the force is perpendicular to the radius of rotation, as it was with the soup can, then the lever arm is the distance from the axis, r. For the door, it is the distance from the hinges to the point where you exert the force, as illustrated in Figure 8-4a, on the next page. If the force is not perpendicular, the perpendicular component of the force must be found. Objectives \u2022 Describe torque and the factors that determine it. \u2022 Calculate net torque. \u2022 Calculate the moment of inertia. Vocabulary lever arm torque moment of inertia Newton\u2019s second law for rotational motion \u25a0", " Figure 8-3 When opening a door that is free to rotate about its hinges, the greatest torque is produced when the force is applied farthest from the hinges (a), at an angle perpendicular to the door (b). a b The force exerted by the string around the can is perpendicular to the radius. If a force is not exerted perpendicular to the radius, however, the lever arm is reduced. To find the lever arm, extend the line of the force until it forms a right angle with a line from the center of rotation. The distance between the intersection and the axis is the lever arm. Thus, using trigonometry, the lever arm, L, can be calculated by the equation L r sin \u03b8, as shown in Figure 8-4b. In this equation, r is the distance from the axis of rotation to the point where the force is exerted, and \u03b8 is the angle between the force and the radius from the axis of rotation to the point where the force is applied. No effect Little effect Maximum effect No effect Some effect Maximum effect Section 8.2 Rotational Dynamics 201 201-210 CH08-S2-845813 6/3/04 10:58 AM Page 202 \u25a0 Figure 8-4 The lever arm is along the width of the door, from the hinge to the point where the force is exerted (a). The lever arm is equal to r sin, when the angle,, between the force and the radius of rotation is not equal to 90\u00b0 (b). a b Axis of rotation L e v e r a r m Lever arm r sin F r F Torque is a measure of how effectively a force causes rotation. The magnitude of torque is the product of the force and the lever arm. Because force is measured in newtons, and distance is measured in meters, torque is measured in newton-meters (Nm). Torque is represented by the Greek letter tau, \u03c4. The equation for torque is shown below. Torque \u03c4 Fr sin \u03b8 Torque is equal to the force times the lever arm. Lever Arm A bolt on a car engine needs to be tightened with a torque of 35 Nm. You use a 25-cm-long wrench and pull on the end of the wrench at an angle of 60.0\u00b0 from the perpendicular. How long is the lever arm, and how much force do you have to exert? 1 Analyze and Sketch the Problem \u2022 Sketch the situation. Find the lever arm by extending the force vector backwards until a line that is perpendicular to it intersects the axis of rotation. Known: r 0.25 m 60.0\u00b0 35 Nm Unknown: L? F? 2 Solve for the Unknown Solve for the length of the lever arm. L r sin (0.25 m)(sin 60.0\u00b0) 0.22 m Substitute r 0.25 m, 60.0\u00b0 60.0\u00b0 25 cm 60.0\u00b0 L e v e r a r m Solve for the force. Fr sin F n r si 35 Nm (0.25 m)(sin 60.0\u00b0) 1.6102 N Substitute 35 Nm, r 0.25 m, 60.0\u00b0 Math Handbook Trigonometric Ratios page 855 3 Evaluate the Answer \u2022 Are the units correct? Force is measured in newtons. \u2022 Does the sign make sense? Only the magnitude of the force needed to rotate the wrench clockwise is calculated. 202 Chapter 8 Rotational Motion 201-210 CH08-S2-845813 3/12/04 5:51 AM Page 203 11. Consider the wrench in Example Problem 1. What force is needed if it is applied to the wrench at a point perpendicular to the wrench? 12. If a torque of 55.0 Nm is required and the largest force that can be exerted by you is 135 N, what is the length of the lever arm that must be used? 13. You have a 0.234-m-long wrench. A job requires a torque of 32.4 Nm, and you can exert a force of 232 N. What is the smallest angle, with respect to the vertical, at which the force can be exerted? 14. You stand on the pedal of a bicycle. If you have a mass of 65 kg, the pedal makes an angle of 35\u00b0 above the horizontal, and the pedal is 18 cm from the center of the chain ring, how much torque would you exert? 15. If the pedal in problem 14 is horizontal, how much torque would you exert? How much torque would you exert when the pedal is vertical? Finding Net Torque Try the following experiment. Get two pencils, some coins, and some transparent tape. Tape two identical coins to the ends of the pencil and balance it on the second pencil, as shown in Figure 8-5. Each coin exerts a torque that is equal to its weight, Fg, times the distance, r, from the balance point to the", " center of the coin, as follows: Fgr But the torques are equal and opposite in direction. Thus, the net torque is zero: 1 2 0 or Fg2 r2 0 Fg1 r1 How can you make the pencil rotate? You could add a second coin on top of one of the two coins, thereby making the two forces different. You also could slide the balance point toward one end or the other of the pencil, thereby making the two distances different. \u25a0 Figure 8-5 The torque exerted by the first coin, Fg1r1, is equal and opposite in direction to the torque exerted by the second coin, Fg2r2, when the pencil is balanced. r1 r2 Fg1 Fg2 Section 8.2 Rotational Dynamics 203 201-210 CH08-S2-845813 6/3/04 10:58 AM Page 204 Balancing Torques Kariann (56 kg) and Aysha (43 kg) want to balance on a 1.75-m-long seesaw. Where should they place the pivot point? 1 Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Draw and label the vectors. rK rA Known: mK mA rK 56 kg 43 kg rA 1.75 m Unknown:? rK? rA 2 Solve for the Unknown Find the two forces. FgK FgA Kariann: FgK mKg (56 kg)(9.80 m/s2) 5.5102 N Aysha: FgA mAg (43 kg)(9.80 m/s2) 4.2102 N Substitute mK 56 kg, g 9.80 m/s2 Substitute mA 43 kg, g 9.80 m/s2 Define Kariann\u2019s distance in terms of the length of the seesaw and Aysha\u2019s distance. rK 1.75 m rA When there is no rotation, the sum of the torques is zero. FgKrK FgArA FgKrK FgK(1.75 m rA) FgArA FgArA 0.0 Nm 0.0 Nm Substitute rK 1.75 m rA Solve for rA. FgK(1.75 m) FgK(rA) FgArA FgArA FgA) rA FgKrA (FgK 0.0 Nm FgK(1.75 m) FgK(1.75 m) Math Handbook Isolating a Variable page 845 rA K ) m.75 (1 F g A) Fg (F g K (5.5102 N)(1.75 m) (5.5102 N 4.2102 N) 0.99 m Substitute FgK 5.5102 N, FgA 4.2102 N 3 Evaluate the Answer \u2022 Are the units correct? Distance is measured in meters. \u2022 Do the signs make sense? Distances are positive. \u2022 Is the magnitude realistic? Aysha is about 1 m from the center, so Kariann is about 0.75 m away from it. Because Kariann\u2019s weight is greater than Aysha\u2019s weight, the lever arm on Kariann\u2019s side should be shorter. Aysha is farther from the pivot, as expected. 204 Chapter 8 Rotational Motion 201-210 CH08-S2-845813 6/3/04 10:59 AM Page 205 16. Ashok, whose mass is 43 kg, sits 1.8 m from the center of a seesaw. Steve, whose mass is 52 kg, wants to balance Ashok. How far from the center of the seesaw should Steve sit? 17. A bicycle-chain wheel has a radius of 7.70 cm. If the chain exerts a 35.0-N force on the wheel in the clockwise direction, what torque is needed to keep the wheel from turning? 18. Two baskets of fruit hang from strings going around pulleys of different diameters, as shown in Figure 8-6. What is the mass of basket A? 19. Suppose the radius of the larger pulley in problem 18 was increased to 6.0 cm. What is the mass of basket A now? 20. A bicyclist, of mass 65.0 kg, stands on the pedal of a bicycle. The crank, which is 0.170 m long, makes a 45.0\u00b0 angle with the vertical, as shown in Figure 8-7. The crank is attached to the chain wheel, which has a radius of 9.70 cm. What force must the chain exert to keep the wheel from turning? 4.5 cm 1.1 cm \u25a0 Figure 8-6 (Not to scale) A", " 9.70 cm \u25a0 Figure 8-7 0.23 kg 45.0\u00b0 0.170 m The Moment of Inertia If you exert a force on a point mass, its acceleration will be inversely proportional to its mass. How does an extended object rotate when a torque is exerted on it? To observe firsthand, recover the pencil, the coins, and the transparent tape that you used earlier in this chapter. First, tape the coins at the ends of the pencil. Hold the pencil between your thumb and forefinger, and wiggle it back and forth. Take note of the forces that your thumb and forefinger exert. These forces create torques that change the angular velocity of the pencil and coins. Now move the coins so that they are only 1 or 2 cm apart. Wiggle the pencil as before. Did the amount of torque and force need to be changed? The torque that was required was much less this time. Thus, the amount of mass is not the only factor that determines how much torque is needed to change angular velocity; the location of that mass also is relevant. The resistance to rotation is called the moment of inertia, which is represented by the symbol I and has units of mass times the square of the distance. For a point object located at a distance, r, from the axis of rotation, the moment of inertia is given by the following equation. Moment of Inertia of a Point Mass I mr2 The moment of inertia of a point mass is equal to the mass of the object times the square of the object\u2019s distance from the axis of rotation. Section 8.2 Rotational Dynamics 205 201-210 CH08-S2-845813 3/12/04 5:53 AM Page 206 Table 8-2 Moments of Inertia for Various Objects Object Thin hoop of radius r Solid, uniform cylinder of radius r Location of Axis Through central diameter Through center Uniform sphere of radius r Through center Long, uniform rod of length l Through center Long, uniform rod of length l Through end \u25a0 Figure 8-8 The moment of inertia of a book depends on the axis of rotation. The moment of inertia of the book in (a) is larger than the moment of inertia of the book in (b) because the average distance of the book\u2019s mass from the rotational axis is larger. Thin, rectangular plate of length l and width w Through center w Diagram r r r Axis Axis Axis Axis l Axis l Axis l Moment of Inertia mr 2 1 mr 2 2 2 mr 2 5 1 ml 2 2 1 1 ml 2 3 1 m(l 2 w 2) 1 2 a b Axis of rotation Axis of rotation As you have seen, the moment of inertia for complex objects, such as the pencil and coins, depends on how far the coins are from the axis of rotation. A bicycle wheel, for example, has almost all of its mass in the rim and tire. Its moment of inertia is almost exactly equal to mr 2, where r is the radius of the wheel. For most objects, however, the mass is distributed continuously and so the moment of inertia is less than mr 2. For example, as shown in Table 8-2, for a solid cylinder of radius r, mr 2, while for a solid sphere, I 2 I 1 mr 2. 5 2 The moment of inertia also depends on the location of the rotational axis, as illustrated in Figure 8-8. To observe this firsthand, hold a book in the upright position, by placing your hands at the bottom of the book. Feel the torque needed to rock the book towards you, and then away from you. Now put your hands in the middle of the book and feel the torque needed to rock the book toward you and then away from you. Note that much less torque is needed when your hands are placed in the middle of the book because the average distance of the book\u2019s mass from the rotational axis is much less in this case. 206 Chapter 8 Rotational Motion 201-210 CH08-S2-845813 6/3/04 11:00 AM Page 207 Moment of Inertia A simplified model of a twirling baton is a thin rod with two round objects at each end. The length of the baton is 0.65 m, and the mass of each object is 0.30 kg. Find the moment of inertia of the baton if it is rotated about the midpoint between the round objects. What is the moment of inertia of the baton when it is rotated around one end? Which is greater? Neglect the mass of the rod. 1 Analyze and Sketch the Problem \u2022 Sketch the situation. Show the baton with the two different axes of rotation and the distances from the axes of rotation to the masses. 0.33 m 0.33 m 0.30 kg 0.30 kg Axis 0.65 m Math Handbook Operations with Significant Digits pages 835\u2013836 Known: m 0.30", " kg l 0.65 m Unknown: I? 2 Solve for the Unknown Calculate the moment of inertia of each mass separately. Rotating about the center of the rod: r 1 l 2 1 (0.65 m) 2 0.33 m Isingle mass mr 2 (0.30 kg)(0.33 m)2 0.033 kgm2 Substitute l = 0.65 m Substitute m 0.30 kg, r 0.33 m Find the moment of inertia of the baton. I 2Isingle mass 2(0.033 kgm2) 0.066 kgm2 Rotating about one end of the rod: Isingle mass mr 2 (0.30 kg)(0.65 m)2 0.13 kgm2 Substitute Isingle mass 0.033 kgm2 Substitute m 0.30 kg, r 0.65 m Find the moment of inertia of the baton. I Isingle mass 0.13 kgm2 The moment of inertia is greater when the baton is swung around one end. 3 Evaluate the Answer \u2022 Are the units correct? Moment of inertia is measured in kgm2. \u2022 Is the magnitude realistic? Masses and distances are small, and so are the moments of inertia. Doubling the distance increases the moment of inertia by a factor of 4. Thus, doubling the distance overcomes having only one mass contributing. Section 8.2 Rotational Dynamics 207 201-210 CH08-S2-845813 3/12/04 5:54 AM Page 208 21. Two children of equal masses sit 0.3 m from the center of a seesaw. Assuming that their masses are much greater than that of the seesaw, by how much is the moment of inertia increased when they sit 0.6 m from the center? 22. Suppose there are two balls with equal diameters and masses. One is solid, and the other is hollow, with all its mass distributed at its surface. Are the moments of inertia of the balls equal? If not, which is greater? 23. Figure 8-9 shows three massive spheres on a rod of very small mass. Consider the moment of inertia of the system, first when it is rotated about sphere A, and then when it is rotated about sphere C. Are the moments of inertia the same or different? Explain. If the moments of inertia are different, in which case is the moment of inertia greater? A C 24. Each sphere in the previous problem has a mass of 0.10 kg. The distance between spheres A and C is 0.20 m. Find the moment of inertia in the following instances: rotation about sphere A, rotation about sphere C. \u25a0 Figure 8-9 Newton\u2019s Second Law for Rotational Motion Newton's second law for linear motion is expressed as a Fnet/m. If you rewrite this equation to represent rotational motion, acceleration is replaced by angular acceleration,, force is replaced by net torque, net, and mass is replaced by moment of inertia, I. Thus, Newton\u2019s second law for rotational motion states that angular acceleration is directly proportional to the net torque and inversely proportional to the moment of inertia. This law is expressed by the following equation. Newton\u2019s Second Law for Rotational Motion et n I The angular acceleration of an object is equal to the net torque on the object, divided by the moment of inertia. Recall the coins taped on the pencil. To change the direction of rotation of the pencil\u2014to give it angular acceleration\u2014you had to apply torque to the pencil. The greater the moment of inertia, the more torque needed to produce the same angular acceleration. Rank the objects shown in the diagram according to their moments of inertia about the indicated axes. All spheres have equal masses and all separations are the same. A B C D 208 Chapter 8 Rotational Motion 201-210 CH08-S2-845813 3/12/04 5:55 AM Page 209 Torque A solid steel wheel has a mass of 15 kg and a diameter of 0.44 m. It starts at rest. You want to make it rotate at 8.0 rev/s in 15 s. a. What torque must be applied to the wheel? b. If you apply the torque by wrapping a strap around the outside of the wheel, how much force should you exert on the strap? F 1 Analyze and Sketch the Problem \u2022 Sketch the situation. The torque must be applied in a counterclockwise direction; force must be exerted as shown. Known: m 15 kg r 1 (0.44 m) = 0.22 m 2 i f t 15 s 0.0 rad/s 2(8.0 rev/s) Unknown:? I?? F? \u03c4 0.44 m 2 Solve for the Unknown a. Solve for angular acceleration. t 2(8.0 rev/s) (0.0 rad/s) 15 s 3.4 rad/s", "2 Solve for the moment of inertia. I 1 mr 2 2 1 (15 kg)(0.22 m)2 2 0.36 kgm2 Solve for torque. I (0.36 kgm2)(3.4 rad/s2) 1.2 kgm2/s2 1.2 Nm b. Solve for force. Fr F r N m.2 1 0 m.22 5.5 N Substitute 2(8.0 rev/s), i f 0.0 rad/s Substitute m 15 kg, r 0.22 m Substitute I 0.36 kgm2, 3.4 rad/s2 Math Handbook Operations with Significant Digits pages 835\u2013836 Substitute 1.2 Nm, r 0.22 m 3 Evaluate the Answer \u2022 Are the units correct? Torque is measured in Nm and force is measured in N. \u2022 Is the magnitude realistic? Despite its large mass, the small size of the wheel makes it relatively easy to spin. Section 8.2 Rotational Dynamics 209 201-210 CH08-S2-845813 6/3/04 11:01 AM Page 210 25. Consider the wheel in Example Problem 4. If the force on the strap were twice as great, what would be the speed of rotation of the wheel after 15 s? 26. A solid wheel accelerates at 3.25 rad/s2 when a force of 4.5 N exerts a torque on it. If the wheel is replaced by a wheel with all of its mass on the rim, the moment of inertia is given by I mr 2. If the same angular velocity were desired, what force would have to be exerted on the strap? 27. A bicycle wheel can be accelerated either by pulling on the chain that is on the gear or by pulling on a string wrapped around the tire. The wheel\u2019s radius is 0.38 m, while the radius of the gear is 0.14 m. If you obtained the needed acceleration with a force of 15 N on the chain, what force would you need to exert on the string? 28. The bicycle wheel in problem 27 is used with a smaller gear whose radius is 0.11 m. The wheel can be accelerated either by pulling on the chain that is on the gear or by pulling string that is wrapped around the tire. If you obtained the needed acceleration with a force of 15 N on the chain, what force would you need to exert on the string? 29. A disk with a moment of inertia of 0.26 kgm2 is attached to a smaller disk mounted on the same axle. The smaller disk has a diameter of 0.180 m and a mass of 2.5 kg. A strap is wrapped around the smaller disk, as shown in Figure 8-10. Find the force needed to give this system an angular acceleration of 2.57 rad/s2. F \u25a0 Figure 8-10 In summary, changes in the amount of torque applied to an object, or changes in the moment of inertia, affect the rate of rotation. In this section, you learned how Newton\u2019s second law of motion applies to rotational motion. In the next section, you will learn how to keep objects from rotating. 8.2 Section Review 30. Torque Vijesh enters a revolving door that is not moving. Explain where and how Vijesh should push to produce a torque with the least amount of force. 31. Lever Arm You try to open a door, but you are unable to push at a right angle to the door. So, you push the door at an angle of 55\u00b0 from the perpendicular. How much harder would you have to push to open the door just as fast as if you were to push it at 90\u00b0? 32. Net Torque Two people are pulling on ropes wrapped around the edge of a large wheel. The wheel has a mass of 12 kg and a diameter of 2.4 m. One person pulls in a clockwise direction with a 43-N force, while the other pulls in a counterclockwise direction with a 67-N force. What is the net torque on the wheel? 33. Moment of Inertia Refer to Table 8-2 on page 206 and rank the moments of inertia from least to greatest of the following objects: a sphere, a wheel with almost all of its mass at the rim, and a solid disk. All have equal masses and diameters. Explain the advantage of using the one with the least moment of inertia. 34. Newton\u2019s Second Law for Rotational Motion A rope is wrapped around a pulley and pulled with a force of 13.0 N. The pulley\u2019s radius is 0.150 m. The pulley\u2019s rotational speed goes from 0.0 to 14.0 rev/min in 4.50 s. What is the moment of inertia of the pulley? 35. Critical Thinking A ball on an extremely lowfriction", ", tilted surface, will slide downhill without rotating. If the surface is rough, however, the ball will roll. Explain why, using a free-body diagram. 210 Chapter 8 Rotational Motion physicspp.com/self_check_quiz 211-217 CH08-S3-845813 6/3/04 11:05 AM Page 211 8.3 Equilibrium Why are some vehicles more likely than others to roll over when involved in an accident? What causes a vehicle to roll over? The answer lies in the design of the vehicle. In this section, you will learn some of the factors that cause an object to tip over. The Center of Mass How does an object rotate around its center of mass? A wrench may spin about its handle or end-over-end. Does any single point on the wrench follow a straight path? Figure 8-11a shows the path of the wrench. You can see that there is a single point whose path traces a straight line, as if the wrench could be replaced by a point particle at that location. The center of mass of an object is the point on the object that moves in the same way that a point particle would move. Locating the center of mass How can you locate the center of mass of an object? First, suspend the object from any point. When the object stops swinging, the center of mass is along the vertical line drawn from the suspension point as shown in Figure 8-11b. Draw the line. Then, suspend the object from another point. Again, the center of mass must be below this point. Draw a second vertical line. The center of mass is at the point where the two lines cross, as shown in Figure 8-11c. The wrench, racket, and all other freely rotating objects rotate about an axis that goes through their center of mass. Where is the center of mass of a person located? Objectives \u2022 Define center of mass. \u2022 Explain how the location of the center of mass affects the stability of an object. \u2022 Define the conditions for equilibrium. \u2022 Describe how rotating frames of reference give rise to apparent forces. Vocabulary center of mass centrifugal \u201cforce\u201d Coriolis \u201cforce\u201d a b c Center of mass \u25a0 Figure 8-11 The path of the center of mass of a wrench is a straight line (a). The center of mass of an object such as a tennis racket can be found by first suspending it from any point (b). The point where the strings intersect is the location of the racket\u2019s center of mass (c). 211 (t)Richard Megna/Fundamental Photographs, (others)Hutchings Photography Section 8.3 Equilibrium 211-217 CH08-S3-845813 6/3/04 11:06 AM Page 212 \u25a0 Figure 8-12 The upward motion of the ballet dancer\u2019s head is less than the upward motion of the center of mass. Thus, the head and torso move in a nearly horizontal path. This creates an illusion of floating. Path of the dancer, s head P a t h o f the dancer, s center of mass The Fosbury-Flop In high jumping, a technique called the Fosbury-Flop allows a high jumper to clear the bar when it is placed at the highest position. This is possible because the athlete\u2019s center of mass passes below the bar as he or she somersaults over the bar, with his or her back toward it. The center of mass of a human body For a person who is standing with his or her arms hanging straight down, the center of mass is a few centimeters below the navel, midway between the front and back of the person\u2019s body. It is slightly higher in young children, because of their relatively larger heads. Because the human body is flexible, however, its center of mass is not fixed. If you raise your hands above your head, your center of mass rises 6 to 10 cm. A ballet dancer, for example, can appear to be floating on air by changing her center of mass in a leap. By raising her arms and legs while in the air, as shown in Figure 8-12, the dancer moves her center of mass closer to her head. The path of the center of mass is a parabola, so the dancer\u2019s head stays at almost the same height for a surprisingly long time. Center of Mass and Stability What factors determine whether a vehicle is stable or prone to roll over in an accident? To understand the problem, think about tipping over a box. A tall, narrow box, standing on end, tips more easily than a low, broad box. Why? To tip a box, as shown in Figure 8-13, you must rotate it about a corner. You pull at the top with a force, F, applying a torque, F. The weight of the box, acting on the center of mass, Fg, applies an opposing torque, w", ". When the center of mass is directly above the point of support, w is zero. The only torque is the one applied by you. As the box rotates farther, its center of mass is no longer above its base of support, and both torques act in the same direction. At this point, the box tips over rapidly. \u25a0 Figure 8-13 The bent arrows show the direction of the torque produced by the force exerted to tip over a box. \u03c4 F F \u03c4 F F \u03c4w Fg \u03c4F F \u03c4w Fg Fg 212 Chapter 8 Rotational Motion 211-217 CH08-S3-845813 3/12/04 6:06 AM Page 213 Spinning Tops 1. Cut out two cardboard disks of 10-cm and 15-cm diameter. 2. Use a pencil with an eraser that has rounded edges from use. If it is new, rub it on paper to round it. 3. Spin the pencil and try to make it stand on the eraser. Repeat several times and record your observations. 4. Carefully push the pencil through the center of the 10-cm disk. 5. Spin the pencil with the disk and try to make it stand on the eraser. 6. Move the disk to different points on the pencil. Spin and record your observations. 7. Repeat steps 4-6 with the 15-cm disk. Analyze and Conclude 8. Sequence the three trials in order from least to most stable. 9. Describe the location of the pencil\u2019s center of mass. 10. Analyze the placement of the disk and its effect on stability. Stability An object is said to be stable if an external force is required to tip it. The box in Figure 8-13 is stable as long as the direction of the torque due to its weight, w tends to keep it upright. This occurs as long as the box\u2019s center of mass lies above its base. To tip the box over, you must rotate its center of mass around the axis of rotation until it is no longer above the base of the box. To rotate the box, you must lift its center of mass. The broader the base, the more stable the object is. For this reason, if you are standing on a bus that is weaving through traffic and you want to avoid falling down, you need to stand with your feet spread apart. Why do vehicles roll over? Figure 8-14 shows two vehicles rolling over. Note that the one with the higher center of mass does not have to be tilted very far for its center of mass to be outside its base\u2014its center of mass does not have to be raised as much as the other vehicle\u2019s. The lower the location of an object\u2019s center of mass, the greater its stability. You are stable when you stand flat on your feet. When you stand on tiptoe, however, your center of mass moves forward directly above the balls of your feet, and you have very little stability. A small person can use torque, rather than force, to defend himself or herself against a stronger person. In judo, aikido, and other martial arts, the fighter uses torque to rotate the opponent into an unstable position, where the opponent\u2019s center of mass does not lie above his or her feet. In summary, if the center of mass is outside the base of an object, it is unstable and will roll over without additional torque. If the center of mass is above the base of the object, it is stable. If the base of the object is very narrow and the center of mass is high, then the object is stable, but the slightest force will cause it to tip over. Conditions for Equilibrium If your pen is at rest, what is needed to keep it at rest? You could either hold it up or place it on a desk or some other surface. An upward force must be exerted on the pen to balance the downward force of gravity. You must also hold the pen so that it will not rotate. An object is said to be in static equilibrium if both its velocity and angular velocity are zero or constant. Thus, for an object to be in static equilibrium, it must meet two conditions. First, it must be in translational equilibrium; that is, the net force exerted on the object must be zero. Second, it must be in rotational equilibrium; that is, the net torque exerted on the object must be zero. \u25a0 Figure 8-14 Larger vehicles have a higher center of mass than smaller ones. The higher the center of mass, the smaller the tilt needed to cause the vehicle\u2019s center of mass to move outside its base and cause the vehicle to roll over. 1 2 Section 8.3 Equilibrium 213 211-217 CH08-S3-845813 3/18/04 1:55 PM Page 214 Static Equilibrium A 5.8-kg ladder, 1.80 m long, rests on two sawhorses. Sawhorse A is 0", ".60 m from one end of the ladder, and sawhorse B is 0.15 m from the other end of the ladder. What force does each sawhorse exert on the ladder? 1 Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Choose the axis of rotation at the point where FA acts on the ladder. Thus, the torque due to FA is zero. Known: m 5.8 kg l 1.80 m 0.60 m lA 0.15 m lB Unknown:? FA? FB 1.80 m 0.15 m 0.60 m 0.30 m 0.75 m A FA B FB Fg rB rg FB 2 Solve for the Unknown For a ladder that has a constant density, the center of mass is at the center rung. The net force is the sum of all forces on the ladder. Axis of rotation Fg FA Fnet 0.0 N FA FB FB (Fg) Fg The ladder is in translational equilibrium, so the net force exerted on it is zero. Solve for FA. Fg FA FB Find the torques due to Fg and FB. g is in the clockwise direction. B is in the counterclockwise direction. The net torque is the sum of all torques on the object. g B rgFg rBFB g g B net 0.0 Nm B g B rgFg rBFB Solve for FB. FB F rg g r B m g rg r B FA Fg Fg Using the expression FA FB m g rg r B m g rg mg r B r g) mg (1 r B 214 Chapter 8 Rotational Motion The ladder is in rotational equilibrium, so net 0.0 Nm. Substitute B rBFB, g rgFg Math Handbook Isolating a Variable page 845 Substitute Fg mg Fg FB, substitute in the expressions for FB and Fg. Substitute FB m g rg r B Substitute Fg mg 211-217 CH08-S3-845813 3/12/04 6:07 AM Page 215 rg Solve for rg. 1 lA 2 0.90 m 0.60 m 0.30 m Solve for rB. For a ladder, which has a constant density, the center of mass is at the center rung. l 0.90 m, lA Substitute 2 0.60 m rB (0.90 m lB) (0.90 m lA) (0.90 m 0.15 m) (0.90 m 0.60 m) 0.75 m 0.30 m 1.05 m Substitute lB 0.15 m, lA 0.60 m Calculate FB. m g rg r B FB (0.30 m)(5.8 kg)(9.80 m/s2) (1.05 m) 16 N Calculate FA. FA r g mg 1 )(5.8 kg)(9.80 m/s2) 1 ( 41 ( Substitute rg 0.30 m, m 5.8 kg, g 9.80 m/s2, rB 1.05 m Substitute rg 0.30 m, m 5.8 kg, g 9.80 m/s2, rB 1.05 m 3 Evaluate the Answer \u2022 Are the units correct? Forces are measured in newtons. \u2022 Do the signs make sense? Both forces are upward. \u2022 Is the magnitude realistic? The forces add up to the weight of the ladder, and the force exerted by the sawhorse closer to the center of mass is greater, which is correct. 36. What would be the forces exerted by the two sawhorses if the ladder in Example Problem 5 had a mass of 11.4 kg? 37. A 7.3-kg ladder, 1.92 m long, rests on two sawhorses, as shown in Figure 8-15. Sawhorse A, on the left, is located 0.30 m from the end, and sawhorse B, on the right, is located 0.45 m from the other end. Choose the axis of rotation to be the center of mass of the ladder. 1.92 m 0.30 m 0.66 m 0.51 m 0.45 m a. What are the torques acting on the ladder? b. Write the equation for rotational equilibrium. c. Solve the equation for FA in terms of Fg. d. How would the forces exerted by the two sawhorses change if A were moved very close to, but not directly under, the center of mass? A \u25a0 Figure 8-15 B 38. A 4.5-m-long wooden plank with a 24-kg mass is supported in two places. One support is directly under the center of the board, and the other is at one end. What are the forces exerted by the two supports? 39", ". A 85-kg diver walks to the end of a diving board. The board, which is 3.5 m long with a mass of 14 kg, is supported at the center of mass of the board and at one end. What are the forces on the two supports? Section 8.3 Equilibrium 215 211-217 CH08-S3-845813 3/12/04 6:08 AM Page 216 Rotating Frames of Reference When you are on a on a rapidly spinning amusement-park ride, it feels like a strong force is pushing you to the outside. A pebble on the floor of the ride would accelerate outward without a horizontal force being exerted on it in the same direction. The pebble would not move in a straight line. In other words, Newton\u2019s laws would not apply. This is because rotating frames of reference are accelerated frames. Newton\u2019s laws are valid only in inertial or nonaccelerated frames. Motion in a rotating reference frame is important to us because Earth rotates. The effects of the rotation of Earth are too small to be noticed in the classroom or lab, but they are significant influences on the motion of the atmosphere and therefore, on climate and weather. Centrifugal \u201cForce\u201d Suppose you fasten one end of a spring to the center of a rotating platform. An object lies on the platform and is attached to the other end of the spring. As the platform rotates, an observer on the platform sees the object stretch the spring. The observer might think that some force toward the outside of the platform is pulling on the object. This apparent force is called centrifugal \u201cforce.\u201d It is not a real force because there is no physical outward push on the object. Still, this \u201cforce\u201d seems real, as anyone who has ever been on an amusement-park ride can attest. As the platform rotates, an observer on the ground sees things differently. This observer sees the object moving in a circle. The object accelerates toward the center because of the force of the spring. As you know, the v2/r. It also can acceleration is centripetal acceleration and is given by ac 2/r. Centripetal acceleration be written in terms of angular velocity, as ac is proportional to the distance from the axis of rotation and depends on the square of the angular velocity. Thus, if you double the rotational frequency, the acceleration increases by a factor of 4. The Coriolis \u201cForce\u201d A second effect of rotation is shown in Figure 8-16. Suppose a person standing at the center of a rotating disk throws a ball toward the edge of the disk. Consider the horizontal motion of the ball as seen by two observers and ignore the vertical motion of the ball as it falls. a Viewed from fixed frame b Coriolis \"force\" viewed from rotating frame Center Observed positions of ball Observed positions of ball \u25a0 Figure 8-16 The Coriolis \u201cforce\u201d exists only in rotating reference frames. 216 Chapter 8 Rotational Motion 211-217 CH08-S3-845813 6/3/04 11:07 AM Page 217 An observer standing outside the disk, as shown in Figure 8-16a, sees the ball travel in a straight line at a constant speed toward the edge of the disk. However, the other observer, who is stationed on the disk and rotating with it, as shown in Figure 8-16b, sees the ball follow a curved path at a constant speed. A force seems to be acting to deflect the ball. This apparent force is called the Coriolis \u201cforce.\u201d Like the centrifugal \u201cforce,\u201d the Coriolis \u201cforce\u201d is not a real force. It seems to exist because we observe a deflection in horizontal motion when we are in a rotating frame of reference. Coriolis \u201cforce\u201d due to Earth Suppose a cannon is fired from a point on the equator toward a target due north of it. If the projectile were fired directly northward, it would also have an eastward velocity component because of the rotation of Earth. This eastward speed is greater at the equator than at any other latitude. Thus, as the projectile moves northward, it also moves eastward faster than points on Earth below it do. The result is that the projectile lands east of the target as shown in Figure 8-17. While an observer in space would see Earth\u2019s rotation, an observer on Earth could claim that the projectile missed the target because of the Coriolis \u201cforce\u201d on the rocket. Note that for objects moving toward the equator, the direction of the apparent force is westward. A projectile will land west of the target when fired due south. The direction of winds around high- and low-pressure areas results from the Coriolis \u201cforce.\u201d Winds flow from areas of high to low pressure.", " Because of the Coriolis \u201cforce\u201d in the northern hemisphere, winds from the south go to the east of low-pressure areas. Winds from the north, however, end up west of low-pressure areas. Therefore, winds rotate counterclockwise around low-pressure areas in the northern hemisphere. In the southern hemisphere however, winds rotate clockwise around low-pressure areas. Most amusement-park rides thrill the riders because they are in accelerated reference frames while on the ride. The \u201cforces\u201d felt by roller-coaster riders at the tops and bottoms of hills, and when moving almost vertically downward, are mostly related to linear acceleration. On Ferris wheels, rotors, other circular rides, and on the curves of roller coasters, centrifugal \u201cforces\u201d provide most of the excitement. Actual path Equator Expected path \u25a0 Figure 8-17 An observer on Earth sees the Coriolis \u201cforce\u201d cause a projectile fired due north to deflect to the right of the intended target. Meteorology Connection Meteorology Connection 8.3 Section Review 40. Center of Mass Can the center of mass of an object be located in an area where the object has no mass? Explain. 41. Stability of an Object Why is a modified vehicle with its body raised high on risers less stable than a similar vehicle with its body at normal height? 42. Conditions for Equilibrium Give an example of an object for each of the following conditions. a. rotational equilibrium, but not translational equilibrium b. translational equilibrium, but not rotational equilibrium 43. Center of Mass Where is the center of mass of a roll of masking tape? 44. Locating the Center of Mass Describe how you would find the center of mass of this textbook. 45. Rotating Frames of Reference A penny is placed on a rotating, old-fashioned record turntable. At the highest speed, the penny starts sliding outward. What are the forces acting on the penny? 46. Critical Thinking You have learned why the winds around a low-pressure area move in a counterclockwise direction. Would the winds move in the same or opposite direction in the southern hemisphere? Explain. physicspp.com/self_check_quiz Section 8.3 Equilibrium 217 218-219 CH08-LAB-845813 6/3/04 11:10 AM Page 218 Alternate CBL instructions can be found on the Web site. physicspp.com Translational and Rotational Equilibrium For maintenance on large buildings, scaffolding can be hung on the outside. In order for the scaffolding to support workers, it must be in translational and rotational equilibrium. If two or more forces act on the scaffolding, each can produce a rotation about either end. Scaffolding with uniform mass distribution acts as though all of the mass is concentrated at its center. In translational equilibrium the object is not accelerating; thus, the upward and downward forces are equal. In order to achieve rotational equilibrium, the sum of all the clockwise torques must equal the sum of all the counterclockwise torques as measured from a pivot point. That is, the net torque must be zero. In this lab you will model scaffolding hung from two ropes using a meterstick and spring scales, and use numbers to measure the forces on the scaffolding. QUESTION What conditions are required for equilibrium when parallel forces act on an object? Objectives Procedure \u25a0 Collect and organize data about the forces acting on the scaffolding. \u25a0 Describe clockwise and counterclockwise The left spring scale will be considered a pivot point for the purposes of this lab. Therefore, the lever arm will be measured from this point. torque. \u25a0 Compare and contrast translational and rotational equilibrium. Safety Precautions 1. Place the ring stands 80 cm apart. 2. Attach a Buret clamp to each of the ring stands. 3. Verify that the scales are set to zero before use. If the scales need to be adjusted, ask your teacher for assistance. 4. Hang a spring scale from each Buret clamp \u25a0 Use care to avoid dropping masses. attached to a ring stand. Materials meterstick two 0-5 N spring scales two ring stands two Buret clamps 500-g hooked mass 200-g hooked mass 5. Hook the meterstick onto the spring scale in such a manner that the 10-cm mark is supported by one hook and the 90-cm mark is supported by the other hook. 6. Read each spring scale and record the force in Data Table 1. 7. Hang a 500-g mass on the meterstick at the 30-cm mark. This point should be 20-cm from the left scale. 8. Read each spring scale and record the force in Data Table 1. 9. Hang a 200-g mass on the meterstick at the 70-cm mark. This point should be 60 cm from the left scale. 10. Read each spring scale and record the force in Data", " Table 1. 218 218-219 CH08-LAB-845813 6/3/04 11:11 AM Page 219 Data Table 1 Object Added Distance From Left Scale (m) Left Scale Reading (N) Right Scale Reading (N) Meterstick 500-g mass 200-g mass 0.4 0.2 0.6 c cc Lever Arm (m) Force (N) c (Nm) cc (Nm) Data Table 2 Object Added Meterstick 500-g mass 200-g mass Right scale Data Table 3 Object Added Meterstick 500-g mass 200-g mass Right scale Analyze 1. Calculate Find the mass of the meterstick. 2. Calculate Find the force, or weight, that results from each object and record it in Data Table 2. For the right scale, read the force it exerts and record it in Data Table 2. 3. Using the point where the left scale is attached as a pivot point, identify the forces located elsewhere that cause the scaffold to rotate clockwise or counterclockwise. Mark these in Data Table 2 with an x. 4. Record the lever arm distance of each force from the pivot point in Data Table 2. 5. Use Numbers Calculate the torque for each object by multiplying the force and lever arm distance. Record these values in Data Table 3. Conclude and Apply 1. Is the system in translational equilibrium? How do you know? 2. Draw a free-body diagram of your system, showing all the forces. 3. Compare and contrast the sum of the clockwise c, and the counterclockwise torques, torques, cc. 4. What is the percent difference between c and cc? Going Further Use additional masses at locations of your choice with your teacher\u2019s permission and record your data. Real-World Physics Research the safety requirements in your area for putting up, using, and dismantling scaffolding. To find out more about rotational motion, visit the Web site: physicspp.com 219 220 CH08-FEATURE-845813 3/12/04 6:23 AM Page 220 The Stability of Sport-Utility Vehicles The Stability of Sport-Utility Vehicles Why are sport-utility vehicles more flippable? Many believe that the large size of the sport-utility vehicle makes it more stable and secure. But, a sport-utility vehicle, as well as other tall vehicles such as vans, is much more likely to roll over than a car. The Problem A sport-utility vehicle has a high center of mass which makes it more likely to topple. Another factor that affects rollover is the static stability factor, which is the ratio of the track width to the center of mass. Track width is defined as half the distance between the two front wheels. The higher the static stability factor, the more likely a vehicle will stay upright. Many sport-utility vehicles have a center of mass 13 or 15 cm higher than passenger cars. Their track width, however, is about the same as that of passenger cars. Suppose the stability factor for a sport-utility vehicle is 1.06 and 1.43 for a car. Statistics show that in a single-vehicle crash, the sport-utility vehicle has a 37 percent chance of rolling over, while the car has a 10.6 percent chance of rolling over. However, the static stability factor oversimpli- fies the issue. Weather and driver behavior are also contributers to rollover crashes. Vehicle factors, such as tires, suspension systems, inertial properties, and advanced handling systems all play a role as well. It is true that most rollover crashes occur when a vehicle swerves off the road and hits a rut, soft soil, or other surface irregularity. This usually occurs when a driver is not paying proper attention or is speeding. Safe drivers greatly reduce their chances of being involved in a rollover accident by paying attention and driving at the correct speed. Still, weather and driver behavior being equal, the laws of physics indicate that sport-utility vehicles carry an increased risk. What Is Being Done? Some models are being built with wider track widths or stronger roofs. Optional side-curtain air bags have sensors to keep the bags inflated for up to 6 s, rather than the usual fraction of a second. This will cushion passengers if the vehicle should flip several times. 220 Technology and Society Engine control unit Wheel speed sensors Yaw sensor ESC hydraulic unit Steering angle sensor Wheel speed sensors An ESC system processes information from the sensors and automatically applies the brakes to individual wheels when instability is detected. A promising new technology called Electronic Stability Control (ESC) can be used to prevent rollover accidents. An ESC system has electronic sensors that detect when a vehicle begins to spin due to oversteering, and also when it begins to slide in a plowlike manner because of understeering. In these instances, an ESC system automatically applies the brakes at one or more wheels, thereby reorienting the vehicle in", " the right direction. Safe driving is the key to preventing many automobile accidents. Knowledge of the physics behind rollover accidents and the factors that affect rollover accidents may help make you an informed, safe driver. Going Further 1. Hypothesize In a multi-vehicle acci- dent, sport-utility vehicles generally fare better than the passenger cars involved in the accident. Why is this so? 2. Debate the Issue ESC is a life-saving technology. Should it be mandatory in all sport-utility vehicles? Why or why not? 221-227 CH08-SG CA STP-845813 3/12/04 6:25 AM Page 221 8.1 Describing Rotational Motion Vocabulary \u2022 radian (p. 197) \u2022 angular displacement (p. 198) \u2022 angular velocity (p. 198) \u2022 angular acceleration (p. 199) Key Concepts \u2022 Angular position and its changes are measured in radians. One complete revolution is 2 rad. \u2022 Angular velocity is given by the following equation. t \u2022 Angular acceleration is given by the following equation. t \u2022 For a rotating, rigid object, the angular displacement, velocity, and acceleration can be related to the linear displacement, velocity, and acceleration for any point on the object. d r v r a r 8.2 Rotational Dynamics Vocabulary \u2022 lever arm (p. 201) \u2022 torque (p. 202) \u2022 moment of inertia (p. 205) \u2022 Newton\u2019s second law for rotational motion (p. 208) Key Concepts \u2022 When torque is exerted on an object, its angular velocity changes. \u2022 Torque depends on the magnitude of the force, the distance from the axis of rotation at which it is applied, and the angle between the force and the radius from the axis of rotation to the point where the force is applied. Fr sin \u2022 The moment of inertia of an object depends on the way the object\u2019s mass is distributed about the rotational axis. For a point object: I mr2 \u2022 Newton\u2019s second law for rotational motion states that angular acceleration is directly proportional to the net torque and inversely proportional to the moment of inertia. et n I Key Concepts \u2022 The center of mass of an object is the point on the object that moves in the same way that a point particle would move. \u2022 An object is stable against rollover if its center of mass is above its base. \u2022 An object is in equilibrium if there are no net forces exerted on it and if there are no net torques acting on it. \u2022 Centrifugal \u201cforce\u201d and the Coriolis \u201cforce\u201d are two apparent forces that appear when a rotating object is analyzed from a coordinate system that rotates with it. 8.3 Equilibrium Vocabulary \u2022 center of mass (p. 211) \u2022 centrifugal \u201cforce\u201d (p. 216) \u2022 Coriolis \u201cforce\u201d (p. 217) physicspp.com/vocabulary_puzzlemaker 221 221-227 CH08-SG CA STP-845813 3/12/04 6:26 AM Page 222 Concept Mapping 47. Complete the following concept map using the following terms: angular acceleration, radius, tangential acceleration, centripetal acceleration. Angular velocity Mastering Concepts 48. A bicycle wheel rotates at a constant 25 rev/min. Is its angular velocity decreasing, increasing, or constant? (8.1) 49. A toy rotates at a constant 5 rev/min. Is its angular acceleration positive, negative, or zero? (8.1) 50. Do all parts of Earth rotate at the same rate? Explain. (8.1) 51. A unicycle wheel rotates at a constant 14 rev/min. Is the total acceleration of a point on the tire inward, outward, tangential, or zero? (8.1) 52. Think about some possible rotations of your textbook. Are the moments of inertia about these three axes the same or different? Explain. (8.2) 53. Torque is important when tightening bolts. Why is force not important? (8.2) 54. Rank the torques on the five doors shown in Figure 8-18 from least to greatest. Note that the magnitude of all the forces is the same. (8.2) A C B D E \u25a0 Figure 8-18 55. Explain how you can change an object\u2019s angular frequency. (8.2) 56. To balance a car\u2019s wheel, it is placed on a vertical shaft and weights are added to make the wheel horizontal. Why is this equivalent to moving the center of mass until it is at the center of the wheel? (8.3) 57. A stunt driver maneuvers a monster truck so that it is traveling on only two wheels. Where is the center of mass of the truck? (8.3) 58. Suppose you stand flat-footed, then you rise and balance on t", "iptoe. If you stand with your toes touching a wall, you cannot balance on tiptoe. Explain. (8.3) 59. Why does a gymnast appear to be floating on air when she raises her arms above her head in a leap? (8.3) 60. Why is a vehicle with wheels that have a large diameter more likely to roll over than a vehicle with wheels that have a smaller diameter? (8.3) Applying Concepts 61. Two gears are in contact and rotating. One is larger than the other, as shown in Figure 8-19. Compare their angular velocities. Also compare the linear velocities of two teeth that are in contact. \u25a0 Figure 8-19 62. Videotape When a videotape is rewound, why does it wind up fastest towards the end? 63. Spin Cycle What does a spin cycle of a washing machine do? Explain in terms of the forces on the clothes and water. 64. How can you experimentally find the moment of inertia of an object? 65. Bicycle Wheels Three bicycle wheels have masses that are distributed in three different ways: mostly at the rim, uniformly, and mostly at the hub. The wheels all have the same mass. If equal torques are applied to them, which one will have the greatest angular acceleration? Which one will have the least? 66. Bowling Ball When a bowling ball leaves a bowler\u2019s hand, it does not spin. After it has gone about half the length of the lane, however, it does spin. Explain how its rotation rate increased and why it does not continue to increase. 222 Chapter 8 Rotational Motion For more problems, go to Additional Problems, Appendix B. 221-227 CH08-SG CA STP-845813 6/10/04 6:43 PM Page 223 67. Flat Tire Suppose your car has a flat tire. You get 73. The outer edge of a truck tire that has a radius of out your tools and find a lug wrench to remove the nuts off the bolt studs. You find it impossible to turn the nuts. Your friend suggests ways you might produce enough torque to turn them. What three ways might your friend suggest? 45 cm has a velocity of 23 m/s. What is the angular velocity of the tire in rad/s? 74. A steering wheel is rotated through 128\u00b0, as shown in Figure 8-22. Its radius is 22 cm. How far would a point on the steering wheel\u2019s edge move? 68. Tightrope Walkers Tightrope walkers often carry long poles that sag so that the ends are lower than the center as shown in Figure 8-20. How does such a pole increase the tightrope walker\u2019s stability? Hint: Consider both center of mass and moment of inertia. 22 cm 128\u00b0 \u25a0 Figure 8-20 69. Merry-Go-Round While riding a merry-go-round, you toss a key to a friend standing on the ground. For your friend to be able to catch the key, should you toss it a second or two before you reach the spot where your friend is standing or wait until your friend is directly behind you? Explain. 70. Why can you ignore forces that act on the axis of rotation of an object in static equilibrium when determining the net torque? 71. In solving problems about static equilibrium, why is the axis of rotation often placed at a point where one or more forces are acting on the object? Mastering Problems 8.1 Describing Rotational Motion 72. A wheel is rotated so that a point on the edge moves through 1.50 m. The radius of the wheel is 2.50 m, as shown in Figure 8-21. Through what angle (in radians) is the wheel rotated? 2.50 m 1.50 m \u25a0 Figure 8-22 75. Propeller A propeller spins at 1880 rev/min. a. What is its angular velocity in rad/s? b. What is the angular displacement of the propeller in 2.50 s? 76. The propeller in the previous problem slows from 475 rev/min to 187 rev/min in 4.00 s. What is its angular acceleration? 77. An automobile wheel with a 9.00 cm radius, as shown in Figure 8-23, rotates at 2.50 rad/s. How fast does a point 7.00 cm from the center travel? m 0 c 9.0 7.00 cm \u25a0 Figure 8-23 78. Washing Machine A washing machine\u2019s two spin cycles are 328 rev/min and 542 rev/min. The diameter of the drum is 0.43 m. a. What is the ratio of the centripetal accelerations for the fast and slow spin cycles? Recall that ac 2 v and v rw r b. What is the ratio of the linear velocity of an \u25a0 Figure 8-21 physicspp.com/chapter_test object at", " the surface of the drum for the fast and slow spin cycles? 79. Find the maximum centripetal acceleration in terms of g for the washing machine in problem 78. Chapter 8 Assessment 223 Mark D. Phillips/AFP/CORBIS 221-227 CH08-SG CA STP-845813 6/9/04 4:23 PM Page 224 80. A laboratory ultracentrifuge is designed to produce a centripetal acceleration of 0.35106 g at a distance of 2.50 cm from the axis. What angular velocity in rev/min is required? 8.2 Rotational Dynamics 81. Wrench A bolt is to be tightened with a torque of 8.0 Nm. If you have a wrench that is 0.35 m long, what is the least amount of force you must exert? 82. What is the torque on a bolt produced by a 15-N force exerted perpendicular to a wrench that is 25 cm long, as shown in Figure 8-24? F 90\u00b0 25 cm \u25a0 Figure 8-24 83. A toy consisting of two balls, each 0.45 kg, at the ends of a 0.46-m-long, thin, lightweight rod is shown in Figure 8-25. Find the moment of inertia of the toy. The moment of inertia is to be found about the center of the rod. 0.45 kg 0.45 kg 0.46 m \u25a0 Figure 8-25 84. A bicycle wheel with a radius of 38 cm is given an angular acceleration of 2.67 rad/s2 by applying a force of 0.35 N on the edge of the wheel. What is the wheel\u2019s moment of inertia? 85. Toy Top A toy top consists of a rod with a diameter of 8.0-mm and a disk of mass 0.0125 kg and a diameter of 3.5 cm. The moment of inertia of the rod can be neglected. The top is spun by wrapping a string around the rod and pulling it with a velocity that increases from zero to 3.0 m/s over 0.50 s. a. What is the resulting angular velocity of the top? b. What force was exerted on the string? 8.3 Equilibrium 86. A 12.5-kg board, 4.00 m long, is being held up on one end by Ahmed. He calls for help, and Judi responds. a. What is the least force that Judi could exert to lift the board to the horizontal position? What part of the board should she lift to exert this force? b. What is the greatest force that Judi could exert to lift the board to the horizontal position? What part of the board should she lift to exert this force? 87. Two people are holding up the ends of a 4.25-kg wooden board that is 1.75 m long. A 6.00-kg box sits on the board, 0.50 m from one end, as shown in Figure 8-26. What forces do the two people exert? 6.00 kg 1.25 m 0.50 m \u25a0 Figure 8-26 88. A car\u2019s specifications state that its weight distribution is 53 percent on the front tires and 47 percent on the rear tires. The wheel base is 2.46 m. Where is the car\u2019s center of mass? Mixed Review 89. A wooden door of mass, m, and length, l, is held horizontally by Dan and Ajit. Dan suddenly drops his end. a. What is the angular acceleration of the door just after Dan lets go? b. Is the acceleration constant? Explain. 90. Topsoil Ten bags of topsoil, each weighing 175 N, are placed on a 2.43-m-long sheet of wood. They are stacked 0.50 m from one end of the sheet of wood, as shown in Figure 8-27. Two people lift the sheet of wood, one at each end. Ignoring the weight of the wood, how much force must each person exert? 1.93 m 0.50 m \u25a0 Figure 8-27 224 Chapter 8 Rotational Motion For more problems, go to Additional Problems, Appendix B. 221-227 CH08-SG CA STP-845813 6/3/04 11:16 AM Page 225 91. Basketball A basketball is rolled down the court. A regulation basketball has a diameter of 24.1 cm, a mass of 0.60 kg, and a moment of inertia of 5.8103 kgm2. The basketball\u2019s initial velocity is 2.5 m/s. a. What is its initial angular velocity? b. The ball rolls a total of 12 m. How many revolutions does it make? c. What is its total angular displacement? 92. The basketball in the previous problem stops rolling after traveling 12 m. 97. The second hand on a watch is 12 mm long. What is the velocity of its tip? 98. L", "umber You buy a 2.44-m-long piece of 10 cm 10 cm lumber. Your friend buys a piece of the same size and cuts it into two lengths, each 1.22 m long, as shown in Figure 8-29. You each carry your lumber on your shoulders. a. Which load is easier to lift? Why? b. Both you and your friend apply a torque with your hands to keep the lumber from rotating. Which load is easier to keep from rotating? Why? a. If its acceleration was constant, what was its angular acceleration? b. What torque was acting on it as it was slowing down? 93. A cylinder with a 50 m diameter, as shown in Figure 8-28, is at rest on a surface. A rope is wrapped around the cylinder and pulled. The cylinder rolls without slipping. a. After the rope has been pulled a distance of 2.50 m at a constant speed, how far has the center of mass of the cylinder moved? b. If the rope was pulled a distance of 2.50 m in 1.25 s, how fast was the center of mass of the cylinder moving? c. What is the angular velocity of the cylinder? 50 m \u25a0 Figure 8-28 94. Hard Drive A hard drive on a modern computer spins at 7200 rpm (revolutions per minute). If the drive is designed to start from rest and reach operating speed in 1.5 s, what is the angular acceleration of the disk? 95. Speedometers Most speedometers in automobiles measure the angular velocity of the transmission and convert it to speed. How will increasing the diameter of the tires affect the reading of the speedometer? 96. A box is dragged across the floor using a rope that is a distance h above the floor. The coefficient of friction is 0.35. The box is 0.50 m high and 0.25 m wide. Find the force that just tips the box. 2.44 m 1.22 m 1.22 m \u25a0 Figure 8-29 99. Surfboard Harris and Paul carry a surfboard that is 2.43 m long and weighs 143 N. Paul lifts one end with a force of 57 N. a. What force must Harris exert? b. What part of the board should Harris lift? 100. A steel beam that is 6.50 m long weighs 325 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending from each end. Suki, who weighs 575 N, stands on the beam in the center and then walks toward one end. How close to the end can she come before the beam begins to tip? Thinking Critically 101. Apply Concepts Consider a point on the edge of a rotating wheel. a. Under what conditions can the centripetal acceleration be zero? b. Under what conditions can the tangential (linear) acceleration be zero? c. Can the tangential acceleration be nonzero while the centripetal acceleration is zero? Explain. d. Can the centripetal acceleration be nonzero while the tangential acceleration is zero? Explain. 102. Apply Concepts When you apply the brakes in a car, the front end dips. Why? physicspp.com/chapter_test Chapter 8 Assessment 225 221-227 CH08-SG CA STP-845813 3/12/04 6:29 AM Page 226 103. Analyze and Conclude A banner is suspended from a horizontal, pivoted pole, as shown in Figure 8-30. The pole is 2.10 m long and weighs 175 N. The banner, which weighs 105 N, is suspended 1.80 m from the pivot point or axis of rotation. What is the tension in the cable supporting the pole? 105. Analyze and Conclude Gerald and Evelyn carry the following objects up a flight of stairs: a large mirror, a dresser, and a television. Evelyn is at the front end, and Gerald is at the bottom end. Assume that both Evelyn and Gerald exert only upward forces. Cable Axis of rotation Center of mass 25.0\u00b0 1.05 m 1.80 m 2.10 m \u25a0 Figure 8-30 104. Analyze and Conclude A pivoted lamp pole is shown in Figure 8-31. The pole weighs 27 N, and the lamp weighs 64 N. a. What is the torque caused by each force? b. Determine the tension in the rope supporting the lamp pole. y Axis of rotation x Rope 105.0\u00b0 0.33 m 0.44 m Lamp \u25a0 Figure 8-31 a. Draw a free-body diagram showing Gerald and Evelyn exerting the same force on the mirror. b. Draw a free-body diagram showing Gerald exerting more force on the bottom of the dresser. c. Where would the center of mass of the television have to be so that Gerald carries all the weight? Writing in Physics 106. Astronomers know that if a satellite is too close to a planet", ", it will be torn apart by tidal forces. That is, the difference in the gravitational force on the part of the satellite nearest the planet and the part farthest from the planet is stronger than the forces holding the satellite together. Do research on the Roche limit and determine how close the Moon would have to orbit Earth to be at the Roche limit. 107. Automobile engines are rated by the torque that they produce. Research and explain why torque is an important quantity to measure. Cumulative Review 108. Two blocks, one of mass 2.0 kg and the other of mass 3.0 kg, are tied together with a massless rope. This rope is strung over a massless, resistance-free pulley. The blocks are released from rest. Find the following. (Chapter 4) a. the tension in the rope b. the acceleration of the blocks. 109. Eric sits on a see-saw. At what angle, relative to the vertical, will the component of his weight parallel to the plane be equal to one-third the perpendicular component of his weight? (Chapter 5) 110. The pilot of a plane wants to reach an airport 325 km due north in 2.75 hours. A wind is blowing from the west at 30.0 km/h. What heading and airspeed should be chosen to reach the destination on time? (Chapter 6) 111. A 60.0-kg speed skater with a velocity of 18.0 m/s comes into a curve of 20.0-m radius. How much friction must be exerted between the skates and ice to negotiate the curve? (Chapter 6) 226 Chapter 8 Rotational Motion For more problems, go to Additional Problems, Appendix B. 221-227 CH08-SG CA STP-845813 6/3/04 11:17 AM Page 227 Multiple Choice 1. The illustration below shows two boxes on opposite ends of a board that is 3.0 m long. The board is supported in the middle by a fulcrum. The box on the left has a mass, m1, of 25 kg, and the box on the right has a mass, m2, of 15 kg. How far should the fulcrum be positioned from the left side of the board in order to balance the masses horizontally? 0.38 m 0.60 m 1.1 m 1.9 m m 2 m 1 2. A force of 60 N is exerted on one end of a 1.0-m-long lever. The other end of the lever is attached to a rotating rod that is perpendicular to the lever. By pushing down on the end of the lever, you can rotate the rod. If the force on the lever is exerted at an angle of 30\u00b0, what torque is exerted on the lever? (sin 30\u00b0 0.5; cos 30\u00b0 0.87; tan 30\u00b0 0.58) 30 N 52 N 60 N 69 N 3. A child attempts to use a wrench to remove a nut on a bicycle. Removing the nut requires a torque of 10 N\u00b7m. The maximum force the child is capable of exerting at a 90\u00b0 angle is 50 N. What is the length of the wrench the child must use to remove the nut? 0.1 m 0.15 m 0.2 m 0.25 m 4. A car moves a distance of 420 m. Each tire on the car has a diameter of 42 cm. Which of the following shows how many revolutions each tire makes as they move that distance? 5.0 101 rev 1.0 102 rev 1.5 102 rev 1.0 103 rev 5. A thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center. A force of 25 N is exerted tangentially to the hoop. If the hoop\u2019s radius is 2.0 m, what is its angular acceleration? 1.3 rad/s 2.5 rad/s 5.0 rad/s 6.3 rad/s 6. Two of the tires on a farmer\u2019s tractor have diameters of 1.5 m. If the farmer drives the tractor at a linear velocity of 3.0 m/s, what is the angular velocity of each tire? 2.0 rad/s 2.3 rad/s 4.0 rad/s 4.5 rad/s Extended Answer 7. You use a 25-cm long wrench to remove the lug nuts on a car wheel, as shown in the illustration below. If you pull up on the end of the wrench with a force of 2.0102 N at an angle of 30\u00b0, what is the torque on the wrench? (sin 30\u00b0 0.5, cos 30\u00b0 0.87) F 30\u00b0 25 cm When Eliminating, Cross It Out Consider each answer choice individually and cross out the ones you have eliminated. If you cannot write in the test booklet, use the scratch paper to list and cross off the answer choices. You will save time", " and stop yourself from choosing an answer you have mentally eliminated. physicspp.com/standardized_test Chapter 8 Standardized Test Practice 227 228-235 CH09-S1-845813 3/19/04 4:59 PM Page 228 What You\u2019ll Learn \u2022 You will describe momentum and impulse and apply them to the interactions between objects. \u2022 You will relate Newton\u2019s third law of motion to conservation of momentum. \u2022 You will explore the momentum of rotating objects. Why It\u2019s Important Momentum is the key to success in many sporting events, including baseball, football, ice hockey, and tennis. Baseball Every baseball player dreams of hitting a home run. When a player hits the ball, at the moment of collision, the ball and the bat are deformed by the collision. The resulting change in momentum determines the batter\u2019s success. Think About This What is the force on a baseball bat when a home run is hit out of the park? physicspp.com 228 fotobankyokohama/firstlight.ca 228-235 CH09-S1-845813 3/10/04 10:25 PM Page 229 What happens when a hollow plastic ball strikes a bocce ball? Question What direction will a hollow plastic ball and a bocce ball move after a head-on collision? Critical Thinking What factor(s) would cause the bocce ball to move backward after colliding with the hollow plastic ball? Procedure 1. Roll a bocce ball and a hollow plastic ball toward each other on a smooth surface. 2. Observe the direction each one moves after the collision. 3. Repeat the experiment, this time keeping the bocce ball stationary, while rolling the hollow plastic ball toward it. 4. Observe the direction each one moves after the collision. 5. Repeat the experiment one more time, but keep the hollow plastic ball stationary, while rolling the bocce ball toward it. 6. Observe the direction each one moves after the collision. Analysis What factors affect how fast the balls move after the collision? What factors determine the direction each one moves after the collision? 9.1 Impulse and Momentum It is always exciting to watch a baseball player hit a home run. The pitcher fires the baseball toward the plate. The batter swings at the baseball and the baseball recoils from the impact of the bat at high speed. Rather than concentrating on the force between the baseball and bat and their resulting accelerations, as in previous chapters, you will approach this collision in a different way in this chapter. The first step in analyzing this type of interaction is to describe what happens before, during, and after the collision between the baseball and bat. You can simplify the collision between the baseball and the bat by making the assumption that all motion is in the horizontal direction. Before the collision, the baseball moves toward the bat. During the collision, the baseball is squashed against the bat. After the collision, however, the baseball moves at a higher velocity away from the bat, and the bat continues in its path, but at a slower velocity. Objectives \u2022 Define the momentum of an object. \u2022 Determine the impulse given to an object. \u2022 Define the angular momentum of an object. Vocabulary impulse momentum impulse-momentum theorem angular momentum angular impulse-angular momentum theorem Section 9.1 Impulse and Momentum 229 Horizons Companies 228-235 CH09-S1-845813 6/3/04 12:12 PM Page 230 Force on a Baseball Impulse and Momentum 1.5104 1.0104 5.0103 ) N ( e c r o F 0.0 1.5 1.0 0.5 0.0 Time (ms) \u25a0 Figure 9-1 The force acting on a baseball increases, then rapidly decreases during a collision, as shown in this force-time graph. \u2022 Momentum and impulse vectors are orange. \u2022 Force vectors are blue. \u2022 Acceleration vectors are violet. \u2022 Velocity vectors are red. \u2022 Displacement vectors are green. 1.0 0.5 1.5 How are the velocities of the ball, before and after the collision, related to the force acting on it? Newton\u2019s second law of motion describes how the velocity of an object is changed by a net force acting on it. The change in velocity of the ball must have been caused by the force exerted by the bat on the ball. The force changes over time, as shown in Figure 9-1. Just after contact is made, the ball is squeezed, and the force increases. After the force reaches its maximum, which is more than 10,000 times the weight of the ball, the ball recovers its shape and snaps away from the bat. The force rapidly returns to zero. This whole event takes place within about 3.0 ms. How can you calculate the change in velocity of the baseball? Impulse Newton\u2019s second law of motion, F ma, can", " be rewritten by using the definition of acceleration as the change in velocity divided by the time needed to make that change. It can be represented by the following equation: F ma m Multiplying both sides of the equation by the time interval, t, results in the following equation: v t Ft mv Impulse, or Ft, is the product of the average force on an object and the time interval over which it acts. Impulse is measured in newton-seconds. For instances in which the force varies with time, the magnitude of an impulse is found by determining the area under the curve of a force-time graph, such as the one shown in Figure 9-1. vi. Therefore, mv mvf The right side of the equation, mv, involves the change in velocity: v vf mvi. The product of the object\u2019s mass, m, and the object\u2019s velocity, v, is defined as the momentum of the object. Momentum is measured in kgm/s. An object\u2019s momentum, also known as linear momentum, is represented by the following equation. Momentum p mv The momentum of an object is equal to the mass of the object times the object\u2019s velocity. mvi. Because mvf Recall the equation Ft mv mvf pi, this equation can be rewritten as follows: Ft mv pf pf and pi. mvi pi, describes the change in momentum The right side of this equation, pf of an object. Thus, the impulse on an object is equal to the change in its momentum, which is called the impulse-momentum theorem. The impulse-momentum theorem is represented by the following equation. Impulse-Momentum Theorem Ft pf The impulse on an object is equal to the object\u2019s final momentum minus the object\u2019s initial momentum. pi 230 Chapter 9 Momentum and Its Conservation 228-235 CH09-S1-845813 6/29/04 2:30 PM Page 231 If the force on an object is constant, the impulse is the product of the force multiplied by the time interval over which it acts. Generally, the force is not constant, however, and the impulse is found by using an average force multiplied by the time interval over which it acts, or by finding the area under a force-time graph. Because velocity is a vector, momentum also is a vector. Similarly, impulse is a vector because force is a vector. This means that signs will be important for motion in one dimension. Using the Impulse-Momentum Theorem What is the change in momentum of a baseball? From the impulsemomentum theorem, you know that the change in momentum is equal to the impulse acting on it. The impulse on a baseball can be calculated by using a force-time graph. In Figure 9-1, the area under the curve is approximately 13.1 Ns. The direction of the impulse is in the direction of the force. Therefore, the change in momentum of the ball also is 13.1 Ns. Because 1 Ns is equal to 1 kgm/s, the momentum gained by the ball is 13.1 kgm/s in the direction of the force acting on it. Assume that a batter hits a fastball. Before the collision of the ball and bat, the ball, with a mass of 0.145 kg, has a velocity of 38 m/s. Assume that the positive direction is toward the pitcher. Therefore, the baseball\u2019s (0.145 kg)(38 m/s) 5.5 kgm/s. momentum is pi What is the momentum of the ball after the collision? Solve the Ft. impulse-momentum theorem for the final momentum: pf The ball\u2019s final momentum is the sum of the initial momentum and the impulse. Thus, the ball\u2019s final momentum is calculated as follows. pi pf 13.1 kgm/s pi 5.5 kgm/s 13.1 kgm/s 7.6 kgm/s What is the baseball\u2019s final velocity? Because pf mvf, solving for vf yields the following: kg /s m.6 pf 52 m/s m g k 45 0 m 7.1 /s vf A speed of 52 m/s is fast enough to clear most outfield fences if the base- ball is hit in the correct direction. Using the Impulse-Momentum Theorem to Save Lives A large change in momentum occurs only when there is a large impulse. A large impulse can result either from a large force acting over a short period of time or from a smaller force acting over a long period of time. What happens to the driver when a crash suddenly stops a car? An impulse is needed to bring the driver\u2019s momentum to zero. According to the impulse-momentum equation, Ft pf pi.", " The final momentum, pf, is zero. The initial momentum, pi, is the same with or without an air bag. Thus, the impulse, Ft, also is the same. An air bag, such as the one shown in Figure 9-2, reduces the force by increasing the time interval during which it acts. It also exerts the force over a larger area of the person\u2019s body, thereby reducing the likelihood of injuries. Running Shoes Running is hard on the feet. When a runner\u2019s foot strikes the ground, the force exerted by the ground on it is as much as four times the runner\u2019s weight. The cushioning in an athletic shoe is designed to reduce this force by lengthening the time interval over which the force is exerted. \u25a0 Figure 9-2 An air bag is inflated during a collision when the force due to the impact triggers the sensor. The chemicals in the air bag\u2019s inflation system react and produce a gas that rapidly inflates the air bag. Section 9.1 Impulse and Momentum 231 Rick Fischer/Masterfile 228-235 CH09-S1-845813 6/1/04 4:15 AM Page 232 Average Force A 2200-kg vehicle traveling at 94 km/h (26 m/s) can be stopped in 21s by gently applying the brakes. It can be stopped in 3.8 s if the driver slams on the brakes, or in 0.22 s if it hits a concrete wall. What average force is exerted on the vehicle in each of these stops? 1 Analyze and Sketch the Problem \u2022 Sketch the system. \u2022 Include a coordinate axis and select the positive direction x 2200 kg to be the direction of the velocity of the car. 94 km/h \u2022 Draw a vector diagram for momentum and impulse. Known: m 2200 kg +26 m/s vi +0.0 m/s vf tgentle braking thard braking thitting a wall 21 s 3.8 s 0.22 s Unknown: Fgentle braking Fhard braking Fhitting a wall??? 2 Solve for the Unknown Determine the initial momentum, pi. pi mvi (2200 kg)(26 m/s) 5.7104 kgm/s Substitute m 2200 kg, vi 26 m/s Vector diagram pi pf Impulse Determine the final momentum, pf. pf mvf (2200 kg)(0.0 m/s) 0.0 kgm/s Substitute m 2200 kg, vf 0.0 m/s Apply the impulse-momentum theorem to obtain the force needed to stop the vehicle. Ft pf Ft (0.0 kgm/s) (5.7104 kgm/s) pi Substitute pf 0.0 kgm/s, pi 5.7104 kgm/s 5.7104 kgm/s 5.7104 kgm/s t F Fgentle braking Fhard braking F hitting a wall 5.7104 kgm/s 21 s 2.7103 N 5.7104 kgm/s 3.8 s 1.5104 N 5.7104 kgm/s 0.22 s 2.6105 N Substitute tgentle braking 21 s Substitute thard braking 3.8 s Substitute thitting a wall 0.22 s Math Handbook Operations with Significant Digits pages 835\u2013836 3 Evaluate the Answer \u2022 Are the units correct? Force is measured in newtons. \u2022 Does the direction make sense? Force is exerted in the direction opposite to the velocity of the car and thus, is negative. \u2022 Is the magnitude realistic? People weigh hundreds of newtons, so it is reasonable that the force needed to stop a car would be in the thousands of newtons. The impulse is the same for all three stops. Thus, as the stopping time is shortened by more than a factor of 10, the force is increased by more than a factor of 10. 232 Chapter 9 Momentum and Its Conservation 228-235 CH09-S1-845813 6/1/04 4:15 AM Page 233 1. A compact car, with mass 725 kg, is moving at 115 km/h toward the east. Sketch the moving car. a. Find the magnitude and direction of its momentum. Draw an arrow on your sketch showing the momentum. b. A second car, with a mass of 2175 kg, has the same momentum. What is its velocity? 2. The driver of the compact car in the previous problem suddenly applies the brakes hard for 2.0 s. As a result, an average force of 5.0103 N is exerted on the car to slow it down. a. What is the change in momentum; that is, the magnitude and direction of the impulse, on the car? b. Complete the \u201cbefore\u201d and \u201cafter\u201d sketches, and determine the momentum and", " the velocity of the car now. 3. A 7.0-kg bowling ball is rolling down the alley with a velocity of 2.0 m/s. For each impulse, shown in Figures 9-3a and 9-3b, find the resulting speed and direction of motion of the bowling ball. 4. The driver accelerates a 240.0-kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s to 28.0 m/s over a time interval of 60.0 s. a. Sketch the event, showing the initial and final situations. b. What is the snowmobile\u2019s change in momentum? What is the impulse on the snowmobile Time (s Time (s) \u25a0 Figure 9-3 c. What is the magnitude of the average force that is exerted on the snowmobile? 5. Suppose a 60.0-kg person was in the vehicle that hit the concrete wall in Example Problem 1. The velocity of the person equals that of the car both before and after the crash, and the velocity changes in 0.20 s. Sketch the problem. a. What is the average force exerted on the person? b. Some people think that they can stop their bodies from lurching forward in a vehicle that is suddenly braking by putting their hands on the dashboard. Find the mass of an object that has a weight equal to the force you just calculated. Could you lift such a mass? Are you strong enough to stop your body with your arms? Angular Momentum As you learned in Chapter 8, the angular velocity of a rotating object changes only if torque is applied to it. This is a statement of Newton\u2019s law for rotational motion, I/t. This equation can be rearranged in the same way as Newton\u2019s second law of motion was, to produce t I. The left side of this equation, t, is the angular impulse of the rotating object. The right side can be rewritten as i. The product of a f rotating object\u2019s moment of inertia and angular velocity is called angular momentum, which is represented by the symbol L. The angular momentum of an object can be represented by the following equation. Angular Momentum L = I The angular momentum of an object is equal to the product of the object\u2019s moment of inertia and the object\u2019s angular velocity. Section 9.1 Impulse and Momentum 233 228-235 CH09-S1-845813 3/10/04 10:36 PM Page 234 Angular momentum is measured in kgm2/s. Just as the linear momentum of an object changes when an impulse acts on it, the angular momentum of an object changes when an angular impulse acts on it. Thus, the angular impulse on the object is equal to the change in the object\u2019s angular momentum, which is called the angular impulse-angular momentum theorem. The angular impulse-angular momentum theorem is represented by the following equation. Angular Impulse-Angular Momentum Theorem t Lf The angular impulse on an object is equal to the object\u2019s final angular momentum minus the object\u2019s initial angular momentum. Li If there are no forces acting on an object, its linear momentum is constant. If there are no torques acting on an object, its angular momentum is also constant. Because an object\u2019s mass cannot be changed, if its momentum is constant, then its velocity is also constant. In the case of angular momentum, however, the object\u2019s angular velocity does not remain constant. This is because the moment of inertia depends on the object\u2019s mass and the way it is distributed about the axis of rotation or revolution. Thus, the angular velocity of an object can change even if no torques are acting on it. Consider, for example, a planet orbiting the Sun. The torque on the planet is zero because the gravitational force acts directly toward the Sun. Therefore, the planet\u2019s angular momentum is constant. When the distance between the planet and the Sun decreases, however, the planet\u2019s moment of inertia of revolution in orbit about the Sun also decreases. Thus, the planet\u2019s angular velocity increases and it moves faster. This is an explanation of Kepler\u2019s second law of planetary motion, based on Newton\u2019s laws of motion. \u25a0 Figure 9-4 The diver\u2019s center of mass is in front of her feet as she gets ready to dive (a). As the diver changes her moment of inertia by moving her arms and legs to increase her angular momentum, the location of the center of mass changes, but the path of the center of mass remains a parabola (b). a b Astronomy Connection Astronomy Connection Center of mass Fg F 234 Tim Fuller Chapter 9 Momentum and Its Conservation 228-235 CH09-S1-845813 6/1/04 1:26 AM Page 235 Consider the diver in Figure 9-4.", " How does she start rotating her body? She uses the diving board to apply an external torque to her body. Then, she moves her center of mass in front of her feet and uses the board to give a final upward push to her feet. This torque acts over time, t, and thus increases the angular momentum of the diver. Before the diver reaches the water, she can change her angular velocity by changing her moment of inertia. She may go into a tuck position, grabbing her knees with her hands. By moving her mass closer to the axis of rotation, the diver decreases her moment of inertia and increases her angular velocity. When she nears the water, she stretches her body straight, thereby increasing the moment of inertia and reducing the angular velocity. As a result, she goes straight into the water. An ice-skater uses a similar method to spin. To begin rotating on one foot, the ice-skater applies an external torque to her body by pushing a portion of the other skate into the ice, as shown in Figure 9-5. If she pushes on the ice in one direction, the ice will exert a force on her in the opposite direction. The force results in a torque if the force is exerted some distance away from the pivot point, and in a direction that is not toward it. The greatest torque for a given force will result if the push is perpendicular to the lever arm. The ice-skater then can control his angular velocity by changing her moment of inertia. Both arms and one leg can be extended from the body to slow the rotation, or pulled in close to the axis of rotation to speed it up. To stop spinning, another torque must be exerted by using the second skate to create a way for the ice to exert the needed force. 9.1 Section Review \u25a0 Figure 9-5 To spin on one foot, an ice-skater extends one leg and pushes on the ice. The ice exerts an equal and opposite force on her body and produces an external torque. 6. Momentum Is the momentum of a car traveling south different from that of the same car when it travels north at the same speed? Draw the momentum vectors to support your answer. 7. Impulse and Momentum When you jump from a height to the ground, you let your legs bend at the knees as your feet hit the floor. Explain why you do this in terms of the physics concepts introduced in this chapter. 8. Momentum Which has more momentum, a supertanker tied to a dock or a falling raindrop? 9. Impulse and Momentum A 0.174-kg softball is pitched horizontally at 26.0 m/s. The ball moves in the opposite direction at 38.0 m/s after it is hit by the bat. a. Draw arrows showing the ball\u2019s momentum before and after the bat hits it. b. What is the change in momentum of the ball? c. What is the impulse delivered by the bat? d. If the bat and softball are in contact for 0.80 ms, what is the average force that the bat exerts on the ball? 10. Momentum The speed of a basketball as it is dribbled is the same when the ball is going toward the floor as it is when the ball rises from the floor. Is the basketball\u2019s change in momentum equal to zero when it hits the floor? If not, in which direction is the change in momentum? Draw the basketball\u2019s momentum vectors before and after it hits the floor. 11. Angular Momentum An ice-skater spins with his arms outstretched. When he pulls his arms in and raises them above his head, he spins much faster than before. Did a torque act on the iceskater? If not, how could his angular velocity have increased? 12. Critical Thinking An archer shoots arrows at a target. Some of the arrows stick in the target, while others bounce off. Assuming that the masses of the arrows and the velocities of the arrows are the same, which arrows produce a bigger impulse on the target? Hint: Draw a diagram to show the momentum of the arrows before and after hitting the target for the two instances. physicspp.com/self_check_quiz Section 9.1 Impulse and Momentum 235 Rick Stewart/Getty Images 236-245 CH09-S2-845813 3/10/04 11:22 PM Page 236 9.2 Conservation of Momentum Objectives \u2022 Relate Newton\u2019s third law to conservation of momentum. \u2022 Recognize the conditions under which momentum is conserved. \u2022 Solve conservation of momentum problems. Vocabulary closed system isolated system law of conservation of momentum law of conservation of angular momentum Before Collision (initial) C D pCi pDi During Collision C D FD on C FC on D After Collision (final) C D pCf pDf \u25a0 Figure 9-6 When two balls collide, they exert forces on each other that change", " their momenta. In the first section of this chapter, you learned how a force applied during a time interval changes the momentum of a baseball. In the discussion of Newton\u2019s third law of motion, you learned that forces are the result of interactions between two objects. The force of a bat on a ball is accompanied by an equal and opposite force of the ball on the bat. Does the momentum of the bat, therefore, also change? Two-Particle Collisions The bat, the hand and arm of the batter, and the ground on which the batter is standing are all objects that interact when a batter hits the ball. Thus, the bat cannot be considered a single object. In contrast to this complex system, examine for a moment the much simpler system shown in Figure 9-6, the collision of two balls. During the collision of the two balls, each one briefly exerts a force on the other. Despite the differences in sizes and velocities of the balls, the forces that they exert on each other are equal and opposite, according to Newton\u2019s third law of motion. These forces are represented by the following equation: FD on C FC on D How do the impulses imparted by both balls compare? Because the time intervals over which the forces are exerted are the same, the impulses must be equal in magnitude but opposite in direction. How did the momenta of the balls change as a result of the collision? According to the impulse-momentum theorem, the change in momentum is equal to the impulse. Compare the changes in the momenta of the two balls. For ball C: pCf For ball D: pDf pCi pDi FD on C FC on D t t pCf pCi Because the time interval over which the forces were exerted is the same, the impulses are equal in magnitude, but opposite in direction. According to Newton\u2019s third law of motion, FC on D pDi), or pCf This equation states that the sum of the momenta of the balls is the same before and after the collision. That is, the momentum gained by ball D is equal to the momentum lost by ball C. If the system is defined as the two balls, the momentum of the system is constant, and therefore, momentum is conserved for the system. FD on C. Thus, pDf (pDf pDi. pCi Momentum in a Closed, Isolated System Under what conditions is the momentum of the system of two balls conserved? The first and most obvious condition is that no balls are lost and no balls are gained. Such a system, which does not gain or lose mass, is said to be a closed system. The second condition required to conserve the momentum of a system is that the forces involved are internal forces; that is, there are no forces acting on the system by objects outside of it. 236 Chapter 9 Momentum and Its Conservation 236-245 CH09-S2-845813 3/10/04 11:23 PM Page 237 When the net external force on a closed system is zero, the system is described as an isolated system. No system on Earth can be said to be absolutely isolated, however, because there will always be some interactions between a system and its surroundings. Often, these interactions are small enough to be ignored when solving physics problems. Systems can contain any number of objects, and the objects can stick together or come apart in a collision. Under these conditions, the law of conservation of momentum states that the momentum of any closed, isolated system does not change. This law will enable you to make a connection between conditions, before and after an interaction, without knowing any of the details of the interaction. Speed A 1875-kg car going 23 m/s rear-ends a 1025-kg compact car going 17 m/s on ice in the same direction. The two cars stick together. How fast do the two cars move together immediately after the collision? 1 Analyze and Sketch the Problem \u2022 Define the system. \u2022 Establish a coordinate system. \u2022 Sketch the situation showing the \u201cbefore\u201d and \u201cafter\u201d states. \u2022 Draw a vector diagram for the momentum. Unknown: vf =? Known: mC = 1875 kg vCi = +23 m/s mD = 1025 kg vDi = +17 m/s 2 Solve for the Unknown Momentum is conserved because the ice makes the total external force on the cars nearly zero. x Before (initial) After (final) C D C D vCi vDi vCf vDf vf Vector diagram pCi pDi pf pi pCi pDi pf pDi pi pCi mCvCi pCf mDvDi pDf mCvCf mDvDf Because the two cars stick together, their velocities after the collision, denoted as vf, are equal. vf", " mDvDi mD)vf (mC vDf Math Handbook Order of Operations page 843 vCf mCvCi Solve for vf. vDi) vC (mC D ) (m C D m m vf i 21 m/s (1875 kg)(23 m/s) (1025 kg)(17 m/s) (1875 kg 1025 kg) Substitute mC mD 1025 kg, vDi 1875 kg, vCi 17 m/s 23 m/s, 3 Evaluate the Answer \u2022 Are the units correct? Velocity is measured in m/s. \u2022 Does the direction make sense? vi and vf are in the positive direction; therefore, vf should be positive. \u2022 Is the magnitude realistic? The magnitude of vf is between the initial speeds of the two cars, but closer to the speed of the more massive one, so it is reasonable. Section 9.2 Conservation of Momentum 237 236-245 CH09-S2-845813 3/11/04 2:05 AM Page 238 13. Two freight cars, each with a mass of 3.0105 kg, collide and stick together. One was initially moving at 2.2 m/s, and the other was at rest. What is their final speed? 14. A 0.105-kg hockey puck moving at 24 m/s is caught and held by a 75-kg goalie at rest. With what speed does the goalie slide on the ice? 15. A 35.0-g bullet strikes a 5.0-kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and bullet fly off together at 8.6 m/s. What was the original speed of the bullet? 16. A 35.0-g bullet moving at 475 m/s strikes a 2.5-kg bag of flour that is on ice, at rest. The bullet passes through the bag, as shown in Figure 9-7, and exits it at 275 m/s. How fast is the bag moving when the bullet exits? 17. The bullet in the previous problem strikes a 2.5-kg steel ball that is at rest. The bullet bounces backward after its collision at a speed of 5.0 m/s. How fast is the ball moving when the bullet bounces backward? 18. A 0.50-kg ball that is traveling at 6.0 m/s collides head-on with a 1.00-kg ball moving in the opposite direction at a speed of 12.0 m/s. The 0.50-kg ball bounces backward at 14 m/s after the collision. Find the speed of the second ball after the collision. 275 m/s \u25a0 Figure 9-7 Recoil It is very important to define a system carefully. The momentum of a baseball changes when the external force of a bat is exerted on it. The baseball, therefore, is not an isolated system. On the other hand, the total momentum of two colliding balls within an isolated system does not change because all forces are between the objects within the system. Can you find the final velocities of the two in-line skaters in Figure 9-8? Assume that they are skating on a smooth surface with no external forces. They both start at rest, one behind the other. a b \u25a0 Figure 9-8 The internal forces exerted by Skater C, the boy, and Skater D, the girl, cannot change the total momentum of the system. 238 Laura Sifferlin Chapter 9 Momentum and Its Conservation 236-245 CH09-S2-845813 3/11/04 2:08 AM Page 239 Skater C, the boy, gives skater D, the girl, a push. Now, both skaters are moving, making this situation similar to that of an explosion. Because the push was an internal force, you can use the law of conservation of momentum to find the skaters\u2019 relative velocities. The total momentum of the system was zero before the push. Therefore, it must be zero after the push. Before pDi pCi 0 pCf mCvCf pCf pCf After pDf pDf pDf mDvDf The coordinate system was chosen so that the positive direction is to the left. The momenta of the skaters after the push are equal in magnitude but opposite in direction. The backward motion of skater C is an example of recoil. Are the skaters\u2019 velocities equal and opposite? The last equation shown above, for the velocity of skater C, can be rewritten as follows: \u25a0 Figure 9-9 The xenon atoms in the ion engine are ionized by bombarding them with electrons. Then, the positively charged xenon ions are accelerated to high speeds. vCf m m DvDf C The velocities depend", " on the skaters\u2019 relative masses. If skater C has a mass of 68.0 kg and skater D\u2019s mass is 45.4 kg, then the ratio of their velocities will be 68.0 : 45.4, or 1.50. The less massive skater moves at the greater velocity. Without more information about how hard skater C pushed skater D, however, you cannot find the velocity of each skater. Propulsion in Space How does a rocket in space change its velocity? The rocket carries both fuel and oxidizer. When the fuel and oxidizer combine in the rocket motor, the resulting hot gases leave the exhaust nozzle at high speed. If the rocket and chemicals are the system, then the system is a closed system. The forces that expel the gases are internal forces, so the system is also an isolated system. Thus, objects in space can accelerate by using the law of conservation of momentum and Newton\u2019s third law of motion. A NASA space probe, called Deep Space 1, performed a flyby of an asteroid a few years ago. The most unusual of the 11 new technologies on board was an ion engine that exerts as much force as a sheet of paper resting on a person\u2019s hand. The ion engine shown in Figure 9-9, operates differently from a traditional rocket engine. In a traditional rocket engine, the products of the chemical reaction taking place in the combustion chamber are released at high speed from the rear. In the ion engine, however, xenon atoms are expelled at a speed of 30 km/s, producing a force of only 0.092 N. How can such a small force create a significant change in the momentum of the probe? Instead of operating for only a few minutes, as the traditional chemical rockets do, the ion engine can run continuously for days, weeks, or months. Therefore, the impulse delivered by the engine is large enough to increase the momentum of the 490-kg spacecraft until it reaches the speed needed to complete its mission. Rebound Height An object\u2019s momentum is the product of its mass and velocity. 1. Drop a large rubber ball from about 15 cm above a table. 2. Measure and record the ball\u2019s rebound height. 3. Repeat steps 1\u20132 with a small rubber ball. 4. Hold the small rubber ball on top of, and in contact with, the large rubber ball. 5. Release the two rubber balls from the same height, so that they fall together. 6. Measure the rebound heights of both rubber balls. Analyze and Conclude 7. Describe the rebound height of each rubber ball dropped by itself. 8. Compare and contrast the rebound heights from number 7 with those from number 6. 9. Explain your observations. Section 9.2 Conservation of Momentum 239 NASA 236-245 CH09-S2-845813 3/10/04 11:16 PM Page 240 Speed An astronaut at rest in space fires a thruster pistol that expels 35 g of hot gas at 875 m/s. The combined mass of the astronaut and pistol is 84 kg. How fast and in what direction is the astronaut moving after firing the pistol? 1 Analyze and Sketch the Problem \u2022 Define the system. \u2022 Establish a coordinate axis. \u2022 Sketch the \u201cbefore\u201d and \u201cafter\u201d conditions. \u2022 Draw a vector diagram showing momenta. + x Before (initial) After (final) vDf D vCf Known: mC mD vCi vDf 84 kg 0.035 kg vDi 875 m/s +0.0 m/s Unknown:? vCf vi 0.0 m/s \u2022 p i Vector Diagram p Df p Cf 2 Solve for the Unknown The system is the astronaut, the gun, and the chemicals that produce the gas. pi pCi pDi = 0.0 kgm/s Before the pistol is fired, all parts of the system are at rest; thus, the initial momentum is zero. Use the law of conservation of momentum to find pf. p \u2022 p f Cf p Df pi pf 0.0 kgm/s pCf pCf pDf pDf mCvCf vCf mDvDf m vDf) ( D m C Solve for the final velocity of the astronaut, vCf. The momentum of the astronaut is equal in magnitude, but opposite in direction to the momentum of the gas leaving the pistol. Math Handbook Isolating a Variable page 845 (0.035 kg)(875 m/s) 84 kg Substitute mD 0.035 kg, vDf 875 m/s, mC 84 kg +0.36 m/s 3 Evaluate the Answer \u2022 Are the units correct? The velocity is measured in m/s. \u2022 Does the direction make sense? The velocity of the astronaut is in the opposite direction", " to that of the expelled gas. \u2022 Is the magnitude realistic? The astronaut\u2019s mass is much larger than that of the gas, so the velocity of the astronaut is much less than that of the expelled gas. 19. A 4.00-kg model rocket is launched, expelling 50.0 g of burned fuel from its exhaust at a speed of 625 m/s. What is the velocity of the rocket after the fuel has burned? Hint: Ignore the external forces of gravity and air resistance. 20. A thread holds a 1.5-kg cart and a 4.5-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giving the 1.5-kg cart a speed of 27 cm/s to the left. What is the velocity of the 4.5-kg cart? 21. Carmen and Judi dock a canoe. 80.0-kg Carmen moves forward at 4.0 m/s as she leaves the canoe. At what speed and in what direction do the canoe and Judi move if their combined mass is 115 kg? 240 Chapter 9 Momentum and Its Conservation 236-245 CH09-S2-845813 3/10/04 11:17 PM Page 241 Two-Dimensional Collisions Up until now, you have looked at momentum in only one dimension. The law of conservation of momentum holds for all closed systems with no external forces. It is valid regardless of the directions of the particles before or after they interact. But what happens in two or three dimensions? Figure 9-10 shows the result of billiard ball C striking stationary billiard ball D. Consider the two billiard balls to be the system. The original momentum of the moving ball is pCi and the momentum of the stationary ball is zero. Therefore, the momentum of the system before the collision is equal to pCi. pDf. After the collision, both billiard balls are moving and have momenta. As long as the friction with the tabletop can be ignored, the system is closed and isolated. Thus, the law of conservation of momentum can be used. The initial momentum equals the vector sum of the final momenta, so pCi pCf The equality of the momenta before and after the collision also means that the sum of the components of the vectors before and after the collision must be equal. Suppose the x-axis is defined to be in the direction of the initial momentum, then the y-component of the initial momentum is equal to zero. Therefore, the sum of the final y-components also must be zero: pCf, y pDf, y 0 The y-components are equal in magnitude but are in the opposite direction and, thus, have opposite signs. The sum of the horizontal components also is equal: pCi pCf, x pDf, x \u25a0 Figure 9-10 The law of conservation of momentum holds for all isolated, closed systems, regardless of the directions of objects before and after a collision. C C p Ci p Cf C D p Df D Section 9.2 Conservation of Momentum 241 236-245 CH09-S2-845813 6/10/04 7:43 PM Page 242 Speed A 1325-kg car, C, moving north at 27.0 m/s, collides with a 2165-kg car, D, moving east at 11.0 m/s. The two cars are stuck together. In what direction and with what speed do they move after the collision? 1 Analyze and Sketch the Problem \u2022 Define the system. \u2022 Sketch the \u201cbefore\u201d and \u201cafter\u201d states. \u2022 Establish the coordinate axis with the y-axis north and the x-axis east. \u2022 Draw a momentum-vector diagram. Known: mC mD vCi, y vDi, x 1325 kg 2165 kg 27.0 m/s 11.0 m/s Unknown:? vf, x? vf, y? 2 Solve for the Unknown Determine the initial momenta of the cars and the momentum of the system. Before (initial) y x D vDi 11.0 m/ pCi pDi mCvCi, y (1325 kg)(27.0 m/s) 3.58104 kgm/s (north) mDvDi, x (2165 kg)(11.0 m/s) 2.38104 kgm/s (east) Substitute mC = 1325 kg, vCi, y 27.0 m/s Substitute mD 2165 kg, vDi, x 11.0 m/s After (final) vf D C Vector Diagram pDi 90\u00b0 pCi pf Use the law of conservation of momentum to find pf. pf, x pf, y pi, x pi, y 2.38104 kgm/s", " 3.58104 kgm/s pDi pCi Use the diagram to set up equations for pf, x and pf, y. Substitute pi, x Substitute pi, y pf y )2 (pf, x)2 (pf, (2.38104 kgm/s)2 (3.58104 kgm/s)2 4.30104 kgm/s p, f p y) Solve for. tan1 ( tan1 ( 56.4.38104 kgm/s 3.58104 kgm/s Substitute pf, x pf, y 3.58104 kgm/s 2.38104 kgm/s, Math Handbook Inverses of Sine, Cosine, and Tangent page 856 Substitute pf, y 3.58104 kgm/s, pf, x 2.38104 kgm/s Determine the final speed. pf mD) (mC vf 4.30104 kgm/s (1325 kg 2165 kg) 12.3 m/s Substitute pf 4.30104 kgm/s, mC 1325 kg, mD 2165 kg 3 Evaluate the Answer \u2022 Are the units correct? The correct unit for speed is m/s. \u2022 Do the signs make sense? Answers are both positive and at the appropriate angles. \u2022 Is the magnitude realistic? The cars stick together, so vf must be smaller than vCi. 242 Chapter 9 Momentum and Its Conservation 236-245 CH09-S2-845813 6/1/04 1:42 AM Page 243 22. A 925-kg car moving north at 20.1 m/s collides with a 1865-kg car moving west at 13.4 m/s. The two cars are stuck together. In what direction and at what speed do they move after the collision? 23. A 1383-kg car moving south at 11.2 m/s is struck by a 1732-kg car moving east at 31.3 m/s. The cars are stuck together. How fast and in what direction do they move immediately after the collision? 24. A stationary billiard ball, with a mass of 0.17 kg, is struck by an identical ball moving at 4.0 m/s. After the collision, the second ball moves 60.0\u00b0 to the left of its original direction. The stationary ball moves 30.0\u00b0 to the right of the moving ball\u2019s original direction. What is the velocity of each ball after the collision? 25. A 1345-kg car moving east at 15.7 m/s is struck by a 1923-kg car moving north. They are stuck together and move with an initial velocity of 14.5 m/s at = 63.5\u00b0. Was the north-moving car exceeding the 20.1 m/s speed limit? Conservation of Angular Momentum Like linear momentum, angular momentum can be conserved. The law of conservation of angular momentum states that if no net external torque acts on an object, then its angular momentum does not change. This is represented by the following equation. Law of Conservation of Angular Momentum L1 An object\u2019s initial angular momentum is equal to its final angular momentum. L2 For example, Earth spins on its axis with no external torques. Its angular momentum is constant. Thus, Earth\u2019s angular momentum is conserved. As a result, the length of a day does not change. A spinning ice-skater also demonstrates conservation of angular momentum. Figure 9-11a shows an ice-skater spinning with his arms extended. When he pulls in his arms, as shown in Figure 9-11b, he begins spinning faster. Without an external torque, his angular momentum does not change; that is, L I is constant. Thus, the ice-skater\u2019s increased angular velocity must be accompanied by a decreased moment of inertia. By pulling his arms close to his body, the ice-skater brings more mass closer to the axis of rotation, thereby decreasing the radius of rotation and decreasing his moment of inertia. You can calculate changes in angular velocity using the law of conservation of angular momentum. thus, Ii Lf Li If i I i f I i f f \u25a0 Figure 9-11 When the iceskater\u2019s arms are extended, the moment of inertia increases and his angular velocity decreases (a). When his arms are closer to his body the moment of inertia decreases and results in an increased angular velocity (b). a b Because frequency is f /2, the above equation can be rewritten as follows) f ( 2 f I f i thus, f I f i f Section 9.2 Conservation of Momentum 243 F. Scott Grant/IMAGE Communications 236-245 CH09-S2-845813 3/10/04 11:24 PM Page 244", " Rotational axis Vertical Spin Precession due to torque Center of mass Gravitational force \u25a0 Figure 9-12 The upper end of the top precesses due to the torque acting on the top. Notice that because f,, and I appear as ratios in these equations, any units may be used, as long as the same unit is used for both values of the quantity. If a torque-free object starts with no angular momentum, it must continue to have no angular momentum. Thus, if part of an object rotates in one direction, another part must rotate in the opposite direction. For example, if you switch on a loosely held electric drill, the drill body will rotate in the direction opposite to the rotation of the motor and bit. Pivot point Angle of rotational axis with the vertical Consider a ball thrown at a weather vane. The ball, moving in a straight line, can start the vane rotating. Consider the ball and vane to be a system. With no external torques, angular momentum is conserved. The vane spins faster if the ball has a large mass, m, a large velocity, v, and hits at right angles as far as possible from the pivot of the vane. The angular momentum of a moving object, such as the ball, is given by L mvr, where r is the perpendicular distance from the axis of rotation. Tops and Gyroscopes Because of the conservation of angular momentum, the direction of rotation of a spinning object can be changed only by applying a torque. If you played with a top as a child, you may have spun it by pulling the string wrapped around its axle. When a top is vertical, there is no torque on it, and the direction of its rotation does not change. If the top is tipped, as shown in Figure 9-12, a torque tries to rotate it downward. Rather than tipping over, however, the upper end of the top revolves, or precesses slowly about the vertical axis. Because Earth is not a perfect sphere, the Sun exerts a torque on it, causing it to precess. It takes about 26,000 years for Earth\u2019s rotational axis to go through one cycle of precession. Your friend was driving her 1265-kg car north on Oak Street when she was hit by a 925-kg compact car going west on Maple Street. The cars stuck together and slid 23.1 m at 42\u00b0 north of west. The speed limit on both streets is 22 m/s (50 mph). Assume that momentum was conserved during the collision and that acceleration was constant during the skid. The coefficient of kinetic friction between the tires and the pavement is 0.65. 1. Your friend claims that she wasn\u2019t speeding, but that the driver of other car was. How fast was your friend driving before the crash? 2. How fast was the other car moving before the crash? Can you support your friend\u2019s case in court? y x 925 kg 42\u00b0 1265 kg 244 Chapter 9 Momentum and Its Conservation 236-245 CH09-S2-845813 6/1/04 1:43 AM Page 245 gyroscope, such as the one shown in Figure 9-13, is a wheel or disk that spins rapidly around one axis while being free to rotate around one or two other axes. The direction of its large angular momentum can be changed only by applying an appropriate torque. Without such a torque, the direction of the axis of rotation does not change. Gyroscopes are used in airplanes, submarines, and spacecraft to keep an unchanging reference direction. Giant gyroscopes are used in cruise ships to reduce their motion in rough water. Gyroscopic compasses, unlike magnetic compasses, maintain direction even when they are not on a level surface. A football quarterback uses the gyroscope effect to make an accurate forward pass. As he throws, he spins, or spirals the ball. If the quarterback throws the ball in the direction of its spin axis of rotation, the ball keeps its pointed end forward, thereby reducing air resistance. Thus, the ball can be thrown far and accurately. If its spin direction is slightly off, the ball wobbles. If the ball is not spun, it tumbles end over end. The flight of a plastic disk also is stabilized by spin. A well-spun plastic disk can fly many meters through the air without wobbling. You are able to perform tricks with a yo-yo because its fast rotational speed keeps it rotating in one plane. \u25a0 Figure 9-13 Because the orientation of the spin axis of the gyroscope does not change even when it is moved, the gyroscope can be used to fix direction. 9.2 Section Review 26. Angular Momentum The outer rim of a plastic disk is thick and heavy. Besides making it easier to catch, how does this affect the rotational properties of the plastic disk? 27. Speed A cart, weighing 24.5 N, is released", " from rest on a 1.00-m ramp, inclined at an angle of 30.0\u00b0 as shown in Figure 9-14. The cart rolls down the incline and strikes a second cart weighing 36.8 N. a. Calculate the speed of the first cart at the bot- tom of the incline. b. If the two carts stick together, with what initial speed will they move along? 24.5 N 1.00 m 36.8 N 30.0\u00b0 \u25a0 Figure 9-14 28. Conservation of Momentum During a tennis serve, the racket of a tennis player continues forward after it hits the ball. Is momentum conserved in the collision? Explain, making sure that you define the system. 29. Momentum A pole-vaulter runs toward the launch point with horizontal momentum. Where does the vertical momentum come from as the athlete vaults over the crossbar? 30. Initial Momentum During a soccer game, two players come from opposite directions and collide when trying to head the ball. They come to rest in midair and fall to the ground. Describe their initial momenta. 31. Critical Thinking You catch a heavy ball while you are standing on a skateboard, and then you roll backward. If you were standing on the ground, however, you would be able to avoid moving while catching the ball. Explain both situations using the law of conservation of momentum. Explain which system you use in each case. physicspp.com/self_check_quiz Section 9.2 Conservation of Momentum 245 file photo 246-247 CH09-LAB-845813 6/1/04 2:14 AM Page 246 Alternate CBL instructions can be found on the Web site. physicspp.com Sticky Collisions In this activity, one moving cart will strike a stationary cart. During the collision, the two carts will stick together. You will measure mass and velocity, both before and after the collision. You then will calculate the momentum both before and after the collision. QUESTION How is the momentum of a system affected by a sticky collision Objectives Procedure \u25a0 Describe how momentum is transferred during a collision. \u25a0 Calculate the momenta involved. \u25a0 Interpret data from a collision. \u25a0 Draw conclusions that support the law of conservation of momentum. Safety Precautions Materials Internet access required 1. View Chapter 9 lab video clip 1 at physicspp.com/internet_lab to determine the mass of the carts. 2. Record the mass of each cart. 3. Watch video clip 2: Cart 1 strikes Cart 2. 4. In the video, three frames represent 0.1 s. Record in the data table the distance Cart 1 travels in 0.1 s before the collision. 5. Observe the collision. Record in the data table the distance the Cart 1-Cart 2 system travels in 0.1 s after the collision. 6. Repeat steps 3\u20135 for video clip 3: Carts 1 and 3 strike Cart 2. 7. Repeat steps 3\u20135 for video clip 4: Carts 1, 3, and 4 strike Cart 2. 8. Repeat steps 3-5 for video clip 5: Carts 1 and 3 strike Carts 2 and 4. 9. Repeat steps 3\u20135 for video clip 6: Cart 1 strikes Carts 2, 3, and 4. 246 246-247 CH09-LAB-845813 6/7/04 6:06 AM Page 247 Data Tables Cart Mass (kg) 1 2 3 4 Time of Approach (s) Distance Covered in Approach (cm) Initial Velocity (cm/s) Mass of Approaching Cart(s) (g) Initial Momentum (gcm/s) Time of Departure (s) Distance Covered in Departure (cm) Final Velocity (cm/s) Mass of Departing Cart(s) (g) Final Momentum (gcm/s) 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 Analyze Real-World Physics 1. Calculate the initial and final velocities for each of the cart systems. 2. Calculate the initial and final momentum for each of the cart systems. 3. Make and Use Graphs Make a graph showing final momentum versus initial momentum for all the video clips. 1. Suppose a linebacker collides with a stationary quarterback and they become entangled. What will happen to the velocity of the linebackerquarterback system if momentum is conserved? 2. If a car rear-ends a stationary car so that the two cars become attached, what will happen to the velocity of the first car? The second car? Conclude and Apply 1. What is the relationship between the initial momentum and the final momentum of the cart systems in a sticky collision? 2. In theory, what should be the slope of the line in your graph? 3. The initial and final data numbers may not be the same due to the precision of the instruments,", " friction, and other variables. Is the initial momentum typically greater or less than the final momentum? Explain. Going Further 1. Describe what the velocity and momentum data might look like if the carts did not stick together, but rather, bounced off of each other? 2. Design an experiment to test the impact of friction during the collision of the cart systems. Predict how the slope of the line in your graph will change with your experiment, and then try your experiment. Interpret Data Visit physicspp.com/ internet_lab to post your findings from the experiment testing the impact of friction during the collisions of the cart systems. Examine the data and graph of the final momentum versus initial momentum on the Web site. Notice how close to or far off the slope is from 1.00. To find out more about momentum, visit the Web site: physicspp.com 247 248 CH09-FEATURE-845813 3/11/04 12:53 AM Page 248 Solar Sailing Solar Sailing Nearly 400 years ago, Johannes Kepler observed that comet tails appeared to be blown by a solar breeze. He suggested that ships would be able to travel in space with sails designed to catch this breeze. Thus, the idea for solar sails was born. How Does a Solar Sail Work? A solar sail is a spacecraft without an engine. A solar sail works like a giant fabric mirror that is free to move. Solar sails usually are made of 5-micron-thick aluminized polyester film or polyimide film with a 100-nm-thick aluminum layer deposited on one side to form the reflective surface. Reflected sunlight, rather than rocket fuel, provides the force. Sunlight is made up of individual particles called photons. Photons have momentum, and when a photon bounces off a solar sail, it transfers its momentum to the sail, which propels the spacecraft along. but solar sails only require photons from the Sun. Thus, solar sails may be a superior way to move large masses over great distances in outer space. Future Journeys The Cosmos 1, the first solar-sail prototype, is scheduled for a launch in the very near future. The Cosmos-1 mission is an international, privately funded venture. The spacecraft looks like a flower with eight huge, solar-sail petals. Being the first solar sail, goals are modest. The mission will be considered successful if the Cosmos 1 operates for just a few days, accelerating under sunlight pressure. Solar sails are important, not only for travel, but also for creating new types of space and Earth weather monitoring stations. These stations would be able to provide greater coverage of Earth and more advanced warning of solar storms that cause problems to communication and electric power grids. It is This artist\u2019s rendering shows Cosmos 1, the first solar sail scheduled for launch in the near future. The force of impacting photons is small in comparison to the force rocket fuel can supply. So, small sails experience only a small amount of force from sunlight, while larger sails experience a greater force. Thus, solar sails may be a kilometer or so across. What speeds can a solar sail achieve? This depends on the momentum transferred to the sail by photons, as well as the sail\u2019s mass. To travel quickly through the solar system, a sail and the spacecraft should be lightweight. Photons supplied by the Sun are constant. They impact the sail every second of every hour of every day during a space flight. The Sun\u2019s continuous supply of photons over time allows the sail to build up huge velocities and enables the spacecraft to travel great distances within a convenient time frame. Rockets require enormous amounts of fuel to move large masses, hoped that in the next few decades, solar sails will be used as interplanetary shuttles because of their ability to travel great distances in convenient time frames. Vast distances could someday be traversed by vehicles that do not consume any fuel. Going Further 1. Research how solar sails can help provide advanced warning of solar storms. 2. Critical Thinking A certain solar-sail model is predicted to take more time to reach Mars than a rocket-propelled spacecraft would, but less time to go to Pluto than a rocket-propelled spacecraft would. Explain why this is so. 248 Future Technology 249-255 CH09-SG CA STP-845813 3/11/04 1:12 AM Page 249 9.1 Impulse and Momentum Vocabulary \u2022 impulse (p. 230) \u2022 momentum (p. 230) \u2022 impulse-momentum theorem (p. 230) \u2022 angular momentum (p. 233) \u2022 angular impulse-angular momentum theorem (p. 234) Key Concepts \u2022 When doing a momentum problem, first examine the system before and after the event. \u2022 The momentum of an object is the product of its mass and velocity and is a vector quantity. p mv \u2022 The impulse on an object is the average net force exerted on the object multiplied by the time interval over which the force acts. Impluse Ft \u2022 The impulse", " on an object is equal to the change in momentum of the object. Ft pf pi \u2022 The angular momentum of a rotating object is the product of its moment of inertia and its angular velocity. L I \u2022 The angular impulse-angular momentum theorem states that the angular impulse on an object is equal to the change in the object\u2019s angular momentum. t Lf Li 9.2 Conservation of Momentum Vocabulary \u2022 closed system (p. 236) \u2022 isolated system (p. 237) \u2022 law of conservation of momentum (p. 237) \u2022 law of conservation of angular momentum (p. 243) Key Concepts \u2022 According to Newton\u2019s third law of motion and the law of conservation of momentum, the forces exerted by colliding objects on each other are equal in magnitude and opposite in direction. \u2022 Momentum is conserved in a closed, isolated system. pf pi \u2022 The law of conservation of momentum can be used to explain the propulsion of rockets. \u2022 Vector analysis is used to solve momentum-conservation problems in two dimensions. \u2022 The law of conservation of angular momentum states that if there are no external torques acting on a system, then the angular momentum is conserved. Lf Li \u2022 Because angular momentum is conserved, the direction of rotation of a spinning object can be changed only by applying a torque. physicspp.com/vocabulary_puzzlemaker 249 249-255 CH09-SG CA STP-845813 6/1/04 2:38 AM Page 250 Concept Mapping 32. Complete the following concept map using the following terms: mass, momentum, average force, time over which the force is exerted. Product is the impulse Produces a change in Product is the velocity Mastering Concepts 33. Can a bullet have the same momentum as a truck? Explain. (9.1) 34. A pitcher throws a curve ball to the catcher. Assume that the speed of the ball doesn\u2019t change in flight. (9.1) a. Which player exerts the larger impulse on the ball? b. Which player exerts the larger force on the ball? 35. Newton\u2019s second law of motion states that if no net force is exerted on a system, no acceleration is possible. Does it follow that no change in momentum can occur? (9.1) 36. Why are cars made with bumpers that can be pushed in during a crash? (9.1) 37. An ice-skater is doing a spin. (9.1) a. How can the skater\u2019s angular momentum be changed? b. How can the skater\u2019s angular velocity be changed without changing the angular momentum? 38. What is meant by \u201can isolated system?\u201d (9.2) 39. A spacecraft in outer space increases its velocity by firing its rockets. How can hot gases escaping from its rocket engine change the velocity of the craft when there is nothing in space for the gases to push against? (9.2) 40. A cue ball travels across a pool table and collides with the stationary eight ball. The two balls have equal masses. After the collision, the cue ball is at rest. What must be true regarding the speed of the eight ball? (9.2) 41. Consider a ball falling toward Earth. (9.2) a. Why is the momentum of the ball not conserved? b. In what system that includes the falling ball is the momentum conserved? 42. A falling basketball hits the floor. Just before it hits, the momentum is in the downward direction, and after it hits the floor, the momentum is in the upward direction. (9.2) a. Why isn\u2019t the momentum of the basketball conserved even though the bounce is a collision? b. In what system is the momentum conserved? 43. Only an external force can change the momentum of a system. Explain how the internal force of a car\u2019s brakes brings the car to a stop. (9.2) 44. Children\u2019s playgrounds often have circular-motion rides. How could a child change the angular momentum of such a ride as it is turning? (9.2) Applying Concepts 45. Explain the concept of impulse using physical ideas rather than mathematics. 46. Is it possible for an object to obtain a larger impulse from a smaller force than it does from a larger force? Explain. 47. Foul Ball You are sitting at a baseball game when a foul ball comes in your direction. You prepare to catch it bare-handed. To catch it safely, should you move your hands toward the ball, hold them still, or move them in the same direction as the moving ball? Explain. 48. A 0.11-g bullet leaves a pistol at 323 m/s, while a similar bullet leaves a rifle at 396 m/s. Explain the difference in exit speeds of the two bullets, assuming that the forces exerted on the bullets by the", " expanding gases have the same magnitude. 49. An object initially at rest experiences the impulses described by the graph in Figure 9-15. Describe the object\u2019s motion after impulses A, B, and C 10 11 B C Time (s) \u25a0 Figure 9-15 250 Chapter 9 Momentum and Its Conservation For more problems, go to Additional Problems, Appendix B. 249-255 CH09-SG CA STP-845813 6/1/04 2:38 AM Page 251 50. During a space walk, the tether connecting an astronaut to the spaceship breaks. Using a gas pistol, the astronaut manages to get back to the ship. Use the language of the impulse-momentum theorem and a diagram to explain why this method was effective. 51. Tennis Ball As a tennis ball bounces off a wall, its momentum is reversed. Explain this action in terms of the law of conservation of momentum. Define the system and draw a diagram as a part of your explanation. 52. Imagine that you command spaceship Zeldon, which is moving through interplanetary space at high speed. How could you slow your ship by applying the law of conservation of momentum? 53. Two trucks that appear to be identical collide on an icy road. One was originally at rest. The trucks are stuck together and move at more than half the original speed of the moving truck. What can you conclude about the contents of the two trucks? 54. Explain, in terms of impulse and momentum, why it is advisable to place the butt of a rifle against your shoulder when first learning to shoot. 55. Bullets Two bullets of equal mass are shot at equal speeds at blocks of wood on a smooth ice rink. One bullet, made of rubber, bounces off of the wood. The other bullet, made of aluminum, burrows into the wood. In which case does the block of wood move faster? Explain. Mastering Problems 9.1 Impulse and Momentum 56. Golf Roc\u00edo strikes a 0.058-kg golf ball with a force of 272 N and gives it a velocity of 62.0 m/s. How long was Roc\u00edo\u2019s club in contact with the ball? 57. A 0.145-kg baseball is pitched at 42 m/s. The batter hits it horizontally to the pitcher at 58 m/s. a. Find the change in momentum of the ball. b. If the ball and bat are in contact for 4.6104 s, what is the average force during contact? 58. Bowling A force of 186 N acts on a 7.3-kg bowling ball for 0.40 s. What is the bowling ball\u2019s change in momentum? What is its change in velocity? 59. A 5500-kg freight truck accelerates from 4.2 m/s to 7.8 m/s in 15.0 s by the application of a constant force. a. What change in momentum occurs? b. How large of a force is exerted? 60. In a ballistics test at the police department, Officer Rios fires a 6.0-g bullet at 350 m/s into a container that stops it in 1.8 ms. What is the average force that stops the bullet? 61. Volleyball A 0.24-kg volleyball approaches Tina with a velocity of 3.8 m/s. Tina bumps the ball, giving it a speed of 2.4 m/s but in the opposite direction. What average force did she apply if the interaction time between her hands and the ball was 0.025 s? 62. Hockey A hockey player makes a slap shot, exerting a constant force of 30.0 N on the hockey puck for 0.16 s. What is the magnitude of the impulse given to the puck? 63. Skateboarding Your brother\u2019s mass is 35.6 kg, and he has a 1.3-kg skateboard. What is the combined momentum of your brother and his skateboard if they are moving at 9.50 m/s? 64. A hockey puck has a mass of 0.115 kg and is at rest. A hockey player makes a shot, exerting a constant force of 30.0 N on the puck for 0.16 s. With what speed does it head toward the goal? 65. Before a collision, a 25-kg object was moving at 12 m/s. Find the impulse that acted on the object if, after the collision, it moved at the following velocities. a. 8.0 m/s b. 8.0 m/s 66. A 0.150-kg ball, moving in the positive direction at 12 m/s, is acted on by the impulse shown in the graph in Figure 9-16. What is the ball\u2019s speed at 4.0 s Time (s) \u25a0 Figure 9-16 67. Baseball A 0.145-kg baseball is moving at 35 m/s when it is caught by a player.", " a. Find the change in momentum of the ball. b. If the ball is caught with the mitt held in a stationary position so that the ball stops in 0.050 s, what is the average force exerted on the ball? c. If, instead, the mitt is moving backward so that the ball takes 0.500 s to stop, what is the average force exerted by the mitt on the ball? physicspp.com/chapter_test Chapter 9 Assessment 251 249-255 CH09-SG CA STP-845813 3/11/04 1:15 AM Page 252 68. Hockey A hockey puck has a mass of 0.115 kg and strikes the pole of the net at 37 m/s. It bounces off in the opposite direction at 25 m/s, as shown in Figure 9-17. a. What is the impulse on the puck? b. If the collision takes 5.0104 s, what is the average force on the puck? 0.115 kg 25 m/s \u25a0 Figure 9-17 69. A nitrogen molecule with a mass of 4.71026 kg, moving at 550 m/s, strikes the wall of a container and bounces back at the same speed. 72. Accident A car moving at 10.0 m/s crashes into a barrier and stops in 0.050 s. There is a 20.0-kg child in the car. Assume that the child\u2019s velocity is changed by the same amount as that of the car, and in the same time period. a. What is the impulse needed to stop the child? b. What is the average force on the child? c. What is the approximate mass of an object whose weight equals the force in part b? d. Could you lift such a weight with your arm? e. Why is it advisable to use a proper restraining seat rather than hold a child on your lap? 9.2 Conservation of Momentum 73. Football A 95-kg fullback, running at 8.2 m/s, collides in midair with a 128-kg defensive tackle moving in the opposite direction. Both players end up with zero speed. a. Identify the \u201cbefore\u201d and \u201cafter\u201d situations and draw a diagram of both. b. What was the fullback\u2019s momentum before the collision? a. What is the impulse the molecule delivers to the c. What was the change in the fullback\u2019s wall? b. If there are 1.51023 collisions each second, what is the average force on the wall? 70. Rockets Small rockets are used to make tiny adjustments in the speeds of satellites. One such rocket has a thrust of 35 N. If it is fired to change the velocity of a 72,000-kg spacecraft by 63 cm/s, how long should it be fired? 71. An animal rescue plane flying due east at 36.0 m/s drops a bale of hay from an altitude of 60.0 m, as shown in Figure 9-18. If the bale of hay weighs 175 N, what is the momentum of the bale the moment before it strikes the ground? Give both magnitude and direction. 36.0 m/s 175 N 60.0 m \u25a0 Figure 9-18 momentum? d. What was the change in the defensive tackle\u2019s momentum? e. What was the defensive tackle\u2019s original momentum? f. How fast was the defensive tackle moving originally? 74. Marble C, with mass 5.0 g, moves at a speed of 20.0 cm/s. It collides with a second marble, D, with mass 10.0 g, moving at 10.0 cm/s in the same direction. After the collision, marble C continues with a speed of 8.0 cm/s in the same direction. a. Sketch the situation and identify the system. Identify the \u201cbefore\u201d and \u201cafter\u201d situations and set up a coordinate system. b. Calculate the marbles\u2019 momenta before the collision. c. Calculate the momentum of marble C after the collision. d. Calculate the momentum of marble D after the collision. e. What is the speed of marble D after the collision? 75. Two lab carts are pushed together with a spring mechanism compressed between them. Upon release, the 5.0-kg cart repels one way with a velocity of 0.12 m/s, while the 2.0-kg cart goes in the opposite direction. What is the velocity of the 2.0-kg cart? 252 Chapter 9 Momentum and Its Conservation For more problems, go to Additional Problems, Appendix B. 249-255 CH09-SG CA STP-845813 6/8/04 3:54 PM Page 253 76. A 50.0-g projectile is launched with a horizontal velocity of 647 m/s from a 4.65-kg launcher moving in the same", " direction at 2.00 m/s. What is the launcher\u2019s velocity after the launch? 80. A 2575-kg van runs into the back of an 825-kg compact car at rest. They move off together at 8.5 m/s. Assuming that the friction with the road is negligible, calculate the initial speed of the van. 77. A 12.0-g rubber bullet travels at a velocity of 150 m/s, hits a stationary 8.5-kg concrete block resting on a frictionless surface, and ricochets in the opposite direction with a velocity of 1.0102 m/s, as shown in Figure 9-19. How fast will the concrete block be moving? 1.0102 m/s 8.5 kg 12.0 g \u25a0 Figure 9-19 78. Skateboarding Kofi, with mass 42.00 kg, is riding a skateboard with a mass of 2.00 kg and traveling at 1.20 m/s. Kofi jumps off and the skateboard stops dead in its tracks. In what direction and with what velocity did he jump? 79. Billiards A cue ball, with mass 0.16 kg, rolling at 4.0 m/s, hits a stationary eight ball of similar mass. If the cue ball travels 45\u00b0 above its original path and the eight ball travels 45\u00b0 below the horizontal, as shown in Figure 9-20, what is the velocity of each ball after the collision? 45\u00b0 45\u00b0 \u25a0 Figure 9-20 81. In-line Skating Diego and Keshia are on in-line skates and stand face-to-face, then push each other away with their hands. Diego has a mass of 90.0 kg and Keshia has a mass of 60.0 kg. a. Sketch the event, identifying the \u201cbefore\u201d and \u201cafter\u201d situations, and set up a coordinate axis. b. Find the ratio of the skaters\u2019 velocities just after their hands lose contact. c. Which skater has the greater speed? d. Which skater pushed harder? 82. A 0.200-kg plastic ball moves with a velocity of 0.30 m/s. It collides with a second plastic ball of mass 0.100 kg, which is moving along the same line at a speed of 0.10 m/s. After the collision, both balls continue moving in the same, original direction. The speed of the 0.100-kg ball is 0.26 m/s. What is the new velocity of the 0.200-kg ball? Mixed Review 83. A constant force of 6.00 N acts on a 3.00-kg object for 10.0 s. What are the changes in the object\u2019s momentum and velocity? 84. The velocity of a 625-kg car is changed from 10.0 m/s to 44.0 m/s in 68.0 s by an external, constant force. a. What is the resulting change in momentum of the car? b. What is the magnitude of the force? 85. Dragster An 845-kg dragster accelerates on a race track from rest to 100.0 km/h in 0.90 s. a. What is the change in momentum of the dragster? b. What is the average force exerted on the dragster? c. What exerts that force? 86. Ice Hockey A 0.115-kg hockey puck, moving at 35.0 m/s, strikes a 0.365-kg jacket that is thrown onto the ice by a fan of a certain hockey team. The puck and jacket slide off together. Find their velocity. 87. A 50.0-kg woman, riding on a 10.0-kg cart, is moving east at 5.0 m/s. The woman jumps off the front of the cart and lands on the ground at 7.0 m/s eastward, relative to the ground. a. Sketch the \u201cbefore\u201d and \u201cafter\u201d situations and assign a coordinate axis to them. b. Find the cart\u2019s velocity after the woman jumps off. physicspp.com/chapter_test Chapter 9 Assessment 253 249-255 CH09-SG CA STP-845813 3/11/04 1:18 AM Page 254 88. Gymnastics Figure 9-21 shows a gymnast 92. Analyze and Conclude Two balls during a performing a routine. First, she does giant swings on the high bar, holding her body straight and pivoting around her hands. Then, she lets go of the high bar and grabs her knees with her hands in the tuck position. Finally, she straightens up and lands on her feet. a. In the second and final parts of the gymnast\u2019s routine, around what axis does she spin? b. Rank in order, from greatest to least, her", " moments of inertia for the three positions. collision are shown in Figure 9-22, which is drawn to scale. The balls enter from the left, collide, and then bounce away. The heavier ball, at the bottom of the diagram, has a mass of 0.600 kg, and the other has a mass of 0.400 kg. Using a vector diagram, determine whether momentum is conserved in this collision. Explain any difference in the momentum of the system before and after the collision. c. Rank in order, from greatest to least, her angular velocities in the three positions. \u25a0 Figure 9-21 89. A 60.0-kg male dancer leaps 0.32 m high. a. With what momentum does he reach the ground? b. What impulse is needed to stop the dancer? c. As the dancer lands, his knees bend, lengthening the stopping time to 0.050 s. Find the average force exerted on the dancer\u2019s body. d. Compare the stopping force with his weight. Thinking Critically 90. Apply Concepts A 92-kg fullback, running at 5.0 m/s, attempts to dive directly across the goal line for a touchdown. Just as he reaches the line, he is met head-on in midair by two 75-kg linebackers, both moving in the direction opposite the fullback. One is moving at 2.0 m/s and the other at 4.0 m/s. They all become entangled as one mass. a. Sketch the event, identifying the \u201cbefore\u201d and \u201cafter\u201d situations. b. What is the velocity of the football players after the collision? c. Does the fullback score a touchdown? \u25a0 Figure 9-22 Writing in Physics 93. How can highway barriers be designed to be more effective in saving people\u2019s lives? Research this issue and describe how impulse and change in momentum can be used to analyze barrier designs. 94. While air bags save many lives, they also have caused injuries and even death. Research the arguments and responses of automobile makers to this statement. Determine whether the problems involve impulse and momentum or other issues. Cumulative Review 95. A 0.72-kg ball is swung vertically from a 0.60-m string in uniform circular motion at a speed of 3.3 m/s. What is the tension in the cord at the top of the ball\u2019s motion? (Chapter 6) 96. You wish to launch a satellite that will remain above the same spot on Earth\u2019s surface. This means the satellite must have a period of exactly one day. Calculate the radius of the circular orbit this satellite must have. Hint: The Moon also circles Earth and both the Moon and the satellite will obey Kepler\u2019s third law. The Moon is 3.9108 m from Earth and its period is 27.33 days. (Chapter 7) 91. Analyze and Conclude A student, holding a bicycle wheel with its axis vertical, sits on a stool that can rotate without friction. She uses her hand to get the wheel spinning. Would you expect the student and stool to turn? If so, in which direction? Explain. 97. A rope is wrapped around a drum that is 0.600 m in diameter. A machine pulls with a constant 40.0 N force for a total of 2.00 s. In that time, 5.00 m of rope is unwound. Find, at 2.00 s, and I. (Chapter 8) 254 Chapter 9 Momentum and Its Conservation For more problems, go to Additional Problems, Appendix B. 249-255 CH09-SG CA STP-845813 6/1/04 2:39 AM Page 255 Multiple Choice 1. When a star that is much larger than the Sun nears the end of its lifetime, it begins to collapse, but continues to rotate. Which of the following describes the conditions of the collapsing star\u2019s moment of inertia (I), angular momentum (L), and angular velocity (\u03c9)? I increases, L stays constant, \u03c9decreases I decreases, L stays constant, \u03c9increases I increases, L increases, \u03c9increases I increases, L increases, \u03c9stays constant 2. A 40.0-kg ice-skater glides with a speed of 2.0 m/s toward a 10.0-kg sled at rest on the ice. The iceskater reaches the sled and holds on to it. The ice-skater and the sled then continue sliding in the same direction in which the ice-skater was originally skating. What is the speed of the ice-skater and the sled after they collide? 0.4 m/s 0.8 m/s 1.6 m/s 3.2 m/s 3. A bicyclist applies the brakes and slows the motion of the wheels. The angular momentum of each wheel then decreases from 7.0 kgm2/s to", " 3.5 kgm2/s over a period of 5.0 s. What is the angular impulse on each wheel? 0.7 kgm2/s 1.4 kgm2/s 2.1 kgm2/s 3.5 kgm2/s 4. A 45.0-kg ice-skater stands at rest on the ice. A friend tosses the skater a 5.0-kg ball. The skater and the ball then move backwards across the ice with a speed of 0.50 m/s. What was the speed of the ball at the moment just before the skater caught it? 2.5 m/s 3.0 m/s 4.0 m/s 5.0 m/s 5. What is the difference in momentum between a 50.0-kg runner moving at a speed of 3.00 m/s and a 3.00103-kg truck moving at a speed of only 1.00 m/s? 1275 kgm/s 2550 kgm/s 2850 kgm/s 2950 kgm/s 6. When the large gear in the diagram rotates, it turns the small gear in the opposite direction at the same linear speed. The larger gear has twice the radius and four times the mass of the smaller gear. What is the angular momentum of the larger gear as a function of the angular momentum of the smaller gear? Hint: The moment of inertia for a disk is 1 mr 2, where m 2 is mass and r is the radius of the disk. 8Lsmall 16Lsmall 2Lsmall 4Lsmall 7. A force of 16 N exerted against a rock with an impulse of 0.8 kgm/s causes the rock to fly off the ground with a speed of 4.0 m/s. What is the mass of the rock? 0.2 kg 0.8 kg 1.6 kg 4.0 kg Extended Answer 8. A 12.0-kg rock falls to the ground. What is the impulse on the rock if its velocity at the moment it strikes the ground is 20.0 m/s? If It Looks Too Good To Be True Beware of answer choices in multiple-choice questions that seem ready-made and obvious. Remember that only one answer choice for each question is correct. The rest are made up by testmakers to distract you. This means that they might look very appealing. Check each answer choice carefully before making your final selection. physicspp.com/standardized_test Chapter 9 Standardized Test Practice 255 256-265 CH10-S1-845813 3/21/04 10:21 PM Page 256 What You\u2019ll Learn \u2022 You will recognize that work and power describe how the external world changes the energy of a system. \u2022 You will relate force to work and explain how machines ease the load. Why It\u2019s Important Simple machines and the compound machines formed from them make many everyday tasks easier to perform. Mountain Bikes A multispeed mountain bicycle with shock absorbers allows you to match the ability of your body to exert forces, to do work, and to deliver power climbing steep hills, traversing flat terrain at high speeds, and safely descending hills. Think About This How does a multispeed mountain bicycle enable a cyclist to ride over any kind of terrain with the least effort? physicspp.com 256 Colorstock/Getty Images 256-265 CH10-S1-845813 6/10/04 4:11 PM Page 257 What factors affect energy? Question What factors affect the energy of falling objects and their ability to do work? Procedure Analysis Compare your data for the different craters. Is there an overall trend to your data? Explain. Critical Thinking As the balls are dropped into the sand, they do work on the sand. Energy can be defined as the ability of an object to do work on itself or its surroundings. Relate the trend(s) you found in this lab to the energy of the balls. How can the energy of a ball be increased? 1. Place about 2 cm of fine sand in the bottom of a pie plate or baking pan. 2. Obtain a variety of metal balls or glass marbles of different sizes. 3. Hold a meterstick vertically in one hand, with one end just touching the surface of the sand. With the other hand, drop one of the balls into the sand. Record the height from which you dropped the ball. 4. Carefully remove the ball from the sand, so as not to disturb the impact crater it made. Measure the depth of the crater and how far sand was thrown from the crater. 5. Record the mass of the ball. 6. Smooth out the sand in the pie plate and perform steps 3\u20135 with different sizes of balls and drop them from varying heights. Be sure to drop different sizes of balls from the same height, as well as the same ball from different heights. 10.1 Energy and Work In Chapter", " 9, you learned about the conservation of momentum. You learned that you could examine the state of a system before and after an impulse acted on it without knowing the details about the impulse. The law of conservation of momentum was especially useful when considering collisions, during which forces sometimes changed dramatically. Recall the discussion in Chapter 9 of the two skaters who push each other away. While momentum is conserved in this situation, the skaters continue to move after pushing each other away; whereas before the collision, they were at rest. When two cars crash into each other, momentum is conserved. Unlike the skaters, however, the cars, which were moving prior to the collision, became stationary after the crash. The collision probably resulted in a lot of twisted metal and broken glass. In these types of situations, some other quantity must have been changed as a result of the force acting on each system. Objectives \u2022 Describe the relationship between work and energy. \u2022 Calculate work. \u2022 Calculate the power used. Vocabulary work energy kinetic energy work-energy theorem joule power watt Section 10.1 Energy and Work 257 256-265 CH10-S1-845813 6/1/04 2:57 AM Page 258 Work and Energy d F Recall that change in momentum is the result of an impulse, which is the product of the average force exerted on an object and the time of the interaction. Consider a force exerted on an object while the object moves a certain distance. Because there is a net force, the object will be accelerated, a F/m, and its velocity will increase. Examine Table 3-3 in Chapter 3, on page 68, which lists equations describing the relationships among position, velocity, and time for motion under constant acceleration. Consider the equation involving acceleration, velocity, and distance: 2ad vf 2. If you use Newton\u2019s second law to replace a with F/m and multiply both sides by m/2, you obtain Fd 1 mvf 2 2 1 mvi 2 2 vi 2. \u25a0 Figure 10-1 Work is done when a constant force, F, is exerted on the backpack in the direction of motion and the backpack moves a distance, d. Work The left side of the equation describes something that was done to the system by the external world (the environment). A force, F, was exerted on an object while the object moved a distance, d, as shown in Figure 10-1. If F is a constant force, exerted in the direction in which the object is moving, then work, W, is the product of the force and the object\u2019s displacement. Work W Fd Work is equal to a constant force exerted on an object in the direction of motion, times the object\u2019s displacement. 2 1 mvi 2 2 describes a property of the system. You probably have used the word work in many other ways. For example, a computer might work well, learning physics can be hard work, and you might work at an after-school job. To physicists, however, work has a very precise meaning. Recall that Fd 1 2. Rewriting the equation W Fd results mvf 2 in W 1 2 1 2. The right side of the equation involves the object\u2019s mvi mvf 2 2 mass and its velocities after and before the force was exerted. The quantity 1 mvi 2 Kinetic energy What property of a system does 1 2 describe? A mvi 2 massive, fast-moving vehicle can do damage to objects around it, and a baseball hit at high speed can rise high into the air. That is, an object with this property can produce a change in itself or the world around it. This property, the ability of an object to produce a change in itself or the world around it, is called energy. The fast-moving vehicle and the baseball possess energy that is associated with their motion. This energy resulting from motion is called kinetic energy and is represented by the symbol KE. Kinetic Energy KE 1 mv2 2 The kinetic energy of an object is equal to 1 times the mass of the object 2 multiplied by the speed of the object squared. v F Sun Planet \u25a0 Figure 10-2 If a planet is in a circular orbit, then the force is perpendicular to the direction of motion. Consequently, the gravitational force does no work on the planet. Substituting KE into the equation W 1 2 results in W KEf mvf 2 KEi. The right side is the difference, or change, in kinetic energy. The work-energy theorem states that when work is done on an object, the result is a change in kinetic energy. The work-energy theorem can be represented by the following equation. 2 1 mvi 2 258 Chapter 10 Energy, Work, and Simple Machines 256-265 CH10-S1-845813 3/21/04 11:43 PM Page 259 Work-Energy Theorem W KE Work is equal to the change in kinetic", " energy. The relationship between work done and the change in energy that results was established by nineteenth-century physicist James Prescott Joule. To honor his work, a unit of energy is called a joule (J). For example, if a 2-kg object moves at 1 m/s, it has a kinetic energy of 1 kgm2/s2, or 1 J. Recall that a system is the object of interest and the external world is everything else. For example, one system might be a box in a warehouse and the external world might consist of yourself, Earth\u2019s mass, and anything else external to the box. Through the process of doing work, energy can move between the external world and the system. Notice that the direction of energy transfer can go both ways. If the external world does work on a system, then W is positive and the energy of the system increases. If, however, a system does work on the external world, then W is negative and the energy of the system decreases. In summary, work is the transfer of energy by mechanical means. Calculating Work The first equation used to calculate work is W Fd. This equation, however, holds only for constant forces exerted in the direction of motion. What happens if the force is exerted perpendicular to the direction of motion? An everyday example of this is the motion of a planet around the Sun, as shown in Figure 10-2. If the orbit is circular, then the force is always perpendicular to the direction of motion. Recall from Chapter 6 that a perpendicular force does not change the speed of an object, only its direction. Consequently, the speed of the planet doesn\u2019t change. Therefore, its kinetic energy also is constant. Using the equation W KE, you can see that when KE is constant, KE 0 and thus, W 0. This means that if F and d are at right angles, then W 0. Because the work done on an object equals the change in energy, work also is measured in joules. One joule of work is done when a force of 1 N acts on an object over a displacement of 1 m. An apple weighs about 1 N. Thus, when you lift an apple a distance of 1 m, you do 1 J of work on it. Constant force exerted at an angle You\u2019ve learned that a force exerted in the direction of motion does an amount of work given by W Fd. A force exerted perpendicular to the motion does no work. What work does a force exerted at an angle do? For example, what work does the person pushing the car in Figure 10-3a do? You know that any force can be replaced by its components. If the coordinate system shown in Figure 10-3b is used, the 125-N force, F, exerted in the direction of the person\u2019s arm, has two components. The magnitude of the horizontal component, Fx, is related to the magnitude of the force, F, by a cosine function: cos 25.0\u00b0 Fx/F. By F cos 25.0\u00b0 (125 N)(cos 25.0\u00b0) 113 N. solving for Fx, you obtain Fx F sin 25.0\u00b0 Using the same method, the vertical component Fy (125 N)(sin 25.0\u00b0) 52.8 N, where the negative sign shows that the force is downward. Because the displacement is in the x direction, only the x-component does work. The y-component does no work. a b y Fx 25.0\u00b0 F 125 N x Fy \u25a0 Figure 10-3 If a force is applied to a car at an angle, the net force doing the work is the component that acts in the direction of the displacement. Section 10.1 Energy and Work 259 Hutchings Photography 256-265 CH10-S1-845813 3/21/04 10:26 PM Page 260 The work you do when you exert a force on an object, at an angle to the direction of motion, is equal to the component of the force in the direction of the displacement, multiplied by the distance moved. The magnitude of the component force acting in the direction of displacement is found by multiplying the magnitude of force, F, by the cosine of the angle F cos. Thus, the between F and the direction of the displacement: Fx work done is represented by the following equation. Work (Angle Between Force and Displacement) W Fd cos Work is equal to the product of force and displacement, times the cosine of the angle between the force and the direction of the displacement. Other agents exert forces on the pushed car as well. Which of these agents do work? Earth\u2019s gravity acts downward, the ground exerts a normal force upward, and friction exerts a horizontal force opposite the direction of motion. The upward and downward forces are perpendicular to the direction of motion and do no work. For these forces, 90\u00b0, which makes cos", " 0, and thus, W 0. The work done by friction acts in the direction opposite that of motion\u2014at an angle of 180\u00b0. Because cos 180\u00b0 1, the work done by friction is negative. Negative work done by a force exerted by something in the external world reduces the kinetic energy of the system. If the person in Figure 10-3a were to stop pushing, the car would quickly stop moving\u2014 its energy of motion would be reduced. Positive work done by a force increases the energy, while negative work decreases it. Use the problemsolving strategies below when you solve problems related to work. Work When solving work-related problems, use the following strategies. 1. Sketch the system and show the force that is doing the work. 2. Draw the force and displacement vectors of the system. 3. Find the angle,, between each force and displacement. 4. Calculate the work done by each force using W Fd cos. 5. Calculate the net work done. Check the sign of the work using the direction of energy transfer. If the energy of the system has increased, the work done by that force is positive. If the energy has decreased, then the work done by that force is negative. 260 Chapter 10 Energy, Work, and Simple Machines Work Diagram d FN Fg FA d A FA d d N FN g Fg 256-265 CH10-S1-845813 3/21/04 10:26 PM Page 261 d F F Work and Energy A 105-g hockey puck is sliding across the ice. A player exerts a constant 4.50-N force over a distance of 0.150 m. How much work does the player do on the puck? What is the change in the puck\u2019s energy? 1 Analyze and Sketch the Problem \u2022 Sketch the situation showing initial conditions. \u2022 Establish a coordinate system with x to the right. \u2022 Draw a vector diagram. x Known: m 105 g F 4.50 N d 0.150 m Unknown: W? KE? 2 Solve for the Unknown Use the equation for work when a constant force is exerted in the same direction as the object\u2019s displacement. W Fd (4.50 N)(0.150 m) 0.675 Nm 0.675 J Substitute F 4.50 N, d 0.150 m 1 J 1 Nm Math Handbook Operations with Significant Digits pages 835\u2014836 Use the work-energy theorem to determine the change in energy of the system. W KE KE 0.675 J Substitute W 0.675 J 3 Evaluate the Answer \u2022 Are the units correct? Work is measured in joules. \u2022 Does the sign make sense? The player (external world) does work on the puck (the system). So the sign of work should be positive. 1. Refer to Example Problem 1 to solve the following problem. a. If the hockey player exerted twice as much force, 9.00 N, on the puck, how would the puck\u2019s change in kinetic energy be affected? b. If the player exerted a 9.00 N-force, but the stick was in contact with the puck for only half the distance, 0.075 m, what would be the change in kinetic energy? 2. Together, two students exert a force of 825 N in pushing a car a distance of 35 m. a. How much work do the students do on the car? b. If the force was doubled, how much work would they do pushing the car the same distance? 3. A rock climber wears a 7.5-kg backpack while scaling a cliff. After 30.0 min, the climber is 8.2 m above the starting point. a. How much work does the climber do on the backpack? b. If the climber weighs 645 N, how much work does she do lifting herself and the backpack? c. What is the average power developed by the climber? Section 10.1 Energy and Work 261 256-265 CH10-S1-845813 3/21/04 10:27 PM Page 262 Force and Displacement at an Angle A sailor pulls a boat a distance of 30.0 m along a dock using a rope that makes a 25.0\u00b0 angle with the horizontal. How much work does the sailor do on the boat if he exerts a force of 255 N on the rope? 1 Analyze and Sketch the Problem \u2022 Establish coordinate axes. \u2022 Sketch the situation showing the boat with initial conditions. \u2022 Draw a vector diagram showing the force and its component in the direction of the displacement. y x Unknown: W? Known: F 255 N d 30.0 m 25.0\u00b0 F 25.0\u00b0 d F F cos 25.0\u00b0 2 3 Solve for the Unknown Use the equation for work done when there is an angle between the force and displacement. W Fd cos (255 N)(30.0 m)(cos 25.0", "\u00b0) 6.93103 J Substitute F 255 N, d 30.0 m, 25.0\u00b0 Evaluate the Answer \u2022 Are the units correct? Work is measured in joules. \u2022 Does the sign make sense? The sailor does work on the boat, which agrees with a positive sign for work. Math Handbook Trigonometric Ratios page 855 4. If the sailor in Example Problem 2 pulled with the same force, and along the same distance, but at an angle of 50.0\u00b0, how much work would he do? 5. Two people lift a heavy box a distance of 15 m. They use ropes, each of which makes an angle of 15\u00b0 with the vertical. Each person exerts a force of 225 N. How much work do they do? 6. An airplane passenger carries a 215-N suitcase up the stairs, a displacement of 4.20 m vertically, and 4.60 m horizontally. a. How much work does the passenger do? b. The same passenger carries the same suitcase back down the same set of stairs. How much work does the passenger do now? 7. A rope is used to pull a metal box a distance of 15.0 m across the floor. The rope is held at an angle of 46.0\u00b0 with the floor, and a force of 628 N is applied to the rope. How much work does the force on the rope do? 8. A bicycle rider pushes a bicycle that has a mass of 13 kg up a steep hill. The incline is 25\u00b0 and the road is 275 m long, as shown in Figure 10-4. The rider pushes the bike parallel to the road with a force of 25 N. a. How much work does the rider do on the bike? b. How much work is done by the force of gravity on the bike? 262 Chapter 10 Energy, Work, and Simple Machines 5 m 7 2 5 N 2 25\u00b0 \u25a0 Figure 10-4 (Not to scale) 256-265 CH10-S1-845813 6/1/04 2:58 AM Page 263 Finding work done when forces change A graph of force versus displacement lets you determine the work done by a force. This graphical method can be used to solve problems in which the force is changing. Figure 10-5a shows the work done by a constant force of 20.0 N that is exerted to lift an object a distance of 1.50 m. The work done by this constant force is represented by W Fd (20.0 N)(1.50 m) 30.0 J. The shaded area under the graph is equal to (20.0 N)(1.50 m), or 30.0 J. The area under a force-displacement graph is equal to the work done by that force, even if the force changes. Figure 10-5b shows the force exerted by a spring, which varies linearly from 0.0 to 20.0 N as it is compressed 1.50 m. The work done by the force that compressed the spring is the area under the graph, which is the area of a triangle, 1 (base)(altitude), or 2 W 1 (20.0 N)(1.50 m) 15.0 J. 2 Work done by many forces Newton\u2019s second law of motion relates the net force on an object to its acceleration. In the same way, the work-energy theorem relates the net work done on a system to its energy change. If several forces are exerted on a system, calculate the work done by each force, and then add the results. Power Until now, none of the discussions of work has mentioned the time it takes to move an object. The work done by a person lifting a box of books is the same whether the box is lifted onto a shelf in 2 s or each book is lifted separately so that it takes 20 min to put them all on the shelf. Although the work done is the same, the rate at which it is done is different. Power is the work done, divided by the time taken to do the work. In other words, power is the rate at which the external force changes the energy of the system. It is represented by the following equation. Power P W t Power is equal to the work done, divided by the time taken to do the work. \u25a0 Figure 10-5 Work can be obtained graphically by finding the area under a force-displacement graph 20.0 0.0 1.50 Displacement (m) 20.0 0.0 1.50 Displacement (m) \u25a0 Figure 10-6 These students are doing work at different rates while climbing the stairs. Consider the three students in Figure 10-6. The girl hurrying up the stairs is more powerful than the boy who is walking up the stairs. Even though the same work is accomplished by both, the girl accomplishes it in less time and thus develops more power. In the case of the two students walking", " up the stairs, both accomplish work in the same amount of time. Power is measured in watts (W). One watt is 1 J of energy transferred in 1 s. A watt is a relatively small unit of power. For example, a glass of water weighs about 2 N. If you lift it 0.5 m to your mouth, you do 1 J of work. If you lift the glass in 1 s, you are doing work at the rate of 1 W. Because a watt is such a small unit, power often is measured in kilowatts (kW). One kilowatt is equal to 1000 W. Section 10.1 Energy and Work 263 Laura Sifferlin 256-265 CH10-S1-845813 3/21/04 10:28 PM Page 264 Power An electric motor lifts an elevator 9.00 m in 15.0 s by exerting an upward force of 1.20104 N. What power does the motor produce in kW? 1 Analyze and Sketch the Problem \u2022 Sketch the situation showing the elevator with initial conditions. \u2022 Establish a coordinate system with up as positive. \u2022 Draw a vector diagram for the force and displacement. y F d F Known: d 9.00 m t 15.0 s F 1.20104 N Unknown: P? 2 Solve for the Unknown Solve for power. P W t d F t (1.20104 N)(9.00 m) (15.0 s) 7.20 kW Substitute W Fd Substitute F 1.20104 N, d 9.00 m, t 15.0 s Math Handbook Operations with Scientific Notation pages 842\u2014843 3 Evaluate the Answer \u2022 Are the units correct? Power is measured in J/s. \u2022 Does the sign make sense? The positive sign agrees with the upward direction of the force. 9. A box that weighs 575 N is lifted a distance of 20.0 m straight up by a cable attached to a motor. The job is done in 10.0 s. What power is developed by the motor in W and kW? 10. You push a wheelbarrow a distance of 60.0 m at a constant speed for 25.0 s, by exerting a 145-N force horizontally. a. What power do you develop? b. If you move the wheelbarrow twice as fast, how much power is developed? 11. What power does a pump develop to lift 35 L of water per minute from a depth of 110 m? (1 L of water has a mass of 1.00 kg.) 12. An electric motor develops 65 kW of power as it lifts a loaded elevator 17.5 m in 35 s. How much force does the motor exert? 13. A winch designed to be mounted on a truck, as shown in Figure 10-7, is advertised as being able to exert a 6.8103-N force and to develop a power of 0.30 kW. How long would it take the truck and the winch to pull an object 15 m? 14. Your car has stalled and you need to push it. You notice as the car gets going that you need less and less force to keep it going. Suppose that for the first 15 m, your force decreased at a constant rate from 210.0 N to 40.0 N. How much work did you do on the car? Draw a force-displacement graph to represent the work done during this period. 264 Warn Industries Inc. Chapter 10 Energy, Work, and Simple Machines \u25a0 Figure 10-7 256-265 CH10-S1-845813 3/21/04 10:29 PM Page 265 \u25a0 Figure 10-8 When riding a multispeed bicycle, if the muscles in your body exert a force of 400 N and the speed is 2.6 m/s, the power output is over 1000 W. Maximizing Power on a Multispeed Bicycle Output power 1000 Velocity 200 400 600 Force (N) 500 You may have noticed in Example Problem 3 that when the force and displacement are in the same direction, P Fd/t. However, because the ratio d/t is the speed, power also can be calculated using P Fv. When you are riding a multispeed bicycle, how do you choose the correct gear? You want to get your body to deliver the largest amount of power. By considering the equation P Fv you can see that either zero force or zero speed results in no power delivered. The muscles cannot exert extremely large forces, nor can they move very fast. Thus, some combination of moderate force and moderate speed will produce the largest amount of power. Figure 10-8 shows that in this particular situation, the maximum power output is over 1000 W when the force is about 400 N and speed is about 2.6 m/s. All engines\u2014not just humans\u2014have these limitations. Simple machines often are designed to match the force and speed that the engine can deliver to the needs of", " the job. You will learn more about simple machines in the next section. Tour de France A bicyclist in the Tour de France rides at about 8.94 m/s for more than 6 h a day. The power output of the racer is about 1 kW. One-fourth of that power goes into moving the bike against the resistance of the air, gears, and tires. Three-fourths of the power is used to cool the racer\u2019s body. 10.1 Section Review 15. Work Murimi pushes a 20-kg mass 10 m across a floor with a horizontal force of 80 N. Calculate the amount of work done by Murimi. 16. Work A mover loads a 185-kg refrigerator into a moving van by pushing it up a 10.0-m, frictionfree ramp at an angle of inclination of 11.0\u00b0. How much work is done by the mover? 17. Work and Power Does the work required to lift a book to a high shelf depend on how fast you raise it? Does the power required to lift the book depend on how fast you raise it? Explain. 18. Power An elevator lifts a total mass of 1.1103 kg a distance of 40.0 m in 12.5 s. How much power does the elevator generate? 19. Work A 0.180-kg ball falls 2.5 m. How much work does the force of gravity do on the ball? 20. Mass A forklift raises a box 1.2 m and does 7.0 kJ of work on it. What is the mass of the box? 21. Work You and a friend each carry identical boxes from the first floor of a building to a room located on the second floor, farther down the hall. You choose to carry the box first up the stairs, and then down the hall to the room. Your friend carries it down the hall on the first floor, then up a different stairwell to the second floor. Who does more work? 22. Work and Kinetic Energy If the work done on an object doubles its kinetic energy, does it double its velocity? If not, by what ratio does it change the velocity? 23. Critical Thinking Explain how to find the change in energy of a system if three agents exert forces on the system at once. physicspp.com/self_check_quiz Section 10.1 Energy and Work 265 266-273 CH10-S2-845813 3/21/04 10:33 PM Page 266 10.2 Machines Objectives \u2022 Demonstrate a knowledge of the usefulness of simple machines. \u2022 Differentiate between ideal and real machines in terms of efficiency. \u2022 Analyze compound machines in terms of combinations of simple machines. \u2022 Calculate efficiencies for simple and compound machines. Vocabulary machine effort force resistance force mechanical advantage ideal mechanical advantage efficiency compound machine Everyone uses machines every day. Some are simple tools, such as bottle openers and screwdrivers, while others are complex, such as bicycles and automobiles. Machines, whether powered by engines or people, make tasks easier. A machine eases the load by changing either the magnitude or the direction of a force to match the force to the capability of the machine or the person. Benefits of Machines Consider the bottle opener in Figure 10-9. When you use the opener, you lift the handle, thereby doing work on the opener. The opener lifts the cap, doing work on it. The work that you do is called the input work, Wi. The work that the machine does is called the output work, Wo. Recall that work is the transfer of energy by mechanical means. You put work into a machine, such as the bottle opener. That is, you transfer energy to the opener. The opener, in turn, does work on the cap, thereby transferring energy to it. The opener is not a source of energy, and therefore, the cap cannot receive more energy than the amount of energy that you put into the opener. Thus, the output work can never be greater than the input work. The machine simply aids in the transfer of energy from you to the bottle cap. Mechanical advantage The force exerted by a person on a machine is called the effort force, Fe. The force exerted by the machine is called the resistance force, Fr. As shown in Figure 10-9a, Fe is the upward force exerted by the person using the bottle opener and Fr is the upward force exerted by the bottle opener. The ratio of resistance force to effort force, Fr/Fe, is called the mechanical advantage, MA, of the machine. F Mechanical Advantage MA r F e The mechanical advantage of a machine is equal to the resistance force divided by the effort force. a b Fr Fe dr de \u25a0 Figure 10-9 A bottle opener is an example of a simple machine. It makes opening a bottle easier, but it does not lessen the work required to do so. 266 Hutchings Photography Chapter 10 Energy, Work, and Simple Machines 266-273 CH10-S2-845813 3", "/21/04 10:34 PM Page 267 a b \u25a0 Figure 10-10 A fixed pulley has a mechanical advantage equal to 1 (a). A pulley system with a movable pulley has a mechanical advantage greater than 1 (b). F F In a fixed pulley, such as the one shown in Figure 10-10a, the forces, Fe and Fr, are equal, and consequently MA is 1. What is the advantage of this machine? The fixed pulley is useful, not because the effort force is lessened, but because the direction of the effort force is changed. Many machines, such as the bottle opener shown in Figure 10-9 and the pulley system shown in Figure 10-10b, have a mechanical advantage greater than 1. When the mechanical advantage is greater than 1, the machine increases the force applied by a person. You can write the mechanical advantage of a machine in another way using the definition of work. The input work is the product of the effort force that a person exerts, Fe, and the distance his or her hand moved, de. In the same way, the output work is the product of the resistance force, Fr, and the displacement of the load, dr. A machine can increase force, but it cannot increase energy. An ideal machine transfers all the energy, so the output work equals the input work: Wo This equation can be rewritten Fr /Fe Wi or Frdr Fede. de/dr. Recall that mechanical advantage is given by MA Fr/Fe. Therefore, for an ideal machine, ideal mechanical advantage, IMA, is equal to the displacement of the effort force, divided by the displacement of the load. The ideal mechanical advantage can be represented by the following equation. d Ideal Mechanical Advantage IMA e d r The ideal mechanical advantage of an ideal machine is equal to the displacement of the effort force, divided by the displacement of the load. Note that you measure the distances moved to calculate the ideal mechanical advantage, but you measure the forces exerted to find the actual mechanical advantage. Section 10.2 Machines 267 266-273 CH10-S2-845813 3/21/04 10:35 PM Page 268 Efficiency In a real machine, not all of the input work is available as output work. Energy removed from the system means that there is less output work from the machine. Consequently, the machine is less efficient at accomplishing the task. The efficiency of a machine, e, is defined as the ratio of output work to input work. W 100 Efficiency e o W i The efficiency of a machine (in %) is equal to the output work, divided by the input work, multiplied by 100. 1, and its An ideal machine has equal output and input work, Wo/Wi efficiency is 100 percent. All real machines have efficiencies of less than 100 percent. Efficiency can be expressed in terms of the mechanical advantage and ideal mechanical advantage. Efficiency, e Wo/Wi, can be rewritten as follows: F rd W r o F d W i e e Because MA Fr/Fe and IMA de/dr, the following expression can be written for efficiency. A M 100 Efficiency e M A I The efficiency of a machine (in %) is equal to its mechanical advantage, divided by the ideal mechanical advantage, multiplied by 100. A machine\u2019s design determines its ideal mechanical advantage. An efficient machine has an MA almost equal to its IMA. A less-efficient machine has a small MA relative to its IMA. To obtain the same resistance force, a greater force must be exerted in a machine of lower efficiency than in a machine of higher efficiency. An electric pump pulls water at a rate of 0.25 m3/s from a well that is 25 m deep. The water leaves the pump at a speed of 8.5 m/s. 1. What power is needed to lift the water to the surface? 2. What power is needed to increase the pump\u2019s kinetic energy? 3. If the pump\u2019s efficiency is 80 percent, how much power must be delivered to the pump? 268 Chapter 10 Energy, Work, and Simple Machines 8.5 m/s 25 m (Not to scale) 266-273 CH10-S2-845813 6/7/04 6:08 AM Page 269 Fr a d b c Fe Fr Fr Fe e Fe Fr Fe Fe f Fe 1_ 2 Fr 1_ 2 Fr Fr Compound Machines Most machines, no matter how complex, are combinations of one or more of the six simple machines: the lever, pulley, wheel and axle, inclined plane, wedge, and screw. These machines are shown in Figure 10-11. The IMA of all the machines shown in Figure 10-11 is the ratio of distances moved. For machines, such as the lever and the wheel and axle, this ratio can be replaced by the ratio of the distance between the place where the force is applied and the pivot point. A common version of the wheel and", " axle is a steering wheel, such as the one shown in Figure 10-12. The IMA is the ratio of the radii of the wheel and axle. A machine consisting of two or more simple machines linked in such a way that the resistance force of one machine becomes the effort force of the second is called a compound machine. \u25a0 Figure 10-12 The IMA for the steering wheel is re/rr. rr Wheel Pivot point Fr Axle re Fe \u25a0 Figure 10-11 Simple machines include the lever (a), pulley (b), wheel and axle (c), inclined plane (d), wedge (e), and screw (f). Section 10.2 Machines 269 266-273 CH10-S2-845813 3/21/04 10:36 PM Page 270 \u25a0 Figure 10-13 A series of simple machines combine to transmit the force that the rider exerts on the pedal to the road. Fchain on gear Fgear on chain Frider on road Frider on pedal In a bicycle, the pedal and front gear act like a wheel and axle. The effort force is the force that the rider exerts on the pedal, Frider on pedal. The resistance is the force that the front gear exerts on the chain, Fgear on chain, as shown in Figure 10-13. The chain exerts an effort force on the rear gear, Fchain on gear, equal to the force exerted on the chain. This gear and the rear wheel act like another wheel and axle. The resistance force is the force that the wheel exerts on the road, Fwheel on road. According to Newton\u2019s third law, the ground exerts an equal forward force on the wheel, which accelerates the bicycle forward. The MA of a compound machine is the product of the MAs of the simple machines from which it is made. For example, in the case of the bicycle illustrated in Figure 10-13, the following is true. Wheel and Axle The gear mechanism on your bicycle multiplies the distance that you travel. What does it do to the force? 1. Mount a wheel and axle system on a sturdy support rod. 2. Wrap a 1-m-long piece of string clockwise around the axle. 3. Wrap another piece of 1-m-long string counterclockwise around the large diameter wheel. 4. Hang a 500-g mass from the end of the string on the larger wheel. CAUTION: Avoid dropping the mass. 5. Pull the string from the axle down so that the mass is lifted by about 10 cm. Analyze and Conclude 6. What did you notice about the force on the string in your hand? 7. What did you notice about the distance that your hand needed to move to lift the mass? Explain the results in terms of the work done on both strings. MA MAmachine 1 )( F g w F n i a h c MAmachine 2 ) w F el ch rid r e MA ( The IMA of each wheel-and-axle machine is the ratio of the distances moved. s iu ad r l a d e p For the pedal gear, IMA g a ius d ra r e t n fro us i d a r ar e g For the rear wheel, IMA re ra s u i d l e he ar w For the bicycle, then, s iu ad r l a d e p ius d ra r a e g t n fro IMA ( ( )(re )( pe ar e g l e he ar us) i d a r s u i d ra Because both gears use the same chain and have teeth of the same size, you can count the number of teeth to find the IMA, as follows. IMA ( teeth on rear gear teeth on front gear )(pe g n le m r a l da s u i rad l e e h w th) Shifting gears on a bicycle is a way of adjusting the ratio of gear radii to obtain the desired IMA. You know that if the pedal of a bicycle is at the top or bottom of its circle, no matter how much downward force you exert, the pedal will not turn. The force of your foot is most effective when the force is exerted perpendicular to the arm of the pedal; that is, when the torque is largest. Whenever a force on a pedal is specified, assume that it is applied perpendicular to the arm. 270 Chapter 10 Energy, Work, and Simple Machines Laura Sifferlin 266-273 CH10-S2-845813 6/1/04 3:13 AM Page 271 Mechanical Advantage You examine the rear wheel on your bicycle. It has a radius of 35.6 cm and has a gear with a radius of 4.00 cm. When the chain is pulled with a force of 155 N, the wheel rim moves 14.0 cm. The efficiency of this part of the bicycle is 95.0 percent. a. What is the IMA", " of the wheel and gear? b. What is the MA of the wheel and gear? c. What is the resistance force? d. How far was the chain pulled to move the rim 14.0 cm? 1 2 Analyze and Sketch the Problem \u2022 Sketch the wheel and axle. \u2022 Sketch the force vectors. 35.6 cm Fe Known: 4.00 cm 35.6 cm 155 N re rr Fe e 95.0% dr 14.0 cm Solve for the Unknown a. Solve for IMA. r IMA.112 Unknown: Fr de IMA? MA??? 4.00 cm Fr For a wheel-and-axle machine, IMA is equal to the ratio of radii. Substitute re 4.00 cm, rr 35.6 cm b. Solve for MA. A M 100 e M A I ) IMA MA ( 0) 0.112 (9. 5 0 0 1 0.106 e 00 1 c. Solve for force. F MA r F e Fr (MA)(Fe) (0.106)(155 N) 16.4 N d. Solve for distance. d IMA e d r de (IMA)(dr) (0.112)(14.0 cm) 1.57 cm Substitute e 95.0%, IMA 0.112 Substitute MA 0.106, Fe 155 N Substitute IMA 0.112, dr 14.0 cm Math Handbook Isolating a Variable page 845 3 Evaluate the Answer \u2022 Are the units correct? Force is measured in newtons and distance in centimeters. \u2022 Is the magnitude realistic? IMA is low for a bicycle because a greater Fe is traded for a greater dr. MA is always smaller than IMA. Because MA is low, Fr also will be low. The small distance the axle moves results in a large distance covered by the wheel. Thus, de should be very small. Section 10.2 Machines 271 266-273 CH10-S2-845813 3/21/04 10:38 PM Page 272 24. If the gear radius in the bicycle in Example Problem 4 is doubled, while the force exerted on the chain and the distance the wheel rim moves remain the same, what quantities change, and by how much? 25. A sledgehammer is used to drive a wedge into a log to split it. When the wedge is driven 0.20 m into the log, the log is separated a distance of 5.0 cm. A force of 1.7104 N is needed to split the log, and the sledgehammer exerts a force of 1.1104 N. a. What is the IMA of the wedge? b. What is the MA of the wedge? c. Calculate the efficiency of the wedge as a machine. 26. A worker uses a pulley system to raise a 24.0-kg carton 16.5 m, as shown in Figure 10-14. A force of 129 N is exerted, and the rope is pulled 33.0 m. a. What is the MA of the pulley system? b. What is the efficiency of the system? 27. You exert a force of 225 N on a lever to raise a 1.25103-N rock a distance of 13 cm. If the efficiency of the lever is 88.7 percent, how far did you move your end of the lever? 28. A winch has a crank with a 45-cm radius. A rope is wrapped around a drum with a 7.5-cm radius. One revolution of the crank turns the drum one revolution. a. What is the ideal mechanical advantage of this machine? b. If, due to friction, the machine is only 75 percent efficient, how much force would have to be exerted on the handle of the crank to exert 750 N of force on the rope? 33.0 m 24.0 kg 16.5 m 129 N \u25a0 Figure 10-14 Multi-gear bicycle On a multi-gear bicycle, the rider can change the MA of the machine by choosing the size of one or both gears. When accelerating or climbing a hill, the rider increases the ideal mechanical advantage to increase the force that the wheel exerts on the road. To increase the IMA, the rider needs to make the rear gear radius large compared to the front gear radius (refer to the IMA equation on page 270). For the same force exerted by the rider, a larger force is exerted by the wheel on the road. However, the rider must rotate the pedals through more turns for each revolution of the wheel. On the other hand, less force is needed to ride the bicycle at high speed on a level road. The rider needs to choose a gear that has a small rear gear and a large front gear that will result in a smaller IMA. Thus, for the same force exerted by the rider, a smaller force is exerted by the wheel on the road. However, in return, the", " rider does not have to move the pedals as far for each revolution of the wheel. An automobile transmission works in the same way. To accelerate a car from rest, large forces are needed and the transmission increases the IMA. At high speeds, however, the transmission reduces the IMA because smaller forces are needed. Even though the speedometer shows a high speed, the tachometer indicates the engine\u2019s low angular speed. 272 Chapter 10 Energy, Work, and Simple Machines 266-273 CH10-S2-845813 3/24/04 8:12 AM Page 273 The Human Walking Machine Movement of the human body is explained by the same principles of force and work that describe all motion. Simple machines, in the form of levers, give humans the ability to walk and run. The lever systems of the human body are complex. However each system has the following four basic parts. 1. a rigid bar (bone) 2. a source of force (muscle contraction) 3. a fulcrum or pivot (movable joints between bones) 4. a resistance (the weight of the body or an object being lifted or moved) Figure 10-15 shows the parts of the lever system in a human leg. Lever systems of the body are not very efficient, and mechanical advantages are low. This is why walking and jogging require energy (burn calories) and help people lose weight. When a person walks, the hip acts as a fulcrum and moves through the arc of a circle, centered on the foot. The center of mass of the body moves as a resistance around the fulcrum in the same arc. The length of the radius of the circle is the length of the lever formed by the bones of the leg. Athletes in walking races increase their velocity by swinging their hips upward to increase this radius. A tall person\u2019s body has lever systems with less mechanical advantage than a short person\u2019s does. Although tall people usually can walk faster than short people can, a tall person must apply a greater force to move the longer lever formed by the leg bones. How would a tall person do in a walking race? What are the factors that affect a tall person\u2019s performance? Walking races are usually 20 or 50 km long. Because of the inefficiency of their lever systems and the length of a walking race, very tall people rarely have the stamina to win. Biology Connection Biology Connection 2 3 4 1 \u25a0 Figure 10-15 The human walking machine. 10.2 Section Review 29. Simple Machines Classify the tools below as a lever, a wheel and axle, an inclined plane, a wedge, or a pulley. a. screwdriver b. pliers c. chisel d. nail puller 30. IMA A worker is testing a multiple pulley system to estimate the heaviest object that he could lift. The largest downward force he could exert is equal to his weight, 875 N. When the worker moves the rope 1.5 m, the object moves 0.25 m. What is the heaviest object that he could lift? 31. Compound Machines A winch has a crank on a 45-cm arm that turns a drum with a 7.5-cm radius through a set of gears. It takes three revolutions of the crank to rotate the drum through one revolution. What is the IMA of this compound machine? 32. Efficiency Suppose you increase the efficiency of a simple machine. Do the MA and IMA increase, decrease, or remain the same? 33. Critical Thinking The mechanical advantage of a multi-gear bicycle is changed by moving the chain to a suitable rear gear. a. To start out, you must accelerate the bicycle, so you want to have the bicycle exert the greatest possible force. Should you choose a small or large gear? b. As you reach your traveling speed, you want to rotate the pedals as few times as possible. Should you choose a small or large gear? c. Many bicycles also let you choose the size of the front gear. If you want even more force to accelerate while climbing a hill, would you move to a larger or smaller front gear? physicspp.com/self_check_quiz Section 10.2 Machines 273 274-275 CH10-LAB-845813 6/1/04 3:23 AM Page 274 Stair Climbing and Power Can you estimate the power you develop as you climb a flight of stairs? Climbing stairs requires energy. As the weight of the body moves through a distance, work is done. Power is a measure of the rate at which work is done. In this activity you will try to maximize the power you develop by applying a vertical force up a flight of stairs over a period of time. QUESTION What can you do to increase the power you develop as you climb a flight of stairs? Objectives Safety Precautions \u25a0 Predict the factors that affect power. \u25a0 Calculate the power developed. \u25a0 Define power operationally. \u25a0 Interpret force, distance, work, time and", " power data. \u25a0 Make and use graphs of work versus time, power versus force, and power versus time. \u25a0 Avoid wearing loose clothing. Materials meterstick (or tape measure) stopwatch bathroom scale Procedure 1. Measure and record the mass of each person in your group using a bathroom scale. If the scale does not have kilogram units, convert the weight in pounds to kilograms. Recall that 2.2 lbs 1 kg. 2. Measure the distance from the floor to the top of the flight of stairs you will climb. Record it in the data table. 3. Have each person in your group climb the flight of stairs in a manner that he or she thinks will maximize the power developed. 4. Use your stopwatch to measure the time it takes each person to perform this task. Record your data in the data table. 274 Horizons Companies 274-275 CH10-LAB-845813 6/1/04 3:24 AM Page 275 Data Table Mass (kg) Weight (N) Distance (m) Work Done (J) Time (s) Power Generated (W) Analyze 1. Calculate Find each person\u2019s weight in newtons and record it in the data table. 2. Calculate the work done by each person. 3. Calculate the power developed by each person in your group as he or she climbs the flight of stairs. 4. Make and Use Graphs Use the data you calculated to draw a graph of work versus time and draw the best-fit line. 5. Draw a graph of power versus work and draw the best-fit line. 6. Draw a graph of power versus time and draw the best-fit line. Conclude and Apply 1. Did each person in your group have the same power rating? Why or why not? 2. Which graph(s) showed a definite relationship between the two variables? 3. Explain why this relationship exists. 4. Write an operational definition of power. Going Further 1. What three things can be done to increase the power you develop while climbing the flight of stairs? 2. Why were the fastest climbers not necessarily the ones who developed the most power? 3. Why were the members of your group with more mass not necessarily the ones who developed the most power? 4. Compare and contrast your data with those of other groups in your class. Real-World Physics 1. Research a household appliance that has a power rating equal to or less than the power you developed by climbing the stairs. 2. Suppose an electric power company in your area charges $0.06/kWh. If you charged the same amount for the power you develop climbing stairs, how much money would you earn by climbing stairs for 1 h? 3. If you were designing a stair climbing machine for the local health club, what information would you need to collect? You decide that you will design a stair climbing machine with the ability to calculate the power developed. What information would you have the machine collect in order to let the climber know how much power he or she developed? To find out more about energy, work, and simple machines, visit the Web site: physicspp.com 275 276 CH10-FEATURE-845813 6/1/04 3:28 AM Page 276 Bicycle Gear Shifters In a multispeed bicycle with two or three front gears and from five to eight rear gears, front and rear derailleurs (shifters) are employed to position the chain. Changing the combination of front and rear gears varies the IMA of the system. A larger IMA reduces effort in climbing hills. A lower IMA allows for greater speed on level ground, but more effort is required. IMA number of teeth on rear gear number of teeth on front gear Rear derailleur Upper gear Lower gear 1 The left-hand shift lever adjusts the position of the front derailleur by means of a connecting cable. 2 The right-hand shift lever adjusts the position of the rear derailleur by means of a wire cable routed along the frame from the shift lever to the derailleur. Front derailleur Thinking Critically 1. Calculate What is the IMA of a multispeed bicycle in the following instances? a. when the chain is set on a front gear with 52 teeth and a rear gear with 14 teeth b. when the chain is set on a front gear with 42 teeth and a rear gear with 34 teeth 2. Apply Which setting in the previous problem, a or b, would you select to race with a friend on level ground? To climb a steep hill? 3 The front derailleur lifts the chain off of the gear teeth and moves it to another gear near it in response to the tightening or loosening of the cable that connects it to the left-hand shift lever. Gear-shift cable 4 The rear derailleur, responding to movements of the right-hand shift lever, positions the chain on the rear gears. The rear derailleur assembly includes two small gears. The lower gear, tensioned by a", " spring, keeps the chain tight as it moves from larger to smaller gears. The upper gear moves in or out to position the chain on the gears next to it. 276 How It Works 277-283 CH10-SG CA STP-845813 6/7/04 6:13 AM Page 277 10.1 Energy and Work Vocabulary \u2022 work (p. 258) \u2022 energy (p. 258) \u2022 kinetic energy (p. 258) \u2022 work-energy theorem (p. 258) \u2022 joule (p. 259) \u2022 power (p. 263) \u2022 watt (p. 263) 10.2 Machines Vocabulary \u2022 machine (p. 266) \u2022 effort force (p. 266) \u2022 resistance force (p. 266) \u2022 mechanical advantage (p. 266) \u2022 ideal mechanical advantage (p. 267) \u2022 efficiency (p. 268) \u2022 compound machine (p. 269) Key Concepts \u2022 Work is the transfer of energy by mechanical means. \u2022 A moving object has kinetic energy. W Fd KE 1 mv2 2 \u2022 The work done on a system is equal to the change in energy of the system. W KE \u2022 Work is the product of the force exerted on an object and the distance the object moves in the direction of the force. W Fd cos \u2022 The work done can be determined by calculating the area under a force- displacement graph. \u2022 Power is the rate of doing work, that is the rate at which energy is transferred. P W t Key Concepts \u2022 Machines, whether powered by engines or humans, do not change the amount of work done, but they do make the task easier. \u2022 A machine eases the load, either by changing the magnitude or the direction of the force exerted to do work. \u2022 The mechanical advantage, MA, is the ratio of resistance force to effort force. F MA r F e \u2022 The ideal mechanical advantage, IMA, is the ratio of the distances moved. d IMA e d r \u2022 The efficiency of a machine is the ratio of output work to input work. W 100 e o W i \u2022 In all real machines, MA is less than IMA. \u2022 The efficiency of a machine can by found from the real and ideal mechanical advantages. A M 100 e M A I physicspp.com/vocabulary_puzzlemaker 277 277-283 CH10-SG CA STP-845813 3/21/04 11:54 PM Page 278 Concept Mapping 34. Create a concept map using the following terms: force, displacement, direction of motion, work, change in kinetic energy. Mastering Concepts 35. In what units is work measured? (10.1) 36. Suppose a satellite revolves around Earth in a circular orbit. Does Earth\u2019s gravity do any work on the satellite? (10.1) 37. An object slides at constant speed on a frictionless surface. What forces act on the object? What work is done by each force? (10.1) 38. Define work and power. (10.1) 39. What is a watt equivalent to in terms of kilograms, meters, and seconds? (10.1) 40. Is it possible to get more work out of a machine than you put into it? (10.2) 41. Explain how the pedals of a bicycle are a simple machine. (10.2) Applying Concepts 42. Which requires more work, carrying a 420-N backpack up a 200-m-high hill or carrying a 210-N backpack up a 400-m-high hill? Why? 43. Lifting You slowly lift a box of books from the floor and put it on a table. Earth\u2019s gravity exerts a force, magnitude mg, downward, and you exert a force, magnitude mg, upward. The two forces have equal magnitudes and opposite directions. It appears that no work is done, but you know that you did work. Explain what work was done. 44. You have an after-school job carrying cartons of new copy paper up a flight of stairs, and then carrying recycled paper back down the stairs. The mass of the paper does not change. Your physics teacher says that you do not work all day, so you should not be paid. In what sense is the physics teacher correct? What arrangement of payments might you make to ensure that you are properly compensated? 45. You carry the cartons of copy paper down the stairs, and then along a 15-m-long hallway. Are you working now? Explain. 46. Climbing Stairs Two people of the same mass climb the same flight of stairs. The first person climbs the stairs in 25 s; the second person does so in 35 s. a. Which person does more work? Explain your answer. b. Which person produces more power? Explain your answer. 47. Show that power delivered can be written as P Fv cos. 48. How can you increase the ideal mechanical advantage of a machine? 49. Wed", "ge How can you increase the mechanical advantage of a wedge without changing its ideal mechanical advantage? 50. Orbits Explain why a planet orbiting the Sun does not violate the work-energy theorem. 51. Claw Hammer A claw hammer is used to pull a nail from a piece of wood, as shown in Figure 10-16. Where should you place your hand on the handle and where should the nail be located in the claw to make the effort force as small as possible? \u25a0 Figure 10-16 Mastering Problems 10.1 Energy and Work 52. The third floor of a house is 8 m above street level. How much work is needed to move a 150-kg refrigerator to the third floor? 53. Haloke does 176 J of work lifting himself 0.300 m. What is Haloke\u2019s mass? 54. Football After scoring a touchdown, an 84.0-kg wide receiver celebrates by leaping 1.20 m off the ground. How much work was done by the wide receiver in the celebration? 55. Tug-of-War During a tug-of-war, team A does 2.20105 J of work in pulling team B 8.00 m. What force was team A exerting? 56. To keep a car traveling at a constant velocity, a 551-N force is needed to balance frictional forces. How much work is done against friction by the car as it travels from Columbus to Cincinnati, a distance of 161 km? 278 Hutchings Photography Chapter 10 Energy, Work, and Simple Machines For more problems, go to Additional Problems, Appendix B. 277-283 CH10-SG CA STP-845813 6/1/04 3:38 AM Page 279 57. Cycling A cyclist exerts a force of 15.0 N as he rides a bike 251 m in 30.0 s. How much power does the cyclist develop? 58. A student librarian lifts a 2.2-kg book from the floor to a height of 1.25 m. He carries the book 8.0 m to the stacks and places the book on a shelf that is 0.35 m above the floor. How much work does he do on the book? 59. A force of 300.0 N is used to push a 145-kg mass 30.0 m horizontally in 3.00 s. a. Calculate the work done on the mass. b. Calculate the power developed. 60. Wagon A wagon is pulled by a force of 38.0 N exerted on the handle at an angle of 42.0\u00b0 with the horizontal. If the wagon is pulled in a circle of radius 25.0 m, how much work is done? 61. Lawn Mower Shani is pushing a lawn mower with a force of 88.0 N along a handle that makes an angle of 41.0\u00b0 with the horizontal. How much work is done by Shani in moving the lawn mower 1.2 km to mow the yard? 62. A 17.0-kg crate is to be pulled a distance of 20.0 m, requiring 1210 J of work to be done. If the job is done by attaching a rope and pulling with a force of 75.0 N, at what angle is the rope held? 63. Lawn Tractor A 120-kg lawn tractor, shown in Figure 10-17, goes up a 21\u00b0 incline that is 12.0 m long in 2.5 s. Calculate the power that is developed by the tractor 21\u00b0 \u25a0 Figure 10-17 64. You slide a crate up a ramp at an angle of 30.0\u00b0 by exerting a 225-N force parallel to the ramp. The crate moves at a constant speed. The coefficient of friction is 0.28. How much work did you do on the crate as it was raised a vertical distance of 1.15 m? 65. Piano A 4.2103-N piano is to be slid up a 3.5-m frictionless plank at a constant speed. The plank makes an angle of 30.0\u00b0 with the horizontal. Calculate the work done by the person sliding the piano up the plank. 66. Sled Diego pulls a 4.5-kg sled across level snow with a force of 225 N on a rope that is 35.0\u00b0 above the horizontal, as shown in Figure 10-18. If the sled moves a distance of 65.3 m, how much work does Diego do? 4.5 kg 5 N 2 2 35.0\u00b0 \u25a0 Figure 10-18 67. Escalator Sau-Lan has a mass of 52 kg. She rides up the escalator at Ocean Park in Hong Kong. This is the world\u2019s longest escalator, with a length of 227 m and an average inclination of 31\u00b0. How much work does the escalator do on Sau-Lan? 68. Lawn Roller A lawn roller is pushed across a lawn by a force of 115 N along the direction of the handle", ", which is 22.5\u00b0 above the horizontal. If 64.6 W of power is developed for 90.0 s, what distance is the roller pushed? 69. John pushes a crate across the floor of a factory with a horizontal force. The roughness of the floor changes, and John must exert a force of 20 N for 5 m, then 35 N for 12 m, and then 10 N for 8 m. a. Draw a graph of force as a function of distance. b. Find the work John does pushing the crate. 70. Maricruz slides a 60.0-kg crate up an inclined ramp that is 2.0-m long and attached to a platform 1.0 m above floor level, as shown in Figure 10-19. A 400.0-N force, parallel to the ramp, is needed to slide the crate up the ramp at a constant speed. a. How much work does Maricruz do in sliding the crate up the ramp? b. How much work would be done if Maricruz simply lifted the crate straight up from the floor to the platform.0 m \u25a0 Figure 10-19 physicspp.com/chapter_test Chapter 10 Assessment 279 277-283 CH10-SG CA STP-845813 6/1/04 3:38 AM Page 280 71. Boat Engine An engine moves a boat through the water at a constant speed of 15 m/s. The engine must exert a force of 6.0 kN to balance the force that the water exerts against the hull. What power does the engine develop? 75. Oil Pump In 35.0 s, a pump delivers 0.550 m3 of oil into barrels on a platform 25.0 m above the intake pipe. The oil\u2019s density is 0.820 g/cm3. a. Calculate the work done by the pump. b. Calculate the power produced by the pump. 72. In Figure 10-20, the magnitude of the force necessary to stretch a spring is plotted against the distance the spring is stretched. a. Calculate the slope of the graph, k, and show that F kd, where k 25 N/m. b. Find the amount of work done in stretching the spring from 0.00 m to 0.20 m by calculating the area under the graph from 0.00 m to 0.20 m. c. Show that the answer to part b can be calculated 1kd2, where W is the using the formula W 2 work, k 25 N/m (the slope of the graph), and d is the distance the spring is stretched (0.20 m). ) N ( e c r o F 8.00 6.00 4.00 2.00 0.00 \u25a0 Figure 10-20 0.10 0.20 0.30 Elongation (m) 73. Use the graph in Figure 10-20 to find the work needed to stretch the spring from 0.12 m to 0.28 m. 74. A worker pushes a crate weighing 93 N up an inclined plane. The worker pushes the crate horizontally, parallel to the ground, as illustrated in Figure 10-21. a. The worker exerts a force of 85 N. How much work does he do? b. How much work is done by gravity? (Be careful with the signs you use.) c. The coefficient of friction is 0.20. How much work is done by friction? (Be careful with the signs you use.) 85 N 3.0 m 5.0 m 4.0 m \u25a0 Figure 10-21 93 N 76. Conveyor Belt A 12.0-m-long conveyor belt, inclined at 30.0\u00b0, is used to transport bundles of newspapers from the mail room up to the cargo bay to be loaded onto delivery trucks. Each newspaper has a mass of 1.0 kg, and there are 25 newspapers per bundle. Determine the power that the conveyor develops if it delivers 15 bundles per minute. 77. A car is driven at a constant speed of 76 km/h down a road. The car\u2019s engine delivers 48 kW of power. Calculate the average force that is resisting the motion of the car. 78. The graph in Figure 10-22 shows the force and displacement of an object being pulled. a. Calculate the work done to pull the object 7.0 m. b. Calculate the power that would be developed if the work was done in 2.0 s. ) N ( e c r o F 40.0 20.0 0.0 2.0 4.0 6.0 Displacement (m) \u25a0 Figure 10-22 10.2 Machines 79. Piano Takeshi raises a 1200-N piano a distance of 5.00 m using a set of pulleys. He pulls in 20.0 m of rope. a. How much effort force would Takeshi apply if this were an ideal machine? b", ". What force is used to balance the friction force if the actual effort is 340 N? c. What is the output work? d. What is the input work? e. What is the mechanical advantage? 80. Lever Because there is very little friction, the lever is an extremely efficient simple machine. Using a 90.0percent-efficient lever, what input work is required to lift an 18.0-kg mass through a distance of 0.50 m? 280 Chapter 10 Energy, Work, and Simple Machines For more problems, go to Additional Problems, Appendix B. 277-283 CH10-SG CA STP-845813 3/21/04 10:51 PM Page 281 81. A pulley system lifts a 1345-N weight a distance of 0.975 m. Paul pulls the rope a distance of 3.90 m, exerting a force of 375 N. a. What is the ideal mechanical advantage of the system? b. What is the mechanical advantage? c. How efficient is the system? 82. A force of 1.4 N is exerted through a distance of 40.0 cm on a rope in a pulley system to lift a 0.50-kg mass 10.0 cm. Calculate the following. a. the MA b. the IMA c. the efficiency 83. A student exerts a force of 250 N on a lever, through a distance of 1.6 m, as he lifts a 150-kg crate. If the efficiency of the lever is 90.0 percent, how far is the crate lifted? 84. What work is required to lift a 215-kg mass a distance of 5.65 m, using a machine that is 72.5 percent efficient? 85. The ramp in Figure 10-23 is 18 m long and 4.5 m high. a. What force, parallel to the ramp (FA), is required to slide a 25-kg box at constant speed to the top of the ramp if friction is disregarded? b. What is the IMA of the ramp? c. What are the real MA and the efficiency of the 87. Crane A motor with an efficiency of 88 percent operates a crane with an efficiency of 42 percent. If the power supplied to the motor is 5.5 kW, with what constant speed does the crane lift a 410-kg crate of machine parts? 88. A compound machine is constructed by attaching a lever to a pulley system. Consider an ideal compound machine consisting of a lever with an IMA of 3.0 and a pulley system with an IMA of 2.0. a. Show that the IMA of this compound machine is 6.0. b. If the compound machine is 60.0 percent efficient, how much effort must be applied to the lever to lift a 540-N box? c. If you move the effort side of the lever 12.0 cm, how far is the box lifted? Mixed Review 89. Ramps Isra has to get a piano onto a 2.0-m-high platform. She can use a 3.0-m-long frictionless ramp or a 4.0-m-long frictionless ramp. Which ramp should Isra use if she wants to do the least amount of work? 90. Brutus, a champion weightlifter, raises 240 kg of weights a distance of 2.35 m. a. How much work is done by Brutus lifting the weights? ramp if a parallel force of 75 N is actually required? b. How much work is done by Brutus holding the 1 8 m F A weights above his head? c. How much work is done by Brutus lowering them back to the ground? d. Does Brutus do work if he lets go of the weights 4.5 m and they fall back to the ground? \u25a0 Figure 10-23 Fg 86. Bicycle Luisa pedals a bicycle with a gear radius of 5.00 cm and a wheel radius of 38.6 cm, as shown in Figure 10-24. If the wheel revolves once, what is the length of the chain that was used? 38.6 cm 5.00 cm \u25a0 Figure 10-24 e. If Brutus completes the lift in 2.5 s, how much power is developed? 91. A horizontal force of 805 N is needed to drag a crate across a horizontal floor with a constant speed. You drag the crate using a rope held at an angle of 32\u00b0. a. What force do you exert on the rope? b. How much work do you do on the crate if you move it 22 m? c. If you complete the job in 8.0 s, what power is developed? 92. Dolly and Ramp A mover\u2019s dolly is used to transport a refrigerator up a ramp into a house. The refrigerator has a mass of 115 kg. The ramp is 2.10 m long and rises 0.850 m. The mover pulls the dolly with a force of", " 496 N up the ramp. The dolly and ramp constitute a machine. a. What work does the mover do? b. What is the work done on the refrigerator by the machine? c. What is the efficiency of the machine? physicspp.com/chapter_test Chapter 10 Assessment 281 277-283 CH10-SG CA STP-845813 3/21/04 10:52 PM Page 282 93. Sally does 11.4 kJ of work dragging a wooden crate 25.0 m across a floor at a constant speed. The rope makes an angle of 48.0\u00b0 with the horizontal. a. How much force does the rope exert on the 97. Apply Concepts A sprinter of mass 75 kg runs the 50.0-m dash in 8.50 s. Assume that the sprinter\u2019s acceleration is constant throughout the race. a. What is the average power of the sprinter over crate? b. What is the force of friction acting on the crate? c. What work is done by the floor through the force the 50.0 m? b. What is the maximum power generated by the sprinter? of friction between the floor and the crate? c. Make a quantitative graph of power versus time 94. Sledding An 845-N sled is pulled a distance of 185 m. The task requires 1.20104 J of work and is done by pulling on a rope with a force of 125 N. At what angle is the rope held? 95. An electric winch pulls a 875-N crate up a 15\u00b0 incline at 0.25 m/s. The coefficient of friction between the crate and incline is 0.45. a. What power does the winch develop? b. If the winch is 85 percent efficient, what is the electrical power that must be delivered to the winch? Thinking Critically 96. Analyze and Conclude You work at a store, carrying boxes to a storage loft that is 12 m above the ground. You have 30 boxes with a total mass of 150 kg that must be moved as quickly as possible, so you consider carrying more than one up at a time. If you try to move too many at once, you know that you will go very slowly, resting often. If you carry only one box at a time, most of the energy will go into raising your own body. The power (in watts) that your body can develop over a long time depends on the mass that you carry, as shown in Figure 10-25. This is an example of a power curve that applies to machines as well as to people. Find the number of boxes to carry on each trip that would minimize the time required. What time would you spend doing the job? Ignore the time needed to go back down the stairs and to lift and lower each box. Power v. Mass ) W ( r e w o P 25 20 15 10 5 0 5 10 15 20 25 30 Mass (kg) \u25a0 Figure 10-25 for the entire race. 98. Apply Concepts The sprinter in the previous problem runs the 50.0-m dash in the same time, 8.50 s. However, this time the sprinter accelerates in the first second and runs the rest of the race at a constant velocity. a. Calculate the average power produced for that first second. b. What is the maximum power that the sprinter now generates? Writing in Physics 99. Just as a bicycle is a compound machine, so is an automobile. Find the efficiencies of the component parts of the power train (engine, transmission, wheels, and tires). Explore possible improvements in each of these efficiencies. 100. The terms force, work, power, and energy often mean the same thing in everyday use. Obtain examples from advertisements, print media, radio, and television that illustrate meanings for these terms that differ from those used in physics. Cumulative Review 101. You are helping your grandmother with some gardening and have filled a garbage can with weeds and soil. Now you have to move the garbage can across the yard and realize it is so heavy that you will need to push it, rather than lift it. If the can has a mass of 24 kg, the coefficient of kinetic friction between the can\u2019s bottom and the muddy grass is 0.27, and the static coefficient of friction between those same surfaces is 0.35, how hard do you have to push horizontally to get the can to just start moving? (Chapter 5) 102. Baseball If a major league pitcher throws a fastball horizontally at a speed of 40.3 m/s (90 mph) and it travels 18.4 m (60 ft, 6 in), how far has it dropped by the time it crosses home plate? (Chapter 6) 103. People sometimes say that the Moon stays in its orbit because the \u201ccentrifugal force just balances the centripetal force, giving no net force.\u201d Explain why this idea is", " wrong. (Chapter 8) 282 Chapter 10 Energy, Work, and Simple Machines For more problems, go to Additional Problems, Appendix B. 277-283 CH10-SG CA STP-845813 3/21/04 10:53 PM Page 283 Multiple Choice 1. A pulley system consists of two fixed pulleys and two movable pulleys that lift a load that has a weight of 300 N. If the effort force used to lift the load is 100 N, what is the mechanical advantage of the system? 1 3 3 4 3 6 2. The box in the diagram is being pushed up the ramp with a force of 100.0 N. If the height of the ramp is 3.0 m, what is the work done on the box? (sin 30\u00b0 0.50, cos 30\u00b0 0.87, tan 30\u00b0 0.58) 150 J 260 J 450 J 600 J 5. A 20.0-N block is attached to the end of a rope, and the rope is looped around a pulley system. If you pull the opposite end of the rope a distance of 2.00 m, the pulley system raises the block a distance of 0.40 m. What is the pulley system\u2019s ideal mechanical advantage? 2.5 4.0 5.0 10.0 6. Two people carry identical 40.0-N boxes up a ramp. The ramp is 2.00 m long and rests on a platform that is 1.00 m high. One person walks up the ramp in 2.00 s, and the other person walks up the ramp in 4.00 s. What is the difference in power the two people use to carry the boxes up the ramp? 5.00 W 10.0 W 20.0 W 40.0 W 0 F 1 0. 0 N 30\u00b0 3.0 m 7. A 4-N soccer ball sits motionless on a field. A player\u2019s foot exerts a force of 5 N on the ball for a distance of 0.1 m, and the ball rolls a distance of 10 m. How much kinetic energy does the ball gain from the player? 0.5 J 0.9 J 9 J 50 J 3. A compound machine used to raise heavy boxes consists of a ramp and a pulley. The efficiency of pulling a 100-kg box up the ramp is 50%. If the efficiency of the pulley is 90%, what is the overall efficiency of the compound machine? 40% 45% 50% 70% 4. A skater with a mass of 50.0 kg slides across an icy pond with negligible friction. As he approaches a friend, both he and his friend hold out their hands, and the friend exerts a force in the direction opposite to the skater\u2019s movement, which slows the skater\u2019s speed from 2.0 m/s2 to 1.0 m/s2. What is the change in the skater\u2019s kinetic energy? 25 J 75 J 100 J 150 J Extended Answer 8. The diagram shows a box being pulled by a rope with a force of 200.0 N along a horizontal surface. The angle the rope makes with the horizontal is 45\u00b0. Calculate the work done on the box and the power required to pull it a distance of 5.0 m in 10.0 s. (sin 45\u00b0 cos 45\u00b0 0.7) 45\u00b0 Beat the Clock and then Go Back As you take a practice test, pace yourself to finish each section just a few minutes early so you can go back and check over your work. physicspp.com/standardized_test Chapter 10 Standardized Test Practice 283 284-292 CH11-S1-845813 6/1/04 5:10 AM Page 284 What You\u2019ll Learn \u2022 You will learn that energy is a property of an object that can change the object\u2019s position, motion, or its environment. \u2022 You will learn that energy changes from one form to another, and that the total amount of energy in a closed system remains constant. Why It\u2019s Important Energy turns the wheels of our world. People buy and sell energy to operate electric appliances, automobiles, and factories. Skiing The height of the ski jump determines the energy the skier has at the bottom of the ramp before jumping into the air and flying many meters down the slope. The distance that the ski jumper travels depends on his or her use of physical principles such as air resistance, balance, and energy. Think About This How does the height of the ski ramp affect the distance that the skier can jump? physicspp.com 284 David Madison Sports Images 284-292 CH11-S1-845813 3/11/04 2:45 AM Page 285 How can you analyze the energy of a bouncing basketball? Question What is the relationship between the height a basketball is dropped from and the height it reaches when it bounces back? Procedure 1. Place a meter", "stick against a wall. Choose an initial height from which to drop a basketball. Record the height in the data table. 2. Drop the ball and record how high the ball bounced. 3. Repeat steps 1 and 2 by dropping the basketball from three other heights. 4. Make and Use Graphs Construct a graph of bounce height ( y) versus drop height (x). Find the best-fit line. Analysis Use the graph to find how high a basketball would bounce if it were dropped from a height of 10.0 m. When the ball is lifted and ready to drop, it possesses energy. What are the factors that influence this energy? Critical Thinking Why doesn\u2019t the ball bounce back to the height from which it was dropped? 11.1 The Many Forms of Energy The word energy is used in many different ways in everyday speech. Some fruit-and-cereal bars are advertised as energy sources. Athletes use energy in sports. Companies that supply your home with electricity, natural gas, or heating fuel are called energy companies. Scientists and engineers use the term energy much more precisely. As you learned in the last chapter, work causes a change in the energy of a system. That is, work transfers energy between a system and the external world. In this chapter, you will explore how objects can have energy in a variety of ways. Energy is like ice cream\u2014it comes in different varieties. You can have vanilla, chocolate, or peach ice cream. They are different varieties, but they are all ice cream and serve the same purpose. However, unlike ice cream, energy can be changed from one variety to another. In this chapter, you will learn how energy is transformed from one variety (or form) to another and how to keep track of the changes. Objectives \u2022 Use a model to relate work and energy. \u2022 Calculate kinetic energy. \u2022 Determine the gravitational potential energy of a system. \u2022 Identify how elastic potential energy is stored. Vocabulary rotational kinetic energy gravitational potential energy reference level elastic potential energy Section 11.1 The Many Forms of Energy 285 Horizons Companies 284-292 CH11-S1-845813 3/11/04 2:47 AM Page 286 \u25a0 Figure 11-1 When you earn money, the amount of cash that you have increases (a). When you spend money, the amount of cash that you have decreases (b). a b $before Cash flow $after $before Cash flow $after \u25a0 Figure 11-2 The kinetic energy after throwing or catching a ball is equal to the kinetic energy before plus the input work. A Model of the Work-Energy Theorem In the last chapter, you were introduced to the work-energy theorem. You learned that when work is done on a system, the energy of that system increases. On the other hand, if the system does work, then the energy of the system decreases. These are abstract ideas, but keeping track of energy is much like keeping track of your spending money. If you have a job, the amount of money that you have increases every time you are paid. This process can be represented with a bar graph, as shown in Figure 11-1a. The orange bar represents how much money you had to start with, and the blue bar represents the amount that you were paid. The green bar is the total amount that you possess after the payment. An accountant would say that your cash flow was positive. What happens when you spend the money that you have? The total amount of money that you have decreases. As shown in Figure 11-1b, the bar that represents the amount of money that you had before you bought that new CD is higher than the bar that represents the amount of money remaining after your shopping trip. The difference is the cost of the CD. Cash flow is shown as a bar below the axis because it represents money going out, which can be shown as a negative number. Energy is similar to your spending money. The amount of money that you have changes only when you earn more or spend it. Similarly, energy can be stored, and when energy is spent, it affects the motion of a system. Throwing a ball Gaining and losing energy also can be illustrated by throwing and catching a ball. In Chapter 10, you learned that when you exert a constant force, F, on an object through a distance, d, in the direction of the force, you do an amount of work, represented by W Fd. The work is positive because the force and motion are in the same direction, and the energy of the object increases by an amount equal to W. Suppose the object is a ball, and you exert a force to throw the ball. As a result of the force you apply, the ball gains kinetic energy. This process is shown in Figure 11-2a. You can again use a bar graph to represent the process. This time, the height of the bar represents the amount of work, or energy, measured in joules. The kinetic energy after the work is done is equal to the", " sum of the initial kinetic energy plus the work done on the ball. a Begin Throwing a ball End b Begin Catching a ball vball Stopped d W 0 F vball d W 0 F End Stopped KEbefore W KEafter KEbefore W KEafter 286 Chapter 11 Energy and Its Conservation 284-292 CH11-S1-845813 6/1/04 5:12 AM Page 287 Catching a ball What happens when you catch a ball? Before hitting your hands or glove, the ball is moving, so it has kineticc energy. In catching it, you exert a force on the ball in the direction opposite to its motion. Therefore, you do negative work on it, causing it to stop. Now that the ball is not moving, it has no kinetic energy. This process and the bar graph that represents it are shown in Figure 11-2b. Kinetic energy is always positive, so the initial kinetic energy of the ball is positive. The work done on the ball is negative and the final kinetic energy is zero. Again, the kinetic energy after the ball has stopped is equal to the sum of the initial kinetic energy plus the work done on the ball. Kinetic Energy Recall that kinetic energy, KE 1 mv2, where m is the mass of the object 2 and v is the magnitude of its velocity. The kinetic energy is proportional to the object\u2019s mass. A 7.26-kg shot put thrown through the air has much more kinetic energy than a 0.148-kg baseball with the same velocity, because the shot put has a greater mass. The kinetic energy of an object is also proportional to the square of the object\u2019s velocity. A car speeding at 20 m/s has four times the kinetic energy of the same car moving at 10 m/s. Kinetic energy also can be due to rotational motion. If you spin a toy top in one spot, does it have kinetic energy? You might say that it does not because the top is not moving anywhere. However, to make the top rotate, someone had to do work on it. Therefore, the top has rotational kinetic energy. This is one of the several varieties of energy. Rotational kinetic 1 I2, where I is the object\u2019s energy can be calculated using KErot 2 moment of inertia and is the object\u2019s angular velocity. The diver, shown in Figure 11-3a, does work as she pushes off of the diving board. This work produces both linear and rotational kinetic energies. When the diver\u2019s center of mass moves as she leaps, linear kinetic energy is produced. When she rotates about her center of mass, as shown in Figure 11-3b, rotational kinetic energy is produced. Because she is moving toward the water and rotating at the same time while in the tuck position, she has both linear and rotational kinetic energy. When she slices into the water, as shown in Figure 11-3c, she has linear kinetic energy. a b c 1. A skater with a mass of 52.0 kg moving at 2.5 m/s glides to a stop over a distance of 24.0 m. How much work did the friction of the ice do to bring the skater to a stop? How much work would the skater have to do to speed up to 2.5 m/s again? 2. An 875.0-kg compact car speeds up from 22.0 m/s to 44.0 m/s while passing another car. What are its initial and final energies, and how much work is done on the car to increase its speed? 3. A comet with a mass of 7.851011 kg strikes Earth at a speed of 25.0 km/s. Find the kinetic energy of the comet in joules, and compare the work that is done by Earth in stopping the comet to the 4.21015 J of energy that was released by the largest nuclear weapon ever built. \u25a0 Figure 11-3 The diver does work as she pushes off of the diving board (a). This work produces rotational kinetic energy as she rotates about her center of mass (b) and she has linear kinetic energy when she slices into the water (c). Section 11.1 The Many Forms of Energy 287 Ken Redmond/Ken Redmond Photography 284-292 CH11-S1-845813 6/7/04 5:33 AM Page 288 \u25a0 Figure 11-4 Money in the form of bills, quarters, and pennies are different forms of the same thing. 5 $1 Bills 20 Quarters 500 Pennies $5 $5 $5 $5 $1.00 5 $5 $0.25 20 $5 $0.01 500 \u25a0 Figure 11-5 Kinetic and potential energy are constantly being exchanged when juggling. Stored Energy Imagine a group of boulders high on a hill. These boulders have been lifted up by geological processes against the force", " of gravity; thus, they have stored energy. In a rock slide, the boulders are shaken loose. They fall and pick up speed as their stored energy is converted to kinetic energy. In the same way, a small, spring-loaded toy, such as a jack-in-the-box, has stored energy, but the energy is stored in a compressed spring. While both of these examples represent energy stored by mechanical means, there are many other means of storing energy. Automobiles, for example, carry their energy stored in the form of chemical energy in the gasoline tank. Energy is made useful or causes motion when it changes from one form to another. How does the money model that was discussed earlier illustrate the transformation of energy from one form to another? Money, too, can come in different forms. You can have one five-dollar bill, 20 quarters, or 500 pennies. In all of these cases, you still have five dollars. The height of the bar graph in Figure 11-4 represents the amount of money in each form. In the same way, you can use a bar graph to represent the amount of energy in various forms that a system has. Gravitational Potential Energy Look at the oranges being juggled in Figure 11-5. If you consider the system to be only one orange, then it has several external forces acting on it. The force of the juggler\u2019s hand does work, giving the orange its original kinetic energy. After the orange leaves the juggler\u2019s hand, only the force of gravity acts on it. How much work does gravity do on the orange as its height changes? Work done by gravity Let h represent the orange\u2019s height measured from the juggler\u2019s hand. On the way up, its displacement is upward, but the force on the orange, Fg, is downward, so the work done by gravity is nega mgh. On the way back down, the force and displacement are tive: Wg mgh. in the same direction, so the work done by gravity is positive: Wg Thus, while the orange is moving upward, gravity does negative work, slowing the orange to a stop. On the way back down, gravity does positive work, increasing the orange\u2019s speed and thereby increasing its kinetic energy. The orange recovers all of the kinetic energy it originally had when it returns to the height at which it left the juggler\u2019s hand. It is as if the orange\u2019s kinetic energy is stored in another form as the ball rises and is transformed back to kinetic energy as the ball falls. Consider a system that consists of an object plus Earth. The gravitational attraction between the object and Earth is a force that always does work on the object as it moves. If the object moves away from Earth, energy is stored in the system as a result of the gravitational force between the object and Earth. This stored energy is called gravitational potential energy and is represented by the symbol PE. The height to which the object has risen is determined by using a reference level, the position where PE is defined to be zero. For an object with mass, m, that has risen to a height, h, above the reference level, gravitational potential energy is represented by the following equation. 288 Hutchings Photography Chapter 11 Energy and Its Conservation 284-292 CH11-S1-845813 3/11/04 2:49 AM Page 289 Gravitational Potential Energy PE mgh The gravitational potential energy of an object is equal to the product of its mass, the acceleration due to gravity, and the distance from the reference level. In the equation for gravitational potential energy, g is the acceleration due to gravity. Gravitational potential energy, like kinetic energy, is measured in joules. Kinetic energy and potential energy of a system Consider the energy of a system consisting of an orange used by the juggler plus Earth. The energy in the system exists in two forms: kinetic energy and gravitational potential energy. At the beginning of the orange\u2019s flight, all the energy is in the form of kinetic energy, as shown in Figure 11-6a. On the way up, as the orange slows down, energy changes from kinetic energy to potential energy. At the highest point of the orange\u2019s flight, the velocity is zero. Thus, all the energy is in the form of gravitational potential energy. On the way back down, potential energy changes back into kinetic energy. The sum of kinetic energy and potential energy is constant at all times because no work is done on the system by any external forces. Reference levels In Figure 11-6a, the reference level is the juggler\u2019s hand. That is, the height of the orange is measured from the juggler\u2019s hand. Thus, at the juggler\u2019s hand, h 0 m and PE 0 J. You can set the reference level at any height that is convenient for solving a given problem. Suppose the reference level is set at the highest point of the", " orange\u2019s flight. Then, h 0 m and the system\u2019s PE 0 J at that point, as illustrated in Figure 11-6b. The potential energy of the system is negative at the beginning of the orange\u2019s flight, zero at the highest point, and negative at the end of the orange\u2019s flight. If you were to calculate the total energy of the system represented in Figure 11-6a, it would be different from the total energy of the system represented in Figure 11-6b. This is because the reference levels are different in each case. However, the total energy of the system in each situation would be constant at all times during the flight of the orange. Only changes in energy determine the motion of a system. Potential Energy of an Atom It is interesting to consider the relative sizes of potential energy per atom. For instance, a carbon atom has a mass of about 21026 kg. If you lift it 1 m above the ground, its gravitational potential energy is about 21025 J. The electrostatic energy that holds the electrons on the atom has a value of about 1019 J, and the nuclear potential energy that holds the nucleus together is greater than 1012 J. The nuclear potential energy is at least a million million times greater than the gravitational potential energy. \u25a0 Figure 11-6 The energy of an orange is converted from one form to another in various stages of its flight (a). Note that the choice of a reference level is arbitrary, but that the total energy remains constant (b). a Begin Middle End b Begin Middle End h 0 0 Reference level Reference level h KE PE KE KE KE PE KE PE PE KE PE PE Section 11.1 The Many Forms of Energy 289 284-292 CH11-S1-845813 6/10/04 7:44 PM Page 290 Gravitational Potential Energy You lift a 7.30-kg bowling ball from the storage rack and hold it up to your shoulder. The storage rack is 0.610 m above the floor and your shoulder is 1.12 m above the floor. a. When the bowling ball is at your shoulder, what is the bowling ball\u2019s gravitational potential energy relative to the floor? b. When the bowling ball is at your shoulder, what is its gravitational potential energy relative to the storage rack? c. How much work was done by gravity as you lifted the ball from the rack to shoulder level? 1 Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Choose a reference level. \u2022 Draw a bar graph showing the gravitational potential energy with the floor as the reference level. 1.12 m 0.61 m Known: m 7.30 kg hr hs g 9.80 m/s2 0.610 m (relative to the floor) 1.12 m (relative to the floor) Unknown:? PEs rel f? PEs rel r PEr rel f Wby you PEs rel f 2 Solve for the Unknown a. Set the reference level to be at the floor. Solve for the potential energy of the ball at shoulder level. PEs rel f mghs (7.30 kg)(9.80 m/s2)(1.12 m) 80.1 J Substitute m 7.30 kg, g 9.80 m/s2, hshoulder 1.12 m b. Set the reference level to be at the rack height. Solve for the height of your shoulder relative to the rack. h hs hr Solve for the potential energy of the ball. Math Handbook Order of Operations page 843 PEs rel r mgh mg(hs (7.30 kg)(9.80 m/s2)(1.12 m 0.610 m) Substitute m 7.3 kg, g 9.80 m/s2, Substitute h hs hr) hr 36.5 J 1.12 m, hr 0.610 m hs This also is equal to the work done by you. c. The work done by gravity is the weight of the ball times the distance the ball was lifted. W Fd (mg)h (mg)(hs (7.30 kg)(9.80 m/s2)(1.12 m 0.610 m) hr) 36.5 J Because the weight opposes the motion of lifting, the work is negative. Substitute m 7.30 kg, g 9.80 m/s2, 0.610 m hs 1.12 m, hr 3 Evaluate the Answer \u2022 Are the units correct? The potential energy and work are both measured in joules. \u2022 Is the magnitude realistic? The ball should have a greater potential energy relative to the floor than relative to the rack, because the ball\u2019s distance above the reference level is greater. 290 Chapter 11 Energy and Its Conservation 284-292 CH11-S1-845813 6/7/04 11:28 PM Page 291 4. 5. In Example Problem 1, what", " is the potential energy of the bowling ball relative to the rack when it is on the floor? If you slowly lower a 20.0-kg bag of sand 1.20 m from the trunk of a car to the driveway, how much work do you do? 6. A boy lifts a 2.2-kg book from his desk, which is 0.80 m high, to a bookshelf that is 2.10 m high. What is the potential energy of the book relative to the desk? 7. If a 1.8-kg brick falls to the ground from a chimney that is 6.7 m high, what is the change in its potential energy? 8. A warehouse worker picks up a 10.1-kg box from the floor and sets it on a long, 1.1-m-high table. He slides the box 5.0 m along the table and then lowers it back to the floor. What were the changes in the energy of the box, and how did the total energy of the box change? (Ignore friction.) Elastic Potential Energy When the string on the bow shown in Figure 11-7 is pulled, work is done on the bow, storing energy in it. Thus, the energy of the system increases. Identify the system as the bow, the arrow, and Earth. When the string and arrow are released, energy is changed into kinetic energy. The stored energy in the pulled string is called elastic potential energy, which is often stored in rubber balls, rubber bands, slingshots, and trampolines. Energy also can be stored in the bending of an object. When stiff metal or bamboo poles were used in pole-vaulting, the poles did not bend easily. Little work was done on the poles, and consequently, the poles did not store much potential energy. Since flexible fiberglass poles were introduced, however, record pole-vaulting heights have soared. \u25a0 Figure 11-7 Elastic potential energy is stored in the string of this bow. Before the string is released, the energy is all potential (a). As the string is released, the energy is transferred to the arrow as kinetic energy (b). a b Section 11.1 The Many Forms of Energy 291 (r)Getty Images, (l)Luis Romero/AP Wide World Photos 284-292 CH11-S1-845813 6/1/04 5:16 AM Page 292 A pole-vaulter runs with a flexible pole and plants its end into the socket in the ground. When the pole-vaulter bends the pole, as shown in Figure 11-8, some of the pole-vaulter\u2019s kinetic energy is converted to elastic potential energy. When the pole straightens, the elastic potential energy is converted to gravitational potential energy and kinetic energy as the pole-vaulter is lifted as high as 6 m above the ground. Unlike stiff metal poles or bamboo poles, fiberglass poles have an increased capacity for storing elastic potential energy. Thus, pole-vaulters are able to clear bars that are set very high. Mass Albert Einstein recognized yet another form of potential energy: mass itself. He said that mass, by its very nature, is energy. This energy, E0, is called rest energy and is represented by the following famous formula. mc2 Rest Energy E0 The rest energy of an object is equal to the object\u2019s mass times the speed of light squared. \u25a0 Figure 11-8 When a pole-vaulter jumps, elastic potential energy is changed into kinetic energy and gravitational potential energy. According to this formula, stretching a spring or bending a vaulting pole causes the spring or pole to gain mass. In these cases, the change in mass is too small to be detected. When forces within the nucleus of an atom are involved, however, the energy released into other forms, such as kinetic energy, by changes in mass can be quite large. 11.1 Section Review 9. Elastic Potential Energy You get a spring-loaded toy pistol ready to fire by compressing the spring. The elastic potential energy of the spring pushes the rubber dart out of the pistol. You use the toy pistol to shoot the dart straight up. Draw bar graphs that describe the forms of energy present in the following instances. a. The dart is pushed into the gun barrel, thereby compressing the spring. b. The spring expands and the dart leaves the gun barrel after the trigger is pulled. c. The dart reaches the top of its flight. 10. Potential Energy A 25.0-kg shell is shot from a cannon at Earth\u2019s surface. The reference level is Earth\u2019s surface. What is the gravitational potential energy of the system when the shell is at 425 m? What is the change in potential energy when the shell falls to a height of 225 m? 11. Rotational Kinetic Energy Suppose some children push a merry-go-round so that it turns twice as fast as it did before they pushed it. What", " are the relative changes in angular momentum and rotational kinetic energy? 12. Work-Energy Theorem How can you apply the work-energy theorem to lifting a bowling ball from a storage rack to your shoulder? 13. Potential Energy A 90.0-kg rock climber first climbs 45.0 m up to the top of a quarry, then descends 85.0 m from the top to the bottom of the quarry. If the initial height is the reference level, find the potential energy of the system (the climber and Earth) at the top and at the bottom. Draw bar graphs for both situations. 14. Critical Thinking Karl uses an air hose to exert a constant horizontal force on a puck, which is on a frictionless air table. He keeps the hose aimed at the puck, thereby creating a constant force as the puck moves a fixed distance. a. Explain what happens in terms of work and energy. Draw bar graphs. b. Suppose Karl uses a different puck with half the mass of the first one. All other conditions remain the same. How will the kinetic energy and work differ from those in the first situation? c. Explain what happened in parts a and b in terms of impulse and momentum. 292 Bob Daemmrich/The Image Works Chapter 11 Energy and Its Conservation physicspp.com/self_check_quiz 293-301 CH11-S2-845813 6/3/04 12:14 PM Page 293 11.2 Conservation of Energy Consider a ball near the surface of Earth. The sum of gravitational potential energy and kinetic energy in that system is constant. As the height of the ball changes, energy is converted from kinetic energy to potential energy, but the total amount of energy stays the same. Conservation of Energy In our everyday world, it may not seem as if energy is conserved. A hockey puck eventually loses its kinetic energy and stops moving, even on smooth ice. A pendulum stops swinging after some time. The money model can again be used to illustrate what is happening in these cases. Suppose you have a total of $50 in cash. One day, you count your money and discover that you are $3 short. Would you assume that the money just disappeared? You probably would try to remember whether you spent it, and you might even search for it. In other words, rather than giving up on the conservation of money, you would try to think of different places where it might have gone. Law of conservation of energy Scientists do the same thing as you would if you could not account for a sum of money. Whenever they observe energy leaving a system, they look for new forms into which the energy could have been transferred. This is because the total amount of energy in a system remains constant as long as the system is closed and isolated from external forces. The law of conservation of energy states that in a closed, isolated system, energy can neither be created nor destroyed; rather, energy is conserved. Under these conditions, energy changes from one form to another while the total energy of the system remains constant. Conservation of mechanical energy The sum of the kinetic energy and gravitational potential energy of a system is called mechanical energy. In any given system, if no other forms of energy are present, mechanical energy is represented by the following equation. Mechanical Energy of a System E KE PE The mechanical energy of a system is equal to the sum of the kinetic energy and potential energy if no other forms of energy are present. Objectives \u2022 Solve problems using the law of conservation of energy. \u2022 Analyze collisions to find the change in kinetic energy. Vocabulary law of conservation of energy mechanical energy thermal energy elastic collision inelastic collision \u25a0 Figure 11-9 A decrease in potential energy is equal to the increase in kinetic energy. 10.0 N 0.0 J 20.0 J Imagine a system consisting of a 10.0-N ball and Earth, as shown in Figure 11-9. Suppose the ball is released from 2.00 m above the ground, which you set to be the reference level. Because the ball is not yet moving, it has no kinetic energy. Its potential energy is represented by the following equation: 2.00 m 1.00 m PE mgh (10.0 N)(2.00 m) 20.0 J KE PE 10.0 J 10.0 J KE PE 20.0 J 0.0 J The ball\u2019s total mechanical energy, therefore, is 20.0 J. As the ball falls, it loses potential energy and gains kinetic energy. When the ball is 1.00 m above Earth\u2019s surface: PE mgh (10.0 N)(1.00 m) 10.0 J. 0.00 m Ground KE PE Section 11.2 Conservation of Energy 293 293-301 CH11-S2-845813 3/11/04 3:20 AM Page 294 Weight 10.0 N PE 20.0 J 4. 0 m 2.0 m KE 20.0 J KE 20.0 J \u25a0", " Figure 11-10 The path that an object follows in reaching the ground does not affect the final kinetic energy of the object. What is the ball\u2019s kinetic energy when it is at a height of 1.00 m? The system consisting of the ball and Earth is closed and isolated because no external forces are acting upon it. Hence, the total energy of the system, E, remains constant at 20.0 J. E KE PE, so KE E PE KE 20.0 J 10.0 J = 10.0 J When the ball reaches ground level, its potential energy is zero, and its kinetic energy is 20.0 J. The equation that describes conservation of mechanical energy can be written as follows. Conservation of Mechanical Energy PEafter PEbefore KEafter KEbefore When mechanical energy is conserved, the sum of the kinetic energy and potential energy present in the system before the event is equal to the sum of the kinetic energy and potential energy in the system after the event. What happens if the ball does not fall down, but rolls down a ramp, as shown in Figure 11-10? If there is no friction, there are no external forces acting on the system. Thus, the system remains closed and isolated. The ball still moves down a vertical distance of 2.00 m, so its loss of potential energy is 20.0 J. Therefore, it gains 20.0 J of kinetic energy. As long as there is no friction, the path that the ball takes does not matter. Roller coasters In the case of a roller coaster that is nearly at rest at the top of the first hill, the total mechanical energy in the system is the coaster\u2019s gravitational potential energy at that point. Suppose some other hill along the track were higher than the first one. The roller coaster would not be able to climb the higher hill because the energy required to do so would be greater than the total mechanical energy of the system. Skiing Suppose you ski down a steep slope. When you begin from rest at the top of the slope, your total mechanical energy is simply your gravitational potential energy. Once you start skiing downhill, your gravitational potential energy is converted to kinetic energy. As you ski down the slope, your speed increases as more of your potential energy is converted to kinetic energy. In ski jumping, the height of the ramp determines the amount of energy that the jumper has to convert into kinetic energy at the beginning of his or her flight. Pendulums The simple oscillation of a pendulum also demonstrates conservation of energy. The system is the pendulum bob and Earth. Usually, the reference level is chosen to be the height of the bob at the lowest point, when it is at rest. If an external force pulls the bob to one side, the force does work that gives the system mechanical energy. At the instant the bob is released, all the energy is in the form of potential energy, but as the bob swings downward, the energy is converted to kinetic energy. Figure 11-11 shows a graph of the changing potential and kinetic energies of a pendulum. When the bob is at the lowest point, its gravitational potential energy is zero, and its kinetic energy is equal to the total mechanical \u25a0 Figure 11-11 For the simple harmonic motion of a pendulum bob (a), the mechanical energy\u2014 the sum of the potential and kinetic energies\u2014is a constant (b). Energy v. Position PE + KE PE KE A B C Horizontal position 294 Chapter 11 Energy and Its Conservation 293-301 CH11-S2-845813 3/11/04 3:21 AM Page 295 energy in the system. Note that the total mechanical energy of the system is constant if we assume that there is no friction. You will learn more about pendulums in Chapter 14. Loss of mechanical energy The oscillations of a pendulum eventually come to a stop, a bouncing ball comes to rest, and the heights of rollercoaster hills get lower and lower. Where does the mechanical energy in such systems go? Any object moving through the air experiences the forces of air resistance. In a roller coaster, there are frictional forces between the wheels and the tracks. When a ball bounces off of a surface, all of the elastic potential energy that is stored in the deformed ball is not converted back into kinetic energy after the bounce. Some of the energy is converted into thermal energy and sound energy. As in the cases of the pendulum and the roller coaster, some of the original mechanical energy in the system is converted into another form of energy within members of the system or transmitted to energy outside the system, as in air resistance. Usually, this new energy causes the temperature of objects to rise slightly. You will learn more about this form of energy, called thermal energy, in Chapter 12. The following strategies will be helpful to you when solving problems related to conservation of energy. Conservation of Energy When solving problems related to the conservation of energy, use the following strategies. 1. Carefully identify the system. Make sure it is closed. In a closed system", ", no objects enter or leave the system. 2. Identify the forms of energy in the system. 3. Identify the initial and final states of the system. 4. Is the system isolated? a. If there are no external forces acting on the system, then the system is isolated and the total energy of the system is constant. Eafter Ebefore b. If there are external forces, then the following is true. Ebefore W Eafter 5. If mechanical energy is conserved, decide on the reference level for potential energy. Draw bar graphs showing initial and final energy like the bar graphs shown to the right. Energy Bar Graphs Initial Final Total energy KEi PEf PEi KEf Section 11.2 Conservation of Energy 295 293-301 CH11-S2-845813 6/3/04 12:17 PM Page 296 Conservation of Mechanical Energy During a hurricane, a large tree limb, with a mass of 22.0 kg and a height of 13.3 m above the ground, falls on a roof that is 6.0 m above the ground. a. Ignoring air resistance, find the kinetic energy of the limb when it reaches the roof. b. What is the speed of the limb when it reaches the roof? 1 Analyze and Sketch the Problem \u2022 Sketch the initial and final conditions. \u2022 Choose a reference level. \u2022 Draw a bar graph. Known: Before (initial) Ground reference After (final) vi 0.0 m/s hi 13.3 m vf hf 6.0 m m 22.0 kg 13.3 m 6.0 m hlimb hroof g 9.80 m/s2 0.0 m/s vi 0.0 J KEi Unknown: KEf vf?? PEi PEf?? PEi KEi PEf KEf 2 Solve for the Unknown a. Set the reference level as the height of the roof. Solve for the initial height of the limb relative to the roof. Bar Graph h hlimb hroof 13.3 m 6.0 m 7.3 m Substitute hlimb 13.3 m, hroof 6.0 m Solve for the initial potential energy of the limb. PEi mgh (22.0 kg)(9.80 m/s2)(7.3 m) 1.6103 J Substitute m 22.0 kg, g 9.80 m/s2, h 7.3 m Identify the initial kinetic energy of the limb. KEi 0.0 J The tree limb is initially at rest. The kinetic energy of the limb when it reaches the roof is equal to its initial potential energy because energy is conserved. PEf 0.0 J because h 0.0 m at the reference level. b. Solve for the speed of the limb. 2 KEf PEi 1.6103 J KEf vf 1 mvf 2 Ef 2K m.0 2(1 0 1.6 g k 2 2 12 m/s 3 J) Substitute KEf 1.6103 J, m 22.0 kg Math Handbook Square and Cube Roots pages 839\u2013840 3 Evaluate the Answer \u2022 Are the units correct? Velocity is measured in m/s and energy is measured in kgm2/s2 J. \u2022 Do the signs make sense? KE and the magnitude of velocity are always positive. 296 Chapter 11 Energy and Its Conservation 293-301 CH11-S2-845813 7/20/04 2:34 PM Page 297 15. A bike rider approaches a hill at a speed of 8.5 m/s. The combined mass of the bike and the rider is 85.0 kg. Choose a suitable system. Find the initial kinetic energy of the system. The rider coasts up the hill. Assuming there is no friction, at what height will the bike come to rest? 16. Suppose that the bike rider in problem 15 pedaled up the hill and never came to a stop. In what system is energy conserved? From what form of energy did the bike gain mechanical energy? 17. A skier starts from rest at the top of a 45.0-m-high hill, skis down a 30\u00b0 incline into a valley, and continues up a 40.0-m-high hill. The heights of both hills are measured from the valley floor. Assume that you can neglect friction and the effect of the ski poles. How fast is the skier moving at the bottom of the valley? What is the skier\u2019s speed at the top of the next hill? Do the angles of the hills affect your answers? 18. In a belly-flop diving contest, the winner is the diver who makes the biggest splash upon hitting the water. The size of the splash depends not only on the diver\u2019s style, but also on the amount of kinetic energy that the diver has. Consider a contest in which each diver", " jumps from a 3.00-m platform. One diver has a mass of 136 kg and simply steps off the platform. Another diver has a mass of 102 kg and leaps upward from the platform. How high would the second diver have to leap to make a competitive splash? Analyzing Collisions A collision between two objects, whether the objects are automobiles, hockey players, or subatomic particles, is one of the most common situations analyzed in physics. Because the details of a collision can be very complex during the collision itself, the strategy is to find the motion of the objects just before and just after the collision. What conservation laws can be used to analyze such a system? If the system is isolated, then momentum and energy are conserved. However, the potential energy or thermal energy in the system may decrease, remain the same, or increase. Therefore, you cannot predict whether or not kinetic energy is conserved. Figure 11-12 and Figure 11-13 on the next page show three different kinds of collisions. In case 1, the momentum of the system before and after the collision is represented by the following: pi pCi pDi pf pCf pDf (1.00 kg)(1.00 m/s) (1.00 kg)(0.00 m/s) 1.00 kgm/s (1.00 kg)(0.20 m/s) (1.00 kg)(1.20 m/s) 1.00 kgm/s Thus, in case 1, the momentum is conserved. Look again at Figure 11-13 and verify for yourself that momentum is conserved in cases 2 and 3. Before (initial) Case 1 After (final) mC 1.00 kg mD 1.00 kg C D C D \u25a0 Figure 11-12 Two moving objects can have different types of collisions. Case 1: the two objects move apart in opposite directions. vCi 1.00 m/s vDi 0.00 m/s vCf 0.20 m/s vDf 1.20 m/s Section 11.2 Conservation of Energy 297 293-301 CH11-S2-845813 6/10/04 7:45 PM Page 298 \u25a0 Figure 11-13 Case 2: the moving object comes to rest and the stationary object begins to move. Case 3: the two objects are stuck together and move as one. Before (initial) Case 2 After (final) mC 1.00 kg mD 1.00 kg C D C D vCi 1.00 m/s vDi 0.00 m/s vCf 0.00 m/s vDf 1.00 m/s Case 3 C D C D vCi 1.00 m/s vDi 0.00 m/s vCf vDf 0.50 m/s Next, consider the kinetic energy of the system in each of these cases. For case 1 the kinetic energy of the system before and after the collision is represented by the following equations: (1.00 kg)(1.00 m/s)2 1 1 (1.00 kg)(0.00 m/s)2 2 2 0.50 J 1 (1.00 kg)(0.20 m/s)2 1 (1.00 kg)(1.20 m/s)2 2 2 0.74 J KEDf KEDi KECf KECi In case 1, the kinetic energy of the system increased. If energy in the system is conserved, then one or more of the other forms of energy must have decreased. Perhaps when the two carts collided, a compressed spring was released, adding kinetic energy to the system. This kind of collision is sometimes called a superelastic or explosive collision. After the collision in case 2, the kinetic energy is equal to: KEDf (1.0 kg)(0.00 m/s)2 1 (1.0 kg)(1.0 m/s)2 0.50 J 2 KECf Kinetic energy remained the same after the collision. This type of collision, in which the kinetic energy does not change, is called an elastic collision. Collisions between hard, elastic objects, such as those made of steel, glass, or hard plastic, often are called nearly elastic collisions. After the collision in case 3, the kinetic energy is equal to: (1.00 kg)(0.50 m/s)2 1 1 (1.00 kg)(0.50 m/s)2 0.25 J 2 2 KEDf KECf Kinetic energy decreased and some of it was converted to thermal energy. This kind of collision, in which kinetic energy decreases, is called an inelastic collision. Objects made of soft, sticky material, such as clay, act in this way. The three kinds of collisions can be represented using bar graphs, such as those shown in Figure 11-14. Although the", " kinetic energy before and after the collisions can be calculated, only the change in other forms of energy can be found. In automobile collisions, kinetic energy is transferred into other forms of energy, such as heat and sound. \u25a0 Figure 11-14 Bar graphs can be drawn to represent the three kinds of collisions. Before KE Other energy After KE Other energy Case 1: KE increases Case 2: KE is constant Case 3: KE decreases 298 Chapter 11 Energy and Its Conservation 293-301 CH11-S2-845813 7/13/04 1:18 PM Page 299 Kinetic Energy In an accident on a slippery road, a compact car with a mass of 575 kg moving at 15.0 m/s smashes into the rear end of a car with mass 1575 kg moving at 5.00 m/s in the same direction. a. What is the final velocity if the wrecked cars lock together? b. How much kinetic energy was lost in the collision? c. What fraction of the original kinetic energy was lost? 1 Analyze and Sketch the Problem \u2022 Sketch the initial and final conditions. \u2022 Sketch the momentum diagram. Known: Before (initial) mAvAi mBvBi vAi vBi After (final) mB)vf (mA mA vAi 575 kg 15.0 m/s 1575 kg 5.00 m/s vf vBf mB vBi vAf Unknown: KE KEf vf Fraction of KEi lost, KE/KEi Solve for the Unknown a. Use the conservation of momentum equation to find the final velocity. KEi??? 2 pAi mAvAi pBi pBf pAf (mA mBvBi vBi) m vA (m B A ) m m ( B A vf i mB)vf vf Math Handbook Isolating a Variable page 845 (575 kg)(15.0 m/s) (1575 kg)(5.00 m/s) (575 kg 1575 kg) 1575 kg, vBi 7.67 m/s, in the direction of the motion before the collision Substitute mA mB 575 kg, vAi 15.0 m/s, 5.00 m/s b. To determine the change in kinetic energy of the system, KEf and KEi are needed. KEf KEi 2 mB)vf 1 mv2 2 1 (mA 2 1 (575 kg + 1575 kg)(7.67 m/s)2 2 6.32104 J KEAi KEBi 2 1 1 mBvBi mAvAi 2 2 2 (575 kg)(15.0 m/s)2 1 1 (1575 kg)(5.00 m/s)2 2 2 8.44104 J Substitute KEAi Substitute m mA mB Substitute mA 575 kg, mB 1575 kg, vf 7.67 m/s 1 mAvAi 2 2, KEBi 2 1 mBvBi 2 575 kg, mB 5.00 m/s Substitute mA vAi 15.0 m/s, vBi 1575 kg, Solve for the change in kinetic energy of the system. KE KEf KEi 6.32104 J 8.44104 J 2.12104 J Substitute KEf 6.32104 J, KEi 8.44104 J c. Calculate the fraction of the original kinetic energy that is lost. 4 J 0 1 2.1 2 K E 4 0 1 4.4 8 J E K i 0.251 25.1% of the original kinetic energy in the system was lost. Substitute KE 2.11104 J, KEi 8.44104 J 3 Evaluate the Answer \u2022 Are the units correct? Velocity is measured in m/s; energy is measured in J. \u2022 Does the sign make sense? Velocity is positive, consistent with the original velocities. Section 11.2 Conservation of Energy 299 293-301 CH11-S2-845813 6/10/04 7:46 PM Page 300 19. An 8.00-g bullet is fired horizontally into a 9.00-kg block of wood on an air table and is embedded in it. After the collision, the block and bullet slide along the frictionless surface together with a speed of 10.0 cm/s. What was the initial speed of the bullet? 20. A 0.73-kg magnetic target is suspended on a string. A 0.025-kg magnetic dart, shot horizontally, strikes the target head-on. The dart and the target together, acting like a pendulum, swing 12.0 cm above the initial level before instantaneously coming to rest. a. Sketch the situation and choose a system. b. Decide what is conserved in each part and explain your decision. c.", " What was the initial velocity of the dart? 21. A 91.0-kg hockey player is skating on ice at 5.50 m/s. Another hockey player of equal mass, moving at 8.1 m/s in the same direction, hits him from behind. They slide off together. a. What are the total energy and momentum in the system before the collision? b. What is the velocity of the two hockey players after the collision? c. How much energy was lost in the collision? In collisions, you can see how momentum and energy are really very different. Momentum is almost always conserved in a collision. Energy is conserved only in elastic collisions. Momentum is what makes objects stop. A 10.0-kg object moving at 5.00 m/s will stop a 20.0-kg object moving at 2.50 m/s if they have a head-on collision. However, in this case, the smaller object has much more kinetic energy. The kinetic energy of the smaller object is KE 1 (10.0 kg)(5.0 m/s)2 125 J. The kinetic energy of 2 the larger object is KE 1 (20.0 kg)(2.50 m/s)2 62.5 J. Based on the 2 work-energy theorem, you can conclude that it takes more work to make the 10.0-kg object move at 5.00 m/s than it does to move the 20.0-kg object at 2.50 m/s. It sometimes is said that in automobile collisions, the momentum stops the cars but it is the energy in the collision that causes the damage. It also is possible to have a collision in which nothing collides. If two lab carts sit motionless on a table, connected by a compressed spring, their total momentum is zero. If the spring is released, the carts will be forced to move away from each other. The potential energy of the spring will be transformed into the kinetic energy of the carts. The carts will still move away from each other so that their total momentum is zero. A bullet of mass m, moving at speed v1, goes through a motionless wooden block and exits with speed v2. After the collision, the block, which has mass mB, is moving. 1. What is the final speed, vB, of the block? 2. How much energy was lost to the bullet? v1 3. How much energy was lost to friction inside the block? Initial Final v2 vB 300 Chapter 11 Energy and Its Conservation 293-301 CH11-S2-845813 6/3/04 12:18 PM Page 301 It is useful to remember two simple examples of collisions. One is the elastic collision between two objects of equal mass, such as when a cue ball with velocity, v, hits a motionless billiard ball head-on. In this case, after the collision, the cue ball is motionless and the other ball rolls off at velocity, v. It is easy to prove that both momentum and energy are conserved in this collision. The other simple example is to consider a skater of mass m, with velocity v, running into another skater of equal mass who happens to be standing motionless on the ice. If they hold on to each other after the collision, they will slide off at a velocity of 1 v because of the conservation 2 of momentum. The final kinetic energy of the pair would be equal to KE v)2 1 (2m)(1 1 mv2, which is half the initial kinetic energy. This is 4 2 2 because the collision was inelastic. You have investigated examples in which the conservation of energy, and sometimes the conservation of momentum, can be used to calculate the motions of a system of objects. These systems would be too complicated to comprehend using only Newton\u2019s second law of motion. The understanding of the forms of energy and how energy flows from one form to another is one of the most useful concepts in science. The term energy conservation appears in everything from scientific papers to electric appliance commercials. Scientists use the concept of energy to explore topics much more complicated than colliding billiard balls. Energy Exchange 1. Select different-sized steel balls and determine their masses. 2. Stand a spring-loaded laboratory cart on end with the spring mechanism pointing upward. 3. Place a ball on top of the spring mechanism and press down until the ball is touching the cart. 4. Quickly release the ball so that the spring shoots it upward. CAUTION: Stay clear of the ball when launching. 5. Repeat the process several times, and measure the average height. 6. Estimate how high the other sizes of steel balls will rise. Analyze and Conclude 7. Classify the balls by height attained. What can you conclude? 11.2 Section Review 22. Closed Systems Is Earth a closed, isolated sys- tem? Support your answer. 23. Energy A child jumps on a trampoline.", " Draw bar graphs to show the forms of energy present in the following situations. a. The child is at the highest point. b. The child is at the lowest point. 24. Kinetic Energy Suppose a glob of chewing gum and a small, rubber ball collide head-on in midair and then rebound apart. Would you expect kinetic energy to be conserved? If not, what happens to the energy? 25. Kinetic Energy In table tennis, a very light but hard ball is hit with a hard rubber or wooden paddle. In tennis, a much softer ball is hit with a racket. Why are the two sets of equipment designed in this way? Can you think of other ball-paddle pairs in sports? How are they designed? 26. Potential Energy A rubber ball is dropped from a height of 8.0 m onto a hard concrete floor. It hits the floor and bounces repeatedly. Each time it hits the floor, it loses 1 of its total energy. How many 5 times will it bounce before it bounces back up to a height of only about 4 m? 27. Energy As shown in Figure 11-15, a 36.0-kg child slides down a playground slide that is 2.5 m high. At the bottom of the slide, she is moving at 3.0 m/s. How much energy was lost as she slid down the slide? 36.0 kg 2.5 m \u25a0 Figure 11-15 28. Critical Thinking A ball drops 20 m. When it has fallen half the distance, or 10 m, half of its energy is potential and half is kinetic. When the ball has fallen for half the amount of time it takes to fall, will more, less, or exactly half of its energy be potential energy? physicspp.com/self_check_quiz Section 11.2 Conservation of Energy 301 302-303 CH11-LAB-845813 3/11/04 3:50 AM Page 302 Alternate CBL instructions can be found on the Web site. physicspp.com Conservation of Energy There are many examples of situations where energy is conserved. One such example is a rock falling from a given height. If the rock starts at rest, at the moment the rock is dropped, it only has potential energy. As it falls, its potential energy decreases as its height decreases, but its kinetic energy increases. The sum of potential energy and kinetic energy remains constant if friction is neglected. When the rock is about to hit the ground, all of its potential energy has been converted to kinetic energy. In this experiment, you will model a falling object and calculate its speed as it hits the ground. QUESTION How does the transfer of an object\u2019s potential energy to kinetic energy demonstrate conservation of energy? Objectives Materials \u25a0 Calculate the speed of a falling object as it hits the ground by using a model. \u25a0 Interpret data to find the relationship between potential energy and kinetic energy of a falling object. grooved track (two sections) marble or steel ball stopwatch block of wood electronic balance metric ruler graphing calculator Safety Precautions 302 Procedure 1. Place the two sections of grooved track together, as shown in Figure 1. Raise one end of the track and place the block under it, about 5 cm from the raised end. Make sure the ball can roll smoothly across the junction of the two tracks. 2. Record the length of the level portion of the track in the data table. Place a ball on the track directly above the point supported by the block. Release the ball. Start the stopwatch when the ball reaches the level section of track. Stop timing when the ball reaches the end of the level portion of the track. Record the time required for the ball to travel that distance in the data table. 3. Move the support block so that it is under the midsection of the inclined track, as shown in Figure 2. Place the ball on the track just above the point supported by the block. Release the ball and measure the time needed for the ball to roll the length of the level portion of the track and record it in the data table. Notice that even though the incline is steeper, the ball is released from the same height as in step 2. 4. Calculate the speed of the ball on the level portion of the track in steps 2 and 3. Move the support block to a point about three-quarters down the length of the inclined track, as shown in Figure 3. Figure 1 Figure 2 Figure 3 302-303 CH11-LAB-845813 7/20/04 2:48 PM Page 303 Data Table Release Height (m) Distance (m) Time (s) Speed (m/s) 0.05 0.05 0.05 0.01 0.02 0.03 5. Predict the amount of time the ball will take to travel the length of the level portion of the track. Record your prediction. Test your prediction. 6. Place the support block at the midpoint of the inclined track (Figure 2). Measure", " a point on the inclined portion of the track that is 1.0 cm above the level portion of the track. Be sure to measure 1.0 cm above the level portion, and not 1.0 cm above the table. 7. Release the ball from this point and measure the time required for the ball to travel on the level portion of the track and record it in the data table. 8. Use a ruler to measure a point that is 2.0 cm above the level track. Release the ball from this point and measure the time required for the ball to travel on the level portion of the track. Record the time in the data table. 9. Repeat step 8 for 3.0 cm, 4.0 cm, 5.0 cm, 6.0 cm, 7.0 cm, and 8.0 cm. Record the times. Analyze 1. Infer What effect did changing the slope of the inclined plane in steps 2\u20136 have on the speed of the ball on the level portion of the track? 2. Analyze Perform a power law regression for this graph using your graphing calculator. Record the equation of this function. Graph this by inputting the equation into Y=. Draw a sketch of the graph. 3. Using the data from step 9 for the release point of 8.0 cm, find the potential energy of the ball before it was released. Use an electronic balance to find the mass of the ball. Note that height must be in m, and mass in kg. 4. Using the speed data from step 9 for the release point of 8.0 cm calculate the kinetic energy of the ball on the level portion of the track. Remember, speed must be in m/s and mass in kg. Conclude and Apply 1. Solve for speed, y, in terms of height, x. Begin by setting PEi KEf. 2. How does the equation found in the previous question relate to the power law regression calculated earlier? 3. Suppose you want the ball to roll twice as fast on the level part of the track as it did when you released it from the 2-cm mark. Using the power law regression performed earlier, calculate the height from which you should release the ball. 4. Explain how this experiment only models dropping a ball and finding its kinetic energy just as it hits the ground. 5. Compare and Contrast Compare the potential energy of the ball before it is released (step 8) to the kinetic energy of the ball on the level track (step 9). Explain why they are the same or why they are different. 6. Draw Conclusions Does this experiment demonstrate conservation of energy? Explain. Going Further What are potential sources of error in this experiment, and how can they be reduced? Real-World Physics How does your favorite roller coaster demonstrate the conservation of energy by the transfer of potential energy to kinetic energy? To find out more about energy, visit the Web site: physicspp.com 303 304 CH11-FEATURE-845813 6/3/04 12:23 PM Page 304 Running Smarter Running Smarter The Physics of Running Shoes Today\u2019s running shoes are high-tech marvels. They enhance performance and protect your body by acting as shock absorbers. How do running shoes help you win a race? They reduce your energy consumption, as well as allow you to use energy more efficiently. Good running shoes must be flexible enough to bend with your feet as you run, support your feet, and hold them in place. They must be lightweight and provide traction to prevent slipping. Insert Upper Midsole Outsole Running Shoes as Shock Absorbers Today, much of the focus of running shoe technology centers on the cushioned midsole that plays a key role as a shockabsorber and performance enhancer. Each time a runner\u2019s foot hits the ground, the ground exerts an equal and opposite force on the runner\u2019s foot. This force can be nearly four times a runner\u2019s weight, causing aches and pains, shin splints, and damage to knees and ankles over long distances. Cushioning is used in running shoes to decrease the force absorbed by the runner. As a runner\u2019s foot hits the ground and comes to a stop, its momentum changes. The change in momentum is p Ft, where F is the force on that object and t is the time during which the force acts. The cushioning causes the change of momentum to occur over an extended time and reduces the force of the foot on the ground. The decreased force reduces the damage to the runner\u2019s body. Running Shoes Boost Performance A shoe\u2019s cushioning system also affects energy consumption. The bones, muscles, ligaments, 304 Horizons Companies Technology and Society and tendons of the foot and leg are a natural cushioning system. But operating this system requires the body to use stored energy to contract muscles. So if a shoe can be worn that assists a runner\u2019s natural cushioning system, the runner does not expend as much of his", " or her own stored energy. The energy the runner saved can be spent to run farther or faster. The cushioned midsole uses the law of conservation of energy to return as much of the energy to the runner as possible. The runner\u2019s kinetic energy transforms to elastic potential energy, plus heat, when the runner\u2019s foot hits the running surface. If the runner can reduce the amount of energy that is lost as heat, the runner\u2019s elastic potential energy can be converted back to useful kinetic energy. Bouncy, springy, elastic materi- als that resist crushing over time commonly are used to create the cushioned midsole. Options now range from silicon gel pads to complex fluid-filled systems and even springs to give a runner extra energy efficiency. Going Further 1. Use Scientific Explanations Use physics to explain why manufacturers put cushioned midsoles in running shoes. 2. Analyze Which surface would provide more cushioning when running: a grassy field or a concrete sidewalk? Explain why that surface provides better cushioning. 3. Research Some people prefer to run barefoot, even in marathon races. Why might this be so? 305-311 CH11-SG CA STP-845813 3/11/04 5:14 AM Page 305 11.1 The Many Forms of Energy Vocabulary \u2022 rotational kinetic energy Key Concepts \u2022 The kinetic energy of an object is proportional to its mass and the square of (p. 287) \u2022 gravitational potential energy (p. 288) \u2022 reference level (p. 288) \u2022 elastic potential energy (p. 291) its velocity. \u2022 The rotational kinetic energy of an object is proportional to the object\u2019s moment of inertia and the square of its angular velocity. \u2022 When Earth is included in a system, the work done by gravity is replaced by gravitational potential energy. \u2022 The gravitational potential energy of an object depends on the object\u2019s weight and its distance from Earth\u2019s surface. PE mgh \u2022 The reference level is the position where the gravitational potential energy is defined to be zero. \u2022 Elastic potential energy may be stored in an object as a result of its change in shape. \u2022 Albert Einstein recognized that mass itself has potential energy. This energy is called rest energy. mc2 E0 11.2 Conservation of Energy Vocabulary \u2022 law of conservation of energy (p. 293) \u2022 mechanical energy (p. 293) \u2022 thermal energy (p. 295) \u2022 elastic collision (p. 298) \u2022 inelastic collision (p. 298) Key Concepts \u2022 The sum of kinetic and potential energy is called mechanical energy. E KE PE \u2022 If no objects enter or leave a system, the system is considered to be a closed system. \u2022 If there are no external forces acting on a system, the system is considered to be an isolated system. \u2022 The total energy of a closed, isolated system is constant. Within the system, energy can change form, but the total amount of energy does not change. Thus, energy is conserved. KEbefore PEbefore KEafter PEafter \u2022 The type of collision in which the kinetic energy after the collision is less than the kinetic energy before the collision is called an inelastic collision. \u2022 The type of collision in which the kinetic energy before and after the collision is the same is called an elastic collision. \u2022 Momentum is conserved in collisions if the external force is zero. The mechanical energy may be unchanged or decreased by the collision, depending on whether the collision is elastic or inelastic. physicspp.com/vocabulary_puzzlemaker 305 305-311 CH11-SG CA STP-845813 6/3/04 12:28 PM Page 306 Concept Mapping 29. Complete the concept map using the following terms: gravitational potential energy, elastic potential energy, kinetic energy. Energy 39. You throw a clay ball at a hockey puck on ice. The smashed clay ball and the hockey puck stick together and move slowly. (11.2) a. Is momentum conserved in the collision? Explain. b. Is kinetic energy conserved? Explain. 40. Draw energy bar graphs for the following processes. (11.2) a. An ice cube, initially at rest, slides down a potential frictionless slope. linear rotational Mastering Concepts Unless otherwise directed, assume that air resistance is negligible. 30. Explain how work and a change in energy are related. (11.1) 31. What form of energy does a wound-up watch spring have? What form of energy does a functioning mechanical watch have? When a watch runs down, what has happened to the energy? (11.1) 32. Explain how energy change and force are related. (11.1) 33. A ball is dropped from the top of a building. You choose the top of the building to be the reference level, while your friend chooses the bottom. Explain whether the energy calculated using these two reference levels is the same or different for the following situations. (11.1) a", ". the ball\u2019s potential energy at any point b. the change in the ball\u2019s potential energy as a result of the fall b. An ice cube, initially moving, slides up a frictionless slope and instantaneously comes to rest. 41. Describe the transformations from kinetic energy to potential energy and vice versa for a roller-coaster ride. (11.2) 42. Describe how the kinetic energy and elastic potential energy are lost in a bouncing rubber ball. Describe what happens to the motion of the ball. (11.2) Applying Concepts 43. The driver of a speeding car applies the brakes and the car comes to a stop. The system includes the car but not the road. Apply the work-energy theorem to the following situations. a. The car\u2019s wheels do not skid. b. The brakes lock and the car\u2019s wheels skid. 44. A compact car and a trailer truck are both traveling at the same velocity. Did the car engine or the truck engine do more work in accelerating its vehicle? 45. Catapults Medieval warriors used catapults to assault castles. Some catapults worked by using a tightly wound rope to turn the catapult arm. What forms of energy are involved in catapulting a rock to the castle wall? c. the kinetic energy of the ball at any point 46. Two cars collide and come to a complete stop. 34. Can the kinetic energy of a baseball ever be negative? (11.1) 35. Can the gravitational potential energy of a baseball ever be negative? Explain without using a formula. (11.1) 36. If a sprinter\u2019s velocity increases to three times the original velocity, by what factor does the kinetic energy increase? (11.1) 37. What energy transformations take place when an athlete is pole-vaulting? (11.2) 38. The sport of pole-vaulting was drastically changed when the stiff, wooden poles were replaced by flexible, fiberglass poles. Explain why. (11.2) Where did all of their energy go? 47. During a process, positive work is done on a system, and the potential energy decreases. Can you determine anything about the change in kinetic energy of the system? Explain. 48. During a process, positive work is done on a system, and the potential energy increases. Can you tell whether the kinetic energy increased, decreased, or remained the same? Explain. 49. Skating Two skaters of unequal mass have the same speed and are moving in the same direction. If the ice exerts the same frictional force on each skater, how will the stopping distances of their bodies compare? 306 Chapter 11 Energy and Its Conservation For more problems, go to Additional Problems, Appendix B. 305-311 CH11-SG CA STP-845813 6/3/04 12:31 PM Page 307 50. You swing a 55-g mass on the end of a 0.75-m 57. Tony has a mass of 45 kg and is moving with a string around your head in a nearly horizontal circle at constant speed, as shown in Figure 11-16. a. How much work is done on the mass by the tension of the string in one revolution? b. Is your answer to part a in agreement with the work-energy theorem? Explain. speed of 10.0 m/s. a. Find Tony\u2019s kinetic energy. b. Tony\u2019s speed changes to 5.0 m/s. Now what is his kinetic energy? c. What is the ratio of the kinetic energies in parts a and b? Explain. 0.75 m 55 g \u25a0 Figure 11-16 51. Give specific examples that illustrate the following processes. a. Work is done on a system, thereby increasing kinetic energy with no change in potential energy. b. Potential energy is changed to kinetic energy with no work done on the system. c. Work is done on a system, increasing potential energy with no change in kinetic energy. d. Kinetic energy is reduced, but potential energy is unchanged. Work is done by the system. 52. Roller Coaster You have been hired to make a roller coaster more exciting. The owners want the speed at the bottom of the first hill doubled. How much higher must the first hill be built? 53. Two identical balls are thrown from the top of a cliff, each with the same speed. One is thrown straight up, the other straight down. How do the kinetic energies and speeds of the balls compare as they strike the ground? Mastering Problems Unless otherwise directed, assume that air resistance is negligible. 11.1 The Many Forms of Energy 54. A 1600-kg car travels at a speed of 12.5 m/s. What is its kinetic energy? 58. Katia and Angela each have a mass of 45 kg, and they are moving together with a speed of 10.0 m/s. a. What is their combined kinetic energy?", " b. What is the ratio of their combined mass to Katia\u2019s mass? c. What is the ratio of their combined kinetic energy to Katia\u2019s kinetic energy? Explain. 59. Train In the 1950s, an experimental train, which had a mass of 2.50104 kg, was powered across a level track by a jet engine that produced a thrust of 5.00105 N for a distance of 509 m. a. Find the work done on the train. b. Find the change in kinetic energy. c. Find the final kinetic energy of the train if it started from rest. d. Find the final speed of the train if there had been no friction. 60. Car Brakes A 14,700-N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop, as shown in Figure 11-17. The average braking force between the tires and the road is 7100 N. How far will the car slide once the brakes are applied? Before (initial) After (final) v 25 m/s v 0.0 m/s m 14,700 N \u25a0 Figure 11-17 61. A 15.0-kg cart is moving with a velocity of 7.50 m/s down a level hallway. A constant force of 10.0 N acts on the cart, and its velocity becomes 3.20 m/s. a. What is the change in kinetic energy of the cart? b. How much work was done on the cart? c. How far did the cart move while the force acted? 62. How much potential energy does DeAnna with a 55. A racing car has a mass of 1525 kg. What is its kinetic energy if it has a speed of 108 km/h? mass of 60.0 kg, gain when she climbs a gymnasium rope a distance of 3.5 m? 56. Shawn and his bike have a combined mass of 45.0 kg. Shawn rides his bike 1.80 km in 10.0 min at a constant velocity. What is Shawn\u2019s kinetic energy? 63. Bowling A 6.4-kg bowling ball is lifted 2.1 m into a storage rack. Calculate the increase in the ball\u2019s potential energy. physicspp.com/chapter_test Chapter 11 Assessment 307 305-311 CH11-SG CA STP-845813 3/11/04 5:17 AM Page 308 64. Mary weighs 505 N. She walks down a flight of stairs to a level 5.50 m below her starting point. What is the change in Mary\u2019s potential energy? 65. Weightlifting A weightlifter raises a 180-kg barbell to a height of 1.95 m. What is the increase in the potential energy of the barbell? Before (initial) vi 12.0 m/s After (final) vf 0.0 m/s 66. A 10.0-kg test rocket is fired vertically from Cape m 2.00103 kg Canaveral. Its fuel gives it a kinetic energy of 1960 J by the time the rocket engine burns all of the fuel. What additional height will the rocket rise? 67. Antwan raised a 12.0-N physics book from a table 75 cm above the floor to a shelf 2.15 m above the floor. What was the change in the potential energy of the system? 68. A hallway display of energy is constructed in which several people pull on a rope that lifts a block 1.00 m. The display indicates that 1.00 J of work is done. What is the mass of the block? 69. Tennis It is not uncommon during the serve of a professional tennis player for the racket to exert an average force of 150.0 N on the ball. If the ball has a mass of 0.060 kg and is in contact with the strings of the racket, as shown in Figure 11-18, for 0.030 s, what is the kinetic energy of the ball as it leaves the racket? Assume that the ball starts from rest. 150.0 N \u25a0 Figure 11-18 70. Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 45 kg. The rocket supplies a constant force for 22.0 m, and Pam acquires a speed of 62.0 m/s. a. What is the magnitude of the force? b. What is Pam\u2019s final kinetic energy? 71. Collision A 2.00103-kg car has a speed of 12.0 m/s. The car then hits a tree. The tree doesn\u2019t move, and the car comes to rest, as shown in Figure 11-19. a. Find the change in kinetic energy of the car. b. Find the amount of work done as the front of the car crashes into the tree. c. Find the size of the force that pushed in the front \u25a0 Figure 11-19 72", ". A constant net force of 410 N is applied upward to a stone that weighs 32 N. The upward force is applied through a distance of 2.0 m, and the stone is then released. To what height, from the point of release, will the stone rise? 11.2 Conservation of Energy 73. A 98.0-N sack of grain is hoisted to a storage room 50.0 m above the ground floor of a grain elevator. a. How much work was done? b. What is the increase in potential energy of the sack of grain at this height? c. The rope being used to lift the sack of grain breaks just as the sack reaches the storage room. What kinetic energy does the sack have just before it strikes the ground floor? 74. A 20-kg rock is on the edge of a 100-m cliff, as shown in Figure 11-20. a. What potential energy does the rock possess relative to the base of the cliff? b. The rock falls from the cliff. What is its kinetic energy just before it strikes the ground? c. What speed does the rock have as it strikes the ground? 20 kg 100 m \u25a0 Figure 11-20 75. Archery An archer puts a 0.30-kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m. a. Assuming that all the energy goes into the arrow, with what speed does the arrow leave the bow? b. If the arrow is shot straight up, how high does of the car by 50.0 cm. it rise? 308 Chapter 11 Energy and Its Conservation For more problems, go to Additional Problems, Appendix B. 305-311 CH11-SG CA STP-845813 6/3/04 12:32 PM Page 309 76. A 2.0-kg rock that is initially at rest loses 407 J of potential energy while falling to the ground. Calculate the kinetic energy that the rock gains while falling. What is the rock\u2019s speed just before it strikes the ground? 77. A physics book of unknown mass is dropped 4.50 m. What speed does the book have just before it hits the ground? 83. A person weighing 635 N climbs up a ladder to a height of 5.0 m. Use the person and Earth as the system. a. Draw energy bar graphs of the system before the person starts to climb the ladder and after the person stops at the top. Has the mechanical energy changed? If so, by how much? b. Where did this energy come from? 78. Railroad Car A railroad car with a mass of 5.0105 kg collides with a stationary railroad car of equal mass. After the collision, the two cars lock together and move off at 4.0 m/s, as shown in Figure 11-21. a. Before the collision, the first railroad car was moving at 8.0 m/s. What was its momentum? b. What was the total momentum of the two cars after the collision? c. What were the kinetic energies of the two cars before and after the collision? d. Account for the loss of kinetic energy. m 5.0105 kg v 4.0 m/s \u25a0 Figure 11-21 79. From what height would a compact car have to be dropped to have the same kinetic energy that it has when being driven at 1.00102 km/h? 80. Kelli weighs 420 N, and she is sitting on a playground swing that hangs 0.40 m above the ground. Her mom pulls the swing back and releases it when the seat is 1.00 m above the ground. a. How fast is Kelli moving when the swing passes through its lowest position? b. If Kelli moves through the lowest point at 2.0 m/s, how much work was done on the swing by friction? 81. Hakeem throws a 10.0-g ball straight down from a height of 2.0 m. The ball strikes the floor at a speed of 7.5 m/s. What was the initial speed of the ball? 82. Slide Lorena\u2019s mass is 28 kg. She climbs the 4.8-m ladder of a slide and reaches a velocity of 3.2 m/s at the bottom of the slide. How much work was done by friction on Lorena? Mixed Review 84. Suppose a chimpanzee swings through the jungle on vines. If it swings from a tree on a 13-m-long vine that starts at an angle of 45\u00b0, what is the chimp\u2019s velocity when it reaches the ground? 85. An 0.80-kg cart rolls down a frictionless hill of height 0.32 m. At the bottom of the hill, the cart rolls on a flat surface, which exerts a frictional force of 2.0 N on the cart. How far does the cart roll on the flat surface before it comes to a stop?", " 86. High Jump The world record for the men\u2019s high jump is about 2.45 m. To reach that height, what is the minimum amount of work that a 73.0-kg jumper must exert in pushing off the ground? 87. A stuntwoman finds that she can safely break her fall from a one-story building by landing in a box filled to a 1-m depth with foam peanuts. In her next movie, the script calls for her to jump from a fivestory building. How deep a box of foam peanuts should she prepare? 88. Football A 110-kg football linebacker has a head-on collision with a 150-kg defensive end. After they collide, they come to a complete stop. Before the collision, which player had the greater momentum and which player had the greater kinetic energy? 89. A 2.0-kg lab cart and a 1.0-kg lab cart are held together by a compressed spring. The lab carts move at 2.1 m/s in one direction. The spring suddenly becomes uncompressed and pushes the two lab carts apart. The 2-kg lab cart comes to a stop, and the 1.0-kg lab cart moves ahead. How much energy did the spring add to the lab carts? 90. A 55.0-kg scientist roping through the top of a tree in the jungle sees a lion about to attack a tiny antelope. She quickly swings down from her 12.0-m-high perch and grabs the antelope (21.0 kg) as she swings. They barely swing back up to a tree limb out of reach of the lion. How high is this tree limb? physicspp.com/chapter_test Chapter 11 Assessment 309 305-311 CH11-SG CA STP-845813 6/3/04 12:33 PM Page 310 91. An 0.80-kg cart rolls down a 30.0\u00b0 hill from a 96. Analyze and Conclude In a perfectly elastic vertical height of 0.50 m as shown in Figure 11-22. The distance that the cart must roll to the bottom of the hill is 0.50 m/sin 30.0\u00b0 1.0 m. The surface of the hill exerts a frictional force of 5.0 N on the cart. Does the cart roll to the bottom of the hill? collision, both momentum and mechanical energy are conserved. Two balls, with masses mA and mB, are moving toward each other with speeds vA and vB, respectively. Solve the appropriate equations to find the speeds of the two balls after the collision. m 0.80 kg F 5.0 N 0.50 m 30.0\u00b0 \u25a0 Figure 11-22 92. Object A, sliding on a frictionless surface at 3.2 m/s, hits a 2.0-kg object, B, which is motionless. The collision of A and B is completely elastic. After the collision, A and B move away from each other at equal and opposite speeds. What is the mass of object A? 93. Hockey A 90.0-kg hockey player moving at 5.0 m/s collides head-on with a 110-kg hockey player moving at 3.0 m/s in the opposite direction. After the collision, they move off together at 1.0 m/s. How much energy was lost in the collision? Thinking Critically 94. Apply Concepts A golf ball with a mass of 0.046 kg rests on a tee. It is struck by a golf club with an effective mass of 0.220 kg and a speed of 44 m/s. Assuming that the collision is elastic, find the speed of the ball when it leaves the tee. 95. Apply Concepts A fly hitting the windshield of a moving pickup truck is an example of a collision in which the mass of one of the objects is many times larger than the other. On the other hand, the collision of two billiard balls is one in which the masses of both objects are the same. How is energy transferred in these collisions? Consider an elastic collision in which billiard ball m1 has velocity v1 and ball m2 is motionless. a. If m1 m2, what fraction of the initial energy is b. If m1 m2, what fraction of the initial energy transferred to m2? is transferred to m2? c. In a nuclear reactor, neutrons must be slowed down by causing them to collide with atoms. (A neutron is about as massive as a proton.) Would hydrogen, carbon, or iron atoms be more desirable to use for this purpose? 97. Analyze and Conclude A 25-g ball is fired with an initial speed of v1 toward a 125-g ball that is hanging motionless from a 1.25-m string. The balls have a perfectly elastic collision. As a result, the 125-g ball swings out until the string makes an angle of 37.0", "\u00b0 with the vertical. What is v1? Writing in Physics 98. All energy comes from the Sun. In what forms has this solar energy come to us to allow us to live and to operate our society? Research the ways that the Sun\u2019s energy is turned into a form that we can use. After we use the Sun\u2019s energy, where does it go? Explain. 99. All forms of energy can be classified as either kinetic or potential energy. How would you describe nuclear, electric, chemical, biological, solar, and light energy, and why? For each of these types of energy, research what objects are moving and how energy is stored in those objects. Cumulative Reveiw 100. A satellite is placed in a circular orbit with a radius of 1.0107 m and a period of 9.9103 s. Calculate the mass of Earth. Hint: Gravity is the net force on such a satellite. Scientists have actually measured the mass of Earth this way. (Chapter 7) 101. A 5.00-g bullet is fired with a velocity of 100.0 m/s toward a 10.00-kg stationary solid block resting on a frictionless surface. (Chapter 9) a. What is the change in momentum of the bullet if it is embedded in the block? b. What is the change in momentum of the bullet if it ricochets in the opposite direction with a speed of 99 m/s? c. In which case does the block end up with a greater speed? 102. An automobile jack must exert a lifting force of at least 15 kN. (Chapter 10) a. If you want to limit the effort force to 0.10 kN, what mechanical advantage is needed? b. If the jack is 75% efficient, over what distance must the effort force be exerted in order to raise the auto 33 cm? 310 Chapter 11 Energy and Its Conservation For more problems, go to Additional Problems, Appendix B. 305-311 CH11-SG CA STP-845813 6/3/04 12:35 PM Page 311 Multiple Choice 1. A bicyclist increases her speed from 4.0 m/s to 6.0 m/s. The combined mass of the bicyclist and the bicycle is 55 kg. How much work did the bicyclist do in increasing her speed? 11 J 28 J 55 J 550 J 2. The illustration below shows a ball swinging freely in a plane. The mass of the ball is 4.0 kg. Ignoring friction, what is the maximum kinetic energy of the ball as it swings back and forth? 0.14 m/s 21 m/s 7.0 m/s 49 m/s h 2.5 m 3. You lift a 4.5-kg box from the floor and place it on a shelf that is 1.5 m above the ground. How much energy did you use in lifting the box? 9.0 J 49 J 11 J 66 J 4. You drop a 6.0102-kg ball from a height of 1.0 m above a hard, flat surface. The ball strikes the surface and loses 0.14 J of its energy. It then bounces back upward. How much kinetic energy does the ball have just after it bounces off the flat surface? 0.20 J 0.59 J 0.45 J 0.73 J 5. You move a 2.5-kg book from a shelf that is 1.2 m above the ground to a shelf that is 2.6 m above the ground. What is the change in the book\u2019s potential energy? 1.4 J 25 J 3.5 J 34 J 6. A ball of mass m rolls along a flat surface with a speed of v1. It strikes a padded wall and bounces back in the opposite direction. The energy of the ball after striking the wall is half its initial energy. Ignoring friction, which of the following expressions gives the ball\u2019s new speed as a function of its initial speed? 1 v1 2 2 (v1) 2 2(v1) 2v1 7. The illustration below shows a ball on a curved track. The ball starts with zero velocity at the top of the track. It then rolls from the top of the track to the horizontal part at the ground. Ignoring friction, its velocity just at the moment it reaches the ground is 14 m/s. What is the height, h, from the ground to the top of the track? 10 m 20 m 7 m 14 m h Extended Answer 8. A box sits on a platform supported by a compressed spring. The box has a mass of 1.0 kg. When the spring is released, it gives 4.9 J of energy to the box, and the box flies upward. What will be the maximum height above the platform reached by the box before it begins to fall? Use the Process of Elimination On any multiple-choice test, there are two ways to find the correct answer", " to each question. Either you can choose the right answer immediately or you can eliminate the answers that you know are wrong. physicspp.com/standardized_test Chapter 11 Standardized Test Practice 311 312-322 CH12-S1-845813 6/3/04 2:42 PM Page 312 What You\u2019ll Learn \u2022 You will learn how temperature relates to the potential and kinetic energies of atoms and molecules. \u2022 You will distinguish heat from work. \u2022 You will calculate heat transfer and the absorption of thermal energy. Why It\u2019s Important Thermal energy is vital for living creatures, chemical reactions, and the working of engines. Solar Energy A strategy used to produce electric power from sunlight concentrates the light with many mirrors onto one collector that becomes very hot. The energy collected at a high temperature is then used to drive an engine, which turns an electric generator. Think About This What forms of energy does light from the Sun take in the process of converting solar energy into useful work through an engine? physicspp.com 312 CORBIS 312-322 CH12-S1-845813 6/3/04 2:49 PM Page 313 What happens when you provide thermal energy by holding a glass of water? Question What happens to the temperature of water when you hold a glass of water in your hand? Procedure Analysis Calculate the change in temperature of the water. If you had more water in the beaker, would it affect the change in temperature? Critical Thinking Explain what caused the water temperature to change. 1. You will need to use a 250-mL beaker and 150 mL of water. 2. Fill the beaker with the 150 mL of water. 3. Record the initial temperature of the water by holding a thermometer in the water in the beaker. Note that the bulb end of the thermometer must not touch the bottom or sides of the beaker, nor should it touch a table or your hands. 4. Remove the thermometer and hold the beaker of water for 2 min by cupping it with both hands, as shown in the figure. 5. Have your lab partner record the final temperature of the water by placing the thermometer in the beaker. Be sure that the bulb end of the thermometer is not touching the bottom or sides of the beaker. 12.1 Temperature and Thermal Energy The study of heat transformations into other forms of energy, called thermodynamics, began with the eighteenth-century engineers who built the first steam engines. These steam engines were used to power trains, factories, and water pumps for coal mines, and thus they contributed greatly to the Industrial Revolution in Europe and in the United States. In learning to design more efficient engines, the engineers developed new concepts about how heat is related to useful work. Although the study of thermodynamics began in the eighteenth century, it was not until around 1900 that the concepts of thermodynamics were linked to the motions of atoms and molecules in solids, liquids, and gases. Today, the concepts of thermodynamics are widely used in various applications that involve heat and temperature. Engineers use the laws of thermodynamics to continually develop higher performance refrigerators, automobile engines, aircraft engines, and numerous other machines. Objectives \u2022 Describe thermal energy and compare it to potential and kinetic energies. \u2022 Distinguish temperature from thermal energy. \u2022 Define specific heat and calculate heat transfer. Vocabulary conduction thermal equilibrium heat convection radiation specific heat Section 12.1 Temperature and Thermal Energy 313 Horizons Companies 312-322 CH12-S1-845813 6/3/04 2:50 PM Page 314 Thermal Energy You already have studied how objects collide and trade kinetic energies. For example, the many molecules present in a gas have linear and rotational kinetic energies. The molecules also may have potential energy in their vibrations and bending. The gas molecules collide with each other and with the walls of their container, transferring energy among each other in the process. There are numerous molecules moving freely in a gas, resulting in many collisions. Therefore, it is convenient to discuss the total energy of the molecules and the average energy per molecule. The total energy of the molecules is called thermal energy, and the average energy per molecule is related to the temperature of the gas. Hot objects What makes an object hot? When you fill up a balloon with helium, the rubber in the balloon is stretched by the repeated pounding from helium atoms. Each of the billions of helium atoms in the balloon collides with the rubber wall, bounces back, and hits the other side of the balloon, as shown in Figure 12-1. If you put a balloon in sunlight, you might notice that the balloon gets slightly larger. The energy from the Sun makes each of the gas atoms move faster and bounce off the rubber walls of the balloon more often. Each atomic collision with the balloon wall puts a greater force on the balloon and stretches the rubber. Thus, the balloon expands. On the other hand, if you refrigerate a balloon, you will find that it shrinks slightly. Lowering the temperature", " slows the movement of the helium atoms. Hence, their collisions do not transfer enough momentum to stretch the balloon quite as much. Even though the balloon contains the same number of atoms, the balloon shrinks. Solids The atoms in solids also have kinetic energy, but they are unable to move freely as gas atoms do. One way to illustrate the molecular structure of a solid is to picture a number of atoms that are connected to each other by springs. Because of the springs, the atoms bounce back and forth, with some bouncing more than others. Each atom has some kinetic energy and some potential energy from the springs that are attached to it. If a solid has N number of atoms, then the total thermal energy in the solid is equal to the average kinetic and potential energy per atom times N. Helium balloon \u25a0 Figure 12-1 Helium atoms in a balloon collide with the rubber wall and cause the balloon to expand. Hot object KEhot KEcold Thermal Energy and Temperature According to the previous discussion of gases and solids, a hot object has more thermal energy than a similar cold object, as shown in Figure 12-2. This means that, as a whole, the particles in a hot object have greater thermal energy than the particles in a cold object. This does not mean that all the particles in an object have exactly the same amount of energy; they have a wide range of energies. However, the average energy of the particles in a hot object is higher than the average energy of the particles in a cold object. To understand this, consider the heights of students in a twelfth-grade class. Although the students\u2019 heights vary, you can calculate the average height of the students in the class. This average is likely to be greater than the average height of students in a ninth-grade class, even though some ninth-grade students may be taller than some twelfth-grade students. Cold object \u25a0 Figure 12-2 Particles in a hot object have greater kinetic and potential energies than particles in a cold object do. 314 Chapter 12 Thermal Energy 312-322 CH12-S1-845813 6/10/04 4:54 PM Page 315 Temperature Temperature depends only on the average kinetic energy of the particles in the object. Because temperature depends on average kinetic energy, it does not depend on the number of atoms in an object. To understand this, consider two blocks of steel. The first block has a mass of 1 kg, and the second block has a mass of 2 kg. If the 1-kg block is at the same temperature as the 2-kg block, the average kinetic energy of the particles in each block is the same. However, the 2-kg block has twice the mass of the 1-kg block. Hence, the 2-kg block has twice the amount of particles as the 1kg block. Thus, the total amount of kinetic energy of the particles in the 2kg block is twice that of the 1-kg mass. Total kinetic energy is divided by the total number of particles in an object to calculate its average kinetic energy. Therefore, the thermal energy in an object is proportional to the number of particles in it. Temperature, however, is not dependent on the number of particles in an object. Equilibrium and Thermometry How do you measure your body temperature? For example, if you suspect that you have a fever, you might place a thermometer in your mouth and wait for a few minutes before checking the thermometer for your temperature reading. The microscopic process involved in measuring temperature involves collisions and energy transfers between the thermometer and your body. Your body is hot compared to the thermometer, which means that the particles in your body have greater thermal energy and are moving faster than the particles in the thermometer. When the cold glass tube of the thermometer touches your skin, which is warmer than the glass, the faster-moving particles in your skin collide with the slower-moving particles in the glass. Energy is then transferred from your skin to the glass particles by the process of conduction, which is the transfer of kinetic energy when particles collide. The thermal energy of the particles that make up the thermometer increases, while at the same time, the thermal energy of the particles in your skin decreases. Thermal equilibrium As the particles in the glass gain more energy, they begin to give some of their energy back to the particles in your body. At some point, the rate of transfer of energy between the glass and your body becomes equal, and your body and the thermometer are then at the same temperature. At this point, your body and the thermometer are said to have reached thermal equilibrium, the state in which the rate of energy flow between two objects is equal and the objects are at the same temperature, as shown in Figure 12-3. The operation of a thermometer depends on some property, such as volume, which changes with temperature. Many household thermometers contain colored alcohol that expands when heated and rises in a narrow tube. The hotter the thermometer, the more the alcohol expands and the", " higher it rises in the tube. In liquid-crystal thermometers, such as the one shown in Figure 12-4, a set of different kinds of liquid crystals is used. Each crystal\u2019s molecules rearrange at a specific temperature, which causes the color of the crystal to change and indicates the temperature by color. Medical thermometers and the thermometers that monitor automobile engines use very small, temperature-sensitive electronic circuits to take rapid measurements. Before Thermal Equilibrium Hot object (A) Cold object (B) KEA KEB After Thermal Equilibrium KEA KEB \u25a0 Figure 12-3 Thermal energy is transferred from a hot object to a cold object. When thermal equilibrium is reached, the transfer of energy between objects is equal. \u25a0 Figure 12-4 Thermometers use a change in physical properties to measure temperature. A liquidcrystal thermometer changes color with a temperature change. Section 12.1 Temperature and Thermal Energy 315 Tom Pantages 312-322 CH12-S1-845813 3/22/04 12:13 AM Page 316 Interstellar space Human body Surface of the Sun Nuclear bomb Flames Center of the Sun Supernova explosions 108 106 104 102 1 10 100 103 104 105 106 107 108 109 1010 Lowest temperature in laboratory Helium liquefies Superconductivity below this temperature Life exists Uncharged atoms exist below this temperature Nuclei exist below this temperature \u25a0 Figure 12-5 There is an extremely wide range of temperatures throughout the universe. Note that the scale has been expanded in areas of particular interest. K 380 370 360 350 340 330 320 310 300 290 280 270 260 20 10 0 \u00b0C \u00b0F 373.15 110 100 100.00 90 80 70 60 50 40 30 20 10 0 0.00 273.15 212.00 32.00 210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 10 250 260 270 0.00 420 430 440 450 460 273.15 459.67 \u25a0 Figure 12-6 The three mostcommon temperature scales are Kelvin, Celsius, and Fahrenheit. 316 (cl)Getty Images, (r)FPG/Getty Images, (others)CORBIS Chapter 12 Thermal Energy Temperature (K) Temperature Scales: Celsius and Kelvin Over the years, scientists developed temperature scales so that they could compare their measurements with those of other scientists. A scale based on the properties of water was devised in 1741 by Swedish astronomer and physicist Anders Celsius. On this scale, now called the Celsius scale, the freezing point of pure water is defined to be 0\u00b0C. The boiling point of pure water at sea level is defined to be 100\u00b0C. Temperature limits The wide range of temperatures present in the universe is shown in Figure 12-5. Temperatures do not appear to have an upper limit. The interior of the Sun is at least 1.5107\u00b0C. Temperatures do, however, have a lower limit. Generally, materials contract as they cool. If an ideal gas, such as the helium in a balloon is cooled, it contracts in such a way that it occupies a volume that is only the size of the helium atoms at 273.15\u00b0C. At this temperature, all the thermal energy that can be removed has been removed from the gas. It is impossible to reduce the temperature any further. Therefore, there can be no temperature lower than 273.15\u00b0C, which is called absolute zero. The Celsius scale is useful for day-to-day measurements of temperature. It is not conducive for working on science and engineering problems, however, because it has negative temperatures. Negative temperatures suggest a molecule could have negative kinetic energy, which is not possible because kinetic energy is always positive. The solution to this issue is to use a temperature scale based on absolute zero. The zero point of the Kelvin scale is defined to be absolute zero. On the Kelvin scale, the freezing point of water (0\u00b0C) is about 273 K and the boiling point of water is about 373 K. Each interval on this scale, called a kelvin, is equal to 1\u00b0C. Thus, TC 273 TK. Figure 12-6 shows representative temperatures on the three most-common scales: Fahrenheit, Celsius, and Kelvin. Very cold temperatures are reached by liquefying gases. Helium liquefies at 4.2 K, or 269\u00b0C. Even colder temperatures can be reached by making use of special properties of solids, helium isotopes, and atoms and lasers. 312-322 CH12-S1-845813 6/10/04 4:56 PM Page 317 1. Convert the following Kelvin temperatures to Celsius temperatures. a. 115 K b. 172 K c. 125 K d. 402 K e. 425 K f. 212 K 2. Find the Celsius and Kelvin temperatures for the following. a. room temperature c. a hot summer day in North Carolina b. a typical refrigerator d. a winter night in Minnesota Heat and the Flow of Thermal Energy When two objects come in", " contact with each other, they transfer energy. This energy that is transferred between the objects is called heat. Heat is described as the energy that always flows from the hotter object to the cooler object. Left to itself heat never flows from a colder object to a hotter object. The symbol Q is used to represent an amount of heat, which, like other forms of energy, is measured in joules. If Q has a negative value, heat has left the object; if Q has a positive value, heat has been absorbed by the object. Conduction If you place one end of a metal rod in a flame, the hot gas particles in the flame conduct heat to the rod. The other end of the rod also becomes warm within a short period of time. Heat is conducted because the particles in the rod are in direct contact with each other. Convection Thermal energy transfer can occur even if the particles in an object are not in direct contact with each other. Have you ever looked into a pot of water just about to boil? The water at the bottom of the pot is heated by conduction and rises to the top, while the colder water at the top sinks to the bottom. Heat flows between the rising hot water and the descending cold water. This motion of fluid in a liquid or gas caused by temperature differences is called convection. Atmospheric turbulence is caused by convection of gases in the atmosphere. Thunderstorms are excellent examples of large-scale atmospheric convection. Ocean currents that cause changes in weather patterns also result from convection. Radiation The third method of thermal transfer, unlike the first two, does not depend on the presence of matter. The Sun warms Earth from over 150 million km away via radiation, which is the transfer of energy by electromagnetic waves. These waves carry the energy from the hot Sun through the vacuum of space to the much cooler Earth. Specific Heat Some objects are easier to heat than others. On a bright summer day, the Sun warms the sand on a beach and the ocean water. However, the sand on the beach gets quite hot, while the ocean water stays relatively cool. When heat flows into an object, its thermal energy and temperature increase. The amount of the increase in temperature depends on the size of the object and on the material from which the object is made. Steam Heating In a steam heating system of a building, water is turned into steam in a boiler located in a maintenance area or the basement. The steam then flows through insulated pipes to each room in the building. In the radiator, the steam is condensed as liquid water and then flows back through pipes to the boiler to be revaporized. The hot steam physically carries the heat from the boiler, and that energy is released when the steam condenses in the radiator. Some disadvantages of steam heating are that it requires expensive boilers and pipes must contain steam under pressure. Meteorology Connection Meteorology Connection Section 12.1 Temperature and Thermal Energy 317 312-322 CH12-S1-845813 7/21/04 2:55 PM Page 318 Table 12-1 Specific Heat of Common Substances Specific Heat (J/kgK) Material Specific Heat (J/kgK) 897 376 710 385 840 2060 450 Lead Methanol Silver Steam Water Zinc 130 2450 235 2020 4180 388 Material Aluminum Brass Carbon Copper Glass Ice Iron The specific heat of a material is the amount of energy that must be added to the material to raise the temperature of a unit mass by one temperature unit. In SI units, specific heat, represented by C, is measured in J/kgK. Table 12-1 provides values of specific heat for some common substances. For example, 897 J must be added to 1 kg of aluminum to raise its temperature by 1 K. The specific heat of aluminum is therefore 897 J/kgK. The heat gained or lost by an object as its temperature changes depends on the mass, the change in temperature, and the specific heat of the substance. By using the following equation, you can calculate the amount of heat, Q, that must be transferred to change the temperature of an object. Heat Transfer Q mCT mC(Tf Heat transfer is equal to the mass of an object times the specific heat of the object times the difference between the final and initial temperatures. Ti) Liquid water has a high specific heat compared to the other substance in Table 12-1. When the temperature of 10.0 kg of water is increased by 5.0 K, the heat absorbed is Q (10.0 kg)(4180 J/kgK)(5.0 K) = 2.1105 J. Remember that the temperature interval for kelvins is the same as that for Celsius degrees. For this reason, you can calculate T in kelvins or in degrees Celsius. Heat Transfer A 5.10-kg cast-iron skillet is heated on the stove from 295 K to 450 K. How much heat had to be transferred to the iron? 1 Analyze and Sketch the Problem \u2022 Sketch the flow of", " heat into the skillet from the stove top. Known: m = 5.10 kg Ti = 295 K C = 450 J/kgK Tf = 450 K Unknown: Q =? Solve for the Unknown Ti) Q mC(Tf (5.10 kg)(450 J/kgK)(450 K 295 K) 3.6105 J 2 3 Substitute m 5.10 kg, C 450 J/kgK, Tf 450 K, Ti 295 K Evaluate the Answer \u2022 Are the units correct? Heat is measured in J. \u2022 Does the sign make sense? Temperature increased, so Q is positive. Math Handbook Order of Operations page 843 m 5.10 kg Q 295 K 450 K Ti Tf 318 Chapter 12 Thermal Energy 312-322 CH12-S1-845813 6/3/04 2:52 PM Page 319 3. When you turn on the hot water to wash dishes, the water pipes have to heat up. How much heat is absorbed by a copper water pipe with a mass of 2.3 kg when its temperature is raised from 20.0\u00b0C to 80.0\u00b0C? 4. The cooling system of a car engine contains 20.0 L of water (1 L of water has a mass of 1 kg). a. What is the change in the temperature of the water if the engine operates until 836.0 kJ of heat is added? b. Suppose that it is winter, and the car\u2019s cooling system is filled with methanol. The density of methanol is 0.80 g/cm3. What would be the increase in temperature of the methanol if it absorbed 836.0 kJ of heat? c. Which is the better coolant, water or methanol? Explain. 5. Electric power companies sell electricity by the kWh, where 1 kWh 3.6106 J. Suppose that it costs $0.15 per kWh to run an electric water heater in your neighborhood. How much does it cost to heat 75 kg of water from 15\u00b0C to 43\u00b0C to fill a bathtub? Stirrer Thermometer Insulated outer container Lid Calorimetry: Measuring Specific Heat A simple calorimeter, such as the one shown in Figure 12-7, is a device used to measure changes in thermal energy. A calorimeter is carefully insulated so that heat transfer to the external world is kept to a minimum. A measured mass of a substance that has been heated to a high temperature is placed in the calorimeter. The calorimeter also contains a known mass of cold water at a measured temperature. The heat released by the substance is transferred to the cooler water. The change in thermal energy of the substance is calculated using the resulting increase in the water temperature. More sophisticated types of calorimeters are used to measure chemical reactions and the energy content of various foods. Inner container The operation of a calorimeter depends on the conservation of energy in an isolated, closed system. Energy can neither enter nor leave this system. As a result, if the energy of one part of the system increases, the energy of another part of the system must decrease by the same amount. Consider a system composed of two blocks of metal, block A and block B, shown in Figure 12-8a on the next page. The total energy of the system is constant, as represented by the following equation. EB Conservation of Energy EA In an isolated, closed system, the thermal energy of object A plus the thermal energy of object B is constant. constant Water Test substance \u25a0 Figure 12-7 A calorimeter provides an isolated, closed system in which to measure energy transfer. Section 12.1 Temperature and Thermal Energy 319 312-322 CH12-S1-845813 3/22/04 12:16 AM Page 320 \u25a0 Figure 12-8 A system is composed of two model blocks at different temperatures that initially are separated (a). When the blocks are brought together, heat flows from the hotter block to the colder block (b). Total energy remains constant. a b A B EA Insulation EB AB EAB EA EB Suppose that the two blocks initially are separated but can be placed in contact with each other. If the thermal energy of block A changes by an amount EA, then the change in thermal energy of block B, EB, must be related by the equation, EA EB. The change in energy of one block is positive, while the change in energy of the other block is negative. For the block whose thermal energy change is positive, the temperature of the block rises. For the block whose thermal energy change is negative, the temperature falls. 0. Thus, EA EB Assume that the initial temperatures of the two blocks are different. When the blocks are brought together, heat flows from the hotter block to the colder block, as shown in Figure 12-8b. The heat flow continues until the blocks are in thermal equilibrium, which is when the blocks have the same temperature. In", " an isolated, closed system, the change in thermal energy is equal to the heat transferred because no work is done. Therefore, the change in energy for each block can be expressed by the following equation: E Q mCT The increase in thermal energy of block A is equal to the decrease in thermal energy of block B. Thus, the following relationship is true: 0 TB TA mACA mBCB The change in temperature is the difference between the final and initial Ti. If the temperature of a block increases, Ti, and T is positive. If the temperature of the block decreases, Ti, and T is negative. The final temperatures of the two blocks are temperatures; that is, T Tf Tf Tf equal. The following is the equation for the transfer of energy: mACA(Tf TA) mBCB(Tf TB) 0 To solve for Tf, expand the equation. mACATf mACATA mBCBTf mBCBTB 0 Tf(mACA mBCB) mACATA mBCBTB Tf mBCBTB mACATA mACA mBCB 320 Chapter 12 Thermal Energy 312-322 CH12-S1-845813 6/10/04 8:28 PM Page 321 Transferring Heat in a Calorimeter A calorimeter contains 0.50 kg of water at 15\u00b0C. A 0.040-kg block of zinc at 115\u00b0C is placed in the water. What is the final temperature of the system? 1 Analyze and Sketch the Problem \u2022 Let zinc be sample A and water be sample B. \u2022 Sketch the transfer of heat from the hotter zinc to the cooler water. Before block of zinc is placed After block of zinc is placed Water Zinc Unknown:? Tf mB 0.50 kg TB 15\u00b0C mA 0.040 kg TA 115\u00b0C Tf? Known: mA CA TA mB CB TB 0.040 kg 388 J/kg\u00b0C 115\u00b0C 0.50 kg 4180 J/kg\u00b0C 15.0\u00b0C 2 Solve for the Unknown Determine the final temperature using the following equation. Tf mBCBTB mACATA mACA mBCB (0.040 kg)(388 J/kg\u00b0C)(115\u00b0C) (0.50 kg)(4180 J/kg\u00b0C)(15.0\u00b0C) (0.040 kg)(388 J/kg\u00b0C) (0.50 kg)(4180 J/kg\u00b0C) 16\u00b0C Math Handbook Operations with Significant Digits pages 835\u2014836 0.040 kg, Substitute mA CA mB TB 388 J/kg\u00b0C, TA 0.50 kg, CB 15\u00b0C 115\u00b0C, 4180 J/kg\u00b0C, 3 Evaluate the Answer \u2022 Are the units correct? Temperature is measured in Celsius. \u2022 Is the magnitude realistic? The answer is between the initial temperatures of the two samples, as is expected when using a calorimeter. 6. A 2.00102-g sample of water at 80.0\u00b0C is mixed with 2.00102 g of water at 10.0\u00b0C. Assume that there is no heat loss to the surroundings. What is the final temperature of the mixture? 7. A 4.00102-g sample of methanol at 16.0\u00b0C is mixed with 4.00102 g of water at 85.0\u00b0C. Assume that there is no heat loss to the surroundings. What is the final temperature of the mixture? 8. Three lead fishing weights, each with a mass of 1.00102 g and at a temperature of 100.0\u00b0C, are placed in 1.00102 g of water at 35.0\u00b0C. The final temperature of the mixture is 45.0\u00b0C. What is the specific heat of the lead in the weights? 9. A 1.00102-g aluminum block at 100.0\u00b0C is placed in 1.00102 g of water at 10.0\u00b0C. The final temperature of the mixture is 25.0\u00b0C. What is the specific heat of the aluminum? Section 12.1 Temperature and Thermal Energy 321 312-322 CH12-S1-845813 6/3/04 2:52 PM Page 322 \u25a0 Figure 12-9 A lizard regulates its body temperature by hiding under a rock when the atmosphere is hot (a) and sunbathing when the atmosphere gets cold (b). a b Biology Connection Biology Connection Animals can be divided into two groups based on their body temperatures. Most are cold-blooded animals whose body temperatures depend on the environment. The others are warm-blooded animals whose body temperatures are controlled internally. That is, a warm-blooded animal\u2019s body temperature remains stable regardless of the temperature of the environment. In contrast, when the temperature of the environment is high, the body temperature of a cold-blooded animal also becomes", " high. A cold-blooded animal, such as the lizard shown in Figure 12-9, regulates this heat flow by hiding under a rock or crevice, thereby reducing its body temperature. Humans are warm-blooded and have a body temperature of about 37\u00b0C. To regulate its body temperature, a warm-blooded animal increases or decreases the level of its metabolic processes. Thus, a warm-blooded animal may hibernate in winter and reduce its body temperature to approach the freezing point of water. 12.1 Section Review 10. Temperature Make the following conversions. a. 5\u00b0C to kelvins b. 34 K to degrees Celsius c. 212\u00b0C to kelvins d. 316 K to degrees Celsius 11. Conversions Convert the following Celsius tem- peratures to Kelvin temperatures. a. 28\u00b0C b. 154\u00b0C c. 568\u00b0C d. 55\u00b0C e. 184\u00b0C 12. Thermal Energy Could the thermal energy of a bowl of hot water equal that of a bowl of cold water? Explain your answer. 13. Heat Flow On a dinner plate, a baked potato always stays hot longer than any other food. Why? 14. Heat The hard tile floor of a bathroom always feels cold to bare feet even though the rest of the room is warm. Is the floor colder than the rest of the room? 15. Specific Heat If you take a plastic spoon out of a cup of hot cocoa and put it in your mouth, you are not likely to burn your tongue. However, you could very easily burn your tongue if you put the hot cocoa in your mouth. Why? 16. Heat Chefs often use cooking pans made of thick aluminum. Why is thick aluminum better than thin aluminum for cooking? 17. Heat and Food It takes much longer to bake a whole potato than to cook french fries. Why? 18. Critical Thinking As water heats in a pot on a stove, the water might produce some mist above its surface right before the water begins to roll. What is happening, and where is the coolest part of the water in the pot? 322 (l)John Cancalosi/Peter Arnold, Inc., (r)Jenny Hager/The Image Works Chapter 12 Thermal Energy physicspp.com/self_check_quiz 323-331 CH12-S2-845813 6/3/04 2:57 PM Page 323 12.2 Changes of State and the Laws of Thermodynamics Eighteenth-century steam-engine builders used heat to turn liquid water into steam. The steam pushed a piston to turn the engine, and then the steam was cooled and condensed into a liquid again. Adding heat to the liquid water changed not only its temperature, but also its structure. You will learn that changing state means changing form as well as changing the way in which atoms store thermal energy. Changes of State The three most common states of matter are solids, liquids, and gases. As the temperature of a solid is raised, it usually changes to a liquid. At even higher temperatures, it becomes a gas. How can these changes be explained? Consider a material in a solid state. When the thermal energy of the solid is increased, the motion of the particles also increases, as does the temperature. Figure 12-10 diagrams the changes of state as thermal energy is added to 1.0 g of water starting at 243 K (ice) and continuing until it reaches 473 K (steam). Between points A and B, the ice is warmed to 273 K. At some point, the added thermal energy causes the particles to move rapidly enough that their motion overcomes the forces holding the particles together in a fixed location. The particles are still touching each other, but they have more freedom of movement. Eventually, the particles become free enough to slide past each other. Melting point At this point, the substance has changed from a solid to a liquid. The temperature at which this change occurs is the melting point of the substance. When a substance is melting, all of the added thermal energy goes to overcome the forces holding the particles together in the solid state. None of the added thermal energy increases the kinetic energy of the particles. This can be observed between points B and C in Figure 12-10, where the added thermal energy melts the ice at a constant 273 K. Because the kinetic energy of the particles does not increase, the temperature does not increase between points B and C. Boiling point Once a solid is completely melted, there are no more forces holding the particles in the solid state. Adding more thermal energy again increases the motion of the particles, and the temperature of the liquid rises. In the diagram, this process occurs between points C and D. As the temperature increases further, some particles in the liquid acquire enough energy to break free from the other particles. At a specific temperature, known as the boiling point, further addition of energy causes the substance to undergo another change of state. All the added thermal energy converts the substance from the liquid state to the gaseous state", " 373 323 273 243 A Objectives \u2022 Define heats of fusion and vaporization. \u2022 State the first and second laws of thermodynamics. \u2022 Distinguish between heat and work. \u2022 Define entropy. Vocabulary heat of fusion heat of vaporization first law of thermodynamics heat engine entropy second law of thermodynamics \u25a0 Figure 12-10 A plot of temperature versus heat added when 1.0 g of ice is converted to steam. Note that the scale is broken between points D and E. D E Water Steam Steam B C Water Ice Water 61.7 395.7 813.7 3073 Heat (J) Section 12.2 Changes of State and the Laws of Thermodynamics 323 323-331 CH12-S2-845813 6/3/04 2:59 PM Page 324 Table 12-2 Heats of Fusion and Vaporization of Common Substances Heat of Fusion Hf (J/kg) Heat of Vaporization Hv (J/kg) Material Copper Mercury Gold Methanol Iron Silver Lead Water (ice) 2.05105 1.15104 6.30104 1.09105 2.66105 1.04105 2.04104 3.34105 5.07106 2.72105 1.64106 8.78105 6.29106 2.36106 8.64105 2.26106 As in melting, the temperature does not rise while a liquid boils. In Figure 12-10, this transition is represented between points D and E. After the material is entirely converted to gas, any added thermal energy again increases the motion of the particles, and the temperature rises. Above point E, steam is heated to temperatures greater than 373 K. Heat of fusion The amount of energy needed to melt 1 kg of a substance is called the heat of fusion of that substance. For example, the heat of fusion of ice is 3.34105 J/kg. If 1 kg of ice at its melting point, 273 K, absorbs 3.34105 J, the ice becomes 1 kg of water at the same temperature, 273 K. The added energy causes a change in state but not in temperature. The horizontal distance in Figure 12-10 from point B to point C represents the heat of fusion. Heat of vaporization At normal atmospheric pressure, water boils at 373 K. The thermal energy needed to vaporize 1 kg of a liquid is called the heat of vaporization. For water, the heat of vaporization is 2.26106 J/kg. The distance from point D to point E in Figure 12-10 represents the heat of vaporization. Every material has a characteristic heat of vaporization. Between points A and B, there is a definite slope to the line as the temperature is raised. This slope represents the specific heat of the ice. The slope between points C and D represents the specific heat of water, and the slope above point E represents the specific heat of steam. Note that the slope for water is less than those of both ice and steam. This is because water has a greater specific heat than does ice or steam. The heat, Q, required to melt a solid of mass m is given by the following equation. Heat Required to Melt a Solid Q mHf The heat required to melt a solid is equal to the mass of the solid times the heat of fusion of the solid. Similarly, the heat, Q, required to vaporize a mass, m, of liquid is given by the following equation. Heat Required to Vaporize a Liquid Q mHv The heat required to vaporize a liquid is equal to the mass of the liquid times the heat of vaporization of the liquid. When a liquid freezes, an amount of heat, Q mHf, must be removed from the liquid to turn it into a solid. The negative sign indicates that the heat is transferred from the sample to the external world. In the same way, when a vapor condenses to a liquid, an amount of heat, Q mHv, must be removed from the vapor. The values of some heats of fusion, Hf, and heats of vaporization, Hv, are shown in Table 12-2. Melting 1. Label two foam cups A and B. 2. Measure and pour 75 mL of room-temperature water into each cup. Wipe up any spilled liquid. 3. Add an ice cube to cup A, and add ice water to cup B until the water levels are equal. 4. Measure the temperature of the water in each cup at 1-min intervals until the ice has melted. 5. Record the temperatures in a data table and plot a graph. Analyze and Conclude 6. Do the samples reach the same final temperature? Why? 324 Chapter 12 Thermal Energy 323-331 CH12-S2-845813 6/3/04 3:00 PM Page 325 Heat Suppose that you are camping in the mountains. You need to melt 1.50 kg of snow at 0.0\u00b0C and heat it to 70.0", "\u00b0C to make hot cocoa. How much heat will be needed? 1 Analyze and Sketch the Problem \u2022 Sketch the relationship between heat and water in its solid and liquid states. \u2022 Sketch the transfer of heat as the temperature of the water increases. m 1.50 kg 0.0\u00b0C Ti C 4180 J/kg\u00b0C Known: Hf Tf 3.34105 J/kg 70.0\u00b0C Unknown:? Qmelt ice Qheat liquid? Qtotal? 2 Solve for the Unknown Calculate the heat needed to melt ice. Qmelt ice mHf (1.50 kg)(3.34105 J/kg) 5.01105 J 5.01102 kJ Calculate the temperature change. T Tf Ti 70.0\u00b0C 0.0\u00b0C 70.0\u00b0C Substitute m 1.50 kg, Hf 3.34105 J/kg Substitute Tf 70.0\u00b0C, Ti 0.0\u00b0C 1.5 kg Snow Q Hf 0.0\u00b0C Ti 70.0\u00b0C Tf Math Handbook Operations with Scientific Notation pages 842\u2014843 Calculate the heat needed to raise the water temperature. Qheat liquid mCT (1.50 kg)(4180 J/kg\u00b0C)(70.0\u00b0C) 4.39105 J 4.39102 kJ Calculate the total amount of heat needed. Substitute m 1.50 kg, C = 4180 J/kg\u00b0C, T 70.0\u00b0C Qtotal Qmelt ice + Qheat liquid 5.01102 kJ + 4.39102 kJ 9.40102 kJ Substitute Qmelt ice 5.01102 kJ, Qheat liquid 4.39102 kJ 3 Evaluate the Answer \u2022 Are the units correct? Energy units are in joules. \u2022 Does the sign make sense? Q is positive when heat is absorbed. \u2022 Is the magnitude realistic? The amount of heat needed to melt the ice is greater than the amount of heat needed to increase the water temperature by 70.0\u00b0C. It takes more energy to overcome the forces holding the particles in the solid state than to raise the temperature of water. 19. How much heat is absorbed by 1.00102 g of ice at 20.0\u00b0C to become water at 0.0\u00b0C? 20. A 2.00102-g sample of water at 60.0\u00b0C is heated to steam at 140.0\u00b0C. How much heat is absorbed? 21. How much heat is needed to change 3.00102 g of ice at 30.0\u00b0C to steam at 130.0\u00b0C? Section 12.2 Changes of State and the Laws of Thermodynamics 325 323-331 CH12-S2-845813 6/10/04 4:58 PM Page 326 The First Law of Thermodynamics Before thermal energy was linked to the motion of atoms, the study of heat and temperature was considered to be a separate science. The first law developed for this science was a statement about what thermal energy is and where it can go. As you know, you can heat a nail by holding it over a flame or by pounding it with a hammer. That is, you can increase the nail\u2019s thermal energy by adding heat or by doing work on it. We do not normally think that the nail does work on the hammer. However, the work done by the nail on the hammer is equal to the negative of the work done by the hammer on the nail. The first law of thermodynamics states that the change in thermal energy, U, of an object is equal to the heat, Q, that is added to the object minus the work, W, done by the object. Note that U, Q, and W are all measured in joules, the unit of energy. The First Law of Thermodynamics U Q W The change in thermal energy of an object is equal to the heat added to the object minus the work done by the object. Thermodynamics also involves the study of the changes in thermal properties of matter. The first law of thermodynamics is merely a restatement of the law of conservation of energy, which states that energy is neither created nor destroyed, but can be changed into other forms. Another example of changing the amount of thermal energy in a system is a hand pump used to inflate a bicycle tire. As a person pumps, the air and the hand pump become warm. The mechanical energy in the moving piston is converted into thermal energy of the gas. Similarly, other forms of energy, such as light, sound, and electric energy, can be changed into thermal energy. For example, a toaster converts electric energy into heat when it toasts bread, and the Sun warms Earth with light from a distance of over 150 million km away. Heat engines The warmth that you experience when you rub your hands together is a result of", " the conversion of mechanical energy into thermal energy. The conversion of mechanical energy into thermal energy occurs easily. However, the reverse process, the conversion of thermal energy into mechanical energy, is more difficult. A device that is able to continuously convert thermal energy to mechanical energy is called a heat engine. A heat engine requires a high-temperature source from which thermal energy can be removed; a low-temperature receptacle, called a sink, into which thermal energy can be delivered; and a way to convert the thermal energy into work. A diagram of a heat engine is shown in Figure 12-11. An automobile internal-combustion engine, such as the one shown in Figure 12-12, is one example of a heat engine. In the engine, a mixture of air and gasoline vapor is ignited and produces a high-temperature flame. Input heat, QH, flows from the flame to the air in the cylinder. The hot air expands and pushes on a piston, thereby changing thermal energy into mechanical energy. To obtain continuous mechanical energy, the engine must be returned to its starting condition. The heated air is expelled and replaced by new air, and the piston is returned to the top of the cylinder. Hot reservoir at TH QH W Heat engine QL QH W QL Cold reservoir at TL \u25a0 Figure 12-11 A heat engine transforms heat at high temperature into mechanical energy and low-temperature waste heat. 326 Chapter 12 Thermal Energy 323-331 CH12-S2-845813 3/22/04 12:24 AM Page 327 Spark plug Air and gasoline vapor Piston Intake Compression Spark Power Exhaust Exhaust \u25a0 Figure 12-12 The heat produced by burning gasoline causes the gases that are produced to expand and to exert force and do work on the piston. The entire cycle is repeated many times each minute. The thermal energy from the burning of gasoline is converted into mechanical energy, which eventually results in the movement of the car. Not all of the thermal energy from the high-temperature flame in an automobile engine is converted into mechanical energy. When the automobile engine is functioning, the exhaust gases and the engine parts become hot. As the exhaust comes in contact with outside air and transfers heat to it, the temperature of the outside air is raised. In addition, heat from the engine is transferred to a radiator. Outside air passes through the radiator and the air temperature is raised. All of this energy, QL, transferred out of the automobile engine is called waste heat, that is, heat that has not been converted into work. When the engine is working continuously, the internal energy of the engine does not change, or U 0 Q W. The net heat going into the engine is Q QH QL. In an automobile engine, the thermal energy in the flame produces the mechanical energy and the waste heat that is expelled. All heat engines generate waste heat, and therefore no engine can ever convert all of the energy into useful motion or work. QL. Thus, the work done by the engine is W QH Efficiency Engineers and car salespeople often talk about the fuel efficiency of automobile engines. They are referring to the amount of the input heat, QH, that is turned into useful work, W. The actual efficiency of an engine is given by the ratio W/QH. The efficiency could equal 100 percent only if all of the input heat were turned into work by the engine. Because there is always waste heat, even the most efficient engines fall short of 100-percent efficiency. In solar collectors, heat is collected at high temperatures and used to drive engines. The Sun\u2019s energy is transmitted as electromagnetic waves and increases the internal energy of the solar collectors. This energy is then transmitted as heat to the engine, where it is turned into useful work and waste heat. Refrigerators Heat flows spontaneously from a warm object to a cold object. However, it is possible to remove thermal energy from a colder object and add it to a warmer object if work is done. A refrigerator is a common example of a device that accomplishes this transfer with the use of mechanical work. Electric energy runs a motor that does work on a gas and compresses it. Section 12.2 Changes of State and the Laws of Thermodynamics 327 323-331 CH12-S2-845813 3/22/04 12:24 AM Page 328 Hot reservoir at TH QH Refrigerator W QL QH W QL Cold reservoir at TL The gas draws heat from the interior of the refrigerator, passes from the compressor through the condenser coils on the outside of the refrigerator, and cools into a liquid. Thermal energy is transferred into the air in the room. The liquid reenters the interior, vaporizes, and absorbs thermal energy from its surroundings. The gas returns to the compressor and the process is repeated. The overall change in the thermal energy of the gas is zero. Thus, according to the first law of thermodynamics, the sum of the heat removed from the refrigerator\u2019s contents", " and the work done by the motor is equal to the heat expelled, as shown in Figure 12-13. Heat pumps A heat pump is a refrigerator that can be run in two directions. In the summer, the pump removes heat from a house and thus cools the house. In the winter, heat is removed from the cold outside air and transferred into the warmer house. In both cases, mechanical energy is required to transfer heat from a cold object to a warmer one. \u25a0 Figure 12-13 A refrigerator absorbs heat, QL, from the cold reservoir and gives off heat, QH, to the hot reservoir. Work, W, is done on the refrigerator. 328 Chapter 12 Thermal Energy 22. A gas balloon absorbs 75 J of heat. The balloon expands but stays at the same temperature. How much work did the balloon do in expanding? 23. A drill bores a small hole in a 0.40-kg block of aluminum and heats the aluminum by 5.0\u00b0C. How much work did the drill do in boring the hole? 24. How many times would you have to drop a 0.50-kg bag of lead shot from a height of 1.5 m to heat the shot by 1.0\u00b0C? 25. When you stir a cup of tea, you do about 0.050 J of work each time you circle the spoon in the cup. How many times would you have to stir the spoon to heat a 0.15-kg cup of tea by 2.0\u00b0C? 26. How can the first law of thermodynamics be used to explain how to reduce the temperature of an object? The Second Law of Thermodynamics Many processes that are consistent with the first law of thermodynamics have never been observed to occur spontaneously. Three such processes are presented in Figure 12-14. For example, the first law of thermodynamics does not prohibit heat flowing from a cold object to a hot object. However, when hot objects have been placed in contact with cold objects, the hot objects have never been observed to become hotter. Similarly, the cold objects have never been observed to become colder. Entropy If heat engines completely converted thermal energy into mechanical energy with no waste heat, then the first law of thermodynamics would be obeyed. However, waste heat is always generated, and randomly distributed particles of a gas are not observed to spontaneously arrange themselves in specific ordered patterns. In the nineteenth century, French engineer Sadi Carnot studied the ability of engines to convert thermal energy into mechanical energy. He developed a logical proof that even an ideal engine would generate some waste heat. Carnot\u2019s result is best described in terms of a quantity called entropy, which is a measure of the disorder in a system. 323-331 CH12-S2-845813 6/3/04 3:09 PM Page 329 Consistent with the First Law of Thermodynamics but do not occur spontaneously Occur spontaneously Hot Cold Hot Cold Q Q \u25a0 Figure 12-14 Many processes that do not violate the first law of thermodynamics do not occur spontaneously. The spontaneous processes obey both the first and second law of thermodynamics. Heat engine QH W Heat engine QH W QL Time Time When a baseball is dropped and falls due to gravity, it possesses potential and kinetic energies that can be recovered to do work. However, when the baseball falls through the air, it collides with many air molecules that absorb some of its energy. This causes air molecules to move in random directions and at random speeds. The energy absorbed from the baseball causes more disorder among the molecules. The greater the range of speeds exhibited by the molecules, the greater the disorder, which in turn increases the entropy. It is highly unlikely that the molecules that have been dispersed in all directions will come back together, give their energies back to the baseball, and cause it to rise. Entropy, like thermal energy, is contained in an object. If heat is added to an object, entropy is increased. If heat is removed from an object, entropy is decreased. If an object does work with no change in temperature, the entropy does not change, as long as friction is ignored. The change in entropy, S, is expressed by the following equation, in which entropy has units of J/K and the temperature is measured in kelvins. Change in Entropy S Q T The change in entropy of an object is equal to the heat added to the object divided by the temperature of the object in kelvins. Entropy has some interesting properties. Compare the following situations. Explain how and why these changes in entropy are different. 1. Heating 1.0 kg of water from 273 K to 274 K. 2. Heating 1.0 kg of water from 353 K to 354 K. 3. Completely melting 1.0 kg of ice at 273 K. 4. Heating 1.0 kg of lead from 273 K to 274 K. Q 1 kg 1 kg Ti Tf Section 12.2 Changes of State and the Laws of Ther", "modynamics 329 323-331 CH12-S2-845813 3/22/04 12:25 AM Page 330 \u25a0 Figure 12-15 The spontaneous mixing of the food coloring and water is an example of the second law of thermodynamics. The second law of thermodynamics states that natural processes go in a direction that maintains or increases the total entropy of the universe. That is, all things will become more and more disordered unless some action is taken to keep them ordered. The increase in entropy and the second law of thermodynamics can be thought of as statements of the probability of events happening. Figure 12-15 illustrates an increase in entropy as food-coloring molecules, originally separate from the clear water, are thoroughly mixed with the water molecules over time. Figure 12-16 shows an example of the second law of thermodynamics that might be familiar to many teenagers. The second law of thermodynamics predicts that heat flows spontaneously only from a hot object to a cold object. Consider a hot iron bar and a cold cup of water. On the average, the particles in the iron will be moving very fast, whereas the particles in the water will be moving slowly. When the bar is plunged into the water and thermal equilibrium is eventually reached, the average kinetic energy of the particles in the iron and the water will be the same. More particles now have an increased random motion than was true for the initial state. This final state is less ordered than the initial state. The fast particles are no longer confined solely to the iron, and the slower particles are no longer confined only to the water; all speeds are evenly distributed. The entropy of the final state is greater than that of the initial state. Violations of the second law We take for granted many daily events that occur spontaneously, or naturally, in one direction. We would be shocked, however, if the reverse of the same events occurred spontaneously. You are not surprised when a metal spoon, heated at one end, soon becomes uniformly hot. Consider your reaction, however, if a spoon lying on a table suddenly, on its own, became red hot at one end and icy cold at the other. If you dive into a swimming pool, you take for granted that you push the water molecules away as you enter the water. However, you would be amazed if you were swimming in the pool and all the water molecules spontaneously threw you up onto the diving board. Neither of these imagined reverse processes would violate the first law of thermodynamics. They are simply examples of the countless events that do not occur because their processes would violate the second law of thermodynamics. \u25a0 Figure 12-16 If no work is done on a system, entropy spontaneously reaches a maximum. 330 (t)Doug Martin, (others)Richard Hutchings/CORBIS Chapter 12 Thermal Energy 323-331 CH12-S2-845813 6/3/04 3:10 PM Page 331 The second law of thermodynamics and the increase in entropy also give new meaning to what has been commonly called the energy crisis. The energy crisis refers to the continued use of limited resources of fossil fuels, such as natural gas and petroleum. When you use a resource, such as natural gas to heat your home, you do not use up the energy in the gas. As the gas ignites, the internal chemical energy contained in the molecules of the gas is converted into thermal energy of the flame. The thermal energy of the flame is then transferred to thermal energy in the air of your home. Even if this warm air leaks to the outside, the energy is not lost. Energy has not been used up. The entropy, however, has increased. The chemical structure of natural gas is very ordered. As you have learned, when a substance becomes warmer, the average kinetic energy of the particles in the substance increases. In contrast, the random motion of warmed air is very disordered. While it is mathematically possible for the original chemical order to be reestablished, the probability of this occurring is essentially zero. For this reason, entropy often is used as a measure of the unavailability of useful energy. The energy in the warmed air in a home is not as available to do mechanical work or to transfer heat to other objects as the original gas molecules were. The lack of usable energy is actually a surplus of entropy. 12.2 Section Review 27. Heat of Vaporization Old-fashioned heating systems sent steam into radiators in each room of a house. In the radiators, the steam condensed back to water. Analyze this process and explain how it heated a room. 28. Heat of Vaporization How much heat is needed to change 50.0 g of water at 80.0\u00b0C to steam at 110.0\u00b0C? 29. Heat of Vaporization The specific heat of mercury is 140 J/kg\u00b0C. Its heat of vaporization is 3.06105 J/kg. How much energy is needed to heat 1.0 kg of mercury metal from 10.0\u00b0C to its boiling point and vaporize it completely? The", " boiling point of mercury is 357\u00b0C. 30. Mechanical Energy and Thermal Energy James Joule carefully measured the difference in temperature of water at the top and bottom of a waterfall. Why did he expect a difference? 32. Mechanical Energy and Thermal Energy Water flows over a fall that is 125.0 m high, as shown in Figure 12-17. If the potential energy of the water is all converted to thermal energy, calculate the temperature difference between the water at the top and the bottom of the fall. 125.0 m \u25a0 Figure 12-17 33. Entropy Evaluate why heating a home with natural gas results in an increased amount of disorder. 31. Mechanical Energy and Thermal Energy A man uses a 320-kg hammer moving at 5.0 m/s to smash a 3.0-kg block of lead against a 450-kg rock. When he measured the temperature he found that it had increased by 5.0\u00b0C. Explain how this happened. 34. Critical Thinking A new deck of cards has all the suits (clubs, diamonds, hearts, and spades) in order, and the cards are ordered by number within the suits. If you shuffle the cards many times, are you likely to return the cards to their original order? Explain. Of what physical law is this an example? physicspp.com/self_check_quiz Section 12.2 Changes of State and the Laws of Thermodynamics 331 332-333 CH12-LAB-845813 6/3/04 3:15 PM Page 332 Alternate CBL instructions can be found on the Web site. physicspp.com Heating and Cooling When a beaker of water is set on a hot plate and the hot plate is turned on, heat is transferred. It first is transferred to the beaker and then to the water at the bottom of the beaker by conduction. The water then transfers heat from the bottom to the top by moving hot water to the top through convection. Once the heat source is removed or shut off, the water radiates thermal energy until it reaches room temperature. How quickly the water heats up is a function of the amount of heat added, the mass of the water, and the specific heat of water. QUESTION How does the constant supply of thermal energy affect the temperature of water? Objectives Procedure \u25a0 Measure, in SI, temperature and mass. \u25a0 Make and use graphs to help describe the change in temperature of water as it heats up and cools down. \u25a0 Explain any similarities and differences in these two changes. Safety Precautions \u25a0 Be careful when using a hot plate. It can burn the skin. Materials hot plate (or Bunsen burner) 250-mL ovenproof glass beaker 50\u2013200 g of water two thermometers (non-mercury) stopwatch (or timer) 1. Set the hot plate to the highest setting, or as recommended by your teacher. Allow a few minutes for the plate to heat up. 2. Measure the mass of the empty beaker. 3. Pour 150 mL of water into the beaker and measure the combined mass of the water and the beaker. 4. Calculate and record the mass of the water in the beaker. 5. Create a data and observations table. 6. Record the initial temperature of the water and the air in the classroom. Note that the bulb end of the thermometers must not touch the bottom or sides of the beaker, nor should it touch a table or your hands. 7. Place the beaker on the hot plate and record the temperature every minute for 5 min. 8. Carefully remove the beaker from the hot plate and record the temperature every minute for the next 10 min. 9. At the end of 10 min, record the temperature of the air. 10. Turn off the hot plate. 11. When finished, allow the equipment to cool and dispose of the water as instructed by your teacher. 332 Horizons Companies 332-333 CH12-LAB-845813 6/3/04 3:16 PM Page 333 Data Table Mass of water Initial air temperature Final air temperature Change in air temperature Time (min) Temperature (\u00b0C) Heating or Cooling Analyze Going Further 1. Calculate the change in air temperature to determine if air temperature may be an extraneous variable. 2. Make a scatter-plot graph of temperature (vertical axis) versus time (horizontal axis). Use a computer or a calculator to construct the graph, if possible. 1. Does placing your thermometer at the top of the water in your beaker result in different readings than if it is placed at the bottom of the beaker? Explain. 2. Hypothesize what the temperature changes might look like if you had the following amounts of water in the beaker: 50 mL, 250 mL. 3. Calculate What was the change in water tem- 3. Suppose you insulated the beaker you were perature as the water heated up? 4. Calculate What was the drop in", " water temperature when the heat source was removed? 5. Calculate the average slope for the temperature increase by dividing change in temperature by the amount of time the water was heating up. 6. Calculate the average slope for the temperature decrease by dividing change in temperature by the amount of time the heat source was removed. Conclude and Apply 1. Summarize What was the change in water temperature when a heat source was applied? 2. Summarize What was the change in water temperature once the heat source was removed? 3. What would happen to the water temperature after the next 10 min? Would it continue cooling down forever? 4. Did the water appear to heat up or cool down quicker? Why do you think this is so? Hint: Examine the slopes you calculated. 5. Hypothesize Where did the thermal energy in the water go once the water began to cool down? Support your hypothesis. using. How would the beaker\u2019s ability to heat up and cool down be affected? Real-World Physics 1. Suppose you were to use vegetable oil in the beaker instead of water. Hypothesize what the temperature changes might look like if you were to follow the same steps and perform the experiment. 2. If you were to take soup at room temperature and cook it in a microwave oven for 3 min, would the soup return to room temperature in 3 min? Explain your answer. To find out more about thermal energy, visit the Web site: physicspp.com 333 334 CH12-FEATURE-845813 6/4/04 4:28 AM Page 334 The Heat Pump Heat pumps, also called reversible air conditioners, were invented in the 1940s. They are used to heat and cool homes and hotel rooms. Heat pumps change from heaters to air conditioners by reversing the flow of refrigerant through the system. 1 Cooling The thin capillary tube sprays liquid refrigerant into a larger coil inside. Inside Outside Receiver: tank stores refrigerant Air grating 5 The fan cools the coil during cooling and warms the coil during heating. Airflow to rooms Fan motor Airflow from rooms Fan motor Unit cabinet Air grating 1 3 4 2 Compressor pump refrigerant 4 Heating Valves 3 and 4 are opened and valves 1 and 2 are closed for heating. The refrigerant flows upward. The inside coil functions as a condenser and the outside coil functions as an evaporator. 3 Heating The thin capillary tube sprays liquid refrigerant into a larger diameter pipe in an outer coil for heating. 2 Cooling Valves 1 and 2 are opened and valves 3 and 4 are closed for cooling. The refrigerant flows downward. The inside coil functions as an evaporator and the outside coil functions as a condenser. Thinking Critically 1. Observe Trace the flow of refrigerant through the entire system for both heating and cooling. Start at the compressor. 2. Analyze Would a heat pump be able to heat an entire house when the outside temperature drops to extremely cold levels? 334 How It Works 335-339 CH12-SG CA STP-845813 3/22/04 12:37 AM Page 335 12.1 Temperature and Thermal Energy Vocabulary \u2022 conduction (p. 315) \u2022 thermal equilibrium (p. 315) \u2022 heat (p. 317) \u2022 convection (p. 317) \u2022 radiation (p. 317) \u2022 specific heat (p. 318) Key Concepts \u2022 The temperature of a gas is proportional to the average kinetic energy of its particles. \u2022 Thermal energy is a measure of the internal motion of an object\u2019s particles. \u2022 A thermometer reaches thermal equilibrium with the object that it comes in contact with, and then a temperature-dependent property of the thermometer indicates the temperature. \u2022 The Celsius and Kelvin temperature scales are used in scientific work. The magnitude of 1 K is equal to the magnitude of 1\u00b0C. \u2022 At absolute zero, no more thermal energy can be removed from a substance. \u2022 Heat is energy transferred because of a difference in temperature. Q mCT mC(Tf Ti) \u2022 Specific heat is the quantity of heat required to raise the temperature of 1 kg of a substance by 1 K. \u2022 In a closed, isolated system, heat may flow and change the thermal energy of parts of the system, but the total energy of the system is constant. EA EB constant 12.2 Changes of State and the Laws of Thermodynamics Vocabulary \u2022 heat of fusion (p. 324) \u2022 heat of vaporization (p. 324) \u2022 first law of thermodynamics (p. 326) \u2022 heat engine (p. 326) \u2022 entropy (p. 328) \u2022 second law of thermodynamics (p. 330) Key Concepts \u2022 The heat of fusion is the quantity of heat needed to change 1 kg of a substance from its solid to liquid state at its melting point. Q mHf \u2022 The heat of vaporization is the quantity of heat needed to change 1 kg of a substance from its liquid to", " gaseous state at its boiling point. Q mHv \u2022 Heat transferred during a change of state does not change the temperature of a substance. \u2022 The change in energy of an object is the sum of the heat added to it minus the work done by the object. U Q W \u2022 A heat engine continuously converts thermal energy to mechanical energy. \u2022 A heat pump and a refrigerator use mechanical energy to transfer heat from a region of lower temperature to one of higher temperature. \u2022 Entropy is a measure of the disorder of a system. \u2022 The change in entropy of an object is defined to be the heat added to the object divided by the temperature of the object. S Q T physicspp.com/vocabulary_puzzlemaker 335 335-339 CH12-SG CA STP-845813 3/22/04 12:37 AM Page 336 Concept Mapping 35. Complete the following concept map using the following terms: heat, work, internal energy. First law of thermodynamics entropy temperature external forces Mastering Concepts 36. Explain the differences among the mechanical energy of a ball, its thermal energy, and its temperature. (12.1) 37. Can temperature be assigned to a vacuum? Explain. (12.1) 38. Do all of the molecules or atoms in a liquid have the same speed? (12.1) 39. Is your body a good judge of temperature? On a cold winter day, a metal doorknob feels much colder to your hand than a wooden door does. Explain why this is true. (12.1) 40. When heat flows from a warmer object in contact with a colder object, do the two have the same temperature changes? (12.1) 41. Can you add thermal energy to an object without increasing its temperature? Explain. (12.2) 42. When wax freezes, does it absorb or release energy? (12.2) 43. Explain why water in a canteen that is surrounded by dry air stays cooler if it has a canvas cover that is kept wet. (12.2) 44. Which process occurs at the coils of a running air conditioner inside a house, vaporization or condensation? Explain. (12.2) Applying Concepts 45. Cooking Sally is cooking pasta in a pot of boiling water. Will the pasta cook faster if the water is boiling vigorously or if it is boiling gently? 46. Which liquid would an ice cube cool faster, water or methanol? Explain. 47. Equal masses of aluminum and lead are heated to the same temperature. The pieces of metal are placed on a block of ice. Which metal melts more ice? Explain. 48. Why do easily vaporized liquids, such as acetone and methanol, feel cool to the skin? 49. Explain why fruit growers spray their trees with water when frost is expected to protect the fruit from freezing. 50. Two blocks of lead have the same temperature. Block A has twice the mass of block B. They are dropped into identical cups of water of equal temperatures. Will the two cups of water have equal temperatures after equilibrium is achieved? Explain. 51. Windows Often, architects design most of the windows of a house on the north side. How does putting windows on the south side affect the heating and cooling of the house? Mastering Problems 12.1 Temperature and Thermal Energy 52. How much heat is needed to raise the temperature of 50.0 g of water from 4.5\u00b0C to 83.0\u00b0C? 53. A 5.00102-g block of metal absorbs 5016 J of heat when its temperature changes from 20.0\u00b0C to 30.0\u00b0C. Calculate the specific heat of the metal. 54. Coffee Cup A 4.00102-g glass coffee cup is 20.0\u00b0C at room temperature. It is then plunged into hot dishwater at a temperature of 80.0\u00b0C, as shown in Figure 12-18. If the temperature of the cup reaches that of the dishwater, how much heat does the cup absorb? Assume that the mass of the dishwater is large enough so that its temperature does not change appreciably. 20.0\u00b0C 80.0\u00b0C 4.00102 g \u25a0 Figure 12-18 55. A 1.00102-g mass of tungsten at 100.0\u00b0C is placed in 2.00102 g of water at 20.0\u00b0C. The mixture reaches equilibrium at 21.6\u00b0C. Calculate the specific heat of tungsten. 336 Chapter 12 Thermal Energy For more problems, go to Additional Problems, Appendix B. 335-339 CH12-SG CA STP-845813 6/3/04 3:20 PM Page 337 56. A 6.0102-g sample of water at 90.0\u00b0C is mixed with 4.00102 g of water at 22.0\u00b0C. Assume that there is no heat loss to the surroundings. What is the final temperature of", " the mixture? 57. A 10.0-kg piece of zinc at 71.0\u00b0C is placed in a container of water, as shown in Figure 12-19. The water has a mass of 20.0 kg and a temperature of 10.0\u00b0C before the zinc is added. What is the final temperature of the water and the zinc? 20.0 kg 10.0\u00b0C \u25a0 Figure 12-19 g 0.0 k 1 58. The kinetic energy of a compact car moving at 100 km/h is 2.9105 J. To get a feeling for the amount of energy needed to heat water, what volume of water (in liters) would 2.9105 J of energy warm from room temperature (20.0\u00b0C) to boiling (100.0\u00b0C)? 59. Water Heater A 3.0102-W electric immersion heater is used to heat a cup of water, as shown in Figure 12-20. The cup is made of glass, and its mass is 3.00102 g. It contains 250 g of water at 15\u00b0C. How much time is needed to bring the water to the boiling point? Assume that the temperature of the cup is the same as the temperature of the water at all times and that no heat is lost to the air. 3.00102 W 15\u00b0C 250 g 3.00102 g \u25a0 Figure 12-20 60. Car Engine A 2.50102-kg cast-iron car engine contains water as a coolant. Suppose that the engine\u2019s temperature is 35.0\u00b0C when it is shut off, and the air temperature is 10.0\u00b0C. The heat given off by the engine and water in it as they cool to air temperature is 4.40106 J. What mass of water is used to cool the engine? 12.2 Changes of State and the Laws of Thermodynamics 61. Years ago, a block of ice with a mass of about 20.0 kg was used daily in a home icebox. The temperature of the ice was 0.0\u00b0C when it was delivered. As it melted, how much heat did the block of ice absorb? 62. A 40.0-g sample of chloroform is condensed from a vapor at 61.6\u00b0C to a liquid at 61.6\u00b0C. It liberates 9870 J of heat. What is the heat of vaporization of chloroform? 63. A 750-kg car moving at 23 m/s brakes to a stop. The brakes contain about 15 kg of iron, which absorbs the energy. What is the increase in temperature of the brakes? 64. How much heat is added to 10.0 g of ice at 20.0\u00b0C to convert it to steam at 120.0\u00b0C? 65. A 4.2-g lead bullet moving at 275 m/s strikes a steel plate and comes to a stop. If all its kinetic energy is converted to thermal energy and none leaves the bullet, what is its temperature change? 66. Soft Drink A soft drink from Australia is labeled \u201cLow-Joule Cola.\u201d The label says \u201c100 mL yields 1.7 kJ.\u201d The can contains 375 mL of cola. Chandra drinks the cola and then wants to offset this input of food energy by climbing stairs. How high would Chandra have to climb if she has a mass of 65.0 kg? Mixed Review 67. What is the efficiency of an engine that produces 2200 J/s while burning enough gasoline to produce 5300 J/s? How much waste heat does the engine produce per second? 68. Stamping Press A metal stamping machine in a factory does 2100 J of work each time it stamps out a piece of metal. Each stamped piece is then dipped in a 32.0-kg vat of water for cooling. By how many degrees does the vat heat up each time a piece of stamped metal is dipped into it? 69. A 1500-kg automobile comes to a stop from 25 m/s. All of the energy of the automobile is deposited in the brakes. Assuming that the brakes are about 45 kg of aluminum, what would be the change in temperature of the brakes? physicspp.com/chapter_test Chapter 12 Assessment 337 335-339 CH12-SG CA STP-845813 6/3/04 3:33 PM Page 338 70. Iced Tea To make iced tea, you start by brewing the tea with hot water. Then you add ice. If you start with 1.0 L of 90\u00b0C tea, what is the minimum amount of ice needed to cool it to 0\u00b0C? Would it be better to let the tea cool to room temperature before adding the ice? 71. A block of copper at 100.0\u00b0C comes in contact with a block of aluminum at 20.0\u00b0C, as shown in Figure 12-21. The final temperature", " of the blocks is 60.0\u00b0C. What are the relative masses of the blocks? 100.0\u00b0C 20.0\u00b0C Copper Aluminum 60.0\u00b0C 60.0\u00b0C Copper Aluminum \u25a0 Figure 12-21 72. A 0.35-kg block of copper sliding on the floor hits an identical block moving at the same speed from the opposite direction. The two blocks come to a stop together after the collision. Their temperatures increase by 0.20\u00b0C as a result of the collision. What was their velocity before the collision? 73. A 2.2-kg block of ice slides across a rough floor. Its initial velocity is 2.5 m/s and its final velocity is 0.50 m/s. How much of the ice block melted as a result of the work done by friction? Thinking Critically 74. Analyze and Conclude A certain heat engine removes 50.0 J of thermal energy from a hot reservoir at temperature TH 40.0 J of heat to a colder reservoir at temperature 325 K. In the process, it also transfers entropy TL from one reservoir to the other. a. How does the operation of the engine change 545 K and expels the total entropy of the reservoirs? b. What would be the total entropy change in the reservoirs if TL 205 K? 75. Analyze and Conclude During a game, the metabolism of basketball players often increases by as much as 30.0 W. How much perspiration must a player vaporize per hour to dissipate this extra thermal energy? 76. Analyze and Conclude Chemists use calorimeters to measure the heat produced by chemical reactions. For instance, a chemist dissolves 1.01022 molecules of a powdered substance into a calorimeter containing 0.50 kg of water. The molecules break up and release their binding energy to the water. The water temperature increases by 2.3\u00b0C. What is the binding energy per molecule for this substance? 77. Apply Concepts All of the energy on Earth comes from the Sun. The surface temperature of the Sun is approximately 104 K. What would be the effect on our world if the Sun\u2019s surface temperature were 103 K? Writing in Physics 78. Our understanding of the relationship between heat and energy was influenced by a soldier named Benjamin Thompson, Count Rumford; and a brewer named James Prescott Joule. Both relied on experimental results to develop their ideas. Investigate what experiments they did and evaluate whether or not it is fair that the unit of energy is called the Joule and not the Thompson. 79. Water has an unusually large specific heat and large heats of fusion and vaporization. Our weather and ecosystems depend upon water in all three states. How would our world be different if water\u2019s thermodynamic properties were like other materials, such as methanol? Cumulative Review 80. A rope is wound around a drum with a radius of 0.250 m and a moment of inertia of 2.25 kg m2. The rope is connected to a 4.00-kg block. (Chapter 8) a. Find the linear acceleration of the block. b. Find the angular acceleration of the drum. c. Find the tension, FT, in the rope. d. Find the angular velocity of the drum after the block has fallen 5.00 m. 81. A weight lifter raises a 180-kg barbell to a height of 1.95 m. How much work is done by the weight lifter in lifting the barbell? (Chapter 10) 82. In a Greek myth, the man Sisyphus is condemned by the gods to forever roll an enormous rock up a hill. Each time he reaches the top, the rock rolls back down to the bottom. If the rock has a mass of 215 kg, the hill is 33 m in height, and Sisyphus can produce an average power of 0.2 kW, how many times in 1 h can he roll the rock up the hill? (Chapter 11) 338 Chapter 12 Thermal Energy For more problems, go to Additional Problems, Appendix B. 335-339 CH12-SG CA STP-845813 7/14/04 1:34 PM Page 339 Multiple Choice 1. Which of the following temperature conversions is incorrect? 273\u00b0C 0 K 273\u00b0C 546 K 298 K 571\u00b0C 88 K 185\u00b0C 2. What are the units of entropy? J/K K/J J kJ 3. Which of the following statements about thermal equilibrium is false? When two objects are at equilibrium, heat radiation between the objects continues to occur. Thermal equilibrium is used to create energy in a heat engine. The principle of thermal equilibrium is used for calorimetry calculations. When two objects are not at equilibrium, heat will flow from the hotter object to the cooler object. 4. How much heat is required to heat 87 g of methanol ice at 14 K to vapor at 340 K? (melting point 97.6\u00b0C, boiling", " point 64.6\u00b0C) 17 kJ 69 kJ 1.4102 kJ 1.5102 kJ 5. Which statement is true about energy, entropy, and changes of state? Freezing ice increases in energy as it gains molecular order as a solid. The higher the specific heat capacity of a substance, the higher its melting point will be. States of matter with increased kinetic energy have higher entropy. Energy and entropy cannot increase at the same time. 6. How much heat is needed to warm 363 mL of water in a baby bottle from 24\u00b0C to 38\u00b0C? 21 kJ 36 kJ 121 kJ 820 kJ 7. Why is there always some waste heat in a heat engine? Heat cannot flow from a cold object to a hot object. Friction slows the engine down. The entropy increases at each stage. The heat pump uses energy. 8. How much heat is absorbed from the surroundings when 81 g of 0.0\u00b0C ice in a beaker melts and warms to 10\u00b0C? 0.34 kJ 27 kJ 30 kJ 190 kJ Ice m 81 g 0.0\u00b0C Ti 9. You do 0.050 J of work on the coffee in your cup each time you stir it. What would be the increase in entropy in 125 mL of coffee at 65\u00b0C when you stir it 85 times? 0.013 J/K 0.050 J 0.095 J/K 4.2 J Extended Answer 10. What is the difference in heat required to melt 454 g of ice at 0.00\u00b0C, and to turn 454 g of water at 100.0\u00b0C into steam? Is the amount of this difference greater or less than the amount of energy required to heat the 454 g of water from 0.00\u00b0C to 100.0\u00b0C? Your Mistakes Can Teach You The mistakes you make before the test are helpful because they show you areas in which you need more work. When calculating the heat needed to melt and warm a substance, remember to calculate the heat needed for melting as well as the heat needed for raising the temperature of the substance. physicspp.com/standardized_test Chapter 12 Standardized Test Practice 339 340-348 CH13-S1-845813 3/30/04 3:00 AM Page 340 What You\u2019ll Learn \u2022 You will explain the expansion and contraction of matter caused by changes in temperature. \u2022 You will apply Pascal\u2019s, Archimedes\u2019, and Bernoulli\u2019s principles in everyday situations. Why It\u2019s Important Fluids and the forces that they exert enable us to swim and dive, balloons to float, and planes to fly. Thermal expansion affects the designs of buildings, roads, bridges, and machines. Submarines A nuclear submarine is designed to maneuver at all levels in the ocean. It must withstand great differences in pressure and temperature as it moves deeper under water. Think About This How is the submarine able to float at the surface of the ocean and to dive far beneath it? physicspp.com 340 Steve Kaufman/Peter Arnold, Inc. 340-348 CH13-S1-845813 6/3/04 3:37 PM Page 341 Does it float or sink? Question How can you measure the buoyancy of objects? Procedure Analysis 1. Obtain a small vial (with a cap or a seal) and a 500-mL graduated cylinder. Attach a rubber band to the vial in order to suspend it from a spring scale. 2. Use the spring scale to find the weight of the vial. Then, use the graduated cylinder to find the volume of water displaced by the sealed vial when it is floating. Record both of these figures. Immediately wipe up any spilled liquid. 3. Place one nickel into the vial and close the lid. Repeat the procedures in step 2, recording the weight of the vial and the nickel, as well as the volume of water displaced. Also, record whether the vial floats or sinks. 4. Repeat steps 2 and 3, each time adding one nickel until the vial no longer floats. When the vial sinks, use the spring scale to find the apparent weight of the vial. Be sure that the vial is not touching the graduated cylinder when it is suspended under water. Use the information you recorded to calculate the density of the vial-and-nickel system in each of your trials. Also, calculate the mass of the water displaced by the system in each trial. How does density appear to be related to floating? Critical Thinking How does the mass of the vialnickel system appear to be related to the mass of the water displaced by the system? Does this relationship hold regardless of whether the system is floating? 13.1 Properties of Fluids Water and air are probably two of the most common substances in the everyday lives of people. We feel their effects when we drink, when we bathe, and literally", " with every breath we take. In your everyday experience, it might not seem that water and air have a great deal in common. If you think further about them, however, you will recognize that they have common properties. Both water and air flow, and unlike solids, neither one of them has a definite shape. Gases and liquids are two states of matter in which atoms and molecules have great freedom to move. In this chapter, you will explore states of matter. Beginning with gases and liquids, you will learn about the principles that explain how matter responds to changes in temperature and pressure, how hydraulic systems can multiply forces, and how huge metallic ships can float on water. You also will investigate the properties of solids, discovering how they expand and contract, why some solids are elastic, and why some solids seem to straddle the line between solid and liquid. Objectives \u2022 Describe how fluids create pressure. \u2022 Calculate the pressure, volume, and number of moles of a gas. \u2022 Compare gases and plasma. Vocabulary fluids pressure pascal combined gas law ideal gas law thermal expansion plasma Section 13.1 Properties of Fluids 341 Horizons Companies 340-348 CH13-S1-845813 6/10/04 8:33 PM Page 342 Pressure Suppose that you put an ice cube in an empty glass. The ice cube has a certain mass and shape, and neither of these quantities depends on the size or shape of the glass. What happens, however, when the ice melts? Its mass remains the same, but its shape changes. The water flows to take the shape of its container and forms a definite, flat, upper surface, as in Figure 13-1. If you boiled the water, it would change into a gas in the form of water vapor, and it also would flow and expand to fill the room. However, the water vapor would not have any definite surface. Both liquids and gases are fluids, which are materials that flow and have no definite shape of their own. For now, you can assume that you are dealing with ideal fluids, whose particles take up no space and have no intermolecular attractive forces. Pressure in fluids You have applied the law of conservation of energy to solid objects. Can this law also be applied to fluids? Work and energy can be defined if we introduce the concept of pressure, which is the force on a surface, divided by the area of the surface. Since pressure is force exerted over a surface, anything that exerts pressure is capable of producing change and doing work. F Pressure P A Pressure equals force divided by surface area. Pressure (P) is a scalar quantity. In the SI system, the unit of pressure is the pascal (Pa), which is 1 N/m2. Because the pascal is a small unit, the kilopascal (kPa), equal to 1000 Pa, is more commonly used. The force, F, on a surface is assumed to be perpendicular to the surface area, A. Figure 13-2 illustrates the relationships between force, area, and pressure. Table 13-1 shows how pressures vary in different situations. Solids, liquids, and pressure Imagine that you are standing on the surface of a frozen lake. The forces that your feet exert on the ice are spread over the area of your shoes, resulting in pressure on the ice. Ice is a solid that is made up of vibrating water molecules, and the forces that hold the water molecules in place cause the ice to exert upward forces on your feet that equal your weight. If the ice melted, most of the bonds between the water molecules would be weakened. Although the molecules would continue to vibrate and remain close to each other, they also would slide past one another, and you would break through the surface. The moving water molecules would continue to exert forces on your body. Gas particles and pressure The pressure exerted by a gas can be understood by applying the kinetic-molecular theory of gases. The kineticmolecular theory explains the properties of an ideal gas. In reality, the particles of a gas take up space and have intermolecular attractive forces, but an ideal gas is an accurate model of a real gas under most conditions. According to the kinetic-molecular theory, the particles in a gas are in random motion at high speeds and undergoing elastic collisions with each other. When a gas particle hits a container\u2019s surface, it rebounds, which changes its momentum. The impulses exerted by many of these collisions result in gas pressure on the surface. \u25a0 Figure 13-1 The ice cubes, which are solids, have definite shapes. However, the liquid water, a fluid, takes the shape of its container. What fluid is filling the space above the water? \u25a0 Figure 13-2 The astronaut and the landing module both exert pressure on the lunar surface. If the lunar module had a mass of approximately 7300 kg and rested on four pads that were each 91 cm in diameter, what pressure did it exert on the Moon\u2019s surface?", " How could you estimate the pressure exerted by the astronaut? 342 (t)Gerard Photography, (b)NASA\u2019s Goddard Space Flight Center Chapter 13 States of Matter 340-348 CH13-S1-845813 3/30/04 3:05 AM Page 343 Table 13-1 Some Typical Pressures Location The center of Earth The center of the Sun Atmospheric pressure On every square centimeter of Earth\u2019s surface at sea level, the atmospheric gas exerts a force of approximately 10 N, about the weight of a 1-kg object. The pressure of Earth\u2019s atmosphere on your body is so well balanced by your body\u2019s outward forces that you seldom notice it. You probably become aware of this pressure only when your ears pop as the result of pressure changes, as when you ride an elevator in a tall building or fly in an airplane. Atmospheric pressure is about 10 N per 1 cm2 (104 m2), which is about 1.0105 N/m2, or 100 kPa. Other planets in our solar system also have atmospheres. The pressure exerted by these atmospheres, however, varies widely. For example, the pressure at the surface of Venus is about 92 times the pressure at the surface of Earth, while the pressure at the surface of Mars is less than 1 percent of Earth\u2019s. Standard atmosphere The best vacuum Blood pressure The deepest ocean trench Air pressure on top of Mt. Everest Pressure (Pa) 31016 41011 1.1108 1.01325105 1.6104 3104 11013 Calculating Pressure A child weighs 364 N and sits on a three-legged stool, which weighs 41 N. The bottoms of the stool\u2019s legs touch the ground over a total area of 19.3 cm2. a. What is the average pressure that the child and stool exert on the ground? b. How does the pressure change when the child leans over so that only two legs of the stool touch the floor? 1 Analyze and Sketch the Problem \u2022 Sketch the child and the stool, labeling the total force that they exert on the ground. \u2022 List the variables, including the force that the child and stool exert on the ground and the areas for parts a and b. AA AB 19.3 cm2 2 19.3 cm2 3 12.9 cm2 Unknown:? PA? PB Known: Fg child Fg stool Fg total 364 N 41 N Fg stool Fg child 364 N 41 N 405 N 2 Solve for the Unknown Find each pressure. P F/A Fg 405 N Math Handbook Dimensional Calculations pages 846\u2014847 a. PA b. PB )((1 ( N 5 0 4 0 0 m2 c 3. 19 1 ( 2.10102 kPa )((1 ( N 5 0 4 0 0 m2 c 9. 12 1 ( 3.14102 kPa )2 ) cm )2 m )2 ) cm )2 m Substitute F Fg total 405 N, A AA 19.3 cm2 Substitute F Fg total 405 N, A AB 12.9 cm2 3 Evaluate the Answer \u2022 Are the units correct? The units for pressure should be Pa, and 1 N/m2 1 Pa. Section 13.1 Properties of Fluids 343 340-348 CH13-S1-845813 3/30/04 3:06 AM Page 344 1. The atmospheric pressure at sea level is about 1.0105 Pa. What is the force at sea level that air exerts on the top of a desk that is 152 cm long and 76 cm wide? 2. A car tire makes contact with the ground on a rectangular area of 12 cm by 18 cm. If the car\u2019s mass is 925 kg, what pressure does the car exert on the ground as it rests on all four tires? 3. A lead brick, 5.0 cm 10.0 cm 20.0 cm, rests on the ground on its smallest face. Lead has a density of 11.8 g/cm3. What pressure does the brick exert on the ground? 4. 5. In a tornado, the pressure can be 15 percent below normal atmospheric pressure. Suppose that a tornado occurred outside a door that is 195 cm high and 91 cm wide. What net force would be exerted on the door by a sudden 15 percent drop in normal atmospheric pressure? In what direction would the force be exerted? In industrial buildings, large pieces of equipment must be placed on wide steel plates that spread the weight of the equipment over larger areas. If an engineer plans to install a 454-kg device on a floor that is rated to withstand additional pressure of 5.0104 Pa, how large should the steel support plate be? The Gas Laws As scientists first studied gases and pressure, they began to notice some interesting relationships. The first relationship to emerge was named Boyle\u2019s law, after seventeenth-century chemist and physicist Robert Boyle. Boyle\u2019s", " law states that for a fixed sample of gas at constant temperature, the volume of the gas varies inversely with the pressure. Because the product of inversely related variables is a constant, Boyle\u2019s law can be written PV constant, or P1V1 P2V2. The subscripts that you see in the gas laws will help you keep track of different variables, such as pressure and volume, as they change throughout a problem. These variables can be rearranged to solve for an unknown pressure or volume. As shown in Figure 13-3, the relationship between the pressure and the volume of a gas is critical to the sport of scuba diving. A second relationship was discovered about 100 years after Boyle\u2019s work 1 by Jacques Charles. When Charles cooled a gas, the volume shrank by 73 2 of its original volume for every degree cooled, which is a linear relationship. At the time, Charles could not cool gases to the extremely low temperatures achieved in modern laboratories. In order to see what lower limits might be possible, he extended, or extrapolated, the graph of his data to these temperatures. This extrapolation suggested that if the temperature were reduced to 273\u00b0C, a gas would have zero volume. The temperature at which a gas would have zero volume is now called absolute zero, which is represented by the zero of the Kelvin temperature scale. These experiments indicated that under constant pressure, the volume of a sample of gas varies directly with its Kelvin temperature, a result that is now called Charles\u2019s law. Charles\u2019s law can be written V/T constant, or V1/T1 V2/T2. \u25a0 Figure 13-3 The gas in the tank on the diver\u2019s back is at high pressure. This pressure is reduced by the regulator so that the pressure of the gas the diver breathes is equal to the water pressure. In the photo, you can see bubbles coming from the regulator. 344 Timothy O\u2019Keefe/Bruce Coleman Chapter 13 States of Matter 340-348 CH13-S1-845813 3/30/04 3:06 AM Page 345 Combining Boyle\u2019s law and Charles\u2019s law relates the pressure, temperature, and volume of a fixed amount of ideal gas, which leads to the equation called the combined gas law. 2V P 1V P 1 2 = constant Combined Gas Law T T 2 1 For a fixed amount of an ideal gas, the pressure times the volume, divided by the Kelvin temperature equals a constant. As shown in Figure 13-4, the combined gas law reduces to Boyle\u2019s law under conditions of constant temperature and to Charles\u2019s law under conditions of constant pressure. The ideal gas law You can use the kinetic-molecular theory to discover how the constant in the combined gas law depends on the number of particles, N. Suppose that the volume and temperature of an ideal gas are held constant. If the number of particles increases, the number of collisions that the particles make with the container will increase, thereby increasing the pressure. Removing particles decreases the number of collisions, and thus, decreases the pressure. You can conclude that the constant in the combined gas law equation is proportional to N. V kN P T The constant, k, is called Boltzmann\u2019s constant, and its value is 1.381023 Pam3/K. Of course, N, the number of particles, is a very large number. Instead of using N, scientists often use a unit called a mole. One mole (abbreviated mol and represented in equations by n) is similar to one dozen, except that instead of representing 12 items, one mole represents 6.0221023 particles. This number is called Avogadro\u2019s number, after Italian scientist Amedeo Avogadro. Pressure How much pressure do you exert when standing on one foot? Have a partner trace your foot, and then use the outline to estimate its area. 1. Determine your weight in newtons and the area of the outline in m2. 2. Calculate the pressure. 3. Compare and contrast the pressure you exert on the ground with the pressure exerted by various objects. For example, you could weigh a brick and determine the pressure it exerts when resting on different faces. Analyze and Conclude 4. How do shoes with high heels affect the pressure that a person exerts on the ground? \u25a0 Figure 13-4 You can use the combined gas law to derive both Boyle\u2019s and Charles\u2019s laws. What happens if you hold volume constant? Avogadro\u2019s number is numerically equal to the number of particles in a sample of matter whose mass equals the molar mass of the substance. You can use this relationship to convert between mass and n, the number of moles present. Using moles instead of the number of particles changes Boltzmann\u2019s constant. This new constant is abbreviated R, and it has the", " value 8.31 Pam3/molK. Rearranging, you can write the ideal gas law in its most familiar form. 2V P 1V P 1 2 T T 2 1 Combined gas law If temperature is constant If pressure is constant Ideal Gas Law PV nRT For an ideal gas, the pressure times the volume is equal to the number of moles multiplied by the constant R and the Kelvin temperature. Note that with the given value of R, volume must be expressed in m3, temperature in K, and pressure in Pa. In practice, the ideal gas law predicts the behavior of gases remarkably well, except under conditions of high pressures or low temperatures. P1V1 P2V2 Boyle\u2019s law V V 1 2 T2 T1 Charles\u2019s law Section 13.1 Properties of Fluids 345 340-348 CH13-S1-845813 3/30/04 3:07 AM Page 346 Gas Laws A 20.0-L sample of argon gas at 273 K is at atmospheric pressure, 101.3 kPa. The temperature is lowered to 120 K, and the pressure is increased to 145 kPa. a. What is the new volume of the argon sample? b. Find the number of moles of argon atoms in the argon sample. c. Find the mass of the argon sample. The molar mass, M, of argon is 39.9 g/mol. 1 Analyze and Sketch the Problem \u2022 Sketch the situation. Indicate the conditions in the container of argon before and after the change in temperature and pressure. \u2022 List the known and unknown variables. 273 K 101.3 kPa 20.0 L T1 P1 V1 T2 120 K P2 145 kPa V2? Unknown:? V2 moles of argon? mass of argon sample? 20.0 L 101.3 kPa 273 K 145 kPa 120 K Known: V1 P1 T1 P2 T2 R 8.31 Pam3/molK Margon 39.9 g/mol 2 Solve for the Unknown a. Use the combined gas law and solve for V2. 2V P 1V P 1 2 T T 2 1 1T 1V P 2 2T P 1 V2 (101.3 kPa)(20.0 L)(120 K) (145 kPa)(273 K) 6.1 L b. Use the ideal gas law and solve for n. PV nRT V n P R T (101.3103 Pa)(0.0200 m3) (8.31 Pam3/molK)(273 K) 0.893 mol Math Handbook Isolating a Variable page 845 145 kPa, Substitute P1 V1 20.0 L, T1 101.3 kPa, P2 273 K, T2 120 K Substitute P 101.3103 Pa, V 0.0200 m3, R 8.31 m3/molK, T 273 K c. Use the molar mass to convert from moles of argon in the sample to mass of the sample. m Mn margon sample (39.9 g/mol)(0.893 mol) 35.6 g Substitute M 39.9 g/mol, n 0.893 mol 3 Evaluate the Answer \u2022 Are the units correct? The volume, V2, is in liters, and the mass of the sample is in grams. \u2022 Is the magnitude realistic? The change in volume is consistent with an increase in pressure and decrease in temperature. The calculated mass of the argon sample is reasonable. 346 Chapter 13 States of Matter 340-348 CH13-S1-845813 3/30/04 3:07 AM Page 347 6. A tank of helium gas used to inflate toy balloons is at a pressure of 15.5106 Pa and a temperature of 293 K. The tank\u2019s volume is 0.020 m3. How large a balloon would it fill at 1.00 atmosphere and 323 K? 7. What is the mass of the helium gas in the previous problem? The molar mass of helium gas is 4.00 g/mol. 8. A tank containing 200.0 L of hydrogen gas at 0.0\u00b0C is kept at 156 kPa. The temperature is raised to 95\u00b0C, and the volume is decreased to 175 L. What is the new pressure of the gas? 9. The average molar mass of the components of air (mainly diatomic oxygen gas and diatomic nitrogen gas) is about 29 g/mol. What is the volume of 1.0 kg of air at atmospheric pressure and 20.0\u00b0C? Thermal Expansion As you applied the combined gas law, you discovered how gases expand as their temperatures increase. When heated, all forms of matter\u2014solids, liquids, and gases\u2014generally become less dense and expand to fill more space. This property, known as thermal expansion", ", has many useful applications, such as circulating air in a room. When the air near the floor of a room is warmed, gravity pulls the denser, colder air near the ceiling down, which pushes the warmer air upward. This circulation of air within a room is called a convection current. Figure 13-5 shows convection currents starting as the hot air above the flames rises. You also can see convection currents in a pot of hot, but not boiling, water on a stove. When the pot is heated from the bottom, the colder and denser water sinks to the bottom where it is warmed and then pushed up by the continuous flow of cooler water from the top. This thermal expansion occurs in most liquids. A good model for all liquids does not exist, but it is useful to think of a liquid as a finely ground solid. Groups of two, three, or more particles move together as if they were tiny pieces of a solid. When a liquid is heated, particle motion causes these groups to expand in the same way that particles in a solid are pushed apart. The spaces between groups increase. As a result, the whole liquid expands. With an equal change in temperature, liquids expand considerably more than solids, but not as much as gases. Why ice floats Because matter expands as it is heated, you might predict that ice would be more dense than water, and therefore, it should sink. However, when water is heated from 0\u00b0C to 4\u00b0C, instead of expanding, it contracts as the forces between particles increase and the ice crystals collapse. These forces between water molecules are strong, and the crystals that make up ice have a very open structure. Even when ice melts, tiny crystals remain. These remaining crystals are melting, and the volume of the water decreases until the temperature reaches 4\u00b0C. However, once the temperature of water moves above 4\u00b0C, its volume increases because of greater molecular motion. The practical result is that water is most dense at 4\u00b0C and ice floats. This fact is very important to our lives and environment. If ice sank, lakes would freeze from the bottom each winter and many would never melt completely in the summer. \u25a0 Figure 13-5 This image was made by a special technique that enables you to see different densities in the air. Convection currents are set up as warmer, less dense air rises and cooler, denser air sinks. Section 13.1 Properties of Fluids 347 Gary Settles/Pennsylvania State University 340-348 CH13-S1-845813 3/30/04 3:08 AM Page 348 \u25a0 Figure 13-6 The colorful lighting effects in neon signs are caused by luminous plasmas formed in the glass tubing. Plasma If you heat a solid, it melts to form a liquid. Further heating results in a gas. What happens if you increase the temperature still further? Collisions between the particles become violent enough to tear the electrons off the atoms, thereby producing positively charged ions. The gaslike state of negatively charged electrons and positively charged ions is called plasma. Plasma is considered to be another fluid state of matter. The plasma state may seem to be uncommon; however, most of the matter in the universe is plasma. Stars consist mostly of plasma at extremely high temperatures. Much of the matter between stars and galaxies consists of energetic hydrogen that has no electrons. This hydrogen is in the plasma state. The primary difference between gas and plasma is that plasma can conduct electricity, whereas gas cannot. Lightning bolts are in the plasma state. Neon signs, such as the one shown in Figure 13-6, fluorescent bulbs, and sodium vapor lamps all contain glowing plasma. 13.1 Section Review 10. Pressure and Force Suppose that you have two boxes. One is 20 cm 20 cm 20 cm. The other is 20 cm 20 cm 40 cm. a. How does the pressure of the air on the outside of the two boxes compare? b. How does the magnitude of the total force of the air on the two boxes compare? 11. Meteorology A weather balloon used by meteorologists is made of a flexible bag that allows the gas inside to freely expand. If a weather balloon containing 25.0 m3 of helium gas is released from sea level, what is the volume of gas when the balloon reaches a height of 2100 m, where the pressure is 0.82105 Pa? Assume that the temperature is unchanged. 12. Gas Compression In a certain internal-combustion engine, 0.0021 m3 of air at atmospheric pressure and 303 K is rapidly compressed to a pressure of 20.1105 Pa and a volume of 0.0003 m3. What is the final temperature of the compressed gas? 13. Density and Temperature Starting at 0\u00b0C, how will the density of water change if it is heated to 4\u00b0C? To 8\u00b0C? 14. The Standard Molar Volume What is the volume of 1 mol of a gas at atmospheric pressure and a temperature of 273 K? 15. The Air in", " a Refrigerator How many moles of air are in a refrigerator with a volume of 0.635 m3 at a temperature of 2.00\u00b0C? If the average molar mass of air is 29 g/mol, what is the mass of the air in the refrigerator? 16. Critical Thinking Compared to the particles that make up carbon dioxide gas, the particles that make up helium gas are very small. What can you conclude about the number of particles in a 2.0-L sample of carbon dioxide gas compared to the number of particles in a 2.0-L sample of helium gas if both samples are at the same temperature and pressure? 348 Norman Tomalin/Bruce Coleman Chapter 13 States of Matter physicspp.com/self_check_quiz 349-351 CH13-S2-845813 3/30/04 3:11 AM Page 349 13.2 Forces Within Liquids The liquids considered thus far have been ideal liquids, in which the particles are totally free to slide past one another. The unexpected behavior of water between 0\u00b0C and 4\u00b0C, however, illustrates that in real fluids, particles exert electromagnetic forces of attraction, called cohesive forces, on each other. These and other forces affect the behavior of fluids. Cohesive Forces Have you ever noticed that dewdrops on spiderwebs and falling drops of oil are nearly spherical? What happens when rain falls on a freshly washed and waxed car? The water drops bead up into rounded shapes, as shown in the spiderweb in Figure 13-7. All of these phenomena are examples of surface tension, which is the tendency of the surface of a liquid to contract to the smallest possible area. Surface tension is a result of the cohesive forces among the particles of a liquid. Notice that beneath the surface of the liquid shown in Figure 13-8a on the next page, each particle of the liquid is attracted equally in all directions by neighboring particles, and even to the particles of the wall of the container. As a result, no net force acts on any of the particles beneath the surface. At the surface, however, the particles are attracted downward and to the sides, but not upward. There is a net downward force, which acts on the top layers and causes the surface layer to be slightly compressed. The surface layer acts like a tightly stretched rubber sheet or a film that is strong enough to support the weight of very light objects, such as the water strider in Figure 13-8b on the next page. The surface tension of water also can support a steel paper clip, even though the density of steel is nine times greater than that of water. Try it! Why does surface tension produce spherical drops? The force pulling the surface particles into a liquid causes the surface to become as small as possible, and the shape that has the least surface for a given volume is a sphere. The higher the surface tension of the liquid, the more resistant the liquid is to having its surface broken. For example, liquid mercury has much stronger cohesive forces than water does. Thus, liquid mercury forms spherical drops, even when it is placed on a smooth surface. On the other hand, liquids such as alcohol and ether have weaker cohesive forces. A drop of either of these liquids flattens out on a smooth surface. Viscosity In nonideal fluids, the cohesive forces and collisions between fluid molecules cause internal friction that slows the fluid flow and dissipates mechanical energy. The measure of this internal friction is called the viscosity of the liquid. Water is not very viscous, but motor oil is very viscous. As a result of its viscosity, motor oil flows slowly over the parts of an engine to coat the metal and reduce rubbing. Lava, molten rock that flows from a volcano or vent in Earth\u2019s surface, is one of the most viscous fluids. There are several types of lava, and the viscosity of each type varies with composition and temperature. Objectives \u2022 Explain how cohesive forces cause surface tension. \u2022 Explain how adhesive forces cause capillary action. \u2022 Discuss evaporative cooling and the role of condensation in cloud formation. Vocabulary cohesive forces adhesive forces \u25a0 Figure 13-7 Rainwater beads up on a spider\u2019s web because water drops have surface tension. Geology Connection Geology Connection Section 13.2 Forces Within Liquids 349 Frank Cezus 349-351 CH13-S2-845813 3/30/04 3:12 AM Page 350 a b \u25a0 Figure 13-8 Molecules in the interior of a liquid are attracted in all directions (a). A water strider can walk on water because molecules at the surface have a net inward attraction that results in surface tension (b). Plants Cohesive forces in liquids actually allow them to be stretched just like rubber bands. This stretching is difficult to achieve in the laboratory, but it is common in plants. The strength of the cohesive forces in water keeps the water from breaking and bubbling as it goes through the plant tissue to the leaves.", " If not for these forces, trees could not grow higher than about 10 m. Adhesive Forces Similar to cohesive forces, adhesive forces are electromagnetic attractive forces that act between particles of different substances. If a glass tube with a small inner diameter is placed in water, the water rises inside the tube. The water rises because the adhesive forces between glass and water molecules are stronger than the cohesive forces between water molecules. This phenomenon is called capillary action. The water continues to rise until the weight of the water that is lifted balances the total adhesive force between the glass and water molecules. If the radius of the tube increases, the volume and the weight of the water will increase proportionally faster than the surface area of the tube. Thus, water is lifted higher in a narrow tube than in a wider one. Capillary action causes molten wax to rise in a candle\u2019s wick and water to move up through the soil and into the roots of plants. When a glass tube is placed in a beaker of water, the surface of the water climbs the outside of the tube, as shown in Figure 13-9a. The adhesive forces between the glass molecules and water molecules are greater than the cohesive forces between the water molecules. In contrast, the cohesive forces between mercury molecules are greater than the adhesive forces between the mercury and glass molecules, so the liquid does not climb the tube. These forces also cause the center of the mercury\u2019s surface to depress, as shown in Figure 13-9b. Evaporation and Condensation Why does a puddle of water disappear on a hot, dry day? As you learned in Chapter 12, the particles in a liquid are moving at random speeds. If a fast-moving particle can break through the surface layer, it will escape from the liquid. Because there is a net downward cohesive force at the surface, however, only the most energetic particles escape. This escape of particles is called evaporation. a b \u25a0 Figure 13-9 Water climbs the outside wall of this glass tube (a), while the mercury is depressed by the rod (b). The forces of attraction between mercury atoms are stronger than any adhesive forces between the mercury and the glass. 350 Chapter 13 States of Matter (t)Runk/Schoenberger/Grant Heilman Photography, (others)Matt Meadows 349-351 CH13-S2-845813 6/3/04 3:40 PM Page 351 Evaporative cooling Evaporation has a cooling effect. On a hot day, your body perspires, and the evaporation of your sweat cools you down. In a puddle of water, evaporation causes the remaining liquid to cool down. Each time a particle with higher-than-average kinetic energy escapes from the water, the average kinetic energy of the remaining particles decreases. As you learned in Chapter 12, a decrease in average kinetic energy is a decrease in temperature. You can test this cooling effect by pouring a small amount of rubbing alcohol in the palm of your hand. Alcohol molecules evaporate easily because they have weak cohesive forces. As the molecules evaporate, the cooling effect is quite noticeable. A liquid that evaporates quickly is called a volatile liquid. Have you ever wondered why humid days feel warmer than dry days at the same temperature? On a day that is humid, the water vapor content of the air is high. Because there are already many water molecules in the air, the water molecules in perspiration are less likely to evaporate from the skin. Evaporation is the body\u2019s primary cooling mechanism, so the body is not able to cool itself as effectively on a humid day. Particles of liquid that have evaporated into the air can also return to the liquid phase if the kinetic energy or temperature decreases, a process called condensation. What happens if you bring a cold glass into a hot, humid area? The outside of the glass soon becomes coated with condensed water. Water molecules moving randomly in the air surrounding the glass strike the cold surface, and if they lose enough energy, the cohesive forces become strong enough to prevent their escape. The air above any body of water, as shown in Figure 13-10, contains evaporated water vapor, which is water in the form of gas. If the temperature is reduced, the water vapor condenses around tiny dust particles in the air and produces droplets only 0.01 mm in diameter. A cloud of these droplets is called fog. Fog often forms when moist air is chilled by the cold ground. Fog also can form in your home. When a carbonated drink is opened, the sudden decrease in pressure causes the temperature of the gas in the container to drop, which condenses the water vapor dissolved in that gas. \u25a0 Figure 13-10 Warm, moist, surface air rises until it reaches a height where the temperature is at the point at which water vapor condenses and forms clouds. 13.2 Section Review 17. Evaporation and Cooling In the past, when a baby had a high fever, the doctor might", " have suggested gently sponging off the baby with rubbing alcohol. Why would this help? 18. Surface Tension A paper clip, which has a density greater than that of water, can be made to stay on the surface of water. What procedures must you follow for this to happen? Explain. 19. Language and Physics The English language includes the terms adhesive tape and working as a cohesive group. In these terms, are adhesive and cohesive being used in the same context as their meanings in physics? 20. Adhesion and Cohesion In terms of adhesion and cohesion, explain why alcohol clings to the surface of a glass rod but mercury does not. 21. Floating How can you tell that the paper clip in problem 18 was not floating? 22. Critical Thinking On a hot, humid day, Beth sat on the patio with a glass of cold water. The outside of the glass was coated with water. Her younger sister, Jo, suggested that the water had leaked through the glass from the inside to the outside. Suggest an experiment that Beth could do to show Jo where the water came from. physicspp.com/self_check_quiz Section 13.2 Forces Within Liquids 351 Orville Andrews 352-358 CH13-S3-845813 3/30/04 3:16 AM Page 352 13.3 Fluids at Rest and in Motion Objectives \u2022 Relate Pascal\u2019s principle to simple machines and occurrences. \u2022 Apply Archimedes\u2019 principle to buoyancy. \u2022 Apply Bernoulli\u2019s principle to airflow. Vocabulary Pascal\u2019s principle buoyant force Archimedes\u2019 principle Bernoulli\u2019s principle streamlines \u25a0 Figure 13-11 The pressure exerted by the force of the small piston is transmitted throughout the fluid and results in a multiplied force on the larger piston. F2 F1 A1 A2 Piston 1 Piston 2 352 Chapter 13 States of Matter You have learned how fluids exert pressure, the force per unit area. You also know that the pressure exerted by fluids changes; for example, atmospheric pressure drops as you climb a mountain. In this section, you will learn about the forces exerted by resting and moving fluids. Fluids at Rest If you have ever dived deep into a swimming pool or lake, you know that your body, especially your ears, is sensitive to changes in pressure. You may have noticed that the pressure you felt on your ears did not depend on whether your head was upright or tilted, but that if you swam deeper, the pressure increased. Pascal\u2019s principle Blaise Pascal, a French physician, noted that the pressure in a fluid depends upon the depth of the fluid and has nothing to do with the shape of the fluid\u2019s container. He also discovered that any change in pressure applied at any point on a confined fluid is transmitted undiminished throughout the fluid, a fact that is now known as Pascal\u2019s principle. Every time you squeeze a tube of toothpaste, you demonstrate Pascal\u2019s principle. The pressure that your fingers exert at the bottom of the tube is transmitted through the toothpaste and forces the paste out at the top. Likewise, if you squeeze one end of a helium balloon, the other end of the balloon expands. When fluids are used in machines to multiply forces, Pascal\u2019s principle is being applied. In a common hydraulic system, a fluid is confined to two connecting chambers, as shown in Figure 13-11. Each chamber has a piston that is free to move, and the pistons have different surface areas. If a force, F1, is exerted on the first piston with a surface area of A1, the pressure, P1, exerted on the fluid can be determined by using the following equation. P1 F1 A 1 This equation is simply the definition of pressure: pressure equals the force per unit area. The pressure exerted by the fluid on the second piston, with a surface area A2, can also be determined. F2 A 2 P2 According to Pascal\u2019s principle, pressure is transmitted without change throughout a fluid, so pressure P2 is equal in value to P1. You can deter F2/A2 and mine the force exerted by the second piston by using F1/A1 solving for F2. This force is shown by the following equation. Force Exerted by a Hydraulic Lift F2 1A F 2 A 1 The force exerted by the second piston is equal to the force exerted by the first piston multiplied by the ratio of the area of the second piston to the area of the first piston. 352-358 CH13-S3-845813 3/30/04 3:18 AM Page 353 23. Dentists\u2019 chairs are examples of hydraulic-lift systems. If a chair weighs 1600 N and rests on a piston with a cross-sectional area of 1440 cm2, what force must be applied to the smaller piston, with a cross-sectional area of 72 cm2, to lift the", " chair? 24. A mechanic exerts a force of 55 N on a 0.015 m2 hydraulic piston to lift a small automobile. The piston that the automobile sits on has an area of 2.4 m2. What is the weight of the automobile? 25. By multiplying a force, a hydraulic system serves the same purpose as a lever or seesaw. If a 400-N child standing on one piston is balanced by a 1100-N adult standing on another piston, what is the ratio of the areas of their pistons? 26. In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a cross-sectional area of 7.0102 m2 and a large piston with a cross-sectional area of 2.1101 m2. An engine weighing 2.7103 N rests on the large piston. a. What force must be applied to the small piston to lift the engine? b. If the engine rises 0.20 m, how far does the smaller piston move? Swimming Under Pressure When you are swimming, you feel the pressure of the water increase as you dive deeper. This pressure is actually a result of gravity; it is related to the weight of the water above you. The deeper you go, the more water there is above you, and the greater the pressure. The pressure of the water is equal to the weight, Fg, of the column of water above you divided by the column\u2019s cross-sectional area, A. Even though gravity pulls only in the downward direction, the fluid transmits the pressure in all directions: up, down, and to the sides. You can find the pressure of the water by applying the following equation. F g P A The weight of the column of water is Fg mg, and the mass is equal to the density,, of the water times its volume, m V. You also know that the volume of the water is the area of the base of the column times its height, V Ah. Therefore, Fg Ahg. Substituting Ahg for Fg in the equation for water pressure gives P Fg/A Ahg/A. Divide A from the numerator and denominator to arrive at the simplest form of the equation for the pressure exerted by a column of water on a submerged body. Pressure of Water on a Body P hg The pressure that a column of water exerts on a body is equal to the density of water times the height of the column times the acceleration due to gravity. This formula works for all fluids, not just water. The pressure of a fluid on a body depends on the density of the fluid, its depth, and g. If there were water on the Moon, the pressure of the water at any depth would be one-sixth as great as on Earth. As illustrated in Figure 13-12, submersibles, both crewed and robotic, have explored the deepest ocean trenches and encountered pressures in excess of 1000 times standard air pressure. \u25a0 Figure 13-12 In 1960, the Trieste, a crewed submersible, descended to the bottom of the Marianas Trench, a depth of over 10,500 m. The crewed submersible Alvin, shown below, can safely dive to a depth of 4500 m. Section 13.3 Fluids at Rest and in Motion 353 Ralph White/CORBIS 352-358 CH13-S3-845813 6/4/04 8:58 AM Page 354 h h l Ftop Fbottom Buoyancy What produces the upward force that allows you to swim? The increase in pressure with increasing depth creates an upward force called the buoyant force. By comparing the buoyant force on an object with its weight, you can predict whether the object will sink or float. Suppose that a box is immersed in water. It has a height of l and its top and bottom each have a surface area of A. Its volume, then, is V lA. Water pressure exerts forces on all sides, as shown in Figure 13-13. Will the box sink or float? As you know, the pressure on the box depends on its depth, h. To find out whether the box will float in water, you will need to determine how the pressure on the top of the box compares with the pressure from below the box. Compare these two equations: Ftop Fbottom Ptop A hgA Pbottom A (l h)gA On the four vertical sides, the forces are equal in all directions, so there is no net horizontal force. The upward force on the bottom is larger than the downward force on the top, so there is a net upward force. The buoyant force can now be determined. Fbuoyant Ftop Fbottom (l h)gA hgA lgA Vg These calculations show the net upward force to be proportional to the volume of the box. This volume equals the volume of the fluid displaced, or pushed", " out of the way, by the box. Therefore, the magnitude of the buoyant force, Vg, equals the weight of the fluid displaced by the object. Buoyant Force Fbuoyant fluidVg The buoyant force on an object is equal to the weight of the fluid displaced by the object, which is equal to the density of the fluid in which the object is immersed multiplied by the object\u2019s volume and the acceleration due to gravity. This relationship was discovered in the third century B.C. by Greek scientist Archimedes. Archimedes\u2019 principle states that an object immersed in a fluid has an upward force on it that is equal to the weight of the fluid displaced by the object. The force does not depend on the weight of the object, only on the weight of the displaced fluid. Sink or float? If you want to know whether an object sinks or floats, you have to take into account all of the forces acting on the object. The buoyant force pushes up, but the weight of the object pulls it down. The difference between the buoyant force and the object\u2019s weight determines whether an object sinks or floats. Suppose that you submerge three objects in a tank filled with water ( 1.00103 kg/m3). Each of the objects has a volume of 100 cm3, or 1.00104 m3. The first object is a steel block with a mass of 0.90 kg. water \u25a0 Figure 13-13 A fluid exerts a greater upward force on the bottom of an immersed object than the downward force on the top of the object. The net upward force is called the buoyant force. 354 Chapter 13 States of Matter 352-358 CH13-S3-845813 3/30/04 3:19 AM Page 355 The second is an aluminum soda can with a mass of 0.10 kg. The third is an ice cube with a mass of 0.090 kg. How will each item move when it is immersed in water? The upward force on all three objects, as shown in Figure 13-14, is the same, because all displace the same weight of water. This buoyant force can be calculated as follows. Fbuoyant waterVg (1.00103 kg/m3)(1.00104 m3)(9.80 m/s2) 0.980 N The weight of the block of steel is 8.8 N, much greater than the buoyant force. There is a net downward force, so the block will sink to the bottom of the tank. The net downward force, its apparent weight, is less than its real weight. All objects in a liquid, even those that sink, have an apparent weight that is less than when the object is in air. The apparent weight can Fbuoyant. For the block of be expressed by the equation Fapparent steel, the apparent weight is 8.8 N 0.98 N, or 7.8 N. Fg The weight of the soda can is 0.98 N, the same as the weight of the water displaced. There is, therefore, no net force, and the can will remain wherever it is placed in the water. It has neutral buoyancy. Objects with neutral buoyancy are described as being weightless; their apparent weight is zero. This property is similar to that experienced by astronauts in orbit, which is why astronaut training sometimes takes place in swimming pools. The weight of the ice cube is 0.88 N, less than the buoyant force, so there is a net upward force, and the ice cube will rise. At the surface, the net upward force will lift part of the ice cube out of the water. As a result, less water will be displaced, and the upward force will be reduced. The ice cube will float with enough volume in the water so that the weight of water displaced equals the weight of the ice cube. An object will float if its density is less than the density of the fluid in which it is immersed. Ships Archimedes\u2019 principle explains why ships can be made of steel and still float; if the hull is hollow and large enough so that the average density of the ship is less than the density of water, the ship will float. You may have noticed that a ship loaded with cargo rides lower in the water than a ship with an empty cargo hold. You can demonstrate this effect by fashioning a small boat out of folded aluminum foil. The boat should float easily, and it will ride lower in the water if you add a cargo of paper clips. If the foil is crumpled into a tight ball, the boat will sink because of its increased density. Similarly, the continents of Earth float upon a denser material below the surface. The drifting motion of these continental plates is responsible for the present shapes and locations of the continents. Other examples of Archimedes\u2019 principle in action include submarines and fishes. Submarines take advantage of Archimedes\u2019 principle", " as water is pumped into or out of a number of different chambers to change the submarine\u2019s average density, causing it to rise or sink. Fishes that have swim bladders also use Archimedes\u2019 principle to control their depths. Such a fish can expand or contract its swim bladder, just like you can puff up your cheeks. To move upward in the water, the fish expands its swim bladder to displace more water and increase the buoyant force. The fish moves downward by contracting the volume of its swim bladder. a b c Fbuoyant Fnet Fg Fbuoyant Fnet 0 Fg Fbuoyant Fnet Fg \u25a0 Figure 13-14 A block of steel (a), an aluminum can of soda (b), and an ice cube (c) all have the same volume, displace the same amount of water, and experience the same buoyant force. However, because their weights are different, the net forces on the three objects are also different. Geology Connection Geology Connection Section 13.3 Fluids at Rest and in Motion 355 352-358 CH13-S3-845813 3/30/04 3:20 AM Page 356 Archimedes\u2019 Principle A cubic decimeter, 1.00103 m3, of a granite building block is submerged in water. The density of granite is 2.70103 kg/m3. a. What is the magnitude of the buoyant force acting on the block? b. What is the apparent weight of the block? 1 Analyze and Sketch the Problem \u2022 Sketch the cubic decimeter of granite immersed in water. \u2022 Show the upward buoyant force and the downward force due to gravity Fbuoyant Fg Math Handbook Operations with Scientific Notation pages 842\u2014843 acting on the granite. Known: V 1.00103 m3 granite water 2.70103 kg/m3 1.00103 kg/m3 Unknown: Fbuoyant Fapparent?? 2 Solve for the Unknown a. Calculate the buoyant force on the granite block. Fbuoyant waterVg (1.00103 kg/m3)(1.00103 m3)(9.80 m/s2) 9.80 N Substitute V 1.00103 m3, g 9.80 m/s2 1.00103 kg/m3, water b. Calculate the granite\u2019s weight and then find its apparent weight. Fg graniteVg mg (2.70103 kg/m3)(1.00103 m3)(9.80 m/s2) 26.5 N Fg 26.5 N 9.80 N 16.7 N Fbuoyant Fapparent Substitute V 1.00103 m3, g 9.80 m/s2 2.70103 kg/m3, granite Substitute Fg 26.5 N, Fbuoyant 9.80 N 3 Evaluate the Answer \u2022 Are the units correct? The forces and apparent weight are in newtons, as expected. \u2022 Is the magnitude realistic? The buoyant force is about one-third the weight of the granite, a sensible answer because the density of water is about one-third that of granite. 27. Common brick is about 1.8 times denser than water. What is the apparent weight of a 0.20 m3 block of bricks under water? 28. A girl is floating in a freshwater lake with her head just above the water. If she weighs 610 N, what is the volume of the submerged part of her body? 29. What is the tension in a wire supporting a 1250-N camera submerged in water? The volume of the camera is 16.5103 m3. 30. Plastic foam is about 0.10 times as dense as water. What weight of bricks could you stack on a 1.0 m 1.0 m 0.10 m slab of foam so that the slab of foam floats in water and is barely submerged, leaving the bricks dry? 31. Canoes often have plastic foam blocks mounted under the seats for flotation in case the canoe fills with water. What is the approximate minimum volume of foam needed for flotation for a 480-N canoe? 356 Chapter 13 States of Matter 352-358 CH13-S3-845813 6/3/04 8:41 AM Page 357 Fluids in Motion: Bernoulli\u2019s Principle Try the experiment shown in Figure 13-15. Hold a strip of notebook paper just under your lower lip. Then blow hard across the top surface. Why does the strip of paper rise? The blowing of the air has decreased the air pressure above the paper. Because this pressure decreases, the pressure in the still air below the paper pushes the paper upward. The relationship between the velocity and pressure exerted by a moving fluid is named for Swiss scientist Daniel Bernoulli. Bernoulli\u2019s principle states that as the velocity of ", "a fluid increases, the pressure exerted by that fluid decreases. This principle is a statement of work and energy conservation as applied to fluids. One instance in which fluid velocity can increase is when it flows through a constriction. The nozzles on some garden hoses can be opened or narrowed so that the velocity of the water spray can be changed. You may have seen the water in a stream speed up as it passed through narrowed sections of the stream bed. As the nozzle of the hose and the stream channel become wider or narrower, the velocity of the fluid changes to maintain the overall flow of water. In addition to streams and hoses, the pressure of blood in our circulatory systems depends partly on Bernoulli\u2019s principle. Treatments of heart disease involve removing obstructions in the arteries and veins and preventing clots in the blood. Consider a horizontal pipe completely filled with a smoothly flowing ideal fluid. If a certain mass of the fluid enters one end of the pipe, then an equal mass must come out the other end. Now consider a section of pipe with a cross section that becomes narrower, as shown in Figure 13-16a. To keep the same mass of fluid moving through the narrow section in a fixed amount of time, the velocity of the fluid must increase. As the fluid\u2019s velocity increases, so does its kinetic energy. This means that net work has been done on the swifter fluid. This net work comes from the difference between the work that was done to move the mass of fluid into the pipe and the work that was done by the fluid pushing the same mass out of the pipe. The work is proportional to the force on the fluid, which, in turn, depends on the pressure. If the net work is positive, the pressure at the input end of the section, where the velocity is lower, must be larger than the pressure at the output end, where the velocity is higher. Applications of Bernoulli\u2019s principle There are many common applications of Bernoulli\u2019s principle, such as paint sprayers and perfume bottles. The simple atomizer on a perfume bottle works by blowing air across the top of a tube sunk into perfume, which creates lower pressure at the top of the tube than in the bottle. As a result, perfume is forced into the air flow. A gasoline engine\u2019s carburetor, which is where air and gas are mixed, is another common application of Bernoulli\u2019s principle. Part of the carburetor is a tube with a constriction, as shown in the diagram in Figure 13-16b. The pressure on the gasoline in the fuel supply is the same as the pressure in the thicker part of the tube. Air flowing through the narrow section of the tube, which is attached to the fuel supply, is at a lower pressure, so fuel is forced into the air flow. By regulating the flow of air in the tube, the amount of fuel mixed into the air can be varied. Newer cars tend to have fuel injectors rather than carburetors, but carburetors are common in older cars and the motors of small gasoline-powered machines, such as lawn mowers. \u25a0 Figure 13-15 Blowing across the surface of a sheet of paper demonstrates Bernoulli\u2019s principle. \u25a0 Figure 13-16 Pressure P1 is greater than P2 because v1 is less than v2 (a). In a carburetor, low pressure in the narrow part of the tube draws fuel into the air flow (b). a b Air P1 v1 P2 v2 Air/gasoline mixture Gasoline Section 13.3 Fluids at Rest and in Motion 357 Horizons Companies 352-358 CH13-S3-845813 6/4/04 9:01 AM Page 358 \u25a0 Figure 13-17 The smooth streamlines show the air flowing above a car that is being tested in a wind tunnel. Streamlines Automobile and aircraft manufacturers spend a great deal of time and money testing new designs in wind tunnels to ensure the greatest efficiency of movement through air. The flow of fluids around objects is represented by streamlines, as shown in Figure 13-17. Objects require less energy to move through a smooth streamlined flow. Streamlines can best be illustrated by a simple demonstration. Imagine carefully squeezing tiny drops of food coloring into a smoothly flowing fluid. If the colored lines that form stay thin and well defined, the flow is said to be streamlined. Notice that if the flow narrows, the streamlines move closer together. Closely spaced streamlines indicate greater velocity and therefore reduced pressure. If streamlines swirl and become diffused, the flow of the fluid is said to be turbulent. Bernoulli\u2019s principle does not apply to turbulent flow. 13.3 Section Review 32. Floating and Sinking Does a full soda pop can float or sink in water? Try it. Does it matter whether or not the drink is diet? All soda pop cans contain the same volume of liquid, 354 mL, and dis", "place the same volume of water. What is the difference between a can that sinks and one that floats? 33. Floating and Density A fishing bobber made of cork floats with one-tenth of its volume below the water\u2019s surface. What is the density of cork? 34. Floating in Air A helium balloon rises because of the buoyant force of the air lifting it. The density of helium is 0.18 kg/m3, and the density of air is 1.3 kg/m3. How large a volume would a helium balloon need to lift a 10-N lead brick? 35. Transmission of Pressure A toy rocket launcher is designed so that a child stomps on a rubber cylinder, which increases the air pressure in a launching tube and pushes a foam rocket into the sky. If the child stomps with a force of 150 N on a 2.5103 m2 area piston, what is the additional force transmitted to the 4.0104 m2 launch tube? 36. Pressure and Force An automobile weighing 2.3104 N is lifted by a hydraulic cylinder with an area of 0.15 m2. a. What is the pressure in the hydraulic cylinder? b. The pressure in the lifting cylinder is produced by pushing on a 0.0082 m2 cylinder. What force must be exerted on this small cylinder to lift the automobile? 37. Displacement Which of the following displaces more water when it is placed in an aquarium? a. A 1.0-kg block of aluminum or a 1.0-kg block of lead? b. A 10-cm3 block of aluminum or a 10-cm3 block of lead? 38. Critical Thinking As you discovered in Practice Problem 4, a tornado passing over a house sometimes makes the house explode from the inside out. How might Bernoulli\u2019s principle explain this phenomenon? What could be done to reduce the danger of a door or window exploding outward? 358 Andy Sacks/Getty Images Chapter 13 States of Matter physicspp.com/self_check_quiz 359-363 CH13-S4-845813 4/2/04 6:30 PM Page 359 13.4 Solids How do solids and liquids differ? Solids are stiff, they can be cut in pieces, and they retain their shapes. You can push on solids. Liquids flow and if you push your finger on water, your finger will move through it. However, if you have ever watched butter warm and begin to lose its shape, you may have wondered if the line between solids and liquids is always distinct. Solid Bodies Under certain conditions, solids and liquids are not easily distinguished. As bottle glass is heated through the molten state, the change from solid to liquid is so gradual that it is difficult to tell which is which. Some solids, such as crystalline quartz, are made of particles that are lined up in orderly patterns. Other solids, such as glass, are made of jumbled arrangements of particles, just like a liquid. As shown in Figure 13-18, quartz and fused quartz (also called quartz glass) are the same chemically, but their physical properties are quite different. When the temperature of a liquid is lowered, the average kinetic energy of the particles decreases. As the particles slow down, the cohesive forces have more effect, and for many solids, the particles become frozen into a fixed pattern called a crystal lattice, shown in Figure 13-19 on the next page. Although the cohesive forces hold the particles in place, the particles in a crystalline solid do not stop moving completely. Rather, they vibrate around their fixed positions. In other materials, such as butter and glass, the particles do not form a fixed crystalline pattern. Such a substance, which has no regular crystal structure but does have a definite volume and shape, is called an amorphous solid. Amorphous solids also are classified as viscous, or slowly flowing, liquids. Pressure and freezing As a liquid becomes a solid, its particles usually fit more closely together than in the liquid state, making solids more dense than liquids. As you have learned, however, water is an exception because it is most dense at 4\u00b0C. Water is also an exception to another general rule. For most liquids, an increase in the pressure on the surface of the liquid increases its freezing point. Because water expands as it freezes, an increase in pressure forces the molecules closer together and opposes the freezing. Therefore, higher pressure lowers the freezing point of water very slightly. a b Objectives \u2022 Relate the properties of solids to their structures. \u2022 Explain why solids expand and contract when the temperature changes. \u2022 Calculate the expansion of solids. \u2022 Explain the importance of thermal expansion. Vocabulary crystal lattice amorphous solid coefficient of linear expansion coefficient of volume expansion \u25a0 Figure 13-18 In crystalline quartz, the particles are arranged in an orderly pattern (a). A crystalline solid", " melts at a specific temperature. Fused quartz is the same chemical as crystalline quartz, but the particles are jumbled randomly in the solid. When fused quartz melts, its properties change slowly over a range of temperatures, allowing it to be worked in a fashion similar to everyday glass (b). Section 13.4 Solids 359 (l)Craig Kramer, (r)James L. Amos/Peter Arnold, Inc. 359-363 CH13-S4-845813 3/30/04 3:25 AM Page 360 \u25a0 Figure 13-19 Ice, the solid form of water, has a larger volume than an equal mass of its liquid form (a). The crystalline structure of ice is in the form of a lattice (b). a b O H It has been hypothesized that the drop in water\u2019s freezing point caused by the pressure of an ice-skater\u2019s blades produces a thin film of liquid between the ice and the blades. Calculations of the pressure caused by even the sharpest blade show that the ice is still too cold to melt. More recent measurements, however, have shown that the friction between the blade and the ice generates enough thermal energy to melt the ice and create a thin layer of water. This explanation is supported by measurements of the spray of ice particles, which are considerably warmer than the ice itself. The same process of melting occurs during snow skiing. Elasticity of solids External forces applied to a solid object may twist or bend it out of shape. The ability of a solid object to return to its original form when the external forces are removed is called the elasticity of the solid. If too much deformation occurs, the object will not return to its original shape because its elastic limit has been exceeded. Elasticity depends on the electromagnetic forces that hold the particles of a substance together. Malleability and ductility are two properties that depend on the structure and elasticity of a substance. Because gold can be flattened and shaped into thin sheets, it is said to be malleable. Copper is a ductile metal because it can be pulled into thin strands of wire. Thermal Expansion of Solids It is standard practice for engineers to design small gaps, called expansion joints, into concrete-and-steel highway bridges to allow for the expansion of parts in the heat of summer. Objects expand only a small amount when they are heated, but that small amount could be several centimeters in a 100-m-long bridge. If expansion gaps were not present, the bridge could buckle or parts of it could break. High temperatures can also damage railroad tracks that are laid without expansion joints, as shown in Figure 13-20. Some materials, such as the ovenproof glass that is used for cooking and laboratory experiments, are designed to have the least possible thermal expansion. Large telescope mirrors are made of a ceramic material that is designed to undergo essentially no thermal expansion. To understand the expansion of heated solids, picture a solid as a collection of particles connected by springs that represent the attractive forces between the particles. When the particles get too close, the springs push them apart. When a solid is heated, the kinetic energy of the particles increases, and they vibrate rapidly and move farther apart, weakening the \u25a0 Figure 13-20 The extreme temperatures of a summer day caused these railroad tracks to buckle. 360 Chapter 13 States of Matter (t)file photo, (b)courtesy U.S. Department of Transportation 359-363 CH13-S4-845813 3/30/04 3:25 AM Page 361 attractive forces between the particles. As a result, when the particles vibrate more violently with increased temperature, their average separation increases and the solid expands. Lower temperature The change in length of a solid is proportional to the change in temperature, as shown in Figure 13-21. A solid will expand in length twice as much when its temperature is increased by 20\u00b0C than when it is increased by 10\u00b0C. The expansion also is proportional to its length. A 2-m bar will expand twice as much as a 1-m bar with the same change in temperature. The length, L2, of a solid at temperature T2 can be found with the following equation, where L1 is the length at temperature T1 and alpha,, is the coefficient of linear expansion. L1 T1) L2 L1(T2 Using simple algebra, you can solve for. L1(T2 T L1 L L1 L2 T1) L Coefficient of Linear Expansion T L 1 The coefficient of linear expansion is equal to the change in length, divided by the original length and the change in temperature. The unit for the coefficient of linear expansion is 1/\u00b0C, or (\u00b0C)1. Since solids expand in three directions, the coefficient of volume expansion,, is about three times the coefficient of linear expansion. V Coefficient of Volume Expansion T V 1 The coefficient of volume expansion is equal to the change in volume divided by the original volume and", " the change in temperature. Again, the unit for is 1/\u00b0C, or (\u00b0C)1. The two coefficients of thermal expansion for a variety of materials are given in Table 13-2. Table 13-2 Coefficients of Thermal Expansion at 20\u00b0C Material Coefficient of Linear Expansion, (\u00b0C)1 Coefficient of Volume Expansion, (\u00b0C)1 25106 9106 3106 12106 16106 Solids Aluminum Glass (soft) Glass (ovenproof) Concrete Copper Liquids Methanol Gasoline Water 75106 27106 9106 36106 48106 1200106 950106 210106 L \u2206L Higher temperature \u25a0 Figure 13-21 The change in length of a material is proportional to the original length and the change in temperature. Section 13.4 Solids 361 359-363 CH13-S4-845813 3/30/04 3:26 AM Page 362 Linear Expansion A metal bar is 1.60 m long at room temperature, 21\u00b0C. The bar is put into an oven and heated to a temperature of 84\u00b0C. It is then measured and found to be 1.7 mm longer. What is the coefficient of linear expansion of this material? 1 Analyze and Sketch the Problem \u2022 Sketch the bar, which is 1.7 mm longer at 84\u00b0C than at 21\u00b0C. \u2022 Identify the initial length of the bar, L1, and the change in length, L. L1 \u2206L Unknown:? Known: 1.60 m L1 L 1.7103 m T1 T2 21\u00b0C 84\u00b0C 2 Solve for the Unknown Calculate the coefficient of linear expansion using the known length, change in length, and change in temperature. L L T 1 1.7103 m (1.60 m)(84\u00b0C 21\u00b0C) 1.7105\u00b0C1 Substitute L 1.7103 m, L1 T1) 84\u00b0C 21\u00b0C T (T2 1.60 m, Math Handbook Operations with Significant Digits pages 835\u2014836 3 Evaluate the Answer \u2022 Are the units correct? The units are correctly expressed in \u00b0C1. \u2022 Is the magnitude realistic? The magnitude of the coefficient is close to the accepted value for copper. 39. A piece of aluminum house siding is 3.66 m long on a cold winter day of 28\u00b0C. How much longer is it on a very hot summer day at 39\u00b0C? 40. A piece of steel is 11.5 cm long at 22\u00b0C. It is heated to 1221\u00b0C, close to its melting temperature. How long is it? 41. A 400-mL glass beaker at room temperature is filled to the brim with cold water at 4.4\u00b0C. When the water warms up to 30.0\u00b0C, how much water will spill from the beaker? 42. A tank truck takes on a load of 45,725 L of gasoline in Houston, where the temperature is 28.0\u00b0C. The truck delivers its load in Minneapolis, where the temperature is 12.0\u00b0C. a. How many liters of gasoline does the truck deliver? b. What happened to the gasoline? 43. A hole with a diameter of 0.85 cm is drilled into a steel plate. At 30.0\u00b0C, the hole exactly accommodates an aluminum rod of the same diameter. What is the spacing between the plate and the rod when they are cooled to 0.0\u00b0C? 44. A steel ruler is marked in millimeters so that the ruler is absolutely correct at 30.0\u00b0C. By what percentage would the ruler be incorrect at 30.0\u00b0C? 362 Chapter 13 States of Matter 359-363 CH13-S4-845813 6/4/04 9:04 AM Page 363 You need to make a 1.00-m-long bar that expands with temperature in the same way as a 1.00-m-long bar of copper would. As shown in the figure at the right, your bar must be made from a bar of iron and a bar of aluminum attached end to end. How long should each of them be? Liron Laluminum Applications of thermal expansion Different materials expand at different rates, as indicated by the different coefficients of expansion given in Table 13-2. Engineers must consider these different expansion rates when designing structures. Steel bars are often used to reinforce concrete, and therefore the steel and concrete must have the same expansion coefficient. Otherwise, the structure could crack on a hot day. Similarly, a dentist must use filling materials that expand and contract at the same rate as tooth enamel. Different rates of expansion have useful applications. For example, engineers have taken advantage of these differences to construct a useful device called a bimetallic strip, which is used in thermostats. A bimetallic strip consists of two strips of different metals welded or riveted together. Usually, one", " strip is brass and the other is iron. When heated, brass expands more than iron does. Thus, when the bimetallic strip of brass and iron is heated, the brass part of the strip becomes longer than the iron part. The bimetallic strip bends with the brass on the outside of the curve. If the bimetallic strip is cooled, it bends in the opposite direction. The brass is then on the inside of the curve. In a home thermostat, shown in Figure 13-22, the bimetallic strip is installed so that it bends toward an electric contact as the room cools. When the room cools below the setting on the thermostat, the bimetallic strip bends enough to make electric contact with the switch, which turns on the heater. As the room warms, the bimetallic strip bends in the other direction. When the room\u2019s temperature reaches the setting on the thermostat, the electric circuit is broken and the heater switches off. \u25a0 Figure 13-22 In this thermostat, a coiled bimetallic strip controls the flow of mercury for opening and closing electrical switches. 13.4 Section Review 45. Relative Thermal Contraction On a hot day, you are installing an aluminum screen door in a concrete door frame. You want the door to fit well on a cold winter day. Should you make the door fit tightly in the frame or leave extra room? 46. States of Matter Why could candle wax be considered a solid? Why might it also be considered a viscous liquid? 47. Thermal Expansion Can you heat a piece of cop- per enough to double its length? 48. States of Matter Does Table 13-2 provide a way to distinguish between solids and liquids? 49. Solids and Liquids A solid can be defined as a material that can be bent and will resist bending. Explain how these properties relate to the binding of atoms in a solid, but do not apply to a liquid. 50. Critical Thinking The iron ring in Figure 13-23 was made by cutting a small piece from a solid ring. If the ring in the figure is heated, will the gap become wider or narrower? Explain your answer. \u25a0 Figure 13-23 physicspp.com/self_check_quiz Section 13.4 Solids 363 Doug Martin 364-365 CH13-LAB-845813 3/30/04 3:50 AM Page 364 Alternate CBL instructions can be found on the Web site. physicspp.com Evaporative Cooling If you have ever spilled a small amount of rubbing alcohol on your skin, you probably noticed how cool it felt. You have learned that this coolness is caused by evaporation. In this experiment, you will test the rates at which different types of alcohol evaporate. An alcohol is a substance that has a hydroxyl functional group (\u2013OH) attached to a carbon or a chain of carbons. From your observations of evaporative cooling, you will infer the relative strength of the cohesive forces in the tested alcohols. QUESTION How do the rates of evaporation compare for different alcohols? Objectives Materials \u25a0 Collect and organize data for the evaporation of alcohols. \u25a0 Compare and contrast the rates of evapora- tion for various alcohols. \u25a0 Analyze why some alcohols evaporate faster than others. \u25a0 Infer the relationship between cohesive forces and the rate of evaporation. methanol (methyl alcohol) ethanol (ethyl alcohol) 2-propanol (isopropyl alcohol) masking tape (two pieces) thermometer (non-mercury) filter paper (three pieces, 2.5 cm 2.5 cm) small rubber bands Safety Precautions \u25a0 The chemicals used in this experiment are flammable and poisonous. Do not inhale the fumes from these chemicals. Do not have any open flame near these chemicals. Use in a well-ventilated room or fume hood. \u25a0 Avoid contact with the chemicals on your skin or clothing. Notify your teacher immediately if an accident or spill occurs. \u25a0 Wash your hands after the lab is over. Procedure 1. Wrap the thermometer with a square piece of filter paper fastened by a small rubber band. To do this, first slip the rubber band onto the thermometer. Then, wrap the paper around the thermometer and roll the rubber band over the wrapped paper. The paper should fit snugly over the thermometer\u2019s end. 2. Obtain a small beaker of methanol. Place the paper-covered end of the thermometer in the container of methanol. Do not let the container fall over. Keep the thermometer in the container for 1 min. 364 364-365 CH13-LAB-845813 3/30/04 3:50 AM Page 365 Data Table Liquid Methyl alcohol Ethyl alcohol Isopropyl alcohol T2 (\u00b0C) T1 (\u00b0C) T (\u00b0C) 3. After 1 min", " has elapsed, record the temperature reading on the thermometer in the data table under T1. This is the initial temperature of the methanol. 4. Remove the thermometer from the methanol. Place the thermometer over the edge of a table top so that the thermometer\u2019s tip extends about 5 cm beyond the edge of the table. Use the masking tape to anchor the thermometer in place. 5. Observe the temperature during the experiment. After 4 min have elapsed, observe and record the temperature in the data table in the column marked T2. 6. Roll the rubber band up the thermometer and dispose of the filter paper as directed by your teacher. 7. Repeat steps 1\u20136, but use ethanol as the liquid. Record your results in the data table. 8. Repeat steps 1\u20136, but use isopropyl alcohol as the liquid. Record your results in the data table. Analyze 1. Interpret Data Did the thermometer show a temperature increase or decrease for your trials? Why? 2. Calculate T for each of your liquids by finding the difference between the ending temperatures and the initial temperatures of the liquids (T2 T1). 3. Using the chemical formulas for methanol (CH3OH), ethanol (C2H5OH), and isopropyl alcohol (C3H7OH), determine the molar mass of each of the liquids you tested. You will need to refer to the periodic table to determine the molar masses. 4. Infer What can the T for each trial tell you about the rates of evaporation of the alcohols? 5. Think Critically Why was paper used on the thermometer instead of using only the thermometer? Conclude and Apply 1. Using the rates of evaporation of the alcohols you studied, how can you determine which alcohol had the strongest cohesive forces? 2. Which alcohol had the weakest cohesive forces? 3. What general trend did you find between the change in temperature (T ) and the molar mass of an alcohol? 4. Hypothesize Would a fan blowing in the lab room change the room\u2019s air temperature? Would it change the T that you observed? Explain. Going Further Predict the size of T for 1-butanol, which has the formula C4H9OH, relative to the alcohols that you tested. Real-World Physics The National Weather Service began using a new windchill index in 2001. The old chart was based on data derived from water-freezing experiments done in Antarctica in the 1940s. Explain how windchill relates to evaporative cooling, why this phenomenon is important in cold weather, and how the new chart improves upon the old chart. To find out more about states of matter, visit the Web site: physicspp.com 365 366 CH13-FEATURE-845813 6/4/04 9:08 AM Page 366 A Strange Matter You now are familiar with the four most common states of matter: solid, liquid, gas, and plasma. But did you know that there is a fifth state of matter? Meet the Bose-Einstein Condensate (BEC). What is a Bose-Einstein Condensate? The origins of the BEC are in the 1920s in Satyendra Nath Bose\u2019s studies of the quantum rules governing photon energies. Einstein applied Bose\u2019s equations to atoms. The equations showed that if the temperature of certain atoms were low enough, most of the atoms would be in the same quantum level. In other words, at extremely low temperatures, atoms that had occupied different energy states suddenly fall into the lowest possible energy state. At these temperatures, which are not found in nature, but are created in the lab through some ingenious technology, the atoms of a BEC cannot be distinguished. Even their positions are identical. How is a BEC created? The first BEC was created in 1995 by Eric Cornell and Carl Wieman of Boulder, Colorado. To make their BEC, Cornell and Wieman used rubidium atoms. They had to figure out how to cool these atoms to a lower temperature than had ever been achieved. You might be surprised to learn that one important step in reaching the necessary temperatures was to use lasers to cool the rubidium atoms. Lasers can burn through metal, but they also can cool a sample of atoms if they are tuned so that their photons bounce off the atoms. In this case, the photons will carry off some of the atoms\u2019 energy, lowering the temperature of the sample. But the laser will not cool the sample unless it is precisely tuned. When the laser is tuned to the proper fre- quency, the result is a sample of very cold atoms known as \u201coptical molasses.\u201d How can the optical molasses be contained? Thermal contact between the optical molasses and a 366 courtesy JILA BEC group/University of Colorado Extreme Physics material container surely would warm the chilled atoms because any container would", " have a higher temperature than the atoms. So the container of choice for the BEC is a nonmaterial one\u2014a container formed by combining lasers with a magnetic field. While optical molasses is cold (about 1/10,000 K), it is not cold enough to form a BEC. Scientists use evaporative cooling to make the final step to the required temperatures. In evaporative cooling, the optical molasses is contained in a stronger magnetic container that allows the highest energy atoms to escape. The atoms with the lowest possible energy are left behind. These are the atoms that suddenly condense to form a BEC. The images below show the final cooling of a sample of atoms to form a BEC (the central lump) at fifty-billionths of a kelvin. In these three images, the central peak forms as atoms condense to a BEC. Going Further 1. Evaluate What sorts of difficulties might scientists have to overcome in finding applications for the BEC? 2. Compare and Contrast Is the evaporative cooling process that creates the BEC the same process that helps keep you cool on a hot day? Explain. 367-373 CH13-SG CA STP-845813 3/30/04 3:58 AM Page 367 13.1 Properties of Fluids Vocabulary \u2022 fluids (p. 342) \u2022 pressure (p. 342) \u2022 pascal (p. 342) \u2022 combined gas law (p. 345) \u2022 ideal gas law (p. 345) \u2022 thermal expansion (p. 347) \u2022 plasma (p. 348) Key Concepts \u2022 Matter in the fluid state flows and has no definite shape of its own. \u2022 Pressure is the force divided by the area on which it is exerted. F P A \u2022 The combined gas law can be used to calculate changes in the volume, temperature, and pressure of an ideal gas. 2V P 1V P 1 2 T T 2 1 \u2022 The ideal gas law can be written as follows. PV nRT 13.2 Forces Within Liquids Vocabulary \u2022 cohesive forces (p. 349) \u2022 adhesive forces (p. 350) Key Concepts \u2022 Cohesive forces are the attractive forces that like particles exert on one another. Surface tension and viscosity both result from cohesive forces. \u2022 Adhesive forces are the attractive forces that particles of different substances exert on one another. Capillary action results from adhesive forces. 13.3 Fluids at Rest and in Motion Vocabulary \u2022 Pascal\u2019s principle (p. 352) \u2022 buoyant force (p. 354) \u2022 Archimedes\u2019 principle (p. 354) \u2022 Bernoulli\u2019s principle (p. 357) \u2022 streamlines (p. 358) 13.4 Solids Vocabulary \u2022 crystal lattice (p. 359) \u2022 amorphous solid (p. 359) \u2022 coefficient of linear expansion (p. 361) \u2022 coefficient of volume expansion (p. 361) Key Concepts \u2022 According to Pascal\u2019s principle, an applied pressure change is transmitted undiminished throughout a fluid. F2 1A F 2 A 1 \u2022 Pressure at any depth is proportional to the fluid\u2019s weight above that depth. P hg \u2022 According to Archimedes\u2019 principle, the buoyant force equals the weight of the fluid displaced by an object. Fbuoyant fluidVg \u2022 Bernoulli\u2019s principle states that the pressure exerted by a fluid decreases as its velocity increases. Key Concepts \u2022 A crystalline solid has a regular pattern of particles, and an amorphous solid has an irregular pattern of particles. \u2022 The thermal expansion of a solid is proportional to the temperature change and original size, and it depends on the material. L T L 1 V T V 1 physicspp.com/vocabulary_puzzlemaker 367 367-373 CH13-SG CA STP-845813 6/4/04 9:11 AM Page 368 Concept Mapping 51. Complete the concept map below using the 59. According to Pascal\u2019s principle, what happens to the following terms: density, viscosity, elasticity, pressure. A term may be used more than once. pressure at the top of a container if the pressure at the bottom is increased? (13.3) States of Matter fluids solids Mastering Concepts 52. How are force and pressure different? (13.1) 53. A gas is placed in a sealed container, and some liquid is placed in a container of the same size. The gas and liquid both have definite volume. How do they differ? (13.1) 54. In what way are gases and plasmas similar? In what way are they different? (13.1) 55. The Sun is made of plasma. How is this plasma different from the plasmas on Earth? (13.1) 56. Lakes A frozen lake melts in the spring. What effect does this have on the temperature of the air above the lake? (13.2) 57. H", "iking Canteens used by hikers often are covered with canvas bags. If you wet the canvas bag covering a canteen, the water in the canteen will be cooled. Explain. (13.2) 58. What do the equilibrium tubes in Figure 13-24 tell you about the pressure exerted by a liquid? (13.3) 60. How does the water pressure 1 m below the surface of a small pond compare with the water pressure the same distance below the surface of a lake? (13.3) 61. Does Archimedes\u2019 principle apply to an object inside a flask that is inside a spaceship in orbit? (13.3) 62. A stream of water goes through a garden hose into a nozzle. As the water speeds up, what happens to the water pressure? (13.3) 63. How does the arrangement of atoms in a crystalline substance differ from that in an amorphous substance? (13.4) 64. Does the coefficient of linear expansion depend on the unit of length used? Explain. (13.4) Applying Concepts 65. A rectangular box with its largest surface resting on a table is rotated so that its smallest surface is now on the table. Has the pressure on the table increased, decreased, or remained the same? 66. Show that a pascal is equivalent to a kg/ms2. 67. Shipping Cargo Compared to an identical empty ship, would a ship filled with table-tennis balls sink deeper into the water or rise in the water? Explain. 68. Drops of mercury, water, ethanol, and acetone are placed on a smooth, flat surface, as shown in Figure 13-25. From this figure, what can you conclude about the cohesive forces in these liquids? \u25a0 Figure 13-24 \u25a0 Figure 13-25 69. How deep would a water container have to be to have the same pressure at the bottom as that found at the bottom of a 10.0-cm deep beaker of mercury, which is 13.55 times as dense as water? 368 Chapter 13 States of Matter (r)Tim Courlas, (l)Richard Megna/Fundamental Photographs For more problems, go to Additional Problems, Appendix B. 367-373 CH13-SG CA STP-845813 6/4/04 9:12 AM Page 369 70. Alcohol evaporates more quickly than water does at 78. Soft Drinks Sodas are made fizzy by the carbon the same temperature. What does this observation allow you to conclude about the properties of the particles in the two liquids? 71. Suppose you use a hole punch to make a circular hole in aluminum foil. If you heat the foil, will the size of the hole decrease or increase? Explain. 72. Equal volumes of water are heated in two narrow tubes that are identical, except that tube A is made of soft glass and tube B is made of ovenproof glass. As the temperature increases, the water level rises higher in tube B than in tube A. Give a possible explanation. 73. A platinum wire easily can be sealed in a glass tube, but a copper wire does not form a tight seal with the glass. Explain. 74. Five objects with the following densities are put into a tank of water. a. 0.85 g/cm3 b. 0.95 g/cm3 c. 1.05 g/cm3 d. 1.15 g/cm3 e. 1.25 g/cm3 The density of water is 1.00 g/cm3. The diagram in Figure 13-26 shows six possible positions of these objects. Select a position, from 1 to 6, for each of the five objects. Not all positions need to be selected. dioxide (CO2) dissolved in the liquid. An amount of carbon dioxide equal to about 8.0 L of carbon dioxide gas at atmospheric pressure and 300.0 K can be dissolved in a 2-L bottle of soda. The molar mass of CO2 is 44 g/mol. a. How many moles of carbon dioxide are in the 2-L bottle? (1 L 0.001 m3) b. What is the mass of the carbon dioxide in the 2-L bottle of soda? 79. As shown in Figure 13-27, a constant-pressure thermometer is made with a cylinder containing a piston that can move freely inside the cylinder. The pressure and the amount of gas enclosed in the cylinder are kept constant. As the temperature increases or decreases, the piston moves up or down in the cylinder. At 0\u00b0C, the height of the piston is 20 cm. What is the height of the piston at 100\u00b0C? Cylinder Piston Gas 20 cm \u25a0 Figure 13-27 1 2 3 4 5 \u25a0 Figure 13-26 Mastering Problems 13.1 Properties of Fluids 80. A piston with an area of 0.015 m2 encloses a constant amount of gas in a cylinder with a volume of 0.23 m3.", " The initial pressure of the gas is 1.5105 Pa. A 150-kg mass is then placed on the piston, and the piston moves downward to a new position, as shown in Figure 13-28. If the temperature is constant, what is the new volume of the gas in the cylinder? Volume 0.23 m3 Piston area 0.015 m2 Volume? 6 75. Textbooks A 0.85-kg physics book with dimensions of 24.0 cm 20.0 cm is at rest on a table. a. What force does the book apply to the table? b. What pressure does the book apply? 76. A 75-kg solid cylinder that is 2.5 m long and has an end radius of 7.0 cm stands on one end. How much pressure does it exert? 77. What is the total downward force of the atmosphere on the top of your head right now? Assume that the top of your head has an area of about 0.025 m2. Gas 150 kg Gas \u25a0 Figure 13-28 physicspp.com/chapter_test Chapter 13 Assessment 369 367-373 CH13-SG CA STP-845813 3/30/04 4:00 AM Page 370 81. Automobiles A certain automobile tire is specified to be used at a gauge pressure of 30.0 psi, or 30.0 pounds per square inch. (One pound per square inch equals 6.90103 Pa.) The term gauge pressure means the pressure above atmospheric pressure. Thus, the actual pressure in the tire is 1.01105 Pa (30.0 psi)(6.90103 Pa/psi) 3.08105 Pa. As the car is driven, the tire\u2019s temperature increases, and the volume and pressure increase. Suppose you filled a car\u2019s tire to a volume of 0.55 m3 at a temperature of 280 K. The initial pressure was 30.0 psi, but during the drive, the tire\u2019s temperature increased to 310 K and the tire\u2019s volume increased to 0.58 m3. a. What is the new pressure in the tire? b. What is the new gauge pressure? 13.3 Fluids at Rest and in Motion 82. Reservoirs A reservoir behind a dam is 17-m deep. What is the pressure of the water at the following locations? a. the base of the dam b. 4.0 m from the top of the dam 83. A test tube standing vertically in a test-tube rack contains 2.5 cm of oil ( 0.81 g/cm3) and 6.5 cm of water. What is the pressure exerted by the two liquids on the bottom of the test tube? 84. Antiques An antique yellow metal statuette of a bird is suspended from a spring scale. The scale reads 11.81 N when the statuette is suspended in air, and it reads 11.19 N when the statuette is completely submerged in water. a. Find the volume of the statuette. b. Is the bird made of gold ( 19.3103 kg/m3) or gold-plated aluminum ( 2.7103 kg/m3)? 85. During an ecology experiment, an aquarium halffilled with water is placed on a scale. The scale shows a weight of 195 N. a. A rock weighing 8 N is added to the aquarium. If the rock sinks to the bottom of the aquarium, what will the scale read? b. The rock is removed from the aquarium, and the amount of water is adjusted until the scale again reads 195 N. A fish weighing 2 N is added to the aquarium. What is the scale reading with the fish in the aquarium? 86. What is the size of the buoyant force on a 26.0-N ball that is floating in fresh water? 87. What is the apparent weight of a rock submerged 88. What is the maximum weight that a balloon filled with 1.00 m3 of helium can lift in air? Assume that the density of air is 1.20 kg/m3 and that of helium is 0.177 kg/m3. Neglect the mass of the balloon. 89. If a rock weighs 54 N in air and has an apparent weight of 46 N when submerged in a liquid with a density twice that of water, what will be its apparent weight when it is submerged in water? 90. Oceanography As shown in Figure 13-29, a large buoy used to support an oceanographic research instrument is made of a cylindrical, hollow iron tank. The tank is 2.1 m in height and 0.33 m in diameter. The total mass of the buoy and the research instrument is about 120 kg. The buoy must float so that one end is above the water to support a radio transmitter. Assuming that the mass of the buoy is evenly distributed, how much of the buoy will be above the waterline when it is floating?? m 0.", "33 m 2.1 m \u25a0 Figure 13-29 13.4 Solids 91. A bar of an unknown metal has a length of 0.975 m at 45\u00b0C and a length of 0.972 m at 23\u00b0C. What is its coefficient of linear expansion? 92. An inventor constructs a thermometer from an aluminum bar that is 0.500 m in length at 273 K. He measures the temperature by measuring the length of the aluminum bar. If the inventor wants to measure a 1.0-K change in temperature, how precisely must he measure the length of the bar? 93. Bridges How much longer will a 300-m steel bridge be on a 30\u00b0C day in August than on a 10\u00b0C night in January? in water if the rock weighs 45 N in air and has a volume of 2.1103 m3? 94. What is the change in length of a 2.00-m copper pipe if its temperature is raised from 23\u00b0C to 978\u00b0C? 370 Chapter 13 States of Matter For more problems, go to Additional Problems, Appendix B. 367-373 CH13-SG CA STP-845813 6/4/04 9:15 AM Page 371 95. What is the change in volume of a 1.0-m3 concrete block if its temperature is raised 45\u00b0C? Mixed Review 103. What is the pressure on the hull of a submarine at 96. Bridges Bridge builders often use rivets that are a depth of 65 m? larger than the rivet hole to make the joint tighter. The rivet is cooled before it is put into the hole. Suppose that a builder drills a hole 1.2230 cm in diameter for a steel rivet 1.2250 cm in diameter. To what temperature must the rivet be cooled if it is to fit into the rivet hole, which is at 20\u00b0C? 104. Scuba Diving A scuba diver swimming at a depth of 5.0 m under water exhales a 4.2106 m3 bubble of air. What is the volume of that bubble just before it reaches the surface of the water? 105. An 18-N bowling ball floats with about half of the 97. A steel tank filled with methanol is 2.000 m in diameter and 5.000 m in height. It is completely filled at 10.0\u00b0C. If the temperature rises to 40.0\u00b0C, how much methanol (in liters) will flow out of the tank, given that both the tank and the methanol will expand? 98. An aluminum sphere is heated from 11\u00b0C to 580\u00b0C. If the volume of the sphere is 1.78 cm3 at 11\u00b0C, what is the increase in volume of the sphere at 580\u00b0C? 99. The volume of a copper sphere is 2.56 cm3 after being heated from 12\u00b0C to 984\u00b0C. What was the volume of the copper sphere at 12\u00b0C? 100. A square of iron plate that is 0.3300 m on each side is heated from 0\u00b0C to 95\u00b0C. a. What is the change in the length of the sides of the square? b. What is the relative change in area of the square? ball submerged. a. What is the diameter of the bowling ball? b. What would be the approximate apparent weight of a 36-N bowling ball? 106. An aluminum bar is floating in a bowl of mercury. When the temperature is increased, does the aluminum float higher or sink deeper into the mercury? 107. There is 100.0 mL of water in an 800.0-mL softglass beaker at 15.0\u00b0C. How much will the water level have dropped or risen when the bottle and water are heated to 50.0\u00b0C? 108. Auto Maintenance A hydraulic jack used to lift cars for repairs is called a three-ton jack. The large piston is 22 mm in diameter, and the small one is 6.3 mm in diameter. Assume that a force of 3 tons is 3.0104 N. a. What force must be exerted on the small piston to lift a 3-ton weight? 101. An aluminum cube with a volume of 0.350 m3 at b. Most jacks use a lever to reduce the force 350.0 K is cooled to 270.0 K. a. What is its volume at 270.0 K? b. What is the length of a side of the cube at 270.0 K? 102. Industry A machinist builds a rectangular mechanical part for a special refrigerator system from two rectangular pieces of steel and two rectangular pieces of aluminum. At 293 K, the part is a perfect square, but at 170K, the part becomes warped, as shown in Figure 13-30. Which parts were made of steel and which were made of aluminum? 2 2 1 T = 293K 3 1 T = 170K 3 4 \u25a0", " Figure 13-30 4 needed on the small piston. If the resistance arm is 3.0 cm, how long must the effort arm of an ideal lever be to reduce the force to 100.0 N? 109. Ballooning A hot-air balloon contains a fixed volume of gas. When the gas is heated, it expands and pushes some gas out at the lower, open end. As a result, the mass of the gas in the balloon is reduced. Why would the air in a balloon have to be hotter to lift the same number of people above Vail, Colorado, which has an altitude of 2400 m, than above the tidewater flats of Virginia, which have an altitude of 6 m? 110. The Living World Some plants and animals are able to live in conditions of extreme pressure. a. What is the pressure exerted by the water on the skin of a fish or worm that lives near the bottom of the Puerto Rico Trench, 8600 m below the surface of the Atlantic Ocean? Use 1030 kg/m3 for the density of seawater. b. What would be the density of air at that pressure, relative to its density above the surface of the ocean? physicspp.com/chapter_test Chapter 13 Assessment 371 367-373 CH13-SG CA STP-845813 6/3/04 9:08 AM Page 372 Thinking Critically 111. Apply Concepts You are washing dishes in the sink. A serving bowl has been floating in the sink. You fill the bowl with water from the sink, and it sinks to the bottom. Did the water level in the sink go up or down when the bowl was submerged? Writing in Physics 116. Some solid materials expand when they are cooled. Water between 4\u00b0 and 0\u00b0C is the most common example, but rubber bands also expand in length when cooled. Research what causes this expansion. 112. Apply Concepts Persons confined to bed are less 117. Research Joseph Louis Gay-Lussac and his likely to develop bedsores if they use a waterbed rather than an ordinary mattress. Explain. 113. Analyze and Conclude One method of measuring the percentage of body fat is based on the fact that fatty tissue is less dense than muscle tissue. How can a person\u2019s average density be assessed with a scale and a swimming pool? What measurements does a physician need to record to find a person\u2019s average percentage of body fat? 114. Analyze and Conclude A downward force of 700 N is required to fully submerge a plastic foam sphere, as shown in Figure 13-31. The density of the foam is 95 kg/m3. a. What percentage of the sphere would be submerged if the sphere were released to float freely? b. What is the weight of the sphere in air? c. What is the volume of the sphere? contributions to the gas laws. How did GayLussac\u2019s work contribute to the discovery of the formula for water? Cumulative Review 118. Two blocks are connected by a string over a frictionless, massless pulley such that one is resting on an inclined plane and the other is hanging over the top edge of the plane, as shown in Figure 13-33. The hanging block has a mass of 3.0 kg and the block on the plane has a mass of 2.0 kg. The coefficient of kinetic friction between the block and the inclined plane is 0.19. Answer the following questions assuming the blocks are released from rest. (Chapter 5) a. What is the acceleration of the blocks? b. What is the tension in the string connecting the blocks? 700 N 2.0 kg 45\u00b0 3.0 kg \u25a0 Figure 13-33 \u25a0 Figure 13-31 119. A compact car with a mass of 875 kg, moving 115. Apply Concepts Tropical fish for aquariums are often transported home from pet shops in transparent plastic bags filled mostly with water. If you placed a fish in its unopened transport bag in a home aquarium, which of the cases in Figure 13-32 best represents what would happen? Explain your reasoning. Bag water level Bag water level Aquarium water level Bag water level Aquarium water level Aquarium water level \u25a0 Figure 13-32 south at 15 m/s, is struck by a full-sized car with a mass of 1584 kg, moving east at 12 m/s. The two cars stick together, and momentum is conserved. (Chapter 9) a. Sketch the situation, assigning coordinate axes and identifying \u201cbefore\u201d and \u201cafter.\u201d b. Find the direction and speed of the wreck immediately after the collision, remembering that momentum is a vector quantity. c. The wreck skids along the ground and comes to a stop. The coefficient of kinetic friction while the wreck is skidding is 0.55. Assume that the acceleration is constant. How far does the wreck skid after impact? 120. A 188-W motor will lift a load at the rate (speed) of 6.50 cm", "/s. How great a load can the motor lift at this rate? (Chapter 10) 372 Chapter 13 States of Matter For more problems, go to Additional Problems, Appendix B. 367-373 CH13-SG CA STP-845813 6/4/04 9:18 AM Page 373 Multiple Choice 1. Gas with a volume of 10.0 L is trapped in an expandable cylinder. If the pressure is tripled and the temperature is increased by 80.0 percent (as measured on the Kelvin scale), what will be the new volume of the gas? 2.70 L 6.00 L 16.7 L 54.0 L 2. Nitrogen gas at standard atmospheric pressure, 101.3 kPa, has a volume of 0.080 m3. If there are 3.6 mol of the gas, what is the temperature? 0.27 K 270 K 0.27\u00b0C 270\u00b0C 3. As diagrammed below, an operator applies a force of 200.0 N to the first piston of a hydraulic lift, which has an area of 5.4 cm2. What is the pressure applied to the hydraulic fluid? 3.7101 Pa 2.0103 Pa 3.7103 Pa 3.7105 Pa 200.0 N 6. What is the buoyant force on a 17-kg object that displaces 85 L of water? 1.7102 N 8.3102 N 1.7105 N 8.3105 N 7. Which one of the following items does not contain matter in the plasma state? neon lighting stars lightning incandescent lighting 8. What is the mass of 365 mL of carbon dioxide gas at 3.0 atm pressure (1 atm 101.3 kPa) and 24\u00b0C? The molar mass for carbon dioxide is 44.0 g/mol. 0.045 g 2.0 g 45 g 2.0 kg Extended Answer 9. A balloon has a volume of 125 mL of air at standard atmospheric pressure, 101.3 kPa. If the balloon is anchored 1.27 m under the surface of a swimming pool, as illustrated in the diagram below, what is the new volume of the balloon? Piston 1 Piston 2 1.27 m 4. If the second piston in the lift diagrammed above exerts a force of 41,000 N, what is the area of the second piston? 0.0049 m2 0.026 m2 0.11 m2 11 m2 5. The density of cocobolo wood from Costa Rica is 1.10 g/cm3. What is the apparent weight of a figurine that displaces 786 mL when submerged in a freshwater lake? 0.770 N 0.865 N 7.70 N 8.47 N Work Weak Muscles; Maintain Strong Ones If you\u2019re preparing for a standardized test that covers many topics, it\u2019s sometimes difficult to focus on all the topics that require your attention. Focus most of your energy on your weaker areas and review your stronger topics frequently. physicspp.com/standardized_test Chapter 13 Standardized Test Practice 373 374-380 CH14-S1-845813 3/30/04 5:30 AM Page 374 What You\u2019ll Learn \u2022 You will examine vibrational motion and learn how it relates to waves. \u2022 You will determine how waves transfer energy. \u2022 You will describe wave behavior and discuss its practical significance. Why It\u2019s Important Knowledge of the behavior of vibrations and waves is essential to the understanding of resonance and how safe buildings and bridges are built, as well as how communications through radio and television are achieved. \u201cGalloping Gertie\u201d Shortly after it was opened to traffic, the Tacoma Narrows Bridge near Tacoma, Washington, began to vibrate whenever the wind blew (see inset). One day, the oscillations became so large that the bridge broke apart and collapsed into the water below. Think About This How could a light wind cause the bridge in the inset photo to vibrate with such large waves that it eventually collapsed? physicspp.com 374 Michael T. Sedam/CORBIS, (inset)CORBIS 374-380 CH14-S1-845813 6/4/04 8:53 AM Page 375 How do waves behave in a coiled spring? Question How do pulses that are sent down a coiled spring behave when the other end of the spring is stationary? Procedure Analysis What happens to the pulses as they travel through the spring? What happens as they hit the end of the spring? How did the pulse in step 1 compare to that generated in step 2? Critical Thinking What are some properties that seem to control how a pulse moves through the spring? 1. Stretch out a coiled spiral spring, but do not overstretch it. One person should hold one end still, while the other person generates a sideways pulse in the spring. Observe the pulse while it travels along the", " spring and when it hits the held end. Record your observations. 2. Repeat step 1 with a larger pulse. Record your observations. 3. Generate a different pulse by compressing the spring at one end and letting go. Record your observations. 4. Generate a third type of pulse by twisting one end of the spring and then releasing it. Record your observations. 14.1 Periodic Motion You\u2019ve probably seen a clock pendulum swing back and forth. You would have noticed that every swing followed the same path, and each trip back and forth took the same amount of time. This action is an example of vibrational motion. Other examples include a metal block bobbing up and down on a spring and a vibrating guitar string. These motions, which all repeat in a regular cycle, are examples of periodic motion. In each example, the object has one position at which the net force on it is zero. At that position, the object is in equilibrium. Whenever the object is pulled away from its equilibrium position, the net force on the system becomes nonzero and pulls the object back toward equilibrium. If the force that restores the object to its equilibrium position is directly proportional to the displacement of the object, the motion that results is called simple harmonic motion. Two quantities describe simple harmonic motion. The period, T, is the time needed for an object to repeat one complete cycle of the motion, and the amplitude of the motion is the maximum distance that the object moves from equilibrium. Objectives \u2022 Describe the force in an elastic spring. \u2022 Determine the energy stored in an elastic spring. \u2022 Compare simple harmonic motion and the motion of a pendulum. Vocabulary periodic motion simple harmonic motion period amplitude Hooke\u2019s law pendulum resonance Section 14.1 Periodic Motion 375 Horizons Companies 374-380 CH14-S1-845813 6/4/04 8:54 AM Page 376 a b c The Mass on a Spring 0 m x m 2x m \u25a0 Figure 14-1 The force exerted by a spring is directly proportional to the distance the spring is stretched. How does a spring react to a force that is applied to it? Figure 14-1a shows a spring hanging from a support with nothing attached to it. The spring does not stretch because no external force is exerted on it. Figure 14-1b shows the same spring with an object of weight mg hanging from it. The spring has stretched by distance x so that the upward force it exerts balances the downward force of gravity acting on the object. Figure 14-1c shows the same spring stretched twice as far, 2x, to support twice the weight, 2mg, hanging from it. Hooke\u2019s law states that the force exerted by a spring is directly proportional to the amount that the spring is stretched. A spring that acts in this way is said to obey Hooke\u2019s law, which can be expressed by the following equation. 2mg mg Hooke\u2019s Law F kx The force exerted by a spring is equal to the spring constant times the distance the spring is compressed or stretched from its equilibrium position. In this equation, k is the spring constant, which depends on the stiffness and other properties of the spring, and x is the distance that the spring is stretched from its equilibrium position. Not all springs obey Hooke\u2019s law, but many do. Those that do are called elastic. Potential energy When a force is applied to stretch a spring, such as by hanging an object on its end, there is a direct linear relationship between the exerted force and the displacement, as shown by the graph in Figure 14-2. The slope of the graph is equal to the spring constant, given in units of newtons per meter. The area under the curve represents the work done to stretch the spring, and therefore equals the elastic potential energy that is stored in the spring as a result of that work. The base of the triangle is x, and the height is the force, which, according to the equation for Hooke\u2019s law, is equal to kx, so the potential energy in the spring is given by the following equation. \u25a0 Figure 14-2 The spring constant of a spring can be determined from the graph of force versus displacement of the spring. Potential Energy in a Spring PEsp The potential energy in a spring is equal to one-half times the product of the spring constant and the square of the displacement. 1 kx2 2 The units of the area, and thus, of the potential energy, are newton\u00b7meters, or joules. How does the net force depend upon position? When an object hangs on a spring, the spring stretches until its upward force, Fsp, balances the object\u2019s weight, Fg, as shown in Figure 14-3a. The block is then in its equilibrium position. If you pull the object down, as in Figure 14-3b, the spring force increases, producing a net force upward. When you let", " go of the object, it accelerates in the upward direction, as in Figure 14-3c. However, as the stretch of the spring is reduced, the upward force decreases. F (N) 0 x (m) 376 Chapter 14 Vibrations and Waves 374-380 CH14-S1-845813 6/4/04 8:55 AM Page 377 /s2 Fsp Fg Fsp a a = 0 m/s2 Fg Fsp Fg Fsp a a = 0 m/s2 Fg Fsp Fg In Figure 14-3d, the upward force of the spring and the object\u2019s weight are equal\u2014there is no acceleration. Because there is no net force, the object continues its upward velocity, moving above the equilibrium position. In Figure 14-3e, the net force is in the direction opposite the displacement of the object and is directly proportional to the displacement, so the object moves with a simple harmonic motion. The object returns to the equilibrium position, as in Figure 14-3f. \u25a0 Figure 14-3 Simple harmonic motion is demonstrated by the vibrations of an object hanging on a spring. The Spring Constant and the Energy in a Spring A spring stretches by 18 cm when a bag of potatoes weighing 56 N is suspended from its end. a. Determine the spring constant. b. How much elastic potential energy is stored in the spring when it is stretched this far? 1 Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Show and label the distance that the spring has stretched and its equilibrium position. Known: x 18 cm F 56 N Unknown: k? PEsp? 2 Solve for the Unknown a. Use F kx and solve for k. 0 m 18 cm 56 N The minus sign can be dropped because it just means that the force is restoring. k F x N 6 5 Substitute F 56 N, x 0.18 m.1 8 m 0 310 N/m 1 kx2 2 1 (310 N/m)(0.18 m)2 2 5.0 J b. PEsp Substitute k 310 N/m, x 0.18 m Math Handbook Operations with Significant Digits pages 835\u2014836 3 Evaluate the Answer \u2022 Are the units correct? N/m are the correct units for the spring constant. (N/m)(m2) Nm J, which is the correct unit for energy. \u2022 Is the magnitude realistic? The spring constant is consistent with a scale used, for example, to weigh groceries. The energy of 5.0 J is equal to the value obtained from W Fx mgh, when the average force of 28 N is applied. Section 14.1 Periodic Motion 377 374-380 CH14-S1-845813 3/30/04 5:35 AM Page 378 1. How much force is necessary to stretch a spring 0.25 m when the spring constant is 95 N/m? 2. A spring has a spring constant of 56 N/m. How far will it stretch when a block weighing 18 N is hung from its end? 3. What is the spring constant of a spring that stretches 12 cm when an object weighing 24 N is hung from it? 4. A spring with a spring constant of 144 N/m is compressed by a distance of 16.5 cm. How much elastic potential energy is stored in the spring? 5. A spring has a spring constant of 256 N/m. How far must it be stretched to give it an elastic potential energy of 48 J? When the external force holding the object is released, as in Figure 14-3c, the net force and the acceleration are at their maximum, and the velocity is zero. As the object passes through the equilibrium point, Figure 14-3d, the net force is zero, and so is the acceleration. Does the object stop? No, it would take a net downward force to slow the object, and that will not exist until the object rises above the equilibrium position. When the object comes to the highest position in its oscillation, the net force and the acceleration are again at their maximum, and the velocity is zero. The object moves down through the equilibrium position to its starting point and continues to move in this vibratory manner. The period of oscillation, T, depends upon the mass of the object and the strength of the spring. Automobiles Elastic potential energy is an important part of the design and building of today\u2019s automobiles. Every year, new models of cars are tested to see how well they withstand damage when they crash into barricades at low speeds. A car\u2019s ability to retain its integrity depends upon how much of the kinetic energy it had before the crash can be converted into the elastic potential energy of the frame after the crash. Many bumpers are modified springs that store energy as a car hits a barrier in a slow-speed collision. After the car stops and the spring is compressed, the spring returns to its equilibrium", " position, and the car recoils from the barrier. Pendulums Simple harmonic motion also can be demonstrated by the swing of a pendulum. A simple pendulum consists of a massive object, called the bob, suspended by a string or light rod of length l. After the bob is pulled to one side and released, it swings back and forth, as shown in Figure 14-4. The string or rod exerts a tension force, FT, and gravity exerts a force, Fg, on the bob. The vector sum of the two forces produces the net force, shown at three positions in Figure 14-4. At the left and right positions shown in Figure 14-4, the net force and acceleration are maximum, and the velocity is zero. At the middle position in Figure 14-4, the net force and acceleration are zero, and the velocity is maximum. You can see that the net force is a restoring force; that is, it is opposite the direction of the displacement of the bob and is trying to restore the bob to its equilibrium position. \u25a0 Figure 14-4 Fnet, the vector sum of Ft and Fg, is the restoring force for the pendulum. F T FT Fnet Fg Fnet FT Fg Fg 378 Chapter 14 Vibrations and Waves 374-380 CH14-S1-845813 3/30/04 5:35 AM Page 379 For small angles (less than about 15\u00b0) the restoring force is proportional to the displacement, so the movement is simple harmonic motion. The period of a pendulum is given by the following equation. l Period of a Pendulum T 2 g The period of a pendulum is equal to two pi times the square root of the length of the pendulum divided by the acceleration due to gravity. Notice that the period depends only upon the length of the pendulum and the acceleration due to gravity, not on the mass of the bob or the amplitude of oscillation. One application of the pendulum is to measure g, which can vary slightly at different locations on Earth. Finding g Using a Pendulum A pendulum with a length of 36.9 cm has a period of 1.22 s. What is the acceleration due to gravity at the pendulum\u2019s location? 1 2 Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Label the length of the pendulum. Known: l 36.9 cm T 1.22 s Unknown: g? Solve for the Unknown l T 2 g Solve for g. )2l g (2 2 T 4 9 36. 0 ( m) Substitute l 0.369 m, T 1.22 s )2 ( 2 s 2. 1 9.78 m/s2 2 3 Evaluate the Answer \u2022 Are the units correct? m/s2 are the correct units for acceleration. \u2022 Is the magnitude realistic? The calculated value of g is quite close to the standard value of g, 9.80 m/s2. This pendulum could be at a high elevation above sea level. 36.9 cm Math Handbook Isolating a Variable page 845 6. What is the period on Earth of a pendulum with a length of 1.0 m? 7. How long must a pendulum be on the Moon, where g 1.6 m/s2, to have a period of 2.0 s? 8. On a planet with an unknown value of g, the period of a 0.75-m-long pendulum is 1.8 s. What is g for this planet? Section 14.1 Periodic Motion 379 374-380 CH14-S1-845813 6/4/04 8:55 AM Page 380 A car of mass m rests at the top of a hill of height h before rolling without friction into a crash barrier located at the bottom of the hill. The crash barrier contains a spring with a spring constant, k, which is designed to bring the car to rest with minimum damage. 1. Determine, in terms of m, h, k, and g, the maximum distance, x, that the spring will be compressed when the car hits it. 2. If the car rolls down a hill that is twice as high, how much farther will the spring be compressed? 3. What will happen after the car has been brought to rest? Foucault Pendulum A Foucault pendulum has a long wire, about 16 m in length, with a heavy weight of about 109 kg attached to one end. According to Newton\u2019s first law of motion, a swinging pendulum will keep swinging in the same direction unless it is pushed or pulled in another direction. However, because Earth rotates every 24 h underneath the pendulum, to an observer it would seem as though the pendulum\u2019s direction of swing has changed. To demonstrate this, pegs are arranged in a circle on the floor beneath so that the swinging pendulum will knock them down as the floor", " rotates. At the north pole, this apparent rotation would be 15\u00b0/h. Resonance To get a playground swing going, you \u201cpump\u201d it by leaning back and pulling the chains at the same point in each swing, or a friend gives you repeated pushes at just the right times. Resonance occurs when small forces are applied at regular intervals to a vibrating or oscillating object and the amplitude of the vibration increases. The time interval between applications of the force is equal to the period of oscillation. Other familiar examples of resonance include rocking a car to free it from a snowbank and jumping rhythmically on a trampoline or a diving board. The largeamplitude oscillations caused by resonance can create stresses. Audiences in theater balconies, for example, sometimes damage the structures by jumping up and down with a period equal to the natural oscillation period of the balcony. Resonance is a special form of simple harmonic motion in which the additions of small amounts of force at specific times in the motion of an object cause a larger and larger displacement. Resonance from wind, combined with the design of the bridge supports, may have caused the original Tacoma Narrows Bridge to collapse. 14.1 Section Review 9. Hooke\u2019s Law Two springs look alike but have different spring constants. How could you determine which one has the greater spring constant? 10. Hooke\u2019s Law Objects of various weights are hung from a rubber band that is suspended from a hook. The weights of the objects are plotted on a graph against the stretch of the rubber band. How can you tell from the graph whether or not the rubber band obeys Hooke\u2019s law? 12. Energy of a Spring What is the difference between the energy stored in a spring that is stretched 0.40 m and the energy stored in the same spring when it is stretched 0.20 m? 13. Resonance If a car\u2019s wheel is out of balance, the car will shake strongly at a specific speed, but not when it is moving faster or slower than that speed. Explain. 11. Pendulum How must the length of a pendulum be changed to double its period? How must the length be changed to halve the period? 14. Critical Thinking How is uniform circular motion similar to simple harmonic motion? How are they different? 380 Chapter 14 Vibrations and Waves physicspp.com/self_check_quiz 381-386 CH14-S2-845813 3/30/04 5:41 AM Page 381 14.2 Wave Properties Both particles and waves carry energy, but there is an important differ- ence in how they do this. Think of a ball as a particle. If you toss the ball to a friend, the ball moves from you to your friend and carries energy. However, if you and your friend hold the ends of a rope and you give your end a quick shake, the rope remains in your hand. Even though no matter is transferred, the rope still carries energy through the wave that you created. A wave is a disturbance that carries energy through matter or space. You have learned how Newton\u2019s laws of motion and principles of conservation of energy govern the behavior of particles. These laws and principles also govern the motion of waves. There are many kinds of waves that transmit energy, including the waves you cannot see. Mechanical Waves Water waves, sound waves, and the waves that travel down a rope or spring are types of mechanical waves. Mechanical waves require a medium, such as water, air, ropes, or a spring. Because many other waves cannot be directly observed, mechanical waves can serve as models. Transverse waves The two disturbances shown in Figure 14-5a are called wave pulses. A wave pulse is a single bump or disturbance that travels through a medium. If the wave moves up and down at the same rate, a periodic wave is generated. Notice in Figure 14-5a that the rope is disturbed in the vertical direction, but the pulse travels horizontally. A wave with this type of motion is called a transverse wave. A transverse wave is one that vibrates perpendicular to the direction of the wave\u2019s motion. Longitudinal waves In a coiled-spring toy, you can create a wave pulse in a different way. If you squeeze together several turns of the coiled-spring toy and then suddenly release them, pulses of closely-spaced turns will move away in both directions, as shown in Figure 14-5b. This is called a longitudinal wave. The disturbance is in the same direction as, or parallel to, the direction of the wave\u2019s motion. Sound waves are longitudinal waves. Fluids usually transmit only longitudinal waves. a b Objectives \u2022 Identify how waves transfer energy without transferring matter. \u2022 Contrast transverse and longitudinal waves. \u2022 Relate wave speed, wavelength, and frequency. Vocabulary wave wave pulse periodic wave transverse wave longitudinal wave surface wave trough crest wavelength frequency \u25a0 Figure 14-5", " A quick shake of a rope sends out transverse wave pulses in both directions (a). The squeeze and release of a coiledspring toy sends out longitudinal wave pulses in both directions (b). Section 14.2 Wave Properties 381 381-386 CH14-S2-845813 3/30/04 5:41 AM Page 382 a b Crest Wave motion Trough Surface waves Waves that are deep in a lake or ocean are longitudinal; at the surface of the water, however, the particles move in a direction that is both parallel and perpendicular to the direction of wave motion, as shown in Figure 14-6. Each of the waves is a surface wave, which has characteristics of both transverse and longitudinal waves. The energy of water waves usually comes from distant storms, whose energy initially came from the heating of Earth by solar energy. This energy, in turn, was carried to Earth by transverse electromagnetic waves from the Sun. Measuring a Wave There are many ways to describe or measure a wave. Some characteristics depend on how the wave is produced, whereas others depend on the medium through which the wave travels. Speed How fast does a wave move? The speed of the pulse shown in Figure 14-7 can be found in the same way as the speed of a moving car is determined. First, measure the displacement of the wave peak, d, then divide this by the time interval, t, to find the speed, given by v d/t. The speed of a periodic wave can be found in the same way. For most mechanical waves, both transverse and longitudinal, the speed depends only on the medium through which the waves move. Amplitude How does the pulse generated by gently shaking a rope differ from the pulse produced by a violent shake? The difference is similar to the difference between a ripple in a pond and an ocean breaker: they have different amplitudes. You have learned that the amplitude of a wave is the maximum displacement of the wave from its position of rest, or equilibrium. Two similar waves having different amplitudes are shown in Figure 14-8. A wave\u2019s amplitude depends on how it is generated, but not on its speed. More work must be done to generate a wave with a greater amplitude. For example, strong winds produce larger water waves than those formed by gentle breezes. Waves with greater amplitudes transfer more energy. \u25a0 Figure 14-6 Surface waves have properties of both transverse and longitudinal waves (a). The paths of the individual particles are circular (b). \u25a0 Figure 14-7 These two photographs were taken 0.20 s apart. During that time, the crest moved 0.80 m. The velocity of the wave is 4.0 m/s. 382 (t)CORBIS, (others)Tom Pantages Chapter 14 Vibrations and Waves 381-386 CH14-S2-845813 6/3/04 8:57 AM Page 383 Whereas a small wave might move sand on a beach a few centimeters, a giant wave can uproot and move a tree. For waves that move at the same speed, the rate at which energy is transferred is proportional to the square of the amplitude. Thus, doubling the amplitude of a wave increases the amount of energy it transfers each second by a factor of 4. Wave A Crest Amplitude Trough Wavelength Rather than focusing on one point on a wave, imagine taking a snapshot of the wave so that you can see the whole wave at one instant in time. Figure 14-8 shows each low point, called a trough, and each high point, called a crest, of a wave. The shortest distance between points where the wave pattern repeats itself is called the wavelength. Crests are spaced by one wavelength. Each trough also is one wavelength from the next. The Greek letter lambda,, represents wavelength. Wave B Phase Any two points on a wave that are one or more whole wavelengths apart are in phase. Particles in the medium are said to be in phase with one another when they have the same displacement from equilibrium and the same velocity. Particles in the medium with opposite displacements and velocities are 180\u00b0 out of phase. A crest and a trough, for example, are 180\u00b0 out of phase with each other. Two particles in a wave can be anywhere from 0\u00b0 to 180\u00b0 out of phase with one another. Period and frequency Although wave speed and amplitude can describe both pulses and periodic waves, period, T, and frequency, f, apply only to periodic waves. You have learned that the period of a simple harmonic oscillator, such as a pendulum, is the time it takes for the motion of the oscillator to complete one cycle. Such an oscillator is usually the source, or cause, of a periodic wave. The period of a wave is equal to the period of the source. In Figures 14-9a through 14-9d, the period, T, equals 0.04 s, which is the time it takes the source to complete one cycle", ". The same time is taken by P, a point on the rope, to return to its initial phase. a c Vibrating blade b d P P P P \u25a0 Figure 14-8 The amplitude of wave A is larger than that of wave B. \u25a0 Figure 14-9 One end of a string, with a piece of tape at point P, is attached to a blade vibrating 25 times per second. Note the change in position of point P over time. Section 14.2 Wave Properties 383 381-386 CH14-S2-845813 3/30/04 5:43 AM Page 384 The frequency of a wave, f, is the number of complete oscillations it makes each second. Frequency is measured in hertz. One hertz (Hz) is one oscillation per second. The frequency and period of a wave are related by the following equation. Frequency of a Wave f 1 T The frequency of a wave is equal to the reciprocal of the period. Both the period and the frequency of a wave depend only on its source. They do not depend on the wave\u2019s speed or the medium. Although you can directly measure a wavelength, the wavelength depends on both the frequency of the oscillator and the speed of the wave. In the time interval of one period, a wave moves one wavelength. Therefore, the wavelength of a wave is the speed multiplied by the period, vT. Because the frequency is usually more easily found than the period, this equation is most often written in the following way. Wavelength v f The wavelength of a wave is equal to the velocity divided by the frequency. Picturing waves If you took a snapshot of a transverse wave on a spring, it might look like one of the waves shown in Figure 14-8. This snapshot could be placed on a graph grid to show more information about the wave, as in Figure 14-10a. Similarly, if you record the motion of a single particle, such as point P in Figure 14-9, that motion can be plotted on a displacement-versus-time graph, as in Figure 14-10b. The period is found using the time axis of the graph. Longitudinal waves can also be depicted by graphs, where the y-axis could represent pressure, for example. a b y (m) y (m) 0.3 0.2 0.1 0.0 0.1 0.2 0.3 0.3 0.2 0.1 0.0 0.1 0.2 0.3 2.0 4.0 6.0 8.0 10.0 x (m) 1.0 2.0 3.0 4.0 5.0 t (s) \u25a0 Figure 14-10 Waves can be represented by graphs. The wavelength of this wave is 4.0 m (a). The period is 2.0 s (b). The amplitude, or displacement, is 0.2 m in both graphs. If these graphs represent the same wave, what is its speed? 384 Chapter 14 Vibrations and Waves 381-386 CH14-S2-845813 6/10/04 8:35 PM Page 385 Characteristics of a Wave A sound wave has a frequency of 192 Hz and travels the length of a football field, 91.4 m, in 0.271 s. a. What is the speed of the wave? b. What is the wavelength of the wave? c. What is the period of the wave? d. If the frequency was changed to 442 Hz, what would be the new wavelength and period? 1 2 Analyze and Sketch the Problem \u2022 Draw a model of the football field. \u2022 Diagram a velocity vector. Known: f 192 Hz d 91.4 m t 0.271 s Unknown: v?? T? Solve for the Unknown a. Solve for v 337 m/s Substitute d 91.4 m, t 0.271 s b. Solve for. v f s Substitute v 337 m/s, f 192 Hz /.76 m c. Solve for T. T 1 f 1 Hz 192 0.00521 s d.762 m T 1 f 1 Hz 442 0.00226 s Substitute f 192 Hz Substitute v 337 m/s, f 442 Hz Substitute f 442 Hz v Math Handbook Operations with Significant Digits pages 835\u2014836 3 Evaluate the Answer \u2022 Are the units correct? Hz has the units s1, so (m/s)/Hz (m/s)s m, which is correct. \u2022 Are the magnitudes realistic? A typical sound wave travels approximately 343 m/s, so 337 m/s is reasonable. The frequencies and periods are reasonable for sound waves. 442 Hz is close to a 440-Hz A above middle-C on a piano. Section 14.2 Wave Properties 385 381-386 CH14-S2-845813 3/30/04", " 5:44 AM Page 386 15. A sound wave produced by a clock chime is heard 515 m away 1.50 s later. a. What is the speed of sound of the clock\u2019s chime in air? b. The sound wave has a frequency of 436 Hz. What is the period of the wave? c. What is the wave\u2019s wavelength? 16. A hiker shouts toward a vertical cliff 465 m away. The echo is heard 2.75 s later. a. What is the speed of sound of the hiker\u2019s voice in air? b. The wavelength of the sound is 0.750 m. What is its frequency? c. What is the period of the wave? 17. If you want to increase the wavelength of waves in a rope, should you shake it at a higher or lower frequency? 18. What is the speed of a periodic wave disturbance that has a frequency of 3.50 Hz and a wavelength of 0.700 m? 19. The speed of a transverse wave in a string is 15.0 m/s. If a source produces a disturbance that has a frequency of 6.00 Hz, what is its wavelength? 20. Five pulses are generated every 0.100 s in a tank of water. What is the speed of propagation of the wave if the wavelength of the surface wave is 1.20 cm? 21. A periodic longitudinal wave that has a frequency of 20.0 Hz travels along a coil spring. If the distance between successive compressions is 0.600 m, what is the speed of the wave? You probably have been intuitively aware that waves carry energy that can do work. You may have seen the massive damage done by the huge storm surge of a hurricane or the slower erosion of cliffs and beaches done by small, everyday waves. It is important to remember that while the amplitude of a mechanical wave determines the amount of energy it carries, only the medium determines the wave\u2019s speed. 14.2 Section Review 22. Speed in Different Media If you pull on one end of a coiled-spring toy, does the pulse reach the other end instantaneously? What happens if you pull on a rope? What happens if you hit the end of a metal rod? Compare and contrast the pulses traveling through these three materials. 23. Wave Characteristics You are creating transverse waves in a rope by shaking your hand from side to side. Without changing the distance that your hand moves, you begin to shake it faster and faster. What happens to the amplitude, wavelength, frequency, period, and velocity of the wave? 24. Waves Moving Energy Suppose that you and your lab partner are asked to demonstrate that a transverse wave transports energy without transferring matter. How could you do it? 25. Longitudinal Waves Describe longitudinal waves. What types of media transmit longitudinal waves? 26. Critical Thinking If a raindrop falls into a pool, it creates waves with small amplitudes. If a swimmer jumps into a pool, waves with large amplitudes are produced. Why doesn\u2019t the heavy rain in a thunderstorm produce large waves? 386 Chapter 14 Vibrations and Waves physicspp.com/self_check_quiz 387-391 CH14-S3-845813 3/30/04 5:48 AM Page 387 14.3 Wave Behavior When a wave encounters the boundary of the medium in which it is traveling, it often reflects back into the medium. In other instances, some or all of the wave passes through the boundary into another medium, often changing direction at the boundary. In addition, many properties of wave behavior result from the fact that two or more waves can exist in the same medium at the same time\u2014quite unlike particles. Waves at Boundaries Recall from Section 14.2 that the speed of a mechanical wave depends only on the properties of the medium it passes through, not on the wave\u2019s amplitude or frequency. For water waves, the depth of the water affects wave speed. For sound waves in air, the temperature affects wave speed. For waves on a spring, the speed depends upon the spring\u2019s tension and mass per unit length. Examine what happens when a wave moves across a boundary from one medium into another, as in two springs of different thicknesses joined end-to-end. Figure 14-11 shows a wave pulse moving from a large spring into a smaller one. The wave that strikes the boundary is called the incident wave. One pulse from the larger spring continues in the smaller spring, but at the specific speed of waves traveling through the smaller spring. Note that this transmitted wave pulse remains upward. Some of the energy of the incident wave\u2019s pulse is reflected backward into the larger spring. This returning wave is called the reflected wave. Whether or not the reflected wave is upright or inverted depends on the characteristics of the two springs. For example, if the waves in the smaller spring have a higher speed because the spring is heavier or stiffer, then the reflected", " wave will be inverted. a b Objectives \u2022 Relate a wave\u2019s speed to the medium in which the wave travels. \u2022 Describe how waves are reflected and refracted at boundaries between media. \u2022 Apply the principle of superposition to the phenomenon of interference. Vocabulary incident wave reflected wave principle of superposition interference node antinode standing wave wave front ray normal law of reflection refraction \u25a0 Figure 14-11 The junction of the two springs is a boundary between two media. A pulse reaching the boundary (a) is partially reflected and partially transmitted (b). Section 14.3 Wave Behavior 387 Tom Pantages 387-391 CH14-S3-845813 3/30/04 5:49 AM Page 388 a b \u25a0 Figure 14-12 A pulse approaches a rigid wall (a) and is reflected back (b). Note that the amplitude of the reflected pulse is nearly equal to the amplitude of the incident pulse, but it is inverted. \u25a0 Figure 14-13 When two equal pulses meet, there is a point, called the node (N), where the medium remains undisturbed (a). Constructive interference results in maximum interference at the antinode (A) (b). If the opposite pulses have unequal amplitudes, cancellation is incomplete (c). What happens if the boundary is a wall rather than another spring? When a wave pulse is sent down a spring connected to a rigid wall, the energy transmitted is reflected back from the wall, as shown in Figure 14-12. The wall is the boundary of a new medium through which the wave attempts to pass. Instead of passing through, the pulse is reflected from the wall with almost exactly the same amplitude as the pulse of the incident wave. Thus, almost all the wave\u2019s energy is reflected back. Very little energy is transmitted into the wall. Also note that the pulse is inverted. If the spring were attached to a loose ring around a pole, a free-moving boundary, the wave would not be inverted. Superposition of Waves Suppose a pulse traveling down a spring meets a reflected pulse coming back. In this case, two waves exist in the same place in the medium at the same time. Each wave affects the medium independently. The principle of superposition states that the displacement of a medium caused by two or more waves is the algebraic sum of the displacements caused by the individual waves. In other words, two or more waves can combine to form a new wave. If the waves move in opposite directions, they can cancel or form a new wave of lesser or greater amplitude. The result of the superposition of two or more waves is called interference 388 Tom Pantages Chapter 14 Vibrations and Waves 1 2 3 4 N N N N 387-391 CH14-S3-845813 6/4/04 9:20 AM Page 389 Wave interference Wave interference can be either constructive or destructive. When two pulses with equal but opposite amplitudes meet, the displacement of the medium at each point in the overlap region is reduced. The superposition of waves with equal but opposite amplitudes causes destructive interference, as shown in Figure 14-13a. When the pulses meet and are in the same location, the displacement is zero. Point N, which does not move at all, is called a node. The pulses continue to move and eventually resume their original form. Constructive interference occurs when wave displacements are in the same direction. The result is a wave that has an amplitude greater than those of any of the individual waves. Figure 14-13b shows the constructive interference of two equal pulses. A larger pulse appears at point A when the two waves meet. Point A has the largest displacement and is called the antinode. The two pulses pass through each other without changing their shapes or sizes. If the pulses have unequal amplitudes, the resultant pulse at the overlap is the algebraic sum of the two pulses, as shown in Figure 14-13c. Standing waves You can apply the concept of superimposed waves to the control of the formation of large amplitude waves. If you attach one end of a rope or coiled spring to a fixed point, such as a doorknob, and then start to vibrate the other end, the wave leaves your hand, is reflected at the fixed end, is inverted, and returns to your hand. When it reaches your hand, the reflected wave is inverted and travels back down the rope. Thus, when the wave leaves your hand the second time, its displacement is in the same direction as it was when it left your hand the first time. a What if you want to increase the amplitude of the wave that you create? Suppose you adjust the motion of your hand so that the period of vibration equals the time needed for the wave to make one round-trip from your hand to the door and back. Then, the displacement given by your hand to the rope each time will add to the displacement of the reflected wave. As a result, the oscillation of the rope in one segment will be much greater than", " the motion of your hand. You would expect this based on your knowledge of constructive interference. This large-amplitude oscillation is an example of mechanical resonance. The nodes are at the ends of the rope and an antinode is in the middle, as shown in Figure 14-14a. Thus, the wave appears to be standing still and is called a standing wave. You should note, however, that the standing wave is the interference of the two traveling waves moving in opposite directions. If you double the frequency of vibration, you can produce one more node and one more antinode in the rope. Then it appears to vibrate in two segments. Further increases in frequency produce even more nodes and antinodes, as shown in Figures 14-14b and c. b c Wave Interaction With a coiled-spring toy, you can create pressure waves, as well as transverse waves of various amplitudes, speeds, and orientations. 1. Design an experiment to test what happens when waves from different directions meet. 2. Perform your experiment and record your observations. Analyze and Conclude 3. Does the speed of either wave change? 4. Do the waves bounce off each other, or do they pass through each other? \u25a0 Figure 14-14 Interference produces standing waves in a rope. As the frequency is increased, as shown from top to bottom, the number of nodes and antinodes increases. Section 14.3 Wave Behavior 389 Richard Megna/Fundamental Photographs 387-391 CH14-S3-845813 3/30/04 5:50 AM Page 390 \u25a0 Figure 14-15 Circular waves spread outward from their source (a). The wave can be represented by circles drawn at their crests (b). Notice that the rays are perpendicular to the wave fronts. a b Ray Wave front Waves in Two Dimensions You have studied waves on a rope and on a spring reflecting from rigid supports, where the amplitude of the waves is forced to be zero by destructive interference. These mechanical waves move in only one dimension. However, waves on the surface of water move in two dimensions, and sound waves and electromagnetic waves will later be shown to move in three dimensions. How can two-dimensional waves be demonstrated? Picturing waves in two dimensions When you throw a small stone into a calm pool of water, you see the circular crests and troughs of the resulting waves spreading out in all directions. You can sketch those waves by drawing circles to represent the wave crests. If you dip your finger into water with a constant frequency, the resulting sketch would be a series of concentric circles, called wave fronts, centered on your finger. A wave front is a line that represents the crest of a wave in two dimensions, and it can be used to show waves of any shape, including circular waves and straight waves. Figure 14-15a shows circular waves in water, and Figure 14-15b shows the wave fronts that represent those water waves. Wave fronts drawn to scale show the wavelengths of the waves, but not their amplitudes. Whatever their shape, two-dimensional waves always travel in a direction that is perpendicular to their wave fronts. That direction can be represented by a ray, which is a line drawn at a right angle to the crest of the wave. When all you want to show is the direction in which a wave is traveling, it is convenient to draw rays instead of wave fronts. Reflection of waves in two dimensions A ripple tank can be used to show the properties of two-dimensional waves. A ripple tank contains a thin layer of water. Vibrating boards produce wave pulses, or, in the case of b Barrier r i Normal Reflected ray Incident ray \u25a0 Figure 14-16 A wave pulse in a ripple tank is reflected by a barrier (a). The ray diagram models the wave in time sequence as it approaches the barrier and is then reflected to the right (b). a 390 (t)Fundamental Photographs, (b)Runk/Shoenberger from Grant Heilman Chapter 14 Vibrations and Waves 387-391 CH14-S3-845813 6/4/04 9:23 AM Page 391 Figure 14-16a, traveling waves of water with constant frequency. A lamp above the tank produces shadows below the tank that show the locations of the crests of the waves. The wave pulse travels toward a rigid barrier that reflects the wave: the incident wave moves upward, and the reflected wave moves to the right. The direction of wave motion can be modeled by a ray diagram. Figure 14-16b shows the ray diagram for the waves in the ripple tank. The ray representing the incident wave is the arrow pointing upward. The ray representing the reflected wave points to the right. The direction of the barrier also is shown by a line, which is drawn at a right angle, or perpendicular, to the barrier, called the normal. The angle between the incident ray and the normal is called the angle of incidence. The angle between the normal", " and the reflected ray is called the angle of reflection. The law of reflection states that the angle of incidence is equal to the angle of reflection. a b Refraction of waves in two dimensions A ripple tank also can be used to model the behavior of waves as they move from one medium into another. Figure 14-17a shows a glass plate placed in a ripple tank. The water above the plate is shallower than the water in the rest of the tank and acts like a different medium. As the waves move from deep to shallow water, their speed decreases, and the direction of the waves changes. Because the waves in the shallow water are generated by the waves in the deep water, their frequency is not changed. Based on the equation \u03bb v/f, the decrease in the speed of the waves means that the wavelength is shorter in the shallower water. The change in the direction of waves at the boundary between two different media is known as refraction. Figure 14-17b shows a wave front and ray model of refraction. When you study the reflection and refraction of light in Chapter 17, you will learn the law of refraction, called Snell\u2019s law. You may not be aware that echoes are caused by the reflection of sound off hard surfaces, such as the walls of a large warehouse or a distant cliff face. Refraction is partly responsible for rainbows. When white light passes through a raindrop, refraction separates the light into its individual colors. \u25a0 Figure 14-17 As the water waves move over a shallower region of the ripple tank where a glass plate is placed, they slow down and their wavelength decreases (a). Refraction can be represented by a diagram of wave fronts and rays (b). 14.3 Section Review 27. Waves at Boundaries Which of the following wave characteristics remain unchanged when a wave crosses a boundary into a different medium: frequency, amplitude, wavelength, velocity, and/or direction? 28. Refraction of Waves Notice in Figure 14-17a how the wave changes direction as it passes from one medium to another. Can two-dimensional waves cross a boundary between two media without changing direction? Explain. 29. Standing Waves In a standing wave on a string fixed at both ends, how is the number of nodes related to the number of antinodes? 30. Critical Thinking As another way to understand wave reflection, cover the right-hand side of each drawing in Figure 14-13a with a piece of paper. The edge of the paper should be at point N, the node. Now, concentrate on the resultant wave, shown in darker blue. Note that it acts like a wave reflected from a boundary. Is the boundary a rigid wall, or is it open-ended? Repeat this exercise for Figure 14-13b. physicspp.com/self_check_quiz Section 14.3 Wave Behavior 391 Tom Pantages 392-393 CH14-LAB-845813 6/4/04 9:30 AM Page 392 Pendulum Vibrations A pendulum can provide a simple model for the investigation of wave properties. In this experiment, you will design a procedure to use the pendulum to examine amplitude, period, and frequency of a wave. You also will determine the acceleration due to gravity from the period and length of the pendulum. QUESTION How can a pendulum demonstrate the properties of waves? Objectives Procedure \u25a0 Determine what variables affect a pendulum\u2019s period. \u25a0 Investigate the frequency and period amplitude of a pendulum. \u25a0 Measure g, the acceleration due to gravity, using the period and length of a pendulum. Safety Precautions Possible Materials string (1.5 m) three sinkers paper clip ring stand with ring stopwatch 1. Design a pendulum using a ring stand, a string with a paper clip, and a sinker attached to the paper clip. Be sure to check with your teacher and have your design approved before you proceed with the lab. 2. For this investigation, the length of the pendulum is the length of the string plus half the length of the bob. The amplitude is how far the bob is pulled from its equilibrium point. The frequency is the cycles/s of the bob. The period is the time for the bob to travel back and forth (one cycle). When collecting data for the period, find the time it takes to make ten cycles, and then calculate the period in s/cycles. When finding frequency, count how many cycles occur in 10 s, and then convert your value to cycles/s. 3. Design a procedure that keeps the mass of the bob and the amplitude constant, but varies the length. Determine the frequency and period of the pendulum. Record your results in the data table. Use several trials at several lengths to collect your data. 4. Design a procedure that keeps length and amplitude constant, but varies the mass of the bob. Determine the frequency and period of the pendulum. Record your results in the data table. Use several trials to collect", " your data. 5. Design a procedure that keeps length and mass of the bob constant, but varies the amplitude of the pendulum. Determine the frequency and period of the pendulum. Record your results in the data table. Use several trials to collect your data. 6. Design a procedure using the pendulum to calculate g, the acceleration due to gravity, using the equation T 2/g. T is the period, and is the length of the pendulum string. Remember to use several trials to collect your data. 392 Horizons Companies 392-393 CH14-LAB-845813 3/30/04 6:24 AM Page 393 Data Table 1 This data table format can be used for steps 2\u20135. Trial 1 Trial 2 Trial 3 Average Length 1 Length 2 Length 3 Mass 1 Mass 2 Mass 3 Amplitude 1 Amplitude 2 Amplitude 3 Period (s/cycle) \u2014\u2014\u2014 \u2014\u2014\u2014 \u2014\u2014\u2014 Frequency (cycles/s) \u2014\u2014\u2014 \u2014\u2014\u2014 \u2014\u2014\u2014 Data Table 2 This data table format can be used for step 6, finding g. Trial 1 Trial 2 Trial 3 Average Period (s/cycle) Length of String (m) Length 1 Length 2 Length 3 Analyze 1. Summarize What is the relationship between the pendulum\u2019s amplitude and its period? 4. Analyze and Conclude When does the pendulum bob have the greatest kinetic energy? 2. Summarize What is the relationship between the pendulum\u2019s bob mass and its period? 5. Analyze and Conclude When does the pendulum bob have the greatest potential energy? 3. Compare and Contrast How are the period and length of a pendulum related? 4. Analyze Calculate g from your data in step 6. 5. Error Analysis What is the percent error of your experimental g value? What are some possible reasons for the difference between your experimental value of g and the accepted value of g? Conclude and Apply 1. Infer What variable(s) affects a pendulum\u2019s period? 2. Analyze Why is it better to run three or more trials to obtain the frequency and period of each pendulum? 3. Compare How is the motion of a pendulum like that of a wave? Going Further Suppose you had a very long pendulum. What other observations could be made, over the period of a day, of this pendulum\u2019s motion? Real-World Physics Pendulums are used to drive some types of clocks. Using the observations from your experiments, what design problems are there in using your pendulum as a time-keeping instrument? To find out more about the behavior of waves, visit the Web site: physicspp.com 393 394 CH14-FEATURE-845813 3/30/04 6:26 AM Page 394 Earthquake Protection Earthquake Protection An earthquake is the equivalent of a violent explosion somewhere beneath the surface of Earth. The mechanical waves that radiate from an earthquake are both transverse and longitudinal. Transverse waves shake a structure horizontally, while longitudinal waves cause vertical shaking. Earthquakes cannot be predicted or prevented, so we must construct our buildings to withstand them. and its foundation. To minimize vertical shaking of a building, springs are inserted into the vertical members of the framework. These springs are made of a strong rubber compound compressed within heavy structural steel cylinders. Sideways shaking is diminished by placing sliding supports beneath the building columns. These allow the structure to remain stationary if the ground beneath it moves sideways. Foundation wall Flexible grating \u201dMoat\u201c allows movement between building and surrounding earth. Columns (extensions of internal support beams) Braces prevent columns from bending. Structural rubber cushions allow 24 inches of vertical movement. Special pads support the building, yet allow sliding if earth moves horizontally. Column ends can slide 22 inches in any direction on smooth support pads. New building designs reduce damage by earthquakes. As our knowledge of earthquakes increases, existing buildings must be retrofitted to withstand newly discovered types of earthquakerelated failures. Reducing Damage Most bridges and parking ramps were built by stacking steelreinforced concrete sections atop one another. Gravity keeps them in place. These structures are immensely strong under normal conditions, but they can be shaken apart by a strong earthquake. New construction codes dictate that their parts must be bonded together by heavy steel straps. Earthquake damage to buildings also can be reduced by allowing a small amount of controlled movement between the building frame 394 Technology and Society Long structures, like tunnels and bridges, must be constructed to survive vertical or horizontal shearing fractures of the earth beneath. The Bay Area Rapid Transit tunnel that runs beneath San Francisco Bay has flexible couplings for stability should the bay floor buckle. Going Further 1. Research What is the framework of your school made of and how were the foundations built? 2. Observe Find a brick building that has a crack in one of its walls. See if you can tell why the crack formed and why it took the path that it did. What might this have to do with earthquakes", "? 395-401 CH14-SG CA STP-845813 3/30/04 6:31 AM Page 395 14.1 Periodic Motion Vocabulary \u2022 periodic motion (p. 375) \u2022 simple harmonic motion (p. 375) \u2022 period (p. 375) \u2022 amplitude (p. 375) \u2022 Hooke\u2019s law (p. 376) \u2022 pendulum (p. 378) \u2022 resonance (p. 380) 14.2 Wave Properties Vocabulary \u2022 wave (p. 381) \u2022 wave pulse (p. 381) \u2022 periodic wave (p. 381) \u2022 transverse wave (p. 381) \u2022 longitudinal wave (p. 381) \u2022 surface wave (p. 382) \u2022 trough (p. 383) \u2022 crest (p. 383) \u2022 wavelength (p. 383) \u2022 frequency (p. 384) 14.3 Wave Behavior Vocabulary \u2022 incident wave (p. 387) \u2022 reflected wave (p. 387) \u2022 principle of superposition (p. 388) \u2022 interference (p. 388) \u2022 node (p. 389) \u2022 antinode (p. 389) \u2022 standing wave (p. 389) \u2022 wave front (p. 390) \u2022 ray (p. 390) \u2022 normal (p. 391) \u2022 law of reflection (p. 391) \u2022 refraction (p. 391) Key Concepts \u2022 Periodic motion is any motion that repeats in a regular cycle. \u2022 Simple harmonic motion results when the restoring force on an object is directly proportional to the object\u2019s displacement from equilibrium. Such a force obeys Hooke\u2019s law. F kx \u2022 The elastic potential energy stored in a spring that obeys Hooke\u2019s law is expressed by the following equation. PEsp 1 kx2 2 \u2022 The period of a pendulum can be found with the following equation. l T 2 g Key Concepts \u2022 Waves transfer energy without transferring matter. \u2022 In transverse waves, the displacement of the medium is perpendicular to the direction of wave motion. In longitudinal waves, the displacement is parallel to the direction of wave motion. \u2022 Frequency is the number of cycles per second and is related to period by: f 1 T \u2022 The wavelength of a continuous wave can be found by using the following equation. v f Key Concepts \u2022 When a wave crosses a boundary between two media, it is partially transmitted and partially reflected. \u2022 The principle of superposition states that the displacement of a medium resulting from two or more waves is the algebraic sum of the displacements of the individual waves. \u2022 Interference occurs when two or more waves move through a medium at the same time. \u2022 When two-dimensional waves are reflected from boundaries, the angles of incidence and reflection are equal. \u2022 The change in direction of waves at the boundary between two different media is called refraction. physicspp.com/vocabulary_puzzlemaker 395 395-401 CH14-SG CA STP-845813 6/4/04 9:34 AM Page 396 Concept Mapping 31. Complete the concept map using the following terms and symbols: amplitude, frequency, v,, T. Waves 45. Suppose you send a pulse along a rope. How does the position of a point on the rope before the pulse arrives compare to the point\u2019s position after the pulse has passed? (14.2) 46. What is the difference between a wave pulse and a periodic wave? (14.2) 47. Describe the difference between wave frequency and speed period wavelength wave velocity. (14.2) A f Mastering Concepts 32. What is periodic motion? Give three examples of periodic motion. (14.1) 33. What is the difference between frequency and period? How are they related? (14.1) 34. What is simple harmonic motion? Give an example of simple harmonic motion. (14.1) 35. If a spring obeys Hooke\u2019s law, how does it behave? (14.1) 36. How can the spring constant of a spring be determined from a graph of force versus displacement? (14.1) 37. How can the potential energy in a spring be determined from the graph of force versus displacement? (14.1) 38. Does the period of a pendulum depend on the mass of the bob? The length of the string? Upon what else does the period depend? (14.1) 39. What conditions are necessary for resonance to occur? (14.1) 40. How many general methods of energy transfer are there? Give two examples of each. (14.2) 41. What is the primary difference between a mechanical wave and an electromagnetic wave? (14.2) 42. What are the differences among transverse, longitudinal, and surface waves? (14.2) 43. Waves are sent along a spring of fixed length. (14.2) a. Can the speed of the waves in the spring be changed?", " Explain. b. Can the frequency of a wave in the spring be changed? Explain. 44. What is the wavelength of a wave? (14.2) 48. Suppose you produce a transverse wave by shaking one end of a spring from side to side. How does the frequency of your hand compare with the frequency of the wave? (14.2) 49. When are points on a wave in phase with each other? When are they out of phase? Give an example of each. (14.2) 50. What is the amplitude of a wave and what does it represent? (14.2) 51. Describe the relationship between the amplitude of a wave and the energy it carries. (14.2) 52. When a wave reaches the boundary of a new medium, what happens to it? (14.3) 53. When a wave crosses a boundary between a thin and a thick rope, as shown in Figure 14-18, its wavelength and speed change, but its frequency does not. Explain why the frequency is constant. (14.3) \u25a0 Figure 14-18 54. How does a spring pulse reflected from a rigid wall differ from the incident pulse? (14.3) 55. Describe interference. Is interference a property of only some types of waves or all types of waves? (14.3) 56. What happens to a spring at the nodes of a standing wave? (14.3) 57. Violins A metal plate is held fixed in the center and sprinkled with sugar. With a violin bow, the plate is stroked along one edge and made to vibrate. The sugar begins to collect in certain areas and move away from others. Describe these regions in terms of standing waves. (14.3) 58. If a string is vibrating in four parts, there are points where it can be touched without disturbing its motion. Explain. How many of these points exist? (14.3) 59. Wave fronts pass at an angle from one medium into a second medium, where they travel with a different speed. Describe two changes in the wave fronts. What does not change? (14.3) 396 Chapter 14 Vibrations and Waves For more problems, go to Additional Problems, Appendix B. 395-401 CH14-SG CA STP-845813 7/14/04 1:45 PM Page 397 Applying Concepts 60. A ball bounces up and down on the end of a spring. Mastering Problems 14.1 Periodic Motion Describe the energy changes that take place during one complete cycle. Does the total mechanical energy change? 61. Can a pendulum clock be used in the orbiting International Space Station? Explain. 62. Suppose you hold a 1-m metal bar in your hand and hit its end with a hammer, first, in a direction parallel to its length, and second, in a direction at right angles to its length. Describe the waves produced in the two cases. 63. Suppose you repeatedly dip your finger into a sink full of water to make circular waves. What happens to the wavelength as you move your finger faster? 64. What happens to the period of a wave as the frequency increases? 65. What happens to the wavelength of a wave as the frequency increases? 66. Suppose you make a single pulse on a stretched spring. How much energy is required to make a pulse with twice the amplitude? 67. You can make water slosh back and forth in a shallow pan only if you shake the pan with the correct frequency. Explain. 68. In each of the four waves in Figure 14-19, the pulse on the left is the original pulse moving toward the right. The center pulse is a reflected pulse; the pulse on the right is a transmitted pulse. Describe the rigidity of the boundaries at A, B, C, and D. A B C D \u25a0 Figure 14-19 69. A spring stretches by 0.12 m when some apples weighing 3.2 N are suspended from it, as shown in Figure 14-20. What is the spring constant of the spring? 3.2 N \u25a0 Figure 14-20 70. Car Shocks Each of the coil springs of a car has a spring constant of 25,000 N/m. How much is each spring compressed if it supports one-fourth of the car\u2019s 12,000-N weight? 71. How much potential energy is stored in a spring with a spring constant of 27 N/m if it is stretched by 16 cm? 72. Rocket Launcher A toy rocket-launcher contains a spring with a spring constant of 35 N/m. How far must the spring be compressed to store 1.5 J of energy? 73. Force-versus-length data for a spring are plotted on the graph in Figure 14-21. a. What is the spring constant of the spring? b. What is the energy stored in the spring when it is stretched to a length of 0.50 m? ) N", " ( e c r o F 12.0 8.0 4.0 0.0 0.20 0.40 0.60 Length (m) \u25a0 Figure 14-21 physicspp.com/chapter_test Chapter 14 Assessment 397 395-401 CH14-SG CA STP-845813 3/30/04 6:32 AM Page 398 74. How long must a pendulum be to have a period of 2.3 s on the Moon, where g 1.6 m/s2? 14.2 Wave Properties 75. Building Motion The Sears Tower in Chicago, shown in Figure 14-22, sways back and forth in the wind with a frequency of about 0.12 Hz. What is its period of vibration? \u25a0 Figure 14-22 76. Ocean Waves An ocean wave has a length of 12.0 m. A wave passes a fixed location every 3.0 s. What is the speed of the wave? 77. Water waves in a shallow dish are 6.0-cm long. At one point, the water moves up and down at a rate of 4.8 oscillations/s. a. What is the speed of the water waves? b. What is the period of the water waves? 78. Water waves in a lake travel 3.4 m in 1.8 s. The period of oscillation is 1.1 s. a. What is the speed of the water waves? b. What is their wavelength? 79. Sonar A sonar signal of frequency 1.00106 Hz has a wavelength of 1.50 mm in water. a. What is the speed of the signal in water? b. What is its period in water? c. What is its period in air? 80. A sound wave of wavelength 0.60 m and a velocity of 330 m/s is produced for 0.50 s. a. What is the frequency of the wave? b. How many complete waves are emitted in this time interval? 81. The speed of sound in water is 1498 m/s. A sonar signal is sent straight down from a ship at a point just below the water surface, and 1.80 s later, the reflected signal is detected. How deep is the water? 82. Pepe and Alfredo are resting on an offshore raft after a swim. They estimate that 3.0 m separates a trough and an adjacent crest of each surface wave on the lake. They count 12 crests that pass by the raft in 20.0 s. Calculate how fast the waves are moving. 83. Earthquakes The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A seismograph records the arrival of the transverse waves 68 s before the arrival of the longitudinal waves. How far away is the earthquake? 14.3 Wave Behavior 84. Sketch the result for each of the three cases shown in Figure 14-23, when the centers of the two approaching wave pulses lie on the dashed line so that the pulses exactly overlap. 1 2 3 \u25a0 Figure 14-23 85. If you slosh the water in a bathtub at the correct frequency, the water rises first at one end and then at the other. Suppose you can make a standing wave in a 150-cm-long tub with a frequency of 0.30 Hz. What is the velocity of the water wave? 86. Guitars The wave speed in a guitar string is 265 m/s. The length of the string is 63 cm. You pluck the center of the string by pulling it up and letting go. Pulses move in both directions and are reflected off the ends of the string. a. How long does it take for the pulse to move to the string end and return to the center? b. When the pulses return, is the string above or below its resting location? c. After 0.50 s, how far is the front of the wave from the source of the sound? c. If you plucked the string 15 cm from one end of the string, where would the two pulses meet? 398 Bruce Leighty/Image Finders Chapter 14 Vibrations and Waves For more problems, go to Additional Problems, Appendix B. 395-401 CH14-SG CA STP-845813 6/10/04 8:38 PM Page 399 87. Sketch the result for each of the four cases shown in 92. Bungee Jumper A high-altitude bungee jumper Figure 14-24, when the centers of each of the two wave pulses lie on the dashed line so that the pulses exactly overlap. 1 2 3 4 \u25a0 Figure 14-24 Mixed Review 88. What is the period of a pendulum with a length of 1.4 m? 89. The frequency of yellow light is 5.11014 Hz. Find the wavelength of yellow light. The speed of light is 3.00108 m/s. 90. Radio", " Wave AM-radio signals are broadcast at frequencies between 550 kHz (kilohertz) and 1600 kHz and travel 3.0108 m/s. a. What is the range of wavelengths for these signals? b. FM frequencies range between 88 MHz (megahertz) and 108 MHz and travel at the same speed. What is the range of FM wavelengths? 91. You are floating just offshore at the beach. Even though the waves are steadily moving in toward the beach, you don\u2019t move any closer to the beach. a. What type of wave are you experiencing as you float in the water? b. Explain why the energy in the wave does not move you closer to shore. c. In the course of 15 s you count ten waves that pass you. What is the period of the waves? d. What is the frequency of the waves? e. You estimate that the wave crests are 3 m apart. What is the velocity of the waves? f. After returning to the beach, you learn that the waves are moving at 1.8 m/s. What is the actual wavelength of the waves? jumps from a hot-air balloon using a 540-m-bungee cord. When the jump is complete and the jumper is just suspended from the cord, it is stretched 1710 m. What is the spring constant of the bungee cord if the jumper has a mass of 68 kg? 93. The time needed for a water wave to change from the equilibrium level to the crest is 0.18 s. a. What fraction of a wavelength is this? b. What is the period of the wave? c. What is the frequency of the wave? 94. When a 225-g mass is hung from a spring, the spring stretches 9.4 cm. The spring and mass then are pulled 8.0 cm from this new equilibrium position and released. Find the spring constant of the spring and the maximum speed of the mass. 95. Amusement Ride You notice that your favorite amusement-park ride seems bigger. The ride consists of a carriage that is attached to a structure so it swings like a pendulum. You remember that the carriage used to swing from one position to another and back again eight times in exactly 1 min. Now it only swings six times in 1 min. Give your answers to the following questions to two significant digits. a. What was the original period of the ride? b. What is the new period of the ride? c. What is the new frequency? d. How much longer is the arm supporting the carriage on the larger ride? e. If the park owners wanted to double the period of the ride, what percentage increase would need to be made to the length of the pendulum? 96. Clocks The speed at which a grandfather clock runs is controlled by a swinging pendulum. a. If you find that the clock loses time each day, what adjustment would you need to make to the pendulum so it will keep better time? b. If the pendulum currently is 15.0 cm, by how much would you need to change the length to make the period lessen by 0.0400 s? 97. Bridge Swinging In the summer over the New River in West Virginia, several teens swing from bridges with ropes, then drop into the river after a few swings back and forth. a. If Pam is using a 10.0-m length of rope, how long will it take her to reach the peak of her swing at the other end of the bridge? b. If Mike has a mass that is 20 kg more than Pam, how would you expect the period of his swing to differ from Pam\u2019s? c. At what point in the swing is KE at a maximum? d. At what point in the swing is PE at a maximum? e. At what point in the swing is KE at a minimum? f. At what point in the swing is PE at a minimum? physicspp.com/chapter_test Chapter 14 Assessment 399 395-401 CH14-SG CA STP-845813 6/4/04 9:36 AM Page 400 98. You have a mechanical fish scale that is made with a spring that compresses when weight is added to a hook attached below the scale. Unfortunately, the calibrations have completely worn off of the scale. However, you have one known mass of 500.0 g that displaces the spring 2.0 cm. a. What is the spring constant for the spring? b. If a fish displaces the spring 4.5 cm, what is the mass of the fish? 99. Car Springs When you add a 45-kg load to the trunk of a new small car, the two rear springs compress an additional 1.0 cm. a. What is the spring constant for each of the springs? b. How much additional potential energy is stored in each of the car springs after loading the trunk? a. Make a graph of the force applied to the spring versus the spring", " length. Plot the force on the y-axis. b. Determine the spring constant from the graph. c. Using the graph, find the elastic potential energy stored in the spring when it is stretched to 0.50 m. 103. Apply Concepts Gravel roads often develop regularly spaced ridges that are perpendicular to the road, as shown in Figure 14-25. This effect, called washboarding, occurs because most cars travel at about the same speed and the springs that connect the wheels to the cars oscillate at about the same frequency. If the ridges on a road are 1.5 m apart and cars travel on it at about 5 m/s, what is the frequency of the springs\u2019 oscillation? 100. The velocity of a wave on a string depends on how tightly the string is stretched, and on the mass per unit length of the string. If FT is the tension in the string, and is the mass/unit length, then the velocity, v, can be determined by the following equation. T v F A piece of string 5.30-m long has a mass of 15.0 g. What must the tension in the string be to make the wavelength of a 125-Hz wave 120.0 cm? Thinking Critically 101. Analyze and Conclude A 20-N force is required to stretch a spring by 0.5 m. a. What is the spring constant? b. How much energy is stored in the spring? c. Why isn\u2019t the work done to stretch the spring equal to the force times the distance, or 10 J? 102. Make and Use Graphs Several weights were suspended from a spring, and the resulting extensions of the spring were measured. Table 14-1 shows the collected data. Table 14-1 Weights on a Spring Force, F (N) Extension, x (m) 2.5 5.0 7.5 10.0 12.5 15.0 0.12 0.26 0.35 0.50 0.60 0.71 \u25a0 Figure 14-25 Writing in Physics 104. Research Christiaan Huygens\u2019 work on waves and the controversy between him and Newton over the nature of light. Compare and contrast their explanations of such phenomena as reflection and refraction. Whose model would you choose as the best explanation? Explain why. Cumulative Review 105. A 1400-kg drag racer automobile can complete a one-quarter mile (402 m) course in 9.8 s. The final speed of the automobile is 250 mi/h (112 m/s). (Chapter 11) a. What is the kinetic energy of the automobile? b. What is the minimum amount of work that was done by its engine? Why can't you calculate the total amount of work done? c. What was the average acceleration of the automobile? 106. How much water would a steam engine have to evaporate in 1 s to produce 1 kW of power? Assume that the engine is 20 percent efficient. (Chapter 12) 400 Darrell Gulin/CORBIS Chapter 14 Vibrations and Waves For more problems, go to Additional Problems, Appendix B. 395-401 CH14-SG CA STP-845813 6/4/04 9:38 AM Page 401 Multiple Choice 1. What is the value of the spring constant of a spring with a potential energy of 8.67 J when it\u2019s stretched 247 mm? 70.2 N/m 71.1 N/m 142 N/m 284 N/m 2. What is the force acting on a spring with a spring constant of 275 N/m that is stretched 14.3 cm? 2.81 N 19.2 N 39.3 N 3.931030 N 3. A mass stretches a spring as it hangs from the spring. What is the spring constant? 0.25 N/m 0.35 N/m 26 N/m 3.5102 N/m 0.85 m 30.4 g 4. A spring with a spring constant of 350 N/m pulls a door closed. How much work is done as the spring pulls the door at a constant velocity from an 85.0-cm stretch to a 5.0-cm stretch? 112 Nm 130 J 224 Nm 1.12103 J 5. What is the correct rearrangement of the formula for the period of a pendulum to find the length of the pendulum? l 4 2g T2 g T l 2 4 2g T l )2 (2 T g l 2 6. What is the frequency of a wave with a period of 3 s? 0.3 Hz 3 Hz c Hz 3 3 Hz 7. Which option describes a standing wave? Waves Direction Medium Identical Same Same Nonidentical Opposite Different Identical Opposite Same Nonidentical Same Different 8. A 1.2-m wave travels 11.2 m to a wall and back again in 4 s. What is the frequency of", " the wave? 0.2 Hz 2 Hz 1.2 m 5 Hz 9 Hz 11.2 m 9. What is the length of a pendulum that has a period of 4.89 s? 5.94 m 11.9 m 24.0 m 37.3 m Extended Answer 10. Use dimensional analysis of the equation kx mg to derive the units of k. Practice, Practice, Practice Practice to improve your performance on standardized tests. Don\u2019t compare yourself to anyone else. physicspp.com/standardized_test Chapter 14 Standardized Test Practice 401 402-410 CH15-S1-845813 3/30/04 6:53 AM Page 402 What You\u2019ll Learn \u2022 You will describe sound in terms of wave properties and behavior. \u2022 You will examine some of the sources of sound. \u2022 You will explain properties that differentiate between music and noise. Why It\u2019s Important Sound is an important means of communication and, in the form of music, cultural expression. Musical Groups A small musical group might contain two or three instruments, while a marching band can contain 100 or more. The instruments in these groups form sounds in different ways, but they can create exciting music when they are played together. Think About This How do the instruments in a musical group create the sounds that you hear? Why do various instruments sound different even when they play the same note? physicspp.com 402 Tim Fuller 402-410 CH15-S1-845813 3/30/04 6:56 AM Page 403 How can glasses produce musical notes? Question How can you use glasses to produce different musical notes, and how do glasses with stems compare to those without stems? Procedure Analysis 1. Select a stemmed glass with a thin rim. 2. Prepare Carefully inspect the top edge of the glass for sharp edges. Notify your teacher if you observe any sharp edges. Be sure to repeat this inspection every time you select a different glass. 3. Place the glass on the table in front of you. Firmly hold the base with one hand. Wet your finger and slowly rub it around the top edge of the glass. CAUTION: Glass is fragile. Handle carefully. 4. Record your observations. Increase or decrease the speed a little. What happens? 5. Select a stemmed glass that is larger or smaller than the first glass. Repeat steps 2-4. 6. Select a glass without a stem and repeat steps 2\u20134. Summarize your observations. Which glasses\u2014 stemmed, not stemmed, or both\u2014were able to produce ringing tones? What factors affected the tones produced? Critical Thinking Propose a method for producing different notes from the same glass. Test your proposed method. Suggest a test to further investigate the properties of glasses that can produce ringing tones. 15.1 Properties and Detection of Sound Sound is an important part of existence for many living things. Animals can use sound to hunt, attract mates, and warn of the approach of predators. In humans, the sound of a siren can heighten our awareness of our surroundings, while the sound of music can soothe and relax us. From your everyday experiences, you already are familiar with several of the characteristics of sound, including volume, tone, and pitch. Without thinking about it, you can use these, and other characteristics, to categorize many of the sounds that you hear; for example, some sound patterns are characteristic of speech, while others are characteristic of a musical group. In this chapter, you will study the physical principles of sound, which is a type of wave. In Chapter 14, you learned how to describe waves in terms of speed, frequency, wavelength, and amplitude. You also discovered how waves interact with each other and with matter. Knowing that sound is a type of wave allows you to describe some of its properties and interactions. First, however, there is a question that you need to answer: exactly what type of wave is sound? Objectives \u2022 Demonstrate the properties that sound shares with other waves. \u2022 Relate the physical properties of sound waves to our perception of sound. \u2022 Identify some applications of the Doppler effect. Vocabulary sound wave pitch loudness sound level decibel Doppler effect Section 15.1 Properties and Detection of Sound 403 Horizons Companies 402-410 CH15-S1-845813 6/10/04 5:15 PM Page 404 \u25a0 Figure 15-1 Before the bell is struck, the air around it is a region of average pressure (a). Once the bell is struck, however, the vibrating edge creates regions of high and low pressure. The dark areas represent regions of higher pressure; the light areas represent regions of lower pressure (b). For simplicity, the diagram shows the regions moving in one direction; in reality, the waves move out from the bell in all directions. a b Sound Waves Put your fingers against your throat as you hum or speak. Can you feel the vibrations? Have you ever put your hand on the loudspeaker of a boom box? Figure 15-", "1 shows a vibrating bell that also can represent your vocal cords, a loudspeaker, or any other sound source. As it moves back and forth, the edge of the bell strikes the particles in the air. When the edge moves forward, air particles are driven forward; that is, the air particles bounce off the bell with a greater velocity. When the edge moves backward, air particles bounce off the bell with a lower velocity. The result of these velocity changes is that the forward motion of the bell produces a region where the air pressure is slightly higher than average. The backward motion produces slightly below-average pressure. Collisions among the air particles cause the pressure variations to move away from the bell in all directions. If you were to focus at one spot, you would see the value of the air pressure rise and fall, not unlike the behavior of a pendulum. In this way, the pressure variations are transmitted through matter. c b Describing sound A pressure variation that is transmitted through matter is a sound wave. Sound waves move through air because a vibrating source produces regular variations, or oscillations, in air pressure. The air particles collide, transmitting the pressure variations away from the source of the sound. The pressure of the air oscillates about the mean air pressure, as shown in Figure 15-2. The frequency of the wave is the number of oscillations in pressure each second. The wavelength is the distance between successive regions of high or low pressure. Because the motion of the particles in air is parallel to the direction of the wave\u2019s motion, sound is a longitudinal wave. The speed of sound in air depends on the temperature, with the speed increasing by about 0.6 m/s for each 1\u00b0C increase in air temperature. At room temperature (20\u00b0C), sound moves through air at sea level at a speed of 343 m/s. Sound also travels through solids and liquids. In general, the speed of sound is greater in \u25a0 Figure 15-2 A coiled spring models the compressions and rarefactions of a sound wave (a). The pressure of the air rises and falls as the sound wave propagates through the atmosphere (b). You can use a sine curve alone to model changes in pressure. Note that the positions of x, y, and z show that the wave, not matter, moves forward (c). a 404 Chapter 15 Sound 402-410 CH15-S1-845813 6/4/04 9:44 AM Page 405 solids and liquids than in gases. Table 15-1 lists the speeds of sound waves in various media. Sound cannot travel in a vacuum because there are no particles to collide. Sound waves share the general properties of other waves. For example, they reflect off hard objects, such as the walls of a room. Reflected sound waves are called echoes. The time required for an echo to return to the source of the sound can be used to find the distance between the source and the reflective object. This principle is used by bats, by some cameras, and by ships that employ sonar. Two sound waves can interfere, causing dead spots at nodes where little sound can be heard. As you learned in Chapter 14, the frequency and wavelength of a wave are related to the speed of the wave by the equation v/f. Table 15-1 Speed of Sound in Various Media Medium Air (0\u00b0) Air (20\u00b0) Helium (0\u00b0) Water (25\u00b0) Seawater (25\u00b0) Copper (25\u00b0) Iron (25\u00b0) m/s 331 343 972 1493 1533 3560 5130 1. Find the wavelength in air at 20\u00b0C of an 18-Hz sound wave, which is one of the lowest frequencies that is detectable by the human ear. 2. What is the wavelength of an 18-Hz sound wave in seawater at 25\u00b0C? 3. Find the frequency of a sound wave moving through iron at 25\u00b0C with a wavelength of 1.25 m. 4. If you shout across a canyon and hear the echo 0.80 s later, how wide is the canyon? 5. A 2280-Hz sound wave has a wavelength of 0.655 m in an unknown medium. Identify the medium. Detection of Pressure Waves Sound detectors convert sound energy\u2014the kinetic energy of the vibrating air particles\u2014into another form of energy. A common detector is a microphone, which converts sound waves into electrical energy. A microphone consists of a thin disk that vibrates in response to sound waves and produces an electrical signal. You will learn about this transformation process in Chapter 25, during your study of electricity and magnetism. The human ear As shown in Figure 15-3, the human ear is a detector that receives pressure waves and converts them to electrical impulses. Sound waves entering the auditory canal cause vibrations of the tympanic membrane. Three tiny bones then transfer these vibrations to fluid in the cochlea. Tiny hairs lining the spiral-shaped cochlea detect certain", " frequencies in the vibrating fluid. These hairs stimulate nerve cells, which send impulses to the brain and produce the sensation of sound. The ear detects sound waves over a wide range of frequencies and is sensitive to an enormous range of amplitudes. In addition, human hearing can distinguish many different qualities of sound. Knowledge of both physics and biology is required to understand the complexities of the ear. The interpretation of sounds by the brain is even more complex, and it is not totally understood. \u25a0 Figure 15-3 The human ear is a complex sense organ that translates sound vibrations into nerve impulses that are sent to the brain for interpretation. The malleus, incus, and stapes are the three bones of the middle ear that sometimes are referred to as the hammer, anvil, and stirrup. Auditory canal Malleus Stapes Incus Auditory nerve Tympanic membrane (eardrum) Cochlea Section 15.1 Properties and Detection of Sound 405 402-410 CH15-S1-845813 3/30/04 6:58 AM Page 406 Perceiving Sound Pitch Marin Mersenne and Galileo first determined that the pitch we hear depends on the frequency of vibration. Pitch can be given a name on the musical scale. For instance, the middle C note has a frequency of 262 Hz. The ear is not equally sensitive to all frequencies. Most people cannot hear sounds with frequencies below 20 Hz or above 16,000 Hz. Older people are less sensitive to frequencies above 10,000 Hz than are young people. By age 70, most people cannot hear sounds with frequencies above 8000 Hz. This loss affects the ability to understand speech. Loudness Frequency and wavelength are two physical characteristics of sound waves. Another physical characteristic of sound waves is amplitude. Amplitude is the measure of the variation in pressure along a wave. In humans, sound is detected by the ear and interpreted by the brain. The loudness of a sound, as perceived by our sense of hearing, depends primarily on the amplitude of the pressure wave. The human ear is extremely sensitive to pressure variations in sound waves, which is the amplitude of the wave. Recall from Chapter 13 that 1 atm of pressure equals 1.01105 Pa. The ear can detect pressure-wave amplitudes of less than one-billionth of an atmosphere, or 2105 Pa. At the other end of the audible range, pressure variations of approximately 20 Pa or greater cause pain. It is important to remember that the ear detects only pressure variations at certain frequencies. Driving over a mountain pass changes the pressure on your ears by thousands of pascals, but this change does not take place at audible frequencies. Because humans can detect a wide range in pressure variations, these amplitudes are measured on a logarithmic scale called the sound level. The unit of measurement for sound level is the decibel (dB). The sound level depends on the ratio of the pressure variation of a given sound wave to the pressure variation in the most faintly heard sound, 2105 Pa. Such an amplitude has a sound level of 0 dB. A sound with a pressure amplitude ten times larger (2104 Pa) is 20 dB. A pressure amplitude ten times larger than this is 40 dB. Most people perceive a 10-dB increase in sound level as about twice as loud as the original level. Figure 15-4 shows the sound level for a variety of sounds. In addition to pressure variations, power and intensity of sound waves can be described by decibel scales. Exposure to loud sounds, in the form of noise or music, has been shown to cause the ear to lose its sensitivity, especially to high frequencies. The longer a person is exposed to loud sounds, the greater the effect. A person can recover from short-term exposure in a period of hours, but the effects \u25a0 Figure 15-4 This decibel scale shows the sound levels of some familiar sounds. 10 dB Barely audible 50 dB Casual conversation 80 dB Alarm clock 110 dB Rock concert 30 dB Whisper, 1 m away 70 dB Heavy traffic 100 dB Siren 140 dB Jet engine 406 Chapter 15 Sound 402-410 CH15-S1-845813 3/30/04 6:59 AM Page 407 of long-term exposure can last for days or weeks. Long exposure to 100-dB or greater sound levels can produce permanent damage. Many rock musicians have suffered serious hearing loss, some as much as 40 percent. Hearing loss also can result from loud music being transmitted to stereo headphones from personal radios and CD players. In some cases, the listeners are unaware of just how high the sound levels really are. Cotton earplugs reduce the sound level only by about 10 dB. Special ear inserts can provide a 25-dB reduction. Specifically designed earmuffs and inserts as shown in Figure 15-5, can reduce the sound level by up to 45 dB. Loudness, as perceived by the human ear, is not directly proportional to the pressure variations in a sound wave. The ear\u2019s sensitivity depends on both pitch and amplitude", ". Also, perception of pure tones is different from perception of a mixture of tones. The Doppler Effect Have you ever noticed that the pitch of an ambulance, fire, or police siren changed as the vehicle sped past you? The pitch was higher when the vehicle was moving toward you, then it dropped to a lower pitch as the source moved away. This frequency shift is called the Doppler effect and is shown in Figure 15-6. The sound source, S, is moving to the right with a speed of vs. The waves that it emits spread in circles centered on the source at the time it produced the waves. As the source moves toward the sound detector, Observer A in Figure 15-6a, more waves are crowded into the space between them. The wavelength is shortened to A. Because the speed of sound is not changed, more crests reach the ear per second, which means that the frequency of the detected sound increases. When the source is moving away from the detector, Observer B in Figure 15-6a, the wavelength is lengthened to B and the detected frequency is lower. Figure 15-6b illustrates the Doppler effect for a moving source of sound on water waves in a ripple tank. A Doppler shift also occurs if the detector is moving and the source is stationary. In this case, the Doppler shift results from the relative velocity of the sound waves and the detector. As the detector approaches the stationary source, the relative velocity is larger, resulting in an increase in the wave crests reaching the detector each second. As the detector recedes from the source, the relative velocity is smaller, resulting in a decrease in the wave crests reaching the detector each second. a b \u03bb B vs S Observer B \u03bb A Observer A \u25a0 Figure 15-5 Continuous exposure to loud sounds can cause serious hearing loss. In many occupations, workers, such as this flight controller, must wear ear protection. A; the wavelength \u25a0 Figure 15-6 As a soundproducing source moves toward an observer, the wavelength is shortened to is B for waves produced by a source moving away from an observer (a). A moving waveproducing source illustrates the Doppler effect in a ripple tank (b). Section 15.1 Properties and Detection of Sound 407 (t)Willie Maldonad/Getty Images, (b)Max Kurtz/Ztek 402-410 CH15-S1-845813 3/30/04 6:59 AM Page 408 For both a moving source and a moving observer, the frequency that the observer hears can be calculated using the equation below. Doppler Effect fd fs v v v d v s The frequency perceived by a detector is equal to the velocity of the detector relative to the velocity of the wave, divided by the velocity of the source relative to the velocity of the wave, multiplied by the wave\u2019s frequency. In the Doppler effect equation, v is the velocity of the sound wave, vd is the velocity of the detector, vs is the velocity of the sound\u2019s source, fs is the frequency of the wave emitted by the source, and fd is the frequency received by the detector. This equation applies when the source is moving, when the observer is moving, and when both are moving. As you solve problems using the above equation, be sure to define the coordinate system so that the positive direction is from the source to the detector. The sound waves will be approaching the detector from the source, so the velocity of sound is always positive. Try drawing diagrams to confirm that the term (v vd)/(v vs) behaves as you would predict based on what you have learned about the Doppler effect. Notice that for a source moving toward the detector (positive direction, which results in a smaller denominator compared to a stationary source) and for a detector moving toward the source (negative direction and increased numerator compared to a stationary detector), the detected frequency, fd, increases. Similarily, if the source moves away from the detector or if the detector moves away from the source, then fd decreases. Read the Connecting Math to Physics feature below to see how the Doppler effect equation reduces when the source or observer is stationary. Reducing Equations When an element in a complex equation is equal to zero, the equation might reduce to a form that is easier to use. vs v v fd v d v s 0 v v ) ) Stationary detector, source in motion: vd fs( fs( fs( fs fd v d v s 0 v v ) vd) Stationary source, detector in motion: vs fs( fs( fs( fs( fs(1 ) ) d 408 Chapter 15 Sound 402-410 CH15-S1-845813 3/30/04 7:00 AM Page 409 The Doppler Effect A trumpet player sounds C above middle C (524 Hz) while traveling in a convertible at 24.6 m/s. If the", " car is coming toward you, what frequency would you hear? Assume that the temperature is 20\u00b0C. 1 Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Establish a coordinate axis. Make sure that the positive direction is from the source to the detector. \u2022 Show the velocities of the source and detector. Unknown:? fd Known: v 343 m/s 24.6 m/s vs 0 m/s vd 524 Hz fs 2 Solve for the Unknown v d v s Use fd v v 0 m/s. ) with vd ) fd fs( fs( 524 Hz( 1 v 1 s v v vs Math Handbook Fractions page 837 ) Substitute v 343 m/s, vs 24.6 m/s, and fs 524 Hz 564 Hz 3 Evaluate the Answer \u2022 Are the units correct? Frequency is measured in hertz. \u2022 Is the magnitude realistic? The source is moving toward you, so the frequency should be increased. 6. Repeat Example Problem 1, but with the car moving away from you. What frequency would you hear? 7. You are in an auto traveling at 25.0 m/s toward a pole-mounted warning siren. If the siren\u2019s frequency is 365 Hz, what frequency do you hear? Use 343 m/s as the speed of sound. 8. You are in an auto traveling at 55 mph (24.6 m/s). A second auto is moving toward you at the same speed. Its horn is sounding at 475 Hz. What frequency do you hear? Use 343 m/s as the speed of sound. 9. A submarine is moving toward another submarine at 9.20 m/s. It emits a 3.50-MHz ultrasound. What frequency would the second sub, at rest, detect? The speed of sound in water is 1482 m/s. 10. A sound source plays middle C (262 Hz). How fast would the source have to go to raise the pitch to C sharp (271 Hz)? Use 343 m/s as the speed of sound. Section 15.1 Properties and Detection of Sound 409 402-410 CH15-S1-845813 3/30/04 7:00 AM Page 410 \u25a0 Figure 15-7 In a process called echolocation, bats use the Doppler effect to locate prey. The Doppler effect occurs in all wave motion, both mechanical and electromagnetic. It has many applications. Radar detectors use the Doppler effect to measure the speed of baseballs and automobiles. Astronomers observe light from distant galaxies and use the Doppler effect to measure their speeds and infer their distances. Physicians can detect the speed of the moving heart wall in a fetus by means of the Doppler effect in ultrasound. Bats use the Doppler effect to detect and catch flying insects. When an insect is flying faster than a bat, the reflected frequency is lower, but when the bat is catching up to the insect, as in Figure 15-7, the reflected frequency is higher. Not only do bats use sound waves to navigate and locate their prey, but they often must do so in the presence of other bats. This means they must discriminate their own calls and reflections against a background of many other sounds of many frequencies. Scientists continue to study bats and their amazing abilities to use sound waves. Biology Connection Biology Connection 15.1 Section Review 11. Graph The eardrum moves back and forth in response to the pressure variations of a sound wave. Sketch a graph of the displacement of the eardrum versus time for two cycles of a 1.0-kHz tone and for two cycles of a 2.0-kHz tone. 12. Effect of Medium List two sound characteristics that are affected by the medium through which the sound passes and two characteristics that are not affected. 13. Sound Properties What physical characteristic of a sound wave should be changed to change the pitch of the sound? To change the loudness? 14. Decibel Scale How much greater is the sound pressure level of a typical rock band\u2019s music (110 dB) than a normal conversation (50 dB)? 15. Early Detection In the nineteenth century, people put their ears to a railroad track to get an early warning of an approaching train. Why did this work? 16. Bats A bat emits short pulses of high-frequency sound and detects the echoes. a. In what way would the echoes from large and small insects compare if they were the same distance from the bat? b. In what way would the echo from an insect flying toward the bat differ from that of an insect flying away from the bat? 17. Critical Thinking Can a trooper using a radar detector at the side of the road determine the speed of a car at the instant the car passes the trooper? Explain. 410 Stephen Dalton/Animals Animals Chapter 15 Sound physicspp.com/self_check_quiz 411-419 CH15-S2-845813", " 3/30/04 7:06 AM Page 411 Objectives \u2022 Describe the origin of sound. \u2022 Demonstrate an understanding of resonance, especially as applied to air columns and strings. \u2022 Explain why there are variations in sound among instruments and among voices. Vocabulary closed-pipe resonator open-pipe resonator fundamental harmonics dissonance consonance beat 15.2 The Physics of Music In the middle of the nineteenth century, German physicist Hermann Helmholtz studied sound production in musical instruments and the human voice. In the twentieth century, scientists and engineers developed electronic equipment that permits not only a detailed study of sound, but also the creation of electronic musical instruments and recording devices that allow us to listen to music whenever and wherever we wish. Sources of Sound Sound is produced by a vibrating object. The vibrations of the object create particle motions that cause pressure oscillations in the air. A loudspeaker has a cone that is made to vibrate by electrical currents. The surface of the cone creates the sound waves that travel to your ear and allow you to hear music. Musical instruments such as gongs, cymbals, and drums are other examples of vibrating surfaces that are sources of sound. The human voice is produced by vibrations of the vocal cords, which are two membranes located in the throat. Air from the lungs rushing through the throat starts the vocal cords vibrating. The frequency of vibration is controlled by the muscular tension placed on the vocal cords. In brass instruments, such as the trumpet and tuba, the lips of the performer vibrate, as shown in Figure 15-8a. Reed instruments, such as the clarinet and saxophone, have a thin wooden strip, or reed, that vibrates as a result of air blown across it, as shown in Figure 15-8b. In flutes and organ pipes, air is forced across an opening in a pipe. Air moving past the opening sets the column of air in the instrument into vibration. In stringed instruments, such as the piano, guitar, and violin, wires or strings are set into vibration. In the piano, the wires are struck; in the guitar, they are plucked; and in the violin, the friction of the bow causes the strings to vibrate. Often, the strings are attached to a sounding board that vibrates with the strings. The vibrations of the sounding board cause the pressure oscillations in the air that we hear as sound. Electric guitars use electronic devices to detect and amplify the vibrations of the guitar strings. a b Brass instrument Mouthpiece Woodwind instrument Mouthpiece Reed \u25a0 Figure 15-8 The shapes of the mouthpieces of a brass instrument (a) and a reed instrument (b) help determine the characteristics of the sound each instrument produces. Section 15.2 The Physics of Music 411 411-419 CH15-S2-845813 6/4/04 9:47 AM Page 412 Tuning fork Air column Hammer Tube Water \u25a0 Figure 15-9 Raising or lowering the tube changes the length of the air column. When the column is in resonance with the tuning fork, the sound is loudest. \u25a0 Figure 15-10 A tube placed in water is a closed-pipe resonator. In closed pipes, high pressure waves reflect as high pressure (a). In open pipes, the reflected waves are inverted (b). Resonance in Air Columns If you have ever used just the mouthpiece of a brass or reed instrument, you know that the vibration of your lips or the reed alone does not make a sound with any particular pitch. The long tube that makes up the instrument must be attached if music is to result. When the instrument is played, the air within this tube vibrates at the same frequency, or in resonance, with a particular vibration of the lips or reed. Remember that resonance increases the amplitude of a vibration by repeatedly applying a small external force at the same natural frequency. The length of the air column determines the frequencies of the vibrating air that will be set into resonance. For many instruments, such as flutes, saxophones, and trombones, changing the length of the column of vibrating air varies the pitch of the instrument. The mouthpiece simply creates a mixture of different frequencies, and the resonating air column acts on a particular set of frequencies to amplify a single note, turning noise into music. A tuning fork above a hollow tube can provide resonance in an air column, as shown in Figure 15-9. The tube is placed in water so that the bottom end of the tube is below the water surface. A resonating tube with one end closed to air is called a closed-pipe resonator. The length of the air column is changed by adjusting the height of the tube above the water. If the tuning fork is struck with a rubber hammer and the length of the air column is varied as the tube is lifted up and down in the water, the sound alternately becomes louder and softer. The sound is loud when the air column is in resonance with the tuning", " fork. A resonating air column intensifies the sound of the tuning fork. Standing pressure wave How does resonance occur? The vibrating tuning fork produces a sound wave. This wave of alternate high- and lowpressure variations moves down the air column. When the wave hits the water surface, it is reflected back up to the tuning fork, as indicated in Figure 15-10a. If the reflected high-pressure wave reaches the tuning fork at the same moment that the fork produces another high-pressure wave, then the emitted and returning waves reinforce each other. This reinforcement of waves produces a standing wave, and resonance is achieved. An open-pipe resonator is a resonating tube with both ends open that also will resonate with a sound source. In this case, the sound wave does not reflect off a closed end, but rather off an open end. The pressure of the reflected wave is inverted; for example, if a high-pressure wave strikes the open end, a low-pressure wave will rebound, as shown in Figure 15-10b. a b Time Closed pipes: high pressure reflects as high pressure Time Open pipes: high pressure reflects as low pressure 412 Chapter 15 Sound Resonance lengths A standing sound wave in a pipe can be represented by a sine wave, as shown in Figure 15-11. Sine waves can represent either the air pressure or the displacement of the air particles. You can see that standing waves have nodes and antinodes. In the pressure graphs, the nodes are regions of mean atmospheric pressure, and at the antinodes, the pressure oscillates between its maximum and minimum values. 411-419 CH15-S2-845813 6/4/04 9:48 AM Page 413 Closed Pipe Pressure node (average-pressure region) Displacement antinode \u25a0 Figure 15-11 Sine waves represent standing waves in pipes. Pressure antinode (high- or low-pressure region) Air pressure Displacement node Displacement of air Open Pipe Pressure node (average-pressure region) Pressure antinode (high- or low-pressure region) Pressure node (average-pressure region) Displacement antinode Displacement node Displacement antinode Air pressure Displacement of air In the case of the displacement graph, the antinodes are regions of high displacement and the nodes are regions of low displacement. In both cases, two antinodes (or two nodes) are separated by one-half wavelength. Resonance frequencies in a closed pipe The shortest column of air that can have an antinode at the closed end and a node at the open end is onefourth of a wavelength long, as shown in Figure 15-12. As the frequency is increased, additional resonance lengths are found at half-wavelength intervals. Thus, columns of length /4, 3/4, 5/4, 7/4, and so on will all be in resonance with a tuning fork. In practice, the first resonance length is slightly longer than one-fourth of a wavelength. This is because the pressure variations do not drop to zero exactly at the open end of the pipe. Actually, the node is approximately 0.4 pipe diameters beyond the end. Additional resonance lengths, however, are spaced by exactly one-half of a wavelength. Measurements of the spacing between resonances can be used to find the velocity of sound in air, as shown in the next Example Problem. Hearing and Frequency The human auditory canal acts as a closed-pipe resonator that increases the ear\u2019s sensitivity for frequencies between 2000 and 5000 Hz, but the full range of frequencies that people hear extends from 20 to 20,000 Hz. A dog\u2019s hearing extends to frequencies as high as 45,000 Hz, and a cat\u2019s extends to frequencies as high as 100,000 Hz. Pressure node (averagepressure region) Pressure antinode (highor low-pressure region) Node Antinode Node Antinode \u03bb 1 4L \u03bb 3 4 \u2014 L 3 f1 v \u2014 \u03bb 1 v \u2014 4L f3 \u2014 3f1 3v 4L L \u03bb 5 4 \u2014 L 5 f5 \u2014 5f1 5v 4L \u25a0 Figure 15-12 A closed pipe resonates when its length is an odd number of quarter wavelengths. Section 15.2 The Physics of Music 413 411-419 CH15-S2-845813 3/30/04 7:07 AM Page 414 \u25a0 Figure 15-13 An open pipe resonates when its length is an even number of quarter wavelengths. Pressure node (averagepressure region) Pressure antinode (highor low-pressure region) Pressure node (averagepressure region) Node Antinode Node Antinode Node L \u03bb 1 2L f1 v v \u2014 \u2014 \u03bb 1 2L \u03bb L 2 f2 \u2014 2f1 v L \u03bb 3 2L \u2014 3 f3 \u2014 3f1 3v 2L Resonance frequencies in an", " open pipe The shortest column of air that can have nodes at both ends is one-half of a wavelength long, as shown in Figure 15-13. As the frequency is increased, additional resonance lengths are found at half-wavelength intervals. Thus, columns of length /2,, 3/2, 2, and so on will be in resonance with a tuning fork. If open and closed pipes of the same length are used as resonators, the wavelength of the resonant sound for the open pipe will be half as long as that for the closed pipe. Therefore, the frequency will be twice as high for the open pipe as for the closed pipe. For both pipes, resonance lengths are spaced by half-wavelength intervals. Hearing resonance Musical instruments use resonance to increase the loudness of particular notes. Open-pipe resonators include flutes and saxophones. Clarinets and the hanging pipes under marimbas and xylophones are examples of closed-pipe resonators. If you shout into a long tunnel, the booming sound you hear is the tunnel acting as a resonator. The seashell in Figure 15-14 acts as a closed-pipe resonator. Resonance on Strings Although the waveforms on vibrating strings vary in shape, depending upon how they are produced, such as by plucking, bowing, or striking, they have many characteristics in common with standing waves on springs and ropes, which you studied in Chapter 14. A string on an instrument is clamped at both ends, and therefore, the string must have a node at each end when it vibrates. In Figure 15-15, you can see that the first mode of vibration has an antinode at the center and is one-half of a wavelength long. The next resonance occurs when one wavelength fits on the string, and additional standing waves arise when the string length is 3/2, 2, 5/2, and so on. As with an open pipe, the resonant frequencies are whole-number multiples of the lowest frequency. The speed of a wave on a string depends on the tension of the string, as well as its mass per unit length. This makes it possible to tune a stringed instrument by changing the tension of its strings. The tighter the string, the faster the wave moves along it, and therefore, the higher the frequency of its standing waves. \u25a0 Figure 15-14 A seashell acts as a closed-pipe resonator to amplify certain frequencies from the background noise. 414 David Mechlin/Index Stock Images Chapter 15 Sound 411-419 CH15-S2-845813 3/30/04 7:08 AM Page 415 \u25a0 Figure 15-15 A string resonates with standing waves when its length is a whole number of half wavelengths. L 1 2L v v f1 L 2 1 2 L v 2f1 f2 L 3 f3 L 2 3 3 v 3f1 2 L Because strings are so small in cross-sectional area, they move very little air when they vibrate. This makes it necessary to attach them to a sounding board, which transfers their vibrations to the air and produces a stronger sound wave. Unlike the strings themselves, the sounding board should not resonate at any single frequency. Its purpose is to convey the vibrations of all the strings to the air, and therefore it should vibrate well at all frequencies produced by the instrument. Because of the complicated interactions among the strings, the sounding board, and the air, the design and construction of stringed instruments are complex processes, considered by many to be as much an art as a science. Sound Quality A tuning fork produces a soft and uninteresting sound. This is because its tines vibrate like simple harmonic oscillators and produce the simple sine wave shown in Figure 15-16a. Sounds made by the human voice and musical instruments are much more complex, like the wave in Figure 15-16b. Both waves have the same frequency, or pitch, but they sound very different. The complex wave is produced by using the principle of superposition to add waves of many frequencies. The shape of the wave depends on the relative amplitudes of these frequencies. In musical terms, the difference between the two waves is called timbre, tone color, or tone quality Time Pure sound Clarinet \u25a0 Figure 15-16 A graph of pure sound versus time (a) and a graph of clarinet sound waves versus time (b) are shown. Time Section 15.2 The Physics of Music 415 411-419 CH15-S2-845813 3/30/04 7:08 AM Page 416 Finding the Speed of Sound Using Resonance When a tuning fork with a frequency of 392 Hz is used with a closed-pipe resonator, the loudest sound is heard when the column is 21.0 cm and 65.3 cm long. What is the speed of sound in this case? Is the temperature warmer or cooler than normal room temperature, which is 20\u00b0C? Explain your answer. 1 2 3 Analy", "ze and Sketch the Problem \u2022 Sketch the closed-pipe resonator. \u2022 Mark the resonance lengths. Unknown: v? Known: f 392 Hz LA LB 21.0 cm 65.3 cm Solve for the Unknown Solve for the length of the wave using the length-wavelength relationship for a closed pipe. 1 2 2(LB LA) LA LB Rearrange the equation for. 2(0.653 m 0.210 m) 0.886 m Use v. f v f (392 Hz)(0.886 m) 347 m/s Substitute LB 0.653 m, LA 0.210 m Rearrange the equation for v. Substitute f 392 Hz, 0.886 m LA LB Math Handbook Order of Operations page 843 The speed is slightly greater than the speed of sound at 20\u00b0C, indicating that the temperature is slightly higher than normal room temperature. Evaluate the Answer \u2022 Are the units correct? (Hz)(m) (1/s)(m) m/s. The answer\u2019s units are correct. \u2022 Is the magnitude realistic? The speed is slightly greater than 343 m/s, which is the speed of sound at 20\u00b0C. 18. A 440-Hz tuning fork is held above a closed pipe. Find the spacing between the resonances when the air temperature is 20\u00b0C. 19. A 440-Hz tuning fork is used with a resonating column to determine the velocity of sound in helium gas. If the spacings between resonances are 110 cm, what is the velocity of sound in helium gas? 20. The frequency of a tuning fork is unknown. A student uses an air column at 27\u00b0C and finds resonances spaced by 20.2 cm. What is the frequency of the tuning fork? Use the speed calculated in Example Problem 2 for the speed of sound in air at 27\u00b0C. 21. A bugle can be thought of as an open pipe. If a bugle were straightened out, it would be 2.65-m long. a. If the speed of sound is 343 m/s, find the lowest frequency that is resonant for a bugle (ignoring end corrections). b. Find the next two resonant frequencies for the bugle. 416 Chapter 15 Sound 411-419 CH15-S2-845813 6/4/04 9:51 AM Page 417 Violin Clarinet Piano 1000 2000 3000 1000 2000 3000 1000 2000 3000 Frequency (Hz) Frequency (Hz) Frequency (Hz) \u25a0 Figure 15-17 A violin, a clarinet, and a piano produce characteristic sound spectra. Each spectrum is unique, as is the timbre of the instrument. The sound spectrum: fundamental and harmonics The complex sound wave in Figure 15-16b was made by a clarinet. Why does the clarinet produce such a sound wave? The air column in a clarinet acts as a closed pipe. Look back at Figure 15-12, which shows three resonant frequencies for a closed pipe. Because the clarinet acts as a closed pipe, for a clarinet of length L the lowest frequency, f1, that will be resonant is v/4L. This lowest frequency is called the fundamental. A closed pipe also will resonate at 3f1, 5f1, and so on. These higher frequencies, which are odd-number multiples of the fundamental frequency, are called harmonics. It is the addition of these harmonics that gives a clarinet its distinctive timbre. Some instruments, such as an oboe, act as open-pipe resonators. Their v/2L with fundamental frequency, which is also the first harmonic, is f1 subsequent harmonics at 2f1, 3f1, 4f1, and so on. Different combinations and amplitudes of these harmonics give each instrument its own unique timbre. A graph of the amplitude of a wave versus its frequency is called a sound spectrum. The spectra of three instruments are shown in Figure 15-17. 1. Determine the tension, FT, in a violin string of mass m and length L that will play the fundamental note at the same frequency as a closed pipe also of length L. Express your answer in terms of m, L, and the speed of sound in air, v. The equation for the speed of a wave T, where FT is the tension in the on a string is u string and is the mass per unit length of the string. F 2. What is the tension in a string of mass 1.0 g and 40.0 cm long that plays the same note as a closed pipe of the same length? L L Section 15.2 The Physics of Music 417 (l)Geoff Butler, (c)HIRB/Index Stock Images, (r)Photodisc/Artbase 411-419 CH15-S2-845813 3/30/04 9:11 AM Page 418 a b 1:2 2:3 \u25a0 Figure 15", "-18 These time graphs show the superposition of two waves having the ratios of 1:2, 2:3, 3:4, and 4:5. Sounds Good Sometimes it can be pretty tough to tell just by looking whether an instrument will act as an openpipe resonator or as a closed-pipe resonator. Ask a musician who plays a wind instrument to bring it to class. 1. Measure the length of the instrument. 2. Have the musician play the lowest possible note on the instrument. 3. Determine the frequency of the note using a frequency generator. Analyze and Conclude 4. Draw Conclusions Did the tested instrument behave most like a closed-pipe resonator or most like an open-pipe resonator? Was the frequency the fundamental or one of the subsequent harmonics? 5. Determine the frequencies of the notes that would form an octave, a perfect fifth, a perfect fourth, and a major third with your observed note. 418 Chapter 15 Sound 3:4 Perfect fourth Octave c d Perfect fifth 4:5 Major third Consonance and dissonance When sounds that have two different pitches are played at the same time, the resulting sound can be either pleasant or jarring. In musical terms, several pitches played together are called a chord. An unpleasant set of pitches is called dissonance. If the combination is pleasant, the sounds are said to be in consonance. What makes a sound pleasant to listen to? Different cultures have different definitions, but most Western cultures accept the definitions of Pythagoras, who lived in ancient Greece. Pythagoras experimented by plucking two strings at the same time. He noted that pleasing sounds resulted when the strings had lengths in small, whole-number ratios, for example 1:2, 2:3, or 3:4. This means that their pitches (frequencies) will also have small, whole-number ratios. Musical intervals Two notes with frequencies related by the ratio 1:2 are said to differ by an octave. For example, if a note has a frequency of 440 Hz, a note that is one octave higher has a frequency of 880 Hz. The fundamental and its harmonics are related by octaves; the first harmonic is one octave higher than the fundamental, the second is two octaves higher, and so on. The sum of the fundamental and the first harmonic is shown in Figure 15-18a. It is the ratio of two frequencies, not the size of the interval between them, that determines the musical interval. In other musical intervals, two pitches may be close together. For example, the ratio of frequencies for a \u201cmajor third\u201d is 4:5. A typical major third is made up of the notes C and E. The note C has a frequency of 262 Hz, so E has a frequency of (5/4)(262 Hz) = 327 Hz. In the same way, notes in a \u201cfourth\u201d (C and F) have a frequency ratio of 3:4, and those in a \u201cfifth\u201d (C and G) have a ratio of 2:3. Graphs of these pleasant sounds are shown in Figure 15-18. More than two notes sounded together also can produce consonance. The three notes called do, mi, and sol make a major chord. For at least 2500 years, this has been recognized as the sweetest of the threenote chords; it has the frequency ratio of 4:5:6. Beats You have seen that consonance is defined in terms of the ratio of frequencies. When the ratio becomes nearly 1:1, the frequencies become very close. Two frequencies that are nearly identical interfere to produce high and low sound levels, as illustrated in Figure 15-19. This oscillation of wave amplitude is called a beat. The frequency of a beat is the magnitude \uf8e6. of difference between the frequencies of the two waves, fbeat When the difference is less than 7 Hz, the ear detects this as a pulsation of loudness. Musical instruments often are tuned by sounding one against another and adjusting the frequency of one until the beat disappears. \uf8e6fA fB 411-419 CH15-S2-845813 3/30/04 7:12 AM Page 419 T2 T1 e d u t i l p m A \u25a0 Figure 15-19 Beats occur as a result of the superposition of two sound waves of slightly different frequencies. Tbeat Time Sound Reproduction and Noise How often do you listen to music produced directly by a human voice or musical instrument? Most of the time, the music has been recorded and played through electronic systems. To reproduce the sound faithfully, the system must accommodate all frequencies equally. A good stereo system keeps the amplitudes of all frequencies between 20 and 20,000 Hz the same to within 3 dB. A telephone system, on the other hand, needs only to transmit the information in spoken language. Frequencies between 300 and 3000 Hz are sufficient. Reducing the number of", " frequencies present helps reduce the noise. A noise wave is shown in Figure 15-20. Many frequencies are present with approximately the same amplitude. While noise is not helpful in a telephone system, some people claim that listening to noise has a calming effect. For this reason, some dentists use noise to help their patients relax. \u25a0 Figure 15-20 Noise is composed of several frequencies and involves random changes in frequency and amplitude. 15.2 Section Review 22. Origins of Sound What is the vibrating object that produces sounds in each of the following? a. a human voice b. a clarinet c. a tuba d. a violin 23. Resonance in Air Columns Why is the tube from which a tuba is made much longer than that of a cornet? 26. Resonance in Closed Pipes One closed organ pipe has a length of 2.40 m. a. What is the frequency of the note played by this pipe? b. When a second pipe is played at the same time, a 1.40-Hz beat note is heard. By how much is the second pipe too long? 27. Timbre Why do various instruments sound differ- ent even when they play the same note? 24. Resonance in Open Tubes How must the length of an open tube compare to the wavelength of the sound to produce the strongest resonance? 28. Beats A tuning fork produces three beats per second with a second, 392-Hz tuning fork. What is the frequency of the first tuning fork? 25. Resonance on Strings A violin sounds a note of F sharp, with a pitch of 370 Hz. What are the frequencies of the next three harmonics produced with this note? 29. Critical Thinking Strike a tuning fork with a rubber hammer and hold it at arm\u2019s length. Then press its handle against a desk, a door, a filing cabinet, and other objects. What do you hear? Why? physicspp.com/self_check_quiz Section 15.2 The Physics of Music 419 420-421 CH15-LAB-845813 3/30/04 7:20 AM Page 420 Alternate CBL instructions can be found on the Web site. physicspp.com Speed of Sound If a vibrating tuning fork is held above a closed pipe of the proper length, the air in the pipe will vibrate at the same frequency, f, as the tuning fork. By placing a glass tube in a large, water-filled graduated cylinder, the length of the glass tube can be changed by raising or lowering it in the water. The shortest column of air that will resonate occurs when the tube is one-fourth of a wavelength long. This resonance will produce the loudest sound, and the wavelength at this resonance is described by 4L, where L is the length from the water to the open end of the pipe. In this lab, you will determine L, calculate, and calculate the speed of sound. QUESTION How can you use a closed-pipe resonator to determine the speed of sound? Objectives \u25a0 Collect and organize data to obtain resonant points in a closed pipe. \u25a0 Measure the length of a closed-pipe resonator. \u25a0 Analyze the data to determine the speed of sound. Safety Precautions \u25a0 Immediately wipe up any spilled liquids. \u25a0 Glass is fragile. Handle with care. Materials three tuning forks of known frequencies graduated cylinder (1000-mL) water tuning fork mallet metric ruler thermometer (non-mercury) glass tube (approximately 40 cm in length and 3.5 cm in diameter) Procedure 1. Put on your safety goggles. Fill the graduated cylinder nearly to the top with water. 2. Measure the room air temperature and record it in Data Table 1. 3. Select a tuning fork and record its frequency in Data Tables 2 and 3. 4. Measure and record the diameter of the glass tube in Data Table 2. 5. Carefully place the glass tube into the water- filled graduated cylinder. 6. Hold the tuning fork by the base. Swiftly strike it on the side with the tuning fork mallet. Do not strike the tuning fork on the laboratory table or other hard surface. 7. Hold the vibrating fork over the open end of the glass tube and slowly raise the tube and the fork until the loudest sound is heard. Once this point is located, move the tube up and down slightly to determine the exact point of resonance. Measure the distance from the water to the top of the glass tube and record this distance in Data Table 2. 8. Repeat steps 3, 6, and 7 for two additional tuning forks and record your results as trials 2 and 3. The three tuning forks that you test should resonate at three different frequencies. 9. Empty the water from the graduated cylinder. 420 420-421 CH15-LAB-845813 6/10/04 8:40 PM Page 421 Data Table 1 Trial Temperature (\u00b0C) Accepted Speed of Sound (m/s) Experimental Speed of Sound (m/s) Data", " Table 2 Trial Tuning Fork Frequency (Hz) Diameter (m) Length of Tube Above Water (m) Calculated Wavelength (m) 1 2 3 1 2 3 Analyze 1. Calculate the accepted speed of sound using the relationship v 331 m/s 0.60T, where v is the speed of sound at temperature T, and T is the air temperature in degrees Celsius. Record this as the accepted speed of sound in Data Tables 1 and 3 for all the trials. Data Table 3 Trial Tuning Fork Frequency (Hz) Accepted Speed of Sound (m/s) Corrected Calculated Wavelength (m) Corrected Experimental Speed of Sound (m/s) 1 2 3 2. Since the first resonant point is located when the tube is one-fourth of a wavelength above the water, use the measured length of the tube to determine the calculated wavelength for each trial. Record the calculated wavelengths in Data Table 2. 3. Multiply the values in Data Table 2 of wavelength and frequency to determine the experimental speed of sound and record this in Data Table 1 for each of the trials. 4. Error Analysis For each trial in Data Table 1, determine the relative error between the experimental and accepted speed of sound. Relative error \uf8e6Accepted value Experimental value\uf8e6 Accepted value 100 5. Critique To improve the accuracy of your calculations, the tube diameter must be taken into consideration. The following relationship provides a more accurate calculation of wavelength: 4(L 0.4d ), where is the wavelength, L is the length of the tube above the water, and d is the inside diameter of the tube. Using the values in Data Table 1 for length and diameter, recalculate and record it in Data Table 3 as the corrected wavelength. Calculate the corrected experimental speed of sound by multiplying the tuning fork frequency and corrected wavelength and record the new value for the corrected experimental speed of sound in Data Table 3. 6. Error Analysis For each trial in Data Table 3, determine the relative error between the corrected experimental speed and the accepted speed of sound. Use the same formula that you used in step 4, above. Conclude and Apply 1. Infer In general, the first resonant point occurs when the tube length /4. What are the next two lengths where resonance will occur? 2. Think Critically If you had a longer tube, would it be possible to locate another position where resonance occurs? Explain your answer. Going Further Which result produced the more accurate speed of sound? Real-World Physics Explain the relationship between the size of organ pipes and their resonant frequencies. To find out more about the properties of sound waves, visit the Web site: physicspp.com 421 422 CH15-FEATURE-845813 3/30/04 7:23 AM Page 422 Sound Waves in the Sun The study of wave oscillations in the Sun is called helioseismology. Naturally occurring sound waves (p waves), gravity waves, and surface gravity waves all occur in the Sun. All of these waves are composed of oscillating particles, but different forces cause the oscillations. For sound waves, pressure differences cause the particles to oscillate. In the Sun, sound waves travel through the convective zone, which is just under the surface, or photosphere. The sound waves do not travel in a straight line, as shown in the image. The motion of the surface of the Sun is measured by observing Doppler shifts in sunlight. The measured vibrations are a complicated pattern that equals the sum of all of the standing waves present in the Sun. Just like a ringing bell, many overtones are present in the Sun. Through careful analysis, the individual standing waves in the Sun and their intensities can be calculated. Photosphere p Ringing like a Bell The sound waves in the Sun cause the surface of the Sun to vibrate in the radial direction, much like a ringing bell vibrates. When a bell is rung, a clapper hits the bell in one place and standing waves are created. The surface of the Sun does have standing waves, but they are not caused by one large event. Instead, scientists hypothesize that many smaller disruptions in the convective zone start most of the sound waves in the Sun. Just like boiling water in a pot can be noisy, bubbles that are larger than the state of Texas form on the surface of the Sun and start sound waves. Unlike a pot of boiling water, the sound coming from the Sun is much too low for us to hear. The A above middle C on the piano has a period of 0.00227 s (f 440 Hz). The middle mode of oscillation of the waves in the Sun has a period of 5 min (f 0.003 Hz). Because we cannot hear the sound waves from the Sun, scientists measure the motion of the surface of the Sun to learn about its sound waves. Because a sound wave takes 2 h to travel from one side of the Sun to the other, the Sun must be observed for long time", " periods. This necessity makes observations from Earth difficult because the Sun is not visible during the night. In 1995, the Solar and Heliospheric Observatory (SOHO) was launched by NASA. This satellite orbits Earth such that it always can observe the Sun. 422 Extreme Physics Core Radiative zone Convective zone Chromosphere Sound waves (p waves) travel through the Sun\u2019s convective zone. Results Because composition, temperature, and density affect the propagation of sound waves, the Sun\u2019s wave oscillations provide information about its interior. SOHO results have given insight into the rotation rate of the Sun as a function of latitude and depth, as well as the density and temperature of the Sun. These results are compared to theoretical calculations to improve our understanding of the Sun. Going Further 1. Hypothesize How do scientists separate the surface motion due to sound waves from the motion due to the rotation of the Sun? 2. Critical Thinking Would sound waves in another star, similar to the Sun but different in size, have the same wavelength as sound waves in the Sun? 423-429 CH15-SG CA STP-845813 3/30/04 7:28 AM Page 423 15.1 Properties and Detection of Sound Vocabulary \u2022 sound wave (p. 404) \u2022 pitch (p. 406) \u2022 loudness (p. 406) \u2022 sound level (p. 406) \u2022 decibel (p. 406) \u2022 Doppler effect (p. 407) Key Concepts \u2022 Sound is a pressure variation transmitted through matter as a longitudinal wave. \u2022 A sound wave has frequency, wavelength, speed, and amplitude. Sound waves reflect and interfere. \u2022 The speed of sound in air at room temperature (20\u00b0C) is 343 m/s. The speed increases roughly 0.6 m/s with each 1\u00b0C increase in temperature. \u2022 Sound detectors convert the energy carried by a sound wave into another form of energy. The human ear is a highly efficient and sensitive detector of sound waves. \u2022 The frequency of a sound wave is heard as its pitch. \u2022 The pressure amplitude of a sound wave can be measured in decibels (dB). \u2022 The loudness of sound as perceived by the ear and brain depends mainly on its amplitude. \u2022 The Doppler effect is the change in frequency of sound caused by the motion of either the source or the detector. It can be calculated with the following equation. fs v v v d v s fd 15.2 The Physics of Music Vocabulary \u2022 closed-pipe resonator (p. 412) Key Concepts \u2022 Sound is produced by a vibrating object in a material medium. \u2022 Most sounds are complex waves that are composed of more than one \u2022 open-pipe resonator frequency. (p. 412) \u2022 fundamental (p. 417) \u2022 harmonics (p. 417) \u2022 dissonance (p. 418) \u2022 consonance (p. 418) \u2022 beat (p. 418) \u2022 An air column can resonate with a sound source, thereby increasing the amplitude of its resonant frequency. \u2022 A closed pipe resonates when its length is /4, 3/4, 5/4, and so on. Its resonant frequencies are odd-numbered multiples of the fundamental. \u2022 An open pipe resonates when its length is /2, 2/2, 3/2, and so on. Its resonant frequencies are whole-number multiples of the fundamental. \u2022 A clamped string has a node at each end and resonates when its length is /2, 2/2, 3/2, and so on, just as with an open pipe. The string\u2019s resonant frequencies are also whole-number multiples of the fundamental. \u2022 The frequencies and intensities of the complex waves produced by a musical instrument determine the timbre that is characteristic of that instrument. \u2022 The fundamental frequency and harmonics can be described in terms of resonance. \u2022 Notes on a musical scale differ in frequency by small, whole-number ratios. An octave has a frequency ratio of 1:2. \u2022 Two waves with almost the same frequency interfere to produce beats. physicspp.com/vocabulary_puzzlemaker 423 423-429 CH15-SG CA STP-845813 3/30/04 7:29 AM Page 424 Concept Mapping 30. Complete the concept map below using the following terms: amplitude, perception, pitch, speed. Sound properties frequency loudness 41. Trombones Explain how the slide of a trombone, shown in Figure 15-21, changes the pitch of the sound in terms of a trombone being a resonance tube. (15.2) Mastering Concepts 31. What are the physical characteristics of sound waves? (15.1) 32. When timing the 100-m run, officials at the finish line are instructed to start their stopwatches at the sight of smoke from the starter\u2019s pistol and not at the sound of its firing. Explain.", " What would happen to the times for the runners if the timing started when sound was heard? (15.1) 33. Name two types of perception of sound and the physical characteristics of sound waves that correspond to them. (15.1) 34. Does the Doppler shift occur for only some types of waves or for all types of waves? (15.1) 35. Sound waves with frequencies higher than can be heard by humans, called ultrasound, can be transmitted through the human body. How could ultrasound be used to measure the speed of blood flowing in veins or arteries? Explain how the waves change to make this measurement possible. (15.1) 36. What is necessary for the production and transmission of sound? (15.2) 37. Singing How can a certain note sung by an opera singer cause a crystal glass to shatter? (15.2) 38. Marching In the military, as marching soldiers approach a bridge, the command \u201croute step\u201d is given. The soldiers then walk out-of-step with each other as they cross the bridge. Explain. (15.2) 39. Musical Instruments Why don\u2019t most musical instruments sound like tuning forks? (15.2) \u25a0 Figure 15-21 Applying Concepts 42. Estimation To estimate the distance in kilometers between you and a lightning flash, count the seconds between the flash and the thunder and divide by 3. Explain how this rule works. Devise a similar rule for miles. 43. The speed of sound increases by about 0.6 m/s for each degree Celsius when the air temperature rises. For a given sound, as the temperature increases, what happens to the following? a. the frequency b. the wavelength 44. Movies In a science-fiction movie, a satellite blows up. The crew of a nearby ship immediately hears and sees the explosion. If you had been hired as an advisor, what two physics errors would you have noticed and corrected? 45. The Redshift Astronomers have observed that the light coming from distant galaxies appears redder than light coming from nearer galaxies. With the help of Figure 15-22, which shows the visible spectrum, explain why astronomers conclude that distant galaxies are moving away from Earth. 4107 m 5107 m 6107 m 7107 m \u25a0 Figure 15-22 40. Musical Instruments What property distinguishes notes played on both a trumpet and a clarinet if they have the same pitch and loudness? (15.2) 46. Does a sound of 40 dB have a factor of 100 (102) times greater pressure variation than the threshold of hearing, or a factor of 40 times greater? 424 (t)Tim Fuller, (b)Carolina Biological/Visuals Unlimited Chapter 15 Sound For more problems, go to Additional Problems, Appendix B. 423-429 CH15-SG CA STP-845813 3/30/04 7:30 AM Page 425 47. If the pitch of sound is increased, what are the changes in the following? a. the frequency b. the wavelength c. the wave velocity d. the amplitude of the wave 56. Photography As shown in Figure 15-24, some cameras determine the distance to the subject by sending out a sound wave and measuring the time needed for the echo to return to the camera. How long would it take the sound wave to return to such a camera if the subject were 3.00 m away? 48. The speed of sound increases with temperature. Would the pitch of a closed pipe increase or decrease when the temperature of the air rises? Assume that the length of the pipe does not change. 49. Marching Bands Two flutists are tuning up. If the conductor hears the beat frequency increasing, are the two flute frequencies getting closer together or farther apart? 50. Musical Instruments A covered organ pipe plays a certain note. If the cover is removed to make it an open pipe, is the pitch increased or decreased? 51. Stringed Instruments On a harp, Figure 15-23a, long strings produce low notes and short strings produce high notes. On a guitar, Figure 15-23b, the strings are all the same length. How can they produce notes of different pitches? a b \u25a0 Figure 15-23 Mastering Problems 15.1 Properties and Detection of Sound 52. You hear the sound of the firing of a distant cannon 5.0 s after seeing the flash. How far are you from the cannon? 53. If you shout across a canyon and hear an echo 3.0 s later, how wide is the canyon? 54. A sound wave has a frequency of 4700 Hz and travels along a steel rod. If the distance between compressions, or regions of high pressure, is 1.1 m, what is the speed of the wave? 55. Bats The sound emitted by bats has a wavelength of 3.5 mm. What is the sound\u2019s frequency in air? 3.00 m \u25a0 Figure 15-24 57. Sound with a frequency of 261", ".6 Hz travels through water at 25\u00b0C. Find the sound\u2019s wavelength in water. Do not confuse sound waves moving through water with surface waves moving through water. 58. If the wavelength of a 4.40102-Hz sound in freshwater is 3.30 m, what is the speed of sound in freshwater? 59. Sound with a frequency of 442 Hz travels through an iron beam. Find the wavelength of the sound in iron. 60. Aircraft Adam, an airport employee, is working near a jet plane taking off. He experiences a sound level of 150 dB. a. If Adam wears ear protectors that reduce the sound level to that of a typical rock concert, what decrease in dB is provided? b. If Adam then hears something that sounds like a barely audible whisper, what will a person not wearing the ear protectors hear? 61. Rock Music A rock band plays at an 80-dB sound level. How many times greater is the sound pressure from another rock band playing at each of the following sound levels? a. 100 dB b. 120 dB 62. A coiled-spring toy is shaken at a frequency of 4.0 Hz such that standing waves are observed with a wavelength of 0.50 m. What is the speed of propagation of the wave? 63. A baseball fan on a warm summer day (30\u00b0C) sits in the bleachers 152 m away from home plate. a. What is the speed of sound in air at 30\u00b0C? b. How long after seeing the ball hit the bat does the fan hear the crack of the bat? physicspp.com/chapter_test Chapter 15 Assessment 425 Getty Images 423-429 CH15-SG CA STP-845813 3/30/04 7:30 AM Page 426 64. On a day when the temperature is 15\u00b0C, a person stands some distance, d, as shown in Figure 15-25, from a cliff and claps his hands. The echo returns in 2.0 s. How far away is the cliff? d \u25a0 Figure 15-25 (Not to scale) 65. Medical Imaging Ultrasound with a frequency of 4.25 MHz can be used to produce images of the human body. If the speed of sound in the body is the same as in salt water, 1.50 km/s, what is the length of a 4.25-MHz pressure wave in the body? 66. Sonar A ship surveying the ocean bottom sends sonar waves straight down into the seawater from the surface. As illustrated in Figure 15-26, the first reflection, off of the mud at the sea floor, is received 1.74 s after it was sent. The second reflection, from the bedrock beneath the mud, returns after 2.36 s. The seawater is at a temperature of 25\u00b0C, and the speed of sound in mud is 1875 m/s. a. How deep is the water? b. How thick is the mud? t 1.74 s t 2.36 s Seawater Mud Bedrock \u25a0 Figure 15-26 (Not to scale) 69. A train moving toward a sound detector at 31.0 m/s blows a 305-Hz whistle. What frequency is detected on each of the following? a. a stationary train b. a train moving toward the first train at 21.0 m/s 70. The train in the previous problem is moving away from the detector. What frequency is now detected on each of the following? a. a stationary train b. a train moving away from the first train at a speed of 21 m/s 15.2 The Physics of Music 71. A vertical tube with a tap at the base is filled with water, and a tuning fork vibrates over its mouth. As the water level is lowered in the tube, resonance is heard when the water level has dropped 17 cm, and again after 49 cm of distance exists from the water to the top of the tube. What is the frequency of the tuning fork? 72. Human Hearing The auditory canal leading to the eardrum is a closed pipe that is 3.0 cm long. Find the approximate value (ignoring end correction) of the lowest resonance frequency. 73. If you hold a 1.2-m aluminum rod in the center and hit one end with a hammer, it will oscillate like an open pipe. Antinodes of pressure correspond to nodes of molecular motion, so there is a pressure antinode in the center of the bar. The speed of sound in aluminum is 5150 m/s. What would be the bar\u2019s lowest frequency of oscillation? 74. One tuning fork has a 445-Hz pitch. When a second fork is struck, beat notes occur with a frequency of 3 Hz. What are the two possible frequencies of the second fork? 75. Flutes A flute acts as an open pipe. If a flute sounds a note with a 370-Hz pitch, what are the frequencies of", " the second, third, and fourth harmonics of this pitch? 76. Clarinets A clarinet sounds the same note, with a pitch of 370 Hz, as in the previous problem. The clarinet, however, acts as a closed pipe. What are the frequencies of the lowest three harmonics produced by this instrument? 67. Determine the variation in sound pressure of a conversation being held at a sound level of 60 dB. 68. A fire truck is moving at 35 m/s, and a car in front of the truck is moving in the same direction at 15 m/s. If a 327-Hz siren blares from the truck, what frequency is heard by the driver of the car? 77. String Instruments A guitar string is 65.0 cm long and is tuned to produce a lowest frequency of 196 Hz. a. What is the speed of the wave on the string? b. What are the next two higher resonant frequencies for this string? 426 Chapter 15 Sound For more problems, go to Additional Problems, Appendix B. 423-429 CH15-SG CA STP-845813 6/4/04 9:56 AM Page 427 78. Musical Instruments The lowest note on an organ is 16.4 Hz. a. What is the shortest open organ pipe that will resonate at this frequency? b. What is the pitch if the same organ pipe is closed? Mixed Review 85. An open organ pipe is 1.65 m long. What fundamental frequency note will it produce if it is played in helium at 0\u00b0C? 79. Musical Instruments Two instruments are playing musical A (440.0 Hz). A beat note with a frequency of 2.5 Hz is heard. Assuming that one instrument is playing the correct pitch, what is the frequency of the pitch played by the second instrument? 80. A flexible, corrugated, plastic tube, shown in Figure 15-27, is 0.85 m long. When it is swung around, it creates a tone that is the lowest pitch for an open pipe of this length. What is the frequency? 0.85 m \u25a0 Figure 15-27 81. The tube from the previous problem is swung faster, producing a higher pitch. What is the new frequency? 82. During normal conversation, the amplitude of a pressure wave is 0.020 Pa. a. If the area of an eardrum is 0.52 cm2, what is the force on the eardrum? b. The mechanical advantage of the three bones in the middle ear is 1.5. If the force in part a is transmitted undiminished to the bones, what force do the bones exert on the oval window, the membrane to which the third bone is attached? c. The area of the oval window is 0.026 cm2. What is the pressure increase transmitted to the liquid in the cochlea? 83. Musical Instruments One open organ pipe has a length of 836 mm. A second open pipe should have a pitch that is one major third higher. How long should the second pipe be? 84. As shown in Figure 15-28, a music box contains a set of steel fingers clamped at one end and plucked on the other end by pins on a rotating drum. What is the speed of a wave on a finger that is 2.4 cm long and plays a note of 1760 Hz? Steel fingers \u25a0 Figure 15-28 86. If you drop a stone into a well that is 122.5 m deep, as illustrated in Figure 15-29, how soon after you drop the stone will you hear it hit the bottom of the well? 122.5 m \u25a0 Figure 15-29 (Not to scale) 87. A bird on a newly discovered planet flies toward a surprised astronaut at a speed of 19.5 m/s while singing at a pitch of 945 Hz. The astronaut hears a tone of 985 Hz. What is the speed of sound in the atmosphere of this planet? 88. In North America, one of the hottest outdoor temperatures ever recorded is 57\u00b0C and one of the coldest is 62\u00b0C. What are the speeds of sound at those two temperatures? 89. A ship\u2019s sonar uses a frequency of 22.5 kHz. The speed of sound in seawater is 1533 m/s. What is the frequency received on the ship that was reflected from a whale traveling at 4.15 m/s away from the ship? Assume that the ship is at rest. 90. When a wet finger is rubbed around the rim of a glass, a loud tone of frequency 2100 Hz is produced. If the glass has a diameter of 6.2 cm and the vibration contains one wavelength around its rim, what is the speed of the wave in the glass? 91. History of Science In 1845, Dutch scientist Christoph Buys-Ballot developed a test of the Doppler effect. He had a trumpet player sound an A note at 440 Hz while riding on a flat", "car pulled by a locomotive. At the same time, a stationary trumpeter played the same note. Buys-Ballot heard 3.0 beats per second. How fast was the train moving toward him? physicspp.com/chapter_test Chapter 15 Assessment 427 Horizons Companies 423-429 CH15-SG CA STP-845813 7/13/04 1:11 PM Page 428 92. You try to repeat Buys-Ballot\u2019s experiment from the previous problem. You plan to have a trumpet played in a rapidly moving car. Rather than listening for beat notes, however, you want to have the car move fast enough so that the moving trumpet sounds one major third above a stationary trumpet. a. How fast would the car have to move? b. Should you try the experiment? Explain. F 93. Guitar Strings The equation for the speed of a wave T, where FT is the tension in the on a string is v string and is the mass per unit length of the string. A guitar string has a mass of 3.2 g and is 65 cm long. What must be the tension in the string to produce a note whose fundamental frequency is 147 Hz? 94. A train speeding toward a tunnel at 37.5 m/s sounds its horn at 327 Hz. The sound bounces off the tunnel mouth. What is the frequency of the reflected sound heard on the train? Hint: Solve the problem in two parts. First, assume that the tunnel is a stationary observer and find the frequency. Then, assume that the tunnel is a stationary source and find the frequency measured on the train. Thinking Critically 95. Make and Use Graphs The wavelengths of the sound waves produced by a set of tuning forks with given frequencies are shown in Table 15-2 below. a. Plot a graph of wavelength versus the frequency (controlled variable). What type of relationship does the graph show? b. Plot a graph of wavelength versus the inverse of the frequency (1/f). What kind of graph is this? Determine the speed of sound from this graph. Table 15-2 Tuning Forks 97. Analyze and Conclude Describe how you could use a stopwatch to estimate the speed of sound if you were near the green on a 200-m golf hole as another group of golfers hit their tee shots. Would your estimate of the speed of sound be too large or too small? 98. Apply Concepts A light wave coming from a point on the left edge of the Sun is found by astronomers to have a slightly higher frequency than light from the right side. What do these measurements tell you about the Sun\u2019s motion? 99. Design an Experiment Design an experiment that could test the formula for the speed of a wave on a string. Explain what measurements you would make, how you would make them, and how you would use them to test the formula. Writing in Physics 100. Research the construction of a musical instrument, such as a violin or French horn. What factors must be considered besides the length of the strings or tube? What is the difference between a quality instrument and a cheaper one? How are they tested for tone quality? 101. Research the use of the Doppler effect in the study of astronomy. What is its role in the big bang theory? How is it used to detect planets around other stars? To study the motions of galaxies? Cumulative Review 102. Ball A, rolling west at 3.0 m/s, has a mass of 1.0 kg. Ball B has a mass of 2.0 kg and is stationary. After colliding with ball B, ball A moves south at 2.0 m/s. (Chapter 9) Frequency (Hz) Wavelength (m) a. Sketch the system, showing the velocities and 131 147 165 196 220 247 2.62 2.33 2.08 1.75 1.56 1.39 96. Make Graphs Suppose that the frequency of a car horn is 300 Hz when it is stationary. What would the graph of the frequency versus time look like as the car approached and then moved past you? Complete a rough sketch. momenta before and after the collision. b. Calculate the momentum and velocity of ball B after the collision. 103. Chris carries a 10-N carton of milk along a level hall to the kitchen, a distance of 3.5 m. How much work does Chris do? (Chapter 10) 104. A movie stunt person jumps from a five-story building (22 m high) onto a large pillow at ground level. The pillow cushions her fall so that she feels a deceleration of no more than 3.0 m/s2. If she weighs 480 N, how much energy does the pillow have to absorb? How much force does the pillow exert on her? (Chapter 11) 428 Chapter 15 Sound For more problems, go to Additional Problems, Appendix B. 423-429 CH15-SG CA STP-845813 3/", "30/04 7:33 AM Page 429 6. Reba hears 20 beats in 5.0 s when she plays two notes on her piano. She is certain that one note has a frequency of 262 Hz. What are the possible frequencies of the second note? 242 Hz or 282 Hz 258 Hz or 266 Hz 260 Hz or 264 Hz 270 Hz or 278 Hz 7. Which of the following pairs of instruments have resonant frequencies at each wholenumber multiple of the lowest frequency? a clamped string and a closed pipe a clamped string and an open pipe an open pipe and a closed pipe an open pipe and a reed instrument Extended Answer 8. The figure below shows the first resonance length of a closed air column. If the frequency of the sound is 488 Hz, what is the speed of the sound? L 16.8 cm Multiple Choice 1. How does sound travel from its source to your ear? by changes in air pressure by vibrations in wires or strings by electromagnetic waves by infrared waves 2. Paulo is listening to classical music in the speakers installed in his swimming pool. A note with a frequency of 327 Hz reaches his ears while he is under water. What is the wavelength of the sound that reaches Paulo\u2019s ears? Use 1493 m/s for the speed of sound in water. 2.19101 m 4.57 m 2.19 nm 4.88105 m 3. The sound from a trumpet travels at 351 m/s in air. If the frequency of the note is 298 Hz, what is the wavelength of the sound wave? 9.93104 m 0.849 m 1.18 m 1.05105 m 4. The horn of a car attracts the attention of a stationary observer. If the car is approaching the observer at 60.0 km/h and the horn has a frequency of 512 Hz, what is the frequency of the sound perceived by the observer? Use 343 m/s for the speed of sound in air. 488 Hz 512 Hz 538 Hz 600 Hz 5. As shown in the diagram below, a car is receding at 72 km/h from a stationary siren. If the siren is wailing at 657 Hz, what is the frequency of the sound perceived by the driver? Use 343 m/s for the speed of sound. 543 Hz 620 Hz 647 Hz 698 Hz vsound x vcar Write It Down Most tests ask you a large number of questions in a small amount of time. Write down your work whenever possible. Do math on paper, not in your head. Underline and reread important facts in passages and diagrams\u2014do not try to memorize them. physicspp.com/standardized_test Chapter 15 Standardized Test Practice 429 430-438 CH16-S1-845813 3/22/04 1:17 AM Page 430 What You\u2019ll Learn \u2022 You will understand sources of light and how light illuminates the universe around us. \u2022 You will be able to describe the wave nature of light and some phenomena that reveal this nature. Why It\u2019s Important Light is a primary source of information about how the universe behaves. We all use information such as color, brightness, and shadow every day to interpret the events occurring around us. Balloon Race You can tell the difference between the competing balloons because of the different colors visible in the sunlight. You can distinguish the balloons from the backgrounds because of color differences in the grass and sky. Think About This What causes these differences in color? How are these colors related? physicspp.com 430 Getty Images 430-438 CH16-S1-845813 6/4/04 6:56 AM Page 431 How can you determine the path of light through air? Question What path does light take as it travels through the air? Procedure Analysis Describe the image of the reflected light beam that you see on the index card. Describe the path that the light beam takes. Critical Thinking Can you see the light beam in the air? Why or why not? 1. Punch a hole with a pushpin in the center of an index card. 2. Using clay, stand the index card so that its longer edge is on the table top. 3. Turn on a lamp and have one lab partner hold the lamp so that the lightbulb shines through the hole in the card. CAUTION: Lamp can get hot over time. 4. Hold a mirror on the opposite side of the index card so that light coming through the hole strikes the mirror. Darken the room. 5. Angle the mirror so that it reflects the beam of light onto the back of the card. CAUTION: Be careful not to reflect the light beam into someone\u2019s eyes. 6. Write down your observations. 16.1 Illumination Light and sound are two methods by which you can receive information. Of the two, light seems to provide the greater variety of information. The human eye can detect tiny changes in the size, position, brightness, and color of", " an object. Our eyes usually can distinguish shadows from solid objects and sometimes distinguish reflections of objects from the objects themselves. In this section, you will learn where light comes from and how it illuminates the universe around you. One of the first things that you ever discovered about light, although you may not have been conscious of your discovery, is that it travels in a straight line. How do you know this? When a narrow beam of light, such as that of a flashlight or sunlight streaming through a small window, is made visible by dust particles in the air, you see the path of the light as a straight line. When your body blocks sunlight, you see your outline in a shadow. Also, whenever you locate an object with your eyes and walk toward it, you most likely walk in a straight path. These things are possible only because light travels in straight lines. Based on this knowledge of how light travels, models have been developed that describe how light works. Objectives \u2022 Develop the ray model of light. \u2022 Predict the effect of distance on light\u2019s illumination. \u2022 Solve problems involving the speed of light. Vocabulary ray model of light luminous source illuminated source opaque transparent translucent luminous flux illuminance Section 16.1 Illumination 431 Horizons Companies 430-438 CH16-S1-845813 3/22/04 1:20 AM Page 432 a b c Ray Reflected ray \u25a0 Figure 16-1 A ray is a straight line that represents the linear path of a narrow beam of light (a). A light ray can change direction if it is reflected (b) or refracted (c). Obstruction Obstruction Refracted ray \u2022 Light rays are red. Ray Model of Light Isaac Newton, whose laws of motion you studied in Chapter 6, believed that light is a stream of fast-moving, unimaginably tiny particles, which he called corpuscles. However, his model could not explain all of the properties of light. Experiments showed that light also behaves like a wave. In the ray model of light, light is represented as a ray that travels in a straight path, the direction of which can be changed only by placing an obstruction in the path, as shown in Figure 16-1. The ray model of light was introduced as a way to study how light interacts with matter, regardless of whether light is a particle or a wave. This study of light is called ray optics or geometric optics. Sources of light Rays of light come from sources of light. Our major source of light is the Sun. Other natural sources of light include flames, sparks, and even fireflies. In the past 100 years, humans have been able to produce several other kinds of light sources. Incandescent bulbs, fluorescent lamps, television screens, lasers, and tiny, light-emitting diodes (LEDs) are each a result of humans using electricity to produce light. What is the difference between sunlight and moonlight? Sunlight, of course, is much, much brighter. There also is an important fundamental difference between the two. The Sun is a luminous source, an object that emits light. In contrast, the Moon is an illuminated source, an object that becomes visible as a result of the light reflecting off it, as shown in Figure 16-2. An incandescent lamp, such as a common lightbulb, is luminous because electrical energy heats a thin tungsten wire in the bulb and causes it to glow. An incandescent source emits light as a result of its high temperature. A bicycle reflector, on the other hand, works as an illuminated source. It is designed to become highly visible when it is illuminated by luminous automobile headlights. Moon Illuminated source Sun Earth Luminous source \u25a0 Figure 16-2 The Sun acts as a luminous source to Earth and the Moon. The Moon acts as an illuminated source to Earth. (Illustration not to scale) 432 Chapter 16 Fundamentals of Light 430-438 CH16-S1-845813 6/4/04 6:26 AM Page 433 a b c \u25a0 Figure 16-3 The transparent glass allows objects to be seen through it (a). The translucent lamp shade allows light to pass through, although the lightbulb source itself is not visible (b). The opaque tarp covers the statue, preventing the statue from being seen (c). Illuminated sources are visible to you because light is reflecting off or transmitting (passing) through the object to your eyes. Media, such as brick, that do not transmit light, but reflect some light, are opaque media. Media that transmit light, such as air and glass, are transparent media. Media that transmit light, but do not permit objects to be seen clearly through them, are translucent media. Lamp shades and frosted lightbulbs are examples of objects that are made of translucent media. All three types of media are illustrated in Figure 16-3. Transparent or translucent media not only transmit light, but they also can reflect a fraction of the light", ". For example, you often can see your reflection in a glass window. Quantity of light The rate at which light energy is emitted from a luminous source is called the luminous flux, P. The unit of luminous flux is the lumen (lm). A typical 100-W incandescent lightbulb emits approximately 1750 lm. You can think of the luminous flux as a measure of the rate at which light rays come out of a luminous source. Imagine placing a lightbulb at the center of a 1-m-radius sphere, as shown in Figure 16-4. The lightbulb emits light in almost all directions. The 1750 lm of luminous flux characterizes all of the light that strikes the inside surface of the sphere in a given unit of time. Even if the sphere was 2 m in radius, the luminous flux of the lightbulb would be the same as for the 1-m-radius sphere, because the total number of light rays does not increase. Once you know the quantity of light being emitted by a luminous source, you can determine the amount of illumination that the luminous source provides to an object, such as a book. The illumination of a surface, or the rate at which light strikes the surface, is called the illuminance, E. You can think of this as a measure of the number of light rays that strike a surface. Illuminance is measured in lux, lx, which is equivalent to lumens per square meter, lm/m2. Consider the setup shown in Figure 16-4. What is the illuminance of the sphere\u2019s inside surface? The equation for the surface area of a sphere is 4r2, so the surface area of this sphere is 4(1.00 m)2 4 m2. The luminous flux striking each square meter of the sphere is 1750 lm/(4 m2) 139 lx. At a distance of 1.00 m from the bulb, 139 lm strikes each square meter. The illuminance of the inside of the sphere is 139 lx. \u25a0 Figure 16-4 Luminous flux is the rate at which light is emitted from a luminous source, whereas illuminance is the rate at which light falls on a surface. Luminous flux P 1750 Im r 1 m Illuminance 1750 4 lx E1 Section 16.1 Illumination 433 (l)Andrew McKim/Masterfile, (m)(r)Laura Sifferlin 430-438 CH16-S1-845813 6/10/04 8:41 PM Page 434 \u25a0 Figure 16-5 The illuminance, E, produced by a point source of light varies inversely as the square of the distance from the light source E1 E1 1 9 E1 An inverse-square relationship What would happen if the sphere surrounding the lamp were larger? If the sphere had a radius of 2.00 m, the luminous flux still would total 1750 lm, but the area of the sphere would be 4(2.00 m)2 16.0 m2, four times larger than the 1.00-m sphere, as shown in Figure 16-5. The illuminance of the inside of the 2.00-m sphere is 1750 lm/(16.0 m2) 34.8 lx, so 34.8 lm strikes each square meter. The illuminance on the inside surface of the 2.00-m sphere is one-fourth the illuminance on the inside of the 1.00-m sphere. In the same way, the inside of a sphere with a 3.00-m radius has an illuminance only (1/3)2, or 1/9, as large as the 1.00-m sphere. Figure 16-5 shows that the illuminance produced by a point source is proportional to 1/r2, an inverse-square relationship. As the light rays spread out in straight lines in all directions from a point source, the number of light rays available to illuminate a unit of area decreases as the square of the distance from the point source. Luminous intensity Some luminous sources are specified in candela, cd. A candela is not a measure of luminous flux, but of luminous intensity. The luminous intensity of a point source is the luminous flux that falls on 1 m2 of the inside of a 1-m-radius sphere. Thus, luminous intensity is luminous flux divided by 4. A bulb with 1750 lm of flux has an intensity of 1750 lm/4 139 cd. In Figure 16-6, the lightbulb is twice as far away from the screen as the candle. For the lightbulb to provide the same illuminance on the lightbulb side of the screen as the candle does on the candle side of the screen, the lightbulb would", " have to be four times brighter than the candle, and, therefore, the luminous intensity of the lightbulb would have to be four times the luminous intensity of the candle. Lamp Screen Candle 2d d \u25a0 Figure 16-6 The illuminance is the same on both sides of the screen, though the lightbulb is brighter than the candle. 434 Chapter 16 Fundamentals of Light 430-438 CH16-S1-845813 6/4/04 6:33 AM Page 435 How to Illuminate a Surface How would you increase the illuminance of your desktop? You could use a brighter bulb, which would increase the luminous flux, or you could move the light source closer to the surface of your desk, thereby decreasing the distance between the light source and the surface it is illuminating. To make the problem easier, you can use the simplification that the light source is a point source. Thus, the illuminance and distance will follow the inverse-square relationship. The problem is further simplified if you assume that light from the source strikes perpendicular to the surface of the desk. Using these simplifications, the illuminance caused by a point light source is represented by the following equation. P Point Source Illuminance E 4 r2 If an object is illuminated by a point source of light, then the illuminance at the object is equal to the luminous flux of the light source, divided by the surface area of the sphere, whose radius is equal to the distance the object is from the light source. Remember that the luminous flux of the light source is spreading out spherically in all directions, so only a fraction of the luminous flux is available to illuminate the desk. Use of this equation is valid only if the light from the luminous source strikes perpendicular to the surface it is illuminating. It is also only valid for luminous sources that are small enough or far enough away to be considered point sources. Thus, the equation does not give accurate values of illuminance for long, fluorescent lamps or incandescent lightbulbs that are close to the surfaces that they illuminate. Direct and Inverse Relationships The illuminance provided by a source of light has both a direct and an inverse relationship. Math x y z2 a Physics P E 4 r 2 If z is constant, then y is directly proportional to x. If r is constant, then E is directly proportional to P. \u2022 When x increases, y increases. \u2022 When P increases, E increases. \u2022 When x decreases, y decreases. \u2022 When P decreases, E decreases. If x is constant, then y is inversely proportional to z2. If P is constant, then E is inversely proportional to r 2. \u2022 When z2 increases, y decreases. \u2022 When r 2 increases, E decreases. \u2022 When z2 decreases, y increases. \u2022 When r 2 decreases, E increases. Illuminated Minds When deciding how to achieve the correct illuminance on students\u2019 desktops, architects must consider the luminous flux of the lights as well as the distance of the lights above the desktops. In addition, the efficiencies of the light sources are an important economic factor. Section 16.1 Illumination 435 430-438 CH16-S1-845813 6/9/04 4:25 PM Page 436 Illumination of a Surface What is the illuminance at on your desktop if it is lighted by a 1750-lm lamp that is 2.50 m above your desk? 1 2 Analyze and Sketch the Problem \u2022 Assume that the lightbulb is the point source. \u2022 Diagram the position of the bulb and desktop. Label P and r. Known: P 1.75103 lm r 2.50 m Unknown: E? Solve for the Unknown The surface is perpendicular to the direction in which the light ray is traveling, so you can use the point source illuminance equation7 Substitue P 1.75103 lm, r 2.50 m )2 4.5 m 0 2 ( 22.3 lm/m2 22.3 lx P 1.75103 lm 2.50 m Math Handbook Operations with Significant Digits pages 835\u2014836 3 Evaluate the Answer \u2022 Are the units correct? The units of luminance are lm/m2 lx, which the answer agrees with. \u2022 Do the signs make sense? All quantities are positive, as they should be. \u2022 Is the magnitude realistic? The illuminance is less than the luminous flux, which it should be at this distance. 1. A lamp is moved from 30 cm to 90 cm above the pages of a book. Compare the illumination on the book before and after the lamp is moved. 2. What is the illumination on a surface that is 3.0 m below a 150-W incandescent lamp that emits a luminous flux of 2275 lm? 3. Draw a graph of the", " illuminance produced by a 150-W incandescent lamp between 0.50 m and 5.0 m. 4. A 64-cd point source of light is 3.0 m above the surface of a desk. What is the illumination on the desk\u2019s surface in lux? 5. A public school law requires a minimum illuminance of 160 lx at the surface of each student\u2019s desk. An architect\u2019s specifications call for classroom lights to be located 2.0 m above the desks. What is the minimum luminous flux that the lights must produce? 6. A screen is placed between two lamps so that they illuminate the screen equally, as shown in Figure 16-7. The first lamp emits a luminous flux of 1445 lm and is 2.5 m from the screen. What is the distance of the second lamp from the screen if the luminous flux is 2375 lm? 436 Chapter 16 Fundamentals of Light Screen P 2375 lm 2.5 m P 1445 lm \u25a0 Figure 16-7 (Not to scale) 430-438 CH16-S1-845813 6/4/04 10:03 AM Page 437 Engineers who design lighting systems must understand how the light will be used. If an even illumination is needed to prevent dark areas, the common practice is to evenly space normal lights over the area to be illuminated, as was most likely done with the lights in your classroom. Because such light sources do not produce truly even light, however, engineers also design special light sources that control the spread of the light, such that they produce even illuminations over large surface areas. Much work has been done in this field with automobile headlights. The Speed of Light For light to travel from a source to an object to be illuminated, it must travel across some distance. According to classical mechanics, if you can measure the distance and the time it takes to travel that distance, you can calculate a speed. Before the seventeenth century, most people believed that light traveled instantaneously. Galileo was the first to hypothesize that light has a finite speed, and to suggest a method of measuring its speed using distance and time. His method, however, was not precise enough, and he was forced to conclude that the speed of light is too fast to be measured over a distance of a few kilometers. Danish astronomer Ole Roemer was the first to determine that light does travel with a measurable speed. Between 1668 and 1674, Roemer made 70 measurements of the 1.8-day orbital period of Io, one of Jupiter\u2019s moons. He recorded the times when Io emerged from Jupiter\u2019s shadow, as shown in Figure 16-8. He made his measurements as part of a project to improve maps by calculating the longitude of locations on Earth. This is an early example of the needs of technology driving scientific advances. After making many measurements, Roemer was able to predict when the next eclipse of Io would occur. He compared his predictions with the actual measured times and found that Io\u2019s orbital period increased on average by about 13 s per orbit when Earth was moving away from Jupiter and decreased on average by about 13 s per orbit when Earth was approaching Jupiter. Roemer believed that Jupiter\u2019s moons were just as regular in their orbits as Earth\u2019s moon; thus, he wondered what might cause this discrepancy in the measurement of Io\u2019s orbital period. vJupiter vIo o w d a h S Io Jupiter 2 1 Sun 3 4 Earth \u25a0 Figure 16-8 Roemer measured the time between eclipses as Io emerged from Jupiter\u2019s shadow. During successive eclipses, Io\u2019s measured orbital period became increasingly smaller or larger depending on whether Earth was moving toward (from position 3 to 1) or away from (from position 1 to 3) Jupiter. (Illustration not to scale) Section 16.1 Illumination 437 430-438 CH16-S1-845813 6/4/04 10:05 AM Page 438 Measurements of the speed of light Roemer concluded that as Earth moved away from Jupiter, the light from each new appearance of Io took longer to travel to Earth because of the increasing distance to Earth. Likewise, as Earth approached Jupiter, Io\u2019s orbital period seemed to decrease. Roemer noted that during the 182.5 days that it took Earth to travel from position 1 to position 3, as shown in Figure 16-8, there were (185.2 days)(1 Io eclipse/1.8 days) 103 Io eclipses. Thus, for light to travel the diameter of Earth\u2019s orbit, he calculated that it takes (103 eclipses )(13 s/eclipse) 1.3103 s, or 22 min. Using the presently known value of the diameter of Earth\u2019s orbit (2.91011 m), Roemer\u2019s value of 22 min gives a value for the speed of light of 2.", "91011 m/((22 min)(60 s/min)) 2.2108 m/s. Today, the speed of light is known to be closer to 3.0108 m/s. Thus, light takes 16.5 min, not 22 min, to cross Earth\u2019s orbit. Nevertheless, Roemer had successfully proved that light moves at a finite speed. Although many measurements of the speed of light have been made, the most notable were performed by American physicist Albert A. Michelson. Between 1880 and the 1920s, he developed Earth-based techniques to measure the speed of light. In 1926, Michelson measured the time required for light to make a round-trip between two California mountains 35 km apart. Michelson used a set of rotating mirrors to measure such small time intervals. Michelson\u2019s best result was (2.997996 0.00004)108 m/s. For this work, he became the first American to receive a Nobel prize in science. The speed of light in a vacuum is a very important and universal value; thus it has its own special symbol, c. Based on the wave nature of light, which you will study in the next section, the International Committee on Weights and Measurements has measured and defined the speed of light in a vacuum to be c 299,792,458 m/s. For many calculations, the value c 3.00108 m/s is precise enough. At this speed, light travels 9.461012 km in a year. This amount of distance is called a light-year. 16.1 Section Review 7. Use of Material Light Properties Why might you choose a window shade that is translucent? Opaque? 11. Distance of Light Travel How far does light travel in the time it takes sound to travel 1 cm in air at 20\u00b0C? 8. Illuminance Does one lightbulb provide more illuminance than two identical lightbulbs at twice the distance? Explain. 9. Luminous Intensity Two lamps illuminate a screen equally\u2014lamp A at 5.0 m, lamp B at 3.0 m. If lamp A is rated 75 cd, what is lamp B rated? 12. Distance of Light Travel The distance to the Moon can be found with the help of mirrors left on the Moon by astronauts. A pulse of light is sent to the Moon and returns to Earth in 2.562 s. Using the measured value of the speed of light to the same precision, calculate the distance from Earth to the Moon. 10. Distance of a Light Source Suppose that a lightbulb illuminating your desk provides only half the illuminance that it should. If it is currently 1.0 m away, how far should it be to provide the correct illuminance? 13. Critical Thinking Use the correct time taken for light to cross Earth\u2019s orbit, 16.5 min, and the diameter of Earth\u2019s orbit, 2.981011 m, to calculate the speed of light using Roemer\u2019s method. Does this method appear to be accurate? Why or why not? 438 Chapter 16 Fundamentals of Light physicspp.com/self_check_quiz 439-447 CH16-S2-845813 6/4/04 10:10 AM Page 439 Objectives \u2022 Describe how diffraction demonstrates that light is a wave. \u2022 Predict the effect of combining colors of light and mixing pigments. \u2022 Explain phenomena such as polarization and the Doppler effect. Vocabulary diffraction primary color secondary color complementary color primary pigment secondary pigment polarization Malus\u2019s law 16.2 The Wave Nature of Light You probably have heard that light is composed of waves, but what evidence do you have that this is so? Suppose that you walk by the open door of the band-rehearsal room at school. You hear the music as you walk toward the rehearsal-room door long before you can see the band members through the door. Sound seems to have reached you by bending around the edge of the door, whereas the light that enables you to see the band members has traveled only in a straight line. If light is composed of waves, why doesn\u2019t light seem to act in the same way as sound does? In fact, light does act in the same way; however, the effect is much less obvious with light than with sound. Diffraction and the Wave Model of Light In 1665, Italian scientist Francesco Maria Grimaldi observed that the edges of shadows are not perfectly sharp. He introduced a narrow beam of light into a dark room and held a rod in front of the light such that it cast a shadow on a white surface. The shadow cast by the rod on the white surface was wider than the shadow should have been if light traveled in a straight line past the edges of the rod. Grimaldi also noted that the shadow was bordered by colored bands. Grimaldi recognized this phenomenon as diffraction, which is the bending of light around", " a barrier. In 1678, Dutch scientist Christiaan Huygens argued in favor of a wave model to explain diffraction. According to Huygens\u2019 principle, all the points of a wave front of light can be thought of as new sources of smaller waves. These wavelets expand in every direction and are in step with one another. A flat, or plane, wave front of light consists of an infinite number of point sources in a line. As this wave front passes by an edge, the edge cuts the wave front such that each circular wavelet generated by each Huygens\u2019 point will propagate as a circular wave in the region where the original wave front was bent, as shown in Figure 16-9. This is diffraction. \u25a0 Figure 16-9 According to Huygens\u2019 principle, the crest of each wave can be thought of as a series of point sources. Each point source creates a circular wavelet. All the wavelets combine to make a flat wave front, except at the edge where circular wavelets of the Huygens\u2019 points move away from the wave front. Barrier Section 16.2 The Wave Nature of Light 439 439-447 CH16-S2-845813 3/22/04 1:29 AM Page 440 Color In 1666, possibly prompted by Grimaldi\u2019s publication of his diffraction results, Newton performed experiments on the colors produced when a narrow beam of sunlight passed through a glass prism, as shown in Figure 16-10. Newton called the ordered arrangement of colors a spectrum. Using his corpuscle model of light, he believed that particles of light were interacting with some unevenness in the glass to produce the spectrum. To test this assumption, Newton allowed the spectrum from one prism to fall on a second prism. If the spectrum was caused by irregularities in the glass, he reasoned that the second prism would increase the spread in colors. Instead, the second prism reversed the spreading of colors and recombined them to form white light. After more experiments, Newton concluded that white light is composed of colors, and that a property of the glass other than unevenness caused the light to separate into colors. Based on the work of Grimaldi, Huygens, and others, we know that light has wave properties and that each color of light is associated with a wavelength. Light falls within the range of wavelengths from about 400 nm (4.00107 m) to 700 nm (7.00107 m), as shown in Figure 16-11. The longest wavelengths are seen as red light. As wavelength decreases, the color changes to orange, yellow, green, blue, indigo, and finally, violet. Red (7.00107 m) Violet (4.00107 m) \u25a0 Figure 16-11 The spectrum of light ranges from the long, red wavelength to the short, violet wavelength. As white light crosses the boundary from air into glass and back into air in Figure 16-11, its wave nature causes each different color of light to be bent, or refracted, at a different angle. This unequal bending of the different colors causes the white light to be spread into a spectrum. This reveals that different wavelengths of light interact in different but predictable ways with matter. Color by addition of light White light can be formed from colored light in a variety of ways. For example, when the correct intensities of red, green, and blue light are projected onto a white screen, as in Figure 16-12, the region where these three colors overlap on the screen will appear to be white. Thus, red, green, and blue light form white light when they are combined. This is called the additive color process, which is used in color-television tubes. A color-television tube contains tiny, dotlike sources of red, green, and blue light. When all three colors of light have the correct intensities, the screen appears to be white. For this reason, red, green, and blue are each called a primary color. The primary colors can be mixed in pairs to form three additional colors, as shown in Figure 16-12. Red and green light together produce yellow light, blue and green light produce cyan, and red and blue light produce magenta. The colors yellow, cyan, and magenta are each called a secondary color, because each is a combination of two primary colors. \u25a0 Figure 16-10 White light, when passed through a prism, is separated into a spectrum of colors. \u25a0 Figure 16-12 Different combinations of blue, green, and red light can produce yellow, cyan, magenta, or white light. Green Yellow White Cyan Red Blue Magenta 440 (b)Matt Meadows, (t)Kodak Chapter 16 Fundamentals of Light 439-447 CH16-S2-845813 3/22/04 1:30 AM Page 441 Color by Temperature Some artists refer to red and orange as hot colors and green and blue as cool colors. Do colors", " really relate to temperature in this way? 1. Obtain a glass prism from your teacher. 2. Obtain a lamp with a dimmer switch from your teacher. Turn on the lamp and turn off the room light. Set the dimmer to minimum brightness of the lamp. 3. Slowly increase the brightness of the lamp. CAUTION: Lamp can get hot and burn skin. 4. Observe the color of light produced by the prism and how it relates to the warmth of the lightbulb on your hand. Analyze and Conclude 5. What colors appeared first when the light was dim? 6. What colors were the last to appear as you brightened the light? 7. How do these colors relate to the temperature of the filament? As shown in Figure 16-12, yellow light can be made from red light and green light. If yellow light and blue light are projected onto a white screen with the correct intensities, the surface will appear to be white. Complementary colors are two colors of light that can be combined to make white light. Thus, yellow is a complementary color of blue, and vice versa, because the two colors of light combine to make white light. In the same way, cyan and red are complementary colors, as are magenta and green. Yellowish laundry can be whitened with a bluing agent added to detergent. Color by subtraction of light As you learned in the first section of this chapter, objects can reflect and transmit light. They also can absorb light. Not only does the color of an object depend on the wavelengths present in the light that illuminates the object, but an object\u2019s color also depends on what wavelengths are absorbed by the object and what wavelengths are reflected. The natural existence or artificial placement of dyes in the material of an object, or pigments on its surface, give the object color. A dye is a molecule that absorbs certain wavelengths of light and transmits or reflects others. When light is absorbed, its energy is taken into the object that it strikes and is turned into other forms of energy. A red shirt is red because the dyes in it reflect red light to our eyes. When white light falls on the red object shown in Figure 16-13, the dye molecules in the object absorb the blue and green light and reflect the red light. When only blue light falls on the red object, very little light is reflected and the object appears to be almost black. The difference between a dye and a pigment is that pigments usually are made of crushed minerals, rather than plant or insect extracts. Pigment particles can be seen with a microscope. A pigment that absorbs only one primary color and reflects two from white light is called a primary pigment. Yellow pigment absorbs blue light and reflects red and green light. Yellow, cyan, and magenta are the colors of primary pigments. A pigment that absorbs two primary colors and reflects one color is called a secondary pigment. The colors of secondary pigments are red (which absorbs green and blue light), green (which absorbs red and blue light), and blue (which absorbs red and green light). Note that the primary pigment colors are the secondary colors. In the same way, the secondary pigment colors are the primary colors. \u25a0 Figure 16-13 The dyes in the dice selectively absorb and reflect various wavelengths of light. The dice are illuminated by white light (a), red light (b), and blue light (c). a b c Section 16.2 The Wave Nature of Light 441 Tom Pantages 439-447 CH16-S2-845813 6/4/04 10:15 AM Page 442 \u25a0 Figure 16-14 The primary pigments are magenta, cyan, and yellow. Mixing these in pairs produces the secondary pigments: red, green, and blue. The primary and secondary pigments are shown in Figure 16-14. When the primary pigments yellow and cyan are mixed, the yellow absorbs blue light and the cyan absorbs red light. Thus, Figure 16-14 shows yellow and cyan combining to make green pigment. When yellow pigment is mixed with the secondary pigment, blue, which absorbs green and red light, all of the primary colors are absorbed, so the result is black. Thus, yellow and blue are complementary pigments. Cyan and red, as well as magenta and green, are also complementary pigments. A color printer uses yellow, magenta, and cyan dots of pigment to make a color image on paper. Often, pigments that are used are finely ground compounds, such as titanium(IV) oxide (white), chromium(III) oxide (green), and cadmium sulfide (yellow). Pigments mix to form suspensions rather than solutions. Their chemical form is not changed in a mixture, so they still absorb and reflect the same wavelengths. Results in color You can now begin to understand the colors that you see in the photo at the beginning of this chapter. The plants on the hillside look green because of the chlorophy", "ll in them. One type of chlorophyll absorbs red light and the other absorbs blue light, but they both reflect green light. The energy in the red and blue light that is absorbed is used by the plants for photosynthesis, which is the process by which green plants make food. In the same photo, the sky is bluish. Violet and blue light are scattered (repeatedly reflected) much more by molecules in the air than are other wavelengths of light. Green and red light are not scattered much by the air, which is why the Sun looks yellow or orange, as shown in Figure 16-15. However, violet and blue light from the Sun are scattered in all directions, illuminating the sky in a bluish hue. Chemistry Connection Chemistry Connection Biology Connection Biology Connection \u25a0 Figure 16-15 The Sun can appear to be a shade of yellow or orange because of the scattering of violet and blue light. 442 (t)Laura Sifferlin, (b)file photo Chapter 16 Fundamentals of Light 439-447 CH16-S2-845813 6/10/04 5:17 PM Page 443 a Direction of string displacement b Direction of string displacement Direction of wave propagation Direction of wave propagation Polarization of Light Have you ever looked at light reflected off a road through polarizing sunglasses? As you rotate the glasses, the road first appears to be dark, then light, and then dark again. Light from a lamp, however, changes very little as the glasses are rotated. Why is there a difference? Normal lamplight is not polarized. However, the light that is coming from the road is reflected and has become polarized. Polarization is the production of light in a single plane of oscillation. Polarization by filtering Polarization can be understood by considering a rope model of light waves, as shown in Figure 16-16. The transverse mechanical waves in the rope represent transverse light waves. The slot represents what is referred to as the polarizing axis of a polarizing medium. When the rope waves are parallel to the slot, they pass through. When they are perpendicular to the slot, the waves are blocked. Polarizing media contain long molecules in which electrons can oscillate, or move back and forth, all in the same direction. As light travels past the molecules, the electrons can absorb light waves that oscillate in the same direction as the electrons. This process allows light waves vibrating in one direction to pass through, while the waves vibrating in the other direction are absorbed. The direction of a polarizing medium perpendicular to the long molecules is called the polarizing axis. Only waves oscillating parallel to that axis can pass through. Ordinary light actually contains waves vibrating in every direction perpendicular to its direction of travel. If a polarizing medium is placed in a beam of ordinary light, only the components of the waves in the same direction as the polarizing axis can pass through. On average, half of the total light amplitude passes through, thereby reducing the intensity of the light by half. The polarizing medium produces light that is polarized. Such a medium is called a polarizing filter. Polarization by reflection When you look through a polarizing filter at the light reflected by a sheet of glass and rotate the filter, you will see the light brighten and dim. The light is partially polarized along the plane of the glass when it is reflected. That is, the reflected ray contains a great deal of light vibrating parallel to the surface of the glass. The polarization of light reflected by roads is the reason why polarizing sunglasses reduce glare. The fact that the intensity of light reflected off a road varies as polarizing sunglasses are rotated suggests that the reflected light is partially polarized. Photographers can use polarizing filters over camera lenses to block reflected light, as shown in Figure 16-17. \u25a0 Figure 16-16 In this rope model of light, light is a single wave oriented in relation to the vertical plane and thus passes through a vertical polarizer (a). It cannot pass through a horizontal polarizer (b). a b \u25a0 Figure 16-17 This photo of a music store, taken without a polarizing filter, contains the glare of light off of the surface of the window (a). This photo of the same scene was taken with a polarizing filter (b). Section 16.2 The Wave Nature of Light 443 439-447 CH16-S2-845813 6/4/04 10:32 AM Page 444 a b \u25a0 Figure 16-18 When two polarizing filters are arranged with their polarizing axes in parallel, a maximum amount of light passes through (a). When two polarizing filters are arranged with perpendicular axes, no light passes through (b). Polarization analysis Suppose that you produce polarized light with a polarizing filter. What would happen if you place a second polarizing filter in the path of the polarized light? If the polarizing axis of the second filter is parallel to that of the first, the light will pass through, as shown in Figure 16-18", "a. If the polarizing axis of the second filter is perpendicular to that of the first, no light will pass through, as shown in Figure 16-18b. The law that explains the reduction of light intensity as it passes through a second polarizing filter is called Malus\u2019s law. If the light intensity after the first polarizing filter is I1, then a second polarizing filter, with its polarizing axis at an angle,, relative to the polarizing axis of the first, will result in a light intensity, I2, that is equal to or less than I1. I1cos2 Malus\u2019s Law I2 The intensity of light coming out of a second polarizing filter is equal to the intensity of polarized light coming out of a first polarizing filter multiplied by the cosine, squared, of the angle between the polarizing axes of the two filters. Using Malus\u2019s law, you can compare the light intensity coming out of the second polarizing filter to the light intensity coming out of the first polarizing filter, and thereby determine the orientation of the polarizing axis of the first filter relative to the second filter. A polarizing filter that uses Malus\u2019s law to accomplish this is called an analyzer. Analyzers can be used to determine the polarization of light coming from any source. You place an analyzer filter between the two cross-polarized filters, such that its polarizing axis is not parallel to either of the two filters, as shown in the figure to the right. 1. You observe that some light passes through filter 2, though no light passed through filter 2 previous to inserting the analyzer filter. Why does this happen? 2. The analyzer filter is placed at an angle of relative to the polarizing axes of filter 1. Derive an equation for the intensity of light coming out of filter 2 compared to the intensity of light coming out of filter 1. Filter 1 Polarized light Analyzer Polarizing axes Filter 2 90\u00b0 444 Laura Sifferlin Chapter 16 Fundamentals of Light 439-447 CH16-S2-845813 6/4/04 10:35 AM Page 445 The Speed of a Light Wave As you learned in Chapter 14, the wavelength,, of a wave is a function of its speed in the medium in which it is traveling and its constant frequency, f. Because light has wave properties, the same mathematical models used to describe waves in general can be used to describe light. For light of a given frequency traveling through a vacuum, wavelength is a function of the speed of light, c, which can be written as c/f. The development 0 of the laser in the 1960s provided new methods of measuring the speed of light. The frequency of light can be counted with extreme precision using lasers and the time standard provided by atomic clocks. Measurements of wavelengths of light, however, are much less precise. Different colors of light have different frequencies and wavelengths, but in a vacuum, they all travel at c. Because all wavelengths of light travel at the same speed in a vacuum, when you know the frequency of a light wave in a vacuum, you can calculate its wavelength, and vice versa. Thus, using precise measurements of light frequency and light speed, you can calculate a precise value of light wavelength. Relative motion and light What happens if a source of light is traveling toward you or you are moving toward the light source? You learned in Chapter 15 that the frequency of a sound heard by the listener changes if either the source or the listener of the sound is moving. The same is true for light. However, when you consider the velocities of a source of sound and the observer, you are really considering each one\u2019s velocity relative to the medium through which the sound travels. Because light waves are not vibrations of the particles of a mechanical medium, unlike sound waves, the Doppler effect of light can involve only the velocities of the source and the observer relative to each other. The magnitude of the difference between the velocities of the source and observer is called the relative speed. Remember that the only factors in the Doppler effect are the velocity components along the axis between the source and observer, as shown in Figure 16-19. a b lig h t vs vs, axis Source \u25a0 Figure 16-19 The observer and the light source have different velocities (a). The magnitude of the vector subtraction of the velocity components along the axis between the source of light and the observer of the light is referred to as the relative speed along the axis between the source and observer, v (b). (Illustration not to scale) lig h t Source v vo, axis vo Observer Observer v \uf8efvs, axis vo, axis \uf8ef Section 16.2 The Wave Nature of Light 445 439-447 CH16-S2-845813 6/10/04 8:43 PM Page 446 The Doppler effect To study the Do", "ppler effect for light, the problem can be simplified by considering axial relative speeds that are much less than the speed of light (v c). This simplification is used to develop the equation for the obseved light frequency, fobs, which is the frequency of light as seen by an observer. Observed Light Frequency fobs ) f(1 v c The observed frequency of light from a source is equal to the actual frequency of the light generated by the source, times the quantity 1 plus the relative speed along the axis between the source and the observer if they are moving toward each other, or 1 minus the relative speed if they are moving away from each other. Because most observations of the Doppler effect for light have been made in the context of astronomy, the equation for the Doppler effect for light generally is written in terms of wavelength rather than frequency. Using the relationship c/f and the v c simplification, the following equation can be used for the Doppler shift,, which is the difference between the observed wavelength of light and the actual wavelength. Doppler Shift ( obs ) v c The difference between the observed wavelength of light and the actual wavelength of light generated by a source is equal to the actual wavelength of light generated by the source, times the relative speed of the source and observer, divided by the speed of light. This quantity is positive if they are moving away from each other or negative if they are moving toward each other. A positive change in wavelength means that the light is red-shifted. This occurs when the relative velocity of the source is in a direction away from the observer. A negative change in wavelength means that the light is blueshifted. This occurs when the relative velocity of the source is in a direction toward the observer. When the wavelength is red-shifted (lengthens), the observed frequency is lower as a result of the inverse relationship between the two variables, because the speed of light remains constant. When the wavelength is blue-shifted, the observed frequency is higher. Researchers can determine how astronomical objects, such as galaxies, are moving relative to Earth by observing the Doppler shift of light. This is done by observing the spectrum of light coming from stars in the galaxy using a spectrometer, as shown in Figure 16-20. Elements that are present in the stars of galaxies emit specific wavelengths in the lab. A spectrometer is able to measure the Doppler shift of these wavelengths. \u25a0 Figure 16-20 Three emission lines of hydrogen are visibly redshifted in the spectrum of quasar 3C 273, as indicated by the taglines outside the spectra. Their wavelengths are shifted approximately 16% of what they are in a laboratory setting. 3C 273 Comparison spectrum 446 Maarten Schmidt Chapter 16 Fundamentals of Light 439-447 CH16-S2-845813 6/4/04 10:36 AM Page 447 14. What is the frequency of oxygen\u2019s spectral line if its wavelength is 513 nm? 15. A hydrogen atom in a galaxy moving with a speed of 6.55106 m/s away from Earth emits light with a frequency of 6.161014 Hz. What frequency of light from that hydrogen atom would be observed by an astronomer on Earth? 16. A hydrogen atom in a galaxy moving with a speed of 6.551016 m/s away from Earth emits light with a wavelength of 4.86107 m. What wavelength would be observed on Earth from that hydrogen atom? 17. An astronomer is looking at the spectrum of a galaxy and finds that it has an oxygen spectral line of 525 nm, while the laboratory value is measured at 513 nm. Calculate how fast the galaxy would be moving relative to Earth. Explain whether the galaxy is moving toward or away from Earth and how you know. In 1929, Edwin Hubble suggested that the universe is expanding. Hubble reached his conclusion of the expanding universe by analyzing emission spectra from many galaxies. Hubble noticed that the spectral lines of familiar elements were at longer wavelengths than expected. The lines were shifted toward the red end of the spectrum. No matter what area of the skies he observed, the galaxies were sending red-shifted light to Earth. What do you think caused the spectral lines to be red-shifted? Hubble concluded that all galaxies are moving away from Earth. You have learned that some characteristics of light can be explained with a simple ray model of light, whereas others require a wave model of light. In Chapters 17 and 18, you will use both of these models to study how light interacts with mirrors and lenses. In Chapter 19, you will learn about other aspects of light that can be understood only through the use of the wave model of light. 16.2 Section Review Astronomy Connection Astronomy Connection 18. Addition of Light Colors What color of light must be combined with blue light to obtain white light? 19. Combination of Pigments What primary pigment colors must be mixed to produce red? Explain how red results using color", " subtraction for pigment colors. 20. Light and Pigment Interaction What color will a yellow banana appear to be when illuminated by each of the following? a. white light b. green and red light c. blue light 21. Wave Properties of Light The speed of red light is slower in air and water than in a vacuum. The frequency, however, does not change when red light enters water. Does the wavelength change? If so, how? 22. Polarization Describe a simple experiment that you could do to determine whether sunglasses in a store are polarizing. 23. Critical Thinking Astronomers have determined that Andromeda, a neighboring galaxy to our own galaxy, the Milky Way, is moving toward the Milky Way. Explain how they determined this. Can you think of a possible reason why Andromeda is moving toward our galaxy? physicspp.com/self_check_quiz Section 16.2 The Wave Nature of Light 447 448-449 CH16-LAB-845813 6/4/04 11:01 AM Page 448 Alternate CBL instructions can be found on the Web site. physicspp.com Polarization of Light A light source that produces transverse light waves that are all in the same fixed plane is said to be polarized in that plane. A polarizing filter can be used to find light sources that produce polarized light. Some media can rotate the plane of polarization of light as it transmits the light. Such media are said to be optically active. In this activity, you will investigate these concepts of polarized light. QUESTION What types of luminous and illuminated light sources produce polarized light? Objectives Materials \u25a0 Experiment with various sources of light and polarizing filters. \u25a0 Describe the results of your experiment. \u25a0 Recognize possible uses of polarizing filters in everyday life. Safety Precautions two polarizing filter sheets incandescent light source fluorescent light source pieces of white and black paper calculator with a liquid crystal display clear, plastic protractor mirror \u25a0 Minimize the length of time you look directly at bright light sources. \u25a0 Do not do this lab with laser light sources. \u25a0 Do not look at the Sun, even if you are using polarizing filters. \u25a0 Light sources can get hot and burn skin. Procedure 1. Take a polarizing filter and look at an incandescent light source. Rotate the filter. Write your observations in the data table. 2. Use a polarizing filter to look at a fluorescent light source. Rotate the filter. Write your observations in the data table. 3. Use a polarizing filter to observe light reflected off the surface of a mirror at approximately a 45\u00b0 angle. Rotate the filter. Record your observations in the data table. 448 Horizons Companies 448-449 CH16-LAB-845813 6/10/04 8:44 PM Page 449 Data Table Light Source Observations 1 2 3 4 5 6 7 8 4. Use a polarizing filter to observe light reflected off a white piece of paper at approximately a 45\u00ba angle. Rotate the filter. Record your observations in the data table. 5. Use a polarizing filter to observe light reflected off a piece of black paper at approximately a 45\u00b0 angle. Rotate the filter. Record your observations in the data table. 6. Use a polarizing filter to observe a liquid crystal display on a calculator. Rotate the filter. Write your observations in the data table. 7. Place one polarizing filter on top of the other filter. Look at an incandescent light source through this set of the filters. Rotate one of the filters with respect to the other. Make a complete rotation. Record your observations in the data table. 8. Place a clear, plastic protractor between the two polarizing filters. Look at an incandescent light source with this. Do a complete rotation of one of the filters. Position the two filters the same way that produced no light in step 7. Record your observations in the data table. Analyze 1. Interpret Data Does incandescent light produce polarized light? How do you know? 2. Interpret Data Does fluorescent light produce polarized light? How do you know? 3. Interpret Data Does reflection from a mirrored surface produce polarized light? How do you know? 4. Compare and Contrast How does reflected light from white paper compare to reflected light from black paper in terms of polarized light? Why are they different? 5. Interpret Data Is the light from liquid crystal displays polarized? How do you know? Conclude and Apply 1. Analyze and Conclude How can two polarizing filters be used to prevent any light from passing through them? 2. Analyze and Conclude Why can the clear, plastic protractor between the polarizing filters be seen even though nothing else can be seen through the polarizing filters? 3. Draw Conclusions In general, what types of situations produce polarized light? Going Further 1. On a sunny day, look at the polarization of blue sky near and far from the Sun using a polarizing filter. CAUTION: Do not look", " directly at the Sun. What characteristics of polarized light do you observe? 2. Is reflected light from clouds polarized? Make an observation to confirm your answer. Real-World Physics 1. Why are high quality sunglasses made with polarizing lenses? 2. Why are polarizing sunglasses a better option than tinted sunglasses when driving a car? To find out more about light, visit the Web site: physicspp.com 449 450 CH16-FEATURE-845813 6/4/04 11:12 AM Page 450 Advances In Lighting Advances In Lighting The use of a gas-discharge lamp depends upon the type of gas: neon for advertising, xenon for searchlights and camera flashes, and sodium vapor for streetlights. Each type of gas produces a different color, but the construction of each lamp is very similar. History has recorded the use of oil, candles, and gas to provide illumination in the dark hours of the night. However, there has always been inherent danger with the use of open flames to provide light. The invention of electric lighting in the nineteenth century provided brighter light and improved safety to the public. The original form of electric light, which is still in common use, is the incandescent bulb. A tungsten filament is heated by electricity until it glows white. The tungsten does not burn, but it vaporizes, which eventually breaks the filament. Because the light is a result of heating the tungsten, this is not very efficient. Recent pursuits in electric lighting have produced longer-lasting, lower-heat light sources. Clockwise from the upper left, the photos show LEDs, a fluorescent light, a halogen light, and gas-discharge lamps in the form of neon lights. Fluorescent Lamps The glow produced by mercury vapor is almost invisible because most of its spectrum is in the ultraviolet region, which is not visible. A fluorescent lamp is made by painting the inside of a mercury-discharge lamp with phosphor, a chemical that glows brightly when ultraviolet light strikes it. Fluorescent lights can be made in any color by changing the mixture of red, green, and blue phosphors. They have a long life and are economical to use, because they produce little heat and a great deal of light. Quartz-Halogen Lamps To prevent a filament from breaking, the bulb can be made very small and filled with bromine or iodine gas. Tungsten ions from the filament combine with the gas molecules in the cooler parts of the lamp to make a compound, which circulates through the lamp and recombines with the filament. The light is very white and bright, but it also is very hot. An ordinary glass bulb would melt, so fused quartz, which has a very high melting point, is used. Gas-Discharge Lamps This type of lamp is made of a glass tube with a wire electrode sealed into each end. All of the air is extracted and replaced by a very small amount of a specially chosen gas. A high voltage is applied across the electrodes. The electricity ionizes, or strips, some electrons from the gas atoms. An ionized gas is a good conductor, so electric current flows through it, causing the gas to glow brightly. Light-Emitting Diodes The light of the future may be the white lightemitting diode, or LED. The LED produces white light by illuminating a tiny phosphor screen inside the LED with blue light. LEDs are bright enough to read by and produce almost no heat as they operate. They are so efficient that a car battery could power the lamps in a home for days without being recharged. Going Further 1. Observe Novelty stores sell many devices that use lights. Examine some of them to see what types of lamp technology are used. 2. Research Find out about the inner construction, characteristic color, and typical uses of a few types of gasdischarge lamps. 450 (br)Burazin/Masterfile, (bl)David Duran/Fundamental Photographs, (tr)Getty Images, (tl)Jerry Driendl/Getty Images Technology and Society 451-455 CH16-SG CA STP-845813 7/14/04 1:48 PM Page 451 16.1 Illumination Vocabulary \u2022 ray model of light (p. 432) \u2022 luminous source (p. 432) \u2022 illuminated source (p. 432) \u2022 opaque (p. 433) \u2022 transparent (p. 433) \u2022 translucent (p. 433) \u2022 luminous flux (p. 433) \u2022 illuminance (p. 433) Key Concepts \u2022 Light travels in a straight line through any uniform medium. \u2022 Materials can be characterized as being transparent, translucent, or opaque, depending on the amount of light that they reflect, transmit, or absorb. \u2022 The luminous flux of a light source is the rate at which light is emitted. It is measured in lumens (lm). \u2022 Illuminance is the luminous flux per unit area. It is measured in", " lux (lx), or lumens per square meter (lm/m2). \u2022 For a point source, illuminance follows an inverse-square relationship with distance and a direct relationship with luminous flux. P E 4 r2 \u2022 In a vacuum, light has a constant speed of c 3.00108 m/s. 16.2 The Wave Nature of Light Vocabulary \u2022 diffraction (p. 439) \u2022 primary color (p. 440) \u2022 secondary color (p. 440) \u2022 complementary colors (p. 441) \u2022 primary pigment (p. 441) \u2022 secondary pigment (p. 441) \u2022 polarization (p. 443) \u2022 Malus\u2019s law (p. 444) Key Concepts \u2022 Light can have wavelengths between 400 and 700 nm. \u2022 White light is a combination of the spectrum of colors, each color having a different wavelength. \u2022 Combining the primary colors, red, blue, and green, forms white light. Combinations of two primary colors form the secondary colors, yellow, cyan, and magenta. \u2022 The primary pigments, cyan, magenta, and yellow, are used in combinations of two to produce the secondary pigments, red, blue, and green. \u2022 Polarized light consists of waves oscillating in the same plane. \u2022 When two polarizing filters are used to polarize light, the intensity of the light coming out of the last filter is dependent on the angle between the polarizing axes of the two filters. I2 I1cos2 \u2022 Light waves traveling through a vacuum can be characterized in terms of frequency, wavelength, and the speed of light. 0 c f \u2022 Light waves are Doppler shifted based upon the relative speed along the axis of the observer and the source of light. fobs ) f(1 v c ( obs ) v c physicspp.com/vocabulary_puzzlemaker 451 451-455 CH16-SG CA STP-845813 6/4/04 11:17 AM Page 452 Concept Mapping 24. Complete the following concept map using the following terms: wave, c, Doppler effect, polarization. Light Models Applying Concepts 40. A point source of light is 2.0 m from screen A and 4.0 m from screen B, as shown in Figure 16-21. How does the illuminance at screen B compare with the illuminance at screen A? Screen A Screen B ray straight path Source diffraction 2 m 4 m 0 c/f Mastering Concepts 25. Sound does not travel through a vacuum. How do we know that light does? (16.1) 26. Distinguish between a luminous source and an illuminated source. (16.1) 27. Look carefully at an ordinary, frosted, incandescent bulb. Is it a luminous or an illuminated source? (16.1) 28. Explain how you can see ordinary, nonluminous classroom objects. (16.1) 29. Distinguish among transparent, translucent, and opaque objects. (16.1) 30. To what is the illumination of a surface by a light source directly proportional? To what is it inversely proportional? (16.1) 31. What did Galileo assume about the speed of light? (16.1) 32. Why is the diffraction of sound waves more familiar in everyday experience than is the diffraction of light waves? (16.2) 33. What color of light has the shortest wavelength? (16.2) 34. What is the range of the wavelengths of light, from shortest to longest? (16.2) 35. Of what colors does white light consist? (16.2) 36. Why does an object appear to be black? (16.2) 37. Can longitudinal waves be polarized? Explain. (16.2) 38. If a distant galaxy were to emit a spectral line in the green region of the light spectrum, would the observed wavelength on Earth shift toward red light or toward blue light? Explain. (16.2) \u25a0 Figure 16-21 41. Reading Lamp You have a small reading lamp that is 35 cm from the pages of a book. You decide to double the distance. a. Is the illuminance at the book the same? b. If not, how much more or less is it? 42. Why are the insides of binoculars and cameras painted black? 43. Eye Sensitivity The eye is most sensitive to yellowgreen light. Its sensitivity to red and blue light is less than 10 percent as great. Based on this knowledge, what color would you recommend that fire trucks and ambulances be painted? Why? 44. Streetlight Color Some very efficient streetlights contain sodium vapor under high pressure. They produce light that is mainly yellow with some red. Should a community that has these lights buy dark blue police cars? Why or why not? Refer to Figure 16-22 for problems 45 and 46. 45. What happens to", " the illuminance at a book as the lamp is moved farther away from the book? \u25a0 Figure 16-22 39. What happens to the wavelength of light as the 46. What happens to the luminous intensity of the frequency increases? (16.2) lamp as it is moved farther away from the book? 452 Chapter 16 Fundamentals of Light For more problems, go to Additional Problems, Appendix B. 451-455 CH16-SG CA STP-845813 6/4/04 6:51 AM Page 453 47. Polarized Pictures Photographers often put 55. A three-way bulb uses 50, 100, or 150 W of polarizing filters over camera lenses to make clouds in the sky more visible. The clouds remain white, while the sky looks darker. Explain this based on your knowledge of polarized light. 48. An apple is red because it reflects red light and absorbs blue and green light. a. Why does red cellophane look red in reflected light? b. Why does red cellophane make a white lightbulb look red when you hold the cellophane between your eye and the lightbulb? c. What happens to the blue and green light? 49. You put a piece of red cellophane over one flashlight and a piece of green cellophane over another. You shine the light beams on a white wall. What color will you see where the two flashlight beams overlap? 50. You now put both the red and green cellophane pieces over one of the flashlights in Problem 49. If you shine the flashlight beam on a white wall, what color will you see? Explain. 51. If you have yellow, cyan, and magenta pigments, how can you make a blue pigment? Explain. 52. Traffic Violation Suppose that you are a traffic officer and you stop a driver for going through a red light. Further suppose that the driver draws a picture for you (Figure 16-23) and explains that the light looked green because of the Doppler effect when he went through it. Explain to him using the Doppler shift equation just how fast he would have had to be going for the red light ( 645 nm), to appear green ( 545 nm). Hint: For the purpose of this problem, assume that the Doppler shift equation is valid at this speed. Red light Looks green vcar electrical power to deliver 665, 1620, or 2285 lm in its three settings. The bulb is placed 80 cm above a sheet of paper. If an illumination of at least 175 lx is needed on the paper, what is the minimum setting that should be used? 56. Earth\u2019s Speed Ole Roemer found that the average increased delay in the disappearance of Io from one orbit around Jupiter to the next is 13 s. a. How far does light travel in 13 s? b. Each orbit of Io takes 42.5 h. Earth travels the distance calculated in part a in 42.5 h. Find the speed of Earth in km/s. c. Check to make sure that your answer for part b is reasonable. Calculate Earth\u2019s speed in orbit using the orbital radius, 1.5108 km, and the period, 1.0 yr. 57. A student wants to compare the luminous flux of a lightbulb with that of a 1750-lm lamp. The lightbulb and the lamp illuminate a sheet of paper equally. The 1750-lm lamp is 1.25 m away from the sheet of paper; the lightbulb is 1.08 m away. What is the lightbulb\u2019s luminous flux? 58. Suppose that you wanted to measure the speed of light by putting a mirror on a distant mountain, setting off a camera flash, and measuring the time it takes the flash to reflect off the mirror and return to you, as shown in Figure 16-24. Without instruments, a person can detect a time interval of about 0.10 s. How many kilometers away would the mirror have to be? Compare this distance with that of some known distances. Mirror You d \u25a0 Figure 16-23 Mastering Problems 16.1 Illumination \u25a0 Figure 16-24 16.2 The Wave Nature of Light 59. Convert 700 nm, the wavelength of red light, to meters. 53. Find the illumination 4.0 m below a 405-lm lamp. 54. Light takes 1.28 s to travel from the Moon to Earth. What is the distance between them? 60. Galactic Motion How fast is a galaxy moving relative to Earth if a hydrogen spectral line of 486 nm is red-shifted to 491 nm? physicspp.com/chapter_test Chapter 16 Assessment 453 451-455 CH16-SG CA STP-845813 6/4/04 6:46 AM Page 454 61. Suppose that you are facing due east at sunrise. Sunlight is reflected off the surface of", " a lake, as shown in Figure 16-25. Is the reflected light polarized? If so, in what direction? Thinking Critically 70. Research Why did Galileo\u2019s method for measuring the speed of light not work? 71. Make and Use Graphs A 110-cd light source is 1.0 m from a screen. Determine the illumination on the screen originally and for every meter of increasing distance up to 7.0 m. Graph the data. a. What is the shape of the graph? b. What is the relationship between illuminance and distance shown by the graph? 72. Analyze and Conclude If you were to drive at sunset in a city filled with buildings that have glass-covered walls, the setting Sun reflected off the building\u2019s walls might temporarily blind you. Would polarizing glasses solve this problem? Writing in Physics 73. Write an essay describing the history of human understanding of the speed of light. Include significant individuals and the contribution that each individual made. 74. Look up information on the SI unit candela, cd, and explain in your own words the standard that is used to set the value of 1 cd. Cumulative Review 75. A 2.0-kg object is attached to a 1.5-m long string and swung in a vertical circle at a constant speed of 12 m/s. (Chapter 7) a. What is the tension in the string when the object is at the bottom of its path? b. What is the tension in the string when the object \u25a0 Figure 16-25 62. Polarizing Sunglasses In which direction should the transmission axis of polarizing sunglasses be oriented to cut the glare from the surface of a road: vertically or horizontally? Explain. 63. Galactic Motion A hydrogen spectral line that is known to be 434 nm is red-shifted by 6.50 percent in light coming from a distant galaxy. How fast is the galaxy moving away from Earth? 64. For any spectral line, what would be an unrealistic value of the apparent wavelength for a galaxy moving away from Earth. Why? Mixed Review 65. Streetlight Illumination A streetlight contains two identical bulbs that are 3.3 m above the ground. If the community wants to save electrical energy by removing one bulb, how far from the ground should the streetlight be positioned to have the same illumination on the ground under the lamp? 66. An octave in music is a doubling of frequency. is at the top of its path? Compare the number of octaves that correspond to the human hearing range to the number of octaves in the human vision range. 67. A 10.0-cd point-source lamp and a 60.0-cd pointsource lamp cast equal intensities on a wall. If the 10.0-cd lamp is 6.0 m from the wall, how far from the wall is the 60.0-cd lamp? 68. Thunder and Lightning Explain why it takes 5 s to hear thunder when lightning is 1.6 km away. 69. Solar Rotation Because the Sun rotates on its axis, one edge of the Sun moves toward Earth and the other moves away. The Sun rotates approximately once every 25 days, and the diameter of the Sun is 1.4109 m. Hydrogen on the Sun emits light of frequency 6.161014 Hz from the two sides of the Sun. What changes in wavelength are observed? 76. A space probe with a mass of 7.600103 kg is traveling through space at 125 m/s. Mission control decides that a course correction of 30.0\u00b0 is needed and instructs the probe to fire rockets perpendicular to its present direction of motion. If the gas expelled by the rockets has a speed of 3.200 km/s, what mass of gas should be released? (Chapter 9) 77. When a 60.0-cm-long guitar string is plucked in the middle, it plays a note of frequency 440 Hz. What is the speed of the waves on the string? (Chapter 14) 78. What is the wavelength of a sound wave with a frequency of 17,000 Hz in water at 25\u00b0C? (Chapter 15) 454 Chapter 16 Fundamentals of Light For more problems, go to Additional Problems, Appendix B. 451-455 CH16-SG CA STP-845813 6/4/04 11:19 AM Page 455 Multiple Choice 1. In 1987, a supernova was observed in a neighboring galaxy. Scientists believed the galaxy was 1.661021 m away. How many years prior to the observation did the supernova explosion actually occur? 5.53103 yr 1.75105 yr 5.531012 yr 1.741020 yr 2. A galaxy is moving away at 5.8106 m/s. Its light appears to observers to have a frequency of 5.61014 Hz. What is the emitted frequency of the light? 1.11013 Hz 5.51014 Hz 5.71014 Hz 6.", "21014 Hz 6. The illuminance due to a 60.0-W lightbulb at 3.0 m is 9.35 lx. What is the total luminous flux of the bulb? 8.3102 lm 7.4101 lm 1.2102 lm 1.1103 lm 7. Light from the Sun takes about 8.0 min to reach Earth. How far away is the Sun? 2.4109 m 1.41010 m 1.4108 km 2.4109 km 8. What is the frequency of 404 nm of light in a vacuum? 2.48103 Hz 7.43105 Hz 2.48106 Hz 7.431014 Hz 3. Which of the following light color combinations is incorrect? Extended Answer Red plus green produces yellow. Red plus yellow produces magenta. Blue plus green produces cyan. Blue plus yellow produces white. 4. The illuminance of direct sunlight on Earth is about 1105 lx. A light on a stage has an intensity in a certain direction of 5106 cd. At what distance from the stage does a member of the audience experience an illuminance equal to that of sunlight? 1.4101 m 7 m 10 m 5101 m 5. What is meant by the phrase color by subtraction of light? Adding green, red, and blue light produces white light. Exciting phosphors with electrons in a television produces color. Paint color is changed by subtracting certain colors, such as producing blue paint from green by removing yellow. The color that an object appears to be is a result of the material absorbing specific light wavelengths and reflecting the rest. 9. A celestial object is known to contain an element that emits light at a wavelength of 525 nm. The observed spectral line for this element is at 473 nm. Is the object approaching or receding, and at what speed? 10. Nonpolarized light of intensity Io is incident on a polarizing filter, and the emerging light strikes a second polarizing filter, as shown in the figure. What is the light intensity emerging from the second polarizing filter? Io Non-polarized light 45\u00b0 Polarizing filters Ask Questions When you have a question about what will be on a test, the way a test is scored, the time limits placed on each section, or anything else, ask the instructor or the person giving the test. physicspp.com/standardized_test Chapter 16 Standardized Test Practice 455 456-463 CH17-S1-845813 6/4/04 11:29 AM Page 456 What You\u2019ll Learn \u2022 You will learn how light reflects off different surfaces. \u2022 You will learn about the different types of mirrors and their uses. \u2022 You will use ray tracing and mathematical models to describe images formed by mirrors. Why It\u2019s Important How light reflects off a surface into your eyes determines the reflection that you see. When you look down at the surface of a lake, you see an upright reflection of yourself. Mountain Scene When you look across a lake, you might see a scene like the one in this photo. The image of the trees and mountains in the lake appears to you to be upside-down. Think About This Why would the image you see of yourself in the lake be upright, while the image of the mountain is upside-down? physicspp.com 456 George Matchneer 456-463 CH17-S1-845813 6/4/04 11:28 AM Page 457 How is an image shown on a screen? Question What types of mirrors are able to reflect an image onto a screen? Procedure 1. Obtain an index card, a plane mirror, a concave mirror, a convex mirror, and a flashlight from your teacher. 6. Repeat steps 3\u20135 with the concave and convex mirror. 7. Perform step 4 for each mirror with a flashlight and observe the reflection on the index card. 2. Turn off the room lights and stand near the Analysis Which mirror(s) produced a real image? What are some things you notice about the image(s) you see? Critical Thinking Based upon your observation of the flashlight images, propose an explanation of how a real image is formed. window. 3. Hold the index card in one hand. Hold the flat, plane mirror in the other hand. 4. Reflect the light coming through the window onto the index card. CAUTION: Do not look directly at the Sun or at the reflection of the Sun in a mirror. Slowly move the index card closer to and then farther away from the mirror and try to make a clear image of objects outside the window. 5. If you can project a clear image, this is called a real image. If you only see a fuzzy light on the index card then no real image is formed. Record your observations. 17.1 Reflection from Plane Mirrors Undoubtedly, as long as there have been humans, they have seen their faces reflected in the quiet water", " of lakes and ponds. When you look at the surface of a body of water, however, you don\u2019t always see a clear reflection. Sometimes, the wind causes ripples in the water, and passing boats create waves. Disturbances on the surface of the water prevent the light from reflecting in a manner such that a clear reflection is visible. Almost 4000 years ago, Egyptians understood that reflection requires smooth surfaces. They used polished metal mirrors to view their images. Sharp, well-defined, reflected images were not possible until 1857, however, when Jean Foucault, a French scientist, developed a method of coating glass with silver. Modern mirrors are produced using ever-increasing precision. They are made with greater reflecting ability by the evaporation of aluminum or silver onto highly polished glass. The quality of reflecting surfaces is even more important in applications such as lasers and telescopes. More than ever before, clear reflections in modern, optical instruments require smooth surfaces. Objectives \u2022 Explain the law of reflection. \u2022 Distinguish between specular and diffuse reflection. \u2022 Locate the images formed by plane mirrors. Vocabulary specular reflection diffuse reflection plane mirror object image virtual image Section 17.1 Reflection from Plane Mirrors 457 Horizons Companies 456-463 CH17-S1-845813 3/18/04 7:47 PM Page 458 Reflected light Incident light Normal r i Surface r i Plane of travel \u25a0 Figure 17-1 The incident ray and the reflected ray are in the same plane of travel. \u2022 Light rays and wave fronts are red. \u2022 Mirrors are light blue. The Law of Reflection What happens to the light that is striking this book? When you hold the book up to the light, you will see that no light passes through it. Recall from Chapter 16 that an object like this is called opaque. Part of the light is absorbed, and part is reflected. The absorbed light spreads out as thermal energy. The behavior of the reflected light depends on the type of surface and the angle at which the light strikes the surface. Recall from Chapter 14 that when a wave traveling in two dimensions encounters a barrier, the angle of incidence is equal to the angle of reflection of the wave. This same two-dimensional reflection relationship applies to light waves. Consider what happens when you bounce-pass a basketball. The ball bounces in a straight line, as viewed from above, to the other player. Light reflects in the same way as a basketball. Figure 17-1 shows a ray of light striking a reflecting surface. The normal is an imaginary line that is perpendicular to a surface at the location where light strikes the surface. The reflected ray, the incident ray, and the normal to the surface always will be in the same plane. Although the light is traveling in three dimensions, the reflection of the light is planar (two-dimensional). The planar and angle relationships are known together as the law of reflection. Law of Reflection r i The angle that a reflected ray makes as measured from the normal to a reflective surface equals the angle that the incident ray makes as measured from the same normal. This law can be explained in terms of the wave model of light. Figure 17-2 shows a wave front of light approaching a reflective surface. As each point along the wave front reaches the surface, it reflects off at the same angle as the preceding point. Because all points are traveling at the same speed, they all travel the same total distance in the same time. Thus, the wave front as a whole leaves the surface at an angle equal to its incident angle. Note that the wavelength of the light does not affect this process. Red, green, and blue light all follow this law. \u25a0 Figure 17-2 A wave front of light approaches a reflective surface. Point P on the wave front strikes the surface first (a). Point Q strikes the surface after point P reflects at an angle equal to the incident angle (b). The process continues with all points reflecting off at angles equal to their incident angles, resulting in a reflected wave front (c). a b Normal c Normal Normal Wave front Wave front Q i P i r P i Q Wave front P Q i i r Reflective surface Reflective surface Reflective surface 458 Chapter 17 Reflection and Mirrors 456-463 CH17-S1-845813 6/4/04 11:32 AM Page 459 a c b d \u25a0 Figure 17-3 When a beam of light strikes a mirrored surface (a), the parallel rays in the beam reflect in parallel and maintain the light as a beam (b). When the light beam strikes a rough surface (c), the parallel rays that make up the beam are reflected from different microscopic surfaces, thereby diffusing the beam (d). Smooth and rough surfaces Consider the beam of light shown in Figure 17-3a. All of the rays in this beam of light reflect off the surface parallel to one another, as shown in Figure 17-3b. This occurs only if the reflecting surface is not rough on the scale of the", " wavelength of light. Such a surface is considered to be smooth relative to the light. A smooth surface, such as a mirror, causes specular reflection, in which parallel light rays are reflected in parallel. C17-06A-845813 What happens when light strikes a surface that appears to be smooth, but actually is rough on the scale of the wavelength of light, such as the page of this textbook or a white wall? Is light reflected? How could you demonstrate this? Figure 17-3c shows a beam of light reflecting off a sheet of paper, which has a rough surface. All of the light rays that make up the beam are parallel before striking the surface, but the reflected rays are not parallel, as shown in Figure 17-3d. This is diffuse reflection, the scattering of light off a rough surface. The law of reflection applies to both smooth and rough surfaces. For a rough surface, the angle that each incident ray makes with the normal equals the angle that its reflected ray makes with the normal. However, on a microscopic scale, the normals to the surface locations where the rays strike are not parallel. Thus, the reflected rays cannot be parallel. The rough surface prevents them from being parallel. In this case, a reflected beam cannot be seen because the reflected rays are scattered in different directions. With specular reflection, as with a mirror, you can see your face. But no matter how much light is reflected off a wall or a sheet of paper, you will never be able to use them as mirrors. Section 17.1 Reflection from Plane Mirrors 459 Henry Leap/James Lehman 456-463 CH17-S1-845813 6/4/04 11:35 AM Page 460 Changing the Angle of Incidence A light ray strikes a plane mirror at an angle of 52.0\u00b0 to the normal. The mirror then rotates 35.0\u00b0 around the point where the beam strikes the mirror so that the angle of incidence of the light ray decreases. The axis of rotation is perpendicular to the plane of the incident and the reflected rays. What is the angle of rotation of the reflected ray? 1 2 Analyze and Sketch the Problem \u2022 Sketch the situation before the rotation of the mirror. \u2022 Draw another sketch with the angle of rotation applied to the mirror. Known: i, initial mirror 52.0\u00b0 35.0\u00b0 Unknown:? r Solve for the Unknown For the angle of incidence to reduce, rotate clockwise. \u03b8 i, final i, initial \u03b8 \u2206\u03b8 52.0\u00b0 35.0\u00b0 17.0\u00b0 clockwise from the new normal Substitute \u03b8 mirror i, initial 52.0\u00b0, \u2206\u03b8 mirror i, initial 52.0\u00b0 r, initial 52.0\u00b0 Initial normal \u2206 mirror 35.0\u00b0 i, final r, final \u2206 mirror 35.0\u00b0 Apply the law of reflection. \u03b8 r, final \u03b8 17.0\u00b0 counterclockwise i, final from the new normal Substitute \u03b8 i, final 17.0\u00b0 Using the two sketches, determine the angle through which the reflected ray has rotated. 52.0\u00b0 35.0\u00b0 17.0\u00b0 70.0\u00b0 clockwise from the original angle \u2206\u03b8 r Math Handbook Operations with Significant Digits pages 835\u2013836 3 Evaluate the Answer \u2022 Is the magnitude realistic? Comparing the final sketch with the initial sketch shows that the angle the light ray makes with the normal decreases as the mirror rotates clockwise toward the light ray. It makes sense, then, that the reflected ray also rotates clockwise. 1. Explain why the reflection of light off ground glass changes from diffuse to specular if you 2. 3. spill water on it. If the angle of incidence of a ray of light is 42\u00b0, what is each of the following? a. the angle of reflection b. the angle the incident ray makes with the mirror c. the angle between the incident ray and the reflected ray If a light ray reflects off a plane mirror at an angle of 35\u00b0 to the normal, what was the angle of incidence of the ray? 4. Light from a laser strikes a plane mirror at an angle of 38\u00b0 to the normal. If the laser is moved so that the angle of incidence increases by 13\u00b0, what is the new angle of reflection? 5. Two plane mirrors are positioned at right angles to one another. A ray of light strikes one mirror at an angle of 30\u00b0 to the normal. It then reflects toward the second mirror. What is the angle of reflection of the light ray off the second mirror? 460 Chapter 17 Reflection and Mirrors 456-463 CH17-S1-845813 6/3/04 9:22 AM Page 461 Objects and Plane-Mirror Images If you looked at yourself in a mirror this morning you saw your reflection in a plane mirror. A plane mirror is a flat, smooth surface from which light", " is reflected by specular reflection. To understand reflection from a mirror, you must consider the object of the reflection and the type of image that is formed. In Chapter 16, the word object was used to refer to sources of light. In describing mirrors, the word object is used in the same way, but with a more specific application. An object is a source of light rays that are to be reflected by a mirrored surface. An object can be a luminous source, such as a lightbulb, or an illuminated source, such as a girl, as shown in Figure 17-4. Most objects that you will work with in this chapter are a source of light that diverges, or spreads out from the source. Consider a single point on the bird in Figure 17-5. Light reflects diffusely from the crest of the bird, the object point. What happens to the light? Some of the light travels from the bird to the mirror and reflects. What does the boy see? Some of the reflected light from the bird hits his eyes. Because his brain processes this information as if the light has traveled a straight path, it seems to the boy as if the light had followed the dashed lines. The light seems to have come from a point behind the mirror, the image point. The boy in Figure 17-5 sees rays of light that come from many points on the bird. The combination of the image points produced by reflected light rays forms the image of the bird. It is a virtual image, which is a type of image formed by diverging light rays. A virtual image is always on the opposite side of the mirror from the object. Images of real objects produced by plane mirrors are always virtual images. Properties of Plane-Mirror Images Looking at yourself in a mirror, you can see that your image appears to be the same distance behind the mirror as you are in front of the mirror. How could you test this? Place a ruler between you and the mirror. Where does the image touch the ruler? You also see that your image is oriented as you are, and it matches your size. This is where the expression mirror image originates. If you move toward the mirror, your image moves toward the mirror. If you move away, your image also moves away. Object point Diverging rays Image point \u25a0 Figure 17-4 The lightbulb is a luminous source that produces diverging light by shining in all directions. The girl is an illuminated source that produces diverging light by the diffused reflection from her body of light that comes from the lightbulb. Diverging rays Mirror \u25a0 Figure 17-5 The reflected rays that enter the eye appear to originate at a point behind the mirror. Section 17.1 Reflection from Plane Mirrors 461 Laura Sifferlin 456-463 CH17-S1-845813 6/4/04 11:36 AM Page 462 O ho do Ray 1 Ray 2 i r P 1 P 2 di I hi \u25a0 Figure 17-6 Light rays (two are shown) leave a point on the object. Some strike the mirror and are reflected into the eye. Sight lines, drawn as dashed lines, extend from the location on the mirror where the rays are reflected back to where they converge. The image is located where the sight lines converge: di do. Virtual Image Position Suppose you are looking at your image in a plane mirror. Can you measure the location of the image? 1. Obtain a camera from your teacher with a focusing ring that has distances marked on it. 2. Stand 1.0 m from a mirror and focus on the edge of the mirror. Check the reading on the focusing ring. It should be 1.0 m. 3. Measure the position of the image by focusing the camera on your image. Check the reading on the focusing ring. Analyze and Conclude 4. How far is the image behind the mirror? 5. Why is the camera able to focus on a virtual image that is behind the mirror even though there is no real object behind the mirror? Image position and height The geometric model of Figure 17-6 demonstrates why the distances are the same. Two rays from point O at the tip of the candle strike the mirror at point P1 and point P2, respectively. Both rays are reflected according to the law of reflection. The reflected rays are extended behind the mirror as sight lines, converging at point I, which is the image of point O. Ray 1, which strikes the mirror at an angle of incidence of 0\u00b0, is reflected back on itself, so the sight line is at 90\u00b0 to the mirror, just as ray 1. Ray 2 is also reflected at an angle equal to the angle of incidence, so the sight line is at the same angle to the mirror as ray 2. \uf8e7 This geometric model reveals that line segments O are corresponding sides of two congruent triangles, OP1P2 and IP1P2. The position of the object with respect to the mirror, or the object position, do,", " has \uf8e7 \uf8e7 \uf8e7 a length equal to the length of O. The apparent position of the image P 1 with respect to the mirror, or the image position, di, has a length equal to \uf8e7 \uf8e7 \uf8e7 the length of I. Using the convention that image position is negative to P 1 indicate that the image is virtual, the following is true. \uf8e7 \uf8e7 \uf8e7 P and I 1 \uf8e7 \uf8e7 P 1 Plane-Mirror Image Position di With a plane mirror, the image position is equal to the negative of the object position. The negative sign indicates that the image is virtual. do You can draw rays from the object to the mirror to determine the size of the image. The sight lines of two rays originating from the bottom of the candle in Figure 17-6 will converge at the bottom of the image. Using the law of reflection and congruent-triangle geometry, the following is true of the object height, ho, and image height, hi, and any other dimension of the object and image. Plane-Mirror Image Height hi With a plane mirror, image height is equal to object height. ho 462 Chapter 17 Reflection and Mirrors 456-463 CH17-S1-845813 6/10/04 5:22 PM Page 463 \u25a0 Figure 17-7 The image formed in a plane mirror is the same size as the object and is the same distance behind the mirror as the object is in front. However, when the boy blinks his right eye, the left eye of his image blinks. Image orientation A plane mirror produces an image with the same orientation as the object. If you are standing on your feet, a plane mirror produces an image of you standing on your feet. If you are doing a headstand, the mirror shows you doing a headstand. However, there is a difference between you and the appearance of your image in a mirror. Follow the sight lines in Figure 17-7. The ray that diverges from the right hand of the boy converges at what appears to be the left hand of his image. Left and right appear to be reversed by a plane mirror. Why, then, are top and bottom not also reversed? This does not happen because a plane mirror does not really reverse left and right. The mirror in Figure 17-7 only reverses the boy\u2019s image such that it is facing in the opposite direction as the boy, or, in other words, it produces a front-to-back reversal. Consider the image of the mountain in the photo at the beginning of the chapter. In this case, the image of the mountain can be described as upside down, but the image is actually a front-to-back reversal of your view of the actual mountain. Because the mirror (the lake surface) is horizontal, rather than vertical, your perspective, or angle of view, makes the image look upside down. Turn your book 90\u00b0 counterclockwise and look at Figure 17-7 again. The actual boy is now facing down, and his image is facing up, upside down relative to the actual boy, just like the image of the mountain. The only thing that has changed is your perspective. 17.1 Section Review 6. Reflection A light ray strikes a flat, smooth, reflecting surface at an angle of 80\u00b0 to the normal. What is the angle that the reflected ray makes with the surface of the mirror? 7. Law of Reflection Explain how the law of reflec- tion applies to diffuse reflection. 8. Reflecting Surfaces Categorize each of the following as a specular or a diffuse reflecting surface: paper, polished metal, window glass, rough metal, plastic milk jug, smooth water surface, and ground glass. 9. Image Properties A 50-cm-tall dog stands 3 m from a plane mirror and looks at its image. What is the image position, height, and type? 10. Image Diagram A car is following another car down a straight road. The first car has a rear window tilted 45\u00b0. Draw a ray diagram showing the position of the Sun that would cause sunlight to reflect into the eyes of the driver of the second car. 11. Critical Thinking Explain how diffuse reflection of light off an object enables you to see an object from any angle. physicspp.com/self_check_quiz Section 17.1 Reflection from Plane Mirrors 463 464-473 CH17-S2-845813 6/4/04 11:39 AM Page 464 17.2 Curved Mirrors Objectives \u2022 Explain how concave and convex mirrors form images. \u2022 Describe properties and uses of spherical mirrors. \u2022 Determine the locations and sizes of spherical mirror images. Vocabulary concave mirror principal axis focal point focal length real image spherical aberration magnification convex mirror If you look at the surface of a shiny spoon, you will notice that your reflection is different from", " what you see in a plane mirror. The spoon acts as a curved mirror, with one side curved inward and the other curved outward. The properties of curved mirrors and the images that they form depend on the shape of the mirror and the object\u2019s position. Concave Mirrors The inside surface of a shiny spoon, the side that holds food, acts as a concave mirror. A concave mirror has a reflective surface, the edges of which curve toward the observer. Properties of a concave mirror depend on how much it is curved. Figure 17-8 shows how a spherical concave mirror works. In a spherical concave mirror, the mirror is shaped as if it were a section of a hollow sphere with an inner reflective surface. The mirror has the same geometric center, C, and radius of curvature, r, as a sphere of radius, r. The line that includes line segment CM is the principal axis, which is the straight line perpendicular to the surface of the mirror that divides the mirror in half. Point M is the center of the mirror where the principal axis intersects the mirror. \u25a0 Figure 17-8 The focal point of a spherical concave mirror is located halfway between the center of curvature and the mirror surface. Rays entering parallel to the principal axis are reflected to converge at the focal point, F. Principal axis C P M i r F r f Mirror surface When you point the principal axis of a concave mirror toward the Sun, all the rays are reflected through a single point. You can locate this point by moving a sheet of paper toward and away from the mirror until the smallest and sharpest spot of sunlight is focused on the paper. This spot is called the focal point of the mirror, the point where incident light rays that are parallel to the principal axis converge after reflecting from the mirror. The Sun is a source of parallel light rays because it is very far away. All of the light that comes directly from the Sun must follow almost parallel paths to Earth, just as all of the arrows shot by an archer must follow almost parallel paths to hit within the circle of a bull\u2019s-eye. When a ray strikes a mirror, it is reflected according to the law of reflection. Figure 17-8 shows that a ray parallel to the principal axis is reflected and crosses the principal axis at point F, the focal point. F is at the halfway point between M and C. The focal length, f, is the position of the focal point with respect to the mirror along the principal axis and can be expressed as f r/2. The focal length is positive for a concave mirror. 464 Chapter 17 Reflection and Mirrors 464-473 CH17-S2-845813 7/14/04 1:49 PM Page 465 a b c O Object C F O Object C F Mirror 2 1 I Mirror Image 2 1 I O Object C F Mirror Image 2 1 I Graphical Method of Finding the Image You have already drawn rays to follow the path of light that reflects off plane mirrors. This method is even more useful when applied to curved mirrors. Not only can the location of the image vary, but so can the orientation and size of the image. You can use a ray diagram to determine properties of an image formed by a curved mirror. Figure 17-9 shows the formation of a real image, an image that is formed by the converging of light rays. The image is inverted and larger than the object. The rays actually converge at the point where the image is located. The point of intersection, I, of the two reflected rays determines the position of the image. You can see the image floating in space if you place your eye so that the rays that form the image fall on your eye, as in Figure 17-9a. As Figure 17-9b shows, however, your eye must be oriented so as to see the rays coming from the image location. You cannot look at the image from behind. If you were to place a movie screen at this point, the image would appear on the screen, as shown in Figure 17-9c. You cannot do this with virtual images. To more easily understand how ray tracing works with curved mirrors, you can use simple, one-dimensional objects, such as the arrow shown in Figure 17-10a. A spherical concave mirror produces an inverted real image if the object position, do, is greater than twice the focal length, f. The object is then beyond the center of curvature, C. If the object is placed between the center of curvature and the focal point, F, as shown in Figure 17-10b, the image is again real and inverted. However, the size of the image is now greater than the size of the object. a O1 do Ray 1 Ray 2 Image F b I1 Object C Image C I1 di Principal plane \u25a0 Figure 17-9 The real image, as seen by the unaided eye (a). The unaided eye cannot see the real", " image if it is not in a location to catch the rays that form the image (b). The real image as seen on a white opaque screen (c). \u25a0 Figure 17-10 When the object is farther from the mirror than C, the image is a real image that is inverted and smaller compared to the object (a). When the object is located between C and F, the real image is inverted, larger than the object, and located beyond C (b). di Ray 2 Object F O1 Ray 1 do Principal plane Section 17.2 Curved Mirrors 465 464-473 CH17-S2-845813 6/9/04 4:27 PM Page 466 \u25a0 Figure 17-11 A Gregorian telescope produces a real image that is upright. Secondary concave mirror Telescope tube Astronomy Connection Astronomy Connection Primary concave mirror How can the inverted real image created by a concave mirror be turned right-side up? In 1663, Scottish astronomer James Gregory developed the Gregorian telescope, shown in Figure 17-11, to resolve this problem. It is composed of a large concave mirror and a small concave mirror arranged such that the smaller mirror is outside of the focal point of the larger mirror. Parallel rays of light from distant objects strike the larger mirror and reflect toward the smaller mirror. The rays then reflect off the smaller mirror and form a real image that is oriented exactly as the object is. Using Ray Tracing to Locate Images Formed by Spherical Mirrors Use the following strategies for spherical-mirror problems. Refer to Figure 17-10. 1. Using lined or graph paper, draw the principal axis of the mirror as a horizontal line from the left side to the right side of your paper, leaving six blank lines above and six blank lines below. 2. Place a point and a label on the principal axis the object, C, and F, as follows. a. If the mirror is a concave mirror and the object is beyond C, away from the mirror, place the mirror at the right side of the page, place the object at the left side of the page, and place C and F to scale. b. If the mirror is a concave mirror and the object is between C and F, place the mirror at the right side of the page, place C at the center of the paper, F halfway between the mirror and C, and the object to scale. c. For any other situation, place the mirror in the center of the page. Place the object or F (whichever is the greatest distance from the mirror) at the left side of the page, and place the other to scale. 3. To represent the mirror, draw a vertical line at the mirror point that extends the full 12 lines of space. This is the principal plane. 4. Draw the object as an arrow and label its top O1. For concave mirrors, objects inside of C should not be higher than three lines high. For all other situations, the objects should be six lines high. The scale for the height of the object will be different from the scale along the principal axis. 5. Draw ray 1, the parallel ray. It is parallel to the principal axis and reflects off the principal plane and passes through F. 6. Draw ray 2, the focus ray. It passes through F, reflects off the principal plane, and is reflected parallel to the principal axis. 7. The image is located where rays 1 and 2 (or their sight lines) cross after reflection. Label the point I1. The image is an arrow perpendicular from the principal axis to I1. 466 Lick Observatory Chapter 17 Reflection and Mirrors 464-473 CH17-S2-845813 6/4/04 11:42 AM Page 467 a b C F Mirror diameter F \u25a0 Figure 17-12 A concave spherical mirror reflects some rays, such that they converge at points other than the focus (a). A parabolic mirror focuses all parallel rays at a point (b). Real image defects in concave mirrors In tracing rays, you have reflected the rays from the principal plane, which is a vertical line representing the mirror. In reality, rays are reflected off the mirror itself, as shown in Figure 17-12a. Notice that only parallel rays that are close to the principal axis, or paraxial rays, are reflected through the focal point. Other rays converge at points closer to the mirror. The image formed by parallel rays reflecting off a spherical mirror with a large mirror diameter and a small radius of curvature is a disk, not a point. This effect, called spherical aberration, makes an image look fuzzy, not sharp. A mirror ground to the shape of a parabola, as in Figure 17-12b, suffers no spherical aberration. Because of the cost of manufacturing large, perfectly parabolic mirrors, many of the newest telescopes use spherical mirrors and smaller, specially-designed secondary mirrors or lenses to correct for spherical aberration. Also, spherical aberration", " is reduced as the ratio of the mirror\u2019s diameter, shown in Figure 17-12a, to its radius of curvature is reduced. Thus, lower-cost spherical mirrors can be used in lower-precision applications. Mathematical Method of Locating the Image The spherical mirror model can be used to develop a simple equation for spherical mirrors. You must use the paraxial ray approximation, which states that only rays that are close to and almost parallel with the principal axis are used to form an image. Using this, in combination with the law of reflection, leads to the mirror equation, relating the focal length, f, object position, do, and image position, di, of a spherical mirror. 1 1 1 Mirror Equation d d f i o The reciprocal of the focal length of a spherical mirror is equal to the sum of the reciprocals of the image position and the object position. When using this equation to solve problems, it is important to remember that it is only approximately correct. It does not predict spherical aberration, because it uses the paraxial ray approximation. In reality, light coming from an object toward a mirror is diverging, so not all of the light is close to or parallel to the axis. When the mirror diameter is small relative to the radius of curvature to minimize spherical aberration, this equation predicts image properties more precisely. Hubble Trouble In 1990, NASA launched the Hubble Space Telescope into orbit around Earth. Hubble was expected to provide clear images without atmospheric distortions. However, soon after it was deployed, Hubble was found to have a spherical aberration. In 1993, corrective optics, called COSTAR, were installed on Hubble to enable it to produce clear images. Section 17.2 Curved Mirrors 467 464-473 CH17-S2-845813 6/4/04 11:44 AM Page 468 Adding and Subtracting Fractions When using the mirror equation, you first use math to move the fraction that contains the quantity you are seeking to the left-hand side of the equation and everything else to the right. Then you combine the two fractions on the right-hand side using a common denominator that results from multiplying the denominators. Math Physics 1 )(z ) (1 ) )( )( d o d o ) ( )( do 1 fd d o i di fd o f do Using this approach, the following relationships can be derived for image position, object position, and focal length: di fd o f do do di f f di ido d f di do Magnification Another property of a spherical mirror is magnification, m, which is how much larger or smaller the image is relative to the object. In practice, this is a simple ratio of the image height to the object height. Using similar-triangle geometry, this ratio can be written in terms of image positon and object position. di h i Magnification m do h o The magnification of an object by a spherical mirror, defined as the image height divided by the object height, is equal to the negative of the image position, divided by the object position. Image position is positive for a real image when using the preceding equations. Thus, the magnification is negative, which means that the image is inverted compared to the object. If the object is beyond point C, the absolute value of the magnification for the real image is less than 1. This means that the image is smaller than the object. If the object is placed between point C and point F, the absolute value of the magnification for the real image is greater than 1. Thus, the image is larger than the object. 468 Chapter 17 Reflection and Mirrors 464-473 CH17-S2-845813 3/18/04 8:53 PM Page 469 Real Image Formation by a Concave Mirror A concave mirror has a radius of 20.0 cm. A 2.0-cm-tall object is 30.0 cm from the mirror. What is the image position and image height? 1 2 Analyze and Sketch the Problem \u2022 Draw a diagram with the object and the mirror. \u2022 Draw two principal rays to locate the image in the diagram. Known: 2.0 cm ho 30.0 cm do r 20.0 cm Unknown:? di? hi Solve for the Unknown Focal length is half the radius of curvature. r f 2 20.0 cm 2 10.0 cm Substitute r 20.0 cm do O1 ho C F hi I1 di Use the mirror equation and solve for image position. 1 1 1 f d d o i di fd o do f (10.0 cm)(30.0 cm) 30.0 cm 10.0 cm Substitute f 10.0 cm, do 30.0 cm Math Handbook Fractions page 837 15.0 cm (real image, in front of the mirror) Use the magnification equation and solve for image height. h di i m h do o iho d d o (15", ".0 cm)(2.0 cm) 30.0 cm hi Substitute di 15.0 cm, ho 2.0 cm, do 30.0 cm 1.0 cm (inverted, smaller image) 3 Evaluate the Answer \u2022 Are the units correct? All positions are in centimeters. \u2022 Do the signs make sense? Positive position and negative height agree with the drawing. 12. Use a ray diagram, drawn to scale, to solve Example Problem 2. 13. An object is 36.0 cm in front of a concave mirror with a 16.0-cm focal length. Determine the image position. 14. A 3.0-cm-tall object is 20.0 cm from a 16.0-cm-radius concave mirror. Determine the image position and image height. 15. A concave mirror has a 7.0-cm focal length. A 2.4-cm-tall object is 16.0 cm from the mirror. Determine the image height. 16. An object is near a concave mirror of 10.0-cm focal length. The image is 3.0 cm tall, inverted, and 16.0 cm from the mirror. What are the object position and object height? Section 17.2 Curved Mirrors 469 464-473 CH17-S2-845813 6/4/04 11:48 AM Page 470 a Ray 2 O1 F Ray 1 Object do I1 b Image di \u25a0 Figure 17-13 When an object is located between the focal point and a spherical concave mirror, a virtual image that is upright and larger compared to the object is formed behind the mirror (a), as shown with the stack of blocks (b). What else do you see in this picture? Virtual Images with Concave Mirrors You have seen that as an object approaches the focal point, F, of a concave mirror, the image moves farther away from the mirror. If the object is at the focal point, all reflected rays are parallel. They never meet, therefore, and the image is said to be at infinity, so the object could never be seen. What happens if the object is moved even closer to the mirror? What do you see when you move your face close to a concave mirror? The image of your face is right-side up and behind the mirror. A concave mirror produces a virtual image if the object is located between the mirror and the focal point, as shown in the ray diagram in Figure 17-13a. Again, two rays are drawn to locate the image of a point on an object. As before, ray 1 is drawn parallel to the principal axis and reflected through the focal point. Ray 2 is drawn as a line from the point on the object to the mirror, along a line defined by the focal point and the point on the object. At the mirror, ray 2 is reflected parallel to the principal axis. Note that ray 1 and ray 2 diverge as they leave the mirror, so there cannot be a real image. However, sight lines extended behind the mirror converge, showing that the virtual image forms behind the mirror. When you use the mirror equation to solve problems involving concave mirrors for which an object is between the mirror and the focal point, you will find that the image position is negative. The magnification equation gives a positive magnification greater than 1, which means that the image is upright and larger compared to the object, like the image in Figure 17-13b. An object of height ho is located at do relative to a concave mirror with focal length f. 1. Draw and label a ray diagram showing the focal length and location of the object if the image is located twice as far from the mirror as the object. Prove your answer mathematically. Calculate the focal length as a function of object position for this placement. 2. Draw and label a ray diagram showing the location of the object if the image is located twice as far from the mirror as the focal point. Prove your answer mathematically. Calculate the image height as a function of the object height for this placement. 3. Where should the object be located so that no image is formed? 470 Horizons Companies Chapter 17 Reflection and Mirrors 464-473 CH17-S2-845813 6/10/04 5:26 PM Page 471 O1 Object Ray 1 Ray 2 do I1 F Image di \u25a0 Figure 17-14 A convex mirror always produces virtual images that are upright and smaller compared to the object. Convex Mirrors In the first part of this chapter, you learned that the inner surface of a shiny spoon acts as a concave mirror. If you turn the spoon around, the outer surface acts as a convex mirror, a reflective surface with edges that curve away from the observer. What do you see when you look at the back of a spoon? You see an upright, but smaller image of yourself. Properties of a spherical convex mirror are shown in Figure 17-14", ". Rays reflected from a convex mirror always diverge. Thus, convex mirrors form virtual images. Points F and C are behind the mirror. In the mirror equation, f and di are negative numbers because they are both behind the mirror. The ray diagram in Figure 17-14 represents how an image is formed by a spherical convex mirror. The figure uses two rays, but remember that there are an infinite number of rays. Ray 1 approaches the mirror parallel to the principal axis. The reflected ray is drawn along a sight line from F through the point where ray 1 strikes the mirror. Ray 2 approaches the mirror on a path that, if extended behind the mirror, would pass through F. The reflected part of ray 2 and its sight line are parallel to the principal axis. The two reflected rays diverge, and the sight lines intersect behind the mirror at the location of the image. An image produced by a single convex mirror is a virtual image that is upright and smaller compared to the object. The magnification equation is useful for determining the apparent dimensions of an object as seen in a spherical convex mirror. If you know the diameter of an object, you can multiply by the magnification fraction to see how the diameter changes. You will find that the diameter is smaller, as are all other dimensions. This is why the objects appear to be farther away than they actually are for convex mirrors. Field of view It may seem that convex mirrors would have little use because the images that they form are smaller than the objects. However, this property of convex mirrors does have practical uses. By forming smaller images, convex mirrors enlarge the area, or field of view, that an observer sees, as shown in Figure 17-15. Also, the center of this field of view is visible from any angle of an observer off the principal axis of the mirror; thus, the field of view is visible from a wide perspective. For this reason, convex mirrors often are used in cars as passenger-side rearview mirrors. \u25a0 Figure 17-15 Convex mirrors produce images that are smaller than the objects. This increases the field of view for observers. Section 17.2 Curved Mirrors 471 Stewart Weir/Eye Ubiquitous/CORBIS 464-473 CH17-S2-845813 3/18/04 8:55 PM Page 472 Image in a Security Mirror A convex security mirror in a warehouse has a 0.50-m focal length. A 2.0-m-tall forklift is 5.0 m from the mirror. What are the image position and image height? 1 2 Analyze and Sketch the Problem \u2022 Draw a diagram with the mirror and the object. \u2022 Draw two principal rays to locate the image in the diagram. Known: 2.0 m ho 5.0 m do f 0.50 m Unknown:? di? hi O1 ho Solve for the Unknown Use the mirror equation and solve for image position. 1 1 1 d d f o i fd o do f ( m) Substitute f 0.50 m, do 0. 5 ( m) 0 0.5 0 5. 0 m 5.0 di m 5.0 m I1 F do di 0.45 m (virtual image, behind the mirror) Use the magnification equation and solve for image height. h di i m h do o iho d d o hi Math Handbook Isolating a Variable page 845 (0.45 m)(2.0 m) (5.0 m) Substitute di 0.45 m, ho 2.0 m, do 5.0 m 0.18 m (upright, smaller image) 3 Evaluate the Answer \u2022 Are the units correct? All positions are in meters. \u2022 Do the signs make sense? A negative position indicates a virtual image; a positive height indicates an image that is upright. These agree with the diagram. 17. An object is located 20.0 cm in front of a convex mirror with a 15.0-cm focal length. Find the image position using both a scale diagram and the mirror equation. 18. A convex mirror has a focal length of 13.0 cm. A lightbulb with a diameter of 6.0 cm is placed 60.0 cm from the mirror. What is the lightbulb\u2019s image position and diameter? 19. A convex mirror is needed to produce an image that is three-fourths the size of an object and located 24 cm behind the mirror. What focal length should be specified? 20. A 7.6-cm-diameter ball is located 22.0 cm from a convex mirror with a radius of curvature of 60.0 cm. What are the ball\u2019s image position and diameter? 21. A 1.8-m-tall girl stands 2.4 m from a store\u2019s security mirror. Her image appears to be 0.36 m tall. What is the focal length of the mirror?", " 472 Chapter 17 Reflection and Mirrors 464-473 CH17-S2-845813 3/18/04 8:56 PM Page 473 m Image Same size Virtual Reduced, inverted Enlarged, inverted Real Real Enlarged Virtual Reduced Virtual Mirror Type f Plane N/A Table 17-1 Single-Mirror System Properties do 0 do r do di \uf8e6 do \uf8e6di (negative) r di f Concave Convex r do f r di f do 0 0 do \uf8e6 do \uf8e6di (negative) \uf8e6f\uf8e6 \uf8e6di \uf8e6 0 (negative) Mirror Comparison How do the various types of mirrors compare? Table 17-1 compares the properties of single-mirror systems with objects that are located on the principal axis of the mirror. Virtual images are always behind the mirror, which means that the image position is negative. When the absolute value of a magnification is between zero and one, the image is smaller than the object. A negative magnification means the image is inverted relative to the object. Notice that the single plane mirror and convex mirror produce only virtual images, whereas the concave mirror can produce real images or virtual images. Plane mirrors give simple reflections, and convex mirrors expand the field of view. A concave mirror acts as a magnifier when an object is within the focal length of the mirror. 17.2 Section Review 22. Image Properties If you know the focal length of a concave mirror, where should you place an object so that its image is upright and larger compared to the object? Will this produce a real or virtual image? 23. Magnification An object is placed 20.0 cm in front of a concave mirror with a focal length of 9.0 cm. What is the magnification of the image? 24. Object Position The placement of an object in front of a concave mirror with a focal length of 12.0 cm forms a real image that is 22.3 cm from the mirror. What is the object position? 25. Image Position and Height A 3.0-cm-tall object is placed 22.0 cm in front of a concave mirror having a focal length of 12.0 cm. Find the image position and height by drawing a ray diagram to scale. Verify your answer using the mirror and magnification equations. 26. Ray Diagram A 4.0-cm-tall object is located 14.0 cm from a convex mirror with a focal length of 12.0 cm. Draw a scale ray diagram showing the image position and height. Verify your answer using the mirror and magnification equations. 27. Radius of Curvature A 6.0-cm-tall object is placed 16.4 cm from a convex mirror. If the image of the object is 2.8 cm tall, what is the radius of curvature of the mirror? 28. Focal Length A convex mirror is used to produce an image that is two-thirds the size of an object and located 12 cm behind the mirror. What is the focal length of the mirror? 29. Critical Thinking Would spherical aberration be less for a mirror whose height, compared to its radius of curvature, is small or large? Explain. physicspp.com/self_check_quiz Section 17.2 Curved Mirrors 473 474-475 CH17-LAB-845813 6/10/04 5:27 PM Page 474 Alternate CBL instructions can be found on the Web site. physicspp.com Concave Mirror Images A concave mirror reflects light rays that arrive parallel to the principal axis through the focal point. Depending on the object position, different types of images can be formed. Real images can be projected onto a screen while virtual images cannot. In this experiment you will investigate how changing the object position affects the image location and type. QUESTION What are the conditions needed to produce real and virtual images using a concave mirror? Objectives Procedure \u25a0 Collect and organize data of object and image positions. \u25a0 Observe real and virtual images. \u25a0 Summarize conditions for production of real and virtual images with a concave mirror. Safety Precautions \u25a0 Do not look at the reflection of the Sun in a mirror or use a concave mirror to focus sunlight. Materials concave mirror flashlight screen support mirror holder two metersticks four meterstick supports screen lamp with a 15-W lightbulb 1. Determine the focal length of your concave mirror by using the following procedure. CAUTION: Do not use the Sun to perform this procedure. Reflect light from a flashlight onto a screen and slowly move the screen closer or farther away from the mirror until a sharp, bright image is visible. Measure the distance between the screen and the mirror along the principal axis. Record this value as the actual focal length of the mirror, f. 2. On the lab table, set up two metersticks on supports in a V orientation. Place the zero measurement ends at the apex of the two meter", "sticks. 3. Place the mirror in a mirror holder and place it at the apex of the two metersticks. 4. Using the lamp as the object of the reflection, place it on one meterstick at the opposite end from the apex. Place the mirror and the screen, supported by a screen support, on the other meterstick at the opposite end from the apex. 5. Turn the room lights off. 6. Turn on the lamp. CAUTION: Do not touch the hot lightbulb. Measure object position, do, and record this as Trial 1. Measure the object height, ho, and record it as Trial 1. This is measured as the actual height of the lightbulb, or glowing filament if the bulb is clear. 7. Adjust the mirror or metersticks, as necessary, such that the reflected light shines on the screen. Slowly move the screen back and forth along the meterstick until a sharp image is seen. Measure image position, di, and the image height, hi, and record these as Trial 1. 474 Horizons Companies 474-475 CH17-LAB-845813 6/4/04 11:56 AM Page 475 Data Table Trial do (cm) di (cm) ho (cm) hi (cm) 1 2 3 4 5 Calculation Table Trial 1 (cm1) d o 1 (cm1) d i 1 1 (cm1) d d o i fcalc (cm) % error 1 2 3 4 5 8. Move the lamp closer to the mirror so that do is twice the focal length, f. Record this as Trial 2. Move the screen until an image is obtained on the screen. Measure di and hi, and record these as Trial 2. 9. Move the lamp closer to the mirror so that do is a few centimeters larger than f. Record this as Trial 3. Move the screen until an image is obtained on the screen. Measure di and hi, and record these as Trial 3. 10. Move the lamp so that do is equal to f. Record this as Trial 4 data. Move the screen back and forth and try to obtain an image. What do you observe? 11. Move the lamp so that do is less than f by a few centimeters. Record this as Trial 5. Move the screen back and forth and try to obtain an image. What do you observe? Analyze 1. Use Numbers Calculate 1/do and 1/di and enter the values in the calculation table. 2. Use Numbers Calculate the sum of 1/do and 1/di and enter the values in the calculation table. Calculate the reciprocal of this number and enter it in the calculation table as fcalc. 3. Error Analysis Compare the experimental focal length, fcalc, with f, the accepted focal length, by finding the percent error. \uf8e6f fcalc 100 percent error = f \uf8e6 Conclude and Apply 1. Classify What type of image was observed in each of the trials? 2. Analyze What conditions cause real images to be formed? 3. Analyze What conditions cause virtual images to be formed? Going Further 1. What are the conditions needed for the image to be larger than the object? 2. Review the methods used for data collection. Identify sources of error and what might be done to improve accuracy. Real-World Physics What advantage would there be in using a telescope with a concave mirror? To find out more about reflection, visit the Web site: physicspp.com 475 476 CH17-FEATURE-845813 3/18/04 9:04 PM Page 476 Adaptive Optical Systems Adaptive Optical Systems Adaptive mirror Light from telescope Distorted wave front Beam splitter Control system Corrected wave front Wave-front sensor High-resolution camera AOS compensates for distortion when viewing Titan, Saturn\u2019s largest moon. Conventional telescope Hubble Space Telescope Keck telescope with AOS The computer looks at this error and calcu- lates how the adaptive mirror should be wrinkled to bring each of the lenslet images back into place. The star image reflected to the observer then will be correct, and a clear image of all other objects (like galaxies and planets) in the vicinity also will be seen clearly. The adaptive mirror is re-shaped about 1000 times per second. Going Further 1. Research What is done if there is not a suitable star for the wave-front sensor to analyze in a region of space under observation? 2. Apply Research how adaptive optics will be used in the future to correct vision. Objects in space are difficult to observe from Earth because they twinkle. Our moving, unevenlyheated atmosphere refracts their light in a chaotic manner. It\u2019s like trying to look at a small object through the bottom of an empty, clear, glass jar while rotating the jar. Flexible Adaptive Mirror An adaptive optical system (AOS) continuously compensates for the distortion of the atmosphere, removing the twinkle from", " star images to allow astronomers to view and photograph steady images of the most distant objects in the visible universe. An AOS directs the magnified image of the stars from the telescope onto a flexible adaptive mirror made of thin glass. This mirror is stretched across 20\u201330 movable pistons that can poke or pull the surface of the mirror into many complicated shapes. Each piston is driven by a fast, computer-controlled motor. When the mirror surface is shaped into just the right pattern at just the right time, it will compensate for the convective movement of the atmosphere between the telescope and the star, and reflect a clear image to the observer or camera. Wave-Front Sensor To detect the atmospheric distortion at each instant of time, a wave-front sensor looks at a single star through the telescope. This device has an array of tiny lenses (lenslets) in several rows. Each lenslet forms an image of the star on a sensitive screen behind it. The position of each image can be read by the computer. If each image is not directly behind its lenslet, then the computer knows that the star\u2019s light waves are being distorted by the atmosphere. A star is a distant point source of light, so it should produce plane waves. Distorted images of a star are non-planar light waves, and these uneven waves cause the images of the star behind some of the lenslets to be displaced. 476 Future Technology 477-483 CH17-SG CA STP-845813 6/4/04 12:09 PM Page 477 17.1 Reflection from Plane Mirrors Vocabulary \u2022 specular reflection (p. 459) \u2022 diffuse reflection (p. 459) \u2022 plane mirror (p. 461) \u2022 object (p. 461) \u2022 image (p. 461) \u2022 virtual image (p. 461) 17.2 Curved Mirrors Vocabulary \u2022 concave mirror (p. 464) \u2022 principal axis (p. 464) \u2022 focal point (p. 464) \u2022 focal length (p. 464) \u2022 real image (p. 465) \u2022 spherical aberration (p. 467) \u2022 magnification (p. 468) \u2022 convex mirror (p. 471) Key Concepts \u2022 According to the law of reflection, the angle that an incident ray makes with the normal equals the angle that the reflected ray makes with the normal. r i \u2022 The law of reflection applies to both smooth and rough surfaces. A rough surface, however, has normals that go in lots of different directions, which means that parallel incident rays will not be reflected in parallel. \u2022 A smooth surface produces specular reflection. A rough surface produces diffuse reflection. \u2022 Specular reflection results in the formation of images that appear to be behind plane mirrors. \u2022 An image produced by a plane mirror is always virtual, is the same size as the object, has the same orientation, and is the same distance from the mirror as the object. di do hi ho Key Concepts \u2022 You can locate the image created by a spherical mirror by drawing two rays from a point on the object to the mirror. The intersection of the two reflected rays is the image of the object point. \u2022 The mirror equation gives the relationship among image position, object position, and focal length of a spherical mirror. 1 1 1 d d f i o \u2022 The magnification of a mirror image is given by equations relating either the positions or the heights of the image and the object. h di i m h do o \u2022 A single concave mirror produces a real image that is inverted when the object position is greater than the focal length. \u2022 A single concave mirror produces a virtual image that is upright when the object position is less than the focal length. \u2022 A single convex mirror always produces a virtual image that is upright and smaller compared to the object. \u2022 By forming smaller images, convex mirrors make images seem farther away and produce a wide field of view. \u2022 Mirrors can be used in combinations to produce images of any size, orientation, and location desired. The most common use of combinations of mirrors is as telescopes. physicspp.com/vocabulary_puzzlemaker 477 477-483 CH17-SG CA STP-845813 6/4/04 12:11 PM Page 478 Concept Mapping 30. Complete the following concept map using the following terms: convex, upright, inverted, real, virtual. Mirrors plane concave Applying Concepts 44. Wet Road A dry road is more of a diffuse reflector than a wet road. Based on Figure 17-16, explain why a wet road appears blacker to a driver than a dry road does. virtual virtual Wet asphalt upright upright Mastering Concepts 31. How does specular reflection differ from diffuse reflection? (17.1) 32. What is meant by the phrase \u201cnormal to the surface\u201d? (17.1) 33. Where is the image produced by a plane mirror located? (17", ".1) 34. Describe the properties of a plane mirror. (17.1) 35. A student believes that very sensitive photographic film can detect a virtual image. The student puts photographic film at the location of a virtual image. Does this attempt succeed? Explain. (17.1) 36. How can you prove to someone that an image is a real image? (17.1) 37. An object produces a virtual image in a concave mirror. Where is the object located? (17.2) 38. What is the defect that all concave spherical mirrors have and what causes it? (17.2) 39. What is the equation relating the focal point, object position, and image position? (17.2) 40. What is the relationship between the center of curvature and the focal length of a concave mirror? (17.2) 41. If you know the image position and object position relative to a curved mirror, how can you determine the mirror\u2019s magnification? (17.2) 42. Why are convex mirrors used as rearview mirrors? (17.2) 43. Why is it impossible for a convex mirror to form a real image? (17.2) Dry asphalt \u25a0 Figure 17-16 45. Book Pages Why is it desirable that the pages of a book be rough rather than smooth and glossy? 46. Locate and describe the physical properties of the image produced by a concave mirror when the object is located at the center of curvature. 47. An object is located beyond the center of curvature of a spherical concave mirror. Locate and describe the physical properties of the image. 48. Telescope You have to order a large concave mirror for a telescope that produces high-quality images. Should you order a spherical mirror or a parabolic mirror? Explain. 49. Describe the properties of the image seen in the single convex mirror in Figure 17-17. \u25a0 Figure 17-17 478 Chapter 17 Reflection and Mirrors For more problems, go to Additional Problems, Appendix B. 477-483 CH17-SG CA STP-845813 3/18/04 9:10 PM Page 479 50. List all the possible arrangements in which you could use a spherical mirror, either concave or convex, to form a real image. 51. List all possible arrangements in which you could use a spherical mirror, either concave or convex, to form an image that is smaller compared to the object. 52. Rearview Mirrors The outside rearview mirrors of cars often carry the warning \u201cObjects in the mirror are closer than they appear.\u201d What kind of mirrors are these and what advantage do they have? Mastering Problems 17.1 Reflection from Plane Mirrors 53. A ray of light strikes a mirror at an angle of 38\u00b0 to the normal. What is the angle that the reflected angle makes with the normal? 54. A ray of light strikes a mirror at an angle of 53\u00b0 to the normal. a. What is the angle of reflection? b. What is the angle between the incident ray and the reflected ray? 55. A ray of light incident upon a mirror makes an angle of 36\u00b0 with the mirror. What is the angle between the incident ray and the reflected ray? 56. Picture in a Mirror Penny wishes to take a picture of her image in a plane mirror, as shown in Figure 17-18. If the camera is 1.2 m in front of the mirror, at what distance should the camera lens be focused? \u25a0 Figure 17-18 physicspp.com/chapter_test 57. Two adjacent plane mirrors form a right angle, as shown in Figure 17-19. A light ray is incident upon one of the mirrors at an angle of 30\u00b0 to the normal. a. What is the angle at which the light ray is reflected from the other mirror? b. A retroreflector is a device that reflects incoming light rays back in a direction opposite to that of the incident rays. Draw a diagram showing the angle of incidence on the first mirror for which the mirror system acts as a retroreflector. 30\u00b0 \u25a0 Figure 17-19 58. Draw a ray diagram of a plane mirror to show that if you want to see yourself from your feet to the top of your head, the mirror must be at least half your height. 59. Two plane mirrors are connected at their sides so that they form a 45\u00b0 angle between them. A light ray strikes one mirror at an angle of 30\u00b0 to the normal and then reflects off the second mirror. Calculate the angle of reflection of the light ray off the second mirror. 60. A ray of light strikes a mirror at an angle of 60\u00b0 to the normal. The mirror is then rotated 18\u00b0 clockwise, as shown in Figure 17-20. What is the angle that the reflected ray makes with the mirror? Incident light 18\u00b0 Normal 60\u00b0 Mirror \u25a0 Figure 17-20 Chapter 17 Assessment", " 479 Hutchings Photography 477-483 CH17-SG CA STP-845813 6/10/04 5:35 PM Page 480 17.2 Curved Mirrors 61. A concave mirror has a focal length of 10.0 cm. What is its radius of curvature? 62. An object located 18 cm from a convex mirror produces a virtual image 9 cm from the mirror. What is the magnification of the image? 63. Fun House A boy is standing near a convex mirror in a fun house at a fair. He notices that his image appears to be 0.60 m tall. If the magnification of the mirror is 1, what is the boy\u2019s height? 3 64. Describe the image produced by the object in Figure 17-21 as real or virtual, inverted or upright, and smaller or larger than the object. C F \u25a0 Figure 17-21 65. Star Image Light from a star is collected by a concave mirror. How far from the mirror is the image of the star if the radius of curvature is 150 cm? 66. Find the image position and height for the object shown in Figure 17-22. 68. An object is 30.0 cm from a concave mirror of 15.0 cm focal length. The object is 1.8 cm tall. Use the mirror equation to find the image position. What is the image height? 69. Dental Mirror A dentist uses a small mirror with a radius of 40 mm to locate a cavity in a patient\u2019s tooth. If the mirror is concave and is held 16 mm from the tooth, what is the magnification of the image? 70. A 3.0-cm-tall object is 22.4 cm from a concave mirror. If the mirror has a radius of curvature of 34.0 cm, what are the image position and height? 71. Jeweler\u2019s Mirror A jeweler inspects a watch with a diameter of 3.0 cm by placing it 8.0 cm in front of a concave mirror of 12.0-cm focal length. a. Where will the image of the watch appear? b. What will be the diameter of the image? 72. Sunlight falls on a concave mirror and forms an image that is 3.0 cm from the mirror. An object that is 24 mm tall is placed 12.0 cm from the mirror. a. Sketch the ray diagram to show the location of the image. b. Use the mirror equation to calculate the image position. c. How tall is the image? 73. Shiny spheres that are placed on pedestals on a lawn are convex mirrors. One such sphere has a diameter of 40.0 cm. A 12-cm-tall robin sits in a tree that is 1.5 m from the sphere. Where is the image of the robin and how tall is the image? Mixed Review 74. A light ray strikes a plane mirror at an angle of 28\u00b0 to the normal. If the light source is moved so that the angle of incidence increases by 34\u00b0, what is the new angle of reflection? 75. Copy Figure 17-23 on a sheet of paper. Draw rays on the diagram to determine the height and location of the image. 3.8 cm F 16 cm 31 cm \u25a0 Figure 17-22 67. Rearview Mirror How far does the image of a car appear behind a convex mirror, with a focal length of 6.0 m, when the car is 10.0 m from the mirror? 3.0 cm F 8.0 cm 4.0 cm \u25a0 Figure 17-23 480 Chapter 17 Reflection and Mirrors For more problems, go to Additional Problems, Appendix B. 477-483 CH17-SG CA STP-845813 6/10/04 5:36 PM Page 481 76. An object is located 4.4 cm in front of a concave 84. A 1.6-m-tall girl stands 3.2 m from a convex mirror. mirror with a 24.0-cm radius. Locate the image using the mirror equation. What is the focal length of the mirror if her image appears to be 0.28 m tall? 77. A concave mirror has a radius of curvature of 26.0 cm. An object that is 2.4 cm tall is placed 30.0 cm from the mirror. a. Where is the image position? b. What is the image height? 78. What is the radius of curvature of a concave mirror that magnifies an object by a factor of 3.2 when the object is placed 20 cm from the mirror? 79. A convex mirror is needed to produce an image one-half the size of an object and located 36 cm behind the mirror. What focal length should the mirror have? 80. Surveillance Mirror A convenience store uses a surveillance mirror to monitor the store\u2019s aisles. Each mirror has a radius of curvature", " of 3.8 m. a. What is the image position of a customer who stands 6.5 m in front of the mirror? b. What is the image height of a customer who is 1.7 m tall? 81. Inspection Mirror A production-line inspector wants a mirror that produces an image that is upright with a magnification of 7.5 when it is located 14.0 mm from a machine part. a. What kind of mirror would do this job? b. What is its radius of curvature? 82. The object in Figure 17-24 moves from position 1 to position 2. Copy the diagram onto a sheet of paper. Draw rays showing how the image changes. 1 C 2 F 1.0 m 1.5 m 2.0 m 2.5 m 85. Magic Trick A magician uses a concave mirror with a focal length of 8.0 m to make a 3.0-m-tall hidden object, located 18.0 m from the mirror, appear as a real image that is seen by his audience. Draw a scale ray diagram to find the height and location of the image. 86. A 4.0-cm-tall object is placed 12.0 cm from a convex mirror. If the image of the object is 2.0 cm tall, and the image is located at 6.0 cm, what is the focal length of the mirror? Draw a ray diagram to answer the question. Use the mirror equation and the magnification equation to verify your answer. Thinking Critically 87. Apply Concepts The ball in Figure 17-25 slowly rolls toward the concave mirror on the right. Describe how the size of the ball\u2019s image changes as it rolls along. C F \u25a0 Figure 17-25 88. Analyze and Conclude The object in Figure 17-26 is located 22 cm from a concave mirror. What is the focal length of the mirror? \u25a0 Figure 17-24 22 cm 83. A ball is positioned 22 cm in front of a spherical mirror and forms a virtual image. If the spherical mirror is replaced with a plane mirror, the image appears 12 cm closer to the mirror. What kind of spherical mirror was used? physicspp.com/chapter_test \u25a0 Figure 17-26 Chapter 17 Assessment 481 CORBIS 477-483 CH17-SG CA STP-845813 6/10/04 5:39 PM Page 482 89. Use Equations Show that as the radius of curvature of a concave mirror increases to infinity, the mirror equation reduces to the relationship between the object position and the image position for a plane mirror. 90. Analyze and Conclude An object is located 6.0 cm from a plane mirror. If the plane mirror is replaced with a concave mirror, the resulting image is 8.0 cm farther behind the mirror. Assuming that the object is located between the focal point and the concave mirror, what is the focal length of the concave mirror? 91. Analyze and Conclude The layout of the twomirror system shown in Figure 17-11 is that of a Gregorian telescope. For this question, the larger concave mirror has a radius of curvature of 1.0 m, and the smaller mirror is located 0.75 m away. Why is the secondary mirror concave? 92. Analyze and Conclude An optical arrangement used in some telescopes is the Cassegrain focus, shown in Figure 17-27. This telescope uses a convex secondary mirror that is positioned between the primary mirror and the focal point of the primary mirror. Convex secondary mirror Concave primary mirror F Telescope tube Eyepiece \u25a0 Figure 17-27 a. A single convex mirror produces only virtual images. Explain how the convex mirror in this telescope functions within the system of mirrors to produce real images. b. Are the images produced by the Cassegrain focus upright or inverted? How does this relate to the number of times that the light crosses? Writing in Physics 93. Research a method used for grinding, polishing, and testing mirrors used in reflecting telescopes. You may report either on methods used by amateur astronomers who make their own telescope optics, or on a method used by a project at a national laboratory. Prepare a one-page report describing the method, and present it to the class. 94. Mirrors reflect light because of their metallic coating. Research and write a summary of one of the following: a. the different types of coatings used and the advantages and disadvantages of each b. the precision optical polishing of aluminum to such a degree of smoothness that no glass is needed in the process of making a mirror Cumulative Review 95. A child runs down the school hallway and then slides on the newly waxed floor. He was running at 4.7 m/s before he started sliding and he slid 6.2 m before stopping. What was the coefficient of friction of the waxed floor? (Chapter 11) 96. A 1.0 g piece of copper falls", " from a height of 1.0104 m from an airplane to the ground. Because of air resistance it reaches the ground moving at a velocity of 70.0 m/s. Assuming that half of the energy lost by the piece was distributed as thermal energy to the copper, how much did it heat during the fall? (Chapter 12) 97. It is possible to lift a person who is sitting on a pillow made from a large sealed plastic garbage bag by blowing air into the bag through a soda straw. Suppose that the cross-sectional area of the person sitting on the bag is 0.25 m2 and the person\u2019s weight is 600 N. The soda straw has a cross-sectional area of 2105 m2. With what pressure must you blow into the straw to lift the person that is sitting on the sealed garbage bag? (Chapter 13) 98. What would be the period of a 2.0-m-long pendulum on the Moon\u2019s surface? The Moon\u2019s mass is 7.341022 kg, and its radius is 1.74106 m. What is the period of this pendulum on Earth? (Chapter 14) 99. Organ pipes An organ builder must design a pipe organ that will fit into a small space. (Chapter 15) a. Should he design the instrument to have open pipes or closed pipes? Explain. b. Will an organ constructed with open pipes sound the same as one constructed with closed pipes? Explain. 100. Filters are added to flashlights so that one shines red light and the other shines green light. The beams are crossed. Explain in terms of waves why the light from both flashlights is yellow where the beams cross, but revert back to their original colors beyond the intersection point. (Chapter 16) 482 Chapter 17 Reflection and Mirrors For more problems, go to Additional Problems, Appendix B. 477-483 CH17-SG CA STP-845813 3/18/04 9:16 PM Page 483 Multiple Choice 1. Where is the object located if the image that is produced by a concave mirror is smaller than the object? at the mirror\u2019s focal point between the mirror and the focal point between the focal point and center of curvature past the center of curvature 2. What is the focal length of a concave mirror that magnifies, by a factor of 3.2, an object that is placed 30 cm from the mirror? 23 cm 32 cm 44 cm 46 cm 3. An object is placed 21 cm in front of a concave mirror with a focal length of 14 cm. What is the image position? 42 cm 8.4 cm 8.4 cm 42 cm 4. The light rays in the illustration below do not properly focus at the focal point. This problem occurs with ________. all spherical mirrors all parabolic mirrors only defective spherical mirrors only defective parabolic mirrors 6. A concave mirror produces an inverted image that is 8.5 cm tall, located 34.5 cm in front of the mirror. If the focal point of the mirror is 24.0 cm, then what is the height of the object that is reflected? 2.3 cm 3.5 cm 14 cm 19 cm 7. A concave mirror with a focal length of 16.0 cm produces an image located 38.6 cm from the mirror. What is the distance of the object from the front of the mirror? 2.4 cm 11.3 cm 22.6 cm 27.3 cm 8. A convex mirror is used to produce an image that is three-fourths the size of an object and located 8.4 cm behind the mirror. What is the focal length of the mirror? 34 cm 11 cm 6.3 cm 4.8 cm 9. A cup sits 17 cm from a concave mirror. The image of the book appears 34 cm in front of the mirror. What are the magnification and orientation of the cup\u2019s image? 0.5, inverted 0.5, upright 2.0, inverted 2.0, upright Extended Answer 10. A 5.0-cm-tall object is located 20.0 cm from a convex mirror with a focal length of 14.0 cm. Draw a scale-ray diagram showing the image height. F 5. A ray of light strikes a plane mirror at an angle of 23\u00b0 to the normal. What is the angle between the reflected ray and the mirror? 67\u00b0 134\u00b0 23\u00b0 46\u00b0 Your Answers Are Better Than the Test\u2019s When you know how to solve a problem, solve it before looking at the answer choices. Often, more than one answer choice will look good, so do the calculations first, and arm yourself with your answer before looking. physicspp.com/standardized_test Chapter 17 Standardized Test Practice 483 484-492 CH18-S1-845813 6/4/04 12:47 PM Page 484 What You\u2019ll Learn \u2022 You will learn", " how light changes direction and speed when it travels through different materials. \u2022 You will compare properties of lenses and the images that they form. \u2022 You will learn about different applications of lenses, including how lenses in your eyes enable you to see. Why It\u2019s Important Some light travels in a straight path from objects to your eyes. Some light is reflected before it reaches you. Other light follows a path that appears to be bent. Wavy Trees If you swim underwater, you will notice that things underwater look normal, but objects above the surface of the water appear to be distorted by the waves on the surface. Think About This What causes the images of the trees to be wavy? physicspp.com 484 Ric Frazier/Masterfile 484-492 CH18-S1-845813 3/18/04 2:53 PM Page 485 What does a straw in a liquid look like from the side view? Question Does a straw look different when observed through water, oil, and corn syrup? Critical Thinking Form a hypothesis as to when a solid object appears to be broken and when it does not. Be sure to include an explanation of the amount of break. Procedure 1. Fill one 400-mL beaker with water. 2. Fill a second 400-mL beaker halfway with corn syrup and the rest with water (pour slowly as to avoid mixing the two liquids). 3. Fill a third 400-mL beaker halfway with water and the rest with cooking oil (pour slowly as to avoid mixing the two liquids). 4. Place a straw gently in each beaker and lean it on the spout. 5. Observe each straw through the side of the beaker as you slowly turn the beaker. 6. Make a data table to record descriptions of the straws\u2019 appearance in each solution. Analysis In which containers does the straw appear to be broken? Are all amounts of break the same? When does the straw not appear to be broken? Explain. 18.1 Refraction of Light Looking at the surface of a swimming pool on a summer day, you can see sunlight reflecting off the water. You can see objects that are in the pool because some of the sunlight travels into the water and reflects off the objects. When you look closely at objects in the water, however, you will notice that they look distorted. For example, things beneath the surface look closer than normal, the feet of a person standing still in the pool appear to move back and forth, and lines along the bottom of the pool seem to sway with the movement of the water. These effects occur because light changes direction as it passes from water to air. As you learned in Chapter 16, the path of light is bent as it crosses the boundary between two media due to refraction. The amount of refraction depends on properties of the two media and on the angle at which the light strikes the boundary. As waves travel along the surface of the water, the boundary between the air and water moves up and down, and tilts back and forth. The path of light leaving the water shifts as the boundary moves, causing objects under the surface to appear to waver. Objectives \u2022 Solve problems involving refraction. \u2022 Explain total internal reflection. \u2022 Explain some optical effects caused by refraction. Vocabulary index of refraction Snell\u2019s law of refraction critical angle total internal reflection dispersion Section 18.1 Refraction of Light 485 Horizons Companies 484-492 CH18-S1-845813 6/10/04 8:48 PM Page 486 \u25a0 Figure 18-1 Light bends toward the normal as it moves from air to glass and bends away from the normal as it moves from glass to air (a). The bending of light makes objects appear to be shifted from their actual locations (b). \u2022 Refracting media and lenses are light blue. Table 18-1 Indices of Refraction for Yellow Light ( 589 nm in vacuum) Medium n Vacuum Air Water Ethanol Crown glass Quartz Flint glass Diamond 1.00 1.0003 1.33 1.36 1.52 1.54 1.62 2.42 Air Glass Air Snell\u2019s Law of Refraction What happens when you shine a narrow beam of light at the surface of a piece of glass? As you can see in Figure 18-1, it bends as it crosses the boundary from air to glass. The bending of light, called refraction, was first studied by Ren\u00e9 Descartes and Willebrord Snell around the time of Kepler and Galileo. To discuss the results of Descartes and Snell, you have to define two angles. The angle of incidence, \u03b8 1, is the angle at which the light ray strikes the surface. It is measured from the normal to the surface. The angle of refraction, \u03b8 2, is the angle at which the transmitted light leaves the surface. It also is measured with respect to the normal. In 1621, Snell found that when light passed", " from air into a transparent substance, the sines of the angles were related by the equation sin \u03b8 n. Here n is a constant that depends on the substance, not on the angles, and is called the index of refraction. The indices of refraction for some substances are listed in Table 18-1. The relationship found by Snell is valid when light goes across a boundary between any two materials. This more general equation is known as Snell\u2019s law of refraction. 1/sin \u03b8 2 Snell\u2019s Law of Refraction n1 sin The product of the index of refraction of the first medium and the sine of the angle of incidence is equal to the product of the index of refraction of the second medium and the sine of the angle of refraction. n2 sin 2 1 Figure 18-1 shows how Snell\u2019s law applies when light travels through a piece of glass with parallel surfaces, such as a windowpane. The light is refracted both when it enters the glass and again when it leaves the glass. When light goes from air to glass it moves from material with a lower n2. index of refraction to one with a higher index of refraction. That is, n1 To keep the two sides of the equation equal, one must have sin \u03b8 sin \u03b8 2. 1 The light beam is bent toward the normal to the surface. When light travels from glass to air it moves from material having a n2. higher index of refraction to one with a lower index. In this case, n1 To keep the two sides of the equation equal one must have sin \u03b8. sin \u03b8 2 1 That is, the light is bent away from the normal. Note that the direction of the ray when it leaves the glass is the same as it was before it struck the glass, but it is shifted from its original position. 486 Horizons Companies Chapter 18 Refraction and Lenses 484-492 CH18-S1-845813 6/4/04 12:54 PM Page 487 Angle of Refraction A light beam in air hits a sheet of crown glass at an angle of 30.0\u00b0. At what angle is the light beam refracted? 1 Analyze and Sketch the Problem \u2022 Make a sketch of the air and crown glass boundary. \u2022 Draw a ray diagram. Unknown:? 2 Known: 1 n1 n2 30.0\u00b0 1.00 1.52 2 Solve for the Unknown Use Snell\u2019s law to solve for the sine of the angle of refraction. n1 sin 1 sin 2 2 1 n2 sin 2 ) sin ( n 1 n 2 sin1(( sin1((1 19.2 ) sin 1) ) sin 30.0\u00b0) 0 2 Substitute n1 1.00, n2 1.52, 1 30.0\u00b0 1 n1 2 n2 Math Handbook Inverses of Sine, Cosine, and Tangent page 856 3 Evaluate the Answer \u2022 Are the units correct? Angles are expressed in degrees. \u2022 Is the magnitude realistic? The index of refraction, n2, is greater than the index 2, must be less than the angle of refraction, n1. Therefore, the angle of refraction, of incidence, 1. 1. A laser beam in air is incident upon ethanol at an angle of incidence of 37.0\u00b0. What is the angle of refraction? 2. Light in air is incident upon a piece of crown glass at an angle of incidence of 45.0\u00b0. What is the angle of refraction? 3. Light passes from air into water at 30.0\u00b0 to the normal. Find the angle of refraction. 4. Light is incident upon a diamond facet at 45.0\u00b0. What is the angle of refraction? 5. A block of unknown material is submerged in water. Light in the water is incident on the block at an angle of incidence of 31\u00b0. The angle of refraction of the light in the block is 27\u00b0. What is the index of refraction of the material of the block? Refraction is responsible for the Moon appearing red during a lunar eclipse. A lunar eclipse occurs when Earth blocks sunlight from the Moon. As a result, you might expect the Moon to be completely dark. Instead, light refracts through Earth\u2019s atmosphere and bends around Earth toward the Moon. Recall that Earth\u2019s atmosphere scatters most of the blue and green light. Thus, mostly red light illuminates the Moon. Because the Moon reflects most colors of light equally well, it reflects the red light back to Earth, and therefore the Moon appears to be red. Astronomy Connection Astronomy Connection Section 18.1 Refraction of Light 487 484-492 CH18-S1-845813 6/10/04 8:48 PM Page 488 a b n1 Q 1 R 1 P 2 2 S n", "2 \u25a0 Figure 18-2 Light moves from air to glass to air again (a). Light slows down and bends toward the normal when it enters a region of a higher index of refraction (b). Wave Model of Refraction The wave model of light was developed almost 200 years after Snell published his research. An understanding that light interacts with atoms when traveling through a medium, such that it moves more slowly than in a vacuum, was achieved three hundred years after Snell\u2019s work. The wave relationship that you learned in Chapter 16 for light traveling through a vacuum, c/f, can be rewritten as v/f, where v is the speed of light 0 in any medium and is the wavelength. The frequency of light, f, does not change when it crosses a boundary. That is, the number of oscillations per second that arrive at a boundary is the same as the number that leave the boundary and transmit through the refracting medium. So, the wavelength of light,, must decrease when light slows down. Wavelength in a medium is shorter than wavelength in a vacuum. What happens when light travels from a region with a high speed into one with a low speed, as shown in Figure 18-2a? The diagram in Figure 18-2b shows a beam of light as being made up of a series of parallel, straight wave fronts. Each wave front represents the crest of a wave and is perpendicular to the direction of the beam. The beam strikes the surface at an angle, 1. Consider the triangle PQR. Because the wave fronts are perpendicular to the direction of the beam, PQR is a right angle and QRP is equal to 1 is equal to the distance between P and Q divided by the distance between P and R. 1. Therefore, sin sin 1 \uf8e7 \uf8e7 \uf8e7Q P \uf8e7 R P The angle of refraction, PSR. In this case 2, can be related in a similar way to the triangle \uf8e7 By taking the ratio of the sines of the two angles, P the following equation: \uf8e7 R is canceled, leaving sin 2 \uf8e7 \uf8e7 \uf8e7R \uf8e7S R P \uf8e7 \uf8e7 n i s \uf8e7R \uf8e7 Figure 18-2b is drawn such that the distance between P and Q is equal to \uf8e7 the length of three wavelengths of light in medium 1, or P 1. In a \uf8e7 similar way, R 2. Substituting these two values into the previous equation and canceling the common factor of 3 provides an equation that relates the angles of incidence and refraction with the wavelength of the light in each medium Using v/f in the above equation and canceling the common factor of f, the equation is rewritten as follows Snell\u2019s law also can be written as a ratio of the sines of the angles of incidence and refraction 488 Horizons Companies Chapter 18 Refraction and Lenses 484-492 CH18-S1-845813 6/11/04 3:26 PM Page 489 Index of refraction Using the transitive property of equality, the previous two equations lead to the following equation: v n 2 1 v n 1 2 In a vacuum, n 1 and v c. If either medium is a vacuum, then the equation is simplified to an equation that relates the index of refraction to the speed of light in a medium. c Index of Refraction n v The index of refraction of a medium is equal to the speed of light in a vacuum divided by the speed of light in the medium. This definition of the index of refraction can be used to find the wavelength of light in a medium compared to the wavelength the light would have in a vacuum. In a medium with an index of refraction n, the speed of light is given by v c/n. The wavelength of the light in a vacuum is 0 and v c/n into 0 v/f. (c/n)/(c/ 0) 0/n, and thus the wavelength of light in a medium is smaller than the wavelength in a vacuum. c/f. Solve for frequency, and substitute f c/ Total Internal Reflection The angle of refraction is larger than the angle of incidence when light passes into a medium of a lower index of refraction, as shown in Figure 18-3a. This leads to an interesting phenomenon. As the angle of incidence increases, the angle of refraction increases. At a certain angle of incidence known as the critical angle, c, the refracted light ray lies along the boundary of the two media, as shown in Figure 18-3b. Recall from Chapter 16 that when light strikes a transparent boundary, even though much of the light is transmitted, some is reflected. Total internal reflection occurs when light traveling from a region of a higher index of refraction to a region of a lower index of refraction strikes the boundary at an angle", " greater than the critical angle such that all light reflects back into the region of the higher index of refraction, as shown in Figure 18-3c. To construct an equation for the critical angle of any boundary, you can use Snell\u2019s law and substitute 1 c and 90.0\u00b0. 2 Critical Angle for Total Internal Reflection sin c n 2 n 1 The sine of the critical angle is equal to the index of refraction of the refracting medium divided by the index of refraction of the incident medium. Total internal reflection causes some curious effects. Suppose that you are looking up at the surface from underwater in a calm pool. You might see an upside-down reflection of another nearby object that also is underwater or a reflection of the bottom of the pool itself. The surface of the water acts like a mirror. Likewise, when you are standing on the side of a pool, it is possible for things below the surface of the water to not be visible to you. When a swimmer is underwater, near the surface, and on the opposite side of the pool from you, you might not see him or her. This is because the light from his or her body is reflected. \u25a0 Figure 18-3 Ray A is partially refracted and partially reflected (a). Ray B is refracted along the boundary of the medium and forms the critical angle (b). An angle of incidence greater than the critical angle results in the total internal reflection of Ray C, which follows the law of reflection (c). a Air Water Refracted ray 2 n2 n1 1 Ray A Critical ray c Ray B b Air Water c Air Air Water i r r i Reflected ray Ray 3 Ray C Section 18.1 Refraction of Light 489 484-492 CH18-S1-845813 6/4/04 1:00 PM Page 490 n2 (Cladding) n2 n1 n1 (Glass fiber) \u25a0 Figure 18-4 Light impulses from a source enter one end of the optical fiber. Each time the light strikes the surface, the angle of incidence is larger than the crictical angle, and, therefore, the light is kept within the fiber. Optical fibers are an important technical application of total internal reflection. As shown in Figure 18-4, the light traveling through the transparent fiber always hits the internal boundary of the optical fiber at an angle greater than the critical angle, so all of the light is reflected and none of the light is transmitted through the boundary. Thus, the light maintains its intensity over the distance of the fiber. Mirages On a hot summer day, you sometimes can see the mirage effect shown in Figure 18-5a. As you drive down a road, you see what appears to be the reflection of an oncoming car in a pool of water. The pool, however, disappears as you approach it. The mirage is the result of the Sun heating the road. The hot road heats the air above it and produces a thermal layering of air that causes light traveling toward the road to gradually bend upward. This makes the light appear to be coming from a reflection in a pool, as shown in Figure 18-5b. Figure 18-5c shows how this occurs. As light from a distant object travels downward toward the road, the index of refraction of the air decreases as the air gets hotter, but the temperature change is gradual. Recall from Chapter 16 that light wave fronts are comprised of Huygens\u2019 wavelets. In the case of a mirage, the Huygens\u2019 wavelets closer to the ground travel faster than those higher up, causing the wave fronts to gradually turn upward. A similar phenomenon, called a superior mirage, occurs when a reflection of a distant boat appears above the boat. The water keeps the air that is closer to its surface cooler. \u25a0 Figure 18-5 A mirage is seen on the surface of a road (a). Light from the car bends upward into the eye of the observer (b). The bottom of the wave front moves faster than the top (c). a b Cooler air; larger n Hotter air; smaller n c Slower Ray Wave fronts Wavelets Faster 490 John M. Dunay IV/Fundamental Photographs Chapter 18 Refraction and Lenses 484-492 CH18-S1-845813 6/4/04 1:05 PM Page 491 Dispersion of Light The speed of light in a medium is determined by interactions between the light and the atoms that make up the medium. Recall from Chapters 12 and 13 that temperature and pressure are related to the energy of particles on the atomic level. The speed of light, and therefore, the index of refraction for a gaseous medium, can change slightly with temperature. In addition, the speed of light and the index of refraction vary for different wavelengths of light in the same liquid or solid medium. You learned in Chapter 16 that white light separates into a spectrum of colors when it passes through a glass", " prism, as shown in Figure 18-6a. This phenomenon is called dispersion. If you look carefully at the light that passes through a prism, you will notice that violet is refracted more than red, as shown in Figure 18-6b. This occurs because the speed of violet light through glass is less than the speed of red light through glass. Violet light has a higher frequency than red light, which causes it to interact differently with the atoms of the glass. This results in glass having a slightly higher index of refraction for violet light than it has for red light. Rainbows A prism is not the only means of dispersing light. A rainbow is a spectrum formed when sunlight is dispersed by water droplets in the atmosphere. Sunlight that falls on a water droplet is refracted. Because of dispersion, each color is refracted at a slightly different angle, as shown in Figure 18-7a. At the back surface of the droplet, some of the light undergoes internal reflection. On the way out of the droplet, the light once again is refracted and dispersed. Although each droplet produces a complete spectrum, an observer positioned between the Sun and the rain will see only a certain wavelength of light from each droplet. The wavelength depends on the relative positions of the Sun, the droplet, and the observer, as shown in Figure 18-7b. Because there are many droplets in the sky, a complete spectrum is visible. The droplets reflecting red light make an angle of 42\u00b0 in relation to the direction of the Sun\u2019s rays; the droplets reflecting blue light make an angle of 40\u00b0. a b White Red Orange Yellow Green Blue Violet \u25a0 Figure 18-6 White light directed through a prism is dispersed into bands of different colors (a). Different colors of light bend different amounts when they enter a medium (b). a b Primary rainbow Refraction Incident ray Sunlight Water droplet Reflection Higher raindrop Mid-level raindrop Lower raindrop \u25a0 Figure 18-7 Rainbows form because white light is dispersed as it enters, reflects at the inside boundary, and exits the raindrops (a). Because of dispersion, only one color from each raindrop reaches an observer (b). (Illustration not to scale) Section 18.1 Refraction of Light 491 David Parker/Photo Researchers 484-492 CH18-S1-845813 6/4/04 1:08 PM Page 492 \u25a0 Figure 18-8 A mist across your view allows for light comprising the entire spectrum of colors to reach your eyes in the form of a rainbow. Reflection from the raindrops sometimes enables you to see a second rainbow with the colors reversed. Sometimes, you can see a faint second-order rainbow like the one shown in Figure 18-8. The second rainbow is outside of the first, is fainter, and has the order of the colors reversed. Light rays that are reflected twice inside water droplets produce this effect. Very rarely, a third rainbow is visible outside the second. What is your prediction about how many times light is reflected in the water droplets and the order of appearance of the colors for the third rainbow? 18.1 Section Review 6. Index of Refraction You notice that when a light ray enters a certain liquid from water, it is bent toward the normal, but when it enters the same liquid from crown glass, it is bent away from the normal. What can you conclude about the liquid\u2019s index of refraction? 7. Index of Refraction A ray of light has an angle of incidence of 30.0\u00b0 on a block of unknown material and an angle of refraction of 20.0\u00b0. What is the index of refraction of the material? 8. Speed of Light Could an index of refraction ever be less than 1? What would this imply about the speed of light in that medium? 10. Total Internal Reflection If you were to use quartz and crown glass to make an optical fiber, which would you use for the cladding layer? Why? 11. Angle of Refraction A beam of light passes from water into polyethylene with n 1.50. If 57.5\u00b0, i what is the angle of refraction in the polyethylene? 12. Critical Angle Is there a critical angle for light traveling from glass to water? From water to glass? 13. Dispersion Why can you see the image of the Sun just above the horizon when the Sun itself has already set? 9. Speed of Light What is the speed of light in chlo- roform (n 1.51)? 14. Critical Thinking In what direction can you see a rainbow on a rainy late afternoon? Explain. 492 Gloria H. Chomica/Masterfile Chapter 18 Refraction and Lenses physicspp.com/self_check_quiz 493-499 CH18-S2-845813 7/21/04 3:02 PM Page 493 18.2 Conve", "x and Concave Lenses The refraction of light in nature that forms rainbows and red lunar eclipses is beautiful, but refraction also is useful. In 1303, French physician Bernard of Gordon wrote of the use of lenses to correct eyesight. Around 1610, Galileo used two lenses to make a telescope, with which he discovered the moons of Jupiter. Since Galileo\u2019s time, lenses have been used in many instruments, such as microscopes and cameras. Lenses are probably the most useful of all optical devices. Types of Lenses A lens is a piece of transparent material, such as glass or plastic, that is used to focus light and form an image. Each of a lens\u2019s two faces might be either curved or flat. The lens in Figure 18-9a is called a convex lens because it is thicker at the center than at the edges. A convex lens often is called a converging lens because when surrounded by material with a lower index of refraction it refracts parallel light rays so that the rays meet at a point. The lens in Figure 18-9b is called a concave lens because it is thinner in the middle than at the edges. A concave lens often is called a diverging lens because when surrounded by material with a lower index of refraction rays passing through it spread out. When light passes through a lens, refraction occurs at the two lens surfaces. Using Snell\u2019s law and geometry, you can predict the paths of rays passing through lenses. To simplify such problems, assume that all refraction occurs on a plane, called the principal plane, that passes through the center of the lens. This approximation, called the thin lens model, applies to all the lenses that you will learn about in this chapter section. Lens equations The problems that you will solve involve spherical thin lenses, lenses that have faces with the same curvature as a sphere. Based on the thin lens model, as well as the other simplifications used in solving problems for spherical mirrors, equations have been developed that look exactly like the equations for spherical mirrors. The thin lens equation relates the focal length of a spherical thin lens to the object position and the image position. 1 1 Thin Lens Equation 1 d d f o i The inverse of the focal length of a spherical lens is equal to the sum of the inverses of the image position and the object position. The magnification equation for spherical mirrors used in Chapter 17 also can be used for spherical thin lenses. It is used to determine the height and orientation of the image formed by a spherical thin lens. Objectives \u2022 Describe how real and virtual images are formed by single convex and concave lenses. \u2022 Locate images formed by lenses using ray tracing and equations. \u2022 Explain how chromatic aberration can be reduced. Vocabulary lens convex lens concave lens thin lens equation chromatic aberration achromatic lens a b di h i Magnification m do h o The magnification of an object by a spherical lens, defined as the image height divided by the object height, is equal to the negative of the image position divided by the object position. \u25a0 Figure 18-9 A convex lens causes rays of light to converge (a). A concave lens causes rays of light to diverge (b). Section 18.2 Convex and Concave Lenses 493 Horizons Companies 493-499 CH18-S2-845813 6/4/04 1:13 PM Page 494 Table 18-2 Properties of a Single Spherical Lens System Lens Type f do di m Image 2f do 2f di f Convex 2f do f 2f di f do 0 \uf8e6 do \uf8e6di (negative) Reduced Inverted Enlarged Inverted Real Real Enlarged Virtual Concave 0 do \uf8e6f\uf8e6 \uf8e6di \uf8e6 0 (negative) Reduced Virtual Using the equations for lenses It is important that you use the proper sign conventions when using these equations. Table 18-2 shows a comparison of the image position, magnification, and type of image formed by single convex and concave lenses when an object is placed at various object positions, do, relative to the lens. Notice the similarity of this table to Table 17-1 for mirrors. As with mirrors, the distance from the principal plane of a lens to its focal point is the focal length, f. The focal length depends upon the shape of the lens and the index of refraction of the lens material. Focal lengths and image positions can be negative. For lenses, virtual images are always on the same side of the lens as the object, which means that the image position is negative. When the absolute value of a magnification is between zero and one, the image is smaller than the object. Magnifications with absolute values greater than one represent images that are larger than the objects. A negative magnification means the image is inverted compared to the object. Notice that a concave lens produces only virtual images,", " whereas a convex lens can produce real images or virtual images. Convex Lenses and Real Images As shown in Figure 18-10a, paper can be ignited by producing a real image of the Sun on the paper. Recall from Chapter 17 that the rays of the Sun are almost exactly parallel when they reach Earth. After being refracted by the lens, the rays converge at the focal point, F, of the lens. Figure 1810b shows two focal points, one on each side of the lens. You could turn the lens around, and it will work the same. a b Principal plane Focal point Focal point F Principal axis \u25a0 Figure 18-10 A converging lens can be used to ignite paper (a). Light entering parallel to the principal axis converges at the focal point of the lens, concentrating solar energy (b). 494 David Young-Wolff/PhotoEdit Chapter 18 Refraction and Lenses 493-499 CH18-S2-845813 3/18/04 3:08 PM Page 495 Object Ray 1 Ray 2 2F F F Image 2F 30 cm do 15 cm di Ray diagrams In Figure 18-11, rays are traced from an object located far from a convex lens. For the purpose of locating the image, you only need to use two rays. Ray 1 is parallel to the principal axis. It refracts and passes through F on the other side of the lens. Ray 2 passes through F on its way to the lens. After refraction, its path is parallel to the principal axis. The two rays intersect at a point beyond F and locate the image. Rays selected from other points on the object converge at corresponding points to form the complete image. Note that this is a real image that is inverted and smaller compared to the object. You can use Figure 18-11 to locate the image of an object that is closer to the lens than the object in the figure. If a refracted ray is reversed in direction, it will follow its original path in the reverse direction. This means that the image and object may be interchanged by changing the direction of the rays. Imagine that the path of light through the lens in Figure 18-11 is reversed and the object is at a distance of 15 cm from the right side of the lens. The new image, located 30 cm from the left side of the lens, is a real image that is inverted and larger compared to the object. If the object is placed at twice the focal length from the lens at the point 2F, as shown in Figure 18-12, the image also is found at 2F. Because of symmetry, the image and object have the same size. Thus, you can conclude that if an object is more than twice the focal length from the lens, the image is smaller than the object. If the object is between F and 2F, then the image is larger than the object. \u25a0 Figure 18-12 When an object is placed at a distance equal to twice the focal length from the lens, the image is the same size as the object. Object Ray 1 Ray 2 2F F F 2F Image \u25a0 Figure 18-11 When an object is placed at a distance greater than twice the focal length of the lens, a real image is produced that is inverted and smaller compared to the object. If the object is placed at the location of the image, you could locate the new image by tracing the same rays in the opposite direction. Lens Masking Effects What happens when you mask, or cover, part of a lens? Does this cause only part of a real image to be formed by the lens? 1. Stick the edge of a convex lens into a ball of clay and place the lens on a tabletop. CAUTION: Lenses have sharp edges. Handle carefully. 2. Use a small lamp on one side and a screen on the other side to get a sharp image of the lamp\u2019s lightbulb. CAUTION: Lamps get hot and can burn skin. 3. Predict what will happen to the image if you place your hand over the top half of the lens. This is called masking. 4. Observe the effects of masking more of the lens and masking less of the lens. Analyze and Conclude 5. How much of the lens is needed for a complete image? 6. What is the effect of masking the lens? Section 18.2 Convex and Concave Lenses 495 493-499 CH18-S2-845813 6/4/04 1:15 PM Page 496 An Image Formed by a Convex Lens An object is placed 32.0 cm from a convex lens that has a focal length of 8.0 cm. a. Where is the image? b. c. What is the orientation of the image? If the object is 3.0 cm high, how tall is the image? 1 2 Analyze and Sketch the Problem \u2022 Sketch the situation, locating the object", " and the lens. \u2022 Draw the two principal rays. Known: 32.0 cm do 3.0 cm ho f 8.0 cm Unknown:? di? hi Solve for the Unknown a. Use the thin lens equation to determine di. 1 1 1 d d f o i fd o f do ) Substitute f 8.0 cm, do )( m c.0 32 m c 0. (8 3 m c.0 8 m c 0. 2 di Object Ray 1 di ho Ray 2 2F F do F 2F hi Image 11 cm (11 cm away from the lens on the side opposite the object) b. Use the magnification equation and solve for image height. 32.0 cm hi h di i m h do o iho d d o (11.0 cm) Substitute di m c )( 3.0 2 3 m c 1.0 cm (1.0 cm tall) 11 cm, ho 3.0 cm, do 32.0 cm Math Handbook Operations with Significant Digits pages 835\u2013836 c. The negative sign in part b means that the image is inverted. 3 Evaluate the Answer \u2022 Are the units correct? All are in centimeters. \u2022 Do the signs make sense? Image position is positive (real image) and image height is negative (inverted compared to the object), which make sense for a convex lens. 15. A 2.25-cm-tall object is 8.5 cm to the left of a convex lens of 5.5-cm focal length. Find the image position and height. 16. An object near a convex lens produces a 1.8-cm-tall real image that is 10.4 cm from the lens and inverted. If the focal length of the lens is 6.8 cm, what are the object position and height? 17. An object is placed to the left of a convex lens with a 25-mm focal length so that its image is the same size as the object. What are the image and object positions? 18. Use a scale ray diagram to find the image position of an object that is 30 cm to the left of a convex lens with a 10-cm focal length. 19. Calculate the image position and height of a 2.0-cm-tall object located 25 cm from a convex lens with a focal length of 5.0 cm. What is the orientation of the image? 496 Chapter 18 Refraction and Lenses 493-499 CH18-S2-845813 6/4/04 1:16 PM Page 497 \u25a0 Figure 18-13 The two principal rays show that a convex lens forms a virtual image that is upright and larger compared to the object when the object is located between the lens and the focal point. Because the principal rays are simply part of a model to help locate an image, they do not have to pass through the picture of the lens in a diagram. In reality, the image is formed only by the light that passes through the actual lens. Ray 2 F Ray 1 Object do F Virtual image di Convex Lenses and Virtual Images When an object is placed at the focal point of a convex lens, the refracted rays will emerge in a parallel beam and no image will be seen. When the object is brought closer to the lens, the rays will diverge on the opposite side of the lens, and the rays will appear to an observer to come from a spot on the same side of the lens as the object. This is a virtual image that is upright and larger compared to the object. Figure 18-13 shows how a convex lens forms a virtual image. The object is located between F and the lens. Ray 1, as usual, approaches the lens parallel to the principal axis and is refracted through the focal point, F. Ray 2 travels from the tip of the object, in the direction it would have if it had started at F on the object side of the lens. The dashed line from F to the object shows you how to draw ray 2. Ray 2 leaves the lens parallel to the principal axis. Rays 1 and 2 diverge as they leave the lens. Thus, no real image is possible. Drawing sight lines for the two rays back to their apparent intersection locates the virtual image. It is on the same side of the lens as the object, and it is upright and larger compared to the object. Note that the actual image is formed by light that passes through the lens, but you can still determine the location of the image by drawing rays that do not have to pass through the lens. 20. A newspaper is held 6.0 cm from a convex lens of 20.0-cm focal length. Find the image position of the newsprint image. 21. A magnifying glass has a focal length of 12.0 cm. A coin, 2.0 cm in diameter, is placed 3.4 cm from the lens. Locate the image of the coin. What", " is the diameter of the image? 22. A convex lens with a focal length of 22.0 cm is used to view a 15.0-cm-long pencil located 10.0 cm away. Find the height and orientation of the image. 23. A stamp collector wants to magnify a stamp by 4.0 when the stamp is 3.5 cm from the lens. What focal length is needed for the lens? 24. A magnifier with a focal length of 30 cm is used to view a 1-cm-tall object. Use ray tracing to determine the location and size of the image when the magnifier is positioned 10 cm from the object. Section 18.2 Convex and Concave Lenses 497 493-499 CH18-S2-845813 6/4/04 1:19 PM Page 498 \u25a0 Figure 18-14 Concave lenses produce only virtual images that are upright and smaller compared to their objects. Object Ray 1 Ray 2 do F Virtual image di F Concave Lenses A concave lens causes all rays to diverge. Figure 18-14 shows how such a lens forms a virtual image. Ray 1 approaches the lens parallel to the principal axis. It leaves the lens along a line that extends back through the focal point. Ray 2 approaches the lens as if it is going to pass through the focal point on the opposite side, and leaves the lens parallel to the principal axis. The sight lines of rays 1 and 2 intersect on the same side of the lens as the object. Because the rays diverge, they produce a virtual image. The image is located at the point from where the two rays apparently diverge. The image also is upright and smaller compared to the object. This is true no matter how far from the lens the object is located. The focal length of a concave lens is negative. When solving problems for concave lenses using the thin lens equation, you should remember that the sign convention for focal length is different from that of convex lenses. If the focal point for a concave lens is 24 cm from the lens, you should use the value f 24 cm in the thin lens equation. All images for a concave lens are virtual. Thus, if an image distance is 20 cm. The object given as 20 cm from the lens, then you should use di position always will be positive. Defects of Spherical Lenses Throughout this section, you have studied lenses that produce perfect images at specific positions. In reality, spherical lenses, just like spherical mirrors, have intrinsic defects that cause problems with the focus and color of images. Spherical lenses exhibit an aberration associated with their spherical design, just as mirrors do. In addition, the dispersion of light through a spherical lens causes an aberration that mirrors do not exhibit. Spherical aberration The model that you have used for drawing rays through spherical lenses suggests that all parallel rays focus at the same position. However, this is only an approximation. In reality, parallel rays that pass through the edges of a spherical lens focus at positions different from those of parallel rays that pass through the center. This inability of a spherical lens to focus all parallel rays to a single point is called spherical aberration. Making lens surfaces aspherical, such as in cameras, eliminates spherical aberration. In high-precision instruments, many lenses, often five or more, are used to form sharp, well-defined images. 498 Chapter 18 Refraction and Lenses 493-499 CH18-S2-845813 6/4/04 1:22 PM Page 499 Chromatic aberration Lenses have a second defect that mirrors do not have. A lens is like a prism, so different wavelengths of light are refracted at slightly different angles, as you can see in Figure 18-15a. Thus, the light that passes through a lens, especially near the edges, is slightly dispersed. An object viewed through a lens appears to be ringed with color. This effect is called chromatic aberration. The term chromatic comes from the Greek word chromo, which means \u201ccolor.\u201d White light White light a Lens b Achromatic lens Chromatic aberration is always present when a single lens is used. However, this defect can be greatly reduced by an achromatic lens, which is a system of two or more lenses, such as a convex lens with a concave lens, that have different indices of refraction. Such a combination of lenses is shown in Figure 18-15b. Both lenses in the figure disperse light, but the dispersion caused by the convex lens is almost canceled by the dispersion caused by the concave lens. The index of refraction of the convex lens is chosen so that the combination of lenses still converges the light. \u25a0 Figure 18-15 All simple lenses have chromatic aberration, in which light of different wavelengths is focused at different points (a). An achromatic lens is a combination of lenses, which minim", "izes the chromatic defect (b). 18.2 Section Review 25. Magnification Magnifying glasses normally are used to produce images that are larger than the related objects, but they also can produce images that are smaller than the related objects. Explain. 26. Image Position and Height A 3.0-cm-tall object is located 2.0 cm from a convex lens having a focal length of 6.0 cm. Draw a ray diagram to determine the location and size of the image. Use the thin lens equation and the magnification equation to verify your answer. 27. Types of Lenses The cross sections of four different thin lenses are shown in Figure 18-16. a. Which of these lenses, if any, are convex, or con- verging, lenses? b. Which of these lenses, if any, are concave, or diverging, lenses? 28. Chromatic Aberration All simple lenses have chromatic aberration. Explain, then, why you do not see this effect when you look through a microscope. 29. Chromatic Aberration You shine white light through a convex lens onto a screen and adjust the distance of the screen from the lens to focus the red light. Which direction should you move the screen to focus the blue light? 30. Critical Thinking An air lens constructed of two watch glasses is placed in a tank of water. Copy Figure 18-17 and draw the effect of this lens on parallel light rays incident on the lens. Water Light rays Air \u25a0 Figure 18-16 Water \u25a0 Figure 18-17 physicspp.com/self_check_quiz Section 18.2 Convex and Concave Lenses 499 500-503 CH18-S3-845813 3/18/04 3:15 PM Page 500 18.3 Applications of Lenses Objectives \u2022 Describe how the eye focuses light to form an image. \u2022 Explain nearsightedness and farsightedness and how eyeglass lenses correct these defects. \u2022 Describe the optical systems in some common optical instruments. Vocabulary nearsightedness farsightedness Biology Connection Biology Connection The properties that you have learned for the refraction of light through lenses are used in almost every optical instrument. In many cases, a combination of lenses and mirrors is used to produce clear images of small or faraway objects. Telescopes, binoculars, cameras, microscopes, and even your eyes contain lenses. Lenses in Eyes The eye is a remarkable optical device. As shown in Figure 18-18, the eye is a fluid-filled, almost spherical vessel. Light that is emitted or reflected off an object travels into the eye through the cornea. The light then passes through the lens and focuses onto the retina that is at the back of the eye. Specialized cells on the retina absorb this light and send information about the image along the optic nerve to the brain. Focusing images Because of its name, you might assume that the lens of an eye is responsible for focusing light onto the retina. In fact, light entering the eye is primarily focused by the cornea because the air-cornea surface has the greatest difference in indices of refraction. The lens is responsible for the fine focus that allows you to clearly see both distant and nearby objects. Using a process called accommodation, muscles surrounding the lens can contract or relax, thereby changing the shape of the lens. This, in turn, changes the focal length of the eye. When the muscles are relaxed, the image of distant objects is focused on the retina. When the muscles contract, the focal length is shortened, and this allows images of closer objects to be focused on the retina. Axons connecting to optic nerve Vitreous fluid Lens Cornea Iris Pupil Aqueous fluid Ciliary muscle Rods Cones Retina Optic nerve \u25a0 Figure 18-18 The human eye is complex and has many components that must work together. 500 Chapter 18 Refraction and Lenses 500-503 CH18-S3-845813 6/4/04 1:27 PM Page 501 a c b d Nearsightedness and farsightedness The eyes of many people do not focus sharp images on the retina. Instead, images are focused either in front of the retina or behind it. External lenses, in the form of eyeglasses or contact lenses, are needed to adjust the focal length and move images to the retina. Figure 18-19a shows the condition of nearsightedness, or myopia, whereby the focal length of the eye is too short to focus light on the retina. Images are formed in front of the retina. As shown in Figure 18-19b, concave lenses correct this by diverging light, thereby increasing images\u2019 distances from the lens, and forming images on the retina. You also can see in Figure 18-19c that farsightedness, or hyperopia, is the condition in which the focal length of the eye is too long. Images are therefore formed past the retina. A", " similar result is caused by the increasing rigidity of the lenses in the eyes of people who are more than about 45 years old. Their muscles cannot shorten the focal length enough to focus images of close objects on the retina. For either defect, convex lenses produce virtual images farther from the eye than the associated objects, as shown in Figure 18-19d. The images then become the objects for the eye lens and can be focused on the retina, thereby correcting the defect. \u25a0 Figure 18-19 A nearsighted person cannot see distant objects clearly because images are focused in front of the retina (a). A concave lens corrects this defect (c). A farsighted person cannot see close objects clearly because images are focused behind the retina (b). A convex lens corrects this defect (d). Contacts Contact lenses produce the same results as eyeglasses do. These small, thin lenses are placed directly on the corneas. A thin layer of tears between the cornea and lens keeps the lens in place. Most of the refraction occurs at the air-lens surface, where the difference in indices of refraction is greatest. As light enters the eye, it first encounters the air/cornea interface. Consider a ray of light that strikes the interface between the air and a person\u2019s cornea at an angle of 30.0\u00b0 to the normal. The index of refraction of the cornea is approximately 1.4. 1. Use Snell\u2019s law to calculate the angle of refraction. 2. What would the angle of refraction be if the person was swimming underwater? 3. Is the refraction greater in air or in water? Does this mean that objects under water seem closer or more distant than they would in air? 4. If you want the angle of refraction for the light ray in water to be the same as it is for air, what should the new angle of incidence be? Air Cornea 30.0\u00b0 2 Section 18.3 Applications of Lenses 501 500-503 CH18-S3-845813 6/10/04 6:29 PM Page 502 \u25a0 Figure 18-20 An astronomical refracting telescope creates a virtual image that is inverted compared to the object. (Illustration not to scale) Objective lens Focal point of eyepiece Focal point of objective lens Image of eyepiece Eyepiece Image of objective lens Refracting Telescopes An astronomical refracting telescope uses lenses to magnify distant objects. Figure 18-20 shows the optical system for a Keplerian telescope. Light from stars and other astronomical objects is so far away that the rays can be considered parallel. The parallel rays of light enter the objective convex lens and are focused as a real image at the focal point of the objective lens. The image is inverted compared to the object. This image then becomes the object for the convex lens of the eyepiece. Notice that the eyepiece lens is positioned so that the focal point of the objective lens is between the eyepiece lens and its focal point. This means that a virtual image is produced that is upright and larger than the first image. However, because the first image was already inverted, the final image is still inverted. For viewing astronomical objects, an image that is inverted is acceptable. In a telescope, the convex lens of the eyepiece is almost always an achromatic lens. Recall that an achromatic lens is a combination of lenses that function as one lens. The combination of lenses eliminates the peripheral colors, or chromatic aberration, that can form on images. \u25a0 Figure 18-21 Binoculars are like two side-by-side refracting telescopes. Thumbscrew Eyepiece Prisms Objective lens 502 Chapter 18 Refraction and Lenses Binoculars Binoculars, like telescopes, produce magnified images of faraway objects. Figure 18-21 shows a typical binocular design. Each side of the binoculars is like a small telescope: light enters a convex objective lens, which inverts the image. The light then travels through two prisms that use total internal reflection to invert the image again, so that the viewer sees an image that is upright compared to the object. The prisms also extend the path along which the light travels and direct it toward the eyepiece of the binoculars. Just as the separation of your two eyes gives you a sense of three dimensions and depth, the prisms allow a greater separation of the objective lenses, thereby improving the three-dimensional view of a distant object. 500-503 CH18-S3-845813 6/4/04 1:30 PM Page 503 a Aperture b \u25a0 Figure 18-22 The single-lens reflex camera shown here can divert the image formed by the lens through a prism for viewing (a) or directly to the film (b). Mirror Film Lens Shutter closed Shutter open Cameras Figure 18-22a shows the optical", " system used in a single-lens reflex camera. As light enters the camera, it passes through an achromatic lens. This lens system refracts the light much like a single convex lens would, forming an image that is inverted on the reflex mirror. The image is reflected upward to a prism that inverts and redirects the light to the viewfinder. When the person holding the camera takes a photograph, he or she presses the shutter-release button, which briefly raises the mirror, as shown in Figure 18-22b. The light, instead of being diverted upward to the prism, then travels along a straight path to form an image on the film. Eyepiece \u25a0 Figure 18-23 The objective lens and the eyepiece in this simple microscope produce an image that is inverted and larger compared to the object. Microscopes Like a telescope, a microscope has both an objective convex lens and a convex eyepiece. However, microscopes are used to view small objects. Figure 18-23 shows the optical system used in a simple compound microscope. The object is located between one and two focal lengths from the objective lens. A real image is produced that is inverted and larger than the object. As with a telescope, this image then becomes the object for the eyepiece. This image is between the eyepiece and its focal point. A virtual image is produced that is upright and larger than the image of the objective lens. Thus, the viewer sees an image that is inverted and greatly larger than the original object. 18.3 Section Review Objective lenses Sample on slide Light lens Lamp 31. Refraction Explain why the cornea is the primary 34. Image Why is the image that you observe in a focusing element in the eye. refracting telescope inverted? 32. Lens Types Which type of lens, convex or concave, should a nearsighted person use? Which type should a farsighted person use? 33. Focal Length Suppose your camera is focused on a person who is 2 m away. You now want to focus it on a tree that is farther away. Should you move the lens closer to the film or farther away? 35. Prisms What are three benefits of having prisms in binoculars? 36. Critical Thinking When you use the highest magnification on a microscope, the image is much darker than it is at lower magnifications. What are some possible reasons for the darker image? What could you do to obtain a brighter image? physicspp.com/self_check_quiz Section 18.3 Applications of Lenses 503 504-505 CH18-LAB-845813 6/15/04 11:11 PM Page 504 Convex Lenses and Focal Length The thin lens equation states that the inverse of the focal length is equal to the sum of the inverses of the image position from the lens and the object position from the lens. QUESTION How is the image position with a thin convex lens related to the object position and the focal length? Objectives Materials \u25a0 Make and use graphs to describe the relationship between the image position with a thin convex lens and the object position. \u25a0 Use models to show that no matter the image position, the focal length is a constant. 25-W straight-line filament bulb lamp base thin convex lens meterstick lens holder index card Safety Precautions \u25a0 Ensure the lamp is turned off before plugging and unplugging it from the electrical outlet. \u25a0 Use caution when handling lamps. They get hot and can burn the skin. \u25a0 Lenses have sharp edges. Handle carefully. Procedure 1. Place a meterstick on your lab table so that it is balancing on the thin side and the metric numbers are right side up. 2. Place a convex lens in a lens holder and set it on the meterstick on or between the 10-cm and 40-cm marks on the meterstick. (Distances will vary depending on the focal length of the lens used.) 3. Turn on the lamp and set it next to the meterstick so that the center of the lightbulb is even with the 0-cm end of the meterstick. 4. Hold an index card so that the lens is between the lamp and the index card. 5. Move the index card back and forth until an upside-down image of the lightbulb is as sharp as possible. 6. Record the distance of the lightbulb from the lens (do) and the distance of the image from the lens (di). 7. Move the lens to another spot between 10 cm and 40 cm and repeat steps 5 and 6. (Distances will vary depending on the focal length of the lens used.) 8. Repeat step 7 three more times. 504 504-505 CH18-LAB-845813 6/10/04 8:52 PM Page 505 Data Table Trial do (cm) di (cm) 1 2 3 4 5 Calculation Table Trial 1 (cm1) d o 1 (cm1) d", " i 1 1 (cm1) d d o i f (cm) 1 2 3 4 5 Analyze Going Further 1. Make and Use Graphs Make a scatter-plot graph of the image position (vertical axis) versus the object position (horizontal axis). Use a computer or calculator to construct the graph if possible. 2. Use Numbers Calculate 1/do and 1/di and enter the values in the calculation table. 3. Use Numbers Calculate the sum of 1/do and 1/di and enter the values in the calculation table. Calculate the reciprocal of this number and enter it in the calculation table as f. Conclude and Apply 1. Interpret Data Looking at the graph, describe the relationship between do and di. 2. Interpret Data Find out the actual focal length of the lens from your teacher. How accurate are your calculations of f? 3. Interpret Data Compare the results of your focal length calculations of the five trials. Are your results similar? 4. Lab Techniques Why do you suppose you were instructed not to hold your lens closer than 10 cm or farther than 40 cm? 1. Which measurement is more precise: do or di? Why do you think so? 2. What can you do to make either (or both) measurement(s) more accurate? Real-World Physics 1. If you were to take a picture with a camera, first of a distant scene, then of an object less than a meter away, how should the distance between the lens and film be changed? 2. What are two ways in which the image projected onto your retina differs from the object you look at? (Remember the lens in your eye is also convex.) To find out more about lenses and refraction, visit the Web site: physicspp.com 505 506 CH18-FEATURE-845813 6/4/04 1:32 PM Page 506 Gravitational Lenses In 1979, astronomers at the Jodrell Bank Observatory in Great Britain discovered two quasars that were separated by only 7 arc seconds (seven 0.36th\u2019s of a degree). Measurements showed they should have been 500,000 light years apart. The two quasars seemed to fluctuate in brightness and in rhythm with each other. The most amazing thing, however, was that the quasars had identical spectra. There appeared to be two different objects, but the two objects were the same. In 1919, comparison of starlight before and during a solar eclipse proved Einstein\u2019s theory to be true. In 1936, Einstein proposed the phenomenon of the gravitational lens. Because light can be bent by the gravitational fields of massive objects, virtual images of rings should be seen by observers on Earth when a massive object is between Earth and the object being observed. Einstein never observed such a phenomenon, but his theory of relativity supported the possible existence of gravitational lenses. The illustration shows how light from a distant galaxy is bent around a galaxy cluster before reaching Earth. The blue shapes are multiple images of the same galaxy produced by gravitational lensing from galaxy cluster 0024+1654 in the center of the photo. Further work by astronomers around the world confirmed that there was just a single quasar, and that its light was being distorted by a cluster of galaxies dominated by a massive elliptical galaxy lying in the line of sight between the quasar and Earth. The astronomers realized that they were seeing two images of one quasar. The galaxy acted like an imperfect convex lens, focusing the deflected light in such a way that two images were formed from one object. Why would they think that the light was bent? Gravity and Light The astronomers remembered the work of Albert Einstein and his theory of relativity. Einstein proposed that light would be bent by the gravitational fields of massive objects. In the classical theory of space, known as Euclidean space, light travels in a straight line. According to Einstein, light bends when it comes near a massive object. 506 NASA Extreme Physics The Evidence As often occurs in science, once someone discovers something for the first time, many more supporting discoveries are made soon after. Since Einstein\u2019s proposals, and the discovery in 1979 of the double-image quasar, many more gravitational lenses have been observed. Both Einstein\u2019s rings and multiple images have been observed. Einstein\u2019s rings result when the gravitational lens and the light from the object are in near-perfect alignment. Multiple images are formed when the gravitational lens and the light from the object are not in perfect alignment. Over 50 gravitational lenses have been discovered. Going Further 1. Infer Why was the discovery of gravita- tional lenses important? 2. Compare and Contrast How are gravitational lenses similar to convex lenses? How are they different? 507-513 CH18-SG CA STP-845813 6/4/04 1:39 PM Page 507 18.1 Refraction of Light Vocabulary \u2022 index of refraction (p. 486) \u2022 Snell\u2019s law of refraction", " (p. 486) \u2022 critical angle (p. 489) \u2022 total internal reflection (p. 489) \u2022 dispersion (p. 491) Key Concepts \u2022 The path of travel of light bends when it passes from a medium with an index of refraction, n1, into a medium with a different index of refraction, n2. n1 sin 1 n2 sin 2 \u2022 The ratio of the speed of light in a vacuum, c, to the speed of light in any medium, v, is the index of refraction, n, of the medium. c n v \u2022 When light traveling through a medium hits a boundary of a medium with a smaller index of refraction, if the angle of incidence exceeds the critical angle, internal reflection. c, the light will be reflected back into the original medium by total sin c n 2 n 1 18.2 Convex and Concave Lenses Vocabulary \u2022 lens (p. 493) \u2022 convex lens (p. 493) \u2022 concave lens (p. 493) \u2022 thin lens equation (p. 493) \u2022 chromatic aberration (p. 499) Key Concepts \u2022 The focal length, f; the object position, do; and the image position, di, for a lens are related by the thin lens equation. 1 1 1 d d f i o \u2022 The magnification, m, of an image by a lens is defined in the same way as \u2022 achromatic lens (p. 499) the magnification of an image by a mirror. h di i m h do o \u2022 A single convex lens produces a real image that is inverted when the object position is greater than the focal length. The image is reduced or enlarged, depending on the object position. \u2022 A single convex lens produces a virtual image that is upright and larger than the object when the object is located between the lens and the focal point. \u2022 A single concave lens always produces a virtual image that is upright and smaller than the object. \u2022 All simple lenses have chromatic aberration. All lenses made with spherical surfaces have spherical aberration. 18.3 Applications of Lenses Vocabulary \u2022 nearsightedness (p. 501) \u2022 farsightedness (p. 501) Key Concepts \u2022 Differences in indices of refraction between air and the cornea are primarily responsible for focusing light in the eye. \u2022 Optical instruments use combinations of lenses to obtain clear images of small or distant objects. physicspp.com/vocabulary_puzzlemaker 507 507-513 CH18-SG CA STP-845813 3/18/04 3:30 PM Page 508 Concept Mapping 37. Complete the following concept map using the following terms: inverted, larger, smaller, virtual. Lenses 48. Describe how the eye focuses light. (18.3) 49. What is the condition in which the focal length of the eye is too short to focus light on the retina? (18.3) 50. What type of image is produced by the objective lens in a refracting telescope? (18.3) convex concave 51. The prisms in binoculars increase the distance real virtual between the objective lenses. Why is this useful? (18.3) 52. What is the purpose of a camera\u2019s reflex mirror? (18.3) upright upright Applying Concepts 53. Which substance, A or B, in Figure 18-24 has a larger index of refraction? Explain. unchanged size larger Mastering Concepts 38. How does the angle of incidence compare with the angle of refraction when a light ray passes from air into glass at a nonzero angle? (18.1) 39. How does the angle of incidence compare with the angle of refraction when a light ray leaves glass and enters air at a nonzero angle? (18.1) 40. Regarding refraction, what is the critical angle? (18.1) A B 41. Although the light coming from the Sun is refracted while passing through Earth\u2019s atmosphere, the light is not separated into its spectrum. What does this indicate about the speeds of different colors of light traveling through air? (18.1) 42. Explain why the Moon looks red during a lunar eclipse. (18.1) 43. How do the shapes of convex and concave lenses differ? (18.2) 44. Locate and describe the physical properties of the image produced by a convex lens when an object is placed some distance beyond 2F. (18.2) 45. What factor, other than the curvature of the surfaces of a lens, determines the location of the focal point of the lens? (18.2) 46. To project an image from a movie projector onto a screen, the film is placed between F and 2F of a converging lens. This arrangement produces an image that is inverted. Why does the filmed scene appear to be upright when the film is viewed", "? (18.2) 47. Describe why precision optical instruments use achromatic lenses. (18.2) \u25a0 Figure 18-24 54. A light ray strikes the boundary between two transparent media. What is the angle of incidence for which there is no refraction? 55. How does the speed of light change as the index of refraction increases? 56. How does the size of the critical angle change as the index of refraction increases? 57. Which pair of media, air and water or air and glass, has the smaller critical angle? 58. Cracked Windshield If you crack the windshield of your car, you will see a silvery line along the crack. The glass has separated at the crack, and there is air in the crack. The silvery line indicates that light is reflecting off the crack. Draw a ray diagram to explain why this occurs. What phenomenon does this illustrate? 59. Legendary Mirage According to legend, Eric the Red sailed from Iceland and discovered Greenland after he had seen the island in a mirage. Describe how the mirage might have occurred. 508 Chapter 18 Refraction and Lenses For more problems, go to Additional Problems, Appendix B. 507-513 CH18-SG CA STP-845813 6/4/04 1:43 PM Page 509 60. A prism bends violet light more than it bends red light. Explain. 61. Rainbows Why would you never see a rainbow in the southern sky if you were in the northern hemisphere? In which direction should you look to see rainbows if you are in the southern hemisphere? 62. Suppose that Figure 18-14 is redrawn with a lens of the same focal length but a larger diameter. Explain why the location of the image does not change. Would the image be affected in any way? 63. A swimmer uses a magnifying glass to observe a small object on the bottom of a swimming pool. She discovers that the magnifying glass does not magnify the object very well. Explain why the magnifying glass is not functioning as it would in air. 64. Why is there chromatic aberration for light that goes through a lens but not for light that reflects from a mirror? 65. When subjected to bright sunlight, the pupils of your eyes are smaller than when they are subjected to dimmer light. Explain why your eyes can focus better in bright light. 66. Binoculars The objective lenses in binoculars form real images that are upright compared to their objects. Where are the images located relative to the eyepiece lenses? Mastering Problems 18.1 Refraction of Light 67. A ray of light travels from air into a liquid, as shown in Figure 18-25. The ray is incident upon the liquid at an angle of 30.0\u00b0. The angle of refraction is 22.0\u00b0. a. Using Snell\u2019s law, calculate the index of refraction of the liquid. b. Compare the calculated index of refraction to those in Table 18-1. What might the liquid be? 30\u00b0 22\u00b0 Air Liquid \u25a0 Figure 18-25 68. Light travels from flint glass into ethanol. The angle of refraction in the ethanol is 25.0\u00b0. What is the angle of incidence in the glass? 69. A beam of light strikes the flat, glass side of a waterfilled aquarium at an angle of 40.0\u00b0 to the normal. For glass, n 1.50. a. At what angle does the beam enter the glass? b. At what angle does the beam enter the water? 70. Refer to Table 18-1. Use the index of refraction of diamond to calculate the speed of light in diamond. 71. Refer to Table 18-1. Find the critical angle for a diamond in air. 72. Aquarium Tank A thick sheet of plastic, n 1.500, is used as the side of an aquarium tank. Light reflected from a fish in the water has an angle of incidence of 35.0\u00b0. At what angle does the light enter the air? 73. Swimming-Pool Lights A light source is located 2.0 m below the surface of a swimming pool and 1.5 m from one edge of the pool, as shown in Figure 18-26. The pool is filled to the top with water. a. At what angle does the light reaching the edge of the pool leave the water? b. Does this cause the light viewed from this angle to appear deeper or shallower than it actually is? 2 m 1.5 m \u25a0 Figure 18-26 (Not to scale) 74. A diamond\u2019s index of refraction for red light, 656 nm, is 2.410, while that for blue light, 434 nm, is 2.450. Suppose that white light is incident on the diamond at 30.0\u00b0. Find the angles of refraction for red and blue light. 75. The index of refraction of crown glass is 1.53 for violet", " light, and it is 1.51 for red light. a. What is the speed of violet light in crown glass? b. What is the speed of red light in crown glass? physicspp.com/chapter_test Chapter 18 Assessment 509 507-513 CH18-SG CA STP-845813 6/10/04 8:53 PM Page 510 76. The critical angle for a special glass in air is 41.0\u00b0. What is the critical angle if the glass is immersed in water? 82. A converging lens has a focal length of 25.5 cm. If it is placed 72.5 cm from an object, at what distance from the lens will the image be? 77. A ray of light in a tank of water has an angle of incidence of 55.0\u00b0. What is the angle of refraction in air? 78. The ray of light shown in Figure 18-27 is incident upon a 60\u00b0-60\u00b0-60\u00b0 glass prism, n 1.5. a. Using Snell\u2019s law of refraction, determine the angle, 2, to the nearest degree. b. Using elementary geometry, determine the value of. 1. c. Determine 2 Air Air 1 45\u00b0 60\u00b0 P Q 2 1 2 R nglass 1.5 Glass 60\u00b0 60\u00b0 \u25a0 Figure 18-27 79. The speed of light in a clear plastic is 1.90108 m/s. A ray of light strikes the plastic at an angle of 22.0\u00b0. At what angle is the ray refracted? 80. A light ray enters a block of crown glass, as illustrated in Figure 18-28. Use a ray diagram to trace the path of the ray until it leaves the glass. 45\u00b0 \u25a0 Figure 18-28 18.2 Convex and Concave Lenses 81. The focal length of a convex lens is 17 cm. A candle is placed 34 cm in front of the lens. Make a ray diagram to locate the image. 83. If an object is 10.0 cm from a converging lens that has a focal length of 5.00 cm, how far from the lens will the image be? 84. A convex lens is needed to produce an image that is 0.75 times the size of the object and located 24 cm from the lens on the other side. What focal length should be specified? 85. An object is located 14.0 cm from a convex lens that has a focal length of 6.0 cm. The object is 2.4 cm high. a. Draw a ray diagram to determine the location, size, and orientation of the image. b. Solve the problem mathematically. 86. A 3.0-cm-tall object is placed 22 cm in front of a converging lens. A real image is formed 11 cm from the lens. What is the size of the image? 87. A 3.0-cm-tall object is placed 15.0 cm in front of a converging lens. A real image is formed 10.0 cm from the lens. a. What is the focal length of the lens? b. If the original lens is replaced with a lens having twice the focal length, what are the image position, size, and orientation? 88. A diverging lens has a focal length of 15.0 cm. An object placed near it forms a 2.0-cm-high image at a distance of 5.0 cm from the lens. a. What are the object position and object height? b. The diverging lens is now replaced by a converging lens with the same focal length. What are the image position, height, and orientation? Is it a virtual image or a real image? 18.3 Applications of Lenses 89. Camera Lenses Camera lenses are described in terms of their focal length. A 50.0-mm lens has a focal length of 50.0 mm. a. A camera with a 50.0-mm lens is focused on an object 3.0 m away. What is the image position? b. A 1000.0-mm lens is focused on an object 125 m away. What is the image position? 90. Eyeglasses To clearly read a book 25 cm away, a farsighted girl needs the image to be 45 cm from her eyes. What focal length is needed for the lenses in her eyeglasses? 91. Copy Machine The convex lens of a copy machine has a focal length of 25.0 cm. A letter to be copied is placed 40.0 cm from the lens. a. How far from the lens is the copy paper? b. How much larger will the copy be? 510 Chapter 18 Refraction and Lenses For more problems, go to Additional Problems, Appendix B. 507-513 CH18-SG CA STP-845813 3/18/04 3:32 PM Page 511 92. Camera A camera lens with a", " focal length of 35 mm is used to photograph a distant object. How far from the lens is the real image of the object? Explain. 93. Microscope A slide of an onion cell is placed 12 mm from the objective lens of a microscope. The focal length of the objective lens is 10.0 mm. a. How far from the lens is the image formed? b. What is the magnification of this image? c. The real image formed is located 10.0 mm beneath the eyepiece lens. If the focal length of the eyepiece is 20.0 mm, where does the final image appear? d. What is the final magnification of this compound system? 94. Telescope The optical system of a toy refracting telescope consists of a converging objective lens with a focal length of 20.0 cm, located 25.0 cm from a converging eyepiece lens with a focal length of 4.05 cm. The telescope is used to view a 10.0-cm-high object, located 425 cm from the objective lens. a. What are the image position, height, and orientation as formed by the objective lens? Is this a real or virtual image? b. The objective lens image becomes the object for the eyepiece lens. What are the image position, height, and orientation that a person sees when looking into the telescope? Is this a real or virtual image? c. What is the magnification of the telescope? Mixed Review 95. A block of glass has a critical angle of 45.0\u00b0. What is its index of refraction? 96. Find the speed of light in antimony trioxide if it has an index of refraction of 2.35. 97. A 3.0-cm-tall object is placed 20 cm in front of a converging lens. A real image is formed 10 cm from the lens. What is the focal length of the lens? 98. Derive n sin 1/sin 2 from the general form of Snell\u2019s law of refraction, n1 sin 1 any assumptions and restrictions. n2 sin 2. State 101. Apparent Depth Sunlight reflects diffusively off the bottom of an aquarium. Figure 18-29 shows two of the many light rays that would reflect diffusively from a point off the bottom of the tank and travel to the surface. The light rays refract into the air as shown. The red dashed line extending back from the refracted light ray is a sight line that intersects with the vertical ray at the location where an observer would see the image of the bottom of the tank. a. Compute the direction that the refracted ray will travel above the surface of the water. b. At what depth does the bottom of the tank appear to be if you look into the water? Divide this apparent depth into the true depth and compare it to the index of refraction. 2 5.0\u00b0 12 cm \u25a0 Figure 18-29 102. It is impossible to see through adjacent sides of a square block of glass with an index of refraction of 1.5. The side adjacent to the side that an observer is looking through acts as a mirror. Figure 18-30 shows the limiting case for the adjacent side to not act like a mirror. Use your knowledge of geometry and critical angles to show that this ray 1.5. configuration is not achievable when nglass nglass 1.5 1 90\u00b0 Air 2 99. Astronomy How many more minutes would it take light from the Sun to reach Earth if the space between them were filled with water rather than a vacuum? The Sun is 1.5108 km from Earth. 100. What is the focal length of the lenses in your eyes when you read a book that is 35.0 cm from them? The distance from each lens to the retina is 0.19 mm. 1 Air 2 90\u00b0 \u25a0 Figure 18-30 physicspp.com/chapter_test Chapter 18 Assessment 511 507-513 CH18-SG CA STP-845813 6/4/04 1:48 PM Page 512 103. Bank Teller Window A 25-mm-thick sheet of plastic, n 1.5, is used in a bank teller\u2019s window. A ray of light strikes the sheet at an angle of 45\u00b0. The ray leaves the sheet at 45\u00b0, but at a different location. Use a ray diagram to find the distance between the ray that leaves and the one that would have left if the plastic were not there. Thinking Critically 104. Recognize Spatial Relationships White light traveling through air (n 1.0003) enters a slab of glass, incident at exactly 45\u00b0. For dense flint glass, n 1.7708 for blue light ( 435.8 nm) and n 1.7273 for red light ( 643.8 nm). What is the angular dispersion of the red and blue light? 105. Compare and Contrast Find the critical angle for ice (n 1.31).", " In a very cold world, would fiberoptic cables made of ice or those made of glass do a better job of keeping light inside the cable? Explain. 106. Recognize Cause and Effect Your lab partner 25 cm and hi used a convex lens to produce an image with 4.0 cm. You are examining a di concave lens with a focal length of 15 cm. You place the concave lens between the convex lens and the original image, 10 cm from the image. To your surprise, you see a real image on the wall that is larger than the object. You are told that the image from the convex lens is now the object for the concave lens, and because it is on the opposite side of the concave lens, it is a virtual object. Use these hints to find the new image position and image height and to predict whether the concave lens changed the orientation of the original image. 107. Define Operationally Name and describe the effect that causes the rainbow-colored fringe commonly seen at the edges of a spot of white light from a slide or overhead projector. 108. Think Critically A lens is used to project the image of an object onto a screen. Suppose that you cover the right half of the lens. What will happen to the image? Writing in Physics 109. The process of accommodation, whereby muscles surrounding the lens in the eye contract or relax to enable the eye to focus on close or distant objects, varies for different species. Investigate this effect for different animals. Prepare a report for the class showing how this fine focusing is accomplished for different eye mechanisms. 110. Investigate the lens system used in an optical instrument such as an overhead projector or a particular camera or telescope. Prepare a graphics display for the class explaining how the instrument forms images. Cumulative Review 111. If you drop a 2.0 kg bag of lead shot from a height of 1.5 m, you could assume that half of the potential energy will be converted into thermal energy in the lead. The other half would go to thermal energy in the floor. How many times would you have to drop the bag to heat it by 10\u00b0C? (Chapter 12) 112. A blacksmith puts an iron hoop or tire on the outer rim of a wooden carriage wheel by heating the hoop so that it expands to a diameter greater than the wooden wheel. When the hoop cools, it contracts to hold the rim in place. If a blacksmith has a wooden wheel with a 1.0000-m diameter and wants to put a rim with a 0.9950-m diameter on the wheel, what is the minimum temperature change the iron 12106/\u00b0C) (Chapter 13) must experience? ( iron 113. A car sounds its horn as it approaches a pedestrian in a crosswalk. What does the pedestrian hear as the car brakes to allow him to cross the street? (Chapter 15) 114. Suppose that you could stand on the surface of the Sun and weigh yourself. Also suppose that you could measure the illuminance on your hand from the Sun\u2019s visible spectrum produced at that position. Next, imagine yourself traveling to a position 1000 times farther away from the center of the Sun as you were when standing on its surface. (Chapter 16) a. How would the force of gravity on you from the Sun at the new position compare to what it was at the surface? b. How would the illuminance on your hand from the Sun at the new position compare to what it was when you were standing on its surface? (For simplicity, assume that the Sun is a point source at both positions.) c. Compare the effect of distance upon the gravitational force and illuminance. 115. Beautician\u2019s Mirror The nose of a customer who is trying some face powder is 3.00-cm high and is located 6.00 cm in front of a concave mirror having a 14.0-cm focal length. Find the image position and height of the customer\u2019s nose by means of the following. (Chapter 17) a. a ray diagram drawn to scale b. the mirror and magnification equations 512 Chapter 18 Refraction and Lenses For more problems, go to Additional Problems, Appendix B. 507-513 CH18-SG CA STP-845813 6/4/04 1:49 PM Page 513 Multiple Choice 1. A flashlight beam is directed at a swimming pool in the dark at an angle of 46\u00b0 with respect to the normal to the surface of the water. What is the angle of refraction of the beam in the water? (The refractive index for water is 1.33.) 18\u00b0 30\u00b0 33\u00b0 44\u00b0 2. The speed of light in diamond is 1.24108 m/s. What is the index of refraction of diamond? 0.0422 0.413 1.24 2.42 3. Which one of the items below is not involved in the formation of rainbows? diffraction dispersion reflection refraction 4.", " George\u2019s picture is being taken by Cami, as shown in the figure, using a camera which has a convex lens with a focal length of 0.0470 m. Determine George\u2019s image position. 4.82 cm 20.7 cm 1.86 m 1.86 cm 4.70 cm 1.91 m 5. What is the magnification of an object that is 4.15 m in front of a camera that has an image position of 5.0 cm? 0.83 0.012 0.83 1.2 6. Which one of the items below is not involved in the formation of mirages? heating of air near the ground Huygens\u2019 wavelets reflection refraction 7. What is the image position for the situation shown in the figure? 6.00 m 1.20 m 0.167 m 0.833 m Object F 2.00 m F 3.00 m 8. What is the critical angle for total internal reflection when light travels from glass (n 1.52) to water (n 1.33)? 48.8\u00b0 61.0\u00b0 29.0\u00b0 41.2\u00b0 9. What happens to the image formed by a convex lens when half of the lens is covered? half of the image disappears the image dims the image gets blurry the image inverts Extended Answer 10. The critical angle for total internal reflection at a diamond-air boundary is 24.4\u00b0. What is the angle of refraction in the air if light is incident on the boundary at an angle of 20.0\u00b0? 11. An object that is 6.98 cm from a lens produces an image that is 2.95 cm from the lens on the same side of the lens. Determine the type of lens that is producing the image and explain how you know. Use as Much Time as You Can You will not get extra points for finishing a test early. Work slowly and carefully to prevent careless errors that can occur when you are hurrying to finish. physicspp.com/standardized_test Chapter 18 Standardized Test Practice 513 514-523 CH19-S1-845813 3/19/04 3:44 PM Page 514 What You\u2019ll Learn \u2022 You will learn how interference and diffraction patterns demonstrate that light behaves like a wave. \u2022 You will learn how interference and diffraction patterns occur in nature and how they are used. Why It\u2019s Important Interference and diffraction can be seen all around you. Compact discs demonstrate diffraction, bubbles show interference, and the wings of a Morpho butterfly show both. Bubble Solution Bubble solution in a container is transparent. However, if you suspend the solution in a grid of plastic, swirls of color can be seen. These colors are not caused by pigments or dyes in the soap, but by an effect of the wave nature of light. Think About This How does bubble solution produce a rainbow of colors? physicspp.com 514 Adrienne Hart-Davis/Science Photo Library/Photo Researchers 514-523 CH19-S1-845813 3/12/04 7:11 AM Page 515 Why does a compact disc reflect a rainbow of light? Question How is light affected when it reflects off a compact disc? Procedure 1. Obtain a compact disc (CD or DVD), a light projector, and color light filters from your teacher. 2. Lay the compact disc on a table with the reflective surface facing up. 3. Place a color filter on the light projector. 4. Turn the light on and shine it on the compact disc, causing the light to reflect off it onto a white surface. CAUTION: Do not look directly into the projector light. 5. Record your observations of the light on the screen. 6. Turn the light off and change to a different color filter. 7. Repeat steps 4\u20135 with the new color filter. 8. Repeat steps 4\u20135 with white light. Analysis Does the color of light affect the pattern? How is the reflection of white light different from that of singlecolor light? Critical Thinking Look closely at your observations of white light reflecting from the discs. Suggest possible sources for the bands of color. 19.1 Interference In Chapter 16, you learned that light sometimes acts like a wave. Light can be diffracted when it passes by an edge, just like water waves and sound waves can. In Chapters 17 and 18, you learned that reflection and refraction can be explained when light is modeled as a wave. What led scientists to believe that light has wave properties? They discovered that light could be made to interfere, which you will learn about in this section. When you look at objects that are illuminated by a white light source such as a nearby lightbulb, you are seeing incoherent light, which is light with unsynchronized wave fronts. The effect of incoherence in waves can be seen in the example of heavy rain falling on a swimming pool. The surface of the water", " is choppy and does not have any regular pattern of wave fronts or standing waves. Because light waves have such a high frequency, incoherent light does not appear choppy to you. Instead, as light from an incoherent white light source illuminates an object, you see the superposition of the incoherent light waves as an even, white light. Objectives \u2022 Explain how light falling on two slits produces an interference pattern. \u2022 Calculate light wavelengths from interference patterns. \u2022 Apply modeling techniques to thin-film interference. Vocabulary incoherent light coherent light interference fringes monochromatic light thin-film interference Section 19.1 Interference 515 Horizons Companies 514-523 CH19-S1-845813 6/4/04 3:36 PM Page 516 \u25a0 Figure 19-1 Smooth wave fronts of light are created by point sources (a) and by lasers (b). a Circular wave fronts Straight line wave fronts b Rays Rays \u25a0 Figure 19-2 These are doubleslit interference patterns for blue light (a), for red light (b), and for white light (c). a b Interference of Coherent Light The opposite of incoherent light is coherent light, which is light from two or more sources that add together in superposition to produce smooth wave fronts. A smooth wave front can be created by a point source, as shown in Figure 19-1a. A smooth wave front also can be created by multiple point sources when all point sources are synchronized, such as with a laser, as shown in Figure 19-1b. Only the superposition of light waves from coherent light sources can produce the interference phenomena that you will examine in this section. English physician Thomas Young proved that light has wave properties when he produced an interference pattern by shining light from a single coherent source through two slits. Young directed coherent light at two closely spaced, narrow slits in a barrier. When the overlapping light from the two slits fell on an observing screen, the overlap did not produce even illumination, but instead created a pattern of bright and dark bands that Young called interference fringes. He explained that these bands must be the result of constructive and destructive interference of light waves from the two slits in the barrier. c In a double-slit interference experiment that uses monochromatic light, which is light of only one wavelength, constructive interference produces a bright central band of the given color on the screen, as well as other bright bands of near-equal spacing and near-equal width on either side, as shown in Figures 19-2a and 19-2b. The intensity of the bright bands decreases the farther the band is from the central band, as you can easily see in Figure 19-2a. Between the bright bands are dark areas where destructive interference occurs. The positions of the constructive and destructive interference bands depend on the wavelength of the light. When white light is used in a double-slit experiment, however, interference causes the appearance of colored spectra instead of bright and dark bands, as shown in Figure 19-2c. All wavelengths interfere constructively in the central bright band, and thus that band is white. The positions of the other colored bands result from the overlap of the interference fringes that occur where wavelengths of each separate color interfere constructively. 516 Tom Pantages Chapter 19 Interference and Diffraction 514-523 CH19-S1-845813 6/4/04 3:36 PM Page 517 \u25a0 Figure 19-3 The coherent source that is created by a narrow single slit produces coherent, nearly cylindrical waves that travel to the two slits in the second barrier. Two coherent, nearly cylindrical waves leave the double slit. Double-slit interference To create coherent light from incoherent light, Young placed a light barrier with a narrow slit in front of a monochromatic light source. Because the width of the slit is very small, only a coherent portion of the light passes through and is diffracted by the slit, producing nearly cylindrical diffracted wave fronts, as shown in Figure 19-3. Because a cylinder is symmetrical, the two portions of the wave front arriving at the second barrier with two slits will be in phase. The two slits at the second barrier then produce coherent, nearly cylindrical wave fronts that can then interfere, as shown in Figure 19-3. Depending on their phase relationship, the two waves can undergo constructive or destructive interference, as shown in Figure 19-4. \u25a0 Figure 19-4 A pair of in-phase waves is created at the two slits. At some locations, the waves might undergo constructive interference to create a bright band (a), or destructive interference to create a dark band (b). a Slit 1 Slit 2 P b Slit 1 P Constructive interference Slit 2 Destructive interference Source 1 Source 2 Superposition Source 1 Source 2 Superposition Section 19.1 Interference 517 514-523 CH19-S1-8", "45813 7/22/04 2:41 AM Page 518 Destructive interference Constructive interference Source S1 S2 P1 x P0 Screen Dark Bright Dark Bright Dark Bright Dark S2 d Q S1 R L a b \u25a0 Figure 19-5 The interference of monochromatic light that passes through the double slit produces bright and dark bands on a screen (a). This diagram (b) represents an analysis of the first bright band. The distance from the slits to the screen, L, is about 105 times longer than the separation, d, between the two slits. (Illustrations not to scale) Measuring the wavelength of light A top view of nearly cylindrical wave fronts and Young\u2019s double slit experiment are shown in Figure 19-5a. The wave fronts interfere constructively and destructively to form a pattern of light and dark bands. A typical diagram that is used to analyze Young\u2019s experiment is shown in Figure 19-5b. Light that reaches point P0 travels the same distance from each slit. Because the waves are in phase, they interfere constructively on the screen to create the central bright band at P0. There is also constructive interference at the first bright band, P1, on either side of the central band, because line segment P1S1 is one wavelength,, longer than the line segment P1S2. Thus, the waves arrive at P1 in phase. There are two triangles shaded in the figure. The larger triangle is a right triangle, so tan x/L. In the smaller triangle RS1S2, the side S1R is the length difference of the two light paths, which is one wavelength. There are now two simplifications that make the problem easier to solve. 1. If L is much larger than d, then line segments S1P1 and S2P1 are nearly parallel to each other and to line segment QP1, and triangle RS1S2 is very nearly a right triangle. Thus, sin /d. 2. If the angle is small, then sin is very nearly equal to tan. With the above simplifications, the relationships tan x/L, sin /d, and sin tan combine to form the equation x/L /d. Solving for gives the following. d Wavelength from Double-Slit Experiment x L The wavelength of light, as measured by a double slit, is equal to the distance on the screen from the central bright band to the first bright band, multiplied by the distance between the slits, divided by the distance to the screen. Constructive interference from two slits occurs at locations, xm, on either side of the central bright band, which are determined using the equation m xmd/L, where m 0, 1, 2, etc., as limited by the small angle simplification. The central bright band occurs at m 0. Frequently, the band given by m 1 is called the first-order band, and so on. 518 Chapter 19 Interference and Diffraction 514-523 CH19-S1-845813 3/12/04 7:15 AM Page 519 Wavelength of Light A double-slit experiment is performed to measure the wavelength of red light. The slits are 0.0190 mm apart. A screen is placed 0.600 m away, and the first-order bright band is found to be 21.1 mm from the central bright band. What is the wavelength of the red light? 1 Analyze and Sketch the Problem \u2022 Sketch the experiment, showing the slits and the screen. \u2022 Draw the interfence pattern with bands in appropriate locations. d x Screen Slits L Math Handbook Operations with Scientific Notation pages 842\u2013843 Known: d 1.90105 m x 2.11102 m L 0.600 m Unknown:? 2 Solve for the Unknown d x L (2.11102 m)(1.90105 m) (0.600 m) 6.68107 nm 668 nm Substitute x 2.11102 m, d 1.90105 m, L 0.600 m 3 Evaluate the Answer \u2022 Are the units correct? The answer is in units of length, which is correct for wavelength. \u2022 Is the magnitude realistic? The wavelength of red light is near 700 nm, and that of blue is near 400 nm. Thus, the answer is reasonable for red light. 1. Violet light falls on two slits separated by 1.90105 m. A first-order bright band appears 13.2 mm from the central bright band on a screen 0.600 m from the slits. What is? 2. Yellow-orange light from a sodium lamp of wavelength 596 nm is aimed at two slits that are separated by 1.90105 m. What is the distance from the central band to the first-order yellow band if the screen is 0.600 m from the slits? In a double", "-slit experiment, physics students use a laser with 632.8 nm. A student places the screen 1.000 m from the slits and finds the first-order bright band 65.5 mm from the central line. What is the slit separation? 3. 4. Yellow-orange light with a wavelength of 596 nm passes through two slits that are separated by 2.25105 m and makes an interference pattern on a screen. If the distance from the central line to the first-order yellow band is 2.00102 m, how far is the screen from the slits? Young presented his findings in 1803, but was ridiculed by the scientific community. His conclusions did not begin to gain acceptance until 1820 after Jean Fresnel proposed a mathematical solution for the wave nature of light in a competition. One of the judges, Sim\u00e9on Denis Poisson, showed that, if Fresnel was correct, a shadow of a circular object illuminated with coherent light would have a bright spot at the center of the shadow. Another judge, Jean Arago, proved this experimentally. Before this, both Poisson and Arago were skeptics of the wave nature of light. Section 19.1 Interference 519 514-523 CH19-S1-845813 6/4/04 3:37 PM Page 520 \u25a0 Figure 19-6 Each wavelength is reinforced where the soap film thickness is /4, 3/4, 5/4 (a). Because each color has a different wavelength, a series of color bands is reflected from the soap film (b). a b 1 4 3 4 Ray 1 Ray 2 Ray 1 Ray 2 Nonreflective Eyeglasses A thin film can be placed on the lenses of eyeglasses to keep them from reflecting wavelengths of light that are highly visible to the human eye. This prevents the glare of reflected light. Thin-Film Interference Have you ever seen a spectrum of colors produced by a soap bubble or by the oily film on a water puddle in a parking lot? These colors were not the result of separation of white light by a prism or of absorption of colors in a pigment. The spectrum of colors was a result of the constructive and destructive interference of light waves due to reflection in a thin film, a phenomenon called thin-film interference. If a soap film is held vertically, as in Figure 19-6, its weight makes it thicker at the bottom than at the top. The thickness varies gradually from top to bottom. When a light wave strikes the film, it is partially reflected, as shown by ray 1, and partially transmitted. The reflected and transmitted waves have the same frequency as the original. The transmitted wave travels through the film to the back surface, where, again, part is reflected, as shown by ray 2. This act of splitting each light wave from an incoherent source into a matched pair of waves means that the reflected light from a thin film is coherent. Color reinforcement How is the reflection of one color enhanced? This happens when the two reflected waves are in phase for a given wavelength. If the thickness of the soap film in Figure 19-6 is one-fourth of the wavelength of the wave in the film, /4, then the round-trip path length in the film is /2. In this case, it would appear that ray 2 would return to the front surface one-half wavelength out of phase with ray 1, and that the two waves would cancel each other based on the superposition principle. However, when a transverse wave is reflected from a medium with a slower wave speed, the wave is inverted. With light, this happens at a medium with a larger index of refraction. As a result, ray 1 is inverted on reflection; whereas ray 2 is reflected from a medium with a smaller index of refraction (air) and is not inverted. Thus, ray 1 and ray 2 are in phase. If the film thickness, d, satisfies the requirement, d /4, then the color of light with that wavelength will be most strongly reflected. Note that because the wavelength of light in the film is shorter than the wavelength in air, d film/4, or, in terms of the wavelength in air, d vacuum/4nfilm. The two waves reinforce each other as they leave the film. Light with other wavelengths undergoes destructive interference. 520 Tom Pantages Chapter 19 Interference and Diffraction 514-523 CH19-S1-845813 3/12/04 7:16 AM Page 521 As you know, different colors of light have different wavelengths. For a film of varying thickness, such as the one shown in Figure 19-6, the wavelength requirement will be met at different thicknesses for different colors. The result is a rainbow of color. Where the film is too thin to produce constructive interference for any wavelength of color, the film appears to be black. Notice in Figure 19-6b that the spectrum repeats. When the thickness of the film is 3/4, the round-trip distance is", " 3/2, and constructive interference occurs again. Any thickness equal to 1/4, 3/4, 5/4, and so on, satisfies the conditions for constructive interference for a given wavelength. Applications of thin-film interference The example of a film of soapy water in air involves constructive interference with one of two waves inverted upon reflection. In the chapter opener example of bubble solution, as the thickness of the film changes, the wavelength undergoing constructive interference changes. This creates a shifting spectrum of color on the surface of the film soap when it is under white light. In other examples of thinfilm interference, neither wave or both waves might be inverted. You can develop a solution for any problem involving thin-film interference using the following strategies. Thin-Film Interference When solving thin-film interference problems, construct an equation that is specific to the problem by using the following strategies. 1. Make a sketch of the thin film and the two coherent waves. For simplicity, draw the waves as rays. 2. Read the problem. Is the reflected light enhanced or reduced? When it is enhanced, the two reflected waves undergo constructive interference. When the reflected light is reduced, the waves undergo destructive interference. 3. Are either or both waves inverted on reflection? If the index of refraction changes from a lower to a higher value, then the wave is inverted. If it changes from a higher to a lower value, there is no inversion. 4. Find the extra distance that the second wave must travel through the thin film to create the needed interference. a. If you need constructive interference and one wave is inverted OR you need destructive interference and either both waves or none are inverted, then the difference in distance is an odd number of half wavelengths: (m 1/2) film, where m 0, 1, 2, etc. b. If you need constructive interference and either both waves or none are inverted OR you need destructive interference and one wave is inverted, then the difference is an integer number of wavelengths: m film, where m 1, 2, 3, etc. 5. Set the extra distance traveled by the second ray to twice the film thickness, 2d. 6. Recall from Chapter 18 that film vacuum/nfilm. Reflection from a Thin Film Ray 1 Ray 2 Medium 1 d Film Medium 2 n1 nfilm n2 Section 19.1 Interference 521 514-523 CH19-S1-845813 6/4/04 3:38 PM Page 522 Oil and Water You observe colored rings on a puddle and conclude that there must be an oil slick on the water. You look directly down on the puddle and see a yellow\u2013green ( 555 nm) region. If the refractive index of oil is 1.45 and that of water is 1.33, what is the minimum thickness of oil that could cause this color? Analyze and Sketch the Problem \u2022 Sketch the thin film and layers above and below it. \u2022 Draw rays showing reflection off the top of the film as well as the bottom. Unknown: d? Known: nwater noil 1.33 1.45 555 nm 1 2 Ray 1 Ray 2 555 nm Air nair 1 d? Oil noil 1.45 Water nwater 1.33 Solve for the Unknown Because noil there is no phase inversion on the second reflection. Thus, there is one wave inversion. The wavelength in oil is less than it is in air. nair, there is a phase inversion on the first reflection. Because nwater noil, Follow the problem-solving strategy to construct the equation. 2d (m 1 Because you want the minimum thickness, m 0. ) n oil 2 Substitute m 0 d 4n oil m n 5 5 5 Substitute 555 nm, noil 1. 5) 4 )( 4 ( 95.7 nm 1.45 Math Handbook Operations with Significant Digits pages 835\u2013836 3 Evaluate the Answer \u2022 Are the units correct? The answer is in nm, which is correct for thickness. \u2022 Is the magnitude realistic? The minimum thickness is smaller than one wavelength, which is what it should be. 5. In the situation in Example Problem 2, what would be the thinnest film that would create a reflected red ( 635 nm) band? 6. A glass lens has a nonreflective coating placed on it. If a film of magnesium fluoride, n 1.38, is placed on the glass, n 1.52, how thick should the layer be to keep yellowgreen light from being reflected? 7. A silicon solar cell has a nonreflective coating placed on it. If a film of sodium monoxide, n 1.45, is placed on the silicon, n 3.5, how thick should the layer be to keep yellowgreen light ( 555 nm) from being reflected? 8. You can observe thin-film interference by dipping a bubble wand into some bubble solution and holding the wand in the air. What is the", " thickness of the thinnest soap film at which you would see a black stripe if the light illuminating the film has a wavelength of 521 nm? Use n 1.33. 9. What is the thinnest soap film (n 1.33) for which light of wavelength 521 nm will constructively interfere with itself? 522 Chapter 19 Interference and Diffraction 514-523 CH19-S1-845813 3/12/04 7:24 AM Page 523 a b c \u25a0 Figure 19-7 The Morpho butterfly has a blue color that shimmers iridescently (a). An electron microscope is used to view a cross section of the wing looking down the length of the ridges (b). The ridges are steplike structures. Matched pairs of light rays from a single step and from multiple steps can interfere (c). Ray 1 Ray 1 Ray 2 Ray 2 n 1.5 Air dair 127 nm n 1.5 dstep 64 nm Thin-film interference also occurs naturally on the wings of the Morpho butterfly, shown in Figure 19-7a. The shimmering blue of the butterfly is caused by ridges that project up from the ground scales of the wings, as shown in Figure 19-7b. Light is reflected from and refracted through a series of steplike structures, as diagrammed in Figure 19-7c, forming a blue interference pattern that appears to shimmer to those who see the butterfly. 19.1 Section Review 10. Film Thickness Lucien is blowing bubbles and holds the bubble wand up so that a soap film is suspended vertically in the air. What is the second thinnest width of the soap film at which he could expect to see a bright stripe if the light illuminating the film has a wavelength of 575 nm? Assume the soap solution has an index of refraction of 1.33. 11. Bright and Dark Patterns Two very narrow slits are cut close to each other in a large piece of cardboard. They are illuminated by monochromatic red light. A sheet of white paper is placed far from the slits, and a pattern of bright and dark bands is seen on the paper. Describe how a wave behaves when it encounters a slit, and explain why some regions are bright while others are dark. 12. Interference Patterns Sketch the pattern described in problem 11. 13. Interference Patterns Sketch what happens to the pattern in problem 11 when the red light is replaced by blue light. 14. Film Thickness A plastic reflecting film (n 1.83) is placed on an auto glass window (n 1.52). a. What is the thinnest film that will reflect yellow- green light? b. Unfortunately, a film this thin cannot be manufactured. What is the next-thinnest film that will produce the same effect? 15. Critical Thinking The equation for wavelength from a double-slit experiment uses the simplification that is small so that sin tan. Up to what angle is this a good approximation when your data has two significant figures? Would the maximum angle for a valid approximation increase or decrease as you increase the precision of your angle measurement? physicspp.com/self_check_quiz Section 19.1 Interference 523 CORBIS 524-531 CH19-S2-845813 6/4/04 3:56 PM Page 524 19.2 Diffraction Objectives \u2022 Explain how diffraction gratings form diffraction patterns. \u2022 Describe how diffraction gratings are used in grating spectrometers. \u2022 Discuss how diffraction limits the ability to distinguish two closely spaced objects with a lens. Vocabulary diffraction pattern diffraction grating Rayleigh criterion \u25a0 Figure 19-8 Notice the wide central band and the narrower bands on either side. A single-slit diffraction pattern for red light would have a wider central band than blue light as long as the same size slit is used for both colors. In Chapter 16, you learned that smooth wave fronts of light spread when they are diffracted around an edge. Diffraction was explained using Huygens\u2019 principle that a smooth wave front is made up of many small point-source wavelets. The cutting of coherent light on two edges spaced closely together produces a diffraction pattern, which is a pattern on a screen of constructive and destructive interference of Huygens\u2019 wavelets. Single-Slit Diffraction When coherent, blue light passes through a single, small opening that is larger than the wavelength of the light, the light is diffracted by both edges, and a series of bright and dark bands appears on a distant screen, as shown in Figure 19-8. Instead of the nearly equally spaced bands produced by two coherent sources in Young\u2019s double-slit experiment, this pattern has a wide, bright central band with dimmer, narrower bands on either side. When using red light instead of blue, the width of the bright central band increases. With", " white light, the pattern is a mixture of patterns of all the colors of the spectrum. To see how Huygens\u2019 wavelets produce the diffraction pattern, imagine a slit of width w as being divided into an even number of Huygens\u2019 points, as shown in Figure 19-9. Each Huygens\u2019 point acts as a point source of Huygens\u2019 wavelets. Divide the slit into two equal parts and choose one source from each part so that the pair is separated by a distance w/2. This pair of sources produces coherent, cylindrical waves that will interfere. For any Huygens\u2019 wavelet produced in the top half, there will be another Huygens\u2019 wavelet in the bottom half, a distance w/2 away, that it will interfere with destructively to create a dark band on the screen. All similar pairings of Huygens\u2019 wavelets interfere destructively at dark bands. Conversely, a bright band on the screen is where pairings of Huygens\u2019 wavelets interfere constructively. In the dim regions between bright and dark bands, partial destructive interference occurs. Top view Perspective view w 2 w \u25a0 Figure 19-9 A slit of width w is divided into pairs of lines that form Huygens\u2019 wavelets, each pair separated by w/2. 524 Tom Pantages Chapter 19 Interference and Diffraction 524-531 CH19-S2-845813 6/4/04 3:57 PM Page 525 r1 r2 w 2 2 \u25a0 Figure 19-10 This diagram represents an analysis of the first dark band. The distance to the screen, L, is much larger than the slit width, w. (Illustration not to scale) P1 (first dark band) x1 P0 (central bright band) Single slit Screen Diffraction pattern When the single slit is illuminated, a central bright band appears at location P0 on the screen, as shown in Figure 19-10. The first dark band is at position P1. At this location, the path lengths r1 and r2 of the two Huygens\u2019 wavelets differ by one-half wavelength, thereby producing a dark band by destructive interference. This model is mathematically similar to that of double-slit interference. A comparison of a single-slit diffraction pattern with a double-slit interference pattern using slits of the same width reveals that all of the bright interference fringes of the doubleslit interference pattern fit within the width of the central bright band of the single-slit diffraction pattern. Double-slit interference results from the interference of the single-slit diffraction patterns of the two slits. An equation now can be developed for the diffraction pattern produced by a single slit using the same simplifications that were used for double-slit interference, assuming that the distance to the screen is much larger than w. The separation distance between the sources of the two interfering waves is now w/2. To find the distance measured on the screen to the first dark band, x1, note that the path length difference is now /2 because at the dark band there is destructive interference. As a result x1/L /w. You can see from Figure 19-10 that it might be difficult to measure the distance to the first dark band from the center of the central bright band. A better method of determining x1 is to measure the width of the central bright band, 2x1. The following equation gives the width of the central bright band from single-slit diffraction. L 2 Width of Bright Band in Single-Slit Diffraction 2x1 w The width of the central bright band is equal to the product of twice the wavelength times the distance to the screen, divided by the width of the slit. Canceling the 2\u2019s out of the above equation gives you the distance from the center of the central bright band to where the first dark band occurs. The location of additional dark bands can be found where the path lengths differ by 3/2, 5/2, and so on. This can be expressed as xm mL/w, where m 1, 2, 3, etc., as limited by the small angle simplication. When m 1, this equation provides the position of the first-order dark band. The second-order dark band occurs at m 2, and so forth. Section 19.2 Diffraction 525 524-531 CH19-S2-845813 3/12/04 7:40 AM Page 526 16. Monochromatic green light of wavelength 546 nm falls on a single slit with a width of 0.095 mm. The slit is located 75 cm from a screen. How wide will the central bright band be? 17. Yellow light with a wavelength of 589 nm passes through a slit of width 0.110 mm and", " makes a pattern on a screen. If the width of the central bright band is 2.60102 m, how far is it from the slits to the screen? 18. Light from a He-Ne laser ( 632.8 nm) falls on a slit of unknown width. A pattern is formed on a screen 1.15 m away, on which the central bright band is 15 mm wide. How wide is the slit? 19. Yellow light falls on a single slit 0.0295 mm wide. On a screen that is 60.0 cm away, the central bright band is 24.0 mm wide. What is the wavelength of the light? 20. White light falls on a single slit that is 0.050 mm wide. A screen is placed 1.00 m away. A student first puts a blue-violet filter ( 441 nm) over the slit, then a red filter ( 622 nm). The student measures the width of the central bright band. a. Which filter produced the wider band? b. Calculate the width of the central bright band for each of the two filters. Single-slit diffraction patterns make the wave nature of light noticeable when the slits are 10 to 100 times the wavelength of the light. Larger openings, however, cast sharp shadows, as Isaac Newton first observed. While the single-slit pattern depends on the wavelength of light, it is only when a large number of slits are put together that diffraction provides a useful tool for measuring wavelength. You have several unknown substances and wish to use a single-slit diffraction apparatus to determine what each one is. You decide to place a sample of an unknown substance in the region between the slit and the screen and use the data that you obtain to determine the identity of each substance by calculating its index of refraction. vacuum, the width of the slit, w, the distance 1. Come up with a general formula for the index of refraction of an unknown substance in terms of the wavelength of the light, from the slit to the screen, L, and the distance between the central bright band and the first dark band, x1. 2. If the source you used had a wavelength of 634 nm, the slit width was 0.10 mm, the distance from the slit to the screen was 1.15 m, and you immersed the apparatus in water (nsubstance the width of the center band to be? 1.33), then what would you expect Unknown substance L x1 Incoming light 526 Chapter 19 Interference and Diffraction 524-531 CH19-S2-845813 6/4/04 3:57 PM Page 527 Diffraction Gratings Although double-slit interference and single-slit diffraction depend on the wavelength of light, diffraction gratings, such as those shown in Figure 19-11, are used to make precision measurements of wavelength. A diffraction grating is a device made up of many single slits that diffract light and form a diffraction pattern that is an overlap of singleslit diffraction patterns. Diffraction gratings can have as many as 10,000 slits per centimeter. That is, the spacing between the slits can be as small as 106 m, or 1000 nm. One type of diffraction grating is called a transmission grating. A transmission grating can be made by scratching very fine lines with a diamond point on glass that transmits light. The spaces between the scratched lines act like slits. A less expensive type of grating is a replica grating. A replica grating is made by pressing a thin plastic sheet onto a glass grating. When the plastic is pulled away, it contains an accurate imprint of the scratches. Jewelry made from replica gratings, shown in Figure 19-12a, produces a spectrum. Reflection gratings are made by inscribing fine lines on metallic or reflective glass surfaces. The color spectra produced when white light reflects off the surface of a CD or DVD is the result of a reflection grating, as shown in Figure 19-12b. If you were to shine monochromatic light on a DVD, the reflected light would produce a diffraction pattern on a screen. Transmission and reflection gratings produce similar diffraction patterns, which can be analyzed in the same manner. Holographic diffraction gratings produce the brightest spectra. They are made by using a laser and mirrors to create an interference pattern consisting of parallel bright and dark lines. The pattern is projected on a piece of metal that is coated with a light-sensitive material. The light produces a chemical reaction that hardens the material. The metal is then placed in acid, which attacks the metal wherever it was not protected by the hardened material. The result is a series of hills and valleys in the metal identical to the original interference pattern. The metal can be used as a reflection grating or a replica transmission grating can be made from it. Because of the", " sinusoidal shape of the hills and valleys, the diffraction patterns are exceptionally bright. a b \u25a0 Figure 19-11 Diffraction gratings are used to create diffraction patterns for the analysis of light sources. \u25a0 Figure 19-12 A transmission grating spectrum is created by jewelry made with replica gratings (a). Compact discs act as reflection gratings, creating a spectrum diffraction pattern when they are placed under white light (b). 527 Section 19.2 Diffraction (t) Central Scientific Company, (bl)Ron Tanaka/Artbase, (br)Mark Thayer 524-531 CH19-S2-845813 6/4/04 3:58 PM Page 528 \u25a0 Figure 19-13 A spectroscope is used to measure the wavelengths of light emitted by a light source. Telescope Collimator Slit Light rays Grating Measuring wavelength An instrument used to measure light wavelengths using a diffraction grating is called a grating spectroscope, as shown in the diagram in Figure 19-13. The source to be analyzed emits light that is directed to a slit. The light from the slit passes through a diffraction grating. The grating produces a diffraction pattern that is viewed through a telescope. The diffraction pattern produced by a diffraction grating has narrow, equally spaced, bright lines, as shown in Figure 19-14. The larger the number of slits per unit length of the grating, the narrower the lines in the diffraction pattern. As a result, the distance between the bright lines can be measured much more precisely with a grating spectroscope than with a double slit. Earlier in this chapter, you found that the interference pattern produced by a double slit could be used to calculate wavelength. An equation for the diffraction grating can be developed in the same way as for the double slit. However, with a diffraction grating, could be large, so the small angle simplification does not apply. Wavelength can be found by measuring the angle,, between the central bright line and the first-order bright line. Wavelength from a Diffraction Grating d sin The wavelength of light is equal to the slit separation distance times the sine of the angle at which the first-order bright line occurs. Constructive interference from a diffraction grating occurs at angles on either side of the central bright line given by the equation m dsin, where m 0, 1, 2, etc. The central bright line occurs at m 0. \u25a0 Figure 19-14 A grating was used to produce diffraction patterns for red light (a) and white light (b). a b 528 Kodansha Chapter 19 Interference and Diffraction 524-531 CH19-S2-845813 7/22/04 2:39 AM Page 529 Using a DVD as a Diffraction Grating A student noticed the beautiful spectrum reflected off a rented DVD. She directed a beam from her teacher\u2019s green laser pointer at the DVD and found three bright spots reflected on the wall. The label on the pointer indicated that the wavelength was 532 nm. The student found that the spacing between the spots was 1.29 m on the wall, which was 1.25 m away. What is the spacing between the rows on the DVD? 1 Analyze and Sketch the Problem \u2022 Sketch the experiment, showing the DVD as a grating and the spots on the wall. \u2022 Identify and label the knowns. Unknown: d? Known: x 1.29 m L 1.25 m 532 nm 2 Solve for the Unknown Find the angle between the central bright spot and one next to it using tan x/L. x tan1 L tan1 1 9. 2 m 5 2. m 45.9\u00b0 1 Substitute x 1.29 m, L 1.25 m Use the diffraction grating wavelength and solve for d. d sin d sin 9 m Substitute 532109 m, 45.9\u00b0 1 0 53 2 \u00b0.9 n 5 4 i s 7.41107 m Spot 1 Spot 2 { { x x Spot 3 Wall L Math Handbook Trigonometric Ratios page 855 3 Evaluate the Answer \u2022 Are the units correct? The answer is in m, which is correct for separation. \u2022 Is the magnitude realistic? With x and L almost the same size, d is close to. 21. White light shines through a grating onto a screen. Describe the pattern that is produced. 22. If blue light of wavelength 434 nm shines on a diffraction grating and the spacing of the resulting lines on a screen that is 1.05 m away is 0.55 m, what is the spacing between the slits in the grating? 23. A diffraction grating with slits separated by 8.60107 m is illuminated by violet light with a wavelength of 421 nm. If the screen is 80.0 cm from the grating, what is the separation of the", " lines in the diffraction pattern? 24. Blue light shines on the DVD in Example Problem 3. If the dots produced on a wall that is 0.65 m away are separated by 58.0 cm, what is the wavelength of the light? 25. Light of wavelength 632 nm passes through a diffraction grating and creates a pattern on a screen that is 0.55 m away. If the first bright band is 5.6 cm from the central bright band, how many slits per centimeter does the grating have? Section 19.2 Diffraction 529 524-531 CH19-S2-845813 6/4/04 3:59 PM Page 530 In thin-film interference, the interference pattern is visible only within a narrow angle of view straight over the film. This would be the case for the Morpho butterfly\u2019s blue, shimmering interference pattern, if not for the layer of glass-like scales on top of the layer of ground scales. This glass-like scale layer acts as a diffraction grating and causes the blue, shimmering interference pattern to be spread to a diffraction pattern with a wider angle of view. Scientists believe that this makes the Morpho butterfly more visible to potential mates. Resolving Power of Lenses The circular lens of a telescope, a microscope, and even your eye acts as a hole, called an aperture, through which light is allowed to pass. An aperture diffracts the light, just as a single slit does. Alternating bright and dark rings occur with a circular aperture, as shown in Figure 19-15. The equation for an aperture is similar to that for a single slit. However, an aperture has a circular edge rather than the two edges of a slit, so slit width, w, is replaced by aperture diameter, D, and an extra geometric factor of 1.22 enters the equation, resulting in x1 1.22L/D. When light from a distant star is viewed through the aperture of a telescope, the image is spread out due to diffraction. If two stars are close enough together, the images may be blurred together, as shown in Figure 19-16. In 1879, Lord Rayleigh, a British physicist, mathematician, and Nobel prize winner, established a criterion for determining whether there is one or two stars in such an image. The Rayleigh criterion states that if the center of the bright spot of one star\u2019s image falls on the first dark ring of the second, the two images are at the limit of resolution. That is, a viewer will be able to tell that there are two stars rather than only one. If two images are at the limit of resolution, how far apart are the objects? Using the Rayleigh criterion, the centers of the bright spots of the two images are a distance of x1 apart. Figure 19-16 shows that similar x1/L. Combining this with triangles can be used to find that xobj/Lobj 1.22L/D to eliminate x1/L, and solving for the separation distance x1 between objects, xobj, the following equation can be derived. Lobj 1.22 Rayleigh Criterion xobj D The separation distance between objects that are at the limit of resolution is equal to 1.22, times the wavelength of light, times the distance from the circular aperture to the objects, divided by the diameter of the circular aperture. Image Telescope aperture x1 xobj L Lobj \u25a0 Figure 19-15 The diffraction pattern of a circular aperture produces alternating dark and bright rings. (Illustration not to scale) Astronomy Connection Astronomy Connection \u25a0 Figure 19-16 Similar-triangle geometry allows you to calculate the actual separation distance of objects. The blue and red colors are used only for the purpose of illustration. (Illustration not to scale) 530 Chapter 19 Interference and Diffraction 524-531 CH19-S2-845813 6/4/04 3:59 PM Page 531 Diffraction in the eye In bright light the eye\u2019s pupil is about 3 mm in diameter. The eye is most sensitive to yellow-green light where 550 nm. 2104 Lobj. The So the Rayleigh criterion applied to the eye gives xobj distance between the pupil and retina is about 2 cm, so two barely resolved point sources would be separated by about 4 m on the retina. The spacing between the light detectors, the cones, in the most sensitive part of the eye, the fovea, is about 2 m. Thus, in the ideal case, the three adjacent cones would record light, dark, and light. It would seem that the eye is ideally constructed. If cones were closer together, they would see details of the diffraction pattern, not of the sources. If cones were farther apart, they would not be able to resolve all possible detail. Applying the Rayleigh criterion to find the ability of the eye to separate two distance sources shows that the eye could separate two automobile", " headlamps (1.5 m apart) at a distance of 7 km. In practice, however, the eye is not limited by diffraction. Imperfections in the lens and the liquid that fills the eye reduce the eye\u2019s resolution to about five times that set by the Rayleigh criterion. Most people use their eyes for purposes other than resolving point sources. For example, the eye seems to have a built-in ability to detect straight edges. Many telescope manufacturers advertise that their instruments are diffraction limited. That is, they claim that their telescopes can separate two point sources at the Rayleigh criterion. To reach this limit they must grind the mirrors and lenses to an accuracy of one-tenth of a wavelength, or about 55 nm. The larger the diameter of the mirror, the greater the resolution of the telescope. Unfortunately, light from planets or stars must go through Earth\u2019s atmosphere. The same variations in the atmosphere that cause stars to twinkle keep telescopes from reaching the diffraction limit. Because the Hubble Space Telescope is above Earth\u2019s atmosphere, the resolution of its images is much better than those of larger telescopes on Earth\u2019s surface. Retinal Projection Screen Did you know that you can use the retina of your eyeball as a screen? CAUTION: Do not do the following with a laser or the Sun. 1. Plug in and turn on an incandescent lamp with a straight filament. Stand about 2 m from the lamp. 2. Hold a diffraction grating in front of your eye so the color spectra are oriented horizontally. 3. Observe the color spectra patterns and draw your observations using colored pencils. Analyze and Conclude 4. Which color is closest to the central bright line (the light filament)? Which is farthest? 5. How many spectra are you able to see on either side of the light? 6. Interpret Data Are your observations consistent with the equation for the wavelength from a diffraction grating? 19.2 Section Review 26. Distance Between First-Order Dark Bands Monochromatic green light of wavelength 546 nm falls on a single slit of width 0.080 mm. The slit is located 68.0 cm from a screen. What is the separation of the first dark bands on each side of the central bright band? 27. Diffraction Patterns Many narrow slits are close to each other and equally spaced in a large piece of cardboard. They are illuminated by monochromatic red light. A sheet of white paper is placed far from the slits, and a pattern of bright and dark bands is visible on the paper. Sketch the pattern that would be seen on the screen. pattern. The dots produced by the first grating are spread out more than those produced by the second. Which grating has more lines per millimeter? 29. Rayleigh Criterion The brightest star in the winter sky in the northern hemisphere is Sirius. In reality, Sirius is a system of two stars that orbit each other. If the Hubble Space Telescope (diameter 2.4 m) is pointed at the Sirius system, which is 8.44 light-years from Earth, what is the minimum separation there would need to be between the stars in order for the telescope to be able to resolve them? Assume that the average light coming from the stars has a wavelength of 550 nm. 28. Line Spacing You shine a red laser light through one diffraction grating and form a pattern of red dots on a screen. Then you substitute a second diffraction grating for the first one, forming a different 30. Critical Thinking You are shown a spectrometer, but do not know whether it produces its spectrum with a prism or a grating. By looking at a whitelight spectrum, how could you tell? physicspp.com/self_check_quiz Section 19.2 Diffraction 531 532-533 CH19-LAB-845813 6/8/04 3:50 PM Page 532 Alternate CBL instructions can be found on the Web site. physicspp.com Double-Slit Interference of Light Light sometimes behaves as a wave. As coherent light strikes a pair of slits that are close together, the light passing through the slits will create a pattern of constructive and destructive interference on a screen. In this investigation you will develop a procedure and measure the wavelength of a monochromatic light source using two slits. QUESTION How can a double-slit interference pattern of light be used to measure the light\u2019s wavelength? Objectives Procedure \u25a0 Observe a double-slit interference pattern of 1. Determine which equation applies to double- monochromatic light. \u25a0 Calculate the wavelength of light using a double-slit interference pattern. Safety Precautions \u25a0 Use laser protective eyewear approved by ANSI. \u25a0 Never look directly into the light of a laser. Possible Materials laser pointer or laser to be tested double-slit plate laser pointer or laser of known wavelength", " clothes pin to hold a laser pointer clay ball to hold the double-slit plate meterstick slit interference. 2. Use a double slit of known slit-separation distance, d, or develop a method to determine d. 3. Sketch how light passes through a double slit to help you determine how x and L can be measured. 4. Using your sketch from step 3 and the list of possible materials provided in this lab, design the lab setup and write a procedure for performing the experiment. 5. Determine the values of m that would be invalid for the equation. 6. CAUTION: Looking directly into laser light could damage your eyes. 7. Be sure to check with your teacher and have approval before you implement your design. 8. Perform your experiment. Write your data in a data table similar to the one on the next page. 532 532-533 CH19-LAB-845813 6/8/04 3:51 PM Page 533 Data Table Source Color Accepted (m) d (m) x (m) L (m) m 1 2 3 4 5 Analyze Going Further 1. Adjust the distance of your slits from the screen. Is there a distance that allows you to collect the most data with the best precision? 2. Calculate the wavelength,, of your light source using m and measurements of x, d, and L. 3. Error Analysis Compare your calculated wavelength to the accepted value by determining the percentage of error. Conclude and Apply 1. Conclude Did your procedure enable you to use a double-slit interference pattern to measure the wavelength of light? Explain. 2. Estimate what results you would get if you used a plate with a smaller slit separation distance, d, and performed the experiment exactly the same? 3. Infer How would your observations change if you used green light, but used the same double-slit plate and screen distance? What would you observe? 1. Use a Scientific Explanation Describe why the double-slit interference pattern dims, brightens, and dims again as distance from the center of the pattern increases. 2. Error Analysis Describe several things you could do in the future to reduce systematic error in your experiment. 3. Evaluate Examine the measuring equipment you used and determine which equipment limited you the most on the precision of your calculations and which equipment gave you more precision than you needed, if any. 4. Lab Techniques What might be done to an experimental setup to use white light from a normal lightbulb to produce a double-slit interference pattern? Real-World Physics 1. When white light shines through slits in a screen door, why is a pattern not visible in the shadow on a wall? 2. Would things look different if all of the light that illuminated your world was coherent? Explain. To find out more about interference patterns, visit the Web site: physicspp.com 533 534 CH19-FEATURE-845813 3/12/04 7:45 AM Page 534 Holography Holography is a form of photography that produces a threedimensional image. Dennis Gabor made the first hologram in 1947, but holography was impractical until the invention of the gas laser in 1960. Holograms are used on credit cards to help prevent counterfeiting, and they may one day be used for ultra-high-density data storage. How is a hologram made? Beam splitter Laser Object beam Mirror Lens 1 Laser light strikes a semitransparent mirror known as a beam splitter. This creates two coherent beams. Reference beam Object Mirror Lens Laser Mirror Virtual image Scattered light Holographic exposure plate 2 The reference and object beams are directed by mirrors, and are made to diverge by lenses. 3 Light scattered by reflection off the object, in this case a basket of pears, interferes with the reference beam. The interference pattern that is formed by the two beams is recorded on the holographic exposure plate. Lens Mirror Transparent developed film 5 A person sees the image as if viewing the original object through a window. Moving his or her head changes the perspective. Thinking Critically 1. Infer A hologram records a complex pattern of constructive and destructive interference fringes. Why do you suppose a vibration-isolated surface is needed for good results? 2. Use Scientific Explanations Identify and explain where the following wave properties occur in the diagrams: reflection, refraction, and interference. 4 When a transparent film of the developed plate is placed in a diverging laser beam, light passing through the film creates a three-dimensional virtual image of the original object with rainbowlike bands of color. 534 How It Works 535-539 CH19-SG CA STP-845813 6/8/04 3:19 PM Page 535 19.1 Interference Vocabulary \u2022 incoherent light (p. 515) \u2022 coherent light (p. 516) \u2022 interference fringes (p. 516", ") \u2022 monochromatic light (p. 516) \u2022 thin-film interference (p. 520) 19.2 Diffraction Vocabulary \u2022 diffraction pattern (p. 524) \u2022 diffraction grating (p. 527) \u2022 Rayleigh criterion (p. 530) Key Concepts \u2022 Incoherent light illuminates an object evenly, just as a lightbulb illuminates your desk. \u2022 Only the superposition of light waves from coherent light sources can produce an interference pattern. \u2022 Interference demonstrates that light has wave properties. \u2022 Light passing through two closely spaced, narrow slits produces a pattern on a screen of dark and light bands called interference fringes. \u2022 Interference patterns can be used to measure the wavelength of light. d x L \u2022 Interference patterns can result from the creation of coherent light at the refractive boundary of a thin film. Key Concepts \u2022 Light passing through a narrow slit is diffracted, or spread out from a straight-line path, and produces a diffraction pattern on a screen. \u2022 The diffraction pattern from a single slit has a bright central band that has a width equal to the distance between the first dark bands on either side of the bright central band. 2x1 L 2 w \u2022 Diffraction gratings consist of large numbers of slits that are very close together and produce narrow lines that result from an overlap of the singleslit diffraction patterns of all of the slits in the grating. \u2022 Diffraction gratings can be used to measure the wavelength of light precisely or to separate light composed of different wavelengths. d sin \u2022 Diffraction limits the ability of an aperture to distinguish two closely spaced objects. Lobj 1.22 D xobj \u2022 If the central bright spot of one image falls on the first dark ring of the second image, the images are at the limit of resolution. physicspp.com/vocabulary_puzzlemaker 535 535-539 CH19-SG CA STP-845813 6/4/04 11:28 PM Page 536 Concept Mapping 31. Monochromatic light of wavelength illuminates two slits in a Young\u2019s double-slit experiment setup that are separated by a distance, d. A pattern is projected onto a screen a distance, L, away from the slits. Complete the following concept map using, L, and d to indicate how you could vary them to produce the indicated change in the spacing between adjacent bright bands, x. d x L to increase x, to decrease x, increase decrease increase decrease Mastering Concepts 32. Why is it important that monochromatic light was used to make the interference pattern in Young\u2019s interference experiment? (19.1) 33. Explain why the position of the central bright band of a double-slit interference pattern cannot be used to determine the wavelength of the light waves. (19.1) 34. Describe how you could use light of a known wave- length to find the distance between two slits. (19.1) 35. Describe in your own words what happens in thinfilm interference when a dark band is produced by light shining on a soap film suspended in air. Make sure you include in your explanation how the wavelength of the light and thickness of the film are related. (19.1) 36. White light shines through a diffraction grating. Are the resulting red lines spaced more closely or farther apart than the resulting violet lines? Why? (19.2) 37. Why do diffraction gratings have large numbers of slits? Why are these slits so close together? (19.2) 38. Why would a telescope with a small diameter not be able to resolve the images of two closely spaced stars? (19.2) 39. For a given diffraction grating, which color of visible light produces a bright line closest to the central bright band? (19.2) Applying Concepts 40. For each of the following examples, indicate whether the color is produced by thin-film interference, refraction, or the presence of pigments. a. soap bubbles b. rose petals c. oil films d. a rainbow 41. How can you tell whether a pattern is produced by a single slit or a double slit? 42. Describe the changes in a single-slit diffraction pattern as the width of the slit is decreased. 43. Science Fair At a science fair, one exhibition is a very large soap film that has a fairly consistent width. It is illuminated by a light with a wavelength of 432 nm, and nearly the entire surface appears to be a lovely shade of purple. What would you see in the following situations? a. the film thickness was doubled b. the film thickness was increased by half a wavelength of the illuminating light c. the film thickness was decreased by one quarter of a wavelength of the illuminating light 44. What are the differences in the characteristics of the diffraction patterns formed by diffraction gratings containing 104 lines/", "cm and 105 lines/cm? 45. Laser Pointer Challenge You have two laser pointers, a red one and a green one. Your friends Mark and Carlos disagree about which has the longer wavelength. Mark insists that red light has a longer wavelength, while Carlos is sure that green has the longer wavelength. You have a diffraction grating handy. Describe what demonstration you would do with this equipment and how you would explain the results to Carlos and Mark to settle their disagreement. 46. Optical Microscope Why is blue light used for illumination in an optical microscope? Mastering Problems 19.1 Interference 47. Light falls on a pair of slits 19.0 m apart and 80.0 cm from a screen, as shown in Figure 19-17. The first-order bright band is 1.90 cm from the central bright band. What is the wavelength of the light? 19.0 m Double slit Screen 80.0 cm \u25a0 Figure 19-17 (Not to scale) 536 Chapter 19 Interference and Diffraction For more problems, go to Additional Problems, Appendix B. 535-539 CH19-SG CA STP-845813 6/4/04 11:28 PM Page 537 48. Oil Slick After a short spring shower, Tom and Ann take their dog for a walk and notice a thin film of oil (n 1.45) on a puddle of water, producing different colors. What is the minimum thickness of a place where the oil creates constructive interference for light with a wavelength of 545 nm? 49. Light of wavelength 542 nm falls on a double slit. First-order bright bands appear 4.00 cm from the central bright band. The screen is 1.20 m from the slits. How far apart are the slits? 50. Insulation Film Winter is approaching and Alejandro is helping to cover the windows in his home with thin sheets of clear plastic (n = 1.81) to keep the drafts out. After the plastic is taped up around the windows such that there is air between the plastic and the glass panes, the plastic is heated with a hair dryer to shrink-wrap the window. The thickness of the plastic is altered during this process. Alejandro notices a place on the plastic where there is a blue stripe of color. He realizes that this is created by thin-film interference. What are three possible thicknesses of the portion of the plastic where the blue stripe is produced if the wavelength of the light is 4.40102 nm? 51. Samir shines a red laser pointer through three different double-slit setups. In setup A, the slits are separated by 0.150 mm and the screen is 0.60 m away from the slits. In setup B, the slits are separated by 0.175 mm and the screen is 0.80 m away. Setup C has the slits separated by 0.150 mm and the screen a distance of 0.80 m away. Rank the three setups according to the separation between the central bright band and the first-order bright band, from least to most separation. Specifically indicate any ties. 19.2 Diffraction 52. Monochromatic light passes through a single slit with a width of 0.010 cm and falls on a screen 100 cm away, as shown in Figure 19-18. If the width of the central band is 1.20 cm, what is the wavelength of the light? 0.010 cm 53. A good diffraction grating has 2.5103 lines per cm. What is the distance between two lines in the grating? 54. Light with a wavelength of 4.5105 cm passes through a single slit and falls on a screen 100 cm away. If the slit is 0.015 cm wide, what is the distance from the center of the pattern to the first dark band? 55. Hubble Space Telescope Suppose the Hubble Space Telescope, 2.4 m in diameter, is in orbit 1.0105 m above Earth and is turned to view Earth, as shown in Figure 19-19. If you ignore the effect of the atmosphere, how large an object can the telescope resolve? Use = 5.1107 m. \u25a0 Figure 19-19 56. Monochromatic light with a wavelength of 425 nm passes through a single slit and falls on a screen 75 cm away. If the central bright band is 0.60 cm wide, what is the width of the slit? 57. Kaleidoscope Jennifer is playing with a kaleidoscope from which the mirrors have been removed. The eyehole at the end is 7.0 mm in diameter. If she can just distinguish two bluishpurple specks on the other end of the kaleidoscope separated by 40 m, what is the length of the kaleidoscope? Use 650 nm and assume that the resolution is diffraction limited through the eyehole. Single slit Screen 100 cm \u25a0 Figure 19-18 (Not to scale) 58. Spectroscope A spectroscope", " uses a grating with 12,000 lines/cm. Find the angles at which red light, 632 nm, and blue light, 421 nm, have first-order bright lines. physicspp.com/chapter_test Chapter 19 Assessment 537 535-539 CH19-SG CA STP-845813 6/4/04 11:28 PM Page 538 Mixed Review 59. Record Marie uses an old 331 rpm record as a 3 diffraction grating. She shines a laser, 632.8 nm, on the record, as shown in Figure 19-20. On a screen 4.0 m from the record, a series of red dots 21 mm apart are visible. a. How many ridges are there in a centimeter along the radius of the record? b. Marie checks her results by noting that the ridges represent a song that lasts 4.01 minutes and takes up 16 mm on the record. How many ridges should there be in a centimeter? 21 mm 21 mm Laser Record Screen 4.0 m \u25a0 Figure 19-20 (Not to scale) 60. An anti-reflective coating, n 1.2, is applied to a lens. If the thickness of the coating is 125 nm, what is (are) the color(s) of light for which complete destructive interference will occur? Hint: Assume the lens is made out of glass. 61. Camera When a camera with a 50-mm lens is set at f/8, its aperture has an opening 6.25 mm in diameter. a. For light with 550 nm, what is the resolution of the lens? The film is 50.0 mm from the lens. b. The owner of a camera needs to decide which film to buy for it. The expensive one, called finegrained film, has 200 grains/mm. The less costly, coarse-grained film has only 50 grains/mm. If the owner wants a grain to be no smaller than the width of the central bright spot calculated in part a, which film should he purchase? Thinking Critically 62. Apply Concepts Yellow light falls on a diffraction grating. On a screen behind the grating, you see three spots: one at zero degrees, where there is no diffraction, and one each at 30\u00b0 and 30\u00b0. You now add a blue light of equal intensity that is in the same direction as the yellow light. What pattern of spots will you now see on the screen? 63. Apply Concepts Blue light of wavelength passes through a single slit of width w. A diffraction pattern appears on a screen. If you now replace the blue light with a green light of wavelength 1.5, to what width should you change the slit to get the original pattern back? 64. Analyze and Conclude At night, the pupil of a human eye has an aperture diameter of 8.0 mm. The diameter is smaller in daylight. An automobile\u2019s headlights are separated by 1.8 m. a. Based upon Rayleigh\u2019s criterion, how far away can the human eye distinguish the two headlights at night? Hint: Assume a wavelength of 525 nm. b. Can you actually see a car\u2019s headlights at the distance calculated in part a? Does diffraction limit your eyes\u2019 sensing ability? Hypothesize as to what might be the limiting factors. Writing in Physics 65. Research and describe Thomas Young\u2019s contributions to physics. Evaluate the impact of his research on the scientific thought about the nature of light. 66. Research and interpret the role of diffraction in medicine and astronomy. Describe at least two applications in each field. Cumulative Review 67. How much work must be done to push a 0.5-m3 block of wood to the bottom of a 4-m-deep swimming pool? The density of wood is 500 kg/m3. (Chapter 13) 68. What are the wavelengths of microwaves in an oven if their frequency is 2.4 GHz? (Chapter 14) 69. Sound wave crests that are emitted by an airplane are 1.00 m apart in front of the plane, and 2.00 m apart behind the plane. (Chapter 15) a. What is the wavelength of the sound in still air? b. If the speed of sound is 330 m/s, what is the frequency of the source? c. What is the speed of the airplane? 70. A concave mirror has a 48.0-cm radius. A 2.0-cmtall object is placed 12.0 cm from the mirror. Calculate the image position and image height. (Chapter 17) 71. The focal length of a convex lens is 21.0 cm. A 2.00-cm-tall candle is located 7.50 cm from the lens. Use the thin-lens equation to calculate the image position and image height. (Chapter 18) 538 Chapter 19 Interference and Diffraction For more problems", ", go to Additional Problems, Appendix B. 535-539 CH19-SG CA STP-845813 6/8/04 3:20 PM Page 539 Multiple Choice 1. What is the best possible explanation for why the colors of a thin film, such as a soap bubble or oil on water, appear to change and move as you watch? because convective heat waves in the air next to the thin film distort the light because the film thickness at any given location changes over time because the wavelengths in sunlight vary over time because your vision varies slightly over time 2. Light at 410 nm shines through a slit and falls on a flat screen as shown in the figure below. The width of the slit is 3.8106 m. What is the width of the central bright band? 0.024 m 0.031 m 0.048 m 0.063 m First-order dark band 0.29 m 3. For the situation in problem 2, what is the angle,, of the first dark band? 3.1\u00b0 6.2\u00b0 12.4\u00b0 17\u00b0 4. Two stars 6.2104 light-years from Earth are 3.1 light-years apart. What is the smallest diameter telescope that could resolve them using 610 nm light? 5.0105 m 6.1105 m 1.5102 m 1.5107 m 5. A grating has slits that are 0.055 mm apart. What is the angle of the first-order bright line for light with a wavelength of 650 nm? 0.012\u00b0 0.68\u00b0 1.0\u00b0 11\u00b0 6. A laser beam at 638 nm illuminates two narrow slits. The third-order band of the resulting pattern is 7.5 cm from the center bright band. The screen is 2.475 m from the slits. How far apart are the slits? 5.8108 m 6.3107 m 2.1105 m 6.3105 m 7. A flat screen is placed 4.200 m from a pair of slits that are illuminated by a beam of monochromatic light. On the screen, the separation between the central bright band and the second-order bright band is 0.082 m. The distance between the slits is 5.3105 m. Determine the wavelength of the light. 2.6107 m 5.2107 m 6.2107 m 1.0106 m 8. A clown is blowing soap bubbles and you notice that the color of one region of a particularly large bubble matches the color of his nose. If the bubble is reflecting 6.5107 m red light waves, and the index of refraction of the soap film is 1.41, what is the minimum thickness of the soap bubble at the location where it is reflecting red? 1.2107 m 3.5107 m 9.2107 m 1.9106 m Extended Answer 9. A diffraction grating that has 6000 slits per cm produces a diffraction pattern that has a firstorder bright line at 20\u00b0 from the central bright line. What is the wavelength of the light? Don\u2019t Be Afraid To Ask For Help If you are practicing for a test and you are having difficulty understanding why you got a question wrong or you are having difficulty even arriving at an answer, ask someone for help. It is important to ask for help before a test because you cannot ask for help during a test. physicspp.com/standardized_test Chapter 19 Standardized Test Practice 539 540-545 CH20-S1-845813 3/22/04 2:29 AM Page 540 What You\u2019ll Learn \u2022 You will observe the behavior of electric charges and analyze how these charges interact with matter. \u2022 You will examine the forces that act between electric charges. Why It\u2019s Important Static electricity enables the operation of devices such as printers and copiers, but it has harmful effects on electronic components and in the form of lightning. Lightning The tiny spark that you experience when you touch a doorknob and the dazzling display of lightning in a storm are both examples of the discharge of static electricity. The charging processes and the means of discharging are vastly different in scale, but they are similar in their fundamental nature. Think About This What causes charge to build up in a thundercloud, and how does it discharge in the form of a spectacular lightning bolt? physicspp.com 540 Kent Wood/Photo Researchers 540-545 CH20-S1-845813 6/7/04 7:43 AM Page 541 Which forces act over a distance? Question What happens when a plastic ruler is rubbed with wool and then brought near a pile of paper scraps? Procedure 1. Place 15-20 scraps of paper from a hole punch on the table. 2. Take a plastic ruler and rub it with a piece of wool. 3. Bring the ruler close to the pieces of paper. Observe the effect the ruler has on the scraps", " of paper. Critical Thinking What forces are acting on the pieces of paper before the ruler is brought close to them? What can you infer about the forces on the paper after the ruler is brought near? Based on your answers to the previous questions, form a hypothesis that explains the effect the ruler has on the scraps of paper. Analysis What happens to the pieces of paper when the ruler is brought close to them? What happens to the pieces of paper that come in contact with the ruler? Did you observe any unexpected results when the ruler was brought close to the paper scraps? If so, describe these results. 20.1 Electric Charge You may have had the experience of rubbing your shoes on a carpet to create a spark when you touched someone. In 1752, Benjamin Franklin set off a flurry of research in the field of electricity when his famous kite experiment showed that lightning is similar to the sparks caused by friction. In his experiment, Franklin flew a kite with a key attached to the string. As a thunderstorm approached, the loose threads of the kite string began to stand up and repel one another, and when Franklin brought his knuckle close to the key, he experienced a spark. Electric effects produced in this way are called static electricity. In this chapter, you will investigate electrostatics, the study of electric charges that can be collected and held in one place. The effects of electrostatics are observable over a vast scale, from huge displays of lightning to the submicroscopic world of atoms and molecules. Current electricity, which is produced by batteries and generators, will be explored in later chapters. Objectives \u2022 Demonstrate that charged objects exert forces, both attractive and repulsive. \u2022 Recognize that charging is the separation, not the creation, of electric charges. \u2022 Describe the differences between conductors and insulators. Vocabulary electrostatics neutral insulator conductor Section 20.1 Electric Charge 541 Horizons Companies 540-545 CH20-S1-845813 6/7/04 7:45 AM Page 542 Charged Objects Have you ever noticed the way that your hair is attracted to the comb when you comb your hair on a dry day or the way that your hair stands on end after it is rubbed with a balloon? Perhaps you also have found that socks sometimes stick together when you take them out of a clothes dryer. If so, you will recognize the attraction of the bits of paper to a plastic ruler demonstrated by the Launch Lab and shown in Figure 20-1. You might have noticed the way the paper pieces jumped up to the ruler as you worked through the Launch Lab. There must be a new, relatively strong force causing this upward acceleration because it is larger than the downward acceleration caused by the gravitational force of Earth. There are other differences between this new force and gravity. Paper is attracted to a plastic ruler only after the ruler has been rubbed; if you wait a while, the attractive property of the ruler disappears. Gravity, on the other hand, does not require rubbing and does not disappear. The ancient Greeks noticed effects similar to that of the ruler when they rubbed amber. The Greek word for amber is elektron, and today this attractive property is called electric. An object that exhibits electric interaction after rubbing is said to be charged. Like charges You can explore electric interactions with simple objects, such as transparent tape. Fold over about 5 mm of the end of a strip of tape for a handle, and then stick the remaining 8- to 12-cm-long part of the tape strip on a dry, smooth surface, such as your desktop. Stick a second, similar piece of tape next to the first. Quickly pull both strips off the desk and bring them near each other. A new property causes the strips to repel each other: they are electrically charged. Because they were prepared in the same way, they must have the same type of charge. Thus, you have demonstrated that two objects with the same type of charge repel each other. You can learn more about this charge by doing some simple experiments. You may have found that the tape is attracted to your hand. Are both sides attracted, or just one? If you wait a while, especially in humid weather, you will find that the electric charge disappears. You can restore it by again sticking the tape to the desk and pulling it off. You also can remove its charge by gently rubbing your fingers down both sides of the tape. Opposite charges Now, stick one strip of tape on the desk and place the second strip on top of the first. As shown in Figure 20-2a, use the handle of the bottom strip of tape to pull the two off the desk together. Rub them with your fingers until they are no longer attracted to your hand. You now have removed all the electric charge. With one hand on the handle of one strip and the other on the handle of the second strip, quickly pull the two strips apart. You will find that they are now both charged. They once again are attracted to your hands.", " Do they still repel each other? No, they now attract each other. They are charged, but they are no longer charged alike. They have opposite charges and therefore attract each other. \u25a0 Figure 20-1 Rubbing a plastic ruler with wool produces a new force of attraction between the ruler and bits of paper. When the ruler is brought close to bits of paper, the attractive electric force accelerates the paper bits upward against the force of gravity. 542 StudiOhio Chapter 20 Static Electricity 540-545 CH20-S1-845813 6/7/04 7:49 AM Page 543 a b Is tape the only object that you can charge? Once again, stick one strip of tape to the desk and the second strip on top. Label the bottom strip B and the top strip T. Pull the pair off together. Discharge them, then pull them apart. Stick the handle end of each strip to the edge of a table, the bottom of a lamp shade, or some similar object. The two should hang down a short distance apart. Finally, rub a comb or pen on your clothing and bring it near one strip of tape and then the other. You will find that one strip will be attracted to the comb, while the other will be repelled by it, as shown in Figure 20-2b. You now can explore the interactions of charged objects with the strips of tape. Experimenting with charge Try to charge other objects, such as glasses and plastic bags. Rub them with different materials, such as silk, wool, and plastic wrap. If the air is dry, scuff your shoes on carpet and bring your finger near the strips of tape. To test silk or wool, slip a plastic bag over your hand before holding the cloth. After rubbing, take your hand out of the bag and bring both the bag and cloth near the strips of tape. Most charged objects will attract one strip and repel the other. You will never find an object that repels both strips of tape, although you might find some that attract both. For example, your finger will attract both strips. You will explore this effect later in this chapter. Types of charge From your experiments, you can make a list of objects labeled B, for bottom, which have the same charge as the tape stuck on the desk. Another list can be made of objects labeled T, which have the same charge as the top strip of tape. There are only two lists, because there are only two types of charge. Benjamin Franklin called them positive and negative charges. Using Franklin\u2019s convention, when hard rubber and plastic are rubbed, they become negatively charged. When materials such as glass and wool are rubbed, they become positively charged. Just as you showed that an uncharged pair of tape strips became oppositely charged, you probably were able to show that if you rubbed plastic with wool, the plastic became negatively charged and the wool positively charged. The two kinds of charges were not created alone, but in pairs. These experiments suggest that matter normally contains both charges, positive and negative. Contact in some way separates the two. To explore this further, you must consider the microscopic picture of matter. A Microscopic View of Charge Electric charges exist within atoms. In 1897, J.J. Thomson discovered that all materials contain light, negatively charged particles that he called electrons. Between 1909 and 1911, Ernest Rutherford, a student of Thomson from New Zealand, discovered that the atom has a massive, positively charged nucleus. When the positive charge of the nucleus equals the negative charge of the surrounding electrons, then the atom is neutral. \u25a0 Figure 20-2 Strips of tape can be given opposite charges (a) and then be used to demonstrate the interactions of like and opposite charges (b). \u2022 Positive charges are shown in red. \u2022 Negative charges are shown in blue. Section 20.1 Electric Charge 543 Tom Pantages 540-545 CH20-S1-845813 3/22/04 2:34 AM Page 544 \u25a0 Figure 20-3 When wool is used to charge a rubber rod, electrons are removed from the wool atoms and cling to the rubber atoms. In this way, both objects become charged. a b + + - + -+ - - + - + - + - + + - Hard rubber rod - + + - + + - - + + + Wool + + With the addition of energy, the outer electrons can be removed from atoms. An atom missing electrons has an overall positive charge, and consequently, any matter made of these electron-deficient atoms is positively charged. The freed electrons can remain unattached or become attached to other atoms, resulting in negatively charged particles. From a microscopic viewpoint, acquiring charge is a process of transferring electrons. Separation of charge If two neutral objects are rubbed together, each can become charged. For instance, when rubber and wool are rubbed together, electrons from atoms on the wool are transferred to the rubber, as shown in Figure 20-3. The extra electrons on the rubber result in a net negative", " charge. The electrons missing from the wool result in a net positive charge. The combined total charge of the two objects remains the same. Charge is conserved, which is one way of saying that individual charges never are created or destroyed. All that happens is that the positive and negative charges are separated through a transfer of electrons. Complex processes that affect the tires of a moving car or truck can cause the tires to become charged. Processes inside a thundercloud can cause the cloud bottom to become negatively charged and the cloud top to become positively charged. In both these cases, charge is not created, but separated. Conductors and Insulators Hold a plastic rod or comb at its midpoint and rub only one end. You will find that only the rubbed end becomes charged. In other words, the charges that you transferred to the plastic stayed where they were put; they did not move. A material through which a charge will not move easily is called an electric insulator. The strips of tape that you charged earlier in this chapter acted in this way. Glass, dry wood, most plastics, cloth, and dry air are all good insulators. Suppose that you support a metal rod on an insulator so that it is isolated, or completely surrounded by insulators. If you then touch the charged comb to one end of the metal rod, you will find that the charge spreads very quickly over the entire rod. A material that allows charges to move about easily is called an electric conductor. Electrons carry, or conduct, electric charge through the metal. Metals are good conductors because at least one electron on each atom of the metal can be removed easily. These electrons act as if they no longer belong to any one atom, but to the metal as a whole; consequently, they move freely throughout the piece of metal. Figure 20-4 contrasts how charges behave when they are Conductor or Insulator? It might be tempting to classify an element as solely a conductor or solely an insulator, but the classification can change depending on the form the element takes. For example, carbon in the form of diamond is an insulator, but carbon in the form of graphite can conduct charge. This is because the carbon atoms in diamonds are tightly bonded to four other carbons, while the carbon atoms in graphite form three stronger bonds and a fourth, weaker bond that allows electrons a limited amount of movement. As a result, graphite is a much better conductor than diamond, even though both are simply carbon atoms. 544 Horizons Companies Chapter 20 Static Electricity 540-545 CH20-S1-845813 3/22/04 2:34 AM Page 545 a Conductor b Insulator \u25a0 Figure 20-4 Charges placed on a conductor will spread over the entire surface (a). Charges placed on an insulator will remain where they are placed (b). placed on a conductor with how they behave on an insulator. Copper and aluminum are both excellent conductors and are used commercially to carry electricity. Plasma, a highly ionized gas, and graphite also are good conductors of electric charge. When air becomes a conductor Air is an insulator; however, under certain conditions, charges move through air as if it were a conductor. The spark that jumps between your finger and a doorknob after you have rubbed your feet on a carpet discharges you. In other words, you have become neutral because the excess charges have left you. Similarly, lightning discharges a thundercloud. In both of these cases, air became a conductor for a brief moment. Recall that conductors must have charges that are free to move. For a spark or lightning to occur, freely moving charged particles must be formed in the normally neutral air. In the case of lightning, excess charges in the cloud and on the ground are great enough to remove electrons from the molecules in the air. The electrons and positively or negatively charged atoms form a plasma, which is a conductor. The discharge of Earth and the thundercloud by means of this conductor forms a luminous arc called lightning. In the case of your finger and the doorknob, the discharge is called a spark. 20.1 Section Review 1. Charged Objects After a comb is rubbed on a wool sweater, it is able to pick up small pieces of paper. Why does the comb lose that ability after a few minutes? 2. Types of Charge In the experiments described earlier in this section, how could you find out which strip of tape, B or T, is positively charged? 3. Types of Charge A pith ball is a small sphere made of a light material, such as plastic foam, often coated with a layer of graphite or aluminum paint. How could you determine whether a pith ball that is suspended from an insulating thread is neutral, is charged positively, or is charged negatively? 4. Charge Separation A rubber rod can be charged negatively when it is rubbed with wool. What happens to the charge of the wool? Why? 5. Conservation of Charge An apple contains trillions of charged particles. Why don", "\u2019t two apples repel each other when they are brought together? 6. Charging a Conductor Suppose you hang a long metal rod from silk threads so that the rod is isolated. You then touch a charged glass rod to one end of the metal rod. Describe the charges on the metal rod. 7. Charging by Friction You can charge a rubber rod negatively by rubbing it with wool. What happens when you rub a copper rod with wool? 8. Critical Thinking It once was proposed that electric charge is a type of fluid that flows from objects with an excess of the fluid to objects with a deficit. Why is the current two-charge model better than the single-fluid model? physicspp.com/self_check_quiz Section 20.1 Electric Charge 545 546-553 CH20-S2-845813 6/7/04 7:52 AM Page 546 20.2 Electric Force Objectives \u2022 Summarize the relationships between electric forces, charges, and distance. \u2022 Explain how to charge objects by conduction and induction. \u2022 Develop a model of how charged objects can attract a neutral object. \u2022 Apply Coulomb\u2019s law to problems in one and two dimensions. Vocabulary electroscope charging by conduction charging by induction grounding Coulomb\u2019s law coulomb elementary charge \u25a0 Figure 20-5 A charged rod, when brought close to another charged and suspended rod, will attract or repel the suspended rod. Electric forces must be strong because they can easily produce acceler- ations larger than the acceleration caused by gravity. You also have learned that they can be either repulsive or attractive, while gravitational forces always are attractive. Over the years, many scientists made attempts to measure electric forces. Daniel Bernoulli, best known for his work with fluids, made some crude measurements in 1760. In the 1770s, Henry Cavendish showed that electric forces must obey an inverse square force law, but being extremely shy, he did not publish his work. His manuscripts were discovered over a century later, after all his work had been duplicated by others. Forces on Charged Bodies The forces that you observed on tape strips also can be demonstrated by suspending a negatively charged, hard rubber rod so that it turns easily, as shown in Figure 20-5. If you bring another negatively charged rod near the suspended rod, the suspended rod will turn away. The negative charges on the rods repel each other. It is not necessary for the rods to make contact. The force, called the electric force, acts at a distance. If a positively charged glass rod is suspended and a similarly charged glass rod is brought close, the two positively charged rods also will repel each other. If a negatively charged rod is brought near a positively charged rod, however, the two will attract each other, and the suspended rod will turn toward the oppositely charged rod. The results of your tape experiments and these actions of charged rods can be summarized in the following way: \u2022 There are two kinds of electric charges: positive and negative. \u2022 Charges exert forces on other charges at a distance. \u2022 The force is stronger when the charges are closer together. \u2022 Like charges repel; opposite charges attract. Neither a strip of tape nor a large rod that is hanging in open air is a very sensitive or convenient way of determining charge. Instead, a device called an electroscope is used. An electroscope consists of a metal knob connected by a metal stem to two thin, lightweight pieces of metal foil, called leaves. Figure 20-6 shows a neutral electroscope. Note that the leaves hang loosely and are enclosed to eliminate stray air currents. a b c + + + + + --- - - + ++ - -- 546 Chapter 20 Static Electricity 546-553 CH20-S2-845813 3/22/04 2:38 AM Page 547 Charging by conduction When a negatively charged rod is touched to the knob of an electroscope, electrons are added to the knob. These charges spread over all the metal surfaces. As shown in Figure 20-7a, the two leaves are charged negatively and repel each other; therefore, they spread apart. The electroscope has been given a net charge. Charging a neutral body by touching it with a charged body is called charging by conduction. The leaves also will spread apart if the electroscope is charged positively. How, then, can you find out whether the electroscope is charged positively or negatively? The type of charge can be determined by observing the leaves when a rod of known charge is brought close to the knob. The leaves will spread farther apart if the rod and the electroscope have the same charge, as shown in Figure 20-7b. The leaves will fall slightly if the electroscope\u2019s charge is opposite that of the rod, as in Figure 20-7c. Separation of charge on neutral objects Earlier in this chapter, when you brought your finger near either charged strip of tape, the tape was attracted", " to your finger. Your finger, however, was neutral\u2014it had equal amounts of positive and negative charge. You know that in conductors, charges can move easily, and that in the case of sparks, electric forces can change insulators into conductors. Given this information, you can develop a plausible model for the force that your finger exerted on the strips of tape. Suppose you move your finger, or any uncharged object, close to a positively charged object. The negative charges in your finger will be attracted to the positively charged object, and the positive charges in your finger will be repelled. Your finger will remain neutral, but the positive and negative charges will be separated. The electric force is stronger for charges that are closer together; therefore, the separation results in an attractive force between your finger and the charged object. The force that a charged ruler exerts on neutral pieces of paper is the result of the same process, the separation of charges. The negative charges at the bottom of thunderclouds also can cause charge separation in Earth. Positive charges in the ground are attracted to Earth\u2019s surface under the cloud. The forces of the charges in the cloud and those on Earth\u2019s surface can break molecules into positively and negatively charged particles. These charged particles are free to move, and they establish a conducting path from the ground to the cloud. The lightning that you observe occurs when a bolt travels at speeds on the order of 500,000 km/h along the conducting path and discharges the cloud. Knob Insulator Leaves \u25a0 Figure 20-6 An electroscope is a device used for detecting charges. In a neutral electroscope, the leaves hang loosely, almost touching one another. \u25a0 Figure 20-7 A negatively charged electroscope will have its leaves spread apart (a). A negatively charged rod pushes electrons down to the leaves, causing them to spread farther apart (b). A positively charged rod attracts some of the electrons, causing the leaves to spread apart less (c). a ---- b ---- -- c + ++ + + ---- ----- -- ----- --- -- Section 20.2 Electric Force 547 546-553 CH20-S2-845813 6/14/04 2:20 PM Page 548 a + + -+ + ++- b ---- c + + + + ++ -- - + + + + ++ -- \u25a0 Figure 20-8 One method of charging by induction begins with neutral spheres that are touching (a). A charged rod is brought near them (b), then the spheres are separated and the charged rod is removed (c). The charges on the separated spheres are equal in magnitude, but opposite in sign. \u25a0 Figure 20-9 A negatively charged rod induces a separation of charges in an electroscope (a). The electroscope is grounded, and negative charges are pushed from the electroscope to the ground (b). The ground is removed before the rod, and the electroscope is left with a positive charge (c). Charging by induction Suppose that two identical, insulated metal spheres are touching, as shown in Figure 20-8a. When a negatively charged rod is brought close to one, as in Figure 20-8b, electrons from the first sphere will be forced onto the sphere farther from the rod and will make it negatively charged. The closer sphere is now positively charged. If the spheres are separated while the rod is nearby, each sphere will have a charge, and the charges will be equal but opposite, as shown in Figure 20-8c. This process of charging an object without touching it is called charging by induction. A single object can be charged by induction through grounding, which is the process of connecting a body to Earth to eliminate excess charge. Earth is a very large sphere, and it can absorb great amounts of charge without becoming noticeably charged itself. If a charged body is touched to Earth, almost any amount of charge can flow to Earth. If a negatively charged rod is brought close to the knob of an electroscope, as in Figure 20-9a, electrons are repelled onto the leaves. If the knob is then grounded on the side opposite the charged rod, electrons will be pushed from the electroscope into the ground until the leaves are neutral, as in Figure 20-9b. Removing the ground before the rod leaves the electroscope with a deficit of electrons, and it will be positively charged, as in Figure 20-9c. Grounding also can be used as a source of electrons. If a positive rod is brought near the knob of a grounded electroscope, electrons will be attracted from the ground, and the electroscope will obtain a negative charge. When this process is employed, the charge induced on the electroscope is opposite that of the object used to charge it. Because the rod never touches the electroscope, its charge is not transferred, and it can be used many times to charge objects by induction. a -- ---- ++ +++ + b -- -- -- - - + - -- + ++ ++ c + + -- ---- + + + + 548 Chapter 20 Static", " Electricity 546-553 CH20-S2-845813 3/22/04 2:40 AM Page 549 Coulomb\u2019s Law You have seen that a force acts between two or more charged objects. In your experiments with tape, you found that the force depends on distance. The closer you brought the charged comb to the tape, the stronger the force was. You also found that the more you charged the comb, the stronger the force was. How can you vary the quantity of charge in a controlled way? This problem was solved in 1785 by French physicist Charles Coulomb. The type of apparatus used by Coulomb is shown in Figure 20-10. An insulating rod with small conducting spheres, A and A, at each end was suspended by a thin wire. A similar sphere, B, was placed in contact with sphere A. When they were touched with a charged object, the charge spread evenly over the two spheres. Because they were the same size, they received equal amounts of charge. The symbol for charge is q. Therefore, the amount of charge on the spheres can be represented by the notation qA and qB. Force depends on distance Coulomb found how the force between the two charged spheres depended on the distance. First, he carefully measured the amount of force needed to twist the suspending wire through a given angle. He then placed equal charges on spheres A and B and varied the distance, r, between them. The force moved A, which twisted the suspending wire. By measuring the deflection of A, Coulomb could calculate the force of repulsion. He showed that the force, F, varied inversely with the square of the distance between the centers of the spheres. 1 F 2 r Force depends on charge To investigate the way in which the force depended on the amount of charge, Coulomb had to change the charges on the spheres in a measured way. He first charged spheres A and B equally, as before. Then he selected an uncharged sphere, C, of the same size as sphere B. When C was placed in contact with B, the spheres shared the charge that had been on B alone. Because the two were the same size, B then had only half of its original charge. Therefore, the charge on B was only one-half the charge on A. After Coulomb adjusted the position of B so that the distance, r, between A and B was the same as before, he found that the force between A and B was half of its former value. That is, he found that the force varied directly with the charge of the bodies. F qAqB After many similar measurements, Coulomb summarized the results in a law now known as Coulomb\u2019s law: the magnitude of the force between charge qA and charge qB, separated by a distance r, is proportional to the magnitude of the charges and inversely proportional to the square of the distance between them. qB qA F r2 The unit of charge: the coulomb The amount of charge that an object has is difficult to measure directly. Coulomb\u2019s experiments, however, showed that the quantity of charge could be related to force. Thus, Coulomb could define a standard quantity of charge in terms of the amount of force that it produces. The SI standard unit of charge is called the coulomb (C). Charged spheres Thin wire \u25a0 Figure 20-10 Coulomb used a similar type of apparatus to measure the force between two spheres, A and B. He observed the deflection of A while varying the distance between A and B. Investigating Induction and Conduction Use a balloon and an electroscope to investigate charging by induction and charging by conduction. 1. Predict what will happen if you charge a balloon by rubbing it with wool and bring it near a neutral electroscope. 2. Predict what will happen if you touch the balloon to the electroscope. 3. Test your predictions. Analyze and Conclude 4. Describe your results. 5. Explain the movements of the leaves in each step of the experiment. Include diagrams. 6. Describe the results if the wool had been used to charge the electroscope. 549 Jose Pessoa-Divisao de Documentacao fotografica do Instituto Portugues de Museus Section 20.2 Electric Force 546-553 CH20-S2-845813 3/22/04 2:40 AM Page 550 a FB on A A FA on B B b A FB on A FA on B B \u25a0 Figure 20-11 The rule for determining the direction of force is: like charges repel; unlike charges attract. 550 Chapter 20 Static Electricity One coulomb is the charge of 6.241018 electrons or protons. A typical lightning bolt can carry 5 C to 25 C of charge. The charge on a single electron is 1.601019 C. The magnitude of the charge of an electron is called the elementary charge. Even small pieces of matter, such as", " coins, contain up to 106 C of negative charge. This enormous amount of negative charge produces almost no external effects because it is balanced by an equal amount of positive charge. If the charge is unbalanced, even as small a charge as 109 C can result in large forces. According to Coulomb\u2019s law, the magnitude of the force on charge qA caused by charge qB a distance r away can be written as follows. qB qA Coulomb\u2019s Law F K r2 The force between two charges is equal to Coulomb\u2019s constant, times the product of the two charges, divided by the square of the distance between them. When the charges are measured in coulombs, the distance in meters, and the force in newtons, the constant, K, is 9.0109 Nm2/C2. The Coulomb\u2019s law equation gives the magnitude of the force that charge qA exerts on qB and also the force that qB exerts on qA. These two forces are equal in magnitude but opposite in direction. You can observe this example of Newton\u2019s third law of motion in action when you bring two strips of tape with like charges together. Each exerts forces on the other. If you bring a charged comb near either strip of tape, the strip, with its small mass, moves readily. The acceleration of the comb and you is, of course, much less because of the much greater mass. The electric force, like all other forces, is a vector quantity. Force vectors need both a magnitude and a direction. However, the Coulomb\u2019s law equation above gives only the magnitude of the force. To determine the direction, you need to draw a diagram and interpret charge relations carefully. If two positively charged objects, A and B, are brought near, the forces they exert on each other are repulsive, as shown in Figure 20-11a. If, instead, B is negatively charged, the forces are attractive, as shown in Figure 20-11b. Electric Force Problems Use these steps to find the magnitude and direction of the force between charges. 1. Sketch the system showing all distances and angles to scale. 2. Diagram the vectors of the system. 3. Use Coulomb\u2019s law to find the magnitude of the force. 4. Use your diagram along with trigonometric relations to find the direction of the force. 5. Perform all algebraic operations on both the numbers and the units. Make sure that the units match the variables in question. 6. Consider the magnitude of your answer. Is it reasonable? 546-553 CH20-S2-845813 6/7/04 7:56 AM Page 551 Coulomb\u2019s Law in Two Dimensions Sphere A, with a charge of 6.0 C, is located near another charged sphere, B. Sphere B has a charge of 3.0 C and is located 4.0 cm to the right of A. a. What is the force of sphere B on sphere A? b. A third sphere, C, with a 1.5-C charge, is added to the configuration. If it is located 3.0 cm directly beneath A, what is the new net force on sphere A? 1 Analyze and Sketch the Problem \u2022 Establish coordinate axes and sketch the spheres. \u2022 Show and label the distances between the spheres. \u2022 Diagram and label the force vectors. y x Known: rAB rAC 6.0 C 3.0 C 1.5 C qA qB qC 4.0 cm 3.0 cm Unknown:? FB on A? FC on A Fnet? 2 Solve for the Unknown a. Find the force of sphere B on sphere A. FB on A q q K B A 2 rA B (9.0109 Nm2/C2) 1.0102 N (6.0106 C)(3.0106 C) (4.0102 m)2 FC on A Fnet FB on A B qB A rAB rAC qA qC C Substitute qA qB 3.0 C, rAB 6.0 C, 4.0 cm Because spheres A and B have unlike charges, the force of B on A is to the right. b. Find the force of sphere C on sphere A. FC on A q q K C A 2 rA C (9.0109 Nm2/C2) 9.0101 N (6.0106 C)(1.5106 C) (3.0102 m)2 Substitute qA qC 1.5 C, rAC 6.0 C, 3.0 cm Spheres A and C have like charges, which repel. The force of C on A is upward. Find the vector sum of FB on A and FC on A to find Fnet on sphere A. Fnet 2 on A 2 FC FB on A (", "1.0102 N)2 (9.0101N)2 Substitute FB on A 130 N F tan.0102 N, FC on A 9.0101 N Math Handbook Inverses of Sine, Cosine, and Tangent page 856 Substitute FC on A 9.0101 N, FB on A 1.0102 N o n A F C F A) tan1 ( ) tan1 (9 42\u00b0 130 N, 42\u00b0 above the x-axis Fnet 3 Evaluate the Answer \u2022 Are the units correct? (Nm2/C2)(C)(C)/m2 N. The units work out to be newtons. \u2022 Does the direction make sense? Like charges repel; unlike charges attract. \u2022 Is the magnitude realistic? The magnitude of the net force is in agreement with the magnitudes of the component forces. Section 20.2 Electric Force 551 546-553 CH20-S2-845813 6/7/04 7:58 AM Page 552 9. A negative charge of 2.0104 C and a positive charge of 8.0104 C are separated by 0.30 m. What is the force between the two charges? 10. A negative charge of 6.0106 C exerts an attractive force of 65 N on a second charge that is 0.050 m away. What is the magnitude of the second charge? 11. The charge on B in Example Problem 1 is replaced by a charge of 3.00 C. Diagram the new situation and find the net force on A. 12. Sphere A is located at the origin and has a charge of 2.0106 C. Sphere B is located at 0.60 m on the x-axis and has a charge of 3.6106 C. Sphere C is located at +0.80 m on the x-axis and has a charge of 4.0106 C. Determine the net force on sphere A. 13. Determine the net force on sphere B in the previous problem. As you use the Coulomb\u2019s law equation, keep in mind that Coulomb\u2019s law is valid only for point charges or uniform spherical charge distributions. That is, a charged sphere may be treated as if all the charge were located at its center if the charge is spread evenly across its entire surface or throughout its volume. If a sphere is a conductor and another charge is brought near it, the charges on the sphere will be attracted or repelled, and the charge no longer will act as if it were at the sphere\u2019s center. Therefore, it is important to consider how large and how far apart two charged spheres are before applying Coulomb\u2019s law. The problems in this textbook assume that charged spheres are small enough and far enough apart to be considered point charges unless otherwise noted. When shapes such as long wires or flat plates are considered, Coulomb\u2019s law must be modified to account for the nonpoint charge distributions. Application of Electrostatic Forces There are many applications of electric forces on particles. For example, these forces can collect soot in smokestacks, thereby reducing air pollution, as shown in Figure 20-12. Tiny paint droplets, charged by As shown in the figure on the right, two spheres of equal mass, m, and equal positive charge, q, are a distance, r, apart. 1. Derive an expression for the charge, q, that must be on each sphere so that the spheres are in equilibrium; that is, so that the attractive and repulsive forces between them are balanced. 2. If the distance between the spheres is doubled, how will that affect the expression for the value of q that you determined in the previous problem? Explain. mass m charge q 3. If the mass of each sphere is 1.50 kg, determine the charge on each sphere needed to maintain the equilibrium. r mass m charge q 552 Chapter 20 Static Electricity 546-553 CH20-S2-845813 3/22/04 2:42 AM Page 553 \u25a0 Figure 20-12 The fly ash being released by these smokestacks is a by-product of burning coal. Static-electricity precipitators can be used to reduce fly ash emissions. induction, can be used to paint automobiles and other objects very uniformly. Photocopy machines use static electricity to place black toner on a page so that a precise reproduction of the original document is made. In other instances, applications are concerned with the control of static charge. For example, static charge can ruin film if it attracts dust, and electronic equipment can be damaged by the discharge of static charge. In these cases, applications are designed to avoid the buildup of static charge and to safely eliminate any charge that does build up. 20.2 Section Review 14. Force and Charge How are electric force and charge related? Describe the force when the charges are like charges and the force when the charges are opposite charges. 19.", " Charging by Induction In an electroscope being charged by induction, what happens when the charging rod is moved away before the ground is removed from the knob? 15. Force and Distance How are electric force and distance related? How would the force change if the distance between two charges were tripled? 16. Electroscopes When an electroscope is charged, the leaves rise to a certain angle and remain at that angle. Why do they not rise farther? 17. Charging an Electroscope Explain how to charge an electroscope positively using a. a positive rod. b. a negative rod. 18. Attraction of Neutral Objects What two properties explain why a neutral object is attracted to both positively and negatively charged objects? 20. Electric Forces Two charged spheres are held a distance, r, apart. One sphere has a charge of 3C, and the other sphere has a charge of 9C. Compare the force of the 3C sphere on the 9C sphere with the force of the 9C sphere on the 3C sphere. 21. Critical Thinking Suppose that you are testing Coulomb\u2019s law using a small, positively charged plastic sphere and a large, positively charged metal sphere. According to Coulomb\u2019s law, the force depends on 1/r 2, where r is the distance between the centers of the spheres. As the spheres get close together, the force is smaller than expected from Coulomb\u2019s law. Explain. physicspp.com/self_check_quiz Section 20.2 Electric Force 553 Mike McClure/Maxximages 554-555 CH20-LAB-845813 6/7/04 8:03 AM Page 554 Charged Objects In this chapter, you observed and studied phenomena that result from the separation of electric charges. You learned that hard rubber and plastic tend to become negatively charged when they are rubbed, while glass and wool tend to become positively charged. But what happens if two objects that tend to become negatively charged are rubbed together? Will electrons be transferred? If so, which material will gain electrons, and which will lose them? In this physics lab, you will design a procedure to further your investigations of positive and negative charges. QUESTION How can you test materials for their ability to hold positive and negative charges? Objectives Materials \u25a0 Observe that different materials tend to become positively or negatively charged. \u25a0 Compare and contrast the ability of materials to acquire and hold positive and negative charges. \u25a0 Interpret data to order a list of materials from strongest tendency to be negatively charged to strongest tendency to be positively charged. 15-cm plastic ruler thread ring stand with ring masking tape materials to be charged, such as rubber rods, plastic rods, glass rods, PVC pipe, copper pipe, steel pipe, pencils, pens, wool, silk, plastic wrap, plastic sandwich bags, waxed paper, and aluminum foil Procedure 1. Use the lab photo as a guide to suspend a 15-cm plastic ruler. It is advisable to wash the ruler in soapy water, then rinse and dry it thoroughly before each use, especially if it is a humid day. The thread should be attached at the midpoint of the ruler with two or three wraps of masking tape between the thread and ruler. 2. Use the following situations as a reference for types of charges a material can have: 1) a plastic ruler rubbed with wool gives the plastic ruler an excess negative charge and the wool an excess positive charge, and 2) a plastic ruler rubbed with plastic wrap gives the plastic ruler an excess positive charge and the plastic wrap an excess negative charge. Safety Precautions 554 Horizons Companies 554-555 CH20-LAB-845813 3/22/04 2:46 AM Page 555 Data Table Material 1 Material 2 Charge on Ruler (,, 0) Observation of Ruler\u2019s Movements Charge on Material 1 (,, 0) Charge on Material 2 (,, 0) 3. Design a procedure to test which objects tend to become negatively charged and which tend to become positively charged. Try various combinations of materials and record your observations in the data table. 4. Develop a test to see if an object is neutral. Remember that a charged ruler may be attracted to a neutral object if it induces a separation of charge in the neutral object. 5. Be sure to check with your teacher and have your procedure approved before you proceed with your lab. Analyze 1. Observe and Infer As you brought charged materials together, could you detect a force between the charged materials? Describe this force. 2. Formulate Models Make a drawing of the charge distribution on the two materials for one of your trials. Use this drawing to explain why the materials acted the way they did during your experiments with them. 3. Draw Conclusions Which materials hold an excess charge? Which materials do not hold a charge very well? 4. Draw Conclusions Which materials tend to become negatively charged? Which tend to become positively charged? 5. Interpret Data Use your data table to list the relative tendencies of materials to be", " positively or negatively charged. Conclude and Apply 1. Explain what is meant by the phrases excess charge and charge imbalance when referring to static electricity. 2. Does excess charge remain on a material or does it dissipate over time? 3. Could you complete this physics lab using a metal rod in place of the suspended plastic ruler? Explain. 4. Clear plastic wrap seals containers of food. Why does plastic wrap cling to itself after it is pulled from its container? Going Further Review the information in your textbook about electroscopes. Redesign the lab using an electroscope, rather than a suspended ruler, to test for the type of charge on an object. Real-World Physics Trucks often have a rubber strap or a chain that drags along the road. Why are they used? To find out more about static charge, visit the Web site: physicspp.com 555 556 CH20-FEATURE-845813 6/7/04 8:07 AM Page 556 Spacecraft and Static Electricity Spacecraft and Static Electricity Most objects on Earth do not build up substantial static-electric charges because a layer of moisture clings to surfaces, allowing charges to migrate to or from the ground. As you learned in this chapter, Earth can absorb almost any amount of charge. However, there is no moisture in space, and Earth is far away. Charged particles ejected from the Sun, or in the ionosphere, strike and cling to spacecraft, charging their surfaces to thousands of volts. Plasma and Charging In Chapter 13, you learned that plasma consists of free electrons and positive ions. Orbiting spacecraft are surrounded by a thin cloud of this plasma. The electrons in plasma can move far more easily than more massive positive ions. Thus, spacecraft surfaces tend to attract electrons and develop a negative charge. This negative charge eventually attracts some heavy positive ions, which strike the spacecraft and can damage its surface. On the International Space Station, an additional difficulty stems from the array of solar panels that convert energy from the Sun into electricity. When the arrays are powering the space station, the voltage on the surface of the craft tends to be close to the voltage of the solar array. As a result, it is possible that an electric arc could form between the space station and the plasma that surrounds it. spacecraft\u2019s components, there is a remote chance that the buildup of charge might endanger astronauts on space walks. To discharge the potential difference and protect craft and crew, the space station\u2019s skin must be connected by a conductor, called a plasma contactor, to the plasma cloud surrounding it. The connection begins on board the station, where a stream of xenon gas from a tank in the Plasma Contactor Unit (PCU) is ionized by an electric current. This ionization takes place in the cathode assembly. The ionized xenon, now in the plasma state, passes out of the craft through the cathode assembly. It is this stream of conductive plasma that connects the craft to the surrounding plasma cloud, thereby reducing the potential difference to safe levels. Plasma Contactor Unit (PCU) Cathode assembly Xenon tank This is a cutaway drawing of the PCU. Future Applications Future spacecraft might integrate the plasma contactor into the propulsion system. For example, the Variable Specific Impulse Magnetoplasma Rocket (VASIMR) could use the plasma exhaust that it produces to provide an electric connection between the spacecraft and the surrounding plasma. Scientists think that this type of rocket could be used in the future to travel between planets. Going Further 1. Apply What is the purpose of a plasma contactor? How is it similar to using your finger to ground an electroscope? 2. Research How could scientists assess the charge on the surface of the International Space Station? Consequences of an Arc Arcs are extremely hot and carry a great deal of current. They can prematurely ignite retro-rockets or explosive bolts and interfere with the operation of the spacecraft\u2019s electronic equipment. The solar panels are particularly susceptible to arc damage. In addition to damage to the 556 NASA Future Technology 557-561 CH20-SG CA STP-845813 6/7/04 8:13 AM Page 557 20.1 Electric Charge Vocabulary \u2022 electrostatics (p. 541) \u2022 neutral (p. 543) \u2022 insulator (p. 544) \u2022 conductor (p. 544) 20.2 Electric Force Vocabulary \u2022 electroscope (p. 546) \u2022 charging by conduction (p. 547) \u2022 charging by induction (p. 548) \u2022 grounding (p. 548) \u2022 Coulomb\u2019s law (p. 549) \u2022 coulomb (p. 549) \u2022 elementary charge (p. 550) Key Concepts \u2022 There are two kinds of electric charge, positive and negative. Interactions of these charges explain the attraction and repulsion that you observed in the strips of tape. \u2022 Electric charge is not created or destroyed; it is conserved.", " Charging is the separation, not creation, of electric charges. \u2022 Objects can be charged by the transfer of electrons. An area with excess electrons has a net negative charge; an area with a deficit of electrons has a net positive charge. \u2022 Charges added to one part of an insulator remain on that part. Insulators include glass, dry wood, plastics, and dry air. \u2022 Charges added to a conductor quickly spread over the surface of the object. In general, examples of conductors include graphite, metals, and matter in the plasma state. \u2022 Under certain conditions, charges can move through a substance that is ordinarily an insulator. Lightning moving through air is one example. Key Concepts \u2022 When an electroscope is charged, electric forces cause its thin metal leaves to spread apart. \u2022 An object can be charged by conduction by touching it with a charged object. \u2022 A charged object will induce a separation of charges within a neutral conductor. This process will result in an attractive force between the charged object and the neutral conductor. \u2022 To charge a conductor by induction, a charged object is first brought near it, causing a separation of charges. Then, the conductor to be charged is separated, trapping opposite charges on the two halves. \u2022 Grounding is the removal of excess charge by touching an object to Earth. Grounding can be used in the process of charging an electroscope by induction. \u2022 Coulomb\u2019s law states that the force between two charged particles varies directly with the product of their charges and inversely with the square of the distance between them. qB qA F K r2 To determine the direction of the force, remember the following rule: like charges repel; unlike charges attract. \u2022 The SI unit of charge is the coulomb. One coulomb (C) is the magnitude of the charge of 6.241018 electrons or protons. The elementary charge, the charge of a proton or electron, is 1.601019 C. physicspp.com/vocabulary_puzzlemaker 557 557-561 CH20-SG CA STP-845813 6/7/04 8:15 AM Page 558 Concept Mapping 22. Complete the concept map below using the following terms: conduction, distance, elementary charge. Electric Force Coulomb\u2019s law charging charge coulomb induction 34. Lightning Lightning usually occurs when a negative charge in a cloud is transported to Earth. If Earth is neutral, what provides the attractive force that pulls the electrons toward Earth? 35. Explain what happens to the leaves of a positively charged electroscope when rods with the following charges are brought close to, but not touching, the electroscope. a. positive b. negative 36. As shown in Figure 20-13, Coulomb\u2019s law and Newton\u2019s law of universal gravitation appear to be similar. In what ways are the electric and gravitational forces similar? How are they different? Mastering Concepts 23. If you comb your hair on a dry day, the comb can become positively charged. Can your hair remain neutral? Explain. (20.1) Law of Universal Gravitation mAmB F G r 2 24. List some insulators and conductors. (20.1) mA mB Coulomb\u2019s Law qAqB F K r 2 qA qB 25. What property makes metal a good conductor and rubber a good insulator? (20.1) 26. Laundry Why do socks taken from a clothes dryer sometimes cling to other clothes? (20.2) 27. Compact Discs If you wipe a compact disc with a clean cloth, why does the CD then attract dust? (20.2) 28. Coins The combined charge of all electrons in a nickel is hundreds of thousands of coulombs. Does this imply anything about the net charge on the coin? Explain. (20.2) 29. How does the distance between two charges impact the force between them? If the distance is decreased while the charges remain the same, what happens to the force? (20.2) 30. Explain how to charge a conductor negatively if you have only a positively charged rod. (20.2) Applying Concepts 31. How does the charge of an electron differ from the charge of a proton? How are they similar? 32. Using a charged rod and an electroscope, how can you find whether or not an object is a conductor? 33. A charged rod is brought near a pile of tiny plastic spheres. Some of the spheres are attracted to the rod, but as soon as they touch the rod, they are flung off in different directions. Explain. r r \u25a0 Figure 20-13 (Not to scale) 37. The constant, K, in Coulomb\u2019s equation is much larger than the constant, G, in the universal gravitation equation. Of what significance is this? 38. The text describes Coulomb\u2019s method for charging two spheres, A and B,", " so that the charge on B was exactly half the charge on A. Suggest a way that Coulomb could have placed a charge on sphere B that was exactly one-third the charge on sphere A. 39. Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance, r, apart. He then made the charge on B one-third the charge on A. How far apart would the two spheres then have had to be for A to have had the same deflection that it had before? 40. Two charged bodies exert a force of 0.145 N on each other. If they are moved so that they are onefourth as far apart, what force is exerted? 41. Electric forces between charges are enormous in comparison to gravitational forces. Yet, we normally do not sense electric forces between us and our surroundings, while we do sense gravitational interactions with Earth. Explain. 558 Chapter 20 Static Electricity For more problems, go to Additional Problems, Appendix B. 557-561 CH20-SG CA STP-845813 3/22/04 2:53 AM Page 559 Mastering Problems 20.2 Electric Force 42. Two charges, qA and qB, are separated by a distance, r, and exert a force, F, on each other. Analyze Coulomb\u2019s law and identify what new force would exist under the following conditions. a. qA is doubled b. qA and qB are cut in half c. r is tripled d. r is cut in half e. qA is tripled and r is doubled 43. Lightning A strong lightning bolt transfers about 25 C to Earth. How many electrons are transferred? 44. Atoms Two electrons in an atom are separated by 1.51010 m, the typical size of an atom. What is the electric force between them? 45. A positive and a negative charge, each of magnitude 2.5105 C, are separated by a distance of 15 cm. Find the force on each of the particles. 46. A force of 2.4102 N exists between a positive charge of 8.0105 C and a positive charge of 3.0105 C. What distance separates the charges? 47. Two identical positive charges exert a repulsive force of 6.4109 N when separated by a distance of 3.81010 m. Calculate the charge of each. 48. A positive charge of 3.0 C is pulled on by two negative charges. As shown in Figure 20-14, one negative charge, 2.0 C, is 0.050 m to the west, and the other, 4.0 C, is 0.030 m to the east. What total force is exerted on the positive charge? 4.0 C 2.0 C 3.0 C 0.050 m 0.030 m \u25a0 Figure 20-14 49. Figure 20-15 shows two positively charged spheres, one with three times the charge of the other. The spheres are 16 cm apart, and the force between them is 0.28 N. What are the charges on the two spheres? q 3q 16 cm \u25a0 Figure 20-15 50. Charge in a Coin How many coulombs of charge are on the electrons in a nickel? Use the following method to find the answer. a. Find the number of atoms in a nickel. A nickel has a mass of about 5 g. A nickel is 75 percent Cu and 25 percent Ni, so each mole of the coin\u2019s atoms will have a mass of about 62 g. b. Find the number of electrons in the coin. On average, each atom has 28.75 electrons. c. Find the coulombs on the electrons. 51. Three particles are placed in a line. The left particle has a charge of 55 C, the middle one has a charge of 45 C, and the right one has a charge of 78 C. The middle particle is 72 cm from each of the others, as shown in Figure 20-16. a. Find the net force on the middle particle. b. Find the net force on the right particle. 55 C 45 C 78 C 72 cm 72 cm \u25a0 Figure 20-16 Mixed Review 52. A small metal sphere with charge 1.2105 C is touched to an identical neutral sphere and then placed 0.15 m from the second sphere. What is the electric force between the two spheres? 53. Atoms What is the electric force between an electron and a proton placed 5.31011 m apart, the approximate radius of a hydrogen atom? 54. A small sphere of charge 2.4 C experiences a force of 0.36 N when a second sphere of unknown charge is placed 5.5 cm from it. What is the charge of the second sphere? 55. Two identically charged spheres placed 12 cm apart have an electric force of 0.28 N between them. What is the charge of each sphere? 56. In an experiment using Coulomb\u2019s apparatus, a", " sphere with a charge of 3.6108 C is 1.4 cm from a second sphere of unknown charge. The force between the spheres is 2.7102 N. What is the charge of the second sphere? 57. The force between a proton and an electron is 3.51010 N. What is the distance between these two particles? physicspp.com/chapter_test Chapter 20 Assessment 559 557-561 CH20-SG CA STP-845813 7/22/04 3:52 PM Page 560 Thinking Critically 58. Apply Concepts Calculate the ratio of the electric force to the gravitational force between the electron and the proton in a hydrogen atom. 59. Analyze and Conclude Sphere A, with a charge of 64 C, is positioned at the origin. A second sphere, B, with a charge of 16 C, is placed at 1.00 m on the x-axis. a. Where must a third sphere, C, of charge 12 C be placed so there is no net force on it? b. If the third sphere had a charge of 6 C, where should it be placed? c. If the third sphere had a charge of 12 C, where should it be placed? 60. Three charged spheres are located at the positions shown in Figure 20-17. Find the total force on sphere B. 4.5 C y A 4.0 cm 8.2 C x B 3.0 cm 6.0 C C \u25a0 Figure 20-17 61. The two pith balls in Figure 20-18 each have a mass of 1.0 g and an equal charge. One pith ball is suspended by an insulating thread. The other is brought to 3.0 cm from the suspended ball. The suspended ball is now hanging with the thread forming an angle of 30.0\u00b0 with the vertical. The ball is in equilibrium with FE, Fg, and FT. Calculate each of the following. a. Fg on the suspended ball b. FE c. the charge on the balls 30.0\u00b0 FE 3.0 cm \u25a0 Figure 20-18 62. Two charges, qA and qB, are at rest near a positive test charge, qT, of 7.2 C. The first charge, qA, is a positive charge of 3.6 C located 2.5 cm away from qT at 35\u00b0; qB is a negative charge of 6.6 C located 6.8 cm away at 125\u00b0. a. Determine the magnitude of each of the forces acting on qT. b. Sketch a force diagram. c. Graphically determine the resultant force acting on qT. Writing in Physics 63. History of Science Research several devices that were used in the seventeenth and eighteenth centuries to study static electricity. Examples that you might consider include the Leyden jar and the Wimshurst machine. Discuss how they were constructed and how they worked. 64. In Chapter 13, you learned that forces exist between water molecules that cause water to be denser as a liquid between 0\u00b0C and 4\u00b0C than as a solid at 0\u00b0C. These forces are electrostatic in nature. Research electrostatic intermolecular forces, such as van der Waals forces and dipole-dipole forces, and describe their effects on matter. Cumulative Review 65. Explain how a pendulum can be used to determine the acceleration of gravity. (Chapter 14) 66. A submarine that is moving 12.0 m/s sends a sonar ping of frequency 1.50103 Hz toward a seamount that is directly in front of the submarine. It receives the echo 1.800 s later. (Chapter 15) a. How far is the submarine from the seamount? b. What is the frequency of the sonar wave that strikes the seamount? c. What is the frequency of the echo received by the submarine? 67. Security Mirror A security mirror is used to produce an image that is three-fourths the size of an object and is located 12.0 cm behind the mirror. What is the focal length of the mirror? (Chapter 17) 68. A 2.00-cm-tall object is located 20.0 cm away from a diverging lens with a focal length of 24.0 cm. What are the image position, height, and orientation? Is this a real or a virtual image? (Chapter 18) 69. Spectrometer A spectrometer contains a grating of 11,500 slits/cm. Find the angle at which light of wavelength 527 nm has a first-order bright band. (Chapter 19) 560 Chapter 20 Static Electricity For more problems, go to Additional Problems, Appendix B. 557-561 CH20-SG CA STP-845813 3/22/04 2:55 AM Page 561 Multiple Choice 1. How many electrons have been removed from a positively charged electroscope if it has a net charge of 7.51011", " C? 7.51011 electrons 2.1109 electrons 1.2108 electrons 4.7108 electrons 2. The force exerted on a particle with a charge of 5.0109 C by a second particle that is 4 cm away is 8.4105 N. What is the charge of the second particle? 4.21013 C 2.0109 C 3.0109 C 6.0105 C 3. Three charges, A, B, and C, are located in a line, as shown below. What is the net force on charge B? 78 N toward A 78 N toward C 130 N toward A 210 N toward C 8.510 6 C 3.110 6 C 6.410 6 C A B C 4.2 cm 2.9 cm 4. What is the charge on an electroscope that has an excess of 4.81010 electrons? 3.31030 C 4.81010 C 7.7109 C 4.81010 C 5. Two charged bodies exert a force of 86 N on each other. If they are moved so that they are six times farther apart, what is the new force that they will exert on each other? 2.4 N 14 N 86 N 5.2102 N 6. Two equally charged bodies exert a force of 90 N on each other. If one of the bodies is exchanged for a body of the same size, but three times as much charge, what is the new force that they will exert on each other? 2.7102 N 8.1102 N 10 N 30 N 7. An alpha particle has a mass of 6.681027 kg and a charge of 3.21019 C. What is the ratio of the electrostatic force to the gravitational force between two alpha particles? 1 4.8107 2.31015 3.11035 8. Charging a neutral body by touching it with a charged body is called charging by _______. conduction induction grounding discharging 9. Macy rubs a balloon with wool, giving the balloon a charge of 8.91014 C. What is the force between the balloon and a metal sphere that is charged to 25 C and is 2 km away? 8.91015 N 5.0109 N 2.21012 N 5.6104 N Extended Answer 10. According to the diagram, what is the net force exerted by charges A and B on charge C? In your answer, include a diagram showing the force vectors FA on C, FB on C, and Fnet. y (m) 0.30 A 2.0 C 0.30 B 2.0 C 4.0 C C 0.40 x (m) Slow Down Check to make sure you are answering the question that each problem is posing. Read the questions and answer choices very carefully. Remember that doing most of the problems and getting them right is always preferable to doing all of the problems and getting a lot of them wrong. physicspp.com/standardized_test Chapter 20 Standardized Test Practice 561 562-568 CH21-S1-845813 3/19/04 3:57 PM Page 562 What You\u2019ll Learn \u2022 You will relate electric fields to electric forces and distinguish between them. \u2022 You will relate electric potential difference to work and energy. \u2022 You will describe how charges are distributed on conductors. \u2022 You will explain how capacitors store electric charges. Why It\u2019s Important Electricity is an essential form of energy for modern societies. High-Energy Discharge A high-voltage generator produces the glow you see inside these discharge spheres. Think About This Why doesn\u2019t an ordinary lightbulb glow in the same way as these discharge spheres connected to a high-voltage generator? physicspp.com 562 Horizons Companies 562-568 CH21-S1-845813 3/18/04 9:56 PM Page 563 How do charged objects interact at a distance? Question How is a charged object affected by interaction with other charged objects at a distance? Procedure Analysis 1. Inflate and tie off two balloons. Attach a 1 -m length of string to each balloon. 2 2. Rub one balloon back and forth on your shirt six to eight times, causing it to become charged. Hang it from a cabinet, table, or other support by the string with a piece of tape. 3. Rub the second balloon the same way and then suspend it from its string. 4. Observe Slowly bring the second balloon toward the suspended one. How do the balloons behave? Tape the second balloon so it hangs by its string next to the first balloon. 5. Observe Bring your hand toward the charged balloons. What happens? What did you observe as the two balloons were brought near each other? What happened as your hand was brought near the balloons? Critical Thinking With what two objects have you previously observed similar behaviors of action at a distance? 21.1 Creating and Measuring Electric Fields Electric force, like gravitational force, which you studied", " in Chapter 8, varies inversely as the square of the distance between two point objects. Both forces can act from great distances. How can a force be exerted across what seems to be empty space? Michael Faraday suggested that because an electrically charged object, A, creates a force on another charged object, B, anywhere in space, object A must somehow change the properties of space. Object B somehow senses the change in space and experiences a force due to the properties of the space at its location. We call the changed property of space an electric field. An electric field means that the interaction is not between two distant objects, but between an object and the field at its location. The forces exerted by electric fields can do work, transferring energy from the field to another charged object. This energy is something you use on a daily basis, whether you plug an appliance into an electric outlet or use a battery-powered, portable device. In this chapter, you will learn more about electric fields, forces, and electric energy. Objectives \u2022 Define an electric field. \u2022 Solve problems relating to charge, electric fields, and forces. \u2022 Diagram electric field lines. Vocabulary electric field electric field line Section 21.1 Creating and Measuring Electric Fields 563 Horizons Companies 562-568 CH21-S1-845813 6/7/04 8:25 AM Page 564 The Electric Field How can you measure an electric field? Place a small charged object at some location. If there is an electric force on it, then there is an electric field at that point. The charge on the object that is used to test the field, called the test charge, must be small enough that it doesn\u2019t affect other charges. Consider Figure 21-1, which illustrates a charged object with a charge of q. Suppose you place the positive test charge at some point, A, and measure a force, F. According to Coulomb\u2019s law, the force is directly proportional to the strength of the test charge, q. That is, if the charge is doubled, so is the force. Therefore, the ratio of the force to the charge is a constant. If you divide the force, F, by the test charge, q, you obtain a vector quantity, F/q. This quantity does not depend on the test charge, only on the force, F, and the location of point A. The electric field at point A, the location of q, is represented by the following equation. Fo q n Electric Field Strengh E q The strength of an electric field is equal to the force on a positive test charge divided by the strength of the test charge. The direction of an electric field is the direction of the force on a positive test charge. The magnitude of the electric field strength is measured in newtons per coulomb, N/C. A picture of an electric field can be made by using arrows to represent the field vectors at various locations, as shown in Figure 21-1. The length of the arrow is used to show the strength of the field. The direction of the arrow shows the field direction. To find the field from two charges, the fields from the individual charges are added vectorially. A test charge can be used to map out the field resulting from any collection of charges. Typical electric field strengths produced by charge collections are shown in Table 21-1. An electric field should be measured only by a very small test charge. This is because the test charge also exerts a force on q. It is important that the force exerted by the test charge does not cause charge to be redistributed on a conductor, thereby causing q to move to another location and thus, changing the force on q as well as the electric field strength being measured. A test charge always should be small enough so that its effect on q is negligible. Table 21-1 Approximate Values of Typical Electric Fields Field Value (N/C) Near a charged, hard-rubber rod In a television picture tube Needed to create a spark in air At an electron\u2019s orbit in a hydrogen atom 1103 1105 3106 51011 \u2022 Electric field lines are indigo. \u2022 Positive charges are red. \u2022 Negative charges are blue. q Electric field q Charge q q \u25a0 Figure 21-1 Arrows can be used to represent the magnitude and direction of the electric field about an electric charge at various locations. 564 Chapter 21 Electric Fields 562-568 CH21-S1-845813 3/18/04 9:58 PM Page 565 Electric Field Strength An electric field is measured using a positive test charge of 3.0106 C. This test charge experiences a force of 0.12 N at an angle of 15\u00b0 north of east. What are the magnitude and direction of the electric field strength at the location of the test charge? 1 Analyze and Sketch the Problem \u2022 Draw and label the test charge, q. \u2022 Show and label the coordinate system centered on the test charge. \u2022 Diagram and label", " the force vector at 15\u00b0 north of east. N Known: q 3.0106 C F 0.12 N at 15\u00b0 N of E Unknown: E? 2 Solve for the Unknown F E q 2.1 0 N Substitute F 0.12 N, q 3.0106 C 6 3.0 0 1 N/C 4.0104 N/C The force on the test charge and the electric field are in the same direction. E 4.0104 N/C at 15\u00b0 N of E q F 15\u00b0 E Math Handbook Operations with Significant Digits pages 835\u2014836 3 Evaluate the Answer \u2022 Are the units correct? Electric field strength is correctly measured in N/C. \u2022 Does the direction make sense? The field direction is in the direction of the force because the test charge is positive. \u2022 Is the magnitude realistic? This field strength is consistent with the values listed in Table 21-1. 1. A positive test charge of 5.0106 C is in an electric field that exerts a force of 2.0104 N on it. What is the magnitude of the electric field at the location of the test charge? 2. A negative charge of 2.0108 C experiences a force of 0.060 N to the right in an electric field. What are the field\u2019s magnitude and direction at that location? 3. A positive charge of 3.0107 C is located in a field of 27 N/C directed toward the south. What is the force acting on the charge? 4. A pith ball weighing 2.1103 N is placed in a downward electric field of 6.5104 N/C. What charge (magnitude and sign) must be placed on the pith ball so that the electric force acting on it will suspend it against the force of gravity? 5. You are probing the electric field of a charge of unknown magnitude and sign. You first map the field with a 1.0106-C test charge, then repeat your work with a 2.0106-C test charge. a. Would you measure the same forces at the same place with the two test charges? Explain. b. Would you find the same field strengths? Explain. Section 21.1 Creating and Measuring Electric Fields 565 562-568 CH21-S1-845813 3/18/04 9:59 PM Page 566 Electric Field Strength What is the electric field strength at a point that is 0.30 m to the right of a small sphere with a charge of 4.0106 C? Analyze and Sketch the Problem \u2022 Draw and label the sphere and its charge, q, and the test charge, q. \u2022 Show and label the distance between the charges. \u2022 Diagram and label the force vector acting on q. q 4.0106 C Known: q 4.0106 C d 0.30 m Unknown: E? d 0.30 m q F 1 2 Math Handbook Operations with Scientific Notation pages 842\u2014843 0 6 C) Substitute K 9.0109 Nm2/C2, q 4.0106 C, d 0.30 m 1.0 )2 0 m.3 (0 Solve for the Unknown The force and the magnitude of the test charge are unknown, so use Coulomb\u2019s law in combination with the electric field strength9.0109 Nm2/C2)(4 4.0105 N/C q Substitute F K q 2 d E 4.0105 N/C toward the sphere, or to the left 3 Evaluate the Answer \u2022 Are the units correct? (Nm2/C2)(C)/m2 N/C. The units work out to be N/C, which is correct for electric field strength. \u2022 Does the direction make sense? The negative sign indicates that the positive test charge is attracted toward the negative point charge. \u2022 Is the magnitude realistic? This field strength is consistent with the values listed in Table 21-1. 6. What is the magnitude of the electric field strength at a position that is 1.2 m from a point charge of 4.2106 C? 7. What is the magnitude of the electric field strength at a distance twice as far from the point charge in problem 6? 8. What is the electric field at a position that is 1.6 m east of a point charge of 7.2106 C? 9. The electric field that is 0.25 m from a small sphere is 450 N/C toward the sphere. What is the charge on the sphere? 10. How far from a point charge of 2.4106 C must a test charge be placed to measure a field of 360 N/C? 566 Chapter 21 Electric Fields 562-568 CH21-S1-845813 3/18/04 9:59 PM Page 567 a b c \u25a0 Figure 21-2 Lines of force are drawn perpendicularly away from", " a positively charged object (a) and perpendicularly into a negatively charged object (b). Electric field lines are shown between like charged and oppositely charged objects (c). \u25a0 Figure 21-3 In the Van de Graaff generator (a), charge is transferred onto a moving belt at A, and from the belt to the metal dome at B. An electric motor does the work needed to increase the electric potential energy. When a person touches a Van de Graaff generator, the results can be dramatic (b). b So far, you have measured an electric field at a single point. Now, imagine moving the test charge to another location. Measure the force on it again and calculate the electric field. Repeat this process again and again until you assign every location in space a measurement of the vector quantity of the electric field strength associated with it. The field is present even if there is no test charge to measure it. Any charge placed in an electric field experiences a force on it resulting from the electric field at that location. The strength of the force depends on the magnitude of the field, E, and the magnitude of the charge, q. Thus, F Eq. The direction of the force depends on the direction of the field and the sign of the charge. Picturing the Electric Field A picture of an electric field is shown in Figure 21-2. Each of the lines used to represent the actual field in the space around a charge is called an electric field line. The direction of the field at any point is the tangent drawn to a field line at that point. The strength of the electric field is indicated by the spacing between the lines. The field is strong where the lines are close together. It is weaker where the lines are spaced farther apart. Although only two-dimensional models can be shown here, remember that electric fields exist in three dimensions. a The direction of the force on a positive test charge near another positive charge is away from the other charge. Thus, the field lines extend radially outward like the spokes of a wheel, as shown in Figure 21-2a. Near a negative charge, the direction of the force on the positive test charge is toward the negative charge, so the field lines point radially inward, as shown in Figure 21-2b. When there are two or more charges, the field is the vector sum of the fields resulting individual charges. The field lines become curved and the pattern is more complex, as shown in Figure 21-2c. Note that field lines always leave a positive charge and enter a negative charge, and that they never cross each other. Grounded base Metal dome Insulator from Motor the Belt A B Section 21.1 Creating and Measuring Electric Fields 567 (r)Horizons Companies, (l)courtesy PASCO Scientific 562-568 CH21-S1-845813 3/18/04 10:00 PM Page 568 \u25a0 Figure 21-4 Lines of force between unlike charges (a, c) and between like charges (b, d) describe the behavior of a positively charged object in a field. The top figures are computer tracings of electric field lines. a c b d Robert Van de Graaff devised the high-voltage electrostatic generator in the 1930s. Van de Graaff\u2019s machine, shown in Figure 21-3a on the previous page, is a device that transfers large amounts of charge from one part of the machine to a metal terminal at the top of the device. Charge is transferred onto a moving belt at the base of the generator, position A, and is transferred off the belt at the metal dome at the top, position B. An electric motor does the work needed to increase the electric potential energy. A person touching the terminal of a Van de Graaff machine is charged electrically. The charges on the person\u2019s hairs repel each other, causing the hairs to follow the field lines, as shown in Figure 21-3b. Another method of visualizing field lines is to use grass seed in an insulating liquid, such as mineral oil. The electric forces cause a separation of charge in each long, thin grass seed. The seeds then turn so that they line up along the direction of the electric field. The seeds thus form a pattern of the electric field lines, as in Figure 21-4. Field lines do not really exist. They are simply a means of providing a model of an electric field. Electric fields, on the other hand, do exist. Although they provide a method of calculating the force on a charged body, they do not explain why charged bodies exert forces on each other. 21.1 Section Review 11. Measuring Electric Fields Suppose you are asked to measure the electric field in space. How do you detect the field at a point? How do you determine the magnitude of the field? How do you choose the magnitude of the test charge? What do you do next? 13. Field Lines In Figure 21-4, can you tell which charges are positive and which are negative", "? What would you add to complete the field lines? 14. Field Versus Force How does the electric field, E, at the test charge differ from the force, F, on it? 12. Field Strength and Direction A positive test charge of magnitude 2.40108 C experiences a force of 1.50103 N toward the east. What is the electric field at the position of the test charge? 15. Critical Thinking Suppose the top charge in Figure 21-2c is a test charge measuring the field resulting from the two negative charges. Is it small enough to produce an accurate measure? Explain. 568 Kodansha Chapter 21 Electric Fields physicspp.com/self_check_quiz 569-579 CH21-S2-845813 6/3/04 10:14 AM Page 569 21.2 Applications of Electric Fields As you have learned, the concept of energy is extremely useful in mechanics. The law of conservation of energy allows us to solve motion problems without knowing the forces in detail. The same is true in the study of electrical interactions. The work performed moving a charged particle in an electric field can result in the particle\u2019s gaining potential, or kinetic energy, or both. Because this chapter investigates charges at rest, only changes in potential energy will be discussed. Energy and Electric Potential Recall the change in gravitational potential energy of a ball when it is lifted, as shown in Figure 21-5. Both the gravitational force, F, and the gravitational field, g F/m, point toward Earth. If you lift a ball against the force of gravity, you do work on it, thereby increasing its potential energy. The situation is similar with two unlike charges: they attract each other, and so you must do work to pull one charge away from the other. When you do the work, you transfer energy to the charge where that energy is stored as potential energy. The larger the test charge, the greater the increase in its potential energy, PE. Although the force on the test charge depends on its magnitude, q, the electric field it experiences does not. The electric field, E F/q, is the force per unit charge. In a similar way, the electric potential difference, V, is defined as the work done moving a positive test charge between two points in an electric field divided by the magnitude of the test charge. W q on Electric Potential Difference V q The difference in electrical potential is the ratio of the work needed to move a charge to the strength of that charge. Electric potential difference is measured in joules per coulomb. One joule per coulomb is called a volt (J/C V). Consider the situation shown in Figure 21-6 on the next page. The negative charge creates an electric field toward itself. Suppose you place a small positive test charge in the field at position A. It will experience a force in the direction of the field. If you now move the test charge away from the negative charge to position B, as in Figure 21-6a, you will have to exert a force, F, on the charge. Because the force that you exert is in the same direction as the displacement, the work that you do on the test charge is positive. Therefore, there also will be a positive change in the electric potential difference. The change in potential difference does not depend on the magnitude of the test charge. It depends only on the field and the displacement. Increase in gravitational potential energy Ball g a Objectives \u2022 Define electric potential difference. \u2022 Calculate potential difference from the work required to move a charge. \u2022 Describe how charges are distributed on solid and hollow conductors. \u2022 Solve problems pertaining to capacitance. Vocabulary electric potential difference volt equipotential capacitor capacitance \u25a0 Figure 21-5 Work is needed to move an object against the force of gravity (a) and against the electric force (b). In both cases, the potential energy of the object is increased. b E B A Increase in electric potential energy Section 21.2 Applications of Electric Fields 569 569-579 CH21-S2-845813 3/18/04 10:05 PM Page 570 \u25a0 Figure 21-6 Electric potential difference is determined by measuring the work per unit charge. If you move unlike charges apart, you increase the electric potential difference (a). If you move unlike charges closer together, you reduce the electric potential difference (b). a B A Fon charge d Fon charge b B A d PE(A) Low V Won charge work by you PE(B) PE(B) High V High V Won charge work by charge PE(A) Low V Suppose you now move the test charge back to position A from position B, as in Figure 21-6b. The force that you exert is now in the direction opposite the displacement, so the work that you do is negative. The electric potential difference is also negative. In fact, it is equal and opposite to the potential difference for the move from position A to ", "position B. The electric potential difference does not depend on the path used to go from one position to another. It does depend on the two positions. Is there always an electric potential difference between the two positions? Suppose you move the test charge in a circle around the negative charge. The force that the electric field exerts on the test charge is always perpendicular to the direction in which you moved it, so you do no work. Therefore, the electric potential difference is zero. Whenever the electric potential difference between two or more positions is zero, those positions are said to be at equipotential. Only differences in potential energy can be measured. The same is true of electric potential; thus, only differences in electric potential are important. The electric potential difference from point A to point B is defined as V VB VA. Electric potential differences are measured with a voltmeter. Sometimes, the electric potential difference is simply called the voltage. Do not confuse electric potential difference, V, with the unit for volts, V. a F F b F F F F High V High V Low V F F Low V Static Electricity Modern electronic devices, such as personal computers, contain components that are easily damaged by static electric discharges. To prevent damage to these sensitive components during repair, a technician will wear a conductive strap around his or her wrist. The other end of this strap is clipped to a grounded piece of metal. The strap conducts charge away from the technician and eliminates any possible potential difference with the grounded equipment. \u25a0 Figure 21-7 Electric potential is smaller when two unlike charges are closer together (a) and larger when two like charges are closer together (b). 570 Chapter 21 Electric Fields 569-579 CH21-S2-845813 3/18/04 10:06 PM Page 571 You have seen that electric potential difference increases as a positive test charge is separated from a negative charge. What happens when a positive test charge is separated from a positive charge? There is a repulsive force between these two charges. Potential energy decreases as the two charges are moved farther apart. Therefore, the electric potential is smaller at points farther from the positive charge, as shown in Figure 21-7. As you learned in Chapter 11, the potential energy of a system can be defined as zero at any reference point. In the same way, the electric potential of any point can be defined as zero. No matter what reference point is chosen, the value of the electric potential difference from point A to point B always will be the same. The Electric Potential in a Uniform Field A uniform electric force and field can be made by placing two large, flat, conducting plates parallel to each other. One is charged positively and the other is charged negatively. The electric field between the plates is constant, except at the edges of the plates, and its direction is from the positive to the negative plate. The pattern formed by the grass seeds pictured in Figure 21-8 represents the electric field between parallel plates. If a positive test charge, q, is moved a distance, d, in the direction opposite the electric field direction, the work done is found by the relationship Won q Fd. Thus, the electric potential difference, the work done per unit charge, is V = Fd/q (F/q)d. Now, the electric field intensity is the force per unit charge, E F/q. Therefore, the electric potential difference, V, between two points a distance, d, apart in a uniform field, E, is represented by the following equation. Electric Potential Difference in a Uniform Field V Ed The electrical potential difference in a uniform field is equal to the product of electric field intensity and the distance moved by a charge. The electric potential increases in the direction opposite the electric field direction. That is, the electric potential is higher near the positively charged plate. By dimensional analysis, the product of the units of E and d is (N/C)(m). This is equivalent to one J/C, which is the definition of 1 V. \u25a0 Figure 21-8 A representation of an electric field between parallel plates is shown. 16. The electric field intensity between two large, charged, parallel metal plates is 6000 N/C. The plates are 0.05 m apart. What is the electric potential difference between them? 17. A voltmeter reads 400 V across two charged, parallel plates that are 0.020 m apart. What is the electric field between them? 18. What electric potential difference is applied to two metal plates that are 0.200 m apart if the electric field between them is 2.50103 N/C? 19. When a potential difference of 125 V is applied to two parallel plates, the field between them is 4.25103 N/C. How far apart are the plates? 20. A potential difference of 275 V is applied to two parallel plates that are 0.35 cm apart. What is the electric field between the plates? Section 21.2 Applications of Electric Fields 571 Kodansha 569-579 CH21-S", "2-845813 3/18/04 10:06 PM Page 572 Work Required to Move a Proton Between Charged Parallel Plates Two charged parallel plates are 1.5 cm apart. The magnitude of the electric field between the plates is 1800 N/C. a. What is the electric potential difference between the plates? b. What work is required to move a proton from the negative plate to the positive plate? 1 Analyze and Sketch the Problem \u2022 Draw the plates separated by 1.5 cm. \u2022 Label one plate with positive charges and the other with negative charges. \u2022 Draw uniformly spaced electric field lines from the positive plate to the negative plate. \u2022 Indicate the electric field strength between the plates. \u2022 Place a proton in the electric field. Known: E 1800 N/C d 1.5 cm q 1.601019 C Unknown: \u2206V? W? E 1800 N/C d 1.5 cm Math Handbook Operations with Scientific Notation pages 842\u2014843 2 Solve for the Unknown V Ed (1800 N/C)(0.015 m) 27 V V W q W qV Substitute E 1800 N/C, d 0.015 m (1.601019 C)(27 V) 4.31018 J Substitute q 1.601019 C, V 27 V 3 Evaluate the Answer \u2022 Are the units correct? (N/C)(m) Nm/C J/C V. The units work out to be volts. CV C(J/C) J, the unit for work. \u2022 Does the sign make sense? Positive work must be done to move a positive charge toward a positive plate. \u2022 Is the magnitude realistic? With such a small charge moved through a potential difference of a few volts, the work performed will be small. 21. What work is done when 3.0 C is moved through an electric potential difference of 1.5 V? 22. A 12-V car battery can store 1.44106 C when it is fully charged. How much work can be done by this battery before it needs recharging? 23. An electron in a television picture tube passes through a potential difference of 18,000 V. How much work is done on the electron as it passes through that potential difference? 24. If the potential difference in problem 18 is between two parallel plates that are 2.4 cm apart, what is the magnitude of the electric field between them? 25. The electric field in a particle-accelerator machine is 4.5105 N/C. How much work is done to move a proton 25 cm through that field? 572 Chapter 21 Electric Fields 569-579 CH21-S2-845813 6/7/04 8:28 AM Page 573 Electric Fields Tie a pith ball on the end of a 20-cm nylon thread and tie the other end to a plastic straw. Holding the straw horizontally, notice that the ball hangs straight down. Now rub a piece of wool on a 30 cm 30 cm square of plastic foam to charge both objects. Stand the foam vertically. Hold the straw and touch the pith ball to the wool. 1. Predict what will happen when the ball is close to the foam. 2. Test your prediction by slowly bringing the hanging ball toward the charged plastic foam. 3. Predict the ball\u2019s behavior at different locations around the foam, and test your prediction. 4. Observe the angle of the thread as you move the pith ball to different regions around the foam. Analyze and Conclude 5. Explain, in terms of the electric field, why the ball swings toward the charged plastic. 6. Compare the angle of the thread at various points around the foam. Why did it change? 7. Infer what the angle of the thread indicates about the strength and the direction of the electric field. Millikan\u2019s Oil-Drop Experiment One important application of the uniform electric field between two parallel plates is the measurement of the charge of an electron. This first was determined by American physicist Robert A. Millikan in 1909. Figure 21-9 shows the method used by Millikan to measure the charge carried by a single electron. First, fine oil drops were sprayed from an atomizer into the air. These drops were charged by friction with the atomizer as they were sprayed. Gravity acting on the drops caused them to fall, and a few of them entered the hole in the top plate of the apparatus. An electric potential difference then was placed across the two plates. The resulting electric field between the plates exerted a force on the charged drops. When the top plate was made positive enough, the electric force caused negatively charged drops to rise. The electric potential difference between the plates was adjusted to suspend a charged drop between the plates. At this point, the downward force of Earth\u2019s gravitational field and the upward force of the electric field were equal in magnitude. The magnitude of the electric field, E, was determined from the electric potential difference between", " the plates. A second measurement had to be made to find the weight of the drop using the relationship mg, which was too tiny to measure by ordinary methods. To make this measurement, a drop first was suspended. Then, the electric field was turned off, and the rate of the fall of the drop was measured. Because of friction with the air molecules, the oil drop quickly reached terminal velocity, which was related to the mass of the drop by a complex equation. Using the measured terminal velocity to calculate mg and knowing E, the charge, q, could be calculated. Charge on an electron Millikan found that there was a great deal of variation in the charges of the drops. When he used X rays to ionize the air and add or remove electrons from the drops, he noted, however, that the changes in the charge on the drops were always a multiple of 1.601019 C. The changes were caused by one or more electrons being added to or removed from the drops. Millikan concluded that the smallest change in charge that could occur was the amount of charge of one electron. Therefore, Millikan proposed that each electron always has the same charge, 1.601019 C. Millikan\u2019s experiment showed that charge is quantized. This means that an object can have only a charge with a magnitude that is some integral multiple of the charge of an electron. Atomizer Charged plate Battery Oil drop Microscope Charged plate \u25a0 Figure 21-9 This illustration shows a cross-sectional view of the apparatus that Millikan used to determine the charge on an electron. Section 21.2 Applications of Electric Fields 573 569-579 CH21-S2-845813 3/18/04 10:07 PM Page 574 Finding the Charge on an Oil Drop In a Millikan oil-drop experiment, a drop has been found to weigh 2.41014 N. The parallel plates are separated by a distance of 1.2 cm. When the potential difference between the plates is 450 V, the drop is suspended, motionless. a. What is the charge on the oil drop? b. If the upper plate is positive, how many excess electrons are on the oil drop? 1 Analyze and Sketch the Problem \u2022 Draw the plates with the oil drop suspended between them. \u2022 Draw and label vectors representing the forces. \u2022 Indicate the potential difference and the distance between the plates. Known: V 450 V Fg d 1.2 cm 2.41014 N qE Fe Unknown: charge on drop, q? number of electrons, n? V 450 V d 1.2 cm Fg mg 2 Solve for the Unknown To be suspended, the electric force and gravitational force must be balanced. Fg Fe qE Fg q V Fg d Solve for q. d F g q V (2.41014 N)(0.012 m) 450 V 6.41019 C Substitute Fe qE V Substitute E d Math Handbook Isolating a Variable page 845 Substitute Fg 2.41014 N, d 0.012 m, V 450 V Solve for the number of electrons on the drop Substitute q 6.41019 C, e 1.61019 C 3 Evaluate the Answer \u2022 Are the units correct? Nm/V J/(J/C) C, the unit for charge. \u2022 Is the magnitude realistic? This is a small whole number of elementary charges. 26. A drop is falling in a Millikan oil-drop apparatus with no electric field. What forces are acting on the oil drop, regardless of its acceleration? If the drop is falling at a constant velocity, describe the forces acting on it. 27. An oil drop weighs 1.91015 N. It is suspended in an electric field of 6.0103 N/C. What is the charge on the drop? How many excess electrons does it carry? 28. An oil drop carries one excess electron and weighs 6.41015 N. What electric field strength is required to suspend the drop so it is motionless? 29. A positively charged oil drop weighing 1.21014 N is suspended between parallel plates separated by 0.64 cm. The potential difference between the plates is 240 V. What is the charge on the drop? How many electrons is the drop missing? 574 Chapter 21 Electric Fields 569-579 CH21-S2-845813 3/18/04 10:08 PM Page 575 a Charged sphere Neutral sphere A B b \u25a0 Figure 21-10 A charged sphere shares charge equally with a neutral sphere of equal size when they are placed in contact with each other. Sharing of Charge All systems come to equilibrium when the energy of the system is at a minimum. For example, if a ball is placed on a hill, it finally will come to rest in a valley where its gravitational potential energy is smallest. This also would be the location where its gravitational potential has been reduced by the largest amount. This same principle explains what happens when an insulated,", " positively charged metal sphere, such as the one shown in Figure 21-10, touches a second, uncharged sphere. The excess charges on sphere A repel each other, so when the neutral sphere, B, touches sphere A, there is a net force on the charges on A toward B. Suppose that you were to physically move the charges, individually, from A to B. When you move the first charge, the other charges on A would push it toward B, so, to control its speed, you would have to exert a force in the opposite direction. Therefore, you do negative work on it, and the electric potential difference from A to B is negative. When the next few charges are moved, they feel a small repulsive force from the charges already on B, but there is still a net positive force in that direction. At some point, the force pushing a charge off A will equal the repulsive force from the charges on B, and the electric potential difference is zero. After this point of equilibrium, work would have to be done to move the next charge to B, so this would not happen by itself and would require an increase in the energy of the system. However, if you did continue to move charges, the electric potential difference from A to B would then be positive. Thus, you can see that charges would move from A to B without external forces until there is no electric potential difference between the two spheres. Different sizes of spheres Suppose that the two spheres have different sizes, as in Figure 21-11. Although the total numbers of charges on the two spheres are the same, the larger sphere has a larger surface area, so the charges can spread farther apart, and the repulsive force between them is reduced. Thus, if the two spheres now are touched together, there will be a net force that will move charges from the smaller to the larger sphere. Again, the charges will move to the sphere with the lower electric potential until there is no electric potential difference between the two spheres. In this case, the larger sphere will have a larger charge when equilibrium is reached. Metal Spheres of Unequal Size a Low V High V b Same V Same q Different q \u25a0 Figure 21-11 Charges are transferred from a sphere with high potential to a sphere with lower potential when they touch. The charges move to create no potential difference. Section 21.2 Applications of Electric Fields 575 569-579 CH21-S2-845813 3/18/04 10:08 PM Page 576 \u25a0 Figure 21-12 The ground wire on a fuel truck prevents ignition of the gasoline vapors. a b c A B C \u25a0 Figure 21-13 On a conducting sphere, (a), the charge is evenly distributed around the surface. The charges on the hollow sphere, (b), are entirely on the outer surface. In irregular shapes, (c), the charges will be closest together at sharp points. 576 Tim Courlas Chapter 21 Electric Fields The same principle explains how charges move on the individual spheres, or on any conductor. They distribute themselves so that the net force on each charge is zero. With no force, there is no electric field along the surface of the conductor. Thus, there is no electric potential difference anywhere on the surface. The surface of a conductor is, therefore, an equipotential surface. If a charged body is grounded by touching Earth, almost any amount of charge can flow to Earth until the electric potential difference between that body and Earth is reduced to zero. Gasoline trucks, for example, can become charged by friction. If the charge on a gasoline truck were to jump to Earth through gasoline vapor, it could cause an explosion. To prevent this, a metal wire on the truck safely conducts the charge to the ground, as shown in Figure 21-12. Similarly, if a computer is not grounded, an electric potential difference between the computer and Earth can occur. If a person then touches the computer, charges could flow through the computer to the person and damage the equipment or hurt the person. Electric Fields Near Conductors The charges on a conductor are spread as far apart as they can be to make the energy of the system as low as possible. The result is that all charges are on the surface of a solid conductor. If the conductor is hollow, excess charges will move to the outer surface. If a closed metal container is charged, there will be no charges on the inside surfaces of the container. In this way, a closed metal container shields the inside from electric fields. For example, people inside a car are protected from the electric fields generated by lightning. Likewise, on an open coffee can, there will be very few charges inside and none near the bottom. Even if the inner surface of an object is pitted or bumpy, giving it a larger surface area than the outer surface, the charge still will be entirely on the outside. On the outside of a conductor, however, the electric field often is not zero. Even though the surface of a conductor is at an equipotential, the electric field", " around the outside of it depends on the shape of the conductor, as well as on the electric potential difference between it and Earth. The charges are closer together at sharp points of a conductor, as indicated in Figure 21-13. Therefore, the field lines are closer together and the field is stronger. This field can become so strong that when electrons are knocked off of atoms by passing cosmic rays, the electrons and resulting ions are accelerated by the field, causing them to strike other atoms, resulting in more ionization of atoms. This chain reaction is what results in the pink glow, 569-579 CH21-S2-845813 6/7/04 8:34 AM Page 577 such as that seen inside a gas-discharge sphere. If the field is strong enough, when the particles hit other molecules they will produce a stream of ions and electrons that form a plasma, which is a conductor. The result is a spark, or, in extreme cases, lightning. To reduce discharges and sparking, conductors that are highly charged or that operate at high potentials are made smooth in shape to reduce the electric fields. In contrast, a lightning rod is pointed so that the electric field will be strong near the end of the rod. As the field accelerates electrons and ions, they form the start of a conducting path from the rod to the clouds. As a result of the rod\u2019s sharply pointed shape, charges in the clouds spark to the rod, rather than to a chimney or other high point on a house or other building. From the rod, a conductor takes the charges safely to the ground. Lightning usually requires a potential difference of millions of volts between Earth and the clouds. Even a small gas-discharge tube operates at several thousand volts. Household wiring, on the other hand, does not normally carry a high enough potential difference to cause such discharges. Storing Charges: The Capacitor When you lift a book, you increase its gravitational potential energy. This can be interpreted as storing energy in a gravitational field. In a similar way, you can store energy in an electric field. In 1746, Dutch physician and physicist Pieter Van Musschenbroek invented a small device that could store a large electric charge. In honor of the city in which he worked, it was called a Leyden jar. Benjamin Franklin used a Leyden jar to store the charge from lightning and in many other experiments. A version of the Leyden jar is still in use today in electric equipment. This new device for storing a charge has a new form, is much smaller in size, and is called a capacitor. As charge is added to an object, the electric potential difference between that object and Earth increases. For a given shape and size of an object, the ratio of charge stored to electric potential difference, q/V, is a constant called the capacitance, C. For a small sphere far from the ground, even a small amount of added charge will increase the electric potential difference. Thus, C is small. A larger sphere can hold more charge for the same increase in electric potential difference, and its capacitance is larger. Capacitors are designed to have specific capacitances. All capacitors are made up of two conductors that are separated by an insulator. The two conductors have equal and opposite charges. Capacitors are used today in electric circuits to store charge. Commercial capacitors, such as those shown in Figure 21-14, typically contain strips of aluminum foil separated by thin plastic that are tightly rolled up to conserve space. The capacitance of a capacitor is independent of the charge on it, and can be measured by first placing charge q on one plate and charge q on the other, and then measuring the electric potential difference, V, that results. The capacitance is found by using the following equation, and is measured in farads, F. q Capacitance C V Capacitance is the ratio of charge on one plate to potential difference. \u25a0 Figure 21-14 Various types of capacitors are pictured below. Section 21.2 Applications of Electric Fields 577 Horizons Companies 569-579 CH21-S2-845813 3/18/04 10:10 PM Page 578 The farad as a unit of measure One farad, F, named after Michael Faraday, is one coulomb per volt, C/V. Just as 1 C is a large amount of charge, 1 F is also a fairly large capacitance. Most capacitors used in modern electronics have capacitances between 10 picofarads (101012 F) and 500 microfarads (500106 F). However, memory capacitors that are used to prevent loss of memory in some computers can have capacitance from 0.5 F to 1.0 F. Note that if the charge is increased, the electric potential difference also increases. The capacitance depends only on the construction of the capacitor, not on the charge, q. Finding Capacitance A sphere has an electric", " potential difference between it and Earth of 40.0 V when it has been charged to 2.4106 C. What is its capacitance? 1 Analyze and Sketch the Problem \u2022 Draw a sphere above Earth and label the charge and q 2.4106 C V 40.0 V potential difference. Known: V 40.0 V q 2.4106 C Unknown: C? 2 Solve for the Unknown q C V 0 6 C Substitute V 40.0 V, q 2.4106 C 1 2.4 V.0 0 4 6.0108 F 0.060 F 3 Evaluate the Answer \u2022 Are the units correct? C/V = F. The units are farads. \u2022 Is the magnitude realistic? A small capacitance would store a small charge at a low voltage. Math Handbook Operations with Significant Digits pages 835\u2014836 30. A 27-F capacitor has an electric potential difference of 45 V across it. What is the charge on the capacitor? 31. Both a 3.3-F and a 6.8-F capacitor are connected across a 24-V electric potential difference. Which capacitor has a greater charge? What is it? 32. The same two capacitors as in problem 31 are each charged to 3.5104 C. Which has the larger electric potential difference across it? What is it? 33. A 2.2-F capacitor first is charged so that the electric potential difference is 6.0 V. How much additional charge is needed to increase the electric potential difference to 15.0 V? 34. When a charge of 2.5105 C is added to a capacitor, the potential difference increases from 12.0 V to 14.5 V. What is the capacitance of the capacitor? 578 Chapter 21 Electric Fields 569-579 CH21-S2-845813 3/18/04 10:11 PM Page 579 The plates of a capacitor attract each other because they carry opposite charges. A capacitor consisting of two parallel plates that are separated by a distance, d, has capacitance, C. 1. Derive an expression for the force between the two plates when the capacitor has charge, q. 2. What charge must be stored on a 22-F capacitor to have a force of 2.0 N between the plates if they are separated by 1.5 mm? d Varieties of capacitors Capacitors have many shapes and sizes, as shown in Figure 21-14. Some are large enough to fill whole rooms and can store enough charge to create artificial lightning or power giant lasers that release thousands of joules of energy in a few billionths of a second. Capacitors in television sets can store enough charge at several hundred volts to be very dangerous if they are touched. These capacitors can remain charged for hours after the televisions have been turned off. This is why you should not open the case of a television or a computer monitor even if it is unplugged. The capacitance of a capacitor is controlled by varying the surface area of the two conductors, or plates, within a capacitor, by the distance between the plates, and by the nature of the insulating material. Capacitors are named for the type of insulator, or dielectric, used to separate the plates, and include ceramic, mica, polyester, paper, and air. Higher capacitance is obtained by increasing the surface area and decreasing the separation of the plates. Certain dielectrics have the ability to effectively offset some of the charge on the plates and allow more charge to be stored. 21.2 Section Review 35. Potential Difference What is the difference between electric potential energy and electric potential difference? 39. Capacitance How much charge is stored on a 0.47-F capacitor when a potential difference of 12 V is applied to it? 36. Electric Field and Potential Difference Show that a volt per meter is the same as a newton per coulomb. 37. Millikan Experiment When the charge on an oil drop suspended in a Millikan apparatus is changed, the drop begins to fall. How should the potential difference on the plates be changed to bring the drop back into balance? 38. Charge and Potential Difference In problem 37, if changing the potential difference has no effect on the falling drop, what does this tell you about the new charge on the drop? 40. Charge Sharing If a large, positively charged, conducting sphere is touched by a small, negatively charged, conducting sphere, what can be said about the following? a. the potentials of the two spheres b. the charges on the two spheres 41. Critical Thinking Referring back to Figure 21-3a, explain how charge continues to build up on the metal dome of a Van de Graaff generator. In particular, why isn\u2019t charge repelled back onto the belt at point B? physicspp.com/self_check_quiz Section 21.2 Applications of Electric Fields 579 580-581 CH21-LAB-", "845813 6/7/04 8:39 AM Page 580 Alternate CBL instructions can be found on the Web site. physicspp.com Charging of Capacitors A capacitor is an electric device that is made from two conductors, or plates, that are separated by an insulator. It is designed to have a specific capacitance. The capacitance depends on the physical characteristics and geometric arrangement of the conductors and the insulator. In the circuit schematic, the capacitor appears to create an open circuit, even when the switch is in the closed position. However, because capacitors store charge, when the switch is closed, charge from the battery will move to the capacitor. The equal, but opposite charges on the two plates within the capacitor establishes a potential difference, or voltage. As charge is added to the capacitor, the electric potential difference increases. In this laboratory activity you will examine the charging of several different capacitors. QUESTION How do the charging times of different capacitors vary with capacitance? Objectives Procedure \u25a0 Collect and organize data on the rate of 1. Before you begin, leave the switch open (off). charge of different capacitors. \u25a0 Compare and contrast the rate of charging for different capacitances. \u25a0 Make and use graphs of potential difference versus time for several capacitors. Safety Precautions Materials 9-V battery 9-V battery clip hook-up wires switch voltmeter 47-k resistor stopwatch capacitors: 1000 F, 500 F, 240 F Do not attach the battery at this time. CAUTION: Be careful to avoid a short circuit, especially by permitting the leads from the battery clip to touch each other. Connect the circuit, as illustrated. Do this by connecting either end of the resistor to one side of the switch. The resistor is used to reduce the charging of the capacitor to a measurable rate. Connect the other end of the resistor to the negative side of the 9-V battery clip. Inspect your 1000-F capacitor to determine whether either end is marked with a negative sign, or an arrow with negative signs on it, that points to the lead that is to be connected to the negative side of the battery. Connect this negative lead to the other side of the switch. Attach the unconnected (positive) lead of the capacitor to the positive lead from the battery clip. 2. Connect the positive terminal of the voltmeter to the positive side of the capacitor and the negative terminal to the negative side of the capacitor. Compare your circuit to the photo to verify your connections. Attach the battery after your teacher has inspected the circuit. 3. Prepare a data table having columns for time and potential difference on each of the three different capacitors. 4. One person should watch the time and another should record potential difference at the designated times. Close the switch and measure the voltage at 5-s intervals. Open the switch after you have collected data. 580 Horizons Companies 580-581 CH21-LAB-845813 6/16/04 10:14 PM Page 581 Data Table Time (s) Voltage (V) across 1000 F Voltage (V) across 500 F Voltage (V) across 240 F Time (s) Voltage (V) across 1000 F Voltage (V) across 500 F Voltage (V) across 240 F 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 5. When you have completed the trial, take a short piece of wire and place it across both ends of the capacitor. This will cause the capacitor to discharge. 6. Replace the 1000-F capacitor with a 500-F capacitor. Repeat steps 4\u20135 and enter data into the appropriate columns of your data table for the 500-F capacitor. 7. Replace the 500-F capacitor with a 240-F capacitor. Repeat steps 4\u20135 and enter data into the appropriate column of your data table for this last capacitor. Analyze 1. Observe and Infer Does each capacitor charge to 9 V? Propose an explanation for the observed behavior. 2. Make and Use Graphs Prepare a graph that plots the time horizontally and the potential difference vertically. Make a separate labeled line for each capacitor. Conclude and Apply 1. Interpret Data Does the voltage on the capacitor immediately jump to the battery\u2019s potential difference (9-V)? Explain the reason for the observed behavior. 2. Infer Does the larger capacitor require a longer time to become fully charged? Explain why or why not. Going Further 1. The time for a capacitor to charge to the voltage of the battery depends upon its capacitance and the opposition to the flow of charge in the circuit. In this lab, the opposition to the flow of charge was controlled by the 47-k resistor that was placed in the circuit. In circuits with a capacitor and resistance, such as in this activity, the time in seconds to charge the capacitor to 63.3 percent of the applied voltage is equal to the product of the capacitor and resistance. This is called the time constant. Therefore, T", " RC, where T is in seconds, R is in ohms, and C is in microfarads. Calculate the time constant for each of the capacitors with the 47-k resistor. 2. Compare your time constants to the values from your graph. Real-World Physics Explain Small, disposable, flash cameras, as well as regular electronic flash units, require time before the flash is ready to be used. A capacitor stores the energy for the flash. Explain what might be going on during the time you must wait to take your next picture. To find out more about electric fields, visit the Web site: physicspp.com 581 582 CH21-FEATURE-845813 3/18/04 10:17 PM Page 582 Lightning Rods Lightning can be very destructive because it creates huge currents in materials that are poor conductors and generates a great deal of heat. In addition to protecting a structure by dissipating some of the charge before lightning strikes, lightning rods are excellent conductors that provide a safe path for the current. Benjamin Franklin is credited with inventing the lightning rod in the 1750s. 3 Positive charges spark out from the lightning rod, meeting the step leader. The conducting path is complete and current neutralizes the separation of charges. Even if the strike does not hit the lightning rod directly, the massive current still can leap to the rod, which is the path of least resistance to the ground. Step leader Positive arc Lightning rod 1 In a thunderstorm, negative charges accumulate in the lowest regions of clouds. The negative electric field from the clouds repels electrons on the ground, inducing a positive charge terminal. 2 The strong electric field 4 The current Thinking Critically accelerates electrons and ions, causing a chain reaction in the air, forming plasma. The ionized air is a conductor, and it branches out from the cloud forming what are called step leaders. travels safely through the conductor to the ground terminal. 1. Hypothesize Along what path would the current travel if a house without a lightning rod were struck by lightning? 2. Evaluate Should the resistance between the ground terminal and Earth be high or low? 3. Infer What are the dangers of an incorrectly installed lightning rod system? 582 How It Works 583-589 CH21-SG CA STP-845813 3/18/04 10:24 PM Page 583 21.1 Creating and Measuring Electric Fields Vocabulary \u2022 electric field (p. 563) \u2022 electric field line (p. 567) Key Concepts \u2022 An electric field exists around any charged object. The field produces forces on other charged objects. \u2022 The electric field is the force per unit charge. F E q \u2022 The direction of the electric field is the direction of the force on a tiny, positive test charge. \u2022 Electric field lines provide a picture of the electric field. They are directed away from positive charges and toward negative charges. They never cross, and their density is related to the strength of the field. 21.2 Applications of Electric Fields Vocabulary \u2022 electric potential difference (p. 569) \u2022 volt (p. 569) \u2022 equipotential (p. 570) \u2022 capacitor (p. 577) \u2022 capacitance (p. 577) Key Concepts \u2022 Electric potential difference is the change in potential energy per unit charge in an electric field. V W q \u2022 Electric potential differences are measured in volts. \u2022 The electric field between two parallel plates is uniform between the plates, except near the edges. In a uniform field, the potential difference is related to the field strength by the following. V Ed \u2022 Robert Millikan\u2019s experiments showed that electric charge is quantized. \u2022 Robert Millikan also showed that the negative charge carried by an electron is 1.601019 C. \u2022 Charges will move in a conductor until the electric potential is the same everywhere on the conductor. \u2022 Grounding makes the potential difference between an object and Earth equal to zero. \u2022 Grounding can prevent sparks resulting from a neutral object making contact with objects that have built-up charge on them. \u2022 Electric fields are strongest near sharply pointed conductors. \u2022 Capacitance is the ratio of the charge on an object to its electric potential difference. q C V \u2022 Capacitance is independent of the charge on an object and the electric potential difference across it. \u2022 Capacitors are used to store charge. physicspp.com/vocabulary_puzzlemaker 583 583-589 CH21-SG CA STP-845813 6/15/04 11:23 PM Page 584 Concept Mapping 42. Complete the concept map below using the following terms: capacitance, field strength, J/C, work. force N/C Electric Field test charge C/V potential difference Mastering Concepts 43. What are the two properties that a test charge must have? (21.1) 44. How is the direction of an electric field defined? (21.1) 45. What are electric field lines? (21.1", ") 46. How is the strength of an electric field indicated with electric field lines? (21.1) 47. Draw some of the electric field lines between each of the following. (21.1) a. two like charges of equal magnitude b. two unlike charges of equal magnitude c. a positive charge and a negative charge having twice the magnitude of the positive charge d. two oppositely charged parallel plates 48. In Figure 21-15, where do the electric field lines leave the positive charge end? (21.1) \u25a0 Figure 21-15 49. What SI unit is used to measure electric potential energy? What SI unit is used to measure electric potential difference? (21.2) 50. Define volt in terms of the change in potential energy of a charge moving in an electric field. (21.2) 51. Why does a charged object lose its charge when it is touched to the ground? (21.2) 52. A charged rubber rod that is placed on a table maintains its charge for some time. Why is the charged rod not discharged immediately? (21.2) 53. A metal box is charged. Compare the concentration of charge at the corners of the box to the charge concentration on the sides of the box. (21.2) 54. Computers Delicate parts in electronic equipment, such as those pictured in Figure 21-16, are contained within a metal box inside a plastic case. Why? (21.2) \u25a0 Figure 21-16 Applying Concepts 55. What happens to the strength of an electric field when the charge on the test charge is halved? 56. Does it require more energy or less energy to move a constant positive charge through an increasing electric field? 57. What will happen to the electric potential energy of a charged particle in an electric field when the particle is released and free to move? 58. Figure 21-17 shows three spheres with charges of equal magnitude, with their signs as shown. Spheres y and z are held in place, but sphere x is free to move. Initially, sphere x is equidistant from spheres y and z. Choose the path that sphere x will begin to follow. Assume that no other forces are acting on the spheres. E D x C A B y z \u25a0 Figure 21-17 584 Horizons Companies Chapter 21 Electric Fields For more problems, go to Additional Problems, Appendix B. 583-589 CH21-SG CA STP-845813 3/18/04 10:25 PM Page 585 59. What is the unit of electric potential difference in 67. A positive charge of 1.0105 C, shown in terms of m, kg, s, and C? 60. What do the electric field lines look like when the electric field has the same strength at all points in a region? 61. Millikan Oil-Drop Experiment When doing a Millikan oil-drop experiment, it is best to work with drops that have small charges. Therefore, when the electric field is turned on, should you try to find drops that are moving rapidly or slowly? Explain. 62. Two oil drops are held motionless in a Millikan oil-drop experiment. a. Can you be sure that the charges are the same? b. The ratios of which two properties of the oil drops have to be equal? 63. Jos\u00e9 and Sue are standing on an insulating platform and holding hands when they are given a charge, as in Figure 21-18. Jos\u00e9 is larger than Sue. Who has the larger amount of charge, or do they both have the same amount? \u25a0 Figure 21-18 64. Which has a larger capacitance, an aluminum sphere with a 1-cm diameter or one with a 10-cm diameter? 65. How can you store different amounts of charge in a capacitor? Mastering Problems 21.1 Creating and Measuring Electric Fields The charge of an electron is 1.601019 C. 66. What charge exists on a test charge that experiences a force of 1.4108 N at a point where the electric field intensity is 5.0104 N/C? Figure 21-19, experiences a force of 0.30 N when it is located at a certain point. What is the electric field intensity at that point? 0.30 N 1.0105 C \u25a0 Figure 21-19 68. A test charge experiences a force of 0.30 N on it when it is placed in an electric field intensity of 4.5105 N/C. What is the magnitude of the charge? 69. The electric field in the atmosphere is about 150 N/C downward. a. What is the direction of the force on a negatively charged particle? b. Find the electric force on an electron with charge 1.61019 C. c. Compare the force in part b with the force of gravity on the same electron (mass 9.11031 kg). 70. Carefully sketch each of the following. a. the electric field produced by a 1.0-C charge b", ". the electric field resulting from a 2.0-C charge (Make the number of field lines proportional to the change in charge.) 71. A positive test charge of 6.0106 C is placed in an electric field of 50.0-N/C intensity, as in Figure 21-20. What is the strength of the force exerted on the test charge? q 6.0106 C E 50.0 N/C \u25a0 Figure 21-20 72. Charges X, Y, and Z all are equidistant from each other. X has a 1.0-C charge, Y has a 2.0-C charge, and Z has a small negative charge. a. Draw an arrow representing the force on charge Z. b. Charge Z now has a small positive charge on it. Draw an arrow representing the force on it. physicspp.com/chapter_test Chapter 21 Assessment 585 583-589 CH21-SG CA STP-845813 6/7/04 8:50 AM Page 586 73. In a television picture tube, electrons are accelerated by an electric field having a value of 1.00105 N/C. a. Find the force on an electron. b. If the field is constant, find the acceleration of the electron (mass 9.111031 kg). 74. What is the electric field strength 20.0 cm from a point charge of 8.0107 C? 75. The nucleus of a lead atom has a charge of 82 protons. a. What are the direction and magnitude of the electric field at 1.01010 m from the nucleus? b. What are the direction and magnitude of the force exerted on an electron located at this distance? 21.2 Applications of Electric Fields 76. If 120 J of work is performed to move 2.4 C of charge from the positive plate to the negative plate shown in Figure 21-21, what potential difference exists between the plates? 2.4 C \u25a0 Figure 21-21 77. How much work is done to transfer 0.15 C of charge through an electric potential difference of 9.0 V? 78. An electron is moved through an electric potential difference of 450 V. How much work is done on the electron? 79. A 12-V battery does 1200 J of work transferring charge. How much charge is transferred? 80. The electric field intensity between two charged plates is 1.5103 N/C. The plates are 0.060 m apart. What is the electric potential difference, in volts, between the plates? 81. A voltmeter indicates that the electric potential difference between two plates is 70.0 V. The plates are 0.020 m apart. What electric field intensity exists between them? 82. A capacitor that is connected to a 45.0-V source contains 90.0 C of charge. What is the capacitor\u2019s capacitance? 83. What electric potential difference exists across a 5.4-F capacitor that has a charge of 8.1104 C? 84. The oil drop shown in Figure 21-22 is negatively charged and weighs 4.51015 N. The drop is suspended in an electric field intensity of 5.6103 N/C. a. What is the charge on the drop? b. How many excess electrons does it carry? 8.51015 N \u25a0 Figure 21-22 85. What is the charge on a 15.0-pF capacitor when it is connected across a 45.0-V source? 86. A force of 0.065 N is required to move a charge of 37 C a distance of 25 cm in a uniform electric field, as in Figure 21-23. What is the size of the electric potential difference between the two points? 37 C 0.053 N 25 cm \u25a0 Figure 21-23 87. Photoflash The energy stored in a capacitor with capacitance C, and an electric potential difference, V, is represented by W 1 CV 2. One application 2 of this is in the electronic photoflash of a strobe light, like the one in Figure 21-24. In such a unit, a capacitor of 10.0 F is charged to 3.0102 V. Find the energy stored. \u25a0 Figure 21-24 88. Suppose it took 25 s to charge the capacitor in problem 87. a. Find the average power required to charge the capacitor in this time. b. When this capacitor is discharged through the strobe lamp, it transfers all its energy in 1.0104 s. Find the power delivered to the lamp. c. How is such a large amount of power possible? 586 Horizons Companies Chapter 21 Electric Fields For more problems, go to Additional Problems, Appendix B. 583-589 CH21-SG CA STP-845813 3/18/04 10:26 PM Page 587 89. Lasers Lasers are used to try to produce controlled fusion reactions. These lasers require brief pulses of energy that are stored in large rooms filled", " with capacitors. One such room has a capacitance of 61103 F charged to a potential difference of 10.0 kV. a. Given that W 1 CV2, find the energy stored 2 in the capacitors. b. The capacitors are discharged in 10 ns (1.0108 s). What power is produced? c. If the capacitors are charged by a generator with a power capacity of 1.0 kW, how many seconds will be required to charge the capacitors? Mixed Review 90. How much work does it take to move 0.25 C between two parallel plates that are 0.40 cm apart if the field between the plates is 6400 N/C? 91. How much charge is stored on a 0.22-F parallel plate capacitor if the plates are 1.2 cm apart and the electric field between them is 2400 N/C? 92. Two identical small spheres, 25 cm apart, carry equal but opposite charges of 0.060 \u00b5C, as in Figure 21-25. If the potential difference between them is 300 V, what is the capacitance of the system? 0.060 C 0.060 C V 300 V 25 cm \u25a0 Figure 21-25 93. The plates of a 0.047 F capacitor are 0.25 cm apart and are charged to a potential difference of 120 V. How much charge is stored on the capacitor? 94. What is the strength of the electric field between the plates of the capacitor in Problem 93 above? 95. An electron is placed between the plates of the capacitor in Problem 93 above, as in Figure 21-26. What force is exerted on that electron? V 120 V C 0.047 F 0.25 cm \u25a0 Figure 21-26 96. How much work would it take to move an additional 0.010 C between the plates at 120 V in Problem 93? 97. The graph in Figure 21-27 represents the charge stored in a capacitor as the charging potential increases. What does the slope of the line represent? Charge Stored on Capacitor ) C ( q 15 10 5 0 10 20 V (V) 30 \u25a0 Figure 21-27 98. What is the capacitance of the capacitor represented by Figure 21-27? 99. What does the area under the graph line in Figure 21-27 represent? 100. How much work is required to charge the capacitor in problem 98 to a potential difference of 25 V? 101. The work found in Problem 100 above is not equal to qV. Why not? 102. Graph the electric field strength near a positive point charge as a function of distance from it. 103. Where is the field of a point charge equal to zero? 104. What is the electric field strength at a distance of zero meters from a point charge? Is there such a thing as a true point charge? Thinking Critically 105. Apply Concepts Although a lightning rod is designed to carry charge safely to the ground, its primary purpose is to prevent lightning from striking in the first place. How does it do that? 106. Analyze and Conclude In an early set of experiments in 1911, Millikan observed that the following measured charges could appear on a single oil drop. What value of elementary charge can be deduced from these data? a. 6.5631019 C b. 8.2041019 C c. 11.501019 C d. 13.131019 C e. 16.481019 C f. 18.081019 C g. 19.711019 C h. 22.891019 C i. 26.131019 C physicspp.com/chapter_test Chapter 21 Assessment 587 583-589 CH21-SG CA STP-845813 6/7/04 8:52 AM Page 588 107. Analyze and Conclude Two small spheres, A and B, lie on the x-axis, as in Figure 21-28. Sphere A has a charge of 3.00106 C. Sphere B is 0.800 m to the right of sphere A and has a charge of 5.00106 C. Find the magnitude and direction of the electric field strength at a point above the x-axis that would form the apex of an equilateral triangle with spheres A and B. Writing in Physics 110. Choose the name of an electric unit, such as coulomb, volt, or farad, and research the life and work of the scientist for whom it was named. Write a brief essay on this person and include a discussion of the work that justified the honor of having a unit named for him. 3.00106 C A 5.00106 C B 0.800 m \u25a0 Figure 21-28 108. Analyze and Conclude In an ink-jet printer, drops of ink are given a certain amount of charge before they move between two large, parallel plates. The purpose of the plates is to deflect the charges so that they are stopped by a gutter and do not reach the paper. This is shown in", " Figure 21-29. The plates are 1.5-cm long and have an electric field of E 1.2106 N/C between them. Drops with a mass m 0.10 ng, and a charge q 1.01016 C, are moving horizontally at a speed, v 15 m/s, parallel to the plates. What is the vertical displacement of the drops when they leave the plates? To answer this question, complete the following steps. a. What is the vertical force on the drops? b. What is their vertical acceleration? c. How long are they between the plates? d. How far are they displaced? 1.5 cm Cumulative Review 111. Michelson measured the speed of light by sending a beam of light to a mirror on a mountain 35 km away. (Chapter 16) a. How long does it take light to travel the distance to the mountain and back? b. Assume that Michelson used a rotating octagon with a mirror on each face of the octagon. Also assume that the light reflects from one mirror, travels to the other mountain, reflects off of a fixed mirror on that mountain, and returns to the rotating mirrors. If the rotating mirror has advanced so that when the light returns, it reflects off of the next mirror in the rotation, how fast is the mirror rotating? c. If each mirror has a mass of 1.0101 g and rotates in a circle with an average radius of 1.0101 cm, what is the approximate centripetal force needed to hold the mirror while is it rotating? 112. Mountain Scene You can see an image of a distant mountain in a smooth lake just as you can see a mountain biker next to the lake because light from each strikes the surface of the lake at about the same angle of incidence and is reflected to your eyes. If the lake is about 100 m in diameter, the reflection of the top of the mountain is about in the middle of the lake, the mountain is about 50 km away from the lake, and you are about 2 m tall, then approximately how high above the lake does the top of the mountain reach? (Chapter 17) q m v E 1.2106 N/C \u25a0 Figure 21-29 109. Apply Concepts Suppose the Moon had a net negative charge equal to q, and Earth had a net positive charge equal to 10q. What value of q would yield the same magnitude of force that you now attribute to gravity? Gutter 113. A converging lens has a focal length of 38.0 cm. If it is placed 60.0 cm from an object, at what distance from the lens will the image be? (Chapter 18) 114. A force, F, is measured between two charges, Q and q, separated by a distance, r. What would the new force be for each of the following? (Chapter 20) a. r is tripled b. Q is tripled c. both r and Q are tripled d. both r and Q are doubled e. all three, r, Q, and q, are tripled 588 Chapter 21 Electric Fields For more problems, go to Additional Problems, Appendix B. 583-589 CH21-SG CA STP-845813 6/7/04 8:56 AM Page 589 Multiple Choice 1. Why is an electric field measured only by a small test charge? 6. How was the magnitude of the field in Millikan\u2019s oil-drop experiment determined? so the charge doesn\u2019t disturb the field because small charges have small momentum so its size doesn\u2019t nudge the charge to be measured aside because an electron always is used as the test charge and electrons are small 2. A force of 14 N exists on charge q, which is 2.1109 C. What is the magnitude of the electric field? 0.15109 N/C 6.7109 N/C 29109 N/C 6.7109 N/C 3. A positive test charge of 8.7 C experiences a force of 8.1106 N at an angle of 24\u00b0 N of E. What are the magnitude and direction of the electric field strength at the location of the test charge? 7.0108 N/C, 24\u00b0 N of E 1.7106 N/C, 24\u00b0 S of W 1.1103 N/C, 24\u00b0 W of S 9.3101 N/C, 24\u00b0 N of E 4. What is the potential difference between two plates that are 18 cm apart with a field of 4.8103 N/C? 27 V 86 V 0.86 kV 27 kV 5. How much work is done on a proton to move it from the negative plate to a positive plate 4.3 cm away if the field is 125 N/C? 5.51023 J 8.61019 J 1.11016 J 5.4 J using a measurable electromagnet from the electric potential between the plates from the magnitude of", " the charge by an electrometer 7. In an oil drop experiment, a drop with a weight of 1.91014 N was suspended motionless when the potential difference between the plates that were 63 mm apart was 0.78 kV. What was the charge on the drop? 1.51018 C 3.91016 C 1.21015 C 9.31013 C 0.78 kV 63 mm 8. A capacitor has a capacitance of 0.093 F. If the charge on the capacitor is 58 C, what is the electrical potential difference? 5.41012 V 1.6106 V 6.2102 V 5.4103 V Extended Answer 9. Assume 18 extra electrons are on an oil drop. Calculate the charge of the oil drop, and calculate the potential difference needed to suspend it if it has a weight of 6.121014 N and the plates are 14.1 mm apart. 125 N/C 4.3 cm Use the Buddy System Study in a group. A small study group works well because it allows you to draw from a broader base of skills and content knowledge. Keep your group small, question each other, and stay on target. physicspp.com/standardized_test Chapter 21 Standardized Test Practice 589 590-600 CH22-S1-845813 3/22/04 3:31 AM Page 590 What You\u2019ll Learn \u2022 You will explain energy transfer in circuits. \u2022 You will solve problems involving current, potential difference, and resistance. \u2022 You will diagram simple electric circuits. Why It\u2019s Important The electric tools and appliances that you use are based upon the ability of electric circuits to transfer energy resulting from potential difference, and thus, perform work. Power Transmission Lines Transmission lines crisscross our country to transfer energy to where it is needed. This transfer is accomplished at high potential differences, often as high as 500,000 V. Think About This Transmission line voltages are too high to use safely in homes and businesses. Why are such high voltages used in transmission lines? physicspp.com 590 Lester Lefkowitz/CORBIS 590-600 CH22-S1-845813 6/10/04 8:59 PM Page 591 Can you get a lightbulb to light? Question Given a wire, a battery, and a lightbulb, can you get the bulb to light? have in common? What do your diagrams of the unlit bulb have in common? From your observations, what conditions seem to be necessary in order for the bulb to light? Critical Thinking What causes electricity to flow through the bulb? Procedure 1. Obtain a lightbulb, a wire, and a battery. Try to find as many ways as possible to get the lightbulb to light. Caution: Wire is sharp and can cut skin. Wire can also get hot if connected across the battery. 2. Diagram two ways in which you are able to get the lightbulb to work. Be sure to label the battery, the wire, and the bulb. 3. Diagram at least three ways in which you are not able to get the bulb to light. Analysis How did you know if electric current was flowing? What do your diagrams of the lit bulb 22.1 Current and Circuits As you learned in Chapter 11, flowing water at the top of a waterfall has both potential and kinetic energy. However, the large amount of natural potential and kinetic energy available from resources such as Niagara Falls are of little use to people or manufacturers who are 100 km away, unless that energy can be transported efficiently. Electric energy provides the means to transfer large quantities of energy great distances with little loss. This transfer usually is done at high potential differences through power lines, such as those shown in the photo on the left. Once this energy reaches the consumer, it can easily be converted into another form or combination of forms, including sound, light, thermal energy, and motion. Because electric energy can so easily be changed into other forms, it has become indispensable in our daily lives. Even quick glances around you will likely generate ample examples of the conversion of electric energy. Inside, lights to help you read at night, microwaves and electric ranges to cook food, computers, and stereos all rely on electricity for power. Outside, street lamps, store signs, advertisements, and the starters in cars all use flowing electric charges. In this chapter, you will learn how potential differences, resistance, and current are related. You also will learn about electric power and energy transfer. Objectives \u2022 Describe conditions that create current in an electric circuit. \u2022 Explain Ohm\u2019s law. \u2022 Design closed circuits. \u2022 Differentiate between power and energy in an electric circuit. Vocabulary electric current conventional current battery electric circuit ampere resistance resistor parallel connection series connection Section 22.1 Current and Circuits 591 Horizons Companies 590-600 CH22-S1-845813 7/22/04 3:38", " PM Page 592 Producing Electric Current In Chapter 21, you learned that when two conducting spheres touch, charges flow from the sphere at a higher potential to the one at a lower potential. The flow continues until there is no potential difference between the two spheres. A flow of charged particles is an electric current. In Figure 22-1a, two conductors, A and B, are connected by a wire conductor, C. Charges flow from the higher potential difference of B to A through C. This flow of positive charge is called conventional current. The flow stops when the potential difference between A, B, and C is zero. You could maintain the electric potential difference between B and A by pumping charged particles from A back to B, as illustrated in Figure 22-1b. Since the pump increases the electric potential energy of the charges, it requires an external energy source to run. This energy could come from a variety of sources. One familiar source, a voltaic or galvanic cell (a common dry cell), converts chemical energy to electric energy. Several galvanic cells connected together are called a battery. A second source of electric energy\u2014a photovoltaic cell, or solar cell\u2014changes light energy into electric energy. Electric Circuits The charges in Figure 22-1b move around a closed loop, cycling from the pump to B, through C, to A and back to the pump. Any closed loop or conducting path allowing electric charges to flow is called an electric circuit. A circuit includes a charge pump, which increases the potential energy of the charges flowing from A to B, and a device that reduces the potential energy of the charges flowing from B to A. The potential energy lost by the charges, qV, moving through the device is usually converted into some other form of energy. For example, electric energy is converted to kinetic energy by a motor, to light energy by a lamp, and to thermal energy by a heater. A charge pump creates the flow of charged particles that make up a current. Consider a generator driven by a waterwheel, such as the one pictured in Figure 22-2a. The water falls and rotates the waterwheel and generator. Thus, the kinetic energy of the water is converted to electric energy by the generator. The generator, like the charge pump, increases the electric potential difference, V. Energy in the amount qV is needed to increase the potential difference of the charges. This energy comes from the change in energy of the water. Not all of the water\u2019s kinetic energy, however, is converted to electric energy, as shown in Figure 22-2b. If the generator attached to the waterwheel is connected to a motor, the charges in the wire flow into the motor. The flow of charges continues through the circuit back to the generator. The motor converts electric energy to kinetic energy. Conservation of charge Charges cannot be created or destroyed, but they can be separated. Thus, the total amount of charge\u2014the number of negative electrons and positive ions\u2014in the circuit does not change. If one coulomb flows through the generator in 1 s, then one coulomb also will flow through the motor in 1 s. Thus, charge is a conserved quantity. Energy also is conserved. The change in electric energy, E, equals qV. Because q is conserved, \u25a0 Figure 22-1 Conventional current is defined as positive charges flowing from the positive plate to the negative plate (a). A generator pumps the positive charges back to the positive plate and maintains the current (b). In most metals, negatively-charged electrons actually flow from the negative to the positive plate, creating the appearance of positive charges that are moving in the opposite direction. a B A b B C Positive charges Current soon ceases C Charge pump A Current maintained 592 Chapter 22 Current Electricity 590-600 CH22-S1-845813 3/22/04 3:35 AM Page 593 a b Waterfall Motor Positive charge flow Positive charge flow Generator Waterwheel Potential energy of water Generator Electric energy Motor Work done by motor Thermal energy \u25a0 Figure 22-2 The potential energy of the waterfall is eventually converted into work done on the bucket (a). The production and use of electric current is not 100 percent efficient. Some thermal energy is produced by the splashing water, friction, and electric resistance (b). the net change in potential energy of the charges going completely around the circuit must be zero. The increase in potential difference produced by the generator equals the decrease in potential difference across the motor. If the potential difference between two wires is 120 V, the waterwheel and the generator must do 120 J of work on each coulomb of charge that is delivered. Every coulomb of charge moving through the motor delivers 120 J of energy to the motor. Rates of Charge Flow and Energy Transfer Power, which is defined in watts, W, measures the rate at which energy is transferred. If a generator transfers 1 J of kinetic energy to electric energy each second, it is transferring energy at the rate", " of 1 J/s, or 1 W. The energy carried by an electric current depends on the charge transferred, q, and the potential difference across which it moves, V. Thus, E qV. Recall from Chapter 20 that the unit for the quantity of electric charge is the coulomb. The rate of flow of electric charge, q/t, called electric current, is measured in coulombs per second. Electric current is represented by I, so I q/t. A flow of 1 C/s is called an ampere, A. The energy carried by an electric current is related to the voltage, E qV. Since current, I q/t, is the rate of charge flow, the power, P E/t, of an electric device can be determined by multiplying voltage and current. To derive the familiar form of the equation for the power delivered to an electric device, you can use P E/t and substitute E qV and q It. Power P IV Power is equal to the current times the potential difference. If the current through the motor in Figure 22-2a is 3.0 A and the potential difference is 120 V, the power in the motor is calculated using the expression P (3.0 C/s)(120 J/C) 360 J/s, which is 360 W. Section 22.1 Current and Circuits 593 590-600 CH22-S1-845813 3/22/04 3:36 AM Page 594 Electric Power and Energy A 6.0-V battery delivers a 0.50-A current to an electric motor connected across its terminals. a. What power is delivered to the motor? b. If the motor runs for 5.0 min, how much electric energy is delivered? 1 Analyze and Sketch the Problem \u2022 Draw a circuit showing the positive terminal of a battery connected to a motor and the return wire from the motor connected to the negative terminal of the battery. \u2022 Show the direction of conventional current. Known: V 6.0 V l 0.50 A t 5.0 min Unknown: P? E? 2 Solve for the Unknown a. Use P IV to find the power. P IV P (0.50 A)(6.0 V) 3.0 W Substitue I 0.50 A, V 6.0 V Motor V I I Battery b. In Chapter 10, you learned that P = E/t. Solve for E to find the energy. E Pt (3.0 W)(5.0 min) (3.0 J/s)(5.0 min)( 9.0102 J s 0 6 n i m 1 ) Substitute P 3.0 W, t 5.0 min Math Handbook Significant Digits page 834 3 Evaluate the Answer \u2022 Are the units correct? Power is measured in watts, and energy is measured in joules. \u2022 Is the magnitude realistic? With relatively low voltage and current, a few watts of power is reasonable. 1. The current through a lightbulb connected across the terminals of a 125-V outlet is 0.50 A. At what rate does the bulb convert electric energy to light? (Assume 100 percent efficiency.) 2. A car battery causes a current of 2.0 A through a lamp and produces 12 V across it. What is the power used by the lamp? 3. What is the current through a 75-W lightbulb that is connected to a 125-V outlet? 4. The current through the starter motor of a car is 210 A. If the battery maintains 12 V across the motor, how much electric energy is delivered to the starter in 10.0 s? 5. A flashlight bulb is rated at 0.90 W. If the lightbulb drops 3.0 V, how much current goes through it? 594 Chapter 22 Current Electricity 590-600 CH22-S1-845813 3/22/04 3:37 AM Page 595 Table 22-1 Changing Resistance Factor How resistance changes Example Length Resistance increases as length increases. L1 L2 Cross-sectional area Resistance increases as cross-sectional area decreases. A1 A2 Temperature Resistance increases as temperature increases. RL1 RL2 RA1 RA2 RT1 RT2 Material Keeping length, cross-sectional area, and temperature constant, resistance varies with the material used. T1 T2 Platinum Iron Aluminum Gold Copper Silver Resistance and Ohm\u2019s Law Suppose two conductors have a potential difference between them. If they are connected with a copper rod, a large current is created. On the other hand, putting a glass rod between them creates almost no current. The property determining how much current will flow is called resistance. Table 22-1 lists some of the factors that impact resistance. Resistance is measured by placing a potential difference across a conductor and dividing the voltage by the current. The resistance, R, is defined as the ratio of electric potential difference, V, to the current", ", I. Resistance R V I Resistance is equal to voltage divided by current. The resistance of the conductor, R, is measured in ohms. One ohm (1 ) is the resistance permitting an electric charge of 1 A to flow when a potential difference of 1 V is applied across the resistance. A simple circuit relating resistance, current, and voltage is shown in Figure 22-3. A 12-V car battery is connected to one of the car\u2019s 3- brake lights. The circuit is completed by a connection to an ammeter, which is a device that measures current. The current carrying the energy to the lights will measure 4 A. \u25a0 Figure 22-3 One ohm,, is defined as 1 V/A. In a circuit with a 3- resistance and a 12-V battery, there is a 4-A current. 12 V I V R 12 V 3 \u2126 4 A 3 \u2126 4 A Section 22.1 Current and Circuits 595 590-600 CH22-S1-845813 3/22/04 3:37 AM Page 596 a A 0.2 A b A 0.1 A c A 0.1 A 6 V 30 3 V 30 6 V 60 I I I \u25a0 Figure 22-4 The current through a simple circuit (a) can be regulated by removing some of the dry cells (b) or by increasing the resistance of the circuit (c). \u25a0 Figure 22-5 A potentiometer can be used to change current in an electric circuit. a 12 V12 V12 V The unit for resistance is named for German scientist Georg Simon Ohm, who found that the ratio of potential difference to current is constant for a given conductor. The resistance for most conductors does not vary as the magnitude or direction of the potential applied to it changes. A device having constant resistance independent of the potential difference obeys Ohm\u2019s law. Most metallic conductors obey Ohm\u2019s law, at least over a limited range of voltages. Many important devices, however, do not. A radio and a pocket calculator contain many devices, such as transistors and diodes, that do not obey Ohm\u2019s law. Even a lightbulb has resistance that depends on its temperature and does not obey Ohm\u2019s law. Wires used to connect electric devices have low resistance. A 1-m length of a typical wire used in physics labs has a resistance of about 0.03. Wires used in home wiring offer as little as 0.004 of resistance for each meter of length. Because wires have so little resistance, there is almost no potential drop across them. To produce greater potential drops, a large resistance concentrated into a small volume is necessary. A resistor is a device designed to have a specific resistance. Resistors may be made of graphite, semiconductors, or wires that are long and thin. There are two ways to control the current in a circuit. Because I V/R, I can be changed by varying V, R, or both. Figure 22-4a shows a simple circuit. When V is 6 V and R is 30, the current is 0.2 A. How could the current be reduced to 0.1 A? According to Ohm\u2019s law, the greater the voltage placed across a resistor, the larger the current passing through it. If the current through a resistor is cut in half, the potential difference also is cut Switch Motor b I Motor Potentiometer Switch Battery I Battery Potentiometer 596 Chapter 22 Current Electricity 590-600 CH22-S1-845813 6/7/04 9:51 AM Page 597 in half. In Figure 22-4b, the voltage applied across the resistor is reduced from 6 V to 3 V to reduce the current to 0.1 A. A second way to reduce the current to 0.1 A is to replace the 30- resistor with a 60- resistor, as shown in Figure 22-4c. Resistors often are used to control the current in circuits or parts of circuits. Sometimes, a smooth, continuous variation of the current is desired. For example, the speed control on some electric motors allows continuous, rather than step-by-step, changes in the rotation of the motor. To achieve this kind of control, a variable resistor, called a potentiometer, is used. A circuit containing a potentiometer is shown in Figure 22-5. Some variable resistors consist of a coil of resistance wire and a sliding contact point. Moving the contact point to various positions along the coil varies the amount of wire in the circuit. As more wire is placed in the circuit, the resistance of the circuit increases; thus, the current changes in accordance with the equation I V/R. In this way, the speed of a motor can be adjusted from fast, with little wire in the circuit, to slow, with a lot of wire in the circuit", ". Other examples of using variable resistors to adjust the levels of electrical energy can be found on the front of a TV: the volume, brightness, contrast, tone, and hue controls are all variable resistors. The human body The human body acts as a variable resistor. When dry, skin\u2019s resistance is high enough to keep currents that are produced by small and moderate voltages low. If skin becomes wet, however, its resistance is lower, and the electric current can rise to dangerous levels. A current as low as 1 mA can be felt as a mild shock, while currents of 15 mA can cause loss of muscle control and currents of 100 mA can cause death. Diagramming Circuits A simple circuit can be described in words. It can also be depicted by photographs or artists\u2019 drawings of the parts. Most frequently, however, an electric circuit is drawn using standard symbols for the circuit elements. Such a diagram is called a circuit schematic. Some of the symbols used in circuit schematics are shown in Figure 22-6. Resistance The resistance of an operating 100-W lightbulb is about 140. When the lightbulb is turned off and at room temperature, its resistance is only about 10. This is because of the great difference between room temperature and the lightbulb\u2019s operating temperature. Biology Connection Biology Connection \u25a0 Figure 22-6 These symbols commonly are used to diagram electric circuits. Conductor Switch Fuse Resistor (fixed) Potentiometer (variable resistor) Inductor Electric connection No electric connection Ground Battery V A Capacitor Lamp DC generator Voltmeter Ammeter Section 22.1 Current and Circuits 597 590-600 CH22-S1-845813 3/22/04 3:38 AM Page 598 Current Through a Resistor A 30.0-V battery is connected to a 10.0- resistor. What is the current in the circuit? 1 Analyze and Sketch the Problem \u2022 Draw a circuit containing a battery, an ammeter, and a resistor. \u2022 Show the direction of the conventional current. Ammeter Known: V 30.0 V R 10.0 Unknown: I =? 2 Solve for the Unknown Use I V /R to determine the current. V I R V Substitute V 30.0 V, R 10..00 A V I Battery Resistor R Math Handbook Operations with Significant Digits pages 835\u2014836 3 Evaluate the Answer \u2022 Are the units correct? Current is measured in amperes. \u2022 Is the magnitude realistic? There is a fairly large voltage and a small resistance, so a current of 3.00 A is reasonable. For all problems, assume that the battery voltage and lamp resistances are constant, no matter what current is present. 6. An automobile panel lamp with a resistance of 33 is placed across a 12-V battery. What is the current through the circuit? 7. A motor with an operating resistance of 32 is connected to a voltage source. The current in the circuit is 3.8 A. What is the voltage of the source? 8. A sensor uses 2.0104 A of current when it is operated by a 3.0-V battery. What is the resistance of the sensor circuit? 9. A lamp draws a current of 0.50 A when it is connected to a 120-V source. a. What is the resistance of the lamp? b. What is the power consumption of the lamp? 10. A 75-W lamp is connected to 125 V. a. What is the current through the lamp? b. What is the resistance of the lamp? 11. A resistor is added to the lamp in the previous problem to reduce the current to half of its original value. a. What is the potential difference across the lamp? b. How much resistance was added to the circuit? c. How much power is now dissipated in the lamp? 598 Chapter 22 Current Electricity 590-600 CH22-S1-845813 3/22/04 3:39 AM Page 599 a Voltmeter b 12 V Resistance 3 \u2126 V 12 V 3 4 A A Ammeter 12 V \u25a0 Figure 22-7 A simple electric circuit is represented pictorially (a) and schematically (b). Battery An artist\u2019s drawing and a schematic of the same circuit are shown in Figures 22-7a and 22-7b. Notice in both the drawing and the schematic that the electric charge is shown flowing out of the positive terminal of the battery. To draw schematic diagrams, use the problem-solving strategy below, and always set up a conventional current. You learned that an ammeter measures current and a voltmeter measures potential differences. Each instrument has two terminals, usually labeled and. A voltmeter measures the potential difference across any component of a circuit. When connecting the voltmeter in a circuit, always connect the terminal to the end of the circuit component that is closer to the positive terminal of the battery", ", and connect the terminal to the other side of the component. Drawing Schematic Diagrams Follow these steps when drawing schematic diagrams. 1. Draw the symbol for the battery or other source of electric energy, such as a generator, on the left side of the page. Put the positive terminal on top. 2. Draw a wire coming out of the positive terminal. When you reach a resistor or other device, draw the symbol for it. 3. If you reach a point where there are two current paths, such as at a voltmeter, draw a path until the two current paths join again. Then draw the second path. in the diagram. Follow one 4. Follow the current path until you reach the negative terminal of the battery. 5. Check your work to make sure that you have included all parts and that there are complete paths for the current to follow. Current Affairs Do you think that current diminishes as it passes through different elements in the circuit? As a scientist, you can test this question. 1. Draw a circuit that includes a power supply and two miniature lamps. 2. Draw the circuit again and include an ammeter to measure the current between the power supply and the lamps. 3. In a third diagram, show the ammeter at a position to measure the current between the lamps. Analyze and Conclude 4. Predict if the current between the lamps will be more than, less than, or the same as the current before the lamps. Explain. 5. Test your prediction by building the circuits. CAUTION: Wire is sharp and can cut skin. Section 22.1 Current and Circuits 599 590-600 CH22-S1-845813 6/7/04 9:54 AM Page 600 12. Draw a circuit diagram to include a 60.0-V battery, an ammeter, and a resistance of 12.5 in series. Indicate the ammeter reading and the direction of the current. 13. Draw a series-circuit diagram showing a 4.5-V battery, a resistor, and an ammeter that reads 85 mA. Determine the resistance and label the resistor. Choose a direction for the conventional current and indicate the positive terminal of the battery. 14. Add a voltmeter to measure the potential difference across the resistors in problems 12 and 13 and repeat the problems. 15. Draw a circuit using a battery, a lamp, a potentiometer to adjust the lamp\u2019s brightness, and an on-off switch. 16. Repeat the previous problem, adding an ammeter and a voltmeter across the lamp. When a voltmeter is connected across another component, it is called a parallel connection because the circuit component and the voltmeter are aligned parallel to each other in the circuit, as diagrammed in Figure 22-8a. Any time the current has two or more paths to follow, the connection is labeled parallel. The potential difference across the voltmeter is equal to the potential difference across the circuit element. Always associate the words voltage across with a parallel connection. An ammeter measures the current through a circuit component. The same current going through the component must go through the ammeter, so there can be only one current path. A connection with only one current path, called a series connection, is shown in Figure 22-8b. To add an ammeter to a circuit, the wire connected to the circuit component must be removed and connected to the ammeter instead. Then, another wire is connected from the second terminal of the ammeter to the circuit component. In a series connection, there can be only a single path through the connection. Always associate the words current through with a series connection. V 12 V I 3 12 V I a b 12 V I 3 I 4 A A \u25a0 Figure 22-8 These schematics show a parallel (a) and a series circuit (b). 22.1 Section Review 17. Schematic Draw a schematic diagram of a circuit that contains a battery and a lightbulb. Make sure the lightbulb will light in this circuit. 18. Resistance Joe states that because R V/I, if he increases the voltage, the resistance will increase. Is Joe correct? Explain. 19. Resistance You want to measure the resistance of a long piece of wire. Show how you would construct a circuit with a battery, a voltmeter, an ammeter, and the wire to be tested to make the measurement. Specify what you would measure and how you would compute the resistance. 20. Power A circuit has 12 of resistance and is connected to a 12-V battery. Determine the change in power if the resistance decreases to 9.0. 21. Energy A circuit converts 2.2103 J of energy when it is operated for 3.0 min. Determine the amount of energy it will convert when it is operated for 1 h. 22. Critical Thinking We say that power is \u201cdissipated\u201d in a resistor. To dissipate is to use, to waste, or to", " squander. What is \u201cused\u201d when charge flows through a resistor? 600 Chapter 22 Current Electricity physicspp.com/self_check_quiz 601-605 CH22-S2-845813 3/22/04 3:43 AM Page 601 22.2 Using Electric Energy Many familiar household appliances convert electric energy to some other form, such as light, kinetic energy, sound, or thermal energy. When you turn on one of these appliances, you complete a circuit and begin converting electric energy. In this section, you will learn to determine the rate of energy conversion and the amount that is converted. Energy Transfer in Electric Circuits Energy that is supplied to a circuit can be used in many different ways. A motor converts electric energy to mechanical energy, and a lamp changes electric energy into light. Unfortunately, not all of the energy delivered to a motor or a lamp ends up in a useful form. Lightbulbs, especially incandescent lightbulbs, become hot. Motors are often far too hot to touch. In each case, some of the electric energy is converted into thermal energy. You will now examine some devices that are designed to convert as much energy as possible into thermal energy. Heating a resistor Current moving through a resistor causes it to heat up because flowing electrons bump into the atoms in the resistor. These collisions increase the atoms\u2019 kinetic energy and, thus, the temperature of the resistor. A space heater, a hot plate, and the heating element in a hair dryer all are designed to convert electric energy into thermal energy. These and other household appliances, such as those pictured in Figure 22-9, act like resistors when they are in a circuit. When charge, q, moves through a resistor, its potential difference is reduced by an amount, V. As you have learned, the energy change is represented by qV. In practical use, the rate at which energy is changed\u2014the power, P E/t\u2014is more important. Earlier, you learned that current is the rate at which charge flows, I q/t, and that power dissipated in a resistor is represented by P IV. For a resistor, V IR. Thus, if you know I and R, you can substitute V IR into the equation for electric power to obtain the following. Objectives \u2022 Explain how electric energy is converted into thermal energy. \u2022 Explore ways to deliver electric energy to consumers near and far. \u2022 Define kilowatt-hour. Vocabulary superconductor kilowatt-hour \u25a0 Figure 22-9 These appliances are designed to change electric energy into thermal energy. Power P I2R Power is equal to current squared times resistance. Thus, the power dissipated in a resistor is proportional both to the square of the current passing through it and to the resistance. If you know V and R, but not I, you can substitute I V/R into P IV to obtain the following equation. 2 Power P V R Power is equal to the voltage squared divided by the resistance. Section 22.2 Using Electric Energy 601 Hutchings Photography 601-605 CH22-S2-845813 3/22/04 3:44 AM Page 602 The power is the rate at which energy is converted from one form to another. Energy is changed from electric to thermal energy, and the temperature of the resistor rises. If the resistor is an immersion heater or burner on an electric stovetop, for example, heat flows into cold water fast enough to bring the water to the boiling point in a few minutes. If power continues to be dissipated at a uniform rate, then after time t, the energy converted to thermal energy will be E Pt. Because P I2R and P V 2/R, the total energy to be converted to thermal energy can be written in the following ways. E Pt Thermal Energy E I 2Rt 2 t E V R Thermal energy is equal to the power dissipated multiplied by the time. It is also equal to the current squared multiplied by resistance and time as well as the voltage squared divided by resistance multiplied by time. Electric Heat A heater has a resistance of 10.0. It operates on 120.0 V. a. What is the power dissipated by the heater? b. What thermal energy is supplied by the heater in 10.0 s? I 1 Analyze and Sketch the Problem \u2022 Sketch the situation. \u2022 Label the known circuit components, which are a 120.0-V potential difference source and a 10.0- resistor. Known: R 10.0 V 120.0 V t 10.0 s Unknown: P? E? 2 Solve for the Unknown a. Because R and V are known, use P V 2/R. V)2 0. 0 2 P (1 1 0. 0 1.44 kW b. Solve for the energy. E Pt (1.44 kW)(10.0 s) 14.4 kJ Substitute V 120.0 V, R 10.0 Substitute P 1", ".44 kW, t 10.0 s 120.0 V 10.0 I Math Handbook Exponents page 839 3 Evaluate the Answer \u2022 Are the units correct? Power is measured in watts, and energy is measured in joules. \u2022 Are the magnitudes realistic? For power, 102 102 101 103, so kilowatts is reasonable. For energy, 103 101 104, so an order of magnitude of 10,000 joules is reasonable. 602 Chapter 22 Current Electricity 601-605 CH22-S2-845813 6/7/04 9:57 AM Page 603 23. A 15- electric heater operates on a 120-V outlet. a. What is the current through the heater? b. How much energy is used by the heater in 30.0 s? c. How much thermal energy is liberated in this time? 24. A 39- resistor is connected across a 45-V battery. a. What is the current in the circuit? b. How much energy is used by the resistor in 5.0 min? 25. A 100.0-W lightbulb is 22 percent efficient. This means that 22 percent of the electric energy is converted to light energy. a. How many joules does the lightbulb convert into light each minute it is in operation? b. How many joules of thermal energy does the lightbulb produce each minute? 26. The resistance of an electric stove element at operating temperature is 11. a. If 220 V are applied across it, what is the current through the stove element? b. How much energy does the element convert to thermal energy in 30.0 s? c. The element is used to heat a kettle containing 1.20 kg of water. Assume that 65 percent of the heat is absorbed by the water. What is the water\u2019s increase in temperature during the 30.0 s? 27. A 120-V water heater takes 2.2 h to heat a given volume of water to a certain temperature. How long would a 240-V unit operating with the same current take to accomplish the same task? Superconductors A superconductor is a material with zero resistance. There is no restriction of current in superconductors, so there is no potential difference, V, across them. Because the power that is dissipated in a conductor is given by the product IV, a superconductor can conduct electricity without loss of energy. At present, almost all superconductors must be kept at temperatures below 100 K. The practical uses of superconductors include MRI magnets and in synchrotrons, which use huge amounts of current and can be kept at temperatures close to 0 K. \u25a0 Figure 22-10 In the year 2000, energy produced by Itaip\u00fa Dam met 24 percent of Brazil\u2019s electric energy needs and 95 percent of Paraguay\u2019s. Transmission of Electric Energy Hydroelectric facilities, such as the one at Itaip\u00fa Dam, shown in Figure 22-10, are capable of producing a great deal of energy. This hydroelectric energy often must be transmitted over long distances to reach homes and industries. How can the transmission occur with as little loss to thermal energy as possible? Thermal energy is produced at a rate represented by P I2R. Electrical engineers call this unwanted thermal energy the joule heating loss, or I2R loss. To reduce this loss, either the current, I, or the resistance, R, must be reduced. All wires have some resistance, even though their resistance is small. The large wire used to carry electric current into a home has a resistance of 0.20 for 1 km. Section 22.2 Using Electric Energy 603 Hans-Jurgen Burkard/Peter Arnold, Inc. 601-605 CH22-S2-845813 3/22/04 3:45 AM Page 604 a b \u25a0 Figure 22-11 Watt-hour meters measure the amount of electric energy used by a consumer (a). Meter readings then are used in calculating the cost of energy (b). Suppose that a farmhouse were connected directly to a power plant 3.5 km away. The resistance in the wires needed to carry a current in a circuit to the home and back to the plant is represented by the following equation: R 2(3.5 km)(0.20 /km) 1.4. An electric stove might cause a 41-A current through the wires. The power dissipated in the wires is represented by the following relationships: P I2R (41 A)2 (1.4 ) 2400 W. All of this power is converted to thermal energy and, therefore, is wasted. This loss could be minimized by reducing the resistance. Cables of high conductivity and large diameter (and therefore low resistance) are available, but such cables are expensive and heavy. Because the loss of energy is also proportional to the square of the current in the conductors, it is even more important to keep the", " current in the transmission lines low. How can the current in the transmission lines be kept low? The electric energy per second (power) transferred over a long-distance transmission line is determined by the relationship P IV. The current is reduced without the power being reduced by an increase in the voltage. Some long-distance lines use voltages of more than 500,000 V. The resulting lower current reduces the I2R loss in the lines by keeping the I2 factor low. Long-distance transmission lines always operate at voltages much higher than household voltages in order to reduce I2R loss. The output voltage from the generating plant is reduced upon arrival at electric substations to 2400 V, and again to 240 V or 120 V before being used in homes. The Kilowatt-Hour While electric companies often are called power companies, they actually provide energy rather than power. Power is the rate at which energy is delivered. When consumers pay their home electric bills, an example of which is shown in Figure 22-11, they pay for electric energy, not power. The amount of electric energy used by a device is its rate of energy consumption, in joules per second (W) times the number of seconds that the device is operated. Joules per second times seconds, (J/s)s, equals the total amount of joules of energy. The joule, also defined as a watt-second, is a relatively small amount of energy, too small for commercial sales use. For this reason, electric companies measure energy sales in a unit of a Use the figure to the right to help you answer the questions below. 1. Initially, the capacitor is uncharged. Switch 1 is closed, and Switch 2 remains open. What is the voltage across the capacitor? 2. Switch 1 is now opened, and Switch 2 remains open. What is the voltage across the capacitor? Why? 3. Next, Switch 2 is closed, while Switch 1 remains open. What is the voltage across the capacitor and the current through the resistor immediately after Switch 2 is closed? 4. As time goes on, what happens to the voltage across the capacitor and the current through the resistor? Switch 2 Switch 1 1.5 F 15 V 1200 604 (t)Bischel Studios, (b)Hutchings Photography Chapter 22 Current Electricity 601-605 CH22-S2-845813 6/7/04 9:59 AM Page 605 large number of joules called a kilowatt-hour, kWh. A kilowatt-hour is equal to 1000 watts delivered continuously for 3600 s (1 h), or 3.6 106 J. Not many household devices other than hot-water heaters, stoves, clothes dryers, microwave ovens, heaters, and hair dryers require more than 1000 W of power. Ten 100-W lightbulbs operating all at once use only 1 kWh of energy when they are left on for one full hour. 28. An electric space heater draws 15.0 A from a 120-V source. It is operated, on the average, for 5.0 h each day. a. How much power does the heater use? b. How much energy in kWh does it consume in 30 days? c. At $0.12 per kWh, how much does it cost to operate the heater for 30 days? 29. A digital clock has a resistance of 12,000 and is plugged into a 115-V outlet. a. How much current does it draw? b. How much power does it use? c. If the owner of the clock pays $0.12 per kWh, how much does it cost to operate the clock for 30 days? 30. An automotive battery can deliver 55 A at 12 V for 1.0 h and requires 1.3 times as much energy for recharge due to its lessthan-perfect efficiency. How long will it take to charge the battery using a current of 7.5 A? Assume that the charging voltage is the same as the discharging voltage. 31. Rework the previous problem by assuming that the battery requires the application of 14 V when it is recharging. You have learned several ways in which power companies solve the problems involved in transmitting electric current over great distances. You also have learned how power companies calculate electric bills and how to predict the cost of running various appliances in the home. The distribution of electric energy to all corners of Earth is one of the greatest engineering feats of the twentieth century. 22.2 Section Review 32. Energy A car engine drives a generator, which produces and stores electric charge in the car\u2019s battery. The headlamps use the electric charge stored in the car battery. List the forms of energy in these three operations. 33. Resistance A hair dryer operating from 120 V has two settings, hot and warm. In which setting is the resistance likely to be smaller? Why? 34. Power Determine the power change in a circuit if the applied voltage is", " decreased by one-half. 35. Efficiency Evaluate the impact of research to improve power transmission lines on society and the environment. 36. Voltage Why would an electric range and an electric hot-water heater be connected to a 240-V circuit rather than a 120-V circuit? 37. Critical Thinking When demand for electric power is high, power companies sometimes reduce the voltage, thereby producing a \u201cbrown-out.\u201d What is being saved? physicspp.com/self_check_quiz Section 22.2 Using Electric Energy 605 606-607 CH22-LAB-845813 3/22/04 4:14 AM Page 606 Voltage, Current, and Resistance In this chapter, you studied the relationships between voltage, current, and resistance in simple circuits. Voltage is the potential difference that pushes current through a circuit, while resistance determines how much current will flow if a potential difference exists. In this activity, you will collect data and make graphs in order to investigate the mathematical relationships between voltage and current and between resistance and current. QUESTION What are the relationships between voltage and current and resistance and current? Objectives Materials \u25a0 Measure current in SI. \u25a0 Describe the relationship between the resistance of a circuit and the total current flowing through a circuit. \u25a0 Describe the relationship between voltage and the total current flowing through a circuit. \u25a0 Make and use graphs to show the relationships between current and resistance and between current and voltage. Safety Precautions \u25a0 CAUTION: Resistors and circuits may become hot. \u25a0 CAUTION: Wires are sharp and can cut skin. four 1.5-V D batteries four D-battery holders one 10-k resistor one 500-A ammeter five wires with alligator clips one 20-k resistor one 30-k resistor one 40-k resistor Procedure Part A 1. Place the D battery in the D-battery holder. 2. Create a circuit containing the D battery, 10-k resistor, and 500-A ammeter. 3. Record the values for resistance and current in Data Table 1. For resistance, use the value of the resistor. For current, read and record the value given by the ammeter. 4. Replace the 10-k resistor with a 20-k resistor. 5. Record the resistance and the current in Data Table 1. 6. Repeat steps 4\u20135, but replace the 20-k resistor with a 30-k resistor. 7. Repeat steps 4\u20135, but replace the 30-k resistor with a 40-k resistor. Part B 8. Recreate the circuit that you made in step 2. Verify the current in the circuit and record the values for voltage and current in Data Table 2. 9. Add a second 1.5-V D battery to the setup and record the values for voltage and current in Data Table 2. When you are using more than one battery, record the sum of the batteries\u2019 voltages as the voltage in Data Table 2. 10. Repeat step 9 with three 1.5-V D batteries. 11. Repeat step 9 with four 1.5-V D batteries. 606 Horizons Companies 606-607 CH22-LAB-845813 6/7/04 10:01 AM Page 607 Data Table 1 Voltage (V) 1.5 1.5 1.5 1.5 Data Table 2 Voltage (V) Resistance (k) Current (A) Resistance (k) Current (A) 10 10 10 10 Analyze Going Further 1. Make and Use Graphs Graph the current versus the resistance. Place resistance on the x-axis and current on the y-axis. 1. What would be the current in a circuit with a voltage of 3.0 V and a resistance of 20 k? How did you determine this? 2. Make and Use Graphs Graph the current versus the voltage. Place voltage on the x-axis and current on the y-axis. 3. Error Analysis Other than the values of the resistors, what factors could have affected the current in Part A? How might the effect of these factors be reduced? 4. Error Analysis Other than the added batteries, what factors could have affected the current in Part B? How might the effect of these factors be reduced? Conclude and Apply 1. Looking at the first graph that you made, describe the relationship between resistance and current? 2. Why do you suppose this relationship between resistance and current exists? 3. Looking at the second graph that you made, how would you describe the relationship between voltage and current? 4. Why do you suppose this relationship between voltage and current exists? 2. Could you derive a formula from your lab data to explain the relationship among voltage, current, and resistance? Hint: Look at the graph of current versus voltage. Assume it is a straight line that goes through the origin. 3. How well does your data match this formula? Explain. Real-World Physics 1. Identify some common appliances that use 240 V rather than 120 V. 2", ". Why do the appliances that you identified require 240 V? What would be the consequences for running such an appliance on a 120-V circuit? To find out more about current electricity, visit the Web site: physicspp.com 607 608 CH22-FEATURE-845813 3/22/04 4:19 AM Page 608 Hybrid Cars Hybrid Cars Meet the hybrid car. It is fuel-efficient, comfortable, safe, quiet, clean, and it accelerates well. Hybrid sales are growing and are expected to exceed 200,000 vehicles in 2005. Why are they called hybrids? A vehicle is called a hybrid if it uses two or more sources of energy. For example, diesel-electric locomotives are hybrids. But the term hybrid vehicle usually refers to a car that uses gas and electricity. Conventional cars have large engines that enable them to accelerate quickly and to drive up steep hills. But the engine\u2019s size makes it inefficient. In a hybrid, a lighter, more efficient gas engine meets most driving needs. When extra energy is needed, it is supplied by electricity from rechargeable batteries. 5 2 4 6 1 3 A hybrid car has a gas engine (1) and an electric motor (2). How do hybrids work? The illustration above shows one type of hybrid, called a parallel hybrid. The small internal combustion engine (1) powers the car during most driving situations. The gas engine and electric motor (2) are connected to the wheels by the same transmission (3). Computerized electronics (4) decide when to use the electric motor, when to use the engine, and when to use both. This type of hybrid has no external power source besides the gas in the fuel tank (5). Unlike an electric car, you don\u2019t need to plug the hybrid into an electric outlet to recharge the batteries (6). Rather, the batteries are recharged by a process called regenerative braking, as shown in the schematic diagram. In conventional vehicles, the brakes apply friction to the wheels, converting a vehicle\u2019s kinetic energy into heat. However, a hybrid\u2019s electric motor 608 Technology and Society Gas tank Battery Transmission Gas engine Electric motor/ generator PE from gas and battery The gas engine and electric motor turn the wheels KE recharges the battery In regenerative breaking, energy from the moving car recharges the batteries. can act as a generator. When the electric motor slows the car, kinetic energy is converted to electric energy, which then recharges the batteries. Can hybrids benefit society? Hybrid cars improve gas mileage and reduce tailpipe emissions. Improved gas mileage saves on the cost of operating the car. Tailpipe emissions include carbon dioxide and carbon monoxide, as well as various hydrocarbons and nitrogen oxides. These emissions can contribute to certain problems, such as smog. Because hybrids improve gas mileage and reduce tailpipe emissions, many people feel that these cars are one viable way to help protect air quality and conserve fuel resources. Going Further 1. Analyze and Conclude What is regenerative braking? 2. Predict Will increased sales of hybrids benefit society? Support your answer. 609-615 CH22-SG CA STP-845813 3/22/04 4:23 AM Page 609 22.1 Current and Circuits Vocabulary \u2022 electric current (p. 592) \u2022 conventional current (p. 592) \u2022 battery (p. 592) \u2022 electric circuit (p. 592) \u2022 ampere (p. 593) \u2022 resistance (p. 595) \u2022 resistor (p. 596) \u2022 parallel connection (p. 600) \u2022 series connection (p. 600) Key Concepts \u2022 Conventional current is defined as current in the direction in which a positive charge would move. \u2022 Generators convert mechanical energy to electric energy. \u2022 A circuit converts electric energy to heat, light, or some other useful output. \u2022 As charge moves through a circuit, resistors cause a drop in potential energy. \u2022 An ampere is equal to one coulomb per second (1 C/s). \u2022 Power can be found by multiplying voltage times current. P IV \u2022 The resistance of a device is given by the ratio of the device\u2019s voltage to its current. R V I \u2022 Ohm\u2019s law states that the ratio of potential difference to current is a constant for a given conductor. Any resistance that does not change with temperature, voltage, or the direction of charge flow obeys Ohm\u2019s law. \u2022 Circuit current can be controlled by changing voltage, resistance, or both. 22.2 Using Electric Energy Vocabulary \u2022 superconductor (p. 603) \u2022 kilowatt-hour (p. 605) Key Concepts \u2022 The power in a circuit is equal to the square of the current times the resistance, or to the voltage squared divided by the resistance. 2 P I2R or P V R \u2022 If power is dissipated at a uniform rate, the thermal energy converted equals power multiplied by time.", " Power also can be represented by I2R and V2/R to give the last two equations. E Pt I2Rt t V 2 R \u2022 Superconductors are materials with zero resistance. At present, the practical uses of superconductors are limited. \u2022 Unwanted thermal energy produced in the transmission of electric energy is called the joule heating loss, or I2R loss. The best way to minimize the joule heating loss is to keep the current in the transmission wires low. Transmitting at higher voltages enables current to be reduced without power being reduced. \u2022 The kilowatt-hour, kWh, is an energy unit. It is equal to 3.6106 J. physicspp.com/vocabulary_puzzlemaker 609 609-615 CH22-SG CA STP-845813 3/22/04 4:24 AM Page 610 Concept Mapping 38. Complete the concept map using the following terms: watt, current, resistance. Electricity rate of flow rate of conversion opposition to flow power 44. Describe the energy conversions that occur in each of the following devices. (22.1) a. an incandescent lightbulb b. a clothes dryer c. a digital clock radio 45. Which wire conducts electricity with the least resistance: one with a large cross-sectional diameter or one with a small cross-sectional diameter? (22.1) 46. A simple circuit consists of a resistor, a battery, and connecting wires. (22.1) a. Draw a circuit schematic of this simple circuit. b. How must an ammeter be connected in a circuit for the current to be correctly read? c. How must a voltmeter be connected to a resistor for ampere ohm the potential difference across it to be read? Mastering Concepts 39. Define the unit of electric current in terms of fundamental MKS units. (22.1) 40. How should a voltmeter be connected in Figure 22-12 to measure the motor\u2019s voltage? (22.1) 1 4 2 3 \u25a0 Figure 22-12 41. How should an ammeter be connected in Figure 22-12 to measure the motor\u2019s current? (22.1) 42. What is the direction of the conventional motor current in Figure 22-12? (22.1) 43. Refer to Figure 22-12 to answer the following questions. (22.1) a. Which device converts electric energy to mechanical energy? b. Which device converts chemical energy to electric energy? 47. Why do lightbulbs burn out more frequently just as they are switched on rather than while they are operating? (22.2) 48. If a battery is short-circuited by a heavy copper wire being connected from one terminal to the other, the temperature of the copper wire rises. Why does this happen? (22.2) 49. What electric quantities must be kept small to transmit electric energy economically over long distances? (22.2) 50. Define the unit of power in terms of fundamental MKS units. (22.2) Applying Concepts 51. Batteries When a battery is connected to a complete circuit, charges flow in the circuit almost instantaneously. Explain. 52. Explain why a cow experiences a mild shock when it touches an electric fence. 53. Power Lines Why can birds perch on high-voltage lines without being injured? 54. Describe two ways to increase the current in a circuit. 55. Lightbulbs Two lightbulbs work on a 120-V circuit. One is 50 W and the other is 100 W. Which bulb has a higher resistance? Explain. 56. If the voltage across a circuit is kept constant and the resistance is doubled, what effect does this have on the circuit's current? c. Which device turns the circuit on and off? d. Which device provides a way to adjust speed? 57. What is the effect on the current in a circuit if both the voltage and the resistance are doubled? Explain. 610 Chapter 22 Current Electricity For more problems, go to Additional Problems, Appendix B. 609-615 CH22-SG CA STP-845813 3/22/04 4:25 AM Page 611 58. Ohm\u2019s Law Sue finds a device that looks like a resistor. When she connects it to a 1.5-V battery, she measures only 45106 A, but when she uses a 3.0-V battery, she measures 25103 A. Does the device obey Ohm\u2019s law? 59. If the ammeter in Figure 22-4a on page 596 were moved to the bottom of the diagram, would the ammeter have the same reading? Explain. 60. Two wires can be placed across the terminals of a 6.0-V battery. One has a high resistance, and the other has a low resistance. Which wire will produce thermal energy at a faster rate? Why?", " Mastering Problems 22.1 Current and Circuits 61. A motor is connected to a 12-V battery, as shown in Figure 22-13. a. How much power is delivered to the motor? b. How much energy is converted if the motor runs for 15 min? Motor 1.5 A 12 V \u25a0 Figure 22-13 62. Refer to Figure 22-14 to answer the following questions. a. What should the ammeter reading be? b. What should the voltmeter reading be? c. How much power is delivered to the resistor? d. How much energy is delivered to the resistor per hour? A 27 V 18 V I \u25a0 Figure 22-14 63. Refer to Figure 22-15 to answer the following questions. a. What should the ammeter reading be? b. What should the voltmeter reading be? c. How much power is delivered to the resistor? d. How much energy is delivered to the resistor per hour? A 27 V 9.0 V I \u25a0 Figure 22-15 64. Refer to Figure 22-16 to answer the following questions. a. What should the ammeter reading be? b. What should the voltmeter reading be? c. How much power is delivered to the resistor? d. How much energy is delivered to the resistor per hour? A 9.0 V 18 V I \u25a0 Figure 22-16 65. Toasters The current through a toaster that is connected to a 120-V source is 8.0 A. What power is dissipated by the toaster? 66. Lightbulbs A current of 1.2 A is measured through a lightbulb when it is connected across a 120-V source. What power is dissipated by the bulb? 67. A lamp draws 0.50 A from a 120-V generator. a. How much power is delivered? b. How much energy is converted in 5.0 min? 68. A 12-V automobile battery is connected to an electric starter motor. The current through the motor is 210 A. a. How many joules of energy does the battery deliver to the motor each second? b. What power, in watts, does the motor use? physicspp.com/chapter_test Chapter 22 Assessment 611 609-615 CH22-SG CA STP-845813 6/7/04 10:04 AM Page 612 69. Dryers A 4200-W clothes dryer is connected to a 220-V circuit. How much current does the dryer draw? 70. Flashlights A flashlight bulb is connected across a 3.0-V potential difference. The current through the bulb is 1.5 A. a. What is the power rating of the bulb? b. How much electric energy does the bulb convert in 11 min? 71. Batteries A resistor of 60.0 has a current of 0.40 A through it when it is connected to the terminals of a battery. What is the voltage of the battery? 72. What voltage is applied to a 4.0- resistor if the current is 1.5 A? 73. What voltage is placed across a motor with a 15operating resistance if there is 8.0 A of current? 74. A voltage of 75 V is placed across a 15- resistor. What is the current through the resistor? 75. Some students connected a length of nichrome wire to a variable power supply to produce between 0.00 V and 10.00 V across the wire. They then measured the current through the wire for several voltages. The students recorded the data for the voltages used and the currents measured, as shown in Table 22-2. a. For each measurement, calculate the resistance. b. Graph I versus V. c. Does the nichrome wire obey Ohm\u2019s law? If not, for all the voltages, specify the voltage range for which Ohm\u2019s law holds. 77. A lamp draws a 66-mA current when connected to a 6.0-V battery. When a 9.0-V battery is used, the lamp draws 75 mA. a. Does the lamp obey Ohm\u2019s law? b. How much power does the lamp dissipate when it is connected to the 6.0-V battery? c. How much power does it dissipate at 9.0 V? 78. Lightbulbs How much energy does a 60.0-W lightbulb use in half an hour? If the lightbulb converts 12 percent of electric energy to light energy, how much thermal energy does it generate during the half hour? 79. The current through a lamp connected across 120 V is 0.40 A when the lamp is on. a. What is the lamp\u2019s resistance when it is on? b. When the lamp is cold, its resistance is 1/5 as great as it is when the lamp is hot. What is the lamp\u2019s cold resistance? c", ". What is the current through the lamp as it is turned on if it is connected to a potential difference of 120 V? 80. The graph in Figure 22-17 shows the current through a device called a silicon diode. a. A potential difference of 0.70 V is placed across the diode. What is the resistance of the diode? b. What is the diode\u2019s resistance when a 0.60-V potential difference is used? c. Does the diode obey Ohm\u2019s law? Current in a Diode Table 22-2 Voltage, V (volts) Current, I (amps) Resistance, R V/I (amps) 2.00 4.00 6.00 8.00 10.00 2.00 4.00 6.00 8.00 10.00 0.0140 0.0270 0.0400 0.0520 0.0630 0.0140 0.0280 0.0390 0.0510 0.0620 _______________ _______________ _______________ _______________ _______________ _______________ _______________ _______________ _______________ _______________ ).02 0.01 0 0.2 0.4 0.6 0.8 Voltage (V) \u25a0 Figure 22-17 81. Draw a schematic diagram to show a circuit including a 90-V battery, an ammeter, and a resistance of 45 connected in series. What is the ammeter reading? Draw arrows showing the direction of conventional current. 76. Draw a series circuit diagram to include a 16resistor, a battery, and an ammeter that reads 1.75 A. Indicate the positive terminal and the voltage of the battery, the positive terminal of the ammeter, and the direction of conventional current. 22.2 Using Electric Energy 82. Batteries A 9.0-V battery costs $3.00 and will deliver 0.0250 A for 26.0 h before it must be replaced. Calculate the cost per kWh. 612 Chapter 22 Current Electricity For more problems, go to Additional Problems, Appendix B. 609-615 CH22-SG CA STP-845813 3/22/04 4:26 AM Page 613 83. What is the maximum current allowed in a 5.0-W, 220- resistor? 84. A 110-V electric iron draws 3.0 A of current. How much thermal energy is developed in an hour? 85. For the circuit shown in Figure 22-18, the maximum safe power is 50 W. Use the figure to find the following: a. the maximum safe current b. the maximum safe voltage V 40 \u25a0 Figure 22-18 I 86. Utilities Figure 22-19 represents an electric furnace. Calculate the monthly (30-day) heating bill if electricity costs $0.10 per kWh and the thermostat is on one-fourth of the time. \u25a0 Figure 22-19 87. Appliances A window air conditioner is estimated to have a cost of operation of $50 per 30 days. This is based on the assumption that the air conditioner will run half of the time and that electricity costs $0.090 per kWh. Determine how much current the air conditioner will take from a 120-V outlet. 88. Radios A transistor radio operates by means of a 9.0-V battery that supplies it with a 50.0-mA current. a. If the cost of the battery is $2.49 and it lasts for 300.0 h, what is the cost per kWh to operate the radio in this manner? b. The same radio, by means of a converter, is plugged into a household circuit by a homeowner who pays $0.12 per kWh. What does it now cost to operate the radio for 300.0 h? Mixed Review 89. If a person has $5, how long could he or she play a 200 W stereo if electricity costs $0.15 per kWh? 90. A current of 1.2 A is measured through a 50.0resistor for 5.0 min. How much heat is generated by the resistor? 91. A 6.0- resistor is connected to a 15-V battery. a. What is the current in the circuit? b. How much thermal energy is produced in 10.0 min? 92. Lightbulbs An incandescent lightbulb with a resistance of 10.0 when it is not lit and a resistance of 40.0 when it is lit has 120 V placed across it. a. What is the current draw when the bulb is lit? b. What is the current draw at the instant the bulb is turned on? c. When does the lightbulb use the most power? 93. A 12-V electric motor\u2019s speed is controlled by a potentiometer. At the motor\u2019s slowest setting, it uses 0.02 A. At its highest setting, the motor", " uses 1.2 A. What is the range of the potentiometer? 94. An electric motor operates a pump that irrigates a farmer\u2019s crop by pumping 1.0104 L of water a vertical distance of 8.0 m into a field each hour. The motor has an operating resistance of 22.0 and is connected across a 110-V source. a. What current does the motor draw? b. How efficient is the motor? on 120 V. a. What is the current in the coil while it is operating? b. What energy is supplied to the coil in 5.0 min? c. If the coil is immersed in an insulated container holding 20.0 kg of water, what will be the increase in the temperature of the water? Assume 100 percent of the heat is absorbed by the water. d. At $0.08 per kWh, how much does it cost to operate the heating coil 30 min per day for 30 days? 96. Appliances An electric heater is rated at 500 W. a. How much energy is delivered to the heater in half an hour? b. The heater is being used to heat a room containing 50 kg of air. If the specific heat of air is 1.10 kJ/kg\u00b0C, and 50 percent of the thermal energy heats the air in the room, what is the change in air temperature in half an hour? c. At $0.08 per kWh, how much does it cost to run the heater 6.0 h per day for 30 days? Thermostat 240.0 V 4.80 95. A heating coil has a resistance of 4.0 and operates physicspp.com/chapter_test Chapter 22 Assessment 613 609-615 CH22-SG CA STP-845813 3/22/04 4:26 AM Page 614 Thinking Critically 97. Formulate Models How much energy is stored in a capacitor? The energy needed to increase the potential difference of a charge, q, is represented by E qV. But in a capacitor, V q/C. Thus, as charge is added, the potential difference increases. As more charge is added, however, it takes more energy to add the additional charge. Consider a 1.0-F \u201csupercap\u201d used as an energy storage device in a personal computer. Plot a graph of V as the capacitor is charged by adding 5.0 C to it. What is the voltage across the capacitor? The area under the curve is the energy stored in the capacitor. Find the energy in joules. Is it equal to the total charge times the final potential difference? Explain. 98. Apply Concepts A microwave oven operates at 120 V and requires 12 A of current. Its electric efficiency (converting AC to microwave radiation) is 75 percent, and its conversion efficiency from microwave radiation to heating water is also 75 percent. a. Draw a block power diagram similar to the energy diagram shown in Figure 22-2b on page 593. Label the function of each block according to total joules per second. b. Derive an equation for the rate of temperature increase (T/s) from the information presented in Chapter 12. Solve for the rate of temperature rise given the rate of energy input, the mass, and the specific heat of a substance. c. Use your equation to solve for the rate of temperature rise in degrees Celsius per second when using this oven to heat 250 g of water above room temperature. d. Review your calculations carefully for the units used and discuss why your answer is in the correct form. e. Discuss, in general terms, different ways in which you could increase the efficiency of microwave heating. f. Discuss, in efficiency terms, why microwave ovens are not useful for heating everything. g. Discuss, in general terms, why it is not a good idea to run microwave ovens when they are empty. 100. Apply Concepts The sizes of 10- resistors range from a pinhead to a soup can. Explain. 101. Make and Use Graphs The diode graph shown in Figure 22-17 on page 612 is more useful than a similar graph for a resistor that obeys Ohm\u2019s law. Explain. 102. Make and Use Graphs Based on what you have learned in this chapter, identify and prepare two parabolic graphs. Writing in Physics 103. There are three kinds of equations encountered in science: (1) definitions, (2) laws, and (3) derivations. Examples of these are: (1) an ampere is equal to one coulomb per second, (2) force is equal to mass times acceleration, (3) power is equal to voltage squared divided by resistance. Write a one-page explanation of where \u201cresistance is equal to voltage divided by current\u201d fits. Before you begin to write, first research the three categories given above. 104. In Chapter 13, you learned that matter expands when it is heated. Research", " the relationship between thermal expansion and high-voltage transmission lines. Cumulative Review 105. A person burns energy at the rate of about 8.4106 J per day. How much does she increase the entropy of the universe in that day? How does this compare to the entropy increase caused by melting 20 kg of ice? (Chapter 12) 106. When you go up the elevator of a tall building, your ears might pop because of the rapid change in pressure. What is the pressure change caused by riding in an elevator up a 30-story building (150 m)? The density of air is about 1.3 kg/m3 at sea level. (Chapter 13) 107. What is the wavelength in air of a 17-kHz sound wave, which is at the upper end of the frequency range of human hearing? (Chapter 15) 99. Analyze and Conclude A salesclerk in an appliance store states that microwave ovens are the most electrically efficient means of heating objects. a. Formulate an argument to refute the clerk\u2019s claim. Hint: Think about heating a specific object. b. Formulate an argument to support the clerk\u2019s claim. Hint: Think about heating a specific object. c. Formulate a diplomatic reply to the clerk. 108. Light of wavelength 478 nm falls on a double slit. First-order bright bands appear 3.00 mm from the central bright band. The screen is 0.91 m from the slits. How far apart are the slits? (Chapter 19) 109. A charge of 3.0106 C is 2.0 m from a second charge of 6.0105 C. What is the magnitude of the force between them? (Chapter 20) 614 Chapter 22 Current Electricity For more problems, go to Additional Problems, Appendix B. 609-615 CH22-SG CA STP-845813 3/22/04 4:27 AM Page 615 Multiple Choice 1. A 100-W lightbulb is connected to a 120-V electric line. What is the current that the lightbulb draws? 0.8 A 1 A 1.2 A 2 A 7. A 50.0- resistance causes a current of 5.00 mA to flow through a circuit connected to a battery. What is the power in the circuit? 1.00102 W 1.00103 W 1.25103 W 2.50103 W 2. A 5.0- resistor is connected to a 9.0-V battery. How much thermal energy is produced in 7.5 min? 8. How much electric energy is delivered to a 60.0-W lightbulb if the bulb is left on for 2.5 hours? 1.2102 J 1.3103 J 3.0103 J 7.3103 J 4.2102 J 2.4101 J 1.5102 J 5.4105 J 3. The current in the flashlight shown below is 0.50 A, and the voltage is the sum of the voltages of the individual batteries. What is the power delivered to the bulb of the flashlight? 0.11 W 1.1 W 2.3 W 4.5 W 1.5 V 1.5 V 1.5 V 4. If the flashlight in the illustration above is left on for 3.0 min, how much electric energy is delivered to the bulb? 6.9 J 14 J 2.0102 J 4.1102 J 5. A current of 2.0 A flows through a circuit containing a motor with a resistance of 12. How much energy is converted if the motor runs for one minute? 4.8101 J 2.0101 J 2.9103 J 1.7105 J Extended Answer 9. The diagram below shows a simple circuit containing a DC generator and a resistor. The table shows the resistances of several small electric devices. If the resistor in the diagram represents a hair dryer, what is the current in the circuit? How much energy does the hair dryer use if it runs for 2.5 min? Device Hair dryer Heater Small motor Resistance () 8.5 10.0 12.0 DC generator 120 V I 6. What is the effect on the current in a simple More Than One Graphic circuit if both the voltage and the resistance are reduced by half? divided by 2 no change multiplied by 2 multiplied by 4 If a test question has more than one table, graph, diagram, or drawing with it, use them all. If you answer based on just one graphic, you probably will miss an important piece of information. physicspp.com/standardized_test Chapter 22 Standardized Test Practice 615 616-626 CH23-S1-845813 3/22/04 5:14 AM Page 616 What You\u2019ll Learn \u2022 You will distinguish among series circuits, parallel circuits, and series-parallel combinations, and solve problems involving them. \u2022 You will explain the function", " of fuses, circuit breakers, and ground-fault interrupters, and describe how ammeters and voltmeters are used in circuits. Why It\u2019s Important Electric circuits are the basis of every electric device, from electric lights to microwave ovens to computers. Learning how circuits work will help you understand how countless electric devices function. Electric Load Centers Electric load centers form the link between the utility company and the circuits in a building. Each circuit breaker protects an individual circuit, which has the various loads connected in parallel. Think About This Why are the building loads connected in parallel? How are the circuit breakers connected? physicspp.com 616 Catherine Karnow/CORBIS, (inset)Horizons Companies 616-626 CH23-S1-845813 3/22/04 5:16 AM Page 617 How do fuses protect electric circuits? Question How does a fuse prevent an electric circuit from drawing too much current and creating a safety hazard? Procedure Analysis Explain how the thickness of a wire is related to how fast the wire will overheat and break apart. Why have circuit breakers replaced fuses in the electric circuit boxes of new homes? Critical Thinking Why is it important to replace a burned-out fuse in a house or car electric circuit with one that has the correct rating? 1. Connect the negative terminal of a 9-V battery to one terminal of a flashlight-bulb socket using a copper wire. CAUTION: Wire ends may be sharp and could cause cuts. 2. Connect the other terminal of the bulb socket to a single strand of steel wool using copper wire. Make sure the strand of steel wool is suspended over a small glass container. 3. Connect the other end of the single strand of steel wool to a switch using another piece of copper wire. Make sure the switch is open (turned off). 4. Connect the other terminal of the switch to the positive terminal of a power supply or a battery. 5. Hypothesize Predict what will happen when the switch is closed (turned on). 6. Observe Close the switch and make observations of the strand of steel wool. 7. Repeat steps 1\u20136 using a thicker strand of steel wool, or twist several strands together to form a single, thicker strand. 23.1 Simple Circuits Although the connection may not immediately be clear to you, a mountain river can be used to model an electric circuit. From its source high in the mountains, the river flows downhill to the plains below. No matter which path the river takes, its change in elevation, from the mountaintop to the plain, is the same. Some rivers flow downhill in a single stream. Other rivers may split into two or more smaller streams as they flow over a waterfall or through a series of rapids. In this case, part of the river follows one path, while other parts of the river follow different paths. No matter how many paths the river takes, however, the total amount of water flowing down the mountain remains unchanged. In other words, the amount of water flowing downhill is not affected by the path it takes. Objectives \u2022 Describe series and parallel circuits. \u2022 Calculate currents, voltage drops, and equivalent resistances in series and parallel circuits. Vocabulary series circuit equivalent resistance voltage divider parallel circuit Section 23.1 Simple Circuits 617 Horizons Companies 616-626 CH23-S1-845813 6/7/04 10:26 AM Page 618 How does the river shown in Figure 23-1 model an electric circuit? The distance that the river drops is similar to the potential difference in a circuit. The amount of water flowing in the river is similar to current in a circuit. Narrow rapids create resistance and are similar to resistors in a circuit. What part of a river is similar to a battery or a generator in an electric circuit? The energy source needed to raise water to the top of the mountain is the Sun. Solar energy evaporates water from lakes and seas leading to the formation of clouds that release rain or snow that falls on the mountaintops. Continue to think about the mountain river model as you read about the current in electric circuits. Series Circuits Three students are connecting two identical lamps to a battery, as illustrated in Figure 23-2. Before they make the final connection to the battery, their teacher asks them to predict the brightnesses of the two lamps. Each student knows that the brightness of a lamp depends on the current through it. The first student predicts that only the lamp close to the positive () terminal of the battery will light because all the current will be used up as thermal and light energy. The second student predicts that only part of the current will be used up, and the second lamp will glow, but more brightly than the first. The third student predicts that the lamps will be of equal brightness because current is a flow of charge and the charge leaving the first lamp has nowhere else to go in the circuit except through the second lamp. The third student reasons that because the", " current will be the same in each lamp, the brightness also will be the same. How do you predict the lights will behave? If you consider the mountain river model for this circuit, you will realize that the third student is correct. Recall from Chapter 20 that charge cannot be created or destroyed. Because the charge in the circuit has only one path to follow and cannot be destroyed, the same amount of charge must leave a circuit as enters the circuit. This means that the current is the same everywhere in the circuit. If you connect three ammeters in the circuit, as shown in Figure 23-3, they all will show the same current. A circuit such as this, in which all current travels through each device, is called a series circuit. If the current is the same throughout the circuit, what is used by the lamp to produce the thermal and light energy? Recall that power, the rate at which electric energy is converted, is represented by P IV. Thus, if there is a potential difference, or voltage drop, across the lamp, then electric energy is being converted into another form. The resistance of the lamp is defined as R V/I. Thus, the potential difference, also called the voltage drop, is V IR. Current and resistance in a series circuit From the river model, you know that the sum of the drops in height is equal to the total drop from the top of the mountain to sea level. In an electric circuit, the increase in voltage provided by the generator or other energy source, Vsource, is equal to the sum of voltage drops across lamps A and B, and is represented by the following equation: Vsource VA VB \u25a0 Figure 23-1 No matter what path a river or a stream takes down a mountain, the amount of water and the drop in elevation are the same. \u25a0 Figure 23-2 What is your prediction about the brightnesses of the two lightbulbs after the circuit is connected? 618 file photo Chapter 23 Series and Parallel Circuits 616-626 CH23-S1-845813 3/22/04 5:17 AM Page 619 Vsource RA RB \u25a0 Figure 23-3 The ammeters show that the current is the same everywhere in a series circuit. To find the potential drop across a resistor, multiply the current in the circuit by the resistance of the individual resistor. Because the IRB. Therefore, current through the lamps is the same, VA RB). The current through the circuit IRB, or Vsource Vsource is represented by the following equation: IRA and VB I(RA IRA V I ce ur so R R B A The same idea can be extended to any number of resistances in series, not just two. The same current would exist in the circuit with a single resistor, R, that has a resistance equal to the sum of the resistances of the two lamps. Such a resistance is called the equivalent resistance of the circuit. For resistors in series, the equivalent resistance is the sum of all the individual resistances, as expressed by the following equation. Equivalent Resistance for Resistors in Series R RA The equivalent resistance of resistors in series equals the sum of the individual resistances of the resistors. RB... Notice that the equivalent resistance is greater than that of any individual resistor. Therefore, if the battery voltage does not change, adding more devices in series always decreases the current. To find the current through a series circuit, first calculate the equivalent resistance and then use the following equation. Current Vso I urce R Current in a series circuit is equal to the potential difference of the source divided by the equivalent resistance. 1. Three 20- resistors are connected in series across a 120-V generator. What is the equivalent resistance of the circuit? What is the current in the circuit? 2. A 10-, 15-, and 5- resistor are connected in a series circuit with a 90-V battery. What is the equivalent resistance of the circuit? What is the current in the circuit? 3. A 9-V battery is in a circuit with three resistors connected in series. a. If the resistance of one of the resistors increases, how will the equivalent resistance change? b. What will happen to the current? c. Will there be any change in the battery voltage? 4. A string of holiday lights has ten bulbs with equal resistances connected in series. When the string of lights is connected to a 120-V outlet, the current through the bulbs is 0.06 A. a. What is the equivalent resistance of the circuit? b. What is the resistance of each bulb? 5. Calculate the voltage drops across the three resistors in problem 2, and verify that their sum equals the voltage of the battery. Section 23.1 Simple Circuits 619 616-626 CH23-S1-845813 3/22/04 5:18 AM Page 620 RA V I RB VB \u25a0 Figure 23-4 In this voltagedivider circuit, the values", " of RA and RB are chosen such that the voltage drop across RB is the desired voltage. Voltage drops in a series circuit As current moves through any circuit, the net change in potential must be zero. This is because the circuit\u2019s electric energy source, the battery or generator, raises the potential an amount equal to the potential drop produced when the current passes through the resistors. Therefore, the net change is zero. An important application of series resistors is a circuit called a voltage divider. A voltage divider is a series circuit used to produce a voltage source of desired magnitude from a higher-voltage battery. For example, suppose you have a 9-V battery but need a 5-V potential source. Consider the circuit shown in Figure 23-4. Two resistors, RA and RB, are connected in series across a battery of magnitude V. The equivalent resistance of the circuit is R RA RB. The current is represented by the following equation: I V R V RB RA The desired voltage, 5 V, is the voltage drop, VB, across resistor RB: VB Into this equation, the earlier equation for current is substituted. IRB. VB IRB RA V RB RB RB V RB RA Voltage dividers often are used with sensors, such as photoresistors. The resistance of a photoresistor depends upon the amount of light that strikes it. Photoresistors are made of semiconductors, such as silicon, selenium, or cadmium sulfide. A typical photoresistor can have a resistance of 400 when light is striking it compared with a resistance of 400,000 when the photoresistor is in the dark. The voltage output of a voltage divider that uses a photoresistor depends upon the amount of light striking the photoresistor sensor. This circuit can be used as a light meter, such as the one shown in Figure 23-5. In this device, an electronic circuit detects the potential difference and converts it to a measurement of illuminance that can be read on the digital display. The amplified voltmeter reading will drop as illuminance increases. \u25a0 Figure 23-5 The voltage output of this voltage divider depends upon the amount of light striking the photoresistor sensor (a). Light meters used in photography make use of a voltage divider (b). a Sensitivity adjustment (potentiometer) b Dry cells Amplified voltmeter V Photoresistor sensor Light 620 Laura Sifferlin Chapter 23 Series and Parallel Circuits 616-626 CH23-S1-845813 6/7/04 10:27 AM Page 621 Voltage Drops in a Series Circuit Two resistors, 47.0 and 82.0, are connected in series across a 45.0-V battery. a. What is the current in the circuit? b. What is the voltage drop across each resistor? c. If the 47.0- resistor is replaced by a 39.0- resistor, will the current increase, decrease, or remain the same? d. What is the new voltage drop across the 82.0- resistor? 1 Analyze and Sketch the Problem \u2022 Draw a schematic of the circuit. 45.0 V Known: Vsource RA RB 47.0 82.0 Unknown: I?? VA? VB 2 Solve for the Unknown a. To determine the current, first find the equivalent resistance. Vso urce and R RA I R RB V A RA RB VA VB Substitute R RA + RB V e urc so R R A B.0 45 V Substitute Vsource 47.0 2.0 8 0.349 A 45.0 V, RA 47.0, RB 82.0 b. Use V IR for each resistor. VA VB IRA (0.349 A)(47.0 ) 16.4 V IRB (0.349 A)(82.0 ) 28.6 V Substitute I 0.349 A, RA 47.0 Substitute I 0.349 A, RB 82.0 Math Handbook Operations with Significant Digits pages 835\u2014836 c. Calculate current, this time using 39.0 as RA. V I e urc so R R B A.0 V 45 Substitute Vsource 39.0 82.0 0.372 A The current will increase. 45.0 V, RA 39.0, RB 82.0 d. Determine the new voltage drop in RB. VB IRB (0.372 A)(82.0 ) 30.5 V Substitute I 0.372 A, RB 82.0 3 Evaluate the Answer \u2022 Are the units correct? Current is A V/; voltage is V A. \u2022 Is the magnitude realistic? For current, if R V, I 1. The voltage drop across any one resistor must be less than the voltage of the circuit. Both values of VB are less than Vsource, which is 45 V. Section 23.1 Simple Circuits 621 616", "-626 CH23-S1-845813 3/22/04 5:19 AM Page 622 6. The circuit shown in Example Problem 1 is producing these symptoms: the ammeter reads 0 A, VA reads 0 V, and VB reads 45 V. What has happened? 7. Suppose the circuit shown in Example Problem 1 has these values: RA 17.0 V. No other information is available. 292, and VA RB a. What is the current in the circuit? 255, b. What is the battery voltage? c. What are the total power dissipation and the individual power dissipations? d. Does the sum of the individual power dissipations in the circuit equal the total power dissipation in the circuit? Explain. 8. Holiday lights often are connected in series and use special lamps that short out when the voltage across a lamp increases to the line voltage. Explain why. Also explain why these light sets might blow their fuses after many bulbs have failed. 9. The circuit in Example Problem 1 has unequal resistors. Explain why the resistor with the lower resistance will operate at a lower temperature. 10. A series circuit is made up of a 12.0-V battery and three resistors. The voltage across one resistor is 1.21 V, and the voltage across another resistor is 3.33 V. What is the voltage across the third resistor? Voltage Divider A 9.0-V battery and two resistors, 390 and 470, are connected as a voltage divider. What is the voltage across the 470- resistor? 1 Analyze and Sketch the Problem \u2022 Draw the battery and resistors in a series circuit. 9.0 V Known: Vsource RA RB 390 470 Unknown:? VB 2 Solve for the Unknown R RA RB Vso I urce R VB RB Substitute R RA V e urc so R R B A IRB R V rce ou s B R R A B ) Substitute Vsource 0 (47 V).0 9 ( 70 4 0 9 3 4.9 V V R A Substitute I e urc so R B 9.0 V, RA 390, RB 470 RA RB I V VB Math Handbook Order of Operations page 843 3 Evaluate the Answer \u2022 Are the units correct? The voltage is V V/. The ohms cancel, leaving volts. \u2022 Is the magnitude realistic? The voltage drop is less than the battery voltage. Because 470 is more than half of the equivalent resistance, the voltage drop is more than half of the battery voltage. 622 Chapter 23 Series and Parallel Circuits 616-626 CH23-S1-845813 3/22/04 5:20 AM Page 623 11. A 22- resistor and a 33- resistor are connected in series and placed across a 120-V potential difference. a. What is the equivalent resistance of the circuit? b. What is the current in the circuit? c. What is the voltage drop across each resistor? d. What is the voltage drop across the two resistors together? 12. Three resistors of 3.3 k, 4.7 k, and 3.9 k are connected in series across a 12-V battery. a. What is the equivalent resistance? b. What is the current through the resistors? c. What is the voltage drop across each resistor? d. Find the total voltage drop across the three resistors. 13. A student makes a voltage divider from a 45-V battery, a 475-k resistor, and a 235-k resistor. The output is measured across the smaller resistor. What is the voltage? 14. Select a resistor to be used as part of a voltage divider along with a 1.2-k resistor. The drop across the 1.2-k resistor is to be 2.2 V when the supply is 12 V. Parallel Circuits Look at the circuit shown in Figure 23-6. How many current paths are there? The current from the generator can go through any of the three resistors. A circuit in which there are several current paths is called a parallel circuit. The three resistors are connected in parallel; both ends of the three paths are connected together. In the mountain river model, such a circuit is illustrated by three paths for the water over a waterfall. Some paths might have a large flow of water, while others might have a small flow. The sum of the flows, however, is equal to the total flow of water over the falls. In addition, regardless of which channel the water flows through, the drop in height is the same. Similarly, in a parallel electric circuit, the total current is the sum of the currents through each path, and the potential difference across each path is the same. \u25a0 Figure 23-6 The parallel paths for current in this diagram are analogous to multiple paths that a river might take down a mountain. RC RB RA Generator Parallel Resistance Hook up a power supply, a resistor, and an ammeter in", " a series circuit. 1. Predict what will happen to the current in the circuit when a second, identical resistor is added in parallel to the first. 2. Test your prediction. 3. Predict the new currents when the circuit contains three and four identical resistors in parallel. 4. Test your prediction. Analyze and Conclude 5. Make a data table to show your results. 6. Explain your results. (Hint: Include the idea of resistance.) Section 23.1 Simple Circuits 623 616-626 CH23-S1-845813 3/22/04 5:20 AM Page 624 120 V 38 A A 5 A A 13 A A 20 A A 24 120 V V 9 120 V V 6 120 V V \u25a0 Figure 23-7 In a parallel circuit, the total current is equal to the sum of the currents in the individual paths. Testing Resistance Ohmmeters, which are used to measure resistance, work by passing a known voltage across a resistor and measuring the current. The resistance is then displayed. Some ohmmeters use potentials of less than 1 V to avoid damaging delicate electronic components, whereas others may use hundreds of volts to check the integrity of insulating materials. What is the current through each resistor in a parallel electric circuit? It depends upon the individual resistances. For example, in Figure 23-7, the potential difference across each resistor is 120 V. The current through a resistor is given by I V/R, so you can calculate the current through the 24- resistor as I (120 V)/(24 ) 5.0 A and then calculate the currents through the other two resistors. The total current through the generator is the sum of the currents through the three paths, in this case, 38 A. What would happen if the 6- resistor were removed from the circuit? Would the current through the 24- resistor change? That current depends only upon the potential difference across it and its resistance; because neither has changed, the current also is unchanged. The same is true for the current through the 9- resistor. The branches of a parallel circuit are independent of each other. The total current through the generator, however, would change. The sum of the currents in the branches would be 18 A if the 6- resistor were removed. Resistance in a parallel circuit How can you find the equivalent resistance of a parallel circuit? In Figure 23-7, the total current through the generator is 38 A. Thus, the value of a single resistor that results in a 38-A current when a 120-V potential difference is placed across it can easily be calculated by using the following equation.2 Notice that this resistance is smaller than that of any of the three resistors in parallel. Placing two or more resistors in parallel always decreases the equivalent resistance of a circuit. The resistance decreases because each new resistor provides an additional path for current, thereby increasing the total current while the potential difference remains unchanged. To calculate the equivalent resistance of a parallel circuit, first note that the total current is the sum of the currents through the branches. If IA, IB, and IC are the currents through the branches and I is the total current, then I IA IC. The potential difference across each resistor is the same, so the current through each resistor, for example, RA, can be found from V/RA. Therefore, the equation for the sum of the currents is as follows: IA IB Dividing both sides of the equation by V provides an equation for the equivalent resistance of the three parallel resistors. Equivalent Resistance for Resistors in Parallel The reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances. This equation can be used for any number of resistors in parallel. 624 Chapter 23 Series and Parallel Circuits 616-626 CH23-S1-845813 6/7/04 10:30 AM Page 625 Equivalent Resistance and Current in a Parallel Circuit Three resistors, 60.0, 30.0, and 20.0, are connected in parallel across a 90.0-V battery. a. Find the current through each branch of the circuit. b. Find the equivalent resistance of the circuit. c. Find the current through the battery. 1 Analyze and Sketch the Problem \u2022 Draw a schematic of the circuit. \u2022 Include ammeters to show where you would measure each of the currents. Known: 60.0 RA 30.0 RB 20.0 RC V 90.0 V Unknown:??? I? R? IA IB IC A IA A IB A IC A 90.0 V 60.0 30.0 20.0 2 Solve for the Unknown V for each branch. a. Because the voltage across each resistor is the same, use I R IA IB IC V R A V Substitute V 90.0 V, RA 0. 0 9 6 0. 0 1.50 A V R B V Substitute V 90.0 V, RB 0. 0 9 0. 0 3", " 3.00 A V R C V Substitute V 90.0 V, RC 0. 0 9 0. 0 2 4.50 A 60.0 30.0 20.0 Math Handbook Fractions page 837 b. Use the equivalent resistance equation for parallel circuits. 60.0, RB 30.0, RC 20. Substitute RA 0 0 0 20. 30. 60. 1 0 10. R 10.0 V to find the total current. c. Use I R V I R V Substitute V 90.0 V, R 10..00 A 3 Evaluate the Answer \u2022 Are the units correct? Current is measured in amps; resistance is measured in ohms. \u2022 Is the magnitude realistic? The equivalent resistance is less than any single resistor. The current for the circuit, I, equals the sum of the current found for each resistor, IA IC. IB Section 23.1 Simple Circuits 625 616-626 CH23-S1-845813 6/15/04 11:31 PM Page 626 15. Three 15.0- resistors are connected in parallel and placed across a 30.0-V battery. a. What is the equivalent resistance of the parallel circuit? b. What is the current through the entire circuit? c. What is the current through each branch of the circuit? 16. A 120.0- resistor, a 60.0- resistor, and a 40.0- resistor are connected in parallel and placed across a 12.0-V battery. a. What is the equivalent resistance of the parallel circuit? b. What is the current through the entire circuit? c. What is the current through each branch of the circuit? 17. Suppose that one of the 15.0- resistors in problem 15 is replaced by a 10.0- resistor. a. Does the equivalent resistance change? If so, how? b. Does the amount of current through the entire circuit change? If so, in what way? c. Does the amount of current through the other 15.0- resistors change? If so, how? 18. A 150- branch in a circuit must be reduced to 93. A resistor will be added to this branch of the circuit to make this change. What value of resistance should be used and how must the resistor be connected? 19. A 12-, 2-W resistor is connected in parallel with a 6.0-, 4-W resistor. Which will become hotter if the voltage across them keeps increasing? Series and parallel connections differ in how they affect a lighting cirucuit. Imagine a 60-W and a 100-W bulb are used in a lighting circuit. Recall that the brightness of a lightbulb is proportional to the power it dissipates, and that P I2R. When the bulbs are connected in parallel, each is connected across 120 V and the 100-W bulb glows more brightly. When connected in series, the current through each bulb is the same. Because the resistance of the 60-W bulb is greater than that of the 100-W bulb, the higher-resistance 60-W bulb dissipates more power and glows more brightly. 23.1 Section Review 20. Circuit Types Compare and contrast the voltages and the currents in series and parallel circuits. 21. Total Current A parallel circuit has four branch currents: 120 mA, 250 mA, 380 mA, and 2.1 A. How much current is supplied by the source? 22. Total Current A series circuit has four resistors. The current through one resistor is 810 mA. How much current is supplied by the source? 23. Circuits A switch is connected in series with a 75-W bulb to a source of 120 V. a. What is the potential difference across the switch when it is closed (turned on)? b. What is the potential difference across the switch if another 75-W bulb is added in series? 24. Critical Thinking The circuit in Figure 23-8 has four identical resistors. Suppose that a wire is added to connect points A and B. Answer the following questions, and explain your reasoning. a. What is the current through the wire? b. What happens to the current through each resistor? c. What happens to the current drawn from the battery? d. What happens to the potential difference across each resistor? R A R R B R \u25a0 Figure 23-8 626 Chapter 23 Series and Parallel Circuits physicspp.com/self_check_quiz 627-631 CH23-S2-845813 3/22/04 5:25 AM Page 627 Objectives \u2022 Explain how fuses, circuit breakers, and ground-fault interrupters protect household wiring. \u2022 Analyze and solve problems involving combined seriesparallel circuits. \u2022 Explain how voltmeters and ammeters are used in circuits. Vocabulary short circuit fuse circuit breaker ground-fault interrupter combination series-parallel circuit ammeter voltmeter 23", ".2 Applications of Circuits You have learned about some of the elements of household wiring cir- cuits. It is important to understand the requirements and limitations of these systems. Above all, you need to be aware of the safety measures that must be followed to prevent accidents and injuries. Safety Devices In an electric circuit, fuses and circuit breakers act as safety devices. They prevent circuit overloads that can occur when too many appliances are turned on at the same time or when a short circuit occurs in one appliance. A short circuit occurs when a circuit with a very low resistance is formed. The low resistance causes the current to be very large. When appliances are connected in parallel, each additional appliance placed in operation reduces the equivalent resistance in the circuit and increases the current through the wires. This additional current might produce enough thermal energy to melt the wiring\u2019s insulation, cause a short circuit, or even begin a fire. A fuse is a short piece of metal that melts when too large a current passes through it. The thickness of the metal used in the fuse is determined by the amount of current that the circuit is designed to handle safely. If a large, unsafe current passes through the circuit, the fuse melts and breaks the circuit. A circuit breaker, shown in Figure 23-9, is an automatic switch that opens when the current reaches a threshold value. If there is a current greater than the rated (threshold) value in the circuit, the circuit becomes overloaded. The circuit breaker opens and stops the current. Current follows a single path from the power source through an electrical appliance and back to the source. Sometimes, a faulty appliance or an accidental drop of the appliance into water might create another current pathway. If this pathway flows through the user, serious injury could result. A current as small as 5 mA flowing through a person could result in electrocution. A ground-fault interrupter in an electric outlet prevents such injuries because it contains an electronic circuit that detects small differences in current caused by an extra current path and opens the circuit. Electric codes for buildings often require ground-fault interrupters to be used in bathroom, kitchen, and exterior outlets. Circuit Breaker On/off reset switch handle Switch contacts Current out to loads Bimetallic strip Latch Current in from central switch \u25a0 Figure 23-9 When too much current flows through the bimetallic strip, the heat that is generated causes the strip to bend and release the latch. The handle moves to the off position, causing the switch to open and break the circuit. Section 23.2 Applications of Circuits 627 627-631 CH23-S2-845813 6/7/04 10:39 AM Page 628 15-A fuse 120 V \u25a0 Figure 23-10 The parallel wiring arrangement used in homes allows the simultaneous use of more than one appliance. However, if too many appliances are used at once, the fuse could melt. Household applications Figure 23-10 diagrams a parallel circuit used in the wiring of homes, and also shows some common appliances that would be connected in parallel. The current in any one circuit does not depend upon the current in the other circuits. Suppose that a 240-W television is plugged into a 120-V outlet. The current is represented by I P/V. For the television, I (240 W)/(120 V) 2.0 A. When a 720-W curling iron is plugged into the outlet, its current draw is I (720 W)/(120 V) 6.0 A. Finally, a 1440-W hair dryer is plugged into the same outlet. The current through the hair dryer is I (1440 W)/(120 V) 12 A. The resistance of each appliance can be calculated using the equation R V/I. The equivalent resistance of the three appliances is as follows. 1 1 1 1 60 10 20 R 1 6 R 6 A fuse is connected in series with the power source so that the entire current passes through it. The current through the fuse is calculated using the equivalent resistance. V 20 A 20 1 I V 6 R If the fuse in the circuit is rated as 15 A, the 20-A current would exceed the rating and cause the fuse to melt, or \u201cblow,\u201d and cut off current. Fuses and circuit breakers also protect against the large currents created by a short circuit. Without a fuse or a circuit breaker, the current caused by a short circuit easily could start a fire. For example, a short circuit could occur if the insulation on a lamp cord became old and brittle. The two wires in the cord might accidentally touch, resulting in a resistance in the wire of only 0.010. This resistance results in a huge current. 0 2 1 I V 12,000 A 1 0.0 0 R V Such a current would cause a fuse or a circuit breaker to open the circuit, thereby preventing the wires from becoming hot enough to start a fire. When the galvanometer, a device used to measure very small currents", " or voltages, in this circuit measures zero, the circuit is said to be balanced. 1. Your lab partner states that the only way to balance this circuit is to make all the resistors equal. Will this balance the circuit? Is there more than one way to balance this circuit? Explain. 2. Derive a general equation for a balanced circuit using the given labels. Hint: Treat the circuit as a voltage divider. 3. Which of the resistors can be replaced with a variable resistor and then used to balance the circuit? 4. Which of the resistors can be replaced with a variable resistor and then used as a sensitivity control? Why would this be necessary? How would it be used in practice? R1 V R2 A R3 R4 R5 B 628 Horizons Companies Chapter 23 Series and Parallel Circuits 627-631 CH23-S2-845813 6/7/04 10:41 AM Page 629 Small resistance from wiring 120 V 120 V \u25a0 Figure 23-11 The small resistance inherent in wires is in series with the parallel resistances of household appliances. Appliances in parallel Combined Series-Parallel Circuits Have you ever noticed the light in your bathroom or bedroom dim when you turned on a hair dryer? The light and the hair dryer are connected in parallel across 120 V. Because of the parallel connection, the current through the light should not have changed when you turned on the hair dryer. Yet the light did dim, so the current must have changed. The dimming occurred because the house wiring had a small resistance. As shown in Figure 23-11, this resistance was in series with the parallel circuit. Such a circuit, which includes series and parallel branches, is called a combination seriesparallel circuit. The following are strategies for analyzing such circuits. Series-Parallel Circuits When analyzing a combination series-parallel circuit, use the following steps to break down the problem. 1. Draw a schematic diagram of the circuit. 2. Find any parallel resistors. Resistors in parallel have separate current paths. They must have the same potential differences across them. Calculate the single equivalent resistance of a resistor that can replace them. Draw a new schematic using that resistor. 3. Are any resistors (including the equivalent resistor) now in series? Resistors in series have one and only one current path through them. Calculate a single new equivalent resistance that can replace them. Draw a new schematic diagram using that resistor. 4. Repeat steps 2 and 3 until you can reduce the circuit to a single resistor. Find the total circuit current. Then go backwards through the circuits to find the currents through and the voltages across individual resistors. Reducing Circuit Diagrams RA 8.0 IA Step 1 60 V IB RB 25 IC RC 15 RA 8.0 Step 2 Step 3 RBC 9.4 R 17.4 Section 23.2 Applications of Circuits 629 627-631 CH23-S2-845813 7/22/04 4:04 PM Page 630 Series-Parallel Circuit A hair dryer with a resistance of 12.0 and a lamp with a resistance of 125 are connected in parallel to a 125-V source through a 1.50- resistor in series. Find the current through the lamp when the hair dryer is on. 1 2 Analyze and Sketch the Problem \u2022 Draw the series-parallel circuit including the hair dryer and lamp. \u2022 Replace RA and RB with a single equivalent resistance, Rp. 125 12.0 Known: 1.50 RC Vsource = 125 V RA RB Unknown: IA Rp I? R??? Solve for the Unknown Find the equivalent resistance for the parallel circuit, then find the equivalent resistance for the entire circuit, and then calculate the current. 1 1 1 1 1 Substitute RA 0 R 12. 125 R R p B A 10.9 RP R RC 1.50 11.0 Substitute RC Rp 12.4 125, RB 12.0 1.50, Rp 10.9 RC IA RA IB RB RC RP I R Substitute Vsource 125 V, R 12.4 Substitute I 10.1 A, RC 1.50 Substitute Vsource 125 V, VC 15.2 V VC (10.1 A)(1.50 ) Vso V 5 2 1 urce I 4 R. 2 1 10.1 A IRC 15.2 V VC Vsource 1.10102 V VA 02 V Substitute VA 1 1.1 0 R 5 2 1 A 0.880 A 125 V 15.2 V VA IA 1.10102 V, RA 125 Math Handbook Operations with Significant Digits pages 835\u2014836 3 Evaluate the Answer \u2022 Are the units correct? Current is measured in amps, and potential drops are measured in volts. \u2022 Is the magnitude realistic? The resistance is greater than the voltage, so the current should be less than 1 A. 25. A series-parallel circuit", " has three resistors: one dissipates 2.0 W, the second 3.0 W, and the third 1.5 W. How much current does the circuit require from a 12-V battery? 26. There are 11 lights in series, and they are in series with two lights in parallel. If the 13 lights are identical, which of them will burn brightest? 27. What will happen to the circuit in problem 26 if one of the parallel lights burns out? 28. What will happen to the circuit in problem 26 if one of the parallel lights shorts out? 630 Chapter 23 Series and Parallel Circuits 627-631 CH23-S2-845813 6/7/04 10:47 AM Page 631 Ammeters and Voltmeters An ammeter is a device that is used to measure the current in any branch or part of a circuit. If, for example, you wanted to measure the current through a resistor, you would place an ammeter in series with the resistor. This would require opening the current path and inserting an ammeter. Ideally, the use of an ammeter should not change the current in the circuit. Because the current would decrease if the ammeter increased the resistance in the circuit, the resistance of an ammeter is designed to be as low as possible. Figure 23-12a shows an ammeter as a meter placed in parallel with a 0.01- resistor. Because the resistance of the ammeter is much less than that of the resistors, the current decrease is negligible. Another instrument, called a voltmeter, is used to measure the voltage drop across a portion of a circuit. To measure the potential drop across a resistor, a voltmeter is connected in parallel with the resistor. Voltmeters are designed to have a very high resistance so as to cause the smallest possible change in currents and voltages in the circuit. Consider the circuit shown in Figure 23-12b. A voltmeter is shown as a meter in series with a 10-k resistor. When the voltmeter is connected in parallel with RB, the equivalent resistance of the combination is smaller than RB alone. Thus, the total resistance of the circuit decreases, and the current increases. The value of RA has not changed, but the current through it has increased, thereby increasing the potential drop across it. The battery, however, holds the potential drop across RA and RB constant. Thus, the potential drop across RB must decrease. The result of connecting a voltmeter across a resistor is to lower the potential drop across it. The higher the resistance of the voltmeter, the smaller the voltage change. Practical meters have resistances of 10 M. 0.01- Ammeter 20 V a RA 10.00 RB 10.00 0.01 10.00 10.00 20.01 b Voltmeter 10 k 10.00 RA 20 V 10.00 RB \u25a0 Figure 23-12 An ammeter is connected in series with two resistors (a). The small resistance of the ammeter slightly alters the current in the circuit. A voltmeter is connected in parallel with a resistor (b). The high resistance of the voltmeter results in a negligible change in the circuit current and voltage. 23.2 Section Review Refer to Figure 23-13 for questions 29\u201333, and 35. The bulbs in the circuit are identical. A I3 B A I1 I2 A A V \u25a0 Figure 23-13 1 2 C 3 31. Circuits in Series The wire at point C is broken and a small resistor is inserted in series with bulbs 2 and 3. What happens to the brightnesses of the two bulbs? Explain. 32. Battery Voltage A voltmeter connected across bulb 2 measures 3.8 V, and a voltmeter connected across bulb 3 measures 4.2 V. What is the battery voltage? 33. Circuits Using the information from problem 32, determine if bulbs 2 and 3 are identical. 34. Circuit Protection Describe three common safety devices associated with household wiring. 29. Brightness How do the bulb brightnesses compare? 30. Current If I3 measures 1.7 A and I1 measures 1.1 A, how much current is flowing in bulb 2? 35. Critical Thinking Is there a way to make the three bulbs in Figure 23-13 burn with equal intensity without using any additional resistors? Explain. physicspp.com/self_check_quiz Section 23.2 Applications of Circuits 631 632-633 CH23-LAB-845813 6/7/04 10:52 AM Page 632 Alternate CBL instructions can be found on the Web site. physicspp.com Series and Parallel Circuits In every circuit there is a relationship among current, potential difference, and resistance in electric circuits. In this experiment, you will investigate how the relationship of current, potential difference, and resistance in series circuits compares to that in parallel circuits. QUESTION How do relationships among current, potential difference, and resistance compare in series and parallel circuits? Objectives Materials", " \u25a0 Describe the relationship among current, potential difference, and resistance in a series circuit. \u25a0 Summarize the relationship among current, potential difference, and resistance in a parallel circuit. \u25a0 Collect data for current and potential differ- ence using electric meters. \u25a0 Calculate resistance in a lightbulb from current low-voltage power supply two light sockets two small lightbulbs ammeter or multimeter (0\u2013500-mA scale) voltmeter or multimeter (0\u201330-V scale) about ten copper wires with alligator clips and potential-difference data. Procedure Safety Precautions \u25a0 Hazard from electric shock is minimal because of the low currents used in this experiment. This experiment should not be carried out using current from an AC circuit, as this current is deadly. \u25a0 Handle wire ends with care as they may be sharp and could cause cuts. 1. Wire two lightbulb sockets in series with an ammeter and a low-voltage power supply. Observe the correct polarity when wiring the ammeter. 2. Screw the lightbulbs into the sockets. Turn on the power supply. Adjust the power control so that the bulbs are dimly lit. 3. Unscrew one of the bulbs. Record your obser- vations in the data table. 4. Screw in the bulb again and find the potential difference across both sets of bulbs by placing the positive probe of the voltmeter on the positive end of the circuit and the negative probe on the negative end of the circuit. Record your data in the data table. 5. Find the potential difference across each individual lightbulb by placing the positive probe of the voltmeter on the positive end of a bulb and the negative probe on the negative end of the bulb. Record your data in the data table. Repeat for the other bulb in series. 6. Place the ammeter at various locations in the series circuit. Record these currents in the data table. 7. Wire the two lightbulb sockets in parallel with the low-voltage power supply and in series with an ammeter. 632 Horizons Companies 632-633 CH23-LAB-845813 6/10/04 9:01 PM Page 633 Data Table Step Observations 3 4 5 6 8 9 10 11 8. Screw the lightbulbs into the sockets. Turn on the power supply. Adjust the power control so that the bulbs are dimly lit. Record the current shown on the ammeter in the data table. 9. Check the potential difference across the entire circuit and across each lightbulb. Record the values in the data table. 10. Place the voltmeter probes across one of the lightbulbs. Now unscrew one of the lightbulbs. Record your observations of both lightbulbs, and record the current and potential difference read by the meters in the data table. 11. Return the lightbulb you removed in the previous step to its socket. Now unscrew the other lightbulb. Record your observations of both lightbulbs, and record the current and potential difference read by the meters in the data table. Analyze 1. Calculate the resistance of the pair of lightbulbs in the series circuit. 2. Calculate the resistance of each lightbulb in the series circuit. 3. How does the resistance of the pair of lightbulbs compare to the individual resistance of each lightbulb? 4. How does the potential difference across the individual lightbulbs compare to the potential difference across the pair of lightbulbs in the series circuit? 5. Calculate the resistance of each of the light- bulbs while they are in the parallel circuit. How does this compare to the resistance calculated for the bulbs in the series circuit? Conclude and Apply 1. Summarize the relationship among current, potential difference, and resistance in a series circuit. 2. Summarize the relationship between current and potential difference in a parallel circuit. Going Further Repeat the experiment using lightbulbs of different voltage ratings (for example: 1.5 V, 3.0 V, and 6.0 V). Real-World Physics 1. The lightbulbs in most homes all are rated for 120 V, no matter how many bulbs there are. How is the ability to use any number of samevoltage bulbs affected by the way in which the bulbs are wired (series or parallel)? 2. Why do lights in a home dim when a large appliance, such as an air conditioner, is turned on? To find out more about series and parallel circuits, visit the Web site: physicspp.com 633 634 CH23-FEATURE-845813 6/7/04 10:55 AM Page 634 Ground Fault Circuit Interrupters (GFCI) A ground fault occurs when electricity takes an incorrect path to ground, such as through a person\u2019s body. Charles Dalziel, an engineering professor at the University of California, was an expert on the effects of electric shock. When he realized that ground faults were the cause", " of many electrocutions, he invented a device to prevent such accidents. How does a ground fault circuit interrupter (GFCI) work? No current here Sensor electronics Trip coil To load Equal current Transformer Neutral No ground current 3 The net current induces a changing magnetic field in the transformer core, which induces a current in the loop with the sensor electronics. The sensor detects the current and energizes an electromagnet, called the trip coil, thus breaking the circuit. This process takes 0.025 s. 1 In normal operation, there are equal and opposite currents through the wires, so the net current through the transformer is zero. Current in sensor loop Coil trips Unequal current Fault current Circuit interrupted 4 A test button on the GFCI connects a small resistor to the circuit, again creating unequal currents in the transformer and opening the circuit. 2 In a ground fault, there is current from the hot conductor through a person and to the ground. The currents in the two wires are unequal, and a net current passes through the transformer. Thinking Critically 1. Hypothesize Ground faults can occur when a person standing in water touches an appliance that is plugged in. How is water a factor in creating the ground fault? 2. Analyze and Conclude Will a GFCI protect a person who touches one wire with one hand and the other wire with the other hand? Explain. 3. Calculate In one GFCI, the test resistor is 14.75 k\u2126. For a voltage of 115 V, calculate the current through this resistor. Is this a lot of current? 634 How It Works 635-641 CH23-SG CA STP-845813 3/22/04 5:38 AM Page 635 23.1 Simple Circuits Vocabulary \u2022 series circuit (p. 618) \u2022 equivalent resistance (p. 619) \u2022 voltage divider (p. 620) \u2022 parallel circuit (p. 623) Key Concepts \u2022 The current is the same everywhere in a simple series circuit. \u2022 The equivalent resistance of a series circuit is the sum of the resistances of its parts. R RA RB... \u2022 The current in a series circuit is equal to the potential difference divided by the equivalent resistance. Vso I urce R \u2022 The sum of the voltage drops across resistors that are in series is equal to the potential difference applied across the combination. \u2022 A voltage divider is a series circuit used to produce a voltage source of desired magnitude from a higher-voltage battery. \u2022 The voltage drops across all branches of a parallel circuit are the same. \u2022 In a parallel circuit, the total current is equal to the sum of the currents in the branches. \u2022 The reciprocal of the equivalent resistance of parallel resistors is equal to the sum of the reciprocals of the individual resistances \u2022 If any branch of a parallel circuit is opened, there is no current in that branch. The current in the other branches is unchanged. 23.2 Applications of Circuits Vocabulary \u2022 short circuit (p. 627) \u2022 fuse (p. 627) \u2022 circuit breaker (p. 627) \u2022 ground-fault interrupter (p. 627) \u2022 combination series- parallel circuit (p. 629) \u2022 ammeter (p. 631) \u2022 voltmeter (p. 631) Key Concepts \u2022 A fuse or circuit breaker, placed in series with appliances, creates an open circuit when dangerously high currents flow. \u2022 A complex circuit is a combination of series and parallel branches. Any parallel branch first is reduced to a single equivalent resistance. Then, any resistors in series are replaced by a single resistance. \u2022 An ammeter is used to measure the current in a branch or part of a circuit. An ammeter always has a low resistance and is connected in series. \u2022 A voltmeter measures the potential difference (voltage) across any part or combination of parts of a circuit. A voltmeter always has a high resistance and is connected in parallel with the part of the circuit being measured. physicspp.com/vocabulary_puzzlemaker 635 635-641 CH23-SG CA STP-845813 6/7/04 11:04 AM Page 636 Concept Mapping 36. Complete the concept map using the following R3, constant terms: series circuit, R R1 current, parallel circuit, constant potential. R2 Resistors in circuits Type of circuit Principle Resistance Mastering Concepts 37. Why is it frustrating when one bulb burns out on a string of holiday tree lights connected in series? (23.1) 38. Why does the equivalent resistance decrease as more resistors are added to a parallel circuit? (23.1) 39. Several resistors with different values are connected in parallel. How do the values of the individual resistors compare with the equivalent resistance? (23.1) 40. Why is household wiring constructed in parallel instead of in series? (23.1) 41. Why is there", " a difference in equivalent resistance between three 60- resistors connected in series and three 60- resistors connected in parallel? (23.1) 42. Compare the amount of current entering a junction in a parallel circuit with that leaving the junction. (A junction is a point where three or more conductors are joined.) (23.1) 43. Explain how a fuse functions to protect an electric circuit. (23.2) 44. What is a short circuit? Why is a short circuit dangerous? (23.2) 45. Why is an ammeter designed to have a very low resistance? (23.2) 46. Why is a voltmeter designed to have a very high resistance? (23.2) Applying Concepts 48. What happens to the current in the other two lamps if one lamp in a three-lamp series circuit burns out? 49. Suppose the resistor, RA, in the voltage divider in Figure 23-4 is made to be a variable resistor. What happens to the voltage output, VB, of the voltage divider if the resistance of the variable resistor is increased? 50. Circuit A contains three 60- resistors in series. Circuit B contains three 60- resistors in parallel. How does the current in the second 60- resistor of each circuit change if a switch cuts off the current to the first 60- resistor? 51. What happens to the current in the other two lamps if one lamp in a three-lamp parallel circuit burns out? 52. An engineer needs a 10- resistor and a 15- resistor, but there are only 30- resistors in stock. Must new resistors be purchased? Explain. 53. If you have a 6-V battery and many 1.5-V bulbs, how could you connect them so that they light but do not have more than 1.5 V across each bulb? 54. Two lamps have different resistances, one larger than the other. a. If the lamps are connected in parallel, which is brighter (dissipates more power)? b. When the lamps are connected in series, which lamp is brighter? 55. For each of the following, write the form of circuit that applies: series or parallel. a. The current is the same everywhere throughout the entire circuit. b. The total resistance is equal to the sum of the individual resistances. c. The voltage drop across each resistor in the circuit is the same. d. The voltage drop in the circuit is proportional to the resistance. e. Adding a resistor to the circuit decreases the total resistance. f. Adding a resistor to the circuit increases the total resistance. g. If the current through one resistor in the circuit goes to zero, there is no current in the entire circuit. h. If the current through one resistor in the circuit goes to zero, the current through all other resistors remains the same. i. This form is suitable for house wiring. 47. How does the way in which an ammeter is connected in a circuit differ from the way in which a voltmeter is connected? (23.2) 56. Household Fuses Why is it dangerous to replace the 15-A fuse used to protect a household circuit with a fuse that is rated at 30 A? 636 Chapter 23 Series and Parallel Circuits For more problems, go to Additional Problems, Appendix B. 635-641 CH23-SG CA STP-845813 6/7/04 11:06 AM Page 637 Mastering Problems 23.1 Simple Circuits 57. Ammeter 1 in Figure 23-14 reads 0.20 A. a. What should ammeter 2 indicate? b. What should ammeter 3 indicate? A1 22 A2 15 V \u25a0 Figure 23-14 A3 58. Calculate the equivalent resistance of these seriesconnected resistors: 680, 1.1 k, and 10 k. 59. Calculate the equivalent resistance of these parallelconnected resistors: 680, 1.1 k, and 10.2 k. 60. A series circuit has two voltage drops: 5.50 V and 6.90 V. What is the supply voltage? 61. A parallel circuit has two branch currents: 3.45 A and 1.00 A. What is the current in the energy source? 62. Ammeter 1 in Figure 23-14 reads 0.20 A. a. What is the total resistance of the circuit? b. What is the battery voltage? c. How much power is delivered to the 22- resistor? d. How much power is supplied by the battery? 63. Ammeter 2 in Figure 23-14 reads 0.50 A. a. Find the voltage across the 22- resistor. b. Find the voltage across the 15- resistor. c. What is the battery voltage? 64. A 22- lamp and a 4.5- lamp are connected in series and placed across a potential difference of 45 V as shown in Figure 23-15. a. What is the equivalent resistance of the circuit", "? b. What is the current in the circuit? c. What is the voltage drop across each lamp? d. What is the power dissipated in each lamp? 65. Refer to Figure 23-16 to answer the following questions. a. What should the ammeter read? b. What should voltmeter 1 read? c. What should voltmeter 2 read? d. How much energy is supplied by the battery per minute? V1 35 A 10 V 15 V2 \u25a0 Figure 23-16 66. For Figure 23-17, the voltmeter reads 70.0 V. a. Which resistor is the hottest? b. Which resistor is the coolest? c. What will the ammeter read? d. What is the power supplied by the battery? A V V 35 15 50 \u25a0 Figure 23-17 67. For Figure 23-18, the battery develops 110 V. a. Which resistor is the hottest? b. Which resistor is the coolest? c. What will ammeter 1 read? d. What will ammeter 2 read? e. What will ammeter 3 read? f. What will ammeter 4 read? 22 4.5 45 V A2 A3 A4 20.0 50.0 10.0 V A1 \u25a0 Figure 23-15 \u25a0 Figure 23-18 physicspp.com/chapter_test Chapter 23 Assessment 637 635-641 CH23-SG CA STP-845813 6/7/04 11:08 AM Page 638 68. For Figure 23-18, ammeter 3 reads 0.40 A. 23.2 Applications of Circuits a. What is the battery voltage? b. What will ammeter 1 read? c. What will ammeter 2 read? d. What will ammeter 4 read? 69. What is the direction of the conventional current in the 50.0- resistor in Figure 23-18? 70. The load across a battery consists of two resistors, with values of 15 and 47, connected in series. a. What is the total resistance of the load? b. What is the voltage of the battery if the current in the circuit is 97 mA? 71. Holiday Lights A string of 18 identical holiday tree lights is connected in series to a 120-V source. The string dissipates 64 W. a. What is the equivalent resistance of the light string? b. What is the resistance of a single light? c. What power is dissipated by each light? 72. One of the lights in problem 71 burns out. The light shorts out the bulb filament when it burns out. This drops the resistance of the lamp to zero. a. What is the resistance of the light string now? b. Find the power dissipated by the string. c. Did the power increase or decrease when the bulb burned out? 73. A 16.0- and a 20.0- resistor are connected in parallel. A difference in potential of 40.0 V is applied to the combination. a. Compute the equivalent resistance of the parallel circuit. b. What is the total current in the circuit? c. What is the current in the 16.0- resistor? 74. Amy needs 5.0 V for an integrated-circuit experiment. She uses a 6.0-V battery and two resistors to make a voltage divider. One resistor is 330. She decides to make the other resistor smaller. What value should it have? 75. Pete is designing a voltage divider using a 12-V battery and a 82- resistor as RB. What resistor should be used as RA if the output voltage across RB is to be 4.0 V? 76. Television A typical television dissipates 275 W when it is plugged into a 120-V outlet. a. Find the resistance of the television. b. The television and 2.5- wires connecting the outlet to the fuse form a series circuit that works like a voltage divider. Find the voltage drop across the television. c. A 12- hair dryer is plugged into the same outlet. Find the equivalent resistance of the two appliances. d. Find the voltage drop across the television and the hair dryer. 77. Refer to Figure 23-19 and assume that all the resistors are 30.0. Find the equivalent resistance. IA IB IC 78. Refer to Figure 23-19 and assume that each resistor dissipates 120 mW. Find the total dissipation. \u25a0 Figure 23-19 and IB 79. Refer to Figure 23-19 and assume that IA 1.7 mA. Find IC. 80. Refer to Figure 23-19 and assume that IB 1.7 mA. Find IA. and IC 13 mA 13 mA 81. Refer to Figure 23-20 to answer the following questions. a. Determine the total resistance. b. Determine the current through the 25- resistor. c. Which resistor is the hottest? Coolest? 25.0 30.0 10.0 25 V 20.0", " 40.0 \u25a0 Figure 23-20 82. A circuit contains six 60-W lamps with a resistance of 240- each and a 10.0- heater connected in parallel. The voltage across the circuit is 120 V. Find the current in the circuit for the following situations. a. Four lamps are turned on. b. All of the lamps are turned on. c. Six lamps and the heater are operating. 83. If the circuit in problem 82 has a 12-A fuse, will the fuse melt if all the lamps and the heater are on? 84. During a laboratory exercise, you are supplied with a battery of potential difference V, two heating elements of low resistance that can be placed in water, an ammeter of very small resistance, a voltmeter of extremely high resistance, wires of negligible resistance, a beaker that is well insulated and has negligible heat capacity, and 0.10 kg of water at 25\u00b0C. By means of a diagram and standard symbols, show how these components should be connected to heat the water as rapidly as possible. 638 Chapter 23 Series and Parallel Circuits For more problems, go to Additional Problems, Appendix B. 635-641 CH23-SG CA STP-845813 6/8/04 5:27 PM Page 639 85. If the voltmeter used in problem 84 holds steady at 45 V and the ammeter reading holds steady at 5.0 A, estimate the time in seconds required to completely vaporize the water in the beaker. Use 4.2 kJ/kg\u00b0C as the specific heat of water and 2.3106 J/kg as the heat of vaporization of water. 86. Home Circuit A typical home circuit is shown in Figure 23-21. The wires to the kitchen lamp each have a resistance of 0.25. The lamp has a resistance of 0.24 k. Although the circuit is parallel, the lead lines are in series with each of the components of the circuit. a. Compute the equivalent resistance of the circuit consisting of just the lamp and the lead lines to and from the lamp. b. Find the current to the lamp. c. Find the power dissipated in the lamp. Kitchen light 240 Power saw Wall outlets 0.25 Switch box 120 V 0.25 \u25a0 Figure 23-21 Mixed Review 87. A series circuit has two voltage drops: 3.50 V and 4.90 V. What is the supply voltage? 88. A parallel circuit has two branch currents: 1.45 A and 1.00 A. What is the current in the energy source? 89. A series-parallel circuit has three resistors, dissipating 5.50 W, 6.90 W, and 1.05 W, respectively. What is the supply power? 90. Determine the maximum safe power in each of three 150-, 5-W resistors connected in series. 91. Determine the maximum safe power in each of three 92-, 5-W resistors connected in parallel. 92. A voltage divider consists of two 47-k resistors connected across a 12-V battery. Determine the measured output for the following. a. an ideal voltmeter b. a voltmeter with a resistance of 85 k c. a voltmeter with a resistance of 10106 physicspp.com/chapter_test 93. Determine the maximum safe voltage that can be applied across the three series resistors in Figure 23-22 if all three are rated at 5.0 W. 92 150 220 \u25a0 Figure 23-22 94. Determine the maximum safe total power for the circuit in problem 93. 95. Determine the maximum safe voltage that can be applied across three parallel resistors of 92, 150, and 220, as shown in Figure 23-23, if all three are rated at 5.0 W. 92 150 220 \u25a0 Figure 23-23 Thinking Critically 96. Apply Mathematics Derive equations for the resistance of two equal-value resistors in parallel, three equal-value resistors in parallel, and N equalvalue resistors in parallel. 97. Apply Concepts Three-way lamps, of the type in Figure 23-24, having a rating of 50 W, 100 W, and 150 W, are common. Draw four partial schematic diagrams that show the lamp filaments and the switch positions for each brightness level, as well as the off position. (You do not need to show the energy source.) Label each diagram. \u25a0 Figure 23-24 Chapter 23 Assessment 639 Horizons Companies 635-641 CH23-SG CA STP-845813 6/7/04 11:15 AM Page 640 98. Apply Concepts Design a circuit that will light one dozen 12-V bulbs, all to the correct (same) intensity, from a 48-V battery. a. Design A requires that should one bulb burn out, all other bulbs continue to produce light. b. Design B requires that should one bulb burn out, those bulbs that continue working must produce the correct intensity.", " c. Design C requires that should one bulb burn out, one other bulb also will go out. d. Design D requires that should one bulb burn out, either two others will go out or no others will go out. 99. Apply Concepts A battery consists of an ideal source of potential difference in series with a small resistance. The electric energy of the battery is produced by chemical reactions that occur in the battery. However, these reactions also result in a small resistance that, unfortunately, cannot be completely eliminated. A flashlight contains two batteries in series, as shown in Figure 23-25. Each has a potential difference of 1.50 V and an internal resistance of 0.200. The bulb has a resistance of 22.0. a. What is the current through the bulb? b. How much power does the bulb dissipate? c. How much greater would the power be if the batteries had no internal resistance? Battery Battery 1.50 V 0.200 1.50 V 0.200 22.0 \u25a0 Figure 23-25 100. Apply Concepts An ohmmeter is made by connecting a 6.0-V battery in series with an adjustable resistor and an ideal ammeter. The ammeter deflects full-scale with a current of 1.0 mA. The two leads are touched together and the resistance is adjusted so that 1.0 mA flows. a. What is the resistance of the adjustable resistor? b. The leads are now connected to an unknown resistance. What resistance would produce a current of half-scale, 0.50 mA? Quarter-scale, 0.25 mA? Three-quarters-scale, 0.75 mA? c. Is the ohmmeter scale linear? Explain. Writing in Physics 101. Research Gustav Kirchhoff and his laws. Write a one-page summary of how they apply to the three types of circuits presented in this chapter. Cumulative Review 102. Airplane An airplane flying through still air produces sound waves. The wave fronts in front of the plane are spaced 0.50 m apart and those behind the plane are spaced 1.50 m apart. The speed of sound is 340 m/s. (Chapter 15) a. What would be the wavelength of the sound waves if the airplane were not moving? b. What is the frequency of the sound waves produced by the airplane? c. What is the speed of the airplane? d. What is the frequency detected by an observer located directly in front of the airplane? e. What is the frequency detected by an observer located directly behind the airplane? 103. An object is located 12.6 cm from a convex mirror with a focal length of 18.0 cm. What is the location of the object\u2019s image? (Chapter 17) 104. The speed of light in a special piece of glass is 1.75108 m/s. What is its index of refraction? (Chapter 18) 105. Monocle An antireflective coating with an index of refraction of 1.4 is applied to a monocle with an index of refraction of 1.52. If the thickness of the coating is 75 nm, what is/are the wavelength(s) of light for which complete destructive interference will occur? (Chapter 19) 106. Two charges of 2.0105 C and 8.0106 C experience a force between them of 9.0 N. How far apart are the two charges? (Chapter 20) 107. A field strength, E, is measured a distance, d, from a point charge, Q. What would happen to the magnitude of E in the following situations? (Chapter 21) a. d is tripled b. Q is tripled c. both d and Q are tripled d. the test charge q is tripled e. all three, d, Q, and q, are tripled 108. The current flow in a 12-V circuit drops from 0.55 A to 0.44 A. Calculate the change in resistance. (Chapter 22) 640 Chapter 23 Series and Parallel Circuits For more problems, go to Additional Problems, Appendix B. 635-641 CH23-SG CA STP-845813 6/7/04 11:20 AM Page 641 Multiple Choice Use the following circuit diagram to answer questions 1\u20133. 6.0 V R1 3.0 R2 12 R3 4.0 1. What is the equivalent resistance of the circuit? 1 9 1 1.0 1.5 19 2. What is the current in the circuit? 0.32 A 0.80 A 3. How much current is in R3? 0.32 A 1.5 A 1.2 A 4.0 A 2.0 A 4.0 A 4. What would a voltmeter placed across R2 read? 0.32 V 1.5 V 3.8 V 6.0 V Use the following circuit diagram to answer questions 5 and 6. 60.0 V RA 12.0 RB", " RC 25.0 15.0 6. What is the current in the circuit? 1.15 A 2.35 A 2.80 A 5.61 A 7. Nina connects eight 12- lamps in series. What is the total resistance of the circuit? 0.67 1.5 12 96 8. Which statement is true? The resistance of a typical ammeter is very high. The resistance of a typical voltmeter is very low. Ammeters have zero resistance. A voltmeter causes a small change in current. Extended Answer 9. Chris is throwing a tailgate party before a nighttime football game. To light the tailgate party, he connects 15 large outdoor lamps to his 12.0-V car battery. Once connected, the lamps do not glow. An ammeter shows that the current through the lamps is 0.350 A. If the lamps require a 0.500-A current in order to work, how many lamps must Chris remove from the circuit? 10. A series circuit has an 8.0-V battery and four 15.0. Calculate the current and 8.0, R3 13.0, 4.0, R2 resistors, R1 and R4 the power in the circuit. 5. What is the equivalent resistance of the circuit? 8.42 10.7 21.4 52.0 Take a Break If you have the opportunity to take a break or get up from your desk during a test, take it. Getting up and moving around will give you extra energy and help you clear your mind. During the break, think about something other than the test so you\u2019ll be able to begin again with a fresh start. physicspp.com/standardized_test Chapter 23 Standardized Test Practice 641 642-651 CH24-S1-845813 3/30/04 8:40 PM Page 642 What You\u2019ll Learn \u2022 You will assign forces of attraction or repulsion between magnetic poles. \u2022 You will relate magnetism to electric charge and electricity. \u2022 You will describe how electromagnetism can be harnessed for practical applications. Why It\u2019s Important Magnetism is the basis for many technologies. Information on the hard drive of a computer is stored as a magnetic pattern. Atom Smashers An accelerator tube, such as the one pictured, is surrounded by superconducting magnets. There is no magnetic field at the center of the tube where high-energy particles travel. If the particles stray from the center, they receive a magnetic push to keep them there. Think About This How do forces applied by magnets cause particles to accelerate? Can any particle be accelerated? physicspp.com 642 Fermilab/Peter Arnold, Inc. 642-651 CH24-S1-845813 7/14/04 1:51 PM Page 643 In which direction do magnetic fields act? Question What would be the direction of force on a magnetized object in a magnetic field? Procedure Analysis What did the red end of the compass needle typically point toward? Away from? Why might some of the arrows not point to either location stated in question 1? Critical Thinking What you have diagrammed with your arrows is called a magnetic field. Recall what a gravitational field and an electric field are, and define magnetic field. 1. Place a bar magnet horizontally in front of you so that the north pole faces left. 2. Place a second bar magnet horizontally next to, and 5.0 cm away from the first (you should be able to place the compass between the magnets). The north pole also should be facing the left. 3. Draw your setup on a sheet of paper. Be sure to label the poles. 4. Place a compass by the two magnets. Draw the direction the arrow is pointing. 5. Continue to move the compass to other positions, each time drawing the direction it points until you have drawn 15\u201320 arrows. 6. Repeat steps 3\u20135, this time with the two north poles facing each other. 24.1 Magnets: Permanent and Temporary The existence of magnets and magnetic fields has been known for more than 2000 years. Chinese sailors employed magnets as navigational compasses approximately 900 years ago. Throughout the world, early scientists studied magnetic rocks, called lodestones. Today, magnets play an increasingly important role in our everyday lives. Electric generators, simple electric motors, television sets, cathode-ray displays, tape recorders, and computer hard drives all depend on the magnetic effects of electric currents. If you have ever used a compass or picked up tacks or paper clips with a magnet, you have observed some effects of magnetism. You even might have made an electromagnet by winding wire around a nail and connecting it to a battery. The properties of magnets become most obvious when you experiment with two of them. To enhance your study of magnetism, you can experiment with magnets, such as those shown in Figure 24-1 on the next page. Objectives \u2022 Describe the properties of magnets and the", " origin of magnetism in materials. \u2022 Compare and contrast various magnetic fields. Vocabulary polarized magnetic fields magnetic flux first right-hand rule solenoid electromagnet second right-hand rule domain Section 24.1 Magnets: Permanent and Temporary 643 Horizons Companies 642-651 CH24-S1-845813 3/30/04 8:42 PM Page 644 \u25a0 Figure 24-1 Common magnets are available in most hardware stores. General Properties of Magnets Suspend a magnet from a thread, as in Figure 24-2a. If you use a bar magnet, you might have to tie a yoke around it to keep it horizontal. When the magnet comes to rest, is it lined up in any particular direction? Now rotate the magnet so that it points in a different direction. When you release the magnet, does it come to rest in the same direction? If so, in which direction does it point? You should have found that the magnet lined up in a north-south direction. Mark the end that points to the north with the letter N for reference. From this simple experiment, you can conclude that a magnet is polarized; that is, it has two distinct and opposite ends. One of the poles is the northseeking pole; the other is the south-seeking pole. A compass is nothing more than a small magnet, mounted so that it is free to turn. Suspend another magnet to determine the north end, and mark it as you did with the first magnet. While one magnet is suspended, observe the interaction of the two magnets by bringing the other magnet near, as in Figure 24-2b. What happens as you bring the two ends that were pointing north, the north poles, toward each other? Now try it with the south poles. Lastly, what happens as you bring opposite poles (the north pole of one magnet and the south pole of the other magnet) toward each other? You should have observed that the two north poles repelled each other, as did the two south poles. However, the north pole of one magnet should have attracted the south pole of the other magnet. Like poles repel; unlike poles attract. Magnets always have two opposite magnetic poles. If you break a magnet in half, you create two smaller magnets, and each will have two poles. Scientists have tried to break magnets into separate north and south poles, called monopoles, but no one has succeeded, not even on the microscopic level. Knowing that magnets always orient themselves in a north-south direction, it may occur to you that Earth itself is a giant magnet. Because opposite poles attract and the north pole of a compass magnet points north, the south pole of the Earth-magnet must be near Earth\u2019s geographic north pole. \u25a0 Figure 24-2 If you suspend a magnet by a thread, it will align itself with magnetic properties in Earth (a). The magnet\u2019s north pole will point north. If you then move the north pole of a second magnet toward the north pole of the suspended magnet, the suspended magnet will move away (b). a b 644 Horizons Companies Chapter 24 Magnetic Fields 642-651 CH24-S1-845813 3/30/04 8:44 PM Page 645 \u25a0 Figure 24-3 A common nail is attracted to a magnet. In the process, the nail itself becomes magnetized. Here you can see when a magnet is touching the nail, the nail is able to attract other metal objects. However, if you separate the magnet from the nail, some of the objects will drop off because the nail will have lost some of its magnetism. How do magnets affect other materials? As you probably discovered as a child, magnets attract things besides other magnets, such as nails, tacks, paper clips, and many other metal objects. Unlike the interaction between two magnets, however, either end of a magnet will attract either end of a piece of metal. How can you explain this behavior? First, you can touch a magnet to a nail and then touch the nail to smaller metal pieces. The nail itself becomes a magnet, as shown in Figure 24-3. The magnet causes the nail to become polarized. The direction of polarization of the nail depends on the polarization of the magnet. If you pull away the magnet, the nail loses some of its magnetization and will no longer exhibit as much attraction for other metal objects. If you repeat the experiment shown in Figure 24-3 with a piece of soft iron (iron with a low carbon content) in place of a nail, you will notice that the iron loses all of its attraction for the other metal objects when the magnet is pulled away. This is because soft iron is a temporary magnet. A nail has other material in it to make it harder and allows it to retain some of its magnetism when a permanent magnet is pulled away. Permanent magnets The magnetism of permanent magnets is produced in the same way in which you created the magnetism of the nail. Because of the microscopic structure of the magnet material,", " the induced magnetism becomes permanent. Many permanent magnets are made of an iron alloy called ALNICO V, that contains a mix of aluminum, nickel, and cobalt. A variety of rare earth elements, such as neodymium and gadolinium, produce permanent magnets that are extremely strong for their size. Magnetic Fields Around Permanent Magnets When you experimented with two magnets, you noticed that the forces between magnets, both attraction and repulsion, occur not only when the magnets touch each other, but also when they are held apart. In the same way that long-range electric and gravitational forces can be described by electric and gravitational fields, magnetic forces can be described by the existence of fields around magnets. These magnetic fields are vector quantities that exist in a region in space where a magnetic force occurs. Section 24.1 Magnets: Permanent and Temporary 645 Matt Meadows 642-651 CH24-S1-845813 3/30/04 8:45 PM Page 646 a b \u25a0 Figure 24-4 The magnetic field of a bar magnet shows up clearly in three dimensions when the magnet is suspended in glycerol with iron filings (a). It is, however, easier to set up a magnet on a sheet of paper covered with iron filings to see the pattern in two dimensions (b). The presence of a magnetic field around a magnet can be shown using iron filings. Each long, thin, iron filing becomes a small magnet by induction. Just like a tiny compass needle, the iron filing rotates until it is parallel to the magnetic field. Figure 24-4a shows filings in a glycerol solution surrounding a bar magnet. The three-dimensional shape of the field is visible. In Figure 24-4b, the filings make up a two-dimensional plot of the field, which can help you visualize magnetic field lines. Filings also can show how the field can be distorted by an object. Magnetic field lines Note that magnetic field lines, like electric field lines, are imaginary. They are used to help us visualize a field, and they also provide a measure of the strength of the magnetic field. The number of magnetic field lines passing through a surface is called the magnetic flux. The flux per unit area is proportional to the strength of the magnetic field. As you can see in Figure 24-4, the magnetic flux is most concentrated at the poles; thus, this is where the magnetic field strength is the greatest. The direction of a magnetic field line is defined as the direction in which the north pole of a compass points when it is placed in the magnetic field. Outside the magnet, the field lines emerge from the magnet at its north pole and enter the magnet at its south pole, as illustrated in Figure 24-5. What happens inside the magnet? There are no isolated poles on which field lines can start or stop, so magnetic field lines always travel inside the magnet from the south pole to the north pole to form closed loops. N S \u2022 Positive charges are red. \u2022 Negative charges are blue. \u2022 Electric field lines are indigo. \u2022 Magnetic field lines are green. \u25a0 Figure 24-5 Magnetic field lines can be visualized as closed loops leaving the north pole of a magnet and entering the south pole of the same magnet. 646 Horizons Companies Chapter 24 Magnetic Fields 642-651 CH24-S1-845813 3/30/04 8:45 PM Page 647 a b \u25a0 Figure 24-6 The magnetic field lines indicated by iron filings on paper clearly show that like poles repel (a) and unlike poles attract (b). The iron filings do not form continuous lines between like poles. Between a north and a south pole, however, the iron filings show that field lines run directly between the two magnets. What kinds of magnetic fields are produced by pairs of bar magnets? You can visualize these fields by placing two magnets on a sheet of paper, and then sprinkling the paper with iron filings. Figure 24-6a shows the field lines between two like poles. In contrast, two unlike poles (north and south) placed close together produce the pattern shown in Figure 24-6b. The filings show that the field lines between two unlike poles run directly from one magnet to the other. Forces on objects in magnetic fields Magnetic fields exert forces on other magnets. The field produced by the north pole of one magnet pushes the north pole of a second magnet away in the direction of the field line. The force exerted by the same field on the south pole of the second magnet is attractive in a direction opposite the field lines. The second magnet attempts to line up with the field, just like a compass needle. When a sample made of iron, cobalt, or nickel is placed in the magnetic field of a permanent magnet, the field lines become concentrated within the sample. Lines leaving the north pole of the magnet enter one end of the sample, pass through it, and leave the other end. Thus, the end of the sample closest to the magnet\u2019s north pole becomes the sample\u2019s south", " pole, and the sample is attracted to the magnet. 1. If you hold a bar magnet in each hand and bring your hands close together, will the force be attractive or repulsive if the magnets are held in the following ways? a. the two north poles are brought close together b. a north pole and a south pole are brought together 2. Figure 24-7 shows five disk magnets floating above each other. The north pole of the top-most disk faces up. Which poles are on the top side of each of the other magnets? 3. A magnet attracts a nail, which, in turn, attracts many small tacks, as shown in Figure 24-3 on page 645. If the north pole of the permanent magnet is the left end, as shown, which end of the nail is the south pole? 4. Why do magnetic compasses sometimes give false readings? \u25a0 Figure 24-7 Section 24.1 Magnets: Permanent and Temporary 647 (b)Doug Martin, (others)Horizons Companies 642-651 CH24-S1-845813 6/7/04 11:29 AM Page 648 \u25a0 Figure 24-8 Using an apparatus similar to the one shown (a), Oersted was able to demonstrate a connection between magnetism and electricity by applying current to the wire (b). a Power supply current AC DC OFF ON b Power supply current AC DC OFF ON Compass Compass Electromagnetism In 1820, Danish physicist Hans Christian Oersted was experimenting with electric currents in wires. Oersted laid a wire across the top of a small compass and connected the ends of the wire to complete an electrical circuit, as shown in Figure 24-8a. He had expected the needle to point toward the wire or in the same direction as the current in the wire. Instead, he was amazed to see that the needle rotated until it pointed perpendicular to the wire, as shown in Figure 24-8b. The forces on the compass magnet\u2019s poles were perpendicular to the direction of current in the wire. Oersted also found that when there was no current in the wire, no magnetic forces existed. If a compass needle turns when placed near a wire carrying an electric current, it must be the result of a magnetic field created by the current. You easily can show the magnetic field around a current-carrying wire by placing a wire vertically through a horizontal piece of cardboard on which iron filings are sprinkled. When there is current through the wire, the filings will form a pattern of concentric circles, around the wire, as shown in Figure 24-9. The circular lines indicate that magnetic field lines around current-carrying wires form closed loops in the same way that field lines about permanent magnets form closed loops. The strength of the magnetic field around a long, straight wire is proportional to the current in the wire. The strength of the field also varies inversely with the distance from the wire. A compass shows the direction of the field lines. If you reverse the direction of the current, the compass needle also reverses its direction, as shown in Figure 24-10a. a I I b Current Right hand Direction of magnetic field \u25a0 Figure 24-9 The magnetic field produced by the current in a wire through a cardboard disk shows up as concentric circles of iron filings around the wire. \u25a0 Figure 24-10 The magnetic field produced by current in a straight-wire conductor reverses when the current in the wire is reversed (a). The first right-hand rule for a straight, current-carrying wire shows the direction of the magnetic field (b). 648 Kodansha Chapter 24 Magnetic Fields 642-651 CH24-S1-845813 3/30/04 8:46 PM Page 649 a I I Magnetic field b I I \u25a0 Figure 24-11 The magnetic field around a circular loop of currentcarrying wire can be modeled with the aid of the first right-hand rule (a). A current in a solenoid creates a magnetic field with the field from each coil adding to all the others (b). The first right-hand rule is a method you can use to determine the direction of a magnetic field relative to the direction of conventional current. Imagine holding a length of insulated wire with your right hand. Keep your thumb pointed in the direction of the conventional (positive) current. The fingers of your hand circle the wire and point in the direction of the magnetic field, as illustrated in Figure 24-10b. Magnetic field near a coil An electric current in a single circular loop of wire forms a magnetic field all around the loop. Applying the right-hand rule to any part of the wire loop, it can be shown that the direction of the field inside the loop is always the same. In Figure 24-11a, the field is always up, or out of the page. Outside the loop, it is always down, or into the page. When a wire is looped several times to form a coil and a current", " is allowed to flow through the coil, the field around all the loops is in the same direction, as shown in Figure 24-11b. A long coil of wire consisting of many loops is called a solenoid. The field from each loop in a solenoid adds to the fields of the other loops and creates a greater total field strength. When there is an electric current in a coil of wire, the coil has a field similar to a permanent magnet. When this current-carrying coil is brought close to a suspended bar magnet, one end of the coil repels the north pole of the magnet. Thus, the current-carrying coil has a north and a south pole and is itself a magnet. This type of magnet, which is created when current flows through a wire coil, is called an electromagnet. The strength of the field is proportional to the current in the coil. The magnetic field produced by each loop is the same. Because these fields are in the same direction, increasing the number of loops increases the strength of the magnetic field. The strength of an electromagnet also can be increased by placing an iron rod or core inside the coil. The core supports the magnetic field better than air does. It increases the magnetic field because the field of the solenoid creates a temporary magnetic field in the core, just as a nearby permanent magnet does when brought near a metal object. The second right-hand rule is a method you can use to determine the direction of the field produced by an electromagnet relative to the flow of conventional current. Imagine holding an insulated coil with your right hand. If you then curl your fingers around the loops in the direction of the conventional (positive) current, as in Figure 24-12, your thumb will point toward the north pole of the electromagnet. Electromagnets Cranes for moving iron and steel in industrial settings frequently use electromagnets. One such magnet, which operates at 230 V and draws 156 A, can lift over 11,300 kg! \u25a0 Figure 24-12 The second right-hand rule can be used to determine the polarity of an electromagnet. N I S I Section 24.1 Magnets: Permanent and Temporary 649 642-651 CH24-S1-845813 6/7/04 11:30 AM Page 650 5. A long, straight, current-carrying wire runs from north to south. a. A compass needle placed above the wire points with its north pole toward the east. In what direction is the current flowing? b. If a compass is put underneath the wire, in which direction will the compass needle point? 6. How does the strength of a magnetic field, 1 cm from a current- carrying wire, compare with each of the following? a. the strength of the field that is 2 cm from the wire b. the strength of the field that is 3 cm from the wire 7. A student makes a magnet by winding wire around a nail and connecting it to a battery, as shown in Figure 24-13. Which end of the nail, the pointed end or the head, will be the north pole? \u25a0 Figure 24-13 8. You have a spool of wire, a glass rod, an iron rod, and an aluminum rod. Which rod should you use to make an electromagnet to pick up steel objects? Explain. 9. The electromagnet in problem 8 works well, but you decide that you would like to make its strength adjustable by using a potentiometer as a variable resistor. Is this possible? Explain. A Microscopic Picture of Magnetic Materials Recall that when you put a piece of iron, nickel, or cobalt next to a magnet, the element also becomes magnetic, and it develops north and south poles. The magnetism, however, is only temporary. The creation of this temporary polarity depends on the direction of the external field. When you take away the external field, the element loses its magnetism. The three elements\u2014iron, nickel, and cobalt\u2014behave like electromagnets in many ways. They have a property called ferromagnetism. In the early nineteenth century, French scientist Andr\u00e9-Marie Amp\u00e8re knew that the magnetic effects of an electromagnet are the result of electric current through its loops. He proposed a theory of magnetism in iron to explain this behavior. Amp\u00e8re reasoned that the effects of a bar magnet must result from tiny loops of current within the bar. Magnetic domains Although the details of Amp\u00e8re\u2019s reasoning were wrong, his basic idea was correct. Each electron in an atom acts like a tiny electromagnet. When the magnetic fields of the electrons in a group of neighboring atoms are all aligned in the same direction, the group is called a domain. Although they may contain 1020 individual atoms, domains are still very small\u2014usually from 10 to 1000 microns. Thus, even a small sample of iron contains a huge number of domains. When a piece of iron is not in", " a magnetic field, the domains point in random directions, and their magnetic fields cancel one another out. If, however, a piece of iron is placed in a magnetic field, the domains tend to align with the external field, as shown in Figure 24-14. In the case of a temporary magnet, after the external field is removed, the domains return to their random arrangement. In a permanent magnet, the iron has been alloyed with other substances to keep the domains aligned after the external magnetic field is removed. 3-D Magnetic Fields Tie a string to the middle of a nail so that the nail will hang horizontally. Put a small piece of tape around the string where it wraps around the nail so that the string will not slip. Insert the nail into a coil and apply a voltage to the coil. Turn off the power and remove the nail from the coil. Now hold the string to suspend the nail. 1. Predict how the nail will behave in the presence of a permanent magnet. 2. Test your prediction. Analyze and Conclude 3. Explain what evidence you have that the nail became magnetized. 4. Make a 3-D drawing that shows the magnetic field around the magnet. 650 Chapter 24 Magnetic Fields 642-651 CH24-S1-845813 3/30/04 8:48 PM Page 651 Recording media Electromagnets make up the recording heads of audiocassette and videotape recorders. Recorders create electrical signals that represent the sounds or pictures being recorded. The electric signals produce currents in the recording head that create magnetic fields. When magnetic recording tape, which has many tiny bits of magnetic material bonded to thin plastic, passes over the recording head, the domains of the bits are aligned by the magnetic fields of the head. The directions of the domains\u2019 alignments depend on the direction of the current in the head and become a magnetic record of the sounds or pictures being recorded. The magnetic material on the tape allows the domains to keep their alignments until a strong enough magnetic field is applied to change them again. On a playback of the tape, the signal, produced by currents generated as the head passes over the magnetic particles, goes to an amplifier and a pair of loudspeakers or earphones. When a previously recorded tape is used to record new sounds, an erase head produces a rapidly alternating magnetic field that randomizes the directions of the domains on the tape. A magnetic history of the Earth Rocks containing iron have recorded the history of the varying directions of Earth\u2019s magnetic field. Rocks on the seafloor were produced when molten rock poured out of cracks in the bottom of the oceans. As they cooled, the rocks were magnetized in the direction of Earth\u2019s field at the time. As a result of seafloor spreading, the rocks farther from the cracks are older than those near the cracks. Scientists who first examined seafloor rocks were surprised to find that the direction of the magnetization in different rocks varied. They concluded from their data that the north and south magnetic poles of Earth have exchanged places many times in Earth\u2019s history. The origin of Earth\u2019s magnetic field is not well understood. How this field might reverse direction is even more of a mystery. a b \u25a0 Figure 24-14 A piece of iron (a) becomes a magnet only when its domains align (b). 24.1 Section Review 10. Magnetic Fields Is a magnetic field real, or is it just a means of scientific modeling? 11. Magnetic Forces Identify some magnetic forces around you. How could you demonstrate the effects of those forces? 12. Magnetic Fields A current-carrying wire is passed through a card on which iron filings are sprinkled. The filings show the magnetic field around the wire. A second wire is close to and parallel to the first wire. There is an identical current in the second wire. If the two currents are in the same direction, how will the first magnetic field be affected? How will it be affected if the two currents are in opposite directions? 13. Direction of a Magnetic Field Describe the righthand rule used to determine the direction of a magnetic field around a straight, current-carrying wire. 14. Electromagnets A glass sheet is placed over an active electromagnet, and iron filings sprinkled on the sheet create a pattern on it. If this experiment is repeated with the polarity of the power supply reversed, what observable differences will result? Explain. 15. Critical Thinking Imagine a toy containing two parallel, horizontal metal rods, one above the other. The top rod is free to move up and down. a. The top rod floats above the lower one. If the top rod\u2019s direction is reversed, however, it falls down onto the lower rod. Explain why the rods could behave in this way. b. Assume that the top rod was lost and replaced with another one. In this case, the top rod falls on top of the bottom rod no matter what its orientation is. What type of replacement rod must have been", " used? physicspp.com/self_check_quiz Section 24.1 Magnets: Permanent and Temporary 651 652-659 CH24-S2-845813 6/10/04 6:39 PM Page 652 24.2 Forces Caused by Magnetic Fields Objectives \u2022 Relate magnetic induction to the direction of the force on a current-carrying wire in a magnetic field. \u2022 Solve problems involving magnetic field strength and the forces on currentcarrying wires, and on moving, charged particles in magnetic fields. \u2022 Describe the design and operation of an electric motor. Vocabulary third right-hand rule galvanometer electric motor armature \u25a0 Figure 24-15 Current-carrying wires experience forces when they are placed in magnetic fields. In this case the force can be down (a), or up (b), depending on the direction of the current. a N F As you learned in the previous section, while Amp\u00e8re was studying the behaviors of magnets, he noted that an electric current produces a magnetic field similar to that of a permanent magnet. Because a magnetic field exerts forces on permanent magnets, Amp\u00e8re hypothesized that there is also a force on a current-carrying wire when it is placed in a magnetic field. Forces on Currents in Magnetic Fields The force on a wire in a magnetic field can be demonstrated using the arrangement shown in Figure 24-15. A battery produces current in a wire directly between two bar magnets. Recall that the direction of the magnetic field between two magnets is from the north pole of one magnet to the south pole of the other magnet. When there is a current in the wire, a force is exerted on the wire. Depending on the direction of the current, the force on the wire either pushes it down, as shown in Figure 24-15a, or pulls it up, as shown in Figure 24-15b. Michael Faraday discovered that the force on the wire is at right angles to both the direction of the magnetic field and the direction of the current. Determining the force\u2019s direction Faraday\u2019s description of the force on a current-carrying wire does not completely describe the direction because the force can be upward or downward. The direction of the force on a current-carrying wire in a magnetic field can be found by using the third right-hand rule. This technique is illustrated in Figure 24-16. The magnetic field is represented by the symbol B, and its direction is represented by a series of arrows. To use the third right-hand rule, point the fingers of your right hand in the direction of the magnetic field, and point your thumb in the direction of the conventional (positive) current in the wire. The palm of your hand will be facing in the direction of the force acting on the wire. When drawing a directional arrow that is into or out of the page, direction is indicated with crosses and dots, respectively. Think of the crosses as the tail feathers of the arrow, and the dots as the arrowhead. Soon after Oersted announced his discovery that the direction of the magnetic field in a wire is perpendicular to the flow of electric current in the wire, Amp\u00e8re was able to demonstrate the forces that current-carrying wires exert on each other. Figure 24-17a shows the direction of the magnetic field around each of the current-carrying wires, which is determined by the first right-hand rule. By applying the third right-hand rule to either wire, you can show why the wires attract each other. Figure 24-17b demonstrates the opposite situation. When currents are in opposite directions, the wires have a repulsive force between them. b F N I S SI 652 Chapter 24 Magnetic Fields 652-659 CH24-S2-845813 7/22/04 3:46 PM Page 653 F S N B I F B I \u25a0 Figure 24-16 The third right-hand rule can be used to determine the direction of force when the current and magnetic field are known. Force on a wire resulting from a magnetic field It is possible to determine the force of magnetism exerted on a current-carrying wire passing through a magnetic field at right angles to the wire. Experiments show that the magnitude of the force, F, on the wire, is proportional to the strength of the field, B, the current, I, in the wire, and the length, L, of the wire in the magnetic field. The relationship of these four factors is as follows: Force on a Current-Carrying Wire in a Magnetic Field F ILB The force on a current-carrying wire in a magnetic field is equal to the product of magnetic field strength, the current, and the length of the wire. The strength of a magnetic field, B, is measured in teslas, T. 1 T is equivalent to 1 N/Am. Note that if the wire is not perpendicular to the magnetic field, a factor of sin is introduced in the above equation, resulting in", " F ILB sin. As the wire becomes parallel to the magnetic field, the angle becomes zero, and the force is reduced to zero. When 90\u00b0, the equation is again F ILB. Loudspeakers One use of the force on a current-carrying wire in a magnetic field is in a loudspeaker. A loudspeaker changes electric energy to sound energy using a coil of fine wire mounted on a paper cone and placed in a magnetic field. The amplifier driving the loudspeaker sends a current through the coil. The current changes direction between 20 and 20,000 times each second, depending on the pitch of the tone it represents. A force exerted on the coil, because it is in a magnetic field, pushes the coil either into or out of the field, depending on the direction of the current. The motion of the coil causes the cone to vibrate, thereby creating sound waves in the air \u25a0 Figure 24-17 Two currentcarrying conductors are attracted when the currents are in the same direction (a), and are repelled when the currents are in opposite directions (b). F F Section 24.2 Forces Caused by Magnetic Fields 653 652-659 CH24-S2-845813 6/7/04 11:33 AM Page 654 Calculate the Strength of a Magnetic Field A straight wire carrying a 5.0-A current is in a uniform magnetic field oriented at right angles to the wire. When 0.10 m of the wire is in the field, the force on the wire is 0.20 N. What is the strength of the magnetic field, B? 1 Analyze and Sketch the Problem \u2022 Sketch the wire and show the direction of the current with an arrow, the magnetic field lines labeled B, and the force on the wire, F. \u2022 Determine the direction of the force using the third right-hand rule. The field, wire, and force are all at right angles. Unknown: B? Known: I 5.0 A L 0.10 m F 0.20 N 2 Solve for the Unknown B is uniform and because B and I are perpendicular to each other, F ILB. F ILB Solve for B. F B I L 0 2. 0 Substitute F 0.20 N, I 5.0 A, L 0.10 m ( 0. ) A (5.0 N 10 m) 0.40 N/Am 0.40 T B is 0.40 T from left to right and perpendicular to I and F. B I L F Math Handbook Operations with Significant Digits pages 835\u2014836 3 Evaluate the Answer \u2022 Are the units correct? The answer is in teslas, the correct unit for a magnetic field. \u2022 Is the magnitude realistic? The current and the length make the magnetic field fairly large, which is realistic. 16. What is the name of the rule used to predict the direction of force on a current-carrying wire at right angles to a magnetic field? Identify what must be known to use this rule. 17. A wire that is 0.50 m long and carrying a current of 8.0 A is at right angles to a 0.40-T magnetic field. How strong is the force that acts on the wire? 18. A wire that is 75 cm long, carrying a current of 6.0 A, is at right angles to a uniform magnetic field. The magnitude of the force acting on the wire is 0.60 N. What is the strength of the magnetic field? 19. A 40.0-cm-long copper wire carries a current of 6.0 A and weighs 0.35 N. A certain magnetic field is strong enough to balance the force of gravity on the wire. What is the strength of the magnetic field? 20. How much current will be required to produce a force of 0.38 N on a 10.0 cm length of wire at right angles to a 0.49-T field? 654 Chapter 24 Magnetic Fields 652-659 CH24-S2-845813 3/30/04 8:55 PM Page 655 a b N S F B Iout Iin Galvanometers The forces exerted on a loop of wire in a magnetic field can be used to measure current. If a small loop of current-carrying wire is placed in the strong magnetic field of a permanent magnet, as in Figure 24-18a, it is possible to measure very small currents. The current passing through the loop goes in one end of the loop and out the other end. Applying the third right-hand rule to each side of the loop, note that one side of the loop is forced down, while the other side of the loop is forced up. The resulting torque rotates the loop, and the magnitude of the torque acting on the loop is proportional to the magnitude of the current. This principle is used in a galvanometer. A galvanometer is a device used to measure very small currents, and therefore, it can", " be used as a voltmeter or an ammeter. A small spring in the galvanometer exerts a torque that opposes the torque that results from the flow of current through the wire loop; thus, the amount of rotation is proportional to the current. The meter is calibrated by finding out how much the coil turns when a known current is sent through it, as shown in Figure 24-18b. The galvanometer can then be used to measure unknown currents. Many galvanometers produce full-scale deflections with as little as 50 A (50106 A) of current. The resistance of the coil of wire in a sensitive galvanometer is about 1000. To measure larger currents, a galvanometer can be converted into an ammeter by placing a resistor with resistance smaller than the galvanometer in parallel with the meter, as shown in Figure 24-19a. Most of the current, Is, passes through the resistor, called the shunt, because the current is inversely proportional to resistance; whereas only a few microamps, Im, flow through the galvanometer. The resistance of the shunt is chosen according to the desired deflection scale. A galvanometer also can be connected as a voltmeter. To make a voltmeter, a resistor, called the multiplier, is placed in series with the meter, as shown in Figure 24-19b. The galvanometer measures the current through the multiplier. The current is represented by I V/R, where V is the voltage across the voltmeter and R is the effective resistance of the galvanometer and the multiplier resistor. Now suppose you want the needle of a voltmeter to move across the entire scale when 10 V is placed across it. The resistor is chosen so that at 10 V, the meter is deflected full-scale by the current through the meter and the resistor. Coil Spring N Soft iron core I S Magnetic torque \u25a0 Figure 24-18 If a wire loop is placed in a magnetic field when there is a current, the loop will rotate (a). The coil in a galvanometer rotates in proportion to the magnitude of the current (b). \u25a0 Figure 24-19 A galvanometer is connected for use as an ammeter (a), and a galvanometer is connected for use as a voltmeter (b). a Ammeter Is Rshunt Im G b Voltmeter R Rmultiplier I G Section 24.2 Forces Caused by Magnetic Fields 655 652-659 CH24-S2-845813 3/30/04 8:56 PM Page 656 \u25a0 Figure 24-20 In an electric motor, split-ring commutators allow the current in the wire loops to change direction and thus enable the loops in the motor to rotate 360\u00b0. N Brush S Commutator Electric motors You have seen how the simple loop of wire used in a galvanometer cannot rotate more than 180\u00b0. The forces push the right side of the loop up and the left side of the loop down until the loop reaches the vertical position. The loop will not continue to turn because the forces are still up and down, now parallel to the loop, and can cause no further rotation. How can you allow the loop to continue to rotate? The current through the loop must reverse direction just as the loop reaches its vertical position. This reversal allows the loop to continue rotating, as illustrated in Figure 24-20. To reverse current direction, an electric connection is made between contacts, called brushes, and a ring that is split into two halves, called a split-ring commutator. Brushes, which are usually pieces of graphite, make contact with the commutator and allow current to flow into the loop. As the loop rotates, so does the commutator. The split ring is arranged so that each half of the commutator changes brushes just as the loop reaches the vertical position. Changing brushes reverse the current in the loop. As a result, the direction of the force on each side of the loop is reversed, and the loop continues to rotate. This process repeats at each half-turn, causing the loop to spin in the magnetic field. The result is an electric motor, which is an apparatus that converts electric energy into rotational kinetic energy. Although only one loop is indicated in Figure 24-20, in an electric motor, the wire coil, called the armature, is made of many loops mounted on a shaft or axle. The total force acting on the armature is proportional to nILB, where n is the total number of turns on the armature, B is the strength of the magnetic field, I is the current, and L is the length of wire in each turn that moves through the magnetic field. The magnetic field is produced either by permanent magnets or by an electromagnet, called a field coil. The torque on the armature, and, as a result, the speed of the motor, is controlled by varying the current through the motor. The figure shows two identical motors with a common shaft. For simplicity, the commutators are not", " shown. Each armature coil consists of 48 turns of wire with rectangular dimensions of 17 cm wide by 35 cm deep. The armature resistance is 12. The red wire travels to the left (along half the width) and then back to the rear of the motor (along the depth). The magnetic field is 0.21 T. The diameter of the pulley is 7.2 cm. A rope fixed to the pulley and the floor prevents the motor shaft from turning. 1. Given F ILB, derive an equation for the torque on the armature for the position shown. 2. With S1 closed and S2 open, determine the torque on the shaft and the force on the spring scale. S Shaft N S2 N 35 V S1 S Pulley 3. With both switches closed, determine the torque on the shaft and the force on the spring scale. Spring scale 120 V 4. What happens to torque if the armature is in a different position? 656 Chapter 24 Magnetic Fields 652-659 CH24-S2-845813 6/7/04 11:35 AM Page 657 Cathodes Anodes Horizontal and vertical deflecting electromagnets Electron beams Glass screen Mask Coating of phosphor strips \u25a0 Figure 24-21 Pairs of magnets deflect the electron beam vertically and horizontally to form pictures for viewing. The Force on a Single Charged Particle Charged particles do not have to be confined to a wire. They also can move in a vacuum where the air particles have been removed to prevent collisions. A picture tube, also called a cathode-ray tube, in a computer monitor or television set uses electrons deflected by magnetic fields to form the pictures on the screen, as illustrated in Figure 24-21. Electric fields pull electrons off atoms in the negative electrode, or cathode. Other electric fields gather, accelerate, and focus the electrons into a narrow beam. Magnetic fields control the motion of the beam back-and-forth and up-and-down across the screen. The screen is coated with a phosphor that glows when it is struck by the electrons, thereby producing the picture. The force produced by a magnetic field on a single electron depends on the velocity of the electron, the strength of the field, and the angle between directions of the velocity and the field. Consider a single electron moving in a wire of length L. The electron is moving perpendicular to the magnetic field. The current, I, is equal to the charge per unit time entering the wire, I q/t. In this case, q is the charge of the electron and t is the time it takes to move the distance, L. The time required for a particle with speed v to travel distance L is found by using the equation of motion, d vt, or, in this case, t L/v. As a result, the equation for the current, I q/t, can be replaced by I qv/L. Therefore, the force on a single electron moving perpendicular to a magnetic field of strength B can be found. Force of a Magnetic Field on a Charged, Moving Particle F qvB The force on a particle moving in a magnetic field is equal to the product of the field strength, the charge of the particle, and its velocity. The particle\u2019s charge is measured in coulombs, C, its velocity in meters per second, m/s, and the strength of the magnetic field in teslas, T. The direction of the force is perpendicular to both the velocity of the particle and the magnetic field. The direction given by the third right-hand rule is for positively charged particles. For electrons, the force is in the opposite direction. Section 24.2 Forces Caused by Magnetic Fields 657 652-659 CH24-S2-845813 6/7/04 11:37 AM Page 658 Force on a Charged Particle in a Magnetic Field A beam of electrons travels at 3.0106 m/s through a uniform magnetic field of 4.0102 T at right angles to the field. How strong is the force acting on each electron? F v B 1 Analyze and Sketch the Problem \u2022 Draw the beam of electrons and its direction of motion; the magnetic field of lines, labeled B; and the force on the electron beam, F. Remember that the force is opposite the force given by the third right-hand rule because of the electron\u2019s negative charge. Known: v 3.0106 m/s B 4.0102 T q 1.601019 C Unknown: F? 2 Solve for the Unknown F qvB (1.601019 C)(3.0106 m/s)(4.0102 T) 1.91014 N Substitute q 1.601019 C, v 3.0106 m/s, B 4.0102 T 3 Evaluate the Answer \u2022 Are the units correct? T N/(Am), and A C/s", "; so T Ns/(Cm). Thus, (TCm)/s N, the unit for force. \u2022 Does the direction make sense? Use the third right-hand rule to verify that the directions of the forces are correct, recalling that the force on the electron is opposite the force given by the third right-hand rule. \u2022 Is the magnitude realistic? Forces on electrons and protons are always small fractions of a newton. Math Handbook Operations with Scientific Notation pages 842\u2014843 21. In what direction does the thumb point when using the third right- hand rule for an electron moving at right angles to a magnetic field? 22. An electron passes through a magnetic field at right angles to the field at a velocity of 4.0106 m/s. The strength of the magnetic field is 0.50 T. What is the magnitude of the force acting on the electron? 23. A stream of doubly ionized particles (missing two electrons, and thus, carrying a net charge of two elementary charges) moves at a velocity of 3.0104 m/s perpendicular to a magnetic field of 9.0102 T. What is the magnitude of the force acting on each ion? 24. Triply ionized particles in a beam carry a net positive charge of three elementary charge units. The beam enters a magnetic field of 4.0102 T. The particles have a speed of 9.0106 m/s. What is the magnitude of the force acting on each particle? 25. Doubly ionized helium atoms (alpha particles) are traveling at right angles to a magnetic field at a speed of 4.0104 m/s. The field strength is 5.0102 T. What force acts on each particle? 658 Chapter 24 Magnetic Fields 652-659 CH24-S2-845813 7/14/04 1:54 PM Page 659 Copper coil wire Read/write head \u25a0 Figure 24-22 Information is written to a computer disk by changing the magnetic field in a read/write head as the media passes beneath it. This causes magnetic particles in the media to align themselves in a pattern that represents the stored information. Disk surface Bit(0) Bit(1) Bit(0) Storing Information with Magnetic Media Data and software commands for computers are processed digitally in bits. Each bit is identified as either a 0 or a 1. How are these bits stored? The surface of a computer storage disk is covered with an even distribution of magnetic particles within a film. The direction of the particles\u2019 domains changes in response to a magnetic field. During recording onto the disk, current is routed to the disk drive\u2019s read/write head, which is an electromagnet composed of a wire-wrapped iron core. The current through the wire induces a magnetic field in the core. When the read/write head passes over the spinning storage disk, as in Figure 24-22, the domains of atoms in the magnetic film line up in bands. The orientation of the domains depends on the direction of the current. Two bands code for one bit of information. Two bands magnetized with the poles oriented in the same direction represent 0. Two bands represent 1 with poles oriented in opposite directions. The recording current always reverses when the read/write head begins recording the next data bit. To retrieve data, no current is sent to the read/write head. Rather, the magnetized bands in the disk induce current in the coil as the disk spins beneath the head. Changes in the direction of the induced current are sensed by the computer and interpreted as 0\u2019s and 1\u2019s. 24.2 Section Review 26. Magnetic Forces Imagine that a current-carrying wire is perpendicular to Earth\u2019s magnetic field and runs east-west. If the current is east, in which direction is the force on the wire? 29. Motors When the plane of the coil in a motor is perpendicular to the magnetic field, the forces do not exert a torque on the coil. Does this mean that the coil does not rotate? Explain. 27. Deflection A beam of electrons in a cathode-ray tube approaches the deflecting magnets. The north pole is at the top of the tube; the south pole is on the bottom. If you are looking at the tube from the direction of the phosphor screen, in which direction are the electrons deflected? 28. Galvanometers Compare the diagram of a galvanometer in Figure 24-18 on page 655 with the electric motor in Figure 24-20 on page 656. How is the galvanometer similar to an electric motor? How are they different? 30. Resistance A galvanometer requires 180 A for full-scale deflection. What is the total resistance of the meter and the multiplier resistor for a 5.0-V full-scale deflection? 31. Critical Thinking How do you know that the forces on parallel current-carrying wires are a result of magnetic attraction between wires, and not a result of electrostatics?", " Hint: Consider what the charges are like when the force is attractive. Then consider what the forces are when three wires carry currents in the same direction. physicspp.com/self_check_quiz Section 24.2 Forces Caused by Magnetic Fields 659 660-661 CH24-LAB-845813 6/7/04 11:44 AM Page 660 Creating an Electromagnet An electromagnet uses the magnetic field generated by a current to magnetize a piece of metal. In this activity, you will construct an electromagnet and test one variable that you think might affect the strength of it. QUESTION What is one variable that determines the strength of an electromagnet? Objectives \u25a0 Hypothesize which variables might affect the strength of an electromagnet. \u25a0 Observe the effects on an electromagnet\u2019s strength. \u25a0 Collect and organize data comparing the chosen variable and magnet strength. \u25a0 Make and use graphs to help identify a relationship between a controlling variable and a responding variable. \u25a0 Analyze and conclude what the effect is of the chosen variable on magnet strength. Possible Materials large paper clips small paper clips steel BBs wire steel nail 6-V lantern batteries 9-V batteries DC power source Procedure Safety Precautions 660 Horizons Companies 1. List the materials you will use to make your electromagnet. 2. List all the possible variables you think could affect the strength of an electromagnet. 3. Choose the one variable you will vary to deter- mine whether it does, in fact, affect the strength of an electromagnet. 4. Determine a method to detect the strength of the magnetic field produced by the electromagnet. 5. Have your teacher approve your lists before continuing. 6. Write a brief procedure for your experiment. Be sure to include all the values for the variables you will be keeping constant. 7. Create a data table like the one on the following page that displays the two quantities you will measure. 8. Build your electromagnet by using a nail and a length of wire. Wrap the wire around the nail. Be sure to leave several inches from both ends of the wire sticking out from your coil to allow attachment to the power source. CAUTION: The end of the nail or wire may be sharp. Exercise care when handling these materials to avoid being cut or scraped. 660-661 CH24-LAB-845813 6/7/04 11:47 AM Page 661 Data Table Number of ________________ Number of ________________ 9. Have your teacher inspect your magnet before 2. If you wanted to increase magnet strength, continuing. 10. Perform your experiment and record your data. CAUTION: If you are using BBs in your experiment, avoid possible injury by immediately picking up any BBs that should happen to fall to the floor. which method seems the most cost effective? Explain. 3. If you need to easily vary the strength of an electromagnet, how would you suggest that be done? Analyze 1. Make and Use Graphs Create a graph showing the relationship between your two variables. 2. What were the variables that you attempted to control in this experiment? Were there any you were unable to control? 3. If you evaluated the strength of the electromagnet by the amount of material it could pick up, how did you try to control any error from the magnet attracting only whole numbers of objects? Conclude and Apply 1. What is the relationship between your chosen variable and the strength of a magnet? 2. What variables did other students in your class find that also affected the strength of an electromagnet? 3. Were there any variables, by any group, that were found not to affect the strength of the electromagnet? Going Further 1. Compare the various variables students found that affected magnet strength. Did any of the variables appear to greatly increase strength without much change in the independent variable? If so, which ones? Real-World Physics 1. If you needed to create a stronger electromagnet for use in a small space, such as inside a laptop computer, what method would you use to increase the electromagnetic strength, given the size constraints? 2. Some buildings have electromagnets to hold fire doors open when the building is occupied. These magnets are mounted to the wall, like a door stop, behind the door. Thinking about the actions a fire alarm system would need to perform to control a fire, what is the advantage of using a system like this to hold the doors? How might a system like this be an advantage, or a disadvantage, in the event of a natural disaster? 3. Some electric bells work by having an arm strike the side of a metal dome-shaped bell. How might an electromagnet be used to make this bell work? How might the bell be wired to allow the arm to strike repeatedly (continual ringing) until the power supply is removed? To find out more about magnetic fields, visit the Web site: physicspp.com 661 662 CH24-FEATURE-8", "45813 6/7/04 11:48 AM Page 662 The Hall Effect Something as simple as magnetic fields deflecting charged particles has led to a revolution in how we measure or detect the movement of things, such as bicycle wheels and automotive crankshafts. It all starts when current passes through a wide, flat conductor, in the presence of a magnetic field. A Useful Sensor Engineers have developed the Hall-effect sensor. These tiny black plastic devices contain a thin film of silicon with wires connected, as shown in the diagram. The Hall voltage wires are connected to a tiny amplifier so that other instruments can detect it. If a permanent magnet is moved near a Hall-effect sensor, the voltage from the amplifier will increase. Thus, the sensor can be used to detect the proximity of the magnet. Magnetic field Thin silicon ribbon Hall voltage Hall sensor on bracket Wires to speedometer Bike fork Current path diverted toward edge of ribbon Voltage source Magnet A magnetic field forces more electrons to the edge of a thin metal strip. This creates the Hall voltage. Bicycle speedometers use a Hall-effect sensor to display the speed at which a bicycle is moving. The magnetic lines of force are perpendicular to the ribbon\u2019s broad surface. This makes the flowing electrons crowd into one side of the ribbon. Because there are more electrons on one edge of the ribbon than on the other, a voltage, called the Hall voltage, is generated across the width of the ribbon. The magnitude of the Hall voltage is dependent upon the strength of the magnetic field. E.H. Hall discovered this effect in 1879. Its industrial and scientific significance were discovered only recently because the Hall voltage is small in ribbons of conventional metals. Now, very thin layers of semiconducting silicon yield substantial Hall voltages. The Hall effect can be used to explore conduction in different types of materials. The sign of the Hall-effect voltage gives the sign of the moving charge, and the magnitude of the voltage tells us about the density and velocity of the charge. Such experiments have shown that in copper and most other metals, electrons carry the charge, but in zinc it is the positive charges that move. 662 Extreme Physics Everyday Applications Bicycle speedometers use a permanent magnet attached to the front wheel. Each revolution of the wheel brings the magnet close to a Hall-effect sensor. The resulting pulses are counted and timed. Halleffect sensors also are used to time the spark in automobile engines. When a magnet mounted on the crankshaft or distributor rotor moves near a sensor, a voltage pulse is produced, and the ignition system instantly fires the spark plug. Going Further 1. Analyze Why are the Hall-voltage electrodes positioned directly across from each other? What if they weren\u2019t? 2. Critical Thinking Might a strong magnetic field applied across a conducting ribbon change the resistance of that ribbon as a result of the Hall effect? Consider what you learned about the cross-sectional areas of wires. 663-669 CH24-SG CA STP-845813 3/30/04 9:03 PM Page 663 24.1 Magnets: Permanent and Temporary Vocabulary \u2022 polarized (p. 644) \u2022 magnetic field (p. 645) \u2022 magnetic flux (p. 646) \u2022 first right-hand rule (p. 649) \u2022 solenoid (p. 649) \u2022 electromagnet (p. 649) \u2022 second right-hand rule (p. 649) \u2022 domain (p. 650) Key Concepts \u2022 Like magnetic poles repel; unlike magnetic poles attract. \u2022 Magnetic fields exit from the north pole of a magnet and enter its south pole. \u2022 Magnetic field lines always form closed loops. \u2022 A magnetic field exists around any carrying-current wire. \u2022 A coil of wire carrying a current has a magnetic field. The field about the coil is like the field about a permanent magnet. 24.2 Forces Caused by Magnetic Fields Vocabulary \u2022 third right-hand rule (p. 652) \u2022 galvanometer (p. 655) \u2022 electric motor (p. 656) \u2022 armature (p. 656) Key Concepts \u2022 The strength of a magnetic field is measured in teslas. \u2022 When a current-carrying wire is placed in a magnetic field, there exists a force on the wire, perpendicular to both the field and the wire. \u2022 The force on a current-carrying wire in a magnetic field is proportional to the field strength, the current flow, and the length of the wire. F ILB \u2022 A galvanometer consists of a loop of wire in a magnetic field, and is used to measure small currents. When current is passed through the loop, a force on the wire loop results in a deflection of the loop. \u2022 A galvanometer can be used as an ammeter by adding a shunt resistor in parallel with the galvanometer. \u2022 A galvanometer can be used as a voltmeter by adding a multiplier resistor in series with the galvanometer. \u2022 A loudspeaker functions by", " varying the current through a coil that is placed in a magnetic field. The coil is attached to a paper cone that moves when the coil moves. As the current varies, the cone vibrates, thereby producing sound. \u2022 An electric motor consists of a coil of wire placed in a magnetic field. When there is a current in the coil, the coil rotates as a result of the force on the wire in the magnetic field. Complete 360\u00b0 rotation is achieved by using a commutator to switch the direction of the current in the coil as the coil rotates. \u2022 The force that a magnetic field exerts on a charged particle depends on three factors: the velocity of the particle, the charge of the particle, and the strength of the field. The direction of the force is perpendicular to both the field and the particle\u2019s velocity. F qvB \u2022 Computer monitors and television screens function by using magnets to focus and direct particles on phosphor screens. When particles strike the screen, light is emitted, and produces images on the screen. physicspp.com/vocabulary_puzzlemaker 663 663-669 CH24-SG CA STP-845813 6/7/04 12:21 PM Page 664 Concept Mapping 32. Complete the following concept map using the following: right-hand rule, F qvB, and F ILB. Force resulting from a Magnetic Field exerted on a current-carrying wire moving charge has a has a has a magnitude direction magnitude given by given by given by Mastering Concepts 33. State the rule for magnetic attraction and repulsion. (24.1) 34. Describe how a temporary magnet differs from a permanent magnet. (24.1) 35. Name the three most important common magnetic elements. (24.1) 36. Draw a small bar magnet and show the magnetic field lines as they appear around the magnet. Use arrows to show the direction of the field lines. (24.1) 37. Draw the magnetic field between two like magnetic poles and then between two unlike magnetic poles. Show the directions of the fields. (24.1) 38. If you broke a magnet in two, would you have isolated north and south poles? Explain. (24.1) 39. Describe how to use the first right-hand rule to determine the direction of a magnetic field around a straight current-carrying wire. (24.1) 40. If a current-carrying wire is bent into a loop, why is the magnetic field inside the loop stronger than the magnetic field outside? (24.1) 41. Describe how to use the second right-hand rule to determine the polarity of an electromagnet. (24.1) 42. Each electron in a piece of iron is like a tiny magnet. The iron, however, may not be a magnet. Explain. (24.1) 43. Why will dropping or heating a magnet weaken it? (24.1) 44. Describe how to use the third right-hand rule to determine the direction of force on a currentcarrying wire placed in a magnetic field. (24.2) 45. A strong current suddenly is switched on in a wire. No force acts on the wire, however. Can you conclude that there is no magnetic field at the location of the wire? Explain. (24.2) 46. What kind of meter is created when a shunt is added to a galvanometer? (24.2) Applying Concepts 47. A small bar magnet is hidden in a fixed position inside a tennis ball. Describe an experiment that you could do to find the location of the north pole and the south pole of the magnet. 48. A piece of metal is attracted to one pole of a large magnet. Describe how you could tell whether the metal is a temporary magnet or a permanent magnet. 49. Is the magnetic force that Earth exerts on a compass needle less than, equal to, or greater than the force that the compass needle exerts on Earth? Explain. 50. Compass Suppose you are lost in the woods but have a compass with you. Unfortunately, the red paint marking the north pole of the compass needle has worn off. You have a flashlight with a battery and a length of wire. How could you identify the north pole of the compass? 51. A magnet can attract a piece of iron that is not a permanent magnet. A charged rubber rod can attract an uncharged insulator. Describe the different microscopic processes producing these similar phenomena. 52. A current-carrying wire runs across a laboratory bench. Describe at least two ways in which you could find the direction of the current. 53. In which direction, in relation to a magnetic field, would you run a current-carrying wire so that the force on it, resulting from the field, is minimized, or even made to be zero? 664 Chapter 24 Magnetic Fields For more problems, go to Additional Problems, Appendix B. 663-669", " CH24-SG CA STP-845813 6/7/04 12:23 PM Page 665 54. Two wires carry equal currents and run parallel to each other. a. If the two currents are in opposite directions, where will the magnetic field from the two wires be larger than the field from either wire alone? b. Where will the magnetic field from both be exactly twice as large as from one wire? c. If the two currents are in the same direction, where will the magnetic field be exactly zero? 55. How is the range of a voltmeter changed when the resistor\u2019s resistance is increased? 56. A magnetic field can exert a force on a charged particle. Can the field change the particle\u2019s kinetic energy? Explain. 57. A beam of protons is moving from the back to the front of a room. It is deflected upward by a magnetic field. What is the direction of the field causing the deflection? 58. Earth\u2019s magnetic field lines are shown in Figure 24-23. At what location, poles or equator, is the magnetic field strength greatest? Explain. Magnetic pole North pole S N S N 61. Refer to Figure 24-26 to answer the following \u25a0 Figure 24-25 questions. a. Where are the poles? b. Where is the north pole? c. Where is the south pole? 1 2 3 4 \u25a0 Figure 24-26 62. Figure 24-27 shows the response of a compass in two different positions near a magnet. Where is the south pole of the magnet located? S N South pole Magnetic pole \u25a0 Figure 24-23 Mastering Problems 24.1 Magnets: Permanent and Temporary 59. As the magnet below in Figure 24-24 moves toward the suspended magnet, what will the magnet suspended by the string do? \u25a0 Figure 24-27 63. A wire that is 1.50 m long and carrying a current of 10.0 A is at right angles to a uniform magnetic field. The force acting on the wire is 0.60 N. What is the strength of the magnetic field? 64. A conventional current flows through a wire, as shown in Figure 24-28. Copy the wire segment and sketch the magnetic field that the current generates. I S N N S \u25a0 Figure 24-28 \u25a0 Figure 24-24 60. As the magnet in Figure 24-25 moves toward the suspended magnet, what will the magnet that is suspended by the string do? 65. The current is coming straight out of the page in Figure 24-29. Copy the figure and sketch the magnetic field that the current generates. \u25a0 Figure 24-29 physicspp.com/chapter_test Chapter 24 Assessment 665 663-669 CH24-SG CA STP-845813 6/7/04 12:41 PM Page 666 66. Figure 24-30 shows the end view of an 69. What is the resistor shown in Figure 24-31 called? electromagnet with current flowing through it. a. What is the direction of the magnetic field inside the loops? b. What is the direction of the magnetic field outside the loops? 70. The arrangement shown in Figure 24-32 is used to convert a galvanometer to what type of device? G I \u25a0 Figure 24-30 67. Ceramic Magnets The repulsive force between two ceramic magnets was measured and found to depend on distance, as given in Table 24-1. a. Plot the force as a function of distance. b. Does this force follow an inverse square law? Table 24-1 Separation, d (cm) Force, F (N) 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.93 0.40 0.13 0.057 0.030 0.018 0.011 0.0076 0.0053 0.0038 0.0028 24.2 Forces Caused by Magnetic Fields 68. The arrangement shown in Figure 24-31 is used to convert a galvanometer to what type of device? G \u25a0 Figure 24-31 \u25a0 Figure 24-32 71. What is the resistor shown in Figure 24-32 called? 72. A current-carrying wire is placed between the poles of a magnet, as shown in Figure 24-33. What is the direction of the force on the wire? I S S N N \u25a0 Figure 24-33 73. A wire that is 0.50 m long and carrying a current of 8.0 A is at right angles to a uniform magnetic field. The force on the wire is 0.40 N. What is the strength of the magnetic field? 74. The current through a wire that is 0.80 m long is 5.0 A. The wire is perpendicular to a 0.60-T magnetic field. What is the magnitude of the force on the wire? 75. A wire that is 25 cm long is at right angles to a 0", ".30-T uniform magnetic field. The current through the wire is 6.0 A. What is the magnitude of the force on the wire? 76. A wire that is 35 cm long is parallel to a 0.53-T uniform magnetic field. The current through the wire is 4.5 A. What force acts on the wire? 77. A wire that is 625 m long is perpendicular to a 0.40-T magnetic field. A 1.8-N force acts on the wire. What current is in the wire? 78. The force on a 0.80-m wire that is perpendicular to Earth\u2019s magnetic field is 0.12 N. What is the current in the wire? Use 5.0105T for Earth\u2019s magnetic field. 79. The force acting on a wire that is at right angles to a 0.80-T magnetic field is 3.6 N. The current in the wire is 7.5 A. How long is the wire? 666 Chapter 24 Magnetic Fields For more problems, go to Additional Problems, Appendix B. 663-669 CH24-SG CA STP-845813 6/7/04 12:46 PM Page 667 80. A power line carries a 225-A current from east to west, parallel to the surface of Earth. a. What is the magnitude of the force resulting from Earth\u2019s magnetic field acting on each meter of the wire? Use BEarth=5.0105T. b. What is the direction of the force? c. In your judgment, would this force be important in designing towers to hold this power line? Explain. 81. Galvanometer A galvanometer deflects full-scale for a 50.0-A current. a. What must be the total resistance of the series resistor and the galvanometer to make a voltmeter with 10.0-V full-scale deflection? b. If the galvanometer has a resistance of 1.0 k, what should be the resistance of the series (multiplier) resistor? 82. The galvanometer in problem 81 is used to make an ammeter that deflects full-scale for 10 mA. a. What is the potential difference across the galvanometer (1.0 k resistance) when a current of 50 A passes through it? b. What is the equivalent resistance of parallel resistors having the potential difference calculated in a circuit with a total current of 10 mA? c. What resistor should be placed parallel with the galvanometer to make the resistance calculated in part b? 83. A beam of electrons moves at right angles to a magnetic field of 6.0102 T. The electrons have a velocity of 2.5106 m/s. What is the magnitude of the force on each electron? 84. Subatomic Particle A muon (a particle with the same charge as an electron) is traveling at 4.21107 m/s at right angles to a magnetic field. The muon experiences a force of 5.001012 N. a. How strong is the magnetic field? b. What acceleration does the muon experience if its mass is 1.881028 kg? 85. A singly ionized particle experiences a force of 4.11013 N when it travels at right angles through a 0.61-T magnetic field. What is the velocity of the particle? 86. A room contains a strong, uniform magnetic field. A loop of fine wire in the room has current flowing through it. Assume that you rotate the loop until there is no tendency for it to rotate as a result of the field. What is the direction of the magnetic field relative to the plane of the coil? 87. A force of 5.781016 N acts on an unknown particle traveling at a 90\u00b0 angle through a magnetic field. If the velocity of the particle is 5.65104 m/s and the field is 3.20102 T, how many elementary charges does the particle carry? Mixed Review 88. A copper wire of insignificant resistance is placed in the center of an air gap between two magnetic poles, as shown in Figure 24-34. The field is confined to the gap and has a strength of 1.9 T. a. Determine the force on the wire (direction and magnitude) when the switch is open. b. Determine the force on the wire (direction and magnitude) when the switch is closed. c. Determine the force on the wire (direction and magnitude) when the switch is closed and the battery is reversed. d. Determine the force on the wire (direction and magnitude) when the switch is closed and the wire is replaced with a different piece having a resistance of 5.5. 7.5 cm N S 5.5 \u2126 24 V \u25a0 Figure 24-34 89. Two galvanometers are available. One has 50.0-A full-scale sensitivity and the other has 500.0-A full-scale sensitivity. Both have", " the same coil resistance of 855. Your challenge is to convert them to measure a current of 100.0 mA, full-scale. a. Determine the shunt resistor for the 50.0-A meter. b. Determine the shunt resistor for the 500.0-A meter. c. Determine which of the two is better for actual use. Explain. 90. Subatomic Particle A beta particle (high-speed electron) is traveling at right angles to a 0.60-T magnetic field. It has a speed of 2.5107 m/s. What size force acts on the particle? 91. The mass of an electron is 9.111031 kg. What is the magnitude of the acceleration of the beta particle described in problem 90? physicspp.com/chapter_test Chapter 24 Assessment 667 663-669 CH24-SG CA STP-845813 3/30/04 9:06 PM Page 668 92. A magnetic field of 16 T acts in a direction due west. An electron is traveling due south at 8.1105 m/s. What are the magnitude and the direction of the force acting on the electron? 93. Loudspeaker The magnetic field in a loudspeaker is 0.15 T. The wire consists of 250 turns wound on a 2.5-cm-diameter cylindrical form. The resistance of the wire is 8.0. Find the force exerted on the wire when 15 V is placed across the wire. 94. A wire carrying 15 A of current has a length of 25 cm in a magnetic field of 0.85 T. The force on a current-carrying wire in a uniform magnetic field can be found using the equation F ILB sin. Find the force on the wire when it makes the following angles with the magnetic field lines of c. 0\u00b0 b. 45\u00b0 a. 90\u00b0 95. An electron is accelerated from rest through a potential difference of 20,000 V, which exists between plates P1 and P2, shown in Figure 24-35. The electron then passes through a small opening into a magnetic field of uniform field strength, B. As indicated, the magnetic field is directed into the page. a. State the direction of the electric field between the plates as either P1 to P2 or P2 to P1. b. In terms of the information given, calculate the electron\u2019s speed at plate P2. c. Describe the motion of the electron through the magnetic field. P1 P2 Electron \u25a0 Figure 24-35 Thinking Critically 96. Apply Concepts A current is sent through a vertical spring, as shown in Figure 24-36. The end of the spring is in a cup filled with mercury. What will happen? Why? Spring Mercury \u25a0 Figure 24-36 97. Apply Concepts The magnetic field produced by a long, current-carrying wire is represented by B (2107 Tm/A)(I/d), where B is the field strength in teslas, I is the current in amps, and d is the distance from the wire in meters. Use this equation to estimate some magnetic fields that you encounter in everyday life. a. The wiring in your home seldom carries more than 10 A. How does the magnetic field that is 0.5 m from such a wire compare to Earth\u2019s magnetic field? b. High-voltage power transmission lines often carry 200 A at voltages as high as 765 kV. Estimate the magnetic field on the ground under such a line, assuming that it is about 20 m high. How does this field compare with a magnetic field in your home? c. Some consumer groups have recommended that pregnant women not use electric blankets in case the magnetic fields cause health problems. Estimate the distance that a fetus might be from such a wire, clearly stating your assumptions. If such a blanket carries 1 A, find the magnetic field at the location of the fetus. Compare this with Earth\u2019s magnetic field. 98. Add Vectors In almost all cases described in problem 97, a second wire carries the same current in the opposite direction. Find the net magnetic field that is a distance of 0.10 m from each wire carrying 10 A. The wires are 0.01 m apart. Make a scale drawing of the situation. Calculate the magnitude of the field from each wire and use a right-hand rule to draw vectors showing the directions of the fields. Finally, find the vector sum of the two fields. State its magnitude and direction. Writing In Physics 99. Research superconducting magnets and write a one-page summary of proposed future uses for such magnets. Be sure to describe any hurdles that stand in the way of the practical application of these magnets. Cumulative Review 100. How much work is required to move a charge of 6.40103 C through a potential difference of 2500 V? (Chapter 21) 101. The current flow in a 120-V circuit increases", " from 1.3 A to 2.3 A. Calculate the change in power. (Chapter 22) 102. Determine the total resistance of three, 55- resistors connected in parallel and then seriesconnected to two 55- resistors connected in series. (Chapter 23) 668 Chapter 24 Magnetic Fields For more problems, go to Additional Problems, Appendix B. 663-669 CH24-SG CA STP-845813 7/14/04 4:41 PM Page 669 Multiple Choice 1. A straight wire carrying a current of 7.2 A has a field of 8.9103 T perpendicular to it. What length of wire in the field will experience a force of 2.1 N? 2.6103 m 3.1102 m 1.3101 m 3.3101 m 2. Assume that a 19-cm length of wire is carrying a current perpendicular to a 4.1-T magnetic field and experiences a force of 7.6 mN. What is the current in the wire? 3.4107 A 9.8103 A 1.0102 A 9.8 A 3. A 7.12-\u00b5C charge is moving at the speed of light in a magnetic field of 4.02 mT. What is the force on the charge? 8.59 N 2.90101 N 8.591012 N 1.001016 N 7.12 C 4. An electron is moving at 7.4105 m/s perpendicular to a magnetic field. It experiences a force of 18 N. What is the strength of the magnetic field? 6.51015 T 2.4105 T 1.3107 T 1.51014 T 6. Which statement about magnetic monopoles is false? A monopole is a hypothetical separate north pole. Research scientists use them for internal medical testing applications. A monopole is a hypothetical separate south pole. They don\u2019t exist. 7. A uniform magnetic field of 0.25 T points vertically downward. A proton enters the field with a horizontal velocity of 4.0106 m/s. What are the magnitude and direction of the instantaneous force exerted on the proton as it enters the magnetic field? 1.61013 N to the left 1.61013 N downward 1.0106 N upward 1.0106 N to the right Extended Answer 8. Derive the units of teslas in kilograms, meters, seconds, and coulombs using dimensional analysis and the formulas F qvB and F ILB. 9. A wire attached to a 5.8-V battery is in a circuit with 18 \u2126. 14 cm of the wire is in a magnetic field of 0.85 T and the force on the wire is 22 mN. What is the angle of the wire in the field given that the formula for angled wires in fields is F ILB sin \u03b8? 7.4105 m/s 18 N 5. Which factor will not affect the strength of a Read the Directions solenoid? number of wraps strength of current thickness of wire core type No matter how many times you\u2019ve taken a particular test or practiced for an exam, it\u2019s always a good idea to read through the directions provided at the beginning of each section. It only takes a moment and could prevent you from making a simple mistake throughout the test that could cause you to do poorly. physicspp.com/standardized_test Chapter 24 Standardized Test Practice 669 670-678 CH25-S1-845813 3/30/04 9:14 PM Page 670 What You\u2019ll Learn \u2022 You will describe how changing magnetic fields can generate electric potential differences. \u2022 You will apply this phenomenon to the construction of generators and transformers. Why It\u2019s Important The relationship between magnetic fields and current makes possible the three cornerstones of electrical technology: motors, generators, and transformers. Hydroelectric Generators Dams commonly are built on rivers to provide a source of power for nearby communities. Within the dam, the potential and kinetic energy of water is turned into electric energy. Think About This How do the generators located inside the dam convert the kinetic and potential energy of the water into electric energy? physicspp.com 670 Tom Pantages, (inset)Robert Cameron/Getty Images 670-678 CH25-S1-845813 3/30/04 9:16 PM Page 671 What happens in a changing magnetic field? Question How does a changing magnetic field affect a coil of wire passed through it? Procedure 1. Place two bar magnets about 8 cm apart. 2. Attach a sensitive galvanometer to either end of a piece of coiled copper wire. 3. Slowly move the wire between the magnets. Note the galvanometer reading. 4. Vary the angle of the movement of the copper wire and the velocity of the wire. Note your results. Analysis What causes the galvanometer to move? What situation makes the galvanometer deflect the most?", " Critical Thinking When the wire is moved between the magnets, what is happening to the wire? 25.1 Electric Current from Changing Magnetic Fields In Chapter 24, you learned how Hans Christian Oersted discovered that Objectives an electric current produces a magnetic field. Michael Faraday thought that the reverse must also be true: that a magnetic field produces an electric current. In 1822, Michael Faraday wrote a goal in his notebook: \u201cConvert magnetism into electricity.\u201d Faraday tried many combinations of magnetic fields and wires without success. After nearly ten years of unsuccessful experiments, Faraday found that he could induce electric current by moving a wire through a magnetic field. In the same year, Joseph Henry, an American high-school teacher, also showed that a changing magnetic field could produce electric current. Henry took an idea developed by another scientist and broadened the application to other educational demonstration devices to make them more sensitive or powerful. Henry\u2019s versions of these devices were not new discoveries, but he made the devices more dramatic and effective as educational aids. However, Henry, unlike Faraday, chose not to publish his discoveries. \u2022 Explain how a changing magnetic field produces an electric current. \u2022 Define electromotive force. \u2022 Solve problems involving wires moving in magnetic fields. Vocabulary electromagnetic induction fourth right-hand rule electromotive force electric generator average power Section 25.1 Electric Current from Changing Magnetic Fields 671 Horizons Companies 670-678 CH25-S1-845813 3/30/04 9:18 PM Page 672 \u25a0 Figure 25-1 When a wire is moved in a magnetic field, there is an electric current in the wire, but only while the wire is moving. The direction of the current depends on the direction in which the wire is moving through the field. The arrows indicate the direction of conventional current. N N Electromagnetic Induction Figure 25-1 shows one of Faraday\u2019s experiments, in which a wire loop that is part of a closed circuit is placed in a magnetic field. When the wire is held stationary or is moved parallel to the magnetic field, there is no current, but when the wire moves up through the field, the current is in one direction. When the wire moves down through the field, the current is in the opposite direction. An electric current is generated in a wire only when the wire cuts magnetic field lines. Faraday found that to generate current, either the conductor can move through a magnetic field or a magnetic field can move past the conductor. It is the relative motion between the wire and the magnetic field that produces the current. The process of generating a current through a circuit in this way is called electromagnetic induction. How can you tell the direction of the current? To find the force on the charges in the wire, use the fourth right-hand rule to hold your right hand so that your thumb points in the direction in which the wire is moving and your fingers point in the direction of the magnetic field. The palm of your hand will point in the direction of the conventional (positive) current, as illustrated in Figure 25-2. \u25a0 Figure 25-2 The fourth right-hand rule can be used to find the direction of the forces on the charges in a conductor that is moving in a magnetic field. F B v F B N S v Motion of the wire I 672 Chapter 25 Electromagnetic Induction 670-678 CH25-S1-845813 3/30/04 9:19 PM Page 673 Electromotive Force When you studied electric circuits, you learned that a source of electrical energy, such as a battery, is needed to produce a continuous current. The potential difference, or voltage, given to the charges by a battery is called the electromotive force, or EMF. Electromotive force, however, is not actually a force; instead, it is a potential difference and is measured in volts. Thus, the term electromotive force is misleading. Like many other historical terms still in use, it originated before the related principles\u2014in this case, those of electricity\u2014were well understood. The EMF is the influence that makes current flow from lower to higher potential, like a water pump in a water fountain. What created the potential difference that caused an induced current in Faraday\u2019s experiment? When you move a wire through a magnetic field, you exert a force on the charges and they move in the direction of the force. Work is done on the charges. Their electrical potential energy, and thus their potential, is increased. The difference in potential is called the induced EMF. EMF depends on the magnetic field, B, the length of the wire in the magnetic field, L, and the velocity of the wire in the field that is perpendicular to the field, v(sin ). Electromotive Force EMF BLv(sin ) Electromotive force is equal to the magnitude of the magnetic field, times the length of the wire times the component of the velocity of the wire in the field perpendicular to", " the field. If a wire moves through a magnetic field at an angle to the field, only the component of the wire\u2019s velocity that is perpendicular to the direction of the magnetic field generates EMF. If the wire moves through the field with a velocity that is exactly perpendicular to the field, then the above equation reduces to EMF BLv, because sin 90\u00b0 1. Checking the units of the EMF equation will help you work algebra correctly in related problems. The unit for measuring EMF is the volt, V. In Chapter 24, B was defined as F/IL; therefore, the units for B are N/Am. Velocity is measured in m/s. Using dimensional analysis, (N/Am)(m)(m/s) Nm/As J/C V. Recall from previous chapters that J Nm, A C/s, and V J/C. Aluminum diaphragm \u25a0 Figure 25-3 In this drawing of a moving coil microphone, the aluminum diaphragm is connected to a coil in a magnetic field. When sound waves vibrate the diaphragm, the coil moves in the magnetic field and generates a current that is proportional to the sound wave. Coil Connecting wires Application of induced EMF A microphone is a simple application that depends on an induced EMF. A dynamic microphone is similar in construction to a loudspeaker. The microphone shown in Figure 25-3 has a diaphragm attached to a coil of wire that is free to move in a magnetic field. Sound waves vibrate the diaphragm, which moves the coil in the magnetic field. The motion of the coil, in turn, induces an EMF across the ends of the coil. The induced EMF varies as the frequency of the sound varies. In this way, the sound wave is converted to an electrical signal. The voltage generated is small, typically 103 V, but it can be increased, or amplified, by electronic devices. S S N Magnet (also serves as the supporting frame) Section 25.1 Electric Current from Changing Magnetic Fields 673 670-678 CH25-S1-845813 6/7/04 12:51 PM Page 674 Induced EMF A straight wire, 0.20 m long, moves at a constant speed of 7.0 m/s perpendicular to a magnetic field of strength 8.0102 T. a. What EMF is induced in the wire? b. The wire is part of a circuit that has a resistance of 0.50. What is the current c. 1 through the wire? If a different metal was used for the wire, which has a resistance of 0.78, what would the new current be? Analyze and Sketch the Problem \u2022 Establish a coordinate system. \u2022 Draw a straight wire of length L. Connect an ammeter to the wire to represent a measurement of current. y Bout of page z x L v A \u2022 Choose a direction for the magnetic field that is perpendicular to the length of wire. \u2022 Choose a direction for the velocity that is perpendicular to both the length of the wire and the magnetic field. Known: v 7.0 m/s L 0.20 m B 8.0102 T 0.50 R1 0.78 R2 Unknown: EMF? I? 2 Solve for the Unknown a. EMF BLv Substitute B 8.0102, L 0.20 m, v 7.0 m/s (8.0102 T )(0.20 m)(7.0 m/s) 0.11 Tm2/s 0.11 V V I R F EM Substitute V EMF R V Substitute EMF 0.11 V, R1 1 1. 0 0 5. 0 0.22 A Using the fourth right-hand rule, the direction of the current is counterclockwise. F I EM R V Substitute EMF 0.11 V, R2.14 A 0.50 0.78 b. c. Math Handbook Rounding page 835 3 The current is counterclockwise. Evaluate the Answer \u2022 Are the units correct? Volt is the correct unit for EMF. Current is measured in amperes. \u2022 Does the direction make sense? The direction obeys the fourth right-hand rule: v is the direction of the thumb, B is the same direction as the fingers, and F is the direction that the palm faces. Current is in the same direction as the force. \u2022 Is the magnitude realistic? The answers are near 101. This agrees with the quantities given and the algebra performed. 674 Chapter 25 Electromagnetic Induction 670-678 CH25-S1-845813 3/30/04 9:21 PM Page 675 1. A straight wire, 0.5 m long, is moved straight up at a speed of 20 m/s through a 0.4-T magnetic field pointed in the horizontal direction. a. What EMF is induced in the", " wire? b. The wire is part of a circuit of total resistance of 6.0. What is the current in the circuit? 2. A straight wire, 25 m long, is mounted on an airplane flying at 125 m/s. The wire moves in a perpendicular direction through Earth\u2019s magnetic field (B 5.0105 T). What EMF is induced in the wire? 3. A straight wire, 30.0 m long, moves at 2.0 m/s in a perpendicular direction through a 1.0-T magnetic field. a. What EMF is induced in the wire? b. The total resistance of the circuit of which the wire is a part is 15.0. What is the current? 4. A permanent horseshoe magnet is mounted so that the magnetic field lines are vertical. If a student passes a straight wire between the poles and pulls it toward herself, the current flow through the wire is from right to left. Which is the north pole of the magnet? Electric Generators The electric generator, invented by Michael Faraday, converts mechanical energy to electrical energy. An electric generator consists of a number of wire loops placed in a strong magnetic field. The wire is wound around an iron core to increase the strength of the magnetic field. The iron and wires are called the armature, which is similar to that of an electric motor. The armature is mounted so that it can rotate freely in the magnetic field. As the armature turns, the wire loops cut through the magnetic field lines and induce an EMF. Commonly called the voltage, the EMF developed by the generator depends on the length of wire rotating in the field. Increasing the number of loops in the armature increases the wire length, thereby increasing the induced EMF. Note that you could have a length of wire with only part of it in the magnetic field. Only the portion within the magnetic field induces an EMF. Current from a generator When a generator is connected in a closed circuit, the induced EMF produces an electric current. Figure 25-4 shows a single-loop generator without an iron core. The direction of the induced current can be found from the third right-hand rule. As the loop rotates, the strength and the direction of the current change. \u25a0 Figure 25-4 An electric current is generated in a wire loop as the loop rotates. S Up N Down Section 25.1 Electric Current from Changing Magnetic Fields 675 670-678 CH25-S1-845813 3/30/04 9:21 PM Page 676 a Top view b Top view c Side view I max 1 3 I max Side view 2 I 0 I 0 4 I max 1 I max I 0 2 I 0 4 Time t n e r r u C 3 I max \u25a0 Figure 25-5 The cross-sectional view of a rotating wire loop shows the position of the loop when maximum current is generated (a). When the loop is vertical, the current is zero (b). The current varies with time as the loop rotates (c). The variation of EMF with time can be shown with a similar graph. \u25a0 Figure 25-6 Only the segments bc and ad have current induced through them. This can be shown using the fourth right-hand rule. The current is greatest when the motion of the loop is perpendicular to the magnetic field; that is, when the loop is in the horizontal position, as shown in Figure 25-5a. In this position, the component of the loop\u2019s velocity perpendicular to the magnetic field is greatest. As the loop rotates from the horizontal to the vertical position, as shown in Figure 25-5b, it moves through the magnetic field lines at an ever-increasing angle. Thus, it cuts through fewer magnetic field lines per unit of time, and the current decreases. When the loop is in the vertical position, the wire segments move parallel to the field and the current is zero. As the loop continues to turn, the segment that was moving up begins to move down and reverses the direction of the current in the loop. This change in direction takes place each time the loop turns through 180\u00b0. The current changes smoothly from zero to some maximum value and back to zero during each half-turn of the loop. Then it reverses direction. A graph of current versus time is shown in Figure 25-5c. Does the entire loop contribute to the induced EMF? Look at Figure 25-6, where all four sides of the loop are depicted in the magnetic field. If the fourth right-hand rule is applied to segment ab, the direction of the induced current is toward the side of the wire. The same applies to segment cd. Thus, no current is induced along the length of the wire in ab or cd. But in segment bc, the direction of the induced current is from b to c, and in segment ad, the current is from d to a. Because the conducting loop is rotating in a circular motion, the relative angle", " between a point on the loop and the magnetic field constantly changes. The electromotive force can be calculated by the electromotive force equation given earlier, EMF BLv(sin ), except that L is now the length of segment bc. The maximum voltage is induced when a conductor is moving perpendicular to the magnetic field and thus 90\u00b0. I (induced) b v \u03b8 \u03b8 c Axis v B d a v I (induced) 676 Chapter 25 Electromagnetic Induction Generators, such as those in the chapter opening image, work in a similar fashion. Potential energy from water stored behind a dam is converted to kinetic energy, which spins the turbines. The turbines, in turn, turn coils of conductors in a magnetic field, thereby inducing an EMF. Generators and motors are almost identical in construction, but \u25a0 Figure 25-7 An alternatingcurrent generator transmits current to an external circuit by way of a brush-slip-ring arrangement (a). The alternating current produced varies with time (b). The resulting power is always positive and also is sinusoidal (c). 670-678 CH25-S1-845813 6/10/04 9:03 PM Page 677 N S a Slip rings Brushes b I max I eff 0 t n e r r u C \u2013I max c Pmax Pave r e w o P 0 Time Time they convert energy in opposite directions. A generator converts mechanical energy to electrical energy, while a motor converts electrical energy to mechanical energy. Alternating-Current Generators An energy source turns the armature of a generator in a magnetic field at a fixed number of revolutions per second. In the United States, electric utilities use a 60-Hz frequency, in which the current goes from one direction to the other and back to the first 60 times per second. Figure 25-7a shows how an alternating current, AC, in an armature is transmitted to the rest of the circuit. The brush-slip-ring arrangement permits the armature to turn freely while still allowing the current to pass into the external circuit. As the armature turns, the alternating current varies between some maximum value and zero, as shown in the graph in Figure 25-7b. Average power The power produced by a generator is the product of the current and the voltage. Because both current and voltage vary, the power associated with an alternating current varies. Figure 25-7c shows a graph of the power produced by an AC generator. Note that power is always positive because I and V are either both positive or both negative. Average power, PAC, is half the maximum power; thus, PAC 1 PAC max. 2 Effective voltage and current It is common to describe alternating current and voltage in terms of effective current and voltage, rather than referring to their maximum values. Recall from Chapter 22 that P I2R. Thus, you can express effective current, Ieff, in terms of the average AC 2R. To determine Ieff in terms of maximum current, Imax, power as PAC 1 PAC max, and substitute in I2R. start with the power relationship, PAC 2 Then solve for Ieff. Ieff Effective Current 2 Imax 2 2 times the maximum current. Effective current is equal to 2 0.707 Imax Ieff Section 25.1 Electric Current from Changing Magnetic Fields 677 670-678 CH25-S1-845813 3/30/04 9:22 PM Page 678 Similarly, the following equation can be used to express effective voltage. Effective Voltage Veff 2 Vmax 2 2 times the maximum voltage. Effective voltage is equal to 2 0.707 Vmax Effective voltage, also is commonly referred to as RMS (root mean square) voltage. In the United States, the voltage generally available at wall outlets is described as 120 V, where 120 V is the magnitude of the effective voltage, not the maximum voltage. The frequency and effective voltage that are used vary in different countries. 5. A generator develops a maximum voltage of 170 V. a. What is the effective voltage? b. A 60-W lightbulb is placed across the generator with an Imax of 0.70 A. What is the effective current through the bulb? c. What is the resistance of the lightbulb when it is working? 6. The RMS voltage of an AC household outlet is 117 V. What is the maximum voltage across a lamp connected to the outlet? If the RMS current through the lamp is 5.5 A, what is the maximum current in the lamp? 7. An AC generator delivers a peak voltage of 425 V. a. What is the Veff in a circuit placed across the generator? b. The resistance is 5.0102. What is the effective current? If the average power dissipated by an electric light is 75 W, what is the peak power? 8. In this section you have explored how moving wires in magnetic fields can induce current. However, as Faraday discovered, changing magnetic fields around a conductor", " also can induce current in the conductor. In the next section, you will explore changing magnetic fields and the applications of induction by changing magnetic fields. 25.1 Section Review 9. Generator Could you make a generator by mounting permanent magnets on a rotating shaft and keeping the coil stationary? Explain. 10. Bike Generator A bike generator lights the headlamp. What is the source of the energy for the bulb when the rider travels along a flat road? 11. Microphone Consider the microphone shown in Figure 25-3. When the diaphragm is pushed in, what is the direction of the current in the coil? 12. Frequency What changes to the generator are required to increase the frequency? 13. Output Voltage Explain why the output voltage of an electric generator increases when the magnetic field is made stronger. What else is affected by strengthening the magnetic field? 14. Generator Explain the fundamental operating principle of an electric generator. 15. Critical Thinking A student asks, \u201cWhy does AC dissipate any power? The energy going into the lamp when the current is positive is removed when the current is negative. The net is zero.\u201d Explain why this reasoning is wrong. 678 Chapter 25 Electromagnetic Induction physicspp.com/self_check_quiz 679-685 CH25-S2-845813 3/30/04 9:42 PM Page 679 25.2 Changing Magnetic Fields Induce EMF In a generator, current is produced when the armature turns through a magnetic field. The act of generating current produces a force on the wires in the armature. In what direction is the force on the wires of an armature? Lenz\u2019s Law Consider a section of one loop that moves through a magnetic field, as shown in Figure 25-8a. An EMF, equal to BLv, will be induced in the wire. If the magnetic field is out of the page and velocity is to the right, then the fourth right-hand rule shows a downward EMF, as illustrated in Figure 258b, and consequently a downward current is produced. In Chapter 24, you learned that a wire carrying a current through a magnetic field will experience a force acting on it. This force results from the interaction between the existing magnetic field and the magnetic field generated around all currents. To determine the direction of this force, use the third right-hand rule: if current, I, is down and the magnetic field, B, is out, then the resulting force is to the left, as shown in Figure 25-8c. This means that the direction of the force on the wire opposes the original motion of the wire, v. That is, the force acts to slow down the rotation of the armature. The method of determining the direction of a force was first demonstrated in 1834 by H.F.E. Lenz and is therefore called Lenz\u2019s law. Lenz\u2019s law states that the direction of the induced current is such that the magnetic field resulting from the induced current opposes the change in the field that caused the induced current. Note that it is the change in the field and not the field itself that is opposed by the induced magnetic effects. Opposing change Figure 25-9, on the next page, is an example of how Lenz\u2019s law works. The north pole of a magnet is moved toward the left end of a coil of wire. To create a force that will oppose the approach of the north pole, the left end of the coil also must become a north pole. In other words, the magnetic field lines must emerge from the left end of the coil. Using the second right-hand rule, which you learned in Chapter 24, you will see that if Lenz\u2019s law is correct, the induced current must be in a counterclockwise direction as viewed from the end of the coil where the magnet is inserted. Experiments have shown that this is so. If the magnet is turned so that a south pole approaches the coil, the induced current will flow in a clockwise direction. Objectives \u2022 Apply Lenz\u2019s law. \u2022 Explain back-EMF and how it affects the operation of motors and generators. \u2022 Explain self-inductance and how it affects circuits. \u2022 Solve transformer problems involving voltage, current, and turn ratios. Vocabulary Lenz\u2019s law eddy current self-inductance transformer primary coil secondary coil mutual inductance step-up transformer step-down transformer \u25a0 Figure 25-8 A wire, length L, moving through a magnetic field, B, induces an electromotive force. If the wire is part of a circuit, then there will be a current, I. This current will interact with the magnetic field and produce a force, F. Notice that the resulting force opposes the motion, v, of the wire. a I EMF = BLv b c Bout of page v L F = BIL I F (EMF) v F = BIL I Section 25.", "2 Changing Magnetic Fields Induce EMF 679 679-685 CH25-S2-845813 3/30/04 9:43 PM Page 680 \u25a0 Figure 25-9 The magnet approaching the coil causes an induced current to flow. Lenz\u2019s law predicts the direction of flow shown. S S N N Induced current If a generator produces only a small current, then the opposing force on the armature will be small, and the armature will be easy to turn. If the generator produces a larger current, the force on the larger current will be greater, and the armature will be more difficult to turn. A generator supplying a large current is producing a large amount of electric energy. The opposing force on the armature means that mechanical energy must be supplied to the generator to produce the electric energy, consistent with the law of conservation of energy. Motors and Lenz\u2019s law Lenz\u2019s law also applies to motors. When a current-carrying wire moves in a magnetic field, an EMF is generated. This EMF, called the back-EMF, is in a direction that opposes the current. When a motor is first turned on, there is a large current because of the low resistance of the motor. As the motor begins to turn, the motion of the wires across the magnetic field induces a back-EMF that opposes the current. Therefore, the net current through the motor is reduced. If a mechanical load is placed on the motor, as in a situation in which work is being done to lift a weight, the rotation of the motor will slow. This slowing down will decrease the back-EMF, which will allow more current through the motor. Note that this is consistent with the law of conservation of energy: if current increases, so does the rate at which electric power is being sent to the motor. This power is delivered in mechanical form to the load. If the mechanical load stops the motor, current can be so high that wires overheat. As current draw varies with the changing speed of an electric motor, the voltage drop across the resistance of the wires supplying the motor also varies. Another device, such as a lightbulb, that is in parallel with the motor, also would experience the drop in voltage. This is why you may have noticed some lights in a house dimming when a large motorized appliance, such as an air conditioner or a table saw, starts operating. When the current to the motor is interrupted by a switch in the circuit being turned off or by the motor\u2019s plug being pulled from a wall outlet, the sudden change in the magnetic field generates a back-EMF. This reverse voltage can be large enough to cause a spark across the switch or between the plug and the wall outlet. Application of Lenz\u2019s law A sensitive balance, such as the kind used in laboratories, uses Lenz\u2019s law to stop its oscillation when an object is placed on the pan. As shown in Figure 25-10, a piece of metal attached to the balance arm is located between the poles of a horseshoe magnet. \u25a0 Figure 25-10 Sensitive balances use eddy-current damping to control oscillations of the balance beam (a). As the metal plate on the end of the beam moves through the magnetic field, a current is generated in the metal. This current, in turn, produces a magnetic field that opposes the motion that caused it, and the motion of the beam is dampened (b). a b Eddy current 680 Matt Meadows Chapter 25 Electromagnetic Induction 679-685 CH25-S2-845813 3/30/04 9:44 PM Page 681 When the balance arm swings, the metal moves through the magnetic field. Currents called eddy currents are generated in the metal and produce a magnetic field that acts to oppose the motion that caused the currents. Thus, the metal piece is slowed down. The force opposes the motion of the metal in either direction but does not act if the metal is still. Thus, the force does not change the mass read by the balance. This effect is called eddy-current damping. A practical motor or transformer core is constructed from thin laminations, or layers, each one insulated from the other, to reduce the circulation of eddy currents. Eddy currents are generated when a piece of metal moves through a magnetic field. The reverse is also true: a current is generated when a metal loop is placed in a changing magnetic field. According to Lenz\u2019s law, the current generated will oppose the changing magnetic field. The current generates a magnetic field of its own in the opposite direction that causes the uncut, aluminum ring in Figure 25-11 to float. An AC current is in the coil, so a constantly changing magnetic field is generated. This changing magnetic field induces an EMF in the rings. If these rings were constructed from a nonconducting material such as nylon or brass, an EM", "F could not be induced. For the uncut ring, the EMF causes a current that produces a magnetic field that will oppose the change in the generating magnetic field. The interaction of these two magnetic fields causes the ring to push away from the coil, similar to the way in which the north poles of two magnets push away from each other. For the lower ring, which has been sawed through, an EMF is generated, but no current can result because of an incomplete path. Hence, no opposing magnetic field is produced by the ring. Self-Inductance Back-EMF can be explained in another way. As Faraday showed, EMF is induced whenever a wire cuts the lines of a magnetic field. The current through the wire shown in Figure 25-12 increases from Figure 25-12a to Figure 25-12c. The current generates a magnetic field, shown by magnetic field lines. As the current and magnetic field increase, new lines are created. As more lines are added, they cut through the coil wires and generate an EMF to oppose the current changes. The EMF will make the potential of the top of the coil more negative than the bottom. This induction of EMF in a wire carrying changing current is called self-inductance. \u25a0 Figure 25-11 Current is induced in the continuous metal ring, while there is no current in the cut ring. a I b I c I \u25a0 Figure 25-12 As the current in the coil increases from (a) on the left to (c) on the right, the magnetic field generated by the current also increases. This increase in the magnetic field produces an EMF that opposes the current direction. EMF induced Section 25.2 Changing Magnetic Fields Induce EMF 681 Horizons Companies 679-685 CH25-S2-845813 3/30/04 9:44 PM Page 682 The size of the induced EMF is proportional to the rate at which field lines cut through the wires. The faster the current is changed, the larger the opposing EMF. If the current reaches a steady value, the magnetic field is constant, and the EMF is zero. When the current is decreased, an EMF is generated that tends to prevent the reduction in the magnetic field and current. Because of self-inductance, work has to be done to increase the current flowing through the coil. Energy is stored in the magnetic field. This is similar to the way in which a charged capacitor stores energy in the electric field between its plates. Transformers Transformers are used to increase or decrease AC voltages. Usage of transformers is common because they change voltages with relatively little loss of energy. In fact, many of the devices in your home, such as game systems, printers, and stereos, have transformers inside their casings or as part of their cords. How transformers work Self-inductance produces an EMF when current changes in a single coil. A transformer has two coils, electrically insulated from each other, but wound around the same iron core. One coil is called the primary coil. The other coil is called the secondary coil. When the primary coil is connected to a source of AC voltage, the changing current creates a changing magnetic field, which is carried through the core to the secondary coil. In the secondary coil, the changing field induces a varying EMF. This effect is called mutual inductance. The EMF induced in the secondary coil, called the secondary voltage, is proportional to the primary voltage. The secondary voltage also depends on the ratio of the number of turns on the secondary coil to the number of turns on the primary coil, as shown by the following expressions. e ry g ta l o v se a nd o c e g a lt o v ry a im r p number of turns on secondary coil number of turns on primary coil N V s s N V p p If the secondary voltage is larger than the primary voltage, the transformer is called a step-up transformer, as shown in Figure 25-13a. If the voltage coming out of the transformer is smaller than the voltage put in, then it is called a step-down transformer, as shown in Figure 25-13b. Motor and Generator Motors and generators differ, mainly in the way they convert energy\u2014electric to mechanical compared to mechanical to electric. 1. Make a series circuit with an efficient DC motor, a miniature lamp, and an ammeter. 2. Rotate the handle, or motor shaft, to try to light the lamp. Analyze and Conclude 3. What happens if you vary the speed at which you rotate the handle? 4. Predict what will happen if you connect your motor to a second motor. \u25a0 Figure 25-13 In a transformer, the ratio of input voltage to output voltage depends upon the ratio of the number of turns on the primary coil to the number of turns on the secondary coil. The output voltage can be the same as the input, greater than the input (a), or less", " than the input (b). a Step-Up Transformer Primary Secondary b Step-Down Transformer Primary Secondary 10.0 A 100 V 5 turns 2.5 A 400 V 1000 W Core 20 turns 1000 W 2.0 A 1000 V 2000 W 50 turns 10.0 A 200 V 10 turns 2000 W Core 682 Chapter 25 Electromagnetic Induction 679-685 CH25-S2-845813 3/30/04 9:45 PM Page 683 In an ideal transformer, the electric power delivered to the secondary circuit equals the power supplied to the primary circuit. An ideal transformer dissipates no power itself, and can be represented by the following equations: Pp Ps VpIp VsIs Rearranging the equation to find the ratio Vp/Vs shows that the current in the primary circuit depends on how much current is required by the secondary circuit. This relationship can be combined with the relationship shown earlier between voltage and the number of turns to result in the following. N V I p p s Transformer Equation I N V s s p The ratio of the current in the secondary coil to the current in the primary coil is equal to the ratio of the voltage in the primary coil to the secondary coil, which is also equal to the ratio of the number of turns on the voltage in the primary coil to the number of turns on the secondary coil. As mentioned previously, a step-up transformer increases voltage. Because transformers cannot increase the power output, there must be a corresponding decrease in current through the secondary circuit. Similarly, in a step-down transformer, the current is greater in the secondary circuit than it is in the primary circuit. A voltage decrease corresponds to a current increase, as shown in the Connecting Math to Physics. Another way to understand this is to consider a transformer as 100 percent efficient, as is typically assumed in industry. Therefore, in most cases, it may be assumed that the input power and the output power are the same. Figure 25-13 illustrates the principles of step-up and step-down transformers. As shown in Figure 25-14, some transformers can function either as step-up transformers or step-down transformers, depending on how they are hooked up. Inequalities Study the following expressions to help you understand the relationships among voltage, current, and the number of coils in step-up and step-down transformers. Step-Up Transformer Step-Down Transformer Vp Vs Ip Is Np Ns Vp Vs Ip Is Np Ns \u25a0 Figure 25-14 If the input voltage is connected to the coils on the left, where there is a larger number of turns, the transformer functions as a step-down transformer. If the input voltage is connected at the right, the transformer functions as a step-up transformer. Common Units Transformers typically are rated in volt-amps reactive (VA, kilo-VA, mega-VA). Technically, only pure resistivetype loads have their power expressed in watts and reactive loads in volt-amps. Section 25.2 Changing Magnetic Fields Induce EMF 683 Horizons Companies 679-685 CH25-S2-845813 3/30/04 9:45 PM Page 684 Step-Up Transformers A step-up transformer has a primary coil consisting of 200 turns and a secondary coil consisting of 3000 turns. The primary coil is supplied with an effective AC voltage of 90.0 V. a. What is the voltage in the secondary circuit? b. The current in the secondary circuit is 2.0 A. What is the current in the primary circuit? 1 2 Analyze and Sketch the Problem \u2022 Draw an iron core with turns of wire. \u2022 Label the variables I, V, and N. Known: Vp Is 200 3000 Np Ns 90.0 V 2.0 A Unknown:? Vs? Ip Ip Vp Np Ns Is Vs Solve for the Unknown a. Solve for Vs. N V s s N V p p sV N p N p 0.0 V) Substitute Ns )( 9 (3000 0 0 2 1350 V Vs 3000, Vp 90.0 V, Np 200 b. The power in the primary and secondary circuits are equal assuming 100 percent efficiency. Pp VpIp Substitute Pp Ps VsIs Solve for Ip. Is V s V p.0 A) Substitute Vs V )(2 (1350 V.0 90 3.0101 A Ip VpIp, Ps VsIs Math Handbook Significant Digits pages 833\u2014834 1350 V, Is 2.0 A, Vp 90.0 V 3 Evaluate the Answer \u2022 Are the units correct? Voltage should be in volts and current in amps. \u2022 Is the magnitude realistic? A large step-up ratio of turns results in a large secondary voltage yet a smaller secondary current. The answers agree with this. For the following problems, effective currents", " and voltages are indicated. 16. A step-down transformer has 7500 turns on its primary coil and 125 turns on its secondary coil. The voltage across the primary circuit is 7.2 kV. What voltage is being applied across the secondary circuit? If the current in the secondary circuit is 36 A, what is the current in the primary circuit? 17. A step-up transformer has 300 turns on its primary coil and 90,000 turns on its secondary coil. The EMF of the generator to which the primary circuit is attached is 60.0 V. What is the EMF in the secondary circuit? The current in the secondary circuit is 0.50 A. What current is in the primary circuit? 684 Chapter 25 Electromagnetic Induction 679-685 CH25-S2-845813 6/8/04 3:59 PM Page 685 A distribution transformer ( T1) has its primary coil connected to a 3.0-kV AC source. The secondary coil is connected to the primary coil of a second transformer ( T2) by copper conductors. Finally, the secondary coil of transformer T2 connects to a load that uses 10.0 kW of power. Transformer T1 has a turn ratio of 5:1, and T2 has a load voltage of 120 V. The transformer efficiencies are 100.0 percent and 97.0 percent, respectively. 1. Calculate the load current. 2. How much power is being dissipated by transformer T2? 3. What is the secondary current of transformer T1? 4. How much current is the AC source supplying to T1? Everyday uses of transformers As you learned in Chapter 22, long-distance transmission of electrical energy is economical only if low currents and very high voltages are used. Step-up transformers are used at power sources to develop voltages as high as 480,000 V. High voltages reduce the current required in the transmission lines, keeping the energy lost to resistance low. When the energy reaches the consumer, step-down transformers, such as those shown in Figure 25-15, provide appropriately low voltages for consumer use. Transformers in home appliances further adjust voltages to useable levels. If you have ever had to charge a toy or operate a personal electronic device, you probably had to plug a large \u201cblock\u201d into the wall outlet. A transformer of the type discussed in this chapter is contained inside of that block. In this case, it is probably reducing the household voltage of about 120 V to something in the 3-V to 26-V range. Not all transformers are step-up or step-down. Transformers can be used to isolate one circuit from another. This is possible because the wire of the primary coil never makes direct contact with the wire of the secondary coil. This type of transformer would most likely be found in some small electronic devices. \u25a0 Figure 25-15 Step-down transformers are used to reduce the high voltages in transmission lines to levels appropriate for consumers at the points of use. 25.2 Section Review 18. Coiled Wire and Magnets You hang a coil of wire with its ends joined so that it can swing easily. If you now plunge a magnet into the coil, the coil will swing. Which way will it swing relative to the magnet and why? 19. Motors If you unplugged a running vacuum cleaner from a wall outlet, you would be much more likely to see a spark than you would be if you unplugged a lighted lamp from the wall. Why? 21. Transformers Frequently, transformer coils that have only a few turns are made of very thick (lowresistance) wire, while those with many turns are made of thin wire. Why? 22. Step-Up Transformers Refer to the step-up transformer shown in Figure 25-13. Explain what will happen to the primary current if the secondary coil is short-circuited. 20. Transformers and Current Explain why a transformer may only be operated on alternating current. 23. Critical Thinking Would permanent magnets make good transformer cores? Explain. physicspp.com/self_check_quiz Section 25.2 Changing Magnetic Fields Induce EMF 685 Matt Meadows 686-687 CH25-LAB-845813 3/30/04 10:57 PM Page 686 Induction and Transformers A transformer is an electric device without any moving components. It is made of two electric circuits interlinked by a magnetic field. A transformer is used to increase or decrease an AC potential difference, which often is called voltage. Transformers can be found everywhere. Every electronic device that plugs into your household electric circuits incorporates a transformer, usually to lower the voltage going to the device. Televisions that have standard cathode-ray picture tubes incorporate high-voltage transformers, which raise the standard household voltage to tens of thousands of volts. This accelerates electrons from the rear of the tube to the screen. In this experiment, you will use two coils with a", " removable iron core. One coil is called the primary coil, the other the secondary coil. When an AC voltage is applied to the primary coil, the changing magnetic field induces a current, and thus, a voltage in the secondary coil. This induced voltage is expressed by Vs/Vp Ns/Np, where N refers to the number of turns in the coils. QUESTION What is the relationship between voltages in the two coils of a transformer? Objectives Procedure \u25a0 Describe how a transformer works. \u25a0 Observe the effect of DC voltage on a transformer. \u25a0 Observe the effect of AC voltage on a transformer. Safety Precautions Materials primary and secondary coil apparatus small AC power supply AC voltmeter DC power supply (0-6V, 0-5A) connecting wires with alligator clips small lightbulb with wires 1. Estimate the number of coils of wire on the primary and secondary coils. Do this by counting the number of coils in 1 cm and multiplying by the coil\u2019s length in centimeters. The primary coil has one layer. The secondary coil has two layers of wire, so double the value for it. Record your results in the data table. 2. Place a small lightbulb across the contacts of the secondary coil. Carefully place the secondary coil into the primary coil. Slowly insert the iron core into the center of the secondary coil. 3. Attach two wires to the output of the DC power supply. Attach the positive wire from the power supply to one of the primary connections. Turn the power supply to nearly its maximum output setting. Holding the free end of the wire attached to the negative connection, gently tap its end to the other primary coil connection. Observe the area where you touch the wire to the connection. Record your observations in the data table. Horizons Companies 686-687 CH25-LAB-845813 3/30/04 10:58 PM Page 687 Data Table Number of primary coils Number of secondary coils Step 3 observation Step 4 observation Step 5 observation Step 6 observation Step 7 coil volts (V) Step 8 observation Step 9 iron core 4. Observe the lightbulb while you are gently tapping the connection. What happens as the wire makes contact and then breaks the electric contact? Record your observation in the data table. 5. Hold the negative wire to the primary coil connection for 5 s and observe the lightbulb. Record your observation in the data table. 6. Disconnect the DC power supply and put it away while leaving the small lightbulb attached to the secondary coil. Attach the AC power supply to the two primary coil connections. Plug in the AC power supply and observe the lightbulb. Record your observations in the data table. 7. Select the AC scale for your voltmeter. Insert the probes into the voltmeter and carefully touch them to the primary coil and measure the applied voltage. Move the probe from the primary coil and measure the secondary coil voltage. Record both readings in the data table. 8. Repeat step 7, but slowly remove the iron core from the secondary coil. What happens to the lightbulb? Measure both primary and secondary coil voltages while the core is being removed. Record your observations in the data table. 9. Carefully feel the iron core. What is your obser- vation? Record it in the data table. Analyze 1. Calculate the ratio of Ns/Np from your data. 2. Calculate the ratio of Vs/Vp from your data. 3. Interpret Data How do the ratios Ns/Np and Vs/Vp compare? 4. Recognize Cause and Effect Based on the data for step 7, is this transformer a step-up or a step-down transformer? What evidence do you have to support this conclusion? Conclude and Apply 1. Infer How can you explain your observation of the lightbulb in step 4? 2. Infer How can you explain the phenomena you observed at the negative connection of the primary coil in step 3? 3. Infer How can you explain your observations of the primary and the secondary coil voltages as you removed the iron core in step 8? 4. Explain Explain the temperature of the iron core you observed in step 9. Going Further Why does the transformer work only with alternating and not direct current? Real-World Physics Discuss the use of transformers to assist in the delivery of electricity from the power plant to your home. To find out more about induction and transformers, visit the Web site: physicspp.com 687 688 CH25-FEATURE-845813 6/8/04 4:01 PM Page 688 How a Credit-Card Reader Works Credit cards have revolutionized the world\u2019s economies by enabling money to be transferred quickly and easily. The credit-card reader, which captures data from a magnetic strip on the back of a card, is one of the most important links in the electronic transfer of money. 1 A permanent magnet that is touched to a strip of plastic", " coated with iron oxide leaves a magnetized region. 2 The magnet can be turned around to produce regions of opposite polarity. These regions are assigned the binary digits 1 and 0 and code information, such as the card holder\u2019s name and card number Credit-card reader Back of credit card 3 The card reader contains a pickup head, which is a tiny, wire-wound iron ring with a gap. and 4 As the card is swiped, the magnetized strip is pulled past the gap in the pickup head, generating a varying voltage across the coil. N S 0 V S N 1 V Card swipe The binary code on the strip is translated into a voltage waveform. The resulting waveform is stored in a computer\u2019s memory and then transmitted to the bank\u2019s verification office. Thinking Critically 1. Observe Why is the order of the binary numbers in step 5 reversed from the magnetic strip to the voltage waveform? 2. Analyze What would happen if the tiny iron ring did not have a gap in it? 688 How It Works 689-695 CH25-SG CA STP-845813 6/8/04 4:06 PM Page 689 25.1 Electric Current from Changing Magnetic Fields Vocabulary \u2022 electromagnetic induction Key Concepts \u2022 Michael Faraday discovered that if a wire moves through a magnetic field, (p. 672) \u2022 fourth right-hand rule (p. 672) \u2022 electromotive force (p. 673) \u2022 electric generator (p. 675) \u2022 average power (p. 677) an electric current can flow. \u2022 The current produced depends upon the angle between the velocity of the wire and the magnetic field. Maximum current occurs when the wire is moving at right angles to the field. \u2022 Electromotive force, EMF, is the potential difference created across the moving wire. EMF is measured in volts. \u2022 The EMF in a straight length of wire moving through a uniform magnetic field is the product of the magnetic field, B, the length of the wire, L, and the component of the velocity of the moving wire through the field that is perpendicular to the field, v(sin ). EMF BLv sin \u2022 Effective current and voltage can be used to describe alternating current and voltage. Ieff 0.707 Imax Veff 0.707 Vmax \u2022 A generator and a motor are similar devices. A generator converts mechanical energy to electric energy, whereas a motor converts electric energy to mechanical energy. 25.2 Changing Magnetic Fields Induce EMF Vocabulary \u2022 Lenz\u2019s law (p. 679) \u2022 eddy current (p. 681) \u2022 self-inductance (p. 681) \u2022 transformer (p. 682) \u2022 primary coil (p. 682) \u2022 secondary coil (p. 682) \u2022 mutual inductance (p. 682) \u2022 step-up transformer (p. 682) \u2022 step-down transformer (p. 682) Key Concepts \u2022 Lenz\u2019s law states that an induced current is always produced in a direction such that the magnetic field resulting from the induced current opposes the change in the magnetic field that is causing the induced current. \u2022 Back-EMF is created by a current-carrying wire moving in a magnetic field. Back-EMF opposes the current. \u2022 Self-inductance is a property of a wire carrying a changing current. The faster the current is changing, the greater the induced EMF that opposes that change. \u2022 A transformer has two coils wound about the same core. An AC current through the primary coil induces an alternating EMF in the secondary coil. The voltages in alternating-current circuits may be increased or decreased by transformers physicspp.com/vocabulary_puzzlemaker 689 689-695 CH25-SG CA STP-845813 3/30/04 11:06 PM Page 690 Concept Mapping 24. Complete the following concept map using the following terms: generator, back-EMF, Lenz\u2019s law. motor EMF Michael Faraday Mastering Concepts 25. What is the armature of an electric generator? (25.1) 26. Why is iron used in an armature? (25.1) For problems 27\u201329, refer to Figure 25-16. 27. A single conductor moves through a magnetic field and generates a voltage. In what direction should the wire be moved, relative to the magnetic field to generate the minimum voltage? (25.1) 28. What is the polarity of the voltage induced in the wire when it passes the south pole of the magnetic field? (25.1) 29. What is the effect of increasing the net conductor length in an electric generator? (25.1) Bout of page Wire \u25a0 Figure 25-16 30. How were Oersted\u2019s and Faraday\u2019s results similar? How were they different? (25.1) 31. You have", " a coil of wire and a bar magnet. Describe how you could use them to generate an electric current. (25.1) 32. What does EMF stand for? Why is the name inaccurate? (25.1) 33. What is the difference between a generator and a motor? (25.1) 34. List the major parts of an AC generator. (25.1) 35. Why is the effective value of an AC current less than its maximum value? (25.1) 36. Hydroelectricity Water trapped behind a dam turns turbines that rotate generators. List all the forms of energy that take part in the cycle that includes the stored water and the electricity produced. (25.1) 37. State Lenz\u2019s law. (25.2) 38. What causes back-EMF in an electric motor? (25.2) 39. Why is there no spark when you close a switch and put current through an inductor, but there is a spark when you open the switch? (25.2) 40. Why is the self-inductance of a coil a major factor when the coil is in an AC circuit but a minor factor when the coil is in a DC circuit? (25.2) 41. Explain why the word change appears so often in this chapter. (25.2) 42. Upon what does the ratio of the EMF in the primary circuit of a transformer to the EMF in the secondary circuit of the transformer depend? (25.2) Applying Concepts 43. Use unit substitution to show that the units of BLv are volts. 44. When a wire is moved through a magnetic field, does the resistance of the closed circuit affect current only, EMF only, both, or neither? 45. Biking As Logan slows his bike, what happens to the EMF produced by his bike\u2019s generator? Use the term armature in your explanation. 46. The direction of AC voltage changes 120 times each second. Does this mean that a device connected to an AC voltage alternately delivers and accepts energy? 690 Chapter 25 Electromagnetic Induction For more problems, go to Additional Problems, Appendix B. 689-695 CH25-SG CA STP-845813 3/30/04 11:38 PM Page 691 47. A wire is moved horizontally between the poles of a magnet, as shown in Figure 25-17. What is the direction of the induced current? N N S S 51. Earth\u2019s Magnetic Field The direction of Earth\u2019s magnetic field in the northern hemisphere is downward and to the north as shown in Figure 25-21. If an east-west wire moves from north to south, in which direction is the current? North magnetic pole North pole \u25a0 Figure 25-17 48. You make an electromagnet by winding wire around a large nail, as shown in Figure 25-18. If you connect the magnet to a battery, is the current larger just after you make the connection or several tenths of a second after the connection is made? Or, is it always the same? Explain. \u25a0 Figure 25-18 49. A segment of a wire loop is moving downward through the poles of a magnet, as shown in Figure 25-19. What is the direction of the induced current? N N S S v \u25a0 Figure 25-19 50. A transformer is connected to a battery through a switch. The secondary circuit contains a lightbulb, as shown in Figure 25-20. Will the lamp be lighted as long as the switch is closed, only at the moment the switch is closed, or only at the moment the switch is opened? Explain. S N South pole South magnetic pole \u25a0 Figure 25-21 52. You move a length of copper wire down through a magnetic field, B, as shown in Figure 25-19. a. Will the induced current move to the right or left in the wire segment in the diagram? b. As soon as the wire is moved in the field, a current appears in it. Thus, the wire segment is a current-carrying wire located in a magnetic field. A force must act on the wire. What will be the direction of the force acting on the wire as a result of the induced current? 53. A physics instructor drops a magnet through a copper pipe, as illustrated in Figure 25-22. The magnet falls very slowly, and the students in the class conclude that there must be some force opposing gravity. a. What is the direction of the current induced in the pipe by the falling magnet if the south pole is toward the bottom? b. The induced current produces a magnetic field. What is the direction of the field? c. How does this field reduce the acceleration of the falling magnet? N S \u25a0 Figure 25-22 Primary Secondary \u25a0 Figure 25-20 54. Generators Why is a generator more difficult to rotate when it is connected to a circuit and supplying current than it is when it is standing", " alone? physicspp.com/chapter_test Chapter 25 Assessment 691 689-695 CH25-SG CA STP-845813 6/8/04 4:09 PM Page 692 55. Explain why the initial start-up current is so high in an electric motor. Also explain how Lenz\u2019s law applies at the instant t 0. 64. An AC generator develops a maximum EMF of 565 V. What effective EMF does the generator deliver to an external circuit? 56. Using Figure 25-10 in conjunction with Lenz\u2019s law, explain why all practical transformer cores incorporate a laminated core. 57. A practical transformer is constructed with a laminated core that is not a superconductor. Because the eddy currents cannot be completely eliminated, there is always a small core loss. This results, in part, in a net loss of power within the transformer. What fundamental law makes it impossible to bring this loss to zero? 58. Explain the process of mutual induction within a transformer. 59. Shawn drops a magnet, north pole down, through a vertical copper pipe. a. What is the direction of the induced current in the copper pipe as the bottom of the magnet passes? b. The induced current produces a magnetic field. What is the direction of the induced magnetic field? Mastering Problems 25.1 Electric Current from Changing Magnetic Fields 60. A wire, 20.0-m long, moves at 4.0 m/s perpendicularly through a magnetic field. An EMF of 40 V is induced in the wire. What is the strength of the magnetic field? 61. Airplanes An airplane traveling at 9.50102 km/h passes over a region where Earth\u2019s magnetic field is 4.5105 T and is nearly vertical. What voltage is induced between the plane\u2019s wing tips, which are 75 m apart? 62. A straight wire, 0.75-m long, moves upward through a horizontal 0.30-T magnetic field, as shown in Figure 25-23, at a speed of 16 m/s. a. What EMF is induced in the wire? b. The wire is part of a circuit with a total resistance of 11. What is the current? N N S S v \u25a0 Figure 25-23 63. At what speed would a 0.20-m length of wire have to move across a 2.5-T magnetic field to induce an EMF of 10 V? 65. An AC generator develops a maximum voltage of 150 V. It delivers a maximum current of 30.0 A to an external circuit. a. What is the effective voltage of the generator? b. What effective current does the generator deliver to the external circuit? c. What is the effective power dissipated in the circuit? 66. Electric Stove An electric stove is connected to an AC source with an effective voltage of 240 V. a. Find the maximum voltage across one of the stove\u2019s elements when it is operating. b. The resistance of the operating element is 11. What is the effective current? 67. You wish to generate an EMF of 4.5 V by moving a wire at 4.0 m/s through a 0.050-T magnetic field. How long must the wire be, and what should be the angle between the field and direction of motion to use the shortest wire? 68. A 40.0-cm wire is moved perpendicularly through a magnetic field of 0.32 T with a velocity of 1.3 m/s. If this wire is connected into a circuit of 10resistance, what is the current? 69. You connect both ends of a copper wire with a total resistance of 0.10 to the terminals of a galvanometer. The galvanometer has a resistance of 875. You then move a 10.0-cm segment of the wire upward at 1.0 m/s through a 2.0102-T magnetic field. What current will the galvanometer indicate? 70. The direction of a 0.045-T magnetic field is 60\u00b0 above the horizontal. A wire, 2.5-m long, moves horizontally at 2.4 m/s. a. What is the vertical component of the magnetic field? b. What EMF is induced in the wire? 71. Dams A generator at a dam can supply 375 MW (375106 W) of electrical power. Assume that the turbine and generator are 85 percent efficient. a. Find the rate at which falling water must supply energy to the turbine. b. The energy of the water comes from a change in potential energy, PE mgh. What is the change in PE needed each second? c. If the water falls 22 m, what is the mass of the water that must pass through the turbine each second to supply this power? 692 Chapter 25 Electromagnetic Induction For more problems, go to Additional Problems, Appendix B. 689-695 CH25-SG CA ST", "P-845813 7/15/04 5:16 PM Page 693 72. A conductor rotating in a magnetic field has a 77. Hair Dryers A hair dryer manufactured for use in length of 20 cm. If the magnetic-flux density is 4.0 T, determine the induced voltage when the conductor is moving perpendicular to the line of force. Assume that the conductor travels at a constant velocity of 1 m/s. 73. Refer to Example Problem 1 and Figure 25-24 to determine the following. a. induced voltage in the conductor b. current flow (I) c. direction of flux rotation around the conductor d. polarity of point A relative to point B R 1.0 A v 3.6 m/s L 0.50 m EMF A B 7.0102 T Bout of page \u25a0 Figure 25-24 the United States uses 10 A at 120 V. It is used with a transformer in England, where the line voltage is 240 V. a. What should be the ratio of the turns of the transformer? b. What current will the hair dryer now draw? 78. A 150-W transformer has an input voltage of 9.0 V and an output current of 5.0 A. a. Is this a step-up or step-down transformer? b. What is the ratio of Voutput to Vinput? 79. Scott connects a transformer to a 24-V source and measures 8.0 V at the secondary circuit. If the primary and secondary circuits were reversed, what would the new output voltage be? Mixed Review 80. A step-up transformer\u2019s primary coil has 500 turns. Its secondary coil has 15,000 turns. The primary circuit is connected to an AC generator having an EMF of 120 V. a. Calculate the EMF of the secondary circuit. b. Find the current in the primary circuit if the current in the secondary circuit is 3.0 A. c. What power is drawn by the primary circuit? What power is supplied by the secondary circuit? 25.2 Changing Magnetic Fields Induce EMF 81. With what speed must a 0.20-m-long wire cut 74. The primary coil of a transformer has 150 turns. It is connected to a 120-V source. Calculate the number of turns on the secondary coil needed to supply the following voltages. a. 625 V b. 35 V c. 6.0 V 75. A step-up transformer has 80 turns on its primary coil and 1200 turns on its secondary coil. The primary circuit is supplied with an alternating current at 120 V. a. What voltage is being applied across the secondary circuit? b. The current in the secondary circuit is 2.0 A. What current is in the primary circuit? c. What are the power input and output of the transformer? 76. Laptop Computers The power supply in a laptop computer requires an effective voltage of 9.0 V from a 120-V line. a. If the primary coil has 475 turns, how many does the secondary coil have? b. A 125-mA current is in the computer. What current is in the primary circuit? across a magnetic field for which B is 2.5 T if it is to have an EMF of 10 V induced in it? 82. At what speed must a wire conductor 50-cm long be moved at right angles to a magnetic field of induction 0.20 T to induce an EMF of 1.0 V in it? 83. A house lighting circuit is rated at 120-V effective voltage. What is the peak voltage that can be expected in this circuit? 84. Toaster A toaster draws 2.5 A of alternating current. What is the peak current through this toaster? 85. The insulation of a capacitor will break down if the instantaneous voltage exceeds 575 V. What is the largest effective alternating voltage that may be applied to the capacitor? 86. Circuit Breaker A magnetic circuit breaker will open its circuit if the instantaneous current reaches 21.25 A. What is the largest effective current the circuit will carry? 87. The electricity received at an electrical substation has a potential difference of 240,000 V. What should the ratio of the turns of the step-down transformer be to have an output of 440 V? physicspp.com/chapter_test Chapter 25 Assessment 693 689-695 CH25-SG CA STP-845813 6/8/04 4:10 PM Page 694 88. An alternating-current electric generator supplies a 45-kW industrial electric heater. If the system voltage is 660 Vrms, what is the peak current supplied? 89. A certain step-down transformer has 100 turns on the primary coil and 10 turns on the secondary coil. If a 2.0-kW resistive load is connected to the transformer, what is the effective primary current that flows? Assume that the secondary voltage is 60.0 Vpk. 90. A transformer rated", " at 100 kVA has an efficiency of 98 percent. a. If the connected load consumes 98 kW of power, what is the input power to the transformer? b. What is the maximum primary current with the transformer consuming its rated reactive power? Assume that VP 600 V. 91. A wire, 0.40-m long, cuts perpendicularly across a magnetic field for which B is 2.0 T at a velocity of 8.0 m/s. a. What EMF is induced in the wire? b. If the wire is in a circuit with a resistance of 6.4, what is the size of the current in the wire? 92. A coil of wire, which has a total length of 7.50 m, is moved perpendicularly to Earth\u2019s magnetic field at 5.50 m/s. What is the size of the current in the wire if the total resistance of the wire is 5.0102 m? Assume Earth\u2019s magnetic field is 5105 T. 93. The peak value of the alternating voltage applied to a 144- resistor is 1.00102 V. What power must the resistor be able to handle? 94. Television The CRT in a television uses a step-up transformer to change 120 V to 48,000 V. The secondary side of the transformer has 20,000 turns and an output of 1.0 mA. a. How many turns does the primary side have? b. What is the input current? Thinking Critically 95. Apply Concepts Suppose that an \u201canti-Lenz\u2019s law\u201d existed that meant a force was exerted to increase the change in a magnetic field. Thus, when more energy was demanded, the force needed to turn the generator would be reduced. What conservation law would be violated by this new \u201claw\u201d? Explain. 96. Analyze Real transformers are not 100 percent efficient. Write an expression for transformer efficiency in percent using power. A step-down transformer that has an efficiency of 92.5 percent is used to obtain 28.0 V from a 125-V household voltage. The current in the secondary circuit is 25.0 A. What is the current in the primary circuit? 97. Analyze and Conclude A transformer that supplies eight homes has an efficiency of 95 percent. All eight homes have operating electric ovens that each draw 35 A from 240-V lines. How much power is supplied to the ovens in the eight homes? How much power is dissipated as heat in the transformer? Writing in Physics 98. Common tools, such as an electric drill, are typically constructed using a universal motor. Using your local library, and other sources, explain how this type of motor may operate on either AC or DC current. Cumulative Review 99. Light is emitted by a distant star at a frequency of 4.561014 Hz. If the star is moving toward Earth at a speed of 2750 km/s, what frequency light will be detected by observers on Earth? (Chapter 16) 100. A distant galaxy emits light at a frequency of 7.291014 Hz. Observers on Earth receive the light at a frequency of 6.141014 Hz. How fast is the galaxy moving, and in what direction? (Chapter 16) 101. How much charge is on a 22-F capacitor with 48 V applied to it? (Chapter 21) 102. Find the voltage across a 22-, 5.0-W resistor operating at half of its rating. (Chapter 22) 103. Determine the total resistance of three, 85- resistors connected in parallel and then seriesconnected to two 85- resistors connected in parallel, as shown in Figure 25-25. (Chapter 23) 85 85 85 85 85 \u25a0 Figure 25-25 104. An electron with a velocity of 2.1106 m/s is at right angles to a 0.81-T magnetic field. What is the force on the electron produced by the magnetic field? What is the electron\u2019s acceleration? The mass of an electron is 9.111031 kg. (Chapter 24) 694 Chapter 25 Electromagnetic Induction For more problems, go to Additional Problems, Appendix B. 689-695 CH25-SG CA STP-845813 3/30/04 11:09 PM Page 695 Multiple Choice 1. Which dimensional analysis is correct for the calculation of EMF? (NAm)(J) (NA/m)(m)(m/s) JC (NmA/s)(1/m)(m/s) 2. 4.20 V of electromotive force are induced on a 427-mm-long piece of wire that is moving at a rate of 18.6 cm/s. What is the magnetic field that induced the EMF? 5.29104 T 1.89102 T 0.334 T 52.9 T 3. Which of the following will fail to induce an electric current in the wire? 4", ". 15 cm of wire is moving at the rate of 0.12 m/s through a perpendicular magnetic field of strength 1.4 T. Calculate the EMF induced in the wire. 0 V 0.018 V 0.025 V 2.5 V 5. A transformer uses a 91-V supply to operate a 13-V device. If the transformer has 130 turns on the primary coil and the device uses 1.9 A of current from the transformer, what is the current supplied to the primary coil? 0.27 A 0.70 A 4.8 A 13.3 A 6. An AC generator that delivers a peak voltage of 202 V connects to an electric heater with a resistance of 480. What is the effective current in the heater? 0.298 A 1.68 A 2.38 A 3.37 A N S 480 202 V Extended Answer 7. Compare the power lost in transmission for an 800-W line at 160 V to the same power on a line at 960 V. Assume the resistance of the line is 2. What conclusion can you draw? N S Investigate Ask your instructor what kinds of questions to expect on the test. Also ask for practice tests so that you can become familiar with the test-taking materials. physicspp.com/standardized_test Chapter 25 Standardized Test Practice 695 696-704 CH26-S1-845813 3/19/04 4:08 PM Page 696 What You\u2019ll Learn \u2022 You will learn how combined electric and magnetic fields can be used to determine the masses of electrons, atoms, and molecules. \u2022 You will explain how electromagnetic waves are created, travel through space, and are detected. Why It\u2019s Important Many electromagnetic waves\u2014from radio and television waves, to visible light, microwaves, and X rays\u2014play vital roles in our lives. Parabolic Receivers This parabolic dish antenna is designed to receive radio waves from satellites orbiting hundreds of kilometers above Earth\u2019s surface and objects well beyond the solar system. Think About This A parabolic dish gets its name from the shape of its reflecting surface\u2014 a parabola. Why are parabolic dish antennas well suited for receiving weak television signals? physicspp.com 696 Rick England/Getty Images 696-704 CH26-S1-845813 6/8/04 4:19 PM Page 697 From where do radio stations broadcast? Question Radio signals are electromagnetic waves. How far away are the transmitters that are used to broadcast the radio station signals that you can tune in on the AM band? Procedure 1. The AM radio band ranges from 540 kHz to 1690 kHz. Make a data table with columns for Frequency (kHz), Station Call Sign, Signal Strength, Location, and Distance (km). 2. Turn on the radio, adjust it to 540 kHz, and set the volume to a moderate level. 3. Collect and Organize Data Slowly adjust the frequency upward until you hear a radio station whose broadcast you can clearly understand. Listen to the broadcast for a short time to hear if the station identifies its call sign. Record the station\u2019s frequency, signal strength (strong, medium, or weak), and call sign in your data table. 4. Repeat step 3 until you have reached the upper end of the AM radio band, 1690 kHz. 5. Determine where each station broadcasts its signal from. Record the city each station broadcasts from in your data table. 6. Measure in SI Using maps, locate the city from which each radio station broadcasts. Determine the distance to each transmitter and record the value in your data table. Analysis How far away was the farthest radio station broadcast? Does the distance from the transmitter affect the station\u2019s signal strength? Critical Thinking Changing the position of the antenna often affects a station\u2019s signal strength. What does this imply about the nature of radio waves? 26.1 Interactions of Electric and Magnetic Fields and Matter Although you may not know what the terms stand for, you have Objectives \u2022 Describe the operation of probably heard of shortwave radio, microwaves, and VHF and UHF television signals. Each of these terms is used to describe one of the many types of electromagnetic waves that are broadcast through the air to provide you with radio, television, and other forms of communication. All of these waves consist of electric and magnetic fields propagating through space. The key to understanding how these waves behave is understanding the nature of the electron. Why? Because electromagnetic waves are produced by accelerating electrons\u2014the electrons\u2019 charge produces electric fields and the electrons\u2019 motion produces magnetic fields. Furthermore, the waves are broadcast and received by antennas, devices made of matter that also contains electrons. Thus, the logical first step in understanding how electromagnetic waves are produced, propagated, received, and used for so many devices is to learn about the properties of the electron. a cathode-ray tube. \u2022 Solve problems involving the interaction of charged particles with the electric and magnetic fields", " in cathode-ray tubes and mass spectrometers. \u2022 Explain how a mass spectrometer separates ions of different masses. Vocabulary isotope mass spectrometer Section 26.1 Interactions of Electric and Magnetic Fields and Matter 697 Horizons Companies 696-704 CH26-S1-845813 6/8/04 4:20 PM Page 698 Mass of an Electron How do you determine the mass of something that cannot be seen with the unaided eye and whose mass is so small that it cannot be measured even by the most sensitive scale? Such was the challenge\u2014that of determining the mass of an electron\u2014facing physicists in the late 1800s. The solution required a series of discoveries. The first piece of the puzzle came from Robert Millikan. As described in Chapter 21, Millikan balanced charged oil droplets in an electric field and was able to determine the charge, q, of an electron (1.6021019 C). Next, British physicist J. J. Thomson was able to determine the charge-to-mass ratio, q/m, of an electron. Knowing both the charge-to-mass ratio, q/m, and the charge of an electron, q, Thomson was able to calculate the mass of an electron. Thomson\u2019s experiments with electrons In 1897, J. J. Thomson performed the first experimental measurement of the charge-to-mass ratio of an electron. Thomson used a cathode-ray tube, a device that generates a stream of electrons. Figure 26-1 shows the setup used in the experiment. In order to minimize collisions between the electron beam and the molecules in the air, Thomson removed virtually all of the air from within the glass tube. V Anodes Charged deflection plates B Deflecting magnet Fluorescent coating Cathode Slits Deflecting magnet E Electron beam Inside the cathode-ray tube, an electric field is produced by a large potential difference between the cathode and anode. Electrons are emitted from the cathode and are accelerated toward the anodes by the electric field. Some of the electrons pass through slits in the anodes, forming a narrow beam. When the electrons reach the end of the tube they strike a fluorescent coating, causing it to glow. Thomson used electric and magnetic fields to exert force on and deflect the electron beam as it passed through the tube. The electric field, which was produced by charged parallel plates, was oriented perpendicular to the beam. The electric field (of strength E) produced a force (equal to qE) that acted on the electrons and deflected them upward, toward the positive plate. The magnetic field was produced by two electromagnets and was oriented at right angles to both the beam and the electric field. Recall from Chapter 24 that the force exerted by a magnetic field is perpendicular to the field and to the direction of motion of the electrons. Thus, the magnetic field (of strength B) produced a force (equal to Bqv, where v is the electron velocity) that acted on the electrons and deflected them downward. \u25a0 Figure 26-1 The charge-to-mass ratio of an electron was first measured with the Thomson adaptation of a cathode-ray tube. The electromagnets and charged deflection plates both are used to alter the path of the electron beam. 698 Chapter 26 Electromagnetism 696-704 CH26-S1-845813 6/8/04 4:21 PM Page 699 The electric and magnetic fields could be adjusted until the beam of electrons followed a straight, or undeflected, path. When this occurred, the forces due to the two fields were equal in magnitude and opposite in direction. Mathematically, this can be represented as follows: Bqv Eq Solving this equation for v yields the following expression: Eq E v B B q This equation shows that the forces are balanced only for electrons that have a specific velocity, v. If the electric field is turned off, only the force due to the magnetic field remains. The magnetic force is perpendicular to the direction of motion of the electrons, causing them to undergo centripetal (center-directed) acceleration. The accelerating electrons follow a circular path of radius r. Using Newton\u2019s second law of motion, the following equation can be written to describe the electron\u2019s path: Solving for q/m results in the following equation. v2 Bqv m r Charge-to-Mass Ratio in a Thomson Tube v q B r m In a Thomson tube, the ratio of an electron\u2019s charge to its mass is equal to the ratio of the electron\u2019s velocity divided by the product of the magnetic field strength and the radius of the electron\u2019s circular path. Thomson calculated the straight trajectory velocity, v, using measured values of E and B. Next, he measured the distance between the spot formed by the undeflected beam and the spot formed when the magnetic field acted on the beam. Using this distance, he", " calculated the radius of the electron\u2019s circular path, r. Knowing the value of r, Thomson was able to calculate q/m. By averaging many experimental trials, he determined that q/m 1.7591011 C/kg. Using this value for q/m and the known value of q, the mass of the electron (m) was calculated. \u25a0 Figure 26-2 This photograph shows the circular tracks of electrons (e) and positrons (e) moving through the magnetic field in a bubble chamber, a type of particle detector used in the early years of high-energy physics. Electrons and positrons curve in opposite directions. 1 q 0 1 2 0.6 1 C 9.1071031 kg m 1 1.7 1 0 9 5 1 m q/ C kg / m 9.111031 kg 9 Thomson\u2019s experiments with protons Thomson also used his cathoderay test apparatus to determine the charge-to-mass ratio for positive ions. He took advantage of the fact that positively charged particles undergo the opposite deflection experienced by electrons moving through an electric or magnetic field. The differing deflection of electrons and positive ions can be seen in Figure 26-2. To accelerate positively charged particles into the deflection region, Thomson reversed the direction of the electric field between the cathode and anodes. He also added a small amount of hydrogen gas to the tube. The electric field pulled electrons off the hydrogen atoms, changing the atoms into positive ions. These positive hydrogen ions, or protons, were then accelerated through a tiny slit in the anode. The resulting proton beam passed through electric and magnetic fields on its way toward the end of the tube. Section 26.1 Interactions of Electric and Magnetic Fields and Matter 699 CERN, P. Loiez/Science Photo Library/Photo Researchers 696-704 CH26-S1-845813 3/12/04 8:10 AM Page 700 Using this technique, the mass of a proton was determined in the same manner as was the mass of the electron. The mass of a proton was found to be 1.671027 kg. Thomson went on to use this technique to determine the masses of heavier ions produced when electrons were stripped from gases, such as helium, neon, and argon. Path Radius An electron with a mass of 9.111031 kg moves through a cathode-ray tube at 2.0105 m/s perpendicular to a magnetic field of 3.5102 T. The electric field is turned off. What is the radius of the circular path that is followed by the electron? B v FB r Math Handbook Operations with Scientific Notation pages 842\u2013843 1 Analyze and Sketch the Problem \u2022 Draw the path of the electron and label the velocity, v. \u2022 Sketch the magnetic field perpendicular to the velocity. \u2022 Diagram the force acting on the electron. Add the radius of the electron\u2019s path to your sketch. Unknown: r? Known: v 2.0105 m/s B 3.5102 T m 9.111031 kg q 1.6021019 C 2 Solve for the Unknown Use Newton\u2019s second law of motion to describe an electron in a cathode-ray tube subjected to a magnetic field. v2 Bqv m r v r = m Bq (9.111031 kg)(2.0105 m/s) (3.5102 T)(1.6021019 C) = = 3.3105 m Substitute m 9.111031 kg, v 2.0105 m/s, B 3.5102 T, q 1.6021019 C 3 Evaluate the Answer \u2022 Are the units correct? The radius of the circular path is a length measurement, given in units of meters. Assume that all charged particles move perpendicular to a uniform magnetic field. 1. A proton moves at a speed of 7.5103 m/s as it passes through a magnetic field of 0.60 T. Find the radius of the circular path. Note that the charge carried by the proton is equal to that of the electron, but is positive. 2. Electrons move through a magnetic field of 6.0102 T balanced by an electric field of 3.0103 N/C. What is the speed of the electrons? 3. Calculate the radius of the circular path that the electrons in problem 2 follow in the absence of the electric field. 4. Protons passing without deflection through a magnetic field of 0.60 T are balanced by an electric field of 4.5103 N/C. What is the speed of the moving protons? 700 Chapter 26 Electromagnetism 696-704 CH26-S1-845813 6/8/04 4:22 PM Page 701 The Mass Spectrometer An interesting thing happened when Thomson put neon gas into the cathode-ray tube\u2014he observed two glowing dots on the screen", " instead of one. Each dot corresponded to a unique charge-to-mass ratio; that is, he was able to calculate two different values for q/m. Thomson concluded that different atoms of the same element could have identical chemical properties, but have different masses. Each of the differing forms of the same atom, with the same chemical properties but different masses, is an isotope. A device similar to Thomson\u2019s cathode-ray tube that is commonly used to study isotopes is the mass spectrometer. The mass spectrometer is able to precisely measure the charge-to-mass ratios of positive ions. From the charge-to-mass ratio, the mass of each isotope can be calculated. The material under investigation is called the ion source, as it is used to produce the positive ions. The ion source must either be a gas or a material that can be heated to form a vapor. The positive ions are formed when accelerated electrons strike the gas or vapor atoms. The collisions knock electrons off the atoms, forming positive ions. A potential difference, V, between the electrodes produces an electric field that is used to accelerate the ions. Figure 26-3 shows one type of mass spectrometer. \u25a0 Figure 26-3 The mass spectrometer is used to analyze isotopes of an element. Inside the spectrometer, a magnet causes the positive ions in a vacuum chamber to be deflected according to their mass (a). In the vacuum chamber, the process is recorded on a photographic plate or a solid-state detector (b). Face of magnetic pole a To vacuum pump Vacuum chamber b Face of magnetic pole To vacuum pump Vacuum chamber Magnet Positive ions Electron beam Gas entry Photographic plate To vacuum pump Section 26.1 Interactions of Electric and Magnetic Fields and Matter 701 696-704 CH26-S1-845813 3/12/04 8:11 AM Page 702 To select ions with a specific velocity, the ions first are made to pass through electric and magnetic deflecting fields, as in Thomson\u2019s cathoderay tube. Ions that pass undeflected through these two fields then move into a region that is subject only to a uniform magnetic field. There, the ions follow a circular path. The radii of the paths can be used to determine the charge-to-mass ratios of the ions. The path radius, r, for an ion can be calculated from Newton\u2019s second law of motion. v2 Bqv m r Solving for r yields the following equation: v r m qB The velocity of an undeflected ion can be calculated from the equation for the kinetic energy of ions accelerated from rest, through a known potential difference, V. KE 1 mv2 qV 2 qV v 2 m Substituting this expression for v in the equation r mv/qB gives the radius of the circular path. v r m qB m B q qV 2 m m 1 2V q B Simplifying this equation by multiplying both sides by B yields the following: V Br 2m q This equation can be used to determine the charge-to-mass ratio of an ion. Charge-to-Mass Ratio of an Ion in a Mass Spectrometer q 2 V 2 r2 B m In a mass spectrometer, the ratio of an ion\u2019s charge to its mass is equal to the ratio of twice the potential difference divided by the product of the square of the magnetic field strength and the square of the radius of the ion\u2019s circular path. As shown in Figure 26-3, in one type of mass spectrometer, the ions strike a plate of photographic film, where they leave a mark. The diameter of the curved path traveled by the ions in the vacuum chamber can be easily measured because it is the distance between the mark made on the film and the slit in the electrode. Therefore, the radius of the path, r, is half of this measured distance. Modeling a Mass Spectrometer Make a ramp by placing a small ball of clay under one end of a grooved ruler. Place a 6-mmdiameter steel ball about halfway up the ramp and release it. 1. Observe the ball as it rolls down the ramp and along the table top. 2. Experiment with the location of a strong magnet near the path of the ball on the table top. Place the magnet close to the path so that it causes the ball\u2019s path to curve, but not so close that it causes the ball to collide with the magnet. Repeat step 1 as needed. 3. Predict what will happen to the ball\u2019s path when the ball is released from a higher or lower point on the ramp. 4. Test your prediction. Analyze and Conclude 5. Explain if the observed results are consistent with those for a charged particle moving through a magnetic field. 702 Chapter 26 Electromagnetism 696-704 CH26-S1-845813 6/8/04 4", ":22 PM Page 703 Mass of a Neon Atom The operator of a mass spectrometer produces a beam of doubly ionized (2) neon atoms. They first are accelerated by a potential difference of 34 V. Then, as the ions pass through a magnetic field of 0.050 T, the radius of their path is 53 mm. Determine the mass of the neon atom to the closest whole number of proton masses. 1 2 Analyze and Sketch the Problem \u2022 Draw the circular path of the ions. Label the radius. \u2022 Draw and label the potential difference between the electrodes. V 34 V B 0.050 T r 0.053 m 1.671027 kg Known: mproton q 2(1.601019 C) 3.201019 C Unknown:? mneon? Nproton Solve for the Unknown Use the equation for the charge-to-mass ratio of an ion in a mass spectrometer. q 2 V r 2 2 B mn eon 2r 2 qB 2V (3.201019 C)(0.050 T)2(0.053 m)2 2(34 V) mneon XXXXX B XXXXX Ion path XXXXX r XXXXX V Math Handbook Isolating a Variable page 845 Substitute q 3.201019 C, B 0.050 T, r 0.053 m, and V 34 V 3.31026 kg Divide the mass of neon by the mass of a proton to find the number of proton masses. Nproton to 20 protons 3.31026 kg 1.671027 kg/proton 3 Evaluate the Answer \u2022 Are the units correct? Mass should be measured in grams or kilograms. The number of protons should not be represented by any units. \u2022 Is the magnitude realistic? Neon has two isotopes, with masses of approximately 20 and 22 proton masses. 5. A beam of singly ionized (1+) oxygen atoms is sent through a mass spectrometer. The values are B 7.2102 T, q 1.601019 C, r 0.085 m, and V 110 V. Find the mass of an oxygen atom. 6. A mass spectrometer analyzes and gives data for a beam of doubly ionized (2) argon atoms. The values are q 2(1.601019 C), B 5.0102 T, r 0.106 m, and V 66.0 V. Find the mass of an argon atom. 7. A stream of singly ionized (1) lithium atoms is not deflected as it passes through a magnetic field of 1.5103 T that is perpendicular to an electric field of 6.0102 N/C. What is the speed of the lithium atoms as they pass through the two fields? 8. In Example Problem 2, the mass of a neon isotope is determined. Another neon isotope is found to have a mass of 22 proton masses. How far apart on the photographic film would these two isotopes land? Section 26.1 Interactions of Electric and Magnetic Fields and Matter 703 696-704 CH26-S1-845813 3/12/04 8:12 AM Page 704 Chemistry Connection Chemistry Connection Atomic mass scale 4.345% 83.789% 9.501% 2.365% 50 51 52 53 54 55 Marks on photographic film plate and isotopic abundances (%) \u25a0 Figure 26-4 Mass spectrometers are widely used to determine the isotopic composition of an element. The above illustration shows the results of analyzing the marks left on a film plate by chromium isotopes. Isotopic analysis The approximate spacing between marks on the film for an ionized chromium (Cr) sample is shown in Figure 26-4. The four distinct red marks indicate that a naturally occurring sample of chromium is composed of four isotopes. The width of the mark corresponds to the abundance of the isotope. Note that the isotope with a mass number of 52 is the most abundant isotope, and that the sum of the percentages for the four isotopes equals 100 percent. As you may recall from chemistry, the mass of each element listed in the periodic table is actually a weighted average of the masses of all of the stable isotopes of that element. Note that all of the chromium ions that hit the film have the same charge. Their charge depends on how many electrons were removed from the neutral chromium atoms used as the ion source. Recall that the ions are formed when accelerated electrons are used to knock electrons off neutral atoms. After the first electron is removed, producing a singly ionized (1) atom, more energy is required to remove the second electron and produce a double ionized (2) atom. This additional energy can be provided by electrons that undergo a greater acceleration because they are subjected to a greater electric field. Thus, higher-energy accelerated electrons can produce both singly and doubly charged ions.", " In this way, the operator of the mass spectrometer can choose the charge on the ion to be studied. Other applications Mass spectrometers have numerous applications. For example, a mass spectrometer can be used to \u201cpurify\u201d a sample of uranium into its component isotopes. Rather than striking a detector to measure relative abundance, the separated isotopes are collected. The different isotopes are, in turn, used in varying applications. Mass spectrometers also are used to detect and identify trace amounts of molecules in a sample, an application extensively used in the environmental and forensic sciences. The device is so sensitive that researchers are able to separate ions with mass differences as small as one ten-thousandth of one percent and are able to identify the presence of a single molecule within a 10 billionmolecule sample. 26.1 Section Review 9. Cathode-Ray Tube Describe how a cathode-ray tube forms an electron beam. 10. Magnetic Field The radius of the circular path of an ion in a mass spectrometer is given by r = (1/B)2Vm/q. Use this equation to explain how a mass spectrometer is able to separate ions of different masses. 11. Magnetic Field A modern mass spectrometer can analyze molecules having masses of hundreds of proton masses. If the singly charged ions of these molecules are produced using the same accelerating voltage, how would the mass spectrometer\u2019s magnetic field have to be changed for the ions to hit the film? 12. Path Radius A proton moves at a speed of 4.2104 m/s as it passes through a magnetic field of 1.20 T. Find the radius of the circular path. 13. Mass A beam of doubly ionized (2) oxygen atoms is accelerated by a potential difference of 232 V. The oxygen then enters a magnetic field of 75 mT and follows a curved path with a radius of 8.3 cm. What is the mass of the oxygen atom? 14. Critical Thinking Regardless of the energy of the electrons used to produce ions, J. J. Thomson never could remove more than one electron from a hydrogen atom. What could he have concluded about the positive charge of a hydrogen atom? 704 Chapter 26 Electromagnetism physicspp.com/self_check_quiz 705-713 CH26-S2-845813 3/12/04 8:16 AM Page 705 26.2 Electric and Magnetic Fields in Space Although you probably do not realize it, you rely on electromagnetic waves every day. Signals broadcast from television and radio stations, orbiting satellites, and even those emanating from distant galaxies are all electromagnetic waves. Electromagnetic waves also are used in common consumer products, such as microwaves ovens, remote-control garage door openers, and cellular phones, to name a few. In this section, you will learn about the fields that make up electromagnetic waves, and how the waves are produced and received. Electromagnetic Waves Great advancements in the understanding of electromagnetic waves were made during the nineteenth century. These advancements led to the development of new devices and technologies that had a huge impact on modern society. A series of breakthroughs In 1821, while performing a demonstration for his students, Danish physicist Hans Christian Oersted noticed that an electric current caused the needle in a nearby compass to deflect. Oersted realized that his observation displayed a fundamental connection between electricity and magnetism. He concluded that an electric current in a conductor produces a magnetic field, and that a changing electric current produces a changing magnetic field. Oersted\u2019s discovery created excitement in the scientific community and led to a flood of new research. Eleven years after Oersted, Englishman Michael Faraday, and American high school physics teacher John Henry independently discovered induction. Induction is the production of an electric field due to a moving magnetic field. Interestingly, induced electric fields exist even if there is not a wire present, as shown in Figure 26-5a. Thus, a changing magnetic field produces a corresponding changing electric field. Notice that the field lines of the induced electric field, shown in Figure 26-5a, are closed loops. This is because, unlike an electrostatic field, there are no charges on which the field lines begin or end. Objectives \u2022 Describe how electromagnetic waves propagate through space. \u2022 Solve problems involving electromagnetic wave properties. \u2022 Describe the factors affecting an antenna\u2019s ability to receive an electromagnetic wave of a specific wavelength. \u2022 Solve problems involving electromagnetic wave propagation through dielectrics. Vocabulary electromagnetic wave dielectrics antenna electromagnetic spectrum electromagnetic radiation piezoelectricity receiver \u25a0 Figure 26-5 These diagrams represent an induced electric field (a), a magnetic field (b), and both electric and magnetic fields (c). Section 26.2 Electric and Magnetic Fields in Space 705 705-713 CH26-S2-845813 3/12/04 8:17 AM Page 706", " In 1860, Scottish physicist James Maxwell postulated that the opposite of induction is also true; that is, that a changing electric field produces a changing magnetic field. This is shown in Figure 26-5b on the previous page. Maxwell also suggested that charges were not necessary\u2014a changing electric field alone would produce the magnetic field. He then predicted that both accelerating charges and changing magnetic fields would produce electric and magnetic fields that move through space. A combined electric and magnetic field that travels through space is an electromagnetic wave, or EM wave. The orientations of the fields making up an electromagnetic wave are shown in Figure 26-5c on the previous page. In 1887, Heinrich Hertz, a German physicist, experimentally confirmed that Maxwell\u2019s theory was correct. Maxwell\u2019s theory led to a complete description of electricity and magnetism. Electromagnetic wave properties The speed of an electromagnetic wave later was found to be approximately 3.00108 m/s, now denoted as c, the speed of light. Light, a type of electromagnetic wave, and all other forms of electromagnetic waves, travel through space at c. The wavelength of an electromagnetic wave, its frequency, and the speed of light all are related. Wavelength-Frequency Relationship for a Wave v f The wavelength of a wave is equal to its speed divided by its frequency. In this equation, the wavelength,, is measured in m; the speed, v, is measured in m/s; and the frequency, f, is measured in Hz. Note that for an electromagnetic wave traveling in air or a vacuum, the speed, v, is equal to c, the speed of light. Thus, for an electromagnetic wave, the equation becomes the following: c f In the equation, c 3.00108 m/s. Note that in the wavelength-frequency equation, the product of frequency and wavelength is constant\u2014equal to c\u2014for any electromagnetic wave. Thus, as wavelength increases, frequency decreases, and vice-versa. In other words, an electromagnetic wave with a long wavelength has a low frequency, whereas an electromagnetic wave with a short wavelength has a high frequency. 15. What is the speed in air of an electromagnetic wave having a frequency of 3.21019 Hz? 16. What is the wavelength of green light having a frequency of 5.701014 Hz? 17. An electromagnetic wave has a frequency of 8.21014 Hz. What is the wavelength of the wave? 18. What is the frequency of an electromagnetic wave having a wavelength of 2.2102 m? 706 Chapter 26 Electromagnetism 705-713 CH26-S2-845813 3/12/04 8:17 AM Page 707 Electromagnetic wave propagation through matter Electromagnetic waves also can travel through matter. Sunlight shining through a glass of water is an example of light waves traveling through three different forms of matter: air, glass, and water. Air, glass, and water are nonconducting materials known as dielectrics. The velocity of an electromagnetic wave through a dielectric is always less than the speed of the wave in a vacuum, and it can be calculated using the following equation: c v K In this equation, the wave velocity, v, is measured in m/s; the speed of light, c, has a value of 3.00108 m/s; and the relative dielectric constant, K, is a dimensionless quantity. In a vacuum, K 1.00000, and the wave velocity is equal to c. In air, K 1.00054, and electromagnetic waves move just slightly slower than c. 19. What is the speed of an electromagnetic wave traveling through the air? Use c 299,792,458 m/s in your calculation. 20. For light traveling through water, the dielectric constant is 1.77. What is the speed of light traveling through water? 21. The speed of light traveling through a material is 2.43108 m/s. What is the dielectric constant of the material? Electromagnetic wave propagation through space The formation of an electromagnetic wave is shown in Figure 26-6. An antenna, which is a wire designed to transmit or receive electromagnetic waves, is connected to an alternating current (AC) source. The AC source produces a varying potential difference in the antenna that alternates at the frequency of the AC source. This varying potential difference generates a corresponding varying electric field that propagates away from the antenna. The changing electric field also generates a varying magnetic field perpendicular to the page. Although the magnetic field is not shown in Figure 26-6, it also propagates away from the antenna. The combined electric and magnetic fields are electromagnetic waves that spread out into space, moving at the speed of light. \u25a0 Figure 26-6 An alternating current source connected to the antenna produces a varying potential difference in the antenna. This varying potential difference generates a varying electric field (a). The varying electric field produces a changing magnetic field", " (not shown), and this magnetic field, in turn, generates an electric field. This process continues and the electromagnetic wave propagates away from the antenna, (b) and (c). a b c Antenna Electric field Section 26.2 Electric and Magnetic Fields in Space 707 705-713 CH26-S2-845813 3/12/04 8:18 AM Page 708 \u25a0 Figure 26-7 Portions of the electric and magnetic fields generated by an antenna might look like this at an instant in time (a). Note how the electric and magnetic fields are perpendicular to each other and to the direction of the wave velocity, v (b). Axis of polarization \u25a0 Figure 26-8 The illustration below provides examples of various types of electromagnetic radiation and their wavelengths. If it were possible to see invisible electromagnetic waves approaching, the changing fields would appear as in Figure 26-7. The electric field oscillates up and down, while the magnetic field oscillates at right angles to the electric field. Both of the fields are at right angles to the wave direction. Note that an electromagnetic wave produced by an antenna is polarized; that is, its electric field is parallel to the antenna\u2019s conductor. Radiation Sources The Electromagnetic Spectrum Type of Radiation Frequency (Hz) Wavelength (m) Detectable Objects 106 Hz (1 MHz) 108 Hz (1 GHz) 1010 Hz Radio waves Microwaves 1012 Hz Infrared 1014 Hz Light 1016 Hz Ultraviolet X rays Gamma rays 1018 Hz 1020 Hz 1022 Hz 1024 Hz 103 m (1 km) 100 m 10 m 1 m 1 cm 104 m (0.1 mm) 106 m (1 m) 109 m (1 nm) 1012 m (1 pm) 1015 House Baseball Bee Cell Virus Molecule Atom Nucleus Proton Radio antenna Klystron or magnetron Lamps and lasers Synchrotron radiation sources X-ray tubes Radioactive sources Particle accelerators 708 Chapter 26 Electromagnetism 705-713 CH26-S2-845813 6/8/04 4:26 PM Page 709 Visible light makes up only a very small portion of the entire electromagnetic spectrum. The wavelengths for some of the colors of visible light are shown in Table 26-1. 1. Which color of light has the longest wavelength? Table 26-1 Wavelengths of Visible Light Color Wavelength (nm) 2. Which color travels the fastest in a vacuum? Violet\u2013Indigo 3. Waves with longer wavelengths diffract around objects in their path more than waves with shorter wavelengths. Which color will diffract the most? The least? 4. Calculate the frequency range for each color of light given in Table 26-1. Blue Green Yellow Orange Red 390 to 455 455 to 492 492 to 577 577 to 597 597 to 622 622 to 700 Producing Electromagnetic Waves Waves from an AC source As you just learned, an AC source connected to an antenna can transmit electromagnetic waves. The wave frequency is equal to the frequency of the rotating AC generator and is limited to about 1 kHz. The range of frequencies and wavelengths that make up all forms of electromagnetic radiation is shown in Figure 26-8 and is called the electromagnetic spectrum. Waves from a coil and a capacitor A common method of generating high-frequency electromagnetic waves is to use a coil and a capacitor connected in a series circuit. If the capacitor is charged by a battery, the potential difference across the capacitor creates an electric field. When the battery is removed, the capacitor discharges as the stored electrons flow through the coil, creating a magnetic field. When the capacitor is discharged, the coil\u2019s magnetic field collapses. A back-EMF then develops and recharges the capacitor in the opposite direction, and the process is repeated. When an antenna is connected across the capacitor, the fields of the capacitor are transmitted into space. One complete oscillation cycle is shown in Figure 26-9. I B strongest I I E strongest D i r e ction of cycle E strongest I I B strongest I \u25a0 Figure 26-9 One complete oscillation cycle for a coil and capacitor circuit is shown. The capacitor and coil sizes determine the oscillations per second of the circuit, which equals the frequency of the waves produced. Section 26.2 Electric and Magnetic Fields in Space 709 705-713 CH26-S2-845813 3/12/04 8:19 AM Page 710 The process occurring in the coil-and-capacitor circuit can be compared with the cyclic oscillations of a swinging pendulum, as shown in Figure 26-10. Assume that the electrons in the coil and capacitor are represented by the pendulum\u2019s bob. The moving bob has the greatest speed at the bottom of its swing, a position at which kinetic energy, KE, is maximized, and potential energy, PE, due to gravity is zero. This point in the pendulum\u2019", "s motion, shown in Figure 26-10a, is similar to the peak electric current flow in the coil when the charge on the capacitor is zero. When the bob reaches the peak of its swing, its vertical displacement and PE are maximized, whereas its KE is zero because the bob\u2019s velocity is zero. As shown in Figure 26-10b, this point in the motion is similar to when the capacitor holds the maximum charge and the current through the coil is zero. Energy in the coil-and-capacitor circuit As you just learned, the PE of the pendulum is largest when its vertical displacement is greatest, and the KE is largest when the velocity is greatest. The sum of the PE and KE\u2014the total energy\u2014is constant throughout the motion of the pendulum. In the coil-and-capacitor circuit, both the magnetic field produced by the coil and the electric field in the capacitor contain energy. When the current is largest, the energy stored in the magnetic field is greatest. When the current is zero, the electric field of the capacitor is largest, and all the energy is contained in the electric field. The total energy of the circuit (the sum of the magnetic field energy, the electric field energy, the thermal losses, and the energy carried away by the generated electromagnetic waves) is constant. Energy that is carried, or radiated, in the form of electromagnetic waves is frequently called electromagnetic radiation. Just as a pendulum eventually stops swinging if it is left alone, the oscillations in a coil and capacitor die out over time due to resistance in the circuit. The oscillations of both systems can be made to continue by adding energy. Gentle pushes, applied at the correct times, will keep a pendulum swinging. The largest amplitude swings occur when the frequency of pushes matches the frequency of swinging motion. This is the condition of resonance, which was discussed in Chapter 14. Similarly, voltage pulses applied to the coil-and-capacitor circuit at the right frequency keep the oscillations in the circuit going. One way of doing this is to add a second coil to the circuit, to form a transformer. A transformer is shown in Figure 26-11. The alternating current induced in the secondary coil is increased by an amplifier and added back to the coil and capacitor. This type of circuit can produce frequencies up to approximately 400 MHz. a PE 0 KE max b PE max KE 0 Energy stored in coil\u2019s magnetic field I B Energy stored in capacitor\u2019s electric field E Frequencies The Federal Communications Commission (FCC) assigns each radio and TV station a carrier wave with a specific frequency. A station broadcasts by varying its carrier wave. When the wave is received by your radio or TV set, the carrier wave is stripped away and the information from the wave is processed so that you can see or hear it. \u25a0 Figure 26-10 The motion of a pendulum is analogous to the action of electrons in a coil-andcapacitor circuit. Motion of the pendulum\u2019s bob is analogous to the current flow in the circuit (a). The point where the pendulum\u2019s motion comes to a stop is analogous to zero current in the circuit (b). 710 Chapter 26 Electromagnetism 705-713 CH26-S2-845813 6/8/04 4:26 PM Page 711 Secondary coil Primary coil Capacitor B Amplifier \u25a0 Figure 26-11 In a transformer, the amplified oscillation from the secondary coil is in resonance with the coil-and-capacitor circuit and keeps the oscillations going. Waves from a resonant cavity The oscillation frequency produced by a coil-and-capacitor circuit can be increased by decreasing the size of the coil and capacitor used. However, above frequencies of 1 GHz, individual coils and capacitors can no longer be used. High frequency microwaves, with frequencies from 1 GHz to 100 GHz, are produced using a resonant cavity. The resonant cavity is a rectangular box that acts as both a coil and a capacitor. The size of the box determines the frequency of oscillation. Microwave ovens have resonant cavities that produce the microwaves used to cook food. To produce even higher frequency infrared waves, the size of the resonant cavity would have to be reduced to molecular size. The oscillating electrons that produce infrared waves are, in fact, within the molecules. Visible and ultraviolet waves are generated by electrons within atoms. X rays and gamma rays are the result of accelerating charges in the nuclei of atoms. All electromagnetic waves arise from accelerated charges, and all travel at the speed of light. Waves from piezoelectricity Coils and capacitors are not the only method of generating oscillation voltages. Quartz crystals deform when a voltage is applied across them, a property known as piezoelectricity. The application of an AC voltage to a cut section of quartz crystal results in sustained oscillations. An inverse linear relationship exists between crystal thickness", " and oscillation frequency. Just as a piece of metal vibrates at a specific frequency when it is bent and released, so does a quartz crystal. A crystal can be cut so that it vibrates at a specific desired frequency. An applied voltage deforms the crystal and starts the vibrations. The piezoelectric property also generates an EMF when the crystal is deformed. Because this EMF is produced at the vibrating frequency of the crystal, it can be amplified and returned to the crystal to keep it vibrating. Because of their nearly constant frequencies of vibration, quartz crystals commonly are used in watches. Section 26.2 Electric and Magnetic Fields in Space 711 Bert Krages/Visuals Unlimited 705-713 CH26-S2-845813 3/12/04 8:20 AM Page 712 E Capacitor Coil Antenna Amplifier \u25a0 Figure 26-12 The changing electric fields from a radio station signal cause electrons in the antenna to accelerate. The information carried by the signal is then decoded and amplified and used to drive a loudspeaker. Biology Connection Biology Connection 712 Chapter 26 Electromagnetism Reception of Electromagnetic Waves Now that you know how electromagnetic waves are produced and transmitted, how do you suppose the waves are detected? As you may have guessed, reception involves an antenna. As shown in Figure 26-12, the wave\u2019s electric fields accelerate the electrons of the material making up the antenna. The acceleration is largest when the antenna is positioned in the same direction as the wave polarization; that is, when it is parallel to the direction of the wave\u2019s electric fields. A potential difference across the terminals of the antenna oscillates at the frequency of the electromagnetic wave. This voltage is largest when the length of the antenna is one-half the wavelength of the wave it is to detect. Thus, an antenna\u2019s length is designed to be one-half of the wavelength of the wave it is supposed to receive. For this reason, an antenna designed to receive radio and television waves is much longer than one designed to receive microwaves. While a simple wire antenna can detect electromagnetic waves, several wires are more effective. A television antenna often consists of two or more wires spaced about one-quarter wavelength apart. Electric fields that are generated in the individual wires form constructive interference patterns that increase the strength of the signal. It is important to realize that all electromagnetic waves, not just visible light waves, undergo reflection, refraction, and diffraction. Thus, it should not be a surprise to learn that dish antennas, like the one shown at the beginning of this chapter, reflect very short wavelength electromagnetic signals, just as parabolic mirrors reflect visible light waves. A dish antenna\u2019s large surface area for collecting and focusing waves makes it well-suited to receive weak radio signals. A parabolic dish antenna works by reflecting and focusing the received signals off its surface and into a device called the horn. The horn, which is supported by a tripod structure over the main dish, contains a short dipole antenna. The horn channels the signals to a receiver, a device consisting of an antenna, a coil-and-capacitor circuit, a detector to decode the signal, and an amplifier. Selection of waves As you know, many different radio and television stations transmit electromagnetic waves at the same time. If the information being broadcast is to be understood, the waves of a particular station must be selected. To select waves of a particular frequency (and reject the others) a tuner uses a coil-and-capacitor circuit connected to an antenna. The capacitance is adjusted until the oscillation frequency of the circuit equals the frequency of the desired wave. When this is done, only waves of the desired frequency can cause significant oscillations of the electrons in the circuit. Energy from waves Waves carry energy as well as information. At microwave and infrared frequencies, waves accelerate electrons in molecules. The energy of the waves is converted to thermal energy in the molecules. This is how microwave ovens cook food. Light waves also can transfer energy to electrons. Photographic film makes use of this fact by using the energy in light waves to drive a chemical reaction within the film. The result is a permanent record of the light from the subject that strikes the film. At higher frequencies, ultraviolet (UV) radiation causes many chemical reactions to occur, including those in living cells that produce sunburn and tanning. 705-713 CH26-S2-845813 3/12/04 8:21 AM Page 713 X Rays In 1895, German physicist Wilhelm Roentgen sent electrons through an evacuated glass tube, similar to the one shown in Figure 26-13. Roentgen used a very high voltage across the tube to give the electrons large kinetic energies. When the electrons struck the metal anode target within the tube, Roentgen noticed a glow on a phosphorescent screen a short distance away. The glow continued even when a piece of wood was placed between the tube and the screen. He", " concluded that some kind of highly penetrating rays were coming from the tube. Because Roentgen did not know what these strange rays were, he called them X rays. A few weeks later, Roentgen found that photographic plates were darkened by X rays. He also discovered that soft body tissue was transparent to the rays, but that bone blocked them. He produced an X-ray picture of his wife\u2019s hand. Within months, doctors recognized the valuable medical uses of this phenomenon. It now is known that an X ray is a high-frequency electromagnetic wave. In an X-ray tube, electrons first are accelerated to high speeds by means of potential differences of 20,000 V or more. When the electrons crash into matter, their kinetic energies are converted into the very high-frequency electromagnetic waves called X rays. Electrons are accelerated to these speeds in cathode-ray tubes, such as the picture tube in a television. When the electrons hit the inside surface of a television screen\u2019s face plate, they come to a sudden stop and cause the colored phosphors to glow. This sudden stopping of the electrons also can produce harmful X rays. Thus, the face-plate glass in a television screen contains lead to stop the X rays and protect the viewers. High-voltage cathode Electrons X rays Metal target anode \u25a0 Figure 26-13 X rays are emitted when high energy electrons strike the metal target inside the X-ray tube. The target can be changed to produce X rays of different wavelengths. 26.2 Section Review 22. Wave Propagation Explain how electromagnetic waves are able to propagate through space. 23. Electromagnetic Waves What are some of the primary characteristics of electromagnetic waves? Do electromagnetic waves behave differently from the way that other waves, such as sound waves, behave? Explain. 24. Frequency An electromagnetic wave is found to have a wavelength of 1.5105 m. What is the frequency of the wave? 25. TV Signals Television antennas normally have metal rod elements that are oriented horizontally. From this information, what can you deduce about the directions of the electric fields in television signals? 26. Parabolic Receivers Why is it important for a parabolic dish\u2019s receiving antenna to be properly aligned with the transmitter? 27. Antenna Design Television channels 2 through 6 have frequencies just below the FM radio band, while channels 7 through 13 have much higher frequencies. Which signals would require a longer antenna: those of channel 7 or those of channel 6? Provide a reason for your answer. 28. Dielectric Constant The speed of light traveling through an unknown material is 1.98108 m/s. Given that the speed of light in a vacuum is 3.00108 m/s, what is the dielectric constant of the unknown material? 29. Critical Thinking Most of the UV radiation from the Sun is blocked by the ozone layer in Earth\u2019s atmosphere. In recent years, scientists have discovered that the ozone layer over both Antarctica and the Arctic Ocean is thinning. Use what you have learned about electromagnetic waves and energy to explain why some scientists are very concerned about the thinning ozone layer. physicspp.com/self_check_quiz Section 26.2 Electric and Magnetic Fields in Space 713 714-715 CH26-LAB-845813 6/8/04 4:30 PM Page 714 Alternate CBL instructions can be found on the Web site. physicspp.com Electromagnetic Wave Shielding The electromagnetic spectrum consists of many types of electromagnetic radiation, each of which can be classified by its frequency or wavelength. Gamma rays have the highest frequency and energy, and have wavelengths that are a fraction of a nanometer. Following gamma rays in order of increasing wavelength (decreasing frequency and energy) are the following forms of electromagnetic radiation: X rays, ultraviolet, visible, infrared, microwaves, and radio waves. Only wavelengths that fall within the range of visible light can be detected by the human eye\u2014all other forms of radiation are invisible. Electromagnetic receivers, such as those used in radios and televisions, detect waves with an antenna. Because every electrical device with a changing or alternating current radiates electromagnetic waves, waves from these sources can interfere with the reception of a desired signal. Some materials are effective at blocking or shielding radio waves. In this lab you will investigate the radio wave shielding effectiveness of various materials. QUESTION What types of materials shield electromagnetic waves? Objectives Materials \u25a0 Experiment with various materials to determine if they are effective at shielding electromagnetic waves. \u25a0 Observe and infer about types of materials that shield radio waves. \u25a0 Collect and organize data on types of shielding. Safety Precautions small battery-operated AM/FM radio two small cardboard boxes metal box or can with lid aluminum foil static shielding bag (type used to protect computer parts) metal screen masking tape leather gloves stapler \u25a0 Always wear safety goggles and a lab apron. \u25a0 Wear gloves when bending or handling Procedure the wire screen. \u25a0 Use caution when working with", " staples to avoid puncturing skin. 1. Prepare the aluminum covered box. Cover the outside of one box and its lid with aluminum foil. Cover the lid separately from the rest of the box so the lid can be removed and replaced. 2. Prepare the wire screen box. Fold a piece of wire screen so that it forms a four-sided box shape with open ends. Use staples to hold connecting edges of the screen together. Make sure the wire box is big enough for the radio to fit inside. Next, cut a piece of wire screen to fit over each open end. Staple one of these pieces in place over one of the ends, making sure to leave no openings. Then, staple one edge of the remaining piece of wire screen to the other end of the wire box. This piece of screen will act as a door that can be opened and closed. 714 Horizons Companies 714-715 CH26-LAB-845813 6/8/04 4:31 PM Page 715 Data Table Band Frequency (Hz) Enclosure Observations Band Frequency (Hz) Enclosure Observations AM AM AM AM AM AM A person\u2019s arms Cardboard box Cardboard box covered with aluminum foil Wire screen box Metal box Static shielding bag FM FM FM FM FM FM A person\u2019s arms Cardboard box Cardboard box covered with aluminum foil Wire screen box Metal box Static shielding bag 3. Turn on the radio and tune it to a strong signal from a station in the AM band. Record the frequency of the station. The frequency can be determined from the radio dial or display, or the station\u2019s broadcast frequency may be announced while you are listening. 4. Hold the radio next to your body and cover it with your arms. Ignoring the fact than the sound is muffled because you are covering the speaker, how has the reception of the signal been affected? Record your observations. 5. Place the radio inside a cardboard box and put the lid on the box. Listen to the radio\u2019s reception and record your observations. 6. Repeat step 5 four more times using the aluminum foil covered box, the wire screen box (with the door closed), the metal box (with lid on), and the static shielding bag, respectively. 7. Change the radio to the FM band and tune in a strong station. Record the frequency of the station and then repeat steps 4\u20136. Analyze 1. Summarize Which materials were effective at shielding radio waves? 2. Use Numbers Calculate the wavelengths for each of the radio frequencies you used. Recall that c f, where the velocity, c, of the electromagnetic waves is 3.00108 m/s. 3. Compare and Contrast How does the wavelength of the tuned-in radio signal compare to the size of holes or openings in the materials used to shield the radio? 4. Interpret Data What did the materials that effectively shielded the radio from receiving signals have in common? Conclude and Apply 1. Explain Offer an explanation of what might be happening to the electric and magnetic fields of the radio wave fields that are blocked from reaching the radio by the shielding materials. 2. Infer Why was covering the radio with your arms not effective at blocking the radio waves? 3. Use Scientific Explanations Ocean water absorbs radio waves, limiting their penetration below the surface to a depth equal to approximately one wavelength. Because of this, very low frequency (40\u201380 Hz) radio waves are used to contact submerged submarines. Why might the site for a high-powered radio transmitter used for submarine communication be located in a remote area far from the ocean? (Hint: Estimate the length if a one-half wavelength antenna were used.) Going Further How does the size of the small holes in the metal screen of a microwave oven door compare to the wavelength of a 2.4-GHz microwave? Real-World Physics Suppose you want to mail some photos or an audio recording stored on a magnetic computer diskette to a friend. What should you do to protect the diskette from electromagnetic waves during shipping? To find out more about electromagnetic waves, visit the Web site: physicspp.com 715 716 CH26-FEATURE-845813 3/12/04 8:28 AM Page 716 Cellular Phones Cellular Phones Network of Base Stations Do you own a cellular phone? Once rare and expensive, cell phones are now commonplace and relatively affordable. Approximately 20 percent of American teenagers have one. Cellular Networks The cell phone gets its name from the way cell-phone companies divide a city into small regions called cells. Each cell is a hexagon-shaped zone within a larger hexagonal grid. Cells are typically about 26 square kilometers in area, though they vary in size depending on terrain and the number of cell phone users. Located within each cell is a base station, consisting of a tall tower-like structure and boxes or buildings that contain radio equipment. When you make a call, the signal from your phone is transmitted to the base station located within your cell. This signal then is transmitted", " from the local base station to the base station where the person you are calling is located. Cell How then, do cell phones communicate with base stations? Cell phones use radio waves to transmit information to and receive information from base stations. A cell phone is essentially a two-way radio containing both a radio transmitter and a radio receiver. The cell-phone transmitter takes the sound of your voice, encodes it onto a radio-frequency wave and then transmits the radio wave to a nearby base station. The base station then relays the signal to the destination base station by broadcasting radio waves through the air. On the receiving end of the call, the cell phone picks up the radio signal, decodes it, and reconverts it into audible sounds you can understand. By making use of two different frequencies (one frequency for talking and one frequency for listening) both people connected by the call can talk at the same time. 716 Technology and Society A cellular phone company\u2019s system of base stations can relay your call all the way across the country, even when both people involved in the call are moving. When you move from cell to cell, the signal is automatically relayed to the correct base station in the system. Risks of Cell Phone Use Using a cell phone is not a risk-free activity. For example, driving while talking on a cell phone is dangerous. One study found that people talking on their cell phones while driving were four times more likely to be involved in an accident than non-cell phone users. Some service stations post warning signs prohibiting cell phone use. Static electricity generated by the cell phone might cause the gasoline fumes to ignite. Another potential risk is more controversial. Because cell phones broadcast radio waves, they emit electromagnetic energy known as radio-frequency (RF) energy. There is some evidence that cell phones emit enough radiation to cause severe health problems, such as brain cancer and Base station Alzheimer\u2019s disease. There is evidence supporting both sides of the debate. Currently, no one knows for certain what the long-term health effects of cell phone use are, if any. Going Further 1. Use Scientific Explanations How did cellular phones get their name? 2. Compare and Contrast How are an AM/FM radio and a cellular phone similar? How are they different? 3. Critical Thinking Explain why the low-power transmitters used by cell phones are important in keeping the phones lightweight. 717-721 CH26-SG CA STP-845813 6/8/04 4:46 PM Page 717 26.1 Interactions of Electric and Magnetic Fields and Matter Vocabulary \u2022 isotope (p. 701) \u2022 mass spectrometer (p. 701) Key Concepts \u2022 The ratio of charge to mass of an electron was measured by J. J. Thomson using balanced electric and magnetic fields in a cathode-ray tube. q v B r m \u2022 An electron\u2019s mass can be found by combining Thomson\u2019s result with Millikan\u2019s measurement of the electron\u2019s charge. \u2022 Atoms of the same element can have different masses. \u2022 The mass spectrometer uses both electric and magnetic fields to measure the masses of ionized atoms and molecules. \u2022 The mass spectrometer can be used to determine the charge-to-mass ratio of an ion. V 2 q 2 r2 B m 26.2 Electric and Magnetic Fields in Space Vocabulary \u2022 electromagnetic wave Key Concepts \u2022 Electromagnetic waves are coupled, changing electric and magnetic fields (p. 706) \u2022 dielectrics (p. 707) \u2022 antenna (p. 707) \u2022 electromagnetic spectrum (p. 709) \u2022 electromagnetic radiation (p. 710) \u2022 piezoelectricity (p. 711) \u2022 receiver (p. 712) that move through space. \u2022 The wavelength of a wave is equal to its speed divided by its frequency. v f For an electromagnetic wave traveling in a vacuum, the speed in the above equation, v, is equal to the speed of light, c. \u2022 The velocity of an electromagnetic wave through a dielectric is less than the speed of light in a vacuum. \u2022 A changing current in a transmitting antenna is used to generate electromagnetic waves. \u2022 Electromagnetic radiation can transmit energy and information through a medium or a vacuum. \u2022 Piezoelectricity is the property of a crystal causing it to bend or deform and produce electrical vibrations when a voltage is applied across it. \u2022 Receiving antennas convert electromagnetic waves to varying electric fields in conductors. \u2022 Electromagnetic waves can be detected by the EMF that they produce in an antenna. Particular frequencies of electromagnetic waves can be selected by using a resonating coil-and-capacitor circuit, known as a tuner. \u2022 A receiver obtains transmitted information from electromagnetic waves. \u2022 The length of the most efficient antenna is one-half the wavelength of the wave to be detected. \u2022 Microwave and infrared waves can accelerate electrons in molecules, thereby producing thermal energy", ". \u2022 X rays are high-frequency electromagnetic waves emitted by rapidly accelerated electrons. physicspp.com/vocabulary_puzzlemaker 717 717-721 CH26-SG CA STP-845813 6/8/04 4:46 PM Page 718 Concept Mapping 30. Complete the following concept map using the following term and symbols: E, c, magnetic field. Electromagnetic waves electric field travel at the speed of light B Mastering Concepts 31. What are the mass and charge of an electron? (26.1) 32. What are isotopes? (26.1) 33. The direction of an induced magnetic field is always at what angle to the changing electric field? (26.2) 34. Why must an AC generator be used to propagate electromagnetic waves? If a DC generator were used, when would it create electromagnetic waves? (26.2) 35. A vertical antenna wire transmits radio waves. Sketch the antenna and the electric and magnetic fields that it creates. (26.2) 36. What happens to a quartz crystal when a voltage is applied across it? (26.2) 37. How does an antenna\u2019s receiving circuit select electromagnetic radio waves of a certain frequency and reject all others? (26.2) Applying Concepts 38. The electrons in a Thomson tube travel from left to right, as shown in Figure 26-14. Which deflection plate should be charged positively to bend the electron beam upward? V \u25a0 Figure 26-14 Upper charged plate Lower charged plate 39. The Thomson tube in question 38 uses a magnetic field to deflect the electron beam. What would the direction of the magnetic field need to be to bend the beam downward? 40. Show that the units of E/B are the same as the units for velocity. 41. The vacuum chamber of a mass spectrometer is shown in Figure 26-15. If a sample of ionized neon is being tested in the mass spectrometer, in what direction must the magnetic field be directed to bend the ions into a clockwise semicircle? Vacuum chamber Deflected paths Film plate Dual slits Ion beam \u25a0 Figure 26-15 42. If the sign of the charge on the particles in question 41 is changed from positive to negative, do the directions of either or both of the fields have to be changed to keep the particles undeflected? Explain. 43. For each of the following properties, identify whether radio waves, light waves, or X rays have the largest value. a. wavelength b. frequency c. velocity 44. TV Waves The frequency of television waves broadcast on channel 2 is about 58 MHz. The waves broadcast on channel 7 are about 180 MHz. Which channel requires a longer antenna? 45. Suppose the eyes of an alien being are sensitive to microwaves. Would you expect such a being to have larger or smaller eyes than yours? Why? 718 Chapter 26 Electromagnetism For more problems, go to Additional Problems, Appendix B. 717-721 CH26-SG CA STP-845813 6/8/04 4:47 PM Page 719 Mastering Problems 26.1 Interactions of Electric and Magnetic Fields and Matter 46. Electrons moving at 3.6104 m/s pass through an electric field with an intensity of 5.8103 N/C. How large a magnetic field must the electrons also experience for their path to be undeflected? 47. A proton moves across a 0.36-T magnetic field, as shown in Figure 26-16. If the proton moves in a circular path with a radius of 0.20 m, what is the speed of the proton \u25a0 Figure 26-16 Path of charged particle X X X X X X 53. Silicon Isotopes In a mass spectrometer, ionized silicon atoms have curvatures, as shown in Figure 26-17. If the smaller radius corresponds to a mass of 28 proton masses, what is the mass of the other silicon isotope? Vacuum chamber 16.23 cm 17.97 cm r1 r2 r2 r1 Film plate Film plate Dual slits Silicon ion beam \u25a0 Figure 26-17 48. A proton enters a 6.0102-T magnetic field with a speed of 5.4104 m/s. What is the radius of the circular path it follows? 49. An electron is accelerated by a 4.5-kV potential difference. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.0 cm? 50. A mass spectrometer yields the following data for a beam of doubly ionized (2) sodium atoms: B 8.0102 T, q 2(1.601019 C), r 0.077 m, and V 156 V. Calculate the mass of a sodium atom. 51. An alpha particle has a mass of approximately 6.61027 kg and has a charge of 2. Such a particle", " is observed to move through a 2.0-T magnetic field along a path of radius 0.15 m. a. What speed does the particle have? b. What is its kinetic energy? c. What potential difference would be required to give it this kinetic energy? 52. A mass spectrometer analyzes carbon-containing molecules with a mass of 175103 proton masses. What percent differentiation is needed to produce a sample of molecules in which only carbon isotopes of mass 12, and none of mass 13, are present? 26.2 Electric and Magnetic Fields in Space 54. Radio Waves The radio waves reflected by a parabolic dish are 2.0 cm long. How long should the antenna be that detects the waves? 55. TV A television signal is transmitted on a carrier frequency of 66 MHz. If the wires on a receiving antenna are placed 1 apart, determine the physical 4 distance between the receiving antenna wires. 56. Bar-Code Scanner A bar-code scanner uses a laser light source with a wavelength of about 650 nm. Determine the frequency of the laser light source. 57. What is the optimum length of a receiving antenna that is to receive a 101.3-MHz radio signal? 58. An EM wave with a frequency of 100-MHz is transmitted through a coaxial cable having a dielectric constant of 2.30. What is the velocity of the wave\u2019s propagation? 59. Cell Phone A certain cellular telephone transmitter operates on a carrier frequency of 8.00108 Hz. What is the optimal length of a cell phone antenna designed to receive this signal? Note that singleended antennas, such as those used by cell phones, generate peak EMF when their length is one-fourth the wavelength of the wave. physicspp.com/chapter_test Chapter 26 Assessment 719 717-721 CH26-SG CA STP-845813 6/8/04 4:48 PM Page 720 Mixed Review 60. The mass of a doubly ionized (2) oxygen atom is found to be 2.71026 kg. If the mass of an atomic mass unit (amu) is equal to 1.671027 kg, how many atomic mass units are in the oxygen atom? 61. Radio An FM radio station broadcasts on a frequency of 94.5 MHz. What is the antenna length that would give the best reception for this station? 62. At what frequency does a cell phone with an 8.3-cm-long antenna send and receive signals? Recall from question 59 that single-ended antennas, such as those used by cell phones, generate peak EMF when their length is one-fourth the wavelength of the wave they are broadcasting or receiving. 63. An unknown particle is accelerated by a potential difference of 1.50102 V. The particle then enters a magnetic field of 50.0 mT, and follows a curved path with a radius of 9.80 cm. What is the ratio of q/m? Thinking Critically 64. Apply Concepts Many police departments use radar guns to catch speeding drivers. A radar gun is a device that uses a high-frequency electromagnetic signal to measure the speed of a moving object. The frequency of the radar gun\u2019s transmitted signal is known. This transmitted signal reflects off of the moving object and returns to the receiver on the radar gun. Because the object is moving relative to the radar gun, the frequency of the returned signal is different from that of the originally transmitted signal. This phenomenon is known as the Doppler shift. When the object is moving toward the radar gun, the frequency of the returned signal is greater than the frequency of the original signal. If the initial transmitted signal has a frequency of 10.525 GHz and the returned signal shows a Doppler shift of 1850 Hz, what is the speed of the moving object? Use the following equation: vtarget c f o r e l p D p 2f m s an r t i t t ed Where, vtarget velocity of target (m/s) c speed of light (m/s) Doppler shift frequency (Hz) frequency of transmitted wave (Hz) fDoppler ftransmitted 65. Apply Concepts H. G. Wells wrote a science-fiction novel called The Invisible Man, in which a man drinks a potion and becomes invisible, although he retains all of his other faculties. Explain why an invisible person would not be able to see. 66. Design an Experiment You are designing a mass spectrometer using the principles discussed in this chapter, but with an electronic detector replacing the photographic film. You want to distinguish singly ionized (1) molecules of 175 proton masses from those with 176 proton masses, but the spacing between adjacent cells in your detector is 0.10 mm. The molecules must have been accelerated by a potential difference of at least 500.0 V to be detected. What are some of the values of V, B, and r that your apparatus should have? Writing in Physics 67. Compose a 1", "\u20132 page report in which you outline the operation of a typical television, DVD, or VCR infrared remote-control unit. Explain why the simultaneous use of multiple remote-control units typically does not cause the units to interfere with each other. Your report should include block diagrams and sketches. Cumulative Review 68. A He\u2013Ne laser ( 633 nm) is used to illuminate a slit of unknown width, forming a pattern on a screen that is located 0.95 m behind the slit. If the first dark band is 8.5 mm from the center of the central bright band, how wide is the slit? (Chapter 19) 69. The force between two identical metal spheres with the charges shown in Figure 26-18 is F. If the spheres are touched together and returned to their original positions, what is the new force between them? (Chapter 20) 4q 2q \u25a0 Figure 26-18 70. What is the electric field strength between two parallel plates spaced 1.2 cm apart if a potential difference of 45 V is applied to them? (Chapter 21) 71. Calculate the daily cost of operating an air compressor that runs one-fourth of the time and draws 12.0 A from a 245-V circuit if the cost is $0.0950 per kWh. (Chapter 22) 72. A 440-cm length of wire carrying 7.7 A is at right angles to a magnetic field. The force on the wire is 0.55 N. What is the strength of the field? (Chapter 24) 73. A north-south wire is moved toward the east through a magnetic field that is pointing down, into Earth. What is the direction of the induced current? (Chapter 25) 720 Chapter 26 Electromagnetism For more problems, go to Additional Problems, Appendix B. 721 CH26-SG CA STP-845813 3/12/04 8:36 AM Page 721 Multiple Choice 1. For a charged particle moving in a circular trajectory, _______. 5. Which one of the following situations does not create an electromagnetic wave? the magnetic force is parallel to the velocity and is directed toward the center of the circular path the magnetic force may be perpendicular to the velocity and is directed away from the center of the circular path the magnetic force always remains parallel to the velocity and is directed away from the center of the circular path the magnetic force always remains perpendicular to the velocity and is directed toward the center of the circular path 2. The radius of the circular path that a proton travels while in a constant 0.10-T magnetic field is 6.6 cm. What is the velocity of the proton? 6.3105 m/s 2.0106 m/s 6.3107 m/s 2.01012 m/s Direct current (DC) voltage is applied to a piezoelectric quartz crystal. Current passes through a wire contained inside a plastic pipe. Current passes through a coil-and-capacitor circuit with a molecular-size resonant cavity. High energy electrons strike a metal target in an X-ray tube. 6. A proton beam has a radius of 0.52 m as it moves perpendicular to a magnetic field of 0.45 T. If the mass of an individual proton is 1.671027 kg, what is the speed of the protons making up the beam? 1.2 m/s 4.7103 m/s 2.2107 m/s 5.8108 m/s x x x x x m 6..10 T x x Path of proton x x x x x 2 cm 0.5 r x x x x x B 0.45 T x x Path of proton beam 3. The dielectric constant of ruby mica is 5.4. What is the speed of light as it passes through ruby mica? 3.2103 m/s 9.4104 m/s 5.6107 m/s 1.3108 m/s 4. A certain radio station broadcasts with waves that are 2.87 m long. What is the frequency of the radio waves? 9.57109 Hz 3.48101 Hz 1.04108 Hz 3.00108 Hz Extended Answer 7. A deuteron (the nucleus of deuterium) has a mass of 3.341027 kg and a charge of e. It travels in a magnetic field of 1.50 T in a circular path, with a radius of 0.0400 m. What is the velocity of the particle? Watch the Little Words Underline words such as never, always, least, not, and except when you see them in test questions. These small words dramatically impact the meaning of a question. physicspp.com/standardized_test Chapter 26 Standardized Test Practice 721 722 Replace CH27-S1-845813 3/22/04 2:57 PM Page 722 What You\u2019ll Learn \u2022 You will understand that light behaves like", " particles having momentum and energy. \u2022 You will learn that small particles of matter behave like waves and are subject to diffraction and interference. Why It\u2019s Important Quantum theory provides the basis for an amazing device called a scanning tunneling microscope (STM). The STM is vital to researchers studying DNA and chemical reaction mechanisms. It also is used in the development of smaller and faster computers. Atomic-Scale Images Two types of silicon atoms, appearing in red and blue, are seen in this STM image of silicon. Think About This An STM was used by The Colorado School of Mines to produce this image of the surface of silicon. The STM uses the ability of electrons to jump across a barrier. How does this jump, which is impossible according to the law of conservation of energy, occur? physicspp.com 722 photo courtsey of Peter Sutter/Colorado School of Mines 722-734 CH27-S1-845813 3/12/04 6:43 AM Page 723 What does the spectrum of a glowing lightbulb look like? Question What colors of visible light are emitted by a glowing, incandescent lightbulb? Procedure 1. Screw a clear, incandescent lightbulb into a lamp socket. 2. Plug the lamp into an electrical outlet that is controlled by a dimmer switch. Turn the lamp on and dim it so that it glows weakly. CAUTION: Do not touch the glowing bulb, as it is very hot and can cause burns. 3. Dim or turn off the other lights in the classroom. 4. Standing about 1-2 m away from the lightbulb, hold a holographic diffraction grating close to your eye. Observe the lightbulb through the diffraction grating. CAUTION: Do not directly view the glowing lightbulb without using the diffraction grating, as damage to your vision may result. 5. Make and Use Scientific Illustrations Use colored pencils to sketch a diagram of what you observe. 6. Turn up the dimmer control to increase the lightbulb to its maximum brightness. 7. Make and Use Scientific Illustrations Use colored pencils to sketch a diagram of what you observe. Analysis Describe the spectrum emitted by the lightbulb. Is it continuous or a series of distinct colored lines? Describe how the observed spectrum changed when the lightbulb glowed brighter. Critical Thinking What is the source of the light emitted by the bulb? What happens to the temperature of the lightbulb\u2019s filament when the bulb glows brighter? 27.1 A Particle Model of Waves James Maxwell\u2019s electromagnetic wave theory, which you learned about in the previous chapter, was proven to be correct by the experiments of Heinrich Hertz in 1889. Light was then firmly established as an electromagnetic wave. All of optics, including phenomena such as interference, diffraction, and polarization, seemed to be explainable in terms of the electromagnetic wave theory. Problems remained for physicists, however, because Maxwell\u2019s notion of light as a purely electromagnetic wave could not explain several other important phenomena. These problems generally involved the absorption or emission of electromagnetic radiation. Two such problems were the emission spectrum given off by a hot body (hot object) and the discharge of electrically charged particles from a metal surface when ultraviolet radiation was incident upon it. As you will learn, these phenomena could be explained once it was understood that electromagnetic waves have particle-like properties in addition to wavelike properties. Objectives \u2022 Describe the spectrum emitted by a hot body. \u2022 Explain the photoelectric and Compton effects. \u2022 Solve problems involving the photoelectric effect. Vocabulary emission spectrum quantized photoelectric effect threshold frequency photon work function Compton effect Section 27.1 A Particle Model of Waves 723 Horizons Companies 722-734 CH27-S1-845813 3/12/04 6:44 AM Page 724 Radiation from Incandescent Bodies Why was the radiation emitted from a hot body puzzling to physicists? The problem had to do with the intensity and frequency of the emitted radiation at different temperatures. Maxwell\u2019s electromagnetic wave theory could not account for the observed radiation emissions of hot bodies. What then, is the nature of the radiation emitted by hot bodies? The lightbulb that you observed in the Launch Lab at the start of this chapter is an example of a hot body. As predicted by electromagnetic theory, light and infrared radiation are emitted by the vibrating charged particles within the filament. The filament glows because it is hot, and it is said to be incandescent; hence, the name incandescent lightbulb. The colors that you see depend upon the relative intensities of the emitted electromagnetic waves of various frequencies, and the sensitivity of your eyes to those waves. When the dimmer control is used to increase the voltage to the bulb, the temperature of the glowing filament increases. As a result, the color changes from deep red to orange to yellow and finally,", " to white. This color change occurs because the higher-temperature filament emits higher-frequency radiation. The higher-frequency radiation comes from the higher-frequency end of the visible spectrum (the violet end) and results in the filament appearing to be whiter. What would you expect to see if you viewed the glowing filament through a diffraction grating? When viewed in this way, all of the colors of the rainbow would be visible. The bulb also emits infrared radiation that you would not see. A plot of the intensity of the light emitted from a hot body over a range of frequencies is known as an emission spectrum. Emission spectra of the incandescent body at temperatures of 4000 K, 5800 K, and 8000 K are shown in Figure 27-1. Note that at each temperature, there is a frequency at which the maximum amount of energy is emitted. If you compare the location of each curve\u2019s maximum, you will see that as the temperature increases, the frequency at which the maximum amount of energy is emitted also increases. \u25a0 Figure 27-1 This graph shows the emission spectra of an incandescent body at three different temperatures. Glows in the Dark Close the shades and turn off the lights in the room. Shine a flashlight at a beaker that contains fluorescein. Now place a red filter over the flashlight so that only red light hits the beaker. 1. Describe the results. 2. Predict how using a green filter instead of the red filter will affect the results. 3. Test your prediction. 4. Explain the results. 5. Predict whether the fluorescein will glow when a blue filter is used and provide an explanation for your prediction. 6. Test your prediction. Analyze and Conclude 7. Write a brief explanation summarizing and explaining your observations 10 724 Chapter 27 Quantum Theory Radiation from an Incandescent Body Infrared Visible Ultraviolet 8000 K 5800 K 4000 K 1 2 3 5 4 8 6 Frequency (1014 Hz) 7 9 10 11 12 722-734 CH27-S1-845813 3/12/04 6:45 AM Page 725 Astronomy Connection Astronomy Connection Temperature of the Universe The universe is filled with the radiation that was emitted when it was a very hot object. Currently, the emission spectrum of the universe matches that of a body with a temperature of 2.7 K. This is very cold. As you may recall, 0 K is the lowest possible temperature on the Kelvin temperature scale and is known as absolute zero. \u25a0 Figure 27-2 This stamp honors Max Planck\u2019s work and refers to the constant that bears his name, Planck\u2019s constant, h. Planck\u2019s constant, which has a value of 6.6261034 JHz, is used in many quantum-related equations. The total power emitted by a hot body also increases with temperature. The power (the energy emitted per second) of an electromagnetic wave is proportional to the hot body\u2019s kelvin temperature raised to the fourth power, T4. Thus, hotter bodies radiate considerably more power than do cooler bodies. Probably the most common example of a hot body radiating a great amount of power is the Sun, a dense ball of gases heated to incandescence by the energy produced within it. The Sun\u2019s surface temperature is 5800 K, and it radiates 41026 W of power, an enormous quantity. On average, each square meter of Earth\u2019s surface receives about 1000 J of energy per second (1000 W), enough to power ten 100-W lightbulbs. The problem with Maxwell\u2019s electromagnetic theory was that it is unable to explain the shape of the spectrum shown in Figure 27-1. Between 1887 and 1900, many physicists used existing classical physics theories to try to explain the shape. They all failed. In 1900, German physicist Max Planck found that he could calculate the spectrum if he introduced a revolutionary hypothesis: that atoms are not able to continuously change their energy. Planck assumed that the vibrational energy of the atoms in a solid could have only specific frequencies, as shown by the following equation. Energy of Vibration E nhf The energy of a vibrating atom is equal to the product of an integer, Planck\u2019s constant, and the frequency of the vibration. In the equation, f is the frequency of vibration of the atom, h is a constant, called Planck\u2019s constant, with a value of 6.6261034 J/Hz, and n is an integer such as 0, 1, 2, 3... n 0: E (0)hf 0 n 1: E (1)hf hf n 2: E (2)hf 2hf n 3: E (3)hf 3hf and so on Thus, the energy, E, can have the values hf, 2hf, 3hf, and so", " on, but never values such as 2 hf or 3 hf. In other words, energy is quantized\u2014it exists only 4 3 in bundles of specific amounts. h is usually rounded to 6.631034 J/Hz for calculations. Planck also proposed that atoms do not always radiate electromagnetic waves when they are vibrating, as predicted by Maxwell. Instead, Planck proposed that atoms emit radiation only when their vibrational energy changes. For example, if the energy of an atom changes from 3hf to 2hf, the atom emits radiation. The energy radiated is equal to the change in energy of the atom, in this case hf. Planck found that the constant, h, has an extremely small value. This means that the energy-changing steps are too small to be noticeable in ordinary bodies. Still, the introduction of quantized energy was extremely troubling to physicists, especially to Planck himself. It was the first hint that the classical physics of Newton and Maxwell might be valid only under certain conditions. Planck was honored for his groundbreaking theory of quantized energy with a Nobel prize in 1918 and, more recently, with the postage stamp shown in Figure 27-2. Section 27.1 A Particle Model of Waves 725 Horizons Companies 722-734 CH27-S1-845813 3/12/04 6:47 AM Page 726 The Photoelectric Effect Physicists of the early 1900s also were challenged by another troubling experimental result that could not be explained by Maxwell\u2019s wave theory. When ultraviolet radiation was incident on a negatively charged zinc plate, the plate discharged. When ordinary visible light was incident on the same charged plate, the plate did not discharge. This result was contrary to electromagnetic theory. Both ultraviolet radiation and visible light are forms of electromagnetic radiation, so why would the zinc plate be discharged by one but not by the other? And why would a positively charged zinc plate not be similarly discharged? Further study showed that the negatively charged zinc plate was discharging by emitting or ejecting electrons. The emission of electrons when electromagnetic radiation falls on an object is called the photoelectric effect. The photoelectric effect can be studied in a photocell, such as the one shown in Figure 27-3. The cell contains two metal electrodes sealed in a tube from which the air has been removed. The evacuated tube keeps the metal surfaces from oxidizing and keeps the electrons from being slowed or stopped by particles in the air. The larger electrode, the cathode, usually is coated with cesium or another alkali metal. The smaller electrode, the anode, is made of a thin wire so that it blocks only a very small amount of radiation. The tube often is made of quartz so as to allow ultraviolet wavelengths to pass through it. A potential difference placed across the electrodes attracts electrons to the anode. When no radiation falls on the cathode, there is no current in the circuit. When radiation falls on the cathode, a current is produced, which is measured by the ammeter, as shown in Figure 27-3. The current is produced because the photoelectric effect causes the ejection of electrons, also called photoelectrons, from the cathode. The flow of electrons is the current in the circuit. The electrons travel to the anode, the positive electrode. Threshold frequency Not all radiation falling on the cathode results in a current. Electrons are ejected from the cathode only if the frequency of the radiation is greater than a certain minimum value, called the threshold frequency, f0. The threshold frequency varies widely, depending on the type of metal. For example, all wavelengths of visible light except red will eject electrons from cesium, but no wavelength of visible light will eject electrons from zinc. Higher-frequency ultraviolet radiation is needed to produce the photoelectric effect in zinc. \u25a0 Figure 27-3 In the photocell shown, electrons ejected from the cathode flow to the anode, completing the circuit and generating an electric current (a). This handheld light meter works because of the photoelectric effect and is used by photographers to measure light levels (b). a UV or light source b Radiation Cathode Anode Photocell Electrons Ammeter 726 Horizons Companies Chapter 27 Quantum Theory 722-734 CH27-S1-845813 3/12/04 7:21 AM Page 727 No matter how intense, radiation with a frequency below f0 will not cause the ejection of electrons from metal. Conversely, even very lowintensity radiation with a frequency at or above the threshold frequency causes the immediate ejection of electrons. When the incident radiation\u2019s frequency is equal to or greater than the threshold frequency, increasing the intensity of the radiation causes an increase in the flow of photoelectrons. How does the electromagnetic wave theory explain the photoelectric effect? It can\u2019t. According to electromagnetic wave theory, the electric field accelerates and ejects the electrons from the metal, and the strength of the electric field is related to", " the intensity of the radiation (not to the radiation\u2019s frequency). Thus, it follows that electrons in the metal would need to absorb energy from a dim light source for a very long time before they gained enough energy to be ejected. As you just learned, however, this is not the case. Observations show that electrons are ejected immediately when even low-intensity radiation at or above the threshold frequency is incident on the metal. Photons and quantized energy In 1905, Albert Einstein published a revolutionary theory that explained the photoelectric effect. According to Einstein, light and other forms of electromagnetic radiation consist of discrete, quantized bundles of energy, each of which was later called a photon. The energy of a photon depends on its frequency. Energy of a Photon E hf The energy of a photon is equal to the product of Planck\u2019s constant and the frequency of the photon. In the above equation, f is frequency in Hz, and h is Planck\u2019s constant. Because the unit Hz 1/s or s1, the J/Hz unit of Planck\u2019s constant is also equivalent to Js. Because the joule is too large a unit of energy to use with atomic-sized systems, the more convenient energy unit of the electron volt (eV) is usually used. One electron volt is the energy of an electron accelerated across a potential difference of 1 V. 1 eV (1.601019 C)(1 V) 1.601019 CV 1.601019 J Using the definition of an electron volt allows the photon energy equa- tion to be rewritten in a simplified form, as shown below. eVnm c 1240 Energy of a Photon E h The energy of a photon is equal to the constant 1240 eVnm divided by the wavelength of the photon. An explanation of the derivation of this equation and how to use it is given in the Problem-Solving Strategies on the next page. Section 27.1 A Particle Model of Waves 727 722-734 CH27-S1-845813 3/12/04 6:48 AM Page 728 Units of hc and Photon Energy Converting the quantity hc to the unit eVnm results in a simplified equation that can be used to solve problems involving photon wavelength. 1. The energy of a photon of wavelength is given by the equation E hf. 2. Because f c/, this equation can be written as E hc/. 3. When using the equation E hc/, if the value of hc in eVnm is divided by in nm, you will obtain the energy in eV. Thus, it is useful to know the value of hc in eVnm. 4. The conversion of hc to the unit eVnm is as follows: hc (6.6261034 J/Hz)(2.998108 m/s)( )(10 V e 1 019 J 1 1.602 1 9 nm) m 1240 eV 5. Substituting hc 1240 eVnm into the equation for the energy of a photon yields the following, where is in nm and E is in eV: eVnm c 1240 E h 6. Use the above equation to solve photon energy problems when energy in eV is desired. It is important to note that Einstein\u2019s theory of the photon goes further than Planck\u2019s theory of radiation from hot bodies. While Planck had proposed that vibrating atoms emit electromagnetic radiation with energy equal to nhf, he did not suggest that light and other forms of electromagnetic radiation act like particles. Einstein\u2019s theory of the photon reinterpreted and extended Planck\u2019s theory of radiation from hot bodies. Einstein\u2019s photoelectric-effect theory is able to explain the existence of a threshold frequency. A photon with a minimum frequency and energy, hf0, is needed to eject an electron from metal. If the photon has a frequency below f0, the photon will not have the energy needed to eject an electron. Because one photon interacts with one electron, an electron cannot simply accumulate subthreshold energy photons until it has enough energy to be ejected. On the other hand, radiation with a frequency greater than f0 has more than enough energy to eject an electron. In fact, the excess energy, hf hf0, becomes the kinetic energy of the ejected electron. Kinetic Energy of an Electron Ejected Due to the Photoelectric Effect KE hf hf0 The kinetic energy of an ejected electron is equal to the difference between the incoming photon energy, hf, and the energy needed to free the electron from the metal, hf0. 728 Chapter 27 Quantum Theory 722-734 CH27-S1-845813 6/8/04 4:51 PM Page 729 UV or light source Anode Photocell Ammeter Radiation Cathode Potentiometer \u25a0 Figure 27-4 The", " maximum kinetic energy of electrons ejected from the cathode can be measured using this apparatus. The ammeter measures the current through the circuit. By adjusting the potentiometer, the experimenter can determine the potential that results in zero current. At the zero current threshold, the maximum possible kinetic energy of the ejected electron can be calculated. Note that hf0 is the minimum energy needed to free the most loosely held electron in an atom. Because not all electrons in an atom have the same energy, some will require more than this minimum energy in order to escape. As a result, the ejected electrons have differing kinetic energies. Thus, it is important to realize that the phrase \u201ckinetic energy of the ejected electrons\u201d refers to the maximum kinetic energy that an ejected electron could have. Some of the ejected electrons will have less. Testing the photoelectric theory How can Einstein\u2019s theory be tested? The kinetic energy of the ejected electrons can be measured indirectly by a device like the one illustrated in Figure 27-4. A variable electric potential difference is used to adjust the voltage across the tube. When the potential difference is adjusted to make the anode negative, the ejected electrons must expend energy to reach the anode. Only electrons ejected from the cathode with sufficient kinetic energy will be able to reach the anode. As shown in Figure 27-4, light of the chosen frequency illuminates the cathode. Gradually, the experimenter increases the opposing potential difference, thereby making the anode more negative. As the opposing potential difference increases, more kinetic energy is needed for the electrons to reach the anode, and fewer electrons arrive there to complete the circuit. At a certain voltage, called the stopping potential, there are no electrons with enough kinetic energy to reach the anode, and the current stops. At the stopping potential, the kinetic energy of the electrons at the cathode equals the work done by the electric field to stop them. This is represented in equation form as KE qV0. In this equation, V0 is the magnitude of the stopping potential in volts (J/C), and q is the charge of the electron (1.601019 C). Note that the negative sign in the equation along with the negative value of q yield a positive value for KE. Applications The photoelectric effect is used in a variety of everyday applications. Solar panels, shown in Figure 27-5, use the photoelectric effect to convert the Sun\u2019s light into electricity. Garage-door openers have safety beams of infrared light that create current in the receiver through the photoelectric effect. If the beam of light is interrupted by an object as the garage door is closing, the current in the receiver stops and triggers the opener to open the door. The photoelectric effect also is used in nightlights and photo eyes that turn lights on and off automatically, depending on whether it is day or night. \u25a0 Figure 27-5 The solar panels on this building use the photoelectric effect to convert the Sun\u2019s light to electricity. Section 27.1 A Particle Model of Waves 729 Ruth Dixon 722-734 CH27-S1-845813 3/12/04 6:50 AM Page 730 Photoelectron Kinetic Energy The stopping potential of a certain photocell is 4.0 V. What is the kinetic energy given to the electrons by the incident light? Give your answer in both joules and electron volts. 1 Analyze and Sketch the Problem \u2022 Draw the cathode and anode, the incident radiation, and the direction of the ejected electron. Note that the stopping potential prevents electrons from flowing across the photocell. Known: V0 q 1.601019 C 4.0 V Unknown: KE (in J and eV)? 2 Solve for the Unknown The electric field does work on the electrons. When the work done, W, equals the negative of the initial kinetic energy, KE, electrons no longer flow across the photocell. KE W 0 J Cathode Anode Incident light Solve for KE. KE W qV0 (1.601019 C)(4.0 V) 6.41019 J Substitute W qV0 Substitute q 1.601019 C, V0 = 4.0 V Convert KE from joules to electron volts. KE (6.41019 J)( V e 1 19 J 1.60 0 1 ) 4.0 eV Math Handbook Operations with Scientific Notation pages 842\u2013843 3 Evaluate the Answer \u2022 Are the units correct? Joules and electron volts are both units of energy. \u2022 Does the sign make sense? Kinetic energy is always positive. \u2022 Is the magnitude realistic? The energy in electron volts is equal in magnitude to the stopping potential difference in volts. 1. An electron has an energy of 2.3 eV. What is the energy of the electron in joules? 2. What is the energy in eV of an electron with a velocity of 6.2106 m/s", "? 3. What is the velocity of the electron in problem 1? 4. The stopping potential for a photoelectric cell is 5.7 V. Calculate the maximum kinetic energy of the emitted photoelectrons in eV. 5. The stopping potential required to prevent current through a photocell is 3.2 V. Calculate the maximum kinetic energy in joules of the photoelectrons as they are emitted. 730 Chapter 27 Quantum Theory 722-734 CH27-S1-845813 6/8/04 4:52 PM Page 731 Suppose a nickel with a mass of 5.0 g vibrates up and down while it is connected to a spring. The maximum velocity of the nickel during the oscillations is 1.0 cm/s. Assume that the vibrating nickel models the quantum vibrations of the electrons within an atom, where the energy of the vibrations is given by the equation E nhf. 1. Find the maximum kinetic energy of the vibrating object. 2. The vibrating object emits energy in the form of light with a frequency of 5.01014 Hz. If the energy is emitted in a single step, find the energy lost by the object. 3. Determine the number of equally sized energy-step reductions that the object would have to make in order to lose all of its energy. Mass 5.0 g Maximum velocity 1.0 cm/s A graph of the kinetic energies of the electrons ejected from a metal versus the frequencies of the incident photons is a straight line, as shown in Figure 27-6. All metals have similar graphs with the same slope. This slope equals the rise/run ratio of the line, which is equal to Planck\u2019s constant, h. e is r slope r u n change in maximum kinetic energy of ejected electrons change in frequency of incident radiation K E h f The graphs of various metals differ only in the threshold frequency needed to free the electrons. In Figure 27-6, the threshold frequency, f0, is the point at which KE 0. In this case, f0, located at the intersection of the line with the x-axis, is approximately 4.41014 Hz. The threshold frequency is related to the work function of the metal. The work function of a metal is the energy needed to free the most weakly bound electron from the metal. The magnitude of the work function is equal to hf0. When a photon of frequency f0 is incident on a metal, the energy of the photon is sufficient to release the electron, but not sufficient to provide the electron with any kinetic energy. 3 \u25a0 Figure 27-6 This graph shows that as the frequency of the incident radiation increases, the kinetic energy of the ejected electrons increases proportionally. Maximum Kinetic Energy of Photoelectrons v. Frequency Between 1905 and 1916, American physicist Robert Millikan performed a brilliant set of experiments in which he attempted to disprove Einstein\u2019s photoelectric theory. While his results confirmed Einstein\u2019s equation, he did not accept Einstein\u2019s \u201cradical\u201d notion of the photon. Millikan\u2019s experiments made it possible for Einstein to receive a Nobel prize for his photoelectric theory in 1921. Two years later, in 1923, Millikan was awarded a Nobel prize for determining the charge of an electron and for his investigations into the photoelectric effect Frequency (1014 Hz) 8 10 11 12 Section 27.1 A Particle Model of Waves 731 722-734 CH27-S1-845813 3/12/04 6:52 AM Page 732 Work Function and Energy A photocell uses a sodium cathode. The sodium cathode has a threshold wavelength of 536 nm. a. Find the work function of sodium in eV. b. If ultraviolet radiation with a wavelength of 348 nm falls on sodium, what is the energy of the ejected electrons in eV? 1 Analyze and Sketch the Problem \u2022 Draw the cathode and anode, the incident radiation, and the direction of the ejected electron. Cathode Anode Known: 536 nm 0 hc 1240 eVnm 348 nm Unknown: W? KE? Incident radiation 2 Solve for the Unknown a. Find the work function using Planck\u2019s constant and the threshold wavelength. c h W hf0 0 nm V e 12 0 4 5 6 m n 3 2.31 eV Substitute hc 1240 eVnm, 0 536 nm b. Use Einstein\u2019s photoelectric-effect equation to determine the energy of the incident radiation. eVnm E 1240 nm V e 12 0 4 m n 8 4 3 3.56 eV Substitute 348 nm Math Handbook Operations with Significant Digits pages 835\u2013836 To calculate the energy of the ejected electron, subtract the work function from the energy of the incident radiation. KE hf hf0 E W 3.56 eV 2.31 eV 1.25 eV Substitute", " h c c E, h W 0 Substitute E 3.56 eV, W 2.31 eV c c h h 0 3 Evaluate the Answer \u2022 Are the units correct? Performing dimensional analysis on the units verifies that eV is the proper unit for KE. \u2022 Does the sign make sense? KE is always positive. \u2022 Are the magnitudes realistic? Energies should be a few electron volts. 6. The threshold wavelength of zinc is 310 nm. Find the threshold frequency, in Hz, and the work function, in eV, of zinc. 7. The work function for cesium is 1.96 eV. What is the kinetic energy, in eV, of photoelectrons ejected when 425-nm violet light falls on the cesium? 8. When a metal is illuminated with 193-nm ultraviolet radiation, electrons with energies of 3.5 eV are emitted. What is the work function of the metal? 9. A metal has a work function of 4.50 eV. What is the longest-wavelength radiation that will cause it to emit photoelectrons? 732 Chapter 27 Quantum Theory 722-734 CH27-S1-845813 3/12/04 6:53 AM Page 733 The Compton Effect The photoelectric effect demonstrates that a photon, even though it has no mass, has kinetic energy just as a particle does. In 1916, Einstein predicted that the photon should have another particle property: momentum. He showed that the momentum of a photon should be equal to E/c. Because E hf and f/c 1/, the photon\u2019s momentum is given by the following equation. f Photon Momentum p h h c The momentum of a photon is equal to Planck\u2019s constant divided by the photon\u2019s wavelength. Experiments done by an American physicist, Arthur Holly Compton, in 1922 tested Einstein\u2019s theory. The results of Compton\u2019s experiments further supported the particle model of light. Compton directed X rays of a known wavelength at a graphite target, as shown in Figure 27-7a, and measured the wavelengths of the X rays scattered by the target. He observed that some of the X rays were scattered without change in wavelength, whereas others had a longer wavelength than that of the original radiation. These results are shown in Figure 27-7b. Note that the peak wavelength for the unscattered X rays corresponds to the wavelength of the original incident X rays, whereas the peak wavelength for the scattered X rays is greater than that of the original incident X rays. Recall that the equation for the energy of a photon, E hf, also can be written as E hc/. This second equation shows that the energy of a photon is inversely proportional to its wavelength. The increase in wavelength that Compton observed meant that the X-ray photons had lost both energy and momentum. The shift in the energy of scattered photons is called the Compton effect. This shift in energy is very small, only about 103 nm, and is a measurable effect only when X rays having wavelengths of 102 nm or less are used. a Unscattered X rays Graphite target 90\u00b0 Scattered X rays X-ray spectrometer X-ray source \u25a0 Figure 27-7 Compton used an apparatus similar to this one to study the nature of photons (a). The increased wavelength of the scattered photons is evidence that the X-ray photons have lost energy (b). 0\u00b0 Unscattered X rays Wavelength 90\u00b0 Scattered X rays Wavelength Section 27.1 A Particle Model of Waves 733 722-734 CH27-S1-845813 3/12/04 6:54 AM Page 734 a b Before Collision After Collision \u25a0 Figure 27-8 Much like the collision between two billiard balls (a), when a photon strikes an electron, the energy and momentum gained by the electron equal the energy and momentum lost by the photon (b). Incident photon Electron Energy hf Energy mc2 Momentum h Recoiling electron Scattered photon Energy Eelectron Momentum mv Energy hf Momentum h In later experiments, Compton observed that electrons were ejected from the graphite block during the experiment. He suggested that the X-ray photons collided with electrons in the graphite target and transferred energy and momentum to them. Compton thought that these photonelectron collisions were similar to the elastic collisions experienced by billiard balls, as shown in Figure 27-8. He tested this idea by measuring the energy of the ejected electrons. Compton found that the energy and momentum gained by the electrons equaled the energy and momentum lost by the photons. Thus, photons obey the laws of conservation of momentum and energy when they are involved in collisions with other particles. 27.1 Section Review 10. Photoelectric Effect Why is high-intensity, lowfrequency light unable to eject electrons from a metal, whereas low-intensity, high-frequency light can? Explain. 11. Frequency and Energy", " of Hot-Body Radiation As the temperature of a body is increased, how does the frequency of peak intensity change? How does the total amount of radiated energy change? 12. Photoelectric and Compton Effects An experimenter sends an X ray into a target. An electron, but no other radiation, emerges from the target. Explain whether this event is a result of the photoelectric effect or the Compton effect. 13. Photoelectric and Compton Effects Distinguish the photoelectric effect from the Compton effect. 14. Photoelectric Effect Green light ( 532 nm) strikes an unknown metal, causing electrons to be ejected. The ejected electrons can be stopped by a potential of 1.44 V. What is the work function, in eV, of the metal? 15. Energy of a Photon What is the energy, in eV, of the photons produced by a laser pointer having a 650-nm wavelength? 16. Photoelectric Effect An X ray is absorbed in a bone and releases an electron. If the X ray has a wavelength of approximately 0.02 nm, estimate the energy, in eV, of the electron. 17. Compton Effect An X ray strikes a bone, collides with an electron, and is scattered. How does the wavelength of the scattered X ray compare to the wavelength of the incoming X ray? 18. Critical Thinking Imagine that the collision of two billiard balls models the interaction of a photon and an electron during the Compton effect. Suppose the electron is replaced by a much more massive proton. Would this proton gain as much energy from the collision as the electron does? Would the photon lose as much energy as it does when it collides with the electron? 734 Richard Megna/Fundamental Photographs Chapter 27 Quantum Theory physicspp.com/self_check_quiz 735-737 CH27-S2-845813 6/8/04 6:29 PM Page 735 27.2 Matter Waves The photoelectric effect and Compton scattering showed that a mass- less electromagnetic wave has momentum and energy, like a particle. If an electromagnetic wave has particlelike properties, could a particle exhibit interference and diffraction, as a wave does? In other words, does a particle have wavelike properties? In 1923, French physicist Louis de Broglie proposed just this, that material particles have wave properties. This proposal was so extraordinary that it was ignored by other scientists until Einstein read de Broglie\u2019s papers and supported his ideas. De Broglie Waves Recall that the momentum of an object is equal to its mass times its velocity, p mv. By analogy with the momentum of a photon, p h/, de Broglie proposed that the momentum of a particle is represented by the following equation: h p mv The wavelength in the above relationship represents that of the moving particle and is known as the de Broglie wavelength. The following equation solves directly for the de Broglie wavelength. h De Broglie Wavelength h v m p The de Broglie wavelength of a moving particle is equal to Planck\u2019s constant divided by the particle\u2019s momentum. According to de Broglie, particles such as electrons and protons should show wavelike properties. Effects such as diffraction and interference had never been observed for particles, so de Broglie\u2019s work was greeted with considerable doubt. In 1927, however, the results of two independent experiments proved that electrons are diffracted just as light is. In one experiment, English physicist George Thomson aimed a beam of electrons at a very thin crystal. Atoms in crystals are arrayed in a regular pattern that acts as a diffraction grating. Electrons diffracted from the crystal formed the same patterns that X rays of a similar wavelength formed. Figure 27-9 shows the pattern made by diffracting electrons. In the United States, Clinton Davisson and Lester Germer performed a similar experiment using electrons reflected and diffracted from thick crystals. The two experiments proved that material particles have wave properties. The wave nature of objects you see and handle every day is not observable because their wavelengths are extremely short. For example, consider the de Broglie wavelength of a 0.145-kg baseball when it leaves a bat with a speed of 38 m/s. h = = m v 6.631034 Js (0.145 kg)(38 m/s) 1.21034 m This wavelength is far too small to have observable effects. As you will see in the following Example Problem, however, an extremely small particle, such as an electron, has a wavelength that can be observed and measured. Objectives \u2022 Describe evidence of the wave nature of matter. \u2022 Solve problems involving the de Broglie wavelength of particles. \u2022 Describe the dual nature of waves and particles, and the importance of the Heisenberg uncertainty principle. Vocabulary de Broglie wavelength Heisenberg uncertainty principle \u25a0 Figure 27-9 Electron diffraction patterns, such", " as this one for a cubic zirconium crystal, demonstrate the wave properties of particles. Section 27.2 Matter Waves 735 Max Kurtz/Ztek 735-737 CH27-S2-845813 6/8/04 6:30 PM Page 736 De Broglie Wavelength An electron is accelerated by a potential difference of 75 V. What is its de Broglie wavelength? 1 Analyze and Sketch the Problem \u2022 Include the positive and negative plates in your drawing. Known: V 75 V q 1.601019 C m 9.111031 kg h 6.631034 Js Unknown:? 2 Solve for the Unknown Write relationships for the kinetic energy of the electron based on potential difference and motion and use them to calculate the electron\u2019s velocity. KE qV, and KE 1 mv 2 2 1 mv 2 qV 2 Equate both forms of KE. 75 V Math Handbook Isolating a Variable page 845 m 2qV 2(1.601019 C)(75 V) 9.111031 kg Solve for v. v Substitute q 1.601019 C, V 75 V, m 9.111031 kg 5.1106 m/s Solve for momentum, p. p mv (9.111031 kg)(5.1106 m/s) 4.61024 kgm/s Solve for the de Broglie wavelength,. Substitute m 9.111031 kg, v 5.1106 m/s h p 63. 6 0 1 Substitute h 6.631034 Js, p 4.61024 kgm/s 0 2 4 1 6 4. s 34 J g /s m k 1.41010 m, which is equivalent to 0.14 nm 3 Evaluate the Answer \u2022 Are the units correct? Dimensional analysis on the units verifies m/s for v and nm for. \u2022 Do the signs make sense? Positive values are expected for both v and. \u2022 Are the magnitudes realistic? The wavelength is close to 0.1 nm, which is in the X-ray region of the electromagnetic spectrum. 19. A 7.0-kg bowling ball rolls with a velocity of 8.5 m/s. a. What is the de Broglie wavelength of the bowling ball? b. Why does the bowling ball exhibit no observable wave behavior? 20. What is the de Broglie wavelength and speed of an electron accelerated by a potential difference of 250 V? 21. What voltage is needed to accelerate an electron so it has a 0.125-nm wavelength? 22. The electron in Example Problem 3 has a de Broglie wavelength of 0.14 nm. What is the kinetic energy, in eV, of a proton (m 1.671027 kg) with the same wavelength? 736 Chapter 27 Quantum Theory 735-737 CH27-S2-845813 6/8/04 6:30 PM Page 737 Particles and Waves a y Is light a particle or a wave? Evidence suggests that both particle and wave models are needed to explain the behavior of light. As you are about to discover, quantum theory and the dual nature of electromagnetic radiation led to fascinating scientific principles and applications. One such application, the scanning tunneling microscope (STM), is discussed in the How it Works feature on page 740. Determining location and momentum It is logical to think that to accurately define the properties of an object, you would need to devise an experiment that directly measures the desired properties. For example, you cannot simply state that a particle is at a certain location moving with a specific speed. Rather, an experiment must be performed that locates the particle and measures its speed. How can you detect the location of a particle? You must touch it or reflect light from it. If light is used, then the reflected light must be collected by an instrument or the human eye. Because of diffraction effects, the light used to detect the particle spreads out and makes it impossible to locate the particle exactly. The use of shorter-wavelength light or radiation decreases diffraction and allows the location of a particle to be more precisely measured. Heisenberg uncertainty principle As a result of the Compton effect, however, when short-wavelength, high-energy radiation strikes a particle, the particle\u2019s momentum is changed, as shown in Figure 27-10. Therefore, the act of precisely measuring the location of a particle has the effect of changing the particle\u2019s momentum. The more precise the determination of a particle\u2019s location, the greater the uncertainty is in its momentum. In the same way, if the momentum of the particle is measured, the position of the particle changes and becomes less certain. This situation is summarized by the Heisenberg uncertainty principle, which states that it is impossible to measure precisely both the position and momentum of a particle at the same time. This principle,", " named for German physicist Werner Heisenberg, is the result of the dual wave and particle properties of light and matter. The Heisenberg uncertainty principle tells us there is a limit to how accurately position and momentum can be measured. 27.2 Section Review Photon Electron x Momentum 0 Before Collision b y Photon wavelength increases x Electron momentum increases After Collision \u25a0 Figure 27-10 A particle can be seen only when light is scattered from it. Thus, the electron remains undetected (a) until an incident photon strikes it (b). The collision scatters the photon and electron and changes their momenta. 23. Wavelike Properties Describe the experiment that confirmed that particles have wavelike properties. 24. Wave Nature Explain why the wave nature of matter is not obvious. 25. De Broglie Wavelength What is the de Broglie wavelength of an electron accelerated through a potential difference of 125 V? 26. Wavelengths of Matter and Radiation When an electron collides with a massive particle, the electron\u2019s velocity and wavelength decrease. How is it possible to increase the wavelength of a photon? 27. Heisenberg Uncertainty Principle When light or a beam of atoms passes through a double slit, an interference pattern forms. Both results occur even when atoms or photons pass through the slits one at a time. How does the Heisenberg uncertainty principle explain this? 28. Critical Thinking Physicists recently made a diffraction grating of standing waves of light. Atoms passing through the grating produce an interference pattern. If the spacing of the slits in the grating were 1 (about 250 nm), what was the 2 approximate de Broglie wavelength of the atoms? physicspp.com/self_check_quiz Section 27.2 Matter Waves 737 738-739 CH27-LAB-845813 6/8/04 6:34 PM Page 738 Modeling the Photoelectric Effect The emission of electrons from an object when electromagnetic radiation is incident upon it is known as the photoelectric effect. Electrons are ejected from the object only when the frequency of the radiation is greater than a certain minimum value, called the threshold frequency. In this investigation you will model the photoelectric effect using steel balls. You will examine why only certain types of electromagnetic radiation result in the emission of photoelectrons. QUESTION How can steel balls be used to model the photoelectric effect? Objectives Materials \u25a0 Formulate a model to investigate the photo- electric effect. \u25a0 Describe how the energy of a photon is related to its frequency. \u25a0 Use scientific explanations to explain why macroscopic phenomena cannot explain the quantum behavior of the atom. steel balls (3) grooved channel (U-channel or shelf bracket) books red, orange, yellow, green, blue, and violet marking pens (or colored stickers) metric ruler isopropyl alcohol Safety Precautions Procedure \u25a0 Keep isopropyl alcohol away from open flame. \u25a0 Do not swallow isopropyl alcohol. \u25a0 Isopropyl alcohol can dry out skin. 738 Horizons Companies 1. Shape the grooved channel, as shown in the photo, and use several books to support the channel, as shown. Make sure the books do not block the ends of the channel. 2. Mark a capital letter R with a red marking pen on the channel 4 cm above the table, as shown. The R represents red. 3. Mark a capital letter V with a violet marking pen on the channel at a distance 14 cm above the table as shown. The V represents violet. Use the other colored pens to place marks for blue, B, green, G, yellow, Y, and orange, O, uniformly between the marks for R and V. 4. Place two steel balls at the lowest point on the channel. These steel balls represent the atom\u2019s valence electrons. 5. Hold a steel ball in place on the channel at the R position. This steel ball represents an incident photon of red light. Note that the red photon has the lowest energy of the six colors of light being modeled. 6. Release the steel ball (photon) and see if it has enough energy to remove a valence electron from the atom; that is, observe if either of the two steel ball electrons escapes from the channel. Record your observations in the data table. 738-739 CH27-LAB-845813 6/8/04 6:35 PM Page 739 Data Table Color or Energy of Photon Observations Red Orange Yellow Green Blue Violet Less than red Greater than violet 7. Remove the steel ball that represents the inci- dent photon from the lower part of the channel. Replace the two steel balls used to represent the valence electrons at the lowest point on the channel. 8. Repeat steps 5\u20137 for each of the colors you marked on the channel. Be sure always to start with the two steel balls at the lowest point on the channel. Note that the violet photon has the greatest", " energy of the six colors of light being modeled. Record your observations in the data table. 9. Repeat steps 5\u20137, but release the steel ball representing the incident photon from a point slightly lower than the R position. Record your observations in the data table. 10. Repeat steps 5\u20137, but release the steel ball representing the incident photon from a point slightly higher than the V position. Record your observations in the data table. 11. Answer question 1 in the Conclude and Apply section and then test your prediction. 12. When you have finished the lab, return all materials to the locations specified by your instructor. Clean off the ink markings on the channel with isopropyl alcohol (or remove the colored stickers placed on the channel). Analyze 1. Interpret Data Which color(s) of photons was able to remove at least one electron in your model? 2. Interpret Data Were any of the photons energetic enough to remove more than one electron? If so, identify the photon\u2019s color. 3. Use Models In step 9, what type of photon does the steel ball represent? 4. Use Models In step 10, what type of photon does the steel ball represent? 5. Explain Should photons of visible light be the only photons considered when investigating the photoelectric effect? Why or why not? 6. Summarize Summarize your observations in terms of the energies of photons. Conclude and Apply 1. Infer What would happen if two red photons hit the two valence electrons at the same time? Test your prediction. 2. Think Critically Some materials hold on to their valence electrons more tightly than others. How could the model be modified to show this? 3. Draw Conclusions In this model, what happens to the photon\u2019s energy when it collides with an electron but does not remove the electron from the atom? Going Further Using the formula E hf, where h is Planck\u2019s constant and f is the frequency of the electromagnetic radiation, calculate the energy of a red photon compared to the energy of a blue photon. Real-World Physics Photographers often have red lights in their darkrooms. Why do they use red light, but not blue light? To find out more about the photoelectric effect and quantum theory, visit the Web site: physicspp.com 739 740 CH27-FEATURE-845813 6/8/04 6:37 PM Page 740 Scanning Tunneling Microscope The scanning tunneling microscope (STM) was invented in 1981 by Gerd Binnig and Heinrich Rohrer. Five years later, they were awarded the Nobel Prize in physics. The atomic-level resolution of an STM allows scientists to form images of atoms, such as the image of silicon atoms shown on the monitor below. How does an STM work? 1 A voltage is applied to the sample to be viewed. The sample must be a conductor. Microscope probe 90% of current occurs within this area Tip Computer- generated image 3 The control system scans the probe back and forth and up and down above the surface of the sample. The distance between the surface and the tip is kept constant, producing a constant current. The upand-down movement of the tip is recorded and turned into an image. Control system V d 1 nm Surface of sample Sample Voltage Thinking Critically 1. Calculate If the tunneling electron current is 1.0109 A, how many electrons flow to the tip in 1 s? 2. Evaluate In an STM, the relationship between the current, I, and the distance, d, between the probe and the sample is I I0ekd, where I0 and k are constants. Use sample values to verify that the current decreases when the distance increases. 3. Design an Experiment What would you do if you wanted to use an STM to study a nonconductive sample? 2 The tip of an STM\u2019s microscope probe is positioned extremely close to the sample (about 1 nm above the surface). As predicted by quantum physics, some electrons jump, or tunnel, between the surface of the sample and the tip of the probe. The moving electrons produce a current (measured in nanoamperes). 740 How It Works 741-745 CH27-SG CA STP-845813 6/8/04 6:53 PM Page 741 27.1 A Particle Model of Waves Vocabulary \u2022 emission spectrum (p. 724) \u2022 quantized (p. 725) \u2022 photoelectric effect (p. 726) \u2022 threshold frequency (p. 726) \u2022 photon (p. 727) \u2022 work function (p. 731) \u2022 Compton effect (p. 733) Key Concepts \u2022 Objects that are hot enough to be incandescent emit light because of the vibrations of the charged particles inside their atoms. \u2022 The spectrum of incandescent objects covers a broad range of wavelengths. The spectrum depends upon the temperature of the incandescent objects. \u2022 Plan", "ck explained the spectrum of an incandescent object by supposing that a particle can have only certain energies that are multiples of a constant, now called Planck\u2019s constant. E nhf \u2022 Einstein explained the photoelectric effect by postulating that light exists in bundles of energy called photons. eVnm c 1240 E hf h \u2022 The photoelectric effect is the emission of electrons by certain metals when they are exposed to electromagnetic radiation. KE hf hf0 \u2022 The photoelectric effect allows the measurement of Planck\u2019s constant, h. \u2022 The work function, which is equivalent to the binding energy of the electron, is measured by the threshold frequency in the photoelectric effect. \u2022 The Compton effect demonstrates that photons have momentum, as predicted by Einstein. f p h h c \u2022 Even though photons, which travel at the speed of light, have zero mass, they do have energy and momentum. Key Concepts \u2022 The wave nature of material particles was suggested by de Broglie and verified experimentally by the diffraction of electrons through crystals. All moving particles have a wavelength, known as the de Broglie wavelength. h h v m p \u2022 The particle and wave aspects are complementary parts of the complete nature of both matter and light. \u2022 The Heisenberg uncertainty principle states that it is not possible to simultaneously measure the precise position and momentum of any particle of light or matter. 27.2 Matter Waves Vocabulary \u2022 de Broglie wavelength (p. 735) \u2022 Heisenberg uncertainty principle (p. 737) physicspp.com/vocabulary_puzzlemaker 741 741-745 CH27-SG CA STP-845813 3/12/04 8:18 AM Page 742 Concept Mapping 29. Complete the following concept map using these terms: dual nature, mass, wave properties, momentum, diffraction. particle properties interference Mastering Concepts 30. Incandescent Light An incandescent lightbulb is controlled by a dimmer. What happens to the color of the light given off by the bulb as the dimmer control is turned down? (27.1) 31. Explain the concept of quantized energy. (27.1) 32. What is quantized in Max Planck\u2019s interpretation of the radiation of incandescent bodies? (27.1) 33. What is a quantum of light called? (27.1) 34. Light above the threshold frequency shines on the metal cathode in a photocell. How does Einstein\u2019s photoelectric effect theory explain the fact that as the light intensity increases, the current of photoelectrons increases? (27.1) 35. Explain how Einstein\u2019s theory accounts for the fact that light below the threshold frequency of a metal produces no photoelectrons, regardless of the intensity of the light. (27.1) 36. Photographic Film Because certain types of black- and-white film are not sensitive to red light, they can be developed in a darkroom that is illuminated by red light. Explain this on the basis of the photon theory of light. (27.1) 37. How does the Compton effect demonstrate that photons have momentum as well as energy? (27.1) 38. The momentum, p, of a particle of matter is given by p mv. Can you calculate the momentum of a photon using the same equation? Explain. (27.2) 39. Explain how each of the following electron properties could be measured. (27.2) a. charge b. mass c. wavelength 40. Explain how each of the following photon properties could be measured. (27.2) a. energy b. momentum c. wavelength Applying Concepts 41. Use the emission spectrum of an incandescent body at three different temperatures shown in Figure 27-1 on page 724 to answer the following questions. a. At what frequency does the peak emission intensity occur for each of the three temperatures? b. What can you conclude about the relationship between the frequency of peak radiation emission intensity and temperature for an incandescent body? c. By what factor does the intensity of the red light given off change as the body\u2019s temperature increases from 4000 K to 8000 K? 42. Two iron bars are held in a fire. One glows dark red, while the other glows bright orange. a. Which bar is hotter? b. Which bar is radiating more energy? 43. Will high-frequency light eject a greater number of electrons from a photosensitive surface than lowfrequency light, assuming that both frequencies are above the threshold frequency? 44. Potassium emits photoelectrons when struck by blue light, whereas tungsten emits photoelectrons when struck by ultraviolet radiation. a. Which metal has a higher threshold frequency? b. Which metal has a larger work function? 45. Compare the de Broglie wavelength of the baseball shown in Figure 27-11 with the diameter of the baseball. 0.10 m 21 m/", "s \u25a0 Figure 27-11 Mastering Problems 27.1 A Particle Model of Waves 46. According to Planck\u2019s theory, how does the frequency of vibration of an atom change if it gives off 5.441019 J while changing its value of n by 1? 47. What potential difference is needed to stop electrons with a maximum kinetic energy of 4.81019 J? 48. What is the momentum of a photon of violet light that has a wavelength of 4.0102 nm? 742 Chapter 27 Quantum Theory For more problems, go to Additional Problems, Appendix B. 741-745 CH27-SG CA STP-845813 3/12/04 8:19 AM Page 743 49. The stopping potential of a certain metal is shown in Figure 27-12. What is the maximum kinetic energy of the photoelectrons in the following units? a. electron volts b. joules 58. A neutron is held in a trap with a kinetic energy of only 0.025 eV. a. What is the velocity of the neutron? b. Find the de Broglie wavelength of the neutron. Cathode Anode 59. The kinetic energy of a hydrogen atom\u2019s electron is 13.65 eV. a. Find the velocity of the electron. b. Calculate the electron\u2019s de Broglie wavelength. c. Given that a hydrogen atom\u2019s radius is 0.519 nm, calculate the circumference of a hydrogen atom and compare it with the de Broglie wavelength for the atom\u2019s electron. 60. An electron has a de Broglie wavelength of 0.18 nm. a. How large a potential difference did it experience if it started from rest? b. If a proton has a de Broglie wavelength of 0.18 nm, how large is the potential difference that it experienced if it started from rest? Mixed Review 61. What is the maximum kinetic energy of photoelectrons ejected from a metal that has a stopping potential of 3.8 V? 62. The threshold frequency of a certain metal is 8.01014 Hz. What is the work function of the metal? 63. If light with a frequency of 1.61015 Hz falls on the metal in the previous problem, what is the maximum kinetic energy of the photoelectrons? 64. Find the de Broglie wavelength of a deuteron (nucleus of 2H isotope) of mass 3.31027 kg that moves with a speed of 2.5104 m/s. 65. The work function of iron is 4.7 eV. a. What is the threshold wavelength of iron? b. Iron is exposed to radiation of wavelength 150 nm. What is the maximum kinetic energy of the ejected electrons in eV? 66. Barium has a work function of 2.48 eV. What is the longest wavelength of light that will cause electrons to be emitted from barium? 67. An electron has a de Broglie wavelength of 400.0 nm, the shortest wavelength of visible light. a. Find the velocity of the electron. b. Calculate the energy of the electron in eV. 68. Electron Microscope An electron microscope is useful because the de Broglie wavelengths of electrons can be made smaller than the wavelength of visible light. What energy in eV has to be given to an electron for it to have a de Broglie wavelength of 20.0 nm? 5.0 V \u25a0 Figure 27-12 50. The threshold frequency of a certain metal is 3.001014 Hz. What is the maximum kinetic energy of an ejected photoelectron if the metal is illuminated by light with a wavelength of 6.50102 nm? 51. The threshold frequency of sodium is 4.41014 Hz. How much work must be done to free an electron from the surface of sodium? 52. If light with a frequency of 1.001015 Hz falls on the sodium in the previous problem, what is the maximum kinetic energy of the photoelectrons? 53. Light Meter A photographer\u2019s light meter uses a photocell to measure the light falling on the subject to be photographed. What should be the work function of the cathode if the photocell is to be sensitive to red light ( 680 nm) as well as to the other colors of light? 54. Solar Energy A home uses about 41011 J of energy each year. In many parts of the United States, there are about 3000 h of sunlight each year. a. How much energy from the Sun falls on one square meter each year? b. If this solar energy can be converted to useful energy with an efficiency of 20 percent, how large an area of converters would produce the energy needed by the home? 27.2 Matter Waves 55. What is the de Broglie wavelength of an electron moving at 3.0106 m/s? 56. What velocity would an electron need to have a de Broglie wavelength of 3", ".01010 m? 57. A cathode-ray tube accelerates an electron from rest across a potential difference of 5.0103 V. a. What is the velocity of the electron? b. What is the wavelength associated with the electron? physicspp.com/chapter_test Chapter 27 Assessment 743 741-745 CH27-SG CA STP-845813 3/12/04 8:20 AM Page 744 69. Incident radiation falls on tin, as shown in Figure 27-13. The threshold frequency of tin is 1.21015 Hz. a. What is the threshold wavelength of tin? b. What is the work function of tin? c. The incident electromagnetic radiation has a wavelength of 167 nm. What is the kinetic energy of the ejected electrons in eV? Cathode Anode 72. Make and Use Graphs A student completed a photoelectric-effect experiment and recorded the stopping potential as a function of wavelength, as shown in Table 27-1. The photocell had a sodium cathode. Plot the data (stopping potential versus frequency) and use your calculator to draw the bestfit straight line (regression line). From the slope and intercept of the line, find the work function, the threshold wavelength, and the value of h/q from this experiment. Compare the value of h/q to the accepted value. 167 nm \u25a0 Figure 27-13 Thinking Critically 70. Apply Concepts A helium-neon laser emits photons with a wavelength of 632.8 nm. a. Find the energy, in joules, of each photon emitted by the laser. b. A typical small laser has a power of 0.5 mW (equivalent to 5104 J/s). How many photons are emitted each second by the laser? 71. Apply Concepts Just barely visible light with an intensity of 1.51011 W/m2 enters a person\u2019s eye, as shown in Figure 27-14. a. If this light shines into the person\u2019s eye and passes through the person\u2019s pupil, what is the power, in watts, that enters the person\u2019s eye? b. Use the given wavelength of the incident light and information provided in Figure 27-14 to calculate the number of photons per second entering the eye. Cornea Table 27-1 Stopping Potential v. Wavelength (nm) V0 (eV) 200 300 400 500 600 4.20 2.06 1.05 0.41 0.03 Writing in Physics 73. Research the most massive particle for which interference effects have been seen. Describe the experiment and how the interference was created. Cumulative Review 74. The spring in a pogo stick is compressed 15 cm when a child who weighs 400.0 N stands on it. What is the spring constant of the spring? (Chapter 14) 75. A marching band sounds flat as it plays on a very cold day. Why? (Chapter 15) 76. A charge of 8.0107 C experiences a force of 9.0 N when placed 0.02 m from a second charge. What is the magnitude of the second charge? (Chapter 20) 550 nm Lens 77. A homeowner buys a dozen identical 120-V light Pupil (diameter 7.0 mm) \u25a0 Figure 27-14 sets. Each light set has 24 bulbs connected in series, and the resistance of each bulb is 6.0. Calculate the total load in amperes if the homeowner operates all the sets from a single exterior outlet. (Chapter 23) 78. The force on a 1.2-m wire is 1.1103 N. The wire is perpendicular to Earth\u2019s magnetic field. How much current is flowing in the wire? (Chapter 24) 744 Chapter 27 Quantum Theory For more problems, go to Additional Problems, Appendix B. 741-745 CH27-SG CA STP-845813 6/8/04 6:54 PM Page 745 Multiple Choice 1. The energy level of an atom changes as it absorbs and emits energy. Which of the following is NOT a possible energy level of an atom? 3 hf 4 hf 3hf 4hf 2. How is the threshold frequency related to the photoelectric effect? It is the minimum frequency of incident radiation needed to cause the ejection of atoms from the anode of a photocell. It is the maximum frequency of incident radiation needed to cause the ejection of atoms from the anode of a photocell. It is the frequency of incident radiation below which electrons will be ejected from an atom. 5. As shown in the diagram below, an electron is accelerated by a potential difference of 95.0 V. What is the de Broglie wavelength of the electron? 5.021022 m 1.261010 m 2.521010 m 5.10106 m 95 V It is the minimum frequency of incident radiation needed to cause the ejection of electrons from an atom", ". 6. What is the de Broglie wavelength of an electron moving at 391 km/s? The mass of an electron is 9.111031 kg. 3. A photon has a frequency of 1.141015 Hz. What is the energy of the photon? 5.821049 J 7.551019 J 8.771016 J 1.091012 J 4. Radiation with an energy of 5.17 eV strikes a photocell as shown below. If the work function of the photocell is 2.31 eV, what is the energy of the ejected photoelectron? 0.00 eV 2.23 eV 2.86 eV 7.48 eV Cathode Anode 3.51025 m 4.791015 m 4.81015 m 1.86109 m 7. What is the work function of a metal? a measure of how much work an electron emitted from the metal can do equal to the threshold frequency the energy needed to free the metal atom\u2019s innermost electron the energy needed to free the most weakly bound electron Extended Answer 8. An object has a de Broglie wavelength of 2.31034 m when its velocity is 45 m/s. What is the mass, in kg, of the object? \u03bb Incident radiation Wear A Watch If you are taking a timed test, make sure to pace yourself. Do not spend too much time on any one question. Skip over difficult questions and return to them after answering the easier questions. physicspp.com/standardized_test Chapter 27 Standardized Test Practice 745 746-759 CH28-S1-845813 3/22/04 6:18 AM Page 746 What You\u2019ll Learn \u2022 You will learn about the discovery of the atom\u2019s composition. \u2022 You will determine energies of the hydrogen atom. \u2022 You will learn how quantum theory led to the modern atomic model. \u2022 You will learn how lasers work and what their applications are. Why It\u2019s Important The quantum model of the atom and the transition of electrons between energy levels explain much of the observed behavior of all matter. Emission Spectra Each of these gas-filled tubes emits a unique spectrum of colors. The bright light is emitted when electrons in the gas make transitions to lower energy states. Think About This Why are the colors of the lights different, and how could you identify what gases are used in each tube? physicspp.com 746 Carl & Ann Purcell/CORBIS 746-759 CH28-S1-845813 3/22/04 6:19 AM Page 747 How can identifying different spinning coins model types of atoms? Question When a quarter, a nickel, a penny, and a dime are spun on a tabletop, what characteristics allow you to identify the type of spinning coin? Procedure 1. Hold a quarter up on its edge and flick it with your index finger to set it spinning. Note the appearance and sound of the spinning coin until it comes to a stop on the table. 2. Repeat step 1 three more times using a dime, a nickel, and a penny, respectively. 3. Have a classmate spin the coins, one at a time, in a random order. Observe each coin only after it is already spinning and then try to identify the type of coin that it is. 4. Repeat step 3, except this time, keep your eyes closed while trying to identify each of the spinning coins. Analysis How successful were you at identifying the individual coins when you were limited to listening to the sounds they made? What are the characteristics of a spinning coin that can be used to identify its type? What instruments might make the identification of the spinning coins easier? Critical Thinking Excited atoms of an element in a highvoltage gas-discharge tube dissipate energy by emitting light. How might the emitted light help you identify the type of atom in the discharge tube? What instruments might help you do this? 28.1 The Bohr Model of the Atom By the end of the nineteenth century, most scientists agreed on the existence of atoms. J. J. Thomson\u2019s discovery of the electron provided convincing evidence that the atom was made up of even smaller, subatomic particles. Every atom tested by Thomson contained negatively charged electrons, and these electrons possessed very little mass. Because atoms were known to be much more massive than the mass accounted for by the electrons they contain, scientists began looking for the missing mass that must be part of each atom. What was the nature of this yet-to-bediscovered massive part of the atom? How was this mass distributed within the atom? Moreover, atoms were known to be electrically neutral, yet, only negatively charged electrons had been identified within the atom. How were the negatively charged electrons arranged in the atom? What was the source of the atom\u2019s neutrality? Were positively charged particles also present in the atom? Knowing that their understanding of the atom was far from complete, scientists began searching", " for answers to numerous and challenging questions. Objectives \u2022 Describe the structure of the nuclear atom. \u2022 Compare and contrast continuous spectra and line-emission spectra. \u2022 Solve problems using orbital-radius and energy-level equations. Vocabulary alpha particles nucleus absorption spectrum energy level ground state excited state principal quantum number Section 28.1 The Bohr Model of the Atom 747 Horizons Companies 746-759 CH28-S1-845813 6/8/04 6:59 PM Page 748 The Nuclear Model Many questions faced the researchers investigating the nature of the atom. What caused the emission of light from atoms? How were the electrons distributed in the atom? Physicists and chemists from many countries searched for the solutions to this puzzle. The results not only provided knowledge about the structure of the atom, but also a totally new approach to both physics and chemistry. The history of the research into the nature of the atom is one of the most exciting stories of the twentieth century. J. J. Thomson believed that a massive, positively charged substance filled the atom. He pictured the negatively charged electrons as being distributed throughout this positively charged substance like raisins in a muffin. Ernest Rutherford, along with laboratory collaborators Hans Geiger and Ernest Marsden, however, performed a series of experiments that showed the atom had a very different structure. Rutherford\u2019s experiments made use of radioactive compounds that emitted penetrating rays. Some of these emissions had been found to be massive, positively charged particles that moved at high speeds. These particles, which were later named alpha particles, are represented by the symbol. The -particles in Rutherford\u2019s experiments could be detected by the small flashes of light that were emitted when the particles collided with a zinc-sulfide-coated screen. As shown in Figure 28-1, Rutherford directed a beam of -particles at an extremely thin sheet of gold foil. Rutherford was aware of Thomson\u2019s model of the atom, and he expected only minor deflections of the -particles as they passed through the thin gold foil. He thought that the paths of the massive, high-speed -particles would be only slightly altered as they passed through the evenly distributed positive charge making up each gold atom. The test results amazed him. While most of the -particles passed through the gold foil either undeflected or only slightly deflected, a few of the particles were scattered through very large angles. Some were even deflected through angles larger than 90\u00b0. A diagram of these results is shown in Figure 28-2. Rutherford compared his amazement to that of firing a 15-inch cannon shell at tissue paper and then having the shell bounce back and hit him. Source of particles \u25a0 Figure 28-1 After bombarding metal foil with alpha particles, Rutherford\u2019s team concluded that most of the mass of the atom was concentrated in the nucleus. Beam of particles Deflected particles Circular fluorescent screen Gold foil 748 Chapter 28 The Atom 746-759 CH28-S1-845813 6/8/04 7:00 PM Page 749 Atoms of gold foil Nucleus \u25a0 Figure 28-2 Most of the alpha particles directed at the thin sheet of gold foil passed through it without deflection. One in 20,000, however, were deflected at large angles. Alpha particles Using Coulomb\u2019s force law and Newton\u2019s laws of motion, Rutherford concluded that the results could be explained only if all of the atom\u2019s positive charge were concentrated in a tiny, massive central core, now called the nucleus. Therefore, Rutherford\u2019s model of the atom is called the nuclear model. Researchers have since determined that all the positive charge and more than 99.9 percent of the mass of the atom are contained in its nucleus. The electrons, which do not contribute a significant amount of mass to the atom, are distributed outside of and far away from the nucleus. Thus, the space occupied by the electrons defines the overall size, or diameter, of the atom. Because the diameter of the atom is about 10,000 times larger than the diameter of the nucleus, the atom mostly is made up of empty space. Emission spectra How are the electrons arranged around the nucleus of the atom? One of the clues that scientists used to answer this question came from the study of the light emitted by atoms. Recall from the previous chapter that the set of electromagnetic wavelengths emitted by an atom is called the atom\u2019s emission spectrum. As shown in Figure 28-3, atoms of a gaseous sample can be made to emit light in a gas-discharge tube. You probably are familiar with the colorful neon signs used by some businesses. These signs work on the same principles as gas-discharge tubes do. A gas-discharge tube consists of a low-pressure gas contained within a glass tube that has metal electrodes attached to each end. The gas glows when high voltage is applied across the tube. What interested scientists the most about this phenomenon was", " the fact that each different gas glowed with a different, unique color. The characteristic glows emitted by several gases are shown in Figure 28-3. a b c \u25a0 Figure 28-3 When high voltage is applied to a gas, the gas emits light, producing a unique glow. Hydrogen gas glows magenta (a), mercury glows bright blue (b), and nitrogen glows rose-orange (c). Section 28.1 The Bohr Model of the Atom 749 Richard Megna/Fundamental Photographs 746-759 CH28-S1-845813 3/22/04 6:22 AM Page 750 Gas-discharge tube containing gas Aligning slits Prism Bright lines Detector y x High voltage a b c \u25a0 Figure 28-4 A prism spectroscope can be used to observe emission spectra (a). The emission spectra of mercury (b) and barium (c) show characteristic lines. When the light emitted by the gas is passed through a prism or a diffraction grating, the emission spectrum of an atom is obtained. An emission spectrum can be studied in greater detail using an instrument called a spectroscope. As shown in Figure 28-4a, light in a prism spectroscope passes through a slit and is then dispersed as it travels through a prism. A lens system (not shown in the diagram) focuses the dispersed light so that it can be viewed, or recorded on a photographic plate or an electronic detector. The spectroscope forms an image of the slit at a different position for each wavelength. The spectrum of a hot body, or incandescent solid, such as the filament in a lightbulb, is a continuous band of colors from red through violet. The spectrum of a gas, however, is a series of distinct lines of different colors. The bright-line emission spectra for mercury gas and barium gas are shown in Figure 28-4b and Figure 28-4c, respectively. Each colored line corresponds to a particular wavelength of light emitted by the atoms of that gas. An emission spectrum is also a useful analytic tool, as it can be used to identify an unknown sample of gas. When the unknown gas is placed in a gas-discharge tube, it can be made to emit light. The emitted light consists of wavelengths that are uniquely characteristic of the atoms of that gas. Thus, the unknown gas can be identified by comparing its wavelengths with the wavelengths present in the spectra of known samples. An emission spectrum also can be used to analyze a mixture of gases. When the emission spectrum of a combination of elements is photographed, an analysis of the lines on the photograph can indicate the identities and the relative concentrations of the elements present. If the material being examined contains a large amount of any particular element, the lines for that element are more intense on the photograph than those of the other elements of lesser quantities. Through a comparison of the line intensities, the percentage composition of the material can be determined. 750 (t)Richard Megna/Fundamental Photographs, (b)Wabash Instrument Corp./Fundamental Photographs Chapter 28 The Atom 746-759 CH28-S1-845813 6/8/04 7:02 PM Page 751 Absorption spectra In 1814, Josef von Fraunhofer observed the presence of several dark lines in the spectrum of sunlight. These dark lines, now called Fraunhofer lines, are seen in Figure 28-5. He reasoned that as sunlight passed through the gaseous atmosphere surrounding the Sun, the gases absorbed certain characteristic wavelengths. These absorbed wavelengths produced the dark lines in the observed spectrum. The set of wavelengths absorbed by a gas is the absorption spectrum of the gas. The composition of the Sun\u2019s atmosphere was determined by comparing the missing lines in the observed spectrum with the known emission spectra of various elements. The compositions of many other stars have been determined using this technique. You can observe an absorption spectrum by passing white light through a gas sample and a spectroscope, as shown in Figure 28-6a. Because the gas absorbs specific wavelengths, the normally continuous spectrum of the white light has dark lines in it after passing through the gas. For a gas, the bright lines of the emission spectrum and the dark lines of the absorption spectrum often occur at the same wavelengths, as shown in Figure 28-6b and Figure 28-6c, respectively. Thus, cool, gaseous elements absorb the same wavelengths that they emit when excited. As you might expect, the composition of a gas can be determined from the wavelengths of the dark lines of the gas\u2019s absorption spectrum. a White light \u25a0 Figure 28-5 Fraunhofer lines appear in the absorption spectrum of the Sun. There are many lines, some of which are faint and others that are very dark, depending on abundances of the elements in the Sun. Astronomy Connection Astronomy Connection \u25a0 Figure 28-6 This apparatus is used to produce the absorption spectrum of", " sodium (a). The emission spectrum of sodium consists of several distinct lines (b), whereas the absorption spectrum of sodium is nearly continuous (c). Dark lines Projector Sodium vapor Prism Continuous spectrum b c 400 nm Sodium emission spectrum 750 nm 400 nm Sodium absorption spectrum 750 nm 751 Images were created by Dr. Donald Mickey, University of Hawaii, Institute for Astronomy, from National Solar Observatory spectra data. Section 28.1 The Bohr Model of the Atom 746-759 CH28-S1-845813 3/22/04 6:23 AM Page 752 Spectroscopy Both emission and absorption spectra are valuable scientific tools. As a result of the elements\u2019 characteristic spectra, scientists are able to analyze, identify, and quantify unknown materials by observing the spectra that they emit or absorb. The emission and absorption spectra of elements are important in industry as well as in scientific research. For example, steel mills reprocess large quantities of scrap iron of varying compositions. The exact composition of a sample of scrap iron can be determined in minutes by spectrographic analysis. The composition of the steel can then be adjusted to suit commercial specifications. Aluminum, zinc, and other metal-processing plants employ the same method. The study of spectra is a branch of science known as spectroscopy. Spectroscopists are employed throughout research and industrial communities. Spectroscopy has proven to be an effective tool for analyzing materials on Earth, and it is the only currently available tool for studying the composition of stars over the vast expanse of space. The Bohr Model of the Atom In the nineteenth century, many physicists tried to use atomic spectra to determine the structure of the atom. Hydrogen was studied extensively because it is the lightest element and has the simplest spectrum. The visible spectrum of hydrogen consists of four lines: red, green, blue, and violet, as shown in Figure 28-7. Any theory that explained the structure of the atom would have to account for these wavelengths and support the nuclear model. However, the nuclear model as proposed by Rutherford was not without its problems. Rutherford had suggested that electrons orbit the nucleus much like the planets orbit the Sun. There was, however, a serious flaw in this planetary model. Problems with the planetary model An electron in an orbit constantly is accelerated toward the nucleus. As you learned in Chapter 26, accelerating electrons radiate energy by emitting electromagnetic waves. At the rate that an orbiting electron would lose energy, it should spiral into the nucleus within 109 s. This, however, must not be happening because atoms are known to be stable. Thus, the planetary model was not consistent with the laws of electromagnetism. In addition, the planetary model predicted that the accelerating electrons would radiate energy at all wavelengths. However, as you just learned, the light emitted by atoms is radiated only at specific wavelengths. Danish physicist Niels Bohr went to England in 1911 and joined Rutherford\u2019s group to work on determining the structure of the atom. He tried to unite the nuclear model with Planck\u2019s quantized energy levels and Einstein\u2019s theory of light. This was a courageous idea because as of 1911, neither of these revolutionary ideas was widely understood or accepted. 410 nm 434 nm 486 nm 656 nm \u25a0 Figure 28-7 The emission spectrum of hydrogen in the visible range has four lines. 752 Chapter 28 The Atom 746-759 CH28-S1-845813 6/8/04 7:02 PM Page 753 Quantized Energy Bohr began with the planetary arrangement of electrons, as diagrammed in Figure 28-8, but then made the bold hypothesis that the laws of electromagnetism do not apply inside the atom. He postulated that an electron in a stable orbit does not radiate energy, even though it is accelerating. Bohr referred to this stable condition as a stationary state. He went on to assume that only stationary states with specific amounts of energy are allowed. In other words, Bohr considered the energy levels in an atom to be quantized. As shown in Figure 28-9, the quantization of energy in atoms can be likened to a flight of stairs with decreasing-height steps. To go up the stairs you must move from one step to the next\u2014it is impossible to stop at a midpoint between steps. Instead of steps, atoms have quantized amounts of energy, each of which is called an energy level. Just as you cannot occupy a position between steps, an atom\u2019s energy cannot have a value between allowed energy levels. An atom with the smallest allowable amount of energy is said to be in the ground state. When an atom absorbs energy, it moves, or makes a transition to, a higher energy level. Any energy level above the ground state is called an excited state. Energy of an Atom What determines the amount of energy an atom has? An atom\u2019s energy equals the sum of the kinetic energy of the electrons and the potential energy from the attractive force between the", " electrons and the nucleus. The energy of an atom with electrons in a nearby orbit is less than that of an atom with electrons in a faraway orbit because work must be done to move the electrons away from the nucleus. Thus, atoms in excited, higher-energy states have electrons in larger, or more distant, orbits. Because energy is quantized and energy is related to the size of the orbit, the size of the orbit also is quantized. The model of an atom just described, that of a central nucleus with orbiting electrons having specific quantized energy levels, is known as the Bohr model of the atom. If Bohr was correct in hypothesizing that stable atoms do not radiate energy, then what is responsible for an atom\u2019s characteristic emission spectrum? To answer this question, Bohr suggested that electromagnetic energy is emitted when the atom changes from one stationary state to another. Incorporating Einstein\u2019s photoelectric theory, Bohr knew that the energy of hf. He then postulated every photon is given by the equation, Ephoton that when an atom absorbs a photon, the atom\u2019s energy increases by an amount equal to that of the photon. This excited atom then makes a transition to a lower energy level by emitting a photon. \u25a0 Figure 28-8 Bohr\u2019s planetary model of the atom was based on the postulation that electrons move in fixed orbits around the nucleus. An atom\u2019s energy levels Steps decreasing in height Fifth Fourth Third Second Excited states First (ground state) Nucleus \u25a0 Figure 28-9 These decreasingheight steps are analogous to the allowed energy levels in an atom. Note how the difference in energy between adjacent energy levels decreases as the energy level increases. Section 28.1 The Bohr Model of the Atom 753 746-759 CH28-S1-845813 6/8/04 7:02 PM Page 754 E1 E1 E2 Ephoton 1 E3 Ephoton 2 E2 Ephoton 3 E3 Ephoton 1 Ephoton 2 1 E3 2 3 1 When the atom makes the transition from its initial energy level, Ei, to its final energy level, Ef, the change in energy, Eatom, is given by the following equation. Ephoton 3 3 Ef As shown in Figure 28-10, the change in energy of the atom equals the energy of the emitted photon. Eatom Ei E2 E1 2 Ground state \u25a0 Figure 28-10 The energy of the emitted photon is equal to the difference in energy between the initial and final energy levels of the atom. 754 Chapter 28 The Atom Ephoton Eatom or Ef The following equations summarize the relationships between the change in energy states of an atom and the energy of the photon emitted. Ephoton Ei Eatom Energy of an Emitted Photon Ephoton The energy of an emitted photon is equal to the product of Planck\u2019s constant and the emitted photon\u2019s frequency. The energy of an emitted photon also is equal to the loss in the atom\u2019s energy. hf, or Ephoton Predictions of the Bohr Model A scientific theory must do more than present postulates; it must allow predictions to be made that can be checked against experimental data. A good theory also can be applied to many different problems, and it ultimately provides a simple, unified explanation of some part of the physical world. Bohr used his theory to calculate the wavelengths of light emitted by a hydrogen atom. The calculations were in excellent agreement with the values measured by other scientists. As a result, Bohr\u2019s model was widely accepted. Unfortunately, the model only worked for the element hydrogen; it could not predict the spectrum of helium, the next-simplest element. In addition, there was not a good explanation as to why the laws of electromagnetism should work everywhere but inside the atom. Not even Bohr believed that his model was a complete theory of the structure of the atom. Despite its shortcomings, however, the Bohr model describes the energy levels and wavelengths of light emitted and absorbed by hydrogen atoms remarkably well. Development of Bohr\u2019s model Bohr developed his model by applying ma, to the electron. The net force Newton\u2019s second law of motion, Fnet is described by Coulomb\u2019s law for the interaction between an electron of charge q that is a distance r from a proton of charge q. That force is given by F Kq2/r2. The acceleration of the electron in a circular orbit about a much more massive proton is given by a v2/r, where the negative sign shows that the direction is inward. Thus, Bohr obtained the following relationship: 2 v2 q K m 2 r r In the equation, K is the constant from Coulomb\u2019s law and has a value of 9.0109 Nm2/C2. 746-", "759 CH28-S1-845813 6/8/04 7:03 PM Page 755 Next, Bohr considered the angular momentum of the orbiting electron, which is equal to the product of an electron\u2019s momentum and the radius of its circular orbit. The angular momentum of the electron is thus given by mvr. Bohr postulated that angular momentum also is quantized; that is, the angular momentum of an electron can have only certain values. He claimed that the allowed values were multiples of h/2, where h is Planck\u2019s constant. Using n to represent an integer, Bohr proposed that mvr nh/2. Using Kq2/r2 mv2/r and rearranging the angular momentum equation, v nh/2mr, Bohr found that the orbital radii of the electrons in a hydrogen atom are given by the following equation. Electron Orbital Radius in Hydrogen rn 2 2 n h 4 q2 2 K m The radius of an electron in orbit n is equal to the product of the square of Planck\u2019s constant and the square of the integer n divided by the quantity four times the square of, times the constant K, times the mass of an electron, times the square of the charge of an electron. You can calculate the radius of the innermost orbit of a hydrogen atom, also known as the Bohr radius, by substituting known values and n 1 into the above equation. r1 (6.6261034 Js)2(1)2 42(9.0109 Nm2/C2)(9.111031 kg)(1.601019 C)2 5.31011 J2s2/Nm2kg 5.31011 m, or 0.053 nm By performing a little more algebra you can show that the total energy of the atom, which is the sum of the kinetic energy of the electron and the potential energy, and is given by Kq2/2r, is represented by the following equation: 2 2mq4 K 1 2 2 n h En By substituting numerical values for the constants, you can calculate the total energy of the atom in joules, which yields the following equation: En 1 2.171018 J 2 n Converting the relationship to units of electron volts yields the following equation. Energy of a Hydrogen Atom En 1 13.6 eV 2 n The total energy of an atom with principal quantum number n is equal to the product of 13.6 eV and the inverse of n2. Both the electron\u2019s orbital radius and the energy of the atom are quantized. The integer, n, that appears in these equations is called the principal quantum number. It is the principal quantum number that determines the quantized values of r and E. In summary, the radius, r, increases as the square of n, whereas the energy, E, depends on 1/n2. Bright-Line Spectra Turn on a gas-discharge tube power supply attached to a gas tube so that the tube glows. CAUTION: Handle gas tube carefully to avoid breaking. Do not touch any exposed metal when the power supply is turned on. Dangerous voltages are present. Always turn off the power supply before changing gas tubes. Turn off the room lights. 1. Describe the color that you observe. 2. Observe the gas-discharge tube through a diffraction grating. 3. Sketch the results of viewing the gas-discharge tube through the diffraction grating. 4. Predict whether the observed spectrum will change when a different gas-discharge tube is viewed through the diffraction grating. 5. Test your prediction. Analyze and Conclude 6. Sketch the results of viewing the new gas-discharge tube through the diffraction grating. 7. Explain why there are differences in the two spectra. Section 28.1 The Bohr Model of the Atom 755 746-759 CH28-S1-845813 6/8/04 7:03 PM Page 756 Energy and electron transitions You may be wondering why the energy of an atom in the Bohr model has a negative value. Recall from Chapter 11 that only energy differences have meaning. The zero energy level can be chosen at will. In this case, zero energy is defined as the energy of the atom when the electron is infinitely far from the nucleus and has no kinetic energy. This condition exists when the atom has been ionized; that is, when the electron has been removed from the atom. Because work has to be done to ionize an atom, the energy of an atom with an orbiting electron is less than zero. The energy of an atom has a negative value. When an atom makes a transition from a lower to a higher energy level, the total energy becomes less negative, but the overall energy change is positive. Some of hydrogen\u2019s energy levels and the possible energy level transitions that it", " can undergo are shown in Figure 28-11. Note that an excited hydrogen atom can emit electromagnetic energy in the infrared, visible, or ultraviolet range depending on the transition that occurs. Ultraviolet light is emitted when the atom drops into its ground state from any excited state. The four visible lines in the hydrogen spectrum are produced when the atom drops from the n 3 or higher energy state into the n 2 energy state. \u25a0 Figure 28-11 The distinct set of color lines that make up a hydrogen atom\u2019s visible spectrum are known as the Balmer series. This visible light is the result of the photons emitted when electrons make transitions to the second energy level, n 2. Other electron transitions in a hydrogen atom result in the emission of ultraviolet (Lyman series) and infrared (Paschen series) electromagnetic energy. Detail of Balmer Series Transitions 0.38 eV 0.54 eV 0.85 eV 1.51 eV 3.40 eV 13.6 eV 756 Chapter 28 The Atom E6 E5 E4 E3 E2 E1 Paschen series (infrared) n 6 n 5 n 4 n 3 Balmer series (visible) Lyman series (ultraviolet) n 2 Ground state n 1 Nucleus 746-759 CH28-S1-845813 3/22/04 6:26 AM Page 757 Energy Levels A hydrogen atom absorbs energy, causing its electron to move from the innermost energy level (n 1), to the second energy level (n 2). Determine the energy of the first and second energy levels, and the energy absorbed by the atom. 1 Analyze and Sketch the Problem \u2022 Diagram the energy levels E1 and E2. \u2022 Indicate the direction of increasing energy on the diagram. Known: Quantum number of innermost energy level, n 1 Quantum number of second energy level, n 2 Unknown: Energy of level E1 Energy of level E2 Energy difference, E??? 2 Solve for the Unknown Use the equation for the energy of an electron in its orbit to calculate the energy of each level E2 E1 En E1 E2 1 13.6 eV 2 n 1 13.6 eV )2 (1 13.6 eV 1 13.6 eV )2 (2 3.40 eV Substitute n 1 Substitute n 2 Math Handbook Operations with Significant Digits pages 835\u2013836 The energy absorbed by the atom, E, is equal to the energy difference between the final energy level of the atom, Ef, and the initial energy level of the atom, Ei. Ei E1 E Ef E2 3.40 eV (13.6 eV) 10.2 eV Energy is absorbed. Substitute Ef Substitute E2 E1 E2, Ei 3.40 eV, E1 13.6 eV 3 Evaluate the Answer \u2022 Are the units correct? Orbital energy values should be measured in electron volts. \u2022 Is the sign correct? The energy difference is positive when electrons move from lower energy levels to higher energy levels. \u2022 Is the magnitude realistic? The energy needed to move an electron from the first energy level to the second energy level should be approximately 10 eV, which it is. 1. Calculate the energies of the second, third, and fourth energy levels in the hydrogen atom. 2. Calculate the energy difference between E3 and E2 in the hydrogen atom. 3. Calculate the energy difference between E4 and E2 in the hydrogen atom. 4. The text shows the solution of the equation rn 2n 2 h for n 1, the innermost orbital 4 2K m radius of the hydrogen atom. Note that with the exception of n2, all factors in the equation are constants. The value of r1 is 5.31011 m, or 0.053 nm. Use this information to calculate the radii of the second, third, and fourth allowable energy levels in the hydrogen atom. 5. The diameter of the hydrogen nucleus is 2.51015 m, and the distance between the nucleus and the first electron is about 51011 m. If you use a ball with a diameter of 7.5 cm to represent the nucleus, how far away will the electron be? q 2 Section 28.1 The Bohr Model of the Atom 757 746-759 CH28-S1-845813 3/22/04 6:27 AM Page 758 Frequency and Wavelength of Emitted Photons An excited hydrogen atom drops from the second energy level (n 2) to the first energy level (n 1). Calculate the energy and the wavelength of the emitted photon. Use E1 and E2 values from Example Problem 1. 1 2 Analyze and Sketch the Problem \u2022 Diagram the energy levels E1 and E2. \u2022 Indicate the direction of increasing energy and show the emission of a photon on the diagram. Known: Energy of", " level E1 Energy of level E2 13.6 eV 3.40 eV Unknown: Frequency, f? Wavelength,? Energy difference, E? Eatom E2 3.40 eV Ephoton E1 13.6 eV Solve for the Unknown The energy of the emitted photon is equal to E, the energy difference between the final energy level of the atom, Ef, and the initial energy level of the atom, Ei. Ei E2 E Ef E1 13.6 eV (3.40 eV) 10.2 eV Energy is emitted. Substitute Ef Substitute E1 E2 E1, Ei 13.6 eV, E2 3.40 eV Math Handbook Isolating a Variable page 845 To determine the wavelength of the photon, use the following equations. E E hf, so f h c f, so c f c (E /h) hc E nm eV 12 0 4 1.2 V e 0 122 nm Solve the photon energy equation for frequency. Solve the wavelength-frequency equation for wavelength. E Substitute f h Substitute hc 1240 eVnm, E 10.2 eV 3 Evaluate the Answer \u2022 Are the units correct? Energy is measured in electron volts. The prefix nano- modifies the base SI unit, the meter, which is the correct unit for the wavelength. \u2022 Are the signs correct? Energy is released when the atom emits a photon during the transition from the second energy level to the first; thus, the energy difference is negative. \u2022 Is the magnitude realistic? Energy released in this transition produces light in the ultraviolet region below 400 nm. 6. Find the wavelength of the light emitted in Practice Problems 2 and 3. Which lines in Figure 28-7 correspond to each transition? 7. For a particular transition, the energy of a mercury atom drops from 8.82 eV to 6.67 eV. a. What is the energy of the photon emitted by the mercury atom? b. What is the wavelength of the photon emitted by the mercury atom? 8. The ground state of a helium ion is 54.4 eV. A transition to the ground state emits a 304-nm photon. What was the energy of the excited state? 758 Chapter 28 The Atom 746-759 CH28-S1-845813 7/16/04 9:08 PM Page 759 Although the Bohr atomic model accurately explained the behavior of a hydrogen atom, it was unable to explain the behavior of any other atom. Verify the limitations of the Bohr model by analyzing an electron transition in a neon atom. Unlike a hydrogen atom, a neon atom has ten electrons. One of these electrons makes a transition between the n 5 and the n 3 energy states, emitting a photon in the process. 1. Assuming that the neon atom\u2019s electron can be treated as an electron in a hydrogen atom, what photon energy does the Bohr model predict? 2. Assuming that the neon atom\u2019s electron can be treated as an electron in a hydrogen atom, what photon wavelength does the Bohr model predict? n 5 n 4 n 3 E5 E4 E3 n 2 E2 3. The actual wavelength of the photon emitted during the transition is 632.8 nm. What is the percent error of the Bohr model\u2019s prediction of photon wavelength? n 1 E1 632.8 nm The Bohr model was a major contribution to scientists\u2019 understanding of the structure of the atom. In addition to calculating the emission spectrum, Bohr and his students were able to calculate the ionization energy of a hydrogen atom. The ionization energy of an atom is the energy needed to completely free an electron from an atom. The calculated ionization value closely agreed with experimental data. The Bohr model further provided an explanation of some of the chemical properties of the elements. The idea that atoms have electron arrangements unique to each element is the foundation of much of our knowledge of chemical reactions and bonding. Niels Bohr, whose accomplishments are commemorated in the stamps shown in Figure 28-12, was awarded a Nobel prize in 1922. \u25a0 Figure 28-12 Honored in these postage stamps from Denmark and Sweden, Niels Bohr\u2019s great contribution to our understanding of the atom earned him worldwide recognition and a Nobel prize. 28.1 Section Review 9. Rutherford\u2019s Nuclear Model Summarize the structure of the atom according to Rutherford\u2019s nuclear model. 13. Absorption Spectrum Explain how the absorption spectrum of a gas can be determined. Describe the reasons for the spectrum\u2019s appearance. 10. Spectra How do the emission spectra of incandescent solids and atomic gases differ? In what ways are they similar? 11. Bohr Model Explain how energy is conserved when an atom absorbs a photon of light. 12. Orbit Radius A helium ion behaves like a hydrogen atom. The radius", " of the ion\u2019s lowest energy level is 0.0265 nm. According to Bohr\u2019s model, what is the radius of the second energy level? 14. Bohr Model Hydrogen has been detected transitioning from the 101st to the 100th energy levels. What is the radiation\u2019s wavelength? Where in the electromagnetic spectrum is this emission? 15. Critical Thinking The nucleus of a hydrogen atom has a radius of about 1.51015 m. If you were to build a model of the hydrogen atom using a softball (r 5 cm) to represent the nucleus, where would you locate an electron in the n 1 Bohr orbit? Would it be in your classroom? On school property? physicspp.com/self_check_quiz Section 28.1 The Bohr Model of the Atom 759 760-765 CH28-S2-845813 3/22/04 6:47 AM Page 760 28.2 The Quantum Model of the Atom Objectives \u2022 Describe the shortcomings of Bohr\u2019s atomic model. \u2022 Describe the quantum model of the atom. \u2022 Explain how a laser works. \u2022 Describe the properties of laser light. Vocabulary quantum model electron cloud quantum mechanics coherent light incoherent light stimulated emission laser \u25a0 Figure 28-13 For an electron to have a stable orbit around the nucleus, the circumference of the orbit must be a whole-number multiple, n, of the de Broglie wavelength. Note that the wholenumber multiples n 3 and n 5 are stable, whereas n 2.9 is not. 760 Chapter 28 The Atom The postulates that Bohr made could not be explained on the basis of accepted physics principles of the time. For example, electromagnetic theory required that the accelerated particles radiate energy, causing the rapid collapse of the atom. In addition, the idea that electron orbits have well-defined radii was in conflict with the Heisenberg uncertainty principle. How could Bohr\u2019s work be put on a firm foundation? From Orbits to an Electron Cloud The first hint to the solution of these problems was provided by Louis de Broglie. Recall from Chapter 27 that de Broglie proposed that particles have wave properties, just as light has particle properties. The de Broglie wavelength of a particle with momentum mv is defined as h/mv. The angular momentum of a particle can be defined as mvr hr/. Thus, Bohr\u2019s required quantized angular-momentum condition, mvr nh/2, can be written in the following way: n h or n 2r r h 2 That is, the circumference of the Bohr orbit, 2r, is equal to a whole number multiple, n, of de Broglie wavelengths,. Figure 28-13 illustrates this relationship. In 1926, Austrian physicist Erwin Schr\u00f6edinger used de Broglie\u2019s wave model to create a quantum theory of the atom based on waves. This theory did not propose a simple planetary model of an atom, as Bohr\u2019s model had. In particular, the radius of the electron orbit was not likened to the radius of the orbit of a planet about the Sun. The Heisenberg uncertainty principle states that it is impossible to know both the position and momentum of an electron at the same time. Thus, the modern quantum model of the atom predicts only the probability that an electron is in a specific region. Interestingly, the quantum model predicts that the most probable distance between the electron and the nucleus for a hydrogen atom is the same as the radius predicted by the Bohr model. n 3 n 5 n 2.9 Stable condition: three complete cycles per orbit Stable condition: five complete cycles per orbit Unstable condition 760-765 CH28-S2-845813 6/8/04 7:10 PM Page 761 n 2 10 Bohr radii 10 Bohr radii n 1 10 Bohr radii 10 Bohr radii \u25a0 Figure 28-14 These plots show the probability of finding the electron in a hydrogen atom for distances up to approximately 10 Bohr radii from the nucleus for the first and second energy levels. The density of the distribution of dots corresponds to the probability of finding the electron. Note that the Bohr radius 0.053 nm. The probability of the electron being at any specific radius can be calculated, and a three-dimensional plot can be constructed showing regions of equal probability. The region in which there is a high probability of finding the electron is called the electron cloud. A slice through the electron cloud for the two lowest energy states of a hydrogen atom is shown in Figure 28-14. Even though the quantum model of the atom is difficult to visualize, quantum mechanics, the study of the properties of matter using its wave properties, makes use of this model and has been extremely successful in predicting many details of atomic structure. These details are very difficult to calculate precisely for all but the simplest atoms. Only very sophisticated computers can make", " highly accurate approximations for the heavier atoms. Quantum mechanics also enables the structures of many molecules to be calculated, allowing chemists to determine the arrangement of atoms in the molecules. Guided by quantum mechanics, chemists have been able to create new and useful molecules that do not occur in nature. Quantum mechanics also is used to analyze the details of the emission and absorption of light by atoms. As a result of quantum mechanical theory, a new source of light was developed. Lasers As you know, light emitted by an incandescent source has a continuum of wavelengths, whereas light produced by an atomic gas consists of only a few distinct wavelengths. Light from both sources travels in all directions. Furthermore, light waves emitted by atoms at one end of a gas-discharge tube are not necessarily in step, or synchronized, with waves from the other end of the tube. That is, the waves are not necessarily all at the same point in their cycle at the same time. Recall from Chapter 19 that waves that are in step, with their minima and maxima coinciding, are said to be coherent. Light waves that are coherent are referred to as coherent light. Out-of-step light waves produce incoherent light. Both types of light waves are shown in Figure 28-15. Light is emitted by atoms that have been excited. So far, you have learned about two ways in which atoms can be excited: thermal excitation and electron collision. Atoms also can be excited by collisions with photons of exactly the right energy. a Incoherent Light b Coherent Light \u25a0 Figure 28-15 Waves of incoherent light (a) and coherent light (b) are shown. Time Time Section 28.2 The Quantum Model of the Atom 761 760-765 CH28-S2-845813 3/22/04 6:48 AM Page 762 \u25a0 Figure 28-16 During spontaneous emission, an electron in an atom drops from the excited state, E2, to the ground state, E1, by spontaneously emitting a photon with energy, hf (a). During stimulated emission, an excited atom is struck by a photon with energy E2 the ground state and emits a photon. Both the incident and emitted photon have the same E1 (b). energy: Ephoton E1. The atom drops to E2 762 Chapter 28 The Atom a Spontaneous Emission b Stimulated Emission hf Ephoton E2 E1 Ephoton hf E2 E1 Ephoton hf Ephoton hf Before After Spontaneous and stimulated emission What happens when an atom is in an excited state? After a very short time, it normally returns to the ground state, giving off a photon of the same energy that it absorbed, as shown in Figure 28-16a. This process is called spontaneous emission. In 1917, Einstein considered what happens to an atom already in an excited state that is struck by a photon with an energy equal to the energy difference between the excited state and the ground state. He showed that the atom, by a process known as stimulated emission, returns to the ground state and emits a photon with an energy equal to the energy difference between the two states. The photon that caused, or stimulated, the emission is not affected. The two photons leaving the atom not only will have the same frequency, but they will be in step, or coherent, as shown in Figure 28-16b. Either of the two photons can now strike other excited atoms, thereby producing additional photons that are in step with the original photons. This process can continue and produce an avalanche of photons, all of the same wavelength and all having their maxima and minima at the same times. For this process to occur, certain conditions must be met. First, there must be other atoms in the excited state. Second, the atoms must remain in the excited state long enough to be struck by a photon. Third, the photons must be contained so that they are able to strike other excited atoms. In 1959, a device called a laser was invented that fulfilled all the conditions needed to produce coherent light. The word laser is an acronym that stands for light amplification by stimulated emission of radiation. An atom that emits light when it is stimulated in a laser is said to lase. Atom excitation The atoms in a laser can be excited, or pumped, as outlined in Figure 28-17. An intense flash of light with a wavelength shorter than that of the laser can be used to pump the atoms. The shorter wavelength, higher energy photons produced by the flash collide with and excite the lasing atoms. When one of the excited atoms decays to a lower energy state by emitting a photon, the avalanche of photons begins. This results in the emission of a brief flash, or pulse, of laser light. Alternatively, the lasing atoms can be excited by collisions with other atoms. In the helium-neon lasers often seen in science classrooms, an electric discharge excites the helium atoms. These excited helium atoms collide with", " the neon atoms, pumping them to an excited state and causing them to lase. The laser light resulting from this process is continuous rather than pulsed. 760-765 CH28-S2-845813 3/22/04 6:48 AM Page 763 Lasing The photons emitted by the lasing atoms are contained by confining the lasing atoms within a glass tube that has parallel mirrors at each end. One of the mirrors is more than 99.9% reflective and reflects nearly all of the light hitting it, whereas the other mirror is partially reflective and allows only about 1 percent of the light hitting it to pass through. Photons that are emitted in the direction of the ends of the tube will be reflected back into the gas by the mirrors. The reflected photons strike more atoms, releasing more photons with each pass between the mirrors. As the process continues, a high intensity of photons builds. The photons that exit the tube through the partially reflecting mirror produce the laser beam. Figure 28-18 shows a laser being used in a laboratory. Excited state Collision occurs Stimulated emission Lasing transition Pumping process Return to ground state Because all the stimulated photons are emitted in step with the photons that struck the atoms, laser light is coherent light. The light is also all of one wavelength, or monochromatic, because the transition of electrons between only one pair of energy levels in one type of atom is involved. The parallel mirrors used in the laser result in the emitted laser light being highly directional. In other words, laser light does not diverge much as it travels. Because a typical laser beam is very small, often only about 2 mm in diameter, the light is very intense. Many solid, liquid, and gas substances can be made to lase. Most substances produce laser light at only one wavelength. The light from some lasers, however, can be tuned, or adjusted, over a range of wavelengths. Ground state \u25a0 Figure 28-17 When a photon strikes an excited atom, it stimulates the atom to emit a second coherent photon and to make a transition to a lower state. \u25a0 Figure 28-18 This argon laser produces a beam of coherent light. Section 28.2 The Quantum Model of the Atom 763 ImageState/Firstlight 760-765 CH28-S2-845813 6/8/04 7:11 PM Page 764 Table 28-1 Common Lasers Medium Wavelength (nm) Type Krypton-fluoride excimer 248 (ultraviolet) Pulsed (KrF gas) Nitrogen (N2 gas) Indium gallium nitride (InGaN crystal) Argon ion (Ar+ gas) 337 (ultraviolet) Pulsed 420 Continuous 476.5, 488.0, 514.5 Continuous Neon (Ne gas) 632.8 Gallium aluminum arsenide 635, 680 Continuous Continuous (GaAlAs crystal) Gallium arsenide (GaAs crystal) 840\u20131350 (infrared) Continuous Neodymium (Nd: YAG crystal) 1064 (infrared) Pulsed Carbon dioxide (CO2 gas) 10,600 (infrared) Continuous Laser Applications If you have used a CD or a DVD player, then you have used a laser. These lasers, as well as the ones used in laser pointers, are made of semiconducting solids. The laser in a CD player is made of layers of gallium arsenide (GaAs) and gallium aluminum and arsenide (GaAlAs). The lasing layer is only 200-nm thick, and each side of the crystal is only 1\u20132 mm long. The atoms in the semiconducting solid are pumped by an electric current, and the resulting photons are amplified as they bounce between the polished ends of the crystal. Table 28-1 shows the wavelength and type, pulsed or continuous, produced by some common lasers. Most lasers are very inefficient. For example, no more than 1 percent of the electrical energy delivered to a gas laser is converted to light energy. Although crystal lasers have efficiencies near 20 percent, they often have much less power than gas lasers do. Despite their inefficiency, the unique properties of laser light have led to many applications. Laser beams are narrow and highly directional\u2014they do not spread out over long distances. For this reason, surveyors use laser beams in applications such as checking the straightness of long tunnels and pipes. When astronauts visited the Moon, they left behind mirrors on the Moon\u2019s surface. Scientists on Earth have used the mirrors to reflect a laser beam transmitted from Earth. The distance between Earth and the Moon was thus accurately determined. By tracking the Moon\u2019s location from different parts of Earth, the movement of Earth\u2019s tectonic plates has been measured. Laser light also is commonly used in fiber-optics communications. A fiber optic cable makes use of total internal reflection to transmit light over distances of many kilometers with little loss of signal energy. The laser", "aracteristic spectra. Samples with extremely small numbers of atoms can be analyzed in this way. In fact, single atoms have been detected and held almost motionless by means of laser excitation. The concentrated power of laser light is used in a variety of ways. In medicine, for example, lasers are used to reshape the corneas of eyes. Lasers also can be used in surgery, as shown in Figure 28-19, in place of a knife to cut flesh with little loss of blood. In industry, lasers are used to cut materials such as steel and to weld materials together. In the future, lasers may be able to produce nuclear fusion to create an almost inexhaustible energy source. A hologram, shown in Figure 28-20, is a photographic recording of both the phase and the intensity of light. Holograms are made possible by the coherent nature of laser light. Holograms form realistic three-dimensional images and can be used, among other applications, in industry to study the vibrations of sensitive equipment and their components. 28.2 Section Review 16. Lasers Which of the lasers in Table 28-1 emits the reddest light (visible light with the longest wavelength)? Which of the lasers emit blue light? Which of the lasers emit beams that are not visible to the human eye? 17. Pumping Atoms Explain whether green light could be used to pump a red laser. Why could red light not be used to pump a green laser? 18. Bohr Model Limitations Although it was able to accurately predict the behavior of hydrogen, in what ways did Bohr\u2019s atomic model have serious shortcomings? 19. Quantum Model Explain why the Bohr model of the atom conflicts with the Heisenberg uncertainty principle, whereas the quantum model does not. 20. Lasers Explain how a laser makes use of stimu- lated emission to produce coherent light. 21. Laser Light What are the four characteristics of laser light that make it useful? 22. Critical Thinking Suppose that an electron cloud were to get so small that the atom was almost the size of the nucleus. Use the Heisenberg uncertainty principle to explain why this would take a tremendous amount of energy. physicspp.com/self_check_quiz Section 28.2 The Quantum Model of the Atom 765 (t)Getty Images, (b)Roger Ressmeyer/CORBIS 766-767 CH28-LAB-845813 6/8/04 7:15 PM Page 766 Finding the Size of an Atom Ernest Rutherford used statistical analysis and probability to help analyze the results of his gold foil experiment. In this experiment, you will model the gold foil experiment using BBs and cups. You will then analyze your results in terms of probability to estimate the size of an object that cannot be seen. QUESTION How can probability be used to determine the size of an object that cannot be seen? Objectives Procedure \u25a0 Interpret data to determine the probability of a BB striking an unseen object. \u25a0 Calculate the size of an unseen object based on probability. Safety Precautions \u25a0 Be sure to immediately pick up BBs that have fallen onto the floor. Materials shoe box three identical small paper cups 200 BBs centimeter ruler large towel or cloth 1. Use the centimeter ruler to measure the length and width of the inside of the shoe box. Record the measurements in the data table. 2. Use the centimeter ruler to measure the diameter of the top of one of the cups. Record the measurement in the data table. 3. Place the shoe box in the center of a folded towel, such that the towel extends at least 30 cm beyond each side of the shoe box. 4. Randomly place the three paper cups in the bottom of the shoe box. 5. Have your lab partner randomly drop 200 BBs into the shoe box. Make sure he or she distributes the BBs evenly over the area of the shoe box. Note that some of the BBs may miss the shoe box and land on the towel. 6. Count the number of BBs in the cups and record the value in the data table. Analyze 1. Calculate the area of the shoe box. The area of a rectangular shape is given by the equation Area Length Width. 2. Calculate the area of a cup using the diameter you measured. The area of a circle is given by (Diam eter)2 the equation Area. 4 3. Calculate the total area of the cups by multiplying the area per cup by the total number of cups. 4. Calculate the percentage of shoe box that is occupied by the three cups by dividing the total area of the cups by the area of the shoe box and then multiplying by 100. 5. Calculate the percentage of BBs that landed in the cups by dividing the number of BBs in the cups by the number of BBs dropped, and then multiplying by 100. 766 Horizons Companies 766-767 CH28-LAB-845813 6/8/04 7:", "16 PM Page 767 Data Table Shoe box length (cm) Shoe box width (cm) Shoe box area (cm2) Measured diameter of cup (cm) Calculated area of a cup (cm2) Your Data Data from Group 2 Data from Group 3 Data from Group 4 Data from Group 5 Class Average Total number of cups 3 3 3 3 3 3 Total calculated area of cups (cm2) Percentage of shoe box occupied by cups (%) Number of BBs dropped 200 200 200 200 200 200 Number of BBs in cups Percent of BBs in cups Percent of shoe box occupied by cups based on probability Total area of cups based on probability (cm2) Number of cups 3 3 3 3 3 3 Area of one cup based on probability (cm2) 6. Determine the percentage of the shoe box occupied by the cups based on probability. Note that this percentage is (ideally) equal to the percentage of BBs that landed in cups. 7. Calculate the total area of the cups based on probability. To calculate this value, multiply the percentage of the shoe box occupied by the cups (based on probability) by the area of the shoe box. 8. Calculate the area of each cup based on probability by dividing the total area of the cups based on probability by three. 9. Record the experimental data from the other groups in the data table and then calculate classroom averages for all of the data. 10. Error Analysis Compare your calculated value for the area of the cup based on probability (experimental value) with the area of the cup calculated from the measured diameter (accepted value). What is the percent error in your value based upon probability? Calculate the percent error using the following equation: Percent error \uf8e6 Accepted value Experimental value \uf8e6 Accepted value 100 Conclude and Apply 1. Were you able to accurately determine the area of a cup based on probability? Explain in terms of the percent error. 2. List the error sources in this experiment and describe their effects on the results. Going Further If larger cups were used in the experiment, do you think you would need fewer, the same, or more BBs to achieve accurate results? Explain. Real-World Physics Your teacher polls the class about postponing a test. Does the accuracy of the poll depend on how many students are surveyed? Explain. To find out more about the atom, visit the Web site: physicspp.com 767 768 CH28-FEATURE-845813 3/22/04 6:56 AM Page 768 Atom Laser Atom Laser The recently developed atom laser is a technology with a promising future. Unlike traditional lasers that emit beams or pulses of coherent photons, atom lasers emit beams or pulses of coherent atoms. As explained below, coherent atoms are different from the incoherent atoms that make up ordinary matter. developed a way to eject small pulses (between 100,000 and 1,000,000 atoms) of coherent atoms from a Bose-Einstein condensate into a beam. In this first atom laser, pulses of coherent atoms could only travel in a single direction. The emitted atoms behaved like projectiles, History In 1923, Louis Victor de Broglie predicted that all particles, including atoms, have wave properties. Wavelength is inversely related to the particle\u2019s mass and velocity, and it is too small to be observed at room temperature. As the atom is cooled, its speed is reduced and its wavelength increases. In the 1920s, Albert Einstein and Satyendra Nath Bose researched particles known as bosons. They predicted that if bosons could be cooled to their lowest possible energy state, then all of the particles would have the same phase and the same wavelength. In other words, the particles would have coherent properties. This unusual phase of matter is called a BoseEinstein condensate. following downward arcing paths due to the influence of gravity. As shown in the photo, the coherent atoms in each pulse tend to spread apart as the beam propagates. In 1999, William Phillips found a way to send pulses of coherent atoms in any direction and how to prevent the atoms from spreading out as the beam propagated. By making a series of very short pulses, Phillips was able to form a continuous beam of coherent atoms. The Future Bose-Einstein condensates and atom lasers will be used to study the basic properties of quantum mechanics and matter waves. Scientists anticipate that atom lasers will be useful in making more-precise atomic clocks and in building small electronic circuits. The atom laser also may be used in atomic interferometry to precisely measure gravitational forces and to test relativity. An atom laser emits pulses of coherent sodium atoms. Each pulse contains 105 to 106 atoms, and the pulses are accelerated downward due to gravity. The spreading of the pulses is a result of repulsive forces. The first Bose-Einstein condensates were formed in 1995 by Eric Cornell and Carl Wieman, and independently, by Wolfgang Ketterle. In further research, Ketterle brought two", " separate Bose-Einstein condensate samples near each other. Ketterle observed wavelike interference patterns from the atoms in the condensates. He went on to prove that all of the atoms in the condensate had the same wavelength and were in phase. The atoms in the condensate were coherent, just as Bose and Einstein predicted. The First Atom Laser In 1997, Ketterle and his collaborators announced the first steps in the development of an atom laser. They 768 Wolfgang Ketterle/Massachusetts Institute of Technology Future Technology Going Further 1. Research Investigate what fermions are and if they can form Bose-Einstein condensates. (Hint: See how the Pauli exclusion principle applies to fermions.) 2. Critical Thinking Atom lasers operate in an ultrahigh vacuum environment. Why do you think this is so? 769-773 CH28-SG CA STP-845813 3/22/04 6:59 AM Page 769 28.1 The Bohr Model of the Atom Vocabulary \u2022 alpha particles (p. 748) \u2022 nucleus (p. 749) \u2022 absorption spectrum (p. 751) \u2022 energy level (p. 753) \u2022 ground state (p. 753) \u2022 excited state (p. 753) \u2022 principal quantum number (p. 755) Key Concepts \u2022 Ernest Rutherford directed positively charged, high-speed alpha particles at thin metal foils. By studying the paths of the reflected particles, he showed that an atom is mostly empty space with a tiny, massive, positively charged nucleus at its center. \u2022 The spectrum produced by atoms of an element can be used to identify an unknown sample of that element. \u2022 If white light passes through a gas, the gas absorbs the same wavelengths that it would emit if it were excited. If the light is then passed through a prism, the absorption spectrum of the gas is visible. \u2022 Niels Bohr\u2019s model of the atom correctly showed that the energy of an atom can have only certain values; thus, it is quantized. He showed that the energy of a hydrogen atom in energy level n is equal to the product of 13.6 eV and the inverse of n2. En 1 13.6 eV 2 n \u2022 According to the Bohr model, atoms make transitions between allowable energy levels, absorbing or emitting energy in the form of photons (electromagnetic waves). The energy of the photon is equal to the difference between the initial and final states of the atom. Ephoton Ef Ei \u2022 According to the Bohr model, the radius of an electron\u2019s orbit can have only certain (quantized) values. The radius of the electron in energy level n of a hydrogen atom is given by the following equation. rn 2 2 n h 4 q2 2 K m 28.2 The Quantum Model of the Atom Vocabulary \u2022 quantum model (p. 760) \u2022 electron cloud (p. 761) \u2022 quantum mechanics (p. 761) \u2022 coherent light (p. 761) \u2022 incoherent light (p. 761) \u2022 stimulated emission (p. 762) \u2022 laser (p. 762) Key Concepts \u2022 In the quantum-mechanical model of the atom, the atom\u2019s energy has only specific, quantized values. \u2022 In the quantum-mechanical model of the atom, only the probability of finding the electron in a specific region can be determined. In the hydrogen atom, the most probable distance of the electron from the nucleus is the same as the electron\u2019s orbital radius in the Bohr model. \u2022 Quantum mechanics is extremely successful in calculating the properties of atoms, molecules, and solids. \u2022 Lasers produce light that is directional, powerful, monochromatic, and coherent. Each property gives the laser useful applications. physicspp.com/vocabulary_puzzlemaker 769 769-773 CH28-SG CA STP-845813 3/22/04 7:00 AM Page 770 Concept Mapping 23. Complete the following concept map using these terms: energy levels, fixed electron radii, Bohr model, photon emission and absorption, energy-level difference. Applying Concepts 34. As the complexity of energy levels changes from atom to atom, what do you think happens to the spectra that they produce? has results that include 35. Northern Lights The northern lights are caused by high-energy particles from the Sun striking atoms high in Earth\u2019s atmosphere. If you looked at these lights through a spectrometer, would you see a continuous or line spectrum? Explain. 36. If white light were emitted from Earth\u2019s surface and observed by someone in space, would its spectrum appear to be continuous? Explain. 37. Is money a good example of quantization? Is water? Explain. with energy equal to 38. Refer to Figure 28-21. A photon with energy of 6.2 eV enters", " a mercury atom in the ground state. Will it be absorbed by the atom? Explain. Energy Level Diagram for a Mercury Atom Mastering Concepts 24. Describe how Rutherford determined that the positive charge in an atom is concentrated in a tiny region, rather than spread throughout the atom. (28.1) 25. How does the Bohr model explain why the absorption spectrum of hydrogen contains exactly the same frequencies as its emission spectrum? (28.1) 26. Review the planetary model of the atom. What are some of the problems with a planetary model of the atom? (28.1) 27. Analyze and critique the Bohr model of the atom. What three assumptions did Bohr make in developing his model? (28.1) 28. Gas-Discharge Tubes Explain how line spectra from gas-discharge tubes are produced. (28.1) 29. How does the Bohr model account for the spectra emitted by atoms? (28.1) 30. Explain why line spectra produced by hydrogen gasdischarge tubes are different from those produced by helium gas-discharge tubes. (28.1) 31. Lasers A laboratory laser has a power of only 0.8 mW (8104 W). Why does it seem more powerful than the light of a 100-W lamp? (28.2) 32. A device similar to a laser that emits microwave radiation is called a maser. What words likely make up this acronym? (28.2) ) Ionization (0.00) E9 (1.56) E8 (1.57) E7 (2.48) E6 (2.68) E5 (3.71) E4 (4.95) E3 (5.52) E2 (5.74 E1 (10.38) 10.38 8.82 8.81 7.90 7.70 6.67 5.43 4.86 4.64 ).00 \u25a0 Figure 28-21 39. A certain atom has four energy levels, with E4 being the highest and E1 being the lowest. If the atom can make transitions between any two levels, how many spectral lines can the atom emit? Which transition produces the photon with the highest energy? 40. A photon is emitted when an electron in an excited hydrogen atom drops through energy levels. What is the maximum energy that the photon can have? If this same amount of energy were given to the atom in the ground state, what would happen? 41. Compare the quantum mechanical theory of the atom with the Bohr model. 33. What properties of laser light led to its use in light 42. Given a red, green, and blue laser, which produces shows? (28.2) photons with the highest energy? 770 Chapter 28 The Atom For more problems, go to Additional Problems, Appendix B. 769-773 CH28-SG CA STP-845813 6/8/04 7:19 PM Page 771 Mastering Problems 28.1 The Bohr Model of the Atom 43. A calcium atom drops from 5.16 eV above the ground state to 2.93 eV above the ground state. What is the wavelength of the photon emitted? 44. A calcium atom in an excited state, E2, has an energy level 2.93 eV above the ground state. A photon of energy 1.20 eV strikes the calcium atom and is absorbed by it. To what energy level is the calcium atom raised? Refer to Figure 28-22. Energy Level Diagram for a Calcium Atom 6.08 (Ionization) 5.58 5.49 5.18 5.32 5.16 4.62 4.55 E10 E9E8 E7E6 E5E4E3 E2 E1 4.13 2.93 ) \u25a0 Figure 28-22 45. A calcium atom is in an excited state at the E6 energy level. How much energy is released when the atom drops down to the E2 energy level? Refer to Figure 28-22. 46. A photon of orange light with a wavelength of 6.00102 nm enters a calcium atom in the E6 excited state and ionizes the atom. What kinetic energy will the electron have as it is ejected from the atom? 47. Calculate the energy associated with the E7 and the E2 energy levels of the hydrogen atom. 48. Calculate the difference in energy levels in the previous problem. Refer to Figure 28-21 for Problems 49 and 50. 49. A mercury atom is in an excited state at the E6 energy level. a. How much energy would be needed to ionize the atom? b. How much energy would be released if the atom dropped down to the E2 energy level instead? 50. A mercury atom in an excited state has an energy of 4.95 eV. It absorbs a photon that raises it to the next-higher energy level. What is the energy", " and the frequency of the photon? 51. What energies are associated with a hydrogen atom\u2019s energy levels of E2, E3, E4, E5, and E6? 52. Using the values calculated in problem 51, calculate the following energy differences. a. E6 b. E6 E5 E3 E2 E2 c. E4 d. E5 e. E5 E3 53. Use the values from problem 52 to determine the frequencies of the photons emitted when an electron in a hydrogen atom makes the energy level changes listed. 54. Determine the wavelengths of the photons having the frequencies that you calculated in problem 53. 55. A hydrogen atom emits a photon with a wavelength of 94.3 nm when its falls to the ground state. From what energy level did the electron fall? 56. For a hydrogen atom in the n 3 Bohr orbital, find the following. a. the radius of the orbital b. the electric force acting between the proton and the electron c. the centripetal acceleration of the electron d. the orbital speed of the electron (Compare this speed with the speed of light.) 28.2 The Quantum Model of the Atom 57. CD Players Gallium arsenide lasers are commonly used in CD players. If such a laser emits at 840 nm, what is the difference in eV between the two lasing energy levels? 58. A GaInNi laser lases between energy levels that are separated by 2.90 eV. a. What wavelength of light does it emit? b. In what part of the spectrum is this light? 59. A carbon-dioxide laser emits very high-power infrared radiation. What is the energy difference in eV between the two lasing energy levels? Consult Table 28-1. 60. The power in a laser beam is equal to the energy of each photon times the number of photons per second that are emitted. a. If you want a laser at 840 nm to have the same power as one at 427 nm, how many times more photons per second are needed? b. Find the number of photons per second in a 5.0-mW 840-nm laser. 61. HeNe Lasers The HeNe lasers used in many classrooms can be made to lase at three wavelengths: 632.8 nm, 543.4 nm, and 1152.3 nm. a. Find the difference in energy between the two states involved in the generation of each wavelength. b. Identify the color of each wavelength. physicspp.com/chapter_test Chapter 28 Assessment 771 769-773 CH28-SG CA STP-845813 3/22/04 7:01 AM Page 772 Mixed Review 62. A photon with an energy of 14.0 eV enters a hydrogen atom in the ground state and ionizes it. With what kinetic energy will the electron be ejected from the atom? 63. Calculate the radius of the orbital associated with the energy levels E5 and E6 of the hydrogen atom. 64. A hydrogen atom is in the n 2 level. a. If a photon with a wavelength of 332 nm strikes the atom, show that the atom will be ionized. b. When the atom is ionized, assume that the electron receives the excess energy from the ionization. What will be the kinetic energy of the electron in joules? 65. A beam of electrons is directed onto a sample of atomic hydrogen gas. What minimum energy of the electrons is needed for the hydrogen atoms to emit the red light produced when the atom goes from the n 3 to the n 2 state? 66. The most precise spectroscopy experiments use \u201ctwo-photon\u201d techniques. Two photons with identical wavelengths are directed at the target atoms from opposite directions. Each photon has half the energy needed to excite the atoms from the ground state to the desired energy level. What laser wavelength would be needed to make a precise study of the energy difference between n 1 and n 2 in hydrogen? Thinking Critically 67. Apply Concepts The result of projecting the spectrum of a high-pressure mercury vapor lamp onto a wall in a dark room is shown in Figure 28-23. What are the differences in energy levels for each of the three visible lines? 436 nm 546 nm 579 nm \u25a0 Figure 28-23 68. Interpret Scientific Illustrations After the emission of the visible photons described in problem 67, the mercury atom continues to emit photons until it reaches the ground state. From an inspection of Figure 28-21, determine whether or not any of these photons would be visible. Explain. 69. Anaylze and Conclude A positronium atom consists of an electron and its antimatter relative, the positron, bound together. Although the lifetime of this \u201catom\u201d is very short\u2014on the average it lives one-seventh of a microsecond\u2014its energy levels can be measured. The Bohr model can be used to calculate energies with the mass of the", " electron replaced by one-half its mass. Describe how the radii of the orbits and the energy of each level would be affected. What would be the wavelength of the E2 to E1 transition? Writing in Physics 70. Do research on the history of models of the atom. Briefly describe each model and identify its strengths and weaknesses. 71. Green laser pointers emit light with a wavelength of 532 nm. Do research on the type of laser used in this type of pointer and describe its operation. Indicate whether the laser is pulsed or continuous. Cumulative Review 72. The force on a test charge of 3.00107 C is 0.027 N. What is the electric field strength at the position of the test charge? (Chapter 21) 73. A technician needs a 4- resistor but only has 1resistors of that value. Is there a way to combine what she has? Explain. (Chapter 23) 74. A 1.0-m-long wire is moved at right angles to Earth\u2019s magnetic field where the magnetic induction is 5.0105 T at a speed of 4.0 m/s. What is the EMF induced in the wire? (Chapter 25) 75. The electrons in a beam move at 2.8108 m/s in an electric field of 1.4104 N/C. What value must the magnetic field have if the electrons pass through the crossed fields undeflected? (Chapter 26) 76. Consider the modifications that J. J. Thomson would need to make to his cathode-ray tube so that it could accelerate protons (rather than electrons), then answer the following questions. (Chapter 26) a. To select particles of the same velocity, would the ratio E/B have to be changed? Explain. b. For the deflection caused by the magnetic field alone to remain the same, would the B field have to be made smaller or larger? Explain. 77. The stopping potential needed to return all the electrons ejected from a metal is 7.3 V. What is the maximum kinetic energy of the electrons in joules? (Chapter 27) 772 Chapter 28 The Atom For more problems, go to Additional Problems, Appendix B. 769-773 CH28-SG CA STP-845813 3/22/04 7:02 AM Page 773 Multiple Choice 1. Which model of the atom was based on the results of Rutherford\u2019s gold foil experiment? Questions 5 and 6 refer to the diagram showing the Balmer series electron transitions in a hydrogen atom. Bohr model nuclear model plum pudding model quantum mechanical model 2. A mercury atom emits light with a wavelength of 405 nm. What is the energy difference between the two energy levels involved with this emission? 0.22 eV 2.14 eV 3.06 eV 4.05 eV 3. The diagram below shows the energy levels for a mercury atom. What wavelength of light is emitted when a mercury atom makes a transition from E7 to E4? 167 nm 251 nm 500 nm 502 nm ) Ionization (0.00) E9 (1.56) E8 (1.57) E7 (2.48) E6 (2.68) E5 (3.71) E4 (4.95) E3 (5.52) E2 (5.74 E1 (10.38) 10.38 8.82 8.81 7.90 7.70 6.67 5.43 4.86 4.64 0.00 ).38 eV 0.54 eV 0.85 eV 1.51 eV 3.40 eV 13.6 eV E6 E5 E4 E3 E2 E1 5. Which energy level transition is responsible for the emission of light with the greatest frequency? E2 to E5 E3 to E2 E3 to E6 E6 to E2 6. What is the frequency of the Balmer series line related to the energy level transition from E4 to E2? (Note that 1 eV 1.601019 J.) 2.551014 Hz 4.321014 Hz 6.151014 Hz 1.081015 Hz Extended Answer 7. Determine the wavelength of light emitted when a hydrogen atom makes a transition from the n 5 energy level to the n 2 energy level. 4. Which statement about the quantum model of the atom is false? The possible energy levels of the atom are quantized. The locations of the electrons around the nucleus are known precisely. The electron cloud defines the area where electrons are likely to be located. Stable electron orbits are related to the de Broglie wavelength. Stumbling Is Not Failing Occasionally, you will encounter a question that you have no idea how to answer. Even after reading the question several times, it still may not make sense. If the question is multiple choice, focus on the part of the question that you know something about. Eliminate as", " many choices as you can. Then take your best guess and move on. physicspp.com/standardized_test Chapter 28 Standardized Test Practice 773 774-783 CH29-S1-845813 3/22/04 7:07 AM Page 774 What You\u2019ll Learn \u2022 You will be able to distinguish among electric conductors, semiconductors, and insulators. \u2022 You will examine how pure semiconductors are modified to produce desired electric properties. \u2022 You will compare diodes and transistors. Why It\u2019s Important Semiconductors have electric properties that allow them to act as oneway conductors to amplify weak signals in many common electronic devices. Fast Math Computers and electronic devices use the controlled movement of electrons and holes in semiconductors to do quick calculations and logical operations. Think About This A silicon microchip might be small, but it may contain the equivalent of millions of resistors, diodes, and transistors. How can this level of complexity be produced in such a tiny structure? physicspp.com 774 774-783 CH29-S1-845813 6/8/04 7:26 PM Page 775 How can you show conduction in a diode? Question Which way does a two-color light-emitting diode (LED) conduct? viewed through the stroboscopic disk? Critical Thinking Suggest a possible explanation for your observations. Procedure 1. Obtain a bi-color (red-green) LED and a 9- to 12-V AC power supply or transformer. 2. Wire a 100-\u2126 resistor and the LED in series with the AC source. 3. Be careful when plugging in the AC source so you are not shocked. Do not touch the resistor as it may become hot. Plug the AC source into a GFCI receptacle. 4. Record your observations of the LED. 5. Hold a stroboscopic disk in front of the LED and spin it. Record your observations of the LED as viewed through the disk. Analysis What color was the LED after you plugged in the power supply? What color was the LED as 29.1 Conduction in Solids Electronic devices depend not only on natural conductors and insula- tors, but also on materials that have been designed and produced by scientists and engineers working together. This brief investigation into electronics begins with a study of how materials conduct electricity. All electronic devices owe their origins to the vacuum tubes of the early 1900s. In vacuum tubes, electron beams flow through space to amplify and control faint electric signals. Vacuum tubes are big, require lots of electric power, and generate considerable heat. They have heated filaments, which require the replacement of the tubes after one to five years. In the late 1940s, solid-state devices were invented that could do the jobs of vacuum tubes. These devices are made of materials, such as silicon and germanium, known as semiconductors. The devices amplify and control very weak electric signals through the movement of electrons within a tiny crystalline space. Because very few electrons flow in them and they have no filaments, devices made from semiconductors operate with a low power input. They are very small, don\u2019t generate much heat, and are inexpensive to manufacture. The estimated useful life of these devices is 20 years or more. Objectives \u2022 Describe electron motion in conductors and semiconductors. \u2022 Compare and contrast n-type and p-type semiconductors. Vocabulary semiconductors band theory intrinsic semiconductors dopants extrinsic semiconductors Section 29.1 Conduction in Solids 775 Horizons Companies 774-783 CH29-S1-845813 3/22/04 7:11 AM Page 776 Band Theory of Solids You have learned about electric conductors and insulators. In conductors, electric charges can move easily, but not in insulators. When you examine these two types of materials at the atomic level, the difference in the way they are able to carry charges becomes apparent. You learned in Chapter 13 that crystalline solids consist of atoms bound together in regular arrangements. You also know from Chapters 27 and 28 that an atom consists of a dense, positively charged nucleus surrounded by a cloud of negatively charged electrons. These electrons can occupy only certain allowed energy levels. Under most conditions, the electrons in an atom occupy the lowest possible energy levels. This condition is referred to as the ground state. Because the electrons can have only certain energies, any energy changes that occur are quantized; that is, the energy changes occur in specific amounts. Energy bands Suppose you could construct a solid by assembling atoms together, one by one. You would start with an atom in the ground state. At large interatomic spacings ( 0.8 nm) with no very near neighbors, the graph in Figure 29-1 shows two discrete energy levels for the atom. As the solid crystal forms by moving atoms closer to the atom,", " the electric fields of these other neighboring atoms affect the energy levels of its electrons. In the solid crystal, the result is that the ground state energy levels in each atom are split into multiple levels by the electric fields of all of its neighbors. There are so many of these levels, and they are so close together, that they no longer appear as distinct levels, but as the energy bands shown in Figure 29-1. The lower energy or valence bands are occupied by bonding electrons in the crystal, and the higher energy or conduction bands are available for electrons to move from atom to atom. Notice in Figure 29-1 that the atomic separations for crystalline silicon and crystalline carbon (diamond) translate to valence bands and conduction bands that are separated by energy gaps. These gaps have no energy levels available for electrons. They are called forbidden energy regions. This description of valence and conduction bands, separated by forbidden energy gaps, is known as the band theory of solids and can be used to better understand electric conduction. For example, the band diagram in \u25a0 Figure 29-1 Energy levels of an atom are split apart when other atoms are brought closer, resulting in an energy gap between the valence and conduction bands. Carbon Silicon Conduction band E 5.5 eV Valence band Conduction band E 1.1 eV Valence band C Si y g r e n E 776 Chapter 29 Solid-State Electronics 0.2 0.4 0.6 0.8 Atomic separation (nm) 774-783 CH29-S1-845813 3/22/04 7:11 AM Page 777 Figure 29-1 suggests that a lot more energy will be required to move valence electrons from the valence band to the conduction band in the case of crystalline carbon (diamond structure) compared to silicon. Carbon in graphite form is a much better conductor because the structure of the atom arrangement in graphite gives it a smaller energy gap than that of diamond. Crystalline silicon has a smaller energy gap than diamond does. At absolute zero, the valence band of silicon would be completely full and the conduction band would be completely empty. At room temperature, some number of valence electrons have enough thermal energy to jump the 1.1-eV gap to the conduction band and serve as charge carriers. As the temperature increases and more electrons gain enough energy to jump the gap, the conductivity of silicon will increase. Germanium has an energy gap of 0.7 eV, which is smaller than that of silicon. This means that germanium is a better conductor than silicon at any given temperature. However, it also means that germanium is too sensitive to heat for many electronic applications. Relatively small changes in temperature cause large changes in the conductivity of germanium, making control and stability of circuits troublesome. Lead has an interatomic spacing of 0.27 nm. Figure 29-1 shows that this would translate to a band-gap diagram in which the conduction band overlaps the valence band. One would, therefore, expect lead to be a good conductor, and it is. Materials with overlapping, partially filled bands are conductors, as indicated in Figure 29-2. Conductors When a potential difference is placed across a material, the resulting electric field exerts a force on the electrons. The electrons accelerate and gain energy and the field does work on them. If there are bands within the material that are only partially filled, then there are energy levels available that are only slightly higher than the electrons\u2019 ground state levels. As a result, the electrons that gain energy from the field can move from one atom to the next. Such movement of electrons from one atom to the next is an electric current, and the entire process is known as electric conduction. Materials with partially filled bands, such as the metals aluminum, lead, and copper, conduct electricity easily. Random motion The free electrons in conductors move about rapidly in a random way, changing directions when they collide with the cores of the atoms. However, if an electric field is put across a length of wire, there will be a net force pushing the electrons in one direction. Although their motion is not greatly affected, they have a slow overall movement dictated by the electric field, as shown in Figure 29-3. Electrons continue to move rapidly with speeds of 106 m/s in random directions, and they drift very slowly at speeds of 105 m/s or slower toward the positive end of the wire. This model of conductors is called the electron-gas model. If the temperature is increased, the speeds of the electrons increase, and, consequently, they collide more frequently with atomic cores. Thus, as the temperature rises, the conductivity of metals is reduced. Conductivity is the reciprocal of resistivity. As conductivity is reduced, a material\u2019s resistance rises. \u25a0 Figure 29-2 In a material that is a good conductor, the conduction band is partially filled.", " The blueshaded area shows energies occupied by electrons. Conductor y g r e n E Conduction band Valence band V \u25a0 Figure 29-3 The electrons move rapidly and randomly in a conductor. If a field is applied across the wire, the electrons drift toward one end. Electron flow is opposite in direction to conventional current. Section 29.1 Conduction in Solids 777 774-783 CH29-S1-845813 3/22/04 7:12 AM Page 778 The Free-Electron Density of a Conductor How many free electrons exist in a cubic centimeter of copper? Each atom contributes one electron. The density, atomic mass, and number of atoms per mole of copper can be found in Appendix D. 1 Analyze the Problem \u2022 Identify the knowns using Appendix D. Known: For copper: 1 free e per atom 8.96 g/cm3 M 63.54 g/mol NA 6.021023 atoms/mol Unknown: free e/cm3? 2 Solve for the Unknown e)(NA)( 1)( ) (fr e e e e fre 3 om t a m c atoms)( )(6.02 (1 e 1 e e fr 3. 6 om t a 1 8.491022 free e/cm3 in copper 10 m 1 M ol 23 ol m g 54 )(8 g 6.9 3 m c 1 ) Cu Cu Cu e e Cu Cu Cu e e Cu Cu Cu e e e e e Substitute free e/1 atom 1 free e/1 atom, 6.021023 atoms/mol, M 63.54 g/mol, NA 8.96 g/cm3 3 Evaluate the Answer \u2022 Are the units correct? Dimensional analysis on the units confirms the number of free electrons per cubic centimeter. \u2022 Is the magnitude realistic? One would expect a large number of electrons in a cubic centimeter. Math Handbook Dimensional Calculations pages 846\u2014847 1. Zinc, with a density of 7.13 g/cm3 and an atomic mass of 65.37 g/mol, has two free electrons per atom. How many free electrons are there in each cubic centimeter of zinc? 2. Silver has 1 free electron per atom. Use Appendix D and determine the number of free electrons in 1 cm3 of silver. 3. Gold has 1 free electron per atom. Use Appendix D and determine the number of free electrons in 1 cm3 of gold. 4. Aluminum has 3 free electrons per atom. Use Appendix D and determine the number of free electrons in 1 cm3 of aluminum. 5. The tip of the Washington Monument was made of 2835 g of aluminum because it was a rare and costly metal in the 1800s. Use problem 4 and determine the number of free electrons in the tip of the Washington Monument. Insulators In an insulating material, the valence band is filled to capacity and the conduction band is empty. As shown in Figure 29-4, an electron must gain a large amount of energy to go to the next energy level. In an insulator, the lowest energy level in the conduction band is 5\u201310 eV above the highest energy level in the valence band, as shown in Figure 29-4a. There is at least a 5-eV gap of energies that no electrons can possess. 778 Chapter 29 Solid-State Electronics 774-783 CH29-S1-845813 3/22/04 7:13 AM Page 779 a Insulator Conduction band E 5 eV Forbidden gap Valence band b Semiconductor Conduction band E 1 eV Forbidden gap Valence band \u25a0 Figure 29-4 Compare the valence and conduction bands in an insulator (a) and in a semiconductor (b). Compare these diagrams with the one shown in Figure 29-2. Although electrons have some kinetic energy as a result of their thermal energy, the average kinetic energy of electrons at room temperature is not sufficient for them to jump the forbidden gap. If a small electric field is placed across an insulator, almost no electrons gain enough energy to reach the conduction band, so there is no current. Electrons in an insulator must be given a large amount of energy to be pulled into the conduction band. As a result, the electrons in an insulator tend to remain in place, and the material does not conduct electricity. Semiconductors Electrons can move more freely in semiconductors than in insulators, but not as easily as in conductors. As shown in Figure 29-4b, the energy gap between the valence band and the conduction band is approximately 1 eV. How does the structure of a semiconductor explain its electronic characteristics? Atoms of the most common semiconductors, silicon (Si) and germanium (Ge), each have four valence electrons. These four electrons are involved in binding the atoms together into the solid crystal. The val", "ence electrons form a filled band, as in an insulator, but the forbidden gap between the valence and conduction bands is much smaller than in an insulator. Not much energy is needed to pull one of the electrons from a silicon atom and put it into the conduction band, as illustrated in Figure 29-5a. Indeed, the gap is so small that some electrons reach the conduction band as a result of their thermal kinetic energy alone. That is, the random motion of atoms and electrons gives some electrons enough energy to break free of their home atoms and wander around the silicon crystal. If an electric field is applied to a semiconductor, electrons in the conduction band move through the solid according to the direction of the applied electric field. In contrast to the effect in a metal, the higher the temperature of a semiconductor, the more able electrons are to reach the conduction band, and the higher the conductivity. An atom from which an electron has broken free is said to contain a hole. As shown in Figure 29-5b, a hole is an empty energy level in the valence band. The atom now has a net positive charge. An electron from the conduction band can jump into this hole and become bound to an atom once again. When a hole and a free electron recombine, their opposite charges neutralize each other. \u25a0 Figure 29-5 Some electrons in semiconductors have enough thermal kinetic energy to break free and wander through the crystal, as shown in the crystal structure (a) and in the bands (b). a b e Atom core Electron Hole e Free electron Conduction band e Hole Valence band Section 29.1 Conduction in Solids 779 774-783 CH29-S1-845813 6/8/04 7:27 PM Page 780 The electron, however, has left behind a hole at its previous location. Thus, as in a game of musical chairs, the negatively charged, free electrons move in one direction and the positively charged holes move in the opposite direction. Pure semiconductors that conduct as a result of thermally freed electrons and holes are called intrinsic semiconductors. Because so few electrons or holes are available to carry charge, conduction in intrinsic semiconductors is very low, making their resistances very high. Fraction of Free Electrons in an Intrinsic Semiconductor Because of the thermal kinetic energy of solid silicon at room temperature, there are 1.451010 free electrons/cm3. What is the number of free electrons per atom of silicon at room temperature? 1 Analyze the Problem \u2022 Identify the knowns and unknowns. Known: 2.33 g/cm3 M 28.09 g/mol 6.021023 atoms/mol NA For Si: 1.451010 free e/cm3 Unknown: free e/atom of Si? 2 Solve for the Unknown 1 N A e ( e fr e om t a )(1.451010 free e/cm3 for Si) )(M )(1 )(1.451 g)( )(2 ( 3 m c 1 g 3.3 2 atoms 2.901013 free e/atom of Si m 1 ol 6.02 23 10 09 ol m 8. 1 free e 10 0 3 m c ) Si Si Si Si Si Si e Si Si Si Si Atom core Electron Hole e Free electron 6.021023 atoms/mol, Substitute NA M 28.09 g/mol, 2.33 g/cm3, free e/cm3 Si 1.451010 free e/cm3 or, roughly 1 out of 3 trillion Si atoms has a free electron 3 Evaluate the Answer \u2022 Are the units correct? Using dimensional analysis confirms the correct units. \u2022 Is the magnitude realistic? In an intrinsic semiconductor, such as silicon at room temperature, very few atoms have free electrons. Math Handbook Operations with Scientific Notation pages 842\u2014843 6. In pure germanium, which has a density of 5.23 g/cm3 and an atomic mass of 72.6 g/mol, there are 2.251013 free electrons/cm3 at room temperature. How many free electrons are there per atom? 7. At 200.0 K, silicon has 1.89105 free electrons/cm3. How many free electrons are there per atom at this temperature? What does this temperature represent on the Celsius scale? 8. At 100.0 K, silicon has 9.231010 free electrons/cm3. How many free electrons are there per atom at this temperature? What does this temperature represent on the Celsius scale? 9. At 200.0 K, germanium has 1.161010 free electrons/cm3. How many free electrons are there per atom at this temperature? 10. At 100.0 K, germanium has 3.47 free electrons/cm3. How many free electrons are there per atom at", " this temperature? 780 Chapter 29 Solid-State Electronics 774-783 CH29-S1-845813 4/1/04 8:55 PM Page 781 Doped Semiconductors The conductivity of intrinsic semiconductors must be increased greatly to make practical devices. Dopants are electron donor or acceptor atoms that can be added in low concentrations to intrinsic semiconductors. Dopants increase conductivity by making extra electrons or holes available. The doped semiconductors are known as extrinsic semiconductors. n-type semiconductors If an electron donor with five valence electrons, such as arsenic (As), is used as a dopant for silicon, the product is called an n-type semiconductor. Figure 29-6a shows a location in a silicon crystal where a dopant atom has replaced one of the silicon atoms. Four of the five As valence electrons bind to neighboring silicon. The fifth electron is called the donor electron. The energy of this donor electron is so close to the conduction band that thermal energy can easily move the electron from the dopant atom into the conduction band, as shown in Figure 29-7a. Conduction in n-type semiconductors is increased by the availability of these extra donor electrons to the conduction band. p-type semiconductors If an electron acceptor with three valence electrons, such as gallium (Ga), is used as a dopant for silicon, the product is called a p-type semiconductor. When a gallium atom replaces a silicon atom, one binding electron is missing, creating a hole in the silicon crystal, as shown in Figure 29-6b. Electrons in the conduction band can easily drop into these holes, creating new holes. Conduction in p-type semiconductors is enhanced by the availability of the extra holes provided by the acceptor dopant atoms, as shown in Figure 29-7b. Both p-type and n-type semiconductors are electrically neutral. Adding dopant atoms of either type does not add any net charge to a semiconductor. Both types of semiconductor use electrons and holes in conduction. Only a few dopant atoms per million silicon atoms are needed to increase the conductivity of semiconductors by a factor of 1000 or more. Silicon is doped by putting a pure silicon crystal in a vacuum with a sample of the dopant material. The dopant is heated until it is vaporized, and the atoms condense on the cold silicon. The dopant diffuses into the silicon on warming, and a thin layer of aluminum or gold is evaporated onto the doped crystal. A wire is welded to this metal layer, allowing the user to apply a potential difference across the doped silicon. a n-type b p-type Electron Donor levels Acceptor levels a Arsenic Donor Si Si Si Si As Si Si Si Si Excess electron free to move b Gallium Acceptor Si Si Si Si Ga Si Si Si Si Excess hole free to move \u25a0 Figure 29-6 A donor atom of arsenic with five valence electrons replaces a silicon atom and provides an unbound electron in the silicon crystal (a). An acceptor atom of gallium with three valence electrons creates a hole in the crystal (b). Hole Conduction bands Forbidden gap Electron Valence bands Hole \u25a0 Figure 29-7 In an n-type semiconductor (a), donor energy levels place electrons in the conduction band. In a p-type semiconductor (b), acceptor energy levels result in holes in the valence band. Section 29.1 Conduction in Solids 781 774-783 CH29-S1-845813 3/22/04 7:14 AM Page 782 Thermistors The electric conductivity of intrinsic and extrinsic semiconductors is sensitive to both temperature and light. Unlike metals in which conductivity is reduced when the temperature rises, an increase in temperature of a semiconductor allows more electrons to reach the conduction band, and conductivity increases and resistance decreases. One semiconductor device, the thermistor, is designed so that its resistance depends very strongly on temperature. The thermistor can be used as a sensitive thermometer and to compensate for temperature variations of other components in an electric circuit. Thermistors also can be used to detect radio waves, infrared radiation, and other forms of radiation. The Conductivity of Doped Silicon Silicon is doped with arsenic so that one in every million silicon atoms is replaced by an arsenic atom. Each arsenic atom donates one electron to the conduction band. a. What is the density of free electrons? b. By what ratio is this density greater than that of intrinsic Arsenic Donor c. 1 2 silicon with 1.451010 free e/cm3? Is conduction mainly by the electrons of the silicon or the arsenic? Analyze the Problem \u2022 Identify the knowns and unknowns. Known: 1", " As atom/106 Si atoms 1 free e/As atom 4.991022 Si atoms/cm3 1.451010 free e/cm3 in intrinsic Si Unknown: free e/cm3 donated by As? ratio of As-donated free e to intrinsic free e? Solve for the Unknown from As) ( a. (fre e e 3 m c e ee fr o at s A )()(Si to a 3 m c ms) m Excess electron free to move Si Si Si Si As Si Si Si Si Substitute free e/As atom 1 free e/1 As atom, As atoms/Si atoms 1 As atom/1106 Si atoms, Si atoms/cm3 4.991022 Si atoms/cm3 m 1 1 e e 3 m c (fre m o t oms t a e ee fr o at s A A 1 a s 1 6 10 i S ) ( )( 4.991022 Si atoms cm3 ) )( 4.991016 free e/cm3 from As donor in doped Si ) free e/cm3 in doped Si free e/cm3 in intrinsic Si 4.991016 free e/cm3 in doped Si 1.451010 free e/cm3 in intrinsic Si b. Ratio ( ( 3.44106 As-donated electron per instrinsic Si electron ) Substitute 4.991016 free e/cm3 in doped Si, 1.451010 free e/cm3 in intrinsic Si c. Because there are over 3 million arsenic-donated electrons for every intrinsic electron, conduction is mainly by the arsenic-donated electrons. Math Handbook Ratios page 838 3 Evaluate the Answer \u2022 Are the units correct? Using dimensional analysis confirms the correct units. \u2022 Is the magnitude realistic? The ratio is large enough so that intrinsic electrons make almost no contribution to conductivity. 782 Chapter 29 Solid-State Electronics 774-783 CH29-S1-845813 3/22/04 7:15 AM Page 783 11. 12. If you wanted to have 1104 as many electrons from arsenic doping as thermally free electrons in silicon at room temperature, how many arsenic atoms should there be per silicon atom? If you wanted to have 5103 as many electrons from arsenic doping as thermally free electrons in the germanium semiconductor described in problem 6, how many arsenic atoms should there be per germanium atom? 13. Germanium at 400.0 K, has 1.131015 thermally liberated carriers/cm3. If it is doped with 1 As atom per 1 million Ge atoms, what is the ratio of doped carriers to thermal carriers? 14. Silicon at 400.0 K, has 4.541012 thermally liberated carriers/cm3. If it is doped with 1 As atom per 1 million Si, what is the ratio of doped carriers to thermal carriers? 15. Based on problem 14, draw a conclusion about the behavior of germanium devices as compared to silicon devices at temperatures in excess of the boiling point of water. Light meters Other useful applications of semiconductors depend on their light sensitivity. When light falls on a semiconductor, the light can excite electrons from the valence band to the conduction band in the same way that other energy sources excite atoms. Thus, the resistance decreases as the light intensity increases. Extrinsic semiconductors can be tailored to respond to specific wavelengths of light. These include the infrared and visible regions of the spectrum. Materials such as silicon and cadmium sulfide serve as light-dependent resistors in light meters used by lighting engineers to design the illumination of stores, offices, and homes; and by photographers to adjust their cameras to capture the best images, as shown in Figure 29-8. \u25a0 Figure 29-8 Photographers use light meters to measure the intensity of incident light on an object. 29.1 Section Review 16. Carrier Mobility In which type of material, a conductor, a semiconductor, or an insulator, are electrons most likely to remain with the same atom? 19. Conductor or Insulator? Magnesium oxide has a forbidden gap of 8 eV. Is this material a conductor, an insulator, or a semiconductor? 17. Semiconductors If the temperature increases, the number of free electrons in an intrinsic semiconductor increases. For example, raising the temperature by 8\u00b0C doubles the number of free electrons in silicon. Is it more likely that an intrinsic semiconductor or a doped semiconductor will have a conductivity that depends on temperature? Explain. 18. Insulator or Conductor? Silicon dioxide is widely used in the manufacture of solid-state devices. Its energy-band diagram shows a gap of 9 eV between the valence band and the conduction band. Is it more useful as an insulator or a conductor? 20. Intrinsic and Extrinsic", " Semiconductors You are designing an integrated circuit using a single crystal of silicon. You want to have a region with relatively good insulating properties. Should you dope this region or leave it as an intrinsic semiconductor? 21. Critical Thinking Silicon produces a doubling of thermally liberated carriers for every 8\u00b0C increase in temperature, and germanium produces a doubling of thermally liberated carriers for every 13\u00b0C increase. It would seem that germanium would be superior for high-temperature applications, but the opposite is true. Explain. physicspp.com/self_check_quiz Section 29.1 Conduction in Solids 783 784-789 CH29-S2-845813 3/22/04 7:16 AM Page 784 29.2 Electronic Devices Today\u2019s electronic instruments, such as radios, televisions, CD players, and microcomputers, rely on semiconductor devices that are combined on chips of silicon a few millimeters wide. In these devices, current and voltage vary in more complex ways than are described by Ohm\u2019s law. As a result, semiconductor devices can change current from AC to DC and amplify voltages. Diodes The simplest semiconductor device is the diode. A diode consists of a sandwich of p-type and n-type semiconductors. Rather than two separate pieces of doped silicon being joined, a single sample of intrinsic silicon is treated first with a p-dopant, then with an n-dopant. Metal contacts are coated on each region so that wires can be attached, as shown in Figure 29-9a. The boundary between the p-type and the n-type regions is called the junction. The resulting device, therefore, is called a pn-junction diode. The free electrons on the n-side of the junction are attracted to the positive holes on the p-side. The electrons readily move into the p-side and recombine with the holes. Holes from the p-side similarly move into the n-side, where they recombine with electrons. As a result of this flow, the n-side has a net positive charge, and the p-side has a net negative charge. These charges produce forces in the opposite direction that stop further movement of charge carriers. The region around the junction is left with neither holes nor free electrons. This region, depleted of charge carriers, is called the depletion layer. Because it has no charge carriers, it is a poor conductor of electricity. Thus, a junction diode consists of relatively good conductors at the ends that surround a poor conductor. a Junction Diode Metal p-type n-type Metal Objectives \u2022 Describe how diodes limit current to motion in only one direction. \u2022 Explain how a transistor can amplify or increase voltage changes. Vocabulary diode depletion layer transistor microchip \u25a0 Figure 29-9 A diagram of the pn-junction diode (a) shows the depletion layer, where there are no charge carriers. Compare the magnitude of current in a reverse-biased diode (b) and a forward-biased diode (c). Reverse-Biased Diode b Forward-Biased Diode c R Electrons and holes recombine at junction R Holes Depletion layer Junction Electrons p-type n-type p-type n-type Holes filled I Electrons leave New holes created I New electrons added 784 Chapter 29 Solid-State Electronics 784-789 CH29-S2-845813 6/8/04 7:30 PM Page 785 When a diode is connected into a circuit in the way shown in Figure 29-9b, both the free electrons in the n-type semiconductor and the holes in the p-type semiconductor are attracted toward the battery. The width of the depletion layer is increased, and no charge carriers meet. Almost no current passes through the diode: it acts like a very large resistor, almost an insulator. A diode oriented in this manner is a reverse-biased diode. If the battery is connected in the opposite direction, as shown in Figure 29-9c, charge carriers are pushed toward the junction. If the voltage of the battery is large enough\u20140.6 V for a silicon diode\u2014electrons reach the p-end and fill the holes. The depletion layer is eliminated, and a current passes through the diode. The battery continues to supply electrons for the n-end. It removes electrons from the p-end, which is the same as supplying holes. With further increases in voltage from the battery, the current increases. A diode in this kind of circuit is a forward-biased diode. The graph shown in Figure 29-10 shows the current through a silicon diode as a function of voltage across it. If the applied voltage is negative, the reverse-biased diode acts like a very high-value resistor and only a tiny current passes (", "about 1011 A for a silicon diode). If the voltage is positive, the diode is forward-biased and acts like a low-value resistor, but not, however, one that obeys Ohm\u2019s law. One major use of a diode is to convert AC voltage to DC voltage with only one polarity. When a diode is used in a circuit that does this, it is called a rectifier. The arrow in the symbol for the diode, which you\u2019ll see in Example Problem 4, shows the direction of conventional current. Diode Current v. Voltage 2 1 0 1 Voltage (V) 20 15 10 \u25a0 Figure 29-10 The graph indicates current-voltage characteristics for a silicon junction diode. A Diode in a Simple Circuit A silicon diode, with I/V characteristics like those shown in Figure 29-10, is connected to a power supply through a 470- resistor. The power supply forward-biases the diode, and its voltage is adjusted until the diode current is 12 mA. What is the power supply voltage? 1 Analyze and Sketch the Problem \u2022 Draw a circuit diagram connecting a diode, a 470- resistor, and a power supply. Indicate the direction of current. Known: I 0.012 A 0.70 V Vd R 470 Unknown:? Vb R I Vb Diode Vd 2 Solve for the Unknown The voltage drop across the resistor is known from V IR, and the power supply voltage is the sum of the resistor and the diode voltage drops. Vb IR Vd (0.012 A)(470 ) 0.70 V 6.3 V Substitute I 0.012 A, R 470, Vd 0.70 V 3 Evaluate the Answer \u2022 Are the units correct? The power supply\u2019s potential difference is in volts. \u2022 Is the magnitude realistic? It is in accord with the current and the resistance. Math Handbook Order of Operations page 843 Section 29.2 Electronic Devices 785 784-789 CH29-S2-845813 6/8/04 7:31 PM Page 786 \u25a0 Figure 29-11 Diode lasers are used as both light emitters and detectors in bar-code scanners. 22. What battery voltage would be needed to produce a current of 2.5 mA in the diode in Example Problem 4? 23. What battery voltage would be needed to produce a current of 2.5 mA if another identical diode were added in series with the diode in Example Problem 4? 24. Describe how the diodes in the previous problem should be connected. 25. Describe what would happen in problem 23 if the diodes were connected in series but with improper polarity. 26. A germanium diode has a voltage drop of 0.40 V when 12 mA passes through it. If a 470- resistor is used in series, what battery voltage is needed? Light-emitting diodes Diodes made from combinations of gallium and aluminum with arsenic and phosphorus emit light when they are forwardbiased. When electrons reach the holes in the junction, they recombine and release the excess energy at the wavelengths of light. These diodes are called light-emitting diodes, or LEDs. Some LEDs are configured to emit a narrow beam of coherent, monochromatic laser light. Such diode lasers are compact, powerful light sources. They are used in CD players, laser pointers, and supermarket bar-code scanners, as shown in Figure 29-11. Diodes can detect light as well as emit it. Light falling on the junction of a reverse-biased pn-junction diode creates electrons and holes, resulting in a current that depends on the light intensity. Approximations often are used in diode circuits because diode resistance is not constant. For diode circuits, the first approximation ignores the forward voltage drop across the diode. The second approximation takes into account a typical value for the diode voltage drop. A third approximation uses additional information about the diode, often in the form of a graph, as shown in the illustration to the right. The curve is the characteristic current-volatage curve for the diode. The straight line shows current-voltage conditions for all possible diode voltage drops for a 180- resistor, a 1.8-V battery, and a diode, from a zero diode voltage drop and 10.0 mA at one end, to a 1.8-V drop, 0.0 mA at the other end. Use the diode circuit in Example Problem 4 with Vb but with R 180 : 1. Determine the diode current using the first approximation. 1.8 V, 2. Determine the diode current using the second approximation and assuming a 0.70-V diode drop. 3. Determine the diode current using the third approximation by using", " the accompanying diode graph. 4. Estimate the error for all three approximations, ignoring the ) 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 Diode Current v. Voltage Graphic solution 0.2 0.6 1.0 1.4 1.8 Voltage (V) battery and resistor. Discuss the impact of greater battery voltages on the errors. 786 Getty Images Chapter 29 Solid-State Electronics 784-789 CH29-S2-845813 6/8/04 7:31 PM Page 787 a pnp-transistor b npn-transistor Circuit symbols B Transistors base C collector E E emitter Transistors and Integrated Circuits A transistor is a simple device made of doped semiconductor material. An npn-transistor consists of layers of n-type semiconductor on either side of a thin p-type layer. The central layer is called the base and the regions on either side are the emitter and the collector. The schematic symbols for the two transistor types are shown in Figure 29-12. The arrow on the emitter shows the direction of conventional current. The operation of an npn-transistor is illustrated in Figure 29-13. The two pn-junctions in the transistor can be thought of as initially forming two back-to-back diodes. The battery on the right, VC, keeps the collector more positive than the emitter. The base-collector diode is reverse-biased, with a wide depletion layer, so there is no current from the collector to the base. When the battery on the left, VB, is connected, the base is more positive than the emitter. That makes the base-emitter diode forward-biased, allowing current IB from the base to the emitter. The very thin base region is part of both diodes in the transistor. The charges injected by IB reduce the reverse bias of the base-collector diode, permitting charge to flow from the collector to the emitter. A small change in IB thus produces a large change in IC. The collector current causes a voltage drop across resistor RC. Small changes in the voltage, VB, applied to the base produce large changes in the collector current and thus changes in the voltage drop across RC. As a result, the transistor amplifies small voltage changes into much larger changes. If instead the center layer is an n-type region, then the device is called a pnp-transistor. A pnp-transistor works the same way, except that the potentials of both batteries are reversed. Current gain The current gain from the base circuit to the collector circuit is a useful indicator of the performance of a transistor. Although the base current is quite small, it is dependent on the base-emitter voltage that is controlling the collector current. For example, if VB in Figure 29-13 is removed, the collector current will drop to zero. If VB is increased, the base current, IB, increases. The collector current, IC, will also increase, but many times more (perhaps 100 times or so). The current gain from the base to the collector ranges from 50 to 300 for general-purpose transistors. \u25a0 Figure 29-12 Compare the circuit symbols used to represent a pnp-transistor (a) and an npntransistor (b). Diode Laser A typical diode laser emits light at 800 nm, which is the near infrared. The beam is output from a small spot on a GaAlAs chip, and when powered by 80 mA, the diode has a forward voltage drop of about 2 V. Diode lasers commonly are used in optical fiber transmissions. \u25a0 Figure 29-13 A circuit using an npn-transistor demonstrates how voltage can be amplified. IC RC VC B RB IB VB Section 29.2 Electronic Devices 787 784-789 CH29-S2-845813 3/22/04 7:18 AM Page 788 Red Light Make a series circuit with a DC power supply, a 470- resistor, and a red LED. Connect the short lead of the LED to the negative side of the power supply which is plugged into a GFCI-protected receptacle. Attach the other lead to the resistor. Connect the remaining resistor lead to the positive side of the power supply. Slowly increase the voltage until the LED glows. Note the voltage setting on the power supply. 1. Hypothesize what will happen if you reverse the direction of current. 2. Experiment by reversing the connections to the battery. Analyze and Conclude 3. Explain your observations in terms of LED characteristics. In a tape player, the small voltage variations from the voltage induced in a coil by magnetized regions on the tape are amplified to move the speaker coil. In computers, small currents in the base-emitter circuits can turn on or turn off large currents in the collector-emitter", " circuits. In addition, several transistors can be connected together to perform logic operations or to add numbers together. In these cases, they act as fast switches rather than as amplifiers. Microchips An integrated circuit, called a microchip, consists of thousands of transistors, diodes, resistors, and conductors, each less than a micrometer across. All these components can be made by doping silicon with donor or acceptor atoms. A microchip begins as an extremely pure single crystal of silicon, 10\u201330 cm in diameter and 1\u20132 m long, as shown in Figure 29-14. The silicon is sliced by a diamond-coated saw into wafers less than 1-mm thick. The circuit is then built layer by layer on the surface of this wafer. By a photographic process, most of the wafer\u2019s surface is covered by a protective layer, with a pattern of selected areas left uncovered so that they can be doped appropriately. The wafer is then placed in a vacuum chamber. Vapors of a dopant such as arsenic enter the machine, doping the wafer in the unprotected regions. By controlling the amount of exposure, the engineer can control the conductivity of the exposed regions of the chip. This process creates resistors, as well as one of the two layers of a diode or one of the three layers of a transistor. The protective layer is removed, and another one with a different pattern of exposed areas is applied. Then the wafer is exposed to another dopant, often gallium, producing pn-junctions. If a third layer is added, npn-transistors can be formed. The wafer also may be exposed to oxygen to produce areas of silicon dioxide insulation. A layer exposed to aluminum vapors can produce a pattern of thin conducting pathways among the resistors, diodes, and transistors. \u25a0 Figure 29-14 A technician prepares a large silicon crystal to be sliced into wafers for microchips. 788 784-789 CH29-S2-845813 6/8/04 7:32 PM Page 789 \u25a0 Figure 29-15 Microchips form the heart of the central processing unit of computers. A penny is shown in the picture to represent scale. Thousands of identical circuits, usually called chips, are produced at one time on a single wafer. The chips are then tested, sliced apart, and mounted in a carrier; wires are attached to the contacts; and the final assembly is then sealed into a protective plastic body. The tiny size of microchips, shown in Figure 29-15, allows the placement of complicated circuits in a small space. Because electronic signals need only travel tiny distances, this miniaturization has increased the speed of computers. Chips now are used in appliances and automobiles as well as in computers. Semiconductor electronics requires that physicists, chemists, and engineers work together. Physicists contribute their understanding of the motion of electrons and holes in semiconductors. Physicists and chemists together add precisely controlled amounts of dopants to extremely pure silicon. Engineers develop the means of mass-producing chips containing thousands of miniaturized diodes and transistors. Together, their efforts have brought our world into this electronic age. 29.2 Section Review 27. Transistor Circuit The emitter current in a transistor circuit is always equal to the sum of the base IC. If the current and the collector current: IE current gain from the base to the collector is 95, what is the ratio of emitter current to base current? IB 28. Diode Voltage Drop If the diode characterized in Figure 29-10 is forward-biased by a battery and a series resistor so that there is more than 10 mA of current, the voltage drop is always about 0.70 V. Assume that the battery voltage is increased by 1 V. a. By how much does the voltage across the diode or the voltage across the resistor increase? b. By how much does the current through the resistor increase? 29. Diode Resistance Compare the resistance of a pn-junction diode when it is forward-biased and when it is reverse-biased. 30. Diode Polarity In a light-emitting diode, which terminal should be connected to the p-end to make the diode light? 31. Current Gain The base current in a transistor circuit measures 55 A and the collector current measures 6.6 mA. What is the current gain from base to collector? 32. Critical Thinking Could you replace an npntransistor with two separate diodes connected by their p-terminals? Explain. physicspp.com/self_check_quiz Section 29.2 Electronic Devices 789 Horizons Companies 790-791 CH29-LAB-845813 6/8/04 7:34 PM Page 790 Alternate CBL instructions can be found on the Web site. physicspp.com Diode Current and Voltage S", "emiconductor devices, such as diodes and transistors, are fabricated using a semiconductor that is made of partly p-type material and partly n-type material. A semiconductor doped with donor atoms is called an n-type semiconductor, while a semiconductor doped with an element leaving a vacancy or a hole in the lattice structure is referred to as a p-type semiconductor. A diode is made by doping adjacent regions of a semiconductor with donor and acceptor atoms, forming a p-n junction. In this lab, you will investigate the voltage and current characteristics of a diode that is placed in a direct current circuit and compare the response with your knowledge of resistors. QUESTION How do the current-voltage characteristics of a diode, an LED, and a resistor compare? Objectives Materials \u25a0 Collect and organize data of voltage drop and current for a diode and an LED. \u25a0 Measure the current passing through a diode and an LED as a function of voltage drop. \u25a0 Compare and contrast the current-voltage characteristics of a resistor with diodes. DC power supply, variable, 0\u201312 VDC 100-\u2126 resistor, 1 - or 1-W 2 1N4002 diode LED, red ammeter, DC, 0\u2013100 mA voltmeter, 0\u20135 VDC hook-up wire Safety Precautions \u25a0 Use caution with electric connections. Avoid contact with the resistor, which may become hot. \u25a0 Plug power supplies into only GFCI-protected receptacles to prevent shock hazard. Procedure 1. Prepare a data table similar to the one shown on page 791. 2. As indicated on the schematic diagram below, wire the negative terminal of the power supply to the negative side of the ammeter using the hook-up wire provided. 100 012 VDC Diode V a mA b LED 3. Locate the end of the diode with the silver band around it. Attach this end to the positive side of the ammeter. 4. Attach one end of the 100-\u2126 resistor to the free end of the diode. 790 790-791 CH29-LAB-845813 6/8/04 7:35 PM Page 791 Data Table Voltage (V) Drop Across Diode Diode Current (mA) LED Current (mA) 0 0.1 0.2 0.3 1.7 1.8 1.9 2.0 \u2014\u2014\u2014\u2014 \u2014\u2014\u2014\u2014 \u2014\u2014\u2014\u2014 \u2014\u2014\u2014\u2014 5. Attach a wire from the free end of the 100-\u2126 resistor to the positive lead on the power supply. 6. As shown in the schematic, the voltmeter is in parallel with the diode. Attach a wire from the positive side of the voltmeter to the end of the diode attached to the resistor. Connect the negative side of the voltmeter to the end of the diode with the silver band, which is attached to the ammeter. 7. The diode circuit should look like part a of the schematic. Make sure the power supply is turned to zero and plug it in. Slowly turn up the power supply to increase the voltage drop across the diode from 0 up to 0.8 V, in 0.1-V increments. Record the corresponding current at each voltage. CAUTION: If your current goes higher than the capacity of your ammeter, do not increase the voltage any higher, and discontinue taking readings. Turn the power supply to zero and unplug it. 8. Observe the LED leads. One should be shorter than the other. Replace the 1N4002 diode with the LED so that it corresponds with part b of the schematic. To find out more about solid-state electronics, visit the Web site: physicspp.com 9. Connect the shorter lead on the LED to the positive side of the ammeter (negative side of the voltmeter) where the silver banded end of the diode had been connected. Connect the longer lead of the LED to the resistor and to the positive side of the voltmeter. 10. Plug in the power supply. Slowly turn up the power supply to increase the voltage drop across the LED from 0 up to 2.0 V, in 0.1-V increments. Record the corresponding current at each voltage. Additionally, observe the LED and record your observations of it. Analyze 1. Make and Use Graphs On one chart, sketch and label graphs of current versus voltage drop for both the diode and the LED. Place current on the y-axis and voltage on the x-axis. What are the shapes of these curves? 2. Formulate Models Using Ohm\u2019s law, compute and plot on the same graph the voltage-current relationship for a 100-\u2126 resistor from 0 to 2 V. Label this line 100 \u2126. What is the form of this plot? Conclude and Apply 1. Compare and Contrast How do", " the currentvoltage curves for a diode, an LED, and a resistor compare? 2. Which of these devices follow Ohm\u2019s law? 3. Analyze and Conclude Diodes are described as having a turn-on voltage. What is the turnon voltage for a silicon diode? For the LED you used? 4. Explain Why would the specifications for an LED give a light output at a specific current, such as 20 mA? Going Further What could be done to get better measurements of current for the diode? Real-World Physics Small incandescent lightbulbs typically draw 75\u2013150 mA of current at a particular voltage. Why might manufacturers prefer using LEDs in a battery-powered CD or MP3 player? 791 792 CH29-FEATURE-845813 6/8/04 7:44 PM Page 792 Artificial Intelligence The phrase artificial intelligence was first used in 1955. It is defined as \u201cthe scientific understanding of the mechanisms underlying thought and intelligent behavior and their embodiment in machines.\u201d Sometimes, a task needs artificial intelligence to be very logical. At other times, it may need artificial intelligence to think and behave with human biases. The goals in the field of artificial intelligence are to develop systems that can do both. Artificial intelligence also is used to create expert systems in computers that are programmed with knowledge about specific topics. Humans can tell the computer the details of a specific situation, and the computer calculates the most logical course of action. In a medical environment, an expert system can be used to accurately diagnose disorders. Artificial intelligence weighs the facts of the situation and then infers which actions are most appropriate. However, artificial intelligence can operate only with facts that have been taught to the computer. Users must constantly be aware of this limitation of expert systems. A prototype Mars rover decides how to navigate obstacles. The robot, Kismet, displays human facial expressions. Applications Artificial intelligence already is used in many areas, and it will do even more for us in the future. When a computer plays chess, it searches through hundreds of thousands of possible moves before selecting the best one. Research is being done to improve the efficiency of search algorithms. Artificial intelligence currently is used for speech recognition to allow hands-free dialing of cell phones and for some interactive telephone transactions. It is not yet fully capable of understanding natural language, but that is a goal. Three-dimensional computer vision is another future application. To mimic the sensory input and behaviors of humans, computers need to extract three-dimensional reality from two-dimensional images. Progress has been made, but humans are still much better than computers at this. With improved vision, artificial intelligence may control automobiles on Earth, or robots exploring another planet, with no human navigators needed. 792 Extreme Physics (l)NASA, (r)Sam Ogden/Photo Researchers Careers Studying mathematics, mathematical logic, and computer programming languages is important for developing systems that can make rational decisions. Knowledge of psychology assures that these decisions also can have a human character. Going Further 1. Debate the Issue Are there ethical limits to the development of artificial intelligence? 2. Recognize Cause and Effect What problems might cause an expert system to make a poor decision? 3. Critical Thinking In what situations must artificial intelligence be absolutely rational, and in what situations should it include human biases? 793-797 CH29-SG CA STP-845813 6/8/04 7:47 PM Page 793 29.1 Conduction in Solids Vocabulary \u2022 semiconductors (p. 775) \u2022 band theory (p. 776) \u2022 intrinsic semiconductors (p. 780) \u2022 dopants (p. 781) \u2022 extrinsic semiconductors (p. 781) Key Concepts \u2022 Electric conduction may be explained by the band theory of solids. \u2022 In solids, the allowed energy levels for outer electrons in an atom are spread into broad bands by the electric fields of electrons on neighboring atoms. \u2022 The valence and conduction bands are separated by forbidden energy gaps; that is, by regions of energy levels that electrons may not possess. \u2022 In conductors, electrons can move through the solid because the conduction band is partially filled. \u2022 Electrons in metals have a fast random motion. A potential difference across the metal causes a slow drift of electrons, called an electric current. \u2022 In insulators, more energy is needed to move electrons into the conduction band than is generally available. \u2022 Conduction in semiconductors is enhanced by doping pure crystals with small amounts of other kinds of atoms, called dopants. \u2022 n-type semiconductors are doped with electron donor atoms, and they conduct by the response of these donor electrons to applied potential differences. \u2022 Arsenic, with five valence electrons, is an example of a donor atom. \u2022 p-type semiconductors are doped with electron acceptor atoms, and they conduct by making holes available to electrons in the conduction band. \u2022 Gall", "ium, with three valence electrons, is an example of an acceptor atom. 29.2 Electronic Devices Vocabulary \u2022 diode (p. 784) \u2022 depletion layer (p. 784) \u2022 transistor (p. 787) \u2022 microchip (p. 788) Key Concepts \u2022 A pn-junction diode consists of a layer of a p-type semiconductor joined with a layer of an n-type semiconductor. \u2022 Diodes conduct charges in one direction only. They can be used in rectifier circuits to convert AC to DC. \u2022 Electrons and holes near either side of the diode junction combine to produce a region without charge carriers known as the depletion layer. \u2022 Applying a potential difference of the proper polarity across the diode makes the depletion layer even wider, no current is observed, and the diode is said to be reverse-biased. \u2022 Reversing the polarity of the applied potential across the diode greatly reduces the depletion layer, current is observed, and the diode is said to be forward-biased. \u2022 A transistor is a sandwich of three layers of semiconductor material, configured as either npn- or pnp-layers. The center base layer is very thin compared to the other layers, the emitter and collector. \u2022 A transistor can act as an amplifier to convert a weak signal into a much stronger one. \u2022 The ratio of the collector-emitter current to the base current is known as the current gain and is a useful measure of transistor amplification. \u2022 Conductivity of semiconductors increases with increasing temperature or illumination, making them useful as thermometers or light meters. Diodes that emit light when a potential is applied are used in optical devices. physicspp.com/vocabulary_puzzlemaker 793 793-797 CH29-SG CA STP-845813 6/8/04 10:05 PM Page 794 Concept Mapping 33. Complete the concept map using the following terms: transistor, silicon diode, emits light, conducts both ways. Circuit Components copper wire LED 41. For the energy-band diagrams shown in Figure 29-16, which have half-full conduction bands? 42. For the energy-band diagrams shown in Figure 29-16, which ones represent semiconductors? 43. The resistance of graphite decreases as temperature rises. Does graphite conduct electricity more like copper or more like silicon does? 44. Which of the following materials would make a better insulator: one with a forbidden gap 8-eV wide, one with a forbidden gap 3-eV wide, or one with no forbidden gap? conducts one way amplifies 45. Consider atoms of the three materials in problem 44. From which material would it be most difficult to remove an electron? Mastering Concepts 34. How do the energy levels in a crystal of an element differ from the energy levels in a single atom of that element? (29.1) 35. Why does heating a semiconductor increase its conductivity? (29.1) 36. What is the main current carrier in a p-type semiconductor? (29.1) 37. An ohmmeter is an instrument that places a potential difference across a device to be tested, measures the current, and displays the resistance of the device. If you connect an ohmmeter across a diode, will the current you measure depend on which end of the diode was connected to the positive terminal of the ohmmeter? Explain. (29.2) 38. What is the significance of the arrowhead at the emitter in a transistor circuit symbol? (29.2) 39. Describe the structure of a forward-biased diode, and explain how it works. (29.2) Applying Concepts 40. For the energy-band diagrams shown in Figure 29-16, which one represents a material with an extremely high resistance? a b c \u25a0 Figure 29-16 46. State whether the bulb in each of the circuits of Figure 29-17 (a, b, and c) is lighted. a b c \u25a0 Figure 29-17 47. In the circuit shown in Figure 29-18, state whether lamp L1, lamp L2, both, or neither is lighted. L1 L2 \u25a0 Figure 29-18 48. Use the periodic table to determine which of the following elements could be added to germanium to make a p-type semiconductor: B, C, N, P, Si, Al, Ge, Ga, As, In, Sn, or Sb. 49. Does an ohmmeter show a higher resistance when a pn-junction diode is forward-biased or reverse-biased? 50. If the ohmmeter in problem 49 shows the lower resistance, is the ohmmeter lead on the arrow side of the diode at a higher or lower potential than the lead connected to the other side? 51. If you dope pure germanium", " with gallium alone, do you produce a resistor, a diode, or a transistor? 794 Chapter 29 Solid-State Electronics For more problems, go to Additional Problems, Appendix B. 793-797 CH29-SG CA STP-845813 6/8/04 10:06 PM Page 795 52. Draw the time-versus-amplitude waveform for point A in Figure 29-19a assuming an input AC waveform, as shown in Figure 29-19b \u25a0 Figure 29-19 Mastering Problems 29.1 Conduction in Solids 53. How many free electrons exist in a cubic centimeter of sodium? Its density is 0.971 g/cm3, its atomic mass is 22.99 g/mol, and there is 1 free electron per atom. 54. At a temperature of 0\u00b0C, thermal energy frees 1.55109 e/cm3 in pure silicon. The density of silicon is 2.33 g/cm3, and the atomic mass of silicon is 28.09 g/mol. What is the fraction of atoms that have free electrons? 29.2 Electronic Devices 55. LED The potential drop across a glowing LED is about 1.2 V. In Figure 29-20, the potential drop across the resistor is the difference between the battery voltage and the LED\u2019s potential drop. What is the current through each of the following? a. the LED b. the resistor R 240 Battery V 6.0 V 1.2 V LED \u25a0 Figure 29-20 56. Jon wants to raise the current through the LED in problem 55 up to 3.0101 mA so that it glows brighter. Assume that the potential drop across the LED is still 1.2 V. What resistor should be used? 57. Diode A silicon diode with I/V characteristics, as shown in Figure 29-10, is connected to a battery through a 270- resistor. The battery forward-biases the diode, and the diode current is 15 mA. What is the battery voltage? 58. Assume that the switch shown in Figure 29-21 is off. a. Determine the base current. b. Determine the collector current. c. Determine the voltmeter reading. Time 1500 120,000 A 3.5 V A 15 V V \u25a0 Figure 29-21 59. Assume that the switch shown in Figure 29-21 is on, and that there is a 0.70-V drop across the baseemitter junction and a current gain from base to collector of 220. a. Determine the base current. b. Determine the collector current. c. Determine the voltmeter reading. Mixed Review 60. The forbidden gap in silicon is 1.1 eV. Electromagnetic waves striking the silicon cause electrons to move from the valence band to the conduction band. What is the longest wavelength of radiation that could excite an electron in this way? Recall that E 1240 eVnm/. 61. Si Diode A particular silicon diode at 0\u00b0C shows a current of 1.0 nA when it is reverse-biased. What current can be expected if the temperature increases to 104\u00b0C? Assume that the reverse-bias voltage remains constant. (The thermal carrier production of silicon doubles for every 8\u00b0C increase in temperature.) 62. Ge Diode A particular germanium diode at 0\u00b0C shows a current of 1.5 A when it is reverse-biased. What current can be expected if the temperature increases to 104\u00b0C? Assume that the reverse-biasing voltage remains constant. (The thermal charge-carrier production of germanium doubles for every 13\u00b0C increase in temperature.) physicspp.com/chapter_test Chapter 29 Assessment 795 793-797 CH29-SG CA STP-845813 3/22/04 7:39 AM Page 796 63. LED A light-emitting diode (LED) produces green 66. Apply Concepts The I/V characteristics of two light with a wavelength of 550 nm when an electron moves from the conduction band to the valence band. Find the width of the forbidden gap in eV in this diode. LEDs that glow with different colors are shown in Figure 29-24. Each is to be connected through a resistor to a 9.0-V battery. If each is to be run at a current of 0.040 A, what resistors should be chosen for each? 64. Refer to Figure 29-22. a. Determine the voltmeter reading. b. Determine the reading of A1. c. Determine the reading of A2. LED Current v. Voltage ).04 0.02 220 All diodes are silicon. 10.0 V \u25a0 Figure 29-22 V 0 0.5 1 1.5 2 2.5 A1 A2 Voltage (V) \u25a0 Figure 29-24 Thinking", " Critically 65. Apply Concepts A certain motor, in Figure 29-23, runs in one direction with a given polarity applied and reverses direction with the opposite polarity. a. Which circuit (a, b, or c) will allow the motor to run in only one direction? b. Which circuit will cause a fuse to blow if the incorrect polarity is applied? c. Which circuit produces the correct direction of rotation regardless of the applied polarity? d. Discuss the advantages and disadvantages of all three circuits. a M c M b M \u25a0 Figure 29-23 67. Apply Concepts Suppose that the two LEDs in problem 66 are now connected in series. If the same battery is to be used and a current of 0.035 A is desired, what resistor should be used? Writing in Physics 68. Research the Pauli exclusion principle and the life of Wolfgang Pauli. Highlight his outstanding contributions to science. Describe the application of the exclusion principle to the band theory of conduction, especially in semiconductors. 69. Write a one-page paper discussing the Fermi energy level as it applies to energy-band diagrams for semiconductors. Include at least one drawing. Cumulative Review 70. An alpha particle, a doubly ionized (2) helium atom, has a mass of 6.71027 kg and is accelerated by a voltage of 1.0 kV. If a uniform magnetic field of 6.5102 T is maintained on the alpha particle, what will be the particle\u2019s radius of curvature? (Chapter 26) 71. What is the potential difference needed to stop photoelectrons that have a maximum kinetic energy of 8.01019 J? (Chapter 27) 72. Calculate the radius of the orbital associated with the energy level E4 of the hydrogen atom. (Chapter 28) 796 Chapter 29 Solid-State Electronics For more problems, go to Additional Problems, Appendix B. 793-797 CH29-SG CA STP-845813 6/8/04 10:12 PM Page 797 Multiple Choice 1. Which statement about diodes is false? Diodes can ___________. amplify voltage detect light emit light rectify AC 2. Cadmium has two free electrons per atom. How many free electrons are there per cm3 of cadmium? The density of cadmium is 8650 kg/m3. 1.241021 9.261022 9.261024 1.171027 3. The base current in a transistor circuit measures 45 A and the collector current measures 8.5 mA. What is the current gain from base to collector? 110 190 205 240 4. In problem 3, if the base current is increased by 5 A, how much is the collector current increased? 5 A 1 mA 10 mA 190 A 5. A transistor circuit shows a collector current of 4.75 mA, and the base to collector current gain is 250. What is the base current? 1.19 A 18.9 A 4.75 mA 1190 mA 6. Which line in the following table best describes both n- and p-type silicon semiconductors? 7. Which line in the following table best describes the behavior of intrinsic silicon semiconductors to increasing temperature? Effect of Increasing Temperature on Intrinsic Silicon Semiconductors Conductivity Resistance Increases Increases Decreases Decreases Increases Decreases Increases Decreases 8. Thermal electron production in silicon doubles for every 8\u00b0C increase in temperature. A silicon diode at 0\u00b0C shows a current of 2.0 nA when reverse-biased. What will be the current at 112\u00b0C if the reverse-bias voltage is constant? 11 A 33 A 44 A 66 A Extended Answer 9. A silicon diode is connected in the forwardbiased direction to a power supply though a 485- resistor, as shown below. If the diode voltage drop is 0.70 V, what is the power supply voltage when the diode current is 14 mA? R 485 I 14 mA Vb Vd 0.70 V n-type p-type Gallium-doped Added electrons Added electrons Arsenic-doped Arsenic-doped Added holes Added holes Gallium-doped Focus If students near you are talking during a test, you should move. Respond only to the instructor when taking a test. Talking is a distraction and the instructor might think that you are cheating. Don\u2019t take the chance. Focus on the test. physicspp.com/standardized_test Chapter 29 Standardized Test Practice 797 798-805 CH30-S1-845813 3/31/04 12:46 AM Page 798 What You\u2019ll Learn \u2022 You will describe the components of a nucleus and how radioactive decay affects these components. \u2022 You will calculate the energy released in nuclear reactions. \u2022 You will examine how radioactive isotopes and nuclear energy are produced and used. \u2022", " You will understand the building blocks of matter. Why It\u2019s Important Nuclear physics has many applications, including medical research, energy production, and studying the structure of matter. Medicine Radioactive isotopes are used to create images of the brain and other organs at work for medical diagnoses and research. Think About This How does the radiation that is emitted by radioactive isotopes allow scientists and doctors to trace the operations of the human body? physicspp.com 798 Kay Chernush 798-805 CH30-S1-845813 6/15/04 11:17 PM Page 799 How can you model the nucleus? Question How is the force exerted by double-sided tape similar to the strong force? diameters. How does that compare with the range of the force of the tape? The strong force is the same for neutrons and protons. Does this model behave this way? Critical Thinking Stable nuclei usually have more neutrons than protons. Why would this model behave the same way? Procedure 1. Cover the outer edges of 3\u20136 disk magnets with double-sided tape. Do the same thing to 3\u20136 disks of wood or aluminum of the same size. The magnets represent protons. The other disks represent neutrons. 2. Arrange the magnets with all the north poles facing up. 3. Describe the force exerted on one \u201cproton\u201d when another \u201cproton\u201d is brought closer, until they finally touch. 4. Describe the force exerted on a \u201cneutron\u201d by either another \u201cneutron\u201d or a \u201cproton\u201d when they are brought together, until they finally touch. Analysis The strong force drops to zero when the centers of two nucleons are farther apart than their 30.1 The Nucleus Ernest Rutherford not only established the existence of the nucleus, but he also conducted some of the early experiments to discover its structure. It is important to realize that neither Rutherford\u2019s experiments, nor the experiments of those who followed, offered the opportunity to observe the atom directly. Instead, inferences were drawn from the observations that researchers made. Recall from Chapter 28 that Rutherford\u2019s team made careful measurements of the deflection of alpha particles as they hit gold foil. These deflections could be explained if the atom was mostly empty space. The experiment showed that the atom had a small, dense, positively charged center surrounded by nearly massless electrons. After the discovery of radioactivity by Becquerel in 1896, he and others looked at the effects produced when a nucleus breaks apart in natural radioactive decay. Marie and Pierre Curie discovered the new element radium and made it available to researchers all over the world, thereby increasing the study of radioactivity. Scientists discovered that through radioactivity one kind of atom could change into another kind, and thus, atoms must be made up of smaller parts. Ernest Rutherford and Fredrick Soddy used radioactivity to probe the center of the atom, the nucleus. Objectives \u2022 Determine the number of neutrons and protons in nuclides. \u2022 Define the binding energy of the nucleus. \u2022 Relate the energy released in a nuclear reaction to the change in binding energy during the reaction. Vocabulary atomic number atomic mass unit mass number nuclide strong nuclear force nucleons binding energy mass defect Section 30.1 The Nucleus 799 798-805 CH30-S1-845813 6/10/04 9:04 PM Page 800 Description of the Nucleus Are nuclei made up only of positively charged particles? At first, only the mass of the nucleus and the fact that the charge was positive were known. The magnitude of the charge of the nucleus was found as a result of X-ray scattering experiments done by Henry Moseley, a member of Rutherford\u2019s team. The results showed that the positively charged protons accounted for roughly half the mass of the nucleus. One hypothesis was that the extra mass was the result of protons, and that electrons in the nucleus reduced the charge to the observed value. This hypothesis, however, had some fundamental problems. In 1932, English physicist James Chadwick solved the problem when he discovered a neutral particle that had a mass approximately that of the proton. This particle, known as the neutron, accounted for the missing mass of the nucleus without increasing its charge. Mass and charge of the nucleus The only charged particle in the nucleus is the proton. The atomic number, Z, of an atom is the number of protons. The total charge of the nucleus is the number of protons times the elementary charge, e. nuclear charge Ze Both the proton and the neutron have a mass that is about 1800 times the mass of an electron. The proton and neutron masses are approximately equal to 1 u, where u is the atomic mass unit, 1.661027 kg. To determine the approximate mass of the nucleus, multiply the number of neutrons and protons, or mass number, A,", " by u. nuclear mass A(u) Size of the nucleus Rutherford\u2019s experiments produced the first measurements of the size of the nucleus. He found that the nucleus has a diameter of about 1014 m. A typical atom might have a radius 10,000 times larger than the size of the nucleus. Although the nucleus contains nearly all of the mass of an atom, proportionally the nucleus occupies less space in the atom than the Sun does in the solar system. The nucleus is incredibly dense\u2014about 1.41018 kg/m3. If a nucleus could be one cubic centimeter, it would have a mass of about one billion tons. \u25a0 Figure 30-1 The nuclides of hydrogen (a) and helium (b) illustrate that all the nuclides of an element have the same numbers of protons, but have different numbers of neutrons. Protons are red and neutrons are gray in this drawing. a Hydrogen isotopes Helium isotopes 2 p 1 n 2 p 2 n 800 Chapter 30 Nuclear Physics 798-805 CH30-S1-845813 3/31/04 12:50 AM Page 801 Do all elements have the same mass numbers? Looking at the periodic table on page 916, you might notice that the first four elements all have an atomic mass, A, near a whole number. Boron, on the other hand, has a mass of 10.8 u. If the nucleus is made up of only protons and neutrons, each with a mass of approximately 1 u, shouldn\u2019t the total mass of any atom be near a whole number? The puzzle of atomic masses that were not whole numbers was solved with the mass spectrometer. You learned in Chapter 26 how the mass spectrometer demonstrated that an element could have atoms with different masses. For example, in an analysis of a pure sample of neon, not one, but two spots appeared on the film of the spectrometer. The two spots were produced by neon atoms of different masses. One variety of the neon atom was found to have a mass of 20 u, while the second type had a mass of 22 u. All neutral neon atoms have 10 protons in the nucleus and 10 electrons in the atom. One kind of neon atom, however, has 10 neutrons in its nucleus, while the other has 12 neutrons. The two kinds of atoms are called isotopes of neon. The nucleus of an isotope is called a nuclide. All nuclides of an element have the same number of protons, but have different numbers of neutrons, as illustrated by the hydrogen and helium nuclides shown in Figure 30-1. All isotopes of a neutral element have the same number of electrons around the nucleus and behave chemically in the same way. Average mass The measured mass of neon gas is 20.183 u. This figure is now understood to be the average mass of the naturally occurring isotopes of neon. Thus, while the mass of an individual atom of neon is close to a whole number of mass units, the atomic mass determined from an average sample of neon atoms is not. Most elements have several isotopic forms that occur naturally. The mass of one isotope of carbon, carbon-12, is now 1 used to define the mass unit. One u is defined to be the mass of the 2 1 carbon-12 isotope. A special method of notation is used to describe an isotope. A subscript representing the atomic number, or charge, Z, is written to the left of the symbol for the element. A superscript written to the left of the symbol is the mass number, A. This notation takes the form A ZX, where X is any element. For example, carbon-12 is 12 6C, and the two isotopes of neon, 10Ne and 22 with atomic number 10, are written as 20 10Ne. 1. Three isotopes of uranium have mass numbers of 234, 235, and 238. The atomic number of uranium is 92. How many neutrons are in the nuclei of each of these isotopes? 2. An isotope of oxygen has a mass number of 15. How many neutrons are in the nucleus of this isotope? 3. How many neutrons are in the mercury isotope 200 4. Write the symbols for the three isotopes of hydrogen that have zero, 80Hg? one, and two neutrons in the nucleus. Section 30.1 The Nucleus 801 798-805 CH30-S1-845813 6/8/04 7:52 PM Page 802 What holds the nucleus together? The negatively charged electrons that surround the positively charged nucleus of an atom are held in place by the attractive electromagnetic force. Because the nucleus consists of positively charged protons and neutral neutrons, the repulsive electromagnetic force among the protons might be expected to cause them to fly apart. Because this does not happen, an even stronger attractive force must exist within the nucleus. The", " Strong Nuclear Force The strong nuclear force, also called the strong force, acts between protons and neutrons that are close together, as they are in a nucleus. This force is more than 100 times stronger than the electromagnetic force. The range of the strong force is short, only about the radius of a proton, 1.41015 m. It is attractive and is of the same strength among protons and protons, protons and neutrons, and neutrons and neutrons. Both neutrons and protons are called nucleons. The strong nuclear force holds the nucleons in the nucleus. If a nucleon were to be pulled out of a nucleus, work would have to be done to overcome the attractive force. Doing work adds energy to the system. Thus, the assembled nucleus has less energy than the separate protons and neutrons that make it up. The difference is the binding energy of the nucleus. Because the assembled nucleus has less energy, all binding energies are negative. Binding Energy of the Nucleus Einstein showed that mass and energy are equivalent, so the binding energy can be expressed in the form of an equivalent amount of mass, according to the following equation. Energy Equivalent of Mass E mc2 The energy contained in matter is equal to the mass times the square of the speed of light in a vacuum. Because energy has to be added to take a nucleus apart, the mass of the assembled nucleus is less than the sum of the masses of the nucleons that compose it. For example, the helium nucleus, 4 2He, consists of two protons and two neutrons. The mass of a proton is 1.007276 u. The mass of a neutron is 1.008665 u. If the mass of the helium nucleus were equal to the sum of the masses of the two protons and the two neutrons, you would expect that the mass of the nucleus would be 4.031882 u. Careful measurement, however, shows that the mass of a helium nucleus is only 4.002603 u. The actual mass of the helium nucleus is less than the mass of its constituent parts by 0.029279 u. The difference between the sum of the masses of the individual nucleons and the actual mass is called the mass defect. Mass spectrometers usually measure the masses of isotopes; that is, the nuclides with all their electrons. When calculating the mass defect of a nucleus, one must make sure that the mass of the electrons is accounted for properly. Thus, the mass of hydrogen (one proton and one electron) usually is given in mass-defect problems. Forces A positron is a positively charged electron. The electromagnetic attraction of an electron and a positron is 4.21042 times stronger than their gravitational attraction. 802 Chapter 30 Nuclear Physics 798-805 CH30-S1-845813 3/31/04 12:50 AM Page 803 Masses normally are measured in atomic mass units. It will be useful, then, to determine the energy equivalent of 1 u (1.66051027 kg). To determine the energy, you must multiply the mass by the square of the speed of light in a vacuum (2.9979108 m/s). This is expressed to five significant digits. E mc2 (1.66051027 kg)(2.9979108 m/s)2 1.49241010 kgm2/s2 1.49241010 J The most convenient unit of energy to use is the electron volt. E (1.49241010 J)(1 eV/1.602171019 J) 9.3149108 eV 931.49 MeV Hence, 1 u of mass is equivalent to 931.49 MeV of energy. Figure 30-2 shows how the binding energy per nucleon depends on the mass of the nucleus. Heavier nuclei are bound more strongly than lighter nuclei. Except for a few nuclei, the binding energy per nucleon becomes more negative as the mass number, A, increases to a value of 56, which is that of iron, Fe. The most tightly bound nucleus is 56 26Fe; thus, nuclei become more stable as their mass numbers approach that of iron. Nuclei whose mass numbers are larger than that of iron are less strongly bound, and are therefore less stable. A nuclear reaction will occur naturally if energy is released by the reaction; that is, if the nucleus is transformed into one closer to the lowpoint of the graph at A 56. At lower mass numbers, below 56 26Fe, a nuclear reaction will occur naturally if it increases the atomic mass. In the Sun and other stars, hydrogen is converted into helium, carbon, and other heavier elements in a reaction that releases energy, causing the electromagnetic radiation that you experience as visible light. \u25a0 Figure 30-2 The binding energy per nucleon depends on the number of nucleons, A Average Binding Energy per Nucleon v. Number of Nucleons", " 0 2 4 6 8 4He 10 16O 12C 56Fe 20 40 60 80 100 120 140 160 180 200 220 240 Number of nucleons (A) Section 30.1 The Nucleus 803 798-805 CH30-S1-845813 6/8/04 7:52 PM Page 804 At mass numbers above 56, a nuclear reaction will occur naturally if it decreases the atomic mass. When uranium-238 decays to thorium-234, the resulting thorium nucleus is more stable than the uranium nucleus. Energy is released in the form of a radioactive particle with mass and kinetic energy. Thorium will not spontaneously transform into uranium because energy would have to be added to the nucleus. The heaviest nuclei in the periodic table are created in this way, by slamming together smaller nuclei in particle accelerators. In general, such heavy elements can exist only for fractions of a second before the nuclei decay into smaller, more stable nuclei. In contrast, when small nuclei gain nucleons, the binding energy of the larger nucleus is more negative, and thus, more stable than the sum of the binding energies of the smaller nuclei. In the next section, calculations of binding energy will be used to understand nuclear reactions. The binding energy explains the energy released when small nuclei fuse, as in stars, and large nuclei split, as in the decay of radioactive elements. Mass Defect and Nuclear Binding Energy Find the mass defect and binding energy of tritium, 3 1H. The mass of the tritium isotope is 3.016049 u, the mass of a hydrogen atom is 1.007825 u, and the mass of a neutron is 1.008665 u. 1 Analyze the Problem Known: mass of 1 hydrogen atom 1.007825 u mass of 1 neutron 1.008665 u mass of tritium 3.016049 u binding energy of 1 u 931.49 MeV Unknown: total mass of nucleons and electron? mass defect? binding energy of tritium? 2 Solve for the Unknown Add the masses of the hydrogen atom (one proton and one electron) and two neutrons. mass of 1 hydrogen atom: plus mass of 2 neutrons: total mass of nucleons: 1.007825 u 2.017330 u 3.025155 u The mass defect is equal to the actual mass of tritium less the mass of the sum of its parts. mass of tritium: less mass of nucleons: mass defect: 3.016049 u 3.025155 u 0.009106 u The binding energy is the energy equivalent of the mass defect. Math Handbook Operations with Significant Digits pages 835\u2014836 E (mass defect in u)(binding energy of 1 u) E (0.009106 u)(931.49 MeV/u) 8.4821 MeV Substitute mass defect 0.009106 u, binding energy per u 931.49 MeV 3 Evaluate the Answer \u2022 Are the units correct? Mass is measured in u, and energy is measured in MeV. \u2022 Does the sign make sense? Binding energy should be negative. \u2022 Is the magnitude realistic? According to Figure 30-2, binding energies per nucleon in this range are between 2 MeV and 3 MeV, so the answer for three nucleons is reasonable. 804 Chapter 30 Nuclear Physics 798-805 CH30-S1-845813 6/8/04 7:53 PM Page 805 Use these values to solve the following problems: mass of hydrogen 1.007825 u, mass of neutron 1.008665 u, 1 u 931.49 MeV. 5. The carbon isotope 12 6C has a mass of 12.0000 u. a. Calculate its mass defect. b. Calculate its binding energy in MeV. 6. The isotope of hydrogen that contains one proton and one neutron is called deuterium. The mass of the atom is 2.014102 u. a. What is its mass defect? b. What is the binding energy of deuterium in MeV? 7. A nitrogen isotope, 15 7N, has seven protons and eight neutrons. It has a mass of 15.010109 u. a. Calculate the mass defect of this nucleus. b. Calculate the binding energy of the nucleus. 8. An oxygen isotope, 16 8O, has a nuclear mass of 15.994915 u. a. What is the mass defect of this isotope? b. What is the binding energy of its nucleus? In few areas of physics has basic knowledge led to applications as quickly as in the field of nuclear physics. The medical use of the radioactive element radium began within 20 years of its discovery. Proton accelerators were used for medical applications less than one year after being invented. In the case of nuclear fission (", "splitting nuclei), the military application was under development before the basic physics was even known. Peaceful applications followed in less than ten years. 30.1 Section Review 9. Nuclei Consider these two pairs of nuclei: 12 6C and 11 13 alike? In which way are they different? 5B and 11 6C and 6C. In which way are the two 10. Binding Energy When tritium, 3 1H, decays, it emits a beta particle and becomes 3 2He. Which nucleus would you expect to have a more negative binding energy? 11. Strong Nuclear Force The range of the strong nuclear force is so short that only nucleons that are adjacent to each other are affected by the force. Use this fact to explain why, in large nuclei, the repulsive electromagnetic force can overcome the strong attractive force and make the nucleus unstable. 12. Mass Defect Which of the two nuclei in problem 10 has the larger mass defect? 13. Mass Defect and Binding Energy The radioac6C has a mass of 14.003074 u. tive carbon isotope 14 a. What is the mass defect of this isotope? b. What is the binding energy of its nucleus? 14. Critical Thinking In old stars, not only are helium and carbon produced by joining more tightly bound nuclei, but so are oxygen ( Z 8) and silicon (Z 14). What is the atomic number of the heaviest nucleus that could be formed in this way? Explain. physicspp.com/self_check_quiz Section 30.1 The Nucleus 805 806-814 CH30-S2-845813 7/29/04 3:53 PM Page 806 30.2 Nuclear Decay and Reactions Objectives \u2022 Describe three forms of radioactive decay. \u2022 Solve nuclear equations. \u2022 Calculate the amount remaining and activity of radioactive material after a given time. \u2022 Define nuclear fission and fusion. \u2022 Describe the operation of a nuclear reactor. Vocabulary radioactive alpha decay beta decay gamma decay nuclear reaction half-life activity fission chain reaction fusion In 1896, Henri Becquerel was working with compounds containing the element uranium. To his surprise, he found that photographic plates that had been covered to keep out light became fogged, or partially exposed, when these uranium compounds were anywhere near the plates. This fogging suggested that some kind of ray from the uranium had passed through the plate coverings. Several materials other than uranium or its compounds also were found to emit these penetrating rays. Materials that emit this kind of radiation are now said to be radioactive. Because particles are emitted from the material, it is said to decay. Nuclei decay from a less stable form to a more stable form. Radioactivity is a natural process. Radioactive Decay In 1899, Ernest Rutherford and his colleagues discovered that the element radon spontaneously splits into a lighter nucleus and a light helium nucleus. That same year, Rutherford also discovered that uranium compounds produce three different kinds of radiation. He separated the types of radiation according to their penetrating ability and named them (alpha), (beta), and (gamma) radiation. Alpha radiation can be stopped by a thick sheet of paper, while 6 mm of aluminum is needed to stop most beta particles. Several centimeters of lead are required to stop gamma rays. Alpha decay An alpha particle is the nucleus of a helium atom, 4 2He. The emission of an particle from a nucleus is a process called alpha decay. The mass number of an particle, 4 2He, is 4, and the atomic number is 2. When a nucleus emits an particle, the mass number, A, of the decaying nucleus is reduced by 4, and the atomic number of the nucleus, Z, is reduced by 2. The element changes, or transmutes, into a different element. For example, 238 92U transmutes into 234 90Th through alpha decay. Beta decay Beta particles are electrons emitted by the nucleus. The nucleus contains no electrons, however, so where do these electrons come from? Beta decay occurs when a neutron is changed to a proton within the nucleus. In all reactions, charge must be conserved. That is, the charge before the reaction must equal the charge after the reaction. In beta decay, when a neutron, charge 0, changes to a proton, charge 1, an electron, charge 1, also appears. In beta decay, a nucleus with N neutrons and Z protons ends up as a nucleus of N1 neutrons and Z1 protons. Another particle, an antineutrino, also is emitted in beta decay. Gamma decay A redistribution of the energy within the nucleus results in gamma decay. The ray is a high-energy photon. Neither the mass number nor the atomic number is changed in gamma decay. Gamma radiation often accompanies alpha and beta decay. The three types of radiation are summarized in Table 30-1. Radioactive elements often go through a series of successive decays until 92U undergo", "es 14 separate they form a stable nucleus. For example, 238 decays before the stable lead isotope 206 82Pb is produced. 806 Chapter 30 Nuclear Physics 806-814 CH30-S2-845813 3/31/04 12:56 AM Page 807 Table 30-1 The Three Types of Radiation Alpha Particle Beta Particle Gamma Ray Charge 2 Least penetration Transmutes nucleus Charge 1 Middle energy Transmutes nucleus No charge Highest penetration Changes only energy Nuclear Reactions and Equations A nuclear reaction occurs whenever the energy or number of neutrons or protons in a nucleus changes. Just as in chemical reactions, some nuclear reactions occur with a release of energy, while others occur only when energy is added to a nucleus. One form of nuclear reaction is the emission of particles by radioactive nuclei. The reaction releases excess energy in the form of the kinetic energy of the emitted particles. Two such reactions are shown in Figure 30-3. Nuclear reactions can be described by words, diagrams, or equations. The symbols used for the nuclei in nuclear equations make the calculation of atomic number and mass number in nuclear reactions simpler. For example, the equation for the reaction in Figure 30-3a is as follows. 238 92U \u2192 234 90Th 4 2He The total number of nuclear particles stays the same during the reaction, so the sum of the superscripts on each side must be equal: 238 234 4. The total charge also is conserved, so the sum of the subscripts on each side must be equal: 92 90 2. In beta decay, an electron, 1, are produced. 0e, and an antineutrino, 0 0 (The symbol for the antineutrino is the Greek letter nu with a bar over it; the bar denotes a particle of antimatter.) The transmutation of a thorium atom by the emission of a particle is shown in Figure 30-3b. 234 90Th \u2192 234 91Pa 1 0e 0 0 Note that the sum of the superscripts on the left-hand side of the equation equals the sum of the superscripts on the right-hand side. Equality also exists between the subscripts on the two sides of the equation. \u25a0 Figure 30-3 The emission of an alpha particle by uranium-238 results in the formation of thorium-234 (a). The emission of a beta particle by thorium-234 results in the formation of protactinium-234 (b). a particle 4 2 He b 238 92 U 234 90 Th antineutrino \uf8e7 0 0 particle 0 e 1 234 90 Th 234 91 Pa Section 30.2 Nuclear Decay and Reactions 807 806-814 CH30-S2-845813 6/8/04 7:55 PM Page 808 Alpha and Beta Decay Write the nuclear equation for each radioactive process. a. A radioactive radium isotope, 226 88Ra, emits an particle and becomes the radon isotope 222 86Rn. b. A radioactive lead isotope, 209 82Pb, decays into the bismuth isotope, 209 83Bi, by the emission of a particle and an antineutrino. 1 Analyze the Problem Known: a. 226 88Ra \u2192 particle 222 particle 4 86Rn 2He 83Bi particle antineutrino b. 209 82Pb \u2192 209 particle 1 0e antineutrino 0 0 2 Solve for the Unknown Unknown: Is the decay possible? Is the decay possible? a. 226 b. 209 88Ra \u2192 4 82Pb \u2192 209 2He 222 86Rn 83Bi 1 0e 0 0 Substitute 4 2He for particle Substitute 1 for antineutrino 0e for particle and 0 0 3 Evaluate the Answer \u2022 Is the number of nucleons conserved? a. 226 222 4, so mass number is conserved. b. 209 209 0 0, so mass number is conserved. \u2022 Is charge conserved? a. 88 86 2, so charge is conserved. b. 82 83 1 0, so charge is conserved. 15. Write the nuclear equation for the transmutation of a radioactive uranium isotope, 234 into a thorium isotope, 230 90Th, by the emission of an particle. 16. Write the nuclear equation for the transmutation of a radioactive thorium isotope, 230 into a radioactive radium isotope, 226 88Ra, by the emission of an particle. 17. Write the nuclear equation for the transmutation of a radioactive radium isotope, 226 92U, 90Th, 88Ra, into a radon isotope, 222 18. A radioactive lead isotope, 214 86Rn, by decay. the emission of a particle and an antineutrino. Write the nuclear equation. 82Pb, can change to a radioactive bismuth isotope, 214 83Bi, by 19. A radioactive carbon isotope", ", 14 14 7N. Write the nuclear equation. 6C, undergoes decay to become the nitrogen isotope In alpha and beta decay, one nucleus, shown on the left of the equation, decays to another nucleus and one or more radioactive particles, shown on the right of the equation. Another example of transmutation occurs when a particle collides with the nucleus, often resulting in the emission 1H \u2192 13 of other particles, as in 12 7N. Such reactions are illustrated in the next Example Problem, and in the discussion of fission later in the section. 6C 1 808 Chapter 30 Nuclear Physics 806-814 CH30-S2-845813 3/31/04 12:58 AM Page 809 Solve Nuclear Equations When nitrogen gas is bombarded with particles, high-energy protons are emitted. What new isotope is created? 1 Analyze the Problem Known: nitrogen 14 proton 1 1H 7N, 4 2He Unknown: What isotope belongs on the right side of the equation? 2 Solve for the Unknown Write the equation for the nuclear reaction. 7N \u2192 1 2He 14 4 1H A Z X Solve for Z and for A. Z 2 7 1 8 A 4 14 1 17 Math Handbook Solving Equations page 844 Use the periodic table on page 916. The element with Z 8 is oxygen. The isotope must be 17 8O. 3 Evaluate the Answer \u2022 Does the equation balance? The number of nucleons is conserved: 4 14 1 17. The charge is conserved: 2 7 1 8. 20. Use the periodic table on page 916 to complete the following. a. 14 b. 55 6C \u2192? 1 24Cr \u2192? 1 \u03bd\uf8e7 0e 0 0 \u03bd\uf8e7 0e 0 0 21. Write the nuclear equation for the transmutation of a seaborgium isotope, 263 106Sg, into a rutherfordium isotope, 259 104Rf, by the emission of an alpha particle. 22. A proton collides with the nitrogen isotope 15 7N, forming a new isotope and an alpha particle. What is the isotope? Write the nuclear equation. 23. Write the nuclear equations for the beta decay of the following isotopes. a. 210 80Pb b. 210 83Bi c. 234 90Th d. 239 93Np Half-Life The time required for half of the atoms in any given quantity of a radioactive isotope to decay is the half-life of that element. After each half-life, the number of undecayed nuclei is cut in half, as shown in Figure 30-4. Each particular isotope has its own half-life. For example, the half-life of the radium isotope 226 88Ra is 1600 years. That is, in 1600 years, half of a given quantity of 226 88Ra decays into another element, radon. In a second 1600 years, half of the remaining sample of radium will have decayed. In other words, one-fourth of the original amount still will remain after 3200 years. In contrast, a sample of polonium-210 will decay to one-fourth the original amount in just 276 days.0 3/4 1/2 1/4 0 \u25a0 Figure 30-4 Radioactive nuclei decay to more stable forms. Remaining Nuclei as a Function of Time 1 2 3 4 5 Number of half-lives Section 30.2 Nuclear Decay and Reactions 809 806-814 CH30-S2-845813 6/8/04 7:55 PM Page 810 Table 30-2 Half-Lives of Selected Isotopes Element Isotope Half-Life Radiation Produced Hydrogen Carbon Cobalt Iodine Lead Polonium Polonium Uranium Uranium Plutonium Plutonium 3 1H 14 6C 60 27Co 131 53I 212 82Pb 194 84Po 210 84Po 235 92U 238 92U 236 94Pu 242 94Pu 12.3 years 5730 years 30 years 8.07 days 10.6 hours 0.7 seconds 138 days 7.1108 years 4.51109 years 2.85 years 3.79105 years,,,,,, The half-lives of selected isotopes are shown in Table 30-2. If you know the original amount of a radioactive substance and its half-life, you can calculate the amount remaining after a given number of half-lives. Half-Life t remaining original1 2 The amount of a radioactive isotope remaining in a sample equals the original amount times 1 to the t, where t is the 2 number of half-lives that have passed. Half-lives of radioactive isotopes are used to date objects. The age of a sample of organic material can be found by measuring the amount of carbon-14 remaining. The age of Earth was calculated", " based on the decay of uranium into lead. The decay rate, or number of decays per second, of a radioactive substance is called its activity. Activity is proportional to the number of radioactive atoms present. Therefore, the activity of a particular sample also is reduced by one-half in one half-life. Consider 131 53I, with a half-life of 8.07 days. If the activity of a certain sample of iodine-131 is 8105 decays/s, then 8.07 days later, its activity will be 4105 decays/s. After another 8.07 days, its activity will be 2105 decays/s. The activity of a sample also is related to its half-life. The shorter the half-life, the higher the activity. Consequently, if you know the activity of a substance and the amount of that substance, you can determine its half-life. The SI unit for decays per second is a Becquerel (Bq). Refer to Figure 30-4 and Table 30-2 to solve the following problems. 24. A sample of 1.0 g of tritium, 3 1H, is produced. What will be the mass of the tritium remaining after 24.6 years? 25. The isotope 238 93Np has a half-life of 2.0 days. If 4.0 g of neptunium is produced on Monday, what will be the mass of neptunium remaining on Tuesday of the next week? 26. A sample of polonium-210 is purchased for a physics class on September 1. Its activity is 2106 Bq. The sample is used in an experiment on June 1. What activity can be expected? 27. Tritium, 3 1H, once was used in some watches to produce a fluorescent glow so that the watches could be read in the dark. If the brightness of the glow is proportional to the activity of the tritium, what would be the brightness of such a watch, in comparison to its original brightness, when the watch is six years old? 810 Chapter 30 Nuclear Physics 806-814 CH30-S2-845813 3/31/04 12:58 AM Page 811 Artificial Radioactivity Radioactive isotopes can be formed from stable isotopes by bombardment with particles, protons, neutrons, electrons, or gamma rays. The resulting unstable nuclei emit radiation until they are converted into stable isotopes. The radioactive nuclei may emit alpha, beta, and gamma radiation, as well as neutrinos, antineutrinos, and positrons. (A positron is a positively charged electron, 1 0e.) Artificially produced radioactive isotopes often are used in medicine and medical research. In many medical applications, patients are given radioactive isotopes of elements that are absorbed by specific parts of the body. A physician uses a radiation counter to monitor the activity in the region in question. A radioactive isotope also can be attached to a molecule that will be absorbed in the area of interest, as is done in positron emission tomography, better known as the PET scan. A PET scan of a brain is shown in Figure 30-5. Radiation often is used to destroy cancer cells. These cells are more sensitive to the damaging effects of radiation because they divide more often than normal cells. Gamma rays from the isotope 60 27Co are used to treat cancer patients. Radioactive iodine is injected to target thyroid cancer. In a third method, particles produced in a particle accelerator are beamed into tissue in such a way that they decay in the cancerous tissue and destroy the cells. Nuclear Fission The possibility of obtaining useful forms of energy from nuclear reactions was discussed in the 1930s. The most promising results came from bombarding substances with neutrons. In Italy, in 1934, Enrico Fermi and Emilio Segr\u00e9 produced many new radioactive isotopes by bombarding uranium with neutrons. German chemists Otto Hahn and Fritz Strassmann showed in 1939 that the resulting atoms acted chemically like barium. One week later, Lise Meitner and Otto Frisch proposed that the neutrons had caused a division of the uranium into two smaller nuclei, resulting in a large release of energy. Such a division of a nucleus into two or more fragments is called fission. The possibility that fission could be not only a source of energy, but also an explosive weapon, was immediately realized by many scientists. Fission occurs when a nucleus is divided into two or more fragments, releasing neutrons and energy. The uranium isotope, 235 92U, undergoes fission when it is bombarded with neutrons. The elements barium and krypton are typical results of fission. The reaction is illustrated with the following equation. 1 0n 235 92U \u2192 92 36Kr 141 56Ba 31 0n 200 MeV The energy released by each fission can be found by calculating the masses of the atoms", " on each side of the equation. In the uranium-235 reaction, the total mass on the right side of the equation is 0.215 u smaller than that on the left. The energy equivalent of this mass is 3.211011 J, or 2.00102 MeV. This energy appears as the kinetic energy of the products of the fission. Radiation Treatment Gamma rays destroy both cancerous cells and healthy cells; thus, the beams of radiation must be directed only at cancerous cells. 9F decays, \u25a0 Figure 30-5 For a PET scan, physicians inject a solution in which a radioactive isotope, such as 18 9F, is attached to a molecule that will concentrate in the target tissues. When the 18 it produces positrons, which annihilate nearby electrons, producing gamma rays. The PET scanner detects the gamma rays. A computer then makes a threedimensional map of the isotope distribution. Here, a normal brain (top), and the brain of a person suffering from dementia (bottom) are contrasted. Section 30.2 Nuclear Decay and Reactions 811 CNRI/Photo Researchers 806-814 CH30-S2-845813 3/31/04 12:59 AM Page 812 \u25a0 Figure 30-6 The nuclear fission chain reaction of uranium-235 takes place in the core of a nuclear reactor. 235 92U 92 36Kr 1 0n 235 92U 1 0n 1 0n 1 0n 92 36Kr 141 56Ba 235 92U 235 92U 141 56Ba 92 36Kr 141 56Ba 92 36Kr 141 56Ba 235 92U Control rod 235 92U 235 92U 235 92U Control rod 1 0n 1 0n 1 0n 1 0n 1 0n 1 0n 1 0n 1 0n 1 0n The neutrons needed to cause the fission of additional 235 92U nuclei can be the neutrons produced once the fission process is started. If one or more of the neutrons cause a fission, that fission releases three more neutrons, each of which can cause more fission. This continual process of repeated fission reactions caused by the release of neutrons from previous fission reactions is called a chain reaction. The process is illustrated in Figure 30-6. Nuclear Reactors \u25a0 Figure 30-7 The glow is due to the Cerenkov effect, which occurs when high speed particles entering the water exceed the speed of light in water. The electrons emit photons, which cause the water to glow when fuel rods are placed in it. This glow is not the result of radioactivity. 812 Science VU/Visuals Unlimited Chapter 30 Nuclear Physics To create a controlled chain reaction and make use of the energy produced, the neutrons need to interact with the fissionable uranium at the right rate. Most of the neutrons released by the fission of 235 92U atoms are moving at high speeds. These are called fast neutrons. In addition, naturally occurring uranium consists of less than one percent 235 92U and more 92U. When a 238 than 99 percent 238 92U nucleus absorbs a fast neutron, it does not undergo fission, but becomes a new isotope, 239 92U. The absorption of neutrons by 238 92U keeps most of the neutrons from reaching the fissionable 92U atoms. Thus, most neutrons released by the fission of 235 235 92U are unable to cause the fission of another 235 92U atom. To control the reaction, the uranium is broken up into small pieces and placed in a moderator, a material that can slow down, or moderate, the fast neutrons. When a neutron collides with a light atom it transfers momentum and energy to the atom. In this way, the neutron loses energy. The moderator thus slows many fast neutrons to speeds at which they can be absorbed more easily by 235 92U. The larger number of slow neutrons greatly increases the probability that a neutron released by the 92U nucleus will cause another 235 fission of a 235 92U nucleus to fission. If there is enough 235 92U in the sample, a chain reaction can occur. To increase the amount of fissionable uranium, the uranium may be enriched by adding more 235 92U. Both types of uranium are used in nuclear reactors. 92U than by 238 806-814 CH30-S2-845813 3/31/04 12:59 AM Page 813 \u25a0 Figure 30-8 In a nuclear power plant, the thermal energy released in nuclear reactions is converted to electric energy. Containment structure Control rods Steam Reactor Pump Steam generator Heat exchanger Pump Steam turbine (generates electricity) Condenser Pump Large body of water The type of nuclear reactor used in the United States, the pressurized water reactor, contains about 200 metric tons of uranium sealed in hundreds of metal rods. The rods are immersed in water, as shown in Figure 30-7. Water not only is the moderator, but also transfers thermal energy", " away from the fission of uranium. Rods of cadmium metal are placed between the uranium rods. Cadmium absorbs neutrons easily and also acts as a moderator. The cadmium rods are moved in and out of the reactor to control the rate of the chain reaction. Thus, the rods are called control rods. When the control rods are inserted completely into the reactor, they absorb enough of the neutrons released by the fission reactions to prevent any further chain reaction. As the control rods are removed from the reactor, the rate of energy release increases, with more free neutrons available to continue the chain reaction. Energy released by the fission heats the water surrounding the uranium rods. The water itself doesn\u2019t boil because it is under high pressure, which increases its boiling point. As shown in Figure 30-8, this water is pumped to a heat exchanger, where it causes other water to boil, producing steam that turns turbines. The turbines are connected to generators that produce electrical energy. Fission of 235 92U nuclei produces Kr, Ba, and other atoms in the fuel rods. Most of these atoms are radioactive. About once a year, some of the uranium fuel rods must be replaced. The old rods no longer can be used in the reactor, but they are still extremely radioactive and must be stored in a location that can be secured. Methods of permanently storing these radioactive waste products currently are being developed. Nuclear Fusion In nuclear fusion, nuclei with small masses combine to form a nucleus with a larger mass, as shown in Figure 30-9 on the next page. In the process, energy is released. You learned earlier in this chapter that the larger nucleus is more tightly bound, so its mass is less than the sum of the masses of the smaller nuclei. This loss of mass corresponds to a release of energy. Modeling Radioactive Decay You will need 50 pennies to represent 50 atoms of a radioactive isotope. In this simulation, heads indicates that a nucleus has not decayed. 1. Record 50 heads as the starting point. 2. Shake all the pennies in a large cup and pour them out. Remove all the tails and set them aside. Count and record the number of heads. 3. Repeat step 2 with the pennies that were heads in the last throw. Each throw simulates one half-life. Analyze and Conclude 4. Graph the number of pennies as a function of the number of half-lives. 5. Collect the results from other students and use the totals to make a new graph. 6. Compare this graph with the individual ones. Which more closely matches the theoretical graph in Figure 30-4? Section 30.2 Nuclear Decay and Reactions 813 806-814 CH30-S2-845813 3/31/04 12:59 AM Page 814 2 1H 3 1H \u25a0 Figure 30-9 The fusion of deuterium and tritium produces helium. Protons are red and neutrons are gray in the figure. 4 2He An example of fusion is the process that occurs in the Sun. Four hydrogen nuclei (protons) fuse in several steps to form one helium nucleus. The mass of the four protons is greater than the mass of the helium nucleus that is produced. The energy equivalent of this mass difference appears as the kinetic energy of the resultant particles. The energy released by the fusion that produces one helium-4 nucleus is 25 MeV. In comparison, the energy released when one dynamite molecule reacts chemically is about 20 eV, about 1 million times smaller. 1 0n There are several processes by which fusion occurs in the Sun. The most important process is the proton-proton chain. 0e 0 0 1 1 1H 1 1H 2 2He 3 1H \u2192 2 1H \u2192 3 2He \u2192 4 1H 1 2He 2He 21 3 1H The first two reactions must occur twice to produce the two 3 2He particles needed for the final reaction. The net result (subtracting out the two protons produced in the final step) is that four protons produce one 4 2He, two positrons, and two neutrinos. The repulsive force between the charged nuclei requires the fusing nuclei to have high energies. Thus, fusion reactions take place only when the nuclei have large amounts of thermal energy. The proton-proton chain requires a temperature of about 2107 K, such as that found in the center of the Sun. Fusion reactions also occur in a hydrogen, or thermonuclear, bomb. In this device, the high temperature necessary to produce the fusion reaction is produced by exploding a uranium fission, or atomic, bomb. 30.2 Section Review 28. Beta Decay How can an electron be expelled from a nucleus in beta decay if the nucleus has no electrons? 29. Nuclear Reactions The polonium isotope 210 84Po undergoes alpha decay. Write the equation for the reaction.", " 30. Half-Life Use Figure 30-4 and Table 30-2 to esti53I would mate in how many days a sample of 131 have three-eighths its original activity. 31. Nuclear Reactor Lead often is used as a radiation shield. Why is it not a good choice for a moderator in a nuclear reactor? 32. Fusion One fusion reaction involves two deuterium nuclei, 2 1H. A deuterium molecule contains two deuterium atoms. Why doesn\u2019t this molecule undergo fusion? 33. Energy Calculate the energy released in the first. fusion reaction in the Sun, 1 1H 1 1H \u2192 2 1H 1 0e 0 0 34. Critical Thinking Alpha emitters are used in smoke detectors. An emitter is mounted on one plate of a capacitor, and the particles strike the other plate. As a result, there is a potential difference across the plates. Explain and predict which plate has the more positive potential. 814 Chapter 30 Nuclear Physics physicspp.com/self_check_quiz 815-823 CH30-S3-845813 3/31/04 1:04 AM Page 815 30.3 The Building Blocks of Matter The first physicists who studied the nucleus with high speed particles had to make do with particles from radioactive sources. Other experimenters used cosmic rays, produced by not-yet-understood processes in stars and galaxies. In the early 1930s, the first laboratory devices that could accelerate protons and particles to energies high enough to penetrate the nucleus were developed. Two devices, the linear accelerator and the synchrotron, are regularly used today. Linear Accelerators A linear accelerator can be used to accelerate protons or electrons. It consists of a series of hollow tubes within a long, evacuated chamber. The tubes are connected to a source of high-frequency alternating voltage, as illustrated in Figure 30-10. Protons are produced in an ion source similar to that described in Chapter 26. When the first tube has a negative potential, protons are accelerated into it. There is no electric field within the tube, so the protons move at a constant velocity. The length of the tube and the frequency of the voltage are adjusted so that when the protons have reached the far end of the tube, the potential of the second tube is negative in relation to that of the first. The resulting electric field in the gap between the tubes accelerates the protons into the second tube. This process continues, with the protons receiving an acceleration between each pair of tubes. The energy of the protons is increased by 105 eV with each acceleration. The protons ride along the crest of an electric field wave, much as a surfboard moves on the ocean. At the end of the accelerator, the protons can have energies of many millions or billions of electron volts. A similar method is used to accelerate electrons. Note that one can accelerate only charged particles. a b Ion source Alternating voltage Beam of particles Target Objectives \u2022 Describe the operation of particle accelerators and particle detectors. \u2022 Describe the Standard Model of matter and explain the role of force carriers. Vocabulary quarks leptons Standard Model force carriers pair production weak nuclear force \u25a0 Figure 30-10 The linear accelerator at Stanford University is 3.3 km long and accelerates electrons to energies of 20 GeV (a). Protons in a linear accelerator are accelerated by changing the charge on the tubes as the protons move (b). (Not to scale) Section 30.3 The Building Blocks of Matter 815 Stanford Linear Accelerator Center/Photo Researchers 815-823 CH30-S3-845813 3/31/04 1:30 AM Page 816 a The Synchrotron An accelerator may be made smaller by using a magnetic field to bend the path of the particles into a circle. In a synchrotron, the bending magnets are separated by accelerating regions, as shown in Figure 30-11b. In the straight regions, high-frequency alternating voltage accelerates the particles. The strength of the magnetic field and the length of the path are chosen so that the particles reach the location of the alternating electric field precisely when the field\u2019s polarity will accelerate them. One of the largest synchrotrons in operation is at the Fermi National Accelerator Laboratory near Chicago, shown in Figure 30-11a. Protons there reach energies of 1 TeV (1012 eV). A proton beam and an antiproton beam travel the circle in opposite directions. (An antiproton is the antimatter pair to a proton; it has the same mass and opposite charge.) The beams collide in several interaction regions and the results are studied. Particle Detectors Once particles are produced, the results of the collision need to be detected. In other words, they need to interact with matter in such a way that we can sense them with our relatively limited human senses. Your hand will stop an particle,", " yet you will have no idea that the particle struck you. And as you read this sentence, billions of solar neutrinos pass through your body without so much as a twitch from you. Over the past century, scientists have devised tools to detect and distinguish the products of nuclear reactions. In the last section, you learned that uranium samples fogged photographic plates. When particles, particles, or gamma rays strike photographic film, the film becomes fogged, or exposed. Thus, photographic film can be used to detect radiation. Many other devices are used to detect charged particles and gamma rays. Most of these devices make use of the fact that a collision with a high-speed particle will remove electrons from atoms. That is, the high-speed particles ionize the matter that they bombard. In addition, some substances fluoresce, or emit photons, when they are exposed to certain types of radiation. Thus, fluorescent substances also can be used to detect radiation. These three means of detecting radiation are illustrated in Figure 30-12. \u25a0 Figure 30-12 Particles can be detected when they interact with matter, exposing film, charging the matter, or causing the matter to emit a photon. Expose film Particle source or Ionize Photon Fluoresce b Injector Drift tube Accelerating units Electromagnets To target \u25a0 Figure 30-11 Fermi Laboratory\u2019s synchrotron has a diameter of 2 km (a). The synchrotron is a circular accelerator. Magnets are used to control the path and acceleration of the particles (b). 816 courtesy of Fermilab Chapter 30 Nuclear Physics 815-823 CH30-S3-845813 6/8/04 7:57 PM Page 817 Geiger counter A Geiger-Mueller tube contains a copper cylinder with a negative charge. Down the center of this cylinder runs a rigid wire with a positive charge. The voltage across the wire and cylinder is kept just below the point at which a spontaneous discharge, or spark, occurs. When a charged particle or gamma ray enters the tube, it ionizes a gas atom between the copper cylinder and the wire. The positive ion that is produced is accelerated toward the copper cylinder by the potential difference, and the electron is accelerated toward the positive wire. As these charged particles move toward the electrodes, they create an avalanche of charged particles, and a pulse of current goes through the tube. Condensation trails A device once used to detect particles was the Wilson cloud chamber. The chamber contained an area supersaturated with water vapor or ethanol vapor. When charged particles traveled through the chamber, leaving a trail of ions in their paths, vapor condensed into small droplets on the ions. In this way, visible trails of droplets, or fog, were formed. In a similar detector still used today, the bubble chamber, charged particles pass through a liquid held just above the boiling point. In this case, the trails of ions cause small vapor bubbles to form, marking the particles\u2019 trajectories, as shown in Figure 30-13. Recent technology has produced detection chambers called wire chambers that are like giant Geiger-Mueller tubes. Huge plates are separated by a small gap filled with a low-pressure gas. A discharge is produced in the path of a particle passing through the chamber. A computer locates the discharge and records its position for later analysis. Neutral particles do not leave trails because they do not produce discharges. The laws of conservation of energy and momentum in collisions can be used to tell if any neutral particles were produced. Other detectors measure the energy of the particles. The entire array of detectors used in high-energy accelerator experiments, such as the Collider Detector at Fermilab (CDF), can be up to three stories high, as shown in Figure 30-14a. The CDF is designed to monitor a quarter-million particle collisions each second, as though the detector functioned as a 5000-ton camera, creating a computer picture of the collision events, as shown in Figure 30-14b. a b \u25a0 Figure 30-13 This false-color bubble chamber photograph shows the tracks of charged particles. \u25a0 Figure 30-14 The Collider Detector at Fermilab (CDF) records the tracks from billions of collisions (a). A CDF computer image of a top quark event is shown (b). 817 Section 30.3 The Building Blocks of Matter (t)Science VU/Visuals Unlimited, (l)David Parker/Photo Researchers, (r)Fermilab/Photo Researchers 815-823 CH30-S3-845813 3/31/04 8:07 AM Page 818 Elementary Particles The model of the atom in 1930 was fairly simple: protons and neutrons surrounded by electrons. There also was a particle with no mass called a gamma particle (photon). Different combinations of these four elementary particles seemed to describe the contents of the universe quite well. More detailed studies of radioactive decay disturbed this", " simple picture. While the particles and gamma rays emitted by radioactive nuclei have single energies that depend on the decaying nucleus, particles are emitted with a wide range of energies. One might expect the energy of the particles to be equal to the difference between the energy of the nucleus before decay and the energy of the nucleus produced by the decay. In fact, the wide range of energies of electrons emitted during beta decay suggested to Niels Bohr that energy might not be conserved in nuclear reactions. Wolfgang Pauli in 1931 and Enrico Fermi in 1934 suggested that an unseen neutral particle was emitted with the particle. Named the neutrino (\u201clittle neutral one\u201d in Italian) by Fermi, the particle, which is actually an antineutrino, was not directly observed until 1956. Studies with cosmic rays discovered more particles. A positive electron, or positron, was found in 1932, and what seemed to be a heavy electron, the muon, in 1937. In 1935, a remarkable hypothesis by Japanese physicist Hideki Yukawa spurred much research in the years to follow. Yukawa hypothesized the existence of a new particle that could carry the nuclear force through space, just as the photon carries the electromagnetic force. In 1947 a possible particle, the pion, was discovered. While it was not the carrier of the strong force, it was a new type of matter. Experiments with particle accelerators resulted in the identification of more and more particles, some with intermediate masses, others much more massive than the proton. They had positive and negative charges, or none at all. Some lifetimes were 1023 s, while others had no detectable decays. Most mysterious were the chargeless, massless neutrinos. At one point Enrico Fermi, asked to identify a particle track, replied \u201cIf I could remember the names of all these particles, I\u2019d be a botanist!\u201d The Standard Model How could all these particles be called \u201celementary?\u201d By the late 1960s it became clear that neither protons, neutrons, nor pions were \u201celementary.\u201d They themselves are made up of particles called quarks, as shown in Figure 30-15. Electrons and neutrinos belong to a different family, called leptons. Physicists now believe that all particles can be grouped into three families: quarks, leptons, and force carriers. This model of the building blocks of matter is called the Standard Model. Quarks make up the threequark baryons, the two-quark mesons, and the recently discovered fivequark pentaquarks. Leptons are particles like electrons and neutrinos. There are six different quarks and six different leptons. Quarks and leptons form matter, while force carriers are particles that carry, or transmit, forces between matter. For example, photons carry the electromagnetic interaction. Eight particles, called gluons, carry the strong nuclear interaction that binds quarks into protons and the protons and neutrons into nuclei. Three particles, the weak bosons, are involved in beta decay. \u25a0 Figure 30-15 Even though quarks have fractional charges, all the particles they make have whole-number charges Proton Neutron Pion 818 Chapter 30 Nuclear Physics 815-823 CH30-S3-845813 3/31/04 1:06 AM Page 819 Quarks and Leptons Bosons Top quark Z W Charm quark Bottom quark Tau Muon Strange quark Down quark Up quark Electron Tau neutrino Muon neutrino Electron neutrino Massless Bosons Photon Gluons Graviton ) V e M ( s s a M 106 105 104 103 102 101 100 101 107 108 109 The graviton is the name given to the yet-undetected carrier of the gravitational interaction. The properties of particles that comprise the Standard Model are summarized in Figure 30-16. Protons and Neutrons The quark model describes nucleons, the proton and the neutron, as an assembly of quarks. The nucleons each are made up of three quarks. The proton has two up quarks, u, (charge 2 e) and one down quark, d, 3 (charge 1 e). A proton is described as p uud. The charge on the proton 3 is the sum of the charges of the three quarks, (2 1 2 )e e. The 3 3 3 neutron is made up of one up quark and two down quarks, n = udd. The 1 1 charge of the neutron is zero, (2 )e 0. 3 3 3 Individual free quarks cannot be observed, because the strong force that holds them together becomes larger as the quarks are pulled farther apart. In this case, the strong force acts like the force of a spring. It is", " unlike the electric force, which becomes weaker as charged particles are moved farther apart. In the quark model, the strong force is transmitted by gluons. Antimatter The positive electron, or positron, is an example of an antiparticle, a particle of antimatter. The electron and positron have the same mass and charge magnitude; however, the signs of their charges are opposite. When a positron and an electron collide, the two can annihilate each other, resulting in energy in the form of gamma rays, as shown in Figure 30-17. \u25a0 Figure 30-16 The known quarks and leptons are divided into three families. The everyday world is made from particles in the lefthand family (u, d, e). Particles in the middle group (c, s, ) are found in cosmic rays and are routinely produced in particle accelerators. Particles in the righthand family (b, t, ) are believed to have existed briefly during the earliest moments of the Big Bang and are created in high-energy collisions. The bosons carry the weak, electromagnetic, strong, and gravitational forces. Masses are stated as energy equivalents, given by Einstein\u2019s formula E mc 2. \u25a0 Figure 30-17 The collision of a positron and an electron results in gamma ray production. 0e 1 0e 1 rays Section 30.3 The Building Blocks of Matter 819 815-823 CH30-S3-845813 3/31/04 1:06 AM Page 820 Conversions between mass and energy The amount of energy created in the annihilation can be calculated using Einstein\u2019s equation for the energy equivalent of mass, E mc2. The mass of the electron is 9.111031 kg. The mass of the positron is the same. Therefore, the energy equivalent of the positron and the electron together can be calculated as follows. E 2(9.111031 kg)(3.00108 m/s)2 E (1.641013 J)(1 eV / 1.601019 J) E 1.02106 eV or 1.02 MeV When a positron and an electron at rest annihilate each other, the sum of the energies of the gamma rays emitted is 1.02 MeV. The inverse of annihilation also can occur. That is, energy can be converted directly into matter. If a gamma ray with at least 1.02 MeV of energy passes close by a nucleus, a positron and electron pair can be produced. \u2192 e e The conversion of energy into a matter-antimatter pair of particles is called pair production. Individual reactions, such as \u2192 e and \u2192 e, however, cannot occur, because such events would violate the law of conservation of charge. Reactions such as \u2192 e proton also do not occur; the pair must be a particle and its corresponding antiparticle. Matter and antimatter particles always come in pairs. The production of a positron-electron pair is shown in Figure 30-18. A magnetic field around the bubble chamber causes the oppositely charged particles\u2019 tracks to curve in opposite directions. The gamma ray that produced the pair produced no track. If the energy of the gamma ray is larger than 1.02 MeV, the excess energy goes into the kinetic energy of the positron and electron. The positron soon collides with another electron and they are both annihilated, resulting in the production of two or three gamma rays with a total energy of no less than 1.02 MeV. Particle conservation Each quark and each lepton also has its antiparticle. The antiparticles are identical to the particles except that for charged particles, an antiparticle will have the opposite charge. For example, an up quark, u, has a charge of 2, while an anti-up quark, u, has a 3 charge of 2. A proton, uud, has a charge of 1 and the antiproton, uudd, 3 has a charge of 2 1 2 1. When a particle and its antiparticle 3 3 3 collide, they annihilate each other and are transformed into photons, or lighter particle-antiparticle pairs and energy. The total number of quarks and the total number of leptons in the universe is constant. That is, quarks and leptons are created or destroyed only in particle-antiparticle pairs. On the other hand, force carriers such as gravitons, photons, gluons, and weak bosons can be created or destroyed if there is enough energy. Antiprotons also can be created. An antiproton has a mass equal to that of the proton but is negatively charged. Protons have 1836 times as much mass as electrons. Thus, the energy needed to create proton-antiproton pairs is comparably larger. The first proton-antiproton pair was produced and observed at Berkeley, California in 1955", ". Neutrons also have an antiparticle, called an antineutron. \u25a0 Figure 30-18 When a particle is produced, its corresponding antimatter particle is also produced. Here, a gamma ray decays into an electron-positron pair. 820 Science Photo Library/Photo Researchers Chapter 30 Nuclear Physics 815-823 CH30-S3-845813 6/8/04 7:57 PM Page 821 35. The mass of a proton is 1.671027 kg. a. Find the energy equivalent of the proton\u2019s mass in joules. b. Convert this value to eV. c. Find the smallest total -ray energy that could result in a proton-antiproton pair. 36. A positron and an electron can annihilate and form three gammas. Two gammas are detected. One has an energy of 225 keV, the other 357 keV. What is the energy of the third gamma? 37. The mass of a neutron is 1.008665 u. a. Find the energy equivalent of the neutron\u2019s mass in MeV. b. Find the smallest total -ray energy that could result in the production of a neutron- antineutron pair. 38. The mass of a muon is 0.1135 u. It decays into an electron and two neutrinos. What is the energy released in this decay? Beta Decay and the Weak Interaction The high energy electrons emitted in the beta decay of a radioactive nucleus do not exist in the nucleus. From where do they come? In the decay process a neutron is transformed into a proton. While in a stable nucleus, the neutron does not decay. A free neutron, or one in an unstable nucleus, however, can decay into a proton by emitting a particle. Sharing the outgoing energy with the proton and particle is an antineutrino, 0. 0 The antineutrino has a very small mass and is uncharged, but like the photon, it carries momentum and energy. The neutron decay equation is written as follows. 1 0n \u2192 1 1p 1 0e 0 0 When an isotope decays by emission of a positron, or antielectron, a process like beta decay occurs. A proton within the nucleus changes into a neutron with the emission of a positron, 1 0e 0 0 0e, and a neutrino, 0 0 0n 1 1p \u2192 1. 1 The decay of neutrons into protons and protons into neutrons cannot be explained by the strong force. The existence of beta decay indicates that there must be another interaction, the weak nuclear force, acting in the nucleus. This second type of nuclear force is much weaker than the strong nuclear force. 92U decays by emission and two successive emissions back into 238 uranium again. 1. Show the three nuclear decay equations. 2. Predict the atomic mass number of the uranium formed. Section 30.3 The Building Blocks of Matter 821 815-823 CH30-S3-845813 3/31/04 1:07 AM Page 822 \u25a0 Figure 30-19 Beta decay of a proton to a neutron can be represented with the quark model: d \u2192 u W, then then W e \uf8e7 Quark model of beta decay The difference between a proton, uud, and a neutron, udd, is only one quark. Beta decay in the quark model occurs in two steps, as shown in Figure 30-19. First, one d quark in a neutron changes to a u quark with the emission of a W boson. The W boson is one of the three carriers of the weak force. In the second step, the W boson decays into an electron and an antineutrino. Similarly, in the decay of a proton, a neutron and a W boson are emitted. The W boson then decays into a positron and a neutrino. The emission of the third weak force carrier, the Z0 boson, is not accompanied by a change from one quark to another. The Z0 boson produces an interaction between the nucleons and the electrons in atoms that is similar to, but much weaker than, the electromagnetic force holding the atom together. The interaction first was detected in 1979. The W, W, and Z0 bosons first were observed directly in 1983. While neutrinos and antineutrinos have long been thought to be massless, recent experiments that detect neutrinos from the Sun and from distant accelerators show that neutrinos do have mass, although these masses are much smaller than those of any other known particles. Testing the Standard Model You can see in Figure 30-16 that the quarks and leptons are separated into three families. The everyday world is made from particles in the left-hand family: protons, neutrons, and", " electrons. Particles in the middle group are found in cosmic rays and are routinely produced in particle accelerators. Particles in the right-hand family are believed to have existed briefly during the earliest moments of the Big Bang and are created in high-energy collisions. What determines the masses of the quarks and leptons? The Higgs boson, which has been hypothesized as the particle that determines the masses of the leptons and quarks, has not yet been discovered. The Standard Model is not a theory; it does not explain the masses of particles, nor why there are three families of quarks and leptons. Why are there four forces? The differences among the four fundamental interactions are evident: the forces may act on different quantities such as charge or mass, they may have different dependencies on distance, and the force carriers have different properties. There are, however, some similarities among the interactions. For instance, the force between charged particles, the electromagnetic interaction, is carried by photons in much the same way as weak bosons carry the weak interaction. The electric force acts over a long range because the photon has zero mass, while the weak force acts over short distances because the W and Z bosons are relatively massive. The mathematical structures of the theories of the weak interaction and electromagnetic interaction, however, are similar. 822 Chapter 30 Nuclear Physics 815-823 CH30-S3-845813 6/10/04 9:06 PM Page 823 a b c \u25a0 Figure 30-20 In a supernova, the electromagnetic and weak forces are indistinguishable. The increased light and neutrinos from supernova 1987A, shown here, reached Earth at the same time, demonstrating that neutrinos travel near the speed of light and are produced in supernovas, as predicted. Supernova 1987A is shown before exploding (a), during explosion (b), and in a close-up from Hubble (c). Astrophysical theories of supernovae indicate that during massive stellar explosions, such as the one shown in Figure 30-20, the two interactions are identical. Present theories of the origin of the universe suggest that the two forces were identical during the early moments of the cosmos as well. For this reason, the electromagnetic and weak forces are said to be unified into a single force, called the electroweak force. In the same way that the electromagnetic and weak forces were unified into the electroweak force during the 1970s, physicists presently are developing theories that include the strong force as well. Work is still incomplete. Theories are being improved, and experiments to test these theories are being planned. A fully unified theory that includes gravitation will require even more work. Even more perplexing are the results of studies of the galaxies that suggest that matter described by the Standard Model makes up only a small fraction of the mass in the universe. A much larger amount is formed of dark matter, so called because it doesn\u2019t interact with photons or ordinary matter, except through the gravitational force. In addition, there appears to be dark energy, an unknown force that accelerates the expansion of the universe. Thus the studies of the tiniest particles that make up nuclei are connected directly to investigations of the largest systems, the galaxies that make up the universe. Elementary particle physicists and cosmologists used to be at opposite ends of the length scale; now they question together \u201cWhat IS the world made of?\u201d Perhaps you will be able to answer that question. 30.3 Section Review 39. Nucleus Bombardment Why would a proton require more energy than a neutron when used to bombard a nucleus? 42. The Standard Model Research the limitations of the Standard Model and possible replacements. 43. Critical Thinking Consider the following equa- 40. Particle Accelerator Protons the Fermi Laboratory accelerator, Figure 30-11, move counterclockwise. In what direction is the magnetic field of the bending magnets? in 41. Pair Production Figure 30-18 shows the production of two electron/positron pairs. Why does the bottom set of tracks curve less than the top pair of tracks? tions. u \u2192 d W and W \u2192 e How could they be used to explain the radioactive decay of a nucleon that results in the emission of a positron and a neutrino? Write the equation involving nucleons rather than quarks. physicspp.com/self_check_quiz Section 30.3 The Building Blocks of Matter 823 (r)NASA, (others)David Malin/Anglo-Australian Observatory 824-825 CH30-LAB-845813 6/8/04 7:59 PM Page 824 Alternate CBL instructions can be found on the Web site. physicspp.com Exploring Radiation Radiation detectors use various means to detect radiation. One common type of detector in use is a Geiger-Mueller tube. It consists of a metal tube filled with gas at a low pressure and a long wire along the tube\u2019s axis", ". The wire is at a high potential difference, such as 400\u2013800 V, relative to the metal tube. At one end of the tube is a thin, fragile window. When a high-energy photon or charged particle enters the tube through the window, some of the gas becomes ionized. The ionized electrons are attracted to the wire and speed up. Then they ionize additional atoms creating a pulse of charge striking the wire. This charge pulse is converted to a voltage pulse, amplified, and counted or sent to a speaker. Previously, you learned that light and other electromagnetic radiation travels in all directions in straight lines from a source, such as the Sun. In this lab you will explore the relationship between distance from a radioactive gamma and beta source, and the measured radiation intensity. QUESTION What is the relationship between distance and radiation intensity from a gamma and a beta source? Objectives Safety Precautions \u25a0 Measure radiation. \u25a0 Use variables, constants, and controls to design your experiment. \u25a0 Collect and organize data from gamma and beta radiation activity compared to the distance from the source. \u25a0 Compare and contrast beta and gamma radia- tion activity. \u25a0 If a Geiger counter is used, keep hands, pencils, etc. away from the end of the Geiger tube as the tube window is very thin and fragile. \u25a0 Plug equipment into only GFCI-protected receptacles to prevent shock hazard. \u25a0 Do not eat, drink, or apply makeup when working with radioactive materials. \u25a0 Be careful not to crack open the protective plastic case over the radioactive material. Inform your teacher immediately if this exposure happens. Materials gamma and beta sources radiation counter or student radiation monitor meterstick masking tape stopwatch 824 Horizons Companies 824-825 CH30-LAB-845813 6/10/04 9:08 PM Page 825 Data Table Background Radiation (cpm counts per minute) cpm Distance (cm) Beta-Measured Count Rate (cpm) Beta-Corrected Count Rate (cpm) Gamma-Measured Count Rate (cpm) Gamma-Corrected Count Rate (cpm) 2 4 6 8 10 12 14 Procedure 1. The type of radiation counter or Geiger-Mueller tube and counter that is available in schools varies dramatically. Your procedure should take into consideration how to assemble and handle the type of equipment that is available for your use\u2014both the detector and the radioactive material. 2. With the detector at least 1 m away from the radioactive materials, turn on the detector and measure the radiation. This is called background radiation. Record your data in the data table. 3. Measure the beta and gamma radiation from your sources at various distances. 4. Subtract the background count rate from the count rate recorded to obtain the corrected activity. 5. Be sure to check with your teacher and have your design approved before you proceed with the lab. 3. Make and Use Graphs Make a plot of the corrected sample count rate versus 1/d 2 for the beta and gamma data. Conclude and Apply 1. Explain how the two graphs compare. What relationship exists between distance and count rate? 2. Explain how the background count rate would compare if you were at sea level, such as along the coast, compared to the level in the Rocky Mountains. 3. Describe what happens to the beta count rate when the Geiger-Mueller tube is moved back to three times the initial distance; for example, 18 cm as compared to 6 cm. Going Further What other physics phenomena follow similar patterns? Analyze Real-World Physics 1. Observe and Infer What is the background radiation source in this experiment? Explain how closeness to radioactive materials is a potential hazard for you or others. 2. Make and Use Graphs Make a plot of gamma count rate versus distance, placing distance on the horizontal axis and corrected sample count rate on the vertical axis. If the count rate levels are similar, plot the beta count rate on the same graph, and label the graph for each set of data. To find out more about radiation, visit the Web site: physicspp.com 825 826 CH30-FEATURE-845813 6/8/04 8:01 PM Page 826 Thermonuclear Fusion Thermonuclear Fusion For several decades, physicists have been seeking to create and sustain a fusion reaction that will generate more energy than it consumes. A thermonuclear reactor would generate great heat from small amounts of deuterium, 2 and tritium, 3 seawater. 1H, 1H, which can be extracted from To initiate a fusion reaction, a mixture of deuterium and tritium must be heated and compressed under conditions typical of those in the Sun. The required temperature would destroy the sort of containers used in fission plants. Confining the plasma is one of the chief design problems for fusion reactors. Magnetic Confinement In a magnetic confinement reactor, a strong current is passed through a container of deuterium and tritium gas so that the plasma is compressed within the arc.", " Additional magnetic fields shape the plasma stream, as shown in the diagram, confining it away from the container walls. One promising configuration keeps the plasma in a toroid, or donut shape, which has the great advantage of having no ends to seal. Magnetic confinement: the hot plasma is compressed and confined by the magnetic field. Inertial Confinement If you look at the rapid, string-like motion of a continuous electric arc, you\u2019ll see that it\u2019s very difficult to form plasma into a stable shape. In the inertial confinement reactor, a microscopic pellet of frozen deuterium-tritium is illuminated on all sides by powerful laser beams. These lasers heat the 826 Future Technology outer layer of the pellet so quickly that it explodes. Simultaneously the remainder of the pellet is compressed and heated so greatly that a fusion reaction starts. The energy from the fusion of the pellet exceeds the energy used to heat the pellet. A stream of pellets is fused one after another to obtain a sustained reaction, and the obtained heat is captured to create steam for turbines. In inertial confinement, beams of light or X rays from a laser rapidly heat the surface of the pellet, forming a surrounding plasma envelope. The rest of the fuel is compressed by the blowoff of the hot surface material. The Future While thermonuclear fusion has been sustained in both types of reactors, researchers have had trouble achieving a breakeven reaction (one in which the energy produced in the reaction exceeds the energy needed to sustain the reaction). Progress toward a practical thermonuclear reactor has been expensive and slow, but the promise is great. A fusion reactor is not completely free of radiation hazards because neutrons are produced in fusion reactions. But because the fuel itself is not radioactive, the amount of nuclear waste would be negligible. Going Further 1. Analyze Why does the thermonuclear reactor appear to be such an attractive source of energy? 2. Compare You have seen three types of \u2018thermal\u2019 electric power plants. What features do they all have in common? 827-831 CH30-SG CA STP-845813 3/31/04 2:03 AM Page 827 30.1 The Nucleus Vocabulary \u2022 atomic number (p. 800) \u2022 atomic mass unit (p. 800) \u2022 mass number (p. 800) \u2022 nuclide (p. 801) \u2022 strong nuclear force (p. 802) \u2022 nucleon (p. 802) \u2022 binding energy (p. 802) \u2022 mass defect (p. 802) Key Concepts \u2022 The number of protons in a nucleus is given by the atomic number, Z. \u2022 The sum of the numbers of protons and neutrons in a nucleus is equal to the mass number, A. \u2022 Atoms having nuclei with the same number of protons but different numbers of neutrons are called isotopes. \u2022 The strong nuclear force binds the nucleus together. \u2022 The energy released in a nuclear reaction can be calculated by finding the mass defect, the difference in mass of the particles before and after the reaction. \u2022 The binding energy is the energy equivalent of the mass defect. E mc2 30.2 Nuclear Decay and Reactions Vocabulary \u2022 radioactive (p. 806) \u2022 alpha decay (p. 806) \u2022 beta decay (p. 806) \u2022 gamma decay (p. 806) \u2022 nuclear reaction (p. 807) \u2022 half-life (p. 809) \u2022 activity (p. 810) \u2022 fission (p. 811) \u2022 chain reaction (p. 812) \u2022 fusion (p. 813) Key Concepts \u2022 An unstable nucleus decays, transmuting into another element. \u2022 Radioactive decay produces three kinds of particles. Alpha,, particles are helium nuclei, beta,, particles are high-speed electrons, and gamma,, rays are high-energy photons. \u2022 In nuclear reactions, the sums of the mass number, A, and the total charge, Z, are not changed. \u2022 The half-life of a radioactive isotope is the time required for half of the nuclei to decay. After t half-lives: t remaining original 1 2 \u2022 The number of decays of a radioactive sample per second is the activity. \u2022 In nuclear fission, the uranium nucleus is split into two smaller nuclei with a release of neutrons and energy. \u2022 Nuclear reactors use the energy released in fission to generate electrical energy. \u2022 The fusion of hydrogen nuclei into a helium nucleus releases the energy that causes stars to shine. 30.3 The Building Blocks of Matter Vocabulary \u2022 quarks (p. 818) \u2022 leptons (p. 818) \u2022 Standard Model (p. 818) \u2022 force carriers (p. 818) \u2022 pair production (p. 820) \u2022 weak nuclear force (p. 821) Key Concepts \u2022 Linear accelerators and synchrotrons produce high-energy", " particles. \u2022 The Geiger-Mueller counter, cloud chamber, and other particle detectors use the ionization caused by charged particles passing through matter. \u2022 All matter appears to be made up of quarks and leptons. \u2022 Matter interacts with other matter through particles called force carriers. \u2022 The Standard Model includes the quarks, leptons, and force carriers. \u2022 When corresponding antimatter and matter particles combine, their mass and energy are converted into energy or lighter matter-antimatter particle pairs. \u2022 By pair production, energy is transformed into a matter-antimatter pair. physicspp.com/vocabulary_puzzlemaker 827 827-831 CH30-SG CA STP-845813 6/8/04 8:02 PM Page 828 Concept Mapping 44. Organize the following terms into the concept map: Standard Model, quarks, gamma rays, force carriers, protons, neutrons, leptons, W bosons, neutrinos, electrons, gluons. Includes Which compose Examples Examples 56. Forces In which of the four interactions (strong, weak, electromagnetic, and gravitational) do the following particles take part? (30.3) a. electron b. proton c. neutrino 57. What happens to the atomic number and mass number of a nucleus that emits a positron? (30.3) 58. Antimatter What would happen if a meteorite made of antiprotons, antineutrons, and positrons landed on Earth? (30.3) Applying Concepts 59. Fission A Web site claims that scientists have been able to cause iron nuclei to undergo fission. Is the claim likely to be true? Explain. 60. Use the graph of binding energy per nucleon in Figure 30-2 to determine whether the reaction 1H 1 2 2He is energetically possible. 1H \u2192 3 Mastering Concepts 45. What force inside a nucleus acts to push the nucleus apart? What force inside the nucleus acts to hold the nucleus together? (30.1) 46. Define the mass defect of a nucleus. To what is it related? (30.1) 47. Which are generally more unstable, small or large nuclei? (30.1) 48. Which isotope has the greater number of protons, uranium-235 or uranium-238? (30.1) 49. Define the term transmutation as used in nuclear physics and give an example. (30.2) 50. Radiation What are the common names for an particle, particle, and radiation? (30.2) 51. What two quantities must always be conserved in any nuclear equation? (30.2) 52. Nuclear Power What sequence of events must occur for a chain reaction to take place? (30.2) 53. Nuclear Power What role does a moderator play in a fission reactor? (30.2) 54. Fission and fusion are opposite processes. How can each release energy? (30.2) 55. High-Energy Physics Why would a linear accelerator not work with a neutron? (30.3) 61. Isotopes Explain the difference between naturally and artificially produced radioactive isotopes. 62. Nuclear Reactor In a nuclear reactor, water that passes through the core of the reactor flows through one loop, while the water that produces steam for the turbines flows through a second loop. Why are there two loops? 63. The fission of a uranium nucleus and the fusion of four hydrogen nuclei to produce a helium nucleus both produce energy. a. Which produces more energy? b. Does the fission of a kilogram of uranium nuclei or the fusion of a kilogram of hydrogen produce more energy? c. Why are your answers to parts a and b different? Mastering Problems 30.1 The Nucleus 64. What particles, and how many of each, make up an atom of 109 47Ag? 65. What is the isotopic symbol (the one used in nuclear equations) of a zinc atom composed of 30 protons and 34 neutrons? 16S has a nuclear mass 66. The sulfur isotope 32 of 31.97207 u. a. What is the mass defect of this isotope? b. What is the binding energy of its nucleus? c. What is the binding energy per nucleon? 828 Chapter 30 Nuclear Physics For more problems, go to Additional Problems, Appendix B. 827-831 CH30-SG CA STP-845813 7/29/04 5:02 PM Page 829 67. A nitrogen isotope, 12 7N, has a nuclear mass of 12.0188 u. a. What is the binding energy per nucleon? b. Does it require more energy to separate 7N nucleus or from a 7N has a mass of 14.00307 u. 68. The two positively charged protons in a helium a nucleon from a 14 7N nucleus? 14", " 12 nucleus are separated by about 2.01015 m. Use Coulomb\u2019s law to find the electric force of repulsion between the two protons. The result will give you an indication of the strength of the strong nuclear force. 69. The binding energy for 4 2He is 28.3 MeV. Calculate the mass of a helium isotope in atomic mass units. 30.2 Nuclear Decay and Reactions 70. Write the complete nuclear equation for the alpha decay of 222 86Rn. 71. Write the complete nuclear equation for the beta decay of 89 36Kr. 72. Complete each nuclear reaction. a. 225 b. 227 89Ac \u2192 4 88Ra \u2192 1 29Cu 1 235U 1 c. 65 d. 92 2He _____ 0e _____ _____ 0n \u2192 _____ \u2192 1 40Zr 3(1 0n \u2192 96 1p _____ 0n) _____ 73. An isotope has a half-life of 3.0 days. What percent of the original material will be left after a. 6.0 days? b. 9.0 days? c. 12 days? 74. In an accident in a research laboratory, a radioactive isotope with a half-life of three days is spilled. As a result, the radiation is eight times the maximum permissible amount. How long must workers wait before they can enter the room? 75. When a boron isotope, 11 5B, is bombarded with protons, it absorbs a proton and emits a neutron. a. What element is formed? b. Write the nuclear equation for this reaction. c. The isotope formed is radioactive and decays by emitting a positron. Write the complete nuclear equation for this reaction. 76. The first atomic bomb released an energy equivalent of 2.0101 kilotons of TNT. One kiloton of TNT is equivalent to 5.01012 J. Uranium-235 releases 3.211011 J/atom. What was the mass of the uranium-235 that underwent fission to produce the energy of the bomb? physicspp.com/chapter_test 77. During a fusion reaction, two deuterons, 2 1H, combine to form a helium isotope, 3 is produced? 2He. What other particle 78. 209 84Po has a half-life of 103 years. How long would it take for a 100-g sample to decay so that only 3.1 g of Po-209 was left? 30.3 The Building Blocks of Matter 79. What would be the charge of a particle composed of three up quarks? 80. The charge of an antiquark is opposite that of a quark. A pion is composed of an up quark and an anti-down quark, ud. What would be the charge of this pion? 81. Pions are composed of a quark and an antiquark. Find the charge of a pion made up of the following. a. uu b. du c. dd 82. Baryons are particles that are made of three quarks. Find the charge on each of the following baryons. a. neutron: ddu b. antiproton: uud 83. The synchrotron at the Fermi Laboratory has a diameter of 2.0 km. Protons circling in it move at approximately the speed of light in a vacuum. a. How long does it take a proton to complete one revolution? b. The protons enter the ring at an energy of 8.0 GeV. They gain 2.5 MeV each revolution. How many revolutions must they travel before they reach 400.0 GeV of energy? c. How long does it take the protons to be accelerated to 400.0 GeV? d. How far do the protons travel during this acceleration? 84. Figure 30-21 shows tracks in a bubble chamber. What are some reasons one track might curve more than another? \u25a0 Figure 30-21 Chapter 30 Assessment 829 Fermilab/Peter Arnold, Inc. 827-831 CH30-SG CA STP-845813 7/29/04 4:02 PM Page 830 Mixed Review 85. Each of the following nuclei can absorb an particle. Assume that no secondary particles are emitted by the nucleus. Complete each equation. a. 14 7N 4 13Al 4 2He \u2192 _____ 2He \u2192 _____ b. 27 86. 211 86Rn has a half-life of 15 h. What fraction of a sample would be left after 60 h? 87. One of the simplest fusion reactions involves the 1H(2.014102 u), from production of deuterium, 2 a neutron and a proton. Write the complete fusion reaction and find the amount of energy released. 88. A 232 92U nucleus, mass 232.0372 u, decays to 228 mass 228.0287 u, by emitting an", " particle, mass 4.0026 u, with a kinetic energy of 5.3 MeV. What must be the kinetic energy of the recoiling thorium nucleus? 90Th, Thinking Critically 89. Infer Gamma rays carry momentum. The momentum of a gamma ray of energy E is equal to E/c, where c is the speed of light. When an electron-positron pair decays into two gamma rays, both momentum and energy must be conserved. The sum of the energies of the gamma rays is 1.02 MeV. If the positron and electron are initially at rest, what must be the magnitude and direction of the momentum of the two gamma rays? 90. Infer An electron-positron pair, initially at rest, also can decay into three gamma rays. If all three gamma rays have equal energies, what must be their relative directions? Make a sketch. 91. Estimate One fusion reaction in the Sun releases about 25 MeV of energy. Estimate the number of such reactions that occur each second from the luminosity of the Sun, which is the rate at which it releases energy, 41026 W. 92. Interpret Data An isotope undergoing radioactive decay is monitored by a radiation detector. The number of counts in each five-minute interval is recorded. The results are shown in Table 30-3. The sample is then removed and the radiation detector records 20 counts resulting from cosmic rays in 5 min. Find the half-life of the isotope. Note that you should first subtract the 20-count background reading from each result. Then plot the counts as a function of time. From your graph, determine the half-life. Table 30-3 Radioactive Decay Measurements Time (min) Counts (per 5 min) 0 5 10 15 20 25 30 987 375 150 70 40 25 18 Writing in Physics 93. Research the present understanding of dark matter in the universe. Why is it needed by cosmologists? Of what might it be made? 94. Research the hunt for the top quark. Why did physicists hypothesize its existence? Cumulative Review 95. An electron with a velocity of 1.7106 m/s is at right angles to a 0.91-T magnetic field. What is the force on the electron produced by the magnetic field? (Chapter 24) 96. An EMF of 2.0 mV is induced in a wire that is 0.10 m long when it is moving perpendicularly across a uniform magnetic field at a velocity of 4.0 m/s. What is the magnetic induction of the field? (Chapter 25) 97. An electron has a de Broglie wavelength of 400.0 nm, the shortest wavelength of visible light. (Chapter 27) a. Find the velocity of the electron. b. Calculate the energy of the electron in eV. 98. A photon with an energy of 14.0 eV enters a hydrogen atom in the ground state and ionizes it. With what kinetic energy will the electron be ejected from the atom? (Chapter 28) 99. A silicon diode (V 0.70 V)that is conducting 137 mA is in series with a resistor and a 6.67-V power source. (Chapter 29) a. What is the voltage drop across the resistor? b. What is the value of the resistor? 830 Chapter 30 Nuclear Physics For more problems, go to Additional Problems, Appendix B. 827-831 CH30-SG CA STP-845813 3/31/04 2:05 AM Page 831 Multiple Choice 1. How many protons, neutrons, and electrons are in the isotope nickel-60, 60 28Ni? Protons Neutrons Electrons 28 28 32 32 32 28 32 28 28 32 28 28 2. What has occurred in the following reaction? 212 82Pb \u2192 212 83Bi e alpha decay beta decay gamma decay loss of a proton 3. What is the product when pollonium-210, 210 84Po, undergoes alpha decay? 206 82Pb 208 82Pb 210 85Pb 210 80Pb 4. A sample of radioactive iodine-131 emits beta particles at the rate of 2.5108 Bq. The half-life is 8 days. What will be the activity after 16 days? 1.6107 Bq 6.2107 Bq 1.2108 Bq 2.5108 Bq 5. Identify the unknown isotope in this reaction: 6C? neutron 14 7N \u2192 14 1 1H 2 1H 3 1H 4 2He 6. Which type of decay does not change the number of protons or neutrons in the nucleus? positron alpha beta gamma 7. Polonium-210 has a half-life of 138 days. How much of a 2.34-kg sample will remain after four years? 0.644 mg 1.50 mg 1.51 g 10.6 g 8.", " An electron and a positron collide, annihilate one another, and release their energy as a gamma ray. What is the minimum energy of the gamma ray? (The energy equivalent of the mass of an electron is 0.51 MeV.) 0.51 MeV 1.02 MeV 931.49 MeV 1863 MeV 9. The illustration below shows the tracks in a bubble chamber produced when a gamma ray decays into a positron and an electron. Why doesn\u2019t the gamma ray leave a track? Gamma rays move too quickly for their tracks to be detected. Only pairs of particles can leave tracks in a bubble chamber. A particle must have mass to interact with the liquid and leave a track, and the gamma ray is virtually massless. The gamma ray is electrically neutral, so it does not ionize the liquid in the bubble chamber. Extended Answer 10. The fission of a uranium-235 nucleus releases about 3.21011 J. One ton of TNT releases about 4109 J. How many uranium-235 nuclei are in a nuclear fission weapon that releases energy equivalent to 20,000 tons of TNT? Do Some Reconnaissance Find out what the conditions will be for taking the test. Is it timed or untimed? Can you use a calculator or other tools? Will those tools be provided? Will mathematical constants be given? Know these things in advance so that you can practice taking tests under the same conditions. physicspp.com/standardized_test Chapter 30 Standardized Test Practice 831 832 EM TOC-845813 6/8/04 10:22 PM Page 832 Math Handbook I. Symbols 833 II. Measurements and Significant Digits Significant Digits Rounding Operations with Significant Digits 833\u2013836 III. Fractions, Ratios, Rates, and Proportions Fractions Ratios Rates Proportions 837\u2013838 IV. Exponents, Powers, Roots, 839\u2013841 and Absolute Value Exponents Square and Cube Roots Operations with Exponents Absolute Value V. Scientific Notation Large Numbers\u2014 841\u2013843 Using Positive Exponents Small Numbers\u2014 Using Negative Exponents Operations with Scientific Notation 833 Isolating a Variable Square Root Property Quadratic Equations Quadratic Formula Dimensional Calculations Dimensional Analysis VII. Graphs of Relations The Coordinate Plane Graphing Data to 848\u2013852 Determine a Relationship Interpolating and Extrapolating Interpreting Line Graphs Linear Equations Slope Direct Variation Inverse Variation Quadratic Graphs VIII. Geometry and 853\u2013856 Trigonometry Perimeter, Area, and Volume Area Under a Graph Right Triangles Trigonometric Ratios Law of Cosines and Law of Sines Inverses of Sine, Cosine, and Tangent Graphs of Trigonometric Functions VI. Equations 843\u2013847 IX. Logarithms 857 Order of Operations Solving Equations Logarithms with Base b Common Logarithms Antilogarithms or Inverse Logarithms Additional Problems Solutions for Practice Problems Tables Color Conventions Electric Circuit Symbols SI Base Units SI Derived Units Useful Conversions Physical Constants SI Prefixes Moments of Inertia for Various Objects Densities of Some Common Substances Melting and Boiling Points of Some Substances Specific Heats of Some Common Substances 910 910 911 911 911 912 912 912 913 913 913 832 858 890 910 913 914 Heats of Fusion and Vaporization of Some Common Substances Coefficients of Thermal Expansion at 20\u00b0C Speed of Sound in Various Media 914 914 Wavelengths of Visible Light Dielectric Constants, K (20\u00b0C) 914 915 The Planets 915 The Moon 915 The Sun Periodic Table of the Elements 916 917 The Elements 918 Safety Symbols 919 927 833-857 EM MHB-845813 6/8/04 10:24 PM Page 833 I. Symbols \u221d change in quantity plus or minus a quantity is proportional to is equal to is approximately equal to is approximately equal to is less than or equal to is greater than or equal to is much less than \u2261 is defined as } } a b ab a(b) a b a/b a b a \uf8e6a\uf8e6 logb x a multiplied by b a divided by b square root of a absolute value of a log to the base, b, of x II. Measurements and Significant Digits Connecting Math to Physics Math is the language of physics. Using math, physicists are able to describe relationships among the measurements that they make using equations. Each measurement is associated with a symbol that is used in physics equations. The symbols are called variables. Significant Digits All measured quantities are approximated and have significant digits. The number of significant digits indicates", " the precision of the measurement. Precision is a measure of exactness. The number of significant digits in a measurement depends on the smallest unit on the measuring tool. The digit farthest to the right in a measurement is estimated. Example: What is the estimated digit for each of the measuring sticks in the figure below used to measure the length of the rod? Using the lower measuring tool, the length is between 9 and 10 cm. The measurement would be estimated to the nearest tenth of a centimeter. If the length was exactly on the 9-cm or 10-cm mark, record it as 9.0 cm or 10.0 cm. Using the upper measuring tool, the length is between 9.5 and 9.6 cm. The measurement would be estimated to the nearest hundredth of a centimeter. If the length was exactly on the 9.5-cm or 9.6-cm mark, record it as 9.50 cm or 9.60 cm. 0 mm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 cm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 833 833-857 EM MHB-845813 3/22/04 8:43 AM Page 834 All nonzero digits in a measurement are significant digits. Some zeros are significant and some are not. All digits between and including the first nonzero digit from the left through the last digit on the right are significant. Use the following rules when determining the number of significant digits. 1. Nonzero digits are significant. 2. Final zeros after a decimal point are significant. 3. Zeros between two significant digits are significant. 4. Zeros used only as placeholders are not significant. Example: State the number of significant digits in each measurement. 5.0 g has two significant digits. 14.90 g has four significant digits. 0.0 has one significant digit. 300.00 mm has five significant digits. 5.06 s has three significant digits. 304 s has three significant digits. Using rules 1 and 2 Using rules 1 and 2 Using rules 2 and 4 Using rules 1, 2, and 3 Using rules 1 and 3 Using rules 1 and 3 0.0060 mm has two significant digits (6 and the last 0). Using rules 1, 2, and 4 140 mm has two significant digits (just 1 and 4). Using rules 1 and 4 1. State the number of significant digits in each measurement. a. 1405 m b. 2.50 km c. 0.0034 m d. 12.007 kg e. 5.8106 kg f. 3.03105 mL There are two cases in which numbers are considered exact, and thus, have an infinite number of significant digits. 1. Counting numbers have an infinite number of significant digits. 2. Conversion factors have an infinite number of significant digits. Examples: The factor \u201c2\u201d in 2mg has an infinite number. The number 2 is a counting number. It is an exact integer. The number \u201c4\u201d in 4 electrons has an infinite number. Because you cannot have a partial electron, the number 4, a counting number, is considered to have an infinite number of significant digits. 60 s/1 min has an infinite number. There are exactly 60 seconds in 1 minute, thus there are an infinite number of significant digits in the conversion factor. 834 Appendix A Math Handbook 833-857 EM MHB-845813 6/8/04 10:26 PM Page 835 Rounding A number can be rounded to a specific place value (like hundreds or tenths) or to a specific number of significant digits. To do this, determine the place being rounded, and then use the following rules. 1. When the leftmost digit to be dropped is less than 5, that digit and any digits that follow are dropped. Then the last digit in the rounded number remains unchanged. 2. When the leftmost digit to be dropped is greater than 5, that digit and any digits that follow are dropped, and the last digit in the rounded number is increased by one. 3. When the leftmost digit to be dropped is 5 followed by a nonzero number, that digit 4. and any digits that follow are dropped. The last digit in the rounded number increases by one. If the digit to the right of the last significant digit is equal to 5, and 5 is followed by a zero or no other digits, look at the last significant digit. If it is odd, increase it by one; if it is even, do not round up. Examples: Round the following numbers to the stated number of significant digits.7645 rounded to 3 significant digits is 8.76. 8.7676 rounded to 3 significant digits is 8.77. 8.7519 rounded to 2 significant digits is 8.8. 92.350 rounded to 3 significant digits is 92.4. 92.25 rounded to 3 significant digits is 92.2.", " Using rule 1 Using rule 2 Using rule 3 Using rule 4 Using rule 4 2. Round each number to the number of significant digits shown in parentheses. a. 1405 m (2) b. 2.50 km (2) c. 0.0034 m (1) d. 12.007 kg (3) Operations with Significant Digits When using a calculator, do all of the operations with as much precision as the calculator allows, and then round the result to the correct number of significant digits. The correct number of significant digits in the result depends on the measurements and on the operation. Addition and subtraction Look at the digits to the right of the decimal point. Round the result to the least precise value among the measurements\u2014the smallest number of digits to the right of the decimal points. Example: Add 1.456 m, 4.1 m, and 20.3 m. The least precise values are 4.1 m and 20.3 m because they have only one digit to the right of the decimal points. 1.456 m 4.1 m 20.3 m 25.856 m Add the numbers The sum is only as precise as the least precise number being added. 25.9 m Round the result to the estimated place with the largest place value Math Handbook 835 833-857 EM MHB-845813 3/22/04 8:44 AM Page 836 Multiplication and division Look at the number of significant digits in each measurement. Perform the calculation. Round the result so that it has the same number of significant digits as the measurement with the least number of significant digits. Example: Multiply 20.1 m by 3.6 m. (20.1 m)(3.6 m) 72.36 m2 The least precise value is 3.6 m with two significant digits. The product can only have as many digits as the least precise of the multiplied numbers. 72 m Round the result to two significant digits 3. Simplify the following expressions using the correct number of significant digits. a. 5.012 km 3.4 km 2.33 km b. 45 g 8.3 g c. 3.40 cm 7.125 cm d. 54 m 6.5 s Combination When doing a calculation that requires a combination of addition/ subtraction and multiplication/division, use the multiplication/division rule. Examples: d 19 m (25.0 m/s)(2.50 s) 1 (10.0 m/s2 )(2.50 s)2 2 5.0101 m 19 m only has two significant digits, so the answer should only have two significant digits m.0 0 1 m 0 slope 70. s 1 1 s 9 2 3.3 m/s 26 s and 11 s only have two significant digits each, so the answer should only have two significant digits Multistep calculations Do not round to significant digits in the middle of a multistep calculation. Instead, round to a reasonable number of decimal places that will not cause you to lose significance in your answer. When you get to your final step where you are solving for the answer asked for in the question, you should then round to the correct number of significant digits. Example: F (24 N)2 (36 N)2 576 N2 1296 N2 Do not round to 580 N2 and 1300 N2 1872 N2 43 N Do not round to 1800 N2 Final answer, so it should be rounded to two significant digits 836 Appendix A Math Handbook 833-857 EM MHB-845813 6/8/04 10:28 PM Page 837 III. Fractions, Ratios, Rates, and Proportions Fractions A fraction names a part of a whole or a part of a group. A fraction also can express a ratio (see page 838). It consists of a numerator, a division bar, and a denominator. er r to a m u n r to a in om n e d number of parts chosen total number of parts Simplification Sometimes, it is easier to simplify an expression before substituting the known values of the variables. Variables often cancel out of the expression. n. p Example: Simplify w p n (p n) )( w p n) (1)( w n w Factor out the p in the numerator and the denominator, and break the fraction into the product of two fractions Substitute ( p/p) Multiply the numerators and the denominators Multiplication and division To multiply fractions, multiply the numerators and multiply the denominators. t. s and Example: Multiply the fractions b a t s b a s)( ( a To divide fractions, multiply the first fraction by the reciprocal of the second fraction. To find the reciprocal of a fraction, invert it\u2014switch the numerator and the denominator. t. s by Example: Divide the fraction b a t ( s)(b s ) Multip", "ly the first fraction by the reciprocal of the second fraction a b a t b s t a Multiply the numerators and the denominators Addition and subtraction To add or subtract two fractions, first write them as fractions with a common denominator; that is, with the same denominator. To find a common denominator, multiply the denominators of both fractions. Then, add or subtract the numerators and use the common denominator. 2. Example: Add the fractions 1 and b a 1 b a 2 (1 ) 2)(a b a b )( 2a ab Multiply each fraction by a fraction equal to 1 Multiply the numerators and the denominators Write a single fraction with the common denominator 4. Perform the indicated operation. Write the answer in simplest form. y a. 1 3 x 3 a b. b b 2 c. (3 ) )(1 x y a 1 d. 2 2 5 Math Handbook 837 833-857 EM MHB-845813 6/8/04 8:06 PM Page 838 Ratios A ratio is a comparison between two numbers by division. Ratios can be written in several different ways. The ratio of 2 and 3 can be written in four different ways: 2 to 3 2 out of 3 2:3 2 3 Rates A rate is a ratio that compares two quantities with different measurement units. A unit rate is a rate that has been simplified so that the denominator is 1. Example: Write 98 km in 2.0 hours as a unit rate. m k 8 9 urs o h 0 2. ( m k 8 9 98 km in 2.0 hours is a ratio of h urs o 0 2. )( ) ) (49)( 49 km per hour or 49 km/h m k ur o h m k ur o h 8 9 0. 2 Split the fraction into the product of a number fraction and a unit fraction Simplify the number fraction Example: Write 16 Swedish crowns in 2 U.S. dollars as a unit rate. 16 Swedish crowns in $2.00 American currency is a ratio of 16 Swedish crowns 2 U.S. dollars 2 ns) ns) 6)(Sw (1 = (8)(Sw w o r c sh i d e s r a ll do. S. U w o r c sh i d e s r a ll do. S. U = 8 Swedish crowns per U.S. dollar or 8 Swedish crowns/U.S. dollar 16 Swedish crowns 2 U.S. dollars Proportions A proportion is an equation that states that two ratios are equal: c, where b and d are not zero. a d b Proportions used to solve ratio problems often include three numbers and one variable. You can solve the proportion to find the value of the variable. To solve a proportion, use cross multiplication. c for a. a Example: Solve the proportion d b c a Cross multiply: d b ad bc c a b d Write the equation resulting from cross multiplying Solve for a 5. Solve the following proportions. 2 a. 4 x 3 b. 3 6 1 2 1.5 7. 5 d. 2 w 0. 5 838 Appendix A Math Handbook 833-857 EM MHB-845813 3/22/04 8:47 AM Page 839 IV. Exponents, Powers, Roots, and Absolute Value Exponents An exponent is a number that tells how many times a number, a, is used as a factor. An exponent is written as a superscript. In the term a n, a is the base and n is the exponent. a n is called the nth power of a or a raised to the nth power. Exponent an Base Connecting Math to Physics A subscript is not an exponent. In physics, a subscript is used to further describe the variable. For example, v0 can be used to represent the velocity at time 0. A subscript is a part of the variable. Positive exponent For any nonzero number, a, and any integer, n, an (a1)(a2)(a3)...(an) Example: Simplify the following exponent terms. 104 (10)(10)(10)(10) 10,000 23 (2)(2)(2) 8 Zero exponent For any nonzero number, a, Examples: Simplify the following zero exponent terms. a 0 1 20 1 130 1 Negative exponent For any nonzero number, a, and any integer, n, 1 an n a Examples: Write the following negative exponent terms as fractions Square and Cube Roots A square root of a number is one of its two equal factors. A radical sign,, indicates a square root. A square root can be shown as exponent 1, as in b b 2 can use a calculator to find square roots. 1. You 2 Examples: Simplify the following square root terms. a 2 (a)(a) a", " 9 (3)(3) 3 64 (8.0)(8.0) 8.0 38.44 6.200 39 6.244997... 6.2 The answer has a zero to the right of the decimal to keep two significant digits Place two zeros to the right of the calculator answer to keep four significant digits Round the calculator answer to keep two significant digits Math Handbook 839 833-857 EM MHB-845813 6/8/04 10:31 PM Page 840 A cube root of a number is one of its three equal factors. A radical sign with the 3, indicates a cube root. A cube root also can be shown as exponent 1, 3 b b 1. 3 number 3, as in 3 Example: Simplify the following cube root terms. 125 3 (5.00)(5.00)(5.00) 5.00 39.304 3.4000. Find each root. Round the answer to the nearest hundredth. a. 22 729 b. 3 c. 676 d. 3 7. Simplify by writing without a radical sign. 46.656 a. 16a 2b4 8. Write using exponents. a. n3 b. 9t 6 1 b. a Operations with Exponents In the following operations with exponents, a and b can be numbers or variables. Product of powers To multiply terms with the same base, add the exponents, as in (a m) (a n) a mn. Quotient of powers To divide terms with the same base, subtract the bottom exponent from the top exponent, as in am/an amn. Power of a power To calculate the power of a power, use the same base and multiply the exponents, as in (am )n a mn. n th root of a power To calculate the root of a power, use the same base and divide the power exponent by the root exponent, as in n a m a m n. Power of a product To calculate the power of a product of a and b, raise both to the power and find their product, as in (ab)n a nb n. 9. Write an equivalent form using the properties of exponents. 2 t 3 a. x x 10. Simplify m q qv 2 m b. t 3 c. (d 2n)2 d. x 2x 840 Appendix A Math Handbook 833-857 EM MHB-845813 3/22/04 8:48 AM Page 841 Absolute Value The absolute value of a number, n, is its magnitude, regardless of its sign. The absolute value of n is written as \uf8e6n\uf8e6. Because magnitudes cannot be less than zero, absolute values always are greater than or equal to zero. Examples: \uf8e63\uf8e6 3 \uf8e63\uf8e6 3 3 units 3 units. Scientific Notation A number of the form a10n is written in scientific notation, where 1 a 10, and n is an integer. The base, 10, is raised to a power, n. The term a must be less than 10. Power a10n Term Base 10 Connecting Math to Physics Physicists use scientific notation to express, compare, and calculate with measurements that are greater than 10 or less than 1. For example, the mass of a proton is written as 6.731028 kg. The density of water is written as 1.000103 kg/m3. This shows, using significant digit rules, that this measurement is exactly 1000 to four significant digits. However, writing the density of water as 1000 kg/m3 would imply that it has only one significant digit, which is incorrect. Scientific notation helps physicists keep accurate track of significant digits. Large Numbers\u2014Using Positive Exponents Multiplying by a power of 10 is like moving the decimal point that same number of places to the left (if the power is negative) or right (if the power is positive). To express a large number in scientific notation, first determine the value for a, 1 a 10. Count the number of decimal places from the decimal point in a to the decimal point in the number. Use that count as the power of 10. A calculator shows scientific notation with e for exponent, as in 2.4e11 2.41011. Some calculators use an E to show the exponent, or there is often a place on the display where the calculator can show smaller-sized digits representing the exponent. Example: Write 7,530,000 in scientific notation. The value for a is 7.53. ( The decimal point is to the right of the first nonzero digit.) So the form will be 7.5310n. 7,530,000 7.53106 There are six decimal places, so the power is 6 To write the standard form of a number expressed in scientific notation, write the value of a, and place extra zeros to the right of", " the number. Use the power and move the decimal point in a that many places to the right. Example: Write the following number in standard form. 2.389105 2.38900105 238,900 Math Handbook 841 833-857 EM MHB-845813 3/22/04 8:49 AM Page 842 Small Numbers\u2014Using Negative Exponents To express a small number in scientific notation, first determine the value for a, 1 a 10. Then count the number of decimal places from the decimal point in a to the decimal point in the number. Use that number as the power of 10. Multiplying by a number with a negative power is the same as dividing by that number with the corresponding positive power. Example: Write 0.000000285 in scientific notation. The value for a is 2.85. (The decimal point is to the right of the first nonzero digit.) So the form will be 2.8510n. 0.000000285 2.85107 There are seven decimal places, so the power is 7 To express a small number in standard form, write the value for a and place extra zeros to the left of a. Use the power and move the decimal point in a that many places to the left. Example: 1.6104 00001.6104 0.00016 11. Express each number in scientific notation. a. 456,000,000 b. 0.000020 12. Express each number in standard notation. a. 3.03107 b. 9.71010 Operations with Scientific Notation Calculating with numbers written in scientific notation uses the properties of expo- nents (see page 840). Multiplication Multiply the terms and add the powers of 10. Example: Simplify. (4.0108)(1.2105) (4.0 1.2)(108 105) Group terms and bases of 10 (4.8)(1085) (4.8)(103) 4.8103 Multiply terms Add powers of 10 Recombine in scientific notation Division Divide the base numbers and subtract the exponents of 10. Example: Simplify. 7 0 ) (1 6 0. (.001073 6.00104 7 0 3 0 842 Appendix A Math Handbook Group terms and bases of 10 Divide terms and subtract powers of 10 833-857 EM MHB-845813 6/8/04 10:32 PM Page 843 Addition and subtraction Adding and subtracting numbers in scientific notation is more challenging, because the powers of 10 must be the same in order to add or subtract the numbers. This means that one of the numbers may need to be rewritten with a different power of 10. If the powers of 10 are equal, then use the distributive property. Example: Simplify. (3.2105) (4.8105) (3.2 4.8)105 8.0105 Group terms Add terms Example: Simplify. (3.2105) (4.8104) (3.2105) (0.48105) Rewrite 4.8104 as 0.48105 (3.2 0.48)105 3.68105 3.7105 Group terms Add terms Round using addition/subtraction rule for significant digits 13. Evaluate each expression. Express the result in scientific notation. a. (5.2104)(4.0 108) b. (2.4103) (8.0104) VI. Equations Order of Operations Scientists and mathematicians have agreed on a set of steps or rules, called the order of operations, so that everyone interprets mathematical symbols in the same way. Follow these steps in order when you evaluate an expression or use a formula. 1. Simplify the expressions inside grouping symbols, like parentheses ( ), brackets [ ], braces { }, and fraction bars. 2. Evaluate all powers and roots. 3. Do all multiplications and/or divisions from left to right. 4. Do all additions and/or subtractions from left to right. Example: Simplify the following expression. 4 3(4 1) 23 4 3(3) 23 4 3(3) 8 4 9 8 8 Order of operations step 1 Order of operations step 2 Order of operations step 3 Order of operations step 4 Connecting Math to Physics The previous example was shown step-by-step to demonstrate order of operations. When solving a physics problem, do not round to the correct number of siginificant digits until after the final calculation. In calculations involving an expression in a numerator and an expression in a denominator, the numerator and the denominator are separate groups and are to be calculated before dividing the numerator by the denominator. So, the multiplication/division rule is used to determine the final number of significant digits. Math Handbook 843 833-857 EM MHB-845813 3/22/04 8:51 AM Page 844 Sol", "ving Equations To solve an equation means to find the value of the variable that makes the equation a true statement. To solve equations, apply the distributive property and the properties of equality. Any properties of equalities that you apply on one side of an equation, you also must apply on the other side. Distributive property For any numbers a, b, and c, a(b c) ab ac a(b c) ab ac Example: Use the distributive property to expand the following expression. 3(x 2) 3x (3)(2) 3x 6 Addition and subtraction properties of equality If two quantities are equal and the same number is added to (or subtracted from) each, then the resulting quantities also are equal. Example: Solve x 3 7 using the addition property 10 Example: Solve t 2 5 using the subtraction property Multiplication and division properties of equality If two equal quantities each are multiplied by (or divided by) the same number, then the resulting quantities also are equal. ac bc b a, for c 0 c c Example: Solve 1 a 3 using the multiplication property. 4 1 a 3 4 a)(4) 3(4) (1 a 12 4 Example: Solve 6n 8 using the division property. 6n 8 8 n 1 6 6 6 n 3 Example: Solve 2t 8 5t 4 for t. 2t 8 5t 4 8 4 5t 2t 12 3t 4 t 844 Appendix A Math Handbook 833-857 EM MHB-845813 3/22/04 8:51 AM Page 845 Isolating a Variable Suppose an equation has more than one variable. To isolate a variable\u2014that is, to solve the equation for a variable\u2014write an equivalent equation so that one side contains only that variable with a coefficient of 1. Connecting Math to Physics Isolate the variable P ( pressure) in the ideal gas law equation. PV nRT P T V nR V V ) nR P(V T V V T P nR V Divide both sides by V Group v v Substitute 14. Solve for x. a. 2 3x 17 b. x 4 2 3x c. t 1 x 4 3 d. a b x c 3 6 e. 2x x f. ax bx c d Square Root Property If a and n are real numbers, n 0, and a 2 n, then a n. Multiply both sides by r Connecting Math to Physics Solve for v in Newton\u2019s second law for a satellite orbiting Earth. Em Gm v 2 m r 2 r Em rGm v 2 rm r 2 r Em Gm mv 2 r Gm v 2 m rm m Gm v 2 E r E v 2 E v Substitute m 1 m Use the positive value for speed Substitute r 1 r Divide both sides by m Take the square root Gm r Em Gm r When using the square root property, it is important to consider for what you are solving. Because we solved for speed in the above example, it did not make sense to use the negative value of the square root. Also, you need to consider if the negative or positive value gives you a realistic solution. For example, when using the square root property to solve for time, a negative value may give you a time before the situation even started. Math Handbook 845 833-857 EM MHB-845813 3/22/04 8:52 AM Page 846 Quadratic Equations A quadratic equation has the form ax 2 bx c 0, where a 0. A quadratic equation has one variable with a power (exponent) of 2. It also may include that same variable to the first power. The solutions of a quadratic equation can be estimated by graphing on a graphing calculator. If b = 0, then there is no x-term in the quadratic equation. The equation can be solved by isolating the squared variable and finding the square root of each side of the equation using the square root property. Quadratic Formula The solutions of any quadratic equation can be calculated by using the quadratic formula. The solutions of ax 2 bx c 0, where a 0, are given by b2 4ac b x 2a Like in the square root property, it is important to consider if the solutions to the quadratic formula give you a realistic answer to the problem you are solving. Usually, you can throw out one of the solutions because it is unrealistic. Projectile motion often requires the use of the quadratic formula when solving equations, so keep the realism of the solution in mind when solving. 15. Solve for x. a. 4x 2 19 17 b. 12 3x 2 9 c. x 2 2x 24 0 d. 24x 2 14x 6 0 Dimensional Calculations When doing calculations, you must include", " the units of each measurement that is written in the calculation. All operations that are performed on the number also are performed on its units. Connecting Math to Physics The acceleration due to gravity, a, is given by the equation x. A free-falling object near the Moon drops 20.5 m in 5.00 s. Find the acceleration, a. a 2 t2 Acceleration is measured in meters per second squared. x a 2 t2 ) m.5 0 2 ( a 2 )2.0 ( 0 s 5 m or 1.64 m/s2 4 a 1.6 2 s The number two is an exact number, so it does not affect the determination of significant digits Calculate and round to three significant digits 846 Appendix A Math Handbook 833-857 EM MHB-845813 3/22/04 8:53 AM Page 847 Unit conversion Use a conversion factor to convert from one measurement unit to another of the same type, such as from minutes to seconds. This is equivalent to multiplying by one. t. Connecting Math to Physics Find x when v0 = 67 m/s and t = 5.0 min. Use the equation, x v0 ds 1 n o c 60 e s te u n i m 1 x v0 t ) m(5.0 x 67 s 1 x 20100 m 2.0104 m s 0 6 1 Multiply by the conversion factor m n i 1 min)( s 0 6 n i m 1 Calculate and round to two significant digits. The numbers 60 s and 1 min are exact numbers, so they do not affect the determination of significant digits. 1 0 16. Simplify t 4.0 s / m 6 1 17. Find the velocity of a dropped brick after 5.0 s using v at and a 9.80 m/s2. 18. Calculate the product: (32 19. An Olympic record for 100.0 m is 9.87 s. What is the speed in)( )(60 m)( m 1 cm 0 10 in kilometers per hour? Dimensional Analysis Dimensional analysis is a method of doing algebra with the units. It often is used to check the validity of the units of a final result and the equation being used, without completely redoing the calculation. Physics Example Verify that the final answer of df di vit 1 at 2 will have the units m. 2 df di is measured in m. t is measured in s. vi is measured in m/s. a is measured in m/s2. m )(s) 1 m (m )(s)2 ( 2 s 2 m (m)(s (m)(s ) 1 ) 2 2 s m (m)(1) 1 (m)(1 Substitute the units for each variable Simplify the fractions using the distributive property Substitute s/s = 1, s2/s2 = 1 Everything simplifies to m, thus df is in m The factor of 1 in the above does not apply to the units. It applies only to any number 2 values that would be inserted for the variables in the equation. It is easiest to remove number factors like the 1 when first setting up the dimensional analysis. 2 Math Handbook 847 833-857 EM MHB-845813 6/9/04 4:35 PM Page 848 VII. Graphs of Relations The Coordinate Plane Points are located in reference to two perpendicular number lines, called axes. The horizontal number line is called the x-axis; the vertical number line is called the y-axis. The x-axis represents the independent variable. The y-axis represents the dependent variable. A point is represented by two coordinates ( x, y), which also is called an ordered pair. The value of the independent variable, x, always is listed first in the ordered pair. The ordered pair (0, 0) represents the origin, which is the point where the two axes intersect. The vertical axis also is called the y-axis. y 4 2 (4, 3) 4 2 0 2 4 x The origin, at (0, 0), is the point where the axes intersect. 2 4 The coordinate system also is called the coordinate plane. Each point is named by an ordered pair. The horizontal axis also is called the x-axis. Graphing Data to Determine a Relationship Use the following steps to graph data. 1. Draw two perpendicular axes. 2. Identify the independent and the dependent variables. Label each axis using the variable names. 3. Determine the range of data for each variable. Use the ranges to decide on a conven- ient scale for each axis. Mark and number the scales. 4. Plot each data point. 5. When the points seem to lie approximately in a line, draw a \u201cbest-fit\u201d line through the points. When the points do not lie in a line, draw a smooth curve through as many data points as possible. When there does", " not appear to be a trend or a pattern to the dots, do not draw any line or curve. U.S. Dollars and British Pounds 70 6. Write a title that clearly describes what the graph represents. Expense Pounds Dollars Hotel Meals Entertainment Transportation 40 30 15 6 58 43 22 9 848 Appendix A Math Handbook s r a l l o D 60 50 40 30 20 10 0 10 20 30 40 50 60 70 Pounds 833-857 EM MHB-845813 3/22/04 8:59 AM Page 849 Interpolating and Extrapolating Interpolation is a process used to estimate a value for a relation that lies between two known values. Extrapolation is a process used to estimate a value for a relation that lies beyond the known values. The equation of a line through the points (the known values) helps you interpolate and extrapolate. Example: Using the data and the graph, estimate the value, in dollars, of 20 pounds. Use interpolation. Locate two points on either side of 20 pounds\u2014 15 pounds and 30 pounds. Draw a line (or place a straight edge) through the graphed points. Draw a line segment from 20 on the x-axis to that line. Draw a line segment from that intersection point to the y-axis. Read the scale on the y-axis. 20 pounds is about 28 or 29 dollars. Example: Use extrapolation to estimate the value of 50 pounds. Draw a line segment from 50 on the x-axis to the line through the points. Extend the line if necessary. Read the corresponding value on the y-axis. Extend the axis scale if necessary. 50 pounds is a little more than 70 dollars. s r a l l o D 70 60 50 40 30 20 10 0 U.S.Dollars and British Pounds 10 20 30 40 50 60 Pounds Interpreting Line Graphs A line graph shows the linear relationship between two variables. Two types of line graphs that describe motion are used frequently in physics. Connecting Math to Physics The following line graph shows a changing relationship between the two graphed variables. Move back toward starting point n o i t i s o P 0 Move ahead quickly Move ahead slowly Maintain position Time This line graph shows a constant relationship between the two graphed variables Constant velocity Time Math Handbook 849 833-857 EM MHB-845813 3/22/04 8:59 AM Page 850 Linear Equations A linear equation can be written as a relation (or a function), y mx b, where m and b are real numbers, m represents the slope of the line, and b represents the y-intercept, the point at which the line crosses the y-axis. Dependent variable Independent variable y mx b Slope y-intercept The graph of a linear equation is a line. The line represents all of the solutions of the linear equation. To graph a linear equation, choose three values for the independent variable. (Only two points are needed, but the third point serves as a check.) Calculate the corresponding values for the dependent variable. Plot each ordered pair (x, y) as a point. Draw a best-fit line through the points. Example: Graph y 1 x3 2 Calculate three ordered pairs to obtain points to plot. Plot of Ordered Pairs y Ordered Pairs x 0 2 6 y 3 2 0 x Slope The slope of a line is the ratio of the change in y-coordinates to the change in x-coordinates. It is the ratio of the vertical change (rise) to the horizontal change (run). This number tells how steep the line is. It can be a positive or a negative number. To find the slope of a line, select two points, (x1, y1) and (x2, y2). Calculate the run, which is the difference (change) between the two x-coordinates, x2 Calculate the rise, which is the difference (change) between the two y-coordinates, y2 y. Form the ratio. y1 y2 y y, e s i Slope x2 where x1 x1 x. \u2206y y y2 y1 0 (x2, y2) (x1, y1) x1 x2 x \u2206x 850 Appendix A Math Handbook 833-857 EM MHB-845813 6/15/04 11:41 PM Page 851 Direct Variation If there is some nonzero constant, m, such that y mx, then y varies directly as x. That means as the independent variable x increases, the dependent variable y increases. The variables x and y also are said to be proportional. This is a linear equation of the form y mx b in which the value of b is zero. The graph passes through the origin, (0, 0). Connecting Math to Physics In the force equation for an", " ideal spring, F kx, where F is the spring force, k is the spring constant, and x is the spring displacement, spring force varies directly as ( is proportional to) spring displacement. That is, the spring force increases as the spring displacement increases Inverse Variation If there is some nonzero constant, m, such that y m, then y varies inversely as x. x That means as the independent variable x increases, the dependent variable y decreases. The variables x and y also are said to be inversely proportional. This is not a linear equation because it contains the product of two variables. The graph of an inverse relationship is a hyperbola. This relationship can be written as y m x y m 1 x xy m Example: Graph the equation xy 90. Ordered Pairs y x 9 10 15 6 30 3 45 2 45 2 3 6 10 30 15 9 Graph of an Inverse Variation y 40 20 8 4 0 20 x 4 8 Connecting Math to Physics In the equation for the speed of a wave, v, where f is wavelength, f is frequency, and v is wave speed, wavelength varies inversely as ( is inversely proportional to) frequency. That is, as the frequency of a wave increases, the wavelength decreases. v is constant. Math Handbook 851 833-857 EM MHB-845813 3/25/04 11:12 AM Page 852 Quadratic Graphs A quadratic relationship is a relationship of the form y ax 2 bx c, where a 0. A quadratic relationship includes the square of the independent variable, x. The graph of a quadratic relationship is a parabola. Whether the parabola opens upward or downward depends on whether the value of the coefficient of the squared term, a, is positive or negative. Example: Graph the equation y x 2 4x 1. Ordered Pairs Graph of a Quadratic Equation y 0 x y x2 4x 1 Connecting Math to Physics A position-time graph in the shape of a quadratic relation means that the object is moving at a constant acceleration. Ordered Pairs Time (s) Position (m) 1 2 3 4 3 6 11 18 Quadratic Graph of Constant Acceleration ) 20 16 12 8 4 0 2 4 Time (s) 852 Appendix A Math Handbook 833-857 EM MHB-845813 3/22/04 9:02 AM Page 853 VIII. Geometry and Trigonometry Perimeter, Area, and Volume Perimeter, Circumference Linear units Area Squared units Surface Area Squared units Volume Cubic units P 4a A a 2 P 2l 2w A lw A 1 bh 2 C 2r A r 2 Square, side a Rectangle, length l width w Triangle base b height h Cube side a Circle radius r Cylinder radius r height h Sphere radius r SA 6a2 V a3 SA 2rh 2r 2 V r 2h SA 4r 2 V 4 r 3 3 Connecting Math to Physics Look for geometric shapes in your physics problems. They could be in the form of objects or spaces. For example, two-dimensional shapes could be formed by velocity vectors, as well as position vectors. Area Under a Graph To calculate the approximate area under a graph, cut the area into smaller pieces and find the area of each piece using the formulas shown above. To approximate the area under a line, cut the area into a rectangle and a triangle, as shown in (a). To approximate the area under a curve, draw several rectangles from the x-axis to the curve, as in (b). Using more rectangles with a smaller base will provide a closer approximation of the area Time Total area Area of the rectangle Area of the triangle n o i t i s o P 0 Time Total area Area 1 Area 2 Area 3\u2026 Math Handbook 853 833-857 EM MHB-845813 3/22/04 9:03 AM Page 854 Right Triangles The Pythagorean theorem states that if a and b are the measures of the legs of a right triangle and c is the measure of the hypotenuse, then c 2 a2 b 2. To determine the length of the hypotenuse, use the square root property. Because distance is positive, the negative value does not have meaning. c a2 b2 Example: In the triangle, a 4 cm and b 3 cm. Find c. c a2 b2 (4 cm)2 (3cm)2 16 cm2 9 cm2 25 cm2 5 cm Hypotenuse c Leg a b Leg 4 c 3 45\u00b0-45\u00b0-90\u00b0 triangles The length of the hypotenuse is 2 times the length of a leg. 45\u00b0 - 45\u00b0 - 90\u00b0 Right Triangle (2)x 45\u00b0 x 30\u00b0-60\u00b0-90\u00b0 triangles The length of the hypotenuse", " is twice the length of the shorter leg. The length of the longer leg is 3 times the length of the shorter leg. 30\u00b0- 60\u00b0 - 90\u00b0 Right Triangle 2x 30\u00b0 (3)x 854 Appendix A Math Handbook 45\u00b0 x 60\u00b0 x 833-857 EM MHB-845813 3/22/04 9:04 AM Page 855 Trigonometric Ratios A trigonometric ratio is a ratio of the lengths of sides of a right triangle. The most common trigonometric ratios are sine, cosine, and tangent.To memorize these ratios, learn the acronym SOH-CAH-TOA. SOH stands for Sine, Opposite, Hypotenuse. CAH stands for Cosine, Adjacent, Hypotenuse. TOA stands for Tangent, Opposite, Adjacent. Words The sine is the ratio of the length of the side opposite to the angle over the length of the hypotenuse. Memory Aid Symbols SOH o te i s p p o sin a sin c u n e t o p hy se The cosine is the ratio of the length of the side adjacent to the angle over the length of the hypotenuse. CAH t n e c ja d a cos b cos c te e s u n o p hy The tangent is the ratio of the length of the side opposite to the angle over the length of the side adjacent to the angle. TOA p te i s o p tan o t n e c a j ad a tan b Example: In right triangle ABC, if a 3 cm, b 4 cm, and c 5 cm, find sin and cos. c m sin 3 0.6 cm 5 m c m c cos 4 0.8 cm c m 5 Example: In right triangle ABC, if 30.0\u00b0 and c 20.0 cm, find a and b. b a cos 30.0\u00b0 sin 30.0\u00b0 cm 20.0 cm 20.0 a (20.0 cm)(sin 30.0\u00b0) 10.0 b (20.0 cm)(cos 30.0\u00b0) 17.3 Law of Cosines and Law of Sines B a C Side opposite Hypotenuse c A b Side adjacent The laws of cosines and sines let you calculate sides and angles in any triangle. Law of cosines The law of cosines looks like the Pythagorean theorem, except for the last term. is the angle opposite side c. If the angle is 90\u00b0, the cos 0 and the last term equals zero. If is greater than 90\u00b0, its cosine is a negative number. c 2 a2 b2 2ab cos a Example: Find the length of the third side of a triangle with a 10.0 cm, b 12.0 cm, 110.0\u00b0. c 2 a2 b2 2ab cos c a2 b2 2ab cos (10.0 cm)2 (12.0cm)2 2(10.0 cm)(12.0 cm)(cos 110.0\u00b0) 1.00102 cm2 144 cm2 (60.0 cm2)(cos 110.0\u00b0) 16.3 b c Math Handbook 855 833-857 EM MHB-845813 3/22/04 9:05 AM Page 856 Law of sines The law of sines is an equation of three ratios, where a, b, and c are the sides opposite angles A, B, and C, respectively. Use the law of sines when you know the measures of two angles and any side of a triangle. C B sin A sin sin c b a Example: In a triangle, C 60.0\u00b0, a 4.0 cm, c 4.6 cm. Find the measure of angle A. C A sin sin c a n C sin A a si c (4.0 cm) (sin 60.0\u00b0) 4.6 49\u00b0 C b a A c B Inverses of Sine, Cosine, and Tangent The inverses of sine, cosine, and tangent allow you to do the reverse of the sine, cosine, and tangent functions and find the angle. The trigonometric functions and their inverses are as follows: Trigonometric Function y sin x y cos x y tan x Inverse x sin1 y or x arc sin y x cos1 y or x arc cos y x tan1 x or x arc tan y Graphs of Trigonometric Functions The sine function, y sin x, and the cosine function, y cos x are periodic functions. The period for each function is 2. x can be any real number. y is any real numbers between 1 and 1, inclusive. y sin x y 1 2 0 x 2 y cos x 2 1 y 1 0 1 x 2 856 Appendix A Math Handbook 833", "-857 EM MHB-845813 6/8/04 10:41 PM Page 857 IX. Logarithms Logarithms with Base b Let b and x be positive integers, b 1. The logarithm of x with base b, written logb x, equals y, where y is the exponent that makes the equation by x true. The log to the base b of x is the number to which you can raise b to get x. logb x y if and only if b y x Memory aid: \u201cthe log is the exponent Examples: Calculate the following logarithms. 1 4 log2 6 1 log10 1000 3 1 Because 24 6 1 Because 103 1000 When you want to find the log of a number, you also can use a calculator. Connecting Math to Physics Physicists use logarithms to work with measurements that extend over several orders of magnitude, or powers of 10. Geophysicists use the Richter scale, a logarithmic scale that allows them to rate earthquakes from 0 to 7, or larger, though the power of earthquakes differ by 7 or more powers of 10. Common Logarithms Base 10 logarithms are called common logarithms. They often are written without the subscript 10. log10 x log x x 0 Antilogarithms or Inverse Logarithms An antilogarithm is the inverse of a logarithm. An antilogarithm is the number which has a given logarithm. Example: Solve log x 4 for x. log x 4 x 104 104 is the antilogarithm of 4 Connecting Math to Physics An equation for loudness, L, in decibels, is L 10 log10 R, where R is the relative intensity of the sound. Calculate R for a fireworks display with a loudness of 130 decibels. 130 10 log10 R 13 log10 R R 1013 Use logarithm rule Divide by 10 When you know the logarithm of a number and want to know the number itself, use a calculator to find the antilogarithm. 20. Write in exponential form. log3 81 4 21. Write in logarithmic form. 103 0.001 22. Find x. log x 3.125 Math Handbook 857 858-889 EM APP B-845813 6/8/04 8:07 PM Page 858 B x i d n e p p A Chapter 1 1. The density,, of an object is given by the ratio of the object\u2019s mass, m, and volume, V, according to the equation m/V. What is the density of a cube that is 1.2 cm on each side and has a mass of 25.6 g? 2. An object that is moving in a straight line with speed v covers a distance, d vt, in time t. Rewrite the equation to find t in terms of d and v. How long does it take a plane that is traveling at 350 km/h to travel 1750 km? 3. Convert 523 kg to milligrams. 4. The liquid measure milliliter, mL, is the same as 1 cm3. How many milliliters of liquid can be held in a 2.5-m3 container? 5. Part of the label from a vitamin container is shown below. The abbreviation \u201cmcg\u201d stands for micrograms. Convert the values to milligrams. Each Tablet Contains Folic Acid 400 mcg Vitamin B12 6 mcg Biotin 30 mcg %DV 100% 100% 10% 6. What type of relationship is shown in the scatter plot? Write an equation to model the data. 1.2 1.0 0.8 0.6 0.4 0.2 0 858 Appendix B 7. How many significant digits are there in each of the following measurements? a. 100 m b. 0.0023 m/s c. 100.1 m d. 2.0023 8. The buoyant (upward) force exerted by water on a submerged object is given by the formula F Vg, where is the density of water (1.0103 kg/m3), V is the volume of the object in m3, and g is the acceleration due to gravity (9.80 m/s2). The force is in newtons. Rewrite the equation to find V in terms of F. Use this to find the volume of a submerged barrel if the buoyant force on it is 9200 N. 9. What is 2.3 kg 0.23 g? 10. Solve the following problems. a. 15.5 cm 12.1 cm m 8 67. 4 b. 1 m s / 2. 3 11. An experiment was performed to determine the period of a pend", "ulum as a function of the length of its string. The data in the table below were measured. a. Plot period, T, versus length, l. b. Are the data linear? c. Plot the period versus the square root of the length. d. What is the relationship between the period and the square root of the length? Length (m) Period (s) 0.1 0.2 0.4 0.6 0.8 1.0 0.6 0.9 1.3 1.6 1.8 2.0 1 2 3 4 5 6 12. Based on the previous problem, what should be the period of a pendulum whose length is 0.7 m? 858-889 EM APP B-845813 6/8/04 10:47 PM Page 859 Chapter 2 1. A position-time graph for a bicycle is shown in the figure below. a. What is the position of the bicycle at 1.00 min? b. What is the position of the bicycle at 3.50 min? c. What is the displacement of the bicycle between the times 1.00 min and 5.00 min? d. Describe the motion of the bicycle. 40.0 30.0 20.0 10..00 1.00 2.00 3.00 4.00 5.00 Time (min) 2. The position of an automobile is plotted as a function of time in the accompanying figure. a. What is the position of the car at 0.00 s? b. What is the position of the automobile after 2.00 s has elapsed? c. How far did the automobile travel between the times 1.00 s and 3.00 s 50.0 40.0 30.0 20.0 10.0 0.. A jogger runs at a constant rate of 10.0 m every 2.0 s. The jogger starts at the origin, and runs in the positive direction for 3600.0 s. The figure below is a positiontime graph showing the position of the jogger from time t 0.0 s to time t 20.0 s. Where is the runner at time t 5.0 s? t 15.0 s 100.0 75.0 50.0 25.0 0.0 4.0 8.0 12.0 16.0 20.0 Time (s) 4. Two trains simultaneously leave the same train station at noon. One train travels north and the other travels south. The position-time graph for both trains is shown in the accompanying figure. a. What is the position of the train traveling north at 6.0 h? b. What is the position of the train traveling south at 6.0 h? c. What is the distance between the trains at 6.0 h? What is the distance at 10.0 h? d. At what time are the trains 600.0 km apart? e. Which train is moving more quickly? North 1000..0 4.0 6.0 8.0 10.0 Time 1.00 2.00 3.00 4.00 5.00 Time (s) South 1000.0 Appendix B 859 858-889 EM APP B-845813 3/31/04 4:14 AM Page 860 5. Two cars head out in the same direction. Car A starts 1.0 min before car B. The position-time graphs for both cars are shown in the accompanying figure. a. How far apart are the two cars when car B starts out at t 1.0 min? b. At what time do the cars meet? c. How far apart are the cars at time t 3.0 min? 7. A child\u2019s toy train moves at a constant speed of 2.0 cm/s. a. Draw the position-time graph showing the position of the toy for 1.0 min. b. What is the slope of the line represent- ing the motion of the toy? 8. The position of an airplane as a function of time is shown in the figure below. a. What is the average velocity of the airplane? Car A b. What is the average speed of the airplane? Car B Time (h) 0.0 1.0 2.0 3.0 4.0 5..0 1.5 1.0 0..0 1.0 2.0 3.0 4.0 5.0 Time (min) 6. The position-time graph for two joggers, A and B, is shown in the accompanying figure. a. How far apart are the two runners at 10.0 min? b. At what time are they 1.00 km apart? c. How far apart are they at 50.0 min? d. At what time do they meet? e. What distance does jogger B cover between 30.0 min and 50.0 min? f. What distance does jogger A cover between 30.0", " min and 50.0 min 200.0 400.0 600.0 800.0 9. The position-time graph for a hot-air balloon that is in flight is shown in the accompanying figure. a. What is the average velocity of the balloon? b. What is the average speed of the balloon? Jogger B Time (h) 0.0 1.0 2.0 3.0 4.0 5.0 2.00 1.50 1.00 0.50 ) Jogger A 0.00 10.0 20.0 30.0 40.0 50.0 Time (min) 860 Appendix 10.0 20.0 30.0 40.0 858-889 EM APP B-845813 6/8/04 10:55 PM Page 861 Chapter 3 1. Jason and his sister, Tara, are riding bicycles. Jason tries to catch up to Tara, who has a 10.0-s head start. a. What is Jason\u2019s acceleration? b. What is Tara\u2019s acceleration? c. At what time do they have the same Jason Tara velocity? 8.0 6.0 4.0 2..0 10.0 20.0 30.0 40.0 Time (s) 2. A dragster starts from rest and accelerates for 4.0 s at a rate of 5.0 m/s2. It then travels at a constant speed for 2.5 s. A parachute opens, stopping the vehicle at a constant rate in 2.0 s. Plot the v-t graph representing the entire motion of the dragster. 3. A car traveling at 21 m/s misses the turnoff on the road and collides into the safety guard rail. The car comes to a complete stop in 0.55 s. a. What is the average acceleration of the car? b. If the safety rail consisted of a section of rigid rail, the car would stop in 0.15 s. What would be the acceleration in this case? 4. On the way to school, Jamal realizes that he left his physics homework at home. His car was initially heading north at 24.0 m/s. It takes him 35.5 s to turn his car around and head south at 15.0 m/s. If north is designated to be the positive direction, what is the average acceleration of the car during this 35.5-s interval? 5. A cheetah can reach a top speed of 27.8 m/s in 5.2 s. What is the cheetah\u2019s average acceleration? 6. After being launched, a rocket attains a speed of 122 m/s before the fuel in the motor is completely used. If you assume that the acceleration of the rocket is constant at 32.2 m/s2, how much time does it take for the fuel to be completely consumed? 7. An object in free fall has an acceleration of 9.80 m/s2 assuming that there is no air resistance. What is the speed of an object dropped from the top of a tall cliff 3.50 s after it has been released, if you assume the effect of air resistance against the object is negligible? 8. A train moving with a velocity of 51 m/s east undergoes an acceleration of 2.3 m/s2 as it approaches a town. What is the velocity of the train 5.2 s after it has begun to decelerate? 9. The v-t graph of a runner is shown in the accompanying figure. a. What is the displacement of the runner between t 0.00 s and t 20.0 s? b. What is the displacement of the runner between t 20.0 s and t 50.0 s? c. What is the displacement of the runner between t 50.0 s and t 60.0 s.0 6.0 4.0 2..00 10.0 20.0 30.0 40.0 50.0 60.0 Time (s) 10. Draw the v-t graph of an automobile that accelerates uniformly from rest at t 0.00 s and covers a distance of 180.0 m in 12.0 s. Appendix B 861 858-889 EM APP B-845813 3/31/04 4:16 AM Page 862 17. A cyclist rides at a constant speed of 12.0 m/s for 1.20 min and then coasts to a stop with uniform acceleration 21.2 s later. If the total distance traveled is 1321 m, then what is the acceleration while the bike coasts to a stop? 18. A hiker tosses a rock into a canyon. He hears it strike water 4.78 s later. How far down is the surface of the water? 19. A rock is thrown upward with a speed of 26 m/s. How long after it is thrown will the rock have a velocity of 48 m/s toward", " the ground? 20. The high-dive board at most pools is 3.00 m above the water. A diving instructor dives off the board and strikes the water 1.18 s later. a. What was the initial velocity of the diver? b. How high above the board did the diver rise? Chapter 4 1. Draw a free-body diagram for the space shuttle just after it leaves the ground. Identify the forces acting on the shuttle. Make sure that you do not neglect air resistance. Also be sure that you indicate the direction of the acceleration, as well as the net force. 2. Draw a free-body diagram for a goldfish that is motionless in the middle of a fishbowl. Identify the forces acting on the fish. Indicate the direction of the net force on the fish and the direction of the acceleration of the fish. 11. The v-t graph of a car is shown in the accompanying figure. What is the displacement of the car from t 0.00 s to t 15.0 s 30.0 15.0 0.00 15.0 30..00 8.00 12.00 16.00 Time (s) 12. Suppose a car rolls down a 52.0-m-long inclined parking lot and is stopped by a fence. If it took the car 11.25 s to roll down the hill, what was the acceleration of the car before striking the fence? 13. A sky diver in free fall reaches a speed of 65.2 m/s when she opens her parachute. The parachute quickly slows her down to 7.30 m/s at a constant rate of 29.4 m/s2. During this period of acceleration, how far does she fall? 14. A child rolls a ball up a hill at 3.24 m/s. If the ball experiences an acceleration of 2.32 m/s2, how long will it take for the ball to have a velocity of 1.23 m/s down the hill? 15. A cheetah can accelerate from rest to a speed of 27.8 m/s in 5.20 s. The cheetah can maintain this speed for 9.70 s before it quickly runs out of energy and stops. What distance does the cheetah cover during this 14.9-s run? 16. A cab driver in a hurry is sitting at a red light. When the light turns green she rapidly accelerates for 3.50 s at 6.80 m/s2. The next light is still red. She then slams on the brakes, accelerating at a rate of 9.60 m/s2 before coming to rest at the stop light. What was her total distance for this trip? 862 Appendix B 858-889 EM APP B-845813 3/31/04 4:17 AM Page 863 3. Draw a free-body diagram for a submerged beach ball as it rises toward the surface just after being released. Identify the forces acting on the beach ball and indicate the direction of the net force and the acceleration. 7. Two dogs fight over a bone. The larger of the two pulls on the bone to the right with a force of 42 N. The smaller one pulls to the left with a force of 35 N. a. Draw the free-body diagram for the bone. b. What is the net force acting on the bone? c. If the bone has a mass of 2.5 kg, what is its acceleration? 4. Muturi is rearranging some furniture. He pushes the dresser with a force of 143 N, and there is opposing frictional force of 112 N. What is the net force? 5. One of the floats in a Thanksgiving Day parade requires four people pulling on ropes to maintain a constant speed of 3.0 km/h for the float. Two people pull with a force of 210 N each, and the other two pull with a force of 140 N each. a. Draw a free-body diagram. b. What is the force of friction between the float and the ground? Dog 1 Dog 2 8. A large model rocket engine can produce a thrust of 12.0 N upon ignition. This engine is part of a rocket with a total mass of 0.288 kg when launched. a. Draw a free-body diagram of the rocket just after launch. b. What is the net force that is acting on the model rocket just after it leaves the ground? c. What is the initial acceleration of the rocket. Five people are playing tug-of-war. Anders and Alyson pull to the right with 45 N and 35 N, respectively. Calid and Marisol pull to the left with 53 N and 38 N, respectively. With what force and in what direction does Benito pull if the game is tied? Appendix B 863 858-889 EM APP B-845813 6/8/04 11:01 PM Page 864. Erika is on an elevator and presses the", " button to go down. When the elevator first starts moving, it has an acceleration of 2.5 m/s2 downward. Erika and the elevator have a combined mass of 1250 kg. a. Draw a free-body diagram for the elevator. b. What is the tension in the cable that provides the upward force on the elevator car? 14. A 7.25-g bullet is fired from a gun. The muzzle velocity of the bullet is 223 m/s. Assume that the bullet accelerates at a constant rate along the barrel of the gun before it emerges with constant speed. The barrel of the gun is 0.203 m long. What average force does the bullet exert on the gun? 15. A 15.2-kg police battering ram exerts an average force of 125 N on a 10.0-kg door. a. What is the average acceleration of the 10. Ngan has a weight of 314.5 N on Mars and door? a weight of 833.0 N on Earth. a. What is Ngan\u2019s mass on Mars? b. What is the acceleration due to gravity b. What is the average acceleration of the battering ram? 16. As a demonstration, a physics teacher attaches a 7.5-kg object to the ceiling by a nearly massless string. This object then supports a 2.5-kg object below it by another piece of string. Finally, another piece of string hangs off the bottom of the lower object to be pulled with ever increasing force until the string breaks somewhere. The string will break when the tension reaches 156 N. a. Which length of string will break first? b. What is the maximum downward force the physics teacher can apply before the string breaks? 7.5 kg 2.5 kg on Mars, gMars? 11. Alex is on the wrestling team and has a mass of 85.3 kg. Being a whiz at physics, he realizes that he is over the 830.0-N cutoff for his weight class. If he can convince the trainers to measure his weight in the elevator, what must be the acceleration of the elevator so that he just makes his weight class? 12. During a space launch, an astronaut typically undergoes an acceleration of 3 gs, which means he experiences an acceleration that is three times that of gravity alone. What would be the apparent weight of a 205-kg astronaut that experiences a 3-g liftoff? 13. Alfonso and Sarah like to go sky diving together. Alfonso has a mass of 88 kg, and Sarah has a mass of 66 kg. While in free fall together, Alfonso pushes Sarah horizontally with a force of 12.3 N. a. What is Alfonso\u2019s horizontal acceleration? b. What is Sarah\u2019s horizontal acceleration? 17. A 10.0-kg object is held up by a string that will break when the tension exceeds 1.00102 N. At what upward acceleration will the string break? 864 Appendix B 858-889 EM APP B-845813 3/31/04 4:19 AM Page 865 18. A large sculpture is lowered into place by a crane. The sculpture has a mass of 2225 kg. When the sculpture makes contact with the ground, the crane slowly releases the tension in the cable as workers make final adjustments to the sculpture\u2019s position on the ground. a. Draw a free-body diagram of the sculp- ture when it is in contact with the ground, and there is still tension in the cable while the workers make the final adjustments. b. What is the normal force on the sculpture when the tension in the cable is 19,250 N? Chapter 5 1. A soccer ball is kicked from a 22-m-tall platform. It lands 15 m from the base of the platform. What is the net displacement of the ball? 2. If the net displacement is 32 m for the same situation, as described in problem 1, how far from the platform base must the ball land? 3. For any single force vector, there is only one angle for which its x- and y-components are equal in size. a. What is that angle? b. How many times bigger is a vector at this particular angle than either of its components? 4. A cue ball on a billiards table travels at 1.0 m/s for 2.0 s. After striking another ball, it travels at 0.80 m/s for 2.5 s at an angle of 60.0\u00b0 from its original path. a. How far does the cue ball travel before and after it strikes the other ball? b. What is the net displacement of the cue ball for the entire 4.5-s time interval? 120.00\u00b0 5. The table below represents a set of force vectors. These vectors begin at the origin of a coordinate system, and end at the coordinates given in the table. Vector # x-", "value (N) y-value (N) A B C 0.0 5.0 0.0 6.0 0.0 10.0 a. What is the magnitude of the resultant of the sum of these three vectors? b. What is the size of the angle,, that the resultant makes with the horizontal? c. Into which quadrant does the resultant point? y A B R C x Appendix B 865 858-889 EM APP B-845813 6/8/04 11:05 PM Page 866. A 9.0-kg crate sits on a level, rough floor. A 61-N force is needed just to start it moving. What is the size of the coefficient of maximum static friction? 7. Given the graph below answer the follow- ing questions. a. What is the value of b. If the frictional force is 1.5 N, what is k for this system? FN? c. Does tripling FN triple Fapplied? d. Do Fapplied and FN act in the same direction? Explain why or why not FN (N) 8. A wooden block sits on a level lab table. A string draped over a pulley connects to a bucket that can be filled with lead pellets. Maggie wants to measure how much applied mass (pellets bucket) is needed to move the block along the table at a constant speed. a. If the applied mass of the bucket and the lead pellets is 0.255 kg, and the block has a weight of 12 N, what is the value of k? b. If some extra pellets are added, describe the behavior of the block. 866 Appendix B 9. A boat travels 75 km southeast, then 56 km due east, then 25 km 30.0\u00b0 north of east. a. Sketch the vector set on a N-E-S-W grid. b. Find its net E-W component of displacement and net N-S component of displacement. c. Find its net displacement. d. Find its net angle relative to an E-W axis. 10. A 1.25-kg box is being pulled across a level k is 0.80. If the string is surface where suddenly cut, causing the applied force to immediately go to 0.0 N, what is the rate of acceleration of the block? Hint: The acceleration will be a negative value. 11. If the velocity of the box in problem 10 was 5.0 m/s at time zero, what will be its speed after 0.50 s? How far will it travel in that time interval? 12. A 0.17-kg hockey puck leaves the stick on a slap shot traveling 21 m/s. If no other forces act on the puck, friction eventually will bring it to rest 62 m away. a. Based on these data, determine the value of k for this hockey puck on ice. b. Will the answer change if a puck of different mass, but same material and shape is used? Explain why or why not. 13. A 105-N suitcase sits on a rubber ramp at an airport carousel at a 25\u00b0 angle. The weight vector can be broken into two perpendicular components. a. What is the magnitude of the component parallel to the ramp surface? b. What is the magnitude of the compo- nent at a right angle to the ramp surface? c. Which of those components is the normal force? 105 N Fx Fy Fg 858-889 EM APP B-845813 6/8/04 11:10 PM Page 867 14. In problem 13, which force offsets the kinetic friction force? If Fx exceeds Ff, will the suitcase accelerate down the ramp? 15. Dalila decides to determine the coefficient of maximum static friction for wood against wood by conducting an experiment. First she places a block of wood on a small plank; next, she slowly lifts one end of the plank upward. She notes the angle at which the block just begins to slide, and claims tan. She is correct. Show why. that s 16. Jonathan, with a mass of 81 kg, starts at rest from the top of a water slide angled at 42\u00b0 from the ground. He exits at the bottom 3.0 s later going 15 m/s. What is k? 17. A 64-N box is pulled by a rope at a constant speed across a rough horizontal surface. If the coefficient of kinetic friction is 0.81, what is the magnitude of the applied force if that force is directed parallel to the floor? 18. Suppose that in problem 17 the applied force is directed by the rope at an angle,, to the floor. a. The normal force is no longer simply Fg. Show how to compute net vertical force. b. The frictional force still opposes the horizontal motion of the box. Show how to compute net horizontal force. Chapter 6 1. A football player kicks a field goal from a distance of 45 m from the goal", "post. The football is launched at a 35\u00b0 angle above the horizontal. What initial velocity is required so that the football just clears the goalpost crossbar that is 3.1 m above the ground? Ignore air resistance and the dimensions of the football. 2. In a certain cathode-ray tube, a beam of electrons, moving at a constant velocity, enters a region of constant electric force midway between two parallel plates that are 10.0 cm long and 1.0 cm apart. In this region, the electrons experience an acceleration, a, toward the upper plate. If the electrons enter this region at a velocity of 3.0106 m/s, what is the acceleration that needs to be applied so that the electrons just miss the upper deflection plate? y d d Electron beam x Deflector plates 3. A skateboard track has a horizontal segment followed by a ramp that declines at a 45\u00b0 angle, as shown. a. How long would the ramp need to be to provide a landing for a skateboarder who launches from the horizontal segment at a velocity of 5.0 m/s? b. If the initial velocity is doubled, what happens to the required length of the ramp 45\u00b0 4. In an attempt to make a 3-point shot from a distance of 6.00 m, a basketball player lofts the ball at an angle of 68\u00b0 above the horizontal. The ball has an initial velocity of 10.0 m/s and an initial distance from the floor of 1.50 m. a. What is the maximum height reached by the ball? b. If the basket rim height is 3.05 m, how far above the rim is the ball? Appendix B 867 858-889 EM APP B-845813 6/8/04 11:25 PM Page 868. How far does a baseball that is thrown horizontally at 42.5 m/s drop over a horizontal distance of 18.4 m? 6. A projectile is launched from zero height with an initial angle,, above the horizontal and with an initial velocity, vi. a. Show that the range of the projectile\u2014the distance from the launch point at which the height is again zero\u2014is given by: 2 v sin 2 R i g b. What launch angle,, results in the largest range? 7. If a ring were constructed as part of a space station, how fast must a 50.0-m-radius ring rotate to simulate Earth\u2019s gravity? 8. A turntable for vinyl records works by constraining the needle to track inside a groove in a very close approximation of uniform circular motion. If the turntable rotational speed is 33 1 rpm, what is the 3 needle\u2019s centripetal acceleration when it is 14.6 cm from the center? 9. A park ride is designed so that the rider is on the edge of a revolving platform, 3.5 m from the platform\u2019s center. This platform is mounted 8.0 m from the center of a larger revolving platform. The smaller platform makes one revolution in 6.0 s and completes 2.0 rev for each revolution of the larger platform. All rotations are counterclockwise. At the instant shown, what are the magnitude and the direction of the rider\u2019s velocity with respect to the ground? 10. Two objects are placed on a flat turntable at 10.0 cm and 20.0 cm from the center, respectively. The coefficient of static friction with the turntable is 0.50. The turntable\u2019s rotational speed is gradually increased. a. Which of the two objects will begin to slide first? Why? b. At what rotational speed does the inner object begin to slide? 11. An airplane\u2019s airspeed is 2.0102 km/h due east. Because of a wind blowing to the north, it is approaching its destination 15\u00b0 north of east. a. What is the wind speed? b. How fast is the airplane approaching its destination? 12. A river is flowing 4.0 m/s to the east. A boater on the south shore plans to reach a dock on the north shore 30.0\u00b0 downriver by heading directly across the river. a. What should be the boat\u2019s speed relative to the water? b. What is the boat\u2019s speed relative to the dock? 30.0\u00b0 Q 8.0 m vRP P vPQ R 3.5 m Chapter 7 1. A year is defined as the time it takes for a planet to travel one full revolution around the Sun. One Earth year is 365 days. Using the data in Table 7-1, calculate the number of Earth days in one Neptune year. If Neptune takes about 16 h to complete one of its days, how many Neptunian days long is Neptune\u2019s year? 868 Appendix B 858-889 EM APP B-845813 3/31/", "04 4:24 AM Page 869 2. Suppose a new planet was discovered to orbit the Sun with a period 5 times that of Pluto. Using the data in Table 7-1, calculate the average distance from the Sun for this new planet. 3. If a meteorite hit the Earth and moved it 2.411010 m closer to Venus, how many days would there be in an Earth year? Use the data in Table 7-1. 4. The Moon is a satellite of the Earth. The Moon travels one full revolution around the Earth in 27.3 days. Given that the mass of the Earth is 5.971024 kg, what is the average distance from the Earth to the Moon? 5. A satellite travels 7.18107 m from the center of one of the planets in our solar system at a speed of 4.20104 m/s. Using the data in Table 7-1, identify the planet. 6. Earth\u2019s atmosphere is divided into four layers: the troposphere (0\u201310 km), the stratosphere (10\u201350 km), the mesosphere (50\u201380 km), and the thermosphere (80\u2013 500 km). What is the minimum velocity an object must have to enter the thermosphere? Chapter 8 1. The rotational velocity of a merry-go-round is increased at a constant rate from 1.5 rad/s to 3.5 rad/s in a time of 9.5 s. What is the rotational acceleration of the merrygo-round? 2. A record player\u2019s needle is 6.5 cm from the center of a 45-rpm record. What is the velocity of the needle? 3. Suppose a baseball rolls 3.2 m across the floor. If the ball\u2019s angular displacement is 82 rad, what is the circumference of the ball? 4. A painter uses a 25.8-cm long screwdriver to pry the lid off of a can of paint. If a force of 85 N is applied to move the screwdriver 60.0\u00b0 from the perpendicular, calculate the torque. 5. A force of 25 N is applied vertically at the end of a wrench handle that is 45 cm long to tighten a bolt in the clockwise direction. What torque is needed by the bolt to keep the wrench from turning. A 92-kg man uses a 3.05-m board to attempt to move a boulder, as shown in the diagram below. He pulls the end of the board and is able to move it to 45\u00b0 from the perpendicular. Calculate the torque applied. 45\u00b0 7. If a 25-kg child tries to apply the same torque as in the previous question, what would the length of the lever arm need to be? 8. Logan, whose mass is 18 kg, sits 1.20 m from the center of a seesaw. If Shiro must sit 0.80 m from the center to balance Logan, what is Shiro\u2019s mass? 9. Two forces\u2014a 55-N force in the clockwise direction and a 35-N force in the counterclockwise direction\u2014are applied to a merrygo-round with a diameter of 4.5 m. What is the net torque on the merry-go-round? 4. 5 m 35 N 55 N 10. A student sits on a stool holding a 5.0-kg dumbbell in each hand. He extends his arms such that each dumbbell is 0.60 m from the axis of rotation. The student\u2019s moment of inertia is 5.0 kgm2. What is the moment of inertia of the student and the dumbbells? Appendix B 869 858-889 EM APP B-845813 6/10/04 9:10 PM Page 870 B x i d n e p p A 11. A basketball player spins a basketball with a radius of 15 cm on his finger. The mass of the ball is 0.75 kg. What is the moment of inertia about the basketball? 12. A merry-go-round in the park has a radius of 2.6 m and a moment of inertia of 1773 kgm2. What is the mass of the merry-go-round? 13. The merry-go-round described in the previous problem is pushed with a constant force of 53 N. What is the angular acceleration? 14. What is the angular velocity of the merry- go-round described in problems 12 and 13 after 85 s, if it started from rest? 15. An ice-skater with a moment of inertia of 1.1 kgm2 begins to spin with her arms extended. After 25 s, she has an angular velocity of 15 rev/s. What is the net torque acting on the skater? 16. A board that is 1.5 m long is supported in two places. If the force exerted by the first support is 25 N and the forced exerted by the second is 62 N, what is the mass", " of the board? 17. A child begins to build a house of cards by laying an 8.5-cm-long playing card with a mass of 0.75 g across two other playing cards: support card A and support card B. If support card A is 2.0 cm from the end and exerts a force of 1.5103 N, how far from the end is support card B located? Let the axis of rotation be at the point support card A comes in contact with the top card. 18. If support card A in the previous problem was moved so that it now is 2.5 cm from the end, how far from the other end does support card B need to be to reestablish equilibrium? Chapter 9 2. If the ball in the previous problem interacts with the wall for a time interval of 0.22 s, what is the average force exerted on the wall? 3. A 42.0-kg skateboarder traveling at 1.50 m/s hits a wall and bounces off of it. If the magnitude of the impulse is 150.0 kgm/s, calculate the final velocity of the skateboarder. 4. A 50.0-g toy car traveling with a velocity of 3.00 m/s due north collides head-on with an 180.0-g fire truck traveling with a velocity of 0.50 m/s due south. The toys stick together after the collision. What are the magnitude and direction of their velocity after the collision? 5. A 0.040-kg bullet is fired into a 3.50-kg block of wood, which was initially at rest. The bullet remains embedded within the block of wood after the collision. The bullet and the block of wood move at a velocity of 7.40 m/s. What was the original velocity of the bullet? 6. Ball A, with a mass of 0.20 kg, strikes ball B, with a mass of 0.30 kg. The initial velocity of ball A is 0.95 m/s. Ball B is initially at rest. What are the final speed and direction of ball A and B after the collision if they stick together? 7. An ice-skater with a mass of 75.0 kg pushes off against a second skater with a mass of 42.0 kg. Both skaters are initially at rest. After the push, the larger skater moves off with a speed of 0.75 m/s eastward. What is the velocity (magnitude and direction) of the smaller skater after the push? 8. Suppose a 55.0-kg ice-skater, who was initially at rest, fires a 2.50-kg gun. The 0.045-kg bullet leaves the gun at a velocity of 565.0 m/s. What is the velocity of the ice-skater after she fires the gun? 1. A ball with an initial momentum of 9. A 1200-kg cannon is placed at rest on 6.00 kgm/s bounces off a wall and travels in the opposite direction with a momentum of 4.00 kgm/s. What is the magnitude of the impulse acting on the ball? an ice rink. A 95.0-kg cannonball is shot from the cannon. If the cannon recoils at a speed of 6.80 m/s, what is the speed of the cannonball? 870 Appendix B 858-889 EM APP B-845813 6/9/04 3:54 PM Page 871 10. An 82-kg receiver, moving 0.75 m/s north, is tackled by a 110.0-kg defensive lineman moving 0.15 m/s east. The football players hit the ground together. Calculate their final velocity (magnitude and direction). 11. A 985-kg car traveling south at 29.0 m/s hits a truck traveling 18.0 m/s west, as shown in the figure below. After the collision, the vehicles stick together and travel with a final momentum of 4.0104 kgm/s at an angle of 45\u00b0. What is the mass of the truck? 29.0 m/s 18.0 m/s 45\u00b0 p 4.0104 kgm/s 12. A 77.0-kg woman is walking 0.10 m/s east in the gym. A man throws a 15.0-kg ball south and accidentally hits the woman. The woman and the ball move together with a velocity of 0.085 m/s. Calculate the direction the woman and the ball move. Chapter 10 1. A toy truck is pushed across a table 0.80 m north, and pulled back across the table 0.80 m south. If a constant horizontal force of 15 N was applied in both directions, what is the net work? 2. A 15-kg child experiences an acceleration of 0.25 m/s2 as she is pulled 1.", "7 m horizontally across the floor by her sister. Calculate the change in the child\u2019s kinetic energy. 3. A man pushes a couch a distance of 0.75 m. If 113 J of work is done, what is the magnitude of the horizontal force applied? 4. Two blocks are tied together by a horizontal string and pulled a distance of 2.7 m across an air hockey table with a constant force of 35 N. The force is directed at an upward angle of 35\u00b0 from the 9.0-kg block, as shown in the figure. What is the change in kinetic energy in the two-block system? 35 N 35\u00b0 6.0 kg 9.0 kg 5. If the two-block system described in the previous problem was initially at rest, what is the final velocity? 6. A toy car with a mass of 0.75 kg is pulled 3.2 m across the floor with a constant force of 110 N. If 67 J of work is done, what is the upward angle of the force? 7. If a 75-W lightbulb is left on for 2.0 h, how much work is done? 8. A 6.50-horsepower (hp) self-propelled lawn mower is able to go from 0.00 m/s to 0.56 m/s in 0.050 s. If the mass of the lawn mower is 48.0 kg, what distance does the lawn mower travel in this time? (Use 1 hp 746 W.) 9. A winch that\u2019s powered by a 156-W motor lifts a crate 9.8 m in 11 s. What is the mass of the crate? 10. A man exerts a force of 310 N on a lever to raise a crate with a mass of 910 kg. If the efficiency of the lever is 78 percent, what is the lever\u2019s IMA? 11. A worker uses a pulley to lift a 45-kg object. If the mechanical advantage of the pulley is 5.2, what is the effort force exerted by the worker? 12. When the chain on a bicycle is pulled 0.95 cm, the rear wheel rim moves a distance of 14 cm. If the gear has a radius of 3.5 cm, what is the radius of the rear wheel? Appendix B 871 858-889 EM APP B-845813 6/9/04 3:55 PM Page 872 B x i d n e p p A Chapter 11 1. A crate with a mass of 210 kg is horizontally accelerated by a force of 95 N. The force is directed at an upward angle so that the vertical part of the force is 47.5 N and the horizontal part of the force is 82.3 N (see the diagram below). If the crate is pulled 5.5 m across the floor, what is the change in kinetic energy of the crate? 47.5 N 95 N 82.3 N 210 kg Smooth floor 2. Assuming that the crate described in problem 1 was initially at rest, what is the final velocity of the crate? 3. If the crate described in problem 1 experienced a frictional force of 15 N, what is the final kinetic energy of the crate? 4. A 150-kg roller-coaster car climbs to the top of a 91-m-high hill. How much work is done against gravity as the car is lifted to the top of the hill? 5. A pendulum bob with a mass of 0.50 kg swings to a maximum height of 1.0 m. What is the kinetic energy when the pendulum bob is at a height of 0.40 m? 6. A sled and its rider, with a total mass of 95 kg, are perched on top of a 25-m-tall hill. A second hill is 12 m tall (see the diagram below). The rider is given an initial push providing 3674 J of kinetic energy. Neglecting friction, what is the velocity at the top of the top of the 12-m-tall hill? 25 m 12 m 7. A 35-kg child is riding on a swing that rises to a maximum height of 0.80 m. Neglecting friction, what is the child\u2019s gravitational potential energy at the top of the swing? What is the child\u2019s kinetic energy at the top of the swing? 8. A sled and its rider are perched at the top of a hill that is 35 m tall. If the gravitational potential energy is 3.0104 J, what is the weight of the sled and rider? 9. The big hill on a roller-coaster ride is 91 m tall. If the mass of the roller-coaster car and its two riders is 314 kg and the maximum velocity reached by this roller-coaster ride is 28 m/s, how much energy was lost to friction? 10. A dartboard with a mass of 2.20 kg", " is suspended from the ceiling such that it is initially at rest. A 0.030-kg dart is thrown at the dartboard with a velocity of 1.3 m/s. After the dart hits the dartboard, the dart and the board initially move with a velocity of 0.025 m/s. How much kinetic energy is lost in the system? 11. In a physics laboratory, students crash carts together on a frictionless track. According to the following data, was kinetic energy conserved? Mass (kg) Cart A Cart B 0.25 0.36 vi (m/s) 0.18 0.20 vf (m/s) 0.21 0.11 12. A 0.150-kg ball that is thrown at a velocity of 30.0 m/s hits a wall and bounces back in the opposite direction with a speed of 25.0 m/s. How much work was done by the ball? Chapter 12 1. Convert the following Celsius tempera- tures to Kelvin temperatures. a. 196\u00b0C b. 32\u00b0C c. 212\u00b0C d. 273\u00b0C e. 273\u00b0C f. 27\u00b0C 872 Appendix B 858-889 EM APP B-845813 6/9/04 3:56 PM Page 873 2. Find the Celsius and the Kelvin tempera- tures for the following objects. a. average body temperature b. hot coffee c. iced tea d. boiling water 3. Convert the following Kelvin temperatures 8. A 220-g iron horseshoe is heated to 825\u00b0C and then plunged into a bucket filled with 20.0 kg of water at 25.0\u00b0C. Assuming that no energy is transferred to the surroundings, what is the final temperature of the water at equilibrium? to Celsius temperatures. a. 4 K b. 25 K c. 272 K d. 373 K e. 298 K f. 316 K 4. A 9.8-g lead bullet with a muzzle velocity of 3.90102 m/s is stopped by a wooden block. What is the change in temperature of the bullet if one-fourth of its original kinetic energy goes into heating the bullet? v 3.90102 m/s Bullet m 9.8 g Wooden block 5. What is the change in temperature of 2.2 kg of the following substances if 8.5103 J of thermal energy is added to each of the substances? a. ice b. water c. steam d. aluminum e. silver f. copper 6. A 2350-kg granite tombstone absorbs 2.8107 J of energy from the Sun to change its temperature from 5.0\u00b0C at night to 20.0\u00b0C during an autumn day. Determine the specific heat of granite from this information. 7. A 2.00103-g sample of water at 100.0\u00b0C is mixed with a 4.00103-g sample of water at 0.0\u00b0C in a calorimeter. What is the equilibrium temperature of the mixture Water T 25.0\u00b0C m 20.0 kg Horseshoe T 825\u00b0C 9. A cook adds 1.20 kg of soup bones to 12.5 kg of hot broth for flavoring. The temperatures of the bones and the broth are 20.0\u00b0C and 90.0\u00b0C, respectively. Assume that the specific heat of the broth is the same as that of water. Also assume that no heat is lost to the environment. If the equilibrium temperature is 87.2\u00b0C, what is the specific heat of the bones? Bones m 1.20 kg t 20.0C Broth T 90.0C 10. A 150.0-g copper cylinder at 425\u00b0C is placed on a large block of ice at 0.00\u00b0C. Assume that no energy is transferred to the surroundings. What is the mass of the ice that will melt? 11. How much energy is needed to melt one troy ounce, 31.1 g, of gold at its melting point? Appendix B 873 858-889 EM APP B-845813 6/9/04 3:57 PM Page 874 4. As shown in the figure below, a bubble of gas with a volume of 1.20 cm3 is released under water. As it rises to the surface, the temperature of the bubble increases from 27\u00b0C to 54\u00b0C, and the pressure is cut to one-third of its initial value. What is the volume of the bubble at the surface? 5. A sample of ethane gas (molar mass 30.1 g/mol) occupies 1.2102 m3 at 46\u00b0C and 2.4105 Pa. How many moles of ethane are present in the sample? What is the mass of the sample? 6. The constant R equals 8.31 Pam3/molK as it is presented in the ideal gas law in this chapter. One mole", " of an ideal gas occupies 22.4 L at standard temperature and pressure, STP, which is defined as 0.00\u00b0C and 1.00 atm. Given this information, deduce the value of R in Latm/molK. 7. Suppose that two linked pistons are both cylindrical in shape. Show that the ratio of forces generated is directly proportional to the square of the radii of the two crosssectional circular areas. 8. The cross-sectional areas of the pistons in the system shown below have a ratio of 25 to 1. If the maximum force that can be applied to the small piston is 12 N, what is the maximum weight that can be lifted 12. Trying to make an interesting pattern, an artist slowly pours 1.00 cm3 of liquid gold onto a large block of ice at 0.00\u00b0C. While being poured, the liquid gold is at its melting point of 1064\u00b0C. The density of gold is 19.3 g/cm3, and its specific heat is 128 J/kgK. What mass of ice will melt after all the gold cools to 0.00\u00b0C? 13. A cylinder containing 1.00 g of water at the boiling point is heated until all the water turns into steam. The expanding steam pushes a piston 0.365 m. There is a 215-N frictional force acting against the piston. What is the change in the thermal energy of the water? 14. What is the change in temperature of water after falling over a 50.0-m-tall waterfall. Assume that the water is at rest just before falling off and just after it reaches the bottom of the falls. 15. Two 6.35-kg lead bricks are rubbed against each other until their temperature rises by 1.50 K. How much work was done on the bricks? Chapter 13 1. Use Table 13.1 to estimate the pressure in atmospheres (atm) on a climber standing atop Mt. Everest. Is this more or less than half standard atmospheric pressure (1.0 atm)? 2. A woman wearing high heels briefly supports all her weight on the heels. If her mass is 45 kg and each heel has an area of 1.2 cm2, what pressure does she exert on the floor? 3. A typical brick has dimensions of 20.0 cm 10.0 cm 5.0 cm and weighs 20.0 N. How does the pressure exerted by a typical brick when it is resting on its smallest side compare to the pressure exerted when it is resting on its largest side? 874 Appendix B 858-889 EM APP B-845813 6/9/04 3:58 PM Page 875 9. A car of mass 1.35103 kg sits on a large piston that has a surface area of 1.23 m2. The large piston is linked to a small piston of area 144 cm2. What is the weight of the car? What force must a mechanic exert on the small piston to raise the car? 10. At what depth in freshwater does the water exert a pressure of 1.00 atm (1 atm 1.013105 Pa) on a scuba diver? 11. An iceberg floats in seawater, partly under water and partly exposed. Show that the value of Vsubmerged/Vtotal equals seawater. What percentage of the iceberg is exposed? Use 1.03103 kg/m3 for the density of seawater, and 0.92103 kg/m3 for the density of ice. ice/ 12. A concrete block ( 36106 \u00b0C1) of volume 0.035 m3 at 30.0\u00b0C is cooled to 10.0\u00b0C. What is the change in volume? 13. A glass mirror used in a mountaintop telescope is subject to temperatures ranging from 15\u00b0C to 45\u00b0C. At the lowest temperature, it has a diameter of 5.1 m. If the coefficient of linear expansion for this glass is 3.0106 \u00b0C1, what is the maximum change in diameter the mirror undergoes due to thermal expansion? 14. As shown in the figure below, a bimetallic strip is made from a piece of copper ( 16106 \u00b0C1) and a piece of steel ( 8106 \u00b0C1). The two pieces are equal in length at room temperature. a. As the strip is heated from room tem- perature, what is the ratio of the change in the length of the copper to that of the steel? b. How will the strip bend when it is heated above room temperature? When it is cooled below room temperature? Copper 15. A fishing line\u2019s lead sinker has a volume of 1.40105 m3. The density of lead is 1.2104 kg/m3. What is the apparent weight of the sinker when immersed in freshwater? Seawater is slightly denser than freshwater. Is the sinker\u2019s apparent weight bigger", " or smaller in seawater? Explain. Chapter 14 1. What is the mass of the watermelon shown below, if the spring constant is 128 N/m? 0.48 m 2. How many centimeters will a spring stretch when a 2.6-kg block is hung vertically from a spring with a spring constant of 89 N/m? 3. How much elastic potential energy does a spring with a constant of 54 N/m have when it is stretched 18 cm? 4. What is the period of the pendulum of the clock below XII I XI X IX VIII II III IV VII VI V 62 cm 0.52 kg Wooden handle Steel 5. A clock pendulum has a period of 0.95 s. How much longer will it have to be to have a period of 1.0 s? Appendix B 875 858-889 EM APP B-845813 3/31/04 4:38 AM Page 876 6. Boat A is traveling at 4.6 m/s. Boat B is moving away from boat A at 9.2 m/s, as shown in the figure below. The captain of boat B blows an air horn with a frequency of 550 Hz. What frequency does boat A hear? Use 343 m/s for the speed of sound. vd vs Boat A Boat B 7. A submarine is traveling toward a stationary detector. If the submarine emits a 260-Hz sound that is received by the detector as a 262-Hz sound, how fast is the submarine traveling? Use 1533 m/s for the speed of sound in seawater. 8. The end of a pipe is inserted into water. A tuning fork is held over the pipe. If the pipe resonates at lengths of 15 cm and 35 cm, what is the frequency of the tuning fork? Use 343 m/s for the speed of sound. 9. A 350-Hz tuning fork is held over the end of a pipe that is inserted into water. What is the spacing between the resonance lengths of the pipe if the speed of sound is 348 m/s? Chapter 16 1. What is the distance, r, between the lightbulb and the table in the figure below? P 2152 lm r E 180 lx. What is the length of a pendulum with a period of 89.4 ms? 7. Al is chopping wood across a clearing from Su. Su sees the axe come down and hears the sound of the impact 1.5 s later. How wide is the clearing? 8. When an orchestra is tuning up, the first violinist plays a note at 256 Hz. a. What is the wavelength of that sound wave if the speed of sound in the concert hall is 340 m/s? b. What is the period of the wave? 9. Geo is standing on a breakwater and he notices that one wave goes by every 4.2 s. The distance between crests is 12.3 m. a. What is the frequency of the wave? b. What is the speed of the wave? Chapter 15 1. Sound waves are being used to determine the depth of a freshwater lake, as shown in the figure below. If the water is 25\u00b0C and it takes 1.2 s for the echo to return to the sensor, how deep is the lake? Detector Sound source 2. Find the wavelength of an 8300-Hz wave in copper. 3. Janet is standing 58.2 m from Ninovan. If Ninovan shouts, how long will it take Janet to hear her? Use 343 m/s for the speed of sound. 4. The engine on a motorcycle hums at 85 Hz. If the motorcycle travels toward a resting observer at a velocity of 29.6 m/s, what frequency does the observer hear? Use 343 m/s for the speed of sound. 5. If the speed of a wave on a 78-cm-long guitar string is known to be 370 m/s, what is the fundamental frequency? 876 Appendix B 858-889 EM APP B-845813 3/31/04 4:39 AM Page 877. What is the luminous flux of a flashlight that provides an illuminance of 145 lx to the surface of water when held 0.50 m above the water? 3. An overhead light fixture holds three lightbulbs. Each lightbulb has a luminous flux of 1892 lm. The light fixture is 1.8 m above the floor. What is the illuminance on the floor? 4. What is the wavelength in air of light that has a frequency of 4.61014 Hz? 5. A helium atom is in a galaxy traveling at a speed of 4.89106 m/s away from the Earth. An astronomer on Earth observes a frequency from the helium atom of 6.521014 Hz. What frequency of light is emitted from the helium atom? 6. An astronomer observes that a molecule in a galaxy traveling toward Earth emits light with a wavelength of", " 514 nm. The astronomer identifies the molecule as one that actually emits light with a wavelength of 525 nm. At what velocity is the galaxy moving? Chapter 17 1. A light ray is reflected off a plane mirror at an angle of 25\u00b0 from the normal. A light ray from another source is reflected 54\u00b0 from the normal. What is the difference in the angles of incidence for the two light sources? 2. A light ray is reflected off of a plane mirror, as shown. What is the angle of reflection? 13\u00b0 3. The angle between an incident and a reflected ray is 70.0\u00b0. What is the angle of reflection? 4. An image is produced by a concave mirror, as shown. What is the object\u2019s position? 25 cm 15 cm C Image 5. What is the magnification of the object in problem 4? 6. If the object in problem 4 is 3.5 cm tall, how tall is the image? 7. A 6.2-m-tall object is 2.3 m from a convex mirror with a 0.8-m focal length. What is the magnification? 8. How tall is the image in problem 7? 9. A ball is 6.5 m from a convex mirror with a magnification of 0.75. If the image is 0.25 m in diameter, what is the diameter of the actual ball? Chapter 18 1. A piece of flint glass is lying on top of a container of water (see figure below). When a red beam of light in air is incident upon the flint glass at an angle of 28\u00b0, what is the angle of refraction in the flint glass? Air Flint glass Water 2. When the angle of refraction in the flint glass of problem 1 is 22\u00b0, what is the angle of refraction in the water? 3. When the beam of light in problem 1 enters the flint glass from the water, what is the maximum angle of incidence in the water such that light will transmit into the air above the flint glass? Hint: Use an angle of refraction of the light beam in air that is almost 90 \u00b0. 4. An object that is 24 cm from a convex lens produces a real image that is 13 cm from the lens. What is the focal length of the lens? 5. A 5.0-cm-tall object is placed 16 cm from a convex lens with a focal length of 8.4 cm. What are the image height and orientation? Appendix B 877 858-889 EM APP B-845813 6/9/04 3:59 PM Page 878 6. An object is 185 cm from a convex lens with a focal length of 25 cm. When an inverted image is 12 cm tall, how tall is the associated object? 7. What are the image height and orientation produced by the setup in the following figure Object 5.0 cm F 10.0 cm 15.0 cm 8. A convex lens can be used as a magnifying glass. When an object that is 15.0 cm from the lens has an image that is exactly 55 times the size of the object, what is the focal length of the lens? 9. A concave lens with a focal length of 220 cm produces a virtual image that is 36 cm tall. When the object is placed 128 cm from the lens, what is the magnification? Chapter 19 1. A physics student performs a double-slit diffraction experiment on an optical bench, as shown in the figure below. The light from a helium-neon laser has a wavelength of 632.8 nm. The laser light passes through two slits separated by 0.020 mm. What is the distance between the centers of the central band and the first-order band? Screen Laser Diffraction grating 50.0 cm 150.0 cm 878 Appendix B 2. A laser of unknown wavelength replaces the helium-neon laser in the experiment described in problem 1. To obtain the best diffraction pattern, the screen had to be moved to 104.0 cm. The distance between the centers of the central band and the first-order band is 1.42 cm. What wavelength of light is produced by the laser? 3. Light with a wavelength of 454.5 nm passes through two slits that are 95.2 cm from a screen. The distance between the centers of the central band and the first-order band is 15.2 mm. What is the slit separation? 4. What color of light would be reflected from the soapy water film shown in the figure below? Air Soapy water nSW 1.33 Air 76.1 nm 5. A 95.7-nm film of an unknown substance is able to prevent 555-nm light from being reflected when surrounded by air. What is the index of refraction of the substance? 6. An oil film (n 1.45) on the surface of a puddle of water on the street is 118 nm", " thick. What frequency of light would be reflected? 7. Violet light ( 415 nm) falls on a slit that is 0.040 mm wide. The distance between the centers of the central bright band and the third-order dark band is 18.7 cm. What is the distance from the slit to the screen? 8. Red light ( 685 nm) falls on a slit that is 0.025 mm wide. The distance between the centers of the central bright band and the second-order dark band is 6.3 cm. What is the width of the central band? 9. The width of the central bright band of a diffraction pattern is 2.9 cm. Laser light of unknown wavelength passes through a single slit that is 0.042 mm wide and onto a screen that is 1.5 m from the slit. What is the wavelength of the light? 858-889 EM APP B-845813 6/9/04 3:59 PM Page 879 10. A diffraction grating has 13,400 lines per inch. What is the slit separation distance? (Use 1 in 2.54 cm) 11. Light from a helium-neon laser ( 632.8 nm) passes through the diffraction grating described in problem 10. What is the angle between the central bright line and the first-order bright line? 12. A diffraction grating with slits separated by 3.40106 m is illuminated by light with a wavelength of 589 nm. The separation between lines in the diffraction pattern is 0.25 m. What is the distance between the diffraction grating and the screen? Chapter 20 1. The magnitude of a charge, q, is to be determined by transferring the charge equally to two pith balls. Each of the pith balls has a mass of m, and is suspended by an insulating thread of length l. When the charge is transferred, the pith balls separate to form an equilibrium in which each thread forms an angle,, with the vertical. a. Draw a force diagram showing the forces that are acting on the rightmost pith ball. b. Derive an expression for q as a function of, m, and l. c. Use your derived expression to determine the value of q when 5.00\u00b0, m 2.00 g, and l 10.0 cm. As shown in the figure below, four charges, each with charge q, are distributed symmetrically around the origin, O(0, 0), at A(1.000, 0), B(0, 1.000), C(1.000, 0), and D(0, 1.000). Find the force on a fifth charge, qT, located at T(5.000, 0). y (m) B 0 C A D T x (m) 3. Consider that the four charges in the previous problem are now combined into a single charge, 4q, located at the origin. What is the force on charge qT? Chapter 21 1. What are the magnitude and the sign of a point charge that experiences a force of 0.48 N east when placed in an electric field of 1.6105 N/C west? 2. A test charge of 1.0106 C, located at the point T(0, 1) m, experiences a force of 0.19 N directed toward the origin, along the y-axis, due to two identical point charges located at A(1, 0) m and B(1, 0) m. a. What is the sign of the charges at A and B? b. What are the magnitude and direction of the electric field at T? y qT T(0, 1) qA A(1, 0) qB B(1, 0) x Appendix B 879 858-889 EM APP B-845813 6/9/04 4:00 PM Page 880. A test charge of 0.5107 C is placed in an electric field of 6.2104 N/C, directed 15\u00b0 north of east. What is the force experienced by the charge? 4. By what percent must the distance from a point charge increase in order to have a reduction in the electric field strength by 40 percent? 5. A particle of mass m 2.0106 kg is in a circular orbit about a point charge. The charge, q, is 3.0105 C and is at a distance of r 20.0 cm. a. What is the electric field strength at all points on the orbit around the point charge? b. Considering only electrostatic forces, what charge, q, on the particle is required to sustain an orbital period of 3.0103 s? FE 20.0 cm q 6. Two charges of equal magnitude and opposite sign are placed 0.50 m apart. The electric field strength midway between them is 4.8104 N/C toward the negative charge. What is the magnitude of each", " charge? q E q r 2 r 2 7. A pith ball weighing 3.0102 N carries a charge of 1.0106 C. The pith ball is placed between two large, parallel, metal plates, that are separated by 0.050 m. What is the potential difference, V, that must be applied in order to suspend the pith ball between the plates? 8. The electric field strength, defined as a force per unit charge, has the units newtons per coulomb, N/C. The formula for electric potential difference, V Ed, however, suggests that E also can be expressed as volts per meter, V/m. a. By analysis of the units, show that these two expressions for the units of electric field strength are equivalent. b. Suggest a reason why V/m is often a more useful way to express electric field strength. 9. Suppose you have two parallel, metal plates that have an electric field between them of strength 3.0104 N/C, and are 0.050 m apart. Consider a point, P, located 0.030 m from plate A, the negatively charged plate, when answering the following questions. a. What is the electric potential at P relative to plate A? b. What is the electric potential at P relative to plate B, the positively charged plate? 10. A certain 1.5-V size-AA battery has a storage capacity of 2500 C. How much work can this battery perform? 11. In a vacuum tube, electrons accelerate from the cathode element to the plate element, which is maintained at a positive potential with respect to the cathode. If the plate voltage is 240 V, how much kinetic energy has an electron acquired when it reaches the plate? 12. An oil drop with five excess electrons is suspended in an electric field of 2.0103 N/C. What is the mass of the oil drop? 13. An oil drop weighing 7.51015 N carries three excess electrons. a. What potential difference is required to suspend the drop between parallel plates separated by 2.3 cm? b. If the oil drop picks up another elec- tron, by how much must the potential difference between the plates be reduced to maintain the oil drop in suspension? 880 Appendix B 858-889 EM APP B-845813 6/9/04 4:01 PM Page 881 14. A charge of 2.00106 C is moved against a constant electric field. If 4.50104 J of work is done on the charge, what is the potential difference between the initial and the final locations of the charge? 15. Capacitors C1 220 F and C2 470 F are connected across a 48.0-V electric potential difference. a. What are the charges, q1 and q2, on each of the capacitors? b. What is the total charge, qT, on both capacitors? c. Repeat steps a and b for a new potential difference, v 96.0 V. d. Now consider the two capacitors as a system. What would be a single equivalent capacitor, Ceq, that could replace C1 and C2, and be capable of yielding the same results? e. Based on your responses to the above, make a conjecture concerning the equivalent capacitance of a system of capacitors\u2014all of which are connected across the same potential difference. 16. A new 90.0-V battery with a storage capacity of 2.5104 C charges a 6800-F capacitor with the switch in position A. Then the switch is thrown to position B to discharge the capacitor. a. How many times can this cycle be repeated before the battery is completely discharged? b. If the capacitor discharge occurs in 120 ms, what is the average power dissipated in the discharge circuit during one cycle? B A Discharge circuit 90.0 17. A 0.68-F capacitor carries a charge on one plate of 1.36105 C. What is the potential difference across the leads of this capacitor Chapter 22 1. A decorative lightbulb rated at 7.50 W draws 60.0 mA when lit. What is the voltage drop across the bulb? 2. A 1.2-V nickel-cadmium battery has a rated storage capacity of 4.0103 mAh (milliamp-hours). a. What is the battery charge capacity in coulombs? Hint: 1 C 1 As. b. How long can this battery supply a current of 125 mA? 3. A heating element of an electric furnace consumes 5.0103 W when connected across a 240-V source. What current flows through the element? 4. The cold filament resistance of a lightbulb is 20.0. The bulb consumes 75 W when it is operating from a 120-V source. By what factor does the start-up current in the bulb exceed the operating current? 5. In the circuit shown below, a potentiometer", " is used to vary the current to the lamp. If the only resistance is due to the lamp, what is the current in the circuit? I 12 V Rlamp 20.0 6. An electric toaster consumes 1875 W in operation. If it is plugged into a 125-V source, what is its resistance? 7. Draw a circuit diagram to include a 90.0-V battery, a 220- resistor, and a 680resistor in series. Show the direction of conventional current. 8. Modify the diagram of the previous problem to include an ammeter and a voltmeter to measure the voltage drop across the 680- resistor. 9. If the total resistance in the circuit in problem 8 is 9.0102, what would the ammeter and the voltmeter indicate? Appendix B 881 858-889 EM APP B-845813 6/9/04 4:02 PM Page 882 B x i d n e p p A 10. An electric motor with a load delivers 5.2 hp to its shaft (1 hp 746 W). Under these conditions, it operates at 82.8 percent efficiency. (Efficiency is defined as the ratio of power output to power input.) a. How much current does the motor draw from a 240-V source? b. What happens to the remaining 17.2 percent of the input power? 11. An industrial heating process uses a current of 380 A supplied at 440-V potential. a. What is the effective resistance of the heating element? b. How much energy is used by this process during an 8-h shift? 12. An 8.0- electric heater operates from a 120-V source. a. How much current does the heater require? b. How much time does the heater need to generate 2.0104 J of thermal energy? 13. A manufacturer of lightbulbs advertises that its 55-W bulb, which produces 800.0 lm of light output, provides almost the same light as a 60.0-W bulb with an energy savings. The 60.0-W bulb produces 840.0 lm. a. Which bulb most efficiently converts electric energy to light? b. Assume a bulb has a lifetime of 1.0103 h and an electric energy cost of $0.12/kWh. How much less does the 55-W bulb cost to operate over its lifetime? c. Is most of the cost savings due to higher bulb efficiency, or due to the consumer\u2019s willingness to accept lower light output? 14. By what factor would the I2R loss in transmission wires be reduced if the transmission voltage were boosted from 220 V to 22 kV? Assume that the rate of energy delivered is unchanged. 15. The electric-utility invoice for a household shows a usage of 1245 kWh during a certain 30-day period. What is the average power consumption during this period? Chapter 23 1. A series circuit contains a 47- resistor, an 82- resistor, and a 90.0-V battery. What resistance, R3, must be added in series to reduce the current to 350 mA? 2. What is the minimum number of 100.0resistors that must be connected in series together with a 12.0-V battery to ensure that the current does not exceed 10.0 mA? 3. A 120.0-V generator is connected in series with a 100.0- resistor, a 400.0- resistor, and a 700.0- resistor. a. How much current is flowing in the circuit? b. What is the voltage drop in the 400- resistor? 4. Show that the total power dissipated in a circuit of series-connected resistors is P I2R, where R is the equivalent resistance. 5. In an experiment, three identical lightbulbs are connected in series across a 120.0-V source, as shown. When the switch is closed, all of the bulbs are illuminated. However, when the experiment is repeated the next day, bulbs A and B are illuminated brighter than normal, and bulb C is dark. A voltmeter is used to measure the voltages in the circuit as shown. The voltmeter readings are: V1 V2 V3 120.0 V 60.0 V 0.0 V a. What has happened in this circuit? b. Explain why bulbs A and B are brighter than normal. c. Is there more or less current flowing now in the circuit? (Bulb A) RA (Bulb B) RB (Bulb C) RC 120.0 V V1 V2 V3 882 Appendix B 858-889 EM APP B-845813 6/9/04 4:03 PM Page 883. A string of holiday lights has 25 identical bulbs connected in series. Each bulb dissipates 1.00 W when the string is connected to a 125-V outlet. a. How much power must be supplied by the 125", "-V source? b. What is the equivalent resistance of this circuit? c. What is the resistance of each bulb? d. What is the voltage drop across each bulb? 7. Two resistors are connected in series across a 12.0-V battery. The voltage drop across one of the resistors is 5.5 V. a. What is the voltage drop across the other resistor? b. If the current in the circuit is 5.0 mA, what are the two resistor values? 8. In problem 7, the specifications for the resistors state that their resistance may vary from the listed nominal value. If the possible ranges of actual resistance values are as follows, 1050 R1 1240 R2 1370, what is the possible minimum and maximum value of the nominal 5.5-V voltage drop? 1160, and 9. A voltage-divider circuit is constructed with a potentiometer, as shown below. There are two fixed resistances of 113 k and 294 k. The resistance of the potentiometer can range from 0.0 to 100.0 k. a. What is Vout when the potentiometer is at its minimum setting of 0.0? b. What is Vout when the potentiometer is at its maximum setting of 100.0 k? c. What potentiometer setting is required to adjust Vout to exactly 65.0 V? 113 k 100.0 V 100.0 k 294 k Vout 10. Two lightbulbs, one rated at 25.0 W and one rated at 75.0 W, are connected in parallel across a 125-V source. a. Which bulb is the brightest? b. What is the operating resistance of each of the bulbs? c. The bulbs are rewired in series. What is the dissipated power in each bulb? Which bulb is the brightest? 11. A 10.0-V battery has an internal resistance of 0.10. The internal resistance can be modeled as a series resistor, as shown. a. Derive an expression for the battery terminal voltage, V, as a function of the current, I. b. Create a graph of voltage versus current for a current range of 0.0 to 1.0 A. c. What value of load resistor RL needs to be placed across the battery terminals to give a current of 1.0 A? d. How does the V-I function differ from that of an ideal voltage source? RI 0.10 10.0 V V RL Battery Load 12. Show that the total power dissipated in a circuit of parallel-connected resistors is 2 P V, where R is the equivalent resistance. R 13. A holiday light string of ten bulbs is equipped with shunts that short out the bulbs when the voltage drop increases to line voltage, which happens when a bulb burns out. Each bulb has a resistance of 200.0. The string is connected to a household circuit at 120.0 V. If the string is protected by a 250.0-mA fuse, how many bulbs can fail without blowing the fuse? Appendix B 883 858-889 EM APP B-845813 6/9/04 4:04 PM Page 884 14. A 60.0-W lightbulb and a 75.0-W lightbulb are connected in series with a 120-V source. a. What is the equivalent resistance of the circuit? b. Suppose an 1875-W hair dryer is now plugged into the parallel circuit with the lightbulbs. What is the new equivalent resistance of the circuit? 15. What is the equivalent resistance of the following resistor network 29.4 k 25.5 k 6.34 k 7.87 k 47.5 k 3. Jingdan turned a screwdriver into a magnet by rubbing it with a strong bar magnet to pick up a screw that has fallen into an inaccessible spot. How could he demagnetize his screwdriver after picking up the screw? 4. A long, straight, current-carrying wire carries a current from west to east. A compass is held above the wire. a. Which direction does the north pole of the compass point? b. When the compass is moved under- neath the wire, in which direction does the north pole point? 5. Consider the sketch of the electromagnet in the figure below. Which components could you change to increase or decrease the strength of the electromagnet? Explain your answer. Chapter 24 1. The magnetic field of Earth resembles the field of a bar magnet. The north pole of a compass needle, used for navigation, generally points toward the geographic north pole. Which magnetic pole of the Earth is the compass needle pointing toward? 2. A magnet is used to collect some spilled paper clips. What is the magnetic pole at the end of the paper clip that is indicated in the figure? N S 884 Appendix B 6. Sketch a graph showing the relationship between the magnetic field around a straight, current-carrying", " wire, and the distance from the wire. 7. How long is a wire in a 0.86-T field that carries a current of 1.4 A and experiences a force of 13 N? 8. A 6.0-T magnetic field barely prevents a 0.32-m length of copper wire with a current of 1.8 A from dropping to the ground. What is the mass of the wire? 9. How much current will be needed to produce a force of 1.1 N on a 21-cm-long piece of wire at right angles to a 0.56-T field? 10. Alpha particles (particles containing two protons, two neutrons, but no electrons) are traveling at right angles to a 47-T field with a speed of 36 cm/s. What is the force on each particle? 858-889 EM APP B-845813 6/9/04 4:05 PM Page 885 11. A force of 7.11012 N is exerted on some Al3 ions (an atom missing three electrons) that are traveling at 430 km/s perpendicular to a magnetic field. What is the magnetic field? 12. Electrons traveling at right angles to a magnetic field experience a force of 8.31013 N when they are in a magnetic field of 6.2101 T. How fast are the electrons moving? Chapter 25 1. A 72-cm wire is moved at an angle of 72\u00b0 through a magnetic field of 1.7102 T and experiences an EMF of 1.2 mV. How fast is the wire moving? 2. A 14.2-m wire moves 3.12 m/s perpendi- cular to a 4.21-T field. a. What EMF is induced in the wire? b. Assume that the resistance in the wire is 0.89. What is the amount of current in the wire? 3. A 3.1-m length of straight wire has a resistance of 3.1. The wire moves at 26 cm/s at an angle of 29\u00b0 through a magnetic field of 4.1 T. What is the induced current in the wire? 4. A generator delivers an effective current of 75.2 A to a wire that has a resistance of 0.86. a. What is the effective voltage? b. What is the peak voltage of the generator? 5. What is the RMS voltage of a household outlet if the peak voltage is 165 V? 6. An outlet has a peak voltage of 170 V. a. What is the effective voltage? b. What effective current is delivered to an 11- toaster? 7. A step-up transformer has a primary coil consisting of 152 turns and a secondary coil with 3040 turns. The primary coil receives a peak voltage of 98 V. a. What is the effective voltage in the primary coil? 8. A step-down transformer has 9000 turns on the primary coil and 150 turns on the secondary coil. The EMF in the primary coil is 16 V. What is the voltage being applied to the secondary coil? 9. A transformer has 124 turns on the primary coil and 18,600 turns on the secondary coil. a. Is this a step-down or a step-up transformer? b. If the effective voltage in the secondary coil is 3.2 kV, what is the peak voltage being delivered to the primary coil? Chapter 26 1. A stream of singly ionized (1) fluorine atoms passes undeflected through a magnetic field of 2.5103 T that is balanced by an electric field of 3.5103 V/m. The mass of the fluorine atoms is 19 times that of a proton. a. What is the speed of the fluorine ions? b. If the electric field is switched off, what is the radius of the circular path followed by the ions? 2. An electron moves perpendicular to Earth\u2019s magnetic field with a speed of 1.78106 m/s. If the strength of Earth\u2019s magnetic field is about 5.00105 T, what is the radius of the electron\u2019s circular path? 3. A proton with a velocity of 3.98104 m/s perpendicular to the direction of a magnetic field follows a circular path with a diameter of 4.12 cm. If the mass of a proton is 1.671027 kg, what is the strength of the magnetic field? 4. A beam of doubly ionized (2) calcium atoms is analyzed by a mass spectrometer. If B 4.5103 T, r 0.125 m, and the mass of the calcium ions is 6.681026 kg, what is the voltage of the mass spectrometer? 5. The speed of light in crown glass is 1.97108 m/s. What is the dielectric constant of crown glass? b. What is the effective voltage in the secondary coil? 6. The dielect", "ric constant of diamond is 6.00. What is the speed of light in diamond? Appendix B 885 858-889 EM APP B-845813 6/9/04 4:07 PM Page 886. By curving different isotopes through paths with different radii, a mass spectrometer can be used to purify a sample of mixed uranium-235 and uranium-238 isotopes. Assume that B 5.00103 T, V 55.0 V, and that each uranium isotope has a 5 ionization state. Uranium-235 has a mass that is 235 times that of a proton, while uranium-238 has a mass that is 238 times that of a proton. By what distance will the two isotopes be separated by the mass spectrometer? R1 Uranium-235 R2 Uranium-238 R2 R1 Uranium ion beam 8. A mass spectrometer often is used in carbon dating to determine the ratio of C-14 isotopes to C-12 isotopes in a biological sample. This ratio then is used to estimate how long ago the once-living organism died. Because a mass spectrometer is sensitive to the charge-to-mass ratio, it is possible for a contaminant particle to alter the value measured for the C-14/C-12 ratio, and thus, yield erroneous results. When ionized, C-14 forms an ion with a 4 charge, and the mass of C-14 is 14 times that of a proton. Consider a contaminant lithium particle. If the most common lithium isotope has a mass that is seven times that of a proton, what must be the charge of the lithium ion needed to contaminate a carbon-14 experiment? 9. What is the wavelength of a radio wave with a frequency of 90.7 MHz? 10. What is the frequency of a microwave with a wavelength of 3.27 mm? 11. What is the frequency of an X ray with a wavelength of 1.001010 m? 12. In recent years, physicists have slowed the speed of light passing through a material to about 1.20 mm/s. What is the dielectric constant of this material? Chapter 27 1. If the maximum kinetic energy of emitted photoelectrons is 1.41018 J, what is the stopping potential of a certain photocell? 2. The stopping potential of a photocell is 2.3 V. What is the initial velocity of an emitted photoelectron that is brought to a stop by the photocell? 3. If a photoelectron traveling at 8.7105 m/s is stopped by a photocell, what is the photocell\u2019s stopping potential? 4. Light with a frequency of 7.51014 Hz is able to eject electrons from the metal surface of a photocell that has a threshold frequency of 5.21014 Hz. What stopping potential is needed to stop the emitted photoelectrons? 5. A metal has a work function of 4.80 eV. Will ultraviolet radiation with a wavelength of 385 nm be able to eject a photoelectron from the metal? 6. When a metal is illuminated with radiation with a wavelength of 152 nm, photoelectrons are ejected with a velocity of 7.9105 m/s. What is the work function, in eV, of the metal? Cathode Anode v 7.9105 m/s 152 nm Incident radiation 886 Appendix B 858-889 EM APP B-845813 6/10/04 9:11 PM Page 887 7. The de Broglie wavelength for an electron traveling at 9.6105 m/s is 7.61010 m. What is the mass of the electron? 8. What is the de Broglie wavelength of a 68-kg man moving with a kinetic energy of 8.5 J? 9. An electron has a de Broglie wavelength of 5.21010 m. What potential difference is responsible for this wavelength? Chapter 28 1. An electron in a hydrogen atom makes a transition from E3 to E1. How much energy does the atom lose? 2. An electron in the hydrogen atom loses 3.02 eV as it falls to energy level E2. From which energy level did the atom fall? 3. Which energy level in a hydrogen atom has a radius of 7.63109 m? 4. When an electron falls from E4 to E1, what is the frequency of the emitted photon? 5. If an electron moves from E3 to E5, what is the wavelength of the photon absorbed by the atom? 6. If a hydrogen atom in its ground state absorbs a photon with a wavelength of 93 nm, it jumps to an excited state. What is the value of the energy in that excited state? Chapter 29 1. Indium has 3 free electrons per atom. Use Appendix D and determine the number of free electrons in 1.0 kg of indium. 2. Cadm", "ium has 2 free electrons per atom. Use Appendix D and determine the number of free electrons in 1.0 dm3 of Cd. 3. Copper has 1 free electron per atom. What length of 1.00-mm diameter copper wire contains 7.811024 free electrons? Use Appendix D for physical constants. 4. At 400.0 K, germanium has 1.131015 free electrons/cm3. How many free electrons per Ge atom are there at this temperature. At 400.0 K, silicon has 4.541012 free electrons/cm3. How many free electrons per Si atom are there at this temperature? 6. At 200.0 K, silicon has 3.791018 free electrons per atom. How many free electrons/cm3 are there in silicon at this temperature? 7. Silicon has 1.451010 free electrons/cm3 at room temperature. If you wanted to have 3106 as many electrons from arsenic doping as thermal free electrons from silicon at room temperature, how many arsenic atoms should there be for each silicon atom? Each arsenic atom provides 1 free electron. Use Appendix D for physical constants. 8. At 200.0 K, germanium has 1.161010 thermally liberated charge carriers/cm3. If it is doped with 1 As atom to 525,000 Ge atoms, what is the ratio of doped carriers to thermal carriers at this temperature? See Appendix D for physical constants. 9. At 200.0 K, silicon has 1.89105 thermally liberated charge carriers/cm3. If it is doped with 1 As atom to 3.75 million Si atoms, what is the ratio of doped carriers to thermal carriers at this temperature? See Appendix D for physical constants. 10. The diode shown below has a voltage drop, Vd, of 0.45 V when I 11 mA. If a 680- resistor, R, is connected in series, what power supply voltage, Vb is needed? R I Diode Vd Vb 11. A diode in a circuit similar to the one in Figure 29-26 has a voltage drop, Vd, of 0.95 V when I 18 mA. If a 390- resistor, R, is connected in series, what power supply voltage, Vb is needed? Appendix B 887 858-889 EM APP B-845813 6/24/04 11:36 AM Page 888 12. What power supply voltage would be 9. The graph below shows a sequence of B x i d n e p p A needed to produce a current of 27 mA in the circuit in problem 10? Assume the diode voltage is unchanged. Chapter 30 1. Carbon-14, or 14C, is an isotope of the 6C, and is used in dating common 12 ancient artifacts. What is the composition of its nucleus? 2. An isotope of iodine (Z 53) is used to treat thyroid conditions. Its mass number is 131. How many neutrons are in its nucleus? 3. The only nonradioactive isotope of fluorine has nine protons and ten neutrons. a. What is its mass number? b. The atomic mass unit, u, is equal to 1.661027 kg. What is fluorine-19\u2019s approximate mass in kilograms? c. Write the full symbol of this atom. 4. The magnesium isotope 25 12Mg has a mass of 24.985840 u. a. Calculate its mass defect. b. Calculate its binding energy in MeV. 5. The isotope 10 5B has a mass of 10.012939 u. a. Calculate the mass defect. b. Calculate its binding energy in MeV. c. Calculate its binding energy per nucleon. 6. The most stable isotope of all is 56 Its binding energy per nucleon is 8.75 MeV/nucleon. a. What is the binding energy of this 26Fe. isotope? b. What is the mass defect of this isotope? 94Pu can be transmuted to an 7. The isotope 239 isotope of uranium, 235 a. Write the nuclear equation for this 92U. transmutation. b. Identify the particle that is ejected. 8. The radioisotope 222 84Po undergoes alpha decay to form an isotope of lead (lead has atomic number 82). Determine what the mass number of that isotope must be by writing a nuclear equation. 888 Appendix B alpha and beta decays, labeled 1, 2, 3, and 4. Consult Table 30-1 as needed. a. Which represent alpha decays, and which represent beta decays? b. What is the overall change in mass in the sequence? In the number of neutrons? A 218 216 214 212 210 80 82 84 Z Atomic number 10. Use Appendix D to complete the two nuclear equations. Include correct", " subscripts and superscripts for each of the particles. a. 32 b. 92 15P \u2192? antineutrino 235U \u2192? 11. Write the complete nuclear equation for 16S. the beta decay of 34 12. A positron is identical to a beta particle, except that its charge is 1 instead of 1. 37 19K undergoes spontaneous positron decay to form an isotope of argon. Identify the isotope of argon by writing a nuclear equation. 13. Iodine-131 has a half-life of 8.0 days. If there are 60.0 mg of this isotope at time zero, how much remains 24 days later? 14. Refer to Table 30-2. Given a sample of cobalt-60, a. how long a time is needed for it to go through four half-lives? b. what fraction remains at the end of that time? 858-889 EM APP B-845813 3/31/04 4:51 AM Page 889 15. The graph shows the activity of a certain radioisotope over time. Deduce its half-life from this data. 17. One source of a star\u2019s energy is the fusion of two deuterons to form an alpha particle plus a gamma ray 48 36 24 12 0 3 6 9 12 Time (h) 16. The mass of a proton and of an antiproton is 1.00728 u. Recall that the conversion of exactly 1 u into energy yields 931.5 MeV. a. Calculate the mass used up when a proton and an antiproton annihilate one another. b. Calculate the energy released here. Particle Mass in u 2 1H 4 2He 0 0 2.0136 4.0026 0.0000 a. Write the nuclear equation for this fusion. b. Calculate the mass \u201clost\u201d in this process. c. Calculate the energy released in MeV. 18. When 1 mol (235 g) of uranium-235 undergoes fission, about 2.01010 kJ of energy are released. When 4.0 g of hydrogen undergoes fusion, about 2.0109 kJ are released. a. For each, calculate the energy yield per gram of fuel. b. Which process produces more energy per gram of fuel Appendix B 889 890-909 EM APP C-845813 6/15/04 10:39 PM Page 890 Chapter 1 0. 9 V 1. I V 0.18 0 4 3. t 10.0 s s2 0 / m 0.4 0 a M 1 H z z) ( H 0 00 5. 750 kHz (1 ) 0.75 MHz,0 0 1,00 00 k z H 1 Hz b. 9. a. s 0 6 h 24 in) ( m ) (60 ) 31,622,400 s 7. 366 days ( i m n 1 da y 1 h 1 1.6 km 1600 m 1.62 m 1.62 m 1200 m 12 m 1613.62 m 1600 m or 1.6 km rounded to two significant digits 6.201 cm cm 7.4 0.68 cm 12.0 cm 26.281 cm cm 26.3 11. a. 320 cm2 or 3.2102 cm2 b. 13.6 km2 Chapter 2 27. Because the bicycle is moving in the positive direction, the average speed and average velocity are the same. Using the points (0.00 min, 0.0 km) and (15.0 min, 10.0 km), d \uf8e6 v \uf8e6 t 1\uf8e6 \uf8e6 d t1 \uf8e6 0.67 km/min d 2 t 2 10.0 km 0.0 km 15.0 min 0.0 min \uf8e6 The bicycle is moving in the positive direction at a speed of 0.67 km/min. Chapter 3 1. Time interval Velocity Position Start 2 v2 3 v3 Stop v2 1 v1 v1 a 9. The car begins at a position of 125.0 m and moves toward the origin, arriving at the origin 5.0 s after it begins moving. The car continues beyond the origin. 3. a. 5.0 to 15.0 s b. 0.0 to 5.0 s c. 15.0 to 20. 11. a. 4.0 s 13. a. 19 s c. 100.0 b. 100.0 m b. 58 80.0 60.0 40.0 20.0 0.00 10.0 30.0 50.0 70.0 Time (s) 15. runner B 17. approximately 30 m d \uf8e6 25. a. v \uf8e6 t 1\uf8e6 d t1 d 2 t 2 1.0 m 0.0 m 3.0 s 0.0 s \uf8e6", " \uf8e6 \uf8e60.33 m/s\uf8e6 0.33 m/s \uf8e6 b. The average velocity is the slope of the line, including the sign, so it is 0.33 m/s or 0.33 m/s north. 890 Appendix C 5. 1..0 5.0 10.0 15.0 20.0 Time (s) 36 m/s 7.0 m/s2 15 m/s v 7. a 0 s 3. t v 9. a. a t 0.0 m/s 25 m/s 3.0 s 8.3 m/s2 b. Half as great (4.2 m/s2) 0.5 cm/y 1.0 cm/y 1.0 y v 11. a t vi m k 1 0 s) 0 )(36 30.0 km/h (3.5 m/s2)(6.8 s)( 0 00 1 m h 1 120 km/h 0.5 cm/y2 at 19. vf 21. vf at vi vi vf so t a m/s 2 2 3.0 /s m s2 / m.1 2 9.0 s 890-909 EM APP C-845813 6/15/04 10:40 PM Page 891 5 23. (7 1 k m 1 0 s)( 0 m)(36 ) 2.7 102 km/h 0 00 1 m h 1 s 31.0 1 2 6.0 12.0 Time (s Time (s) Part 1: Constant acceleration: 1 (3.0 m/s)(6.0 s) 2 9.0 m d1 Part 2: Constant velocity: d2 (3.0 m/s)(12.0 s 6.0 s) 18 m thus, d d1 d2 9.0 m 18 m 27 m 25. a 25 20 15 10 5 0 5 10 15 20 25 Distance the driver walked to the gas station: 33. Part 1: Constant velocity: d vt (1.5 m/s)(20.0 min)( 1800.8 km Time to walk back to the car: 00 m 1 8 t d 1500 s 25 min /s m.2 1 v b. 20,000 ) 15,000 10,000 5000 0 1000 2000 3000 4000 5000 Time (s) 27. 29. vi) (vf v v 2 2 vi)t d vt 2 (22 m/s 44 m/s)(11 s) 2 (vf 1.2102 m vi) (vf v v 2 2 vi)t d vt 2 ) 7.5 m/s 0.94 m/s 9 2( 1 m d vf 2 t 5. 4 s so vi (vf d vt (4.3 m/s)(19 min)( Part 2: Constant acceleration: s 0 6 n i m 1 ) 4902 m df di vit 1 at2 2 vit) d 2(df i a 2 t 2(5.0103 m 4902 m (4.3 m/s)(19.4 s)) (19.4 s)2 0.077 m/s2 43. a. Now the positive direction is downward. vf vf at, where a g 9.80 m/s2 vi 0.0 m/s (9.80 m/s2)(4.0 s) 39 m/s when the downward direction is positive b. d vit 1 at2 2 (0.0 m/s)(4.0 s) 1 (9.80 m/s2)(4.0 s)2 2 78 m The brick still falls 78 m 45. a. a g, and at the maximum height, vf 2 2ad becomes 2 vi 2 2gd vf vi 0 2 2 ) s / m.5 2 2 ( v d 25.8 9 ( 2 g d 2 b. Calculate time to rise using vf 0. with a g and vf.30 vi at, The time to fall equals the time to rise, so the time to remain in the air is (2)(2.30 s) 4.60 s tair 2trise Appendix C 891 890-909 EM APP C-845813 6/15/04 10:41 PM Page 892 Chapter 4 1. 3. System y FEarth\u2019s mass on flowerpot v a Fnet 19. a. The scale reads 585 N. Since there is no accelera- tion, your weight equals the downward force of gravity: mg Fg F 5 N 8 g so m 59.7 kg /s2 m 0 9. g 5 8 b. On the Moon, g changes: Fg mgMoon (59.7 kg)(1.60 m/s2) 95.5 N System 29. The", " only force acting on the brick is the gravitational attraction of Earth\u2019s mass. The brick exerts an equal and opposite force on Earth. 31. Suitcase Cart Fcart on suitcase Fsurface on cart FEarth\u2019s mass on cart Fsuitcase on cart FEarth\u2019s mass on suitcase 33. Identify the tire as the system and the direction of pulling as positive. Fwheel on tire ma 0 Fnet FMika on tire FDiego on tire Fwheel on tire FDiego on tire FMika on tire 23 N 31 N 54 N Chapter 5 1. 141 km 125.0 km R2 A2 B2 R A2 B2 65.0 km (65.0km)2 (125.0 km)2 141 km 3. R2 A2 B2 2AB cos R A2 B2 2AB cos (4.5 km)2 (6.4 km)2 2(4.5 km)(6.4 km)(cos 135\u00b0) 1.0101 km Fnet 0 Ffriction on crate Fpull on crate 5. y System Frope on bucket a 0 Fnet 0 v v v v FEarth\u2019s mass on bucket 7. Fnet 225 N 165 N 6.0101 N in the direction of the larger force 15. The scale reads the weight of the watermelon: mg (4.0 kg)(9.80 m/s2) 39 N Fg 17. Identify Reiko\u2019s direction as positive and the rope as the system. FReiko on rope Fnet FReiko on rope ma FTaru on rope ma FTaru on rope (0.75 kg)(1.25 m/s2) 16.0 N 17 N 892 Appendix C 890-909 EM APP C-845813 6/15/04 10:42 PM Page 893 5. Identify north and west as the positive directions. d1(sin ) (0.40 km)(sin 60.0\u00b0) 0.35 km d1W d1(cos ) (0.40 km)(cos 60.0\u00b0) 0.20 km d1N 0.50 km d2N d2W d2W d1W RW d2N d1N RN 2 2 RN R RW 0.35 km 0.50 km 0.85 km 0.20 km 0.00 km 0.20 km 0.00 km (0.85km)2 (0.20 km)2 0.87 km W R m) ) tan1(0 tan1 77\u00b0 N 0.87 km at 77\u00b0 west of north 7. The resultant is 10.0 km. Using the Pythagorean theorem, the distance east is R2 A2 B2, so B R2 A2 (10.0km)2 (8.0 km)2 6.0 km 9. It could never be shorter than one of its components, but if it lies along either the x- or y-axis, then one of its components equals its length. 17. FN mg 52 N Since the speed is constant, the friction force equals the force exerted by the girl, 36 N. Ff kFN F 6 f 3 N 0.69 F 2 5 N N so k 19. FAmes on box Ffriction sFN smg (0.55)(134 N) 74 N 21. Ff, before so FN Ff, after k, before FN Ff N 1.0101 N 8 5. f e b o 58., after FN e r, (0.06)(1.0101 N) 0.6 N 23. Fnet k F F ma mg kFN = F kmg ma 65 N (41 kg)(0.12 m/s2) (41 kg)(9.80 m/s2) 0.15 25.The initial velocity is 1.0 m/s, the final velocity is 2.0 m/s, and the acceleration is 2.0 m/s2, so v v a i ; let ti f tf ti vi vf a tf 0 and solve for tf. 2.0 m/s 1.0 m/s 2.0 m/s2 0.50 s 33. Fg 60.0\u00b0 y FN x Ff Fgy Fgx Fg 35. Fg, perpendicular Fg cos mg cos Fg, per cos1 ndicular pe g m cos1 449 N (50.0 kg)(9.80 m/s2) 23.6\u00b0 37. Fg, parallel Fg(sin ), when the angle is with respect to the horizontal. Fg(cos ), when the angle is with Fg, perpendicular respect to the horizontal. 2Fg, parallel Fg, perpendicular s o c F Fg, 1 ular g n e p r e p 2 n ta F in ll c i", " d, tan1(1 ) 26.6\u00b0 relative to the horizontal, 2 or 63.4\u00b0 relative to the vertical. 39. Since a g(sin k cos ), a (9.80 m/s2)(sin 31\u00b0 (0.15)(cos 31\u00b0)) 3.8 m/s2 41. a g(sin k cos ) (cos ) k(cos ) a g(sin ) g If a 0, 0 g(sin ) g k(cos ) sin in s k s o c \u00b0 3 7 in s \u00b0 c 7 3 s o 0.75 Appendix C 893 890-909 EM APP C-845813 6/15/04 10:44 PM Page 894 Chapter 6 0, y yt 1 gt2 becomes 2 1. a. Since y y 1 gt2 2 y or t2 2 g y t 2 g 07 2 ( 9.8 4.00 s m) /s2 8.4 m b. x vxt (5.0 m/s)(4.00 s) 2.0101 m c. vx 5.0 m/s. This is the same as the initial horizontal speed because the acceleration due to gravity influences only the vertical motion. For the vertical gt with v vy and vi, the component, use v vi initial vertical component of velocity, zero. At t 4.00 s vy gt (9.80 m/s2)(4.00 s) 39.2 m/. x vxt; x t v x y 1 gt2 2 g( 1 2 )2 (9.80 m/s2)(0 1 0 m 7 0. 2 /s m 0. 2 6.0103 m or 0.60 cm )2 x v x 5. Following the method of Practice Problem 4, Hangtime: in 2vi s t g 2(27.0 m/s)(sin 60.0\u00b0) 9.80 m/s2 4.77 s Distance: x (vi cos )t (27.0 m/s)(cos 60.0\u00b0)(4.77 s) 64.4 m Maximum height: at t 1 (4.77 s) 2.38 s 2 y (vi sin )t 1 gt2 2 (27.0 m/s)(sin 60.0\u00b0)(2.38 s) 1 (9.80 m/s2)(2.38 s)2 2 27.9 m 894 Appendix C 13.ac /s)2 2 m (22 v 8.6 m/s2 m 56 r FN. The friction force must supply Recall that Ff the centripetal force, so Ff of the normal force is FN friction must be at least F a m a.6 8 f c c 0.8 9 F g g m N mac. The magnitude mg. The coefficient of 2 m s / 0.88 2 s / m 15. Ff Fc v2 m r 1 m/s)2 4. ( (45 kg ) 3 m. 6 120 23. vc/g w/g vc/w 0.75 m/s 0.020 m/s 0.73 m/s 25. v vp 2 2 vw (150 km/h)2 (75 km/h)2 1.7102 km/h T 1. ( Chapter 7 )2 )3 ( (4.2 units)3( 3 r G r 1 G T I rG ) 23.4103 units3 3 29 units 3. ( T )2 M T E Thus, TM r M r E with rM 1.52rE )3 ( 2 ( )3 TE (365 days)2 ( 2rE)3 1.5 rE r M r E 4.68105 days2 684 days 5. ( T T s)2 M Thus, rs r r M s)3 ( s)2 3 3( )2 da (3.90105km)3( rM d M 3 7.961013 km3 4.30104 km 890-909 EM APP C-845813 3/31/04 6:22 AM Page 895 13. a. v mE h G r E (6.671011 Nm2/kg2)(5.971024 kg) (6.38106 m 1.5105 m) 7.8103 m/s mh b. T 2 (rE G )3 E (6.38106 m 1.5105 m)3 (6.671011 Nm2/kg2)(5.971024 kg) 2 5.3103 s 88 min Chapter 8 1. a. (60)(2rad) 120rad or 377 rad b. 2rad or 6.28 rad 1 )(2rad) c. ( rad or 0.524 rad 1", " 2 6 Vertical: 0.0\u00b0 Fr sin mgr sin (65 kg)(9.80 m/s2)(0.18 m)(sin 0.0\u00b0) 0.0 Nm 17. F gr (35.0 N)(0.0770 m) 2.70 Nm chain Thus, a torque of 2.70 Nm must be exerted to balance this torque. 19. m1 m1 r2 m 2 r 1 (0.23 kg)(1.1 cm) 6.0 cm 0.042 kg 21. For the mass of the two children, I mr 2 + mr 2 2mr 2 When r is doubled, I is multiplied by a factor of 4. 23. The moments of inertia are different. If the spacing between spheres is r and each sphere has mass m, then rotation about sphere A is I mr 2 m(2r)2 5mr 2 Rotation about sphere C is I mr 2 mr 2 2mr 2 3. a. The changes in velocity are the same, so the linear accelerations are the same. The moment of inertia is greater when rotating around sphere A. b. Because the radius of the wheel is reduced from 35.4 cm to 24 cm, the angular acceleration will be increased. 1 2 5.23 rad/s2 a 2 r 1. 5 8. 24 0 7.7 rad/s2 /s2 m m 11. Fr sin so F n r si 35 Nm (0.25 m)(sin 90.0\u00b0) 1.4102 N 13. Fr sin so sin1( ) F r 32.4 Nm (232 N)(0.234 m) sin1( 36.6\u00b0 ) 15. Fr sin Horizontal: 90.0\u00b0 Fr sin 25. Torque is now twice as great. The angular acceleration is also twice as great, so the change in angular velocity is twice as great. Thus, the final angular velocity is 32rad/s, or 16 rev/s. 27. The torque on the wheel comes from either the chain or the string. Iwheel chain Iwheel wheel Thus, chain wheel wheel wheel Fchain rgear Fstring rwheel Fstring r Fch ear g n ai rw h l e e 14 m) 0. ( (15 N ) 8 3 0. m 5.5 N r F 29. I I so F I r Ilarge) (Ismall r 1 2 Ilarge msmall rsmall mgr sin (65 kg)(9.80 m/s2)(0.18 m)(sin 90.0\u00b0) 1.1102 Nm 1 (2.5 kg)(0.090 m)2 0.26 kgm2(2.57 rad/s2) 2 0.090 m 7.7 N Appendix C 895 890-909 EM APP C-845813 3/31/04 6:23 AM Page 896 37. a. clockwise: A counterclockwise: B FArA FA(0.96 m 0.30 m) (0.66 m)FA FBrB FB(0.96 m 0.45 m) (0.51 m)FB Chapter 9 1. E v p 0 a. p mv b. B net A so B A (0.51 m)FB c. Fg FA thus, FA FB Fg or FA (0.66 m)FA FB )FA m 6 (0 1 m.5 Fg.6 0 Fg ma 7.3 kg)(9.80 m/s2 31 N d. FA would become greater, and FB would be less. 39. Choose the center of mass of the board as the pivot. The force of Earth\u2019s gravity on the board is exerted totally on the support under the center of mass. end Fendrend diver Thus, Fend rdiver Fdiverrdiver Fd r e iv r n e mdi r e v r n e d d grdiver (85 kg)(9.80 m/s2)(1.75 m) 1.75 m 8.3102 N Fcenter Fend Thus, Fcenter To find the force on the center support, notice that because the board is not moving. Fg Fdiver Fg Fdiver 2Fdiver Fg 2mdiverg mboardg mboard) g(2mdiver (9.80 m/s2)(2(85 kg) 14 kg) Fend 1.8103 N 896 Appendix C (725 kg)(115 km/h)(1 0 00 m k 1 2.32104 kgm/s eastward m)( 1 36 h 0 s 0 ) p b. v m 0 s (2.32104 kgm/s)36 k m 1 0 0 00 1 m 1 h 2175 kg 38.4 km/h eastward 3. a", ". Ft pf pi Ft mvf m vf mvf mvi (5.0 N)(1.0 s) (7.0 kg)(2.0 m/s) 7.0 kg 2.7 m/s in the same direction as the original velocity Ft mvf m b. vf (5.0 N)(1.0 s)(7.0 kg)(2.0 m/s) 7.0 kg 1.3 m/s in the same direction as the original velocity 5. Before After vf 0 vi x pi a. Ft p = pf pi pf F t mvi pf F t m (0.0 kgm/s)(60.0 kg)(94 km/h)1 h 1 0 00 0 s 0 36 k m 1 0.20 s 7.8103 N opposite to the direction of motion b. Fg mg F 03 N 1 g 7 8. 8.0102 kg m 2 /s m 0 8. 9 g Such a mass is too heavy to lift. You cannot safely stop yourself with your arms. 890-909 EM APP C-845813 6/15/04 10:47 PM Page 897 13. pi pf 2mvf mvAi mvBi mvBi mvAi 2 vf 2.2 m/s 0.0 m/s 2 1.1 m/s 15. mbvbi (mb mwvwi mw)vf where vf is the common final speed of the bullet and piece of lumber. 0.0 m/s, Because vwi mw)vf (mb mb vbi (0.0350 kg 5.0 kg)(8.6 m/s) 0.0350 kg 1.2103 m/s 17. The system is the bullet and the ball. mbulletvbullet, i vball, i mballvball, i 0.0 m/s and vbullet, f vbullet, f) mbullet(vbullet, i mball so vball, f mbulletvbullet, f 5.0 m/s mballvball, f (0.0350 kg)(475 m/s (5.0 m/s)) 2.5 kg 6.7 m/s 19. pri prf pfuel, f pfuel, f 0.0 kgm/s pfuel, i where prf If the initial mass of the rocket (including fuel) is 4.00 kg, then the final mass of the rocket is mr 4.00 kg 0.0500 kg 3.95 kg mrf 0.0 kgm/s mrfvrf mfu v fuel, f el vrf m f r mfuelvfuel, f (0.0500 kg)(625 m/s) 3.95 kg 7.91 m/s pJf 0.0 kgm/s 21. pCi pJi where pCi mCvCf pCf pJi mJvJf m vCf C m J so vJf pf vf y)2 pi (pi, x)2 (pi, )2 (m1v1i)2 (m1v2i pf m1 m2 )2 (m1v1i)2 (m1v2i m2 m1 ((1732 kg)(31.3 m/s))2 ((1383 kg)(11.2 m/s))2 1383 kg 1732 kg 18.1 m/s p y) tan1(, i p x i, v m ) tan1( 1 1 m v 2 2 (1383 kg)(11.2 m/s) (1732 kg)(31.3 m/s) tan1 i i 15.9\u00b0 south of east 25. Before: pi, x pf pf, y pf, y v2i m1v1i (1345 kg)(15.7 m/s) 2.11104 kgm/s pi m2)vf (m1 (1345 kg + 1923 kg)(14.5 m/s) 4.74104 kgm/s pf sin (4.74104 kgm/s)(sin 63.5\u00b0) 4.24104 kgm/s m2v2i pi, y p gm/s y 4.24 4 k 0 1 f, k g 3 2 19 m 2 22.1 m/s Yes, it was exceeding the speed limit. Chapter 10 1. a. Because W Fd and KE W, doubling the force would double the work, which would double the change in kinetic energy to 1.35 J. b. Because W Fd, halving the distance would cut in half the work", ", which also would cut the change in kinetic energy in half, to 0.68 J80.0 kg)(4.0 m/s) 115 kg 2.8 m/s in the opposite direction 23. Car 1 is moving south. Car 2 is moving east. Before: pi, x pi, y p1, x p1, y p2, x p2, y 0 m2v2i 0 m1v1i 3. a. W mgd (7.5 kg)(9.80 m/s2)(8.2 m) 6.0102 J b. W Fd 6.0102 J (645 N)(8.2 m) 6.0102 J 5.9103 J Appendix C 897 890-909 EM APP C-845813 7/29/04 4:05 PM Page 898 c.3 W J(1 i m s 0 6 n) 5. W Fd cos 2(255 N)(15 m)(cos 15\u00b0) 6.5103 J 7. W Fd cos (628 N)(15.0 m)(cos 46.0\u00b0) 6.54103 J 9. P W t.0 m) 20 ( (575 N ). s 0 10 1.15103 W 1.15 kW gd m 11. P W t t m (35 L/min)(1.00 kg/L) t 35 kg/min Thus, P m gd t (35 L/min)(1 min/60 s)(9.80 m/s2)(110 m) 0.63 kW d 13. P F t d t F P 340 s (6.8103 N)(15 m) 0.30103 W 5.7 min d 0.2 0 ( m ) e 25. a. IMA 4.0 5 0... MA 4 F ) N 10 1. 1 ( e M A. 5 100 1 c. efficiency 100 38 27. efficiency 100 W i F rd r 100 cy) n e i i f f e Fe( (1.25103 N)(0.13 m)(100) (225 N)(88.7) So de 0.81 m 898 Appendix C Chapter 11 1. To bring the skater to a stop: KEi 2 1 mvi 2 2 W KEf 1 mvf 2 (52.0 kg)(0.00 m/s)2 1 1 (52.0 kg)(2.5 m/s)2 2 2 160 J To speed up again: KEi W KEf 1 2 1 mvi mvf 2 2 2 (52.0 kg)(2.5 m/s)2 1 1 (52.0 kg)(0.00 m/s)2 2 2 160 J 3. KE 1 mv2 2 1 (7.851011 kg)(2.50104 m/s)2 2 2.451020.8104 bombs would be required to produce the same amount of energy used by Earth in stopping the comet. 5. W Fd hi) mg(hf (20.0 kg)(9.80 m/s2)(0.00 m 1.20 m) 2.35102 J 7. Choose the ground as the reference level. PE mg(hf hi) (1.8 kg)(9.80 m/s2)(0.00 m 6.7 m) 1.2102 J 15. The system is the bike the rider Earth. There are no external forces, so total energy is conserved. KE 1 mv2 2 1 (85.0 kg)(8.5 m/s)2 2 3.1103 J KEf PEi PEf KEi 1 mv2 0 0 mgh 2 2 v2 m s) /.5 8 h 3.7 m s2) / m 0.8 (9 (2 2 g ( ) 890-909 EM APP C-845813 6/15/04 10:50 PM Page 899 17. Bottom of the valley: PEf PEi KEi 0 mgh 1 mv2 0 2 v 2gh KEf 2(9.80 m/s2)(45.0m) 29.7 m/s Top of the next hill: PEi PEf KEf KEi 1 0 mghi mv 2 mghf 2 hf) v 2g(hi 2(9.80 m/s2)(45.0m 40.0 m) 9.90 m/s No, the angles of the hills do not affect the answers. 19. Conservation of momentum: mv (m M)V, or v (m M)V m (0.00800 kg 9.00 kg)(0.100 m/s) 0.00800 kg 1.13102 m/s Chapter 12 273 115 273 158\u00b0C 273 172 273 101\u00b0", "C 273 125 273 148\u00b0C 273 402 273 129\u00b0C 273 425 273 152\u00b0C 273 212 273 61\u00b0C 1. a. TC b. TC c. TC d. TC e. TC f. TC TK TK TK TK TK TK 3. Q mCT (2.3 kg)(385 J/kgK)(80.0\u00b0C 20.0\u00b0C) 5.3104 J 5. Q mCT (75 kg)(4180 J/kgK)(43\u00b0C 15\u00b0C) 8.8106 J 6 J.8 8 0 1 2.4 kWh 06 3.6 Wh k 1 / J (2.4 kWh)($0.15 per kWh) = $0.36 9. Heat gained by the water: Q mCT (0.100 kg)(4180 J/kg\u00b0C)(15.0\u00b0C) 6.27 kJ Thus, heat lost by the aluminum block T 6.27 kJ maluminumCaluminum hence, Caluminum Q T malum inum 6.27 kJ (0.100 kg)(75.0\u00b0C) 8.36102 J/kg\u00b0C 19. Q mCT mHf (0.100 kg)(2060 J/kg\u00b0C)(20.0\u00b0C) (0.100 kg)(3.34105 J/kg) 3.75104 J 21. Q mCice T mHf mCsteam mCwater (0.300 kg)(2060 J/kg\u00b0C)(0.0\u00b0C (30.0\u00b0C)) T mHv 0.300 kg)(3.34105 J/kg) (0.300 kg)(4180 J/kg\u00b0C)(100.0\u00b0C 0.0\u00b0C) (0.300 kg)(2.26106 J/kg) (0.300 kg) (2020 J/kg\u00b0C)(130.0\u00b0C 100.0\u00b0C) 9.40102 kJ 23. U Q Wblock; since Wdrill Wblock and assume no heat added to drill: mCT 0 Wdrill (0.40 kg)(897 J/kg\u00b0C)(5.0\u00b0C) 1.8103 J 25. U mCT (0.15 kg)(4180 J/kg\u00b0C)(2.0\u00b0C) 1.3103 J The number of stirs is 03 J 2.6104 1 1.3 J 0 5.0 0 Chapter 13 1. F PA Plw (1.0105 Pa)(1.52 m)(0.76 m) 1.2105 N 7. mACA(Tf TAi) mWCW(Tf TWi) 0 Since in this particular case, mA cancel and TWi CAT C i A C C W A W Tf mW, the masses (2450 J/kgK)(16.0\u00b0C) (4180 J/kgK)(85.0\u00b0C) 2450 J/kgK 4180 J/kgK 59.5\u00b0C 3. mbrick V lwh ickg mb Fgb l hg hg w F r rick P w l w l A A (11.8 g/cm3)(1 cm3 )( 3 m g k 1 1 g 0 00 6 10 1 )(0.200 m)(9.80 m/s2) 23.1 kPa Appendix C 899 890-909 EM APP C-845813 6/15/04 10:51 PM Page 900 F g. g m 5. The maximum pressure is P A A Therefore g A m P (454 kg)(9.80 m/s2) 5.0104 Pa 8.9102 m2 7. PV nRT; so V n P R T m nM V m P M R T (15.5106 Pa)(0.020 m3) (8.31 Pam3/molK)(293 K) 5.1102 g (4.00 g/mol) 9. PV nRT; so T V nR P 0 m 1 03 1 g. where n M l o m g/ 9 2 T 20.0\u00b0C 273 293K (8.31 Pam3/molk)(293 K) 1 0 03 g 1. l o m g/ 9 2 (1.013105 Pa.83 m3 23. F2 1A F 2 A 1 (1600 N)(72 cm2) 1440 cm2 8.0101 N 25. F2 1A F 2. Thus.4 The adult stands on the larger piston. 27. Fapparent Fg Fbuoyant brickVg ( (1.8103 kg/m3 brick waterVg water)Vg 1.00103 kg/m3)(0.20 m3)(9", ".80 m/s2) 1.6103 N 29. To hold the camera in place, the tension in the wire must equal the apparent weight of the camera. T Fapparent Fg Fg 1250 N (1.00103 kg/m3)(16.5103 m3)(9.80 m/s2) 1.09103 N Fbuoyant waterVg 900 Appendix C 31. The buoyant force on the foam must equal 480 N. We are assuming the canoe is made of dense material. waterVg Fbuoyant F nt a y o u b therefore, V rg a w t e 480 N (1.00103 kg/m3)(9.80 m/s2) 4.9102 m3 39. L2 L1 L L1(T2 L1(T2 T1) T1) ; so (25106 \u00b0C1)(3.66 m)(39\u00b0C (28\u00b0C)) 6.1103 m 6.1 mm 41. At the beginning, 400 mL of 4.4\u00b0C water is in the beaker. Find the change in volume at 30.0\u00b0C. V VT (210106 \u00b0C1)(400106 m3)(30.0\u00b0C 4.4\u00b0C) 2106 m3 2 mL 43. The aluminum shrinks more than the steel. Let L be the diameter of the rod. Laluminum LT (25106 \u00b0C1)(0.85 cm)(0.0\u00b0C 30.0\u00b0C) 6.4104 cm The diameter of the steel hole shrinks by Lsteel LT (12106 \u00b0C1)(0.85 cm)(0.0\u00b0C 30.0\u00b0C) 3.1104 cm The spacing between the rod and the hole will be 1 (6.4104 cm 3.1104 cm) 1.6104 cm 2 Chapter 14 1. F kx (95 Nm)(0.25 m) 24 N 3. F kx k F x N 4 2 2 m.1 0 2.0102 N/m 5. PEsp 1 kx2 2 k 2P Esp.61 m J m 890-909 EM APP C-845813 6/15/04 10:52 PM Page 901 7. T 2 l g )2 T l g( 2 s)2 (1.6 m/s2)(2.0 2 0.16 m m 5 5 1 15. a. v d 343 m/s 0 5 t s 1. m 1 b. T 1 0.00229 s Hz 43 6 f s 0.787 m / m 3 3 4 c. v z H 6 3 4 f 17. Lower s 2.50 m 1.0 5 m / 19. v f, so v 0.0 6 z H f 21. v f (0.600 m)(20.0 Hz) 12.0 m/s Chapter 15 1. v 3 3 4 f 8 1 /s 19 m m H z 3. /s 4.10103 Hz. v, f so v f (0.655 m)(2280 Hz) 1490 m/s This speed corresponds to water at 25\u00b0C. Chapter 16 E r e t f a 1 4d after P 4db efore Therefore, after the lamp is moved the illumination is one-ninth of the original illumination. 3. Illuminance of a 150-W bulb P 2275, d 0.50, 0.75,..., 5.0 P E(d) 4 d2 720 ) x l ( E P 2275 lm 7.2 0.50 5.0 r (m vs 7. v 343 m/s, fs 0 m/s, vd v v ) fs ( d v v s 365 Hz( 392 Hz fd vd fd 9. v 1482 m/s, fs 0 m/s v v ) fs ( d v v s 3.50 MHz( 3.52 MHz 365 Hz, 25.0 m/s 343 m/s 25.0 m/s 343 m/s P 5. E 4 d2 P 4Ed2 4(160 lm/m2)(2.0 m)2 8.0103 lm ) 3.50 MHz, vs 9.20 m/s, 15. The relative speed along the axis is much less than the speed of light. Thus, you can use the observed light frequency equation. Because the astronomer and the galaxy are moving away from each other, use the negative form of the observed light frequency equation. 1482 m/s 1482 m/s 9.20 m/s ) fobs f (1 v ) c 6 (6.161014 Hz)(1 6 s) / m 1 0 8 m s / 10 6.031014 Hz 5. 0. 3", " 5 0 19. Resonance spacing 1.1 m, so 2.2 m 2 v f (2.2 m)(440 Hz) = 970 m/s 21. a. 1 2L 2(2.65 m) 5.30 m f1 The lowest frequency is /s 64.7 Hz m 3 3 4 v m 5 0.3 1 /s 129 Hz 6 2 2 ) 194 Hz 6 2 ( 2 5 3 /s ) m f3 b. f2 17. Assume that the relative speed along the axis is much less than the speed of light. Thus, you can use the Doppler shift equation. ) v c ( obs The observed (apparent) wavelength appears to be longer than the known (actual) wavelength of the oxygen spectral line. This means that the astronomer and the galaxy are moving away from each other. So use the positive form of the Doppler shift equation. Appendix C 901 890-909 EM APP C-845813 6/15/04 10:54 PM Page 902 obs ( ) v c Solve for the unknown variable. ( ) obs v c (3.00108 m/s) 7.02106 m/s (525 nm 513 nm) 513 nm Chapter 17 1. Polishing makes the surface smoother. 3. i r 35\u00b0 5. r1 i2 30\u00b0 i1 90\u00b0 r1 90\u00b0 30\u00b0 60\u00b0 1 1 13. 1 d d f o i f do do f (36.0 cm)(16.0 cm) 36.0 cm 16.0 cm di 28.8 cm 15. 1 d d f o i f do f do (16.0 cm)(7.0 cm) 16.0 cm 7.0 cm di 12.4 cm h di m i h do o iho d d hi o (12.4 cm)(2.4 cm) 16.0 cm 1.9 cm 17. O1 Ray 1 1 1 19. 1 d d f o i di do so f di do di, so do and m do 24 cm and m 0.75, so ( cm) 32 cm 4 2.7 5 0 di m di do and f (32 cm)(24 cm) 32 cm (24 cm) 96 cm h di i 21. m h do o ohi d h o di (2.4 m)(0.36 m) 1.8 m 0.48 m 1 1 1 d d f o i ido d f di do (0.48 m)(2.4 m) 0.48 m 2.4 m 0.60 m Chapter 18 1. n1 sin 1 2 2 n2 sin 2 n i n1 s sin1( 1) n 2 n i s sin1((1.00)( 6 3. 1 26.3\u00b0 n2 sin 2 n i n1 s 1) sin1( n 2 n i (s sin1((1.00) 3 3. 1 22.1\u00b0 37.0\u00b0)) 30.0\u00b0)) 3. n1 sin 1 Ray 2 Horizontal scale: 1 block 1.0 cm I1 F di 8.6 cm n2 5. n1 sin 1 2 n2 sin in s n1 1 si n 2 31\u00b0) (s n i (1.. do do f (20.0 cm)(15.0 cm) 20.0 cm (15.0 cm) di 1 1 15. 1 d d f o i f do do f ) m c.5 )(5 ( m c.5 8 m c.5 5 m c.5 8 di 15.6 cm, or 16 cm to two significant digits 8.57 cm 902 Appendix C 890-909 EM APP C-845813 6/15/04 10:55 PM Page 903 h di i m h do o iho d d (15.6 cm)(2.25 cm) 8.5 cm hi o 4.1 cm 1 1 17. 1 d d f i o di because with do di and m 1 m do Therefore, 2 1 d f i di do 2f 2(25 mm) 5.0101 mm di 5.0101 mm 1 1 19. 1 d d f o i f do do f ) m c.0 )(5 ( m c 5 2 m c.0 5 m c 5 2 di 6.2 cm h di m i h do o iho d d hi o (6.2 cm)(2.0 cm) 25 cm 0.50 cm (inverted image) 1 1 21. 1 d d f o i f do do f (3.4 cm)(12.0 cm) 3.4 cm 12.0 cm so di 4.7 cm odi h d o hi (2.0 cm)(4.7 cm) 3.4 cm 2.8 cm (4.0)(3.5", " cm) 14 cm di 23. m do mdo so di 1 1 1 d d f i o di do so f do di (3.5 cm)(14 cm) 3.5 cm (14 cm) 4.7 cm Chapter 19 d 1. x L (13.2103 m)(1.90105 m) 0.600 m 418 nm d 3. x L L d x (632.8109 m)(1.000 m) 65.5103 m 9.66106 m 9.66 m 5. 2t (m 1 ) n 2 oil For the thinnest film, m 0. 5 n 3 6 109 nm t 4.4 1 )( ( 4n oil nair, there is a phase inversion on nfilm, there is the first reflection. Because nsilicon a phase inversion on the second reflection. 7. Because nfilm m 5 For destructive interference to keep yellow-green from being reflected: 2t m 1 nf 2 ilm For the thinnest film, m 0. 5 n 5 5 95.7 nm t.4 1 )( 4 ( 4n film m 5) 9. For constructive interference 2t (m 1 ) nf 2 ilm For the thinnest film, m 0. 1 m n 2 5 t 97.9 nm 3).3 1 )( 4 ( 4n film 17. 2x1 L 2 w x1 )w (2 L 2 (2.60102 m)(0.110103 m) 2(589109 m) 2.43 m 19. 2x1 L 2 w )w x1 (2 L 2 (2.40103 m)(0.0295103m) 2(60.0102 m) 5.90102 nm 21. A full spectrum of color is seen. Because of the variety of wavelength, dark fringes of one wavelength would be filled by bright fringes of another color. Appendix C 903 890-909 EM APP C-845813 7/29/04 4:17 PM Page 904 23. d sin sin d sin1 d x tan L x L tan (9.0109 Nm2/C2) (3.6106 C)(4.0106 C) (0.20 m)2 (9.0109 Nm2/C2) (2.0106 C)(3.6106 C) (0.60 m)2 3.1 N toward the right Chapter 21 L tan sin1 d (0.800 m) tansin1.449 m 25. d sin There is one slit per distance, d, so 1 gives slits d per centimeter. d sin x 1 sintan L 9m 63210 m 10 sintan 6 5.0 m 5.5 0 6 m 6.210 6.210 4 cm li 1 t 6.2 4cm 0 1 1.6103 slits/cm Chapter 20 q q K 9. F B A 2 d A B (9.0109 Nm2/C2)(2.0104 C)(8.0104 C) (0.30 m)2 1.6104 N The force is attractive. 11. y x Fnet FC on A FB on A B qB A rAB rAC qA qC C Magnitudes of all forces remain the same. The direction changes to 42\u00b0 above the x axis, or 138\u00b0. 13. FA on B FC on B Fnet FC on B FA on 904 Appendix C 4 10 0. 2 1. E F 4.0101 N/. E F q so F Eq (27 N/C)(3.0107 C) 8.1106 N 5. a. No. The force on the 2.0-C charge would be twice that on the 1.0-C charge. b. Yes. You would divide the force by the strength of the test charge, so the results would be the same. 7. Because the field strength varies as the square of the distance from the point charge, the new field strength will be one-fourth of the old field strength, or 6.5103 N/C. q F 9. E K 2 q d \u2019 d2 so q E K (450 N/C)(0.25 m)2 9.0109 Nm2/C2 3.1109 C The charge is negative, because the field is directed toward it. 17. V Ed 0 0 4 V V E 2104 N/C 2 0..94102 m 19. V Ed, so d 03 1 4.25 E N/C 21. W qV (3.0 C)(1.5 V) 4.5 J 23. W qV (1.601019 C)(1.8104 V) 2.91015 J 25. W qV qEd (1.601019 C", ")(4.5105 N/C)(0.25 m) 1.81014 J 27. Fg 5 1 N C / N Eq F 10 9. 1 3.21019 # electrons 890-909 EM APP C-845813 6/15/04 10:57 PM Page 905 1 4 0 4 2 V V 3.8104 N/C 29. E 0 3 m 1 6.4 d 10 2. 1 E F F 3.21019 C, so # electrons 31. q CV, so the larger capacitor has a greater charge. 15. Lamp Battery Switch Potentiometer V 8.0 A 0 1 2 23. a. I V \u2126 5 1 R q (6.8106 F)(24 V) 1.6104 C b. E I2Rt (8.0 A)2(15 \u2126)(30.0 s) 2.9104 J 33. q CV so q C(V2 V1) (2.2106 F)(15.0 V 6.0 V) 2.0105 C Chapter 22 1. P IV (0.50 A)(125 V) 63 J/s 63 W 3. P IV W P 5 7 0.60 A I 5 V 2 1 V 5. P IV P 0 W 0.30 A 0.9 I V.0 3 V 7. V IR (3.8 A)(32 \u2126) 1.2102 V 0 2 1 9. a. R V 2.4102 \u2126 5 0. 0 I V A b. P IV (0.50 A)(120 V) 6.0101 W 11. a. The new value of the current is c. 2.9104 J, because all electrical energy is converted to thermal energy. 25. a. E Pt (0.22)(100.0 J/s)(1.0 min) s 0 6 m n i 1 1.3103 J b. E Pt (0.78)(100.0 J/s)(1.0 min) s 0 6 m n i 1 4.7103 J t) 27. E IVt = I(2V)( 2 For a given amount of energy, doubling the voltage will divide the time by 2. h 1.1 h t 2.2 2 15 1 29. a. I V,0 12 R V 9.6103 A 00 \u2126 b. P VI (115 V)(9.6103 A) 1.1 W c. Cost (1.1103 kW)($0.12/kWh) (30 days)(24 h/day) $0.10 (1.3)IVt (1.3)(55 A)(12 V)(1.0 h) 858 Wh 31. Echarge 8 W 85 E 8.2 h t 1 ) ( A (7.5 V I h 4 V.30 A 0.6 2 So V IR (0.30 A)(2.1102 \u2126) 6.3101 V Chapter 23 b. The total resistance of the circuit is now V A 5 2 1 V 4.2102 \u2126 Rtotal 3 0. 0 I Therefore, Rtotal 4.2102 \u2126 2.1102 \u2126 2.1102 \u2126 Rlamp Rres c. P IV (0.30 A)(6.3101 V) 19 W 13. 4.5 V A 85 mA I V 5. 4 R V 53 \u2126 8 5 0 0. I A 53 1. R R1 R2 R3 20 \u2126 20 \u2126 20 \u2126 60. a. It will increase. b. I V, so it will decrease. R c. No. It does not depend on the resistance. 5. V1 V2 V3 V1 IR1 IR2 IR3 V2 (3 A)(10 \u2126) 30 V (3 A)(15 \u2126) 45 V (3 A)(5 \u2126) 15 V V3 30 V 45 V 15 V 90 V voltage of battery 0. 7 1 7. a. I V 66.7 mA. 0 5 5 2 R V \u2126 Appendix C 905 890-909 EM APP C-845813 7/29/04 4:18 PM Page 906 b. First, find the total resistance, then solve for voltage. 19. Neither. They both reach maximum dissipation at RB 255 \u2126 292 \u2126 547 \u2126 R RA V IR (66.7 mA)(547 \u2126) 36.5 V c. P IV (66.7 mA)(36.5 V) 2.43 W PA PB I2RA I2RB (66.7 mA)2(255 \u2126) 1.13 W (", "66.7 mA)2(292 \u2126) 1.30 W d. Yes. The law of conservation of energy states that energy cannot be created or destroyed; therefore, the rate at which energy is converted, or power is dissipated, will equal the sum of all parts. 9. The resistor with the lower resistance will dissipate less power; thus, it will be cooler. 11. a. R R1 R2 22 \u2126 33 \u2126 55 \u2126 V 2.2 A 0 1 2 b. I V \u2126 5 5 R c. V IR Across a 22-\u2126 resistor, V V1 R IR1 R1 V(22 \u2126) 48 V 1 0 2 \u2126 5 5 Across a 33-\u2126 resistor, V V2 R IR2 R2 V(33 \u2126) 72 V 1 0 2 \u2126 5 5 d. V 48 V 72 V 1.20102 V 13. VB RB V RB RA (45 V)(235 k\u2126) 475 k\u2126 235 k\u2126 15 V 1 1 1 1 15. a 15. 15. 15. 3 0 \u2126 15. R 5.00 \u2126 V 6.00 A 0. 0 3 b.00 A 0. 0 3 V c. I \u2126 0. 5 1 R 1 17. a. Yes, it gets smaller. b. Yes, it gets larger. No, it remains the same. Currents are independent. 906 Appendix C the same voltage. 2 P V R V PR The voltage is equal across parallel resistors, so: V P1R1 P2R2 (2 W)(12 \u2126) (4 W)(6.0 \u2126) 5 V maximum 25. By conservation of energy (and power): PT P2 P1 P3 2.0 W 3.0 W 1.5 W 6.5 W IV PT P W T 6.5 I 0.54 A V 2 V 1 27. Then, all of the working lights are in series. The 12 working lights will burn with equal intensity. Chapter 24 1. a. repulsive b. attractive 3. the bottom (the point) 5. a. from south to north b. west 7. the pointed end 9. Yes. Connect the potentiometer in series with the power supply and the coil. Adjusting the potentiometer for more resistance will decrease the current flow and the strength of the field. 17. F BIL (0.40 N/Am)(8.0 A)(0.50 m) 1.6 N 19. F BIL, F weight of wire N 5 3. 0 F 0.15 T B 0 00 m).4 ( ) (6.0 A L I 21. Opposite to the direction of the electron motion 23. F Bqv (9.0102 T)(2)(1.601019 C)(3.0104 m/s) 8.61016 N 25. F Bqv (5.0102 T)(2)(1.601019 C)(4.0104 m/s) 6.41016 N 890-909 EM APP C-845813 6/15/04 11:00 PM Page 907 Chapter 25 1. a. EMF BLv (0.4 T)(0.5 m)(20 m/s) 4 V F b. I EM R V 4 \u2126.0 6 0.7 A 3. a. EMF BLv (1.0 T)(30.0 m)(2.0 m/s) 6.0101V 1 V 4.0 A 10 F 6. 0 b. I EM \u2126.0 5 1 R 5. a. Veff (0.707)Vmax (0.707)(170 V) 1.2102 V b. Ieff (0.707)Imax (0.707)(0.70 A) 0.49 A V f f e c. R Ie f f V x ma 2 Ima x 2 V x m a Im.4102 \u2126 7. a. Veff V x ma 2 V 25 4 2 3.01102 V b. Ieff V eff.60 A N V P P 17 90,000) ( (60.0 V ) 0 0 3 VS VPIP 1.80104 V VSIS IS V S V P IP (1.80104 V)(0.50 A) 60.0 V 1.5102 V Chapter 26 v 2 1. Bqv m r v r m Bq (1.671027 kg)(7.5103 m/s) (0.60 T)(1.601019 C) 1.3104 m v 2 3. Bqv m r v r m Bq (9.111031 kg)(5", ".0104 m/s) (6.0102 T)(1.601019 C) 4.7106 m 2q 2r 5. m B 2 V (7.2102 T)2(0.085 m)2(1.601019 C) 2(110 V) 2.71026 kg 7. Bqv Eq 0 3 1 T 0.5 1 4.0105 m/s 15. All electromagnetic waves travel through air or a vacuum at c, 3.00108 m/s 17.7107 m c 19. v K m/s 8 5,4 299 2,79 54.0 1 0 0 2.99712108 m/s c 21. v K c)2 so K ( v 0 0. (3 3 4. 2 1.52 8 0 1 8 10 s)2 / m s / m Appendix C 907 890-909 EM APP C-845813 6/15/04 11:01 PM Page 908 019 J) 3.71019 J 1 eV Chapter 27 1. (2.3 eV)(1.60 1 3. m 9.111031 kg, KE 1 mv2 2 KE so v 2 1 2.7 3 ( 3 1.1 1 9 9.0105 m/s ) 9k 0 1 0 J g 1 m 5. KE qV0 (1.601019 C)(3.2 J/C) 5.11019 J 7. KEmax eVnm hf0 1240 nm 1.96 eV e V 12 0 4 4 m n 5 2 0.960 eV 9. hf0 hc 4.50 eV 4.50 eV, so 0 nm eV 12 0 4 V e 50. 4 276 nm Thus 19. a. v m J 34 s 0 1 63. 6 ( s) / m 8 g) k 0. (7 1.11035 m.5 b. The wavelength is too small to show observable effects. 21. h p h so p KE 1 mv2 2 2 p 2 m h2 2m s 6 34 10 (9.111031 kg) J m eV 1 (1.5441017 J)( ) 019 J 1 1.60 96.5 eV, so it would have to be accelerated through 96.5 V. 908 Appendix C E2 E3 1. En Chapter 28 13 6 eV. 2 n 1 eV 3.40 eV 6. 3 2 2 ) ( 1 eV 1.51 eV 6. 3 2 3 ) ( 1 eV 0.850 eV 6. 3 2 ( ) 4 3. E E4 E2 1 1 (13.6 eV)( ) 2 2 4 2 1 1 (13.6 eV)( ) 1 6 4 E4 2.55 eV 1 1 0 1 5 x m 5. 5 1 0 1.5 2 5 m 0.07 m x 2103 m 7. a. E 8.82 eV 6.67 eV 2.15 eV c h b. E (6.631034 Js)(3.00108 m/s) (2.15 eV)(1.601019 JeV) 5.78107 m 578 nm Chapter 29 1. free e/cm3 (2 e/atom)(6.021023 atoms/mol)( 1 5. 6 ) ol m g 37 (7.13 g/cm3) 1.311023 3. free e/cm3 (1 e/atom)(6.021023 atoms/mol)( l o m 1 7.9 6 19 ) g (19.32 g/cm3) 5.901022 5. free e (1.811023 free e/cm3)( ) g 5 83 2 m3 /c g 0 2.7 1.901026 free e in the tip 7. free e/atom ( m 1 ol 6.02 23 10 3.7810 8. 1 atoms )(2 09 ol m 18 free e/atom g)( 3 cm 1 3 g.3 2 free e )(1.891 05 ) 3 cm TK TC 273 TC TK 273 200.0 273 73\u00b0C 890-909 EM APP C-845813 7/29/04 4:20 PM Page 909 9. free e/atom ( m 1 ol 6.02 23 10 )(7 atoms 2. m g)( 3 m c 3.2 5 6 l o )( g 1.161010 free e cm3 ) b. E (0.113986 u)(931.49 MeV/u) 106.18 MeV 15. 234 92U \u2192 230 90Th 4 2He 2.671013 11. ratio free e/cm3 in doped Si free e", " Hz lx T V W Pa m/s m3 m/s2 m2 A2s4/kgm2 kg/m3 As kgm/Cs2 kgm2/A2s3 kgm2/As3 kgm2/s2 kgm/s2 s1 cd/m2 kg/As2 kgm2/As3 kgm2/s3 kg/ms2 m/s m3 V/A Nm Ns/Cm W/A or J/C J/s N/m2 T a b e s l Useful Conversions 1 kg 6.021026 u 1 oz \u2194 28.4 g 1 kg \u2194 2.21 lb 1 lb 4.45 N 1 atm 14.7 lb/in2 1 atm 1.01105 N/m2 1 atm 101 kPa 1 cal 4.184 J 1 eV 1.601019 J 1 kWh 3.60 MJ 1 hp 746 W 1 mol 6.0221023 items Appendix D 911 910-918 EM APP D-845813 6/9/04 4:11 PM Page 912 Quantity Symbol Value Approximate Value Physical Constants Atomic mass unit Avogadro\u2019s number Boltzmann\u2019s constant Constant in Coulomb\u2019s law Elementary charge Gas constant Gravitational constant Mass of an electron Mass of a proton Mass of a neutron Planck\u2019s constant Speed of light in a vacuum u NA k K e R G me mp mn h c 1.660538861027 kg 6.02214151023 mol1 1.38065051023 Pam3/K 8.987551788109 Nm2/C2 1.602176531019 C 8.314472 Pam3/molK 6.67421011 Nm2/kg2 9.10938261031 kg 1.672621711027 kg 1.674927281027 kg 6.62606931034 Js 2.99792458108 m/s 1.661027 kg 6.0221023 mol1 1.381023 Pam3/K 9.0109 Nm2/C2 1.6021019 C 8.31 Pam3/molK 6.671011 Nm2/kg2 9.111031 kg 1.671027 kg 1.671027 kg 6.631034 Js 3.00108 m/s SI Prefixes Moments of Inertia for Various Objects f c p n m milli pico centi nano micro femto Prefix Symbol Scientific Notation 1015 1012 109 106 103 102 101 101 102 103 106 109 1012 1015 mega hecto deka peta giga deci tera kilo da 912 Appendix D Object Thin hoop of radius r Location of Axis Through central diameter Solid, uniform cylinder of radius r Through center Uniform sphere of radius r Through center Long, uniform rod of length l Through center Long, uniform rod of length l Through end Thin, rectangular plate of length l and width w Through center Diagram Moment of Inertia Axis mr 2 r r r Axis Axis Axis l Axis l Axis l w 1 mr 2 2 2 mr 2 5 1 ml 2 2 1 1 ml 2 3 1 m(l 2 w 2) 1 2 910-918 EM APP D-845813 6/9/04 4:12 PM Page 913 Densities of Some Common Substances Substance Aluminum Cadmium Copper Germanium Gold Hydrogen Indium Iron Lead Mercury Oxygen Silicon Silver Water (4\u00b0C) Zinc Material Aluminum Brass Carbon Copper Glass Ice Iron Melting and Boiling Points of Some Substances Substance Melting point (\u00b0C) Boiling point (\u00b0C) Aluminum Copper Germanium Gold Indium Iron Lead Silicon Silver Water Zinc 660.37 1083 937.4 1064.43 156.61 1535 327.5 1410 961.93 0.000 419.58 2467 2567 2830 2808 2080 2750 1740 2355 2212 100.000 907 Density (g/cm3) 2.702 8.642 8.92 5.35 19.31 8.99105 7.30 7.86 11.34 13.546 1.429103 2.33 10.5 1.000 7.14 Specific Heats of Some Common Substances Specific Heat (J/kgK) Material Specific Heat (J/kgK) 897 376 710 385 840 2060 450 Lead Methanol Silver Steam Water Zinc 130 2450 235 2020 4180 388 Heats of Fusion and Vaporization of Some Common Substances Material Copper Gold Iron Lead Mercury Methanol Silver Water (ice) Heat of Fusion, Hf (J/kg) 2.05105 6.30104 2.66105 2.04104 1.", "15104 1.09105 1.04105 3.34105 Heat of Vaporization, Hv (J/kg) 5.07106 1.64106 6.29106 8.64105 2.72105 8.78105 2.36106 2.26106 T a b e s l Appendix D 913 910-918 EM APP D-845813 6/9/04 4:13 PM Page 914 Coefficients of Thermal Expansion at 20\u00b0C Material Coefficient of Linear Expansion, (\u00b0C)1 Coefficient of Volume Expansion, (\u00b0C)1 25106 19106 12106 17106 9106 3106 12106 9106 Solids Aluminum Brass Concrete Copper Glass (soft) Glass (ovenproof) Iron, steel Platinum Liquids Gasoline Mercury Methanol Water Gases Air (and most other gases) 75106 56106 36106 48106 27106 9106 35106 27106 950106 180106 1100106 210106 3400106 Speed of Sound in Various Media Wavelengths of Visible Light Medium Air (0\u00b0) Air (20\u00b0) Helium (0\u00b0) Hydrogen (0\u00b0) Water (25\u00b0) Seawater (0\u00b0) Rubber Copper (25\u00b0) Iron (25\u00b0) Ovenproof glass Diamond /s 331 343 972 1286 1493 1533 1600 3560 5130 5640 12,000 914 Appendix D Color Wavelength in Nanometers (nm) Violet light Indigo light Blue light Cyan light Green light Yellow light Orange light Red light 380\u2013430 nm 430\u2013450 nm 450\u2013500 nm 500\u2013520 nm 520\u2013565 nm 565\u2013590 nm 590\u2013625 nm 625\u2013740 nm Dielectric Constants, K (20\u00b0C) Vacuum Air (1 atm) Neon (1 atm) Glass Quartz Fused quartz Water 1.0000 1.00059 1.00013 4\u20137 4.3 3.75 80 910-918 EM APP D-845813 3/31/04 7:22 AM Page 915 The Planets Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto 0.3302 4.8685 5.9736 0.64185 1898.6 568.46 86.832 102.43 0.0125 2439.7 6051.8 6378.1 3397 71,492 60,268 25,559 24,764 1195 5427 5234 5515 3933 1326 687 1270 1638 1750 Mass (kg1024) Equatorial radius (km) Mean density (kg/m3) Albedo 0.119 0.750 0.306 0.250 0.343 0.342 0.300 0.290 0.4\u20130.6 Semimajor axis (km106) Orbital period (Earth days) Orbital inclination (degrees) Orbital eccentricity Rotational period (hours) Axial tilt (degrees) Average surface temperature (K) 57.91 108.21 149.60 227.92 778.57 1433.53 2872.46 4495.06 5869.66 87.969 224.701 365.256 686.980 4332.589 10,759.22 30,685.4 60,189 90,465 7.00 3.39 0.000 1.850 1.304 2.485 0.772 1.769 17.16 0.2056 0.0067 0.0167 0.0935 0.0489 0.0565 0.0457 0.0113 0.2444 1407.6 5832.5R 23.9345 24.6229 9.9250 10.656 17.24R 16.11 153.2928R 0.01 177.36 23.45 25.19 3.13 26.73 97.77 28.32 122.53 440 737 288 210 112 84 53 55 50 R indicates retrograde motion. The Moon The Sun Mass Equatorial radius Mean density Albedo Semimajor axis Revolution period Synodic (lunar) period Orbital inclination 0.073491024 kg 1738.1 km 3350 kg/m3 0.11 0.3844 km106 27.3217 Earth days 29.53 Earth days 18.28\u00b0\u201328.58\u00b0 Mass Equatorial radius Mean density Absolute magnitude Luminosity Spectral type Rotational period Mean energy production 1,989,1001024 kg 696000 km 1408 kg/m3 4.83 384.61024 J/s G2 V 609.12 h 0.1937103 J/kg Orbital eccentricity 0.0549 Average temperature 5778 K Rotational period 655.728 h T a b e s l Appendix D 915 910-918 EM APP D-845813 3/31/04 7:23 AM Page 916 B8 8", " B7 7 B6 6 B5 5 B4 4 B3 A2 A1 916 Appendix D 910-918 EM APP D-845813 3/31/04 7:24 AM Page 917 Element Symbol Actinium Aluminum Americium Antimony Argon Arsenic Astatine Barium Berkelium Beryllium Bismuth Bohrium Boron Bromine Cadmium Calcium Californium Carbon Cerium Cesium Chlorine Chromium Cobalt Copper Curium Darmstadtium Dubnium Dysprosium Einsteinium Erbium Europium Fermium Fluorine Francium Gadolinium Gallium Germanium Gold Hafnium Hassium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium Lead Lithium Lutetium Magnesium Manganese Meitnerium Mendelevium Ac Al Am Sb Ar As At Ba Bk Be Bi Bh B Br Cd Ca Cf C Ce Cs Cl Cr Co Cu Cm Ds Db Dy Es Er Eu Fm F Fr Gd Ga Ge Au Hf Hs He Ho H In I Ir Fe Kr La Lr Pb Li Lu Mg Mn Mt Md Atomic Number 89 13 95 51 18 33 85 56 97 4 83 107 5 35 48 20 98 6 58 55 17 24 27 29 96 110 105 66 99 68 63 100 9 87 64 31 32 79 72 108 2 67 1 49 53 77 26 36 57 103 82 3 71 12 25 109 101 The Elements Atomic Mass (227) 26.982 (243) 121.760 39.948 74.922 (210) 137.327 (247) 9.012 208.980 (264) 10.811 79.904 112.411 40.078 (251) 12.011 140.116 132.905 35.453 51.996 58.933 63.546 (247) (281) (262) 162.500 (252) 167.259 151.964 (257) 18.998 (223) 157.25 69.723 72.64 196.967 178.49 (277) 4.003 164.930 1.008 114.818 126.904 192.217 55.845 83.798 138.906 (262) 207.2 6.941 174.967 24.305 54.938 (268) (258) Element Symbol Mercury Molybdenum Neodymium Neon Neptunium Nickel Niobium Nitrogen Nobelium Osmium Oxygen Palladium Phosphorus Platinum Plutonium Polonium Potassium Praseodymium Promethium Protactinium Radium Radon Rhenium Rhodium Rubidium Ruthenium Rutherfordium Samarium Scandium Seaborgium Selenium Silicon Silver Sodium Strontium Sulfur Tantalum Technetium Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium Element 111[*] Element 112[*] Element 114[*] Hg Mo Nd Ne Np Ni Nb N No Os O Pd P Pt Pu Po K Pr Pm Pa Ra Rn Re Rh Rb Ru Rf Sm Sc Sg Se Si Ag Na Sr S Ta Tc Te Tb Tl Th Tm Sn Ti W U V Xe Yb Y Zn Zr Uuu Uub Uuq Atomic Number 80 42 60 10 93 28 41 7 102 76 8 46 15 78 94 84 19 59 61 91 88 86 75 45 37 44 104 62 21 106 34 14 47 11 38 16 73 43 52 65 81 90 69 50 22 74 92 23 54 70 39 30 40 111 112 114 Atomic Mass 200.59 95.94 144.24 20.180 (237) 58.693 92.906 14.007 (259) 190.23 15.999 106.42 30.974 195.078 (244) (209) 39.098 140.908 (145) 231.036 (226) (222) 186.207 102.906 85.468 101.07 (261) 150.36 44.956 (266) 78.96 28.086 107.868 22.990 87.62 32.065 180.948 (98) 127.60 158.925 204.383 232.038 168.934 118.710 47.867 183.84 238.029 50.942 131.293 173.04 88.906 65.409 91.224 (272) (285) (289) T a b e s l * Names have not yet been approved by IUPAC. Appendix D 917 910-918 EM APP D-845813 3/31/04 7:24 AM", " Page 918 SAFETY SYMBOLS HAZARD EXAMPLES PRECAUTION REMEDY DISPOSAL BIOLOGICAL Special disposal procedures need to be followed. Organisms or other biological materials that might be harmful to humans Certain chemicals, living organisms Do not dispose of these materials in the sink or trash can. Dispose of wastes as directed by your teacher. Bacteria, fungi, blood, unpreserved tissues, plant materials Avoid skin contact with these materials. Wear mask or gloves. Notify your teacher if you suspect contact with material. Wash hands thoroughly. EXTREME TEMPERATURE Objects that can burn skin by being too cold or too hot Boiling liquids, hot plates, dry ice, liquid nitrogen Use proper protection when handling. Go to your teacher for first aid. SHARP OBJECT FUME ELECTRICAL IRRITANT CHEMICAL Use of tools or glassware that can easily puncture or slice skin Razor blades, pins, scalpels, pointed tools, dissecting probes, broken glass Practice commonsense behavior and follow guidelines for use of the tool. Go to your teacher for first aid. Possible danger to respiratory tract from fumes Ammonia, acetone, nail polish remover, heated sulfur, moth balls Make sure there is good ventilation. Never smell fumes directly. Wear a mask. Leave foul area and notify your teacher immediately. Possible danger from electrical shock or burn Improper grounding, liquid spills, short circuits, exposed wires Double-check setup with teacher. Check condition of wires and apparatus. Do not attempt to fix electrical problems. Notify your teacher immediately. Substances that can irritate the skin or mucous membranes of the respiratory tract Pollen, moth balls, steel wool, fiberglass, potassium permanganate Wear dust mask and gloves. Practice extra care when handling these materials. Go to your teacher for first aid. Chemicals that can react with and destroy tissue and other materials Bleaches such as hydrogen peroxide; acids such as sulfuric acid, hydrochloric acid; bases such as ammonia, sodium hydroxide Wear goggles, gloves, and an apron. Immediately flush the affected area with water and notify your teacher TOXIC Substance may be poisonous if touched, inhaled, or swallowed. Mercury, many metal compounds, iodine, poinsettia plant parts Follow your teacher\u2019s instructions. FLAMMABLE Flammable chemicals may be ignited by open flame, spark, or exposed heat. Alcohol, kerosene, potassium permanganate Avoid open flames and heat when using flammable chemicals. OPEN FLAME Open flame in use, may cause fire. Hair, clothing, paper, synthetic materials Tie back hair and loose clothing. Follow teacher\u2019s instruction on lighting and extinguishing flames. Always wash hands thoroughly after use. Go to your teacher for first aid. Notify your teacher immediately. Use fire safety equipment if applicable. Notify your teacher immediately. Use fire safety equipment if applicable. Eye Safety Proper eye protection should be worn at all times by anyone performing or observing science activities. Clothing Protection This symbol appears when substances could stain or burn clothing. Radioactivity This symbol appears when radioactive materials are used. Handwashing After the lab, wash hands with soap and water before removing goggles. 918 Appendix D 919-926 EM GLOSSARY-845813 3/31/04 7:45 AM Page 919 absorption spectrum (751): The characteristic set of wavelengths absorbed by a gas, which can be used to identify that gas. acceleration (59): The rate at which the velocity of an object changes. acceleration due to gravity (72): The acceleration of an object in free fall, resulting from the influence of Earth\u2019s gravity; acceleration due to gravity on Earth, g, is 9.80 m/s2 toward the center of Earth. accuracy (13): A characteristic of a measured value that describes how well the results of a measurement agree with the \u201creal\u201d value, which is the accepted value, as measured by competent experimenters. achromatic lens (499): A combination of two or more lenses with different indices of refraction (such as a concave lens with a convex lens) that is used to minimize a chromatic aberration. activity (810): The number of decays per second of a radioac- tive substance. adhesive forces (350): The electromagnetic forces of attraction that particles of different substances exert on one another; responsible for capillary action. alpha decay (806): The radioactive decay process in which an alpha () particle is emitted from a nucleus. alpha particles (748): Massive, positively charged atomic particles that move at high speed; are represented by the symbol. ammeter (631): A low-resistance device connected in series that is used to measure the electric current in any branch or part of a circuit. amorphous solid (359): A substance having a definite shape and volume, but lacking a regular crystal structure. ampere (593): A flow of electric", " charge, or electric current, equal to one coulomb per second (1 C/s). amplitude (375): In any periodic motion, the maximum dis- tance an object moves from equilibrium. angular acceleration (199): The change in angular velocity divided by the time needed to make the change; is measured in rad/s2. angular displacement (198): The change in the angle as an object rotates. angular impulse\u2013angular momentum theorem (234): States that the angular impulse on an object is equal to the change in the object\u2019s angular momentum. angular momentum (233): The product of a rotating object\u2019s moment of inertia and its angular velocity; is measured in kgm2/s. angular velocity (198): The angular displacement of an object divided by the time needed to make the displacement. antenna (707): A wire designed to transmit or receive electro- magnetic waves. antinode (389): The point with the largest displacement when two wave pulses meet. apparent weight (98): The force experienced by an object, resulting from all the forces acting on it, giving the object an acceleration. Archimedes\u2019 principle (354): States that an object immersed in a fluid has an upward force on it that equals the weight of the fluid displaced by the object. armature (656): The wire coil of an electric motor, made up of many loops mounted on an axle or shaft; torque on an armature, and the motor\u2019s resultant speed, is controlled by varying the current through the motor. atomic mass unit (800): A unit of mass, u, where u is equal to 1.661027 kg. atomic number (800): The number of protons in an atom\u2019s nucleus. average acceleration (59): The change in an object\u2019s velocity during a measurable time interval, divided by that specific time interval; is measured in m/s2. average power (677): Half the maximum power associated with an alternating current. average speed (44): How fast an object is moving; is the absolute value of the slope of an object\u2019s position-time graph. average velocity (44): The change in position, divided by the time during which the change occurred; is the slope of an object\u2019s position-time graph. l G o s s a r y band theory (776): The theory that electric conduction in solids can be better understood in terms of valance and conduction bands separated by forbidden energy gaps. battery (592): A device made up of several galvanic cells connected together that converts chemical energy to electric energy. beat (418): The oscillation of wave amplitude that results from the superposition of two sound waves with almost identical frequencies. Bernoulli\u2019s principle (357): States that as the velocity of a fluid increases, the pressure exerted by that fluid decreases. beta decay (806): The radioactive decay process in which a neutron is changed to a proton within the nucleus and a beta () particle and antineutrino are emitted. binding energy (802): The energy equivalent of the mass defect; is always negative. buoyant force (354): The upward force exerted on an object immersed in a fluid. capacitance (577): The ratio of an object\u2019s stored charge to its electric potential difference. capacitor (577): An electric device used to store charge that is made up of two conductors separated by an insulator. center of mass (211): The point on the object that moves in the same way that a point particle would move. centrifugal \u201cforce\u201d (216): The apparent force that seems to pull on a moving object, but does not exert a physical outward push on it, and is observed only in rotating frames of reference. centripetal acceleration (153): The center-seeking acceleration of an object moving in a circle at a constant speed. centripetal force (154): The net force exerted toward the center of the circle that causes an object to have a centripetal acceleration. chain reaction (812): Continual process of repeated fission reactions caused by the release of neutrons from previous fission reactions. charging by conduction (547): The process of charging a neutral object by touching it with a charged object. charging by induction (548): The process of charging an object without touching it, which can be accomplished by bringing a charged object close to a neutral object, causing a separation of charges, then separating the object to be charged, trapping opposite but equal charges. chromatic aberration (499): A spherical lens defect in which light passing through a lens is focused at different points, causing an object viewed through a lens to seem to be ringed with color. circuit breaker (627): An automatic switch that acts as a safety device in an electric circuit by opening and stopping the current flow when too much current flows through a circuit. closed system (236): A system that does not gain or lose mass. Glossary 919 9", "19-926 EM GLOSSARY-845813 6/9/04 4:15 PM Page 920 y r a s s o G l closed-pipe resonator (412): A resonating tube with one end closed to air; its resonant frequencies are odd-numbered multiples of the fundamental. coefficient of kinetic friction (127): The slope of a line, k, between two surfaces, relating frictional force to normal force. coefficient of linear expansion (361): The change in length divided by the original length and the change in temperature. coefficient of static friction (127): A dimensionless constant depending on the two surfaces in contact. It is used to calculate the maximum static frictional force that needs to be overcome before motion begins. coefficient of volume expansion (361): The change in volume divided by the original volume and the change in temperature; is about three times the coefficient of linear expansion because solids expand in three directions. coherent light (516): Light from two or more sources, whose additive superposition produces smooth wave fronts. (761): Light whose waves are in step, with coinciding maxima and minima. cohesive forces (349): The electromagnetic forces of attraction that like particles exert on one another; responsible for surface tension and viscosity. combination series-parallel circuit (629): A complex electric circuit that includes both series and parallel branches. combined gas law (345): For a fixed amount of an ideal gas, the pressure times the volume, divided by the Kelvin temperature equals a constant; reduces to Boyle\u2019s law if temperature is constant and to Charles\u2019s law if pressure is constant. complementary color (441): A color of light, which when combined with another color of light, produces white light. components (122): Projections of the component vectors. compound machine (269): A machine consisting of two or more simple machines that are connected so that the resistance force of one machine becomes the effort force of the second machine. Compton effect (733): The shift in the energy of scattered photons. concave lens (493): A diverging lens, thinner at its middle than at its edges, that spreads out light rays passing through it when surrounded by material with a lower index of refraction; produces a smaller, virtual, upright image. concave mirror (464): A mirror that reflects light from its inwardly curving surface and can produce either an upright, virtual image or an inverted, real image. conduction (315): The process by which kinetic energy is transferred when particles collide. conductor (544): A material, such as copper, through which a charge will move easily. consonance (418): A pleasant set of pitches. convection (317): A type of thermal energy transfer that occurs from the motion of fluid in liquid or gas that is caused by differences in temperature. conventional current (592): A flow of positive charges that move from higher potential to lower potential. convex lens (493): A converging lens, thicker at its center than at its edges, that refracts parallel light rays so the rays meet at a point when surrounded by material with a lower index of refraction; can produce a smaller, inverted, real image, or a larger, upright, virtual image. convex mirror (471): A mirror that reflects light from its outwardly curving surface and produces an upright, reduced, virtual image. coordinate system (34): A system used to describe motion that gives the zero point location of the variable being studied and the direction in which the values of the variable increase. Coriolis \u201cforce\u201d (217): The apparent force that seems to deflect a moving object from its path and is observed only in rotating frames of reference. coulomb (549): The SI standard unit of charge; one coulomb, C, is the magnitude of the charge of 6.241018 electrons or protons. Coulomb\u2019s law (549): States that the force between two charges varies directly with the product of their charge and inversely with the square of the distance between them. crest (383): The high point of a wave. critical angle (489): The certain angle of incidence in which the refracted light ray lies along the boundary between two media. crystal lattice (359): A fixed, regular pattern formed when the temperature of a liquid is lowered, the average kinetic energy of its particles decreases and, for many solids, the particles become frozen but do not stop moving and instead, vibrate around their fixed positions. de Broglie wavelength (735): The wavelength associated with a moving particle. decibel (406): The unit of measurement for sound level; also can describe the power and intensity of sound waves. dependent variable (15): The factor in an experiment that depends on the independent variable. depletion layer (784): The region around a pn-junction diode where there are no charge carriers and electricity is poorly conducted. dielectrics (707): Nonconducting materials\u2014glass, air, and water\u2014", "through which electromagnetic waves move at less than the speed of light in a vacuum. diffraction (439): The bending of light around a barrier. diffraction grating (527): A device consisting of large numbers of single slits that are quite close together, diffract light, and form a diffraction pattern that is an overlap of single-slit diffraction patterns; can be used to precisely measure light wavelength or to separate light of different wavelengths. diffraction pattern (524): A pattern on a screen of construc- tive and destructive interference of Huygens\u2019 wavelets. diffuse reflection (459): A scattered, fuzzy reflection pro- duced by a rough surface. dimensional analysis (6): A method of treating units as algebraic quantities, which can be cancelled; can be used to check that an answer will be in the correct units. diode (784): The simplest semiconductor device; conducts charges in one direction only and consists of a sandwich of p-type and n-type semiconductors. dispersion (491): The separation of white light into a spectrum of colors by such means as a glass prism or water droplets in the atmosphere. displacement (36): A change in position having both magnitude and direction; is equal to the final position minus the initial position. dissonance (418): An unpleasant, jarring set of pitches. distance (34): A scalar quantity that describes how far an object is from the origin. domain (650): A very small group, usually 10\u20131000, that is formed when the magnetic fields of the electrons in a group of neighboring atoms are aligned in the same direction. dopants (781): Electron donor or acceptor atoms that can be added in low concentration to intrinsic semiconductors, increasing their conductivity by making either extra electrons or holes available. Doppler effect (407): The change in the frequency of sound caused by the movement of either the source, the detector, or both the detector and the source. Doppler shift (446): The difference between the observed wavelength of light and actual wavelength of light based on the relative speed of the observer and the source of the light. drag force (100): The force exerted by a fluid on an object moving through the fluid; depends on the object\u2019s motion and properties and the fluid\u2019s properties. 920 Glossary 919-926 EM GLOSSARY-845813 3/31/04 7:47 AM Page 921 eddy current (681): A current generated in a piece of metal that is moving through a changing magnetic field, producing a magnetic field that opposes the motion that caused the current. efficiency (268): The ratio of output work to input work. effort force (266): The force a person exerts on a machine. elastic collision (298): A type of collision in which the kinetic energy before and after the collision remains the same. elastic potential energy (291): The potential energy that may be stored in an object, such as a rubber band, as a result of its change in shape. electric circuit (592): A closed loop or pathway that allows electric charges to flow. electric current (592): A flow of charged particles. electric field (563): The field that exists around any charged object; produces forces that can do work, transferring energy from the field to another charged object. electric field lines (567): Lines that provide a picture of an electric field, indicate the field\u2019s strength by the spacing between the lines, never cross, and are directed toward negative charges and away from positive charges. electric generator (675): A device that converts mechanical energy to electric energy and consists of a number of wire loops placed in a strong magnetic field. electric motor (656): An apparatus that converts electric ener- gy into rotational kinetic energy. electric potential difference (569): The change in potential energy per unit charge in an electric field. electromagnet (649): A magnet created when current flows through a wire coil. electromagnetic induction (672): The process of generating a current through a circuit due to the relative motion between a wire and a magnetic field when the wire is moved through the magnetic field or the magnetic field moves past the wire. electromagnetic radiation (710): Energy that is carried, or radiated, in the form of electromagnetic waves. electromagnetic spectrum (709): The entire range of frequencies and wavelengths that make up all forms of electromagnetic radiation, including radio waves, microwaves, visible light, and X rays. electromagnetic wave (706): Coupled, changing electric and magnetic field that travels through space. electromotive force (673): The potential difference, measured in volts, given to the charges by a battery; is abbreviated EMF. electron cloud (761) The region in which there is a high prob- ability of finding an electron. electroscope (546): A device that is used to detect electric charges and consists of a metal knob connected by a metal stem to two thin metal leaves. electrostatics (541): The", " study of electric charges that can be collected and held in one place. elementary charge (550): The magnitude of the charge of an electron. emission spectrum (724): A plot of the intensity of light emitted from a hot body over a range of frequencies. energy (258): The ability of an object to produce a change in itself or in the world around it. energy level (753): The quantized amount of energy that an atom may have in each level. entropy (328): A measure of the disorder in a system. equilibrant (131): A force that places an object in equilibrium; is the same magnitude as the resultant, but opposite in direction. equilibrium (95): The condition in which the net force on an object is zero. equipotential (570): The electric potential difference of zero between two or more positions in an electric field. equivalent resistance (619): For resistors in a series, is the sum of all the individual resistances. excited state (753): Any energy level of an atom that is high- er than its ground state. extrinsic semiconductors (781): Semiconductors with greatly enhanced conductivity resulting from the addition of dopants. l G o s s a r y farsightedness (501): A vision defect in which a person cannot see close objects clearly because images are focused behind the retina; can be corrected with a convex lens. first law of thermodynamics (326): States that the change in thermal energy of an object is equal to the heat that is added to the object, minus the work done by the object. first right-hand rule (648): A method used to determine the direction of a magnetic field relative to the direction of conventional current. fission (811): The process in which a nucleus is divided into two or more fragments, and neutrons and energy are released. fluids (342): Materials (liquids and gases) that flow and have no definite shape of their own. focal length (464): The position of the focal point with respect to the mirror along the principal axis. focal point (464): The point where incident light rays that are parallel to the principal axis converge after reflecting from the mirror. force (88): A push or pull exerted on an object that causes a change in motion; has both direction and magnitude and may be a contact or a field force. force carriers (818): Particles that transmit, or carry forces between matter. fourth right-hand rule (672): A method used to determine the direction of the forces on the charges in a conductor that is moving in a magnetic field. free-body diagram (89): A physical model that represents the forces acting on a system. free fall (72): The motion of a body when air resistance is negligible and the motion can be considered due to the force of gravity alone. frequency (384): The number of complete oscillations that a wave makes each second; is measured in hertz, Hz. fundamental (417): For a musical instrument, the lowest fre- quency of sound that will resonate. fuse (627): A short piece of metal that acts as a safety device in an electric circuit by melting and stopping the current from flowing if a dangerously high current passes through the circuit. fusion (813): The process in which nuclei with small masses combine to form a nucleus with a larger mass and energy is released. galvanometer (655): A device that is used to measure very small currents; can be used as a voltmeter or an ammeter. gamma decay (806): The radioactive decay process in which there is a redistribution of energy within the nucleus, but no change in atomic mass or charge. gravitational field (183): The field that surrounds any object with mass; equals the universal gravitational constant, times the mass of the object, divided by the square of the distance from the object\u2019s center. Glossary 921 919-926 EM GLOSSARY-845813 3/31/04 7:48 AM Page 922 y r a s s o G l gravitational force (175): The attractive force between two objects that is directly proportional to the mass of the objects. impulse-momentum theorem (230): States that the impulse on an object equals the object\u2019s final momentum minus the object\u2019s initial momentum. gravitational mass (184): The size of the gravitational force incident wave (387): A wave that strikes a boundary between between two objects. two media. gravitational potential energy (288): The stored energy in a system resulting from the gravitational force between Earth and the object. ground-fault interrupter (627): A device that contains an electronic circuit that detects small current differences caused by an extra current path; it opens the circuit, prevents electrocution, and often is required as a safety measure for bathroom, kitchen, and exterior outlets. grounding (548): The process of removing excess charge by touching an object to Earth. ground state (753): State of an atom with the", " smallest allow- able amount of energy. incoherent light (515): Light with unsynchronized wave fronts that illuminates objects with an even, white light. (761): Light whose waves are out of step, with their maxima and minima not coinciding. independent variable (15): The factor that is changed or manipulated during an experiment. index of refraction (486): For a medium, is the ratio of the speed of light in a vacuum to the speed of light in that medium. inelastic collision (298): A type of collision in which the kinetic energy after the collision is less than the kinetic energy before the collision. inertia (95): The tendency of an object to resist change. inertial mass (183): A measure of an object\u2019s resistance to any type of force. instantaneous acceleration (59): The change in an object\u2019s velocity at a specific instant of time. instantaneous position (40): The position of an object at any half-life (809): The time required for half the atoms in a given particular instant in time. quantity of a radioactive isotope to decay. harmonics (417): Higher frequencies, which are odd-numbered multiples of the fundamental frequency; give certain musical instruments their own unique timbre. heat (317): Energy transferred between two objects in contact with one another and always flows from the hotter object to the cooler object. heat engine (326): A device that continuously converts thermal energy to mechanical energy; requires a high-temperature thermal energy source, a low-temperature receptacle (a sink), and a way to convert the thermal energy into work. heat of fusion (324): The amount of heat required to change 1 kg of a substance from a solid state to a liquid state at its melting point. heat of vaporization (324): The amount of heat required to change 1 kg of a substance from a liquid state to a gaseous state at its boiling point. Heisenberg uncertainty principle (737): States that it is impossible to simultaneously measure the exact position and momentum of a particle of light or matter. Hooke\u2019s law (376): States that the force acting on a spring is directly proportional to the amount that the spring is stretched. hypothesis (8): An educated, testable guess about how vari- ables are related. instantaneous velocity (46): A measure of motion that tells the speed and direction of an object at a specific instant in time. insulator (544): A material, such as glass, through which a charge will not move easily. interaction pair (102): A pair of forces that are equal in strength, but opposite in direction. interference (388): Results from the superposition of two or more waves; can be constructive (wave displacements in the same direction) or destructive (waves with equal but opposite amplitudes). interference fringes (516): A pattern of light and dark bands on a screen, resulting from the constructive and destructive interference of light waves passing through two narrow, closely spaced slits in a barrier. intrinsic semiconductors (780): Pure semiconductors that conduct charge as a result of thermally freed electrons and holes. inverse relationship (18): A hyperbolic relationship that exists when one variable depends on the inverse of the other variable. isolated system (237): A closed system on which the net exter- nal force is zero. isotope (701): Each of the differing forms of the same atom that have different masses, but have the same chemical properties. ideal gas law (345): For an ideal gas, the pressure times the volume is equal to the number of moles, times the constant, R, and the Kelvin temperature; predicts the behavior of gases remarkably well unless under high-pressure or lowtemperature conditions. ideal mechanical advantage (267): For an ideal machine, is equal to the displacement of the effort force, divided by displacement of the load. illuminance (433): The rate at which light strikes a surface, or falls on a unit area; is measured in lumens per square meter, lm/m2, or lux, lx. illuminated source (432): An object, such as the Moon, that becomes visible as a result of the light reflecting off it. image (461): The combination of image points in a plane mir- ror from which the reflected object seems to originate. impulse (230): The product of the average net force on an object and the time interval over which the force acts. joule (259): Unit of energy, J; 1 J of work is done when a force of 1 N acts on an object over a displacement of 1 m. Kepler\u2019s first law (172): States that the planets move in ellip- tical paths, with the Sun at one focus. Kepler\u2019s second law (173): States that an imaginary line from the Sun to a planet sweeps out equal areas in equal time intervals. Kepler\u2019s third law (173): States that the square", " of the ratio of the periods of any two planets is equal to the cube of the ratio of their average distances from the Sun. 922 Glossary 919-926 EM GLOSSARY-845813 6/9/04 4:16 PM Page 923 l G o s s a r y kilowatt-hour (604): An energy unit used by electric companies to measure energy sales; 1 kWh is equal to 1000 watts, W, delivered continuously for 3600 s (1 h). kinetic energy (258): The energy of an object, resulting from its motion. kinetic friction (126): The force exerted on one surface by a second surface when the two surfaces rub against one another because one or both of the surfaces are moving. laser (762): A device that produces powerful, coherent, directional, monochromatic light that can be used to excite other atoms; the acronym stands for light amplification by stimulated emission of radiation. law of conservation of angular momentum (243): States that if there are no net external torques on an object, then its angular momentum is conserved. law of conservation of energy (293): States that in a closed, isolated system, energy is not created or destroyed, but rather, is conserved. law of conservation of momentum (237): States that the momentum of any closed, isolated system does not change. law of reflection (391): States that the angle of incidence is equal to the angle of reflection. (458): States that the angle a reflected ray makes, as measured from the normal to a reflective surface, equals the angle the incident ray makes, as measured from the same normal. law of universal gravitation (175): States that gravitational force between any two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. lens (493): A piece of transparent material, such as glass or plastic, that is used to focus light and form an image. Lenz\u2019s law (679): States that an induced current always is produced in a direction such that the magnetic field resulting from the induced current opposes the change in the magnetic field that is causing the induced current. leptons (818): Any of a family of particles like electrons and neutrinos. lever arm (201): The perpendicular distance from the axis of rotation to the point where force is exerted. linear relationship (16): A type of relationship that exists between two variables whose graphed data points lie on a straight line. line of best fit (15): A line that best passes through or near graphed data; used to describe data and predict where new data will appear on the graph. longitudinal wave (381): A mechanical wave in which the disturbance is in the same direction, or parallel to, the direction of wave motion. loudness (406): Sound intensity as sensed by the ear and interpreted by the brain; depends mainly on the pressure wave\u2019s amplitude. luminous flux (433): The rate at which light energy is emitted from a luminous source; is measured in lumens, lm. luminous source (432): An object, such as the Sun or an incandescent lamp, that emits light. machine (266): A tool that makes work easier (but does not change the amount of work) by changing the magnitude or the direction of the force exerted to do work. magnetic field (645): The area around a magnet, or around any current-carrying wire or coil of wire, where a magnetic force exists. magnetic flux (646): The number of magnetic field lines that pass through a surface. magnification (468): The amount that an image is enlarged or reduced in size, relative to the object. magnitude (35): A measure of size. Malus\u2019s law (444): States that the intensity of light coming out of a second polarizing filter equals the intensity of polarized light coming out of a first polarizing filter, times the cosine, squared, of the angle between the polarizing axes of the two filters. mass defect (802): The difference between the sum of the masses of individual nucleons and the actual mass. mass number (800): The sum of the numbers of neutrons and protons in an atom\u2019s nucleus. mass spectrometer (701): Device that uses both electric and magnetic fields to measure the masses of ionized atoms and molecules and can determine the charge-to-mass ratio of an ion. measurement (11): A comparison between an unknown quantity and a standard. mechanical advantage (266): The ratio of resistance force to effort force. mechanical energy (293): The sum of kinetic and gravitation- al potential energy of a system. microchip (788): An integrated circuit consisting of thousands of diodes, transistors, resistors, and conductors. mirror equation (467): Relates the focal length, the object position, and the image position of a spherical mirror. moment of inertia (205): The resistance to rotation. momentum (230):", " The product of the object\u2019s mass and the object\u2019s velocity; is measured in kgm/s. monochromatic light (516): Light having only one wave- length. motion diagram (33): A series of images showing the positions of a moving object taken at regular (equal) time intervals. mutual inductance (682): Effect in which a changing current through a transformer\u2019s primary coil creates a changing magnetic field, which is carried through the core to the transformer\u2019s secondary coil, where the changing field induces a varying EMF. nearsightedness (501): A vision defect in which a person cannot see distant objects clearly because images are focused in front of the retina; can be corrected with a concave lens. net force (92): The vector sum of all the forces on an object. neutral (543): An atom whose positively charged nucleus exactly balances the negative charge of the surrounding electrons. Newton\u2019s first law (94): States that an object at rest will remain at rest, and a moving object will continue moving in a straight line with constant speed, if and only if the net force acting on that object is zero. Newton\u2019s second law (93): States that the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object being accelerated. Newton\u2019s second law for rotational motion (208): States that the angular acceleration of an object is directly proportional to the net torque and inversely proportional to the moment of inertia. Newton\u2019s third law (102): States that all forces come in pairs and that the two forces in a pair act on different objects and are equal in strength and opposite in direction. node (389): The stationary point where two equal wave pulses meet and are in the same location, having a displacement of zero. Glossary 923 919-926 EM GLOSSARY-845813 6/9/04 4:16 PM Page 924 y r a s s o G l normal (391): The line in a ray diagram that shows the direction of the barrier and is drawn at a right angle, or perpendicular, to the barrier. normal force (107): The perpendicular contact force exerted by a surface on another object. nuclear reaction (807): Occurs when the number of neutrons or protons in a nucleus changes; can be produced by bombardment of nuclei by gamma rays, other nuclei, protons, neutrons, alpha particles, or electrons. nucleons (802): Protons and neutrons. nucleus (749): The tiny, massive, positively charged central core of an atom. nuclide (801): The nucleus of an isotope. object (461): A luminous or illuminated source of light rays that are to be reflected by a mirrored surface. observed light frequency (446): The frequency of light as seen by an observer. opaque (433): A medium that absorbs light and reflects some light rather than transmitting it, preventing objects from being seen through it. open-pipe resonator (412): A resonating tube with both ends open that also will resonate with a sound source; its resonant frequencies are whole-number multiples of the fundamental. origin (34): The point at which both variables in a coordinate system have the value zero. pair production (820): The conversion of energy into a matter- antimatter pair of particles. parallel circuit (623): A type of electric circuit in which there are several current paths; its total current is equal to the sum of the currents in the individual branches, and if any branch is opened, the current in the other branches remains unchanged. parallel connection (600): A type of connection in which the circuit component and the voltmeter are aligned parallel to one another in the circuit, the potential difference across the voltmeter equals the potential difference across the circuit element, and there are two or more current paths to follow. particle model (33): A simplified version of a motion diagram in which the moving object is replaced by a series of single points. pascal (342): The SI unit of pressure. Pascal\u2019s principle (352): States that any change in pressure applied at any point on a confined fluid is transmitted undiminished throughout the fluid. pendulum (378): A device that can demonstrate simple harmonic motion when its bob (a massive ball or weight), suspended by a string or light rod, is pulled to one side and released, causing it to swing back and forth. period (375): In any periodic motion, the amount of time required for an object to repeat one complete cycle of motion. periodic motion (375): Any motion that repeats in a regular cycle. periodic wave (381): A mechanical wave that moves up and down at the same rate. photoelectric effect (726): The emission of electrons by certain metals that is produced when they are exposed to electromagnetic radiation. photon (727): A discrete, quantized bundle of radiation that travels at the speed of light, has zero mass, and", " has energy and momentum. 924 Glossary physics (3): The study of matter and energy and their relation- ships. piezoelectricity (711): The property of a crystal that causes it to bend or deform, producing electric vibrations, when a voltage is applied across it. pitch (406): The highness or lowness of a sound wave, which depends on the frequency of vibration. plane mirror (461): A flat, smooth surface from which light is reflected by regular reflection, producing a virtual image that is the same size as the object, has the same orientation, and is the same distance from the mirror as the object. plasma (348): A gaslike, fluid state of matter made up of negatively charged electrons and positively charged ions that can conduct electricity; makes up most of the matter in the universe, such as stars. polarization (443): Light whose waves oscillate only in a single plane. (644): For a magnet, describes the property of having two distinct, opposite ends, one of which is a northseeking pole and the other of which is a south-seeking pole. position (34): The separation between an object and the origin; it can be either positive or negative. position-time graph (38): A graph that can be used to determine an object\u2019s velocity and position, as well as where and when two objects meet, by plotting the time data on a horizontal axis and the position data on a vertical axis. power (263): The work done, divided by the time needed to do the work. precision (12): A characteristic of a measured value describing the degree of exactness of a measurement. pressure (342): The force on a surface, divided by the surface\u2019s area. primary coil (682): An electrically insulated transformer coil, which, when connected to an AC voltage source, induces an alternating EMF in the secondary coil. primary color (440): Red, green, and blue, which can be combined to form white light and mixed in pairs to produce the secondary colors: yellow, cyan, and magenta. primary pigment (441): Cyan, magenta, and yellow, each of which absorbs one primary color from white light and reflects two primary colors; can be mixed in pairs to produce the secondary pigments: red, green, and blue. principal axis (464): A straight line perpendicular to the surface of a mirror that divides the mirror in half. principle of superposition (388): States that the displacement of a medium caused by two or more waves is the algebraic sum of the displacements of the individual waves. principal quantum number (755): The integer, n, that determines the quantized values of r and E for an electron\u2019s orbital radius in hydrogen and the energy of a hydrogen atom\u2014the radius, r, increases as the square of n, whereas the energy, E, depends on 1/n2. projectile (147): An object shot through the air, such as a football, that has independent vertical and horizontal motions and, after receiving an initial thrust, travels through the air only under the force of gravity. quadratic relationship (17): A parabolic relationship that results when one variable depends on the square of another variable. quantized (725): Energy exists only in bundles of specific amounts. quantum mechanics (761): The study of the properties of matter, using its wave properties. quantum model (760): A model that predicts only the proba- bility that an electron is in a particular region. quarks (818): Tiny particles that make up protons, neutrons, and pions. 919-926 EM GLOSSARY-845813 3/31/04 8:00 AM Page 925 l G o s s a r y radian (197): 1 of a revolution; abbreviated \u201crad.\u201d 2 radiation (317): The thermal transfer of energy by electromag- netic waves through the vacuum of space. radioactive (806): Materials that undergo radioactive decay and emit penetrating rays. ray (390): A line that can show the direction a wave is travel- ing and is drawn at a right angle to a wave\u2019s crest. ray model of light (432): A model that represents light as a ray that travels in a straight path, whose direction can be changed only by putting an obstruction in the path. Rayleigh criterion (530): States that if the central bright spot of one image falls on the first dark ring of the second image, the images are at the limit of resolution. real image (465): An inverted optical image that is smaller than the object and is formed by the converging of light rays. receiver (712): A device used to obtain information from electromagnetic waves; consists of an antenna, a coil-and-capacitor circuit, a detector to decode the signal, and an amplifier. reference level (288): The position where gravitational poten- tial energy is defined as zero. reflected wave (", "387): An erect or inverted returning wave that results from some of the energy of the incident wave\u2019s pulse being reflected backward. refraction (391): The change in direction of waves at the boundary between two different media. resistance (595): A property that determines how much cur- rent will flow; is equal to voltage divided by current. resistance force (266): The force exerted by a machine. resistor (596): A device with a specific resistance; may be made of long, thin wires; graphite; or semiconductors and often is used to control the current in circuits or parts of circuits. resonance (380): A special form of simple harmonic motion that occurs when small forces are applied at regular intervals to an oscillating or vibrating object and the amplitude of the vibration increases. resultant (35): A vector that results from the sum of two other vectors; it always points from the first vector\u2019s tail to the last vector\u2019s tip. rotational kinetic energy (287): The kinetic energy of an object, proportional to the object\u2019s moment of inertia and the square of its angular velocity. scalars (35): Quantities, such as temperature or distance, that are just numbers without any direction. scientific law (9): A well-established rule about the natural world that sums up, but doesn\u2019t explain, a pattern in nature. scientific method (8): A systematic method of observing, experimenting, and analyzing to answer questions about the natural world. scientific theory (10): An explanation based on numerous observations, supported by experimental results, that may explain why things work the way they do. secondary coil (682): An electrically insulated transformer coil in which an alternating EMF is induced by an AC current through the primary coil. secondary color (440): Yellow, cyan, and magenta, each of which is produced by combining two primary colors. secondary pigment (441): Red, green, and blue, each of which absorbs two primary colors from white light and reflects one primary color; can be produced by mixing pairs of cyan, magenta, and yellow pigments. second law of thermodynamics (330): States that natural processes go in a direction that maintains or increases the total entropy of the universe. second right-hand rule (649): A method used to determine the direction of the field produced by an electromagnet, relative to the flow of conventional current. self-inductance (682): The inductance of EMF in a wire carry- ing a changing current. semiconductors (775): Conductive materials, such as silicon or germanium, which, when made into solid-state devices, can amplify and control weak electric signals through electron movement within a tiny crystalline space. series circuit (618): A type of electric circuit in which all current travels through each device and is the same everywhere; its current is equal to the potential difference divided by the equivalent resistance. series connection (600): A type of connection in which there is only a single current path. short circuit (627): Occurs when a very low resistance circuit is formed, causing a very large current that could easily start a fire from overheated wires. significant digits (7): All the valid digits in a measurement, the number of which indicates the measurement\u2019s precision. simple harmonic motion (375): A motion that occurs when the restoring force on an object is directly proportional to the object\u2019s displacement from equilibrium. Snell\u2019s law of refraction (486): States that the product of the index of refraction of a medium and the sine of the angle of incidence equals the product of the index of refraction of a second medium and the sine of the angle of refraction. solenoid (649): A long coil of wire with many loops; fields from each loop add to the fields of the other loops, creating a greater total field strength. sound level (406): A logarithmic scale that measures amplitudes; depends on the ratio of the pressure variation of a particular sound wave to the pressure variation in the most faintly heard sound; unit of measurement is the decibel, dB. sound wave (404): A pressure variation transmitted through matter as a longitudinal wave; it reflects and interferes and has frequency, wavelength, speed, and amplitude. specific heat (318): The amount of energy that must be added to a material to raise the temperature of a unit mass by one temperature unit; is measured in J/kgK. specular reflection (459): A reflection produced by a smooth surface in which parallel light rays are reflected in parallel. spherical aberration (467): The image defect of a spherical mirror that does not allow parallel light rays far from the principal axis to converge at the focal point, and produces an image that is fuzzy, not sharp. Standard Model (818): A model of the building blocks of matter in which all particles can be grouped into three families: quarks, leptons, and force carriers. standing wave (389): A", " wave that appears to be standing still, produced by the interference of two traveling waves moving in opposite directions. static friction (126): The force exerted on one surface by a second surface when there is no motion between the two surfaces. step-down transformer (682): A type of transformer in which the voltage coming out of the transformer is smaller than the voltage put into the transformer. step-up transformer (682): A type of transformer in which the secondary voltage coming out of the transformer is larger than the primary voltage going into the transformer. stimulated emission (762): The process that occurs when an excited atom is struck by a photon having energy equal to the energy difference between the excited state and the ground state\u2014the atom drops to the ground state and emits a photon with energy equal to the energy difference between the two states. streamlines (358): Lines representing the flow of fluids around objects. Glossary 925 919-926 EM GLOSSARY-845813 3/31/04 8:02 AM Page 926 vector resolution (122): The process of breaking a vector into its components. vectors (35): Quantities, such as position, that have both mag- nitude and direction. velocity-time graph (58): A graph that can be used to plot the velocity of an object versus time and to determine the sign of an object\u2019s acceleration. virtual image (461): The image formed of diverging light rays; is always on the opposite side of the mirror from the object. volt (569): The unit equal to one joule per coulomb, 1 J/C. voltage divider (620): A series circuit that is used to produce a voltage source of desired magnitude from a higher-voltage battery; often is used with sensors such as photoresistors. voltmeter (631): A high-resistance device used to measure the voltage drop across any portion or a combination of portions of a circuit and is connected in parallel with the part of the circuit being measured. watt (263): Unit of power, W; 1 J of energy transferred in 1 s. wave (381): A disturbance that carries energy through matter or space; transfers energy without transferring matter. wave front (390): A line representing the crest of a wave in two dimensions that can show the wavelength, but not the amplitude, of the wave when drawn to scale. wavelength (383): The shortest distance between points where the wave pattern repeats itself, such as from crest to crest or from trough to trough. wave pulse (381): A single disturbance or pulse that travels through a medium. weak nuclear force (821): A weak force acting in the nucleus in beta,, decay. weightlessness (98): An object\u2019s apparent weight of zero that results when there are no contact forces pushing up on the object. work (258): The transfer of energy by mechanical means; is done when a constant force is exerted on an object in the direction of motion, times the object\u2019s displacement. work-energy theorem (258): States that when work is done on an object, a change in kinetic energy occurs. work function (731): The energy required to free the most weakly bound electron from a metal; measured by the threshold frequency in the photoelectric effect. X ray (713): High-frequency electromagnetic wave emitted by rapidly accelerated electrons. strong nuclear force (802): An attractive force that binds the nucleus together; is of the same strength between protons and protons, protons and neutrons, and neutrons and neutrons. superconductor (603): A material with zero resistance that can conduct electricity without a loss of energy. surface wave (382): A mechanical wave in which the particles move both parallel and perpendicular to the direction of wave motion. y r a s s o G l tension (105): The specific name for the force exerted by a rope or a string. terminal velocity (101): The constant velocity of an object that is reached when the drag force equals the force of gravity. thermal energy (295): A measure of the internal motion of an object\u2019s particles. thermal equilibrium (315): The state in which the rate of energy flow between two objects is equal and the objects are at the same temperature. thermal expansion (347): A property of all forms of matter that causes the matter to expand, becoming less dense, when heated. thin-film interference (520): A phenomenon in which a spectrum of colors is produced due to the constructive and destructive interference of light waves reflected in a thin film. thin lens equation (493): States that the inverse of the focal length of a sperical lens equals the sum of the inverses of the image position and the object position. third right-hand rule (652): A method that can be used to determine the direction of the force on a current-carrying wire in a magnetic field. threshold frequency (726): The certain minimum value at or above which the frequency of radiation causes the ejection of electrons", " from a metal. time interval (36): The difference between two times. torque (202): A measure of how effectively a force causes rota- tion; is equal to the force times the lever arm. total internal reflection (489): Occurs when light traveling through an area with a higher index of refraction to an area with a lower index of refraction hits a boundary at an angle that exceeds the critical angle and all light reflects back into the area with the higher index of refraction. trajectory (147): The path of a projectile through space. transformer (682): A device that can decrease or increase the voltages in AC circuits with relatively little energy loss. transistor (787): A simple device made of doped semiconducting material that can act as an amplifier, converting a weak signal to a much stronger one. translucent (433): A medium that transmits light and also can reflect a fraction of the light, but does not allow objects to be seen clearly through it. transparent (433): A medium that transmits light and also can reflect a fraction of the light, allowing objects to be seen clearly through it. transverse wave (381): A mechanical wave that vibrates per- pendicular to the direction of the wave\u2019s motion. trough (383): The low point of a wave. uniform circular motion (153): The movement of an object or particle trajectory at a constant speed around a circle with a fixed radius. 926 Glossary 927-942 EM INDEX-845813 6/10/04 9:15 PM Page 927 I n d e x Aberration, chromatic, 499; in Hubble Space Telescope, 467; spherical, 467, 498 Absolute value, 841 Absolute zero, 316 Absorption spectrum, 751\u2013752 Acceleration, 59\u201364; angular, 199, 199 table; average (a), 59\u201364, 63 prob., 65, 68; centripetal, 153, 154\u2013155; constant, 57 lab, 65\u201371, 68 table; direction of, 63; of Earth, 104 prob.; force and, 88, 90\u201391; gravity and free fall and, 72\u201375; instantaneous, 59; linear vs. angular, 199 table; motion diagrams and, 60, 60 prob.; Newton\u2019s second law and. See Newton\u2019s second law; positive and negative, 61\u201362; velocity-time graphs and, 62; velocity vs., 64 Acceleration due to gravity (g), 72\u201375, 74 prob., 76\u201377 lab; amusement park rides, 74; ball thrown upwards, 72\u201373; pendulums and, 378\u2013379, 379 prob., 392\u2013393 lab; from universal gravitation and Newton\u2019s second law of motion, 182; weightlessness of water in falling cup and, 182 lab Accelerators, linear, 815; synchrotrons, 816 Accommodation, 500 Accuracy, 13\u201314 Achromatic lenses, 499, 502 Activities. See Launch Labs; MiniLabs; Physics Labs Activity, 810 Adaptive optical system (AOS), 476 Additive color process, 440 Adhesive forces, 350 Agent, 88 Air bags, 231 Air columns, resonance in, 412\u2013414 Air, conduction by, 545 Air resistance, drag force and, 100\u2013101; projectile motion and, 152; streamlines and, 358 Aircraft, streamlining of, 358 Alcohol molecules, evaporation of, 351, 364\u2013365 lab Algorithms, 22 ALNICO V, 645 Alpha decay, 806, 808 prob. Alpha particles, 748, 806, 807 table Alpha radiation, 806 Alternating current (AC), 677, 682, 707\u2013708, 709 Alternating current (AC) generators, 677\u2013678 AM radio waves, 697 lab Ammeter, 595, 599, 600, 631, 655, 729 illus. Amorphous solid, 359 Ampere (A), 5 table, 593 Amp\u00e8re, Andr\u00e9-Marie, 650 Amplifier, 788 Amplitude, 375, 382\u2013383, 392\u2013393 lab, 405\u2013407 Amusement-park rides, accelerated frames of reference and, 216; acceleration of free-fall rides, 74; centrifugal \u201cforce\u201d and, 216; confusion of body\u2019s systems by, 138; conservation of mechanical energy by, 294 Analyzers, 444 Angle of incidence, 486, 488; critical, 489; law of reflection and, 458, 459, 460 prob.; total internal reflection and, 489; two-dimensional waves and, 391 Angle of reflection, 391 Angle of refraction, 486, 487 prob.; total internal reflection and, 489\u2013490; wave model of light and, 488 Angle of revolution, 198 Angle of the resultant vector, 123 Angular acceleration, 199 Angular displacement,", " 198 Angular frequency, 200 Angular impulse-angular momentum theorem, 234\u2013235 Angular momentum, 233\u2013235; conservation of, 243\u2013245; of electrons, 755 Angular velocity, 198\u2013199; affect of force on, 201; conservation of angular momentum and, 243\u2013245 Animals, body temperature of, 322 Annihilation, positron/electron, 819\u2013820 Anodes, 726 Antenna, 707\u2013708, 712 Antimatter, 819\u2013820; pair production and, 820; particle conservation and, 820 Antineutrino, 818, 822 Antineutrons, 820 Antinodes, 389 Antiparticles, 819 Antiprotons, 820 Aperture, 530 Apparent weight, 98, 99 prob., 182 Applying Physics, carbon conductors, 544; contact lenses, 501; dragster records, 68; electromagnets, 649; Fosbury-Flop, 212; friction, causes of, 130; geosynchronous orbits, 180; hearing and frequency, 413; illuminance of desktops, 435; nonreflective eyeglasses, 520; ohmmeters, 624; potential energy of an atom, 289; power output of Tour de France racer, 265; relative size of potential energy of atom, 289; resistance, 597; running shoes, 231; speed records, 44; spherical aberration on Hubble Space Telescope, 467; static electricity, damage to electronic devices by, 570; steam heat, 317; transformer ratings, 683 Arago, Jean, 519 Archimedes, 354 Archimedes\u2019 principle, 354\u2013355, 356 prob. Arcs, electric, 556 Area, force and pressure and, 342, 343 prob. Aristotle, 3 lab Arithmetic, significant digits and, 7, 8 prob. Armatures, 656 Arsenic, semiconductors and, 781 Artificial intelligence, 792 Artificial radioactivity, 811 Artificial satellites, Kepler\u2019s laws and orbit of, 173; Landsat 7, 180; limits on mass of, 180; orbital speed and period of, 180, 181 prob. Astrolabe, 172 illus. Astronauts, weightlessness of in space, 182 Astronomy, expansion of the universe, 447; Hubble Space Telescope, 467; use of Doppler effect by, 410 Atmosphere, convection in, 317 Atmospheric pressure, 343, 343 table Atom laser, 768 Atom smashing, 799 Atomic clocks, 50, 445 Atomic mass unit, 800 Atomic number (Z), 800 Atomic particles, 748 Atomic spectra, absorption spectrum, 751\u2013752; Balmer series and, 756 illus.; determining atomic structure from, 752; emission spectra, 749\u2013750, 752, 755 lab Atomic structure, Bohr model of, 752\u2013756, 757 prob., 758 prob., 759, 759 prob.; nuclear model of, 748\u2013752, 766\u2013767 lab; quantum model of, 760\u2013761 Atomizers, Bernoulli\u2019s principle and, 357 Atoms, 747\u2013765; absorption spectrum and, 751\u2013752; charge of, 543\u2013544; electron orbital radius, 755; emission spectra from, 749\u2013750, 752, 755 lab; energy of emitted photons from, 753\u2013754; energy of, total, 755, 757 prob.; energy of vibration of (E), 725; excitation of, 761\u2013762; lasers from excited. See Lasers; mass of, 749; neutral, 543; quantized energy levels in, 753\u2013754; relative size of potential energy of, 289; spinning coins as model of, 747 lab Audiocassette recorders, 651 Auditory canal, 405 Automobile carburetors, Bernoulli\u2019s principle and, 357 Automobile engines, first law of thermodynamics and, 326\u2013327; spark plugs in, 662 Automobile transmissions, ideal mechanical advantage (IMA) of, 272 Automobiles, collisions of, 231; elastic potential energy of bumpers, 378; hybrid cars, 608; speed of, 20\u201321 lab; streamlining of, 358 Average acceleration (a), 59\u201364, 63 prob., 65, 68 Average power, 677 Average speed, 44, 45 prob. Average velocity, 44, 45 prob.; equation for, 44; motion diagrams and, 46\u201347 Avogadro, Amedeo, 345 Avogadro\u2019s number, 345 Axis, principle, 464 Axles, 269 Index 927 927-942 EM INDEX-845813 6/10/04 9:16 PM Page 928 x e d n I Back-EMF, 680\u2013681 Balanced electric circuits, 628 prob. Balances, Lenz\u2019s law and, 680", "\u2013681 Balloons, charging of, 563 lab Balls, acceleration of one thrown upward, 72\u201373; change in momentum of after collision, 231; energy change from throwing and catching (workenergy theorem), 286\u2013287; flight of one thrown at an initial angle, 151 prob.; momentum of two colliding, 236 Balmer series, 756 illus. Band theory of solids, 776\u2013777 Bands, conduction, 776\u2013777; valence, 776\u2013777 Base, transistor, 787 Base units, SI, 5, 5 table Bathroom scales, 96, 98, 110 Bats, Doppler effect and, 410 Batteries, 592 Beats, 418 BEC. See Bose-Einstein condensates Becquerel, Henri, 799, 806 Bernoulli, Daniel, 357, 546 Bernoulli\u2019s principle (velocity and pressure of moving fluid), 357 Beta decay, 806, 808 prob. See also Radioactive decay; weak interactions and, 821\u2013822 Beta particles, 806, 807 table Beta radiation, 806. See also Radiation; distance and intensity of, 824\u2013825 lab Bicycle gears, ideal mechanical advantage (IMA), 270, 270 lab, 271 prob., 272, 276 Bicycles, 270, 276; ideal mechanical advantage (IMA) of, 270, 270 lab, 271 prob., 272; mechanical advantage (MA) of, 270, 272; multi-gear, 272; power output of Tour de France racer, 265 Big Bang Theory, 9 illus. Billiard ball collisions, 301 Bimetallic strips, 363 Binary numbers, 22 Binding energy, nuclear, 799 lab, 802\u2013804, 804 prob. Binoculars, lenses in, 502 Black holes, 185, 188 Blood pressure, 343 table Blue-shifted light, 446 Bob, pendulum, 378 Body temperature, regulation of, 322 Bohr model of the atom, 752\u2013756, 759; energy of emitted photons, 753\u2013754, 758 prob.; ionization energy and, 759; Newton\u2019s second law and, 754\u2013755; orbital energy values, 756, 757 prob.; orbital radius and, 755, 760; problems with, 754, 759 prob.; quantized energy levels in, 753\u2013754 Bohr, Niels, 752\u2013753, 754\u2013755, 759 Bohr radius, 755 Boiling point, 323\u2013324 Boltzmann\u2019s constant (k), 345 Bose-Einstein condensates, 366, 768 Bose, Satyendra Nath, 366, 768 Bosons, 768 Bottom quarks, 819 illus. Bow and arrow, elastic potential energy and, 291 Boyle, Robert, 344 Boyle\u2019s law (volume and pressure of a gas), 344, 346 prob. Brahe, Tycho, 171, 172 Brass instruments, source of sound from, 411 Bridges, earthquake-proof, 394 Bright-line emission spectra, 750, 755 lab Brushes, 656 Bubble chambers, 817 Buildings, earthquake-proof, 394 Buoyant force, 341 lab, 354\u2013355, 356 prob. Butterflies, thin-film interference on wing of, 523, 530 Cadmium, nuclear reactors and, 813 Callisto, distance of from Jupiter, 174 prob. Calorimeter, 319, 321 prob. Calorimetry, 319\u2013320, 321 prob. Cameras, lenses in, 503 Cancer, from cellular phones, 716; radiation treatments for, 811 Candela (cd), 5 table, 434 Capacitance (C), 577; charging of capacitors, 580\u2013581 lab; determining, 577, 578, 578 prob.; unit of (farad), 578 Capacitors, 577\u2013579; capacitance (C) of, 577, 578, 578 prob.; charging of, 580\u2013581 lab; controlling capacitance, 579; electromagnetic waves from, 709\u2013710; types of, 577 illus., 579 Capillary action, 350 Carbon, 544 Carbon-14 dating, 11, 810 Carburetors, Bernoulli\u2019s principle and, 357 Careers, physics-related, 3 Carnot, Sadi, 328 Carriers, force, 818 Cars. See Automobiles Cassette tapes, 651 Cathode-ray tube, 657; charge-to-mass ratio of electrons and, 698\u2013699; charge-to-mass ratio of protons and, 699\u2013700 Cathodes, 726 Cavendish, Henry, 177\u2013178, 546 CD players, 764 Cells, photovoltaic, 592 Cellular phones, 716 Celsius, Anders, 316 Celsius scale, 316 Center of mass, 211\u2013212; of", " human body, 212; locating, 211, 213 lab; stability and, 212\u2013213 Centi-, 6 table Centrifugal \u201cforce\u201d, 156, 216 Centripetal acceleration, 153, 154\u2013155, 216 Centripetal force, 154, 155 Centripetal motion, space station design and, 162 Cerenkov effect, 812 illus. Cesium clocks, 50 Chadwick, James, 800 Chain reactions, 812 Challenge Problems, Bohr model of the atom, 759 prob.; capacitance (C), 579 prob.; coiled springs, 380 prob.; collisions and conservation of energy, 300 prob.; concave mirror, locating image on, 470 prob.; Coulomb\u2019s law and electric force, 552 prob.; currentvoltage curve of a diode, 786 prob.; distribution transformers, 685 prob.; electric circuits, balanced, 628 prob.; entropy, 329 prob.; force and motion in one dimension, 100 prob.; free-fall objects, 75 prob.; graphing, 18 prob.; kinetic energy of photoelectron, 731 prob.; lenses and human vision, 501 prob.; linear expansion and, 363 prob.; moment of inertia (I), 208 prob.; motion on inclined planes, 132 prob.; physics of music, 417 prob.; planetary motion, 176 prob.; power and efficiency, 268 prob.; reduction of light intensity when passed through second filter (Malus\u2019s law), 444 prob.; relative-velocity, 157 prob.; single-slit diffraction, 526 prob.; torque of electric motors, 656 prob.; two-dimensional collisions and, 244 prob.; uniform straight-line motion of multiple objects, 40 prob.; voltage across a capacitor, 604 prob. Chandra X-ray Observatory, 188 Changes of state, 323\u2013324; boiling point, 323\u2013324; heat of fusion, 324, 324 table, 325 prob.; heat of vaporization, 324, 324 table; melting point, 323, 324 lab Charge, atomic, 543\u2013544; electrostatic. See Electrostatic charge; elementary, 550; neutral, 543 Charge pumps, 592 Charge-to-mass ratio, of electrons, 698\u2013699; of ion in a mass spectrometer, 702, 703 prob.; of protons, 699\u2013700 Charged objects, 542\u2013544, 554\u2013555 lab. See also Neutral objects; charging by conduction, 547; charging by induction, 548; forces on, 546\u2013548; like charges, 542; opposite charges, 542\u2013543 Charges, 542\u2013544. See also Electric currents; Electrostatics; elementary, 550; like, 542; neutral, 543; opposite, 542\u2013543; separation of, 544, 547; sharing of, 575\u2013576; storing of, 577; unit of (C), 549, 578 Charging by conduction, 547, 549 lab Charging by induction, 548, 549 lab Charles, Jacques, 344 Charles\u2019s law (volume and tempera- ture of a gas), 344 Charm quarks, 819 illus. Chips, 22, 789 Chlorophyll, 442 Chromatic aberration, 499 Circuit breakers, 627\u2013628 Circuits. See Electric circuits Circular motion, uniform. See Uniform circular motion Circular waves, 390\u2013391 928 Index 927-942 EM INDEX-845813 6/10/04 9:17 PM Page 929 Clarinets, sound from, 411, 414, 417 Clocks, cesium, 50; light, 78 Closed-pipe resonators, 412, 413, 414, 416 prob., 420\u2013421 lab Closed systems, 236 Cloud chamber, Wilson, 817 Cochlea, 405 Coefficient of friction (), 127, 129\u2013130, 136\u2013137 lab; kinetic ( 127, 129 table, 136\u2013137 lab; static ( 127, 129 table, 136\u2013137 lab; typical values, 129 table k), s), Coefficient of kinetic friction ( k), 127, 129 table, 136\u2013137 lab Coefficient of linear expansion, 361, 361 table Coefficient of static friction ( s), 127, 129 table, 136\u2013137 lab Coefficient of volume expansion, 361, 361 table Coefficients of thermal expansion, 361, 361 table Coherent light, 516\u2013519, 761; from incoherent with double-slit experiment, 517; interference of, 516\u2013519; from lasers, 763. See also Lasers Cohesive forces, 349; in plants, 350; surface tension and, 349, 350 illus.; viscosity and, 349 Coil-and-capacitor circuit, electromagnetic waves from, 709\u2013710 Coiled springs. See Springs Coils, primary, 682; secondary, 682 Cold-blooded animals, 322 Collider Det", "ector at Fermilab (CDF), 817 Collisions, baseballs, 231; billiard balls, 301; car crashes, 231; conservation of energy and, 297\u2013298, 299 prob., 300 prob., 300\u2013301; conservation of momentum and, 300; elastic, 298, 301 lab; ice skating and, 301; impulse and momentum and, 230\u2013231, 232 prob.; inelastic, 298; superelastic (explosive), 298; two-dimensional, 241, 242 prob.; two-particle, 236, 246\u2013247 lab Color printers, additive pigment process, 442 Color television tubes, additive color process and, 440 Colors, 440\u2013442; additive color process, 440\u2013441, 442; complementary, 441; green plants and chlorophyll, 442; hot and cool, 441 lab; Newton\u2019s study of, 440; primary, 440; secondary, 440; subtractive color process, 441\u2013442; thin-film interference and reinforcement of, 520\u2013521 Combination series-parallel circuits, 629, 630 prob. Combined gas law, 345 Comets, 172 Commutators, 656 Compact discs (CDs), lasers in CD players, 764; as reflection gratings, 515 lab, 527 Compasses, 648 Complementary colors, 441 Complementary pigments, 442 Components, vector, 122. See also Vectors I n d e x Compound machines, 269\u2013270; Convex lenses, 493; focal length and, bicycles, 270, 271 prob., 272, 276; ideal mechanical advantage (IMA) of, 270, 271 prob., 272; mechanical advantage (MA) of, 270, 271 prob. Compton, Arthur Holly, 733, 734 Compton effect, 733\u2013734 Computer animation, 19 illus. Computer storage disks, 659 Computers, 22, 659 Concave lenses, 493, 498; properties of, 494, 494 table; thin lens equation and, 494, 498 Concave mirrors, 464\u2013468, 470; locating image on, 465\u2013468, 469 prob., 470 prob.; vs. other single-mirror systems, 473 table; Problem-Solving Strategies, 466; real image defects on, 467; real image formed by, 466, 474\u2013475 lab; virtual image formed by, 470, 474\u2013475 lab Condensation, 351 Condensation trails, 817 Conduction, 315; band theory of solids and, 776\u2013777; charging by, 547, 549 lab; electrical, 777; heat transfer by, 315, 317; in solids, 775\u2013783 Conduction bands, 776\u2013777 Conductivity, 777 Conductors, electrical, 544\u2013545; air as, 545; electric fields near, 576\u2013577; free electrons in solid, 777, 778 prob. Connecting Math to Physics, 47, 68, 123, 521, 728; adding and subtracting fractions, 468; direct and inverse relationships, 175, 435; inequalities, 683; reducing equations, 408 Conservation of angular momentum, 243\u2013245; angular velocity and, 243\u2013244; tops and gyroscopes and, 244\u2013245 Conservation of energy, 285 lab, 293\u2013301; in a calorimeter, 319\u2013320; collisions between two objects and, 297\u2013298, 299 prob., 300\u2013301; conservation of mechanical energy and, 293\u2013295, 296 prob.; loss of mechanical energy and, 295; ProblemSolving Strategies, 295; transfer of potential energy to kinetic energy and, 301 lab, 302\u2013303 lab 495, 504\u2013505 lab; masking of, 495 lab; properties of, 494, 494 table; real images and, 494\u2013495, 496 prob.; virtual images and, 494, 497 Convex mirrors, 471; field of view of, 471; location of image on, 471, 472 prob.; vs. other single-mirror systems, 473 table; virtual images formed by, 471 Coordinate systems, 34\u201335, 122, 135, 155, 848 Copernicus, Nicholas, 171 Copper, ductility of, 360 Coriolis \u201cforce\u201d, 216\u2013217 Cornell, Eric, 366, 768 Cosines, law of, 120 COSTAR, 467 Coulomb (C), 549\u2013550 Coulomb, Charles, 549 Coulomb\u2019s law, 549\u2013550, 551 prob., 552, 552 prob.; direction of magnitude of force and, 550; equation for, 550; Problem-Solving Strategies, 550 c), 489 Credit card readers, 688 Crest, wave, 383 Critical angle ( Crystal lasers, 764 Crystal lattice, 359 Cube", " roots, 840 Curie, Marie, 3, 799 Curie, Pierre, 799 Currents. See Electric currents Curved mirrors, comparing, 473, 473 table; concave, 464\u2013470; convex, 471, 472 prob.; mathematical method to locate image on, 467, 469 prob.; magnification by, 468; parabolic, 467; ray tracing to locate images formed by, 465\u2013467 Cymbals, source of sound from, 411 Dalziel, Harles, 634 Dams, 676\u2013677 Dark matter, 823 Davisson, Clinton, 735 de Broglie, Louis, 735, 760, 768 de Broglie wavelength, 735, 736 prob., Conservation of momentum, law of, 760 237. See also Momentum Consonance, 418 Constant acceleration, 65\u201371; displacement and, 67 prob.; equations for, 68 table; position and, 66, 68, 68 table; velocity and, 65, 68, 68 table Constructive interference, 516; from diffraction gratings, 528; double-slit interference and, 516, 517, 518; mechanical waves and, 389; thin-film interference and, 520 Contact forces, 88 Contact lenses, 501 Control rods, 813 Controlled chain reactions, 812\u2013813 Convection, 317 Conventional currents, 592 Converging lenses. See Convex lenses Conversion factors, 6, 847 Dead spots, 405 Decay. See also Radioactive decay; alpha, 806, 808 prob.; beta, 806, 808 prob.; gamma, 806; nuclear reactions and, 807\u2013808, 809 prob. Deci-, 6 table Decibel (dB), 406 Deep Space 1, 239 Deflection of light, Einstein\u2019s theory of relativity, 185 Density, free-electron, 777, 778 prob. Dependent variables, 15, 16, 848 Depletion layer, 784 Derived units, SI, 5 Descartes, R\u00e9ne, 486 Destructive interference, 516; doubleslit interference and, 516, 517, 518; mechanical waves and, 389; thin-film interference and, 520 Index 929 927-942 EM INDEX-845813 6/10/04 9:17 PM Page 930 x e d n I Diagrams, coordinate systems, 34\u201335, DVDs, lasers in DVD players, 764; 122, 135, 155; electric circuit, 597\u2013599, 599 lab; free-body, 89; motion, 32\u201333; ray. See Ray tracing Diamonds, 544 Dielectrics, 707 Diffraction, 524\u2013528, 530\u2013531; doubleslit, 516\u2013519, 532\u2013533 lab; in the eye, 531, 531 lab; resolving power of lenses and, 530\u2013531; single-slit, 524\u2013526, 526 prob.; wave model of light and, 439 Diffraction gratings, 527\u2013528, 530, 531 lab; DVDs, spacing between rows of, 529 prob.; grating spectroscopes, 528; holographic, 527; reflection, 515 lab, 527, 529 prob.; replica, 527; retinal projection screens, 531 lab; transmission, 527; wavelength of light from, 528, 529 prob. Diffraction patterns, 524\u2013526 Diffuse reflections, 459 Digital virtual discs (DVDs), lasers in DVD players, 764; as reflection gratings, 515 lab, 527, 529 prob. Digits, significant. See Significant digits Dimensional analysis, 6\u20137, 847 Diodes, 784\u2013786. See also Transistors; current and voltage of, 785, 785 prob., 786 prob., 790\u2013791 lab; light-emitting, 450, 786, 788 lab Direct relationships, 175, 435, 851 Dispersion, 491\u2013492 Displacement (d), 36\u201337; angular, 198; equation for, 36; length and direction of, 37; from a v-t graph, 67 prob., 68 Dissonance, 418 Distance, coordinate system, 34 Divers, angular momentum (L) and, 235; pressure of water on a body, 353; rotational kinetic energy (KErot) and, 287 Dividers, voltage. See Voltage divider circuits Dogs, hearing of, 413 Domains, magnetic, 650 Dopants, 781, 782 prob. Doped semiconductors, 781; conductivity, 782 prob.; diodes, 784\u2013786, 785 prob., 786 prob.; light meters and, 783; microchips (integrated circuits) and, 788\u2013789; n-type, 781;", " p-type, 781; thermistors, 782; transistors and, 787\u2013788 Doppler effect, 407\u2013408, 410; equation for, 408; frequency and, 409 prob.; light waves and, 445\u2013446; measuring motion on surface of the Sun by, 422; reducing equation for, 408; uses of, 410 Double-image quasars, gravitational lenses and, 506 Double-slit interference, 516\u2013519; interference patterns, 516; wavelength of light from, 518\u2013519, 519 prob., 532\u2013533 lab Down quark, 819 illus. Drag force, 100\u2013101 Drums, source of sound from, 411 Dry cells, 592 Ductility, 360 as reflection gratings, 515 lab, 527, 529 prob. Dyes, absorption and reflection of light by, 441 Ear, as closed-pipe resonator, 413; damage to from loud sounds, 406\u2013407; detection of sound by, 405; frequencies detected by, 413; sensitivity to amplitude of sound waves, 406 Earth, acceleration of, 104 prob.; acceleration of objects due to gravity of. See Acceleration due to gravity; continents, floating of and Archimedes\u2019 principle, 355; Coriolis \u201cforce\u201d, 217; gravitational field, strength of, 183; heating of by radiation, 317; magnetic field of, 651; mass of, determination of, 178; Newton\u2019s thought experiments on satellite orbits and, 179; period of satellite orbiting, 180, 181 prob.; positive/negative rotation of, 198; pressure at center of, 343 table; radius, mass, and distance from the Sun, 173 table; speed of object on equator, 199; speed of satellite orbiting, 180, 181 prob.; the Sun as source of light for, 432; weight (Fg) on, 96 Earthquakes, building construction and, 394 Echolocation, 410 illus. Eclipses, lunar, 487 Eddy currents, 681 Effective current, 677 Effective voltage, 677, 678 Efficiency, 268 Effort force (Fe), 266 Einstein, Albert, 10 illus.; Bose-Einstein condensates and, 366, 768; energy equivalent of mass (E mc2), 802, 820; general theory of relativity, 10 illus., 184\u2013185; gravitational lenses and, 506; photoelectric theory of, 727\u2013729; photon momentum and, 733; rest energy (E0 stimulated emission of atoms and, 762 mc2), 292; Ejected electron, kinetic energy of, 728\u2013729, 730 prob., 732 prob. Elastic collisions, 298, 299 prob., 301 lab Elastic potential energy, 291\u2013292 Elasticity, 360 Electric arcs, 556 Electric bills, 604 Electric circuits, 592\u2013593; applications of, 620, 627\u2013628; balanced, 628 prob.; diagramming, 591 lab, 597, 599 lab, 599\u2013600; fuses and, 617 lab; measuring current and voltage, 631; parallel, 623\u2013624, 623 lab, 625 prob., 626, 632\u2013633 lab; rates of charge flow and energy transfer, 593, 594 prob.; release of thermal energy by, 601\u2013602; resistance and Ohm\u2019s law, 595\u2013597, 598 prob.; series, 618\u2013620, 621\u2013623 prob., 632\u2013633 lab; series-parallel, 629, 630 prob.; short, 627; voltage divider, 620, 622 prob. Electric conduction, 777 Electric currents (I), 592; ammeters and, 631, 655; from changing magnetic fields, 671 lab, 671\u2013673, 674\u2013675 prob.; controlling in a circuit with resistors, 596\u2013597, 598 prob.; effective, 677; from electric generators, 675\u2013678; energy carried by, 592, 593, 594 prob.; forces on in magnetic fields, 652\u2013653, 654 prob.; galvanometers and, 628 prob., 655; in parallel circuits, 623\u2013624, 625\u2013626 prob.; power of, 593, 594 prob.; producing, 591 lab, 592; resistance-current relationships, 4 lab, 18, 606\u2013607 lab; in series circuits, 618\u2013620, 621\u2013623 prob.; in seriesparallel circuits, 629, 630 prob.; transmitting over long distances, 603\u2013604; unit of (ampere), 5 table, 593; voltagecurrent relationships, 606\u2013607 lab Electric field lines, 567 Electric fields (E), 563\u2013579. See also Electric forces; affect of on charged object, 563 lab; capacitors and, 577\u2013579", ", 578 prob., 579 prob.; conductors and, 576\u2013577; electric potential and, 569\u2013571, 572 prob.; generating with Van de Graaff generator, 567 illus., 568; Millikan\u2019s oil-drop experiment and, 573, 574 prob.; sharing of charges and, 575\u2013576; strength of, 564, 564 table, 565 prob., 566 prob., 567, 573 lab; typical values, 564 table; visualizing, 564, 567\u2013568 Electric forces, 541 lab, 546\u2013553. See also Electric fields; applications of, 552\u2013553; charging by conduction, 547, 549 lab; charging by induction, 548, 549 lab; Coulomb\u2019s law and, 549\u2013550, 551 prob., 552; ProblemSolving Strategies, 550; separation of charges on neutral objects, 547 Electric generators, 675\u2013678 Electric guitars, 411 Electric heaters, thermal energy released by, 601\u2013602, 602 prob. Electric motors, 656, 656 prob., 680 Electric potential, 569\u2013571, 572 prob. Electric potential difference (V), 4 prob., 569\u2013571, 572 prob.; capacitors and, 577\u2013579, 578 prob., 579 prob.; conductors and, 576\u2013577; Millikan\u2019s oil-drop experiment and, 573, 574 prob.; sharing of charges and, 575\u2013576; in a uniform field, 571, 572 prob., 573, 574 prob.; unit of (volt), 569; of zero (equipotential positions), 570 Electric power transmission, 603\u2013605 Electric transmission lines, 604 Electrical conduction in solids, 775\u2013783 Electricity, static. See Static electricity; transmission of, 603\u2013605 Electromagnetic force, 802 Electromagnetic induction, 671 lab, 672; electric generators and, 675\u2013678; electromotive force (EMF) and, 673, 674\u2013675 prob.; Lenz\u2019s law and, 679\u2013681; self-inductance and, 930 Index 927-942 EM INDEX-845813 6/10/04 9:18 PM Page 931 I n d e x 681\u2013682; transformers and, 682\u2013683, 684 prob., 685 Electromagnetic receivers, 712 Electromagnetic spectrum, 708 illus., 709 Electromagnetic wave theory, 706, 723; problems with, 723, 724, 725, 726 Electromagnetic waves, 706\u2013713; from AC sources, 707\u2013708, 709; from a coil and capacitor, 709\u2013710; de Broglie wavelength and, 735, 736 prob.; discovery of, 705\u2013706; electromagnetic spectrum, 708 illus., 709; emission spectra from, 724\u2013725; energy of (photoelectric theory), 726\u2013729, 730 prob., 731, 732 prob., 738\u2013739 lab; materials that shield, 714\u2013715 lab; momentum of (Compton effect), 733\u2013734; from piezoelectricity, 711; power of, 725; receiving, 697 lab, 712; from a resonant cavity, 711; speed of, 706; travel of through matter and space, 707\u2013708; velocity of through dielectric, 707; wavelength-frequency relationship for, 706 Electromagnetism, 648\u2013649; charge-tomass ratio of electrons and, 698\u2013699; charge-to-mass ratio of protons and, 699\u2013700; determination of path of electron and, 699, 700 prob.; electromagnetic waves and, 714\u2013715 lab; magnetic field near current-carrying wires, 648\u2013649; mass spectrometers and, 701\u2013702, 703 prob., 704 Electromagnets, 649; audiocassette/videocassette recorders and, 651; computer storage disks and, 659; creating, 660\u2013661 lab; magnetic domains and, 650 Electromotive force (EMF), 673, 674\u2013675 prob.; back-EMF, 680\u2013682; induced in electric generators, 675\u2013677; transformers and, 682\u2013683 Electron clouds, 761 Electron diffraction pattern, wave properties of particles and, 735 Electron neutrino, 819 illus. Electron volt, energy of an electron and, 727 Electronic devices, diodes, 784\u2013786, 785 prob., 786 prob., 788 lab; transistors and integrated devices, 787", "\u2013789 Electrons, angular momentum of, 755; charge-to-mass ratio of, 698\u2013699; discovery of, 747; electromagnetic force in nucleus and, 802; force of magnetic fields on, 657, 658 prob.; free in semiconductors, 779\u2013780, 780 prob., 782 prob.; frequency and wavelength of emitted, 758 prob.; kinetic energy of ejected due to photoelectric effect, 728\u2013729, 730 prob., 732 prob.; magnitude of charge of, 550; mass of from charge-to-mass ratio, 698, 699; measurement of charge of in Millikan\u2019s oil-drop experiment, 573, 574 prob.; orbital energy values, 755\u2013756, 757 prob.; orbital radius of, 755, 760; path of, determining, 699, 700 prob. Electroscope, 546, 548; charging by conduction, 547, 549 lab; charging by induction, 548, 549 lab Electrostatic charge, 541 lab, 541\u2013545. See also Electric force; charged objects, 542\u2013543, 554\u2013555 lab; conductors and insulators and, 544\u2013545; microscopic view of, 543\u2013544; separation of, 544, 592\u2013593; unit of (C), 549\u2013550 Electrostatic generator, 568 Electrostatics, 541\u2013553; applications of, 552\u2013553; charged objects, 542\u2013543, 554\u2013555 lab; charging by conduction, 547, 549 lab; charging by induction, 548, 549 lab; conductors and insulators and, 544\u2013545; Coulomb\u2019s law and, 549\u2013550, 551 prob., 552, 552 prob.; forces on charged bodies, 546\u2013548; grounding and, 548; microscopic view of charge, 543\u2013544; separation of charge on neutral objects, 547 Electroweak force, 823 Elementary charge, 550 Elevators, 98, 99 prob., 108\u2013109 lab Elliptical orbits, Kepler\u2019s first law and, 172 EM wave. See Electromagnetic wave EMF. See Electromotive force (EMF) Emission spectra, 724; of atoms, 749\u2013750, 752, 755 lab; of light from hot bodies, 724 lab, 724; photoelectric effect and, 726\u2013729, 730 prob., 731, 732 prob.; shape of, 725 Emissions, spontaneous, 762; stimulated, 762 Energy, 258\u2013259, 285; conduction and, 315; conservation of. See Conservation of energy; direction of force and, 257 lab; elastic potential, 291\u2013292; of electric currents, 592, 593, 594 prob.; gravitational potential (PE), 288\u2013289, 290 prob.; kinetic (KE), 258\u2013259, 287, 289; mechanical, 293\u2013295; quantized, 725; mc2), 292; thermal. See Therrest (E0 mal energy; unit of (Joule), 259; work and change in (work-energy theorem), 258\u2013259, 261 prob., 286\u2013287 Energy crisis, 331 Energy equivalent of mass (E mc2), 802\u2013803, 820 Energy levels, electron, 753; energies of, 757 prob.; frequency and wavelength of emitted photons and, 758 prob.; transitions in, 756 Engines, heat, 326\u2013327; internal- combustion, 326\u2013327 Entropy, 328\u2013331, 329 prob.; equation for change in, 329; second law of thermodynamics and, 330\u2013331 Equations, angle of resultant vector, 123; angular acceleration, 199; angular impulse-angular momentum theorem, 234; angular momentum, 233; angular velocity, 198; average acceleration, 62; average velocity, 44; buoyant force, 354; capacitance, 577; centripetal acceleration, 154; charge-to-mass ratio in a Thomson tube, 699; charge-to-mass ratio of an ion in mass spectrometer, 702; coefficient of linear expansion, 361; coefficient of volume expansion, 361; combined gas law (pressure, volume, and temperature), 345; conservation of angular momentum, 243; conservation of energy in a calorimeter, 319; cosines, law of, 120; Coulomb\u2019s law, 550; critical angle for total internal reflection, 489; current in series circuit, 619; de Broglie wavelength, 735; displacement, 36; Doppler effect, 408; Doppler shift, 446; effective current, 677; efficiency, 268; electric field strength, 564; electric potential difference, 569, 571; electromotive force (EMF), 673; electron orbital radius, 755; energy equivalent of mass (E", " mc2), 802; energy of a photon, 727; energy of an atom, 755; energy of an emitted photon, 754; energy of vibration, 725; equivalent for resistors in parallel, 624; equivalent resistance for resistors in series, 619; final velocity with average acceleration, 65; first law of thermodynamics, 326; force of magnetic field on charged particles, 657; force on current-carrying wire in a magnetic field, 653; frequency of a wave, 384; gravitational field strength, 183; gravitational mass, 184; gravitational potential energy, 289; half-life, 810; heat required to melt a solid, 324; heat required to vaporize a liquid, 324; heat transfer, 318; Hooke\u2019s law, 376; hydraulic lift, force exerted by, 352; ideal gas law, 345; ideal mechanical advantage (IMA), 267; impulse-momentum theorem, 230; index of refraction, 489; inertial mass, 183; inverse relationship between two variables, 18; Kepler\u2019s third law, 173; kinetic energy, 258; kinetic friction force, 127; linear relationship between two variables, 16; magnification by a spherical mirror, 468; Malus\u2019s law, 444; mechanical advantage (MA), 266; mirror equation, 467; moment of inertia of a point of mass, 205; motion for average velocity, 47; Newton\u2019s second law for linear motion, 93; Newton\u2019s second law for rotational motion, 208; Newton\u2019s third law, 102; observed light frequency, 446; order of operation in, 843; period of a pendulum, 379; period of satellite orbiting Earth, 180; photon momentum, 733; plane-mirror image height, 462; plane-mirror image position, 462; point source of illuminance, 435; position with average acceleration, 68; potential energy in a spring, 376; power, 263; power dissipated by a resistor, 601; power of electric circuit, 593; pressure, 342; pressure of water on a body, 353; Pythagorean theorem, 120; quadratic relationship between two variables, 17; Rayleigh criterion, 530; reducing form of, 408; reflection, law of, 458; resistance, 595; mc2), 292; sines, law rest energy (E0 of, 120; slope of a straight line, 17; Snell\u2019s law of refraction, 486; solving, 843\u2013845; speed of a satellite orbiting Earth, 180; static friction force, 127; thermal energy dissipated by a resistor, 602; thin lens, 493; time interval, 36; torque, 202; transformer, 683; velocity with constant acceleration, 68; wavelength-frequency relationship for a Index 931 927-942 EM INDEX-845813 6/10/04 9:18 PM Page 932 x e d n I wave, 706; wavelength of light from double-slit experiment, 518; wavelength of light from diffraction grating, 528; width of bright bands on singleslit diffraction pattern, 525; work, 258, 260; work-energy theorem, 259 Equator, speed of object on, 199 Equilibrant, 131 Equilibrium, 95, 211\u2013217; center of mass and, 211\u2013213; force and motion in two dimensions and, 131; rotational, 213, 218\u2013219 lab; static, 213, 214\u2013215 prob.; translational, 213, 218\u2013219 lab Equipotential positions, 570 Equivalent resistance, in parallel circuits, 624, 625\u2013626 prob., 628; in series circuits, 619\u2013620 Error, measurement, 11\u201312, 13 Evaporation, 350\u2013351, 364\u2013365 lab Evaporative cooling, 351; of alcohols, 364\u2013365 lab; production of Bose-Einstein condensates, 366 Example Problems, acceleration, direction and magnitude of, 63 prob.; acceleration on motion diagram, 60 prob.; alpha and beta decay, 808 prob.; angle of refraction, 487 prob.; Archimedes\u2019 principle, 356 prob.; average velocity and speed, 45 prob.; balanced friction forces, 128 prob.; calorimeter, transfer of heat in, 321 prob.; capacitance, 578 prob.; coefficient of linear expansion, 362 prob.; collisions, kinetic energy lost in, 299 prob.; concave mirror, locating image on, 469 prob.; conductivity of doped silicon, 782 prob.; conservation of mechanical energy, 296 prob.; convex lens, image formed by, 496 prob.; convex mirror, locating image on, 472 prob.; Coulomb\u2019s law and electric force, 551 prob.; current through a resistor, 598 prob.; de Broglie wavelength, 736 prob.; diode", " in a single circuit, 785 prob.; displacement from a motion diagram, 69 prob.; displacement from a v-t graph, 67 prob.; Doppler effect, 409 prob.; electric potential difference in a uniform field, 572 prob.; electric power and energy, 594 prob.; electron, path of 700 prob.; energy levels, 757 prob.; force and displacement at an angle, 262 prob.; force, average, 232 prob.; force of magnetic field on charged particles, 658 prob.; free-electron density in a conductor, 778 prob.; free electrons in intrinsic semiconductor, 780 prob.; gas laws, 346 prob.; gravitational potential energy, 290 prob.; heat transfer, 318 prob.; horizontal acceleration, 97 prob.; illumination of a surface, 436 prob.; inclined planes, 133 prob., 134 prob.; induced electromotive force, 674 prob.; inertia, moment of, 207 prob.; lever arm and torque, 202 prob.; magnetic fields, strength of, 654 prob.; mass defect and nuclear binding energy, 804 prob.; mass spectrometer, mass of a neon atom from, 703 prob.; mechanical advantage (MA)/ideal mechanical advantage (IMA), 271 prob.; nuclear equations, 808 prob., 809 prob.; oil drop, charge on (Millikan\u2019s experi- ment), 574 prob.; orbital sizes of Jupiter\u2019s moons, 174 prob.; orbital speed and period of satellite orbiting Earth, 181 prob.; pendulum, acceleration due to gravity, 379 prob.; photoelectron kinetic energy, 730 prob.; power, 264 prob.; power and energy released by a heater, 602 prob.; pressure, 343 prob.; real and apparent weight, 99 prob.; reflection and angle of incidence, 460 prob.; relative-velocity, 158 prob.; resistance-current relationships, 4 prob.; series-parallel circuits, 630 prob.; space propulsion, 240 prob.; speed of sound using resonance, 416 prob.; spring constant and energy of a spring, 377 prob.; step-up transformers, 684 prob.; thinfilm interference, 522 prob.; torque, 204 prob., 209 prob.; two-dimensional collisions, 242 prob.; two-part motion, 70 prob.; two-particle collisions, 237 prob.; uniform circular motion, 155 prob.; uniform straight-line motion of multiple objects, 41 prob.; uniform straight-line motion of single object, 39 prob.; vector addition/subtraction, 121 prob., 124 prob.; voltage dividers and, 622 prob.; voltage drops in series circuits, 621 prob.; wavelength of light from doubleslit experiment, 519 prob.; wavelength of light from diffraction grating, 529 prob.; work and energy, 261 prob.; work function and kinetic energy, 732 prob. Excimer lasers, 764, 765 Excited state, 753 Expansion joints, 360 Experiments. See also Launch Labs; MiniLabs; Physics Labs; comparing results of, 11\u201312; scientific method and, 8 Explosive collisions, 298 Exponential growth, 18 Exponents, 839, 840 External world, 88 Extrapolation, 849 Extreme Physics, artificial intelligence, 792; black holes, 188; Bose-Einstein condensates, 366; cesium clocks, 50; gravitational lenses, 506; Hall voltage, 662; sound waves in the Sun, 422; time dilation, 78 Extrinsic semiconductors, 781, 782, 783 Eye, 500\u2013501; diffraction on retina, 531, 531 lab; focusing of images, 500; nearsightedness and farsightedness, 501; refraction from, 501 prob.; resolving power of, 531 Eye surgery, lasers and, 764, 765 Eyeglasses, 501, 520 Fahrenheit scale, 316 Farad (F), 578 Faraday, Michael, 183, 578, 652, 675, 705 Farsightedness, 501 Fast neutrons, 812 Fermato-, 6 table Fermi, Enrico, 811, 818 Fermi National Accelerator laboratory, 816 Ferromagnetism, 650 Fiber-optic communications, lasers and, 764 Field coils, 656 Field forces, contact forces vs., 88 Field lines, electric, 567; magnetic, 646 Fields, electric. See Electric fields; gravitational, 182\u2013183; magnetic. See Magnetic fields Filtering, polarization by, 443 Final time (tf), 36 First law of thermodynamics, 326\u2013328; heat engines and, 326\u2013327; heat pumps and, 328, 334; refrigerators and, 327\u2013328 First right-hand rule, 649 Fish, swim bladders and Archimedes\u2019 principle, 355 Fission, nuclear, 811\u2013812 Fluids, 342. See also Gases; Liquids; Archimedes\u2019 principle and, 354\u2013355, 356 prob.; Bernoulli\ufffd", "\ufffds principle and, 357\u2013358; buoyancy and, 341 lab, 354\u2013355, 356 prob.; hydraulic systems and, 352; ideal, 342; Pascal\u2019s principle and, 352; plasma, 348; pressure in, 342; pressure of on a body, 353; velocity of moving, 357 Fluorescein, 724 lab Fluorescent lamps, 432, 450 Flutes, 411, 412 Flux, luminous, 433; magnetic, 646 Focal length ( f ), 464; of concave mirror, 464; of convex lenses and, 494\u2013495, 504\u2013505 lab; of thin lens, 494 Focal point, 464; of lenses, 494; of mirrors, 464 Foci, 172 Fog, 351 Forbidden energy regions, 776 Force, 88. See also Force in one dimension; Force in two dimensions; adhesive, 350; angular velocity and, 201; area and pressure and, 342, 343 prob.; buoyant, 341 lab, 354\u2013355, 356 prob.; centrifugal, 156, 216; centripetal, 154, 155; on charged bodies, 546\u2013550, 551 prob.; cohesive, 349; contact, 88; drag, 100\u2013101; effort, 266; electric. See Electric forces; field, 88; friction. See Friction; gravitational. See Gravitational force; interaction, 102\u2013103, 104 prob.; magnitude of common, 91 table; net. See Net force; normal, 107; resistance, 266; strong nuclear, 802; types of, 94 table; unit of (N), 91; weak nuclear, 821, 823; work and (angle between force and displacement), 260, 262 prob. Force-acceleration graphs, 90\u201391 Force carriers, 818 Force in one dimension, 87\u201395. See also Force; Force in two dimensions; acceleration and, 88, 90\u201393; combining forces, 92; contact vs. field forces, 88; drag force and terminal velocity, 100\u2013101; elevators and, 98, 108\u2013109 lab; free-body diagrams of, 89; interaction pairs and, 102\u2013103, 104 prob.; 932 Index 927-942 EM INDEX-845813 6/10/04 9:19 PM Page 933 I n d e x Newton\u2019s first law and, 94\u201395; Newton\u2019s second law and, 96, 97 prob., 98; Newton\u2019s third law and, 102\u2013103, 103 lab, 104 prob.; normal force and, 107; Problem-Solving Strategies, 98; real and apparent weight and, 98, 99 prob.; ropes and strings and, 87 lab, 105, 106 prob.; tension and, 105, 106 prob. Force in two dimensions, 131\u2013135. See also Force; Force in one dimension; equilibrium and, 119 lab, 131; inclined planes, 132, 133 prob., 134 prob., 135, 135 lab Fosbury-Flop, 212 Fossil fuels, energy crisis and, 331 Foucault, Jean, 457 Fourth right-hand rule, 672 Fractions, 468, 837 Fractions of revolutions (radians), 197, 198 illus. Frames of reference, rotating, 216 Franklin, Benjamin, 541, 577, 582 Fraunhofer, Josef von, 751 Fraunhofer lines, 751 Free-body diagrams, 89 Free-electron density, 777, 778 prob. Free-fall objects, 72\u201375; acceleration of due to gravity, 72, 182; amusement park rides, 74; balls thrown upwards, 72\u201373 Frequency ( f ), 384, 385 prob.; funda- mental, 417; threshold, 726\u2013727, 728, 731, 732 prob.; unit of (Hz), 384; wavelength-frequency relationship, 706; of waves from a pendulum, 392\u2013393 lab Fresnel, Jean, 519 Friction (Ff ), 94 table, 126\u2013130; balanced friction forces, 127, 128 prob.; causes of, 130; coefficients of, 127, 129 table; kinetic, 126, 127; model for, 126\u2013127; static, 126, 127; unbalanced friction forces, 127; 129 prob. Fringes, interference, 516 Frisch, Otto, 811 Fuel efficiency, 327 Fundamental frequency, 417 Fuses, 617 lab, 627, 628 Fusion, nuclear, 813\u2013814, 826 Future Technology, adaptive optical systems, 476; atom lasers, 768; computers, 22; rotating space stations, 162; solar sails, 248; spacecraft and static electricity, 556; thermonuclear fusion, 826 Gabor, Dennis, 534 Gadolinium, 645 Galilei, Galileo, 10 illus., 72, 174 prob., 405, 437 Gallium, 781 Gallium aluminum, 764 Gallium", " arsenide, 764 Galvanic cells, 592 Galvanometers, 628 prob., 655, 671 lab Gamma decay, 806 Gamma particles, 818 Gamma radiation, 806, 807 table, 811, 824\u2013825 lab Gas-discharge lamps, 450 Gas lasers, 764 Gas laws, 344\u2013345, 347 prob.; Boyle\u2019s law (volume and pressure of a gas), 344, 346 prob.; Charles\u2019s law (volume and temperature of a gas), 344; combined gas law, 345; ideal gas law, 345 Gases, 323; absorption spectrum of, 751\u2013752; convection currents and, 347; emission spectra from, 750, 752, 755 lab; gas laws and. See Gas laws; pressure exerted by, 342 Geiger, Hans, 748 Geiger counters, 817, 824\u2013825 lab Geiger-Mueller tubes, 817 General theory of relativity, Einstein\u2019s, 184\u2013185 Generators, 675\u2013678; alternating- current, 677\u2013678; electric, 592 illus., 675\u2013677 Geometric optics, 432 Geometry, 853\u2013856 Germanium, 777, 779 Germer, Lessmer, 735 Giga-, 6 table Glasses, eye. See Eyeglasses Global Positioning System (GPS), 14 Gluons, 818, 819 illus., 820 Gold foil experiment, Rutherford\u2019s, 748, 766\u2013767 lab, 799 Gold, malleability of, 360 Gongs, source of sound from, 411 GPS systems. See Global Positioning System (GPS) Graduated cylinder, precision of measurements in, 12 illus. Graphite, 544, 777 Graphs, 15\u201319, 18 prob., 848\u2013852; force-acceleration, 90\u201391; line, 15\u201317; non-linear relationships on, 17\u201318; position-time, 38\u201342, 58; predicting values from, 19; slope of lines on. See Slope of a line; variables on, 15, 16; velocity-time, 58\u201359; of waves, 384, 390 Grass, green color of, 442 Grass seeds, visualizing electric field lines and, 568 Grating spectroscopes, 528 Gratings, diffraction. See Diffraction gratings Gravitational field (g), 182\u2013183 Gravitational force (Fg), 175; Newton\u2019s law of universal gravitation and, 175\u2013178; orbital speed and period of satellite orbiting Earth and, 180, 181 prob.; planetary motion and, 176, 178; universal gravitation constant (G) and, 176, 177\u2013178; weightlessness and, 182 Gravitational lenses, 506 Gravitational mass (mgrav), 184 Gravitational potential energy (PE), 288\u2013289, 290 prob.; of an atom, 289; kinetic energy and potential energy of a system and, 289; mechanical energy and, 293\u2013294; reference level and, 288, 289; work and, 288 Graviton, 819, 820 Gravity, acceleration due to, 72\u201375, 76\u201377 lab, 88; development of theory of, 10 illus.; Einstein\u2019s general theory of relativity and, 184\u2013185 Gravity waves (g waves), 422 Grimaldi, Francesco Maria, 439 Ground-fault interrupters, 627, 634 Ground faults, 634 Ground state, 753 Grounding, 548 Guitars, sound from, 411 Gyroscopes, 245 Hahn, Otto, 811 Hale-Bopp comet, 172 Half-life, 809\u2013810, 810 table, 813 lab Hall, E. H., 662 Hall voltage, 662 Halley\u2019s comet, 172 Harmonic motion. See Simple harmonic motion Harmonics, 417. See also Simple harmonic motion hc, units of and photon energy, 727, 728 Hearing, damage to from loud sounds, 406\u2013407; sensitivity to amplitude of sound waves, 406; sound detection and, 405 Heat, 317; calorimetry and, 319\u2013320, 321 prob.; conduction, 317; convection, 317; radiation, 317; release by resistors, 601\u2013602; specific, 317\u2013318, 319\u2013320, 321 prob.; transfer of, 317\u2013318, 318 prob., 319 prob. Heat engines, 326\u2013327 Heat of fusion, 324, 324 table, 325 prob. Heat of vaporization, 324, 324 table Heat pumps, 328, 334 Heisenberg uncertainty principle, 737, 760 Heisenberg, Werner, 737 Helioseismology, 422 Helium, nuclear fusion and, 814 Helmholtz, Hermann, 411 Henry, John, 671 Hertz (Hz), 384 Hertz, Heinrich, 723 Holes, 779\u2013780 Holograms,", " 534, 765 Holographic diffraction gratings, 527 Holography, 534 Home electric bills, 604 Home thermostats, 363 Home wiring, 627\u2013628 Hooke\u2019s law, 376, 377 prob. Horns, 712 Hot bodies, radiation from, 723 lab, 724 lab, 724\u2013725 Hot plates, release of thermal energy by, 601 Household appliances, equivalent resistance and, 628; kilowatt power of, 605; thermal energy released by, 601 How It Works, bathroom scales, 110; bicycles, 276; credit card readers, 688; ground-fault interrupters (GFCI), 634; heat pumps, 334; lightning rods, 582; scanning tunneling microscope, 740 Index 933 927-942 EM INDEX-845813 6/10/04 9:19 PM Page 934 x e d n I Hubble, Edwin, 447 Hubble Space Telescope, 467 Human body, body temperature of, 322; center of mass of, 212; hearing and. See Hearing; as simple machine, 273; speech and, 411; as a variable resistor, 597; vision and. See Vision Hunter, John H., 110 Huygens, Christiaan, 439 Huygen\u2019s wavelets, 439, 490, 524\u2013525 Hybrid cars, 608 Hydraulic lift systems, 352 Hydroelectric electricity, 603 Hydrogen, atomic spectra, 752; electron cloud, 761 illus.; electron orbital radius, 755; ionization energy, 759; orbital energy values, 756, 757 prob. Hyperbola, 18 Hyperopia, 501 Hypothesis, 8 Ice, floating of, 347, 355; pressure and freezing of, 359\u2013360 Ice skaters, angular momentum (L) and, 235; collisions between two, 301 Ideal fluids, 342 Ideal gas law, 345 Ideal machines, efficiency of, 268; ideal mechanical advantage (IMA) of, 267 Ideal mechanical advantage (IMA), 267; of automobile transmissions, 272; of bicycle gears, 270, 271 prob., 272, 276; of compound machines, 270, 271 prob.; efficiency and, 268; of simple machines, 269 Illuminance (E), 433\u2013434; equation for point source, 435; inverse-square relationship and, 434; of a surface, 435, 436 prob., 437 Illuminated source, 432 Illumination, from illuminated sources, 432; luminous flux (P) and, 433, 434; luminous intensity and, 434; from luminous sources, 432; of a surface (illuminance, E), 433\u2013434, 435, 436 prob., 437 IMA. See Ideal mechanical advantage (IMA) Images, 461; from concave lenses, 498; from concave mirrors, 465\u2013468, 469 prob., 470, 474\u2013475 lab; from convex lenses and, 494\u2013495, 496 prob., 497; from convex mirrors, 465\u2013468, 470 prob., 471, 472 prob.; mirror, 461; from plane mirrors, 461\u2013463, 462 lab; real, 465, 474\u2013475 lab, 494\u2013495, 496 prob.; spherical thin lenses and, 494; virtual. See Virtual images Impulse (Ft), 230; impulse-momentum theorem and, 230\u2013231, 232 prob.; of two particles in closed, isolated system, 236 Impulse-momentum theorem, 230\u2013231, 232 prob.; angular impulseangular momentum theorem, 234\u2013235; baseballs and, 231; car crashes and, 231, 232 prob. In-line skaters, recoil of, 238\u2013239 Incandescent bodies, radiation from, 723 lab, 724\u2013725 Incandescent bulbs, 432, 531 lab, 723 lab Incidence, angle of. See Angle of incidence Incident wave, 387 Inclined planes, 135 lab; ideal mechanical advantage (IMA) of, 269; motion along, 132, 133 prob., 134 prob., 135; as simple machines, 269 Incoherent light, 515, 761 Independent variables, 15, 16, 848 Index of refraction, 486, 486 table, 489 Induced electromotive force, 673, 674 prob., 676 Inductance, mutual, 682; self, 681\u2013682 Induction, charging by, 548, 549 lab, 554\u2013555 lab, 705 Inelastic collision, 298 Inequalities, 683 Inertia, 95; moment of (I), 205\u2013206, 206 table, 207 prob., 208 prob. Inertial mass (minertial), 183 Initial time (ti), 36 Instantaneous acceleration, 59 Instantaneous angular acceleration, 199 Instantaneous angular velocity, 199 Instant", "aneous position (d), 40 Instantaneous velocity, 46, 46 lab Instruments, accuracy of, 13 Insulators, 544, 778\u2013779 Intensity, luminous. See Luminous intensity Interaction pairs, 102; Earth\u2019s acceleration and, 104 prob.; normal force and, 107; Problem-Solving Strategies, 103; tension forces and, 105, 106 prob. Interference, 388\u2013389, 516\u2013521; double-slit, 516, 517\u2013519, 519 prob., 532\u2013533 lab; thin-film, 520\u2013521, 522 prob., 523 Interference fringes, 516 Interference patterns, double-slit, 516; single-slit, 525 Internal-combustion engine, 326\u2013327 Internal reflection, total, 489\u2013490 International Bureau of Weights and Measures, 6 International Prototype Meter, 6 illus. International Space Station (ISS), 162, 556 International System of Units. See SI Internet Physics Labs, forces in an elevator, 108\u2013109 lab; measurement of speeding objects, 20\u201321 lab; momentum in a collision, 246\u2013247 lab Interpolation, 849 Intrinsic semiconductors, 780, 782 Inverse relationships, 18, 175, 851 Inverse square law, 175 Inverse square relationship, illumi- nance and, 434 Io (Jupiter\u2019s moon), 174 prob., 437 Iodine, radioactive, 811 Ion engines, 239 Ionization energy, 759 Ions, charge-to-mass ratio of, 702, 703 prob.; sources of, 704 Isolated systems, 237 Isotopes, 701, 801. See also Radioactive isotopes; activity of, 810; analysis of with mass spectrometer, 701, 704; artificial, 811; half-life of, 809\u2013810, 810 table, 813 lab Jodrell Bank Observatory, 506 Joule (J), 259 Joule heating loss, 603, 604 Joule, James Prescott, 259 Junction, diode, 784 Jupiter, determination of speed of light and, 437\u2013438; orbital sizes of Jupiter\u2019s moons, 174 prob.; radius, mass, and distance from the Sun, 173 table Kelvin (K), 5 table, 316 Kelvin scale, 316 Kepler, Johannes, 172\u2013173 Keplerian telescope, 502 Kepler\u2019s first law, 172 Kepler\u2019s second law, 173, 174 prob. Kepler\u2019s third law, 173, 176 Ketterle, Wolfgang, 768 Kilo-, 6 table Kilogram (kg), 5 table Kilopascal (kPa), 342 Kilowatt-hour (kWh), 605 Kinetic energy (KE), 258\u2013259, 287; collision between two objects and, 297\u2013298, 299 prob., 300; mechanical energy and, 293\u2013294; of photoelectrons, 728\u2013729, 730 prob.; potential energy and, 289; rotational (KErot), 287; work and change in (work-energy theorem), 258\u2013259, 261 prob. Kinetic friction (Ff, kinetic), 126\u2013127, 129; balanced friction forces and, 128 prob; coefficient of, 127, 136\u2013137 lab; unbalanced friction forces and, 129 prob. Labs. See Launch Labs; MiniLabs; Physics Labs Lamps, types of, 450 Landsat 7, 180 Lascaux cave, radiocarbon dating and, 11 Lasers, 445, 762\u2013765; atom, 768; CD and DVD players and, 764; eye surgery with, 764, 765; fiber-optic communi- 934 Index 927-942 EM INDEX-845813 6/10/04 9:20 PM Page 935 I n d e x cations and, 764; holograms and, 765; inefficiency of, 764; production of Bose-Einstein condensates and, 366; production of coherent light by, 763; spectroscopy with, 765 Launch Labs. See also Mini Labs; Physics Labs; atoms, spinning penny model of, 747 lab; balloons, charging of, 563 lab; buoyancy of objects, 341 lab; conservation of momentum, 285 lab; currents and circuits, 591 lab; electric current from changing magnetic fields, 671 lab; electric fields and charged objects, 563 lab; electromagnetic (radio) waves, 697 lab; energy, factors affecting, 257 lab; force on suspended object, 87 lab; fuses and electric circuits, 617 lab; graphs of constant speed and acceleration, 57 lab; LED lights through stroboscope, 775 lab; light, path of through air, 431 lab; light reflected off a compact disc, 515 lab; magnetic fields, direction of, 643 lab; modeling nucleus, 799 lab", "; momentum, 229 lab; music, production of, 403 lab; planetary orbits, 171 lab; projectile motion, 147 lab; rate of fall of objects, 3 lab; reflection of image onto screen, 457 lab; rotational motion, 197 lab; spectrum of incandescent light bulb, 723 lab; static electricity and electric force, 541 lab; straight-line motion, 31 lab; sum of forces, 119 lab; temperature changes, 313 lab; waves in a coiled spring, 375 lab Lava, viscosity of, 349 Law of conservation of energy, 293\u2013301, 301 lab; mechanical energy and, 293\u2013295, 296 prob.; ProblemSolving Strategies, 295; two-object collisions and, 297\u2013298, 299 prob., 300\u2013301 Law of conservation of momentum, 237, 241, 242 prob. Law of Cosines, 120, 855 Law of inertia. See Newton\u2019s first law Law of reflection, 391, 458\u2013459; angle of incidence and, 458, 459, 460 prob.; rough surfaces and, 459; twodimensional waves and, 391 Law of Sines, 120, 856 Law of universal gravitation, 175\u2013178, 182; gravitational mass (mgrav) and, 184; inverse square law and, 175; Kepler\u2019s third law and, 176; universal gravitation constant (G), 176, 177\u2013178 Laws, 9. See also specific laws Laws of thermodynamics, 326\u2013331 LEDs. See Light-emitting diodes (LEDs) Length, SI base quantity (m), 5 table Lenses, 493\u2013499, 494 table, 500\u2013503; achromatic, 499, 502; binoculars and, 502; camera, 503; chromatic aberration on, 499; concave, 493, 494 table, 498; convex, 493, 494\u2013495, 496 prob., 497, 504\u2013505 lab; focal length (f) and, 494; focal point, 494; gravitational, 506; human eye and vision and, 500\u2013501, 501 prob.; magnification, 493, 494; masking of, 495 lab; microscope, 503; path of rays passing through, 493; refracting telescopes and, 502; resolving power, 530\u2013531; spherical aberration on, 498 Lenz\u2019s law, 679\u2013681 Leptons, 818, 819 illus., 820, 822 Lever arm, 201, 202 prob. Levers, 269 Leyden jar, 577 Light. See also Illumination; blue-shifted, 446; coherent, 516, 761; color and, 440\u2013442; diffraction, 439, 515 lab, 524\u2013528, 529 prob., 530\u2013531, 531 lab; dispersion, 491\u2013492; Doppler effect for, 446\u2013447; Einstein\u2019s general theory of relativity and, 185; incoherent, 515, 761; interference, 515\u2013521, 522 prob., 523; lasers and. See Lasers; luminous flux (P) and, 433, 434; luminous intensity and, 434; monochromatic, 516; observed light frequency, 446; opaque media and, 433; path of through air, 431 lab; polarization, 443\u2013444, 448\u2013449 lab; ray model of, 432\u2013434; red-shifted, 446; reflection. See Reflection; refraction. See Refraction; rope model of, 443; speed of, 437\u2013438, 445\u2013447; translucent media and, 433; transparent media and, 433; wave nature of, 439\u2013447, 441 lab; wavelength of, 518\u2013519, 519 prob., 528, 532\u2013533 lab Light clock, 78 Light-emitting diodes (LEDs), 432, 450, 785 lab, 786, 788 lab, 790\u2013791 lab Light meters, 620, 783 Light spectrum, 440 Light bulbs, lighting with an electric current, 591 lab; parallel circuits and, 626; radiation from incandescent, 723 lab, 724\u2013725; resistance of, 595 Lighting, advances in, 450 Lightning, 545, 547, 577, 582 Lightning rods, 577, 582 Like charges, 542 Line graphs, 15\u201317, 849\u2013851 Linear accelerators, 815 Linear expansion of a solid, 361, 361 table, 362 prob., 363 Linear relationships, 16\u201317, 850 Liquid-crystal thermometers, 315 Liquid mercury, 349 Liquids, 323. See also Fluids; adhesive forces in, 350; boiling point of, 323\u2013324; buoyancy and, 341 lab, 354\u2013355, 356", " prob.; cohesive forces in, 349; condensation of, 351; evaporation of, 350\u2013351, 364\u2013365 lab; heat of vaporization of, 324, 324 table; pressure and freezing of water, 342, 359\u2013360; pressure of water on a body, 353; surface tension of, 349, 350 illus.; thermal expansion of, 347; viscosity of, 349; volatile, 351 Logarithms, 857 Long-period comets, 172 Longitudinal wave, 381 Loudness, 406 Loudspeakers, 653 Lumen (lm), 433 Luminous flux (P), 433 Luminous intensity, 434; unit of (candela), 5 table, 434 Luminous source, 432; rate of light from (luminous flux (P)), 433, 434 Lunar eclipses, refraction and, 487 Lux (lx), 433 Lyman series, 756 illus. m (slope). See Slope of a line MA. See Mechanical advantage (MA) Machines, 266\u2013273; compound, 269\u2013270, 271 prob., 272; efficiency of, 268; effort force on (Fe), 266; human body as, 273; ideal mechanical advantage (IMA), 267, 269, 270, 271 prob., 272; mechanical advantage (MA), 266\u2013267, 270, 271 prob.; resistance force of (Fr), 266; simple, 269 Magnetic domains, 650 Magnetic field lines, 646 Magnetic fields, 645\u2013647; audio- cassette/videocassette recorders and, 651; direction of, 643 lab, 646; Earth\u2019s, 651; electric current from changing. See Electromagnetic induction; electric motors and, 656; forces on charged particles, 657, 658 prob.; forces on electric currents in, 652\u2013653; forces on objects in, 647; galvanometers and, 655; loudspeakers and, 653; magnetic domains and, 650; near a coil, 649; near current-carrying wires, 648\u2013649; strength of, 653, 654 prob., 660\u2013661 lab; three-dimensional, 650 lab Magnetic flux, 646 Magnetic poles, 644 Magnetic storage media, 651, 659 Magnetic strips, credit card, 688 Magnets, 644\u2013647; electromagnets, 649, 660\u2013661 lab; Hall voltage and, 662; magnetic fields around, 645\u2013647, 647 prob.; permanent, 645; poles of, 644; temporary, 645, 650 Magnification, 468, 472 prob.; convex mirrors and, 471; spherical concave mirrors and, 468; spherical thin lenses and, 493, 494 Magnitude, 35 Malleability, 360 Malus\u2019s law, 444, 444 prob. Marianas Trench, 353 illus. Marimbas, 414 Mars, hypothesis of canals on, 9; radius, mass, and distance from the Sun, 173 table; surface of, 9 illus. Marsden, Ernest, 748 Mass (m), affect of on space (Einstein\u2019s theory of relativity), 184\u2013185; of an atom, 800, 801; of Earth, 178; energy of (rest energy, E0), 292; gravitational (mgrav), 184; inertial (minertial), 183; kinetic energy (KE) of an object and, 287; projectile motion and, 148 lab; on a spring. See Springs; unit of (kg), 5 table Mass defect, 802\u2013803, 804 prob. Mass number, 800, 801 Index 935 927-942 EM INDEX-845813 6/10/04 9:20 PM Page 936 x e d n I Mass spectrometers, 701\u2013702, 704; Millikan\u2019s oil-drop experiment, 573, identification of trace materials with, 704; ions charge-to-mass ratio from, 702; isotope analysis with, 704; mass of an atom from, 703 prob.; modeling of, 702 lab; purification of samples with, 704 Math, 4\u20137, 833\u2013857. See also Connecting Math to Physics; absolute value, 841; dimensional analysis, 6\u20137, 847; equations, solving, 843\u2013847; exponents, 839\u2013840; fractions, 837; geometry, 853\u2013855; logarithms, 857; proportions, 838; rates, 838; rounding, 835; scientific notation, 841\u2013843; significant digits, 7, 833\u2013834, 835; square and cube roots, 839\u2013840; symbols, 833; trigonometry, 855\u2013856 Matter, energy contained in (E mc2), 802; propagation of waves through, 7", "07; states of. See States of matter; wavelength of, 735, 736 prob. Matter waves, 735, 736 prob. Maximum static friction force (Ff, static), 127, 129 Maxwell, James, 706, 723 Measurement error, 11\u201312, 13 Measurements, 8 lab, 11; accuracy of, 13; comparing, 11\u201312; errors in reading, 13; margin of error, 11\u201312; precision of, 8 lab, 12; SI units, 5; significant digits and, 7 Mechanical advantage (MA), 266\u2013267; of compound machines, 270, 271 prob.; ideal. See Ideal mechanical advantage (IMA) Mechanical energy (E), 293\u2013295; conservation of, 293\u2013295, 296 prob.; loss of, 295 Mechanical waves, 381\u2013382; amplitude of, 382\u2013383; behavior of at boundaries, 387\u2013388; depicting on graphs, 384; frequency of, 384; interference between, 388\u2013389, 389 lab; period of, 383, 385 prob.; phase of, 383; resonance and, 389; speed of, 382, 384, 385 prob., 387; superposition of, 388\u2013389; two-dimensional, 390\u2013391; wavelength of, 383, 385 prob. Mega-, 6 table Meitner, Lise, 811 Melting point, 323, 324 lab Memory, computer, 22 Mercury (element), 350 Mercury (planet), radius, mass, and distance from the Sun, 173 table; shape of orbit of, 171 lab Mesons, 818 Metals, kinetic energies of electron ejected from, 731; work function of, 731, 732 prob. Meter (m), 5 table Michelson, Albert, 438 Micro-, 6 table Microchips, 788\u2013789 Microphones, 673 Microscopes, lenses in, 503; scanning tunneling microscope (STM), 740 Microwaves, 711, 712 Milli-, 6 table Millikan, Robert A., 573, 698, 731 574 prob., 698 Mini Labs. See also Launch Labs; Physics Labs; bright-line emission spectra, 755 lab; change, measuring, 8 lab; changing velocity and, 58 lab; convex lenses, masking of, 495 lab; diffraction on the human retina, 531 lab; electric field strength, 573 lab; emission spectrum from hot body, 724 lab; energy, exchange, 301 lab; hot and cool colors, 441 lab; inclined plane, 135 lab; induction and conduction, 549 lab; instantaneous velocity vectors, 46 lab; light-emitting diodes, 788 lab; magnetic fields, three-dimensional, 650 lab; mass spectrometer, modeling of, 702 lab; melting point, 324 lab; parallel resistance, 623 lab; pressure of a solid, 345 lab; projectile motion, 148 lab; rebound heights, 239 lab; retinal projection screen, 531 lab; schematic diagrams, 599 lab; spinning tops, 213 lab; virtual image position, 462 lab; weightlessness of water, 182 lab; wheels and axles on bicycle gears, 270 lab; wind instruments, 418 lab Mirages, 490 Mirror equation, 467 Mirror image, 461 Mirrors, comparison of types, 473, 473 table; concave, 464\u2013468, 469 prob., 470, 474\u2013475 lab; convex, 471, 472 prob., 473 table; focal length (f ), 464; focal point, 464; in lasers, 763; locating image on curved, 465\u2013468, 469 prob.; magnification by spherical, 468; mirror equation, 467; plane, 460 prob., 461\u2013463, 462 lab; production of, 457; reflection of image onto screen by, 457 lab Models, 19 Moderator, 812 Mole (mol), 5 table, 345 Moment of inertia (I), 205\u2013206, 207 prob., 208 prob.; equation for, 205; of various objects, 206 table Momentum (p), 230; angular, 233\u2013235, 243\u2013245; conservation of, 237, 243\u2013244; impulse-momentum theorem and, 230\u2013231, 232 prob.; of a particle, 737; rebound heights, 239 lab; recoil and, 238\u2013239; space propulsion and, 239, 240 prob.; two-dimensional collisions and, 241, 242 prob., two-particle collisions and, 229 lab, 236\u2013237, 237 prob., 246\u2013247 lab Monochromatic light, 516 Moon (Earth\u2019s), distance to, 13; as illuminated source, 432; lunar eclipses and, 487 Moonlight, 432 Moons, orbital sizes of Jupiter\u2019s, 174 prob. Moore, Gordon, 22 Morpho butterfly, thin-film interfer- ence on wing of, 523, 530 Motion, accelerated. See", " Acceleration; circular. See Circular motion; force and. See Force; periodic. See Periodic motion; planetary. See Planetary motion; rotational. See Rotational motion; two-dimensional. See Two-dimensional motion; uniform straight line. See Straight-line motion Motion diagrams, 32\u201333, 48\u201349 lab; acceleration on, 60, 60 prob.; average velocity, 46\u201347; changing velocity and, 58; coordinate systems and, 34\u201335; particle model, 33; time intervals and displacement on, 36\u201337 Motor oil, viscosity of, 349 Motors, electric, 656, 656 prob. MRI magnets, 603 Multi-gear bicycles, 272, 276 Multiple objects, straight-line motion of, 40, 41 prob., 42 Multipliers, 655 Muon neutrinos, 819 illus. Muons, 818, 819 illus. Musical intervals, 418 Music, hearing loss from loud, 406\u2013407 Music, physics of, 411\u2013419; beats, 418, 419 illus.; consonance and dissonance, 418; fundamental frequency, 417; harmonics, 417; loudness of notes, 414; musical intervals, 418; percussion instruments, 411; reed instruments, 411; reproduction of music and noise, 419; resonance in air columns, 412\u2013414, 446 prob.; resonance on strings, 414\u2013415; sound spectrum and, 417; sources of sound, 403 lab, 411; stringed instruments, 411, 414\u2013415, 417 prob.; timbre (tone quality or color), 415; wind instruments, 418 lab Mutual inductance, 682 Myopia, 501 n-type semiconductors, 781. See also Diodes; Transistors Nano-, 6 table National Institute of Science and Technology (NIST), 6 Natural gas, 331 Nearsightedness, 501 Negative acceleration, 61\u201362 Negative exponents, 842 Negative position, 35 Neodymium, 645 Neon signs, plasma in, 348 Neptune, discovery of, 179; radius, mass, and distance from the Sun, 173 table Net force, 92; centripetal force and, 155; of a coiled spring, 376\u2013378; of a pendulum, 378\u2013379 Net torque, 203 Neutral objects, 543\u2013544, 547. See also Charged objects Neutrinos, 818, 822 Neutrons, mass of nucleus and, 800; strong nuclear forces and, 802 Newton (N), 91 Newton, Sir Isaac, 10 illus., 175, 179, 440 Newton\u2019s first law, 94\u201395 936 Index 927-942 EM INDEX-845813 7/29/04 4:32 PM Page 937 I n d e x Newton\u2019s second law, 93; acceleration due to Earth\u2019s gravitation and, 182; apparent weight and weightlessness, 98; bathroom scales and, 96; Bohr model of the atom and, 754\u2013755; centripetal force and, 155; drag force and terminal velocity and, 100\u2013101; horizontal acceleration and, 97 prob.; Problem-Solving Strategies, 98 Newton\u2019s second law for rotational motion, 208, 209 prob. Newton\u2019s third law, 102\u2013103, 104 prob. NIST-F1 cesium clock, 50 Node, 389 Noise, 419 Nonideal fluids, viscosity of, 349 Nonlinear relationships, 17\u201318; inverse relationship between two variables, 18, 851; quadratic relationship between two variables, 17, 852 Nonreflective eyeglasses, 520 Normal, 391 Normal force (FN ), 94 table, 107, 127 npn-transistor, 787 Nuclear equations, 807, 808 prob., 809 prob. Nuclear fission, 811\u2013812 Nuclear fusion, 813\u2013814, 826 Nuclear model of the atom, 748\u2013752; absorption spectrum and, 751\u2013752; emission spectra and, 749\u2013750, 752; problems with, 752; Rutherford\u2019s gold foil experiments and, 748\u2013749 Nuclear reactions, 807; equations for, 807, 808 prob., 809 prob.; nuclear fission and, 811\u2013812; spontaneity of, 803\u2013804 Nuclear reactors, 812\u2013813 Nucleons, 802, 819 Nucleus, 749, 799\u2013805; binding energy of, 799 lab, 802\u2013804, 804 prob.; charge of, 800; discovery of, 799; mass defect of, 802\u2013803, 804 prob.; mass of (mass number), 800, 801; Rutherford\u2019s experiments on. See Gold foil experiment, Rutherford\u2019s; size of, 800; strong nuclear forces and, 802 Nuclides, 801 Object, 461 Oboe, 417 Obs", "erved light frequency (fobs), 446 Ocean trench, pressure at deepest, 343 table Oersted, Hans Christian, 648, 671, 705 Ohm (\u2126), 595 Ohm, George Simon, 596 Ohm\u2019s law, 595\u2013597 Oil-drop experiment, Millikan\u2019s, 573, 574 prob., 698 Oil, thin-film interference and, 522 prob. Opaque media, 433, 458 Open-pipe resonators, 412, 414 Opposite charges, 542\u2013543 Optical fibers, total internal reflec- tion and, 490 Optical systems, adaptive, 476 Optics. See Lenses; Mirrors Orbital energy values, 755\u2013756, 757 prob. Orbital periods, 172; Kepler\u2019s first law and, 172; Kepler\u2019s second law and, 173; Kepler\u2019s third law and, 173, 174 prob.; Newton\u2019s law of universal gravitation and, 176; of a planet orbiting the Sun, 173, 176; of satellite orbiting Earth, 180, 181 prob. Orbital radius, 755, 760 Orbits, early studies of, 171; geosynchronous, 180; Kepler\u2019s laws and, 172\u2013173, 174 prob.; Newton\u2019s law of universal gravitation and, 175\u2013178; Newton\u2019s thought experiments on, 179; orbital speed and period of a planet orbiting Earth, 180, 181 prob.; period of a planet orbiting the Sun, 176; shape of, 171 lab, 172 Order of operations, 843 Organ pipes, source of sound from, 411 Origin, coordinate system, 34 Oscillation cycle, coil and capacitor circuit, 709\u2013710 Outward force. See Centrifugal force Ovenproof glass, 360 p-type semiconductors, 781. See also Diodes; Transistors Pair production, 820 Parabola, 17 illus., 147 Parabolic dish antennas, 712 Parabolic mirrors, 467 Parallax, 13 Parallel circuits, 623\u2013624, 624 lab, 625\u2013626 prob., 626, 632\u2013633 lab. See also Series circuits; Series-parallel circuits Parallel connections, 600 Paraxial ray simplification, 467 Particle accelerators, 815\u2013816 Particle detectors, 816\u2013817 Particle model of light, 737; Compton effect and, 733\u2013734; photoelectric theory and, 726\u2013729, 730 prob., 731, 732 prob., 738\u2013739 lab; radiation from, 724 lab; incandescent bodies and, 724\u2013725 Particle model of motion, 33 changing velocity and, 58; uniform motion and, 33 Particles, Compton effect and, 733\u2013734; de Broglie wavelength and, 735, 736 prob.; Heisenberg uncertainty principle and, 737; location of, 737; magnetic forces on, 657, 658 prob.; momentum of, 735, 737; photoelectric theory of, 726\u2013729, 730 prob., 731, 732 prob.; wave properties of, 735, 736 prob., 737 Pascal (Pa), 342 Pascal, Blaise, 352 Pascal\u2019s principle, 352 Paschen series, 756 illus. Pauli, Wolfgang, 818 Pendulums, 378\u2013379; acceleration of due to gravity, 378\u2013379, 379 prob., 392\u2013393 lab; conservation of mechanical energy and, 294\u2013295; examining wave characteristics with, 392\u2013393 lab; period of, 379 Penzias, Arno, 9 illus. Percussion instruments, source of sound from, 411 Period (T), 375; of a coiled spring, 378; harmonic, 375; of a pendulum, 379, 392\u2013393 lab; of a wave, 383, 385 prob. Periodic motion, 375\u2013380, 392\u2013393 lab Periodic wave, 381, 383 Permanent magnets, 645; magnetic fields around, 645\u2013647 Perspiration, evaporative cooling by, 351 Phase, mechanical wave, 383 Phosphor, 657 Photocells, 726; photoelectron kinetic energy, 728\u2013729, 730 prob., 732 prob.; threshold frequency, 726\u2013727; work function of a metal and, 731, 732 prob. Photocopiers, 553 Photoelectric effect, 726; Einstein\u2019s theory of, 727\u2013729; energy of a photon and, 727\u2013728, 731 prob.; kinetic energy of an ejected electron, 728\u2013729, 730 prob.; Millikan\u2019s experiments proving, 731; modeling, 738\u2013739 lab; threshold frequency and, 726\u2013727; work function of a metal and, 731, 732 prob. Photon energy (E), 7", "27, 753\u2013754; kinetic energy of electron, 728\u2013729, 730 prob.; Problem-Solving Strategies, 728; work function of a metal and, 731, 732 prob. Photons, 727, 820; energy of, 727\u2013729, 731 prob., 753\u2013754; frequency and wavelength of, 758 prob.; momentum of, 733\u2013734 Photoresistors, 620 Photosynthesis, chlorophyll and, 442 Photovoltaic cells, 592 Physicists, 3 Physics, 3\u20134. See also Extreme Physics Physics Labs, acceleration due to gravity, 76\u201377 lab; capacitors, charging of, 580\u2013581 lab; charged objects, 554\u2013555 lab; coefficient of friction, 136\u2013137 lab; concave mirror images, 474\u2013475 lab; conservation of energy, 302\u2013303 lab; convex lenses and focal length, 504\u2013505 lab; double-slit interference, 532\u2013533 lab; electric circuits, 632\u2013633 lab; electromagnetic wave shielding, 714\u2013715 lab; electromagnet, creating, 660\u2013661 lab; evaporative cooling of alcohols, 364\u2013365 lab; forces in an elevator, 108\u2013109 lab; measurement of speeding objects, 20\u201321 lab; motion diagrams, creating, 48\u201349 lab; orbits of planets and satellites, 186\u2013187 lab; pendulums and waves, 392\u2013393 lab; photoelectric effect, 738\u2013739 lab; polarization of light, 448\u2013449 lab; power (P) on a stair climber, 274\u2013275 lab; projectile motion, 160\u2013161 lab; Index 937 927-942 EM INDEX-845813 6/10/04 9:22 PM Page 938 x e d n I radiation intensity and distance, 824\u2013825 lab; speed of sound, 420\u2013421 lab; thermal energy and water temperature, 332\u2013333 lab; transformers, induced voltage of, 686\u2013687 lab; translational and rotational equilibrium, 218\u2013219 lab; two-particle collisions, 246\u2013247 lab; voltage, current, and resistance relationships, 606\u2013607 lab Physics of music. See Music, physics of Piano, 411, 417 illus. Pico-, 6 table Piezoelectricity, 711 Pigments, 441\u2013442; chlorophyll, 442; complementary, 442; primary, 441, 442; secondary, 441 Pion, 818 Pitch, 406; Doppler effect and, 407 Planck, Max, 725, 728 Planck\u2019s constant (h), 725, 727 Plane mirrors, 460 prob., 461\u2013463, 462 lab, 473 table Planetary model of the atom. See Nuclear model of the atom Planetary motion, angular momentum (L) and, 234; Brahe\u2019s study of, 171, 172; Copernicus\u2019s study of, 171; Einstein\u2019s general theory of relativity and, 185; gravitational fields and, 182\u2013183; Kepler\u2019s laws and, 172\u2013173, 174 prob.; modeling, 186\u2013187 lab; Newton\u2019s law of universal gravitation and, 175\u2013178; Newton\u2019s thought experiments on satellites and, 179; orbital period, 176, 180, 181 prob.; orbital shape, 171 lab, 172; orbital size, 174 prob.; orbital speed, 180, 181 prob. Planets, motion; radius, mass, and distance of from the Sun, 173 table; See also Planetary motion Plants, capillary action and, 350; chlorophyll in, 442; cohesive forces in, 350 Plasma, 348, 545, 556 Plasma contactor, 556 Pluto, 173 table pn-junction diode, 784\u2013786, 785 prob., 786 prob. pnp-transistor, 787 Point source, illuminance of, 435, 436 prob.; luminous intensity of, 434 Poisson, Simeon Denis, 519 Polarization of light, 443\u2013444, 444 prob., 448\u2013449 lab Polarized, 644, 645 Polarized light, 443 Polarizing filters, 443\u2013444, 444 prob., 448\u2013449 lab Pole-vaulting, elastic potential energy and, 291\u2013292 Polonium, 799 Position (d), 34; constant acceleration and, 66, 68, 68 table; instantaneous, 40; negative, 35 Position-time graphs, 38\u201342; average speed and, 44; constant acceleration and, 66; motion of multiple objects, 40, 40 prob., 41 prob., 42; motion of single object, 38, 39 prob., 40; object velocity and, 43\u201344; straight-line formula vs., 46, 47 table Positive acceleration, 61\u201362 Positive exponents, 841 Positron Emission", " Tomography (PET), 811 Positrons, 819 Potential energy, of an atom, 289; elastic, 291\u2013292; gravitational, 288\u2013289, 290 prob., 293\u2013294; in a spring 376\u2013378, work and, 288 Potentiometer, 596 illus., 597 Power (P), 263, 265, 268 prob.; average, 677; of electric currents, 593, 594 prob., 677; of an electric motor, 264 prob.; of electromagnetic waves, 725; from a resistor, 601\u2013602, 602 prob.; on a stair climber, 274\u2013275 lab; unit for (W), 263 Power (mathematics), 840 Power companies, 604\u2013605 Practice Problems, acceleration (Chapter 3), 61 prob., 64 prob., 65 prob., 67 prob., 69 prob., 71 prob., 74 prob.; atoms (Chapter 28), 757 prob., 758 prob.; conservation of energy (Chapter 11), 287 prob., 291 prob., 297 prob., 300 prob.; current electricity (Chapter 22), 594 prob., 598 prob., 600 prob., 603 prob., 605 prob.; electric fields (Chapter 21), 565 prob., 566 prob., 571 prob., 572 prob., 574 prob., 578 prob.; electromagnetic induction (Chapter 25), 675 prob., 678 prob., 684 prob.; electromagnetism (Chapter 26), 700 prob., 703 prob., 706 prob., 707 prob.; force and motion (Chapter 4), 89 prob., 93 prob., 97 prob., 100 prob., 104 prob., 106 prob.; force in two-dimensions (Chapter 5), 121 prob., 125 prob., 128 prob., 130 prob., 133 prob., 135 prob.; fundamentals of light (Chapter 16), 436 prob., 447 prob.; illumination (Chapter 16), 436 prob., 447 prob.; interference and diffraction (Chapter 19), 519 prob., 522 prob., 526 prob., 529 prob.; magnetic fields (Chapter 24), 647 prob., 650 prob., 654 prob., 658 prob.; momentum (Chapter 9), 233 prob., 238 prob., 240 prob.; motion in two dimensions (Chapter 6), 150 prob., 152 prob., 156 prob., 159 prob.; the nucleus (Chapter 30), 801 prob., 805 prob., 808 prob., 809 prob., 810 prob.; physics toolkit (Chapter 1), 5 prob., 7 prob., 8 prob., 18 prob.; planetary motion (Chapter 7), 174 prob., 181 prob.; quantum theory (Chapter 27), 730 prob., 732 prob., 736 prob.; reflection and mirrors (Chapter 17), 460 prob., 469 prob., 472 prob.; refraction and lenses (Chapter 18), 487 prob., 496 prob., 497 prob.; representing motion (Chapter 2), 39 prob., 41 prob., 45 prob.; rotational motion (Chapter 8), 200 prob., 203 prob., 205 prob., 208 prob., 210 prob., 215 prob.; series and parallel circuits (Chapter 23), 619 prob., 622 prob., 623 prob., 626 prob., 630 prob.; solid-state electronics (Chapter 29), 778 prob., 780 prob., 783 prob., 786 prob.; sound (Chapter 15), 405 prob., 409 prob., 416 prob.; states of matter (Chapter 13), 344 prob., 356 prob., 362 prob.; static electricity and electric force (Chapter 20), 552 prob.; thermal energy (Chapter 12), 317 prob., 319 prob., 321 prob., 325 prob., 328 prob.; vibration and waves (Chapter 14), 379 prob., 386 prob.; work, energy, and simple machines (Chapter 10), 261 prob., 262 prob., 264 prob., 272 prob. Precision, 8 lab, 12, 13\u201314 Predictions, graphs and, 19 Prefixes, SI, 6 table Pressure (P), 342; atmospheric, 343; Boyle\u2019s law (volume and pressure of a gas) and, 344, 346 prob.; calculation of, 342, 343 prob., 345 lab; combined gas law (pressure, volume, and temperature), 345; of fluids, 342, 352, 353, 357\u2013358; Pascal\u2019s principle and, 352; relationship with force and area, 342; typical values, 343 table; unit of (Pa), 342 Pressure waves, damage to human ear by, 406\u2013407; detection of by human ear, 405; loudness and, 406\u2013407; pitch and, 406; resonance length and, 412\u2013414; sensitivity of human ear to, 406; sound as, 404\u2013405 Pressurized water reactors, 813 Primary coils, 682 Primary colors, 440 Primary pigments, 441, 442 Principle axis, 464 Principle of superposition, 388 Principle quantum number, 755 Prism, dispersion of light and, 491 Problem-Solving Strategies, conservation of energy, 295; electric force, 550; force and motion, 98; interaction pairs", ", 103; line-graphs, plotting, 16; motion in two dimensions (projectile motion), 149; ray tracing to locate images formed by curved mirrors, 466; schematic diagrams, 599; series-parallel circuits, 629; thin-film interference, 521; units of hc and photon energy, 728; vector addition/subtraction, 123; work, 260 Projectile launcher, designing, 160\u2013161 lab Projectile motion, 147 lab, 147\u2013152; air resistance and, 152; factors affecting path of, 160\u2013161 lab; mass and, 148 lab; Problem-Solving Strategies, 149; of projectiles launched at an angle, 150, 151 prob., 152 prob. Projectiles, 147 Proportions, 838 Protons, atomic number (Z) and, 800; charge-to-mass ratio of, 699\u2013700; electromagnetic force in nucleus and, 802; mass of from charge-to-mass ratio, 700; mass of nucleus and, 800; strong nuclear forces and, 802 Pulleys, 269; ideal mechanical advantage (IMA) of, 269; mechanical advantage (MA) of, 267; Pythagorean theorem, 120 Quadratic equations, 846 Quadratic formula, 846 938 Index 927-942 EM INDEX-845813 6/10/04 9:22 PM Page 939 I n d e x Quadratic graphs, 852 Quadratic relationships, 17, 850 Quantized energy, 725 Quantum mechanics, 761 Quantum number, principle, 755 Quantum theory of the atom, 760\u2013761 Quarks, 818, 819 table, 820, 822 Quartz crystal, piezoelectricity and, 711 Quartz-halogen lamps, 450 Quasars, double-image, 506 Radar detectors, 410 Radian (rad), 197, 198 illus. Radiation, 317; alpha, 806, 807 table; beta, 806, 807 table; distance and intensity of, 824\u2013825 lab; gamma, 806, 807 table; from incandescent bodies, 723 lab, 724\u2013725; medical uses of, 811 Radiation detectors, 817, 824\u2013825 lab Radio waves, receiving, 697 lab, 712; shielding from, 714\u2013715 lab Radioactive, 806 Radioactive dating, 11, 810 Radioactive decay, 18, 799, 806; activity and, 810; alpha, 806, 808 prob.; beta, 806, 808 prob.; gamma, 806; half-life and, 809\u2013810, 813 lab; nuclear equations showing, 807, 808 prob., 809 prob. Radioactive iodine, 811 Radioactive isotopes, activity of, 810; analysis with mass spectrometer, 701, 704; artificial, 811; half-life of, 809\u2013810, 810 table, 813 lab; medical uses of, 811; nuclear equations and, 807, 808 prob., 809 prob. Radioactive waste, 813 Radioactivity, activity and, 810; artificial, 811; cancer treatment with, 811; discovery of, 799; distance and intensity of, 824\u2013825 lab; half-life and, 809\u2013810, 810 table, 813 lab; nuclear equations and, 807, 808 prob., 809 prob.; types of radiation, 806, 807 table Radiocarbon dating, margin of error, 11 Rainbows, 491\u2013492 Rates, 838 Ratios, 838 Ray model of light, 432\u2013434; illumination of a surface and, 435, 436 prob., 437; luminous flux (P) and, 433; luminous intensity and, 434; speed of light and, 437\u2013438 Ray optics, 432 Ray tracing, concave mirrors and, 465\u2013467, 470 prob.; convex lenses and, 495, 496 prob., 497; convex mirrors and, 471, 472 prob.; plane mirrors and, 462; two-dimensional waves and, 391 Rayleigh criterion, 530\u2013531 Rayleigh, Lord, 530 Reactions, chain, 812; fusion, 813\u2013814, 826 Real image, 465; concave mirrors and, 465\u2013467, 474\u2013475 lab; convex lenses and, 494\u2013495, 496 prob. Real weight, 98, 99 prob. Receivers, 712 Recoil, 238\u2013239 Red-shifted light, 446 Reducing equations, 408 Reed instruments, 411 Reeds, 411 Reference level, 288, 289, 290 prob. Reflected wave, 387 Reflection, 457\u2013473; angle of, 391; concave mirrors and, 464\u2013468, 469 prob., 470; convex mirrors, 471", ", 472 prob.; diffuse, 459; of image onto movie screen, 457 lab; law of, 391, 458\u2013459, 460 prob.; plane mirrors and, 461\u2013463; polarization of light by, 443; regular, 459; total internal, 489\u2013490 Reflection gratings, 515 lab, 527, 529 prob. Refracting telescopes, 502 Refraction, 391, 485 lab, 485\u2013492; angle of, 486, 487 prob., 488; dispersion of light by, 491\u2013492; index of, 486, 486 table, 489; lenses and. See Lenses; lunar eclipses and, 487; mirages and, 490; Snell\u2019s law of, 486\u2013487, 488; total internal reflection and, 489\u2013490; two-dimensional waves and, 391; wave model of, 488\u2013489 Refrigerators, 327\u2013328 Relative velocity, 157\u2013159 Relativity, Einstein\u2019s theory of, 184\u2013185 Replica gratings, 527 Resistance (R), 595\u2013597; factors affecting, 595 table; in parallel circuits, 624, 625\u2013626 prob.; reducing loss of energy by, 604; resistance-current relationships, 4 lab, 18, 606\u2013607 lab; in series circuits, 619\u2013620, 621 prob. 622 prob.; in series-parallel circuits and, 629, 630 prob.; zero in superconductors, 603 Resistance force (Fr), 266 Resistivity, 777 Resistors, 596\u2013597, 598 prob.; currentvoltage characteristics, 790\u2013791 lab; in parallel circuits, 623\u2013624, 625\u2013626 prob.; release of thermal energy by, 601\u2013602; in series circuits, 619\u2013620, 621 prob. Resonance, 380; in air columns, 412\u2013414; hearing and, 414; speed of sound from, 413, 416 prob., 420\u2013421 lab; on strings, 414\u2013415 Resonance length, 412\u2013413, 416 prob. Resonant cavities, 711 Resonators, 412\u2013414; closed-pipe, 412; open-pipe, 412; resonance length and, 412\u2013413; standing pressure waves and, 412 Rest energy (E0), 292 Resultant force, 131 Resultants, 35 Retinal projection screen, 531 lab Reversible air conditioners. See Heat pumps Revolution, angle of (), 198; fractions of (radians), 197, 198 illus. Right-hand rules, first, 649; fourth, 672; second, 649; third, 652 Ripple tanks, 390\u2013391 Rise, 17 Rockets, affect of Coriolis force on, 217; space propulsion and, 239, 240 prob. Rocks, history of Earth\u2019s magnetic field in, 651 Rod, pendulum, 378 Roemer, Ole, 437\u2013438 Roentgen, Wilhelm, 713 Roller coasters, confusion of body\u2019s systems by, 138; conservation of mechanical energy and, 294 Rope model of light waves, 443 Ropes, forces of, 105, 106 prob. Rossi X-ray Timing Explorer, 188 Rotating frames of reference, 216 Rotational equilibrium, 213, 218\u2013219 lab Rotational kinetic energy (KErot), 287 Rotational motion, 197\u2013217; effect of force on angular velocity, 201; angle of revolution, 198; angular acceleration, 199; angular displacement, 198; angular frequency, 200; angular momentum (L), 233\u2013235; angular velocity, 198\u2013199; center of mass and, 211\u2013213, 213 lab; centrifugal \u201cforce\u201d and, 216; Coriolis \u201cforce\u201d and, 216\u2013217; distance of a point on a rotating object, 198; equilibrium and, 213, 214\u2013215 prob., 218\u2013219 lab; lever arms, 201\u2013202, 202 prob.; measurement of (radians), 197, 198 illus.; moment of inertia (I) and, 205\u2013206, 207 prob., 208 prob.; Newton\u2019s second law of, 208, 209 prob.; rotating frames of reference, 216; rotational kinetic energy (KErot), 287; sign and direction of, 198; stability of objects and, 213; torque and, 202\u2013203, 204 prob. Rounding, 835 Rubbing alcohol, evaporation of, 351, 364\u2013365 lab Run, 17 Running shoes, physics of, 231, 304 Rutherford, Ernest, 543, 748, 800; discovery of three types of radiation, 806; gold foil experiments of, 748\u2013749, 766\u2013767 lab, 799 Satellites, mass of, 180; orbits of, 173, 180, 181 prob., 186\u2013187 lab; speed of, 180, 181 prob. Saturn, 173 table", " Saxophone, 411, 412, 414 Scalars, 35 Scales, weight (Fg) and, 96, 98, 99 prob., 110 Scanning tunneling microscope (STM), 737, 740 Schematic diagrams, 597, 597 table, 598 prob., 599, 599 lab Schr\u00f6dinger, Erwin, 760 Schwarzschild radius, 188 Scientific law, 9 Index 939 927-942 EM INDEX-845813 6/10/04 9:23 PM Page 940 x e d n I Scientific method, 8\u201310 Scientific notation, 841\u2013843 Scientific theory, 10 Scuba diving, regulation of pressure, 344 illus. Seafloor spreading, 651 Second (s), 5 table Second law of thermodynamics, 329 prob., 330\u2013331 Second right-hand rules, 649 Secondary coils, 682 Secondary colors, 440 Secondary pigments, 441 Segr\u00e9, Emilio, 811 Self-inductance, 681\u2013682 Semiconductors, 775, 779\u2013783; diodes, 784\u2013786, 785 prob., 786 prob., 788 lab, 790\u2013791 lab; doped, 781\u2013782, 782 prob.; extrinsic, 781; free electrons in, 779\u2013780, 780 prob., 782 prob.; intrinsic, 780, 780 prob.; light meters using, 783; n-type, 781; p-type, 781; thermistors, 782; in transistors and integrated devices, 787\u2013789 Separation of charges, 547 Series circuits, 618\u2013620, 619 prob., 632\u2013633 lab. See also Parallel circuits; Series-parallel circuits; current in, 619; equivalent resistance and, 619; voltage drops in, 620, 621 prob. Series connections, 600 Series-parallel circuits, 629, 630 prob. Sextant, 172 illus. Sharing of charges, 575\u2013576 Ships, floating of, 355 Shock absorbers, running shoes as, 304 Shock, electric, 597 Short circuits, 627, 628 Short-period comets, 172 SI units, 5\u20136; base units, 5, 5 table; converting between, 6; prefixes, 6 table Significant digits, 7, 8 prob., 12, 833\u2013834, 835\u2013836 Silicon, semiconductors and, 779, 781, 782 prob. Simple circuits. See Series circuits Simple harmonic motion, 375; amplitude of, 375; pendulums and, 378\u2013379, 379 prob., 392\u2013393 lab; period of (T), 375; resonance and, 380; springs and, 376\u2013378, 377 prob., 380 prob. Simple machines, 269; human body as, 273; ideal mechanical advantage (IMA), 269; six types of, 269 Sine waves, resonance length and, 412\u2013414 Sines, law of, 120, 856 Single-slit diffraction, 524\u2013526; index of refraction of unknown substance, 526 prob.; width of bright bands on diffraction pattern, 525 Sinusoids, 18 Skiing, conservation of mechanical energy and, 294 Skin, resistance of, 597 Sky, color of, 442 Slope of a line (m), 17, 850; on force- acceleration graphs, 91; motion diagrams and, 46; position-time graphs and, 43\u201344 Snell, Willebrord, 486 Snell\u2019s law of refraction, 486\u2013487, 501 prob. Soap, spectrum produced by, 520 Soddy, Frederick, 799 Solar and Heliospheric Observatory (SOHO), 422 Solar breeze, 248 Solar cells, 592 Solar collectors, 327 Solar sails, 248 Solenoids, 649 Solid-state electronics, 775\u2013789; conductors and, 777, 778 prob.; diodes, 784\u2013786, 785 prob., 786 prob., 790\u2013791 lab; insulators, 778\u2013779; semiconductors, 775, 779\u2013783; transistors and integrated devices, 787\u2013789 Solids, 323; amorphous, 359; band theory of, 776\u2013777; conduction of electricity by, 775\u2013783; crystalline, 359; elasticity of, 360; heat of fusion of, 324, 324 table, 325 prob.; insulating, 778\u2013779; melting point of, 323, 324 lab; pressure and, 342, 343 prob., 359\u2013360; thermal expansion of, 347, 360\u2013361, 362 prob., 363, 363 prob. Songbird migration, relative velocity and, 159 Sound. See also Sound waves; music. See Music, physics of; properties and detection of, 403\u2013407 Sound detectors, 405 Sound level,", " 406 Sound quality, 415, 417\u2013418 Sound spectra, 417 Sound waves, 387, 403, 404\u2013410; detection of by human ear, 405; Doppler effect and, 407\u2013408, 409 prob., 410; loudness and, 406\u2013407; from musical instruments. See Music, physics of; pitch and, 406; properties of, 404\u2013405; reproduction of, 419; sound detectors and, 405; speed of, 404\u2013405, 405 table, 413, 416 prob., 420\u2013421 lab; in the Sun, 422; wavelength of, 404 Space, affect of mass on (Einstein\u2019s theory of relativity), 184\u2013185 Space heaters, release of thermal energy by, 601, 602 prob. Space stations, rotating, 162 Space telescopes, 476 Space travel, affect on humans, 162; solar sails and, 248; space propulsion, 239, 240 prob.; static electricity and, 556; time dilation and, 78; weightlessness and, 182 Specific heat, 318; of common substances, 318 table; measuring by calorimetry, 319\u2013320, 321 prob. Spectra, 440; absorption, 751\u2013752; electromagnetic, 708 illus., 709; emission. See Emission spectra; of incandescent body, 723 lab; rainbows and, 491\u2013492 Spectroscopes, 750, 752, 765 Spectroscopy, 752, 765 Specular reflection, 459 Speech, 411 Speed, average, 20\u201321 lab, 44, 45 prob. Speed of light, ray model of light and, 437\u2013438; wave model of light and, 445\u2013447; wavelength-frequency relationship and, 706 Speed of sound, 404\u2013405, 405 table, 413, 416 prob., 420\u2013421 lab Speedometers, 662 Spherical aberration, 467, 498 Spherical mirrors, 464; aberration on, 467; locating image on, 467, 469 prob.; magnification by, 468; mirror equation for, 467; ray tracing and, 465\u2013467 Spiderwebs, surface tension and, 349 Spontaneous emissions, 762 Sport utility vehicles, stability of, 220 Spring constant (k), 376, 377 prob. Spring force (Fsp ), 94 table Springs, elastic, 375 lab, 376\u2013378, 380 prob.; graphing elongation of, 17 prob.; mass on (Hooke\u2019s law), 376, 377 prob.; potential energy of, 376\u2013378, 377 prob.; waves in, 375 lab, 381, 389, 389 lab Square root property, 845 Square roots, 839 Stability, 212\u2013213, 213 lab, 220 Stair climbers, work and power and, 274\u2013275 lab Standard model of matter, 818\u2013823; antimatter and, 819\u2013820; leptons and quarks, 818\u2013819, 819 table; protons and neutrons, 819; testing of, 822\u2013823; weak nuclear force and, 821\u2013822 Standardized Test Practice, 29, 55, 85, 117, 145, 169, 227, 283, 311, 339, 373, 401, 429, 455, 483, 513, 539, 561, 589, 615, 641, 669, 695, 721, 745, 773, 797, 831 Standing waves, 389, 412\u2013413 Stars, plasma in, 348 States of matter, Bose-Einstein condensates, 366, 768; changes in, 323\u2013324, 324 lab, 325 prob.; fluids. See Fluids; gases. See Gases; solids. See Solids Static electricity, applications of, 552\u2013553; charging by conduction, 547, 549 lab; charging by induction, 548, 549 lab; conductors and insulators and, 544\u2013545; Coulomb\u2019s law, 549\u2013550, 551 prob., 552, 552 prob.; damage to electronic devices by, 570; electric charge and, 541\u2013544, 554\u2013555 lab; electric force and, 541 lab, 546\u2013550, 551 prob., 552 prob.; separation of charge on neutral objects, 547; spacecraft and, 556 Static equilibrium, 213, 214\u2013215 prob. Static friction (Ff, static), 126, 127, 129; coefficient of, 127, 136\u2013137 lab Steam heat, 317 Step-down transformers, 682, 683 Step-up transformers, 682, 683, 684 prob. Stimulated emission, 762 STM. See Scanning tunneling microscope Stored energy. See Potential energy Straight line equation, 16\u201317, 46, 47 table, 850 940 Index 927-942 EM INDEX-845813 6/10/", "04 9:23 PM Page 941 I n d e x Straight-line motion, 31\u201347, 31 lab; average speed and, 44, 45 prob.; average velocity and, 43\u201344, 45 prob., 46\u201347; coordinate systems analyzing, 34\u201335; displacement from, 36\u201337; instantaneous velocity and, 46; motion diagrams for, 32\u201333, 48\u201349 lab; of multiple objects, 40, 40 prob., 41 prob., 42; nonuniform vs. uniform, 58; particle model of, 33; position-time graphs of, 38\u201342; of single objects, 38, 39 prob., 40 Strange quarks, 819 illus. Strassmann, Fritz, 811 Streamlines, 358 Stringed instruments, 411, 414\u2013415 Strings, forces of, 105, 106 prob. Stroboscope, 775 lab Strong nuclear forces, 802 Submersibles, Archimedes\u2019 principle and, 355; pressure of water on, 353 Subtractive color process, 441 Sun, color of, 442; fusion in, 814; Kepler\u2019s laws of planetary motion and, 172\u2013173; as luminous source, 432; mass of, 173 table; period of a planet orbiting, 176; power of waves from, 725; pressure at center of, 343 table; radiation from, 317, 432; radius of, 173 table; sound waves in, 422 Superconductors, 603 Superelastic collisions, 298 Superior mirages, 490 Surface gravity waves, 422 Surface tension, 349, 350 illus. Surface waves, 382. See also Mechanical waves Swim bladders, 355 Swimming, pressure and, 353 Symbols, 833 Synchrotrons, 816 System\u00e9 Internationale d\u2019Unites. See SI units Systems, 88; closed, 236; hydraulic, 352; isolated, 236 Tape players, 788 Tau neutrino, 819 illus. Technology. See Extreme Physics; Future Technology; Technology and Society Technology and Society, amusement park rides, 138; cellular phones, 716; hybrid cars, 608; lighting, advances in, 450; running shoes, 304; sport utility vehicles, 220 Telescopes, adaptive optical systems (AOSs) on, 476; refracting, 502; resolving power, 530, 531 Television, 579, 657, 712 Temperature, Charles\u2019s law (volume and temperature of a gas) and, 344; combined gas law (pressure, volume, and temperature), 345, 346 prob.; limits of, 316; power of incandescent bodies and, 724; regulation of internal by animals, 322; SI base quantity (kelvin), 5 table; speed of sound and, 404, 405 table; thermal energy and, 314\u2013315; thermometers and, 315; of the universe, 725 Temperature scales, 316, 317 prob. Tension (FT), 94 table, 105, 106 prob. Tera-, 6 table Terminal velocity, 101 Teslas (T), 653 Test-Taking Tips, 29, 55, 85, 117, 145, 169, 227, 283, 311, 339, 373, 401, 429, 455, 483, 513, 539, 561, 589, 615, 641, 669, 695, 721, 745, 773, 797, 831 Theories, 10 Thermal energy, 295; calorimetry and, 319\u2013320, 321 prob.; changes of state and, 323\u2013324, 324 lab, 325 prob.; entropy and the second law of thermodynamics, 328\u2013331, 329 prob.; first law of thermodynamics and, 326\u2013328; joule heating loss, 603, 604; kinetic and potential energy and, 314; release by resistors, 601\u2013602, 602 prob.; temperature and, 314\u2013315, 332\u2013333 lab; thermometers and thermal equilibrium, 315; transfer of, 313 lab, 317, 318 prob., 319 prob. Thermal equilibrium, 315 Thermal expansion, 347; coefficients of, 361, 361 table; of gases, 347; of liquids, 347; of solids, 360\u2013361, 362 prob., 363, 363 prob. Thermistors, 782 Thermodynamics, 313. See also Thermal energy; first law of, 326\u2013328; second law of, 329 prob., 330\u2013331 Thermometers, 315 Thermonuclear fusion, 826 Thermostats, 363 Thin-film interference, 520\u2013521, 522 prob., 523; applications of, 521; on butterfly wings, 523, 530; ProblemSolving Strategies, 521 Thin lens equation, 493, 498 Third right-hand rule, 652 Thomson, George, 735 Thomson, J. J., 543, 698\u2013699, 747, 748 Thomson tube, charge-to-mass ratio of electrons", ", 698\u2013699; charge-to-mass ratio of protons, 699\u2013700 Thorium-234, 807 illus. Threshold frequency, 726\u2013727; Einstein\u2019s photoelectric-effect theory and, 728; work function of a metal and, 731, 732 prob. Thrust (Fthrust), 94 table Timbre, 415 Time dilation, 78 Time interval, 36 Time, SI base quantity (s), 5 table Tone color, 415 Tone quality, 415 Top quarks, 819 illus. Tops, 213 lab, 244 Torque, 202\u2013203; applying to a steel wheel, 209 prob.; balancing, 204 prob.; of electric motors, 656, 656 prob.; net, 203 Total internal reflection, 489\u2013490 Trajectory, 147, 148; factors affecting, 160\u2013161 lab; perspective of the viewer and, 152; of projectiles launched at angles, 150, 151 prob. Transformers, 682\u2013683; relationship between voltages in two coils of, 686\u2013687 lab; step-down, 682, 683; step-up, 682, 683, 684 prob.; uses of, 685 Transistors, 787\u2013789; computers and, 22; npn, 787; pnp, 787 Translational equilibrium, 213, 218\u2013219 lab Translucent media, 433 Transmission gratings, 527 Transparent media, 433 Transverse wave, 381 Trigonometry, 855\u2013856 Trombone, 412 Trough, wave, 383 Trumpet, 411 Tuba, 411 Tug-of-war, 103 lab, 105 Turbines, 676\u2013677 Twin paradox, 78 Two-dimensional collisions, 241, 242 prob., 244 prob. Two-dimensional motion, ProblemSolving Strategies, 149; projectile motion, 147 lab, 147\u2013152, 160\u2013161 lab; relative velocity and, 157\u2013159, 158 prob.; uniform circular motion and, 153\u2013156 Two-dimensional waves, 390\u2013391; graphing, 390; reflection and, 390\u2013391; refraction and, 391 Two-particle collisions, 236, 237 prob., 246\u2013247 lab Two-point calibration, 13 Two-slit interference. See Double-slit interference Tympanic membrane, 405 Ultrasound, 410 Ultraviolet waves, 711 Uniform circular motion, 153\u2013156; centripetal acceleration, 153, 154\u2013155, 155 prob.; centripetal force, 155 Uniform field, electric potential difference in, 571, 572 prob.; equation for, 571; Millikan\u2019s oil-drop experiment and, 573, 574 prob. Units, converting, 847 Universal gravitation, 175\u2013178; gravitational mass (mgrav) and, 184; Kepler\u2019sthird law and, 176; Newton\u2019s discovery of, 175; orbital period and, 176, 180, 181 prob.; orbital speed and, 180, 181 prob.; universal gravitation constant (G), 176, 177\u2013178 Universal gravitation constant (G), 176, 177\u2013178 Universe, expansion of, 446\u2013447; origins of, 823; temperature of, 725 Up quark, 819 illus. Uranium, 799, 807 illus., 812\u2013813 Uranus, 173 table, 179 Index 941 927-942 EM INDEX-845813 6/10/04 9:24 PM Page 942 mechanical advantage of, 269, 271 prob. White light, 440\u2013441 Wieman, Carl, 366, 768 Wilson cloud chamber, 817 Wind instruments, 418 lab Wire loops, magnetic field near, 649 Wires, force on from magnetic field, 653, 654 prob.; power dissipated in, 604; resistance of, 596, 604 Work (W), 258; calculating, 259\u2013260, 261 prob.; efficiency and, 268; energy change and (work-energy theorem), 258\u2013259, 286\u2013287; equation for, 258; force and displacement at an angle, 259\u2013260, 262 prob.; gravitational potential energy (PE) and, 288; ideal mechanical advantage (IMA) from, 267; kinetic energy and, 258\u2013259; power (P) and, 263, 264 prob., 265, 274\u2013275 lab; Problem-Solving Strategies, 260; wheels and axles on bicycle gears, 270 lab Work-energy theorem, 258\u2013259, 261 prob., 286\u2013287 Work function of a metal, 731, 732 prob. X rays, 713, 733\u2013734 y-intercept (b), 17, 848 Young, Thomas, 516 Yukawa, Hideki, 818 Z0 boson, 822 Zero-g, 182 Zeros, significant, 7 x e d n I v-t graph. See Velocity-time graph", " Vacuum, pressure in, 343 table; speed of light in (c), 438, 445 Vacuum tubes, 775 Valence bands, 776\u2013777 Van de Graaff generator, 567 illus., 568 Van de Graaff, Robert, 568 Van Musschenbroek, Pieter, 577 Variable resistors, 597 Variable Specific Impulse Magnetoplasma Rocket (VASIMR), 556 Variables, 15, 16; dependent, 15, 16; independent, 15, 16; isolating in equations, 845; linear relationships between, 16\u201317, 850; quadratic relationships between, 17, 852 Vector components, 122 Vector resolution, 122 Vectors, 35, 119\u2013125; adding single- dimensional, 35; adding twodimensional, 120, 121 prob., 123, 124 prob.; components of (vector resolution), 122; equilibrant of, 131; length and direction of displacement, 37; subtracting singledimensional, 37 Velocity, 43\u201347; acceleration vs., 64; angular, 198\u2013199; average, 44, 45 prob., 46\u201347; with average acceleration, 65; changing, 58, 58 lab; with constant acceleration, 68; instantaneous, 46; position and constant acceleration and, 68, 68 table; relative, 157\u2013159; time dilation at high, 78 Velocity-time graphs, 58\u201359; acceleration and, 62, 63 prob.; displacement from, 66, 67 prob., 68 Venus, 173 table Vibrating atom, energy of, 725 Vibrational motion. See Periodic motion Videotape recorders, 651 Violin, 411, 417 illus. Virtual images, 461, 473, 473 table; from concave lenses, 494; from concave mirrors, 470, 474\u2013475 lab; from convex lenses, 494, 497; from convex mirrors, 471; from plane mirrors, 462, 462 lab Viscosity, 349 Vision, 500\u2013501, 501 prob.; focusing of images, 500; nearsightedness and farsightedness, 501; resolving power of eye and Rayleigh criterion, 531 Voice, human, 411 Volatile liquids, 351 Volt (V), 569 Voltage, 570, 606\u2013607 lab; effective, 677, 678; Hall, 662; measurement with voltmeter, 631; in parallel circuits, 624; in series circuits, 618\u2013619, 621\u2013623 prob.; of transformers, 683, 684 prob.; voltage dividers and, 620, 621 prob. Voltage divider circuits, 620, 621 prob., 622 prob. Voltage drops, 620, 621 prob., 631 Voltaic cells, 592 Voltmeter, 599, 600, 631, 655 Volume, combined gas law and, 345 W boson, 822 W bosons, 822 Warm-blooded animals, 322 Water, capillary action of, 350; heat of fusion, 324; heat of vaporization, 324; pressure and freezing of, 359\u2013360; pressure of column of on a body, 353; specific heat of, 318; surface tension of, 349, 350 illus.; thermal energy and temperature of, 332\u2013333 lab; waves, 381, 382, 387, 390\u2013391 Watt (W), 263 Watt-hour meters, 604 illus. Watt-second, 604\u2013605 Wave fronts, 390 Wave model of light, 439\u2013447, 737; color and, 440\u2013442, 441 lab; de Broglie wavelength and, 735, 736 prob.; Doppler effect and, 445\u2013446; expansion of the universe and, 446\u2013447; interference and, 515\u2013521; observed light frequency, 446; polarization of light, 442\u2013444, 444 prob., 448\u2013449 lab; reflection and, 458\u2013459; refraction and, 488\u2013489 Wave pulse, 381 Wavelength, 383; of light waves, 445, 518, 519 prob., 528, 532\u2013533 lab; of mechanical waves, 383\u2013384, 385 prob.; refraction and, 488\u2013489; of sound waves, 404; wavelength-frequency relationship, 706 Waves, 381\u2013386; amplitude of, 382\u2013383; behavior of at boundaries, 387\u2013388; crest of, 383; depicting on graphs, 384; electromagnetic. See Electromagnetic waves; frequency of, 384; incident, 387; interference between, 388\u2013389, 389 lab; longitudinal, 381\u2013382; mechanical, 381\u2013382; period of, 383, 385 prob.; periodic, 381; phase of, 383; reflected, 387; resonance and, 389; sound. See Sound waves; speed of, 382, 385 prob., 387; springs and. See Springs, elastic", "; standing, 389; superposition of, 388\u2013389; transverse, 381; trough of, 383; two-dimensional, 390\u2013391; wavelength of. See Wavelength Weak bosons, 818, 820 Weak nuclear forces, 821 Weather, Coriolis force and, 217 Wedge, 269 \u201cWeighing Earth\u201d experiment, Cavendish\u2019s, 177\u2013178 Weight (Fg), 94 table; of astronauts in space, 182; determination of, 96; real and apparent, 98, 99 prob.; unit for (newton), 96 Weightlessness, 98, 182, 182 lab Wheel-and-axle machines, 269\u2013270; bicycle gears, 270 lab; ideal mechanical advantage (IMA) of, 269, 271 prob.; 942 Index 943-944 EM CREDITS-845813 6/10/04 7:13 PM Page 943 COVER (t)Michael Kevin Daly/CORBIS, (bl)Don Farrell/Getty Images, (br)Volker Mohrke/CORBIS, (inset)Myron Jay Dorf/CORBIS; 2 NASA; 3 Horizons Companies; 4 FOXTROT \u00a91998 Bill Amend. Reprinted with permission of UNIVERSAL PRESS SYNDICATE. All rights reserved.; 5 Mark E. 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A one-semester course on quantum information could be based upon a selection of material from Chapters 1 through 3, depending on the background of the class. Following that, Chapters 8 through 10 on quantum error-correction, followed by Chapters 11 and 12 on quantum entropy and quantum information theory, respectively. A full year class could cover all material in the book, with time for additional readings selected from the \u2018History and further reading\u2019 section of several chapters. Quantum computation and quantum information also lend themselves ideally to independent research projects for students. Aside from classes on quantum computation and quantum information, there is another way we hope the book will be used, which is as the text for an introductory class in quantum mechanics for physics students. Conventional introductions to quantum mechanics rely heavily on the mathematical machinery of partial differential equations. We believe this often obscures the fundamental ideas. Quantum computation and quantum informa- Preface xxv tion offers an excellent conceptual laboratory for understanding the basic concepts and unique aspects of quantum mechanics, without the use of heavy mathematical machinery. Such a class would focus on the introduction to quantum mechanics in Chapter 2, basic material on quantum circuits in Chapter 4, a selection of material on quantum algorithms from Chapters 5 and 6, Chapter 7 on physical implementations of quantum computation, and then almost any selection of material from Part III of the book, depending upon taste. Note to the student We have written the book to be as self-contained as possible. The main exception is that occasionally we have omitted arguments that one really needs to work through oneself to believe; these are usually given as exercises. Let us suggest that you should at least attempt all the exercises as you work through the book. With few exceptions the exercises can be worked out in a few minutes. If you are having a lot of dif\ufb01culty with many of the exercises it may be a sign that you need to go back and pick up one or more key concepts. Further reading As already noted, each chapter concludes with a \u2018History and further reading\u2019 section. There are also a few broad-ranging references that might be of interest to readers. Preskill\u2019s[Pre98b] superb lecture notes approach quantum computation and quantum information from a somewhat different point of view than this book. Good overview articles on speci\ufb01c subjects include (in order of their appearance in this book): Aharonov\u2019s review of quantum computation[Aha99b],", " Kitaev\u2019s review of algorithms and error-correction[Kit97b], Mosca\u2019s thesis on quantum algorithms[Mos99], Fuchs\u2019 thesis[Fuc96] on distinguishability and distance measures in quantum information, Gottesman\u2019s thesis on quantum errorcorrection[Got97], Preskill\u2019s review of quantum error-correction[Pre97], Nielsen\u2019s thesis on quantum information theory[Nie98], and the reviews of quantum information theory by Bennett and Shor[BS98] and by Bennett and DiVincenzo[BD00]. Other useful references include Gruska\u2019s book[Gru99], and the collection of review articles edited by Lo, Spiller, and Popescu[LSP98]. Errors Any lengthy document contains errors and omissions, and this book is surely no exception to the rule. If you \ufb01nd any errors or have other comments to make about the book, please email them to: qci@squint.org. As errata are found, we will add them to a list maintained at the book web site: http://www.squint.org/qci/. Acknowledgements A few people have decisively in\ufb02uenced how we think about quantum computation and quantum information. For many enjoyable discussions which have helped us shape and re\ufb01ne our views, MAN thanks Carl Caves, Chris Fuchs, Gerard Milburn, John Preskill and Ben Schumacher, and ILC thanks Tom Cover, Umesh Vazirani, Yoshi Yamamoto, and Bernie Yurke. An enormous number of people have helped in the construction of this book, both directly and indirectly. A partial list includes Dorit Aharonov, Andris Ambainis, Nabil Amer, Howard Barnum, Dave Beckman, Harry Buhrman, the Caltech Quantum Optics Foosballers, Andrew Childs, Fred Chong, Richard Cleve, John Conway, John Cortese, Michael DeShazo, Ronald de Wolf, David DiVincenzo, Steven van Enk, Henry Everitt, Ron Fagin, Mike Freedman, Michael Gagen, Neil Gershenfeld, Daniel Gottesman, Jim Harris, Alexander Holevo, Andrew Huibers, Julia Kempe, Alesha Kitaev, Manny Knill, Shing Kong, Raymond La\ufb02amme, Andrew Landahl, Ron Legere, Debbie Leung, Daniel Lidar, Elliott Lieb, Theresa Lynn, Hideo Mabuchi, Yu Manin, Mike Mosca, Alex Pines, Sridhar Rajagopalan, Bill Risk, Beth Ruskai, Sara Schneider, Robert Schrader, Peter Shor, Sheri Stoll, Volker Strassen, Armin Uhlmann, Lieven Vandersypen, Anne Verhulst, Debby Wallach, Mike Westmoreland, Dave Wineland, Howard Wiseman, John Yard, Xinlan Zhou, and Wojtek Zurek. Thanks to the folks at Cambridge University Press for their help turning this book from an idea into reality. Our especial thanks go to our thoughtful and enthusiastic editor Simon Capelin, who shepherded this project along for more than three years, and to Margaret Patterson, for her timely and thorough copy-editing of the manuscript. Parts of this book were completed while MAN was a Tolman Prize Fellow at the California Institute of Technology, a member of the T-6 Theoretical Astrophysics Group at the Los Alamos National Laboratory, and a member of the University of New Mexico Center for Advanced Studies, and while ILC was a Research Staff Member at the IBM Almaden Research Center, a consulting Assistant Professor of Electrical Engineering at Stanford University, a visiting researcher at the University of California Berkeley Department of Computer Science, a member of the Los Alamos National Laboratory T-6 Theoretical Astrophysics Group, and a visiting researcher at the University of California Santa Barbara Institute for Theoretical Physics. We also appreciate the warmth and hospitality of the Aspen Center for Physics, where the \ufb01nal page proofs of this book were \ufb01nished. MAN and ILC gratefully acknowledge support from DARPA under the NMRQC research initiative and the QUIC Institute administered by the Army Research Of\ufb01ce. We also thank the National Science Foundation, the National Security Agency, the Of\ufb01ce of Naval Research, and IBM for their generous support. Nomenclature and notation There are several items of nomenclature and notation which have two or more meanings in common use in the \ufb01eld of quantum computation and quantum information. To prevent confusion from arising, this section collects many of the more frequently used of these items, together with the conventions that will be adhered to in this book. Linear algebra and quantum mechanics All", " vector spaces are assumed to be \ufb01nite dimensional, unless otherwise noted. In many instances this restriction is unnecessary, or can be removed with some additional technical work, but making the restriction globally makes the presentation more easily comprehensible, and doesn\u2019t detract much from many of the intended applications of the results. A positive operator A is one for which \u03c8|A|\u03c8 \u2265 0 for all |\u03c8. A positive de\ufb01nite operator A is one for which \u03c8|A|\u03c8 > 0 for all |\u03c8 = 0. The support of an operator is de\ufb01ned to be the vector space orthogonal to its kernel. For a Hermitian operator, this means the vector space spanned by eigenvectors of the operator with non-zero eigenvalues. The notation U (and often but not always V ) will generically be used to denote a unitary operator or matrix. H is usually used to denote a quantum logic gate, the Hadamard gate, and sometimes to denote the Hamiltonian for a quantum system, with the meaning clear from context. Vectors will sometimes be written in column format, as for example,, 1 2 (0.1) and sometimes for readability in the format (1, 2). The latter should be understood as shorthand for a column vector. For two-level quantum systems used as qubits, we shall usually identify the state |0 with the vector (1, 0), and similarly |1 with (0, 1). We also de\ufb01ne the Pauli sigma matrices in the conventional way \u2013 see \u2018Frequently used quantum gates and circuit symbols\u2019, below. Most signi\ufb01cantly, the convention for the Pauli sigma z matrix is that \u03c3z|0 = |0 and \u03c3z|1 = \u2212|1, which is reverse of what some physicists (but usually not computer scientists or mathematicians) intuitively expect. The origin of this dissonance is that the +1 eigenstate of \u03c3z is often identi\ufb01ed by physicists with a so-called \u2018excited state\u2019, and it seems natural to many to identify this with |1, rather than with |0 as is done in this book. Our choice is made in order to be consistent with the usual indexing of matrix elements in linear algebra, which makes it natural to identify the \ufb01rst column of \u03c3z with the action of \u03c3z on |0, and the second column with the action on |1. This choice is also in use throughout the quantum computation and quantum information community. In addition to the conventional notations \u03c3x, \u03c3y and \u03c3z for the Pauli sigma matrices, it will also be convenient to use the notations \u03c31, \u03c32, \u03c33 for these xxx Nomenclature and notation three matrices, and to de\ufb01ne \u03c30 as the 2\u00d72 identity matrix. Most often, however, we use the notations I, X, Y and Z for \u03c30, \u03c31, \u03c32 and \u03c33, respectively. Information theory and probability As be\ufb01ts good information theorists, logarithms are always taken to base two, unless otherwise noted. We use log(x) to denote logarithms to base 2, and ln(x) on those rare occasions when we wish to take a natural logarithm. The term probability distribution x px = 1. The is used to refer to a \ufb01nite set of real numbers, px, such that px \u2265 0 and relative entropy of a positive operator A with respect to a positive operator B is de\ufb01ned by S(A||B) \u2261 tr(A log A) \u2212 tr(A log B). Miscellanea \u2295 denotes modulo two addition. Throughout this book \u2018z\u2019 is pronounced \u2018zed\u2019. Frequently used quantum gates and circuit symbols Certain schematic symbols are often used to denote unitary transforms which are useful in the design of quantum circuits. For the reader\u2019s convenience, many of these are gathered together below. The rows and columns of the unitary transforms are labeled from left to right and top to bottom as 00... 0, 00... 1 to 11... 1 with the bottom-most wire being the least signi\ufb01cant bit. Note that ei\u03c0/4 is the square root of i, so that the \u03c0/8 gate is the square root of the phase gate, which itself is the square root of the Pauli-Z gate. Hadamard Pauli-X Pauli-Y Pauli-Z Phase \u03c0/8 1\u221a 2 1 1 1 \u22121 1 0 0 1", " 0 \u2212i i 0 1 0 0 \u22121 0 1 i 0 1 0 0 ei\u03c0/4 controlled- swap controlled-Z controlled-phase Toffoli Fredkin (controlled-swap) measurement qubit classical bit n qubits \u2022 Z = \u2022 \u2022 \u2295 \u2022 \u00d7 \u00d7 Nomenclature and notation xxxi \u23a1 \u23a2 \u23a3 \u23a1 \u23a2 \u23a3 \u23a1 \u23a2 \u23a3 Projection onto |0 and |1 wire carrying a single qubit (time goes left to right) wire carrying a single classical bit wire carrying n qubits I Fundamental concepts 1 Introduction and overview Science offers the boldest metaphysics of the age. It is a thoroughly human construct, driven by the faith that if we dream, press to discover, explain, and dream again, thereby plunging repeatedly into new terrain, the world will somehow come clearer and we will grasp the true strangeness of the universe. And the strangeness will all prove to be connected, and make sense. \u2013 Edward O. Wilson Information is physical. \u2013 Rolf Landauer What are the fundamental concepts of quantum computation and quantum information? How did these concepts develop? To what uses may they be put? How will they be presented in this book? The purpose of this introductory chapter is to answer these questions by developing in broad brushstrokes a picture of the \ufb01eld of quantum computation and quantum information. The intent is to communicate a basic understanding of the central concepts of the \ufb01eld, perspective on how they have been developed, and to help you decide how to approach the rest of the book. Our story begins in Section 1.1 with an account of the historical context in which quantum computation and quantum information has developed. Each remaining section in the chapter gives a brief introduction to one or more fundamental concepts from the \ufb01eld: quantum bits (Section 1.2), quantum computers, quantum gates and quantum circuits (Section 1.3), quantum algorithms (Section 1.4), experimental quantum information processing (Section 1.5), and quantum information and communication (Section 1.6). Along the way, illustrative and easily accessible developments such as quantum teleportation and some simple quantum algorithms are given, using the basic mathematics taught in this chapter. The presentation is self-contained, and designed to be accessible even without a background in computer science or physics. As we move along, we give pointers to more in-depth discussions in later chapters, where references and suggestions for further reading may also be found. If as you read you\u2019re \ufb01nding the going rough, skip on to a spot where you feel more comfortable. At points we haven\u2019t been able to avoid using a little technical lingo which won\u2019t be completely explained until later in the book. Simply accept it for now, and come back later when you understand all the terminology in more detail. The emphasis in this \ufb01rst chapter is on the big picture, with the details to be \ufb01lled in later. 1.1 Global perspectives Quantum computation and quantum information is the study of the information processing tasks that can be accomplished using quantum mechanical systems. Sounds pretty 2 Introduction and overview simple and obvious, doesn\u2019t it? Like many simple but profound ideas it was a long time before anybody thought of doing information processing using quantum mechanical systems. To see why this is the case, we must go back in time and look in turn at each of the \ufb01elds which have contributed fundamental ideas to quantum computation and quantum information \u2013 quantum mechanics, computer science, information theory, and cryptography. As we take our short historical tour of these \ufb01elds, think of yourself \ufb01rst as a physicist, then as a computer scientist, then as an information theorist, and \ufb01nally as a cryptographer, in order to get some feel for the disparate perspectives which have come together in quantum computation and quantum information. 1.1.1 History of quantum computation and quantum information Our story begins at the turn of the twentieth century when an unheralded revolution was underway in science. A series of crises had arisen in physics. The problem was that the theories of physics at that time (now dubbed classical physics) were predicting absurdities such as the existence of an \u2018ultraviolet catastrophe\u2019 involving in\ufb01nite energies, or electrons spiraling inexorably into the atomic nucleus. At \ufb01rst such problems were resolved with the addition of ad hoc hypotheses to classical physics, but as a better understanding of atoms and radiation was gained these attempted explanations became more and more convoluted. The crisis came to a head in the early 1920s after a quarter century of turmoil, and resulted in the creation of the modern theory of quantum mechanics. Quantum mechanics has been an indispensable part of science ever since, and has been applied with enormous success to everything under and inside the Sun, including the structure of the atom,", " nuclear fusion in stars, superconductors, the structure of DNA, and the elementary particles of Nature. What is quantum mechanics? Quantum mechanics is a mathematical framework or set of rules for the construction of physical theories. For example, there is a physical theory known as quantum electrodynamics which describes with fantastic accuracy the interaction of atoms and light. Quantum electrodynamics is built up within the framework of quantum mechanics, but it contains speci\ufb01c rules not determined by quantum mechanics. The relationship of quantum mechanics to speci\ufb01c physical theories like quantum electrodynamics is rather like the relationship of a computer\u2019s operating system to speci\ufb01c applications software \u2013 the operating system sets certain basic parameters and modes of operation, but leaves open how speci\ufb01c tasks are accomplished by the applications. The rules of quantum mechanics are simple but even experts \ufb01nd them counterintuitive, and the earliest antecedents of quantum computation and quantum information may be found in the long-standing desire of physicists to better understand quantum mechanics. The best known critic of quantum mechanics, Albert Einstein, went to his grave unreconciled with the theory he helped invent. Generations of physicists since have wrestled with quantum mechanics in an effort to make its predictions more palatable. One of the goals of quantum computation and quantum information is to develop tools which sharpen our intuition about quantum mechanics, and make its predictions more transparent to human minds. For example, in the early 1980s, interest arose in whether it might be possible to use quantum effects to signal faster than light \u2013 a big no-no according to Einstein\u2019s theory of relativity. The resolution of this problem turns out to hinge on whether it is possible to clone an unknown quantum state, that is, construct a copy of a quantum state. If cloning were possible, then it would be possible to signal faster than light using quantum effects. Global perspectives 3 However, cloning \u2013 so easy to accomplish with classical information (consider the words in front of you, and where they came from!) \u2013 turns out not to be possible in general in quantum mechanics. This no-cloning theorem, discovered in the early 1980s, is one of the earliest results of quantum computation and quantum information. Many re\ufb01nements of the no-cloning theorem have since been developed, and we now have conceptual tools which allow us to understand how well a (necessarily imperfect) quantum cloning device might work. These tools, in turn, have been applied to understand other aspects of quantum mechanics. A related historical strand contributing to the development of quantum computation and quantum information is the interest, dating to the 1970s, of obtaining complete control over single quantum systems. Applications of quantum mechanics prior to the 1970s typically involved a gross level of control over a bulk sample containing an enormous number of quantum mechanical systems, none of them directly accessible. For example, superconductivity has a superb quantum mechanical explanation. However, because a superconductor involves a huge (compared to the atomic scale) sample of conducting metal, we can only probe a few aspects of its quantum mechanical nature, with the individual quantum systems constituting the superconductor remaining inaccessible. Systems such as particle accelerators do allow limited access to individual quantum systems, but again provide little control over the constituent systems. Since the 1970s many techniques for controlling single quantum systems have been developed. For example, methods have been developed for trapping a single atom in an \u2018atom trap\u2019, isolating it from the rest of the world and allowing us to probe many different aspects of its behavior with incredible precision. The scanning tunneling microscope has been used to move single atoms around, creating designer arrays of atoms at will. Electronic devices whose operation involves the transfer of only single electrons have been demonstrated. Why all this effort to attain complete control over single quantum systems? Setting aside the many technological reasons and concentrating on pure science, the principal answer is that researchers have done this on a hunch. Often the most profound insights in science come when we develop a method for probing a new regime of Nature. For example, the invention of radio astronomy in the 1930s and 1940s led to a spectacular sequence of discoveries, including the galactic core of the Milky Way galaxy, pulsars, and quasars. Low temperature physics has achieved its amazing successes by \ufb01nding ways to lower the temperatures of different systems. In a similar way, by obtaining complete control over single quantum systems, we are exploring untouched regimes of Nature in the hope of discovering new and unexpected phenomena. We are just now taking our \ufb01rst steps along these lines, and already a few interesting surprises have been discovered in this regime. What else shall we discover as we obtain more complete control over single quantum systems, and extend it to more complex systems? Quantum computation and quantum information \ufb01t naturally into this program. They provide a useful series of challenges at varied levels of dif\ufb01culty for people devising methods to", " better manipulate single quantum systems, and stimulate the development of new experimental techniques and provide guidance as to the most interesting directions in which to take experiment. Conversely, the ability to control single quantum systems is essential if we are to harness the power of quantum mechanics for applications to quantum computation and quantum information. Despite this intense interest, efforts to build quantum information processing systems 4 Introduction and overview have resulted in modest success to date. Small quantum computers, capable of doing dozens of operations on a few quantum bits (or qubits) represent the state of the art in quantum computation. Experimental prototypes for doing quantum cryptography \u2013 a way of communicating in secret across long distances \u2013 have been demonstrated, and are even at the level where they may be useful for some real-world applications. However, it remains a great challenge to physicists and engineers of the future to develop techniques for making large-scale quantum information processing a reality. Let us turn our attention from quantum mechanics to another of the great intellectual triumphs of the twentieth century, computer science. The origins of computer science are lost in the depths of history. For example, cuneiform tablets indicate that by the time of Hammurabi (circa 1750 B.C.) the Babylonians had developed some fairly sophisticated algorithmic ideas, and it is likely that many of those ideas date to even earlier times. The modern incarnation of computer science was announced by the great mathematician Alan Turing in a remarkable 1936 paper. Turing developed in detail an abstract notion of what we would now call a programmable computer, a model for computation now known as the Turing machine, in his honor. Turing showed that there is a Universal Turing Machine that can be used to simulate any other Turing machine. Furthermore, he claimed that the Universal Turing Machine completely captures what it means to perform a task by algorithmic means. That is, if an algorithm can be performed on any piece of hardware (say, a modern personal computer), then there is an equivalent algorithm for a Universal Turing Machine which performs exactly the same task as the algorithm running on the personal computer. This assertion, known as the Church\u2013Turing thesis in honor of Turing and another pioneer of computer science, Alonzo Church, asserts the equivalence between the physical concept of what class of algorithms can be performed on some physical device with the rigorous mathematical concept of a Universal Turing Machine. The broad acceptance of this thesis laid the foundation for the development of a rich theory of computer science. Not long after Turing\u2019s paper, the \ufb01rst computers constructed from electronic components were developed. John von Neumann developed a simple theoretical model for how to put together in a practical fashion all the components necessary for a computer to be fully as capable as a Universal Turing Machine. Hardware development truly took off, though, in 1947, when John Bardeen, Walter Brattain, and Will Shockley developed the transistor. Computer hardware has grown in power at an amazing pace ever since, so much so that the growth was codi\ufb01ed by Gordon Moore in 1965 in what has come to be known as Moore\u2019s law, which states that computer power will double for constant cost roughly once every two years. Amazingly enough, Moore\u2019s law has approximately held true in the decades since the 1960s. Nevertheless, most observers expect that this dream run will end some time during the \ufb01rst two decades of the twenty-\ufb01rst century. Conventional approaches to the fabrication of computer technology are beginning to run up against fundamental dif\ufb01culties of size. Quantum effects are beginning to interfere in the functioning of electronic devices as they are made smaller and smaller. One possible solution to the problem posed by the eventual failure of Moore\u2019s law is to move to a different computing paradigm. One such paradigm is provided by the theory of quantum computation, which is based on the idea of using quantum mechanics to perform computations, instead of classical physics. It turns out that while an ordinary computer can be used to simulate a quantum computer, it appears to be impossible to Global perspectives 5 perform the simulation in an ef\ufb01cient fashion. Thus quantum computers offer an essential speed advantage over classical computers. This speed advantage is so signi\ufb01cant that many researchers believe that no conceivable amount of progress in classical computation would be able to overcome the gap between the power of a classical computer and the power of a quantum computer. What do we mean by \u2018ef\ufb01cient\u2019 versus \u2018inef\ufb01cient\u2019 simulations of a quantum computer? Many of the key notions needed to answer this question were actually invented before the notion of a quantum computer had even arisen. In particular, the idea of ef\ufb01cient and inef\ufb01cient algorithms was made mathematically precise by the \ufb01eld of computational complexity. Roughly speaking, an ef\ufb01cient algorithm is one which runs in time polynomial in the size of the problem solved. In contrast, an inef\ufffd", "\ufffd\ufffdcient algorithm requires superpolynomial (typically exponential) time. What was noticed in the late 1960s and early 1970s was that it seemed as though the Turing machine model of computation was at least as powerful as any other model of computation, in the sense that a problem which could be solved ef\ufb01ciently in some model of computation could also be solved ef\ufb01ciently in the Turing machine model, by using the Turing machine to simulate the other model of computation. This observation was codi\ufb01ed into a strengthened version of the Church\u2013 Turing thesis: Any algorithmic process can be simulated ef\ufb01ciently using a Turing machine. The key strengthening in the strong Church\u2013Turing thesis is the word ef\ufb01ciently. If the strong Church\u2013Turing thesis is correct, then it implies that no matter what type of machine we use to perform our algorithms, that machine can be simulated ef\ufb01ciently using a standard Turing machine. This is an important strengthening, as it implies that for the purposes of analyzing whether a given computational task can be accomplished ef\ufb01ciently, we may restrict ourselves to the analysis of the Turing machine model of computation. One class of challenges to the strong Church\u2013Turing thesis comes from the \ufb01eld of analog computation. In the years since Turing, many different teams of researchers have noticed that certain types of analog computers can ef\ufb01ciently solve problems believed to have no ef\ufb01cient solution on a Turing machine. At \ufb01rst glance these analog computers appear to violate the strong form of the Church\u2013Turing thesis. Unfortunately for analog computation, it turns out that when realistic assumptions about the presence of noise in analog computers are made, their power disappears in all known instances; they cannot ef\ufb01ciently solve problems which are not ef\ufb01ciently solvable on a Turing machine. This lesson \u2013 that the effects of realistic noise must be taken into account in evaluating the ef\ufb01ciency of a computational model \u2013 was one of the great early challenges of quantum computation and quantum information, a challenge successfully met by the development of a theory of quantum error-correcting codes and fault-tolerant quantum computation. Thus, unlike analog computation, quantum computation can in principle tolerate a \ufb01nite amount of noise and still retain its computational advantages. The \ufb01rst major challenge to the strong Church\u2013Turing thesis arose in the mid 1970s, when Robert Solovay and Volker Strassen showed that it is possible to test whether an integer is prime or composite using a randomized algorithm. That is, the Solovay\u2013Strassen test for primality used randomness as an essential part of the algorithm. The algorithm did not determine whether a given integer was prime or composite with certainty. Instead, the algorithm could determine that a number was probably prime or else composite with 6 Introduction and overview certainty. By repeating the Solovay\u2013Strassen test a few times it is possible to determine with near certainty whether a number is prime or composite. The Solovay-Strassen test was of especial signi\ufb01cance at the time it was proposed as no deterministic test for primality was then known, nor is one known at the time of this writing. Thus, it seemed as though computers with access to a random number generator would be able to ef\ufb01ciently perform computational tasks with no ef\ufb01cient solution on a conventional deterministic Turing machine. This discovery inspired a search for other randomized algorithms which has paid off handsomely, with the \ufb01eld blossoming into a thriving area of research. Randomized algorithms pose a challenge to the strong Church\u2013Turing thesis, suggesting that there are ef\ufb01ciently soluble problems which, nevertheless, cannot be ef\ufb01ciently solved on a deterministic Turing machine. This challenge appears to be easily resolved by a simple modi\ufb01cation of the strong Church\u2013Turing thesis: Any algorithmic process can be simulated ef\ufb01ciently using a probabilistic Turing machine. This ad hoc modi\ufb01cation of the strong Church\u2013Turing thesis should leave you feeling rather queasy. Might it not turn out at some later date that yet another model of computation allows one to ef\ufb01ciently solve problems that are not ef\ufb01ciently soluble within Turing\u2019s model of computation? Is there any way we can \ufb01nd a single model of computation which is guaranteed to be able to ef\ufb01ciently simulate any other model of computation? Motivated by this question, in 1985 David Deutsch asked whether the laws of physics could be use to derive an even stronger version of the Church\u2013Turing thesis. Instead of adopting ad hoc hypotheses", ", Deutsch looked to physical theory to provide a foundation for the Church\u2013Turing thesis that would be as secure as the status of that physical theory. In particular, Deutsch attempted to de\ufb01ne a computational device that would be capable of ef\ufb01ciently simulating an arbitrary physical system. Because the laws of physics are ultimately quantum mechanical, Deutsch was naturally led to consider computing devices based upon the principles of quantum mechanics. These devices, quantum analogues of the machines de\ufb01ned forty-nine years earlier by Turing, led ultimately to the modern conception of a quantum computer used in this book. At the time of writing it is not clear whether Deutsch\u2019s notion of a Universal Quantum Computer is suf\ufb01cient to ef\ufb01ciently simulate an arbitrary physical system. Proving or refuting this conjecture is one of the great open problems of the \ufb01eld of quantum computation and quantum information. It is possible, for example, that some effect of quantum \ufb01eld theory or an even more esoteric effect based in string theory, quantum gravity or some other physical theory may take us beyond Deutsch\u2019s Universal Quantum Computer, giving us a still more powerful model for computation. At this stage, we simply don\u2019t know. What Deutsch\u2019s model of a quantum computer did enable was a challenge to the strong form of the Church\u2013Turing thesis. Deutsch asked whether it is possible for a quantum computer to ef\ufb01ciently solve computational problems which have no ef\ufb01cient solution on a classical computer, even a probabilistic Turing machine. He then constructed a simple example suggesting that, indeed, quantum computers might have computational powers exceeding those of classical computers. This remarkable \ufb01rst step taken by Deutsch was improved in the subsequent decade by many people, culminating in Peter Shor\u2019s 1994 demonstration that two enormously important problems \u2013 the problem of \ufb01nding the prime factors of an integer, and the so- Global perspectives 7 called \u2018discrete logarithm\u2019 problem \u2013 could be solved ef\ufb01ciently on a quantum computer. This attracted widespread interest because these two problems were and still are widely believed to have no ef\ufb01cient solution on a classical computer. Shor\u2019s results are a powerful indication that quantum computers are more powerful than Turing machines, even probabilistic Turing machines. Further evidence for the power of quantum computers came in 1995 when Lov Grover showed that another important problem \u2013 the problem of conducting a search through some unstructured search space \u2013 could also be sped up on a quantum computer. While Grover\u2019s algorithm did not provide as spectacular a speedup as Shor\u2019s algorithms, the widespread applicability of search-based methodologies has excited considerable interest in Grover\u2019s algorithm. At about the same time as Shor\u2019s and Grover\u2019s algorithms were discovered, many people were developing an idea Richard Feynman had suggested in 1982. Feynman had pointed out that there seemed to be essential dif\ufb01culties in simulating quantum mechanical systems on classical computers, and suggested that building computers based on the principles of quantum mechanics would allow us to avoid those dif\ufb01culties. In the 1990s several teams of researchers began \ufb02eshing this idea out, showing that it is indeed possible to use quantum computers to ef\ufb01ciently simulate systems that have no known ef\ufb01cient simulation on a classical computer. It is likely that one of the major applications of quantum computers in the future will be performing simulations of quantum mechanical systems too dif\ufb01cult to simulate on a classical computer, a problem with profound scienti\ufb01c and technological implications. What other problems can quantum computers solve more quickly than classical computers? The short answer is that we don\u2019t know. Coming up with good quantum algorithms seems to be hard. A pessimist might think that\u2019s because there\u2019s nothing quantum computers are good for other than the applications already discovered! We take a different view. Algorithm design for quantum computers is hard because designers face two dif\ufb01cult problems not faced in the construction of algorithms for classical computers. First, our human intuition is rooted in the classical world. If we use that intuition as an aid to the construction of algorithms, then the algorithmic ideas we come up with will be classical ideas. To design good quantum algorithms one must \u2018turn off\u2019 one\u2019s classical intuition for at least part of the design process, using truly quantum effects to achieve the desired algorithmic end. Second, to be truly interesting it is not enough to design an algorithm that is merely quantum mechanical. The algorithm must be better than any existing classical algorithm! Thus, it is possible that one may \ufb01nd an algorithm which makes use", " of truly quantum aspects of quantum mechanics, that is nevertheless not of widespread interest because classical algorithms with comparable performance characteristics exist. The combination of these two problems makes the construction of new quantum algorithms a challenging problem for the future. Even more broadly, we can ask if there are any generalizations we can make about the power of quantum computers versus classical computers. What is it that makes quantum computers more powerful than classical computers \u2013 assuming that this is indeed the case? What class of problems can be solved ef\ufb01ciently on a quantum computer, and how does that class compare to the class of problems that can be solved ef\ufb01ciently on a classical computer? One of the most exciting things about quantum computation and quantum information is how little is known about the answers to these questions! It is a great challenge for the future to understand these questions better. Having come up to the frontier of quantum computation, let\u2019s switch to the history 8 Introduction and overview of another strand of thought contributing to quantum computation and quantum information: information theory. At the same time computer science was exploding in the 1940s, another revolution was taking place in our understanding of communication. In 1948 Claude Shannon published a remarkable pair of papers laying the foundations for the modern theory of information and communication. Perhaps the key step taken by Shannon was to mathematically de\ufb01ne the concept of information. In many mathematical sciences there is considerable \ufb02exibility in the choice of fundamental de\ufb01nitions. Try thinking naively for a few minutes about the following question: how would you go about mathematically de\ufb01ning the notion of an information source? Several different answers to this problem have found widespread use; however, the de\ufb01nition Shannon came up with seems to be far and away the most fruitful in terms of increased understanding, leading to a plethora of deep results and a theory with a rich structure which seems to accurately re\ufb02ect many (though not all) real-world communications problems. Shannon was interested in two key questions related to the communication of information over a communications channel. First, what resources are required to send information over a communications channel? For example, telephone companies need to know how much information they can reliably transmit over a given telephone cable. Second, can information be transmitted in such a way that it is protected against noise in the communications channel? Shannon answered these two questions by proving the two fundamental theorems of information theory. The \ufb01rst, Shannon\u2019s noiseless channel coding theorem, quanti\ufb01es the physical resources required to store the output from an information source. Shannon\u2019s second fundamental theorem, the noisy channel coding theorem, quanti\ufb01es how much information it is possible to reliably transmit through a noisy communications channel. To achieve reliable transmission in the presence of noise, Shannon showed that error-correcting codes could be used to protect the information being sent. Shannon\u2019s noisy channel coding theorem gives an upper limit on the protection afforded by errorcorrecting codes. Unfortunately, Shannon\u2019s theorem does not explicitly give a practically useful set of error-correcting codes to achieve that limit. From the time of Shannon\u2019s papers until today, researchers have constructed more and better classes of error-correcting codes in their attempts to come closer to the limit set by Shannon\u2019s theorem. A sophisticated theory of error-correcting codes now exists offering the user a plethora of choices in their quest to design a good error-correcting code. Such codes are used in a multitude of places including, for example, compact disc players, computer modems, and satellite communications systems. Quantum information theory has followed with similar developments. In 1995, Ben Schumacher provided an analogue to Shannon\u2019s noiseless coding theorem, and in the process de\ufb01ned the \u2018quantum bit\u2019 or \u2018qubit\u2019 as a tangible physical resource. However, no analogue to Shannon\u2019s noisy channel coding theorem is yet known for quantum information. Nevertheless, in analogy to their classical counterparts, a theory of quantum error-correction has been developed which, as already mentioned, allows quantum computers to compute effectively in the presence of noise, and also allows communication over noisy quantum channels to take place reliably. Indeed, classical ideas of error-correction have proved to be enormously important in developing and understanding quantum error-correcting codes. In 1996, two groups working independently, Robert Calderbank and Peter Shor, and Andrew Steane, discov- Global perspectives 9 ered an important class of quantum codes now known as CSS codes after their initials. This work has since been subsumed by the stabilizer codes, independently discovered by Robert Calderbank, Eric Rains, Peter Shor and Neil Sloane, and by Daniel Gottesman. By building upon the basic ideas of classical linear coding theory, these discoveries greatly facilitated a rapid understanding of quantum error-correcting codes and their application to quantum", " computation and quantum information. The theory of quantum error-correcting codes was developed to protect quantum states against noise. What about transmitting ordinary classical information using a quantum channel? How ef\ufb01ciently can this be done? A few surprises have been discovered in this arena. In 1992 Charles Bennett and Stephen Wiesner explained how to transmit two classical bits of information, while only transmitting one quantum bit from sender to receiver, a result dubbed superdense coding. Even more interesting are the results in distributed quantum computation. Imagine you have two computers networked, trying to solve a particular problem. How much communication is required to solve the problem? Recently it has been shown that quantum computers can require exponentially less communication to solve certain problems than would be required if the networked computers were classical! Unfortunately, as yet these problems are not especially important in a practical setting, and suffer from some undesirable technical restrictions. A major challenge for the future of quantum computation and quantum information is to \ufb01nd problems of real-world importance for which distributed quantum computation offers a substantial advantage over distributed classical computation. Let\u2019s return to information theory proper. The study of information theory begins with the properties of a single communications channel. In applications we often do not deal with a single communications channel, but rather with networks of many channels. The subject of networked information theory deals with the information carrying properties of such networks of communications channels, and has been developed into a rich and intricate subject. By contrast, the study of networked quantum information theory is very much in its infancy. Even for very basic questions we know little about the information carrying abilities of networks of quantum channels. Several rather striking preliminary results have been found in the past few years; however, no unifying theory of networked information theory exists for quantum channels. One example of networked quantum information theory should suf\ufb01ce to convince you of the value such a general theory would have. Imagine that we are attempting to send quantum information from Alice to Bob through a noisy quantum channel. If that channel has zero capacity for quantum information, then it is impossible to reliably send any information from Alice to Bob. Imagine instead that we consider two copies of the channel, operating in synchrony. Intuitively it is clear (and can be rigorously justi\ufb01ed) that such a channel also has zero capacity to send quantum information. However, if we instead reverse the direction of one of the channels, as illustrated in Figure 1.1, it turns out that sometimes we can obtain a non-zero capacity for the transmission of information from Alice to Bob! Counter-intuitive properties like this illustrate the strange nature of quantum information. Better understanding the information carrying properties of networks of quantum channels is a major open problem of quantum computation and quantum information. Let\u2019s switch \ufb01elds one last time, moving to the venerable old art and science of cryptography. Broadly speaking, cryptography is the problem of doing communication or 10 Introduction and overview Figure 1.1. Classically, if we have two very noisy channels of zero capacity running side by side, then the combined channel has zero capacity to send information. Not surprisingly, if we reverse the direction of one of the channels, we still have zero capacity to send information. Quantum mechanically, reversing one of the zero capacity channels can actually allow us to send information! computation involving two or more parties who may not trust one another. The best known cryptographic problem is the transmission of secret messages. Suppose two parties wish to communicate in secret. For example, you may wish to give your credit card number to a merchant in exchange for goods, hopefully without any malevolent third party intercepting your credit card number. The way this is done is to use a cryptographic protocol. We\u2019ll describe in detail how cryptographic protocols work later in the book, but for now it will suf\ufb01ce to make a few simple distinctions. The most important distinction is between private key cryptosystems and public key cryptosystems. The way a private key cryptosystem works is that two parties, \u2018Alice\u2019 and \u2018Bob\u2019, wish to communicate by sharing a private key, which only they know. The exact form of the key doesn\u2019t matter at this point \u2013 think of a string of zeroes and ones. The point is that this key is used by Alice to encrypt the information she wishes to send to Bob. After Alice encrypts she sends the encrypted information to Bob, who must now recover the original information. Exactly how Alice encrypts the message depends upon the private key, so that to recover the original message Bob needs to know the private key, in order to undo the transformation Alice applied. Unfortunately, private key cryptosystems have some severe problems in many contexts. The most basic problem is how to distribute the keys? In many ways, the key distribution problem is just as dif\ufb01cult as the original problem of communicating in private \u2013 a malevolent third party may be eavesdropping", " on the key distribution, and then use the intercepted key to decrypt some of the message transmission. One of the earliest discoveries in quantum computation and quantum information was that quantum mechanics can be used to do key distribution in such a way that Alice and Bob\u2019s security can not be compromised. This procedure is known as quantum cryptography or quantum key distribution. The basic idea is to exploit the quantum mechanical principle that observation in general disturbs the system being observed. Thus, if there is an eavesdropper listening in as Alice and Bob attempt to transmit their key, the presence of the eavesdropper will be visible as a disturbance of the communications channel Alice and Bob are using to establish the key. Alice and Bob can then throw out the key bits established while the eavesdropper was listening in, and start over. The \ufb01rst quantum cryptographic ideas were proposed by Stephen Wiesner in the late 1960s, but unfortu- Global perspectives 11 nately were not accepted for publication! In 1984 Charles Bennett and Gilles Brassard, building on Wiesner\u2019s earlier work, proposed a protocol using quantum mechanics to distribute keys between Alice and Bob, without any possibility of a compromise. Since then numerous quantum cryptographic protocols have been proposed, and experimental prototypes developed. At the time of this writing, the experimental prototypes are nearing the stage where they may be useful in limited-scale real-world applications. The second major type of cryptosystem is the public key cryptosystem. Public key cryptosystems don\u2019t rely on Alice and Bob sharing a secret key in advance. Instead, Bob simply publishes a \u2018public key\u2019, which is made available to the general public. Alice can make use of this public key to encrypt a message which she sends to Bob. What is interesting is that a third party cannot use Bob\u2019s public key to decrypt the message! Strictly speaking, we shouldn\u2019t say cannot. Rather, the encryption transformation is chosen in a very clever and non-trivial way so that it is extremely dif\ufb01cult (though not impossible) to invert, given only knowledge of the public key. To make inversion easy, Bob has a secret key matched to his public key, which together enable him to easily perform the decryption. This secret key is not known to anybody other than Bob, who can therefore be con\ufb01dent that only he can read the contents of Alice\u2019s transmission, to the extent that it is unlikely that anybody else has the computational power to invert the encryption, given only the public key. Public key cryptosystems solve the key distribution problem by making it unnecessary for Alice and Bob to share a private key before communicating. Rather remarkably, public key cryptography did not achieve widespread use until the mid-1970s, when it was proposed independently by Whit\ufb01eld Dif\ufb01e and Martin Hellman, and by Ralph Merkle, revolutionizing the \ufb01eld of cryptography. A little later, Ronald Rivest, Adi Shamir, and Leonard Adleman developed the RSA cryptosystem, which at the time of writing is the most widely deployed public key cryptosystem, believed to offer a \ufb01ne balance of security and practical usability. In 1997 it was disclosed that these ideas \u2013 public key cryptography, the Dif\ufb01e\u2013Hellman and RSA cryptosystems \u2013 were actually invented in the late 1960s and early 1970s by researchers working at the British intelligence agency GCHQ. The key to the security of public key cryptosystems is that it should be dif\ufb01cult to invert the encryption stage if only the public key is available. For example, it turns out that inverting the encryption stage of RSA is a problem closely related to factoring. Much of the presumed security of RSA comes from the belief that factoring is a problem hard to solve on a classical computer. However, Shor\u2019s fast algorithm for factoring on a quantum computer could be used to break RSA! Similarly, there are other public key cryptosystems which can be broken if a fast algorithm for solving the discrete logarithm problem \u2013 like Shor\u2019s quantum algorithm for discrete logarithm \u2013 were known. This practical application of quantum computers to the breaking of cryptographic codes has excited much of the interest in quantum computation and quantum information. We have been looking at the historical antecedents for quantum computation and quantum information. Of course, as the \ufb01eld has grown and matured, it has sprouted its own sub\ufb01elds of research, whose antecedents lie mainly within quantum computation and quantum information. Perhaps the most striking of these is the study of quantum entanglement. Entanglement is a uniquely quantum mechanical resource that plays a key role in many of the most interesting applications of quantum computation and quantum information; entanglement is iron to the", " classical world\u2019s bronze age. In recent years there has been a 12 Introduction and overview tremendous effort trying to better understand the properties of entanglement considered as a fundamental resource of Nature, of comparable importance to energy, information, entropy, or any other fundamental resource. Although there is as yet no complete theory of entanglement, some progress has been made in understanding this strange property of quantum mechanics. It is hoped by many researchers that further study of the properties of entanglement will yield insights that facilitate the development of new applications in quantum computation and quantum information. 1.1.2 Future directions We\u2019ve looked at some of the history and present status of quantum computation and quantum information. What of the future? What can quantum computation and quantum information offer to science, to technology, and to humanity? What bene\ufb01ts does quantum computation and quantum information confer upon its parent \ufb01elds of computer science, information theory, and physics? What are the key open problems of quantum computation and quantum information? We will make a few very brief remarks about these overarching questions before moving onto more detailed investigations. Quantum computation and quantum information has taught us to think physically about computation, and we have discovered that this approach yields many new and exciting capabilities for information processing and communication. Computer scientists and information theorists have been gifted with a new and rich paradigm for exploration. Indeed, in the broadest terms we have learned that any physical theory, not just quantum mechanics, may be used as the basis for a theory of information processing and communication. The fruits of these explorations may one day result in information processing devices with capabilities far beyond today\u2019s computing and communications systems, with concomitant bene\ufb01ts and drawbacks for society as a whole. Quantum computation and quantum information certainly offer challenges aplenty to physicists, but it is perhaps a little subtle what quantum computation and quantum information offers to physics in the long term. We believe that just as we have learned to think physically about computation, we can also learn to think computationally about physics. Whereas physics has traditionally been a discipline focused on understanding \u2018elementary\u2019 objects and simple systems, many interesting aspects of Nature arise only when things become larger and more complicated. Chemistry and engineering deal with such complexity to some extent, but most often in a rather ad hoc fashion. One of the messages of quantum computation and information is that new tools are available for traversing the gulf between the small and the relatively complex: computation and algorithms provide systematic means for constructing and understanding such systems. Applying ideas from these \ufb01elds is already beginning to yield new insights into physics. It is our hope that this perspective will blossom in years to come into a fruitful way of understanding all aspects of physics. We\u2019ve brie\ufb02y examined some of the key motivations and ideas underlying quantum computation and quantum information. Over the remaining sections of this chapter we give a more technical but still accessible introduction to these motivations and ideas, with the hope of giving you a bird\u2019s-eye view of the \ufb01eld as it is presently poised. Quantum bits 13 1.2 Quantum bits The bit is the fundamental concept of classical computation and classical information. Quantum computation and quantum information are built upon an analogous concept, the quantum bit, or qubit for short. In this section we introduce the properties of single and multiple qubits, comparing and contrasting their properties to those of classical bits. What is a qubit? We\u2019re going to describe qubits as mathematical objects with certain speci\ufb01c properties. \u2018But hang on\u2019, you say, \u2018I thought qubits were physical objects.\u2019 It\u2019s true that qubits, like bits, are realized as actual physical systems, and in Section 1.5 and Chapter 7 we describe in detail how this connection between the abstract mathematical point of view and real systems is made. However, for the most part we treat qubits as abstract mathematical objects. The beauty of treating qubits as abstract entities is that it gives us the freedom to construct a general theory of quantum computation and quantum information which does not depend upon a speci\ufb01c system for its realization. What then is a qubit? Just as a classical bit has a state \u2013 either 0 or 1 \u2013 a qubit also has a state. Two possible states for a qubit are the states |0 and |1, which as you might guess correspond to the states 0 and 1 for a classical bit. Notation like \u2018| \u2019 is called the Dirac notation, and we\u2019ll be seeing it often, as it\u2019s the standard notation for states in quantum mechanics. The difference between bits and qubits is that a qubit can be in a state other than |0 or |1. It is also possible to form linear combinations of states, often called superpositions: |\u03c8 = \u03b1 |0 + \u03b2 |1. (1.1) The", " numbers \u03b1 and \u03b2 are complex numbers, although for many purposes not much is lost by thinking of them as real numbers. Put another way, the state of a qubit is a vector in a two-dimensional complex vector space. The special states |0 and |1 are known as computational basis states, and form an orthonormal basis for this vector space. We can examine a bit to determine whether it is in the state 0 or 1. For example, computers do this all the time when they retrieve the contents of their memory. Rather remarkably, we cannot examine a qubit to determine its quantum state, that is, the values of \u03b1 and \u03b2. Instead, quantum mechanics tells us that we can only acquire much more restricted information about the quantum state. When we measure a qubit we get either the result 0, with probability |\u03b1|2, or the result 1, with probability |\u03b2|2. Naturally, |\u03b1|2 + |\u03b2|2 = 1, since the probabilities must sum to one. Geometrically, we can interpret this as the condition that the qubit\u2019s state be normalized to length 1. Thus, in general a qubit\u2019s state is a unit vector in a two-dimensional complex vector space. This dichotomy between the unobservable state of a qubit and the observations we can make lies at the heart of quantum computation and quantum information. In most of our abstract models of the world, there is a direct correspondence between elements of the abstraction and the real world, just as an architect\u2019s plans for a building are in correspondence with the \ufb01nal building. The lack of this direct correspondence in quantum mechanics makes it dif\ufb01cult to intuit the behavior of quantum systems; however, there is an indirect correspondence, for qubit states can be manipulated and transformed in ways which lead to measurement outcomes which depend distinctly on the different properties of the state. Thus, these quantum states have real, experimentally veri\ufb01able consequences, which we shall see are essential to the power of quantum computation and quantum information. 14 Introduction and overview The ability of a qubit to be in a superposition state runs counter to our \u2018common sense\u2019 understanding of the physical world around us. A classical bit is like a coin: either heads or tails up. For imperfect coins, there may be intermediate states like having it balanced on an edge, but those can be disregarded in the ideal case. By contrast, a qubit can exist in a continuum of states between |0 and |1 \u2013 until it is observed. Let us emphasize again that when a qubit is measured, it only ever gives \u20180\u2019 or \u20181\u2019 as the measurement result \u2013 probabilistically. For example, a qubit can be in the state 1\u221a 2 |0 + 1\u221a 2 |1, (1.2) 2|2) of the time, and the which, when measured, gives the result 0 \ufb01fty percent (|1/ result 1 \ufb01fty percent of the time. We will return often to this state, which is sometimes denoted |+. \u221a Despite this strangeness, qubits are decidedly real, their existence and behavior extensively validated by experiments (discussed in Section 1.5 and Chapter 7), and many different physical systems can be used to realize qubits. To get a concrete feel for how a qubit can be realized it may be helpful to list some of the ways this realization may occur: as the two different polarizations of a photon; as the alignment of a nuclear spin in a uniform magnetic \ufb01eld; as two states of an electron orbiting a single atom such as shown in Figure 1.2. In the atom model, the electron can exist in either the so-called \u2018ground\u2019 or \u2018excited\u2019 states, which we\u2019ll call |0 and |1, respectively. By shining light on the atom, with appropriate energy and for an appropriate length of time, it is possible to move the electron from the |0 state to the |1 state and vice versa. But more interestingly, by reducing the time we shine the light, an electron initially in the state |0 can be moved \u2018halfway\u2019 between |0 and |1, into the |+ state. Figure 1.2. Qubit represented by two electronic levels in an atom. Naturally, a great deal of attention has been given to the \u2018meaning\u2019 or \u2018interpretation\u2019 that might be attached to superposition states, and of the inherently probabilistic nature of observations on quantum systems. However, by and large, we shall not concern ourselves with such discussions in this book. Instead, our intent will be to develop mathematical and conceptual pictures which are predictive. One picture useful in thinking about qubits is the following geometric representation. Because |\u03b1|2 + |\u03b2|2 = 1, we", " may rewrite Equation (1.1) as |\u03c8 = ei\u03b3 cos \u03b8 2 |0 + ei\u03d5 sin \u03b8 2 Quantum bits 15 |1, (1.3) where \u03b8, \u03d5 and \u03b3 are real numbers. In Chapter 2 we will see that we can ignore the factor of ei\u03b3 out the front, because it has no observable effects, and for that reason we can effectively write |\u03c8 = cos \u03b8 2 |0 + ei\u03d5 sin \u03b8 2 |1. (1.4) The numbers \u03b8 and \u03d5 de\ufb01ne a point on the unit three-dimensional sphere, as shown in Figure 1.3. This sphere is often called the Bloch sphere; it provides a useful means of visualizing the state of a single qubit, and often serves as an excellent testbed for ideas about quantum computation and quantum information. Many of the operations on single qubits which we describe later in this chapter are neatly described within the Bloch sphere picture. However, it must be kept in mind that this intuition is limited because there is no simple generalization of the Bloch sphere known for multiple qubits. |\u03c8 y z \u03b8 \u03d5 x Figure 1.3. Bloch sphere representation of a qubit. How much information is represented by a qubit? Paradoxically, there are an in\ufb01nite number of points on the unit sphere, so that in principle one could store an entire text of Shakespeare in the in\ufb01nite binary expansion of \u03b8. However, this conclusion turns out to be misleading, because of the behavior of a qubit when observed. Recall that measurement of a qubit will give only either 0 or 1. Furthermore, measurement changes the state of a qubit, collapsing it from its superposition of |0 and |1 to the speci\ufb01c state consistent with the measurement result. For example, if measurement of |+ gives 0, then the post-measurement state of the qubit will be |0. Why does this type of collapse occur? Nobody knows. As discussed in Chapter 2, this behavior is simply one of the fundamental postulates of quantum mechanics. What is relevant for our purposes is that from a single measurement one obtains only a single bit of information about the state of the qubit, thus resolving the apparent paradox. It turns out that only if in\ufb01nitely many 16 Introduction and overview identically prepared qubits were measured would one be able to determine \u03b1 and \u03b2 for a qubit in the state given in Equation (1.1). But an even more interesting question to ask might be: how much information is represented by a qubit if we do not measure it? This is a trick question, because how can one quantify information if it cannot be measured? Nevertheless, there is something conceptually important here, because when Nature evolves a closed quantum system of qubits, not performing any \u2018measurements\u2019, she apparently does keep track of all the continuous variables describing the state, like \u03b1 and \u03b2. In a sense, in the state of a qubit, Nature conceals a great deal of \u2018hidden information\u2019. And even more interestingly, we will see shortly that the potential amount of this extra \u2018information\u2019 grows exponentially with the number of qubits. Understanding this hidden quantum information is a question that we grapple with for much of this book, and which lies at the heart of what makes quantum mechanics a powerful tool for information processing. 1.2.1 Multiple qubits Hilbert space is a big place. \u2013 Carlton Caves Suppose we have two qubits. If these were two classical bits, then there would be four possible states, 00, 01, 10, and 11. Correspondingly, a two qubit system has four computational basis states denoted |00, |01, |10, |11. A pair of qubits can also exist in superpositions of these four states, so the quantum state of two qubits involves associating a complex coef\ufb01cient \u2013 sometimes called an amplitude \u2013 with each computational basis state, such that the state vector describing the two qubits is |\u03c8 = \u03b100|00 + \u03b101|01 + \u03b110|10 + \u03b111|11. (1.5) Similar to the case for a single qubit, the measurement result x (= 00, 01, 10 or 11) occurs with probability |\u03b1x|2, with the state of the qubits after the measurement being |x. The condition that probabilities sum to one is therefore expressed by the normalization x\u2208{0,1}2 |\u03b1x|2 = 1, where the notation \u2018{0, 1}2\u2019 means \u2018the set of strings condition that of length two with each letter being either zero or one\u2019. For a two qubit system, we could measure just a", " subset of the qubits, say the \ufb01rst qubit, and you can probably guess how this works: measuring the \ufb01rst qubit alone gives 0 with probability |\u03b100|2 + |\u03b101|2, leaving the post-measurement state |\u03c8 = \u03b100|00 + \u03b101|01 |\u03b100|2 + |\u03b101|2. (1.6) |\u03b100|2 + |\u03b101|2 Note how the post-measurement state is re-normalized by the factor so that it still satis\ufb01es the normalization condition, just as we expect for a legitimate quantum state. An important two qubit state is the Bell state or EPR pair, |00 + |11 \u221a 2. (1.7) This innocuous-looking state is responsible for many surprises in quantum computation Quantum computation 17 and quantum information. It is the key ingredient in quantum teleportation and superdense coding, which we\u2019ll come to in Section 1.3.7 and Section 2.3, respectively, and the prototype for many other interesting quantum states. The Bell state has the property that upon measuring the \ufb01rst qubit, one obtains two possible results: 0 with probability 1/2, leaving the post-measurement state |\u03d5 = |00, and 1 with probability 1/2, leaving |\u03d5 = |11. As a result, a measurement of the second qubit always gives the same result as the measurement of the \ufb01rst qubit. That is, the measurement outcomes are correlated. Indeed, it turns out that other types of measurements can be performed on the Bell state, by \ufb01rst applying some operations to the \ufb01rst or second qubit, and that interesting correlations still exist between the result of a measurement on the \ufb01rst and second qubit. These correlations have been the subject of intense interest ever since a famous paper by Einstein, Podolsky and Rosen, in which they \ufb01rst pointed out the strange properties of states like the Bell state. EPR\u2019s insights were taken up and greatly improved by John Bell, who proved an amazing result: the measurement correlations in the Bell state are stronger than could ever exist between classical systems. These results, described in detail in Section 2.6, were the \ufb01rst intimation that quantum mechanics allows information processing beyond what is possible in the classical world. More generally, we may consider a system of n qubits. The computational basis states of this system are of the form |x1x2... xn, and so a quantum state of such a system is speci\ufb01ed by 2n amplitudes. For n = 500 this number is larger than the estimated number of atoms in the Universe! Trying to store all these complex numbers would not be possible on any conceivable classical computer. Hilbert space is indeed a big place. In principle, however, Nature manipulates such enormous quantities of data, even for systems containing only a few hundred atoms. It is as if Nature were keeping 2500 hidden pieces of scratch paper on the side, on which she performs her calculations as the system evolves. This enormous potential computational power is something we would very much like to take advantage of. But how can we think of quantum mechanics as computation? 1.3 Quantum computation Changes occurring to a quantum state can be described using the language of quantum computation. Analogous to the way a classical computer is built from an electrical circuit containing wires and logic gates, a quantum computer is built from a quantum circuit containing wires and elementary quantum gates to carry around and manipulate the quantum information. In this section we describe some simple quantum gates, and present several example circuits illustrating their application, including a circuit which teleports qubits! 1.3.1 Single qubit gates Classical computer circuits consist of wires and logic gates. The wires are used to carry information around the circuit, while the logic gates perform manipulations of the information, converting it from one form to another. Consider, for example, classical single bit gate, whose operation logic gates. The only non-trivial member of this class is the is de\ufb01ned by its truth table, in which 0 \u2192 1 and 1 \u2192 0, that is, the 0 and 1 states are interchanged. Can an analogous quantum gate for qubits be de\ufb01ned? Imagine that we had some process which took the state |0 to the state |1, and vice versa. Such a process 18 Introduction and overview gate. However, would obviously be a good candidate for a quantum analogue to the specifying the action of the gate on the states |0 and |1 does not tell us what happens to superpositions of the states |0 and |1, without further knowledge about the properties gate acts linearly, that is, it takes the state of quantum gates. In fact, the quantum \u03b1", "|0 + \u03b2|1 to the corresponding state in which the role of |0 and |1 have been interchanged, \u03b1|1 + \u03b2|0. (1.8) (1.9) gate acts linearly and not in some nonlinear fashion is a very Why the quantum interesting question, and the answer is not at all obvious. It turns out that this linear behavior is a general property of quantum mechanics, and very well motivated empirically; moreover, nonlinear behavior can lead to apparent paradoxes such as time travel, fasterthan-light communication, and violations of the second laws of thermodynamics. We\u2019ll explore this point in more depth in later chapters, but for now we\u2019ll just take it as given. gate in matrix form, which follows directly from the linearity of quantum gates. Suppose we de\ufb01ne a matrix X to represent the quantum There is a convenient way of representing the quantum gate as follows: X \u2261 0 1 1 0. (1.10) (The notation X for the quantum state \u03b1|0 + \u03b2|1 is written in a vector notation as is used for historical reasons.) If the quantum, \u03b1 \u03b2 (1.11) with the top entry corresponding to the amplitude for |0 and the bottom entry the amplitude for |1, then the corresponding output from the quantum gate is X =. \u03b2 \u03b1 \u03b1 \u03b2 (1.12) gate is to take the state |0 and replace it by the state Notice that the action of the corresponding to the \ufb01rst column of the matrix X. Similarly, the state |1 is replaced by the state corresponding to the second column of the matrix X. So quantum gates on a single qubit can be described by two by two matrices. Are there any constraints on what matrices may be used as quantum gates? It turns out that there are. Recall that the normalization condition requires |\u03b1|2 + |\u03b2|2 = 1 for a quantum state \u03b1|0 + \u03b2|1. This must also be true of the quantum state |\u03c8 = \u03b1|0 + \u03b2|1 after the gate has acted. It turns out that the appropriate condition on the matrix representing the gate is that the matrix U describing the single qubit gate be unitary, that is U \u2020U = I, where U \u2020 is the adjoint of U (obtained by transposing and then complex conjugating U ), and I is the two by two identity matrix. For example, for the gate it is easy to verify that X \u2020X = I. Amazingly, this unitarity constraint is the only constraint on quantum gates. Any unitary matrix speci\ufb01es a valid quantum gate! The interesting implication is that in contrast to the classical case, where only one non-trivial single bit gate exists \u2013 the Quantum computation 19 z z y y x x z x Figure 1.4. Visualization of the Hadamard gate on the Bloch sphere, acting on the input state (|0 + |1)/ \u221a y 2. gate \u2013 there are many non-trivial single qubit gates. Two important ones which we shall use later are the Z gate: Z \u2261 1 0 0 \u22121, (1.13) which leaves |0 unchanged, and \ufb02ips the sign of |1 to give \u2212|1, and the Hadamard gate, H \u2261 1\u221a 2 1 1 1 \u22121. (1.14) \u221a This gate is sometimes described as being like a \u2018square-root of a |0 into (|0 + |1)/ \u221a |1 into (|0 \u2212 |1)/ |1. Note, however, that H 2 is not a thus applying H twice to a state does nothing to it. \u2019 gate, in that it turns 2 (\ufb01rst column of H), \u2018halfway\u2019 between |0 and |1, and turns 2 (second column of H), which is also \u2018halfway\u2019 between |0 and gate, as simple algebra shows that H 2 = I, and The Hadamard gate is one of the most useful quantum gates, and it is worth trying to visualize its operation by considering the Bloch sphere picture. In this picture, it turns out that single qubit gates correspond to rotations and re\ufb02ections of the sphere. The Hadamard operation is just a rotation of the sphere about the \u02c6y axis by 90\u25e6, followed by a rotation about the \u02c6x axis by 180\u25e6, as illustrated in Figure 1.4. Some important single qubit gates are shown in Figure 1.5, and contrasted with the classical case. Figure 1.5. Single bit (left) and qubit (right) logic gates. There are in\ufb01nitely many two by two unitary matrices, and thus in\ufb01n", "itely many single 20 Introduction and overview qubit gates. However, it turns out that the properties of the complete set can be understood from the properties of a much smaller set. For example, as explained in Box 1.1, an arbitrary single qubit unitary gate can be decomposed as a product of rotations cos \u03b3 2 sin \u03b3 2 and a gate which we\u2019ll later understand as being a rotation about the \u02c6z axis, \u2212 sin \u03b3 2 cos \u03b3 2,, e\u2212i\u03b2/2 0 0 ei\u03b2/2 (1.15) (1.16) together with a (global) phase shift \u2013 a constant multiplier of the form ei\u03b1. These gates can be broken down further \u2013 we don\u2019t need to be able to do these gates for arbitrary \u03b1, \u03b2 and \u03b3, but can build arbitrarily good approximations to such gates using only certain special \ufb01xed values of \u03b1, \u03b2 and \u03b3. In this way it is possible to build up an arbitrary single qubit gate using a \ufb01nite set of quantum gates. More generally, an arbitrary quantum computation on any number of qubits can be generated by a \ufb01nite set of gates that is said to be universal for quantum computation. To obtain such a universal set we \ufb01rst need to introduce some quantum gates involving multiple qubits. Box 1.1: Decomposing single qubit operations In Section 4.2 starting on page 174 we prove that an arbitrary 2\u00d72 unitary matrix may be decomposed as 0 ei\u03b2/2 U = ei\u03b1 e\u2212i\u03b2/2 0 cos \u03b3 2 sin \u03b3 2 where \u03b1, \u03b2, \u03b3, and \u03b4 are real-valued. Notice that the second matrix is just an ordinary rotation. It turns out that the \ufb01rst and last matrices can also be understood as rotations in a different plane. This decomposition can be used to give an exact prescription for performing an arbitrary single qubit quantum logic gate. \u2212 sin \u03b3 2 cos \u03b3 2 e\u2212i\u03b4/2 0 0 ei\u03b4/2, (1.17),,, (exclusive- gates alone, which is thus known as a universal gate. By contrast, the 1.3.2 Multiple qubit gates Now let us generalize from one to multiple qubits. Figure 1.6 shows \ufb01ve notable multiple gates. An important bit classical gates, the theoretical result is that any function on bits can be computed from the composition of alone or is not universal. One way of seeing this is to note that applying even together with gate does not change the total parity of the bits. As a result, any circuit involving an gates will, if two inputs x and y have the same parity, give outputs only with the same parity, restricting the class of functions which may be computed, and thus precluding universality. and and ), The prototypical multi-qubit quantum logic gate is the controlled- gate. This gate has two input qubits, known as the control qubit and the target qubit, respecis shown in the top right of Figure 1.6; tively. The circuit representation for the the top line represents the control qubit, while the bottom line represents the target or Quantum computation 21 Figure 1.6. On the left are some standard single and multiple bit gates, while on the right is the prototypical multiple qubit gate, the controlledrespect to the amplitudes for |00, |01, |10, and |11, in that order.. The matrix representation of the controlled-, UCN, is written with qubit. The action of the gate may be described as follows. If the control qubit is set to 0, then the target qubit is left alone. If the control qubit is set to 1, then the target qubit is \ufb02ipped. In equations: |00 \u2192 |00; |01 \u2192 |01; |10 \u2192 |11; |11 \u2192 |10. (1.18) gate, since Another way of describing the the action of the gate may be summarized as |A, B \u2192 |A, B \u2295 A, where \u2295 is addition gate does. That is, the control qubit and the modulo two, which is exactly what the target qubit are is as a generalization of the classical ed and stored in the target qubit. Yet another way of describing the action of the is to give a matrix representation, as shown in the bottom right of Figure 1.6. You can easily verify that the \ufb01rst column of UCN describes the transformation that occurs to |00, and similarly for the other computational basis states, |01, |10, and |11. As for the single qubit case, the requirement that probability be conserved is expressed in the fact that", " UCN is a unitary matrix, that is, U \u2020 CN UCN = I. We noticed that the can be regarded as a type of generalized- or the regular gate represents the classical other classical gates such as the gates in a sense similar to the way the quantum gates gate? It turns out that this is not possible. The reason is because the are essentially irreversible or non-invertible. For example, given the output A \u2295 B from gate, it is not possible to determine what the inputs A and B were; there is an an irretrievable loss of information associated with the irreversible action of the gate. On the other hand, unitary quantum gates are always invertible, since the inverse of a unitary matrix is also a unitary matrix, and thus a quantum gate can always be inverted by another quantum gate. Understanding how to do classical logic in this reversible or invertible sense will be a crucial step in understanding how to harness the power of and gate. Can gate be understood as unitary 22 Introduction and overview quantum mechanics for computation. We\u2019ll explain the basic idea of how to do reversible computation in Section 1.4.1. Of course, there are many interesting quantum gates other than the controlled-. and single qubit gates are the prototypes for all However, in a sense the controlledother gates because of the following remarkable universality result: Any multiple qubit and single qubit gates. The proof is given in logic gate may be composed from Section 4.5, and is the quantum parallel of the universality of the gate. 1.3.3 Measurements in bases other than the computational basis We\u2019ve described quantum measurements of a single qubit in the state \u03b1|0 + \u03b2|1 as yielding the result 0 or 1 and leaving the qubit in the corresponding state |0 or |1, with respective probabilities |\u03b1|2 and |\u03b2|2. In fact, quantum mechanics allows somewhat more versatility in the class of measurements that may be performed, although certainly nowhere near enough to recover \u03b1 and \u03b2 from a single measurement! Note that the states |0 and |1 represent just one of many possible choices of basis 2 and |\u2212 \u2261 2. An arbitrary state |\u03c8 = \u03b1|0 + \u03b2|1 can be re-expressed in terms of the states for a qubit. Another possible choice is the set |+ \u2261 (|0 + |1)/ (|0 \u2212 |1)/ states |+ and |\u2212: \u221a \u221a |\u03c8 = \u03b1|0 + \u03b2|1 = \u03b1 |+ + |\u2212 \u221a 2 + \u03b2 |+ \u2212 |\u2212 \u221a 2 = \u03b1 + \u03b2 \u221a 2 |+ + \u03b1 \u2212 \u03b2 \u221a 2 |\u2212. (1.19) It turns out that it is possible to treat the |+ and |\u2212 states as though they were the computational basis states, and measure with respect to this new basis. Naturally, measuring with respect to the |+, |\u2212 basis results in the result \u2018+\u2019 with probability |\u03b1 + \u03b2|2/2 and the result \u2018\u2212\u2019 with probability |\u03b1 \u2212 \u03b2|2/2, with corresponding post-measurement states |+ and |\u2212, respectively. More generally, given any basis states |a and |b for a qubit, it is possible to express an arbitrary state as a linear combination \u03b1|a + \u03b2|b of those states. Furthermore, provided the states are orthonormal, it is possible to perform a measurement with respect to the |a, |b basis, giving the result a with probability |\u03b1|2 and b with probability |\u03b2|2. The orthonormality constraint is necessary in order that |\u03b1|2 + |\u03b2|2 = 1 as we expect for probabilities. In an analogous way it is possible in principle to measure a quantum system of many qubits with respect to an arbitrary orthonormal basis. However, just because it is possible in principle does not mean that such a measurement can be done easily, and we return later to the question of how ef\ufb01ciently a measurement in an arbitrary basis can be performed. There are many reasons for using this extended formalism for quantum measurements, but ultimately the best one is this: the formalism allows us to describe observed experimental results, as we will see in our discussion of the Stern\u2013Gerlach experiment in Section 1.5.1. An even more sophisticated and convenient (but essentially equivalent) formalism for describing quantum measurements is described in the next chapter, in Section 2.2.3. 1.3.4 Quantum circuits We\u2019ve already met a few simple quantum circuits. Let\u2019s look in a little more detail at the elements of a quantum circuit. A simple quantum circuit containing three quantum gates is shown in Figure 1.7. The circuit is to be read from left-to-right. Each line Quantum computation", " 23 in the circuit represents a wire in the quantum circuit. This wire does not necessarily correspond to a physical wire; it may correspond instead to the passage of time, or perhaps to a physical particle such as a photon \u2013 a particle of light \u2013 moving from one location to another through space. It is conventional to assume that the state input to the circuit is a computational basis state, usually the state consisting of all |0s. This rule is broken frequently in the literature on quantum computation and quantum information, but it is considered polite to inform the reader when this is the case. The circuit in Figure 1.7 accomplishes a simple but useful task \u2013 it swaps the states of the two qubits. To see that this circuit accomplishes the swap operation, note that the sequence of gates has the following sequence of effects on a computational basis state |a, b, |a, b \u2212\u2192 |a, a \u2295 b \u2212\u2192 |a \u2295 (a \u2295 b), a \u2295 b = |b, a \u2295 b \u2212\u2192 |b, (a \u2295 b) \u2295 b = |b, a, (1.20) where all additions are done modulo 2. The effect of the circuit, therefore, is to interchange the state of the two qubits. Figure 1.7. Circuit swapping two qubits, and an equivalent schematic symbol notation for this common and useful circuit. There are a few features allowed in classical circuits that are not usually present in quantum circuits. First of all, we don\u2019t allow \u2018loops\u2019, that is, feedback from one part of the quantum circuit to another; we say the circuit is acyclic. Second, classical circuits allow, with the resulting single wire wires to be \u2018joined\u2019 together, an operation known as of the inputs. Obviously this operation is not reversible and containing the bitwise in our quantum circuits. Third, the inverse therefore not unitary, so we don\u2019t allow, whereby several copies of a bit are produced is also not allowed in operation, quantum circuits. In fact, it turns out that quantum mechanics forbids the copying of a operation impossible! We\u2019ll see an example of this in the next qubit, making the section when we attempt to design a circuit to copy a qubit. As we proceed we\u2019ll introduce new quantum gates as needed. It\u2019s convenient to introduce another convention about quantum circuits at this point. This convention is illustrated in Figure 1.8. Suppose U is any unitary matrix acting on some number n of qubits, so U can be regarded as a quantum gate on those qubits. Then we can de\ufb01ne a controlled-U gate which is a natural extension of the controlledgate. Such a gate has a single control qubit, indicated by the line with the black dot, and n target qubits, indicated by the boxed U. If the control qubit is set to 0 then nothing happens to the target qubits. If the control qubit is set to 1 then the gate U is applied to the target qubits. The prototypical example of the controlled-U gate is the controlledgate, which is a controlled-U gate with U = X, as illustrated in Figure 1.9. Another important operation is measurement, which we represent by a \u2018meter\u2019 symbol, 24 Introduction and overview Figure 1.8. Controlled-U gate. Figure 1.9. Two different representations for the controlled-. as shown in Figure 1.10. As previously described, this operation converts a single qubit state |\u03c8 = \u03b1|0 + \u03b2|1 into a probabilistic classical bit M (distinguished from a qubit by drawing it as a double-line wire), which is 0 with probability |\u03b1|2, or 1 with probability |\u03b2|2. Figure 1.10. Quantum circuit symbol for measurement. We shall \ufb01nd quantum circuits useful as models of all quantum processes, including but not limited to computation, communication, and even quantum noise. Several simple examples illustrate this below. 1.3.5 Qubit copying circuit? The gate is useful for demonstrating one particularly fundamental property of quantum information. Consider the task of copying a classical bit. This may be done gate, which takes in the bit to copy (in some unknown state x) using a classical and a \u2018scratchpad\u2019 bit initialized to zero, as illustrated in Figure 1.11. The output is two bits, both of which are in the same state x. Suppose we try to copy a qubit in the unknown state |\u03c8 = a |0 + b |1 in the same manner by using a gate. The input state of the two qubits may be written as a |0 + b |1 |0 = a |00 + b |10, (1.21) is to negate the second qubit when the \ufb01r", "st qubit is 1, and thus The function of the output is simply a |00 + b |11. Have we successfully copied |\u03c8? That is, have we created the state |\u03c8|\u03c8? In the case where |\u03c8 = |0 or |\u03c8 = |1 that is indeed what this circuit does; it is possible to use quantum circuits to copy classical information encoded as a |0 or a |1. However, for a general state |\u03c8 we see that |\u03c8|\u03c8 = a2|00 + ab|01 + ab|10 + b2|11. (1.22) Quantum computation 25 Figure 1.11. Classical and quantum circuits to \u2018copy\u2019 an unknown bit or qubit. Comparing with a|00 + b|11, we see that unless ab = 0 the \u2018copying circuit\u2019 above does not copy the quantum state input. In fact, it turns out to be impossible to make a copy of an unknown quantum state. This property, that qubits cannot be copied, is known as the no-cloning theorem, and it is one of the chief differences between quantum and classical information. The no-cloning theorem is discussed at more length in Box 12.1 on page 532; the proof is very simple, and we encourage you to skip ahead and read the proof now. There is another way of looking at the failure of the circuit in Figure 1.11, based on the intuition that a qubit somehow contains \u2018hidden\u2019 information not directly accessible to measurement. Consider what happens when we measure one of the qubits of the state a|00 + b|11. As previously described, we obtain either 0 or 1 with probabilities |a|2 and |b|2. However, once one qubit is measured, the state of the other one is completely determined, and no additional information can be gained about a and b. In this sense, the extra hidden information carried in the original qubit |\u03c8 was lost in the \ufb01rst measurement, and cannot be regained. If, however, the qubit had been copied, then the state of the other qubit should still contain some of that hidden information. Therefore, a copy cannot have been created. 1.3.6 Example: Bell states Let\u2019s consider a slightly more complicated circuit, shown in Figure 1.12, which has a, and transforms the four computational basis states Hadamard gate followed by a according to the table given. As an explicit example, the Hadamard gate takes the input |00 to (|0 + |1)|0/ 2. Note how this works: \ufb01rst, the Hadamard transform puts the top qubit in a superposition;, and the target gets inverted only when the this then acts as a control input to the control is 1. The output states gives the output state (|00 + |11)/ 2, and then the \u221a \u221a |\u03b200 = |\u03b201 = |\u03b210 = |\u03b211 = |00 + |11 \u221a 2 |01 + |10 \u221a 2 |00 \u2212 |11 \u221a 2 |01 \u2212 |10 \u221a 2 ; ; ; and, (1.23) (1.24) (1.25) (1.26) are known as the Bell states, or sometimes the EPR states or EPR pairs, after some of the people \u2013 Bell, and Einstein, Podolsky, and Rosen \u2013 who \ufb01rst pointed out the strange properties of states like these. The mnemonic notation |\u03b200, |\u03b201, |\u03b210, |\u03b211 may be 26 Introduction and overview understood via the equations |\u03b2xy \u2261 |0, y + (\u22121)x|1, \u00afy \u221a 2, (1.27) where \u00afy is the negation of y. In |00 |01 |10 |11 \u221a Out \u221a (|00 + |11)/ (|01 + |10)/ (|00 \u2212 |11)/ (|01 \u2212 |10)/ \u221a \u221a 2 \u2261 |\u03b200 2 \u2261 |\u03b201 2 \u2261 |\u03b210 2 \u2261 |\u03b211 Figure 1.12. Quantum circuit to create Bell states, and its input\u2013ouput quantum \u2018truth table\u2019. 1.3.7 Example: quantum teleportation We will now apply the techniques of the last few pages to understand something nontrivial, surprising, and a lot of fun \u2013 quantum teleportation! Quantum teleportation is a technique for moving quantum states around, even in the absence of a quantum communications channel linking the sender of the quantum state to the recipient. Here\u2019s how quantum teleportation works. Alice and Bob met long ago but now live far apart. While together they generated an EPR pair, each taking one qubit of the EPR pair when they separated. Many years later, Bob", " is in hiding, and Alice\u2019s mission, should she choose to accept it, is to deliver a qubit |\u03c8 to Bob. She does not know the state of the qubit, and moreover can only send classical information to Bob. Should Alice accept the mission? Intuitively, things look pretty bad for Alice. She doesn\u2019t know the state |\u03c8 of the qubit she has to send to Bob, and the laws of quantum mechanics prevent her from determining the state when she only has a single copy of |\u03c8 in her possession. What\u2019s worse, even if she did know the state |\u03c8, describing it precisely takes an in\ufb01nite amount of classical information since |\u03c8 takes values in a continuous space. So even if she did know |\u03c8, it would take forever for Alice to describe the state to Bob. It\u2019s not looking good for Alice. Fortunately for Alice, quantum teleportation is a way of utilizing the entangled EPR pair in order to send |\u03c8 to Bob, with only a small overhead of classical communication. In outline, the steps of the solution are as follows: Alice interacts the qubit |\u03c8 with her half of the EPR pair, and then measures the two qubits in her possession, obtaining one of four possible classical results, 00, 01, 10, and 11. She sends this information to Bob. Depending on Alice\u2019s classical message, Bob performs one of four operations on his half of the EPR pair. Amazingly, by doing this he can recover the original state |\u03c8! The quantum circuit shown in Figure 1.13 gives a more precise description of quantum teleportation. The state to be teleported is |\u03c8 = \u03b1|0+\u03b2|1, where \u03b1 and \u03b2 are unknown amplitudes. The state input into the circuit |\u03c80 is |\u03c80 = |\u03c8|\u03b200 (1.28) Quantum computation 27 Figure 1.13. Quantum circuit for teleporting a qubit. The two top lines represent Alice\u2019s system, while the bottom line is Bob\u2019s system. The meters represent measurement, and the double lines coming out of them carry classical bits (recall that single lines denote qubits). \u03b1|0(|00 + |11) + \u03b2|1(|00 + |11), (1.29) = 1\u221a 2 where we use the convention that the \ufb01rst two qubits (on the left) belong to Alice, and the third qubit to Bob. As we explained previously, Alice\u2019s second qubit and Bob\u2019s qubit gate, obtaining start out in an EPR state. Alice sends her qubits through a \u03b1|0(|00 + |11) + \u03b2|1(|10 + |01) (1.30) |\u03c81 =. 1\u221a 2 She then sends the \ufb01rst qubit through a Hadamard gate, obtaining |\u03c82 = 1 2 \u03b1(|0 + |1)(|00 + |11) + \u03b2(|0 \u2212 |1)(|10 + |01). This state may be re-written in the following way, simply by regrouping terms: |\u03c82 = |00 1 2 + |10 \u03b1|0 + \u03b2|1 \u03b1|0 \u2212 \u03b2|1 + |01 + |11 \u03b1|1 + \u03b2|0 \u03b1|1 \u2212 \u03b2|0. (1.31) (1.32) This expression naturally breaks down into four terms. The \ufb01rst term has Alice\u2019s qubits in the state |00, and Bob\u2019s qubit in the state \u03b1|0 + \u03b2|1 \u2013 which is the original state |\u03c8. If Alice performs a measurement and obtains the result 00 then Bob\u2019s system will be in the state |\u03c8. Similarly, from the previous expression we can read off Bob\u2019s postmeasurement state, given the result of Alice\u2019s measurement: 00 \u2212\u2192 |\u03c83(00) \u2261 01 \u2212\u2192 |\u03c83(01) \u2261 10 \u2212\u2192 |\u03c83(10) \u2261 11 \u2212\u2192 |\u03c83(11) \u2261 \u03b1|0 + \u03b2|1 \u03b1|1 + \u03b2|0 \u03b1|0 \u2212 \u03b2|1 \u03b1|1 \u2212 \u03b2|0 (1.33) (1.34) (1.35) (1.36). Depending on Alice\u2019s measurement outcome, Bob\u2019s qubit will end up in one of these four possible states. Of course, to know which state it is in, Bob must be told the result of Alice\u2019s measurement \u2013 we will show later that it is this fact which prevents teleportation 28 Introduction and overview from being used to transmit information faster than light. Once Bob", " has learned the measurement outcome, Bob can \u2018\ufb01x up\u2019 his state, recovering |\u03c8, by applying the appropriate quantum gate. For example, in the case where the measurement yields 00, Bob doesn\u2019t need to do anything. If the measurement is 01 then Bob can \ufb01x up his state by applying the X gate. If the measurement is 10 then Bob can \ufb01x up his state by applying the Z gate. If the measurement is 11 then Bob can \ufb01x up his state by applying \ufb01rst an X and then a Z gate. Summing up, Bob needs to apply the transformation Z M1X M2 (note how time goes from left to right in circuit diagrams, but in matrix products terms on the right happen \ufb01rst) to his qubit, and he will recover the state |\u03c8. There are many interesting features of teleportation, some of which we shall return to later in the book. For now we content ourselves with commenting on a couple of aspects. First, doesn\u2019t teleportation allow one to transmit quantum states faster than light? This would be rather peculiar, because the theory of relativity implies that faster than light information transfer could be used to send information backwards in time. Fortunately, quantum teleportation does not enable faster than light communication, because to complete the teleportation Alice must transmit her measurement result to Bob over a classical communications channel. We will show in Section 2.4.3 that without this classical communication, teleportation does not convey any information at all. The classical channel is limited by the speed of light, so it follows that quantum teleportation cannot be accomplished faster than the speed of light, resolving the apparent paradox. A second puzzle about teleportation is that it appears to create a copy of the quantum state being teleported, in apparent violation of the no-cloning theorem discussed in Section 1.3.5. This violation is only illusory since after the teleportation process only the target qubit is left in the state |\u03c8, and the original data qubit ends up in one of the computational basis states |0 or |1, depending upon the measurement result on the \ufb01rst qubit. What can we learn from quantum teleportation? Quite a lot! It\u2019s much more than just a neat trick one can do with quantum states. Quantum teleportation emphasizes the interchangeability of different resources in quantum mechanics, showing that one shared EPR pair together with two classical bits of communication is a resource at least the equal of one qubit of communication. Quantum computation and quantum information has revealed a plethora of methods for interchanging resources, many built upon quantum teleportation. In particular, in Chapter 10 we explain how teleportation can be used to build quantum gates which are resistant to the effects of noise, and in Chapter 12 we show that teleportation is intimately connected with the properties of quantum error-correcting codes. Despite these connections with other subjects, it is fair to say that we are only beginning to understand why it is that quantum teleportation is possible in quantum mechanics; in later chapters we endeavor to explain some of the insights that make such an understanding possible. 1.4 Quantum algorithms What class of computations can be performed using quantum circuits? How does that class compare with the computations which can be performed using classical logical circuits? Can we \ufb01nd a task which a quantum computer may perform better than a classical computer? In this section we investigate these questions, explaining how to perform classical computations on quantum computers, giving some examples of problems for Quantum algorithms 29 which quantum computers offer an advantage over classical computers, and summarizing the known quantum algorithms. 1.4.1 Classical computations on a quantum computer Can we simulate a classical logic circuit using a quantum circuit? Not surprisingly, the answer to this question turns out to be yes. It would be very surprising if this were not the case, as physicists believe that all aspects of the world around us, including classical logic circuits, can ultimately be explained using quantum mechanics. As pointed out earlier, the reason quantum circuits cannot be used to directly simulate classical circuits is because unitary quantum logic gates are inherently reversible, whereas many classical gate are inherently irreversible. logic gates such as the Any classical circuit can be replaced by an equivalent circuit containing only reversible elements, by making use of a reversible gate known as the Toffoli gate. The Toffoli gate has three input bits and three output bits, as illustrated in Figure 1.14. Two of the bits are control bits that are unaffected by the action of the Toffoli gate. The third bit is a target bit that is \ufb02ipped if both control bits are set to 1, and otherwise is left alone. Note that applying the Toffoli gate twice to a set of bits has the effect (a, b, c) \u2192 (a, b, c \u2295 ab) \u2192 (a, b, c), and thus the Toffoli gate is a reversible gate, since it has an inverse \u2013 itself", ". Inputs Outputs Figure 1.14. Truth table for the Toffoli gate, and its circuit representation. The Toffoli gate can be used to simulate gates, as shown in Figure 1.15, and, as shown in Figure 1.16. With these two operations it can also be used to do becomes possible to simulate all other elements in a classical circuit, and thus an arbitrary classical circuit can be simulated by an equivalent reversible circuit. The Toffoli gate has been described as a classical gate, but it can also be implemented as a quantum logic gate. By de\ufb01nition, the quantum logic implementation of the Toffoli gate simply permutes computational basis states in the same way as the classical Toffoli gate. For example, the quantum Toffoli gate acting on the state |110 \ufb02ips the third qubit because the \ufb01rst two are set, resulting in the state |111. It is tedious but not dif\ufb01cult to write this transformation out as an 8 by 8 matrix, U, and verify explicitly that U is a unitary matrix, and thus the Toffoli gate is a legitimate quantum gate. The quantum Toffoli gate can be used to simulate irreversible classical logic gates, just as the classical 30 Introduction and overview Figure 1.15. Classical circuit implementing a to the output from the is on the third bit., while the third bit is prepared in the standard state 1, sometimes known as an ancilla state. The gate using a Toffoli gate. The top two bits represent the input Figure 1.16. bits standard ancilla states), and the output from with the Toffoli gate, with the second bit being the input to the (and the other two appearing on the second and third bits. Toffoli gate was, and ensures that quantum computers are capable of performing any computation which a classical (deterministic) computer may do. What if the classical computer is non-deterministic, that is, has the ability to generate random bits to be used in the computation? Not surprisingly, it is easy for a quantum computer to simulate this. To perform such a simulation it turns out to be suf\ufb01cient to produce random fair coin tosses, which can be done by preparing a qubit in the state |0, sending it through a Hadamard gate to produce (|0 + |1)/ 2, and then measuring the state. The result will be |0 or |1 with 50/50 probability. This provides a quantum computer with the ability to ef\ufb01ciently simulate a non-deterministic classical computer. Of course, if the ability to simulate classical computers were the only feature of quantum computers there would be little point in going to all the trouble of exploiting quantum effects! The advantage of quantum computing is that much more powerful functions may be computed using qubits and quantum gates. In the next few sections we explain how to do this, culminating in the Deutsch\u2013Jozsa algorithm, our \ufb01rst example of a quantum algorithm able to solve a problem faster than any classical algorithm. \u221a 1.4.2 Quantum parallelism Quantum parallelism is a fundamental feature of many quantum algorithms. Heuristically, and at the risk of over-simplifying, quantum parallelism allows quantum computers to evaluate a function f (x) for many different values of x simultaneously. In this section we explain how quantum parallelism works, and some of its limitations. Suppose f (x) : {0, 1} \u2192 {0, 1} is a function with a one-bit domain and range. A Quantum algorithms 31 convenient way of computing this function on a quantum computer is to consider a two qubit quantum computer which starts in the state |x, y. With an appropriate sequence of logic gates it is possible to transform this state into |x, y \u2295 f (x), where \u2295 indicates addition modulo 2; the \ufb01rst register is called the \u2018data\u2019 register, and the second register the \u2018target\u2019 register. We give the transformation de\ufb01ned by the map |x, y \u2192 |x, y \u2295 f (x) a name, Uf, and note that it is easily shown to be unitary. If y = 0, then the \ufb01nal state of the second qubit is just the value f (x). (In Section 3.2.5 we show that given a classical circuit for computing f there is a quantum circuit of comparable ef\ufb01ciency which computes the transformation Uf on a quantum computer. For our purposes it can be considered to be a black box.) Figure 1.17. Quantum circuit for evaluating f (0) and f (1) simultaneously. Uf is the quantum circuit which takes inputs like |x, y to |x, y \u2295 f", " (x). \u221a Consider the circuit shown in Figure 1.17, which applies Uf to an input not in the computational basis. Instead, the data register is prepared in the superposition (|0 + 2, which can be created with a Hadamard gate acting on |0. Then we apply Uf, |1)/ resulting in the state: |0, f (0) + |1, f (1) \u221a 2. (1.37) This is a remarkable state! The different terms contain information about both f (0) and f (1); it is almost as if we have evaluated f (x) for two values of x simultaneously, a feature known as \u2018quantum parallelism\u2019. Unlike classical parallelism, where multiple circuits each built to compute f (x) are executed simultaneously, here a single f (x) circuit is employed to evaluate the function for multiple values of x simultaneously, by exploiting the ability of a quantum computer to be in superpositions of different states. This procedure can easily be generalized to functions on an arbitrary number of bits, by using a general operation known as the Hadamard transform, or sometimes the Walsh\u2013 Hadamard transform. This operation is just n Hadamard gates acting in parallel on n qubits. For example, shown in Figure 1.18 is the case n = 2 with qubits initially prepared as |0, which gives |0 + |1 \u221a 2 |0 + |1 \u221a 2 = |00 + |01 + |10 + |11 2 (1.38) as output. We write H \u22972 to denote the parallel action of two Hadamard gates, and read \u2018\u2297\u2019 as \u2018tensor\u2019. More generally, the result of performing the Hadamard transform on n 32 Introduction and overview qubits initially in the all |0 state is 1\u221a 2n x |x, (1.39) where the sum is over all possible values of x, and we write H \u2297n to denote this action. That is, the Hadamard transform produces an equal superposition of all computational basis states. Moreover, it does this extremely ef\ufb01ciently, producing a superposition of 2n states using just n gates. Figure 1.18. The Hadamard transform H \u22972 on two qubits. Quantum parallel evaluation of a function with an n bit input x and 1 bit output, f (x), can thus be performed in the following manner. Prepare the n + 1 qubit state |0\u2297n|0, then apply the Hadamard transform to the \ufb01rst n qubits, followed by the quantum circuit implementing Uf. This produces the state 1\u221a 2n x |x|f (x). (1.40) In some sense, quantum parallelism enables all possible values of the function f to be evaluated simultaneously, even though we apparently only evaluated f once. However, this parallelism is not immediately useful. In our single qubit example, measurement of the state gives only either |0, f (0) or |1, f (1)! Similarly, in the general case, measurement of x |x, f (x) would give only f (x) for a single value of x. Of course, a classical the state computer can do this easily! Quantum computation requires something more than just quantum parallelism to be useful; it requires the ability to extract information about more than one value of f (x) from superposition states like x |x, f (x). Over the next two sections we investigate examples of how this may be done. 1.4.3 Deutsch\u2019s algorithm A simple modi\ufb01cation of the circuit in Figure 1.17 demonstrates how quantum circuits can outperform classical ones by implementing Deutsch\u2019s algorithm (we actually present a simpli\ufb01ed and improved version of the original algorithm; see \u2018History and further reading\u2019 at the end of the chapter). Deutsch\u2019s algorithm combines quantum parallelism with a property of quantum mechanics known as interference. As before, let us use the Hadamard gate to prepare the \ufb01rst qubit as the superposition (|0 + |1)/ 2, but now let us prepare the second qubit y as the superposition (|0 \u2212 |1)/ 2, using a Hadamard gate applied to the state |1. Let us follow the states along to see what happens in this circuit, shown in Figure 1.19. \u221a \u221a The input state |\u03c80 = |01 (1.41) Quantum algorithms 33 Figure 1.19. Quantum circuit implementing Deutsch\u2019s algorithm. is sent through two Hadamard gates to give |0 + |1 \u221a 2 |\u03c81 =. |0 \u2212 |1 \u221a", " 2 (1.42) \u221a A little thought shows that if we apply Uf to the state |x(|0 \u2212 |1)/ the state (\u22121)f (x)|x(|0 \u2212 |1)/ two possibilities: 2 then we obtain 2. Applying Uf to |\u03c81 therefore leaves us with one of \u221a |\u03c82 = \u23a7 \u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23a9 \u00b1 \u00b1 |0 + |1 \u221a 2 |0 \u2212 |1 \u221a 2 |0 \u2212 |1 \u221a 2 |0 \u2212 |1 \u221a 2 if f (0) = f (1) if f (0) = f (1). The \ufb01nal Hadamard gate on the \ufb01rst qubit thus gives us \u23a7 |\u03c83 = \u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23a9 \u00b1|0 \u00b1|1 |0 \u2212 |1 \u221a 2 |0 \u2212 |1 \u221a 2 if f (0) = f (1) if f (0) = f (1). (1.43) (1.44) Realizing that f (0) \u2295 f (1) is 0 if f (0) = f (1) and 1 otherwise, we can rewrite this result concisely as |\u03c83 = \u00b1|f (0) \u2295 f (1), (1.45) |0 \u2212 |1 \u221a 2 so by measuring the \ufb01rst qubit we may determine f (0) \u2295 f (1). This is very interesting indeed: the quantum circuit has given us the ability to determine a global property of f (x), namely f (0) \u2295 f (1), using only one evaluation of f (x)! This is faster than is possible with a classical apparatus, which would require at least two evaluations. This example highlights the difference between quantum parallelism and classical randomized algorithms. Naively, one might think that the state |0|f (0) + |1|f (1) corresponds rather closely to a probabilistic classical computer that evaluates f (0) with probability one-half, or f (1) with probability one-half. The difference is that in a classical computer these two alternatives forever exclude one another; in a quantum computer it is 34 Introduction and overview possible for the two alternatives to interfere with one another to yield some global property of the function f, by using something like the Hadamard gate to recombine the different alternatives, as was done in Deutsch\u2019s algorithm. The essence of the design of many quantum algorithms is that a clever choice of function and \ufb01nal transformation allows ef\ufb01cient determination of useful global information about the function \u2013 information which cannot be attained quickly on a classical computer. 1.4.4 The Deutsch\u2013Jozsa algorithm Deutsch\u2019s algorithm is a simple case of a more general quantum algorithm, which we shall refer to as the Deutsch\u2013Jozsa algorithm. The application, known as Deutsch\u2019s problem, may be described as the following game. Alice, in Amsterdam, selects a number x from 0 to 2n \u2212 1, and mails it in a letter to Bob, in Boston. Bob calculates some function f (x) and replies with the result, which is either 0 or 1. Now, Bob has promised to use a function f which is of one of two kinds; either f (x) is constant for all values of x, or else f (x) is balanced, that is, equal to 1 for exactly half of all the possible x, and 0 for the other half. Alice\u2019s goal is to determine with certainty whether Bob has chosen a constant or a balanced function, corresponding with him as little as possible. How fast can she succeed? In the classical case, Alice may only send Bob one value of x in each letter. At worst, Alice will need to query Bob at least 2n/2 + 1 times, since she may receive 2n/2 0s before \ufb01nally getting a 1, telling her that Bob\u2019s function is balanced. The best deterministic classical algorithm she can use therefore requires 2n/2 + 1 queries. Note that in each letter, Alice sends Bob n bits of information. Furthermore, in this example, physical distance is being used to arti\ufb01cially elevate the cost of calculating f (x), but this is not needed in the general problem, where f (x) may be inherently dif\ufb01cult to calculate. If Bob and Alice were able to exchange qubits, instead of just classical bits, and if Bob agreed to calculate f (", "x) using a unitary transform Uf, then Alice could achieve her goal in just one correspondence with Bob, using the following algorithm. Analogously to Deutsch\u2019s algorithm, Alice has an n qubit register to store her query in, and a single qubit register which she will give to Bob, to store the answer in. She begins by preparing both her query and answer registers in a superposition state. Bob will evaluate f (x) using quantum parallelism and leave the result in the answer register. Alice then interferes states in the superposition using a Hadamard transform on the query register, and \ufb01nishes by performing a suitable measurement to determine whether f was constant or balanced. The speci\ufb01c steps of the algorithm are depicted in Figure 1.20. Let us follow the states through this circuit. The input state |\u03c80 = |0\u2297n|1 (1.46) is similar to that of Equation (1.41), but here the query register describes the state of n qubits all prepared in the |0 state. After the Hadamard transform on the query register and the Hadamard gate on the answer register we have |\u03c81 = x\u2208{0,1}n |x \u221a 2n |0 \u2212 |1 \u221a 2. (1.47) The query register is now a superposition of all values, and the answer register is in an Quantum algorithms 35 Figure 1.20. Quantum circuit implementing the general Deutsch\u2013Jozsa algorithm. The wire with a \u2018/\u2019 through it represents a set of n qubits, similar to the common engineering notation. evenly weighted superposition of 0 and 1. Next, the function f is evaluated (by Bob) using Uf : |x, y \u2192 |x, y \u2295 f (x), giving |\u03c82 = x (\u22121)f (x)|x \u221a 2n. |0 \u2212 |1 \u221a 2 (1.48) Alice now has a set of qubits in which the result of Bob\u2019s function evaluation is stored in the amplitude of the qubit superposition state. She now interferes terms in the superposition using a Hadamard transform on the query register. To determine the result of the Hadamard transform it helps to \ufb01rst calculate the effect of the Hadamard transform on a state |x. By checking the cases x = 0 and x = 1 separately we see that for a single qubit H|x = z(\u22121)xz|z/ \u221a 2. Thus H \u2297n|x1,..., xn = z1,...,zn(\u22121)x1z1+\u00b7\u00b7 +xnzn|z1,..., zn 2n \u221a This can be summarized more succinctly in the very useful equation H \u2297n|x = \u221a z(\u22121)x\u00b7z|z 2n, where x \u00b7 z is the bitwise inner product of x and z, modulo 2. Using this equation and (1.48) we can now evaluate |\u03c83, |\u03c83 = z x (\u22121)x\u00b7z+f (x)|z 2n |0 \u2212 |1 \u221a 2. (1.51) Alice now observes the query register. Note that the amplitude for the state |0\u2297n is x(\u22121)f (x)/2n. Let\u2019s look at the two possible cases \u2013 f constant and f balanced \u2013 to discern what happens. In the case where f is constant the amplitude for |0\u2297n is +1 or \u22121, depending on the constant value f (x) takes. Because |\u03c83 is of unit length it follows that all the other amplitudes must be zero, and an observation will yield 0s for all qubits in the query register. If f is balanced then the positive and negative contributions to the amplitude for |0\u2297n cancel, leaving an amplitude of zero, and a measurement must yield a result other than 0 on at least one qubit in the query register. Summarizing, if Alice. (1.49) (1.50) 36 Introduction and overview measures all 0s then the function is constant; otherwise the function is balanced. The Deutsch\u2013Jozsa algorithm is summarized below. Algorithm: Deutsch\u2013Jozsa Inputs: (1) A black box Uf which performs the transformation |x|y \u2192 |x|y \u2295 f (x), for x \u2208 {0,..., 2n \u2212 1} and f (x) \u2208 {0, 1}. It is promised that f (x) is either constant for all values of x, or else f (x) is balanced", ", that is, equal to 1 for exactly half of all the possible x, and 0 for the other half. Outputs: 0 if and only if f is constant. Runtime: One evaluation of Uf. Always succeeds. Procedure: |0\u2297n|1 1. 2. \u2192 1\u221a 2n 3. \u2192 x 4. \u2192 z x 5. \u2192 z 2n\u22121 x=0 (\u22121)f (x)|x |x |0 \u2212 |1 \u221a 2 |0 \u2212 |1 \u221a 2 (\u22121)x\u00b7z+f (x)|z \u221a 2n initialize state create superposition using Hadamard gates |0 \u2212 |1 \u221a 2 calculate function f using Uf perform Hadamard transform measure to obtain \ufb01nal output z We\u2019ve shown that a quantum computer can solve Deutsch\u2019s problem with one evaluation of the function f compared to the classical requirement for 2n/2 + 1 evaluations. This appears impressive, but there are several important caveats. First, Deutsch\u2019s problem is not an especially important problem; it has no known applications. Second, the comparison between classical and quantum algorithms is in some ways an apples and oranges comparison, as the method for evaluating the function is quite different in the two cases. Third, if Alice is allowed to use a probabilistic classical computer, then by asking Bob to evaluate f (x) for a few randomly chosen x she can very quickly determine with high probability whether f is constant or balanced. This probabilistic scenario is perhaps more realistic than the deterministic scenario we have been considering. Despite these caveats, the Deutsch\u2013Jozsa algorithm contains the seeds for more impressive quantum algorithms, and it is enlightening to attempt to understand the principles behind its operation. Exercise 1.1: (Probabilistic classical algorithm) Suppose that the problem is not to distinguish between the constant and balanced functions with certainty, but rather, with some probability of error < 1/2. What is the performance of the best classical algorithm for this problem? 1.4.5 Quantum algorithms summarized The Deutsch\u2013Jozsa algorithm suggests that quantum computers may be capable of solving some computational problems much more ef\ufb01ciently than classical computers. Unfortunately, the problem it solves is of little practical interest. Are there more interesting Quantum algorithms 37 problems whose solution may be obtained more ef\ufb01ciently using quantum algorithms? What are the principles underlying such algorithms? What are the ultimate limits of a quantum computer\u2019s computational power? Broadly speaking, there are three classes of quantum algorithms which provide an advantage over known classical algorithms. First, there is the class of algorithms based upon quantum versions of the Fourier transform, a tool which is also widely used in classical algorithms. The Deutsch\u2013Jozsa algorithm is an example of this type of algorithm, as are Shor\u2019s algorithms for factoring and discrete logarithm. The second class of algorithms is quantum search algorithms. The third class of algorithms is quantum simulation, whereby a quantum computer is used to simulate a quantum system. We now brie\ufb02y describe each of these classes of algorithms, and then summarize what is known or suspected about the computational power of quantum computers. Quantum algorithms based upon the Fourier transform The discrete Fourier transform is usually described as transforming a set x0,..., xN \u22121 of N complex numbers into a set of complex numbers y0,..., yN \u22121 de\ufb01ned by yk \u2261 1\u221a N N \u22121 j=0 e2\u03c0ijk/N xj. (1.52) Of course, this transformation has an enormous number of applications in many branches of science; the Fourier transformed version of a problem is often easier than the original problem, enabling a solution. The Fourier transform has proved so useful that a beautiful generalized theory of Fourier transforms has been developed which goes beyond the de\ufb01nition (1.52). This general theory involves some technical ideas from the character theory of \ufb01nite groups, and we will not attempt to describe it here. What is important is that the Hadamard transform used in the Deutsch\u2013Jozsa algorithm is an example of this generalized class of Fourier transforms. Moreover, many of the other important quantum algorithms also involve some type of Fourier transform. The most important quantum algorithms known, Shor\u2019s fast algorithms for factoring and discrete logarithm, are two examples of algorithms based upon the Fourier transform de\ufb01ned in Equation (1.52). The Equation (1.52) does not appear terribly quantum mechanical in the form we have written it. Imagine, however, that we de\ufb01ne a linear transformation U on n qubits by its action on computational basis states |j, where 0 \u2264 j \u2264 2n \u2212 1, |j", " \u2212\u2192 1\u221a 2n 2n\u22121 k=0 e2\u03c0ijk/2n |k. (1.53) It can be checked that this transformation is unitary, and in fact can be realized as a quantum circuit. Moreover, if we write out its action on superpositions, 2n\u22121 j=0 xj|j \u2212\u2192 1\u221a 2n 2n\u22121 2n\u22121 \u23a1 \u23a3 k=0 j=0 \u23a4 e2\u03c0ijk/2n xj \u23a6 |k = 2n\u22121 k=0 yk|k, (1.54) we see that it corresponds to a vector notation for the Fourier transform (1.52) for the case N = 2n. 38 Introduction and overview How quickly can we perform the Fourier transform? Classically, the fast Fourier transform takes roughly N log(N ) = n2n steps to Fourier transform N = 2n numbers. On a quantum computer, the Fourier transform can be accomplished using about log2(N ) = n2 steps, an exponential saving! The quantum circuit to do this is explained in Chapter 5. This result seems to indicate that quantum computers can be used to very quickly compute the Fourier transform of a vector of 2n complex numbers, which would be fantastically useful in a wide range of applications. However, that is not exactly the case; the Fourier transform is being performed on the information \u2018hidden\u2019 in the amplitudes of the quantum state. This information is not directly accessible to measurement. The catch, of course, is that if the output state is measured, it will collapse each qubit into the state |0 or |1, preventing us from learning the transform result yk directly. This example speaks to the heart of the conundrum of devising a quantum algorithm. On the one hand, we can perform certain calculations on the 2n amplitudes associated with n qubits far more ef\ufb01ciently than would be possible on a classical computer. But on the other hand, the results of such a calculation are not available to us if we go about it in a straightforward manner. More cleverness is required in order to harness the power of quantum computation. Fortunately, it does turn out to be possible to utilize the quantum Fourier transform to ef\ufb01ciently solve several problems that are believed to have no ef\ufb01cient solution on a classical computer. These problems include Deutsch\u2019s problem, and Shor\u2019s algorithms for discrete logarithm and factoring. This line of thought culminated in Kitaev\u2019s discovery of a method to solve the Abelian stabilizer problem, and the generalization to the hidden subgroup problem, Let f be a function from a \ufb01nitely generated group G to a \ufb01nite set X such that f is constant on the cosets of a subgroup K, and distinct on each coset. Given a quantum black box for performing the unitary transform U |g|h = |g|h\u2295f (g), for g \u2208 G, h \u2208 X, and \u2295 an appropriately chosen binary operation on X, \ufb01nd a generating set for K. The Deutsch\u2013Jozsa algorithm, Shor\u2019s algorithms, and related \u2018exponentially fast\u2019 quantum algorithms can all be viewed as special cases of this algorithm. The quantum Fourier transform and its applications are described in Chapter 5. Quantum search algorithms A completely different class of algorithms is represented by the quantum search algorithm, whose basic principles were discovered by Grover. The quantum search algorithm solves the following problem: Given a search space of size N, and no prior knowledge about the structure of the information in it, we want to \ufb01nd an element of that search space satisfying a known property. How long does it take to \ufb01nd an element satisfying that property? Classically, this problem requires approximately N operations, but the quantum search algorithm allows it to be solved using approximately N operations. \u221a The quantum search algorithm offers only a quadratic speedup, as opposed to the more impressive exponential speedup offered by algorithms based on the quantum Fourier transform. However, the quantum search algorithm is still of great interest, since searching heuristics have a wider range of application than the problems solved using the quantum Fourier transform, and adaptations of the quantum search algorithm may have utility Quantum algorithms 39 for a very wide range of problems. The quantum search algorithm and its applications are described in Chapter 6. Quantum simulation Simulating naturally occurring quantum mechanical systems is an obvious candidate for a task at which quantum computers may excel, yet which is believed to be dif\ufb01cult on a classical computer. Classical computers have dif\ufb01culty simulating general quantum systems for much the same reasons they have dif\ufffd", "\ufffd\ufffdculty simulating quantum computers \u2013 the number of complex numbers needed to describe a quantum system generally grows exponentially with the size of the system, rather than linearly, as occurs in classical systems. In general, storing the quantum state of a system with n distinct components takes something like cn bits of memory on a classical computer, where c is a constant which depends upon details of the system being simulated, and the desired accuracy of the simulation. By contrast, a quantum computer can perform the simulation using kn qubits, where k is again a constant which depends upon the details of the system being simulated. This allows quantum computers to ef\ufb01ciently perform simulations of quantum mechanical systems that are believed not to be ef\ufb01ciently simulatable on a classical computer. A signi\ufb01cant caveat is that even though a quantum computer can simulate many quantum systems far more ef\ufb01ciently than a classical computer, this does not mean that the fast simulation will allow the desired information about the quantum system to be obtained. When measured, a kn qubit simulation will collapse into a de\ufb01nite state, giving only kn bits of information; the cn bits of \u2018hidden information\u2019 in the wavefunction is not entirely accessible. Thus, a crucial step in making quantum simulations useful is development of systematic means by which desired answers can be ef\ufb01ciently extracted; how to do this is only partially understood. Despite this caveat, quantum simulation is likely to be an important application of quantum computers. The simulation of quantum systems is an important problem in many \ufb01elds, notably quantum chemistry, where the computational constraints imposed by classical computers make it dif\ufb01cult to accurately simulate the behavior of even moderately sized molecules, much less the very large molecules that occur in many important biological systems. Obtaining faster and more accurate simulations of such systems may therefore have the welcome effect of enabling advances in other \ufb01elds in which quantum phenomena are important. In the future we may discover a physical phenomenon in Nature which cannot be ef\ufb01ciently simulated on a quantum computer. Far from being bad news, this would be wonderful! At the least, it will stimulate us to extend our models of computation to encompass the new phenomenon, and increase the power of our computational models beyond the existing quantum computing model. It also seems likely that very interesting new physical effects will be associated with any such phenomenon! Another application for quantum simulation is as a general method to obtain insight into other quantum algorithms; for example, in Section 6.2 we explain how the quantum search algorithm can be viewed as the solution to a problem of quantum simulation. By approaching the problem in this fashion it becomes much easier to understand the origin of the quantum search algorithm. Finally, quantum simulation also gives rise to an interesting and optimistic \u2018quantum corollary\u2019 to Moore\u2019s law. Recall that Moore\u2019s law states that the power of classical 40 Introduction and overview computers will double once every two years or so, for constant cost. However, suppose we are simulating a quantum system on a classical computer, and want to add a single qubit (or a larger system) to the system being simulated. This doubles or more the memory requirements needed for a classical computer to store a description of the state of the quantum system, with a similar or greater cost in the time needed to simulate the dynamics. The quantum corollary to Moore\u2019s law follows from this observation, stating that quantum computers are keeping pace with classical computers provided a single qubit is added to the quantum computer every two years. This corollary should not be taken too seriously, as the exact nature of the gain, if any, of quantum computation over classical is not yet clear. Nevertheless, this heuristic statement helps convey why we should be interested in quantum computers, and hopeful that they will one day be able to outperform the most powerful classical computers, at least for some applications. The power of quantum computation How powerful are quantum computers? What gives them their power? Nobody yet knows the answers to these questions, despite the suspicions fostered by examples such as factoring, which strongly suggest that quantum computers are more powerful than classical computers. It is still possible that quantum computers are no more powerful than classical computers, in the sense that any problem which can be ef\ufb01ciently solved on a quantum computer can also be ef\ufb01ciently solved on a classical computer. On the other hand, it may eventually be proved that quantum computers are much more powerful than classical computers. We now take a brief look at what is known about the power of quantum computation. Computational complexity theory is the subject of classifying the dif\ufb01culty of various computational problems, both classical and quantum, and to understand the power of quantum computers we will \ufb01rst examine some general ideas from computational complexity. The most basic idea is that of a complexity", " class. A complexity class can be thought of as a collection of computational problems, all of which share some common feature with respect to the computational resources needed to solve those problems. Two of the most important complexity classes go by the names P and NP. Roughly speaking, P is the class of computational problems that can be solved quickly on a classical computer. NP is the class of problems which have solutions which can be quickly checked on a classical computer. To understand the distinction between P and NP, consider the problem of \ufb01nding the prime factors of an integer, n. So far as is known there is no fast way of solving this problem on a classical computer, which suggests that the problem is not in P. On the other hand, if somebody tells you that some number p is a factor of n, then we can quickly check that this is correct by dividing p into n, so factoring is a problem in NP. It is clear that P is a subset of NP, since the ability to solve a problem implies the ability to check potential solutions. What is not so clear is whether or not there are problems in NP that are not in P. Perhaps the most important unsolved problem in theoretical computer science is to determine whether these two classes are different:? = NP. P (1.55) Most researchers believe that NP contains problems that are not in P. In particular, there is an important subclass of the NP problems, the NP-complete problems, that are Quantum algorithms 41 of especial importance for two reasons. First, there are thousands of problems, many highly important, that are known to be NP-complete. Second, any given NP-complete problem is in some sense \u2018at least as hard\u2019 as all other problems in NP. More precisely, an algorithm to solve a speci\ufb01c NP-complete problem can be adapted to solve any other problem in NP, with a small overhead. In particular, if P = NP, then it will follow that no NP-complete problem can be ef\ufb01ciently solved on a classical computer. It is not known whether quantum computers can be used to quickly solve all the problems in NP, despite the fact that they can be used to solve some problems \u2013 like factoring \u2013 which are believed by many people to be in NP but not in P. (Note that factoring is not known to be NP-complete, otherwise we would already know how to ef\ufb01ciently solve all problems in NP using quantum computers.) It would certainly be very exciting if it were possible to solve all the problems in NP ef\ufb01ciently on a quantum computer. There is a very interesting negative result known in this direction which rules out using a simple variant of quantum parallelism to solve all the problems in NP. Speci\ufb01cally, one approach to the problem of solving problems in NP on a quantum computer is to try to use some form of quantum parallelism to search in parallel through all the possible solutions to the problem. In Section 6.6 we will show that no approach based upon such a search-based methodology can yield an ef\ufb01cient solution to all the problems in NP. While it is disappointing that this approach fails, it does not rule out that some deeper structure exists in the problems in NP that will allow them all to be solved quickly using a quantum computer. P and NP are just two of a plethora of complexity classes that have been de\ufb01ned. Another important complexity class is PSPACE. Roughly speaking, PSPACE consists of those problems which can be solved using resources which are few in spatial size (that is, the computer is \u2018small\u2019), but not necessarily in time (\u2018long\u2019 computations are \ufb01ne). PSPACE is believed to be strictly larger than both P and NP although, again, this has never been proved. Finally, the complexity class BPP is the class of problems that can be solved using randomized algorithms in polynomial time, if a bounded probability of error (say 1/4) is allowed in the solution to the problem. BPP is widely regarded as being, even more so than P, the class of problems which should be considered ef\ufb01ciently soluble on a classical computer. We have elected to concentrate here on P rather than BPP because P has been studied in more depth, however many similar ideas and conclusions arise in connection with BPP. What of quantum complexity classes? We can de\ufb01ne BQP to be the class of all computational problems which can be solved ef\ufb01ciently on a quantum computer, where a bounded probability of error is allowed. (Strictly speaking this makes BQP more analogous to the classical complexity class BPP than to P, however we will ignore this subtlety for the purposes of the present discussion, and treat it as the analogue of P.) Exactly where BQP \ufb01ts with respect to P", ", NP and PSPACE is as yet unknown. What is known is that quantum computers can solve all the problems in P ef\ufb01ciently, but that there are no problems outside of PSPACE which they can solve ef\ufb01ciently. Therefore, BQP lies somewhere between P and PSPACE, as illustrated in Figure 1.21. An important implication is that if it is proved that quantum computers are strictly more powerful than classical computers, then it will follow that P is not equal to PSPACE. Proving this latter result has been attempted without success by many computer scientists, suggesting that it may be non-trivial to prove that quantum computers are more powerful than classical computers, despite much evidence in favor of this proposition. 42 Introduction and overview Figure 1.21. The relationship between classical and quantum complexity classes. Quantum computers can quickly solve any problem in P, and it is known that they can\u2019t solve problems outside of PSPACE quickly. Where quantum computers \ufb01t between P and PSPACE is not known, in part because we don\u2019t even know whether PSPACE is bigger than P! We won\u2019t speculate further on the ultimate power of quantum computation now, preferring to wait until after we have better understood the principles on which fast quantum algorithms are based, a topic which occupies us for most of Part II of this book. What is already clear is that the theory of quantum computation poses interesting and signi\ufb01cant challenges to the traditional notions of computation. What makes this an important challenge is that the theoretical model of quantum computation is believed to be experimentally realizable, because \u2013 to the best of our knowledge \u2013 this theory is consistent with the way Nature works. If this were not so then quantum computation would be just another mathematical curiosity. 1.5 Experimental quantum information processing Quantum computation and quantum information is a wonderful theoretical discovery, but its central concepts, such as superpositions and entanglement, run counter to the intuition we garner from the everyday world around us. What evidence do we have that these ideas truly describe how Nature operates? Will the realization of large-scale quantum Experimental quantum information processing 43 computers be experimentally feasible? Or might there be some principle of physics which fundamentally prohibits their eventual scaling? In the next two sections we address these questions. We begin with a review of the famous \u2018Stern\u2013Gerlach\u2019 experiment, which provides evidence for the existence of qubits in Nature. We then widen our scope, addressing the broader problem of how to build practical quantum information processing systems. 1.5.1 The Stern\u2013Gerlach experiment The qubit is a fundamental element for quantum computation and quantum information. How do we know that systems with the properties of qubits exist in Nature? At the time of writing there is an enormous amount of evidence that this is so, but in the early days of quantum mechanics the qubit structure was not at all obvious, and people struggled with phenomena that we may now understand in terms of qubits, that is, in terms of two level quantum systems. A decisive (and very famous) early experiment indicating the qubit structure was conceived by Stern in 1921 and performed with Gerlach in 1922 in Frankfurt. In the original Stern\u2013Gerlach experiment, hot atoms were \u2018beamed\u2019 from an oven through a magnetic \ufb01eld which caused the atoms to be de\ufb02ected, and then the position of each atom was recorded, as illustrated in Figure 1.22. The original experiment was done with silver atoms, which have a complicated structure that obscures the effects we are discussing. What we describe below actually follows a 1927 experiment done using hydrogen atoms. The same basic effect is observed, but with hydrogen atoms the discussion is easier to follow. Keep in mind, though, that this privilege wasn\u2019t available to people in the early 1920s, and they had to be very ingenious to think up explanations for the more complicated effects they observed. Hydrogen atoms contain a proton and an orbiting electron. You can think of this electron as a little \u2018electric current\u2019 around the proton. This electric current causes the atom to have a magnetic \ufb01eld; each atom has what physicists call a \u2018magnetic dipole moment\u2019. As a result each atom behaves like a little bar magnet with an axis corresponding to the axis the electron is spinning around. Throwing little bar magnets through a magnetic \ufb01eld causes the magnets to be de\ufb02ected by the \ufb01eld, and we expect to see a similar de\ufb02ection of atoms in the Stern\u2013Gerlach experiment. How the atom is de\ufb02ected depends upon both the atom\u2019s magnetic dipole moment \u2013 the axis the electron is spinning around \u2013 and the magnetic \ufb01eld generated by the Stern\u2013 Gerlach device. We won\u2019t go through the details, but suf", "\ufb01ce to say that by constructing the Stern\u2013Gerlach device appropriately, we can cause the atom to be de\ufb02ected by an amount that depends upon the \u02c6z component of the atom\u2019s magnetic dipole moment, where \u02c6z is some \ufb01xed external axis. Two major surprises emerge when this experiment is performed. First, since the hot atoms exiting the oven would naturally be expected to have their dipoles oriented randomly in every direction, it would follow that there would be a continuous distribution of atoms seen at all angles exiting from the Stern\u2013Gerlach device. Instead, what is seen is atoms emerging from a discrete set of angles. Physicists were able to explain this by assuming that the magnetic dipole moment of the atoms is quantized, that is, comes in discrete multiples of some fundamental amount. This observation of quantization in the Stern\u2013Gerlach experiment was surprising to physicists of the 1920s, but not completely astonishing because evidence for quantization 44 Introduction and overview effects in other systems was becoming widespread at that time. What was truly surprising was the number of peaks seen in the experiment. The hydrogen atoms being used were such that they should have had zero magnetic dipole moment. Classically, this is surprising in itself, since it corresponds to no orbital motion of the electron, but based on what was known of quantum mechanics at that time this was an acceptable notion. Since the hydrogen atoms would therefore have zero magnetic moment, it was expected that only one beam of atoms would be seen, and this beam would not be de\ufb02ected by the magnetic \ufb01eld. Instead, two beams were seen, one de\ufb02ected up by the magnetic \ufb01eld, and the other de\ufb02ected down! This puzzling doubling was explained after considerable effort by positing that the electron in the hydrogen atom has associated with it a quantity called spin. This spin is not in any way associated to the usual rotational motion of the electron around the proton; it is an entirely new quantity to be associated with an electron. The great physicist Heisenberg labeled the idea \u2018brave\u2019 at the time it was suggested, and it is a brave idea, since it introduces an essentially new physical quantity into Nature. The spin of the electron is posited to make an extra contribution to the magnetic dipole moment of a hydrogen atom, in addition to the contribution due to the rotational motion of the electron. Figure 1.22. Abstract schematic of the Stern\u2013Gerlach experiment. Hot hydrogen atoms are beamed from an oven through a magnetic \ufb01eld, causing a de\ufb02ection either up (| + Z) or down (| \u2212 Z). What is the proper description of the spin of the electron? As a \ufb01rst guess, we might hypothesize that the spin is speci\ufb01ed by a single bit, telling the hydrogen atom to go up or down. Additional experimental results provide further useful information to determine if this guess needs re\ufb01nement or replacement. Let\u2019s represent the original Stern\u2013Gerlach apparatus as shown in Figure 1.22. Its outputs are two beams of atoms, which we shall call | + Z and | \u2212 Z. (We\u2019re using suggestive notation which looks quantum mechanical, but of course you\u2019re free to use whatever notation you prefer.) Now suppose we cascade two Stern\u2013Gerlach apparatus together, as shown in Figure 1.23. We arrange it so that the second apparatus is tipped sideways, so the magnetic \ufb01eld de\ufb02ects atoms along the \u02c6x axis. In our thought-experiment we\u2019ll block off the | \u2212 Z output from the \ufb01rst Stern\u2013Gerlach apparatus, while the | + Z output is sent through a second apparatus oriented along the \u02c6x axis. A detector is placed at the \ufb01nal output to measure the distribution of atoms along the \u02c6x axis. A classical magnetic dipole pointed in the + \u02c6z direction has no net magnetic moment in the \u02c6x direction, so we might expect that the \ufb01nal output would have one central peak. However, experimentally it is observed that there are two peaks of equal intensity! So perhaps these atoms are peculiar, and have de\ufb01nite magnetic moments along each axis, independently. That is, maybe each atom passing through the second apparatus can be Experimental quantum information processing 45 Figure 1.23. Cascaded Stern\u2013Gerlach measurements. described as being in a state we might write as | + Z| + X or | + Z| \u2212 X, to indicate the two values for spin that might be observed. Figure 1.24. Three stage cascaded Stern\u2013Gerlach measurements. Another experiment, shown in Figure 1.24,", " can test this hypothesis by sending one beam of the previous output through a second \u02c6z oriented Stern\u2013Gerlach apparatus. If the atoms had retained their | + Z orientation, then the output would be expected to have only one peak, at the | + Z output. However, again two beams are observed at the \ufb01nal output, of equal intensity. Thus, the conclusion would seem to be that contrary to classical expectations, a | + Z state consists of equal portions of | + X and | \u2212 X states, and a | + X state consists of equal portions of | + Z and | \u2212 Z states. Similar conclusions can be reached if the Stern\u2013Gerlach apparatus is aligned along some other axis, like the \u02c6y axis. The qubit model provides a simple explanation of this experimentally observed be- havior. Let |0 and |1 be the states of a qubit, and make the assignments | + Z \u2190 |0 | \u2212 Z \u2190 |1 | + X \u2190 (|0 + |1)/ | \u2212 X \u2190 (|0 \u2212 |1)/ 2. \u221a 2 \u221a (1.56) (1.57) (1.58) (1.59) \u221a Then the results of the cascaded Stern\u2013Gerlach experiment can be explained by assuming that the \u02c6z Stern\u2013Gerlach apparatus measures the spin (that is, the qubit) in the computational basis |0, |1, and the \u02c6x Stern\u2013Gerlach apparatus measures the spin with respect to 2. For example, in the cascaded \u02c6z- \u02c6x- \u02c6z experiment, the basis (|0 + |1)/ if we assume that the spins are in the state | + Z = |0 = (| + X + | \u2212 X)/ 2 after exiting the \ufb01rst Stern\u2013Gerlach experiment, then the probability for obtaining | + X out of the second apparatus is 1/2, and the probability for | \u2212 X is 1/2. Similarly, the probability for obtaining | + Z out of the third apparatus is 1/2. A qubit model thus properly predicts results from this type of cascaded Stern\u2013Gerlach experiment. 2, (|0 \u2212 |1)/ \u221a \u221a 46 Introduction and overview This example demonstrates how qubits could be a believable way of modeling systems in Nature. Of course it doesn\u2019t establish beyond all doubt that the qubit model is the correct way of understanding electron spin \u2013 far more experimental corroboration is required. Nevertheless, because of many experiments like these, we now believe that electron spin is best described by the qubit model. What is more, we believe that the qubit model (and generalizations of it to higher dimensions; quantum mechanics, in other words) is capable of describing every physical system. We now turn to the question of what systems are especially well adapted to quantum information processing. 1.5.2 Prospects for practical quantum information processing Building quantum information processing devices is a great challenge for scientists and engineers of the third millennium. Will we rise to meet this challenge? Is it possible at all? Is it worth attempting? If so, how might the feat be accomplished? These are dif\ufb01cult and important questions, to which we essay brief answers in this section, to be expanded upon throughout the book. The most fundamental question is whether there is any point of principle that prohibits us from doing one or more forms of quantum information processing? Two possible obstructions suggest themselves: that noise may place a fundamental barrier to useful quantum information processing; or that quantum mechanics may fail to be correct. Noise is without a doubt a signi\ufb01cant obstruction to the development of practical quantum information processing devices. Is it a fundamentally irremovable obstruction that will forever prevent the development of large-scale quantum information processing devices? The theory of quantum error-correcting codes strongly suggests that while quantum noise is a practical problem that needs to be addressed, it does not present a fundamental problem of principle. In particular, there is a threshold theorem for quantum computation, which states, roughly speaking, that provided the level of noise in a quantum computer can be reduced below a certain constant \u2018threshold\u2019 value, quantum error-correcting codes can be used to push it down even further, essentially ad in\ufb01nitum, for a small overhead in the complexity of the computation. The threshold theorem makes some broad assumptions about the nature and magnitude of the noise occurring in a quantum computer, and the architecture available for performing quantum computation; however, provided those assumptions are satis\ufb01ed, the effects of noise can be made essentially negligible for quantum information processing. Chapters 8, 10 and 12 discuss quantum noise, quantum error-correction and the threshold theorem in detail. A second possibility that may preclude quantum information processing is if quantum mechanics is incorrect.", " Indeed, probing the validity of quantum mechanics (both relativistic and non-relativistic) is one reason for being interested in building quantum information processing devices. Never before have we explored a regime of Nature in which complete control has been obtained over large-scale quantum systems, and perhaps Nature may reveal some new surprises in this regime which are not adequately explained by quantum mechanics. If this occurs, it will be a momentous discovery in the history of science, and can be expected to have considerable consequences in other areas of science and technology, as did the discovery of quantum mechanics. Such a discovery might also impact quantum computation and quantum information; however, whether the impact would enhance, detract or not affect the power of quantum information processing cannot be predicted in advance. Until and unless such effects are found we have no way of knowing how they might affect information processing, so for the remainder of this book Experimental quantum information processing 47 we go with all the evidence to date and assume that quantum mechanics is a complete and correct description of the world. Given that there is no fundamental obstacle to building quantum information processing devices, why should we invest enormous amounts of time and money in the attempt to do so? We have already discussed several reasons for wanting to do so: practical applications such as quantum cryptography and the factoring of large composite numbers; and the desire to obtain fundamental insights into Nature and into information processing. These are good reasons, and justify a considerable investment of time and money in the effort to build quantum information processing devices. However, it is fair to say that a clearer picture of the relative power of quantum and classical information processing is needed in order to assess their relative merits. To obtain such a picture requires further theoretical work on the foundations of quantum computation and quantum information. Of particular interest is a decisive answer to the question \u2018Are quantum computers more powerful than classical computers?\u2019 Even if the answer to such a question eludes us for the time being, it would be useful to have a clear path of interesting applications at varying levels of complexity to aid researchers aiming to experimentally realize quantum information processing. Historically, the advance of technology is often hastened by the use of short- to medium-term incentives as a stepping-stone to long-term goals. Consider that microprocessors were initially used as controllers for elevators and other simple devices, before graduating to be the fundamental component in personal computers (and then on to who-knows-what). Below we sketch out a path of short- to medium-term goals for people interested in achieving the long-term goal of large-scale quantum information processing. Surprisingly many small-scale applications of quantum computation and quantum information are known. Not all are as \ufb02ashy as cousins like the quantum factoring algorithm, but the relative ease of implementing small-scale applications makes them extremely important as medium-term goals in themselves. Quantum state tomography and quantum process tomography are two elementary processes whose perfection is of great importance to quantum computation and quantum information, as well as being of independent interest in their own right. Quantum state tomography is a method for determining the quantum state of a system. To do this, it has to overcome the \u2018hidden\u2019 nature of the quantum state \u2013 remember, the state can\u2019t be directly determined by a measurement \u2013 by performing repeated preparations of the same quantum state, which is then measured in different ways in order to build up a complete description of the quantum state. Quantum process tomography is a more ambitious (but closely related) procedure to completely characterize the dynamics of a quantum system. Quantum process tomography can, for example, be used to characterize the performance of an alleged quantum gate or quantum communications channel, or to determine the types and magnitudes of different noise processes in a system. Beside obvious applications to quantum computation and quantum information, quantum process tomography can be expected to have signi\ufb01cant applications as a diagnostic tool to aid in the evaluation and improvement of primitive operations in any \ufb01eld of science and technology where quantum effects are important. Quantum state tomography and quantum process tomography are described in more detail in Chapter 8. Various small-scale communications primitives are also of great interest. We have already mentioned quantum cryptography and quantum teleportation. The former is likely to be useful in practical applications involving the distribution of a small amount of key 48 Introduction and overview material that needs to be highly secure. The uses of quantum teleportation are perhaps more open to question. We will see in Chapter 12 that teleportation may be an extremely useful primitive for transmitting quantum states between distant nodes in a network, in the presence of noise. The idea is to focus one\u2019s efforts on distributing EPR pairs between the nodes that wish to communicate. The EPR pairs may be corrupted during communication, but special \u2018entanglement distillation\u2019 protocols can then be used to \u2018clean up\u2019 the EPR pairs, enabling them to be used to teleport quantum states from one location to another. In fact, procotols", " based upon entanglement distillation and teleportation offer performance superior to more conventional quantum error-correction techniques in enabling noise free communication of qubits. What of the medium-scale? A promising medium-scale application of quantum information processing is to the simulation of quantum systems. To simulate a quantum system containing even a few dozen \u2018qubits\u2019 (or the equivalent in terms of some other basic system) strains the resources of even the largest supercomputers. A simple calculation is instructive. Suppose we have a system containing 50 qubits. To describe the state of such a system requires 250 \u2248 1015 complex amplitudes. If the amplitudes are stored to 128 bits of precision, then it requires 256 bits or 32 bytes in order to store each amplitude, for a total of 32 \u00d7 1015 bytes of information, or about 32 thousand terabytes of information, well beyond the capacity of existing computers, and corresponding to about the storage capacity that might be expected to appear in supercomputers during the second decade of the twenty-\ufb01rst century, presuming that Moore\u2019s law continues on schedule. 90 qubits at the same level of precision requires 32 \u00d7 1027 bytes, which, even if implemented using single atoms to represent bits, would require kilograms (or more) of matter. How useful will quantum simulations be? It seems likely that conventional methods will still be used to determine elementary properties of materials, such as bond strengths and basic spectroscopic properties. However, once the basic properties are well understood, it seems likely that quantum simulation will be of great utility as a laboratory for the design and testing of properties of novel molecules. In a conventional laboratory setup, many different types of \u2018hardware\u2019 \u2013 chemicals, detectors, and so on \u2013 may be required to test a wide variety of possible designs for a molecule. On a quantum computer, these different types of hardware can all be simulated in software, which is likely to be much less expensive and much faster. Of course, \ufb01nal design and testing must be performed with real physical systems; however, quantum computers may enable a much larger range of potential designs to be explored and evaluated en route to a better \ufb01nal design. It is interesting to note that such ab initio calculations to aid in the design of new molecules have been attempted on classical computers; however, they have met with limited success due to the enormous computational resources needed to simulate quantum mechanics on a classical computer. Quantum computers should be able to do much better in the relatively near future. What of large-scale applications? Aside from scaling up applications like quantum simulation and quantum cryptography, relatively few large-scale applications are known: the factoring of large numbers, taking discrete logarithms, and quantum searching. Interest in the \ufb01rst two of these derives mainly from the negative effect they would have of limiting the viability of existing public key cryptographic systems. (They might also be of substantial practical interest to mathematicians interested in these problems simply for their own sake.) So it does not seem likely that factoring and discrete logarithm Experimental quantum information processing 49 will be all that important as applications for the long run. Quantum searching may be of tremendous use because of the wide utility of the search heuristic, and we discuss some possible applications in Chapter 6. What would really be superb are many more large-scale applications of quantum information processing. This is a great goal for the future! Given a path of potential applications for quantum information processing, how can it be achieved in real physical systems? At the small scale of a few qubits there are already several working proposals for quantum information processing devices. Perhaps the easiest to realize are based upon optical techniques, that is, electromagnetic radiation. Simple devices like mirrors and beamsplitters can be used to do elementary manipulations of photons. Interestingly, a major dif\ufb01culty has been producing single photons on demand; experimentalists have instead opted to use schemes which produce single photons \u2018every now and then\u2019, at random, and wait for such an event to occur. Quantum cryptography, superdense coding, and quantum teleportation have all been realized using such optical techniques. A major advantage of the optical techniques is that photons tend to be highly stable carriers of quantum mechanical information. A major disadvantage is that photons don\u2019t directly interact with one another. Instead, the interaction has to be mediated by something else, like an atom, which introduces additional noise and complications into the experiment. An effective interaction between two photons is set up, which essentially works in two steps: photon number one interacts with the atom, which in turn interacts with the second photon, causing an overall interaction between the two photons. An alternative scheme is based upon methods for trapping different types of atom: there is the ion trap, in which a small number of charged atoms are trapped in a con\ufb01ned space; and neutral atom traps, for trapping uncharged atoms in a con\ufb01ned space. Quantum information processing schemes", " based upon atom traps use the atoms to store qubits. Electromagnetic radiation also shows up in these schemes, but in a rather different way than in what we referred to as the \u2018optical\u2019 approach to quantum information processing. In these schemes, photons are used to manipulate the information stored in the atoms themselves, rather than as the place the information is stored. Single qubit quantum gates can be performed by applying appropriate pulses of electromagnetic radiation to individual atoms. Neighboring atoms can interact with one another via (for example) dipole forces that enable quantum gates to be accomplished. Moreover, the exact nature of the interaction between neighboring atoms can be modi\ufb01ed by applying appropriate pulses of electromagnetic radiation to the atoms, giving the experimentalist control over what gates are performed in the system. Finally, quantum measurement can be accomplished in these systems using the long established quantum jumps technique, which implements with superb accuracy the measurements in the computational basis used for quantum computation. Another class of quantum information processing schemes is based upon Nuclear Magnetic Resonance, often known by its initials, NMR. These schemes store quantum information in the nuclear spin of atoms in a molecule, and manipulate that information using electromagnetic radiation. Such schemes pose special dif\ufb01culties, because in NMR it is not possible to directly access individual nuclei. Instead, a huge number (typically around 1015) of essentially identical molecules are stored in solution. Electromagnetic pulses are applied to the sample, causing each molecule to respond in roughly the same way. You should think of each molecule as being an independent computer, and the sample as containing a huge number of computers all running in parallel (classically). 50 Introduction and overview NMR quantum information processing faces three special dif\ufb01culties that make it rather different from other quantum information processing schemes. First, the molecules are typically prepared by letting them equilibrate at room temperature, which is so much higher than typical spin \ufb02ip energies that the spins become nearly completely randomly oriented. This fact makes the initial state rather more \u2018noisy\u2019 than is desirable for quantum information processing. How this noise may be overcome is an interesting story that we tell in Chapter 7. A second problem is that the class of measurements that may be performed in NMR falls well short of the most general measurements we would like to perform in quantum information processing. Nevertheless, for many instances of quantum information processing the class of measurements allowed in NMR is suf\ufb01cient. Third, because molecules cannot be individually addressed in NMR you might ask how it is that individual qubits can be manipulated in an appropriate way. Fortunately, different nuclei in the molecule can have different properties that allow them to be individually addressed \u2013 or at least addressed at a suf\ufb01ciently \ufb01ne-grained scale to allow the operations essential for quantum computation. Many of the elements required to perform large-scale quantum information processing can be found in existing proposals: superb state preparation and quantum measurements can be performed on a small number of qubits in the ion trap; superb dynamics can be performed in small molecules using NMR; fabrication technology in solid state systems allows designs to be scaled up tremendously. A single system having all these elements would be a long way down the road to a dream quantum computer. Unfortunately, all these systems are very different, and we are many, many years from having large-scale quantum computers. However, we believe that the existence of all these properties in existing (albeit different) systems does bode well for the long-term existence of largescale quantum information processors. Furthermore, it suggests that there is a great deal of merit to pursuing hybrid designs which attempt to marry the best features of two or more existing technologies. For example, there is much work being done on trapping atoms inside electromagnetic cavities. This enables \ufb02exible manipulation of the atom inside the cavity via optical techniques, and makes possible real-time feedback control of single atoms in ways unavailable in conventional atom traps. To conclude, note that it is important not to assess quantum information processing as though it were just another technology for information processing. For example, it is tempting to dismiss quantum computation as yet another technological fad in the evolution of the computer that will pass in time, much as other fads have passed \u2013 for example, the \u2018bubble memories\u2019 widely touted as the next big thing in memory during the early 1980s. This is a mistake, since quantum computation is an abstract paradigm for information processing that may have many different implementations in technology. One can compare two different proposals for quantum computing as regards their technological merits \u2013 it makes sense to compare a \u2018good\u2019 proposal to a \u2018bad\u2019 proposal \u2013 however even a very poor proposal for a quantum computer is of a different qualitative nature from a superb design for a classical computer. 1.6 Quantum information The term \u2018quantum information\u2019 is used in two distinct ways in the \ufb01eld of quantum", " computation and quantum information. The \ufb01rst usage is as a broad catch-all for all manner of operations that might be interpreted as related to information processing Quantum information 51 using quantum mechanics. This use encompasses subjects such as quantum computation, quantum teleportation, the no-cloning theorem, and virtually all other topics in this book. The second use of \u2018quantum information\u2019 is much more specialized: it refers to the study of elementary quantum information processing tasks. It does not typically include, for example, quantum algorithm design, since the details of speci\ufb01c quantum algorithms are beyond the scope of \u2018elementary\u2019. To avoid confusion we will use the term \u2018quantum information theory\u2019 to refer to this more specialized \ufb01eld, in parallel with the widely used term \u2018(classical) information theory\u2019 to describe the corresponding classical \ufb01eld. Of course, the term \u2018quantum information theory\u2019 has a drawback of its own \u2013 it might be seen as implying that theoretical considerations are all that matter! Of course, this is not the case, and experimental demonstration of the elementary processes studied by quantum information theory is of great interest. The purpose of this section is to introduce the basic ideas of quantum information theory. Even with the restriction to elementary quantum information processing tasks, quantum information theory may look like a disordered zoo to the beginner, with many apparently unrelated subjects falling under the \u2018quantum information theory\u2019 rubric. In part, that\u2019s because the subject is still under development, and it\u2019s not yet clear how all the pieces \ufb01t together. However, we can identify a few fundamental goals uniting work on quantum information theory: (1) Identify elementary classes of static resources in quantum mechanics. An example is the qubit. Another example is the bit; classical physics arises as a special case of quantum physics, so it should not be surprising that elementary static resources appearing in classical information theory should also be of great relevance in quantum information theory. Yet another example of an elementary class of static resources is a Bell state shared between two distant parties. (2) Identify elementary classes of dynamical processes in quantum mechanics. A simple example is memory, the ability to store a quantum state over some period of time. Less trivial processes are quantum information transmission between two parties, Alice and Bob; copying (or trying to copy) a quantum state, and the process of protecting quantum information processing against the effects of noise. (3) Quantify resource tradeoffs incurred performing elementary dynamical processes. For example, what are the minimal resources required to reliably transfer quantum information between two parties using a noisy communications channel? Similar goals de\ufb01ne classical information theory; however, quantum information theory is broader in scope than classical information theory, for quantum information theory includes all the static and dynamic elements of classical information theory, as well as additional static and dynamic elements. The remainder of this section describes some examples of questions studied by quantum information theory, in each case emphasizing the fundamental static and dynamic elements under consideration, and the resource tradeoffs being considered. We begin with an example that will appear quite familiar to classical information theorists: the problem of sending classical information through a quantum channel. We then begin to branch out and explore some of the new static and dynamic processes present in quantum mechanics, such as quantum error-correction, the problem of distinguishing quantum states, and entanglement transformation. The chapter concludes with some re\ufb02ections on how the 52 Introduction and overview tools of quantum information theory can be applied elsewhere in quantum computation and quantum information. 1.6.1 Quantum information theory: example problems Classical information through quantum channels The fundamental results of classical information theory are Shannon\u2019s noiseless channel coding theorem and Shannon\u2019s noisy channel coding theorem. The noiseless channel coding theorem quanti\ufb01es how many bits are required to store information being emitted by a source of information, while the noisy channel coding theorem quanti\ufb01es how much information can be reliably transmitted through a noisy communications channel. What do we mean by an information source? De\ufb01ning this notion is a fundamental problem of classical and quantum information theory, one we\u2019ll re-examine several times. For now, let\u2019s go with a provisional de\ufb01nition: a classical information source is described by a set of probabilities pj, j = 1, 2,..., d. Each use of the source results in the \u2018letter\u2019 j being emitted, chosen at random with probability pj, independently for each use of the source. For instance, if the source were of English text, then the numbers j might correspond to letters of the alphabet and punctuation, with the probabilities pj giving the relative frequencies with which the different letters appear in regular English text. Although it is not true that the letters in English appear in an independent fashion, for our purposes", " it will be a good enough approximation. Regular English text includes a considerable amount of redundancy, and it is possible to exploit that redundancy to compress the text. For example, the letter \u2018e\u2019 occurs much more frequently in regular English text than does the letter \u2018z\u2019. A good scheme for compressing English text will therefore represent the letter \u2018e\u2019 using fewer bits of information than it uses to represent \u2018z\u2019. Shannon\u2019s noiseless channel coding theorem quanti\ufb01es exactly how well such a compression scheme can be made to work. More precisely, the noiseless channel coding theorem tells us that a classical source described by probabilities pj can be compressed so that on average each use of the source can be represented using H(pj) bits of information, where H(pj) \u2261 \u2212 j pj log(pj) is a function of the source probability distribution known as the Shannon entropy. Moreover, the noiseless channel coding theorem tells us that to attempt to represent the source using fewer bits than this will result in a high probability of error when the information is decompressed. (Shannon\u2019s noiseless channel coding theorem is discussed in much greater detail in Chapter 12.) Shannon\u2019s noiseless coding theorem provides a good example where the goals of information theory listed earlier are all met. Two static resources are identi\ufb01ed (goal number 1): the bit and the information source. A two-stage dynamic process is identi\ufb01ed (goal 2), compressing an information source, and then decompressing to recover the information source. Finally a quantitative criterion for determining the resources consumed (goal 3) by an optimal data compression scheme is found. Shannon\u2019s second major result, the noisy channel coding theorem, quanti\ufb01es the amount of information that can be reliably transmitted through a noisy channel. In particular, suppose we wish to transfer the information being produced by some information source to another location through a noisy channel. That location may be at another point in space, or at another point in time \u2013 the latter is the problem of storing information in the presence of noise. The idea in both instances is to encode the information being produced using error-correcting codes, so that any noise introduced by the channel can be corrected at the other end of the channel. The way error-correcting codes achieve this Quantum information 53 is by introducing enough redundancy into the information sent through the channel so that even after some of the information has been corrupted it is still possible to recover the original message. For example, suppose the noisy channel is for the transmission of single bits, and the noise in the channel is such that to achieve reliable transmission each bit produced by the source must be encoded using two bits before being sent through the channel. We say that such a channel has a capacity of half a bit, since each use of the channel can be used to reliably convey roughly half a bit of information. Shannon\u2019s noisy channel coding theorem provides a general procedure for calculating the capacity of an arbitrary noisy channel. Shannon\u2019s noisy channel coding theorem also achieves the three goals of information theory we stated earlier. Two types of static resources are involved (goal 1), the information source, and the bits being sent through the channel. Three dynamical processes are involved (goal 2). The primary process is the noise in the channel. To combat this noise we perform the dual processes of encoding and decoding the state in an error-correcting code. For a \ufb01xed noise model, Shannon\u2019s theorem tells us how much redundancy must be introduced by an optimal error-correction scheme if reliable information transmission is to be achieved (goal 3). For both the noiseless and noisy channel coding theorems Shannon restricted himself to storing the output from an information source in classical systems \u2013 bits and the like. A natural question for quantum information theory is what happens if the storage medium is changed so that classical information is transmitted using quantum states as the medium. For example, it may be that Alice wishes to compress some classical information produced by an information source, transmitting the compressed information to Bob, who then decompresses it. If the medium used to store the compressed information is a quantum state, then Shannon\u2019s noiseless channel coding theorem cannot be used to determine the optimal compression and decompression scheme. One might wonder, for example, if using qubits allows a better compression rate than is possible classically. We\u2019ll study this question in Chapter 12, and prove that, in fact, qubits do not allow any signi\ufb01cant saving in the amount of communication required to transmit information over a noiseless channel. Naturally, the next step is to investigate the problem of transmitting classical information through a noisy quantum channel. Ideally, what we\u2019d like is a result that quanti\ufb01es the capacity of such a channel for the transmission of information. Evaluating the capacity is", " a very tricky job for several reasons. Quantum mechanics gives us a huge variety of noise models, since it takes place in a continuous space, and it is not at all obvious how to adapt classical error-correction techniques to combat the noise. Might it be advantageous, for example, to encode the classical information using entangled states, which are then transmitted one piece at a time through the noisy channel? Or perhaps it will be advantageous to decode using entangled measurements? In Chapter 12 we\u2019ll prove the HSW (Holevo\u2013Schumacher\u2013Westmoreland) theorem, which provides a lower bound on the capacity of such a channel. Indeed, it is widely believed that the HSW theorem provides an exact evaluation of the capacity, although a complete proof of this is not yet known! What remains at issue is whether or not encoding using entangled states can be used to raise the capacity beyond the lower bound provided by the HSW theorem. All evidence to date suggests that this doesn\u2019t help raise the capacity, but it is still a fascinating open problem of quantum information theory to determine the truth or falsity of this conjecture. 54 Introduction and overview Quantum information through quantum channels Classical information is, of course, not the only static resource available in quantum mechanics. Quantum states themselves are a natural static resource, even more natural than classical information. Let\u2019s look at a different quantum analogue of Shannon\u2019s coding theorems, this time involving the compression and decompression of quantum states. To begin, we need to de\ufb01ne some quantum notion of an information source, analogous to the classical de\ufb01nition of an information source. As in the classical case, there are several different ways of doing this, but for the sake of de\ufb01niteness let\u2019s make the provisional de\ufb01nition that a quantum source is described by a set of probabilities pj and corresponding quantum states |\u03c8j. Each use of the source produces a state |\u03c8j with probability pj, with different uses of the source being independent of one another. Is it possible to compress the output from such a quantum mechanical source? Consider the case of a qubit source which outputs the state |0 with probability p and the state |1 with probability 1 \u2212 p. This is essentially the same as a classical source emitting single bits, either 0 with probability p, or 1 with probability 1 \u2212 p, so it is not surprising that similar techniques can be used to compress the source so that only H(p, 1 \u2212 p) qubits are required to store the compressed source, where H(\u00b7) is again the Shannon entropy function. \u221a What if the source had instead been producing the state |0 with probability p, and 2 with probability 1 \u2212 p? The standard techniques of classical the state (|0 + |1)/ data compression no longer apply, since in general it is not possible for us to distinguish the states |0 and (|0 + |1)/ 2. Might it still be possible to perform some type of compression operation? \u221a It turns out that a type of compression is still possible, even in this instance. What is interesting is that the compression may no longer be error-free, in the sense that the quantum states being produced by the source may be slightly distorted by the compression\u2013 decompression procedure. Nevertheless, we require that this distortion ought to become very small and ultimately negligible in the limit of large blocks of source output being compressed. To quantify the distortion we introduce a \ufb01delity measure for the compression scheme, which measures the average distortion introduced by the compression scheme. The idea of quantum data compression is that the compressed data should be recovered with very good \ufb01delity. Think of the \ufb01delity as being analogous to the probability of doing the decompression correctly \u2013 in the limit of large block lengths, it should tend towards the no error limit of 1. Schumacher\u2019s noiseless channel coding theorem quanti\ufb01es the resources required to do quantum data compression, with the restriction that it be possible to recover the source with \ufb01delity close to 1. In the case of a source producing orthogonal quantum states |\u03c8j with probabilities pj Schumacher\u2019s theorem reduces to telling us that the source may be compressed down to but not beyond the classical limit H(pj). However, in the more general case of non-orthogonal states being produced by the source, Schumacher\u2019s theorem tells us how much a quantum source may be compressed, and the answer is not the Shannon entropy H(pj)! Instead, a new entropic quantity, the von Neumann entropy, turns out to be the correct answer. In general, the von Neumann entropy agrees with the Shannon entropy if and only if the states |\u03c8j are orthogonal. Otherwise, the von Neumann entropy for the source pj, |\u03c8j", " is in general strictly smaller than the Shannon entropy H(pj). Thus, for example, a source producing the state |0 with probability p \u221a and (|0 + |1)/ H(p, 1 \u2212 p) qubits per use of the source! 2 with probability 1 \u2212 p can be reliably compressed using fewer than Quantum information 55 The basic intuition for this decrease in resources required can be understood quite easily. Suppose the source emitting states |0 with probability p and (|0 + |1)/ 2 with probability 1 \u2212 p is used a large number n times. Then by the law of large numbers, with high probability the source emits about np copies of |0 and n(1 \u2212 p) copies of (|0 + |1)/ \u221a \u221a 2. That is, it has the form \u2297n(1\u2212p) |0\u2297np |0 + |1 \u221a 2 up to re-ordering of the systems involved. Suppose we expand the product of |0 + |1 terms on the right hand side. Since n(1 \u2212 p) is large, we can again use the law of large numbers to deduce that the terms in the product will be roughly one-half |0s and onehalf |1s. That is, the |0 + |1 product can be well approximated by a superposition of states of the form (1.60), |0\u2297n(1\u2212p)/2|1\u2297n(1\u2212p)/2. (1.61) Thus the state emitted by the source can be approximated as a superposition of terms of the form |0\u2297n(1+p)/2|1\u2297n(1\u2212p)/2. (1.62) How many states of this form are there? Roughly n choose n(1 + p)/2, which by Stirling\u2019s approximation is equal to N \u2261 2nH[(1+p)/2,(1\u2212p)/2]. A simple compression method then is to label all states of the form (1.62) |c1 through |cN. It is possible to perform a unitary transform on the n qubits emitted from the source that takes |cj to |j|0\u2297n\u2212nH[(1+p)/2,(1\u2212p)/2], since j is an nH[(1 + p)/2, (1 \u2212 p)/2] bit number. The compression operation is to perform this unitary transformation, and then drop the \ufb01nal n\u2212nH[(1+p)/2, (1\u2212p)/2] qubits, leaving a compressed state of nH[(1+p)/2, (1\u2212p)/2] qubits. To decompress we append the state |0\u2297n\u2212nH[(1+p)/2,(1\u2212p)/2] to the compressed state, and perform the inverse unitary transformation. This procedure for quantum data compression and decompression results in a storage requirement of H[(1 + p)/2, (1 \u2212 p)/2] qubits per use of the source, which whenever p \u2265 1/3 is an improvement over the H(p, 1 \u2212 p) qubits we might naively have expected from Shannon\u2019s noiseless channel coding theorem. In fact, Schumacher\u2019s noiseless channel coding theorem allows us to do somewhat better even than this, as we will see in Chapter 12; however, the essential reason in that construction is the same as the reason we were able to compress here: we exploited the fact that |0 and (|0 + |1)/ 2 are not orthogonal. Intuitively, the states contain some redundancy since both have a component in the |0 direction, which results in more physical similarity than would be obtained from orthogonal states. It is this redundancy that we have exploited in the coding scheme just described, and which is used in the full proof of Schumacher\u2019s noiseless channel coding theorem. Note that the restriction p \u2265 1/3 arises because when p < 1/3 this particular scheme doesn\u2019t exploit the redundancy in the states: we end up effectively increasing the redundancy present in the problem! Of course, this is an artifact of the particular scheme we have chosen, and the general solution exploits the redundancy in a much more sensible way to achieve data compression. \u221a Schumacher\u2019s noiseless channel coding theorem is an analogue of Shannon\u2019s noiseless 56 Introduction and overview channel coding theorem for the compression and decompression of quantum states. Can we \ufb01nd an analogue of Shannon\u2019s noisy channel coding theorem? Considerable progress on this important question has been made, using the theory of quantum error-correcting codes; however,", " a fully satisfactory analogue has not yet been found. We review some of what is known about the quantum channel capacity in Chapter 12. Quantum distinguishability Thus far all the dynamical processes we have considered \u2013 compression, decompression, noise, encoding and decoding error-correcting codes \u2013 arise in both classical and quantum information theory. However, the introduction of new types of information, such as quantum states, enlarges the class of dynamical processes beyond those considered in classical information theory. A good example is the problem of distinguishing quantum states. Classically, we are used to being able to distinguish different items of information, at least in principle. In practice, of course, a smudged letter \u2018a\u2019 written on a page may be very dif\ufb01cult to distinguish from a letter \u2018o\u2019, but in principle it is possible to distinguish between the two possibilities with perfect certainty. \u221a On the other hand, quantum mechanically it is not always possible to distinguish between arbitrary states. For example, there is no process allowed by quantum mechanics that will reliably distinguish between the states |0 and (|0 + |1)/ 2. Proving this rigorously requires tools we don\u2019t presently have available (it is done in Chapter 2), but by considering examples it\u2019s pretty easy to convince oneself that it is not possible. Suppose, for example, that we try to distinguish the two states by measuring in the computational basis. Then, if we have been given the state |0, the measurement will yield 0 with probability 1. However, when we measure (|0 + |1)/ 2 the measurement yields 0 with probability 1/2 and 1 with probability 1/2. Thus, while a measurement result of 1 implies that state must have been (|0 + |1)/ 2, since it couldn\u2019t have been |0, we can\u2019t infer anything about the identity of the quantum state from a measurement result of 0. \u221a \u221a This indistinguishability of non-orthogonal quantum states is at the heart of quantum computation and quantum information. It is the essence of our assertion that a quantum state contains hidden information that is not accessible to measurement, and thus plays a key role in quantum algorithms and quantum cryptography. One of the central problems of quantum information theory is to develop measures quantifying how well non-orthogonal quantum states may be distinguished, and much of Chapters 9 and 12 is concerned with this goal. In this introduction we\u2019ll limit ourselves to pointing out two interesting aspects of indistinguishability \u2013 a connection with the possibility of fasterthan-light communication, and an application to \u2018quantum money.\u2019 \u221a Imagine for a moment that we could distinguish between arbitrary quantum states. We\u2019ll show that this implies the ability to communicate faster than light, using entanglement. Suppose Alice and Bob share an entangled pair of qubits in the state (|00 + |11)/ 2. Then, if Alice measures in the computational basis, the post-measurement states will be |00 with probability 1/2, and |11 with probability 1/2. Thus Bob\u2019s system is either in the state |0, with probability 1/2, or in the state |1, with probability 1/2. Suppose, however, that Alice had instead measured in the |+, |\u2212 basis. Recall that |0 = (|+ + |\u2212)/ 2. A little algebra shows that the initial state of Alice and Bob\u2019s system may be rewritten as (| + + + | \u2212 \u2212)/ 2. Therefore, if Alice measures in the |+, |\u2212 basis, the state of Bob\u2019s system after the measurement 2 and |1 = (|+ \u2212 |\u2212)/ \u221a \u221a \u221a Quantum information 57 will be |+ or |\u2212 with probability 1/2 each. So far, this is all basic quantum mechanics. But if Bob had access to a device that could distinguish the four states |0, |1, |+, |\u2212 from one another, then he could tell whether Alice had measured in the computational basis, or in the |+, |\u2212 basis. Moreover, he could get that information instantaneously, as soon as Alice had made the measurement, providing a means by which Alice and Bob could achieve faster-than-light communication! Of course, we know that it is not possible to distinguish non-orthogonal quantum states; this example shows that this restriction is also intimately tied to other physical properties which we expect the world to obey. \u221a The indistinguishability of non-orthogonal quantum states need not always be a handicap. Sometimes it can be a boon. Imagine that a bank produces banknotes imprinted with a (classical) serial number, and a sequence of qubits each in either the state |0 or (|0 + |1)/ 2. Nobody but the bank knows what sequence of these two states is embedded in the", " note, and the bank maintains a list matching serial numbers to embedded states. The note is impossible to counterfeit exactly, because it is impossible for a would-be counterfeiter to determine with certainty the state of the qubits in the original note, without destroying them. When presented with the banknote a merchant (of certi\ufb01able repute) can verify that it is not a counterfeit by calling the bank, telling them the serial number, and then asking what sequence of states were embedded in the note. They can then check that the note is genuine by measuring the qubits in the |0, |1 or (|0 + |1)/ 2 basis, as directed by the bank. With probability which increases exponentially to one with the number of qubits checked, any would-be counterfeiter will be detected at this stage! This idea is the basis for numerous other quantum cryptographic protocols, and demonstrates the utility of the indistinguishability of non-orthogonal quantum states. 2, (|0 \u2212 |1)/ \u221a \u221a Exercise 1.2: Explain how a device which, upon input of one of two non-orthogonal quantum states |\u03c8 or |\u03d5 correctly identi\ufb01ed the state, could be used to build a device which cloned the states |\u03c8 and |\u03d5, in violation of the no-cloning theorem. Conversely, explain how a device for cloning could be used to distinguish non-orthogonal quantum states. Creation and transformation of entanglement Entanglement is another elementary static resource of quantum mechanics. Its properties are amazingly different from those of the resources most familiar from classical information theory, and they are not yet well understood; we have at best an incomplete collage of results related to entanglement. We don\u2019t yet have all the language needed to understand the solutions, but let\u2019s at least look at two information-theoretic problems related to entanglement. Creating entanglement is a simple dynamical process of interest in quantum information theory. How many qubits must two parties exchange if they are to create a particular entangled state shared between them, given that they share no prior entanglement? A second dynamical process of interest is transforming entanglement from one form into another. Suppose, for example, that Alice and Bob share between them a Bell state, and wish to transform it into some other type of entangled state. What resources do they need to accomplish this task? Can they do it without communicating? With classical communication only? If quantum communication is required then how much quantum communication is required? 58 Introduction and overview Answering these and more complex questions about the creation and transformation of entanglement forms a fascinating area of study in its own right, and also promises to give insight into tasks such as quantum computation. For example, a distributed quantum computation may be viewed as simply a method for generating entanglement between two or more parties; lower bounds on the amount of communication that must be done to perform such a distributed quantum computation then follow from lower bounds on the amount of communication that must be performed to create appropriate entangled states. 1.6.2 Quantum information in a wider context We have given but the barest glimpse of quantum information theory. Part III of this book discusses quantum information theory in much greater detail, especially Chapter 11, which deals with fundamental properties of entropy in quantum and classical information theory, and Chapter 12, which focuses on pure quantum information theory. Quantum information theory is the most abstract part of quantum computation and quantum information, yet in some sense it is also the most fundamental. The question driving quantum information theory, and ultimately all of quantum computation and quantum information, is what makes quantum information processing tick? What is it that separates the quantum and the classical world? What resources, unavailable in a classical world, are being utilized in a quantum computation? Existing answers to these questions are foggy and incomplete; it is our hope that the fog may yet lift in the years to come, and we will obtain a clear appreciation for the possibilities and limitations of quantum information processing. Problem 1.1: (Feynman-Gates conversation) Construct a friendly imaginary discussion of about 2000 words between Bill Gates and Richard Feynman, set in the present, on the future of computation. (Comment: You might like to try waiting until you\u2019ve read the rest of the book before attempting this question. See the \u2018History and further reading\u2019 below for pointers to one possible answer for this question.) Problem 1.2: What is the most signi\ufb01cant discovery yet made in quantum computation and quantum information? Write an essay of about 2000 words for an educated lay audience about the discovery. (Comment: As for the previous problem, you might like to try waiting until you\u2019ve read the rest of the book before attempting this question.) History and further reading Most of the material in this chapter is revisited in more depth in later chapters. Therefore the historical references", " and further reading below are limited to material which does not recur in later chapters. Piecing together the historical context in which quantum computation and quantum information have developed requires a broad overview of the history of many \ufb01elds. We have tried to tie this history together in this chapter, but inevitably much background material was omitted due to limited space and expertise. The following recommendations attempt to redress this omission. History and further reading 59 The history of quantum mechanics has been told in many places. We recommend especially the outstanding works of Pais[Pai82, Pai86, Pai91]. Of these three, [Pai86] is most directly concerned with the development of quantum mechanics; however, Pais\u2019 biographies of Einstein[Pai82] and of Bohr[Pai91] also contain much material of interest, at a less intense level. The rise of technologies based upon quantum mechanics has been described by Milburn[Mil97, Mil98]. Turing\u2019s marvelous paper on the foundations of computer science[Tur36] is well worth reading. It can be found in the valuable historical collection of Davis[Dav65]. Hofstadter[Hof79] and Penrose[Pen89] contain entertaining and informative discussions of the foundations of computer science. Shasha and Lazere\u2019s biography of \ufb01fteen leading computer scientists[SL98] gives considerable insight into many different facets of the history of computer science. Finally, Knuth\u2019s awesome series of books[Knu97, Knu98a, Knu98b] contain an amazing amount of historical information. Shannon\u2019s brilliant papers founding information theory make excellent reading[Sha48] (also reprinted in [SW49]). MacWilliams and Sloane[MS77] is not only an excellent text on error-correcting codes, but also contains an enormous amount of useful historical information. Similarly, Cover and Thomas[CT91] is an excellent text on information theory, with extensive historical information. Shannon\u2019s collected works, together with many useful historical items have been collected in a large volume[SW93] edited by Sloane and Wyner. Slepian has also collected a useful set of reprints on information theory[Sle74]. Cryptography is an ancient art with an intricate and often interesting history. Kahn[Kah96] is a huge history of cryptography containing a wealth of information. For more recent developments we recommend the books by Menezes, van Oorschot, and Vanstone[MvOV96], Schneier[Sch96a], and by Dif\ufb01e and Landau[DL98]. Quantum teleportation was discovered by Bennett, Brassard, Cr\u00b4epeau, Jozsa, Peres, and Wootters[BBC+93], and later experimentally realized in various different forms by Boschi, Branca, De Martini, Hardy and Popescu[BBM+98] using optical techniques, by Bouwmeester, Pan, Mattle, Eibl, Weinfurter, and Zeilinger[BPM+97] using photon polarization, by Furusawa, S\u00f8rensen, Braunstein, Fuchs, Kimble, and Polzik using \u2018squeezed\u2019 states of light[FSB+98], and by Nielsen, Knill, and La\ufb02amme using NMR[NKL98]. Deutsch\u2019s problem was posed by Deutsch[Deu85], and a one-bit solution was given in the same paper. The extension to the general n-bit case was given by Deutsch and Jozsa[DJ92]. The algorithms in these early papers have been substantially improved subsequently by Cleve, Ekert, Macchiavello, and Mosca[CEMM98], and independently in unpublished work by Tapp. In this chapter we have given the improved version of the algorithm, which \ufb01ts very nicely into the hidden subgroup problem framework that will later be discussed in Chapter 5. The original algorithm of Deutsch only worked probabilistically; Deutsch and Jozsa improved this to obtain a deterministic algorithm, but their method required two function evaluations, in contrast to the improved algorithms presented in this chapter. Nevertheless, it is still conventional to refer to these algorithms as Deutsch\u2019s algorithm and the Deutsch\u2013Jozsa algorithm in honor of two huge leaps forward: the concrete demonstration by Deutsch that a quantum computer could do something faster than a classical computer; and the extension by Deutsch and Jozsa which demonstrated for the \ufb01rst time a similar gap for the scaling of the time required to solve a problem. Excellent discussions of the Stern\u2013Gerlach experiment can be found in standard quantum mechanics textbooks such as the texts by Sakurai[Sak95], Volume III of Feynman, Leighton and Sands[FLS65a], and Cohen-", "Tannoudji, Diu and Lalo\u00a8e[CTDL77a, CTDL77b]. Problem 1.1 was suggested by the lovely article of Rahim[Rah99]. 2 Introduction to quantum mechanics I ain\u2019t no physicist but I know what matters. \u2013 Popeye the Sailor Quantum mechanics: Real Black Magic Calculus \u2013 Albert Einstein Quantum mechanics is the most accurate and complete description of the world known. It is also the basis for an understanding of quantum computation and quantum information. This chapter provides all the necessary background knowledge of quantum mechanics needed for a thorough grasp of quantum computation and quantum information. No prior knowledge of quantum mechanics is assumed. Quantum mechanics is easy to learn, despite its reputation as a dif\ufb01cult subject. The reputation comes from the dif\ufb01culty of some applications, like understanding the structure of complicated molecules, which aren\u2019t fundamental to a grasp of the subject; we won\u2019t be discussing such applications. The only prerequisite for understanding is some familiarity with elementary linear algebra. Provided you have this background you can begin working out simple problems in a few hours, even with no prior knowledge of the subject. Readers already familiar with quantum mechanics can quickly skim through this chapter, to become familiar with our (mostly standard) notational conventions, and to assure themselves of familiarity with all the material. Readers with little or no prior knowledge should work through the chapter in detail, pausing to attempt the exercises. If you have dif\ufb01culty with an exercise, move on, and return later to make another attempt. The chapter begins with a review of some material from linear algebra in Section 2.1. This section assumes familiarity with elementary linear algebra, but introduces the notation used by physicists to describe quantum mechanics, which is different to that used in most introductions to linear algebra. Section 2.2 describes the basic postulates of quantum mechanics. Upon completion of the section, you will have understood all of the fundamental principles of quantum mechanics. This section contains numerous simple exercises designed to help consolidate your grasp of this material. The remaining sections of the chapter, and of this book, elucidate upon this material, without introducing fundamentally new physical principles. Section 2.3 explains superdense coding, a surprising and illuminating example of quantum information processing which combines many of the postulates of quantum mechanics in a simple setting. Sections 2.4 and 2.5 develop powerful mathematical tools \u2013 the density operator, puri\ufb01cations, and the Schmidt decomposition \u2013 which are especially useful in the study of quantum computation and quantum information. Understanding these tools will also help you consolidate your understanding of elementary quantum mechanics. Finally, Section 2.6 examines the question of how quantum mechanics goes beyond the usual \u2018classical\u2019 understanding of the way the world works. Linear algebra 61 2.1 Linear algebra This book is written as much to disturb and annoy as to instruct. \u2013 The \ufb01rst line of About Vectors, by Banesh Hoffmann. Life is complex \u2013 it has both real and imaginary parts. \u2013 Anonymous Linear algebra is the study of vector spaces and of linear operations on those vector spaces. A good understanding of quantum mechanics is based upon a solid grasp of elementary linear algebra. In this section we review some basic concepts from linear algebra, and describe the standard notations which are used for these concepts in the study of quantum mechanics. These notations are summarized in Figure 2.1 on page 62, with the quantum notation in the left column, and the linear-algebraic description in the right column. You may like to glance at the table, and see how many of the concepts in the right column you recognize. In our opinion the chief obstacle to assimilation of the postulates of quantum mechanics is not the postulates themselves, but rather the large body of linear algebraic notions required to understand them. Coupled with the unusual Dirac notation adopted by physicists for quantum mechanics, it can appear (falsely) quite fearsome. For these reasons, we advise the reader not familiar with quantum mechanics to quickly read through the material which follows, pausing mainly to concentrate on understanding the absolute basics of the notation being used. Then proceed to a careful study of the main topic of the chapter \u2013 the postulates of quantum mechanics \u2013 returning to study the necessary linear algebraic notions and notations in more depth, as required. The basic objects of linear algebra are vector spaces. The vector space of most interest to us is Cn, the space of all n-tuples of complex numbers, (z1,..., zn). The elements of a vector space are called vectors, and we will sometimes use the column matrix notation \u23a4 \u23a5 \u23a6 \u23a1 \u23a2 \u23a3 z1... zn (2.1) to indicate a vector. There is an addition operation de\ufb01ned which takes pairs of vectors to other vectors. In Cn the addition operation for", " vectors is de\ufb01ned by \u23a1 \u23a4 \u23a1 \u23a1 \u23a4 \u23a4 \u23a2 \u23a3 \u23a5 \u23a6 + \u23a2 \u23a3 \u23a5 \u23a6 \u2261 \u23a2 \u23a3 \u23a5 \u23a6, (2.2) z1... zn z 1... z n z1 + z 1... zn + z n where the addition operations on the right are just ordinary additions of complex numbers. Furthermore, in a vector space there is a multiplication by a scalar operation. In Cn this operation is de\ufb01ned by \u23a1 \u23a2 \u23a3 z z1... zn \u23a4 \u23a1 \u23a5 \u23a6 \u2261 \u23a2 \u23a3 \u23a4 \u23a5 \u23a6, zz1... zzn (2.3) 62 Introduction to quantum mechanics where z is a scalar, that is, a complex number, and the multiplications on the right are ordinary multiplication of complex numbers. Physicists sometimes refer to complex numbers as c-numbers. Quantum mechanics is our main motivation for studying linear algebra, so we will use the standard notation of quantum mechanics for linear algebraic concepts. The standard quantum mechanical notation for a vector in a vector space is the following: |\u03c8. (2.4) \u03c8 is a label for the vector (any label is valid, although we prefer to use simple labels like \u03c8 and \u03d5). The |\u00b7 notation is used to indicate that the object is a vector. The entire object |\u03c8 is sometimes called a ket, although we won\u2019t use that terminology often. A vector space also contains a special zero vector, which we denote by 0. It satis\ufb01es the property that for any other vector |v, |v + 0 = |v. Note that we do not use the ket notation for the zero vector \u2013 it is the only exception we shall make. The reason for making the exception is because it is conventional to use the \u2018obvious\u2019 notation for the zero vector, |0, to mean something else entirely. The scalar multiplication operation is such that z0 = 0 for any complex number z. For convenience, we use the notation (z1,..., zn) to denote a column matrix with entries z1,..., zn. In Cn the zero element is (0, 0,..., 0). A vector subspace of a vector space V is a subset W of V such that W is also a vector space, that is, W must be closed under scalar multiplication and addition. Notation Description z\u2217 |\u03c8 \u03c8| \u03d5|\u03c8 Complex conjugate of the complex number z. (1 + i)\u2217 = 1 \u2212 i Vector. Also known as a ket. Vector dual to |\u03c8. Also known as a bra. Inner product between the vectors |\u03d5 and |\u03c8. |\u03d5 \u2297 |\u03c8 Tensor product of |\u03d5 and |\u03c8. |\u03d5|\u03c8 A\u2217 AT A\u2020 \u03d5|A|\u03c8 Abbreviated notation for tensor product of |\u03d5 and |\u03c8. Complex conjugate of the A matrix. Transpose of the A matrix. Hermitian conjugate or adjoint of the A matrix, A\u2020 = (AT )\u2217. \u2020 = a b c d a\u2217 b\u2217 Inner product between |\u03d5 and A|\u03c8. Equivalently, inner product between A\u2020|\u03d5 and |\u03c8. c\u2217 d\u2217. Figure 2.1. Summary of some standard quantum mechanical notation for notions from linear algebra. This style of notation is known as the Dirac notation. 2.1.1 Bases and linear independence A spanning set for a vector space is a set of vectors |v1,..., |vn such that any vector i ai|vi of vectors |v in the vector space can be written as a linear combination |v = Linear algebra 63 in that set. For example, a spanning set for the vector space C2 is the set |v1 \u2261 since any vector 1 0 ; |v2 \u2261 0 1 |v = a1 a2, (2.5) (2.6) in C2 can be written as a linear combination |v = a1|v1 + a2|v2 of the vectors |v1 and |v2. We say that the vectors |v1 and |v2 span the vector space C2. Generally, a vector space may have many different spanning sets. A second spanning set for the vector space C2 is the set |v1 \u2261 1\u221a 2 ; 1 1 |v2 \u2261 1\u221a", " 2, 1 \u22121 (2.7) since an arbitrary vector |v = (a1, a2) can be written as a linear combination of |v1 and |v2, |v = a1 + a2\u221a 2 |v1 + a1 \u2212 a2\u221a 2 |v2. (2.8) A set of non-zero vectors |v1,..., |vn are linearly dependent if there exists a set of complex numbers a1,..., an with ai = 0 for at least one value of i, such that a1|v1 + a2|v2 + \u00b7 \u00b7 \u00b7 + an|vn = 0. (2.9) A set of vectors is linearly independent if it is not linearly dependent. It can be shown that any two sets of linearly independent vectors which span a vector space V contain the same number of elements. We call such a set a basis for V. Furthermore, such a basis set always exists. The number of elements in the basis is de\ufb01ned to be the dimension of V. In this book we will only be interested in \ufb01nite dimensional vector spaces. There are many interesting and often dif\ufb01cult questions associated with in\ufb01nite dimensional vector spaces. We won\u2019t need to worry about these questions. Exercise 2.1: (Linear dependence: example) Show that (1, \u22121), (1, 2) and (2, 1) are linearly dependent. 2.1.2 Linear operators and matrices A linear operator between vector spaces V and W is de\ufb01ned to be any function A : V \u2192 W which is linear in its inputs, A i ai|vi = |vi. aiA i (2.10) Usually we just write A|v to denote A(|v). When we say that a linear operator A is de\ufb01ned on a vector space, V, we mean that A is a linear operator from V to V. An important linear operator on any vector space V is the identity operator, IV, de\ufb01ned by the equation IV |v \u2261 |v for all vectors |v. Where no chance of confusion arises we drop the subscript V and just write I to denote the identity operator. Another important linear operator is the zero operator, which we denote 0. The zero operator maps all vectors to 64 Introduction to quantum mechanics the zero vector, 0|v \u2261 0. It is clear from (2.10) that once the action of a linear operator A on a basis is speci\ufb01ed, the action of A is completely determined on all inputs. Suppose V, W, and X are vector spaces, and A : V \u2192 W and B : W \u2192 X are linear operators. Then we use the notation BA to denote the composition of B with A, de\ufb01ned by (BA)(|v) \u2261 B(A(|v)). Once again, we write BA|v as an abbreviation for (BA)(|v). The most convenient way to understand linear operators is in terms of their matrix representations. In fact, the linear operator and matrix viewpoints turn out to be completely equivalent. The matrix viewpoint may be more familiar to you, however. To see the connection, it helps to \ufb01rst understand that an m by n complex matrix A with entries Aij is in fact a linear operator sending vectors in the vector space Cn to the vector space Cm, under matrix multiplication of the matrix A by a vector in Cn. More precisely, the claim that the matrix A is a linear operator just means that A i ai|vi = i aiA|vi (2.11) is true as an equation where the operation is matrix multiplication of A by column vectors. Clearly, this is true! We\u2019ve seen that matrices can be regarded as linear operators. Can linear operators be given a matrix representation? In fact they can, as we now explain. This equivalence between the two viewpoints justi\ufb01es our interchanging terms from matrix theory and operator theory throughout the book. Suppose A : V \u2192 W is a linear operator between vector spaces V and W. Suppose |v1,..., |vm is a basis for V and |w1,..., |wn is a basis for W. Then for each j in the range 1,..., m, there exist complex numbers A1j through Anj such that A|vj = i Aij|wi. (2.12) The matrix whose entries are the values Aij is said to form a matrix representation of the operator A. This matrix representation of A is completely equivalent to the operator A, and we will use the matrix representation and abstract operator viewpoints interchangeably. Note that to make the connection between matrices and linear operators", " we must specify a set of input and output basis states for the input and output vector spaces of the linear operator. Exercise 2.2: (Matrix representations: example) Suppose V is a vector space with basis vectors |0 and |1, and A is a linear operator from V to V such that A|0 = |1 and A|1 = |0. Give a matrix representation for A, with respect to the input basis |0, |1, and the output basis |0, |1. Find input and output bases which give rise to a different matrix representation of A. Exercise 2.3: (Matrix representation for operator products) Suppose A is a linear operator from vector space V to vector space W, and B is a linear operator from vector space W to vector space X. Let |vi, |wj, and |xk be bases for the vector spaces V, W, and X, respectively. Show that the matrix representation for the linear transformation BA is the matrix product of the matrix representations for B and A, with respect to the appropriate bases. Linear algebra 65 Exercise 2.4: (Matrix representation for identity) Show that the identity operator on a vector space V has a matrix representation which is one along the diagonal and zero everywhere else, if the matrix representation is taken with respect to the same input and output bases. This matrix is known as the identity matrix. 2.1.3 The Pauli matrices Four extremely useful matrices which we shall often have occasion to use are the Pauli matrices. These are 2 by 2 matrices, which go by a variety of notations. The matrices, and their corresponding notations, are depicted in Figure 2.2. The Pauli matrices are so useful in the study of quantum computation and quantum information that we encourage you to memorize them by working through in detail the many examples and exercises based upon them in subsequent sections. \u03c30 \u2261 I \u2261 \u03c32 \u2261 \u03c3y \u2261 Y \u2261 1 0 0 1 0 \u2212i i 0 \u03c31 \u2261 \u03c3x \u2261 X \u2261 \u03c33 \u2261 \u03c3z \u2261 1 Figure 2.2. The Pauli matrices. Sometimes I is omitted from the list with just X, Y and Z known as the Pauli matrices. 2.1.4 Inner products An inner product is a function which takes as input two vectors |v and |w from a vector space and produces a complex number as output. For the time being, it will be convenient to write the inner product of |v and |w as (|v, |w). This is not the standard quantum mechanical notation; for pedagogical clarity the (\u00b7, \u00b7) notation will be useful occasionally in this chapter. The standard quantum mechanical notation for the inner product (|v, |w) is v|w, where |v and |w are vectors in the inner product space, and the notation v| is used for the dual vector to the vector |v; the dual is a linear operator from the inner product space V to the complex numbers C, de\ufb01ned by v|(|w) \u2261 v|w \u2261 (|v, |w). We will see shortly that the matrix representation of dual vectors is just a row vector. A function (\u00b7, \u00b7) from V \u00d7 V to C is an inner product if it satis\ufb01es the requirements that: (1) (\u00b7, \u00b7) is linear in the second argument, |v, i \u03bbi|wi = |v, |wi. \u03bbi i (2) (|v, |w) = (|w, |v)\u2217. (3) (|v, |v) \u2265 0 with equality if and only if |v = 0. For example, Cn has an inner product de\ufb01ned by ((y1,..., yn), (z1,..., zn)) \u2261 y\u2217 i zi = y\u2217 1... y\u2217 n i \u23a4 \u23a5 \u23a6. \u23a1 \u23a2 \u23a3 z1... zn (2.13) (2.14) 66 Introduction to quantum mechanics We call a vector space equipped with an inner product an inner product space. Exercise 2.5: Verify that (\u00b7, \u00b7) just de\ufb01ned is an inner product on Cn. Exercise 2.6: Show that any inner product (\u00b7, \u00b7) is conjugate-linear in the \ufb01rst argument, \u03bbi|wi, |v = i i \u03bb\u2217 i (|wi, |v). (2.15) Discussions of quantum mechanics often refer to Hilbert space. In the \ufb01nite dimensional complex vector spaces that come up in quantum computation and quantum information, a Hilbert space is exactly the same thing as an inner product space. From now on", " we use the two terms interchangeably, preferring the term Hilbert space. In in\ufb01nite dimensions Hilbert spaces satisfy additional technical restrictions above and beyond inner product spaces, which we will not need to worry about. Vectors |w and |v are orthogonal if their inner product is zero. For example, |w \u2261 (1, 0) and |v \u2261 (0, 1) are orthogonal with respect to the inner product de\ufb01ned by (2.14). We de\ufb01ne the norm of a vector |v by |v \u2261 v|v. (2.16) A unit vector is a vector |v such that |v = 1. We also say that |v is normalized if |v = 1. It is convenient to talk of normalizing a vector by dividing by its norm; thus |v/|v is the normalized form of |v, for any non-zero vector |v. A set |i of vectors with index i is orthonormal if each vector is a unit vector, and distinct vectors in the set are orthogonal, that is, i|j = \u03b4ij, where i and j are both chosen from the index set. Exercise 2.7: Verify that |w \u2261 (1, 1) and |v \u2261 (1, \u22121) are orthogonal. What are the normalized forms of these vectors? Suppose |w1,..., |wd is a basis set for some vector space V with an inner product. There is a useful method, the Gram\u2013Schmidt procedure, which can be used to produce an orthonormal basis set |v1,..., |vd for the vector space V. De\ufb01ne |v1 \u2261 |w1/ |w1, and for 1 \u2264 k \u2264 d \u2212 1 de\ufb01ne |vk+1 inductively by |vk+1 \u2261 |wk+1 \u2212 |wk+1 \u2212 k k i=1 i=1 vi|wk+1|vi vi|wk+1|vi. (2.17) It is not dif\ufb01cult to verify that the vectors |v1,..., |vd form an orthonormal set which is also a basis for V. Thus, any \ufb01nite dimensional vector space of dimension d has an orthonormal basis, |v1,..., |vd. Exercise 2.8: Prove that the Gram\u2013Schmidt procedure produces an orthonormal basis for V. From now on, when we speak of a matrix representation for a linear operator, we mean a matrix representation with respect to orthonormal input and output bases. We also use the convention that if the input and output spaces for a linear operator are the same, then the input and output bases are the same, unless noted otherwise. Linear algebra 67 With these conventions, the inner product on a Hilbert space can be given a convenient j vj|j be representations of i wi|i and |v = matrix representation. Let |w = vectors |w and |v with respect to some orthonormal basis |i. Then, since i|j = \u03b4ij, \u239b v|w = \u239d i \u239e vi|i, wj|j \u23a0 = = v\u2217 1... v\u2217 n j \u23a1 \u23a2 \u23a3 w1... wn \u23a4 \u23a5 \u23a6. ij v\u2217 i wj\u03b4ij = i v\u2217 i wi (2.18) (2.19) That is, the inner product of two vectors is equal to the vector inner product between two matrix representations of those vectors, provided the representations are written with respect to the same orthonormal basis. We also see that the dual vector v| has a nice interpretation as the row vector whose components are complex conjugates of the corresponding components of the column vector representation of |v. There is a useful way of representing linear operators which makes use of the inner product, known as the outer product representation. Suppose |v is a vector in an inner product space V, and |w is a vector in an inner product space W. De\ufb01ne |wv| to be the linear operator from V to W whose action is de\ufb01ned by |wv| |v \u2261 |w v|v = v|v|w. (2.20) This equation \ufb01ts beautifully into our notational conventions, according to which the expression |wv|v could potentially have one of two meanings: we will use it to denote the result when the operator |wv| acts on |v, and it has an existing interpretation as the result of multiplying |w by the complex number v|v. Our de\ufb01nitions are chosen so that these two", " potential meanings coincide. Indeed, we de\ufb01ne the former in terms of the latter! By de\ufb01nition We can take linear combinations of outer product operators |wv| in the obvious way. i ai|wivi| is the linear operator which, when acting on |v, produces i ai|wivi|v as output. The usefulness of the outer product notation can be discerned from an important result known as the completeness relation for orthonormal vectors. Let |i be any orthonormal i vi|i for basis for the vector space V, so an arbitrary vector |v can be written |v = some set of complex numbers vi. Note that i|v = vi and therefore |ii| |v = |ii|v = vi|i = |v. i Since the last equation is true for all |v it follows that i i |ii| = I. i This equation is known as the completeness relation. One application of the completeness relation is to give a means for representing any operator in the outer product notation. Suppose A : V \u2192 W is a linear operator, |vi is an orthonormal basis for V, and |wj an orthonormal basis for W. Using the completeness relation twice we obtain A = IW AIV (2.23) (2.21) (2.22) 68 Introduction to quantum mechanics = = ij ij |wjwj|A|vivi| wj|A|vi|wjvi|, (2.24) (2.25) which is the outer product representation for A. We also see from this equation that A has matrix element wj|A|vi in the ith column and jth row, with respect to the input basis |vi and output basis |wj. A second application illustrating the usefulness of the completeness relation is the Cauchy\u2013Schwarz inequality. This important result is discussed in Box 2.1, on this page. Exercise 2.9: (Pauli operators and the outer product) The Pauli matrices (Figure 2.2 on page 65) can be considered as operators with respect to an orthonormal basis |0, |1 for a two-dimensional Hilbert space. Express each of the Pauli operators in the outer product notation. Exercise 2.10: Suppose |vi is an orthonormal basis for an inner product space V. What is the matrix representation for the operator |vjvk|, with respect to the |vi basis? Box 2.1: The Cauchy-Schwarz inequality The Cauchy\u2013Schwarz inequality is an important geometric fact about Hilbert spaces. It states that for any two vectors |v and |w, |v|w|2 \u2264 v|vw|w. To see this, use the Gram\u2013Schmidt procedure to construct an orthonormal basis |i w|w. for the vector space such that the \ufb01rst member of the basis |i is |w/ i |ii| = I, and dropping some non-negative Using the completeness relation terms gives v|vw|w = v|ii|vw|w i v|ww|v w|w \u2265 w|w = v|ww|v = |v|w|2, (2.26) (2.27) (2.28) as required. A little thought shows that equality occurs if and only if |v and |w are linearly related, |v = z|w or |w = z|v, for some scalar z. 2.1.5 Eigenvectors and eigenvalues An eigenvector of a linear operator A on a vector space is a non-zero vector |v such that A|v = v|v, where v is a complex number known as the eigenvalue of A corresponding to |v. It will often be convenient to use the notation v both as a label for the eigenvector, and to represent the eigenvalue. We assume that you are familiar with the elementary properties of eigenvalues and eigenvectors \u2013 in particular, how to \ufb01nd them, via the characteristic equation. The characteristic function is de\ufb01ned to be c(\u03bb) \u2261 det |A \u2212 \u03bbI|, Linear algebra 69 where det is the determinant function for matrices; it can be shown that the characteristic function depends only upon the operator A, and not on the speci\ufb01c matrix representation used for A. The solutions of the characteristic equation c(\u03bb) = 0 are the eigenvalues of the operator A. By the fundamental theorem of algebra, every polynomial has at least one complex root, so every operator A has at least one eigenvalue, and a corresponding eigenvector. The eigenspace corresponding to an", " eigenvalue v is the set of vectors which have eigenvalue v. It is a vector subspace of the vector space on which A acts. A diagonal representation for an operator A on a vector space V is a representation i \u03bbi|ii|, where the vectors |i form an orthonormal set of eigenvectors for A, A = with corresponding eigenvalues \u03bbi. An operator is said to be diagonalizable if it has a diagonal representation. In the next section we will \ufb01nd a simple set of necessary and suf\ufb01cient conditions for an operator on a Hilbert space to be diagonalizable. As an example of a diagonal representation, note that the Pauli Z matrix may be written 1 0 0 \u22121 Z = = |00| \u2212 |11|, (2.29) where the matrix representation is with respect to orthonormal vectors |0 and |1, respectively. Diagonal representations are sometimes also known as orthonormal decompositions. When an eigenspace is more than one dimensional we say that it is degenerate. For example, the matrix A de\ufb01ned by \u23a1 \u23a3 2.30) has a two-dimensional eigenspace corresponding to the eigenvalue 2. The eigenvectors (1, 0, 0) and (0, 1, 0) are said to be degenerate because they are linearly independent eigenvectors of A with the same eigenvalue. Exercise 2.11: (Eigendecomposition of the Pauli matrices) Find the eigenvectors, eigenvalues, and diagonal representations of the Pauli matrices X, Y, and Z. Exercise 2.12: Prove that the matrix is not diagonalizable. 1 1 0 1 (2.31) 2.1.6 Adjoints and Hermitian operators Suppose A is any linear operator on a Hilbert space, V. It turns out that there exists a unique linear operator A\u2020 on V such that for all vectors |v, |w \u2208 V, (|v, A|w) = (A\u2020|v, |w). (2.32) This linear operator is known as the adjoint or Hermitian conjugate of the operator A. From the de\ufb01nition it is easy to see that (AB)\u2020 = B\u2020A\u2020. By convention, if |v is a vector, then we de\ufb01ne |v\u2020 \u2261 v|. With this de\ufb01nition it is not dif\ufb01cult to see that (A|v)\u2020 = v|A\u2020. 70 Introduction to quantum mechanics Exercise 2.13: If |w and |v are any two vectors, show that (|wv|)\u2020 = |vw|. Exercise 2.14: (Anti-linearity of the adjoint) Show that the adjoint operation is anti-linear, \u2020 aiAi = i i i A\u2020 a\u2217 i. (2.33) Exercise 2.15: Show that (A\u2020)\u2020 = A. In a matrix representation of an operator A, the action of the Hermitian conjugation operation is to take the matrix of A to the conjugate-transpose matrix, A\u2020 \u2261 (A\u2217)T, where the \u2217 indicates complex conjugation, and T indicates the transpose operation. For example, we have 1 + 3i 1 + i 2i 1 \u2212 4i \u2020 = 1 \u2212 3i \u22122i 1 \u2212 i 1 + 4i. (2.34) An operator A whose adjoint is A is known as a Hermitian or self-adjoint operator. An important class of Hermitian operators is the projectors. Suppose W is a k-dimensional vector subspace of the d-dimensional vector space V. Using the Gram\u2013 Schmidt procedure it is possible to construct an orthonormal basis |1,..., |d for V such that |1,..., |k is an orthonormal basis for W. By de\ufb01nition, k |ii| P \u2261 (2.35) i=1 is the projector onto the subspace W. It is easy to check that this de\ufb01nition is independent of the orthonormal basis |1,..., |k used for W. From the de\ufb01nition it can be shown that |vv| is Hermitian for any vector |v, so P is Hermitian, P \u2020 = P. We will often refer to the \u2018vector space\u2019 P, as shorthand for the vector space onto which P is a projector. The orthogonal complement of P is the operator Q \u2261 I \u2212 P. It is easy to see that Q is", " a projector onto the vector space spanned by |k + 1,..., |d, which we also refer to as the orthogonal complement of P, and may denote by Q. Exercise 2.16: Show that any projector P satis\ufb01es the equation P 2 = P. An operator A is said to be normal if AA\u2020 = A\u2020A. Clearly, an operator which is Hermitian is also normal. There is a remarkable representation theorem for normal operators known as the spectral decomposition, which states that an operator is a normal operator if and only if it is diagonalizable. This result is proved in Box 2.2 on page 72, which you should read closely. Exercise 2.17: Show that a normal matrix is Hermitian if and only if it has real eigenvalues. A matrix U is said to be unitary if U \u2020U = I. Similarly an operator U is unitary if U \u2020U = I. It is easily checked that an operator is unitary if and only if each of its matrix representations is unitary. A unitary operator also satis\ufb01es U U \u2020 = I, and therefore U is normal and has a spectral decomposition. Geometrically, unitary operators are important because they preserve inner products between vectors. To see this, let |v and |w be any Linear algebra 71 two vectors. Then the inner product of U |v and U |w is the same as the inner product of |v and |w, U |v, U |w = v|U \u2020U |w = v|I|w = v|w. (2.36) This result suggests the following elegant outer product representation of any unitary U. Let |vi be any orthonormal basis set. De\ufb01ne |wi \u2261 U |vi, so |wi is also an orthonormal i |wivi|. basis set, since unitary operators preserve inner products. Note that U = Conversely, if |vi and |wi are any two orthonormal bases, then it is easily checked that the operator U de\ufb01ned by U \u2261 i |wivi| is a unitary operator. Exercise 2.18: Show that all eigenvalues of a unitary matrix have modulus 1, that is, can be written in the form ei\u03b8 for some real \u03b8. Exercise 2.19: (Pauli matrices: Hermitian and unitary) Show that the Pauli matrices are Hermitian and unitary. Exercise 2.20: (Basis changes) Suppose A and A are matrix representations of an operator A on a vector space V with respect to two different orthonormal bases, |vi and |wi. Then the elements of A and A are A A ij = wi|A|wj. Characterize the relationship between A and A. ij = vi|A|vj and A special subclass of Hermitian operators is extremely important. This is the positive operators. A positive operator A is de\ufb01ned to be an operator such that for any vector |v, (|v, A|v) is a real, non-negative number. If (|v, A|v) is strictly greater than zero for all |v = 0 then we say that A is positive de\ufb01nite. In Exercise 2.24 on this page you will show that any positive operator is automatically Hermitian, and therefore by the spectral i \u03bbi|ii|, with non-negative eigenvalues \u03bbi. decomposition has diagonal representation Exercise 2.21: Repeat the proof of the spectral decomposition in Box 2.2 for the case when M is Hermitian, simplifying the proof wherever possible. Exercise 2.22: Prove that two eigenvectors of a Hermitian operator with different eigenvalues are necessarily orthogonal. Exercise 2.23: Show that the eigenvalues of a projector P are all either 0 or 1. Exercise 2.24: (Hermiticity of positive operators) Show that a positive operator is necessarily Hermitian. (Hint: Show that an arbitrary operator A can be written A = B + iC where B and C are Hermitian.) Exercise 2.25: Show that for any operator A, A\u2020A is positive. 2.1.7 Tensor products The tensor product is a way of putting vector spaces together to form larger vector spaces. This construction is crucial to understanding the quantum mechanics of multiparticle systems. The following discussion is a little abstract, and may be dif\ufb01cult to follow if you\u2019re not already familiar with the tensor product, so feel free to skip ahead now and revisit later when you come to the discussion of tensor products in quantum mechanics. Suppose V and W are vector spaces of dimension m", " and n respectively. For convenience we also suppose that V and W are Hilbert spaces. Then V \u2297 W (read \u2018V tensor 72 Introduction to quantum mechanics Box 2.2: The spectral decomposition \u2013 important! The spectral decomposition is an extremely useful representation theorem for normal operators. Theorem 2.1: (Spectral decomposition) Any normal operator M on a vector space V is diagonal with respect to some orthonormal basis for V. Conversely, any diagonalizable operator is normal. Proof The converse is a simple exercise, so we prove merely the forward implication, by induction on the dimension d of V. The case d = 1 is trivial. Let \u03bb be an eigenvalue of M, P the projector onto the \u03bb eigenspace, and Q the projector onto the orthogonal complement. Then M = (P + Q)M (P + Q) = P M P + QM P + P M Q + QM Q. Obviously P M P = \u03bbP. Furthermore, QM P = 0, as M takes the subspace P into itself. We claim that P M Q = 0 also. To see this, let |v be an element of the subspace P. Then M M \u2020|v = M \u2020M |v = \u03bbM \u2020|v. Thus, M \u2020|v has eigenvalue \u03bb and therefore is an element of the subspace P. It follows that QM \u2020P = 0. Taking the adjoint of this equation gives P M Q = 0. Thus M = P M P + QM Q. Next, we prove that QM Q is normal. To see this, note that QM = QM (P + Q) = QM Q, and QM \u2020 = QM \u2020(P + Q) = QM \u2020Q. Therefore, by the normality of M, and the observation that Q2 = Q, QM Q QM \u2020Q = QM QM \u2020Q = QM M \u2020Q = QM \u2020M Q = QM \u2020QM Q = QM \u2020Q QM Q, (2.37) (2.38) (2.39) (2.40) (2.41) so QM Q is normal. By induction, QM Q is diagonal with respect to some orthonormal basis for the subspace Q, and P M P is already diagonal with respect to some orthonormal basis for P. It follows that M = P M P + QM Q is diagonal with respect to some orthonormal basis for the total vector space. In terms of the outer product representation, this means that M can be written as M = i \u03bbi|ii|, where \u03bbi are the eigenvalues of M, |i is an orthonormal basis for V, and each |i an eigenvector of M with eigenvalue \u03bbi. In terms of projectors, i \u03bbiPi, where \u03bbi are again the eigenvalues of M, and Pi is the projector M = onto the \u03bbi eigenspace of M. These projectors satisfy the completeness relation i Pi = I, and the orthonormality relation PiPj = \u03b4ijPi. W \u2019) is an mn dimensional vector space. The elements of V \u2297 W are linear combinations of \u2018tensor products\u2019 |v \u2297 |w of elements |v of V and |w of W. In particular, if |i and |j are orthonormal bases for the spaces V and W then |i \u2297 |j is a basis for V \u2297 W. We often use the abbreviated notations |v|w, |v, w or even |vw for the tensor product |v \u2297 |w. For example, if V is a two-dimensional vector space with basis vectors |0 and |1 then |0 \u2297 |0 + |1 \u2297 |1 is an element of V \u2297 V. By de\ufb01nition the tensor product satis\ufb01es the following basic properties: Linear algebra 73 (1) For an arbitrary scalar z and elements |v of V and |w of W, |v \u2297 |w z z|v = \u2297 |w = |v \u2297. (2.42) z|w (2) For arbitrary |v1 and |v2 in V and |w in W, |v1 + |v2 \u2297 |w = |v1 \u2297 |w + |v2 \u2297 |w. (3) For arbitrary |v in V and |w1 and |w2 in W, |v \u2297 |", "w1 + |w2 = |v \u2297 |w1 + |v \u2297 |w2. (2.43) (2.44) What sorts of linear operators act on the space V \u2297 W? Suppose |v and |w are vectors in V and W, and A and B are linear operators on V and W, respectively. Then we can de\ufb01ne a linear operator A \u2297 B on V \u2297 W by the equation (A \u2297 B)(|v \u2297 |w) \u2261 A|v \u2297 B|w. (2.45) The de\ufb01nition of A \u2297 B is then extended to all elements of V \u2297 W in the natural way to ensure linearity of A \u2297 B, that is, (A \u2297 B) ai|vi \u2297 |wi \u2261 i i aiA|vi \u2297 B|wi. (2.46) It can be shown that A \u2297 B de\ufb01ned in this way is a well-de\ufb01ned linear operator on V \u2297 W. This notion of the tensor product of two operators extends in the obvious way to the case where A : V \u2192 V and B : W \u2192 W map between different vector spaces. Indeed, an arbitrary linear operator C mapping V \u2297 W to V \u2297 W can be represented as a linear combination of tensor products of operators mapping V to V and W to W, ciAi \u2297 Bi, (2.47) where by de\ufb01nition i ciAi \u2297 Bi |v \u2297 |w \u2261 i ciAi|v \u2297 Bi|w. (2.48) The inner products on the spaces V and W can be used to de\ufb01ne a natural inner product on V \u2297 W. De\ufb01ne \u239b \u239d i ai|vi \u2297 |wi, j \u239e bj|v j \u2297 |w j \u23a0 \u2261 ij i bjvi|v a\u2217 jwi|w j. (2.49) It can be shown that the function so de\ufb01ned is a well-de\ufb01ned inner product. From this inner product, the inner product space V \u2297W inherits the other structure we are familiar with, such as notions of an adjoint, unitarity, normality, and Hermiticity. All this discussion is rather abstract. It can be made much more concrete by moving i C = 74 Introduction to quantum mechanics to a convenient matrix representation known as the Kronecker product. Suppose A is an m by n matrix, and B is a p by q matrix. Then we have the matrix representation: \u23a1 \u23a2 \u23a2 \u23a2 \u23a3 A \u2297 B \u2261 nq!\" A11B A12B... A1nB A21B A22B... A2nB............ # \u23ab \u23a4 \u23aa\u23aa\u23aa\u23ac \u23a5 \u23a5 \u23a5 \u23aa\u23aa\u23aa\u23ad \u23a6 mp. (2.50) Am1B Am2B... AmnB In this representation terms like A11B denote p by q submatrices whose entries are proportional to B, with overall proportionality constant A11. For example, the tensor product of the vectors (1, 2) and (2, 3) is the vector \u2297 The tensor product of the Pauli matrices X and Y is i i 0 0 \u2212i i 0 0 0 0 0 (2.51) (2.52) Finally, we mention the useful notation |\u03c8\u2297k, which means |\u03c8 tensored with itself k times. For example |\u03c8\u22972 = |\u03c8 \u2297 |\u03c8. An analogous notation is also used for operators on tensor product spaces. Exercise 2.26: Let |\u03c8 = (|0 + |1)/ \u221a 2. Write out |\u03c8\u22972 and |\u03c8\u22973 explicitly, both in terms of tensor products like |0|1, and using the Kronecker product. Exercise 2.27: Calculate the matrix representation of the tensor products of the Pauli operators (a) X and Z; (b) I and X; (c) X and I. Is the tensor product commutative? Exercise 2.28: Show that the transpose, complex conjugation, and adjoint operations distribute over the tensor product, (A \u2297 B)\ufffd", "\ufffd = A\u2217 \u2297 B\u2217; (A \u2297 B)T = AT \u2297 BT ; (A \u2297 B)\u2020 = A\u2020 \u2297 B\u2020.(2.53) Exercise 2.29: Show that the tensor product of two unitary operators is unitary. Exercise 2.30: Show that the tensor product of two Hermitian operators is Hermitian. Exercise 2.31: Show that the tensor product of two positive operators is positive. Exercise 2.32: Show that the tensor product of two projectors is a projector. Exercise 2.33: The Hadamard operator on one qubit may be written as (|0 + |1)0| + (|0 \u2212 |1)1| H =. 1\u221a 2 (2.54) Linear algebra 75 Show explicitly that the Hadamard transform on n qubits, H \u2297n, may be written as H \u2297n = 1\u221a 2n (\u22121)x\u00b7y|xy|. x,y (2.55) Write out an explicit matrix representation for H \u22972. 2.1.8 Operator functions There are many important functions which can be de\ufb01ned for operators and matrices. Generally speaking, given a function f from the complex numbers to the complex numbers, it is possible to de\ufb01ne a corresponding matrix function on normal matrices (or some subclass, such as the Hermitian matrices) by the following construca a|aa| be a spectral decomposition for a normal operator A. De\ufb01ne tion. Let A = a f (a)|aa|. A little thought shows that f (A) is uniquely de\ufb01ned. This prof (A) \u2261 cedure can be used, for example, to de\ufb01ne the square root of a positive operator, the logarithm of a positive-de\ufb01nite operator, or the exponential of a normal operator. As an example, e\u03b8 0 0 e\u2212\u03b8, (2.56) exp(\u03b8Z) = since Z has eigenvectors |0 and |1. Exercise 2.34: Find the square root and logarithm of the matrix. 4 3 3 4 (2.57) Exercise 2.35: (Exponential of the Pauli matrices) Let v be any real, three-dimensional unit vector and \u03b8 a real number. Prove that where v \u00b7 \u03c3 \u2261 3 i=1 exp(i\u03b8v \u00b7 \u03c3) = cos(\u03b8)I + i sin(\u03b8)v \u00b7 \u03c3, (2.58) vi\u03c3i. This exercise is generalized in Problem 2.1 on page 117. Another important matrix function is the trace of a matrix. The trace of A is de\ufb01ned to be the sum of its diagonal elements, tr(A) \u2261 i Aii. (2.59) The trace is easily seen to be cyclic, tr(AB) = tr(BA), and linear, tr(A + B) = tr(A) + tr(B), tr(zA) = z tr(A), where A and B are arbitrary matrices, and z is a complex number. Furthermore, from the cyclic property it follows that the trace of a matrix is invariant under the unitary similarity transformation A \u2192 U AU \u2020, as tr(U AU \u2020) = tr(U \u2020U A) = tr(A). In light of this result, it makes sense to de\ufb01ne the trace of an operator A to be the trace of any matrix representation of A. The invariance of the trace under unitary similarity transformations ensures that the trace of an operator is well de\ufb01ned. As an example of the trace, suppose |\u03c8 is a unit vector and A is an arbitrary operator. To evaluate tr(A|\u03c8\u03c8|) use the Gram\u2013Schmidt procedure to extend |\u03c8 to an 76 Introduction to quantum mechanics orthonormal basis |i which includes |\u03c8 as the \ufb01rst element. Then we have tr(A|\u03c8\u03c8|) = i|A|\u03c8\u03c8|i i = \u03c8|A|\u03c8. (2.60) (2.61) This result, that tr(A|\u03c8\u03c8|) = \u03c8|A|\u03c8 is extremely useful in evaluating the trace of an operator. Exercise 2.36: Show that the Pauli matrices except for I have trace zero. Exercise 2.37: (Cyclic property of the trace) If A and B are two linear operators show that Exercise 2.38: (Linearity of the trace) If A and B are two", " linear operators, show tr(AB) = tr(BA). (2.62) that tr(A + B) = tr(A) + tr(B) and if z is an arbitrary complex number show that tr(zA) = ztr(A). (2.63) (2.64) Exercise 2.39: (The Hilbert\u2013Schmidt inner product on operators) The set LV of linear operators on a Hilbert space V is obviously a vector space \u2013 the sum of two linear operators is a linear operator, zA is a linear operator if A is a linear operator and z is a complex number, and there is a zero element 0. An important additional result is that the vector space LV can be given a natural inner product structure, turning it into a Hilbert space. (1) Show that the function (\u00b7, \u00b7) on LV \u00d7 LV de\ufb01ned by (A, B) \u2261 tr(A\u2020B) (2.65) is an inner product function. This inner product is known as the Hilbert\u2013Schmidt or trace inner product. (2) If V has d dimensions show that LV has dimension d2. (3) Find an orthonormal basis of Hermitian matrices for the Hilbert space LV. The commutator between two operators A and B is de\ufb01ned to be 2.1.9 The commutator and anti-commutator [A, B] \u2261 AB \u2212 BA. (2.66) If [A, B] = 0, that is, AB = BA, then we say A commutes with B. Similarly, the anti-commutator of two operators A and B is de\ufb01ned by {A, B} \u2261 AB + BA; (2.67) we say A anti-commutes with B if {A, B} = 0. It turns out that many important properties of pairs of operators can be deduced from their commutator and anti-commutator. Perhaps the most useful relation is the following connection between the commutator and the property of being able to simultaneously diagonalize Hermitian operators A and B, that is, write A = set of eigenvectors for A and B. i ai|ii|, B = Linear algebra 77 i bi|ii|, where |i is some common orthonormal Theorem 2.2: (Simultaneous diagonalization theorem) Suppose A and B are Hermitian operators. Then [A, B] = 0 if and only if there exists an orthonormal basis such that both A and B are diagonal with respect to that basis. We say that A and B are simultaneously diagonalizable in this case. This result connects the commutator of two operators, which is often easy to compute, to the property of being simultaneously diagonalizable, which is a priori rather dif\ufb01cult to determine. As an example, consider that \u2212 0 \u2212i i 0 0 1 1 0 [X, Y ] = 0 1 = 2i 0 \u2212i i 0 1 0 1 0 0 \u22121 = 2iZ, (2.68) (2.69) (2.70) so X and Y do not commute. You have already shown, in Exercise 2.11, that X and Y do not have common eigenvectors, as we expect from the simultaneous diagonalization theorem. Proof You can (and should!) easily verify that if A and B are diagonal in the same orthonormal basis then [A, B] = 0. To show the converse, let |a, j be an orthonormal basis for the eigenspace Va of A with eigenvalue a; the index j is used to label possible degeneracies. Note that AB|a, j = BA|a, j = aB|a, j, (2.71) and therefore B|a, j is an element of the eigenspace Va. Let Pa denote the projector onto the space Va and de\ufb01ne Ba \u2261 PaBPa. It is easy to see that the restriction of Ba to the space Va is Hermitian on Va, and therefore has a spectral decomposition in terms of an orthonormal set of eigenvectors which span the space Va. Let\u2019s call these eigenvectors |a, b, k, where the indices a and b label the eigenvalues of A and Ba, and k is an extra index to allow for the possibility of a degenerate Ba. Note that B|a, b, k is an element of Va, so B|a, b, k = PaB|a, b, k. Moreover we have Pa|a, b, k = |a, b, k, so B|a, b, k = PaBPa|a, b, k = b|a, b, k. (2.", "72) It follows that |a, b, k is an eigenvector of B with eigenvalue b, and therefore |a, b, k is an orthonormal set of eigenvectors of both A and B, spanning the entire vector space on which A and B are de\ufb01ned. That is, A and B are simultaneously diagonalizable. Exercise 2.40: (Commutation relations for the Pauli matrices) Verify the commutation relations [X, Y ] = 2iZ; [Y, Z] = 2iX; [Z, X] = 2iY. (2.73) There is an elegant way of writing this using jkl, the antisymmetric tensor on 78 Introduction to quantum mechanics three indices, for which jkl = 0 except for 123 = 231 = 312 = 1, and 321 = 213 = 132 = \u22121: [\u03c3j, \u03c3k] = 2i 3 l=1 jkl\u03c3l. (2.74) Exercise 2.41: (Anti-commutation relations for the Pauli matrices) Verify the anti-commutation relations where i = j are both chosen from the set 1, 2, 3. Also verify that (i = 0, 1, 2, 3) {\u03c3i, \u03c3j} = 0 (2.75) Exercise 2.42: Verify that \u03c32 i = I. AB = [A, B] + {A, B} 2. Exercise 2.43: Show that for j, k = 1, 2, 3, \u03c3j\u03c3k = \u03b4jkI + i 3 l=1 jkl\u03c3l. (2.76) (2.77) (2.78) Exercise 2.44: Suppose [A, B] = 0, {A, B} = 0, and A is invertible. Show that B must be 0. Exercise 2.45: Show that [A, B]\u2020 = [B\u2020, A\u2020]. Exercise 2.46: Show that [A, B] = \u2212[B, A]. Exercise 2.47: Suppose A and B are Hermitian. Show that i[A, B] is Hermitian. 2.1.10 The polar and singular value decompositions The polar and singular value decompositions are useful ways of breaking linear operators up into simpler parts. In particular, these decompositions allow us to break general linear operators up into products of unitary operators and positive operators. While we don\u2019t understand the structure of general linear operators terribly well, we do understand unitary operators and positive operators in quite some detail. The polar and singular value decompositions allow us to apply this understanding to better understand general linear operators. Theorem 2.3: (Polar decomposition) Let A be a linear operator on a vector space V. Then there exists unitary U and positive operators J and K such that A = U J = KU, (2.79) where the unique positive operators J and K satisfying these equations are de\ufb01ned by J \u2261 unique. AA\u2020. Moreover, if A is invertible then U is A\u2020A and K \u2261 \u221a \u221a Linear algebra 79 We call the expression A = U J the left polar decomposition of A, and A = KU the right polar decomposition of A. Most often, we\u2019ll omit the \u2018right\u2019 or \u2018left\u2019 nomenclature, and use the term \u2018polar decomposition\u2019 for both expressions, with context indicating which is meant. \u221a Proof J \u2261 A\u2020A is a positive operator, so it can be given a spectral decomposition, J = i \u03bbi|ii| (\u03bbi \u2265 0). De\ufb01ne |\u03c8i \u2261 A|i. From the de\ufb01nition, we see that \u03c8i|\u03c8i = \u03bb2 i. Consider for now only those i for which \u03bbi = 0. For those i de\ufb01ne |ei \u2261 |\u03c8i/\u03bbi, so the |ei are normalized. Moreover, they are orthogonal, since if i = j then ei|ej = i|A\u2020A|j/\u03bbi\u03bbj = i|J 2|j/\u03bbi\u03bbj = 0. We have been considering i such that \u03bbi = 0. Now use the Gram\u2013Schmidt procedure to extend the orthonormal set |ei so it forms an orthonormal basis, which we also label |ei. De\ufb01ne a unitary operator U \u2261 i |eii|. When \u03bbi = 0 we have U J|i = \u03bbi|ei = |\u03c8i = A|i. When \u03bbi = 0 we have U J|", "i = 0 = |\u03c8i. We have proved that the action of A and U J agree on the basis |i, and thus that A = U J. J is unique, since multiplying A = U J on the left by the adjoint equation A\u2020 = JU \u2020 \u221a gives J 2 = A\u2020A, from which we see that J = A\u2020A, uniquely. A little thought shows that if A is invertible, then so is J, so U is uniquely determined by the equation U = AJ \u22121. The proof of the right polar decomposition follows, since A = U J = U JU \u2020U = KU, where K \u2261 U JU \u2020 is a positive operator. Since AA\u2020 = KU U \u2020K = K 2 we must have K = AA\u2020, as claimed. \u221a The singular value decomposition combines the polar decomposition and the spectral theorem. Corollary 2.4: (Singular value decomposition) Let A be a square matrix. Then there exist unitary matrices U and V, and a diagonal matrix D with non-negative entries such that The diagonal elements of D are called the singular values of A. A = U DV. (2.80) Proof By the polar decomposition, A = SJ, for unitary S, and positive J. By the spectral theorem, J = T DT \u2020, for unitary T and diagonal D with non-negative entries. Setting U \u2261 ST and V \u2261 T \u2020 completes the proof. Exercise 2.48: What is the polar decomposition of a positive matrix P? Of a unitary matrix U? Of a Hermitian matrix, H? Exercise 2.49: Express the polar decomposition of a normal matrix in the outer product representation. Exercise 2.50: Find the left and right polar decompositions of the matrix 1 1 0 1. (2.81) 80 Introduction to quantum mechanics 2.2 The postulates of quantum mechanics All understanding begins with our not accepting the world as it appears. \u2013 Alan Kay The most incomprehensible thing about the world is that it is comprehensible. \u2013 Albert Einstein Quantum mechanics is a mathematical framework for the development of physical theories. On its own quantum mechanics doesn\u2019t tell you what laws a physical system must obey, but it does provide a mathematical and conceptual framework for the development of such laws. In the next few sections we give a complete description of the basic postulates of quantum mechanics. These postulates provide a connection between the physical world and the mathematical formalism of quantum mechanics. The postulates of quantum mechanics were derived after a long process of trial and (mostly) error, which involved a considerable amount of guessing and fumbling by the originators of the theory. Don\u2019t be surprised if the motivation for the postulates is not always clear; even to experts the basic postulates of quantum mechanics appear surprising. What you should expect to gain in the next few sections is a good working grasp of the postulates \u2013 how to apply them, and when. 2.2.1 State space The \ufb01rst postulate of quantum mechanics sets up the arena in which quantum mechanics takes place. The arena is our familiar friend from linear algebra, Hilbert space. Postulate 1: Associated to any isolated physical system is a complex vector space with inner product (that is, a Hilbert space) known as the state space of the system. The system is completely described by its state vector, which is a unit vector in the system\u2019s state space. Quantum mechanics does not tell us, for a given physical system, what the state space of that system is, nor does it tell us what the state vector of the system is. Figuring that out for a speci\ufb01c system is a dif\ufb01cult problem for which physicists have developed many intricate and beautiful rules. For example, there is the wonderful theory of quantum electrodynamics (often known as QED), which describes how atoms and light interact. One aspect of QED is that it tells us what state spaces to use to give quantum descriptions of atoms and light. We won\u2019t be much concerned with the intricacies of theories like QED (except in so far as they apply to physical realizations, in Chapter 7), as we are mostly interested in the general framework provided by quantum mechanics. For our purposes it will be suf\ufb01cient to make some very simple (and reasonable) assumptions about the state spaces of the systems we are interested in, and stick with those assumptions. The simplest quantum mechanical system, and the system which we will be most concerned with, is the qubit. A qubit has a two-dimensional state space. Suppose |0 and |1 form an orthonormal basis for that state space. Then an arbitrary state vector in the state space can be written |\u03c8 = a|0 + b|1, (2.82) The post", "ulates of quantum mechanics 81 where a and b are complex numbers. The condition that |\u03c8 be a unit vector, \u03c8|\u03c8 = 1, is therefore equivalent to |a|2 + |b|2 = 1. The condition \u03c8|\u03c8 = 1 is often known as the normalization condition for state vectors. We will take the qubit as our fundamental quantum mechanical system. Later, in Chapter 7, we will see that there are real physical systems which may be described in terms of qubits. For now, though, it is suf\ufb01cient to think of qubits in abstract terms, without reference to a speci\ufb01c realization. Our discussions of qubits will always be referred to some orthonormal set of basis vectors, |0 and |1, which should be thought of as being \ufb01xed in advance. Intuitively, the states |0 and |1 are analogous to the two values 0 and 1 which a bit may take. The way a qubit differs from a bit is that superpositions of these two states, of the form a|0 + b|1, can also exist, in which it is not possible to say that the qubit is de\ufb01nitely in the state |0, or de\ufb01nitely in the state |1. We conclude with some useful terminology which is often used in connection with i \u03b1i|\u03c8i is a the description of quantum states. We say that any linear combination superposition of the states |\u03c8i with amplitude \u03b1i for the state |\u03c8i. So, for example, the state |0 \u2212 |1 \u221a 2 is a superposition of the states |0 and |1 with amplitude 1/ amplitude \u22121/ 2 for the state |1. \u221a (2.83) \u221a 2 for the state |0, and 2.2.2 Evolution How does the state, |\u03c8, of a quantum mechanical system change with time? The following postulate gives a prescription for the description of such state changes. Postulate 2: The evolution of a closed quantum system is described by a unitary transformation. That is, the state |\u03c8 of the system at time t1 is related to the state |\u03c8 of the system at time t2 by a unitary operator U which depends only on the times t1 and t2, |\u03c8 = U |\u03c8. (2.84) Just as quantum mechanics does not tell us the state space or quantum state of a particular quantum system, it does not tell us which unitary operators U describe realworld quantum dynamics. Quantum mechanics merely assures us that the evolution of any closed quantum system may be described in such a way. An obvious question to ask is: what unitary operators are natural to consider? In the case of single qubits, it turns out that any unitary operator at all can be realized in realistic systems. Let\u2019s look at a few examples of unitary operators on a single qubit which are important in quantum computation and quantum information. We have already seen several examples of such unitary operators \u2013 the Pauli matrices, de\ufb01ned in Section 2.1.3, and the quantum gates described in Chapter 1. As remarked in Section 1.3.1, the X matrix is gate. The X and often known as the quantum Z Pauli matrices are also sometimes referred to as the bit \ufb02ip and phase \ufb02ip matrices: the X matrix takes |0 to |1, and |1 to |0, thus earning the name bit \ufb02ip; and the Z matrix leaves |0 invariant, and takes |1 to \u2212|1, with the extra factor of \u22121 added known as a phase factor, thus justifying the term phase \ufb02ip. We will not use the term phase \ufb02ip for gate, by analogy to the classical 82 Introduction to quantum mechanics Z very often, since it is easily confused with the phase gate to be de\ufb01ned in Chapter 4. (Section 2.2.7 contains more discussion of the many uses of the term \u2018phase\u2019.) Another interesting unitary operator is the Hadamard gate, which we denote H. This 2, H|1 \u2261 (|0 \u2212 |1)/ 2, and corresponding matrix \u221a \u221a has the action H|0 \u2261 (|0 + |1)/ representation H = 1\u221a 2 1 1 1 \u22121. (2.85) Exercise 2.51: Verify that the Hadamard gate H is unitary. Exercise 2.52: Verify that H 2 = I. Exercise 2.53: What are the eigenvalues and eigenvectors of H? Postulate 2 requires that the system being described be closed. That is, it is not interacting in any way with other systems. In reality, of course, all", " systems (except the Universe as a whole) interact at least somewhat with other systems. Nevertheless, there are interesting systems which can be described to a good approximation as being closed, and which are described by unitary evolution to some good approximation. Furthermore, at least in principle every open system can be described as part of a larger closed system (the Universe) which is undergoing unitary evolution. Later, we\u2019ll introduce more tools which allow us to describe systems which are not closed, but for now we\u2019ll continue with the description of the evolution of closed systems. Postulate 2 describes how the quantum states of a closed quantum system at two different times are related. A more re\ufb01ned version of this postulate can be given which describes the evolution of a quantum system in continuous time. From this more re\ufb01ned postulate we will recover Postulate 2. Before we state the revised postulate, it is worth pointing out two things. First, a notational remark. The operator H appearing in the following discussion is not the same as the Hadamard operator, which we just introduced. Second, the following postulate makes use of the apparatus of differential equations. Readers with little background in the study of differential equations should be reassured that they will not be necessary for much of the book, with the exception of parts of Chapter 7, on real physical implementations of quantum information processing. Postulate 2: The time evolution of the state of a closed quantum system is described by the Schr\u00a8odinger equation, i d|\u03c8 dt = H|\u03c8. In this equation, is a physical constant known as Planck\u2019s constant whose value must be experimentally determined. The exact value is not important to us. In practice, it is common to absorb the factor into H, effectively setting = 1. H is a \ufb01xed Hermitian operator known as the Hamiltonian of the closed system. (2.86) If we know the Hamiltonian of a system, then (together with a knowledge of ) we understand its dynamics completely, at least in principle. In general \ufb01guring out the Hamiltonian needed to describe a particular physical system is a very dif\ufb01cult problem \u2013 much of twentieth century physics has been concerned with this problem \u2013 which requires substantial input from experiment in order to be answered. From our point of The postulates of quantum mechanics 83 view this is a problem of detail to be addressed by physical theories built within the framework of quantum mechanics \u2013 what Hamiltonian do we need to describe atoms in such-and-such a con\ufb01guration \u2013 and is not a question that needs to be addressed by the theory of quantum mechanics itself. Most of the time in our discussion of quantum computation and quantum information we won\u2019t need to discuss Hamiltonians, and when we do, we will usually just posit that some matrix is the Hamiltonian as a starting point, and proceed from there, without attempting to justify the use of that Hamiltonian. Because the Hamiltonian is a Hermitian operator it has a spectral decomposition H = E|EE|, (2.87) E with eigenvalues E and corresponding normalized eigenvectors |E. The states |E are conventionally referred to as energy eigenstates, or sometimes as stationary states, and E is the energy of the state |E. The lowest energy is known as the ground state energy for the system, and the corresponding energy eigenstate (or eigenspace) is known as the ground state. The reason the states |E are sometimes known as stationary states is because their only change in time is to acquire an overall numerical factor, |E \u2192 exp(\u2212iEt/)|E. As an example, suppose a single qubit has Hamiltonian H = \u03c9X. (2.88) (2.89) In this equation \u03c9 is a parameter that, in practice, needs to be experimentally determined. We won\u2019t worry about the parameter overly much here \u2013 the point is to give you a feel for the sort of Hamiltonians that are sometimes written down in the study of quantum computation and quantum information. The energy eigenstates of this Hamiltonian are 2 and (|0 \u2212 |1)/ obviously the same as the eigenstates of X, namely (|0 + |1)/ 2, with corresponding energies \u03c9 and \u2212\u03c9. The ground state is therefore (|0 \u2212 |1)/ 2, and the ground state energy is \u2212\u03c9. \u221a \u221a \u221a What is the connection between the Hamiltonian picture of dynamics, Postulate 2, and the unitary operator picture, Postulate 2? The answer is provided by writing down the solution to Schr\u00a8odinger\u2019s equation, which is easily veri\ufb01ed to be: |\u03c8(t2) = exp |\u03c8(t1) = U (t1, t2)|\ufffd", "\ufffd(t1), (2.90) \u2212iH(t2 \u2212 t1) where we de\ufb01ne U (t1, t2) \u2261 exp \u2212iH(t2 \u2212 t1). (2.91) You will show in the exercises that this operator is unitary, and furthermore, that any unitary operator U can be realized in the form U = exp(iK) for some Hermitian operator K. There is therefore a one-to-one correspondence between the discrete-time description of dynamics using unitary operators, and the continuous time description using Hamiltonians. For most of the book we use the unitary formulation of quantum dynamics. Exercise 2.54: Suppose A and B are commuting Hermitian operators. Prove that exp(A) exp(B) = exp(A + B). (Hint: Use the results of Section 2.1.9.) 84 Introduction to quantum mechanics Exercise 2.55: Prove that U (t1, t2) de\ufb01ned in Equation (2.91) is unitary. Exercise 2.56: Use the spectral decomposition to show that K \u2261 \u2212i log(U ) is Hermitian for any unitary U, and thus U = exp(iK) for some Hermitian K. In quantum computation and quantum information we often speak of applying a unitary operator to a particular quantum system. For example, in the context of quantum circuits we may speak of applying the unitary gate X to a single qubit. Doesn\u2019t this contradict what we said earlier, about unitary operators describing the evolution of a closed quantum system? After all, if we are \u2018applying\u2019 a unitary operator, then that implies that there is an external \u2018we\u2019 who is interacting with the quantum system, and the system is not closed. An example of this occurs when a laser is focused on an atom. After a lot of thought and hard work it is possible to write down a Hamiltonian describing the total atom\u2013 laser system. The interesting thing is that when we write down the Hamiltonian for the atom\u2013laser system and consider the effects on the atom alone, the behavior of the state vector of the atom turns out to be almost but not quite perfectly described by another Hamiltonian, the atomic Hamiltonian. The atomic Hamiltonian contains terms related to laser intensity, and other parameters of the laser, which we can vary at will. It is as if the evolution of the atom were being described by a Hamiltonian which we can vary at will, despite the atom not being a closed system. More generally, for many systems like this it turns out to be possible to write down a time-varying Hamiltonian for a quantum system, in which the Hamiltonian for the system is not a constant, but varies according to some parameters which are under an experimentalist\u2019s control, and which may be changed during the course of an experiment. The system is not, therefore, closed, but it does evolve according to Schr\u00a8odinger\u2019s equation with a time-varying Hamiltonian, to some good approximation. The upshot is that to begin we will often describe the evolution of quantum systems \u2013 even systems which aren\u2019t closed \u2013 using unitary operators. The main exception to this, quantum measurement, will be described in the next section. Later on we will investigate in more detail possible deviations from unitary evolution due to the interaction with other systems, and understand more precisely the dynamics of realistic quantum systems. 2.2.3 Quantum measurement We postulated that closed quantum systems evolve according to unitary evolution. The evolution of systems which don\u2019t interact with the rest of the world is all very well, but there must also be times when the experimentalist and their experimental equipment \u2013 an external physical system in other words \u2013 observes the system to \ufb01nd out what is going on inside the system, an interaction which makes the system no longer closed, and thus not necessarily subject to unitary evolution. To explain what happens when this is done, we introduce Postulate 3, which provides a means for describing the effects of measurements on quantum systems. Postulate 3: Quantum measurements are described by a collection {Mm} of measurement operators. These are operators acting on the state space of the system being measured. The index m refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is |\u03c8 immediately before the measurement then the probability that result m occurs is The postulates of quantum mechanics 85 given by p(m) = \u03c8|M \u2020 mMm|\u03c8, and the state of the system after the measurement is Mm|\u03c8. \u03c8|M \u2020 mMm|\u03c8 The measurement operators satisfy the completeness equation, m M \u2020 mMm = I. The completeness equation expresses the fact that probabilities sum to one:", " 1 = p(m) = \u03c8|M \u2020 mMm|\u03c8. (2.92) (2.93) (2.94) (2.95) m m This equation being satis\ufb01ed for all |\u03c8 is equivalent to the completeness equation. However, the completeness equation is much easier to check directly, so that\u2019s why it appears in the statement of the postulate. A simple but important example of a measurement is the measurement of a qubit in the computational basis. This is a measurement on a single qubit with two outcomes de\ufb01ned by the two measurement operators M0 = |00|, M1 = |11|. Observe that 1 = M1. Thus the each measurement operator is Hermitian, and that M 2 completeness relation is obeyed, I = M \u2020 M1 = M0 + M1. Suppose the state being measured is |\u03c8 = a|0 + b|1. Then the probability of obtaining measurement outcome 0 is 0 = M0, M 2 M0 + M \u2020 1 0 p(0) = \u03c8|M \u2020 0 M0|\u03c8 = \u03c8|M0|\u03c8 = |a|2. (2.96) Similarly, the probability of obtaining the measurement outcome 1 is p(1) = |b|2. The state after measurement in the two cases is therefore M0|\u03c8 |a| = M1|\u03c8 |b| = a |a| b |b| |0 |1. (2.97) (2.98) We will see in Section 2.2.7 that multipliers like a/|a|, which have modulus one, can effectively be ignored, so the two post-measurement states are effectively |0 and |1, just as described in Chapter 1. The status of Postulate 3 as a fundamental postulate intrigues many people. Measuring devices are quantum mechanical systems, so the quantum system being measured and the measuring device together are part of a larger, isolated, quantum mechanical system. (It may be necessary to include quantum systems other than the system being measured and the measuring device to obtain a completely isolated system, but the point is that this can be done.) According to Postulate 2, the evolution of this larger isolated system can be described by a unitary evolution. Might it be possible to derive Postulate 3 as a consequence of this picture? Despite considerable investigation along these lines there is still disagreement between physicists about whether or not this is possible. We, however, are going to take the very pragmatic approach that in practice it is clear when to apply 86 Introduction to quantum mechanics Postulate 2 and when to apply Postulate 3, and not worry about deriving one postulate from the other. Over the next few sections we apply Postulate 3 to several elementary but important measurement scenarios. Section 2.2.4 examines the problem of distinguishing a set of quantum states. Section 2.2.5 explains a special case of Postulate 3, the projective or von Neumann measurements. Section 2.2.6 explains another special case of Postulate 3, known as POVM measurements. Many introductions to quantum mechanics only discuss projective measurements, omitting a full discussion of Postulate 3 or of POVM elements. For this reason we have included Box 2.5 on page 91 which comments on the relationship between the different classes of measurement we describe. Exercise 2.57: (Cascaded measurements are single measurements) Suppose {Ll} and {Mm} are two sets of measurement operators. Show that a measurement de\ufb01ned by the measurement operators {Ll} followed by a measurement de\ufb01ned by the measurement operators {Mm} is physically equivalent to a single measurement de\ufb01ned by measurement operators {Nlm} with the representation Nlm \u2261 MmLl. 2.2.4 Distinguishing quantum states An important application of Postulate 3 is to the problem of distinguishing quantum states. In the classical world, distinct states of an object are usually distinguishable, at least in principle. For example, we can always identify whether a coin has landed heads or tails, at least in the ideal limit. Quantum mechanically, the situation is more complicated. In Section 1.6 we gave a plausible argument that non-orthogonal quantum states cannot be distinguished. With Postulate 3 as a \ufb01rm foundation we can now give a much more convincing demonstration of this fact. Distinguishability, like many ideas in quantum computation and quantum information, is most easily understood using the metaphor of a game involving two parties, Alice and Bob. Alice chooses a state |\u03c8i (1 \u2264 i \u2264 n) from some \ufb01xed set of states known to both parties. She gives the state |\u03c8i to Bob, whose task it is to", " identify the index i of the state Alice has given him. Suppose the states |\u03c8i are orthonormal. Then Bob can do a quantum measurement to distinguish these states, using the following procedure. De\ufb01ne measurement operators Mi \u2261 |\u03c8i\u03c8i|, one for each possible index i, and an additional measurement operator M0 de\ufb01ned as the positive square root of the positive operator I \u2212 |\u03c8i\u03c8i|. These operators satisfy the completeness relation, and if the state |\u03c8i is prepared then p(i) = \u03c8i|Mi|\u03c8i = 1, so the result i occurs with certainty. Thus, it is possible to reliably distinguish the orthonormal states |\u03c8i. i = 0 By contrast, if the states |\u03c8i are not orthonormal then we can prove that there is no quantum measurement capable of distinguishing the states. The idea is that Bob will do a measurement described by measurement operators Mj, with outcome j. Depending on the outcome of the measurement Bob tries to guess what the index i was using some rule, i = f (j), where f (\u00b7) represents the rule he uses to make the guess. The key to why Bob can\u2019t distinguish non-orthogonal states |\u03c81 and |\u03c82 is the observation that |\u03c82 can be decomposed into a (non-zero) component parallel to |\u03c81, and a component orthogonal to |\u03c81. Suppose j is a measurement outcome such that f (j) = 1, that is, Bob guesses that the state was |\u03c81 when he observes j. But because of the component of |\u03c82 parallel The postulates of quantum mechanics 87 to |\u03c81, there is a non-zero probability of getting outcome j when |\u03c82 is prepared, so sometimes Bob will make an error identifying which state was prepared. A more rigorous argument that non-orthogonal states can\u2019t be distinguished is given in Box 2.3, but this captures the essential idea. Box 2.3: Proof that non-orthogonal states can\u2019t be reliably distinguished A proof by contradiction shows that no measurement distinguishing the nonorthogonal states |\u03c81 and |\u03c82 is possible. Suppose such a measurement is possible. If the state |\u03c81 (|\u03c82) is prepared then the probability of measuring j such that f (j) = 1 (f (j) = 2) must be 1. De\ufb01ning Ei \u2261 j Mj, these observations may be written as: j:f (j)=i M \u2020 \u03c81|E1|\u03c81 = 1; \u03c82|E2|\u03c82 = 1. (2.99) i\u03c81|Ei|\u03c81 = 1, and since \u03c81|E1|\u03c81 = 1 i Ei = I it follows that Since E2|\u03c81 = 0. Suppose we decompose we must have \u03c81|E2|\u03c81 = 0, and thus |\u03c82 = \u03b1|\u03c81 + \u03b2|\u03d5, where |\u03d5 is orthonormal to |\u03c81, |\u03b1|2 + |\u03b2|2 = 1, and |\u03b2| < 1 \u221a since |\u03c81 and |\u03c82 are not orthogonal. Then E2|\u03d5, which implies a contradiction with (2.99), as E2|\u03c82 = \u03b2 \u221a \u221a \u03c82|E2|\u03c82 = |\u03b2|2\u03d5|E2|\u03d5 \u2264 |\u03b2|2 < 1, where the second last inequality follows from the observation that \u03d5|E2|\u03d5 \u2264 \u03d5|Ei|\u03d5 = \u03d5|\u03d5 = 1. (2.100) (2.101) i 2.2.5 Projective measurements In this section we explain an important special case of the general measurement postulate, Postulate 3. This special class of measurements is known as projective measurements. For many applications of quantum computation and quantum information we will be concerned primarily with projective measurements. Indeed, projective measurements actually turn out to be equivalent to the general measurement postulate, when they are augmented with the ability to perform unitary transformations, as described in Postulate 2. We will explain this equivalence in detail in Section 2.2.8, as the statement of the measurement postulate for projective measurements is super\ufb01cially rather different from the general postulate, Postulate 3. Projective measurements: A projective measurement is described by an observable, M, a Hermitian operator on the state space of the system being observed. The observable has a spectral decomposition, M = m", "Pm, (2.102) m where Pm is the projector onto the eigenspace of M with eigenvalue m. The possible outcomes of the measurement correspond to the eigenvalues, m, of the observable. Upon measuring the state |\u03c8, the probability of getting result m is 88 Introduction to quantum mechanics given by p(m) = \u03c8|Pm|\u03c8. (2.103) Given that outcome m occurred, the state of the quantum system immediately after the measurement is Pm|\u03c8 \u221a p(m). (2.104) Projective measurements can be understood as a special case of Postulate 3. Suppose the measurement operators in Postulate 3, in addition to satisfying the completeness relation mMm = I, also satisfy the conditions that Mm are orthogonal projectors, that is, the Mm are Hermitian, and MmMm = \u03b4m,mMm. With these additional restrictions, Postulate 3 reduces to a projective measurement as just de\ufb01ned. m M \u2020 Projective measurements have many nice properties. In particular, it is very easy to calculate average values for projective measurements. By de\ufb01nition, the average (see Appendix 1 for elementary de\ufb01nitions and results in probability theory) value of the measurement is E(M ) = = m m = \u03c8| m p(m) m\u03c8|Pm|\u03c8 mPm |\u03c8 m = \u03c8|M |\u03c8. (2.110) (2.111) (2.112) (2.113) This is a useful formula, which simpli\ufb01es many calculations. The average value of the observable M is often written M \u2261 \u03c8|M |\u03c8. From this formula for the average follows a formula for the standard deviation associated to observations of M, [\u0394(M )]2 = (M \u2212 M )2 = M 2 \u2212 M 2. (2.114) (2.115) The standard deviation is a measure of the typical spread of the observed values upon measurement of M. In particular, if we perform a large number of experiments in which the state |\u03c8 is prepared and the observable M is measured, then the standard deviation \u0394(M ) M 2 \u2212 M 2. This forof the observed values is determined by the formula \u0394(M ) = mulation of measurement and standard deviations in terms of observables gives rise in an elegant way to results such as the Heisenberg uncertainty principle (see Box 2.4). Exercise 2.58: Suppose we prepare a quantum system in an eigenstate |\u03c8 of some observable M, with corresponding eigenvalue m. What is the average observed value of M, and the standard deviation? Two widely used nomenclatures for measurements deserve emphasis. Rather than giving an observable to describe a projective measurement, often people simply list a comm Pm = I and PmPm = plete set of orthogonal projectors Pm satisfying the relations The postulates of quantum mechanics 89 Box 2.4: The Heisenberg uncertainty principle Perhaps the best known result of quantum mechanics is the Heisenberg uncertainty principle. Suppose A and B are two Hermitian operators, and |\u03c8 is a quantum state. Suppose \u03c8|AB|\u03c8 = x + iy, where x and y are real. Note that \u03c8|[A, B]|\u03c8 = 2iy and \u03c8|{A, B}|\u03c8 = 2x. This implies that |\u03c8|[A, B]|\u03c8|2 + |\u03c8|{A, B}|\u03c8|2 = 4 |\u03c8|AB|\u03c8|2. (2.105) By the Cauchy\u2013Schwarz inequality |\u03c8|AB|\u03c8|2 \u2264 \u03c8|A2|\u03c8\u03c8|B2|\u03c8, (2.106) which combined with Equation (2.105) and dropping a non-negative term gives |\u03c8|[A, B]|\u03c8|2 \u2264 4\u03c8|A2|\u03c8\u03c8|B2|\u03c8. (2.107) Suppose C and D are two observables. Substituting A = C \u2212C and B = D\u2212D into the last equation, we obtain Heisenberg\u2019s uncertainty principle as it is usually stated: \u0394(C)\u0394(D) \u2265 |\u03c8|[C, D]|\u03c8| 2. (2.108) You should be wary of a common misconception about the uncertainty principle, that measuring an observable C to some \u2018accuracy\u2019 \u0394(C) causes the value of D to be \u2018disturbed\u2019 by an amount \u0394(D) in such a", " way that some sort of inequality similar to (2.108) is satis\ufb01ed. While it is true that measurements in quantum mechanics cause disturbance to the system being measured, this is most emphatically not the content of the uncertainty principle. The correct interpretation of the uncertainty principle is that if we prepare a large number of quantum systems in identical states, |\u03c8, and then perform measurements of C on some of those systems, and of D in others, then the standard deviation \u0394(C) of the C results times the standard deviation \u0394(D) of the results for D will satisfy the inequality (2.108). As an example of the uncertainty principle, consider the observables X and Y when measured for the quantum state |0. In Equation (2.70) we showed that [X, Y ] = 2iZ, so the uncertainty principle tells us that \u0394(X)\u0394(Y ) \u2265 0|Z|0 = 1. (2.109) One elementary consequence of this is that \u0394(X) and \u0394(Y ) must both be strictly greater than 0, as can be veri\ufb01ed by direct calculation. \u03b4mmPm. The corresponding observable implicit in this usage is M = m mPm. Another widely used phrase, to \u2018measure in a basis |m\u2019, where |m form an orthonormal basis, simply means to perform the projective measurement with projectors Pm = |mm|. Let\u2019s look at an example of projective measurements on single qubits. First is the measurement of the observable Z. This has eigenvalues +1 and \u22121 with corresponding eigenvectors |0 and |1. Thus, for example, measurement of Z on the state |\u03c8 = (|0 + |1)/ 2 gives the result +1 with probability \u03c8|00|\u03c8 = 1/2, and similarly the \u221a 90 Introduction to quantum mechanics result \u22121 with probability 1/2. More generally, suppose v is any real three-dimensional unit vector. Then we can de\ufb01ne an observable: v \u00b7 \u03c3 \u2261 v1\u03c31 + v2\u03c32 + v3\u03c33. (2.116) Measurement of this observable is sometimes referred to as a \u2018measurement of spin along the v axis\u2019, for historical reasons. The following two exercises encourage you to work out some elementary but important properties of such a measurement. Exercise 2.59: Suppose we have qubit in the state |0, and we measure the observable X. What is the average value of X? What is the standard deviation of X? Exercise 2.60: Show that v \u00b7 \u03c3 has eigenvalues \u00b11, and that the projectors onto the corresponding eigenspaces are given by P\u00b1 = (I \u00b1 v \u00b7 \u03c3)/2. Exercise 2.61: Calculate the probability of obtaining the result +1 for a measurement of v \u00b7 \u03c3, given that the state prior to measurement is |0. What is the state of the system after the measurement if +1 is obtained? 2.2.6 POVM measurements The quantum measurement postulate, Postulate 3, involves two elements. First, it gives a rule describing the measurement statistics, that is, the respective probabilities of the different possible measurement outcomes. Second, it gives a rule describing the postmeasurement state of the system. However, for some applications the post-measurement state of the system is of little interest, with the main item of interest being the probabilities of the respective measurement outcomes. This is the case, for example, in an experiment where the system is measured only once, upon conclusion of the experiment. In such instances there is a mathematical tool known as the POVM formalism which is especially well adapted to the analysis of the measurements. (The acronym POVM stands for \u2018Positive Operator-Valued Measure\u2019, a technical term whose historical origins we won\u2019t worry about.) This formalism is a simple consequence of the general description of measurements introduced in Postulate 3, but the theory of POVMs is so elegant and widely used that it merits a separate discussion here. Suppose a measurement described by measurement operators Mm is performed upon a quantum system in the state |\u03c8. Then the probability of outcome m is given by p(m) = \u03c8|M \u2020 mMm|\u03c8. Suppose we de\ufb01ne Em \u2261 M \u2020 mMm. (2.117) Then from Postulate 3 and elementary linear algebra, Em is a positive operator such m Em = I and p(m) = \u03c8|Em|\u03c8. Thus the set of operators Em are suf\ufb01cient to that determine the probabilities of the different measurement outcomes. The operators Em are known as the POVM elements associated with the measurement. The complete set {Em} is", " known as a POVM. As an example of a POVM, consider a projective measurement described by measurement operators Pm, where the Pm are projectors such that PmPm = \u03b4mmPm and m Pm = I. In this instance (and only this instance) all the POVM elements are the same as the measurement operators themselves, since Em \u2261 P \u2020 mPm = Pm. The postulates of quantum mechanics 91 Box 2.5: General measurements, projective measurements, and POVMs Most introductions to quantum mechanics describe only projective measurements, and consequently the general description of measurements given in Postulate 3 may be unfamiliar to many physicists, as may the POVM formalism described in Section 2.2.6. The reason most physicists don\u2019t learn the general measurement formalism is because most physical systems can only be measured in a very coarse manner. In quantum computation and quantum information we aim for an exquisite level of control over the measurements that may be done, and consequently it helps to use a more comprehensive formalism for the description of measurements. Of course, when the other axioms of quantum mechanics are taken into account, projective measurements augmented by unitary operations turn out to be completely equivalent to general measurements, as shown in Section 2.2.8. So a physicist trained in the use of projective measurements might ask to what end we start with the general formalism, Postulate 3? There are several reasons for doing so. First, mathematically general measurements are in some sense simpler than projective measurements, since they involve fewer restrictions on the measurement operators; there is, for example, no requirement for general measurements analogous to the condition PiPj = \u03b4ijPi for projective measurements. This simpler structure also gives rise to many useful properties for general measurements that are not possessed by projective measurements. Second, it turns out that there are important problems in quantum computation and quantum information \u2013 such as the optimal way to distinguish a set of quantum states \u2013 the answer to which involves a general measurement, rather than a projective measurement. A third reason for preferring Postulate 3 as a starting point is related to a property of projective measurements known as repeatability. Projective measurements are repeatable in the sense that if we perform a projective measurement once, and obtain the outcome m, repeating the measurement gives the outcome m again and does not change the state. To see this, suppose |\u03c8 was the initial state. After the \u2019 \ufb01rst measurement the state is |\u03c8m = \u03c8|Pm|\u03c8. Applying Pm to Pm|\u03c8 |\u03c8m does not change it, so we have \u03c8m|Pm|\u03c8m = 1, and therefore repeated measurement gives the result m each time, without changing the state. This repeatability of projective measurements tips us off to the fact that many important measurements in quantum mechanics are not projective measurements. For instance, if we use a silvered screen to measure the position of a photon we destroy the photon in the process. This certainly makes it impossible to repeat the measurement of the photon\u2019s position! Many other quantum measurements are also not repeatable in the same sense as a projective measurement. For such measurements, the general measurement postulate, Postulate 3, must be employed. Where do POVMs \ufb01t in this picture? POVMs are best viewed as a special case of the general measurement formalism, providing the simplest means by which one can study general measurement statistics, without the necessity for knowing the post-measurement state. They are a mathematical convenience that sometimes gives extra insight into quantum measurements. ( / 92 Introduction to quantum mechanics Exercise 2.62: Show that any measurement where the measurement operators and the POVM elements coincide is a projective measurement. m Em = I. Above we noticed that the POVM operators are positive and satisfy m Em = I. Suppose now that {Em} is some arbitrary set of positive operators such that We will show that there exists a set of measurement operators Mm de\ufb01ning a measurement described by the POVM {Em}. De\ufb01ning Mm \u2261 mMm = m Em = I, and therefore the set {Mm} describes a measurement with POVM {Em}. For this reason it is convenient to de\ufb01ne a POVM to be any set of operators {Em} such m Em = I is that: (a) each operator Em is positive; and (b) the completeness relation obeyed, expressing the fact that probabilities sum to one. To complete the description of POVMs, we note again that given a POVM {Em}, the probability of outcome m is given by p(m) = \u03c8|Em|\u03c8. Em we see that m M \u2020 \u221a We\u2019ve looked at projective measurements as an example of the use of POVMs, but it wasn\u2019", "t very exciting since we didn\u2019t learn much that was new. The following more sophisticated example illustrates the use of the POVM formalism as a guide for our intuition in quantum computation and quantum information. Suppose Alice gives Bob a qubit prepared in one of two states, |\u03c81 = |0 or |\u03c82 = (|0 + |1)/ 2. As explained in Section 2.2.4 it is impossible for Bob to determine whether he has been given |\u03c81 or |\u03c82 with perfect reliability. However, it is possible for him to perform a measurement which distinguishes the states some of the time, but never makes an error of mis-identi\ufb01cation. Consider a POVM containing three elements, \u221a \u221a E1 \u2261 E2 \u2261 2 \u221a 1 + \u221a 2 2 \u221a |11|, |0 \u2212 |1 1 + 2 E3 \u2261 I \u2212 E1 \u2212 E2. 0| \u2212 1| 2, (2.118) (2.119) (2.120) It is straightforward to verify that these are positive operators which satisfy the completeness relation m Em = I, and therefore form a legitimate POVM. Suppose Bob is given the state |\u03c81 = |0. He performs the measurement described by the POVM {E1, E2, E3}. There is zero probability that he will observe the result E1, since E1 has been cleverly chosen to ensure that \u03c81|E1|\u03c81 = 0. Therefore, if the result of his measurement is E1 then Bob can safely conclude that the state he received must have been |\u03c82. A similar line of reasoning shows that if the measurement outcome E2 occurs then it must have been the state |\u03c81 that Bob received. Some of the time, however, Bob will obtain the measurement outcome E3, and he can infer nothing about the identity of the state he was given. The key point, however, is that Bob never makes a mistake identifying the state he has been given. This infallibility comes at the price that sometimes Bob obtains no information about the identity of the state. This simple example demonstrates the utility of the POVM formalism as a simple and intuitive way of gaining insight into quantum measurements in instances where only the measurement statistics matter. In many instances later in the book we will only be concerned with measurement statistics, and will therefore use the POVM formalism rather than the more general formalism for measurements described in Postulate 3. Exercise 2.63: Suppose a measurement is described by measurement operators Mm. The postulates of quantum mechanics 93 Show that there exist unitary operators Um such that Mm = Um Em is the POVM associated to the measurement. \u221a Em, where Exercise 2.64: Suppose Bob is given a quantum state chosen from a set |\u03c81,..., |\u03c8m of linearly independent states. Construct a POVM {E1, E2,..., Em+1} such that if outcome Ei occurs, 1 \u2264 i \u2264 m, then Bob knows with certainty that he was given the state |\u03c8i. (The POVM must be such that \u03c8i|Ei|\u03c8i > 0 for each i.) 2.2.7 Phase \u2018Phase\u2019 is a commonly used term in quantum mechanics, with several different meanings dependent upon context. At this point it is convenient to review a couple of these meanings. Consider, for example, the state ei\u03b8|\u03c8, where |\u03c8 is a state vector, and \u03b8 is a real number. We say that the state ei\u03b8|\u03c8 is equal to |\u03c8, up to the global phase factor ei\u03b8. It is interesting to note that the statistics of measurement predicted for these two states are the same. To see this, suppose Mm is a measurement operator associated to some quantum measurement, and note that the respective probabilities for outcome m occurring are \u03c8|M \u2020 mMm|\u03c8. Therefore, from an observational point of view these two states are identical. For this reason we may ignore global phase factors as being irrelevant to the observed properties of the physical system. mMm|\u03c8 and \u03c8|e\u2212i\u03b8M \u2020 mMmei\u03b8|\u03c8 = \u03c8|M \u2020 There is another kind of phase known as the relative phase, which has quite a different meaning. Consider the states |0 + |1 \u221a 2 and \u221a |0 \u2212 |1 \u221a 2. (2.121) \u221a In the \ufb01rst state the amplitude of |1 is 1/ 2. For the second state the amplitude is \u22121/ 2. In each case the magnitude of the amplitudes is the same, but they differ in sign. More generally, we say that two amplitudes,", " a and b, differ by a relative phase if there is a real \u03b8 such that a = exp(i\u03b8)b. More generally still, two states are said to differ by a relative phase in some basis if each of the amplitudes in that basis is related by such a phase factor. For example, the two states displayed above are the same up to a relative phase shift because the |0 amplitudes are identical (a relative phase factor of 1), and the |1 amplitudes differ only by a relative phase factor of \u22121. The difference between relative phase factors and global phase factors is that for relative phase the phase factors may vary from amplitude to amplitude. This makes the relative phase a basis-dependent concept unlike global phase. As a result, states which differ only by relative phases in some basis give rise to physically observable differences in measurement statistics, and it is not possible to regard these states as physically equivalent, as we do with states differing by a global phase factor Exercise 2.65: Express the states (|0 + |1)/ \u221a 2 and (|0 \u2212 |1)/ \u221a 2 in a basis in which they are not the same up to a relative phase shift. 2.2.8 Composite systems Suppose we are interested in a composite quantum system made up of two (or more) distinct physical systems. How should we describe states of the composite system? The following postulate describes how the state space of a composite system is built up from the state spaces of the component systems. 94 Introduction to quantum mechanics Postulate 4: The state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Moreover, if we have systems numbered 1 through n, and system number i is prepared in the state |\u03c8i, then the joint state of the total system is |\u03c81 \u2297 |\u03c82 \u2297 \u00b7 \u00b7 \u00b7 \u2297 |\u03c8n. Why is the tensor product the mathematical structure used to describe the state space of a composite physical system? At one level, we can simply accept it as a basic postulate, not reducible to something more elementary, and move on. After all, we certainly expect that there be some canonical way of describing composite systems in quantum mechanics. Is there some other way we can arrive at this postulate? Here is one heuristic that is sometimes used. Physicists sometimes like to speak of the superposition principle of quantum mechanics, which states that if |x and |y are two states of a quantum system, then any superposition \u03b1|x + \u03b2|y should also be an allowed state of a quantum system, where |\u03b1|2 + |\u03b2|2 = 1. For composite systems, it seems natural that if |A is a state of system A, and |B is a state of system B, then there should be some corresponding state, which we might denote |A|B, of the joint system AB. Applying the superposition principle to product states of this form, we arrive at the tensor product postulate given above. This is not a derivation, since we are not taking the superposition principle as a fundamental part of our description of quantum mechanics, but it gives you the \ufb02avor of the various ways in which these ideas are sometimes reformulated. A variety of different notations for composite systems appear in the literature. Part of the reason for this proliferation is that different notations are better adapted for different applications, and we will also \ufb01nd it convenient to introduce some specialized notations on occasion. At this point it suf\ufb01ces to mention a useful subscript notation to denote states and operators on different systems, when it is not clear from context. For example, in a system containing three qubits, X2 is the Pauli \u03c3x operator acting on the second qubit. Exercise 2.66: Show that the average value of the observable X1Z2 for a two qubit \u221a system measured in the state (|00 + |11)/ 2 is zero. In Section 2.2.5 we claimed that projective measurements together with unitary dynamics are suf\ufb01cient to implement a general measurement. The proof of this statement makes use of composite quantum systems, and is a nice illustration of Postulate 4 in action. Suppose we have a quantum system with state space Q, and we want to perform a measurement described by measurement operators Mm on the system Q. To do this, we introduce an ancilla system, with state space M, having an orthonormal basis |m in one-to-one correspondence with the possible outcomes of the measurement we wish to implement. This ancilla system can be regarded as merely a mathematical device appearing in the construction, or it can be interpreted physically as an extra quantum system introduced into the problem, which we assume has a state space with the required properties. Letting |0 be any \ufb01xed state of", " M, de\ufb01ne an operator U on products |\u03c8|0 of states |\u03c8 from Q with the state |0 by U |\u03c8|0 \u2261 m Mm|\u03c8|m. Using the orthonormality of the states |m and the completeness relation (2.122) m M \u2020 mMm = The postulates of quantum mechanics 95 I, we can see that U preserves inner products between states of the form |\u03c8|0, \u03d5|0|U \u2020U |\u03c8|0 = = \u03d5|M \u2020 mMm|\u03c8 m|m m,m \u03d5|M \u2020 mMm|\u03c8 m = \u03d5|\u03c8. (2.123) (2.124) (2.125) By the results of Exercise 2.67 it follows that U can be extended to a unitary operator on the space Q \u2297 M, which we also denote by U. Exercise 2.67: Suppose V is a Hilbert space with a subspace W. Suppose U : W \u2192 V is a linear operator which preserves inner products, that is, for any |w1 and |w2 in W, w1|U \u2020U |w2 = w1|w2. Prove that there exists a unitary operator U : V \u2192 V which extends U. That is, U |w = U |w for all |w in W, but U is de\ufb01ned on the entire space V. Usually we omit the prime symbol and just write U to denote the extension. (2.126) Next, suppose we perform a projective measurement on the two systems described by projectors Pm \u2261 IQ \u2297 |mm|. Outcome m occurs with probability p(m) = \u03c8|0|U \u2020PmU |\u03c8|0 = m,m = \u03c8|M \u2020 mMm|\u03c8, \u03c8|M \u2020 mm|(IQ \u2297 |mm|)Mm|\u03c8|m (2.127) (2.128) (2.129) just as given in Postulate 3. The joint state of the system QM after measurement, conditional on result m occurring, is given by PmU |\u03c8|0 \u03c8|U \u2020PmU |\u03c8 = Mm|\u03c8|m \u03c8|M \u2020 mMm|\u03c8. (2.130) It follows that the state of system M after the measurement is |m, and the state of system Q is Mm|\u03c8, \u03c8|M \u2020 mMm|\u03c8 (2.131) just as prescribed by Postulate 3. Thus unitary dynamics, projective measurements, and the ability to introduce ancillary systems, together allow any measurement of the form described in Postulate 3 to be realized. Postulate 4 also enables us to de\ufb01ne one of the most interesting and puzzling ideas associated with composite quantum systems \u2013 entanglement. Consider the two qubit state |\u03c8 = |00 + |11 \u221a 2. (2.132) This state has the remarkable property that there are no single qubit states |a and |b such that |\u03c8 = |a|b, a fact which you should now convince yourself of: 96 Introduction to quantum mechanics Exercise 2.68: Prove that |\u03c8 = |a|b for all single qubit states |a and |b. We say that a state of a composite system having this property (that it can\u2019t be written as a product of states of its component systems) is an entangled state. For reasons which nobody fully understands, entangled states play a crucial role in quantum computation and quantum information, and arise repeatedly through the remainder of this book. We have already seen entanglement play a crucial role in quantum teleportation, as described in Section 1.3.7. In this chapter we give two examples of the strange effects enabled by entangled quantum states, superdense coding (Section 2.3), and the violation of Bell\u2019s inequality (Section 2.6). 2.2.9 Quantum mechanics: a global view We have now explained all the fundamental postulates of quantum mechanics. Most of the rest of the book is taken up with deriving consequences of these postulates. Let\u2019s quickly review the postulates and try to place them in some kind of global perspective. Postulate 1 sets the arena for quantum mechanics, by specifying how the state of an isolated quantum system is to be described. Postulate 2 tells us that the dynamics of closed quantum systems are described by the Schr\u00a8odinger equation, and thus by unitary evolution. Postulate 3 tells us how to extract information from our quantum systems by giving a prescription for the description of measurement. Postulate 4 tells us how the state spaces of different quantum", " systems may be combined to give a description of the composite system. What\u2019s odd about quantum mechanics, at least by our classical lights, is that we can\u2019t directly observe the state vector. It\u2019s a little bit like a game of chess where you can never \ufb01nd out exactly where each piece is, but only know the rank of the board they are on. Classical physics \u2013 and our intuition \u2013 tells us that the fundamental properties of an object, like energy, position, and velocity, are directly accessible to observation. In quantum mechanics these quantities no longer appear as fundamental, being replaced by the state vector, which can\u2019t be directly observed. It is as though there is a hidden world in quantum mechanics, which we can only indirectly and imperfectly access. Moreover, merely observing a classical system does not necessarily change the state of the system. Imagine how dif\ufb01cult it would be to play tennis if each time you looked at the ball its position changed! But according to Postulate 3, observation in quantum mechanics is an invasive procedure that typically changes the state of the system. What conclusions should we draw from these strange features of quantum mechanics? Might it be possible to reformulate quantum mechanics in a mathematically equivalent way so that it had a structure more like classical physics? In Section 2.6 we\u2019ll prove Bell\u2019s inequality, a surprising result that shows any attempt at such a reformulation is doomed to failure. We\u2019re stuck with the counter-intuitive nature of quantum mechanics. Of course, the proper reaction to this is glee, not sorrow! It gives us an opportunity to develop tools of thought that make quantum mechanics intuitive. Moreover, we can exploit the hidden nature of the state vector to do information processing tasks beyond what is possible in the classical world. Without this counter-intuitive behavior, quantum computation and quantum information would be a lot less interesting. We can also turn this discussion about, and ask ourselves: \u2018If quantum mechanics is so different from classical physics, then how come the everyday world looks so classical?\u2019 Why do we see no evidence of a hidden state vector in our everyday lives? It turns out Application: superdense coding 97 that the classical world we see can be derived from quantum mechanics as an approximate description of the world that will be valid on the sort of time, length and mass scales we commonly encounter in our everyday lives. Explaining the details of how quantum mechanics gives rise to classical physics is beyond the scope of this book, but the interested reader should check out the discussion of this topic in \u2018History and further reading\u2019at the end of Chapter 8. 2.3 Application: superdense coding Superdense coding is a simple yet surprising application of elementary quantum mechanics. It combines in a concrete, non-trivial way all the basic ideas of elementary quantum mechanics, as covered in the previous sections, and is therefore an ideal example of the information processing tasks that can be accomplished using quantum mechanics. Superdense coding involves two parties, conventionally known as \u2018Alice\u2019 and \u2018Bob\u2019, who are a long way away from one another. Their goal is to transmit some classical information from Alice to Bob. Suppose Alice is in possession of two classical bits of information which she wishes to send Bob, but is only allowed to send a single qubit to Bob. Can she achieve her goal? Superdense coding tells us that the answer to this question is yes. Suppose Alice and Bob initially share a pair of qubits in the entangled state |\u03c8 = |00 + |11 \u221a 2. (2.133) Alice is initially in possession of the \ufb01rst qubit, while Bob has possession of the second qubit, as illustrated in Figure 2.3. Note that |\u03c8 is a \ufb01xed state; there is no need for Alice to have sent Bob any qubits in order to prepare this state. Instead, some third party may prepare the entangled state ahead of time, sending one of the qubits to Alice, and the other to Bob. Figure 2.3. The initial setup for superdense coding, with Alice and Bob each in possession of one half of an entangled pair of qubits. Alice can use superdense coding to transmit two classical bits of information to Bob, using only a single qubit of communication and this preshared entanglement. By sending the single qubit in her possession to Bob, it turns out that Alice can communicate two bits of classical information to Bob. Here is the procedure she uses. If she wishes to send the bit string \u201800\u2019 to Bob then she does nothing at all to her qubit. If she wishes to send \u201801\u2019 then she applies the phase \ufb02ip Z to her qubit. If she wishes to gate, X, to her qubit. If she wishes to send send \u201810\u2019 then she applies the quantum", " \u201811\u2019 then she applies the iY gate to her qubit. The four resulting states are easily seen 98 Introduction to quantum mechanics to be: 00 : |\u03c8 \u2192 01 : |\u03c8 \u2192 10 : |\u03c8 \u2192 11 : |\u03c8 \u2192 |00 + |11 \u221a 2 |00 \u2212 |11 \u221a 2 |10 + |01 \u221a 2 |01 \u2212 |10 \u221a 2. (2.134) (2.135) (2.136) (2.137) As we noted in Section 1.3.6, these four states are known as the Bell basis, Bell states, or EPR pairs, in honor of several of the pioneers who \ufb01rst appreciated the novelty of entanglement. Notice that the Bell states form an orthonormal basis, and can therefore be distinguished by an appropriate quantum measurement. If Alice sends her qubit to Bob, giving Bob possession of both qubits, then by doing a measurement in the Bell basis Bob can determine which of the four possible bit strings Alice sent. Summarizing, Alice, interacting with only a single qubit, is able to transmit two bits of information to Bob. Of course, two qubits are involved in the protocol, but Alice never need interact with the second qubit. Classically, the task Alice accomplishes would have been impossible had she only transmitted a single classical bit, as we will show in Chapter 12. Furthermore, this remarkable superdense coding protocol has received partial veri\ufb01cation in the laboratory. (See \u2018History and further reading\u2019 for references to the experimental veri\ufb01cation.) In later chapters we will see many other examples, some of them much more spectacular than superdense coding, of quantum mechanics being harnessed to perform information processing tasks. However, a key point can already be seen in this beautiful example: information is physical, and surprising physical theories such as quantum mechanics may predict surprising information processing abilities. Exercise 2.69: Verify that the Bell basis forms an orthonormal basis for the two qubit state space. Exercise 2.70: Suppose E is any positive operator acting on Alice\u2019s qubit. Show that \u03c8|E \u2297 I|\u03c8 takes the same value when |\u03c8 is any of the four Bell states. Suppose some malevolent third party (\u2018Eve\u2019) intercepts Alice\u2019s qubit on the way to Bob in the superdense coding protocol. Can Eve infer anything about which of the four possible bit strings 00, 01, 10, 11 Alice is trying to send? If so, how, or if not, why not? 2.4 The density operator We have formulated quantum mechanics using the language of state vectors. An alternate formulation is possible using a tool known as the density operator or density matrix. This alternate formulation is mathematically equivalent to the state vector approach, but it provides a much more convenient language for thinking about some commonly encountered scenarios in quantum mechanics. The next three sections describe the density operator formulation of quantum mechanics. Section 2.4.1 introduces the density operator using the concept of an ensemble of quantum states. Section 2.4.2 develops some general The density operator 99 properties of the density operator. Finally, Section 2.4.3 describes an application where the density operator really shines \u2013 as a tool for the description of individual subsystems of a composite quantum system. 2.4.1 Ensembles of quantum states The density operator language provides a convenient means for describing quantum systems whose state is not completely known. More precisely, suppose a quantum system is in one of a number of states |\u03c8i, where i is an index, with respective probabilities pi. We shall call {pi, |\u03c8i} an ensemble of pure states. The density operator for the system is de\ufb01ned by the equation \u03c1 \u2261 pi|\u03c8i\u03c8i|. (2.138) i The density operator is often known as the density matrix; we will use the two terms interchangeably. It turns out that all the postulates of quantum mechanics can be reformulated in terms of the density operator language. The purpose of this section and the next is to explain how to perform this reformulation, and explain when it is useful. Whether one uses the density operator language or the state vector language is a matter of taste, since both give the same results; however it is sometimes much easier to approach problems from one point of view rather than the other. Suppose, for example, that the evolution of a closed quantum system is described by the unitary operator U. If the system was initially in the state |\u03c8i with probability pi then after the evolution has occurred the system will be in the state U |\u03c8i with probability pi. Thus, the evolution of the density operator is described by the equation \u03c1 = i pi|\u03c8i\u03c8i| U\u2212\u2192 i piU", " |\u03c8i\u03c8i|U \u2020 = U \u03c1U \u2020. (2.139) Measurements are also easily described in the density operator language. Suppose we perform a measurement described by measurement operators Mm. If the initial state was |\u03c8i, then the probability of getting result m is p(m|i) = \u03c8i|M \u2020 mMm|\u03c8i = tr(M \u2020 mMm|\u03c8i\u03c8i|), (2.140) where we have used Equation (2.61) to obtain the last equality. By the law of total probability (see Appendix 1 for an explanation of this and other elementary notions of probability theory) the probability of obtaining result m is p(m) = p(m|i)pi i = pitr(M \u2020 mMm|\u03c8i\u03c8i|) i = tr(M \u2020 mMm\u03c1). (2.141) (2.142) (2.143) What is the density operator of the system after obtaining the measurement result m? If the initial state was |\u03c8i then the state after obtaining the result m is |\u03c8m i = Mm|\u03c8i. \u03c8i|M \u2020 mMm|\u03c8i (2.144) 100 Introduction to quantum mechanics Thus, after a measurement which yields the result m we have an ensemble of states |\u03c8m i with respective probabilities p(i|m). The corresponding density operator \u03c1m is therefore \u03c1m = i p(i|m)|\u03c8m i \u03c8m i | = i p(i|m) Mm|\u03c8i\u03c8i|M \u2020 m \u03c8i|M \u2020 mMm|\u03c8i. (2.145) But by elementary probability theory, p(i|m) = p(m, i)/p(m) = p(m|i)pi/p(m). Substituting from (2.143) and (2.140) we obtain pi \u03c1m = Mm|\u03c8i\u03c8i|M \u2020 m mMm\u03c1) tr(M \u2020 i Mm\u03c1M \u2020 m tr(M \u2020 mMm\u03c1). = (2.146) (2.147) What we have shown is that the basic postulates of quantum mechanics related to unitary evolution and measurement can be rephrased in the language of density operators. In the next section we complete this rephrasing by giving an intrinsic characterization of the density operator that does not rely on the idea of a state vector. Before doing so, however, it is useful to introduce some more language, and one more fact about the density operator. First, the language. A quantum system whose state |\u03c8 is known exactly is said to be in a pure state. In this case the density operator is simply \u03c1 = |\u03c8\u03c8|. Otherwise, \u03c1 is in a mixed state; it is said to be a mixture of the different pure states in the ensemble for \u03c1. In the exercises you will be asked to demonstrate a simple criterion for determining whether a state is pure or mixed: a pure state satis\ufb01es tr(\u03c12) = 1, while a mixed state satis\ufb01es tr(\u03c12) < 1. A few words of warning about the nomenclature: sometimes people use the term \u2018mixed state\u2019 as a catch-all to include both pure and mixed quantum states. The origin for this usage seems to be that it implies that the writer is not necessarily assuming that a state is pure. Second, the term \u2018pure state\u2019 is often used in reference to a state vector |\u03c8, to distinguish it from a density operator \u03c1. Finally, imagine a quantum system is prepared in the state \u03c1i with probability pi. It is not dif\ufb01cult to convince yourself that the system may be described by the density matrix i pi\u03c1i. A proof of this is to suppose that \u03c1i arises from some ensemble {pij, |\u03c8ij} (note that i is \ufb01xed) of pure states, so the probability for being in the state |\u03c8ij is pipij. The density matrix for the system is thus \u03c1 = ij = i pipij|\u03c8ij\u03c8ij| pi\u03c1i, (2.148) (2.149) j pij|\u03c8ij\u03c8ij|. We say that \u03c1 is a mixture where we have used the de\ufb01nition \u03c1i = of the states \u03c1i with probabilities pi. This concept of a mixture comes up repeatedly in the analysis of problems like quantum noise, where the effect of the noise is to introduce ignorance into our knowledge of the quantum state.", " A simple example is provided by the measurement scenario described above. Imagine that, for some reason, our record of the result m of the measurement was lost. We would have a quantum system in the state \u03c1m with probability p(m), but would no longer know the actual value of m. The state of The density operator 101 such a quantum system would therefore be described by the density operator \u03c1 = p(m)\u03c1m m m m = = tr(M \u2020 mMm\u03c1) Mm\u03c1M \u2020 m tr(M \u2020 mMm\u03c1) Mm\u03c1M \u2020 m, (2.150) (2.151) (2.152) a nice compact formula which may be used as the starting point for analysis of further operations on the system. 2.4.2 General properties of the density operator The density operator was introduced as a means of describing ensembles of quantum states. In this section we move away from this description to develop an intrinsic characterization of density operators that does not rely on an ensemble interpretation. This allows us to complete the program of giving a description of quantum mechanics that does not take as its foundation the state vector. We also take the opportunity to develop numerous other elementary properties of the density operator. The class of operators that are density operators are characterized by the following useful theorem: Theorem 2.5: (Characterization of density operators) An operator \u03c1 is the density operator associated to some ensemble {pi, |\u03c8i} if and only if it satis\ufb01es the conditions: (1) (Trace condition) \u03c1 has trace equal to one. (2) (Positivity condition) \u03c1 is a positive operator. Proof Suppose \u03c1 = i pi|\u03c8i\u03c8i| is a density operator. Then tr(\u03c1) = pitr(|\u03c8i\u03c8i|) = pi = 1, (2.153) i i so the trace condition tr(\u03c1) = 1 is satis\ufb01ed. Suppose |\u03d5 is an arbitrary vector in state space. Then \u03d5|\u03c1|\u03d5 = pi\u03d5|\u03c8i\u03c8i|\u03d5 pi|\u03d5|\u03c8i|2 (2.154) (2.155) (2.156) so the positivity condition is satis\ufb01ed. Conversely, suppose \u03c1 is any operator satisfying the trace and positivity conditions. Since \u03c1 is positive, it must have a spectral decomposition where the vectors |j are orthogonal, and \u03bbj are real, non-negative eigenvalues of \u03c1. j \u03c1 = \u03bbj|jj|, (2.157) i = i \u2265 0, 102 Introduction to quantum mechanics j \u03bbj = 1. Therefore, a system in state |j with From the trace condition we see that probability \u03bbj will have density operator \u03c1. That is, the ensemble {\u03bbj, |j} is an ensemble of states giving rise to the density operator \u03c1. This theorem provides a characterization of density operators that is intrinsic to the operator itself: we can de\ufb01ne a density operator to be a positive operator \u03c1 which has trace equal to one. Making this de\ufb01nition allows us to reformulate the postulates of quantum mechanics in the density operator picture. For ease of reference we state all the reformulated postulates here: Postulate 1: Associated to any isolated physical system is a complex vector space with inner product (that is, a Hilbert space) known as the state space of the system. The system is completely described by its density operator, which is a positive operator \u03c1 with trace one, acting on the state space of the system. If a quantum system is in the state \u03c1i with probability pi, then the density operator for the system is i pi\u03c1i. Postulate 2: The evolution of a closed quantum system is described by a unitary transformation. That is, the state \u03c1 of the system at time t1 is related to the state \u03c1 of the system at time t2 by a unitary operator U which depends only on the times t1 and t2, \u03c1 = U \u03c1U \u2020. (2.158) Postulate 3: Quantum measurements are described by a collection {Mm} of measurement operators. These are operators acting on the state space of the system being measured. The index m refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is \u03c1 immediately before the measurement then the probability that result m occurs is given by and the state of the system after the measurement is p(m) = tr(M \u2020 mMm\u03c1), Mm\u03c1M \u2020 m tr(M \u2020 mMm\u03c1). The measurement operators satisfy the completeness equation, M \u2020 mMm = I. (2.159) (", "2.160) (2.161) m Postulate 4: The state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Moreover, if we have systems numbered 1 through n, and system number i is prepared in the state \u03c1i, then the joint state of the total system is \u03c11 \u2297 \u03c12 \u2297... \u03c1n. These reformulations of the fundamental postulates of quantum mechanics in terms of the density operator are, of course, mathematically equivalent to the description in terms of the state vector. Nevertheless, as a way of thinking about quantum mechanics, the density operator approach really shines for two applications: the description of quantum systems whose state is not known, and the description of subsystems of a composite The density operator 103 quantum system, as will be described in the next section. For the remainder of this section we \ufb02esh out the properties of the density matrix in more detail. Exercise 2.71: (Criterion to decide if a state is mixed or pure) Let \u03c1 be a density operator. Show that tr(\u03c12) \u2264 1, with equality if and only if \u03c1 is a pure state. It is a tempting (and surprisingly common) fallacy to suppose that the eigenvalues and eigenvectors of a density matrix have some special signi\ufb01cance with regard to the ensemble of quantum states represented by that density matrix. For example, one might suppose that a quantum system with density matrix \u03c1 = 3 4 |00| + |11|. 1 4 (2.162) must be in the state |0 with probability 3/4 and in the state |1 with probability 1/4. However, this is not necessarily the case. Suppose we de\ufb01ne ) ) |a \u2261 |b \u2261 ) |0 + |1 |1, (2.163) (2.164) and the quantum system is prepared in the state |a with probability 1/2 and in the state |b with probability 1/2. Then it is easily checked that the corresponding density matrix is \u03c1 = 1 2 |aa| + |bb| = 1 2 3 4 |00| + |11|. 1 4 (2.165) That is, these two different ensembles of quantum states give rise to the same density matrix. In general, the eigenvectors and eigenvalues of a density matrix just indicate one of many possible ensembles that may give rise to a speci\ufb01c density matrix, and there is no reason to suppose it is an especially privileged ensemble. A natural question to ask in the light of this discussion is what class of ensembles does give rise to a particular density matrix? The solution to this problem, which we now give, has surprisingly many applications in quantum computation and quantum information, notably in the understanding of quantum noise and quantum error-correction (Chapters 8 and 10). For the solution it is convenient to make use of vectors | \u02dc\u03c8i which may not be i | \u02dc\u03c8i \u02dc\u03c8i|, normalized to unit length. We say the set | \u02dc\u03c8i generates the operator \u03c1 \u2261 and thus the connection to the usual ensemble picture of density operators is expressed pi|\u03c8i. When do two sets of vectors, | \u02dc\u03c8i and | \u02dc\u03d5j generate the by the equation | \u02dc\u03c8i = same operator \u03c1? The solution to this problem will enable us to answer the question of what ensembles give rise to a given density matrix. \u221a Theorem 2.6: (Unitary freedom in the ensemble for density matrices) The sets | \u02dc\u03c8i and | \u02dc\u03d5j generate the same density matrix if and only if | \u02dc\u03c8i = j uij| \u02dc\u03d5j, (2.166) where uij is a unitary matrix of complex numbers, with indices i and j, and we 104 Introduction to quantum mechanics \u2018pad\u2019 whichever set of vectors | \u02dc\u03c8i or | \u02dc\u03d5j is smaller with additional vectors 0 so that the two sets have the same number of elements. As a consequence of the theorem, note that \u03c1 = i pi|\u03c8i\u03c8i| = j qj|\u03d5j\u03d5j| for normalized states |\u03c8i, |\u03d5j and probability distributions pi and qj if and only if \u221a pi|\u03c8i = \u221a qj|\u03d5j, uij (2.167) j for some unitary matrix uij, and we may pad the smaller ensemble with entries having probability zero in order to make the two ensembles the same size. Thus, Theorem 2.6 characterizes the freedom in ensembles {pi, |\u03c8", "i} giving rise to a given density matrix \u03c1. Indeed, it is easily checked that our earlier example of a density matrix with two different decompositions, (2.162), arises as a special case of this general result. Let\u2019s turn now to the proof of the theorem. Proof Suppose | \u02dc\u03c8i = j uij| \u02dc\u03d5j for some unitary uij. Then | \u02dc\u03c8i \u02dc\u03c8i| = uiju\u2217 i ik| \u02dc\u03d5j \u02dc\u03d5k| ijk u\u2020 kiuij | \u02dc\u03d5j \u02dc\u03d5k| = = = jk jk j i \u03b4kj| \u02dc\u03d5j \u02dc\u03d5k| | \u02dc\u03d5j \u02dc\u03d5j|, which shows that | \u02dc\u03c8i and | \u02dc\u03d5j generate the same operator. Conversely, suppose A = i | \u02dc\u03c8i \u02dc\u03c8i| = j | \u02dc\u03d5j \u02dc\u03d5j|. (2.168) (2.169) (2.170) (2.171) (2.172) Let A = k \u03bbk|kk| be a decomposition for A such that the states |k are orthonormal, and the \u03bbk are strictly positive. Our strategy is to relate the states | \u02dc\u03c8i to the states \u221a \u03bbk|k, and similarly relate the states | \u02dc\u03d5j to the states |\u02dck. Combining the two | \u02dck \u2261 relations will give the result. Let |\u03c8 be any vector orthonormal to the space spanned by the | \u02dck, so \u03c8| \u02dck \u02dck|\u03c8 = 0 for all k, and thus we see that 0 = \u03c8|A|\u03c8 = \u03c8| \u02dc\u03c8i \u02dc\u03c8i|\u03c8 = |\u03c8| \u02dc\u03c8i|2. (2.173) i i Thus \u03c8| \u02dc\u03c8i = 0 for all i and all |\u03c8 orthonormal to the space spanned by the | \u02dck. It follows that each | \u02dc\u03c8i can be expressed as a linear combination of the | \u02dck, | \u02dc\u03c8i = k cik| \u02dck. Since A = k | \u02dck \u02dck| = i | \u02dc\u03c8i \u02dc\u03c8i| we see that | \u02dck \u02dck| = cikc\u2217 il | \u02dck\u02dcl|. (2.174) The operators |\u02dck\u02dcl| are easily seen to be linearly independent, and thus it must be that k kl i The density operator 105 i cikc\u2217 il = \u03b4kl. This ensures that we may append extra columns to c to obtain a unitary k vik| \u02dck, where we have appended zero vectors to the list k wjk| \u02dck. Thus matrix v such that | \u02dc\u03c8i = of | \u02dck. Similarly, we can \ufb01nd a unitary matrix w such that | \u02dc\u03d5j = | \u02dc\u03c8i = j uij| \u02dc\u03d5j, where u = vw\u2020 is unitary. Exercise 2.72: (Bloch sphere for mixed states) The Bloch sphere picture for pure states of a single qubit was introduced in Section 1.2. This description has an important generalization to mixed states as follows. (1) Show that an arbitrary density matrix for a mixed state qubit may be written as 2.175) where r is a real three-dimensional vector such that r \u2264 1. This vector is known as the Bloch vector for the state \u03c1. (2) What is the Bloch vector representation for the state \u03c1 = I/2? (3) Show that a state \u03c1 is pure if and only if r = 1. (4) Show that for pure states the description of the Bloch vector we have given coincides with that in Section 1.2. Exercise 2.73: Let \u03c1 be a density operator. A minimal ensemble for \u03c1 is an ensemble {pi, |\u03c8i} containing a number of elements equal to the rank of \u03c1. Let |\u03c8 be any state in the support of \u03c1. (The support of a Hermitian operator A is the vector space spanned by the eigenvectors of A with non-zero eigenvalues.) Show that there is a minimal ensemble for \u03c1 that contains |\u03c8, and moreover that in any such ensemble |\u03c8 must appear with probability pi = 1 \u03c8i|\u03c1\u22121|\u03c8i,", " (2.176) where \u03c1\u22121 is de\ufb01ned to be the inverse of \u03c1, when \u03c1 is considered as an operator acting only on the support of \u03c1. (This de\ufb01nition removes the problem that \u03c1 may not have an inverse.) 2.4.3 The reduced density operator Perhaps the deepest application of the density operator is as a descriptive tool for subsystems of a composite quantum system. Such a description is provided by the reduced density operator, which is the subject of this section. The reduced density operator is so useful as to be virtually indispensable in the analysis of composite quantum systems. Suppose we have physical systems A and B, whose state is described by a density operator \u03c1AB. The reduced density operator for system A is de\ufb01ned by \u03c1A \u2261 trB(\u03c1AB), where trB is a map of operators known as the partial trace over system B. The partial trace is de\ufb01ned by (2.177) |a1a2| \u2297 |b1b2| trB (2.178) where |a1 and |a2 are any two vectors in the state space of A, and |b1 and |b2 are any two vectors in the state space of B. The trace operation appearing on the right hand side \u2261 |a1a2| tr(|b1b2|), 106 Introduction to quantum mechanics is the usual trace operation for system B, so tr(|b1b2|) = b2|b1. We have de\ufb01ned the partial trace operation only on a special subclass of operators on AB; the speci\ufb01cation is completed by requiring in addition to Equation (2.178) that the partial trace be linear in its input. It is not obvious that the reduced density operator for system A is in any sense a description for the state of system A. The physical justi\ufb01cation for making this identi\ufb01cation is that the reduced density operator provides the correct measurement statistics for measurements made on system A. This is explained in more detail in Box 2.6 on page 107. The following simple example calculations may also help understand the reduced density operator. First, suppose a quantum system is in the product state \u03c1AB = \u03c1 \u2297 \u03c3, where \u03c1 is a density operator for system A, and \u03c3 is a density operator for system B. Then \u03c1A = trB(\u03c1 \u2297 \u03c3) = \u03c1 tr(\u03c3) = \u03c1, (2.184) which is the result we intuitively expect. Similarly, \u03c1B = \u03c3 for this state. A less trivial example is the Bell state (|00 + |11)/ |00 + |11 \u221a 2 2. This has density operator 00| + 11| \u221a 2 (2.185) \u03c1 = \u221a = |0000| + |1100| + |0011| + |1111| 2. (2.186) Tracing out the second qubit, we \ufb01nd the reduced density operator of the \ufb01rst qubit, \u03c11 = tr2(\u03c1) = = = = tr2(|0000|) + tr2(|1100|) + tr2(|0011|) + tr2(|1111|) 2 |00|0|0 + |10|0|1 + |01|1|0 + |11|1|1 2 |00| + |11| 2 I 2. (2.187) (2.188) (2.189) (2.190) (2.191) Notice that this state is a mixed state, since tr((I/2)2) = 1/2 < 1. This is quite a remarkable result. The state of the joint system of two qubits is a pure state, that is, it is known exactly; however, the \ufb01rst qubit is in a mixed state, that is, a state about which we apparently do not have maximal knowledge. This strange property, that the joint state of a system can be completely known, yet a subsystem be in mixed states, is another hallmark of quantum entanglement. Exercise 2.74: Suppose a composite of systems A and B is in the state |a|b, where |a is a pure state of system A, and |b is a pure state of system B. Show that the reduced density operator of system A alone is a pure state. Exercise 2.75: For each of the four Bell states, \ufb01nd the reduced density operator for each qubit. Quantum teleportation and the reduced density operator A useful application of the reduced density operator is to the analysis of quantum teleportation. Recall from Section 1.3.7 that quantum teleportation is a procedure for sending The density operator 107 Box 2.6: Why", " the partial trace? Why is the partial trace used to describe part of a larger quantum system? The reason for doing this is because the partial trace operation is the unique operation which gives rise to the correct description of observable quantities for subsystems of a composite system, in the following sense. Suppose M is any observable on system A, and we have some measuring device \u02dcM denote the corresponding which is capable of realizing measurements of M. Let observable for the same measurement, performed on the composite system AB. \u02dcM is necessarily equal to M \u2297 IB. Note that Our immediate goal is to argue that if the system AB is prepared in the state |m|\u03c8, where |m is an eigenstate of M with eigenvalue m, and |\u03c8 is any state of B, then the measuring device must yield the result m for the measurement, with probability one. Thus, if Pm is the projector onto the m eigenspace of the observable M, then the corresponding projector for \u02dcM is Pm \u2297 IB. We therefore have \u02dcM = mPm \u2297 IB = M \u2297 IB. (2.179) m The next step is to show that the partial trace procedure gives the correct measurement statistics for observations on part of a system. Suppose we perform a measurement on system A described by the observable M. Physical consistency requires that any prescription for associating a \u2018state\u2019, \u03c1A, to system A, must have the property that measurement averages be the same whether computed via \u03c1A or \u03c1AB, tr(M \u03c1A) = tr( \u02dcM \u03c1AB) = tr((M \u2297 IB)\u03c1AB). This equation is certainly satis\ufb01ed if we choose \u03c1A \u2261 trB(\u03c1AB). In fact, the partial trace turns out to be the unique function having this property. To see this uniqueness property, let f (\u00b7) be any map of density operators on AB to density operators on A such that (2.180) tr(M f (\u03c1AB)) = tr((M \u2297 IB)\u03c1AB), (2.181) for all observables M. Let Mi be an orthonormal basis of operators for the space of Hermitian operators with respect to the Hilbert\u2013Schmidt inner product (X, Y ) \u2261 tr(XY ) (compare Exercise 2.39 on page 76). Then expanding f (\u03c1AB) in this basis gives f (\u03c1AB) = i = i Mitr(Mif (\u03c1AB)) Mitr((Mi \u2297 IB)\u03c1AB). (2.182) (2.183) It follows that f is uniquely determined by Equation (2.180). Moreover, the partial trace satis\ufb01es (2.180), so it is the unique function having this property. quantum information from Alice to Bob, given that Alice and Bob share an EPR pair, and have a classical communications channel. 108 Introduction to quantum mechanics At \ufb01rst sight it appears as though teleportation can be used to do faster than light communication, a big no-no according to the theory of relativity. We surmised in Section 1.3.7 that what prevents faster than light communication is the need for Alice to communicate her measurement result to Bob. The reduced density operator allows us to make this rigorous. Recall that immediately before Alice makes her measurement the quantum state of the three qubits is (Equation (1.32)): |\u03c82 = |00 1 2 +|10 \u03b1|0 + \u03b2|1 \u03b1|0 \u2212 \u03b2|1 + |01 + |11 \u03b1|1 + \u03b2|0 \u03b1|1 \u2212 \u03b2|0. (2.192) Measuring in Alice\u2019s computational basis, the state of the system after the measurement is: \u03b1|0 + \u03b2|1 \u03b1|1 + \u03b2|0 \u03b1|0 \u2212 \u03b2|1 \u03b1|1 \u2212 \u03b2|0 |00 |01 |10 |11 with probability with probability with probability with probability 2.193) (2.194) (2.195) (2.196) The density operator of the system is thus \u03c1 = 1 |0000|(\u03b1|0 + \u03b2|1)(\u03b1\u22170| + \u03b2\u22171|) + |0101|(\u03b1|1 + \u03b2|0)(\u03b1\u22171| + \u03b2\u22170|) 4 +|1010|(\u03b1|0 \u2212 \u03b2|1)(\u03b1\u22170| \u2212 \u03b2\u22171|) + |1111|(\u03b1|1 \u2212 \u03b2|0)(\u03b1\u22171| \u2212 \u03b2\u22170|). (2.197) Tracing out Alice\u2019s system, we see that the reduced density operator of Bob\u2019s system is \u03c1B = = = = 1 (\u03b1|0 + \u03b2", "|1)(\u03b1\u22170| + \u03b2\u22171|) + (\u03b1|1 + \u03b2|0)(\u03b1\u22171| + \u03b2\u22170|) 4 +(\u03b1|0 \u2212 \u03b2|1)(\u03b1\u22170| \u2212 \u03b2\u22171|) + (\u03b1|1 \u2212 \u03b2|0)(\u03b1\u22171| \u2212 \u03b2\u22170|) 2(|\u03b1|2 + |\u03b2|2)|00| + 2(|\u03b1|2 + |\u03b2|2)|11| 4 |00| + |11| 2 I 2, (2.198) (2.199) (2.200) (2.201) where we have used the completeness relation in the last line. Thus, the state of Bob\u2019s system after Alice has performed the measurement but before Bob has learned the measurement result is I/2. This state has no dependence upon the state |\u03c8 being teleported, and thus any measurements performed by Bob will contain no information about |\u03c8, thus preventing Alice from using teleportation to transmit information to Bob faster than light. The Schmidt decomposition and puri\ufb01cations 109 2.5 The Schmidt decomposition and puri\ufb01cations Density operators and the partial trace are just the beginning of a wide array of tools useful for the study of composite quantum systems, which are at the heart of quantum computation and quantum information. Two additional tools of great value are the Schmidt decomposition and puri\ufb01cations. In this section we present both these tools, and try to give the \ufb02avor of their power. Theorem 2.7: (Schmidt decomposition) Suppose |\u03c8 is a pure state of a composite system, AB. Then there exist orthonormal states |iA for system A, and orthonormal states |iB of system B such that |\u03c8 = i \u03bbi|iA|iB, where \u03bbi are non-negative real numbers satisfying co-ef\ufb01cients. (2.202) i \u03bb2 i = 1 known as Schmidt i \u03bb2 i|iAiA| and \u03c1B = This result is very useful. As a taste of its power, consider the following consequence: let |\u03c8 be a pure state of a composite system, AB. Then by the Schmidt decomposition \u03c1A = i|iBiB|, so the eigenvalues of \u03c1A and \u03c1B are i \u03bb2 identical, namely \u03bb2 i for both density operators. Many important properties of quantum systems are completely determined by the eigenvalues of the reduced density operator of the system, so for a pure state of a composite system such properties will be the same for both systems. As an example, consider the state of two qubits, (|00 + |01 + |11)/ 3. (\u03c1B)2 This has no obvious symmetry property, yet if you calculate tr you will discover that they have the same value, 7/9 in each case. This is but one small consequence of the Schmidt decomposition. (\u03c1A)2 and tr \u221a Proof We give the proof for the case where systems A and B have state spaces of the same dimension, and leave the general case to Exercise 2.76. Let |j and |k be any \ufb01xed orthonormal bases for systems A and B, respectively. Then |\u03c8 can be written |\u03c8 = jk ajk|j|k, (2.203) for some matrix a of complex numbers ajk. By the singular value decomposition, a = udv, where d is a diagonal matrix with non-negative elements, and u and v are unitary matrices. Thus |\u03c8 = ujidiivik|j|k. (2.204) ijk De\ufb01ning |iA \u2261 j uji|j, |iB \u2261 k vik|k, and \u03bbi \u2261 dii, we see that this gives |\u03c8 = \u03bbi|iA|iB. (2.205) It is easy to check that |iA forms an orthonormal set, from the unitarity of u and the orthonormality of |j, and similarly that the |iB form an orthonormal set. i 110 Introduction to quantum mechanics Exercise 2.76: Extend the proof of the Schmidt decomposition to the case where A and B may have state spaces of different dimensionality. Exercise 2.77: Suppose ABC is a three component quantum system. Show by example that there are quantum states |\u03c8 of such systems which can not be written in the form |\u03c8 = \u03bbi|iA|iB|iC, (2.206) where \u03bbi are real numbers, and |iA, |iB", ", |iC are orthonormal bases of the respective systems. i The bases |iA and |iB are called the Schmidt bases for A and B, respectively, and the number of non-zero values \u03bbi is called the Schmidt number for the state |\u03c8. The Schmidt number is an important property of a composite quantum system, which in some sense quanti\ufb01es the \u2018amount\u2019 of entanglement between systems A and B. To get some idea of why this is the case, consider the following obvious but important property: the Schmidt number is preserved under unitary transformations on system A or system i \u03bbi|iA|iB is the Schmidt decomposition for |\u03c8 B alone. To see this, notice that if i \u03bbi(U |iA)|iB is the Schmidt decomposition for U |\u03c8, where U is a unitary then operator acting on system A alone. Algebraic invariance properties of this type make the Schmidt number a very useful tool. Exercise 2.78: Prove that a state |\u03c8 of a composite system AB is a product state if and only if it has Schmidt number 1. Prove that |\u03c8 is a product state if and only if \u03c1A (and thus \u03c1B) are pure states. A second, related technique for quantum computation and quantum information is puri\ufb01cation. Suppose we are given a state \u03c1A of a quantum system A. It is possible to introduce another system, which we denote R, and de\ufb01ne a pure state |AR for the joint system AR such that \u03c1A = trR(|ARAR|). That is, the pure state |AR reduces to \u03c1A when we look at system A alone. This is a purely mathematical procedure, known as puri\ufb01cation, which allows us to associate pure states with mixed states. For this reason we call system R a reference system: it is a \ufb01ctitious system, without a direct physical signi\ufb01cance. To prove that puri\ufb01cation can be done for any state, we explain how to construct a system R and puri\ufb01cation |AR for \u03c1A. Suppose \u03c1A has orthonormal decomposition i pi|iAiA|. To purify \u03c1A we introduce a system R which has the same state \u03c1A = space as system A, with orthonormal basis states |iR, and de\ufb01ne a pure state for the combined system |AR \u2261 pi|iA|iR. (2.207) \u221a i We now calculate the reduced density operator for system A corresponding to the state |AR: trR(|ARAR|) = \u221a pipj|iAjA| tr(|iRjR|) ij \u221a = ij pipj|iAjA| \u03b4ij (2.208) (2.209) EPR and the Bell inequality 111 pi|iAiA| = i = \u03c1A. (2.210) (2.211) Thus |AR is a puri\ufb01cation of \u03c1A. Notice the close relationship of the Schmidt decomposition to puri\ufb01cation: the procedure used to purify a mixed state of system A is to de\ufb01ne a pure state whose Schmidt basis for system A is just the basis in which the mixed state is diagonal, with the Schmidt coef\ufb01cients being the square root of the eigenvalues of the density operator being puri\ufb01ed. In this section we\u2019ve explained two tools for studying composite quantum systems, the Schmidt decomposition and puri\ufb01cations. These tools will be indispensable to the study of quantum computation and quantum information, especially quantum information, which is the subject of Part III of this book. Exercise 2.79: Consider a composite system consisting of two qubits. Find the Schmidt decompositions of the states |00 + |11 \u221a 2 ; |00 + |01 + |10 + |11 2 ; and |00 + |01 + |10 \u221a 3. (2.212) Exercise 2.80: Suppose |\u03c8 and |\u03d5 are two pure states of a composite quantum system with components A and B, with identical Schmidt coef\ufb01cients. Show that there are unitary transformations U on system A and V on system B such that |\u03c8 = (U \u2297 V )|\u03d5. Exercise 2.81: (Freedom in puri\ufb01cations) Let |AR1 and |AR2 be two puri\ufb01cations of a state \u03c1A to a composite system AR. Prove that there exists a unitary transformation UR acting on system R such that |AR1 =", " (IA \u2297 UR)|AR2. Exercise 2.82: Suppose {pi, |\u03c8i} is an ensemble of states generating a density matrix i pi|\u03c8i\u03c8i| for a quantum system A. Introduce a system R with \u03c1 = orthonormal basis |i. \u221a (1) Show that (2) Suppose we measure R in the basis |i, obtaining outcome i. With what pi|\u03c8i|i is a puri\ufb01cation of \u03c1. i probability do we obtain the result i, and what is the corresponding state of system A? (3) Let |AR be any puri\ufb01cation of \u03c1 to the system AR. Show that there exists an orthonormal basis |i in which R can be measured such that the corresponding post-measurement state for system A is |\u03c8i with probability pi. 2.6 EPR and the Bell inequality Anybody who is not shocked by quantum theory has not understood it. \u2013 Niels Bohr 112 Introduction to quantum mechanics I recall that during one walk Einstein suddenly stopped, turned to me and asked whether I really believed that the moon exists only when I look at it. The rest of this walk was devoted to a discussion of what a physicist should mean by the term \u2018to exist\u2019. \u2013 Abraham Pais...quantum phenomena do not occur in a Hilbert space, they occur in a laboratory. \u2013 Asher Peres...what is proved by impossibility proofs is lack of imagination. \u2013 John Bell This chapter has focused on introducing the tools and mathematics of quantum mechanics. As these techniques are applied in the following chapters of this book, an important recurring theme is the unusual, non-classical properties of quantum mechanics. But what exactly is the difference between quantum mechanics and the classical world? Understanding this difference is vital in learning how to perform information processing tasks that are dif\ufb01cult or impossible with classical physics. This section concludes the chapter with a discussion of the Bell inequality, a compelling example of an essential difference between quantum and classical physics. When we speak of an object such as a person or a book, we assume that the physical properties of that object have an existence independent of observation. That is, measurements merely act to reveal such physical properties. For example, a tennis ball has as one of its physical properties its position, which we typically measure using light scattered from the surface of the ball. As quantum mechanics was being developed in the 1920s and 1930s a strange point of view arose that differs markedly from the classical view. As described earlier in the chapter, according to quantum mechanics, an unobserved particle does not possess physical properties that exist independent of observation. Rather, such physical properties arise as a consequence of measurements performed upon the system. For example, according to quantum mechanics a qubit does not possess de\ufb01nite properties of \u2018spin in the z direction, \u03c3z\u2019, and \u2018spin in the x direction, \u03c3x\u2019, each of which can be revealed by performing the appropriate measurement. Rather, quantum mechanics gives a set of rules which specify, given the state vector, the probabilities for the possible measurement outcomes when the observable \u03c3z is measured, or when the observable \u03c3x is measured. Many physicists rejected this new view of Nature. The most prominent objector was Albert Einstein. In the famous \u2018EPR paper\u2019, co-authored with Nathan Rosen and Boris Podolsky, Einstein proposed a thought experiment which, he believed, demonstrated that quantum mechanics is not a complete theory of Nature. The essence of the EPR argument is as follows. EPR were interested in what they termed \u2018elements of reality\u2019. Their belief was that any such element of reality must be represented in any complete physical theory. The goal of the argument was to show that quantum mechanics is not a complete physical theory, by identifying elements of reality that were not included in quantum mechanics. The way they attempted to do this was by introducing what they claimed was a suf\ufb01cient condition for a physical property to be an element of reality, namely, that it be possible to predict with certainty the value that property will have, immediately before measurement. EPR and the Bell inequality 113 Box 2.7: Anti-correlations in the EPR experiment Suppose we prepare the two qubit state |\u03c8 = |01 \u2212 |10 \u221a 2, (2.213) a state sometimes known as the spin singlet for historical reasons. It is not dif\ufb01cult to show that this state is an entangled state of the two qubit system. Suppose we perform a measurement of spin along the v axis on both qubits, that is, we measure the observable v \u00b7 \u03c3 (de\ufb01ned in Equation (2.116) on page 90) on each qubit, getting a result of +1 or \u22121 for each qubit", ". It turns out that no matter what choice of v we make, the results of the two measurements are always opposite to one another. That is, if the measurement on the \ufb01rst qubit yields +1, then the measurement on the second qubit will yield \u22121, and vice versa. It is as though the second qubit knows the result of the measurement on the \ufb01rst, no matter how the \ufb01rst qubit is measured. To see why this is true, suppose |a and |b are the eigenstates of v \u00b7 \u03c3. Then there exist complex numbers \u03b1, \u03b2, \u03b3, \u03b4 such that |0 = \u03b1|a + \u03b2|b |1 = \u03b3|a + \u03b4|b. (2.214) (2.215) Substituting we obtain |01 \u2212 |10 \u221a 2 = (\u03b1\u03b4 \u2212 \u03b2\u03b3) But \u03b1\u03b4 \u2212 \u03b2\u03b3 is the determinant of the unitary matrix to a phase factor ei\u03b8 for some real \u03b8. Thus |ab \u2212 |ba \u221a 2. (2.216), and thus is equal \u03b1 \u03b2 \u03b4 \u03b3 |01 \u2212 |10 \u221a 2 = |ab \u2212 |ba \u221a, 2 (2.217) up to an unobservable global phase factor. As a result, if a measurement of v \u00b7 \u03c3 is performed on both qubits, then we can see that a result of +1 (\u22121) on the \ufb01rst qubit implies a result of \u22121 (+1) on the second qubit. Consider, for example, an entangled pair of qubits belonging to Alice and Bob, re- spectively: |01 \u2212 |10 \u221a 2. (2.218) Suppose Alice and Bob are a long way away from one another. Alice performs a measurement of spin along the v axis, that is, she measures the observable v \u00b7 \u03c3 (de\ufb01ned in Equation (2.116) on page 90). Suppose Alice receives the result +1. Then a simple quantum mechanical calculation, given in Box 2.7, shows that she can predict with certainty 114 Introduction to quantum mechanics that Bob will measure \u22121 on his qubit if he also measures spin along the v axis. Similarly, if Alice measured \u22121, then she can predict with certainty that Bob will measure +1 on his qubit. Because it is always possible for Alice to predict the value of the measurement result recorded when Bob\u2019s qubit is measured in the v direction, that physical property must correspond to an element of reality, by the EPR criterion, and should be represented in any complete physical theory. However, standard quantum mechanics, as we have presented it, merely tells one how to calculate the probabilities of the respective measurement outcomes if v \u00b7 \u03c3 is measured. Standard quantum mechanics certainly does not include any fundamental element intended to represent the value of v \u00b7 \u03c3, for all unit vectors v. The goal of EPR was to show that quantum mechanics is incomplete, by demonstrating that quantum mechanics lacked some essential \u2018element of reality\u2019, by their criterion. They hoped to force a return to a more classical view of the world, one in which systems could be ascribed properties which existed independently of measurements performed on those systems. Unfortunately for EPR, most physicists did not accept the above reasoning as convincing. The attempt to impose on Nature by \ufb01at properties which she must obey seems a most peculiar way of studying her laws. Indeed, Nature has had the last laugh on EPR. Nearly thirty years after the EPR paper was published, an experimental test was proposed that could be used to check whether or not the picture of the world which EPR were hoping to force a return to is valid or not. It turns out that Nature experimentally invalidates that point of view, while agreeing with quantum mechanics. The key to this experimental invalidation is a result known as Bell\u2019s inequality. Bell\u2019s inequality is not a result about quantum mechanics, so the \ufb01rst thing we need to do is momentarily forget all our knowledge of quantum mechanics. To obtain Bell\u2019s inequality, we\u2019re going to do a thought experiment, which we will analyze using our common sense notions of how the world works \u2013 the sort of notions Einstein and his collaborators thought Nature ought to obey. After we have done the common sense analysis, we will perform a quantum mechanical analysis which we can show is not consistent with the common sense analysis. Nature can then be asked, by means of a real experiment, to decide between our common sense notions of how the world works, and quantum mechanics. Imagine we perform the following experiment, illustrated in Figure 2.4. Charlie prepares two particles. It doesn\u2019t matter how he prepares the particles, just that he is capable of repeating the experimental procedure which he uses. Once he has performed the preparation,", " he sends one particle to Alice, and the second particle to Bob. Once Alice receives her particle, she performs a measurement on it. Imagine that she has available two different measurement apparatuses, so she could choose to do one of two different measurements. These measurements are of physical properties which we shall label PQ and PR, respectively. Alice doesn\u2019t know in advance which measurement she will choose to perform. Rather, when she receives the particle she \ufb02ips a coin or uses some other random method to decide which measurement to perform. We suppose for simplicity that the measurements can each have one of two outcomes, +1 or \u22121. Suppose Alice\u2019s particle has a value Q for the property PQ. Q is assumed to be an objective property of Alice\u2019s particle, which is merely revealed by the measurement, much as we imagine the position of a tennis ball to be revealed by the particles of light being scattered off it. Similarly, let R denote the value revealed by a measurement of the property PR. EPR and the Bell inequality 115 Similarly, suppose that Bob is capable of measuring one of two properties, PS or PT, once again revealing an objectively existing value S or T for the property, each taking value +1 or \u22121. Bob does not decide beforehand which property he will measure, but waits until he has received the particle and then chooses randomly. The timing of the experiment is arranged so that Alice and Bob do their measurements at the same time (or, to use the more precise language of relativity, in a causally disconnected manner). Therefore, the measurement which Alice performs cannot disturb the result of Bob\u2019s measurement (or vice versa), since physical in\ufb02uences cannot propagate faster than light. Q R S T Figure 2.4. Schematic experimental setup for the Bell inequalities. Alice can choose to measure either Q or R, and Bob chooses to measure either S or T. They perform their measurements simultaneously. Alice and Bob are assumed to be far enough apart that performing a measurement on one system can not have any effect on the result of measurements on the other. We are going to do some simple algebra with the quantity QS + RS + RT \u2212 QT. Notice that QS + RS + RT \u2212 QT = (Q + R)S + (R \u2212 Q)T. (2.219) Because R, Q = \u00b11 it follows that either (Q + R)S = 0 or (R \u2212 Q)T = 0. In either case, it is easy to see from (2.219) that QS + RS + RT \u2212 QT = \u00b12. Suppose next that p(q, r, s, t) is the probability that, before the measurements are performed, the system is in a state where Q = q, R = r, S = s, and T = t. These probabilities may depend on how Charlie performs his preparation, and on experimental noise. Letting E(\u00b7) denote the mean value of a quantity, we have E(QS + RS + RT \u2212 QT ) = \u2264 qrst qrst = 2. p(q, r, s, t)(qs + rs + rt \u2212 qt) (2.220) p(q, r, s, t) \u00d7 2 (2.221) (2.222) Also, E(QS + RS + RT \u2212 QT ) = p(q, r, s, t)qs + p(q, r, s, t)rs qrst + qrst qrst p(q, r, s, t)rt \u2212 qrst p(q, r, s, t)qt (2.223) = E(QS) + E(RS) + E(RT ) \u2212 E(QT ). (2.224) Comparing (2.222) and (2.224) we obtain the Bell inequality, E(QS) + E(RS) + E(RT ) \u2212 E(QT ) \u2264 2. (2.225) 116 Introduction to quantum mechanics This result is also often known as the CHSH inequality after the initials of its four discoverers. It is part of a larger set of inequalities known generically as Bell inequalities, since the \ufb01rst was found by John Bell. By repeating the experiment many times, Alice and Bob can determine each quantity on the left hand side of the Bell inequality. For example, after \ufb01nishing a set of experiments, Alice and Bob get together to analyze their data. They look at all the experiments where Alice measured PQ and Bob measured PS. By multiplying the results of their experiments together, they get a sample of values for QS. By averaging over this sample, they can estimate E(QS) to an accuracy only limited by the number of experiments which they perform. Similarly, they can estimate all the other quantities on the", " left hand side of the Bell inequality, and thus check to see whether it is obeyed in a real experiment. It\u2019s time to put some quantum mechanics back in the picture. Imagine we perform the following quantum mechanical experiment. Charlie prepares a quantum system of two qubits in the state |\u03c8 = |01 \u2212 |10 \u221a 2. (2.226) He passes the \ufb01rst qubit to Alice, and the second qubit to Bob. They perform measurements of the following observables: Q = Z1 R = X1 S = T = \u2212Z2 \u2212 X2\u221a 2 Z2 \u2212 X2\u221a 2. (2.227) (2.228) Simple calculations show that the average values for these observables, written in the quantum mechanical \u00b7 notation, are: QS = 1\u221a 2 ; RS = 1\u221a 2 ; RT = 1\u221a 2 ; QT = \u2212 1\u221a 2. Thus, QS + RS + RT \u2212 QT = 2 \u221a 2. (2.229) (2.230) Hold on! We learned back in (2.225) that the average value of QS plus the average value of RS plus the average value of RT minus the average value of QT can never exceed two. Yet here, quantum mechanics predicts that this sum of averages yields 2 \u221a 2! Fortunately, we can ask Nature to resolve the apparent paradox for us. Clever experiments using photons \u2013 particles of light \u2013 have been done to check the prediction (2.230) of quantum mechanics versus the Bell inequality (2.225) which we were led to by our common sense reasoning. The details of the experiments are outside the scope of the book, but the results were resoundingly in favor of the quantum mechanical prediction. The Bell inequality (2.225) is not obeyed by Nature. What does this mean? It means that one or more of the assumptions that went into the derivation of the Bell inequality must be incorrect. Vast tomes have been written analyzing the various forms in which this type of argument can be made, and analyzing the subtly different assumptions which must be made to reach Bell-like inequalities. Here we merely summarize the main points. There are two assumptions made in the proof of (2.225) which are questionable: Chapter problems 117 (1) The assumption that the physical properties PQ, PR, PS, PT have de\ufb01nite values Q, R, S, T which exist independent of observation. This is sometimes known as the assumption of realism. (2) The assumption that Alice performing her measurement does not in\ufb02uence the result of Bob\u2019s measurement. This is sometimes known as the assumption of locality. These two assumptions together are known as the assumptions of local realism. They are certainly intuitively plausible assumptions about how the world works, and they \ufb01t our everyday experience. Yet the Bell inequalities show that at least one of these assumptions is not correct. What can we learn from Bell\u2019s inequality? For physicists, the most important lesson is that their deeply held commonsense intuitions about how the world works are wrong. The world is not locally realistic. Most physicists take the point of view that it is the assumption of realism which needs to be dropped from our worldview in quantum mechanics, although others have argued that the assumption of locality should be dropped instead. Regardless, Bell\u2019s inequality together with substantial experimental evidence now points to the conclusion that either or both of locality and realism must be dropped from our view of the world if we are to develop a good intuitive understanding of quantum mechanics. What lessons can the \ufb01elds of quantum computation and quantum information learn from Bell\u2019s inequality? Historically the most useful lesson has perhaps also been the most vague: there is something profoundly \u2018up\u2019 with entangled states like the EPR state. A lot of mileage in quantum computation and, especially, quantum information, has come from asking the simple question: \u2018what would some entanglement buy me in this problem?\u2019 As we saw in teleportation and superdense coding, and as we will see repeatedly later in the book, by throwing some entanglement into a problem we open up a new world of possibilities unimaginable with classical information. The bigger picture is that Bell\u2019s inequality teaches us that entanglement is a fundamentally new resource in the world that goes essentially beyond classical resources; iron to the classical world\u2019s bronze age. A major task of quantum computation and quantum information is to exploit this new resource to do information processing tasks impossible or much more dif\ufb01cult with classical resources. Problem 2.1: (Functions of the Pauli matrices) Let f (\u00b7) be any function from complex numbers to complex numbers. Let n be a normalized vector in three dimensions, and let \u03b8 be real. Show that f (\u03b8) + f (\u2212\u03b8) 2", " f (\u03b8) \u2212 f (\u2212\u03b8) 2 f (\u03b8n \u00b7 \u03c3) = n \u00b7 \u03c3. I + (2.231) Problem 2.2: (Properties of the Schmidt number) Suppose |\u03c8 is a pure state of a composite system with components A and B. (1) Prove that the Schmidt number of |\u03c8 is equal to the rank of the reduced density matrix \u03c1A \u2261 trB(|\u03c8\u03c8|). (Note that the rank of a Hermitian operator is equal to the dimension of its support.) (2) Suppose |\u03c8 = j |\u03b1j|\u03b2j is a representation for |\u03c8, where |\u03b1j and |\u03b2j are (un-normalized) states for systems A and B, respectively. Prove that the 118 Introduction to quantum mechanics number of terms in such a decomposition is greater than or equal to the Schmidt number of |\u03c8, Sch(\u03c8). (3) Suppose |\u03c8 = \u03b1|\u03d5 + \u03b2|\u03b3. Prove that Sch(\u03c8) \u2265 |Sch(\u03d5) \u2212 Sch(\u03b3)|. (2.232) Problem 2.3: (Tsirelson\u2019s inequality) Suppose Q = q \u00b7 \u03c3, R = r \u00b7 \u03c3, S = s \u00b7 \u03c3, T = t \u00b7 \u03c3, where q, r, s and t are real unit vectors in three dimensions. Show that (2 = 4I + [Q, R] \u2297 [S, T ]. (2.233) Use this result to prove that, (2.234) so the violation of the Bell inequality found in Equation (2.230) is the maximum possible in quantum mechanics. History and further reading There are an enormous number of books on linear algebra at levels ranging from High School through to Graduate School. Perhaps our favorites are the two volume set by Horn and Johnson[HJ85, HJ91], which cover an extensive range of topics in an accessible manner. Other useful references include Marcus and Minc[MM92], and Bhatia[Bha97]. Good introductions to linear algebra include Halmos[Hal58], Perlis[Per52], and Strang[Str76]. There are many excellent books on quantum mechanics. Unfortunately, most of these books focus on topics of tangential interest to quantum information and computation. Perhaps the most relevant in the existing literature is Peres\u2019 superb book[Per93]. Beside an extremely clear exposition of elementary quantum mechanics, Peres gives an extensive discussion of the Bell inequalities and related results. Good introductory level texts include Sakurai\u2019s book[Sak95], Volume III of the superb series by Feynman, Leighton, and Sands[FLS65a], and the two volume work by Cohen-Tannoudji, Diu and Lalo\u00a8e[CTDL77a, CTDL77b]. All three of these works are somewhat closer in spirit to quantum computation and quantum information than are most other quantum mechanics texts, although the great bulk of each is still taken up by applications far removed from quantum computation and quantum information. As a result, none of these texts need be read in detail by someone interested in learning about quantum computation and quantum information. However, any one of these texts may prove handy as a reference, especially when reading articles by physicists. References for the history of quantum mechanics may be found at the end of Chapter 1. Many texts on quantum mechanics deal only with projective measurements. For applications to quantum computing and quantum information it is more convenient \u2013 and, we believe, easier for novices \u2013 to start with the general description of measurements, of which projective measurements can be regarded as a special case. Of course, ultimately, as we have shown, the two approaches are equivalent. The theory of generalized measurements which we have employed was developed between the 1940s and 1970s. Much of the history can be distilled from the book of Kraus[Kra83]. Interesting discussion of quantum measurements may be found in Section 2.2 of Gardiner[Gar91], and in the book by Braginsky and Khahili[BK92]. The POVM measurement for distinguishing History and further reading 119 non-orthogonal states described in Section 2.2.6 is due to Peres[Per88]. The extension described in Exercise 2.64 appeared in Duan and Guo[DG98]. Superdense coding was invented by Bennett and Wiesner[BW92]. An experiment implementing a variant of superdense coding using entangled photon pairs was performed by Mattle, Weinfurter, Kwiat, and Zeilinger[MWKZ96]. The density operator formalism was introduced independently by Landau[Lan27] and by von Neumann[von27]. The unitary freedom in the ensemble for density matrices, Theorem 2.6,", " was \ufb01rst pointed out by Schro\u00a8dinger[Sch36], and was later rediscovered and extended by Jaynes[Jay57] and by Hughston, Jozsa and Wootters[HJW93]. The result of Exercise 2.73 is from the paper by Jaynes, and the results of Exercises 2.81 and 2.82 appear in the paper by Hughston, Jozsa and Wootters. The class of probability distributions which may appear in a density matrix decomposition for a given density matrix has been studied by Uhlmann[Uhl70] and by Nielsen[Nie99b]. Schmidt\u2019s eponymous decomposition appeared in[Sch06]. The result of Exercise 2.77 was noted by Peres[Per95]. The EPR thought experiment is due to Einstein, Podolsky and Rosen[EPR35], and was recast in essentially the form we have given here by Bohm[Boh51]. It is sometimes misleadingly referred to as the EPR \u2018paradox\u2019. The Bell inequality is named in honour of Bell[Bel64], who \ufb01rst derived inequalities of this type. The form we have presented is due to Clauser, Horne, Shimony, and Holt[CHSH69], and is often known as the CHSH inequality. This inequality was derived independently by Bell, who did not publish the result. Part 3 of Problem 2.2 is due to Thapliyal (private communication). Tsirelson\u2019s in- equality is due to Tsirelson[Tsi80]. 3 Introduction to computer science In natural science, Nature has given us a world and we\u2019re just to discover its laws. In computers, we can stuff laws into it and create a world. \u2013 Alan Kay Our \ufb01eld is still in its embryonic stage. It\u2019s great that we haven\u2019t been around for 2000 years. We are still at a stage where very, very important results occur in front of our eyes. \u2013 Michael Rabin, on computer science Algorithms are the key concept of computer science. An algorithm is a precise recipe for performing some task, such as the elementary algorithm for adding two numbers which we all learn as children. This chapter outlines the modern theory of algorithms developed by computer science. Our fundamental model for algorithms will be the Turing machine. This is an idealized computer, rather like a modern personal computer, but with a simpler set of basic instructions, and an idealized unbounded memory. The apparent simplicity of Turing machines is misleading; they are very powerful devices. We will see that they can be used to execute any algorithm whatsoever, even one running on an apparently much more powerful computer. The fundamental question we are trying to address in the study of algorithms is: what resources are required to perform a given computational task? This question splits up naturally into two parts. First, we\u2019d like to understand what computational tasks are possible, preferably by giving explicit algorithms for solving speci\ufb01c problems. For example, we have many excellent examples of algorithms that can quickly sort a list of numbers into ascending order. The second aspect of this question is to demonstrate limitations on what computational tasks may be accomplished. For example, lower bounds can be given for the number of operations that must be performed by any algorithm which sorts a list of numbers into ascending order. Ideally, these two tasks \u2013 the \ufb01nding of algorithms for solving computational problems, and proving limitations on our ability to solve computational problems \u2013 would dovetail perfectly. In practice, a signi\ufb01cant gap often exists between the best techniques known for solving a computational problem, and the most stringent limitations known on the solution. The purpose of this chapter is to give a broad overview of the tools which have been developed to aid in the analysis of computational problems, and in the construction and analysis of algorithms to solve such problems. Why should a person interested in quantum computation and quantum information spend time investigating classical computer science? There are three good reasons for this effort. First, classical computer science provides a vast body of concepts and techniques which may be reused to great effect in quantum computation and quantum information. Many of the triumphs of quantum computation and quantum information have come by combining existing ideas from computer science with novel ideas from quantum Introduction to computer science 121 mechanics. For example, some of the fast algorithms for quantum computers are based upon the Fourier transform, a powerful tool utilized by many classical algorithms. Once it was realized that quantum computers could perform a type of Fourier transform much more quickly than classical computers this enabled the development of many important quantum algorithms. Second, computer scientists have expended great effort understanding what resources are required to perform a given computational task on a classical computer. These results can be used as the basis for a comparison with quantum computation and quantum information. For example, much attention has been focused on the problem of \ufb01nding the prime", " factors of a given number. On a classical computer this problem is believed to have no \u2018ef\ufb01cient\u2019 solution, where \u2018ef\ufb01cient\u2019 has a meaning we\u2019ll explain later in the chapter. What is interesting is that an ef\ufb01cient solution to this problem is known for quantum computers. The lesson is that, for this task of \ufb01nding prime factors, there appears to be a gap between what is possible on a classical computer and what is possible on a quantum computer. This is both intrinsically interesting, and also interesting in the broader sense that it suggests such a gap may exist for a wider class of computational problems than merely the \ufb01nding of prime factors. By studying this speci\ufb01c problem further, it may be possible to discern features of the problem which make it more tractable on a quantum computer than on a classical computer, and then act on these insights to \ufb01nd interesting quantum algorithms for the solution of other problems. Third, and most important, there is learning to think like a computer scientist. Computer scientists think in a rather different style than does a physicist or other natural scientist. Anybody wanting a deep understanding of quantum computation and quantum information must learn to think like a computer scientist at least some of the time; they must instinctively know what problems, what techniques, and most importantly what problems are of greatest interest to a computer scientist. The structure of this chapter is as follows. In Section 3.1 we introduce two models for computation: the Turing machine model, and the circuit model. The Turing machine model will be used as our fundamental model for computation. In practice, however, we mostly make use of the circuit model of computation, and it is this model which is most useful in the study of quantum computation. With our models for computation in hand, the remainder of the chapter discusses resource requirements for computation. Section 3.2 begins by overviewing the computational tasks we are interested in as well as discusing some associated resource questions. It continues with a broad look at the key concepts of computational complexity, a \ufb01eld which examines the time and space requirements necessary to solve particular computational problems, and provides a broad classi\ufb01cation of problems based upon their dif\ufb01culty of solution. Finally, the section concludes with an examination of the energy resources required to perform computations. Surprisingly, it turns out that the energy required to perform a computation can be made vanishingly small, provided one can make the computation reversible. We explain how to construct reversible computers, and explain some of the reasons they are important both for computer science and for quantum computation and quantum information. Section 3.3 concludes the chapter with a broad look at the entire \ufb01eld of computer science, focusing on issues of particular relevance to quantum computation and quantum information. 122 Introduction to computer science 3.1 Models for computation...algorithms are concepts that have existence apart from any programming language. \u2013 Donald Knuth What does it mean to have an algorithm for performing some task? As children we all learn a procedure which enables us to add together any two numbers, no matter how large those numbers are. This is an example of an algorithm. Finding a mathematically precise formulation of the concept of an algorithm is the goal of this section. Historically, the notion of an algorithm goes back centuries; undergraduates learn Euclid\u2019s two thousand year old algorithm for \ufb01nding the greatest common divisor of two positive integers. However, it wasn\u2019t until the 1930s that the fundamental notions of the modern theory of algorithms, and thus of computation, were introduced, by Alonzo Church, Alan Turing, and other pioneers of the computer era. This work arose in response to a profound challenge laid down by the great mathematician David Hilbert in the early part of the twentieth century. Hilbert asked whether or not there existed some algorithm which could be used, in principle, to solve all the problems of mathematics. Hilbert expected that the answer to this question, sometimes known as the entscheidungsproblem, would be yes. Amazingly, the answer to Hilbert\u2019s challenge turned out to be no: there is no algorithm to solve all mathematical problems. To prove this, Church and Turing had to solve the deep problem of capturing in a mathematical de\ufb01nition what we mean when we use the intuitive concept of an algorithm. In so doing, they laid the foundations for the modern theory of algorithms, and consequently for the modern theory of computer science. In this chapter, we use two ostensibly different approaches to the theory of computation. The \ufb01rst approach is that proposed by Turing. Turing de\ufb01ned a class of machines, now known as Turing machines, in order to capture the notion of an algorithm to perform a computational task. In Section 3.1.1, we describe Turing machines, and then discuss some of the simpler variants of the Turing machine model.", " The second approach is via the circuit model of computation, an approach that is especially useful as preparation for our later study of quantum computers. The circuit model is described in Section 3.1.2. Although these models of computation appear different on the surface, it turns out that they are equivalent. Why introduce more than one model of computation, you may ask? We do so because different models of computation may yield different insights into the solution of speci\ufb01c problems. Two (or more) ways of thinking about a concept are better than one. 3.1.1 Turing machines The basic elements of a Turing machine are illustrated in Figure 3.1. A Turing machine contains four main elements: (a) a program, rather like an ordinary computer; (b) a \ufb01nite state control, which acts like a stripped-down microprocessor, co-ordinating the other operations of the machine; (c) a tape, which acts like a computer memory; and (d) a readwrite tape-head, which points to the position on the tape which is currently readable or writable. We now describe each of these four elements in more detail. The \ufb01nite state control for a Turing machine consists of a \ufb01nite set of internal states, Models for computation 123!\" \" %&\u2019 ( #$ Figure 3.1. Main elements of a Turing machine. In the text, blanks on the tape are denoted by a \u2018b\u2019. Note the marking the left hand end of the tape. q1,..., qm. The number m is allowed to be varied; it turns out that for m suf\ufb01ciently large this does not affect the power of the machine in any essential way, so without loss of generality we may suppose that m is some \ufb01xed constant. The best way to think of the \ufb01nite state control is as a sort of microprocessor, co-ordinating the Turing machine\u2019s operation. It provides temporary storage off-tape, and a central place where all processing for the machine may be done. In addition to the states q1,..., qm, there are also two special internal states, labelled qs and qh. We call these the starting state and the halting state, respectively. The idea is that at the beginning of the computation, the Turing machine is in the starting state qs. The execution of the computation causes the Turing machine\u2019s internal state to change. If the computation ever \ufb01nishes, the Turing machine ends up in the state qh to indicate that the machine has completed its operation. The Turing machine tape is a one-dimensional object, which stretches off to in\ufb01nity in one direction. The tape consists of an in\ufb01nite sequence of tape squares. We number the tape squares 0, 1, 2, 3,.... The tape squares each contain one symbol drawn from some alphabet, \u0393, which contains a \ufb01nite number of distinct symbols. For now, it will be convenient to assume that the alphabet contains four symbols, which we denote by 0, 1, b (the \u2018blank\u2019 symbol), and, to mark the left hand edge of the tape. Initially, the tape contains a at the left hand end, a \ufb01nite number of 0s and 1s, and the rest of the tape contains blanks. The read-write tape-head identi\ufb01es a single square on the Turing machine tape as the square that is currently being accessed by the machine. Summarizing, the machine starts its operation with the \ufb01nite state control in the state qs, and with the read-write head at the leftmost tape square, the square numbered 0. The computation then proceeds in a step by step manner according to the program, to be de\ufb01ned below. If the current state is qh, then the computation has halted, and the output of the computation is the current (non-blank) contents of the tape. A program for a Turing machine is a \ufb01nite ordered list of program lines of the form q, x, q, x, s. The \ufb01rst item in the program line, q, is a state from the set of internal states of the machine. The second item, x, is taken from the alphabet of symbols which may appear on the tape, \u0393. The way the program works is that on each machine cycle, the Turing machine looks through the list of program lines in order, searching for a line q, x, \u00b7, \u00b7, \u00b7, such that the current internal state of the machine is q, and the symbol 124 Introduction to computer science being read on the tape is x. If it doesn\u2019t \ufb01nd such a program line, the internal", " state of the machine is changed to qh, and the machine halts operation. If such a line is found, then that program line is executed. Execution of a program line involves the following steps: the internal state of the machine is changed to q; the symbol x on the tape is overwritten by the symbol x, and the tape-head moves left, right, or stands still, depending on whether s is \u22121, +1, or 0, respectively. The only exception to this rule is if the tape-head is at the leftmost tape square, and s = \u22121, in which case the tape-head stays put. Now that we know what a Turing machine is, let\u2019s see how it may be used to compute a simple function. Consider the following example of a Turing machine. The machine starts with a binary number, x, on the tape, followed by blanks. The machine has three internal states, q1, q2, and q3, in addition to the starting state qs and halting state qh. The program contains the following program lines (the numbers on the left hand side are for convenience in referring to the program lines in later discussion, and do not form part of the program): 1 : qs,, q1,, +1 2 : q1, 0, q1, b, +1 3 : q1, 1, q1, b, +1 4 : q1, b, q2, b, \u22121 5 : q2, b, q2, b, \u22121 6 : q2,, q3,, +1 7 : q3, b, qh, 1, 0. What function does this program compute? Initially the machine is in the state qs and at the left-most tape position so line 1, qs,, q1,, +1, is executed, which causes the tape-head to move right without changing what is written on the tape, but changing the internal state of the machine to q1. The next three lines of the program ensure that while the machine is in the state q1 the tape-head will continue moving right while it reads either 0s (line 2) or 1s (line 3) on the tape, over-writing the tape contents with blanks as it goes and remaining in the state q1, until it reaches a tape square that is already blank, at which point the tape-head is moved one position to the left, and the internal state is changed to q2 (line 4). Line 5 then ensures that the tape-head keeps moving left while blanks are being read by the tape-head, without changing the contents of the tape. This keeps up until the tape-head returns to its starting point, at which point it reads a on the tape, changes the internal state to q3, and moves one step to the right (line 6). Line 7 completes the program, simply printing the number 1 onto the tape, and then halting. The preceding analysis shows that this program computes the constant function f (x) = 1. That is, regardless of what number is input on the tape the number 1 is output. More generally, a Turing machine can be thought of as computing functions from the nonnegative integers to the non-negative integers; the initial state of the tape is used to represent the input to the function, and the \ufb01nal state of the tape is used to represent the output of the function. It seems as though we have gone to a very great deal of trouble to compute this simple function using our Turing machines. Is it possible to build up more complicated functions using Turing machines? For example, could we construct a machine such that when two numbers, x and y, are input on the tape with a blank to demarcate them, it will Models for computation 125 output the sum x + y on the tape? More generally, what class of functions is it possible to compute using a Turing machine? It turns out that the Turing machine model of computation can be used to compute an enormous variety of functions. For example, it can be used to do all the basic arithmetical operations, to search through text represented as strings of bits on the tape, and many other interesting operations. Surprisingly, it turns out that a Turing machine can be used to simulate all the operations performed on a modern computer! Indeed, according to a thesis put forward independently by Church and by Turing, the Turing machine model of computation completely captures the notion of computing a function using an algorithm. This is known as the Church\u2013Turing thesis: The class of functions computable by a Turing machine corresponds exactly to the class of functions which we would naturally regard as being computable by an algorithm. The Church\u2013Turing thesis asserts an equivalence between a rigorous mathematical concept \u2013 function computable by a Turing machine \u2013 and the intuitive concept of what it means for a function to be computable by an algorithm. The thesis derives its", " importance from the fact that it makes the study of real-world algorithms, prior to 1936 a rather vague concept, amenable to rigorous mathematical study. To understand the signi\ufb01cance of this point it may be helpful to consider the de\ufb01nition of a continuous function from real analysis. Every child can tell you what it means for a line to be continuous on a piece of paper, but it is far from obvious how to capture that intuition in a rigorous de\ufb01nition. Mathematicians in the nineteenth century spent a great deal of time arguing about the merits of various de\ufb01nitions of continuity before the modern de\ufb01nition of continuity came to be accepted. When making fundamental de\ufb01nitions like that of continuity or of computability it is important that good de\ufb01nitions be chosen, ensuring that one\u2019s intuitive notions closely match the precise mathematical de\ufb01nition. From this point of view the Church\u2013Turing thesis is simply the assertion that the Turing machine model of computation provides a good foundation for computer science, capturing the intuitive notion of an algorithm in a rigorous de\ufb01nition. A priori it is not obvious that every function which we would intuitively regard as computable by an algorithm can be computed using a Turing machine. Church, Turing and many other people have spent a great deal of time gathering evidence for the Church\u2013Turing thesis, and in sixty years no evidence to the contrary has been found. Nevertheless, it is possible that in the future we will discover in Nature a process which computes a function not computable on a Turing machine. It would be wonderful if that ever happened, because we could then harness that process to help us perform new computations which could not be performed before. Of course, we would also need to overhaul the de\ufb01nition of computability, and with it, computer science. Exercise 3.1: (Non-computable processes in Nature) How might we recognize that a process in Nature computes a function not computable by a Turing machine? Exercise 3.2: (Turing numbers) Show that single-tape Turing machines can each be given a number from the list 1, 2, 3,... in such a way that the number uniquely speci\ufb01es the corresponding machine. We call this number the Turing number of the corresponding Turing machine. (Hint: Every positive integer has 126 Introduction to computer science... pak a unique prime factorization pa1 1 and a1,..., ak are non-negative integers.) pa2 2 k, where pi are distinct prime numbers, In later chapters, we will see that quantum computers also obey the Church\u2013Turing thesis. That is, quantum computers can compute the same class of functions as is computable by a Turing machine. The difference between quantum computers and Turing machines turns out to lie in the ef\ufb01ciency with which the computation of the function may be performed \u2013 there are functions which can be computed much more ef\ufb01ciently on a quantum computer than is believed to be possible with a classical computing device such as a Turing machine. Demonstrating in complete detail that the Turing machine model of computation can be used to build up all the usual concepts used in computer programming languages is beyond the scope of this book (see \u2018History and further reading\u2019 at the end of the chapter for more information). When specifying algorithms, instead of explicitly specifying the Turing machine used to compute the algorithm, we shall usually use a much higher level pseudocode, trusting in the Church\u2013Turing thesis that this pseudocode can be translated into the Turing machine model of computation. We won\u2019t give any sort of rigorous de\ufb01nition for pseudocode. Think of it as a slightly more formal version of English or, if you like, a sloppy version of a high-level programming language such as C++ or BASIC. Pseudocode provides a convenient way of expressing algorithms, without going into the extreme level of detail required by a Turing machine. An example use of pseudocode may be found in Box 3.2 on page 130; it is also used later in the book to describe quantum algorithms. There are many variants on the basic Turing machine model. We might imagine Turing machines with different kinds of tapes. For example, one could consider two-way in\ufb01nite tapes, or perhaps computation with tapes of more than one dimension. So far as is presently known, it is not possible to change any aspect of the Turing model in a way that is physically reasonable, and which manages to extend the class of functions computable by the model. As an example consider a Turing machine equipped with multiple tapes. For simplicity we consider the two-tape case, as the generalization to more than two tapes is clear from this example. Like the basic Turing machine, a two", "-tape Turing machine has a \ufb01nite number of internal states q1,..., qm, a start state qs, and a halt state qh. It has two tapes, each of which contain symbols from some \ufb01nite alphabet of symbols, \u0393. As before we \ufb01nd it convenient to assume that the alphabet contains four symbols, 0, 1, b and, where marks the left hand edge of each tape. The machine has two tape-heads, one for each tape. The main difference between the two-tape Turing machine and the basic Turing machine is in the program. Program lines are of the form q, x1, x2, q, x, s1, s2, 1 meaning that if the internal state of the machine is q, tape one is reading x1 at its current position, and tape two is reading x2 at its current position, then the internal state of the machine should be changed to q, x1 overwritten with x 2, and the tape-heads for tape one and tape two moved according to whether s1 or s2 are equal to +1, \u22121 or 0, respectively. 1, x2 overwritten with x, x 2 In what sense are the basic Turing machine and the two-tape Turing machine equivalent models of computation? They are equivalent in the sense that each computational model is able to simulate the other. Suppose we have a two-tape Turing machine which takes as input a bit string x on the \ufb01rst tape and blanks on the remainder of both tapes, Models for computation 127 except the endpoint marker. This machine computes a function f (x), where f (x) is de\ufb01ned to be the contents of the \ufb01rst tape after the Turing machine has halted. Rather remarkably, it turns out that given a two-tape Turing machine to compute f, there exists an equivalent single-tape Turing machine that is also able to compute f. We won\u2019t explain how to do this explicitly, but the basic idea is that the single-tape Turing machine simulates the two-tape Turing machine, using its single tape to store the contents of both tapes of the two-tape Turing machine. There is some computational overhead required to do this simulation, but the important point is that in principle it can always be done. In fact, there exists a Universal Turing machine (see Box 3.1) which can simulate any other Turing machine! Another interesting variant of the Turing machine model is to introduce randomness into the model. For example, imagine that the Turing machine can execute a program line whose effect is the following: if the internal state is q and the tape-head reads x, then \ufb02ip an unbiased coin. If the coin lands heads, change the internal state to qiH, and if it lands tails, change the internal state to qiT, where qiH and qiT are two internal. states of the Turing machine. Such a program line can be represented as q, x, qiH However, even this variant doesn\u2019t change the essential power of the Turing machine model of computation. It is not dif\ufb01cult to see that we can simulate the effect of the above algorithm on a deterministic Turing machine by explicitly \u2018searching out\u2019 all the possible computational paths corresponding to different values of the coin tosses. Of course, this deterministic simulation may be far less ef\ufb01cient than the random model, but the key point for the present discussion is that the class of functions computable is not changed by introducing randomness into the underlying model., qiT Exercise 3.3: (Turing machine to reverse a bit string) Describe a Turing machine which takes a binary number x as input, and outputs the bits of x in reverse order. (Hint: In this exercise and the next it may help to use a multi-tape Turing machine and/or symbols other than, 0, 1 and the blank.) Exercise 3.4: (Turing machine to add modulo 2) Describe a Turing machine to add two binary numbers x and y modulo 2. The numbers are input on the Turing machine tape in binary, in the form x, followed by a single blank, followed by a y. If one number is not as long as the other then you may assume that it has been padded with leading 0s to make the two numbers the same length. Let us return to Hilbert\u2019s entscheidungsproblem, the original inspiration for the founders of computer science. Is there an algorithm to decide all the problems of mathematics? The answer to this question was shown by Church and Turing to be no. In Box 3.2, we explain Turing\u2019s proof of this remarkable fact. This phenomenon of undecidability is now known to extend far beyond the examples", " which Church and Turing constructed. For example, it is known that the problem of deciding whether two topological spaces are topologically equivalent (\u2018homeomorphic\u2019) is undecidable. There are simple problems related to the behavior of dynamical systems which are undecidable, as you will show in Problem 3.4. References for these and other examples are given in the end of chapter \u2018History and further reading\u2019. Besides its intrinsic interest, undecidability foreshadows a topic of great concern in computer science, and also to quantum computation and quantum information: the dis- 128 Introduction to computer science Box 3.1: The Universal Turing Machine We\u2019ve described Turing machines as containing three elements which may vary from machine to machine \u2013 the initial con\ufb01guration of the tape, the internal states of the \ufb01nite state control, and the program for the machine. A clever idea known as the Universal Turing Machine (UTM) allows us to \ufb01x the program and \ufb01nite state control once and for all, leaving the initial contents of the tape as the only part of the machine which needs to be varied. The Universal Turing Machine (see the \ufb01gure below) has the following property. Let M be any Turing machine, and let TM be the Turing number associated to machine M. Then on input of the binary representation for TM followed by a blank, followed by any string of symbols x on the remainder of the tape, the Universal Turing Machine gives as output whatever machine M would have on input of x. Thus, the Universal Turing Machine is capable of simulating any other Turing machine! )#* $ )#* #* * The Universal Turing Machine is similar in spirit to a modern programmable computer, in which the action to be taken by the computer \u2013 the \u2018program\u2019 \u2013 is stored in memory, analogous to the bit string TM stored at the beginning of the tape by the Universal Turing Machine. The data to be processed by the program is stored in a separate part of memory, analogous to the role of x in the Universal Turing Machine. Then some \ufb01xed hardware is used to run the program, producing the output. This \ufb01xed hardware is analogous to the internal states and the (\ufb01xed) program being executed by the Universal Turing Machine. Describing the detailed construction of a Universal Turing Machine is beyond the scope of this book. (Though industrious readers may like to attempt the construction.) The key point is the existence of such a machine, showing that a single \ufb01xed machine can be used to run any algorithm whatsoever. The existence of a Universal Turing Machine also explains our earlier statement that the number of internal states in a Turing machine does not matter much, for provided that number m exceeds the number needed for a Universal Turing Machine, such a machine can be used to simulate a Turing machine with any number of internal states. tinction between problems which are easy to solve, and problems which are hard to solve. Undecidability provides the ultimate example of problems which are hard to solve \u2013 so hard that they are in fact impossible to solve. Exercise 3.5: (Halting problem with no inputs) Show that given a Turing Models for computation 129 machine M there is no algorithm to determine whether M halts when the input to the machine is a blank tape. Exercise 3.6: (Probabilistic halting problem) Suppose we number the probabilistic Turing machines using a scheme similar to that found in Exercise 3.2 and de\ufb01ne the probabilistic halting function hp(x) to be 1 if machine x halts on input of x with probability at least 1/2 and 0 if machine x halts on input of x with probability less than 1/2. Show that there is no probabilistic Turing machine which can output hp(x) with probability of correctness strictly greater than 1/2 for all x. Exercise 3.7: (Halting oracle) Suppose a black box is made available to us which takes a non-negative integer x as input, and then outputs the value of h(x), where h(\u00b7) is the halting function de\ufb01ned in Box 3.2 on page 130. This type of black box is sometimes known as an oracle for the halting problem. Suppose we have a regular Turing machine which is augmented by the power to call the oracle. One way of accomplishing this is to use a two-tape Turing machine, and add an extra program instruction to the Turing machine which results in the oracle being called, and the value of h(x) being printed on the second tape, where x is the current contents of the second tape. It is clear that this model for computation is more powerful than the conventional Turing machine model, since it can be used to compute the halting function. Is the halting problem for this model of computation undecidable? That", " is, can a Turing machine aided by an oracle for the halting problem decide whether a program for the Turing machine with oracle will halt on a particular input? 3.1.2 Circuits Turing machines are rather idealized models of computing devices. Real computers are \ufb01nite in size, whereas for Turing machines we assumed a computer of unbounded size. In this section we investigate an alternative model of computation, the circuit model, that is equivalent to the Turing machine in terms of computational power, but is more convenient and realistic for many applications. In particular the circuit model of computation is especially important as preparation for our investigation of quantum computers. A circuit is made up of wires and gates, which carry information around, and perform simple computational tasks, respectively. For example, Figure 3.2 shows a simple circuit which takes as input a single bit, a. This bit is passed through a gate, which \ufb02ips gate serve merely to the bit, taking 1 to 0 and 0 to 1. The wires before and after the gate; they can represent movement of the bit through carry the bit to and from the space, or perhaps just through time. More generally, a circuit may involve many input and output bits, many wires, and many logical gates. A logic gate is a function f : {0, 1}k \u2192 {0, 1}l from some \ufb01xed number k of input bits to some \ufb01xed number l of output bits. For example, the gate is a gate with one input bit and one output bit which computes the function f (a) = 1 \u2295 a, where a is a single bit, and \u2295 is modulo 2 addition. It is also usual to make the convention that no loops are allowed in the circuit, to avoid possible instabilities, as illustrated in Figure 3.3. We say such a circuit is acyclic, and we adhere to the convention that circuits in the circuit model of computation be acyclic. 130 Introduction to computer science Box 3.2: The halting problem In Exercise 3.2 you showed that each Turing machine can be uniquely associated with a number from the list 1, 2, 3,.... To solve Hilbert\u2019s problem, Turing used this numbering to pose the halting problem: does the machine with Turing number x halt upon input of the number y? This is a well posed and interesting mathematical problem. After all, it is a matter of some considerable interest to us whether our algorithms halt or not. Yet it turns out that there is no algorithm which is capable of solving the halting problem. To see this, Turing asked whether there is an algorithm to solve an even more specialized problem: does the machine with Turing number x halt upon input of the same number x? Turing de\ufb01ned the halting function, * h(x) \u2261 0 1 if machine number x does not halt upon input of x if machine number x halts upon input of x. If there is an algorithm to solve the halting problem, then there surely is an algorithm to evaluate h(x). We will try to reach a contradiction by supposing such an algorithm exists, denoted by HALT(x). Consider an algorithm computing the function TURING(x), with pseudocode TURING(x) y = HALT(x) if y = 0 then halt else end if loop forever Since HALT is a valid program, TURING must also be a valid program, with some Turing number, t. By de\ufb01nition of the halting function, h(t) = 1 if and only if TURING halts on input of t. But by inspection of the program for TURING, we see that TURING halts on input of t if and only if h(t) = 0. Thus h(t) = 1 if and only if h(t) = 0, a contradiction. Therefore, our initial assumption that there is an algorithm to evaluate h(x) must have been wrong. We conclude that there is no algorithm allowing us to solve the halting problem. Figure 3.2. Elementary circuit performing a single gate on a single input bit. partial list includes the There are many other elementary logic gates which are useful for computation. A gate, and the gate, the gate. Each of these gates takes two bits as input, and produces a single bit as output. gate outputs 1 if gate outputs 1 if and only if both of its inputs are 1. The gate, the gate, the The Models for computation 131 Figure 3.3. Circuits containing cycles can be unstable, and are not usually permitted in the circuit model of computation. gate outputs the sum, modulo 2, of and only if at least one of its inputs is 1. The and, respectively, of their inputs, and gates take the its inputs. The and then apply a to whatever is output. The action of these gates is illustrated in Figure 3.", "4. Figure 3.4. Elementary circuits performing the,,,, and gates. There are two important \u2018gates\u2019 missing from Figure 3.4, namely the gate and gate. In circuits we often allow bits to \u2018divide\u2019, replacing a bit with two the copies of itself, an operation referred to as, that is, the value of two bits are interchanged. A third operation missing from Figure 3.4, not really a logic gate at all, is to allow the preparation of extra ancilla or work bits, to allow extra working space during the computation.. We also allow bits to These simple circuit elements can be put together to perform an enormous variety of computations. Below we\u2019ll show that these elements can be used to compute any function whatsoever. In the meantime, let\u2019s look at a simple example of a circuit which adds two n bit integers, using essentially the same algorithm taught to school-children around the 132 Introduction to computer science world. The basic element in this circuit is a smaller circuit known as a half-adder, shown in Figure 3.5. A half-adder takes two bits, x and y, as input, and outputs the sum of the bits x \u2295 y modulo 2, together with a carry bit set to 1 if x and y are both 1, or 0 otherwise. Figure 3.5. Half-adder circuit. The carry bit c is set to 1 when x and y are both 1, otherwise it is 0. Two cascaded half-adders may be used to build a full-adder, as shown in Figure 3.6. A full-adder takes as input three bits, x, y, and c. The bits x and y should be thought of as data to be added, while c is a carry bit from an earlier computation. The circuit outputs two bits. One output bit is the modulo 2 sum, x \u2295 y \u2295 c of all three input bits. The second output bit, c, is a carry bit, which is set to 1 if two or more of the inputs is 1, and is 0 otherwise. ( ( Figure 3.6. Full-adder circuit. By cascading many of these full-adders together we obtain a circuit to add two n-bit integers, as illustrated in Figure 3.7 for the case n = 3.!! ( Figure 3.7. Addition circuit for two three-bit integers, x = x2x1x0 and y = y2y1y0, using the elementary algorithm taught to school-children. We claimed earlier that just a few \ufb01xed gates can be used to compute any function f : {0, 1}n \u2192 {0, 1}m whatsoever. We will now prove this for the simpli\ufb01ed case of a function f : {0, 1}n \u2192 {0, 1} with n input bits and a single output bit. Such a function Models for computation 133 is known as a Boolean function, and the corresponding circuit is a Boolean circuit. The general universality proof follows immediately from the special case of Boolean functions. The proof is by induction on n. For n = 1 there are four possible functions: the identity, which has a circuit consisting of a single wire; the bit \ufb02ip, which is implemented using gate; the function which replaces the input bit with a 0, which can be a single ing the input with a work bit initially in the 0 state; and the function obtained by ing the input with a work which replaces the input with a 1, which can be obtained by bit initially in the 1 state. To complete the induction, suppose that any function on n bits may be computed by a circuit, and let f be a function on n + 1 bits. De\ufb01ne n-bit functions f0 and f1 by f0(x1,..., xn) \u2261 f (0, x1,..., xn) and f1(x1,..., xn) \u2261 f (1, x1,..., xn). These are both n-bit functions, so by the inductive hypothesis there are circuits to compute these functions. It is now an easy matter to design a circuit which computes f. The circuit computes both f0 and f1 on the last n bits of the input. Then, depending on whether the \ufb01rst bit of the input was a 0 or a 1 it outputs the appropriate answer. A circuit to do this is shown in Figure 3.8. This completes the induction. +#,, -+% Figure 3.8. Circuit to compute an arbitrary function f on n + 1 bits, assuming by induction that there are circuits to compute the n-bit functions f0 and f1. Five elements may be identi\ufb01ed in the universal circuit construction:", " (1) wires, which preserve the states of the bits; (2) ancilla bits prepared in standard states, used in the n = 1 case of the proof; (3) the operation, which takes a single bit as input operation, which interchanges and outputs two copies of that bit; (4) the gates. In Chapter 4 we\u2019ll de\ufb01ne the value of two bits; and (5) the the quantum circuit model of computation in a manner analogous to classical circuits. It is interesting to note that many of these \ufb01ve elements pose some interesting challenges when extending to the quantum case: it is not necessarily obvious that good quantum wires for the preservation of qubits can be constructed, even in principle, the, and, 134 Introduction to computer science operation cannot be performed in a straightforward manner in quantum mechanics, due gates to the no-cloning theorem (as explained in Section 1.3.5), and the are not invertible, and thus can\u2019t be implemented in a straightforward manner as unitary quantum gates. There is certainly plenty to think about in de\ufb01ning a quantum circuit model of computation! and Exercise 3.8: (Universality of simulate the are available., and ) Show that the gates, provided wires, ancilla bits and gate can be used to Let\u2019s return from our brief quantum digression, to the properties of classical circuits. We claimed earlier that the Turing machine model is equivalent to the circuit model of computation. In what sense do we mean the two models are equivalent? On the face of it, the two models appear quite different. The unbounded nature of a Turing machine makes them more useful for abstractly specifying what it is we mean by an algorithm, while circuits more closely capture what an actual physical computer does. The two models are connected by introducing the notion of a uniform circuit family. A circuit family consists of a collection of circuits, {Cn}, indexed by a positive integer n. The circuit Cn has n input bits, and may have any \ufb01nite number of extra work bits, and output bits. The output of the circuit Cn, upon input of a number x of at most n bits in length, is denoted by Cn(x). We require that the circuits be consistent, that is, if m < n and x is at most m bits in length, then Cm(x) = Cn(x). The function computed by the circuit family {Cn} is the function C(\u00b7) such that if x is n bits in length then C(x) = Cn(x). For example, consider a circuit Cn that squares an n-bit number. This de\ufb01nes a family of circuits {Cn} that computes the function, C(x) = x2, where x is any positive integer. It\u2019s not enough to consider unrestricted families of circuits, however. In practice, we need an algorithm to build the circuit. Indeed, if we don\u2019t place any restrictions on the circuit family then it becomes possible to compute all sorts of functions which we do not expect to be able to compute in a reasonable model of computation. For example, let hn(x) denote the halting function, restricted to values of x which are n bits in length. Thus hn is a function from n bits to 1 bit, and we have proved there exists a circuit Cn to compute hn(\u00b7). Therefore the circuit family {Cn} computes the halting function! However, what prevents us from using this circuit family to solve the halting problem is that we haven\u2019t speci\ufb01ed an algorithm which will allow us to build the circuit Cn for all values of n. Adding this requirement results in the notion of a uniform circuit family. That is, a family of circuits {Cn} is said to be a uniform circuit family if there is some algorithm running on a Turing machine which, upon input of n, generates a description of Cn. That is, the algorithm outputs a description of what gates are in the circuit Cn, how those gates are connected together to form a circuit, any ancilla bits needed by operations, and where the output from the circuit the circuit, should be read out. For example, the family of circuits we described earlier for squaring n-bit numbers is certainly a uniform circuit family, since there is an algorithm which, given n, outputs a description of the circuit needed to square an n-bit number. You can think of this algorithm as the means by which an engineer is able to generate a description of (and thus build) the circuit for any n whatsoever. By contrast, a circuit family that is not uniform is said to be a non-uniform circuit family. There is no algorithm to construct and The analysis of computational problems 135 the circuit for arbitrary n, which prevents our engineer from building circuits to compute functions", " like the halting function. Intuitively, a uniform circuit family is a family of circuits that can be generated by some reasonable algorithm. It can be shown that the class of functions computable by uniform circuit families is exactly the same as the class of functions which can be computed on a Turing machine. With this uniformity restriction, results in the Turing machine model of computation can usually be given a straightforward translation into the circuit model of computation, and vice versa. Later we give similar attention to issues of uniformity in the quantum circuit model of computation. 3.2 The analysis of computational problems The analysis of computational problems depends upon the answer to three fundamental questions: (1) What is a computational problem? Multiplying two numbers together is a computational problem; so is programming a computer to exceed human abilities in the writing of poetry. In order to make progress developing a general theory for the analysis of computational problems we are going to isolate a special class of problems known as decision problems, and concentrate our analysis on those. Restricting ourselves in this way enables the development of a theory which is both elegant and rich in structure. Most important, it is a theory whose principles have application far beyond decision problems. (2) How may we design algorithms to solve a given computational problem? Once a problem has been speci\ufb01ed, what algorithms can be used to solve the problem? Are there general techniques which can be used to solve wide classes of problems? How can we be sure an algorithm behaves as claimed? (3) What are the minimal resources required to solve a given computational problem? Running an algorithm requires the consumption of resources, such as time, space, and energy. In different situations it may be desirable to minimize consumption of one or more resource. Can we classify problems according to the resource requirements needed to solve them? In the next few sections we investigate these three questions, especially questions 1 and 3. Although question 1, \u2018what is a computational problem?\u2019, is perhaps the most fundamental of the questions, we shall defer answering it until Section 3.2.3, pausing \ufb01rst to establish some background notions related to resource quanti\ufb01cation in Section 3.2.1, and then reviewing the key ideas of computational complexity in Section 3.2.2. Question 2, how to design good algorithms, is the subject of an enormous amount of ingenious work by many researchers. So much so that in this brief introduction we cannot even begin to describe the main ideas employed in the design of good algorithms. If you are interested in this beautiful subject, we refer you to the end of chapter \u2018History and further reading\u2019. Our closest direct contact with this subject will occur later in the book, when we study quantum algorithms. The techniques involved in the creation of quantum algorithms have typically involved a blend of deep existing ideas in algorithm design for classical computers, and the creation of new, wholly quantum mechanical techniques for algorithm design. For this reason, and because the spirit of quantum algorithm design 136 Introduction to computer science is so similar in many ways to classical algorithm design, we encourage you to become familiar with at least the basic ideas of algorithm design. Question 3, what are the minimal resources required to solve a given computational problem, is the main focus of the next few sections. For example, suppose we are given two numbers, each n bits in length, which we wish to multiply. If the multiplication is performed on a single-tape Turing machine, how many computational steps must be executed by the Turing machine in order to complete the task? How much space is used on the Turing machine while completing the task? These are examples of the type of resource questions we may ask. Generally speaking, computers make use of many different kinds of resources, however we will focus most of our attention on time, space, and energy. Traditionally in computer science, time and space have been the two major resource concerns in the study of algorithms, and we study these issues in Sections 3.2.2 through 3.2.4. Energy has been a less important consideration; however, the study of energy requirements motivates the subject of reversible classical computation, which in turn is a prerequisite for quantum computation, so we examine energy requirements for computation in some considerable detail in Section 3.2.5. 3.2.1 How to quantify computational resources Different models of computation lead to different resource requirements for computation. Even something as simple as changing from a single-tape to a two-tape Turing machine may change the resources required to solve a given computational problem. For a computational task which is extremely well understood, like addition of integers, for example, such differences between computational models may be of interest. However, for a \ufb01rst pass at understanding a problem, we would like a way of quantifying resource requirements that is independent of relatively trivial changes in the computational model. One of the tools which has been developed to do this is the asymptotic notation, which can be used to summarize the essential behavior", " of a function. This asymptotic notation can be used, for example, to summarize the essence of how many time steps it takes a given algorithm to run, without worrying too much about the exact time count. In this section we describe this notation in detail, and apply it to a simple problem illustrating the quanti\ufb01cation of computational resources \u2013 the analysis of algorithms for sorting a list of names into alphabetical order. Suppose, for example, that we are interested in the number of gates necessary to add together two n-bit numbers. Exact counts of the number of gates required obscure the big picture: perhaps a speci\ufb01c algorithm requires 24n + 2log n + 16 gates to perform this task. However, in the limit of large problem size the only term which matters is the 24n term. Furthermore, we disregard constant factors as being of secondary importance to the analysis of the algorithm. The essential behavior of the algorithm is summed up by saying that the number of operations required scales like n, where n is the number of bits in the numbers being added. The asymptotic notation consists of three tools which make this notion precise. The O (\u2018big O\u2019) notation is used to set upper bounds on the behavior of a function. Suppose f (n) and g(n) are two functions on the non-negative integers. We say \u2018f (n) is in the class of functions O(g(n))\u2019, or just \u2018f (n) is O(g(n))\u2019, if there are constants c and n0 such that for all values of n greater than n0, f (n) \u2264 cg(n). That is, for suf\ufb01ciently large n, the function g(n) is an upper bound on f (n), up to an unimportant constant The analysis of computational problems 137 factor. The big O notation is particularly useful for studying the worst-case behavior of speci\ufb01c algorithms, where we are often satis\ufb01ed with an upper bound on the resources consumed by an algorithm. When studying the behaviors of a class of algorithms \u2013 say the entire class of algorithms which can be used to multiply two numbers \u2013 it is interesting to set lower bounds on the resources required. For this the \u03a9 (\u2018big Omega\u2019) notation is used. A function f (n) is said to be \u03a9(g(n)) if there exist constants c and n0 such that for all n greater than n0, cg(n) \u2264 f (n). That is, for suf\ufb01ciently large n, g(n) is a lower bound on f (n), up to an unimportant constant factor. Finally, the \u0398 (\u2018big Theta\u2019) notation is used to indicate that f (n) behaves the same as g(n) asymptotically, up to unimportant constant factors. That is, we say f (n) is \u0398(g(n)) if it is both O(g(n)) and \u03a9(g(n)). Asymptotic notation: examples Let\u2019s consider a few simple examples of the asymptotic notation. The function 2n is in the class O(n2), since 2n \u2264 2n2 for all positive n. The function 2n is \u03a9(n3), since n3 \u2264 2n for suf\ufb01ciently large n. Finally, the function 7n2 + n log(n) is \u0398(n2), since 7n2 \u2264 7n2 + n log(n) \u2264 8n2 for all suf\ufb01ciently large values of n. In the following few exercises you will work through some of the elementary properties of the asymptotic notation that make it a useful tool in the analysis of algorithms. \u221a \u221a Exercise 3.9: Prove that f (n) is O(g(n)) if and only if g(n) is \u03a9(f (n)). Deduce that f (n) is \u0398(g(n)) if and only if g(n) is \u0398(f (n)). Exercise 3.10: Suppose g(n) is a polynomial of degree k. Show that g(n) is O(nl) for any l \u2265 k. Exercise 3.11: Show that log n is O(nk) for any k > 0. Exercise 3.12: (nlog n is super-polynomial) Show that nk is O(nlog n) for any k, but that nlog n is never O(nk). Exercise 3.13: (nlog n is sub-exponential) Show that cn is \u03a9(nlog n) for any c > 1, but that n", "log n is never \u03a9(cn). Exercise 3.14: Suppose e(n) is O(f (n)) and g(n) is O(h(n)). Show that e(n)g(n) is O(f (n)h(n)). An example of the use of the asymptotic notation in quantifying resources is the following simple application to the problem of sorting an n element list of names into alphabetical order. Many sorting algorithms are based upon the \u2018compare-and-swap\u2019 operation: two elements of an n element list are compared, and swapped if they are in the wrong order. If this compare-and-swap operation is the only means by which we can access the list, how many such operations are required in order to ensure that the list has been correctly sorted? A simple compare-and-swap algorithm for solving the sorting problem is as follows: (note that compare-and-swap(j,k) compares the list entries numbered j and k, and swaps them if they are out of order) 138 Introduction to computer science for j = 1 to n-1 for k = j+1 to n compare-and-swap(j,k) end k end j It is clear that this algorithm correctly sorts a list of n names into alphabetical order. Note that the number of compare-and-swap operations executed by the algorithm is (n \u2212 1) + (n \u2212 2) + \u00b7 \u00b7 \u00b7 + 1 = n(n \u2212 1)/2. Thus the number of compare-and-swap operations used by the algorithm is \u0398(n2). Can we do better than this? It turns out that we can. Algorithms such as \u2018heapsort\u2019 are known which run using O(n log n) compare-andswap operations. Furthermore, in Exercise 3.15 you\u2019ll work through a simple counting argument that shows any algorithm based upon the compare-and-swap operation requires \u03a9(n log n) such operations. Thus, the sorting problem requires \u0398(n log n) compare-andswap operations, in general. Exercise 3.15: (Lower bound for compare-and-swap based sorts) Suppose an n element list is sorted by applying some sequence of compare-and-swap operations to the list. There are n! possible initial orderings of the list. Show that after k of the compare-and-swap operations have been applied, at most 2k of the possible initial orderings will have been sorted into the correct order. Conclude that \u03a9(n log n) compare-and-swap operations are required to sort all possible initial orderings into the correct order. 3.2.2 Computational complexity The idea that there won\u2019t be an algorithm to solve it \u2013 this is something fundamental that won\u2019t ever change \u2013 that idea appeals to me. \u2013 Stephen Cook Sometimes it is good that some things are impossible. I am happy there are many things that nobody can do to me. \u2013 Leonid Levin It should not come as a surprise that our choice of polynomial algorithms as the mathematical concept that is supposed to capture the informal notion of \u2018practically ef\ufb01cient computation\u2019 is open to criticism from all sides. [... ] Ultimately, our argument for our choice must be this: Adopting polynomial worst-case performance as our criterion of ef\ufb01ciency results in an elegant and useful theory that says something meaningful about practical computation, and would be impossible without this simpli\ufb01cation. \u2013 Christos Papadimitriou What time and space resources are required to perform a computation? In many cases these are the most important questions we can ask about a computational problem. Problems like addition and multiplication of numbers are regarded as ef\ufb01ciently solvable because we have fast algorithms to perform addition and multiplication, which consume The analysis of computational problems 139 little space when running. Many other problems have no known fast algorithm, and are effectively impossible to solve, not because we can\u2019t \ufb01nd an algorithm to solve the problem, but because all known algorithms consume such vast quantities of space or time as to render them practically useless. Computational complexity is the study of the time and space resources required to solve computational problems. The task of computational complexity is to prove lower bounds on the resources required by the best possible algorithm for solving a problem, even if that algorithm is not explicitly known. In this and the next two sections, we give an overview of computational complexity, its major concepts, and some of the more important results of the \ufb01eld. Note that computational complexity is in a sense complementary to the \ufb01eld of algorithm design; ideally, the most ef\ufb01cient algorithms we could design would match perfectly with the lower bounds proved by", " computational complexity. Unfortunately, this is often not the case. As already noted, in this book we won\u2019t examine classical algorithm design in any depth. One dif\ufb01culty in formulating a theory of computational complexity is that different computational models may require different resources to solve the same problem. For instance, multiple-tape Turing machines can solve many problems substantially faster than single-tape Turing machines. This dif\ufb01culty is resolved in a rather coarse way. Suppose a problem is speci\ufb01ed by giving n bits as input. For instance, we might be interested in whether a particular n-bit number is prime or not. The chief distinction made in computational complexity is between problems which can be solved using resources which are bounded by a polynomial in n, or which require resources which grow faster than any polynomial in n. In the latter case we usually say that the resources required are exponential in the problem size, abusing the term exponential, since there are functions like nlog n which grow faster than any polynomial (and thus are \u2018exponential\u2019 according to this convention), yet which grow slower than any true exponential. A problem is regarded as easy, tractable or feasible if an algorithm for solving the problem using polynomial resources exists, and as hard, intractable or infeasible if the best possible algorithm requires exponential resources. As a simple example, suppose we have two numbers with binary expansions x1... xm1 and y1... ym2, and we wish to determine the sum of the two numbers. The total size of the input is n \u2261 m1 + m2. It\u2019s easy to see that the two numbers can be added using a number of elementary operations that scales as \u0398(n); this algorithm uses a polynomial (indeed, linear) number of operations to perform its tasks. By contrast, it is believed (though it has never been proved!) that the problem of factoring an integer into its prime factors is intractable. That is, the belief is that there is no algorithm which can factor an arbitrary n-bit integer using O(p(n)) operations, where p is some \ufb01xed polynomial function of n. We will later give many other examples of problems which are believed to be intractable in this sense. The polynomial versus exponential classi\ufb01cation is rather coarse. In practice, an algorithm that solves a problem using 2n/1000 operations is probably more useful than one which runs in n1000 operations. Only for very large input sizes (n \u2248 108) will the \u2018ef\ufb01cient\u2019 polynomial algorithm be preferable to the \u2018inef\ufb01cient\u2019 exponential algorithm, and for many purposes it may be more practical to prefer the \u2018inef\ufb01cient\u2019 algorithm. Nevertheless, there are many reasons to base computational complexity primarily on the polynomial versus exponential classi\ufb01cation. First, historically, with few exceptions, polynomial resource algorithms have been much faster than exponential algorithms. We 140 Introduction to computer science might speculate that the reason for this is lack of imagination: coming up with algorithms requiring n, n2 or some other low degree polynomial number of operations is often much easier than \ufb01nding a natural algorithm which requires n1000 operations, although examples like the latter do exist. Thus, the predisposition for the human mind to come up with relatively simple algorithms has meant that in practice polynomial algorithms usually do perform much more ef\ufb01ciently than their exponential cousins. A second and more fundamental reason for emphasizing the polynomial versus exponential classi\ufb01cation is derived from the strong Church\u2013Turing thesis. As discussed in Section 1.1, it was observed in the 1960s and 1970s that probabilistic Turing machines appear to be the strongest \u2018reasonable\u2019 model of computation. More precisely, researchers consistently found that if it was possible to compute a function using k elementary operations in some model that was not the probabilistic Turing machine model of computation, then it was always possible to compute the same function in the probabilistic Turing machine model, using at most p(k) elementary operations, where p(\u00b7) is some polynomial function. This statement is known as the strong Church\u2013Turing thesis: Strong Church\u2013Turing thesis: Any model of computation can be simulated on a probabilistic Turing machine with at most a polynomial increase in the number of elementary operations required. The strong Church\u2013Turing thesis is great news for the theory of computational complexity, for it implies that attention may be restricted to the probabilistic Turing machine model of computation. After all, if a problem has no polynomial resource solution on a probabilistic Turing machine, then the strong Church\u2013Turing thesis implies that it", " has no ef\ufb01cient solution on any computing device. Thus, the strong Church\u2013Turing thesis implies that the entire theory of computational complexity will take on an elegant, model-independent form if the notion of ef\ufb01ciency is identi\ufb01ed with polynomial resource algorithms, and this elegance has provided a strong impetus towards acceptance of the identi\ufb01cation of \u2018solvable with polynomial resources\u2019 and \u2018ef\ufb01ciently solvable\u2019. Of course, one of the prime reasons for interest in quantum computers is that they cast into doubt the strong Church\u2013Turing thesis, by enabling the ef\ufb01cient solution of a problem which is believed to be intractable on all classical computers, including probabilistic Turing machines! Nevertheless, it is useful to understand and appreciate the role the strong Church\u2013Turing thesis has played in the search for a model-independent theory of computational complexity. Finally, we note that, in practice, computer scientists are not only interested in the polynomial versus exponential classi\ufb01cation of problems. This is merely the \ufb01rst and coarsest way of understanding how dif\ufb01cult a computational problem is. However, it is an exceptionally important distinction, and illustrates many broader points about the nature of resource questions in computer science. For most of this book, it will be our central concern in evaluating the ef\ufb01ciency of a given algorithm. Having examined the merits of the polynomial versus exponential classi\ufb01cation, we now have to confess that the theory of computational complexity has one remarkable outstanding failure: it seems very hard to prove that there are interesting classes of problems which require exponential resources to solve. It is quite easy to give non-constructive proofs that most problems require exponential resources (see Exercise 3.16, below), and furthermore many interesting problems are conjectured to require exponential resources for their solution, but rigorous proofs seem very hard to come by, at least with the present The analysis of computational problems 141 state of knowledge. This failure of computational complexity has important implications for quantum computation, because it turns out that the computational power of quantum computers can be related to some major open problems in classical computational complexity theory. Until these problems are resolved, it cannot be stated with certainty how computationally powerful a quantum computer is, or even whether it is more powerful than a classical computer! Exercise 3.16: (Hard-to-compute functions exist) Show that there exist Boolean functions on n inputs which require at least 2n/ log n logic gates to compute. 3.2.3 Decision problems and the complexity classes P and NP Many computational problems are most cleanly formulated as decision problems \u2013 problems with a yes or no answer. For example, is a given number m a prime number or not? This is the primality decision problem. The main ideas of computational complexity are most easily and most often formulated in terms of decision problems, for two reasons: the theory takes its simplest and most elegant form in this form, while still generalizing in a natural way to more complex scenarios; and historically computational complexity arose primarily from the study of decision problems. Although most decision problems can easily be stated in simple, familiar language, discussion of the general properties of decision problems is greatly helped by the terminology of formal languages. In this terminology, a language L over the alphabet \u03a3 is a subset of the set \u03a3\u2217 of all (\ufb01nite) strings of symbols from \u03a3. For example, if \u03a3 = {0, 1}, then the set of binary representations of even numbers, L = {0, 10, 100, 110,...} is a language over \u03a3. Decision problems may be encoded in an obvious way as problems about languages. For instance, the primality decision problem can be encoded using the binary alphabet \u03a3 = {0, 1}. Strings from \u03a3\u2217 can be interpreted in a natural way as non-negative integers. For example, 0010 can be interpreted as the number 2. The language L is de\ufb01ned to consist of all binary strings such that the corresponding number is prime. To solve the primality decision problem, what we would like is a Turing machine which, when started with a given number n on its input tape, eventually outputs some equivalent of \u2018yes\u2019 if n is prime, and outputs \u2018no\u2019 if n is not prime. To make this idea precise, it is convenient to modify our old Turing machine de\ufb01nition (of Section 3.1.1) slightly, replacing the halting state qh with two states qY and qN to represent the answers \u2018yes\u2019 and \u2018no\u2019 respectively. In all other ways the machine behaves in the same way, and it still halts when it enters the state qY or qN. More generally, a language L", " is decided by a Turing machine if the machine is able to decide whether an input x on its tape is a member of the language of L or not, eventually halting in the state qY if x \u2208 L, and eventually halting in the state qN if x \u2208 L. We say that the machine has accepted or rejected x depending on which of these two cases comes about. How quickly can we determine whether or not a number is prime? That is, what is the fastest Turing machine which decides the language representing the primality decision problem? We say that a problem is in TIME(f (n)) if there exists a Turing machine which decides whether a candidate x is in the language in time O(f (n)), where n is the length of x. A problem is said to be solvable in polynomial time if it is in TIME(nk) for some \ufb01nite k. The collection of all languages which are in TIME(nk), for some k, is denoted P. P is our \ufb01rst example of a complexity class. More generally, a complexity 142 Introduction to computer science class is de\ufb01ned to be a collection of languages. Much of computational complexity theory is concerned with the de\ufb01nition of various complexity classes, and understanding the relationship between different complexity classes. Not surprisingly, there are problems which cannot be solved in polynomial time. Unfortunately, proving that any given problem can\u2019t be solved in polynomial time seems to be very dif\ufb01cult, although conjectures abound! A simple example of an interesting decision problem which is believed not to be in P is the factoring decision problem: : Given a composite integer m and l < m, does m have a non-trivial factor less than l? An interesting property of factoring is that if somebody claims that the answer is \u2018yes, m does have a non-trivial factor less than l\u2019 then they can establish this by exhibiting such a factor, which can then be ef\ufb01ciently checked by other parties, simply by doing long-division. We call such a factor a witness to the fact that m has a factor less than l. This idea of an easily checkable witness is the key idea in the de\ufb01nition of the complexity class NP, below. We have phrased factoring as a decision problem, but you can easily verify that the decision problem is equivalent to \ufb01nding the factors of a number: Exercise 3.17: Prove that a polynomial-time algorithm for \ufb01nding the factors of a number m exists if and only if the factoring decision problem is in P. Factoring is an example of a problem in an important complexity class known as NP. What distinguishes problems in NP is that \u2018yes\u2019 instances of a problem can easily be veri\ufb01ed with the aid of an appropriate witness. More rigorously, a language L is in NP if there is a Turing machine M with the following properties: (1) If x \u2208 L then there exists a witness string w such that M halts in the state qY after a time polynomial in |x| when the machine is started in the state x-blank-w. (2) If x \u2208 L then for all strings w which attempt to play the role of a witness, the machine halts in state qN after a time polynomial in |x| when M is started in the state x-blank-w. There is an interesting asymmetry in the de\ufb01nition of NP. While we have to be able to quickly decide whether a possible witness to x \u2208 L is truly a witness, there is no such need to produce a witness to x \u2208 L. For instance, in the factoring problem, we have an easy way of proving that a given number has a factor less than m, but exhibiting a witness to prove that a number has no factors less than m is more daunting. This suggests de\ufb01ning coNP, the class of languages which have witnesses to \u2018no\u2019 instances; obviously the languages in coNP are just the complements of languages in NP. How are P and NP related? It is clear that P is a subset of NP. The most famous open problem in computer science is whether or not there are problems in NP which are not in P, often abbreviated as the P = NP problem. Most computer scientists believe that P = NP, but despite decades of work nobody has been able to prove this, and the possibility remains that P = NP. Exercise 3.18: Prove that if coNP = NP then P = NP. Upon \ufb01rst acquaintance it\u2019s tempting to conclude that the conjecture P = NP ought to be pretty easy to resolve. To see why it\u2019s actually rather subtle it helps to", " see couple of The analysis of computational problems 143 examples of problems that are in P and NP. We\u2019ll draw the examples from graph theory, a rich source of decision problems with surprisingly many practical applications. A graph is a \ufb01nite collection of vertices {v1,..., vn} connected by edges, which are pairs (vi, vj) of vertices. For now, we are only concerned with undirected graphs, in which the order of the vertices (in each edge pair) does not matter; similar ideas can be investigated for directed graphs in which the order of vertices does matter. A typical graph is illustrated in Figure 3.9. Figure 3.9. A graph. A cycle in a graph is a sequence v1,..., vm of vertices such that each pair (vj, vj+1) is an edge, as is (v1, vm). A simple cycle is a cycle in which none of the vertices is repeated, except for the \ufb01rst and last vertices. A Hamiltonian cycle is a simple cycle which visits every vertex in the graph. Examples of graphs with and without Hamiltonian cycles are shown in Figure 3.10. Figure 3.10. The graph on the left contains a Hamiltonian cycle, 0, 1, 2, 3, 4, 5, 0. The graph on the right contains no Hamiltonian cycle, as can be veri\ufb01ed by inspection. The Hamiltonian cycle problem ( has no known polynomial time algorithm. Indeed, ) is to determine whether a given graph contains is a decision problem in NP, since if a given graph has a a Hamiltonian cycle or not. Hamiltonian cycle, then that cycle can be used as an easily checkable witness. Moreover, is a problem in the class of so-called NP-complete problems, which can be thought of as the \u2018hardest\u2019 problems in in time t allows any other problem in NP to be solved NP, in the sense that solving in time O(poly(t)). This also means that if any NP-complete problem has a polynomial time solution then it will follow that P = NP. There is a problem, the Euler cycle decision problem, which is super\ufb01cially similar to, but which has astonishingly different properties. An Euler cycle is an ordering of the edges of a graph G so that every edge in the graph is visited exactly once. The Euler 144 Introduction to computer science ) is to determine, given a graph G on n vertices, whether that cycle decision problem (, only graph contains an Euler cycle or not. the path visits edges, rather than vertices. Consider the following remarkable theorem, to be proven in Exercise 3.20: is, in fact, exactly the same problem as Theorem 3.1: (Euler\u2019s theorem) A connected graph contains an Euler cycle if and only if every vertex has an even number of edges incident upon it.. First, check to see whether Euler\u2019s theorem gives us a method for ef\ufb01ciently solving the graph is connected; this is easily done with O(n2) operations, as shown in Exercise 3.19. If the graph is not connected, then obviously no Euler cycle exists. If the graph is connected then for each vertex check whether there is an even number of edges incident upon the vertex. If a vertex is found for which this is not the case, then there is no Euler cycle, otherwise an Euler cycle exists. Since there are n vertices, and at most n(n \u2212 1)/2 edges, this algorithm requires O(n3) elementary operations. Thus is in P! Somehow, there is a structure present in the problem of visiting each edge that can be exploited, yet which does not seem to be re\ufb02ected in the to provide an ef\ufb01cient algorithm for problem of visiting each vertex; it is not at all obvious why such a structure should be present in one case, but not in the other, if indeed it is absent for the problem. Exercise 3.19: The problem is to determine whether there is a path can be between two speci\ufb01ed vertices in a graph. Show that solved using O(n) operations if the graph has n vertices. Use the solution to to show that it is possible to decide whether a graph is connected in O(n2) operations. Exercise 3.20: (Euler\u2019s theorem) Prove Euler\u2019s theorem. In particular, if each vertex has an even number of incident edges, give a constructive procedure for \ufb01nding an Euler cycle. The equivalence between the factoring decision problem and the factoring problem proper is a special instance of one of the most important ideas in computer science, an idea known as reduction. Intuitive", "ly, we know that some problems can be viewed as special instances of other problems. A less trivial example of reduction is the reduction of ). The traveling salesman decision problem is as follows: we are given n cities 1, 2,..., n and a non-negative integer distance dij between each pair of cities. Given a distance d the problem is to determine if there is a tour of all the cities of distance less than d. to the traveling salesman decision problem ( to The reduction of goes as follows. Suppose we have a graph containing n by thinking of each vertex of the graph as a vertices. We turn this into an instance of \u2018city\u2019 and de\ufb01ning the distance dij between cities i and j to be one if vertices i and j are connected, and the distance to be two if the vertices are unconnected. Then a tour of the cities of distance less than n + 1 must be of distance n, and be a Hamiltonian cycle for the graph. Conversely, if a Hamiltonian cycle exists then a tour of the cities of distance less than n + 1 must exist. In this way, given an algorithm for solving, we can convert without much overhead. Two consequences can be it into an algorithm for solving is also tractable. Second, if inferred from this. First, if must also be hard. This is an example of a general technique known is a tractable problem, then is hard then The analysis of computational problems 145 as reduction: we\u2019ve reduced the problem will use repeatedly throughout this book. to the problem. This is a technique we A more general notion of reduction is illustrated in Figure 3.11. A language B is said to be reducible to another language A if there exists a Turing machine operating in polynomial time such that given as input x it outputs R(x), and x \u2208 B if and only if R(x) \u2208 A. Thus, if we have an algorithm for deciding A, then with a little extra overhead we can decide the language B. In this sense, the language B is essentially no more dif\ufb01cult to decide than the language A. Is x \u2208 B Compute R x ( ) in polynomial time Is R x \u2208 A ( )? \u2018\u2018Yes\u2019\u2019 or \u2018\u2018No\u2019\u2019 Figure 3.11. Reduction of B to A. Exercise 3.21: (Transitive property of reduction) Show that if a language L1 is reducible to the language L2 and the language L2 is reducible to L3 then the language L1 is reducible to the language L3. Some complexity classes have problems which are complete with respect to that complexity class, meaning there is a language L in the complexity class which is the \u2018most dif\ufb01cult\u2019 to decide, in the sense that every other language in the complexity class can be reduced to L. Not all complexity classes have complete problems, but many of the complexity classes we are concerned with do have complete problems. A trivial example is provided by P. Let L be any language in P which is not empty or equal to the set of all words. That is, there exists a string x1 such that x1 \u2208 L and a string x2 such that x2 \u2208 L. Then any other language L in P can be reduced to L using the following reduction: given an input x, use the polynomial time decision procedure to determine whether x \u2208 L or not. If it is not, then set R(x) = x1, otherwise set R(x) = x2. Exercise 3.22: Suppose L is complete for a complexity class, and L is another language in the complexity class such that L reduces to L. Show that L is complete for the complexity class. Less trivially, NP also contains complete problems. An important example of such a problem and the prototype for all other NP-complete problems is the circuit satis\ufb01ability gates, is there problem or an assignment of values to the inputs to the circuit that results in the circuit outputting 1, that is, is the circuit satis\ufb01able for some input? The NP-completeness of is known as the Cook\u2013Levin theorem, for which we now outline a proof. : given a Boolean circuit composed of and, 146 Introduction to computer science Theorem 3.2: (Cook\u2013Levin) is NP-complete. Proof is in NP, and The proof has two parts. The \ufb01rst part of the proof is to show that the second part is to show that any language in NP can be reduced to. Both parts of the proof are based on simulation techniques: the \ufb01rst part of the proof is essentially showing that a Turing machine can ef\ufb01ciently simulate a circuit, while the second part of the proof is essentially showing", " that a circuit can ef\ufb01ciently simulate a Turing machine. Both parts of the proof are quite straightforward; for the purposes of illustration we give the second part in some detail. The \ufb01rst part of the proof is to show that is in NP. Given a circuit containing n circuit elements, and a potential witness w, it is obviously easy to check in polynomial time on a Turing machine whether or not w satis\ufb01es the circuit, which establishes that is in NP. The second part of the proof is to show that any language L \u2208 NP can be reduced to. That is, we aim to show that there is a polynomial time computable reduction R such that x \u2208 L if and only if R(x) is a satis\ufb01able circuit. The idea of the reduction is to \ufb01nd a circuit which simulates the action of the machine M which is used to check instance-witness pairs, (x, w), for the language L. The input variables for the circuit will represent the witness; the idea is that \ufb01nding a witness which satis\ufb01es the circuit is equivalent to M accepting (x, w) for some speci\ufb01c witness w. Without loss of generality we may make the following assumptions about M to simplify the construction: (1) M \u2019s tape alphabet is,0,1 and the blank symbol. (2) M runs using time at most t(n) and total space at most s(n) where t(n) and s(n) are polynomials in n. (3) Machine M can actually be assumed to run using time exactly t(n) for all inputs of size n. This is done by adding the lines qY, x, qY, x, 0, and qN, x, qN, x, 0 for each of x =, 0, 1 and the blank, arti\ufb01cially halting the machine after exactly t(n) steps. The basic idea of the construction to simulate M is outlined in Figure 3.12. Each internal state of the Turing machine is represented by a single bit in the circuit. We name the corresponding bits \u02dcqs, \u02dcq1,..., \u02dcqm, \u02dcqY, \u02dcqN. Initially, \u02dcqs is set to one, and all the other bits representing internal states are set to zero. Each square on the Turing machine tape is represented by three bits: two bits to represent the letter of the alphabet (, 0, 1 or blank) currently residing on the tape, and a single \u2018\ufb02ag\u2019 bit which is set to one if the read-write head is pointing to the square, and set to zero otherwise. We denote the bits representing the tape contents by (u1, v1),..., (us(n), vs(n)) and the corresponding \ufb02ag bits by f1,..., fs(n). Initially the uj and vj bits are set to represent the inputs x and w, as appropriate, while f1 = 1 and all other fj = 0. There is also a lone extra \u2018global \ufb02ag\u2019 bit, F, whose function will be explained later. F is initially set to zero. We regard all the bits input to the circuit as \ufb01xed, except for those representing the witness w, which are the variable bits for the circuit. The action of M is obtained by repeating t(n) times a \u2018simulation step\u2019 which simulates the execution of a single program line for the Turing machine. Each simulation step may be broken up into a sequence of steps corresponding in turn to the respective program lines, with a \ufb01nal step which resets the global \ufb02ag F to zero, as The analysis of computational problems 147 m + 3 fixed input bits 3n + 6 fixed input bits 3w(n) variable input bits 3 ( ) s n fixed input bits 1 fixed input bit \u23a7 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 \u23a7 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 \u23a7 \u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\ufffd", "\ufffd\u23a9 \u23a7 \u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23a9 \u23a7 \u23a8 \u23a9 qs \u02dc q1 \u02dc... qm \u02dc qY \u02dc qN \u02dc x b /m+3 /3n+6 w /3w(n) b... b F /3s(n) \" Simulation Step Simulation Step output bit \u02dcqY Simulation Step \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 # t n ( ) simulation steps! Figure 3.12. Outline of the procedure used to simulate a Turing machine using a circuit. illustrated in Figure 3.13. To complete the simulation, we only need to simulate a program line of the form qi, x, qj, x, s. For convenience, we assume qi = qj, but a similar construction works in the case when qi = qj. The procedure is as follows: (1) Check to see that \u02dcqi = 1, indicating that the current state of the machine is qi. (2) For each tape square: (a) Check to see that the global \ufb02ag bit is set to zero, indicating that no action has yet been taken by the Turing machine. (b) Check that the \ufb02ag bit is set to one, indicating that the tape head is at this tape square. (c) Check that the simulated tape contents at this point are x. (d) If all conditions check out, then perform the following steps: 1. Set \u02dcqi = 0 and \u02dcqj = 1. 2. Update the simulated tape contents at this tape square to x. 3. Update the \ufb02ag bit of this and adjacent \u2018squares\u2019 as appropriate, depending on whether s = +1, 0, \u22121, and whether we are at the left hand end of the tape. 4. Set the global \ufb02ag bit to one, indicating that this round of computation has been completed. 148 Introduction to computer science This is a \ufb01xed procedure which involves a constant number of bits, and by the universality result of Section 3.1.2 can be performed using a circuit containing a constant number of gates. Figure 3.13. Outline of the simulation step used to simulate a Turing machine using a circuit. The total number of gates in the entire circuit is easily seen to be O(t(n)(s(n) + n)), which is polynomial in size. At the end of the circuit, it is clear that \u02dcqY = 1 if and only if the machine M accepts (x, w). Thus, the circuit is satis\ufb01able if and only if there exists w such that machine M accepts (x, w), and we have found the desired reduction from L to. gives us a foot in the door which enables us to easily prove that many other problems are NP-complete. Instead of directly proving that a problem is NP-complete, we can instead prove that it is in NP and that reduces to it, so by Exercise 3.22 the problem must be NP-complete. A small sample of NP-complete problems is discussed in Box 3.3. An example of another NP-complete problem is the satis\ufb01ability problem ), which is phrased in terms of a Boolean formula. Recall that a Boolean formula ( \u03d5 is composed of the following elements: a set of Boolean variables, x1, x2,...; Boolean connectives, that is, a Boolean function with one or two inputs and one output, such as \u2227 (AND), \u2228 (OR), and \u00ac (NOT); and parentheses. The truth or falsity of a Boolean formula for a given set of Boolean variables is decided according to the usual laws of Boolean algebra. For example, the formula \u03d5 = x1 \u2228 \u00acx2 has the satisfying assignment x1 = 0 and x2 = 0, while x1 = 0 and x2 = 1 is not a satisfying assignment. The satis\ufb01ability problem is to determine, given a Boolean formula \u03d5, whether or not it is satis\ufb01able by any set of possible inputs. Exercise 3.23: Show that is NP-complete by \ufb01rst showing that reduces to. (Hint: for the reduction it may help to is in NP, and then showing that represent each distinct wire in an instance of Boolean formula.) by different variables in a An important restricted case of is also NP-complete, the 3-satis\ufb01ability problem ), which is concerned with formulae in 3-conjunctive normal form. A formula is ( said to be in conjunctive normal form if it is the AND of a collection", " of clauses, each of which is the OR of one or more literals, where a literal is an expression is of the form x or \u00acx. For example, the formula (x1 \u2228 \u00acx2) \u2227 (x2 \u2228 x3 \u2228 \u00acx4) is in conjunctive normal form. A formula is in 3-conjunctive normal form or 3-CNF if each clause has exactly three literals. For example, the formula (\u00acx1 \u2228x2 \u2228\u00acx2)\u2227(\u00acx1 \u2228x3 \u2228\u00acx4)\u2227(x2 \u2228x3 \u2228x4) is in 3-conjunctive normal form. The 3-satis\ufb01ability problem is to determine whether a formula in 3-conjunctive normal form is satis\ufb01able or not. The proof that justify inclusion in this overview. Even more than is NP-complete is straightforward, but is a little too lengthy to is in some sense and, The analysis of computational problems 149 the NP-complete problem, and it is the basis for countless proofs that other problems are NP-complete. We conclude our discussion of NP-completeness with the surprising in which every clause has two literals, can be solved fact that in polynomial time:, the analogue of Exercise 3.24: ( has an ef\ufb01cient solution) Suppose \u03d5 is a Boolean formula in conjunctive normal form, in which each clause contains only two literals. (1) Construct a (directed) graph G(\u03d5) with directed edges in the following way: the vertices of G correspond to variables xj and their negations \u00acxj in \u03d5. There is a (directed) edge (\u03b1, \u03b2) in G if and only if the clause (\u00ac\u03b1 \u2228 \u03b2) or the clause (\u03b2 \u2228 \u00ac\u03b1) is present in \u03d5. Show that \u03d5 is not satis\ufb01able if and only if there exists a variable x such that there are paths from x to \u00acx and from \u00acx to x in G(\u03d5). (2) Show that given a directed graph G containing n vertices it is possible to determine whether two vertices v1 and v2 are connected in polynomial time. (3) Find an ef\ufb01cient algorithm to solve. Box 3.3: A zoo of NP-complete problems The importance of the class NP derives, in part, from the enormous number of computational problems that are known to be NP-complete. We can\u2019t possibly hope to survey this topic here (see \u2018History and further reading\u2019), but the following examples, taken from many distinct areas of mathematics, give an idea of the delicious melange of problems known to be NP-complete. \u2022 \u2022 \u2022 \u2022 (graph theory): A clique in an undirected graph G is a subset of vertices, each pair of which is connected by an edge. The size of a clique is the number of vertices it contains. Given an integer m and a graph G, does G have a clique of size m? (arithmetic): Given a \ufb01nite collection S of positive integers and a target t, is there any subset of S which sums to t? (linear programming): Given an integer m \u00d7 n matrix A and an m-dimensional vector y with integer values, does there exist an n-dimensional vector x with entries in the set {0, 1} such that Ax \u2264 y? (graph theory): A vertex cover for an undirected graph G is a set of vertices V such that every edge in the graph has one or both vertices contained in V. Given an integer m and a graph G, does G have a vertex cover V containing m vertices? Assuming that P = NP it is possible to prove that there is a non-empty class of problems NPI (NP intermediate) which are neither solvable with polynomial resources, nor are NP-complete. Obviously, there are no problems known to be in NPI (otherwise we would know that P = NP) but there are several problems which are regarded as being likely candidates. Two of the strongest candidates are the factoring and graph isomorphism problems: 150 Introduction to computer science : Suppose G and G are two undirected graphs over the vertices V \u2261 {v1,..., vn}. Are G and G isomorphic? That is, does there exist a one-to-one function \u03d5 : V \u2192 V such that the edge (vi, vj) is contained in G if and only if (\u03d5(vi), \u03d5(vj )) is contained in G? Problems in NPI are interesting to researchers in quantum computation and quantum information for two reasons. First", ", it is desirable to \ufb01nd fast quantum algorithms to solve problems which are not in P. Second, many suspect that quantum computers will not be able to ef\ufb01ciently solve all problems in NP, ruling out NP-complete problems. Thus, it is natural to focus on the class NPI. Indeed, a fast quantum algorithm for factoring has been discovered (Chapter 5), and this has motivated the search for fast quantum algorithms for other problems suspected to be in NPI. 3.2.4 A plethora of complexity classes We have investigated some of the elementary properties of some important complexity classes. A veritable pantheon of complexity classes exists, and there are many non-trivial relationships known or suspected between these classes. For quantum computation and quantum information, it is not necessary to understand all the different complexity classes that have been de\ufb01ned. However, it is useful to have some appreciation for the more important of the complexity classes, many of which have natural analogues in the study of quantum computation and quantum information. Furthermore, if we are to understand how powerful quantum computers are, then it behooves us to understand how the class of problems solvable on a quantum computer \ufb01ts into the zoo of complexity classes which may be de\ufb01ned for classical computers. There are essentially three properties that may be varied in the de\ufb01nition of a complexity class: the resource of interest (time, space,... ), the type of problem being considered (decision problem, optimization problem,... ), and the underlying computational model (deterministic Turing machine, probabilistic Turing machine, quantum computer,... ). Not surprisingly, this gives us an enormous range to de\ufb01ne complexity classes. In this section, we brie\ufb02y review a few of the more important complexity classes and some of their elementary properties. We begin with a complexity class de\ufb01ned by changing the resource of interest from time to space. The most natural space-bounded complexity class is the class PSPACE of decision problems which may be solved on a Turing machine using a polynomial number of working bits, with no limitation on the amount of time that may be used by the machine (see Exercise 3.25). Obviously, P is included in PSPACE, since a Turing machine that halts after polynomial time can only traverse polynomially many squares, but it is also true that NP is a subset of PSPACE. To see this, suppose L is any language in NP. Suppose problems of size n have witnesses of size at most p(n), where p(n) is some polynomial in n. To determine whether or not the problem has a solution, we may sequentially test all 2p(n) possible witnesses. Each test can be run in polynomial time, and therefore polynomial space. If we erase all the intermediate working between tests then we can test all the possibilities using polynomial space. Unfortunately, at present it is not even known whether PSPACE contains problems which are not in P! This is a pretty remarkable situation \u2013 it seems fairly obvious that having unlimited time and polynomial spatial resources must be more powerful than having only a polynomial amount of time. However, despite considerable effort and in- The analysis of computational problems 151 genuity, this has never been shown. We will see later that the class of problems solvable on a quantum computer in polynomial time is a subset of PSPACE, so proving that a problem ef\ufb01ciently solvable on a quantum computer is not ef\ufb01ciently solvable on a classical computer would establish that P = PSPACE, and thus solve a major outstanding problem of computer science. An optimistic way of looking at this result is that ideas from quantum computation might be useful in proving that P = PSPACE. Pessimistically, one might conclude that it will be a long time before anyone rigorously proves that quantum computers can be used to ef\ufb01ciently solve problems that are intractable on a classical computer. Even more pessimistically, it is possible that P = PSPACE, in which case quantum computers offer no advantage over classical computers! However, very few (if any) computational complexity theorists believe that P = PSPACE. Exercise 3.25: (PSPACE \u2286 EXP) The complexity class EXP (for exponential time) contains all decision problems which may be decided by a Turing machine running in exponential time, that is time O(2nk that PSPACE \u2286 EXP. (Hint: If a Turing machine has l internal states, an m letter alphabet, and uses space p(n), argue that the machine can exist in one of at most lmp(n) different states, and that if the Turing machine is to avoid in\ufb01nite loops then it must halt before revisiting a state.) ), where k is", " any constant. Prove Exercise 3.26: (L \u2286 P) The complexity class L (for logarithmic space) contains all decision problems which may be decided by a Turing machine running in logarithmic space, that is, in space O(log(n)). More precisely, the class L is de\ufb01ned using a two-tape Turing machine. The \ufb01rst tape contains the problem instance, of size n, and is a read-only tape, in the sense that only program lines which don\u2019t change the contents of the \ufb01rst tape are allowed. The second tape is a working tape which initially contains only blanks. The logarithmic space requirement is imposed on the second, working tape only. Show that L \u2286 P. Does allowing more time or space give greater computational power? The answer to this question is yes in both cases. Roughly speaking, the time hierarchy theorem states that TIME(f (n)) is a proper subset of TIME(f (n) log2(f (n))). Similarly, the space hierarchy theorem states that SPACE(f (n)) is a proper subset of SPACE(f (n) log(f (n))), where SPACE(f (n)) is, of course, the complexity class consisting of all languages that can be decided with spatial resources O(f (n)). The hierarchy theorems have interesting implications with respect to the equality of complexity classes. We know that L \u2286 P \u2286 NP \u2286 PSPACE \u2286 EXP. (3.1) Unfortunately, although each of these inclusions is widely believed to be strict, none of them has ever been proved to be strict. However, the time hierarchy theorem implies that P is a strict subset of EXP, and the space hierarchy theorem implies that L is a strict subset of PSPACE! So we can conclude that at least one of the inclusions in (3.1) must be strict, although we do not know which one. What should we do with a problem once we know that it is NP-complete, or that some other hardness criterion holds? It turns out that this is far from being the end of the story in problem analysis. One possible line of attack is to identify special cases of the problem which may be amenable to attack. For example, in Exercise 3.24 we saw. that the problem has an ef\ufb01cient solution, despite the NP-completeness of 152 Introduction to computer science Another approach is to change the type of problem which is being considered, a tactic which typically results in the de\ufb01nition of new complexity classes. For example, instead of \ufb01nding exact solutions to an NP-complete problem, we can instead try to \ufb01nd good algorithms for \ufb01nding approximate solutions to a problem. For example, the problem is an NP-complete problem, yet in Exercise 3.27 we show that it is possible to ef\ufb01ciently \ufb01nd an approximation to the minimal vertex cover which is correct to within a factor two! On the other hand, in Problem 3.6 we show that it is not possible correct to within any factor, unless P = NP! to \ufb01nd approximations to solutions of Exercise 3.27: (Approximation algorithm for ) Let G = (V, E) be an undirected graph. Prove that the following algorithm \ufb01nds a vertex cover for G that is within a factor two of being a minimal vertex cover: V C = \u2205 E = E do until E = \u2205 let (\u03b1, \u03b2) be any edge of E V C = V C \u222a {\u03b1, \u03b2} remove from E every edge incident on \u03b1 or \u03b2 return V C. Why is it possible to approximate the solution of one NP-complete problem, but not another? After all, isn\u2019t it possible to ef\ufb01ciently transform from one problem to another? This is certainly true, however it is not necessarily true that this transformation preserves the notion of a \u2018good approximation\u2019 to a solution. As a result, the computational complexity theory of approximation algorithms for problems in NP has a structure that goes beyond the structure of NP proper. An entire complexity theory of approximation algorithms exists, which unfortunately is beyond the scope of this book. The basic idea, however, is to de\ufb01ne a notion of reduction that corresponds to being able to ef\ufb01ciently reduce one approximation problem to another, in such a way that the notion of good approximation is preserved. With such a notion, it is possible to de\ufb01ne complexity classes such as MAXSNP by analogy to the class NP, as the set of problems for which it is possible to ef\ufb01ciently verify approximate solutions to the problem. Complete problems exist for MAXSN", "P, just as for NP, and it is an interesting open problem to determine how the class MAXSNP compares to the class of approximation problems which are ef\ufb01ciently solvable. We conclude our discussion with a complexity class that results when the underlying model of computation itself is changed. Suppose a Turing machine is endowed with the ability to \ufb02ip coins, using the results of the coin tosses to decide what actions to take during the computation. Such a Turing machine may only accept or reject inputs with a certain probability. The complexity class BPP (for bounded-error probabilistic time) contains all languages L with the property that there exists a probabilistic Turing machine M such that if x \u2208 L then M accepts x with probability at least 3/4, and if x \u2208 L, then M rejects x with probability at least 3/4. The following exercise shows that the choice of the constant 3/4 is essentially arbitrary: The analysis of computational problems 153 Exercise 3.28: (Arbitrariness of the constant in the de\ufb01nition of BPP) Suppose k is a \ufb01xed constant, 1/2 < k \u2264 1. Suppose L is a language such that there exists a Turing machine M with the property that whenever x \u2208 L, M accepts x with probability at least k, and whenever x \u2208 L, M rejects x with probability at least k. Show that L \u2208 BPP. Indeed, the Chernoff bound, discussed in Box 3.4, implies that with just a few repetitions of an algorithm deciding a language in BPP the probability of success can be ampli\ufb01ed to the point where it is essentially equal to one, for all intents and purposes. For this reason, BPP even more than P is the class of decision problems which is usually regarded as being ef\ufb01ciently solvable on a classical computer, and it is the quantum analogue of BPP, known as BQP, that is most interesting in our study of quantum algorithms. 3.2.5 Energy and computation Computational complexity studies the amount of time and space required to solve a computational problem. Another important computational resource is energy. In this section, we study the energy requirements for computation. Surprisingly, it turns out that computation, both classical and quantum, can in principle be done without expending any energy! Energy consumption in computation turns out to be deeply linked to the reversibility of the computation. Consider a gate like the gate, which takes as input two bits, and produces a single bit as output. This gate is intrinsically irreversible because, given the output of the gate, the input is not uniquely determined. For example, gate is 1, then the input could have been any one of 00, 01, if the output of the gate is an example of a reversible logic gate because, or 10. On the other hand, the gate, it is possible to infer what the input must have been. given the output of the Another way of understanding irreversibility is to think of it in terms of information erasure. If a logic gate is irreversible, then some of the information input to the gate is lost irretrievably when the gate operates \u2013 that is, some of the information has been erased by the gate. Conversely, in a reversible computation, no information is ever erased, because the input can always be recovered from the output. Thus, saying that a computation is reversible is equivalent to saying that no information is erased during the computation. What is the connection between energy consumption and irreversibility in computation? Landauer\u2019s principle provides the connection, stating that, in order to erase information, it is necessary to dissipate energy. More precisely, Landauer\u2019s principle may be stated as follows: Landauer\u2019s principle (\ufb01rst form): Suppose a computer erases a single bit of information. The amount of energy dissipated into the environment is at least kBT ln 2, where kB is a universal constant known as Boltzmann\u2019s constant, and T is the temperature of the environment of the computer. According to the laws of thermodynamics, Landauer\u2019s principle can be given an alternative form stated not in terms of energy dissipation, but rather in terms of entropy: Landauer\u2019s principle (second form): Suppose a computer erases a single bit of information. The entropy of the environment increases by at least kB ln 2, where kB is Boltzmann\u2019s constant. Justifying Landauer\u2019s principle is a problem of physics that lies beyond the scope of this 154 Introduction to computer science Box 3.4: BPP and the Chernoff bound Suppose we have an algorithm for a decision problem which gives the correct answer with probability 1/2 +, and the wrong answer with probability 1/2 \u2212. If we run the algorithm n times, then it seems reasonable to guess that the", " correct answer is whichever appeared most frequently. How reliably does this procedure work? The Chernoff bound is a simple result from elementary probability which answers this question. Theorem 3.3: (The Chernoff bound) Suppose X1,..., Xn are independent and identically distributed random variables, each taking the value 1 with probability 1/2 +, and the value 0 with probability 1/2 \u2212. Then n p Xi \u2264 n/2 \u2264 e\u221222n. (3.2) i=1 Proof Consider any sequence (x1,..., xn) containing at most n/2 ones. The probability of such a sequence occurring is maximized when it contains n/2 ones, so p(X1 = x1,..., Xn = xn3.3) (3.4) 2 n 2. = (1 \u2212 42) 2n There can be at most 2n such sequences, so i p Xi \u2264 n/2 \u2264 2n \u00d7 (1 \u2212 42) 2n n 2 = (1 \u2212 42) n 2. (3.5) Finally, by calculus, 1 \u2212 x \u2264 exp(\u2212x), so p Xi \u2264 n/2 \u2264 e\u221242n/2 = e\u221222n. (3.6) i What this tells us is that for \ufb01xed, the probability of making an error decreases exponentially quickly in the number of repetitions of the algorithm. In the case of BPP we have = 1/4, so it takes only a few hundred repetitions of the algorithm to reduce the probability of error below 10\u221220, at which point an error in one of the computer\u2019s components becomes much more likely than an error due to the probabilistic nature of the algorithm. book \u2013 see the end of chapter \u2018History and further reading\u2019 if you wish to understand why Landauer\u2019s principle holds. However, if we accept Landauer\u2019s principle as given, then it raises a number of interesting questions. First of all, Landauer\u2019s principle only provides a lower bound on the amount of energy that must be dissipated to erase information. The analysis of computational problems 155 How close are existing computers to this lower bound? Not very, turns out to be the answer \u2013 computers circa the year 2000 dissipate roughly 500kBT ln 2 in energy for each elementary logical operation. Although existing computers are far from the limit set by Landauer\u2019s principle, it is still an interesting problem of principle to understand how much the energy consumption can be reduced. Aside from the intrinsic interest of the problem, a practical reason for the interest follows from Moore\u2019s law: if computer power keeps increasing then the amount of energy dissipated must also increase, unless the energy dissipated per operation drops at least as fast as the rate of increase in computing power. If all computations could be done reversibly, then Landauer\u2019s principle would imply no lower bound on the amount of energy dissipated by the computer, since no bits at all are erased during a reversible computation. Of course, it is possible that some other physical principle might require that energy be dissipated during the computation; fortunately, this turns out not to be the case. But is it possible to perform universal computation without erasing any information? Physicists can cheat on this problem to see in advance that the answer to this question must be yes, because our present understanding of the laws of physics is that they are fundamentally reversible. That is, if we know the \ufb01nal state of a closed physical system, then the laws of physics allow us to work out the initial state of the system. If we believe that those laws are correct, then we must conclude that, there must be some underlying hidden in the irreversible logic gates like reversible computation. But where is this hidden reversibility, and can we use it to construct manifestly reversible computers? and We will use two different techniques to give reversible circuit-based models capable of universal computation. The \ufb01rst model, a computer built entirely of billiard balls and mirrors, gives a beautiful concrete realization of the principles of reversible computation. The second model, based on a reversible logic gate known as the Toffoli gate (which we \ufb01rst encountered in Section 1.4.1), is a more abstract view of reversible computation that will later be of great use in our discussion of quantum computation. It is also possible to build reversible Turing machines that are universal for computation; however, we won\u2019t study these here, since the reversible circuit models turn out to be much more useful for quantum computation. The basic idea of the billiard ball computer is illustrated in Figure 3.14. Billiard ball \u2018inputs\u2019 enter the computer from the left hand side, bouncing off mirrors and each other, before exiting as \u2018outputs\u2019 on the right hand side. The presence or absence of", " a billiard ball at a possible input site is used to indicate a logical 1 or a logical 0, respectively. The fascinating thing about this model is that it is manifestly reversible, insofar as its operation is based on the laws of classical mechanics. Furthermore, this model of computation turns out to be universal in the sense that it can be used to simulate an arbitrary computation in the standard circuit model of computation. Of course, if a billiard ball computer were ever built it would be highly unstable. As any billiards player can attest, a billiard ball rolling frictionlessly over a smooth surface is easily knocked off course by small perturbations. The billiard ball model of computation depends on perfect operation, and the absence of external perturbations such as those caused by thermal noise. Periodic corrections can be performed, but information gained by doing this would have to be erased, requiring work to be performed. Expenditure of energy thus serves the purpose of reducing this susceptibility to noise, which is necessary for a practical, real-world computational machine. For the purposes of this introduction, 156 Introduction to computer science. Figure 3.14. A simple billiard ball computer, with three input bits and three output bits, shown entering on the left and leaving on the right, respectively. The presence or absence of a billiard ball indicates a 1 or a 0, respectively. Empty circles illustrate potential paths due to collisions. This particular computer implements the Fredkin classical reversible logic gate, discussed in the text. we will ignore the effects of noise on the billiard ball computer, and concentrate on understanding the essential elements of reversible computation. The billiard ball computer provides an elegant means for implementing a reversible universal logic gate known as the Fredkin gate. Indeed, the properties of the Fredkin gate provide an informative overview of the general principles of reversible logic gates and circuits. The Fredkin gate has three input bits and three output bits, which we refer to as a, b, c and a, b, c, respectively. The bit c is a control bit, whose value is not changed by the action of the Fredkin gate, that is, c = c. The reason c is called the control bit is because it controls what happens to the other two bits, a and b. If c is set to 0 then a and b are left alone, a = a, b = b. If c is set to 1, a and b are swapped, a = b, b = a. The explicit truth table for the Fredkin gate is shown in Figure 3.15. It is easy to see that the Fredkin gate is reversible, because given the output a, b, c, we can determine the inputs a, b, c. In fact, to recover the original inputs a, b and c we need only apply another Fredkin gate to a, b, c: Exercise 3.29: (Fredkin gate is self-inverse) Show that applying two consecutive Fredkin gates gives the same outputs as inputs. Examining the paths of the billiard balls in Figure 3.14, it is not dif\ufb01cult to verify that this billiard ball computer implements the Fredkin gate: Exercise 3.30: Verify that the billiard ball computer in Figure 3.14 computes the Fredkin gate. In addition to reversibility, the Fredkin gate also has the interesting property that the number of 1s is conserved between the input and output. In terms of the billiard ball computer, this corresponds to the number of billiard balls going into the Fredkin gate being equal to the number coming out. Thus, it is sometimes referred to as being a conservative reversible logic gate. Such reversibility and conservative properties are interesting to a physicist because they can be motivated by fundamental physical princi- The analysis of computational problems 157 Inputs Outputs Figure 3.15. Fredkin gate truth table and circuit representation. The bits a and b are swapped if the control bit c is set, and otherwise are left alone. ples. The laws of Nature are reversible, with the possible exception of the measurement postulate of quantum mechanics, discussed in Section 2.2.3 on page 84. The conservative property can be thought of as analogous to properties such as conservation of mass, or conservation of energy. Indeed, in the billiard ball model of computation the conservative property corresponds exactly to conservation of mass. Figure 3.16. Fredkin gate con\ufb01gured to perform the elementary gates operation, since it routing function, the produces two copies of x at the output. Note that each of these con\ufb01gurations requires the use of extra \u2018ancilla\u2019 bits gate \u2013 and in general the output prepared in standard states \u2013 for example, the 0 input on the \ufb01rst line of the contains \u2018garbage\u2019 not needed for the remainder of the computation. (right). The middle gate also serves to perform the (middle), and a primitive (left), The Fredkin", " gate is not only reversible and conservative, it\u2019s a universal logic gate as well! As illustrated in Figure 3.16, the Fredkin gate can be con\ufb01gured to simulate functions, and thus can be cascaded to simulate any,, and classical circuit whatsoever. To simulate irreversible gates such as using the Fredkin gate, we made use of two ideas. First, we allowed the input of \u2018ancilla\u2019 bits to the Fredkin gate, in specially prepared states, either 0 or 1. Second, the output of the Fredkin gate contained extraneous \u2018garbage\u2019 not needed for the remainder of the computation. These ancilla and garbage bits are not directly important to the computation. Their importance lies in the fact that they make may the computation reversible. Indeed the irreversibility of gates like the be viewed as a consequence of the ancilla and garbage bits being \u2018hidden\u2019. Summarizing, given any classical circuit computing a function f (x), we can build a reversible circuit made entirely of Fredkin gates, which on input of x, together with some ancilla bits and 158 Introduction to computer science in a standard state a, computes f (x), together with some extra \u2018garbage\u2019 output, g(x). Therefore, we represent the action of the computation as (x, a) \u2192 (f (x), g(x)). We now know how to compute functions reversibly. Unfortunately, this computation produces unwanted garbage bits. With some modi\ufb01cations it turns out to be possible to perform the computation so that any garbage bits produced are in a standard state. This construction is crucial for quantum computation, because garbage bits whose value depends upon x will in general destroy the interference properties crucial to quantum gate computation. To understand how this works it is convenient to assume that the is available in our repertoire of reversible gates, so we may as well assume that the ancilla bits a all start out as 0s, with gates being added where necessary to turn the ancilla gate is 0s into 1s. It will also be convenient to assume that the classical controlledavailable, de\ufb01ned in a manner analogous to the quantum de\ufb01nition of Section 1.3.2, that is, the inputs (c, t) are taken to (c, t \u2295 c), where \u2295 denotes addition modulo 2. Notice that t = 0 gives (c, 0) \u2192 (c, c), so the controlledcan be thought of as a reversible copying gate or, which leaves no garbage bits at the output. With the additional gates appended at the beginning of the circuit, the action of the computation may be written as (x, 0) \u2192 (f (x), g(x)). We could also have added gates to the beginning of the circuit, in order to create a copy of x which is not changed during the subsequent computation. With this modi\ufb01cation, the action of the circuit may be written (x, 0, 0) \u2192 (x, f (x), g(x)). (3.7) Equation (3.7) is a very useful way of writing the action of the reversible circuit, because it allows an idea known as uncomputation to be used to get rid of the garbage bits, for a small cost in the running time of the computation. The idea is the following. Suppose we start with a four register computer in the state (x, 0, 0, y). The second register is used to store the result of the computation, and the third register is used to provide workspace for the computation, that is, the garbage bits g(x). The use of the fourth register is described shortly, and we assume it starts in an arbitrary state y. We begin as before, by applying a reversible circuit to compute f, resulting in the state (x, f (x), g(x), y). Next, we use s to add the result f (x) bitwise to the fourth register, leaving the machine in the state (x, f (x), g(x), y \u2295 f (x)). However, all the steps used to compute f (x) were reversible and did not affect the fourth register, so by applying the reverse of the circuit used to compute f we come to the state (x, 0, 0, y \u2295f (x)). Typically, we omit the ancilla 0s from the description of the function evaluation, and just write the action of the circuit as (x, y) \u2192 (x, y \u2295 f (x)). (3.8) In general we refer to this modi\ufb01ed circuit computing f as the reversible circuit computing f, even though in principle there are many other reversible circuits which could be used to compute f. What resource overhead is involved in doing reversible computation? To analyze this", " question, we need to count the number of extra ancilla bits needed in a reversible circuit, and compare the gate counts with classical models. It ought to be clear that the number of gates in a reversible circuit is the same as in an irreversible circuit to within the constant factor which represents the number of Fredkin gates needed to simulate a single element of the irreversible circuit, and an additional factor of two for uncomputation, with an The analysis of computational problems 159 operations used in reversible computation which is linear in overhead for the extra the number of bits involved in the circuit. Similarly, the number of ancilla bits required scales at most linearly with the number of gates in the irreversible circuit, since each element in the irreversible circuit can be simulated using a constant number of ancilla bits. As a result, natural complexity classes such as P and NP are the same no matter whether a reversible or irreversible model of computation is used. For more elaborate complexity classes like PSPACE the situation is not so immediately clear; see Problem 3.9 and \u2018History and further reading\u2019 for a discussion of some such subtleties. Exercise 3.31: (Reversible half-adder) Construct a reversible circuit which, when two bits x and y are input, outputs (x, y, c, x \u2295 y), where c is the carry bit when x and y are added. The Fredkin gate and its implementation using the billiard ball computer offers a beautiful paradigm for reversible computation. There is another reversible logic gate, the Toffoli gate, which is also universal for classical computation. While the Toffoli gate does not have quite the same elegant physical simplicity as the billiard ball implementation of the Fredkin gate, it will be more useful in the study of quantum computation. We have already met the Toffoli gate in Section 1.4.1, but for convenience we review its properties here. The Toffoli gate has three input bits, a, b and c. a and b are known as the \ufb01rst and second control bits, while c is the target bit. The gate leaves both control bits unchanged, \ufb02ips the target bit if both control bits are set, and otherwise leaves the target bit alone. The truth table and circuit representation for the Toffoli gate are shown in Figure 3.17. Inputs Outputs Figure 3.17. Truth table and circuit representation of the Toffoli gate. How can the Toffoli gate be used to do universal computation? Suppose we wish to the bits a and b. To do this using the Toffoli gate, we input a and b as control bits, and send in an ancilla bit set to 1 as the target bit, as shown in Figure 3.18. The of a and b is output as the target bit. As expected from our study of the Fredkin requires the use of a special ancilla input, gate, the Toffoli gate simulation of a and some of the outputs from the simulation are garbage bits. The Toffoli gate can also be used to implement the operation by inputting an ancilla 1 to the \ufb01rst control bit, and a to the second control bit, producing the output 1, a, a. This is illustrated in Figure 3.19. Recalling that are together and 160 Introduction to computer science Figure 3.18. Implementing a, while the third bit is prepared in the standard state 1, sometimes known as an ancilla state. The output from the gate using a Toffoli gate. The top two bits represent the input to the is on the third bit. universal for computation, we see that an arbitrary circuit can be ef\ufb01ciently simulated using a reversible circuit consisting only of Toffoli gates and ancilla bits, and that useful additional techniques such as uncomputation may be achieved using the same methods as were employed with the Fredkin gate. Figure 3.19. bits standard ancilla states. The output from with the Toffoli gate, with the second bit being the input to the, and the other two appears on the second and third bits. Our interest in reversible computation was motivated by our desire to understand the energy requirements for computation. It is clear that the noise-free billiard ball model of computation requires no energy for its operation; what about models based upon the Toffoli gate? This can only be determined by examining speci\ufb01c models for the computation of the Toffoli gate. In Chapter 7, we examine several such implementations, and it turns out that, indeed, the Toffoli gate can be implemented in a manner which does not require the expenditure of energy. There is a signi\ufb01cant caveat attached to the idea that computation can be done without the expenditure of energy. As we noted earlier, the billiard ball model of computation is highly sensitive to noise, and this is true of many other models of reversible computation.", " To nullify the effects of noise, some form of error-correction needs to be done. Such error-correction typically involves the performance of measurements on the system to determine whether the system is behaving as expected, or if an error has occurred. Because the computer\u2019s memory is \ufb01nite, the bits used to store the measurement results utilized in error-correction must eventually be erased to make way for new measurement results. According to Landauer\u2019s principle, this erasure carries an associated energy cost Perspectives on computer science 161 that must be accounted for when tallying the total energy cost of the computation. We analyze the energy cost associated with error-correction in more detail in Section 12.4.4. What can we conclude from our study of reversible computation? There are three key ideas. First, reversibility stems from keeping track of every bit of information; irreversibility occurs only when information is lost or erased. Second, by doing computation reversibly, we obviate the need for energy expenditure during computation. All computations can be done, in principle, for zero cost in energy. Third, reversible computation can be done ef\ufb01ciently, without the production of garbage bits whose value depends upon the input to the computation. That is, if there is an irreversible circuit computing a function f, then there is an ef\ufb01cient simulation of this circuit by a reversible circuit with action (x, y) \u2192 (x, y \u2295 f (x)). What are the implications of these results for physics, computer science, and for quantum computation and quantum information? From the point of view of a physicist or hardware engineer worried about heat dissipation, the good news is that, in principle, it is possible to make computation dissipation-free by making it reversible, although in practice energy dissipation is required for system stability and immunity from noise. At an even more fundamental level, the ideas leading to reversible computation also lead to the resolution of a century-old problem in the foundations of physics, the famous problem of Maxwell\u2019s demon. The story of this problem and its resolution is outlined in Box 3.5 on page 162. From the point of view of a computer scientist, reversible computation validates the use of irreversible elements in models of computation such as the Turing machine (since using them or not gives polynomially equivalent models). Moreover, since the physical world is fundamentally reversible, one can argue that complexity classes based upon reversible models of computation are more natural than complexity classes based upon irreversible models, a point revisited in Problem 3.9 and \u2018History and further reading\u2019. From the point of view of quantum computation and quantum information, reversible computation is enormously important. To harness the full power of quantum computation, any classical subroutines in a quantum computation must be performed reversibly and without the production of garbage bits depending on the classical input. Exercise 3.32: (From Fredkin to Toffoli and back again) What is the smallest number of Fredkin gates needed to simulate a Toffoli gate? What is the smallest number of Toffoli gates needed to simulate a Fredkin gate? 3.3 Perspectives on computer science In a short introduction such as this chapter, it is not remotely possible to cover in detail all the great ideas of a \ufb01eld as rich as computer science. We hope to have conveyed to you something of what it means to think like a computer scientist, and provided a basic vocabulary and overview of some of the fundamental concepts important in the understanding of computation. To conclude this chapter, we brie\ufb02y touch on some more general issues, in order to provide some perspective on how quantum computation and quantum information \ufb01ts into the overall picture of computer science. Our discussion has revolved around the Turing machine model of computation. How does the computational power of unconventional models of computation such as massively parallel computers, DNA computers and analog computers compare with the standard 162 Introduction to computer science Box 3.5: Maxwell\u2019s demon The laws of thermodynamics govern the amount of work that can be performed by a physical system at thermodynamic equilibrium. One of these laws, the second law of thermodynamics, states that the entropy in a closed system can never decrease. In 1871, James Clerk Maxwell proposed the existence of a machine that apparently violated this law. He envisioned a miniature little \u2018demon\u2019, like that shown in the \ufb01gure below, which could reduce the entropy of a gas cylinder initially at equilibrium by individually separating the fast and slow molecules into the two halves of the cylinder. This demon would sit at a little door at the middle partition. When a fast molecule approaches from the left side the demon opens a door between the partitions, allowing the molecule through, and then closes the door. By doing this many times the total entropy of the cylinder can be decreased, in apparent violation of the second law of thermodynamics. The resolution to the Maxwell\u2019", "s demon paradox lies in the fact that the demon must perform measurements on the molecules moving between the partitions, in order to determine their velocities. The result of this measurement must be stored in the demon\u2019s memory. Because any memory is \ufb01nite, the demon must eventually begin erasing information from its memory, in order to have space for new measurement results. By Landauer\u2019s principle, this act of erasing information increases the total entropy of the combined system \u2013 demon, gas cylinder, and their environments. In fact, a complete analysis shows that Landauer\u2019s principle implies that the entropy of the combined system is increased at least as much by this act of erasing information as the entropy of the combined system is decreased by the actions of the demon, thus ensuring that the second law of thermodynamics is obeyed. Turing machine model of computation and, implicitly, with quantum computation? Let\u2019s begin with parallel computing architectures. The vast majority of computers in existence are serial computers, processing instructions one at a time in some central processing unit. By contrast, parallel computers can process more than one instruction at a time, leading to a substantial savings in time and money for some applications. Nevertheless, parallel processing does not offer any fundamental advantage over the standard Turing machine model when issues of ef\ufb01ciency are concerned, because a Turing machine can simulate a parallel computer with polynomially equivalent total physical resources \u2013 the total space and time used by the computation. What a parallel computer gains in time, Perspectives on computer science 163 it loses in the total spatial resources required to perform the computation, resulting in a net of no essential change in the power of the computing model. An interesting speci\ufb01c example of massively parallel computing is the technique of DNA computing. A strand of DNA, deoxyribonucleic acid, is a molecule composed of a sequence (a polymer) of four kinds of nucleotides distinguished by the bases they carry, denoted by the letter A (adenine), C (cytosine), G (guanine) and T (thymine). Two strands, under certain circumstances, can anneal to form a double strand, if the respective base pairs form complements of each other (A matches T and G matches C). The ends are also distinct and must match appropriately. Chemical techniques can be used to amplify the number of strands beginning or ending with speci\ufb01c sequences (polymerase chain reaction), separate the strands by length (gel electrophoresis), dissolve double strands into single strands (changing temperature and pH), read the sequence on a strand, cut strands at a speci\ufb01c position (restriction enzymes), and detect if a certain sequence of DNA is in a test tube. The procedure for using these mechanisms in a robust manner is rather involved, but the basic idea can be appreciated from an example. The directed Hamiltonian path problem is a simple and equivalently hard variant of the Hamiltonian cycle problem of Section 3.2.2, in which the goal is to determine if a path exists or not between two speci\ufb01ed vertices j1 and jN in a directed graph G of N vertices, entering each vertex exactly once, and following only allowed edge directions. This problem can be solved with a DNA computer using the following \ufb01ve steps, in which xj are chosen to be unique sequences of bases (and \u00afxj their complements), DNA strands xjxk encode edges, and strands \u00afxj \u00afxj encode vertices. (1) Generate random paths through G, by combining a mixture of all possible vertex and edge DNA strands, and waiting for the strands to anneal. (2) Keep only the paths beginning with j1 and ending with jN, by amplifying only the double strands beginning with \u00afxj1 and ending with \u00afxjN. (3) Select only paths of length N, by separating the strands according to their length. (4) Select only paths which enter each vertex at least once, by dissolving the DNA into single strands, and annealing with all possible vertex strands one at a time and \ufb01ltering out only those strands which anneal. And (5) detect if any strands have survived the selection steps; if so, then a path exists, and otherwise, it does not. To ensure the answer is correct with suf\ufb01ciently high probability, xj may be chosen to contain many (\u2248 30) bases, and a large number (\u2248 1014 or more are feasible) of strands are used in the reaction. Heuristic methods are available to improve upon this basic idea. Of course, exhaustive search methods such as this only work as long as all possible paths can be generated ef\ufb01ciently, and thus the number of molecules used must grow exponentially as the size of the problem (the number of vertices", " in the example above). DNA molecules are relatively small and readily synthesized, and the huge number of DNA combinations one can \ufb01t into a test tube can stave off the exponential complexity cost increase for a while \u2013 up to a few dozen vertices \u2013 but eventually the exponential cost limits the applicability of this method. Thus, while DNA computing offers an attractive and physically realizable model of computation for the solution of certain problems, it is a classical computing technique and offers no essential improvement in principle over a Turing machine. Analog computers offer a yet another paradigm for performing computation. A computer is analog when the physical representation of information it uses for computation is based on continuous degrees of freedom, instead of zeroes and ones. For example, a thermometer is an analog computer. Analog circuitry, using resistors, capacitors, and ampli\ufb01ers, is also said to perform analog computation. Such machines have an in\ufb01nite 164 Introduction to computer science resource to draw upon in the ideal limit, since continuous variables like position and voltage can store an unlimited amount of information. But this is only true in the absence of noise. The presence of a \ufb01nite amount of noise reduces the number of distinguishable states of a continuous variable to a \ufb01nite number \u2013 and thus restricts analog computers to the representation of a \ufb01nite amount of information. In practice, noise reduces analog computers to being no more powerful than conventional digital computers, and through them Turing machines. One might suspect that quantum computers are just analog computers, because of the use of continuous parameters in describing qubit states; however, it turns out that the effects of noise on a quantum computer can effectively be digitized. As a result, their computational advantages remain even in the presence of a \ufb01nite amount of noise, as we shall see in Chapter 10. What of the effects of noise on digital computers? In the early days of computation, noise was a very real problem for computers. In some of the original computers a vacuum tube would malfunction every few minutes. Even today, noise is a problem for computational devices such as modems and hard drives. Considerable effort was devoted to the problem of understanding how to construct reliable computers from unreliable components. It was proven by von Neumann that this is possible with only a polynomial increase in the resources required for computation. Ironically, however, modern computers use none of those results, because the components of modern computers are fantastically reliable. Failure rates of 10\u221217 and even less are common in modern electronic components. For this reason, failures happen so rarely that the extra effort required to protect against them is not regarded as being worth making. On the other hand, we shall \ufb01nd that quantum computers are very delicate machines, and will likely require substantial application of error-correction techniques. Different architectures may change the effects of noise. For example, if the effect of noise is ignored, then changing to a computer architecture in which many operations are performed in parallel may not change the number of operations which need to be done. However, a parallel system may be substantially more resistant to noise, because the effects of noise have less time to accumulate. Therefore, in a realistic analysis, the parallel version of an algorithm may have some substantial advantages over a serial implementation. Architecture design is a well developed \ufb01eld of study for classical computers. Hardly anything similar has been developed along the same lines for quantum computers, but the study of noise already suggests some desirable traits for future quantum computer architectures, such as a high level of parallelism. A fourth model of computation is distributed computation, in which two or more spatially separated computational units are available to solve a computational problem. Obviously, such a model of computation is no more powerful than the Turing machine model in the sense that it can be ef\ufb01ciently simulated on a Turing machine. However, distributed computation gives rise to an intriguing new resource challenge: how best to utilize multiple computational units when the cost of communication between the units is high. This problem of distributed computation becomes especially interesting as computers are connected through high speed networks; although the total computational capacity of all the computers on a network might be extremely large, utilization of that potential is dif\ufb01cult. Most interesting problems do not divide easily into independent chunks that can be solved separately, and may frequently require global communication between different computational subsystems to exchange intermediate results or synchronize status. The \ufb01eld of communication complexity has been developed to address such issues, by Chapter problems 165 quantifying the cost of communication requirements in solving problems. When quantum resources are available and can be exchanged between distributed computers, the communication costs can sometimes be greatly reduced. A recurring theme through these concluding thoughts and through the entire book is that despite the traditional independence of computer science from physical constraints, ultimately physical laws have tremendous impact not only upon how computers are realized, but also the class of problems they are capable of solving. The success of quantum computation and quantum information as", " a physically reasonable alternative model of computation questions closely held tenets of computer science, and thrusts notions of computer science into the forefront of physics. The task of the remainder of this book is to stir together ideas from these disparate \ufb01elds, and to delight in what results! Problem 3.1: (Minsky machines) A Minsky machine consists of a \ufb01nite set of registers, r1, r2,..., rk, each capable of holding an arbitrary non-negative integer, and a program, made up of orders of one of two types. The \ufb01rst type has the form: The interpretation is that at point m in the program register rj is incremented by one, and execution proceeds to point n in the program. The second type of order has the form: The interpretation is that at point m in the program, register rj is decremented if it contains a positive integer, and execution proceeds to point n in the program. If register rj is zero then execution simply proceeds to point p in the program. The program for the Minsky machine consists of a collection of such orders, of a form like: The starting and all possible halting points for the program are conventionally labeled zero. This program takes the contents of register r1 and adds them to register r2, while decrementing r1 to zero. 166 Introduction to computer science (1) Prove that all (Turing) computable functions can be computed on a Minsky machine, in the sense that given a computable function f (\u00b7) there is a Minsky machine program that when the registers start in the state (n, 0,..., 0) gives as output (f (n), 0,..., 0). (2) Sketch a proof that any function which can be computed on a Minsky machine, in the sense just de\ufb01ned, can also be computed on a Turing machine. Problem 3.2: (Vector games) A vector game is speci\ufb01ed by a \ufb01nite list of vectors, all of the same dimension, and with integer co-ordinates. The game is to start with a vector x of non-negative integer co-ordinates and to add to x the \ufb01rst vector from the list which preserves the non-negativity of all the components, and to repeat this process until it is no longer possible. Prove that for any computable function f (\u00b7) there is a vector game which when started with the vector (n, 0,..., 0) reaches (f (n), 0,..., 0). (Hint: Show that a vector game in k + 2 dimensions can simulate a Minsky machine containing k registers.) Problem 3.3: (Fractran) A Fractran program is de\ufb01ned by a list of positive rational numbers q1,..., qn. It acts on a positive integer m by replacing it by qim, where i is the least number such that qim is an integer. If there is ever a time when there is no i such that qim is an integer, then execution stops. Prove that for any computable function f (\u00b7) there is a Fractran program which when started with 2n reaches 2f (n) without going through any intermediate powers of 2. (Hint: use the previous problem.) Problem 3.4: (Undecidability of dynamical systems) A Fractran program is essentially just a very simple dynamical system taking positive integers to positive integers. Prove that there is no algorithm to decide whether such a dynamical system ever reaches 1. Problem 3.5: (Non-universality of two bit reversible logic) Suppose we are trying to build circuits using only one and two bit reversible logic gates, and ancilla bits. Prove that there are Boolean functions which cannot be computed in this fashion. Deduce that the Toffoli gate cannot be simulated using one and two bit reversible gates, even with the aid of ancilla bits. Problem 3.6: (Hardness of approximation of ) Let r \u2265 1 and suppose that which is guaranteed to \ufb01nd the there is an approximation algorithm for shortest tour among n cities to within a factor r. Let G = (V, E) be any graph on by identifying cities with vertices in V, and n vertices. De\ufb01ne an instance of de\ufb01ning the distance between cities i and j to be 1 if (i, j) is an edge of G, and to be r|V | + 1 otherwise. Show that if the approximation algorithm is applied to then it returns a Hamiltonian cycle for G if one exists, and this instance of otherwise returns a tour of length more than r|V |. From the NP-completeness it follows that no such approximation algorithm", " can exist unless P = NP. of Problem 3.7: (Reversible Turing machines) (1) Explain how to construct a reversible Turing machine that can compute the same class of functions as is computable on an ordinary Turing machine. (Hint: It may be helpful to use a multi-tape construction.) History and further reading 167 (2) Give general space and time bounds for the operation of your reversible Turing machine, in terms of the time t(x) and space s(x) required on an ordinary single-tape Turing machine to compute a function f (x). Problem 3.8: (Find a hard-to-compute class of functions (Research)) Find a natural class of functions on n inputs which requires a super-polynomial number of Boolean gates to compute. Problem 3.9: (Reversible PSPACE = PSPACE) It can be shown that the problem \u2018quanti\ufb01ed satis\ufb01ability\u2019, or language in PSPACE can be reduced to, is PSPACE-complete. That is, every other in polynomial time. The language is de\ufb01ned to consist of all Boolean formulae \u03d5 in n variables x1,..., xn, and in conjunctive normal form, such that: \u2203x1 \u2203x1 \u2200x2 \u2200x2 \u2203x3 \u2203x3... \u2200xn... \u2203xn \u03d5 if n is even; \u03d5 if n is odd. (3.9) (3.10) Prove that a reversible Turing machine operating in polynomial space can be used to solve operating reversibly in polynomial space is equal to PSPACE.. Thus, the class of languages decidable by a computer Problem 3.10: (Ancilla bits and ef\ufb01ciency of reversible computation) Let pm be the mth prime number. Outline the construction of a reversible circuit which, upon input of m and n such that n > m, outputs the product pmpn, that is (m, n) \u2192 (pmpn, g(m, n)), where g(m, n) is the \ufb01nal state of the ancilla bits used by the circuit. Estimate the number of ancilla qubits your circuit requires. Prove that if a polynomial (in log n) size reversible circuit can be found that uses O(log(log n)) ancilla bits then the problem of factoring a product of two prime numbers is in P. History and further reading Computer science is a huge subject with many interesting sub\ufb01elds. We cannot hope for any sort of completeness in this brief space, but instead take the opportunity to recommend a few titles of general interest, and some works on subjects of speci\ufb01c interest in relation to topics covered in this book, with the hope that they may prove stimulating. Modern computer science dates to the wonderful 1936 paper of Turing[Tur36]. The Church\u2013Turing thesis was \ufb01rst stated by Church[Chu36] in 1936, and was then given a more complete discussion from a different point of view by Turing. Several other researchers found their way to similar conclusions at about the same time. Many of these contributions and a discussion of the history may be found in a volume edited by Davis[Dav65]. Provocative discussions of the Church\u2013Turing thesis and undecidability may be found in Hofstadter[Hof79] and Penrose[Pen89]. There are many excellent books on algorithm design. We mention only three. First, there is the classic series by Knuth[Knu97, Knu98a, Knu98b] which covers an enormous portion of computer science. Second, there is the marvelous book by Cormen, Leiserson, and Rivest[CLR90]. This huge book contains a plethora of well-written material on many areas 168 Introduction to computer science of algorithm design. Finally, the book of Motwani and Raghavan[MR95] is an excellent survey of the \ufb01eld of randomized algorithms. The modern theory of computational complexity was especially in\ufb02uenced by the papers of Cook[Coo71] and Karp[Kar72]. Many similar ideas were arrived at independently in Russia by Levin[Lev73], but unfortunately took time to propagate to the West. The classic book by Garey and Johnson[GJ79] has also had an enormous in\ufb02uence on the \ufb01eld. More recently, Papadimitriou[Pap94] has written a beautiful book that surveys many of the main ideas of computational complexity theory. Much of the material in this chapter is based upon Papadimitriou\u2019s book. In this chapter we considered only", " one type of reducibility between languages, polynomial time reducibility. There are many other notions of reductions between languages. An early survey of these notions was given by Ladner, Lynch and Selman[LLS75]. The study of different notions of reducibility later blossomed into a sub\ufb01eld of research known as structural complexity, which has been reviewed by Balc\u00b4azar, Diaz, and Gabarr\u00b4o[BDG88a, BDG88b]. The connection between information, energy dissipation, and computation has a long history. The modern understanding is due to a 1961 paper by Landauer[Lan61], in which Landauer\u2019s principle was \ufb01rst formulated. A paper by Szilard[Szi29] and a 1949 lecture by von Neumann[von66] (page 66) arrive at conclusions close to Landauer\u2019s principle, but do not fully grasp the essential point that it is the erasure of information that requires dissipation. Reversible Turing machines were invented by Lecerf[Lec63] and later, but independently, in an in\ufb02uential paper by Bennett[Ben73]. Fredkin and Toffoli[FT82] introduced reversible circuit models of computation. Two interesting historical documents are Barton\u2019s May, 1978 MIT 6.895 term paper[Bar78], and Ressler\u2019s 1981 Master\u2019s thesis[Res81], which contain designs for a reversible PDP-10! Today, reversible logic is potentially important in implementations of low-power CMOS circuitry[YK95]. Maxwell\u2019s demon is a fascinating subject, with a long and intricate history. Maxwell proposed his demon in 1871[Max71]. Szilard published a key paper in 1929[Szi29] which anticipated many of the details of the \ufb01nal resolution of the problem of Maxwell\u2019s demon. In 1965 Feynman[FLS65b] resolved a special case of Maxwell\u2019s demon. Bennett, building on Landauer\u2019s work[Lan61], wrote two beautiful papers on the subject[BBBW82, Ben87] which completed the resolution of the problem. An interesting book about the history of Maxwell\u2019s demon and its exorcism is the collection of papers by Leff and Rex[LR90]. DNA computing was invented by Adleman, and the solution of the directed Hamiltonian path problem we describe is his[Adl94]. Lipton has also shown how and circuit satis\ufb01ability can be solved in this model[Lip95]. A good general article is Adleman\u2019s Scienti\ufb01c American article[Adl98]; for an insightful look into the universality of DNA operations, see Winfree[Win98]. An interesting place to read about performing reliable computation in the presence of noise is the book by Winograd and Cowan[WC67]. This topic will be addressed again in Chapter 10. A good textbook on computer architecture is by Hennessey, Goldberg, and Patterson.[HGP96]. Problems 3.1 through 3.4 explore a line of thought originated by Minsky (in his beautiful book on computational machines[Min67]) and developed by Conway[Con72, Con86]. The Fractran programming language is certainly one of the most beautiful and elegant universal computational models known, as demonstrated by the following example, known History and further reading 169 as PRIMEGAME[Con86]. PRIMEGAME is de\ufb01ned by the list of rational numbers: 17 91 ; 78 85 ; 19 51 ; 23 38 ; 29 33 ; 77 29 ; 95 23 ; 77 19 ; 1 17 ; 11 13 ; 13 11 ; 15 2 ; 1 7 ;. 55 1 (3.11) Amazingly, when PRIMEGAME is started at 2, the other powers of 2 that appear, namely, 22, 23, 25, 27, 211, 213,..., are precisely the prime powers of 2, with the powers stepping through the prime numbers, in order. Problem 3.9 is a special case of the more general subject of the spatial requirements for reversible computation. See the papers by Bennett[Ben89], and by Li, Tromp and Vitanyi[LV96, LTV98]. II Quantum computation 4 Quantum circuits The theory of computation has traditionally been studied almost entirely in the abstract, as a topic in pure mathematics. This is to miss the point of it. Computers are physical objects, and computations are physical processes. What computers can or cannot compute is determined by the laws of physics alone, and not by pure mathematics. \u2013 David Deutsch Like mathematics, computer science will be somewhat different from the other sciences, in that it deals with arti\ufb01cial laws that can be proved, instead of natural laws that are never known with certainty. \u2013 Donald Knuth The", " opposite of a profound truth may well be another profound truth. \u2013 Niels Bohr This chapter begins Part II of the book, in which we explore quantum computation in detail. The chapter develops the fundamental principles of quantum computation, and establishes the basic building blocks for quantum circuits, a universal language for describing sophisticated quantum computations. The two fundamental quantum algorithms known to date are constructed from these circuits in the following two chapters. Chapter 5 presents the quantum Fourier transform and its applications to phase estimation, order-\ufb01nding and factoring. Chapter 6 describes the quantum search algorithm, and its applications to database search, counting and speedup of solutions to NP-complete problems. Chapter 7 concludes Part II with a discussion of how quantum computation may one day be experimentally realized. Two other topics of great interest for quantum computation, quantum noise and quantum error-correction, are deferred until Part III of the book, in view of their wide interest also outside quantum computation. There are two main ideas introduced in this chapter. First, we explain in detail the fundamental model of quantum computation, the quantum circuit model. Second, we demonstrate that there exists a small set of gates which are universal, that is, any quantum computation whatsoever can be expressed in terms of those gates. Along the way we also have occasion to describe many other basic results of quantum computation. Section 4.1 begins the chapter with an overview of quantum algorithms, focusing on what algorithms are known, and the unifying techniques underlying their construction. Section 4.2 is a detailed study of single qubit operations. Despite their simplicity, single qubit operations offer a rich playground for the construction of examples and techniques, and it is essential to understand them in detail. Section 4.3 shows how to perform multi-qubit controlled unitary operations, and Section 4.4 discusses the description of measurement in the quantum circuits model. These elements are then brought together in Section 4.5 for the statement and proof of the universality theorem. We summarize all the basic elements 172 Quantum circuits of quantum computation in Section 4.6, and discuss possible variants of the model, and the important question of the relationship in computational power between classical and quantum computers. Section 4.7 concludes the chapter with an important and instructive application of quantum computation to the simulation of real quantum systems. This chapter is perhaps the most reader-intensive of all the chapters in the book, with a high density of exercises for you to complete, and it is worth explaining the reason for this intensity. Obtaining facility with the basic elements of the quantum circuit model of computation is quite easy, but requires assimilating a large number of simple results and techniques that must become second nature if one is to progress to the more dif\ufb01cult problem of designing quantum algorithms. For this reason we take an example-oriented approach in this chapter, and ask you to \ufb01ll in many of the details, in order to acquire such a facility. A less intensive, but somewhat super\ufb01cial overview of the basic elements of quantum computation may be obtained by skipping to Section 4.6. 4.1 Quantum algorithms What is a quantum computer good for? We\u2019re all familiar with the frustration of needing more computer resources to solve a computational problem. Practically speaking, many interesting problems are impossible to solve on a classical computer, not because they are in principle insoluble, but because of the astronomical resources required to solve realistic cases of the problem. The spectacular promise of quantum computers is to enable new algorithms which render feasible problems requiring exorbitant resources for their solution on a classical computer. At the time of writing, two broad classes of quantum algorithms are known which ful\ufb01ll this promise. The \ufb01rst class of algorithms is based upon Shor\u2019s quantum Fourier transform, and includes remarkable algorithms for solving the factoring and discrete logarithm problems, providing a striking exponential speedup over the best known classical algorithms. The second class of algorithms is based upon Grover\u2019s algorithm for performing quantum searching. These provide a less striking but still remarkable quadratic speedup over the best possible classical algorithms. The quantum searching algorithm derives its importance from the widespread use of search-based techniques in classical algorithms, which in many instances allows a straightforward adaptation of the classical algorithm to give a faster quantum algorithm. Figure 4.1 sketches the state of knowledge about quantum algorithms at the time of writing, including some sample applications of those algorithms. Naturally, at the core of the diagram are the quantum Fourier transform and the quantum searching algorithm. Of particular interest in the \ufb01gure is the quantum counting algorithm. This algorithm is a clever combination of the quantum searching and Fourier transform algorithms, which can be used to estimate the number of solutions to a search problem more quickly than is possible on a classical computer. The quantum searching algorithm has many potential applications, of which but a few are illustrated. It can be used to extract statistics, such as the minimal element, from an unordered data set,", " more quickly than is possible on a classical computer. It can be used to speed up algorithms for some problems in NP \u2013 speci\ufb01cally, those problems for which a straightforward search for a solution is the best algorithm known. Finally, it can be used to speed up the search for keys to cryptosystems such as the widely used Data Encryption Standard (DES). These and other applications are explained in Chapter 6. Quantum counting Quantum search Speed up for some NP problems Quantum algorithms 173 Fourier transform Hidden subgroup problem Discrete log Order-finding Factoring Statistics mean,median,min Search for crypto keys Break cryptosystems (RSA) Figure 4.1. The main quantum algorithms and their relationships, including some notable applications. The quantum Fourier transform also has many interesting applications. It can be used to solve the discrete logarithm and factoring problems. These results, in turn, enable a quantum computer to break many of the most popular cryptosystems now in use, including the RSA cryptosystem. The Fourier transform also turns out to be closely related to an important problem in mathematics, \ufb01nding a hidden subgroup (a generalization of \ufb01nding the period of a periodic function). The quantum Fourier transform and several of its applications, including fast quantum algorithms for factoring and discrete logarithm, are explained in Chapter 5. Why are there so few quantum algorithms known which are better than their classical counterparts? The answer is that coming up with good quantum algorithms seems to be a dif\ufb01cult problem. There are at least two reasons for this. First, algorithm design, be it classical or quantum, is not an easy business! The history of algorithms shows us that considerable ingenuity is often required to come up with near optimal algorithms, even for apparently very simple problems, like the multiplication of two numbers. Finding good quantum algorithms is made doubly dif\ufb01cult because of the additional constraint that we want our quantum algorithms to be better than the best known classical algorithms. A second reason for the dif\ufb01culty of \ufb01nding good quantum algorithms is that our intuitions are much better adapted to the classical world than they are to the quantum world. If we think about problems using our native intuition, then the algorithms which we come up with are going to be classical algorithms. It takes special insights and special tricks to come up with good quantum algorithms. Further study of quantum algorithms will be postponed until the next chapter. In this chapter we provide an ef\ufb01cient and powerful language for describing quantum algorithms, the language of quantum circuits \u2013 assemblies of discrete sets of components which describe computational procedures. This construction will enable us to quantify the cost of an algorithm in terms of things like the total number of gates required, or the circuit depth. The circuit language also comes with a toolbox of tricks that simpli\ufb01es algorithm design and provides ready conceptual understanding. 174 Quantum circuits 4.2 Single qubit operations The development of our quantum computational toolkit begins with operations on the simplest quantum system of all \u2013 a single qubit. Single qubit gates were introduced in Section 1.3.1. Let us quickly summarize what we learned there; you may \ufb01nd it useful to refer to the notes on notation on page xxiii as we go along. A single qubit is a vector |\u03c8 = a|0 + b|1 parameterized by two complex numbers satisfying |a|2 + |b|2 = 1. Operations on a qubit must preserve this norm, and thus are described by 2\u00d72 unitary matrices. Of these, some of the most important are the Pauli matrices; it is useful to list them again here1. (4.1) Three other quantum gates will play a large part in what follows, the Hadamard gate (denoted H), phase gate (denoted S), and \u03c0/8 gate (denoted T ): H = 1\u221a 2 1 1 1 \u2212 exp(i\u03c0/4) \u221a. (4.2) 2 and S = T 2. A couple of useful algebraic facts to keep in mind are that H = (X + Z)/ You might wonder why the T gate is called the \u03c0/8 gate when it is \u03c0/4 that appears in the de\ufb01nition. The reason is that the gate has historically often been referred to as the \u03c0/8 gate, simply because up to an unimportant global phase T is equal to a gate which has exp(\u00b1i\u03c0/8) appearing on its diagonals. T = exp(i\u03c0/8) exp(\u2212i\u03c0/8) 0 0 exp(i\u03c0/8). (4.3) Nevertheless, the nomenclature is in some respects rather unfortunate, and we often refer to this gate as the T gate. Recall also that a single qubit", " in the state a|0 + b|1 can be visualized as a point (\u03b8, \u03d5) on the unit sphere, where a = cos(\u03b8/2), b = ei\u03d5 sin(\u03b8/2), and a can be taken to be real because the overall phase of the state is unobservable. This is called the Bloch sphere representation, and the vector (cos \u03d5 sin \u03b8, sin \u03d5 sin \u03b8, cos \u03b8) is called the Bloch vector. We shall return to this picture often as an aid to intuition. Exercise 4.1: In Exercise 2.11, which you should do now if you haven\u2019t already done it, you computed the eigenvectors of the Pauli matrices. Find the points on the Bloch sphere which correspond to the normalized eigenvectors of the different Pauli matrices. The Pauli matrices give rise to three useful classes of unitary matrices when they are exponentiated, the rotation operators about the \u02c6x, \u02c6y, and \u02c6z axes, de\ufb01ned by the equations: Rx(\u03b8) \u2261 e\u2212i\u03b8X/2 = cos Ry(\u03b8) \u2261 e\u2212i\u03b8Y /2 = cos Rz(\u03b8) \u2261 e\u2212i\u03b8Z/2 = cos \u03b8 2 \u03b8 2 \u03b8 2 I \u2212 i sin I \u2212 i sin I \u2212 i sin = cos \u03b8 2 \u2212i sin \u03b8 2 \u2212 sin \u03b8 cos \u03b8 2 2 sin \u03b8 cos \u03b8 2 2 e\u2212i\u03b8/2 0 ei\u03b8/2 0 \u2212i sin \u03b8 2 cos \u03b8 2. (4.4) (4.5) (4.6) Single qubit operations 175 Exercise 4.2: Let x be a real number and A a matrix such that A2 = I. Show that exp(iAx) = cos(x)I + i sin(x)A. (4.7) Use this result to verify Equations (4.4) through (4.6). Exercise 4.3: Show that, up to a global phase, the \u03c0/8 gate satis\ufb01es T = Rz(\u03c0/4). Exercise 4.4: Express the Hadamard gate H as a product of Rx and Rz rotations and ei\u03d5 for some \u03d5. If \u02c6n = (nx, ny, nz) is a real unit vector in three dimensions then we generalize the previous de\ufb01nitions by de\ufb01ning a rotation by \u03b8 about the \u02c6n axis by the equation \u03b8 \u03b8 R \u02c6n(\u03b8) \u2261 exp(\u2212i\u03b8 \u02c6n \u00b7 \u03c3/2) = cos I \u2212 i sin (nxX + nyY + nzZ), (4.8) 2 2 where \u03c3 denotes the three component vector (X, Y, Z) of Pauli matrices. Exercise 4.5: Prove that ( \u02c6n \u00b7 \u03c3)2 = I, and use this to verify Equation (4.8). Exercise 4.6: (Bloch sphere interpretation of rotations) One reason why the R \u02c6n(\u03b8) operators are referred to as rotation operators is the following fact, which you are to prove. Suppose a single qubit has a state represented by the Bloch vector \u03bb. Then the effect of the rotation R \u02c6n(\u03b8) on the state is to rotate it by an angle \u03b8 about the \u02c6n axis of the Bloch sphere. This fact explains the rather mysterious looking factor of two in the de\ufb01nition of the rotation matrices. Exercise 4.7: Show that XY X = \u2212Y and use this to prove that XRy(\u03b8)X = Ry(\u2212\u03b8). Exercise 4.8: An arbitrary single qubit unitary operator can be written in the form for some real numbers \u03b1 and \u03b8, and a real three-dimensional unit vector \u02c6n. U = exp(i\u03b1)R \u02c6n(\u03b8) (4.9) 1. Prove this fact. 2. Find values for \u03b1, \u03b8, and \u02c6n giving the Hadamard gate H. 3. Find values for \u03b1, \u03b8, and \u02c6n giving the phase gate 1 0 0 i S =. (4.10) An arbitrary unitary operator on a single qubit can be written in many ways as a combination of rotations, together with global phase shifts on the qubit. The following theorem provides a means of expressing an arbitrary single qubit rotation that will be particularly useful in later applications to", " controlled operations. Theorem 4.1: (Z-Y decomposition for a single qubit) Suppose U is a unitary operation on a single qubit. Then there exist real numbers \u03b1, \u03b2, \u03b3 and \u03b4 such that U = ei\u03b1Rz(\u03b2)Ry(\u03b3)Rz(\u03b4). (4.11) 176 Quantum circuits Proof Since U is unitary, the rows and columns of U are orthonormal, from which it follows that there exist real numbers \u03b1, \u03b2, \u03b3,and \u03b4 such that U = ei(\u03b1\u2212\u03b2/2\u2212\u03b4/2) cos \u03b3 2 ei(\u03b1+\u03b2/2\u2212\u03b4/2) sin \u03b3 2 \u2212ei(\u03b1\u2212\u03b2/2+\u03b4/2) sin \u03b3 2 ei(\u03b1+\u03b2/2+\u03b4/2) cos \u03b3 2. (4.12) Equation (4.11) now follows immediately from the de\ufb01nition of the rotation matrices and matrix multiplication. Exercise 4.9: Explain why any single qubit unitary operator may be written in the form (4.12). Exercise 4.10: (X-Y decomposition of rotations) Give a decomposition analogous to Theorem 4.1 but using Rx instead of Rz. Exercise 4.11: Suppose \u02c6m and \u02c6n are non-parallel real unit vectors in three dimensions. Use Theorem 4.1 to show that an arbitrary single qubit unitary U may be written U = ei\u03b1R \u02c6n(\u03b2)R \u02c6m(\u03b3)R \u02c6n(\u03b4), (4.13) for appropriate choices of \u03b1, \u03b2, \u03b3 and \u03b4. The utility of Theorem 4.1 lies in the following mysterious looking corollary, which is the key to the construction of controlled multi-qubit unitary operations, as explained in the next section. Corollary 4.2: Suppose U is a unitary gate on a single qubit. Then there exist unitary operators A, B, C on a single qubit such that ABC = I and U = ei\u03b1AXBXC, where \u03b1 is some overall phase factor. Proof In the notation of Theorem 4.1, set A \u2261 Rz(\u03b2)Ry(\u03b3/2), B \u2261 Ry(\u2212\u03b3/2)Rz(\u2212(\u03b4 + \u03b2)/2) and C \u2261 Rz((\u03b4 \u2212 \u03b2)/2). Note that ABC = Rz(\u03b2)Ry \u2019 \u03b3 2 ( \u2019 Ry \u2212 ( \u03b3 2 Rz \u2212 Rz 4.14) Since X 2 = I, and using Exercise 4.7, we see that \u2019 ( XBX = XRy \u2212 XXRz \u2212 \u03b3 2 \u03b4 + \u03b2 2 X = Ry ( \u2019 \u03b3 2 Rz. \u03b4 + \u03b2 2 Thus AXBXC = Rz(\u03b2)Ry ( \u2019 \u03b3 2 Ry ( \u2019 \u03b3 2 Rz \u03b4 + \u03b2 2 Rz \u03b4 \u2212 \u03b2 2 = Rz(\u03b2)Ry(\u03b3)Rz(\u03b4). Thus U = ei\u03b1AXBXC and ABC = I, as required. Exercise 4.12: Give A, B, C, and \u03b1 for the Hadamard gate. (4.15) (4.16) (4.17) Controlled operations 177 Exercise 4.13: (Circuit identities) It is useful to be able to simplify circuits by inspection, using well-known identities. Prove the following three identities: HXH = Z; HY H = \u2212Y ; HZH = X. (4.18) Exercise 4.14: Use the previous exercise to show that HT H = Rx(\u03c0/4), up to a global phase. Exercise 4.15: (Composition of single qubit operations) The Bloch representation gives a nice way to visualize the effect of composing two rotations. (1) Prove that if a rotation through an angle \u03b21 about the axis \u02c6n1 is followed by a rotation through an angle \u03b22 about an axis \u02c6n2, then the overall rotation is through an angle \u03b212 about an axis \u02c6n12 given by c12 = c1c2 \u2212 s1s2 \u02c6n1 \u00b7 \u02c6n2 (4.19) s12 \u02c6n12 = s1c2 \u02c6n1 + c1s2 \u02c6n2 \u2212 s1s2 \u02c6n2 \u00d7 \u02c6n1, (4.20) where ci = cos(\u03b2i/2), si = sin(\u03b2i/2), c12 = cos(\u03b212", "/2), and s12 = sin(\u03b212/2). (2) Show that if \u03b21 = \u03b22 and \u02c6n1 = \u02c6z these equations simplify to c12 = c2 \u2212 s2 \u02c6z \u00b7 \u02c6n2 s12 \u02c6n12 = sc( \u02c6z + \u02c6n2) \u2212 s2 \u02c6n2 \u00d7 \u02c6z, (4.21) (4.22) where c = c1 and s = s1. Symbols for the common single qubit gates are shown in Figure 4.2. Recall the basic properties of quantum circuits: time proceeds from left to right; wires represent qubits, and a \u2018/\u2019 may be used to indicate a bundle of qubits. Hadamard Pauli-X Pauli-Y Pauli-Z Phase \u03c0/8 1\u221a 2 1 1 1 \u22121 1 0 1 0 0 \u2212i i 0 1 0 0 \u22121 0 1 i 0 1 0 0 ei\u03c0/4 Figure 4.2. Names, symbols, and unitary matrices for the common single qubit gates. 4.3 Controlled operations \u2018If A is true, then do B\u2019. This type of controlled operation is one of the most useful in computing, both classical and quantum. In this section we explain how complex controlled operations may be implemented using quantum circuits built from elementary operations. 178 Quantum circuits, which we met in SecThe prototypical controlled operation is the controlled, is a quantum gate tion 1.2.1. Recall that this gate, which we\u2019ll often refer to as with two input qubits, known as the control qubit and target qubit, respectively. It is drawn as shown in Figure 4.3. In terms of the computational basis, the action of the is given by |c|t \u2192 |c|t \u2295 c; that is, if the control qubit is set to |1 then the target qubit is \ufb02ipped, otherwise the target qubit is left alone. Thus, in the computational basis |control, target the matrix representation of \u23a4 is \u23a1 \u23a2 \u23a2 \u23a3 4.23) Figure 4.3. Circuit representation for the controlledbottom line the target qubit. gate. The top line represents the control qubit, the More generally, suppose U is an arbitrary single qubit unitary operation. A controlledU operation is a two qubit operation, again with a control and a target qubit. If the control qubit is set then U is applied to the target qubit, otherwise the target qubit is left alone; that is, |c|t \u2192 |cU c|t. The controlled-U operation is represented by the circuit shown in Figure 4.4. Figure 4.4. Controlled-U operation. The top line is the control qubit, and the bottom line is the target qubit. If the control qubit is set then U is applied to the target, otherwise it is left alone. Exercise 4.16: (Matrix representation of multi-qubit gates) What is the 4\u00d74 unitary matrix for the circuit in the computational basis? What is the unitary matrix for the circuit in the computational basis? Controlled operations 179 Exercise 4.17: (Building gate from one controlled-Z gate, that is, the gate whose action in the computational basis is speci\ufb01ed by the unitary matrix from controlled-Z gates) Construct 1 \u23a4 \u23a5 \u23a5 \u23a6, and two Hadamard gates, specifying the control and target qubits. Exercise 4.18: Show that Exercise 4.19: ( action on density matrices) The gate is a simple permutation whose action on a density matrix \u03c1 is to rearrange the elements in the matrix. Write out this action explicitly in the computational basis. Exercise 4.20: ( basis transformations) Unlike ideal classical gates, ideal quantum gates do not have (as electrical engineers say) \u2018high-impedance\u2019 inputs. In fact, the role of \u2018control\u2019 and \u2018target\u2019 are arbitrary \u2013 they depend on what basis behaves you think of a device as operating in. We have described how the with respect to the computational basis, and in this description the state of the control qubit is not changed. However, if we work in a different basis then the control qubit does change: we will show that its phase is \ufb02ipped depending on the state of the \u2018target\u2019 qubit! Show that Introducing basis states |\u00b1 \u2261 (|0 \u00b1 |1)/ that the effect of a target is as follows: 2, use this circuit identity to show with the \ufb01rst qubit as control and the second qubit as \u221a |+|+ \u2192 |+", "|+ |\u2212|+ \u2192 |\u2212|+ |+|\u2212 \u2192 |\u2212|\u2212 |\u2212|\u2212 \u2192 |+|\u2212. (4.24) (4.25) (4.26) (4.27) Thus, with respect to this new basis, the state of the target qubit is not changed, while the state of the control qubit is \ufb02ipped if the target starts as |\u2212, otherwise 180 Quantum circuits it is left alone. That is, in this basis, the target and control have essentially interchanged roles! Our immediate goal is to understand how to implement the controlled-U operation for arbitrary single qubit U, using only single qubit operations and the gate. Our strategy is a two-part procedure based upon the decomposition U = ei\u03b1AXBXC given in Corollary 4.2 on page 176. Our \ufb01rst step will be to apply the phase shift exp(i\u03b1) on the target qubit, controlled by the control qubit. That is, if the control qubit is |0, then the target qubit is left alone, while if the control qubit is |1, a phase shift exp(i\u03b1) is applied to the target. A circuit implementing this operation using just a single qubit unitary gate is depicted on the right hand side of Figure 4.5. To verify that this circuit works correctly, note that the effect of the circuit on the right hand side is |00 \u2192 |00, |01 \u2192 |01, |10 \u2192 ei\u03b1|10, |11 \u2192 ei\u03b1|11, (4.28) which is exactly what is required for the controlled operation on the left hand side. Figure 4.5. Controlled phase shift gate and an equivalent circuit for two qubits. We may now complete the construction of the controlled-U operation, as shown in Figure 4.6. To understand why this circuit works, recall from Corollary 4.2 that U may be written in the form U = ei\u03b1AXBXC, where A, B and C are single qubit operations such that ABC = I. Suppose that the control qubit is set. Then the operation ei\u03b1AXBXC = U is applied to the second qubit. If, on the other hand, the control qubit is not set, then the operation ABC = I is applied to the second qubit; that is, no change is made. That is, this circuit implements the controlled-U operation. Now that we know how to condition on a single qubit being set, what about conditioning on multiple qubits? We\u2019ve already met one example of multiple qubit conditioning, the Toffoli gate, which \ufb02ips the third qubit, the target qubit, conditioned on the \ufb01rst two qubits, the control qubits, being set to one. More generally, suppose we have n + k qubits, and U is a k qubit unitary operator. Then we de\ufb01ne the controlled operation Cn(U ) by the equation Cn(U )|x1x2... xn|\u03c8 = |x1x2... xnU x1x2...xn|\u03c8, where x1x2... xn in the exponent of U means the product of the bits x1, x2,..., xn. That is, the operator U is applied to the last k qubits if the \ufb01rst n qubits are all equal to one, and otherwise, nothing is done. Such conditional operations are so useful that we (4.29) Controlled operations 181 Figure 4.6. Circuit implementing the controlled-U operation for single qubit U. \u03b1, A, B and C satisfy U = exp(i\u03b1)AXBXC, ABC = I. introduce a special circuit notation for them, illustrated in Figure 4.7. For the following we assume that k = 1, for simplicity. Larger k can be dealt with using essentially the same methods, however for k \u2265 2 there is the added complication that we don\u2019t (yet) know how to perform arbitrary operations on k qubits. Figure 4.7. Sample circuit representation for the Cn(U ) operation, where U is a unitary operator on k qubits, for n = 4 and k = 3. Suppose U is a single qubit unitary operator, and V is a unitary operator chosen so that V 2 = U. Then the operation C2(U ) may be implemented using the circuit shown in Figure 4.8. Exercise 4.21: Verify that Figure 4.8 implements the C2(U ) operation. Exercise 4.22: Prove that a C2(U ) gate (for any single qubit unitary U ) can be s. constructed using", " at most eight one-qubit gates, and six controlled- Exercise 4.23: Construct a C1(U ) gate for U = Rx(\u03b8) and U = Ry(\u03b8), using only and single qubit gates. Can you reduce the number of single qubit gates needed in the construction from three to two? The familiar Toffoli gate is an especially important special case of the C2(U ) operation, 182 Quantum circuits Figure 4.8. Circuit for the C2(U ) gate. V is any unitary operator satisfying V 2 = U. The special case V \u2261 (1 \u2212 i)(I + iX)/2 corresponds to the Toffoli gate. the case C2(X). De\ufb01ning V \u2261 (1 \u2212 i)(I + iX)/2 and noting that V 2 = X, we see that Figure 4.8 gives an implementation of the Toffoli gate in terms of one and two qubit operations. From a classical viewpoint this is a remarkable result; recall from Problem 3.5 that one and two bit classical reversible gates are not suf\ufb01cient to implement the Toffoli gate, or, more generally, universal computation. By contrast, in the quantum case we see that one and two qubit reversible gates are suf\ufb01cient to implement the Toffoli gate, and will eventually prove that they suf\ufb01ce for universal computation. Ultimately we will show that any unitary operation can be composed to an arbitrarily and \u03c0/8 gates. good approximation from just the Hadamard, phase, controlledBecause of the great usefulness of the Toffoli gate it is interesting to see how it can be built from just this gate set. Figure 4.9 illustrates a simple circuit for the Toffoli gate made up of just Hadamard, phase, controlled- and \u03c0/8 gates Figure 4.9. Implementation of the Toffoli gate using Hadamard, phase, controlled- and \u03c0/8 gates. Exercise 4.24: Verify that Figure 4.9 implements the Toffoli gate. Exercise 4.25: (Fredkin gate construction) Recall that the Fredkin (controlled-swap) gate performs the transform \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 4.30) Controlled operations 183 (1) Give a quantum circuit which uses three Toffoli gates to construct the Fredkin gate (Hint: think of the swap gate construction \u2013 you can control each gate, one at a time). (2) Show that the \ufb01rst and last Toffoli gates can be replaced by (3) Now replace the middle Toffoli gate with the circuit in Figure 4.8 to obtain gates. a Fredkin gate construction using only six two-qubit gates. (4) Can you come up with an even simpler construction, with only \ufb01ve two-qubit gates? Exercise 4.26: Show that the circuit: \u2022 \u2022 \u2022 Ry \u03c0/ \u2295 Ry \u03c0/ \u2295 Ry \u2212\u03c0/ \u2295 Ry \u2212\u03c0/ differs from a Toffoli gate only by relative phases. That is, the circuit takes |c1, c2, t to ei\u03b8(c1,c2,t)|c1, c2, t \u2295 c1 \u00b7 c2, where ei\u03b8(c1,c2,t) is some relative phase factor. Such gates can sometimes be useful in experimental implementations, where it may be much easier to implement a gate that is the same as the Toffoli up to relative phases than it is to do the Toffoli directly. Exercise 4.27: Using just s and Toffoli gates, construct a quantum circuit to perform the transformation \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 4.31) This kind of partial cyclic permutation operation will be useful later, in Chapter 7. How may we implement Cn(U ) gates using our existing repertoire of gates, where U is an arbitrary single qubit unitary operation? A particularly simple circuit for achieving this task is illustrated in Figure 4.10. The circuit divides up into three stages, and makes use of a small number (n \u2212 1) of working qubits, which all start and end in the state |0. Suppose the control qubits are in the computational basis state |c1, c2,..., cn. The all the control bits c1,..., cn together to \ufb01rst stage of the circuit is to reversibly s c1 and produce the product c1 \u00b7 c2... cn.", " To do this, the \ufb01rst gate in the circuit c2 together, using a Toffoli gate, changing the state of the \ufb01rst work qubit to |c1 \u00b7 c2. s c3 with the product c1 \u00b7 c2, changing the state of the second The next Toffoli gate work qubit to |c1 \u00b7 c2 \u00b7 c3. We continue applying Toffoli gates in this fashion, until the \ufb01nal work qubit is in the state |c1 \u00b7 c2... cn. Next, a U operation on the target qubit is 184 Quantum circuits performed, conditional on the \ufb01nal work qubit being set to one. That is, U is applied if and only if all of c1 through cn are set. Finally, the last part of the circuit just reverses the steps of the \ufb01rst stage, returning all the work qubits to their initial state, |0. The combined result, therefore, is to apply the unitary operator U to the target qubit, if and only if all the control bits c1 through cn are set, as desired. Figure 4.10. Network implementing the Cn(U ) operation, for the case n = 5. Exercise 4.28: For U = V 2 with V unitary, construct a C5(U ) gate analogous to that in Figure 4.10, but using no work qubits. You may use controlled-V and controlled-V \u2020 gates. Exercise 4.29: Find a circuit containing O(n2) Toffoli, and single qubit gates which implements a Cn(X) gate (for n > 3), using no work qubits. Exercise 4.30: Suppose U is a single qubit unitary operation. Find a circuit containing O(n2) Toffoli, Cn(U ) gate (for n > 3), using no work qubits. and single qubit gates which implements a In the controlled gates we have been considering, conditional dynamics on the target qubit occurs if the control bits are set to one. Of course, there is nothing special about one, and it is often useful to consider dynamics which occur conditional on the control bit being set to zero. For instance, suppose we wish to implement a two qubit gate in which the second (\u2018target\u2019) qubit is \ufb02ipped, conditional on the \ufb01rst (\u2018control\u2019) qubit being set to zero. In Figure 4.11 we introduce a circuit notation for this gate, together with an equivalent circuit in terms of the gates we have already introduced. Generically we shall use the open circle notation to indicate conditioning on the qubit being set to zero, while a closed circle indicates conditioning on the qubit being set to one. A more elaborate example of this convention, involving three control qubits, is illustrated in Figure 4.12. The operation U is applied to the target qubit if the \ufb01rst and third qubits are set to zero, and the second qubit is set to one. It is easy to verify by inspection that the circuit on the right hand side of the \ufb01gure implements the desired operation. More generally, it is easy to move between circuits which condition on qubits being set Measurement 185 Figure 4.11. Controlled operation with a qubit being set to zero. gate being performed on the second qubit, conditional on the \ufb01rst to one and circuits which condition on qubits being set to zero, by insertion of X gates in appropriate locations, as illustrated in Figure 4.12. Another convention which is sometimes useful is to allow controlled- gates to have multiple targets, as shown in Figure 4.13. This natural notation means that when the control qubit is 1, then all the qubits marked with a \u2295 are \ufb02ipped, and otherwise nothing happens. It is convenient to use, for example, in constructing classical functions such as permutations, or in encoders and decoders for quantum error-correction circuits, as we shall see in Chapter 10. Exercise 4.31: (More circuit identities) Let subscripts denote which qubit an operator acts on, and let C be a the target qubit. Prove the following identities: with qubit 1 the control qubit and qubit 2 CX1C = X1X2 CY1C = Y1X2 CZ1C = Z1 CX2C = X2 CY2C = Z1Y2 CZ2C = Z1Z2 Rz,1(\u03b8)C = CRz,1(\u03b8) Rx,2(\u03b8)C = CRx,2(\u03b8). 4.4 Measurement (4.32) (4.33) (4.", "34) (4.35) (4.36) (4.37) (4.38) (4.39) A \ufb01nal element used in quantum circuits, almost implicitly sometimes, is measurement. In our circuits, we shall denote a projective measurement in the computational basis (Section 2.2.5) using a \u2018meter\u2019 symbol, illustrated in Figure 4.14. In the theory of quantum circuits it is conventional to not use any special symbols to denote more general measurements, because, as explained in Chapter 2, they can always be represented by unitary transforms with ancilla qubits followed by projective measurements. There are two important principles that it is worth bearing in mind about quantum circuits. Both principles are rather obvious; however, they are of such great utility that they are worth emphasizing early. The \ufb01rst principle is that classically conditioned operations can be replaced by quantum conditioned operations: 186 Quantum circuits Figure 4.12. Controlled-U operation and its equivalent in terms of circuit elements we already know how to implement. The fourth qubit has U applied if the \ufb01rst and third qubits are set to zero, and the second qubit is set to one. \u2022 \u2295 \u2295 \u2261 \u2022 \u2022 \u2295 \u2295 Figure 4.13. Controlled- gate with multiple targets. Principle of deferred measurement: Measurements can always be moved from an intermediate stage of a quantum circuit to the end of the circuit; if the measurement results are used at any stage of the circuit then the classically controlled operations can be replaced by conditional quantum operations. Often, quantum measurements are performed as an intermediate step in a quantum circuit, and the measurement results are used to conditionally control subsequent quantum gates. This is the case, for example, in the teleportation circuit of Figure 1.13 on page 27. However, such measurements can always be moved to the end of the circuit. Figure 4.15 illustrates how this may be done by replacing all the classical conditional operations by corresponding quantum conditional operations. (Of course, some of the interpretation of this circuit as performing \u2018teleportation\u2019 is lost, because no classical information is transmitted from Alice to Bob, but it is clear that the overall action of the two quantum circuits is the same, which is the key point.) The second principle is even more obvious \u2013 and surprisingly useful! Figure 4.14. Symbol for projective measurement on a single qubit. In this circuit nothing further is done with the measurement result, but in more general quantum circuits it is possible to change later parts of the quantum circuit, conditional on measurement outcomes in earlier parts of the circuit. Such a usage of classical information is depicted using wires drawn with double lines (not shown here). Measurement 187 Figure 4.15. Quantum teleportation circuit in which measurements are done at the end, instead of in the middle of the circuit. As in Figure 1.13, the top two qubits belong to Alice, and the bottom one to Bob. Principle of implicit measurement: Without loss of generality, any unterminated quantum wires (qubits which are not measured) at the end of a quantum circuit may be assumed to be measured. To understand why this is true, imagine you have a quantum circuit containing just two qubits, and only the \ufb01rst qubit is measured at the end of the circuit. Then the measurement statistics observed at this time are completely determined by the reduced density matrix of the \ufb01rst qubit. However, if a measurement had also been performed on the second qubit, then it would be highly surprising if that measurement could change the statistics of measurement on the \ufb01rst qubit. You\u2019ll prove this in Exercise 4.32 by showing that the reduced density matrix of the \ufb01rst qubit is not affected by performing a measurement on the second. As you consider the role of measurements in quantum circuits, it is important to keep in mind that in its role as an interface between the quantum and classical worlds, measurement is generally considered to be an irreversible operation, destroying quantum information and replacing it with classical information. In certain carefully designed cases, however, this need not be true, as is vividly illustrated by teleportation and quantum error-correction (Chapter 10). What teleportation and quantum error-correction have in common is that in neither instance does the measurement result reveal any information about the identity of the quantum state being measured. Indeed, we will see in Chapter 10 that this is a more general feature of measurement \u2013 in order for a measurement to be reversible, it must reveal no information about the quantum state being measured! Exercise 4.32: Suppose \u03c1 is the density matrix describing a two qubit system. Suppose we perform a projective measurement in the computational basis of the second qubit. Let P0 = |00| and P1 = |11| be the projectors onto the |0 and |1 states of the second qubit, respectively. Let", " \u03c1 be the density matrix which would be assigned to the system after the measurement by an observer who did not learn the measurement result. Show that \u03c1 = P0\u03c1P0 + P1\u03c1P1. (4.40) Also show that the reduced density matrix for the \ufb01rst qubit is not affected by the measurement, that is, tr2(\u03c1) = tr2(\u03c1). Exercise 4.33: (Measurement in the Bell basis) The measurement model we have speci\ufb01ed for the quantum circuit model is that measurements are performed only 188 Quantum circuits in the computational basis. However, often we want to perform a measurement in some other basis, de\ufb01ned by a complete set of orthonormal states. To perform this measurement, simply unitarily transform from the basis we wish to perform the measurement in to the computational basis, then measure. For example, show that the circuit performs a measurement in the basis of the Bell states. More precisely, show that this circuit results in a measurement being performed with corresponding POVM elements the four projectors onto the Bell states. What are the corresponding measurement operators? Exercise 4.34: (Measuring an operator) Suppose we have a single qubit operator U with eigenvalues \u00b11, so that U is both Hermitian and unitary, so it can be regarded both as an observable and a quantum gate. Suppose we wish to measure the observable U. That is, we desire to obtain a measurement result indicating one of the two eigenvalues, and leaving a post-measurement state which is the corresponding eigenvector. How can this be implemented by a quantum circuit? Show that the following circuit implements a measurement of U : 0 | H \u2022 H |\u03c8in U |\u03c8out Exercise 4.35: (Measurement commutes with controls) A consequence of the principle of deferred measurement is that measurements commute with quantum gates when the qubit being measured is a control qubit, that is: \u2022 U = \u2022 U = U (Recall that the double lines represent classical bits in this diagram.) Prove the \ufb01rst equality. The rightmost circuit is simply a convenient notation to depict the use of a measurement result to classically control a quantum gate. 4.5 Universal quantum gates,, A small set of gates (e.g. ) can be used to compute an arbitrary classical function, as we saw in Section 3.1.2. We say that such a set of gates is universal for classical computation. In fact, since the Toffoli gate is universal for classical computation, quantum circuits subsume classical circuits. A similar universality result is true for quantum computation, where a set of gates is said to be universal for quantum computation if any unitary operation may be approximated to arbitrary accuracy by a quantum circuit Universal quantum gates 189 involving only those gates. We now describe three universality constructions for quantum computation. These constructions build upon each other, and culminate in a proof that any unitary operation can be approximated to arbitrary accuracy using Hadamard, phase,, and \u03c0/8 gates. You may wonder why the phase gate appears in this list, since it can be constructed from two \u03c0/8 gates; it is included because of its natural role in the fault-tolerant constructions described in Chapter 10. The \ufb01rst construction shows that an arbitrary unitary operator may be expressed exactly as a product of unitary operators that each acts non-trivially only on a subspace spanned by two computational basis states. The second construction combines the \ufb01rst construction with the results of the previous section to show that an arbitrary unitary operator may be expressed exactly using single qubit and gates. The third construction combines the second construction with a proof that single qubit operation may be approximated to arbitrary accuracy using the Hadamard, phase, and \u03c0/8 gates. This in turn implies that any unitary operation can be approximated to arbitrary accuracy using Hadamard, phase,, and \u03c0/8 gates. Our constructions say little about ef\ufb01ciency \u2013 how many (polynomially or exponentially many) gates must be composed in order to create a given unitary transform. In Section 4.5.4 we show that there exist unitary transforms which require exponentially many gates to approximate. Of course, the goal of quantum computation is to \ufb01nd interesting families of unitary transformations that can be performed ef\ufb01ciently. Exercise 4.36: Construct a quantum circuit to add two two-bit numbers x and y modulo 4. That is, the circuit should perform the transformation |x, y \u2192 |x, x + y mod 4. 4.5.1 Two-level unitary gates are universal Consider a unitary matrix U which acts on a d-dimensional Hilbert space. In this section we explain how U may", " be decomposed into a product of two-level unitary matrices; that is, unitary matrices which act non-trivially only on two-or-fewer vector components. The essential idea behind this decomposition may be understood by considering the case when U is 3\u00d73, so suppose that U has the form \u23a1 \u23a3 We will \ufb01nd two-level unitary matrices U1,..., U3 such that It follows that U3U2U1U = 4.41) (4.42) (4.43) U1, U2 and U3 are all two-level unitary matrices, and it is easy to see that their inverses, U \u2020 3 are also two-level unitary matrices. Thus, if we can demonstrate (4.42), 1 then we will have shown how to break U up into a product of two-level unitary matrices. 2 and U \u2020, U \u2020 190 Quantum circuits Use the following procedure to construct U1: if b = 0 then set \u23a1 U1 \u2261 \u23a3 \u23a4 \u23a6. 1 0 0 0 1 0 0 0 1 If b = 0 then set \u23a1 \u23a2 \u23a2 \u23a3 U1 \u2261 a\u2217\u221a |a|2+|b|2 b\u221a |a|2+|b|2 0 b\u2217\u221a |a|2+|b|2 \u2212a\u221a |a|2+|b|4.44) (4.45) Note that in either case U1 is a two-level unitary matrix, and when we multiply the matrices out we get U1U = \u23a2 \u23a3 \u23a1 \u23a4 4.46) The key point to note is that the middle entry in the left hand column is zero. We denote the other entries in the matrix with a generic prime ; their actual values do not matter. Now apply a similar procedure to \ufb01nd a two-level matrix U2 such that U2U1U has no entry in the bottom left corner. That is, if c a \u2217 0 0 U2 \u2261 \u23a2 \u23a3 \u23a1 = 0 we set \u23a4 \u23a5 \u23a6, 0 1 0 0 0 1 while if c = 0 then we set U2 \u2261 \u23a1 \u23a2 \u23a2 \u23a2 \u23a3 \u2217 a \u221a |a |2+|c |2 0 \u221a c |a |2+|c |a |2+|c |2 0 \u221a \u2212a |a |2+|c |2 In either case, when we carry out the matrix multiplication we \ufb01nd that \u23a1 U2U1U = \u23a2 \u23a3 \u23a4 \u23a5 \u23a6. 1 d 0 0 f g e h j Since U, U1 and U2 are unitary, it follows that U2U1U is unitary, and thus d since the \ufb01rst row of U2U1U must have norm 1. Finally, set \u23a1 \u23a4 U3 \u2261 \u23a2 \u23a3 4.47) (4.48) (4.49) = g = 0, (4.50) It is now easy to verify that U3U2U1U = I, and thus U = U \u2020 position of U into two-level unitaries. 1 U \u2020 2 U \u2020 3, which is a decom- More generally, suppose U acts on a d-dimensional space. Then, in a similar fashion to the 3\u00d73 case, we can \ufb01nd two-level unitary matrices U1,..., Ud\u22121 such that the matrix Universal quantum gates 191 Ud\u22121Ud\u22122... U1U has a one in the top left hand corner, and all zeroes elsewhere in the \ufb01rst row and column. We then repeat this procedure for the d \u2212 1 by d \u2212 1 unitary submatrix in the lower right hand corner of Ud\u22121Ud\u22122... U1U, and so on, with the end result that an arbitrary d\u00d7d unitary matrix may be written U = V1... Vk, (4.51) where the matrices Vi are two-level unitary matrices, and k \u2264 (d \u2212 1) + (d \u2212 2) + \u00b7 \u00b7 \u00b7 + 1 = d(d \u2212 1)/2. Exercise 4.37: Provide a decomposition of the transform \u23a4 \u23a1 \u23a2 \u23a2 \u23a3 1 2 1 1 1 \u22121 1 \u2212i \u22121 1 1 1 i \u22121 \u2212i 1 \u22121 i", " \u23a5 \u23a5 \u23a6 (4.52) into a product of two-level unitaries. This is a special case of the quantum Fourier transform, which we study in more detail in the next chapter. A corollary of the above result is that an arbitrary unitary matrix on an n qubit system may be written as a product of at most 2n\u22121(2n \u2212 1) two-level unitary matrices. For speci\ufb01c unitary matrices, it may be possible to \ufb01nd much more ef\ufb01cient decompositions, but as you will now show there exist matrices which cannot be decomposed as a product of fewer than d \u2212 1 two-level unitary matrices! Exercise 4.38: Prove that there exists a d\u00d7d unitary matrix U which cannot be decomposed as a product of fewer than d \u2212 1 two-level unitary matrices. 4.5.2 Single qubit and gates are universal We have just shown that an arbitrary unitary matrix on a d-dimensional Hilbert space may be written as a product of two-level unitary matrices. Now we show that single gates together can be used to implement an arbitrary two-level unitary qubit and operation on the state space of n qubits. Combining these results we see that single qubit gates can be used to implement an arbitrary unitary operation on n qubits, and and therefore are universal for quantum computation. Suppose U is a two-level unitary matrix on an n qubit quantum computer. Suppose in particular that U acts non-trivially on the space spanned by the computational basis states |s and |t, where s = s1... sn and t = t1... tn are the binary expansions for s and t. Let \u02dcU be the non-trivial 2\u00d72 unitary submatrix of U ; \u02dcU can be thought of as a unitary operator on a single qubit. Our immediate goal is to construct a circuit implementing U, built from single qubit gates. To do this, we need to make use of Gray codes. Suppose we have and distinct binary numbers, s and t. A Gray code connecting s and t is a sequence of binary numbers, starting with s and concluding with t, such that adjacent members of the list differ in exactly one bit. For instance, with s = 101001 and t = 110011 we have the Gray 192 Quantum circuits code 0 0 0 0 Let g1 through gm be the elements of a Gray code connecting s and t, with g1 = s and gm = t. Note that we can always \ufb01nd a Gray code such that m \u2264 n + 1 since s and t can differ in at most n locations4.53) The basic idea of the quantum circuit implementing U is to perform a sequence of gates effecting the state changes |g1 \u2192 |g2 \u2192... \u2192 |gm\u22121, then to perform a controlled- \u02dcU operation, with the target qubit located at the single bit where gm\u22121 and gm differ, and then to undo the \ufb01rst stage, transforming |gm\u22121 \u2192 |gm\u22122 \u2192... \u2192 |g1. Each of these steps can be easily implemented using operations developed earlier in this chapter, and the \ufb01nal result is an implementation of U. A more precise description of the implementation is as follows. The \ufb01rst step is to swap the states |g1 and |g2. Suppose g1 and g2 differ at the ith digit. Then we accomplish the swap by performing a controlled bit \ufb02ip on the ith qubit, conditional on the values of the other qubits being identical to those in both g1 and g2. Next we use a controlled operation to swap |g2 and |g3. We continue in this fashion until we swap |gm\u22122 with |gm\u22121. The effect of this sequence of m \u2212 2 operations is to achieve the operation |g1 \u2192 |gm\u22121 |g2 \u2192 |g1 |g3 \u2192 |g2......... |gm\u22121 \u2192 |gm\u22122. (4.54) (4.55) (4.56) (4.57) All other computational basis states are left unchanged by this sequence of operations. Next, suppose gm\u22121 and gm differ in the jth bit. We apply a controlled- \u02dcU operation with the jth qubit as target, conditional on the other qubits having the same values as appear in both gm and gm\u22121. Finally, we complete the U operation by undoing the swap operations: we swap |gm\u22121 with |gm\u22122, then |gm\u22122 with |gm\u22123", " and so on, until we swap |g2 with |g1. A simple example illuminates the procedure further. Suppose we wish to implement the two-level unitary transformation \u23a1 Here, a, b, c and d are any complex numbers such that \u02dc4.58) a c b d is a unitary matrix. Notice that U acts non-trivially only on the states |000 and |111. We write a Gray code connecting 000 and 111: Universal quantum gates 193 4.59) From this we read off the required circuit, shown in Figure 4.16. The \ufb01rst two gates shuf\ufb02e the states so that |000 gets swapped with |011. Next, the operation \u02dcU is applied to the \ufb01rst qubit of the states |011 and |111, conditional on the second and third qubits being in the state |11. Finally, we unshuf\ufb02e the states, ensuring that |011 gets swapped back with the state |000. Figure 4.16. Circuit implementing the two-level unitary operation de\ufb01ned by (4.58). Returning to the general case, we see that implementing the two-level unitary operation U requires at most 2(n \u2212 1) controlled operations to swap |g1 with |gm\u22121 and then back again. Each of these controlled operations can be realized using O(n) single qubit and gates; the controlled- \u02dcU operation also requires O(n) gates. Thus, implementing U requires O(n2) single qubit and gates. We saw in the previous section that an arbitrary unitary matrix on the 2n-dimensional state space of n qubits may be written as a product of O(22n) = O(4n) two-level unitary operations. Combining these results, we see that an arbitrary unitary operation on n qubits can be implemented using a circuit containing O(n24n) single qubit and gates. Obviously, this construction does not provide terribly ef\ufb01cient quantum circuits! However, we show in Section 4.5.4 that the construction is close to optimal in the sense that there are unitary operations that require an exponential number of gates to implement. Thus, to \ufb01nd fast quantum algorithms we will clearly need a different approach than is taken in the universality construction. Exercise 4.39: Find a quantum circuit using single qubit operations and s to implement the transformation \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 4.60) 194 Quantum circuits where \u02dcU = a c b d is an arbitrary 2\u00d72 unitary matrix. 4.5.3 A discrete set of universal operations and single qubit unitaries together form In the previous section we proved that the a universal set for quantum computation. Unfortunately, no straightforward method is known to implement all these gates in a fashion which is resistant to errors. Fortunately, in this section we\u2019ll \ufb01nd a discrete set of gates which can be used to perform universal quantum computation, and in Chaper 10 we\u2019ll show how to perform these gates in an error-resistant fashion, using quantum error-correcting codes. Approximating unitary operators Obviously, a discrete set of gates can\u2019t be used to implement an arbitrary unitary operation exactly, since the set of unitary operations is continuous. Rather, it turns out that a discrete set can be used to approximate any unitary operation. To understand how this works, we \ufb01rst need to study what it means to approximate a unitary operation. Suppose U and V are two unitary operators on the same state space. U is the target unitary operator that we wish to implement, and V is the unitary operator that is actually implemented in practice. We de\ufb01ne the error when V is implemented instead of U by E(U, V ) \u2261 max |\u03c8 (U \u2212 V )|\u03c8, (4.61) where the maximum is over all normalized quantum states |\u03c8 in the state space. In Box 4.1 on page 195 we show that this measure of error has the interpretation that if E(U, V ) is small, then any measurement performed on the state V |\u03c8 will give approximately the same measurement statistics as a measurement of U |\u03c8, for any initial state |\u03c8. More precisely, we show that if M is a POVM element in an arbitrary POVM, and PU (or PV ) is the probability of obtaining this outcome if U (or V ) were performed with a starting state |\u03c8, then |PU \u2212 PV | \u2264 2E(U, V ). (4.62) Thus, if E(U", ", V ) is small, then measurement outcomes occur with similar probabilities, regardless of whether U or V were performed. Also shown in Box 4.1 is that if we perform a sequence of gates V1,..., Vm intended to approximate some other sequence of gates U1,..., Um, then the errors add at most linearly, m E(UmUm\u22121... U1, VmVm\u22121... V1) \u2264 E(Uj, Vj). (4.63) j=1 The approximation results (4.62) and (4.63) are extremely useful. Suppose we wish to perform a quantum circuit containing m gates, U1 through Um. Unfortunately, we are only able to approximate the gate Uj by the gate Vj. In order that the probabilities of different measurement outcomes obtained from the approximate circuit be within a tolerance \u0394 > 0 of the correct probabilities, it suf\ufb01ces that E(Uj, Vj) \u2264 \u0394/(2m), by the results (4.62) and (4.63). Universality of Hadamard + phase + + \u03c0/8 gates We\u2019re now in a good position to study the approximation of arbitrary unitary operations by discrete sets of gates. We\u2019re going to consider two different discrete sets of gates, both Universal quantum gates 195 Box 4.1: Approximating quantum circuits Suppose a quantum system starts in the state |\u03c8, and we perform either the unitary operation U, or the unitary operation V. Following this, we perform a measurement. Let M be a POVM element associated with the measurement, and let PU (or PV ) be the probability of obtaining the corresponding measurement outcome if the operation U (or V ) was performed. Then |PU \u2212 PV | = + +\u03c8|U \u2020M U |\u03c8 \u2212 \u03c8|V \u2020M V |\u03c8 + +. Let |\u0394 \u2261 (U \u2212 V )|\u03c8. Simple algebra and the Cauchy\u2013Schwarz inequality show that |PU \u2212 PV | = + + +. +\u03c8|U \u2020M |\u0394 + \u0394|M V |\u03c8 \u2264 |\u03c8|U \u2020M |\u0394| + |\u0394|M V |\u03c8| \u2264 |\u0394 + |\u0394 \u2264 2E(U, V ). (4.64) (4.65) (4.66) (4.67) (4.68) The inequality |PU \u2212 PV | \u2264 2E(U, V ) gives quantitative expression to the idea that when the error E(U, V ) is small, the difference in probabilities between measurement outcomes is also small. Suppose we perform a sequence V1, V2,..., Vm of gates intended to approximate some other sequence of gates, U1, U2,..., Um. Then it turns out that the error caused by the entire sequence of imperfect gates is at most the sum of the errors in the individual gates, E(UmUm\u22121... U1, VmVm\u22121... V1) \u2264 m j=1 E(Uj, Vj). (4.69) To prove this we start with the case m = 2. Note that for some state |\u03c8 we have E(U2U1, V2V1) = (U2U1 \u2212 V2V1)|\u03c8 = (U2U1 \u2212 V2U1)|\u03c8 + (V2U1 \u2212 V2V1)|\u03c8. Using the triangle inequality |a + |b \u2264 |a + |b, we obtain E(U2U1, V2V1) \u2264 (U2 \u2212 V2)U1|\u03c8 + V2(U1 \u2212 V1)|\u03c8 \u2264 E(U2, V2) + E(U1, V1), (4.70) (4.71) (4.72) (4.73) which was the desired result. The result for general m follows by induction. of which are universal. The \ufb01rst set, the standard set of universal gates, consists of the and \u03c0/8 gates. We provide fault-tolerant constructions Hadamard, phase, controlledfor these gates in Chapter 10; they also provide an exceptionally simple universality construction. The second set of gates we consider consists of the Hadamard gate, phase gate, and the Toffoli gate. These gates can also all be done faultgate, the controlledtolerantly; however, the universality proof and fault-tolerance construction for these gates is a little less appealing. We begin the universality proof by showing that the Hadamard and \u03c0/8 gates can be 196 Quantum circuits used", " to approximate any single qubit unitary operation to arbitrary accuracy. Consider the gates T and HT H. T is, up to an unimportant global phase, a rotation by \u03c0/4 radians around the \u02c6z axis on the Bloch sphere, while HT H is a rotation by \u03c0/4 radians around the \u02c6x axis on the Bloch sphere (Exercise 4.14). Composing these two operations gives, up to a global phase, exp \u2212i Z exp \u2212i \u03c0 8 \u03c0 8 X = cos = cos2 I \u2212 i sin I \u2212 i cos \u03c0 8 \u03c0 8 \u03c0 Z 8 \u03c0 8 cos I \u2212 i sin \u03c0 8 (X + Z) + sin \u03c0 8 Y \u03c0 8 X (4.74) sin \u03c0 8. (4.75), cos \u03c0 This is a rotation of the Bloch sphere about an axis along n = (cos \u03c0 8 ) with 8 corresponding unit vector \u02c6n, and through an angle \u03b8 de\ufb01ned by cos(\u03b8/2) \u2261 cos2 \u03c0 8. That is, using only the Hadamard and \u03c0/8 gates we can construct R \u02c6n(\u03b8). Moreover, this \u03b8 can be shown to be an irrational multiple of 2\u03c0. Proving this latter fact is a little beyond our scope; see the end of chapter \u2018History and further reading\u2019., sin \u03c0 8 Next, we show that repeated iteration of R \u02c6n(\u03b8) can be used to approximate to arbitrary accuracy any rotation R \u02c6n(\u03b1). To see this, let \u03b4 > 0 be the desired accuracy, and let N be an integer larger than 2\u03c0/\u03b4. De\ufb01ne \u03b8k so that \u03b8k \u2208 [0, 2\u03c0) and \u03b8k = (k\u03b8)mod 2\u03c0. Then the pigeonhole principle implies that there are distinct j and k in the range 1,..., N such that |\u03b8k \u2212 \u03b8j| \u2264 2\u03c0/N < \u03b4. Without loss of generality assume that k > j, so we have |\u03b8k\u2212j| < \u03b4. Since j = k and \u03b8 is an irrational multiple of 2\u03c0 we must have \u03b8k\u2212j = 0. It follows that the sequence \u03b8l(k\u2212j) \ufb01lls up the interval [0, 2\u03c0) as l is varied, so that adjacent members of the sequence are no more than \u03b4 apart. It follows that for any > 0 there exists an n such that E(R \u02c6n(\u03b1), R \u02c6n(\u03b8)n) < 3. (4.76) Exercise 4.40: For arbitrary \u03b1 and \u03b2 show that E(R \u02c6n(\u03b1), R \u02c6n(\u03b1 + \u03b2)) = |1 \u2212 exp(i\u03b2/2)|, (4.77) and use this to justify (4.76). We are now in position to verify that any single qubit operation can be approximated to arbitrary accuracy using the Hadamard and \u03c0/8 gates. Simple algebra implies that for any \u03b1 HR \u02c6n(\u03b1)H = R \u02c6m(\u03b1), (4.78) where \u02c6m is a unit vector in the direction (cos \u03c0 8 that, \u2212 sin \u03c0 8, cos \u03c0 8 ), from which it follows E(R \u02c6m(\u03b1), R \u02c6m(\u03b8)n) < 3. But by Exercise 4.11 an arbitrary unitary U on a single qubit may be written as U = R \u02c6n(\u03b2)R \u02c6m(\u03b3)R \u02c6n(\u03b4), (4.79) (4.80) up to an unimportant global phase shift. The results (4.76) and (4.79), together with the Universal quantum gates 197 chaining inequality (4.63) therefore imply that for suitable positive integers n1, n2, n3, E(U, R \u02c6n(\u03b8)n1HR \u02c6n(\u03b8)n2 HR \u02c6n(\u03b8)n3 ) <. That is, given any single qubit unitary operator U and any > 0 it is possible to approximate U to within using a circuit composed of Hadamard gates and \u03c0/8 gates alone. (4.81) Since the \u03c0/8 and Hadamard gates allow us to approximate any single qubit unitary operator, it follows from the arguments of Section 4.5.2 that we can approximate any m gate quantum circuit, as follows. Given a quantum circuit containing m", " gates, either s or single qubit unitary gates, we may approximate it using Hadamard, and \u03c0/8 gates (later, we will \ufb01nd that phase gates make it possible to do controlledthe appoximation fault-tolerantly, but for the present universality argument they are not strictly necessary). If we desire an accuracy of for the entire circuit, then this may be achieved by approximating each single qubit unitary using the above procedure to within /m and applying the chaining inequality (4.63) to obtain an accuracy of for the entire circuit. How ef\ufb01cient is this procedure for approximating quantum circuits using a discrete set of gates? This is an important question. Suppose, for example, that approximating an arbitrary single qubit unitary to within a distance were to require \u03a9(21/) gates from the discrete set. Then to approximate the m gate quantum circuit considered in the previous paragraph would require \u03a9(m2m/) gates, an exponential increase over the original circuit size! Fortunately, the rate of convergence is much better than this. Intuitively, it is plausible that the sequence of angles \u03b8k \u2018\ufb01lls in\u2019 the interval [0, 2\u03c0) in a more or less uniform fashion, so that to approximate an arbitrary single qubit gate ought to take roughly \u0398(1/) gates from the discrete set. If we use this estimate for the number of gates required to approximate an arbitrary single qubit gate, then the number required to approximate an m gate circuit to accuracy becomes \u0398(m2/). This is a quadratic increase over the original size of the circuit, m, which for many applications may be suf\ufb01cient. Rather remarkably, however, a much faster rate of convergence can be proved. The Solovay\u2013Kitaev theorem, proved in Appendix 3, implies that an arbitrary single qubit gate may be approximated to an accuracy using O(logc(1/)) gates from our discrete set, where c is a constant approximately equal to 2. The Solovay\u2013Kitaev theorem therefore implies that to approximate a circuit containing m s and single qubit unitaries to an accuracy requires O(m logc(m/)) gates from the discrete set, a polylogarithmic increase over the size of the original circuit, which is likely to be acceptable for virtually all applications. To sum up, we have shown that the Hadamard, phase, controlled- and \u03c0/8 gates s are universal for quantum computation in the sense that given a circuit containing and arbitrary single qubit unitaries it is possible to simulate this circuit to good accuracy using only this discrete set of gates. Moreover, the simulation can be performed ef\ufb01ciently, in the sense that the overhead required to perform the simulation is polynomial in log(m/), where m is the number of gates in the original circuit, and is the desired accuracy of the simulation. Exercise 4.41: This and the next two exercises develop a construction showing that the Hadamard, phase, controlled- and Toffoli gates are universal. Show that 198 Quantum circuits Figure 4.17. Provided both measurement outcomes are 0 this circuit applies Rz(\u03b8) to the target, where cos \u03b8 = 3/5. If some other measurement outcome occurs then the circuit applies Z to the target. the circuit in Figure 4.17 applies the operation Rz(\u03b8) to the third (target) qubit if the measurement outcomes are both 0, where cos \u03b8 = 3/5, and otherwise applies Z to the target qubit. Show that the probability of both measurement outcomes being 0 is 5/8, and explain how repeated use of this circuit and Z = S2 gates may be used to apply a Rz(\u03b8) gate with probability approaching 1. Exercise 4.42: (Irrationality of \u03b8) Suppose cos \u03b8 = 3/5. We give a proof by contradiction that \u03b8 is an irrational multiple of 2\u03c0. (1) Using the fact that ei\u03b8 = (3 + 4i)/5, show that if \u03b8 is rational, then there must exist a positive integer m such that (3 + 4i)m = 5m. (2) Show that (3 + 4i)m = 3 + 4i (mod 5) for all m > 0, and conclude that no m such that (3 + 4i)m = 5m can exist. Exercise 4.43: Use the results of the previous two exercises to show that the Hadamard, phase, controlledcomputation. and Toffoli gates are universal for quantum Exercise 4.44: Show that the three qubit gate G de\ufb01ned by the circuit: \u2022 \u2022 iRx \u03c0\u03b1 is universal for quantum computation whenever \u03b1 is irrational. Exercise 4.", "45: Suppose U is a unitary transform implemented by an n qubit quantum circuit constructed from H, S, form 2\u2212k/2M, for some integer k, where M is a 2n\u00d72n matrix with only complex integer entries. Repeat this exercise with the Toffoli gate replaced by the \u03c0/8 gate. and Toffoli gates. Show that U is of the 4.5.4 Approximating arbitrary unitary gates is generically hard We\u2019ve seen that any unitary transformation on n qubits can be built up out of a small set of elementary gates. Is it always possible to do this ef\ufb01ciently? That is, given a unitary transformation U on n qubits does there always exist a circuit of size polynomial in n approximating U? The answer to this question turns out to be a resounding no: in fact, most unitary transformations can only be implemented very inef\ufb01ciently. One way to see Universal quantum gates 199 this is to consider the question: how many gates does it take to generate an arbitrary state of n qubits? A simple counting argument shows that this requires exponentially many operations, in general; it immediately follows that there are unitary operations requiring exponentially many operations. To see this, suppose we have g different types of gates available, and each gate works on at most f input qubits. These numbers, f and g, are \ufb01xed by the computing hardware we have available, and may be considered to be constants. Suppose we have a quantum circuit containing m gates, starting from the computational basis state |0\u2297n. For any particular gate in the circuit there are therefore = O(nf g) possible choices. It follows that at most O(nf gm) different g n f at most states may be computed using m gates. Figure 4.18. Visualization of covering the set of possible states with patches of constant radius. Suppose we wish to approximate a particular state, |\u03c8, to within a distance. The idea of the proof is to cover the set of all possible states with a collection of \u2018patches,\u2019 each of radius (Figure 4.18), and then to show that the number of patches required rises doubly exponentially in n; comparing with the exponential number of different states that may be computed using m gates will imply the result. The \ufb01rst observation we need is that the space of state vectors of n qubits can be regarded as just the unit (2n+1 \u22121)-sphere. To see this, suppose the n qubit state has amplitudes \u03c8j = Xj + iYj, where Xj and Yj are the real and imaginary parts, respectively, of the jth amplitude. The normalization condition j + Y 2 j ) = 1, which is just the condition for a for quantum states can be written point to be on the unit sphere in 2n+1 real dimensions, that is, the unit (2n+1 \u2212 1)-sphere. Similarly, the surface area of radius near |\u03c8 is approximately the same as the volume of a (2n+1 \u2212 2)-sphere of radius. Using the formula Sk(r) = 2\u03c0(k+1)/2rk/\u0393((k + 1)/2) for the surface area of a k-sphere of radius r, and Vk(r) = 2\u03c0(k+1)/2rk+1/[(k + 1)\u0393((k + 1)/2)] for the volume of a k-sphere of radius r, we see that the number of patches needed to j(X 2 200 Quantum circuits cover the state space goes like S2n+1\u22121(1) V2n+1\u22122() = \u221a \u03c0\u0393(2n \u2212 1 2 )(2n+1 \u2212 1) \u0393(2n)2n+1\u22121, (4.82) where \u0393 is the usual generalization of the factorial function. But \u0393(2n \u22121/2) \u2265 \u0393(2n)/2n, so the number of patches required to cover the space is at least \u03a9 1 2n+1\u22121. (4.83) Recall that the number of patches which can be reached in m gates is O(nf gm), so in order to reach all the -patches we must have nf gm O \u2265 \u03a9 1 2n+1\u22121 which gives us m = \u03a9. 2n log(1/) log(n) (4.84) (4.85) That is, there are states of n qubits which take \u03a9(2n log(1/)/ log(n)) operations to approximate to within a distance. This is exponential in n, and thus is", " \u2018dif\ufb01cult\u2019, in the sense of computational complexity introduced in Chapter 3. Furthermore, this immediately implies that there are unitary transformations U on n qubits which take \u03a9(2n log(1/)/ log(n)) operations to approximate by a quantum circuit implementing an operation V such that E(U, V ) \u2264. By contrast, using our universality constructions and the Solovay\u2013Kitaev theorem it follows that an arbitrary unitary operation U on n qubits may be approximated to within a distance using O(n24n logc(n24n/)) gates. Thus, to within a polynomial factor the construction for universality we have given is optimal; unfortunately, it does not address the problem of determining which families of unitary operations can be computed ef\ufb01ciently in the quantum circuits model. 4.5.5 Quantum computational complexity In Chapter 3 we described a theory of \u2018computational complexity\u2019 for classical computers that classi\ufb01ed the resource requirements to solve computational problems on classical computers. Not surprisingly there is considerable interest in developing a theory of quantum computational complexity, and relating it to classical computational complexity theory. Although only \ufb01rst steps have been taken in this direction, it will doubtless be an enormously fruitful direction for future researchers. We content ourselves with presenting one result about quantum complexity classes, relating the quantum complexity class BQP to the classical complexity class PSPACE. Our discussion of this result is rather informal; for more details you are referred to the paper of Bernstein and Vazirani referenced in the end of chapter \u2018History and further reading\u2019. Recall that PSPACE was de\ufb01ned in Chapter 3 as the class of decision problems which can be solved on a Turing machine using space polynomial in the problem size and an arbitrary amount of time. BQP is an essentially quantum complexity class consisting of those decision problems that can be solved with bounded probability of error using a polynomial size quantum circuit. Slightly more formally, we say a language L is in BQP if there is a family of polynomial size quantum circuits which decides the language, Universal quantum gates 201 accepting strings in the language with probability at least 3/4, and rejecting strings which aren\u2019t in the language with probability at least 3/4. In practice, what this means is that the quantum circuit takes as input binary strings, and tries to determine whether they are elements of the language or not. At the conclusion of the circuit one qubit is measured, with 0 indicating that the string has been accepted, and 1 indicating rejection. By testing the string a few times to determine whether it is in L, we can determine with very high probability whether a given string is in L. Of course, a quantum circuit is a \ufb01xed entity, and any given quantum circuit can only decide whether strings up to some \ufb01nite length are in L. For this reason, we use an entire family of circuits in the de\ufb01nition of BQP; for every possible input length there is a different circuit in the family. We place two restrictions on the circuit in addition to the acceptance / rejection criterion already described. First, the size of the circuits should only grow polynomially with the size of the input string x for which we are trying to determine whether x \u2208 L. Second, we require that the circuits be uniformly generated, in a sense similar to that described in Section 3.1.2. This uniformity requirement arises because, in practice, given a string x of some length n, somebody will have to build a quantum circuit capable of deciding whether x is in L. To do so, they will need to have a clear set of instructions \u2013 an algorithm \u2013 for building the circuit. For this reason, we require that our quantum circuits be uniformly generated, that is, there is a Turing machine capable of ef\ufb01ciently outputting a description of the quantum circuit. This restriction may seem rather technical, and in practice is nearly always satis\ufb01ed trivially, but it does save us from pathological examples such as that described in Section 3.1.2. (You might also wonder if it matters whether the Turing machine used in the uniformity requirement is a quantum or classical Turing machine; it turns out that it doesn\u2019t matter \u2013 see \u2018History and further reading\u2019.) One of the most signi\ufb01cant results in quantum computational complexity is that BQP \u2286 PSPACE. It is clear that BPP \u2286 BQP, where BPP is the classical complexity class of decision problems which can be solved with bounded probability of error using polynomial time on a classical Turing machine. Thus we have the chain of inclusions BPP \u2286 BQP \u2286 PSPACE. Proving that BQP = BPP \u2013", " intuitively the statement that quantum computers are more powerful than classical computers \u2013 will therefore imply that BPP = PSPACE. However, it is not presently known whether BPP = PSPACE, and proving this would represent a major breakthrough in classical computer science! So proving that quantum computers are more powerful than classical computers would have some very interesting implications for classical computational complexity! Unfortunately, it also means that providing such a proof may be quite dif\ufb01cult. Why is it that BQP \u2286 PSPACE? Here is an intuitive outline of the proof (a rigorous proof is left to the references in \u2018History and further reading\u2019). Suppose we have an n qubit quantum computer, and do a computation involving a sequence of p(n) gates, where p(n) is some polynomial in n. Supposing the quantum circuit starts in the state |0 we will explain how to evaluate in polynomial space on a classical computer the probability that it ends up in the state |y. Suppose the gates that are executed on the quantum computer are, in order, U1, U2,..., Up(n). Then the probability of ending up in the state |y is the modulus squared of y|Up(n) \u00b7 \u00b7 \u00b7 U2U1|0. (4.86) This quantity may be estimated in polynomial space on a classical computer. The basic 202 Quantum circuits idea is to insert the completeness relation obtaining y|Up(n) \u00b7 \u00b7 \u00b7 U2U1|0 = x1,...,xp(n)\u22121 x |xx| = I between each term in (4.86), y|Up(n)|xp(n)\u22121xp(n)\u22121|Up(n)\u22122... U2|x1x1|U1|0. (4.87) Given that the individual unitary gates appearing in this sum are operations such as the, and so on, it is clear that each term in the sum can be calculated Hadamard gate, to high accuracy using only polynomial space on a classical computer, and thus the sum as a whole can be calculated using polynomial space, since individual terms in the sum can be erased after being added to the running total. Of course, this algorithm is rather slow, since there are exponentially many terms in the sum which need to be calculated and added to the total; however, only polynomially much space is consumed, and thus BQP \u2286 PSPACE, as we set out to show. A similar procedure can be used to simulate an arbitrary quantum computation on a classical computer, no matter the length of the quantum computation. Therefore, the class of problems solvable on a quantum computer with unlimited time and space resources is no larger than the class of problems solvable on a classical computer. Stated another way, this means that quantum computers do not violate the Church\u2013Turing thesis that any algorithmic process can be simulated ef\ufb01ciently using a Turing machine. Of course, quantum computers may be much more ef\ufb01cient than their classical counterparts, thereby challenging the strong Church\u2013Turing thesis that any algorithmic process can be simulated ef\ufb01ciently using a probabilistic Turing machine. 4.6 Summary of the quantum circuit model of computation In this book the term \u2018quantum computer\u2019 is synonymous with the quantum circuit model of computation. This chapter has provided a detailed look at quantum circuits, their basic elements, universal families of gates, and some applications. Before we move on to more sophisticated applications, let us summarize the key elements of the quantum circuit model of computation: (1) Classical resources: A quantum computer consists of two parts, a classical part and a quantum part. In principle, there is no need for the classical part of the computer, but in practice certain tasks may be made much easier if parts of the computation can be done classically. For example, many schemes for quantum error-correction (Chapter 10) are likely to involve classical computations in order to maximize ef\ufb01ciency. While classical computations can always be done, in principle, on a quantum computer, it may be more convenient to perform the calculations on a classical computer. (2) A suitable state space: A quantum circuit operates on some number, n, of qubits. The state space is thus a 2n-dimensional complex Hilbert space. Product states of the form |x1,..., xn, where xi = 0, 1, are known as computational basis states of the computer. |x denotes a computational basis state, where x is the number whose binary representation is x1... xn. (3) Ability to prepare states in the computational basis: It is assumed that any computational basis state |x1,..., xn can be prepared in at most n steps. Summary", " of the quantum circuit model of computation 203 (4) Ability to perform quantum gates: Gates can be applied to any subset of qubits as desired, and a universal family of gates can be implemented. For example, it gate to any pair of qubits in the quantum should be possible to apply the computer. The Hadamard, phase, which any unitary operation can be approximated, and thus is a universal set of gates. Other universal families exist. and \u03c0/8 gates form a family of gates from (5) Ability to perform measurements in the computational basis: Measurements may be performed in the computational basis of one or more of the qubits in the computer. The quantum circuit model of quantum computation is equivalent to many other models of computation which have been proposed, in the sense that other models result in essentially the same resource requirements for the same problems. As a simple example which illustrates the basic idea, one might wonder whether moving to a design based on three-level quantum systems, rather than the two-level qubits, would confer any computational advantage. Of course, although there may be some slight advantage in using three-level quantum systems (qutrits) over two-level systems, any difference will be essentially negligible from the theoretical point of view. At a less trivial level, the \u2018quantum Turing machine\u2019 model of computation, a quantum generalization of the classical Turing machine model, has been shown to be equivalent to the model based upon quantum circuits. We do not consider that model of computation in this book, but the reader interested in learning more about quantum Turing machines may consult the references given in the end of chapter \u2018History and further reading\u2019. Despite the simplicity and attraction of the quantum circuit model, it is useful to keep in mind possible criticisms, modi\ufb01cations, and extensions. For example, it is by no means clear that the basic assumptions underlying the state space and starting conditions in the quantum circuit model are justi\ufb01ed. Everything is phrased in terms of \ufb01nite dimensional state spaces. Might there be anything to be gained by using systems whose state space is in\ufb01nite dimensional? Assuming that the starting state of the computer is a computational basis state is also not necessary; we know that many systems in Nature \u2018prefer\u2019 to sit in highly entangled states of many systems; might it be possible to exploit this preference to obtain extra computational power? It might be that having access to certain states allows particular computations to be done much more easily than if we are constrained to start in the computational basis. Likewise, the ability to ef\ufb01ciently perform entangling measurements in multi-qubit bases might be as useful as being able to perform just entangling unitary operations. Indeed, it may be possible to harness such measurements to perform tasks intractable within the quantum circuit model. A detailed examination and attempted justi\ufb01cation of the physics underlying the quantum circuit model is outside the scope of the present discussion, and, indeed, outside the scope of present knowledge! By raising these issues we wish to introduce the question of the completeness of the quantum circuit model, and re-emphasize the fundamental point that information is physical. In our attempts to formulate models for information processing we should always attempt to go back to fundamental physical laws. For the purposes of this book, we shall stay within the quantum circuit model of computation. It offers a rich and powerful model of computation that exploits the properties of quantum mechanics to perform amazing feats of information processing, without classical prece- 204 Quantum circuits dent. Whether physically reasonable models of computation exist which go beyond the quantum circuit model is a fascinating question which we leave open for you. 4.7 Simulation of quantum systems Perhaps [...] we need a mathematical theory of quantum automata. [...] the quantum state space has far greater capacity than the classical one: for a classical system with N states, its quantum version allowing superposition accommodates cN states. When we join two classical systems, their number of states N1 and N2 are multiplied, and in the quantum case we get the exponential growth cN1N2. [...] These crude estimates show that the quantum behavior of the system might be much more complex than its classical simulation. \u2013 Yu Manin (1980)[Man80], as translated in [Man99] The quantum-mechanical computation of one molecule of methane requires 1042 grid points. Assuming that at each point we have to perform only 10 elementary operations, and that the computation is performed at the extremely low temperature T = 3 \u00d7 10\u22123K, we would still have to use all the energy produced on Earth during the last century. \u2013 R. P. Poplavskii (1975)[Pop75], as quoted by Manin Can physics be simulated by a universal computer? [...] the physical world is quantum mechanical, and therefore the proper problem is the simulation of quantum physics [...] the full description of quantum mechanics for a large system with R particles", " [...] has too many variables, it cannot be simulated with a normal computer with a number of elements proportional to R [... but it can be simulated with ] quantum computer elements. [...] Can a quantum system be probabilistically simulated by a classical (probabilistic, I\u2019d assume) universal computer? [...] If you take the computer to be the classical kind I\u2019ve described so far [..] the answer is certainly, No! \u2013 Richard P. Feynman (1982)[Fey82] Let us close out this chapter by providing an interesting and useful application of the quantum circuit model. One of the most important practical applications of computation is the simulation of physical systems. For example, in the engineering design of a new building, \ufb01nite element analysis and modeling is used to ensure safety while minimizing cost. Cars are made lightweight, structurally sound, attractive, and inexpensive, by using computer aided design. Modern aeronautical engineering depends heavily on computational \ufb02uid dynamics simulations for aircraft designs. Nuclear weapons are no longer exploded (for the most part), but rather, tested by exhaustive computational modeling. Examples abound, because of the tremendous practical applications of predictive simulations. We begin by describing some instances of the simulation problem, then we present a quantum algorithm for simulation and an illustrative example, concluding with some perspective on this application. 4.7.1 Simulation in action The heart of simulation is the solution of differential equations which capture the physical laws governing the dynamical behavior of a system. Some examples include Newton\u2019s Simulation of quantum systems 205 law, Poisson\u2019s equation, m dx dt d dt = F, \u2212\u2207 \u00b7 (k \u2207u) = Q, the electromagnetic vector wave equation, and the diffusion equation, \u2207 \u00b7 \u2207 E = 0\u03bc0 \u22022 E \u2202t2, \u22072\u03c8 = 1 a2 \u2202\u03c8 \u2202t, (4.88) (4.89) (4.90) (4.91) just to name a very few. The goal is generally: given an initial state of the system, what is the state at some other time and/or position? Solutions are usually obtained by approximating the state with a digital representation, then discretizing the differential equation in space and time such that an iterative application of a procedure carries the state from the initial to the \ufb01nal conditions. Importantly, the error in this procedure is bounded, and known not to grow faster than some small power of the number of iterations. Furthermore, not all dynamical systems can be simulated ef\ufb01ciently: generally, only those systems which can be described ef\ufb01ciently can be simulated ef\ufb01ciently. Simulation of quantum systems by classical computers is possible, but generally only very inef\ufb01ciently. The dynamical behavior of many simple quantum systems is governed by Schr\u00a8odinger\u2019s equation, i d dt |\u03c8 = H|\u03c8. (4.92) We will \ufb01nd it convenient to absorb into H, and use this convention for the rest of this section. For a typical Hamiltonian of interest to physicists dealing with real particles in space (rather than abstract systems such as qubits, which we have been dealing with!), this reduces to i \u2202 \u2202t \u03c8(x) = \u2212 1 2m \u22022 \u2202x2 + V (x) \u03c8(x), (4.93) using a convention known as the position representation x|\u03c8 = \u03c8(x). This is an elliptical equation very much like Equation (4.91). So just simulating Schr\u00a8odinger\u2019s equation is not the especial dif\ufb01culty faced in simulating quantum systems. What is the dif\ufb01culty? The key challenge in simulating quantum systems is the exponential number of differential equations which must be solved. For one qubit evolving according to the Schr\u00a8odinger equation, a system of two differential equations must be solved; for two qubits, four equations; and for n qubits, 2n equations. Sometimes, insightful approximations can be made which reduce the effective number of equations involved, thus making classical simulation of the quantum system feasible. However, there are many physically interesting quantum systems for which no such approximations are known. Exercise 4.46: (Exponential complexity growth of quantum systems) Let \u03c1 be a density matrix describing the state of n qubits. Show that describing \u03c1 requires 4n \u2212 1 independent real numbers. 206 Quantum circuits The reader with a physics background may appreciate that there are many important quantum systems for which classical simulation is intractable. These include the Hubbard model, a model of interacting fermionic particles with the Hamiltonian H = n k=1 V0nk\u2191nk\u2193 +", " t0c\u2217 k\u03c3cj\u03c3, k,j neighbors,\u03c3 (4.94) which is useful in the study of superconductivity and magnetism, the Ising model, H = n k=1 \u03c3k \u00b7 \u03c3k+1, (4.95) and many others. Solutions to such models give many physical properties such as the dielectric constant, conductivity, and magnetic susceptibility of materials. More sophisticated models such as quantum electrodynamics (QED) and quantum chromodynamics (QCD) can be used to compute constants such as the mass of the proton. Quantum computers can ef\ufb01ciently simulate quantum systems for which there is no known ef\ufb01cient classical simulation. Intuitively, this is possible for much the same reason any quantum circuit can be constructed from a universal set of quantum gates. Moreover, just as there exist unitary operations which cannot be ef\ufb01ciently approximated, it is possible in principle to imagine quantum systems with Hamiltonians which cannot be ef\ufb01ciently simulated on a quantum computer. Of course, we believe that such systems aren\u2019t actually realized in Nature, otherwise we\u2019d be able to exploit them to do information processing beyond the quantum circuit model. 4.7.2 The quantum simulation algorithm Classical simulation begins with the realization that in solving a simple differential equation such as dy/dt = f (y), to \ufb01rst order, it is known that y(t + \u0394t) \u2248 y(t) + f (y)\u0394t. Similarly, the quantum case is concerned with the solution of id|\u03c8/dt = H|\u03c8, which, for a time-independent H, is just |\u03c8(t) = e\u2212iHt|\u03c8(0). (4.96) Since H is usually extremely dif\ufb01cult to exponentiate (it may be sparse, but it is also exponentially large), a good beginning is the \ufb01rst order solution |\u03c8(t + \u0394t) \u2248 (I \u2212 iH\u0394t)|\u03c8(t). This is tractable, because for many Hamiltonians H it is straightforward to compose quantum gates to ef\ufb01ciently approximate I \u2212 iH\u0394t. However, such \ufb01rst order solutions are generally not very satisfactory. Ef\ufb01cient approximation of the solution to Equation (4.96), to high order, is possible for many classes of Hamiltonian. For example, in most physical systems, the Hamiltonian can be written as a sum over many local interactions. Speci\ufb01cally, for a system of n particles, L Hk, H = (4.97) k=1 where each Hk acts on at most a constant c number of systems, and L is a polynomial in n. For example, the terms Hk are often just two-body interactions such as XiXj and onebody Hamiltonians such as Xi. Both the Hubbard and Ising models have Hamiltonians of this form. Such locality is quite physically reasonable, and originates in many systems Simulation of quantum systems 207 from the fact that most interactions fall off with increasing distance or difference in energy. There are sometimes additional global symmetry constraints such as particle statistics; we shall come to those shortly. The important point is that although e\u2212iHt is dif\ufb01cult to compute, e\u2212iHkt acts on a much smaller subsystem, and is straightforward to approximate k e\u2212iHkt! How, using quantum circuits. But because [Hj, Hk] = 0 in general, e\u2212iHt = then, can e\u2212iHkt be useful in constructing e\u2212iHt?, Exercise 4.47: For H = k Hk, prove that e\u2212iHt = e\u2212iH1te\u2212iH2t... e\u2212iHLt for all t L if [Hj, Hk] = 0, for all j, k. Exercise 4.48: Show that the restriction of Hk to involve at most c particles implies that in the sum (4.97), L is upper bounded by a polynomial in n. The heart of quantum simulation algorithms is the following asymptotic approximation theorem: Theorem 4.3: (Trotter formula) Let A and B be Hermitian operators. Then for any real t, lim n\u2192\u221e (eiAt/neiBt/n)n = ei(A+B)t. (4.98) Note that (4.98) is true even if A and B do not commute. Even more interestingly, perhaps, it can be generalized to hold for A and B which are generators of certain kinds of semigroups, which correspond", " to general quantum operations; we shall describe such generators (the \u2018Lindblad form\u2019) in Section 8.4.1 of Chapter 8. For now, we only consider the case of A and B being Hermitian matrices. Proof By de\ufb01nition, and thus eiAt/n = I + eiAt/neiBt/n = I + iAt + O 1 n2, i(A + B)t + O 1 n 1 n. 1 n2 Taking products of these gives us (eiAt/neiBt/n)n = I + n and since n k 1 nk = 1 + O k=1 1 n 1 nk n k i(A + B)t k + O, 1 n /k!, this gives (4.99) (4.100) (4.101) lim n\u2192\u221e (eiAt/neiBt/n)n = lim n\u2192\u221e n k=0 (i(A + B)t)k k = ei(A+B)t. (4.102) Modi\ufb01cations of the Trotter formula provide the methods by which higher order 208 Quantum circuits approximations can be derived for performing quantum simulations. For example, using similar reasoning to the proof above, it can be shown that Similarly, ei(A+B)\u0394t = eiA\u0394teiB\u0394t + O(\u0394t2). ei(A+B)\u0394t = eiA\u0394t/2eiB\u0394teiA\u0394t/2 + O(\u0394t3). (4.103) (4.104) An overview of the quantum simulation algorithm is given below, and an explicit example of simulating the one-dimensional non-relativistic Schr\u00a8odinger equation is shown in Box 4.2. Algorithm: Quantum simulation k Hk acting on an N -dimensional system, Inputs: (1) A Hamiltonian H = where each Hk acts on a small subsystem of size independent of N, (2) an initial state |\u03c80, of the system at t = 0, (3) a positive, non-zero accuracy \u03b4, and (3) a time tf at which the evolved state is desired. Outputs: A state | \u02dc\u03c8(tf ) such that | \u02dc\u03c8(tf )|e\u2212iHtf |\u03c80|2 \u2265 1 \u2212 \u03b4. Runtime: O(poly(1/\u03b4)) operations. Procedure: Choose a representation such that the state | \u02dc\u03c8 of n = poly(log N ) qubits approximates the system and the operators e\u2212iHk\u0394t have ef\ufb01cient quantum circuit approximations. Select an approximation method (see for example Equations (4.103)\u2013(4.105)) and \u0394t such that the expected error is acceptable (and j\u0394t = tf for an integer j), construct the corresponding quantum circuit U\u0394t for the iterative step, and do: 1. 2. 3. 4. | \u02dc\u03c80 \u2190 |\u03c80 ; j = 0 \u2192 | \u02dc\u03c8j+1 = U\u0394t| \u02dc\u03c8j \u2192 j = j + 1 ; goto 2 until j\u0394t \u2265 tf \u2192 | \u02dc\u03c8(tf ) = | \u02dc\u03c8j initialize state iterative update loop \ufb01nal result Exercise 4.49: (Baker\u2013Campbell\u2013Hausdorf formula) Prove that e(A+B)\u0394t = eA\u0394teB\u0394te\u2212 1 2 [A,B]\u0394t2 + O(\u0394t3), (4.105) and also prove Equations (4.103) and (4.104). L Exercise 4.50: Let H = k Hk, and de\ufb01ne e\u2212iH1\u0394te\u2212iH2\u0394t... e\u2212iHL\u0394t U\u0394t = e\u2212iHL\u0394te\u2212iHL\u22121\u0394t... e\u2212iH1\u0394t. (4.106) (a) Prove that U\u0394t = e\u22122iH\u0394t + O(\u0394t3). (b) Use the results in Box 4.1 to prove that for a positive integer m, E(U m \u0394t, e\u22122miH\u0394t) \u2264 m\u03b1\u0394t3, (4.107) for some constant \u03b1. Simulation of quantum systems 209 Box 4.2: Quantum simulation of Schr\u00a8odinger\u2019s equation The methods and", " limitations of quantum simulation may be illustrated by the following example, drawn from the conventional models studied by physicists, rather than the abstract qubit model. Consider a single particle living on a line, in a onedimensional potential V (x), governed by the Hamiltonian H = p2 2m + V (x), (4.108) where p is the momentum operator and x is the position operator. The eigenvalues of x are continuous, and the system state |\u03c8 resides in an in\ufb01nite dimensional Hilbert space; in the x basis, it can be written as - \u221e |\u03c8 = \u2212\u221e |xx|\u03c8 dx. (4.109) In practice, only some \ufb01nite region is of interest, which we may take to be the range \u2212d \u2264 x \u2264 d. Furthermore, it is possible to choose a differential step size \u0394x suf\ufb01ciently small compared to the shortest wavelength in the system such that d/\u0394x | \u02dc\u03c8 = ak|k\u0394x (4.110) k=\u2212d/\u0394x provides a good physical approximation of |\u03c8. This state can be represented using n = log(2d/\u0394x + 1) qubits; we simply replace the basis |k\u0394x (an eigenstate of the x operator) with |k, a computational basis state of n qubits. Note that only n qubits are required for this simulation, whereas classically 2n complex numbers would have to be kept track of, thus leading to an exponential resource saving when performing the simulation on a quantum computer. Computation of | \u02dc\u03c8(t) = e\u2212iHt| \u02dc\u03c8(0) must utilize one of the approximations of Equations (4.103)\u2013(4.105) because in general H1 = V (x) does not commute with H0 = p2/2m. Thus, we must be able to compute e\u2212iH1\u0394t and e\u2212iH0\u0394t. Because | \u02dc\u03c8 is expressed in the eigenbasis of H1, e\u2212iH1\u0394t is a diagonal transformation of the form |k \u2192 e\u2212iV (k\u0394x)\u0394t|k. (4.111) It is straightforward to compute this, since we can compute V (k\u0394x)\u0394t. (See also Problem 4.1.) The second term is also simple, because x and p are conjugate variables related by a quantum Fourier transform UFFTxU \u2020 FFT = p, and thus e\u2212iH0\u0394t = UFFTe\u2212ix2\u0394t/2mU \u2020 FFT; to compute e\u2212iH0\u0394t, do |k \u2192 UFFTe\u2212ix2/2mU \u2020 |k. FFT The construction of UFFT is discussed in Chapter 5. (4.112) 4.7.3 An illustrative example The procedure we have described for quantum simulations has concentrated on simulating Hamiltonians which are sums of local interations. However, this is not a fundamental 210 Quantum circuits requirement! As the following example illustrates, ef\ufb01cient quantum simulations are possible even for Hamiltonians which act non-trivially on all or nearly all parts of a large system. Suppose we have the Hamiltonian (4.113) H = Z1 \u2297 Z2 \u2297 \u00b7 \u00b7 \u00b7 \u2297 Zn, which acts on an n qubit system. Despite this being an interaction involving all of the system, indeed, it can be simulated ef\ufb01ciently. What we desire is a simple quantum circuit which implements e\u2212iH\u0394t, for arbitrary values of \u0394t. A circuit doing precisely this, for n = 3, is shown in Figure 4.19. The main insight is that although the Hamiltonian involves all the qubits in the system, it does so in a classical manner: the phase shift applied to the system is e\u2212i\u0394t if the parity of the n qubits in the computational basis is even; otherwise, the phase shift should be ei\u0394t. Thus, simple simulation of H is possible by \ufb01rst classically computing the parity (storing the result in an ancilla qubit), then applying the appropriate phase shift conditioned on the parity, then uncomputing the parity (to erase the ancilla). This strategy clearly works not only for n = 3, but also for arbitrary values of n\u2212i\u0394tZ \u2295 \u2295 \u2295 0 | Figure 4.19. Quantum circuit for simulating the Hamiltonian H = Z1 \u2297 Z2 \u2297 Z3 for time \u0394t. Furthermore,", " extending the same procedure allows us to simulate more complicated extended Hamiltonians. Speci\ufb01cally, we can ef\ufb01ciently simulate any Hamiltonian of the form n. H = \u03c3k c(k), (4.114) k=1 where \u03c3k c(k) is a Pauli matrix (or the identity) acting on the kth qubit, with c(k) \u2208 {0, 1, 2, 3} specifying one of {I, X, Y, Z}. The qubits upon which the identity operation is performed can be disregarded, and X or Y terms can be transformed by single qubit gates to Z operations. This leaves us with a Hamiltonian of the form of (4.113), which is simulated as described above. Exercise 4.51: Construct a quantum circuit to simulate the Hamiltonian performing the unitary transform e\u2212i\u0394tH for any \u0394t. H = X1 \u2297 Y2 \u2297 Z3, (4.115) Using this procedure allows us to simulate a wide class of Hamiltonians containing terms which are not local. In particular, it is possible to simulate a Hamiltonian of the form Simulation of quantum systems 211 L k=1 Hk where the only restriction is that the individual Hk have a tensor product H = structure, and that L is polynomial in the total number of particles n. More generally, all that is required is that there be an ef\ufb01cient circuit to simulate each Hk separately. As an Xk + Z \u2297n can easily be simulated using the above example, the Hamiltonian H = techniques. Such Hamiltonians typically do not arise in Nature. However, they provide a new and possibly valuable vista on the world of quantum algorithms. n k=1 4.7.4 Perspectives on quantum simulation The quantum simulation algorithm is very similar to classical methods, but also differs in a fundamental way. Each iteration of the quantum algorithm must completely replace the old state with the new one; there is no way to obtain (non-trivial) information from an intermediate step without signi\ufb01cantly changing the algorithm, because the state is a quantum one. Furthermore, the \ufb01nal measurement must be chosen cleverly to provide the desired result, because it disturbs the quantum state. Of course, the quantum simulation can be repeated to obtain statistics, but it is desirable to repeat the algorithm only at most a polynomial number of times. It may be that even though the simulation can be performed ef\ufb01ciently, there is no way to ef\ufb01ciently perform a desired measurement. Also, there are Hamiltonians which simply can\u2019t be simulated ef\ufb01ciently. In Section 4.5.4, we saw that there exist unitary transformations which quantum computers cannot ef\ufb01ciently approximate. As a corollary, not all Hamiltonian evolutions can be ef\ufb01ciently simulated on a quantum computer, for if this were possible, then all unitary transformations could be ef\ufb01ciently approximated! Another dif\ufb01cult problem \u2013 one which is very interesting \u2013 is the simulation of equilibration processes. A system with Hamiltonian H in contact with an environment at temperature T will generally come to thermal equilibrium in a state known as the Gibbs state, \u03c1therm = e\u2212H/kBT /Z, where kB is Boltzmann\u2019s constant, and Z = tr e\u2212H/kBT is the usual partition function normalization, which ensures that tr(\u03c1) = 1. The process by which this equilibration occurs is not very well understood, although certain requirements are known: the environment must be large, it must have non-zero population in states with energies matching the eigenstates of H, and its coupling with the system should be weak. Obtaining \u03c1therm for arbitrary H and T is generally an exponentially dif\ufb01cult problem for a classical computer. Might a quantum computer be able to solve this ef\ufb01ciently? We do not yet know. On the other hand, as we discussed above many interesting quantum problems can indeed be simulated ef\ufb01ciently with a quantum computer, even when they have extra constraints beyond the simple algorithms presented here. A particular class of these involve global symmetries originating from particle statistics. In the everyday world, we are used to being able to identify different particles; tennis balls can be followed around a tennis court, keeping track of which is which. This ability to keep track of which object is which is a general feature of classical objects \u2013 by continuously measuring the position of a classical particle it can be tracked at all times, and thus uniquely distinguished from other particles. However, this breaks down in quantum mechanics, which prevents us from following the", " motion of individual particles exactly. If the two particles are inherently different, say a proton and an electron, then we can distinguish them by measuring the sign of the charge to tell which particle is which. But in the case of identical particles, like two electrons, it is found that they are truly indistinguishable. Indistinguishability of particles places a constraint on the state vector of a system which 212 Quantum circuits manifests itself in two ways. Experimentally, particles in Nature are found to come in two distinct \ufb02avors, known as bosons and fermions. The state vector of a system of bosons remains unchanged under permutation of any two constituents, re\ufb02ecting their fundamental indistinguishability. Systems of fermions, in contrast, experience a sign change in their state vector under interchange of any two constituents. Both kinds of systems can be simulated ef\ufb01ciently on a quantum computer. The detailed description of how this is done is outside the scope of this book; suf\ufb01ce it to say the procedure is fairly straightforward. Given an initial state of the wrong symmetry, it can be properly symmetrized before the simulation begins. And the operators used in the simulation can be constructed to respect the desired symmetry, even allowing for the effects of higher order error terms. The reader who is interested in pursuing this and other topics further will \ufb01nd pointers to the literature in \u2018History and further reading,\u2019 at the end of the chapter. Problem 4.1: (Computable phase shifts) Let m and n be positive integers. Suppose f : {0,..., 2m \u2212 1} \u2192 {0,..., 2n \u2212 1} is a classical function from m to n bits which may be computed reversibly using T Toffoli gates, as described in Section 3.2.5. That is, the function (x, y) \u2192 (x, y \u2295 f (x)) may be implemented using T Toffoli gates. Give a quantum circuit using 2T + n (or fewer) one, two, and three qubit gates to implement the unitary operation de\ufb01ned by |x \u2192 exp \u22122i\u03c0f (x) 2n |x. (4.116) Problem 4.2: Find a depth O(log n) construction for the Cn(X) gate. (Comment: The depth of a circuit is the number of distinct timesteps at which gates are applied; the point of this problem is that it is possible to parallelize the Cn(X) construction by applying many gates in parallel during the same timestep.) Problem 4.3: (Alternate universality construction) Suppose U is a unitary matrix on n qubits. De\ufb01ne H \u2261 i ln(U ). Show that (1) H is Hermitian, with eigenvalues in the range 0 to 2\u03c0. (2) H can be written H = hgg, (4.117) g where hg are real numbers and the sum is over all n-fold tensor products g of the Pauli matrices {I, X, Y, Z}. (3) Let \u0394 = 1/k, for some positive integer k. Explain how the unitary operation exp(\u2212ihgg\u0394) may be implemented using O(n) one and two qubit operations. (4) Show that exp(\u2212iH\u0394) = / g exp(\u2212ihgg\u0394) + O(4n\u03942), (4.118) where the product is taken with respect to any \ufb01xed ordering of the n-fold tensor products of Pauli matrices, g. Chapter problems 213 (5) Show that 0 / g U = 1k exp(\u2212ihgg\u0394) + O(4n\u0394). (4.119) (6) Explain how to approximate U to within a distance > 0 using O(n16n/) one and two qubit unitary operations. Problem 4.4: (Minimal Toffoli construction) (Research) (1) What is the smallest number of two qubit gates that can be used to implement the Toffoli gate? (2) What is the smallest number of one qubit gates and gates that can be used to implement the Toffoli gate? (3) What is the smallest number of one qubit gates and controlled-Z gates that can be used to implement the Toffoli gate? Problem 4.5: (Research) Construct a family of Hamiltonians, {Hn}, on n qubits, such that simulating Hn requires a number of operations super-polynomial in n. (Comment: This problem seems to be quite dif\ufb01cult.) Problem 4.6: (Universality with prior entanglement) Controlled-", " gates and single qubit gates form a universal set of quantum logic gates. Show that an alternative universal set of resources is comprised of single qubit unitaries, the ability to perform measurements of pairs of qubits in the Bell basis, and the ability to prepare arbitrary four qubit entangled states. Summary of Chapter 4: Quantum circuits \u2022 Universality: Any unitary operation on n qubits may be implemented exactly by composing single qubit and controlled- gates. \u2022 Universality with a discrete set: The Hadamard gate, phase gate, controlledgate, and \u03c0/8 gate are universal for quantum computation, in the sense that an arbitrary unitary operation on n qubits can be approximated to an arbitrary accuracy > 0 using a circuit composed of only these gates. Replacing the \u03c0/8 gate in this list with the Toffoli gate also gives a universal family. \u2022 Not all unitary operations can be ef\ufb01ciently implemented: There are unitary operations on n qubits which require \u03a9(2n log(1/)/ log(n)) gates to approximate to within a distance using any \ufb01nite set of gates. \u2022 Simulation: For a Hamiltonian H = k Hk which is a sum of polynomially many terms Hk such that ef\ufb01cient quantum circuits for Hk can be constructed, a quantum computer can ef\ufb01ciently simulate the evolution e\u2212iHt and approximate |\u03c8(t) = e\u2212iHt|\u03c8(0), given |\u03c8(0). 214 Quantum circuits History and further reading The gate constructions in this chapter are drawn from a wide variety of sources. The paper by Barenco, Bennett, Cleve, DiVincenzo, Margolus, Shor, Sleator, Smolin, and Weinfurter[BBC+95] was the source of many of the circuit constructions in this chapter, gates. Another useful and for the universality proof for single qubit and controlledsource of insights about quantum circuits is the paper by Beckman, Chari, Devabhaktuni, and Preskill[BCDP96]. A gentle and accessible introduction has been provided by DiVincenzo[DiV98]. The fact that measurements commute with control qubit terminals was pointed out by Grif\ufb01ths and Niu[GN96]. The universality proof for two-level unitaries is due to Reck, Zeilinger, Bernstein, and Bertani[RZBB94]. The universality of the controlledand single qubit gates was proved by DiVincenzo[DiV95b]. The universal gate G in Exercise 4.44 is sometimes known as the Deutsch gate[Deu89]. Deutsch, Barenco, and Ekert[DBE95] and Lloyd[Llo95] independently proved that almost any two qubit quantum logic gate is universal. That errors caused by sequences of gates is at most the sum of the errors of the individual gates was proven by Bernstein and Vazirani [BV97]. The speci\ufb01c universal set of gates we have focused on \u2013 the and \u03c0/8 gates, was proved universal in Boykin, Mor, Hadamard, phase, controlledPulver, Roychowdhury, and Vatan[BMP+99], which also contains a proof that \u03b8 de\ufb01ned by cos(\u03b8/2) \u2261 cos2(\u03c0/8) is an irrational multiple of \u03c0. The bound in Section 4.5.4 is based on a paper by Knill[Kni95], which does a much more detailed investigation of the hardness of approximating arbitrary unitary operations using quantum circuits. In particular, Knill obtains tighter and more general bounds than we do, and his analysis applies also to cases where the universal set is a continuum of gates, not just a \ufb01nite set, as we have considered. The quantum circuit model of computation is due to Deutsch[Deu89], and was further developed by Yao[Yao93]. The latter paper showed that the quantum circuit model of computation is equivalent to the quantum Turing machine model. Quantum Turing machines were introduced in 1980 by Benioff[Ben80], further developed by Deutsch[Deu85] and Yao[Yao93], and their modern de\ufb01nition given by Bernstein and Vazirani[BV97]. The latter two papers also take \ufb01rst steps towards setting up a theory of quantum computational complexity, analogous to classical computational complexity theory. In particular, the inclusion BQP \u2286 PSPACE and some slightly stronger results was proved by Bernstein and Vazirani. Knill and La\ufb02amme[KL99] develop some fascinating connections between quantum and classical computational complexity. Other interesting work on quantum computational complexity includes the paper", " by Adleman, Demarrais and Huang[ADH97], and the paper by Watrous[Wat99]. The latter paper gives intriguing evidence to suggest that quantum computers are more powerful than classical computers in the setting of \u2018interactive proof systems\u2019. The suggestion that non-computational basis starting states may be used to obtain computational power beyond the quantum circuits model was made by Daniel Gottesman and Michael Nielson. That quantum computers might simulate quantum systems more ef\ufb01ciently than classical computers was intimated by Manin[Man80] in 1980, and independently developed in more detail by Feynman[Fey82] in 1982. Much more detailed investigations were subsequently carried out by Abrams and Lloyd[AL97], Boghosian and Taylor[BT97], Sornborger and Stewart[SS99], Wiesner[Wie96], and Zalka[Zal98]. The Trotter formula is attributed to Trotter[Tro59], and was also proven by Chernoff[Che68], although the simpler form for History and further reading 215 unitary operators is much older, and goes back to the time of Sophus Lie. The third order version of the Baker\u2013Campbell\u2013Hausdorff formula, Equation (4.104), was given by Sornborger and Stewart[SS99]. Abrams and Lloyd[AL97] give a procedure for simulating many-body Fermi systems on a quantum computer. Terhal and DiVincenzo address the problem of simulating the equilibration of quantum systems to the Gibbs state[TD98]. The method used to simulate the Schr\u00a8odinger equation in Box 4.2 is due to Zalka[Zal98] and Wiesner[Wie96]. Exercise 4.25 is due to Vandersypen, and is related to work by Chau and Wilczek[CW95]. Exercise 4.45 is due to Boykin, Mor, Pulver, Roychowdhury, and Vatan[BMP+99]. Problem 4.2 is due to Gottesman. Problem 4.6 is due to Gottesman and Chuang[GC99]. 5 The quantum Fourier transform and its applications If computers that you build are quantum, Then spies everywhere will all want \u2019em. Our codes will all fail, And they\u2019ll read our email, Till we get crypto that\u2019s quantum, and daunt \u2019em. \u2013 Jennifer and Peter Shor To read our E-mail, how mean of the spies and their quantum machine; be comforted though, they do not yet know how to factorize twelve or \ufb01fteen. \u2013 Volker Strassen Computer programming is an art form, like the creation of poetry or music. \u2013 Donald Knuth The most spectacular discovery in quantum computing to date is that quantum computers can ef\ufb01ciently perform some tasks which are not feasible on a classical computer. For example, \ufb01nding the prime factorization of an n-bit integer is thought to require exp(\u0398(n1/3 log2/3 n)) operations using the best classical algorithm known at the time of writing, the so-called number \ufb01eld sieve. This is exponential in the size of the number being factored, so factoring is generally considered to be an intractable problem on a classical computer: it quickly becomes impossible to factor even modest numbers. In contrast, a quantum algorithm can accomplish the same task using O(n2 log n log log n) operations. That is, a quantum computer can factor a number exponentially faster than the best known classical algorithms. This result is important in its own right, but perhaps the most exciting aspect is the question it raises: what other problems can be done ef\ufb01ciently on a quantum computer which are infeasible on a classical computer? In this chapter we develop the quantum Fourier transform, which is the key ingredient for quantum factoring and many other interesting quantum algorithms. The quantum Fourier transform, with which we begin in Section 5.1, is an ef\ufb01cient quantum algorithm for performing a Fourier transform of quantum mechanical amplitudes. It does not speed up the classical task of computing Fourier transforms of classical data. But one important task which it does enable is phase estimation, the approximation of the eigenvalues of a unitary operator under certain circumstances, as described in Section 5.2. This allows us to solve several other interesting problems, including the order-\ufb01nding problem and the factoring problem, which are covered in Section 5.3. Phase estimation can also be combined with the quantum search algorithm to solve the problem of counting solutions to a search problem, as described in the next chapter. Section 5.4 concludes the chapter with a discussion of how the quantum Fourier transform may be used to solve", " the hidden The quantum Fourier transform 217 subgroup problem, a generalization of the phase estimation and order-\ufb01nding problems that has among its special cases an ef\ufb01cient quantum algorithm for the discrete logarithm problem, another problem thought to be intractable on a classical computer. 5.1 The quantum Fourier transform A good idea has a way of becoming simpler and solving problems other than that for which it was intended. \u2013 Robert Tarjan One of the most useful ways of solving a problem in mathematics or computer science is to transform it into some other problem for which a solution is known. There are a few transformations of this type which appear so often and in so many different contexts that the transformations are studied for their own sake. A great discovery of quantum computation has been that some such transformations can be computed much faster on a quantum computer than on a classical computer, a discovery which has enabled the construction of fast algorithms for quantum computers. One such transformation is the discrete Fourier transform. In the usual mathematical notation, the discrete Fourier transform takes as input a vector of complex numbers, x0,..., xN \u22121 where the length N of the vector is a \ufb01xed parameter. It outputs the transformed data, a vector of complex numbers y0,..., yN \u22121, de\ufb01ned by yk \u2261 1\u221a N N \u22121 j=0 xje2\u03c0ijk/N. (5.1) The quantum Fourier transform is exactly the same transformation, although the conventional notation for the quantum Fourier transform is somewhat different. The quantum Fourier transform on an orthonormal basis |0,..., |N \u2212 1 is de\ufb01ned to be a linear operator with the following action on the basis states, |j \u2212\u2192 1\u221a N N \u22121 k=0 e2\u03c0ijk/N |k. Equivalently, the action on an arbitrary state may be written N \u22121 j=0 xj|j \u2212\u2192 N \u22121 k=0 yk|k, (5.2) (5.3) where the amplitudes yk are the discrete Fourier transform of the amplitudes xj. It is not obvious from the de\ufb01nition, but this transformation is a unitary transformation, and thus can be implemented as the dynamics for a quantum computer. We shall demonstrate the unitarity of the Fourier transform by constructing a manifestly unitary quantum circuit computing the Fourier transform. It is also easy to prove directly that the Fourier transform is unitary: Exercise 5.1: Give a direct proof that the linear transformation de\ufb01ned by Equation (5.2) is unitary. Exercise 5.2: Explicitly compute the Fourier transform of the n qubit state |00... 0. 218 The quantum Fourier transform and its applications In the following, we take N = 2n, where n is some integer, and the basis |0,..., |2n \u2212 1 is the computational basis for an n qubit quantum computer. It is helpful to write the state |j using the binary representation j = j1j2... jn. More formally, j = j12n\u22121 + j22n\u22122 + \u00b7 \u00b7 \u00b7 + jn20. It is also convenient to adopt the notation 0.jljl+1... jm to represent the binary fraction jl/2 + jl+1/4 + \u00b7 \u00b7 \u00b7 + jm/2m\u2212l+1. With a little algebra the quantum Fourier transform can be given the following useful product representation: |j1,..., jn \u2192 |0 + e2\u03c0i0.jn|1 |0 + e2\u03c0i0.jn\u22121jn|1 \u00b7 \u00b7 \u00b7 2n/2 |0 + e2\u03c0i0.j1j2\u00b7\u00b7\u00b7jn|1. (5.4) This product representation is so useful that you may even wish to consider this to be the de\ufb01nition of the quantum Fourier transform. As we explain shortly this representation allows us to construct an ef\ufb01cient quantum circuit computing the Fourier transform, a proof that the quantum Fourier transform is unitary, and provides insight into algorithms based upon the quantum Fourier transform. As an incidental bonus we obtain the classical fast Fourier transform, in the exercises! The equivalence of the product representation (5.4) and the de\ufb01nition (5.2) follows from some elementary algebra: |j \u2192 1 2n/2 2n\u22121 e2\u03c0ijk/2n |k 1 n l=1 kl2\u2212l e2\u03c0ij... |k1... kn k=0 1 k1=0 1.", ".. k1=0 n e2\u03c0ijkl 2\u2212l|kl kn=0 1 n kn=0 l=1 1 e2\u03c0ijkl 2\u2212l|kl l=1 n l=1 kl=0 |0 + e2\u03c0ij2\u2212l |1 |0 + e2\u03c0i0.jn|1 |0 + e2\u03c0i0.jn\u22121jn|1 \u00b7 \u00b7 \u00b7 2n/2 1 2n/2 1 2n/2 1 2n/2 1 2n/2 = = = = = (5.5) (5.6) (5.7) (5.8) (5.9) |0 + e2\u03c0i0.j1j2\u00b7\u00b7\u00b7jn|1.(5.10) The product representation (5.4) makes it easy to derive an ef\ufb01cient circuit for the quantum Fourier transform. Such a circuit is shown in Figure 5.1. The gate Rk denotes the unitary transformation Rk \u2261 1 0 0 e2\u03c0i/2k. (5.11) To see that the pictured circuit computes the quantum Fourier transform, consider what happens when the state |j1... jn is input. Applying the Hadamard gate to the \ufb01rst bit produces the state 1 21/2 |0 + e2\u03c0i0.j1|1 |j2... jn, (5.12) The quantum Fourier transform 219 Figure 5.1. Ef\ufb01cient circuit for the quantum Fourier transform. This circuit is easily derived from the product representation (5.4) for the quantum Fourier transform. Not shown are swap gates at the end of the circuit which reverse the order of the qubits, or normalization factors of 1/ 2 in the output. \u221a since e2\u03c0i0.j1 = \u22121 when j1 = 1, and is +1 otherwise. Applying the controlled-R2 gate produces the state \u2019 1 21/2 |0 + e2\u03c0i0.j1j2 |1 ( |j2... jn. (5.13) We continue applying the controlled-R3, R4 through Rn gates, each of which adds an extra bit to the phase of the co-ef\ufb01cient of the \ufb01rst |1. At the end of this procedure we have the state \u2019 |0 + e2\u03c0i0.j1j2...jn|1 ( 1 21/2 |j2... jn. (5.14) Next, we perform a similar procedure on the second qubit. The Hadamard gate puts us in the state |0 + e2\u03c0i0.j1j2...jn |1 |0 + e2\u03c0i0.j2 |1 |j3... jn, (5.15) ( \u2019 ( \u2019 1 22/2 and the controlled-R2 through Rn\u22121 gates yield the state \u2019 |0 + e2\u03c0i0.j1j2...jn|1 ( \u2019 |0 + e2\u03c0i0.j2...jn |1 ( |j3... jn. (5.16) 1 22/2 We continue in this fashion for each qubit, giving a \ufb01nal state \u2019 ( \u2019 \u2019 ( |0 + e2\u03c0i0.j1j2...jn|1 |0 + e2\u03c0i0.j2...jn|1... |0 + e2\u03c0i0.jn |1 1 2n/2 (. (5.17) Swap operations (see Section 1.3.4 for a description of the circuit), omitted from Figure 5.1 for clarity, are then used to reverse the order of the qubits. After the swap operations, the state of the qubits is \u2019 |0 + e2\u03c0i0.jn |1 ( \u2019 1 2n/2 |0 + e2\u03c0i0.jn\u22121jn|1 ( \u2019... |0 + e2\u03c0i0.j1j2\u00b7\u00b7\u00b7jn |1 (. (5.18) Comparing with Equation (5.4) we see that this is the desired output from the quantum Fourier transform. This construction also proves that the quantum Fourier transform is unitary, since each gate in the circuit is unitary. An explicit example showing a circuit for the quantum Fourier transform on three qubits is given in Box 5.1. How many gates does this circuit use? We start by doing a Hadamard gate and n \u2212 1 conditional rot", "ations on the \ufb01rst qubit \u2013 a total of n gates. This is followed by a Hadamard gate and n \u2212 2 conditional rotations on the second qubit, for a total of n + (n \u2212 1) gates. Continuing in this way, we see that n + (n \u2212 1) + \u00b7 \u00b7 \u00b7 + 1 = n(n + 1)/2 gates are required, 220 The quantum Fourier transform and its applications Box 5.1: Three qubit quantum Fourier transform For concreteness it may help to look at the explicit circuit for the three qubit quantum Fourier transform: Recall that S and T are the phase and \u03c0/8 gates (see page xxiii). As a matrix the quantum Fourier transform in this instance may be written out explicitly, using \u03c9 = e2\u03c0i/8 = i, as \u221a \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 1 \u03c92 \u03c93 \u03c94 \u03c95 \u03c96 \u03c97 1 \u03c92 \u03c94 \u03c96 \u03c92 \u03c94 \u03c96 1 \u03c93 \u03c96 \u03c91 \u03c94 \u03c97 \u03c92 \u03c95 1 \u03c94 \u03c94 1 \u03c95 \u03c92 \u03c97 \u03c94 \u03c91 \u03c96 \u03c93 1 \u03c96 \u03c94 \u03c92 \u03c96 \u03c94 \u03c92 1 \u03c97 \u03c96 \u03c95 \u03c94 \u03c93 \u03c92 \u03c91 \u03c94 \u03c94 5.19) plus the gates involved in the swaps. At most n/2 swaps are required, and each swap gates. Therefore, this circuit provides a can be accomplished using three controlled\u0398(n2) algorithm for performing the quantum Fourier transform. In contrast, the best classical algorithms for computing the discrete Fourier transform on 2n elements are algorithms such as the Fast Fourier Transform (FFT), which compute the discrete Fourier transform using \u0398(n2n) gates. That is, it requires exponentially more operations to compute the Fourier transform on a classical computer than it does to implement the quantum Fourier transform on a quantum computer. At face value this sounds terri\ufb01c, since the Fourier transform is a crucial step in so many real-world data processing applications. For example, in computer speech recognition, the \ufb01rst step in phoneme recognition is to Fourier transform the digitized sound. Can we use the quantum Fourier transform to speed up the computation of these Fourier transforms? Unfortunately, the answer is that there is no known way to do this. The problem is that the amplitudes in a quantum computer cannot be directly accessed by measurement. Thus, there is no way of determining the Fourier transformed amplitudes of the original state. Worse still, there is in general no way to ef\ufb01ciently prepare the original state to be Fourier transformed. Thus, \ufb01nding uses for the quantum Fourier transform is more subtle than we might have hoped. In this and the next chapter we develop several algorithms based upon a more subtle application of the quantum Fourier transform. Phase estimation 221 Exercise 5.3: (Classical fast Fourier transform) Suppose we wish to perform a Fourier transform of a vector containing 2n complex numbers on a classical computer. Verify that the straightforward method for performing the Fourier transform, based upon direct evaluation of Equation (5.1) requires \u0398(22n) elementary arithmetic operations. Find a method for reducing this to \u0398(n2n) operations, based upon Equation (5.4). Exercise 5.4: Give a decomposition of the controlled-Rk gate into single qubit and gates. Exercise 5.5: Give a quantum circuit to perform the inverse quantum Fourier transform. Exercise 5.6: (Approximate quantum Fourier transform) The quantum circuit construction of the quantum Fourier transform apparently requires gates of exponential precision in the number of qubits used. However, such precision is never required in any quantum circuit of polynomial size. For example, let U be the ideal quantum Fourier transform on n qubits, and V be the transform which results if the controlled-Rk gates are performed to a precision \u0394 = 1/p(n) for some polynomial p(n). Show that the error E(U, V ) \u2261 max|\u03c8 (U \u2212 V )|\u03c8 scales as \u0398(n2/p(n)), and thus polynomial precision in each gate is suf\ufb01cient to guarantee polynomial accuracy in the output state. 5.2 Phase estimation The Fourier transform is the key to a general procedure known as phase estimation,", " which in turn is the key for many quantum algorithms. Suppose a unitary operator U has an eigenvector |u with eigenvalue e2\u03c0i\u03d5, where the value of \u03d5 is unknown. The goal of the phase estimation algorithm is to estimate \u03d5. To perform the estimation we assume that we have available black boxes (sometimes known as oracles) capable of preparing the state |u and performing the controlled-U 2j operation, for suitable non-negative integers j. The use of black boxes indicates that the phase estimation procedure is not a complete quantum algorithm in its own right. Rather, you should think of phase estimation as a kind of \u2018subroutine\u2019 or \u2018module\u2019 that, when combined with other subroutines, can be used to perform interesting computational tasks. In speci\ufb01c applications of the phase estimation procedure we shall do exactly this, describing how these black box operations are to be performed, and combining them with the phase estimation procedure to do genuinely useful tasks. For the moment, though, we will continue to imagine them as black boxes. The quantum phase estimation procedure uses two registers. The \ufb01rst register contains t qubits initially in the state |0. How we choose t depends on two things: the number of digits of accuracy we wish to have in our estimate for \u03d5, and with what probability we wish the phase estimation procedure to be successful. The dependence of t on these quantities emerges naturally from the following analysis. The second register begins in the state |u, and contains as many qubits as is necessary to store |u. Phase estimation is performed in two stages. First, we apply the circuit shown in Figure 5.2. The circuit begins by applying a Hadamard transform to the \ufb01rst register, followed by application of controlled-U operations on the second register, with U raised 222 The quantum Fourier transform and its applications to successive powers of two. The \ufb01nal state of the \ufb01rst register is easily seen to be: \u2019 1 2t/2 |0 + e2\u03c0i2t\u22121\u03d5|1 |0 + e2\u03c0i2t\u22122\u03d5|1... |0 + e2\u03c0i20\u03d5|1 ( \u2019 ( \u2019 ( We omit the second register from this description, since it stays in the state |u throughout the computation. = 1 2t/2 t\u22121 2 k=0 e2\u03c0i\u03d5k|k. (5.20) \u23a7 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 \u23a7 \u23aa\u23a8 \u23a9 First register qubits t Second register | e2\u03c0i(2t\u22121\u03d5)| | e2\u03c0i(22\u03d5)| 1 e2\u03c0i(21\u03d5)| 1 e2\u03c0i(20\u03d5)| 1 |u U 20 U 21 U 22 \u00b7 \u00b7 \u00b7 U 2t\u22121 |u Figure 5.2. The \ufb01rst stage of the phase estimation procedure. Normalization factors of 1/ the right. \u221a 2 have been omitted, on Exercise 5.7: Additional insight into the circuit in Figure 5.2 may be obtained by showing, as you should now do, that the effect of the sequence of controlled-U operations like that in Figure 5.2 is to take the state |j|u to |jU j|u. (Note that this does not depend on |u being an eigenstate of U.) The second stage of phase estimation is to apply the inverse quantum Fourier transform on the \ufb01rst register. This is obtained by reversing the circuit for the quantum Fourier transform in the previous section (Exercise 5.5), and can be done in \u0398(t2) steps. The third and \ufb01nal stage of phase estimation is to read out the state of the \ufb01rst register by doing a measurement in the computational basis. We will show that this provides a pretty good estimate of \u03d5. An overall schematic of the algorithm is shown in Figure 5.3. To sharpen our intuition as to why phase estimation works, suppose \u03d5 may be expressed exactly in t bits, as \u03d5 = 0.\u03d51... \u03d5t. Then the state (5.20) resulting from the \ufb01rst stage of phase estimation may be rewritten \u2019 |0 + e2\u03c0i0.\u03d5t|1 ( \ufffd", "\ufffd 1 2t/2 |0 + e2\u03c0i0.\u03d5t\u22121\u03d5t|1 ( \u2019... |0 + e2\u03c0i0.\u03d51\u03d52\u00b7\u00b7\u00b7\u03d5t|1 (. (5.21) The second stage of phase estimation is to apply the inverse quantum Fourier transform. But comparing the previous equation with the product form for the Fourier transform, Equation (5.4), we see that the output state from the second stage is the product state |\u03d51... \u03d5t. A measurement in the computational basis therefore gives us \u03d5 exactly! Phase estimation 223 Figure 5.3. Schematic of the overall phase estimation procedure. The top t qubits (the \u2018/\u2019 denotes a bundle of wires, as usual) are the \ufb01rst register, and the bottom qubits are the second register, numbering as many as required to perform U. |u is an eigenstate of U with eigenvalue e2\u03c0i\u03d5. The output of the measurement is an 3 approximation to \u03d5 accurate to t \u2212 bits, with probability of success at least 1 \u2212. 2 log 2 + 1 2 Summarizing, the phase estimation algorithm allows one to estimate the phase \u03d5 of an eigenvalue of a unitary operator U, given the corresponding eigenvector |u. An essential feature at the heart of this procedure is the ability of the inverse Fourier transform to perform the transformation 1 2t/2 2t\u22121 j=0 e2\u03c0i\u03d5j|j|u \u2192 | \u02dc\u03d5|u, (5.22) where | \u02dc\u03d5 denotes a state which is a good estimator for \u03d5 when measured. 5.2.1 Performance and requirements The above analysis applies to the ideal case, where \u03d5 can be written exactly with a t bit binary expansion. What happens when this is not the case? It turns out that the procedure we have described will produce a pretty good approximation to \u03d5 with high probability, as foreshadowed by the notation used in (5.22). Showing this requires some careful manipulations. Let b be the integer in the range 0 to 2t \u2212 1 such that b/2t = 0.b1... bt is the best t bit approximation to \u03d5 which is less than \u03d5. That is, the difference \u03b4 \u2261 \u03d5 \u2212 b/2t between \u03d5 and b/2t satis\ufb01es 0 \u2264 \u03b4 \u2264 2\u2212t. We aim to show that the observation at the end of the phase estimation procedure produces a result which is close to b, and thus enables us to estimate \u03d5 accurately, with high probability. Applying the inverse quantum Fourier transform to the state (5.20) produces the state 1 2t 2t\u22121 k,l=0 e \u22122\u03c0ikl 2t e2\u03c0i\u03d5k|l. Let \u03b1l be the amplitude of |(b + l)(mod 2t), \u03b1l \u2261 1 2t 2t\u22121 \u2019 e2\u03c0i(\u03d5\u2212(b+l)/2t) (k. k=0 This is the sum of a geometric series, so \u03b1l = 1 2t 1 \u2212 e2\u03c0i(2t\u03d5\u2212(b+l)) 1 \u2212 e2\u03c0i(\u03d5\u2212(b+l)/2t) (5.23) (5.24) (5.25) 224 The quantum Fourier transform and its applications = 1 2t 1 \u2212 e2\u03c0i(2t\u03b4\u2212l) 1 \u2212 e2\u03c0i(\u03b4\u2212l/2t). (5.26) Suppose the outcome of the \ufb01nal measurement is m. We aim to bound the probability of obtaining a value of m such that |m \u2212 b| > e, where e is a positive integer characterizing our desired tolerance to error. The probability of observing such an m is given by p(|m \u2212 b| > e) = |\u03b1l|2 + |\u03b1l|2. But for any real \u03b8, |1 \u2212 exp(i\u03b8)| \u2264 2, so \u22122t\u22121<l\u2264\u2212(e+1) e+1\u2264l\u22642t\u22121 |\u03b1l| \u2264 2 2t |1 \u2212 e2\u03c0i(\u03b4\u2212l/2t)|. By elementary geometry or calculus |1 \u2212 exp(i\u03b8)| \u2265 2|\u03b8|/\u03c0 whenever \u2212\u03c0 \u2264 \u03b8 \u2264 \u03c0. But when \u22122t\u22121 < l \u2264 2t\u22121 we have \u2212\u03c0 \u2264 2\u03c0(\u03b4 \u2212 l/2t) \u2264 \u03c0. Thus |\u03b1l| \u2264 1 2t", "+1(\u03b4 \u2212 l/2t). Combining (5.27) and (5.29) gives \u23a1 p(|m \u2212 b| > e) \u2264 1 4 \u23a3 \u2212(e+1) l=\u22122t\u22121+1 1 (l \u2212 2t\u03b4)2 + 2t\u22121 l=e+1 1 (l \u2212 2t\u03b4)2 \u23a4 \u23a6. Recalling that 0 \u2264 2t\u03b4 \u2264 1, we obtain p(|m \u2212 b| > e) \u2264 1 4 \u2264 1 2 \u23a1 \u23a3 \u2212(e+1) l=\u22122t\u22121+1 1 l2 + 2t\u22121 l=e+1 1 (l \u2212 1)2 \u23a4 \u23a6 2t\u22121\u22121 1 l2 l=e 2t\u22121\u22121 - \u2264 1 2 = e\u22121 1 2(e \u2212 1). dl 1 l2 Suppose we wish to approximate \u03d5 to an accuracy 2\u2212n, that is, we choose e = 2t\u2212n \u2212 1. By making use of t = n + p qubits in the phase estimation algorithm we see from (5.34) that the probability of obtaining an approximation correct to this accuracy is at least 1 \u2212 1/2(2p \u2212 2). Thus to successfully obtain \u03d5 accurate to n bits with probability of success at least 1 \u2212 we choose t = n + log 2 +. (5.35) 4 5 1 2 In order to make use of the phase estimation algorithm, we need to be able to prepare an eigenstate |u of U. What if we do not know how to prepare such an eigenstate? Suppose that we prepare some other state |\u03c8 in place of |u. Expanding this state in terms of u cu|u. Suppose the eigenstate |u has eigenvalue eigenstates |u of U gives |\u03c8 = e2\u03c0i\u03d5u. Intuitively, the result of running the phase estimation algorithm will be to give (5.27) (5.28) (5.29) (5.30) (5.31) (5.32) (5.33) (5.34) Phase estimation 225 u cu| 6\u03d5u|u, where 6\u03d5u is a pretty good approximation to the as output a state close to phase \u03d5u. Therefore, we expect that reading out the \ufb01rst register will give us a good approximation to \u03d5u, where u is chosen at random with probability |cu|2. Making this argument rigorous is left for Exercise 5.8. This procedure allows us to avoid preparing a (possibly unknown) eigenstate, at the cost of introducing some additional randomness into the algorithm. Exercise 5.8: Suppose the phase estimation algorithm takes the state |0|u to the, the algorithm outputs state | 6\u03d5u|u, so that given the input |0 u cu|u u cu| 6\u03d5u|u. Show that if t is chosen according to (5.35), then the probability for measuring \u03d5u accurate to n bits at the conclusion of the phase estimation algorithm is at least |cu|2(1 \u2212 ). Why is phase estimation interesting? For its own sake, phase estimation solves a problem which is both non-trivial and interesting from a physical point of view: how to estimate the eigenvalue associated to a given eigenvector of a unitary operator. Its real use, though, comes from the observation that other interesting problems can be reduced to phase estimation, as will be shown in subsequent sections. The phase estimation algorithm is summarized below. Algorithm: Quantum phase estimation Inputs: (1) A black box wich performs a controlled-U j operation, for integer j, 3 (2) an eigenstate |u of U with eigenvalue e2\u03c0i\u03d5u, and (3) t = n + qubits initialized to |0. 2 + 1 2 2 log Outputs: An n-bit approximation 6\u03d5u to \u03d5u. Runtime: O(t2) operations and one call to controlled-U j black box. Succeeds with probability at least 1 \u2212. Procedure: 1. 2. 3. 4. 5. |0|u \u2192 1\u221a 2t \u2192 1\u221a 2t 2t\u22121 j=0 2t\u22121 j=0 |j|u |jU j|u initial state create superposition apply black box = 1\u221a 2t 2t\u22121 j=0 \u2192 | 6\u03d5u|u \u2192 6\u03d5u e2\u03c0ij\u03d5u |j|u result of black box apply inverse Fourier transform measure \ufb01rst register Exercise 5", ".9: Let U be a unitary transform with eigenvalues \u00b11, which acts on a state |\u03c8. Using the phase estimation procedure, construct a quantum circuit to collapse |\u03c8 into one or the other of the two eigenspaces of U, giving also a 226 The quantum Fourier transform and its applications classical indicator as to which space the \ufb01nal state is in. Compare your result with Exercise 4.34. 5.3 Applications: order-\ufb01nding and factoring The phase estimation procedure can be used to solve a variety of interesting problems. We now describe two of the most interesting of these problems: the order-\ufb01nding problem, and the factoring problem. These two problems are, in fact, equivalent to one another, so in Section 5.3.1 we explain a quantum algorithm for solving the order-\ufb01nding problem, and in Section 5.3.2 we explain how the order-\ufb01nding problem implies the ability to factor as well. To understand the quantum algorithms for factoring and order-\ufb01nding requires a little background in number theory. All the required materials are collected together in Appendix 4. The description we give over the next two sections focuses on the quantum aspects of the problem, and requires only a little familiarity with modular arithmetic to be readable. Detailed proofs of the number-theoretic results we quote here may be found in Appendix 4. The fast quantum algorithms for order-\ufb01nding and factoring are interesting for at least three reasons. First, and most important in our opinion, they provide evidence for the idea that quantum computers may be inherently more powerful than classical computers, and provide a credible challenge to the strong Church\u2013Turing thesis. Second, both problems are of suf\ufb01cient intrinsic worth to justify interest in any novel algorithm, be it classical or quantum. Third, and most important from a practical standpoint, ef\ufb01cient algorithms for order-\ufb01nding and factoring can be used to break the RSA public-key cryptosystem (Appendix 5). 5.3.1 Application: order-\ufb01nding For positive integers x and N, x < N, with no common factors, the order of x modulo N is de\ufb01ned to be the least positive integer, r, such that xr = 1(mod N ). The order-\ufb01nding problem is to determine the order for some speci\ufb01ed x and N. Order-\ufb01nding is believed to be a hard problem on a classical computer, in the sense that no algorithm is known to solve the problem using resources polynomial in the O(L) bits needed to specify the problem, where L \u2261 log(N ) is the number of bits needed to specify N. In this section we explain how phase estimation may be used to obtain an ef\ufb01cient quantum algorithm for order-\ufb01nding. Exercise 5.10: Show that the order of x = 5 modulo N = 21 is 6. Exercise 5.11: Show that the order of x satis\ufb01es r \u2264 N. The quantum algorithm for order-\ufb01nding is just the phase estimation algorithm applied to the unitary operator U |y \u2261 |xy(mod N ), (5.36) with y \u2208 {0, 1}L. (Note that here and below, when N \u2264 y \u2264 2L \u2212 1, we use the convention that xy(mod N ) is just y again. That is, U only acts non-trivially when Applications: order-\ufb01nding and factoring 227 0 \u2264 y \u2264 N \u2212 1.) A simple calculation shows that the states de\ufb01ned by r\u22121 |us \u2261 1\u221a r exp k=0 \u22122\u03c0isk r |xk mod N, for integer 0 \u2264 s \u2264 r \u2212 1 are eigenstates of U, since U |us = 1\u221a r \u22122\u03c0isk r |xk+1 mod N r\u22121 exp k=0 2\u03c0is r = exp |us. (5.37) (5.38) (5.39) Using the phase estimation procedure allows us to obtain, with high accuracy, the corresponding eigenvalues exp(2\u03c0is/r), from which we can obtain the order r with a little bit more work. Exercise 5.12: Show that U is unitary (Hint: x is co-prime to N, and therefore has an inverse modulo N ). There are two important requirements for us to be able to use the phase estimation procedure: we must have ef\ufb01cient procedures to implement a controlled-U 2j operation for any integer j, and we must be able to ef\ufb01ciently prepare an eigen", "state |us with a nontrivial eigenvalue, or at least a superposition of such eigenstates. The \ufb01rst requirement is satis\ufb01ed by using a procedure known as modular exponentiation, with which we can implement the entire sequence of controlled-U 2j operations applied by the phase estimation procedure using O(L3) gates, as described in Box 5.2. The second requirement is a little tricker: preparing |us requires that we know r, so this is out of the question. Fortunately, there is a clever observation which allows us to circumvent the problem of preparing |us, which is that 1\u221a r r\u22121 s=0 |us = |1. (5.44) 2 log 2 + 1 2 3 In performing the phase estimation procedure, if we use t = 2L + 1 + qubits in the \ufb01rst register (referring to Figure 5.3), and prepare the second register in the state |1 \u2013 which is trivial to construct \u2013 it follows that for each s in the range 0 through r \u2212 1, we will obtain an estimate of the phase \u03d5 \u2248 s/r accurate to 2L + 1 bits, with probability at least (1 \u2212 )/r. The order-\ufb01nding algorithm is schematically depicted in Figure 5.4. Exercise 5.13: Prove (5.44). (Hint: r\u22121 s=0 exp(\u22122\u03c0isk/r) = r\u03b4k0.) In fact, prove that 1\u221a r r\u22121 s=0 e2\u03c0isk/r|us = |xk mod N. (5.45) Exercise 5.14: The quantum state produced in the order-\ufb01nding algorithm, before the inverse Fourier transform, is |\u03c8 = 2t\u22121 j=0 |jU j|1 = 2t\u22121 j=0 |j|xj mod N, (5.46) 228 The quantum Fourier transform and its applications Box 5.2: Modular exponentiation How can we compute the sequence of controlled-U 2j operations used by the phase estimation procedure as part of the order-\ufb01nding algorithm? That is, we wish to compute the transformation |z|y \u2192 |zU zt2t\u22121... U z120|y = |z|xzt2t\u22121 \u00d7 \u00b7 \u00b7 \u00b7 \u00d7 xz120y(mod N ) = |z|xzy(mod N ). (5.40) (5.41) (5.42) Thus the sequence of controlled-U 2j operations used in phase estimation is equivalent to multiplying the contents of the second register by the modular exponential xz(mod N ), where z is the contents of the \ufb01rst register. This operation may be accomplished easily using the techniques of reversible computation. The basic idea is to reversibly compute the function xz(mod N ) of z in a third register, and then to reversibly multiply the contents of the second register by xz(mod N ), using the trick of uncomputation to erase the contents of the third register upon completion. The algorithm for computing the modular exponential has two stages. The \ufb01rst stage uses modular multiplication to compute x2(mod N ), by squaring x modulo N, then computes x4(mod N ) by squaring x2(mod N ), and continues in this way, computing x2j (mod N ) for all j up to t \u2212 1. We use t = 2L + 1 + log(2 + 1/(2)) = O(L), so a total of t \u2212 1 = O(L) squaring operations is performed at a cost of O(L2) each (this cost assumes the circuit used to do the squaring implements the familiar algorithm we all learn as children for multiplication), for a total cost of O(L3) for the \ufb01rst stage. The second stage of the algorithm is based upon the observation we\u2019ve already noted, \u2019 ( \u2019 xz(mod N ) = xzt2t\u22121 (mod N ) xzt\u221212t\u22122 ( (mod N ) \u2019... xz120 ( (mod N ). (5.43) Performing t \u2212 1 modular multiplications with a cost O(L2) each, we see that this product can be computed using O(L3) gates. This is suf\ufb01ciently ef\ufb01cient for our purposes, but more ef\ufb01cient algorithms are possible based on more ef\ufb01cient algorithms for multiplication (see \u2018History and further reading\u2019). Using the techniques of Section 3.2.5, it is now straightforward to construct a reversible circuit", " with a t bit register and an L bit register which, when started in the state (z, y) outputs (z, xzy (mod N )), using O(L3) gates, which can be translated into a quantum circuit using O(L3) gates computing the transformation |z|y \u2192 |z|xzy (mod N ). if we initialize the second register as |1. Show that the same state is obtained if we replace U j with a different unitary transform V, which computes V |j|k = |j|k + xj mod N, (5.47) and start the second register in the state |0. Also show how to construct V using O(L3) gates. Applications: order-\ufb01nding and factoring 229 Register 1 qubits t | / H \u2297t 0 |j \u2022 F T \u2020 Register 2 L qubits | / 1 xj mod N Figure 5.4. Quantum circuit for the order-\ufb01nding algorithm. The second register is shown as being initialized to the |1 state, but if the method of Exercise 5.14 is used, it can be initialized to |0 instead. This circuit can also be used for factoring, using the reduction given in Section 5.3.2. The continued fraction expansion The reduction of order-\ufb01nding to phase estimation is completed by describing how to obtain the desired answer, r, from the result of the phase estimation algorithm, \u03d5 \u2248 s/r. We only know \u03d5 to 2L + 1 bits, but we also know a priori that it is a rational number \u2013 the ratio of two bounded integers \u2013 and if we could compute the nearest such fraction to \u03d5 we might obtain r. Remarkably, there is an algorithm which accomplishes this task ef\ufb01ciently, known as the continued fractions algorithm. An example of how this works is described in Box 5.3. The reason this algorithm satis\ufb01es our needs is the following theorem, which is proved in Appendix 4: Theorem 5.1: Suppose s/r is a rational number such that + + + s r \u2212 \u03d5 + + + \u2264 1 2r2. (5.48) Then s/r is a convergent of the continued fraction for \u03d5, and thus can be computed in O(L3) operations using the continued fractions algorithm. Since \u03d5 is an approximation of s/r accurate to 2L + 1 bits, it follows that |s/r \u2212 \u03d5| \u2264 2\u22122L\u22121 \u2264 1/2r2, since r \u2264 N \u2264 2L. Thus, the theorem applies. Summarizing, given \u03d5 the continued fractions algorithm ef\ufb01ciently produces numbers s and r with no common factor, such that s/r = s/r. The number r is our candidate for the order. We can check to see whether it is the order by calculating xr mod N, and seeing if the result is 1. If so, then r is the order of x modulo N, and we are done! Performance How can the order-\ufb01nding algorithm fail? There are two possibilities. First, the phase estimation procedure might produce a bad estimate to s/r. This occurs with probability at most, and can be made small with a negligible increase in the size of the circuit. More seriously, it might be that s and r have a common factor, in which case the number r returned by the continued fractions algorithm be a factor of r, and not r itself. Fortunately, there are at least three ways around this problem. Perhaps the most straightforward way is to note that for randomly chosen s in the range 0 through r \u2212 1, it\u2019s actually pretty likely that s and r are co-prime, in which case the continued fractions algorithm must return r. To see that this is the case, note that by Problem 4.1 on page 638 the number of prime numbers less than r is at least 230 The quantum Fourier transform and its applications Box 5.3: The continued fractions algorithm The idea of the continued fractions algorithm is to describe real numbers in terms of integers alone, using expressions of the form [a0,..., aM ] \u2261 a0 + 1 a1 + a2+ 1 1...+ 1 aM, (5.49) where a0,..., aM are positive integers. (For applications to quantum computing it is convenient to allow a0 = 0 as well.) We de\ufb01ne the mth convergent (0 \u2264 m \u2264 M ) to this continued fraction to be [a0,..., am]. The continued fractions algorithm is a method for determining the continued fraction expansion of an arbitrary real number. It is easily understood by example. Suppose we are trying to decompose 31/", "13 as a continued fraction. The \ufb01rst step of the continued fractions algorithm is to split 31/13 into its integer and fractional part, Next we invert the fractional part, obtaining 31 13 = 2 +. 5 13 31 13 = 2 +. 1 13 5 These steps \u2013 split then invert \u2013 are now applied to 13/5, giving 31 13 = Next we split and invert 5/3: 31 13 = 2 + 1 2 + 1 1+ 1 3 2 The decomposition into a continued fraction now terminates, since 3 2 = 1 + 1 2 (5.50) (5.51) (5.52) (5.53) (5.54) may be written with a 1 in the numerator without any need to invert, giving a \ufb01nal continued fraction representation of 31/13 as 31 13 = 2 +. 1 2 + 1 1+ 1 1+ 1 2 (5.55) It\u2019s clear that the continued fractions algorithm terminates after a \ufb01nite number of \u2018split and invert\u2019 steps for any rational number, since the numerators which appear (31, 5, 3, 2, 1 in the example) are strictly decreasing. How quickly does this termination occur? It turns out that if \u03d5 = s/r is a rational number, and s and r are L bit integers, then the continued fraction expansion for \u03d5 can be computed using O(L3) operations \u2013 O(L) \u2018split and invert\u2019 steps, each using O(L2) gates for elementary arithmetic. Applications: order-\ufb01nding and factoring 231 r/2 log r, and thus the chance that s is prime (and therefore, co-prime to r) is at least 1/2 log(r) > 1/2 log(N ). Thus, repeating the algorithm 2 log(N ) times we will, with high probability, observe a phase s/r such that s and r are co-prime, and therefore the continued fractions algorithm produces r, as desired. A second method is to note that if r = r, then r is guaranteed to be a factor of r, unless s = 0, which possibility occurs with probability 1/r \u2264 1/2, and which can be discounted further by a few repetitions. Suppose we replace a by a \u2261 ar (mod N ). Then the order of a is r/r. We can now repeat the algorithm, and try to compute the order of a, which, if we succeed, allows us to compute the order of a, since r = r \u00d7 r/r. If we fail, then we obtain r which is a factor of r/r, and we now try to compute the order of a \u2261 (a)r (mod N ). We iterate this procedure until we determine the order of a. At most log(r) = O(L) iterations are required, since each repetition reduces the order of the current candidate a... by a factor of at least two. The third method is better than the \ufb01rst two methods, in that it requires only a constant number of trials, rather than O(L) repetitions. The idea is to repeat the phase, s estimation-continued fractions procedure twice, obtaining r, s 2 1 the second time. Provided s 2 have no common factors, r may be extracted by taking the least common multiple of r1 and r2. The probability that s 2 have no common factors is given by 1 the \ufb01rst time, and r 1 and s 1 and s 2 1 \u2212 p(q|s 1)p(q|s 2), (5.56) q where the sum is over all prime numbers q, and p(x|y) here means the probability of x dividing y. If q divides s 1 then it must also divide the true value of s, s1, on the \ufb01rst iteration, so to upper bound p(q|s 1) it suf\ufb01ces to upper bound p(q|s1), where s1 is chosen uniformly at random from 0 through r \u2212 1. It is easy to see that p(q|s1) \u2264 1/q, and thus p(q|s 2 have no common factors satis\ufb01es 2) \u2264 1/q, and thus the probability that s 1) \u2264 1/q. Similarly, p(q|s 1 and s 1 \u2212 q p(q|s 1)p(q|s 2) \u2265 1 \u2212 q 1 q2. (5.57) The right hand side can be upper bounded in a number of ways; a simple technique is provided in Exercise 5.16, which gives 1 \u2212 q p(q|s 1)p(q|s 2) \u2265 1 4, (5.58)", " and thus the probability of obtaining the correct r is at least 1/4. Exercise 5.15: Show that the least common multiple of positive integers x and y is xy/ gcd(x, y), and thus may be computed in O(L2) operations if x and y are L bit numbers. Exercise 5.16: For all x \u2265 2 prove that 1/y2 dy \u2265 2/3x2. Show that 7 x+1 x - \u221e 1 y2 dy =, 3 4 (5.59) 2 q 1 q2 \u2264 3 2 and thus that (5.58) holds. 232 The quantum Fourier transform and its applications What resource requirements does this algorithm consume? The Hadamard transform requires O(L) gates, and the inverse Fourier transform requires O(L2) gates. The major cost in the quantum circuit proper actually comes from the modular exponentiation, which uses O(L3) gates, for a total of O(L3) gates in the quantum circuit proper. The continued fractions algorithm adds O(L3) more gates, for a total of O(L3) gates to obtain r. Using the third method for obtaining r from r we need only repeat this procedure a constant number of times to obtain the order, r, for a total cost of O(L3). The algorithm is summarized below. Algorithm: Quantum order-\ufb01nding Inputs: (1) A black box Ux,N which performs the transformation |j|k \u2192 |j|xjk mod N, for x co-prime to the L-bit number N, (2) t = 2L + 1 + to the state |1. qubits initialized to |0, and (3) L qubits initialized 2 + 1 2 2 log 3 Outputs: The least integer r > 0 such that xr = 1 (mod N ). Runtime: O(L3) operations. Succeeds with probability O(1). Procedure: 1. 2. 3. 4. 5. 6. |0|1 \u2192 1\u221a 2t \u2192 1\u221a 2t 2t\u22121 j=0 2t\u22121 j=0 \u2248 1\u221a |j|1 initial state create superposition |j|xj mod N apply Ux,N r\u22121 2t\u22121 e2\u03c0isj/r|j|us r2t r\u22121 s=0 s=0 j=0 |8s/r|us \u2192 1\u221a r \u2192 8s/r \u2192 r apply inverse Fourier transform to \ufb01rst register measure \ufb01rst register apply continued fractions algorithm 5.3.2 Application: factoring The problem of distinguishing prime numbers from composites, and of resolving composite numbers into their prime factors, is one of the most important and useful in all of arithmetic. [... ] The dignity of science seems to demand that every aid to the solution of such an elegant and celebrated problem be zealously cultivated. \u2013 Carl Friedrich Gauss, as quoted by Donald Knuth Given a positive composite integer N, what prime numbers when multiplied together equal it? This factoring problem turns out to be equivalent to the order-\ufb01nding problem Applications: order-\ufb01nding and factoring 233 we just studied, in the sense that a fast algorithm for order-\ufb01nding can easily be turned into a fast algorithm for factoring. In this section we explain the method used to reduce factoring to order-\ufb01nding, and give a simple example of this reduction. The reduction of factoring to order-\ufb01nding proceeds in two basic steps. The \ufb01rst step is to show that we can compute a factor of N if we can \ufb01nd a non-trivial solution x = \u00b1 1(mod N ) to the equation x2 = 1(mod N ). The second step is to show that a randomly chosen y co-prime to N is quite likely to have an order r which is even, and such that yr/2 = \u00b1 1(mod N ), and thus x \u2261 yr/2(mod N ) is a non-trivial solution to x2 = 1(mod N ). These two steps are embodied in the following theorems, whose proofs may be found in Section A4.3 of Appendix 4. Theorem 5.2: Suppose N is an L bit composite number, and x is a non-trivial solution to the equation x2 = 1(mod N ) in the range 1 \u2264 x \u2264 N, that is, neither x = 1(mod N ) nor x = N \u2212 1 = \u22121(mod N ). Then at least one of gcd(x \u2212 1, N ) and gcd(x + 1, N ) is a non-trivial factor of N that can", " be computed using O(L3) operations. Theorem 5.3: Suppose N = p\u03b11 1... p\u03b1m m is the prime factorization of an odd composite positive integer. Let x be an integer chosen uniformly at random, subject to the requirements that 1 \u2264 x \u2264 N \u2212 1 and x is co-prime to N. Let r be the order of x modulo N. Then p(r is even and xr/2 = \u2212 1(mod N )) \u2265 1 \u2212 1 2m. (5.60) Theorems 5.2 and 5.3 can be combined to give an algorithm which, with high probability, returns a non-trivial factor of any composite N. All the steps in the algorithm can be performed ef\ufb01ciently on a classical computer except (so far as is known today) an order-\ufb01nding \u2018subroutine\u2019 which is used by the algorithm. By repeating the procedure we may \ufb01nd a complete prime factorization of N. The algorithm is summarized below. Algorithm: Reduction of factoring to order-\ufb01nding Inputs: A composite number N Outputs: A non-trivial factor of N. Runtime: O((log N )3) operations. Succeeds with probability O(1). Procedure: 1. 2. 3. 4. If N is even, return the factor 2. Determine whether N = ab for integers a \u2265 1 and b \u2265 2, and if so return the factor a (uses the classical algorithm of Exercise 5.17). Randomly choose x in the range 1 to N \u22121. If gcd(x, N ) > 1 then return the factor gcd(x, N ). Use the order-\ufb01nding subroutine to \ufb01nd the order r of x modulo N. 234 The quantum Fourier transform and its applications 5. If r is even and xr/2 = \u2212 1(mod N ) then compute gcd(xr/2 \u2212 1, N ) and gcd(xr/2 + 1, N ), and test to see if one of these is a non-trivial factor, returning that factor if so. Otherwise, the algorithm fails. Steps 1 and 2 of the algorithm either return a factor, or else ensure that N is an odd integer with more than one prime factor. These steps may be performed using O(1) and O(L3) operations, respectively. Step 3 either returns a factor, or produces a randomly chosen element x of {0, 1, 2,..., N \u2212 1}. Step 4 calls the order-\ufb01nding subroutine, computing the order r of x modulo N. Step 5 completes the algorithm, since Theorem 5.3 guarantees that with probability at least one-half r will be even and xr/2 = \u2212 1(mod N ), and then Theorem 5.2 guarantees that either gcd(xr/2 \u2212 1, N ) or gcd(xr/2 + 1, N ) is a non-trivial factor of N. An example illustrating the use of this algorithm with the quantum order-\ufb01nding subroutine is shown in Box 5.4. Exercise 5.17: Suppose N is L bits long. The aim of this exercise is to \ufb01nd an ef\ufb01cient classical algorithm to determine whether N = ab for some integers a \u2265 1 and b \u2265 2. This may be done as follows: (1) Show that b, if it exists, satis\ufb01es b \u2264 L. (2) Show that it takes at most O(L2) operations to compute log2 N, x = y/b for b \u2264 L, and the two integers u1 and u2 nearest to 2x. (3) Show that it takes at most O(L2) operations to compute ub repeated squaring) and check to see if either is equal to N. 1 and ub 2 (use (4) Combine the previous results to give an O(L3) operation algorithm to determine whether N = ab for integers a and b. Exercise 5.18: (Factoring 91) Suppose we wish to factor N = 91. Con\ufb01rm that steps 1 and 2 are passed. For step 3, suppose we choose x = 4, which is co-prime to 91. Compute the order r of x with respect to N, and show that xr/2 mod 91 = 64 = \u2212 1(mod 91), so the algorithm succeeds, giving gcd(64 \u2212 1, 19) = 7. It is unlikely that this is the most ef\ufb01cient method you\u2019ve seen for factoring 91. Indeed, if all computations had to be carried out on a classical computer, this", " reduction would not result in an ef\ufb01cient factoring algorithm, as no ef\ufb01cient method is known for solving the order-\ufb01nding problem on a classical computer. Exercise 5.19: Show that N = 15 is the smallest number for which the order-\ufb01nding subroutine is required, that is, it is the smallest composite number that is not even or a power of some smaller integer. 5.4 General applications of the quantum Fourier transform The main applications of the quantum Fourier transform we have described so far in this chapter are phase estimation and order-\ufb01nding. What other problems can be solved with these techniques? In this section, we de\ufb01ne a very general problem known as the hidden subgroup problem, and describe an ef\ufb01cient quantum algorithm for solving it. This problem, which encompasses all known \u2018exponentially fast\u2019 applications of the quantum Fourier transform, can be thought of as a generalization of the task of \ufb01nding the unknown period of a periodic function, in a context where the structure of the domain and range General applications of the quantum Fourier transform 235 Box 5.4: Factoring 15 quantum-mechanically The use of order-\ufb01nding, phase estimation, and continued fraction expansions in the quantum factoring algorithm is illustrated by applying it to factor N = 15. First, we choose a random number which has no common factors with N ; suppose we choose x = 7. Next, we compute the order r of x with respect to N, using the quantum order-\ufb01nding algorithm: begin with the state |0|0 and create the state 1\u221a 2t t\u22121 2 k=0 |k|0 = 1\u221a 2t |0 + |1 + |2 + \u00b7 \u00b7 \u00b7 + |2t \u2212 1 |0 (5.61) by applying t = 11 Hadamard transforms to the \ufb01rst register. Choosing this value of t ensures an error probability of at most 1/4. Next, compute f (k) = xk mod N, leaving the result in the second register, 1\u221a 2t = 2t\u22121 k=0 1\u221a 2t |k|xk mod N (5.62) |0|1 + |1|7 + |2|4 + |3|13 + |4|1 + |5|7 + |6|4 + \u00b7 \u00b7 \u00b7. We now apply the inverse Fourier transform F T \u2020 to the \ufb01rst register and measure it. One way of analyzing the distribution of outcomes obtained is to calculate the reduced density matrix for the \ufb01rst register, and apply F T \u2020 to it, and calculate the measurement statistics. However, since no further operation is applied to the second register, we can instead apply the principle of implicit measurement (Section 4.4) and assume that the second register is measured, obtaining a random result from 1, 7, 4, or 13. Suppose we get 4 (any of the results works); this means the state input. After applying F T \u2020 to F T \u2020 would have been |2 + |6 + |10 + |14 + \u00b7 \u00b7 \u00b7 we obtain some state \u03b1|, with the probability distribution 4 2t 2 | \u03b1 | 0.25 0.2 0.15 0.1 0.05 0 200 400 600 800 1000 1200 1400 1600 1800 2000 shown for 2t = 2048. The \ufb01nal measurement therefore gives either 0, 512, 1024, or 1536, each with probability almost exactly 1/4. Suppose we obtain = 1536 from the measurement; computing the continued fraction expansion thus gives 1536/2048 = 1/(1 + (1/3)), so that 3/4 occurs as a convergent in the expansion, giving r = 4 as the order of x = 7. By chance, r is even, and moreover, xr/2 mod N = 72 mod 15 = 4 = \u2212 1 mod 15, so the algorithm works: computing the greatest common divisor gcd(x2 \u2212 1, 15) = 3 and gcd(x2 + 1, 15) = 5 tells us that 15 = 3\u00d75. 236 The quantum Fourier transform and its applications of the function may be very intricate. In order to present this problem in the most approachable manner, we begin with two more speci\ufb01c applications: period-\ufb01nding (of a one-dimensional function), and discrete logarithms. We then return to the general hidden subgroup problem. Note that the presentation in this section is rather more schematic and conceptual than earlier sections in this chapter; of necessity, this means that the reader interested in understanding all the details will have to work much harder! 5.4.", "1 Period-\ufb01nding Consider the following problem. Suppose f is a periodic function producing a single bit as output and such that f (x + r) = f (x), for some unknown 0 < r < 2L, where x, r \u2208 {0, 1, 2,...}. Given a quantum black box U which performs the unitary transform U |x|y \u2192 |x|y \u2295 f (x) (where \u2295 denotes addition modulo 2) how many black box queries and other operations are required to determine r? Note that in practice U operates on a \ufb01nite domain, whose size is determined by the desired accuracy for r. Here is a quantum algorithm which solves this problem using one query, and O(L2) other operations: Algorithm: Period-\ufb01nding Inputs: (1) A black box which performs the operation U |x|y = |x|y \u2295 f (x), (2) a state to store the function evaluation, initialized to |0, and (3) t = O(L + log(1/)) qubits initialized to |0. Outputs: The least integer r > 0 such that f (x + r) = f (x). Runtime: One use of U, and O(L2) operations. Succeeds with probability O(1). Procedure: 1. 2. 3. 4. 5. 6. 2t\u22121 |0|0 \u2192 1\u221a 2t \u2192 1\u221a 2t x=0 \u2248 1\u221a x=0 2t\u22121 initial state create superposition apply U |x|0 |x|f (x) r\u22121 t\u22121 2 e2\u03c0ix/r|x| \u02c6f () r2t r\u22121 =0 =0 x=0 |8/r| \u02c6f () \u2192 1\u221a r \u2192 8/r \u2192 r apply inverse Fourier transform to \ufb01rst register measure \ufb01rst register apply continued fractions algorithm The key to understanding this algorithm, which is based on phase estimation, and is nearly identical to the algorithm for quantum order-\ufb01nding, is step 3, in which we introduce the state | \u02c6f () \u2261 1\u221a r r\u22121 x=0 e\u22122\u03c0ix/r|f (x), (5.63) General applications of the quantum Fourier transform 237 the Fourier transform of |f (x). The identity used in step 3 is based on |f (x) = e2\u03c0ix/r| \u02c6f (), (5.64) e2\u03c0ix/r = r for x an integer multiple of r, which is easy to verify by noting that and zero otherwise. The approximate equality in step 3 is required because 2t may not be an integer multiple of r in general (it need not be: this is taken account of by the phase estimation bounds). By Equation (5.22), applying the inverse Fourier transform to the \ufb01rst register, in step 4, gives an estimate of the phase /r, where is chosen randomly. r can be ef\ufb01ciently obtained in the \ufb01nal step using a continued fraction expansion. r\u22121 1\u221a r =0 r\u22121 =0 Box 5.5: The shift-invariance property of the Fourier transform The Fourier transform, Equation (5.1), has an interesting and very useful property, known as shift invariance. Using notation which is useful in describing the general application of this property, let us describe the quantum Fourier transform as \u03b1h|h \u2192 \u02dc\u03b1g|g, (5.65) h\u2208H g\u2208G where \u02dc\u03b1g = h\u2208H \u03b1h exp(2\u03c0igh/|G|), H is some subset of G, and G indexes the states in an orthonormal basis of the Hilbert space. For example, G may be the set of numbers from 0 to 2n \u2212 1 for an n qubit system. |G| denotes the number of elements in G. Suppose we apply to the initial state an operator Uk which performs the unitary transform Uk|g = |g + k, (5.66) then apply the Fourier transform. The result, Uk \u03b1h|h = \u03b1h|h + k \u2192 h\u2208H h\u2208H g\u2208G e2\u03c0igk/|G| \u02dc\u03b1g|g (5.67) has the property that the magnitude of the amplitude for |g does not change, no matter what k is, that is: | exp(2\u03c0igk/|G|) \u02dc\u03b1g| = | \u02dc\u03b1g|. In the language of group theory", ", G is a group, H a subgroup of G, and we say that if a function f on G is constant on cosets of H, then the Fourier transform of f is invariant over cosets of H. Why does this work? One way to understand this is to realize that (5.63) is approximately the Fourier transform over {0, 1,..., 2L \u2212 1} of |f (x) (see Exercise 5.20), and the Fourier transform has an interesting and very useful property, known as shift invariance, described in Box 5.5. Another is to realize that what the order-\ufb01nding algorithm does is just to \ufb01nd the period of the function f (k) = xk mod N, so the ability to \ufb01nd the period of a general periodic function is not unexpected. Yet another way is to realize that the implementation of the black box U is naturally done using a certain unitary operator whose eigenvectors are precisely | \u02c6f (), as described in Exercise 5.21 below, so that the phase estimation procedure of Section 5.2 can be applied. 238 The quantum Fourier transform and its applications Exercise 5.20: Suppose f (x + r) = f (x), and 0 \u2264 x < N, for N an integer multiple of r. Compute \u02c6f () \u2261 1\u221a N N \u22121 x=0 e\u22122\u03c0ix/N f (x), (5.68) and relate the result to (5.63). You will need to use the fact that * e2\u03c0ik/N = 0 N/r if is an integer multiple of N/r otherwise. k\u2208{0,r,2r,...,N \u2212r} (5.69) Exercise 5.21: (Period-\ufb01nding and phase estimation) Suppose you are given a unitary operator Uy which performs the transformation Uy|f (x) = |f (x + y), for the periodic function described above. (1) Show that the eigenvectors of Uy are | \u02c6f (), and calculate their eigenvalues. (2) Show that given |f (x0) for some x0, Uy can be used to realize a black box which is as useful as U in solving the period-\ufb01nding problem. 5.4.2 Discrete logarithms The period \ufb01nding problem we just considered is a simple one, in that the domain and range of the periodic function were integers. What happens when the function is more complex? Consider the function f (x1, x2) = asx1+x2 mod N, where all the variables are integers, and r is the smallest positive integer for which ar mod N = 1. This function is periodic, since f (x1 +, x2 \u2212 s) = f (x1, x2), but now the period is a 2-tuple, (, \u2212s), for integer. This may seem to be a strange function, but it is very useful in cryptography, since determining s allows one to solve what is known as the discrete logarithm problem: given a and b = as, what is s? Here is a quantum algorithm which solves this problem using one query of a quantum black box U which performs the unitary transform U |x1|x2|y \u2192 |x1|x2|y \u2295 f (x) (where \u2295 denotes bitwise addition modulo 2), and O(log r2) other operations. We assume knowledge of the minimum r > 0 such that ar mod N = 1, which can be obtained using the order-\ufb01nding algorithm described previously. Algorithm: Discrete logarithm Inputs: (1) A black box which performs the operation U |x1|x2|y = |x1|x2|y \u2295 f (x1, x2), for f (x1, x2) = bx1ax2, (2) a state to store the function evaluation, initialized to |0, and (3) two t = O(log r + log(1/)) qubit registers initialized to |0. Outputs: The least positive integer s such that as = b. Runtime: One use of U, and O(log r2) operations. Succeeds with probability O(1). Procedure: 1. |0|0|0 initial state General applications of the quantum Fourier transform 239 2t\u22121 2t\u22121 x1=0 2t\u22121 x2=0 2t\u22121 |x1|x2|0 create superposition |x1|x2|f (x1, x2) apply U \u2192 1 2t \u2192", " 1 2t x2=0 r\u22121 2t\u22121 2t\u22121 2=0 r\u22121 x1=0 \u23a1 x2=0 2t\u22121 \u23a3 x1=0 \u2248 1 \u221a 2t r = 1 \u221a r 2t \u2192 1\u221a r \u2019 \u2192 \u2192 s 2=0 8s2/r, 82/r ( e2\u03c0i(s2x1+2x2)/r|x1|x2| \u02c6f (s2, 2) \u23a4 \u23a1 \u23a4 2t\u22121 e2\u03c0i(s2x1)/r|x1 \u23a6 \u23a3 e2\u03c0i(2x2)/r|x2 \u23a6 | \u02c6f (s2, 2) 2=0 x1=0 x2=0 r\u22121 | 8s2/r| 82/r| \u02c6f (s2, 2) apply inverse Fourier transform to \ufb01rst two registers measure \ufb01rst two registers apply generalized continued fractions algorithm 2. 3. 4. 5. 6. Again, the key to understanding this algorithm is step 3, in which we introduce the state | \u02c6f (1, 2) = 1\u221a r r\u22121 j=0 e\u22122\u03c0i2j/r|f (0, j), (5.70) the Fourier transform of |f (x1, x2) (see Exercise 5.22). In this equation, the values of 1 and 2 must satisfy r\u22121 k=0 e2\u03c0ik(1/s\u22122)/r = r. (5.71) Otherwise, the amplitude of | \u02c6f (1, 2) is nearly zero. The generalized continued fraction expansion used in the \ufb01nal step to determine s is analogous to the procedures used in Section 5.3.1, and is left as a simple exercise for you to construct. Exercise 5.22: Show that | \u02c6f (1, 2) = r\u22121 r\u22121 x1=0 x2=0 e\u22122\u03c0i(1x1+2x2)/r|f (x1, x2) = 1\u221a r r\u22121 j=0 e\u22122\u03c0i2j/r|f (0, j), (5.72) and we are constrained to have 1/s \u2212 2 be an integer multiple of r for this expression to be non-zero. Exercise 5.23: Compute 1 r r\u22121 r\u22121 1=0 2=0 e\u22122\u03c0i(1x1+2x2)/r| \u02c6f (1, 2) (5.73) using (5.70), and show that the result is f (x1, x2). Exercise 5.24: Construct the generalized continued fractions algorithm needed in 240 The quantum Fourier transform and its applications step 6 of the discrete logarithm algorithm to determine s from estimates of s2/r and 2/r. Exercise 5.25: Construct a quantum circuit for the black box U used in the quantum discrete logarithm algorithm, which takes a and b as parameters, and performs the unitary transform |x1|x2|y \u2192 |x1|x2|y \u2295 bx1ax2. How many elementary operations are required? 5.4.3 The hidden subgroup problem By now, a pattern should be coming clear: if we are given a periodic function, even when the structure of the periodicity is quite complicated, we can often use a quantum algorithm to determine the period ef\ufb01ciently. Importantly, however, not all periods of periodic functions can be determined. The general problem which de\ufb01nes a broad framework for these questions can be succinctly expressed in the language of group theory (see Appendix 2 for a quick review) as follows: Let f be a function from a \ufb01nitely generated group G to a \ufb01nite set X such that f is constant on the cosets of a subgroup K, and distinct on each coset. Given a quantum black box for performing the unitary transform U |g|h = |g|h\u2295f (g), for g \u2208 G, h \u2208 X, and \u2295 an appropriately chosen binary operation on X, \ufb01nd a generating set for K. Order-\ufb01nding, period-\ufb01nding, discrete logarithms, and many other problems are instances of this hidden subgroup problem; some interesting ones are listed in Figure 5.5. For G a \ufb01nite Abelian group, a quantum computer can solve the hidden subgroup problem using a", " number of operations polynomial in log |G|, and one use of the black box function evaluation, using an algorithm very similar to the others in this section. (In fact, solution for a \ufb01nitely generated Abelian group is also possible, along similar lines, but we\u2019ll stick to the \ufb01nite case here.) We shall leave detailed speci\ufb01cation of the algorithm to you as an exercise, which should be simple after we explain the basic idea. Many things remain essentially the same, because \ufb01nite Abelian groups are isomorphic to products of additive groups over the integers in modular arithmetic. This means that the quantum Fourier transform of f over G is well de\ufb01ned (see Section A2.3), and can still be done ef\ufb01ciently. The \ufb01rst non-trivial step of the algorithm is to use a Fourier transform (generalizing the Hadamard operation) to create a superposition over group elements, which is then transformed by applying the quantum black box for f in the next step, to give 1 |G| g\u2208G |g|f (g). (5.74) As before, we would now like to rewrite |f (g) in the Fourier basis. We start with |f (g) = 1 |G| |G|\u22121 =0 e2\u03c0ig/|G|| \u02c6f (), (5.75) where we have chosen exp[\u22122\u03c0ig/|G|] as a representation (see Exercise A2.13) of g \u2208 G indexed by (the Fourier transform maps between group elements and representations: see Exercise A2.23). The key is to recognize that this expression can be simpli\ufb01ed because General applications of the quantum Fourier transform 241 Name G X K Function Deutsch {0, 1}, \u2295 {0, 1} {0} or {0, 1} K = {0, 1} : * K = {0} : * f (x) = 0 f (x) = 1 f (x) = x f (x) = 1 \u2212 x Simon {0, 1}n, \u2295 Period\ufb01nding Order\ufb01nding Z, + Z, + Discrete logarithm Zr \u00d7 Zr + (mod r) any \ufb01nite set any \ufb01nite set {aj} j \u2208 Zr ar = 1 {aj} j \u2208 Zr ar = 1 {0, s} s \u2208 {0, 1}n f (x \u2295 s) = f (x) {0, r, 2r,...} r \u2208 G f (x + r) = f (x) {0, r, 2r,...} r \u2208 G f (x) = ax f (x + r) = f (x) (, \u2212s), s \u2208 Zr f (x1, x2) = akx1+x2 f (x1 +, x2 \u2212 s) = f (x1, x2) Order of a permutation Z2m \u00d7 Z2n + (mod 2m) Z2n {0, r, 2r,...} r \u2208 X Z \u00d7 Z, + ZN (, \u2212s), s \u2208 X f (x, y) = \u03c0x(y) f (x + r, y) = f (x, y) \u03c0 = permutation on X f (x1, x2) = \u03c0(sx1 + x2 mod N ) \u03c0 = permutation on X (H, X) H = any Abelian group any \ufb01nite set {s \u2208 H | f (s, x) = x, \u2200x \u2208 X} f (gh, x) = f (g, f (h, x)) f (gs, x) = f (g, x) Hidden linear function Abelian stabilizer Figure 5.5. Hidden subgroup problems. The function f maps from the group G to the \ufb01nite set X, and is promised to be constant on cosets of the hidden subgroup K \u2286 G. ZN represents the set {0, 1,..., N \u2212 1} in this table, and Z is the integers. The problem is to \ufb01nd K (or a generating set for it), given a black box for f. f is constant and distinct on cosets of the subgroup K, so that | \u02c6f () = 1 |G| g\u2208G e\u22122\u03c0ig/|G||f (g) has nearly zero amplitude for all values of", " except those which satisfy e\u22122\u03c0ih/|G| = |K|. h\u2208K (5.76) (5.77) 242 The quantum Fourier transform and its applications If we can determine, then using the linear constraints given by this expression allows us to determine elements of K, and since K is Abelian, this allows us to eventually determine a generating set for the whole hidden subgroup, solving the problem. However, life is not so simple. An important reason why the period-\ufb01nding and discrete logarithm algorithms work is because of the success of the continued fraction expansion in obtaining from /|G|. In those problems, and |G| are arranged to not have any common factors, with high probability. In the general case, however, this may not be true, since |G| is free to be a composite number with many factors, and we have no useful prior information about. Fortunately, this problem can be solved: as mentioned above, any \ufb01nite Abelian group \u00d7 G is isomorphic to a product of cyclic groups of prime power order, that is, G = Zp1 \u00d7\u00b7 \u00b7 \u00b7\u00d7ZpM, where pi are primes, and Zpi is the group over integers {0, 1,..., pi \u22121} Zp2 with addition modulo pi being the group operation. We can thus re-express the phase which appears in (5.75) as e2\u03c0ig/|G| = e2\u03c0i igi/pi M/ i=1 (5.78) for gi \u2208 Zpi. The phase estimation procedure now gives us, and thus, sample K as described above, to solve the hidden subgroup problem. i, from which we determine Exercise 5.26: Since K is a subgroup of G, when we decompose G into a product of cyclic groups of prime power order, this also decomposes K. Re-express (5.77) to show that determining i allows one to sample from the corresponding cyclic subgroup Kpi of K. Exercise 5.27: Of course, the decomposition of a general \ufb01nite Abelian group G into a product of cyclic groups of prime power order is usually a dif\ufb01cult problem (at least as hard as factoring integers, for example). Here, quantum algorithms come to the rescue again: explain how the algorithms in this chapter can be used to ef\ufb01ciently decompose G as desired. Exercise 5.28: Write out a detailed speci\ufb01cation of the quantum algorithm to solve the hidden subgroup problem, complete with runtime and success probability estimates, for \ufb01nite Abelian groups. Exercise 5.29: Give quantum algorithms to solve the Deutsch and Simon problems listed in Figure 5.5, using the framework of the hidden subgroup problem. 5.4.4 Other quantum algorithms? One of the most intriguing aspects of this framework for describing quantum algorithms in terms of the hidden subgroup problem is the suggestion that more dif\ufb01cult problems might be solvable by considering various groups G and functions f. We have only described the solution of this problem for Abelian groups. What about non-Abelian groups? They are quite interesting (see Appendix 2 for a discussion of general Fourier transforms over non-Abelian groups): for example, the problem of graph isomorphism is to determine if two given graphs are the same under some permutation of the labels of the n vertices (Section 3.2.3). These permutations can be described as transformations under the symmetric group Sn, and algorithms for performing fast Fourier transforms Chapter problems 243 over these groups exists. However, a quantum algorithm for ef\ufb01ciently solving the graph isomporphism problem remains unknown. Even if more general cases of the hidden subgroup problem remain unsolvable by quantum computers, having this unifying framework is useful, because it allows us to ask questions about how one might be able to step outside its limitations. It is dif\ufb01cult to believe that all fast quantum algorithms that will ever be discovered will be just ways to solve the hidden subgroup problem. If one thinks of these problems as being based on the coset invariance property of the Fourier transform, in searching for new algorithms, perhaps the thing to do then is to investigate other transforms with different invariances. Going in another direction, one might ask: what dif\ufb01cult hidden subgroup problems might be ef\ufb01ciently solved given an arbitrary (but speci\ufb01ed independently of the problem) quantum state as a helper? After all, as discussed in Chapter 4, most quantum states are actually exponentially hard to construct. Such a state might be a useful resource (a real \u2018quantum oracle\u2019), if quantum algorithms", " existed to utilize them to solve hard problems! The hidden subgroup problem also captures an important constraint underlying the class of quantum algorithms which are exponentially faster than their (known) classical counterparts: this is a promise problem, meaning that it is of the form \u2018F (X) is promised to have such and such property: characterize that property.\u2019 Rather disappointingly, perhaps, we shall show at the end of the next chapter that, in solving problems without some sort of promise, quantum computers cannot achieve an exponential speedup over classical computers; the best speedup is polynomial. On the other hand, this gives us an important clue as to what kinds of problems quantum computers might be good at: in retrospect, the hidden subgroup problem might be thought of as a natural candidate for quantum computation. What other natural problems are there? Think about it! Problem 5.1: Construct a quantum circuit to perform the quantum Fourier transform |j \u2212\u2192 1 \u221a p p\u22121 k=0 e2\u03c0ijk/p|k (5.79) where p is prime. Problem 5.2: (Measured quantum Fourier transform) Suppose the quantum Fourier transform is performed as the last step of a quantum computation, followed by a measurement in the computational basis. Show that the combination of quantum Fourier transform and measurement is equivalent to a circuit consisting entirely of one qubit gates and measurement, with classical control, and no two qubit gates. You may \ufb01nd the discussion of Section 4.4 useful. Problem 5.3: (Kitaev\u2019s algorithm) Consider the quantum circuit where |u is an eigenstate of U with eigenvalue e2\u03c0i\u03d5. Show that the top qubit is 244 The quantum Fourier transform and its applications measured to be 0 with probability p \u2261 cos2(\u03c0\u03d5). Since the state |u is unaffected by the circuit it may be reused; if U can be replaced by U k, where k is an arbitrary integer under your control, show that by repeating this circuit and increasing k appropriately, you can ef\ufb01ciently obtain as many bits of p as desired, and thus, of \u03d5. This is an alternative to the phase estimation algorithm. Problem 5.4: The runtime bound O(L3) we have given for the factoring algorithm is not tight. Show that a better upper bound of O(L2 log L log log L) operations can be achieved. Problem 5.5: (Non-Abelian hidden subgroups \u2013 Research) Let f be a function on a \ufb01nite group G to an arbitrary \ufb01nite range X, which is promised to be constant and distinct on distinct left cosets of a subgroup K. Start with the state 1 |G|m g1,...,gm |g1,..., gm|f (g1),..., f (gm), (5.80) and prove that picking m = 4 log |G| + 2 allows K to be identi\ufb01ed with probability at least 1 \u2212 1/|G|. Note that G does not necessarily have to be Abelian, and being able to perform a Fourier transform over G is not required. This result shows that one can produce (using only O(log |G|) oracle calls) a \ufb01nal result in which the pure state outcomes corresponding to different possible hidden subgroups are nearly orthogonal. However, it is unknown whether a POVM exists or not which allows the hidden subgroup to be identi\ufb01ed ef\ufb01ciently (i.e. using poly(log |G|) operations) from this \ufb01nal state. Problem 5.6: (Addition by Fourier transforms) Consider the task of constructing a quantum circuit to compute |x \u2192 |x + y mod 2n, where y is a \ufb01xed constant, and 0 \u2264 x < 2n. Show that one ef\ufb01cient way to do this, for values of y such as 1, is to \ufb01rst perform a quantum Fourier transform, then to apply single qubit phase shifts, then an inverse Fourier transform. What values of y can be added easily this way, and how many operations are required? History and further reading 245 Summary of Chapter 5: The quantum Fourier transform and its applications \u2022 When N = 2n the quantum Fourier transform |j = |j1,..., jn \u2212\u2192 1\u221a N e2\u03c0i jk N |k N \u22121 k=0 (5.81) may be written in the form |j \u2192 1 |0 + e2\u03c0i0.jn |1 2n/2 |0 + e2\u03c0i0.jn\u22121jn |1 |0 + e2\u03c0i", "0.j1j2...jn|1,... (5.82) and may be implemented using \u0398(n2) gates. \u2022 Phase estimation: Let |u be an eigenstate of the operator U with eigenvalue e2\u03c0i\u03d5. Starting from the initial state |0\u2297t|u, and given the ability to ef\ufb01ciently perform U 2k for integer k, this algorithm (shown in Figure 5.3) can be used to ef\ufb01ciently obtain the state | \u02dc\u03d5|u, where \u02dc\u03d5 accurately approximates \u03d5 to t \u2212 3 2 2 + 1 log 2 bits with probability at least 1 \u2212. \u2022 Order-\ufb01nding: The order of x modulo N is the least positive integer r such that xr mod N = 1. This number can be computed in O(L3) operations using the quantum phase estimation algorithm, for L-bit integers x and N. \u2022 Factoring: The prime factors of an L-bit integer N can be determined in O(L3) operations by reducing this problem to \ufb01nding the order of a random number x co-prime with N. \u2022 Hidden subgroup problem: All the known fast quantum algorithms can be described as solving the following problem: Let f be a function from a \ufb01nitely generated group G to a \ufb01nite set X such that f is constant on the cosets of a subgroup K, and distinct on each coset. Given a quantum black box for performing the unitary transform U |g|h = |g|h \u2295 f (g), for g \u2208 G and h \u2208 X, \ufb01nd a generating set for K. History and further reading The de\ufb01nition of the Fourier transform may be generalized beyond what we have considered in this chapter. In the general scenario a Fourier transform is de\ufb01ned on a set of complex numbers \u03b1g, where the index g is chosen from some group, G. In this chapter we have chosen G to be the additive group of integers modulo 2n, often denoted Z2n. Deutsch[Deu85] showed that the Fourier transform over the group Zn 2 could be implemented ef\ufb01ciently on a quantum computer \u2013 this is the Hadamard transform of earlier chapters. Shor [Sho94] realized to spectacular effect that quantum computers could ef\ufb01ciently implement the quantum Fourier transform over groups Zm for certain special values of m. Inspired by this result Coppersmith[Cop94], Deutsch (unpublished), and Cleve (unpublished) gave the simple quantum circuits for computing the quantum Fourier transform over Z2n which we have used in this chapter. Cleve, Ekert, Mac- 246 The quantum Fourier transform and its applications chiavello and Mosca[CEMM98] and Grif\ufb01ths and Niu[GN96] independently discovered the product formula (5.4); in fact, this result had been realized much earlier by Danielson and Lanczos. The simpli\ufb01ed proof starting in Equation (5.5) was suggested by Zhou. Grif\ufb01ths and Niu[GN96] are responsible for the measured quantum Fourier transform found in Problem 5.2. The Fourier transform over Z2n was generalized to obtain a Fourier transform over an arbitrary \ufb01nite Abelian group by Kitaev[Kit95], who also introduced the phase estimation procedure in the form given in Problem 5.3. Cleve, Ekert, Macchiavello and Mosca[CEMM98] also integrated several of the techniques of Shor and Kitaev into one nice picture, upon which Section 5.2 is based. A good description of the phase estimation algorithm can be found in Mosca\u2019s Ph.D. thesis[Mos99]. Shor announced the quantum order-\ufb01nding algorithm in a seminal paper in 1994[Sho94], and noted that the problems of performing discrete logarithms and factoring could be reduced to order-\ufb01nding. The \ufb01nal paper, including extended discussion and references, was published in 1997[Sho97]. This paper also contains a discussion of clever multiplication methods that may be used to speed up the algorithm even further than in our description, which uses relatively naive multiplication techniques. With these faster multiplication methods the resources required to factor a composite integer n scale as O(n2 log n log log n), as claimed in the introduction to the chapter. In 1995 Kitaev[Kit95] announced an algorithm for \ufb01nding the stabilizer of a general Abelian group, which he showed could be used to solve", " discrete logarithm and factoring as special cases. In addition, this algorithm contained several elements not present in Shor\u2019s algorithm. A good review of the factoring algorithm was written by Ekert and Jozsa [EJ96]; also see DiVincenzo [DiV95a]. The discussion of continued fractions is based upon Chapter 10 of Hardy and Wright[HW60]. At the time of writing, the most ef\ufb01cient classical algorithm for factoring on a classical computer is the number \ufb01eld sieve. This is described in a collection edited by A. K. Lenstra and H. W. Lenstra, Jr.[LL93]. The generalization of quantum algorithms to solving the hidden subgroup problem has been considered by many authors. Historically, Simon was \ufb01rst to note that a quantum computer could \ufb01nd a hidden period of a function satisfying f (x\u2295s) = f (x)[Sim94, Sim97]. In fact, Shor found his result by generalizing Simon\u2019s result, and by applying a Fourier transform over ZN instead of Simon\u2019s Hadamard transforms (a Fourier transform over Zk 2 ). Boneh and Lipton then noted the connection to the hidden subgroup problem, and described a quantum algorithm for solving the hidden linear function problem[BL95]. Jozsa was the \ufb01rst to explicitly provide a uniform description of the Deutsch\u2013Jozsa, Simon, and Shor algorithms in terms of the hidden subgroup problem[Joz97]. Ekert and Jozsa\u2019s work in studying the role of the Abelian and non-Abelian Fast Fourier Transform algorithms in speedup of quantum algorithms[EJ98] has also been insightful. Our description of the hidden subgroup problem in Section 5.4 follows the framework of Mosca and Ekert[ME99, Mos99]. Cleve has proven that the problem of \ufb01nding an order of a permutation requires an exponential number of queries for a bounded-error probabilistic classical computer[Cle99]. Generalizations of this method to beyond Abelian groups have been attempted by Ettinger and H\u00f8yer[EH99], by Roetteler and Beth[RB98] and Pueschel, Roetteler, and Beth[PRB98], by Beals, who also described constructions of quantum Fourier transforms over the symmetric group[BBC+98], and by Ettinger, H\u00f8yer, and Knill[EHK99]. These results have shown, so far, that there exists a quantum algorithm to solve the hidden subgroup problem for non-Abelian groups using only O(log |G|) oracle calls, but whether this can be realized in polynomial time is unknown (Problem 5.5). History and further reading 247 6 Quantum search algorithms Suppose you are given a map containing many cities, and wish to determine the shortest route passing through all cities on the map. A simple algorithm to \ufb01nd this route is to search all possible routes through the cities, keeping a running record of which route has the shortest length. On a classical computer, if there are N possible routes, it obviously takes O(N ) operations to determine the shortest route using this method. Remarkably, there is a quantum search algorithm, sometimes known as Grover\u2019s algorithm, which N ) operations. enables this search method to be sped up substantially, requiring only O( Moreover, the quantum search algorithm is general in the sense that it can be applied far beyond the route-\ufb01nding example just described to speed up many (though not all) classical algorithms that use search heuristics. \u221a In this chapter we explain the fast quantum search algorithm. The basic algorithm is described in Section 6.1. In Section 6.2 we derive the algorithm from another point of view, based on the quantum simulation algorithm of Section 4.7. Three important applications of this algorithm are also described: quantum counting in Section 6.3, speedup of solution of NP-complete problems in Section 6.4, and search of unstructured databases in Section 6.5. One might hope to improve upon the search algorithm to do even better than a square root speedup but, as we show in Section 6.6, it turns out this is not possible. We conclude in Section 6.7 by showing that this speed limit applies to most unstructured problems. 6.1 The quantum search algorithm Let us begin by setting the stage for the search algorithm in terms of an oracle, similar to that encountered in Section 3.1.1. This allows us to present a very general description of the search procedure, and a geometric way to visualize its action and see how it performs. 6.1.1 The oracle Suppose we wish to search through a search space of N elements", ". Rather than search the elements directly, we concentrate on the index to those elements, which is just a number in the range 0 to N \u2212 1. For convenience we assume N = 2n, so the index can be stored in n bits, and that the search problem has exactly M solutions, with 1 \u2264 M \u2264 N. A particular instance of the search problem can conveniently be represented by a function f, which takes as input an integer x, in the range 0 to N \u2212 1. By de\ufb01nition, f (x) = 1 if x is a solution to the search problem, and f (x) = 0 if x is not a solution to the search problem. Suppose we are supplied with a quantum oracle \u2013 a black box whose internal workings we discuss later, but which are not important at this stage \u2013 with the ability to recognize solutions to the search problem. This recognition is signalled by making use of an oracle The quantum search algorithm 249 qubit. More precisely, the oracle is a unitary operator, O, de\ufb01ned by its action on the computational basis: |x|q O\u2192 |x|q \u2295 f (x), where |x is the index register, \u2295 denotes addition modulo 2, and the oracle qubit |q is a single qubit which is \ufb02ipped if f (x) = 1, and is unchanged otherwise. We can check whether x is a solution to our search problem by preparing |x|0, applying the oracle, and checking to see if the oracle qubit has been \ufb02ipped to |1. (6.1) \u221a In the quantum search algorithm it is useful to apply the oracle with the oracle qubit initially in the state (|0 \u2212 |1)/ 2, just as was done in the Deutsch\u2013Jozsa algorithm of Section 1.4.4. If x is not a solution to the search problem, applying the oracle to the state \u221a 2 does not change the state. On the other hand, if x is a solution to the |x(|0 \u2212 |1)/ search problem, then |0 and |1 are interchanged by the action of the oracle, giving a \ufb01nal state \u2212|x(|0 \u2212 |1)/ \u221a 2. The action of the oracle is thus: |0 \u2212 |1 \u221a 2 O\u2212\u2192 (\u22121)f (x)|x |0 \u2212 |1 \u221a 2 |x. (6.2) \u221a Notice that the state of the oracle qubit is not changed. It turns out that this remains (|0 \u2212 |1)/ 2 throughout the quantum search algorithm, and can therefore be omitted from further discussion of the algorithm, simplifying our description. With this convention, the action of the oracle may be written: |x O\u2212\u2192 (\u22121)f (x)|x. (6.3) We say that the oracle marks the solutions to the search problem, by shifting the phase of the solution. For an N item search problem with M solutions, it turns out that we need only apply the search oracle O( N/M ) times in order to obtain a solution, on a quantum computer. This discussion of the oracle without describing how it works in practice is rather abstract, and perhaps even puzzling. It seems as though the oracle already knows the answer to the search problem; what possible use could it be to have a quantum search algorithm based upon such oracle consultations?! The answer is that there is a distinction between knowing the solution to a search problem, and being able to recognize the solution; the crucial point is that it is possible to do the latter without necessarily being able to do the former. A simple example to illustrate this is the problem of factoring. Suppose we have been given a large number, m, and told that it is a product of two primes, p and q \u2013 the same sort of situation as arises in trying to break the RSA public key cryptosystem (Appendix 5). To determine p and q, the obvious method on a classical computer is to search all numbers from 2 through m1/2 for the smaller of the two prime factors. That is, we successively do a trial division of m by each number in the range 2 to m1/2, until we \ufb01nd the smaller prime factor. The other prime factor can then be found by dividing m by the smaller prime. Obviously, this search-based method requires roughly m1/2 trial divisions to \ufb01nd a factor on a classical computer. The quantum search algorithm can be used to speed up this process. By de\ufb01nition, the action of the oracle upon input of the state |x is to divide m by x, and check to see if the division is exact, \ufb02ipping the oracle qubit if", " this is so. Applying the quantum search algorithm with this oracle yields the smaller of the two prime factors with high probability. 250 Quantum search algorithms But to make the algorithm work, we need to construct an ef\ufb01cient circuit implementing the oracle. How to do this is an exercise in the techniques of reversible computation. We begin by de\ufb01ning the function f (x) \u2261 1 if x divides m, and f (x) = 0 otherwise; f (x) tells us whether the trial division is successful or not. Using the techniques of reversible computation discussed in Section 3.2.5, construct a classical reversible circuit which takes (x, q) \u2013 representing an input register initially set to x and a one bit output register initially set to q \u2013 to (x, q \u2295 f (x)), by modifying the usual (irreversible) classical circuit for doing trial division. The resource cost of this reversible circuit is the same to within a factor two as the irreversible classical circuit used for trial division, and therefore we regard the two circuits as consuming essentially the same resources. Furthermore, the classical reversible circuit can be immediately translated into a quantum circuit that takes |x|q to |x|q \u2295 f (x), as required of the oracle. The key point is that even without knowing the prime factors of m, we can explicitly construct an oracle which recognizes a solution to the search problem when it sees one. Using this oracle and the quantum search algorithm we can search the range 2 to m1/2 using O(m1/4) oracle consultations. That is, we need only perform the trial division roughly m1/4 times, instead of m1/2 times, as with the classical algorithm! The factoring example is conceptually interesting but not practical: there are classical algorithms for factoring which work much faster than searching through all possible divisors. However, it illustrates the general way in which the quantum search algorithm may be applied: classical algorithms which rely on search-based techniques may be sped up using the quantum search algorithm. Later in this chapter we examine scenarios where the quantum search algorithm offers a genuinely useful aid in speeding up the solution of NP-complete problems. 6.1.2 The procedure Schematically, the search algorithm operates as shown in Figure 6.1. The algorithm proper makes use of a single n qubit register. The internal workings of the oracle, including the possibility of it needing extra work qubits, are not important to the description of the quantum search algorithm proper. The goal of the algorithm is to \ufb01nd a solution to the search problem, using the smallest possible number of applications of the oracle. The algorithm begins with the computer in the state |0\u2297n. The Hadamard transform is used to put the computer in the equal superposition state, |\u03c8 = 1 N 1/2 N \u22121 x=0 |x. (6.4) The quantum search algorithm then consists of repeated application of a quantum subroutine, know as the Grover iteration or Grover operator, which we denote G. The Grover iteration, whose quantum circuit is illustrated in Figure 6.2, may be broken up into four steps: (1) Apply the oracle O. (2) Apply the Hadamard transform H \u2297n. (3) Perform a conditional phase shift on the computer, with every computational basis state except |0 receiving a phase shift of \u22121, |x \u2192 \u2212(\u22121)\u03b4x0|x. (6.5) The quantum search algorithm 251 (4) Apply the Hadamard transform H \u2297n. Exercise 6.1: Show that the unitary operator corresponding to the phase shift in the Grover iteration is 2|00| \u2212 I. Figure 6.1. Schematic circuit for the quantum search algorithm. The oracle may employ work qubits for its implementation, but the analysis of the quantum search algorithm involves only the n qubit register. Figure 6.2. Circuit for the Grover iteration, G. Each of the operations in the Grover iteration may be ef\ufb01ciently implemented on a quantum computer. Steps 2 and 4, the Hadamard transforms, require n = log(N ) operations each. Step 3, the conditional phase shift, may be implemented using the techniques of Section 4.3, using O(n) gates. The cost of the oracle call depends upon the speci\ufb01c application; for now, we merely need note that the Grover iteration requires only a single oracle call. It is useful to note that the combined effect of steps 2, 3, and 4 is H \u2297n(2|00| \u2212 I)H \u2297n = 2|\u03c8\u03c8| \u2212 I, (6.6) where |\u03c8 is the equally weighted superposition of states, (", "6.4). Thus the Grover iteration, G, may be written G = (2|\u03c8\u03c8| \u2212 I)O. Exercise 6.2: Show that the operation (2|\u03c8\u03c8| \u2212 I) applied to a general state k \u03b1k|k produces \u2212\u03b1k + 2\u03b1 |k, k (6.7) 252 Quantum search algorithms where \u03b1 \u2261 (2|\u03c8\u03c8| \u2212 I) is sometimes referred to as the inversion about mean operation. k \u03b1k/N is the mean value of the \u03b1k. For this reason, 6.1.3 Geometric visualization What does the Grover iteration do? We have noted that G = (2|\u03c8\u03c8| \u2212 I)O. In fact, we will show that the Grover iteration can be regarded as a rotation in the two-dimensional space spanned by the starting vector |\u03c8 and the state consisting of a uniform superposition of solutions to the search problem. To see this it is useful to adopt the convention x indicates a sum over all x which are solutions to the search problem, and x inthat dicates a sum over all x which are not solutions to the search problem. De\ufb01ne normalized states |\u03b1 \u2261 1\u221a N \u2212 M |\u03b2 \u2261 1\u221a M x |x x |x. (6.8) (6.9) Simple algebra shows that the initial state |\u03c8 may be re-expressed as ) ) |\u03c8 = |\u03b1 + |\u03b2, (6.10) N \u2212 M N M N so the initial state of the quantum computer is in the space spanned by |\u03b1 and |\u03b2. The effect of G can be understood in a beautiful way by realizing that the oracle operation O performs a re\ufb02ection about the vector |\u03b1 in the plane de\ufb01ned by |\u03b1 and |\u03b2. That is, O(a|\u03b1 + b|\u03b2) = a|\u03b1 \u2212 b|\u03b2. Similarly, 2|\u03c8\u03c8| \u2212 I also performs a re\ufb02ection in the plane de\ufb01ned by |\u03b1 and |\u03b2, about the vector |\u03c8. And the product of two re\ufb02ections is a rotation! This tells us that the state Gk|\u03c8 remains in the space spanned (N \u2212 M )/N, by |\u03b1 and |\u03b2 for all k. It also gives us the rotation angle. Let cos \u03b8/2 = so that |\u03c8 = cos \u03b8/2|\u03b1 + sin \u03b8/2|\u03b2. As Figure 6.3 shows, the two re\ufb02ections which comprise G take |\u03c8 to G|\u03c8 = cos 3\u03b8 2 |\u03b1 + sin 3\u03b8 2 |\u03b2, (6.11) so the rotation angle is in fact \u03b8. It follows that continued application of G takes the state to Gk|\u03c8 = cos \u03b8 |\u03b1 + sin \u03b8 |\u03b2. (6.12) 2k + 1 2 2k + 1 2 Summarizing, G is a rotation in the two-dimensional space spanned by |\u03b1 and |\u03b2, rotating the space by \u03b8 radians per application of G. Repeated application of the Grover iteration rotates the state vector close to |\u03b2. When this occurs, an observation in the computational basis produces with high probability one of the outcomes superposed in |\u03b2, that is, a solution to the search problem! An example illustrating the search algorithm with N = 4 is given in Box 6.1. Exercise 6.3: Show that in the |\u03b1, |\u03b2 basis, we may write the Grover iteration as G = cos \u03b8 \u2212 sin \u03b8 cos \u03b8 sin \u03b8, (6.13) The quantum search algorithm 253 Figure 6.3. The action of a single Grover iteration, G: the state vector is rotated by \u03b8 towards the superposition |\u03b2 of all solutions to the search problem. Initially, it is inclined at angle \u03b8/2 from |\u03b1, a state orthogonal to |\u03b2. An oracle operation O re\ufb02ects the state about the state |\u03b1, then the operation 2|\u03c8\u03c8| \u2212 I re\ufb02ects it about |\u03c8. In the \ufb01gure |\u03b1 and |\u03b2 are lengthened slightly to reduce clutter (all states should be unit vectors). After repeated Grover iterations, the state vector gets close to |\u03b2, at which point an observation in the computational basis outputs a solution to the search problem with high probability. The remarkable ef\ufb01ciency of the algorithm occurs because \u03b8 behaves like \u03a9( N/M ) applications of G are required to rotate the state vector close to |\u03b2. M/N ), so only O", "( where \u03b8 is a real number in the range 0 to \u03c0/2 (assuming for simplicity that M \u2264 N/2; this limitation will be lifted shortly), chosen so that 2 sin \u03b8 = \u221a M (N \u2212 M ) N. (6.14) 6.1.4 Performance How many times must the Grover iteration be repeated in order to rotate |\u03c8 near |\u03b2? The initial state of the system is |\u03c8 = M/N |\u03b2, so rotating (N \u2212 M )/N |\u03b1 + M/N radians takes the system to |\u03b2. Let CI(x) denote the integer through arccos closest to the real number x, where by convention we round halves down, CI(3.5) = 3, for example. Then repeating the Grover iteration R = CI arccos \u03b8 M/N (6.15) times rotates |\u03c8 to within an angle \u03b8/2 \u2264 \u03c0/4 of |\u03b2. Observation of the state in the computational basis then yields a solution to the search problem with probability at least one-half. Indeed, for speci\ufb01c values of M and N it is possible to achieve a much higher M/N, and probability of success. For example, when M N we have \u03b8 \u2248 sin \u03b8 \u2248 2 M/N, giving a probability thus the angular error in the \ufb01nal state is at most \u03b8/2 \u2248 of error of at most M/N. Note that R depends on the number of solutions M, but not 254 Quantum search algorithms on the identity of those solutions, so provided we know M we can apply the quantum search algorithm as described. In Section 6.3 we will explain how to remove even the need for a knowledge of M in applying the search algorithm. The form (6.15) is useful as an exact expression for the number of oracle calls used to perform the quantum search algorithm, but it would be useful to have a simpler expression summarizing the essential behavior of R. To achieve this, note from (6.15) that R \u2264 \u03c0/2\u03b8, so a lower bound on \u03b8 will give an upper bound on R. Assuming for the moment that M \u2264 N/2, we have \u03b8 \u03b8 \u2265 sin = from which we obtain an elegant upper bound on the number of iterations required, 2 (6.16). (6.17) N/M ) Grover iterations (and thus oracle calls) must be performed That is, R = O( in order to obtain a solution to the search problem with high probability, a quadratic improvement over the O(N/M ) oracle calls required classically. The quantum search algorithm is summarized below, for the case M = 1. Algorithm: Quantum search Inputs: (1) a black box oracle O which performs the transformation O|x|q = |x|q \u2295 f (x), where f (x) = 0 for all 0 \u2264 x < 2n except x0, for which f (x0) = 1; (2) n + 1 qubits in the state |0. Outputs: x0. Runtime: O( 2n) operations. Succeeds with probability O(1). \u221a Procedure: |0\u2297n|0 1. 2. \u2192 1\u221a 2n 3. \u2192 \u2248 |x0 4. \u2192 x0 2n\u22121 |x x=0 (2|\u03c8\u03c8| \u2212 I)O |0 \u2212 |1 \u221a 2 |0 \u2212 |1 \u221a 2 R 1\u221a 2n 2n\u22121 |x x=0 |0 \u2212 |1 \u221a 2 initial state apply H \u2297n to the \ufb01rst n qubits, and HX to the last qubit \u221a apply the Grover iteration R \u2248 \u03c0 2n/4 times. measure the \ufb01rst n qubits Exercise 6.4: Give explicit steps for the quantum search algorithm, as above, but for the case of multiple solutions (1 < M < N/2). What happens when more than half the items are solutions to the search problem, that M (N \u2212 M )/N ) (compare (6.14)) we is, M \u2265 N/2? From the expression \u03b8 = arcsin(2 see that the angle \u03b8 gets smaller as M varies from N/2 to N. As a result, the number of iterations needed by the search algorithm increases with M, for M \u2265 N/2. Intuitively, \u221a Quantum search as a quantum simulation 255 this is a silly property for a search algorithm to have: we expect that it should become easier to \ufb01nd a solution to the problem as the number of solutions increases. There are at least two", " ways around this problem. If M is known in advance to be larger than N/2 then we can just randomly pick an item from the search space, and then check that it is a solution using the oracle. This approach has a success probability at least one-half, and only requires one consultation with the oracle. It has the disadvantage that we may not know the number of solutions M in advance. In the case where it isn\u2019t known whether M \u2265 N/2, another approach can be used. This approach is interesting in its own right, and has a useful application to simplify the analysis of the quantum algorithm for counting the number of solutions to the search problem, as presented in Section 6.3. The idea is to double the number of elements in the search space by adding N extra items to the search space, none of which are solutions. As a consequence, less than half the items in the new search space are solutions. This is effected by adding a single qubit |q to the search index, doubling the number of items to be searched to 2N. A new augmented oracle O is constructed which marks an item only if it is a solution to the search problem and the extra bit is set to zero. In Exercise 6.5 you will explain how the oracle O may be constructed using one call to O. The new search problem has only M solutions out of 2N entries, so running the search algorithm with the new oracle O we see that at most R = \u03c0/4 2N/M calls to O are required, and it follows that O( N/M ) calls to O are required to perform the search. Exercise 6.5: Show that the augmented oracle O may be constructed using one application of O, and elementary quantum gates, using the extra qubit |q. The quantum search algorithm may be used in a wide variety of ways, some of which will be explored in subsequent sections. The great utility of the algorithm arises because we do not assume any particular structure to the search problems being performed. This is the great advantage of posing the problem in terms of a \u2018black box\u2019 oracle, and we adopt this point of view whenever convenient through the remainder of this chapter. In practical applications, of course, it is necessary to understand how the oracle is being implemented, and in each of the practical problems we concern ourselves with an explicit description of the oracle implementation is given. Exercise 6.6: Verify that the gates in the dotted box in the second \ufb01gure of Box 6.1 perform the conditional phase shift operation 2|0000| \u2212 I, up to an unimportant global phase factor. 6.2 Quantum search as a quantum simulation The correctness of the quantum search algorithm is easily veri\ufb01ed, but it is by no means obvious how one would dream up such an algorithm from a state of ignorance. In this section we sketch a heuristic means by which one can \u2018derive\u2019 the quantum search algorithm, in the hope of lending some intuition as to the tricky task of quantum algorithm design. As a useful side effect we also obtain a deterministic quantum search algorithm. Because our goal is to obtain insight rather than generality, we assume for the sake of simplicity that the search problem has exactly one solution, which we label x. Our method involves two steps. First, we make a guess as to a Hamiltonian which 256 Quantum search algorithms Box 6.1: Quantum search: a two-bit example Here is an explicit example illustrating how the quantum search algorithm works on a search space of size N = 4. The oracle, for which f (x) = 0 for all x except x = x0, in which case f (x0) = 1, can be taken to be one of the four circuits corresponding to x0 = 0, 1, 2, or 3 from left to right, where the top two qubits carry the query x, and the bottom qubit carries the oracle\u2019s response. The quantum circuit which performs the initial Hadamard transforms and a single Grover iteration G is Initially, the top two qubits are prepared in the state |0, and the bottom one as |1. The gates in the dotted box perform the conditional phase shift operation 2|0000| \u2212 I. How many times must we repeat G to obtain x0? From Equation (6.15), using M = 1, we \ufb01nd that less than one iteration is required. It turns out that because \u03b8 = \u03c0/3 in (6.14), only exactly one iteration is required, to perfectly obtain x0, in this special case. In the geometric picture of Figure 6.3, our initial state |\u03c8 = (|00 + |01 + |10 + |11)/2 is 30\u25e6 from |\u03b1, and a single rotation by \u03b8 = 60\u25e6 moves |\u03c8 to |\u03b2. You can con\ufffd", "\ufffd\ufffdrm for yourself directly, using the quantum circuits, that measurement of the top two qubits gives x0, after using the oracle only once. In contrast, a classical computer \u2013 or classical circuit \u2013 trying to differentiate between the four oracles would require on average 2.25 oracle queries! solves the search problem. More precisely, we write down a Hamiltonian H which depends on the solution x and an initial state |\u03c8 such that a quantum system evolving according to H will change from |\u03c8 to |x after some prescribed time. Once we\u2019ve found such a Hamiltonian and initial state, we can move on to the second step, which is to attempt to simulate the action of the Hamiltonian using a quantum circuit. Amazingly, following this procedure leads very quickly to the quantum search algorithm! We have already met this two-part procedure while studying universality in quantum circuits, in Problem 4.3, and it also serves well in the study of quantum searching. We suppose that the algorithm starts with the quantum computer in a state |\u03c8. We\u2019ll tie down what |\u03c8 should be later on, but it is convenient to leave |\u03c8 undetermined until we understand the dynamics of the algorithm. The goal of quantum searching is to Quantum search as a quantum simulation 257 change |\u03c8 into |x or some approximation thereof. What Hamiltonians might we guess do a good job of causing such an evolution? Simplicity suggests that we should guess a Hamiltonian constructed entirely from the terms |\u03c8 and |x. Thus, the Hamiltonian must be a sum of terms like |\u03c8\u03c8|, |xx|, |\u03c8x| and |x\u03c8|. Perhaps the simplest choices along these lines are the Hamiltonians: H = |xx| + |\u03c8\u03c8| H = |x\u03c8| + |\u03c8x|. (6.18) (6.19) It turns out that both these Hamiltonians lead to the quantum search algorithm! For now, however, we restrict ourselves to analyzing the Hamiltonian in Equation (6.18). Recall from Section 2.2.2 that after a time t, the state of a quantum system evolving according to the Hamiltonian H and initially in the state |\u03c8 is given by exp(\u2212iHt)|\u03c8. (6.20) Intuitively it looks pretty good: for small t the effect of the evolution is to take |\u03c8 to (I \u2212 itH)|\u03c8 = (1 \u2212 it)|\u03c8 \u2212 itx|\u03c8|x. That is, the |\u03c8 vector is rotated slightly, into the |x direction. Let\u2019s actually do a full analysis, with the goal being to determine whether there is a t such that exp(\u2212iHt)|\u03c8 = |x. Clearly we can restrict the analysis to the two-dimensional space spanned by |x and |\u03c8. Performing the Gram\u2013Schmidt procedure, we can \ufb01nd |y such that |x, |y forms an orthonormal basis for this space, and |\u03c8 = \u03b1|x + \u03b2|y, for some \u03b1, \u03b2 such that \u03b12 + \u03b22 = 1, and for convenience we have chosen the phases of |x and |y so that \u03b1 and \u03b2 are real and non-negative. In the |x, |y basis we have 1 0 0 0 + \u03b12 \u03b1\u03b2 \u03b22 \u03b1\u03b2 1 + \u03b12 \u03b1\u03b2 \u03b1\u03b2 1 \u2212 \u03b12 = I + \u03b1(\u03b2X + \u03b1Z). (6.21) H = Thus = exp(\u2212iHt)|\u03c8 = exp(\u2212it) cos(\u03b1t)|\u03c8 \u2212 i sin(\u03b1t) (\u03b2X + \u03b1Z) |\u03c8. (6.22) The global phase factor exp(\u2212it) can be ignored, and simple algebra shows that (\u03b2X + \u03b1Z)|\u03c8 = |x, so the state of the system after a time t is cos(\u03b1t)|\u03c8 \u2212 i sin(\u03b1t)|x. (6.23) Thus, observation of the system at time t = \u03c0/2\u03b1 yields the result |x with probability one: we have found a solution to the search problem! Unfortunately, the time of the observation depends on \u03b1, the component of |\u03c8 in the |x direction, and thus on x, which is what we are trying to determine. The obvious solution is to attempt to arrange \u03b1 to be the same for all |x, that is, to choose |\u03c8 to be the uniform superposition state x |x \u221a N (6.24). |\u03c8 = \u221a \u221a Making this choice gives \u03b1 = 1/ N for all x, and thus the time of observation t = N /2 does not depend on knowing the value of", " x. Furthermore, the state (6.24) has \u03c0 the obvious advantage that we already know how to prepare such a state by doing a Hadamard transform. We now know that the Hamiltonian (6.18) rotates the vector |\u03c8 to |x. Can we \ufb01nd 258 Quantum search algorithms a quantum circuit to simulate the Hamiltonian (6.18), and thus obtain a quantum search algorithm? Applying the method of Section 4.7, we see that a natural way of simulating H is to alternately simulate the Hamiltonians H1 \u2261 |xx| and H2 \u2261 |\u03c8\u03c8| for short time increments \u0394t. These Hamiltonians are easily simulated using the methods of Chapter 4, as illustrated in Figures 6.4 and 6.5. Exercise 6.7: Verify that the circuits shown in Figures 6.4 and 6.5 implement the operations exp(\u2212i|xx|\u0394t) and exp(\u2212i|\u03c8\u03c8|\u0394t), respectively, with |\u03c8 as in (6.24). Figure 6.4. Circuit implementing the operation exp(\u2212i|xx|\u0394t) using two oracle calls. Figure 6.5. Circuit implementing the operation exp(\u2212i|\u03c8\u03c8|\u0394t), for |\u03c8 as in (6.24). \u221a \u221a \u221a N /\u0394t), and thus the cumulative error is O(\u0394t2 \u00d7 The number of oracle calls required by the quantum simulation is determined by how small a time-step is required to obtain reasonably accurate results. Suppose we use a simulation step of length \u0394t that is accurate to O(\u0394t2). The total number of steps required N ). is t/\u0394t = \u0398( To obtain a reasonably high success probability we need the error to be O(1), which means that we must choose \u0394t = \u0398(1/ N ) which results in a number of oracle calls that scales like O(N ) \u2013 no better than the classical solution! What if we use a more accurate method of quantum simulation, say one that is accurate to O(\u0394t3)? The cumulative error in this N ), and thus to achieve a reasonable success probability we need to choose case is O(\u0394t2 \u0394t = \u0398(N \u22121/4), resulting in a total number of oracle calls O(N 3/4), which is a distinct improvement over the classical situation, although still not as good as achieved by the quantum search algorithm of Section 6.1! In general going to a more accurate quantum simulation technique results in a reduction in the number of oracle calls required to perform the simulation: N /\u0394t) = O(\u0394t \u221a \u221a Exercise 6.8: Suppose the simulation step is performed to an accuracy O(\u0394tr). Show Quantum search as a quantum simulation 259 that the number of oracle calls required to simulate H to reasonable accuracy is O(N r/2(r\u22121)). Note that as r becomes large the exponent of N approaches 1/2. We have been analyzing the accuracy of the quantum simulation of the Hamiltonian (6.18) using general results on quantum simulation from Section 4.7. Of course, in this instance we are dealing with a speci\ufb01c Hamiltonian, not the general case, which suggests that it might be interesting to calculate explicitly the effect of a simulation step of time \u0394t, rather than relying on the general analysis. We can do this for any speci\ufb01c choice of simulation method \u2013 it can be a little tedious to work out the effect of the simulation step, but it is essentially a straightforward calculation. The obvious starting point is to explicitly calculate the action of the lowest-order simulation techniques, that is, to calculate one or both of exp(\u2212i|xx|\u0394t) exp(\u2212i|\u03c8\u03c8|\u0394t) or exp(\u2212i|\u03c8\u03c8|\u0394t) exp(\u2212i|xx|\u0394t). The results are essentially the same in both instances; we will focus on the study of U (\u0394t) \u2261 exp(\u2212i|\u03c8\u03c8|\u0394t) exp(\u2212i|xx|\u0394t). U (\u0394t) clearly acts non-trivially only in the space spanned by |xx| and |\u03c8\u03c8|, so we restrict ourselves to that space, working in the basis |x, |y, where |y is de\ufb01ned as before. Note that in this representation |xx| = (I + Z)/2 = (I + \u02c6z \u00b7 \u03c3)/2, where \u02c6z \u2261 (0, 0, 1) is the unit vector in the z direction, and |", "\u03c8\u03c8| = (I + \u03c8 \u00b7 \u03c3)/2, where \u03c8 = (2\u03b1\u03b2, 0, (\u03b12 \u2212 \u03b22)) (recall that this is the Bloch vector representation; see Section 4.2). A simple calculation shows that up to an unimportant global phase factor, U (\u0394t) = cos2 \u22122i sin \u0394t 2 \u0394t 2 \u2212 sin2 cos \u0394t 2 \u0394t 2 \u03c8 \u00b7 \u02c6z \u03c8 + \u02c6z 2 I \u03c8 \u00d7 \u02c6z 2 \u0394t 2 + sin \u00b7 \u03c3. (6.25) Exercise 6.9: Verify Equation (6.25). (Hint: see Exercise 4.15.) Equation (6.25) implies that U (\u0394t) is a rotation on the Bloch sphere about an axis of rotation r de\ufb01ned by \u0394t 2 and through an angle \u03b8 de\ufb01ned by r = cos \u03c8 + \u02c6z 2 \u03c8 \u00d7 \u02c6z 2 \u0394t 2, + sin \u03b8 2 cos = cos2 \u0394t 2 \u2212 sin2 \u0394t 2 \u03c8 \u00b7 \u02c6z, which simpli\ufb01es upon substitution of \u03c8 \u00b7 \u02c6z = \u03b12 \u2212 \u03b22 = (2/N \u2212 1) to \u03b8 cos 2 = 1 \u2212 2 N sin2 \u0394t 2. (6.26) (6.27) (6.28) Note that \u03c8 \u00b7 r = \u02c6z \u00b7 r, so both |\u03c8\u03c8| and |xx| lie on the same circle of revolution about the r axis on the Bloch sphere. Summarizing, the action of U (\u0394t) is to rotate |\u03c8\u03c8| about the r axis, through an angle \u03b8 for each application of U (\u0394t), as illustrated in Figure 6.6. We terminate the procedure when enough rotations have been performed to rotate |\u03c8\u03c8| near to the solution |xx|. Now initially we imagined that \u0394t was small, since we were considering the case of quantum simulation, but Equation (6.28) shows 260 Quantum search algorithms that the smart thing to do is to choose \u0394t = \u03c0, in order to maximize the rotation angle \u03b8. If we do this, then we obtain cos(\u03b8/2) = 1 \u2212 2/N, which for large N corresponds to N ), \u03b8 \u2248 4/ just as for the original quantum search algorithm. N, and the number of oracle calls required to \ufb01nd the solution |x is O( \u221a \u221a Figure 6.6. Bloch sphere diagram showing the initial state \u03c8 rotating around the axis of rotation r going toward the \ufb01nal state \u02c6z. Indeed, if we make the choice \u0394t = \u03c0, then this \u2018quantum simulation\u2019 is in fact identical with the original quantum search algorithm, since the operators applied in the quantum simulation are exp(\u2212i\u03c0|\u03c8\u03c8|) = I \u2212 2|\u03c8\u03c8| and exp(\u2212i\u03c0|xx|) = I \u2212 2|xx|, and up to a global phase shift these are identical to the steps making up the Grover iteration. Viewed this way, the circuits shown in Figures 6.2 and 6.3 for the quantum search algorithm are simpli\ufb01cations of the circuits shown in Figures 6.4 and 6.5 for the simulation, in the special case \u0394t = \u03c0! Exercise 6.10: Show that by choosing \u0394t appropriately we can obtain a quantum \u221a search algorithm which uses O( exactly, that is, the algorithm works with probability 1, rather than with some smaller probability. N ) queries, and for which the \ufb01nal state is |x We have re-derived the quantum search algorithm from a different point of view, the point of view of quantum simulation. Why did this approach work? Might it be used to \ufb01nd other fast quantum algorithms? We can\u2019t answer these questions in any de\ufb01nitive way, but the following few thoughts may be of some interest. The basic procedure used is four-fold: (1) specify the problem to be solved, including a description of the desired input and output from the quantum algorithm; (2) guess a Hamiltonian to solve the problem, and verify that it does in fact work; (3) \ufb01nd a procedure to simulate the Hamiltonian; and (4) analyze the resource costs of the simulation. This is different from the more conventional approach in two respects: we guess a Hamiltonian, rather than a quantum circuit, and there is no analogue to the simulation step in the conventional approach. The more important of these two differences is", " the \ufb01rst. There is a great deal of freedom in specifying a quantum circuit to solve a problem. While that freedom is, in part, responsible Quantum counting 261 for the great power of quantum computation, it makes searching for good circuits rather dif\ufb01cult. By contrast, specifying a Hamiltonian is a much more constrained problem, and therefore affords less freedom in the solution of problems, but those same constraints may in fact make it much easier to \ufb01nd an ef\ufb01cient quantum algorithm to solve a problem. We\u2019ve seen this happen for the quantum search algorithm, and perhaps other quantum algorithms will be discovered by this method; we don\u2019t know. What seems certain is that this \u2018quantum algorithms as quantum simulations\u2019 point of view offers a useful alternative viewpoint to stimulate in the development of quantum algorithms. Exercise 6.11: (Multiple solution continuous quantum search) Guess a Hamiltonian with which one may solve the continuous time search problem in the case where the search problem has M solutions. Exercise 6.12: (Alternative Hamiltonian for quantum search) Suppose H = |x\u03c8| + |\u03c8x|. (6.29) (1) Show that it takes time O(1) to rotate from the state |\u03c8 to the state |x, given an evolution according to the Hamiltonian H. (2) Explain how a quantum simulation of the Hamiltonian H may be performed, and determine the number of oracle calls your simulation technique requires to obtain the solution with high probability. 6.3 Quantum counting How quickly can we determine the number of solutions, M, to an N item search problem, if M is not known in advance? Clearly, on a classical computer it takes \u0398(N ) consultations with an oracle to determine M. On a quantum computer it is possible to estimate the number of solutions much more quickly than is possible on a classical computer by combining the Grover iteration with the phase estimation technique based upon the quantum Fourier transform (Chapter 5). This has some important applications. First, if we can estimate the number of solutions quickly then it is also possible to \ufb01nd a solution quickly, even if the number of solutions is unknown, by \ufb01rst counting the number of solutions, and then applying the quantum search algorithm to \ufb01nd a solution. Second, quantum counting allows us to decide whether or not a solution even exists, depending on whether the number of solutions is zero, or non-zero. This has applications, for example, to the solution of NP-complete problems, which may be phrased in terms of the existence of a solution to a search problem. Exercise 6.13: Consider a classical algorithm for the counting problem which samples uniformly and independently k times from the search space, and let X1,..., Xk be the results of the oracle calls, that is, Xj = 1 if the jth oracle call revealed a solution to the problem, and Xj = 0 if the jth oracle call did not reveal a solution to the problem. This algorithm returns the estimate S \u2261 N \u00d7 for the number of solutions to the search problem. Show that the standard deviation in S is \u0394S = least 3/4 of estimating M correctly to within an accuracy M we must have k = \u03a9(N ). M (N \u2212 M )/k. Prove that to obtain a probability at M for all values of j Xj/k \u221a 262 Quantum search algorithms Exercise 6.14: Prove that any classical counting algorithm with a probability at least M for some constant c 3/4 for estimating M correctly to within an accuracy c and for all values of M must make \u03a9(N ) oracle calls. \u221a Quantum counting is an application of the phase estimation procedure of Section 5.2 to estimate the eigenvalues of the Grover iteration G, which in turn enables us to determine the number of solutions M to the search problem. Suppose |a and |b are the two eigenvectors of the Grover iteration in the space spanned by |\u03b1 and |\u03b2. Let \u03b8 be the angle of rotation determined by the Grover iteration. From Equation (6.13) we see that the corresponding eigenvalues are ei\u03b8 and ei(2\u03c0\u2212\u03b8). For ease of analysis it is convenient to assume that the oracle has been augmented, as described in Section 6.1, expanding the size of the search space to 2N, and ensuring that sin2(\u03b8/2) = M/2N. The phase estimation circuit used for quantum counting is shown in Figure 6.7. The function of the circuit is to estimate \u03b8 to m bits of accuracy, with a probability of success at least 1 \u2212. The \ufb01rst register contains t \u2261 m + log(2 + 1/2) qubits, as per the phase estimation", " algorithm, and the second register contains n + 1 qubits, enough to implement the Grover iteration on the augmented search space. The state of the second register x |x by a Hadamard is initialized to an equal superposition of all possible inputs transform. As we saw in Section 6.1 this state is a superposition of the eigenstates |a and |b, so by the results of Section 5.2 the circuit in Figure 6.7 gives us an estimate of \u03b8 or 2\u03c0 \u2212 \u03b8 accurate to within |\u0394\u03b8| \u2264 2\u2212m, with probability at least 1 \u2212. Furthermore, an estimate for 2\u03c0 \u2212 \u03b8 is clearly equivalent to an estimate of \u03b8 with the same level of accuracy, so effectively the phase estimation algorithm determines \u03b8 to an accuracy 2\u2212m with probability 1 \u2212. Figure 6.7. Circuit for performing approximate quantum counting on a quantum computer. Speeding up the solution of NP-complete problems 263 Using the equation sin2(\u03b8/2) = M/2N and our estimate for \u03b8 we obtain an estimate of the number of solutions, M. How large an error, \u0394M, is there in this estimate? |\u0394M | 2N = + + + +sin2 = sin sin \u2212 sin2 + sin \u03b8 + \u0394\u03b8 2 \u03b8 + \u0394\u03b8 2 \u03b8 + \u0394\u03b8 2 \u03b8 + + + +. \u2212 sin (6.30) (6.31) 2 Calculus implies that | sin((\u03b8 +\u0394\u03b8)/2)\u2212sin(\u03b8/2)| \u2264 |\u0394\u03b8|/2, and elementary trigonometry that | sin((\u03b8 + \u0394\u03b8)/2)| < sin(\u03b8/2) + |\u0394\u03b8|/2, so 2 |\u0394M | 2N < 2 sin \u03b8 2 + |\u0394\u03b8| 2 |\u0394\u03b8| 2. (6.32) Substituting sin2(\u03b8/2) = M/2N and |\u0394\u03b8| \u2264 2\u2212m gives our \ufb01nal estimate for the error in our estimate of M, |\u0394M | < \u221a 2M N + N 2m+1 2\u2212m. (6.33) As an example, suppose we choose m = n/2 + 1, and = 1/6. Then we have t = N ) Grover iterations, and thus \u0398( N ) oracle n/2 + 3, so the algorithm requires \u0398( M ). Compare this with calls. By (6.33) our accuracy is |\u0394M | < Exercise 6.14, according to which it would have required O(N ) oracle calls to obtain a similar accuracy on a classical computer. M/2 + 1/4 = O( \u221a \u221a \u221a Indeed, the example just described serves double duty as an algorithm for determining whether a solution to the search problem exists at all, that is, whether M = 0 or M = 0. If M = 0 then we have |\u0394M | < 1/4, so the algorithm must produce the estimate zero with probability at least 5/6. Conversely, if M = 0 then it is easy to verify that the estimate for M is not equal to 0 with probability at least 5/6. Another application of quantum counting is to \ufb01nd a solution to a search problem when the number M of solutions is unknown. The dif\ufb01culty in applying the quantum search algorithm as described in Section 6.1 is that the number of times to repeat the Grover iteration, Equation (6.15), depends on knowing the number of solutions M. This problem can be alleviated by using the quantum counting algorithm to \ufb01rst estimate \u03b8 and M to high accuracy using phase estimation, and then to apply the quantum search algorithm as in Section 6.1, repeating the Grover iteration a number of times determined by (6.15), with the estimates for \u03b8 and M obtained by phase estimation substituted to determine R. The angular error in this case is at most \u03c0/4(1 + |\u0394\u03b8|/\u03b8), so choosing m = n/2 + 1 as before gives an angular error at most \u03c0/4 \u00d7 3/2 = 3\u03c0/8, which corresponds to a success probability of at least cos2(3\u03c0/8) = 1/2 \u2212 1/2 2 \u2248 0.15 for the search algorithm. If the probability of obtaining an estimate of \u03b8 this accurate is 5/6, as in our earlier example, then the total probability of obtaining a solution to the search problem is 5/6 \u00d7 cos2(3\u03c0/8) \u2248 0.12, a probability which may quickly be boosted close to 1 by", " a few repetitions of the combined counting\u2013search procedure. \u221a 6.4 Speeding up the solution of NP-complete problems Quantum searching may be used to speed up the solution to problems in the complexity class NP (Section 3.2.2). We already saw, in Section 6.1.1, how factoring can be sped 264 Quantum search algorithms up; here, we illustrate how quantum search can be applied to assist the solution of the ). Recall that a Hamiltonian cycle of a graph is a simple Hamiltonian cycle problem ( problem is to determine whether a cycle which visits every vertex of the graph. The given graph has a Hamiltonian cycle or not. This problem belongs to the class of NPcomplete problems, widely believed (but not yet proved) to be intractable on a classical computer. A simple algorithm to solve is to perform a search through all possible orderings of the vertices: (1) Generate each possible ordering (v1,..., vn) of vertices for the graph. Repetitions will be allowed, as they ease the analysis without affecting the essential result. (2) For each ordering check to see whether it is a Hamiltonian cycle for the graph. If not, continue checking the orderings. Since there are nn = 2n log n possible orderings of the vertices which must be searched, this algorithm requires 2n log n checks for the Hamiltonian cycle property in the worst case. Indeed, any problem in NP may be solved in a similar way: if a problem of size n has witnesses which can be speci\ufb01ed using w(n) bits, where w(n) is some polynomial in n, then searching through all 2w(n) possible witnesses will reveal a solution to the problem, if one exists. The quantum search algorithm may be used to speed up this algorithm by increasing the speed of the search. Speci\ufb01cally, we use the algorithm described in Section 6.3 to determine whether a solution to the search problem exists. Let m \u2261 log n. The search space for the algorithm will be represented by a string of mn qubits, each block of m qubits being used to store the index to a single vertex. Thus we can write the computational basis states as |v1,..., vn, where each |vi is represented by the appropriate string of m qubits, for a total of nm qubits. The oracle for the search algorithm must apply the transformation: * O|v1,..., vn = |v1,..., vn \u2212|v1,..., vn if v1,..., vn is not a Hamiltonian cycle if v1,..., vn is a Hamiltonian cycle (6.34) Such an oracle is easy to design and implement when one has a description of the graph. One takes a polynomial size classical circuit recognizing Hamiltonian cycles in the graph, and converts it to a reversible circuit, also of polynomial size, computing the transformation (v1,..., vn, q) \u2192 (v1,..., vn, q \u2295 f (v1,..., vn)), where f (v1,..., vn) = 1 if v1,..., vn is a Hamiltonian cycle, and is 0 otherwise. Implementing the corresponding \u221a circuit on a quantum computer with the \ufb01nal qubit starting in the state (|0 \u2212 |1)/ 2 gives the desired transformation. We won\u2019t explicitly describe the details here, except to note the key point: the oracle requires a number of gates polynomial in n, as a direct consequence of the fact that Hamiltonian cycles can be recognized using polynomially many gates classically. Applying the variant of the search algorithm which determines whether a solution to the search problem exists (Section 6.3) we see that it takes O(2mn/2) = O(2nlog n/2) applications of the oracle to determine whether a Hamiltonian cycle exists. When one does exist it is easy to apply the combined counting\u2013search algorithm to \ufb01nd an example of such a cycle, which can then be exhibited as a witness for the problem. To summarize: \u2022 The classical algorithm requires O p(n)2nlog n operations to determine whether a Quantum search of an unstructured database 265 Hamiltonian cycle exists, where the polynomial factor p(n) is overhead predominantly due to the implementation of the oracle, that is, the gates checking whether a candidate path is Hamiltonian or not. The dominant effect in determining the resources required is the exponent in 2nlog n. The classical algorithm is deterministic, that is, it succeeds with probability 1. p(n)2", "nlog n/2 operations to determine whether \u2022 The quantum algorithm requires O a Hamiltonian cycle exists. Once again, the polynomial p(n) is overhead predominantly due to implementation of the oracle. The dominant effect in determining the resources required is the exponent in 2nlog n/2. There is a constant probability (say, 1/6) of error for the algorithm, which may be reduced to 1/6r by r repetitions of the algorithm. \u2022 Asymptotically the quantum algorithm requires the square root of the number of operations the classical algorithm requires. 6.5 Quantum search of an unstructured database Suppose somebody gives you a list containing one thousand \ufb02ower names, and asks you where \u2018Perth Rose\u2019 appears on the list. If the \ufb02ower appears exactly once on the list, and the list is not ordered in any obvious way, then you will need to examine \ufb01ve hundred names, on average, before you \ufb01nd the \u2018Perth Rose\u2019. Might it be possible to speed up this kind of database searching using the quantum search algorithm? Indeed, the quantum search algorithm is sometimes referred to as a database search algorithm, but its usefulness for that application is limited, and based on certain assumptions. In this section we take a look at how the quantum search algorithm can conceptually be used to search an unstructured database, in a setting rather like that found on a conventional computer. The picture we construct will clarify what resources are required to enable a quantum computer to search classical databases. Suppose we have a database containing N \u2261 2n items, each of length l bits. We will label these items d1,..., dN. We want to determine where a particular l bit string, s, is in the database. A classical computer, used to solve this problem, is typically split into two parts, illustrated in Figure 6.8. One is the central processing unit, or CPU, where data manipulation takes place, using a small amount of temporary memory. The second part is a large memory which stores the database in a string of 2n blocks of l bit cells. The memory is assumed to be passive, in the sense that it is not capable of processing data on its own. What is possible is to LOAD data from memory into the CPU, and STORE data from the CPU in memory, and to do manipulations of the data stored temporarily in the CPU. Of course, classical computers may be designed along different lines, but this CPU\u2013memory split is a popular and common architecture. To \ufb01nd out where a given string s is in the unstructured database, the most ef\ufb01cient classical algorithm is as follows. First, an n-bit index to the database elements is set up in the CPU. We assume that the CPU is large enough to store the n \u2261 log N bit index. The index starts out at zero, and is incremented by one on each iteration of the algorithm. At each iteration, the database entry corresponding to the index is loaded into the CPU, and compared to the string which is being searched for. If they are the same, the algorithm outputs the value of the index and halts. If not, the algorithm continues incrementing the index. Obviously, this algorithm requires that items be loaded from 266 Quantum search algorithms ) / / / Figure 6.8. Classical database searching on a computer with distinct central processing unit (CPU) and memory. Only two operations may be directly performed on the memory \u2013 a memory element may be LOADed into the CPU, or an item from the CPU may be STOREd in memory. memory 2n times in the worst case. It is also clear that this is the most ef\ufb01cient possible algorithm for solving the problem in this model of computation. How ef\ufb01ciently can an analogous algorithm be implemented on a quantum computer? And, even if a quantum speedup is possible, how useful is such an algorithm? We show \ufb01rst that a speedup is possible, and then return to the question of the utility of such an algorithm. Suppose our quantum computer consists of two units, just like the classical computer, a CPU and a memory. We assume that the CPU contains four registers: (1) an n qubit \u2018index\u2019 register initialized to |0; (2) an l qubit register initialized to |s and remaining in that state for the entire computation; (3) an l qubit \u2018data\u2019 register initialized to |0; and (4) a 1 qubit register initialized to (|0 \u2212 |1)/ \u221a 2. The memory unit can be implemented in one of two ways. The simplest is a quantum memory containing N = 2n cells of l qubits each, containing the database entries |dx. The second implementation is to implement the memory as a classical memory with N = 2n cells of", " l bits each, containing the database entries dx. Unlike a traditional classical memory, however, it can be addressed by an index x which can be in a superposition of multiple values. This quantum index allows a superposition of cell values to be LOADed from memory. Memory access works in the following way: if the CPU\u2019s index register is in the state |x and the data register is in the state |d, then the contents dx of the xth memory cell are added to the data register: |d \u2192 |d \u2295 dx, where the addition is done bitwise, modulo 2. First, let us see how this capability is used to perform quantum search, then we shall discuss how such a memory might be physically constructed. The key part of implementing the quantum search algorithm is realization of the oracle, which must \ufb02ip the phase of the index which locates s in the memory. Suppose the CPU is in the state |x|s|0 |0 \u2212 |1 \u221a 2. Applying the LOAD operation puts the computer in the state |x|s|dx |0 \u2212 |1 \u221a 2. (6.35) (6.36) Now the second and third registers are compared, and if they are the same, then a bit Quantum search of an unstructured database 267 \ufb02ip is applied to register 4; otherwise nothing is changed. The effect of this operation is |x|s|dx |0 \u2212 |1 \u221a 2 \u2192 \u23a7 \u23aa\u23aa\u23a8 \u23aa\u23aa\u23a9 \u2212|x|s|dx |x|s|dx |0 \u2212 |1 \u221a 2 |0 \u2212 |1 \u221a 2 if dx = s if dx = s. (6.37) The data register is then restored to the state |0 by performing the LOAD operation again. The total action of the oracle thus leaves registers 2, 3 and 4 unaffected, and unentangled with register 1. Thus, the overall effect is to take the state of register 1 from |x to \u2212|x if dx = s, and to leave the register alone otherwise. Using the oracle implemented in this way, we may apply the quantum search algorithm to determine the location of s in the database, using O( N ) LOAD operations, compared to the N LOAD operations that were required classically. \u221a In order for the oracle to function correctly on superpositions it seems at \ufb01rst glance as though the memory needs to be quantum mechanical. In fact, as we noted above, with some caveats the memory can actually be implemented classically, which likely makes it much more resistant to the effects of noise. But a quantum addressing scheme is still needed; a conceptual picture illustrating how this might be done is shown in Figure 6.9. The principle of operation is a means by which the binary encoded state of the quantum index (where 0 to 2n \u2212 1 is represented by n qubits) is translated into a unary encoding (where 0 to 2n \u2212 1 is represented by the position of a single probe within 2n possible locations) which addresses the classical database. The database effects a change on a degree of freedom within the probe which is unrelated to its position. The binary to unary encoding is then reversed, leaving the data register with the desired contents. Are there practical instances in which the quantum search algorithm could be useful for searching classical databases? Two distinct points may be made. First, databases are not ordinarily unstructured. Simple databases, like one containing \ufb02ower names discussed in the introduction to this section, may be maintained in alphabetical order, such that a binary search can be used to locate an item in time which is O(log(N )) for an N -element database. However, some databases may require a much more complex structure, and although sophisticated techniques exist to optimize classical searches, given queries of a suf\ufb01ciently complex or unanticipated nature, a predetermined structure may not be of assistance, and the problem can be regarded as being essentially the unstructured database search problem we discussed. Second, for a quantum computer to be able to search a classical database, a quantum addressing scheme is required. The scheme we depicted requires O(N log N ) quantum switches \u2013 about the same amount of hardware as would be required to store the database itself. Presumably, these switches may someday be as simple and inexpensive as classical memory elements, but if that is not the case, then building a quantum computer to perform a quantum search may not be economically advantageous, compared with using classical computing hardware distributed over the memory elements. Given these considerations, it appears that the principle use of quantum search algorithms will not be in searching classical databases. Rather, their use will probably be in searching for solutions to hard problems, as discussed in the last section, such as the Hamiltonian cycle, traveling salesman, and satis\ufffd", "\ufffdability problems. 268 Quantum search algorithms 01 03 03 02 02 //// 2 3 1 7 8 24 25 26 3 3 32 02 02 03 03 01 \u221a Figure 6.9. Conceptual diagram of a 32 cell classical memory with a \ufb01ve qubit quantum addressing scheme. Each circle represents a switch, addressed by the qubit inscribed within. For example, when |x4 = |0, the corresponding switch routes the input qubit towards the left; when |x4 = |1 the switch routes the input qubit to the right. If |x4 = (|0 + |1)/ enter at the top of the tree, and are routed down to the database, which changes their state according to the contents of the memory. The qubits are then routed back into a de\ufb01nite position, leaving them with the retrieved information. Physically, this could be realized using, for example, single photons for the data register qubits, which are steered using nonlinear interferometers (Chapter 7). The classical database could be just a simple sheet of plastic in which a \u2018zero\u2019 (illustrated as white squares) transmits light unchanged, and a \u2018one\u2019 (shaded squares) shifts the polarization of the incident light by 90\u25e6. 2, then an equal superposition of both routes is taken. The data register qubits Optimality of the search algorithm 269 6.6 Optimality of the search algorithm \u221a We have shown that a quantum computer can search N items, consulting the search N ) times. We now prove that no quantum algorithm can perform this oracle only O( N ) accesses to the search oracle, and thus the algorithm we task using fewer than \u03a9( have demonstrated is optimal. \u221a Suppose the algorithm starts in the state |\u03c8. For simplicity, we prove the lower bound for the case where the search problem has a single solution, x. To determine x we are allowed to apply an oracle Ox which gives a phase shift of \u22121 to the solution |x and leaves all other states invariant, Ox = I \u2212 2|xx|. We suppose the algorithm starts in a state |\u03c8 and applies the oracle Ox exactly k times, with unitary operations U1, U2,..., Uk interleaved between the oracle operations. De\ufb01ne k \u2261 UkOxUk\u22121Ox... U1Ox|\u03c8 |\u03c8x |\u03c8k \u2261 UkUk\u22121... U1|\u03c8. (6.38) (6.39) That is, |\u03c8k is the state that results when the sequence of unitary operations U1,..., Uk is carried out, without the oracle operations. Let |\u03c80 = |\u03c8. Our goal will be to bound the quantity Dk \u2261 \u03c8x k \u2212 \u03c8k 2, (6.40) x where we use the notation \u03c8 for |\u03c8 as a convenience to simplify formulas. Intuitively, Dk is a measure of the deviation after k steps caused by the oracle, from the evolution that would otherwise have ensued. If this quantity is small, then all the states |\u03c8x k are roughly the same, and it is not possible to correctly identify x with high probability. The strategy of the proof is to demonstrate two things: (a) a bound on Dk that shows it can grow no faster than O(k2); and (b) a proof that Dk must be \u03a9(N ) if it is to be possible to distinguish N alternatives. Combining these two results gives the desired lower bound. First, we give an inductive proof that Dk \u2264 4k2. This is clearly true for k = 0, where Dk = 0. Note that Dk+1 = = x x Ox\u03c8x k \u2212 \u03c8k 2 Ox(\u03c8x k \u2212 \u03c8k) + (Ox \u2212 I)\u03c8k 2. (6.41) (6.42) Applying b+c2 \u2264 b2 +2b c+c2 with b \u2261 Ox(\u03c8x \u22122x|\u03c8k|x, gives k \u2212\u03c8k) and c \u2261 (Ox \u2212I)\u03c8k = Dk+1 \u2264 x \u03c8x k \u2212 \u03c8k2 + 4\u03c8x k \u2212 \u03c8k |x|\u03c8k| + 4|\u03c8k|x|2. (6.43) Applying the Cauchy\u2013Schwarz inequality to the second term on the right hand side, and noting that x |x|\u03c8k|2 = 1 gives Dk+1 \u2264 Dk + 4 \u03c8x k \u2212 \u03c8k2 \u2264 Dk + 4 x Dk + 4. 1 2 x 1 2 |\u03c8", "k|x|2 + 4 (6.44) (6.45) 270 Quantum search algorithms By the inductive hypothesis Dk \u2264 4k2 we obtain Dk+1 \u2264 4k2 + 8k + 4 = 4(k + 1)2, (6.46) which completes the induction. To complete the proof we need to show that the probability of success can only be high if Dk is \u03a9(N ). We suppose |x|\u03c8x k |2 \u2265 1/2 for all x, so that an observation yields a solution to the search problem with probability at least one-half. Replacing |x by ei\u03b8|x does not change the probability of success, so without loss of generality we may assume that x|\u03c8x k |, and therefore k = |x|\u03c8x \u03c8x k \u2212 x2 = 2 \u2212 2|x|\u03c8x k \u2212 x2 we see that Ek \u2264 (2 \u2212 k | \u2264 2 \u2212 \u221a x \u03c8x De\ufb01ning Ek \u2261 prove that Dk is \u03a9(N ). De\ufb01ning Fk \u2261 Dk = (\u03c8x k \u2212 x) + (x \u2212 \u03c8k)2 x x \u2212 \u03c8k2 we have \u221a 2. (6.47) 2)N. We are now in position to (6.48) x x \u2212 \u03c8k2 (6.49) \u2265 x x \u03c8x k \u2212 x2 \u2212 2 \u03c8x k \u2212 x x \u2212 \u03c8k + = Ek + Fk \u2212 2 x \u03c8x k \u2212 x x \u2212 \u03c8k. x (6.50) \u221a EkFk, so (6.51) Applying the Cauchy\u2013Schwarz inequality gives we have x \u03c8x k \u2212 x x \u2212 \u03c8k \u2264 Dk \u2265 Ek + Fk \u2212 2 EkFk = ( Fk \u2212 \u221a Ek)2. In Exercise 6.15 you will show that Fk \u2265 2N \u2212 2 Ek \u2264 (2 \u2212 \u221a 2 \u2212 than ( N. Combining this with the result \u221a 2)N gives Dk \u2265 cN for suf\ufb01ciently large N, where c is any constant less 2 \u2212 2)2 \u2248 0.42. Since Dk \u2264 4k2 this implies that \u221a k \u2265 cN/4. (6.52) Summarizing, to achieve a probability of success at least one-half for \ufb01nding a solution to the search problem we must call the oracle \u03a9( N ) times. \u221a Exercise 6.15: Use the Cauchy\u2013Schwarz inequality to show that for any normalized state vector |\u03c8 and set of N orthonormal basis vectors |x, \u221a \u03c8 \u2212 x2 \u2265 2N \u2212 2 N. (6.53) x Exercise 6.16: Suppose we merely require that the probability of an error being made is less than 1/2 when averaged uniformly over the possible values for x, instead \u221a of for all values of x. Show that O( N ) oracle calls are still required to solve the search problem. This result, that the quantum search algorithm is essentially optimal, is both exciting and disappointing. It is exciting because it tells us that for this problem, at least, we have fully plumbed the depths of quantum mechanics; no further improvement is possible. The disappointment arises because we might have hoped to do much better than the square root speedup offered by the quantum search algorithm. The sort of dream result Black box algorithm limits 271 we might have hoped for a priori is that it would be possible to search an N item search space using O(log N ) oracle calls. If such an algorithm existed, it would allow us to solve NP-complete problems ef\ufb01ciently on a quantum computer, since it could search all 2w(n) possible witnesses using roughly w(n) oracle calls, where the polynomial w(n) is the length of a witness in bits. Unfortunately, such an algorithm is not possible. This is useful information for would-be algorithm designers, since it indicates that a naive search-based method for attacking NP-complete problems is guaranteed to fail. Venturing into the realm of opinion, we note that many researchers believe that the essential reason for the dif\ufb01culty of NP-complete problems is that their search space has essentially no structure, and that (up to polynomial factors) the best possible method for solving such a problem is to adopt a search method. If one takes this point of view, then it is bad news for quantum computing, indicating that the class of problems ef\ufb01ciently soluble on a quantum computer, BQP, does not contain the NP-complete problems", ". Of course, this is merely opinion, and it is still possible that the NP-complete problems contain some unknown structure that allows them to be ef\ufb01ciently solved on a quantum computer, or perhaps even on a classical computer. A nice example to illustrate this point is the problem of factoring, widely believed to be in the class NPI of problems intermediate in dif\ufb01culty between P and the NP-complete problems. The key to the ef\ufb01cient quantum mechanical solution of the factoring problem was the exploitation of a structure \u2018hidden\u2019 within the problem \u2013 a structure revealed by the reduction to order\ufb01nding. Even with this amazing structure revealed, it has not been found possible to exploit the structure to develop an ef\ufb01cient classical algorithm for factoring, although, of course, quantum mechanically the structure can be harnessed to give an ef\ufb01cient factoring algorithm! Perhaps a similar structure lurks in other problems suspected to be in NPI, such as the graph isomorphism problem, or perhaps even in the NP-complete problems themselves. Exercise 6.17: (Optimality for multiple solutions) Suppose the search problem N/M ) oracle applications are required to \ufb01nd a has M solutions. Show that O( solution. 6.7 Black box algorithm limits We conclude this chapter with a generalization of the quantum search algorithm which provides insightful bounds on the power of quantum computation. At the beginning of the chapter, we described the search problem as \ufb01nding an n-bit integer x such that the function f : {0, 1}n \u2192 {0, 1} evaluates to f (x) = 1. Related to this is the decision problem of whether or not there exists x such that f (x) = 1. Solving this decision problem is equivalently dif\ufb01cult, and can be expressed as computing the Boolean function F (X) = X0 \u2228 X1 \u2228 \u00b7 \u00b7 \u00b7 \u2228 XN \u22121, where \u2228 denotes the binary OR operation, Xk \u2261 f (k), and X denotes the set {X0, X1,..., XN \u22121}. More generally, we may wish to compute some function other than OR. For example, F (X) could be the AND, PARITY (sum modulo two), or MAJORITY (F (X) = 1 if and only if more Xk = 1 than not) functions. In general, we can consider F to be any Boolean function. How fast (measured in number of queries) can a computer, classical or quantum, compute these functions, given an oracle for f? It might seem dif\ufb01cult to answer such questions without knowing something about the 272 Quantum search algorithms function f, but in fact a great deal can be determined even in this \u2018black box\u2019 model, where the means by which the oracle accomplishes its task is taken for granted, and complexity is measured only in terms of the number of required oracle queries. The analysis of the search algorithm in the previous sections demonstrated one way to approach such problems, but a more powerful approach for obtaining query complexities is the method of polynomials, which we now brie\ufb02y describe. Let us begin with some useful de\ufb01nitions. The deterministic query complexity D(F ) is the minimum number of oracle queries a classical computer must perform to compute F with certainty. The quantum equivalent, QE(F ), is the minimum number of oracle queries a quantum computer requires to compute F with certainty. Since a quantum computer produces probabilistic outputs by nature, a more interesting quantity is the bounded error complexity Q2(F ), the minimum number of oracle queries a quantum computer requires to produce an output which equals F with probability at least 2/3. (The 2/3 is an arbitrary number \u2013 the probability need only be bounded \ufb01nitely away from 1/2 in order to be boosted close to 1 by repetitions.) A related measure is the zeroerror complexity Q0(F ), the minimum number of oracle queries a quantum computer requires to produce an output which either equals F with certainty, or, with probability less than 1/2, an admission of an inconclusive result. All these bounds must hold for any oracle function f (or in other words, any input X into F ). Note that Q2(F ) \u2264 Q0(F ) \u2264 QE(F ) \u2264 D(F ) \u2264 N. The method of polynomials is based upon the properties of minimum-degree multilinear polynomials (over the real numbers) which represent Boolean functions. All the polynomials we shall consider below are functions of Xk \u2208 {0, 1} and are thus multik = Xk. We say that a", " polynomial p : RN \u2192 R represents F if p(X) = linear, since X 2 F (X) for all X \u2208 {0, 1}N (where R denotes the real numbers). Such a polynomial p always exists, since we can explicitly construct a suitable candidate: p(X) = N \u22121/ 1 \u2212 (Yk \u2212 Xk)2. F (Y ) Y \u2208{0,1}N k=0 (6.54) That the minimum degree p is unique is left as Exercise 6.18 for the reader. The minimum degree of such a representation for F, denoted as deg(F ), is a useful measure of the complexity of F. For example, it is known that deg(OR), deg(AND), and deg(PARITY) are all equal to N. In fact, it is known that the degree of most functions is of order N. Moreover, it has also been proven that D(F ) \u2264 2 deg(F )4. (6.55) This result places an upper bound on the performance of deterministic classical computation in calculating most Boolean functions. Extending this concept, if a polynomial satis\ufb01es |p(X)\u2212F (X)| \u2264 1/3 for all X \u2208 {0, 1}N, we say p approximates F, and 8deg(F ) denotes the minimum degree of such an approximating polynomial. Such measures are important in randomized classical computation and, as we shall see, in describing the quantum case. It is known that 8deg(PARITY) = N, N ) and 8deg(AND) \u2208 \u0398( N ), (6.56) \u221a 8deg(OR) \u2208 \u0398( and \u221a D(F ) \u2264 216 8deg(F )6. (6.57) Black box algorithm limits 273 The bounds of Equations (6.55) and (6.57) are only the best known at the time of writing; their proof is outside the scope of this book, but you may \ufb01nd further information about them in \u2018History and further reading\u2019. It is believed that tighter bounds are possible, but these will be good enough for our purposes. Exercise 6.18: Prove that the minimum degree polynomial representing a Boolean function F (X) is unique. Exercise 6.19: Show that P (X) = 1 \u2212 (1 \u2212 X0)(1 \u2212 X1)... (1 \u2212 XN \u22121) represents OR. Polynomials naturally arise in describing the results of quantum algorithms. Let us write the output of a quantum algorithm Q which performs T queries to an oracle O as 2n\u22121 k=0 ck|k. (6.58) We will show that the amplitudes ck are polynomials of degree at most T in the variables X0, X1,..., XN \u22121. Any Q can be realized using the quantum circuit shown in Figure 6.10. The state |\u03c80 right before the \ufb01rst oracle query can be written as \u2019 |\u03c80 = ij ai0j|i|0 + ai1j|i|1 |j, (6.59) ( where the \ufb01rst label corresponds to the n qubit oracle query, the next to a single qubit in which the oracle leaves its result, and the last to the m \u2212 n \u2212 1 working qubits used by Q. After the oracle query, we obtain the state |\u03c81 = ij ( \u2019 ai0j|i|Xi + ai1j|i|Xi \u2295 1 |j, (6.60) but since Xi is either 0 or 1, we can re-express this as \u2019 |\u03c81 = ij (1 \u2212 Xi)ai0j + Xiai1j |i0 + ( \u2019 (1 \u2212 Xi)ai1j + Xiai0j ( |i1 |j. (6.61) Note that in |\u03c80, the amplitudes of the computational basis states were of degree 0 in X, while those of |\u03c81 are of degree 1 (linear in X). The important observation is that any unitary operation which Q performs before or after the oracle query cannot change the degree of these polynomials, but each oracle call can increase the degree by at most one. Thus, after T queries, the amplitudes are polynomials of at most degree T. Moreover, measuring the \ufb01nal output (6.58) in the computational basis produces a result k with probability Pk(X) = |ck|2, which are real-valued polynomials in X of degree at", " most 2T. Figure 6.10. General quantum circuit for a quantum algorithm which performs T queries to an oracle O. U0, U1,..., UT are arbitrary unitary transforms on m qubits, and the oracle acts on n + 1 qubits. 274 Quantum search algorithms The total probability P (X) of obtaining a one as the output from the algorithm is a sum over some subset of the polynomials Pk(X), and thus also has degree at most 2T. In the case that Q produces the correct answer with certainty we must have P (X) = F (X), and thus deg(F ) \u2264 2T, from which we deduce QE(F ) \u2265 deg(F ) 2. (6.62) In the case where Q produces an answer with bounded probability of error it follows that P (X) approximates F (X), and thus 8deg(F ) \u2264 2T, from which we deduce Q2(F ) \u2265 8deg(F ) 2. Combining (6.55) and (6.62), we \ufb01nd that QE(F ) \u2265 1/4 D(F ) 32 Similarly, combining (6.57) and (6.63), we \ufb01nd that Q2(F ) \u2265 1/6 D(F ) 13 824.. (6.63) (6.64) (6.65) This means that in computing Boolean functions using a black box, quantum algorithms may only provide a polynomial speedup over classical algorithms, at best \u2013 and even that is not generally possible (since deg(F ) is \u03a9(N ) for most functions). On the other hand, it is known that for F = OR, D(F ) = N, and the randomized classical query complexity R(F ) \u2208 \u0398(N ), whereas combining (6.63) and (6.56), and the known performance of the quantum search algorithm, shows that Q2(F ) \u2208 \u0398( N ). This square root speedup is just what the quantum search algorithm achieves, and the method of polynomials indicates that the result can perhaps be generalized to a somewhat wider class of problems, but without extra information about the structure of the black box oracle function f, no exponential speedup over classical algorithms is possible. \u221a Exercise 6.20: Show that Q0(OR) \u2265 N by constructing a polynomial which represents the OR function from the output of a quantum circuit which computes OR with zero error. Problem 6.1: (Finding the minimum) Suppose x1,..., xN is a database of \u221a numbers held in memory, as in Section 6.5. Show that only O(log(N ) accesses to the memory are required on a quantum computer, in order to \ufb01nd the smallest element on the list, with probability at least one-half. N ) Problem 6.2: (Generalized quantum searching) Let |\u03c8 be a quantum state, and de\ufb01ne U|\u03c8 \u2261 I \u2212 2|\u03c8\u03c8|. That is, U|\u03c8 gives the state |\u03c8 a \u22121 phase, and leaves states orthogonal to |\u03c8 invariant. (1) Suppose we have a quantum circuit implementing a unitary operator U such that U |0\u2297n = |\u03c8. Explain how to implement U|\u03c8. Chapter problems 275 (2) Let |\u03c81 = |1, |\u03c82 = (|0 \u2212 |1)/ 2, |\u03c83 = (|0 \u2212 i|1)/ unknown oracle O is selected from the set U|\u03c81, U|\u03c82, U|\u03c83. Give a quantum algorithm which identi\ufb01es the oracle with just one application of the oracle. (Hint: consider superdense coding.) \u221a 2. Suppose an \u221a (3) Research: More generally, given k states |\u03c81,..., |\u03c8k, and an unknown oracle O selected from the set U|\u03c81,..., U|\u03c8k, how many oracle applications are required to identify the oracle, with high probability? Problem 6.3: (Database retrieval) Given a quantum oracle which returns |k, y \u2295 Xk given an n qubit query (and one scratchpad qubit) |k, y, show that with high probability, all N = 2n bits of X can be obtained using only N/2 + for any F. N queries. This implies the general upper bound Q2(F ) \u2264 N/2 + \u221a \u221a N Problem 6.4: (Quantum searching and cryptography) Quantum searching can, potentially, be used", " to speed up the search for cryptographic keys. The idea is to search through the space of all possible keys for decryption, in each case trying the key, and checking to see whether the decrypted message makes \u2018sense\u2019. Explain why this idea doesn\u2019t work for the Vernam cipher (Section 12.6). When might it work for cryptosystems such as DES? (For a description of DES see, for example, [MvOV96] or [Sch96a].) Summary of Chapter 6: Quantum search algorithms \u2022 Quantum search algorithm: For a search problem with M solutions out of x |x and then repeat G \u2261 H \u2297nU H \u2297nO a total N = 2n possibilities, prepare N/M ) times, where O is the search oracle, |x \u2192 \u2212|x if x is a solution, of O( no change otherwise, and U takes |0 \u2192 \u2212|0 and leaves all other computational basis states alone. Measuring yields a solution to the search problem with high probability. \u221a \u2022 Quantum counting algorithm: Suppose a search problem has an unknown number M of solutions. G has eigenvalues exp(\u00b1i\u03b8) where sin2(\u03b8/2) = M/N. The Fourier transform based phase estimation procedure enables us to estimate N ) oracle applications. Quantum counting, in turn, M to high accuracy using O( allows us to determine whether a given search problem has any solutions, and to \ufb01nd one if there are, even if the number of solutions is not known in advance. \u2022 Polynomial bounds: For problems which are described as evaluations of total functions F (as opposed to partial functions, or \u2018promise\u2019 problems), quantum algorithms can give no more than a polynomial speedup over classical algorithms. Speci\ufb01cally, Q2(F ) \u2265 tum search is optimal: it is \u0398( D(F )/13 824 \u221a N ).. Moreover, the performance of the quan- 1/6 276 Quantum search algorithms History and further reading The quantum search algorithm and much of its further development and elaboration is due to Grover[Gro96, Gro97]. Boyer, Brassard, H\u00f8yer and Tapp[BBHT98] wrote an in\ufb02uential paper in which they developed the quantum search algorithm for cases where the number of solutions M is greater than one, and outlined the quantum counting algorithm, later developed in more detail by Brassard, H\u00f8yer, and Tapp[BHT98], and from the point of view of phase estimation by Mosca[Mos98]. That the Grover iteration can be understood as a product of two re\ufb02ections was \ufb01rst pointed out in a review by Aharonov[Aha99b]. The continuous-time Hamiltonian (6.18) was \ufb01rst investigated by Farhi and Gutmann[FG98], from a rather different point of view than we take in Section 6.2. That Grover\u2019s algorithm is the best possible oracle-based search algorithm was proved by Bennett, Bernstein, Brassard and Vazirani[BBBV97]. The version of this proof we have presented is based upon that given by Boyer, Brassard, H\u00f8yer and Tapp[BBHT98]. Zalka[Zal99] has re\ufb01ned these proofs to show that the quantum search algorithm is, asymptotically, exactly optimal. The method of polynomials for bounding the power of quantum algorithms was introduced into quantum computing by Beals, Buhrman, Cleve, Mosca, and de Wolf[BBC+98]. An excellent discussion is also available in Mosca\u2019s Ph.D. thesis[Mos99], on which much of the discussion in Section 6.7 is based. A number of results are quoted in that section without proof; here are the citations: Equation (6.55) is attributed to Nisan and Smolensky in [BBC+98], but otherwise is presently unpublished, (6.56) is derived from a theorem by Paturi[Pat92] and (6.57) is derived in [BBC+98]. A better bound than (6.65) is given in [BBC+98], but requires concepts such as block sensitivity which are outside the scope of this book. A completely different approach for bounding quantum black box algorithms, using arguments based on entanglement, was presented by Ambainis[Amb00]. Problem 6.1 is due to D\u00a8urr and H\u00f8yer[DH96]. Problem 6.3 is due to van Dam[van98a]. 7 Quantum computers: physical realization Computers in the future may weigh no more than 1.5 tons. \u2013 Popular Mechanics, forecasting the relentless march of", " science, 1949 I think there is a world market for maybe \ufb01ve computers. \u2013 Thomas Watson, chairman of IBM, 1943 Quantum computation and quantum information is a \ufb01eld of fundamental interest because we believe quantum information processing machines can actually be realized in Nature. Otherwise, the \ufb01eld would be just a mathematical curiosity! Nevertheless, experimental realization of quantum circuits, algorithms, and communication systems has proven extremely challenging. In this chapter we explore some of the guiding principles and model systems for physical implementation of quantum information processing devices and systems. We begin in Section 7.1 with an overview of the tradeoffs in selecting a physical realization of a quantum computer. This discussion provides perspective for an elaboration of a set of conditions suf\ufb01cient for the experimental realization of quantum computation in Section 7.2. These conditions are illustrated in Sections 7.3 through 7.7, through a series of case studies, which consider \ufb01ve different model physical systems: the simple harmonic oscillator, photons and nonlinear optical media, cavity quantum electrodynamics devices, ion traps, and nuclear magnetic resonance with molecules. For each system, we brie\ufb02y describe the physical apparatus, the Hamiltonian which governs its dynamics, means for controlling the system to perform quantum computation, and its principal drawbacks. We do not go into much depth in describing the physics of these systems; as each of these are entire \ufb01elds unto themselves, that would be outside the scope of this book! Instead, we summarize just the concepts relevant to quantum computation and quantum information such that both the experimental challenge and theoretical potential can be appreciated. On the other hand, analyzing these systems from the standpoint of quantum information also provides a fresh perspective which we hope you will \ufb01nd insightful and useful, as it also allows strikingly simple derivations of some important physics. We conclude the chapter in Section 7.8 by discussing aspects of some other physical systems \u2013 quantum dots, superconducting gates, and spins in semiconductors \u2013 which are also of interest to this \ufb01eld. For the bene\ufb01t of the reader wishing to catch just the highlights of each implementation, a summary is provided at the end of each section. 7.1 Guiding principles What are the experimental requirements for building a quantum computer? The elementary units of the theory are quantum bits \u2013 two-level quantum systems; in Section 1.5 we took a brief look at why it is believed that qubits exist in Nature, and what physical forms they may take on. To realize a quantum computer, we must not only give qubits 278 Quantum computers: physical realization some robust physical representation (in which they retain their quantum properties), but also select a system in which they can be made to evolve as desired. Furthermore, we must be able to prepare qubits in some speci\ufb01ed set of initial states, and to measure the \ufb01nal output state of the system. The challenge of experimental realization is that these basic requirements can often only be partially met. A coin has two states, and makes a good bit, but a poor qubit because it cannot remain in a superposition state (of \u2018heads\u2019 and \u2018tails\u2019) for very long. A single nuclear spin can be a very good qubit, because superpositions of being aligned with or against an external magnetic \ufb01eld can last a long time \u2013 even for days. But it can be dif\ufb01cult to build a quantum computer from nuclear spins because their coupling to the world is so small that it is hard to measure the orientation of single nuclei. The observation that the constraints are opposing is general: a quantum computer has to be well isolated in order to retain its quantum properties, but at the same time its qubits have to be accessible so that they can be manipulated to perform a computation and to read out the results. A realistic implementation must strike a delicate balance between these constraints, so that the relevant question is not how to build a quantum computer, but rather, how good a quantum computer can be built. System Nuclear spin Electron spin Ion trap (In+) Electron \u2013 Au Electron \u2013 GaAs Quantum dot Optical cavity Microwave cavity \u03c4Q 10\u22122 \u2212 108 10\u22123 10\u22121 10\u22128 10\u221210 10\u22126 10\u22125 100 \u03c4op 10\u22123 \u2212 10\u22126 10\u22127 10\u221214 10\u221214 10\u221213 10\u22129 10\u221214 10\u22124 nop = \u03bb\u22121 105 \u2212 1014 104 1013 106 103 103 109 104 Figure 7.1. Crude estimates for decoherence times \u03c4Q (seconds), operation times \u03c4op (seconds), and maximum number of operations nop = \u03bb\u22121 = \u03c4Q/\u03c4op for various candidate physical realizations of interacting systems of quantum bits. Despite the number of entries in this table, only three fundamentally different qubit representations are given:", " spin, charge, and photon. The ion trap utilizes either \ufb01ne or hyper\ufb01ne transitions of a trapped atom (Section 7.6), which correspond to electron and nuclear spin \ufb02ips. The estimates for electrons in gold and GaAs, and in quantum dots are given for a charge representation, with an electrode or some con\ufb01ned area either containing an electron or not. In optical and microwave cavities, photons (of frequencies from gigahertz to hundreds of terahertz) populating different modes of the cavities represent the qubit. Take these estimates with a grain of salt: they are only meant to give some perspective on the wide range of possibilities. What physical systems are potentially good candidates for handling quantum information? A key concept in understanding the merit of a particular quantum computer realization is the notion of quantum noise (sometimes called decoherence), the subject of Chapter 8: processes corrupting the desired evolution of the system. This is because the length of the longest possible quantum computation is roughly given by the ratio of \u03c4Q, the time for which a system remains quantum-mechanically coherent, to \u03c4op, the time it takes to perform elementary unitary transformations (which involve at least two qubits). These two times are actually related to each other in many systems, since they are both Conditions for quantum computation 279 determined by the strength of coupling of the system to the external world. Nevertheless, \u03bb = \u03c4op/\u03c4Q can vary over a surprisingly wide range, as shown in Figure 7.1. These estimates give some insight into the merits of different possible physical realizations of a quantum information processing machine, but many other important sources of noise and imperfections arise in actual implementations. For example, manipulations of a qubit represented by two electronic levels of an atom by using light to cause transitions between levels would also cause transitions to other electronic levels with some probability. These would also be considered noise processes, since they take the system out of the two states which de\ufb01ne the qubit. Generally speaking, anything which causes loss of (quantum) information is a noise process \u2013 later, in Chapter 8, we discuss the theory of quantum noise in more depth. 7.2 Conditions for quantum computation Let us return to discuss in detail the four basic requirements for quantum computation which were mentioned at the beginning of the previous section. These requirements are the abilities to: 1. Robustly represent quantum information 2. Perform a universal family of unitary transformations 3. Prepare a \ufb01ducial initial state 4. Measure the output result 7.2.1 Representation of quantum information Quantum computation is based on transformation of quantum states. Quantum bits are two-level quantum systems, and as the simplest elementary building blocks for a quantum computer, they provide a convenient labeling for pairs of states and their physical realizations. Thus, for example, the four states of a spin-3/2 particle, |m = +3/2, |m = +1/2, |m = \u22121/2, |m = \u22123/2, could be used to represent two qubits. For the purpose of computation, the crucial realization is that the set of accessible states should be \ufb01nite. The position x of a particle along a one-dimensional line is not generally a good set of states for computation, even though the particle may be in a quantum state x cx|x. This is because x has a continuous range |x, or even some superposition of possibilities, and the Hilbert space has in\ufb01nite size, so that in the absence of noise the information capacity is in\ufb01nite. For example, in a perfect world, the entire texts of Shakespeare could be stored in (and retrieved from) the in\ufb01nite number of digits in the binary fraction x = 0.010111011001.... This is clearly unrealistic; what happens in reality is that the presence of noise reduces the number of distinguishable states to a \ufb01nite number. In fact, it is generally desirable to have some aspect of symmetry dictate the \ufb01niteness of the state space, in order to minimize decoherence. For example, a spin-1/2 particle lives in a Hilbert space spanned by the | \u2191 and | \u2193 states; the spin state cannot be anything outside this two-dimensional space, and thus is a nearly ideal quantum bit when well isolated. If the choice of representation is poor, then decoherence will result. For example, as described in Box 7.1, a particle in a \ufb01nite square well which is just deep enough to contain two bound states would make a mediocre quantum bit, because transitions from 280 Quantum computers: physical realization the bound states to the continuum of unbound states would be possible. These would lead to decoherence since they could destroy qubit superposition states. For single qubits, the \ufb01", "gure of merit is the minimum lifetime of arbitrary superposition states; a good measure, used for spin states and atomic systems, is T2, the (\u2018transverse\u2019) relaxation time of states 2. Note that T1, the (\u2018longitudinal\u2019) relaxation time of the higher such as (|0 + |1)/ energy |1 state, is just a classical state lifetime, which is usually longer than T2. \u221a Box 7.1: Square wells and qubits A prototypical quantum system is known as the \u2018square well,\u2019 which is a particle in a one-dimensional box, behaving according to Schr\u00a8odinger\u2019s equation, (2.86). The Hamiltonian for this system is H = p2/2m + V (x), where V (x) = 0 for 0 < x < L, and V (x) = \u221e otherwise. The energy eigenstates, expressed as wavefunctions in the position basis, are ( \u2019 n\u03c0 L |\u03c8n = 2 L sin where n is an integer, and |\u03c8n(t) = e\u2212iEnt|\u03c8n, with En = n2\u03c02m/2L2. These states have a discrete spectrum. In particular, suppose that we arrange matters such that only the two lowest energy levels need be considered in an experiment. We de\ufb01ne an arbitrary wavefunction of interest as |\u03c8 = a |\u03c81 + b |\u03c82. Since (7.1) x, |\u03c8(t) = e\u2212i(E1+E2)/2t ae\u2212i\u03c9t|\u03c81 + bei\u03c9t|\u03c82, (7.2) ) where \u03c9 = (E1 \u2212 E2)/2, we can just forget about everything except a and b, and write our state abstractly as the two-component vector |\u03c8 =. This two-level a b system represents a qubit! Does our two-level system transform like a qubit? Under time evolution, this qubit evolves under the effective Hamiltonian H = \u03c9Z, which can be disregarded by moving into the rotating frame. To perform operations to this qubit, we perturb H. Consider the effect of adding the additional term \u03b4V (x) = \u2212V0(t) 9\u03c02 16L x L \u2212 1 2 (7.3) to V (x). In the basis of our two-level system, this can be rewritten by taking the matrix elements Vnm = \u03c8n|\u03b4V (x)|\u03c8m, giving V11 = V22 = 0, and V12 = V21 = V0, such that, to lowest order in V0, the perturbation to H is H1 = V0(t)X. This generates rotations about the \u02c6x axis. Similar techniques can be used to perform other single qubit operations, by manipulating the potential function. This shows how a single qubit can be represented by the two lowest levels in a square well potential, and how simple perturbations of the potential can effect computational operations on the qubit. However, perturbations also introduce higher order effects, and in real physical systems boxes are not in\ufb01nitely deep, other levels begin to enter the picture, and our two-level approximation begins to fail. Also, in reality, the controlling system is just another quantum system, and it couples to the one we are trying to do quantum computation with. These problems lead to decoherence. Conditions for quantum computation 281 7.2.2 Performance of unitary transformations Closed quantum systems evolve unitarily as determined by their Hamiltonians, but to perform quantum computation one must be able to control the Hamiltonian to effect an arbitrary selection from a universal family of unitary transformations (as described in Section 4.5). For example, a single spin might evolve under the Hamiltonian H = Px(t)X + Py(t)Y, where P{x,y} are classically controllable parameters. From Exercise 4.10, we know that by manipulating Px and Py appropriately, one can perform arbitrary single spin rotations. According to the theorems of Section 4.5, any unitary transform can be composed from single spin operations and controlledgates, and thus realization of those two kinds of quantum logic gates are natural goals for experimental quantum computation. However, implicitly required also is the ability to address individual qubits, and to apply these gates to select qubits or pairs of qubits. This is not simple to accomplish in many physical systems. For example, in an ion trap, one can direct a laser at one of many individual ions to selectively excite it, but only as long as the ions are spatially separated by a wavelength or more. Unrecorded imperfections", " in unitary transforms can lead to decoherence. In Chapter 8 we shall see how the average effect of random kicks (small rotations to a single spin about its \u02c6z axis) leads to loss of quantum information which is represented by the relative phases in a quantum state. Similarly, the cumulative effect of systematic errors is decoherence, when the information needed to be able to reverse them is lost. Furthermore, the control parameters in the Hamiltonian are only approximately classical controls: in reality, the controlling system is just another quantum system, and the true Hamiltonian should include the back-action of the control system upon the quantum computer. For example, instead of Px(t) in the above example, one actually has a Jaynes\u2013Cummings type atom\u2013 photon interaction Hamiltonian (Section 7.5.2), with Px(t) = k) or something similar being the cavity photon \ufb01eld. After interacting with a qubit, a photon can carry away information about the state of the qubit, and this is thus a decoherence process. k \u03c9k(t)(ak + a\u2020 Two important \ufb01gures of merit for unitary transforms are the minimum achievable \ufb01delity F (Chapter 9), and the maximum time top required to perform elementary operations such as single spin rotations or a controlled- gate. 7.2.3 Preparation of \ufb01ducial initial states One of the most important requirements for being able to perform a useful computation, even classically, is to be able to prepare the desired input. If one has a box which can perform perfect computations, what use is it if numbers cannot be input? With classical machines, establishing a de\ufb01nite input state is rarely a dif\ufb01culty \u2013 one merely sets some switches in the desired con\ufb01guration and that de\ufb01nes the input state. However, with quantum systems this can be very dif\ufb01cult, depending on the realization of qubits. Note that it is only necessary to be able to (repeatedly) produce one speci\ufb01c quantum state with high \ufb01delity, since a unitary transform can turn it into any other desired input state. For example, being able to put n spins into the |00... 0 state is good enough. The fact that they may not stay there for very long due to thermal heating is a problem with the choice of representation. Input state preparation is a signi\ufb01cant problem for most physical systems. For example, ions can be prepared in good input states by physically cooling them into their ground state 282 Quantum computers: physical realization (Section 7.6), but this is challenging. Moreover, for physical systems in which ensembles of quantum computers are involved, extra concerns arise. In nuclear magnetic resonance (Section 7.7), each molecule can be thought of as a single quantum computer, and a large number of molecules is needed to obtain a measurable signal strength. Although qubits can remain in arbitrary superposition states for relatively long times, it is dif\ufb01cult to put all of the qubits in all of the molecules into the same state, because the energy difference \u03c9 between the |0 and |1 states is much smaller than kBT. On the other hand, simply letting the system equilibrate establishes it in a very well-known state, the thermal one, with the density matrix \u03c1 \u2248 e\u2212H/kBT /Z, where Z is a normalization factor required to maintain tr(\u03c1) = 1. Two \ufb01gures of merit are relevant to input state preparation: the minimum \ufb01delity with which the initial state can be prepared in a given state \u03c1in, and the entropy of \u03c1in. The entropy is important because, for example, it is very easy to prepare the state \u03c1in = I/2n with high \ufb01delity, but that is a useless state for quantum computation, since it is invariant under unitary transforms! Ideally, the input state is a pure state, with zero entropy. Generally, input states with non-zero entropy reduce the accessibility of the answer from the output result. 7.2.4 Measurement of output result What measurement capability is required for quantum computation? For the purpose of the present discussion, let us think of measurement as a process of coupling one or more qubits to a classical system such that after some interval of time, the state of the qubits is indicated by the state of the classical system. For example, a qubit state a|0 + b|1, represented by the ground and excited states of a two-level atom, might be measured by pumping the excited state and looking for \ufb02uorescence. If an electrometer indicates that \ufb02uorescence had been detected by a photomultiplier tube,", " then the qubit would collapse into the |1 state; this would happen with probability |b|2. Otherwise, the electrometer would detect no charge, and the qubit would collapse into the |0 state. An important characteristic of the measurement process for quantum computation is the wavefunction collapse which describes what happens when a projective measurement is performed (Section 2.2.5). The output from a good quantum algorithm is a superposition state which gives a useful answer with high probability when measured. For example, one step in Shor\u2019s quantum factoring algorithm is to \ufb01nd an integer r from the measurement result, which is an integer close to qc/r, where q is the dimension of a Hilbert space. The output state is actually in a nearly uniform superposition of all possible values of c, but a measurement collapses this into a single, random integer, thus allowing r to be determined with high probability (using a continued fraction expansion, as was described in Chapter 5). Many dif\ufb01culties with measurement can be imagined; for example, inef\ufb01cient photon counters and ampli\ufb01er thermal noise can reduce the information obtained about measured qubit states in the scheme just described. Furthermore, projective measurements (sometimes called \u2018strong\u2019 measurements) are often dif\ufb01cult to implement. They require that the coupling between the quantum and classical systems be large, and switchable. Measurements should not occur when not desired; otherwise they can be a decoherence process. Surprisingly, however, strong measurements are not necessary; weak measurements which are performed continuously and never switched off, are usable for quantum com- Harmonic oscillator quantum computer 283 putation. This is made possible by completing the computation in time short compared with the measurement coupling, and by using large ensembles of quantum computers. These ensembles together give an aggregate signal which is macroscopically observable and indicative of the quantum state. Use of an ensemble introduces additional problems. For example, in the factoring algorithm, if the measurement output is qc/r, the algorithm would fail because c, the average value of c, is not necessarily an integer (and thus the continued fraction expansion would not be possible). Fortunately, it is possible to modify quantum algorithms to work with ensemble average readouts. This will be discussed further in Section 7.7. A good \ufb01gure of merit for measurement capability is the signal to noise ratio (SNR). This accounts for measurement inef\ufb01ciency as well as inherent signal strength available from coupling a measurement apparatus to the quantum system. 7.3 Harmonic oscillator quantum computer Before continuing on to describe a complete physical model for a realizable quantum computer, let us pause for a moment to consider a very elementary system \u2013 the simple harmonic oscillator \u2013 and discuss why it does not serve as a good quantum computer. The formalism used in this example will also serve as a basis for studying other physical systems. 7.3.1 Physical apparatus An example of a simple harmonic oscillator is a particle in a parabolic potential well, V (x) = m\u03c92x2/2. In the classical world, this could be a mass on a spring, which oscillates back and forth as energy is transfered between the potential energy of the spring and the kinetic energy of the mass. It could also be a resonant electrical circuit, where the energy sloshes back and forth between the inductor and the capacitor. In these systems, the total energy of the system is a continuous parameter. In the quantum domain, which is reached when the coupling to the external world becomes very small, the total energy of the system can only take on a discrete set of values. An example is given by a single mode of electromagnetic radiation trapped in a high Q cavity; the total amount of energy (up to a \ufb01xed offset) can only be integer multiples of \u03c9, an energy scale which is determined by the fundamental constant and the frequency of the trapped radiation, \u03c9. The set of discrete energy eigenstates of a simple harmonic oscillator can be labeled as |n, where n = 0, 1,..., \u221e. The relationship to quantum computation comes by taking a \ufb01nite subset of these states to represent qubits. These qubits will have lifetimes determined by physical parameters such as the cavity quality factor Q, which can be made very large by increasing the re\ufb02ectivity of the cavity walls. Moreover, unitary transforms can be applied by simply allowing the system to evolve in time. However, there are problems with this scheme, as will become clear below. We begin by studying the system Hamiltonian, then discuss how one might implement simple quantum logic gates such as the controlled-. 284 Quantum computers: physical realization The Hamiltonian for a particle in a one-dimensional parabolic potential is 7.3.2 The Hamiltonian H =", " p2 2m + 1 2 m\u03c92x2, (7.4) where p is the particle momentum operator, m is the mass, x is the position operator, and \u03c9 is related to the potential depth. Recall that x and p are operators in this expression (see Box 7.2), which can be rewritten as H = \u03c9 a\u2020a +, 1 2 where a\u2020 and a are creation and annihilation operators, de\ufb01ned as \u2019 \u2019 a = a\u2020 = 1\u221a 2m\u03c9 1\u221a 2m\u03c9 m\u03c9x + ip ( ( m\u03c9x \u2212 ip. (7.5) (7.6) (7.7) The zero point energy \u03c9/2 contributes an unobservable overall phase factor, which can be disregarded for our present purpose. The eigenstates |n of H, where n = 0, 1,..., have the properties \u221a a\u2020a|n = n|n a\u2020|n = a|n = \u221a n + 1 |n + 1 n |n \u2212 1. (7.10) (7.11) (7.12) Later, we will \ufb01nd it convenient to express interactions with a simple harmonic oscillator by introducing additional terms involving a and a\u2020, and interactions between oscillators with terms such as a\u2020 2. For now, however, we con\ufb01ne our attention to a single oscillator. a2 + a1a\u2020 1 Exercise 7.1: Using the fact that x and p do not commute, and that in fact [x, p] = i, explicitly show that a\u2020a = H/\u03c9 \u2212 1/2. Exercise 7.2: Given that [x, p] = i, compute [a, a\u2020]. Exercise 7.3: Compute [H, a] and use the result to show that if |\u03c8 is an eigenstate of H with energy E \u2265 n\u03c9, then an|\u03c8 is an eigenstate with energy E \u2212 n\u03c9. Exercise 7.4: Show that |n = (a\u2020)n n! \u221a |0. Exercise 7.5: Verify that Equations (7.11) and (7.12) are consistent with (7.10) and the normalization condition n|n = 1. Time evolution of the eigenstates is given by solving the Schr\u00a8odinger equation, (2.86), from which we \ufb01nd that the state |\u03c8(0) = n cn(0)|n evolves in time to become |\u03c8(t) = e\u2212iHt/|\u03c8(0) = cne\u2212in\u03c9t|n. (7.13) We will assume for the purpose of discussion that an arbitrary state can be perfectly prepared, and that the state of the system can be projectively measured (Section 2.2.3), n Harmonic oscillator quantum computer 285 Box 7.2: The quantum harmonic oscillator The harmonic oscillator is an extremely important and useful concept in the quantum description of the physical world, and a good way to begin to understand its properties is to determine the energy eigenstates of its Hamiltonian, (7.4). One way to do this is simply to solve the Schr\u00a8odinger equation 2 2m d2\u03c8n(x) dx2 + 1 2 m\u03c92x2\u03c8n(x) = E\u03c8n(x) for \u03c8n(x) and the eigenenergies E, subject to \u03c8(x) \u2192 0 at x = \u00b1\u221e, and 1; the \ufb01rst \ufb01ve solutions are sketched here: (7.8) 7 |\u03c8(x)|2 = These wavefunctions describe the probability amplitudes that a particle in the harmonic oscillator will be found at different positions within the potential. Although these pictures may give some intuition about what a physical system is doing in co-ordinate space, we will generally be more interested in the abstract algebraic properties of the states. Speci\ufb01cally, suppose |\u03c8 satis\ufb01es (7.8) with energy E. Then de\ufb01ning operators a and a\u2020 as in (7.6)\u2013(7.7), we \ufb01nd that since [H, a\u2020] = \u03c9a\u2020, \u2019 ( Ha\u2020|\u03c8 = [H, a\u2020] + a\u2020H |\u03c8 = (\u03c9 + E)a\u2020|\u03c8, (7.9) that is, a\u2020|\u03c8 is an eigenstate of H, with energy E + \u03c9! Similarly, a|\u03c8 is an eigenstate with energy E \u2212 \u03c9. Because of this, a\u2020 and a are called raising", " and lowering operators. It follows that a\u2020n|\u03c8 are eigenstates for any integer n, with energies E + n\u03c9. There are thus an in\ufb01nite number of energy eigenstates, whose energies are equally spaced apart, by \u03c9. Moreover, since H is positive de\ufb01nite, there must be some |\u03c80 for which a|\u03c80 = 0; this is the ground state \u2013 the eigenstate of H with lowest energy. These results ef\ufb01ciently capture the essence of the quantum harmonic oscillator, and allow us to use a compact notation |n for the eigenstates, where n is an integer, and H|n = (n + 1/2)|n. We shall often work with |n, a, and a\u2020 in this chapter, as harmonic oscillators arise in the guise of many different physical systems. 286 Quantum computers: physical realization but otherwise, there are no interactions with the external world, so that the system is perfectly closed. 7.3.3 Quantum computation Suppose we want to perform quantum computation with the single simple harmonic oscillator described above. What can be done? The most natural choice for representation of qubits are the energy eigenstates |n. This choice allows us to perform a controlledgate in the following way. Recall that this transformation performs the mapping |00L \u2192 |00L |01L \u2192 |01L |10L \u2192 |11L |11L \u2192 |10L, on two qubit states (here, the subscript L is used to clearly distinguish \u2018logical\u2019 states in contrast to the harmonic oscillator basis states). Let us encode these two qubits using the mapping (7.14) |00L = |0 |01L = |2 |10L = (|4 + |1)/ |11L = (|4 \u2212 |1)/ \u221a 2 \u221a 2. (7.15) Now suppose that at t = 0 the system is started in a state spanned by these basis states, and we simply evolve the system forward to time t = \u03c0/\u03c9. This causes the energy eigenstates to undergo the transformation |n \u2192 exp(\u2212i\u03c0a\u2020a)|n = (\u22121)n|n, such that |0, |2, and |4 stay unchanged, but |1 \u2192 \u2212|1. As a result, we obtain the desired controlled- gate transformation. In general, a necessary and suf\ufb01cient condition for a physical system to be able to perform a unitary transform U is simply that the time evolution operator for the system, T = exp(\u2212iHt), de\ufb01ned by its Hamiltonian H, has nearly the same eigenvalue spectrum as U. In the case above, the controlledgate was simple to implement because it only has eigenvalues +1 and \u22121; it was straightforward to arrange an encoding to obtain the same eigenvalues from the time evolution operator for the harmonic oscillator. The Hamiltonian for an oscillator could be perturbed to realize nearly any eigenvalue spectrum, and any number of qubits could be represented by simply mapping them into the in\ufb01nite number of eigenstates of the system. This suggests that perhaps one might be able to realize an entire quantum computer in a single simple harmonic oscillator! 7.3.4 Drawbacks Of course, there are many problems with the above scenario. Clearly, one will not always know the eigenvalue spectrum of the unitary operator for a certain quantum computation, even though one may know how to construct the operator from elementary gates. In fact, for most problems addressed by quantum algorithms, knowledge of the eigenvalue spectrum is tantamount to knowledge of the solution! Another obvious problem is that the technique used above does not allow one computation to be cascaded with another, because in general, cascading two unitary transforms results in a new transform with unrelated eigenvalues. Finally, the idea of using a single harmonic oscillator to perform quantum computation Optical photon quantum computer 287 is \ufb02awed because it neglects the principle of digital representation of information. A Hilbert space of 2n dimensions mapped into the state space of a single harmonic oscillator would have to allow for the possibility of states with energy 2n\u03c9. In contrast, the same Hilbert space could be obtained by using n two-level quantum systems, which has an energy of at most n\u03c9. Similar comparisons can be made between a classical dial with 2n settings, and a register of n classical bits. Quantum computation builds upon digital computation, not analog computation. The main features of the harmonic oscillator quantum computer are summarized below (each system we consider will be summarized similarly, at the end of the corresponding section). With this, we leave behind us the study of single oscillators, and turn next to systems of harmonic oscillators, made of photons and atoms. Harm", "onic oscillator quantum computer \u2022 Qubit representation: Energy levels |0, |1,..., |2n of a single quantum oscillator give n qubits. \u2022 Unitary evolution: Arbitrary transforms U are realized by matching their eigenvalue spectrums to that given by the Hamiltonian H = a\u2020a. \u2022 Initial state preparation: Not considered. \u2022 Readout: Not considered. \u2022 Drawbacks: Not a digital representation! Also, matching eigenvalues to realize transformations is not feasible for arbitrary U, which generally have unknown eigenvalues. 7.4 Optical photon quantum computer An attractive physical system for representing a quantum bit is the optical photon. Photons are chargeless particles, and do not interact very strongly with each other, or even with most matter. They can be guided along long distances with low loss in optical \ufb01bers, delayed ef\ufb01ciently using phase shifters, and combined easily using beamsplitters. Photons exhibit signature quantum phenomena, such as the interference produced in two-slit experiments. Furthermore, in principle, photons can be made to interact with each other, using nonlinear optical media which mediate interactions. There are problems with this ideal scenario; nevertheless, many things can be learned from studying the components, architecture, and drawbacks of an optical photon quantum information processor, as we shall see in this section. 7.4.1 Physical apparatus Let us begin by considering what single photons are, how they can represent quantum states, and the experimental components useful for manipulating photons. The classical behavior of phase shifters, beamsplitters, and nonlinear optical Kerr media is described. Photons can represent qubits in the following manner. As we saw in the discussion of the simple harmonic oscillator, the energy in an electromagnetic cavity is quantized in units of \u03c9. Each such quantum is called a photon. It is possible for a cavity to contain a superposition of zero or one photon, a state which could be expressed as a qubit c0|0 + c1|1, but we shall do something different. Let us consider two cavities, whose total energy is \u03c9, and take the two states of a qubit as being whether the photon is in 288 Quantum computers: physical realization one cavity (|01) or the other (|10). The physical state of a superposition would thus be written as c0|01 + c1|10; we shall call this the dual-rail representation. Note that we shall focus on single photons traveling as a wavepacket through free space, rather than inside a cavity; one can imagine this as having a cavity moving along with the wavepacket. Each cavity in our qubit state will thus correspond to a different spatial mode. One scheme for generating single photons in the laboratory is by attenuating the output of a laser. A laser outputs a state known as a coherent state, |\u03b1, de\ufb01ned as |\u03b1 = e\u2212|\u03b1|2/2 \u221e n=0 \u03b1n \u221a n! |n, (7.16) where |n is an n-photon energy eigenstate. This state, which has been the subject of thorough study in the \ufb01eld of quantum optics, has many beautiful properties which we shall not describe here. It suf\ufb01ces to understand just that coherent states are naturally radiated from driven oscillators such as a laser when pumped high above its lasing threshold. Note that the mean energy is \u03b1|n|\u03b1 = |\u03b1|2. When attenuated, a coherent state just becomes a weaker coherent state, and a weak coherent state can be made to have just one photon, with high probability. Exercise 7.6: (Eigenstates of photon annihilation) Prove that a coherent state is an eigenstate of the photon annihilation operator, that is, show a|\u03b1 = \u03bb|\u03b1 for some constant \u03bb. \u221a \u221a \u221a \u221a For example, for \u03b1 = 0.1, we obtain the state 0.002 |2+\u00b7 \u00b7 \u00b7. Thus if light ever makes it through the attenuator, one knows it is a single photon with probability better than 95%; the failure probability is thus 5%. Note also that 90% of the time, no photons come through at all; this source thus has a rate of 0.1 photons per unit time. Finally, this source does not indicate (by means of some classical readout) when a photon has been output or not; two of these sources cannot be synchronized. 0.09 |1+ 0.90 |0+ Better synchronicity can be achieved using parametric down-conversion. This involves sending photons of frequency \u03c90 into a nonlinear optical medium such as KH2PO4 to generate photon pairs at frequencies \u03c91 + \u03c92 = \u03c90. Momentum is also conserved, such that k1 +k2 =", " k3, so that when a single \u03c92 photon is (destructively) detected, then a single \u03c91 photon is known to exist (Figure 7.2). By coupling this to a gate, which is opened only when a single photon (as opposed to two or more) is detected, and by appropriately delaying the outputs of multiple down-conversion sources, one can, in principle, obtain multiple single photons propagating in time synchronously, within the time resolution of the detector and gate. Single photons can be detected with high quantum ef\ufb01ciency for a wide range of wavelengths, using a variety of technologies. For our purposes, the most important characteristic of a detector is its capability of determining, with high probability, whether zero or one photon exists in a particular spatial mode. For the dual-rail representation, this translates into a projective measurement in the computational basis. In practice, imperfections reduce the probability of being able to detect a single photon; the quantum ef\ufb01ciency \u03b7 (0 \u2264 \u03b7 \u2264 1) of a photodetector is the probability that a single photon incident on the detector generates a photocarrier pair that contributes to detector current. Other important characteristics of a detector are its bandwidth (time responsivity), noise, and \u2018dark counts\u2019 which are photocarriers generated even when no photons are incident. Optical photon quantum computer 289 9: ;: Figure 7.2. Parametric down-conversion scheme for generation of single photons. Three of the most experimentally accessible devices for manipulating photon states are mirrors, phase shifters and beamsplitters. High re\ufb02ectivity mirrors re\ufb02ect photons and change their propagation direction in space. Mirrors with 0.01% loss are not unusual. We shall take these for granted in our scenario. A phase shifter is nothing more than a slab of transparent medium with index of refraction n different from that of free space, n0; for example, ordinary borosilicate glass has n \u2248 1.5n0 at optical wavelengths. Propagation in such a medium through a distance L changes a photon\u2019s phase by eikL, where k = n\u03c9/c0, and c0 is the speed of light in vacuum. Thus, a photon propagating through a phase shifter will experience a phase shift of ei(n\u2212n0)L\u03c9/c0 compared to a photon going the same distance through free space. Another useful component, the beamsplitter, is nothing more than a partially silvered piece of glass, which re\ufb02ects a fraction R of the incident light, and transmits 1\u2212R. In the laboratory, a beamsplitter is usually fabricated from two prisms, with a thin metallic layer sandwiched in-between, schematically drawn as shown in Figure 7.3. It is convenient to de\ufb01ne the angle \u03b8 of a beamsplitter as cos \u03b8 = R; note that the angle parameterizes the amount of partial re\ufb02ection, and does not necessarily have anything to do with the physical orientation of the beamsplitter. The two inputs and two outputs of this device are related by aout = ain cos \u03b8 + bin sin \u03b8 bout = \u2212ain sin \u03b8 + bin cos \u03b8, (7.17) (7.18) where classically we think of a and b as being the electromagnetic \ufb01elds of the radiation at the two ports. Note that in this de\ufb01nition we have chosen a non-standard phase convention convenient for our purposes. In the special case of a 50/50 beamsplitter, \u03b8 = 45\u25e6. Nonlinear optics provides one \ufb01nal useful component for this exercise: a material Figure 7.3. Schematic of an optical beamsplitter, showing the two input ports, the two output ports, and the phase conventions for a 50/50 beamsplitter (\u03b8 = \u03c0/4). The beamsplitter on the right is the inverse of the one on the left (the two are distinguished by the dot drawn inside). The input-output relations for the mode operators a and b are given for \u03b8 = \u03c0/4. 290 Quantum computers: physical realization whose index of refraction n is proportional to the total intensity I of light going through it: n(I) = n + n2I. (7.19) This is known as the optical Kerr effect, and it occurs (very weakly) in materials as mundane as glass and sugar water. In doped glasses, n2 ranges from 10\u221214 to 10\u22127 cm2/W, and in semiconductors, from 10\u221210 to 102. Experimentally, the relevant behavior is that when two beams of light of equal intensity are nearly co-propagated through a Kerr medium, each beam will experience an extra phase", " shift of ein2IL\u03c9/c0 compared to what happens in the single beam case. This would be ideal if the length L could be arbitrarily long, but unfortunately that fails because most Kerr media are also highly absorptive, or scatter light out of the desired spatial mode. This is the primary reason why a single photon quantum computer is impractical, as we shall discuss in Section 7.4.3. We turn next to a quantum description of these optical components. 7.4.2 Quantum computation Arbitrary unitary transforms can be applied to quantum information, encoded with single photons in the c0|01+c1|10 dual-rail representation, using phase shifters, beamsplitters, and nonlinear optical Kerr media. How this works can be understood in the following manner, by giving a quantum-mechanical Hamiltonian description of each of these devices. The time evolution of a cavity mode of electromagnetic radiation is modeled quantummechanically by a harmonic oscillator, as we saw in Section 7.3.2. |0 is the vacuum state, |1 = a\u2020|0 is a single photon state, and in general, |n = a\u2020 n |0 is an n-photon state, n! where a\u2020 is the creation operator for the mode. Free space evolution is described by the Hamiltonian \u221a H = \u03c9a\u2020a, (7.20) and applying (7.13), we \ufb01nd that the state |\u03c8 = c0|0 + c1|1 evolves in time to become |\u03c8(t) = c0|0 + c1e\u2212i\u03c9t|1. Note that the dual-rail representation is convenient because free evolution only changes |\u03d5 = c0|01 + c1|10 by an overall phase, which is undetectable. Thus, for that manifold of states, the evolution Hamiltonian is zero. Phase shifter. A phase shifter P acts just like normal time evolution, but at a different rate, and localized to only the modes going through it. That is because light slows down in a medium with larger index of refraction; speci\ufb01cally, it takes \u0394 \u2261 (n \u2212 n0)L/c0 more time to propagate a distance L in a medium with index of refraction n than in vacuum. For example, the action of P on the vacuum state is to do nothing: P |0 = |0, but on a single photon state, one obtains P |1 = ei\u0394|1. P performs a useful logical operation on a dual-rail state. Placing a phase shifter in one mode retards its phase evolution with respect to another mode, which travels the same distance but without going through the shifter. For dual-rail states this transforms c0|01 + c1|10 to c0e\u2212i\u0394/2|01 + c1ei\u0394/2|10, up to an irrelevant overall phase. Recall from Section 4.2 that this operation is nothing more than a rotation, Rz(\u0394) = e\u2212iZ\u0394/2, (7.21) where we take as the logical zero |0L = |01 and one |1L = |10, and Z is the usual Optical photon quantum computer 291 Pauli operator. One can thus think of P as resulting from time evolution under the Hamiltonian H = (n0 \u2212 n)Z, (7.22) where P = exp(\u2212iHL/c0). Exercise 7.7: Show that the circuit below transforms a dual-rail state by |\u03c8out = |\u03c8in, (7.23) ei\u03c0 0 0 1 if we take the top wire to represent the |01 mode, and |10 the bottom mode, and the boxed \u03c0 to represent a phase shift by \u03c0: Note that in such \u2018optical circuits\u2019, propagation in space is explicitly represented by putting in lumped circuit elements such as in the above, to represent phase evolution. In the dual-rail representation, evolution according to (7.20) changes the logical state only by an unobservable global phase, and thus we are free to disregard it and keep only relative phase shifts. Exercise 7.8: Show that P |\u03b1 = |\u03b1ei\u0394 where |\u03b1 is a coherent state (note that, in general, \u03b1 is a complex number!). Beamsplitter. A similar Hamiltonian description of the beamsplitter also exists, but instead of motivating it phenomenologically, let us begin with the Hamiltonian and show how the expected classical behavior, Equations (7.17)\u2013(7.18) arises from it. Recall that the beamsplitter acts on two modes, which we shall describe by the creation (annihilation) operators a (a\u2020) and b (b\u2020). The Hamiltonian is Hbs", " = i\u03b8 ab\u2020 \u2212 a\u2020b, and the beamsplitter performs the unitary operation B = exp \u03b8 a\u2020b \u2212 ab\u2020. (7.24) (7.25) The transformations effected by B on a and b, which will later be useful, are found to be BaB\u2020 = a cos \u03b8 + b sin \u03b8 and BbB\u2020 = \u2212a sin \u03b8 + b cos \u03b8. (7.26) We verify these relations using the Baker\u2013Campbell\u2013Hausdorf formula (also see Exercise 4.49) e\u03bbGAe\u2212\u03bbG = \u221e n=0 \u03bbn n! Cn, (7.27) where \u03bb is a complex number, A, G, and Cn are operators, and Cn is de\ufb01ned recursively as the sequence of commutators C0 = A, C1 = [G, C0], C2 = [G, C1], C3 = [G, C2],..., Cn = [G, Cn\u22121]. Since it follows from [a, a\u2020] = 1 and [b, b\u2020] = 1 that [G, a] = \u2212b and [G, b] = a, for G \u2261 a\u2020b\u2212ab\u2020, we obtain for the expansion of BaB\u2020 the series coef\ufb01cients 292 Quantum computers: physical realization C0 = a, C1 = [G, a] = \u2212b, C2 = [G, C1] = \u2212a, C3 = [G, C2] = \u2212[G, C0] = b, which in general are Cn even = ina Cn odd = in+1b. From this, our desired result follows straightforwardly: (7.28) (7.29) (7.30) (7.31) (7.32) (i\u03b8)n n! b BaB\u2020 = e\u03b8Gae\u2212\u03b8G \u221e = Cn n=0 \u03b8n n! (i\u03b8)n n! = a cos \u03b8 \u2212 b sin \u03b8. a + i n even = n odd (7.33) The transform BbB\u2020 is trivially found by swapping a and b in the above solution. Note that the beamsplitter operator arises from a deep relationship between the beamsplitter and the algebra of SU (2), as explained in Box 7.3. In terms of quantum logic gates, B performs a useful operation. First note that B|00 = |00, that is, when no photons in either input mode exist, no photons will exist in either output mode. When one photon exists in mode a, recalling that |1 = a\u2020|0, we \ufb01nd that B|01 = Ba\u2020|00 = Ba\u2020B\u2020B|00 = (a\u2020 cos \u03b8 + b\u2020 sin \u03b8)|00 = cos \u03b8|01 + sin \u03b8|10. (7.34) Similarly, B|10 = cos \u03b8|10 \u2212 sin \u03b8|01. Thus, on the |0L and |1L manifold of states, we may write B as B = cos \u03b8 \u2212 sin \u03b8 cos \u03b8 sin \u03b8 = ei\u03b8Y. (7.35) Phase shifters and beamsplitters together allow arbitrary single qubit operations to be performed to our optical qubit. This a consequence of Theorem 4.1 on page 175, which states that all single qubit operations can be generated from \u02c6z-axis rotations Rz(\u03b1) = exp(\u2212i\u03b1Z/2), and \u02c6y-axis rotations, Ry(\u03b1) = exp(\u2212i\u03b1Y /2). A phase shifter performs Rz rotations, and a beamsplitter performs Ry rotations. Exercise 7.9: (Optical Hadamard gate) Show that the following circuit acts as a Hadamard gate on dual-rail single photon states, that is, |01 \u2192 (|01 + |10)/ and |10 \u2192 (|01 \u2212 |10)/ 2 up to an overall phase: \u221a \u221a 2 Exercise 7.10: (Mach\u2013Zehnder interferometer) Interferometers are optical tools used to measure small phase shifts, which are constructed from two beamsplitters. Their basic principle of operation can be understood by this simple exercise. 1. Note that this circuit performs the identity operation: Optical photon quantum computer 293 Box 7.3: SU (2) Symmetry and quantum beamsplitters There is an interesting connection between the Lie group SU (2) and the algebra of two coupled harmonic oscillators, which is useful for understanding the quantum beamsplitter transformation. Identify a\u2020a \u2212 b\u2020b \u2192 Z", " a\u2020b \u2192 \u03c3+ ab\u2020 \u2192 \u03c3\u2212, (7.36) (7.37) (7.38) where Z is the Pauli operator, and \u03c3\u00b1 = (X \u00b1 iY )/2 are raising and lowering operators de\ufb01ned in terms of Pauli X and Y. From the commutation relations for a, a\u2020, b, and b\u2020, it is easy to verify that these de\ufb01nitions satisfy the usual commutation relations for the Pauli operators, (2.40). Also note that the total number operator, a\u2020a + b\u2020b, commutes with \u03c3z, \u03c3+, and \u03c3\u2212, as it should, being an invariant quantity under rotations in the SU (2) space. Using X = a\u2020b + ab\u2020 and Y = \u2212i(a\u2020b \u2212 ab\u2020) in the traditional SU (2) rotation operator R( \u02c6n, \u03b8) = e\u2212i\u03b8\u03c3\u00b7 \u02c6n/2 (7.39) gives us the desired beamsplitter operator when \u02c6n is taken to be the \u2212 \u02c6y-axis. 2. Compute the rotation operation (on dual-rail states) which this circuit performs, as a function of the phase shift \u03d5: Exercise 7.11: What is B|2, 0 for \u03b8 = \u03c0/4? Exercise 7.12: (Quantum beamsplitter with classical inputs) What is B|\u03b1|\u03b2 where |\u03b1 and |\u03b2 are two coherent states as in Equation (7.16)? (Hint: recall that |n = (a\u2020)n n! |0.) \u221a Nonlinear Kerr media. The most important effect of a Kerr medium is the cross phase modulation it provides between two modes of light. That is classically described by the n2 term in (7.19), which is effectively an interaction between photons, mediated by atoms in the Kerr medium. Quantum-mechanically, this effect is described by the Hamiltonian Hxpm = \u2212\u03c7a\u2020ab\u2020b, (7.40) where a and b describe two modes propagating through the medium, and for a crystal of 294 Quantum computers: physical realization length L we obtain the unitary transform K = ei\u03c7La\u2020ab\u2020b. (7.41) \u03c7 is a coef\ufb01cient related to n2, and the third order nonlinear susceptibility coef\ufb01cient usually denoted as \u03c7(3). That the expected classical behavior arises from this Hamiltonian is left as Exercise 7.14 for the reader. By combining Kerr media with beamsplitters, a controlledin the following manner. For single photon states, we \ufb01nd that gate can be constructed K|00 = |00 K|01 = |01 K|10 = |10 K|11 = ei\u03c7L|11, (7.42) (7.43) (7.44) (7.45) and let us take \u03c7L = \u03c0, such that K|11 = \u2212|11. Now consider two dual-rail states, that is, four modes of light. These live in a space spanned by the four basis states |e00 = |1001, |e01 = |1010, |e10 = |0101, |e11 = |0110. Note that we have \ufb02ipped the usual order of the two modes for the \ufb01rst pair, for convenience (physically, the two modes are easily swapped using mirrors). Now, if a Kerr medium is applied to act upon the two middle modes, then we \ufb01nd that K|ei = |ei for all i except K|e11 = \u2212|e11. This is useful because the controlled- operation can be factored into \u23a1 \u23a2 \u23a2 \u23a3 \" # UCN \u23a1 \u23a2 \u23a2 \u23a3 \u23a4 \u23a5 \u23a5 \u23a6! 0 0 1 0 = 1\u221a 2 \" 1 1 1 \u22127.46) where H is the single qubit Hadamard transform (simply implemented with beamsplitters and phase shifters), and K is the Kerr interaction we just considered, with \u03c7L = \u03c0. Such an apparatus has been considered before, for constructing a reversible classical optical logic gate, as described in Box 7.4; in the single photon regime, it also functions as a quantum logic gate. Summarizing, the can be constructed from Kerr media, and arbitrary single qubit operations realized using beamsplitters and phase shifters. Single photons can be created using attenuated lasers, and detected with photodetectors. Thus, in theory, a quantum computer can be implemented using these optical components! Exercise", " 7.13: (Optical Deutsch\u2013Jozsa quantum circuit) In Section 1.4.4 (page 34), we described a quantum circuit for solving the one-bit Deutsch\u2013Jozsa problem. Here is a version of that circuit for single photon states (in the dual-rail representation), using beamsplitters, phase shifters, and nonlinear Kerr media: Optical photon quantum computer 295 Box 7.4: The quantum optical Fredkin gate An optical Fredkin gate can be built using two beamsplitters and a nonlinear Kerr medium as shown in this schematic diagram: Kerr c b a c' b' a' This performs the unitary transform U = B\u2020KB, where B is a 50/50 beamsplitter, K is the Kerr cross phase modulation operator K = ei\u03be b\u2020b c\u2020c, and \u03be = \u03c7L is the product of the coupling constant and the interaction distance. This simpli\ufb01es to give b\u2020 \u2212 a\u2020 b \u2212 a U=exp i\u03bec\u2020c =ei \u03c0 2 b\u2020b e \u03be 2 2 c\u2020c(a\u2020b\u2212b\u2020a) e\u2212i \u03c0 2 2 b\u2020b ei \u03be 2 a\u2020a c\u2020c ei \u03be 2 b\u2020b c\u2020c. (7.47) (7.48) The \ufb01rst and third exponentials are constant phase shifts, and the last two phase shifts come from cross phase modulation. All those effects are not fundamental, and can be compensated for. The interesting term is the second exponential, which is de\ufb01nes the quantum Fredkin operator \u03be c\u2020c (a\u2020b \u2212 b\u2020a). (7.49) F (\u03be) = exp 2 The usual (classical) Fredkin gate operation is obtained for \u03be = \u03c0, in which case when no photons are input at c, then a = a and b = b, but when a single photon is input at c, then a = b and b = a. This can be understood by realizing that F (\u03c7) is like a controlled-beamsplitter operator, where the rotation angle is \u03bec\u2020c. Note that this description does not use the dual-rail representation; in that representation, this Fredkin gate corresponds to a controlled- gate. 1. Construct circuits for the four possible classical functions Uf using Fredkin gates and beamsplitters. 2. Why are no phase shifters necessary in this construction? 3. For each Uf show explicitly how interference can be used to explain how the quantum algorithm works. 296 Quantum computers: physical realization 4. Does this implementation work if the single photon states are replaced by coherent states? Exercise 7.14: (Classical cross phase modulation) To see that the expected classical behavior of a Kerr medium is obtained from the de\ufb01nition of K, Equation (7.41), apply it to two modes, one with a coherent state and the other in state |n; that is, show that K|\u03b1|n = |\u03b1ei\u03c7Ln|n. Use this to compute \u03c1a = Trb K|\u03b1|\u03b2\u03b2|\u03b1|K \u2020 = e\u2212|\u03b2|2 m |\u03b2|2m m! |\u03b1ei\u03c7Lm\u03b1ei\u03c7Lm|, and show that the main contribution to the sum is for m = |\u03b2|2. (7.50) (7.51) (7.52) 7.4.3 Drawbacks The single photon representation of a qubit is attractive. Single photons are relatively simple to generate and measure, and in the dual-rail representation, arbitrary single qubit operations are possible. Unfortunately, interacting photons is dif\ufb01cult \u2013 the best nonlinear Kerr media available are very weak, and cannot provide a cross phase modulation of \u03c0 between single photon states. In fact, because a nonlinear index of refraction is usually obtained by using a medium near an optical resonance, there is always some absorption associated with the nonlinearity, and it can theoretically be estimated that in the best such arrangement, approximately 50 photons must be absorbed for each photon which experiences a \u03c0 cross phase modulation. This means that the outlook for building quantum computers from traditional nonlinear optics components is slim at best. Nevertheless, from studying this optical quantum computer, we have gained some valuable insight into the nature of the architecture and system design of a quantum computer. We now can see what an actual quantum computer might look like in the laboratory (if only suf\ufb01ciently good components were available to construct it), and a striking feature is that it is constructed nearly completely from optical interferometers. In the apparatus, information is encoded both in the photon number and the phase of the photon, and interferometers are used to convert between the two representations", ". Although it is feasible to construct stable optical interferometers, if an alternate, massive representation of a qubit were chosen, then it could rapidly become dif\ufb01cult to build stable interferometers because of the shortness of typical de Broglie wavelengths. Even with the optical representation, the multiple interlocked interferometers which would be needed to realize a large quantum algorithm would be a challenge to stabilize in the laboratory. Historically, optical classical computers were once thought to be promising replacements for electronic machines, but they ultimately failed to live up to expectations when suf\ufb01ciently nonlinear optical materials were not discovered, and when their speed and parallelism advantages did not suf\ufb01ciently outweigh their alignment and power disadvantages. On the other hand, optical communications is a vital and important area; one reason for this is that for distances longer than one centimeter, the energy needed to transmit Optical cavity quantum electrodynamics 297 a bit using a photon over a \ufb01ber is smaller than the energy required to charge a typical 50 ohm electronic transmission line covering the same distance. Similarly, it may be that optical qubits may \ufb01nd a natural home in communication of quantum information, such as in quantum cryptography, rather than in computation. Despite the drawbacks facing optical quantum computer realizations, the theoretical formalism which describes them is absolutely fundamental in all the other realizations we shall study in the remainder of this chapter. In fact, you may think of what we shall turn to next as being just another kind of optical quantum computer, but with a different (and better!) kind of nonlinear medium. Optical photon quantum computer \u2022 Qubit representation: Location of single photon between two modes, |01 and |10, or polarization. \u2022 Unitary evolution: Arbitrary transforms are constructed from phase shifters (Rz rotations), beamsplitters (Ry rotations), and nonlinear Kerr media, which allow two single photons to cross phase modulate, performing exp i\u03c7L|1111|. \u2022 Initial state preparation: Create single photon states (e.g. by attenuating laser light). \u2022 Readout: Detect single photons (e.g. using a photomultipler tube). \u2022 Drawbacks: Nonlinear Kerr media with large ratio of cross phase modulation strength to absorption loss are dif\ufb01cult to realize. 7.5 Optical cavity quantum electrodynamics Cavity quantum electrodynamics (QED) is a \ufb01eld of study which accesses an important regime involving coupling of single atoms to only a few optical modes. Experimentally, this is made possible by placing single atoms within optical cavities of very high Q; because only one or two electromagnetic modes exist within the cavity, and each of these has a very high electric \ufb01eld strength, the dipole coupling between the atom and the \ufb01eld is very high. Because of the high Q, photons within the cavity have an opportunity to interact many times with the atoms before escaping. Theoretically, this technique presents a unique opportunity to control and study single quantum systems, opening many opportunities in quantum chaos, quantum feedback control, and quantum computation. In particular, single-atom cavity QED methods offer a potential solution to the dilemma with the optical quantum computer described in the previous section. Single photons can be good carriers of quantum information, but they require some other medium in order to interact with each other. Because they are bulk materials, traditional nonlinear optical Kerr media are unsatisfactory in satisfying this need. However, well isolated single atoms might not necessarily suffer from the same decoherence effects, and moreover, they could also provide cross phase modulation between photons. In fact, what if the state of single photons could be ef\ufb01ciently transfered to and from single atoms, whose interactions could be controlled? This potential scenario is the topic of this section. 298 Quantum computers: physical realization 7.5.1 Physical apparatus The two main experimental components of a cavity QED system are the electromagnetic cavity and the atom. We begin by describing the basic physics of cavity modes, and then summarize basic ideas about atomic structure and the interaction of atoms with light. Fabry\u2013Perot cavity The main interaction involved in cavity QED is the dipolar interaction d \u00b7 E between an electric dipole moment d and an electric \ufb01eld E. How large can this interaction be? It is dif\ufb01cult in practice to change the size of d; however, | E| is experimentally accessible, and one of the most important tools for realizing a very large electric \ufb01eld in a narrow band of frequencies and in a small volume of space, is the Fabry\u2013Perot cavity. In the approximation that the electric \ufb01eld is monochromatic and occupies a single spatial mode, it can be given a very simple quantum-mechanical description: E(r) = i E0 aeik", "r \u2212 a\u2020e\u2212ikr. (7.56) As described in Box 7.5, these approximations are appropriate for the \ufb01eld in a FabryPerot cavity. Here, k = \u03c9/c is the spatial frequency of the light, E0 is the \ufb01eld strength, is the polarization, and r is the position at which the \ufb01eld is desired. a and a\u2020 are creation and annihilation operators for photons in the mode, and behave as described in Section 7.4.2. Note that the Hamiltonian governing the evolution of the \ufb01eld in the cavity is simply H\ufb01eld = \u03c9a\u2020a, (7.57) and this is consistent with the semiclassical notion that the energy is the volume integral of | E|2 in the cavity. Exercise 7.15: Plot (7.55) as a function of \ufb01eld detuning \u03d5, for R1 = R2 = 0.9. Two-level atoms Until this section of the chapter, we have discussed only photons, or interactions such as the cross phase modulation between photons mediated by a semiclassical medium. Now, let us turn our attention to atoms, their electronic structure, and their interactions with photons. This is, of course, a very deep and well-developed \ufb01eld of study; we shall only describe a small part of it that touches upon quantum computation. The electronic energy eigenstates of an atom can be very complicated (see Box 7.6), but for our purposes modeling an atom as having only two states is an excellent approximation. This two-level atom approximation can be valid because we shall be concerned with the interaction with monochromatic light and, in this case, the only relevant energy levels are those satisfying two conditions: their energy difference matches the energy of the incident photons, and symmetries (\u2018selection rules\u2019) do not inhibit the transition. These conditions arise from basic conservation laws for energy, angular momentum, and parity. Energy conservation is no more than the condition that \u03c9 = E2 \u2212 E1, where E2 and E1 are two eigenenergies of the atom. Angular momentum and parity conservation requirements can be illustrated by considering the matrix element of \u02c6r between two orbital wavefunctions, l1, m1| \u02c6r|l2, m2. Without loss of generality, we can (7.58) Optical cavity quantum electrodynamics 299 Box 7.5: The Fabry\u2013Perot cavity A basic component of a Fabry\u2013Perot cavity is a partially silvered mirror, off which incident light Ea and Eb partially re\ufb02ect and partially transmit, producing the output \ufb01elds Ea and Eb. These are related by the unitary transform Ea Eb, Ea Eb = (7.53) where R is the re\ufb02ectivity of the mirror, and the location of the \u2018\u2212\u2019 sign is a convention chosen as given here for convenience. /// A Fabry\u2013Perot cavity is made from two plane parallel mirrors of re\ufb02ectivities R1 and R2, upon which light Ein is incident from the outside, as shown in the \ufb01gure. Inside the cavity, light bounces back and forth between the two mirrors, such that the \ufb01eld acquires a phase shift ei\u03d5 on each round-trip; \u03d5 is a function of the path length and the frequency of the light. Thus, using (7.53), we \ufb01nd the cavity internal \ufb01eld to be Ecav = Ek = k \u221a 1 \u2212 R1Ein \u221a R1R2 1 + ei\u03d5 \u221a, (7.54) \u221a where E0 = \u221a ei\u03d5/2 1 \u2212 R1Ein, and Ek = \u2212ei\u03d5 \u221a \u221a R1R2Ek\u22121. Similarly, we \ufb01nd Eout = \u221a 1 \u2212 R2, and Ere\ufb02 = R1Ein + 1 \u2212 R1 R2ei\u03d5Ecav. One of the most important characteristics of a Fabry\u2013Perot cavity for our purpose is the power in the cavity internal \ufb01eld as a function of the input power and \ufb01eld frequency, Pcav Pin = + + + + Ecav Ein + + + + 2 = 1 \u2212 R1 \u221a |1 + ei\u03d5 R1R2|2. (7.55) Two aspects are noteworthy. First, frequency selectivity is given by the fact that \u03d5 = \u03c9d/c, where d is the mirror separation, c is the speed of light, and \u03c9 is the frequency of the", " \ufb01eld. Physically, it comes about because of constructive and destructive interference between the cavity \ufb01eld and the front surface re\ufb02ected light. And second, on resonance, the cavity \ufb01eld achieves a maximum value which is approximately 1/(1 \u2212 R) times the incident \ufb01eld. This property is invaluable for cavity QED. 300 Quantum computers: physical realization take \u02c6r to be in the \u02c6x \u2212 \u02c6y plane, such that it can be expressed in terms of spherical harmonics (Box 7.6) as ) \u02c6r = 3 8\u03c0 (\u2212rx + iry)Y1,+1 + (rx + iry)Y1,\u22121 In this basis, the relevant terms in l, m1| \u02c6r|l, m2 are - Y \u2217 l1m1 Y1mYl2m2 d\u03a9. (7.59) (7.60) Recall that m = \u00b11; this integral is non-zero only when m2\u2212m1 = \u00b11 and \u0394l = \u00b11. The \ufb01rst condition is the conservation of angular momentum, and the second, parity, under the dipole approximation where l1, m1| \u02c6r|l2, m2 becomes relevant. These conditions are selection rules which are important in the two-level atom approximation. Exercise 7.16: (Electric dipole selection rules) Show that (7.60) is non-zero only when m2 \u2212 m1 = \u00b11 and \u0394l = \u00b11. In reality, light is never perfectly monochromatic; it is generated from some source such as a laser, in which longitudinal modes, pump noise, and other sources give rise to a \ufb01nite linewidth. Also, an atom coupled to the external world never has perfectly de\ufb01ned energy eigenstates; small perturbations such as nearby \ufb02uctuating electric potentials, or even interaction with the vacuum, cause each energy level to be smeared out and become a distribution with \ufb01nite width. Nevertheless, by choosing an atom and excitation energy carefully, and by taking advantage of the selection rules, it is possible to arrange circumstances such that the two-level atom approximation is superb. The whole point of this procedure is that in this approximation, if |\u03c81 and |\u03c82 are the two selected levels, then the matrix elements of the \u02c6r are rij = \u03c8i| \u02c6r|\u03c8j \u2248 r0Y, where r0 is some constant, and Y is a Pauli operator (Section 2.1.3; that we obtain Y as opposed to X doesn\u2019t really matter \u2013 it is a matter of convention, and convenience, for later calculations). This will be relevant in describing interactions between the atom and incident electric \ufb01elds. The Hamiltonian of the atom itself, in this two-level subspace, is simply (7.65) Hatom = \u03c90 2 Z, (7.66) where \u03c90 is the difference of the energies of the two levels, since the two states are energy eigenstates. 7.5.2 The Hamiltonian The d \u00b7 E interaction between an atom and a cavity con\ufb01ned electric \ufb01eld can be approximated quite well by a much simpler model, in the two-level approximation of the atom, using the quantization of the \ufb01eld in the cavity, and the minute size of the electron compared to the wavelength of the \ufb01eld. Using the fact that d \u221d \u02c6r (the electric dipole size is charge times distance), we can combine (7.56) with (7.65) to obtain the interaction HI = \u2212igY (a \u2212 a\u2020), (7.67) Optical cavity quantum electrodynamics 301 Box 7.6: Energy levels of an atom The electrons of an atom behave like particles in a three dimensional box, with a Hamiltonian of the form HA = k |pk|2 2m \u2212 Ze2 rk + Hrel + Hee + Hso + Hhf, (7.61) where the \ufb01rst two terms describe the balance of the electrons\u2019 kinetic energy with the Coulomb attraction of the negatively charged electrons to the positively charged nucleus, Hrel is a relativistic correction term, Hee describes electron\u2013electron couplings and contributions from the fermionic nature of the electrons, Hso is the spin orbit interaction, which can be interpreted as the spin of the electron interacting with a magnetic \ufb01eld generated by its orbit around the atom, and Hhf is the hyper\ufb01ne interaction: the electron spin interacting with the magnetic", " \ufb01eld generated by the nucleus. The energy eigenstates of HA are generally pretty well categorized according to three integers or half-integers (quantum numbers): n, the principle quantum number; l, the orbital angular momentum; and m, its \u02c6z component. In addition, S, the total electron spin, and I, the nuclear spin, are often important. The eigenvalues of H are roughly determined to order \u03b12 by n, to slightly smaller order by Hee, to order \u03b14 by Hrel and Hso, and to order \u2248 10\u22123\u03b14 by Hhf, where \u03b1 = 1/137 is the dimensionless \ufb01ne structure constant. The derivation of n is simple and follows the usual one-dimensional Schr\u00a8odinger equation solutions for a particle in a box, since the Coulomb con\ufb01ning potential is dependent on radial distance only. However, orbital angular momentum is a feature of being in three dimensions which deserves some explanation. The essential properties arise from the angular dependence of the coordinate representation of HA, in which p becomes the Laplacian operator \u22072, giving the Schr\u00a8odinger equation d\u03a6(\u03d5) d\u03d52 + l(l + 1)\u0398(\u03b8)\u03a6(\u03d5) = 0, d\u0398 d\u03b8 d d\u03b8 (7.62) sin \u03b8 + \u03a6(\u03d5) sin \u03b8 \u0398(\u03b8) sin2 \u03b8 where \u03b8 and \u03d5 are the usual spherical coordinates, and \u03a6 and \u0398 are the eigenfunctions we desire. The solutions Ylm(\u03b8, \u03d5) = \u0398lm(\u03b8)\u03a6m(\u03d5) are the spherical harmonics 9 Ylm(\u03b8, \u03d5) \u2261 (\u22121)m 2l + 1 4\u03c0 (l \u2212 m)! (l + m)! Plm(cos \u03b8)eim\u03d5, (7.63) where Plm are the usual Legendre functions (1 \u2212 x2)m/2 2ll! Plm(x) = dm+l dxm+l (x2 \u2212 1)l. (7.64) In these equations, \u2212l \u2264 m \u2264 l, and it can be shown that m and l must be either integer or half-integer. l is known as the orbital angular momentum, and m is its component along the \u02c6z axis. Similarly, the electron spins S and the nuclear spin I have components ms and mi. As you can see, the description of the energy states of an atom can be quite complicated! Summarizing: for our purposes we may think of the eigenenergies of an atom as being determined by seven numbers: n, l, m, S, ms, I, and mi. 302 Quantum computers: physical realization where we have chosen r = 0 as the point to place the atom (and thus evaluate E), and also oriented the atom such that \u02c6r is aligned properly with the electric \ufb01eld vector. g is some constant (we need not be concerned about the speci\ufb01c values here, just the forms) which describes the strength of the interaction. The i is present simply to allow g to be real, since HI must be Hermitian. HI can be simpli\ufb01ed further, by recognizing that it contains terms which are generally small; to see these, it is useful to de\ufb01ne the Pauli raising and lowering operators, \u03c3\u00b1 = X \u00b1 iY 2, such that we can re-express HI as HI = g(\u03c3+ \u2212 \u03c3\u2212)(a \u2212 a\u2020). (7.68) (7.69) The terms containing \u03c3+a\u2020 and \u03c3\u2212a oscillate at twice the frequencies of interest, which are \u03c9 and \u03c90, and dropping them is a fairly good approximation (the rotating wave approximation) which leads us to the total Hamiltonian H = Hatom + H\ufb01eld + HI, H = \u03c90 2 Z + \u03c9a\u2020a + g(a\u2020\u03c3\u2212 + a\u03c3+). (7.70) where, again, just to recap: the Pauli operators act on the two-level atom, a\u2020, a are raising and lowering operators on the single mode \ufb01eld, \u03c9 is the frequency of the \ufb01eld, \u03c90 is the frequency of the atom, and g is the coupling constant for the interaction between atom and \ufb01eld. This is the fundamental theoretical tool in the study of cavity QED, the Jaynes\u2013Cummings Hamiltonian, which describes interactions between two-level atoms and an electromagnetic \ufb01eld. This Hamiltonian can be written in another convenient form", " by noting that N = a\u2020a + Z/2 is a constant of the motion, that is [H, N ] = 0, so that we \ufb01nd H = \u03c9N + \u03b4Z + g(a\u2020\u03c3\u2212 + a\u03c3+), (7.71) where \u03b4 = (\u03c90 \u2212 \u03c9)/2 is known as the detuning \u2013 the frequency difference between the \ufb01eld and atomic resonance. This Hamiltonian, the Jaynes\u2013Cummings Hamiltonian, is very important, and we shall be spending nearly all of the rest of the chapter studying its properties and guises in different physical systems. Exercise 7.17: (Eigenstates of the Jaynes\u2013Cummings Hamiltonian) Show that |n, 1 + |n + 1, 0 |n, 1 \u2212 |n + 1, 0 |\u03c7n = |\u03c7n = (7.73) (7.72) 1\u221a 2 1\u221a 2 are eigenstates of the Jaynes\u2013Cummings Hamiltonian (7.71) for \u03c9 = \u03b4 = 0, with the eigenvalues \u221a H|\u03c7n = g H|\u03c7n = \u2212g n + 1|\u03c7n \u221a n + 1|\u03c7n, (7.74) (7.75) where the labels in the ket are |\ufb01eld, atom. Optical cavity quantum electrodynamics 303 7.5.3 Single-photon single-atom absorption and refraction The most interesting regime in cavity QED, for our purposes, is that in which single photons interact with single atoms. This is an unusual regime, in which traditional concepts (such as index of refraction and permittivity) in classical theories of electromagnetism break down. In particular, we would like to utilize a single atom to obtain a nonlinear interaction between photons. Let us begin by showing one striking and general characteristic of the atom\u2013\ufb01eld system known as Rabi oscillations. Without loss of generality we may neglect N, since it only contributes a \ufb01xed phase. Recalling that time evolution is given by U = e\u2212iHt (here and in the following, it will often be convenient to drop, and we shall do so freely), and focusing on the case of at most a single excitation in the \ufb01eld mode, where \u23a1 \u23a4 H = \u2212 \u23a3 \u23a6, (7.76\u03b4 (the basis states are |00, |01, |10, from left to right and top to bottom, where the left label corresponds to the \ufb01eld, and the right one to the atom), we \ufb01nd that U = e\u2212i\u03b4t|0000| + (cos \u03a9t + i sin \u03a9t)|0101| \u03b4 \u03a9 \u03b4 sin \u03a9t)|1010| + (cos \u03a9t \u2212 i \u03a9 ( \u2019 |0110| + |1001| sin \u03a9t \u2212 i g \u03a9. (7.77) The interesting behavior is in the last line of this equation, which shows that the atom and \ufb01eld oscillate back and forth exchanging a quantum of energy, at the Rabi frequency \u03a9 = g2 + \u03b42. Exercise 7.18: (Rabi oscillations) Show that (7.77) is correct by using ein\u00b7\u03c3 = sin |n| + i \u02c6n \u00b7 \u03c3 cos |n| (7.78) to exponentiate H. This is an unusually simple derivation of the Rabi oscillations and the Rabi frequency; ordinarily, one solves coupled differential equations to obtain \u03a9, but here we obtain the essential dynamics just by focusing on the single-atom, single-photon subspace! The transformation of the photon, in interacting with a single atom, can be obtained by tracing over the atom\u2019s state (Section 2.4.3). The probability that an initial photon |1 is absorbed by the atom (which we assume starts in its ground state, |0) is simply \u03c7r = k |0k|U |10|2 = g2 g2 + \u03b42 sin2 \u03a9t. (7.79) This has the usual Lorentzian pro\ufb01le expected for absorption as a function of detuning \u03b4 from resonance. The refractive index (of the single atom!) is given by the matrix elements of U in which the atom stays in the ground state. The phase shift experienced by the photon is 304 Quantum computers: physical realization the difference in the angle of rotation experienced by the |1 and the |0 states of the \ufb01eld, tracing over the atom. This is found to be \u03c7i = arg e", "i\u03b4t cos \u03a9t \u2212 i sin \u03a9t. \u03b4 \u03a9 (7.80) For \ufb01xed non-zero \u03b4, as the coupling g is decreased, the absorption probability \u03c7r decreases as g2, but the phase shift \u03c7i remains nearly constant. This is the origin of materials which can perform phase shifts without scattering much light. Exercise 7.19: (Lorentzian absorption pro\ufb01le) Plot (7.79) for t = 1 and g = 1.2, as a function of the detuning \u03b4, and (if you know it) the corresponding classical result. What are the oscillations due to? Exercise 7.20: (Single photon phase shift) Derive (7.80) from U, and plot it for t = 1 and g = 1.2, as a function of the detuning \u03b4. Compare with \u03b4/\u03a92. 2 Figure 7.4. Three level atom (with levels 0, 1, and 2) interacting with two orthogonal polarizations of light, described by the operators a and b. The atom\u2013photon couplings are respectively ga and gb. The energy differences between 0 and 1, and between 0 and 2 are assumed to be nearly equal. A natural application of the atom\u2013photon interaction is to study what happens when two different photon modes (each containing at most one photon) interact with the same atom. This can give rise to a nonlinear interaction between the two modes. Recall from Section 7.4.2 that nonlinear Kerr media can be described phenomenologically as media which induce a cross phase modulation with Hamiltonian of the form H = \u03c7a\u2020ab\u2020b. There, we did not see how that effect arises from fundamental interactions. Using the present formalism, the origin of the Kerr effect can be illustrated using a simple model, in which two polarizations of light interact with a three-level atom, as shown in Figure 7.4. This is described by a modi\ufb01ed version of the Jaynes\u2013Cummings Hamiltonian \u23a3 \u23a6 + ga \u239da \u23a3 \u22121 0 0 0 1 0 0 0 1 \u239b \u23a1 + gb \u239d\u2020 \u23a3 \u23a6 + a7.81) where the basis elements for the 3\u00d73 atom operators are |0, |1, and |2. In matrix form, Optical cavity quantum electrodynamics 305 (7.82) (7.83) the relevant terms in H are found to be the block-diagonal matrix H = \u23a3 where \u23a1 \u23a4 \u23a6, H0 0 0 H1 0 0 0 0 H2 H0 = \u22127.84) H2 = H1 = 0 0 gb \u03b4 \u2212\u03b4 ga 0 0 \u2212\u03b4 ga gb ga 0 \u03b4 0 0 \u2212\u03b4 gb 0 gb ga \u03b4 0 \u03b4 0 in the basis |a, b, atom = |000 for H0, |100, |001, |010, |002 for H1, and |110, |011, |102 for H2, across the columns from left to right. Exponentiating to give U = exp(iHt) allows one to \ufb01nd the single photon phase shifts \u03d5a = arg(100|U |100) \u2212 arg(000|U |000) and \u03d5b = arg(010|U |010) \u2212 arg(000|U |000) and the two photon phase shift \u03d5ab = arg(110|U |110) \u2212 arg(000|U |000). For linear media, one would expect that \u03d5ab = \u03d5a + \u03d5b, that is, the two photon state has twice the phase shift of the single photon state, since exp[\u2212i\u03c9(a\u2020a + b\u2020b)]|11 = exp(\u22122i\u03c9)|11. However, this system behaves nonlinearly, and gives \u03c73 \u2261 \u03d5ab \u2212 \u03d5a \u2212 \u03d5b as shown in Figure 7.5. In this physical system, this Kerr effect arises from the slight amplitude for the atom to exchange quanta between the two optical modes. (7.85 20 10 0 -10 -20 -2 -1 0 Detuning delta 1 2 Figure 7.5. Kerr phase shift \u03c73 in degrees, for t = 0.98 and ga = gb = 1, plotted as a function of the detuning \u03b4, computed from (7.82) for single photons interacting with a single three-level atom. Exercise 7.21: Explicitly exponentiate (7.82) and show that \u03d5ab = arg ei\u03b4t cos \u03a9t \u2212 i, (7.86) \u03b4", " \u03a9 sin \u03a9t 306 Quantum computers: physical realization a + g2 \u03b42 + g2 where \u03a9 = b. Use this to compute \u03c73, the nonlinear Kerr phase shift. This is a very simple way to model and understand the Kerr interaction, which sidesteps much of the complication typically involved in classical nonlinear optics. Exercise 7.22: Associated with the cross phase modulation is also a certain amount of loss, which is given by the probability that a photon is absorbed by the atom. Compute this probability, 1 \u2212 110|U |110, where U = exp(\u2212iHt) for H as in (7.82); compare with 1 \u2212 100|U |100 as a function of \u03b4, ga, gb, and t. 7.5.4 Quantum computation Broadly speaking, cavity QED techniques can be used to perform quantum computation in a number of different ways, two of which are the following: quantum information can be represented by photon states, using cavities with atoms to provide nonlinear interactions between photons; or quantum information can be represented using atoms, using photons to communicate between the atoms. Let us now close out this subject by describing an experiment which demonstrates the \ufb01rst of these methods to realize a quantum logic gate. As we saw in Section 7.4.2, a quantum computer can be constructed using single photon states, phase shifters, beamsplitters, and nonlinear Kerr media, but the \u03c0 cross gate is nearly infeasible with phase modulation required to produce a controlledstandard bulk nonlinear optics techniques. Cavity QED can be used to implement a Kerr interaction, as shown in Section 7.5.3; unlike for bulk media, this can have a very strong effect even at the single photon level, because of the strong \ufb01eld provided by a Fabry\u2013Perot type cavity. Figure 7.6 illustrates a cavity QED experiment which was performed (see \u2018History and further reading\u2019 at the end of the chapter) to demonstrate the potential for realizing a logic gate with the unitary transform \u23a4 \u23a1 \u23a2 \u23a2 \u23a3 1 0 0 0 0 ei\u03d5a 0 0 0 0 ei\u03d5b 0 0 0 0 ei(\u03d5a+\u03d5b+\u0394) \u23a5 \u23a5 \u23a6, (7.87) where \u0394 = 16\u25e6, using single photons. In the experiment, two modes of light (distinguished by a very small frequency difference) with weak coherent states are prepared, one linearly polarized (the probe), and one circularly polarized (the pump), as input to the cavity. This state can be expressed as |\u03c8 in = |\u03b2+ |\u03b1+ + |\u03b1\u2212 \u221a 2, (7.88) recalling that linearly polarized light is an equal superposition of the two possible circularly polarized states, + and \u2212. Approximating the weak coherent states as |\u03b1 \u2248 |0+\u03b1|1 and similarly for |\u03b2 (and leaving out normalizations for the moment) gives |0+ + \u03b2|1+ |\u03c8 in \u2248 |0+ + \u03b1|1+ + |0\u2212 + \u03b1|1\u2212. (7.89) These photons pass through the optical cavity and interact with the atom, which is modeled as causing a different phase shift to occur to states depending on the total number of photons in each polarization (independent of which mode the photons are Optical cavity quantum electrodynamics 307 in). Speci\ufb01cally, we assume that a photon in the |1+ state experiences a ei\u03d5a phase shift if it is in the probe beam, and ei\u03d5b for the pump. In addition to this single photon phase shift, the state |1+1+ experiences an additional Kerr phase shift \u0394, so it becomes ei(\u03d5a+\u03d5b+\u0394)|1+1+. Other states (and in particular, other polarizations) remain unchanged. The physics which leads to this behavior is similar to that described in Section 7.5.3, and the end effect is the same: a cross phase modulation between the pump and the probe light. The output from the cavity is thus |\u03c8out \u2248 |0+ |0+ + \u03b1ei\u03d5a |1+ + |0\u2212 + \u03b1|1\u2212 + ei\u03d5b\u03b2|1+ |0+ + \u03b1ei(\u03d5a+\u0394)|1+ + |0\u2212 + \u03b1|1\u2212 \u2248 |0+|\u03b1, \u03d5a/2 + ei\u03d5b \u03b2|1+|\u03b1, (\u03d5a + \u0394)/2, (7.90) (7.91) where |\u03b1, \u03d5a/2 denotes a linearly polarized probe \ufb01eld rotated", " from the vertical by \u03d5a/2. The \ufb01eld polarizations are measured by the detector, giving \u03d5a \u2248 17.5\u25e6, \u03d5b \u2248 12.5\u25e6 and \u0394 \u2248 16\u25e6. Since \u0394 is a non-trivial value, this result suggests that a universal two qubit logic gate (Exercise 7.23) is possible using single photons, and a single atom in a cavity as a nonlinear optical Kerr medium to interact photons. Probe Pump \u03a9 a \u03a9 b Heterodyne Local oscillator M1 M2 \u03bb/2 PBS Optical pumping Cs beam Figure 7.6. Schematic of an experimental apparatus used to demonstrate the possibility of using a single atom to provide cross phase modulation between single photons, as an elementary quantum logic gate. A linearly polarized weak probe beam of light \u03a9a, and a stronger circularly polarized pump beam \u03a9b are prepared and shone on an optical cavity with high re\ufb02ectivity mirrors M1 and M2. Cesium atoms prepared in the electronic state 6S1/2, F = 4, m = 4 by optical pumping fall (the \ufb01gure shows the atoms upside down) such that the average number of atoms in the cavity is around one. The light traverses the cavity, interacting with the atom; \u03c3+ polarized light causes strong transitions to the 6P3/2, F = 5, m = 5 state, and the orthogonal \u03c3\u2212 polarized light causes weak transitions to the 6P3/2, F = 5, m = 3 state. The polarization of the output light is then measured, using a half wave plate, a polarizing beamsplitter (PBS), and a sensitive balanced heterodyne detector (which selectively detects light at a speci\ufb01c frequency, as determined by the local oscillator). Figure courtesy of Q. Turchette. Several important caveats must be kept in mind in interpreting these experimental results. The incident photons are absorbed with non-trivial probability when traversing the cavity and atom, and thus the true quantum operation performed is not unitary; this 308 Quantum computers: physical realization problem would be aggravated if multiple gates were cascaded, which would be required, gate (which requires \u0394 = \u03c0). In fact, re\ufb02ection for example, to realize a controlledlosses of the cavity arrangement used in this experiment would signi\ufb01cantly impede cascading; to understand how to get around this, a proper time-dependent model would have to be developed and studied. Also, although the cross phase modulation model is consistent with data measured, the photon\u2013atom interaction model used is an ansatz, and other models are not ruled out by the experiment. In fact, it would be possible in principle to use single photon states (as opposed to attenuated coherent states) in the experiment, and measurements of the resulting entanglement of the two modes in |\u03c8out would be a good test. At the time this experiment was carried out, no general procedure was known for fully characterizing a quantum operation and its suitability as a quantum logic gate. However, a method for doing this, known as process tomography, is now well understood (Chapter 8), and remarkably it even allows full characterization of dissipation and other non-unitary behavior. Performing such a test would unambiguously determine exactly the extent to which the experiment described here actually re\ufb02ects a quantum computation. Despite these drawbacks, the experiment does demonstrate fundamental concepts required for quantum information processing. It certi\ufb01es that nonlinear optical behavior such as the Kerr interaction really does occur at the single photon level, thus validating the essence of the Jaynes\u2013Cummings model. Also, this experiment is performed in what is called the bad cavity regime, where the atom\u2019s coherent coupling rate g2/\u03ba to the cavity mode dominates incoherent emission rate \u03b3 into free space, but this coupling is weaker than the rate \u03ba at which input photons enter and leave the cavity. The strong coupling operating regime, in which g > \u03ba > \u03b3, offers an alternative in which larger conditional phase shifts \u0394 may be obtained. Most importantly, perhaps, cavity QED opens the door to a wealth of additional interactions which are valuable for quantum information processing. We have also seen how the quantum information perspective \u2013 focusing on single photons and single atoms \u2013 has allowed us take the Jaynes\u2013Cummings Hamiltonian, the basic cavity QED interaction, and construct from it some of the most fundamental physics of the interaction of electromagnetic waves with matter. We now leave the subject of cavity QED, but as we continue on next to ion traps, and then to magnetic resonance, we shall keep with us these notions of photon\u2013atom interactions, single atoms and photons, and the Jaynes\u2013Cummings Hamiltonian. Exercise 7.23", ": Show that the two qubit gate of (7.87) can be used to realize a gate, when augmented with arbitrary single qubit operations, for controlledany \u03d5a and \u03d5b, and \u0394 = \u03c0. It turns out that for nearly any value of \u0394 this gate is universal when augmented with single qubit unitaries. Optical cavity quantum electrodynamics \u2022 Qubit representation: Location of single photon between two modes, |01 and |10, or polarization. \u2022 Unitary evolution: Arbitrary transforms are constructed from phase shifters (Rz Ion traps 309 rotations), beamsplitters (Ry rotations), and a cavity QED system, comprised of a Fabry\u2013Perot cavity containing a few atoms, to which the optical \ufb01eld is coupled. \u2022 Initial state preparation: Create single photon states (e.g. by attenuating laser light). \u2022 Readout: Detect single photons (e.g. using a photomultipler tube). \u2022 Drawbacks: The coupling of two photons is mediated by an atom, and thus it is desirable to increase the atom\u2013\ufb01eld coupling. However, coupling the photon into and out of the cavity then becomes dif\ufb01cult, and limits cascadibility. 7.6 Ion traps Thus far in this chapter, we have focused mainly on representing qubits using photons. Let us now turn to representations which use atomic and nuclear states. Speci\ufb01cally, as we saw in Section 7.1, electron and nuclear spins provide potentially good representations for qubits. Spin is a strange (but very real!) concept (Box 7.7), but since the energy difference between different spin states is typically very small compared with other energy scales (such as the kinetic energy of typical atoms at room temperature), the spin states of an atom are usually dif\ufb01cult to observe, and even more dif\ufb01cult to control. In carefully crafted environments, however, exquisite control is possible. Such circumstances are provided by isolating and trapping small numbers of charged atoms in electromagnetic traps, then cooling the atoms until their kinetic energy is much lower than the spin energy contribution. After doing this, incident monochromatic light can be tuned to selectively cause transitions which change certain spin states depending on other spin states. This is the essence of how trapped ions can be made to perform quantum computation, as we describe in this section. We begin with an overview of the experimental apparatus and its main components, then we present a Hamiltonian modeling the system. We describe gate with an experiment which has been performed to demonstrate a controlledtrapped 9Be ions, and then close with a few comments on the potential and limitations of the method. Exercise 7.24: The energy of a nuclear spin in a magnetic \ufb01eld is approximately \u03bcNB, where \u03bcN = eh/4\u03c0mp \u2248 5\u00d710\u221227 joules per tesla is the nuclear Bohr magneton. Compute the energy of a nuclear spin in a B = 10 tesla \ufb01eld, and compare with the thermal energy kBT at T = 300 K. 7.6.1 Physical apparatus An ion trap quantum computer has as its main components an electromagnetic trap with lasers and photodetectors, and ions. Trap geometry and lasers The main experimental apparatus, an electromagnetic trap constructed from four cylindrical electrodes, is shown in Figure 7.7. The end segments of the electrodes are biased at a different voltage U0 than the middle, so that the ions are axially con\ufb01ned by a static potential \u03a6dc = \u03baU0 /2 along the \u02c6z axis (\u03ba is a geometrical factor). However, a result known as Earnshaw\u2019s theorem states that a charge cannot be con\ufb01ned in three dimensions by static potentials. Thus, to provide con\ufb01nement, two of the electrodes z2 \u2212 (x2 + y2) 310 Quantum computers: physical realization Box 7.7: Spin Spin is a strange concept. When a particle has spin, it possesses a magnetic moment as if it were a composite particle with some current running around in a loop. But electrons are elementary particles, and the quarks which compose a nucleon are not known to produce spin by orbital motion. Furthermore, the spin of a particle is only ever either integer or half-integer. Spin is nevertheless quite real, and an important part of everyday physics. Integer spin particles, known as bosons, include the photon. Being massless, it is somewhat special and only has spin \u00b11 (and no spin zero) components; these correspond to the two familiar orthogonal polarization states. Sunglasses made from cheap plastic polarizers are effective when driving because sunlight becomes partially polarized in the opposite direction after re\ufb02ecting off of surfaces such as roadways (light polarized with the electric \ufb01eld", " transverse to the interface always partially re\ufb02ects no matter the angle of incidence, in contrast with the transverse magnetic polarization which does not re\ufb02ect when the angle of incidence is at Brewster\u2019s angle). Halfinteger spin particles, known as fermions, include the electron, proton, and neutron. These are \u2018spin-1/2\u2019 particles, in that their spin component can either be +1/2 (spin \u2018up\u2019) or \u22121/2 (spin \u2018down\u2019). When we say \u2018spin\u2019 often what is meant is a spin-1/2 particle. The energy eigenstates of an atom intimately involve spin, and the combination of multiple spins. For example, the nucleus of 9Be has spin 3/2. Spins interact with a magnetic \ufb01eld just as magnetic moments do; in a magnetic \ufb01eld B, an electron with I \u00b7 B. Pictorially, S \u00b7 B, and similarly, a nucleus I has energy gn spin S has energy ge for example, the spin contribution to an atom\u2019s energy levels can be viewed as: \"; &2 &2 3&2 &2 &2 3&2 3&2 &2 &2 \" :$\" \" :$\" 3&2 \" = \" :$\" where we have assumed a spin-1/2 electron, and a spin-3/2 nucleus. By tuning the frequency of an incident laser just right, any of these transitions could be selected, as long as conservation laws (Section 7.5.1) are satis\ufb01ed. In particular, angular momentum conversation implies that when a photon is absorbed by an atom, one unit of angular momentum or spin must change between the initial and \ufb01nal states. These states thus must have de\ufb01nite values of angular momenta; this can be taken into account. Unlike continuous variables such as position and momentum, and other in\ufb01nite Hilbert space systems which must be arti\ufb01cially truncated to represent quantum bits, spin states provide good representations for quantum information because they live in an inherently \ufb01nite state space. Ion traps 311 are grounded, while the other two are driven by a fast oscillating voltage which creates a radiofrequency (RF) potential \u03a6rf = (V0 cos \u03a9T t + Ur)(1 + (x2 \u2212 y2)/R2)/2, where R is a geometrical factor. The segments of the electrodes are capacitively coupled such that the RF potential is constant across them. The combination of \u03a6dc and \u03a6rf creates, on average (over \u03a9T ), a harmonic potential in x, y, and z. Together with the Coulomb repulsion of the ions, this gives a Hamiltonian governing the motion of the N ions in the trap, H = N M 2 i=1 xx2 \u03c92 i + \u03c92 yy2 i + \u03c92 zz2 i + + |pi|2 M 2 N i=1 j>i e2 4\u03c00|ri \u2212 rj|, (7.92) where M is the mass of each ion. Typically, \u03c9x, \u03c9y # \u03c9z by design, so that the ions all lie generally along the \u02c6z axis. As the number of ions becomes large, the geometrical con\ufb01guration of the ions can become quite complicated, forming zig-zag and other patterns, but we shall focus on the simple case where just a few ions are trapped, in a string-like con\ufb01guration. 9: *: ; 0? >: Figure 7.7. Schematic drawing (not to scale) of an ion trap quantum computer, depicting four ions trapped in the center of a potential created by four cylindrical electrodes. The apparatus is typically contained in a high vacuum (\u2248 10\u22128 Pa), and the ions are loaded from a nearby oven. Modulated laser light incident on the ions through windows in vacuum chamber perform operations on and are used to readout the atomic states. Just as a mass on a spring can behave as a quantum system when the coupling to the external world becomes suf\ufb01ciently small, the motion of the electromagnetically con\ufb01ned ion becomes quantized when it is suf\ufb01ciently well isolated. Let us \ufb01rst understand what the quantization means, then consider the isolation criteria. As we saw in Section 7.3, the energy levels of a harmonic oscillator are equally spaced, in units of \u03c9z. In the ion trap, in the regime which concerns us, these energy eigenstates represent different vibrational modes of the entire linear chain of ions moving together as one body, with mass N M.", " These are called the center of mass modes. Each \u03c9z quantum of vibrational energy is called a phonon, and can be thought of as a particle, just as a quantum of electromagnetic radiation in a cavity is a photon. For the above phonon description to hold, certain criteria must hold. First, the coupling to the environment must be suf\ufb01ciently small such that thermalization does not randomize the state of the system (and thus cause it to behave classically). Physically, 312 Quantum computers: physical realization what can happen is that nearby \ufb02uctuating electric and magnetic \ufb01elds push on the ions, causing their motional state to randomly transition between energy eigenstates. Such noise sources are nearly inevitable, in a technical sense, since, for example, one cannot drive the con\ufb01ning electrodes from a perfect voltage source; the source will always have a \ufb01nite resistance, and this resistance gives rise to Johnson noise, which has \ufb02uctuations on time scales the ions are sensitive to. The electric \ufb01eld on local patches of the electrodes can also \ufb02uctuate, randomly driving the ions\u2019 motion. As the randomness increases, the quantum properties of the ions\u2019 state is lost, and their behavior becomes well described by classical statistical averages. For example, both their momentum and position become well de\ufb01ned, which cannot be simultaneously true for a quantum system. Nevertheless, in practice, most technical noise sources can indeed be controlled quite well, to the extent that they do not heat or dephase the trapped ions too much on the time scale of most experiments. In part, one important reason this is possible is that as long as the harmonic approximation holds, the trapped ions are very selective about the frequency of the noise they are sensitive to; just as transitions between atomic levels can be selected by radiation tuned only to the correct frequency, only \ufb02uctuations which have high spectral power density around \u03c9z will affect the ions. It is also quite important for the ions to be suf\ufb01ciently cool so as to make the onedimensional harmonic approximation valid. The true potential is non-quadratic for large displacements along any direction away from the trap center. And higher order vibrational modes in which the ions move relative to each other (instead of moving together) must have energies much higher than the center of mass mode. When this holds, and the ions are cooled to their motional ground state, their transition to the next higher energy state is through absorption of a center of mass phonon; this process is related to the Mossba\u00a8uer effect, in which a photon is absorbed by atoms in a crystal without generating local phonons because the entire crystal recoils together. How are the ions cooled to their motional ground state? The goal is to satisfy kBT \u03c9z, where T is the temperature re\ufb02ecting the kinetic energy of the ions. Essentially, this can be done by using the fact that photons carry not only energy, but also momentum p = h/\u03bb, where \u03bb is the wavelength of the light. Just as the whistle of an approaching train has a higher pitch than a departing train, an atom moving toward a laser beam has transition frequencies which are slightly higher in energy than an atom moving away. If the laser is tuned such that it is absorbed only by approaching atoms, then the atoms slow down because the photons kick them in the opposite direction. This method is known as Doppler cooling. Shining a properly tuned laser (which has momentum vector components along each axis) at trapped atoms thus can cool the atoms down to the limit kBT \u2248 \u0393/2, where \u0393 is the radiative width of the transition used for the cooling. To cool beyond this limit, another method, known as sideband cooling, is then applied, as illustrated in Figure 7.8. This allows one to reach the kBT \u03c9z limit. Another criterion which must be satis\ufb01ed is that the width of the ion oscillation in the trap potential should be small compared to the wavelength of the incident light. This Lamb\u2013Dicke criterion is conveniently expressed in terms of the Lamb-Dicke parameter /2N M \u03c9 is the characteristic length \u03b7 \u2261 2\u03c0z0/\u03bb, where \u03bb is the wavelength, and z0 = scale of the spacing between ions in the trap. The Lamb\u2013Dicke criterion requires that \u03b7 1; this does not strictly have to be met in order for ion traps to be useful for quantum computation, but it is desired to have that \u03b7 \u2248 1 at least, in order that the individual Ion traps 313 Figure 7.8. Sideband cooling method, showing transitions between |0, n and |1, m, where 0 and 1 are two electronic levels, and n and m are phonon levels representing motional states of the", " ion. Laser light is tuned to have energy one phonon less than the electronic transition, such that, for example, the |0, 3 state transitions to the |1, 2 state, as shown. The atom then spontaneously decays into the lower energy 0 state (wiggly lines), randomly going to either |0, 1, |0, 2, or |0, 3 (with nearly equal probabilities). Note that the laser light actually causes all possible transitions between |0, n and |1, n \u2212 1, since these all have the same energy. However, this process does not touch the |0, 0 state, and eventually that is the state in which the atom will be left. ions can be resolved by different laser beams, but without making their motional state too dif\ufb01cult to optically excite in order to perform logic operations. Atomic structure The purpose of the trap apparatus described above is to allow ions to be cooled to the extent that their vibrational state is suf\ufb01ciently close to having zero phonons (|0), an appropriate initial state for computation. Similarly, the internal states of the ions must be initialized appropriately, so they may be used to store quantum information. Let us now consider what these internal states are, and understand why they are good qubit representations by estimating their coherent lifetime. The internal atomic states relevant to the trapped ion we shall consider result from the combination F of electron spin S and nuclear spin I, giving F = S + I. The formal piece of theory which describes this \u2013 known as the addition of angular momenta \u2013 not only describes important physics for understanding atomic structure, but also is an interesting mechanism for quantum information. A single photon interacting with an atom can provide or carry away one unit of angular momentum, as we saw in Section 7.5.1. But there are numerous possible sources of angular momenta in an atom: orbital, electron spin, and nuclear spin. Where it comes from is partly determined by the energy levels selected by the energy of the photon, but beyond that, the photon cannot distinguish between different sources, and to describe what happens we must select a basis in which total angular momentum becomes a uniquely de\ufb01ned property of the state. Consider, for example, two spin-1/2 spins. The \u2018computational\u2019 basis for this two qubit space is |00, |01, |10, |11, but to span the state space we could equally well 314 Quantum computers: physical realization choose the basis |0, 0J = |01 \u2212 |10 \u221a 2 |1, \u22121J = |00 |1, 0J = |01 + |10 \u221a 2 |1, 1J = |11. (7.93) (7.94) (7.95) (7.96) These basis states are special, because they are eigenstates of the total momentum operator, de\ufb01ned by jx = (X1 + X2)/2, jy = (Y1 + Y2)/2, jz = (Z1 + Z2)/2, and J 2 = j2 x + j2 z. y + j2 (7.97) The states |j, mjJ are eigenstates of J 2 with eigenvalue j(j + 1), and simultaneously eigenstates of jz, with eigenvalue mj. These states are the natural ones selected by many physical interactions; for example, in a \u02c6z oriented magnetic \ufb01eld the magnetic moment \u03bc in the Hamiltonian \u03bcBz is proportional to mj, the component of the total angular momentum in the \u02c6z direction. The theory of addition of angular momenta is sophisticated and well developed, and we have but scratched its surface (for the interested reader, some relevant exercises are provided below, and pointers to the literature are given in the \u2018History and further reading\u2019 section at the end of the chapter). Nevertheless, some interesting observations which concern quantum information can already be drawn from the above examples. Normally, we think of entangled states such as the Bell states (Section 1.3.6) as being unnatural states of matter, because they have strange, non-local properties. However, the state |0, 0J is a Bell state! Why does Nature prefer this state here? It is because of a symmetry under which the interaction involving the magnetic moment is invariant under interchange of the two spins. Such symmetries actually occur widely in Nature, and are potentially quite useful for performing entangling measurements and operations. Exercise 7.25: Show that the total angular momenta operators obey the commutation relations for SU (2), that is, [ji, jk] = iikljl. Exercise 7.26: Verify the properties of |j, mjJ by explicitly writing the 4\u00d74 matrices J 2 and jz in the basis de\ufffd", "\ufffdned by |j, mjJ. Exercise 7.27: (Three spin angular momenta states) Three spin-1/2 spins can combine together to give states of total angular momenta with j = 1/2 and j = 3/2. Show that the states |3/2, 1/2 = |3/2, 3/2 = |111 |011 + |101 + |110 |100 + |010 + |001 |3/2, \u22121/2 = 1\u221a 3 1\u221a 3 |3/2, \u22123/2 = |000 \u2212|001 + |100 |1/2, 1/21 = 1\u221a 2 (7.98) (7.99) (7.100) (7.101) (7.102) Ion traps 315 |1/2, \u22121/21 = |1/2, 1/22 = |1/2, \u22121/22 = 1\u221a 2 1\u221a 6 1\u221a 6 |110 \u2212 |011 |001 \u2212 2|010 + |100 \u2212|110 + 2|101 \u2212 |011 (7.103) (7.104) (7.105) form a basis for the space, satisfying J 2|j, mj = j(j + 1)|j, mj and jz|j, mj = mj|j, mj, for jz = (Z1 + Z2 + Z3)/2 (similarly for jx and jy) and J 2 = j2 straightforward brute-force method is simply to simultaneously diagonalize the 8\u00d78 matrices J 2 and jz. z. There are sophisticated ways to obtain these states, but a x + j2 y + j2 Exercise 7.28: (Hyper\ufb01ne states) We shall be taking a look at beryllium in Section 7.6.4 \u2013 the total angular momenta states relevant there involve a nuclear spin I = 3/2 combining with an electron spin S = 1/2 to give F = 2 or F = 1. For a spin-3/2 particle, the angular momenta operators are \u23a1 \u221a \u23a4 \u23a5 \u23a5 \u23a6 0 0 \u221a 3 0 0 2i 0 \u221a \u2212i 3 \u23a4 0 0 0 3 ix = \u23a2 \u23a2 \u23a3 7.106) \u23a4 i 3 0 \u23a1 1 2 \u23a2 \u23a2 \u23a3 \u23a5 \u23a5 \u23a6 \u23a2 \u23a2 \u23a3 iz = iy = (7.107) 0 0 \u221a 3 0 \u22122i 0 0 0 1 0 0 \u221a \u2212i 3 0 0 \u22123 0 0 \u22121 0 0 0 0 1. Show that ix, iy, and iz satisfy SU (2) commutation rules. 2. Give 8\u00d78 matrix representations of fz = iz \u2297 I + I \u2297 Z/2 (where I here represents the identity operator on the appropriate subspace) and similarly fx and fy, and, F 2 = f 2 obtain basis states |F, mF for which F 2|F, mF = F (F + 1)|F, mF and fz|F, mF = mF |F, mF. z. Simultaneously diagonalize fz and F 2 to x + f 2 y + f 2 (7.108) \u23a5 \u23a5 \u23a6 1 2 How long can a superposition of different spin states exist? The limiting process, known as spontaneous emission, occurs when an atom transitions from its excited state to its ground state by emitting a photon. This happens at some random time, at a rate which we shall estimate. It might seem that spontaneously emitting a photon is a strange thing for an atom to do, if it is simply sitting in free space with nothing apparently disturbing it. But this process is actually a very natural consequence of the coupling of the atom to electromagnetic \ufb01elds, described simply by the Jaynes\u2013Cummings interaction, HI = g(a\u2020\u03c3\u2212 + a\u03c3+), (7.109) as we recall from Section 7.5.2. Previously, we used this model to describe how a laser interacts with an atom, but the model also describes what happens to an atom even 316 Quantum computers: physical realization when no optical \ufb01eld is present! Consider an atom in its excited state coupled to a single mode which contains no photon, the state |01 (using |\ufb01eld, atom). This is not an eigenstate of HI, and thus it cannot remain stationary as time evolves. What happens is described by the unitary operator U in (7.77), by which we \ufb01nd that there is a probability", " pdecay = |10|U |01|2 for the atom to decay into its ground state and emit a photon, where pdecay = g2 4 sin2 1 2 (\u03c9 \u2212 \u03c90)2t (\u03c9 \u2212 \u03c90)2, (7.110) to lowest order in g, the atom\u2013\ufb01eld coupling. \u03c9 is the frequency of the photon, and \u03c90 the energy difference between the two levels of the atom. An atom sitting in free space interacts with many different optical modes; inserting the coupling g2 = \u03c92 0 2\u03c90c2 |0|\u03bc|1|2, (7.111) where \u03bc is the atomic dipole operator, integrating over all the optical modes (Exercise 7.29) and taking a time derivative gives the probability per second of decay, \u03b3rad = \u03c93 |0|\u03bc|1|2 0 3\u03c00c5. (7.112) If we make the approximation that |0|\u03bc|1| \u2248 \u03bcB \u2248 9\u00d710\u221224 J/T, the Bohr magneton, and assume that \u03c90/2\u03c0 \u2248 10 GHz, then \u03b3rad \u2248 10\u221215 sec\u22121, a spontaneous emission rate of less than one decay every 3 000 000 years. This calculation is representative of those done to estimate lifetimes of atomic states; as you can see, the hyper\ufb01ne states can have remarkably long coherence times in theory, and this is generally consistent with experiments, in which lifetimes of tens of seconds to tens of hours have been observed. Exercise 7.29: (Spontaneous emission) The spontaneous emission rate (7.112) can be derived from (7.110)\u2013(7.111) by the following steps. 1. Integrate 1 (2\u03c0c)3 8\u03c0 3 - \u221e 0 \u03c92 pdecayd\u03c9, (7.113) where the 8\u03c0/3 comes from summing over polarizations and integrating over the solid angle d\u03a9, and \u03c92/(2\u03c0c)3 comes from the mode density in three-dimensional space. (Hint: you may want to extend the lower limit of the integral to \u2212\u221e.) 2. Differentiate the result with respect to t, to obtain \u03b3rad. The form of g2 is a result of quantum electrodynamics; taking this for granted, the remainder of the calculation as presented here really stems from just the Jaynes\u2013Cummings interaction. Again, we see how considering its properties in the single atom, single photon regime gives us a fundamental property of atoms, without resorting to perturbation theory! Exercise 7.30: (Electronic state lifetimes) A calculation similar to that for \u03b3red can be done to estimate the lifetimes expected for electronic transitions, that is, those which involve energy level changes \u0394n = 0. For such transitions, the relevant Ion traps 317 interaction couples the atom\u2019s electric dipole moment to the electromagnetic \ufb01eld, giving \u03c92 0 2\u03c90 This gives a spontaneous emission rate g2 ed = |0|\u03bced|1|2. \u03b3ed red = \u03c93 0 |0|\u03bced|1|2 3\u03c00c3. (7.114) (7.115) Give a value for \u03b3ed red, taking |0|\u03bced|1| \u2248 qa0, where q is the electric charge, and a0 the Bohr radius, and assuming \u03c90/2\u03c0 \u2248 1015 Hz. The result show how much faster electronic states can decay compared with hyper\ufb01ne states. 7.6.2 The Hamiltonian Combining the simpli\ufb01ed models given in the previous section for the harmonic electromagnetic trap and the atomic structure provides us with the following simpli\ufb01ed toy model for an ion trap quantum information processor. Imagine a single two-level spin interacting via the usual magnetic dipole interaction HI = \u2212\u03bc \u00b7 B with an electromagS is proportional to the spin operator S, netic \ufb01eld, where the dipole moment \u03bc = \u03bcm and the magnetic \ufb01eld is B = B1 \u02c6x cos(kz \u2212 \u03c9t + \u03d5), and B1 is the \ufb01eld strength, k its momentum in the \u02c6z direction, \u03c9 its frequency, and \u03d5 its phase. Note that in this section, we shall use Sx = X/2, Sy = Y /2, and Sz = Z/2 as the spin operators; they are related to the Pauli operators by a factor of two. In addition to the usual electromagnetic interaction, there are interactions with the vibrational modes. The spin is physically con\ufb01ned within a harmonic", " potential of energy scale \u03c9z (Figure 7.9), such that its position becomes quantized and we must describe it by an operator z = z0(a\u2020 + a), where a\u2020, a are raising and lowering operators for the vibrational modes of the particle, representing creation and annihilation of phonons. Figure 7.9. Toy model of a trapped ion: a single particle in a harmonic potential with two internal states, interacting with electromagnetic radiation. Let us assume that the particle is cooled to near its lowest vibrational mode, such that the width of its oscillation in the well is small compared to the wavelength of the incident light, that is, the Lamb\u2013Dicke parameter \u03b7 \u2261 kz0 is small. De\ufb01ning the Rabi frequency 318 Quantum computers: physical realization of the spin as \u03a9 = \u03bcmB1/2, and recalling that Sx = (S+ + S\u2212)/2, we \ufb01nd that the interaction Hamiltonian simpli\ufb01es in the small \u03b7 limit to become HI = \u2212\u03bc \u00b7 B \u2019 S+ei(\u03d5\u2212\u03c9t) + S\u2212e\u2212i(\u03d5\u2212\u03c9t) \u2248 ( \u03a9 2 i + : \u03b7\u03a9 2 S+a + S\u2212a\u2020 + S+a\u2020 + S\u2212a ei(\u03d5\u2212\u03c9t) \u2212 e\u2212i(\u03d5\u2212\u03c9t) ; \u2019 (7.116) (. (7.117) The \ufb01rst term in brackets results from the usual Jaynes\u2013Cummings Hamiltonian as we saw in Section 7.5.2, which occurs when the location z of the spin is a constant. However, it is simpli\ufb01ed and does not contain photon operators because it turns out that as long as B1 is a strong coherent state, we can neglect its quantum properties and leave ourselves with a Hamiltonian which describes just the evolution of the internal atomic state. It is in fact quite remarkable that a coherent state of the \ufb01eld does not become entangled with an atom after interacting with it (to an excellent degree of approximation); this is a deep result which you may explore further by looking at Problem 7.3 at the end of the chapter. We shall also touch on this fact in describing resonance in Section 7.7.2. The second term in brackets describes the coupling of the motional state of the ion to its spin state, through the fact that the magnetic \ufb01eld it sees is dependent on its position. The four terms in braces correspond to four transitions (two up and two down) which are known as the red and blue motional sidebands, illustrated in Figure 7.10. % Figure 7.10. Energy levels of the toy model trapped ion showing the red and blue motional sideband transitions, which correspond to creation or annihilation of a single phonon. There is an in\ufb01nite ladder of additional motional states, which are usually not involved. The states are labeled as |n, m where n represents the spin state, and m the number of phonons. Why these sideband transitions have frequencies \u03c90 \u00b1 \u03c9z is easy to see, by including the free particle Hamiltonian H0 = \u03c90Sz + \u03c9za\u2020a, which causes the spin and phonon operators to evolve as S+(t) = S+ei\u03c90t a\u2020(t) = a\u2020ei\u03c9zt S\u2212(t) = S\u2212e\u2212i\u03c90t a(t) = ae\u2212i\u03c9zt. (7.118) (7.119) (7.120) Ion traps 319 Thus, in the frame of reference of H0, the dominant terms of H are found to be I = eiH0t/HIe\u2212iH0t/ \u23a7 \u23a8 H I = \u23a9 S+a\u2020ei\u03d5 \u2212 S\u2212ae\u2212i\u03d5 S+aei\u03d5 \u2212 S\u2212a\u2020e\u2212i\u03d5 i \u03b7\u03a9 2 i \u03b7\u03a9 2 \u03c9 = \u03c90 + \u03c9z \u03c9 = \u03c90 \u2212 \u03c9z (7.121) where the frequency of the electromagnetic \ufb01eld, \u03c9, is as shown on the right. Extending the above model from one spin to N spins con\ufb01ned within the same harmonic potential is simple if we assume that they share a single center of mass vibrational mode, whose energy is much lower than any other vibrational mode of the system. A straightforward extension of the theory shows that the only required modi\ufb01cation is replacement of \u03a9 by \u03a9/ N, since all N particles move together collectively. \u221a 7.6.3 Quantum computation Quantum computation with trapped ions requires one to be able to construct arbitrary", " unitary transforms on the internal states of the atoms. We now show how this is done, in three steps: we describe (1) how arbitrary single qubit operations are performed on the internal atomic (spin) state, (2) a method for performing a controlled two qubit gate between the spin and the phonon state, and (3) a way to swap quantum information between the spin and the phonon. Given these building blocks, we then describe an experiment which was performed to demonstrate a controlledgate, complete with state preparation and readout. Single qubit operations Applying an electromagnetic \ufb01eld tuned to frequency \u03c90 turns on the internal Hamiltonian term H internal I = \u03a9 2 S+ei\u03d5 + S\u2212e\u2212i\u03d5. (7.122) By choosing \u03d5 and the duration of the interaction appropriately, this allows us to perform rotation operations Rx(\u03b8) = exp(\u2212i\u03b8Sx) and Ry(\u03b8) = exp(\u2212i\u03b8Sy), which, by Theorem 4.1 on page 175, thereby allow us to perform any single qubit operation on the spin state. We shall denote rotations on the jth ion by a subscript, for example, Rxj(\u03b8). Exercise 7.31: Construct a Hadamard gate from Ry and Rx rotations. Controlled phase-\ufb02ip gate Suppose, now, that one qubit is stored in the atom\u2019s internal spin state, and another qubit is stored using the |0 and |1 phonon states. If this is the case, we can perform a controlled phase-\ufb02ip gate, with the unitary transform \u23a4 \u23a1 \u23a2 \u23a2 \u23a3 1 \u23a5 \u23a5 \u23a6. (7.123) It is easiest to explain how to do this with an atom that has a third energy level, as shown in Figure 7.11 (the extra level is not fundamentally necessary; see Problem 7.4). A laser 320 Quantum computers: physical realization is tuned to the frequency \u03c9aux + \u03c9z, to cause transitions between the |20 and |11 states; this turns on a Hamiltonian of the form \u03b7\u03a9 2 +ei\u03d5 + S S Haux = i \u2212e\u2212i\u03d5 (7.124), + and S where S \u2212 denote transitions between |20 and |11, and we assume that higher order motional states are unoccupied. Note that because of the uniqueness of this frequency, no other transitions are excited. We apply the laser with phase and duration to perform a 2\u03c0 pulse, that is, the rotation Rx(2\u03c0) on the space spanned by |11 and |20, which is just the unitary transform |11 \u2192 \u2212|11. All the other states remain unchanged, assuming that undesired states such as |1, 2 have no probability amplitude. This realizes the transform of (7.123), as desired. We shall write this gate as Cj(Z) (denoting a controlled-Z operation), where j indicates which ion the gate is applied to. Note that the same phonon is shared by all the ions, since it is a center-of-mass phonon; because of this, adopting engineering terminology, this has been called the phonon \u2018bus\u2019 qubit in the literature. Figure 7.11. Energy levels of a three-level atom in an ion trap, with two phonon states each. The labels |n, m indicate the atom\u2019s state n and the phonon state m. The |20 \u2194 |11 transition is used to perform a controlled phase-\ufb02ip gate. Swap gate Finally, we need some way to swap qubits between the atom\u2019s internal spin state and the phonon state. This can be done by tuning a laser to the frequency \u03c90 \u2212 \u03c9z, and arranging for the phase to be such that we perform the rotation Ry(\u03c0) on the subspace spanned by |01 and |10, which is just the unitary transform \u23a4 \u23a1 \u23a2 \u23a2 \u23a3 0 1 0 0 0 \u22127.125) on the |00, |01, |10, |11 space. If the initial state is a|00 + b|10 (that is, the phonon is initially |0), then the state after the swap is a|00 + b|01, so this accomplishes the j when acting on ion j; the inverse desired swap operation. We shall write this as j corresponds to Ry(\u2212\u03c0). Technically, because of the minus sign in the operation |1001| entry of Ry(\u03c0), this is not a perfect swap operation, but it is equivalent to one up relative phases (see Exercise 4.26). Thus, this is sometimes referred to as being a \u2018", "mapping operation\u2019 instead of as a swap. Ion traps 321 Putting these gates together allows us to construct a and k (target) using the sequence of operations gate acting on ions j (control) Controlled- gate jk = Hk k Cj(Z) k Hk, (7.126) (time going from right to left, as usual for matrices) where Hk is a Hadamard gate (constructed from Ry and Rx rotations on ion k). This is very similar to how a controlledgate was constructed using beamsplitters and optical Kerr media, as in Equation (7.46). 7.6.4 Experiment Figure 7.12. Photograph of a microfabricated elliptical electrode ion trap, in which ions have been con\ufb01ned. The ions in this trap are barium ions, rather than beryllium, but the basic principles are the same as described in the text. Reproduced courtesy of R. Devoe and C. Kurtsiefer, IBM Almaden Research Center. A controlled- gate using a single trapped ion has been demonstrated (see \u2018History and further reading\u2019 at the end of the chapter); precisely how this experiment is done is insightful. In the experiment, a single ion of 9Be+ is trapped in a coaxial resonator RF ion trap, different in geometry from the linear ion trap of Figure 7.7, but functionally equivalent, and similar to the photograph of an actual ion trap shown in Figure 7.12. Beryllium was chosen for its convenient hyper\ufb01ne and electronic level structure, shown in Figure 7.13. The 2S1/2(1, 1) and 2S1/2(2, 2) energy levels (Exercise 7.28) are used as the atom\u2019s internal qubit state, and the |0 and |1 phonon states as another qubit (labeled 322 Quantum computers: physical realization Figure 7.13. Energy levels of 9Be+ used in the ion trap experiment. Figure courtesy of C. Monroe at NIST. in the \ufb01gure as n = 0 and n = 1). The \u2248 313 nm transition between the 2S1/2(1, 1) and 2S1/2(2, 2) levels is accomplished not by tuning a single laser to the transition frequency, but rather two lasers whose difference frequency is that of the transition. This Raman transition method simpli\ufb01es requirements for laser phase stability. The 2S1/2(2, 0) state is used as the auxiliary level; the 2S1/2 states have different energies by virtue of a 0.18 millitesla magnetic \ufb01eld applied to the system. The trapped ion has vibrational frequencies (\u03c9x, \u03c9y, \u03c9z)/2\u03c0 = (11.2, 18.2, 29.8) MHz in the trap, and a ground state nx = 0 wavefunction spread of about 7 nm, giving a Lamb\u2013Dicke parameter of about \u03b7x = 0.2. The Rabi frequency of the on-resonance transition is \u03a9/2\u03c0 = 140 kHz, the two motional sidebands, \u03b7x\u03a9/2\u03c0 = 30 kHz, and the auxiliary transition \u03b7x\u03a9/2\u03c0 = 12 kHz. The state of the ion is initialized using Doppler and sideband cooling to obtain, with approximately 95% probability, the state |00 = |2S1/2(2, 2)|nx = 0. The internal and motional states of the ion are then prepared in one of the four basis states |00, |01, |10, or |11 using single qubit operations, then a controlledgate is performed using three pulses, which implement a Ry(\u03c0/2) rotation on the internal state qubit, a controlled-Z operation between the two qubits, then a Ry(\u2212\u03c0/2) rotation on the internal state qubit. It is simple to show (Exercise 7.32) that this circuit, drawn in Figure 7.14, realizes a controlled- gate. Readout of the computational output is performed with two measurements. The \ufb01rst is to collect the \ufb02uorescence from the ion which occurs when + circularly polarized light tuned to the 2S1/2(2, 2) \u2013 2P3/2(3, 3) \u2018cycling\u2019 transition is applied. The light does not couple appreciably to the 2S1/2(1, 1) state, and thus the intensity of the observed \ufb02uorescence is proportional to the probability of the internal state qubit being in the |0 Ion traps 323 Figure 7.14. Quantum circuit modeling the ion trap controlledphonon state, and the bottom, the ion\u2019s internal hyper\ufffd", "\ufffd\ufffdne state. experiment. The top wire represents the state; it is a projective measurement. This measurement technique is powerful because the transition cycles many times \u2013 the ion absorbs a photon, jumping to the 2P3/2(3, 3) state, then emits a photon, decaying back into the 2S1/2(2, 2) state where it started. Thousands or more cycles are possible, allowing good statistics to be accumulated. The second measurement is similar to the \ufb01rst, but a swap pulse is applied \ufb01rst to exchange the motional and internal state qubits; this projectively measures the motional state qubit. The experiment as performed veri\ufb01es the classical truth table of the controlledoperation and, in principle, by preparing superposition input states and measuring output density matrices, the unitary transform could be completely characterized using process gate requires about 50 microseconds to tomography (Chapter 8). The controlledperform with the optical power used in the experiment. On the other hand, the coherence time was measured to be somewhere around hundreds to thousands of microseconds. The dominant decoherence mechanisms included instabilities in the laser beam power and the RF ion trap drive frequency and voltage amplitude, and \ufb02uctuating external magnetic \ufb01elds. Moreover, the experiment involved only a single ion, and only two qubits, and gate should generally thus was not useful for computation; to be useful, a controlledbe applied between different ions, and not just between a single ion and the motional state. However, the technical limitations can probably be overcome, and lifetimes can be extended by using the short-lived motional state only intermittently, capitalizing on the much longer coherence times of the internal atomic states. And scaling to larger numbers of ions is conceptually viable. Shown in Figure 7.15 is a string of 40 mercury ions which have been trapped. There are many hurdles to making such systems behave as useful quantum information processing machines, but technological surprises are a neverending saga. Someday, perhaps, trapped ions such as these could be registers of qubits in a quantum computer. Figure 7.15. Image of \ufb02uorescence from about 40 trapped mercury (199Hg+) atomic ions. The ions are spaced by approximately 15 micrometers, and the two apparent gaps are different isotopes of mercury which do not respond to the probe laser. Reprinted courtesy of D. Wineland, at NIST. 324 Quantum computers: physical realization Exercise 7.32: Show that the circuit in Figure 7.14 is equivalent (up to relative phases) to a controlled- gate, with the phonon state as the control qubit. Ion trap quantum computer \u2022 Qubit representation: Hyper\ufb01ne (nuclear spin) state of an atom, and lowest level vibrational modes (phonons) of trapped atoms. \u2022 Unitary evolution: Arbitrary transforms are constructed from application of laser pulses which externally manipulate the atomic state, via the Jaynes\u2013Cummings interaction. Qubits interact via a shared phonon state. \u2022 Initial state preparation: Cool the atoms (by trapping and using optical pumping) into their motional ground state, and hyper\ufb01ne ground state. \u2022 Readout: Measure population of hyper\ufb01ne states. \u2022 Drawbacks: Phonon lifetimes are short, and ions are dif\ufb01cult to prepare in their motional ground states. 7.7 Nuclear magnetic resonance Nuclear spin systems would be nearly ideal for quantum computation if only spin\u2013 spin couplings could be large and controllable; this is an important observation from our study of ion traps in the last section. The principal drawback of ion trap quantum computers is the weakness of the phonon mediated spin\u2013spin coupling technique and its susceptibility to decoherence. One way this limitation could be circumvented would be to trap molecules instead of single atoms \u2013 the magnetic dipole and electron mediated Fermi contact interactions between neighboring nuclei would provide strong natural couplings. However, with their many vibrational modes, single molecules have been dif\ufb01cult to trap and cool, and thus optical manipulation and detection of nuclear spins in trapped molecules has not been feasible except in special circumstances. On the other hand, direct manipulation and detection of nuclear spin states using radiofrequency electromagnetic waves is a well-developed \ufb01eld known as nuclear magnetic resonance (NMR). These techniques are widely used in chemistry, for example, to measure properties of liquids, solids, and gases, to determine the structure of molecules, and to image materials and even biological systems. These many applications have lead the technology of NMR to become quite sophisticated, allowing control and observation of tens to hundreds and thousands of nuclei in an experiment. However, two problems arise in using NMR for quantum computation. First, because of the smallness of the nuclear magnetic moment, a large number (more than", " \u2248 108) molecules must be present in order to produce a measurable induction signal. Conceptually, a single molecule might be a \ufb01ne quantum computer, but how can this be true of an ensemble of molecules? In particular, the output of an NMR measurement is an average over all the molecule\u2019s signals; can the average output of an ensemble of quantum computers be meaningful? Second, NMR is typically applied to physical systems in equilibrium at room temperature, where the spin energy \u03c9 is much less than kBT. This means that the initial state of the spins is nearly completely random. Traditional quantum computation requires the system be prepared in a pure state; how can quantum computation be performed with a system which is in a high entropy mixed state? Nuclear magnetic resonance 325 Solutions to these two problems have made NMR a particularly attractive and insightful method for implementing quantum computation, despite stringent limitations which arise from the thermal nature of typical systems. Many lessons can be learned from NMR: for example, techniques for controlling realistic Hamiltonians to perform arbitrary unitary transforms, methods for characterizing and circumventing decoherence (and systematic errors), and considerations which arise in assembling components in implementing full quantum algorithms on entire systems. We begin with a description of the physical apparatus and the main Hamiltonian involved, then we discuss how quantum information processing with NMR is possible despite the thermal input state and ensemble problems, concluding with some experiments which have been performed demonstrating quantum algorithms, and the drawbacks of this method. 7.7.1 Physical apparatus Let us begin with a general description of the apparatus, whose workings will be mathematically modeled in detail later. The two main parts of a pulsed NMR system for liquid samples, which we shall focus on here, are the sample and the NMR spectrometer. A typical molecule which might be used would contain a number n of protons which have spin 1/2 (other possible nuclei include 13C, 19F, 15N, and 31P), and produce an NMR signal at about 500 MHz when placed in a magnetic \ufb01eld of about 11.7 tesla. The frequencies of different nuclei within a molecule can differ by a few kHz to hundreds of kHz because of differences in the local magnetic \ufb01elds due to chemical environment shielding effects. The molecules are typically dissolved in a solvent, reducing the concentration to the extent that inter-molecular interactions become negligible, leaving a system that might well be described as an ensemble of n qubit quantum \u2018computers.\u2019 The spectrometer itself is constructed from radiofrequency (RF) electronics and a large superconducting magnet, within the bore of which is held the sample in a glass tube, as shown in Figure 7.16. There, the static \u02c6z oriented magnetic \ufb01eld B0 is carefully trimmed to be uniform over approximately 1 cm3 to better than one part in 109. Orthogonal saddle or Helmholtz coils lying in the transverse plane allow small, oscillating magnetic \ufb01elds to be applied along the \u02c6x and \u02c6y directions. These \ufb01elds can be rapidly pulsed on and off to manipulate nuclear spin states. The same coils are also part of tuned circuits which are used to pick up the RF signal generated by the precessing nuclei (much like how a spinning magnet inductively generates an alternating current in a nearby coil). A typical experiment begins with a long waiting period in which the nuclei are allowed to thermalize to equilibrium; this can require several minutes for well-prepared liquid samples. Under control of a (classical) computer, RF pulses are then applied to effect the desired transformation on the state of the nuclei. The high power pulse ampli\ufb01ers are then quickly switched off and a sensitive pre-ampli\ufb01er is enabled, to measure the \ufb01nal state of the spins. This output, called the free induction decay, is Fourier transformed to obtain a frequency spectrum with peaks whose areas are functions of the spin states (Figure 7.17). There are many important practical issues which lead to observable imperfections. For example, spatial inhomogeneities in the static magnetic \ufb01eld cause nuclei in different parts of the \ufb01elds to precess at different frequencies. This broadens lines in the spectrum. An even more challenging problem is the homogeneity of the RF \ufb01eld, which is generated by a coil which must be orthogonal to the B0 magnet; this geometric constraint and the requirement to simultaneously maintain high B0 homogeneity usually forces the RF 326 Quantum computers: physical realization Figure 7.16. Schematic diagram of an NMR apparatus. \ufb01eld to be inhomogeneous and generated by a small coil, leading to imperfect control of the nuclear system. Also, pulse timing, and stability of power, phase, and frequency are", " important issues; however, unlike the ion traps, because of the lower frequencies, good control of these parameters is more tractable. We shall return to imperfections in Section 7.7.4, after understanding the basic mathematical description of the system and the methodology for performing quantum information processing with NMR. 7.7.2 The Hamiltonian The basic theory of NMR can be understood from an ideal model of one and two spins, which we describe here. The \ufb01rst step is to describe how electromagnetic radiation interacts with a single magnetic spin. We then consider the physical nature of couplings between spins which arise in molecules. These tools enable us to model readout of the magnetization which results from transformation of an initial state which is in thermal equilibrium. Finally, we describe a phenomenological model of decoherence, and how its T1 and T2 parameters can be experimentally determined. Single spin dynamics The magnetic interaction of a classical electromagnetic \ufb01eld with a two-state spin is described by the Hamiltonian H = \u2212\u03bc \u00b7 B, where \u03bc is the spin, and B = B0 \u02c6z + B1( \u02c6x cos \u03c9t + \u02c6y sin \u03c9t) is a typical applied magnetic \ufb01eld. B0 is static and very large, and B1 is usually time varying and several orders of magnitude smaller than B0 in strength, so that perturbation theory is traditionally employed to study this system. However, the Schr\u00a8odinger equation for this system can be solved straightforwardly without perturbation theory, using the Pauli matrix techniques of Chapter 2, in terms of which the Hamiltonian can be written as H = \u03c90 2 Z + g(X cos \u03c9t + Y sin \u03c9t), (7.127) Nuclear magnetic resonance 327 where g is related to the strength of the B1 \ufb01eld, and \u03c90 to B0, and X, Y, Z are the Pauli matrices as usual. De\ufb01ne |\u03d5(t) = ei\u03c9tZ/2|\u03c7(t), such that the Schr\u00a8odinger equation i\u2202t|\u03c7(t) = H|\u03c7(t) can be re-expressed as i\u2202t|\u03d5(t) = ei\u03c9Zt/2He\u2212i\u03c9Zt/2 \u2212 Z |\u03d5(t). \u03c9 2 Since ei\u03c9Zt/2Xe\u2212i\u03c9Zt/2 = (X cos \u03c9t \u2212 Y sin \u03c9t), (7.129) simpli\ufb01es to become i\u2202t|\u03d5(t) = \u03c90 \u2212 \u03c9 2 Z + gX |\u03d5(t), (7.128) (7.129) (7.130) (7.131) where the terms on the right multiplying the state can be identi\ufb01ed as the effective \u2018rotating frame\u2019 Hamiltonian. The solution to this equation is i \u03c9 0 \u2212\u03c9 2 Z+gX t |\u03d5(t) = e |\u03d5(0). (7.132) The concept of resonance arises from the behavior of this solution, which can be understood using (4.8) to be a single qubit rotation about the axis \u02c6n = \u02c6z + 2g \u03c90\u2212\u03c9 \u02c6x 1 + 2g \u03c90\u2212\u03c9 2 by an angle |n| = t 2 \u03c90 \u2212 \u03c9 2 + g2. (7.133) (7.134) When \u03c9 is far from \u03c90, the spin is negligibly affected by the B1 \ufb01eld; the axis of its rotation is nearly parallel with \u02c6z, and its time evolution is nearly exactly that of the free B0 Hamiltonian. On the other hand, when \u03c90 \u2248 \u03c9, the B0 contribution becomes negligible, and a small B1 \ufb01eld can cause large changes in the state, corresponding to rotations about the \u02c6x axis. The enormous effect a small perturbation can have on the spin system, when tuned to the appropriate frequency, is responsible for the \u2018resonance\u2019 in nuclear magnetic resonance. The same effect, of course, is also at the heart of the selectivity of two-level atoms for speci\ufb01cally tuned laser \ufb01elds that was used (but not explained) in Section 7.5.1. In general, when \u03c9 = \u03c90, the single spin rotating frame Hamiltonian can be written as H = g1(t)X + g2(t)Y, (7.135) where g1 and g2 are functions of the", " applied transverse RF \ufb01elds. Exercise 7.33: (Magnetic resonance) Show that (7.128) simpli\ufb01es to become (7.129). What laboratory frame Hamiltonian gives rise to the rotating frame Hamiltonian (7.135)? 328 Quantum computers: physical realization Exercise 7.34: (NMR frequencies) Starting with the nuclear Bohr magneton, compute the precession frequency of a proton in a magnetic \ufb01eld of 11.8 tesla. How many gauss should B1 be to accomplish a 90\u25e6 rotation in 10 microseconds? Spin\u2013spin couplings More than one spin is usually present in systems of interest; 1H, 13C, 19F, and 15N all have nuclear spin 1/2. These spins interact through two dominant mechanisms: direct dipolar coupling, and indirect through-bond electron mediated interactions. Throughspace dipolar coupling is described by an interaction Hamiltonian of the form H D 1,2 = \u03b31\u03b32 4r3 \u03c31 \u00b7 \u03c32 \u2212 3(\u03c31 \u00b7 \u02c6n)(\u03c32 \u00b7 \u02c6n), (7.136) where \u02c6n is the unit vector in the direction joining the two nuclei, and \u03c3 is the magnetic moment vector (times two). In a low viscosity liquid, dipolar interactions are rapidly averaged away; mathematically this is calculated by showing that the spherical average of H D 1,2 over \u02c6n goes to zero as the averaging becomes rapid compared to the dipolar coupling energy scale. Through-bond interactions, also known simply as \u2018J-coupling,\u2019 are indirect interactions mediated by electrons shared through a chemical bond; the magnetic \ufb01eld seen by one nucleus is perturbed by the state of its electronic cloud, which interacts with another nucleus through the overlap of the electronic wavefunction with the nucleus (a Fermi contact interaction). This coupling has the form H J 1,2 = J 4 \u03c31 \u00b7 \u03c32 = J 4 Z1Z2 + J 8 \u03c3+\u03c3\u2212 + \u03c3\u2212\u03c3+. (7.137) We shall be interested in the case where J is a scalar (in general it may be a tensor), which is an excellent approximation in liquids and when couplings are weak, or when the interacting nuclear species have vastly different precession frequencies. This case is described by H J 12 \u2248 4 JZ1Z2. (7.138) Exercise 7.35: (Motional narrowing) Show that the spherical average of H D 1,2 over \u02c6n is zero. Thermal equilibrium NMR differs signi\ufb01cantly from the other physical systems we have studied previously in this chapter in that it uses an ensemble of systems, and the primary measurement is an ensemble average. Furthermore, no extensive procedures are employed to prepare the initial state in a special state such as the ground state; in fact, to do so is challenging with present technology. Rather, the initial state is the thermal equilibrium state, e\u2212\u03b2H Z where \u03b2 = 1/kBT, and Z = tr e\u2212\u03b2H is the usual partition function normalization, which ensures that tr(\u03c1) = 1. Since \u03b2 \u2248 10\u22124 at modest \ufb01elds for typical nuclei at room (7.139) \u03c1 =, Nuclear magnetic resonance 329 temperature, the high temperature approximation \u03c1 \u2248 2\u2212n 1 \u2212 \u03b2H (7.140) is appropriate, for a system of n spins. Since spin\u2013spin couplings are very small compared with the precession frequencies, the thermal state density matrix is very nearly diagonal in the Z basis, and thus it can be interpreted as being a mixture of the pure states |00... 0, |00... 01,..., |11... 1. What is actually the true physical state of each ensemble member is a matter of debate, because an in\ufb01nite number of unravelings exist for a given density matrix. In principle, with NMR the true physical state can be measured if the ensemble members (individual molecules) are accessible, but this is experimentally dif\ufb01cult. Exercise 7.36: (Thermal equilibrium NMR state) For n = 1 show that the thermal equilibrium state is \u03c1 \u2248 1 \u2212 and for n = 2 (and \u03c9A \u2248 4\u03c9B), \u03c9 2kBT 1 0 0 \u22121, \u03c1 \u2248 1 \u2212 \u03c9B 4kBT \u23a1 \u23a2 \u23a2 \u23a3 3 0 0 0 0 0 \u22125 (7.141) (7.142) Magnetization readout The principal output of an experiment is the free induction decay signal, mathematically given as V (t) = V0tr", " e\u2212iHt\u03c1eiHt(iXk + Yk), (7.143) where Xk and Yk operate only on the kth spin, and V0 is a constant factor dependent on coil geometry, quality factor, and maximum magnetic \ufb02ux from the sample volume. This signal originates from the pickup coils detecting the magnetization of the sample in the \u02c6x \u2212 \u02c6y plane. In the laboratory frame, this signal will oscillate at a frequency equal to the precession frequency \u03c90 of the nuclei; however, V (t) is usually mixed down with an oscillator locked at \u03c90, then Fourier transformed, such that the \ufb01nal signal appears as shown in Figure 7.17. Exercise 7.37: (NMR spectrum of coupled spins) Calculate V (t) for H = JZ1Z2 4 [1 \u2212 \u03b2\u03c90(Z1 + Z2)]e\u2212i\u03c0Y1/4. How many lines would there be in and \u03c1 = ei\u03c0Y1/4 1 the spectrum of the \ufb01rst spin if the Hamiltonian were H = JZ1(Z2 + Z3 + Z4) (with a similar initial density matrix) and what would their relative magnitudes be? Decoherence A prominent characteristic of the free induction decay whose description lies outside the simple models presented so far for NMR is the exponential decay of the magnetization signal. One cause of this is inhomogeneity in the static magnetic \ufb01eld, which leads to 330 Quantum computers: physical realization 1 0..6 0.4 0.2 0 \u2212500 \u2212400 \u2212300 \u2212200 \u2212100 0 Frequency from 125.77 MHz in Hz 100 200 300 400 500 Figure 7.17. Carbon spectrum of 13C labeled trichloroethylene. The four lines on the left come from the carbon nucleus directly bound to the proton; four lines appear because of couplings to the proton and to the second carbon nucleus, whose own signal gives the closely spaced four lines on the right. The second carbon nucleus is further away from the proton than the \ufb01rst, and thus has a much smaller coupling to it. precessing spins in one part of the sample getting out of phase with those in another part. Effects due to inhomogeneities are reversible in principle, but there are other sources of phase randomization which are fundamentally irreversible, such as those originating from spin\u2013spin couplings. Another irreversible mechanism is the thermalization of the spins to equilibrium at the temperature of their environment, a process which involves exchange of energy. For a single qubit state, these effects may be phenomenologically characterized with a density matrix transformation model, a b\u2217 b 1 \u2212 a \u2192 (a \u2212 a0)e\u2212t/T1 + a0 b\u2217e\u2212t/T2 be\u2212t/T2 (a0 \u2212 a)e\u2212t/T1 + 1 \u2212 a0, (7.144) where T1 and T2 are known as the spin\u2013lattice (or \u2018longitudinal\u2019) and spin\u2013spin (or \u2018transverse\u2019) relaxation rates, respectively, and a0 characterizes the thermal equilibrium state. They de\ufb01ne important time scales for the lifetimes of non-equilibrium classical states and quantum superpositions. Theoretical tools for calculating T1 and T2 in NMR systems are well-developed, and, in fact, measurements of these rates play an important role in using NMR to distinguish between different chemical species. Experimental methods for measuring T1 and T2 are well known in NMR. Let Rx = e\u2212i\u03c0X/4 be a 90\u25e6 pulse about the \u02c6x axis. To measure T1, apply R2 x, wait time \u03c4, then Rx. The \ufb01rst pulse \ufb02ips the spin by 180\u25e6, after which it relaxes for time \u03c4 back towards equilibrium (visualize this as the Bloch vector shrinking back towards the top of the Bloch sphere, the ground state), then the \ufb01nal 90\u25e6 pulse puts the magnetization in the \u02c6x \u2212 \u02c6y plane, where it is detected. The measured magnetization M from this inversion\u2013recovery experiment is found to decay exponentially with \u03c4 as M = M0. To 1 \u2212 2 exp(\u2212\u03c4 /T1) Nuclear magnetic resonance 331 measure T2, one can, to \ufb01rst order, simply measure the linewidth of a peak. A better way, the Carr\u2013Purcell\u2013Meiboom\u2013Gill technique, is to apply an Rx operation, followed x\u2019. This train of 180\u25e6 by k iterations of \u2018wait time \u03c4 /2, apply R2 pulses \ufffd", "\ufffdrefocuses\u2019 couplings (Section 7.7.3) and partially cancels B0 \ufb01eld inhomogeneities, so that one can better estimate the true T2 of the system. The observed magnetization decays as M = M0e\u2212k\u03c4 /T2. x, wait time \u03c4 /2, apply R2 Multiple spin Hamiltonian Summarizing our discussion of the NMR Hamiltonian, we can write H for an n spin coupled system as H = \u03c9kZk + k j,k H J j,k + H RF + j,k H D j,k + H env, (7.145) where the \ufb01rst term is the free precession of the spins in the ambient magnetic \ufb01eld, H D is the magnetic dipole coupling of (7.136), H J is the \u2018J\u2019 coupling of (7.137), H RF describes the effect of the externally applied radiofrequency magnetic \ufb01elds of (7.135), and H env describes interactions with the environment which lead to decoherence, as in (7.144). For the sake of understanding the basic principles about how this Hamiltonian can be manipulated, we shall \ufb01nd it suf\ufb01cient to consider the simpli\ufb01ed Hamiltonian H = \u03c9kZk + k j,k Zj \u2297 Zk + k k (t)Xk + gy gx k(t)Yk, (7.146) in much of the following discussion. The treatment of the more general (7.145) follows the same ideas. 7.7.3 Quantum computation Quantum information processing requires performing unitary transforms to a system prepared in a proper initial state. Three questions arise for the present system: First, how can arbitrary unitary transforms be implemented in a system of n coupled spins described by the Hamiltonian of (7.146)? And second, how can the thermal state (7.140) of an NMR system be used as a proper initial state for computation? Third, the quantum algorithms we have studied in the last three chapters ask for projective measurements to obtain output results, whereas with NMR, we can only easily perform ensemble average measurements. How can we deal with this ensemble readout problem? We answer these questions in this section. Refocusing Perhaps one of the most interesting techniques available to us in performing arbitrary unitary transforms using Hamiltonians of the sort of (7.146) is refocusing, as it is known in the art of NMR. Consider the simple two spin Hamiltonian H = H sys + H RF where H sys = aZ1 + bZ2 + cZ1Z2. (7.147) As was shown in Section 7.7.2, when a large RF \ufb01eld is applied at the proper frequency, to a good approximation, we can approximate e\u2212iHt/ \u2248 e\u2212iHRFt/. (7.148) 332 Quantum computers: physical realization This allows arbitrary single qubit operations to be performed with excellent \ufb01delity. Let us de\ufb01ne Rx1 = e\u2212i\u03c0X1/4 (7.149) as a 90\u25e6 rotation about \u02c6x on spin 1, and similarly for spin 2. The 180\u25e6 rotation R2 the special property that x1 has R2 x1e\u2212iaZ1tR2 x1 = eiaZ1t, (7.150) as can be easily veri\ufb01ed. This is known as refocusing, because of the way it reverses time evolution such that different frequency spins starting together at some point on the Bloch sphere come back to the same point on the Bloch sphere. 180\u25e6 pulses applied in this manner are known as refocusing pulses. Note that in the above expression, a can be an operator as well as a constant (as long as it contains no operators which act on spin 1), and thus e\u2212iHsys t/R2 x1e\u2212iHsys t/R2 x1 = e\u22122ibZ2t/. (7.151) Using another set of refocusing pulses applied to spin 2 would remove even this remaining term. Refocusing is thus a useful technique for removing coupled evolution between spins, and for removing all evolution entirely. Exercise 7.38: (Refocusing) Explicitly show that (7.150) is true (use the anti-commutativity of the Pauli matrices). Exercise 7.39: (Three-dimensional refocusing) What set of pulses can be used to refocus evolution under any single spin Hamiltonian H sys = k ck\u03c3k? Exercise 7.40: (Refocusing dipolar interactions) Give a sequence of pulses which 1,2 into the much simpler form can be used to turn", " two spin dipolar coupling H D of (7.138). ControlledRealization of a controlledgate is simple using refocusing pulses and single qubit pulses. Let us show how this is done for a two spin system with the Hamiltonian of (7.147). From the construction of (7.46), we know that being able to realize the unitary transform gate \u23a1 UCZ = \u23a2 \u23a2 \u23a3 \u23a4 1 \u23a5 \u23a5 \u23a6 (7.152) \u221a is suf\ufb01cient. Since from one evolution period of time \u03c0/4c together with several single qubit pulses is straightforward. i eiZ1Z2\u03c0/4e\u2212iZ1\u03c0/4e\u2212iZ2\u03c0/4 = UCZ, getting a controlled- Exercise 7.41: (NMR controlled- ) Give an explicit sequence of single qubit rotations which realize a controlledHamiltonian of (7.147). You may start with (7.46), but the result can be simpli\ufb01ed to reduce the number of single qubit rotations. between two spins evolving under the Nuclear magnetic resonance 333 Temporal, spatial, and logical labeling Being able to realize arbitrary unitary transforms on a spin system to good \ufb01delity using RF pulses is one of the most attractive aspects of NMR for quantum computation. However, the major drawback is the fact that the initial state is usually the thermal state of (7.140). Despite the high entropy of this state, quantum computation can nevertheless be done, with some cost. Two techniques for achieving this are called temporal and logical labeling. Temporal labeling, also sometimes called temporal averaging, is based on two important facts: quantum operations are linear, and the observables measured in NMR are traceless (see Section 2.2.5 for background on quantum measurements). Suppose a two spin system starts out with the density matrix \u23a1 \u23a2 \u23a2 \u23a3 \u03c11 = (7.153) where a, b, c, and d are arbitrary positive numbers satisfying a+b+c+d = 1. We can use a circuit P constructed from controlledgates to obtain a state with the permuted populations and similarly, \u03c12 = P \u03c11P \u2020 = \u03c13 = P \u2020\u03c11P = \u23a2 \u23a2 \u23a3 \u23a1 \u23a2 \u23a2 \u23a3 \u23a1 \u23a4 \u23a5 \u23a5 \u23a6, 7.154) (7.155) A unitary quantum computation U is applied to each of these states, to obtain (in three separate experiments, which may be performed at different times) Ck = U \u03c1kU \u2020. By linearity, Ck = k=1,2,3 k U \u03c1kU \u2020 (7.156) (7.157) (7.1581 \u2212 a \u03c1k k U \u2020 \u23a1 = (4a \u2212 1)U \u23a2 \u23a2 \u23a3 1 0 0 0 In NMR, observables M (such as Pauli X and Y ) for which tr(M ) = 0 are the only ones ever measured; thus, tr CkM = tr CkM k k (7.159) 334 Quantum computers: physical realization \u239b \u23a1 \u23a2 \u23a2 \u23a3 0 0 0 0 \u239c \u239c \u239d0000|7.160) (7.161) = (4a \u2212 1) tr = (4a \u2212 1) tr We \ufb01nd that the sum of the measured signals from the three experiments gives us a result which is proportional to what we would have obtained had the original system been prepared in a pure state |0000| instead of in the arbitrary state of (7.153). This is always possible to accomplish for arbitrarily prepared systems of any size, given enough experiments which are summed over, and suf\ufb01ciently long coherence time for unitary operations to be performed before decoherence sets in. Note that the different Ck experiments can actually also be done simultaneously with three different systems, or in different parts of a single system; this is experimentally feasible by applying gradient magnetic \ufb01elds which vary systematically over a single sample, and that variant of this technique is called spatial labeling. Exercise 7.42: (Permutations for temporal labeling) Give a quantum circuit to accomplish the permutations P and P \u2020 necessary to transform \u03c11 of (7.153) to \u03c12 of (7.154). Logical labeling is based on similar observations, but does not require multiple experiments to be performed. Suppose we have a system of three nearly identical spins in the state \u23a1 \u23a2 \u23a2 \u23a2 \u23a2 \ufffd", "\ufffd\ufffd \u23a2 \u23a2 \u23a2 \u23a2 \u23a2 \u23a3 6 0 0 0 0 0 0 0 \u03c1 = \u03b4I + \u03b1 \u2248 \u03b4I + 7.162\u2212 6, (7.163) 2 0 0 \u22122 where \u03b4I refers to a background population which is unobservable (because of traceless measurement observables), and \u03b1 \u03b4 is a small constant. We may then apply a unitary operation which performs a permutation P, giving \u03c1 = P \u03c1P \u2020 = \u03b4I + 7.164) Nuclear magnetic resonance 335 \u23a1 Note that the upper 4\u00d74 block of this matrix has the form 6 0 0 \u22122 0 0 0 0 0\u2212 2 0 0 0 0 0\u2212 2 \u23a5 \u23a5 \u23a6 = 8|0000| \u2212 2I, \u23a2 \u23a2 \u23a3 \u23a4 (7.165) where I here denotes the 4\u00d74 identity matrix. Just as for temporal labeling, we \ufb01nd that if a computation is performed on such a state, in this case the |000, |001, |010, |011 manifold, then it produces a result which is proportional to what we would have obtained had the original system been prepared in a pure state |0000|! Experimentally, it is possible to perform \u02dcP, and also to isolate the signal from just this manifold of states. States which are of the form \u03c1 = 2\u2212n(1 \u2212 )I + U |00... 000... 0|U \u2020 (where U is any unitary operator), are called \u2018effective pure states\u2019, or \u2018pseudopure\u2019 states. n is the number of qubits, but the dimension of the Hilbert space need not be a power of two in general. There are many strategies for preparing such states, and in general they all incur some cost. We shall return to discuss this later, in Section 7.7.4. Effective pure states make it possible to observe zero temperature dynamics from a system which equilibrates to a high temperature state, as long as the coupling of the system to its high temperature environment is suf\ufb01ciently small. This is the way it is used in NMR quantum computation. Exercise 7.43: (Permutations for logical labeling) Give a quantum circuit to accomplish the permutations P necessary to transform \u03c1 of (7.163) to \u03c1 of (7.165). Exercise 7.44: (Logical labeling for n spins) Suppose we have a system of n nearly identical spins of Zeeman frequency \u03c9 in thermal equilibrium at temperature T with state \u03c1. What is the largest effective pure state that you can construct from \u03c1 using logical labeling? (Hint: take advantage of states whose labels have Hamming weight of n/2.) Ensemble readout of quantum algorithm results We have seen how arbitrary unitary transforms can be performed on an n spin system with a Hamiltonian of (7.146), and we have learned how to prepare from a thermal state a well-de\ufb01ned input which behaves as a low-entropy ground state. However, to complete the requirements for quantum computation, we must have a way of performing measurements on the system to read out computational results. The dif\ufb01culty is that the output of a typical quantum algorithm is a random number, whose distribution gives information which allows the problem to be solved. Unfortunately, the average value of the random variable does not necessarily give any relevant information. This would be the output if the quantum algorithm were executed without modi\ufb01cation on an NMR quantum computer, since it is performed with a large ensemble of molecules, rather than with a single n spin molecule. This dif\ufb01culty is illustrated by the following example. The quantum factoring algorithm produces as its output a random rational number c/r, where c is an unknown integer, and r is the desired result (also an integer). Normally, a projective measurement is used to obtain c/r, then a classical continued fractions algorithm is performed to obtain c with high probability (Section 5.3.1). The answer is then checked by plugging in the 336 Quantum computers: physical realization result to the original problem, and if it fails the entire algorithm is repeated. Unfortunately, however, if only the ensemble average is available, since c is nearly uniformly distributed the average value c/r gives no meaningful information. A simple resolution to this problem, which works for all quantum algorithms based on the hidden subgroup problem (Chapter 5), is simply to append any required classical post-processing step to the quantum computation. This is always possible because quantum computation subsumes classical computation. In the example given above, we simply ask each individual quantum computer (each molecule) to perform a continued fractions algorithm. The result is then checked on", " each quantum computer, then only the computers which succeed in the veri\ufb01cation give an output. The \ufb01nal ensemble average measurement thus gives r. 7.7.4 Experiment One of the most exciting aspects of the NMR approach is the ready experimental realization of small instances of quantum computation and quantum information tasks. In this concluding section on NMR, we brie\ufb02y describe three experiments which have been performed, demonstrating state tomography, elementary logic gates, and the quantum search algorithm. We also summarize the drawbacks of this method. State tomography How does one debug a quantum computer? A classical computer is analyzed by measuring its internal state at different points in time. Analogously, for a quantum computer, an essential technique is the ability to measure its density matrix \u2013 this is called state tomography. Recall that the density matrix of a single qubit can be expressed as \u03c1 = 1 2 1 + rk\u03c3k, k (7.166) where \u03c3k are the Pauli matrices and rb is a real, three-component vector. Because of the trace orthogonality property of Pauli matrices, tr(\u03c3k\u03c3j) = 2\u03b4kj, it follows that \u03c1 can be reconstructed from the three measurement results rk = \u03c3k = tr(\u03c1\u03c3k). (7.167) (7.168) Measurement of the usual observable in NMR, (7.143), preceded by the appropriate single qubit pulses, allows us to determine \u03c3k, and thus obtain \u03c1. Similar results hold for larger numbers of spins. In practice, it is convenient to measure just the traceless deviation of \u03c1 from the identity; this is called the deviation density matrix. Example results for two and three spin systems are shown in Figure 7.18. e\u2212iHtMk\u03c1M \u2020 Exercise 7.45: (State tomography with NMR) Let the voltage measurement k eiHt(iXk + Yk) be the result of experiment k. Vk(t) = V0tr Show that for two spins, nine experiments, with M0 = I, M1 = Rx1, M2 = Ry1, M3 = Rx2, M4 = Rx2Rx1, M5 = Rx2Ry1 etc. provide suf\ufb01cient data from which \u03c1 can be reconstructed. 30 20 10 0 10 20 30 Nuclear magnetic resonance 337 80 60 40 20 0 20 40 Figure 7.18. Experimentally measured deviation density matrices. Vertical scales are arbitrary, and only the real part is shown; the imaginary components are all relatively small. (top left) The two qubit thermal equilibrium state of the proton and the carbon nucleus in molecules of chloroform (13CHCl3) in a 11.78 tesla magnetic \ufb01eld. The 0.5 milliliter, 200 millimolar sample was diluted in acetone-d6, degassed, and \ufb02ame sealed in a thin walled, 5 mm glass tube. (top right) Two qubit effective pure state created using temporal labeling with the chloroform, as in Equation (7.161). (bottom left) The three qubit thermal equilibrium state of three \ufb02uorine nuclei in tri\ufb02uoroethylene. (bottom right) An effective pure state created from the three spin system using logical labeling, as in Equation (7.164). Exercise 7.46: How many experiments are suf\ufb01cient for three spins? Necessary? Quantum logic gates The two qubit proton\u2013carbon system in chloroform presents an excellent system for demonstration of single qubit and two qubit gates, for many reasons. At \u2248 500 and \u2248 125 MHz in an \u2248 11.8 tesla \ufb01eld, the frequencies of the two spins are well separated and easily addressed. The 215 Hz J-coupling frequency of the two nuclei is also convenient; it is much slower than the time scale required for single qubit RF pulses, but much faster than the relaxation time scales. In typical experiments, the T1 of the proton and carbon are approximately 18 and 25 seconds, respectively, while T2 are 7 and 0.3 seconds. The carbon T2 is short because of interactions with the three quadrupolar chlorine nuclei, but taking the product of the shortest T2 lifetime and the J-coupling indicates that roughly 60 gates should still be realizable before quantum coherence is lost. The Hamiltonian of the two spin system is very well approximated by the expression in (7.147), but it can be simpli\ufb01ed signi\ufb01cantly using an experimental trick. By tuning two oscillators to exactly the rotating frequencies of the proton and carbon, we obtain, in", " the rotating frame de\ufb01ned by the oscillators, the simpli\ufb01ed Hamiltonian H = 2\u03c0JZ1Z2, (7.169) 338 Quantum computers: physical realization where J = 215 Hz. This Hamiltonian makes the realization of the controlledgate transform equivalent up to single qubit quite simple. A circuit which performs a phases is shown in Figure 7.19, as well as a circuit for creating a Bell state, and experimentally measured outputs. e\u2212iH/2\u00afhJ e\u2212iH/2\u00afhJ Rx Rx Rx Ry Ry Figure 7.19. Quantum circuits implemented with NMR and the real part of the experimentally measured output deviation density matrices. In these circuits, Rx and Ry denote single qubit gates which perform 90\u25e6 rotations about \u02c6x and \u02c6y, implemented with RF pulses about 10 microseconds long, and the two qubit box with e\u2212iH/2J is a free evolution period of time 1/2J \u2248 2.3 milliseconds. (top) Controlledfor a thermal state input, showing the exchange of the |10 and |11 diagonal elements, as expected from the 2, and its classical truth table for the output, when a |00 effective pure state is prepared as an input. gate. (bottom) Circuit for creating the Bell state (|00 \u2212 |11)/ circuit, and the output measured \u221a Exercise 7.47: (NMR controlled- gate) Verify that the circuit shown in the top left of Figure 7.19 performs a controlledthat is, it acts properly on classical input states, and furthermore can be turned into a proper controlledrotations. Give another circuit using the same building blocks to realize a proper gate by applying additional single qubit Rz gate, up to single qubit phases; gate. Exercise 7.48: Verify that the circuit shown in the bottom left of Figure 7.19 creates 2 as advertised. the Bell state (|00 \u2212 |11)/ \u221a Exercise 7.49: (NMR swap gate) An important chemical application of NMR is measurement of connectivity of spins, i.e. what protons, carbons, and phosphorus atoms are nearest neighbors in a molecule. One pulse sequence to do this is known as INADEQUATE (incredible natural abundance double quantum transfer experiment \u2013 the art of NMR is full of wonderfully creative acronyms). In the language of quantum computation, it can be understood as simply trying works, the two nuclei to apply a must be neighbors. Another building block which is used in sequences such as between any two resonances; if the Nuclear magnetic resonance 339 TOCSY (total correlation spectroscopy) is a swap operation, but not quite in the perfect form we can describe simply with quantum gates! Construct a quantum circuit using only e\u2212iH/2J, Rx, and Ry operations to implement a swap gate (you may start from the circuit in Figure 1.7). Quantum algorithms Grover\u2019s quantum search algorithm provides another simple example for NMR quantum computation. For a problem size of four elements (n = 2 qubits), we are given the set x = {0, 1, 2, 3} for which f (x) = 0 except at one value x0, where f (x0) = 1. The goal is to \ufb01nd x0, which can be classically accomplished by evaluating f (x) an average of 2.25 times. In comparison, the quantum algorithm \ufb01nds x0 by evaluating f (x) only once (Chapter 6; see, in particular, Box 6.1). Three operators are required: the oracle operator O (which performs a phase \ufb02ip based on the function f (x)), the Hadamard operator on two qubits H \u22972, and the conditional phase shift operator P. The oracle O \ufb02ips the sign of the basis element corresponding to x0; for x0 = 3, this is \u23a1 \u23a2 \u23a2 \u23a3 1 O = (7.170) Denoting a t = 1/2J (2.3 millisecond for the chloroform) period evolution e\u2212iH/2J as \u00afRy2 \u03c4 (up to an irrelevant overall phase factor) \u03c4, we \ufb01nd that O = Ry1 for the x0 = 3 case. H \u22972 is just two single qubit Hadamard operations, H1 \u2297 H2, where Hk = R2 xk \u00afRyk. And the operator P, \u00afRy1Ry2 \u00afRx1 \u00afRx2 \u23a1 \u23a2 \u23a2 \u23a3 P = 1 0 0 \u22121 0 0 0 0 0 \u22121 0 0 0 0 0 \u22121 \u23a4 \u23a5 \u23a5 \u23a6 (7.171)", " \u00afRy1Ry2Rx2 \u00afRy2 \u03c4. From these, we construct the Grover is simply realized as P = Ry1Rx1 iteration G = H \u22972P H \u22972O. This operator can be simpli\ufb01ed straightforwardly by eliminating unnecessary operations which obviously cancel (see Exercise 7.51). Let |\u03c8k = Gk|00 be the state after k applications of the Grover iteration to the initial state. We \ufb01nd that the amplitude x0|\u03c8k \u2248 sin((2k + 1)\u03b8), where \u03b8 = arcsin(1/ 2); this periodicity is a fundamental property of the quantum search algorithm, and is a natural feature to test in an experiment. For the two qubit case, and x0 = 3, we expect |11 = |\u03c81 = \u2212|\u03c84 = |\u03c87 = \u2212|\u03c810, a period of 3 if the overall sign is disregarded. \u221a Exercise 7.50: Find quantum circuits using just single qubit rotations and e\u2212iH/2J to implement the oracle O for x0 = 0, 1, 2. Exercise 7.51: Show that the Grover iteration can be simpli\ufb01ed, by canceling adjacent 340 Quantum computers: physical realization single qubit rotations appropriately, to obtain \u00afRx1 \u00afRx1 \u00afRx1 \u00afRx1 \u00afRx2 \u00afRx2 \u00afRx2 \u00afRx2 for the four possible cases of x0. \u00afRy2 \u03c4 Rx1 \u00afRy2 \u03c4 Rx1 \u00afRy2 \u03c4 \u00afRx1 \u00afRy2 \u03c4 \u00afRx1 \u00afRy1 \u00afRy1 \u00afRy1 \u00afRy1 \u23a7 \u23aa\u23aa\u23a8 \u23aa\u23aa\u23a9 \u00afRy1Rx2 \u00afRy1 \u00afRx2 \u00afRy1Rx2 \u00afRx2 \u00afRy1 G = \u00afRy2 \u03c4 (x0 = 3) \u00afRy2 \u03c4 (x0 = 2) \u00afRy2 \u03c4 (x0 = 1) \u00afRy2 \u03c4 (x0 = 0), (7.172) Figure 7.20 shows the theoretical and measured deviation density matrices \u03c1\u0394n = |\u03c8n\u03c8n| \u2212 tr(|\u03c8n\u03c8n|)/4 for the \ufb01rst seven iterations of U. As expected, \u03c1\u03941 clearly reveals the |11 state corresponding to x0 = 3. Analogous results were obtained for experiments repeated for the other possible values of x0. Measuring each density matrix required 9 \u00d7 3 = 27 experimental repetitions, 9 for the tomographic reconstruction and 3 for the pure state preparation. The longest computation, for n = 7, took less than 35 milliseconds, which was well within the coherence time. The periodicity of Grover\u2019s algorithm is clearly seen in Figure 7.20, with good agreement between theory and experiment. The large signal-to-noise (typically better than 104 to 1) was obtained with just single-shot measurements. Numerical simulations indicate that the 7\u201344% errors are primarily due to inhomogeneity of the magnetic \ufb01eld, magnetization decay during the measurement, and imperfect calibration of the rotations (in order of importance). Drawbacks The bulk-ensemble NMR implementation of quantum computation has been successful in demonstrating quantum algorithms and quantum information tasks with systems up to seven qubits, which is quite impressive. However, there are important limitations which arise from the temporal, spatial, and logical labeling techniques which are at the heart of the method. The essential objective of these labeling techniques is to isolate the signal from the subset of spins which happen to be in the |00... 0 component (or any other standard, high probability state) of the thermal equilibrium state. Temporal and spatial labeling do this by adding up signals to cancel all but the desired state; logical labeling trades off Hilbert space for purity. Irrespective of the method used, however, nothing can increase the probability of the |00... 0 component of the thermal state, p00...0 = 1 Z 00... 0|e\u2212\u03b2H|00... 0. (7.173) k \u03c9Zk, we \ufb01nd that p00...0 is proportional to n2\u2212n, for an n spin molecule. Taking H = This means that the total signal decreases exponentially as the number of qubits \u2018distilled\u2019 into an effective pure state using labeling techniques, for constant initial state temperature. Another limitation comes from using molecules as quantum computers. The structure of the molecule plays the role of the architecture of the computer, determining what pairs (or groups) of qubits interact with each other (analogously, the RF", " pulses serve as the software). Not all qubits are necessarily well connected! This is doubly important since interactions cannot be switched off, except by performing refocusing. Furthermore qubits are addressed by distinguishing them in frequency, but this rapidly becomes dif\ufb01cult as the number of nuclei is increased. A solution to this exists, which is to use a cellular automata style architecture, such as the one-dimensional chain X \u2212 A \u2212 B \u2212 C \u2212 A \u2212 Nuclear magnetic resonance 341 50 0 50 20 10 0 10 20 20 10 0 10 20 50 0 50 50 0 50 20 10 0 10 20 20 10 0 10 20 40 20 0 20 40 50 0 50 20 10 0 10 20 50 0 50 20 10 0 10 20 20 10 0 10 20 50 0 50 20 10 0 10 20 50 0 50 20 10 0 10 20 20 10 0 10 20 1 2 3 4 1 2 3 4 Figure 7.20. Theoretical and experimental deviation density matrices (in arbitrary units) for seven steps of Grover\u2019s algorithm performed on the hydrogen and carbon spins in chloroform. Three full cycles, with a periodicity of four iterations are clearly seen. Only the real component is plotted (the imaginary portion is theoretically zero and was found to contribute less than 12% to the experimental results). Relative errors ||\u03c1theory \u2212 \u03c1expt||/||\u03c1theory|| are shown, in which the ends are distinguished but the middle is composed of a repeating regular sequence. In this architecture, only distinct letters are addressable, and it might seem as if this is a highly restrictive model of computation. However, it has been shown that in fact it is universal, with only polynomial slowdown. The precise amount of slowdown required will of course be important when performing tasks such as the quantum search algorithm, which only has a square root speedup. Methods for circumventing the limitations of labeling techniques also exist. One possibility is to polarize the nuclei through some physical mechanism; this has been done by using optical pumping (similar to how ions are cooled as shown in Figure 7.8) to polarize the electronic state of rubidium atoms, which then transfer their polarization to the nuclei of xenon atoms through formation of short-lived van der Waals molecules. This has also been done for helium. Doing similarly for molecules is conceivable, albeit technologically challenging. Another possibility is the use of a different labeling scheme; logical labeling is essentially a compression algorithm, which increases the relative probability of one state in an ensemble by discarding ensemble members. An improved version of this method has been developed which achieves the entropic limit, giving nH(p) pure qubits from 342 Quantum computers: physical realization n-spin molecules originally at temperature T such that p = (1 \u2212 e\u2212\u0394E/kBT )/2, where \u0394E is the spin \ufb02ip energy. This scheme does not have any exponential cost; the compression can be achieved using only a polynomial number of basic operations. However, it is inef\ufb01cient unless p is relatively small, and today p \u2248 0.4999 in good solenoid magnet systems. Figure 7.21. A selection of simple molecules which have been used to demonstrate various quantum computation and quantum information tasks with NMR. (a) Chloroform: two qubits, proton and carbon, have been used to implement the Deutsch\u2013Jozsa algorithm, and a two qubit quantum search. (b) Alanine: three qubits, composed of the carbon backbone, have been used to demonstrate error-correction. Note how the three carbon nuclei have distinguishable frequencies because their surrounding chemical environments are different (for example, the electronegativity of the oxygen causes it to draw much of the nearby electrons away from the neighboring carbon). (c) 2,3-dibromothiophene: two qubits, composed of the two protons, have been used to simulate four levels of a truncated simple harmonic oscillator. Here, the two protons are at different distances from the sulphur atom, and thus have distinguishable frequencies. (d) Tri\ufb02uorobromoethylene: three qubits, the three \ufb02uorines, have been used to demonstrate logical labeling and the creation of a (|000 + |111)/ 2 superposition state. (e) Trichloroethylene: three qubits, the proton and two carbons, were used to demonstrate teleportation, with the proton\u2019s state being teleported to the rightmost carbon. (f) Sodium formate: two qubits, proton and carbon, used to demonstrate the two qubit quantum error detection code. In this molecule, the sodium radical is used to tune the T2 times of the two qubits to be nearly equal, by changing the ambient temperature to modify its exchange rate with the solvent. \u221a Despite these drawbacks,", " NMR provides a testbed for quantum algorithms and illustrates basic techniques which other realizations will have to implement to perform quantum computation. Some of the molecules which have been used to demonstrate quantum computation and quantum information tasks are shown in Figure 7.21. The NMR idea is also a very rich area for innovation, combining chemistry, physics, engineering, and mathematics, and undoubtedly continued innovation between \ufb01elds will further this technique. Other implementation schemes 343 NMR quantum computer \u2022 Qubit representation: Spin of an atomic nucleus. \u2022 Unitary evolution: Arbitrary transforms are constructed from magnetic \ufb01eld pulses applied to spins in a strong magnetic \ufb01eld. Couplings between spins are provided by chemical bonds between neighboring atoms. \u2022 Initial state preparation: Polarize the spins by placing them in a strong magnetic \ufb01eld, then use \u2018effective pure state\u2019 preparation techniques. \u2022 Readout: Measure voltage signal induced by precessing magnetic moment. \u2022 Drawbacks: Effective pure state preparation schemes reduce the signal exponentially in the number of qubits, unless the initial polarization is suf\ufb01ciently high. 7.8 Other implementation schemes In this chapter we have described but a fraction of the number of ideas people have considered for implementation of quantum computers. Our selection illustrates basic requirements and challenges common to all implementations: robust representation of quantum information, application of unitary transforms, preparation of a \ufb01ducial input state, and measurement of the output. The simple harmonic oscillator example emphasizes how a digital representation is crucial: each unit (qubit, qutrit, qudit, or whatever) of quantum information should reside in physically separate degrees of freedom; otherwise, some resource (such as energy) is inef\ufb01ciently utilized. That example also provides the mathematical basis for studying representations of qubits through the remainder of the chapter. Single photons are nearly ideal qubits, but the nonlinear optical materials required to get them to interact are dif\ufb01cult to realize without causing coherence loss. Cavity-QED techniques can address this problem by using single atoms to interact photons, but even more importantly, they introduce the notion of two-level atoms, and the idea of guarding qubit representations using selection rules imposed by physical symmetries such as dipole selection. A natural extension of this idea is to represent qubits using spin-1/2 particles, which inherently only have two states. This is the tack taken with trapped ions, which store qubits in electron and nuclear spins; the dif\ufb01culty with this method, though, is that the center of mass oscillations \u2013 phonons \u2013 used to mediate interactions between spins have short coherence times. Molecules, in which nuclear spins can be strongly coupled by chemical bonds, can solve this dif\ufb01culty, but the spin resonance signal from single molecules is too small to detect with present technology. NMR quantum computing solves this by creating \u2018effective pure states\u2019 with bulk ensembles of O(1018) molecules, thereby demonstrating simple quantum algorithms in the laboratory. But without providing initial polarization, this capability comes at the cost of signal strength, which decreases exponentially with the number of qubits. As these examples demonstrate, coming up with a good physical realization for a quantum computer is tricky business, fraught with tradeoffs. All of the above schemes are unsatisfactory, in that none allow a large-scale quantum computer to be realized anytime in the near future. However, that does not preclude the possibility, and in fact 344 Quantum computers: physical realization many other implementation schemes have been proposed, some of which we brie\ufb02y touch upon in this \ufb01nal section here. A good way to categorize realizations is in terms of the physical degree of freedom used to represent qubits. Recall Figure 7.1: just about anything which comes in quantum units could be a qubit, but, as we have seen, fundamental physical quanta such as photons and spin are particularly good choices. Another fundamental quantum unit which could serve as a qubit representation is electric charge. Modern electronics provides excellent techniques to create, control, and measure charges, even at the level of single electrons. For example, quantum dots, fabricated from semiconductor materials, metals, or even small molecules, can serve as threedimensional boxes with electrostatic potentials which con\ufb01ne charge quanta. This is veri\ufb01ed by observation of the Coulomb blockade effect, in which electrical conductance through dots of capacitance C is found to increase in discrete steps as a function of bias voltage across the dot, re\ufb02ecting the e2/2C energy required to add each additional electron. Unlike photons, (net) charge cannot be destroyed; charges can only be moved around, and thus a charge state qubit would have to use something like the dual-rail representation of Section 7.4.2", ", whereby the |0 and |1 states correspond to having charge located in either of two dots, or two states within a single dot. \u221a Just as for single photons, single qubit operations on charge state qubits can be performed using electrostatic gates (the analogue of optical phase shifters) and either single mode waveguide couplers (the analogue of beamsplitters) for moving electrons, or tunnel junctions for quantum dots. Electrical charges experience long-range Coulomb interactions with other charges (the potential created by a single charge at distance r is V (r) = e/4\u03c0r), and thus charges far away can modulate the phase of a local charge, much like the Kerr interaction between photons. Controlled Coulomb interactions can thus be used to perform two qubit operations. Finally, single electron charges are straightforward to measure; modern \ufb01eld effect transistors easily detect movements of single charges in their channels, and single electron transistors operating at \u2248 100 millikelvin temperatures can detect charge to better than 10\u22124e/ Hz at frequencies over 200 MHz. Unfortunately, uncontrolled distant charge motion leads to dephasing; this, and other scattering mechanism such as those due to phonon interactions cause coherence times to be relatively short for charge states, on the order of hundreds of femtoseconds to hundreds of picoseconds. Charge carriers in superconductors have also been suggested as qubit representations. At low temperatures in certain metals, two electrons can bind together through a phonon interaction to form a Cooper pair, which has charge 2e. And just as electrons can be con\ufb01ned within quantum dots, Cooper pairs can be con\ufb01ned within an electrostatic box, such that the number of Cooper pairs in the box becomes a good quantum number, and can be used to represent quantum information. Single qubit operations can be realized by using electrostatic gates to modulate the box potential, and Josephson junctions between coupled boxes. These junctions can also be used to couple qubits, and their strength can be modulated using an external magnetic \ufb01eld by coupling appropriately to superconducting interferometric loops. Finally, qubits can be measured simply by measuring electric charge. This superconductor qubit representation is interesting because of the relative robustness of Cooper pairs; it is estimated that the main decoherence mechanism is spontaneous emission of electromagnetic photons, which may allow coherence times exceeding one microsecond, long compared with typical dynamical time scales of hundreds of picoseconds. Unfortunately, just as with the electronic charge representation, Other implementation schemes 345 a \ufb02uctuating background of extraneous charges (\u2018quasiparticles\u2019) is highly deleterious to qubit coherence. One means around this problem, using superconductor technology, is to choose instead a magnetic \ufb02ux qubit representation, in which qubit states correspond to left and right hand orientations of \ufb02ux localized through a superconducting loop device. Here, decoherence is caused by background magnetic \ufb02uctuations, which are expected to be much quieter than the electrostatic case. The locality of magnetic interactions is a good feature for qubit representations, and thus we return to spin, for which schemes have also been proposed to take advantage of solid state technology. A fairly large quantum dot, even one containing many electrons, can behave as a spin-1/2 object, where the spin is carried by a single excess electron. This spin state can be prepared by equilibrating in a strong magnetic \ufb01eld, at low temperatures, such that the spin \ufb02ip energy \u0394E is much larger than kBT. Manipulating a single spin, as we have seen in Section 7.7, can be done by applying pulsed local magnetic \ufb01elds, and coupled qubit operations can be realized with a controlled Heisenberg coupling,, X1X2 + Y1Y2 + Z1Z2 (7.174) H(t) = J(t)S1 \u00b7 S2 = 1 4 where S are the spin operators (Pauli operators divided by two), and J(t) = 4\u03c4 2 0 (t)/u, u being the charging energy of the dot, and \u03c40(t) being the tunneling matrix element controlled by local electrostatic gates placed between dots. This interaction is universal, gate (see exercise below). Spin in the sense that it is equivalent to the controlledstates may, in theory, be measured by allowing the spin-carrying electron to tunnel to a readout paramagnetic dot, or to spin-dependently tunnel through a \u2018spin valve\u2019 to a readout electrometer. The challenge is to realize such measurements in practice; high \ufb01delity single spin measurement in semiconductors has not been accomplished with present technology. Exercise 7.52: (Universality of He", "isenberg Hamiltonian) Show that a swap operation U can be implemented by turning on J(t) for an appropriate amount of time in the Heisenberg coupling Hamiltonian of (7.174), to obtain U = exp(\u2212i\u03c0 S1 \u00b7 S2). The \u2018 \u2019 gate obtained by turning on the interaction for half this time is universal; compute this transform and show how to obtain a gate by composing it with single qubit operations. controlled- \u221a Eventually, with suf\ufb01ciently advanced technology, it may be possible to place, control, and measure single nuclear spins in semiconductors, making possible the following vision. Imagine being able to precisely place single phosphorus atoms (nuclear spin-1/2) within a crystalline wafer of 28Si (nuclear spin 0), positioned beneath lithographically patterned electrostatic gates. These gates allow manipulation of the electron cloud surrounding the 31P dopants, to perform single qubit operations via modulation of the magnetic \ufb01eld seen by the 31P nuclei. Additional gates located above the region separating the 31P dopants can be used to arti\ufb01cially create electron distributions connecting adjacent 31P, much like a chemical bond, thus allowing two qubit operations to be performed. Although fabrication constraints of such a scheme are extremely challenging \u2013 for example, the gates should be separated by 100 \u02daA or less, and the 31P dopants must be registered precisely and in an ordered array \u2013 at least this vision articulates a possible venue for marrying quantum computing with more conventional computing technologies. 346 Quantum computers: physical realization Of the schemes we have described for realizing quantum computers, the ones which have most captivated the attention of technologists are the ones based on solid state technologies. Of course, atomic, molecular, and optical quantum computing schemes continue to be proposed, using systems such as optical lattices (arti\ufb01cial crystals made from atoms con\ufb01ned by crossed beams of light) and Bose condensates which are at the forefront of those \ufb01elds; someday, we may even see quantum computing proposals using mesons, quarks, and gluons, or even black holes. But the motivation to envision some kind of solid state quantum computer is enormous. It has been estimated that the world has invested over US$1 trillion in silicon technology since the invention of the transistor in the late 1940s. Condensed matter systems have also been rich in new physics, such as superconductivity, the quantum Hall effect, and the Coulomb blockade (a classical effect discovered long after it was widely thought that everything about classical physics was well known!). This chapter has concentrated mainly on the implementation of quantum computing machines, but the basic components which were presented are also useful in many other quantum applications. Quantum cryptography and its experimental realization are described in Section 12.6. And pointers to experimental work on quantum teleportation and superdense coding are given in \u2018History and further reading\u2019 at the end of this chapter. The general interface between quantum communication and computation includes challenges such as distributed quantum computation; development of new algorithms and experimental implementations of such systems will certainly continue to be a rich area of research for the future. Much of the allure of quantum computing and communication machines is certainly their potential economic rami\ufb01cations, as novel information technologies. But as we have seen in this chapter, quantum computation and quantum information also motivate new questions about physical systems, and provide different ways to understand their properties. These new ideas embody a need to move away from traditional many-body, statistical, and thermodynamic studies of physical systems, all the way from atoms to condensed matter systems. They represent a new opportunity to focus instead on dynamical properties of physical systems at the single quantum level. Hopefully, by giving a \ufb02avor of the richness of this approach, this chapter will motivate you to continue \u2018thinking algorithmically\u2019 about Physics. Problem 7.1: (Ef\ufb01cient temporal labeling) Can you construct ef\ufb01cient circuits (which require only O(poly(n)) gates) to cyclically permute all diagonal elements in a 2n\u00d72n diagonal density matrix except the |0n0n| term? Problem 7.2: (Computing with linear optics) In performing quantum computation with single photons, suppose that instead of the dual rail representation of Section 7.4.1 we use a unary representation of states, where |00... 01 is 0, |00... 010 is 1, |00... 0100 is 2, and so on, up to |10... 0 being 2n \u2212 1. 1. Show that an arbitrary unitary transformation on these states can be constructed completely from just beamsplitters and phase shifters (and no nonlinear media). Chapter problems 347 2. Construct a circuit of beamsplitters and phase shifters to perform the one qubit Deutsch", "\u2013Jozsa algorithm. 3. Construct a circuit of beamsplitters and phase shifters to perform the two qubit quantum search algorithm. 4. Prove that an arbitrary unitary transform will, in general, require an exponential number (in n) of components to realize. Problem 7.3: (Control via Jaynes\u2013Cummings interactions) Robust and accurate control of small quantum systems \u2013 via an external classical degree of freedom \u2013 is important to the ability to perform quantum computation. It is quite remarkable that atomic states can be controlled by applying optical pulses, without causing superpositions of atomic states to decohere very much! In this problem, we see what approximations are necessary for this to be the case. Let us begin with the Jaynes\u2013Cummings Hamiltonian for a single atom coupled to a single mode of an electromagnetic \ufb01eld, H = a\u2020\u03c3\u2212 + a\u03c3+, where \u03c3\u00b1 act on the atom, and a, a\u2020 act on the \ufb01eld. 1. For U = ei\u03b8H, compute An = n|U |\u03b1, (7.175) (7.176) where |\u03b1 and |n are coherent states and number eigenstates of the \ufb01eld, respectively; An is an operator on atomic states, and you should obtain \u221a \u221a \u221a An = e\u2212|\u03b12| |\u03b1|2 n! cos(\u03b8 sin(\u03b8 n) n + 1) i\u03b1\u221a n+1 \u221a i n \u03b1 sin(\u03b8 \u221a cos(\u03b8 n) n + 1). (7.177) The results of Exercise 7.17 may be helpful. 2. It is useful to make an approximation that \u03b1 is large (without loss of generality, we may choose \u03b1 real). Consider the probability distribution pn = e\u2212x xn n!, (7.178) $ \u221a which has mean n = x, and standard deviation change variables to n = x \u2212 L \u221a x, and use Stirling\u2019s approximation n2 \u2212 n2 = x. Now n! \u2248 to obtain \u221a 2\u03c0n nne\u2212n pL \u2248 e\u2212L2/2 \u221a 2\u03c0. (7.179) (7.180) 3. The most important term is An for n = |\u03b1|2. De\ufb01ne n = \u03b12 + L\u03b1, and for a = y 1 y2 + L y and b = y 1 y2 + L y + 1, (7.181) where y = 1/\u03b1, show that AL \u2248 e\u2212L2/4 (2\u03c0)1/4 cos a\u03d5 (i/b) sin b\u03d5 ia sin a\u03d5 cos b\u03d5, (7.182) 348 Quantum computers: physical realization using \u03b8 = \u03d5/\u03b1. Also verify that % \u221e \u2212\u221e A\u2020 LAL dL = I as expected. 4. The ideal unitary transform which occurs to the atom is U = cos \u03b1\u03b8 i sin \u03b1\u03b8 i sin \u03b1\u03b8 cos \u03b1\u03b8. (7.183) (7.184) How close is AL to U? See if you can estimate the \ufb01delity % \u221e F = min|\u03c8 \u2212\u221e as a Taylor series in y. |\u03c8|U \u2020AL|\u03c8|2 dL (7.185) Problem 7.4: (Ion trap logic with two-level atoms) The controlled- gate described in Section 7.6.3 used a three-level atom for simplicity. It is possible to do without this third level, with some extra complication, as this problem demonstrates. Let \u03a5blue,j \u02c6n sideband frequency, \u03c9 = \u03a9 + \u03c9z, of the jth particle for time \u03b8 similarly for the red sideband. \u02c6n denotes the axis of the rotation in the \u02c6x- \u02c6y plane, controlled by setting the phase of the incident light. The superscript j may be omitted when it is clear which ion is being addressed. Speci\ufb01cally, (\u03b8) denote the operation accomplished by pulsing light at the blue N /\u03b7\u03a9, and \u221a \u03a5blue \u02c6n (\u03b8) = exp ei\u03d5|0011| + e\u2212i\u03d5|1100| \u221a \u221a + ei\u03d5 2|0112| + e\u2212i\u03d5 2|1201| + \u00b7 \u00b7 \u00b7 i\u03b8 2, (7.186) where \u02c6n = \u02c6x cos \u03d5 + \u02c6y sin \u03d5, and the two labels in the ket represent the internal and the motional states", ", respectively, from left to right. The from the fact that a\u2020|n = n + 1|n + 1 for bosonic states. (1) Show that Sj = \u03a5red,j (\u03c0) performs a swap between the internal and motional 2 factor comes \u221a \u221a \u02c6y states of ion j when the motional state is initially |0. (2) Find a value of \u03b8 such that \u03a5blue \u02c6n (\u03b8) acting on any state in the computational subspace, spanned by |00, |01, |10, and |11, leaves it in that subspace. This should work for any axis \u02c6n. (3) Show that if \u03a5blue \u02c6n (\u03d5) stays within the computational subspace, then \u03b1 (\u2212\u03b2)\u03a5blue \u02c6n (\u03b8)\u03a5blue U = \u03a5blue for any choice of rotation angle \u03b2 and axis \u03b1. \u03b1 (\u03b2) also stays within the computational subspace, (4) Find values of \u03b1 and \u03b2 such that U is diagonal. Speci\ufb01cally, it is useful to obtain an operator such as \u221a 2 \u23a1 \u23a2 \u23a2 \u23a3 e\u2212i\u03c0 ei\u03c0/ \u221a 2 (7.187) History and further reading 349 (5) Show that (7.187) describes a non-trivial gate, in that a controlled- gate between the internal states of two ions can be constructed from it and single qubit operations. Can you come up with a composite pulse sequence for performing a without requiring the motional state to initially be |0? Summary of Chapter 7: Quantum computers: physical realization \u2022 There are four basic requirements for implementation of quantum computation: (1) Representation of qubits, (2) Controllable unitary evolution, (3) Preparation of initial qubit states, and (4) Measurement of \ufb01nal qubit states. \u2022 Single photons can serve as good qubits, using |01 and |10 as logical 0 and 1, but conventional nonlinear optical materials which are suf\ufb01ciently strong to allow single photons to interact inevitably absorb or scatter the photons. \u2022 Cavity-QED is a technique by which single atoms can be made to interact strongly with single photons. It provides a mechanism for using an atom to mediate interactions between single photons. \u2022 Trapped ions can be cooled to the extent that their electronic and nuclear spin states can be controlled by applying laser pulses. By coupling spin states through center-of-mass phonons, logic gates between different ions can be performed. \u2022 Nuclear spins are nearly ideal qubits, and single molecules would be nearly ideal quantum computers if their spin states could only be controlled and measured. Nuclear magnetic resonance makes this possible using large ensembles of molecules at room temperature, but at the expense of signal loss due to an inef\ufb01cient preparation procedure. History and further reading For an excellent discussion of why building a quantum computer is dif\ufb01cult, see the article by DiVincenzo[DiV95a], on which Figure 7.1 is based. DiVincenzo also presents \ufb01ve criteria for realizing a quantum computer, which are similar to those discussed in Section 7.2. The quantum simple harmonic oscillator of Section 7.3 is a staple of quantum mechanics, and is treated in any standard textbook, such as [Sak95]. The general necessary and suf\ufb01cient conditions for quantum computation given in Section 7.3.3 were discussed by Lloyd[Llo94]. The optical quantum computer of Section 7.4 uses as its main theoretical tools the formalism of quantum optics, which is described in many textbooks; two excellent ones are [Lou73, Gar91]. For more on basic optics and optical technology, such as polarizers, beamsplitters, photon detectors, and the like, see for example, the textbook [ST91]. The beamsplitter in the single photon regime was studied by Campos, Saleh, and Tiech[CST89], and related to that context, the elegant connection between SU (2) and two coupled harmonic oscillators was \ufb01rst described by Schwinger[Sak95]. The concept of a \u2018dual-rail\u2019 representation of a qubit was suggested by Yurke and used by Chuang and Yamamoto[CY95] to describe a complete quantum computer (using nonlinear Kerr media) to perform the 350 Quantum computers: physical realization Deutsch\u2013Jozsa algorithm (as in Exercise 7.13). The quantum-optical Fredkin gate was \ufb01rst described by Yamamoto, Kitagawa, and Igeta[YKI88] and Milburn[Mil89a]. The single photon generation and detection technology required for an optical quantum", " computer has been discussed by Imamoglu and Yamamoto[IY94] and by Kwiat, Steinberg, Chiao, Eberhard, and Petroff[KSC+94]. An analogous mechanism using electron optics and the Coulomb interaction in place of the Kerr interaction has been discussed by Kitagawa and Ueda[KU91]. The fundamental limits of traditional, off-resonance nonlinear optical materials at single quanta levels have been studied by Watanabe and Yamamoto[WY90]. Cerf, Adami, and Kwiat have studied the simulation of quantum logic using (exponentially many) linear optical components[CAK98]. An in\ufb02uential earlier paper by Reck, Zeilinger, Bernstein and Bertani[RZBB94] described similar constructions, but did not make the explicit connection to quantum computation. Kwiat, Mitchell, Schwindt, and White have constructed an optical simulation of Grover\u2019s quantum search algorithm which uses linear optics, but appears to require exponential resources when scaling[KMSW99]. See Miller[Mil89b] for a discussion of the energetics of optical communication over different distances. Allen and Eberly[AE75] have written a beautiful treatise on two-level atoms and optical resonance. The experiment described in Section 7.5.4 was performed by Turchette, Hood, Lange, Mabuchi, and Kimble at Caltech in 1995[THL+95]. A detailed explanation is also given in Turchette\u2019s Ph.D. thesis[Tur97]. The single photons used in this experiment were called \u2018\ufb02ying qubits\u2019. A different approach, in which atomic states are used as qubits and the atoms travel through optical cavities, was proposed by Domokos, Raimond, Brune, and Haroche[DRBH95]. It is based on the idea of using a single atom to switch a coherent state into a cavity, as described by Davidovich, Maali, Brune, Raimond, and Haroche[DRBH87, DMB+93]. The idea of ion trap quantum compution was proposed by Cirac and Zoller[CZ95]. Our discussion of this idea in Section 7.6.1 bene\ufb01ted greatly from the articles by Steane[Ste97], and by Wineland, Monroe, Itano, Leibfried, King, and Meekof[WMI+98]. Earnshaw\u2019s theorem can be derived from Laplace\u2019s equation, as described in his original paper[Ear42], or a modern electromagnetics textbook such as the one by Ramo, Whinnery, and van Duzer[RWvD84]. Figure 7.8 is drawn after Figure 6 in [Ste97]. Figure 7.7 is drawn after [WMI+98]. The experiment described in Section 7.6.4 was performed by Monroe, Meekhof, King, Itano, and Wineland at the National Institute of Standards and Technology in Boulder, Colorado[MMK+95]. Figure 7.15 is reprinted courtesy of Wineland[WMI+98]. Brewer, DeVoe, and Kallenbach have envisioned using large arrays of planar ion microtraps[BDK92] as a scalable quantum computer; this is the kind of trap shown in Figure 7.12. James has extensively studied the theory of heating and other decoherence mechanisms in ion traps[Jam98]. The impact of decoherence on ion trap quantum computation has also been studied in some depth by Plenio and Knight, who also consider effects such as the failure of the two-level approximation[PK96]. DiVincenzo \ufb01rst suggested the use of nuclear spins in quantum computation[DiV95b], and pointed out that the well known, and very old ENDOR (electron nucleon dougate. ble resonance) pulse sequence is essentially an instance of the controlledHowever, the problem of how to use an ensemble of nuclei at room temperature for quantum computation was not solved until Cory, Fahmy, and Havel[CFH97], and Gershenfeld and Chuang[GC97] realized that effective pure states could be prepared. It was also necessary to realize that quantum algorithms can be modi\ufb01ed to allow ensemble History and further reading 351 readout; a solution to this problem was provided in [GC97], and is presented in Section 7.7.3. Excellent textbooks on NMR have been written by Ernst, Bodenhausen, and Wokaun[EBW87], and by Slichter[Sli96]. Warren has written a criticism of NMR quantum computation[War97]; interestingly, in the same paper he advocates using electron spin resonance (ESR) to do quantum computation. Temporal labeling was proposed by Knill, Chuang,", " and La\ufb02amme[KCL98]. The NMR logic gates, and Bell state preparation circuit of Section 7.7.4 were discussed by Chuang, Gershenfeld, Kubinec, and Leung[CGKL98]. The realization of Grover\u2019s algorithm in Section 7.7.4 was by Chuang, Gershenfeld, and Kubinec[CGK98]; Figure 7.20 is taken from their paper. Linden, Kupce, and Freeman note that the swap gate may be a useful contribution of quantum computing to NMR, and provides a pulse sequence for it[LKF99]. The three spin data in Figure 7.18 showing logical labeling are from the article by Vandersypen, Yannoni, Sherwood and Chuang[VYSC99]. Yamaguchi and Yamamoto have creatively extended the NMR idea to use a crystal lattice[YY99]. The molecules shown in Figure 7.21 are attributed to (a) Chuang, Vandersypen, Zhou, Leung, and Lloyd[CVZ+98] (b) Cory, Mass, Price, Knill, La\ufb02amme, Zurek, and Havel[CMP+98] (c) Somaroo, Tseng, Havel, La\ufb02amme, and Cory[STH+99] (d) Vandersypen, Yannoni, Sherwood and Chuang[VYSC99] (e) Nielsen, Knill, and La\ufb02amme[NKL98] (f) Leung, Vandersypen, Zhou, Sherwood, Yannoni, Kubinec and Chuang[LVZ+99]. Jones, Mosca, and Hansen have also realized various quantum algorithms on small molecules[JM98, JMH98]. The optimal labeling scheme which achieves the entropic limit was devised by Schulman and Vazirani[SV99]. Various criticisms of the NMR approach to quantum information processing have been leveled. Perhaps the most comprehensive discussion is due to Schack and Caves[SC99], building on earlier work by Braunstein, Caves, Jozsa, Linden, Popescu and Schack[BCJ+99], whose main technical conclusions (though not the possible connection with NMR) were obtained by Vidal and Tarrach[Vid99] and also by Zyczkowski, Horodecki, Sanpera and Lewenstein[ZHSL99]. See also the discussion by Linden and Popescu[LP99]. There are far too many proposals for implementations of quantum computers to all be mentioned here, so only a few are given; references in the given citations should provide good leads to additional articles. Lloyd has envisioned many implementations of quantum computers, including polymer systems[Llo93]. Nakamura, Pashkin, and Tsai have demonstrated control of single Cooper-pair qubits and observation of their Rabi oscillations[NPT99]. Mooij, Orlando, Levitov, Tian, van der Waal, and Lloyd have studied the representation of a qubit using a superconducting \ufb02ux representation[MOL+ 99]. Platzman and Dykman have creatively proposed the use of electrons bound to the surface of liquid helium as qubits[PD99]. The description of the spin based quantum dot qubit realization in Section 7.8 was proposed by Loss and DiVincenzo[LD98]; Exercise 7.52 is due to them. An interesting entree into the literature on the coherence times of quantum dots is the article by Huibers, Switkes, Marcus, Campman, and Gossard[HSM+98]. Imamoglu, Awschalom, Burkard, DiVincenzo, Loss, Sherwin, and Small have proposed a quantum computer implementation using electron spins in quantum dots manipulated with cavity QED techniques[IAB+99]. The silicon-based nuclear spin quantum computer with 31P dopants was proposed by Kane[Kan98], and Vrijen, Yablonovitch, Wang, Jiang, Balandin, Roychowdhury, Mor, and DiVincenzo[VYW+99] have extended this to use electron spins buried in a silicon-germanium heterostructure. Finally, Brennen, Caves, Jessen, and Deutsch[BCJD99] have proposed an implementation of quantum computation using neutral atoms trapped in a far off-resonance optical lattice. 352 Quantum computers: physical realization Quantum teleportation has been experimentally realized using single photons and nuclear spins as qubits, as was described in \u2018History and further reading\u2019 at the end of Chapter 1. One of these implementations, by Furusawa, S\u00f8rensen, Braunstein, Fuchs, Kimble, and", " Polzik[FSB+98], is particularly of note in the context of this chapter, because it eschews the use of a \ufb01nite Hilbert space representation of quantum information (such as qubits)! Instead, it utilizes in\ufb01nite dimensional Hilbert spaces, where continuous variables (such as position and momentum, as in Section 7.3.2) parameterize quantum states. This approach to teleportation was originally suggested by Vaidman[Vai94], then further developed by Braunstein and Kimble[BK98a]. The continuous variable representation has also been extended to superdense coding by Braunstein and Kimble[BK99]; to quantum error correction, independently by Braunstein[Bra98] and by Lloyd and Slotine[LS98]; and to computation, by Lloyd and Braunstein[LB99]. III Quantum information 8 Quantum noise and quantum operations Until now we have dealt almost solely with the dynamics of closed quantum systems, that is, with quantum systems that do not suffer any unwanted interactions with the outside world. Although fascinating conclusions can be drawn about the information processing tasks which may be accomplished in principle in such ideal systems, these observations are tempered by the fact that in the real world there are no perfectly closed systems, except perhaps the universe as a whole. Real systems suffer from unwanted interactions with the outside world. These unwanted interactions show up as noise in quantum information processing systems. We need to understand and control such noise processes in order to build useful quantum information processing systems. This is a central topic of the third part of this book, which begins in this chapter with the description of the quantum operations formalism, a powerful set of tools enabling us to describe quantum noise and the behavior of open quantum systems. What is the distinction between an open and a closed system? A swinging pendulum like that found in some mechanical clocks can be a nearly ideal closed system. A pendulum interacts only very slightly with the rest of the world \u2013 its environment \u2013 mainly through friction. However, to properly describe the full dynamics of the pendulum and why it eventually ceases to move one must take into account the damping effects of air friction and imperfections in the suspension mechanism of the pendulum. Similarly, no quantum systems are ever perfectly closed, and especially not quantum computers, which must be delicately programmed by an external system to perform some desired set of operations. For example, if the state of a qubit is represented by two positions of a electron, then that electron will interact with other charged particles, which act as a source of uncontrolled noise affecting the state of the qubit. An open system is nothing more than one which has interactions with some other environment system, whose dynamics we wish to neglect, or average over. The mathematical formalism of quantum operations is the key tool for our description of the dynamics of open quantum systems. This tool is very powerful, in that it simultaneously addresses a wide range of physical scenarios. It can be used to describe not only nearly closed systems which are weakly coupled to their environments, but also systems which are strongly coupled to their environments, and closed systems that are opened suddenly and subject to measurement. Another advantage of quantum operations in applications to quantum computation and quantum information is that they are especially well adapted to describe discrete state changes, that is, transformations between an initial state \u03c1 and \ufb01nal state \u03c1, without explicit reference to the passage of time. This discrete-time analysis is rather different to the tools traditionally used by physicists for the description of open quantum systems (such as \u2018master equations\u2019, \u2018Langevin equations\u2019, and \u2018stochastic differential equations\u2019), which tend to be continuous-time descriptions. 354 Quantum noise and quantum operations The chapter is structured as follows. We begin in Section 8.1 with a discussion of how noise is described in classical systems. The intuition gained by understanding classical noise is invaluable in learning how to think about quantum operations and quantum noise. Section 8.2 introduces the quantum operations formalism from three different points of view, enabling us to become thoroughly familiar with the basic theory of quantum operations. Section 8.3 illustrates several important examples of noise using quantum operations. These include such examples as depolarization, amplitude damping, and phase damping. A geometric approach to understanding quantum noise on single qubits is explained, using the Bloch sphere. Section 8.4 explains some miscellaneous applications of quantum operations: the connection between quantum operations and other tools conventionally used by physicists to describe quantum noise, such as master equations; how to experimentally determine the dynamics a quantum system undergoes using a procedure known as quantum process tomography; and an explanation of how quantum operations can be used to understand the fact that the world around us appears to obey the rules of classical physics, while it really follows quantum mechanical laws. The chapter concludes in Section 8.5 with a discussion of the limitations of the quantum operations formalism as a general approach to the description of noise in quantum systems. 8.1 Classical", " noise and Markov processes To understand noise in quantum systems it is helpful to build some intuition by understanding noise in classical systems. How should we model noise in a classical system? Let\u2019s look at some simple examples to understand how this is done, and what it can teach us about noise in quantum systems. Imagine a bit is being stored on a hard disk drive attached to an ordinary classical computer. The bit starts out in the state 0 or 1, but after a long time it becomes likely that stray magnetic \ufb01elds will cause the bit to become scrambled, possibly \ufb02ipping its state. We can model this by a probability p for the bit to \ufb02ip, and a probability 1 \u2212 p for the bit to remain the same. This process is illustrated in Figure 8.1. Figure 8.1. After a long time a bit on a hard disk drive may \ufb02ip with probability p. What is really going on, of course, is that the environment contains magnetic \ufb01elds which can cause the bit to \ufb02ip. To \ufb01gure out the probability p for the bit to \ufb02ip we need to understand two things. First, we need a model for the distribution of magnetic \ufb01elds in the environment. Assuming that the user of the hard disk drive isn\u2019t doing anything silly like running a strong magnet near the disk drive, we can construct a realistic model by sampling the magnetic \ufb01eld in environments similar to the one in which the drive will be running. Second, we need a model for how magnetic \ufb01elds in the environment Classical noise and Markov processes 355 will interact with bits on the disk. Fortunately, such a model is already well known to physicists, and goes by the name \u2018Maxwell\u2019s equations\u2019. With these two elements in hand, we can in principle calculate the probability p that a bit \ufb02ip will occur on the drive over some prescribed period of time. This basic procedure \u2013 \ufb01nding a model for the environment and for the system\u2013 environment interaction \u2013 is one we follow repeatedly in our study of noise, both classical and quantum. Interactions with the environment are the fundamental source of noise in both classical and quantum systems. It is often not easy to \ufb01nd exact models for the environment or the system\u2013environment interaction; however, by being conservative in our modeling and closely studying the observed properties of a system to see if it obeys our model, it is possible to attain a high degree of accuracy in the modeling of noise in real physical systems. The behavior of the bit on the hard disk can be succinctly summarized in a single equation. Suppose p0 and p1 are the initial probabilities that the bit is in the states 0 and 1, respectively. Let q0 and q1 be the corresponding probabilities after the noise has occurred. Let X be the initial state of the bit, and Y the \ufb01nal state of the bit. Then the law of total probability (Appendix 1) states that p(Y = y) = x p(Y = y|X = x)p(X = x). (8.1) The conditional probabilities p(Y = y|X = x) are called transition probabilities, since they summarize the changes that may occur in the system. Writing these equations out explicitly for the bit on a hard disk we have q0 q1 = 1 \u2212 p p p 1 \u2212 p. p0 p1 (8.2) Let\u2019s look at a slightly more complicated example of noise in a classical system. Imagine that we are trying to build a classical circuit to perform some computational task. Unfortunately, we\u2019ve been given faulty components to build the circuit. Our rather arti\ufb01cial circuit consists of a single input bit, X, to which are applied two consecutive (faulty) gates, producing an intermediate bit Y, and a \ufb01nal bit Z. It seems reasonable to gate works correctly is independent of whether the assume that whether the second \ufb01rst gate worked correctly. This assumption \u2013 that the consecutive noise processes act independently \u2013 is a physically reasonable assumption to make in many situations. It results in a stochastic process X \u2192 Y \u2192 Z of a special type known as a Markov process. Physically, this assumption of Markovicity corresponds to assuming that the gate acts independently of the environenvironment causing the noise in the \ufb01rst gate, a reasonable assumption given that the ment causing the noise in the second gates are likely to be located a considerable distance apart in space. Summarizing, noise in classical systems can be described using the theory of stochastic processes. Often, in the analysis of multi-stage processes it is a good assumption to use Markov processes. For a single stage process the output probabilities q are related to the input probabilities p by the equation where", " E is a matrix of transition probabilities which we shall refer to as the evolution matrix. Thus, the \ufb01nal state of the system is linearly related to the starting state. This q = Ep, (8.3) 356 Quantum noise and quantum operations feature of linearity is echoed in the description of quantum noise, with density matrices replacing probability distributions. What properties must the evolution matrix E possess? We require that if p is a valid probability distribution, then Ep must also be a valid probability distribution. Satisfying this condition turns out to be equivalent to two conditions on E. First, all the entries of E must be non-negative, a condition known as the positivity requirement. If they weren\u2019t, then it would be possible to obtain negative probabilities in Ep. Second, all the columns of E must sum to one, a condition known as the completeness requirement. Suppose this weren\u2019t true. Imagine, for example, that the \ufb01rst column didn\u2019t sum to one. Letting p contain a one in the \ufb01rst entry and zeroes everywhere else, we see that Ep would not be a valid probability distribution in this case. Summarizing, the key features of classical noise are as follows: there is a linear relationship between input and output probabilities, described by a transition matrix with non-negative entries (positivity) and columns summing to one (completeness). Classical noise processes involving multiple stages are described as Markov processes, provided the noise is caused by independent environments. Each of these key features has important analogues in the theory of quantum noise. Of course, there are also some surprising new features of quantum noise! 8.2 Quantum operations 8.2.1 Overview The quantum operations formalism is a general tool for describing the evolution of quantum systems in a wide variety of circumstances, including stochastic changes to quantum states, much as Markov processes describe stochastic changes to classical states. Just as a classical state is described by a vector of probabilities, we shall describe quantum states in terms of the density operator (density matrix) \u03c1, whose properties you may wish to review by rereading Section 2.4, beginning on page 98, before continuing to read this chapter. And similar to how classical states transform as described by (8.3), quantum states transform as \u03c1 = E(\u03c1). (8.4) The map E in this equation is a quantum operation. Two simple examples of quantum operations which we have encountered previously, in Chapter 2, are unitary transformations and measurements, for which E(\u03c1) = U \u03c1U \u2020, and Em(\u03c1) = Mm\u03c1M \u2020 m, respectively (see Exercises 8.1 and 8.2, below). The quantum operation captures the dynamic change to a state which occurs as the result of some physical process; \u03c1 is the initial state before the process, and E(\u03c1) is the \ufb01nal state after the process occurs, possibly up to some normalization factor. Over the next few sections, we develop a general theory of quantum operations incorporating unitary evolution, measurement, and even more general processes! We shall develop three separate ways of understanding quantum operations, illustrated in Figure 8.2, all of which turn out to be equivalent. The \ufb01rst method is based on the idea of studying dynamics as the result of an interaction between a system and an environment, much as classical noise was described in Section 8.1. This method is concrete and easy to relate to the real world. Unfortunately, it suffers from the drawback of not being mathematically convenient. Our second method of understanding quantum op- Quantum operations 357 erations, completely equivalent to the \ufb01rst, overcomes this mathematical inconvenience by providing a powerful mathematical representation for quantum operations known as the operator-sum representation. This method is rather abstract, but is very useful for calculations and theoretical work. Our third approach to quantum operations, equivalent to the other two, is via a set of physically motivated axioms that we would expect a dynamical map in quantum mechanics to satisfy. The advantage of this approach is that it is exceedingly general \u2013 it shows that quantum dynamics will be described by quantum operations under an amazingly wide range of circumstances. However, it does not offer the calculational convenience of the second approach, nor the concrete nature of the \ufb01rst. Taken together, these three approaches to quantum operations offer a powerful tool with which we can understand quantum noise and its effects. Figure 8.2. Three approaches to quantum operations which are equivalent, but offer different advantages depending upon the intended application. Exercise 8.1: (Unitary evolution as a quantum operation) Pure states evolve under unitary transforms as |\u03c8 \u2192 U |\u03c8. Show that, equivalently, we may write \u03c1 \u2192 E(\u03c1) \u2261 U \u03c1U \u2020, for \u03c1 = |\u03c8\u03c8|. Exercise", " 8.2: (Measurement as a quantum operation) Recall from Section 2.2.3 (on page 84) that a quantum measurement with outcomes labeled by m is described by a set of measurement operators Mm such that Let the state of the system immediately before the measurement be \u03c1. Show that for Em(\u03c1) \u2261 Mm\u03c1M \u2020 measurement is m, the state of the system immediately after the mMm = I. m M \u2020 Em(\u03c1) tr(Em(\u03c1)). (8.5) Also show that the probability of obtaining this measurement result is p(m) = tr(Em(\u03c1)). 8.2.2 Environments and quantum operations The dynamics of a closed quantum system are described by a unitary transform. Conceptually, we can think of the unitary transform as a box into which the input state enters and from which the output exits, as illustrated on the left hand side of Figure 8.3. 358 Quantum noise and quantum operations For our purposes, the interior workings of the box are not of concern to us; it could be implemented by a quantum circuit, or by some Hamiltonian system, or anything else. A natural way to describe the dynamics of an open quantum system is to regard it as arising from an interaction between the system of interest, which we shall call the principal system, and an environment, which together form a closed quantum system, as illustrated on the right hand side of Figure 8.3. In other words, suppose we have a system in state \u03c1, which is sent into a box which is coupled to an environment. In general the \ufb01nal state of the system, E(\u03c1), may not be related by a unitary transformation to the initial state \u03c1. We assume (for now) that the system\u2013environment input state is a product state, \u03c1 \u2297 \u03c1env. After the box\u2019s transformation U the system no longer interacts with the environment, and thus we perform a partial trace over the environment to obtain the reduced state of the system alone: E(\u03c1) = trenv U (\u03c1 \u2297 \u03c1env) U \u2020. (8.6) Of course, if U does not involve any interaction with the environment, then E(\u03c1) = \u02dcU \u03c1 \u02dcU \u2020, where \u02dcU is the part of U which acts on the system alone. Equation (8.6) is our \ufb01rst of three equivalent de\ufb01nitions of a quantum operation. Figure 8.3. Models of closed (left) and open (right) quantum systems. An open quantum system consists of two parts, the principal system and an environment. An important assumption is made in this de\ufb01nition \u2013 we assume that the system and the environment start in a product state. In general, of course, this is not true. Quantum systems interact constantly with their environments, building up correlations. One way this expresses itself is via the exchange of heat between the system and its environment. Left to itself a quantum system will relax to the same temperature as its environment, which causes correlations to exist between the two. However, in many cases of practical interest it is reasonable to assume that the system and its environment start out in a product state. When an experimentalist prepares a quantum system in a speci\ufb01ed state they undo all the correlations between that system and the environment. Ideally, the correlations will be completely destroyed, leaving the system in a pure state. Even if this is not the case, we shall see later that the quantum operations formalism can even describe quantum dynamics when the system and environment do not start out in a product state. Another issue one might raise is: how can U be speci\ufb01ed if the environment has nearly in\ufb01nite degrees of freedom? It turns out, very interestingly, that in order for this model to properly describe any possible transformation \u03c1 \u2192 E(\u03c1), if the principal system has a Hilbert space of d dimensions, then it suf\ufb01ces to model the environment as being in a Hilbert space of no more than d2 dimensions. It also turns out not to be necessary for the environment to start out in a mixed state; a pure state will do. We shall return to these points in Section 8.2.3. Quantum operations 359 As an explicit example of the use of Equation (8.6), consider the two qubit quantum circuit shown in Figure 8.4, in which U is a controlledgate, with the principal system the control qubit, and the environment initially in the state \u03c1env = |00| as the target qubit. Inserting into Equation (8.6), it is easily seen that E(\u03c1) = P0\u03c1P0 + P1\u03c1P1, (8.7) where P0 = |00| and P1 = |11| are projection operators", ". Intuitively, this dynamics occurs because the environment stays in the |0 state only when the system is |0; otherwise the environment is \ufb02ipped to the state |1. In the next section we give a derivation of this equation as an example of the operator-sum representation. Figure 8.4. Controlled- gate as an elementary example of a single qubit quantum operation. We have described quantum operations as arising from the interaction of a principal system with an environment; however, it is convenient to generalize the de\ufb01nition somewhat to allow different input and output spaces. For example, imagine that a single qubit, which we label A, is prepared in an unknown state \u03c1. A three-level quantum system (\u2018qutrit\u2019) labelled B is prepared in some standard state |0, and then interacts with system A via a unitary interaction U, causing the joint system to evolve into the state U (\u03c1 \u2297 |00|)U \u2020. We then discard system A, leaving system B in some \ufb01nal state \u03c1. By de\ufb01nition, the quantum operation E describing this process is E(\u03c1) = \u03c1 = trA(U (\u03c1 \u2297 |00|)U \u2020). (8.8) Notice that E maps density operators of the input system, A, to density operators of the output system, B. Most of our discussion of quantum operations below is concerned with quantum operations \u2018on\u2019 some system A, that is, they map density operators of system A to density operators of system A. However it is occasionally useful in applications to allow a more general de\ufb01nition. Such a de\ufb01nition is provided by de\ufb01ning quantum operations as the class of maps which arise as a result of the following processes: some initial system is prepared in an unknown quantum state \u03c1, and then brought into contact with other systems prepared in standard states, allowed to interact according to some unitary interaction, and then some part of the combined system is discarded, leaving just a \ufb01nal system in some state \u03c1. The quantum operation E de\ufb01ning this process simply maps \u03c1 to \u03c1. We will see that this extension to allow different input and output spaces gels naturally with our treatment of quantum operations via the operator-sum representation, and also our axiomatic study. Nevertheless, for the most part it simpli\ufb01es discussion if we assume that the input and output spaces of a quantum operation are the same, using the convenient distinction between \u2018principal system\u2019 and \u2018environment\u2019 which disappears 360 Quantum noise and quantum operations in the general case, and giving occasional exercises to indicate the necessary extensions when the input and output spaces are different. 8.2.3 Operator-sum representation Quantum operations can be represented in an elegant form known as the operator-sum representation, which is essentially a re-statement of Equation (8.6) explicitly in terms of operators on the principal system\u2019s Hilbert space alone. The main result is motivated by the following simple calculation. Let |ek be an orthonormal basis for the (\ufb01nite dimensional) state space of the environment, and let \u03c1env = |e0e0| be the initial state of the environment. There is no loss of generality in assuming that the environment starts in a pure state, since if it starts in a mixed state we are free to introduce an extra system purifying the environment (Section 2.5). Although this extra system is \u2018\ufb01ctitious\u2019, it makes no difference to the dynamics experienced by the principal system, and thus can be used as an intermediate step in calculations. Equation (8.6) can thus be rewritten as \u03c1 \u2297 |e0e0| U \u2020|ek E(\u03c1) = ek|U (8.9) = k k Ek\u03c1E\u2020 k, (8.10) where Ek \u2261 ek|U |e0 is an operator on the state space of the principal system. Equation (8.10) is known as the operator-sum representation of E. The operators {Ek} are known as operation elements for the quantum operation E. The operator-sum representation is important; it will be used repeatedly for the remainder of the book. The operation elements satisfy an important constraint known as the completeness relation, analogous to the completeness relation for evolution matrices in the description of classical noise. In the classical case, the completeness relation arose from the requirement that probability distributions be normalized to one. In the quantum case the completeness relation arises from the analogous requirement that the trace of E(\u03c1) be equal to one, 1 = tr(E(\u03c1)) = tr = tr k k Ek\u03c1E\u2020 k E\u2020 kEk\u03c1. Since", " this relationship is true for all \u03c1 it follows that we must have k E\u2020 kEk = I. (8.11) (8.12) (8.13) (8.14) This equation is satis\ufb01ed by quantum operations which are trace-preserving. There kEk \u2264 I, but they are also non-trace-preserving quantum operations, for which describe processes in which extra information about what occurred in the process is obtained by measurement, as we explain in more detail shortly. Maps E of the form kEk \u2264 I provide our second de\ufb01nition of a quantum operation. of (8.10) for which We show below that this de\ufb01nition is essentially equivalent to the \ufb01rst, Equation (8.6), and in fact is slightly more general, since it allows for non-trace-preserving operations. We k E\u2020 k E\u2020 Quantum operations 361 will often have occasion to move backwards and forwards between these two de\ufb01nitions; it should be clear from context which de\ufb01nition we are working from at any given moment. Exercise 8.3: Our derivation of the operator-sum representation implicitly assumed that the input and output spaces for the operation were the same. Suppose a composite system AB initially in an unknown quantum state \u03c1 is brought into contact with a composite system CD initially in some standard state |0, and the two systems interact according to a unitary interaction U. After the interaction we discard systems A and D, leaving a state \u03c1 of system BC. Show that the map E(\u03c1) = \u03c1 satis\ufb01es E(\u03c1) = Ek\u03c1E\u2020 k, (8.15) k for some set of linear operators Ek from the state space of system AB to the state space of system BC, and such that k E\u2020 kEk = I. The operator-sum representation is important because it gives us an intrinsic means of characterizing the dynamics of the principal system. The operator-sum representation describes the dynamics of the principal system without having to explicitly consider properties of the environment; all that we need to know is bundled up into the operators Ek, which act on the principal system alone. This simpli\ufb01es calculations and often provides considerable theoretical insight. Furthermore, many different environmental interactions may give rise to the same dynamics on the principal system. If it is only the dynamics of the principal system which are of interest then it makes sense to choose a representation of the dynamics which does not include unimportant information about other systems. In the remainder of this section, we explore the properties of the operator-sum representation, and in particular, three features. First, we give it a physical interpretation, in terms of the operation elements Ek. A natural question which arises from this is how an operator-sum representation can be determined for any open quantum system (given, for example, the system\u2013environment interaction or other speci\ufb01cation). This is answered in the second topic addressed below, and the converse, how to construct a model open quantum system for any operator-sum representation, concludes. Exercise 8.4: (Measurement) Suppose we have a single qubit principal system, interacting with a single qubit environment through the transform U = P0 \u2297 I + P1 \u2297 X, (8.16) where X is the usual Pauli matrix (acting on the environment), and P0 \u2261 |00|, P1 \u2261 |11| are projectors (acting on the system). Give the quantum operation for this process, in the operator-sum representation, assuming the environment starts in the state |0. Exercise 8.5: (Spin \ufb02ips) Just as in the previous exercise, but now let U = X\u221a 2 \u2297 I + Y\u221a 2 \u2297 X, (8.17) Give the quantum operation for this process, in the operator-sum representation. 362 Quantum noise and quantum operations Exercise 8.6: (Composition of quantum operations) Suppose E and F are quantum operations on the same quantum system. Show that the composition F \u25e6 E is a quantum operation, in the sense that it has an operator-sum representation. State and prove an extension of this result to the case where E and F do not necessarily have the same input and output spaces. Physical interpretation of the operator-sum representation There is a nice interpretation that can be given to the operator-sum representation. Imagine that a measurement of the environment is performed in the basis |ek after the unitary transformation U has been applied. Applying the principle of implicit measurement, we see that such a measurement affects only the state of the environment, and does not change the state of the principal system. Let \u03c1k be the state of the principal system given that outcome k occurs, so \u03c1k \u221d trE(|ekek|", "U (\u03c1 \u2297 |e0e0|)U \u2020|ekek|) = ek|U (\u03c1 \u2297 |e0e0|)U \u2020|ek (8.18) (8.19) = Ek\u03c1E\u2020 k. Normalizing \u03c1k, \u03c1k = Ek\u03c1E\u2020 k tr(Ek\u03c1E\u2020 k), we \ufb01nd the probability of outcome k is given by Thus p(k) = tr(|ekek|U (\u03c1 \u2297 |e0e0|)U \u2020|ekek|) = tr(Ek\u03c1E\u2020 k). E(\u03c1) = k p(k)\u03c1k = k Ek\u03c1E\u2020 k. (8.20) (8.21) (8.22) (8.23) This gives us a beautiful physical interpretation of what is going on in a quantum operation with operation elements {Ek}. The action of the quantum operation is equivalent to taking the state \u03c1 and randomly replacing it by Ek\u03c1E\u2020 k), with probability tr(Ek\u03c1E\u2020 k). In this sense, it is very similar to the concept of noisy communication channels used in classical information theory; in this vein, we shall sometimes refer to certain quantum operations which describe quantum noise processes as being noisy quantum channels. k/tr(Ek\u03c1E\u2020 Figure 8.5. Controlled- gate as an elementary model of single qubit measurement. Quantum operations 363 A simple example, based on Figure 8.4, illustrates this interpretation of the operatorsum representation. Suppose we choose the states |ek = |0E and |1E, where we include the E subscript to make it clear that the state is a state of the environment. This can be interpreted as doing a measurement in the computational basis of the environment qubit, as shown in Figure 8.5. Doing such a measurement does not, of course, change the state of the principal system. Using subscripts P to denote the principal system, the controlled- may be expanded as U = |0P 0E0P 0E| + |0P 1E0P 1E| + |1P 1E1P 0E| + |1P 0E1P 1E|. (8.24) Thus and therefore E0 = 0E|U |0E = |0P 0P | E1 = 1E|U |0E = |1P 1P |, E(\u03c1) = E0\u03c1E0 + E1\u03c1E1, (8.25) (8.26) (8.27) in agreement with Equation (8.7). Measurements and the operator-sum representation Given a description of an open quantum system, how do we determine an operator-sum representation for its dynamics? We have already found one answer: given the unitary system\u2013environment transformation operation U, and a basis of states |ek for the environment, the operation elements are Ek \u2261 ek|U |e0. (8.28) It is possible to extend this result even further by allowing the possibility that a measurement is performed on the combined system\u2013environment after the unitary interaction, allowing the acquisition of information about the quantum state. It turns out that this physical possibility is naturally connected to non-trace-preserving quantum operations, that is, maps E(\u03c1) = k E\u2020 kEk \u2264 I. k Ek\u03c1E\u2020 k such that Suppose the principal system is initially in a state \u03c1. For convenience we denote the principal system by the letter Q. Adjoined to Q is an environment system E. We suppose that Q and E are initially independent systems, and that E starts in some standard state, \u03c3. The joint state of the system is thus initially \u03c1QE = \u03c1 \u2297 \u03c3. (8.29) We suppose that the systems interact according to some unitary interaction U. After the unitary interaction a projective measurement is performed on the joint system, described by projectors Pm. The case where no measurement is made corresponds to the special case where there is only a single measurement outcome, m = 0, which corresponds to the projector P0 \u2261 I. The situation is summarized in Figure 8.6. Our aim is to determine the \ufb01nal state of Q as a function of the initial state, \u03c1. The \ufb01nal state of QE is given by PmU (\u03c1 \u2297 \u03c3)U \u2020Pm tr(PmU (\u03c1 \u2297 \u03c3)U \u2020Pm), (8.30) 364 Quantum noise and quantum operations \u03c1Q Q Q PmU E Figure 8.6. Environmental model for a quantum operation. given that measurement outcome m occurred. Tracing out E we see that the \ufb01nal state of", " Q alone is trE(PmU (\u03c1 \u2297 \u03c3)U \u2020Pm) tr(PmU (\u03c1 \u2297 \u03c3)U \u2020Pm) This representation of the \ufb01nal state involves the initial state \u03c3 of the environment, the interaction U and the measurement operators Pm. De\ufb01ne a map (8.31). Em(\u03c1) \u2261 trE(PmU (\u03c1 \u2297 \u03c3)U \u2020Pm), (8.32) so the \ufb01nal state of Q alone is Em(\u03c1)/tr(Em(\u03c1)). Note that tr[Em(\u03c1)] is the probability j qj|jj| be an ensemble of outcome m of the measurement occurring. Let \u03c3 = decomposition for \u03c3. Introduce an orthonormal basis |ek for the system E. Note that Em(\u03c1) = qjtrE(|ekek|PmU (\u03c1 \u2297 |jj|)U \u2020Pm|ekek|) jk jk = Ejk\u03c1E\u2020 jk, where \u221a Ejk \u2261 qjek|PmU |j. (8.33) (8.34) (8.35) This equation is a generalization of Equation (8.10), and gives an explicit means for calculating the operators appearing in an operator-sum representation for Em, given that the initial state \u03c3 of E is known, and the dynamics between Q and E are known. The quantum operations Em can be thought of as de\ufb01ning a kind of measurement process generalizing the description of measurements given in Chapter 2. Exercise 8.7: Suppose that instead of doing a projective measurement on the combined principal system and environment we had performed a general measurement described by measurement operators {Mm}. Find operator-sum representations for the corresponding quantum operations Em on the principal system, and show that the respective measurement probabilities are tr[E(\u03c1)]. Quantum operations 365 System\u2013environment models for any operator-sum representation We have shown that interacting quantum systems give rise in a natural way to an operatorsum representation for quantum operations. What about the converse problem? Given a set of operators {Ek} is there some reasonable model environmental system and dynamics which give rise to a quantum operation with those operation elements? By \u2018reasonable\u2019 we mean that the dynamics must be either a unitary evolution or a projective measurement. Here, we show how to construct such a model. We will only show how to do this for quantum operations mapping the input space to the same output space, although it is mainly a matter of notation to generalize the construction to the more general case. In particular, we show that for any trace-preserving or non-trace-preserving quantum operation, E, with operation elements {Ek}, there exists a model environment, E, starting in a pure state |e0, and model dynamics speci\ufb01ed by a unitary operator U and projector P onto E such that E(\u03c1) = trE(P U (\u03c1 \u2297 |e0e0|)U \u2020P ). (8.36) k E\u2020 To see this, suppose \ufb01rst that E is a trace-preserving quantum operation, with operatorsum representation generated by operation elements {Ek} satisfying the completeness kEk = I, so we are only attempting to \ufb01nd an appropriate unitary operator relation U to model the dynamics. Let |ek be an orthonormal basis set for E, in one-to-one correspondence with the index k for the operators Ek. Note that by de\ufb01nition E has such a basis; we are trying to \ufb01nd a model environment giving rise to a dynamics described by the operation elements {Ek}. De\ufb01ne an operator U which has the following action on states of the form |\u03c8|e0, U |\u03c8|e0 \u2261 Ek|\u03c8|ek, (8.37) k where |e0 is just some standard state of the model environment. Note that for arbitrary states |\u03c8 and |\u03d5 of the principal system, \u03c8|e0|U \u2020U |\u03d5|e0 = \u03c8|E\u2020 kEk|\u03d5 = \u03c8|\u03d5, (8.38) k by the completeness relation. Thus the operator U can be extended to a unitary operator acting on the entire state space of the joint system. It is easy to verify that trE(U (\u03c1 \u2297 |e0e0|)U \u2020) = (8.39) Ek\u03c1E\u2020 k, so this model provides a realization of the quantum operation E with operation elements {Ek}. This result is illustrated in Box 8", ".1. k Non-trace-preserving quantum operations can easily be modeled using a construction along the same lines (Exercise 8.8). A more interesting generalization of this construction is the case of a set of quantum operations {Em} corresponding to possible outcomes Em is trace-preserving, since the from a measurement, so the quantum operation m for probabilities of the distinct outcomes sum to one, 1 = all possible inputs \u03c1. See Exercise 8.9, below. m p(m) = tr (\u03c1) Em m Exercise 8.8: (Non-trace-preserving quantum operations) Explain how to construct a unitary operator for a system\u2013environment model of a 366 Quantum noise and quantum operations non-trace-preserving quantum operation, by introducing an extra operator, E\u221e, into the set of operation elements Ek, chosen so that when summing over the complete set of k, including k = \u221e, one obtains k E\u2020 kEk = I. Exercise 8.9: (Measurement model) If we are given a set of quantum operations m Em is trace-preserving, then it is possible to construct a {Em} such that measurement model giving rise to this set of quantum operations. For each m, let Emk be a set of operation elements for Em. Introduce an environmental system, E, with an orthonormal basis |m, k in one-to-one correspondence with the set of indices for the operation elements. Analogously to the earlier construction, de\ufb01ne an operator U such that U |\u03c8|e0 = Emk|\u03c8|m, k. (8.40) mk Next, de\ufb01ne projectors Pm \u2261 Show that performing U on \u03c1 \u2297 |e0e0|, then measuring Pm gives m with probability tr(Em(\u03c1)), and the corresponding post-measurement state of the principal system is Em(\u03c1)/tr(Em(\u03c1)). |m, km, k| on the environmental system, E. k Box 8.1: Mocking up a quantum operation Given a trace-preserving quantum operation expressed in the operator-sum representation, E(\u03c1) = k, we can construct a physical model for it in the following way. From (8.10), we want U to satisfy k Ek\u03c1E\u2020 Ek = ek|U |e0, where U is some unitary operator, and |ek are orthonormal basis vectors for the environment system. Such a U is conveniently represented as the block matrix (8.41E1] [E2] [E3] [E4]... \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7..................... \u23a5 \u23a5 \u23a5 \u23a5 \u23a5 \u23a6 (8.42) in the basis |ek. Note that the operation elements Ek only determine the \ufb01rst block column of this matrix (unlike elsewhere, here it is convenient to have the \ufb01rst label of the states be the environment, and the second, the principal system). Determination of the rest of the matrix is left up to us; we simply choose the entries such that U is unitary. Note that by the results of Chapter 4, U can be implemented by a quantum circuit. 8.2.4 Axiomatic approach to quantum operations Until now the main motivation for our study of quantum operations has been that they provide an elegant way of studying systems which are interacting with an environment. We\u2019re going to switch to a different viewpoint now, where we try to write down physically Quantum operations 367 motivated axioms which we expect quantum operations to obey. This viewpoint is more abstract than our earlier approach, based on explicit system\u2013environment models, but it is also extremely powerful because of that abstraction. The way we\u2019re going to proceed is as follows. First, we\u2019re going to forget everything we\u2019ve learned about quantum operations, and start over by de\ufb01ning quantum operations according to a set of axioms, which we\u2019ll justify on physical grounds. That done, we\u2019ll prove that a map E satis\ufb01es these axioms if and only if it has an operator-sum representation, thus providing the missing link between the abstract axiomatic formulation, and our earlier discussion. We de\ufb01ne a quantum operation E as a map from the set of density operators of the input space Q1 to the set of density operators for the output space Q2, with the following three axiomatic properties: (note that for notational simplicity in the proofs we take Q1 = Q2 = Q) \u2022 A1: First, tr[E(\u03c1)] is the probability that the process represented by E occurs, when \u03c1 is the initial state. Thus,", " 0 \u2264 tr[E(\u03c1)] \u2264 1 for any state \u03c1. \u2022 A2: Second, E is a convex-linear map on the set of density matrices, that is, for probabilities {pi}, E i pi\u03c1i = i piE(\u03c1i). (8.43) \u2022 A3: Third, E is a completely positive map. That is, if E maps density operators of system Q1 to density operators of system Q2, then E(A) must be positive for any positive operator A. Furthermore, if we introduce an extra system R of arbitrary dimensionality, it must be true that (I \u2297 E)(A) is positive for any positive operator A on the combined system RQ1, where I denotes the identity map on system R. The \ufb01rst property is one of mathematical convenience. To cope with the case of measurements, it turns out that it is extremely convenient to make the convention that E does not necessarily preserve the trace property of density matrices, that tr(\u03c1) = 1. Rather, we make the convention that E is to be de\ufb01ned in such a way that tr[E(\u03c1)] is equal to the probability of the measurement outcome described by E occurring. For example, suppose that we are doing a projective measurement in the computational basis of a single qubit. Then two quantum operations are used to describe this process, de\ufb01ned by E0(\u03c1) \u2261 |00|\u03c1|00| and E1(\u03c1) \u2261 |11|\u03c1|11|. Notice that the probabilities of the respective outcomes are correctly given by tr[E0(\u03c1)] and tr[E1(\u03c1)]. With this convention the correctly normalized \ufb01nal quantum state is therefore E(\u03c1) tr[E(\u03c1)]. (8.44) In the case where the process is deterministic, that is, no measurement is taking place, this reduces to the requirement that tr[E(\u03c1)] = 1 = tr(\u03c1), for all \u03c1. As previously discussed, in this case, we say that the quantum operation is a trace-preserving quantum operation, since on its own E provides a complete description of the quantum process. On the other hand, if there is a \u03c1 such that tr[E(\u03c1)] < 1, then the quantum operation is non-tracepreserving, since on its own E does not provide a complete description of the processes that may occur in the system. (That is, other measurement outcomes may occur, with 368 Quantum noise and quantum operations some probability.) A physical quantum operation is one that satis\ufb01es the requirement that probabilities never exceed 1, tr[E(\u03c1)] \u2264 1. The second property stems from a physical requirement on quantum operations. Suppose the input \u03c1 to the quantum operation is obtained by randomly selecting the state from an ensemble {pi, \u03c1i} of quantum states, that is, \u03c1 = i pi\u03c1i. Then we would expect that the resulting state, E(\u03c1)/tr[E(\u03c1)] = E(\u03c1)/p(E) corresponds to a random selection from the ensemble {p(i|E), E(\u03c1i)/tr[E(\u03c1i)]}, where p(i|E) is the probability that the state prepared was \u03c1i, given that the process represented by E occurred. Thus, we demand that E(\u03c1) = p(E) i p(i|E) E(\u03c1i) tr[E(\u03c1i)], (8.45) where p(E) = tr[E(\u03c1)] is the probability that the process described by E occurs on input of \u03c1. By Bayes\u2019 rule (Appendix 1), p(i|E) = p(E|i) pi p(E) = tr[E(\u03c1i)]pi p(E) (8.46) so (8.45) reduces to (8.43). The third property also originates from an important physical requirement, that not only must E(\u03c1) be a valid density matrix (up to normalization) so long as \u03c1 is valid, but furthermore, if \u03c1 = \u03c1RQ is the density matrix of some joint system of R and Q, if E acts only on Q, then E(\u03c1RQ) must still result in a valid density matrix (up to normalization) of the joint system. An example is given in Box 8.2. Formally, suppose we introduce a second (\ufb01nite dimensional) system R. Let I denote the identity map on system R. Then the map I \u2297 E must take positive operators to positive operators. It is perhaps surprising that these three axioms are suf\ufb01cient to de\ufb01ne quantum operations. However, the following theorem shows that they are equivalent", " to the earlier system-environment models and the de\ufb01nition in terms of an operator-sum representation: Theorem 8.1: The map E satis\ufb01es axioms A1, A2 and A3 if and only if E(\u03c1) = Ei\u03c1E\u2020 i, (8.50) i for some set of operators {Ei} which map the input Hilbert space to the output Hilbert space, and i Ei \u2264 I. i E\u2020 Proof i. E is obviously linear, so to check that E is a quantum Suppose E(\u03c1) = operation we need only prove that it is completely positive. Let A be any positive operator acting on the state space of an extended system, RQ, and let |\u03c8 be some state of RQ. De\ufb01ning |\u03d5i \u2261 (IR \u2297 E\u2020 i Ei\u03c1E\u2020 i )|\u03c8, we have \u03c8|(IR \u2297 Ei)A(IR \u2297 E\u2020 i )|\u03c8 = \u03d5i|A|\u03d5i \u2265 0, by the positivity of the operator A. It follows that \u03c8|(I \u2297 E)(A)|\u03c8 = \u03d5i|A|\u03d5i \u2265 0, i (8.51) (8.52) (8.53) Quantum operations 369 Box 8.2: Complete positivity versus positivity The transpose operation on a single qubit provides an example of why complete positivity is an important requirement for quantum operations. By de\ufb01nition, this map transposes the density operator in the computational basis: a b c d T\u2212\u2192 a c b d. (8.47) This map preserves positivity of a single qubit. However, suppose that qubit is part of a two qubit system initially in the entangled state |00 + |11 \u221a 2, (8.48) and the transpose operation is applied to the \ufb01rst of these two qubits, while the second qubit is subject to trivial dynamics. Then the density operator of the system after the dynamics has been applied is \u23a4 \u23a1 \u23a2 \u23a2 \u23a3 8.49) A calculation shows that this operator has eigenvalues 1/2, 1/2, 1/2 and \u22121/2, so this is not a valid density operator. Thus, the transpose operation is an example of a positive map which is not completely positive, that is, it preserves the positivity of operators on the principal system, but does not continue to preserve positivity when applied to systems which contain the principal system as a subsystem. and thus for any positive operator A, the operator (I \u2297 E)(A) is also positive, as required. i E\u2020 i Ei \u2264 I ensures that probabilities are less than or equal to 1. The requirement This completes the \ufb01rst part of the proof. Suppose next that E satis\ufb01es axioms A1, A2 and A3. Our aim will be to \ufb01nd an operator-sum representation for E. Suppose we introduce a system, R, with the same dimension as the original quantum system, Q. Let |iR and |iQ be orthonormal bases for R and Q. It will be convenient to use the same index, i, for these two bases, and this can certainly be done as R and Q have the same dimensionality. De\ufb01ne a joint state |\u03b1 of RQ by |\u03b1 \u2261 |iR|iQ. (8.54) i The state |\u03b1 is, up to a normalization factor, a maximally entangled state of the systems R and Q. This interpretation of |\u03b1 as a maximally entangled state may help in understanding the following construction. Next, we de\ufb01ne an operator \u03c3 on the state space of RQ by We may think of this as the result of applying the quantum operation E to one half of \u03c3 \u2261 (IR \u2297 E)(|\u03b1\u03b1|). (8.55) 370 Quantum noise and quantum operations a maximally entangled state of the system RQ. It is a truly remarkable fact, which we will now demonstrate, that the operator \u03c3 completely speci\ufb01es the quantum operation E. That is, to know how E acts on an arbitrary state of Q, it is suf\ufb01cient to know how it acts on a single maximally entangled state of Q with another system! The trick which allows us to recover E from \u03c3 is as follows. Let |\u03c8 = j \u03c8j|jQ be any state of system Q. De\ufb01ne a corresponding state | \u02dc\u03c8 of system R by the equation | \u02dc\u03c8 \u2261 \u03c8\u2217 j |jR", ". Notice that j \u239e |iRjR| \u2297 E(|iQjQ|) \u23a0 | \u02dc\u03c8 \u239b \u239d ij \u03c8i\u03c8\u2217 j E(|iQjQ|) \u02dc\u03c8|\u03c3| \u02dc\u03c8 = \u02dc\u03c8| = ij = E(|\u03c8\u03c8|). (8.59) |sisi| be some decomposition of \u03c3, where the vectors |si need not be Let \u03c3 = i normalized. De\ufb01ne a map Ei(|\u03c8) \u2261 \u02dc\u03c8|si. (8.60) A little thought shows that this map is a linear map, so Ei is a linear operator on the state space of Q. Furthermore, we have Ei|\u03c8\u03c8|E\u2020 i = \u02dc\u03c8|sisi| \u02dc\u03c8 Thus i i = \u02dc\u03c8|\u03c3| \u02dc\u03c8 = E(|\u03c8\u03c8|). E(|\u03c8\u03c8|) = i Ei|\u03c8\u03c8|E\u2020 i, for all pure states, |\u03c8, of Q. By convex-linearity it follows that E(\u03c1) = Ei\u03c1E\u2020 i i (8.56) (8.57) (8.58) (8.61) (8.62) (8.63) (8.64) (8.65) in general. The condition the trace of E(\u03c1) with a probability. i E\u2020 i Ei \u2264 I follows immediately from axiom A1 identifying Freedom in the operator-sum representation We have seen that the operator-sum representation provides a very general description of the dynamics of an open quantum system. Is it a unique description? Consider quantum operations E and F acting on a single qubit with the operatork and F (\u03c1) = k, where the operation k Ek\u03c1E\u2020 k Fk\u03c1F \u2020 sum representations E(\u03c1) = elements for E and F are de\ufb01ned by I\u221a 2 1\u221a 2 E1 = 0 1 1 0 = E2 = Z\u221a 2 = 1\u221a 2 1 0 0 \u22121 (8.66) Quantum operations 371 and F1 = |00| = 1 0 0 0 F2 = |11| =. 0 0 0 1 (8.67) These appear to be very different quantum operations. What is interesting is that E and \u221a F are actually the same quantum operation. To see this, note that F1 = (E1 + E2)/ 2 and F2 = (E1 \u2212 E2)/ \u221a 2. Thus, (E1 + E2)\u03c1(E\u2020 F (\u03c1) = 1 + E2\u03c1E\u2020 2 = E1\u03c1E\u2020 = E(\u03c1). 1 + E\u2020 2 ) + (E1 \u2212 E2)\u03c1(E\u2020 1 2 \u2212 E\u2020 2 ) (8.68) (8.69) (8.70) This example shows that the operation elements appearing in an operator-sum representation for a quantum operation are not unique. The freedom in the representation is very interesting. Suppose we \ufb02ipped a fair coin, and, depending on the outcome of the coin toss, applied either the unitary operator I or Z to the quantum system. This process corresponds to the \ufb01rst operator-sum representation for E. The second operator-sum representation for E (labeled F above) corresponds to performing a projective measurement in the {|0, |1} basis, with the outcome of the measurement unknown. These two apparently very different physical processes give rise to exactly the same dynamics for the principal system. When do two sets of operation elements give rise to the same quantum operation? Understanding this question is important for at least two reasons. First, from a physical point of view, understanding the freedom in the representation gives us more insight into how different physical processes can give rise to the same system dynamics. Second, understanding the freedom in operator-sum representation is crucial to a good understanding of quantum error-correction. Intuitively, it is clear that there must be a great deal of freedom in an operatorsum representation. Consider a trace-preserving quantum operation E which describes the dynamics of some system such as that shown in Figure 8.3. We have shown that the operation elements Ek = ek|U |e0 for E may be associated with an orthonormal basis |ek for the environment. Suppose that we supplement the interaction U with an additional unitary action U on the environment alone, as shown in Figure 8.7. Clearly this does not change the state of the principal system. What are the corresponding operation elements to this new process, (I \u2297 U )U? We obtain: F", "k = ek|(I \u2297 U ) U |e0 I \u2297 ek|U |ej = j = j U kjEj, ej|U |e0 (8.71) (8.72) (8.73) kj are the matrix elements where we have used the fact that of U with respect to the basis |ek. It turns out that the freedom in the operator-sum representation yielded by this physically motivated picture captures the essence of the complete freedom available in the operator-sum representation, as proved in the following theorem. |ejej| = I, and U j 372 Quantum noise and quantum operations \u03c1 |e0e0| U U E(\u03c1) Figure 8.7. Origin of the unitary freedom in the operator-sum representation. Theorem 8.2: (Unitary freedom in the operator-sum representation) Suppose {E1,..., Em} and {F1,..., Fn} are operation elements giving rise to quantum operations E and F, respectively. By appending zero operators to the shorter list of operation elements we may ensure that m = n. Then E = F if and only if there exist complex numbers uij such that Ei = m unitary matrix. j uijFj, and uij is an m by Proof The key to the proof is Theorem 2.6, on page 103. Recall that this result tells us that two sets of vectors |\u03c8i and |\u03d5j generate the same operator if and only if |\u03c8i = uij|\u03d5j, (8.74) j where uij is a unitary matrix of complex numbers, and we \u2018pad\u2019 whichever set of states |\u03c8i or |\u03d5j is smaller with additional states 0 so that the two sets have the same number of elements. This result allows us to characterize the freedom in operator-sum representations. Suppose {Ei} and {Fj} are two sets of operation elements for the same quantum j for all \u03c1. De\ufb01ne operation, j Fj\u03c1F \u2020 i Ei\u03c1E\u2020 i = |ei \u2261 |fj \u2261 k k Ei|kQ Fj|kQ |kR |kR. Recall the de\ufb01nition of \u03c3 in Equation (8.55), from which it follows that \u03c3 = j |fjfj|, and thus there exists unitary uij such that |ei = uij|fj. But for arbitrary |\u03c8 we have j Ei|\u03c8 = \u02dc\u03c8|ei = uij \u02dc\u03c8|fj Thus uijFj|\u03c8. j = k Ei = uijFj. j (8.75) (8.76) i |eiei| = (8.77) (8.78) (8.79) (8.80) (8.81) Examples of quantum noise and quantum operations 373 Conversely, supposing Ei and Fj are related by a unitary transformation of the form Ei = ij uijFj, simple algebra shows that the quantum operation with operation elements {Ei} is the same as the quantum operation with operation elements {Fj}. Theorem 8.2 can be used to answer another interesting question: what is the maximum size of an environment that would be needed to mock up a given quantum operation? Theorem 8.3: All quantum operations E on a system of Hilbert space dimension d can be generated by an operator-sum representation containing at most d2 elements, E(\u03c1) = M k=1 Ek\u03c1E\u2020 k, (8.82) where 1 \u2264 M \u2264 d2. The proof of this theorem is simple and is left as an exercise for you. Exercise 8.10: Give a proof of Theorem 8.3 based on the freedom in the operator-sum representation, as follows. Let {Ej} be a set of operation elements for E. De\ufb01ne a matrix Wjk \u2261 tr(E\u2020 j Ek). Show that the matrix W is Hermitian and of rank at most d2, and thus there is unitary matrix u such that uW u\u2020 is diagonal with at most d2 non-zero entries. Use u to de\ufb01ne a new set of at most d2 non-zero operation elements {Fj} for E. Exercise 8.11: Suppose E is a quantum operation mapping a d-dimensional input space to a d-dimensional output space. Show that E can be described using a set of at most dd operation elements {Ek}. The freedom in the operator-sum representation is surprisingly useful. We use it, for example, in our study of quantum error-correction in Chapter 10. In that chapter we will see that certain", " sets of operators in the operator-sum representation give more useful information about the quantum error-correction process, and it behooves us to study quantum error-correction from that point of view. As usual, having multiple ways of understanding a process gives us much more insight into what is going on. 8.3 Examples of quantum noise and quantum operations In this section we examine some concrete examples of quantum noise and quantum operations. These models illustrate the power of the quantum operations formalism we have been developing. They are also important in understanding the practical effects of noise on quantum systems, and how noise can be controlled by techniques such as error-correction. We begin in Section 8.3.1 by considering how measurement can be described as a quantum operation, and in particular we consider the trace and partial trace operations. After that, we turn to noise processes, beginning in Section 8.3.2 with the presentation of a graphical method for understanding quantum operations on a single qubit. This method is used in the remainder of the section to illustrate elementary bit and phase \ufb02ip error processes (in Section 8.3.3), the depolarizing channel (in Section 8.3.4), amplitude damping (in Section 8.3.5), and phase damping (in Section 8.3.6). Amplitude and phase 374 Quantum noise and quantum operations damping are ideal models of noise that capture many of the most important features of the noise occurring in quantum mechanical systems, and we not only consider their abstract mathematical formulation, but also how the processes arise in real-world quantum systems. 8.3.1 Trace and partial trace One of the main uses of the quantum operations formalism is to describe the effects of measurement. Quantum operations can be used to describe both the probability of getting a particular outcome from a measurement on a quantum system, and also the state change in the system effected by the measurement. The simplest operation related to measurement is the trace map \u03c1 \u2192 tr(\u03c1) \u2013 which we can show is indeed a quantum operation, in the following way. Let HQ be any input Hilbert space, spanned by an orthonormal basis |1... |d, and let H Q be a onedimensional output space, spanned by the state |0. De\ufb01ne E(\u03c1) \u2261 d i=1 |0i|\u03c1|i0|, (8.83) so that E is a quantum operation, by Theorem 8.1. Note that E(\u03c1) = tr(\u03c1)|00|, so that, up to the unimportant |00| multiplier, this quantum operation is identical to the trace function. An even more useful result is the observation that the partial trace is a quantum operation. Suppose we have a joint system QR, and wish to trace out system R. Let |j be a basis for system R. De\ufb01ne a linear operator Ei : HQR \u2192 HQ by \u239b \u239e \u239d Ei j \u03bbj|qj|j \u23a0 \u2261 \u03bbi|qi, (8.84) where \u03bbj are complex numbers, and |qj are arbitrary states of system Q. De\ufb01ne E to be the quantum operation with operation elements {Ei}, that is, E(\u03c1) \u2261 Ei\u03c1E\u2020 i. (8.85) i By Theorem 8.1, this is a quantum operation from system QR to system Q. Notice that E(\u03c1 \u2297 |jj|) = \u03c1\u03b4j,j = trR(\u03c1 \u2297 |jj|), where \u03c1 is any Hermitian operator on the state space of system Q, and |j and |j are members of the orthonormal basis for system R. By linearity of E and trR, it follows that E = trR. (8.86) 8.3.2 Geometric picture of single qubit quantum operations There is an elegant geometric method for picturing quantum operations on a single qubit. This method allows one to get an intuitive feel for the behavior of quantum operations in terms of their action on the Bloch sphere. Recall from Exercise 2.72 on page 105 that the state of a single qubit can always be written in the Bloch representation8.87) Examples of quantum noise and quantum operations 375 where r is a three component real vector. Explicitly, this gives us 1 2 \u03c1 = 1 + rz rx + iry rx \u2212 iry 1 \u2212 rz. (8.88) In this representation, it turns out that an arbitrary trace-preserving quantum operation is equivalent to a map of the form E\u2192 r = Mr + c, r (8.89) where M is a 3 \u00d7 3 real matrix, and c is a constant vector. This is an af\ufb01ne map, mapping the Bloch sphere into itself", ". To see this, suppose the operators Ei generating the operator-sum representation for E are written in the form 3 Ei = \u03b1iI + aik\u03c3k. k=1 Then it is not dif\ufb01cult to check that Mjk = l alja\u2217 lk + a\u2217 ljalk + |\u03b1l|2 \u2212 p alpa\u2217 lp \u03b4jk + i p ck = 2i l jp jpkalja\u2217 lp, (8.90) jkp(\u03b1la\u2217 lp \u2212 \u03b1\u2217 l alp) (8.91) (8.92) where we have made use of the completeness relation expression for c. i E\u2020 i Ei = I to simplify the The meaning of the af\ufb01ne map, Equation (8.89), is made clearer by considering the polar decomposition of the matrix M, M = U |M |, where U is unitary. Because M is real, it follows that |M | is real and Hermitian, that is, |M | is a symmetric matrix. Furthermore, because M is real we may assume that U is real, and is thus an orthogonal matrix, that is, U T U = I, where T is the transpose operation. Thus we may write M = OS, (8.93) where O is a real orthogonal matrix with determinant 1, representing a proper rotation, and S is a real symmetric matrix. Viewed this way, Equation (8.89) is just a deformation of the Bloch sphere along principal axes determined by S, followed by a proper rotation due to O, followed by a displacement due to c. Exercise 8.12: Why can we assume that O has determinant 1 in the decomposition (8.93)? Exercise 8.13: Show that unitary transformations correspond to rotations of the Bloch sphere. Exercise 8.14: Show that det(S) need not be positive. 376 Quantum noise and quantum operations 8.3.3 Bit \ufb02ip and phase \ufb02ip channels The geometric picture described above can be used to visualize some important quantum operations on single qubits, which will later be used in the theory of error-correction. The bit \ufb02ip channel \ufb02ips the state of a qubit from |0 to |1 (and vice versa) with probability 1 \u2212 p. It has operation elements \u221a p I = E0 = \u221a = E1 = $ 1 \u2212 p 0 1 1 0. (8.94) The effect of the bit \ufb02ip channel is illustrated in Figure 8.8. 1 0.5 0.5 y y 0 0 -0.5 -0.5 1 0.5 0.5 y y 0 0 -0.5 -0.5 1 -1 1 0.5 z 0 -0.5 -1 -1 -1 =\u21d2 1 -1 1 0.5 z 0 -0.5 -1 -1 -1 -0.5 -0.5 0 0 x x 0.5 0.5 -0.5 -0.5 0 0 x x 0.5 0.5 1 1 Figure 8.8. The effect of the bit \ufb02ip channel on the Bloch sphere, for p = 0.3. The sphere on the left represents the set of all pure states, and the deformed sphere on the right represents the states after going through the channel. Note that the states on the \u02c6x axis are left alone, while the \u02c6y- \u02c6z plane is uniformly contracted by a factor of 1 \u2212 2p. This geometric picture makes it very easy to verify certain facts about this quantum operation. For example, it is easy to verify that the quantity tr(\u03c12) for a single qubit is equal to (1 + |r|2)/2, where |r| is the norm of the Bloch vector. The contraction of the Bloch sphere illustrated in Figure 8.8 cannot increase the norm of the Bloch vector, and therefore we can immediately conclude that tr(\u03c12) can only ever decrease for the bit \ufb02ip channel. This is but one example of the use of the geometric picture; once it becomes suf\ufb01ciently familiar it becomes a great source of insight about the properties of quantum operations on a single qubit. The phase \ufb02ip channel has operation elements \u221a E0 = p I = \u221a p 1 0 0 1 $ E1 = 1. (8.95) The effect of the phase \ufb02ip channel is illustrated in Figure 8.9. As a special case of the phase \ufb02ip channel, consider the quantum operation which arises when we choose p = 1/2. Using the freedom", " in the operator-sum representation this operation may be written \u03c1 \u2192 E(\u03c1) = P0\u03c1P0 + P1\u03c1P1, (8.96) where P0 = |00|, P1 = |11|, which corresponds to a measurement of the qubit in the |0, |1 basis, with the result of the measurement unknown. Using the above prescription Examples of quantum noise and quantum operations 377 it is easy to see that the corresponding map on the Bloch sphere is given by (rx, ry, rz) \u2192 (0, 0, rz). (8.97) Geometrically, the Bloch vector is projected along the z axis, and the x and y components of the Bloch vector are lost. 1 0.5 0.5 y y 0 0 -0.5 -0.5 1 0.5 0.5 y y 0 0 -0.5 -0.5 1 -1 1 0.5 z 0 -0.5 -1 -1 -1 =\u21d2 1 -1 1 0.5 z 0 -0.5 -1 -1 -1 -0.5 -0.5 0 0 x x 0.5 0.5 -0.5 -0.5 0 0 x x 0.5 0.5 1 1 Figure 8.9. The effect of the phase \ufb02ip channel on the Bloch sphere, for p = 0.3. Note that the states on the \u02c6z axis are left alone, while the \u02c6x \u2212 \u02c6y plane is uniformly contracted by a factor of 1 \u2212 2p. The bit\u2013phase \ufb02ip channel has operation elements \u221a p I = E0 = \u221a = E1 = $ 1 \u2212 p 0 \u2212i i 0. (8.98) As the name indicates, this is a combination of a phase \ufb02ip and a bit \ufb02ip, since Y = iXZ. The action of the bit\u2013phase \ufb02ip channel is illustrated in Figure 8.10. 1 0.5 0.5 y y 0 0 -0.5 -0.5 1 0.5 0.5 y y 0 0 -0.5 -0.5 1 -1 1 0.5 z 0 -0.5 -1 -1 -1 =\u21d2 1 -1 1 0.5 z 0 -0.5 -1 -1 -1 -0.5 -0.5 0 0 x x 0.5 0.5 -0.5 -0.5 0 0 x x 0.5 0.5 1 1 Figure 8.10. The effect of the bit\u2013phase \ufb02ip channel on the Bloch sphere, for p = 0.3. Note that the states on the \u02c6y axis are left alone, while the \u02c6x- \u02c6z plane is uniformly contracted by a factor of 1 \u2212 2p. 378 Quantum noise and quantum operations Exercise 8.15: Suppose a projective measurement is performed on a single qubit in \u221a the basis |+, |\u2212, where |\u00b1 \u2261 (|0 \u00b1 |1)/ of the result of the measurement, the density matrix evolves according to the equation 2. In the event that we are ignorant \u03c1 \u2192 E(\u03c1) = |++|\u03c1|++| + |\u2212\u2212|\u03c1|\u2212\u2212|. (8.99) Illustrate this transformation on the Bloch sphere. Exercise 8.16: The graphical method for understanding single qubit quantum operations was derived for trace-preserving quantum operations. Find an explicit example of a non-trace-preserving quantum operation which cannot be described as a deformation of the Bloch sphere, followed by a rotation and a displacement. 8.3.4 Depolarizing channel The depolarizing channel is an important type of quantum noise. Imagine we take a single qubit, and with probability p that qubit is depolarized. That is, it is replaced by the completely mixed state, I/2. With probability 1 \u2212 p the qubit is left untouched. The state of the quantum system after this noise is E(\u03c1) = pI 2 + (1 \u2212 p)\u03c1. (8.100) The effect of the depolarizing channel on the Bloch sphere is illustrated in Figure 8.11. 1 0.5 0.5 y y 0 0 -0.5 -0.5 1 0.5 0.5 y y 0 0 -0.5 -0.5 1 -1 1 0.5 z 0 -0.5 -1 -1 -1 =\u21d2 1 -1 1 0.5 z 0 -0.5 -1 -1 -1 -0.5 -0.5 0 0 x x 0.5 0.5 -0.5 -0.5 0", " 0 x x 0.5 0.5 1 1 Figure 8.11. The effect of the depolarizing channel on the Bloch sphere, for p = 0.5. Note how the entire sphere contracts uniformly as a function of p. A quantum circuit simulating the depolarizing channel is illustrated in Figure 8.12. The top line of the circuit is the input to the depolarizing channel, while the bottom two lines are an \u2018environment\u2019 to simulate the channel. We have used an environment with two mixed state inputs. The idea is that the third qubit, initially a mixture of the state |0 with probability 1 \u2212 p and state |1 with probability p acts as a control for whether or not the completely mixed state I/2 stored in the second qubit is swapped into the \ufb01rst qubit. Examples of quantum noise and quantum operations 379 Figure 8.12. Circuit implementation of the depolarizing channel. The form (8.100) is not in the operator-sum representation. However, if we observe that for arbitrary \u03c1, I 2 = \u03c1 + X\u03c1X + Y \u03c1Y + Z\u03c1Z 4 (8.101) (8.102) and then substitute for I/2 into (8.100) we arrive at the equation 1 \u2212 3p 4 (X\u03c1X + Y \u03c1Y + Z\u03c1Z), E(\u03c1) = \u03c1 + p 4 $ \u221a showing that the depolarizing channel has operation elements { \u221a p X/2, p Z/2}. Note, incidentally, that it is frequently convenient to parametrize the 1 \u2212 3p/4 I, p Y /2, \u221a depolarizing channel in different ways, such as E(\u03c1) = (1 \u2212 p)\u03c1 + p 3 (X\u03c1X + Y \u03c1Y + Z\u03c1Z), (8.103) which has the interpretation that the state \u03c1 is left alone with probability 1 \u2212 p, and the operators X, Y and Z applied each with probability p/3. Exercise 8.17: Verify (8.101) as follows. De\ufb01ne E(A) \u2261 A + XAX + Y AY + ZAZ 4, (8.104) and show that E(I) = I; E(X) = E(Y ) = E(Z) = 0. (8.105) Now use the Bloch sphere representation for single qubit density matrices to verify (8.101). The depolarizing channel can, of course, be generalized to quantum systems of dimension more than two. For a d-dimensional quantum system the depolarizing channel again replaces the quantum system with the completely mixed state I/d with probability p, and leaves the state untouched otherwise. The corresponding quantum operation is E(\u03c1) = pI d + (1 \u2212 p)\u03c1. (8.106) Exercise 8.18: For k \u2265 1 show that tr(\u03c1k) is never increased by the action of the depolarizing channel. Exercise 8.19: Find an operator-sum representation for a generalized depolarizing channel acting on a d-dimensional Hilbert space. 380 Quantum noise and quantum operations 8.3.5 Amplitude damping An important application of quantum operations is the description of energy dissipation \u2013 effects due to loss of energy from a quantum system. What are the dynamics of an atom which is spontaneously emitting a photon? How does a spin system at high temperature approach equilibrium with its environment? What is the state of a photon in an interferometer or cavity when it is subject to scattering and attenuation? Each of these processes has its own unique features, but the general behavior of all of them is well characterized by a quantum operation known as amplitude damping, which we can derive by considering the following scenario. Suppose we have a single optical mode containing the quantum state a|0+b|1, a superposition of zero or one photons. The scattering of a photon from this mode can be modeled by thinking of inserting a partially silvered mirror, a beamsplitter, in the path of the photon. As we saw in Section 7.4.2, this beamsplitter allows the photon to couple to another single optical mode (representing the environment), according to the unitary transformation B = exp, where a, a\u2020 and b, b\u2020 are annihilation and creation operators for photons in the two modes. The output after the beamsplitter, assuming the environment starts out with no photons, is simply B|0(a|0+b|1) = a|00+b(cos \u03b8|01+sin \u03b8|10), using Equation (7.34). Tracing over the environment gives us the quantum operation a\u2020b \u2212 ab\u2020 & \u03b8 \u2019 EAD(\u03c1) = E0\u03c1E\u2020 0 + E1\u03c1E\u2020", " 1, (8.107) where Ek = k|B|0 are 1 0 0 0 E0 = E1 = \u221a \u221a 0 1 \u2212 \u03b3 \u03b3 0, (8.108) the operation elements for amplitude damping. \u03b3 = sin2 \u03b8 can be thought of as the probability of losing a photon. Observe that no linear combination can be made of E0 and E1 to give an operation element proportional to the identity (though compare with Exercise 8.23). The E1 operation changes a |1 state into a |0 state, corresponding to the physical process of losing a quantum of energy to the environment. E0 leaves |0 unchanged, but reduces the amplitude of a |1 state; physically, this happens because a quantum of energy was not lost to the environment, and thus the environment now perceives it to be more likely that the system is in the |0 state, rather than the |1 state. Figure 8.13. Circuit model for amplitude damping Exercise 8.20: (Circuit model for amplitude damping) Show that the circuit in Examples of quantum noise and quantum operations 381 Figure 8.13 models the amplitude damping quantum operation, with sin2(\u03b8/2) = \u03b3. Exercise 8.21: (Amplitude damping of a harmonic oscillator) Suppose that our principal system, a harmonic oscillator, interacts with an environment, modeled as another harmonic oscillator, through the Hamiltonian H = \u03c7(a\u2020b + b\u2020a) (8.109) where a and b are the annihilation operators for the respective harmonic oscillators, as de\ufb01ned in Section 7.3. (1) Using U = exp(\u2212iH\u0394t), denoting the eigenstates of b\u2020b as |kb, and selecting the vacuum state |0b as the initial state of the environment, show that the operation elements Ek = kb|U |0b are found to be Ek = n ( n k (1 \u2212 \u03b3)n\u2212k\u03b3k |n \u2212 kn|, (8.110) where \u03b3 = 1 \u2212 cos2(\u03c7\u0394t) is the probability of loosing a single quantum of energy, and states such as |n are eigenstates of a\u2020a. (2) Show that the operation elements Ek de\ufb01ne a trace-preserving quantum operation. Exercise 8.22: (Amplitude damping of single qubit density matrix) For the general single qubit state \u03c1 = a b\u2217 b c show that amplitude damping leads to EAD(\u03c1) = 1 \u2212 (1 \u2212 \u03b3)(1 \u2212 a) \u221a b\u2217 1 \u2212 \u03b3 (8.111) (8.112) \u221a 1 \u2212 \u03b3 b c(1 \u2212 \u03b3). Exercise 8.23: (Amplitude damping of dual-rail qubits) Suppose that a single qubit state is represented by using two qubits, as |\u03c8 = a |01 + b |10. (8.113) Show that EAD \u2297 EAD applied to this state gives a process which can be described by the operation elements $ \u221a \u03b3 Edr 0 = 1 = Edr 1 \u2212 \u03b3 I |0001| + |0010|, (8.114) (8.115) 0 ) happens to the qubit, or the qubit is transformed 1 ) into the state |00, which is orthogonal to |\u03c8. This is a simple that is, either nothing (Edr (Edr error-detection code, and is also the basis for the robustness of the \u2018dual-rail\u2019 qubit discussed in Section 7.4. Exercise 8.24: (Spontaneous emission is amplitude damping) A single atom coupled to a single mode of electromagnetic radiation undergoes spontaneous emission, as was described in Section 7.6.1. To see that this process is just 382 Quantum noise and quantum operations amplitude damping, take the unitary operation resulting from the Jaynes\u2013Cummings interaction, Equation (7.77), with detuning \u03b4 = 0, and give the quantum operation resulting from tracing over the \ufb01eld. A general characteristic of a quantum operation is the set of states that are left invariant under the operation. For example, we have seen how the phase \ufb02ip channel leaves the \u02c6z axis of the Bloch sphere unchanged; this corresponds to states of the form p|00| + (1 \u2212 p)|11| for arbitrary probability p. In the case of amplitude damping, only the ground state |0 is left invariant. That is a natural consequence of our modeling the environment as starting in the |0 state, as if it were at zero temperature. What quantum operation describes the effect of dissipation to", " an environment at \ufb01nite temperature? This process, EGAD, called generalized amplitude damping, is de\ufb01ned for single qubits by the operation elements \u221a E0 = \u221a E1 = E2 = E3 = 1 \u2212 \u03b3 0 0 1 0 0\u221a \u03b3 0, (8.116) (8.117) (8.118) (8.119) where the stationary state \u03c1\u221e = 0 1 \u2212 p p 0 satis\ufb01es EGAD(\u03c1\u221e) = \u03c1\u221e. Generalized amplitude damping describes the \u2018T1\u2019 relaxation processes due to coupling of spins to their surrounding lattice, a large system which is in thermal equilibrium at a temperature often much higher than the spin temperature. This is the case relevant to NMR quantum computation, where some of the properties of EGAD described in Box 8.3 become important. (8.120), Exercise 8.25: If we de\ufb01ne the temperature T of a qubit by assuming that in equilibrium the probabilities of being in the |0 and |1 states satisfy a Boltzmann distribution, that is p0 = e\u2212E0/kBT /Z and p1 = e\u2212E1/kBT /Z, where E0 is the energy of the state |0, E1 the energy of the state |1, and Z = e\u2212E0/kBT + e\u2212E1/kBT, what temperature describes the state \u03c1\u221e? We can visualize the effect of amplitude damping in the Bloch representation as the Bloch vector transformation (rx, ry, rz) \u2192 rx $ 1 \u2212 \u03b3, ry $ 1 \u2212 \u03b3, \u03b3 + rz(1 \u2212 \u03b3). (8.122) When \u03b3 is replaced with a time-varying function like 1 \u2212 e\u2212t/T1 (t is time, and T1 just some constant characterizing the speed of the process), as is often the case for real physical processes, we can visualize the effects of amplitude damping as a \ufb02ow on the Bloch sphere, which moves every point in the unit ball towards a \ufb01xed point at the north pole, where |0 is located. This is shown in Figure 8.14. Examples of quantum noise and quantum operations 383 Box 8.3: Generalized amplitude damping and effective pure states The notion of \u2018effective pure states\u2019 introduced in Section 7.7 was found to be useful in NMR implementations of quantum computers. These states behave like pure states under unitary evolution and measurement of traceless observables. How do they behave under quantum operations? In general, non-unitary quantum operations ruin the effectiveness of these states, but surprisingly, they can behave properly under generalized amplitude damping. Consider a single qubit effective pure state \u03c1 = (1 \u2212 p)I + (2p \u2212 1)|00|. Clearly, traceless measurement observables acting on U \u03c1U \u2020 produce results which are proportional to those on the pure state U |00|U \u2020. Suppose \u03c1 is the stationary state of EGAD. Interestingly, in this case, EGAD(U \u03c1U \u2020) = (1 \u2212 p)I + (2p \u2212 1)EAD(U \u03c1U \u2020). (8.121) That is, under generalized amplitude damping, an effective pure state can remain such, and moreover, the \u2018pure\u2019 component of \u03c1 behaves as if it were undergoing amplitude damping to a reservoir at zero temperature! Similarly, generalized amplitude damping performs the transformation $ $ (rx, ry, rz) \u2192 rx 1 \u2212 \u03b3, ry 1 \u2212 \u03b3, \u03b3(2p \u2212 1) + rz(1 \u2212 \u03b3). (8.123) Comparing (8.122) and (8.123), it is clear that amplitude damping and generalized amplitude damping differ only in the location of the \ufb01xed point of the \ufb02ow; the \ufb01nal state is along the \u02c6z axis, at the point (2p \u2212 1), which is a mixed state. 1 0.5 0.5 y y 0 0 -0.5 -0.5 1 0.5 0.5 y y 0 0 -0.5 -0.5 1 -1 1 0.5 z 0 -0.5 -1 -1 -1 =\u21d2 1 -1 1 0.5 z 0 -0.5 -1 -1 -1 -0.5 -0.5 0 0 x x 0.5 0.5 -0.5 -0.5 0 0 x x 0.5 0.5 1 1 Figure 8.14. The", " effect of the amplitude damping channel on the Bloch sphere, for p = 0.8. Note how the entire sphere shrinks towards the north pole, the |0 state. 8.3.6 Phase damping A noise process that is uniquely quantum mechanical, which describes the loss of quantum information without loss of energy, is phase damping. Physically it describes, for example, 384 Quantum noise and quantum operations what happens when a photon scatters randomly as it travels through a waveguide, or how electronic states in an atom are perturbed upon interacting with distant electrical charges. The energy eigenstates of a quantum system do not change as a function of time, but do accumulate a phase which is proportional to the eigenvalue. When a system evolves for an amount of time which is not precisely known, partial information about this quantum phase \u2013 the relative phases between the energy eigenstates \u2013 is lost. A very simple model for this kind of quantum noise is the following. Suppose that we have a qubit |\u03c8 = a |0 + b |1 upon which the rotation operation Rz(\u03b8) is applied, where the angle of rotation \u03b8 is random. The randomness could originate, for example, from a deterministic interaction with an environment, which never again interacts with the system and thus is implicitly measured (see Section 4.4). We shall call this random Rz operation a phase kick. Let us assume that the phase kick angle \u03b8 is well represented as a random variable which has a Gaussian distribution with mean 0 and variance 2\u03bb. The output state from this process is given by the density matrix obtained from averaging over \u03b8, \u03c1 = = % \u221e \u2212\u221e 1\u221a 4\u03c0\u03bb |a|2 a\u2217b e\u2212\u03bb Rz(\u03b8)|\u03c8\u03c8|R\u2020 z(\u03b8)e\u2212\u03b82/4\u03bb d\u03b8. ab\u2217 e\u2212\u03bb |b|2 (8.124) (8.125) The random phase kicking causes the expected value of the off-diagonal elements of the density matrix to decay exponentially to zero with time. That is a characteristic result of phase damping. Another way to derive the phase damping quantum operation is to consider an interaction between two harmonic oscillators, in a manner similar to how amplitude damping was derived in the last section, but this time with the interaction Hamiltonian H = \u03c7 a\u2020a(b + b\u2020), Letting U = exp(\u2212iH\u0394t), considering only the |0 and |1 states of the a oscillator as our system, and taking the environment oscillator to initially be |0, we \ufb01nd that tracing over the environment gives the operation elements Ek = kb|U |0b, which are (8.126) E0 = E1 =, where \u03bb = 1 \u2212 cos2(\u03c7\u0394t) can be interpreted as the probability that a photon from the system has been scattered (without loss of energy). As was the case for amplitude damping, E0 leaves |0 unchanged, but reduces the amplitude of a |1 state; unlike amplitude damping, however, the E1 operation destroys |0 and reduces the amplitude of the |1 state, and does not change it into a |0. By applying Theorem 8.2, the unitary freedom of quantum operations, we \ufb01nd that a unitary recombination of E0 and E1 gives a new set of operation elements for phase damping, \u221a \u03b1 \u02dcE0 = 1 0 0 1 (8.127) (8.128) (8.129) Examples of quantum noise and quantum operations 385 \u221a 1 \u2212 \u03b1 \u02dcE1 =, 0 1 0 \u22121 (8.130) \u221a where \u03b1 = (1 + same as the phase \ufb02ip channel which we encountered in Section 8.3.3! 1 \u2212 \u03bb)/2. Thus the phase damping quantum operation is exactly the Since phase damping is the same as the phase \ufb02ip channel, we have already seen how it is visualized on the Bloch sphere, in Figure 8.9. This corresponds to the Bloch vector transformation \u221a \u221a (rx, ry, rz) \u2192 rx 1 \u2212 \u03bb, ry 1 \u2212 \u03bb, rz, (8.131) \u221a which has the effect of shrinking the sphere into ellipsoids. Phase damping is often referred to as a \u2018T2\u2019 (or \u2018spin-spin\u2019) relaxation process, for historical reasons, where e\u2212t/2T2 = 1 \u2212 \u03bb. As a function of time, the amount of damping increases, corresponding to an inwards \ufb02ow of all points in the unit ball towards the \u02c6z axis. Note that states along the \ufffd", "\ufffdz axis remain invariant. Historically, phase damping was a process that was almost always thought of, physically, as resulting from a random phase kick or scattering process. It was not until the connection to the phase \ufb02ip channel was discovered that quantum error-correction was developed, since it was thought that phase errors were continuous and couldn\u2019t be described as a discrete process! In fact, single qubit phase errors can always be thought of as resulting from a process in which either nothing happens to a qubit, with probability \u03b1, or with probability 1 \u2212 \u03b1, the qubit is \ufb02ipped by the Z Pauli operation. Although this might not be the actual microscopic physical process happening, from the standpoint of the transformation occurring to a qubit over a discrete time interval large compared to the underlying random process, there is no difference at all. Phase damping is one of the most subtle and important processes in the study of quantum computation and quantum information. It has been the subject of an immense amount of study and speculation, particularly with regard to why the world around us appears to be so classical, with superposition states not a part of our everyday experience! Perhaps it is phase damping that is responsible for this absence of superposition states from the everyday (Exercise 8.31)? The pioneering quantum physicist Schr\u00a8odinger was perhaps the \ufb01rst to pose this problem, and he did this in a particularly stark form, as discussed in Box 8.4. \u03c1in |0 \u2022 \u03c1out Ry(\u03b8) Figure 8.15. Circuit model for phase damping. The upper wire carries the input qubit with an unknown state, and the lower wire is an ancilla qubit used to model the environment. Exercise 8.26: (Circuit model for phase damping) Show that the circuit in Figure 8.15 can be used to model the phase damping quantum operation, provided \u03b8 is chosen appropriately. Exercise 8.27: (Phase damping = phase \ufb02ip channel) Give the unitary 386 Quantum noise and quantum operations transformation which relates the operation elements of (8.127)\u2013(8.128) to those of (8.129)\u2013(8.130); that is, \ufb01nd u such that \u02dcEk = j ukjEj. Exercise 8.28: (One phase damping model circuit) Show that a single gate can be used as a model for phase damping, if we let the controlledinitial state of the environment be a mixed state, where the amount of damping is determined by the probability of the states in the mixture. Exercise 8.29: (Unitality) A quantum process E is unital if E(I) = I. Show that the depolarizing and phase damping channels are unital, while amplitude damping is not. Exercise 8.30: (T2 \u2264 T1/2) The T2 phase coherence relaxation rate is just the exponential decay rate of the off-diagonal elements in the qubit density matrix, while T1 is the decay rate of the diagonal elements (see Equation (7.144)). Amplitude damping has both nonzero T1 and T2 rates; show that for amplitude damping T2 = T1/2. Also show that if amplitude and phase damping are both applied then T2 \u2264 T1/2. Exercise 8.31: (Exponential sensitivity to phase damping) Using (8.126), show that the element \u03c1nm = n|\u03c1|m in the density matrix of a harmonic oscillator decays exponentially as e\u2212\u03bb(n\u2212m)2 under the effect of phase damping, for some constant \u03bb. 8.4 Applications of quantum operations As be\ufb01ts a powerful tool, the quantum operations formalism has numerous applications. In this section we describe two of these applications. Section 8.4.1 describes the theory of master equations, a picture of quantum noise complementary to the quantum operations formalism. The master equation approach describes quantum noise in continuous time using differential equations, and is the approach to quantum noise most often used by physicists. In Section 8.4.2 we describe quantum process tomography, a procedure to experimentally determine the dynamics of a quantum system. 8.4.1 Master equations Open quantum systems occur in a wide range of disciplines, and many tools other than quantum operations can be employed in their study. In this section, we brie\ufb02y describe one such tool, the approach of master equations. The dynamics of open quantum systems have been studied extensively in the \ufb01eld of quantum optics. The main objective in this context is to describe the time evolution of an open system with a differential equation which properly describes non-unitary behavior. This description is provided by the master equation, which can be written most generally in the Lindblad form as d\u03c1 dt = \u2212 i", " [H, \u03c1] + 2Lj\u03c1L\u2020 j \u2212 {L\u2020 jLj, \u03c1}, j (8.134) where {x, y} = xy +yx denotes an anticommutator, H is the system Hamiltonian, a Hermitian operator representing the coherent part of the dynamics, and Lj are the Lindblad Applications of quantum operations 387 Box 8.4: Schr\u00a8odinger\u2019s cat When I hear about Schr\u00a8odinger\u2019s cat, I reach for my gun. \u2013 Stephen Hawking Schr\u00a8odinger\u2019s infamous cat faces life or death contingent upon an automatic device which breaks a vial of poison and kills the cat if an excited atomic state is observed to decay, as illustrated here: Schr\u00a8odinger asked what happens when the atom is in a superposition state? Is the cat alive or dead? Why do superposition states such as this apparently not occur in the everyday world? This conundrum is resolved by realizing that it is very unlikely to occur in real life, because of extreme sensitivity of macroscopic superposition states to decoherence. Let the atom represent a single qubit. The joint system has the initial state |alive|1. Suppose that after one half-life of the atom, the state is the equal superposition |alive(|0 + |1)/ 2 (this represents a simpli\ufb01cation of the actual physics, which are too involved to go into here). The apparatus kills the cat if the atom is in the |0 state; otherwise, the cat lives. This gives the state |\u03c8 = 2, in which the cat\u2019s state has become entangled with that of the atom. This would seem to indicate the cat is simultaneously alive and dead, but suppose we consider the density matrix of this state, |dead|0 + |alive|1 \u221a \u221a / \u03c1 = |\u03c8\u03c8| 1 2 = |alive, 1alive, 1| + |dead, 0dead, 0| +|alive, 1dead, 0| + |dead, 0alive, 1| (8.132). (8.133) Now, in practice it is impossible to perfectly isolate the cat and the atom in their box, and thus information about this superposition state will leak into the external world. For example, heat from the cat\u2019s body could permeate the wall and give some indication of its state to the outside. Such effects may be modeled as phase damping, which exponentially damps out the \ufb01nal two (off-diagonal) terms in \u03c1. To a \ufb01rst approximation, we may model the cat\u2013atom system as a simple harmonic oscillator. An important result about the decoherence of such a system is that coherence between states of high energy difference decays faster than between states with a lower energy difference (Exercise 8.31). Thus \u03c1 will quickly be transformed into a nearly diagonal state, which represents an ensemble of cat\u2013atom states which correspond to either dead or alive, and are not in a superposition of the two states. 388 Quantum noise and quantum operations operators, representing the coupling of the system to its environment. The differential equation takes on the above form in order that the process be completely positive in a sense similar to that described earlier for quantum operations. It is also generally assumed that the system and environment begin in a product state. Furthermore, in order to derive a master equation for a process, one usually begins with a system\u2013environment model Hamiltonian, and then makes the Born and Markov approximations in order to determine Lj. Note that in the master equation approach, tr[\u03c1(t)] = 1 at all times. As an example of a Lindblad equation, consider a two-level atom coupled to the vacuum, undergoing spontaneous emission. The coherent part of the atom\u2019s evolution is described by the Hamiltonian H = \u2212\u03c9\u03c3z/2. \u03c9 is the energy difference of the atomic levels. Spontaneous emission causes an atom in the excited (|1) state to drop down into the ground (|0) state, emitting a photon in the process. This emission is described by \u03b3\u03c3\u2212, where \u03c3\u2212 \u2261 |01| is the atomic lowering operator, and \u03b3 the Lindblad operator is the rate of spontaneous emission. The master equation describing this process is \u221a d\u03c1 dt = \u2212 i [H, \u03c1] + \u03b3 2\u03c3\u2212\u03c1\u03c3+ \u2212 \u03c3+\u03c3\u2212\u03c1 \u2212 \u03c1\u03c3+\u03c3\u2212, (8.135) where \u03c3+ \u2261 \u03c3\u2020 \u2212 is the atomic raising operator. To solve the equation it is helpful to move to the interaction picture, that is, make the change of variables \u02dc\u03c1(t) \u2261 eiHt\u03c1(t)e", "\u2212iHt. The equation of motion for \u02dc\u03c1 is easily found to be d \u02dc\u03c1 dt 2 \u02dc\u03c3\u2212 \u02dc\u03c1 \u02dc\u03c3+ \u2212 \u02dc\u03c3+ \u02dc\u03c3\u2212 \u02dc\u03c1 \u2212 \u02dc\u03c1 \u02dc\u03c3+ \u02dc\u03c3\u2212 = \u03b3 where \u02dc\u03c3\u2212 \u2261 eiHt\u03c3\u2212e\u2212iHt = e\u2212i\u03c9t\u03c3\u2212 \u02dc\u03c3+ \u2261 eiHt\u03c3+e\u2212iHt = ei\u03c9t\u03c3+. Our \ufb01nal equation of motion is thus = \u03b3 2\u03c3\u2212 \u02dc\u03c1\u03c3+ \u2212 \u03c3+\u03c3\u2212 \u02dc\u03c1 \u2212 \u02dc\u03c1\u03c3+\u03c3\u2212. d \u02dc\u03c1 dt (8.136) (8.137) (8.138) (8.139) (8.140) This equation of motion is easily solved using a Bloch vector representation for \u02dc\u03c1. The solution is \u03bbx = \u03bbx(0)e\u2212\u03b3t \u03bby = \u03bby(0)e\u2212\u03b3t \u03bbz = \u03bbz(0)e\u22122\u03b3t + 1 \u2212 e\u22122\u03b3t. (8.141) (8.142) (8.143) De\ufb01ning \u03b3 = 1 \u2212 exp(\u22122t\u03b3) we can easily check that this evolution is equivalent to \u02dc\u03c1(t) = E( \u02dc\u03c1(0)) \u2261 E0 \u02dc\u03c1(0)E\u2020 0 + E1 \u02dc\u03c1(0)E\u2020 1, where E0 \u2261 E1 \u2261 8.144) (8.145) (8.146) Applications of quantum operations 389 are the operation elements de\ufb01ning the quantum operation E. Note that the effect of E is amplitude damping; compare with Equation (8.108). The example we have considered is an instance of the spin-boson model, in which a small, \ufb01nite dimensional quantum system interacts with a bath of simple harmonic oscillators. Physically, it is important in describing the interaction of atoms with electromagnetic radiation, as in cavity QED, or atom and ion traps. The master equation approach is less general than the quantum operations formalism. Solving a master equation allows one to determine the time dependence of a density matrix. Knowing this, in turn, means that the result can be expressed as a quantum operation in the operator-sum representation, \u03c1(t) = k Ek(t)\u03c1(0)E\u2020 k(t), (8.147) where Ek(t) are time dependent operation elements, determined from the solution to the master equation. However, a quantum process described in terms of an operator-sum representation cannot necessarily be written down as a master equation. For example, quantum operations can describe non-Markovian dynamics, simply because they describe only state changes, not continuous time evolution. Nevertheless, each approach has its own place. In fact, even quantum operations do not provide the most general description; we consider in Section 8.5 some processes which are not described by quantum operations. 8.4.2 Quantum process tomography Quantum operations provide a wonderful mathematical model for open quantum systems, and are conveniently visualized (at least for qubits) \u2013 but how do they relate to experimentally measurable quantities? What measurements should an experimentalist do if they wish to characterize the dynamics of a quantum system? For classical systems, this elementary task is known as system identi\ufb01cation. Here, we show how its analogue, known as quantum process tomography, can be performed for \ufb01nite dimensional quantum systems. To understand process tomography we \ufb01rst need to understand another procedure called quantum state tomography. State tomography is the procedure of experimentally determining an unknown quantum state. Suppose we are given an unknown state, \u03c1, of a single qubit. How can we experimentally determine what the state of \u03c1 is? If we are given just a single copy of \u03c1 then it turns out to be impossible to characterize \u03c1. The basic problem is that there is no quantum measurement which can distinguish non-orthogonal quantum states like |0 and (|0 + |1)/ 2 with certainty. However, it is possible to estimate \u03c1 if we have a large number of copies of \u03c1. For instance, if \u03c1 is the quantum state produced by some experiment, then we simply repeat the experiment many times to produce many copies of the state \u03c1. \u221a \u221a Suppose we have many copies of a single qubit density matrix, \u03c1. The set I/ \u221a 2, 2 forms an orthonormal set of matrices with respect to the Hilbert\u2013 2, Y / 2, Z/ \u221a \u221a X/ Schmidt inner product, so \u03c1 may be expanded as \u03c1 = tr(", "\u03c1)I + tr(X\u03c1)X + tr(Y \u03c1)Y + tr(Z\u03c1)Z 2 Recall, however, that expressions like tr(A\u03c1) have an interpretation as the average value of observables. For example, to estimate tr(Z\u03c1) we measure the observable Z a large number of times, m, obtaining outcomes z1, z2,..., zm, all equal to +1 or \u22121. The empirical (8.148). 390 Quantum noise and quantum operations i zi/m, is an estimate for the true value of tr(Z\u03c1). We can average of these quantities, use the central limit theorem to determine how well this estimate behaves for large m, where it becomes approximately Gaussian with mean equal to tr(Z\u03c1) and with standard m, where \u0394(Z) is the standard deviation for a single measurement of deviation \u0394(Z)/ i zi/m is Z, which is upper bounded by 1, so the standard deviation in our estimate at most 1/ m. \u221a \u221a In a similar way we can estimate the quantities tr(X\u03c1) and tr(Y \u03c1) with a high degree of con\ufb01dence in the limit of a large sample size, and thus obtain a good estimate for \u03c1. Generalizing this procedure to the case of more than one qubit is not dif\ufb01cult, at least in principle! Similar to the single qubit case, an arbitrary density matrix on n qubits can be expanded as \u03c1 = v tr (\u03c3v1 \u2297 \u03c3v2 \u2297 \u00b7 \u00b7 \u00b7 \u2297 \u03c3vn \u03c1) \u03c3v1 \u2297 \u03c3v2 \u2297 \u00b7 \u00b7 \u00b7 \u2297 \u03c3vn 2n, (8.149) where the sum is over vectors v = (v1,..., vn) with entries vi chosen from the set 0, 1, 2, 3. By performing measurements of observables which are products of Pauli matrices we can estimate each term in this sum, and thus obtain an estimate for \u03c1. We\u2019ve described how to do state tomography for systems comprised of qubits. What if non-qubit systems are involved? Not surprisingly, it is easy to generalize the above prescription to such systems. We won\u2019t explicitly do so here, but instead refer you to the end of chapter \u2018History and further reading\u2019 for references. Now that we know how to do quantum state tomography, how can we use it to do quantum process tomography? The experimental procedure may be outlined as follows. Suppose the state space of the system has d dimensions; for example, d = 2 for a single qubit. We choose d2 pure quantum states |\u03c81,..., |\u03c8d2, chosen so that the corresponding density matrices |\u03c81\u03c81|,..., |\u03c8d2\u03c8d2| form a basis set for the space of matrices. We explain in more detail how to choose such a set below. For each state |\u03c8j we prepare the quantum system in that state and then subject it to the process which we wish to characterize. After the process has run to completion we use quantum state tomography to determine the state E(|\u03c8j\u03c8j|) output from the process. From a purist\u2019s point of view we are now done, since in principle the quantum operation E is now determined by a linear extension of E to all states. In practice, of course, we would like to have a way of determining a useful representation of E from experimentally available data. We will explain a general procedure for doing so, worked out explicitly for the case of a single qubit. Our goal is to determine a set of operation elements {Ei} for E, E(\u03c1) = i Ei\u03c1E\u2020 i. (8.150) However, experimental results involve numbers, not operators, which are a theoretical concept. To determine the Ei from measurable parameters, it is convenient to consider an equivalent description of E using a \ufb01xed set of operators \u02dcEi, which form a basis for the set of operators on the state space, so that Ei = m eim \u02dcEm (8.151) Applications of quantum operations 391 for some set of complex numbers eim. Equation (8.150) may thus be rewritten as E(\u03c1) = \u02dcEm\u03c1 \u02dcE\u2020 n\u03c7mn, mn (8.152) i eime\u2217 where \u03c7mn \u2261 in are the entries of a matrix which is positive Hermitian by de\ufb01nition. This expression, known as the chi matrix representation, shows that E can be completely described by a complex number matrix,", " \u03c7, once the set of operators Ei has been \ufb01xed. In general, \u03c7 will contain d4 \u2212 d2 independent real parameters, because a general linear map of d by d complex matrices to d by d matrices is described by d4 independent parameters, but there are d2 additional constraints due to the fact that \u03c1 remains Hermitian with trace one; that is, the completeness relation E\u2020 i Ei = I, (8.153) i is satis\ufb01ed, giving d2 real constraints. We will show how to determine \u03c7 experimentally, and then show how an operator-sum representation of the form (8.150) can be recovered once the \u03c7 matrix is known. Let \u03c1j, 1 \u2264 j \u2264 d2 be a \ufb01xed, linearly independent basis for the space of d \u00d7 d matrices; that is, any d \u00d7 d matrix can be written as a unique linear combination of the \u03c1j. A convenient choice is the set of operators |nm|. Experimentally, the output state E(|nm|) may be obtained by preparing the input states |n, |m, |+ = (|n+|m)/ 2, 2 and forming linear combinations of E(|nn|), E(|mm|), and |\u2212 = (|n + i|m)/ E(|++|), and E(|\u2212\u2212|), as follows: \u221a \u221a E(|nm|) = E(|++|) + iE(|\u2212\u2212|) \u2212 1 + i 2 E(|nn|) \u2212 1 + i 2 E(|mm|). (8.154) Thus, it is possible to determine E(\u03c1j) by state tomography, for each \u03c1j. Furthermore, each E(\u03c1j) may be expressed as a linear combination of the basis states, E(\u03c1j) = \u03bbjk\u03c1k, (8.155) k and since E(\u03c1j) is known from the state tomography, \u03bbjk can be determined by standard linear algebraic algorithms. To proceed, we may write \u02dcEm\u03c1j \u02dcE\u2020 n = \u03b2mn jk \u03c1k, (8.156) k where \u03b2mn jk are complex numbers which can be determined by standard algorithms from linear algebra given the \u02dcEm operators and the \u03c1j operators. Combining the last two expressions and (8.152) we have k mn \u03c7mn\u03b2mn jk \u03c1k = \u03bbjk\u03c1k. k From the linear independence of the \u03c1k it follows that for each k, \u03b2mn jk \u03c7mn = \u03bbjk. mn This relation is a necessary and suf\ufb01cient condition for the matrix \u03c7 to give the correct quantum operation E. One may think of \u03c7 and \u03bb as vectors, and \u03b2 as a d4 \u00d7 d4 matrix (8.157) (8.158) 392 Quantum noise and quantum operations st,xy with columns indexed by mn, and rows by jk. To show how\u03c7 may be obtained, let \u03ba be the generalized inverse for the matrix \u03b2, satisfying the relation \u03b2mn jk = jk\u03baxy \u03b2st st \u03b2mn xy. (8.159) Most computer packages for matrix manipulation are capable of \ufb01nding such generalized inverses. We now prove that \u03c7 de\ufb01ned by \u03c7mn \u2261 \u03bamn jk \u03bbjk (8.160) satis\ufb01es the relation (8.158). jk The dif\ufb01culty in verifying that \u03c7 de\ufb01ned by (8.160) satis\ufb01es (8.158) is that, in general, \u03c7 is not uniquely determined by Equation (8.158). For convenience we rewrite these equations in matrix form as \u03b2\u03c7 = \u03bb \u03c7 \u2261 \u03ba\u03bb. (8.161) (8.162) From the construction that led to Equation (8.152) we know there exists at least one solution to Equation (8.161), which we shall call \u03c7. Thus \u03bb = \u03b2\u03c7. The generalized inverse satis\ufb01es \u03b2\u03ba\u03b2 = \u03b2. Premultiplying the de\ufb01nition of \u03c7 by \u03b2 gives \u03b2\u03c7 = \u03b2\u03ba\u03bb = \u03b2\u03ba\u03b2\u03c7 = \u03b2\u03c7 = \u03bb. (8.163) (8.164) (8.165) (8.166) Thus \u03c7 de\ufb01ned by (8.162) satis\ufb01es the Equation (8.161), as we wanted to show. Having determined \u03c7 one immediately obtains the operator", "-sum representation for E in the following manner. Let the unitary matrix U \u2020 diagonalize \u03c7, \u03c7mn = xy Umxdx\u03b4xyU \u2217 ny. From this it can easily be veri\ufb01ed that $ Ei = di UjiEj (8.167) (8.168) j are operation elements for E. Our algorithm may thus be summarized as follows: \u03bb is experimentally determined using state tomography, which in turn determines \u03c7 via the equation \u03c7 = \u03ba\u03bb, which gives us a complete description of E, including a set of operation elements Ei. In the case of a single qubit quantum process, only 12 parameters must be determined (Box 8.5). The dynamics of a two qubit quantum black box E2 pose an even greater challenge for our understanding. In this case there are 240 parameters which need to be determined in order to completely specify the quantum operation acting on the quantum system! Determining these would obviously be quite a considerable undertaking. However, as for the single qubit case, it is relatively straightforward to implement a numerical routine which will automate the calculation, provided experimental state tomography and state preparation procedures are available in the laboratory. Applications of quantum operations 393 Box 8.5: Process tomography for a single qubit The general method of process tomography can be simpli\ufb01ed in the case of a one qubit operation to provide explicit formulas which may be useful in experimental contexts. This simpli\ufb01cation is made possible by choosing the \ufb01xed operators \u02dcEi to have commutation properties which conveniently allow the \u03c7 matrix to be determined by straightforward matrix multiplication. In the one qubit case, we use: E0 = I \u02dcE1 = X \u02dcE2 = \u2212iY \u02dcE3 = Z. (8.169) (8.170) (8.171) (8.172) There are 12 parameters, speci\ufb01ed by \u03c7, which determine an arbitrary single qubit quantum operation E. These parameters may be measured using four sets of experiments. As a speci\ufb01c example, suppose the input states |0, |1, |+ = (|0 + |1)/ 2 and |\u2212 = (|0 + i |1)/ 2 are prepared, and the four matrices \u221a \u221a \u03c1 1 = E(|00|) \u03c1 4 = E(|11|) \u03c1 2 = E(|++|) \u2212 iE(|\u2212\u2212|) \u2212 (1 \u2212 i)(\u03c1 3 = E(|++|) + iE(|\u2212\u2212|) \u2212 (1 + i)(\u03c1 \u03c1 1 + \u03c1 1 + \u03c1 are determined using state tomography. These correspond to \u03c1 (8.173) (8.174) (8.175) (8.176) 4)/2 4)/2 j = E(\u03c1j), where (8.177), 1 0 0 0 \u03c11 = \u03c12 = \u03c11X, \u03c13 = X\u03c11, and \u03c14 = X\u03c11X. From (8.156) and Equations (8.169)\u2013(8.172) we may determine \u03b2, and similarly \u03c1 j determines \u03bb. However, due to the particular choice of basis, and the Pauli matrix representation of \u02dcEi, we may express the \u03b2 matrix as the Kronecker product \u03b2 = \u039b \u2297 \u039b, where \u039b = 1 2 I X X \u22128.178) (8.179) so that \u03c7 may be expressed conveniently as \u03c7 = \u039b in terms of block matrices. We have shown how a useful representation for the dynamics of a quantum system may be experimentally determined using a systematic procedure. This procedure of quantum process tomography is analogous to the system identi\ufb01cation step performed in classical control theory, and plays a similar role in understanding and controlling noisy quantum systems. 394 Quantum noise and quantum operations Exercise 8.32: Explain how to extend quantum process tomography to the case of non-trace-preserving quantum operations, such as arise in the study of measurement. Exercise 8.33: (Specifying a quantum process) Suppose that one wished to completely specify an arbitrary single qubit operation E by describing how a set of points on the Bloch sphere {rk} transform under E. Prove that the set must contain at least four points. Exercise 8.34: (Process tomography for two qubits) Show that the \u03c72 describing the black box operations on two qubits can be expressed as \u03c72 = \u039b2\u03c1\u039b2, (8.180) where \u039b2 = \u039b \u2297 \u039b, \u039b is as de\ufb01ned in Box", " 8.5, and \u03c1 is a block matrix of 16 measured density matrices, \u23a1 \u23a4 14 \u23a2 \u23a2 \u23a3 \u03c1 = P T 11 \u03c1 \u03c1 21 \u03c1 \u03c1 \u03c1 31 \u03c1 41 \u03c1 \u03c1 nm = E(\u03c1nm), \u03c1nm = Tn|0000|Tm, T1 = I \u2297 I, T2 = I \u2297 X, 12 \u03c1 22 \u03c1 32 \u03c1 42 \u03c1 13 \u03c1 23 \u03c1 33 \u03c1 43 \u03c1 where \u03c1 T3 = X \u2297 I, T4 = X \u2297 X, and P = I \u2297 [(\u03c100 + \u03c112 + \u03c121 + \u03c133) \u2297 I] is a permutation matrix. \u23a5 \u23a5 \u23a6 P, 44 24 34 (8.181) Exercise 8.35: (Process tomography example) Consider a one qubit black box of unknown dynamics E1. Suppose that the following four density matrices are obtained from experimental measurements, performed according to Equations (8.173)\u2013(8.176): 8.182) (8.183) (8.184) (8.185) where \u03b3 is a numerical parameter. From an independent study of each of these input\u2013output relations, one could make several important observations: the ground state |0 is left invariant by E1, the excited state |1 partially decays to the ground state, and superposition states are damped. Determine the \u03c7 matrix for this process. 8.5 Limitations of the quantum operations formalism Are there interesting quantum systems whose dynamics are not described by quantum operations? In this section we will construct an arti\ufb01cial example of a system whose evo- Chapter problems 395 lution is not described by a quantum operation, and try to understand the circumstances under which this is likely to occur. Suppose a single qubit is prepared in some unknown quantum state, which we denote \u03c1. The preparation of this qubit involves certain procedures to be carried out in the laboratory in which the qubit is prepared. Suppose that among the laboratory degrees of freedom is a single qubit which, as a side effect of the state preparation procedure, is left in the state |0 if \u03c1 is a state on the bottom half of the Bloch sphere, and is left in the state |1 if \u03c1 is a state on the top half of the Bloch sphere. That is, the state of the system after preparation is \u03c1 \u2297 |00| \u2297 other degrees of freedom if \u03c1 is a state on the bottom half of the Bloch sphere, and \u03c1 \u2297 |11| \u2297 other degrees of freedom (8.186) (8.187) if \u03c1 is a state on the top half of the Bloch sphere. Once the state preparation is done, the system begins to interact with the environment, in this case all the laboratory degrees of freedom. Suppose the interaction is such that a controlledis performed between the principal system and the extra qubit in the laboratory system. Thus, if the system\u2019s Bloch vector was initially in the bottom half of the Bloch sphere it is left invariant by the process, while if it was initially in the top half of the Bloch sphere it is rotated into the bottom half of the Bloch sphere. Obviously, this process is not an af\ufb01ne map acting on the Bloch sphere, and therefore, by the results of Section 8.3.2, it cannot be a quantum operation. The lesson to be learned from this discussion is that a quantum system which interacts with the degrees of freedom used to prepare that system after the preparation is complete will in general suffer a dynamics which is not adequately described within the quantum operations formalism. This is an important conclusion to have reached, as it indicates that there are physically reasonable circumstances under which the quantum operations formalism may not adequately describe the processes taking place in a quantum system. This should be kept in mind, for example, in applications of the quantum process tomography procedure discussed in the previous section. For the remainder of this book we will, however, work within the quantum operations formalism. It provides a powerful, and reasonably general tool for describing the dynamics experienced by quantum systems. Most of all, it provides a means by which concrete progress can be made on problems related to quantum information processing. It is an interesting problem for further research to study quantum information processing beyond the quantum operations formalism. Problem 8.1: (Lindblad form to quantum operation) In the notation of Section 8.4.1, explicitly work through the steps to solve the differential equation \u02d9\u03c1 = \u2212 \u03bb 2 (\u03c3+\u03c3\u2212\u03c1 + \u03c1\u03c3+\u03c3\u2212 \u2212 2\u03c3\u2212\u03c1\u03c3+) (8.188) for \u03c1", "(t). Express the map \u03c1(0) \u2192 \u03c1(t) as \u03c1(t) = k Ek(t)\u03c1(0)E\u2020 k(t). Problem 8.2: (Teleportation as a quantum operation) Suppose Alice is in possession of a single qubit, denoted as system 1, which she wishes to teleport to 396 Quantum noise and quantum operations Bob. Unfortunately, she and Bob only share an imperfectly entangled pair of qubits. Alice\u2019s half of this pair is denoted system 2, and Bob\u2019s half is denoted system 3. Suppose Alice performs a measurement described by a set of quantum operations Em with result m on systems 1 and 2. Show that this induces an operation \u02dcEm relating the initial state of system 1 to the \ufb01nal state of system 3, and that teleportation is accomplished if Bob can reverse this operation using a trace-preserving quantum operation Rm, to obtain \u02dcEm(\u03c1) tr[ \u02dcEm(\u03c1)] Rm = \u03c1, (8.189) where \u03c1 is the initial state of system 1. Problem 8.3: (Random unitary channels) It is tempting to believe that all unital channels, that is, those for which E(I) = I, result from averaging over random unitary operations, that is, E(\u03c1) = k, where Uk are unitary operators and the pk form a probability distribution. Show that while this is true for single qubits, it is untrue for larger systems. k pkUk\u03c1U \u2020 History and further reading 397 Summary of Chapter 8: Quantum noise and quantum operations \u2022 The operator-sum representation: The behavior of an open quantum system can be modeled as E(\u03c1) = k where Ek are operation elements, satisfying ation is trace-preserving. Ek\u03c1E\u2020 k, (8.190) k E\u2020 kEk = I if the quantum oper- \u2022 Environmental models for quantum operations: A trace-preserving quantum operation can always be regarded as arising from the unitary interaction of a system with an initially uncorrelated environment, and vice versa. Non-tracepreserving quantum operations may be treated similarly, except an additional projective measurement is performed on the composite of system and environment, with the different outcomes corresponding to different non-trace-preserving quantum operations. \u2022 Quantum process tomography: A quantum operation on a d-dimensional quantum system can be completely determined by experimentally measuring the output density matrices produced from d2 pure state inputs. \u2022 Operation elements for important single qubit quantum operations: 1 \u2212 3p 4 depolarizing channel i i 0, amplitude damping phase damping phase \ufb02ip bit \ufb02ip bit\u2013phase \ufb02ip 1 0 1 1 0 0 \u2212i i 0 History and further reading Quantum noise is an important topic in several \ufb01elds, and there is an enormous literature on the subject. We will necessarily be restricted to citing only a small sample of 398 Quantum noise and quantum operations the resources available on the topic. An early treatise on quantum noise from a rather mathematical perspective is due to Davies[Dav76]. Caldeira and Leggett[CL83] did some of the \ufb01rst and most complete studies of an important model known as the spin-boson model, using an approach based upon the Feynman path integral. Gardiner[Gar91] studied quantum noise from the perspective of quantum optics. More recently, the quantum optics community has developed what is known as the quantum trajectories approach to quantum noise. Reviews of this subject may be found in the articles by Zoller and Gardiner[ZG97], and Plenio and Knight[PK98]. A large literature exists on the subject of quantum operations. We mention just a few key references, primarily the book by Kraus[Kra83], which contains references to much earlier work on the subject. In\ufb02uential early papers on the subject include those by Hellwig and Kraus[HK69, HK70], and by Choi[Cho75]. Lindblad[Lin76] connected the quantum operations formalism to the theory of continuous time quantum evolution, introducing what is now known as the Lindblad form. Schumacher[Sch96b] and Caves[Cav99] have written excellent summaries of the quantum operations formalism from the point of view of quantum error-correction. Quantum state tomography was suggested by Vogel and Risken[VR89]. Leonhardt[Leo97] has written a recent review containing references to other work. The need for quantum process tomography was pointed out in a paper by Turchette, Hood, Lange, Mabuchi, and Kimble[THL+95]. The theory was developed independently by Chuang and Nielsen[CN97], and by Poyatos, Cirac and Z", "oller[PCZ97]. Jones[Jon94] had earlier sketched out the main ideas of quantum process tomography. An unfortunate confusion of terms has arisen with the word \u2018decoherence\u2019. Historically, it has been used to refer just to a phase damping process, particularly by Zurek[Zur91]. Zurek and other researchers recognized that phase damping has a unique role in the transition from quantum to classical physics; for certain environmental couplings, it occurs on a time scale which is much faster than any amplitude damping process, and can therefore be much more important in determining the loss of quantum coherence. The major point of these studies has been this emergence of classicality due to environmental interactions. However, by and large, the usage of decoherence in quantum computation and quantum information is to refer to any noise process in quantum processing. In this book, we prefer the more generic term \u2018quantum noise\u2019 and tend towards its usage, although occasionally decoherence \ufb01nds a proper place in the context. A more detailed discussion of some of the limitations of the quantum operations formalism (and in particular, the assumption of a system and environment initially in a product state) is provided by Royer[Roy96]. Problem 8.2 is due to Nielsen and Caves[NC97]. Problem 8.3 is due to Landau and Streater[LS93] as part of an in-depth study of the extremal points of the convex set of doubly stochastic quantum operations. 9 Distance measures for quantum information What does it mean to say that two items of information are similar? What does it mean to say that information is preserved by some process? These questions are central to a theory of quantum information processing, and the purpose of this chapter is the development of distance measures giving quantitative answers to these questions. Motivated by our two questions we will be concerned with two broad classes of distance measures, static measures and dynamic measures. Static measures quantify how close two quantum states are, while dynamic measures quantify how well information has been preserved during a dynamic process. The strategy we take is to begin by developing good static measures of distance, and then to use those static measures as the basis for the development of dynamic measures of distance. There is a certain arbitrariness in the way distance measures are de\ufb01ned, both classically and quantum mechanically, and the community of people studying quantum computation and quantum information has found it convenient to use a variety of distance measures over the years. Two of those measures, the trace distance and the \ufb01delity, have particularly wide currency today, and we discuss both these measures in detail in this chapter. For the most part the properties of both are quite similar, however for certain applications one may be easier to deal with than the other. It is for this reason and because both are widely used within the quantum computation and quantum information community that we discuss both measures. 9.1 Distance measures for classical information The idea of distinguishing probability distributions is slippery business. \u2013 Christopher Fuchs Let\u2019s start out in an arena where we can easily apply our intuition \u2013 distance measures for classical information. What are the objects to be compared in classical information theory? We might consider comparing strings of bits like 00010 and 10011. One way of quantifying the distance between these is the Hamming distance, de\ufb01ned to be the number of places at which two bit strings are not equal. For example, the bit strings 00010 and 10011 differ in the \ufb01rst and last place, so the Hamming distance between them is two. Unfortunately, the Hamming distance between two objects is simply a matter of labeling, and a priori there aren\u2019t any labels in the Hilbert space arena of quantum mechanics! A much better place to launch the study of distance measures for quantum information is with the comparison of classical probability distributions. In fact, in classical information theory an information source is usually modeled as a random variable, that is, as a probability distribution over some source alphabet. For example, an unknown source of English text may be modeled as a sequence of random variables over the Roman alphabet. 400 Distance measures for quantum information Before the text is read we can make a fair guess at the relative frequency of the letters that appear in the text, and certain correlations among them, such as the fact that occurrences of the pair of letters \u2018th\u2019 are much more common than the pair \u2018zx\u2019 in English text. This characterization of information sources as probability distributions over some alphabet encourages us to concentrate on the comparison of probability distributions in our search for measures of distance. What does it mean to say that two probability distributions {px} and {qx} over the same index set, x, are similar to one another? It is dif\ufb01cult to give an answer to this question which is obviously the unique \u2018correct\u2019 answer, so instead we propose two different answers, each of", " which is widely used by the quantum computation and quantum information community. The \ufb01rst measure is the trace distance, de\ufb01ned by the equation: D(px, qx) \u2261 1 2 x |px \u2212 qx|. (9.1) This quantity is sometimes known as the L1 distance or Kolmogorov distance. We prefer the term trace distance because it anticipates the later quantum mechanical analogue of this quantity, which is de\ufb01ned using the trace function. The trace distance turns out to be a metric on probability distributions, (a metric D(x, y) must be symmetric, D(x, y) = D(y, x), and satisfy the triangle inequality, D(x, z) \u2264 D(x, y) + D(y, z)) so the use of the term \u2018distance\u2019 is justi\ufb01ed. Exercise 9.1: What is the trace distance between the probability distribution (1, 0) and the probability distribution (1/2, 1/2)? Between (1/2, 1/3, 1/6) and (3/4, 1/8, 1/8)? Exercise 9.2: Show that the trace distance between probability distributions (p, 1 \u2212 p) and (q, 1 \u2212 q) is |p \u2212 q|. A second measure of distance between probability distributions, the \ufb01delity of the probability distributions {px} and {qx}, is de\ufb01ned by F (px, qx) \u2261 \u221a pxqx. x (9.2) The \ufb01delity is a very different way of measuring distance between probability distributions than is the trace distance. To begin with, it is not a metric, although later we discuss a metric derived from the \ufb01delity. To see that the \ufb01delity is not a metric note that when the distributions {px} and {qx} are identical, F (px, qx) = x px = 1. A better geometric understanding of the \ufb01delity is illustrated in Figure 9.1; the \ufb01delity is just the inner \u221a \u221a product between vectors with components qx, which lie on a unit sphere. px and Exercise 9.3: What is the \ufb01delity of the probability distributions (1, 0) and (1/2, 1/2)? Of (1/2, 1/3, 1/6) and (3/4, 1/8, 1/8)? The trace distance and \ufb01delity are mathematically useful means of de\ufb01ning the notion of a distance between two probability distributions. Do these measures have physically motivated operational meanings? In the case of the trace distance, the answer to this Distance measures for classical information 401 Figure 9.1. Geometric interpretation of the \ufb01delity as the inner product between vectors x( unit sphere. (Since 1 = px)2 = qx)2.) x( \u221a \u221a \u221a px and \u221a qx lying on a question is yes. In particular, it is simple to prove that ) ) ) ) ) x\u2208S |p(S) \u2212 q(S)| = max S D(px, qx) = max S px \u2212 ) ) ) ) ), qx x\u2208S (9.3) where the maximization is over all subsets S of the index set {x}. The quantity being maximized is the difference between the probability that the event S occurs, according to the distribution {px}, and the probability that the event S occurs, according to the distribution {qx}. The event S is thus in some sense the optimal event to examine when trying to distinguish the distributions {px} and {qx}, with the trace distance governing how well it is possible to make this distinction. Unfortunately, a similarly clear interpretation for the \ufb01delity is not known. However, in the next section we show that the \ufb01delity is a suf\ufb01ciently useful quantity for mathematical purposes to justify its study, even without a clear physical interpretation. Moreover, we cannot rule out the possibility that a clear interpretation of the \ufb01delity will be discovered in the future. Finally, it turns out that there are close connections between the \ufb01delity and the trace distance, so properties of one quantity can often be used to deduce properties of the other, a fact which is useful surprisingly often. Exercise 9.4: Prove (9.3). Exercise 9.5: Show that the absolute value signs may be removed from Equation (9.3), that is, D(px, qx) = max S (p(S) \u2212 q(S)) = max S qx. px \u2212 x\u2208", "S x\u2208S (9.4) The trace distance and \ufb01delity are static measures of distance for comparing two \ufb01xed probability distributions. There is a third notion of distance which is a dynamic measure of distance in the sense that it measures how well information is preserved by 402 Distance measures for quantum information some physical process. Suppose a random variable X is sent through a noisy channel, giving as output another random variable Y, to form a Markov process X \u2192 Y. For convenience we assume both X and Y have the same range of values, denoted by x. Then the probability that Y is not equal to X, p(X = Y ), is an obvious and important measure of the degree to which information has been preserved by the process. Surprisingly, this dynamic measure of distance can be understood as a special case of the static trace distance! Imagine that the random variable X is given to you, and you make a copy of X, creating a new random variable \u02dcX = X. The random variable X now passes through the noisy channel, leaving as output the random variable Y, as illustrated in Figure 9.2. How close is the initial perfectly correlated pair, ( \u02dcX, X), to the \ufb01nal pair, ( \u02dcX, Y )? Using the trace distance as our measure of \u2018closeness\u2019, we obtain with some simple algebra, 1 2 1 2 1 2 |\u03b4xxp(X = x) \u2212 p( \u02dcX = x, Y = x)| (9.5) xx x= x x= x p( \u02dcX = x, Y = x) + p( \u02dcX = x, Y = x) + ) ) ) )p(X = x) \u2212 p( \u02dcX = x, Y = x) (9.6) p(X = x) \u2212 p( \u02dcX = x, Y = x) 1 2 1 2 x x D(( \u02dcX, X), (X, Y )) = = = = p( \u02dcX =Y ) + 1 \u2212 p( \u02dcX = Y ) 2 p(X =Y ) + p( \u02dcX =Y ) 2 = = p(X =Y ). (9.7) (9.8) (9.9) (9.10) Thus, as illustrated in Figure 9.3, the probability of an error in the channel is equal to the trace distance between the probability distribution for ( \u02dcX, X) and ( \u02dcX, Y ). This is an important construction, since it will be the basis for analogous quantum constructions. This is necessary because there is no direct quantum analogue of the probability p(X = Y ), since there is no notion in quantum mechanics analogous to the joint probability distribution for variables X and Y that exist at different times. Instead, to de\ufb01ne dynamic measures of quantum distance we use an approach similar to the construction just given, based on the idea that it is quantum entanglement, rather than classical correlation, which is the important thing to preserve during a quantum channel\u2019s dynamics. Figure 9.2. Given a Markov process X \u2192 Y we may \ufb01rst make a copy of X, which turns it into Y. \u02dcX, before subjecting X to the noise How close are two quantum states? 403 Figure 9.3. The probability of an error in the channel is equal to the trace distance between the probability distributions for ( \u02dcX, X) and ( \u02dcX, Y ). 9.2 How close are two quantum states? How close are two quantum states? Over the next few sections we describe quantum generalizations of the classical notions of trace distance and \ufb01delity, and discuss in detail the properties of these quantities. 9.2.1 Trace distance We begin by de\ufb01ning the trace distance between quantum states \u03c1 and \u03c3, D(\u03c1, \u03c3) \u2261 1 2 \u221a tr|\u03c1 \u2212 \u03c3|. A\u2020A to be the positive square root of A\u2020A. Notice where as per usual we de\ufb01ne |A| \u2261 that the quantum trace distance generalizes the classical trace distance in the sense that if \u03c1 and \u03c3 commute then the (quantum) trace distance between \u03c1 and \u03c3 is equal to the classical trace distance between the eigenvalues of \u03c1 and \u03c3. More explicitly, if \u03c1 and \u03c3 commute they are diagonal in the same basis, (9.11) \u03c1 = i ri|ii|; \u03c3 = si|ii|, i for some orthonormal basis |i. Thus D(\u03c1, \u03c3) = ) ) ) ) ) 1 2 tr ) ) ) (ri \u2212 si)|ii| ) ) i = D(ri, si). Exercise 9.6: What is the trace", " distance between the density operators |00| + 3 4 |11|; 1 4 2 3 |00| + Between: 1 |11|; 4 \u221a (Recall that |\u00b1 \u2261 (|0 \u00b1 |1)/ |00| + 3 4 2.) |++| + 2 3 |11|? |\u2212\u2212|? 1 3 1 3 (9.12) (9.13) (9.14) (9.15) (9.16) 404 Distance measures for quantum information A good way of getting a feel for the trace distance is to understand it for the special case of a qubit, in the Bloch sphere representation. Suppose \u03c1 and \u03c3 have respective Bloch vectors r and s9.17) (Recall that \u03c3 denotes a vector of Pauli matrices; it should not be confused with the state \u03c3.) The trace distance between \u03c1 and \u03c3 is easily calculated: D(\u03c1, \u03c3) = = 1 2 1 4 tr|\u03c1 \u2212 \u03c3| tr |(r \u2212 s) \u00b7 \u03c3|. (9.18) (9.19) (r \u2212 s) \u00b7 \u03c3 has eigenvalues \u00b1|r \u2212 s|, so the trace of |(r \u2212 s) \u00b7 \u03c3| is 2|r \u2212 s|, and we see that D(\u03c1, \u03c3) = |r \u2212 s| 2. (9.20) That is, the distance between two single qubit states is equal to one half the ordinary Euclidean distance between them on the Bloch sphere! This intuitive geometric picture of the trace distance for qubits is often useful when trying to understand general properties of the trace distance. Conjectured properties can be suggested, refuted, or gain plausibility by looking at simple examples on the Bloch sphere. For example, rotations of the Bloch sphere leave the Euclidean distance invariant. This suggests that the trace distance might be preserved under unitary transformations in general, D(U \u03c1U \u2020, U \u03c3U \u2020) = D(\u03c1, \u03c3), (9.21) a conjecture which you can easily verify with a moment\u2019s thought. We will come back to the Bloch sphere picture often in our investigation of distance measures. To understand the properties of the trace distance a good starting point is to prove a formula for the trace distance generalizing Equation (9.3) for the classical trace distance: D(\u03c1, \u03c3) = max P tr(P (\u03c1 \u2212 \u03c3)), (9.22) where the maximization may be taken alternately over all projectors, P, or over all positive operators P \u2264 I; the formula is valid in either case. This formula gives rise to an appealing interpretation of the trace distance. Recalling that POVM elements are positive operators P \u2264 I, the trace distance is therefore equal to the difference in probabilities that a measurement outcome with POVM element P may occur, depending on whether the state is \u03c1 or \u03c3, maximized over all possible POVM elements P. We prove Equation (9.22) for the case where the maximization is over projectors; the case of positive operators P \u2264 I follows the same reasoning. The proof is based on the fact that \u03c1 \u2212 \u03c3 can be expressed as \u03c1 \u2212 \u03c3 = Q \u2212 S, where Q and S are positive operators with orthogonal support (see Exercise 9.7). It implies that |\u03c1 \u2212 \u03c3| = Q + S, so D(\u03c1, \u03c3) = (tr(Q) + tr(S))/2. But tr(Q \u2212 S) = tr(\u03c1 \u2212 \u03c3) = 0, so tr(Q) = tr(S), and therefore D(\u03c1, \u03c3) = tr(Q). Let P be the projector onto the support of Q. Then tr(P (\u03c1 \u2212 \u03c3)) = tr(P (Q \u2212 S)) = tr(Q) = D(\u03c1, \u03c3). Conversely, let P be any projector. How close are two quantum states? 405 Then tr(P (\u03c1 \u2212 \u03c3)) = tr(P (Q \u2212 S)) \u2264 tr(P Q) \u2264 tr(Q) = D(\u03c1, \u03c3). This completes the proof. Exercise 9.7: Show that for any states \u03c1 and \u03c3, one may write \u03c1 \u2212 \u03c3 = Q \u2212 S, where Q and S are positive operators with support on orthogonal vector spaces. (Hint: use the spectral decomposition \u03c1 \u2212 \u03c3 = U DU \u2020, and split the diagonal matrix D into positive and negative parts. This fact will continue to be useful later.) There is a closely related way of viewing the quantum trace distance which relates it more closely to the classical trace distance: Theorem 9.1: Let {Em} be a", " POVM, with pm \u2261 tr(\u03c1Em) and qm \u2261 tr(\u03c3Em) as the probabilities of obtaining a measurement outcome labeled by m. Then D(\u03c1, \u03c3) = max {Em} D(pm, qm), (9.23) where the maximization is over all POVMs {Em}. Proof Note that D(pm, qm) = |tr(Em(\u03c1 \u2212 \u03c3))|. (9.24) 1 2 m Using the spectral decomposition we may write \u03c1 \u2212 \u03c3 = Q \u2212 S, where Q and S are positive operators with orthogonal support. Thus |\u03c1 \u2212 \u03c3| = Q + S, and Thus |tr(Em(\u03c1 \u2212 \u03c3))| = |tr(Em(Q \u2212 S))| \u2264 tr(Em(Q + S)) \u2264 tr(Em|\u03c1 \u2212 \u03c3|). D(pm, qm) \u2264 1 2 m tr(Em|\u03c1 \u2212 \u03c3|) = 1 2 tr(|\u03c1 \u2212 \u03c3|) = D(\u03c1, \u03c3), (9.25) (9.26) (9.27) (9.28) (9.29) (9.30) where we have applied the completeness relation for POVM elements, m Em = I. Conversely, by choosing a measurement whose POVM elements include projectors onto the support of Q and S, we see that there exist measurements which give rise to probability distributions such that D(pm, qm) = D(\u03c1, \u03c3). Thus, if two density operators are close in trace distance, then any measurement performed on those quantum states will give rise to probability distributions which are close together in the classical sense of trace distance, giving a second interpretation of the trace distance between two quantum states as an achievable upper bound on the trace distance between probability distributions arising from measurements performed on those quantum states. 406 Distance measures for quantum information We call the trace distance a \u2018distance\u2019, so we should check to see whether it has the property of being a metric on the space of density operators. From our geometric picture for a single qubit this is obviously true for a single qubit; is it true more generally? It is clear that D(\u03c1, \u03c3) = 0 if and only if \u03c1 = \u03c3, and that D(\u00b7, \u00b7) is a symmetric function of its inputs. All that remains to check is that the triangle inequality holds, D(\u03c1, \u03c4 ) \u2264 D(\u03c1, \u03c3) + D(\u03c3, \u03c4 ). To see this, note from Equation (9.22) that there exists a projector P such that D(\u03c1, \u03c4 ) = tr(P (\u03c1 \u2212 \u03c4 )) = tr(P (\u03c1 \u2212 \u03c3)) + tr(P (\u03c3 \u2212 \u03c4 )) \u2264 D(\u03c1, \u03c3) + D(\u03c3, \u03c4 ), (9.31) (9.32) (9.33) (9.34) establishing that the trace distance is a metric. At this stage, we don\u2019t know a whole lot about the trace distance. However, we\u2019re in a good position to prove some genuinely spectacular results, useful in a wide variety of contexts. The most interesting result is that no physical process ever increases the distance between two quantum states, a result illustrated in Figure 9.4. We state this more formally as a theorem: Theorem 9.2: (Trace-preserving quantum operations are contractive) Suppose E is a trace-preserving quantum operation. Let \u03c1 and \u03c3 be density operators. Then D(E(\u03c1), E(\u03c3)) \u2264 D(\u03c1, \u03c3). (9.35) Figure 9.4. Trace-preserving quantum operations cause a contraction on the space of density operators. Proof Use the spectral decomposition to write \u03c1 \u2212 \u03c3 = Q \u2212 S, where Q and S are positive matrices with orthogonal support, and let P be a projector such that D(E(\u03c1), E(\u03c3)) = tr[P (E(\u03c1) \u2212 E(\u03c3))]. Observe that tr(Q) \u2212 tr(S) = tr(\u03c1) \u2212 tr(\u03c3) = 0, so tr(Q) = tr(S) and thus tr(E(Q)) = tr(E(S)). Using this observation we see that How close are two quantum states? 407 D(\u03c1, \u03c3) = = = tr|\u03c1 \u2212 \u03c3| tr|Q \u2212 S| tr(Q) + 1 2 tr(E(Q)) + 1 2 1 2 1 2 1 2 tr(S) tr(E(S)) 1 2 = = tr(E(Q)) \u2265 tr(P E(Q)) \u2265 tr(P (E(Q) \u2212 E(S))) = tr(P (E(", "\u03c1) \u2212 E(\u03c3))) = D(E(\u03c1), E(\u03c3)), (9.36) (9.37) (9.38) (9.39) (9.40) (9.41) (9.42) (9.43) (9.44) which completes the proof. There is an important special case of this result which can be understood by the following analogy. Imagine somebody shows you two different paintings in a gallery. Provided you have reasonably good vision, you shouldn\u2019t have any dif\ufb01culty telling them apart. On the other hand, if somebody covers up most of the two paintings then you might have more dif\ufb01culty telling the two apart, as illustrated in Figure 9.5. Similarly, if we \u2018cover up\u2019 parts of two quantum states then we can show that the distance between those two states is never increased. To prove this, recall from page 374 that the partial trace is a trace-preserving quantum operation. By Theorem 9.2, if we take quantum states \u03c1AB and \u03c3AB of a composite quantum system AB then the distance between \u03c1A = trB(\u03c1AB) and \u03c3A = trB(\u03c3AB) is never more than the distance between \u03c1AB and \u03c3AB, D(\u03c1A, \u03c3A) \u2264 D(\u03c1AB, \u03c3AB). (9.45) Figure 9.5. Objects become less distinguishable when only partial information is available. In many applications we want to estimate the trace distance for a mixture of inputs. Such estimates are greatly aided by the following theorem: Theorem 9.3: (Strong convexity of the trace distance) Let{ pi} and {qi} be probability distributions over the same index set, and \u03c1i and \u03c3i be density operators, also with indices from the same index set. Then D pi\u03c1i, i i qi\u03c3i \u2264 D(pi, qi) + i piD(\u03c1i, \u03c3i), (9.46) 408 Distance measures for quantum information where D(pi, qi) is the classical trace distance between the probability distributions {pi} and {qi}. This result can be used to prove convexity results for the trace distance so we refer to this property as the strong convexity property for trace distance. Proof By Equation (9.22) there exists a projector P such that D pi\u03c1i, qi\u03c3i i i pitr(P \u03c1i) \u2212 i qitr(P \u03c3i) pitr(P (\u03c1i \u2212 \u03c3i)) + (pi \u2212 qi)tr(P \u03c3i) i piD(\u03c1i, \u03c3i) + D(pi, qi), (9.47) (9.48) (9.49) = = \u2264 i i i where D(pi, qi) is the trace distance between the probability distributions {pi} and {qi}, and we used Equation (9.22) in the last line. As a special case of this result, we see that the trace distance is jointly convex in its inputs, D pi\u03c1i, pi\u03c3i i i \u2264 i piD(\u03c1i, \u03c3i). (9.50) Exercise 9.8: (Convexity of the trace distance) Show that the trace distance is convex in its \ufb01rst input, D i pi\u03c1i, \u03c3 \u2264 i piD(\u03c1i, \u03c3). (9.51) By symmetry convexity in the second entry follows from convexity in the \ufb01rst. Exercise 9.9: (Existence of \ufb01xed points) Schauder\u2019s \ufb01xed point theorem is a classic result from mathematics that implies that any continuous map on a convex, compact subset of a Hilbert space has a \ufb01xed point. Use Schauder\u2019s \ufb01xed point theorem to prove that any trace-preserving quantum operation E has a \ufb01xed point, that is, \u03c1 such that E(\u03c1) = \u03c1. Exercise 9.10: Suppose E is a strictly contractive trace-preserving quantum operation, that is, for any \u03c1 and \u03c3, D(E(\u03c1), E(\u03c3)) < D(\u03c1, \u03c3). Show that E has a unique \ufb01xed point. Exercise 9.11: Suppose E is a trace-preserving quantum operation for which there exists a density operator \u03c10 and a trace-preserving quantum operation E such that E(\u03c1) = p\u03c10 + (1 \u2212 p)E (\u03c1), for some p, 0 < p \u2264 1. Physically, this means that with probability p the input state is thrown", " out and replaced with the \ufb01xed state \u03c10, while with probability (9.52) How close are two quantum states? 409 1 \u2212 p the operation E occurs. Use joint convexity to show that E is a strictly contractive quantum operation, and thus has a unique \ufb01xed point. Exercise 9.12: Consider the depolarizing channel introduced in Section 8.3.4 on page 378, E(\u03c1) = pI/2 + (1 \u2212 p)\u03c1. For arbitrary \u03c1 and \u03c3 \ufb01nd D(E(\u03c1), E(\u03c3)) using the Bloch representation, and prove explicitly that the map E is strictly contractive, that is, D(E(\u03c1), E(\u03c3)) < D(\u03c1, \u03c3). Exercise 9.13: Show that the bit \ufb02ip channel (Section 8.3.3) is contractive but not strictly contractive. Find the set of \ufb01xed points for the bit \ufb02ip channel. 9.2.2 Fidelity A second measure of distance between quantum states is the \ufb01delity. The \ufb01delity is not a metric on density operators, but we will see that it does give rise to a useful metric. This section reviews the de\ufb01nition and basic properties of the \ufb01delity. The \ufb01delity of states \u03c1 and \u03c3 is de\ufb01ned to be ( F (\u03c1, \u03c3) \u2261 tr \u03c11/2\u03c3\u03c11/2. (9.53) It is certainly not immediately obvious that this is a useful measure of distance between \u03c1 and \u03c3. It doesn\u2019t even look symmetric! Yet we will see that the \ufb01delity is symmetric in its inputs, and has many of the other properties we expect of a good distance measure. There are two important special cases where it is possible to give more explicit formulae for the \ufb01delity. The \ufb01rst is when \u03c1 and \u03c3 commute, that is, are diagonal in the same basis, \u03c1 = i ri|ii|; \u03c3 = si|ii|, i for some orthonormal basis |i. In this case we see that F (\u03c1, \u03c3) = tr risi|ii| risi|ii| i \u221a = tr i \u221a risi = i = F (ri, si). (9.54) (9.55) (9.56) (9.57) (9.58) That is, when \u03c1 and \u03c3 commute the quantum \ufb01delity F (\u03c1, \u03c3) reduces to the classical \ufb01delity F (ri, si) between the eigenvalue distributions ri and si of \u03c1 and \u03c3. Our second example is to calculate the \ufb01delity between a pure state |\u03c8 and an arbitrary state, \u03c1. From Equation (9.53) we see that ( F (|\u03c8, \u03c1) = tr ( \u03c8|\u03c1|\u03c8 |\u03c8\u03c8| = \u03c8|\u03c1|\u03c8. (9.59) (9.60) That is, the \ufb01delity is equal to the square root of the overlap between |\u03c8 and \u03c1. This is an important result which we will make use of often. For the case of a qubit we were able to explicitly evaluate the trace distance between two 410 Distance measures for quantum information states, and give it a simple geometric interpretation as half the Euclidean distance between points on the Bloch sphere. Unfortunately, no similarly clear geometric interpretation is known for the \ufb01delity between two states of a qubit. However, the \ufb01delity does satisfy many of the same properties as the trace distance. For example, it is invariant under unitary transformations: F (U \u03c1U \u2020, U \u03c3U \u2020) = F (\u03c1, \u03c3). (9.61) Exercise 9.14: (Invariance of \ufb01delity under unitary transforms) Prove (9.61) by \u221a \u221a using the fact that for any positive operator A, U AU \u2020 = U AU \u2020. There is also a useful characterization of the \ufb01delity analogous to the characterization (9.22) for the trace distance. Theorem 9.4: (Uhlmann\u2019s theorem) Suppose \u03c1 and \u03c3 are states of a quantum system Q. Introduce a second quantum system R which is a copy of Q. Then F (\u03c1, \u03c3) = max |\u03c8,|\u03d5 |\u03c8|\u03d5|, (9.62) where the maximization is over all puri\ufffd", "\ufffdcations |\u03c8 of \u03c1 and |\u03d5 of \u03c3 into RQ. Before giving the proof of Uhlmann\u2019s theorem we need an easily proved lemma. Lemma 9.5: Let A be any operator, and U unitary. Then |tr(AU )| \u2264 tr|A|, (9.63) with equality being attained by choosing U = V \u2020, where A = |A|V is the polar decomposition of A. Proof Equality is clearly attained under the conditions stated. Observe that |tr(AU )| = |tr(|A|V U )| = ) ) ) ) )tr(|A|1/2|A|1/2V U ) ). The Cauchy\u2013Schwarz inequality for the Hilbert\u2013Schmidt inner product gives: ( |tr(AU )| \u2264 tr|A| tr U \u2020V \u2020|A|V U = tr|A|, which completes the proof. Proof (Uhlmann\u2019s theorem) (9.64) (9.65) Fix orthonormal bases |iR and |iQ in systems R and Q. Because R and Q are of the same dimension, the index i may be assumed to run over the same set of values. De\ufb01ne |iR|iQ. Let |\u03c8 be any puri\ufb01cation of \u03c1. Then the Schmidt decomposition |m \u2261 and a moment\u2019s thought should convince you that i |\u03c8 = UR \u2297 \u221a |m, \u03c1UQ (9.66) (9.68) (9.69) (9.70) (9.71) How close are two quantum states? 411 for some unitary operators UR and UQ on systems R and Q. Similarly, if |\u03d5 is any puri\ufb01cation of \u03c3 then there exist unitary operators VR and VQ such that VR \u2297 |\u03d5 = (9.67) |m. \u03c3VQ \u221a Taking the inner product gives ) ) )m| |\u03c8|\u03d5| = RVR \u2297 U \u2020 U \u2020 Q \u221a \u221a \u03c1 \u03c3VQ |m ) ) ). Using Exercise 9.16 on this page we see that |\u03c8|\u03d5| = ) ) )tr RURU \u2020 V \u2020 Q \u221a \u221a \u03c1 \u03c3VQ ) ) ). Setting U \u2261 VQV \u2020 RURU \u2020 Q we see that |\u03c8|\u03d5| = ) )tr \u221a \u221a \u03c1 \u03c3U ) ). By Lemma 9.5, |\u03c8|\u03d5| \u2264 tr ) )\u221a \u03c1 \u221a \u03c3 ( ) ) = tr \u221a \u03c11/2\u03c3\u03c11/2. \u221a \u221a \u221a To see that equality may be attained, suppose position of attained. \u03c3|V is the polar decom\u03c3. Choosing UQ = UR = VR = I and VQ = V \u2020 we see that equality is \u03c3 = | \u221a \u221a \u03c1 \u03c1 \u03c1 Exercise 9.15: Show that F (\u03c1, \u03c3) = max |\u03d5 |\u03c8|\u03d5|, (9.72) where |\u03c8 is any \ufb01xed puri\ufb01cation of \u03c1, and the maximization is over all puri\ufb01cations of \u03c3. Exercise 9.16: (The Hilbert\u2013Schmidt inner product and entanglement) Suppose R and Q are two quantum systems with the same Hilbert space. Let |iR and |iQ be orthonormal basis sets for R and Q. Let A be an operator on R and B an operator on Q. De\ufb01ne |m \u2261 |iR|iQ. Show that i tr(A\u2020B) = m|(A \u2297 B)|m, (9.73) where the multiplication on the left hand side is of matrices, and it is understood that the matrix elements of A are taken with respect to the basis |iR and those for B with respect to the basis |iQ. Uhlmann\u2019s formula (9.62) does not provide a calculational tool for evaluating the \ufb01delity, as does Equation (9.53). However, in many instances, properties of the \ufb01delity are more easily proved using Uhlmann\u2019s formula than Equation (9.53). For example, Uhlmann\u2019s formula makes it clear that the \ufb01delity is symmetric in its inputs, F (\u03c1, \u03c3) = F (\u03c3, \u03c1), and that the \ufb01del", "ity is bounded between 0 and 1, 0 \u2264 F (\u03c1, \u03c3) \u2264 1. If \u03c1 = \u03c3 then it is clear that F (\u03c1, \u03c3) = 1, from Uhlmann\u2019s formula. If \u03c1 = \u03c3 then |\u03c8 = |\u03d5 for any puri\ufb01cations |\u03c8 and |\u03d5 of \u03c1 and \u03c3, respectively, so F (\u03c1, \u03c3) < 1. On the other hand, Equation (9.53) is sometimes useful as a means for understanding properties of the \ufb01delity. For instance, we see that F (\u03c1, \u03c3) = 0 if and only if \u03c1 and \u03c3 have support on orthogonal subspaces. Intuitively, when \u03c1 and \u03c3 are supported on orthogonal subspaces 412 Distance measures for quantum information they are perfectly distinguishable, so we should expect the \ufb01delity to be minimized at this point. Summarizing, the \ufb01delity is symmetric in its inputs, 0 \u2264 F (\u03c1, \u03c3) \u2264 1, with equality in the \ufb01rst inequality if and only if \u03c1 and \u03c3 have orthogonal support, and equality in the second inequality if and only if \u03c1 = \u03c3. We saw that the quantum trace distance could be related to the classical trace distance by considering the probability distributions induced by a measurement. In a similar way, we can show that F (\u03c1, \u03c3) = min {Em} F (pm, qm), (9.74) where the minimum is over all POVMs {Em}, and pm \u2261 tr(\u03c1Em), qm \u2261 tr(\u03c3Em) are the probability distributions for \u03c1 and \u03c3 corresponding to the POVM {Em}. To see that $ this is true, apply the polar decomposition \u221a \u03c3U, and note that \u03c11/2\u03c3\u03c11/2 = \u221a \u221a \u03c1 \u221a F (\u03c1, \u03c3) = tr( \u03c1 = m \u03c3U ) $ \u221a $ tr( \u03c1 Em Em \u221a \u03c3U ). The Cauchy\u2013Schwarz inequality and some simple algebra gives $ F (\u03c1, \u03c3) \u2264 tr(\u03c1Em)tr(\u03c3Em) which establishes that m = F (pm, qm), F (\u03c1, \u03c3) \u2264 min {Em} F (pm, qm). To see that equality can be attained in this inequality, we need to \ufb01nd a POVM {Em} such that the Cauchy\u2013Schwarz inequality is satis\ufb01ed with equality for each term in the \u03c3U for some set of complex numbers \u03b1m. But sum, that is, \u221a \u03c1 = \u03b1m Em Em $ \u221a \u221a \u221a \u221a \u221a \u03c1 \u03c3U = \u03c11/2\u03c3\u03c11/2, so for invertible \u03c1, \u221a \u03c3U = \u03c1\u22121/2 ( \u03c11/2\u03c3\u03c11/2. Substituting, we \ufb01nd that the equality conditions are that $ Em (I \u2212 \u03b1mM ) = 0, $ where M \u2261 \u03c1\u22121/2 m \u03b2m|mm| is a spectral decomposition for M then we choose Em = |mm| and \u03b1m = 1/\u03b2m. The case of non-invertible \u03c1 follows from continuity. \u03c11/2\u03c3\u03c11/2\u03c1\u22121/2. If M = We proved three important properties of the trace distance \u2013 the metric property, contractivity, and strong convexity. Rather remarkably, analogous properties all hold for the \ufb01delity. What is more, the proof techniques used for \ufb01delity differ considerably from those used for the trace distance. For that reason it\u2019s worth looking at these results in some detail. The \ufb01delity is not a metric; however, there is a simple way of turning the \ufb01delity into a metric. The basic idea can be gleaned from Figure 9.6, by noticing that the angle between two points on the sphere is a metric. For the quantum case, Uhlmann\u2019s theorem tells us that the \ufb01delity between two states is equal to the maximum inner product between (9.75) (9.76) (9.77) (9.78) (9.79) (9.80) (9.81) How close are two quantum states? 413 puri\ufb01cations of those states. This suggests that we de\ufb01ne the angle between states \u03c1 and \u03c3 by A(\u03c1, \u03c3) \u2261 arccos F (\u03c1, \u03c3). (9.82)", " Obviously the angle is non-negative, symmetric in its inputs, and is equal to zero if and only if \u03c1 = \u03c3. If we can show that the angle obeys the triangle inequality then we will have established that the angle is a metric. We prove the triangle inequality using Uhlmann\u2019s theorem and some obvious facts about vectors in three dimensions. Let |\u03d5 be a puri\ufb01cation of \u03c3, and choose puri\ufb01cations |\u03c8 of \u03c1 and |\u03b3 of \u03c4 such that F (\u03c1, \u03c3) = \u03c8|\u03d5 F (\u03c3, \u03c4 ) = \u03d5|\u03b3, (9.83) (9.84) and \u03c8|\u03b3 is real and positive. (Note that this can always be done, by multiplying |\u03c8, |\u03d5 and |\u03b3 by appropriate phase factors, if necessary.) It is clear from Figure 9.6 that arccos(\u03c8|\u03b3) \u2264 A(\u03c1, \u03c3) + A(\u03c3, \u03c4 ). (9.85) But by Uhlmann\u2019s theorem, F (\u03c1, \u03c4 ) \u2265 \u03c8|\u03b3, so A(\u03c1, \u03c4 ) \u2264 arccos(\u03c8|\u03b3). Combining with the previous inequality gives the triangle inequality, A(\u03c1, \u03c4 ) \u2264 A(\u03c1, \u03c3) + A(\u03c3, \u03c4 ). (9.86) Figure 9.6. The angle between points on the surface of the unit sphere is a metric. Exercise 9.17: Show that 0 \u2264 A(\u03c1, \u03c3) \u2264 \u03c0/2, with equality in the \ufb01rst inequality if and only if \u03c1 = \u03c3. Qualitatively, the \ufb01delity behaves like an \u2018upside-down\u2019 version of the trace distance, decreasing as two states become more distinguishable, and increasing as they become less distinguishable. Therefore, we should not expect the contractivity or non-increasing property of the trace distance to hold for \ufb01delity. Instead, the analogous property of being non-decreasing does hold for \ufb01delity. We will refer to this as the monotonicity of the \ufb01delity under quantum operations. 414 Distance measures for quantum information Theorem 9.6: (Monotonicity of the \ufb01delity) Suppose E is a trace-preserving quantum operation. Let \u03c1 and \u03c3 be density operators. Show that F (E(\u03c1), E(\u03c3)) \u2265 F (\u03c1, \u03c3). (9.87) Proof Let |\u03c8 and |\u03d5 be puri\ufb01cations of \u03c1 and \u03c3 into a joint system RQ such that F (\u03c1, \u03c3) = |\u03c8|\u03d5|. Introduce a model environment E for the quantum operation, E, which starts in a pure state |0, and interacts with the quantum system Q via a unitary interaction U. Note that U |\u03c8|0 is a puri\ufb01cation of E(\u03c1), and U |\u03d5|0 is a puri\ufb01cation of E(\u03c3). By Uhlmann\u2019s theorem it follows that F (E(\u03c1), E(\u03c3)) \u2265 |\u03c8|0|U \u2020U |\u03d5|0| = |\u03c8|\u03d5| = F (\u03c1, \u03c3), (9.88) (9.89) (9.90) establishing the property we set out to prove. Exercise 9.18: (Contractivity of the angle) Let E be a trace-preserving quantum operation. Show that A(E(\u03c1), E(\u03c3)) \u2264 A(\u03c1, \u03c3). (9.91) We \ufb01nish off our study of the elementary properties of \ufb01delity by proving a result for the \ufb01delity analogous to the strong convexity of the trace distance, using Uhlmann\u2019s theorem. Theorem 9.7: (Strong concavity of the \ufb01delity) Let pi and qi be probability distributions over the same index set, and \u03c1i and \u03c3i density operators also indexed by the same index set. Then F pi\u03c1i, qi\u03c3i \u221a \u2265 piqiF (\u03c1i, \u03c3i). (9.92) i i i Not surprisingly, this result may be used to prove concavity results for the \ufb01delity, for which reason we dub it the strong concavity property for the \ufb01delity. This property is not strictly analogous to the strong convexity of the trace distance; however, the similarity in spirit leads us to use similar nomenclature. Proof Let |\u03c8i and |\u03d5i be puri\ufffd", "\ufffdcations of \u03c1i and \u03c3i chosen such that F (\u03c1i, \u03c3i) = \u03c8i|\u03d5i. Introduce an ancillary system which has orthonormal basis states |i corresponding to the index set i for the probability distributions. De\ufb01ne \u221a |\u03c8 \u2261 pi|\u03c8i|i; |\u03d5 \u2261 \u221a qi|\u03d5i|i. i i i pi\u03c1i and |\u03d5 is a puri\ufb01cation of (9.93) i qi\u03c3i, so by Note that |\u03c8 is a puri\ufb01cation of Uhlmann\u2019s formula, F pi\u03c1i, qi\u03c3i i i \u2265 |\u03c8|\u03d5| = \u221a piqi\u03c8i|\u03d5i = i \u221a i piqiF (\u03c1i, \u03c3i), (9.94) How close are two quantum states? 415 which establishes the result we set out to prove. Exercise 9.19: (Joint concavity of \ufb01delity) Prove that the \ufb01delity is jointly concave, F pi\u03c1i, pi\u03c3i i i \u2265 i piF (\u03c1i, \u03c3i). (9.95) Exercise 9.20: (Concavity of \ufb01delity) Prove that the \ufb01delity is concave in the \ufb01rst entry, F i pi\u03c1i, \u03c3 \u2265 i piF (\u03c1i, \u03c3). (9.96) By symmetry the \ufb01delity is also concave in the second entry. 9.2.3 Relationships between distance measures The trace distance and the \ufb01delity are closely related, despite their very different forms. Qualitatively, they may be considered to be equivalent measures of distance for many applications. In this section we quantify more precisely the relationship between trace distance and \ufb01delity. In the case of pure states, the trace distance and the \ufb01delity are completely equivalent to one another. To see this, consider the trace distance between two pure states, |a and |b. Using the Gram\u2013Schmidt procedure we may \ufb01nd orthonormal states |0 and |1 such that |a = |0 and |b = cos \u03b8|0 + sin \u03b8|1. Note that F (|a, |b) = | cos \u03b8|. Furthermore, D(|a, |b) = ) ) ) ) tr 1 2 = |sin \u03b8| ( 1 \u2212 cos2 \u03b8 \u2212 cos \u03b8 sin \u03b8 \u2212 cos \u03b8 sin \u03b8 \u2212 sin2 \u03b8 ) ) ) ) = 1 \u2212 F (|a, |b)2. (9.97) (9.98) (9.99) Thus the trace distance between two pure states is a function of the \ufb01delity of those states, and vice versa. This relationship at the level of pure states can be used to deduce a relationship at the level of mixed states. Let \u03c1 and \u03c3 be any two quantum states, and let |\u03c8 and |\u03d5 be puri\ufb01cations chosen such that F (\u03c1, \u03c3) = |\u03c8|\u03d5| = F (|\u03c8, |\u03d5). Recalling that trace distance is non-increasing under the partial trace, we see that D(\u03c1, \u03c3) \u2264 D(|\u03c8, |\u03d5) $ = 1 \u2212 F (\u03c1, \u03c3)2. (9.100) (9.101) Thus, if the \ufb01delity between two states is close to one, it follows that the states are also close in trace distance. The converse is also true. To see this, let {Em} be a POVM such that \u221a F (\u03c1, \u03c3) = pmqm, (9.102) m where pm \u2261 tr(\u03c1Em), qm \u2261 tr(\u03c3Em) are the probabilities for obtaining outcome m for the states \u03c1 and \u03c3, respectively. Observe \ufb01rst that \u221a ( qm \u2212 2F (\u03c1, \u03c3) qm)2 = pm \u2212 (9.103) pm + \u221a m m m 416 Distance measures for quantum information However, it is also true that \u221a ( pm \u2212 ) )\u221a \u221a pm \u2212 qm)2 \u2264 qm ) ) \u2264 \u221a | ) ), so pm + \u221a qm \u221a pm \u2212 qm| | pm + = 2(1 \u2212 F (\u03c1, \u03c3)). ) \u221a )\u221a \u221a m = m m |pm \u2212 qm| = 2D(pm, qm) \u2264 2D(\u03c1", ", \u03c3). Comparing (9.104) and (9.108) we see that 1 \u2212 F (\u03c1, \u03c3) \u2264 D(\u03c1, \u03c3). Summarizing, we have 1 \u2212 F (\u03c1, \u03c3) \u2264 D(\u03c1, \u03c3) \u2264 $ 1 \u2212 F (\u03c1, \u03c3)2. \u221a qm| (9.104) (9.105) (9.106) (9.107) (9.108) (9.109) (9.110) The implication is that the trace distance and the \ufb01delity are qualitatively equivalent measures of closeness for quantum states. Indeed, for many purposes it does not matter whether the trace distance or the \ufb01delity is used to quantify distance, since results about one may be used to deduce equivalent results about the other. Exercise 9.21: When comparing pure states and mixed states it is possible to make a stronger statement than (9.110) about the relationship between trace distance and \ufb01delity. Prove that 1 \u2212 F (|\u03c8, \u03c3)2 \u2264 D(|\u03c8, \u03c3). (9.111) 9.3 How well does a quantum channel preserve information? Friends come and go, but enemies accumulate \u2013 Jones\u2019 Law, attributed to Thomas Jones How well does a quantum channel preserve information? More precisely, suppose a quantum system is in the state |\u03c8 and some physical process occurs, changing the quantum system to the state E(|\u03c8\u03c8|). How well has the channel E preserved the state |\u03c8 of the quantum system? The static measures of distance discussed in previous sections will be used in this section to develop measures of how well a quantum channel preserves information. This type of scenario occurs often in quantum computation and quantum information. For example, in the memory of a quantum computer, |\u03c8 is the initial state of the memory, and E represents the dynamics that the memory undergoes, including noise processes arising from interaction with the environment. A second example is provided by a quantum communication channel for transmitting the state |\u03c8 from one location to another. No channel is ever perfect, so the action of the channel is described by a quantum operation E. An obvious way of quantifying how well the state |\u03c8 is preserved by the channel is to make use of the static distance measures introduced in the previous section. For How well does a quantum channel preserve information? 417 example, we can compute the \ufb01delity between the starting state |\u03c8 and the ending state E(|\u03c8\u03c8|). For the case of the depolarizing channel, we obtain F (|\u03c8, E(|\u03c8\u03c8|)) = \u03c81 \u2212 p)|\u03c8\u03c8| |\u03c8 (9.112) I 2 (9.113) This result agrees well with our intuition \u2013 the higher the probability p of depolarizing, the lower the \ufb01delity of the \ufb01nal state with the initial state. Provided p is very small the \ufb01delity is close to one, and the state E(\u03c1) is practically indistinguishable from the initial state |\u03c8. There is nothing special about the use of the \ufb01delity in the above expression. We could equally well have used the trace distance. However, for the remainder of this chapter we are going to restrict ourselves to measures of distance based upon the \ufb01delity and derived quantities. Using the properties of the trace distance established in the last section it is not dif\ufb01cult, for the most part, to give a parallel development based upon the trace distance. However, it turns out that the \ufb01delity is an easier tool to calculate with, and for that reason we restrict ourselves to considerations based upon the \ufb01delity. Our prototype measure for information preservation, the \ufb01delity F (|\u03c8, E(|\u03c8\u03c8|)), has some drawbacks which need to be remedied. In a real quantum memory or quantum communications channel, we don\u2019t know in advance what the initial state |\u03c8 of the system will be. However, we can quantify the worst-case behavior of the system by minimizing over all possible initial states, Fmin(E) \u2261 min |\u03c8 F (|\u03c8, E(|\u03c8\u03c8|)). (9.114) 1 \u2212 p/2, as the \ufb01delity of the For example, for the p-depolarizing channel Fmin = channel is the same for all input states |\u03c8. A more interesting example is the phase damping channel, $ E(\u03c1) = p\u03c1 + (1 \u2212 p)Z\u03c1Z. For the phase damping channel the \ufb01delity is given by F (|\u03c8, E(|\u03c8", "\u03c8|)) = ( \u03c8| ( p|\u03c8\u03c8| + (1 \u2212 p)Z|\u03c8\u03c8|Z |\u03c8 = p + (1 \u2212 p)\u03c8|Z|\u03c82. (9.115) (9.116) (9.117) The second term under the square root sign is non-negative, and equal to zero when |\u03c8 = (|0 + |1)/ 2. Thus for the phase damping channel the minimum \ufb01delity is \u221a Fmin(E) = \u221a p. (9.118) You might wonder why we minimized over pure states in the de\ufb01nition of Fmin. After all, might not the quantum system of interest start in a mixed state \u03c1? For example, a quantum memory might be entangled with the rest of the quantum computer, and therefore would start out in a mixed state. Fortunately, the joint concavity of the \ufb01delity can be used to show that allowing mixed states does not change Fmin. To see this, suppose 418 Distance measures for quantum information that \u03c1 = i \u03bbi|ii| is the initial state of the quantum system. Then we have \u03bbiE(|ii|) i \u03bbi|ii|, i \u03bbiF (|i, E(|ii|)). F (\u03c1, E(\u03c1)) = F \u2265 i It follows that F (\u03c1, E(\u03c1)) \u2265 F (|i, E(|ii|)) (9.119) (9.120) (9.121) for at least one of the states |i, and thus F (\u03c1, E(\u03c1)) \u2265 Fmin. Of course, we are interested not only in protecting quantum states as they are transmitted through a quantum communications channel, but also as they dynamically undergo computation. Suppose, for example, that as part of a quantum computation we attempt to implement a quantum gate described by the unitary operator U. As described in the last chapter, any such attempt will inevitably encounter some (hopefully not too severe) noise, so the correct description of the gate is using a trace-preserving quantum operation E. A natural measure of how successful our gate has been is the gate \ufb01delity, F (U, E) \u2261 min |\u03c8 F (U |\u03c8, E(|\u03c8\u03c8|)). (9.122) gate on a single qubit, but instead Suppose, for example, that we try to implement a implement the noisy operation E(\u03c1) = (1 \u2212 p)X\u03c1X + pZ\u03c1Z, for some small noise parameter p. Then the gate \ufb01delity for this operation is given by ( F (X, E) = min |\u03c8 \u03c8|X ( (1 \u2212 p)X|\u03c8\u03c8|X + pZ|\u03c8\u03c8|Z X|\u03c8 (1 \u2212 p) + p\u03c8|Y |\u03c82 = min |\u03c8 $ = (1 \u2212 p). (9.123) (9.124) (9.125) In Exercise 9.22 you will show that performing a sequence of gates each with high \ufb01delity is suf\ufb01cient to ensure that the total operation has high \ufb01delity, and thus for the purposes of quantum computation it is suf\ufb01cient to perform each gate in the computation with high \ufb01delity. (Compare also the similar but less general arguments of Chapter 4 on approximating quantum circuits.) Exercise 9.22: (Chaining property for \ufb01delity measures) Suppose U and V are unitary operators, and E and F are trace-preserving quantum operations meant to approximate U and V. Letting d(\u00b7, \u00b7) be any metric on the space of density matrices satisfying d(U \u03c1U \u2020, U \u03c3U \u2020) = d(\u03c1, \u03c3) for all density matrices \u03c1 and \u03c3 and unitary U (such as the angle arccos(F (\u03c1, \u03c3))), de\ufb01ne the corresponding error E(U, E) by E(U, E) \u2261 max \u03c1 d(U \u03c1U \u2020, E(\u03c1)), (9.126) and show that E(V U, F \u25e6 E) \u2264 E(U, E) + E(V, F ). Thus, to perform a quantum computation with high \ufb01delity it suf\ufb01ces to complete each step of the computation with high \ufb01delity. How well does a quantum channel preserve information? 419 Quantum sources of information and the entanglement \ufb01delity We\u2019ve been talking about dynamic measures of information preservation, without de\ufb01ning exactly", " what we mean by a quantum source of information. We\u2019ll now explain two possible de\ufb01nitions for this notion, and use these de\ufb01nitions to motivate the introduction of some dynamic measures of information preservation. A priori, it is not entirely clear how best to go about de\ufb01ning the notion of a quantum source of information. Classically, the best solution to this de\ufb01nitional problem is not at all obvious, and it is possible to come up with inequivalent de\ufb01nitions, each yielding a rich and useful theory of information. Since quantum information contains classical information as a sub\ufb01eld, it should not be surprising if there are even more ways of de\ufb01ning the notion of an information source quantum mechanically! To conclude this chapter we introduce two possible quantum de\ufb01nitions for the notion of an information source, explain how they motivate corresponding distance measures for the preservation of information, and prove some elementary properties of these distance measures. Further discussion of quantum sources of information is deferred until Chapter 12. One attractive de\ufb01nition for quantum sources is to imagine a stream of identical quantum systems (say, qubits) being produced by some physical process, where the states of the respective systems are given by \u03c1X1, \u03c1X2,..., the Xj are independent and identically distributed random variables, and \u03c1j is some \ufb01xed set of density operators. For example, one might imagine a stream of qubits, each of which is prepared in the state |0 with probability one half, or in the state (|0 + |1)/ 2 with probability one half. \u221a This ensemble notion of a quantum source leads naturally to a notion of ensemble average \ufb01delity which captures the idea that the source is well-preserved under the action of a noisy channel described by a trace-preserving quantum operation E, namely F = j pjF (\u03c1j, E(\u03c1j))2, (9.127) where pj are the respective probabilities for the different possible preparations of the system \u03c1j. Obviously, 0 \u2264 F \u2264 1, and provided F \u2248 1 we can be con\ufb01dent that, on average, the channel E preserves the states emitted by the source with a high degree of accuracy. You may wonder why the \ufb01delity appearing on the right hand side of the de\ufb01nition is squared. There are two answers to this question, one simple, and one complex. The simple answer is that including this square term makes the ensemble \ufb01delity more naturally related to the entanglement \ufb01delity, as de\ufb01ned below. The more complex answer is that quantum information is, at present, in a state of infancy and it is not entirely clear what the \u2018correct\u2019 de\ufb01nitions for notions such as information preservation are! Nevertheless, as we shall see in Chapter 12, the ensemble average \ufb01delity and the entanglement \ufb01delity give rise to a rich theory of quantum information, which leads us to believe that these measures are on the right track, even though a complete theory of quantum information has not yet been developed. Exercise 9.23: Show that \u00afF = 1 if and only if E(\u03c1j) = \u03c1j for all j such that pj > 0. There is a second notion of a quantum source we may consider, motivated by the idea that a channel which preserves information well is a channel that preserves entanglement well. The basic idea comes from the discussion of the classical probability of error in Section 9.1. As noted there, a direct analogue of the probability of error p(X = Y ) 420 Distance measures for quantum information cannot be de\ufb01ned for quantum processes, for there is no direct quantum analogue of a probability distribution de\ufb01ned at two times. Instead, we will use a quantum analogue of the idea illustrated in Figure 9.7, which is that a dynamic measure of distance can be de\ufb01ned by \ufb01rst copying the random variable X to \u02dcX, then subjecting X to noise to obtain Y, and using as our measure of distance some metric quantity D[( \u02dcX, X), ( \u02dcX, Y )] between the joint distributions for ( \u02dcX, X) and ( \u02dcX, Y ). Figure 9.7. The probability of an error in the channel is equal to the trace distance between the probability distributions for ( \u02dcX, X) and ( \u02dcX, Y ). A quantum analogue of this model is as follows. A quantum system, Q, is prepared in a state \u03c1. The state of Q is assumed to be entangled in some way with the external world. This entanglement replaces the correlation between X", " and \u02dcX in the classical model. We represent the entanglement by introducing a \ufb01ctitious system R, such that the joint state of RQ is a pure state. It turns out that all results that we prove do not depend in any way on how this puri\ufb01cation is performed, so we may as well suppose that this is an arbitrary entanglement with the outside world. The system Q is then subjected to a dynamics described by a quantum operation, E. The basic situation is illustrated in Figure 9.8. How well is the entanglement between R and Q preserved by the quantum operation E? We quantify this by the entanglement \ufb01delity F (\u03c1, E) which is a function of \u03c1 and E de\ufb01ned for trace-preserving quantum operations E by F (\u03c1, E) \u2261 F (RQ, RQ)2 = RQ| (IR \u2297 E)(|RQRQ|) |RQ, (9.128) (9.129) where the use of a prime indicates the state of a system after the quantum operation has been applied, and the absence of a prime indicates the state of a system before the How well does a quantum channel preserve information? 421 |RQ R Q Q Figure 9.8. The RQ picture of a quantum channel. The initial state of RQ is a pure state. quantum operation has been applied. The quantity appearing on the right hand side of this de\ufb01nition is the square of the static \ufb01delity between the initial and \ufb01nal states of RQ. The use of the square of the static \ufb01delity is purely a convenience, since it simpli\ufb01es certain properties of the entanglement \ufb01delity. Note that the entanglement \ufb01delity depends only upon \u03c1 and E, and not (as it may appear) upon the details of the puri\ufb01cation |RQ. To see this, we use the fact, proved in Exercise 2.81, that any two puri\ufb01cations |R1Q1 and |R2Q2 of \u03c1 are related by a unitary operation, U, that acts upon R alone, |R2Q2 = U |R1Q1. Thus 2Q 1Q F (|R2Q2, \u03c1R 2) = F (|R1Q1, \u03c1R (9.130) 2), which establishes the result. The entanglement \ufb01delity provides a measure of how well the entanglement between R and Q is preserved by the process E, with values close to 1 indicating that the entanglement has been well preserved, and values close to 0 indicating that most of the entanglement has been destroyed. The choice of whether to use the static \ufb01delity squared or the static \ufb01delity is essentially arbitrary; the present de\ufb01nition results in slightly more attractive mathematical properties. One of the attractive properties of the entanglement \ufb01delity is that there is a very simple formula which enables it to be calculated exactly. Suppose Ei is a set of operation elements for a quantum operation E. Then F (\u03c1, E) = RQ|\u03c1RQ|RQ = |RQ|Ei|RQ|2. (9.131) i Suppose we write |RQ = \u221a j pj|j|j, where \u03c1 = \u221a j pj|jj|. Then RQ|Ei|RQ = pjpkj|kj|Ei|k (9.132) (9.133) (9.134) jk = pjj|Ei|j j = tr(Ei\u03c1). i Substituting this expression into Equation (9.131) we obtain the useful computational formula F (\u03c1, E) = |tr(\u03c1Ei)|2. (9.135) 422 Distance measures for quantum information Thus, for example, the entanglement \ufb01delity for the phase damping channel E(\u03c1) = p\u03c1 + (1 \u2212 p)Z\u03c1Z is given by F (\u03c1, E) = p |tr(\u03c1)|2 + (1 \u2212 p) |tr(\u03c1Z)|2 = p + (1 \u2212 p)tr(\u03c1Z)2, (9.136) so we see that as p decreases, the entanglement \ufb01delity decreases, as we intuitively expect. We\u2019ve de\ufb01ned two notions of a quantum information source and associated distance measures, one based on the idea that we want to preserve an ensemble of quantum states with high average \ufb01delity, the other based on the idea that it is entanglement between the", " source and some reference system that we wish to preserve. Perhaps surprisingly, these two de\ufb01nitions turn out to be closely related! The reason for this lies in two useful properties of the entanglement \ufb01delity. First, the entanglement \ufb01delity is a lower bound on the square of the static \ufb01delity between the input and output to the process, F (\u03c1, E) \u2264 F (\u03c1, E(\u03c1)) 2. (9.137) Intuitively, this result states that it is harder to preserve a state plus entanglement with the outside world than it is to merely preserve the state alone. The proof is an elementary application of the monotonicity of the static \ufb01delity under partial trace, F (\u03c1, E) = F (|RQ, \u03c1RQ )2 \u2264 F (\u03c1Q, \u03c1Q )2. The second property of entanglement \ufb01delity we need to relate it to the ensemble average de\ufb01nition is that it is a convex function of \u03c1. To see this, de\ufb01ne f (x) \u2261 F (x\u03c11 + (1 \u2212 x)\u03c12, E), and using Equation (9.135) and elementary calculus we \ufb01nd f (x) = |tr((\u03c11 \u2212 \u03c12)Ei)|2, (9.138) so that f (x) \u2265 0, which implies the convexity of the entanglement \ufb01delity, as required. i Combining these two results we see that \u239b \u239e \u239d F j pj\u03c1j, E \u23a0 \u2264 \u2264 pjF (\u03c1j, E) pjF (\u03c1j, E(\u03c1j))2, j j \u239e and thus \u239b \u239d j F pj\u03c1j, E \u23a0 \u2264 \u00afF! (9.139) (9.140) (9.141) Thus, any quantum channel E which does a good job of preserving the entanglement between a source described by a density operator \u03c1 and a reference system will automatically do a good job of preserving an ensemble source described by probabilities pj and states \u03c1j such that \u03c1 = j pj\u03c1j. In this sense the notion of a quantum source based on entanglement \ufb01delity is a more stringent notion than the ensemble de\ufb01nition, and for this reason we will prefer the entanglement \ufb01delity based de\ufb01nition in our study of quantum information theory, in Chapter 12. We conclude the chapter with a short list of some easily proved properties of the entanglement \ufb01delity that will be useful in later chapters: (1) 0 \u2264 F (\u03c1, E) \u2264 1. Immediate from properties of the static \ufb01delity. Chapter problems 423 (2) The entanglement \ufb01delity is linear in the quantum operation input. This is immediate from the de\ufb01nition of the entanglement \ufb01delity. (3) For pure state inputs, the entanglement \ufb01delity is equal to the static \ufb01delity squared between input and output, F (|\u03c8, E) = F (|\u03c8, E(|\u03c8\u03c8|))2. (9.142) This is immediate from the observation that the state |\u03c8 is a puri\ufb01cation of itself, and the de\ufb01nition of the entanglement \ufb01delity. (4) F (\u03c1, E) = 1 if and only if for all pure states |\u03c8 in the support of \u03c1, E(|\u03c8\u03c8|) = |\u03c8\u03c8|. (9.143) To prove this, suppose F (\u03c1, E) = 1, and |\u03c8 is a pure state in the support of \u03c1. De\ufb01ne p \u2261 1/\u03c8|\u03c1\u22121|\u03c8 > 0 (compare Exercise 2.73 on page 105) and \u03c3 to be a density operator such that (1 \u2212 p)\u03c3 = \u03c1 \u2212 p|\u03c8\u03c8|. Then by convexity, ( 1 = F (\u03c1, E) \u2264 p F (|\u03c8, E) + (1 \u2212 p), (9.144) and thus F (|\u03c8, E) = 1, establishing the result one way. The converse is a straightforward application of the de\ufb01nition of the entanglement \ufb01delity. (5) Suppose that \u03c8|E(|\u03c8\u03c8|)|\u03c8 \u2265 1 \u2212 \u03b7 for all |\u03c8 in the support", " of \u03c1, for some \u03b7. Then F (\u03c1, E) \u2265 1 \u2212 (3\u03b7/2). (See Problem 9.3.) Problem 9.1: (Alternate characterization of the \ufb01delity) Show that F (\u03c1, \u03c3) = inf P tr(\u03c1P )tr(\u03c3P \u22121), (9.145) where the in\ufb01mum is taken over all invertible positive matrices P. Problem 9.2: Let E be a trace-preserving quantum operation. Show that for each \u03c1 there is a set of operation elements {Ei} for E such that F (\u03c1, E) = |tr(\u03c1E1)|2. (9.146) Problem 9.3: Prove fact (5) on this page. Summary of Chapter 9: Distance measures for quantum information \u2022 Trace distance: D(\u03c1, \u03c3) \u2261 1 2 tr|\u03c1 \u2212 \u03c3|. Doubly convex metric on density opera- tors, contractive under quantum operations. \u2022 Fidelity: ( F (\u03c1, \u03c3) \u2261 tr \u03c11/2\u03c3\u03c11/2 = max |\u03c8,|\u03d5 \u221a |\u03c8|\u03d5|. Strongly concave, F ( i pi\u03c1i, i qi\u03c3i) \u2265 i piqiF (\u03c1i, \u03c3i). \u2022 Entanglement \ufb01delity: F (\u03c1, E). Measure of how well entanglement is preserved during a quantum mechanical process, starting with the state \u03c1 of a system Q, which is assumed to be entangled with another quantum system, R, and applying the quantum operation E to system Q. 424 Distance measures for quantum information History and further reading Readers wishing to learn more about distance measures for quantum information would be well advised to start with Fuchs\u2019 1996 Ph.D. Dissertation[Fuc96]. It contains a wealth of material on distance measures for quantum information, including a list of 528 references on related topics, organized into subject areas. Notably, the proof of Equation (9.74) may be found there, as well as much else of interest. Also see the article by Fuchs and van de Graaf[FvdG99]; this paper is the origin of the inequality (9.110), and is also a good overview of distance measures for quantum information, especially in the context of quantum cryptography. The contractivity of the trace distance was proved by Ruskai[Rus94]. The monotonicity of the \ufb01delity was proved by Barnum, Caves, Fuchs, Jozsa and Schumacher[BCF+96]. In the literature both the quantity we call \ufb01delity and its square have been referred to as the \ufb01delity. Uhlmann\u2019s paper[Uhl76] in which he proves the eponymous theorem also contains an extensive discussion of the elementary properties of \ufb01delity. The proof of Uhlmann\u2019s theorem given here is due to Jozsa[Joz94]. Chaining properties for \ufb01delity measures and the relation to noisy quantum computation are discussed in more detail by Aharonov, Kitaev and Nisan[AKN98]. Schumacher[Sch96b] introduced the entanglement \ufb01delity and proved many elementary properties. Knill and La\ufb02amme[KL97] established the connection between subspace \ufb01delity and entanglement \ufb01delity, Problem 9.3. A more detailed proof of this fact appeared in Barnum, Knill and Nielsen[BKN98]. Problem 9.1 is due to Alberti[Alb83]. 10 Quantum error-correction We have learned that it is possible to \ufb01ght entanglement with entanglement. \u2013 John Preskill To be an Error and to be Cast out is part of God\u2019s Design \u2013 William Blake This chapter explains how to do quantum information processing reliably in the presence of noise. The chapter covers three broad topics: the basic theory of quantum errorcorrecting codes, fault-tolerant quantum computation, and the threshold theorem. We begin by developing the basic theory of quantum error-correcting codes, which protect quantum information against noise. These codes work by encoding quantum states in a special way that make them resilient against the effects of noise, and then decoding when it is wished to recover the original state. Section 10.1 explains the basic ideas of classical error-correction, and some of the conceptual challenges that must be overcome to make quantum error-correction possible. Section 10.2 explains a simple example of a quantum error-correcting code, which we then generalize into a theory of quantum error-correcting codes in Section 10.3. Section 10.4 explains some ideas from", " the classical theory of linear codes, and how they give rise to an interesting class of quantum codes known as Calderbank\u2013Shor\u2013Steane (CSS) codes. Section 10.5 concludes our introductory survey of quantum error-correcting codes with a discussion of stabilizer codes, a richly structured class of codes with a close connection to classical error-correcting codes. Our discussion of quantum error-correction assumes that encoding and decoding of quantum states can be done perfectly, without error. This is useful, for example, if we wish to send quantum states over a noisy communication channel and can use almostnoiseless quantum computers to perform pretty good encoding and decoding of the states at each end of the channel. However, this assumption cannot be made if the quantum gates used to do the encoding and decoding are themselves noisy. Fortunately, the theory of fault-tolerant quantum computation, developed in Section 10.6, allows us to remove the assumption of perfect encoding and decoding. Even more impressively, fault-tolerance allow us to perform logical operations on encoded quantum states, in a manner which tolerates faults in the underlying gate operations. The chapter culminates in Section 10.6.4 with the threshold theorem for quantum computation: provided the noise in individual quantum gates is below a certain constant threshold it is possible to ef\ufb01ciently perform an arbitrarily large quantum computation. Of course there are caveats to this result, which we spend some time discussing. Nevertheless, the threshold theorem is a remarkable result indicating that noise likely poses no fundamental barrier to the performance of large-scale quantum computations. 426 Quantum error-correction 10.1 Introduction Noise is a great bane of information processing systems. Whenever possible we build our systems to avoid noise completely, and where that is not possible, we try to protect against the effects of noise. For example, components in modern computers are extremely reliable, with a failure rate typically below one error in 1017 operations. For most practical purposes we can act as if computer components are completely noiseless. On the other hand, many systems in widespread use do suffer from a substantial noise problem. Modems and CD players both make use of error-correcting codes to protect against the effects of noise. The details of the techniques used to protect against noise in practice are sometimes rather complicated, but the basic principles are easily understood. The key idea is that if we wish to protect a message against the effects of noise, then we should encode the message by adding some redundant information to the message. That way, even if some of the information in the encoded message is corrupted by noise, there will be enough redundancy in the encoded message that it is possible to recover or decode the message so that all the information in the original message is recovered. For example, suppose we wish to send a bit from one location to another through a noisy classical communications channel. The effect of the noise in the channel is to \ufb02ip the bit being transmitted with probability p > 0, while with probability 1 \u2212 p the bit is transmitted without error. Such a channel is known as a binary symmetric channel, and is illustrated in Figure 10.1. A simple means of protecting the bit against the effects of noise in the binary symmetric channel is to replace the bit we wish to protect with three copies of itself: 0 \u2192 000 1 \u2192 111. (10.1) (10.2) The bit strings 000 and 111 are sometimes referred to as the logical 0 and logical 1, since they play the role of 0 and 1, respectively. We now send all three bits through the channel. At the receiver\u2019s end of the channel three bits are output, and the receiver has to decide what the value of the original bit was. Suppose 001 were output from the channel. Provided the probability p of a bit \ufb02ip is not too high, it is very likely that the third bit was \ufb02ipped by the channel, and that 0 was the bit that was sent. Figure 10.1. Binary symmetric channel. This type of decoding is called majority voting, since the decoded output from the channel is whatever value, 0 or 1, appears more times in the actual channel output. Majority voting fails if two or more of the bits sent through the channel were \ufb02ipped, and succeeds otherwise. The probability that two or more of the bits are \ufb02ipped is Introduction 427 3p2(1 \u2212 p) + p3, so the probability of error is pe = 3p2 \u2212 2p3. Without encoding, the probability of an error was p, so the code makes the transmission more reliable provided pe < p, which occurs whenever p < 1/2. The type of code just described is called a repetition code, since we encode the message to be sent by repeating it a number of times. A similar technique has been used for millennia as a part of everyday conversation: if we\u2019re having dif\ufb01culty understanding someone\u2019s spoken", " language, perhaps because they have a foreign accent, we ask them to repeat what they\u2019re saying. We may not catch all the words either time, but we can put the iterations together to comprehend a coherent message. Many clever techniques have been developed in the theory of classical error-correcting codes; however, the key idea is always to encode messages by adding enough redundancy that the original message is recoverable after noise has acted on the encoded message, with the amount of redundancy needing to be added depending on the severity of noise in the channel. 10.1.1 The three qubit bit \ufb02ip code To protect quantum states against the effects of noise we would like to develop quantum error-correcting codes based upon similar principles. There are some important differences between classical information and quantum information that require new ideas to be introduced to make such quantum error-correcting codes possible. In particular, at a \ufb01rst glance we have three rather formidable dif\ufb01culties to deal with: \u2022 No cloning: One might try to implement the repetition code quantum mechanically by duplicating the quantum state three or more times. This is forbidden by the no-cloning theorem discussed in Box 12.1 on page 532. Even if cloning were possible, it would not be possible to measure and compare the three quantum states output from the channel. \u2022 Errors are continuous: A continuum of different errors may occur on a single qubit. Determining which error occurred in order to correct it would appear to require in\ufb01nite precision, and therefore in\ufb01nite resources. \u2022 Measurement destroys quantum information: In classical error-correction we observe the output from the channel, and decide what decoding procedure to adopt. Observation in quantum mechanics generally destroys the quantum state under observation, and makes recovery impossible. Fortunately, none of these problems is fatal, as we shall demonstrate. Suppose we send qubits through a channel which leaves the qubits untouched with probability 1 \u2212 p, and \ufb02ips the qubits with probability p. That is, with probability p the state |\u03c8 is taken to the state X|\u03c8, where X is the usual Pauli sigma x operator, or bit \ufb02ip operator. This channel is called the bit \ufb02ip channel, and we now explain the bit \ufb02ip code, which may be used to protect qubits against the effects of noise from this channel. Suppose we encode the single qubit state a|0 + b|1 in three qubits as a|000 + b|111. A convenient way to write this encoding is |0 \u2192 |0L \u2261 |000 |1 \u2192 |1L \u2261 |111, (10.3) (10.4) where it is understood that superpositions of basis states are taken to corresponding superpositions of encoded states. The notation |0L and |1L indicates that these are 428 Quantum error-correction the logical |0 and logical |1 states, not the physical zero and one states. A circuit performing this encoding is illustrated in Figure 10.2. |\u03c8 \u2022 \u2022 |0 |0 \u2295 \u2295 Figure 10.2. Encoding circuit for the three qubit bit \ufb02ip code. The data to be encoded enters the circuit on the top line. Exercise 10.1: Verify that the encoding circuit in Figure 10.2 works as claimed. Suppose the initial state a|0 + b|1 has been perfectly encoded as a|000 + b|111. Each of the three qubits is passed through an independent copy of the bit \ufb02ip channel. Suppose a bit \ufb02ip occurred on one or fewer of the qubits. There is a simple two stage error-correction procedure which can be used to recover the correct quantum state in this case: (1) Error-detection or syndrome diagnosis: We perform a measurement which tells us what error, if any, occurred on the quantum state. The measurement result is called the error syndrome. For the bit \ufb02ip channel there are four error syndromes, corresponding to the four projection operators: P0 \u2261 |000000| + |111111| no error P1 \u2261 |100100| + |011011| bit \ufb02ip on qubit one P2 \u2261 |010010| + |101101| bit \ufb02ip on qubit two P3 \u2261 |001001| + |110110| bit \ufb02ip on qubit three. (10.5) (10.6) (10.7) (10.8) Suppose for example that a bit \ufb02ip occurs on qubit one, so the corrupted state is a|100 + b|011. Notice that \u03c8|P1|\u03c8 = 1 in this case, so the outcome of the measurement result (the error syndrome) is certainly 1. Furthermore, the syndrome measurement does not", " cause any change to the state: it is a|100 + b|011 both before and after syndrome measurement. Note that the syndrome contains only information about what error has occurred, and does not allow us to infer anything about the value of a or b, that is, it contains no information about the state being protected. This is a generic feature of syndrome measurements, since to obtain information about the identity of a quantum state it is in general necessary to perturb that state. (2) Recovery: We use the value of the error syndrome to tell us what procedure to use to recover the initial state. For example, if the error syndrome was 1, indicating a bit \ufb02ip on the \ufb01rst qubit, then we \ufb02ip that qubit again, recovering the original state a|000 + b|111 with perfect accuracy. The four possible error syndromes and the recovery procedure in each case are: 0 (no error) \u2013 do nothing; 1 (bit \ufb02ip on \ufb01rst qubit) \u2013 \ufb02ip the \ufb01rst qubit again; 2 (bit \ufb02ip on second qubit) \u2013 \ufb02ip the second qubit Introduction 429 again; 3 (bit \ufb02ip on third qubit) \u2013 \ufb02ip the third qubit again. For each value of the error syndrome it is easy to see that the original state is recovered with perfect accuracy, given that the corresponding error occurred. This error-correction procedure works perfectly, provided bit \ufb02ips occur on one or fewer of the three qubits. This occurs with probability (1 \u2212 p)3 + 3p(1 \u2212 p)2 = 1 \u2212 3p2 + 2p3. The probability of an error remaining uncorrected is therefore 3p2 \u2212 2p3, just as for the classical repetition code we studied earlier. Once again, provided p < 1/2 the encoding and decoding improve the reliability of storage of the quantum state. Improving the error analysis This error analysis is not completely adequate. The problem is that not all errors and states in quantum mechanics are created equal: quantum states live in a continuous space, so it is possible for some errors to corrupt a state by a tiny amount, while others mess it up completely. An extreme example is provided by the bit \ufb02ip \u2018error\u2019 X, which does not affect the state (|0 + |1)/ 2 at all, but \ufb02ips the |0 state so it becomes a |1. In the former case we would not be worried about a bit \ufb02ip error occurring, while in the latter case we would obviously be very worried. \u221a To address this problem we make use of the \ufb01delity quantity introduced in Chapter 9. Recall that the \ufb01delity between a pure and a mixed state is given by F (|\u03c8, \u03c1) = $ \u03c8|\u03c1|\u03c8. The object of quantum error-correction is to increase the \ufb01delity with which quantum information is stored (or communicated) up near the maximum possible \ufb01delity of one. Let\u2019s compare the minimum \ufb01delity achieved by the three qubit bit \ufb02ip code with the \ufb01delity when no error-correction is performed. Suppose the quantum state of interest is |\u03c8. Without using the error-correcting code the state of the qubit after being sent through the channel is \u03c1 = (1 \u2212 p)|\u03c8\u03c8| + pX|\u03c8\u03c8|X. (10.9) The \ufb01delity is given by ( ( F = \u03c8|\u03c1|\u03c8 = (1 \u2212 p) + p\u03c8|X|\u03c8\u03c8|X|\u03c8. (10.10) The second term under the square root is non-negative, and equal to zero when |\u03c8 = |0, 1 \u2212 p. Suppose the three qubit errorso we see that the minimum \ufb01delity is F = correcting code is used to protect the state |\u03c8 = a|0L + b|1L. The quantum state after both the noise and error-correction is: \u221a & (1 \u2212 p)3 + 3p(1 \u2212 p)2 \u2019 \u03c1 = |\u03c8\u03c8| + \u00b7 \u00b7 \u00b7. (10.11) The omitted terms represent contributions from bit \ufb02ips on two or three qubits. All the omitted terms are positive operators, so the \ufb01delity we calculate will be a lower bound (1 \u2212 p)3 + 3p(1 \u2212 p)2. That is, on the true \ufb01delity. We see that F = 1 \u2212 3p2 + 2p3, so the \ufb01delity of storage", " for the quantum state is the \ufb01delity is at least improved provided p < 1/2, which is the same conclusion we came to earlier based on a much cruder analysis. \u03c8|\u03c1|\u03c8 \u2265 $ $ $ Exercise 10.2: The action of the bit \ufb02ip channel can be described by the quantum operation E(\u03c1) = (1 \u2212 p)\u03c1 + pX\u03c1X. Show that this may be given an alternate operator-sum representation, as E(\u03c1) = (1 \u2212 2p)\u03c1 + 2pP+\u03c1P+ + 2pP\u2212\u03c1P\u2212 where 430 Quantum error-correction \u221a P+ and P\u2212 are projectors onto the +1 and \u22121 eigenstates of X, (|0 + |1)/ and (|0 \u2212 |1)/ a model in which the qubit is left alone with probability 1 \u2212 2p, and is \u2018measured\u2019 by the environment in the |+, |\u2212 basis with probability 2p. 2, respectively. This latter representation can be understood as \u221a 2 There is a different way of understanding the syndrome measurement that is useful in generalizing the three qubit code. Suppose that instead of measuring the four projectors P0, P1, P2, P3 we performed two measurements, the \ufb01rst of the observable Z1Z2 (that is, Z \u2297 Z \u2297 I), and the second of the observable Z2Z3. Each of these observables has eigenvalues \u00b11, so each measurement provides a single bit of information, for a total of two bits of information \u2013 four possible syndromes, just as in the earlier description. The \ufb01rst measurement, of Z1Z2, can be thought of as comparing the \ufb01rst and second qubits to see if they are the same. To see why this is so, note that Z1Z2 has spectral decomposition Z1Z2 = (|0000| + |1111|) \u2297 I \u2212 (|0101| + |1010|) \u2297 I, (10.12) which corresponds to a projective measurement with projectors (|0000| + |1111|) \u2297 I and (|0101| + |1010|) \u2297 I. Thus, measuring Z1Z2 can be thought of as comparing the values of the \ufb01rst and second qubits, giving +1 if they are the same, and \u22121 if they are different. Similarly, measuring Z2Z3 compares the values of the second and third qubits, giving +1 if they are the same, and \u22121 if they are different. Combining these two measurement results we can determine whether a bit \ufb02ip occurred on one of the qubits or not, and if so, which one: if both measurement results give +1 then with high probability no bit \ufb02ip has occurred; if measuring Z1Z2 gives +1 and measuring Z2Z3 gives \u22121 then with high probability just the third qubit \ufb02ipped; if measuring Z1Z2 gives \u22121 and measuring Z2Z3 gives +1 then with high probability just the \ufb01rst qubit \ufb02ipped; and \ufb01nally if both measurements give \u22121 then with high probability just the second qubit \ufb02ipped. What is crucial to the success of these measurements is that neither measurement gives any information about the amplitudes a and b of the encoded quantum state, and thus neither measurement destroys the superpositions of quantum states that we wish to preserve using the code. Exercise 10.3: Show by explicit calculation that measuring Z1Z2 followed by Z2Z3 is equivalent, up to labeling of the measurement outcomes, to measuring the four projectors de\ufb01ned by (10.5)\u2013(10.8), in the sense that both procedures result in the same measurement statistics and post-measurement states. 10.1.2 Three qubit phase \ufb02ip code The bit \ufb02ip code is interesting, but it does not appear to be that signi\ufb01cant an innovation over classical error-correcting codes, and leaves many problems open (for example, many kinds of errors other than bit \ufb02ips can happen to qubits). A more interesting noisy quantum channel is the phase \ufb02ip error model for a single qubit. In this error model the qubit is left alone with probability 1 \u2212 p, and with probability p the relative phase of the |0 and |1 states is \ufb02ipped. More precisely, the phase \ufb02ip operator Z is applied to the qubit with probability p > 0, so the state a|0+b|1 is taken to the state a|0\u2212b|1", " under the phase \ufb02ip. There is no classical equivalent to the phase \ufb02ip channel, since classical channels don\u2019t have any property equivalent to phase. However, there is an easy way to turn the phase \ufb02ip channel into a bit \ufb02ip channel. Suppose we work in the qubit basis Introduction 431 \u221a \u221a 2, |\u2212 \u2261 (|0 \u2212 |1)/ |+ \u2261 (|0 + |1)/ 2. With respect to this basis the operator Z takes |+ to |\u2212 and vice versa, that is, it acts just like a bit \ufb02ip with respect to the labels + and \u2212! This suggests using the states |0L \u2261 | + ++ and |1L \u2261 | \u2212 \u2212\u2212 as logical zero and one states for protection against phase \ufb02ip errors. All the operations needed for error-correction \u2013 encoding, error-detection, and recovery \u2013 are performed just as for the bit \ufb02ip channel, but with respect to the |+, |\u2212 basis instead of the |0, |1 basis. To accomplish this basis change we simply apply the Hadamard gate and its inverse (also the Hadamard gate) at appropriate points in the procedure, since the Hadamard gate accomplishes the change back and forth between the |0, |1 basis and the |+, |\u2212 basis. Figure 10.3. Encoding circuits for the phase \ufb02ip code. More explicitly, the encoding for the phase \ufb02ip channel is performed in two steps: \ufb01rst, we encode in three qubits exactly as for the bit \ufb02ip channel; second, we apply a Hadamard gate to each qubit (Figure 10.3). Error-detection is achieved by applying the same projective measurements as before, but conjugated by Hadamard gates: Pj \u2192 P \u2261 j H \u22973PjH \u22973. Equivalently, syndrome measurement may be performed by measuring the observables H \u22973Z1Z2H \u22973 = X1X2 and H \u22973Z2Z3H \u22973 = X2X3. It is interesting to interpret these measurements along similar lines to the measurement of Z1Z2 and Z2Z3 for the bit \ufb02ip code. Measurement of the observables X1X2 and X2X3 corresponds to comparing the sign of qubits one and two, and two and three, respectively, in the sense that measurement of X1X2, for example, gives +1 for states like |+|+ \u2297 (\u00b7) or |\u2212|\u2212 \u2297 (\u00b7), and \u22121 for states like |+|\u2212 \u2297 (\u00b7) or |\u2212|+ \u2297 (\u00b7). Finally, error-correction is completed with the recovery operation, which is the Hadamard-conjugated recovery operation from the bit \ufb02ip code. For example, suppose we detected a \ufb02ip in the sign of the \ufb01rst qubit from |+ to |\u2212. Then we recover by applying HX1H = Z1 to the \ufb01rst qubit. Similar procedures apply for other error syndromes. Obviously this code for the phase \ufb02ip channel has the same characteristics as the code for the bit \ufb02ip channel. In particular, the minimum \ufb01delity for the phase \ufb02ip code is the same as that for the bit \ufb02ip code, and we have the same criteria for the code producing an improvement over the case with no error-correction. We say that these two channels are unitarily equivalent, since there is a unitary operator U (in this case the Hadamard gate) such that the action of one channel is the same as the other, provided the \ufb01rst channel is preceded by U and followed by U \u2020. These operations may be trivially incorporated into the encoding and error-correction operations. For general unitary operators these ideas are worked out in Problem 10.1 on page 495. 432 Quantum error-correction Exercise 10.4: Consider the three qubit bit \ufb02ip code. Suppose we had performed the error syndrome measurement by measuring the eight orthogonal projectors corresponding to projections onto the eight computational basis states. (1) Write out the projectors corresponding to this measurement, and explain how the measurement result can be used to diagnose the error syndrome: either no bits \ufb02ipped or bit number j \ufb02ipped, where j is in the range one to three. (2) Show that the recovery procedure works only for computational basis states. (3) What is the minimum \ufb01delity for the error-correction procedure? 10.2 The Shor", " code There is a simple quantum code which can protect against the effects of an arbitrary error on a single qubit! The code is known as the Shor code, after its inventor. The code is a combination of the three qubit phase \ufb02ip and bit \ufb02ip codes. We \ufb01rst encode the qubit using the phase \ufb02ip code: |0 \u2192 | + ++, |1 \u2192 | \u2212 \u2212\u2212. Next, we encode each of these 2 and |\u2212 is qubits using the three qubit bit \ufb02ip code: |+ is encoded as (|000 + |111) encoded as (|000 \u2212 |111) 2. The result is a nine qubit code, with codewords given by: \u221a \u221a |0 \u2192 |0L \u2261 (|000 + |111)(|000 + |111)(|000 + |111) |1 \u2192 |1L \u2261 (|000 \u2212 |111)(|000 \u2212 |111)(|000 \u2212 |111) \u221a 2 \u221a 2 2 2. (10.13) Figure 10.4. Encoding circuit for the Shor nine qubit code. Some of the |0 states appear indented, simply to emphasize the concatenated nature of the encoding.. The quantum circuit encoding the Shor code is shown in Figure 10.4. As described above, the \ufb01rst part of the circuit encodes the qubit using the three qubit phase \ufb02ip code; The Shor code 433 comparison with Figure 10.3 shows that the circuits are identical. The second part of the circuit encodes each of these three qubits using the bit \ufb02ip code, using three copies of the bit \ufb02ip code encoding circuit, Figure 10.2. This method of encoding using a hierarchy of levels is known as concatenation. It\u2019s a great trick for obtaining new codes from old, and we use it again later to prove some important results about quantum error-correction. The Shor code is able to protect against phase \ufb02ip and bit \ufb02ip errors on any qubit. To see this, suppose a bit \ufb02ip occurs on the \ufb01rst qubit. As for the bit \ufb02ip code, we perform a measurement of Z1Z2 comparing the \ufb01rst two qubits, and \ufb01nd that they are different. We conclude that a bit \ufb02ip error occurred on the \ufb01rst or second qubit. Next we compare the second and third qubit by performing a measurement of Z2Z3. We \ufb01nd that they are the same, so it could not have been the second qubit which \ufb02ipped. We conclude that the \ufb01rst qubit must have \ufb02ipped, and recover from the error by \ufb02ipping the \ufb01rst qubit again, back to its original state. In a similar way we can detect and recover from the effects of bit \ufb02ip errors on any of the nine qubits in the code. We cope in a similar manner with phase \ufb02ips on the qubits. Suppose a phase \ufb02ip occurs on the \ufb01rst qubit. Such a phase \ufb02ip \ufb02ips the sign of the \ufb01rst block of qubits, changing |000 + |111 to |000 \u2212 |111, and vice versa. Indeed, a phase \ufb02ip on any of the \ufb01rst three qubits has this effect, and the error-correction procedure we describe works for any of these three possible errors. Syndrome measurement begins by comparing the sign of the \ufb01rst and second blocks of three qubits, just as syndrome measurement for the phase \ufb02ip code began by comparing the sign of the \ufb01rst and second qubits. For example, (|000 \u2212 |111)(|000 \u2212 |111) has the same sign (\u2212) in both blocks of qubits, while (|000 \u2212 |111)(|000 + |111) has different signs. When a phase \ufb02ip occurs on any of the \ufb01rst three qubits we \ufb01nd that the signs of the \ufb01rst and second blocks are different. The second and \ufb01nal stage of syndrome measurement is to compare the sign of the second and third blocks of qubits. We \ufb01nd that these are the same, and conclude that the phase must have \ufb02ipped in the \ufb01rst block of three qubits. We recover from this by \ufb02ipping the sign in the \ufb01rst block of three qubits back to its original value. We can recover from a phase \ufb02ip on any of the nine qubits in a", " similar manner. Exercise 10.5: Show that the syndrome measurement for detecting phase \ufb02ip errors in the Shor code corresponds to measuring the observables X1X2X3X4X5X6 and X4X5X6X7X8X9. Exercise 10.6: Show that recovery from a phase \ufb02ip on any of the \ufb01rst three qubits may be accomplished by applying the operator Z1Z2Z3. Suppose both bit and phase \ufb02ip errors occur on the \ufb01rst qubit, that is, the operator Z1X1 is applied to that qubit. Then it is easy to see that the procedure for detecting a bit \ufb02ip error will detect a bit \ufb02ip on the \ufb01rst qubit, and correct it, and the procedure for detecting a phase \ufb02ip error will detect a phase \ufb02ip on the \ufb01rst block of three qubits, and correct it. Thus, the Shor code also enables the correction of combined bit and phase \ufb02ip errors on a single qubit. Indeed, the Shor code protects against much more than just bit and phase \ufb02ip errors on a single qubit \u2013 we now show that it protects against completely arbitrary errors, provided they only affect a single qubit! The error can be tiny \u2013 a rotation about the z axis of the Bloch sphere by \u03c0/263 radians, say \u2013 or it can be an apparently disastrous error like removing the qubit entirely and replacing it with garbage! The interesting thing 434 Quantum error-correction is, no additional work needs to be done in order to protect against arbitrary errors \u2013 the procedure already described works just \ufb01ne. This is an example of the extraordinary fact that the apparent continuum of errors that may occur on a single qubit can all be corrected by correcting only a discrete subset of those errors; all other possible errors being corrected automatically by this procedure! This discretization of the errors is central to why quantum error-correction works, and should be regarded in contrast to classical error-correction for analog systems, where no such discretization of errors is possible. To simplify the analysis, suppose noise of an arbitrary type is occurring on the \ufb01rst qubit only; we\u2019ll come back to what happens when noise is affecting other qubits as well. Following Chapter 8 we describe the noise by a trace-preserving quantum operation E. It is most convenient to analyze error-correction by expanding E in an operator-sum representation with operation elements {Ei}. Supposing the state of the encoded qubit is |\u03c8 = \u03b1|0L + \u03b2|1L before the noise acts, then after the noise has acted the state is E(|\u03c8\u03c8|) = i. To analyze the effects of error-correction it\u2019s easiest to focus on the effect error-correction has on a single term in this sum, say Ei|\u03c8\u03c8|E\u2020 i. As an operator on the \ufb01rst qubit alone Ei may be expanded as a linear combination of the identity, I, the bit \ufb02ip, X1, the phase \ufb02ip, Z1, and the combined bit and phase \ufb02ip, X1Z1: i Ei|\u03c8\u03c8|E\u2020 Ei = ei0I + ei1X1 + ei2Z1 + ei3X1Z1. (10.14) The (un-normalized) quantum state Ei|\u03c8 can thus be written as a superposition of four terms, |\u03c8, X1|\u03c8, Z1|\u03c8, X1Z1|\u03c8. Measuring the error syndrome collapses this superposition into one of the four states |\u03c8, X1|\u03c8, Z1|\u03c8 or X1Z1|\u03c8 from which recovery may then be performed by applying the appropriate inversion operation, resulting in the \ufb01nal state |\u03c8. The same is true for all the other operation elements Ei. Thus, errorcorrection results in the original state |\u03c8 being recovered, despite the fact that the error on the \ufb01rst qubit was arbitrary. This is a fundamental and deep fact about quantum error-correction, that by correcting just a discrete set of errors \u2013 the bit \ufb02ip, phase \ufb02ip, and combined bit\u2013phase \ufb02ip, in this example \u2013 a quantum error-correcting code is able to automatically correct an apparently much larger (continuous!) class of errors. What happens when noise is affecting more than just the \ufb01rst qubit? Two basic ideas are used to cope with this. First, in many situations it", " is a good approximation to assume that the noise acts on qubits independently. Provided the effect of the noise on one qubit is fairly small, we can expand the total effect of the noise as a sum over terms involving errors on no qubits, on a single qubit, on two qubits, and so on, with the terms with errors on no qubits and on a single qubit dominating the higher order terms. Performing error-correction results in the zeroth and \ufb01rst order terms being corrected properly, and leaves only the much smaller second and higher order errors, achieving a net suppression of error. A more detailed analysis of this idea will be given later. Sometimes, of course, it is not reasonable to assume that the noise acts on qubits independently. When this occurs we use a different idea \u2013 error-correcting codes which can correct errors on more than a single qubit. Such codes may be constructed along similar lines to the Shor code, and we explain the basic ideas behind how this can be done later in this chapter. Theory of quantum error-correction 435 10.3 Theory of quantum error-correction Can we construct a general theory of quantum error-correcting codes? This section develops a general framework for studying quantum error-correction, including the quantum error-correction conditions, a set of equations which must be satis\ufb01ed if quantum errorcorrection is to be possible. Of course, possessing such a framework doesn\u2019t guarantee us that good quantum error-correcting codes exist \u2013 that topic is taken up in Section 10.4 on page 445! But it does provide background to enable us to \ufb01nd good quantum errorcorrecting codes. The basic ideas of the theory of quantum error-correction generalize in a natural way the ideas introduced by the Shor code. Quantum states are encoded by a unitary operation into a quantum error-correcting code, formally de\ufb01ned as a subspace C of some larger Hilbert space. It is useful to have a notation for the projector onto the code space C, so we use the notation P ; for the three qubit bit \ufb02ip code P = |000000| + |111111|. After encoding the code is subjected to noise, following which a syndrome measurement is performed to diagnose the type of error which occurred, that is, the error syndrome. Once this has been determined, a recovery operation is performed, to return the quantum system to the original state of the code. The basic picture is illustrated in Figure 10.5: different error syndromes correspond to undeformed and orthogonal subspaces of the total Hilbert space. The subspaces must be orthogonal, otherwise they couldn\u2019t be reliably distinguished by the syndrome measurement. Furthermore, the different subspaces must be undeformed versions of the original code space, in the sense that the errors mapping to the different subspaces must take the (orthogonal) codewords to orthogonal states, in order to be able to recover from the error. This intuitive picture is essentially the content of the quantum error-correction conditions discussed below. To develop a general theory of quantum error-correction it behooves us to make as few assumptions as possible about the nature of noise and about the procedure used to do the error-correction. That is, we won\u2019t necessarily assume that the error-correction is done via a two-stage detection\u2013recovery method, and we won\u2019t make any assumptions about the noise occurring for qubit systems, or being weak. Instead, we just make two very broad assumptions: the noise is described by a quantum operation E, and the complete error-correction procedure is effected by a trace-preserving quantum operation R, which we call the error-correction operation. This error-correction operation bundles into one piece the two steps that we have called error-detection and recovery above. In order for error-correction to be deemed successful, we require that for any state \u03c1 whose support lies in the code C, (R \u25e6 E) (\u03c1) \u221d \u03c1. (10.15) You may be wondering why we wrote \u221d rather than = in the last equation. If E were a trace-preserving quantum operation then by taking traces of both sides of the equation we see that \u221d would be =. However, sometimes we may be interested in error-correcting non-trace-preserving operations E, such as measurements, for which \u221d is appropriate. Of course, the error-correction step R must succeed with probability 1, which is why we required R to be trace-preserving. The quantum error-correction conditions are a simple set of equations which can be checked to determine whether a quantum error-correcting code protects against a particular type of noise E. We will use these conditions to construct a plethora of quantum 436 Quantum error", "-correction (A) (B) Figure 10.5. The packing of Hilbert spaces in quantum coding: (A) bad code, with non-orthogonal, deformed resultant spaces, and (B) good code, with orthogonal (distinguishable), undeformed spaces. codes, and also to investigate some of the general properties of quantum error-correcting codes. Theorem 10.1: (Quantum error-correction conditions) Let C be a quantum code, and let P be the projector onto C. Suppose E is a quantum operation with operation elements {Ei}. A necessary and suf\ufb01cient condition for the existence of an error-correction operation R correcting E on C is that P E\u2020 i EjP = \u03b1ijP, (10.16) for some Hermitian matrix \u03b1 of complex numbers. We call the operation elements {Ei} for the noise E errors, and if such an R exists we say that {Ei} constitutes a correctable set of errors. Proof We prove suf\ufb01ciency \ufb01rst, by constructing an explicit error-correction operation R whenever (10.16) is satis\ufb01ed. The construction we use is of the two-part form used for the Shor code \u2013 error-detection and then recovery \u2013 so the proof also shows that error-correction can always be accomplished using such a two-part procedure. Suppose {Ei} is a set of operation elements satisfying the quantum error-correction conditions, (10.16). By assumption \u03b1 is a Hermitian matrix, and thus can be diagonalized, d = u\u2020\u03b1u, where u is a unitary matrix and d is diagonal. De\ufb01ne operators Fk \u2261 i uikEi. Recalling Theo- Theory of quantum error-correction 437 rem 8.2, we see that {Fk} is also a set of operation elements for E. By direct substitution, kiujlP E\u2020 u\u2020 k FlP = i EjP. (10.17) P F \u2020 Substituting (10.16) simpli\ufb01es this to P F \u2020 we obtain k FlP = ij ij u\u2020 ki\u03b1ijujlP and since d = u\u2020\u03b1u P F \u2020 k FlP = dklP, (10.18) which can be thought of as a simpli\ufb01cation of the quantum error-correction conditions (10.16), because dkl is diagonal. We use the simpli\ufb01ed conditions (10.18) and the polar decomposition (Section 2.1.10 on page 78) to de\ufb01ne the syndrome measurement. From the polar decomposition, we k FkP = see that FkP = Uk dkkUkP for some unitary Uk. The effect of Fk is therefore to rotate the coding subspace into the subspace de\ufb01ned by the projector Pk \u2261 UkP U \u2020 k = FkP U \u2020 dkk. Equation (10.18) implies that these subspaces are orthogonal, k / since when k = l, \u221a ( P F \u2020 \u221a PlPk = P \u2020 l Pk = UlP F \u2020 \u221a l FkP U \u2020 dlldkk k = 0. (10.19) k Pk = I. Recovery is accomplished simply by applying U \u2020 The syndrome measurement is a projective measurement de\ufb01ned by the projectors Pk, augmented by an additional projector if necessary to satisfy the completeness relation k. To see that this errorcorrection procedure works, note that the combined detection\u2013recovery step corresponds to the quantum operation R(\u03c3) = kPk\u03c3PkUk. For states \u03c1 in the code, simple algebra and the de\ufb01nitions show that: \u221a k U \u2020 \u221a U \u2020 kPkFl \u221a \u03c1 kP \u2020 \u03c1 = U \u2020 k FlP \u03c1 kUkP F \u2020 U \u2020 k FlP \u221a dkk $ \u221a = = \u03b4kl = \u03b4kl $ \u03c1 dkkP \u221a dkk \u03c1. Thus R(E(\u03c1)) = kl kPkFl\u03c1F \u2020 U \u2020 l PkUk = \u03b4kldkk\u03c1 kl \u221d \u03c1, (10.20) (10.21) (10.22) (10.23) (10.24) (10.25) (10.26) as required. To prove necessity of the quantum error-correction conditions (10.16), suppose {Ei} is a set of errors which is perfectly correctable by an error-correction", " operation R with operation elements {Rj}. De\ufb01ne a quantum operation EC by EC(\u03c1) \u2261 E(P \u03c1P ). Since P \u03c1P is in the code space for all \u03c1, it follows that R(EC(\u03c1)) \u221d P \u03c1P, (10.27) for all states \u03c1. Moreover, the proportionality factor must be a constant c, not depending 438 Quantum error-correction on \u03c1, if both right and left hand sides are to be linear. Rewriting the last equation explicitly in terms of the operation elements gives RjEiP \u03c1P E\u2020 i R\u2020 j = cP \u03c1P. (10.28) ij This equation holds for all \u03c1. It follows that the quantum operation with operation elements {RjEi} is identical to the quantum operation with a single operation element \u221a cP. Theorem 8.2 implies that there exist complex numbers cki such that RkEiP = ckiP. k = c\u2217 Taking the adjoint of this equation gives P E\u2020 i R\u2020 c\u2217 kickjP. But R is a trace-preserving operation, so P E\u2020 kickjP over k we deduce that P E\u2020 kRkEjP = c\u2217 i R\u2020 i EjP = \u03b1ijP, k c\u2217 where \u03b1ij \u2261 quantum error-correction conditions. kiP and therefore P E\u2020 k R\u2020 kRkEjP = kRk = I. Summing the equation i R\u2020 (10.29) (10.30) kickj is just a Hermitian matrix of complex numbers. These are the Direct veri\ufb01cation of the quantum error-correction conditions is an easy but timeconsuming task. In Sections 10.4 and 10.5 we describe a theoretical formalism which uses the quantum error-correction conditions as a launching point for the construction of many interesting classes of codes, and which circumvents much of the dif\ufb01culty associated with verifying the quantum error-correction conditions directly. For now, you should work through the following example, which demonstrates the quantum error-correction conditions in action: Exercise 10.7: Consider the three qubit bit \ufb02ip code of Section 10.1.1, with $ $ (1 \u2212 p)3I, corresponding projector P = |000000| + |111111|. The noise process this code protects against has operation elements $ p(1 \u2212 p)2X2, p(1 \u2212 p)2X1, { probability that a bit \ufb02ips. Note that this quantum operation is not trace-preserving, since we have omitted operation elements corresponding to bit \ufb02ips on two and three qubits. Verify the quantum error-correction conditions for this code and noise process. p(1 \u2212 p)2X3}, where p is the $ 10.3.1 Discretization of the errors We have discussed the protection of quantum information against a speci\ufb01c noise process E. In general, however, we don\u2019t know exactly what noise is af\ufb02icting a quantum system. It would be useful if a speci\ufb01c code C and error-correction operation R could be used to protect against an entire class of noise processes. Fortunately, the quantum errorcorrection conditions are easily adapted to provide exactly this sort of protection. Theorem 10.2: Suppose C is a quantum code and R is the error-correction operation constructed in the proof of Theorem 10.1 to recover from a noise process E with operation elements {Ei}. Suppose F is a quantum operation with operation elements {Fj} which are linear combinations of the Ei, that is Fj = i mjiEi for some matrix mji of complex numbers. Then the error-correction operation R also corrects for the effects of the noise process F on the code C. Theory of quantum error-correction 439 Box 10.1: Quantum error-correction without measurement In the main text we have described quantum error-correction as a two stage process: an error-detection step effected using quantum measurement, followed by a recovery step effected using conditioned unitary operations. It is possible to perform quantum error-correction using only unitary operations and ancilla systems prepared in standard states. The advantage of knowing how to do this is that for some real-world quantum systems it is very dif\ufb01cult to perform the quantum measurements needed for quantum error-correction, so an alternate procedure is needed. The techniques we use to do that are essentially the same as those described in Chapter 8 for mocking up an arbitrary quantum operation; we now recap the basic idea in the context of quantum error-correction. Suppose the syndrome measurement on the", " principal system \u2013 the one being errorcorrected \u2013 is described by measurement operators Mi, and the corresponding conditional unitary operation is Ui. Introduce an ancilla system with basis states |i corresponding to the possible error syndromes. The ancilla starts in a standard pure state |0 before error-correction. De\ufb01ne a unitary operator U on the principal system plus ancilla by This can be extended to a unitary operator acting on the whole space since U |\u03c8|0 \u2261 (UiMi|\u03c8)|i. i \u03d5|0|U \u2020U |\u03c8|0 = \u03d5|M \u2020 i Mj|\u03c8\u03b4ij ij = \u03d5|M \u2020 i Mi|\u03c8 i = \u03d5|\u03c8. (10.31) (10.32) (10.33) (10.34) i UiMi\u03c3M \u2020 That is, U preserves inner products, and can thus be extended to a unitary operator on the entire state space. The effect of U is to effect the transformation R(\u03c3) = i on the system being error-corrected, exactly the same quantum operation as described in the main text for the performance of quantum errorcorrection. Note that in order for this error-correction procedure to work it is necessary to use a fresh ancilla each time error-correction is performed. i U \u2020 Proof By Theorem 10.1 the operation elements {Ei} must satisfy the quantum error-correction conditions, P EiE\u2020 j P = \u03b1ijP. As shown in the proof of Theorem 10.1, without loss of generality we may assume that the operation elements for E have been chosen such that \u03b1ij = dij is diagonal with real entries. The error-correction operation R has operation elements U \u2020 kPk, where by Equation (10.23) the Uk and Pk are chosen such that for any \u03c1 in the code space: \u221a U \u2020 kPkEi $ \u03c1 = \u03b4ki dkk \u221a \u03c1. (10.35) 440 Quantum error-correction Substituting Fj = i mjiEi gives U \u2020 k PkFj \u221a \u03c1 = $ dkk \u221a \u03c1 mji\u03b4ki $ \u221a dkk \u03c1, i = mjk and thus as required. R(F (\u03c1)) = kj = kj \u221d \u03c1, k PkFj\u03c1F \u2020 U \u2020 j PkUk |mjk|2dkk\u03c1 (10.36) (10.37) (10.38) (10.39) (10.40) This result enables the introduction of a more powerful language to describe quantum error-correcting codes. Instead of talking about the class of error processes E correctable by a code C and error-correction operation R we can talk about a set of error operators (or simply errors) {Ei} which are correctable. By this, we mean that the quantum errorcorrection conditions hold for these operators, P EiE\u2020 j P = \u03b1ijP. (10.41) Theorems 10.1 and 10.2 together imply that any noise process E whose operation elements are built from linear combinations of these error operators {Ei} will be corrected by the recovery operation R! Let\u2019s look at an example of this powerful new viewpoint in action. Suppose E is a quantum operation acting on a single qubit. Then its operation elements {Ei} can each be written as a linear combination of the Pauli matrices \u03c30, \u03c31, \u03c32, \u03c33. Therefore, to check that the Shor code corrects against arbitrary single qubit errors on the \ufb01rst qubit it is suf\ufb01cient to verify that the equations P \u03c31 i \u03c31 jP = \u03b1ijP, (10.42) are satis\ufb01ed, where \u03c31 i are the Pauli matrices (I, X, Y and Z) acting on the \ufb01rst qubit. Once this is done it is assured that any error process whatsoever on the \ufb01rst qubit may be corrected! (The actual calculation is quite simple, and is done as part of Exercise 10.10.) Indeed this example explains a point that can be somewhat mysterious when \ufb01rst confronted with the literature on quantum error-correction: many authors have what appears to be a suspicious fondness for the depolarizing channel, E(\u03c1) = (1 \u2212 p)\u03c1 + p 3 (X\u03c1X + Y \u03c1Y + Z\u03c1Z). It is tempting to assume that this greatly limits the validity of their models of error-", "correction, but this is not so, for as the discussion just now implies, the ability to error-correct the depolarizing channel automatically implies the ability to error-correct an arbitrary single qubit quantum operation. Summarizing, we have learnt that it is possible to discretize quantum errors, that to \ufb01ght the continuum of errors possible on a single qubit it is suf\ufb01cient merely to win the war against a \ufb01nite set of errors, the four Pauli matrices. Similar results hold for higherdimensional quantum systems. This stands in remarkable contrast to the theory of errorcorrection for analog classical systems. Error-correction in such systems is extremely dif\ufb01cult because in principle there are an in\ufb01nite number of different error syndromes. Theory of quantum error-correction 441 Digital error-correction for classical information processing is much more successful because it involves only a \ufb01nite number of error syndromes. The surprising thing we have learnt is that quantum error-correction seems much more similar to classical digital error-correction than it is to classical analog error-correction. Exercise 10.8: Verify that the three qubit phase \ufb02ip code |0L = | + ++, |1L = | \u2212 \u2212\u2212 satis\ufb01es the quantum error-correction conditions for the set of error operators {I, Z1, Z2, Z3}. Exercise 10.9: Again, consider the three qubit phase \ufb02ip code. Let Pi and Qi be the projectors onto the |0 and |1 states, respectively, of the ith qubit. Prove that the three qubit phase \ufb02ip code protects against the error set {I, P1, Q1, P2, Q2, P3, Q3}. Exercise 10.10: Explicitly verify the quantum error-correction conditions for the Shor code, for the error set containing I and the error operators Xj, Yj, Zj for j = 1 through 9. Exercise 10.11: Construct operation elements for a single qubit quantum operation E that upon input of any state \u03c1 replaces it with the completely randomized state I/2. It is amazing that even such noise models as this may be corrected by codes such as the Shor code! 10.3.2 Independent error models How can we make the connection between quantum error-correction and the criteria for doing reliable quantum information processing introduced in Chapter 9? In this section we explain the basic idea of how this may be done using the assumption of independent errors on different qubits. Intuitively, if a noise process acts independently on the different qubits in the code, then provided the noise is suf\ufb01ciently weak error-correction should improve the storage \ufb01delity of the encoded state over the unencoded state. To illustrate this, we begin with the example of the depolarizing channel, which provides an especially simple demonstration of the basic ideas, and then broaden the ideas to include other important channels. Recall that the depolarizing channel may be described by a single parameter, a probability p. The action of the depolarizing channel on a single qubit is de\ufb01ned by the equation E(\u03c1) = (1 \u2212 p)\u03c1 + p/3[X\u03c1X + Y \u03c1Y + Z\u03c1Z], and can be interpreted as saying that nothing happens to the qubit with probability 1 \u2212 p, and each of the operators X, Y and Z is applied to the qubit with probability p/3. The depolarizing channel is especially easy to analyze in the context of quantum error-correction because it has such a nice interpretation in terms of the four basic errors I, X, Y and Z which are most commonly used in the analysis of quantum codes. We\u2019ll explain how this analysis is performed, and then return to the question of what happens when we consider a process which doesn\u2019t have such a simple interpretation in terms of the I, X, Y and Z operations. A simple calculation shows that the minimum \ufb01delity for states sent through a depolarizing channel is given by F = 1 \u2212 2p/3 = 1 \u2212 p/3 + O(p2). $ Exercise 10.12: Show that the \ufb01delity between the state |0 and E(|00|) is 442 Quantum error-correction $ channel is 1 \u2212 2p/3. 1 \u2212 2p/3, and use this to argue that the minimum \ufb01delity for the depolarizing $ Suppose we encode a single qubit of information in an n qubit quantum code which corrects errors on any single qubit. Suppose the depolarizing channel with parameter p acts independently on each of the qubits, giving rise to a joint", " action on all n qubits of E \u2297n(\u03c1) = (1 \u2212 p)n\u03c1 + n 3 j=1 k=1 (1 \u2212 p)n\u22121 p 3 k \u03c1 \u03c3j \u03c3j k + \u00b7 \u00b7 \u00b7, (10.43) where the \u2018...\u2019 indicates higher-order terms which are positive and will drop out of the analysis. After error-correction has been performed all terms appearing in this sum will be returned to the state \u03c1, provided \u03c1 was in the code originally, R \u2297 E \u2297n & (1 \u2212 p)n + n(1 \u2212 p)n\u22121p \u2019 (\u03c1) = \u03c1 + \u00b7 \u00b7 \u00b7, (10.44) so the \ufb01delity satis\ufb01es $ (1 \u2212 p)n\u22121(1 \u2212 p + np) = 1 \u2212 F \u2265 n 2 2 p2 + O(p3). (10.45) Thus, provided the probability of error p is suf\ufb01ciently small, using the quantum errorcorrecting code leads to an improvement in the \ufb01delity of the quantum states being protected by the code. Not all noisy channels can be interpreted so easily as a random combination of no error, bit \ufb02ips, phase \ufb02ips and combinations of the two. Many naturally occurring quantum channels have no such interpretation. Consider the example of amplitude damping (Section 8.3.5), which has operation elements E0 and E1: E0 = 1 0 \u221a 0 1 \u2212 \u03b3 ; E1 = \u221a \u03b3 0 0 0. (10.46) The parameter \u03b3 is a small positive parameter characterizing the strength of the amplitude damping process \u2013 as \u03b3 gets closer to zero, the strength decreases, until ultimately we end up with an essentially noise-free quantum channel. We might reasonably guess that the amplitude damping channel has an equivalent description in terms of a set of operation elements including a term proportional to the identity, {f (\u03b3)I, E 2,...}, where f (\u03b3) \u2192 1 as \u03b3 \u2192 0. If this were the case then an analysis of error-correction for the amplitude damping channel acting independently on multiple qubits could be done that was similar to the analysis of error-correction performed for the depolarizing channel. Surprisingly, it turns out that no such description is possible! This follows from Theorem 8.2 simply because for \u03b3 > 0 no linear combination of E0 and E1 can ever be proportional to the identity, and thus no set of operation elements for the amplitude damping channel can ever include a term proportional to the identity. 1, E Similarly, many other noise processes in quantum mechanics are close to the identity in a physical sense, yet no operator-sum representation for the process contains a large identity component. Intuitively it seems reasonable that in such a circumstance errorcorrection should result in a net gain in the storage \ufb01delity for quantum information, provided the noise is suf\ufb01ciently weak. We will now show that this is in fact the case, using the speci\ufb01c example of the amplitude damping channel for concreteness. A simple calculation shows that the minimum \ufb01delity for the amplitude damping channel applied Theory of quantum error-correction 443 \u221a 1 \u2212 \u03b3. Suppose the qubit is encoded in an n qubit quantum code to a single qubit is capable of correcting arbitrary errors on a single qubit, and that amplitude damping channels of parameter \u03b3 act independently on each qubit. We will sketch the basic idea showing that the effect of quantum error-correction is to change the \ufb01delity of storage to 1 \u2212 O(\u03b32), so for small \u03b3 encoding the qubit in the quantum code results in a net suppression of error. Exercise 10.13: Show that the minimum \ufb01delity F (|\u03c8, E(|\u03c8\u03c8|)) when E is the amplitude damping channel with parameter \u03b3, is \u221a 1 \u2212 \u03b3. Using Ej,k to denote the action of Ej on the kth qubit, the effect of the noise on the encoded qubits may be written: E \u2297n(\u03c1) = E0,1 \u2297 E0,2 \u2297 \u00b7 \u00b7 \u00b7 \u2297 E0,n \u23a1 \u239b + n j=1 \u23a2 \u23a3E1,(\u03b32). \u03c1 E\u2020 0,1 \u23a4 \u239e \u2297 E\u2020 0,2 \u23a1 \u2297 \u00b7 \u00b7 \u00b7 \u2297 E\u2020 0,n \u239b E0,k \u239f \u23a0 \u23a5", " \u23a6 \u03c1 \u23a2 \u23a3E\u2020 1,j \u239c \u239d \u2297 * k = j E\u2020 0,k \u239e \u23a4 \u239f \u23a0 \u23a5 \u23a6 (10.47) \u221a \u03b3(X +iY )/2. Substituting Suppose we write E0 = (1\u2212\u03b3/4)I +\u03b3Z/4+O(\u03b32), and E1 = these expressions in (10.47) gives E \u2297n(\u03c1) = 1 \u2212 \u03b3 4 2n 2n\u22122 n j=1 2n\u22121 n Zj\u03c1 + \u03c1Zj j=1 Xj + iYj Xj \u2212 iYj \u03c1 + O(\u03b32). (10.48) Suppose \u03c1 is a state of the code. Obviously, the effect of error-correction on \u03c1 is to leave it invariant! The effect of error-correction on terms like Zj\u03c1 and \u03c1Zj is most easily understood by considering the effect on Zj|\u03c8\u03c8|, where |\u03c8 is a state of the code. We suppose the code is such that the error Zj takes |\u03c8 to a subspace which is orthogonal to the code, so that when the syndrome measurement is performed terms like Zj|\u03c8\u03c8| disappear. (Note that even if this orthogonality assumption is not made a similar analysis can still be made, by working in terms of error operators which do take the code to orthogonal subspaces.) Thus terms like Zj\u03c1 vanish after error-correction, as do terms like \u03c1Zj, Xj\u03c1Yj and Yj\u03c1Xj. Furthermore, error-correction takes Xj\u03c1Xj and Yj\u03c1Yj back to \u03c1, since the code can correct errors on one qubit. Thus, after error-correction the state of the system is 2n 1 \u2212 \u03b3 4 \u03b3 4 2n\u22122 \u03b3 4 \u03c1 + 2n 1 \u2212 \u03c1 + O(\u03b32) = \u03c1 + O(\u03b32). (10.49) Thus, to order \u03b32 error-correction returns the quantum system to its original state \u03c1, and for weak noise (small \u03b3) error-correction results in a net suppression of errors, just as for the depolarizing channel. Our analysis here was for the amplitude damping noise model, but it is not dif\ufb01cult to generalize this argument to obtain similar conclusions for other noise models. In general, however, for the remainder of this chapter we work mainly with noise models which can be understood as stochastic application of errors corresponding 444 Quantum error-correction to the Pauli matrices, similar to the depolarizing channel, allowing us to do the analysis using familiar concepts from classical probability theory. Keep in mind that the ideas we describe can be extended beyond this simple error model to apply to a much wider range of error models using principles similar to those we have just outlined. 10.3.3 Degenerate codes In many respects quantum error-correcting codes are quite similar to classical codes \u2013 an error is identi\ufb01ed by measuring the error syndrome, and then corrected as appropriate, just as in the classical case. There is, however, an interesting class of quantum codes known as degenerate codes possessing a striking property unknown in classical codes. The idea is most easily illustrated for the case of the Shor code. Consider the effect of the errors Z1 and Z2 on the codewords for the Shor code. As we have already noted, the effect of these errors is the same on both codewords. For classical error-correcting codes errors on different bits necessarily lead to different corrupted codewords. The phenomenon of degenerate quantum codes is a sort of good news\u2013bad news situation for quantum codes. The bad news is that some of the proof techniques used classically to prove bounds on error-correction fall down because they can\u2019t be applied to degenerate codes. We\u2019ll see an example of this in the next section and the quantum Hamming bound. The good news is that degenerate quantum codes seem to be among the most interesting quantum codes! In some sense they are able to \u2018pack more information in\u2019 than are classical codes, because distinct errors do not necessarily have to take the code space to orthogonal spaces, and it is possible (though has not yet been shown) that this extra ability may lead to degenerate codes that can store quantum information more ef\ufb01ciently than any non-degenerate code. 10.3.4 The quantum Hamming bound For applications we would like to make use of the \u2018best\u2019 possible quantum codes. What \ufffd", "\ufffdbest\u2019 means in any given circumstance depends on the application. For this reason we would like to have criteria for determining whether or not a code with particular characteristics exists or not. In this section we develop the quantum Hamming bound, a simple bound which gives some insights into the general properties of quantum codes. Unfortunately the quantum Hamming bound only applies to non-degenerate codes, but it gives us an idea of what more general bounds may look like. Suppose a non-degenerate code is used to encode k qubits in n qubits in such a way that it can correct errors on any subset of t or fewer qubits. Suppose j errors occur, where j \u2264 t. There are n j sets of locations where errors may occur. With each such set of locations there are three possible errors \u2013 the three Pauli matrices X, Y, Z \u2013 that may occur in each qubit, for a total of 3j possible errors. The total number of errors that may occur on t or fewer qubits is therefore t j=0 3j. n j (10.50) (Note that j = 0 corresponds to the case of no errors on any qubit, the \u2018error\u2019 I.) In order to encode k qubits in a non-degenerate way each of these errors must correspond to an orthogonal 2k-dimensional subspace. All of these subspaces must be \ufb01tted into the Constructing quantum codes 445 total 2n-dimensional space available to n qubits, leading to the inequality t j=0 n j 3j2k \u2264 2n, (10.51) which is the quantum Hamming bound. Consider, for example, the case where we wish to encode one qubit in n qubits in such a way that errors on one qubit are tolerated. In this case the quantum Hamming bound reads: 2(1 + 3n) \u2264 2n. (10.52) Substitution shows that this inequality is not satis\ufb01ed for n \u2264 4, while it is for values of n \u2265 5. Therefore, there is no non-degenerate code encoding one qubit in fewer than \ufb01ve qubits in such a way as to protect from all possible errors on a single qubit. Of course, not all quantum codes are non-degenerate, so the quantum Hamming bound is more useful as a rule of thumb than as a hard and fast bound on the existence of quantum codes. (In particular, at the time of writing no codes are known that violate the quantum Hamming bound, even allowing degenerate codes.) Later, we will have occasion to look at some bounds on quantum codes that apply to all quantum codes, not just non-degenerate codes. For example, in Section 12.4.3 we prove the quantum Singleton bound, which implies that any quantum code encoding k qubits in n qubits and able to correct errors on any t qubits must satisfy n \u2265 4t+k. It follows that the smallest code encoding a single qubit and able to correct an arbitrary error on a single qubit must satisfy n \u2265 4 + 1 = 5, and indeed, we will soon exhibit such a \ufb01ve qubit code. 10.4 Constructing quantum codes We now have a theoretical framework for the study of quantum error-correcting codes, but we don\u2019t as yet have many examples of such codes! We begin to remedy this defect by taking a brief tour into the theory of classical linear codes in Section 10.4.1, and then in Section 10.4.2 explain how ideas from classical linear codes can be used to construct a large class of quantum codes known as Calderbank\u2013Shor\u2013Steane (CSS) codes. Our job is completed in Section 10.5 with the development of the stabilizer codes, a class of codes even more general than the CSS codes which offers a powerful means for constructing a wide variety of quantum codes. 10.4.1 Classical linear codes Classical error-correcting codes have many varied and important technological applications, so it is unsurprising that a powerful theory of such codes has been developed. Our interest in the techniques of classical error-correction is that many of these techniques have important implications for quantum error-correction, especially the theory of classical linear codes, which can be used to develop a wide variety of good quantum error-correcting codes. In this section we review classical linear codes, emphasizing especially those ideas important to quantum error-correction. A linear code C encoding k bits of information into an n bit code space is speci\ufb01ed by an n by k generator matrix G whose entries are all elements of Z2, that is, zeroes and ones. The matrix G maps messages to their encoded equivalent. Thus the k bit message x is encoded as Gx, where the message x is treated as a column vector in the obvious 446 Quantum", " error-correction way. Furthermore, the multiplication operation, and all our other arithmetic operations in this section, are done modulo 2. As a simple example, the repetition code mapping a single bit to three repetitions is speci\ufb01ed by the generator matrix \u23a3 G = \u23a6, (10.53) \u23a1 \u23a4 1 1 1 since G maps the possible messages, 0 and 1, to their encoded form, G[0] = (0, 0, 0) and G[1] = (1, 1, 1). (Recall that (a, b,..., z) is our shorthand notation for column vectors.) We say that a code using n bits to encode k bits of information is an [n, k] code; this example is therefore a [3, 1] code. A slightly more complicated example is to encode two bits using three repetitions of each bit \u2013 a [6, 2] code. This has generator matrix \u23a1 \u23a4 10.54) from which we see that G(0, 0) = (0, 0, 0, 0, 0, 0); G(0, 1) = (0, 0, 0, 1, 1, 1); G(1, 0) = (1, 1, 1, 0, 0, 0); G(1, 1) = (1, 1, 1, 1, 1, 1), (10.55) (10.56) just as we expect. The set of possible codewords for the code corresponds to the vector space spanned by the columns of G, so in order that all messages be uniquely encoded we require that the columns of G be linearly independent, but otherwise place no constraints on G. Exercise 10.14: Write an expression for a generator matrix encoding k bits using r repetitions for each bit. This is an [rk, k] linear code, and should have an rk\u00d7k generator matrix. Exercise 10.15: Show that adding one column of G to another results in a generator matrix generating the same code. A great advantage of linear codes over general error-correcting codes is their compact speci\ufb01cation. A general code encoding k bits in n bits requires 2k codewords each of length n to specify the encoding, a total of n2k bits to specify a description of the code. With a linear code we need only specify the kn bits of the generator matrix, an exponential saving in the amount of memory required! This compact description is mirrored in the ability to do ef\ufb01cient encoding and decoding, important features which classical linear codes share with their quantum cousins, the stabilizer codes. We can already see how to perform ef\ufb01cient encoding of a classical linear code: one simply multiplies the k bit message by the n by k generator matrix G to obtain the n bit encoded message, a procedure which can be done using O(nk) operations. One of the attractive features of the generator matrix de\ufb01nition of linear codes is the transparent connection between the messages we wish to encode and how they are encoded. It is not so clear how to perform error-correction. Error-correction for linear Constructing quantum codes 447 codes is most easily understood by introducing an alternative (but equivalent) formulation of linear codes in terms of parity check matrices. In this de\ufb01nition an [n, k] code is de\ufb01ned to consist of all n-element vectors x over Z2 such that Hx = 0, (10.57) where H is an n \u2212 k by n matrix known as the parity check matrix, with entries all zeroes and ones. Equivalently, but more succinctly, the code is de\ufb01ned to be the kernel of H. A code encoding k bits has 2k possible codewords so the kernel of H must be k-dimensional, and therefore we require that H have linearly independent rows. Exercise 10.16: Show that adding one row of the parity check matrix to another does not change the code. Using Gaussian elimination and swapping of bits it is therefore possible to assume that the parity check matrix has the standard form [A|In\u2212k], where A is an (n \u2212 k)\u00d7k matrix. To connect the parity check picture of linear codes with the generator matrix picture we need to develop a procedure that enables us to convert back and forth between the parity check matrix H and the generator matrix G. To go from the parity check matrix to the generator matrix, pick k linearly independent vectors y1,..., yk spanning the kernel of H, and set G to have columns y1 through yk. To go from the generator matrix to the parity check matrix, pick n \u2212 k linearly independent vectors y1,..., yn\u2212k orthogonal to", " the columns of G, and set the rows of H to be yT n\u2212k. (By orthogonal, we mean that the inner product modulo 2 must be zero.) As an example, consider the [3, 1] repetition code de\ufb01ned by the generator matrix (10.53). To construct H we pick out 3 \u2212 1 = 2 linearly independent vectors orthogonal to the columns of G, say (1, 1, 0) and (0, 1, 1), and de\ufb01ne the parity check matrix as: 1,..., yT 10.58) It is easy to check that Hx = 0 only for the codewords x = (0, 0, 0) and x = (1, 1, 1). Exercise 10.17: Find a parity check matrix for the [6, 2] repetition code de\ufb01ned by the generator matrix in (10.54). Exercise 10.18: Show that the parity check matrix H and generator matrix G for the same linear code satisfy HG = 0. Exercise 10.19: Suppose an [n, k] linear code C has a parity check matrix of the form H = [A|In\u2212k], for some (n \u2212 k)\u00d7k matrix A. Show that the corresponding generator matrix is Ik \u2212A G =. (10.59) (Note that \u2212A = A since we are working modulo 2; however, this equation also holds for linear codes over more general \ufb01elds than Z2.) The parity check matrix makes error-detection and recovery quite transparent. Suppose that we encode the message x as y = Gx, but an error e due to noise corrupts y giving the corrupted codeword y = y + e. (Note that + here denotes bitwise addition 448 Quantum error-correction modulo 2.) Because Hy = 0 for all codewords, it follows that Hy = He. We call Hy the error syndrome, and it plays a role similar to the role played by the error syndrome in quantum error-correction; it is a function Hy of the corrupted state y, just as the quantum error syndrome is determined by measuring the corrupted quantum state, and because of the relation Hy = He, the error syndrome contains information about the error that occurred that hopefully will enable recovery to the original codeword y. To see how this is possible, suppose no errors or just one error occurred. Then the error syndrome Hy is equal to 0 in the no error case and is equal to Hej when an error occurs on the jth bit, where ej is the unit vector with component 1 in the jth component. If we assume that errors occur on at most one bit, it is therefore possible to perform error-correction by computing the error syndrome Hy and comparing it to the different possible values of Hej to determine which (if any) bit needs to be corrected. More generally, insight into how error-correction may be performed with a linear code can be attained using the concept of distance. Suppose x and y are words of n bits each. The (Hamming) distance d(x, y) between x and y is de\ufb01ned to be the number of places at which x and y differ. Thus d((1, 1, 0, 0), (0, 1, 0, 1)) = 2, for example. The (Hamming) weight of a word x is de\ufb01ned to be the distance from the string of all zeroes, wt(x) \u2261 d(x, 0), that is, the number of places at which x is non-zero. Note that d(x, y) = wt(x + y). To understand the connection with error-correction suppose we encode x as y = Gx using a linear error-correcting code. Noise corrupts the encoded bit string producing y = y +e. Provided the probability of a bit \ufb02ip is less than 1/2, the most likely codeword to have been encoded is the codeword y which minimizes the number of bit \ufb02ips needed to get from y to y, that is, which minimizes wt(e) = d(y, y). In principle, error-correction with a linear code may be accomplished by simply replacing y by such a y. In practice this may be rather inef\ufb01cient, since determining the minimal distance d(y, y) in general requires searching all 2k possible codewords y. A great deal of effort in classical coding theory has gone into constructing codes with special structure that enable error-correction to be performed more ef\ufb01ciently, however these constructions are beyond the scope of this book. The global properties of the code can also be understood using the Hamming distance. We de\ufb01ne the distance of a code", " to be the minimum distance between any two codewords, d(C) \u2261 min x,y\u2208C,x = y d(x, y). (10.60) But d(x, y) = wt(x + y). Since the code is linear, x + y is a codeword if x and y are, so we see that d(C) = min x\u2208C,x = 0 wt(x). (10.61) Setting d \u2261 d(C), we say that C is an [n, k, d] code. The importance of the distance is that a code with distance at least 2t + 1 for some integer t is able to correct errors on up to t bits, simply by decoding the corrupted encoded message y as the unique codeword y satisfying d(y, y) \u2264 t. Exercise 10.20: Let H be a parity check matrix such that any d \u2212 1 columns are linearly independent, but there exists a set of d linearly dependent columns. Show that the code de\ufb01ned by H has distance d. Exercise 10.21: (Singleton bound) Show that an [n, k, d] code must satisfy n \u2212 k \u2265 d \u2212 1. Constructing quantum codes 449 A good illustrative class of linear error-correcting codes are the Hamming codes. Suppose r \u2265 2 is an integer and let H be the matrix whose columns are all 2r \u22121 bit strings of length r which are not identically 0. This parity check matrix de\ufb01nes a [2r \u2212 1, 2r \u2212 r \u2212 1] linear code known as a Hamming code. An especially important example for quantum error-correction is the case r = 3, which is a [7, 4] code having parity check matrix10.62) Any two columns of H are different, and therefore linearly independent; the \ufb01rst three columns are linearly dependent, so by Exercise 10.20 the distance of the code is 3. It follows that this code is able to correct an error on any single bit. Indeed, the errorcorrection method is very simple. Suppose an error occurs on the jth bit. Inspection of (10.62) reveals that the syndrome Hej is just a binary representation for j, telling us which bit to \ufb02ip to correct the error. Exercise 10.22: Show that all Hamming codes have distance 3, and thus can correct an error on a single bit. The Hamming codes are therefore [2r \u2212 1, 2r \u2212 r \u2212 1, 3] codes. What can we say more generally about the properties of linear codes? In particular, we would like conditions telling us whether or not codes with particular code parameters exist. Not surprisingly, many techniques for proving such conditions exist. One such set of conditions is known as the Gilbert\u2013Varshamov bound, which states that for large n there exists an [n, k] error-correcting code protecting against errors on t bits for some k such that k n \u2265 1 \u2212 H 2t n, (10.63) where H(x) \u2261 \u2212x log(x) \u2212 (1 \u2212 x) log(1 \u2212 x) is the binary Shannon entropy, studied in detail in Chapter 11. The importance of the Gilbert\u2013Varshamov bound is that it guarantees the existence of good codes, provided one doesn\u2019t try to encode too many bits (k) into too small a number of bits (n). The proof of the Gilbert\u2013Varshamov bound is quite simple, and is left as an exercise. Exercise 10.23: Prove the Gilbert\u2013Varshamov bound. We conclude our survey of classical error-correction by explaining an important construction for codes known as the dual construction. Suppose C is an [n, k] code with generator matrix G and parity check matrix H. Then we can de\ufb01ne another code, the dual of C, denoted C \u22a5, to be the code with generator matrix H T and parity check matrix GT. Equivalently, the dual of C consists of all codewords y such that y is orthogonal to all the codewords in C. A code is said to be weakly self-dual if C \u2286 C\u22a5, and strictly self-dual if C = C\u22a5. Rather remarkably, the dual construction for classical linear codes arises naturally in the study of quantum error-correction, and is the key to the construction of an important class of quantum codes known as CSS codes. 450 Quantum error-correction Exercise 10.24: Show that a code with generator matrix G is weakly self-dual if and only if GT G = 0. Exercise 10.25: Let C be a linear code. Show that if x \u2208 C\u22a5 then", " y\u2208C(\u22121)x\u00b7y = 0. y\u2208C(\u22121)x\u00b7y = |C|, while if x \u2208 C\u22a5 then 10.4.2 Calderbank\u2013Shor\u2013Steane codes Our \ufb01rst example of a large class of quantum error-correcting codes is the Calderbank\u2013 Shor\u2013Steane codes, more usually known as CSS codes, after the initials of the inventors of this class of codes. CSS codes are an important subclass of the more general class of stabilizer codes. Suppose C1 and C2 are [n, k1] and [n, k2] classical linear codes such that C2 \u2282 C1 and C1 and C\u22a5 2 both correct t errors. We will de\ufb01ne an [n, k1 \u2212 k2] quantum code CSS(C1, C2) capable of correcting errors on t qubits, the CSS code of C1 over C2, via the following construction. Suppose x \u2208 C1 is any codeword in the code C1. Then we de\ufb01ne the quantum state |x + C2 by |x + C2 \u2261 1$ |C2| y\u2208C2 |x + y, (10.64) where + is bitwise addition modulo 2. Suppose x is an element of C1 such that x \u2212 x \u2208 C2. Then it is easy to see that |x + C2 = |x + C2, and thus the state |x + C2 depends only upon the coset of C1/C2 which x is in, explaining the coset notation we have used for |x + C2. Furthermore, if x and x belong to different cosets of C2, then for no y, y \u2208 C2 does x + y = x + y, and therefore |x + C2 and |x + C2 are orthonormal states. The quantum code CSS(C1, C2) is de\ufb01ned to be the vector space spanned by the states |x + C2 for all x \u2208 C1. The number of cosets of C2 in C1 is |C1|/|C2| so the dimension of CSS(C1, C2) is |C1|/|C2| = 2k1\u2212k2, and therefore CSS(C1, C2) is an [n, k1 \u2212 k2] quantum code. We can exploit the classical error-correcting properties of C1 and C\u22a5 2 to detect and correct quantum errors! In fact, it is possible to error-correct up to t bit and phase \ufb02ip errors on CSS(C1, C2) by making use of the error-correcting properties of C1 and C\u22a5 2, respectively. Suppose the bit \ufb02ip errors are described by an n bit vector e1 with 1s where bit \ufb02ips occurred, and 0s elsewhere, and the phase \ufb02ip errors are described by an n bit vector e2 with 1s where phase \ufb02ips occurred, and 0s elsewhere. If |x+C2 was the original state then the corrupted state is: 1$ |C2| y\u2208C2 (\u22121)(x+y)\u00b7e2|x + y + e1. (10.65) To detect where bit \ufb02ips occurred it is convenient to introduce an ancilla containing suf\ufb01cient qubits to store the syndrome for the code C1, and initially in the all zero state |0. We use reversible computation to apply the parity matrix H1 for the code C1, taking |x + y + e1|0 to |x + y + e1|H1(x + y + e1) = |x + y + e|H1e1, since (x + y) \u2208 C1 is annihilated by the parity check matrix. The effect of this operation is to produce the state: 1$ |C2| y\u2208C2 (\u22121)(x+y)\u00b7e2|x + y + e1|H1e1. (10.66) Constructing quantum codes 451 Exercise 10.26: Suppose H is a parity check matrix. Explain how to compute the transformation |x|0 \u2192 |x|Hx using a circuit composed entirely of controlled- s. Error-detection for the bit \ufb02ip errors is completed by measuring the ancilla to obtain the result H1e1 and discarding the ancilla, giving the state 1$ (\u22121)(x+y)\u00b7e2|x + y + e1. (10.67) |C2| y\u2208C2 Knowing the error syndrome", " H1e1 we can infer the error e1 since C1 can correct up to t errors, which completes the error-detection. Recovery is performed simply by applying gates to the qubits at whichever positions in the error e1 a bit \ufb02ip occurred, removing all the bit \ufb02ip errors and giving the state 1$ |C2| y\u2208C2 (\u22121)(x+y)\u00b7e2|x + y. (10.68) To detect phase \ufb02ip errors we apply Hadamard gates to each qubit, taking the state to 1$ |C2|2n z y\u2208C2 (\u22121)(x+y)\u00b7(e2+z)|z, (10.69) where the sum is over all possible values for n bit z. Setting z \u2261 z + e2, this state may be rewritten: 1$ |C2|2n (\u22121)(x+y)\u00b7z|z + e2. (10.70) y\u2208C2 (The next step appeared as Exercise 10.25 on page 450.) Supposing z \u2208 C\u22a5 2 = |C2|, while if z \u2208 C\u22a5 to see that 2 state may be rewritten: (\u22121)y\u00b7z (\u22121)y\u00b7z then y\u2208C2 y\u2208C2 z it is easy = 0. Thus the 1$ 2n/|C2| z\u2208C\u22a5 2 (\u22121)x\u00b7z|z + e2, (10.71) which looks just like a bit \ufb02ip error described by the vector e2! As for the error-detection for bit \ufb02ips we introduce an ancilla and reversibly apply the parity check matrix H2 for C\u22a5 2 to obtain H2e2, and correct the \u2018bit \ufb02ip error\u2019 e2, obtaining the state 1$ 2n/|C2| z\u2208C\u22a5 2 (\u22121)x\u00b7z|z. (10.72) The error-correction is completed by again applying Hadamard gates to each qubit; we can either compute the result of these gates directly, or note that the effect is to apply Hadamard gates to the state in (10.71) with e2 = 0; since the Hadamard gate is self-inverse this takes us back to the state in (10.68) with e2 = 0: 1$ |C2| y\u2208C2 |x + y, (10.73) which is the original encoded state! One important application of CSS codes is to prove a quantum version of the Gilbert\u2013 Varshamov bound, guaranteeing the existence of good quantum codes. This states that 452 Quantum error-correction in the limit as n becomes large, an [n, k] quantum code protecting against errors on up to t qubits exists for some k such that k n \u2265 1 \u2212 2H 2t n. (10.74) Thus, good quantum error-correcting codes exist, provided one doesn\u2019t try to pack too many qubits k into an n qubit code. The proof of the Gilbert\u2013Varshamov bound for CSS codes is rather more complex than the proof of the classical Gilbert\u2013Varshamov bound, due to the constraints on the classical codes C1 and C2, and is left as an end of chapter problem. Summarizing, suppose C1 and C2 are [n, k1] and [n, k2] classical linear codes, respectively, such that C2 \u2282 C1, and both C1 and C\u22a5 2 can correct errors on up to t bits. Then CSS(C1, C2) is an [n, k1 \u2212 k2] quantum error-correcting code which can correct arbitrary errors on up to t qubits. Furthermore, the error-detection and correction steps require gates, in each case a number linear only the application of Hadamard and controlledin the size of the code. Encoding and decoding can also be performed using a number of gates linear in the size of the code, but we won\u2019t discuss that here; it\u2019s discussed later in more generality in Section 10.5.8. Exercise 10.27: Show that the codes de\ufb01ned by |x + C2 \u2261 1$ |C2| y\u2208C2 (\u22121)u\u00b7y|x + y + v (10.75) and parameterized by u and v are equivalent to CSS(C1, C2) in the sense that they have the same error-correcting properties. These codes, which we\u2019ll refer to as CSSu,v(C1, C2), will be useful later in our study", " of quantum key distribution, in Section 12.6.5. The Steane code An important example of a CSS code may be constructed using the [7, 4, 3] Hamming code whose parity check matrix we recall here10.76) Suppose we label this code C and de\ufb01ne C1 \u2261 C and C2 \u2261 C\u22a5. To use these codes to de\ufb01ne a CSS code we need \ufb01rst to check that C2 \u2282 C1. By de\ufb01nition the parity check matrix of C2 = C\u22a5 is equal to the transposed generator matrix of C1 = C: H[C2] = G[C1]10.77) Exercise 10.28: Verify that the transpose of the matrix in (10.77) is the generator of the [7, 4, 3] Hamming code. Stabilizer codes 453 Comparing with (10.76) we see that the span of the rows of H[C2] strictly contains the span of the rows of H[C1], and since the corresponding codes are the kernels of H[C2] and H[C1] we conclude that C2 \u2282 C1. Furthermore, C\u22a5 2 = (C\u22a5)\u22a5 = C, so both C1 and C\u22a5 2 are distance 3 codes which can correct errors on 1 bit. Since C1 is a [7, 4] code and C2 is a [7, 3] code it follows that CSS(C1, C2) is a [7, 1] quantum code which can correct errors on a single qubit. This [7, 1] quantum code has nice properties that make it easy to work with, and will be used in many of the examples for the remainder of this chapter. It is known as the Steane code, after its inventor. The codewords of C2 are easily determined from (10.77) and Exercise 10.28. Rather than write them out explicitly, we write them out implicitly as the entries in the logical |0L for the Steane code, |0 + C2: |0L = 1\u221a 8 |0000000 + |1010101 + |0110011 + |1100110 +|0001111 + |1011010 + |0111100 + |1101001. (10.78) To determine the other logical codeword we need to \ufb01nd an element of C1 that is not in C2. An example of such an element is (1, 1, 1, 1, 1, 1, 1), giving: |1L = 1\u221a 8 |1111111 + |0101010 + |1001100 + |0011001 +|1110000 + |0100101 + |1000011 + |0010110. (10.79) 10.5 Stabilizer codes We cannot clone, perforce; instead, we split Coherence to protect it from that wrong That would destroy our valued quantum bit And make our computation take too long. Correct a \ufb02ip and phase \u2013 that will suf\ufb01ce. If in our code another error\u2019s bred, We simply measure it, then God plays dice, Collapsing it to X or Y or zed. We start with noisy seven, nine, or \ufb01ve And end with perfect one. To better spot Those \ufb02aws we must avoid, we \ufb01rst must strive To \ufb01nd which ones commute and which do not. With group and eigenstate, we\u2019ve learned to \ufb01x Your quantum errors with our quantum tricks. \u2013 \u2018Quantum Error Correction Sonnet\u2019, by Daniel Gottesman Stabilizer codes, sometimes known as additive quantum codes, are an important class of quantum codes whose construction is analogous to classical linear codes. In order to understand stabilizer codes it is useful to \ufb01rst develop the stabilizer formalism, a powerful method for understanding a wide class of operations in quantum mechanics. The applications of the stabilizer formalism extend far beyond quantum error-correction; however, in 454 Quantum error-correction this book our main concern is with this speci\ufb01c application. After de\ufb01ning the stabilizer formalism, we explain how unitary gates and measurements may be described using it, and an important theorem which quanti\ufb01es the limitations of stabilizer operations. We then present stabilizer constructions for stabilizer codes, along with explicit examples, a useful standard form, and circuits for encoding, decoding, and correction. 10.5.1 The stabilizer formalism The central insight of the stabilizer formalism is easily illustrated by an example. Consider the EPR state of two qubits |\u03c8 = |00 + |11 \ufffd", "\ufffd 2. (10.80) It is easy to verify that this state satis\ufb01es the identities X1X2|\u03c8 = |\u03c8 and Z1Z2|\u03c8 = |\u03c8; we say that the state |\u03c8 is stabilized by the operators X1X2 and Z1Z2. A little less obviously, the state |\u03c8 is the unique quantum state (up to a global phase) which is stabilized by these operators X1X2 and Z1Z2. The basic idea of the stabilizer formalism is that many quantum states can be more easily described by working with the operators that stabilize them than by working explicitly with the state itself. This claim is perhaps rather surprising at \ufb01rst sight; nevertheless it is true. It turns out that many quantum codes (including CSS codes and the Shor code) can be much more compactly described using stabilizers than in the state vector description. Even more importantly, errors on the qubits and operations such as the Hadamard gate, phase gate, and even the controlledgate and measurements in the computational basis are all easily described using the stabilizer formalism! The key to the power of the stabilizer formalism lies in the clever use of group theory, whose basic elements are reviewed in Appendix 2. The group of principal interest is the Pauli group Gn on n qubits. For a single qubit, the Pauli group is de\ufb01ned to consist of all the Pauli matrices, together with multiplicative factors \u00b11, \u00b1i: G1 \u2261 {\u00b1I, \u00b1iI, \u00b1X, \u00b1iX, \u00b1Y, \u00b1iY, \u00b1Z, \u00b1iZ}. (10.81) This set of matrices forms a group under the operation of matrix multiplication. You might wonder why we don\u2019t omit the multiplicative factors \u00b11 and \u00b1i; the reason these are included is to ensure that G1 is closed under multiplication, and thus forms a legitimate group. The general Pauli group on n qubits is de\ufb01ned to consist of all n-fold tensor products of Pauli matrices, and again we allow multiplicative factors \u00b11, \u00b1i. We can now de\ufb01ne stabilizers a little more precisely. Suppose S is a subgroup of Gn and de\ufb01ne VS to be the set of n qubit states which are \ufb01xed by every element of S. VS is the vector space stabilized by S, and S is said to be the stabilizer of the space VS, since every element of VS is stable under the action of elements in S. You should convince yourself of the truth of the following simple exercise: Exercise 10.29: Show that an arbitrary linear combination of any two elements of VS is also in VS. Therefore, VS is a subspace of the n qubit state space. Show that VS is the intersection of the subspaces \ufb01xed by each operator in S (that is, the eigenvalue one eigenspaces of elements of S). Let\u2019s look at a simple example of the stabilizer formalism in action, a case with n = 3 Stabilizer codes 455 qubits and S \u2261 {I, Z1Z2, Z2Z3, Z1Z3}. The subspace \ufb01xed by Z1Z2 is spanned by |000, |001, |110 and |111, and the subspace \ufb01xed by Z2Z3 is spanned by |000, |100,|011 and |111. Note that the elements |000 and |111 are common to both these lists. With these observations and a little thought one realizes that VS must be the subspace spanned by the states |000 and |111. In this example we determined VS simply by looking at the subspaces stabilized by two of the operators in S. This is a manifestation of an important general phenomenon \u2013 the description of a group by its generators. As explained in Appendix 2 a set of elements g1,..., gl in a group G is said to generate the group G if every element of G can be written as a product of elements from the list g1,..., gl, and we write G = g1,..., gl. In the example S = Z1Z2, Z2Z3 as Z1Z3 = (Z1Z2)(Z2Z3) and I = (Z1Z2)2. The great advantage of using generators to describe groups is that they provide a compact means of describing the group. Indeed, in Appendix 2 we show that a group G with size |G| has a set of at most log(|G|) generators. Furthermore, to see that a particular vector is stabilized by a group S we need only check that the vector is stabilized by the", " generators, since it is then automatically stabilized by products of the generators, making this a most convenient representation. (The notation \u00b7 \u00b7 \u00b7 which we use for group generators may potentially be confused with the notation for observable averages introduced in Section 2.2.5 beginning on page 87; however, in practice, it is always clear from context how the notation is being used.) Not just any subgroup S of the Pauli group can be used as the stabilizer for a nontrivial vector space. For example, consider the subgroup of G1 consisting of {\u00b1I, \u00b1X}. Obviously the only solution to (\u2212I)|\u03c8 = |\u03c8 is |\u03c8 = 0, and thus {\u00b1I, \u00b1X} is the stabilizer for the trivial vector space. What conditions must be satis\ufb01ed by S in order that it stabilize a non-trivial vector space, VS? Two conditions easily seen to be necessary are that (a) the elements of S commute, and (b) \u2212I is not an element of S. We don\u2019t yet have all the tools to prove it, but will show later that these two conditions are also suf\ufb01cient for VS to be non-trivial. Exercise 10.30: Show that \u2212I \u2208 S implies \u00b1iI \u2208 S. To see that these two conditions are necessary, suppose VS is non-trivial, so it contains a non-zero vector |\u03c8. Let M and N be elements of S. Then M and N are tensor products of Pauli matrices, possibly with an overall multiplicative factor; because the Pauli matrices all commute or anti-commute with one another, it follows that M and N must either commute or anti-commute. To establish condition (a), that they commute, we suppose that M and N anti-commute and show that this leads to a contradiction. By assumption \u2212N M = M N so we have \u2212|\u03c8 = \u2212N M |\u03c8 = M N |\u03c8 = |\u03c8, where the \ufb01rst and last equalities follow from the fact that M and N stabilize |\u03c8. So we have \u2212|\u03c8 = |\u03c8, which implies that |\u03c8 is the zero vector, which is the desired contradiction. To establish condition (b), that \u2212I \u2208 S just note that if \u2212I is an element of S then we have \u2212I|\u03c8 = |\u03c8, which again leads to a contradiction. Exercise 10.31: Suppose S is a subgroup of Gn generated by elements g1,..., gl. Show that all the elements of S commute if and only if gi and gj commute for each pair i, j. A beautiful example of the stabilizer formalism is provided by the seven qubit Steane 456 Quantum error-correction Name g1 g2 g3 g4 g5 g6 Operator I I I XXXX I XX I I XX Figure 10.6. Stabilizer generators for the Steane seven qubit code. The entries represent tensor products on the respective qubits; for example, ZIZIZIZ = = Z1Z3Z5Z7. code. It turns out that the six generators g1 through g6 listed in Figure 10.6 generate a stabilizer for the code space of the Steane code. Observe how clean and compact this description is when compared with the rather messy speci\ufb01cation in terms of state vectors, (10.78) and (10.79); even further advantages will manifest when we examine quantum error-correction from this viewpoint. Note also the similarity in structure between the generators in Figure 10.6 and the parity check matrices for the classical linear codes C1 and C\u22a5 2 used in the construction of the Steane code. (Recall that for the Steane code C1 = C\u22a5 is the Hamming [7, 4, 3] code with parity check matrix given by (10.76).) The 2 \ufb01rst three generators of the stabilizer have Xs in locations corresponding to the locations of the 1s in the parity check matrix for C1, while the \ufb01nal three generators g4 through g6 have Zs in locations corresponding to the locations of the 1s in the parity check matrix for C\u22a5 2. With these observations in hand the solution to the following exercise becomes almost self-evident. Exercise 10.32: Verify that the generators in Figure 10.6 stabilize the codewords for the Steane code, as described in Section 10.4.2. This use of the stabilizer formalism to describe a quantum code foreshadows our later use of stabilizers to describe a wide class of quantum codes, but for now it is important to appreciate that there is nothing special about the Steane", " code\u2019s status as a quantum code \u2013 it is merely a subspace of a vector space which happens to have a description using stabilizers. In practice, we want our generators g1,..., gl to be independent in the sense that removing any generator gi makes the group generated smaller, g1,..., gi\u22121, gi+1,..., gl = g1,..., gl. (10.82) Determining whether a particular set of generators is independent or not is rather timeconsuming with our current understanding; fortunately, there is a simple way this can be done based on an idea known as the check matrix, so-named because it plays a role in the theory of stabilizer codes analogous to the parity check matrix in classical linear codes. Suppose S = g1,..., gl. There is an extremely useful way of presenting the generators g1,..., gl of S using the check matrix. This is an l\u00d72n matrix whose rows correspond to the generators g1 through gl; the left hand side of the matrix contains 1s to indicate which generators contain Xs, and the right hand side contains 1s to indicate Stabilizer codes 457 which generators contain Zs; the presence of a 1 on both sides indicates a Y in the generator. More explicitly, the ith row is constructed as follows. If gi contains an I on the jth qubit then the jth and n + jth column elements are 0; if it contains an X on the jth qubit then the jth column element is a 1 and the n + jth column element is a 0; if it contains a Z on the jth qubit then the jth column element is 0 and the n + jth column element is 1; if it contains a Y on the jth qubit then both the jth and n + jth columns are 1. In the case of the Steane seven qubit code we can read the check matrix off from Figure 10.610.83) The check matrix doesn\u2019t contain any information about the multiplicative factors out the front of the generators, but it does contain much other useful information, so much so that we use r(g) to denote the 2n-dimensional row vector representation of an element g of the Pauli group. Suppose we de\ufb01ne a 2n\u00d72n matrix \u039b by 0 I I 0 \u039b = (10.84) where the I matrices on the off-diagonals are n\u00d7n. Elements g and g of the Pauli group are easily seen to commute if and only if r(g)\u039br(g)T = 0; the formula x\u039byT de\ufb01nes a sort of \u2018twisted\u2019 inner product between row matrices x and y expressing whether the elements of the Pauli group corresponding to x and y commute or not. Exercise 10.33: Show that g and g commute if and only if r(g)\u039br(g)T = 0. (In the check matrix representation, arithmetic is done modulo two.) Exercise 10.34: Let S = g1,..., gl. Show that \u2212I is not an element of S if and only if g2 j = I for all j, and gj = \u2212 I for all j. Exercise 10.35: Let S be a subgroup of Gn such that \u2212I is not an element of S. Show that g2 = I for all g \u2208 S, and thus g\u2020 = g. A useful connection between independence of the generators and the check matrix is established by means of the following proposition: Proposition 10.3: Let S = g1,..., gl be such that \u2212I is not an element of S. The generators g1 through gl are independent if and only if the rows of the corresponding check matrix are linearly independent. Proof We prove the contrapositive. Note \ufb01rst that g2 i must equal I for all i, by Exercise 10.35. Observe that r(g) + r(g) = r(gg), so addition in the row vector representation corresponds to multiplication of group elements. Thus the rows of the check matrix are linearly is equal to the dependent, i air(gi) = 0, and aj = 0 for some j, if and only if i gai + i 458 Quantum error-correction identity, up to an overall multiplicative factor. But \u2212I \u2208 S so the multiplicative factor must be 1, and the last condition corresponds to the condition gj = g\u22121 i = j gai i, and thus g1,..., gl are not independent generators. j = + The following innocuous looking proposition is surprisingly useful, and can be immediately leveraged into a proof that VS is of dimension", " 2k when S is generated by l = n\u2212k independent commuting generators, and \u2212I \u2208 S. We will use the proposition repeatedly throughout the remainder of this chapter. Once again the tool of choice in the proof is the check matrix representation. Proposition 10.4: Let S = g1,..., gl be generated by l independent generators and satisfy \u2212I \u2208 S. Fix i in the range 1,..., l. Then there exists g \u2208 Gn such that ggig\u2020 = \u2212gi and ggjg\u2020 = gj for all j = i. Proof Let G be the check matrix associated to g1,..., gl. The rows of G are linearly independent by Proposition 10.3, so there exists a 2n-dimensional vector x such that G\u039bx = ei, where ei is the l-dimensional vector with a 1 in the ith position and 0s elsewhere. Let g be such that r(g) = xT. Then by de\ufb01nition of x we have r(gj)\u039br(g)T = 0 for j = i and r(gi)\u039br(g)T = 1, and thus ggig\u2020 = \u2212gi and ggjg\u2020 = gj for j = i. We conclude our look at the most basic elements of the stabilizer formalism by ful\ufb01lling our earlier promise that VS is non-trivial provided S is generated by independent commuting generators and \u2212I \u2208 S. Indeed, if there are l = n \u2212 k generators, then it is at least plausible (and we will prove) that VS is 2k-dimensional, based on the intuitive argument that each additional generator for the stabilizer cuts the dimension of VS by a factor of 1/2, as we might naively expect because the +1 and \u22121 eigenspaces for a tensor product of Pauli matrices divide the total Hilbert space into two subspaces of equal dimension. Proposition 10.5: Let S = g1..., gn\u2212k be generated by n \u2212 k independent and commuting elements from Gn, and such that \u2212I \u2208 S. Then VS is a 2k-dimensional vector space. In all our later discussion of the stabilizer formalism we use the convention that stabilizers are always described in terms of independent commuting generators such that \u2212I \u2208 S. Proof Let x = (x1,..., xn\u2212k) be a vector of n \u2212 k elements of Z2. De\ufb01ne + P x S \u2261 n\u2212k j=1 (I + (\u22121)xj gj) 2n\u2212k. (10.85) Because (I + gj)/2 is the projector onto the +1 eigenspace of gj, it is easy to see that P (0,...,0) must be the projector onto VS. By Proposition 10.4 for each x there exists gx in S Gn such that gxP (0,...,0) S is the same as the dimension of VS. Furthermore, for distinct x the P x S are easily seen to be orthogonal. S, and therefore the dimension of P x (gx)\u2020 = P x S Stabilizer codes 459 The proof is concluded with the algebraic observation that I = x P x S. (10.86) The left hand side is a projector onto a 2n-dimensional space, while the right hand side is a sum over 2n\u2212k orthogonal projectors of the same dimension as VS, and thus the dimension of VS must be 2k. 10.5.2 Unitary gates and the stabilizer formalism We have been discussing the use of the stabilizer formalism to describe vector spaces. The formalism can also be used to describe the dynamics of those vector spaces in the larger state space, under a variety of interesting quantum operations. Aside from the intrinsic interest of understanding quantum dynamical operations, this goal is especially relevant because we will describe quantum error-correcting codes using the stabilizer formalism, and would like an elegant means for understanding the effects of noise and other dynamical processes on those codes. Suppose we apply a unitary operation U to a vector space VS stabilized by the group S. Let |\u03c8 be any element of VS. Then for any element g of S, U |\u03c8 = U g|\u03c8 = U gU \u2020U |\u03c8, (10.87) and thus the state U |\u03c8 is stabilized by U gU \u2020, from which we deduce that the vector space U VS is stabilized by the group U SU \u2020 \u2261 {U gU \u2020|g \u2208 S}. Furthermore, if g1,..., gl generate S, then U g1U \u2020,..", "., U glU \u2020 generate U SU \u2020, so to compute the change in the stabilizer we need only compute how it affects the generators of the stabilizer. The great advantage of this approach to dynamics is that for certain special unitary operations U this transformation of the generators takes on a particularly appealing form. Suppose, for example, that we apply a Hadamard gate to a single qubit. Note that HXH \u2020 = Z; HY H \u2020 = \u2212Y ; HZH \u2020 = X. (10.88) As a consequence we correctly deduce that after a Hadamard gate is applied to the quantum state stabilized by Z (|0), the resulting state will be stabilized by X (|+). Not very impressive, you may think! Imagine though that we had n qubits in a state whose stabilizer is Z1, Z2,..., Zn. It is easy to see that this is the state |0\u2297n. Applying the Hadamard gate to each of the n qubits we see that the state afterwards has a stabilizer X1, X2,..., Xn; again it is easy to see that this is just the familiar state which is an equal superposition of all computational basis states. What is remarkable about this example is that the usual (state vector) description of the \ufb01nal state requires 2n amplitudes to be speci\ufb01ed, compared with the description provided by the generators: X1,..., Xn, which is linear in n! Still, you might say, after applying the Hadamard gate to each of the n qubits there is no entanglement in the quantum computer, so it is not so surprising that a compact description can be obtained. But much more is possible within, which the stabilizer formalism, including an ef\ufb01cient description of the controlledtogether with the Hadamard gate is capable of generating entanglement. To understand how this works, consider how the operators X1, X2, Z1 and Z2 behave under conjugation gate with qubit 1 as control by the controlled-. Denoting by U the controlled- 460 Quantum error-correction Operation controlled- H S X Y Z Input Output X1X2 X1 X2 X2 Z1 Z1 Z1Z2 Z2 Z Z \u2212 X X \u2212Z Z \u2212 X X Z Z Figure 10.7. Transformation properties of elements of the Pauli group under conjugation by various common operations. The controlled- has qubit 1 as the control and qubit 2 as the target. and qubit 2 as the target, we have \u23a1 \u23a2 \u23a2 \u23a3 U X1U \u2020 = = X1X2. 10.89) Similar calculations show that U X2U \u2020 = X2, U Z1U \u2020 = Z1 and U Z2U \u2020 = Z1Z2. To see how U conjugates other operators in the two qubit Pauli group we need only take products of the facts we already know. For example, to calculate U X1X2U \u2020 we observe that U X1X2U \u2020 = U X1U \u2020U X2U \u2020 = (X1X2)X2 = X1. The Y Pauli matrices may be dealt with similarly, as for example U Y2U \u2020 = iU X2Z2U \u2020 = iU X2U \u2020U Z2U \u2020 = iX2(Z1Z2) = Z1Y2. Exercise 10.36: Explicitly verify that U X1U \u2020 = X1X2, U X2U \u2020 = X2, U Z1U \u2020 = Z1, and U Z2U \u2020 = Z1Z2. These and other useful conjugation relations for the Hadamard, phase, and Pauli gates are summarized in Figure 10.7. Exercise 10.37: What is U Y1U \u2020? As an example of using the stabilizer formalism to understand unitary dynamics, consider the swap circuit introduced in Section 1.3.4 on page 23; for convenience, the circuit is illustrated in Figure 10.8. Consider the way the operators Z1 and Z2 transform Stabilizer codes 461 by conjugation by the gates in the circuit. The operator Z1 transforms through the sequence Z1 \u2192 Z1 \u2192 Z1Z2 \u2192 Z2 and the operator Z2 transforms through the sequence Z2 \u2192 Z1Z2 \u2192 Z1 \u2192 Z1. Similarly, X1 \u2192 X2 and X2 \u2192 X1 under the circuit. Of course, if we take U to be the swap operator then it is obvious that U Z1U \u2020 = Z", "2 and U Z2U \u2020 = Z1, and similarly for X1 and X2, just as for the circuit in Figure 10.8. Proving that this implies that the circuit implements U is left as an exercise: Exercise 10.38: Suppose U and V are unitary operators on two qubits which transform Z1, Z2, X1, and X2 by conjugation in the same way. Show this implies that U = V. Figure 10.8. Circuit swapping two qubits. The example of the swap circuit is interesting but doesn\u2019t do justice to the feature of the stabilizer formalism which makes it truly useful \u2013 the ability to describe certain types of quantum entanglement. We\u2019ve already seen that the stabilizer formalism can be gates, and of course these gates used to describe Hadamard gates and controlledtogether can be used to create entangled states (compare Section 1.3.6). We will see that the stabilizer formalism can in fact be used to describe a wide class of entangled states, including many quantum error-correcting codes. What gates other than the Hadamard and controlled- gates can be described within the stabilizer formalism? The most important addition to this set is the phase gate, a single qubit gate whose de\ufb01nition we now recall, S =. (10.90) 1 0 0 i The action of the phase gate by conjugation on the Pauli matrices is easily computed: SXS\u2020 = Y ; SZS\u2020 = Z. (10.91) Exercise 10.39: Verify (10.91). Indeed, it turns out that any unitary operation taking elements of Gn to elements of Gn under conjugation can be composed from the Hadamard, phase and controlledgates. By de\ufb01nition, we say the set of U such that U GnU \u2020 = Gn is the normalizer of Gn, and denote it by N (Gn), so we are claiming that the normalizer of Gn is generated by the gates, in view of which the Hadamard, phase and Hadamard, phase and controlledcontrolledgates are sometimes referred to simply as the normalizer gates. The proof of this result is simple but instructive, and you will work through it in Exercise 10.40 on page 462. Theorem 10.6: Suppose U is any unitary operator on n qubits with the property that 462 Quantum error-correction if g \u2208 Gn then U gU \u2020 \u2208 Gn. Then up to a global phase U may be composed from O(n2) Hadamard, phase and controlled- gates. Exercise 10.40: Provide an inductive proof of Theorem 10.6 as follows. (1) Prove that the Hadamard and phase gates can be used to perform any normalizer operation on a single qubit. \u221a (2) Suppose U is an n + 1 qubit gate in N (Gn+1) such that U Z1U \u2020 = X1 \u2297 g and U X1U \u2020 = Z1 \u2297 g for some g, g \u2208 Gn. De\ufb01ne U on n qubits by U |\u03c8 \u2261 construction for U in Figure 10.9 may be implemented using O(n2) Hadamard, phase and controlled- 20|U (|0 \u2297 |\u03c8). Use the inductive hypothesis to show that the gates. (3) Show that any gate U \u2208 N (Gn+1) may be implemented using O(n2) Hadamard, phase and controlled- gates. Figure 10.9. Construction used to prove that the Hadamard, phase and controllednormalizer N (Gn). gates generate the We\u2019ve seen that many interesting quantum gates are in the normalizer N (Gn); are there any gates which aren\u2019t? It turns out that most quantum gates are outside the normalizer. Two gates of particular interest which are not in the normalizer are the \u03c0/8 and Toffoli gates. Letting U denote the Toffoli gate with qubits 1 and 2 as controls and qubit 3 as the target, and recalling that T denotes the \u03c0/8 gate, we can easily calculate the action by conjugation of the \u03c0/8 and Toffoli gates on Pauli matrices as X + Y\u221a 2 T ZT \u2020 = Z T XT \u2020 = (10.92) and U Z1U \u2020 = Z1 U X1U \u2020 = X1 \u2297 U Z2U \u2020 = Z2 U X2U \u2020 = X2 \u2297 U X3U \u2020 = X3 U Z3U \u2020 = Z3 \u2297 I + Z2 + X3 \u2212", " Z2X3 2 I + Z1 + X3 \u2212 Z1X3 2 I + Z1 + Z2 \u2212 Z1Z2 2. (10.93) (10.94) (10.95) Unfortunately, this makes analyzing quantum circuits including \u03c0/8 and Toffoli gates via the stabilizer formalism much less convenient than circuits which only contain Hadamard, gates. Fortunately, encoding, decoding, error-detection and phase and controlledrecovery for stabilizer quantum codes can all be accomplished using only such normalizer gates, so the stabilizer formalism is very convenient for the analysis of such codes! Exercise 10.41: Verify Equations (10.92) through (10.95). Stabilizer codes 463 10.5.3 Measurement in the stabilizer formalism We have explained how a limited class of unitary gates may be conveniently described within the stabilizer formalism, but even more is true! Measurements in the computational basis may also be easily described within the stabilizer formalism. To understand how this works, imagine we make a measurement of g \u2208 Gn (recall that g is a Hermitian operator, and can thus be regarded as an observable in the sense of Section 2.2.5). For convenience we assume without loss of generality that g is a product of Pauli matrices with no multiplicative factor of \u22121 or \u00b1i out the front. The system is assumed to be in a state |\u03c8 with stabilizer g1,..., gn. How does the stabilizer of the state transform under this measurement? There are two possibilities: \u2022 g commutes with all the generators of the stabilizer. \u2022 g anti-commutes with one or more of the generators of the stabilizer. Suppose the stabilizer has generators g1,..., gn, and that g anti-commutes with g1. Without loss of generality we may assume that g commutes with g2,..., gn, since if it does not commute with one of these elements (say g2) then it is easy to verify that g does commute with g1g2, and we simply replace the generator g2 by g1g2 in our list of generators for the stabilizer. In the \ufb01rst instance it follows that either g or \u2212g is an element of the stabilizer by the following argument. Since gjg|\u03c8 = ggj|\u03c8 = g|\u03c8 for each stabilizer generator, g|\u03c8 is in VS and is thus a multiple of |\u03c8. Because g2 = I it follows that g|\u03c8 = \u00b1|\u03c8, whence either g or \u2212g must be in the stabilizer. We assume that g is in the stabilizer, with the discussion for \u2212g proceeding analogously. In this instance g|\u03c8 = |\u03c8 and thus a measurement of g yields +1 with probability one, and the measurement does not disturb the state of the system, and thus leaves the stabilizer invariant. What about the second instance, when g anti-commutes with g1 and commutes with all the other generators of the stabilizer? Note that g has eigenvalues \u00b11 and so the projectors for the measurement outcomes \u00b11 are given by (I \u00b1 g)/2, respectively, and thus the measurement probabilities are given by p(+1) = tr p(\u22121) = tr |\u03c8\u03c8| |\u03c8\u03c8 Using the facts that g1|\u03c8 = |\u03c8 and gg1 = \u2212g1g gives p(+1) = tr = tr I + g 2 I \u2212 g 2 g1 g1|\u03c8\u03c8| |\u03c8\u03c8|. (10.96) (10.97) (10.98) (10.99) Applying the cyclic property of trace we may take g1 to the right hand end of the trace and absorb it into \u03c8| using g1 = g\u2020 1 (Exercise 10.35 on page 457), giving p(+1) = tr |\u03c8\u03c8| = p(\u22121). (10.100) I \u2212 g 2 Since p(+1) + p(\u22121) = 1 we deduce that p(+1) = p(\u22121) = 1/2. Suppose the result +1 occurs. In this instance the new state of the system is |\u03c8+ \u2261 (I + g)|\u03c8/ 2, which has \u221a 464 Quantum error-correction stabilizer g, g2,..., gn. Similarly, if the result \u22121 occurs the posterior state is stabilized by \u2212g, g2,..., gn. 10.5.4 The Gottesman\u2013Knill theorem The results about using stabilizers to describe unitary", " dynamics and measurements may be summarized in the remarkable Gottesman\u2013Knill theorem: Theorem 10.7: (Gottesman\u2013Knill theorem) Suppose a quantum computation is performed which involves only the following elements: state preparations in the computational basis, Hadamard gates, phase gates, controlledgates, and measurements of observables in the Pauli group (which includes measurement in the computational basis as a special case), together with the possibility of classical control conditioned on the outcome of such measurements. Such a computation may be ef\ufb01ciently simulated on a classical computer. gates, Pauli We have already implicitly proved the Gottesman\u2013Knill theorem. The way the classical computer performs the simulation is simply to keep track of the generators of the stabilizer as the various operations are being performed in the computation. For example, to simulate a Hadamard gate we simply update each of the n generators describing the quantum state. Similarly, simulation of the state preparation, phase gate, controlledgate, Pauli gates, and measurements of observables in the Pauli group may all be done using O(n2) steps on a classical computer, so that a quantum computation involving m operations from this set can be simulated using O(n2m) operations on a classical computer. The Gottesman\u2013Knill theorem highlights how subtle is the power of quantum computation. It shows that some quantum computations involving highly entangled states may be simulated ef\ufb01ciently on classical computers. Of course, not all quantum computations (and therefore, not all types of entanglement) can be described ef\ufb01ciently within the stabilizer formalism, but an impressive variety can be. Consider that interesting quantum information processing tasks such as quantum teleportation (Section 1.3.7) and superdense coding (Section 2.3) can be performed using only the Hadamard gate, controlledgate, and measurements in the computational basis, and can therefore be ef\ufb01ciently simulated on a classical computer, by the Gottesman\u2013Knill theorem. Moreover, we will shortly see that a wide variety of quantum error-correcting codes can be described within the stabilizer formalism. There is much more to quantum computation than just the power bestowed by quantum entanglement! Exercise 10.42: Use the stabilizer formalism to verify that the circuit of Figure 1.13 on page 27 teleports qubits, as claimed. Note that the stabilizer formalism restricts the class of states being teleported, so in some sense this is not a complete description of teleportation, nevertheless it does allow an understanding of the dynamics of teleportation. 10.5.5 Stabilizer code constructions The stabilizer formalism is ideally suited to the description of quantum codes. In this section we describe how this may be done, and use it to illustrate several important codes, such as Shor\u2019s nine qubit code, CSS codes, and a \ufb01ve qubit code which is the smallest Stabilizer codes 465 code that can be used to protect against the effects of arbitrary errors on a single qubit. The basic idea is very simple: an [n, k] stabilizer code is de\ufb01ned to be the vector space VS stabilized by a subgroup S of Gn such that \u2212I \u2208 S and S has n \u2212 k independent and commuting generators, S = g1,..., gn\u2212k. We denote this code C(S). What are the logical basis states for the code C(S)? In principle, given n\u2212k generators for the stabilizer S we can choose any 2k orthonormal vectors in the code C(S) to act as our logical computational basis states. In practice it makes a great deal more sense to choose the states in a more systematic way. One method is as follows. First, we choose operators \u00afZ1,..., \u00afZk \u2208 Gn such that g1,..., gn\u2212k, \u00afZ1,..., \u00afZk forms an independent and commuting set. (We explain in detail how this may be done a little later.) The \u00afZj operator plays the role of a logical Pauli sigma z operator on logical qubit number j, so the logical computational basis state |x1,..., xkL is therefore de\ufb01ned to be the state with stabilizer g1,..., gn\u2212k, (\u22121)x1 \u00afZ1,..., (\u22121)xk \u00afZk. (10.101) Similarly, we de\ufb01ne \u00afXj to be that product of Pauli matrices which takes \u00afZj to \u2212 \u00afZj under conjugation, and leaves all other \u00afZi and gi alone when acting by conjugation. Clearly \u00afXj gate acting on the jth encoded qubit. The", " operator has the effect of a quantum \u00afXj satis\ufb01es \u00afXjgk \u00afX \u2020 j = gk, and thus commutes with all the generators of the stabilizer. \u00afXj commutes with all the \u00afZi except for \u00afZj, with which it It is also easy to check that anti-commutes. How are the error-correcting properties of a stabilizer code related to the generators of its stabilizer? Suppose we encode a state using an [n, k] stabilizer code C(S) with stabilizer S = g1,..., gn\u2212k, and an error E occurs corrupting the data. In three stages of analysis we are going to determine what types of errors can be detected using C(S) and when recovery can be performed. First, we\u2019ll take a look at the effect different types of errors have on the code space, simply to gain some intuition about what types of errors can be detected and corrected; there will be no proofs, as this stage is simply to build intuition. The second stage is the statement and proof of a general theorem telling us what kinds of errors can be detected and corrected by a stabilizer code, based upon the quantum error-correction conditions. The third stage of our analysis is to give a practical prescription for performing error-detection and recovery, using notions such as error syndrome. Suppose C(S) is a stabilizer code corrupted by an error E \u2208 Gn. What happens to the code space when E anti-commutes with an element of the stabilizer? In this case E takes C(S) to an orthogonal subspace, and the error can in principle be detected (and perhaps corrected after detection) by performing an appropriate projective measurement. If E \u2208 S we don\u2019t need to worry since the \u2018error\u2019 E doesn\u2019t corrupt the space at all. The real danger comes when E commutes with all the elements of S but is not actually in S, that is, Eg = gE for all g \u2208 S. The set of E \u2208 Gn such that Eg = gE for all g \u2208 S is known as the centralizer of S in Gn and is denoted Z(S). In fact, for the stabilizer groups S of concern to us, the centralizer is identical to a more familiar group, the normalizer of S, denoted N (S), which is de\ufb01ned to consist of all elements E of Gn such that EgE\u2020 \u2208 S for all g \u2208 S. Exercise 10.43: Show that S \u2286 N (S) for any subgroup S of Gn. 466 Quantum error-correction Exercise 10.44: Show that N (S) = Z(S) for any subgroup S of Gn not containing \u2212I. These observations about the effect of various types of error operator E motivate the statement and proof of the following theorem, which is essentially just a translation of the quantum error-correction conditions (Theorem 10.1) into the terms of stabilizer codes. Theorem 10.8: (Error-correction conditions for stabilizer codes) Let S be the stabilizer for a stabilizer code C(S). Suppose {Ej} is a set of operators in Gn such that E\u2020 j Ek \u2208 N (S) \u2212 S for all j and k. Then {Ej} is a correctable set of errors for the code C(S). Without loss of generality we can restrict ourselves to considering errors Ej in Gn j = Ej, which reduces the error-correction conditions for stabilizer codes to such that E\u2020 having EjEk \u2208 N (S) \u2212 S for all j and k. Proof Let P be the projector onto the code space C(S). For given j and k there are two possibilities: either E\u2020 j Ek in Gn \u2212 N (S). Consider the \ufb01rst case. Then P E\u2020 j EkP = P since P is invariant under multiplication by elements of S. Suppose j Ek \u2208 Gn \u2212 N (S) so that E\u2020 E\u2020 j Ek must anticommute with some element g1 of S. Let g1,..., gn\u2212k be a set of generators of S, so that j Ek in S or E\u2020 Using the anti-commutativity gives P = + n\u2212k l=1 (I + gl) 2n\u2212k. j EkP = (I \u2212 g1)E\u2020 E\u2020 j Ek + n\u2212k l=2 (I + gl) 2n\u2212k. (10.102) (10.103) j EkP = 0 whenever j Ek \u2208 Gn \u2212 N (S). It follows that the set of errors {Ej} satis\ufb01es the quantum error- But P", " (I \u2212 g1) = 0 since (I + g1)(I \u2212 g1) = 0 and therefore P E\u2020 E\u2020 correction conditions, and thus forms a correctable set of errors. The statement and proof of Theorem 10.8 are wonderful theoretical results, but they don\u2019t explicitly tell us how to perform the error-correction operation when it is in fact possible! To understand how this is achieved, suppose g1,..., gn\u2212k is a set of generators for the stabilizer of an [n, k] stabilizer code, and that {Ej} is a set of correctable errors for the code. Error-detection is performed by measuring the generators of the stabilizer g1 through gn\u2212k in turn, to obtain the error syndrome, which consists of the results of the measurements, \u03b21 through \u03b2n\u2212k. If the error Ej occurred then the error syndrome is given by \u03b2l such that EjglE\u2020 j = \u03b2lgl. In the case when Ej is the unique error operator having this syndrome recovery may be achieved simply by applying E\u2020 j. In the case when there are two distinct errors Ej and Ej giving rise to the same error syndrome, it follows that EjP E\u2020 j, where P is the projector onto the code space, so E\u2020 j Ej P E\u2020 j after the error Ej has occurred results in a successful recovery. Thus, for each possible error syndrome we simply pick out a single error Ej with that syndrome, and apply E\u2020 j to achieve recovery when that syndrome is observed. j Ej \u2208 S, and thus applying E\u2020 jEj = P, whence E\u2020 j = Ej P E\u2020 Stabilizer codes 467 Theorem 10.8 motivates the de\ufb01nition of a notion of distance for a quantum code analogous to the distance for a classical code. We de\ufb01ne the weight of an error E \u2208 Gn to be the number of terms in the tensor product which are not equal to the identity. For example, the weight of X1Z4Y8 is three. The distance of a stabilizer code C(S) is de\ufb01ned to be the minimum weight of an element of N (S) \u2212 S, and if C(S) is an [n, k] code with distance d then we say that C(S) is an [n, k, d] stabilizer code. By Theorem 10.8 a code with distance at least 2t + 1 is able to correct arbitrary errors on any t qubits, just as in the classical case. Exercise 10.45: (Correcting located errors) Suppose C(S) is an [n, k, d] stabilizer code. Suppose k qubits are encoded in n qubits using this code, which is then subjected to noise. Fortunately, however, we are told that only d \u2212 1 of the qubits are affected by the noise, and moreover, we are told precisely which d \u2212 1 qubits have been affected. Show that it is possible to correct the effects of such located errors. 10.5.6 Examples We now give a few simple examples of stabilizer codes, including already familiar codes such as the Shor nine qubit code and CSS codes, but from the new point of view of the stabilizer formalism. In each case, the properties of the codes follow easily by applying Theorem 10.8 to the generators of the stabilizer. With the examples under our belt we will turn our attention to \ufb01nding quantum circuits for performing encoding and decoding. The three qubit bit \ufb02ip code Consider the familiar three qubit bit \ufb02ip code spanned by the states |000 and |111, with stabilizer generated by Z1Z2 and Z2Z3. By inspection we see that every possible product of two elements from the error set {I, X1, X2, X3} \u2013 I, X1, X2, X3, X1X2, X1X3, X2X3 \u2013 anti-commutes with at least one of the generators of the stabilizer (except for I, which is in S), and thus by Theorem 10.8 the set {I, X1, X2, X3} forms a correctable set of errors for the three qubit bit \ufb02ip code with stabilizer Z1Z2, Z2Z3. Error-detection for the bit \ufb02ip code is effected by measuring the stabilizer generators, Z1Z2 and Z2Z3. If, for example, the error X1 occurred, then the stabilizer is transformed to \u2212Z1Z2, Z2Z3, so the syndrome measurement gives the results \u22121 and +1. Similarly, the error X2 gives error syndrome", " \u22121 and \u22121, the error X3 gives error syndrome +1 and \u22121, and the trivial error I gives error syndrome +1 and +1. In each instance recovery is effected in the obvious way simply by applying the inverse operation to the error indicated by the error syndrome. The error-correction operation for the bit \ufb02ip code is summarized in Figure 10.10. Of course, the procedure we have outlined is exactly the same as that described earlier for the three qubit bit \ufb02ip code! All this group-theoretic analysis would hardly be worthwhile if this was all the insight we gained. The real utility of the stabilizer formalism only starts to become apparent as we move to more complex examples. Exercise 10.46: Show that the stabilizer for the three qubit phase \ufb02ip code is generated by X1X2 and X2X3. 468 Quantum error-correction Z1Z2 Z2Z3 +1 +1 \u22121 +1 \u22121 +1 \u22121 \u22121 Error type no error bit 3 \ufb02ipped bit 1 \ufb02ipped bit 2 \ufb02ipped Action no action \ufb02ip bit 3 \ufb02ip bit 1 \ufb02ip bit 2 Figure 10.10. Error-correction for the three qubit bit \ufb02ip code in the language of stabilizer codes. Name g1 g2 g3 g4 g5 g6 g7 g8 \u00afZ \u00afX Operator XXXXXX I I I I I I XXXXXX XXXXXXXXX Z Z Z Z Z Z Z Z Z Figure 10.11. The eight generators for the Shor nine qubit code, and the logical Z and logical X operations. (Yes, they really are the reverse of what one might naively expect!) The nine qubit Shor code The stabilizer for the Shor code has eight generators, as illustrated in Figure 10.11. It is easy to check the conditions of Theorem 10.8 for the error set containing I and all single qubit errors. Consider, for example, single qubit errors like X1 and Y4. The product X1Y4 anti-commutes with Z1Z2, and thus is not in N (S). Similarly, all other products of two errors from this error set are either in S or else anti-commute with at least one element of S and thus are not in N (S), implying that the Shor code can be used to correct an arbitrary single qubit error. Exercise 10.47: Verify that the generators of Figure 10.11 generate the two codewords of Equation (10.13). Exercise 10.48: Show that the operations \u00afZ = X1X2X3X4X5X6X7X8X9 and \u00afX = Z1Z2Z3Z4Z5Z6Z7Z8Z9 act as logical Z and X operations on a Shor-code encoded qubit. That is, show that this \u00afZ is independent of and commutes with the generators of the Shor code, and that with the generators of the Shor code, and anti-commutes with \u00afZ. \u00afX is independent of and commutes The \ufb01ve qubit code What is the minimum size for a quantum code which encodes a single qubit so that any error on a single qubit in the encoded state can be detected and corrected? It turns out that the answer to this question is \ufb01ve qubits. (See Section 12.4.3). The stabilizer for Stabilizer codes 469 Name g1 g2 g3 g4 \u00afZ \u00afX Operator XZ ZX I I XZ ZX X I XZ Z ZX I XZ Z Z Z Z Z XXXXX Figure 10.12. The four generators for the \ufb01ve qubit code, and the logical Z and logical X operations. Note that the last three generators may be obtained by shifting the \ufb01rst right. the \ufb01ve qubit code has the generators given in Figure 10.12. Because the \ufb01ve qubit code is the smallest capable of protecting against a single error it might be thought that it is the most useful code; however, for many applications it is more transparent to use the Steane seven qubit code. Exercise 10.49: Use Theorem 10.8 to verify that the \ufb01ve qubit code can protect against an arbitrary single qubit error. The logical codewords for the \ufb01ve qubit code are |00000 + |10010 + |01001 + |10100 |0L = 1 4 |1L = 1 4 +|01010 \u2212 |11011 \u2212 |00110 \u2212 |11000 \u2212 |11101 \u2212 |00011 \u2212 |11110 \u2212 |01111 \u2212 |10001 \u2212 |01100 \u2212 |101", "11 + |00101 |11111 + |01101 + |10110 + |01011 +|10101 \u2212 |00100 \u2212 |11001 \u2212 |00111 \u2212 |00010 \u2212 |11100 \u2212 |00001 \u2212 |10000 \u2212 |01110 \u2212 |10011 \u2212 |01000 + |11010 (10.104) (10.105) Exercise 10.50: Show that the \ufb01ve qubit code saturates the quantum Hamming bound, that is, it satis\ufb01es the inequality of (10.51) with equality. CSS codes and the seven qubit code The CSS codes are an excellent example of a class of stabilizer codes, demonstrating beautifully how easy the stabilizer formalism makes it to understand quantum code construction. Suppose C1 and C2 are [n, k1] and [n, k2] classical linear codes such that C2 \u2282 C1 and C1 and C\u22a5 2 both correct t errors. De\ufb01ne a check matrix with the form H(C\u22a5 2 ) 0 0 H(C1). (10.106) To see that this de\ufb01nes a stabilizer code, we need the check matrix to satisfy the commutativity condition H(C\u22a5 2 )H(C1)T = [H(C1)G(C2)]T = 0 2 )H(C1)T = 0. But we have H(C\u22a5 470 Quantum error-correction because of the assumption C2 \u2282 C1. Indeed, it\u2019s an easy exercise to see that this code is exactly CSS(C1, C2), and that it is capable of correcting arbitrary errors on any t qubits. The seven qubit Steane code is an example of a CSS code, whose check matrix we have already seen, in Equation (10.83). Encoded Z and X operators may be de\ufb01ned for the Steane code as \u00afZ \u2261 Z1Z2Z3Z4Z5Z6Z7; \u00afX \u2261 X1X2X3X4X5X6X7. (10.107) Exercise 10.51: Verify that the check matrix de\ufb01ned in (10.106) corresponds to the stabilizer of the CSS code CSS(C1, C2), and use Theorem 10.8 to show that arbitrary errors on up to t qubits may be corrected by this code. Exercise 10.52: Verify by direct operation on the codewords that the operators of (10.107) act appropriately, as logical Z and X. 10.5.7 Standard form for a stabilizer code The construction of the logical Z and X operators for a stabilizer code is made much easier to understand if we put the code into standard form. To understand what the standard form is, consider the check matrix for an [n, k] stabilizer code C: & \u2019. G = G1|G2 This matrix has n \u2212 k rows. Swapping rows of this matrix corresponds to relabeling generators, swapping corresponding columns on both sides of the matrix to relabeling qubits, and adding two rows corresponds to multiplying generators; it is easy to see that we may always replace a generator gi by gigj when i = j. Thus there is an equivalent code with a different set of generators whose corresponding check matrix corresponds to the matrix G where Gaussian elimination has been done on G1, swapping qubits as necessary: (10.108) r{ n \u2212 k \u2212 r{ r,-./ I 0 n\u2212r,-./ A 0 r,-./ B D n\u2212r,-./ C E, (10.109) where r is the rank of G1. Next, swapping qubits as necessary we perform a Gaussian elimination on E to obtain r{ n \u2212 k \u2212 r \u2212 s{ s{ \u23a1 \u23a3 r,-./ I 0 0 n\u2212k\u2212r\u2212s,-./ A1 0 0 k+s,-./ A2 0 0 r,-./ B D1 D2 n\u2212k\u2212r\u2212s,-./ C1 I 0 k+s,-./ C2 E2 0 \u23a4 \u23a6. (10.110) The last s generators cannot commute with the \ufb01rst r generators unless D2 = 0, and thus we may assume that s = 0. Furthermore, we may also set C1 = 0 by taking appropriate linear combinations of rows, so our check matrix has the form: n\u2212k\u2212r r,-./,-./ I 0 I 0 r{ n \u2212 k \u2212 r{ n\u2212k\u2212r,-./ A1 0 k,-./ A2 0 k,-./ C E r,-./ B D (10.111), where we have relabeled E2 as E and", " D1 as D. It is not dif\ufb01cult to see that this procedure is not unique; however, we will say that any code with check matrix in the form (10.111) is in standard form. Given the standard form for a quantum code it is easy to de\ufb01ne encoded Z operations Stabilizer codes 471 for the code. That is, we have to pick k operators independent of the generators of the stabilizer and of one another, yet commuting with one another, and also with the generators of the stabilizer. Suppose we write the check matrix for these k encoded Z operators in the form Gz = [F1F2F3|F4F5F6], where all the matrices have k rows, and the respective column sizes are r, n \u2212 k \u2212 r, k, r, n \u2212 k \u2212 r and k. We choose these matrices such that Gz = [000|AT 2 0I]. The commutativity of these encoded Z operations with the elements of the stabilizer follows from the equation I \u00d7 (AT 2 )T + A2 = 0, and it is clear that the encoded Z operations commute with one another since they only contain products of Z operators. The independence of the encoded Z operators from the \ufb01rst r generators of the stabilizer follows from the fact that no X terms appear in the de\ufb01nition of the encoded Z operators, the independence from the set of n\u2212k \u2212r generators follows from the (n \u2212 k \u2212 r)\u00d7(n \u2212 k \u2212 r) identity matrix appearing in the check matrix for those generators, and the lack of any corresponding terms in the check matrix for the encoded Z operators. In a similar way we may pick the encoded X operators, with k\u00d72n check matrix [0ET I|CT 00]. Exercise 10.53: Prove that the encoded Z operators are independent of one another. Exercise 10.54: Prove that with the check matrix for the encoded X operators de\ufb01ned as above, the encoded X operators are independent of one another and of the generators, commute with the generators of the stabilizer, with each other, and \u00afXj commutes with all the \u00afZk except \u00afZj, with which it anti-commutes. As an example we bring the check matrix for the Steane code (Equation 10.83) into standard form. We have n = 7 and k = 1 for this code, and inspection of the check matrix shows that the rank of the \u03c3x part is r = 3. The matrix may be brought into standard form by swapping qubits 1 and 4, 3 and 4, and 6 and 7, then by adding row 6 to row 4, then row 6 to row 5, and \ufb01nally adding rows 4 and 5 to row 6. The resulting standard form is10.112) We read off A2 = (1, 1, 0) and thus the encoded Z has check matrix [0000000|1100001], which corresponds to \u00afZ = Z1Z2Z7. Recalling that qubits 1 and 4, 3 and 4, 6 and 7 were swapped, this corresponds to an encoded Z of \u00afZ = Z2Z4Z6 in the original code. This may appear rather puzzling in the light of Equation (10.107), which states that the encoded Z is \u00afZ = Z1Z2Z3Z4Z5Z6Z7; however, the puzzle is resolved by noting that the two \u2018different\u2019 encoded Z operations differ only by a factor Z1Z3Z5Z7, which is an element of the stabilizer of the Steane code, and thus both have the same effect on Steane code states. Exercise 10.55: Find the \u00afX operator for the standard form of the Steane code. Exercise 10.56: Show that replacing an encoded X or Z operator by g times that 472 Quantum error-correction operator, where g is an element of the stabilizer, does not change the action of the operator on the code. Exercise 10.57: Give the check matrices for the \ufb01ve and nine qubit codes in standard form. 10.5.8 Quantum circuits for encoding, decoding, and correction One of the features of stabilizer codes is that their structure enables systematic construction of procedures for encoding, decoding, and error-correction. We describe the general method \ufb01rst, then present some explicit circuit constructions as examples. Let us begin with a general case, an [n, k] stabilizer code with generators g1,..., gn\u2212k, and logical Z operators Z1,..., \u00afZk. Preparing an encoded |0\u2297k state, which is the standard state for beginning a quantum computation, is quite simple. To do this we can start with any easily-", "prepared state \u2013 say, the state |0\u2297n \u2013 and measure each of the observables g1,..., gn\u2212k, Z1,..., Zk in turn. Depending on the measurement outcomes the resulting quantum state will have stabilizer \u00b1g1,...,\u00b1 gn\u2212k, \u00b1Z1,...,\u00b1 Zk, with the various signs being determined by the respective measurement outcomes. The signs of all the stabilizer generators and the \u00afZj can then be \ufb01xed up by applying products of Pauli operators, as described in the proof of Proposition 10.4, resulting in a state with stabilizer g1,..., gn\u2212k, \u00afZ1,..., \u00afZk, that is, to the encoded |0\u2297k. Once this state is prepared it is possible to change it to an arbitrary encoded computational basis state |x1,..., xk by applying the appropriate operators from the set \u00afX1,..., \u00afXk. Of course, this approach to encoding has the disadvantage that it is not unitary. To obtain fully unitary encoding a different approach based upon the standard form of the check matrix may be used; this approach is outlined in Problem 10.3. Also, if you wish to encode an unknown state, this can also be done systematically, starting from an encoded |0\u2297k state, as explained in Problem 10.4. For our purposes it will be suf\ufb01cient to prepare encoded |0\u2297k states. Decoding quantum codes is also quite simple, however it is worth explaining why, for many purposes, a full decoding is not necessary. It turns out that the techniques of fault-tolerant quantum computation can be used to perform logical operations directly on encoded data, without the need to decode the data. Furthermore, the output of a computation performed in this way can be directly determined simply by measuring the logical Z operators, without the need to decode and measure in the computational basis. Thus, doing a fully unitary decoding which preserves the encoded quantum information is not so important for our purposes. If such a decoding procedure is desired for some reason \u2013 perhaps one is using quantum error-correcting codes to transmit information over a noisy communication channel \u2013 then it may be achieved by running the unitary encoding circuit found in Problem 10.3 backwards. The error-correction procedure for a stabilizer code has already been described in Section 10.5.5, and is much like the encoding procedure: simply measure each of the generators g1,... gn\u2212k in turn, obtaining the error syndrome \u03b21,..., \u03b2n\u2212k. Classical computation is then used to determine from \u03b2j the required recovery operations E\u2020 j. The key to constructing encoding, decoding, and error-correction circuits in each of the above descriptions is understanding how to measure operators. Recall that this is a generalization of the normal projective measurements we have widely used, in which the objective is to project a state into an eigenstate of the operator and to obtain both Stabilizer codes 473 the projected state, and an indicator of the eigenvalue. If this reminds you of the phase estimation algorithm of Chapter 5, it\u2019s no coincidence! Recall from that chapter, and from Exercise 4.34 on page 188, that the circuit shown in Figure 10.13 can be used to measure the single qubit operator M (with eigenvalues \u00b11), given a gate which performs a controlled-M operation. Two useful versions of this circuit, which measure X and Z, are given in Figures 10.14 and 10.15. Figure 10.13. Quantum circuit for measuring a single qubit operator M with eigenvalues \u00b11. The top qubit is the ancilla used for the measurement, and the bottom qubit is being measured. |0 H \u2022 X H |0 \u2295 = H \u2022 H Figure 10.14. Quantum circuit for measuring the X operator. Two equivalent circuits are given; the one on the left is the usual construction (as in Figure 10.13), and the one on the right is a useful equivalent circuit. Figure 10.15. Quantum circuit for measuring the Z operator. Two equivalent circuits are given; the one on the left is the usual construction (as in Figure 10.13), and the one on the right is a useful simpli\ufb01cation. Of course, there is nothing special about the fact that M is a single qubit operator: the circuit in Figure 10.13 works just as well if we replace the second qubit with a bundle of qubits, and M is an arbitrary Hermitian operator with eigenvalues \u00b11. Such operators include, for example, the products of Pauli operators we need to measure during the encoding, decoding and error-correction procedures for stabilizer", " codes. As a concrete example, consider the syndrome measurement and encoding procedures for the seven qubit Steane code. A convenient starting point is the standard form of the check matrix for the code, Equation (10.112), because we can immediately read off the generators we need to measure directly from this matrix. Speci\ufb01cally, recall that the left block corresponds to X generators, and the right, Z, so the quantum circuit shown in Figure 10.16 immediately results. Note how the location of the zeroes and ones in the matrix corresponds to the location of the targets for the gates in the left half (which measure X), and the targets in the right half (which measure Z). This circuit can be used to perform error-correction by following the measurement results with products of Pauli operators on the code qubits to correct the errors. Or, by adding an additional 474 Quantum error-correction measurement of \u00afZ and \ufb01xing the signs in the generators of the stabilizer, as described earlier, the circuit can be used to prepare the encoded logical state |0L. |0 |0 |0 |0 |0 | Figure 10.16. Quantum circuit for measuring the generators of the Steane code, to give the error syndrome. The top six qubits are the ancilla used for the measurement, and the bottom seven are the code qubits. Exercise 10.58: Verify that the circuits in Figures 10.13\u201310.15 work as described, and check the claimed circuit equivalences. Exercise 10.59: Show that by using the identities of Figures 10.14 and 10.15, the syndrome circuit of Figure 10.16 can be replaced with the circuit of Figure 10.17 Exercise 10.60: Construct a syndrome measuring circuit analogous to that in Figure 10.16, but for the nine and \ufb01ve qubit codes. Exercise 10.61: Describe explicit recovery operations E\u2020 j corresponding to the different possible error syndromes that may be measured using the circuit in Figure 10.16. 10.6 Fault-tolerant quantum computation One of the most powerful applications of quantum error-correction is not merely the protection of stored or transmitted quantum information, but the protection of quantum information as it dynamically undergoes computation. Remarkably, it turns out that arbitrarily good quantum computation can be achieved even with faulty logic gates, provided only that the error probability per gate is below a certain constant threshold. Over Fault-tolerant quantum computation 475 \u2295 \u2295 \u2295 \u2295 \u2295 \u2295 \u2295 \u2295 \u2295 \u2295 \u2295 \u2295 |0 |0 |0 |0 |0 | \u2022 Figure 10.17. Quantum circuit equivalent to the one in Figure 10.16. the next few sections we explain the techniques of fault-tolerant quantum computation which are used to achieve this remarkable result. We look \ufb01rst at the big picture in Section 10.6.1, before examining in detail the various elements of fault-tolerant quantum computation in Sections 10.6.2 and 10.6.3, and conclude in Section 10.6.4 with a discussion of some of the limitations and possible extensions of the fault-tolerant constructions. Note that a rigorous discussion of the many subtleties of fault-tolerant quantum computation lies somewhat beyond our scope; the interested reader is referred to the end of chapter \u2018History and further reading\u2019. 10.6.1 Fault-tolerance: the big picture The theory of fault-tolerant quantum computation integrates many different ideas en route to the threshold condition. We now describe each of these ideas in turn. We begin with the notion of computing on encoded data, and explain how the fundamental problems of error propagation and error accumulation require that our circuits for computing on encoded data satisfy certain fault-tolerance criteria. We then introduce our fundamental noise model for quantum circuits, which allows us to give more precise de\ufb01nitions for the notion of a fault-tolerant operation. We work through a speci\ufb01c example of a fault\u2013 explaining how it can be tolerant operation in action \u2013 the fault-tolerant controlledused to prevent the propagation and accumulation of errors. We conclude by explaining how fault-tolerant operations can be combined with a procedure known as concatenation to obtain the threshold theorem for quantum computation, and give a simple estimate for the threshold. 476 Quantum error-correction Fundamental issues The basic idea of fault-tolerant quantum computation is to compute directly on encoded quantum states in such a manner that decoding is never required. Suppose we are given a simple quantum circuit such as that shown in Figure 10.18. Unfortunately, noise af\ufb02icts each of the elements used to build this circuit \u2013 the state preparation procedures, quantum logic gates, measurement of the output, and even", " the simple transmission of quantum information along the quantum wires. To combat the effect of this noise we replace each qubit in the original circuit with an encoded block of qubits, using an error-correcting code such as the seven qubit Steane code, and replace each gate in the original circuit with a procedure for performing an encoded gate acting on the encoded state, as shown in Figure 10.19. By performing error-correction periodically on the encoded state we prevent accumulation of errors in the state. Of course, merely performing error-correction periodically is not suf\ufb01cient to prevent the build-up of errors, even if it is applied after every encoded gate. The reasons for this are two-fold. First, and most importantly, the encoded gates can cause errors to propagate. For example, the encoded controlledillustrated in Figure 10.20 may cause an error on the encoded control qubit to propagate to the encoded target qubit. Thus errors in the qubits forming the encoded control qubit can propagate to become errors in the encoded target qubit. Encoded gates should therefore be designed very carefully so that a failure anywhere during the procedure for performing the encoded gate can only propagate to a small number of qubits in each block of the encoded data, in order that error-correction will be effective at removing the errors. Such procedures for performing encoded gates are referred to as fault-tolerant procedures, and we will show that it is possible to perform a universal set of logical and \u03c0/8 gates \u2013 using fault-tolerant operations \u2013 the Hadamard, phase, controlledprocedures. The second issue that must be addressed is that error-correction itself can introduce errors on the encoded qubits, so we must be careful to design error-correction procedures that do not introduce too many errors into the encoded data. This can be achieved using techniques similar to those used to prevent propagation of errors by encoded gates, by taking care to ensure that failures during the procedure for errorcorrection do not propagate to cause too many errors in the encoded data. Figure 10.18. A simple quantum circuit. If each component in the circuit fails with probability p then the probability of an error at the output is O(p). Fault-tolerant operations: de\ufb01nitions Let\u2019s pin down a little more precisely what it means for a particular procedure implementing an encoded quantum gate to be fault-tolerant. We de\ufb01ne the fault-tolerance of a procedure to be the property that if only one component in the procedure fails then the failure causes at most one error in each encoded block of qubits output from the procedure. For example, the failure of a single component in a fault-tolerant recovery Fault-tolerant quantum computation 477 Figure 10.19. A simulation of the circuit in Figure 10.18, using encoded qubits and encoded logical operations. If fault-tolerant procedures are used to perform all the operations then the probability of error at the output is O(p2), where p is the probability for any individual component in the circuit to fail. An interesting feature is the second error-correction step performed on the second qubit. The \u2018operation\u2019 being error-corrected by this step appears to be trivial: nothing happens to the qubit at all! Nevertheless, simply storing qubits for a period of time introduces errors into the qubits, and should be periodically error-corrected in order to prevent the accumulation of errors. \u00d7 \u2022 \u2295 \u2212\u2192 \u2022 \u00d7 \u2295 \u00d7 Figure 10.20. A controlledaffects two. This is also true when encoded qubits are used, and an encoded controlleddiscussed in the text. gate can cause an error to propagate so that instead of affecting one qubit, it is implemented, as procedure for quantum error-correction results in the recovery procedure being performed correctly, up to an error on a single qubit of the output. By \u2018component\u2019 we mean any of the elementary operations used in the encoded gate, which might include noisy gates, noisy measurements, noisy quantum wires, and noisy state preparations. This de\ufb01nition of a fault-tolerant procedure for a quantum gate is sometimes generalized in the literature to cope with some of the more subtle issues that arise in the theory of fault-tolerant computation, but at our level of detail it is suf\ufb01cient. Of course, encoded quantum gates aren\u2019t all we wish to perform during our quantum computations. It is also useful to de\ufb01ne the notions of a fault-tolerant measurement procedure, and fault-tolerant state preparation. A procedure for measuring an observable on a set of encoded qubits is said to be fault-tolerant if the failure of any single component in the procedure results in an error in at most one qubit in each encoded block of qubits at", " the output of the procedure. Furthermore, we require that if only one component fails then the measurement result reported must have probability of error of O(p2), where p is the (maximum) probability of a failure in any one of the components used to implement the measurement procedure. A procedure for preparing a \ufb01xed encoded state is said to be fault-tolerant if, given that a single component failed during the procedure, there is at most a single qubit in error in each encoded block of qubits output from the procedure. To make these notions of fault-tolerance more precise, we need to be a little more speci\ufb01c about our error model. One of the major simpli\ufb01cations we are going to make in our analysis is to describe errors on qubits as being of one of four types: I, X, Y or Z, occurring stochastically with appropriate probabilities. We allow correlated errors, but again to occur on two qubits, when performing gates such as the controlledwill assume that they are of the form of a tensor product of Pauli matrices, occurring with some probability. This probabilistic analysis enables us to use familiar concepts from classical probability theory to determine the total probability that the output from a circuit is correct or not. In more sophisticated presentations of fault-tolerance (see 478 Quantum error-correction \u2018History and further reading\u2019) much more general error models can be considered, for example, allowing correlated errors of arbitrary type across several qubits. Nevertheless, the techniques used in those more sophisticated analyses are essentially generalizations of those we describe, combined with the deep insight we obtained earlier in the chapter that to perform error-correction of a continuum of possible errors it suf\ufb01ces to correct a discrete set of errors. With our noise model at hand, we can be more precise about what we mean when we gate is gate is denoted U, then the effective action say that an error \u2018propagates\u2019 through the circuit. Consider, for example, the in Figure 10.20. Imagine that an X error on the \ufb01rst qubit occurs just before the applied. If the unitary operator for the of the circuit is U X1 = U X1U \u2020U = X1X2U, that is, it is as though the controlledwas applied correctly, but an X error occurred on both qubits after the. Over the remainder of this chapter we repeatedly use this trick of conjugating errors through gates to study how errors propagate through our circuits. A slightly more challenging example of error propagation is to suppose it is the gate itself which fails. What happens gate implements the quantum operation E. Then this then? Suppose our noisy may be rewritten as E = E \u25e6 U \u22121 \u25e6 U, where U is the quantum operation implementing a gate is equivalent to applying a perfect perfect followed by the operation E \u25e6 U \u22121, which is approximately the identity if our noisy is reasonably good, and can be understood in our usual error model of tensor products such as X \u2297 Z occurring on the two qubits with some small probability p. gate. Thus, the noisy Over the next few sections we explain in detail procedures for performing each class of fault-tolerant operations we have described \u2013 fault-tolerant quantum logic with a universal set of gates, fault-tolerant measurement, and fault-tolerant state preparation. The actual constructions we describe are for the Steane code, but they generalize fairly easily to the case of more general stabilizer codes. For now, however, we imagine that we have all of these procedures at our disposal. How can we put them together to perform quantum computation? Example: fault-tolerant controlled- Let\u2019s examine a procedure for implementing a fault-tolerant controlledgate, followed by a fault-tolerant error-correction step, as illustrated in Figure 10.21. Our analysis of this circuit proceeds in 4 steps. Step 1 is the point of entry into the circuit, step 2 is has been performed, step 3 is after the syndrome measurement, after the encoded and step 4 is after the recovery operation has been applied. Our goal is to show that the probability that this circuit introduces two or more errors in the \ufb01rst encoded block behaves as O(p2), where p is the probability of failure of individual components in the circuit. Because a (hypothetical) perfect decoding of the \ufb01rst block of qubits only fails if there are two or more errors in the block, it follows that the probability that a perfectly decoded state contains errors is at most O(p2) larger after the action of this circuit than it was before. To show that this procedure introduces two errors into the \ufb01rst encoded block with probability O(p2), let\u2019s identify all the possible ways this circuit can introduce", " two or more errors into the \ufb01rst encoded block of qubits at the output: (1) There is a single pre-existing error entering the circuit at step 1 in each encoded block of qubits. This can cause two errors in the output from the \ufb01rst block because, Fault-tolerant quantum computation 479 / / \u2191 1 \u2022 \u2295 \u2191 2 Syndrome measurement Syndrome measurement Recovery Recovery \u2191 3 \u2191 4 Figure 10.21. Block diagram of the fault-tolerant procedure construction, including error-correction. for example, the error on the second block may propagate through the encoded circuit to cause an error on the \ufb01rst block of qubits. Provided the operations up to this stage have been done fault-tolerantly, we can argue that the probability of such an error entering on the \ufb01rst block is at most c0p for some constant c0, since such an error must have occurred during either the syndrome measurement or recovery stages in the previous stage of the quantum circuit. c0 is the total number of places at which a failure may occur during syndrome measurement or recovery in the previous stage of the circuit. If we assume for simplicity that the probability of a single pre-existing error entering at step 1 on the second block is also c0p, and that these two errors occur independently, then the probability of this event is at most c2 0p2. For the Steane code construction described below, there are contributions to c0 from the six separate syndrome measurements, each of which has approximately 101 locations at which a failure may occur, together with a recovery operation involving seven components, for a total of approximately c0 \u2248 70. (2) A single pre-existing error enters in step 1 on either the \ufb01rst or second block of. The qubits, and a single failure occurs during the fault-tolerant controlledprobability of this is c1p2, where c1 is a constant de\ufb01ned to be the number of pairs of points at which a failure may occur. For the Steane code construction described below, we argued previously that there are roughly 70 locations times two blocks where a failure may have occurred causing an error to enter the circuit, for a total of 140 locations. There are a further 7 locations at which a failure may occur during the circuit, for a total of c1 \u2248 7 \u00d7 140 \u2248 103 locations at which a pair of failures may occur. (3) Two failures occur during the fault-tolerant. This happens with probability at most c2p2, where c2 is the number of pairs of points at which a failure may occur. For the Steane code, c2 \u2248 102. (4) A failure occurs during the and during the syndrome measurement. The only way two or more errors can occur at the output is if the syndrome measurement gives the incorrect result, which occurs with probability c3p2 for some constant c3 (for the Steane code, c3 \u2248 102). Another case which appears to be of interest, but which does not actually matter, is when the syndrome measurement gives the correct result, in which case the error introduced by the diagnosed and corrected by recovery, leaving only a single error at the output, introduced during syndrome measurement. is correctly (5) Two or more failures occur during the syndrome measurement. This happens with 480 Quantum error-correction probability at most c4p2, where c4 is the number of pairs of points at which a failure may occur. For the Steane code, c4 \u2248 702 \u2248 5 \u00d7 103. (6) A failure occurs during syndrome measurement and a failure during recovery. This happens with probability at most c5p2, where c5 is the number of pairs of points at which a failure may occur. For the Steane code, c5 \u2248 70 \u00d7 7 \u2248 500. (7) Two or more failures occur during recovery. This happens with probability at most c6p2, where c6 is the number of pairs of points at which a failure may occur. For the Steane code, c6 \u2248 72 \u2248 50. Thus, the probability that this circuit introduces two or more errors into the encoded \ufb01rst block of qubits is at most cp2 for some constant c = c2 0 + c1 + c2 + c3 + c4 + c5 + c6 which is approximately equal to 104 for the Steane code. If a perfect decoding were to be performed at the end of the circuit, the probability of an error would therefore be at most cp2. This is a truly remarkable result: we have managed to \ufb01nd an implementation for with the property that individual components may fail with probability p, but the the encoded procedure succeeds with probability 1 \u2212 cp2, and thus, provided p is small enough, in the example, p < 10\u22124, there", " is a net gain to be had by using the encoded procedure! Similar conclusions can be drawn for all the other operations one might wish to do during a quantum computation, so that by doing any of our operations fault-tolerantly we can reduce our probability of a failure from p to cp2, for some constant c. We have estimated c for the, however estimates for other fault-tolerant operations do not differ all that greatly, and we will continue to use c \u2248 104 in our numerical estimates. Concatenated codes and the threshold theorem There is a beautiful construction based on concatenated codes which can be used to reduce the effective error rate achieved by the computation even further! The idea is to recursively apply the scheme described above for simulating a circuit using an encoded circuit, constructing a hierarchy of quantum circuits C0 (the original circuit we wish to simulate), C1, C2,.... In the \ufb01rst stage of this construction, each qubit in the original circuit is encoded in a quantum code whose qubits are themselves encoded in a quantum code, whose own qubits are encoded yet again, and so forth ad in\ufb01nitum, as illustrated in Figure 10.22. In the second stage of this construction, any given gate in the original circuit C0, such as a Hadamard gate, is replaced in the circuit C1 by a fault-tolerant procedure implementing an encoded Hadamard gate and error-correction. Each of the components used in the circuit C1 is then replaced in the circuit C2 by a fault-tolerant procedure implementing an encoded version of the component and error-correction, and so on, ad in\ufb01nitum. Suppose we do two levels of concatenation, as illustrated. If the failure probability of components at the lowest level of the code \u2013 the actual physical qubits \u2013 is p, then the failure probability at the middle level (one level of encoding) is at most cp2, and at the highest level (two levels of encoding) \u2013 the level at which the circuit must function correctly if the computation is to produce the correct output \u2013 is c(cp2)2. Thus, if we concatenate k times, the failure probability for a procedure at the highest level is (cp)2k /c, while the size of the simulating circuit goes as dk times the size of the original circuit, where d is a constant representing the maximum number of operations used in a fault-tolerant procedure to do an encoded gate and error-correction. Suppose then that we wish to simulate a circuit containing p(n) gates, where n speci\ufb01es the size of some problem, and p(n) is a polynomial function in n. This might be, for Fault-tolerant quantum computation 481 Figure 10.22. A two level concatenated code, encoding a single qubit in nine qubits. We use a three qubit code merely to keep the \ufb01gure simple; in practice a code such as the Steane code which can correct arbitrary errors on one or more qubits would be used. example, the circuit for the quantum factoring algorithm. Suppose we wish to achieve a \ufb01nal accuracy of in our simulation of this algorithm. To do so our simulation of each gate in the algorithm must be accurate to /p(n), so we must concatenate a number of times k such that (cp)2k c \u2264 p(n). (10.113) Provided p < pth \u2261 1/c such a k can be found. This condition \u2013 that p < pth \u2013 is known as the threshold condition for quantum computation, since provided it is satis\ufb01ed we can achieve arbitrary accuracy in our quantum computations. How large a simulating circuit is required to achieve this level of accuracy? Note that we have dk = log(p(n)/c) log(1/pc) log d poly(log p(n)/), = O (10.114) where poly indicates a polynomial of \ufb01xed degree, and thus the simulating circuit contains O(poly(log p(n)/)p(n)) (10.115) gates, which is only polylogarithmically larger than the size of the original circuit. Summarizing, we have the threshold theorem for quantum computation: Threshold theorem for quantum computation: A quantum circuit containing p(n) gates may be simulated with probability of error at most using O(poly(log p(n)/)p(n)) (10.116) gates on hardware whose components fail with probability at most p, provided p is below some constant threshold, p < pth, and given reasonable assumptions about the noise in the underlying hardware. 482 Quantum error-correction What is the value of pth? For the Steane", " code, c \u2248 104 according to our counting, so a very rough estimate gives pth \u2248 10\u22124. It needs to be emphasized that our estimates are (very) far from rigorous, however much more sophisticated calculations for the threshold have typically yielded values in the range 10\u22125\u201310\u22126. Note that the precise value of the threshold depends greatly on the assumptions made about the computational capabilities! For example, if parallel operations are not possible, then the threshold condition is impossible to achieve, because errors accumulate in the circuit too quickly for errorcorrection to cope with. Classical computation is also required in addition to the quantum operations, to process the measured syndromes and determine what quantum gates to apply to correct errors. Some discussion of limitations on estimates of the threshold are given in Section 10.6.4. Exercise 10.62: Show by explicit construction of generators for the stabilizer that concatenating an [n1, 1] stabilizer code with an [n2, 1] stabilizer code gives an [n1n2, 1] stabilizer code. 10.6.2 Fault-tolerant quantum logic A key technique in the construction of fault-tolerant quantum circuits is the method of constructing fault-tolerant operations to do logic on encoded states. In Section 4.5.3 of and \u03c0/8 gates form a Chapter 4 we learned that the Hadamard, phase, controlleduniversal set in terms of which any quantum computation may be expressed. We now explain how each of these gates can be implemented fault-tolerantly. Normalizer operations We begin with fault-tolerant constructions for the normalizer operations \u2013 the Hadamard, phase and controlledgates \u2013 for the speci\ufb01c case of the Steane code. By understanding the basic principles underlying the constructions for this concrete example it is easy to generalize them to any stabilizer code. Recall from Equation (10.107) that for the Steane code the Pauli \u00afZ and \u00afX operators on the encoded states can be written in terms of operators on the unencoded qubits, as Z = Z1Z2Z3Z4Z5Z6Z7; \u00afX = X1X2X3X4X5X6X7. (10.117) An encoded Hadamard gate \u00afH should interchange \u00afZ and \u00afX under conjugation, just as the Hadamard gate interchanges Z and X under conjugation. \u00afH = H1H2H3H4H5H6H7 accomplishes this task, so that a Hadamard on the encoded qubit can be implemented as shown in Figure 10.23. Exercise 10.63: Suppose U is any unitary operation mapping the Steane code into itself, and such that U \u00afZU \u2020 = \u00afX and U \u00afXU \u2020 = \u00afZ. Prove that up to a global phase the action of U on the encoded states |0L and |1L is |0L \u2192 (|0L + |1L)/ 2 and |1L \u2192 (|0L \u2212 |1L)/ \u221a \u221a 2. This is a good \ufb01rst step, but just doing logic on the encoded state is not suf\ufb01cient to make this operation fault-tolerant! We also need to understand how errors propagate. Because the circuit implementing \u00afH = H \u22977 does not involve more than one qubit in the encoded block in interaction, it seems physically reasonable to assume that the failure of a single component in the circuit can cause at most one error in the block of qubits output Fault-tolerant quantum computation 483 a|0L + b|1L \u23a7 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 \u23ab \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23ac \u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\u23aa\ufffd", "\ufffd\ufffd\u23aa\u23aa\u23aa\u23aa\u23ad H H H H H H H a+b\u221a 2 |0L + a\u2212b\u221a 2 |1L Figure 10.23. Transversal Hadamard gate on a qubit encoded in the Steane code. from the procedure. To see that this is true, imagine an error on the \ufb01rst qubit occurred just before the encoded H gate was applied. For the sake of de\ufb01niteness, suppose that the error is a Z error, so the combined operation on the qubit is HZ. As in our earlier gate, inserting the identity H \u2020H = I gives analysis of error propagation for the HZ = HZH \u2020H = XH, so such an error is equivalent to \ufb01rst applying H then the error X occurring. Similarly, a failure during the gate operation itself is equivalent to a perfect gate, followed by a small amount of noise acting on the qubit, which we can think of in terms of our usual model of X, Y and Z all occurring with some small probability. The circuit in Figure 10.23 thus really is a fault-tolerant operation, because a single failure occurring anywhere in the procedure doesn\u2019t propagate to affect other qubits, and thus causes at most one error in the block of qubits output from the procedure. Are there any general principles we can distill from the circuit in Figure 10.23? One useful observation is that encoded gates are automatically fault-tolerant if they can be implemented in a bitwise fashion, since that property ensures that a single failure anywhere in the encoded gate introduces at most one error per block of the code, and thus error probabilities do not grow out of the control of the error-correcting code. This property, that an encoded gate can be implemented in a bitwise fashion, is known as the transversality property of an encoded quantum gate. Transversality is interesting because it offers a general design principle for \ufb01nding fault-tolerant quantum circuits, and we see below that many gates other than the Hadamard gate can be given transversal implementations. Keep in mind, though, that it is possible to \ufb01nd fault-tolerant constructions which aren\u2019t transversal, as we\u2019ll see below with the example of the fault-tolerant \u03c0/8 gate. Using the Steane code, many gates other than the Hadamard gate are easily given transversal (and thus fault-tolerant) implementations. Three of the most interesting, in addition to the Hadamard, are the phase gate and the Pauli X and Z gates. Suppose we apply the X gates bitwise to each of the seven qubits of the Steane code. This transforms each Z operator to \u2212Z under conjugation, so Z \u2192 (\u22121)7Z = \u2212Z and \u00afX \u2192 \u00afX under conjugation by the bitwise application of X, and thus this circuit effects an encoded X operation on the states of the Steane code. This circuit is transversal, and thus is automatically fault-tolerant. In a similar way, applying Z bitwise to the states of the Steane code gives a fault-tolerant implementation of an encoded Z. The transversal 484 Quantum error-correction \u00afS implementation of the phase gate is a little more challenging. Under conjugation, must take \u00afZ to \u00afZ and \u00afX to \u00afY = i \u00afX \u00afZ. However, applying the obvious guess \u00afS = S1S2S3S4S5S6S7 takes \u00afZ to \u00afZ under conjugation, and \u00afX to \u2212 \u00afY. The minus sign in front of the \u2212 \u00afY may be \ufb01xed up by applying \u00afZ. Thus, applying the operation ZS to each qubit in the code effects an encoded phase gate, which is transversal and thus fault-tolerant. In contrast to the Hadamard, Pauli, and phase gates, implementing the controlledfault-tolerantly appears at \ufb01rst to be a challenge, because it involves two separate code blocks of seven qubits. How can we realize a which does not introduce more than one error per block of the code? Fortunately, this turns out to be very simple when using the Steane code, as illustrated in Figure 10.24: it is easily seen to be effected by gates applied pairwise between the seven qubits in the two blocks! You seven might worry that this transversal construction violates our own rules; after all, can\u2019t the gates we are doing cause errors to propagate beyond a single qubit? This controlledis correct, but there is no problem, because the error propagation only ever affects at most one qubit in another block", "; it does not adversely affect qubits within the same block. Remember that affecting qubits in other blocks is okay because each block can handle errors on single qubits! gates fails. What happens then? Suppose our noisy More precisely, suppose that an X error on the \ufb01rst qubit occurs just before the between the \ufb01rst qubit of each block, which we\u2019ll label qubits 1 and 8. If controlledgate is denoted U, then the effective action is U X1 = U X1U \u2020U = this controlledX1X8U, that is, it is as though the controlledwas applied correctly, but an X error occurred on the \ufb01rst qubit of both blocks of encoded qubits! Slightly more challenging, gate suppose one of the implements the quantum operation E. Then this may be rewritten as E = E \u25e6 U \u22121 \u25e6 U, where U is the quantum operation implementing a perfect gate. Thus, the noisy followed by the operation E \u25e6 U \u22121, is reasonably good, and can be which is approximately the identity if our noisy understood in our usual error model of tensor products such as X \u2297 Z occurring on the two qubits with some small probability. Fortunately, while such errors involve two qubits, they only involve a single qubit in each block of encoded qubits. Similar conclusions about error propagation apply to errors at other locations. It follows that the failure of a single component anywhere within the procedure we have described propagates to cause no more than one error in each block of encoded qubits, and thus this implementation of the encoded controlled- gate is equivalent to applying a perfect is fault-tolerant. Having found fault-tolerant implementations of the Hadamard, phase and controlledgates, it follows from Theorem 10.6 that any operation in the normalizer can be realized fault-tolerantly. Of course, normalizer operations do not exhaust the complete set of unitary gates required to do quantum computation, but this is a promising start! Exercise 10.64: (Back propagation of errors) It is clear that an X error on the gate propagates to the target qubit. In addition, it turns control qubit of a out that a Z error on the target propagates back to the control! Show this using the stabilizer formalism, and also directly using quantum circuit identities. You may \ufb01nd Exercise 4.20 on page 179 useful. Fault-tolerant quantum computation 485 Figure 10.24. Transversal controlled- between two qubits encoded in separate blocks with the Steane code. Fault-tolerant \u03c0/8 gate The one remaining gate we require to complete the standard set of gates for universal quantum computation is the \u03c0/8 gate. Alternatively, as was noted in Section 4.5.3, adding a fault-tolerant Toffoli gate to our current set of fault-tolerant Hadamard, phase and gates would also give us a universal set, allowing us to perform all the controlledgates required by a quantum computer in a fault-tolerant manner. It turns out that the fault-tolerant \u03c0/8 gate is very simple to realize, and using a similar but more elaborate construction a fault-tolerant Toffoli gate can be realized. Our basic strategy in constructing the fault-tolerant \u03c0/8 gate is to split the construction into three parts. The \ufb01rst part of the construction is a simple circuit to simulate the \u03c0/8 gate using elements we already know how to do fault-tolerantly, such as the controlled, phase and X gates. There are, however, two parts of this circuit which we don\u2019t yet know how to make fault-tolerant. The \ufb01rst is the preparation of an ancilla state for input into the circuit. In order that this ancilla be adequate, we require that the failure of any component during the ancilla preparation should lead to at most a single error in the block of qubits making up the ancilla. We explain how such fault-tolerant ancilla preparation can be done later in this section. The second operation we need is measurement. In order to make the measurement fault-tolerant, we require that the failure of a single component during the procedure for measurement should not affect the measurement outcome. If it did, then the error would propagate to cause errors on many qubits in the \ufb01rst block, since whether the encoded SX operation is performed or not is determined by the measurement result. How to do such a fault-tolerant measurement is described in the next section. (Strictly speaking, for the fault-tolerant measurement procedure we describe the measurement outcome may actually be incorrect with probability O(p2), where p is the probability of failure of a single", " component. We will ignore this for the purposes of the present discussion; it is easily dealt with by a slightly more sophisticated analysis along similar lines.) Figure 10.25 shows a circuit implementing a \u03c0/8 gate. All elements in the circuit can be performed fault-tolerantly, except perhaps those in the dashed box and the measurement. The circuit starts with two encoded qubits, one of which is the qubit |\u03c8 = a|0+b|1 we wish to operate on (let |0 and |1 denote logical states here). The other qubit is prepared 486 Quantum error-correction Figure 10.25. Quantum circuit which fault-tolerantly implements a \u03c0/8 gate. The dashed box represents a (non-fault-tolerant) preparation procedure for the ancilla state (|0 + exp(i\u03c0/4)|1)/ 2; how to do this preparation fault-tolerantly is explained in the text. The slash on the wire denotes a bundle of seven qubits, and the double-line wire represents the classical bit resulting from the measurement. Note that the \ufb01nal SX operation is controlled by the measurement result. \u221a in the state |\u0398 = |0 + exp(i\u03c0/4) |1 \u221a 2, (10.118) which is the state generated by the circuit in the dotted box in the \ufb01gure. We explain how this ancilla preparation step may be done fault-tolerantly in a moment. Next, perform a fault-tolerant controlled- operation, giving a|1 + b|0 + exp(i\u03c0/4) |1 a|0 + b exp(i\u03c0/4) |1 b|0 + a exp(i\u03c0/4) |1 |1 |0 +. (10.119) 1\u221a 2 |0 a|0 + b|1 = 1\u221a 2 Finally, measure the second qubit, and if it is 0 then we are done. Otherwise, perform the operation SX = 1 0 0 i 0 1 1 0 (10.120) to the remaining qubit. Either way, we are left with the state a|0 + b exp(i\u03c0/4) |1 up to an irrelevant global phase, as required for a \u03c0/8 gate. This wonderful result may seem to have come out of nowhere, but in fact it is the result of a systematic construction which is explained in the exercises below. The same construction is used to realize a fault-tolerant Toffoli gate, as shown in Exercise 10.68. The construction of the fault-tolerant \u03c0/8 gate requires a fault-tolerant method to produce the ancilla state |\u0398. This preparation can be achieved using the techniques for fault-tolerant measurements, as explained in detail in the next section. For now we explain the connection to fault-tolerant measurement. As shown in Figure 10.25, |\u0398 may be produced by applying a Hadamard gate and then a \u03c0/8 gate to the state |0. The state |0 is a +1 eigenstate of Z, so it follows that |\u0398 is a +1 eigenstate of T HZHT \u2020 = T XT \u2020 = e\u2212i\u03c0/4SX. |\u0398 can therefore be prepared by \ufb01rst preparing an encoded |0, and then fault-tolerantly measuring e\u2212i\u03c0/4SX. If the result +1 is obtained we conclude that the state has been correctly prepared. If the result \u22121 is obtained, we have one of two options. We can either start over, repeating the procedure until the fault-tolerant measurement of e\u2212i\u03c0/4SX gives the result +1, or we can use the more elegant and ef\ufb01cient observation that since Z SX Z = \u2212SX, applying a fault-tolerant Z operation changes the state from the \u22121 eigenstate of e\u2212i\u03c0/4SX to the +1 eigenstate of Fault-tolerant quantum computation 487 e\u2212i\u03c0/4SX, |\u0398. Whichever procedure is used, a single failure anywhere in the procedure produces an error in at most one qubit in the ancilla state |\u0398. It is not dif\ufb01cult to see that the procedure we have described is fault-tolerant as a whole; however, it may be useful to look at an explicit example to see this. Suppose a single component failure occurred during ancilla construction, leading to an error on a single qubit in the ancilla. This propagates through the encoded controlledgate to cause one error in each of the \ufb01rst and second blocks of qubits. Fortunately, an error on a single", " qubit in the second encoded qubit doesn\u2019t affect the result of our fault-tolerant measurement procedure, so SX is applied or not applied as appropriate, and thus the error on the \ufb01rst block of qubits propagates through to cause a single error in the output from the encoded gate. Similarly, it is not dif\ufb01cult to convince yourself that a single failure anywhere else in this procedure for an encoded \u03c0/8 gate leads to an error on only a single qubit in the output block of encoded qubits. Exercise 10.65: An unknown qubit in the state |\u03c8 can be swapped with a second gates, qubit which is prepared in the state |0 using only two controlledwith the circuit Show that the two circuits below, which use only a single measurement and a classically controlled single qubit operation, also accomplish the same task: gate, with Exercise 10.66: (Fault-tolerant \u03c0/8 gate construction) One way to implement a \u03c0/8 gate is to \ufb01rst swap the qubit state |\u03c8 you wish to transform with some known state |0, then to apply a \u03c0/8 gate to the resulting qubit. Here is a quantum circuit which does that: Doing this does not seem particularly useful, but actually it leads to something which is! Show that by using the relations T X = exp(\u2212i\u03c0/4)SX and T U = U T (U is the controlledthe circuit of Figure 10.25. gate, and T acts on the control qubit) we may obtain Exercise 10.67: Show that the following circuit identities hold: 488 Quantum error-correction (a) (b) Exercise 10.68: (Fault-tolerant Toffoli gate construction) A procedure similar to the above sequence of exercises for the \u03c0/8 gate gives a fault-tolerant Toffoli gate. (1) First, swap the three qubit state |xyz you wish to transform with some known state |000, then apply a Toffoli gate to the resulting qubits. Show that the following circuit accomplishes this task: (2) Using the commutation rules from Exercise 10.67, show that moving the \ufb01nal Toffoli gate all the way back to the left side gives the circuit |0 |0 |0 |x |y |x |y |z \u2295 xy \u2022 \u2295 \u2022 \u2022 X X \u2295 \u2022 (3) Assuming the ancilla preparation shown in the leftmost dotted box can be done fault-tolerantly, show that this circuit can be used to give a fault-tolerant implementation of the Toffoli gate using the Steane code. Fault-tolerant quantum computation 489 10.6.3 Fault-tolerant measurement An extremely useful and important tool in the construction of fault-tolerant circuits is the ability to measure an operator M. Measurements are used to do encoding, read out the result of a quantum computation, to diagnose the syndrome in error-correction, and to do ancilla state preparation in the construction of the fault-tolerant \u03c0/8 and Toffoli gates, and thus are absolutely crucial to fault-tolerant quantum computation. In order that a procedure for performing an encoded measurement be considered to be fault-tolerant, we require that two things be true in order to prevent the propagation of errors. First, a single failure anywhere in the procedure should lead to at most one error in any block of qubits at the end of the procedure. Second, even if a single failure occurs during the procedure, we require that the measurement result be correct with probability 1 \u2212 O(p2). This latter requirement is extremely important, since the measurement result may be used to control other operations in the quantum computer, and if it is incorrect then it may propagate to affect many qubits in other blocks of encoded qubits. Figure 10.26. Quantum circuit for measuring a single qubit operator M with eigenvalues \u00b11. The top qubit is the ancilla used for the measurement, and the bottom qubit is being measured. Recall that measurement of a single qubit observable M may be performed using the circuit shown in Figure 10.26. Suppose M can be given a transversal encoded implementation on a quantum code as bitwise application of a gate M to each qubit of the code. For example, for the Steane code, M = H can be given a transversal implementation as bitwise application of M = H, while a transversal implementation of M = S uses bitwise application of M = ZS. This suggests a possible circuit for measuring the encoded M on the encoded data, as shown schematically in Figure 10.27. Note that a real quantum code, such as the Steane code, would require more", " qubits. Unfortunately, the circuit in Figure 10.27 is not fault-tolerant. To see this, imagine a single failure occurs at the very beginning of the circuit, on the ancilla qubit. This will propagate forward to affect all the encoded qubits, so the circuit is not fault-tolerant. |0 H \u2022 \u2022 \u2022 H \u23a7 \u23aa\u23aa\u23aa\u23aa\u23a8 Encoded Data \u23aa\u23aa\u23aa\u23aa\u23a9 M M M Figure 10.27. Schematic procedure for performing a measurement of an encoded observable M with a transversal implementation as bitwise application of M. The circuit is not fault-tolerant. Note that a real code would require more than three qubits. 490 Quantum error-correction A nice way to make the measurement circuit fault-tolerant is schematically illustrated in Figure 10.28. For simplicity, this \ufb01gure shows the data to be measured encoded in only three qubits; in practice more qubits will be used, such as the seven qubit Steane code, and for concreteness we imagine that it is the Steane code which is being used here. In addition to the encoded data, the circuit introduces one ancilla qubit for each data qubit, initially each in the state |0. The \ufb01rst step is to prepare the ancilla in a \u2018cat\u2019 state, |00... 0 + |11... 1. (Note that the cat state is not encoded in any code.) The circuit used to do this preparation is not itself fault-tolerant, because a single failure during the circuit can cause errors on multiple qubits in the cat state. Nevertheless, this does not affect the fault-tolerance of the entire procedure, because we follow the ancilla preparation by several veri\ufb01cation steps (only a single veri\ufb01cation step is shown in the Figure). |0 |0 |0 Prepare Verify H \u2022 Controlled-M \u2022 Decode \u2022 H \u2022 \u2295 \u2295 \u2022 |0 Encoded Data \u2295 \u2295 \u23a7 \u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23a9 \u2022 \u2022 \u2295 \u2295 M M M Figure 10.28. Schematic procedure for fault-tolerant measurement of an observable M, which is performed on encoded data. This procedure is repeated three times and a majority vote of the measurement outcomes taken, with the result that the measurement result is wrong with probability O(p2), where p is the failure probability of any individual component, and a single error anywhere in the circuit produces at most one error in the data. The veri\ufb01cation works as follows. The basic idea is that to verify that the state is a cat state it is suf\ufb01cient to show that measurement of ZiZj for all pairs of qubits i and j in the cat state will give 1; that is, the parity of any pair of qubits in the cat state is even. To verify this for a particular pair ZiZj (Z2Z3 in the example) we introduce an extra qubit, initially in the state |0. We compute the parity of two of the qubits in the s with the ancilla qubits as controls, and the ancilla by implementing two controlledextra qubit as the target, before measuring the extra qubit. If the measured parity is 1 then we know that the ancilla is not in the cat state, discard it, and start again. Suppose a single component failure occurs somewhere during this sequence of parity checks. This procedure is not fault-tolerant, because it is easy to show that there are single component failures which lead to more than one phase \ufb02ip in the ancilla state. For example, if there is a Z error on the extra qubit between the gates then this can propagate forward to cause Z errors on two of the ancilla qubits. Fortunately, it is easy to show that multiple Z errors in the ancilla qubits do not propagate to the encoded data, although they may cause the \ufb01nal measurement result to be incorrect. To cope with this problem, and as described in more detail below, we repeat this procedure for measurement three times Fault-tolerant quantum computation 491 and take a majority vote, so the probability that the measurement will be wrong two or more times in this way is at most O(p2), where p is the probability of failure for a single component. What about X or Y errors? Well, these can propagate to cause errors in the encoded data, but it is a fortunate fact that a single failure during the cat state preparation and veri\ufb01cation can cause", " at most one X or Y error in the ancilla after veri\ufb01cation, and thus at most one error in the encoded data, ensuring fault-tolerance! Exercise 10.69: Show that a single failure anywhere in the ancilla preparation and veri\ufb01cation can lead to at most one X or Y error in the ancilla output. Exercise 10.70: Show that Z errors in the ancilla do not propagate to affect the encoded data, but result in an incorrect measurement result being observed. After the cat state has been veri\ufb01ed, controlled-M gates are performed between pairs of ancilla and data qubits, with no ancilla qubit being used more than once. Thus, if the ancilla is in the state |00... 0 this results in nothing being done to the encoded data, while if the ancilla is in the state |11... 1 the encoded M operation is applied to the data. The value of the cat state is that it ensures errors do not propagate from one controlled-M gate to another, so a single error in the veri\ufb01cation stage or the sequence of controlled-M gates results in at most a single error in the encoded data. Finally, the gates and measurement result is obtained by decoding the cat state with a series of a Hadamard; the resulting qubit is 0 or 1 depending on the eigenvalue of the state of the data. These \ufb01nal gates do not involve the data, and thus an error in these gates does not propagate to affect the data at all. But what if an error in this \ufb01nal sequence of gates results in an incorrect measurement result? By repeating the measurement procedure three times and taking a majority vote of the results, we can ensure that the probability of an error in the measurement result is O(p2), where p is the probability of failure in an individual component. We have described a method for performing fault-tolerant measurements such that the measurement gives an erroneous result with probability O(p2), where p is the failure probability for the individual components, and a single failure anywhere in the procedure results in an error on at most one qubit in the encoded data. The construction can be applied for any single qubit observable M which can be implemented in a transversal fashion. For the Steane code, this includes the Hadamard, phase and Pauli gates, and with a slight modi\ufb01cation, the observable M = e\u2212i\u03c0/4SX. To perform the controlledM operation on the Steane code for this choice of M we apply controlled-ZSX gates transversally for each pair of qubits in the ancilla and the code, followed by seven T gates applied transversally to the ancilla qubits. As described in Section 10.6.2, a fault-tolerant measurement of this observable can be used to create the ancilla used in the fault-tolerant circuit for the \u03c0/8 gate. Exercise 10.71: Verify that when M = e\u2212i\u03c0/4SX the procedure we have described gives a fault-tolerant method for measuring M. Exercise 10.72: (Fault-tolerant Toffoli ancilla state construction) Show how to fault-tolerantly prepare the state created by the circuit in the dotted box of 492 Quantum error-correction Exercise 10.68, that is, |000 + |010 + |100 + |111 2. (10.121) You may \ufb01nd it helpful to \ufb01rst give the stabilizer generators for this state. Measurement of stabilizer generators We have described the fault-tolerant measurement procedure when M is an encoded observable for a single qubit, however the techniques immediately generalize to other cases. For our purposes it is suf\ufb01cient to be able to measure stabilizer generators, which take the form of a tensor product of Pauli matrices. Such measurements allow us to perform fault-tolerant error-correction, the initial encoding for the quantum computer, and to measure encoded Z operators for the \ufb01nal readout stage of the computation. As a simple example, suppose we wished to measure an operator like X1Z2X3 on the \ufb01rst three qubits of a block of seven qubits encoded using the Steane code. An obvious generalization of Figure 10.28 can be used to perform this measurement, as shown in Figure 10.29. Once again, we perform veri\ufb01ed cat state preparations before applying transversal controlled operations on the encoded data, in order to achieve a fault-tolerant measurement procedure for the operator X1Z2X3. With the ability to fault-tolerantly measure such observables we automatically", " obtain the ability to perform the steps of encoding, syndrome measurement and measurement in the (logical) computational basis that are required to perform quantum computation. For the purposes of encoding, it suf\ufb01ces for quantum computation to prepare an encoded |0 state. For a stabilizer code such as the Steane code, such a preparation can be achieved by fault-tolerantly measuring all the stabilizer generators and the encoded \u00afZ operator, and then \ufb01xing the signs of the stabilizer generators and the encoded Z by applying appropriate fault-tolerant operations, according to the prescription in the proof of Proposition 10.4 in Section 10.5.1. An example illustrating how the Steane code encoded |0 state may be prepared faulttolerantly is explained in Exercise 10.73. Syndrome measurement for error-correction and \ufb01nal read-out in the encoded computational basis of the quantum computer are realized fault-tolerantly along similar lines. |000 + |111 \u221a 2 \u23a7 \u23aa\u23aa\u23a8 \u23aa\u23aa\u23a9 \u2022 H \u2295 \u2295 \u2022 X \u2022 Z \u2022 X Figure 10.29. Schematic procedure for performing a fault-tolerant measurement of the operator XZX on three qubits. Fault-tolerant quantum computation 493 |0 H \u2022 H |0000 + |1111 \u221a 2 \u2192 \u2295 \u2295 \u2295 \u2295 \u23a7 \u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 \u2022 \u2022 H \u2022 \u2022 \u2295 \u2295 \u2022 \u2295 \u2295 \u2295 \u2295 \u2295 Figure 10.30. One step in fault-tolerantly producing the Steane code encoded |0 state. Exercise 10.73: (Fault-tolerant encoded state construction) Show that the Steane code encoded |0 state can be constructed fault-tolerantly in the following manner. (1) Begin with the circuit of Figure 10.16, and replace the measurement of each generator, as shown in Figure 10.30, with each ancilla qubit becoming a cat state |00... 0 + |11... 1, and the operations rearranged to have their controls on different qubits, so that errors do not propagate within the code block. (2) Add a stage to fault-tolerantly measure Z. (3) Calculate the error probability of this circuit, and of the circuit when the generator measurements are repeated three times and majority voting is done. (4) Enumerate the operations which should be performed conditioned on the measurement results and show that they can be done fault-tolerantly. Exercise 10.74: Construct a quantum circuit to fault-tolerantly generate the encoded |0 state for the \ufb01ve qubit code (Section 10.5.6). 10.6.4 Elements of resilient quantum computation The most spectacular success of quantum error-correcting codes \u2013 the threshold for quantum computation \u2013 is that provided the noise in individual quantum gates is below a certain constant threshold it is possible to ef\ufb01ciently perform an arbitrarily large quantum computation. Stated another way, noise is not a serious problem in principle for quantum computation. The basic idea in the proof of the threshold, as was outlined in Section 10.6.1, is to perform fault-tolerant operations directly on encoded states, interleaved with error-correction steps, resulting in a net reduction in error probability from p to O(p2). By concatenating our codes multiple times and creating hierarchical fault-tolerant procedures, the error probability can be reduced further to O(p4), then O(p8) and so on, ultimately being reduced to as low a level as desired, so long as the the original error p is less than some threshold value pth. Using the procedures we have described, we estimate a threshold of approximately pth \u223c 10\u22125-10\u22126. A bold claim such as the threshold theorem obviously needs quali\ufb01ers. It is not the case that it is possible to protect a quantum computation against the effect of completely 494 Quantum error-correction arbitrary noise. The threshold theorem relies for its functioning on a small number of physically reasonable assumptions about the type of noise occurring in the quantum computer, and the quantum computer architecture available in order to reach its powerful conclusion. The error model we have considered is rather simplistic, and it is certainly the case that real quantum computers will experience more varied types of noise than we have considered here. Nevertheless, it seems plausible that the techniques introduced here when coupled with more sophisticated quantum error-correcting codes and with more sophisticated tools for analysis, can result in a threshold for quantum computation applicable in a much wider variety of", " circumstances than those we have considered. We have not the space here to dive into a more sophisticated analysis, but several observations are in order. First, it is interesting to note that the threshold result requires a high degree of parallelism in our circuits. Even if all we wish to do is store quantum information in a quantum memory this operation will require periodic error-correction that demands a high degree of parallelism. Thus, a desirable goal for would-be designers of quantum computers is to develop architectures which are parallelizable, in order that the techniques of fault-tolerant quantum computation may be applied. Second, we note that our presentation of the threshold has completely neglected the cost of the classical computations and communication that are done during state preparation, syndrome measurement, and recovery. The cost of these could potentially be quite high; for example, to do recovery at the highest levels of the concatenated code requires communication between all parts of the quantum system. If this communication cannot be accomplished much faster than the time scale over which errors occur in the system then errors will begin to creep back in, negating the effect of the error-correction. More sophisticated analyses can deal with this problem; however, as with other complications there is a concomitant cost in the form of a more stringent threshold for quantum computation. Third, our fault-tolerant constructions for measurement and the \u03c0/8 gate made use of ancilla qubits in the state |0, perhaps with some slight additional noise. It can be shown that in fact a constant supply of such fresh ancilla qubits is necessary for the threshold theorem to apply, and thus quantum computer designers must provide architectures which are not only parallelizable, but which also allow fresh ancilla qubits to be brought up on a regular basis. Our presentation has focused on basic principles, not on optimizing the methods used, and it is likely that in practice much more streamlined versions of our constructions would be used. A simple but important guiding principle is to choose your codes well. We have focused on the Steane code because it is easy to work with and demonstrates all the fundamental principles; however, in practice other codes may work much better. For example, it may pay handsome dividends at the \ufb01rst level of concatenation to use a code optimized to protect against the type of noise known to occur in the particular physical system being used for implementation. Although the theoretical ideas behind the threshold theorem can be adapted in a variety of different ways, depending upon the noise prevalent in a speci\ufb01c implementation of quantum computation, a skeptic might still claim that all such noise models for which a threshold may be proved are overly restrictive, and will not be realized within any real physical system. Such skepticism is, \ufb01nally, only answerable in the laboratory with a demonstration of large-scale fault-tolerant quantum computation. The marvel of the present result is that it proves that, to the best of our current knowledge, no principle of physics will limit quantum computers from being realized someday. Chapter problems 495 Summarizing, in this chapter we have outlined the basic principles by which quantum information processing may be performed in a resilient manner, focusing on the speci\ufb01c example of quantum computation. The same basic techniques apply also to any other system in which quantum information processing may be performed, such as quantum communications channels for performing tasks such as quantum cryptography. The extreme fragility of quantum information in all known systems makes it likely that some form of quantum error-correction will need to be used in any practical quantum information processing system, but surprisingly, these techniques work so well that arbitrarily reliable quantum computations can be performed using noisy components, provided the error probability in those components is less than some constant threshold. Problem 10.1: Channels E1 and E2 are said to be equivalent if there exist unitary channels U and V such that E2 = U \u25e6 E1 \u25e6 V. (1) Show that the relation of channel equivalence is an equivalence relation. (2) Show how to turn an error-correcting code for E1 into an error-correcting code for E2. Assume that the error-correction procedure for E1 is performed as a projective measurement followed by a conditional unitary operation, and explain how the error-correction procedure for E2 can be performed in the same fashion. Problem 10.2: (Gilbert\u2013Varshamov bound) Prove the Gilbert\u2013Varshamov bound for CSS codes, namely, that an [n, k] CSS code correcting t errors exists for some k such that k n \u2265 1 \u2212 2H 2t n. (10.122) As a challenge, you may like to try proving the Gilbert\u2013Varshamov bound for a general stabilizer code, namely, that there exists an [n, k] stabilizer code correcting errors on t qubits, with k n \u2265 1 \u2212 2 log(", "3)t n \u2212 H 2t n. (10.123) Problem 10.3: (Encoding stabilizer codes) Suppose we assume that the generators for the code are in standard form, and that the encoded Z and X operators have been constructed in standard form. Find a circuit taking the n\u00d72n check matrix corresponding to a listing of all the generators for the code together with the encoded Z operations from \u23a1 \u23a3 to the standard form \u23a1 \u23a3 I A1 A2 B 0 C2 0 D I E 0 0 AT 10.124) (10.125) 496 Quantum error-correction Problem 10.4: (Encoding by teleportation) Suppose you are given a qubit |\u03c8 to encode in a stabilizer code, but you are not told anything about how |\u03c8 was constructed: it is an unknown state. Construct a circuit to perform the encoding in the following manner: (1) Explain how to fault-tolerantly construct the partially encoded state |0|0L + |1|1L \u221a 2, (10.126) by writing this as a stabilizer state, so it can be prepared by measuring stabilizer generators. (2) Show how to fault-tolerantly perform a Bell basis measurement with |\u03c8 and the unencoded qubit from this state. (3) Give the Pauli operations which you need to \ufb01x up the remaining encoded qubit after this measurement, so that it becomes |\u03c8, as in the usual quantum teleportation scheme. Compute the probability of error of this circuit. Also show how to modify the circuit to perform fault-tolerant decoding. Problem 10.5: Suppose C(S) is an [n, 1] stabilizer code capable of correcting errors gate may be implemented between two logical qubits encoded on a single qubit. Explain how a fault-tolerant implementation of the controlledusing this code, using only fault-tolerant stabilizer state preparation, fault-tolerant measurement of elements of the stabilizer, and normalizer gates applied transversally. History and further reading 497 Summary of Chapter 10: Quantum error-correction \u2022 Quantum error-correcting code: An [n, k, d] quantum error-correcting code uses n qubits to encode k qubits, with distance d. \u2022 Quantum error-correction conditions: Let C be a quantum error-correcting code and P be the projector onto C. Then the code can correct the error set {Ei} if and only if P E\u2020 i EjP = \u03b1ijP, (10.127) for some Hermitian matrix \u03b1 of complex numbers. \u2022 Stabilizer codes: Let S be the stabilizer for a stabilizer code C(S) and suppose j Ek \u2208 N (S) \u2212 S for all j {Ej} is a set of errors in the Pauli group such that E\u2020 and k. Then {Ej} is a correctable set of errors for the code C(S). \u2022 Fault-tolerant quantum computation: A universal set of logical operations on encoded quantum states can be performed in such a way that the effective failure probability in the encoded states scales like O(p2), where p is the failure probability in the underlying gates. \u2022 The threshold theorem: Provided the noise in individual quantum gates is below a certain constant threshold and obeys certain physically reasonable assumptions, it is possible to reliably perform an arbitrarily long quantum computation, with only a small overhead in the size of the circuit necessary to ensure reliability. History and further reading There are many excellent texts on error-correcting codes in classical information theory. We especially recommend the wonderful text of MacWilliams and Sloane[MS77]. This begins at a very elementary level, but quickly and smoothly moves into more advanced topics, covering an enormous range of material. A more recent introduction, also very good, is the text by Welsh[Wel88]. Quantum error-correction was independently discovered by Shor[Sho95], who found the nine qubit code presented in Section 10.2, and by Steane[Ste96a], who used a different approach, in which he studied the interference properties of multiple particle entangled states. The quantum error-correction conditions were proved independently by Bennett, DiVincenzo, Smolin and Wootters[BDSW96], and by Knill and La\ufb02amme[KL97], building upon earlier work by Ekert and Macchiavello[EM96]. The \ufb01ve qubit code was discovered by Bennett, DiVincenzo, Smolin and Wootters[BDSW96], and independently by La\ufb02amme, Miquel, Paz and Zurek[LMPZ96]. Calderbank and Shor[CS96], and Steane[Ste96b] used ideas", " from classical error-correction to develop the CSS (Calderbank\u2013Shor\u2013Steane) codes. Calderbank and Shor also stated and proved the Gilbert\u2013Varshamov bound for CSS codes. Gottesman[Got96] invented the stabilizer formalism, and used it to de\ufb01ne stabilizer codes, and investigated some of the properties of some speci\ufb01c codes. Independently, Calderbank, Rains, Shor and Sloane[CRSS97] invented an essentially equivalent approach to quantum error-correction 498 Quantum error-correction based on ideas from classical coding theory. They were able to classify almost all known quantum codes using a GF (4) orthogonal geometry approach[CRSS98], and also provided the \ufb01rst proof of the quantum Gilbert\u2013Varshamov bound for general stabilizer codes, which had earlier been stated by Ekert and Macchiavello[EM96]. The Gottesman\u2013Knill theorem seems to have \ufb01rst been stated by Gottesman in [Got97], where he attributes the result to Knill, with the proof based upon the stabilizer formalism Gottesman had introduced. Gottesman has applied the stabilizer formalism to a wide variety of problems with considerable success; see for example [Got97] for a sample and further references. Our presentation of the stabilizer formalism is based primarily upon [Got97], wherein may be found most of the results we describe, including the result that the Hadamard, generate the normalizer N (Gn). phase and controlled- Many constructions for speci\ufb01c classes of quantum codes are known; we point to just a few here. Rains, Hardin, Shor and Sloane[RHSS97] have constructed interesting examples of quantum codes lying outside the stabilizer codes we have considered. Many people have considered quantum codes based on systems other than qubits; we mention especially the work of Gottesman[Got98a] and Rains[Rai99b] which construct non-binary codes and consider fault-tolerant computation with such codes. Aharonov and Ben-Or[ABO99] construct non-binary codes using interesting techniques based on polynomials over \ufb01nite \ufb01elds, and also investigate fault-tolerant computation with such codes. Approximate quantum error-correction is another topic we have not touched; that approximate quantum error-correction can lead to improved codes was shown by Leung, Nielsen, Chuang and Yamamoto[LNCY97]. A large and interesting class of quantum error-correcting codes (but beyond the scope of this chapter) are known variously by the names noiseless quantum codes and decoherence free subspaces. A substantial body of work exists on these subjects (and clarifying the connections between them). An entry into the literature may be found through the work of Zanardi and Rasetti[ZR98, Zan99], Lidar, Chuang, and Whaley[LCW98], Bacon, Kempe, Lidar and Whaley[BKLW99, LBW99], and of Knill, La\ufb02amme and Viola[KLV99]. Many bounds on quantum error-correcting codes are known, often adapted from similar classical bounds. Ekert and Macchiavello[EM96] pointed out the possibility of proving a quantum analogue to the Hamming bound; this construction and the role of \u2018degenerate\u2019 quantum codes was subsequently clari\ufb01ed by Gottesman[Got96]. Shor and La\ufb02amme[SL97] proved a quantum analogue of a result in classical coding theory, the MacWilliams identities, which touched off a great deal of work studying the properties of certain polynomials related to quantum codes (the weight enumerators) as well as more general work on the problem of bounds for quantum codes, by Ashikhmin[Ash97], Ashikhmin and Lytsin[AL99], and several papers on the topic by Rains[Rai98, Rai99c, Rai99a]. The theory of fault-tolerant computation for classical computers was worked out by von Neumann[von56], and is discussed in the monograph by Winograd and Cowan[WC67]. Shor[Sho96] introduced the idea of fault-tolerance into quantum computation, and showed how to perform all the basic fault tolerant steps (state preparation, quantum logic, errorcorrection, and measurement). Kitaev[Kit97b, Kit97c] independently developed many similar ideas, including fault-tolerant constructions for many basic quantum logic gates. Cirac, Pellizzari and Zoller[CPZ96], and Zurek and La\ufffd", "\ufffd\ufffdamme[ZL96], also took early steps toward fault-tolerant quantum computation. DiVincenzo and Shor generalized Shor\u2019s original construction to show how fault-tolerant measurement of syndromes for any stabilizer code could be performed[DS96], and Gottesman[Got98b] generalized all the fault-tolerant History and further reading 499 constructions, showing how to perform fault-tolerant computation with any stabilizer code. A general review of this work as well as much other survey material may be found in [Got97]; this includes a construction to solve Problem 10.5. The fault-tolerant \u03c0/8 and Toffoli gate constructions are based on a line of thought developed by Gottesman and Chuang[GC99], and Zhou and Chuang[ZLC00]; the circuit for the fault-tolerant Toffoli described in Exercise 10.68 is actually Shor\u2019s original construction[Sho96]. Steane[Ste99] has developed many ingenious constructions for fault-tolerant procedures. Kitaev[Kit97a, Kit97b] has introduced a beautiful set of ideas for implementing faulttolerance, using topological methods to assist in the performance of quantum errorcorrection. The basic idea is that if information is stored in the topology of a system, then that information will naturally be very robust against the effects of noise. These and many other elegant ideas have been developed in further papers by Bravyi and Kitaev[BK98b] and by Freedman and Meyer[FM98]. Preskill[Pre97] is an excellent review of the \ufb01eld of quantum error-correction as a whole, and contains an especially beautiful description of topological quantum error-correction, as well as a provocative discussion of whether topological error-correction can be used to gain insight into fundamental questions about black holes and quantum gravity! Many different groups proved threshold results for quantum computation. These results hold for a wide variety of assumptions, giving essentially different threshold theorems. Aharonov and Ben-Or[ABO97, ABO99] and Kitaev\u2019s[Kit97c, Kit97b] threshold proofs do not require fast or reliable classical computation. Aharonov and Ben-Or also showed that in order for a threshold result to hold, there must be constant parallelism in the quantum computer at each timestep[ABO97]. In their threshold proofs, Gottesman[Got97] and Preskill[Pre98c, GP10] have an especially detailed optimization of the value of the threshold. Knill, La\ufb02amme and Zurek[KLZ98a, KLZ98b]\u2019s results concentrate on proving the threshold theorem for a wide class of error models. Aharonov, Ben-Or, Impagliazzo, and Nisan have also shown that a supply of fresh qubits is necessary for the threshold[ABOIN96]. Further references and historical material may be found within the cited works. In particular, each group built on Shor\u2019s pioneering work[Sho96] on fault-tolerant quantum computation. Numerous excellent reviews of fault-tolerant quantum computation have been written, developing the basic ideas in much greater detail than we have here, from a variety of different points of view. Aharonov\u2019s thesis[Aha99a] develops the threshold theorem and much material of related interest in a self-contained way. Gottesman\u2019s thesis[Got97] also provides a review of fault-tolerant quantum computation, with more emphasis on properties of quantum codes, and developing fault-tolerant constructions for a wide variety of different codes. Knill, La\ufb02amme and Zurek have written a semi-popular overview of the threshold result[KLZ98a]. Finally, Preskill has written two superb articles[Pre98c, Pre98a] explaining quantum error-correction and fault-tolerant quantum computation. 11 Entropy and information Entropy is a key concept of quantum information theory. It measures how much uncertainty there is in the state of a physical system. In this chapter we review the basic de\ufb01nitions and properties of entropy in both classical and quantum information theory. In places the chapter contains rather detailed and lengthy mathematical arguments. On a \ufb01rst reading these sections may be read lightly and returned to later for reference purposes. 11.1 Shannon entropy The key concept of classical information theory is the Shannon entropy. Suppose we learn the value of a random variable X. The Shannon entropy of X quanti\ufb01es how much information we gain, on average, when we learn the value of X. An alternative view is that the entropy of X measures the amount of uncertainty about X before we", " learn its value. These two views are complementary; we can view the entropy either as a measure of our uncertainty before we learn the value of X, or as a measure of how much information we have gained after we learn the value of X. Intuitively, the information content of a random variable should not depend on the labels attached to the different values that may be taken by the random variable. For example, we expect that a random variable taking the values \u2018heads\u2019 and \u2018tails\u2019 with respective probabilities 1/4 and 3/4 contains the same amount of information as a random variable that takes the values 0 and 1 with respective probabilities 1/4 and 3/4. For this reason, the entropy of a random variable is de\ufb01ned to be a function of the probabilities of the different possible values the random variable takes, and is not in\ufb02uenced by the labels used for those values. We often write the entropy as a function of a probability distribution, p1,..., pn. The Shannon entropy associated with this probability distribution is de\ufb01ned by H(X) \u2261 H(p1,..., pn) \u2261 \u2212 px log px. x (11.1) We justify this de\ufb01nition shortly. Note that in the de\ufb01nition \u2013 and throughout this book \u2013 logarithms indicated by \u2018log\u2019 are taken to base two, while \u2018ln\u2019 indicates a natural logarithm. It is conventional to say that entropies are measured in \u2018bits\u2019 with this convention for the logarithm. You may wonder what happens when px = 0, since log 0 is unde\ufb01ned. Intuitively, an event which can never occur should not contribute to the entropy, so by convention we agree that 0 log 0 \u2261 0. More formally, note that limx\u21920 x log x = 0, which provides further support for our convention. Why is the entropy de\ufb01ned in this way? Later in this section, Exercise 11.2 gives an intuitive justi\ufb01cation for this de\ufb01nition of the entropy, based on certain \u2018reasonable\u2019 axioms which might be expected of a measure of information. This intuitive justi\ufb01cation Shannon entropy 501 is reassuring but it is not the whole story. The best reason for this de\ufb01nition of entropy is that it can be used to quantify the resources needed to store information. More concretely, suppose there is some source (perhaps a radio antenna) which is producing information of some sort, say in the form of a bit string. Let\u2019s consider a very simple model for a source: we model it as producing a string X1, X2,... of independent, identically distributed random variables. Most real information sources don\u2019t behave quite this way, but it\u2019s often a good approximation to reality. Shannon asked what minimal physical resources are required to store the information being produced by the source, in such a way that at a later time the information can be reconstructed? The answer to this question turns out to be the entropy, that is, H(X) bits are required per source symbol, where H(X) \u2261 H(X1) = H(X2) =... is the entropy of each random variable modeling the source. This result is known as Shannon\u2019s noiseless coding theorem, and we prove both classical and quantum versions of it in Chapter 12. For a concrete example of Shannon\u2019s noiseless channel coding theorem, suppose an information source produces one of four symbols, 1, 2, 3 or 4. Without compression two bits of storage space corresponding to the four possible outputs are consumed for each use of the source. Suppose however that the symbol 1 is produced by the source with probability 1/2, the symbol 2 with probability 1/4, and the symbols 3 and 4 both with probability 1/8. We can make use of the bias between the source outputs to compress the source, using fewer bits to store commonly occurring symbols such as 1, and more bits to store rarely occurring symbols like 3 and 4. One possible compression scheme is to encode 1 as the bit string 0, 2 as the bit string 10, 3 as the bit string 110, and 4 as the bit string 111. \u00b73 = 7/4 bits Notice that the average length of the compressed string is 1 2 of information per use of the source. This is less than is required in the naive approach to storing this source! Amazingly, this matches the entropy of the source, H(X) = \u22121/2 log(1/2) \u2212 1/4 log(1/4) \u2212 1/8 log(1/8) \u2212 1/8 log(1/8) = 7/4! Moreover,", " it turns out that any attempt to compress the source further results in data being irretrievably lost; the entropy quanti\ufb01es the optimal compression that may be achieved. \u00b73+ 1 8 \u00b72+ 1 8 \u00b71+ 1 4 This operational motivation for the de\ufb01nition of entropy in terms of data compression expresses a key philosophy of information theory, both quantum and classical: fundamental measures of information arise as the answers to fundamental questions about the physical resources required to solve some information processing problem. Exercise 11.1: (Simple calculations of entropy) What is the entropy associated with the toss of a fair coin? With the roll of a fair die? How would the entropy behave if the coin or die were unfair? Exercise 11.2: (Intuitive justi\ufb01cation for the de\ufb01nition of entropy) Suppose we are trying to quantify how much information is provided by an event E which may occur in a probabilistic experiment. We do this using an \u2018information function\u2019 I(E) whose value is determined by the event E. Suppose we make the following assumptions about this function: (1) I(E) is a function only of the probability of the event E, so we may write I = I(p), where p is a probability in the range 0 to 1. (2) I is a smooth function of probability. (3) I(pq) = I(p) + I(q) when p, q > 0. (Interpretation: The information gained 502 Entropy and information when two independent events occur with individual probabilities p and q is the sum of the information gained from each event alone.) Show that I(p) = k log p, for some constant, k. It follows that the average information gain when one of a mutually exclusive set of events with probabilities p1,..., pn occurs is k entropy, up to a constant factor. i pi log pi, which is just the Shannon 11.2 Basic properties of entropy 11.2.1 The binary entropy The entropy of a two-outcome random variable is so useful that we give it a special name, the binary entropy, de\ufb01ned as Hbin(p) \u2261 \u2212p log p \u2212 (1 \u2212 p) log(1 \u2212 p), where p and 1 \u2212 p are the probabilities of the two outcomes. Where context makes the meaning clear we write H(p) rather than Hbin(p). The binary entropy function is plotted in Figure 11.1. Notice that H(p) = H(1 \u2212 p) and that H(p) attains its maximum value of 1 at p = 1/2. (11.8) The binary entropy is an excellent testing ground for the understanding of more general properties of entropy. One property of particular interest is how the entropy behaves when we mix two or more probability distributions. Imagine, for example, that Alice has in her possession two coins, one a quarter from the US, the other a dollar coin from Australia. Both coins have been altered to exhibit bias, with the probability of heads on the US coin being pU, and the probability of heads on the Australian coin being pA. Suppose Alice \ufb02ips the US coin with probability q and the Australian coin with probability 1 \u2212 q, telling Bob whether the result was heads or tails. How much information does Bob gain on average? Intuitively it is clear that Bob should gain at least as much information as the average of the information he would have gained from a US coin \ufb02ip or an Australian coin \ufb02ip. As an equation this intuition may be expressed as: H(qpU + (1 \u2212 q)pA) \u2265 qH(pU) + (1 \u2212 q)H(pA). (11.9) Sometimes the inequality can be strict, because Bob gains information not only about the value (heads or tails) of the coin, but also some additional information about the identity of the coin. For instance, if pU = 1/3 and pA = 5/6, and heads comes up, then Bob has received a pretty good indicator that the coin was Australian. Equation (11.9) is easily shown to be correct. It is an example of a broader concept, that of concavity, which we met in Chapter 9 when discussing distance measures. Recall that a real-valued function f is said to be concave if for any p in the range 0 to 1, f (px + (1 \u2212 p)y) \u2265 pf (x) + (1 \u2212 p)f (y). (11.10) The binary entropy is easily seen to be concave, as can be grasped visually by examining Figure 11.1 and observing that the graph of the binary entropy goes above any line cutting the graph. We will be much concerned with concavity properties of the entropy, both classical and", " quantum. Don\u2019t let the simplicity of the above intuitive argument beguile you into a false complacency: many of the deepest results in quantum information have their roots in skilful application of concavity properties of classical or quantum entropies. Basic properties of entropy 503 Box 11.1: Entropic quantum uncertainty principle There is an elegant entropic way of reformulating the uncertainty principle of quantum mechanics. Recall the Heisenberg uncertainty principle from Box 2.4 on page 89. This states that the standard deviations \u0394(C) and \u0394(D) for observables C and D must satisfy the relation \u0394(C)\u0394(D) \u2265 |\u03c8|[C, D]|\u03c8| 2, (11.2) c c|cc| and D = for a quantum system in the state |\u03c8. d d|dd| be spectral decompositions for C and Let C = D. De\ufb01ne f (C, D) \u2261 maxc,d |c|d| to be the maximum \ufb01delity between any two \u221a eigenvectors of |c and |d. For example, f (X, Z) = 1/ 2 for the Pauli matrices X and Z. Suppose the quantum system is prepared in a state |\u03c8, and let p(c) be the probability distribution associated with a measurement of C, with associated entropy H(C), and q(d) the probability distribution associated with a measurement of D, with associated entropy H(D). The entropic uncertainty principle states that H(C) + H(D) \u2265 2 log. (11.3) 1 f (C, D) A full proof of this result would take us too far a\ufb01eld (see \u2018History and further reading\u2019 for references); however, we can give a simple proof of the weaker result: H(C) + H(D) \u2265 \u22122 log 1 + f (C, D) 2. (11.4) To prove this, note that H(C) + H(D) = \u2212 cd p(c)q(d) log (p(c)q(d)). (11.5) We aim to bound p(c)q(d) = |c|\u03c8\u03c8|d|2 from above. To do this, let | \u02dc\u03c8 be the projection of |\u03c8 into the plane spanned by |c and |d, so | \u02dc\u03c8 has norm \u03bb less than or equal to one. If \u03b8 is the angle |d makes with |c in this plane, and \u03d5 is the angle | \u02dc\u03c8 makes with |d, then we see p(c)q(d) = |c| \u02dc\u03c8 \u02dc\u03c8|d|2 = \u03bb2 cos2(\u03b8 \u2212 \u03d5) cos2(\u03d5). Calculus implies that the maximum is reached when \u03bb = 1 and \u03d5 = \u03b8/2, at which point p(c)q(d) = cos4(\u03b8/2), which can be put in the form p(c)q(d) = 1 + |c|d| 2 2. Combining this with Equation (11.5) gives H(C) + H(D) \u2265 \u22122 log 1 + f (C, D) 2, as claimed. (11.6) (11.7) Moreover, for quantum entropies it is sometimes rather dif\ufb01cult to justify our intuitive beliefs about what concavity properties entropy ought to have. 504 Entropy and information 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 11.1. Binary entropy function H(p). Exercise 11.3: Prove that the binary entropy Hbin(p) attains its maximum value of one at p = 1/2. Exercise 11.4: (Concavity of the binary entropy) From Figure 11.1 it appears that the binary entropy is a concave function. Prove that this is so, that is: Hbin(px1 + (1 \u2212 p)x2) \u2265 pHbin(x1) + (1 \u2212 p)Hbin(x2), (11.11) where 0 \u2264 p, x1, x2 \u2264 1. Prove in addition that the binary entropy is strictly concave, that is, the above inequality is an equality only for the trivial cases x1 = x2, or p = 0, or p = 1. 11.2.2 The relative entropy The relative entropy is", " a very useful entropy-like measure of the closeness of two probability distributions, p(x) and q(x), over the same index set, x. Suppose p(x) and q(x) are two probability distributions on the same index set, x. De\ufb01ne the relative entropy of p(x) to q(x) by H(p(x)q(x)) \u2261 x p(x) log p(x) q(x) \u2261 \u2212H(X) \u2212 x p(x) log q(x). (11.12) We de\ufb01ne \u22120 log 0 \u2261 0 and \u2212p(x) log 0 \u2261 +\u221e if p(x) > 0. It is not immediately obvious what the relative entropy is good for, or even why it is a good measure of distance between two distributions. The following theorem gives some motivation as to why it is regarded as being like a distance measure. Basic properties of entropy 505 Theorem 11.1: (Non-negativity of the relative entropy) The relative entropy is non-negative, H(p(x)q(x)) \u2265 0, with equality if and only if p(x) = q(x) for all x. Proof A very useful inequality in information theory is log x ln 2 = ln x \u2264 x \u2212 1, for all positive x, with equality if and only if x = 1. Here we need to rearrange the result slightly, to \u2212 log x \u2265 (1 \u2212 x)/ ln 2, and then note that H(p(x)q(x)) = \u2212 q(x) p(x) p(x) log x p(x) 1 \u2212 q(x) p(x) (p(x) \u2212 q(x)) (1 \u2212 1) = 0, \u2265 1 ln 2 = = 1 ln 2 1 ln 2 x x (11.13) (11.14) (11.15) (11.16) which is the desired inequality. The equality conditions are easily deduced by noting that equality occurs in the second line if and only if q(x)/p(x) = 1 for all x, that is, the distributions are identical. The relative entropy is often useful, not in itself, but because other entropic quantities can be regarded as special cases of the relative entropy. Results about the relative entropy then give as special cases results about other entropic quantities. For example, we can use the non-negativity of the relative entropy to prove the following fundamental fact about entropies. Suppose p(x) is a probability distribution for X, over d outcomes. Let q(x) \u2261 1/d be the uniform probability distribution over those outcomes. Then H(p(x)q(x)) = H(p(x)1/d) = \u2212H(X) \u2212 p(x) log(1/d) = log d \u2212 H(X).(11.17) x From the non-negativity of the relative entropy, Theorem 11.1, we see that log d \u2212 H(X) \u2265 0, with equality if and only if X is uniformly distributed. This is an elementary fact, but so important that we restate it formally as a theorem. Theorem 11.2: Suppose X is a random variable with d outcomes. Then H(X) \u2264 log d, with equality if and only if X is uniformly distributed over those d outcomes. We use this technique \u2013 \ufb01nding expressions for entropic quantities in terms of the relative entropy \u2013 often in the study of both classical and quantum entropies. Exercise 11.5: (Subadditivity of the Shannon entropy) Show that H(p(x, y)p(x)p(y)) = H(p(x)) + H(p(y)) \u2212 H(p(x, y)). Deduce that H(X, Y ) \u2264 H(X) + H(Y ), with equality if and only if X and Y are independent random variables. 11.2.3 Conditional entropy and mutual information Suppose X and Y are two random variables. How is the information content of X related to the information content of Y? In this section we introduce two concepts \u2013 the 506 Entropy and information conditional entropy and the mutual information \u2013 which help answer this question. The de\ufb01nitions we give for these concepts are rather formal, and at times you may be confused as to why a particular quantity \u2013 say, the conditional entropy \u2013 is to be interpreted in the way we indicate. Keep in mind that the ultimate justi\ufb01cation for these de\ufb01nitions is that they answer resource questions, which we investigate in more detail in Chapter 12, and that the interpretations given to the quantities depend on the nature of the resource question being answered. We already met the joint", " entropy of a pair of random variables implicitly in the previous section. For clarity, we now make this de\ufb01nition explicit. The joint entropy of X and Y is de\ufb01ned in the obvious way, H(X, Y ) \u2261 \u2212 p(x, y) log p(x, y), (11.18) x,y and may be extended in the obvious way to any vector of random variables. The joint entropy measures our total uncertainty about the pair (X, Y ). Suppose we know the value of Y, so we have acquired H(Y ) bits of information about the pair, (X, Y ). The remaining uncertainty about the pair (X, Y ), is associated with our remaining lack of knowledge about X, even given that we know Y. The entropy of X conditional on knowing Y is therefore de\ufb01ned by H(X|Y ) \u2261 H(X, Y ) \u2212 H(Y ). (11.19) The conditional entropy is a measure of how uncertain we are, on average, about the value of X, given that we know the value of Y. A second quantity, the mutual information content of X and Y, measures how much information X and Y have in common. Suppose we add the information content of X, H(X), to the information content of Y. Information which is common to X and Y will have been counted twice in this sum, while information which is not common will have been counted exactly once. Subtracting off the joint information of (X, Y ), H(X, Y ), we therefore obtain the common or mutual information of X and Y : H(X : Y ) \u2261 H(X) + H(Y ) \u2212 H(X, Y ). (11.20) Note the useful equality H(X : Y ) = H(X) \u2212 H(X|Y ) relating the conditional entropy and mutual information. To get some feeling for how the Shannon entropy behaves, we now give some simple relationships between the different entropies. Theorem 11.3: (Basic properties of Shannon entropy) (1) H(X, Y ) = H(Y, X), H(X : Y ) = H(Y : X). (2) H(Y |X) \u2265 0 and thus H(X : Y ) \u2264 H(Y ), with equality if and only if Y is a function of X, Y = f (X). (3) H(X) \u2264 H(X, Y ), with equality if and only if Y is a function of X. (4) Subadditivity: H(X, Y ) \u2264 H(X) + H(Y ) with equality if and only if X and Y are independent random variables. (5) H(Y |X) \u2264 H(Y ) and thus H(X : Y ) \u2265 0, with equality in each if and only if X and Y are independent random variables. (6) Strong subadditivity: H(X, Y, Z) + H(Y ) \u2264 H(X, Y ) + H(Y, Z), with equality if and only if Z \u2192 Y \u2192 X forms a Markov chain. Basic properties of entropy 507 (7) Conditioning reduces entropy: H(X|Y, Z) \u2264 H(X|Y ). Most of the proofs are either obvious or easy exercises. We give a few hints below. Proof (1) Obvious from the relevant de\ufb01nitions. (2) Since p(x, y) = p(x)p(y|x) we have = \u2212 xy x H(X, Y ) = \u2212 p(x, y) log p(x)p(y|x) (11.21) p(x) log p(x) \u2212 xy p(x, y) log p(y|x) (11.22) = H(X) \u2212 p(x, y) log p(y|x). (11.23) xy Thus H(Y |X) = \u2212 with equality if and only if Y is a deterministic function of X. xy p(x, y) log p(y|x). But \u2212 log p(y|x) \u2265 0, so H(Y |X) \u2265 0 (3) Follows from the previous result. (4) To prove subadditivity and, later, strong subadditivity we again use the fact that log x \u2264 (x \u2212 1)/ ln 2 for all positive x, with equality if and only if x = 1. We \ufb01nd that x,y p(x, y) log p(x)p(y) p(x, y) \u2264 1 ln 2 = 1 ln 2 x,y x,y p(x, y) p(x)p(y) p(x, y)", " \u2212 1 (11.24) p(x)p(y) \u2212 p(x, y) = 1 \u2212 1 ln 2 = 0. (11.25) Subadditivity follows. Notice that equality is achieved if and only if p(x, y) = p(x)p(y) for all x and y. That is, the subadditivity inequality is saturated if and only if X and Y are independent. (5) Follows from subadditivity and the relevant de\ufb01nitions. (6) Strong subadditivity of Shannon entropy follows from the same technique used to prove subadditivity; the dif\ufb01culty level is only slightly higher than that proof. You will be asked to supply this proof as Exercise 11.6. (7) Intuitively, we expect that the uncertainty about X, given that we know the value of Y and Z, is less than our uncertainty about X, given that we only know Y. More formally, inserting the relevant de\ufb01nitions, the result that conditioning reduces entropy is equivalent to H(X, Y, Z) \u2212 H(Y, Z) \u2264 H(X, Y ) \u2212 H(Y ), (11.26) which is a rearranged version of the strong subadditivity inequality. Exercise 11.6: (Proof of classical strong subadditivity) Prove that H(X, Y, Z) + H(Y ) \u2264 H(X, Y ) + H(Y, Z), with equality if and only if Z \u2192 Y \u2192 X forms a Markov chain. Exercise 11.7: In Exercise 11.5 you implicitly showed that the mutual information H(X : Y ) could be expressed as the relative entropy of two probability distributions, H(X : Y ) = H(p(x, y)p(x)p(y)). Find an expression for the 508 Entropy and information conditional entropy H(Y |X) as a relative entropy between two probability distributions. Use this expression to deduce that H(Y |X) \u2265 0, and to \ufb01nd the equality conditions. The various relationships between entropies may mostly be deduced from the \u2018entropy Venn diagram\u2019 shown in Figure 11.2. Such \ufb01gures are not completely reliable as a guide to the properties of entropy, but they provide a useful mnemonic for remembering the various de\ufb01nitions and properties of entropy. Figure 11.2. Relationships between different entropies. We conclude our study of the elementary properties of conditional entropy and mutual information with a simple and useful chaining rule for conditional entropies. Theorem 11.4: (Chaining rule for conditional entropies) Let X1,..., Xn and Y be any set of random variables. Then H(X1,..., Xn|Y ) = n i=1 H(Xi|Y, X1,..., Xi\u22121). (11.27) Proof We prove the result for n = 2, and then induct on n. Using only the de\ufb01nitions and some simple algebra we have H(X1, X2|Y ) = H(X1, X2, Y ) \u2212 H(Y ) = H(X1, X2, Y ) \u2212 H(X1, Y ) + H(X1, Y ) \u2212 H(Y ) = H(X2|Y, X1) + H(X1|Y ), (11.28) (11.29) (11.30) which establishes the result for n = 2. Now we assume the result for general n, and show the result holds for n + 1. Using the already established n = 2 case, we have H(X1,..., Xn+1|Y ) = H(X2,..., Xn+1|Y, X1) + H(X1|Y ). (11.31) Applying the inductive hypothesis to the \ufb01rst term on the right hand side gives H(X1,..., Xn+1|Y ) = n+1 i=2 H(Xi|Y, X1,..., Xi\u22121) + H(X1|Y ) (11.32) Basic properties of entropy 509 = n+1 i=1 H(Xi|Y, X1,..., Xi\u22121), (11.33) and the induction goes through. Exercise 11.8: (Mutual information is not always subadditive) Let X and Y be independent identically distributed random variables taking the values 0 or 1 with probability 1/2. Let Z \u2261 X \u2295 Y, where \u2295 denotes addition modulo 2. Show that the mutual", " information in this case is not subadditive, H(X, Y : Z) \u2264 H(X : Z) + H(Y : Z). (11.34) Exercise 11.9: (Mutual information is not always superadditive) Let X1 be a random variable taking values 0 or 1 with respective probabilities of 1/2 and X2 \u2261 Y1 \u2261 Y2 \u2261 X1. Show that the mutual information in this case is not superadditive, H(X1 : Y1) + H(X2 : Y2) \u2264 H(X1, X2 : Y1, Y2) (11.35) 11.2.4 The data processing inequality In many applications of interest we perform computations on the information we have available, but that information is imperfect, as it has been subjected to noise before it becomes available to us. A basic inequality of information theory, the data processing inequality, states that information about the output of a source can only decrease with time: once information has been lost, it is gone forever. Making this statement more precise is the goal of this section. The intuitive notion of information processing is captured in the idea of a Markov chain of random variables. A Markov chain is a sequence X1 \u2192 X2 \u2192 \u00b7 \u00b7 \u00b7 of random variables such that Xn+1 is independent of X1,..., Xn\u22121, given Xn. More formally, p(Xn+1 = xn+1|Xn = xn,..., X1 = x1) = p(Xn+1 = xn+1|Xn = xn). (11.36) Under what conditions does a Markov chain lose information about its early values, as time progresses? The following data processing inequality gives an information-theoretic way of answering this question. Theorem 11.5: (Data processing inequality) Suppose X \u2192 Y \u2192 Z is a Markov chain. Then H(X) \u2265 H(X : Y ) \u2265 H(X : Z). (11.37) Moreover, the \ufb01rst inequality is saturated if and only if, given Y, it is possible to reconstruct X. This result is intuitively plausible: it tells us that if a random variable X is subject to noise, producing Y, then further actions on our part (\u2018data processing\u2019) cannot be used to increase the amount of mutual information between the output of the process and the original information X. Proof 510 Entropy and information The \ufb01rst inequality was proved in Theorem 11.3 on page 507. From the de\ufb01nitions we see that H(X : Z) \u2264 H(X : Y ) is equivalent to H(X|Y ) \u2264 H(X|Z). From the fact that X \u2192 Y \u2192 Z is a Markov chain it is easy to prove (Exercise 11.10) that Z \u2192 Y \u2192 X is also a Markov chain, and thus H(X|Y ) = H(X|Y, Z). The problem is thus reduced to proving that H(X, Y, Z) \u2212 H(Y, Z) = H(X|Y, Z) \u2264 H(X|Z) = H(X, Z) \u2212 H(Z). This is just the already proved strong subadditivity inequality. Suppose H(X : Y ) < H(X). Then it is not possible to reconstruct X from Y, since if Z is the attempted reconstruction based only on knowledge of Y, then X \u2192 Y \u2192 Z must be a Markov chain, and thus H(X) > H(X : Z) by the data processing inequality. Thus Z = X. On the other hand, if H(X : Y ) = H(X), then we have H(X|Y ) = 0 and thus whenever p(X = x, Y = y) > 0 we have p(X = x|Y = y) = 1. That is, if Y = y then we can infer with certainty that X was equal to x, allowing us to reconstruct X. As noted above, if X \u2192 Y \u2192 Z is a Markov chain, then so is Z \u2192 Y \u2192 X. Thus, as a corollary to the data processing inequality we see that if X \u2192 Y \u2192 Z is a Markov chain then H(Z : Y ) \u2265 H(Z : X). (11.38) We refer to this result as the data pipelining inequality. Intuitively, it says that any information Z shares with X must be information which Z also shares with Y ; the information is \u2018pipelined\u2019 from X through Y to Z. Exercise 11.10: Show that if X \u2192 Y \u2192 Z is a Markov chain then so is Z \u2192 Y \u2192 X. 11.3 Von Neumann entropy The", " Shannon entropy measures the uncertainty associated with a classical probability distribution. Quantum states are described in a similar fashion, with density operators replacing probability distributions. In this section we generalize the de\ufb01nition of the Shannon entropy to quantum states. Von Neumann de\ufb01ned the entropy of a quantum state \u03c1 by the formula S(\u03c1) \u2261 \u2212tr(\u03c1 log \u03c1). (11.39) In this formula logarithms are taken to base two, as usual. If \u03bbx are the eigenvalues of \u03c1 then von Neumann\u2019s de\ufb01nition can be re-expressed S(\u03c1) = \u2212 x \u03bbx log \u03bbx, (11.40) where we de\ufb01ne 0 log 0 \u2261 0, as for the Shannon entropy. For calculations it is usually this last formula which is most useful. For instance, the completely mixed density operator in a d-dimensional space, I/d, has entropy log d. From now on, when we refer to entropy, it will usually be clear from context whether we mean the Shannon or von Neumann entropy. Von Neumann entropy 511 Exercise 11.11: (Example calculations of entropy) Calculate S(\u03c1) for 11.41) (11.42) (11.43) Exercise 11.12: (Comparison of quantum and classical entropies) Suppose \u03c1 = p|00| + (1 \u2212 p) H(p, 1 \u2212 p). (|0+|1)(0|+1|) 2. Evaluate S(\u03c1). Compare the value of S(\u03c1) to 11.3.1 Quantum relative entropy As for the Shannon entropy, it is extremely useful to de\ufb01ne a quantum version of the relative entropy. Suppose \u03c1 and \u03c3 are density operators. The relative entropy of \u03c1 to \u03c3 is de\ufb01ned by S(\u03c1||\u03c3) \u2261 tr(\u03c1 log \u03c1) \u2212 tr(\u03c1 log \u03c3). (11.50) As with the classical relative entropy, the quantum relative entropy can sometimes be in\ufb01nite. In particular, the relative entropy is de\ufb01ned to be +\u221e if the kernel of \u03c3 (the vector space spanned by the eigenvectors of \u03c3 with eigenvalue 0) has non-trivial intersection with the support of \u03c1 (the vector space spanned by the eigenvectors of \u03c1 with non-zero eigenvalue), and is \ufb01nite otherwise. The quantum relative entropy is non-negative, a result sometimes known as Klein\u2019s inequality: Theorem 11.7: (Klein\u2019s inequality) The quantum relative entropy is non-negative, S(\u03c1||\u03c3) \u2265 0, (11.51) with equality if and only if \u03c1 = \u03c3. Proof Let \u03c1 = From the de\ufb01nition of the relative entropy we have i pi|ii| and \u03c3 = j qj|jj| be orthonormal decompositions for \u03c1 and \u03c3. S(\u03c1||\u03c3) = pi log pi \u2212 i|\u03c1 log \u03c3|i. (11.52) We substitute into this equation the equations i|\u03c1 = pii| and \u239b \u239e i i i| log \u03c3|i = i| \u239d j log(qj)|jj| \u23a0 |i = j log(qj)Pij, (11.53) where Pij \u2261 i|jj|i \u2265 0, to give S(\u03c1||\u03c3) = i \u239b pi \u239dlog pi \u2212 j \u239e Pij log(qj) \u23a0. (11.54) Note that Pij satis\ufb01es the equations Pij \u2265 0, i Pij = 1 and j Pij = 1. (Regarding Pij 512 Entropy and information Box 11.2: Continuity of the entropy Suppose we vary \u03c1 by a small amount. How much does S(\u03c1) change? Fannes\u2019 inequality tells us that the answer is \u2018not much\u2019, and even provides a bound on how small the change is. Theorem 11.6: (Fannes\u2019 inequality) Suppose \u03c1 and \u03c3 are density matrices such that the trace distance between them satis\ufb01es T (\u03c1, \u03c3) \u2264 1/e. Then |S(\u03c1) \u2212 S(\u03c3)| \u2264 T (\u03c1, \u03c3) log d + \u03b7(T (\u03c1, \u03c3)), (11.44) where d is the dimension of the Hilbert space, and \u03b7(x) \u2261 \u2212x log x. Removing the restriction that T (\u03c1, \u03c3) \u2264 1/e we can prove the", " weaker inequality |S(\u03c1) \u2212 S(\u03c3)| \u2264 T (\u03c1, \u03c3) log d +. 1 e (11.45) Proof To prove Fannes\u2019 inequality we need a simple result relating the trace distance between two operators to their eigenvalues. Let r1 \u2265 r2 \u2265 \u00b7 \u00b7 \u00b7 \u2265 rd be the eigenvalues of \u03c1, in descending order, and s1 \u2265 s2 \u2265 \u00b7 \u00b7 \u00b7 \u2265 sd be the eigenvalues of \u03c3, again in descending order. By the spectral decomposition we may decompose \u03c1 \u2212 \u03c3 = Q \u2212 R, where Q and R are positive operators with orthogonal support, so T (\u03c1, \u03c3) = tr(R) + tr(Q). De\ufb01ning V \u2261 R + \u03c1 = Q + \u03c3, we have T (\u03c1, \u03c3) = tr(R) + tr(Q) = tr(2V ) \u2212 tr(\u03c1) \u2212 tr(\u03c3). Let t1 \u2265 t2 \u2265 \u00b7 \u00b7 \u00b7 \u2265 td be the eigenvalues of T. Note that ti \u2265 max(ri, si), so 2ti \u2265 ri + si + |ri \u2212 si|, and it follows that T (\u03c1, \u03c3) \u2265 |ri \u2212 si|. (11.46) i By calculus whenever |r \u2212 s| \u2264 1/2 it follows that |\u03b7(r) \u2212 \u03b7(s)| \u2264 \u03b7(|r \u2212 s|). A moment\u2019s thought shows that |ri \u2212 si| \u2264 1/2 for all i, so + + + + + + + + (\u03b7(ri) \u2212 \u03b7(si)) + + |S(\u03c1) \u2212 S(\u03c3)| = \u03b7(|ri \u2212 si|). (11.47) \u2264 i i i |ri \u2212 si| and observing that \u03b7(|ri \u2212 si|) = \u0394\u03b7(|ri \u2212 si|/\u0394) \u2212 |ri \u2212 Setting \u0394 \u2261 si| log(\u0394), we see that |S(\u03c1) \u2212 S(\u03c3)| \u2264 \u0394 \u03b7(|ri \u2212 si|/\u0394) + \u03b7(\u0394) \u2264 \u0394 log d + \u03b7(\u0394), (11.48) where we applied Theorem 11.2 to obtain the second inequality. But \u0394 \u2264 T (\u03c1, \u03c3) by (11.46), so by the monotonicity of \u03b7(\u00b7) on the interval [0, 1/e], |S(\u03c1) \u2212 S(\u03c3)| \u2264 T (\u03c1, \u03c3) log d + \u03b7(T (\u03c1, \u03c3)), (11.49) whenever T (\u03c1, \u03c3) \u2264 1/e, which is Fannes\u2019 inequality. The weaker form of Fannes\u2019 inequality for general T (\u03c1, \u03c3) follows with minor modi\ufb01cations. Von Neumann entropy 513 as a matrix, this property is known as double stochasticity.) Because log(\u00b7) is a strictly j Pijqj, with j Pij log qj \u2264 log ri, where ri \u2261 concave function it follows that equality if and only if there exists a value of j for which Pij = 1. Thus S(\u03c1||\u03c3) \u2265 pi log pi ri, (11.55) i with equality if and only if for each i there exists a value of j such that Pij = 1, that is, if and only if Pij is a permutation matrix. This has the form of the classical relative entropy. From the non-negativity of the classical relative entropy, Theorem 11.1, we deduce that S(\u03c1||\u03c3) \u2265 0, (11.56) with equality if and only if pi = ri for all i, and Pij is a permutation matrix. To simplify the equality conditions further, note that by relabeling the eigenstates of \u03c3 if necessary, we can assume that Pij is the identity matrix, and thus that \u03c1 and \u03c3 are diagonal in the same basis. The condition pi = ri tells us that the corresponding eigenvalues of \u03c1 and \u03c3 are identical, and thus \u03c1 = \u03c3 are the equality conditions. The von Neumann entropy has many interesting and useful properties: 11.3.2 Basic properties of entropy Theorem 11.8: (Basic properties of von Neumann entropy) (1) The entropy is non-negative. The entropy is zero if and only if the state is pure. (2) In a d-dimensional Hilbert space the entropy is at most log d. The entropy is equal to log d if and only if the system is in the completely mixed state I/d. (3) Suppose a composite system AB is in a pure", " state. Then S(A) = S(B). (4) Suppose pi are probabilities, and the states \u03c1i have support on orthogonal subspaces. Then S i pi\u03c1i = H(pi) + i piS(\u03c1i). (11.57) (5) Joint entropy theorem: Suppose pi are probabilities, |i are orthogonal states for a system A, and \u03c1i is any set of density operators for another system, B. Then S pi|ii| \u2297 \u03c1i = H(pi) + i i piS(\u03c1i). (11.58) Proof (1) Clear from the de\ufb01nition. (2) The result follows from the non-negativity of the relative entropy, 0 \u2264 S(\u03c1||I/d) = \u2212S(\u03c1) + log d. (3) From the Schmidt decomposition we know that the eigenvalues of the density operators of systems A and B are the same. (Recall the discussion following Theorem 2.7 on page 109.) The entropy is determined completely by the eigenvalues, so S(A) = S(B). 514 Entropy and information (4) Let \u03bbj i and |ej that pi\u03bbj i and |ej i be the eigenvalues and corresponding eigenvectors of \u03c1i. Observe i pi\u03c1i, and thus i are the eigenvalues and eigenvectors of S pi\u03c1i = \u2212 pi\u03bbj i log pi\u03bbj i i = \u2212 ij i pi log pi \u2212 pi i j i log \u03bbj \u03bbj i = H(pi) + as required. (5) Immediate from the preceding result. i piS(\u03c1i), (11.59) (11.60) (11.61) Exercise 11.13: (Entropy of a tensor product) Use the joint entropy theorem to show that S(\u03c1 \u2297 \u03c3) = S(\u03c1) + S(\u03c3). Prove this result directly from the de\ufb01nition of the entropy. By analogy with the Shannon entropies it is possible to de\ufb01ne quantum joint and conditional entropies and quantum mutual information, for composite quantum systems. The joint entropy S(A, B) for a composite system with two components A and B is de\ufb01ned in the obvious way, S(A, B) \u2261 \u2212tr(\u03c1AB log(\u03c1AB)), where \u03c1AB is the density matrix of the system AB. We de\ufb01ne the conditional entropy and mutual information by: S(A|B) \u2261 S(A, B) \u2212 S(B) S(A : B) \u2261 S(A) + S(B) \u2212 S(A, B) = S(A) \u2212 S(A|B) = S(B) \u2212 S(B|A). (11.62) (11.63) (11.64) Some properties of the Shannon entropy fail to hold for the von Neumann entropy, and this has many interesting consequences for quantum information theory. For instance, for random variables X and Y, the inequality H(X) \u2264 H(X, Y ) holds. This makes intuitive sense: surely we cannot be more uncertain about the state of X than we are about the joint state of X and Y. This intuition fails for quantum states. Consider a system AB of two qubits in the entangled state (|00 + |11)/ 2. This is a pure state, so S(A, B) = 0. On the other hand, system A has density operator I/2, and thus has entropy equal to one. Another way of stating this result is that, for this system, the quantity S(B|A) = S(A, B) \u2212 S(A) is negative. \u221a Exercise 11.14: (Entanglement and negative conditional entropy) Suppose |AB is a pure state of a composite system belonging to Alice and Bob. Show that |AB is entangled if and only if S(B|A) < 0. 11.3.3 Measurements and entropy How does the entropy of a quantum system behave when we perform a measurement on that system? Not surprisingly, the answer to this question depends on the type of measurement which we perform. Nevertheless, there are some surprisingly general assertions we can make about how the entropy behaves. Suppose, for example, that a projective measurement described by projectors Pi is Von Neumann entropy 515 performed on a quantum system, but we never learn the result of the measurement. If the state of the system before the measurement was \u03c1 then the state after is given by \u03c1 = Pi\u03c1Pi. (11.65) The following result shows that the entropy is never decreased by this procedure, and remains constant only if the state is not", " changed by the measurement: i Theorem 11.9: (Projective measurements increase entropy) Suppose Pi is a complete set of orthogonal projectors and \u03c1 is a density operator. Then the entropy of the state \u03c1 \u2261 least as great as the original entropy, i Pi\u03c1Pi of the system after the measurement is at S(\u03c1) \u2265 S(\u03c1), (11.66) with equality if and only if \u03c1 = \u03c1. Proof The proof is to apply Klein\u2019s inequality to \u03c1 and \u03c1, 0 \u2264 S(\u03c1||\u03c1) = \u2212S(\u03c1) \u2212 tr(\u03c1 log \u03c1). (11.67) The result will follow if we can prove that \u2212tr(\u03c1 log \u03c1) = S(\u03c1). To do this, we apply i = Pi, and the cyclic property of the completeness relation the trace, to obtain i Pi = I, the relation P 2 \u2212tr(\u03c1 log \u03c1) = \u2212tr Pi\u03c1 log \u03c1 Pi\u03c1 log \u03c1Pi. (11.68) (11.69) i i = \u2212tr Note that \u03c1Pi = Pi\u03c1Pi = Pi\u03c1. That is, Pi commutes with \u03c1 and thus with log \u03c1, so \u2212tr(\u03c1 log \u03c1) = \u2212tr Pi\u03c1Pi log \u03c1 This completes the proof. i = \u2212tr(\u03c1 log \u03c1) = S(\u03c1). (11.70) (11.71) Exercise 11.15: (Generalized measurements can decrease entropy) Suppose a qubit in the state \u03c1 is measured using the measurement operators M1 = |00| and M2 = |01|. If the result of the measurement is unknown to us then the state of the system afterwards is M1\u03c1M \u2020 can decrease the entropy of the qubit. 2. Show that this procedure 1 + M2\u03c1M \u2020 11.3.4 Subadditivity Suppose distinct quantum systems A and B have a joint state \u03c1AB. Then the joint entropy for the two systems satis\ufb01es the inequalities S(A, B) \u2264 S(A) + S(B) S(A, B) \u2265 |S(A) \u2212 S(B)|. (11.72) (11.73) 516 Entropy and information The \ufb01rst of these inequalities is known as the subadditivity inequality for Von Neumann entropy, and holds with equality if and only if systems A and B are uncorrelated, that is, \u03c1AB = \u03c1A \u2297 \u03c1B. The second is called the triangle inequality, or sometimes the Araki-Lieb inequality; it is the quantum analogue of the inequality H(X, Y ) \u2265 H(X) for Shannon entropies. The proof of subadditivity is a simple application of Klein\u2019s inequality, S(\u03c1) \u2264 \u2212tr(\u03c1 log \u03c3). Setting \u03c1 \u2261 \u03c1AB and \u03c3 \u2261 \u03c1A \u2297 \u03c1B, note that \u2212tr(\u03c1 log \u03c3) = \u2212tr(\u03c1AB(log \u03c1A + log \u03c1B)) = \u2212tr(\u03c1A log \u03c1A) \u2212 tr(\u03c1B log \u03c1B) = S(A) + S(B). (11.74) (11.75) (11.76) Klein\u2019s inequality therefore gives S(A, B) \u2264 S(A) + S(B), as desired. The equality conditions \u03c1 = \u03c3 for Klein\u2019s inequality give equality conditions \u03c1AB = \u03c1A \u2297 \u03c1B for subadditivity. To prove the triangle inequality, introduce a system R which puri\ufb01es systems A and B, as in Section 2.5. Applying subadditivity we have S(R) + S(A) \u2265 S(A, R). (11.77) Since ABR is in a pure state, S(A, R) = S(B) and S(R) = S(A, B). The previous inequality may be rearranged to give S(A, B) \u2265 S(B) \u2212 S(A). (11.78) The equality conditions for this inequality are not quite as easily stated as those for subadditivity. Formally, the equality conditions are that \u03c1AR = \u03c1A\u2297\u03c1R. Intuitively, what this means is that A is already as entangled as it can possibly be with the outside world, given its existing correlations with system B. A more detailed mathematical statement of the equality conditions is given in Exercise 11.16 on this page. By symmetry between the systems A and B we also have S(A, B) \u2265 S(A) \u2212 S(B). Combining this with S(A, B) \u2265 S", "(B) \u2212 S(A) gives the triangle inequality. Exercise 11.16: (Equality conditions for S(A, B) \u2265 S(B) \u2212 S(A)) Let i \u03bbi|ii| is a spectral decomposition for \u03c1AB. Show that \u03c1AB = S(A, B) = S(B) \u2212 S(A) if and only if the operators \u03c1A common eigenbasis, and the \u03c1B i \u2261 trA(|ii|) have orthogonal support. i \u2261 trB(|ii|) have a Exercise 11.17: Find an explicit non-trivial example of a mixed state \u03c1 for AB such that S(A, B) = S(B) \u2212 S(A). 11.3.5 Concavity of the entropy The entropy is a concave function of its inputs. That is, given probabilities pi \u2013 noni pi = 1 \u2013 and corresponding density operators \u03c1i, the negative real numbers such that entropy satis\ufb01es the inequality: S i pi\u03c1i \u2265 i piS(\u03c1i). (11.79) i pi\u03c1i expresses the state of a quantum system which is in an The intuition is that the unknown state \u03c1i with probability pi, and our uncertainty about this mixture of states (11.81) (11.82) (11.83) (11.84) i pi\u03c1i should be higher than the average uncertainty of the states \u03c1i, since the state expresses ignorance not only due to the states \u03c1i, but also a contribution due to our ignorance of the index i. Von Neumann entropy 517 Suppose the \u03c1i are states of a system A. Introduce an auxiliary system B whose state space has an orthonormal basis |i corresponding to the index i on the density operators \u03c1i. De\ufb01ne a joint state of AB by: \u03c1AB \u2261 pi\u03c1i \u2297 |ii|. (11.80) To prove concavity we use the subadditivity of the entropy. Note that for the density matrix \u03c1AB we have: i S(A) = S S(B) = S pi\u03c1i i i pi|ii| S(A, B) = H(pi) + piS(\u03c1i). = H(pi) i Applying the subadditivity inequality S(A, B) \u2264 S(A) + S(B) we obtain i piS(\u03c1i) \u2264 S pi\u03c1i, i which is concavity! Note that equality holds if and only if all the states \u03c1i for which pi > 0 are identical; that is, the entropy is a strictly concave function of its inputs. It\u2019s worth pausing to think about the strategy we\u2019ve employed in the proof of concavity, and the similar strategy used to prove the triangle inequality. We introduced an auxiliary system B in order to prove a result about the system A. Introducing auxiliary systems is something done often in quantum information theory, and we\u2019ll see this trick again and again. The intuition behind the introduction of B in this particular instance is as follows: i pi\u03c1i, where the value of i is not we want to \ufb01nd a system of which part is in the state known. System B effectively stores the \u2018true\u2019 value of i; if A were \u2018truly\u2019 in state \u03c1i, the system B would be in state |ii|, and observing system B in the |i basis would reveal this fact. Using auxiliary systems to encode our intuition in a rigorous way is an art, but it is also an essential part of many proofs in quantum information theory. Exercise 11.18: Prove that equality holds in the concavity inequality (11.79) if and only if all the \u03c1is are the same. Exercise 11.19: Show that there exists a set of unitary matrices Uj and a probability distribution pj such that for any matrix A, i piUiAU \u2020 i = tr(A) I d, (11.85) where d is the dimension of the Hilbert space A lives in. Use this observation and the strict concavity of the entropy to give an alternate proof that the completely mixed state I/d on a space of d dimensions is the unique state of maximal entropy. 518 Entropy and information Exercise 11.20: Let P be a projector and Q = I \u2212 P the complementary projector. Prove that there are unitary operators U1 and U2 and a probability p such that for all \u03c1, P \u03c1P + Q\u03c1Q = pU1\u03c1U \u2020 2. Use this observation to give an alternate proof of Theorem 11.9 based on concavity. 1 + (1 \u2212 p)U2\u03c1U", " \u2020 Exercise 11.21: (Concavity of the Shannon entropy) Use the concavity of the von Neumann entropy to deduce that the Shannon entropy is concave in probability distributions. Exercise 11.22: (Alternate proof of concavity) De\ufb01ne f (p) \u2261 S(p\u03c1 + (1 \u2212 p)\u03c3). Argue that to show concavity it is suf\ufb01cient to prove that f (p) \u2264 0. Prove that f (p) \u2264 0, \ufb01rst for the case where \u03c1 and \u03c3 are invertible, and then for the case where they are not. 11.3.6 The entropy of a mixture of quantum states The \ufb02ip side of concavity is the following useful theorem which provides an upper bound on the entropy of a mixture of quantum states. Taken together the two results imply that i pi\u03c1i of quantum states \u03c1i the following inequality holds: for a mixture piS(\u03c1i) \u2264 S pi\u03c1i \u2264 piS(\u03c1i) + H(pi). (11.86) i i i The intuition behind the upper bound on the right hand side is that our uncertainty about i pi\u03c1i is never more than our average uncertainty about the state \u03c1i, plus an the state additional contribution of H(pi) which represents the maximum possible contribution our uncertainty about the index i contributes to our total uncertainty. We now prove this upper bound. Theorem 11.10: Suppose \u03c1 = \u03c1i are density operators. Then i pi\u03c1i, where pi are some set of probabilities, and the S(\u03c1) \u2264 i piS(\u03c1i) + H(pi), (11.87) with equality if and only if the states \u03c1i have support on orthogonal subspaces. Proof We begin with the pure state case, \u03c1i = |\u03c8i\u03c8i|. Suppose the \u03c1i are states of a system A, and introduce an auxiliary system B with an orthonormal basis |i corresponding to the index i on the probabilities pi. De\ufb01ne \u221a |AB \u2261 pi|\u03c8i|i. (11.88) Since |AB is a pure state we have S(B) = S(A) = S i pi|\u03c8i\u03c8i| = S(\u03c1). (11.89) i Suppose we perform a projective measurement on the system B in the |i basis. After the measurement the state of system B is \u03c1B = i pi|ii|. (11.90) Strong subadditivity 519 But by Theorem 11.9 projective measurements never decrease entropy, so S(\u03c1) = S(B) \u2264 S(B) = H(pi). Observing that S(\u03c1i) = 0 for the pure state case, we have proved that S(\u03c1) \u2264 H(pi) + piS(\u03c1i), (11.91) when the states \u03c1i are pure states. Furthermore, equality holds if and only if B = B, which is easily seen to occur if and only if the states |\u03c8i are orthogonal. j pi j|ei jei j| be orthonormal decompoj|. Applying the pure state result and the sitions for the states \u03c1i, so \u03c1 = observation that The mixed state case is now easy. Let \u03c1i = jei j = 1 for each i, we have ij pipi j|ei j pi S(\u03c1) \u2264 \u2212 pipi j log(pipi j) i = \u2212 ij i pi log pi \u2212 pi i j j log pi pi j = H(pi) + piS(\u03c1i), i (11.92) (11.93) (11.94) which is the desired result. The equality conditions for the mixed state case follow immediately from the equality conditions for the pure state case. 11.4 Strong subadditivity The subadditivity and triangle inequalities for two quantum systems can be extended to three systems. The basic result is known as the strong subadditivity inequality, and it is one of the most important and useful results in quantum information theory. The inequality states that for a trio of quantum systems, A, B, C, S(A, B, C) + S(B) \u2264 S(A, B) + S(B, C). (11.95) Unfortunately, unlike the classical case, all known proofs of the quantum strong subadditivity inequality are quite dif\ufb01cult. However, it is so useful in quantum information theory that we give a full proof of this result. The basic structure of the proof is presented in Section 11.4.1, with some of the details of the proof left to Appendix 6. 11.4.1 Proof of strong", " subadditivity The proof of the strong subadditivity inequality which we shall give is based upon a deep mathematical result known as Lieb\u2019s theorem. We begin with a de\ufb01nition which allows us to state Lieb\u2019s theorem. Suppose f (A, B) is a real-valued function of two matrices, A and B. Then f is said to be jointly concave in A and B if for all 0 \u2264 \u03bb \u2264 1, f (\u03bbA1 + (1 \u2212 \u03bb)A2, \u03bbB1 + (1 \u2212 \u03bb)B2) \u2265 \u03bbf (A1, B1) + (1 \u2212 \u03bb)f (A2, B2). (11.96) Exercise 11.23: (Joint concavity implies concavity in each input) Let f (A, B) be a jointly concave function. Show that f (A, B) is concave in A, with B held \ufb01xed. Find a function of two variables that is concave in each of its inputs, but is not jointly concave. 520 Entropy and information Theorem 11.11: (Lieb\u2019s theorem) Let X be a matrix, and 0 \u2264 t \u2264 1. Then the function f (A, B) \u2261 tr(X \u2020AtXB1\u2212t) (11.97) is jointly concave in positive matrices A and B. Proof See Appendix 6 for the proof of Lieb\u2019s theorem. Lieb\u2019s theorem implies a sequence of results, each interesting in its own right, culminating in the proof of strong subadditivity. We begin with the convexity of the relative entropy. Theorem 11.12: (Convexity of the relative entropy) The relative entropy S(\u03c1\u03c3) is jointly convex in its arguments. Proof For arbitrary matrices A and X acting on the same space de\ufb01ne It(A, X) \u2261 tr(X \u2020AtXA1\u2212t) \u2212 tr(X \u2020XA). The \ufb01rst term in this expression is concave in A, by Lieb\u2019s theorem, and the second term is linear in A. Thus, It(A, X) is concave in A. De\ufb01ne (11.98) I(A, X) \u2261 d dt + + + + It(A, X) = tr(X \u2020(log A)XA) \u2212 tr(X \u2020X(log A)A). (11.99) t=0 Noting that I0(A, X) = 0 and using the concavity of It(A, X) in A we have I(\u03bbA1 + (1 \u2212 \u03bb)A2, X) = lim \u0394\u21920 I\u0394(\u03bbA1 + (1 \u2212 \u03bb)A2, X) \u0394 I\u0394(A1, X) \u0394 \u2265 \u03bb lim + (1 \u2212 \u03bb) lim \u0394\u21920 \u0394\u21920 = \u03bbI(A1, X) + (1 \u2212 \u03bb)I(A2, X). I\u0394(A2, X) \u0394 That is, I(A, X) is a concave function of A. De\ufb01ning the block matrices 11.100) (11.101) (11.102) (11.103) we can easily verify that I(A, X) = \u2212S(\u03c1\u03c3). The joint convexity of S(\u03c1\u03c3) follows from the concavity of I(A, X) in A. Corollary 11.13: (Concavity of the quantum conditional entropy) Let AB be a composite quantum system with components A and B. Then the conditional entropy S(A|B) is concave in the state \u03c1AB of AB. Strong subadditivity 521 Proof Let d be the dimension of system A. Note that < < I < < d \u2297 \u03c1B \u03c1AB = \u2212S(A, B) \u2212 tr S \u03c1AB log \u2297 \u03c1B I d = \u2212S(A, B) \u2212 tr(\u03c1B log \u03c1B) + log d = \u2212S(A|B) + log d. (11.104) (11.105) (11.106) Thus S(A|B) = log d \u2212 S(\u03c1ABI/d \u2297 \u03c1B). The concavity of S(A|B) follows from the joint convexity of the relative entropy. Theorem 11.14: (Strong subadditivity) For any trio of quantum systems, A, B, C, the inequalities hold. S(A) + S(B) \u2264 S(", "A, C) + S(B, C) S(A, B, C) + S(B) \u2264 S(A, B) + S(B, C) (11.107) (11.108) Proof The two inequalities are in fact equivalent. We will use concavity of the conditional entropy to prove the \ufb01rst, and then show that the second follows. De\ufb01ne a function T (\u03c1ABC) of density operators on the system ABC, T (\u03c1ABC) \u2261 S(A) + S(B) \u2212 S(A, C) \u2212 S(B, C) = \u2212S(C|A) \u2212 S(C|B). (11.109) From the concavity of the conditional entropy we see that T (\u03c1ABC) is a convex function of i pi|ii| be a spectral decomposition of \u03c1ABC. From the convexity \u03c1ABC. Let \u03c1ABC = of T, T (\u03c1ABC) \u2264 i piT (|ii|). But T (|ii|) = 0 as for a pure state S(A, C) = S(B) and S(B, C) = S(A). It follows that T (\u03c1ABC) \u2264 0, and thus S(A) + S(B) \u2212 S(A, C) \u2212 S(B, C) \u2264 0, (11.110) which is the \ufb01rst inequality we set out to prove. To obtain the second inequality, introduce an auxiliary system R purifying the system ABC. Then using the just-proved inequality we have S(R) + S(B) \u2264 S(R, C) + S(B, C). (11.111) Since ABCR is a pure state, S(R) = S(A, B, C) and S(R, C) = S(A, B), so (11.111 becomes S(A, B, C) + S(B) \u2264 S(A, B) + S(B, C), (11.112) as we set out to show. Exercise 11.24: We obtained strong subadditivity as a consequence of the inequality S(A) + S(B) \u2264 S(A, C) + S(B, C). Show that this inequality can be obtained as a consequence of strong subadditivity. Exercise 11.25: We obtained strong subadditivity as a consequence of the concavity of the conditional information, S(A|B). Show that the concavity of the conditional 522 Entropy and information entropy may be deduced from strong subadditivity. (Hint: You may need to introduce an auxiliary system into the problem.) 11.4.2 Strong subadditivity: elementary applications Strong subadditivity and related results have many useful implications for quantum information theory. Let\u2019s take a look at a few of the elementary consequences of these results. First, it is worth emphasizing how remarkable it is that the inequality S(A) + S(B) \u2264 S(A, C)+S(B, C) holds. The corresponding inequality holds also for Shannon entropies, but for different reasons. For Shannon entropies it is true that H(A) \u2264 H(A, C) and H(B) \u2264 H(B, C), so the sum of the two inequalities must necessarily be true. In the quantum case, it is possible to have either S(A) > S(A, C) or S(B) > S(B, C), yet somehow Nature conspires in such a way that both of these possibilities are not true simultaneously, in order to ensure that the condition S(A) + S(B) \u2264 S(A, C) + S(B, C) is always satis\ufb01ed. Other ways of rephrasing this are in terms of conditional entropies and mutual informations, 0 \u2264 S(C|A) + S(C|B) S(A : B) + S(A : C) \u2264 2S(A), (11.113) (11.114) both of which are also remarkable inequalities, for similar reasons. Note, however, that the inequality 0 \u2264 S(A|C) + S(B|C), which one might hope to be true based upon (11.114), is not, as can be seen by choosing ABC to be a product of a pure state for A with an EPR state for BC. Exercise 11.26: Prove that S(A : B) + S(A : C) \u2264 2S(A). Note that the corresponding inequality for Shannon entropies holds since H(A : B) \u2264 H(A). Find an example where S(A : B) > S(A).", " For practical applications strong subadditivity is often most easily applied by using a rephrasing in terms of the conditional or mutual informations. The following theorem lists three very simple reformulations of strong subadditivity that provide a powerful intuitive guide to the properties of quantum entropy. Theorem 11.15: (1) Conditioning reduces entropy: Suppose ABC is a composite quantum system. Then S(A|B, C) \u2264 S(A|B). (2) Discarding quantum systems never increases mutual information: Suppose ABC is a composite quantum system. Then S(A : B) \u2264 S(A : B, C). (3) Quantum operations never increase mutual information: Suppose AB is a composite quantum system and E is a trace-preserving quantum operation on system B. Let S(A : B) denote the mutual information between systems A and B before E is applied to system B, and S(A : B) the mutual information after E is applied to system B. Then S(A : B) \u2264 S(A : B). Strong subadditivity 523 Proof (1) The proof is the same as the classical proof (part of Theorem 11.3 on page 506), which we reproduce for convenience: S(A|B, C) \u2264 S(A|B) is equivalent to S(A, B, C) \u2212 S(B, C) \u2264 S(A, B) \u2212 S(B), which is equivalent to S(A, B, C) + S(B) \u2264 S(A, B) + S(B, C), which is strong subadditivity. (2) S(A : B) \u2264 S(A : B, C) is equivalent to S(A) + S(B) \u2212 S(A, B) \u2264 S(A) + S(B, C) \u2212 S(A, B, C), which is equivalent to S(A, B, C) + S(B) \u2264 S(A, B) + S(B, C), which is strong subadditivity. (3) By the constructions of Chapter 8 the action of E on B may be simulated by introducing a third system C, initially in a pure state |0, and a unitary interaction U between B and C. The action of E on B is equivalent to the action of U followed by discarding system C. Letting primes denote the state of the systems after U has acted we have S(A : B) = S(A : B, C) because C starts out in a product state with AB, and clearly S(A : B, C) = S(A : B, C). Discarding systems cannot increase mutual information, so S(A : B) \u2264 S(A : B, C). Putting it all together gives S(A : B) \u2264 S(A : B), as required. There is an interesting set of questions related to the subadditivity properties of quantum conditional entropies. We saw earlier that the Shannon mutual information is not subadditive, and thus the quantum mutual information is not subadditive either. What about the subadditivity of the conditional entropy? That is, is it true that S(A1, A2|B1, B2) \u2264 S(A1|B1) + S(A2|B2), (11.115) for any four quantum systems A1, A2, B1 and B2? It turns out that this inequality is correct. What\u2019s more, the conditional entropy is also subadditive in the \ufb01rst and second entries. Proving these facts is an instructive exercise in the application of strong subadditivity. Theorem 11.16: (Subadditivity of the conditional entropy) Let ABCD be a composite of four quantum systems. Then the conditional entropy is jointly subadditive in the \ufb01rst and second entries: S(A, B|C, D) \u2264 S(A|C) + S(B|D). (11.116) Let ABC be a composite of three quantum systems. Then the conditional entropy is subadditive in each of the \ufb01rst and second entries: S(A, B|C) \u2264 S(A|C) + S(B|C) S(A|B, C) \u2264 S(A|B) + S(A|C). (11.117) (11.118) Proof To prove joint subadditivity in both entries, note that by strong subadditivity S(A, B, C, D) + S(C) \u2264 S(A, C) + S(B, C, D). (11.119) Adding S(D) to each side of this inequality, we obtain S(A, B, C, D) + S(C) + S(", "D) \u2264 S(A, C) + S(B, C, D) + S(D). (11.120) 524 Entropy and information Applying strong subadditivity to the last two terms of the right hand side gives S(A, B, C, D) + S(C) + S(D) \u2264 S(A, C) + S(B, D) + S(C, D). (11.121) Rearranging this inequality gives S(A, B|C, D) \u2264 S(A|C) + S(B|D), (11.122) which is the joint subadditivity of the conditional entropy. Subadditivity of the conditional entropy in the \ufb01rst entry, S(A, B|C) \u2264 S(A|C) + S(B|C), is trivially seen to be equivalent to strong subadditivity. Subadditivity in the second entry is slightly more challenging. We wish to show that S(A|B, C) \u2264 S(A|B) + S(A|C). Note that this is equivalent to demonstrating the inequality S(A, B, C) + S(B) + S(C) \u2264 S(A, B) + S(B, C) + S(A, C). (11.123) To prove this, note that at least one of the inequalities S(C) \u2264 S(A, C) or S(B) \u2264 S(A, B) must be true, as S(A|B)+ S(A|C) \u2265 0 by the \ufb01rst inequality in Theorem 11.14. Suppose S(C) \u2264 S(A, C). Adding to this inequality the strong subadditivity inequality, S(A, B, C) + S(B) \u2264 S(A, B) + S(B, C) gives the result. A similar proof holds in the case when S(B) \u2264 S(A, B). When we introduced the relative entropy it was described as being rather like a measure of distance between probability distributions or between density operators. Imagine that a quantum system consists of two parts, labeled A and B, and that we are given two density operators \u03c1AB and \u03c3AB. In order to ful\ufb01l its distance-like promise, a very desirable property of S(\u00b7\u00b7) is that it decrease when part of the system is ignored, that is: S(\u03c1A\u03c3A) \u2264 S(\u03c1AB\u03c3AB). (11.124) This result is known as the monotonicity of the relative entropy. Intuitively this is a very reasonable property for a measure of distance to have; we expect that ignoring part of a physical system makes it harder to distinguish two states of that system (compare Section 9.2.1), and thus decrease any reasonable measure of distance between them. Theorem 11.17: (Monotonicity of the relative entropy) Let \u03c1AB and \u03c3AB be any two density matrices of a composite system AB. Then S(\u03c1A\u03c3A) \u2264 S(\u03c1AB\u03c3AB). (11.125) Proof Exercise 11.19 on page 517 implies that there exist unitary transformations Uj on the space B and probabilities pj such that \u03c1A \u2297 I d = j pjUj\u03c1ABU \u2020 j (11.126) for all \u03c1AB. From the convexity of the relative entropy we obtain S \u03c1A \u2297 < < < < \u03c3A \u2297 I d I d \u2264 pjS \u2019 Uj \u03c1ABU \u2020 j < < < Uj\u03c3ABU \u2020 j (. (11.127) j Chapter problems 525 But the relative entropy is invariant under unitary conjugation, so this gives < < \u03c3AB \u03c1AB < < \u03c3AB \u03c1AB = S. (11.128) S \u03c1A \u2297 < < < < \u03c3A \u2297 I d I d Combining this with the easily veri\ufb01ed observation that S \u03c1A \u2297 = S(\u03c1A\u03c3A), (11.129) \u2264 pjS j < < < < \u03c3A \u2297 I d I d gives the monotonicity of the relative entropy. Problem 11.1: (Generalized Klein\u2019s inequality) Suppose f (\u00b7) is a convex function from real numbers to real numbers. Then f induces a natural function f (\u00b7) on Hermitian operators, as described in Section 2.1.8 on page 75. Prove that tr(f (A) \u2212 f (B)) \u2265 tr((A \u2212 B)f (B)). (11.130) Use this result to show that the relative entropy is", " non-negative. Problem 11.2: (Generalized relative entropy) The de\ufb01nition of the relative entropy may be extended to apply to any two positive operators r and s, S(rs) \u2261 tr(r log r) \u2212 tr(r log s). (11.131) The earlier argument proving joint convexity of the relative entropy goes directly through for this generalized de\ufb01nition: (1) For \u03b1, \u03b2 > 0 show that S(\u03b1r\u03b2s) = \u03b1S(rs) + \u03b1 tr(r) log(\u03b1/\u03b2). (11.132) (2) Prove that the joint convexity of the relative entropy implies the subadditivity of the relative entropy, S(r1 + r2s1 + s2) \u2264 S(r1s1) + S(r2s2). (11.133) (3) Prove that subadditivity of the relative entropy implies joint convexity of the relative entropy. (4) Let pi and qi be probability distributions over the same set of indices. Show that S piri qisi \u2264 i i i piS(risi) + i pitr(ri) log(pi/qi). (11.134) In the case where the ri are density operators so tr(ri) = 1, this reduces to the pretty formula S piri i i qisi \u2264 i piS(risi) + H(pi||qi), (11.135) where H(\u00b7||\u00b7) is the Shannon relative entropy. Problem 11.3: (Analogue of the triangle inequality for conditional entropy) (1) Show that H(X, Y |Z) \u2265 H(X|Z). (2) Show that it is not always true that S(A, B|C) \u2265 S(A|C). 526 Entropy and information (3) Prove the conditional version of the triangle inequality, S(A, B|C) \u2265 S(A|C) \u2212 S(B|C). (11.136) Problem 11.4: (Conditional forms of strong subadditivity) (1) Prove that S(A, B, C|D) + S(B|D) \u2264 S(A, B|D) + S(B, C|D). (2) Show by explicit example that it is not always true that H(D|A, B, C) + H(D|B) \u2264 H(D|A, B) + H(D|B, C). Problem 11.5: (Strong subadditivity \u2013 Research) Find a simple proof of the strong subadditivity inequality for quantum entropies. Summary of Chapter 11: Entropy and information \u2022 Fundamental measures of information arise as the answers to questions about the quantity of physical resources required to solve some information processing problem. S(A) = \u2212tr(\u03c1A log \u03c1A) S(\u03c1\u03c3) = \u2212S(\u03c1) \u2212 tr(\u03c1 log \u03c3) \u2022 Basic de\ufb01nitions: (entropy) (relative entropy) (conditional entropy) S(A|B) = S(A, B) \u2212 S(B) (mutual information) S(A : B) = S(A) + S(B) \u2212 S(A, B) \u2022 Strong subadditivity: S(A, B, C) + S(B) \u2264 S(A, B) + S(B, C). The other entropy inequalities we discussed are corollaries of this or the joint convexity of the relative entropy. \u2022 The relative entropy is jointly convex in its arguments. \u2022 The relative entropy is monotonic: S(\u03c1A\u03c3A) \u2264 S(\u03c1AB\u03c3AB). History and further reading Historically, the concept of entropy \ufb01rst arose in the study of thermodynamics and statistical mechanics. But the modern information-theoretic foundation for the study of entropy came in Shannon\u2019s wonderful papers on information theory [Sha48]. A good general reference on properties of the Shannon entropy (and much else in information theory) are Chapters 2 and 16 of Cover and Thomas[CT91]. General references on the von Neumann entropy are the review article by Wehrl[Weh78], and the book by Ohya and Petz[OP93]. The entropic uncertainty principle we prove is due to Deutsch[Deu83]. Many other people have worked on entropic uncertainty relations, and we just mention two other papers of interest. Kraus[Kra87] conjectured an entropic uncertainty relation strengthening Deutsch\u2019s for a particular class of measurements, and Maassen and Uf\ufb01nk[MU88] proved Kraus\u2019 conjecture. The relative entropy was introduced by Kullback and", " Leibler[KL51], and its quantum generalization is due to Umegaki[Ume62]. Fannes\u2019 inequality appeared in [Fan73]. Klein\u2019s inequality was proved in [Kle31]. The triangle inequality is due to Araki History and further reading 527 and Lieb[AL70]. Strong subadditivity has an interesting history. Robinson and Ruelle[RR67] \ufb01rst noted the importance of classical strong subadditivity for statistical physics. The quantum version was then conjectured in 1968 by Lanford and Robinson[LR68]. Obtaining a proof of the result was rather dif\ufb01cult however. Finally, in 1973 the theorem was proved in two papers: Lieb\u2019s eponymous theorem in [Lie73], while the surprising connection to strong subadditivity was developed in Lieb and Ruskai[LR73b]; see also[LR73a]. Lieb\u2019s theorem is a generalization of the Wigner\u2013Yanase\u2013Dyson conjecture made in 1963 by Wigner and Yanase[WY63] and subsequently extended by Dyson (unpublished); prior to 1973 it was not known that the Wigner\u2013Yanase\u2013Dyson conjecture and strong subadditivity were related! See Wehrl[Weh78] for a discussion of the Wigner\u2013Yanase\u2013Dyson conjecture. The proof of Lieb\u2019s theorem we have given is due to Simon[Sim79], and is a variant of a proof by Uhlmann[Uhl77]. Other proofs of Lieb\u2019s theorem are also known. See for example, Epstein[Eps73], Ando[And79], and Petz[Pet86]. Subadditivity of the relative entropy in the \ufb01rst and second entries was proved by Lieb[Lie75]. Joint subadditivity of the quantum conditional entropy was proved by Nielsen[Nie98]. The monotonicity of the relative entropy was \ufb01rst noted by Lindblad[Lin75]. Problem 11.2 is due to Ruskai[Rus94]. 12 Quantum information theory Classical information theory is mostly concerned with the problem of sending classical information \u2013 letters in an alphabet, speech, strings of bits \u2013 over communications channels which operate in accordance with the laws of classical physics. How does the picture change if we can build quantum-mechanical communications channels? Can we transmit information more ef\ufb01ciently? Can we make use of quantum mechanics to transmit secret information without being eavesdropped on? These are just two of the questions we may ask when communication channels are allowed to be quantum mechanical. This rede\ufb01nition of what a channel is causes us to go back and re-examine the fundamental questions motivating classical information theory, in the search for new answers. This chapter surveys what is known about quantum information theory, including some surprising and intriguing possibilities made possible by quantum communication channels. Quantum information theory is motivated by the study of communications channels, but it has a much wider domain of application, and it is a thought-provoking challenge to capture in a verbal nutshell the goals of the \ufb01eld. As described in Section 1.6, we can identify three fundamental goals uniting work on quantum information theory: to identify elementary classes of static resources in quantum mechanics (which we identify as types of \u2018information\u2019); to identify elementary classes of dynamical processes in quantum mechanics (identi\ufb01ed as types of \u2018information processing\u2019); and to quantify resource tradeoffs incurred performing elementary dynamical processes. Quantum information theory is fundamentally richer than classical information theory, because quantum mechanics includes so many more elementary classes of static and dynamic resources \u2013 not only does it support all the familiar classical types, but there are entirely new classes such as the static resource of entanglement to make life even more interesting than it is classically. The title of the chapter is \u2018Quantum information theory\u2019, and you may be forgiven for wondering how it is that we can hope to cover all aspects of quantum information theory in a single chapter. In fact, quantum information theory contains many facets other than those described here, including the study of quantum operations, the de\ufb01nition and study of \ufb01delity measures, quantum error-correcting codes, and various notions of entropy \u2013 all topics which we have described in detail in earlier chapters. The purpose of the present chapter is to describe quantum information theory in its \u2018purest\u2019 form; those other chapters are focused on developing speci\ufb01c tools, while we are concerned here with the grand sweep of things, with the most general statements one can make about the properties of quantum information. We begin in Section 12.1 with a discussion of some of the unique properties of quantum states when compared to classical states, in the language of information theory. Not only are quantum", " states impossible to copy, generally, but also they cannot be perfectly distinguished! This is quanti\ufb01ed by the Holevo bound. We then consider, in Section 12.2, an Distinguishing quantum states and the accessible information 529 elementary information-theoretic task, data compression, and show how quantum states can be compressed much as classical ones are. This is done by paralleling the theorem of typical sequences with the typical subspace theorem, to prove Schumacher\u2019s quantum noiseless channel coding theorem, the analogue of Shannon\u2019s noiseless channel coding theorem. A natural generalization of this problem is the capacity of a noisy channel for classical information, and in Section 12.3 we de\ufb01ne and prove the analogue to Shannon\u2019s noisy channel coding theorem, known as the Holevo\u2013Schumacher\u2013Westmoreland theorem. The most dif\ufb01cult challenge is the capacity of a noisy quantum channel for quantum information. This is the subject of Section 12.4, in which the entropy exchange, the quantum Fano inequality, and the quantum data processing inequality are de\ufb01ned, but the open question of the capacity is left unresolved. Two applications of the noisy channel relations, the quantum Singleton bound, and the exorcism of Maxwell\u2019s demon, are presented, and the \ufb01rst half of this chapter is summarized. Two recurring themes throughout this exploration of quantum information are entanglement and non-orthogonality, and these subjects are the focus of the last two sections of the chapter. Section 12.5 describes how entanglement can be thought of as a physical resource, and explains how it can be transformed, distilled, and diluted. And \ufb01nally, in Section 12.6, we present quantum cryptography, a provably secure means of communication whose success arises from the many properties of quantum information considered in this chapter. 12.1 Distinguishing quantum states and the accessible information There is a simple game we can play to illustrate the remarkable differences between quantum and classical information. We will describe this game using two \ufb01ctitious protagonists, Alice and Bob; of course the results can be rephrased in more abstract language, but the anthropocentric language makes the results easier to think (and write!) about. Suppose Alice has a classical information source producing symbols X = 0,..., n according to a probability distribution p0,..., pn. The aim of the game is for Bob to determine the value of X as best he can. To achieve this goal, Alice prepares a quantum state \u03c1X chosen from some \ufb01xed set \u03c10,..., \u03c1n, and gives the state to Bob, who makes a quantum measurement on the state he has been given, and then tries to make the best guess he can as to the identity of X, based on his measurement result Y. A good measure of how much information Bob has gained about X through the measurement is the mutual information H(X : Y ) between X and the measurement outcome Y, as de\ufb01ned in Chapter 11. By the data processing inequality we know that Bob can infer X from Y if and only if H(X : Y ) = H(X), and that in general H(X : Y ) \u2264 H(X); we will see later that the closeness of H(X : Y ) to H(X) actually provides a quantitative measure of how well Bob can determine X. Bob\u2019s goal is to choose a measurement which maximizes H(X : Y ), bringing it as close as possible to H(X). To this end, we de\ufb01ne Bob\u2019s accessible information to be the maximum of the mutual information H(X : Y ) over all possible measurement schemes. The accessible information is a measure of how well Bob can do at inferring the state Alice prepared. In classical information theory, the accessible information is not so interesting; while in practice it may be dif\ufb01cult to distinguish two classical states \u2013 consider the troubles we have reading bad handwriting \u2013 in principle it is always possible. In contrast, in quantum mechanics it is not always possible to distinguish distinct states, even in principle. For example, we saw in Box 2.3 on page 87 that there is no quantum mechanical procedure to 530 Quantum information theory reliably distinguish two non-orthogonal quantum states. Stated in terms of accessible information, if Alice prepares the state |\u03c8 with probability p, and another, non-orthogonal state |\u03d5 with probability 1 \u2212 p, the accessible information of this preparation is strictly less than H(p), as it is not possible for Bob to determine the identity of the state with full reliability. Classically, if Alice prepares one of two classical states \u2013 say a bit in the state 0 with probability p, or in the state 1 with probability 1 \u2212 p \u2013", " then there is no fundamental reason of principle why Bob cannot distinguish between these two states, and thus the accessible information is the same as the entropy of preparation, H(p). There\u2019s an important caveat to this discussion, a context in which the concept of accessible information does make sense classically. The context is that of distinguishing probability distributions. Imagine that Alice prepares the state 0 or 1 according to one of two probability distributions, either (p, 1 \u2212 p) or (q, 1 \u2212 q). Given the state, Bob\u2019s goal is to identify which probability distribution Alice used to prepare the state. Clearly, it is not always possible for Bob to perform this identi\ufb01cation with perfect reliability! Nevertheless, this example (analogous to the accessible information for a quantum system prepared in one of a set of mixed states) is of subsidiary importance. What is most important and remarkable is that the fundamental objects in quantum mechanics \u2013 the pure quantum states \u2013 enjoy distinguishability properties that are markedly different and much richer than the corresponding properties for the fundamental objects of classical information theory, such as \u20180\u2019 or \u20181\u2019. The no-cloning theorem provides another perspective on the lack of accessibility suffered by quantum information in comparison to classical information. Classical information can, of course, be copied. This can be done exactly with digital information, like the multiply backed-up LATEX\ufb01le being used to generate this book, or approximately, as with the analog images appearing on each page of this book, which have been copied by printing press prior to distribution. Surprisingly, the no-cloning theorem states that quantum mechanics does not allow unknown quantum states to be copied exactly, and places severe limitations on our ability to make approximate copies. The no-cloning theorem is proved in Box 12.1. At \ufb01rst sight the no-cloning theorem appears rather peculiar. After all, isn\u2019t classical physics a special case of quantum mechanics? How is it possible that we can copy classical information if we can\u2019t copy quantum states? The answer is that the no-cloning theorem does not prevent all quantum states from being copied, it simply says that nonorthogonal quantum states cannot be copied. More precisely, suppose |\u03c8 and |\u03d5 are two non-orthogonal quantum states. Then the no-cloning theorem implies that it is not possible to build a quantum device that, when input with |\u03c8 or |\u03d5, will output two copies of the input state, |\u03c8|\u03c8 or |\u03d5|\u03d5. On the other hand, if |\u03c8 and |\u03d5 are orthogonal, then the no-cloning theorem doesn\u2019t prohibit their cloning. Indeed, it is rather easy to design quantum circuits which copy such states! This observation resolves the apparent contradiction between the no-cloning theorem and the ability to copy classical information, for the different states of classical information can be thought of merely as orthogonal quantum states. Exercise 12.1: Suppose |\u03c8 and |\u03d5 are two orthogonal quantum states of a single qubit. Design a quantum circuit with two input qubits (the \u2018data\u2019 and the \u2018target\u2019 qubits), with the data qubit in either the state |\u03c8 or |\u03d5, and the target qubit Distinguishing quantum states and the accessible information 531 prepared in the standard state |0, which produces as output |\u03c8|\u03c8 or |\u03d5|\u03d5, depending on whether |\u03c8 or |\u03d5 was input to the data qubit. What is the connection between cloning and accessible information? Suppose Alice prepares one of two non-orthogonal quantum states |\u03c8 and |\u03d5 with respective probabilities p and 1\u2212p. Suppose it were the case that Bob\u2019s accessible information about these states was H(p), that is, the laws of quantum mechanics allowed Bob to obtain enough information by measurement to identify which of the two states |\u03c8 and |\u03d5 Alice had prepared. Then Bob could clone the states in a very simple manner: he would perform a measurement to determine which of |\u03c8 and |\u03d5 had been prepared by Alice, and once he had made the identi\ufb01cation, could prepare at will multiple copies of whichever state |\u03c8 or |\u03d5 Alice had given him. Thus, the no-cloning theorem can be seen as a consequence of the fact that the accessible information for these states is strictly less than H(p). It\u2019s also possible to turn this perspective around, and view the fact that the accessible information is less than H(p) as a consequence of the no-cloning theorem! This is done as follows. Imagine that it is possible to clone non-orthogonal states. After receiving the state |\u03c8 or |\u03d5 from Alice, Bob could repeatedly", " apply such a cloning device to obtain the state |\u03c8\u2297n or |\u03d5\u2297n. In the limit of large n these two states become very nearly orthogonal and it is possible to distinguish them with arbitrarily high reliability by a projective measurement. That is, if it were possible to clone then Bob could identify with arbitrarily high probability of success whether the state |\u03c8 or |\u03d5 had been prepared, and thus the accessible information would be H(p). We can therefore view the no-cloning theorem as being equivalent to the statement than in quantum mechanics the accessible information for non-orthogonal states is in general less than the entropy of preparation. As we have emphasized throughout the book, the hidden nature of quantum information lies at the heart of the power of quantum computation and quantum information, and the accessible information captures in a quantitative way this hidden nature of quantum information. Unfortunately, no general method for calculating the accessible information is known; however, a variety of important bounds can be proved, the most important of which is the Holevo bound. 12.1.1 The Holevo bound The Holevo bound is an exceedingly useful upper bound on the accessible information that plays an important role in many applications of quantum information theory. Theorem 12.1: (The Holevo bound) Suppose Alice prepares a state \u03c1X where X = 0,..., n with probabilities p0,..., pn. Bob performs a measurement described by POVM elements {Ey} = {E0,..., Em} on that state, with measurement outcome Y. The Holevo bound states that for any such measurement Bob may do: H(X : Y ) \u2264 S(\u03c1) \u2212 pxS(\u03c1x), (12.6) where \u03c1 = x px\u03c1x. x The Holevo bound is thus an upper bound on the accessible information. The quantity appearing on the right hand side of the Holevo bound is so useful in quantum information theory that it is given a name, the Holevo \u03c7 quantity, and is sometimes denoted \u03c7. 532 Quantum information theory Box 12.1: The no-cloning theorem Is it possible to make a copy of an unknown quantum state? Surprisingly, it turns out that the answer to this question is no. In this box we describe an elementary proof of this fact that captures the essential reason this is not possible. Suppose we have a quantum machine with two slots labeled A and B. Slot A, the data slot, starts out in an unknown but pure quantum state, |\u03c8. This is the state which is to be copied into slot B, the target slot. We assume that the target slot starts out in some standard pure state, |s. Thus the initial state of the copying machine is |\u03c8 \u2297 |s. Some unitary evolution U now effects the copying procedure, ideally, |\u03c8 \u2297 |s U\u2212\u2192 U |\u03c8 \u2297 |s = |\u03c8 \u2297 |\u03c8. (12.1) (12.2) Suppose this copying procedure works for two particular pure states, |\u03c8 and |\u03d5. Then we have |\u03c8 \u2297 |s |\u03d5 \u2297 |s U U = |\u03c8 \u2297 |\u03c8 = |\u03d5 \u2297 |\u03d5. Taking the inner product of these two equations gives \u03c8|\u03d5 = (\u03c8|\u03d5)2. (12.3) (12.4) (12.5) \u221a 2, since these states are not orthogonal. But x = x2 has only two solutions, x = 0 and x = 1, so either |\u03c8 = |\u03d5 or |\u03c8 and |\u03d5 are orthogonal. Thus a cloning device can only clone states which are orthogonal to one another, and therefore a general quantum cloning device is impossible. A potential quantum cloner cannot, for example, clone the qubit states |\u03c8 = |0 and |\u03d5 = (|0 + |1)/ What we have shown is that it is impossible to perfectly clone an unknown quantum state using unitary evolution. Several questions naturally arise: what if we try to copy a mixed state? What if we allow cloning devices that are not unitary? What if we are willing to allow imperfect copies that nevertheless are \u2018good\u2019 according to some interesting measure of \ufb01delity? These are all good questions which have been the subject of much investigation, as can be seen from the end of chapter \u2018History and further reading.\u2019 The short summary of this work is that even if one allows non-unitary cloning devices, the cloning of non-orthogonal pure states remains impossible unless one is willing to tolerate a \ufb01nite loss of \ufb01delity in the copied states. Similar conclusions hold also for", " mixed states, although a somewhat more sophisticated approach is necessary to even de\ufb01ne what is meant by the notion of cloning a mixed state. Proof The proof of the Holevo bound is via a simple and beautiful construction involving three quantum systems, which we label P, Q and M. The system Q is the quantum system Alice gives to Bob; P and M are \ufb01ctitious auxiliary systems introduced to ease the proof, Distinguishing quantum states and the accessible information 533 just as was done in proving many of the entropic inequalities in Chapter 11. Intuitively, P may be thought of as the \u2018preparation\u2019 system. By de\ufb01nition it has an orthonormal basis |x whose elements correspond to the labels 0,..., n on the possible preparations for the quantum system, Q. M can be thought of intuitively as Bob\u2019s \u2018measuring device\u2019, and it has a basis |y whose elements correspond to the possible outcomes 1,..., n of Bob\u2019s measurement. The initial state of the total system is assumed to be \u03c1P QM = px|xx| \u2297 \u03c1x \u2297 |00|, (12.7) x where we write the tensor product decomposition in the order P QM. Intuitively, this state represents the situation that Alice has chosen a value of x with probability px, prepared a corresponding \u03c1x and given it to Bob, who is about to use his measuring apparatus, initially in the standard state |0, to perform the measurement. To describe the measurement we introduce a quantum operation E acting on the systems Q and M only (not P ), whose action is to perform a measurement with POVM elements {Ey} on the system Q, and to store the result of the measurement in system M : E(\u03c3 \u2297 |00|) \u2261 y Ey\u03c3 Ey \u2297 |yy|, (12.8) where \u03c3 is any state of system Q, and |0 is the initial state of the measuring apparatus. In the following exercise you show that E is a trace-preserving quantum operation. Exercise 12.2: De\ufb01ne Uy to be the unitary operator acting on system M whose action on a basis is Uy|y \u2261 |y + y, where the addition is done modulo n + 1. Show Ey \u2297 Uy} is a set of operation elements de\ufb01ning a trace-preserving that { quantum operation E whose action on states of the form \u03c3 \u2297 |00| agrees with (12.8). The proof of Holevo\u2019s bound now proceeds as follows. Using primes to denote states of P QM after application of E, and unprimed states to indicate the time prior to application of E we have S(P : Q) = S(P : Q, M ), since M is initially uncorrelated with P and Q, and S(P : Q, M ) \u2265 S(P : Q, M ), since applying the quantum operation E to QM can\u2019t increase the mutual information between P and QM (Theorem 11.15 on page 522), and \ufb01nally S(P : Q, M ) \u2265 S(P : M ), since discarding systems can\u2019t increase mutual information (also Theorem 11.15). Putting these results together gives S(P : M ) \u2264 S(P : Q). (12.9) This result, with a little simple algebra, is easily understood to be the Holevo bound! Let\u2019s \ufb01rst calculate the quantity on the right hand side. Note that \u03c1P Q = px|xx| \u2297 \u03c1x, (12.10) x from which it follows that S(P ) = H(px), S(Q) = S(\u03c1), and S(P, Q) = H(px) + x pxS(\u03c1x) (by Theorem 11.10 on page 518), whence S(P : Q) = S(P ) + S(Q) \u2212 S(P, Q) = S(\u03c1) \u2212 x pxS(\u03c1x), (12.11) which is exactly what we want on the right hand side of the Holevo bound! To calculate 534 Quantum information theory the quantity on the left hand side of (12.9), note that \u03c1P QM = xy px|xx| \u2297 Ey\u03c1x Ey \u2297 |yy|. (12.12) Tracing out system Q and using the observation that the joint distribution p(x, y) for the pair (X, Y ) satis\ufb01es p(x, y) = pxp(y|x) =", " pxtr(\u03c1xEy) = pxtr( Ey), gives Ey\u03c1x \u03c1P M = xy p(x, y)|xx| \u2297 |yy|, (12.13) whence S(P : M ) = H(X : Y ), which is exactly what we want on the left hand side of the Holevo bound! This completes the proof of the Holevo bound. 12.1.2 Example applications of the Holevo bound The Holevo bound is a keystone in the proof of many results in quantum information theory. For now, we\u2019ll give just a taste of how this important result may be applied. Recall Theorem 11.10, which implies that S(\u03c1) \u2212 pxS(\u03c1x) \u2264 H(X), (12.14) x with equality if and only if the states \u03c1x have orthogonal support. Suppose that the states \u03c1x do not have orthogonal support, so the inequality in (12.14) is strict. Then the Holevo bound implies that H(X : Y ) is strictly less than H(X), and thus it is impossible for Bob to determine X with perfect reliability based on his measurement result Y, which generalizes our existing understanding that if the states prepared by Alice are not orthogonal then it is not possible for Bob to determine with certainty which state Alice prepared. A concrete example involves Alice preparing a single qubit in one of two quantum states according to the outcome of a fair coin toss. If the coin toss yields heads, then Alice prepares the state |0, and if the coin toss yields tails, then Alice prepares the state cos \u03b8|0 + sin \u03b8|1, where \u03b8 is some real parameter. In the |0, |1 basis it follows that \u03c1 may be written cos2 \u03b8 cos \u03b8 sin \u03b8 cos \u03b8 sin \u03b8 sin2 \u03b8. (12.15) A simple calculation shows that the eigenvalues of \u03c1 are (1 \u00b1 cos \u03b8)/2, and the Holevo bound is therefore given by the binary entropy H((1 + cos \u03b8)/2), as illustrated in Figure 12.1. Notice that the Holevo bound is maximized when \u03b8 = \u03c0/2, attaining a value of 1 bit, corresponding to the case of Alice preparing states chosen from an orthogonal set, at which point it is possible for Bob to determine with surety which state Alice prepared. For other values of \u03b8 the Holevo bound is strictly less than 1 bit, and it is impossible for Bob to determine with surety which state Alice prepared. The Holevo bound may be given more operational meaning by making use of the Fano inequality (see Box 12.2 on page 536 for a derivation). Suppose Bob makes a guess \u02dcX = f (Y ) as to which state Alice prepared, based on the outcome of his measurement Y and some rule for making a guess, encapsulated in the function f (\u00b7). Then according to the Fano inequality and the Holevo bound, H(p( \u02dcX = X)) + p( \u02dcX = X) log(|X| \u2212 1) \u2265 H(X|Y ) Distinguishing quantum states and the accessible information 535 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 12.1. Plot of the Holevo bound \u03c7 as a function of \u03b8 when the states |0 and cos \u03b8|0 + sin \u03b8|1 are prepared with equal probability. Notice that the Holevo bound reaches a maximum when \u03b8 = \u03c0/2, corresponding to orthogonal states. It is only at this point that it is possible for Bob to determine with certainty which state Alice prepared. = H(X) \u2212 H(X : Y ) \u2265 H(X) \u2212 \u03c7, (12.19) which allows us to place bounds on how well Bob may infer the value of X. Heuristically, the smaller \u03c7 is, the harder it is for Bob to determine which state Alice prepared. This is illustrated in Figure 12.2 for the case where Alice prepares |0 with probability onehalf and cos \u03b8|0 + sin \u03b8|1 with probability one-half, for which the bound reduces to H(p( \u02dcX = X)) \u2265 1 \u2212 \u03c7 and \u03c7 = H((1 + cos(\u03b8))/2) as we noted before. Notice that when \u03b8 = \u03c0/2, Bob has some \ufb01nite probability of making an error in his guess. This error gets larger as \u03b8 gets closer to zero. Finally, when", " \u03b8 = 0 and the two states are indistinguishable, the lower bound tells us that Bob\u2019s probability of error is at least one-half \u2013 he can do no better than chance in guessing which state Alice prepared, as we would expect. Exercise 12.3: Use the Holevo bound to argue that n qubits can not be used to transmit more than n bits of classical information. Exercise 12.4: Suppose Alice sends Bob an equal mixture of the four pure states ) |X2 = |X1 = |0 ) 1 3 1 3 1 3 |X3 = |X4 = ) \u221a \u221a \u221a |0 + |0 + |0 + 2|1 2e2\u03c0i/3|1 2e4\u03c0i/3|1. (12.20) (12.21) (12.22) (12.23) 536 Quantum information theory Box 12.2: Fano\u2019s inequality Suppose we wish to infer the value of a random variable X based on knowledge of another random variable Y. Intuitively, we expect that the conditional entropy H(X|Y ) limits how well we may perform this inference. The Fano inequality makes this intuition rigorous, and provides a useful bound on how well we may infer X, given Y. Suppose \u02dcX \u2261 f (Y ) is some function of Y which we are using as our best guess for X. Let pe \u2261 p(X = \u02dcX) be the probability that this guess is incorrect. Then the Fano inequality states that H(pe) + pe log(|X| \u2212 1) \u2265 H(X|Y ), (12.16) where H(\u00b7) is the binary entropy and |X| is the number of values X may assume. Qualitatively, what the inequality tells us is that if H(X|Y ) is large (that is, comparable in size to log(|X| \u2212 1)) then the probability pe of making an error in inference must also be large. To prove the Fano inequality, de\ufb01ne an \u2018error\u2019 random variable, E \u2261 1 if X = \u02dcX, and E \u2261 0 if X = \u02dcX. Notice that H(E) = H(pe). Using the chaining rule for conditional entropies (page 508), we have H(E, X|Y ) = H(E|X, Y ) + H(X|Y ). But E is completely determined once X and Y are known, so H(E|X, Y ) = 0 and thus H(E, X|Y ) = H(X|Y ). Applying the chain rule for entropies again but to different variables, we obtain H(E, X|Y ) = H(X|E, Y ) + H(E|Y ). Conditioning reduces entropy, so H(E|Y ) \u2264 H(E) = H(pe), whence H(X|Y ) = H(E, X|Y ) \u2264 H(X|E, Y ) + H(pe). The proof of the Fano inequality is concluded by bounding H(X|E, Y ) as follows (we have omitted a few simple algebraic details, which you can easily \ufb01ll in): H(X|E, Y ) = p(E = 0)H(X|E = 0, Y ) + p(E = 1)H(X|E = 1, Y )(12.17) \u2264 p(E = 0) \u00d7 0 + pe log(|X| \u2212 1) = pe log(|X| \u2212 1), (12.18) where H(X|E = 1, Y ) \u2264 log(|X| \u2212 1) follows from the fact that when E = 1, X = Y, and X can assume at most |X| \u2212 1 values, bounding its entropy, and thus its conditional entropy by log(|X| \u2212 1). Substituting H(X|E, Y ) \u2264 pe log(|X| \u2212 1) into H(X|Y ) \u2264 H(X|E, Y )+H(pe) gives the Fano inequality H(X|Y ) \u2264 H(pe)+ pe log(|X| \u2212 1). Show that the maximum mutual information between Bob\u2019s measurement and Alice\u2019s transmission is less than one bit. A POVM which achieves \u2248 0.415 bits is known. Can you construct this or, better yet, one which achieves the Holevo bound? 12.2 Data compression Let\u2019s switch tacks now and investigate an elementary dynamical process \u2013 data compression \u2013 which arises in both classical and quantum information theory. In its most general form the problem of data compression is to determine what are the minimal Data compression 537 0.5 0.45 0.4 0", ".35 0.3 0.25 0.2 0.15 0.1 0.05 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 12.2. A lower bound on the probability of Bob making an error in inferring whether Alice prepared the state |0 or cos \u03b8|0 + sin \u03b8|1. This lower bound is obtained by combining Fano\u2019s inequality with the Holevo bound. Observe that the bound decreases to zero as \u03b8 gets close to \u03c0/2, where the states may be reliably distinguished. physical requirements needed to store an information source? It is one of the fundamental problems of information theory, with implications far beyond its immediate scope. In both classical and quantum information theory the techniques utilized in solving this problem turn out to have a far wider range of applicability than mere data compression, yet receive perhaps their simplest and most elegant expression in the solution of the data compression problem. In this section we examine in detail both quantum and classical data compression. 12.2.1 Shannon\u2019s noiseless channel coding theorem Shannon\u2019s noiseless channel coding theorem quanti\ufb01es the extent to which we can compress the information being produced by a classical information source. What do we mean by a classical information source? Many models of such a source are possible. A simple and very fruitful model is that a source consists of a sequence of random variables X1, X2,... whose values represent the output of the source. We will \ufb01nd it convenient to assume that the random variables take values from a \ufb01nite alphabet of symbols, although extensions to in\ufb01nite alphabets also hold. Furthermore, we assume that the different uses of the source are independent and identically distributed; that is, the source is what is known as an i.i.d information source. In the real world sources often don\u2019t behave in this way. It is easy to see, for example, that the letters in the English text in front of you don\u2019t occur in an independent fashion; strong correlations exist between the letters. To take a simple example the letter \u2018t\u2019 is followed by the letter \u2018h\u2019 more frequently than one would expect based upon the overall frequency of the letter \u2018h\u2019 in normal English text; we say 538 Quantum information theory that occurrences of \u2018t\u2019 and \u2018h\u2019 do not occur independently, but are correlated. Nevertheless, for a wide variety of information sources (including English text) the assumption of an i.i.d. source works pretty well in practice, and the ideas introduced to deal with the special case of an i.i.d. source can be generalized to more sophisticated sources. Before we get into the technical details of Shannon\u2019s theorem, let\u2019s use a simple example to understand the intuition behind the result. Suppose an i.i.d. information source is producing bits X1, X2, X3,..., each being equal to zero with probability p, and equal to one with probability 1 \u2212 p. The key idea behind Shannon\u2019s theorem is to divide the possible sequences of values x1,..., xn for the random variables X1,..., Xn up into two types \u2013 sequences which are highly likely to occur, known as typical sequences, and sequences which occur rarely, known as atypical sequences. How is this done? As n gets large, we expect that with high probability a fraction p of the symbols output from the source will be equal to zero, and that a fraction 1 \u2212 p will be equal to one. The sequences x1,..., xn for which this assumption is correct are known as typical sequences. Combining this de\ufb01nition with the independence assumption for the source gives p(x1,..., xn) = p(x1)p(x2)... p(xn) \u2248 pnp(1 \u2212 p)(1\u2212p)n (12.24) for typical sequences. Taking logarithms on both sides gives \u2212 log p(x1,..., xn) \u2248 \u2212np log p \u2212 n(1 \u2212 p) log(1 \u2212 p) = nH(X), (12.25) where X is a random variable distributed according to the source distribution and H(X) = \u2212p log(p) \u2212 (1 \u2212 p) log(1 \u2212 p) is the entropy of the source distribution, also known as the entropy rate of the source. Thus p(x1,..., xn) \u2248 2\u2212nH(X), from which we see that there can be at most 2nH(X) typical", " sequences, since the total probability of all typical sequences cannot be greater than one. We now have the tools to understand a simple scheme for data compression. Suppose the output from the source is x1,..., xn. To compress this output, we check to see whether x1,..., xn is a typical sequence. If it\u2019s not, we give up \u2013 declare an error. Fortunately, as n becomes large this happens very rarely, since nearly all sequences are typical in the limit of large n. If the output is a typical sequence, we record that fact. Since there are at most 2nH(X) typical sequences, it only requires nH(X) bits to uniquely identify a particular typical sequence. We choose some such scheme for identi\ufb01cation, and compress the output from the source to the corresponding string of nH(X) bits describing which typical sequence occurred, which can later be decompressed. As n becomes large this scheme succeeds with probability approaching one. Several criticisms can be made of this scheme: (a) It has a small but \ufb01nite chance of failing. Slightly more sophisticated schemes make use of similar ideas to remove the possibility of an error occurring. (b) To do the compression we have to wait until the source has emitted a large number, n, of symbols. Again, adaptations exist which allow the processing to be done as the symbols are emitted by the source. (c) No explicit scheme mapping outputs from the source to the compressed sequences has been given. Once again, slightly more sophisticated schemes can be developed which address this issue. (d) The exact procedure being used to do the data compression depends on the output distribution of the source. What if this is not known? Clever universal compression algorithms exist which can cope with this possibility. The reader who is interested in Data compression 539 these and other issues is referred to the book of Cover and Thomas listed in the end of chapter \u2018History and further reading\u2019. Let\u2019s generalize the notion of typical sequences beyond the binary case. Suppose X1, X2,... is an i.i.d. information source. Typically, the frequency of occurrence of any given letter x in the sequence output from the source will be close to the probability p(x) of that letter occurring on any given use of the source. With this intuitive understanding in mind we make the following rigorous de\ufb01nition of the notion of a typical sequence. Given > 0 we say that a string of source symbols x1x2... xn is -typical if 2\u2212n(H(X)+) \u2264 p(x1,..., xn) \u2264 2\u2212n(H(X)\u2212), (12.26) and denote the set of all such -typical sequences of length n by T (n, ). A useful equivalent reformulation of the de\ufb01nition is as + + + + 1 n log 1 p(x1,..., xn) + + + + \u2264. \u2212 H(X) (12.27) Using the law of large numbers (stated and proved in Box 12.3 on page 541) we can prove the theorem of typical sequences, which makes rigorous the idea that in the limit of large n most sequences output by an information source are typical. Theorem 12.2: (Theorem of typical sequences) (1) Fix > 0. Then for any \u03b4 > 0, for suf\ufb01ciently large n, the probability that a sequence is -typical is at least 1 \u2212 \u03b4. (2) For any \ufb01xed > 0 and \u03b4 > 0, for suf\ufb01ciently large n, the number |T (n, )| of -typical sequences satis\ufb01es (1 \u2212 \u03b4)2n(H(X)\u2212) \u2264 |T (n, )| \u2264 2n(H(X)+). (12.28) (3) Let S(n) be a collection of size at most 2nR, of length n sequences from the source, where R < H(X) is \ufb01xed. Then for any \u03b4 > 0 and for suf\ufb01ciently large n, x\u2208S(n) p(x) \u2264 \u03b4. (12.29) Proof Part 1: A direct application of the law of large numbers. Notice that \u2212 log p(Xi) are independent, identically distributed random variables. By the law of large numbers for any > 0 and \u03b4 > 0 for suf\ufb01ciently large n we have + + n + + + i=1 p \u2212 log p(Xi) n \u2212 E(\u2212 log p(X)) \u2264 \u2265 1 \u2212 \u03b4. (12.30) + + + + +", " But E (log p(X)) = \u2212H(X) and i=1 log p(Xi) = log (p(X1,..., Xn)). Thus |\u2212 log(p(X1,..., Xn))/n \u2212 H(X)| \u2264 p \u2265 1 \u2212 \u03b4. (12.31) n That is, the probability that a sequence is -typical is at least 1 \u2212 \u03b4. Part 2: Follows from the de\ufb01nition of typicality, and the observation that the sum of 540 Quantum information theory the probabilities of the typical sequences must lie in the range 1 \u2212 \u03b4 (by part 1) to 1 (as probabilities cannot sum to more than 1). Thus 1 \u2265 p(x) \u2265 2\u2212n(H(X)+) = |T (n, )|2\u2212n(H(X)+), (12.32) x\u2208T (n,) x\u2208T (n,) from which we deduce |T (n, )| \u2264 2n(H(X)+), and 1 \u2212 \u03b4 \u2264 p(x) \u2264 x\u2208T (n,) x\u2208T (n,) 2\u2212n(H(X)\u2212) = |T (n, )|2\u2212n(H(X)\u2212), (12.33) from which we deduce |T (n, )| \u2265 (1 \u2212 \u03b4)2n(H(X)\u2212). Part 3: The idea is to split the sequences in S(n) up into typical and atypical sequences. The atypical sequences have small probability in the large n limit. The number of typical sequences in S(n) is obviously no larger than the total number of sequences in S(n), which is at most 2nR, and each typical sequence has probability about 2\u2212nH(X) so the total probability of the typical sequences in S(n) scales like 2n(R\u2212H(X)), which goes to zero when R < H(X). More rigorously, choose so R < H(X) \u2212 \u03b4 and 0 < < \u03b4/2. Split the sequences in S(n) up into the -typical sequences, and the -atypical sequences. By part 1, for n suf\ufb01ciently large the total probability of the atypical sequences can be made less than \u03b4/2. There are at most 2nR typical sequences in S(n), each with probability at most 2\u2212n(H(X)\u2212), so the probability of the typical sequences is at most 2\u2212n(H(X)\u2212\u2212R), which goes to zero as n goes to in\ufb01nity. Thus the total probability of the sequences in S(n) is less than \u03b4 for n suf\ufb01ciently large. Shannon\u2019s noiseless channel coding theorem is an easy application of the theorem of typical sequences. We give here a very simple version of the noiseless channel coding theorem; more sophisticated versions are left to the exercises and the end of chapter \u2018History and further reading\u2019. The basic setting is to suppose that X1, X2,... is an i.i.d. classical information source over some \ufb01nite alphabet containing d symbols. A compression scheme of rate R maps possible sequences x = (x1,..., xn) to a bit string of length nR which we denote by Cn(x) = Cn(x1,..., xn). (Note that nR may not be an integer; our notation is simpli\ufb01ed by agreeing that in this case nR = nR.) The matching decompression scheme takes the nR compressed bits and maps them back to a string of n letters from the alphabet, Dn(Cn(x)). A compression\u2013decompression scheme (Cn, Dn) is said to be reliable if the probability that Dn(Cn(x)) = x approaches one as n approaches \u221e. Shannon\u2019s noiseless channel coding theorem speci\ufb01es for what values of the rate R a reliable compression scheme exists, revealing a remarkable operational interpretation for the entropy rate H(X): it is just the minimal physical resources necessary and suf\ufb01cient to reliably store the output from the source. Theorem 12.4: (Shannon\u2019s noiseless channel coding theorem) Suppose {Xi} is an i.i.d. information source with entropy rate H(X). Suppose R > H(X). Then there exists a reliable compression scheme of rate R for the source. Conversely, if R < H(X) then any compression scheme will not be reliable. Proof Suppose R > H(X). Choose >", " 0 such that H(X) + < R. Consider the set T (n, ) Data compression 541 Box 12.3: The law of large numbers Suppose we repeat an experiment a large number of times, each time measuring the value of some parameter, X. We label the results of the experiments X1, X2,.... Assuming that the results of the experiments are independent, we intuitively expect n that the value of the estimator Sn \u2261 Xi/n of the average E(X), should i=1 approach E(X) as n \u2192 \u221e. The law of large numbers is a rigorous statement of this intuition. Theorem 12.3: (Law of large numbers) Suppose X1, X2,... are independent random variables all having the same distribution as a random variable X with \ufb01nite \ufb01rst and second moments, |E(X)| < \u221e and E(X 2) < \u221e. Then for any > 0, p(|Sn \u2212 E(X)| > ) \u2192 0 as n \u2192 \u221e. Proof To begin we assume that E(X) = 0 and discuss what happens when E(X) = 0 upon completion of the proof. Since the random variables are independent with mean zero, it follows that E(XiXj) = E(Xi)E(Xj) = 0 when i = j, and thus E(S2 n) = n i,j=1 E(XiXj) n2 n i=1 E(X 2 i ) n2 = E(X 2) n, = (12.34) where the \ufb01nal equality follows from the fact that X1,..., Xn are identically distributed to X. By the same token, from the de\ufb01nition of the expectation we have - E(S2 n) = n, dP S2 (12.35) where dP is the underlying probability measure. It is clear that either |Sn| \u2264 or |Sn| >, so we can split this integral into two pieces, and then drop one of these pieces with the justi\ufb01cation that it is non-negative, - - - E(S2 n) = dP S2 n + |Sn|\u2264 |Sn|> dP S2 n \u2265 |Sn|> n. dP S2 (12.36) In the region of integration S2 n > 2, and thus - E(S2 n) \u2265 2 |Sn|> dP = 2p(|Sn| > ). (12.37) Comparing this inequality with (12.34) we see that p(|Sn| > ) \u2264 E(X 2) n2 Letting n \u2192 \u221e completes the proof. In the case when E(X) = 0, it is easy to obtain the result, by de\ufb01ning (12.38). Yi \u2261 Xi \u2212 E(X), Y \u2261 X \u2212 E(X). (12.39) Y and Y1, Y2,... are a sequence of independent, identically distributed random variables with E(Y ) = 0 and E(Y 2) < \u221e. The result follows from the earlier reasoning. 542 Quantum information theory of -typical sequences. For any \u03b4 > 0 and for suf\ufb01ciently large n, there are at most 2n(H(x)+) < 2nR such sequences, and the probability of the source producing such a sequence is at least 1 \u2212 \u03b4. The method of compression therefore is simply to examine the output of the source to see if it is -typical. If it is not, then compress to some \ufb01xed nR bit string which indicates failure; the decompression operation simply outputs a random sequence x1,..., xn as a guess to the information produced by the source; effectively we give up on compression in this case. If the output of the source is typical then we compress the output simply by storing an index for the particular sequence using nR bits in the obvious way, allowing later recovery. Suppose R < H(X). The combined compression\u2013decompression operation has at most 2nR possible outputs, so at most 2nR of the sequences output from the source can be compressed and decompressed without an error occurring. By the theorem of typical sequences, for suf\ufb01ciently large n the probability of a sequence output from the source lying in a subset of 2nR sequences goes to zero, for R < H(X). Thus any such compression scheme cannot be reliable. Exercise 12.5: (Variable-length zero error data compression) Consider the following heuristic for a variable length data compression scheme. Let x1,..., xn be the output from n uses of an", " i.i.d. source with entropy rate H(X). If x1,..., xn is typical, then send a H(X) bit index indicating which typical sequence it is. If x1,..., xn is atypical, send an uncompressed log dn bit index for the sequence (recall that d is the alphabet size). Turn this heuristic into a rigorous argument that the source can be compressed to an average of R bits per source symbol, for any R > H(X), with zero probability of error. 12.2.2 Schumacher\u2019s quantum noiseless channel coding theorem A great conceptual breakthrough of quantum information theory is to realize that we can treat quantum states as if they were information, and ask information-theoretic questions about those quantum states. In this section we de\ufb01ne the notion of a quantum source of information, and study the question: to what extent can the \u2018information\u2019 \u2013 quantum states \u2013 being produced by that source be compressed? How might we de\ufb01ne the notion of a quantum information source? As with the de\ufb01nition of a classical information source it is by no means obvious what the best means of making this de\ufb01nition is, and it is possible to come up with several different de\ufb01nitions, not all of which are equivalent. The de\ufb01nition we are going to use is based on the idea that entanglement is what we are trying to compress and decompress. More formally, an (i.i.d) quantum source will be described by a Hilbert space H, and a density matrix \u03c1 on that Hilbert space. We imagine that the state \u03c1 of the system is merely part of a larger system which is in a pure state, and the mixed nature of \u03c1 is due to entanglement between H and the remainder of the system. A compression scheme of rate R for this source consists of two families of quantum operations Cn and Dn, analogous to the compression and decompression schemes used in the classical case. Cn is the compression operation, taking states in H \u2297n to states in a 2nR-dimensional state space, the compressed space. We may regard the compressed space as representing nR qubits. The operation Dn is a decompression operation, which takes states in the compressed space to states in the original state space. The combined compression\u2013decompression operation is therefore Dn \u25e6Cn. Our criterion for reliability is that in the limit of large n the entanglement \ufb01delity F (\u03c1\u2297n, Dn \u25e6 Cn) should tend towards one. The basic idea of quantum data compression is illustrated in Figure 12.3. Data compression 543 \u03c1 n log d qubits Cn \u03c1 Dn \u03c1 nS(\u03c1) qubits n log d qubits Figure 12.3. Quantum data compression. The compression operation Cn compresses a quantum source \u03c1 stored in n log d qubits into nS(\u03c1) qubits. The source is accurately recovered via the decompression operation Dn. The key technical idea making the quantum noiseless channel coding theorem possible is a quantum version of the idea of typical sequences. Suppose the density operator \u03c1 associated with a quantum source has orthonormal decomposition \u03c1 = p(x)|xx|, (12.40) x where |x is an orthonormal set, and p(x) are the eigenvalues of \u03c1. The eigenvalues p(x) of \u03c1 obey the same rules as a probability distribution: they are non-negative and sum to one. Furthermore, H(p(x)) = S(\u03c1). Therefore, it makes sense to talk of an -typical sequence, x1,..., xn for which + + + + 1 n log 1 p(x1)p(x2)... p(xn) + + + + \u2264, \u2212 S(\u03c1) (12.41) in exactly the same fashion as for the classical de\ufb01nition. An -typical state is a state |x1|x2... |xn for which the sequence x1, x2,..., xn is -typical. De\ufb01ne the -typical subspace to be the subspace spanned by all -typical states, |x1... |xn. We\u2019ll denote the -typical subspace by T (n, ), and the projector onto the -typical subspace by P (n, ). Notice that P (n, ) = |x1x1| \u2297 |x2x2| \u2297... |xnxn|. (12.42) x \u2212typ", "ical The theorem of typical sequences may now be translated into an equivalent quantum form, the typical subspace theorem. Theorem 12.5: (Typical subspace theorem) (1) Fix > 0. Then for any \u03b4 > 0, for suf\ufb01ciently large n, tr(P (n, )\u03c1\u2297n) \u2265 1 \u2212 \u03b4. (12.43) (2) For any \ufb01xed > 0 and \u03b4 > 0, for suf\ufb01ciently large n, the dimension |T (n, )| = tr(P (n, )) of T (n, ) satis\ufb01es: (1 \u2212 \u03b4)2n(S(\u03c1)\u2212) \u2264 |T (n, )| \u2264 2n(S(\u03c1)+). (12.44) (3) Let S(n) be a projector onto any subspace of H \u2297n of dimension at most 2nR, where R < S(\u03c1) is \ufb01xed. Then for any \u03b4 > 0, and for suf\ufb01ciently large n, tr(S(n)\u03c1\u2297n) \u2264 \u03b4. (12.45) 544 Quantum information theory In each case the result can be obtained directly using the law of large numbers, but we prefer to use the theorem of typical sequences to emphasize the close connection to the techniques used in the proof of Shannon\u2019s noiseless channel coding theorem. Proof Part 1: Observe that tr(P (n, )\u03c1\u2297n) = x \u2212typical p(x1)p(x2)... p(xn). (12.46) The result follows immediately from part 1 of the theorem of typical sequences. Part 2: Follows immediately from part 2 of the theorem of typical sequences. Part 3: We split the trace up into a trace over the typical subspace and the atypical subspace, tr(S(n)\u03c1\u2297n) = tr(S(n)\u03c1\u2297nP (n, )) + tr(S(n)\u03c1\u2297n(I \u2212 P (n, ))), (12.47) and bound each term separately. For the \ufb01rst term observe that \u03c1\u2297nP (n, ) = P (n, )\u03c1\u2297nP (n, ), since P (n, ) is a projector which commutes with \u03c1\u2297n. But tr(S(n)P (n, )\u03c1\u2297nP (n, )) \u2264 2nR2\u2212n(S(\u03c1)\u2212), (12.48) (12.49) since the eigenvalues of P (n, )\u03c1\u2297nP (n, ) are bounded above by 2\u2212n(S(\u03c1)\u2212). Letting n \u2192 \u221e we see the \ufb01rst term tends to zero. For the second term note that S(n) \u2264 I. Since S(n) and \u03c1\u2297(I \u2212 P (n, )) are both positive operators it follows that 0 \u2264 tr(S(n)\u03c1\u2297n(I \u2212 P (n, )) \u2264 tr(\u03c1\u2297n(I \u2212 P (n, )) \u2192 0 as n \u2192 \u221e, so the second term also tends to zero as n becomes large, giving the result. With the typical subspace theorem under our belts it is not dif\ufb01cult to prove a quantum analogue of Shannon\u2019s noiseless channel coding theorem. The main ideas of the proof are analogous, but the technical analysis is made a little more dif\ufb01cult by the appearance of non-commuting operators in the proof, which have no classical analogue. Theorem 12.6: (Schumacher\u2019s noiseless channel coding theorem) Let {H, \u03c1} be an i.i.d. quantum source. If R > S(\u03c1) then there exists a reliable compression scheme of rate R for the source {H, \u03c1}. If R < S(\u03c1) then any compression scheme of rate R is not reliable. Proof Suppose R > S(\u03c1) and let > 0 be such that S(\u03c1) + \u2264 R. By the typical subspace theorem, for any \u03b4 > 0 and for all n suf\ufb01ciently large, tr(\u03c1\u2297nP (n, )) \u2265 1 \u2212 \u03b4, and dim(T (n, )) \u2264 2nR. Let H n c be any 2nR-dimensional Hilbert space containing T (n, ). The encoding is done in the following fashion. First a measurement is made, described by the complete set of", " orthogonal projectors P (n, ), I \u2212 P (n, ), with corresponding outcomes we denote 0 and 1. If outcome 0 occurs nothing more is done and the state is left in the typical subspace. If outcome 1 occurs then we replace the state of the system with some standard state \u2018|0\u2019 chosen from the typical subspace; it doesn\u2019t matter Data compression 545 what state is used. It follows that the encoding is a map Cn : H \u2297n \u2192 H n 2nR-dimensional subspace H n c, with operator-sum representation c into the Cn(\u03c3) \u2261 P (n, )\u03c3P (n, ) + Ai\u03c3A\u2020 i, (12.50) where Ai \u2261 |0i| and |i is an orthonormal basis for the orthocomplement of the typical subspace. i The decoding operation Dn : H n c \u2192 H \u2297n is de\ufb01ned to be the identity on H n c, Dn(\u03c3) = \u03c3. With these de\ufb01nitions for the encoding and decoding it follows that F (\u03c1\u2297n, Dn \u25e6 Cn) = |tr(\u03c1\u2297nP (n, ))|2 + i \u2265 |tr(\u03c1\u2297nP (n, ))|2 \u2265 |1 \u2212 \u03b4|2 \u2265 1 \u2212 2\u03b4, |tr(\u03c1\u2297nAi)|2 (12.51) (12.52) (12.53) where the last line follows from the typical subspace theorem. But \u03b4 can be made arbitrarily small for suf\ufb01ciently large n, and thus it follows that there exists a reliable compression scheme {Cn, Dn} of rate R whenever S(\u03c1) < R. To prove the converse, suppose R < S(\u03c1). Without loss of generality we suppose that the compression operation maps from H \u2297n to a 2nR-dimensional subspace with corresponding projector S(n). Let Cj be operation elements for the compression operation Cn, and Dk operation elements for the decompression operation Dn. Then we have F (\u03c1\u2297n, Dn \u25e6 Cn) = + + +2. +tr(DkCj\u03c1\u2297n) (12.54) jk Each of the operators Cj maps to within the subspace with projector S(n) so Cj = S(n)Cj. Let Sk(n) be the projector onto the subspace to which the subspace S(n) is mapped by Dk, so we have Sk(n)DkS(n) = DkS(n) and thus DkCj = DkS(n)Cj = Sk(n)DkS(n)Cj = Sk(n)DkCj, whence F (\u03c1\u2297n, Dn \u25e6 Cn) = + + +2. +tr(DkCj\u03c1\u2297nSk(n)) jk Applying the Cauchy\u2013Schwarz inequality gives F (\u03c1\u2297n, Dn \u25e6 Cn) \u2264 tr(DkCj\u03c1\u2297nC\u2020 j D\u2020 k)tr(Sk(n)\u03c1\u2297n). jk Applying part 3 of the typical subspace theorem we see that for any \u03b4 > 0 and for suf\ufb01ciently large n, tr(Sk(n)\u03c1\u2297n) \u2264 \u03b4. Moreover, the proof of the typical subspace theorem implies that the size n needs to be for this to hold does not depend on k. Thus F (\u03c1\u2297n, Dn \u25e6 Cn) \u2264 \u03b4 tr(DkCj\u03c1\u2297nC\u2020 j D\u2020 k) (12.55) (12.56) (12.57) (12.58) jk = \u03b4, since Cn and Dn are trace-preserving. Since \u03b4 was arbitrary it follows that F (\u03c1\u2297n, Dn \u25e6 Cn) \u2192 0 as n \u2192 \u221e, and thus the compression scheme is not reliable. Schumacher\u2019s theorem not only discusses the existence of a reliable compression scheme, but it also gives clues as to how to actually construct one. The key is to be 546 Quantum information theory c into the 2nR-dimensional typiable to ef\ufb01ciently perform the mapping Cn : H \u2297n \u2192 H n cal subspace H n c. Classical compression techniques such as enumerative coding, Huffman coding, and arithmetic coding can be applied but with one strong restriction: the encoding circuit must be completely", " reversible, and also entirely erase the original state in the process of creating the compressed one! After all, by the no-cloning theorem, it cannot copy the original state, so it cannot leave it behind as normal classical compression schemes typically do. A simple example illustrating how quantum compression works is given in Box 12.4. Exercise 12.6: In the notation of Box 12.4, give an explicit expression for CX in terms of X. Also, describe how to construct a quantum circuit to perform Un for arbitrary n. How many elementary operations do you require, as a function of n? Exercise 12.7: (Data compression circuit) Outline the construction of a circuit to reliably compress a qubit source with \u03c1 = p|00| + (1 \u2212 p)|11| into nR qubits for any R > S(\u03c1) = H(p). Exercise 12.8: (Compression of an ensemble of quantum states) Suppose that instead of adopting the de\ufb01nition of a quantum source based on a single density matrix \u03c1 and the entanglement \ufb01delity, we instead adopted the following ensemble de\ufb01nition, that an (i.i.d.) quantum source is speci\ufb01ed by an ensemble {pj, |\u03c8j} of quantum states, and that consecutive uses of the source are independent and produce a state |\u03c8j with probability pj. A compression\u2013decompression scheme (Cn, Dn) is said to be reliable in this de\ufb01nition if the ensemble average \ufb01delity approaches 1 as n \u2192 \u221e: \u00afF \u2261 pj1... pjn F (\u03c1J, (Dn \u25e6 Cn)(\u03c1J ))2, (12.61) J \u03c8j1 j pj|\u03c8j\u03c8j| and show that provided R > S(\u03c1) there exists a reliable where J = (j1,..., jn) and \u03c1J \u2261 |\u03c8j1 \u03c1 \u2261 compression scheme of rate R with respect to this de\ufb01nition of \ufb01delity. | \u2297 \u00b7 \u00b7 \u00b7 \u2297 |\u03c8jn |. De\ufb01ne \u03c8jn 12.3 Classical information over noisy quantum channels Anything that can go wrong, will \u2013 Attributed to Edward A. Murphy, Jr. We all have dif\ufb01culty talking on the telephone from time to time. We say we have a \u2018bad line\u2019 when we have exceptional dif\ufb01culty understanding the person on the other end of the line. This is an example of the general phenomenon of noise which is present to some extent in all information processing systems. As described in Chapter 10 error-correcting codes can be used to combat the effects of noise, allowing reliable communication and computation to take place even in the presence of quite severe noise. Given a particular noisy communications channel N an interesting question is how much information can be transmitted reliably through that channel. For example, it might be possible that 1000 uses of the channel can be used to transmit 500 bits of information using an appropriate errorcorrecting code, with high probability of recovery from any errors the channel introduces. Classical information over noisy quantum channels 547 Box 12.4: Schumacher compression Consider an i.i.d. quantum source characterized by the single qubit density matrix 12.59) This could originate, for example, as a small part of a of much larger entangled system. An alternate way of viewing this source (compare Section 9.3) is that it is producing the state |\u03c80 = |0 or |\u03c81 = (|0 + |1)/ 2 with equal probabilities one half each (see Exercise 12.8). \u03c1 has orthonormal decomposition p|\u00af0\u00af0| + (1 \u2212 p)|\u00af1\u00af1|, where |\u00af0 = cos \u03c0 |0 + sin \u03c0 |1, and p = 8 8 [3 + tan(\u03c0/8)]/4. In this basis, a block of n qubits can be written as the state |1, |\u00af1 = \u2212 sin \u03c0 8 |0 + cos \u03c0 8 \u221a CX |X. (12.60) X={\u00af0\u00af0...\u00af0, \u00af0...\u00af0\u00af1,...,\u00af1\u00af1...\u00af1} By Theorem 12.6, only |X for which the Hamming weight is approximately equal to np (that is, a basis for the typical subspace) need be transmitted in order to enable reconstruction of the original state with high \ufb01delity. This is easy to appreciate, because |\u00af0|\u03c8k| = cos(\u03c0/8) (for k = {0, 1}) is much larger than", " |\u00af1|\u03c8k| = sin(\u03c0/8), and for X with large Hamming weight, the coef\ufb01cients CX are very small. How do we realize such a compression scheme? One approximate way is the following. Suppose we have quantum circuit Un, which permutes basis states |X such that states are re-ordered lexicographically according to Hamming weight. For example, for n = 4 it does 0000 \u2192 0000 0001 \u2192 0001 0010 \u2192 0010 0100 \u2192 0011 1001 \u2192 1000 1010 \u2192 1001 1100 \u2192 1010 0111 \u2192 1011 1011 \u2192 1100 1101 \u2192 1101 1110 \u2192 1110 1111 \u2192 1111 1000 \u2192 0100 0011 \u2192 0101 0101 \u2192 0110 0110 \u2192 0111 and Toffoli Such a transform, which can be realized using just controlledgates, reversibly packs the typical subspace into the \ufb01rst \u2248 nH(p) qubits (from left to right). To complete the scheme, we also need a quantum gate V, which rotates single qubits into the |\u00af0, |\u00af1 basis. The desired compression scheme is then Cn = (V \u2020)\u2297nUnV \u2297n, and we need send only the \ufb01rst nH(p) qubits output from Cn, to enable a sequence of states from the source to be reconstructed with high \ufb01delity, using a decoder which is the inverse of this circuit. A more ef\ufb01cient coding scheme would pack just the states with Hamming weight \u2248 np into the \ufb01rst nH(p) qubit space; this can be done using a quantum version of arithmetic coding, for example. We say such a code has rate 500/1000 = 1/2. A fundamental problem of information theory is to determine the maximum rate for reliable communication through the channel N, a number known as the capacity of the channel. For noisy classical communications channels the capacity of the channel may be calculated using a beautiful result known as Shannon\u2019s noisy channel coding theorem. We 548 Quantum information theory begin our investigation of the communication of classical information in the presence of noise in Section 12.3.1 with a discussion of some of the main ideas behind Shannon\u2019s noisy channel coding theorem. We don\u2019t get too detailed, however, because in Section 12.3.2 we move on to take a detailed look at a generalization of the problem whereby two parties attempt to communicate classical information by the use of a noisy quantum channel! 12.3.1 Communication over noisy classical channels Many of the main ideas about noisy channel coding, both quantum and classical, can be understood by examining the binary symmetric channel. Recall from Section 10.1 that the binary symmetric channel is a noisy communications channel for a single bit of information, whose effect is to \ufb02ip the bit being transmitted with probability p > 0, while with probability 1 \u2212 p the bit is transmitted without error, as illustrated in Figure 12.4. Figure 12.4. Binary symmetric channel. How much information can we reliably transmit per use of a binary symmetric channel? Using error-correcting codes it is possible to transmit information through the channel, but at an overhead in the number of bits used to accomplish the communication. We will argue that the maximum rate at which information can be reliably transmitted is 1 \u2212 H(p), where H(\u00b7) is the Shannon entropy. What does it mean that the transmission be reliably accomplished? This is a good question, since different answers give rise to different possible rates. We are going to use the following de\ufb01nition for reliability: we assume that inputs to the channel may be encoded in large blocks all at once, and require that the probability for an error in transmission using the code goes to zero as the blocksize is made large. Another possible de\ufb01nition of reliability is to suppose again that the encoding may be performed in blocks, but that as the blocksize becomes large the probability of error becomes exactly zero. Unfortunately, this de\ufb01nition turns out to be too optimistic about what can be achieved with error-correction, and leads to zero capacity for the binary symmetric channel! Similarly, if we don\u2019t allow encoding to be performed in large blocks the capacity turns out to be zero. Indeed, it is rather amazing (and not at all obvious) that even with our less ambitious de\ufb01nition of reliability a non-zero rate of information transmission can be achieved. To show that this is possible several clever ideas are needed. Random coding for the binary symmetric channel Suppose we want to transmit nR bits of information using n uses of our binary symmetric channel; that is, we want to transmit information at a rate R through the channel. We are going to outline a proof that an error", "-correcting code exists that accomplishes this with low probability of error in the limit of large n, and provided R < 1 \u2212 H(p). The Classical information over noisy quantum channels 549 \ufb01rst idea we need is a random coding method for constructing an error-correcting code. Suppose (q, 1 \u2212 q) is any \ufb01xed probability distribution over the possible inputs to the channel (0 and 1). (This distribution is often called the a priori distribution of the code \u2013 the introduction of this distribution is just a technical device to enable the random coding method to work, and the randomness present in the distribution should not be confused with the randomness in the channel.) Then we pick out a codeword x = (x1,..., xn) for our code simply by choosing xj = 0 with probability q and xj = 1 with probability 1 \u2212 q, independently for each j = 1,..., n. We repeat this procedure 2nR times, creating a codebook C of 2nR entries; we denote a generic entry in the codebook by xj. Obviously it\u2019s possible to construct some pretty lousy error-correcting codes using this procedure! We might get really unlucky and construct a code all of whose codewords consist of the string of n zeroes, which obviously is not much use for the transmission of information. Nevertheless, it turns out that on average this random coding procedure gives a pretty good error-correcting code. To understand why this is so, let\u2019s look at what the channel does to a single codeword in the code. Since all the codewords are constructed in the same way, we may as well look at the \ufb01rst, x1. What is the effect of the binary symmetric channel on x1? On a codeword of length n we expect roughly np of the bits to be \ufb02ipped, so with high probability the output from the channel will be have a Hamming distance of about np from the codeword x1, as illustrated in Figure 12.5; we say that such an output is on the Hamming sphere of radius np around x1. How many elements are there in this Hamming sphere? The answer to this question is roughly 2nH(p), since the Hamming sphere consists of all the typically occurring outputs y = x1 \u2295 e from the channel, where e is the error that occurs in the channel, \u2295 denotes bitwise addition modulo 2, and by the theorem of typical sequences the number of such typical errors e is about 2nH(p). We\u2019ve focused on a single codeword, but of course this same type of corruption occurs for all the codewords. We can imagine the space of all the codewords and their surrounding Hamming spheres, as depicted in Figure 12.6. If, as we\u2019ve shown there, the Hamming spheres don\u2019t overlap, then there\u2019s an easy way for Bob to decode the output from the channel. He simply checks to see if the output is in one of the Hamming spheres, outputs the corresponding codeword if so, and outputs \u2018error\u2019 if not. Since we\u2019ve assumed the spheres are non-overlapping, given any codeword as input it\u2019s highly likely that this will result in a successful decoding. Indeed, even if the spheres overlap slightly, it is still possible for Bob to perform the decoding with a good chance of success, provided the overlap is small \u2013 with high probability the output from the channel will belong to one (not zero or two or more) of the Hamming spheres, and will result in a successful decoding. When does this small overlap condition occur? To understand this we need to better understand the structure of the possible outputs from the channel. We obtained the codewords for our code by sampling 2nR times from a set (X1,..., Xn) of random variables which are independent and identically distributed with Xj = 0 with probability q and Xj = 1 with probability 1 \u2212 q. Suppose we let Yj be the result of sending Xj through the binary symmetric channel. The theorem of typical sequences implies that the set of typical values for (Y1,..., Yn) is of size roughly 2nH(Y ), where Y is distributed as each of the Yj. What\u2019s more, each of these typical output values has roughly equal probability. Now, if we sample one hundred times uniformly from a population of size one million, 550 Quantum information theory :$ @ :$ #;$ $: (\" :$> : @ :\" Figure 12.5. Suppose the codeword x1 is sent through n uses of the binary symmetric channel. Then a typical output from the channel is an", " element of the Hamming sphere of radius np around the sequence which has been sent. (This \ufb01gure is a closeup of Figure 12.6.) @ $ Figure 12.6. Randomly chosen codewords for the binary symmetric channel, surrounded by their Hamming spheres of \u2018typical\u2019 outputs. A closeup of an individual codeword may be found in Figure 12.5. it\u2019s not too likely that we\u2019re going to get any repeats. In fact, even if we sample one hundred thousand times the number of repeats is going to be pretty small. It\u2019s not until we get out to about a million samples that the number of repeats is going to start getting large relative to the size of the sample. In a similar fashion, the amount of overlap between our 2nR Hamming spheres of radius np is not going to start getting large until Classical information over noisy quantum channels 551 the combined number of elements in all the spheres approaches the size of the space \u2013 2nH(Y ) \u2013 that we are effectively sampling from. Since each sphere contains roughly 2nH(p) elements, this means that we are very likely to have a good error-correcting code provided 2nR \u00d7 2nH(p) < 2nH(Y ), which corresponds to the condition R < H(Y ) \u2212 H(p). (12.62) (12.63) Now the entropy H(Y ) depends on the a priori distribution (q, 1 \u2212 q) chosen for the Xj. To make the rate as high as possible we try to maximize H(Y ). A simple calculation shows that this is achieved by using the uniform a priori distribution corresponding to q = 1/2, for which H(Y ) = 1, and therefore it is possible to achieve the rate R for any R less than 1 \u2212 H(p). We\u2019ve just outlined a proof that it is possible to reliably transmit information through a binary symmetric channel at any rate up to 1\u2212H(p). The proof is rather sketchy, but in fact contains many of the key ideas needed for a rigorous treatment, even in the quantum case. It turns out that the rates we have shown how to achieve are also the fastest it\u2019s possible to transmit the information through the binary symmetric channel; any faster than a rate 1 \u2212 H(p) and the Hamming spheres start to overlap too much to determine what codeword was sent, no matter how the codewords were chosen! Thus 1 \u2212 H(p) is the capacity of the binary symmetric channel. How practical is random coding as a method to achieve high rate codes for the binary symmetric channel? It is true that if we use a random code then with high probability we can operate at a rate near the capacity. Unfortunately, there is a major dif\ufb01culty with this procedure. In order to do the encoding and decoding, the sender and receiver (\u2018Alice\u2019 and \u2018Bob\u2019) must \ufb01rst agree on a strategy for doing these tasks. In the case of random codes, this means that Alice must send Bob a list of all her random codewords. Doing this takes as much or more communication than Alice and Bob will be able to extract from the noisy channel. Clearly, this is undesirable for many applications! The random coding method is merely a method of demonstrating the existence of high rate codes, it is not a practical method for their construction. For wide practical application, what we would like is a method for achieving rates near the channel capacity which does not incur an unacceptable communication overhead for Alice and Bob. It is quite remarkable that methods for constructing such codes have only recently been discovered even for noisy classical channels, despite many decades of intense effort, and it remains an interesting open problem to \ufb01nd similar constructions for noisy quantum channels. Shannon\u2019s noisy channel coding theorem Shannon\u2019s noisy channel coding theorem generalizes the capacity result for the binary symmetric channel to the case of a discrete memoryless channel. Such a channel has a \ufb01nite input alphabet I, and a \ufb01nite output alphabet O. For the binary symmetric channel, I = O = {0, 1}. The action of the channel is described by a set of conditional probabilities, p(y|x), where x \u2208 I and y \u2208 O. These represent the probabilities of the different outputs y from the channel, given that the input was x, and satisfy the rules p(y|x) \u2265 0 (12.64) 552 Quantum information theory y p(y|x) = 1 for all x. (12.65) The channel is memoryless in the sense that the channel acts the same way each time it is used, and different uses are independent of one another. We shall use the symbol N to denote a classical noisy channel", ". Of course, there are many interesting communications channels which aren\u2019t discrete memoryless channels, such as the telephone line example given earlier, which has a continuous set of inputs and outputs. More general channels may be technically more dif\ufb01cult to understand than discrete memoryless channels, but many of the underlying ideas are the same and we refer you to the end of chapter \u2018History and further reading\u2019 for books containing information on this subject. Let\u2019s look at the actual statement of Shannon\u2019s noisy channel coding theorem. We won\u2019t give the details of the proof, since we prove a more general result for quantum channels in the next section, but it is instructive to look at the statement of the classical result. First, we need to make our notion of reliable information transmission a little more precise. The basic idea is illustrated in Figure 12.7. In the \ufb01rst stage one of 2nR possible messages M is produced by Alice and is encoded using a map Cn : {1,..., 2nR} \u2192 In which assigns to each of Alice\u2019s possible messages an input string which is sent through n uses of the channel to Bob, who decodes the channel output using a map Dn : On \u2192 {1,..., 2nR} which assigns a message to each string for each possible output from the channel. For a given encoding\u2013decoding pair, the probability of error is de\ufb01ned to be the maximum probability over all messages M that the decoded output of the channel D(Y ) is not equal to the message M : p(Cn, Dn) \u2261 max M p(Dn(Y ) = M |X = Cn(M )). (12.66) We say a rate R is achievable if there exists such a sequence of encoding\u2013decoding pairs (Cn, Dn), and require in addition that p(Cn, Dn) \u2192 0 as n \u2192 \u221e. The capacity C(N ) of a given noisy channel N is de\ufb01ned to be the supremum over all achievable rates for the channel. *:: >\"\" A\"$ >\"\" +$ %B \"\" >\"\",\" Figure 12.7. The noisy coding problem for classical messages. We require that every one of the 2nR possible messages should be sent uncorrupted through the channel with high probability. A priori it is not at all obvious how to calculate the capacity of a channel \u2013 a barehands calculation would involve taking a supremum over a very large (in\ufb01nite!) class of possible encoding and decoding methods, and does not appear to be a particularly promising approach. Shannon\u2019s noisy channel coding theorem enormously simpli\ufb01es the calculation of capacity, reducing it to a simple and well-de\ufb01ned optimization problem that can be solved exactly in many cases, and which is computationally quite tractable even when an exact solution is not feasible. Classical information over noisy quantum channels 553 Theorem 12.7: (Shannon\u2019s noisy channel coding theorem) For a noisy channel N the capacity is given by C(N ) = max p(x) where the maximum is taken over all input distributions p(x) for X, for one use of the channel, and Y is the corresponding induced random variable at the output of the channel. H(X : Y ), (12.67) As an example of the noisy channel coding theorem, consider the case of a binary symmetric channel \ufb02ipping bits with probability p, and with input distribution p(0) = q, p(1) = 1 \u2212 q. We have H(X : Y ) = H(Y ) \u2212 H(Y |X) = H(Y ) \u2212 p(x)H(Y |X = x). (12.68) (12.69) x But for each x, H(Y |X = x) = H(p) so H(X : Y ) = H(Y ) \u2212 H(p), which is maximized by choosing q = 1/2, so H(Y ) = 1 and therefore C(N ) = 1 \u2212 H(p) by Shannon\u2019s noisy channel coding theorem, just as our earlier intuitive calculation of the channel capacity of the binary symmetric channel suggested. Exercise 12.9: The erasure channel has two inputs, 0 and 1, and three outputs, 0, 1 and e. With probability 1 \u2212 p the input is left alone. With probability p the input is \u2018erased\u2019, and replaced by e. (1) Show that the capacity of the erasure channel is 1 \u2212 p. (2) Prove that the capacity of the erasure channel is greater than the capacity of the binary symmetric channel. Why is this result intuitively plausible? Exercise 12.", "10: Suppose N1 and N2 are two discrete memoryless channels such that the input alphabet of N2 is equal to the output alphabet of N1. Show that C(N2 \u25e6 N1) \u2264 min(C(N1), C(N2)). (12.70) Find an example where the inequality is strict. A slight peculiarity of the noisy channel coding theorem we have presented is that nowhere does the notion of a classical information source appear! Recall that earlier we de\ufb01ned a classical information source as a sequence of independent and identically distributed random variables. We can combine this notion of an information source in an interesting way with the noisy channel coding theorem to obtain what is known as a source\u2013channel coding theorem. The basic idea is illustrated in Figure 12.8. An information source with entropy rate H(X) is producing information. By Shannon\u2019s noiseless channel coding theorem it is possible to compress the information from the source so that it only requires nH(X) bits to describe; this step is sometimes known as source coding. The compressed output of the source is now used as the input message for the noisy channel. Transmitting at a rate R less than capacity, it requires nH(X)/R uses of the channel to reliably transmit the compressed data to the receiver, who can then decompress it to recover the original output from the source. You might wonder whether a better scheme for transmitting an information source over the noisy channel is possible. Perhaps it is possible to do something more ef\ufb01cient 554 Quantum information theory A\"\" \" *:: A\"\" >\"\" A\"$ >\"\" +$ %B \"\" >\"\",\" Figure 12.8. The noisy coding problem for a classical information source, sometimes known as the source-coding model. than this two stage compress\u2013encode and decode\u2013decompress method? In fact, this turns out not to be the case, and the method of source\u2013channel coding described is in fact optimal, but a proof of this fact is beyond our scope; see the end of chapter \u2018History and further reading\u2019 for more details. 12.3.2 Communication over noisy quantum channels Suppose that instead of using a noisy classical communications channel to communicate Alice and Bob make use of a noisy quantum communications channel. More precisely, Alice has some message M that she wants to send to Bob. She encodes that message, just as she did in the classical case, but now the message is encoded as a quantum state, which is sent over the noisy quantum channel. By performing the encoding in just the right way, we hope that Bob will be able to determine what Alice\u2019s message was, with low probability of failure. Moreover, we\u2019d like the rate at which Alice can send information to Bob to be as high as possible. What we want, in other words, is a procedure for computing the capacity for classical information of a noisy quantum channel. This problem has not yet been completely solved, but a great deal of progress has been made, and in this section we examine this progress. What is known is how to calculate the capacity for a channel E assuming that Alice encodes her messages using product states of the form \u03c11 \u2297 \u03c12 \u2297..., where each of the \u03c11, \u03c12,... are potential inputs for one use of the channel E. We call the capacity with this restriction the product state capacity, and denote it C(1)(E) to indicate that input states cannot be entangled across two or more uses of the channel. Note that this restricted model of communication between Alice and Bob does allow Bob to decode using measurements entangled across multiple uses of the channel; in fact, it turns out that this is essential. The only restriction (and an unfortunate restriction it is) is that Alice can only prepare product state inputs. It is believed by many researchers, but has not been proved, that allowing entangled signals doesn\u2019t increase the capacity. The result which allows us to calculate the product state capacity is known as the Holevo\u2013Schumacher\u2013Westmoreland (HSW) theorem, after its discoverers. As does Shannon\u2019s noisy channel coding theorem for classical noisy channels, the HSW theorem provides an effective means for computing the product state capacity for a speci\ufb01ed noisy channel E, and in some instances may even allow the derivation of an exact expression. Classical information over noisy quantum channels 555 Theorem 12.8: (Holevo\u2013Schumacher\u2013Westmoreland (HSW) theorem) Let E be a trace-preserving quantum operation. De\ufb01ne \u23a1 \u239b \u239b \u239e \u239e \u23a4 \u03c7(E) \u2261 max {pj,\u03c1j } \u23a3S \u239dE \u239d j pj\u03c1j \u23a0 \u23a0 \u2212 j", " pjS(E(\u03c1j)) \u23a6, (12.71) where the maximum is over all ensembles {pj, \u03c1j} of possible input states \u03c1j to the channel. Then \u03c7(E) is the product state capacity for the channel E, that is, \u03c7(E) = C (1)(E). The maximum in (12.71) is potentially over an unbounded set. In practice, we use the results of the following exercise to restrict the maximization to pure state ensembles containing at most d2 elements, where d is the dimension of the input to the channel. Exercise 12.11: Show that the maximum in the expression (12.71) may be achieved using an ensemble of pure states. Show further that it suf\ufb01ces to consider only ensembles of at most d2 pure states, where d is the dimension of the input to the channel. The proof of the HSW theorem involves several different ideas and it is easiest to understand the proof by breaking the discussion up into smaller pieces, and then putting the pieces together to obtain the HSW theorem. \u2297 \u03c1M2 Random coding Suppose \u03c1j is a set of possible inputs to the channel E and let \u03c3j \u2261 E(\u03c1j) be the corresponding outputs. We are going to develop a random coding technique similar to that described earlier for the binary symmetric channel, allowing Alice and Bob to communicate using codewords which are products of the states \u03c1j. We let pj be a probability distribution over the indices j, the a priori distribution. Alice wants to send a message M chosen from the set {1,..., 2nR} to Bob. To each possible message M she associates \u2297... \u2297 \u03c1Mn, where M1,..., Mn are chosen from the index set a codeword \u03c1M1 {j}. (The M1,..., Mn are not meant to be a decimal representation of M or anything of that sort!) For each message M Alice chooses M1 by sampling from the distribution {pj}. She chooses M2 similarly and so on through to Mn, which completes the speci\ufb01cation of the codeword. Abusing notation slightly we write \u03c1M \u2261 \u03c1M1 \u2297 \u00b7 \u00b7 \u00b7 \u2297 \u03c1Mn. The corresponding output states are simply denoted with a \u03c3 instead of a \u03c1, so for example we have \u03c3M1 = E(\u03c1M1 ) and \u03c3M = E \u2297n(\u03c1M ). When Bob receives a particular state \u03c3M (corresponding to Alice trying to communicate the message M ) he performs a measurement in an attempt to determine what the message was. Because we are only interested in the measurement statistics and not in the post-measurement state of Bob\u2019s system, it is suf\ufb01cient to describe this measurement using the POVM formalism. We suppose that for each possible message M Bob has a corresponding POVM element EM. It is possible that Bob might have one (or more) POVM elements that don\u2019t correspond to any speci\ufb01c message sent by Alice; obviously these can all be summed together into a single POVM element E0 satisfying E0 = I \u2212 EM. The probability of Bob successfully identifying M is tr(\u03c3M EM ), and therefore the probability of an error being made for the message M is pe M \u2261 1 \u2212 tr(\u03c3M EM ). M = 0 What we want to do is prove the existence of high rate codes such that the probability 556 Quantum information theory M is small for all messages M. To do this we use a counter-intuitive and rather of error pe clever trick introduced by Shannon for the classical problem. We imagine that Alice is producing the messages M by choosing uniformly from the set {1,..., 2nR}, and analyze the average probability of error pav \u2261 M pe M 2nR = M (1 \u2212 tr(\u03c3M EM )) 2nR. (12.72) The \ufb01rst step of the proof is to show that high rate codes exist with pav tending to zero as n becomes large. After this has been done we will use Shannon\u2019s trick to show that this implies the existence of codes with essentially the same rate for which pe M is close to zero for all M. We begin by constructing a POVM {EM } which represents a pretty good (though perhaps not optimal) method for Bob to decode the outputs \u03c3M from the channel. The key idea in the construction, as for the classical binary symmetric channel, is the idea of typicality. Let > 0. Suppose we de\ufb01ne \u00af\u03c3 \u2261 j pj\u03c3j, and let P be the projector onto the -typical", " subspace of \u00af\u03c3\u2297n. By the theorem of typical sequences it follows that for any \u03b4 > 0 and for suf\ufb01ciently large n, \u00af\u03c3\u2297n(I \u2212 P ) tr \u2264 \u03b4. (12.73) For a given message M we are also going to de\ufb01ne a notion of an -typical subspace for \u03c3M, based on the idea that typically \u03c3M is a tensor product of about np1 copies of \u03c11, np2 copies of \u03c12, and so on. De\ufb01ne \u00afS \u2261 j pjS(\u03c3j). Suppose \u03c3j has spectral decomposition k \u03bbj k|ej kej k|, so \u03c3M = K K |EM \u03bbM K EM K |, (12.74) where K = (K1,..., Kn), and for convenience we de\ufb01ne \u03bbM |EM all |EM Kn and. De\ufb01ne PM to be the projector onto the space spanned by |eM2 K \u2261 |eM1 K2 K1 K such that... |eMn Kn K \u2261 \u03bbM1 K1... \u03bbMn \u03bbM2 K2 + + + + 1 n log 1 \u03bbM K \u2212 \u00afS + + + + \u2264. (12.75) (It will be useful to denote by TM the set of all K such that this condition is satis\ufb01ed.) In a similar manner to the proof of the theorem of typical sequences, the law of large numbers implies that for any \u03b4 > 0 and for suf\ufb01ciently large n we have E(tr(\u03c3M PM )) \u2265 1 \u2212 \u03b4, where the expectation is taken with respect to the distribution over codewords \u03c1M (for a \ufb01xed message M ) induced by random coding, and thus for each M, E [tr (\u03c3M (I \u2212 PM ))] \u2264 \u03b4. (12.76) Also note that by the de\ufb01nition (12.75) the dimension of the subspace onto which PM projects can be at most 2n( \u00afS+), and thus E(tr(PM )) \u2264 2n( \u00afS+). (12.77) We now use the typicality notions to de\ufb01ne Bob\u2019s decoding POVM. We de\ufb01ne EM \u2261 M \u22121/2 P PM P P PM P M \u22121/2 P PM P, (12.78) Classical information over noisy quantum channels 557 where A\u22121/2 denotes the generalized inverse of A1/2, that is, the operator which is inverse to A1/2 on the support of A and is otherwise zero. It follows that M EM \u2264 I, and we can de\ufb01ne one more positive operator E0 \u2261 I \u2212 M EM to complete the POVM. The intuition behind this construction is similar to the decoding method described for the binary symmetric channel. In particular, up to small corrections EM is equal to the projector PM, and Bob\u2019s measurement of {EM } corresponds essentially to checking to see if the output from the channel falls into the space on which PM projects; the space onto which this projector projects can be thought of as analogous to the Hamming sphere of radius np around the codewords used for the binary symmetric channel. The main technical part of the proof that random coding works is to obtain an upper bound on the average probability of error pav. The details of how this is done are given in Box 12.5. The result is \u23a1 \u23a4 pav \u2264 1 2nR M \u23a2 \u23a33tr (\u03c3M (I \u2212 P )) + M = M tr(P \u03c3M P PM ) + tr (\u03c3M (I \u2212 PM )) \u23a5 \u23a6. (12.79) The quantity pav is de\ufb01ned with respect to a speci\ufb01c choice of codewords. We are going to calculate the expectation of this quantity over all random codes. By construction E(\u03c3M ) = \u00af\u03c3\u2297n, and \u03c3M and PM are independent when M = M, so we obtain E(pav) \u2264 3tr( \u00af\u03c3\u2297n(I \u2212 P )) + (2nR \u2212 1)tr(P \u00af\u03c3\u2297nP E(P1)) + E (tr(\u03c31(I \u2212 P1))).(12.80) Substituting (12.73) and (12.76) we obtain E(pav) \u2264 4\u03b4 + (2nR \u2212 1)tr(P \u00af\u03c3\u2297nP E(P1", ")). (12.81) But P \u00af\u03c3\u2297nP \u2264 2\u2212n(S( \u00af\u03c3)\u2212)I and by (12.77) we have E(tr(P1)) \u2264 2n( \u00afS+) whence E(pav) \u2264 4\u03b4 + (2nR \u2212 1)2\u2212n(S( \u00af\u03c3)\u2212 \u00afS\u22122). Provided R < S( \u00af\u03c3) \u2212 \u00afS it follows that E(pav) \u2192 0 as n \u2192 \u221e. Indeed, by choosing the ensemble {pj, \u03c1j} to achieve the maximum in (12.71) we see that this must be true whenever R < \u03c7(E). Thus there must exist a sequence of codes of rate R such that pav \u2192 0 as the block-size n of the code is increased. It follows that for any \ufb01xed > 0 (note that this is a new meaning of to replace the old, which is no longer needed!) for suf\ufb01ciently large n (12.82) pav = M pe M 2nR <. (12.83) M < 2. Obviously in order for this to be true at least half the messages M must satisfy pe So we construct a new code by deleting half the codewords (the codewords with high pe M ) from the code with rate R and pav <, obtaining a new code with 2nR/2 = 2n(R\u22121/n) codewords, and with pe M < 2 for all messages M. Obviously this code also has asymptotic rate R, and the probability of an error can be made arbitrarily small for all codewords, not just on average, as n becomes large. Summing up, we have shown that for any rate R less than \u03c7(E) as de\ufb01ned in (12.71), there exists a code using product state inputs enabling transmission through the channel E at rate R. Our proof suffers the same \ufb02aw as do random coding proofs of Shannon\u2019s 558 Quantum information theory classical noisy channel coding theorem, namely, it does not provide a constructive procedure for performing the coding, but it does at least demonstrate the existence of codes at rates up to capacity. Proof of the upper bound Suppose R is greater than \u03c7(E) as de\ufb01ned in (12.71). We will show that it is impossible for Alice to reliably send information to Bob at this rate through the channel E. Our general strategy is to imagine that Alice is producing messages M uniformly at random from the set {1,..., 2nR} and to show that her average error probability must be bounded away from zero, and therefore the maximum error probability must also be bounded away from zero. Suppose Alice encodes message M as \u03c1M = \u03c1M 1 n with corresponding outputs denoted using \u03c3 instead of \u03c1, and Bob decodes using a POVM {EM } which, without loss of generality, we may suppose contains an element EM for each message, M EM = I and possibly an extra element E0 to ensure that the completeness relation is satis\ufb01ed. This gives an average error probability: \u2297 \u00b7 \u00b7 \u00b7 \u2297 \u03c1M pav = M (1 \u2212 tr(\u03c3M EM )) 2nR. (12.96) From Exercise 12.3 we know that R \u2264 log(d), where d is the dimension of the input to the channel, and thus the POVM {EM } contains at most dn + 1 elements. By Fano\u2019s inequality it follows that H(pav) + pav log(dn) \u2265 H(M |Y ), (12.97) where Y is the measurement outcome from Bob\u2019s decoding, and thus npav log d \u2265 H(M ) \u2212 H(M : Y ) \u2212 H(pav) = nR \u2212 H(M : Y ) \u2212 H(pav). (12.98) Applying \ufb01rst the Holevo bound and then the subadditivity of entropy gives H(M : Y ) \u2264 S ( \u00af\u03c3) \u2212 S(\u03c3M 1 M n \u2264 j=1 S( \u00af\u03c3j) \u2212 M \u2297 \u00b7 \u00b7 \u00b7 \u2297 \u03c3M n ) 2nR S(\u03c3M j ) 2nR, (12.99) (12.100) where \u00af\u03c3j \u2261 greater than \u03c7(E) as de\ufb01ned in (12.71), so M \u03c3M j /2nR. Each of the n terms in the sum on the right hand side is no H(M : Y ) \u2264 n\u03c7(E). (12.101) Substituting into (12.98) gives npav log d \u2265 n(R \u2212 \u03c7", "(E)) \u2212 H(pav), and thus in the limit as n becomes large we obtain pav \u2265 (R \u2212 \u03c7(E)) log(d), (12.102) which is bounded away from zero when R > \u03c7(E), which completes the proof that \u03c7(E) is an upper bound on the product state capacity. Classical information over noisy quantum channels 559 Box 12.5: HSW theorem: the error estimate The most technically complicated part of the proof of the HSW theorem is obtaining an estimate for pav. We outline the details of how this is done here; missing steps should be regarded as exercises to be \ufb01lled in. Suppose we de\ufb01ne | \u02dcEM K. Then K \u2261 P |EM \u239b EM = \u239d | \u02dcEM K \u02dcEM K | \u23a0 | \u02dcEM K \u02dcEM K | \u239d | \u02dcEM K \u02dcEM K | \u23a0 \u239e \u22121/2 \u239b \u239e \u22121/2 M K\u2208TM K\u2208TM M K\u2208TM De\ufb01ning \u03b1(M,K),(M,K ) \u2261 \u02dcEM K | \u239b \u239d M K \u2208TM the average probability of error can be written pav = 1 2nR M 0 1 \u2212 K K \u2208TM | \u02dcEM K \u02dcEM K | \u23a0 (12.84) \u239e \u22121/2 | \u02dcEM K, (12.85) 1 \u03bbM K |\u03b1(M,K),(M,K )|2. (12.86) Using K \u03bbM K = 1 and omitting non-positive terms, we see that \u23a1 \u23a1 pav \u2264 1 2nR \u23a3 M K\u2208TM \u03bbM K (1 \u2212 \u03b12 (M,K),(M,K)) + K\u2208TM \u03bbM K \u23a6. (12.87) K | \u02dcEM De\ufb01ne a matrix \u0393 with entries \u03b3(M,K),(M,K ) \u2261 \u02dcEM K, where the indices are such that K \u2208 TM and K \u2208 TM. It is convenient to work in the matrix space de\ufb01ned by these index conventions, to let E denote the unit matrix with respect to these indices, and to use sp (for spur) to denote the trace operations with respect to these indices. A calculation shows that \u03931/2 = [\u03b1(M,K),(M,K )] \u2264 \u03b3(M,K),(M,K) \u2264 1. Using the observation that and it follows that \u03b12 1 \u2212 x2 = (1 + x)(1 \u2212 x) \u2264 2(1 \u2212 x) when 0 \u2264 x \u2264 1 together with (12.87) gives (M,K),(M,K) pav \u2264 1 2nR \u23a32 M K\u2208TM 1 \u2212 \u03b1(M,K),(M,K) + \u03bbM K K\u2208TM \u03bbM K \u23a6. (12.88) \u23a4 \u23a4 De\ufb01ne the diagonal matrix \u039b \u2261 diag(\u03bbM K ) and observe that 2(E \u2212 \u03931/2) = (E \u2212 \u03931/2)2 + (E \u2212 \u0393) = (E \u2212 \u0393)2(E + \u03931/2)\u22122 + (E \u2212 \u0393) \u2264 (E \u2212 \u0393)2 + (E \u2212 \u0393). K (1 \u2212 \u03b1(M,K),(M,K)) = 2sp(\u039b(E \u2212 \u03931/2) \u03bbM (12.89) (12.90) (12.91) (12.92) \u2264 sp(\u039b(E \u2212 \u0393)2) + sp(\u039b(E \u2212 \u0393)). (12.93) Thus, 2 M K\u2208TM (continued) 560 Quantum information theory Box 12.5 (continued): Calculating the spurs on the right hand side, substituting into (12.88), and doing some simple algebra gives \u23a1 \u239b pav \u2264 1 2nR \u23a2 \u23a3 M K \u03bbM K \u239c \u239d2 \u2212 2\u03b3(M,K),(M,K) + |\u03b3(M,K),(M,K )|,K \u2208TM |\u03b3(M,K),(M,K )|2 \u239f \u23a0 + K\u2208TM \u03bbM K \u23a5 \u23a6. (12.94) Substituting de\ufb01nitions and doing some simple algebra gives \ufffd", "\ufffd\ufffd pav \u2264 1 2nR M \u23a2 \u23a32tr (\u03c3M (I \u2212 P )) + tr(\u03c3M (I \u2212 P )PM (I \u2212 P )) + M = M tr(P \u03c3M P PM ) + tr (\u03c3M (I \u2212 PM )) \u23a4 \u23a5 \u23a6. (12.95) The second term is less than tr(\u03c3M (I \u2212 P )), which gives the desired error estimate, (12.79). Examples An interesting implication of the HSW theorem is that any quantum channel E whatsoever can be used to transmit classical information, provided the channel is not simply a constant. For if the channel is not a constant then there exist pure states |\u03c8 and |\u03d5 such that E(|\u03c8\u03c8|) = E(|\u03d5\u03d5|). Substituting the ensemble made up of these two states with equal probabilities 1/2 into the expression (12.71) for the product state capacity we see that C(1)(E) \u2265 S E(|\u03c8\u03c8|) + E(|\u03d5\u03d5|) 2 \u2212 1 2 E(|\u03c8\u03c8|) \u2212 1 2 E(|\u03d5\u03d5|) > 0, (12.103) where the second inequality follows from the strict concavity of the entropy established in Section 11.3.5. Let\u2019s look at a simple example where the product state capacity can be calculated exactly, the case of the depolarizing channel with parameter p. Let {pj, |\u03c8j} be an ensemble of quantum states. Then we have E(|\u03c8j\u03c8j|) = p|\u03c8j\u03c8j| + (1 \u2212 p), I 2 (12.104) a quantum state which has eigenvalues (1 + p)/2 and (1 \u2212 p)/2 from which it follows that S(E(|\u03c8j\u03c8j|)) = H, (12.105) 1 + p 2 which does not depend on |\u03c8j at all. Thus the maximum in (12.71) is achieved by Quantum information over noisy quantum channels 561 j E(|\u03c8j\u03c8j|)), which may be done by simply choosing the maximizing the entropy S( |\u03c8j to form an orthonormal basis (say |0 and |1) for the state space of a single qubit, giving a value for the entropy of one bit, and a product state capacity of C(E) = 1 \u2212 H for the depolarizing channel with parameter p. 1 + p 2 (12.106) Exercise 12.12: Adapt the proof of the HSW theorem to \ufb01nd a proof of Shannon\u2019s noisy channel coding theorem, simplifying the proof wherever possible. 12.4 Quantum information over noisy quantum channels How much quantum information can be reliably transmitted over a noisy quantum channel? This problem of determining the quantum channel capacity is less well understood than is the problem of determining the capacity for sending classical information through a noisy quantum channel. We now present some of the information-theoretic tools that have been developed to understand the capacity of a quantum channel for quantum information, most notably quantum information-theoretic analogues to the Fano inequality (Box 12.2 on page 536), data processing inequality (Section 11.2.4 on page 509) and Singleton bound (Exercise 10.21 on page 449). As for quantum data compression, our point of view in studying these problems is to regard a quantum source as being a quantum system in a mixed state \u03c1 which is entangled with another quantum system, and the measure of reliability for transmission of quantum information by the quantum operation E is the entanglement \ufb01delity F (\u03c1, E). It is useful to introduce, as in Chapter 9, labels Q for the system \u03c1 lives in and R, the reference system, which initially puri\ufb01es Q. In this picture the entanglement \ufb01delity is a measure of how well the entanglement between Q and R was preserved by the action of E on system Q. 12.4.1 Entropy exchange and the quantum Fano inequality How much noise does a quantum operation cause when applied to the state \u03c1 of a quantum system Q? One measure of this is the extent to which the state of RQ, initially pure, becomes mixed as a result of the quantum operation. We de\ufb01ne the entropy exchange of the operation E upon input of \u03c1 by S(\u03c1, E) \u2261 S(R, Q). (12.107) Suppose that the action of the quantum operation E is mocked up by introducing an environment E, initially in a pure state, and then causing a unitary interaction between Q and E, as described in Chapter", " 8. Then the state of RQE after the interaction is a pure state, whence S(R, Q) = S(E), so the entropy exchange may also be identi\ufb01ed with the amount of entropy introduced by the operation E into an initially pure environment E. Note that the entropy exchange does not depend upon the way in which the initial state of Q, \u03c1, is puri\ufb01ed into RQ. The reason is because any two puri\ufb01cations of Q into RQ are related by a unitary operation on the system R, as shown in Exercise 2.81 on page 111. This unitary operation on R obviously commutes with the action of the 562 Quantum information theory quantum operation on Q, and thus the \ufb01nal states of RQ induced by the two different puri\ufb01cations are related by a unitary transformation on R, and thus give rise to the same value for the entropy exchange. Furthermore, it follows from these results that S(E) does not depend upon the particular environmental model for E which is used, provided the model starts with E in a pure state. A useful explicit formula for the entropy exchange can be given based upon the operator-sum representation for quantum operations. Suppose a trace-preserving quantum operation E has operation elements {Ei}. Then, as shown in Section 8.2.3, a unitary model for this quantum operation is given by de\ufb01ning a unitary operator U on QE such that U |\u03c8|0 = Ei|\u03c8|i, (12.108) i where |0 is the initial state of the environment, and |i is an orthonormal basis for the environment. Note that the state of E after application of E is: \u03c1E tr(Ei\u03c1E\u2020 j )|ij|. (12.109) = i,j That is, tr(Ei\u03c1E\u2020 j ) are the matrix elements of E in the |i basis. Given a quantum operation with operation elements {Ei} it is therefore natural to de\ufb01ne a matrix W (the w-matrix) with matrix elements Wij \u2261 tr(Ei\u03c1E\u2020 j ), that is, W is the matrix of E in an appropriate basis. This representation for \u03c1E gives rise to a formula for the entropy exchange which is useful in making explicit calculations, S(\u03c1, E) = S(W ) \u2261 \u2212tr(W log W ). (12.110) Given a quantum operation E and a state \u03c1, it is always possible to choose operation elements {Fj} for E such that W is diagonal; we say W is in canonical form. To see that such a set of operation elements exists, recall from Chapter 8 that a quantum operation may have many different sets of operation elements. In particular, two sets of operators {Ei} and {Fj} are operation elements for the same quantum operation if and only if j ujiEi, where u is a unitary matrix of complex numbers, and it may be necessary Fj = to append 0 operators to the sets Ei or Fj so that the matrix u is a square matrix. Let W be the w-matrix associated to a particular choice of operation elements {Ei} for E. W is a matrix representation of the environmental density operator, and thus is a positive matrix which may be diagonalized by a unitary matrix v, D = vW v\u2020, where D is a diagonal i vjiEi, matrix with non-negative entries. De\ufb01ne operators Fj by the equation Fj \u2261 so the Fj are also a set of operation elements for E, giving rise to a new w-matrix 6W with matrix elements 6Wkl = tr(Fk\u03c1F \u2020 l ) = vkmv\u2217 lnWmn = Dkl. (12.111) mn Thus, the w-matrix is diagonal if calculated with respect to the operation elements {Fj}. Any such set of operation elements {Fj} for E for which the corresponding w-matrix is diagonal is said to be a canonical representation for E with respect to the input \u03c1. We see later that canonical representations turn out to have a special signi\ufb01cance for quantum error-correction. Many properties of the entropy exchange follow easily from properties of the entropy Quantum information over noisy quantum channels 563 discussed in Chapter 11. For example, working in a canonical representation for a tracepreserving quantum operation E on a d-dimensional space, we see immediately that S(I/d, E) = 0 if and only if E is a unitary quantum operation. Therefore, S(I/d, E) can be thought of as quantifying the extent to which incoherent quantum noise occurs on the system as a whole. A", " second example is that the matrix W is linear in \u03c1, and by the concavity of the entropy it follows that S(\u03c1, E) is concave in \u03c1. Since the system RQ can always be chosen to be at most d2-dimensional, where d is the dimension of Q, it follows that the entropy exchange is bounded above by 2 log d. Exercise 12.13: Show that the entropy exchange is concave in the quantum operation E. Intuitively, if the quantum source Q is subject to noise which results in the entanglement RQ becoming mixed, then the \ufb01delity of the \ufb01nal state RQ with the initial state RQ cannot be perfect. Moreover, the greater the noise the worse the \ufb01delity. In Section 12.1.1 an analogous situation arose in the study of classical channels, where the uncertainty H(X|Y ) about the input of a channel, X, given the output, Y, was related to the probability of being able to recover the state of X from Y by the Fano inequality. There is a very useful quantum analogue of this result, relating the entropy exchange S(\u03c1, E) to the entanglement \ufb01delity F (\u03c1, E). Theorem 12.9: (Quantum Fano inequality) Let \u03c1 be a quantum state and E a trace-preserving quantum operation. Then S(\u03c1, E) \u2264 H(F (\u03c1, E)) + (1 \u2212 F (\u03c1, E)) log(d2 \u2212 1), (12.112) where H(\u00b7) is the binary Shannon entropy. Inspection of the quantum Fano inequality reveals an attractive intuitive meaning: if the entropy exchange for a process is large, then the entanglement \ufb01delity for the process must necessarily be small, indicating that the entanglement between R and Q has not been well preserved. Moreover, we note that in the quantum Fano inequality the entropy exchange S(\u03c1, E) plays a role analogous to the role played by the conditional entropy H(X|Y ) in classical information theory. Proof To prove the quantum Fano inequality, let |i be an orthonormal basis for the system RQ chosen so the \ufb01rst state in the set |1 = |RQ. If we form the quantities pi \u2261 i|\u03c1RQ|i, then from the results of Section 11.3.3 it follows that S(R, Q) \u2264 H(p1,..., pd2), where H(pi) is the Shannon information of the set {pi}. Elementary algebra shows that (12.113) H(p1,..., pd2) = H(p1) + (1 \u2212 p1)H Combining this with the observation that H( p2 1\u2212p1 F (\u03c1, E) by de\ufb01nition gives,,..., pd2 1 \u2212 p1 p2 1 \u2212 p1,..., p d2 1\u2212p1 ) \u2264 log(d2 \u2212 1) and p1 =. (12.114) S(\u03c1, E) \u2264 H(F (\u03c1, E)) + (1 \u2212 F (\u03c1, E)) log(d2 \u2212 1), (12.115) 564 Quantum information theory which is the quantum Fano inequality. 12.4.2 The quantum data processing inequality In Section 11.2.4 we discussed the classical data processing inequality. Recall that the data processing inequality states that for a Markov process X \u2192 Y \u2192 Z, H(X) \u2265 H(X : Y ) \u2265 H(X : Z), (12.116) with equality in the \ufb01rst stage if and only if the random variable X can be recovered from Y with probability one. Thus the data processing inequality provides informationtheoretic necessary and suf\ufb01cient conditions for error-correction to be possible. There is a quantum analogue to the data processing inequality applicable to a two stage quantum process described by quantum operations E1 and E2, \u03c1 E1\u2212\u2192 \u03c1 E2\u2212\u2192 \u03c1. We de\ufb01ne the quantum coherent information by I(\u03c1, E) \u2261 S(E(\u03c1)) \u2212 S(\u03c1, E). (12.117) (12.118) This quantity, coherent information, is suspected (but not known) to play a role in quantum information theory analogous to the role played by the mutual information H(X : Y ) in classical information theory. One reason for this belief is that the coherent information satis\ufb01es a quantum data processing inequality analogous to the classical data processing inequality. Theorem 12.10: (Quantum data processing inequality) Let \u03c1 be a quantum state and E1 and E2 trace-pres", "erving quantum operations. Then S(\u03c1) \u2265 I(\u03c1, E1) \u2265 I(\u03c1, E2 \u25e6 E1), (12.119) with equality in the \ufb01rst inequality if and only if it is possible to perfectly reverse the operation E1, in the sense that there exists a trace-preserving reversal operation R such that F (\u03c1, R \u25e6 E) = 1. Comparison with the classical data processing inequality shows that the coherent information plays a role in the quantum data processing inequality identical to the role played by the mutual information in the classical data processing inequality. Of course, such a heuristic argument cannot be regarded as any sort of a rigorous justi\ufb01cation for the view that the coherent information is the correct quantum analogue of the classical mutual information. In order to obtain such a justi\ufb01cation, the coherent information ought to be related to the quantum channel capacity in a similar way to the relation between classical mutual information and classical channel capacity, and such a relationship has not yet been established. (See the end of chapter \u2018History and further reading\u2019 for some partial progress.) How is the notion of perfect reversibility de\ufb01ned in the quantum data processing inequality connected with more familiar notions such as that arising in the context of quantum error-correction? By de\ufb01nition, we say a trace-preserving quantum operation E is perfectly reversible upon input of \u03c1 if there exists a trace-preserving quantum operation R such that F (\u03c1, R \u25e6 E) = 1. (12.120) Quantum information over noisy quantum channels 565 But from item (4) on page 423, it follows that a quantum operation is perfectly reversible if and only if for every state |\u03c8 in the support of \u03c1, (R \u25e6 E)(|\u03c8\u03c8|) = |\u03c8\u03c8|. (12.121) This observation connects the notion of perfect reversibility to quantum error-correcting codes. Recall that a quantum error-correcting code is a subspace of some larger Hilbert space spanned by logical codewords. To be resilient against the noise induced by a quantum operation E it is necessary that the quantum operation E be reversible by a trace-preserving reversal operation R in the sense that for all states |\u03c8 in the code, (R \u25e6 E)(|\u03c8\u03c8|) = |\u03c8\u03c8|. This condition is equivalent to the criterion of perfect reversibility in the statement of the data processing inequality, that F (\u03c1, R \u25e6 E) = 1, for some \u03c1 whose support is the code space. Proof The quantum data processing inequality is proved using a four system construction: R and Q appear in their familiar roles, while E1 and E2 are systems initially in pure states, chosen such that a unitary interaction between Q and E1 generates the dynamics E1, and a unitary interaction between Q and E2 generates the dynamics E2. The proof of the \ufb01rst stage of the quantum data processing inequality is to apply the subadditivity inequality S(R, E 1) \u2264 S(R) + S(E 1) to obtain I(\u03c1, E1) = S(E1(\u03c1)) \u2212 S(\u03c1, E1) = S(Q) \u2212 S(E 1) 1) \u2212 S(E = S(R, E 1) \u2264 S(R) + S(E 1) \u2212 S(E = S(R) = S(Q) = S(\u03c1). 1) = S(R) (12.122) (12.123) (12.124) (12.125) (12.126) The proof of the second part of the data processing inequality is to apply the strong subadditivity inequality, S(R, E 1 From the purity of the total state of RQE 1 2 it follows that, E 2 ) + S(E 1 ) \u2264 S(R, E E 1 ) + S(E 1, E 2 ). (12.127) S(R, E 1, E 2 ) = S(Q). (12.128) Neither of the systems R or E1 are involved in the second stage of the dynamics in which Q and E2 interact unitarily. Thus, their state does not change during this stage: \u03c1RE 1. But from the purity of RQE1 after the \ufb01rst stage of the dynamics, 1 = \u03c1RE S(R, E 1 ) = S(R, E 1) = S(Q). (12.129) The remaining two terms in the strong subadditivity inequality (12.127) are now recognized as entropy exchanges, S(E 1 ) = S(E 1) = S(\u03c1, E1); S(E 1, E 2 ) = S(\u03c1", ", E2 \u25e6 E1). (12.130) Making these substitutions into (12.127) yields S(Q) + S(\u03c1, E1) \u2264 S(Q) + S(\u03c1, E2 \u25e6 E1), (12.131) 566 Quantum information theory which can be rewritten as the second stage of the data processing inequality, I(\u03c1, E1) \u2265 I(\u03c1, E2 \u25e6 E1). To complete the proof we need to show that E is perfectly reversible upon input of \u03c1 if and only if the \ufb01rst inequality in the quantum data processing inequality is satis\ufb01ed with equality, S(\u03c1) = I(\u03c1, E) = S(\u03c1) \u2212 S(\u03c1, E). (12.132) To prove the necessity of this condition for reversal, suppose that E is perfectly reversible upon input of \u03c1, with reversal operation R. From the second stage of the quantum data processing inequality it follows that S(\u03c1) \u2212 S(\u03c1, E) \u2265 S(\u03c1) \u2212 S(\u03c1, R \u25e6 E). (12.133) From the reversibility requirement it follows that \u03c1 = \u03c1. Furthermore, from the quantum Fano inequality (12.112) and the perfect reversibility requirement F (\u03c1, R \u25e6 E) = 1 it follows that S(\u03c1, R \u25e6 E) = 0. Thus the second stage of the quantum data processing inequality when applied to \u03c1 \u2192 E(\u03c1) \u2192 (R \u25e6 E)(\u03c1) may be rewritten S(\u03c1) \u2212 S(\u03c1, E) \u2265 S(\u03c1). (12.134) Combining this with the \ufb01rst part of the quantum data processing inequality, S(\u03c1) \u2265 S(\u03c1) \u2212 S(\u03c1, E), we deduce that S(\u03c1) = S(\u03c1) \u2212 S(\u03c1, E), (12.135) for any E which is perfectly reversible upon input of \u03c1. Next, we give a constructive proof that satisfaction of the condition S(\u03c1) = S(\u03c1) \u2212 S(\u03c1, E) (12.136) implies that the quantum operation E is reversible upon input of \u03c1. Noting that S(\u03c1) = S(Q) = S(R) = S(R), S(\u03c1) = S(Q) = S(R, E) and S(\u03c1, E) = S(E), we see that S(R) + S(E) = S(R, E), which we saw in Section 11.3.4 is equivalent to the condition that \u03c1RE. Suppose the initial state of Q is \u03c1 = i pi|ii|, and that \u221a pi|i|i, where the \ufb01rst system is R and we purify this state into RQ as |RQ = i the second system is Q. Note that \u03c1R = \u03c1R = i pi|ii|. Furthermore, suppose that \u03c1E j qj|jj| for some orthonormal set |j, so that = \u03c1R \u2297 \u03c1E = \u03c1RE = ij piqj|ii| \u2297 |jj|. (12.137) This has eigenvectors |i|j so by the Schmidt decomposition we may write the total state of RQE after the quantum operation E has been applied as |RQE = \u221a piqj|i|i, j|j, (12.138) ij where |i, j is some orthonormal set of states for system Q. De\ufb01ne projectors Pj by Pj \u2261 i |i, ji, j|. The idea of the restoration operation is to \ufb01rst perform a measurement described by the projectors Pj, which reveals the state |j of the environment, and then do a unitary rotation Uj conditional on j which restores the state |i, j to |i: Uj|i, j \u2261 |i. Quantum information over noisy quantum channels 567 That is, j is the measurement syndrome, and Uj the corresponding recovery operation. The complete recovery operation may be written R(\u03c3) \u2261 j UjPj\u03c3PjU \u2020 j. (12.139) The projectors Pj are orthogonal, by the orthogonality of the states |i, j, but may not be complete. If this is the case, then to ensure that the quantum operation R is tracepreserving, it is necessary to add an extra projector \u02dcP \u2261 I \u2212 j Pj to the set of projectors to make the operation trace-preserving. The \ufb01nal state of the system RQE after the reversal operation is given by UjPj", "|RQERQE|PjU \u2020 j j = = \u221a pi1 pi2 qj|i1i2| \u2297 (Uj|i1, ji2, j|U \u2020 j ) \u2297 |jj| (12.140) i1i2 \u221a pi1 j i1,i2 pi2 |i1i2| \u2297 |i1i2| \u2297 \u03c1E, (12.141) from which we see that \u03c1RQ E is perfectly reversible upon input of the state \u03c1, as we desired to show. = \u03c1RQ, and thus F (\u03c1, R \u25e6 E) = 1, that is, the operation This completes the proof of the information-theoretic reversibility conditions for tracepreserving quantum operations. Intuition about the result may be obtained by imagining that Q is a memory element in a quantum computer, R is the remainder of the quantum computer, and E is an environment whose interaction with Q causes noise. The information-theoretic reversibility condition is most elegantly understood as the statement that the state of the environment E after the noise has occurred should not be correlated with the state of the remainder of the quantum computer, R. To state it in more anthropomorphic terms, error-correction is possible precisely when the environment does not learn anything about the remainder of the quantum computer through interacting with Q! Even more concretely, suppose Q is an n qubit system and C is an [n, k] quantum error-correcting code living in the system Q with orthonormal codewords |x and projector P onto the codespace. Consider the density matrix P/2k, which may be puri\ufb01ed to a pure state of RQ: 1\u221a 2k x |x|x. (12.142) Imagine this code is able to correct arbitrary errors on some subset Q1 of the qubits. Then, in particular, it must be able to correct that error which simply swaps those qubits out into the environment and replaces them with some standard state. The informationtheoretic reversibility condition that \u03c1RE can in this case be rephrased as the condition \u03c1RQ1 = \u03c1R \u2297 \u03c1Q1. Thus the reference system R and subsystem Q1 on which errors can be corrected must initially be uncorrelated if correction is to be possible! = \u03c1R \u2297 \u03c1E Exercise 12.14: Show that the condition \u03c1RQ1 = \u03c1R \u2297 \u03c1Q1 is also suf\ufb01cient to be able to correct errors on the subsystem Q1. 568 Quantum information theory The reasoning used in the proof of the quantum data processing inequality can be adapted to prove a wide variety of other inequalities. For example, suppose we have a quantum system Q in a state \u03c1 that is subjected to the quantum operation E. The \ufb01rst stage of the data processing inequality follows by applying the subadditivity inequality for entropy to the systems RE. What if instead we applied the subadditivity inequality to the systems QE, obtaining S(\u03c1) = S(R) = S(R) = S(Q, E) \u2264 S(Q) + S(E) = S(E(\u03c1)) + S(\u03c1, E). (12.143) That is, \u0394S + S(\u03c1, E) \u2265 0, (12.144) where \u0394S \u2261 S(E(\u03c1)) \u2212 S(\u03c1) is the change in entropy caused by the process E. Loosely speaking, this inequality says that the change in entropy of the system plus the change in entropy of the environment must be non-negative, an eminently reasonable statement in accord with the second law of thermodynamics, and one which will aid us in Section 12.4.4 in our thermodynamic analysis of quantum error-correction! Exercise 12.15: Apply all possible combinations of the subadditivity and strong subadditivity inequalities to deduce other inequalities for the two stage quantum process \u03c1 \u2192 \u03c1 = E1(\u03c1) \u2192 \u03c1 = (E2 \u25e6 E1)(\u03c1), expressing the results whenever possible in terms of entropy exchanges and the entropies S(\u03c1), S(\u03c1), S(\u03c1). When it is not possible to express a quantity appearing in such an inequality in these terms, give a prescription for calculating the quantity using only a knowledge of \u03c1 and operation elements {Ej} for E1 and {Fk} for E2. 12.4.3 Quantum Singleton bound The information-theoretic approach to quantum error-correction can be used to prove a beautiful bound on the ability of quantum error-correcting codes to correct errors, the quantum Singleton bound. Recall that", " an [n, k, d] code uses n qubits to encode k qubits, and is able to correct located errors (Exercise 10.45) on up to d \u2212 1 of the qubits. The quantum Singleton bound states that we must have n \u2212 k \u2265 2(d \u2212 1). Contrast this with the classical Singleton bound, Exercise 10.21 on page 449, which states that for an [n, k, d] classical code we must have n \u2212 k \u2265 d \u2212 1. Because a quantum code to correct errors on up to t qubits must have distance at least 2t + 1 it follows that n \u2212 k \u2265 4t. Thus, for example, a code to encode k = 1 qubits and capable of correcting errors on t = 1 of the qubits must satisfy n \u2212 1 \u2265 4, that is, n must be at least 5, so the \ufb01ve qubit code described in Chapter 10 is the smallest possible code for this task. The proof of the quantum Singleton bound is an extension of the information-theoretic techniques we have been using to analyze quantum error-correction. Suppose the code is a 2k-dimensional subspace associated with the system Q, with orthonormal basis denoted by |x. Introduce a 2k-dimensional reference system R also with 2k orthonormal basis vectors denoted |x, and consider the entangled state of RQ, 1\u221a 2k We divide the n qubits of Q up into three disjoint blocks, the \ufb01rst and second, Q1 and Q2, consisting of d \u2212 1 qubits each, and the third Q3 consisting of the remaining n \u2212 2(d \u2212 1) |RQ = (12.145) |x|x. x Quantum information over noisy quantum channels 569 qubits. Because the code has distance d any set of d \u2212 1 located errors may be corrected, and thus it is possible to correct errors on either Q1 or Q2. It follows that R and Q1 must be uncorrelated, as are R and Q2. By this observation, the overall purity of the state of RQ1Q2Q3, and the subadditivity of entropy we have: S(R) + S(Q1) = S(R, Q1) = S(Q2, Q3) \u2264 S(Q2) + S(Q3) S(R) + S(Q2) = S(R, Q2) = S(Q1, Q3) \u2264 S(Q1) + S(Q3). (12.146) (12.147) Adding these two inequalities gives 2S(R) + S(Q1) + S(Q2) \u2264 S(Q1) + S(Q2) + 2S(Q3). Canceling terms and substituting S(R) = k gives k \u2264 S(Q3). But Q3 is n \u2212 2(d \u2212 1) qubits in size, so S(Q3) \u2264 n\u22122(d\u22121), giving k \u2264 n\u22122(d\u22121), whence 2(d\u22121) \u2264 n\u2212k, the quantum Singleton bound. (12.148) As an example of the quantum Singleton bound in action, consider the depolarizing channel E(\u03c1) = p\u03c1 + (1 \u2212 p)/3(X\u03c1X + Y \u03c1Y + Z\u03c1Z). Suppose the depolarizing channel acts independently on a large number n of qubits. If p < 3/4 then more than one quarter of those qubits will suffer errors, so any code capable of recovering from the errors must have t > n/4. But the quantum Singleton bound implies that n \u2212 k \u2265 4t > n, and thus k must be negative, that is, it is not possible to encode any qubits at all in this case. Thus, when p < 3/4, the quantum Singleton bound implies that the capacity of the depolarizing channel for quantum information is zero! 12.4.4 Quantum error-correction, refrigeration and Maxwell\u2019s demon Quantum error-correction may be thought of as a type of refrigeration process, capable of keeping a quantum system at a constant entropy, despite the in\ufb02uence of noise processes which tend to change the entropy of the system. Indeed, quantum error-correction may even appear rather puzzling from this point of view, as it appears to allow a reduction in entropy of the quantum system in apparent violation of the second law of thermodynamics! To understand why there is no violation of the second law we do an analysis of quantum error-correction similar to that used to analyze Maxwell\u2019s demon in Box 3.5 on page 162. Quantum error-correction is essentially a special type of Maxwell\u2019s demon \u2013", " we can imagine a \u2018demon\u2019 performing syndrome measurements on the quantum system, and then correcting errors according to the result of the syndrome measurement. Just as in the analysis of the classical Maxwell\u2019s demon, the storage of the syndrome in the demon\u2019s memory carries with it a thermodynamic cost, in accordance with Landauer\u2019s principle. In particular, because any memory is \ufb01nite, the demon must eventually begin erasing information from its memory, in order to have space for new measurement results. Landauer\u2019s principle states that erasing one bit of information from the memory increases the total entropy of the system \u2013 quantum system, demon, and environment \u2013 by at least one bit. Rather more precisely, we can consider a four-stage error-correction \u2018cycle\u2019 as depicted in Figure 12.9: (1) The system, starting in a state \u03c1, is subjected to a noisy quantum evolution that takes it to a state \u03c1. In typical scenarios for error-correction, we are interested in cases where the entropy of the system increases, S(\u03c1) > S(\u03c1), although this is not necessary. 570 Quantum information theory (2) A demon performs a (syndrome) measurement on the state \u03c1 described by measurement operators {Mm}, obtaining result m with probability m = Mm\u03c1M \u2020 pm = tr(Mm\u03c1M \u2020 m), resulting in the posterior state \u03c1 m/pm. (3) The demon applies a unitary operation Vm (the recovery operation) that creates a \ufb01nal system state m = Vm\u03c1 \u03c1 mV \u2020 m = VmMm\u03c1M \u2020 mV \u2020 m pm. (12.149) (4) The cycle is restarted. In order that this actually be a cycle and that it be a successful error-correction, we must have \u03c1 m. m = \u03c1 for each measurement outcome Figure 12.9. The quantum error-correction cycle. We now show that any reduction in entropy during the second and third stages \u2013 the error-correction stages \u2013 comes at the expense of entropy production in the environment which is at least as large as the entropy reduction in the quantum system being errorcorrected. After the third stage the only record of the measurement result m is the record kept in the demon\u2019s memory. To reset its memory for the next cycle, the demon must erase its record of the measurement result, causing a net increase in the entropy of the environment by Landauer\u2019s principle. The number of bits that must be erased is determined by the representation the demon uses to store the measurement result m; by Shannon\u2019s noiseless channel coding theorem, at least H(pm) bits are required, on average, to store the measurement result, and thus a single error-correction cycle on average involves the dissipation of H(pm) bits of entropy into the environment when the measurement record is erased. Before error-correction the state of the quantum system is \u03c1. After error-correction the state of the quantum system is \u03c1, so the net change in entropy of the system due to the error-correction is \u0394S \u2261 S(\u03c1) \u2212 S(\u03c1). There is an additional entropic cost of H(pm) (on average) associated with erasing the measurement record, for a total cost of \u0394(S)+H(pm). Our goal is to bound this thermodynamic cost, and in so doing demonstrate that the second law of thermodynamics is never violated. To do so, it helps to introduce two items of notation: let E represent the noise process occurring during stage 1 of the error-correction cycle, \u03c1 \u2192 \u03c1 = E(\u03c1), and let R be the quantum operation representing Entanglement as a physical resource 571 the error-correction operation, R(\u03c3) \u2261 m VmMm\u03c3M \u2020 mV \u2020 m. (12.150) nV \u2020 With input \u03c1 the w-matrix for this process has elements Wmn = tr(VmMm\u03c1M \u2020 n ), and thus has diagonal elements Wmm = tr(VmMm\u03c1M \u2020 m), which is just the probability pm the demon obtains measurement outcome m when measuring the error syndrome. By Theorem 11.9 on page 515 the entropy of the diagonal elements of W is at least as great as the entropy of W, so m) = tr(Mm\u03c1M \u2020 mV \u2020 H(pm) \u2265 S(W ) = S(\u03c1, R), with equality if and only if the operators VmMm are a canonical decomposition of R with respect to \u03c1 so that the off-diagonal terms in W vanish. By Equation (12.144) on page 568 it follows that (", "12.151) \u0394S + S(\u03c1, R) = S(\u03c1) \u2212 S(\u03c1) + S(\u03c1, R) \u2265 0. Combining this result with (12.151) we deduce that \u0394S + H(pm) \u2265 0. But \u0394S + H(pm) was the total entropy change caused by the error-correction procedure. We conclude that error-correction can only ever result in a net increase in total entropy, with any decrease in system entropy due to error-correction being paid for with entropy production when the error syndrome produced during error-correction is erased. (12.152) Exercise 12.16: Show that in the case where R perfectly corrects E for the input \u03c1, the inequality S(\u03c1) \u2212 S(\u03c1) + S(\u03c1, R) \u2265 0 (12.153) must actually be satis\ufb01ed with equality. 12.5 Entanglement as a physical resource Thus far our study of quantum information has been focused on resources that are not too far distant from the resources considered in classical information theory. For your convenience Figure 12.10 summarizes many of these results in both their quantum and classical guises. One of the delights of quantum computation and quantum information is that quantum mechanics also contains essentially new types of resource that differ vastly from the sort of resource traditionally regarded as information in classical information theory. Perhaps the best understood of these is quantum entanglement, and it is to this resource that we now turn. We say \u2018best understood\u2019, but that is not saying a whole lot! We are a long way from having a general theory of quantum entanglement. Nevertheless, some encouraging progress towards such a general theory has been made, revealing an intriguing structure to the entangled states, and some quite remarkable connections between the properties of noisy quantum channels and various types of entanglement transformation. We are just going to take a quick peek at what is known, focusing on the transformation properties of entanglement distributed between two systems (\u2018bi-partite\u2019 entanglement), Alice and Bob. There is, of course, a great deal of interest in developing a general theory of entanglement for multi-partite systems, but how to do this is not well understood. 572 Quantum information theory Information Theory Classical Shannon entropy H(X) = \u2212 p(x) log p(x) Quantum von Neumann entropy S(\u03c1) = \u2212tr(\u03c1 log \u03c1) x Distinguishability and accessible information Letters always distinguishable N = |X| Holevo bound H(X : Y ) \u2264 S(\u03c1) \u2212 Noiseless channel coding pxS(\u03c1x) px\u03c1x x x \u03c1 = Shannon\u2019s theorem nbits = H(X) Schumacher\u2019s theorem nqubits = S px\u03c1x Capacity of noisy channels for classical information x Shannon\u2019s noisy coding theorem C(N ) = max p(x) H(X : Y ) Holevo\u2013Schumacher\u2013Westmoreland theorem 0 S(\u03c1)\u2212 C (1)(E) = max {px,\u03c1x} pxS(\u03c1 x) 1 \u03c1 x = E(\u03c1x), \u03c1 = x x px\u03c1 x Information-theoretic relations Fano inequality H(pe) + pe log(|X| \u2212 1) \u2265 H(X|Y ) Quantum Fano inequality H(F (\u03c1, E)) + (1 \u2212 F (\u03c1, E)) log(d2 \u2212 1) \u2265 S(\u03c1, E) Mutual information Coherent information H(X : Y ) = H(Y ) \u2212 H(Y |X) I(\u03c1, E) = S(E(\u03c1)) \u2212 S(\u03c1, E) Data processing inequality X \u2192 Y \u2192 Z H(X) \u2265 H(X : Y ) \u2265 H(X : Z) Quantum data processing inequality \u03c1 \u2192 E1(\u03c1) \u2192 (E2 \u25e6 E1)(\u03c1) S(\u03c1) \u2265 I(\u03c1, E1) \u2265 I(\u03c1, E2 \u25e6 E1) Figure 12.10. Summary of some important classical information relations, and quantum analogues of those relations. Entanglement as a physical resource 573 12.5.1 Transforming bi-partite pure state entanglement The starting point for our investigation is the following simple question: given that Alice and Bob share an entangled pure state |\u03c8, into what other types of entanglement |\u03d5 may they transform |\u03c8, given that they can each perform arbitrary operations on their local systems, including measurement, but can only communicate using classical communication? No quantum communication between Alice and Bob is allowed, constraining the class of transformations they may achieve. As an example, imagine that Alice and Bob share an entangled pair of qubits in the 2.", " Alice performs a two outcome measurement described by \u221a Bell state (|00 + |11)/ measurement operators M1 and M2: M1 = cos \u03b8 0 0 sin \u03b8 ; M2 = sin \u03b8 0 0 cos \u03b8. (12.154) After the measurement the state is either cos \u03b8|00 + sin \u03b8|11 or cos \u03b8|11 + sin \u03b8|00, depending on the measurement outcome, 1 or 2. In the latter case, Alice applies a gate after the measurement, resulting in the state cos \u03b8|01 + sin \u03b8|10. She then sends the measurement result (1 or 2) to Bob, who does nothing to the state if the measurement gate if the result is 2. The \ufb01nal state of the joint system is result is 1, and performs a therefore cos \u03b8|11 + sin \u03b8|00, regardless of the measurement outcome obtained by Alice. \u221a That is, Alice and Bob have transformed their initial entangled resource (|00 + |11)/ 2 into the state cos \u03b8|00 + sin \u03b8|11 using only local operations on their individual systems, and classical communication. It is perhaps not immediately obvious what is the signi\ufb01cance of the problem of entanglement transformation. There is a certain intrinsic interest to the class of transformations we are allowing \u2013 local operations and classical communication (LOCC) \u2013 however, it is by no means clear a priori that this is truly an interesting problem. It turns out, however, that generalizations of this entanglement transformation problem exhibit deep and unexpected connections to quantum error-correction. Furthermore, the techniques introduced in the solution to the problem are of quite considerable interest, and give unexpected insight into the properties of entanglement. In particular, we will discover a close connection between entanglement and the theory of majorization, an area of mathematics that actually predates quantum mechanics! Before jumping into the study of entanglement transformation, let\u2019s \ufb01rst acquaint ourselves with a few relevant facts about majorization. Majorization is an ordering on d-dimensional real vectors intended to capture the notion that one vector is more or less disordered than another. More precisely, suppose x = (x1,..., xd) and y = (y1,..., yd) are two d-dimensional vectors. We use the notation x\u2193 to mean x re-ordered so the components are in decreasing order so, for example, x\u2193 1 is the largest component of x. y\u2193 x\u2193 j for k = 1,..., d, with j \u2264 We say x is majorized by y, written x \u227a y, if equality instead of inequality when k = d. What this de\ufb01nition has to do with notions of disorder will become clear shortly! k k j=1 j=1 The connection between majorization and entanglement transformation is easily stated yet rather surprising. Suppose |\u03c8 and |\u03d5 are states of the joint Alice\u2013Bob system. De\ufb01ne \u03c1\u03c8 \u2261 trB(|\u03c8\u03c8|), \u03c1\u03d5 \u2261 trB(|\u03d5\u03d5|) to be the corresponding reduced density matrices of Alice\u2019s system, and let \u03bb\u03c8 and \u03bb\u03d5 be the vectors whose entries are the eigenvalues of \u03c1\u03c8 and \u03c1\u03d5. We will show that |\u03c8 may be transformed to |\u03d5 by LOCC if and only if \u03bb\u03c8 \u227a \u03bb\u03d5! To demonstrate this we need \ufb01rst a few simple facts about majorization. 574 Quantum information theory Exercise 12.17: Show that x \u227a y if and only if for all real t, d d d j=1 max(xj \u2212 t, 0) \u2264 j=1 max(yj \u2212 t, 0), and Exercise 12.18: Use the previous exercise to show that the set of x such that x \u227a y is xj = yj. j=1 j=1 d convex. The following proposition gives a more intuitive meaning to the notion of majorization, showing that x \u227a y if and only if x can be written as a convex combination of permutations of y. Intuitively, therefore, x \u227a y if x is more disordered than y in the sense that x can be obtained by permuting the elements of y and mixing the resulting vectors. This representation theorem is one of the most useful results in the study of majorization. Proposition 12.11: x \u227a y if and only if x = distribution pj and permutation matrices Pj. j pjPjy for some probability Proof Suppose x \u227a y. Without loss of generality we may suppose x = x\u2193 and y", " = y\u2193. j pjPjy by induction on the dimension d. For d = 1 the We will prove that x = result is clear. Suppose x and y are d + 1-dimensional vectors such that x \u227a y. Then x1 \u2264 y1. Choose j such that yj \u2264 x1 \u2264 yj\u22121, and de\ufb01ne t in the range [0, 1] such that x1 = ty1 + (1 \u2212 t)yj. De\ufb01ne a convex combination of permutations D \u2261 tI + (1 \u2212 t)T, where T is the permutation matrix which transposes the 1st and jth matrix elements. Then Dy = (x1, y2,..., yj\u22121, (1 \u2212 t)y1 + tyj, yj+1,..., yd+1). (12.155) De\ufb01ne x \u2261 (x2,..., xd+1) and y \u2261 (y2,..., yj\u22121, (1 \u2212 t)y1 + tyj, yj+1,..., yd+1). In Exercise 12.19 on this page you show x \u227a y, so by the inductive hypothesis x = jP j and permutation matrices P Dy, where the P j are extended to d + 1 dimensions by acting trivially on the \ufb01rst entry. Since D = (tI + (1 \u2212 t)T ), and a product of permutation matrices is a permutation matrix, the result follows. jy for probabilities p j, whence x = \u2019 jP j j p j p ( Exercise 12.19: Verify that x \u227a y. Conversely, suppose x = follows that x = j pjPjy \u227a y. j pjPjy. It is clear that Pjy \u227a y and by Exercise 12.18 it Matrices which are convex combinations of permutation matrices have many interesting properties. Notice, for example, that the entries of such a matrix must be non-negative, and that each of the rows and columns must sum to one. A matrix with these properties is known as a doubly stochastic matrix, and there is a result known as Birkhoff\u2019s theorem which implies that the doubly stochastic matrices correspond exactly to the set of matrices which can be written as convex combinations of permutation matrices. We won\u2019t prove Birkhoff\u2019s theorem here (see the end of chapter \u2018History and further reading\u2019), but merely state it: Theorem 12.12: (Birkhoff\u2019s theorem) A d by d matrix D is doubly stochastic (that is, has non-negative entries and each row and column sums to 1) if and only if D j pjPj. can be written as a convex combination of permutation matrices, D = j j Entanglement as a physical resource 575 From Birkhoff\u2019s theorem and Proposition 12.11 it follows that x \u227a y if and only if x = Dy for some doubly stochastic D. This result allows us to prove a striking and useful operator generalization of Proposition 12.11. Suppose H and K are two Hermitian operators. Then we say H \u227a K if \u03bb(H) \u227a \u03bb(K), where we use \u03bb(H) to denote the vector of eigenvalues of a Hermitian operator H. Then we have: Theorem 12.13: Let H and K be Hermitian operators. Then H \u227a K if and only if there is a probability distribution pj and unitary matrices Uj such that H = pjUjKU \u2020 j. (12.156) Proof Suppose H \u227a K. Then \u03bb(H) = j pjPj\u03bb(K) by Proposition 12.11. Let \u039b(H) denote the diagonal matrix whose entries are the eigenvalues of H. Then the vector equation \u03bb(H) = j pjPj\u03bb(K) may be re-expressed as \u039b(H) = pjPj\u039b(K)P \u2020 j. (12.157) But H = V \u039b(H)V \u2020 and \u039b(K) = W KW \u2020 for some unitary matrices V and W, giving j pjUjKU \u2020 j where Uj \u2261 V PjW is a unitary matrix, completing the proof in the H = forward direction. j pjUjKU \u2020 j. Similarly to before, this is equivalent to j for some unitary matrices Vj. Writing the matrix", " components Conversely, suppose H = \u039b(H) = of Vj as Vj,kl, we have: j pjVj\u039b(K)V \u2020 \u03bb(H)k = jl pjVj,kl\u03bb(K)lV \u2020 j,lk = jl pj|Vj,kl|2\u03bb(K)l. (12.158) De\ufb01ne a matrix D with entries Dkl \u2261 j pj|Vj,kl|2 so we have \u03bb(H) = D\u03bb(K). The entries of D are non-negative by de\ufb01nition, and the rows and columns of D all sum to one because the rows and columns of the unitary matrices Vj are unit vectors, so D is doubly stochastic and thus \u03bb(H) \u227a \u03bb(K). We now have in hand all the facts about majorization needed in the study of LOCC transformations of bipartite pure state entanglement. The \ufb01rst step of the argument is to reduce the problem from the study of general protocols, which may involve two way classical communication, to protocols involving only one-way classical communication. Proposition 12.14: Suppose |\u03c8 can be transformed to |\u03d5 by LOCC. Then this transformation can be achieved by a protocol involving just the following steps: Alice performs a single measurement described by measurement operators Mj, sends the result j to Bob, who performs a unitary operation Uj on his system. Proof Without loss of generality we may suppose the protocol consists of Alice performing a measurement, sending the result to Bob, who performs a measurement (whose nature may depend on the information received from Alice), and sends the result back to Alice, who performs a measurement... and so on. The idea of the proof is simply to show that the 576 Quantum information theory effect of any measurement Bob can do may be simulated by Alice (with one small caveat) so all Bob\u2019s actions can actually be replaced by actions by Alice! To see that this is the case, imagine Bob performs a measurement with measurement operators Mj on a pure \u03bbl|lA|lB and state |\u03c8. Suppose this pure state has Schmidt decomposition |\u03c8 = de\ufb01ne operators Nj on Alice\u2019s system to have a matrix representation with respect to Alice\u2019s Schmidt basis which is the same as the matrix representation of Bob\u2019s operators kl Mj,kl|kBlB| then we de\ufb01ne Mj with respect to his Schmidt basis. That is, if Mj = \u221a l Nj \u2261 Mj,kl|kAlA|. (12.159) kl \u221a \u221a kl Mj,kl kl Mj,kl \u03bbl|kA|lB, also with probability Suppose Bob performs the measurement de\ufb01ned by the measurement operators Mj. Then the post-measurement state is |\u03c8j \u221d Mj|\u03c8 = \u03bbl|lA|kB, with probability kl \u03bbl|Mj,kl|2. On the other hand, if Alice had measured Nj then the post-measurement state is |\u03d5j \u221d Nj|\u03c8 = kl \u03bbl|Mj,kl|2. Note furthermore that |\u03c8j and |\u03d5j are the same states up to interchange of Alice and Bob\u2019s systems via the map |kA \u2194 |kB, and therefore must have the same Schmidt components. It follows from Exercise 2.80 on page 111 that there exists a unitary Uj on Alice\u2019s system and Vj on Bob\u2019s system such that |\u03c8j = (Uj \u2297 Vj)|\u03d5j. Therefore, Bob performing a measurement described by measurement operators Mj is equivalent to Alice performing the measurement described by measurement operators UjNj followed by Bob performing the unitary transformation Vj. Summarizing, a measurement by Bob on a known pure state can be simulated by a measurement by Alice, up to a unitary transformation by Bob. Imagine then that Alice and Bob engage in a multi-round protocol transforming |\u03c8 to |\u03d5. Without loss of generality, we may suppose that the \ufb01rst round of the protocol consists of Alice performing a measurement and sending the result to Bob. The second round consists of Bob performing a measurement (perhaps with the type of measurement determined by the result of the \ufb01rst) and sending the result to Alice. Instead, however, we can suppose that this measurement is simulated by a measurement performed by Alice, up to a unitary transformation by Bob. Indeed, we can replace all the measurements by Bob and communication from Bob to Alice by measurements by Alice", ", with a unitary to be done by Bob conditional on Alice\u2019s measurement result. Finally, all the measurements performed by Alice can be combined into one single measurement (Exercise 2.57 on page 86), whose result determines a unitary transformation to be performed by Bob; the net effect of this protocol is exactly the same as the original protocol with two-way communication. Theorem 12.15: A bipartite pure state |\u03c8 may be transformed to another pure state |\u03d5 by LOCC if and only if \u03bb\u03c8 \u227a \u03bb\u03d5. Proof Suppose |\u03c8 may be transformed to |\u03d5 by LOCC. By Proposition 12.14 we may assume that the transformation is effected by Alice performing a measurement with measurement operators Mj, then sending the result to Bob, who performs a unitary transformation Uj. From Alice\u2019s point of view she starts with the state \u03c1\u03c8 and ends with the state \u03c1\u03d5 regardless of the measurement outcome, so we must have Mj\u03c1\u03c8M \u2020 j = pj\u03c1\u03d5, (12.160) Entanglement as a physical resource 577 \u221a \u03c1\u03c8 implies that there where pj is the probability of outcome j. Polar decomposing Mj exists a unitary Vj such that \u221a \u221a Mj \u03c1\u03c8 = Mj\u03c1\u03c8M \u2020 j Vj = pj\u03c1\u03d5Vj. Premultiplying this equation by its adjoint gives Summing on j and using the completeness relation j Mj = I gives \u221a \u221a \u03c1\u03c8M \u2020 j Mj \u03c1\u03c8 = pjV \u2020 j \u03c1\u03d5Vj. j M \u2020 \u03c1\u03c8 = j whence \u03bb\u03c8 \u227a \u03bb\u03d5 by Theorem 12.13. pjV \u2020 j \u03c1\u03d5Vj, (12.161) (12.162) (12.163) The proof of the converse is essentially to run the proof in the forwards direction backwards. Suppose \u03bb\u03c8 \u227a \u03bb\u03d5, so \u03c1\u03c8 \u227a \u03c1\u03d5 and by Theorem 12.13 there exist probabilities pj and unitary operators Uj such that \u03c1\u03c8 = j. Assume for now that \u03c1\u03c8 is invertible (this assumption is easily removed; see Exercise 12.20) and de\ufb01ne operators Mj for Alice\u2019s system by j pjUj\u03c1\u03d5U \u2020 \u221a Mj \u03c1\u03c8 \u2261 \u221a pj\u03c1\u03d5U \u2020 j. (12.164) To see that these operators de\ufb01ne a measurement we need to check the completeness relation. We have Mj = j \u03c1\u22121/2 \u221a \u03c8 pj\u03c1\u03d5U \u2020 \u239b and thus \u239e j M \u2020 j Mj = \u03c1\u22121/2 \u03c8 \u239d j pjUj\u03c1\u03d5U \u2020 j \u23a0 \u03c1\u22121/2 \u03c8 = \u03c1\u22121/2 \u03c8 \u03c1\u03c8\u03c1\u22121/2 \u03c8 = I, (12.165) which is the completeness relation. Suppose Alice performs the measurement described by the operators Mj, obtaining the outcome j and the corresponding state |\u03c8j \u221d Mj|\u03c8. Let \u03c1j denote Alice\u2019s reduced density matrix corresponding to the state |\u03c8j, so \u03c1j \u221d Mj\u03c1\u03c8M \u2020 j = pj\u03c1\u03d5, (12.166) by substitution of (12.164), and thus \u03c1j = \u03c1\u03d5. It follows by Exercise 2.81 that Bob may convert |\u03c8j into |\u03d5 by application of a suitable unitary transformation Vj. Exercise 12.20: Show that the assumption that \u03c1\u03c8 is invertible may be removed from the proof of the converse part of Theorem 12.15. Exercise 12.21: (Entanglement catalysis) Suppose Alice and Bob share a pair of \u221a \u221a \u221a \u221a \u221a \u221a 0.4|00 + 0.4|11 + four level systems in the state |\u03c8 = 0.1|33. Show that it is not possible for them to convert this state by LOCC to the state \u221a 0.25|22. Imagine, however, that a friendly bank |\u03d5 = is willing to offer them the loan of a catalyst, an entangled pair of qubits in the state |c = convert the state |\u03c8|c to |\u03d5|c by local operations and classical communication, returning the catalyst |c to the bank after the transformation is complete. 0.4|11", ". Show that it is possible for Alice and Bob to 0.25|11 + \u221a 0.5|00 + \u221a 0.6|00 + 0.1|22 + Exercise 12.22: (Entanglement conversion without communication) Suppose Alice and Bob are trying to convert a pure state |\u03c8 into a pure state |\u03d5 using local operations only \u2013 no classical communication. Show that this is possible if 578 Quantum information theory and only if \u03bb\u03c8 summing to 1, and \u2018 identical non-zero entries. \u223c = \u03bb\u03d5 \u2297 x, where x is some real vector with non-negative entries \u223c =\u2019 means that the vectors on the left and the right have 12.5.2 Entanglement distillation and dilution Suppose that instead of being supplied with a single copy of a state |\u03c8 Alice and Bob are supplied with a large number of copies. What types of entanglement transformation can they accomplish with all these copies? We are going to focus on two particular types of entanglement transformation, known as entanglement distillation and entanglement dilution. The idea of entanglement distillation is for Alice and Bob to convert some large number of copies of a known pure state |\u03c8 into as many copies of the Bell state (|00+|11)/ 2 as possible using local operations and classical communication, requiring not that they succeed exactly, but only with high \ufb01delity. Entanglement dilution is the reverse process of using LOCC to convert a large number of copies of the Bell state 2 into copies of |\u03c8, again with high \ufb01delity in the limit where a large (|00 + |11)/ number of copies of the Bell state are initially available. \u221a \u221a \u221a What motivates the study of entanglement distillation and dilution? Suppose we take seriously the idea that entanglement is a physical resource and that as such it should be possible to quantify entanglement, much as we quantify other physical resources such as energy or entropy. Suppose we decide to pick the Bell state (|00 + |11)/ 2 as our standard unit of entanglement \u2013 the basic measure, rather like the standard kilogram or the standard meter. We can associate a measure of entanglement to a quantum state |\u03c8 in a similar way to the way we associate a mass to an object. Suppose, for instance, that it takes 15 chocolate biscuits of a particular brand to attain a mass equivalent to the standard kilogram; we say that the chocolate biscuits have a mass of 1/15th of a kilogram. Now, strictly speaking, if the chocolate biscuits had a mass of 1/14.8 kilograms we\u2019d be in a bit of trouble, because no integer number of chocolate biscuits is going to balance the standard kilogram, and it\u2019s not so obvious how to de\ufb01ne a non-integer number of chocolate biscuits. Fortunately, what we do is notice that 148 chocolate biscuits exactly balances 10 standard kilograms, so the mass of the chocolate biscuits is 10/148 kilograms. But what if the real mass is not 1/14.8 kilograms, but is something even more esoteric, like 1/14.7982... kilograms? Well, we simply go to the limit of balancing a large number m of chocolate biscuits with another large number n of standard kilograms, and declare the mass of a chocolate biscuit to be the limiting ratio n/m as both m and n become very large. \u221a In a similar way, a potential approach to de\ufb01ning the amount of entanglement present in a pure state |\u03c8 is to imagine that we are given a large number n of Bell states 2, and are asked to produce as many (high-\ufb01delity) copies of |\u03c8 as (|00 + |11)/ possible using local operations and classical communication. If the number of copies of |\u03c8 that can be produced is m then we de\ufb01ne the limiting ratio of n/m to be the entanglement of formation of the state |\u03c8. Alternately, we might imagine performing the process in reverse, going from m copies of |\u03c8 to n copies of (|00 + |11)/ 2 using LOCC, and de\ufb01ning the limiting ratio n/m to be the distillable entanglement of the state |\u03c8. It is by no means obvious that these two de\ufb01nitions give the same number for the quantity of entanglement; we will see that for pure states |\u03c8 the entanglement of formation and the distillable entanglement are in fact exactly the same! \u221a Let\u2019s take a look at a simple protocol for entanglement dilution, and another for Entanglement as a physical resource 579 entanglement distillation. Suppose an", " entangled state |\u03c8 has Schmidt decomposition |\u03c8 = p(x)|xA|xB. (12.167) x We write the squared Schmidt co-ef\ufb01cients p(x) in the form we usually reserve for probability distributions, both because the co-ef\ufb01cients satisfy the usual rules of probability distributions (non-negative and sum to one), and because ideas from probability theory turn out to be useful in understanding entanglement distillation and dilution. The m-fold tensor product |\u03c8\u2297m may be written p(x1)p(x2)... p(xm)|x1Ax2A... xmA|x1Bx2B... xmB. (12.168) |\u03c8\u2297m = x1,x2,...,xm Suppose that we de\ufb01ned a new quantum state |\u03d5m by omitting all those terms x1,..., xm which are not -typical, in the sense de\ufb01ned in Section 12.2.1: |\u03d5m \u2261 p(x1)p(x2)... p(xm)|x1Ax2A... xmA|x1Bx2B... xmB. (12.169) x \u2212typical m \u2261 |\u03d5m/ The state |\u03d5m is not quite a properly normalized quantum state; to normalize it we de\ufb01ne |\u03d5 \u03d5m|\u03d5m. By part 1 of the theorem of typical sequences the \ufb01delity F (|\u03c8\u2297m, |\u03d5 m) \u2192 1 as m \u2192 \u221e. Furthermore, by part 2 of the theorem of typical sequences the number of terms in the sum (12.169) is at most 2m(H(p(x))+) = 2m(S(\u03c1\u03c8 )+), where \u03c1\u03c8 is the result of tracing out Bob\u2019s part of |\u03c8. Suppose then that Alice and Bob are in joint possession of n = m(S(\u03c1\u03c8) + ) Bell states. Alice prepares \u2018both parts\u2019 of |\u03d5 m locally, and then uses the Bell states shared with Bob to teleport what should be Bob\u2019s half of the state |\u03d5 m over to Bob. In this way, Alice and Bob can dilute their n Bell states to obtain |\u03d5 m, a pretty good approximation to |\u03c8\u2297m. This entanglement dilution procedure has n = m(S(\u03c1\u03c8) + ), so the ratio n/m tends to S(\u03c1\u03c8) +. We may choose as small as we like, so we conclude that the entanglement of formation for the state |\u03c8 is no larger than S(\u03c1\u03c8), since we have just shown that (asymptotically) S(\u03c1\u03c8) Bell states may be converted into a single copy of |\u03c8. An entanglement distillation protocol for converting copies of |\u03c8 into Bell states also follows along similar lines. Suppose that Alice and Bob are in possession of m copies of |\u03c8. By performing a measurement onto the -typical subspace of \u03c1\u03c8 Alice may, with high \ufb01delity, convert the state |\u03c8\u2297m into the state |\u03d5 m. The largest Schmidt co-ef\ufb01cient appearing in |\u03d5m is at most 2\u2212m(S(\u03c1\u03c8)\u2212) by the de\ufb01nition of typical sequences. The renormalized state |\u03d5 (1 \u2212 \u03b4) larger, since m has Schmidt co-ef\ufb01cients at most a factor 1/ the theorem of typical sequences tells us that 1 \u2212 \u03b4 is a lower bound on the probability that a sequence is -typical, and may be made arbitrarily close to 1 for suf\ufb01ciently large m is at most 2\u2212m(S(\u03c1\u03c8 )\u2212)/(1 \u2212 \u03b4). Suppose m. Thus, the largest eigenvalue of the state \u03c1\u03d5 we choose any n such that \u221a 2\u2212m(S(\u03c1\u03c8 )\u2212) 1 \u2212 \u03b4 m is majorized by the vector (2\u2212n, 2\u2212n,..., 2\u2212n), Then the vector of eigenvalues for \u03c1\u03d5 and thus by Theorem 12.15 the state |\u03d5 m may be converted to n Bell states by local operations and classical communication. Examining (12.170) we see that this is possible provided n \u2248 mS(\u03c1\u03c8), and thus the entanglement of distillation is at least S", "(\u03c1\u03c8). \u2264 2\u2212n. (12.170) We have exhibited strategies for distilling |\u03c8 into S(\u03c1\u03c8) Bell states and for diluting 580 Quantum information theory S(\u03c1\u03c8) Bell states into a copy of |\u03c8. In fact, it is not very dif\ufb01cult to see that the procedures we have described are really the optimal methods for doing entanglement dilution and distillation! Suppose, for example, that a more ef\ufb01cient protocol for entanglement dilution existed, that could dilute |\u03c8 into S > S(\u03c1\u03c8) Bell states. Then starting with S(\u03c1\u03c8) Bell states Alice and Bob could produce a copy of |\u03c8 using the protocol already described, and then use the hypothetical protocol to produce S Bell states. Thus, by local operations and classical communication, Alice and Bob have taken S(\u03c1\u03c8) Bell states and turned them into S > S(\u03c1\u03c8) Bell states! It is not very dif\ufb01cult to convince yourself (and see Exercise 12.24) that increasing the number of Bell states present using local operations and classical communication is not possible, so such a hypothetical dilution protocol can not exist. In a similar way we can see that the procedure for entanglement distillation is optimal. Thus, the entanglement of formation and entanglement of distillation for the state |\u03c8 are the same, and both equal S(\u03c1\u03c8)! Exercise 12.23: Prove that the procedure for entanglement distillation we have described is optimal. Exercise 12.24: Recall that the Schmidt number of a bi-partite pure state is the number of non-zero Schmidt components. Prove that the Schmidt number of a pure quantum state cannot be increased by local operations and classical communication. Use this result to argue that the number of Bell states shared between Alice and Bob cannot be increased by local operations and classical communication. We\u2019ve learned how to transform Bell states of a bipartite quantum system into copies of another entangled state, |\u03c8, and back again, in an optimal fashion, motivating us to de\ufb01ne the amount of entanglement present in that quantum state to be the number of Bell states into which copies of |\u03c8 may be inter-converted, that is, S(\u03c1\u03c8). What do we learn from this de\ufb01nition? Below we will see that it is possible to obtain some interesting insights into quantum error-correction by further generalizing the notion of distillable entanglement. Nevertheless, at the time of writing it seems fair to say that the study of entanglement is in its infancy, and it is not yet entirely clear what advances in our understanding of quantum computation and quantum information can be expected as a result of the study of quantitative measures of entanglement. We have a reasonable understanding of the properties of pure states of bi-partite quantum systems, but a very poor understanding of systems containing three or more components, or even of mixed states for bi-partite systems. Developing a better understanding of entanglement and connecting that understanding to topics such as quantum algorithms, quantum error-correction and quantum communication is a major outstanding task of quantum computation and quantum information! 12.5.3 Entanglement distillation and quantum error-correction We de\ufb01ned entanglement distillation for pure states, but there is no reason that de\ufb01nition cannot be extended to mixed states. More precisely, suppose \u03c1 is a general state of a bi-partite quantum system belonging to Alice and Bob. They are supplied with a large number m of copies of these states, and using local operations and classical communication attempt to convert these states to the largest possible number n of Bell states, with high \ufb01delity. The distillable entanglement D(\u03c1) of \u03c1 is the limiting value of the ratio n/m Entanglement as a physical resource 581 for the best possible distillation protocol; for pure states |\u03c8 we have already shown that D(|\u03c8) = S(\u03c1\u03c8), but we do not yet know how to evaluate D(\u03c1) for a mixed state. A considerable number of techniques for doing entanglement distillation have been developed, giving lower bounds on the value for D(\u03c1) for speci\ufb01c classes of states \u03c1. We won\u2019t review these techniques here (see the end of chapter \u2018History and further reading\u2019). What we will describe is a fascinating connection between distillable entanglement and quantum error-correction. Imagine that Alice is attempting to send Bob quantum information through a noisy quantum channel E. We suppose the channel is a qubit channel such as the depolarizing channel, although the same basic ideas adapt easily to non-qubit channels", ". One method for sending quantum information through the channel is as follows. Alice prepares a large number m of Bell states, and sends half of each Bell pair through the channel. Suppose the result of applying E to half a Bell pair is to create the state \u03c1, so Alice and Bob end up sharing m copies of \u03c1. Alice and Bob now perform entanglement distillation, producing mD(\u03c1) Bell pairs. Alice can now prepare an mD(\u03c1) qubit state and teleport it to Bob using the mD(\u03c1) Bell pairs which they share. Thus, entanglement distillation protocols can be used as a type of error-correction for quantum communications channels between two parties Alice and Bob, enabling Alice to reliably send mD(\u03c1) qubits of information to Bob, where D(\u03c1) is the distillable entanglement of \u03c1, the state that results when one half of a Bell pair is sent through the noisy channel E connecting Alice and Bob. What is truly remarkable is that this method of communication using entanglement distillation may sometimes work even when conventional quantum error-correction techniques fail. For example, for the depolarizing channel with p = 3/4 we saw in Section 12.4.3 that no quantum information may be transmitted through the channel. However, entanglement distillation protocols are known which can produce a non-zero rate of transmission D(\u03c1) even for this channel! The reason this is possible is because the entanglement distillation protocols allow classical communication back and forth between Alice and Bob, whereas conventional quantum error-correction does not allow any such classical communication. This example allows us to explain the claim we made way back in Chapter 1, illustrated in Figure 12.11, that there are channels with zero capacity for quantum information which, when one such channel is connecting Alice to Bob, and another connects Bob to Alice, can be used to achieve a net \ufb02ow of quantum information! The way this is done is very simple and is based on entanglement distillation. Now, in order that entanglement distillation be possible, we need Alice and Bob to be able to communicate classically, so we set aside half the forward uses of the channel, and all the backward uses of the channel to be used for the transmission of classical information used by the distillation protocol; these channels have non-zero rate for classical information transmission by the HSW theorem. The remaining forward uses of the channel are used to transmit halves of Bell pairs from Alice to Bob, with entanglement distillation being used to extract good Bell pairs out of the resulting states, and then teleportation with the good Bell pairs to achieve a net transmission of quantum information, providing yet another vivid demonstration of the remarkable properties of quantum information! 582 Quantum information theory Figure 12.11. Classically, if we have two very noisy channels of zero capacity running side by side, then the combined channel has zero capacity to send information. Not surprisingly, if we reverse the direction of one of the channels, we still have zero capacity to send information. Quantum mechanically, reversing one of the zero capacity channels can actually allow us to send information! 12.6 Quantum cryptography A \ufb01tting conclusion to this chapter is provided by a most remarkable application of quantum information. As we saw in Chapter 5, quantum computers can be used to break some of the best public key cryptosystems. Fortunately, however, what quantum mechanics takes away with one hand, it gives back with the other: a procedure known as quantum cryptography or quantum key distribution exploits the principles of quantum mechanics to enable provably secure distribution of private information. In this section we describe this procedure, and discuss its security. We begin by explaining the basic ideas of a classical technique, private key cryptography, in Section 12.6.1. Private key cryptography is a much older form of cryptography than the public key cryptosystems (mentioned in Chapter 5), and the principles of private key cryptography are used in quantum cryptosystems. Two other important classical techniques, privacy ampli\ufb01cation and information reconciliation, which are also used in the quantum systems, are described in Section 12.6.2. Three different protocols for quantum key distribution are then presented in Section 12.6.3. How secure are these protocols? It turns out, as we see in Section 12.6.4, that the coherent information, a measure of quantum information we \ufb01rst encountered in Section 12.4.1, gives an information-theoretic lower bound on the in-principle ability to use a quantum communication channel to send private information! This suggests that ideas of quantum information may be useful in proving the security of speci\ufb01c quantum key distribution protocols, and indeed, they are: we conclude the chapter in Section 12.6.5 with a sketch of how the theory of quantum error-correction provides proof of the security of quantum cryptography. 12.6", ".1 Private key cryptography Until the invention of public key cryptography in the 1970s, all cryptosystems operated on a different principle, now known as private key cryptography. In a private key cryptosystem, if Alice wishes to send messages to Bob then Alice must have an encoding key, which allows her to encrypt her message, and Bob must have a matching decoding key, which allows Bob to decrypt the encrypted message. A simple, yet highly effective private key cryptosystem is the Vernam cipher, sometimes called a one time pad. Alice Quantum cryptography 583 and Bob begin with n-bit secret key strings, which are identical. Alice encodes her n-bit message by adding the message and key together, and Bob decodes by subtracting to invert the encoding, as illustrated in Figure 12.12. The great feature of this system is that as long as the key strings are truly secret, it is provably secure. That is, when the protocol used by Alice and Bob succeeds, it does so with arbitrarily high probability (an eavesdropper Eve can always jam the communication channel, but Alice and Bob can detect this jamming and declare failure). And for any eavesdropping strategy employed by Eve, Alice and Bob can guarantee that Eve\u2019s mutual information with their unencoded message can be made as small as desired. In contrast, the security of public key cryptography (Appendix 5) relies on unproven mathematical assumptions about the dif\ufb01culty of solving certain problems like factoring (with classical computers!), even though it is widely used and more convenient. Original message Encryption key Encrypted message Received message Decryption key Decrypted message Public Channel Figure 12.12. The Vernam cipher. Alice encrypts by adding the random key bits (or in this example, letters of the alphabet) to the original message. Bob decrypts by subtracting the key bits to recover the message. The major dif\ufb01culty of private key cryptosystems is secure distribution of key bits. In particular, the Vernam cipher is secure only as long as the number of key bits is at least as large as the size of the message being encoded, and key bits cannot be reused! Thus, the large amount of key bits needed makes such schemes impractical for general use. Furthermore, key bits must be delivered in advance, guarded assiduously until used, then destroyed afterwards; otherwise, in principle, such classical information can be copied without disturbing the originals, thus compromising the security of the whole protocol. Despite these drawbacks, private key cryptosystems such as the Vernam cipher continue to be used because of their provable security, with key material delivered by clandestine meetings, trusted couriers, or private secured communication links. 584 Quantum information theory Exercise 12.25: Consider a system with n users, any pair of which would like to be able to communicate privately. Using public key cryptography how many keys are required? Using private key cryptography how many keys are required? 12.6.2 Privacy ampli\ufb01cation and information reconciliation The \ufb01rst step in private key cryptography is distribution of the key string. What if Alice and Bob start out with imperfect keys? Speci\ufb01cally, suppose that Alice and Bob share correlated random classical bit strings X and Y, and they also have an upper bound on Eve\u2019s mutual information with X and Y. From these imperfect keys, how can they obtain a good enough key to conduct a secure cryptographic protocol? We now show that by performing two steps, information reconciliation, followed by privacy ampli\ufb01cation, they can systematically increase the correlation between their key strings, while reducing eavesdropper Eve\u2019s mutual information about the result, to any desired level of security. These classical steps will be used in the next section, in the quantum key distribution protocol. Information reconciliation is nothing more than error-correction conducted over a public channel, which reconciles errors between X and Y to obtain a shared bit string W while divulging as little as possible to Eve. After this procedure, suppose Eve has obtained a random variable Z which is partially correlated with W. Privacy ampli\ufb01cation is then used by Alice and Bob to distill from W a smaller set of bits S whose correlation with Z is below a desired threshold. Since this last step is conceptually new, let us consider it \ufb01rst. A detailed proof of why privacy ampli\ufb01cation succeeds is beyond the scope of this book, but we will describe the basic method and present the main theorem. One way to accomplish privacy ampli\ufb01cation uses the class of universal hash functions G, which map the set of n-bit strings A to the set of m-bit strings B, such that for any distinct a1, a2 \u2208 A, when g is chosen uniformly at random from G, then the probability that g(a1) = g(a2) is at most 1/|B", "|, The collision entropy of the random variable X with probability distribution p(x) is 0 de\ufb01ned as 1 Hc(X) = \u2212 log p(x)2. (12.171) x (This is sometimes known as the R`enyi entropy of order 2.) Using the concavity of the log function, it is not dif\ufb01cult to show that the Shannon entropy provides an upper bound on this quantity: H(X) \u2265 Hc(X). Hc is important in the following theorem about universal hash functions: Theorem 12.16: Let X be a random variable on alphabet X with probability distribution p(x) and collision entropy Hc(X), and let G be the random variable corresponding to the random choice (with uniform distribution) of a member of the universal class of hash functions from X to {0, 1}m. Then H(G(X)|G) \u2265 Hc(G(X)|G) \u2265 m \u2212 2m\u2212Hc(X). (12.172) Theorem 12.16 can be applied to privacy ampli\ufb01cation in the following manner. Alice and Bob publicly select g \u2208 G and each apply it to W, giving a new bit string S, which Quantum cryptography 585 they choose as their secret key. If Eve\u2019s uncertainty about W given her knowledge Z = z (about a speci\ufb01c instance of the protocol) is known in terms of the collision entropy to be bounded below by some number, say Hc(W |Z = z) > d, then it follows from Theorem 12.16 that Hc(S|G, Z = z) \u2265 m \u2212 2m\u2212d. (12.173) In other words, m may be chosen small enough such that Hc(S|G, Z = z) is nearly equal to m. This maximizes Eve\u2019s uncertainty about the key S, making it a secure secret. Information reconciliation further reduces the number of bits that Alice and Bob can obtain, but this can be bounded as follows. By computing a series of parity checks on subsets of her bits X, Alice can compose a (classical) message u consisting of subset speci\ufb01cations and parities, which, when transmitted to Bob, allows him to correct the errors in his string Y, after which both have the same string W. Clearly, this will require sending k > H(W |Y ) bits of information in u. However, this procedure gives Eve additional knowledge U = u, thus increasing her collision entropy to Hc(W |Z = z, U = u). On average (over possible reconciliation messages u), this increase is bounded below by Hc(W |Z = z, U = u) \u2265 Hc(W |Z = z) \u2212 H(U ), where H(U ) is the usual Shannon entropy of U, but this bound is too weak, because it implies that the probability that the leaked information U = u decreases Hc by more than mH(U ) is only at most 1/m. A stronger bound is provided by this theorem: Theorem 12.17: Let X and U be random variables with alphabets X and U, respectively, where X has probability distribution p(x), and U is jointly distributed with X according to p(x, u). Also let s > 0 be an arbitrary parameter. Then, with probability at least 1 \u2212 2\u2212s, U takes on a value u for which Hc(X|U = u) \u2265 Hc(X) \u2212 2 log |U | \u2212 2s. (12.174) Here, s is known as a security parameter. Applying this to the reconciliation protocol gives the conclusion that Alice and Bob can choose s such that Eve\u2019s collision entropy is bounded from below by Hc(W |Z = z, U = u) \u2265 d \u2212 2(k + s), with probability better than 1 \u2212 2\u2212s. Following this step with privacy ampli\ufb01cation allows them to distill m secret key bits S, for which Eve\u2019s total information is less than 2m\u2212d+2(k+s) bits. CSS code privacy ampli\ufb01cation & information reconciliation As we noted above, information reconciliation is nothing more than error-correction; it turns out that privacy ampli\ufb01cation is also intimately related to error-correction, and both tasks can be implemented by using classical codes. This viewpoint provides a simple conceptual picture which will be useful in the proof of the security of quantum key distribution, in Section 12.6.5, since we have a well-developed theory of quantum errorcorrection codes. With this in mind, it is useful to observe the following. Decoding from a randomly chosen CSS code (see Section 10.4.2) can", " be thought of as performing information reconciliation and privacy ampli\ufb01cation. Although CSS codes are usually used to encode quantum information, for our present purpose we can just consider their classical properties. Consider two classical linear codes C1 and C2 which satisfy the conditions for a t error-correcting [n, m] CSS code: C2 \u2282 C1 and C1 and C\u22a5 2 586 Quantum information theory both correct t errors. Alice chooses a random n bit string X and transmits it to Bob, who receives Y. Let us assume that a priori it is known that along a communication channel between Alice and Bob, the expected number of errors per code block caused by all noise sources including eavesdropping is less than t; in practice, this can be established by random testing of the channel. Furthermore, suppose that Eve knows nothing about the codes C1 and C2; this can be ensured by Alice choosing the code randomly. Finally, suppose that Alice and Bob have an upper bound on the mutual information between Eve\u2019s data Z, and their own data, X and Y. Bob receives Y = X +, where is some error. Since it is known that less than t errors occurred, if Alice and Bob both correct their states to the nearest codeword in C1, their results X, Y \u2208 C1 are identical, W = X = Y. This step is nothing more than information reconciliation. Of course, Eve\u2019s mutual information with W may still be unacceptably large. To reduce this, Alice and Bob identify which of the 2m cosets of C2 in C1 their state W belongs to; that is, that they compute the coset of W + C2 in C1. The result is their m bit key string, S. By virtue of Eve\u2019s lack of knowledge about C2, and the error-correcting properties of C2, this procedure can reduce Eve\u2019s mutual information with S to an acceptable level, performing privacy ampli\ufb01cation. 12.6.3 Quantum key distribution Quantum key distribution (QKD) is a protocol which is provably secure, by which private key bits can be created between two parties over a public channel. The key bits can then be used to implement a classical private key cryptosystem, to enable the parties to communicate securely. The only requirement for the QKD protocol is that qubits can be communicated over the public channel with an error rate lower than a certain threshold. The security of the resulting key is guaranteed by the properties of quantum information, and thus is conditioned only on fundamental laws of physics being correct! The basic idea behind QKD is the following fundamental observation: Eve cannot gain any information from the qubits transmitted from Alice to Bob without disturbing their state. First of all, by the no-cloning theorem (Box 12.1), Eve cannot clone Alice\u2019s qubit. Second, we have the following proposition: Proposition 12.18: (Information gain implies disturbance) In any attempt to distinguish between two non-orthogonal quantum states, information gain is only possible at the expense of introducing disturbance to the signal. Proof Let |\u03c8 and |\u03d5 be the non-orthogonal quantum states Eve is trying to obtain information about. By the results of Section 8.2, we may assume without loss of generality that the process she uses to obtain information is to unitarily interact the state (|\u03c8 or |\u03d5) with an ancilla prepared in a standard state |u. Assuming that this process does not disturb the states, in the two cases one obtains |\u03c8|u \u2192 |\u03c8|v |\u03d5|u \u2192 |\u03d5|v. (12.175) (12.176) Eve would like |v and |v to be different so that she can acquire information about Quantum cryptography 587 the identity of the state. However, since inner products are preserved under unitary transformations, it must be that v|v \u03c8|\u03d5 = u|u \u03c8|\u03d5 v|v = u|u = 1, (12.177) (12.178) which implies that |v and |v must be identical. Thus, distinguishing between |\u03c8 and |\u03d5 must inevitably disturb at least one of these states. We make use of this idea by transmitting non-orthogonal qubit states between Alice and Bob. By checking for disturbance in their transmitted states, they establish an upper bound on any noise or eavesdropping occurring in their communication channel. These \u2018check\u2019 qubits are interspersed randomly among data qubits (from which key bits are later extracted), so that the upper bound applies to the data qubits as well. Alice and Bob then perform information reconciliation and privacy ampli\ufb01cation to distill a shared secret key string. The threshold for the maximum tolerable error rate is thus determined by the e", "f\ufb01cacy of the best information reconciliation and privacy ampli\ufb01cation protocols. Three different QKD protocols which work in this way are presented below. The BB84 protocol Alice begins with a and b, two strings each of (4 + \u03b4)n random classical bits. She then encodes these strings as a block of (4 + \u03b4)n qubits, (4+\u03b4)n. |\u03c8 = |\u03c8akbk, (12.179) k=1 where ak is the kth bit of a (and similarly for b), and each qubit is one of the four states |\u03c800 = |0 |\u03c810 = |1 |\u03c801 = |+ = (|0 + |1)/ |\u03c811 = |\u2212 = (|0 \u2212 |1)/ \u221a 2 \u221a 2. (12.180) (12.181) (12.182) (12.183) The effect of this procedure is to encode a in the basis X or Z, as determined by b. Note that the four states are not all mutually orthogonal, and therefore no measurement can distinguish between (all of) them with certainty. Alice then sends |\u03c8 to Bob, over their public quantum communication channel. Bob receives E(|\u03c8\u03c8|), where E describes the quantum operation due to the combined effect of the channel and Eve\u2019s actions. He then publicly announces this fact. At this point, Alice, Bob, and Eve each have their own states described by separate density matrices. Note also that at this point, since Alice hasn\u2019t revealed b, Eve has no knowledge of what basis she should have measured in to eavesdrop on the communication; at best, she can only guess, and if her guess was wrong, then she would have disturbed the state received by Bob. Moreover, whereas in reality the noise E may be partially due to the environment (a poor channel) in addition to Eve\u2019s eavesdropping, it doesn\u2019t help Eve to have complete control over the channel, so that she is entirely responsible for E. Of course, Bob also \ufb01nds E(|\u03c8\u03c8|) uninformative at this point, because he does not know anything about b. Nevertheless, he goes ahead and measures each qubit in basis X or Z, as determined by a random (4 + \u03b4)n bit string b which he creates on his own. Let 588 Quantum information theory Bob\u2019s measurement result be a. After this, Alice publicly announces b, and by discussion over a public channel, Bob and Alice discard all bits in {a, a} except those for which corresponding bits of b and b are equal. Their remaining bits satisfy a = a, since for these bits Bob measured in the same basis Alice prepared in. Note that b reveals nothing about either a, or the bits a resulting from Bob\u2019s measurement, but it is important that Alice not publish b until after Bob announces reception of Alice\u2019s qubits. For simplicity in the following explanation, let Alice and Bob keep just 2n bits of their result; \u03b4 can be chosen suf\ufb01ciently large so that this can be done with exponentially high probability. Now Alice and Bob perform some tests to determine how much noise or eavesdropping happened during their communication. Alice selects n bits (of their 2n bits) at random, and publicly announces the selection. Bob and Alice then publish and compare the values of these check bits. If more than t bits disagree, then they abort and re-try the protocol from the start. t is selected such that if the test passes, then they can apply information reconciliation and privacy ampli\ufb01cation algorithms to obtain m acceptably secret shared key bits from the remaining n bits. This protocol, known as BB84 after its inventors (see the end of chapter \u2018History and further reading\u2019), is summarized in Figure 12.13, and an experimental realization is described in Box 12.7. Related versions of this protocol, such as using fewer check bits, are also known by the same name. The BB84 QKD protocol 1: Alice chooses (4 + \u03b4)n random data bits. 2: Alice chooses a random (4 + \u03b4)n-bit string b. She encodes each data bit as {|0, |1} if the corresponding bit of b is 0 or {|+, |\u2212} if b is 1. 3: Alice sends the resulting state to Bob. 4: Bob receives the (4 + \u03b4)n qubits, announces this fact, and measures each qubit in the X or Z basis at random. 5: Alice announces b. 6: Alice and Bob discard any bits where Bob measured a different basis than Alice prepared. With high probability, there are at least 2n bits left (if not, abort", " the protocol). They keep 2n bits. 7: Alice selects a subset of n bits that will to serve as a check on Eve\u2019s interference, and tells Bob which bits she selected. 8: Alice and Bob announce and compare the values of the n check bits. If more than an acceptable number disagree, they abort the protocol. 9: Alice and Bob perform information reconciliation and privacy ampli\ufb01ca- tion on the remaining n bits to obtain m shared key bits. Figure 12.13. The four state quantum key distribution protocol known as BB84. Exercise 12.26: Let a k be Bob\u2019s measurement result of qubit |\u03c8akbk, assuming a noiseless channel with no eavesdropping. Show that when b k = bk, a and completely uncorrelated with ak. But when b k = bk, a k = ak. k is random Quantum cryptography 589 Exercise 12.27: (Random sampling tests) The random test of n of 2n check bits allows Alice and Bob to place an upper bound on the number of errors in their untested bits, with high probability. Speci\ufb01cally, for any \u03b4 > 0, the probability of obtaining less than \u03b4n errors on the check bits, and more than (\u03b4 + )n errors on the remaining n bits is asymptotically less than exp[\u2212O(2n)], for large n. We prove this claim here. (1) Without loss of generality, you may assume that there are \u03bcn errors in the 2n bits, where 0 \u2264 \u03bc \u2264 2. Now, if there are \u03b4n errors on the check bits, and (\u03b4 + )n errors on the rest, then \u03b4 = (\u03bc \u2212 )/2. The two conditional statements in the claim thus imply the following: < \u03b4n errors on check bits > (\u03b4 + )n errors on rest \u21d2 \u21d2 < \u03b4n errors on check bits (12.184) > (\u03bc \u2212 \u03b4)n errors on rest, (12.185) and in fact, the top claim on the right implies the bottom one on the right. Using this, show that the probability p which we would like to bound satis\ufb01es \u22121 2n n p < \u03bcn \u03b4n (2 \u2212 \u03bc)n (1 \u2212 \u03b4)n \u03b4n. (12.186) (2) Show that for large n, you can bound 1 an + 1 2anH(b/a) \u2264 an bn \u2264 2anH(b/a), (12.187) where H(\u00b7) is the binary entropy function, Equation (11.8). Apply this to the above bound for p. (3) Apply the bound H(x) < 1 \u2212 2(x \u2212 1/2)2 to obtain the \ufb01nal result, p < exp[\u2212O(2n)]. You may replace \u03bc by a constant which expresses the worst possible case. (4) Compare the result with the Chernoff bound, Box 3.4. Can you come up with a different way to derive an upper bound on p? The B92 protocol The BB84 protocol can be generalized to use other states and bases, and similar conclusions hold. In fact, a particularly simple protocol exists in which only two states are used. For simplicity, it is suf\ufb01cient to consider what happens to a single bit at a time; the description easily generalizes to block tests just as is done in BB84. Suppose Alice prepares one random classical bit a, and, depending on the result, sends Bob \u23a7 \u23aa\u23a8 |0 if a = 0 |\u03c8 = \u23aa\u23a9 |0 + |1 \u221a 2 Depending on a random classical bit a which he generates, Bob subsequently measures the qubit he receives from Alice in either the Z basis |0, |1 (if a = 0), or in the X basis 2 (if a = 1). From his measurement, he obtains the result b, which |\u00b1 = (|0 \u00b1 |1)/ is 0 or 1, corresponding to the \u22121 and +1 eigenstates of X and Z. Bob then publicly announces b (but keeps a secret), and Alice and Bob conduct a public discussion keeping if a = 1. \u221a (12.188) 590 Quantum information theory Box 12.7: Experimental quantum cryptography Quantum key distribution is particularly interesting and astonishing because it is easily experimentally realized. Here is a schematic diagram of one system employing commercial \ufb01ber-optic components to deliver key bits over a ten kilometer distance, which has been built at IBM: \u03bc \u03bc \u03bc \u03bc Bob initially generates strong coherent states |\u03b1 using a diode laser emitting light at a wavelength of 1.3 \u03bcm, and transmits them to Alice", ", who attenuates them to (approximately) generate a single photon. She also polarizes the photon in one of the four states of the BB84 protocol, using as |0 and |1 states horizontal and vertical polarization. She then returns the photon to Bob, who measures it using a polarization analyzer, in a random basis. By using this special con\ufb01guration in which the photon traverses the same path twice, the apparatus can be made to autocompensate for imperfections (such as slowly \ufb02uctuating path lengths and polarization shifts) along the \ufb01ber link. Alice and Bob then select the subset of results in which they used the same basis, reconcile their information, and perform privacy ampli\ufb01cation, communicating over a public channel with photons (over the same \ufb01ber) of 1.55 \u03bcm wavelength. Key bits can be exchanged at the rate of a few hundred per second. Ultimately, improvements in the light source and detector should allow the rate to be improved by a few orders of magnitude. Quantum key distribution over distances exceeding 40 kilometers, and also in installed telecommunication \ufb01ber (under Lake Geneva) has been demonstrated. only those pairs {a, a} for which b = 1. Note that when a = a, then b = 0 always. Only if a = 1 \u2212 a will Bob obtain b = 1, and that occurs with probability 1/2. The \ufb01nal key is a for Alice, and 1 \u2212 a for Bob. This protocol, known as B92 (see the end of chapter \u2018History and further reading\u2019), highlights how the impossibility of perfect distinction between non-orthogonal states lies at the heart of quantum cryptography. As in BB84, because it is impossible for any eavesdropper to distinguish between Alice\u2019s states without disrupting the correlation between the bits Alice and Bob \ufb01nally keep, this protocol allows Alice and Bob to created shared! \" # $ % * +, - -. Quantum cryptography 591 key bits while also placing an upper bound on the noise and eavesdropping during their communication. They can then apply information reconciliation and privacy ampli\ufb01cation to extract secret bits from their resulting correlated random bit strings. Exercise 12.28: Show that when b = 1, then a and a are perfectly correlated with each other. Exercise 12.29: Give a protocol using six states, the eigenstates of X, Y, and Z, and argue why it is also secure. Discuss the sensitivity of this protocol to noise and eavesdropping, in comparison with that of BB84 and B92. The EPR protocol The key bits generated in the BB84 and B92 protocols may appear to have been originated by Alice. However, it turns out that the key can be seen to arise from a fundamentally random process involving the properties of entanglement. This is illustrated by the following protocol. Suppose Alice and Bob share a set of n entangled pairs of qubits in the state |00 + |11 \u221a 2. (12.189) These states are known as EPR pairs. Obtaining these states could have come about in many different ways; for example, Alice could prepare the pairs and then send half of each to Bob, or vice versa. Alternatively, a third party could prepare the pairs and send halves to Alice and Bob. Or they could have met a long time ago and shared them, storing them until the present. Alice and Bob then select a random subset of the EPR pairs, and test to see if they violate Bell\u2019s inequality (Equation (2.225), on page 115 in Section 2.6), or some other appropriate test of \ufb01delity. Passing the test certi\ufb01es that they continue to hold suf\ufb01ciently pure, entangled quantum states, placing a lower bound on the \ufb01delity of the remaining EPR pairs (and thus any noise or eavesdropping). And when they measure these in jointly determined random bases, Alice and Bob obtain correlated classical bit strings from which they can obtain secret key bits as in the B92 and BB84 protocols. Using an argument based on Holevo\u2019s bound, the \ufb01delity of their EPR pairs can be used to establish an upper bound on Eve\u2019s accessible information about the key bits. Where do the key bits come from in this EPR protocol? Since it is symmetric \u2013 Alice and Bob perform identical tasks on their qubits, even possibly simultaneously \u2013 it cannot be said that either Alice or Bob generates the key. Rather, the key is truly random. In fact the same applies to the BB84 protocol, since it can be reduced to an instance of a generalized version of the EPR protocol. Suppose that Alice prepares a random classical bit b, and according to it, measures her half of the EPR pair in either the |0, |1 basis, or 2, obtaining a", ". Let Bob do identically, measuring in (his in the basis |\u00b1 = (|0 \u00b1 |1)/ randomly chosen) basis b and obtaining a. Now they communicate b and b over a public classical channel, and keep as their key only those {a, a} for which b = b. Note that this key is undetermined until Alice or Bob performs a measurement on their EPR pair half. Similar observations can be made about the B92 protocol. For this reason, quantum cryptography is sometimes thought of not as secret key exchange or transfer, but rather as secret key generation, since fundamentally neither Alice nor Bob can pre-determine the key they will ultimately end up with upon completion of the protocol. \u221a 592 Quantum information theory 12.6.4 Privacy and coherent information Thus far, we have described the basic protocol for QKD and argued that it is secure, but we have not provided quantitative bounds. How secure is it? It turns out there is an interesting and fundamental connection between the basic quantitative measures of quantum information discussed in this chapter, and the in-principle obtainable security of quantum cryptography, which we describe below. The quantum coherent information I(\u03c1, E) gives a lower bound on the ability of a quantum channel to send private information. In the most general circumstance, Alice prepares states \u03c1A k, where k = 0, 1,... denotes the different possible states she may send, each with some probability pk. Bob receives states \u03c1B k ) which may be different from \u03c1A k because of channel noise or an eavesdropper, Eve. The mutual information between the result of any measurement Bob may do, and Alice\u2019s value k, Hbob:alice, is bounded above by Holevo\u2019s bound, (12.6), k = E(\u03c1A Hbob:alice \u2264 \u03c7B = S(\u03c1B) \u2212 pkS(\u03c1B k ), (12.190) where \u03c1B = k pk\u03c1B k. Similarly, Eve\u2019s mutual information is bounded above, Heve:alice \u2264 \u03c7E = S(\u03c1E) \u2212 pkS(\u03c1E k ). (12.191) k k Since any excess information Bob has relative to Eve (at least above a certain threshold) can in principle be exploited by Bob and Alice to distill a shared secret key using techniques such as privacy ampli\ufb01cation, it makes sense to de\ufb01ne the quantity P = sup Hbob:alice \u2212 Heve:alice, (12.192) as the guaranteed privacy of the channel, where the supremum is taken over all strategies Alice and Bob may employ to use the channel. This is the maximum excess classical information that Bob may obtain relative to Eve about Alice\u2019s quantum signal. By the HSW theorem, Alice and Bob can employ a strategy such that Hbob:alice = \u03c7B, while for any strategy Eve may employ, Heve:alice \u2264 \u03c7E. Thus, P \u2265 \u03c7B \u2212 \u03c7E, for a suitable choice of strategy. From Exercise 12.11, it follows that a lower bound on the privacy P can be obtained by assuming that all of Alice\u2019s signal states \u03c1A k = |\u03c8A k | are pure states, which are initially unentangled with Eve, who starts out in some state |0E (that may also be assumed to be pure, without loss of generality). In general, the channel from Alice to Bob will include interactions with some environment other than Eve, but to give Eve the greatest possible advantage, all such interactions may be attributed to her, such that the \ufb01nal joint state received by Eve and Bob after transmission is k \u03c8A |\u03c8EB = U |\u03c8A k |0E. (12.193) Since this is a pure state, the reduced density matrices \u03c1E non-zero eigenvalues, and thus the same entropies, S(\u03c1E k ) = S(\u03c1B k ). Thus, k and \u03c1B k will have the same P \u2265 \u03c7B \u2212 \u03c7E = S(\u03c1B) \u2212 pkS(\u03c1B k ) \u2212 S(\u03c1E) + k = S(\u03c1B) \u2212 S(\u03c1E) k pkS(\u03c1E k ) (12.194) (12.195) (12.196) Quantum cryptography 593 = I(\u03c1, E). (12.197) That is, a lower bound on the guaranteed privacy of the channel E is given by the quantum coherent information I(\u03c1, E), as de\ufb01ned in Equation (12.118). Note that this result is not speci\ufb01c to any protocol (which may have its own security \ufb02", "aws). Also, the protocol must perform tests, which are not considered in this calculation, to actually determine properties of the channel E, after which this bound can be applied. So although the information-theoretic bound we have arrived at here is quite elegant, we still have a way to go before being able to quantify the security of QKD! 12.6.5 The security of quantum key distribution How secure is quantum key distribution? Because of the inevitability of disturbance of the communicated state, upon information gain by an eavesdropper, we have good reason to believe in the security of QKD. What we need, however, to conclude that the protocol truly is secure, is a quanti\ufb01able de\ufb01nition of security which explicitly bounds Eve\u2019s knowledge about the \ufb01nal key, given some measure of Alice and Bob\u2019s effort. The following criterion is acceptable: A QKD protocol is de\ufb01ned as being secure if, for any security parameters s > 0 and > 0 chosen by Alice and Bob, and for any eavesdropping strategy, either the scheme aborts, or it succeeds with probability at least 1 \u2212 O(2\u2212s), and guarantees that Eve\u2019s mutual information with the \ufb01nal key is less than 2\u2212. The key string must also be essentially random. In this \ufb01nal section, we give the main elements of a proof that BB84 is secure. This proof serves as a \ufb01tting conclusion to the chapter, because of its elegant use of many concepts of quantum information in providing an argument that is suf\ufb01ciently simple and transparent so as to be incontestable. The origin of this proof, intriguingly, comes from the observation that after information reconciliation and privacy ampli\ufb01cation are performed, the ultimately obtainable key rate turns out to coincide with the achievable qubit transmission rate for CSS codes (Section 10.4.2) over noisy communication channels! Here is the main idea, in outline. It is relatively straightforward to exhaustively establish that the BB84, B92, and EPR protocols are secure if Eve can only attack the transmission one qubit at a time. The dif\ufb01culty lies in dealing with the possibility of collective attacks, in which Eve manipulates and possibly stores large blocks of the transmitted qubits. To address this, we need a more general and powerful argument. Suppose we somehow know that Eve never introduces more than t qubit errors per block. Then Alice could encode her qubits in a t error-correcting quantum code, such that all of Eve\u2019s meddling could be removed by Bob decoding the code. Two things must be established to make this feasible: \ufb01rst, how can an upper bound be placed on t? This turns out to be possible by sampling the channel in the appropriate manner, leaving us with a protocol which is secure, even against collective attacks! Unfortunately, this protocol generally requires a fault-tolerant quantum computer in order to encode and decode qubits robustly. The second challenge is thus to choose a quantum code such that the full sequence of encoding, decoding, and measurement can be performed using no quantum computation or storage \u2013 just single qubit preparation and measurement. Using CSS codes does the trick (after some simpli\ufb01cation), and in fact gives just the BB84 protocol. Below, we begin with a manifestly secure EPR pair based QKD protocol, and then apply solutions to the two challenges to systematically simplify the initial protocol into BB84. 594 Quantum information theory Suppose that Alice has n pairs of entangled qubits, each in the state Requirements for a secure QKD protocol |\u03b200 = |00 + |11 \u221a 2. (12.198) Denote this state as |\u03b200\u2297n. Alice then transmits half of each pair to Bob; because of noise and eavesdropping on the channel, the resulting state may be impure, and can be described as the density matrix \u03c1. Alice and Bob then perform local measurements to obtain a key, as described previously. The following Lemma can be used to show that the \ufb01delity of \u03c1 with respect to |\u03b200\u2297n places an upper bound on the mutual information Eve has with the key. Lemma 12.19: (High \ufb01delity implies low entropy) If F (\u03c1, |\u03b200\u2297n)2 > 1 \u2212 2\u2212s, then S(\u03c1) < (2n + s + 1/ ln 2)2\u2212s + O(2\u22122s). Proof If F (\u03c1, |\u03b200\u2297n)2 = \u2297n\u03b200|\u03c1|\u03b200\u2297n > 1 \u2212 2\u2212s, then the largest eigenvalue of \u03c1 must be larger than 1 \u2212 2", "\u2212s. Therefore, the entropy of \u03c1 is bounded above by the entropy of a diagonal density matrix \u03c1max with diagonal entries 1 \u2212 2\u2212s, 2\u2212s/(22n \u2212 1), 2\u2212s/(22n \u2212 1),..., 2\u2212s/(22n \u22121), that is, \u03c1max has a large entry 1\u22122\u2212s, and the remaining probability is distributed equally among the remaining 22n \u2212 1 entries. Since S(\u03c1max) = \u2212(1 \u2212 2\u2212s) log(1 \u2212 2\u2212s) \u2212 2\u2212s log 2\u2212s 22n \u2212 1, (12.199) the desired result follows. By Holevo\u2019s bound (12.6), S(\u03c1) is an upper bound on the information accessible to Eve, resulting from Alice and Bob\u2019s measurements of \u03c1. This implies that if a QKD protocol can provide Alice and Bob with EPR pairs of \ufb01delity at least 1 \u2212 2\u2212s (with high probability), then it is secure. Exercise 12.30: Simplify (12.199) to obtain the expression for S(\u03c1) given in the statement of the Lemma. Exercise 12.31: It may be unclear why S(\u03c1) bounds Eve\u2019s mutual information with Alice and Bob\u2019s measurement results. Show that this follows from assuming the worst about Eve, giving her all the control over the channel. Random sampling can upper-bound eavesdropping How can a protocol place a lower bound on the \ufb01delity of Alice and Bob\u2019s EPR pairs? The key idea is a classical argument, based on random sampling, which we encountered in the description of the BB84 protocol (Exercise 12.27). Arguments based on classical probability, however, need not apply when considering outcomes of quantum measurements. This is vividly demonstrated by Bell\u2019s inequality (Section 2.6). On the other hand, quantum experiments do allow a classical interpretation whenever measurement observables that refer to only one basis are considered. And fortunately, it happens that measurements in only one basis are required for Alice and Bob to bound the \ufb01delity of their EPR pairs. Quantum cryptography 595 According to (10.14), a qubit transmitted through a noisy quantum channel can be described as having had one of four things happen to it: nothing (I), a bit \ufb02ip (X), a phase \ufb02ip (Z), or a combined bit and phase \ufb02ip (Y ). Recall that the Bell basis is de\ufb01ned by the four states |\u03b200 = |00 + |11 \u221a 2, |\u03b210 = |00 \u2212 |11 \u221a 2, |\u03b201 = |01 + |10 \u221a 2, |\u03b211 = |01 \u2212 |10 \u221a 2. (12.200) Let the second qubit in each pair be the one that Alice sends to Bob. If a bit \ufb02ip error occurs to this qubit, then |\u03b200 is transformed into |\u03b201. Similarly, a phase \ufb02ip gives |\u03b210, and a combination of the two errors gives |\u03b211 (up to irrelevant overall phases). A natural measurement which detects if a bit \ufb02ip has occurred is given by projectors \u03a0bf = |\u03b201\u03b201| + |\u03b211\u03b211| and I \u2212 \u03a0bf, and likewise, the measurement described by projectors \u03a0pf = |\u03b210\u03b210| + |\u03b211\u03b211| and I \u2212 \u03a0pf detects phase \ufb02ips. Since both these measurements commute with the Bell basis, their outcomes obey classical probability arguments. In fact, any measurement which commutes with the Bell basis will also satisfy the same classical arguments. More precisely, Alice and Bob and can bound the \ufb01delity of their EPR pairs by randomly sampling a subset of them. Suppose Alice sends 2n EPR pair halves to Bob. They subsequently randomly select n of them and check these qubits by jointly measuring either \u03a0bf or \u03a0pf (again, chosen randomly). By the same classical arguments used in the random sampling tests in BB84 (Exercise 12.27), if \u03b4n bit or phase \ufb02ip errors are detected, then the remaining n EPR pairs would be exponentially certain to have the same number of errors, were they also to be measured in the Bell basis. Bell states are nonlocal, and generally measurements in the Bell basis require nonlocal operations, which can be dif\ufb01cult. Fortunately, however, they are not required in the present scheme, because \u03a0bf = (I \u2297 I \u2212 Z \u2297 Z)/2, and \u03a0pf = (I \u2297 I \u2212 X \u2297 X)/2. Thus, Alice and Bob can perform the desired checks", " with local measurements of Pauli operators, either by both measuring Z, or both measuring X. Exercise 12.32: Note that the local measurements that Alice and Bob perform, such as I \u2297 X and X \u2297 I, do not commute with the Bell basis. Show that despite this, the statistics which Alice and Bob compile from their measurements are the same as those which they would have obtained had they actually measured \u03a0bf and \u03a0pf. The modi\ufb01ed Lo\u2013Chau protocol Random sampling in the Bell basis thus provides Alice and Bob with EPR pairs \u03c1 with known \ufb01delity to the ideal state |\u03b200\u2297n, and as previously discussed, this bounds Eve\u2019s mutual information with any measurements that might be performed on \u03c1. For \u03c1 to be useful for key generation, however, Alice and Bob must reduce Eve\u2019s mutual information with their state until it is exponentially small. This task can be achieved by applying classical privacy ampli\ufb01cation to their measurement results. Equivalently, Alice and Bob can \ufb01rst perform entanglement distillation, as introduced in Section 12.5.2, to obtain \u03c1 which is very close to |\u03b200\u2297m for some m < n, then measure the \ufb01nal state. This sort of \u2018quantum privacy ampli\ufb01cation\u2019 will be useful to us. A rough argument is as follows. Entanglement distillation can be accomplished by per- 596 Quantum information theory forming quantum error-correction. Since \u03c1 is nearly certain to have \u03b4n errors, encoding these qubits in a \u03b4n correcting quantum error-correction code allows up to \u03b4n errors to be perfectly corrected. As we saw in Sections 10.5.5 and 10.5.8, if an [n, m] stabilizer code is used, then encoding, syndrome measurement and error recovery can be performed by measurements of Pauli operators determined by the rows of the check matrix for the code. Alice and Bob simply perform identical measurements and recovery operations on their respective n qubit halves of \u03c1, producing an error-corrected state which has a \ufb01delity relative to |\u03b200\u2297m that is on the order of one minus the the probability of more than \u03b4n errors occurring. By construction, the syndrome measurements turn out to commute with the Bell basis, since Alice and Bob perform identical tasks. Putting the random sampling and entanglement distillation pieces together gives us the modi\ufb01ed Lo\u2013Chau protocol, presented in Figure 12.14. A few notes about this protocol are in order. The random Hadamard transforms performed in steps 3 and 7 create a symmetry in Eve\u2019s strategy between detecting information encoded in the X and Z bases (and thus causing X and Z errors). They also enable random selection of a measurement of \u03a0bf or \u03a0pf on the check qubits. The speci\ufb01c procedure employed in step 9 can be justi\ufb01ed for the case of any stabilizer code, as in Exercise 12.34. The Gilbert\u2013Varshamov bound for CSS codes, Equation (10.74), shows that good quantum codes exist for large block lengths, so that n can be chosen suf\ufb01ciently large that for a \u03b4n error-correcting [n, m] quantum code, the criteria for security can be satis\ufb01ed. QKD protocol: modi\ufb01ed Lo\u2013Chau 1: Alice creates 2n EPR pairs in the state |\u03b200\u22972n. 2: Alice randomly selects n of the 2n EPR pairs to serve as checks to check for Eve\u2019s interference. She does not do anything with them yet. 3: Alice selects a random 2n-bit string b, and performs a Hadamard transform on the second qubit of each pair for which b is 1. 4: Alice sends the second qubit of each pair to Bob. 5: Bob receives the qubits and publicly announces this fact. 6: Alice announces b and which n qubits are to provide check bits. 7: Bob performs Hadamards on the qubits where b is 1. 8: Alice and Bob each measure their n check qubits in the |0, |1 basis, and publicly share the results. If more than t of these disagree, they abort the protocol. 9: Alice and Bob measure their remaining n qubits according to the check matrix for a pre-determined [n, m] quantum code correcting up to t errors. They share the results, compute the syndromes for the errors, and then correct their state, obtaining m nearly perfect EPR pairs. 10: Alice and Bob measure the m EPR pairs in the |0, |", "1 basis to obtain a shared secret key. Figure 12.14. A QKD protocol which is secure, by virtue of use of perfect quantum computers, error-correction, and random testing of EPR pairs. Quantum cryptography 597 Exercise 12.33: Let {M1, M2,..., Mn} be a set of measurement observables which produce respective results Xi when an input state \u03c1 is measured. Argue that the random variables Xi obey classical probability arguments if [Mi, Mj] = 0, that is, they commute with each other. Exercise 12.34: (Entanglement distillation by error-correction) In Section 10.5.8, we saw that codewords of an [n, m] qubit stabilizer code can be constructed by measuring its generators g1,..., gn\u2212m on an arbitrary n qubit quantum state, then applying Pauli operations to change the result to be a simultaneous +1 eigenstate of the generators. Using that idea, show that if we start out with n EPR pairs in the state |\u03b200\u2297n, and perform identical generator measurements on the two n qubit halves of the pairs, followed by Pauli operations to correct for differences in the measurement results between the pairs, then we obtain an encoded |\u03b200\u2297m state. Also show that if the stabilizer code corrects up to \u03b4n errors, then even if \u03b4n errors are suffered by an n qubit half, we still obtain |\u03b200\u2297m. A quantum error-correction protocol The modi\ufb01ed Lo\u2013Chau protocol makes use of quantum error-correction to perform entanglement distillation, and is built essentially upon the EPR protocol. Entanglement is a frail resource, and quantum error-correction generally requires robust quantum computers, which are challenging to realize. Fortunately, however, this protocol can be systematically simpli\ufb01ed in a series of steps, each of which provably does not compromise the security of the scheme. Let us begin by removing the need to distribute EPR pairs. Note that the measurements which Alice perform at the end of the modi\ufb01ed Lo\u2013Chau protocol can be performed at the very start, with no change in any of the states held by the rest of the world. Alice\u2019s measurements of her halves of the check EPR pairs in step 8 collapse the pairs into n single qubits, so instead of sending entangled states, Alice can simply send single qubits. This gives us the modi\ufb01ed steps 1: Alice creates n random check bits, and n EPR pairs in the state |\u03b200\u2297n. She also encodes n qubits as |0 or |1 according to the check bits. 2: Alice randomly chooses n positions (out of 2n) and puts the check qubits in these positions and half of each EPR pair in the remaining positions. 8: Bob measures the n check qubits in the |0, |1 basis, and publicly shares the results with Alice. If more than t of these disagree, they abort the protocol. Similarly, Alice\u2019s measurements in steps 9 and 10 collapse EPR pairs into random qubits encoded in a random quantum code. This can be seen in the following manner. A particularly convenient choice of code, which we shall employ for the remainder of this section, is an [n, m] CSS code of C1 over C2, CSS(C1, C2), which encodes m qubits in n qubits and corrects up to t errors. Recall from Section 10.4.2 that for this code, H1 and H \u22a5 2 are the parity check matrices corresponding to the classical codes C1 and C\u22a5 2, in which each of the codeword states is w\u2208C2 for vk being a representative of one of the 2m cosets of C2 in C1 (the notation vk is |vk + w, (12.201) 1 |C2| 598 Quantum information theory chosen to suggest a vector v indexed by a key string k). Also recall that there exists a family of codes equivalent to this one, CSSz,x(C1, C2), with codeword states |\u03bevk,z,x = 1 |C2| w\u2208C2 (\u22121)z\u00b7w|vk + w + x. (12.202) These states form an orthonormal basis for a 2n-dimensional Hilbert space (see Exercise 12.35), and thus we may write Alice\u2019s n EPR pair state as |\u03b200\u2297n = 2n j=0 |j|j = vk,z,x |\u03bevk,z,x|\u03bevk,z,x", ". (12.203) Note that in this expression, we have separated the labels into two kets, where the \ufb01rst denotes the qubits Alice keeps, and the second, the qubits which are sent to Bob. When Alice measures the stabilizer generators corresponding to H1 and H \u22a5 2 on her qubits in step 9, she obtains random values for x and z, and similarly, her \ufb01nal measurement in step 10 gives her a random choice of vk. The remaining n qubits are thus left in the state |\u03bevk,z,x, which is the codeword for vk in CSSz,x(C1, C2). This is just the encoded counterpart of a 2m qubit state |k. Therefore, as claimed above, Alice\u2019s measurements produce random qubits encoded in a random code. Thus, instead of sending halves of EPR pairs, Alice can equivalently randomly choose x, z, and k, then encode |k in the code CSSz,x(C1, C2), and send Bob the encoded n qubits. This gives us the modi\ufb01ed steps 1: Alice creates n random check bits, a random m bit key k, and two random n bit strings x and z. She encodes |k in the code CSSz,x(C1, C2). She also encodes n qubits as |0 or |1 according to the check bits. 2: Alice randomly chooses n positions (out of 2n) and puts the check qubits in these positions and encoded qubits in the remaining positions. 6: Alice announces b, x, z, and which n qubits are to provide check bits. 9: Bob decodes the remaining n qubits from CSSz,x(C1, C2). 10: Bob measures his qubits to obtain the shared secret key k. The resulting scheme, known as the CSS codes protocol, is shown in Figure 12.15. Exercise 12.35: Show that the states |\u03bevk,z,x de\ufb01ned in (12.202) form an orthonormal basis for a 2n-dimensional Hilbert space, that is, |\u03bevk,z,x\u03bevk,z,x| = I. (12.204) vk,z,x Hint: for C1 an [n, k1] code, C2 an [n, k2] code, and m = k1 \u2212 k2, note that there are 2m distinct values of vk, 2n\u2212k1 distinct x, and 2k2 distinct z. Exercise 12.36: Verify Equation (12.203). Exercise 12.37: This is an alternative way to understand why Alice\u2019s measurements in steps 9 and 10 collapse EPR pairs into random qubits encoded in a random quantum code. Suppose Alice has an EPR pair (|00 + |11)/ she measures the \ufb01rst qubit in the X basis, then the second qubit collapses into an eigenstate of X determined by the measurement result. Similarly, show that if 2. Show that if \u221a Quantum cryptography 599 she measures in the Z basis, then the second qubit is left in a Z eigenstate labeled by the measurement result. Using this observation and the results of 2, and \u00afZ on her Section 10.5.8, conclude that Alice\u2019s measurements of H1, H \u22a5 EPR pair halves result in a random codeword of CSSz,x(C1, C2) determined by her measurement results. QKD protocol: CSS codes 1: Alice creates n random check bits, a random m bit key k, and two random n bit strings x and z. She encodes |k in the code CSSz,x(C1, C2). She also encodes n qubits as |0 or |1 according to the check bits. 2: Alice randomly chooses n positions (out of 2n) and puts the check qubits in these positions and the encoded qubits in the remaining positions. 3: Alice selects a random 2n-bit string b, and performs a Hadamard transform on each qubit for which b is 1. 4: Alice sends the resulting qubits to Bob. 5: Bob receives the qubits and publicly announces this fact. 6: Alice announces b, x, z, and which n qubits are to provide check bits. 7: Bob performs Hadamards on the qubits where b is 1. 8: Bob measures the n check qubits in the |0, |1 basis, and publicly shares the results with Alice. If more than t of these disagree, they abort the protocol. 9: Bob decodes the remaining n qubits from CSSz,x(C1, C2). 10:", " Bob measures his qubits to obtain the shared secret key k. Figure 12.15. A QKD protocol which is secure, by virtue of simpli\ufb01cation of the modi\ufb01ed Lo\u2013Chau protocol via CSS codes. Reduction to BB84 The CSS codes QKD protocol is secure by virtue of direct reduction from the modi\ufb01ed Lo\u2013Chau protocol, and is much simpler because it does not make any evident use of EPR pairs. Unfortunately, it is still unsatisfactory, because it requires perfect quantum computation to perform the encoding and decoding (instead of just single qubit preparation and measurements), and Bob needs to temporarily store qubits in a quantum memory while waiting for communication from Alice. The use of CSS codes, however, enables these two requirements to be removed, essentially because they decouple phase \ufb02ip correction from bit \ufb02ip correction. First, note that Bob measures his qubits in the Z basis immediately after decoding; thus, the phase correction information Alice sends as z is unnecessary. Thus, since C1 and C2 are classical codes, instead of decoding then measuring, he can immediately measure to obtain vk + w + x + (where represents some possible error, due to the channel and to Eve), then decode classically: he subtracts Alice\u2019s announced value of x, then corrects the result to a codeword in C1, which is de\ufb01nitely vk + w, if the distance of the code is 600 Quantum information theory not exceeded. The \ufb01nal key k is the coset of vk + w + C2 in C1 (see Appendix 2 for an explanation of cosets, and this notation). This gives us: 9: Bob measures the remaining qubits to get vk + w + x +, and subtracts x from the result, correcting it with code C1 to obtain vk + w. 10: Bob computes the coset of vk + w + C2 in C1 to obtain the key k. Second, since Alice need not reveal z, the state she effectively sends is a mixed state, averaged over random values of z, \u03c1vk,x = 1 2n z = = 1 2n|C2| z w\u2208C2 1 |C2| |\u03bevk,z,x\u03bevk,z,x| (12.205) (\u22121)z\u00b7(w1+w2)|vk + w1 + xvk + w2 + x| (12.206) w1,w2\u2208C2 |vk + w + xvk + w + x|. (12.207) This state is simple to create: Alice need only classically choose w \u2208 C2 at random, and construct |vk + w + x, using her randomly determined x and k. We thus have: 1: Alice creates n random check bits, a random n bit string x, a random vk \u2208 C1/C2, and a random w \u2208 C2. She encodes n qubits in the state |0 or |1 according to vk + w + x, and similarly, n qubits according to the check bits. Steps 1 and 9 can be simpli\ufb01ed further, by changing step 6 slightly. Currently, Alice sends |vk + w + x, Bob receives and measures to obtain vk + w + x +, then Alice sends x, which Bob subtracts to obtain vk + w +. But if Alice chooses vk \u2208 C1 (as opposed to C1/C2), then w is unnecessary. Moreover, vk + x is then a completely random n bit string, and instead of the above protocol, it is equivalent if Alice chooses x at random, sends |x, Bob receives and measures to obtain x +, then Alice sends x \u2212 vk, which Bob subtracts to obtain vk +. Now, there is no difference between the random check bits and the code bits! This gives us: 1: Alice chooses a random vk \u2208 C1, and creates 2n qubits in the state |0 or |1 according to 2n random bits. 2: Alice randomly chooses n positions (out of 2n) and designates these as the check qubits, and the remainder as |x. 6: Alice announces b, x \u2212 vk, and which n qubits are to provide check bits. 9: Bob measures the remaining qubits to get x +, and subtracts x \u2212 vk from the result, correcting it with code C1 to obtain vk. 10: Alice and Bob compute the coset of vk + C2 in C1 to obtain the key k. Next, note that Alice need not perform Hadamard operations (although in practice, single qubit operations are not so", " dif\ufb01cult to accomplish with photons). She can instead encode her qubits directly in either the |0, |1 (Z) basis or in the |+, |\u2212 (X) basis, depending on the bits of b: 1: Alice creates (4 + \u03b4)n random bits. For each bit, she creates a qubit in either the |0, |1 basis, or the |+, |\u2212 basis, according to a random bit string b. Quantum cryptography 601 We are almost done: encoding and decoding are now performed classically. The remaining problem is to remove the need for a quantum memory. To solve this problem, suppose that Bob goes ahead and measures immediately after receiving qubits from Alice, choosing randomly to measure in either the X or Z bases. When Alice subsequently announces b, they can keep only those bits for which their bases happened to be the same. This allows Bob to do away completely with his quantum storage device. Note that with high probability they discard half their bits, so in order to obtain the same number of key bits as before, they should start with a little (say, \u03b4) more than twice the number of original random bits. Of course, Alice now must delay her choice of which bits are to be check bits, until after the discarding step. This gives us the \ufb01nal protocol, shown in Figure 12.16. This protocol is exactly the same as BB84, with only minor cosmetic differences. Note how the use of the classical code C1 performs information reconciliation, and computing the coset of vk + C2 in C1 performs privacy ampli\ufb01cation (see Section 12.6.2). QKD protocol: Secure BB84 1: Alice creates (4 + \u03b4)n random bits. 2: For each bit, she creates a qubit in either the Z basis, or the X basis, according to a random bit string b. 3: Alice sends the resulting qubits to Bob. 4: Alice chooses a random vk \u2208 C1. 5: Bob receives the qubits, publicly announces this fact, and measures each in the Z or X bases at random. 6: Alice announces b. 7: Alice and Bob discard those bits Bob measured in a basis other than b. With high probability, there are at least 2n bits left; if not, abort the protocol. Alice decides randomly on a set of 2n bits to continue to use, randomly selects n of these to be check bits, and announces the selection. 8: Alice and Bob publicly compare their check bits. If more than t of these disagree, they abort the protocol. Alice is left with the n bit string x, and Bob with x +. 9: Alice announces x \u2212 vk. Bob subtracts this from his result, correcting it with code C1 to obtain vk. 10: Alice and Bob compute the coset of vk + C2 in C1 to obtain the key k. Figure 12.16. The \ufb01nal QKD protocol arrived at by systematic reduction of the CSS codes protocol, which is exactly the same as BB84 (up to minor cosmetic differences). For clarity, we have dropped the notation. In summary, we have systematically proven the security of the BB84 quantum key distribution protocol, by starting with a manifestly secure scheme requiring perfect quantum computation and quantum memories, and systematically reducing it to BB84. By virtue of having only made modi\ufb01cations which manifestly leave Eve\u2019s quantum state (conditioned on all revealed classical information) unchanged, we conclude that BB84 is secure. Naturally, there are some caveats. This proof only applies to an ideal situation, where the states sent are those described. In practice, qubit sources are imperfect; for example, 602 Quantum information theory such sources are often lasers attenuated to approximately generate single photons, representing qubits (as described in Section 7.4.1). Furthermore, the proof does not place any bounds on the amount of effort Alice and Bob must go to in decoding; for practical key distribution, C1 must be ef\ufb01ciently decodable. This proof also does not provide an upper bound on the tolerable eavesdropping; it utilizes CSS codes, which are not optimal. It is estimated that a rate of bit and phase errors up to 11% is acceptable using a protocol similar to BB84, but with the aid of quantum computers to encode and decode, higher error rates may be tolerable. The ultimate capability of quantum cryptography is an interesting open issue, and we expect such fundamental questions about the physical limits of computation and communication to continue to intrigue and challenge researchers in the future. Exercise 12.38: Show that if you had the ability to distinguish non-orthogonal states, then it would be possible to compromise the security of BB84, and indeed, all of the QKD protocols we have described. Problem 12.1: In this problem we", " will work through an alternate proof of the Holevo bound. De\ufb01ne the Holevo chi quantity, \u03c7 \u2261 S(\u03c1) \u2212 pxS(\u03c1x). (12.208) (1) Suppose the quantum system consists of two parts, A and B. Show that x \u03c7A \u2264 \u03c7AB. (12.209) (Hint: Introduce an extra system which is correlated with AB, and apply strong subadditivity.) (2) Let E be a quantum operation. Use the previous result to show that \u03c7 \u2261 S(E(\u03c1)) \u2212 pxS(E(\u03c1x)) \u2264 \u03c7 \u2261 S(\u03c1) \u2212 pxS(\u03c1x). (12.210) x x That is, the Holevo chi quantity decreases under quantum operations. This is an important and useful fact in its own right. (3) Let Ey be a set of POVM elements. Augment the quantum system under consideration with an \u2018apparatus\u2019 system, M, with an orthonormal basis |y. De\ufb01ne a quantum operation by E(\u03c1 \u2297 |00|) \u2261 y Ey\u03c1 Ey \u2297 |yy|, (12.211) where |0 is some standard pure state of M. Prove that after the action of E, \u03c7M = H(X : Y ). Use this and the previous two results to show that H(X : Y ) \u2264 S(\u03c1) \u2212 pxS(\u03c1x), (12.212) which is the Holevo bound. x Problem 12.2: This result is an extension of the previous problem. Provide a proof of the no-cloning theorem by showing that a cloning process for non-orthogonal pure states would necessarily increase \u03c7. Chapter problems 603 Problem 12.3: For a \ufb01xed quantum source and rate R > S(\u03c1), design a quantum circuit implementing a rate R compression scheme. Problem 12.4: (Linearity forbids cloning) Suppose we have a quantum machine with two slots, A and B. Slot A, the data slot, starts out in an unknown quantum state \u03c1. This is the state to be copied. Slot B, the target slot, starts out in some standard quantum state, \u03c3. We will assume that any candidate copying procedure is linear in the initial state, \u03c1 \u2297 \u03c3 \u2192 E(\u03c1 \u2297 \u03c3) = \u03c1 \u2297 \u03c1, (12.213) where E is some linear function. Show that if \u03c11 = \u03c12 are density operators such that E(\u03c11 \u2297 \u03c3) = \u03c11 \u2297 \u03c11 E(\u03c12 \u2297 \u03c3) = \u03c12 \u2297 \u03c12, (12.214) (12.215) then any mixture of \u03c11 and \u03c12 is not copied correctly by this procedure. Problem 12.5: (Classical capacity of a quantum channel \u2013 Research) Is the product state capacity (12.71) the true capacity of a noisy quantum channel for classical information, that is, the capacity when entangled inputs to the channel are allowed? Problem 12.6: (Methods for achieving capacity \u2013 Research) Find an ef\ufb01cient construction for codes achieving rates near the product state capacity (12.71) of a noisy quantum channel for classical information. Problem 12.7: (Quantum channel capacity \u2013 Research) Find a method to evaluate the capacity of a given quantum channel E for the transmission of quantum information. 604 Quantum information theory Summary of Chapter 12: Quantum information theory \u2022 No-cloning: No quantum device can be constructed which outputs |\u03c8|\u03c8, given |\u03c8, for arbitrary |\u03c8. \u2022 Holevo\u2019s bound: The maximum accessible classical information when trying to distinguish between quantum states \u03c1x sent with probability distribution px is H(X : Y ) \u2264 \u03c7 \u2261 S px\u03c1x \u2212 pxS(\u03c1x). x \u2022 Schumacher\u2019s quantum noiseless channel coding theorem: S(\u03c1) can be interpreted as the number of qubits needed to faithfully represent a quantum source described by \u03c1. x \u2022 The Holevo\u2013Schumacher\u2013Westmoreland theorem: The capacity of a noisy quantum channel E for classical information is given by: C(E) = max S {px,|\u03c8x} pxE(|\u03c8x\u03c8x|) \u2212 pxS(E(|\u03c8x\u03c8x|)). (12.216) x x \u2022 Majorization condition for entanglement transformation: Alice and Bob can transform |\u03c8 to |\u03d5 by local operations and classical communication if and only if \u03bb\u03c8 \u227a \u03bb\u03d5, where \u03bb\u03c8 is", " a vector of the eigenvalues of the reduced density matrix of |\u03c8 (similarly for \u03bb\u03d5). \u2022 Pure state entanglement distillation and dilution: Alice and Bob can convert between n copies of a joint state |\u03c8 and nS(\u03c1) Bell pairs, by local operations and classical communication alone, as n \u2192 \u221e, where \u03c1 is the reduced density matrix. \u2022 Quantum cryptography: Provably secure key distribution is possible by communicating using non-orthogonal quantum states, with a protocol such as BB84. Eavesdropping on the channel will cause a detectable increase in the error rate, because information gain implies disturbance. History and further reading The book of Cover and Thomas[CT91] is a terri\ufb01c introduction to classical information theory. The reader looking for a more advanced yet still readable treatment of information theory should consult Csisz\u00b4ar and K\u00a8orner[CK81]. Also well worth the read are Shannon\u2019s original papers, among the most in\ufb02uential in twentieth century science. These have been reprinted together in a single volume by Shannon and Weaver[SW49]. Bennett and Shor[BS98] and Bennett and DiVincenzo[BD00] have written excellent review articles on quantum information theory. The no-cloning theorem is due to Dieks[Die82] and to Wootters and Zurek[WZ82]. An enormous amount of work has been done extending these results. By far the majority of the papers consider various schemes for cloners which are interesting in some particular way \u2013 they optimize some measure of cloning \ufb01delity, or some other property one might wish a cloner to have. We will not attempt to give a full review of this work here, but History and further reading 605 note that many of these papers may be found on the internet at http://arXiv.org/ in the quant-ph archive. Some papers of especial interest include the work of Barnum, Caves, Fuchs, Jozsa and Schumacher[BCF+96] extending the range of application of the no-cloning theorem to mixed states and non-unitary cloning devices; by Mor[Mor98] on the cloning of states of composite systems; by Westmoreland and Schumacher[WS98] on a possible equivalence between cloning and faster-than-light communication; and a rebuttal by van Enk[van98b]. The Holevo bound was conjectured by Gordon in 1964[Gor64], and proved by Holevo in 1973[Hol73]. The conceptually simple proof we have given is based upon the dif\ufb01cult-toprove strong subadditivity inequality, however Holevo used a more direct approach which has been simpli\ufb01ed by Fuchs and Caves[FC94]. The approach via strong subadditivity is due to Yuen and Ozawa[YO93]; see also Schumacher, Westmoreland and Wootters[SWW96]. The classical noiseless channel coding theorem is due to Shannon[Sha48, SW49]. The quantum noiseless channel coding theorem is due to Schumacher[Sch95], and is described in a pioneering paper that introduced in an integrated fashion many of the fundamental notions of quantum information theory, including sources, \ufb01delity measures, and the notion of quantum states as a resource that could be treated in information-theoretic terms. This last observation, simple but profound, was driven home by Schumacher\u2019s introduction in the paper of the now-ubiquitous term qubit, attributed to a conversation between Schumacher and Wootters. A paper by Jozsa and Schumacher[JS94] simpli\ufb01ed Schumacher\u2019s original approach; this paper was published earlier than [Sch95], but was was written at a later time. These papers were based on the ensemble-average \ufb01delity measure discussed in Exercise 12.8, rather than the entanglement \ufb01delity based proof we have given here, which is based on the approach of Nielsen[Nie98]. A slight hole in the original papers by Schumacher, and Schumacher and Jozsa was \ufb01lled by the work of Barnum, Fuchs, Jozsa and Schumacher[BFJS96]. M. Horodecki[Hor97] subsequently provided a more powerful proof of the same result that also points the way towards a theory of quantum data compression of ensembles of mixed states. The compression scheme described in Box 12.4, which is the quantum analogue of Cover\u2019s enumerative coding method[CT91], is originally attributed to Schumacher[Sch95], and quantum circuits for it are explicitly given by Cleve and DiVincenzo[CD96]. Braun", "stein, Fuchs, Gottesman, and Lo have generalized this to provide a quantum analogue to Huffman encoding[BFGL98], and Chuang and Modha to arithmetic coding[CM00]. The Holevo\u2013Schumacher\u2013Westmoreland (HSW) theorem has an interesting history. The problem it addresses was \ufb01rst discussed by Holevo[Hol79] in 1979, who made some partial progress on the problem. Unaware of this work, Hausladen, Jozsa, Schumacher, Westmoreland, and Wootters[HJS+96] solved a special case of the problem in 1996. Independently, and shortly thereafter, Holevo[Hol98] and Schumacher and Westmoreland[SW97] proved the HSW theorem giving the product state capacity of a noisy quantum channel for classical information. Fuchs[Fuc97] has described some interesting examples of the product state capacity, where the ensemble of states maximizing the expression (12.71) for the capacity contains non-orthogonal members. King and Ruskai[KR99] have made some promising headway on the problem of showing that the product state capacity is the same as the capacity unrestricted to product states, but the general problem remains open. The entropy exchange was de\ufb01ned by Lindblad[Lin91], and rediscovered by Schumacher[Sch96b], who proved the quantum Fano inequality. The coherent information was 606 Quantum information theory introduced by Lloyd[Llo97] and by Schumacher and Nielsen[SN96] in the context of the capacity of a noisy quantum channel; [SN96] proves the quantum data processing inequality. A table containing the inequalities mentioned in Exercise 12.15 may be found in Nielsen\u2019s Ph.D. dissertation[Nie98]. The as yet unsolved problem of determining the quantum channel capacity (Problem 12.7) has an interesting history. Initial work on the subject came from several different perspectives, as may be seen from the papers of Barnum, Nielsen and Schumacher[BNS98], of Bennett, DiVincenzo, Smolin, and Wootters[BDSW96], of Lloyd[Llo97], of Schumacher[Sch96b] and of Schumacher and Nielsen[SN96]. The equivalence of several of these points of view has been understood through the work of Barnum, Knill and Nielsen[BKN98], and Barnum, Smolin and Terhal[BST98]. The capacity has been established for some speci\ufb01c channels (most notably the quantum erasure channel) by Bennett, DiVincenzo and Smolin[BDS97], and a lower bound on the capacity of the depolarizing channel making intriguing use of degenerate quantum codes has been obtained by Shor and Smolin[SS96], and re\ufb01ned by DiVincenzo, Shor and Smolin[DSS98]. Zurek[Zur89], Milburn[Mil96], and Lloyd[Llo96] analyzed examples of quantum Maxwell\u2019s demons, though not in the context of error-correction. The analysis here is based on the work of Nielsen, Caves, Schumacher and Barnum[NCSB98]. This point of view has also been pursued by Vedral[Ved99] to obtain limits on entanglement distillation procedures. The quantum Singleton bound is due to Knill and La\ufb02amme[KL97]. The proof we give is due to Preskill[Pre98b]. The study of entanglement has blossomed into a major area of research, and there are far too many papers on the subject to even begin to give an account here. Once again, see the quant-ph archive at http://arXiv.org/. The conditions for entanglement transformation based on majorization (Theorem 12.15) are due to Nielsen[Nie99a]. Theorem 12.13 is due to Uhlmann[Uhl71, Uhl72, Uhl73]. Proposition 12.14 is due to Lo and Popescu[LP97]. Entanglement catalysis was discovered by Jonathan and Plenio[JP99]. Marshall and Olkin[MO79] is a comprehensive introduction to majorization, including the proof of Birkhoff\u2019s theorem. The limits for entanglement dilution and distillation are due to Bennett, Bernstein, Popescu and Schumacher[BBPS96]. Entanglement distillation protocols for mixed states were introduced by Bennett, Brassard, Popescu, Schumacher, Smolin and Wootters[BBP+96], and the connection to quantum error-correction developed in a ground-breaking paper by Bennett, DiVincenzo, Smolin and Wootters[BDSW96", "] that has stimulated a lot of subsequent research. The example illustrated in Figure 12.11 was noted by Gottesman and Nielsen (unpublished). We mention just a few more papers on entanglement of exceptional interest that may serve as an entry-point to the literature; unfortunately many papers of note are omitted as a result. A series of papers by members of the Horodecki family (Michal, Pawel and Ryszard) have investigated the properties of entanglement in depth; of especial note are [HHH96, HHH98, HHH99a, HHH99b, HHH99c]. Also of great interest are the papers by Vedral and Plenio[VP98] and by Vidal[Vid98]. For an excellent (early) overview of quantum cryptography at the lay level see the article by Bennett, Brassard, and Ekert in Scienti\ufb01c American[BBE92]. Quantum cryptographic ideas were \ufb01rst put forward by Wiesner in the late 1960s. Unfortunately, Wiesner\u2019s ideas were not accepted for publication, and it wasn\u2019t until the early 1980s that the ideas became known. Wiesner proposed that (entangled) quantum states, if they could be stored for long periods of time, could be used as a kind of counterfeit-proof money[Wie, Wie83]. Bennett developed several further protocols, one of which lead to the \ufb01rst experimental implementation, by Bennett, Bessette, Brassard, Salvail, and Smolin[BBB+92], which is History and further reading 607 of historical interest (in principle) since it transmitted its information less than a meter and, moreover, eavesdropping was facilitated by a loud buzzing sound which emanated from the power supply whenever a \u2018one\u2019 was sent! The concept of privacy ampli\ufb01cation was \ufb01rst introduced by Bennett, Brassard, and Robert[BBR88]. For information reconciliation protocols see [BBB+92] and [BS94]. Theorem 12.16 is stated and proved in Bennett, Brassard, Cr`epeau, and Maurer[BBCM95], in a very readable general treatment of privacy ampli\ufb01cation. Note that the information disclosed during reconciliation has an important impact on the threshold for privacy ampli\ufb01cation, as bounded in Theorem 12.17, proven by Cachin and Maurer[CM97]. Privacy ampli\ufb01cation has applications to classical key generation using distant correlated random sources such as starlight sensed by satellites[Mau93]. The four state protocol known as BB84 is named after the authors, Bennett and Brassard[BB84], and similarly, the two state B92 protocol is named after Bennett[Ben92]. The EPR protocol was devised by Ekert[Eke91]. The proof of the random sampling bound in Exercise 12.27 is due to Ambainis. The limitations and security of quantum cryptography have been discussed in depth in many publications. For a sample, see the works by Barnett and Phoenix[BP93]; Brassard[Bra93]; Ekert, Huttner, Palma, and Peres[EHPP94]; also [Phy92]. The connection between coherent information and privacy was recognized by Schumacher and Westmoreland[SW98]. Numerous papers have been published on experimental implementations of quantum cryptographic systems. For a good introduction, see Hughes, Alde, Dyer, Luther, Morgan, and Schauer[HAD+95]; the demonstration of quantum cryptography under Lake Geneva was by Muller, Zbinden, and Gisin[MZG96]. The experiment described in Box 12.7 was performed at IBM, by Bethune and Risk[BR98, BR00], and we thank them for the schematic of their apparatus. A large number of proofs of the security of various quantum key distribution protocols, under different circumstances, have been presented. Of particular note is a complete (and extensive, but somewhat complicated) proof of the security of QKD with BB84 given by Mayers[May98]. Biham, Boyer, Brassard, van de Graaf, and Mor have also considered attacks against BB84[BBB+98]. A simpler proof, which uses EPR states and requires perfect quantum computation has been given by Lo and Chau[LC99]; this is the protocol we began with in Section 12.6.5. Lo has simpli\ufb01ed it to begin with a test which ascertains the error rate, before transmitting key material[Lo99]. The even simpler (and beautiful!) proof given in Section 12.6.5 is due to Shor and Preskill[SP00], who also give the 11% estimate mentioned in Section 12.6.5. Our presentation of this proof also bene\ufb01ted greatly from conversations with Gottes", "man. Appendix 1: Notes on basic probability theory This appendix collects some elementary de\ufb01nitions and results about probability theory. It is assumed that the reader already has some familiarity with this material. The reader who is not familiar with any of the results should take time out to prove them, as indicated in the exercises. The basic notion of probability theory is that of a random variable. A random variable X may take one of a number of values, x, with probabilities p(X = x). We use upper case to denote the random variable, and x to denote the values that random variable may take. We often use the notation p(x) instead of p(X = x), leaving the \u2018X =\u2019 implicit. In this book we shall only be concerned with random variables which take their value from a \ufb01nite set of values, and we always assume that this is the case. Occasionally it is convenient to consider random variables whose values are vectors taking values, for example, in the set (i, j), i = 1,..., m1, j = 1,..., m2. The conditional probability that Y = y given that X = x is de\ufb01ned by p(Y = y|X = x) \u2261 p(X = x, Y = y) p(X = x), (A1.1) where p(X = x, Y = y) is the probability that X = x and Y = y. When p(X = x) = 0 we make the convention that p(Y = y|X = x) = 0. We often use the notation p(y|x) leaving the \u2018Y =\u2019 and \u2018X =\u2019 implicit. Random variables X and Y are said to be independent if p(X = x, Y = y) = p(X = x)p(Y = y) for all x and y. Note that when X and Y are independent, it follows that p(y|x) = p(y) for all x and y. Bayes\u2019 rule relates the conditional probabilities for Y given X to those for X given Y, p(x|y) = p(y|x) p(x) p(y). (A1.2) The probability p(y) appearing in this expression is often re-expressed using the law of total probability discussed below. Exercise A1.1: Prove Bayes\u2019 rule. One of the most important and frequently used results in probability theory is the law of total probability. It states that if X and Y are two random variables, then the probabilities for Y can be expressed in terms of the probabilities for X, and the conditional probabilities for Y given X, p(y) = p(y|x)p(x), (A1.3) where the sum is over all values x that X can take. x Exercise A1.2: Prove the law of total probability. History and further reading 609 The expectation, average, or mean of a random variable X is de\ufb01ned by E(X) \u2261 p(x)x, (A1.4) where the sum is over all values x which the random variable X can take. x Exercise A1.3: Prove that there exists a value of x \u2265 E(X) such that p(x) > 0. Exercise A1.4: Prove that E(X) is linear in X. Exercise A1.5: Prove that for independent random variables X and Y, E(XY ) = E(X)E(Y ). The variance of a random variable X is de\ufb01ned by the expression var(X) \u2261 E[(X \u2212 E(X))2] = E(X 2) \u2212 E(X)2. (A1.5) \u221a The standard deviation, \u0394(X) \u2261 var(X), is a measure of the spread of a random variable about the average. Chebyshev\u2019s inequality quanti\ufb01es more precisely in what sense the standard deviation is a measure of the spread of values a random variable may take. It states that for any \u03bb > 0 and random variable X with \ufb01nite variance, p(|X \u2212 E(X)| \u2265 \u03bb\u0394(X)) \u2264 1 \u03bb2. (A1.6) Thus, the probability of being more than \u03bb standard deviations away from the mean gets small as \u03bb goes to in\ufb01nity. Exercise A1.6: Prove Chebyshev\u2019s inequality. The main text of the book contains many other results in probability theory, including the Chernoff bound on page 154, Fano\u2019s inequality on page 536, and the law of large numbers on page 541. History and", " further reading Probability theory enjoys a surfeit of superb texts. We especially recommend the text of Grimmett and Stirzaker[GS92] as an excellent introduction to many basic ideas in the theory of probability and stochastic processes. From a more purely mathematical point of view, Williams[Wil91] is a superb introduction to the modern theory of probability, with an emphasis on the beautiful theory of martingales. Finally, the classic two-volume text of Feller[Fel68a, Fel68b] is an excellent indepth introduction to the ideas of probability theory. Appendix 2: Group theory The mathematical theory of groups is useful at several points in the study of quantum computation and quantum information. The generalization of the order-\ufb01nding, factoring, and period \ufb01nding algorithms in Chapter 5 is based on the hidden subgroup problem; the stabilizer formalism for quantum error-correction in Chapter 10 is based on some elementary group theory. The number theory described in Appendix 4 uses the properties of the group Z\u2217 n. And, implicitly, the quantum circuits used throughout the book are an example of the use of Lie groups. In this appendix we review some basic material on group theory. We summarize many of the fundamental concepts and provide important de\ufb01nitions, but do not attempt to explain very much, as group theory is a vast subject! A2.1 Basic de\ufb01nitions A group (G, \u00b7) is a non-empty set G with a binary group multiplication operation \u2018\u00b7\u2019, with the following properties: (closure) g1 \u00b7 g2 \u2208 G for all g1, g2 \u2208 G; (associativity) (g1 \u00b7 g2) \u00b7 g3 = g1 \u00b7 (g2 \u00b7 g3), for all g1, g2, g3 \u2208 G; (identity) there exists e \u2208 G such that \u2200g \u2208 G; (inverses) for all g \u2208 G, there exists g\u22121 \u2208 G such that g \u00b7 g\u22121 = e and g\u22121 \u00b7 g = e. We often leave out \u00b7 and write g1 \u00b7 g2 as simply g1g2. We also often refer to the group G without referring explicitly to its multiplication operation, but it must be de\ufb01ned. A group G is \ufb01nite if the number of elements in G is \ufb01nite. The order of a \ufb01nite group G is the number of elements it contains, denoted as |G|. A group G is said to be Abelian if g1g2 = g2g1 for all g1, g2 \u2208 G. A simple example of a \ufb01nite Abelian group is the additive group Zn of integers modulo n, with \u2018multiplication\u2019 operation the operation of modular addition. It is easily checked that this operation satis\ufb01es the closure and associativity axioms; there is an identity element, 0, since x + 0 = x(mod n) for all x, and every x \u2208 G has an inverse, n \u2212 x, since x + (n \u2212 x) = 0(mod n). The order of an element g \u2208 G is the smallest positive integer r such that gr (g multiplied with itself r times) equals the identity element e. A subgroup H of G is a subset of G which forms a group under the same group multiplication operation as G. Theorem A2.1: (Lagrange\u2019s theorem) If H is a subgroup of a \ufb01nite group G then |H| divides |G|. Exercise A2.1: Prove that for any element g of a \ufb01nite group, there always exists a positive integer r such that gr = e. That is, every element of such a group has an order. Exercise A2.2: Prove Lagrange\u2019s theorem. Basic de\ufb01nitions 611 Exercise A2.3: Show that the order of an element g \u2208 G divides |G|. If g1 and g2 are elements of G, then the conjugate of g2 with respect to g1 is the element g\u22121 g2g1. If H is subgroup of G, then it is known as a normal subgroup if g\u22121Hg = H 1 for all g \u2208 G. The conjugacy class Gx of an element x in a group G is de\ufb01ned by Gx \u2261 {g\u22121xg|g \u2208 G}. Exercise A2.4: Show that if y \u2208 Gx then Gy = Gx. Exercise A2.5: Show that if x is an element of an Abelian group G then Gx = {", "x}. An interesting example of a group which is not Abelian is the Pauli group on n qubits. For a single qubit, the Pauli group is de\ufb01ned to consist of all the Pauli matrices, with multiplicative factors \u00b11, \u00b1i allowed by the de\ufb01nition: G1 \u2261 {\u00b1I, \u00b1iI, \u00b1X, \u00b1iX, \u00b1Y, \u00b1iY, \u00b1Z, \u00b1iZ}. (A2.1) This set of matrices forms a group under the operation of matrix multiplication. You might wonder why we don\u2019t omit the multiplicative factors \u00b11 and \u00b1i; the reason these are included is to ensure that G1 is closed under multiplication, and thus forms a legitimate group. The general Pauli group on n qubits is de\ufb01ned to consist of all n-fold tensor products of Pauli matrices, and again we allow multiplicative factors of \u00b11, \u00b1i. A2.1.1 Generators The study of groups is often greatly simpli\ufb01ed by the use of a set of group generators for the group being studied. A set of elements g1,..., gl in a group G is said to generate the group G if every element of G can be written as a product of (possibly repeated) elements from the list g1,..., gl, and we write G = g1,..., gl. For example, G1 = X, Z, iI, since every element in G can be written as a product of X, Z and iI. On the other hand, X = {I, X}, a subgroup of G1, since not every element of G1 can be written as a power of X. The notation \u00b7 \u00b7 \u00b7 which we use for group generators may potentially be confused with the notation for observable averages introduced in Section 2.2.5 (page 87); however, in practice it is always clear from context how the notation is being used. The great advantage of using generators to describe groups is that they provide a compact means of describing the group. Suppose G has size |G|. Then it is pretty easy to show that there is a set of log(|G|) generators generating G. To see this, suppose g1,..., gl is a set of elements in a group G, and g is not an element of g1,..., gl. Let f \u2208 g1,..., gl. Then f g \u2208 g1,..., gl, since if it were then we would have g = f \u22121f g \u2208 g1,..., gl, which we know is false by assumption. Thus for each element f \u2208 g1,..., gl there is an element f g which is in g1,..., gl, g but not in g1,..., gl. Thus adding the generator g to g1,..., gl doubles (or more) the size of the group being generated, from which we conclude that G must have a set of generators containing at most log(|G|) elements. A2.1.2 Cyclic groups A cyclic group G possesses an element a such that any element g \u2208 G can be expressed as an for some integer n. a is known as a generator of G, and we write G = a. A cyclic subgroup H generated by g \u2208 G is the group formed by {e, g, g2,..., gr\u22121}, where r is the order of g. That is, H = g. 612 Group theory Exercise A2.6: Show that any group of prime order is cyclic. Exercise A2.7: Show that every subgroup of a cyclic group is cyclic. Exercise A2.8: Show that if g \u2208 G has \ufb01nite order r, then gm = gn if and only if m = n(mod r). A2.1.3 Cosets For H a subgroup of G, the left coset of H in G determined by g \u2208 G is the set gH \u2261 {gh|h \u2208 H}. The right coset is de\ufb01ned similarly. Often whether a coset is a \u2018left\u2019 or \u2018right\u2019 coset is implied by context. In the case of a group like Zn where the group operation is addition it is conventional to write cosets of a subgroup H in the form g + H, for g \u2208 Zn. Elements of a particular coset gH are known as coset representatives of that coset. Exercise A2.9: Cosets de\ufb01ne an equivalence relation between elements. Show that g1, g2", " \u2208 G are in the same coset of H in G if and only if there exists some h \u2208 H such that g2 = g1h. Exercise A2.10: How many cosets of H are there in G? A2.2 Representations Let Mn be the set of n\u00d7n complex matrices. A matrix group is a set of matrices in Mn which satisfy the properties of a group under matrix multiplication. We shall denote the identity element in such groups as I. A representation \u03c1 of a group G may be de\ufb01ned as a function which maps G to a matrix group, preserving group multiplication. Speci\ufb01cally, g \u2208 G is mapped to \u03c1(g) \u2208 Mn, such that g1g2 = g3 implies \u03c1(g1)\u03c1(g2) = \u03c1(g3). If the map is many to one, it is known as a homomorphism; if it is one to one, then the map is an isomorphism. A representation \u03c1 which maps into Mn has dimension d\u03c1 = n. The representations we have de\ufb01ned are also referred to as being matrix representations; there are more general ones, but these are suf\ufb01cient for our purposes. For the remainder of this appendix all the groups G which we deal with should be taken to be \ufb01nite groups. A2.2.1 Equivalence and reducibility Two important concepts about representations are equivalence and reducibility. The character of a matrix group G \u2282 Mn is a function on the group de\ufb01ned by \u03c7(g) = tr(g), for g \u2208 G, where tr(\u00b7) is the usual trace function on matrices. It has the following properties: (1) \u03c7(I) = n, (2) |\u03c7(g)| \u2264 n, (3) |\u03c7(g)| = n implies g = ei\u03b8I, (4) \u03c7 is constant on any given conjugacy class of G, (5) \u03c7(g\u22121) = \u03c7\u2217(g), and (6) \u03c7(g) is an algebraic number for all g. Two matrix groups are said to be equivalent if they are isomorphic, and corresponding elements under the isomorphism have the same character. Exercise A2.11: (Characters) Prove the properties of characters given above. Exercise A2.12: (Unitary matrix groups) A unitary matrix group is comprised solely of unitary matrices (those which satisfy U \u2020U = I). Show that every matrix group is equivalent to a unitary matrix group. If a representation of a group Representations 613 consists entirely of unitary matrices, we may refer to it as being a unitary representation. A matrix group G in Mn is said to be completely reducible if it is equivalent to another matrix group H which is of block diagonal form, that is all elements m \u2208 H are of the form diag(m1, m2), for some m1 \u2208 Mn1 and m2 \u2208 Mn2. If no such equivalence exists, then the matrix group is irreducible. The following is a useful property of irreducible matrix groups: Lemma A2.2: (Schur\u2019s lemma) Let G \u2282 Mn and H \u2282 Mk be two matrix groups of the same order, |G| = |H|. If there exists a k by n matrix S such that Sgi = hiS for some ordering of all elements gi \u2208 G and hi \u2208 H, then either S is the zero matrix, or n = k and S is a square nonsingular matrix. Exercise A2.13: Show that every irreducible Abelian matrix group is one dimensional. Exercise A2.14: Prove that if \u03c1 is an irreducible representation of G, then |G|/d\u03c1 is an integer. The following theorem connects irreducibility with characters: Theorem A2.3: A matrix group G is irreducible if and only if 1 |G| g\u2208G |\u03c7(g)|2 = 1. (A2.2) The key theorem of representation theory is the following: A2.2.2 Orthogonality Theorem A2.4: (Fundamental theorem) Every group G has exactly r inequivalent irreducible representations, where r is the number of conjugacy classes of G. And if \u03c1p \u2208 Md\u03c1 and \u03c1q are any two of these, then the matrix elements satisfy the orthogonality relations \u03c1p(g) \u03c1q(g) \u03b4il\u03b4jk\u03b4pq, \u22121", " = (A2.3) |G| d\u03c1 g\u2208G ij kl where \u03b4pq = 1 if \u03c1p = \u03c1q and is zero otherwise. Exercise A2.15: Using the Fundamental Theorem, prove that characters are orthogonal, that is: r ri (\u03c7p i )\u2217\u03c7q i = |G|\u03b4pq i=1 and r p=1 (\u03c7p i )\u2217\u03c7p j = |G| ri \u03b4ij, (A2.4) where p, q, and \u03b4pq have the same meaning as in the theorem, and \u03c7p value the character of the pth irreducible representation takes on the ith conjugacy class of G, and ri is the size of the ith conjugacy class. i is the Exercise A2.16: S3 is the group of permutations of three elements. Suppose we order these as mapping 123 to: 123; 231; 312; 213; 132, and 321, respectively. Show that 614 Group theory there exist two one-dimensional irreducible representations of S3, one of which is trivial, and the other of which is 1, 1, 1, \u22121, \u22121, \u22121, corresponding in order to the six permutations given earlier. Also show that there exists a two-dimensional irreducible representation, with the matrices 1 0 1 0 1 0 0 1 \u22121 0 0 1,, \u22121 \u2212 \u221a \u221a 3 3 \u22121 \u221a 1 1 3 \u221a 3 \u22121 \u221a \u2212 \u221a 3 3 \u22121 \u221a \u2212 3 3 \u22121 1,. (A2.5) Verify that the representations are orthogonal. A2.2.3 The regular representation The number 1 is a valid one-dimensional matrix representation of any group. However, it is trivial. A representation is faithful if the matrix group of the representation is isomorphic to the original group. The regular representation is a faithful representation which exists for any group, and is constructed in the following manner. Let v = [g1, g2,..., g|G|]T be a column vector of elements from G. Multiplying each element of v by an element g \u2208 G permutes the entries of the vector; this permutation can be represented by a |G|\u00d7|G| matrix acting by matrix multiplication on v. The |G| such matrices corresponding to the possible different permutations form a faithful representation of G, under matrix multiplication. Exercise A2.17: Prove that the regular representation is faithful. Exercise A2.18: Show that the character of the regular representation is zero except on the representation of the identity element, for which \u03c7(I) = |G|. The decomposition of arbitrary representations into tensor sums of irreducible repre- sentations obeys the following theorem: Theorem A2.5: If \u03c1 is an arbitrary representation of G with character \u03c7, and \u03c1p are the inequivalent irreducible representations of G with characters \u03c7p, then \u03c1 = \u2295pcp\u03c1p, where \u2295 denotes a direct sum, and cp are the numbers determined by cp = 1 |G| r i=1 ri (\u03c7p i )\u2217\u03c7i. (A2.6) Exercise A2.19: Use Theorem A2.5 to show that the regular representation contains d\u03c1p instances of each irreducible representation \u03c1p. Thus, if R denotes the regular representation, and \u02c6G denotes the set of all inequivalent irreducible representations, then \u03c7R i = d\u03c1\u03c7\u03c1 i. (A2.7) Exercise A2.20: The character of the regular representation is zero except for the \u03c1\u2208 \u02c6G Fourier transforms 615 conjugacy class i containing e, the identity element in G. Show, therefore, that d\u03c1\u03c7\u03c1(g) = N \u03b4ge. (A2.8) Exercise A2.21: Show that \u03c1\u2208 \u02c6G \u03c1\u2208 \u02c6G d2 \u03c1 = |G|. A2.3 Fourier transforms Let G be a \ufb01nite group of order N, and f be a function which maps group elements to complex numbers. For an irreducible representation \u03c1 of G, of dimension d\u03c1, we de\ufb01ne the Fourier transform of f to be \u02c6f, ) \u02c6f (\u03c1) \u2261 d\u03c1 N g\u2208G f (g) \u03c1(g). (A2.9) \u02c6G be a Note that for \u03c1 a matrix", " representation, complete set of inequivalent irreducible representations of G. We de\ufb01ne the inverse Fourier transform of \u02c6f to be \u02c6f (\u03c1) maps matrices to matrices. Let f (g) = 1\u221a N \u03c1\u2208 \u02c6G d\u03c1 tr( \u02c6f (\u03c1)\u03c1(g\u22121)). (A2.10) \u03c1 d2 \u03c1 = N, f and \u02c6f can both be expressed as vectors of complex numbers of Because length N. Note that the coef\ufb01cients in the above equations have been chosen such that if \u02c6G consists of unitary representations, then the Fourier transformations are unitary. The above de\ufb01nitions can be understood by substituting (A2.9) into (A2.10) to obtain f (g\u2208 \u02c6G g\u2208G d\u03c1f (g) tr \u03c1(g)\u03c1(g\u22121) d\u03c1f (g) tr \u03c1(g g\u22121) \u03c1\u2208 \u02c6G g\u2208G f (g) g\u2208G \u03c1\u2208 \u02c6G d\u03c1 \u03c7\u03c1(g g\u22121). Using (A2.8), we may simplify (A2.13) to f (g) = f (g)\u03b4gg, (A2.11) (A2.12) (A2.13) (A2.14) as desired. g\u2208G Exercise A2.22: Substitute (A2.10) into (A2.9) and prove that \u02c6f (\u03c1) is obtained. Exercise A2.23: Let us represent an Abelian group G by g \u2208 [0, N \u2212 1], with addition as the group operation, and de\ufb01ne \u03c1h(g) = exp[\u22122\u03c0igh/N ] as the h representation of g. This representation is one-dimensional, so d\u03c1 = 1. Show that the Fourier transform relations for G are \u02c6f (h) = 1\u221a N N \u22121 g=0 f (g) e\u22122\u03c0igh/N and f (h) = 1\u221a N N \u22121 g=0 \u02c6f (g) e2\u03c0ihg/N. (A2.15) 616 Group theory Exercise A2.24: Using the results of Exercise A2.16, construct the Fourier transform over S3 and express it as a 6\u00d76 unitary matrix. History and further reading There are many outstanding texts on group theory, and virtually any book on algebra has a section devoted to the theory of groups. The discussion here borrowed much notation from the text on \ufb01nite groups by Lomont[Lom87]. Hammermesh is a standard reference for group theory in physics[Ham89]. Discussions of Fourier transforms over groups are not so common. Diaconis and Rockmore have written a good article on the ef\ufb01cient computation of Fourier transforms over groups[DR90]; many of their results are also reviewed in F\u00a8assler and Stiefel[FS92]. Beth independently discovered the fast Fourier transform over groups[Bet84], as did Clausen[Cla89]. Appendix 3: The Solovay\u2013Kitaev theorem In Chapter 4 we showed that an arbitrary unitary operation U may be implemented on gates. a quantum computer using a circuit consisting of single qubit and controlledSuch universality results are important because they ensure the equivalence of apparently different models of quantum computation. For example, the universality results ensure that a quantum computer programmer may design quantum circuits containing gates which have four input and output qubits, con\ufb01dent that such gates can be simulated by a constant number of controlled- and single qubit unitary gates. An unsatisfactory aspect of the universality of controlled- and single qubit unitary gates is that the single qubit gates form a continuum, while the methods for fault-tolerant quantum computation described in Chapter 10 work only for a discrete set of gates. Fortunately, also in Chapter 4 we saw that any single qubit gate may be approximated to gate, Hadamard arbitrary accuracy using a \ufb01nite set of gates, such as the controlledgate H, phase gate S, and \u03c0/8 gate. We also gave a heuristic argument that approximating the chosen single qubit gate to an accuracy required only \u0398(1/) gates chosen from the \ufb01nite set. Furthermore, in Chapter 10 we showed that the controlled, Hadamard, phase and \u03c0/8 gates may be implemented in a fault-tolerant manner. In this appendix we show that a much faster rate of convergence than \u0398", "(1/) may be achieved. The Solovay\u2013Kitaev theorem shows that for any gate U on a single qubit, and given any > 0, it is possible to approximate U to a precision using \u0398(logc(1/)) gates from a \ufb01xed \ufb01nite set, where c is a small constant approximately equal to 2. The best possible value for c isn\u2019t known yet, so we are going to explain the proof of the Solovay\u2013Kitaev theorem for c approximately equal to 4, and then in the end of appendix problems outline a method that may be used to reduce c down closer to 2. We will also prove that c cannot be less than 1; determining the best possible value of c between 1 and 2 is an open problem! To appreciate the importance of the Solovay\u2013Kitaev theorem, imagine a quantum computer programmer designs an algorithm using f (n) gates to solve a problem. Suppose the algorithm he or she comes up with uses many gates outside the usual fault-tolerant, Hadamard, phase and \u03c0/8 gates. How many gates are required set of controlledto implement the algorithm fault-tolerantly? If the tolerance to error for the entire algorithm is to be, then the individual gates must be accurate to a tolerance /f (n). By the heuristic argument of Chapter 4 it would take \u0398(f (n)/) gates in the fault-tolerant set to approximate each one of the gates used in the algorithm, for a total cost \u0398(f (n)2/), a polynomial increase in the number of gates required by the algorithm. Using the Solovay\u2013 Kitaev theorem each gate in the algorithm can be simulated by O(logc(f (n)/)) gates in the fault-tolerant set, for a total cost O(f (n) logc(f (n)/)) for the fault-tolerant algorithm, which is only polylogarithmically more than for the original algorithm. For many 618 The Solovay\u2013Kitaev theorem problems, such a polylogarithmic cost is quite acceptable, whereas the polynomial cost provided by the heuristic argument of Chapter 4 may be much less desirable. To state the Solovay\u2013Kitaev theorem more precisely, we need to de\ufb01ne some notation and nomenclature. Recall that SU (2) is the set of all single qubit unitary matrices with determinant equal to one. We restrict our attention to SU (2), since all single qubit unitary gates may be written as the product of an element of SU (2) with an unimportant global phase factor. Suppose G is a \ufb01nite set of elements of SU (2); G plays the role of the \ufb01nite set of elementary gates our quantum computer programmer is using to simulate all the other gates. For the sake of concreteness, think of G as containing the fault-tolerant set H, S, and T, with appropriate global phases added to ensure the determinants are all equal to one. We suppose for the sake of convenience that G contains its own inverses, that is, if U \u2208 G, then U \u2020 \u2208 G. In the case of the fault-tolerant set, that means adding S\u2020 = S3 and T \u2020 = T 7 to the set, which fortunately can be expressed in terms of gates already in the set. A word of length l from G is a product g1g2... gl \u2208 SU (2), where gi \u2208 G for each i. We de\ufb01ne Gl to be the set of all words of length at most l, and G to be the set of all words of \ufb01nite length. \u221a We need some notion of distance to quantify what we mean by an approximation to a unitary matrix. The exact measure used is not all that important. It is convenient for our purposes to use the trace distance studied in Chapter 9, D(U, V ) \u2261 tr|U \u2212 V |, X \u2020X is the positive square root of X \u2020X. Actually, this de\ufb01nition differs where |X| \u2261 by a factor 2 from the de\ufb01nition used in Chapter 9; the reason for using a different normalization is that it makes geometric visualization of the proof of the Solovay\u2013Kitaev theorem easier, as we shall see (it will also be helpful to think of elements in SU (2) as being points in space). A subset S of SU (2) is said to be dense in SU (2) if for any element U of SU (2) and > 0 there is an element s \u2208 S such that D(s, U ) <.", " Suppose S and W are subsets of SU (2). Then S is said to form an -net for W, where > 0, if every point in W is within a distance of some point in S. Our interest is in how fast Gl \u2018\ufb01lls up\u2019 SU (2) as l is increased. That is, for how small an is Gl an -net for SU (2)? The Solovay\u2013Kitaev theorem says that gets small very quickly indeed as l is increased. Exercise A3.1: In Chapter 4 we made use of the distance measure E(U, V ) = max|\u03c8 (U \u2212 V )|\u03c8, where the maximum is over all pure states |\u03c8. Show that when U and V are single qubit rotations, U = R \u02c6m(\u03b8), V = R \u02c6n(\u03d5), D(U, V ) = 2E(U, V ), and thus it does not matter whether we use the trace distance or the measure E(\u00b7, \u00b7) for the Solovay\u2013Kitaev theorem. Theorem A3.1: (Solovay\u2013Kitaev theorem) Let G be a \ufb01nite set of elements in SU (2) containing its own inverses, such that G is dense in SU (2). Let > 0 be given. Then Gl is an -net in SU (2) for l = O(logc(1/)), where c \u2248 4. As already noted, the best possible value of c is somewhat lower than 4, but it is convenient to give the proof for this particular case. In Problem 3.1 we explain how modi\ufb01cations of the proof can be used to lower c. The \ufb01rst part of the proof is to show that the points of Gl get very dense in a small neighbourhood of the identity matrix I as l is increased, a conclusion encapsulated in the following lemma. To state the lemma, we de\ufb01ne S to be the set of all points U in SU (2) such that D(U, I) \u2264. The Solovay\u2013Kitaev theorem 619 Lemma A3.2: Let G be a \ufb01nite set of elements in SU (2) containing its own inverses, and such that G is dense in SU (2). There exists a universal constant 0 independent of G, such that for any \u2264 0, if Gl is an 2-net for S, then G5l is a C3-net for S\u221a C3/2, for some constant C. We prove Lemma A3.2 shortly, but \ufb01rst let\u2019s see how it implies the Solovay\u2013Kitaev theorem. There are two steps to the proof. The \ufb01rst step is to apply Lemma A3.2 iteratively to show that the neighbourhood of the origin \ufb01lls in very quickly as the word length l is increased. Since G is dense in SU (2) we can \ufb01nd an l0 such that Gl0 is an 2 0-net for SU (2), and thus also for S0. Applying Lemma A3.2 with = 0 and l = l0 implies that G5l0 is a C3 and l = 5l0 implies that G52l0 is a C( k times, we \ufb01nd that G5kl0 is an (k)2-net for S(k), where. Applying Lemma A3.2 again with = C3/2 0 )3/2. Iterating this procedure 0-net for S\u221a )3-net for S\u221a C3/2 0 C3/2 \u221a 0 C3/2 0 \u221a C( \u221a (k) = (C0)(3/2)k C. (A3.1) Without loss of generality we may suppose 0 has been chosen such that C0 < 1, and therefore (k) gets small very fast as k increases. It will also be useful to note that, provided 0 is chosen small enough, (k)2 < (k + 1). The second step is to let U be any element of SU (2) and use the translation idea illustrated in Figure A3.1 to approximate U using products of elements of G. Let U0 \u2208 Gl0 be an (0)2-approximation to U. Now de\ufb01ne V so that V U0 = U, that is, V \u2261 U U \u2020 0. Thus D(V, I) = tr|V \u2212 I| = tr|(U \u2212 U0)U \u2020 | = tr|U \u2212 U0| < (0)2 < (1", "). From the iterated application of Lemma A3.2 discussed above, we can \ufb01nd U1 \u2208 G5l0 which is an (1)2-approximation to V. It follows that U1U0 is an (1)2-approximation to U. Now de\ufb01ne V so that V U1U0 = U, that is, V \u2261 U U \u2020 U \u2020 1. Thus D(V, I) = tr|V \u2212 I| = tr|(U \u2212 U1U0)U \u2020 | = tr|U \u2212 U1U0|/ < (1)2 < (2). It follows from the iterated application of Lemma A3.2 that we can \ufb01nd U2 \u2208 G52l0 which is an (2)2-approximation to V, and thus U2U1U0 is an (2)2-approximation to U. Continuing in this way we construct Uk \u2208 G5kl0 so that UkUk\u22121... U0 is an (k)2-approximation to U. Putting it all together, a sequence of l0 + 5l0 + \u00b7 \u00b7 \u00b7 + 5kl0 < 5 4 5kl0 gates can be used to approximate any unitary gate U to an accuracy (k)2. To approximate to some desired accuracy, we therefore must choose k such that U \u2020 1 0 0 0 Substituting (A3.1) this can be restated as (k)2 <. k 3 2 < log(1/C2) 2 log(1/C0). (A3.2) (A3.3) It follows that the number of gates required to approximate to within satis\ufb01es (c = log 5/ log(3/2) \u2248 4) kc c number of gates < 5 4 3 2 That is, the number of gates required to approximate to within is O(logc(1/)), completing the proof of the Solovay\u2013Kitaev theorem. l0 < 5 4 log(1/C2) 2 log(1/C0) 5kl0 = (A3.4) l0. 5 4 The proof of Lemma A3.2 uses a few elementary facts about multiplication of elements 620 The Solovay\u2013Kitaev theorem Figure A3.1. The translation step used in the proof of the Solovay\u2013Kitaev theorem. To approximate an arbitrary single qubit gate we \ufb01rst approximate to within a distance (0)2 using l0 gates from G. Then we improve the approximation by adding 5l0 more gates, for a total accuracy better than (1)2, and continue on in this way, quickly converging to U. of SU (2), which we now recall. The key idea of the lemma is to work in the neighborhood of the identity, which greatly simpli\ufb01es the rather complicated operation of multiplication in SU (2). More precisely, suppose U and V are elements of SU (2), and de\ufb01ne the group commutator of U and V by [U, V ]gp \u2261 U V U \u2020V \u2020. Suppose U and V are both close to the identity, so that they may be written U = e\u2212iA and V = e\u2212iB, where A and B are Hermitian matrices such that tr|A|, tr|B| \u2264 for some small. Expanding e\u00b1iA and e\u00b1iB to terms quadratic in A and B gives (A3.5) D([U, V ]gp, e\u2212[A,B]) = O(3), where [A, B] = AB \u2212 BA is the usual commutator for matrices (in fact, the commutator for the Lie algebra of SU (2)). Thus, in the neighborhood of the identity we can study the group commutator by studying instead the much simpler matrix commutator. (A3.6) The Solovay\u2013Kitaev theorem 621 (A3.7) (A3.8) Indeed, for qubits the matrix commutator has an especially nice form. An arbitrary element of SU (2) may be written U = u(a) \u2261 exp(\u2212ia \u00b7 \u03c3/2) for some real vector a. Similarly V = u(b) = exp(\u2212ib \u00b7 \u03c3/2) for some real vector b. Recalling from Exercise 2.40 that [a \u00b7 \u03c3,b \u00b7 \u03c3] = 2i ( \u2019 a \u00d7 b \u00b7 \u03c3, we see from (A3.6) that \u2019 D [U, V ]gp, u (( \u2019 a \u00d7 b = O", "(3). The basic idea of the proof of Lemma A3.2 is now easy to understand. The details, most of which relate to approximation issues, are \ufb01lled in below for completeness; for now we just give the main idea, as illustrated in Figure A3.2. Suppose we wish to approximate some element U = u(x) in S2. You will see in Exercise A3.4 that trace distances like D(U, I) are equal (up to small corrections) to the Euclidean distance x, so to a good approximation x \u2264 2. We can always choose y and z of length at most, such that x = y \u00d7z. Pick y0 and z0 such that u(y0) and u(z0) are elements of Gl that 2-approximate u(y) and u(z), respectively. Applying (A3.6) to the commutator [u(y0), u(z0)]gp we get an O(3)-approximation to U. This gives a O(3)-net for S2; to \ufb01nish the proof of the lemma we apply a translation step like that in the main part of the proof of the Solovay\u2013Kitaev theorem, obtaining in 5l gates an O(3)-approximation to any element of SO(3/2). 2 C3 2 Figure A3.2. The main idea in the proof of Lemma A3.2. Taking group commutators of elements U1 and U2 in S \ufb01lls in S2 much more densely. Note that the density of circles appearing on the right hand side ought to be much higher than is shown, as there should be one for each pair of circles on the left; the lower density is merely for clarity. The proof of the lemma is completed by applying a translation step (not shown) to get good approximations to any element of S\u221a 33/2. Exercise A3.2: Suppose A and B are Hermitian matrices such that tr|A|, tr|B| \u2264. Prove that for all suf\ufb01ciently small, \u2019 D e\u2212iA, e\u2212iB (, e\u2212[A,B] gp \u2264 d3, (A3.9) 622 The Solovay\u2013Kitaev theorem for some constant d, establishing Equation (A3.6). (Comment: for practical purposes it may be interesting to obtain good bounds on d.) Exercise A3.3: Let x and y be any two real vectors. Show that D(u(x), u(y)) = 2 \u221a 2 1 \u2212 cos(x/2) cos(y/2) \u2212 sin(x/2) sin(y/2) \u02c6x \u00b7 \u02c6y, (A3.10) where x \u2261 x, y \u2261 y, and \u02c6x and \u02c6y are unit vectors in the x and y directions, respectively. Exercise A3.4: Show that in the case y = 0 the formula for D(u(x), u(y)) reduces to D(u(x), I) = 4 sin Exercise A3.5: Show that when xA3.11) D(u(x), u(y)) = x \u2212 y + O(3). (A3.12) Proof (Lemma A3.2) Suppose Gl is an 2-net in S. The \ufb01rst step of the proof is to show that [Gl, Gl]gp is a C3-net for S2 and some constant C. Let U \u2208 S2 and pick x such that U = u(x). By Exercise A3.4 it follows that x \u2264 2 + O(6). Choose any pair of vectors y and z of length at most + O(5) such that x = y \u00d7 z. Gl is an 2-net for S, so choose U1 and U2 in Gl \u2229 S such that D(U1, u(y)) < 2 + O(5) D(U2, u(z)) < 2 + O(5), (A3.13) (A3.14) and let y0 and z0 be chosen such that U1 = u(y0) and U2 = u(z0). By Exercise A3.4 it follows that y0, z0 \u2264 + O(3). Our goal is to show that D(U, [U1, U2]gp) is smaller than C3. To do this, we use the triangle inequality, D(U, [U1, U2]gp) \u2264 D(U, u(y0 \u00d7 z0)) + D(u(y0 \u00d7 z0", "), [U1, U2]gp). The second term is at most d3, by Exercise A3.2, where d is a constant slightly larger than d because of the possible contribution due to the fact that y0, z0 < + O(3), rather than y0, z0 <. Substituting U = u(x), making use of Exercise A3.5, introducing an appropriate constant d and doing some elementary algebra gives (A3.15) D(U, [U1, U2]gp) \u2264 D(u(x), u(y0 \u00d7 z0) + d3 (A3.16) = x \u2212 y0 \u00d7 z0 + d3 = y \u00d7 z \u2212 y0 \u00d7 z0 + d3. (A3.18) = [(y \u2212 y0) + y0] \u00d7 [(z \u2212 z0) + z0] \u2212 y0 \u00d7 z0 + d3 (A3.19) \u2264 (d + 2)3 + O(4) (A3.20) \u2264 C3, (A3.17) (A3.21) where C is an appropriately chosen constant. The second step in the proof of the lemma is to apply a translation step like that used in the main part of the proof of the Solovay\u2013Kitaev theorem. Speci\ufb01cally, given Chapter problems 623 C3, we can \ufb01nd V in Gl such that D(U, V ) \u2264 2, and thus U V \u2020 \u2208 S2. Then U \u2208 S\u221a \ufb01nd W1 and W2 in Gl such that D([W1, W2]gp, U V \u2020) \u2264 C3, and therefore D([W1, W2]gpV, U ) \u2264 C3, (A3.22) which completes the proof. Exercise A3.6: Fixing the set G of elementary gates, describe an algorithm which, given a description of a single qubit unitary gate U and a desired accuracy > 0, ef\ufb01ciently computes a sequence of gates from G that -approximates U. The analysis in this appendix is rather crude, and a much tighter analysis can be made. One issue of especial interest is the best possible value of the exponent c in the O(logc(1/)) bound. It is not dif\ufb01cult to show that c can be no less than 1. To see this, imagine we have a collection of N little balls, all of radius, in SU (2). The volume of these balls scales like d, for some unimportant constant d. Therefore, if the balls are to cover SU (2), N must be of size \u03a9(1/d). Suppose we consider all possible sequences U1U2... Ug consisting of g gates chosen from G. Clearly such sequences can generate at most |G|g distinct unitary operations. Thus we must have |G|g = \u03a9(1/d), which implies the desired lower bound on the number of gates, g = \u03a9 log. 1 (A3.23) Problem 3.1: The following problem outlines a more elaborate construction that achieves an O(log2(1/) logc(log(1/))) bound on the number of gates required to approximate to within of a desired target, for any c > 2. (1) Suppose N is a \u03b4-net in S, for 0 < \u03b4 < \u2264 0, 0 suf\ufb01ciently small. Show that [N, N ]gp is a d\u03b4-net in S2, for some constant d. (2) Suppose Gl is a \u03b4-net in S, for 0 < \u03b4 < \u2264 0. Show that G4kl is a dk\u03b42k\u22121-net in S2k. (3) Suppose we de\ufb01ne k by 4 k \u2261 log 5, log(1/) log(1/0) and suppose we can \ufb01nd l such that Gl is a \u03b40-net for S0, where dk\u03b40 = 0. (A3.24) (A3.25) Show that G4kl is an -net for S 2k 0. (4) Use the already-proved version of the Solovay\u2013Kitaev theorem to show that choosing l = O(kc) suf\ufb01ces in the previous part of this problem, where c = log(5)/ log(3/2) is the constant appearing in the exponent in the already-proved version of the Solovay\u2013Kitaev theorem. (5) Combine the previous results to prove that O(log2(1/) logc(log(1/))) gates can", " be used to -approximate an arbitrary gate in SU (2). (6) Show that any c > 2 can appear in the conclusion of the previous result. 624 The Solovay\u2013Kitaev theorem Problem 3.2: (Research) If it exists, \ufb01nd an approximation procedure asympoticially faster than the result found in the previous problem. Ideally, a procedure would (a) saturate the \u03a9(log(1/)) lower bound on the number of gates required to perform the approximation, and (b) provide an ef\ufb01cient algorithm for constructing such approximating sequences of gates. Problem 3.3: (Research) Fix a \ufb01nite set of single qubit gates G which can be performed fault-tolerantly and which generate a set dense in the single qubit gates; say the \u03c0/8 gate and the Hadamard gate. Develop an elegant, ef\ufb01cient and reasonably tight method which, given an arbitrary single qubit gate U and some > 0, produces a sequence of gates from the fault-tolerant set giving an -approximation to U, up to global phase. History and further reading The results in this appendix were proved by Solovay in 1995 (unpublished manuscript), and independently by Kitaev, who gave an outline of the proof in [Kit97b]. In the same paper Kitaev observed that the result can be generalized to many Lie groups other than SU (2); roughly speaking, the key fact about SU (2) used in the proof was that [S, S]gp \u2287 S\u03a9(2), and other Lie groups for which this fact holds also obey some version of the Solovay\u2013Kitaev theorem. The Solovay\u2013Kitaev theorem is true, for example, for the Lie group SU (d) of d by d unitary matrices with unit determinant. After hearing of this result, Solovay subsequently generalized his proof in a similar fashion. Our presentation has bene\ufb01ted substantially from a 1999 lecture of Freedman, and by discussions with Freedman, Kitaev and Solovay. Appendix 4: Number theory Understanding some elementary number theory is necessary if we are to understand cryptosystems and how quantum computers can be used to break them. In this appendix we review some basic facts about number theory. A4.1 Fundamentals Let\u2019s start off by agreeing about a few conventions for nomenclature and notation. The set of integers is the set {..., \u22122, \u22121, 0, 1, 2,...}, denoted Z. We may occasionally refer to the natural numbers, meaning non-negative integers, but more often we\u2019ll say nonnegative integer or positive integer, in order to make the distinction between the case when zero is included, and when zero is not included. Suppose n is an integer. An integer d divides n (written d|n) if there exists an integer k such that n = dk. We say in this case that d is a factor or divisor of n. Notice that 1 and n are always factors of n. When d does not divide (is not a factor of) n we write d | n. For example, 3|6 and 3|18, but 3 | 5 and 3 | 7. Exercise A4.1: (Transitivity) Show that if a|b and b|c then a|c. Exercise A4.2: Show that if d|a and d|b then d also divides linear combinations of a and b, ax + by, where x and y are integers. Exercise A4.3: Suppose a and b are positive integers. Show that if a|b then a \u2264 b. Conclude that if a|b and b|a then a = b. A prime number is an integer greater than 1 which has only itself and 1 as factors. The \ufb01rst few prime numbers are 2, 3, 5, 7, 11, 13, 17,.... Perhaps the most important single fact about the positive integers is that they may be represented uniquely as a product of factors which are prime numbers. This result is given an appropriately impressive name, the fundamental theorem of arithmetic: Theorem A4.1: (Fundamental theorem of arithmetic) Let a be any integer greater than 1. Then a has a prime factorization of the form a = pa1 1 where p1,..., pn are distinct prime numbers, and a1,..., an are positive integers. Moreover, this prime factorization is unique, up to the order of the factors.... pan n, pa2 2 (A4.1) 626 Number theory Proof The reader who has never seen a proof of the fundamental theorem of arithmetic is strongly encouraged to attempt to supply it for", " themselves. Failing that, a proof may be found in any elementary number theory text; see the end of appendix \u2018History and further reading\u2019 for references. For small numbers it is easy to \ufb01nd the prime factorization by trial and error, for example 20 = 22 \u00b7 51. For large numbers no ef\ufb01cient algorithm is known to \ufb01nd the prime factorization on a classical computer, despite immense effort aimed at \ufb01nding such an algorithm. Exercise A4.4: Find the prime factorizations of 697 and 36 300. A4.2 Modular arithmetic and Euclid\u2019s algorithm We\u2019re all thoroughly familiar with the techniques of ordinary arithmetic. Another type of arithmetic, modular arithmetic, is extremely useful in understanding the properties of numbers. We assume that you are familiar with the elementary ideas of modular arithmetic, and so will quickly breeze through the basic ideas and notation, before coming to more advanced theory. Modular arithmetic can be thought of as the arithmetic of remainders. If we divide 18 by 7 we get the answer 2, with a remainder of 4. More formally, given any positive integers x and n, x can be written (uniquely) in the form x = kn + r, (A4.2) where k is a non-negative integer, the result of dividing x by n, and the remainder r lies in the range 0 to n \u2212 1, inclusive. Modular arithmetic is simply ordinary arithmetic in which we only pay attention to remainders. We use the notation (mod n) to indicate that we are working in modular arithmetic. For instance, we write 2 = 5 = 8 = 11(mod 3), because 2, 5, 8 and 11 all have the same remainder (2) when divided by 3. The appellation \u2018(mod n)\u2019 reminds us that we are working in modular arithmetic, with respect to the number n. Addition, multiplication, and subtraction operations for modular arithmetic may all be de\ufb01ned in the obvious ways, but it is perhaps not so obvious how to de\ufb01ne a division operation. To understand how this may be done we introduce another key concept from number theory, that of the greatest common divisor of two integers. The greatest common divisor of integers a and b is the largest integer which is a divisor of both a and b. We write this number as gcd(a, b). For example, the greatest common divisor of 18 and 12 is 6. An easy way of seeing this is to enumerate the positive divisors of 18 (1, 2, 3, 6, 9, 18) and 12 (1, 2, 3, 4, 6, 12), and then pick out the largest common element in the two lists. This method is quite inef\ufb01cient and impractical for large numbers. Fortunately, there is a much more ef\ufb01cient way of working out the greatest common divisor, a method known as Euclid\u2019s algorithm, whose explication occupies us for the next few pages. Theorem A4.2: (Representation theorem for the gcd) The greatest common divisor of two integers a and b is the least positive integer that can be written in the form ax + by, where x and y are integers. Modular arithmetic and Euclid\u2019s algorithm 627 Proof Let s = ax + by be the smallest positive integer that can be written in this form. Since gcd(a, b) is a divisor of both a and b it is also a divisor of s. It follows that gcd(a, b) \u2264 s. To complete the proof we demonstrate that s \u2264 gcd(a, b) by showing that s is a divisor of both a and b. The proof is by contradiction. Suppose s is not a divisor of a. Then a = ks + r, where the remainder r is in the range 1 to s \u2212 1. Rearranging this equation and using s = ax + by we see that r = a(1 \u2212 kx) + b(\u2212ky) is a positive integer that can be written as a linear combination of a and b, and which is smaller than s. But this contradicts the de\ufb01nition of s as the smallest positive integer that can be written as a linear combination of a and b. We conclude that s must divide a. By symmetry s must also be a divisor of b, which completes the proof. Corollary A4.3: Suppose c divides both a and b. Then c divides gcd(a, b). Proof By Theorem A4.2, gcd(a, b) = ax + by for some integers x and y. Since c divides a and b it must also divide ax + by. When does a number, a", ", have a multiplicative inverse in modular arithmetic? That is, given a and n, when does there exist a b such that ab = 1(mod n)? For example, note that 2 \u00b7 3 = 1(mod 5), so the number 2 has multiplicative inverse 3 in arithmetic modulo 5. On the other hand, trial and error shows that 2 has no multiplicative inverse modulo 4. Finding multiplicative inverses in modular arithmetic turns out to be related to the gcd by the notion of co-primality: integers a and b are said to be co-prime if their greatest common divisor is 1. For example, 14 and 9 are co-prime, since the positive divisors of 14 are 1, 2, 7 and 14, while 1, 3 and 9 are the positive divisors of 9. The following corollary characterizes the existence of multiplicative inverses in modular arithmetic using co-primality. Corollary A4.4: Let n be an integer greater than 1. An integer a has a multiplicative inverse modulo n if and only if gcd(a, n) = 1, that is, a and n are co-prime. Proof Suppose a has a multiplicative inverse, which we denote a\u22121, modulo n. Then aa\u22121 = 1 + kn for some integer k, and thus aa\u22121 + (\u2212k)n = 1. From Theorem A4.2 we conclude that gcd(a, n) = 1. Conversely, if gcd(a, n) = 1 then there must exist integers a\u22121 and b such that aa\u22121 + bn = 1, and therefore aa\u22121 = 1(mod n). Exercise A4.5: For p a prime prove that all integers in the range 1 to p \u2212 1 have multiplicative inverses modulo p. Which integers in the range 1 to p2 \u2212 1 do not have multiplicative inverses modulo p2? Exercise A4.6: Find the multiplicative inverse of 17 modulo 24. Exercise A4.7: Find the multiplicative inverse of n + 1 modulo n2, where n is any integer greater than 1. Exercise A4.8: (Uniqueness of the inverse) Suppose b and b are multiplicative inverses of a, modulo n. Prove that b = b(mod n). 628 Number theory The next theorem is the key to Euclid\u2019s ef\ufb01cient algorithm for \ufb01nding the greatest common divisor of two positive integers. Theorem A4.5: Let a and b be integers, and let r be the remainder when a is divided by b. Then provided r = 0, gcd(a, b) = gcd(b, r). (A4.3) Proof We prove the equality by showing that each side divides the other. To prove that the left hand side divides the right note that r = a\u2212kb for some integer k. Since gcd(a, b) divides a, b and linear combinations of these it follows that gcd(a, b) divides r. By Corollary A4.3, gcd(a, b) divides gcd(b, r). To prove that the right hand side divides the left note that gcd(b, r) divides b, and since a = r + kb is a linear combination of b and r it follows that gcd(b, r) also divides a. By Corollary A4.3, gcd(b, r) divides gcd(a, b). Exercise A4.9: Explain how to \ufb01nd gcd(a, b) if the prime factorizations of a and b are known. Find the prime factorizations of 6825 and 1430, and use them to compute gcd(6825, 1430) Euclid\u2019s algorithm for \ufb01nding the greatest common divisor of positive integers a and b works as follows. First, order a and b so that a > b. Divide b into a, with result k1 and remainder r1: a = k1b + r1. By Theorem A4.5 gcd(a, b) = gcd(b, r1). Next, we perform a second division with b playing the role of a, and r1 playing the role of b: b = k2r1 + r2. By Theorem A4.5 gcd(a, b) = gcd(b, r1) = gcd(r1, r2). Next, we perform a third division with r1 playing the role of a and r2 the role of b: r1 = k3r2 + r3. By Theorem A4.5 gcd(a, b", ") = gcd(b, r1) = gcd(r1, r2) = gcd(r2, r3). We continue in this manner, each time dividing the most recent remainder by the second most recent remainder, obtaining a new result and remainder. The algorithm halts when we obtain a remainder that is zero, that is, rm = km+1rm+1 for some m. We have gcd(a, b) = gcd(rm, rm+1) = rm+1, so the algorithm returns rm+1. As an example of the use of Euclid\u2019s algorithm we \ufb01nd gcd(6825, 1430): 6825 = 4 \u00d7 1430 + 1105 1430 = 1 \u00d7 1105 + 325 1105 = 3 \u00d7 325 + 130 325 = 2 \u00d7 130 + 65 130 = 2 \u00d7 65. (A4.4) (A4.5) (A4.6) (A4.7) (A4.8) From this we see that gcd(6825, 1430) = 65. An adaptation of Euclid\u2019s algorithm may be used to ef\ufb01ciently \ufb01nd integers x and y such that ax + by = gcd(a, b). The \ufb01rst stage is to run through the steps of Euclid\u2019s algorithm, as before. The second stage begins at the second last line of the running of Euclid\u2019s algorithm, and involves successive substitution of the lines higher up in the algorithm as illustrated by the following example: 65 = 325 \u2212 2 \u00d7 130 (A4.9) Modular arithmetic and Euclid\u2019s algorithm 629 = 325 \u2212 2 \u00d7 (1105 \u2212 3 \u00d7 325) = \u22122 \u00d7 1105 + 7 \u00d7 325 = \u22122 \u00d7 1105 + 7 \u00d7 (1430 \u2212 1 \u00d7 1105) = 7 \u00d7 1430 \u2212 9 \u00d7 1105 = 7 \u00d7 1430 \u2212 9 \u00d7 (6825 \u2212 4 \u00d7 1430) = \u22129 \u00d7 6825 + 37 \u00d7 1430. (A4.10) (A4.11) (A4.12) That is, 65 = 6825 \u00d7 (\u22129) + 1430 \u00d7 37, which is the desired representation. What resources are consumed by Euclid\u2019s algorithm? Suppose a and b may be represented as bit strings of at most L bits each. It is clear that none of the divisors ki or remainders ri can be more than L bits long, so we may assume that all computations are done in L bit arithmetic. The key observation to make in a resource analysis is that ri+2 \u2264 ri/2. To prove this we consider two cases: \u2022 ri+1 \u2264 ri/2. It is clear that ri+2 \u2264 ri+1 so we are done. \u2022 ri+1 > ri/2. In this case ri = 1 \u00d7 ri+1 + ri+2, so ri+2 = ri \u2212 ri+1 \u2264 ri/2. Since ri+2 \u2264 ri/2, it follows that the divide-and-remainder operation at the heart of Euclid\u2019s algorithm need be performed at most 2log a = O(L) times. Each divide-andremainder operation requires O(L2) operations, so the total cost of Euclid\u2019s algorithm is O(L3). Finding x and y such that ax + by = gcd(a, b) incurs a minor additional cost: O(L) substitutions are performed, at a cost of O(L2) per substitution to do the arithmetic involved, for a total resource cost of O(L3). Euclid\u2019s algorithm may also be used to ef\ufb01ciently \ufb01nd multiplicative inverses in modular arithmetic. This is implicit in the proof of Corollary A4.4; we now make it explicit. Suppose a is co-prime to n, and we wish to \ufb01nd a\u22121, modulo n. To do so, use Euclid\u2019s algorithm and the co-primality of a and n to \ufb01nd integers x and y such that ax + ny = 1. (A4.13) Note then that ax = (1 \u2212 ny) = 1(mod n), that is, x is the multiplicative inverse of a, modulo n. Furthermore, this algorithm is computationally ef\ufb01cient, taking only O(L3) steps, where L is the length in bits of n. Now that we know how to ef\ufb01ciently \ufb01nd inverses in modular arithmetic, it is only a short step to solve simple linear equations, such", " as ax + b = c(mod n). (A4.14) Suppose a and n are co-prime. Then using Euclid\u2019s algorithm we may ef\ufb01ciently \ufb01nd the multiplicative inverse a\u22121 of a, modulo n, and thus the solution to the previous equation, x = a\u22121(c \u2212 b)(mod n). (A4.15) An important result known as the Chinese remainder theorem extends the range of equations we may solve much further, allowing us to ef\ufb01ciently solve systems of equations in modular arithmetic. Theorem A4.6: (Chinese remainder theorem) Suppose m1,..., mn are positive integers such that any pair mi and mj (i = j) are co-prime. Then the system of equations x = a1(mod m1) x = a2(mod m2A4.16) (A4.17) 630 Number theory x = an(mod mn) (A4.18) has a solution. Moreover, any two solutions to this system of equations are equal modulo M \u2261 m1m2... mn. Proof The proof is to explicitly construct a solution to the system of equations. De\ufb01ne Mi \u2261 M/mi and observe that mi and Mi are co-prime. It follows that Mi has an inverse i aiMiNi. To see that x is a solution to modulo mi, which we denote Ni. De\ufb01ne x \u2261 the system of equations, note that MiNi = 1(mod mi) and MiNi = 0(mod mj) when i = j, so x = ai(mod mi), which demonstrates the existence of a solution. Suppose x and x are both solutions to the system of equations. It follows that x\u2212x = 0(mod mi) for each i, and thus mi divides x \u2212 x for each i. Since the mi are co-prime, it follows that the product M = m1... mn also divides x \u2212 x, so x = x(mod M ), as we set out to show. Euclid\u2019s algorithm and the Chinese remainder theorem are two of the signal triumphs of algorithmic number theory. How ironic then that they should play a role in the sequence of ideas leading up to the RSA cryptosystem, whose presumed security is based on the dif\ufb01culty of performing certain algorithmic tasks in number theory. Nevertheless, this is indeed the case! We turn now to the number-theoretic background necessary to understand the RSA cryptosystem. The key ideas are a famous result of classical number theory, Fermat\u2019s little theorem \u2013 not to be confused with Fermat\u2019s last theorem \u2013 and a generalization of Fermat\u2019s little theorem due to Euler. The proof of Fermat\u2019s little theorem relies on the following elegant lemma. Lemma A4.7: Suppose p is prime and k is an integer in the range 1 to p \u2212 1. Then p divides p k. Proof Consider the identity p(p \u2212 1) \u00b7 \u00b7 \u00b7 (p \u2212 k + 1) = p k k(k \u2212 1) \u00b7 \u00b7 \u00b7 1. (A4.19) Since k \u2265 1 the left hand side (and thus the right) is divisible by p. Since k \u2264 p \u2212 1 the term k(k \u2212 1) \u00b7 \u00b7 \u00b7 1 is not divisible by p. It follows that must be divisible by p. p k Theorem A4.8: (Fermat\u2019s little theorem) Suppose p is a prime, and a is any integer. Then ap = a(mod p). If a is not divisible by p then ap\u22121 = 1(mod p). Proof The second part of the theorem follows from the \ufb01rst, since if a is not divisible by p then a has an inverse modulo p, so ap\u22121 = a\u22121ap = a\u22121a = 1(mod p). We prove the \ufb01rst part of the theorem for positive a (the case of non-positive a follows easily) by induction on a. When a = 1 we have ap = 1 = a(mod p), as required. Suppose the result holds true Modular arithmetic and Euclid\u2019s algorithm 631 for a, that is, ap = a(mod p) and consider the case of a + 1. By the binomial expansion, p (A4.20) (1 + a)p = p k ak. k=0 p k By Lemma A4.7, p divides whenever 1 \u2264 k \u2264 p \u2212 1, so all terms except the \ufb01rst and last vanish from the sum modulo p, (1 + a)p = (1 + ap)(", "mod p). Applying the inductive hypothesis ap = a(mod p) we see that (1 + a)p = (1 + a)(mod p), as required. There is a remarkable generalization of Fermat\u2019s little theorem due to Euler, based on the Euler \u03d5 function. \u03d5(n) is de\ufb01ned to be the number of positive integers less than n which are co-prime to n. As an example, note that all positive integers less than a prime p are co-prime to p, and thus \u03d5(p) = p \u2212 1. The only integers less than p\u03b1 which are not co-prime to p\u03b1 are the multiples of p: p, 2p, 3p,..., (p\u03b1\u22121 \u2212 1)p, from which we deduce \u03d5(p\u03b1) = (p\u03b1 \u2212 1) \u2212 (p\u03b1\u22121 \u2212 1) = p\u03b1\u22121(p \u2212 1). (A4.21) Furthermore, if a and b are co-prime, then the Chinese remainder theorem can be used to show that \u03d5(ab) = \u03d5(a)\u03d5(b). (A4.22) To see this, consider the system of equations x = xa(mod a), x = xb(mod b). Applying the Chinese remainder theorem to this set of equations we see that there is a one-to-one correspondence between pairs (xa, xb) such that 1 \u2264 xa < a, 1 \u2264 xb < b, gcd(xa, a) = 1, gcd(xb, b) = 1, and integers x such that 1 \u2264 x < ab, gcd(x, ab) = 1. There are \u03d5(a)\u03d5(b) such pairs (xa, xb) and \u03d5(ab) such x, from which we deduce (A4.22). Equations (A4.21) and (A4.22) together imply a formula for \u03d5(n) based on the prime factorization of n, n = p\u03b11 1 \u00b7 \u00b7 \u00b7 p\u03b1k k : \u03d5(n) = k/ j=1 p\u03b1j \u22121 j (pj \u2212 1). Exercise A4.10: What is \u03d5(187)? Exercise A4.11: Prove that n = d|n \u03d5(d), (A4.23) (A4.24) where the sum is over all positive divisors d of n, including 1 and n. (Hint: Prove the result for n = p\u03b1 \ufb01rst, then use the multiplicative property (A4.22) of \u03d5 to complete the proof.) Fermat\u2019s little theorem has the following beautiful generalization, due to Euler: Theorem A4.9: Suppose a is co-prime to n. Then a\u03d5(n) = 1(mod n). Proof We \ufb01rst show by induction on \u03b1 that a\u03d5(p\u03b1) = 1(mod p\u03b1). For \u03b1 = 1 the result is just Fermat\u2019s little theorem. Assume the result is true for \u03b1 \u2265 1, so a\u03d5(p\u03b1) = 1 + kp\u03b1, (A4.25) 632 Number theory for some integer k. Then by (A4.21), a\u03d5(p\u03b1+1) = ap\u03b1(p\u22121) = ap\u03d5(p\u03b1) = (1 + kp\u03b1)p p p j = 1 + j=1 kjpj\u03b1. Using Lemma A4.7 it is easy to see that p\u03b1+1 divides every term in the sum, so a\u03d5(p\u03b1+1) = 1(mod p\u03b1+1), (A4.26) (A4.27) (A4.28) (A4.29) (A4.30) \u00b7 \u00b7 \u00b7 p\u03b1m which completes the induction. The proof of the theorem is completed by noting that for j ) for each j, as \u03d5(n) is a multiple of \u03d5(p\u03b1j arbitrary n = p\u03b11 j ). 1 Applying the construction in the proof of the Chinese remainder theorem we \ufb01nd that any solution to the set of equations x = 1(mod p\u03b1j j ) must satisfy x = 1(mod n), and thus a\u03d5(n) = 1(mod n). m, a\u03d5(n) = 1(mod p\u03b1j De\ufb01ne Z\u2217 n are in Z\u2217 n to be the set of all elements in Zn which have inverses modulo n, that is, the set of all elements in Zn which are co-prime to", " n. Z\u2217 n is easily seen to form a group of size \u03d5(n) under multiplication, that is, it contains the multiplicative identity, products of elements in Z\u2217 n is closed under the multiplicative inverse operation. (For an overview of elementary group theory, see Appendix 2.) What is not so obvious is n has when n is a power of an odd prime p, n = p\u03b1. It turns the remarkable structure Z\u2217 out that Z\u2217 p\u03b1 in the sense that any other element x may be written x = gk(mod n) for some non-negative integer k. p\u03b1 is a cyclic group, that is, there is an element g in Z\u2217 p\u03b1 which generates Z\u2217 n, and Z\u2217 Theorem A4.10: Let p be an odd prime, \u03b1 a positive integer. Then Z\u2217 p\u03b1 is cyclic. Proof The proof of this fact is a little beyond the scope of this book. It can be found in many texts containing any considerable amount of number theory. See for example Section 3.2 of Knuth[Knu98a], especially pages 16 through 23. Exercise A4.12: Verify that Z\u2217 multiplication modulo n. n forms a group of size \u03d5(n) under the operation of Exercise A4.13: Let a be an arbitrary element of Z\u2217 n. Show that S \u2261 {1, a, a2,...} n, and that the size of S is the least value of r such that forms a subgroup of Z\u2217 ar = 1(mod n). Exercise A4.14: Suppose g is a generator for Z\u2217 n. Show that g must have order \u03d5(n). Exercise A4.15: Lagrange\u2019s theorem (Theorem A2.1 on page 610) is an elementary result of group theory stating that the size of a subgroup must divide the order of the group. Use Lagrange\u2019s theorem to provide an alternate proof of Theorem A4.9, that is, show that a\u03d5(n) = 1(mod n) for any a \u2208 Z\u2217 n. Reduction of factoring to order-\ufb01nding 633 A4.3 Reduction of factoring to order-\ufb01nding The problem of factoring numbers on a classical computer turns out to be equivalent to another problem, the order-\ufb01nding problem. This equivalence is important as it turns out that quantum computers are able to quickly solve the order-\ufb01nding problem, and thus can factor quickly. In this section we explain the equivalence between these two problems, focusing on the reduction of factoring to order-\ufb01nding. Suppose N is a positive integer, and x is co-prime to N, 1 \u2264 x < N. The order of x modulo N is de\ufb01ned to be the least positive integer r such that xr = 1(mod N ). The order-\ufb01nding problem is to determine r, given x and N. Exercise A4.16: Use Theorem A4.9 to show that the order of x modulo N must divide \u03d5(N ). The reduction of factoring to order-\ufb01nding proceeds in two basic steps. The \ufb01rst step is to show that we can compute a factor of n if we can \ufb01nd a non-trivial solution x = \u00b1 1(mod N ) to the equation x2 = 1(mod N ). The second step is to show that a randomly chosen y co-prime to N is quite likely to have an order r which is even, and such that yr/2 = \u00b1 1(mod N ), and thus x \u2261 yr/2(mod N ) is a solution to x2 = 1(mod N ). Theorem A4.11: Suppose N is a composite number L bits long, and x is a non-trivial solution to the equation x2 = 1(mod N ) in the range 1 \u2264 x \u2264 N, that is, neither x = 1(mod N ) nor x = N \u2212 1 = \u22121(mod N ). Then at least one of gcd(x \u2212 1, N ) and gcd(x + 1, N ) is a non-trivial factor of N that can be computed using O(L3) operations. Proof Since x2 = 1(mod N ), it must be that N divides x2 \u22121 = (x+1)(x\u22121), and thus N must have a common factor with one or the other of (x + 1) and (x \u2212 1). But 1 < x < N \u2212 1 by assumption, so x \u2212 1 < x + 1 < N, from which we see that the common factor can not be N", " itself. Using Euclid\u2019s algorithm we may compute gcd(x \u2212 1, N ) and gcd(x + 1, N ) and thus obtain a non-trivial factor of N, using O(L3) operations. Lemma A4.12: Let p be an odd prime. Let 2d be the largest power of 2 dividing \u03d5(p\u03b1). Then with probability exactly one-half 2d divides the order modulo p\u03b1 of a randomly chosen element of Z\u2217 p\u03b1. Proof Note that \u03d5(p\u03b1) = p\u03b1\u22121(p\u22121) is even, since p is odd, and thus d \u2265 1. By Theorem A4.10 there exists a generator g for Z\u2217 p\u03b1, so an arbitrary element may be written in the form gk(mod p\u03b1) for some k in the range 1 through \u03d5(p\u03b1). Let r be the order of gk modulo p\u03b1 and consider two cases. The \ufb01rst case is when k is odd. From gkr = 1(mod p\u03b1) we deduce that \u03d5(p\u03b1)|kr, and thus 2d|r, since k is odd. The second case is when k is even. Then \u2019 (k/2 gk\u03d5(p\u03b1)/2 = g\u03d5(p\u03b1) = 1k/2 = 1(mod p\u03b1). (A4.31) Thus r|\u03d5(p\u03b1)/2 from which we deduce that 2d does not divide r. 634 Number theory Summarizing, Z\u2217 p\u03b1 may be partitioned into two sets of equal size: those which may be written gk with k odd, for which 2d|r, where r is the order of gk, and those which may be written gk with k even, for which 2d | r. Thus with probability 1/2 the integer 2d divides the order r of a randomly chosen element of Z\u2217 p\u03b1, and with probability 1/2 it does not. Theorem A4.13: Suppose N = p\u03b11 1 m is the prime factorization of an odd composite positive integer. Let x be chosen uniformly at random from Z\u2217 let r be the order of x, modulo N. Then \u00b7 \u00b7 \u00b7 p\u03b1m N, and p(r is even and xr/2 = \u2212 1(mod N )) \u2265 1 \u2212 1 2m. (A4.32) Proof We show that p(r is odd or xr/2 = \u22121(mod N )) \u2264 1 2m. (A4.33) By the Chinese remainder theorem, choosing x uniformly at random from Z\u2217 N is equivalent to choosing xj independently and uniformly at random from Z\u2217, and requiring \u03b1j p j j ) for each j. Let rj be the order of xj modulo p\u03b1j that x = xj(mod p\u03b1j j. Let 2dj be the largest power of 2 that divides rj and 2d be the largest power of 2 that divides r. We will show that to have r odd or xr/2 = \u22121(mod N ) it is necessary that dj takes the same value for all values of j. The result then follows, as from Lemma A4.12 the probability of this occurring is at most 1/2m. The \ufb01rst case we consider is when r is odd. It is easy to see that rj|r for each j, and therefore rj is odd, so dj = 0 for all i = 1,..., k. The second and \ufb01nal case is when r is even and xr/2 = \u22121(mod N ). Then xr/2 = \u22121(mod p\u03b1j j ), so rj | (r/2). Since rj|r we must have dj = d for all j. Theorems A4.11 and A4.13 can be combined to give an algorithm which, with high probability, returns a non-trivial factor of any composite N. All the steps in the algorithm can be performed ef\ufb01ciently on a classical computer except (as far as is known today) an order-\ufb01nding \u2018subroutine\u2019 which is used by the algorithm. By repeating the algorithm we may \ufb01nd a complete prime factorization of N. The algorithm is summarized below. (1) If N is even, return the factor 2. (2) Use the algorithm of Exercise 5.17 to determine whether N = ab for integers a \u2265 1 and b \u2265 2, and if so return the factor a. (3) Randomly choose x in the range 1 to N \u2212 1. If gcd(x, N ) > 1 then return the", " factor gcd(x, N ). (4) Use the order-\ufb01nding subroutine to \ufb01nd the order r of x, modulo N. (5) If r is even and xr/2 = \u2212 1(mod N ) then compute gcd(xr/2 \u2212 1, N ) and gcd(xr/2 + 1, N ), and test to see which is a non-trivial factor, returning that factor. Otherwise, the algorithm fails. Steps 1 and 2 of the algorithm either return a factor, or else ensure that N is an odd integer with more than one prime factor. These steps may be performed using O(1) and O(L3) operations, respectively. Step 3 either returns a factor, or produces a randomly Continued fractions 635 chosen element x of Z\u2217 N. Step 4 calls the order-\ufb01nding subroutine, computing the order r of x, modulo N. Step 5 completes the algorithm, since Theorem A4.13 guarantees that with probability at least one-half r will be even and xr/2 = \u2212 1(mod N ), and then Theorem A4.11 guarantees that either gcd(xr/2 \u22121, N ) or gcd(xr/2 +1, N ) is a non-trivial factor of N. Exercise A4.17: (Reduction of order-\ufb01nding to factoring) We have seen that an ef\ufb01cient order-\ufb01nding algorithm allows us to factor ef\ufb01ciently. Show that an ef\ufb01cient factoring algorithm would allow us to ef\ufb01ciently \ufb01nd the order modulo N of any x co-prime to N. A4.4 Continued fractions There are many remarkable connections between the continuum of real numbers and the integers. One such connection is the beautiful theory of continued fractions. In this section we develop a few elements of the theory of continued fractions, elements crucial to the application of the fast quantum algorithms for order-\ufb01nding and factoring detailed in Chapter 5. As an example of a continued fraction, consider the number s de\ufb01ned by the expression s \u2261 1 2 + 1 2+ 1 2+\u00b7\u00b7\u00b7. (A4.34) \u221a Informally, note that s = 1/(2+s), from which it is easy to satisfy oneself that s = 2\u22121. The idea of the continued fractions method is to describe real numbers in terms of integers alone, using expressions such as (A4.34). A \ufb01nite simple continued fraction is de\ufb01ned by a \ufb01nite collection a0,..., aN of positive integers, [a0,..., aN ] \u2261 a0 + 1 a1 + a2+ 1 1 \u00b7\u00b7\u00b7+ 1 aN. (A4.35) We de\ufb01ne the nth convergent (0 \u2264 n \u2264 N ) to this continued fraction to be [a0,..., an]. Theorem A4.14: Suppose x is a rational number greater than or equal to one. Then x has a representation as a continued fraction, x = [a0,..., aN ], which may be found by the continued fractions algorithm. Proof The continued fractions algorithm is best understood by example. Suppose we are trying to decompose 31/13 as a continued fraction. The \ufb01rst step of the continued fractions algorithm is to split 31/13 into its integer and fractional part, Next we invert the fractional part, obtaining 31 13 = 2 + 31 13 = 2 +. 5 13. 1 13 5 (A4.36) (A4.37) 636 Number theory These steps \u2013 split then invert \u2013 are now applied to 13/5, giving 31 13 = Next we split and invert 5/3: 31 13 = 2 + 1 2 + 1 1+ 1 3 2 (A4.38) (A4.39) The decomposition into a continued fraction now terminates, since 3/2 = 1 + 1/2 may be written with a 1 in the numerator without any need to invert, giving a \ufb01nal continued fraction representation of 31/13 as 31 13 = 2 +. 1 2 + 1 1+ 1 1+ 1 2 (A4.40) It\u2019s clear that the continued fractions algorithm terminates after a \ufb01nite number of \u2018split and invert\u2019 steps for any rational number, since the numerators which appear (31, 3, 2, 1 in the example) are strictly decreasing. How quickly does this termination occur? We\u2019ll come back to that question shortly. The theorem above", " has been stated for x \u2265 1; however, in practice it is convenient to relax the requirement that a0 to be positive and allow it to be any integer, which results in the restriction x \u2265 1 becoming super\ufb02uous. In particular, if x is in the range 0 through 1 as occurs in applications to quantum algorithms, then the continued fraction expansion has a0 = 0. The continued fractions algorithm provides an unambiguous method for obtaining a continued fraction expansion of a given rational number. The only possible ambiguity comes at the \ufb01nal stage, because it is possible to split an integer in two ways, either an = an, or as an = (an \u2212 1) + 1/1, giving two alternate continued fraction expansions. This ambiguity is actually useful, since it allows us to assume without loss of generality that the continued fraction expansion of a given rational number has either an odd or even number of convergents, as desired. Exercise A4.18: Find the continued fraction expansion for x = 19/17 and x = 77/65. Theorem A4.15: Let a0,..., aN be a sequence of positive numbers. Then [a0,..., an] = pn qn, (A4.41) where pn and qn are real numbers de\ufb01ned inductively by p0 \u2261 a0, q0 \u2261 1 and p1 \u2261 1 + a0a1, q1 \u2261 a1, and for 2 \u2264 n \u2264 N, pn \u2261 anpn\u22121 + pn\u22122 qn \u2261 anqn\u22121 + qn\u22122. (A4.42) (A4.43) In the case where aj are positive integers, so too are the pj and qj. Continued fractions 637 Proof We induct on n. The result is easily checked directly for the cases n = 0, n = 1, 2. By de\ufb01nition, for n \u2265 3, [a0,..., an] = [a0,..., an\u22122, an\u22121 + 1/an]. (A4.44) Applying the inductive hypothesis, let \u02dcpj/ \u02dcqj be the sequence of convergents associated with the continued fraction on the right hand side: [a0,..., an\u22122, an\u22121 + 1/an] = \u02dcpn\u22121 \u02dcqn\u22121. It is clear that \u02dcpn\u22123 = pn\u22123, \u02dcpn\u22122 = pn\u22122 and \u02dcqn\u22123 = qn\u22123, \u02dcqn\u22122 = qn\u22122, so \u02dcpn\u22121 \u02dcqn\u22121 = = (an\u22121 + 1/an)pn\u22122 + pn\u22123 (an\u22121 + 1/an)qn\u22122 + qn\u22123 pn\u22121 + pn\u22122/an qn\u22121 + qn\u22122/an. Multiplying top and bottom of the right hand side by an we see that \u02dcpn\u22121 \u02dcqn\u22121 = pn qn. Combining Equations (A4.48), (A4.45) and (A4.44) gives [a0,..., an] = pn qn, as required. (A4.45) (A4.46) (A4.47) (A4.48) (A4.49) Exercise A4.19: Show that qnpn\u22121 \u2212 pnqn\u22121 = (\u22121)n for n \u2265 1. Use this fact to conclude that gcd(pn, qn) = 1. (Hint: Induct on n.) How many values of an must be determined to obtain a continued fraction expansion for a rational number x = p/q > 1, where p and q are co-prime? Suppose a0,..., aN are positive integers. From the de\ufb01nition of pn and qn it follows that pn and qn are increasing sequences. Therefore pn = anpn\u22121 + pn\u22122 \u2265 2pn\u22122 and similarly qn \u2265 2qn\u22122, from which it follows that pn, qn \u2265 2n/2. Thus the 2N/2 \u2264 q \u2264 p, and so N = O(log(p)). It follows that if x = p/q is a rational number, p and q are L bit integers, then the continued fraction expansion for x can be computed using O(L3) operations \u2013 O(L) \u2018split and invert\u2019 steps, each using O(L2) gates for elementary arithmetic. Theorem A4.16: Let x be", " a rational number and suppose p/q is a rational number such that + + + + p q \u2212 x + + + \u2264 1 + 2q2. (A4.50) Then p/q is a convergent of the continued fraction for x. Proof 638 Number theory Let p/q = [a0,..., an] be the continued fraction expansion for p/q, and de\ufb01ne pj, qj as in Theorem A4.15, so that pn/qn = p/q. De\ufb01ne \u03b4 by the equation x \u2261 pn qn + \u03b4 2q2 n, so |\u03b4| < 1. De\ufb01ne \u03bb by the equation \u03bb \u2261 2 qnpn\u22121 \u2212 pnqn\u22121 \u03b4 \u2212 qn\u22121 qn. (A4.51) (A4.52) The reason we de\ufb01ne \u03bb this way is because with a little algebra we can see that it satis\ufb01es the equation x = \u03bbpn + pn\u22121 \u03bbqn + qn\u22121, (A4.53) and therefore x = [a0,..., an, \u03bb]. Choosing n even we see from Exercise A4.19 that \u03bb = 2 \u03b4 \u2212 qn\u22121 qn. By the increasing property of qn it follows that \u03bb = 2 \u03b4 \u2212 qn\u22121 qn > 2 \u2212 1 > 1. (A4.54) (A4.55) Therefore \u03bb is a rational number greater than 1, and so has a simple \ufb01nite continued fraction, \u03bb = [b0,..., bm], and so x = [a0,..., an, b0,..., bm] is a simple \ufb01nite continued fraction for x with p/q as a convergent. Problem 4.1: (Prime number estimate) Let \u03c0(n) be the number of prime numbers which are less than n. A dif\ufb01cult-to-prove result known as the prime number theorem asserts that limn\u2192\u221e \u03c0(n) log(n)/n = 1 and thus \u03c0(n) \u2248 n/ log(n). This problem gives a poor man\u2019s version of the prime number theorem which gives a pretty good lower bound on the distribution of prime numbers. (1) Prove that n \u2264 log (2) Show that 2n n. 2n n \u2264 log? p\u22642n @ log(2n) log p log p, where the sum is over all primes p less than or equal to 2n. (3) Use the previous two results to show that \u03c0(2n) \u2265 n log(2n). (A4.56) (A4.57) History and further reading 639 History and further reading There are many excellent books on number theory. We have made considerable use of the excellent book by Koblitz[Kob94], which combines much introductory material about number theory, algorithms, and cryptography all in one location. A similar combination forming a small part of a much more comprehensive presentation oriented towards algorithms may be found in Chapter 33 of Cormen, Leiserson and Rivest[CLR90]. Our discussion of continued fractions is based upon Chapter 10 of the classic text on number theory by Hardy and Wright[HW60]. Problem 4.1 is adapted from Papadimitriou[Pap94]. Appendix 5: Public key cryptography and the RSA cryptosystem Cryptography is the art of enabling two parties to communicate in private. For example, a consumer wishing to make a purchase on the internet wants to transmit their credit card number over the internet in such a way that only the company they are purchasing from gains access to the number. Rather more ominously, in wartime each of the warring parties wants the means to carry on private communication. To achieve privacy a cryptographic protocol or cryptosystem is used. Effective cryptosystems make it easy for parties who wish to communicate to do so, but make it very dif\ufb01cult for third parties to \u2018eavesdrop\u2019 on the contents of the conversation. A particularly important class of cryptosystems are the public key cryptosystems. The basic idea of public key cryptography is illustrated by the analogy depicted in Figure A5.1. Alice sets up a mailbox with the property that anybody can send her mail, by putting it into the mailbox, but only she can retrieve mail out of the mailbox. To achieve this she gives the mailbox two doors. On top of the mailbox is a locked trap door. Any person able to open the trap door can drop mail into the box. However, the chute from the trap", " door into the box is one way, so they can\u2019t reach into the box and \ufb01sh mail out. Alice makes the key to the trapdoor freely available to the public \u2013 it is a public key \u2013 so that she can receive mail from absolutely anybody. On the front of the mail box is a second door, from which mail already inside the box can be retrieved. Alice is in possession of the sole key for that door; it is her own secret key. This arrangement \u2013 involving two keys, one secret and one public \u2013 allows anybody in the world to communicate with Alice while maintaining privacy. Public key cryptosystems operate according to similar principles. Suppose Alice wishes to receive messages using a public key cryptosystem. She must \ufb01rst generate two cryptographic keys, one a public key, P, the other a secret key, S. The exact nature of these keys depends on the details of the cryptosystem being used. Some cryptosystems use simple objects like numbers as keys, while other cryptosystems use much more complicated mathematical objects, like elliptic curves, as keys. Once Alice has generated her keys, she publishes the public key so that anybody can obtain access to the key. Now suppose Bob wishes to send Alice a private message. He \ufb01rst obtains a copy of Alice\u2019s public key P, and then encrypts the message he wishes to send Alice, using Alice\u2019s public key to perform the encryption. Exactly how the encryption transformation is performed depends on the details of the cryptosystem in use. The key point is that in order to be secure against eavesdropping the encryption stage needs to be very dif\ufb01cult to reverse, even making use of the public key used to encrypt the message in the \ufb01rst place! It\u2019s like the trap door for mail \u2013 what you can put in you can\u2019t take back out, even if you have the key to the trap door. Since the public key and the encoded message is the only information available to an eavesdropper it won\u2019t be possible for the eavesdropper to recover the message. Alice, however, has an additional piece of information not available Public key cryptography and the RSA cryptosystem 641 Figure A5.1. The key ideas of public key cryptography, illustrated in more familiar terms. Essentially the same scheme is implemented by the Post Of\ufb01ce in many countries. to an eavesdropper, the secret key, S. The secret key determines a second transformation, this time on the encrypted message. This transformation is known as decryption, and is inverse to encryption, allowing Alice to recover the original message. In an ideal world that is how public key cryptography would work. Unfortunately, at the time of writing it is not known whether there are any such secure schemes for doing public key cryptography. There do exist several schemes which are widely believed to be secure, and which are in common use for applications such as internet commerce, but wide belief is not equivalent to a proof of security. The reason these schemes are believed to be secure is because so much effort has been devoted to \ufb01nding a means for breaking these schemes (without success!), a sort of proof by attrition. The most widely used of these public key cryptosystems is the RSA cryptosystem, named RSA for the initials of its creators, Rivest, Shamir, and Adleman. The presumed security of the RSA cryptosystem is based, as we shall now see, on the apparent dif\ufb01culty of factoring on a classical computer. Understanding RSA requires a little background in number theory, which is covered in Appendix 4, notably Sections A4.1 and A4.2. Suppose Alice wishes to create public and private keys for use with the RSA cryptosystem. She uses the following procedure: (1) Select two large prime numbers, p and q. (2) Compute the product n \u2261 pq. (3) Select at random a small odd integer, e, that is relatively prime to \u03d5(n) = (p \u2212 1)(q \u2212 1). (4) Compute d, the multiplicative inverse of e, modulo \u03d5(n). (5) The RSA public key is the pair P = (e, n). The RSA secret key is the pair S = (d, n). 642 Public key cryptography and the RSA cryptosystem Suppose a second party, Bob, wishes to encrypt a message M to send to Alice, using the public key (e, n). We assume the message M has only log n bits, as longer messages may be encrypted by breaking M up into blocks of at most log n bits and then encrypting the blocks separately. The encryption procedure for a single block is to compute: E(M ) = M e(mod n). (A5.1) E(M ) is the encrypted version of the message", " M, which Bob transmits to Alice. Alice can quickly decrypt the message using her secret key S = (d, n), simply by raising the encrypted message to the dth power: E(M ) \u2192 D(E(M )) = E(M )d(mod n). (A5.2) For the decryption to be successful we need D(E(M )) = M (mod n). To see that this is the case, note that by construction ed = 1(mod \u03d5(n)) and thus ed = 1 + k\u03d5(n) for some integer k. The proof now proceeds by considering two different cases. In the \ufb01rst case, M is co-prime to n. By Euler\u2019s generalization of Fermat\u2019s little theorem, Theorem A4.9, it follows that M k\u03d5(n) = 1(mod n) and thus, D(E(M )) = E(M )d(mod n) = M ed(mod n) = M 1+k\u03d5(n)(mod n) = M \u00b7 M k\u03d5(n)(mod n) = M (mod n), (A5.3) (A5.4) (A5.5) (A5.6) (A5.7) which establishes that the decryption is successful when M is co-prime to n. Suppose next that M is not co-prime to n, so that one or both of p and q divide M. To be speci\ufb01c, we consider the case where p divides M and q does not divide M ; the other possible cases requre only minor modi\ufb01cations. Because p divides M we have M = 0(mod p) and thus M ed = 0 = M (mod p). Because q does not divide M we have M q\u22121 = 1(mod q) by Fermat\u2019s little theorem, and thus M \u03d5(n) = 1(mod q), since \u03d5(n) = (p \u2212 1)(q \u2212 1). Using ed = 1 + k\u03d5(n) we see that M ed = M (mod q). By the Chinese remainder theorem it follows that we must have M ed = M (mod n), and thus the decryption is also successful when M is not co-prime to n. Exercise A5.1: Written examples of the application of RSA tend to be rather opaque. It\u2019s better to work through an example yourself. Encode the word \u2018QUANTUM\u2019 (or at least the \ufb01rst few letters!), one letter at a time, using p = 3 and q = 11. Choose appropriate values for e and d, and use a representation of English text involving 5 bits per letter. How ef\ufb01ciently can RSA be implemented? There are two implementation issues to be considered. First is the generation of public and private keys for the cryptosystem. If this can\u2019t be done quickly then RSA won\u2019t be good for much. The main bottleneck is the generation of the prime numbers p and q. The way this is attacked is to randomly select a number of the desired length, and then to apply a primality test to determine if the number is, in fact, prime. Fast primality tests such as the Miller\u2013Rabin test can be used to determine whether a number is prime using roughly O(L3) operations, where L is the desired size of the cryptographic key. If the number is found to be composite Public key cryptography and the RSA cryptosystem 643 then we simply repeat the procedure until a prime is found. The prime number theorem (see Problem 4.1) implies that the probability of any given number being prime is about 1/ log(2L) = 1/L, so with high probability O(L) trials are required to obtain a prime number, for a total cost of O(L4) operations to do key generation. The second issue in the implementation of RSA is the ef\ufb01ciency of the encryption and decryption transformations. These are accomplished by modular exponentiation, which we know can be done ef\ufb01ciently using O(L3) operations \u2013 see Box 5.2 on page 228. Thus all the operations required to use the RSA cryptosystem can be done quite quickly on a classical computer, and in practice modest computing power can quite easily cope with keys up to a few thousand bits in length. How can RSA be broken? We describe two methods by which one might hope to break RSA, one based on order-\ufb01nding, the other based on factoring. Suppose Eve receives an encrypted message M e(mod n), and knows the public key (e, n) used to encrypt the message. Suppose she can \ufb01nd the order of", " the encrypted message, that is, she can \ufb01nd the smallest positive integer r such that (M e)r = 1(mod n). (Without loss of generality, we may suppose such an order exists, that is, M e is co-prime to n. If this is not the case, then M e(mod n) and n have a common factor that may be extracted by Euclid\u2019s algorithm, which would allow us to break RSA, as in the second method described below.) Then Exercise A4.16 implies that r divides \u03d5(n). Since e is co-prime to \u03d5(n) it must also be co-prime to r, and thus has a multiplicative inverse modulo r. Let d be such a multiplicative inverse, so ed = 1 + kr for some integer k. Then Eve can recover the original message M by raising the encrypted message to the dth power: (M e)d (mod n) = M 1+kr(mod n) = M \u00b7 M kr(mod n) = M (mod n). (A5.8) (A5.9) (A5.10) It is interesting that Eve never actually learns the secret key (d, n); she only learns (d, n). Of course, d is closely related to d, since d is the inverse of e modulo r, d is the inverse of e modulo \u03d5(n), and r divides \u03d5(n). Nevertheless, this example shows that it is possible to break RSA without necessarily determining the exact value of the secret key. Of course, this method only works if Eve has an ef\ufb01cient method for order-\ufb01nding, and no such method is currently known for a classical computer. On a quantum computer, however, order-\ufb01nding can be accomplished ef\ufb01ciently, as described in Section 5.3.1, and thus RSA can be broken. Exercise A5.2: Show that d is also an inverse of e modulo r, and thus d = d(mod r). A second method for breaking RSA allows one to determine the secret key completely. Suppose Eve could factor n = pq, extracting p and q, and thus giving a means for ef\ufb01ciently computing \u03d5(n) = (p \u2212 1)(q \u2212 1). It is then an easy matter for Eve to compute d, the inverse of e modulo \u03d5(n), and thus completely determine the secret key (d, n). So, if factoring large numbers were easy then it would be easy to break RSA. The presumed security of RSA rests on the fact that these attacks rely on having algorithms to solve problems which are believed (but not known) to be intractable on a classical computer, the order-\ufb01nding and factoring problems. Unfortunately, it\u2019s not even known to be the case that RSA is secure if these problems are hard. It could be that these problems really are dif\ufb01cult, yet there is some other way of breaking RSA. Despite 644 Public key cryptography and the RSA cryptosystem these caveats, more than two decades of attempts to break RSA have resulted in failure, and it is widely believed that RSA is secure against attacks by classical computers. Problem 5.1: Write a computer program for performing encryption and decryption using the RSA algorithm. Find a pair of 20 bit prime numbers and use them to encrypt a 40 bit message. History and further reading Public key cryptosystems were invented by Dif\ufb01e and Hellman in 1976[DH76], and independently by Merkle at about the same time, although his work was not published until 1978[Mer78]. The RSA cryptosystem was invented shortly after by Rivest, Shamir, and Adleman[RSA78]. In 1997 it was disclosed that these ideas \u2013 public key cryptography, the Dif\ufb01e-Hellman and RSA cryptosystems \u2013 were actually invented in the late 1960s and early 1970s by researchers working at the British intelligence agency GCHQ. An account of this work may be found at \u2018http://www.cesg.gov.uk/about/nsecret/\u2019. Primality tests such as the Miller\u2013Rabin and Solovay\u2013Strassen tests are described in Koblitz\u2019s excellent book[Kob94] on number theory and cryptography, which contains a wealth of additional material on public key cryptography. These primality tests were two of the earliest indicators that randomized algorithms may be more ef\ufb01cient for some purposes than deterministic algorithms. The Solovay\u2013Strassen algorithm is due to Solovay and Strassen[SS76], and the Miller\u2013Rabin test is due jointly to Miller[Mil76] and Rabin", "[Rab80]. Appendix 6: Proof of Lieb\u2019s theorem One of the most important and useful results in quantum information theory is the strong subadditivity inequality for von Neumann entropies. This states that for a trio of quantum systems, A, B, C, S(A, B, C) + S(B) \u2264 S(A, B) + S(B, C). (A6.1) Unfortunately, no transparent proof of strong subadditivity is known. Chapter 11 presents a relatively simple proof, based upon a deep mathematical result known as Lieb\u2019s theorem. In this appendix we prove Lieb\u2019s theorem. We begin with a few simple notations and de\ufb01nitions. Suppose f (A, B) is a real-valued function of two matrices, A and B. Then f is said to be jointly concave in A and B if for all 0 \u2264 \u03bb \u2264 1, f (\u03bbA1 + (1 \u2212 \u03bb)A2, \u03bbB1 + (1 \u2212 \u03bb)B2) \u2265 \u03bbf (A1, B1) + (1 \u2212 \u03bb)f (A2, B2). (A6.2) For matrices A and B, we say A \u2264 B if B \u2212 A is a positive matrix. We say A \u2265 B if B \u2264 A. Let A be an arbitrary matrix. We de\ufb01ne the norm of A by A \u2261 max u|u=1 |u|A|u|. (A6.3) In our proof of Lieb\u2019s theorem we will have occasion to use the following easily veri\ufb01ed observations: Exercise A6.1: (\u2264 is preserved under conjugation) If A \u2264 B, show that XAX \u2020 \u2264 XBX \u2020 for all matrices X. Exercise A6.2: Prove that A \u2265 0 if and only if A is a positive operator. Exercise A6.3: (\u2264 is a partial order) Show that the relation \u2264 is a partial order on operators \u2013 that is, it is transitive (A \u2264 B and B \u2264 C implies A \u2264 C), asymmetric (A \u2264 B and B \u2264 A implies A = B), and re\ufb02exive (A \u2264 A). Exercise A6.4: Suppose A has eigenvalues \u03bbi. De\ufb01ne \u03bb to be the maximum of the set |\u03bbi|. Prove that: (1) A \u2265 \u03bb. (2) When A is Hermitian, A = \u03bb. (3) When A =, 1 1 0 1 (A6.4) A = 3/2 > 1 = \u03bb. 646 Proof of Lieb\u2019s theorem Exercise A6.5: (AB and BA have the same eigenvalues) Prove that AB and BA have the same eigenvalues. (Hint: For invertible A, show that det(xI \u2212 AB) = det(xI \u2212 BA), and thus the eigenvalues of AB and BA are the same. By continuity this holds even when A is not invertible.) Exercise A6.6: Suppose A and B are such that AB is Hermitian. Using the previous two observations show that AB \u2264 BA. Exercise A6.7: Suppose A is positive. Show that A \u2264 1 if and only if A \u2264 I. Exercise A6.8: Let A be a positive matrix. De\ufb01ne a superoperator (linear operator on matrices) by the equation A(X) \u2261 AX. Show that A is positive with respect to the Hilbert\u2013Schmidt inner product. That is, for all X, tr(X \u2020A(X)) \u2265 0. Similarly, show that the superoperator de\ufb01ned by A(X) \u2261 XA is positive with respect to the Hilbert\u2013Schmidt inner product on matrices. With these results in hand, we are now in a position to state and prove Lieb\u2019s theorem. Theorem A6.1: (Lieb\u2019s theorem) Let X be a matrix, and 0 \u2264 t \u2264 1. Then the function f (A, B) \u2261 tr(X \u2020AtXB1\u2212t) (A6.5) is jointly concave in positive matrices A and B. Lieb\u2019s theorem is an easy corollary of the following lemma: Lemma A6.2: Let R1, R2, S1, S2, T1, T2 be positive operators such that 0 = [R1, R2] = [S1, S2] = [T1, T2], and R1 \u2265 S1 + T1 R2 \u2265 S2 + T2", ". Then for all 0 \u2264 t \u2264 1, Rt 1 R1\u2212t 2 \u2265 St 1 S1\u2212t 2 + T t 1 T 1\u2212t 2 is true as a matrix inequality. (A6.6) (A6.7) (A6.8) Proof We begin by proving the result for t = 1/2, and then use this to establish the result for general t. It will be convenient to assume that R1 and R2 are invertible, and it is left as an exercise to make the minor technical modi\ufb01cations to the proof necessary to establish the result when this is not the case. Let |x and |y be any two vectors. Applying the Cauchy\u2013Schwarz inequality twice and performing some straightforward manipulations, we have S1/2 2 + T 1/2 |x|(S1/2 1 S1/2 \u2264 |x|S1/2 2 1 \u2264 S1/2 |x S1/2 1 )\u2019 |y| + |x|T 1/2 |y + T 1/2 T 1/2 |y| 2 |x T 1/2 ( \u2019 T 1/2 2 )|y| |y 1 2 2 1 1 \u2264 S1/2 1 |x2 + T 1/2 |x2 1 S1/2 2 |y2 + T 1/2 2 (A6.9) (A6.10) (A6.11) ( |y2 Proof of Lieb\u2019s theorem 647 (A6.12) = x|(S1 + T1)|xy|(S2 + T2)|y. By hypothesis, S1 + T1 \u2264 R1 and S2 + T2 \u2264 R2, so |x|(S1/2 1 2 + T 1/2 S1/2 1 T 1/2 2 )|y| \u2264 x|R1|xy|R2|y. (A6.13) Let |u be any unit vector. Then applying (A6.13) with |x \u2261 R\u22121/2 1 R\u22121/2 2 |u gives u|R\u22121/2 1 (S1/2 1 2 + T 1/2 S1/2 1 T 1/2 2 |uu|R\u22121/2 )R\u22121/2 2 |u R2R\u22121/2 2 |u 2 \u2264 = u|R\u22121/2 1 R1R\u22121/2 1 u|uu|u = 1. Thus De\ufb01ne R\u22121/2 1 (S1/2 1 2 + T 1/2 S1/2 1 T 1/2 2 )R\u22121/2 2 \u2264 1. A \u2261 R\u22121/4 B \u2261 R1/4 1 R\u22121/4 2 R\u22121/4 1. 2 (S1/2 1 2 + T 1/2 S1/2 1 T 1/2 2 )R\u22121/2 2 Note that AB is Hermitian, so by Exercise A6.6 on page 646, )R\u22121/4 R\u22121/4 2 2 + T 1/2 S1/2 R\u22121/4 1 (S1/2 T 1/2 2 2 1 R\u22121/4 1 1 = AB \u2264 BA 2 + T 1/2 S1/2 = R\u22121/2 \u2264 1, (S1/2 1 1 1 T 1/2 2 )R\u22121/2 2 |u and |y \u2261 (A6.14) (A6.15) (A6.16) (A6.17) (A6.18) (A6.19) (A6.20) (A6.21) where the last inequality is just (A6.16). AB is a positive operator, so by Exercise A6.7 on page 646 and the previous inequality, R\u22121/4 1 2 + T 1/2 S1/2 R\u22121/4 2 R\u22121/4 1 )R\u22121/4 (A6.22) (S1/2 T 1/2 2 \u2264 I. 1 2 1 Finally, by Exercise A6.1 on page 645, and the commutativity of R1 and R2, 2 + T 1/2 S1/2 \u2264 R1/2 1 R1/2 2 T 1/2 2 S1/2 1, 1 (A6.23) which establishes that (A6.8) holds for t = 1/2. Let I be the set of all t such that (A6.8) holds. By inspection, we see that 0 and 1 are elements of I, and we have just shown that 1", "/2 is an element of I. We now use the t = 1/2 case to prove the result for any t such that 0 \u2264 t \u2264 1. Suppose \u03bc and \u03b7 are any two elements of I, so that S1\u2212\u03bc 2 S1\u2212\u03b7 (A6.25) 2 These inequalities are of the form (A6.6) and (A6.7) for which the t = 1/2 case has already been proved. Using the t = 1/2 result we see that T 1\u2212\u03bc 2 T 1\u2212\u03b7. 2 \u2265 S\u03bc 1 \u2265 S\u03b7 1 R1\u2212\u03bc 2 R1\u2212\u03b7 2 + T \u03bc + T \u03b7 R\u03bc 1 R\u03b7 1 (A6.24) 1 1 \u2019 R\u03bc 1 R1\u2212\u03bc 2 (1/2 \u2019 R\u03b7 1 (1/2 \u2019 S\u03bc 1 \u2265 S1\u2212\u03bc 2 R1\u2212\u03b7 2 + (1/2 \u2019 (1/2 S\u03b7 1 S1\u2212\u03b7 2 (1/2 \u2019 \u2019 T \u03bc 1 T 1\u2212\u03bc 2 (1/2 T 1\u2212\u03b7 2. (A6.26) T \u03b7 1 648 Proof of Lieb\u2019s theorem Using the commutativity assumptions 0 = [R1, R2] = [S1, S2] = [T1, T2], we see that for \u03bd \u2261 (\u03bc + \u03b7)/2, R\u03bd 1 R1\u2212\u03bd 2 S1\u2212\u03bd 2 Thus whenever \u03bc and \u03b7 are in I, so is (\u03bc + \u03b7)/2. Since 0 and 1 are in I, it is easy to see that any number t between 0 and 1 with a \ufb01nite binary expansion must be in I. Thus I is dense in [0, 1]. The result now follows from the continuity in t of the conclusion, (A6.8). \u2265 S\u03bd 1 T 1\u2212\u03bd 2 (A6.27) + T \u03bd 1. The proof of Lieb\u2019s theorem is a simple application of Lemma A6.2. The clever idea that makes this possible is to choose the operators in Lemma A6.2 to be superoperators \u2013 linear maps on operators. These will be chosen in such a way as to be positive with respect to the Hilbert\u2013Schmidt inner product (A, B) \u2261 tr(A\u2020B). Proof (Lieb\u2019s theorem) Let 0 \u2264 \u03bb \u2264 1 and de\ufb01ne superoperators S1, S2, T1, T2, R1, R2 as follows: S1(X) \u2261 \u03bbA1X S2(X) \u2261 \u03bbXB1 T1(X) \u2261 (1 \u2212 \u03bb)A2X T2(X) \u2261 (1 \u2212 \u03bb)XB2 R1 \u2261 S1 + T1 R2 \u2261 S2 + T2. Observe that S1 and S2 commute, as do T1 and T2, and R1 and R2. Recall Exercise A6.8 on page 646, that all these operators are positive with respect to the Hilbert\u2013Schmidt inner product. By Lemma A6.2, (A6.31) (A6.29) (A6.32) (A6.28) (A6.30) (A6.33) Rt 1 R1\u2212t 2 \u2265 S t 1 S 1\u2212t 2 + T t 1 T 1\u2212t 2. (A6.34) Using the Hilbert\u2013Schmidt inner product to take the X \u00b7X matrix element of the previous inequality gives tr X \u2020 (\u03bbA1 + (1 \u2212 \u03bb)A2)t X (\u03bbB1 + (1 \u2212 \u03bb)B2)1\u2212t X \u2020(\u03bbA1)tX(\u03bbB1)1\u2212t XB1\u2212t 1 + tr ) + (1 \u2212 \u03bb)tr(X \u2020At 2 \u2265 tr = \u03bbtr(X \u2020At 1 X \u2020((1 \u2212 \u03bb)A2)tX((1 \u2212 \u03bb)B2)1\u2212t XB1\u2212t 2 ), (A6.35) (A6.36) which is the desired statement of joint concavity. History and further reading The history of Lieb\u2019s theorem is tied up with the proof of the strong subadditivity inequality for quantum entropies, and may be found together with the history of the proof of that inequality in the \u2018History and further reading\u2019 for Chapter 11. 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Thermodynamic cost of computation, algorithmic complexity and the information metric. Nature, 341:119, 1989. W. H. Zurek. Decoherence and the transition from quantum to classical. Phys. Today, October 1991. [Zur89] [Zur91] [ZR98] Index Bold page numbers indicate the place where the concept is introduced, explained, or de\ufb01ned. Major theorems are listed together under \u2018theorem\u2019, and end-of-chapter problems are listed together under \u2018problem\u2019. \u03a9(\u00b7), 137 \u0398(\u00b7), 137 -net, 618 \u2265 relation for matrices, 645, 82 \u2264 relation for matrices, 645 \u03c0/8 gate, 174 fault-tolerant, 485 Toffoli construction, 182 * operation, 62 0-1 integer programming, 149, 148 Abelian group, 240, 610 Abelian stabilizer problem, 241 Abrams, D. S., 214, 215, [AL97] accessible information, 529 acyclic circuits, 23 Adami, C., 350, [CAK98] additive quantum codes, 453 adjoint, 62, 69 Adleman, L. M., 11, 168, 214, 641, 644, [ADH97], [Adl94], [Adl98], [RSA78] Aharonov, D., xix, xxi, 276, 424, 498, 499, [ABO97], [ABO99], [ABOIN96], [Aha99a], [Aha99b], [AKN98] Alberti, P. M., 424, [Alb83] Alde, D. M., 607, [HAD+95] algorithm Deutsch\u2013Jozsa, 36 discrete logarithm, 238 period-\ufb01nding, 236 quantum order-\ufb01nding, 232 quantum phase estimation, 225 quantum search, 254 quantum simulation, 208 reduction of factoring to order-\ufb01nding, 233 algorithm design, 135 algorithms, 120, 122 Allen, L., 350, [AE75] alphabet, 141 Ambainis, A., xxi, 276, 607, [Amb00] Amer, N., xxi amplitude, 81 amplitude damping, 380 analog computation, 5, 163, 287 ancilla, 94 ancilla", " bits, 131, 20 gate, 130 Ando, T., 527, [And79] angle between states, 413 angular momentum, 314 anti-commutator, 76 Araki, H., 526, 527, [AL70] Araki-Lieb inequality, 516 architecture, quantum computer, 340 Ashikhmin, A., 498, [AL99], [Ash97] asymptotic notation, 136 atom traps, 3 atypical sequences, 538 auxiliary system, 517 Awschalom, D. D., 351, [IAB+99] B92 protocol for QKD, 589 Bacon, D. A., 498, [BKLW99], [LBW99] Balandin, A., 351, [VYW+99] Balc\u00b4azar, J. L., 168, [BDG88a], [BDG88b] Bardeen, J., 4 Barenco, A., 214, [BBC+95], [DBE95] Barnett, S. M., 607, [BP93] Barnum, H., xxi, 424, 605, 606, [BCF+96], [BFJS96], [BKN98], [BNS98], [BST98], [NCSB98] Barton, E., 168, [Bar78] basis for a vector space, 63 Bayes\u2019 rule, 608 BB84 protocol for QKD, 587 Beals, R., 246, 276, [BBC+98] beamsplitters, 288, 291 Beckman, D., xxi, 214, [BCDP96] Bell basis, 98 Bell inequality, 17, 111, 115, 119 Bell states, 16, 25, 98 Bell, J. S., 17, 25, 112, 116, 119, [Bel64] Ben-Or, M., 498, 499, [ABO97], [ABO99], [ABOIN96] Benioff, P., 214, [Ben80] Bennett, C. H., xix, 9, 11, 59, 119, 168, 169, 214, 276, 497, 604, 606, 607, [BB84], [BBB+92], [BBBV97], [BBCM95], [BBC+93], [BBC+95], [BBE92], [BBPS96], [BBP+96], [BBR88], [BD00], [BDS97], [BDSW96], [Ben73], [Ben82], [Ben87], [Ben89], [Ben92], [BS98], [BW92] Bernstein, E., 200, 214, 276, [BBBV97], [BV97] Bernstein, H. J., 214, 350, 606, [BBPS96], [RZBB94] Bertani, P., 214, 350, [RZBB94] beryllium, 310, 315 666 Index Bessette, F., 606, 607, [BBB+92] Beth, T., 246, 616, [Bet84], [PRB98], [RB98] Bethune, D. S., 607, [BR00], [BR98] Bhatia, R., 118, [Bha97] big \u03a9 notation, 137 big \u0398 notation, 137 big O notation, 136 Biham, E., 607, [BBB+98] billiard ball computer, 155 binary entropy, 502 binary symmetric channel, 426 Birkhoff\u2019s theorem, 574 bit, 13 bit \ufb02ip, 81 bit \ufb02ip channel, 376 bit \ufb02ip code, 427 bit \ufb02ip operator, 427 bit\u2013phase \ufb02ip channel, 377 Bloch sphere, 15, 19, 105, 174 Bloch vector, 105, 174, 259 Bodenhausen, G., 351, [EBW87] Boghosian, B. M., 214, [BT97] Bohm, D., 119, [Boh51] Bohr magneton, 309 Bohr, N., 111, 171 Boltzmann\u2019s constant, 153 Boneh, D., 246, [BL95] Boolean circuit, 133 Boolean function, 133 Boschi, D., 59, [BBM+98] Bose condensate, 346 Bouwmeester, D., 59, [BPM+97] Boyer, M., 276, 607, [BBB+98], [BBHT98] Boykin, P. O., 214, 215, [BMP+99] BPP, 152 BQP, 41, 200 bra, 62 Braginsky, V. B., 118, [BK92] Branca, S., 59, [BBM+98] Brassard, G", "., 11, 59, 276, 606, 607, [BB84], [BBB+92], [BBB+98], [BBBV97], [BBCM95], [BBC+93], [BBE92], [BBHT98], [BBP+96], [BBR88], [BHT98], [Bra93], [BS94] Brattain, W., 4 Braunstein, S. L., 59, 351, 352, 605, [BCJ+99], [BFGL98], [BK98a], [BK99], [Bra98], [FSB+98], [LB99] Bravyi, S. B., 499, [BK98b] Brennen, G. K., 351, [BCJD99] Brewer, R. G., 350, [BDK92] Brewster\u2019s angle, 310 Brune, M., 350, [DMB+93], [DRBH87], [DRBH95] Buhrman, H., xxi, 246, 276, [BBC+98] Burkard, G., 351, [IAB+99] c-numbers, 62 Cachin, C., 607, [CM97] Caldeira, A. O., 398, [CL83] Calderbank, A. R., 8, 9, 450, 497, 498, [CRSS97], [CRSS98], [CS96] Calderbank\u2013Shor\u2013Steane codes, 445, 450 Campman, K., 351, [HSM+98] Campos, R. A., 349, [CST89] canonical form for entropy exchange, 562 Capelin, S., xxi Carr\u2013Purcell\u2013Meiboom\u2013Gill technique, 331 cat state for fault-tolerant measurement, 490 catalyst, 577 Cauchy\u2013Schwarz inequality, 68 Caves, C. M., xxi, 16, 351, 398, 424, 605, 606, [BCF+96], [BCJD99], [BCJ+99], [Cav99], [FC94], [NC97], [NCSB98], [SC99] cavity quantum electrodynamics, 277, 297, 343 cellular automata, 340 centralizer, 465 Cerf, N. J., 350, [CAK98] characteristic function, 68 Chari, A. N., 214, [BCDP96] Chau, H. F., 215, 607, [CW95], [LC99] Chebyshev\u2019s inequality, 609 check matrix, 456 Chernoff bound, 154, 609 Chernoff, P. R., 214, [Che68] chi-matrix representation, 391 Chiao, R. Y., 350, [KSC+94] Childs, A. M., xxi Chinese remainder theorem, 629 Choi, M.-D., 398, [Cho75] Chong, F. T., xxi CHSH inequality, 116, 119 Chuang, I. L., 215, 349\u2013351, 398, 498, 499, 605, [CGK98], [CGKL98], [CM00], [CN97], [CVZ+98], [CY95], [GC97], [GC99], [KCL98], [LCW98], [LNCY97], [LVZ+99], [VYSC99], [ZLC00] Church, A., 4, 122, 125, 167, [Chu36] Church\u2013Turing thesis, 4 strong form of, 5, 6, 140 Church-Turing thesis, 125, 226 Cirac, J. I., 350, 398, 498, [CPZ96], [CZ95], [PCZ97] circuit family, 134 circuit model of computation, 129 classical information over noisy quantum channels, 546 classical noise, 354 classical physics, 2 Clausen, M., 616, [Cla89] Clauser, J. F., 119, [CHSH69] Cleve, R., xxi, 59, 214, 245, 246, 276, 605, [BBC+95], [BBC+98], [CD96], [CEMM98], [Cle99] clique, 149 closed quantum systems, 353 CNF, 148 co-prime, 627 Cohen-Tannoudji, C., 59, 118, [CTDL77a], [CTDL77b] coherent information, 564, 572, 592, 605 collusion entropy, 584 communication complexity, 164 commutator, 76 commuting operators, 76, 597 compare-and-swap based sorts, 137 complete positivity, 368 example of a positive map not completely positive, 368 completeness equation, 85, 102 completeness of a problem for a complexity", " class, 145 completeness relation, 67, 360 complex conjugate, 62, 70 complexity class, 142, 150 composite systems, 93 composition of linear operators, 64 computational complexity, 40, 135, 138 dif\ufb01culty of obtaining results in, 140 concatenated codes, 480 concavity, 504 conditional entropy classical, 506 quantum, 514 conditional probability, 608 conjunctive normal form, 148 coNP, 142 conservative property of the Fredkin gate, 156 continued fraction expansion, 229, 230, 282, 335, 635 continued fractions algorithm, 635 controlled operation, 177 controlled- gate, 20, 178 fault-tolerant, 484 convergent, 230, 635 Conway, J. H., xxi, 168, 169, [Con72], [Con86] Cook, S. A., 138, 168, [Coo71] Cooper pair, 344 Coppersmith, D., 245, [Cop94] Cormen, T. H., 167, 639, [CLR90] correctable errors, 440 correctable set of errors, 436 Cortese, J., xxi Cory, D. G., 350, 351, [CFH97], [CMP+98], [STH+99] coset invariance, 237, 243 cosets, 586, 612 Coulomb blockade, 344 counting problem, 216 Cover, T. M., xxi, 59, 526, 539, 604, 605, [CT91] Cowan, J. D., 168, 498, [WC67] creation and annihilation operators, 284 Cr\u00b4epeau, C., 59, 607, [BBCM95], [BBC+93] gate, 131 cryptography, 9, 582, 640, 145 Csisz\u00b4ar, I., 604, [CK81] CSS codes, 445, 450, 593 cycle, 143 cyclic group, 611 cyclic property of trace, 75 cyclic subgroup, 611 data compression, 536 data pipelining inequality, 510 data processing inequality, 572, 606 quantum, 564, 572 Davidovich, L., 350, [DMB+93], [DRBH87] Davies, E. B., 398, [Dav76] Davis, M. D., 59, 167, 652, [Dav65], 662 de Wolf, R., xxi, 246, 276, [BBC+98] decision problems, 135, 141 decoherence as a stochastic phase kick process, 384 estimates of, 278 decoherence free subspace, 498 degeneracy, 69 degenerate codes, 444 Demarrais, J., 214, [ADH97] density matrix, 99 density operator, 99, 119 depolarized, 378 depolarizing channel, 378 DeShazo, M., xxi deterministic query complexity, 272 Deutsch\u2019s algorithm, 32 Deutsch\u2019s problem, 34, 241 Index 667 Deutsch, D., 6, 32, 34, 59, 171, 214, 245, 526, [DBE95], [Deu83], [Deu85], [Deu89], [DJ92] Deutsch, I. H., 351, [BCJD99] Deutsch\u2013Jozsa algorithm, 34, 59, 249 optical implementation, 294 Devabhaktuni, S., 214, [BCDP96] deviation density matrix, 336 DeVoe, R. G., 321, 350, [BDK92] Diaconis, P., 616, [DR90] diagonal representation, 69 diagonalizable operator, 69 Diaz, J., 168, [BDG88a], [BDG88b] Dieks, D., 604, [Die82] Dif\ufb01e, W., 11, 59, 644, [DH76], [DL98] dimension of a vector space, 63 dipolar coupling, 328 Dirac notation, 13, 62 discrete logarithm problem, 216, 217, 241 quantum algorithm for, 238 discrete memoryless channel, 551 distance measures, 399 distance of a code, 448 distillable entanglement, 578 distributed computation, 164 distributed quantum computation, xvii Diu, B., 59, 118, [CTDL77a], [CTDL77b] DiVincenzo, D. P., xix, xxi, 214, 215, 246, 349\u2013351, 497, 498, 604\u2013606, [BBC+95], [BD00], [BDS97], [BDSW96], [CD96], [DiV95a], [DiV95b], [DiV98], [DS96], [DSS98], [IAB+99], [LD98], [TD98], [VYW+99] divisor, 625", " DNA computing, 163 Domokos, P., 350, [DRBH95] double stochasticity, 511 dual linear code, 449 dual vector, 65 dual-rail representation, 288 Duan, L.-M., 119, [DG98] D\u00a8urr, C., 276, [DH96] Dyer, P., 607, [HAD+95] Dykman, M. I., 351, [PD99] dynamic measures of distance, 399, 401 Dyson, F. J., 527 Earnshaw\u2019s theorem, 309 Earnshaw, S., 350, [Ear42] Eberhard, P. H., 350, [KSC+94] Eberly, J. H., 350, [AE75] edges, 143 ef\ufb01ciency of quantum simulations, 206 Eibl, M., 59, [BPM+97] eigenvalue, 68 eigenvector, 68 Einstein, A., 2, 17, 25, 60, 80, 119, [EPR35] Einstein-Podolsky-Rosen thought experiment, 17 Ekert, A. K., 59, 214, 246, 497, 498, 606, 607, [BBE92], [CEMM98], [DBE95], [EHPP94], [EJ96], [EJ98], [Eke91], [EM96], [ME99] electric dipole selection rules, 300 electron spin, 309 element of reality, 112 ENDOR, 350 energy, 83, 153 energy eigenstates, 83 ensemble of pure states, 99 entanglement, 11, 95 668 Index as a physical resource, 571 catalysis, 577 dilution, 578 distillation, 578 mixed state, 580 of formation, 578 entanglement \ufb01delity, 420 entropy, 500 classical, 500 concavity of, 516 of an ensemble, 518 quantum, 510 rate, 538 strict concavity of, 504 entropy exchange, 561, 605 entropy Venn diagram, 508 entscheidungsproblem, 122 environmental models quantum operations, 365 trace-preserving quantum operations, 363 EPR, 17, 111 EPR pairs, 16, 25, 98, 591 EPR protocol for QKD, 591 EPR states, 25 EPR thought experiment, 119 Epstein, H., 527, [Eps73] equilibration of quantum systems, 211 Ernst, R. R., 351, [EBW87] error propagation in fault-tolerant circuits, 483 error syndrome, 428 classical, 448 error-correcting codes, 8, 425 errors, 436 Ettinger, M., 246, [EH99], [EHK99] Euclid\u2019s algorithm, 122, 626 Euler \u03d5 function, 631 Euler cycle, 143 Everitt, H., xxi evolution matrix, 355 EXP, 151 expectation of a random variable, 609 exponential resources, 139 exponential time, 151 Fabry\u2013Perot cavity, 298 factoring, 142 factoring decision problem, 142 factoring problem, 232 factors, 625 Fagin, R., xxi Fahmy, A. F., 350, [CFH97], 23 Fannes\u2019 inequality, 512 Fannes, M., 526, [Fan73] Fano\u2019s inequality, 534, 536, 563, 572, 609 quantum, see quantum Fano inequality, 23 gate, 131 Farhi, E., 276, [FG98] F\u00a8assler, A., 616, [FS92] fault-tolerant computation, 425 fault-tolerant quantum computation, 474 \u03c0/8 gate, 485 assumptions, 493 de\ufb01nition of operations, 476 error propagation, 478 measurement, 477, 489 quantum logic, 482 the threshold theorem, 480, 493 Toffoli gate, 488 feasible computational problems, 139 Feller, W., 609, [Fel68a], [Fel68b] Feynman path integral, 398 Feynman, R. P., 7, 59, 118, 168, 204, 214, [Fey82], [FLS65a], [FLS65b] \ufb01delity, 281 classical, 400 joint concavity of, 415 quantum, 409 \ufb01ne structure constant, 301 \ufb01nite simple continued fraction, 635 \ufb01nite state control, 122 foosball, xxi formal languages, 141 Fourier transform discrete, 217 over a group, 615 over groups, 240 quantum, 191, 209, 217 shift-invariance property, 237 Fractran, 166 Fredkin gate, 156 optical, 295 Fredkin, E., 168, [FT82] Freedman, M. H.,", " xxi, 499, 624, [FM98] Freeman, R., 351, [LKF99] Fuchs, C. A., xix, xxi, 59, 352, 399, 424, 605, [BCF+96], [BFGL98], [BFJS96], [FC94], [FSB+98], [Fuc96], [Fuc97], [FvdG99] full-adder, 132 fundamental theorem of arithmetic, 625 Furusawa, A., 59, 352, [FSB+98] Gabarr\u00b4o, J., 168, [BDG88a], [BDG88b] Gagen, M., xxi Gardiner, C. W., 118, 349, 398, [Gar91], [ZG97] Garey, M. R., 168, [GJ79] Gauss, K. F., 232 generalized amplitude damping, 382 generalized measurements, 118 generators, 611 for the \ufb01ve qubit code, 469 for the Shor code, 468 of a group, 455 Gerlach, W., 43 Gershenfeld, N., xxi, 350, 351, [CGK98], [CGKL98], [GC97] Gibbs state, 211 Gilbert\u2013Varshamov bound for classical codes, 449 for CSS codes, 451, 495, 596 Gisin, N., 607, [MZG96] global phase, 93 Goldberg, D., 168, [HGP96] Gordon, J. P., 605, [Gor64] Gossard, A. C., 351, [HSM+98] Gottesman, D., xix, xxi, 9, 215, 453, 497\u2013499, 605\u2013607, [BFGL98], [GC99], [Got96], [Got97], [Got98a], [Got98b], [GP10] Gram-Schmidt procedure, 66 graph, 143 graph isomorphism problem, 150, 242 graph theory, 143 Gray code, 191 greatest common divisor, 626 Grif\ufb01ths, R. B., 214, 246, [GN96] Grimmett, G. R., 609, [GS92] ground state, 83 group commutator, 620 group theory, 610 Grover iteration, 250 Grover operator, 250 Grover\u2019s algorithm, 7, 38, 248 Grover, L. K., 7, 38, 276, [Gro96], [Gro97] Gruska, J., xix, [Gru99] Guo, G.-C., 119, [DG98] Gutmann, S., 276, [FG98] Hadamard gate, 19, 174 Hadamard transform, 31 half-adder, 132 Halmos, P. R., 118, [Hal58] halting problem, 130 halting state of a Turing machine, 123 Hamiltonian, 82 Hamiltonian cycle problem, 143, 264 Hamiltonian cycle problem inclusion in NP, 143 Hammermesh, M., 616, [Ham89] Hamming code, 449 Hamming distance, 399, 448 Hamming sphere, 549 Hamming weight, 448, 547 Hammurabi, 4 Hansen, R. H., 351, [JMH98] Hardin, R. H., 498, [RHSS97] Hardy, G. H., 246, 639, [HW60] Hardy, L., 59, [BBM+98] Haroche, S., 350, [DMB+93], [DRBH87], [DRBH95] Harris, J. S., xxi Hausladen, P., 605, [HJS+96] Havel, T. F., 350, 351, [CFH97], [CMP+98], [STH+99], 143, 264 Heisenberg uncertainty principle, 88, 89 Heisenberg, W., 44 Hellman, M., 11, 644, [DH76] Hellwig, K.-E., 398, [HK69], [HK70] Hennessey, J. L., 168, [HGP96] Hermitian conjugate, 62, 69, 70 Hermitian operator, 70 hidden linear function problem, 241 hidden subgroup problem, 38, 217, 234, 336 quantum algorithm for, 240 Hilbert space, 66 Hilbert\u2019s problem, 122 Hilbert, D., 122 Hilbert-Schmidt inner product, 76 Hoffmann, B., 61 Hofstadter, D. R., 59, 167, [Hof79] Holevo \u03c7 quantity, 531 Holevo bound, 531, 592 Holevo, A. S., xxi, 605, [Hol73], [Hol79], [Hol98] Holt, R. A., 119", ", [CHSH69] Hood, C. J., 306, 350, 398, [THL+95] Horn, R. A., 118, [HJ85], [HJ91] Horne, M. A., 119, [CHSH69] Horodecki, M., 605, 606, [HHH96], [HHH98], [HHH99a], [HHH99b], [HHH99c], [Hor97] Index 669 Horodecki, P., 351, 606, [HHH96], [HHH98], [HHH99a], [HHH99b], [HHH99c], [ZHSL99] Horodecki, R., 606, [HHH96], [HHH98], [HHH99a], [HHH99b], [HHH99c] H\u00f8yer, P., 246, 276, [BBHT98], [BHT98], [DH96], [EH99], [EHK99] HSW theorem, 581, 592 Huang, M. A., 214, [ADH97] Hubbard model, 206 Hughes, R. J., 607, [HAD+95] Hughston, L. P., 119, [HJW93] Huibers, A. G., xxi, 351, [HSM+98] Huttner, B., 607, [EHPP94] hyper\ufb01ne states, 315 i.i.d. source, 537 identity matrix, 65 identity operator, 63 Igeta, K., 350, [YKI88] Imamoglu, A., 350, 351, [IAB+99], [IY94] Impagliazzo, R., 499, [ABOIN96] INADEQUATE, 338 independent generators of a group, 456 independent random variables, 608 infeasible computational problems, 139 information gain implies disturbance, 586 information reconciliation, 584 information source, classical, 399 information theory, 7 operational motivation for de\ufb01nitions in, 501 inner product, 62, 65 inner product space, 66 integers, 625 interference, 32 interferometers, 296 internal states of a Turing machine, 122 intractable computational problems, 139 ion trap, 277 cooling, 312 geometry, 309 quantum computer, 309, 343 toy model, 317 irreversible logic gate, 153 Ising model, 206 Itano, W. M., 350, [MMK+95], [WMI+98] J-coupling, 328 James, D., 350, [Jam98] Jaynes, E. T., 119, [Jay57] Jaynes\u2013Cummings Hamiltonian, 281, 300, 302, 308, 315, 318 Jessen, P. S., 351, [BCJD99] Jiang, H. W., 351, [VYW+99] Johnson noise, 312 Johnson, C. R., 118, [HJ85], [HJ91] Johnson, D. S., 168, [GJ79] joint concavity, 519, 645 joint convexity of trace distance, 408 joint entropy, 506 quantum, 514 joint entropy theorem, 513 Jonathan, D., 606, [JP99] Jones, J. A., 351, [JM98], [JMH98] Jones, K. R. W., 398, [Jon94] Josephson junction, 344 Jozsa, R., 59, 119, 246, 351, 424, 605, [BBC+93], [BCF+96], [BCJ+99], [BFJS96], [DJ92], [EJ96], 670 Index [EJ98], [HJS+96], [HJW93], [Joz94], [Joz97], [JS94] Kahn, D., 59, [Kah96] Kallenbach, R., 350, [BDK92] Kane, B. E., 351, [Kan98] Karp, R. M., 168, [Kar72] Kay, A., 80, 120 Kempe, J., xxi, 498, [BKLW99] Kerr effect, optical, 290 Kerr media, nonlinear, 293, 305 ket, 62, 62 Khahili, F. Y., 118, [BK92] Kimble, H. J., 59, 306, 350, 352, 398, [BK98a], [BK99], [FSB+98], [THL+95] King, B. E., 350, [MMK+95], [WMI+98] King, C., 605, [KR99] Kitaev\u2019s algorithm, 243 Kitaev, A. Y., xix, xxi, 38, 246, 424, 498, 499,", " 624, [AKN98], [BK98b], [Kit95], [Kit97a], [Kit97b], [Kit97c] Kitagawa, M., 350, [KU91], [YKI88] Klein\u2019s inequality, 526 Klein, O., 526, [Kle31] Knight, P. L., 350, 398, [PK96], [PK98] Knight, T., 168, [YK95] Knill, E., xxi, 59, 214, 246, 351, 424, 497\u2013499, 606, [BKN98], [CMP+98], [EHK99], [KCL98], [KL97], [KL99], [KLV99], [KLZ98a], [KLZ98b], [Kni95], [NKL98] Knuth, D. E., 59, 122, 167, 171, 216, 232, 632, [Knu97], [Knu98a], [Knu98b] Koblitz, N., 639, 644, [Kob94] Kolmogorov distance, 400 Kong, S., xxi K\u00a8orner, J., 604, [CK81] Kraus, K., 118, 398, 526, [HK69], [HK70], [Kra83], [Kra87] Kronecker product, 74 Kubinec, M. G., 351, [CGK98], [CGKL98], [LVZ+99] Kullback, S., 526, [KL51] Kupce, E., 351, [LKF99] Kurtsiefer, C., 321 Kwiat, P. G., 119, 350, [CAK98], [KMSW99], [KSC+94], [MWKZ96] L, 151 L1 distance, 400 Ladner, R. E., 168, [LLS75] La\ufb02amme, R., xxi, 59, 214, 351, 424, 497\u2013499, 606, [CMP+98], [KCL98], [KL97], [KL99], [KLV99], [KLZ98a], [KLZ98b], [LMPZ96], [NKL98], [SL97], [STH+99], [ZL96] Lalo\u00a8e, F., 59, 118, [CTDL77a], [CTDL77b] Lamb\u2013Dicke parameter, 312 Lamb-Dicke criterion, 312 Landahl, A., xxi Landau, L., 119, [Lan27] Landau, L. J., 398, [LS93] Landau, S., 59, [DL98] Landauer\u2019s principle, 153, 569 Maxwell\u2019s demon and, 162 Landauer, R., 1, 168, [Lan61] Lanford, O. E., 527, [LR68] Lange, W., 306, 350, 398, [THL+95] Langevin equations, 353 language, 141 law of large numbers, 541, 609 law of total probability, 608 Lazere, C., 59, [SL98] Lecerf, Y., 168, [Lec63] Leff, H. S., 168, [LR90] Legere, R., xxi Leggett, A. J., 398, [CL83] Leibfried, D., 350, [WMI+98] Leibler, R. A., 526, [KL51] Leighton, R. B., 59, 118, 168, [FLS65a], [FLS65b] Leiserson, C. E., 167, 639, [CLR90] Lenstra, A. K., 246, [LL93] Lenstra Jr., H. W., 246, [LL93] Leonhardt, U., 398, [Leo97] Leung, D. W., xxi, 351, 498, 499, [CGKL98], [CVZ+98], [LNCY97], [LVZ+99], [ZLC00] Levin, L., 138, 168, [Lev73] Levitov, L., 351, [MOL+99] Lewenstein, M., 351, [ZHSL99] Li, M., 169, [LTV98], [LV96] Lidar, D. A., xxi, 498, [BKLW99], [LBW99], [LCW98] Lie formula composition of Lie operations, 207 Lie, S., 215 Lieb\u2019s theorem, 519, 526, 645, 646", " Lieb, E. H., xxi, 527, [AL70], [Lie73], [Lie75], [LR73a], [LR73b] Lindblad form, 207, 386, 398 Lindblad operators, 388 Lindblad, G., 398, 527, 605, [Lin75], [Lin76], [Lin91] Linden, N., 351, [BCJ+99], [LKF99], [LP99] linear algebra, 61, 118 linear code, 445 linear dependence, 63 linear independence, 63 linear operators, 63 linearity of trace, 75 Lipton, R. J., 168, 246, [BL95], [Lip95] literals, 148 Lloyd, S., 214, 215, 349, 351, 352, 606, [AL97], [CVZ+98], [LB99], [Llo93], [Llo94], [Llo95], [Llo96], [Llo97], [LS98], [MOL+99] Lo, H.-K., xix, 605\u2013607, [BFGL98], [LC99], [Lo99], [LP97], [LSP98] local realism, 116 LOCC, 573 logarithmic space, 151 logic gate, 129 logical labeling, 333 Lomont, J. S., 616, [Lom87] Loss, D., 351, [IAB+99], [LD98] Louisell, W. H., 349, [Lou73] Luther, G. G., 607, [HAD+95] Lynch, N. A., 168, [LLS75] Lynn, T., xxi Lytsin, S., 498, [AL99] Maali, A., 350, [DMB+93] Maassen, H., 526, [MU88] Mabuchi, H., xxi, 306, 350, 398, [THL+95] Macchiavello, C., 59, 246, 497, 498, [CEMM98], [EM96] MacWilliams, F. J., 59, 497, [MS77] magnetic resonance, 326 majorization, 573 Manin, Y. I., xxi, 204, 214, [Man80], [Man99] Index 671 Marcus, C. M., 351, [HSM+98] Marcus, M., 118, [MM92] Margolus, N., 214, [BBC+95] Markov chain, 509 Markov processes, 354, 355 Marshall, A. W., 606, [MO79] Martini, F. D., 59, [BBM+98] Mass, W., 351, [CMP+98] master equations, 353, 386 matrices, 64 matrix representation of an operator, 64 Mattle, K., 59, 119, [BPM+97], [MWKZ96] Maurer, U. M., 607, [BBCM95], [CM97], [Mau93] Maxwell\u2019s demon, 161, 162, 569 Maxwell, J. C., 162, 168, [Max71] Mayers, D., 607, [May98] measurement, 84, 185, 356 fault-tolerant, 489 in the Bell basis, 187 in the stabilizer formalism, 463 of an operator, 188 measurement operators, 84, 102 Meekhof, D. M., 350, [MMK+95], [WMI+98] Menezes, A. J., 59, 275, [MvOV96] Merkle, R., 11, 644, [Mer78] metric, 400 Meyer, D. A., 499, [FM98] Milburn, G. J., xxi, 59, 350, 606, [Mil89a], [Mil96], [Mil97], [Mil98] Miller, D. A. B., 350, [Mil89b] Miller, G. L., 644, [Mil76] Minc, H., 118, [MM92] Minsky machine, 165 program, 165 Minsky, M. L., 168, [Min67] Miquel, C., 497, [LMPZ96] mirror, 288 Mitchell, J. R., 350, [KMSW99] mixed state, 100 Modha, D., 605, [CM00] modular arithmetic, 626 modular exponential, 228 modular exponentiation, 227 monotonicity of the relative entropy, 524 Monroe, C., 322, 350, [MMK+95], [WMI+98] Mooij, J. E., 351, [MOL+99] Moore\u2019s law, 4 Moore, G., 4 Mor, T., 214, 215, 351, 605,", " 607, [BBB+98], [BMP+99], [Mor98], [VYW+99] Morgan, G. L., 607, [HAD+95] Mosca, M., xix, xxi, 59, 246, 276, 351, [BBC+98], [CEMM98], [JM98], [JMH98], [ME99], [Mos98], [Mos99] Mossba\u00a8uer effect, 312 Motwani, R., 168, [MR95] Muller, A., 607, [MZG96] multiplicative inverse, 627 Murphy, E. A., Jr., 546 mutual information, classical, 506 mutual information, quantum, 514 Nakamura, Y., 351, [NPT99], 20 gate, 130 natural numbers, 625 Nielsen, M. A., xix, 59, 119, 351, 398, 424, 498, 527, 605, 606, [BKN98], [BNS98], [CN97], [LNCY97], [NC97], [NCSB98], [Nie98], [Nie99a], [Nie99b], [NKL98], [SN96] Nisan, N., 424, 499, [ABOIN96], [AKN98] Niu, C.-S., 214, 246, [GN96] NMR, 324, 343 no-cloning theorem, 3, 24, 530, 586 noise classical, 354 quantum, 353 noiseless channel coding theorem, 8, 500 noiseless quantum codes, 498 noisy channel coding theorem, 8 noisy classical channels, 548 non-Abelian groups, 242 non-trace-preserving quantum operations, 367 non-uniform circuit family, 134, 20 gate, 130 norm, 66 norm of a matrix, 645 normal operator, 70 normalization condition for state vectors, 81 normalized vectors, 66 normalizer of Gn, 461 normalizer operations, 461 fault-tolerant, 482 gate, 129 NP, 40, 142, 263 NPI, 149 nuclear spin, 309 number \ufb01eld sieve, 216 O(\u00b7), 136 observable, 87 Ohya, M., 526, [OP93] Olkin, I., 606, [MO79] one time pad, 583 open quantum systems, 353 operation elements, 360 operator-sum representation, 360 freedom in, 370 optical computer, classical, 296 optical lattice, 346 optical pumping, 312 xenon, 341, 20 gate, 130 oracle, 221 for quantum searching, 248 models of computation, 129 order of x modulo N, 226, 633 of a group element, 610 of a permutation, 241 order-\ufb01nding problem, 216, 226, 241, 633 Orlando, T. P., 351, [MOL+99] orthogonal complement, 70 orthogonality, 66 orthonormal decomposition, 69 orthonormality, 66 outer product notation for operators, 67 output of a computation, 123 Ozawa, M., 605, [YO93] P, 40, 141 Pais, A., 59, 112, [Pai82], [Pai86], [Pai91] 672 Index Palma, G. M., 607, [EHPP94] Pan, J. W., 59, [BPM+97] Papadimitriou, C. M., 138, 168, 639, [Pap94] parallel computers, 162 parity check matrix, 447 partial trace, 105 Pashkin, Y. A., 351, [NPT99] Patterson, D. A., 168, [HGP96] Patterson, M., xxi Paturi, R., 276, [Pat92] Pauli gates, see quantum logic gate Pauli group, 454, 611 Pauli matrices, 65, 174, 427 Paz, J.-P., 497, [LMPZ96] Pellizzari, T., 498, [CPZ96] Penrose, R., 59, 167, [Pen89] Peres, A., 59, 112, 118, 119, 607, [BBC+93], [EHPP94], [Per88], [Per93], [Per95] period-\ufb01nding problem, 241 period-\ufb01nding, quantum algorithm for, 236 Perlis, S., 118, [Per52] Petroff, M. D., 350, [KSC+94] Petz, D., 526, 527, [OP93], [Pet86] phase, 81, 93 phase estimation, quantum algorithm for, 221 phase \ufb02ip, 81 phase \ufb02ip channel, 376 phase \ufb02ip operator, 430 phase gate, 174 phase kicks, as a", " model of quantum noise, 384 phase shifter, 288 Phoenix, S. J. D., 607, [BP93] phonon, 311 photodetector, 288 physical quantum operations, 367 Pines, A., xxi Planck\u2019s constant, 82 Platzman, P. M., 351, [PD99] Plenio, M. B., 350, 398, 606, [JP99], [PK96], [PK98], [VP98] Podolsky, B., 17, 25, 119, [EPR35] Poisson equation, 205 polar decomposition, 78 polynomial resources, 139 polynomial time, 141 Polzik, E. S., 59, 352, [FSB+98] Popescu, S., xix, 59, 351, 606, [BBM+98], [BBPS96], [BBP+96], [BCJ+99], [LP97], [LP99], [LSP98] Popeye the Sailor, 60 Poplavskii, R. P., 204, [Pop75] positive de\ufb01nite operator, 71 positive operator, 71 Positive Operator-Valued Measures, 90 POVM, 90 POVM measurements, 90, 118 completeness relation for, 92 elements for, 90 Poyatos, J. F., 398, [PCZ97] Preskill, J., xix, xxi, 214, 499, 606, 607, [BCDP96], [GP10], [Pre97], [Pre98a], [Pre98b], [Pre98c], [SP00] Price, M., 351, [CMP+98] primality decision problem, 141 prime number, 625 prime number theorem, 638 principle of deferred measurement, 185 principle of implicit measurement, 186 privacy ampli\ufb01cation, 584 private key cryptography, 582, 640 problem, Addition by Fourier transforms, 244 Alternate characterization of the \ufb01delity, 423 Alternate universality construction, 212 Analogue of the triangle inequality for conditional entropy, 525 Ancilla bits and ef\ufb01ciency of reversible computation, 167 Classical capacity of a quantum channel \u2013 Research, 603 Computable phase shifts, 212 Computing with linear optics, 346 Conditional forms of strong subadditivity, 526 Control via Jaynes\u2013Cummings interactions, 347 Database retrieval, 275 Ef\ufb01cient temporal labeling, 346 Encoding by teleportation, 496 Encoding stabilizer codes, 495 Feynman-Gates conversation, 58 Find a hard-to-compute class of functions (Research), 167 Finding the minimum, 274 Fractran, 166 Functions of the Pauli matrices, 117 Generalized Klein\u2019s inequality, 525 Generalized quantum searching, 274 Generalized relative entropy, 525 Gilbert\u2013Varshamov bound, 495 Hardness of approximation of, 166 Ion trap logic with two-level atoms, 348 Kitaev\u2019s algorithm, 243 Lindblad form to quantum operation, 395 Linearity forbids cloning, 603 Measured quantum Fourier transform, 243 Methods for achieving capacity \u2013 Research, 603 Minimal Toffoli construction, 213 Minsky machines, 165 Non-Abelian hidden subgroups \u2013 Research, 244 Non-universality of two bit reversible logic, 166 Prime number estimate, 638 Properties of the Schmidt number, 117 Quantum channel capacity \u2013 Research, 603 Quantum searching and cryptography, 275 Random unitary channels, 396 Reversible PSPACE = PSPACE, 167 Reversible Turing machines, 166 Strong subadditivity \u2013 Research, 526 Teleportation as a quantum operation, 395 Tsirelson\u2019s inequality, 118 Undecidability of dynamical systems, 166 Universality with prior entanglement, 213 Vector games, 166 process tomography, 308, 389 program for a Turing machine, 123 projective measurements, 87, 282 projectors, 70 promise problems, 243 pseudocode, 126 PSPACE, 150 public key cryptography, 582 public key cryptosystems, 11, 640 Pueschel, M., 246, [PRB98] Pulver, M., 214, 215, [BMP+99] pure state, 100 puri\ufb01cations, 109, 110, 119 QKD, 582, 586 B92 protocol, 589 BB84 protocol, 587, 593 EPR protocol, 591 security of, 593 quantum chromodynamics, 206 quantum circuits, 22 quantum code, 435 quantum corollary to Moore\u2019s law, 39 quantum cryptography, 10, 582 quantum dots, 344 quantum ef\ufb01ciency, 288 quantum electrodynamics, 2, 206 quantum error-correction, 425 quantum factoring, 216 quantum Fano inequality, 563, 572, 605 quantum \ufb01eld theory, 6 quantum Fourier transform,", " 216 quantum circuit for, 219 quantum gravity, 6 quantum Hall effect, 346 quantum Hamming bound, 444 quantum information theory, 528 data processing inequality, 564 entanglement, 571 entropy exchange, 561 Holevo bound, 531 quantum cryptography, 582 quantum Fano inequality, 561 Schumacher\u2019s noiseless coding theorem, 542 Singleton bound, 568 summary of important relations, 572, 20, 178, 321, 482 quantum key distribution, 586 quantum logic gate \u03c0/8, 174, 485 controlled phase-\ufb02ip, 319 controlledHadamard, 174, 483 Pauli, 483 phase, 174, 483 single qubit, 174, 319 swap, 320 Toffoli, 29, 485 quantum money, 56 quantum noise, 353 estimates of, 278 quantum operations, 353, 353 chi-matrix representation, 391 environmental models for, 363, 365 limitations to the formalism, 394 non-trace-preserving, 367 partial trace map, 374 physical, 367 trace map, 374 trace-preserving, 367 quantum process tomography, 389, 398 quantum search algorithm, 248, 339 quantum searching, 263 quantum source, 542 quantum state tomography, 389, 398 quantum teleportation, see teleportation, quantum quantum trajectories, 398 quantum Turing machine, 203 qubit, 13, 80, 605 charge representation, 344 harmonic oscillator representation, 283 implementation, 277 Index 673 photon representation, 287 spin representation, 309, 324, 345 square well representation, 280 superconductor representation, 344 qutrit, 203, 343, 359 Rabi frequency, 318 Rabi oscillations, 303 Rabin, M. O., 120, 644, [Rab80] Raghavan, P., 168, [MR95] Rahim, H. Z., 59, [Rah99] Raimond, J. M., 350, [DMB+93], [DRBH87], [DRBH95] Rains, E. M., 9, 497, 498, [CRSS97], [CRSS98], [Rai98], [Rai99a], [Rai99b], [Rai99c], [RHSS97] Rajagopalan, S., xxi Ramo, S., 350, [RWvD84] random coding, 549 randomized algorithms, 5 rank of a Hermitian operator, 117 Rasetti, M., 498, [ZR98] read-write tape-head for a Turing machine, 123 Reck, M., 214, 350, [RZBB94] reduced density operator, 105 reducibility of one language to another, 145 reduction, 144 reduction of factoring to order-\ufb01nding, 232, 633 refocusing, 331 relative entropy, 526 classical, 504 quantum, 511 relative phase, 93 remainders, 626 R`enyi entropy, 584 Ressler, A., 168, [Res81] reversible logic gates, 153 reversible Turing machine, 155 Rex, R., 168, [LR90] Risk, W., xxi Risk, W. P., 607, [BR00], [BR98] Risken, H., 398, [VR89] Rivest, R. L., 11, 167, 639, 644, [CLR90], [RSA78] Robert, J. M., 607, [BBR88] Robinson, D. W., 527, [LR68], [RR67] Rockmore, D., 616, [DR90] Roetteler, M., 246, [PRB98], [RB98] Rosen, N., 17, 25, 119, [EPR35] rotation operators, 174 Roychowdhury, V., 214, 215, 351, [BMP+99], [VYW+99] Royer, A., 398, [Roy96] RSA cryptosystem, 11 Ruelle, D., 527, [RR67] Ruskai, M. B., xxi, 424, 527, 605, [KR99], [LR73a], [LR73b], [Rus94] Sakurai, J. J., 59, 118, 349, [Sak95] Saleh, B. E. A., 349, [CST89], [ST91] Salvail, L., 606, 607, [BBB+92], [BS94] Sands, M., 59, 118, 168, [FLS65a], [FLS65b] Sanpera, A., 351, [ZHSL99] scalar, 62 scanning tunneling microscope, 3 Schack, R., 351, [BCJ+99], [SC99] Schauer, M., 607, [HAD+95] Schmidt bases, 110 Schmidt", " co-ef\ufb01cients, 109 674 Index Schmidt decomposition, 109, 119 Schmidt number, 110 Schmidt, E., 119, [Sch06] Schneider, S., xxi Schneier, B., 59, 275, [Sch96a] Schrader, R., xxi Schr\u00a8odinger equation, 82, 205, 280, 284, 301 quantum simulation of, 209 Schr\u00a8odinger\u2019s cat, 387 Schr\u00a8odinger, E., 119, [Sch36] Schulman, L. J., 351, [SV99] Schumacher compression, 547 Schumacher\u2019s quantum noiseless coding theorem, 542 Schumacher, B. W., xxi, 8, 398, 424, 605\u2013607, [BBPS96], [BBP+96], [BCF+96], [BFJS96], [BNS98], [HJS+96], [JS94], [NCSB98], [Sch95], [Sch96b], [SN96], [SW97], [SW98], [SWW96], [WS98] Schwindt, P. D. D., 350, [KMSW99] security of quantum key distribution, 593 selection rules, electric dipole, 300 self-adjoint operator, 70 Selman, A. L., 168, [LLS75] Shamir, A., 11, 641, 644, [RSA78] Shannon entropy, 500, 526 Shannon\u2019s noiseless coding theorem, 500, 537 Shannon, C. E., 8, 59, 526, 604, 605, [Sha48], [SW49] Shasha, D., 59, [SL98] Sherwin, M., 351, [IAB+99] Sherwood, M. H., 351, [LVZ+99], [VYSC99] shift-invariance property of the Fourier transform, 237 Shimony, A., 119, [CHSH69] Shockley, W., 4 Shor code, 432 Shor\u2019s algorithm, 7 Shor, J., 216 Shor, P. W., xix, xxi, 6, 8, 9, 214, 216, 245, 246, 265, 432, 450, 497\u2013499, 604, 606, 607, [BBC+95], [BS98], [CRSS97], [CRSS98], [CS96], [DS96], [DSS98], [RHSS97], [Sho94], [Sho95], [Sho96], [Sho97], [SL97], [SP00], [SS96] similarity transformation, 75 Simon\u2019s problem, 241 Simon, B., 527, [Sim79] Simon, D. R., 246, [Sim94], [Sim97] simple cycle, 143 simple harmonic oscillator, 277, 283 simulation of one computational model by another, 126 simulations, quantum and classical, 204 single photons, 287, 296, 343 Singleton bound, 445 quantum, 568 singular value decomposition, 78 singular values, 79 Sleator, T., 214, [BBC+95] Slepian, D., 59, [Sle74] Slichter, C. P., 351, [Sli96] Sloane, N. J. A., 59, 497, 498, [CRSS97], [CRSS98], [MS77], [RHSS97], [SW93] Slotine, J. E., 352, [LS98] Small, A., 351, [IAB+99] Smolin, J. A., 214, 497, 606, 607, [BBB+92], [BBC+95], [BBP+96], [BDS97], [BDSW96], [BST98], [DSS98], [SS96] Solovay, R., 5, 624, 644, [SS76] Solovay\u2013Kitaev theorem, 197, 617 Somaroo, S. S., 351, [CMP+98], [STH+99] S\u00f8rensen, J. L., 59, 352, [FSB+98] Sornborger, A. T., 214, 215, [SS99] sorting, 137 source\u2013channel coding, 553 space hierarchy theorem, 151 space-bounded computation, 150 span of a set of vectors, 63 spanning set, 62 spatial labeling, 333 spectral decomposition, 70, 72 Spiller, T., xix, [LSP98] spin, 310 spin orbit interaction, 301 spin singlet, 113 spin valve, 345 spin\u2013lattice relaxation, 280, 330 spin\u2013spin relaxation, 280, 330 spontaneous emission, 315, 380 spur, 559 square well, 280 stabil", "izer codes, 453, 465 stabilizer formalism, 453 stabilizer of a vector space, 454 standard deviation of a random variable, 609 standard form for a stabilizer code, 470 for parity check matrix, 447 starting state of a Turing machine, 123 state space, 80, 102 state tomography, 389 with NMR, 336 state vector, 80 static measures of distance, 399 stationary states, 83 Steane code, 453 Steane, A. M., 8, 350, 450, 497, 499, [Ste96a], [Ste96b], [Ste97], [Ste99] Steinberg, A. M., 350, [KSC+94] Stern, O., 43 Stern\u2013Gerlach experiment, 43 Stewart, E. D., 214, 215, [SS99] Stiefel, E., 616, [FS92] Stirzaker, D. R., 609, [GS92] stochastic differential equations, 353 stochastic processes, 355 Stoll, S., xxi Strang, G., 118, [Str76] Strassen, V., xxi, 5, 216, 644, [SS76] Streater, R. F., 398, [LS93] strict concavity, 504 strictly self-dual linear codes, 449 string theory, 6 strong concavity of the \ufb01delity, 414 strong convexity of trace distance, 408 strong subadditivity classical, 506 proof of, 519 quantum, 519 structural complexity, 168 subadditivity classical, 506 quantum, 516 subgroup, 610 subset-sum, 149 superconductor, 344 superdense coding, 17, 97, 119, 275, 352 superposition principle, 94 superpositions, 81 support of a Hermitian operator, 105 swap operation in Fourier transform, 219 Switkes, M., 351, [HSM+98] Szilard, L., 168, [Szi29] T1, 280, 330 T2, 280, 330 tape for a Turing machine, 123 Tapp, A., 59, 276, [BBHT98], [BHT98] Tarjan, R., 217 Taylor, W., 214, [BT97] Teich, M. C., 349, [ST91] teleportation, quantum, 17, 26, 352 temporal labeling, 333 tensor product, 62, 71, 72 Terhal, B. M., 215, 606, [BST98], [TD98] Thapliyal, A., 119 theorem Z-Y decomposition for a single qubit, 175 Basic properties of Shannon entropy, 506 Basic properties of von Neumann entropy, 513 Birkhoff\u2019s theorem, 574 Chaining rule for conditional entropies, 508 Characterization of density operators, 101 Chinese remainder theorem, 629 Concavity of the quantum conditional entropy, 520 Convexity of the relative entropy, 520 Cook\u2013Levin, 146 Data processing inequality, 509 Error-correction conditions for stabilizer codes, 466 Euler\u2019s theorem, 144 Fannes\u2019 inequality, 512 Fermat\u2019s little theorem, 630 Fundamental theorem, 613 Fundamental theorem of arithmetic, 625 Gottesman\u2013Knill theorem, 464 High \ufb01delity implies low entropy, 594 Holevo\u2013Schumacher\u2013Westmoreland (HSW) theorem, 555 Information gain implies disturbance, 586 Klein\u2019s inequality, 511 Lagrange\u2019s theorem, 610 Law of large numbers, 541 Lieb\u2019s theorem, 520, 646 Monotonicity of the \ufb01delity, 414 Monotonicity of the relative entropy, 524 Non-negativity of the relative entropy, 505 Polar decomposition, 78 Projective measurements increase entropy, 515 Quantum data processing inequality, 564 Quantum error-correction conditions, 436 Quantum Fano inequality, 563 Representation theorem for the gcd, 626 Schmidt decomposition, 109 Schumacher\u2019s noiseless channel coding theorem, 544 Schur\u2019s lemma, 613 Shannon\u2019s noiseless channel coding theorem, 540 Shannon\u2019s noisy channel coding theorem, 553 Simultaneous diagonalization theorem, 77 Singular value decomposition, 79 Solovay\u2013Kitaev theorem, 618 Spectral decomposition, 72 Strong concavity of the \ufb01delity, 414 Strong convexity of the trace distance, 407 Strong subadditivity, 521 Subadditivity of the conditional entropy, 523 Index 675 The Chernoff bound, 154 The Holevo bound, 531 Theorem of typical sequences, 539 Trace-preserving quantum operations are contractive, 406 Trotter formula, 207 Typical subspace theorem, 543 Uhlmann\u2019s theorem, 410 Un", "itary freedom in the ensemble for density matrices, 103 Unitary freedom in the operator-sum representation, 372 thermal equilibrium, 328 Thomas, J. A., 59, 526, 539, 604, 605, [CT91] threshold condition, 480, 493 threshold for quantum computation, 493 threshold theorem, 425 Tian, L., 351, [MOL+99] Tiech, M. C., 349, [CST89] time hierarchy theorem, 151 TIMEf (n), 141 TOCSY, 339 Toffoli gate, 29, 155, 159 control bits, 159 fault-tolerant, 488 target bit, 159 Toffoli, T., 168, [FT82] tomography, 389 trace distance, 618 classical, 400 operational meaning for, 400 quantum, 403 trace inner product, 76 trace of a matrix, 75 trace-preserving quantum operations, 360, 367 tractable computational problems, 139 transistor, 4 transpose, 62, 70 transpose operation, 368 transversal property of fault-tolerant operations, 483 triangle inequality, 516 Tromp, J., 169, [LTV98] Trotter formula, 207 Trotter, H. F., 214, [Tro59] Tsai, J. S., 351, [NPT99] Tseng, C. H., 351, [STH+99] Tsirelson\u2019s inequality, 118, 119 Tsirelson, B. S., 119, [Tsi80] Turchette, Q. A., 306, 307, 350, 398, [THL+95], [Tur97] Turing machine, 4, 120, 122 quantum, 203 Turing number, 125 Turing, A. M., 4, 59, 122, 125, 167, [Tur36] two-level atom, 298, 327 resonance, 326 two-level unitary matrices, 189 typical sequences, 538 Ueda, M., 350, [KU91] Uf\ufb01nk, J. H. B., 526, [MU88] Uhlmann\u2019s formula, 411 Uhlmann\u2019s theorem, 410 Uhlmann, A., xxi, 119, 424, 527, 606, [Uhl70], [Uhl71], [Uhl72], [Uhl73], [Uhl76], [Uhl77] Umegaki, H., 526, [Ume62] uncomputation, 158 Wigner, E. P., 527, [WY63] Wilczek, F., 215, [CW95] Williams, D., 609, [Wil91] Wilson, E. O., 1 Wineland, D. J., xxi, 323, 350, [MMK+95], [WMI+98] Winfree, E., 168, [Win98] Winograd, S., 168, 498, [WC67] Wiseman, H., xxi witness, 142 Wokaun, A., 351, [EBW87] Wootters, W. K., 59, 119, 497, 604\u2013606, [BBC+93], [BBP+96], [BDSW96], [HJS+96], [HJW93], [SWW96], [WZ82] word of length l from G, 618 work bits, 131 Wright, E. M., 246, 639, [HW60] Wyner, A. D., 59, [SW93], 20 gate, 130 Yablonovitch, E., 351, [VYW+99] Yamaguchi, F., 351, [YY99] Yamamoto, Y., xxi, 349\u2013351, 498, [CY95], [IY94], [LNCY97], [WY90], [YKI88], [YY99] Yanase, M. M., 527, [WY63] Yannoni, C. S., 351, [LVZ+99], [VYSC99] Yao, A. C., 214, [Yao93] Yard, J., xxi Younis, S., 168, [YK95] Yuen, H. P., 605, [YO93] Yurke, B., xxi, 349 Z gate, 19 Zalka, C., 214, 215, 276, [Zal98], [Zal99] Zanardi, P., 498, [Zan99], [ZR98] Zbinden, H., 607, [MZG96] Zeilinger, A., 59, 119, 214, 350, [BPM+97], [MWKZ96], [RZBB94] zero operator, 63 zero vector, 62 Zhou, X. L., x", "xi, 246, 351, 499, [CVZ+98], [LVZ+99], [ZLC00] Zoller, P., 350, 398, 498, [CPZ96], [CZ95], [PCZ97], [ZG97] Zurek, W. H., xxi, 351, 398, 497\u2013499, 604, 606, [CMP+98], [KLZ98a], [KLZ98b], [LMPZ96], [WZ82], [ZL96], [Zur89], [Zur91] Zyczkowski, K., 351, [ZHSL99] 676 Index uniform circuit family, 134 unit vector, 66 unitary evolution, 356 unitary operators, 70 approximation, 194 ef\ufb01cient decompositions, 191, 198 Universal Turing Machine, 4, 128 universality, 281 discrete set of universal operations, 194 family of quantum gates, 188, 281 of the Fredkin gate, 157 proof of, 132 single qubit gates and two-level unitary operators, 189, 191 Vaidman, L., 352, [Vai94] van Dam, W., 276, [van98a] van de Graaf, J, 424 van de Graaf, J., 424, 607, [BBB+98], [FvdG99] van der Waal, C. H., 351, [MOL+99] van Duzer, T., 350, [RWvD84] van Enk, S. J., xxi, 605, [van98b] van Oorschot, P. C., 59, 275, [MvOV96] Vandersypen, L. M. K., xxi, 215, 351, [CVZ+98], [LVZ+99], [VYSC99] Vanstone, S. A., 59, 275, [MvOV96] variance of a random variable, 609 Vatan, F., 214, 215, [BMP+99] Vazirani, U., xxi, 200, 214, 276, 351, [BBBV97], [BV97], [SV99] vector game, 166 vector spaces, 61 vector subspace, 62 vectors, 61 Vedral, V., 606, [Ved99], [VP98] Verhulst, A., xxi Vernam cipher, 583 vertex cover, 149 vertices, 143 Vidal, G., 351, 606, [Vid98], [Vid99] Viola, L., 498, [KLV99] Vitanyi, P., 169, [LTV98], [LV96] Vogel, K., 398, [VR89] von Neumann entropy, 54, 510, 526 von Neumann, J., 4, 119, 164, 168, 498, [von27], [von56], [von66] Vrijen, R., 351, [VYW+99] Wallach, D., xxi Walsh\u2013Hadamard transform, 31 Wang, K., 351, [VYW+99] Warren, W., 351, [War97] Watanabe, K., 350, [WY90] Watrous, J., 214, [Wat99] weakly self-dual linear codes, 449 Weaver, W., 59, 604, 605, [SW49] Wehrl, A., 526, 527, [Weh78] Weinfurter, H., 59, 119, 214, [BBC+95], [BPM+97], [MWKZ96] Welsh, D. J. A., 497, [Wel88] Westmoreland, M. D., xxi, 605, 607, [HJS+96], [SW97], [SW98], [SWW96], [WS98] Whaley, K. B., 498, [BKLW99], [LBW99], [LCW98] Whinnery, J. R., 350, [RWvD84] White, A. G., 350, [KMSW99] Wiesner, S. J., 9, 10, 119, 214, 215, 606, [BW92], 663, [Wie83], [Wie96], [Wie]ent. (The potential terms are invariant even when \u03b1 = \u03b1(x)). 1.4 The Hamiltonian Formalism The link between the Lagrangian formalism and the quantum theory goes via the path integral. In this course we will not discuss path integral methods, and focus instead on canonical quantization. For this we need the Hamiltonian formalism of \ufb01eld theory. We start by de\ufb01ning", " the momentum \u03c0a(x) conjugate to \u03c6a(x), \u03c0a(x) = \u2202L \u2202 \u02d9\u03c6a (1.68) The conjugate momentum \u03c0a(x) is a function of x, just like the \ufb01eld \u03c6a(x) itself. It is not to be confused with the total momentum P i de\ufb01ned in (1.43) which is a single number characterizing the whole \ufb01eld con\ufb01guration. The Hamiltonian density is given by H = \u03c0a(x) \u02d9\u03c6a(x) \u2212 L(x) (1.69) where, as in classical mechanics, we eliminate \u02d9\u03c6a(x) in favour of \u03c0a(x) everywhere in H. The Hamiltonian is then simply H = d3x H (1.70) \u2013 19 \u2013 An Example: A Real Scalar Field For the Lagrangian L = 1 2 \u02d9\u03c62 \u2212 1 2(\u2207\u03c6)2 \u2212 V (\u03c6) the momentum is given by \u03c0 = \u02d9\u03c6, which gives us the Hamiltonian, H = d3x 1 2\u03c02 + 1 2(\u2207\u03c6)2 + V (\u03c6) (1.71) (1.72) Notice that the Hamiltonian agrees with the de\ufb01nition of the total energy (1.45) that we get from applying Noether\u2019s theorem for time translation invariance. In the Lagrangian formalism, Lorentz invariance is clear for all to see since the action is invariant under Lorentz transformations. In contrast, the Hamiltonian formalism is not manifestly Lorentz invariant: we have picked a preferred time. For example, the equations of motion for \u03c6(x) = \u03c6(x, t) arise from Hamilton\u2019s equations, \u02d9\u03c6(x, t) = \u2202H \u2202\u03c0(x, t) and \u02d9\u03c0(x, t) = \u2212 \u2202H \u2202\u03c6(x, t) (1.73) which, unlike the Euler-Lagrange equations (1.6), do not look Lorentz invariant. Nevertheless, even though the Hamiltonian framework doesn\u2019t look Lorentz invariant, the physics must remain unchanged. If we start from a relativistic theory, all \ufb01nal answers must be Lorentz invariant even if it\u2019s not manifest at intermediate steps. We will pause at several points along the quantum route to check that this is indeed the case. \u2013 20 \u2013 2. Free Fields \u201cThe career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.\u201d Sidney Coleman 2.1 Canonical Quantization In quantum mechanics, canonical quantization is a recipe that takes us from the Hamiltonian formalism of classical dynamics to the quantum theory. The recipe tells us to take the generalized coordinates qa and their conjugate momenta pa and promote them to operators. The Poisson bracket structure of classical mechanics morphs into the structure of commutation relations between operators, so that, in units with = 1, [qa, qb] = [pa, pb] = 0 [qa, pb] = i \u03b4b a (2.1) In \ufb01eld theory we do the same, now for the \ufb01eld \u03c6a(x) and its momentum conjugate \u03c0b(x). Thus a quantum \ufb01eld is an operator valued function of space obeying the commutation relations [\u03c6a(x), \u03c6b(y)] = [\u03c0a(x), \u03c0b(y)] = 0 [\u03c6a(x), \u03c0b(y)] = i\u03b4(3)(x \u2212 y) \u03b4b a (2.2) Note that we\u2019ve lost all track of Lorentz invariance since we have separated space x and time t. We are working in the Schr\u00a8odinger picture so that the operators \u03c6a(x) and \u03c0a(x) do not depend on time at all \u2014 only on space. All time dependence sits in the states |\u03c8 which evolve by the usual Schr\u00a8odinger equation i d|\u03c8 dt = H |\u03c8 (2.3) We aren\u2019t doing anything di\ufb00erent from usual quantum mechanics; we\u2019re merely applying the old formalism to \ufb01elds. Be warned however that the notation |\u03c8 for the state is deceptively simple: if you were to write the wavefunction in quantum \ufb01eld", " theory, it would be a functional, that is a function of every possible con\ufb01guration of the \ufb01eld \u03c6. The typical information we want to know about a quantum theory is the spectrum of the Hamiltonian H. In quantum \ufb01eld theories, this is usually very hard. One reason for this is that we have an in\ufb01nite number of degrees of freedom \u2014 at least one for every point x in space. However, for certain theories \u2014 known as free theories \u2014 we can \ufb01nd a way to write the dynamics such that each degree of freedom evolves independently \u2013 21 \u2013 from all the others. Free \ufb01eld theories typically have Lagrangians which are quadratic in the \ufb01elds, so that the equations of motion are linear. For example, the simplest relativistic free theory is the classical Klein-Gordon (KG) equation for a real scalar \ufb01eld \u03c6(x, t), \u2202\u00b5\u2202\u00b5\u03c6 + m2\u03c6 = 0 (2.4) To exhibit the coordinates in which the degrees of freedom decouple from each other, we need only take the Fourier transform, Then \u03c6(p, t) satis\ufb01es \u03c6(x, t) = d3p (2\u03c0)3 eip\u00b7x \u03c6(p, t) \u22022 \u2202t2 + (p 2 + m2) \u03c6(p, t) = 0 (2.5) (2.6) Thus, for each value of p, \u03c6(p, t) solves the equation of a harmonic oscillator vibrating at frequency \u03c9p = + p 2 + m2 (2.7) We learn that the most general solution to the KG equation is a linear superposition of simple harmonic oscillators, each vibrating at a di\ufb00erent frequency with a di\ufb00erent amplitude. To quantize \u03c6(x, t) we must simply quantize this in\ufb01nite number of harmonic oscillators. Let\u2019s recall how to do this. 2.1.1 The Simple Harmonic Oscillator Consider the quantum mechanical Hamiltonian H = 1 2p2 + 1 2\u03c92q2 (2.8) with the canonical commutation relations [q, p] = i. To \ufb01nd the spectrum we de\ufb01ne the creation and annihilation operators (also known as raising/lowering operators, or sometimes ladder operators) \u03c9 2 \u03c9 2 a\u2020 = (2.9 \u221a 2\u03c9 i \u221a 2\u03c9 p which can be easily inverted to give q = 1 \u221a 2\u03c9 (a + a\u2020), p = \u2212i \u03c9 2 (a \u2212 a\u2020) (2.10) \u2013 22 \u2013 Substituting into the above expressions we \ufb01nd [a, a\u2020] = 1 while the Hamiltonian is given by H = 1 2\u03c9(aa\u2020 + a\u2020a) = \u03c9(a\u2020a + 1 2) (2.11) (2.12) One can easily con\ufb01rm that the commutators between the Hamiltonian and the creation and annihilation operators are given by [H, a\u2020] = \u03c9a\u2020 and [H, a] = \u2212\u03c9a (2.13) These relations ensure that a and a\u2020 take us between energy eigenstates. Let |E be an eigenstate with energy E, so that H |E = E |E. Then we can construct more eigenstates by acting with a and a\u2020, Ha\u2020 |E = (E + \u03c9)a\u2020 |E, Ha |E = (E \u2212 \u03c9)a |E (2.14) So we \ufb01nd that the system has a ladder of states with energies..., E \u2212 \u03c9, E, E + \u03c9, E + 2\u03c9,... (2.15) If the energy is bounded below, there must be a ground state |0 which satis\ufb01es a |0 = 0. This has ground state energy (also known as zero point energy), H |0 = 1 2\u03c9 |0 Excited states then arise from repeated application of a\u2020, |n = (a\u2020)n |0 with H |n = (n + 1 2)\u03c9 |n where I\u2019ve ignored the normalization of these states so, n| n = 1. 2.2 The Free Scalar Field (2.16) (2.17) We now apply the quantization of the harmonic oscillator to the free scalar \ufb01eld. We write \u03c6 and \u03c0 as a linear sum of an in\ufb01nite number of creation and annihilation operators a\u2020 p and", " ap, indexed by the 3-momentum p, \u03c6(x) = d3p (2\u03c0)3 1 2\u03c9p ap eip\u00b7x + a\u2020 p e\u2212ip\u00b7x \u03c0(x) = d3p (2\u03c0)3 (\u2212i) \u03c9p 2 ap eip\u00b7x \u2212 a\u2020 p e\u2212ip\u00b7x \u2013 23 \u2013 (2.18) (2.19) Claim: The commutation relations for \u03c6 and \u03c0 are equivalent to the following commutation relations for ap and a\u2020 p [\u03c6(x), \u03c6(y)] = [\u03c0(x), \u03c0(y)] = 0 [\u03c6(x), \u03c0(y)] = i\u03b4(3)(x \u2212 y) \u21d4 [ap, aq] = [a\u2020 p, a\u2020 q] = (2\u03c0)3\u03b4(3)(p \u2212 q) q] = 0 [ap, a\u2020 (2.20) Proof: We\u2019ll show this just one way. Assume that [ap, a\u2020 [\u03c6(x), \u03c0(y)] = d3p d3q (\u2212i) (2\u03c0)6 2 d3p (\u2212i) (2\u03c0)3 2 = i\u03b4(3)(x \u2212 y) = \u03c9q \u03c9p \u2212[ap, a\u2020 q] eip\u00b7x\u2212iq\u00b7y + [a\u2020 \u2212eip\u00b7(x\u2212y) \u2212 eip\u00b7(y\u2212x) q] = (2\u03c0)3\u03b4(3)(p \u2212 q). Then p, aq] e\u2212ip\u00b7x+iq\u00b7y (2.21) The Hamiltonian Let\u2019s now compute the Hamiltonian in terms of ap and a\u2020 p. We have d3x \u03c02 + (\u2207\u03c6)2 + m2\u03c62 d3x d3p d3q (2\u03c0)6 1 \u03c9p\u03c9q m2 \u221a \u03c9p\u03c9q 2 d3p 1 (2\u03c0)3 \u03c9p + \u2212 \u221a \u03c9p\u03c9q 2 (ap eip\u00b7x \u2212 a\u2020 p e\u2212ip\u00b7x)(aq eiq\u00b7x \u2212 a\u2020 q e\u2212iq\u00b7x) (ip ap eip\u00b7x \u2212 ip a\u2020 p e\u2212ip\u00b7x) \u00b7 (iq aq eiq\u00b7x \u2212 iq a\u2020 q e\u2212iq\u00b7x) (ap eip\u00b7x + a\u2020 p e\u2212ip\u00b7x)(aq eiq\u00b7x + a\u2020 q e\u2212iq\u00b7x) (\u2212\u03c92 p + p 2 + m2)(ap a\u2212p + a\u2020 p a\u2020 \u2212p) + (\u03c92 p + p 2 + m2)(ap a\u2020 p + a\u2020 p ap) where in the second line we\u2019ve used the expressions for \u03c6 and \u03c0 given in (2.18) and (2.19); to get to the third line we\u2019ve integrated over d3x to get delta-functions \u03b4(3)(p\u00b1q) which, in turn, allow us to perform the d3q integral. Now using the expression for the frequency \u03c92 p = p 2 + m2, the \ufb01rst term vanishes and we\u2019re left with 1 2 H = = d3p (2\u03c0)3 \u03c9p d3p (2\u03c0)3 \u03c9p ap a\u2020 p + a\u2020 p ap a\u2020 p ap + 1 2(2\u03c0)3 \u03b4(3)(0) (2.22) \u2013 24 \u2013 Hmmmm. We\u2019ve found a delta-function, evaluated at zero where it has its in\ufb01nite spike. Moreover, the integral over \u03c9p diverges at large p. What to do? Let\u2019s start by looking at the ground state where this in\ufb01nity \ufb01rst becomes apparent. 2.3 The Vacuum Following our procedure for the harmonic oscillator, let\u2019s de\ufb01ne the vacuum |0 by insisting that it is annihilated by all ap, ap |0 = 0 \u2200 p (2.23) With this de\ufb01nition, the energy E0 of the ground state comes from the second term in (2.22), \u03c9p \u03b4(3)(0) d3p H |0 \u2261 E0 |0 = 1 2 The subject of quantum \ufb01eld theory is rife with in\ufb01nities. Each tells us something important, usually that we\u2019re doing something wrong, or asking the wrong question. Let\u2019s take some time to explore where this in\ufb01nity comes from and how we should deal with it", ". | 0 = \u221e |0 (2.24) In fact there are two di\ufb00erent \u221e\u2019s lurking in the expression (2.24). The \ufb01rst arises because space is in\ufb01nitely large. (In\ufb01nities of this type are often referred to as infra-red divergences although in this case the \u221e is so simple that it barely deserves this name). To extract out this in\ufb01nity, let\u2019s consider putting the theory in a box with sides of length L. We impose periodic boundary conditions on the \ufb01eld. Then, taking the limit where L \u2192 \u221e, we get (2\u03c0)3 \u03b4(3)(0) = lim L\u2192\u221e L/2 \u2212L/2 d3x eix\u00b7p p=0 = lim L\u2192\u221e L/2 \u2212L/2 d3x = V (2.25) where V is the volume of the box. So the \u03b4(0) divergence arises because we\u2019re computing the total energy, rather than the energy density E0. To \ufb01nd E0 we can simply divide by the volume, E0 = = E0 V d3p (2\u03c0)3 1 2 \u03c9p (2.26) which is still in\ufb01nite. We recognize it as the sum of ground state energies for each harmonic oscillator. But E0 \u2192 \u221e due to the |p| \u2192 \u221e limit of the integral. This is a high frequency \u2014 or short distance \u2014 in\ufb01nity known as an ultra-violet divergence. This divergence arises because of our hubris. We\u2019ve assumed that our theory is valid to arbitrarily short distance scales, corresponding to arbitrarily high energies. This is clearly absurd. The integral should be cut-o\ufb00 at high momentum in order to re\ufb02ect the fact that our theory is likely to break down in some way. \u2013 25 \u2013 We can deal with the in\ufb01nity in (2.24) in a more practical way. In physics we\u2019re only interested in energy di\ufb00erences. There\u2019s no way to measure E0 directly, so we can simply rede\ufb01ne the Hamiltonian by subtracting o\ufb00 this in\ufb01nity, H = d3p (2\u03c0)3 \u03c9p a\u2020 p ap (2.27) so that, with this new de\ufb01nition, H |0 = 0. In fact, the di\ufb00erence between this Hamiltonian and the previous one is merely an ordering ambiguity in moving from the classical theory to the quantum theory. For example, if we de\ufb01ned the Hamiltonian of the harmonic oscillator to be H = (1/2)(\u03c9q \u2212 ip)(\u03c9q + ip), which is classically the same as our original choice, then upon quantization it would naturally give H = \u03c9a\u2020a as in (2.27). This type of ordering ambiguity arises a lot in \ufb01eld theories. We\u2019ll come across a number of ways of dealing with it. The method that we\u2019ve used above is called normal ordering. De\ufb01nition: We write the normal ordered string of operators \u03c61(x1)... \u03c6n(xn) as : \u03c61(x1)... \u03c6n(xn) : (2.28) It is de\ufb01ned to be the usual product with all annihilation operators ap placed to the right. So, for the Hamiltonian, we could write (2.27) as : H := d3p (2\u03c0)3 \u03c9p a\u2020 p ap (2.29) In the remainder of this section, we will normal order all operators in this manner. 2.3.1 The Cosmological Constant Above I wrote \u201cthere\u2019s no way to measure E0 directly\u201d. There is a BIG caveat here: gravity is supposed to see everything! The sum of all the zero point energies should contribute to the stress-energy tensor that appears on the right-hand side of Einstein\u2019s equations. We expect them to appear as a cosmological constant \u039b = E0/V, R\u00b5\u03bd \u2212 1 2Rg\u00b5\u03bd = \u22128\u03c0GT\u00b5\u03bd + \u039bg\u00b5\u03bd (2.30) Current observation suggests that 70% of the energy density in the universe has the properties of a cosmological constant with \u039b \u223c (10\u22123eV )4. This is much smaller than other scales in particle physics. In particular, the Standard Model is valid at least up to 1012 eV. Why don\u2019", "t the zero point energies of these \ufb01elds contribute to \u039b? Or, if they do, what cancels them to such high accuracy? This is the cosmological constant problem. No one knows the answer! \u2013 26 \u2013 2.3.2 The Casimir E\ufb00ect \u201cI mentioned my results to Niels Bohr, during a walk. That is nice, he said, that is something new... and he mumbled something about zero-point energy.\u201d Hendrik Casimir Using the normal ordering prescription we can happily set E0 = 0, while chanting the mantra that only energy di\ufb00erences can be measured. But we should be careful, for there is a situation where di\ufb00erences in the energy of vacuum \ufb02uctuations themselves can be measured. To regulate the infra-red divergences, we\u2019ll make the x1 direction periodic, with size L, and impose periodic boundary conditions such that \u03c6(x) = \u03c6(x + Ln) (2.31) with n = (1, 0, 0). We\u2019ll leave y and z alone, but remember that we should compute all physical quantities per unit area A. We insert two re\ufb02ecting plates, separated by a distance d L in the x1 direction. The plates are such that they impose \u03c6(x) = 0 at the position of the plates. The presence of these plates a\ufb00ects the Fourier decomposition of the \ufb01eld and, in particular, means that the momentum of the \ufb01eld inside the plates is quantized as p = n\u03c0 d, py, pz n \u2208 Z+ (2.32) Figure 3: For a massless scalar \ufb01eld, the ground state energy between the plates is E(d) A = \u221e n=1 dpydpz (2\u03c0)2 n\u03c0 d 1 2 2 + p2 y + p2 z while the energy outside the plates is E(L \u2212 d). The total energy is therefore E = E(d) + E(L \u2212 d) (2.33) (2.34) which \u2013 at least naively \u2013 depends on d. If this naive guess is true, it would mean that there is a force on the plates due to the \ufb02uctuations of the vacuum. This is the Casimir force, \ufb01rst predicted in 1948 and observed 10 years later. In the real world, the e\ufb00ect is due to the vacuum \ufb02uctuations of the electromagnetic \ufb01eld, with the boundary conditions imposed by conducting plates. Here we model this e\ufb00ect with a scalar. \u2013 27 \u2013 Ld But there\u2019s a problem. E is in\ufb01nite! What to do? The problem comes from the arbitrarily high momentum modes. We could regulate this in a number of di\ufb00erent ways. Physically one could argue that any real plate cannot re\ufb02ect waves of arbitrarily high frequency: at some point, things begin to leak. Mathematically, we want to \ufb01nd a way to neglect modes of momentum p a\u22121 for some distance scale a d, known as the ultra-violet (UV) cut-o\ufb00. One way to do this is to change the integral (2.33) to, E(d) A = \u221e n=1 dpydpz (2\u03c0)2 n\u03c0 d 1 2 2 + p2 y + p2 z ( n\u03c0 d )2 e\u2212a +p2 y+p2 z (2.35) which has the property that as a \u2192 0, we regain the full, in\ufb01nite, expression (2.33). However (2.35) is \ufb01nite, and gives us something we can easily work with. Of course, we made it \ufb01nite in a rather ad-hoc manner and we better make sure that any physical quantity we calculate doesn\u2019t depend on the UV cut-o\ufb00 a, otherwise it\u2019s not something we can really trust. The integral (2.35) is do-able, but a little complicated. It\u2019s a lot simpler if we look at the problem in d = 1 + 1 dimensions, rather than d = 3 + 1 dimensions. We\u2019ll \ufb01nd that all the same physics is at play. Now the energy is given by E1+1(d) = \u03c0 2d \u221e n=1 n (2.36) We now regulate this sum by introducing the UV cuto\ufb00 a introduced above. This renders the expression \ufb01nite, allowing us to start manipulating it thus, E1+1", "(d) \u2192 \u03c0 2d = \u2212 = \u2212 1 2 1 2 \u221e n e\u2212an\u03c0/d n=1 \u2202 \u2202a \u221e n=1 e\u2212an\u03c0/d \u2202 \u2202a 1 1 \u2212 e\u2212a\u03c0/d ea\u03c0/d (ea\u03c0/d \u2212 1)2 \u03c0 24d = \u03c0 2d d 2\u03c0a2 \u2212 where, in the last line, we\u2019ve used the fact that a d. We can now compute the full energy, + O(a2) (2.37) = E1+1 = E1+1(d) + E1+1(L \u2212 d) = L 2\u03c0a2 \u2212 \u03c0 24 1 d + 1 L \u2212 d + O(a2) (2.38) \u2013 28 \u2013 This is still in\ufb01nite in the limit a \u2192 0, which is to be expected. However, the force is given by \u2202E1+1 \u2202d = \u03c0 24d2 +... (2.39) where the... include terms of size d/L and a/d. The key point is that as we remove both the regulators, and take a \u2192 0 and L \u2192 \u221e, the force between the plates remains \ufb01nite. This is the Casimir force2. If we ploughed through the analogous calculation in d = 3 + 1 dimensions, and performed the integral (2.35), we would \ufb01nd the result 1 A \u2202E \u2202d = \u03c02 480d4 (2.40) The true Casimir force is twice as large as this, due to the two polarization states of the photon. 2.4 Particles Having dealt with the vacuum, we can now turn to the excitations of the \ufb01eld. It\u2019s easy to verify that [H, a\u2020 p] = \u03c9p a\u2020 p and [H, ap] = \u2212\u03c9p ap (2.41) which means that, just as for the harmonic oscillator, we can construct energy eigenstates by acting on the vacuum |0 with a\u2020 p. Let This state has energy |p = a\u2020 p |0 H |p = \u03c9p |p with p = p 2 + m2 \u03c92 (2.42) (2.43) But we recognize this as the relativistic dispersion relation for a particle of mass m and 3-momentum p, E2 p = p 2 + m2 (2.44) 2The number 24 that appears in the denominator of the one-dimensional Casimir force plays a more famous role in string theory: the same calculation in that context is the reason the bosonic string lives in 26 = 24 + 2 spacetime dimensions. (The +2 comes from the fact the string itself is extended in one space and one time dimension). You will need to attend next term\u2019s \u201cString Theory\u201d course to see what on earth this has to do with the Casimir force. \u2013 29 \u2013 We interpret the state |p as the momentum eigenstate of a single particle of mass m. To stress this, from now on we\u2019ll write Ep everywhere instead of \u03c9p. Let\u2019s check this particle interpretation by studying the other quantum numbers of |p. We may take the classical total momentum P given in (1.46) and turn it into an operator. After normal ordering, it becomes P = \u2212 d3x \u03c0 \u2207\u03c6 = d3p (2\u03c0)3 p a\u2020 p ap Acting on our state |p with P, we learn that it is indeed an eigenstate, P |p = p |p (2.45) (2.46) telling us that the state |p has momentum p. Another property of |p that we can study is its angular momentum. Once again, we may take the classical expression for the total angular momentum of the \ufb01eld (1.55) and turn it into an operator, J i = ijk d3x (J 0)jk (2.47) It\u2019s not hard to show that acting on the one-particle state with zero momentum, J i |p = 0 = 0, which we interpret as telling us that the particle carries no internal angular momentum. In other words, quantizing a scalar \ufb01eld gives rise to a spin 0 particle. Multi-Particle States, Bosonic Statistics and Fock Space We can create multi-particle states by acting multiple times with a\u2020\u2019s. We interpret the state in which n a\u2020\u2019s act on the vacuum as an n-particle state, |p1,..., pn = a\u2020 p1... a", "\u2020 pn |0 (2.48) Because all the a\u2020\u2019s commute among themselves, the state is symmetric under exchange of any two particles. For example, |p, q = |q, p (2.49) This means that the particles are bosons. The full Hilbert space of our theory is spanned by acting on the vacuum with all possible combinations of a\u2020\u2019s, |0, a\u2020 p |0, a\u2020 p a\u2020 q |0, a\u2020 p a\u2020 q a\u2020 r |0... (2.50) \u2013 30 \u2013 This space is known as a Fock space. The Fock space is simply the sum of the n-particle Hilbert spaces, for all n \u2265 0. There is a useful operator which counts the number of particles in a given state in the Fock space. It is called the number operator N N = d3p (2\u03c0)3 a\u2020 p ap (2.51) and satis\ufb01es N |p1,..., pn = n |p1,..., pn. The number operator commutes with the Hamiltonian, [N, H] = 0, ensuring that particle number is conserved. This means that we can place ourselves in the n-particle sector, and stay there. This is a property of interactions free theories, but will no longer be true when we consider interactions: create and destroy particles, taking us between the di\ufb00erent sectors in the Fock space. Operator Valued Distributions Although we\u2019re referring to the states |p as \u201cparticles\u201d, they\u2019re not localized in space in any way \u2014 they are momentum eigenstates. Recall that in quantum mechanics the position and momentum eigenstates are not good elements of the Hilbert space since they are not normalizable (they normalize to delta-functions). Similarly, in quantum \ufb01eld theory neither the operators \u03c6(x), nor ap are good operators acting on the Fock space. This is because they don\u2019t produce normalizable states. For example, 0| ap a\u2020 p |0 = p| p = (2\u03c0)3\u03b4(0) and 0|\u03c6(x) \u03c6(x) |0 = x| x = \u03b4(0) (2.52) They are operator valued distributions, rather than functions. This means that although \u03c6(x) has a well de\ufb01ned vacuum expectation value, 0| \u03c6(x) |0 = 0, the \ufb02uctuations of the operator at a \ufb01xed point are in\ufb01nite, 0| \u03c6(x)\u03c6(x) |0 = \u221e. We can construct well de\ufb01ned operators by smearing these distributions over space. For example, we can create a wavepacket |\u03d5 = d3p (2\u03c0)3 e\u2212ip\u00b7x \u03d5(p) |p (2.53) which is partially localized in both position and momentum space. (A typical state might be described by the Gaussian \u03d5(p) = exp(\u2212p 2/2m2)). 2.4.1 Relativistic Normalization We have de\ufb01ned the vacuum |0 which we normalize as 0| 0 = 1. The one-particle states |p = a\u2020 p |0 then satisfy p| q = (2\u03c0)3 \u03b4(3)(p \u2212 q) (2.54) \u2013 31 \u2013 But is this Lorentz invariant? It\u2019s not obvious because we only have 3-vectors. What could go wrong? Suppose we have a Lorentz transformation p \u00b5 \u2192 (p)\u00b5 = \u039b\u00b5 \u03bdp \u03bd (2.55) such that the 3-vector transforms as p \u2192 p. preferable if the two states are related by a unitary transformation, In the quantum theory, it would be |p \u2192 |p = U (\u039b) |p (2.56) This would mean that the normalizations of |p and |p are the same whenever p and p are related by a Lorentz transformation. But we haven\u2019t been at all careful with normalizations. In general, we could get |p \u2192 \u03bb(p, p ) |p (2.57) for some unknown function \u03bb(p, p ). How do we \ufb01gure this out? The trick is to look at an object which we know is Lorentz invariant. One such object is the identity operator on one-particle states (which is really the projection operator onto one-particle states). With the normalization (2.54) we know this is given by 1 = d3p", " (2\u03c0)3 |p p| (2.58) This operator is Lorentz invariant, but it consists of two terms: the measure d3p and the projector |pp|. Are these individually Lorentz invariant? In fact the answer is no. Claim The Lorentz invariant measure is, d3p 2Ep (2.59) Proof: d4p is obviously Lorentz invariant. And the relativistic dispersion relation for a massive particle, p\u00b5p\u00b5 = m2 \u21d2 p 2 0 = E2 p = p 2 + m2 (2.60) is also Lorentz invariant. Solving for p0, there are two branches of solutions: p0 = \u00b1Ep. But the choice of branch is another Lorentz invariant concept. So piecing everything together, the following combination must be Lorentz invariant, d4p \u03b4(p2 0 \u2212 p 2 \u2212 m2) p0>0 = d3p 2p0 p0=Ep which completes the proof. (2.61) \u2013 32 \u2013 From this result we can \ufb01gure out everything else. For example, the Lorentz invariant \u03b4-function for 3-vectors is which follows because 2Ep \u03b4(3)(p \u2212 q) d3p 2Ep 2Ep \u03b4(3)(p \u2212 q) = 1 (2.62) (2.63) So \ufb01nally we learn that the relativistically normalized momentum states are given by |p = 2Ep |p = 2Ep a\u2020 p |0 (2.64) Notice that our notation is rather subtle: the relativistically normalized momentum state |p di\ufb00ers from |p by the factor 2Ep. These states now satisfy Finally, we can rewrite the identity on one-particle states as p| q = (2\u03c0)3 2Ep \u03b4(3)(p \u2212 q) 1 = d3p (2\u03c0)3 1 2Ep |p p| (2.65) (2.66) Some texts also de\ufb01ne relativistically normalized creation operators by a\u2020(p) = 2Ep a\u2020 p. We won\u2019t make use of this notation here. 2.5 Complex Scalar Fields Consider a complex scalar \ufb01eld \u03c8(x) with Lagrangian L = \u2202\u00b5\u03c8 \u2202\u00b5\u03c8 \u2212 M 2\u03c8\u03c8 (2.67) Notice that, in contrast to the Lagrangian (1.7) for a real scalar \ufb01eld, there is no factor of 1/2 in front of the Lagrangian for a complex scalar \ufb01eld. If we write \u03c8 in terms of real scalar \ufb01elds by \u03c8 = (\u03c61 + i\u03c62)/ 2, we get the factor of 1/2 coming from the 2\u2019s. The equations of motion are 1/ \u221a \u221a \u2202\u00b5\u2202\u00b5 \u03c8 + M 2\u03c8 = 0 \u2202\u00b5\u2202\u00b5\u03c8 + M 2\u03c8 = 0 (2.68) where the second equation is the complex conjugate of the \ufb01rst. We expand the complex \ufb01eld operator as a sum of plane waves as \u03c8 = \u03c8\u2020 = d3p (2\u03c0)3 d3p (2\u03c0)3 1 2Ep 1 2Ep bp e+ip\u00b7x + c\u2020 p e\u2212ip\u00b7x p e\u2212ip\u00b7x + cp e+ip\u00b7x b\u2020 (2.69) \u2013 33 \u2013 Since the classical \ufb01eld \u03c8 is not real, the corresponding quantum \ufb01eld \u03c8 is not hermitian. This is the reason that we have di\ufb00erent operators b and c\u2020 appearing in the positive and negative frequency parts. The classical \ufb01eld momentum is \u03c0 = \u2202L/\u2202 \u02d9\u03c8 = \u02d9\u03c8. We also turn this into a quantum operator \ufb01eld which we write as, p e\u2212ip\u00b7x \u2212 cp e+ip\u00b7x b\u2020 d3p (2\u03c0)3 i d3p (2\u03c0)3 (\u2212i) The commutation relations between \ufb01elds and momenta are given by bp e+ip\u00b7x \u2212 c\u2020 p e\u2212ip\u00b7x Ep 2 Ep 2 (2.70) \u03c0\u2020 = \u03c0 = [\u03c8(x), \u03c0(y)] = i\u03b4(3)(x \u2212 y) and [\u03c8(x), \u03c0\u2020(y)] = 0 (2.71) together with others", " related by complex conjugation, as well as the usual [\u03c8(x), \u03c8(y)] = [\u03c8(x), \u03c8\u2020(y)] = 0, etc. One can easily check that these \ufb01eld commutation relations are equivalent to the commutation relations for the operators bp and cp, [bp, b\u2020 [cp, c\u2020 q] = (2\u03c0)3 \u03b4(3)(p \u2212 q) q] = (2\u03c0)3 \u03b4(3)(p \u2212 q) (2.72) and [bp, bq] = [cp, cq] = [bp, cq] = [bp, c\u2020 q] = 0 (2.73) In summary, quantizing a complex scalar \ufb01eld gives rise to two creation operators, b\u2020 p and c\u2020 p. These have the interpretation of creating two types of particle, both of mass M and both spin zero. They are interpreted as particles and anti-particles. In contrast, for a real scalar \ufb01eld there is only a single type of particle: for a real scalar \ufb01eld, the particle is its own antiparticle. Recall that the theory (2.67) has a classical conserved charge Q = i d3x ( \u02d9\u03c8\u03c8 \u2212 \u03c8 \u02d9\u03c8) = i d3x (\u03c0\u03c8 \u2212 \u03c8\u03c0) (2.74) After normal ordering, this becomes the quantum operator Q = d3p (2\u03c0)3 (c\u2020 p cp \u2212 b\u2020 p bp) = Nc \u2212 Nb (2.75) so Q counts the number of anti-particles (created by c\u2020) minus the number of particles (created by b\u2020). We have [H, Q] = 0, ensuring that Q is conserved quantity in the quantum theory. Of course, in our free \ufb01eld theory this isn\u2019t such a big deal because both Nc and Nb are separately conserved. However, we\u2019ll soon see that in interacting theories Q survives as a conserved quantity, while Nc and Nb individually do not. \u2013 34 \u2013 2.6 The Heisenberg Picture Although we started with a Lorentz invariant Lagrangian, we slowly butchered it as we quantized, introducing a preferred time coordinate t. It\u2019s not at all obvious that the theory is still Lorentz invariant after quantization. For example, the operators \u03c6(x) depend on space, but not on time. Meanwhile, the one-particle states evolve in time by Schr\u00a8odinger\u2019s equation, i d |p(t) dt = H |p(t) \u21d2 |p(t) = e\u2212iEp t |p (2.76) Things start to look better in the Heisenberg picture where time dependence is assigned to the operators O, so that OH = eiHt OS e\u2212iHt dOH dt = i[H, OH] (2.77) (2.78) where the subscripts S and H tell us whether the operator is in the Schr\u00a8odinger or Heisenberg picture. In \ufb01eld theory, we drop these subscripts and we will denote the picture by specifying whether the \ufb01elds depend on space \u03c6(x) (the Schr\u00a8odinger picture) or spacetime \u03c6(x, t) = \u03c6(x) (the Heisenberg picture). The operators in the two pictures agree at a \ufb01xed time, say, t = 0. The commutation relations (2.2) become equal time commutation relations in the Heisenberg picture, [\u03c6(x, t), \u03c6(y, t)] = [\u03c0(x, t), \u03c0(y, t)] = 0 [\u03c6(x, t), \u03c0(y, t)] = i\u03b4(3)(x \u2212 y) (2.79) Now that the operator \u03c6(x) = \u03c6(x, t) depends on time, we can start to study how it evolves. For example, we have \u02d9\u03c6 = i[H, \u03c6] = [ i 2 d3y \u03c0(y)2 + \u2207\u03c6(y)2 + m2\u03c6(y)2, \u03c6(x)] d3y \u03c0(y) (\u2212i) \u03b4(3)(y \u2212 x) = \u03c0(x) (2.80) = i Meanwhile, the equation of motion for \u03c0 reads, \u02d9\u03c0 = i[H, \u03c0] = i 2 [ d3y \u03c0(y)2 + \u2207\u03c6", "(y)2 + m2\u03c6(y)2, \u03c0(x)] \u2013 35 \u2013 = i 2 d3y (\u2207y[\u03c6(y), \u03c0(x)]) \u2207\u03c6(y) + \u2207\u03c6(y) \u2207y[\u03c6(y), \u03c0(x)] = \u2212 +2im2\u03c6(y) \u03b4(3)(x \u2212 y) d3y \u2207y \u03b4(3)(x \u2212 y) \u2207y\u03c6(y) \u2212 m2\u03c6(x) = \u22072\u03c6 \u2212 m2\u03c6 (2.81) where we\u2019ve included the subscript y on \u2207y when there may be some confusion about which argument the derivative is acting on. To reach the last line, we\u2019ve simply integrated by parts. Putting (2.80) and (2.81) together we \ufb01nd that the \ufb01eld operator \u03c6 satis\ufb01es the Klein-Gordon equation \u2202\u00b5\u2202\u00b5\u03c6 + m2\u03c6 = 0 (2.82) Things are beginning to look more relativistic. We can write the Fourier expansion of \u03c6(x) by using the de\ufb01nition (2.77) and noting, eiHt ap e\u2212iHt = e\u2212iEp t ap and eiHt a\u2020 p e\u2212iHt = e+iEp t a\u2020 p (2.83) which follows from the commutation relations [H, ap ] = \u2212Ep ap and [H, a\u2020 This then gives, p ] = +Ep a\u2020 p. \u03c6(x, t) = d3p (2\u03c0)3 1 2Ep ap e\u2212ip\u00b7x + a\u2020 p e+ip\u00b7x (2.84) which looks very similar to the previous expansion (2.18) except that the exponent is now written in terms of 4vectors, p \u00b7 x = Ep t \u2212 p \u00b7 x. (Note also that a sign has \ufb02ipped in the exponent due to our Minkowski metric contraction). It\u2019s simple to check that (2.84) indeed satis\ufb01es the Klein-Gordon equation (2.82). 2.6.1 Causality We\u2019re approaching something Lorentz invariant in the Heisenberg picture, where \u03c6(x) now satis\ufb01es the KleinGordon equation. But there\u2019s still a hint of non-Lorentz invariance because \u03c6 and \u03c0 satisfy equal time commutation relations, Figure 4: [\u03c6(x, t), \u03c0(y, t)] = i\u03b4(3)(x \u2212 y) (2.85) \u2013 36 \u2013 txOO21 But what about arbitrary spacetime separations? In particular, for our theory to be causal, we must require that all spacelike separated operators commute, [O1(x), O2(y)] = 0 \u2200 (x \u2212 y)2 < 0 (2.86) This ensures that a measurement at x cannot a\ufb00ect a measurement at y when x and y are not causally connected. Does our theory satisfy this crucial property? Let\u2019s de\ufb01ne \u2206(x \u2212 y) = [\u03c6(x), \u03c6(y)] (2.87) The objects on the right-hand side of this expression are operators. However, it\u2019s easy to check by direct substitution that the left-hand side is simply a c-number function with the integral expression \u2206(x \u2212 y) = d3p (2\u03c0)3 1 2Ep e\u2212ip\u00b7(x\u2212y) \u2212 eip\u00b7(x\u2212y) (2.88) What do we know about this function? \u2022 It\u2019s Lorentz invariant, thanks to the appearance of the Lorentz invariant measure d3p/2Ep that we introduced in (2.59). \u2022 It doesn\u2019t vanish for timelike separation. For example, taking x \u2212 y = (t, 0, 0, 0) gives [\u03c6(x, 0), \u03c6(x, t)] \u223c e\u2212imt \u2212 e+imt. \u2022 It vanishes for space-like separations. This follows by noting that \u2206(x \u2212 y) = 0 at equal times for all (x \u2212 y)2 = \u2212(x \u2212 y)2 < 0, which we can see explicitly by writing [\u03c6(x, t), \u03c6(y, t)] = 1 2 d3p (2\u03c0)3 1 p 2 + m2 eip\u00b7(x\u2212y) \u2212 e\u2212ip\u00b7(", "x\u2212y) (2.89) and noticing that we can \ufb02ip the sign of p in the last exponent as it is an integration variable. But since \u2206(x \u2212 y) Lorentz invariant, it can only depend on (x \u2212 y)2 and must therefore vanish for all (x \u2212 y)2 < 0. We therefore learn that our theory is indeed causal with commutators vanishing outside the lightcone. This property will continue to hold in interacting theories; indeed, it is usually given as one of the axioms of local quantum \ufb01eld theories. I should mention however that the fact that [\u03c6(x), \u03c6(y)] is a c-number function, rather than an operator, is a property of free \ufb01elds only. \u2013 37 \u2013 2.7 Propagators We could ask a di\ufb00erent question to probe the causal structure of the theory. Prepare a particle at spacetime point y. What is the amplitude to \ufb01nd it at point x? We can calculate this: 0| \u03c6(x)\u03c6(y) |0 = = d3p d3p (2\u03c0)6 d3p (2\u03c0)3 1 2Ep 1 4EpEp 0| ap a\u2020 p | 0 e\u2212ip\u00b7x+ip\u00b7y e\u2212ip\u00b7(x\u2212y) \u2261 D(x \u2212 y) (2.90) The function D(x \u2212 y) is called the propagator. For spacelike separations, (x \u2212 y)2 < 0, one can show that D(x \u2212 y) decays like D(x \u2212 y) \u223c e\u2212m|x\u2212y| (2.91) So it decays exponentially quickly outside the lightcone but, nonetheless, is non-vanishing! The quantum \ufb01eld appears to leak out of the lightcone. Yet we\u2019ve just seen that spacelike measurements commute and the theory is causal. How do we reconcile these two facts? We can rewrite the calculation (2.89) as [\u03c6(x), \u03c6(y)] = D(x \u2212 y) \u2212 D(y \u2212 x) = 0 if (x \u2212 y)2 < 0 (2.92) There are words you can drape around this calculation. When (x \u2212 y)2 < 0, there is no Lorentz invariant way to order events. If a particle can travel in a spacelike direction from x \u2192 y, it can just as easily travel from y \u2192 x. In any measurement, the amplitudes for these two events cancel. With a complex scalar \ufb01eld, it is more interesting. We can look at the equation [\u03c8(x), \u03c8\u2020(y)] = 0 outside the lightcone. The interpretation now is that the amplitude for the particle to propagate from x \u2192 y cancels the amplitude for the antiparticle to travel from y \u2192 x. In fact, this interpretation is also there for a real scalar \ufb01eld because the particle is its own antiparticle. 2.7.1 The Feynman Propagator As we will see shortly, one of the most important quantities in interacting \ufb01eld theory is the Feynman propagator, \u2206F (x \u2212 y) = 0| T \u03c6(x)\u03c6(y) |0 = D(x \u2212 y) D(y \u2212 x) x0 > y0 y0 > x0 (2.93) \u2013 38 \u2013 where T stands for time ordering, placing all operators evaluated at later times to the left so, T \u03c6(x)\u03c6(y) = \u03c6(x)\u03c6(y) \u03c6(y)\u03c6(x) x0 > y0 y0 > x0 (2.94) Claim: There is a useful way of writing the Feynman propagator in terms of a 4momentum integral. \u2206F (x \u2212 y) = d4p (2\u03c0)4 i p2 \u2212 m2 e\u2212ip\u00b7(x\u2212y) (2.95) Notice that this is the \ufb01rst time in this course that we\u2019ve integrated over 4-momentum. Until now, we integrated only over 3-momentum, with p0 \ufb01xed by the mass-shell condition to be p0 = Ep. In the expression (2.95) for \u2206F, we have no such condition on p0. However, as it stands this integral is ill-de\ufb01ned because, for each value of p, the denominator p2 \u2212m2 = (p0)2 \u2212 p 2 \u2212m2 produces a pole when p0 = \u00b1Ep = \u00b1p 2 + m2.", " We need a prescription for avoiding these singularities in the p0 integral. To get the Feynman propagator, we must choose the contour to be Figure 5: The contour for the Feynman propagator. Proof: 1 p2 \u2212 m2 = 1 (p0)2 \u2212 E2 p = 1 (p0 \u2212 Ep)(p0 + Ep) (2.96) so the residue of the pole at p0 = \u00b1Ep is \u00b11/2Ep. When x0 > y0, we close the contour in the lower half plane, where p0 \u2192 \u2212i\u221e, ensuring that the integrand vanishes since e\u2212ip0(x0\u2212y0) \u2192 0. The integral over p0 then picks up the residue at p0 = +Ep which is \u22122\u03c0i/2Ep where the minus sign arises because we took a clockwise contour. Hence when x0 > y0 we have \u2206F (x \u2212 y) = d3p (2\u03c0)4 \u22122\u03c0i 2Ep i e\u2212iEp(x0\u2212y0)+ip\u00b7(x\u2212y) \u2013 39 \u2013 Im(p )Re(p )Ep\u2212E+p00 = d3p (2\u03c0)3 1 2Ep e\u2212ip\u00b7(x\u2212y) = D(x \u2212 y) (2.97) which is indeed the Feynman propagator for x0 > y0. In contrast, when y0 > x0, we close the contour in an anti-clockwise direction in the upper half plane to get, \u2206F (x \u2212 y) = = = d3p (2\u03c0)4 d3p (2\u03c0)3 d3p (2\u03c0)3 2\u03c0i (\u22122Ep) i e+iEp(x0\u2212y0)+ip\u00b7(x\u2212y) 1 2Ep 1 2Ep e\u2212iEp(y0\u2212x0)\u2212ip\u00b7(y\u2212x) e\u2212ip\u00b7(y\u2212x) = D(y \u2212 x) (2.98) where to go to from the second line to the third, we have \ufb02ipped the sign of p which is valid since we integrate over d3p and all other quantities depend only on p 2. Once again we reproduce the Feynman propagator. Instead of specifying the contour, it is standard to write the Feynman propagator as \u2206F (x \u2212 y) = d4p (2\u03c0)4 ie\u2212ip\u00b7(x\u2212y) p2 \u2212 m2 + i (2.99) with > 0, and in\ufb01nitesimal. This has the e\ufb00ect of shifting the poles slightly o\ufb00 the real axis, so the integral along the real p0 axis is equivalent to the contour shown in Figure 5. This way of writing the propagator is, for obvious reasons, called the \u201ci prescription\u201d. 2.7.2 Green\u2019s Functions Figure 6: There is another avatar of the propagator: it is a Green\u2019s function for the Klein-Gordon operator. If we stay away from the singularities, we have (\u22022 t \u2212 \u22072 + m2)\u2206F (x \u2212 y) = = \u2212i d4p (2\u03c0)4 i p2 \u2212 m2 (\u2212p2 + m2) e\u2212ip\u00b7(x\u2212y) d4p (2\u03c0)4 e\u2212ip\u00b7(x\u2212y) = \u2212i \u03b4(4)(x \u2212 y) (2.100) Note that we didn\u2019t make use of the contour anywhere in this derivation. For some purposes it is also useful to pick other contours which also give rise to Green\u2019s functions. \u2013 40 \u2013 Im(p )0+i\u03b5i\u03b5\u2212Re(p )0 Figure 7: The retarded contour Figure 8: The advanced contour For example, the retarded Green\u2019s function \u2206R(x \u2212 y) is de\ufb01ned by the contour shown in Figure 7 which has the property \u2206R(x \u2212 y) = D(x \u2212 y) \u2212 D(y \u2212 x) 0 x0 > y0 y0 > x0 (2.101) The retarded Green\u2019s function is useful in classical \ufb01eld theory if we know the initial value of some \ufb01eld con\ufb01guration and want to \ufb01gure out what it evolves into in the presence of a source, meaning that we want to know the solution to the inhomogeneous Klein-Gordon equation, \u2202\u00b5\u2202\u00b5\u03c6 + m2\u03c6 = J(x) (2.102) for some \ufb01xed background", " function J(x). Similarly, one can de\ufb01ne the advanced Green\u2019s function \u2206A(x \u2212 y) which vanishes when y0 < x0, which is useful if we know the end point of a \ufb01eld con\ufb01guration and want to \ufb01gure out where it came from. Given that next term\u2019s course is called \u201cAdvanced Quantum Field Theory\u201d, there is an obvious name for the current course. But it got shot down in the sta\ufb00 meeting. In the quantum theory, we will see that the Feynman Green\u2019s function is most relevant. 2.8 Non-Relativistic Fields Let\u2019s return to our classical complex scalar \ufb01eld obeying the Klein-Gordon equation. We\u2019ll decompose the \ufb01eld as \u03c8(x, t) = e\u2212imt \u02dc\u03c8(x, t) (2.103) Then the KG-equation reads t \u03c8 \u2212 \u22072\u03c8 + m2\u03c8 = e\u2212imt \u00a8\u02dc\u03c8 \u2212 2im \u02d9\u02dc\u03c8 \u2212 \u22072 \u02dc\u03c8 \u22022 = 0 (2.104) with the m2 term cancelled by the time derivatives. The non-relativistic limit of a particle is |p| m. Let\u2019s look at what this does to our \ufb01eld. After a Fourier transform, \u2013 41 \u2013 Im(p )Re(p )E+p00Ep\u2212Im(p )Re(p )Ep\u221200Ep+ this is equivalent to saying that |\u00a8\u02dc\u03c8| m| \u02d9\u02dc\u03c8|. In this limit, we drop the term with two time derivatives and the KG equation becomes, \u2202 \u02dc\u03c8 \u2202t i = \u2212 1 2m \u22072 \u02dc\u03c8 (2.105) This looks very similar to the Schr\u00a8odinger equation for a non-relativistic free particle of mass m. Except it doesn\u2019t have any probability interpretation \u2014 it\u2019s simply a classical \ufb01eld evolving through an equation that\u2019s \ufb01rst order in time derivatives. We wrote down a Lagrangian in section 1.1.2 which gives rise to \ufb01eld equations which are \ufb01rst order in time derivatives. In fact, we can derive this from the relativistic Lagrangian for a scalar \ufb01eld by again taking the limit \u2202t\u03c8 m\u03c8. After losing the \u02dctilde, so \u02dc\u03c8 \u2192 \u03c8, the non-relativistic Lagrangian becomes L = +i\u03c8 \u02d9\u03c8 \u2212 1 2m \u2207\u03c8 \u2207\u03c8 (2.106) where we\u2019ve divided by 1/2m. This Lagrangian has a conserved current arising from the internal symmetry \u03c8 \u2192 ei\u03b1\u03c8. The current has time and space components \u2212\u03c8\u03c8, j\u00b5 = i 2m (\u03c8\u2207\u03c8 \u2212 \u03c8\u2207\u03c8) To move to the Hamiltonian formalism we compute the momentum \u03c0 = \u2202L \u2202 \u02d9\u03c8 = i\u03c8 (2.107) (2.108) This means that the momentum conjugate to \u03c8 is i\u03c8. The momentum does not depend on time derivatives at all! This looks a little disconcerting but it\u2019s fully consistent for a theory which is \ufb01rst order in time derivatives. In order to determine the full trajectory of the \ufb01eld, we need only specify \u03c8 and \u03c8 at time t = 0: no time derivatives on the initial slice are required. Since the Lagrangian already contains a \u201cp \u02d9q\u201d term (instead of the more familiar 1 2p \u02d9q term), the time derivatives drop out when we compute the Hamiltonian. We get, To quantize we impose (in the Schr\u00a8odinger picture) the canonical commutation relations H = 1 2m \u2207\u03c8 \u2207\u03c8 (2.109) [\u03c8(x), \u03c8(y)] = [\u03c8\u2020(x), \u03c8\u2020(y)] = 0 [\u03c8(x), \u03c8\u2020(y)] = \u03b4(3)(x \u2212 y) (2.110) \u2013 42 \u2013 We may expand \u03c8(x) as a Fourier transform \u03c8(x) = d3p (2\u03c0)3 ap eip\u00b7x where the commutation relations (2.110) require [ap, a\u2020 q ] = (2\u03c0)3\u03b4(", "3)(p \u2212 q) (2.111) (2.112) The vacuum satis\ufb01es ap |0 = 0, and the excitations are a\u2020 p1 states have energy... a\u2020 pn |0. The one-particle H |p = p 2 2m |p (2.113) which is the non-relativistic dispersion relation. We conclude that quantizing the \ufb01rst order Lagrangian (2.106) gives rise to non-relativistic particles of mass m. Some comments: \u2022 We have a complex \ufb01eld but only a single type of particle. The anti-particle is not in the spectrum. The existence of anti-particles is a consequence of relativity. \u2022 A related fact is that the conserved charge Q = d3x : \u03c8\u2020\u03c8 : is the particle number. This remains conserved even if we include interactions in the Lagrangian of the form \u2206L = V (\u03c8\u03c8) (2.114) So in non-relativistic theories, particle number is conserved. It is only with relativity, and the appearance of anti-particles, that particle number can change. \u2022 There is no non-relativistic limit of a real scalar \ufb01eld. In the relativistic theory, the particles are their own anti-particles, and there can be no way to construct a multi-particle theory that conserves particle number. 2.8.1 Recovering Quantum Mechanics In quantum mechanics, we talk about the position and momentum operators X and P. In quantum \ufb01eld theory, position is relegated to a label. How do we get back to quantum mechanics? We already have the operator for the total momentum of the \ufb01eld P = d3p (2\u03c0)3 p a\u2020 p ap (2.115) \u2013 43 \u2013 which, on one-particle states, gives P |p = p |p. It\u2019s also easy to construct the position operator. Let\u2019s work in the non-relativistic limit. Then the operator d3p (2\u03c0)3 a\u2020 creates a particle with \u03b4-function localization at x. We write |x = \u03c8\u2020(x) |0. A natural position operator is then \u03c8\u2020(x) = p e\u2212ip\u00b7x (2.116) X = d3x x \u03c8\u2020(x) \u03c8(x) (2.117) so that X |x = x |x. Let\u2019s now construct a state |\u03d5 by taking superpositions of one-particle states |x, |\u03d5 = d3x \u03d5(x) |x (2.118) The function \u03d5(x) is what we would usually call the Schr\u00a8odinger wavefunction (in the position representation). Let\u2019s make sure that it indeed satis\ufb01es all the right properties. Firstly, it\u2019s clear that acting with the position operator X has the right action of \u03d5(x), X i |\u03d5 = d3x xi \u03d5(x) |x but what about the momentum operator P? We will now show that P i |\u03d5 = d3x \u2212i \u2202\u03d5 \u2202xi |x (2.119) (2.120) which tells us that P i acts as the familiar derivative on wavefunctions |\u03d5. To see that this is the case, we write P i |\u03d5 = = d3xd3p (2\u03c0)3 pia\u2020 (2\u03c0)3 pia\u2020 d3xd3p pap \u03d5(x) \u03c8\u2020(x) |0 p e\u2212ip\u00b7x \u03d5(x) |0 (2.121) where we\u2019ve used the relationship [ap, \u03c8\u2020(x)] = e\u2212ip\u00b7x which can be easily checked. Proceeding with our calculation, we have P i |\u03d5 = = = d3xd3p d3xd3p p (2\u03c0)3 a\u2020 (2\u03c0)3 e\u2212ip\u00b7x \u2202\u03d5 d3x \u2202xi \u2212i i \u2202 \u2202xi e\u2212ip\u00b7x \u2202\u03d5 \u2202xi \u2212i \u03d5(x) |0 a\u2020 p |0 |x (2.122) \u2013 44 \u2013 which con\ufb01rms (2.120). So we learn that when acting on one-particle states, the operators X and P act as position and momentum operators in quantum mechanics, with [X i, P j] |\u03d5 = i\u03b4ij |\u03d5. But what about dynamics? How does the wavefunction \u03d5(x, t", ") change in time? The Hamiltonian (2.109) can be rewritten as H = d3x 1 2m \u2207\u03c8 \u2207\u03c8 = d3p (2\u03c0)3 p 2 2m a\u2020 p ap so we \ufb01nd that i \u2202\u03d5 \u2202t = \u2212 1 2m \u22072\u03d5 (2.123) (2.124) But this is the same equation obeyed by the original \ufb01eld (2.105)! Except this time, it really is the Schr\u00a8odinger equation, complete with the usual probabilistic interpretation for the wavefunction \u03d5. Note in particular that the conserved charge arising from the Noether current (2.107) is Q = d3x |\u03d5(x)|2 which is the total probability. Historically, the fact that the equation for the classical \ufb01eld (2.105) and the oneIt was thought that particle wavefunction (2.124) coincide caused some confusion. perhaps we are quantizing the wavefunction itself and the resulting name \u201csecond quantization\u201d is still sometimes used today to mean quantum \ufb01eld theory. It\u2019s important to stress that, despite the name, we\u2019re not quantizing anything twice! We simply quantize a classical \ufb01eld once. Nonetheless, in practice it\u2019s useful to know that if we treat the one-particle Schr\u00a8odinger equation as the equation for a quantum \ufb01eld then it will give the correct generalization to multi-particle states. Interactions Often in quantum mechanics, we\u2019re interested in particles moving in some \ufb01xed background potential V (x). This can be easily incorporated into \ufb01eld theory by working with a Lagrangian with explicit x dependence, L = i\u03c8 \u02d9\u03c8 \u2212 1 2m \u2207\u03c8 \u2207\u03c8 \u2212 V (x) \u03c8\u03c8 (2.125) Note that this Lagrangian doesn\u2019t respect translational symmetry and we won\u2019t have the associated energy-momentum tensor. While such Lagrangians are useful in condensed matter physics, we rarely (or never) come across them in high-energy physics, where all equations obey translational (and Lorentz) invariance. \u2013 45 \u2013 One can also consider interactions between particles. Obviously these are only important for n particle states with n \u2265 2. We therefore expect them to arise from additions to the Lagrangian of the form \u2206L = \u03c8(x) \u03c8(x) \u03c8(x) \u03c8(x) (2.126) which, in the quantum theory, is an operator which destroys two particles before creating two new ones. Such terms in the Lagrangian will indeed lead to inter-particle forces, both in the non-relativistic and relativistic setting. In the next section we explore these types of interaction in detail for relativistic theories. \u2013 46 \u2013 3. Interacting Fields The free \ufb01eld theories that we\u2019ve discussed so far are very special: we can determine their spectrum, but nothing interesting then happens. They have particle excitations, but these particles don\u2019t interact with each other. Here we\u2019ll start to examine more complicated theories that include interaction terms. These will take the form of higher order terms in the Lagrangian. We\u2019ll start by asking what kind of small perturbations we can add to the theory. For example, consider the Lagrangian for a real scalar \ufb01eld, L = 1 2 \u2202\u00b5\u03c6 \u2202\u00b5\u03c6 \u2212 1 2 m2\u03c62 \u2212 n\u22653 \u03bbn n! \u03c6n (3.1) The coe\ufb03cients \u03bbn are called coupling constants. What restrictions do we have on \u03bbn to ensure that the additional terms are small perturbations? You might think that we need simply make \u201c\u03bbn 1\u201d. But this isn\u2019t quite right. To see why this is the case, let\u2019s do some dimensional analysis. Firstly, note that the action has dimensions of angular momentum or, equivalently, the same dimensions as. Since we\u2019ve set = 1, using the convention described in the introduction, we have [S] = 0. With S = d4x L, and [d4x] = \u22124, the Lagrangian density must therefore have [L] = 4 (3.2) What does this mean for the Lagrangian (3.1)? Since [\u2202\u00b5] = 1, we can read o\ufb00 the mass dimensions of all the factors to \ufb01nd, [\u03c6] = 1, [m] = 1", ", [\u03bbn] = 4 \u2212 n (3.3) So now we see why we can\u2019t simply say we need \u03bbn 1, because this statement only makes sense for dimensionless quantities. The various terms, parameterized by \u03bbn, fall into three di\ufb00erent categories \u2022 [\u03bb3] = 1: For this term, the dimensionless parameter is \u03bb3/E, where E has dimensions of mass. Typically in quantum \ufb01eld theory, E is the energy scale of the process of interest. This means that \u03bb3 \u03c63/3! is a small perturbation at high energies E \u03bb3, but a large perturbation at low energies E \u03bb3. Terms that we add to the Lagrangian with this behavior are called relevant because they\u2019re most relevant at low energies (which, after all, is where most of the physics we see lies). In a relativistic theory, E > m, so we can always make this perturbation small by taking \u03bb3 m. \u2013 47 \u2013 \u2022 [\u03bb4] = 0: this term is small if \u03bb4 1. Such perturbations are called marginal. \u2022 [\u03bbn] < 0 for n \u2265 5: The dimensionless parameter is (\u03bbnEn\u22124), which is small at low-energies and large at high energies. Such perturbations are called irrelevant. As you\u2019ll see later, it is typically impossible to avoid high energy processes in quantum \ufb01eld theory. (We\u2019ve already seen a glimpse of this in computing the vacuum energy). This means that we might expect problems with irrelevant operators. Indeed, these lead to \u201cnon-renormalizable\u201d \ufb01eld theories in which one cannot make sense of the in\ufb01nities at arbitrarily high energies. This doesn\u2019t necessarily mean that the theory is useless; just that it is incomplete at some energy scale. Let me note however that the naive assignment of relevant, marginal and irrelevant is not always \ufb01xed in stone: quantum corrections can sometimes change the character of an operator. An Important Aside: Why QFT is Simple Typically in a quantum \ufb01eld theory, only the relevant and marginal couplings are important. This is basically because, as we\u2019ve seen above, the irrelevant couplings become small at low-energies. This is a huge help: of the in\ufb01nite number of interaction terms that we could write down, only a handful are actually needed (just two in the case of the real scalar \ufb01eld described above). Let\u2019s look at this a little more. Suppose that we some day discover the true superduper \u201ctheory of everything unimportant\u201d that describes the world at very high energy scales, say the GUT scale, or the Planck scale. Whatever this scale is, let\u2019s call it \u039b. It is an energy scale, so [\u039b] = 1. Now we want to understand the laws of physics down at our puny energy scale E \u039b. Let\u2019s further suppose that down at the energy scale E, the laws of physics are described by a real scalar \ufb01eld. (They\u2019re not of course: they\u2019re described by non-Abelian gauge \ufb01elds and fermions, but the same argument applies in that case so bear with me). This scalar \ufb01eld will have some complicated interaction terms (3.1), where the precise form is dictated by all the stu\ufb00 that\u2019s going on in the high energy superduper theory. What are these interactions? Well, we could write our dimensionful coupling constants \u03bbn in terms of dimensionless couplings gn, multiplied by a suitable power of the relevant scale \u039b, \u03bbn = gn \u039bn\u22124 (3.4) The exact values of dimensionless couplings gn depend on the details of the high-energy superduper theory, but typically one expects them to be of order 1: gn \u223c O(1). This \u2013 48 \u2013 means that for experiments at small energies E \u039b, the interaction terms of the form \u03c6n with n > 4 will be suppressed by powers of (E/\u039b)n\u22124. This is usually a (e.g for the energies E explored at the suppression by many orders of magnitude. LHC, E/Mpl \u223c 10\u221216). It is this simple argument, based on dimensional analysis, that ensures that we need only focus on the \ufb01rst few terms in the interaction: those which are relevant and marginal. It also means that if we only have access to low-energy experiments (which we do!), it\u2019s going to be very di\ufffd", "\ufffdcult to \ufb01gure out the high energy theory (which it is!), because its e\ufb00ects are highly diluted except for the relevant and marginal interactions. The discussion given above is a poor man\u2019s version of the ideas of e\ufb00ective \ufb01eld theory and Wilson\u2019s renormalization group, about which you can learn more in the \u201cStatistical Field Theory\u201d course. Examples of Weakly Coupled Theories In this course we\u2019ll study only weakly coupled \ufb01eld theories i.e. ones that can truly be considered as small perturbations of the free \ufb01eld theory at all energies. In this section, we\u2019ll look at two types of interactions 1) \u03c64 theory: L = 1 2 \u2202\u00b5\u03c6\u2202\u00b5\u03c6 \u2212 1 2 m2\u03c62 \u2212 \u03bb 4! \u03c64 (3.5) with \u03bb 1. We can get a hint for what the e\ufb00ects of this extra term will be. Expanding out \u03c64 in terms of ap and a\u2020 p, we see a sum of interactions that look like p a\u2020 a\u2020 p a\u2020 p a\u2020 p and a\u2020 p a\u2020 p a\u2020 p ap etc. (3.6) These will create and destroy particles. This suggests that the \u03c64 Lagrangian describes a theory in which particle number is not conserved. Indeed, we could check that the number operator N now satis\ufb01es [H, N ] = 0. 2) Scalar Yukawa Theory L = \u2202\u00b5\u03c8\u2202\u00b5\u03c8 + 1 2 \u2202\u00b5\u03c6\u2202\u00b5\u03c6 \u2212 M 2\u03c8\u03c8 \u2212 1 2 m2\u03c62 \u2212 g\u03c8\u03c8\u03c6 (3.7) with g M, m. This theory couples a complex scalar \u03c8 to a real scalar \u03c6. While the individual particle numbers of \u03c8 and \u03c6 are no longer conserved, we do still have a symmetry rotating the phase of \u03c8, ensuring the existence of the charge Q de\ufb01ned in (2.75) such that [Q, H] = 0. This means that the number of \u03c8 particles minus the number of \u03c8 anti-particles is conserved. It is common practice to denote the antiparticle as \u00af\u03c8. \u2013 49 \u2013 The scalar Yukawa theory has a slightly worrying aspect: the potential has a stable local minimum at \u03c6 = \u03c8 = 0, but is unbounded below for large enough \u2212g\u03c6. This means we shouldn\u2019t try to push this theory too far. A Comment on Strongly Coupled Field Theories In this course we restrict attention to weakly coupled \ufb01eld theories where we can use perturbative techniques. The study of strongly coupled \ufb01eld theories is much more di\ufb03cult, and one of the major research areas in theoretical physics. For example, some of the amazing things that can happen include \u2022 Charge Fractionalization: Although electrons have electric charge 1, under the right conditions the elementary excitations in a solid have fractional charge 1/N (where N \u2208 2Z + 1). For example, this occurs in the fractional quantum Hall e\ufb00ect. \u2022 Con\ufb01nement: The elementary excitations of quantum chromodynamics (QCD) are quarks. But they never appear on their own, only in groups of three (in a baryon) or with an anti-quark (in a meson). They are con\ufb01ned. \u2022 Emergent Space: There are \ufb01eld theories in four dimensions which at strong coupling become quantum gravity theories in ten dimensions! The strong coupling e\ufb00ects cause the excitations to act as if they\u2019re gravitons moving in higher dimensions. This is quite extraordinary and still poorly understood. It\u2019s called the AdS/CFT correspondence. 3.1 The Interaction Picture There\u2019s a useful viewpoint in quantum mechanics to describe situations where we have small perturbations to a well-understood Hamiltonian. Let\u2019s return to the familiar ground of quantum mechanics with a \ufb01nite number of degrees of freedom for a moment. In the Schr\u00a8odinger picture, the states evolve as i d|\u03c8S dt = H |\u03c8S (3.8) while the operators OS are independent of time. In contrast, in the Heisenberg picture the states are \ufb01xed and the operators change in time OH(t) = eiHtOS e\u2212iHt |\u03c8H = eiHt |\u03c8S \u2013 50 \u2013 (3.9)", " The interaction picture is a hybrid of the two. We split the Hamiltonian up as H = H0 + Hint (3.10) The time dependence of operators is governed by H0, while the time dependence of states is governed by Hint. Although the split into H0 and Hint is arbitrary, it\u2019s useful when H0 is soluble (for example, when H0 is the Hamiltonian for a free \ufb01eld theory). The states and operators in the interaction picture will be denoted by a subscript I and are given by, |\u03c8(t)I = eiH0t |\u03c8(t)S OI(t) = eiH0tOSe\u2212iH0t (3.11) This last equation also applies to Hint, which is time dependent. The interaction Hamiltonian in the interaction picture is, HI \u2261 (Hint)I = eiH0t(Hint)S e\u2212iH0t (3.12) The Schr\u00a8odinger equation for states in the interaction picture can be derived starting from the Schr\u00a8odinger picture i d|\u03c8S dt = HS |\u03c8S \u21d2 i \u21d2 i So we learn that d dt d|\u03c8I dt e\u2212iH0t |\u03c8I = (H0 + Hint)S e\u2212iH0t |\u03c8I = eiH0t(Hint)S e\u2212iH0t |\u03c8I (3.13) i d|\u03c8I dt = HI(t) |\u03c8I (3.14) 3.1.1 Dyson\u2019s Formula \u201cWell, Birmingham has much the best theoretical physicist to work with, Peierls; Bristol has much the best experimental physicist, Powell; Cambridge has some excellent architecture. You can make your choice.\u201d Oppenheimer\u2019s advice to Dyson on which university position to accept. We want to solve (3.14). Let\u2019s write the solution as |\u03c8(t)I = U (t, t0) |\u03c8(t0)I (3.15) \u2013 51 \u2013 where U (t, t0) is a unitary time evolution operator such that U (t1, t2)U (t2, t3) = U (t1, t3) and U (t, t) = 1. Then the interaction picture Schr\u00a8odinger equation (3.14) requires that i dU dt = HI(t) U If HI were a function, then we could simply solve this by U (t, t0)?= exp t \u2212i t0 HI(t) dt (3.16) (3.17) But there\u2019s a problem. Our Hamiltonian HI is an operator, and we have ordering issues. Let\u2019s see why this causes trouble. The exponential of an operator is de\ufb01ned in terms of the expansion, exp \u2212i t t0 HI(t) dt = 1 \u2212 i t t0 HI(t) dt + (\u2212i)2 2 t t0 2 HI(t) dt +...(3.18) But when we try to di\ufb00erentiate this with respect to t, we \ufb01nd that the quadratic term gives us \u2212 1 2 t t0 HI(t) dt HI(t) \u2212 1 2 HI(t) t t0 HI(t) dt (3.19) Now the second term here looks good, since it will give part of the HI(t)U that we need on the right-hand side of (3.16). But the \ufb01rst term is no good since the HI(t) sits the wrong side of the integral term, and we can\u2019t commute it through because [HI(t), HI(t)] = 0 when t = t. So what\u2019s the way around this? Claim: The solution to (3.16) is given by Dyson\u2019s Formula. (Essentially \ufb01rst \ufb01gured out by Dirac, although the compact notation is due to Dyson). U (t, t0) = T exp \u2212i HI(t) dt t t0 (3.20) where T stands for time ordering where operators evaluated at later times are placed to the left T (O1(t1) O2(t2)) = O1(t1) O2(t2) O2(t2) O1(t1) t1 > t2 t2 > t1 (3.21) Expanding out the expression (3.20), we now have U (t, t0) = 1 \u2212 i t t0", " dt HI(t) + (\u2212i)2 2 t t0 t t t dt + dt HI(t)HI(t) t dt HI(t)HI(t) dt +... t0 t0 \u2013 52 \u2013 Actually these last two terms double up since t t dt t0 t dt HI(t)HI(t) = = t t0 t dt dt t dt HI(t)HI(t) t0 t dt HI(t)HI(t) (3.22) t0 t0 where the range of integration in the \ufb01rst expression is over t \u2265 t, while in the second expression it is t \u2264 t which is, of course, the same thing. The \ufb01nal expression is the same as the second expression by a simple relabelling. This means that we can write U (t, t0) = 1 \u2212 i t t0 dt HI(t) + (\u2212i)2 t t dt t0 t0 dt HI(t)HI(t) +... (3.23) Proof: The proof of Dyson\u2019s formula is simpler than explaining what all the notation means! Firstly observe that under the T sign, all operators commute (since their order is already \ufb01xed by the T sign). Thus i \u2202 \u2202t T exp \u2212i t t0 dt HI(t) = T HI(t) exp \u2212i t dt HI(t) = HI(t) T exp \u2212i t0 t dt HI(t) (3.24) t0 since t, being the upper limit of the integral, is the latest time so HI(t) can be pulled out to the left. Before moving on, I should confess that Dyson\u2019s formula is rather formal. It is typically very hard to compute time ordered exponentials in practice. The power of the formula comes from the expansion which is valid when HI is small and is very easily computed. 3.2 A First Look at Scattering Let us now apply the interaction picture to \ufb01eld theory, starting with the interaction Hamiltonian for our scalar Yukawa theory, Hint = g d3x \u03c8\u2020\u03c8\u03c6 (3.25) Unlike the free theories discussed in Section 2, this interaction doesn\u2019t conserve particle number, allowing particles of one type to morph into others. To see why this is, we use \u2013 53 \u2013 the interaction picture and follow the evolution of the state: |\u03c8(t) = U (t, t0) |\u03c8(t0), where U (t, t0) is given by Dyson\u2019s formula (3.20) which is an expansion in powers of Hint. But Hint contains creation and annihilation operators for each type of particle. In particular, \u2022 \u03c6 \u223c a + a\u2020: This operator can create or destroy \u03c6 particles. Let\u2019s call them mesons. \u2022 \u03c8 \u223c b + c\u2020: This operator can destroy \u03c8 particles through b, and create antiparticles through c\u2020. Let\u2019s call these particles nucleons. Of course, in reality nucleons are spin 1/2 particles, and don\u2019t arise from the quantization of a scalar \ufb01eld. But we\u2019ll treat our scalar Yukawa theory as a toy model for nucleons interacting with mesons. \u2022 \u03c8\u2020 \u223c b\u2020 + c: This operator can create nucleons through b\u2020, and destroy anti- nucleons through c. Importantly, Q = Nc \u2212 Nb remains conserved in the presence of Hint. At \ufb01rst order in perturbation theory, we \ufb01nd terms in Hint like c\u2020b\u2020a. This kills a meson, producing a nucleon-anti-nucleon pair. It will contribute to meson decay \u03c6 \u2192 \u03c8 \u00af\u03c8. At second order in perturbation theory, we\u2019ll have more complicated terms in (Hint)2, for example (c\u2020b\u2020a)(cba\u2020). This term will give contributions to scattering processes \u03c8 \u00af\u03c8 \u2192 \u03c6 \u2192 \u03c8 \u00af\u03c8. The rest of this section is devoted to computing the quantum amplitudes for these processes to occur. To calculate amplitudes we make an important, and slightly dodgy, assumption: Initial and \ufb01nal states are eigenstates of the free theory This means that we take the initial state |i at t \u2192 \u2212\u221e, and the \ufb01nal state |f at t \u2192 +\u221e, to be eigenstates of the free Hamiltonian H0. At some level, this sounds plausible: at t \u2192 \u2212\u221e, the particles in a", " scattering process are far separated and don\u2019t feel the e\ufb00ects of each other. Furthermore, we intuitively expect these states to be eigenstates of the individual number operators N, which commute with H0, but not Hint. As the particles approach each other, they interact brie\ufb02y, before departing again, each going on its own merry way. The amplitude to go from |i to |f is lim t\u00b1\u2192\u00b1\u221e f | U (t+, t\u2212) |i \u2261 f | S |i (3.26) where the unitary operator S is known as the S-matrix. (S is for scattering). There are a number of reasons why the assumption of non-interacting initial and \ufb01nal states is shaky: \u2013 54 \u2013 \u2022 Obviously we can\u2019t cope with bound states. For example, this formalism can\u2019t describe the scattering of an electron and proton which collide, bind, and leave as a Hydrogen atom. It\u2019s possible to circumvent this objection since it turns out that bound states show up as poles in the S-matrix. \u2022 More importantly, a single particle, a long way from its neighbors, is never alone in \ufb01eld theory. This is true even in classical electrodynamics, where the electron sources the electromagnetic \ufb01eld from which it can never escape. In quantum electrodynamics (QED), a related fact is that there is a cloud of virtual photons surrounding the electron. This line of thought gets us into the issues of renormalization \u2014 more on this next term in the \u201cAQFT\u201d course. Nevertheless, motivated by this problem, after developing scattering theory using the assumption of noninteracting asymptotic states, we\u2019ll mention a better way. 3.2.1 An Example: Meson Decay Consider the relativistically normalized initial and \ufb01nal states, |i = 2Ep a\u2020 p |0 |f = 4Eq1Eq2 b\u2020 q1 c\u2020 q2 |0 (3.27) The initial state contains a single meson of momentum p; the \ufb01nal state contains a nucleon-anti-nucleon pair of momentum q1 and q2. We may compute the amplitude for the decay of a meson to a nucleon-anti-nucleon pair. To leading order in g, it is f | S |i = \u2212ig f | d4x \u03c8\u2020(x)\u03c8(x)\u03c6(x) |i (3.28) Let\u2019s go slowly. We \ufb01rst expand out \u03c6 \u223c a + a\u2020 using (2.84). (Remember that the \u03c6 in this formula is in the interaction picture, which is the same as the Heisenberg picture of the free theory). The a piece will turn |i into something proportional to |0, while the a\u2020 piece will turn |i into a two meson state. But the two meson state will have zero overlap with f |, and there\u2019s nothing in the \u03c8 and \u03c8\u2020 operators that lie between them to change this fact. So we have f | S |i = \u2212ig f | = \u2212ig f | d4x \u03c8\u2020(x)\u03c8(x) d3k (2\u03c0)3 2Ep 2Ek ak a\u2020 p e\u2212ik\u00b7x |0 d4x \u03c8\u2020(x)\u03c8(x)e\u2212ip\u00b7x |0 (3.29) p, picking up a \u03b4(3)(p \u2212 k) deltawhere, in the second line, we\u2019ve commuted ak past a\u2020 function which kills the d3k integral. We now similarly expand out \u03c8 \u223c b + c\u2020 and \u2013 55 \u2013 \u03c8\u2020 \u223c b\u2020 + c. To get non-zero overlap with f |, only the b\u2020 and c\u2020 contribute, for they create the nucleon and anti-nucleon from |0. We then have f | S |i = \u2212ig 0| d4xd3k1d3k2 (2\u03c0)6 Eq1Eq2 Ek1 Ek2 cq2bq1c\u2020 k1 b\u2020 k2 |0 ei(k1+k2\u2212p)\u00b7x = \u2212ig (2\u03c0)4 \u03b4(4)(q1 + q2 \u2212 p) (3.30) and so we get our \ufb01rst quantum \ufb01eld theory amplitude. Notice that the \u03b4-function puts constraints on the possible decays. In particular, the decay only happens at all if m \u2265 2M. To see this, we may always boost ourselves to a reference frame where the meson", " is stationary, so p = (m, 0, 0, 0). Then the delta function imposes momentum conservation, telling us that q1 = \u2212q2 and m = 2M 2 + |q|2. Later you will learn how to turn this quantum amplitude into something more physical, namely the lifetime of the meson. The reason this is a little tricky is that we must square the amplitude to get the probability for decay, which means we get the square of a \u03b4-function. We\u2019ll explain how to deal with this in Section 3.6 below, and again in next term\u2019s \u201cStandard Model\u201d course. 3.3 Wick\u2019s Theorem From Dyson\u2019s formula, we want to compute quantities like f | T {HI(x1)... HI(xn)} |i, where |i and |f are eigenstates of the free theory. The ordering of the operators is \ufb01xed by T, time ordering. However, since the HI\u2019s contain certain creation and annihilation operators, our life will be much simpler if we can start to move all annihilation operators to the right where they can start killing things in |i. Recall that this is the de\ufb01nition of normal ordering. Wick\u2019s theorem tells us how to go from time ordered products to normal ordered products. 3.3.1 An Example: Recovering the Propagator Let\u2019s start simple. Consider a real scalar \ufb01eld which we decompose in the Heisenberg picture as \u03c6(x) = \u03c6+(x) + \u03c6\u2212(x) (3.31) where \u03c6+(x) = \u03c6\u2212(x) = d3p (2\u03c0)3 d3p (2\u03c0)3 1 2Ep 1 2Ep ap e\u2212ip\u00b7x a\u2020 p e+ip\u00b7x \u2013 56 \u2013 (3.32) where the \u00b1 signs on \u03c6\u00b1 make little sense, but apparently you have Pauli and Heisenberg to blame. (They come about because \u03c6+ \u223c e\u2212iEt, which is sometimes called the positive frequency piece, while \u03c6\u2212 \u223c e+iEt is the negative frequency piece). Then choosing x0 > y0, we have T \u03c6(x)\u03c6(y) = \u03c6(x)\u03c6(y) = (\u03c6+(x) + \u03c6\u2212(x))(\u03c6+(y) + \u03c6\u2212(y)) (3.33) = \u03c6+(x)\u03c6+(y) + \u03c6\u2212(x)\u03c6+(y) + \u03c6\u2212(y)\u03c6+(x) + [\u03c6+(x), \u03c6\u2212(y)] + \u03c6\u2212(x)\u03c6\u2212(y) where the last line is normal ordered, and for our troubles we have picked up the extra term D(x\u2212y) = [\u03c6+(x), \u03c6\u2212(y)] which is the propagator we met in (2.90). So for x0 > y0 we have T \u03c6(x)\u03c6(y) =: \u03c6(x)\u03c6(y) : + D(x \u2212 y) (3.34) Meanwhile, for y0 > x0, we may repeat the calculation to \ufb01nd T \u03c6(x)\u03c6(y) =: \u03c6(x)\u03c6(y) : + D(y \u2212 x) (3.35) So putting this together, we have the \ufb01nal expression T \u03c6(x)\u03c6(y) =: \u03c6(x)\u03c6(y) : + \u2206F (x \u2212 y) (3.36) where \u2206F (x \u2212 y) is the Feynman propagator de\ufb01ned in (2.93), for which we have the integral representation \u2206F (x \u2212 y) = d4k (2\u03c0)4 ieik\u00b7(x\u2212y) k2 \u2212 m2 + i (3.37) Let me reiterate a comment from Section 2: although T \u03c6(x)\u03c6(y) and : \u03c6(x)\u03c6(y) : are both operators, the di\ufb00erence between them is a c-number function, \u2206F (x \u2212 y). De\ufb01nition: We de\ufb01ne the contraction of a pair of \ufb01elds in a string of operators... \u03c6(x1)... \u03c6(x2)... to mean replacing those operators with the Feynman propagator, leaving all other operators untouched. We use the notation, \u03c6(x1)... \u03c6(x2", ")...... to denote contraction. So, for example, \u03c6(x)\u03c6(y) = \u2206F (x \u2212 y) \u2013 57 \u2013 (3.38) (3.39) A similar discussion holds for complex scalar \ufb01elds. We have T \u03c8(x)\u03c8\u2020(y) =: \u03c8(x)\u03c8\u2020(y) : +\u2206F (x \u2212 y) prompting us to de\ufb01ne the contraction \u03c8(x)\u03c8\u2020(y) = \u2206F (x \u2212 y) and \u03c8(x)\u03c8(y) = \u03c8\u2020(x)\u03c8\u2020(y) = 0 (3.40) (3.41) 3.3.2 Wick\u2019s Theorem For any collection of \ufb01elds \u03c61 = \u03c6(x1), \u03c62 = \u03c6(x2), etc, we have T (\u03c61... \u03c6n) =: \u03c61... \u03c6n : + : all possible contractions : (3.42) To see what the last part of this equation means, let\u2019s look at an example. For n = 4, the equation reads T (\u03c61\u03c62\u03c63\u03c64) = : \u03c61\u03c62\u03c63\u03c64 : + + \u03c61\u03c62 : \u03c63\u03c64 : + \u03c63\u03c64 + \u03c61\u03c62 \u03c61\u03c63 \u03c61\u03c63 : \u03c62\u03c64 : + four similar terms \u03c62\u03c63 \u03c62\u03c64 + \u03c61\u03c64 (3.43) Proof: The proof of Wick\u2019s theorem proceeds by induction and a little thought. It\u2019s true for n = 2. Suppose it\u2019s true for \u03c62... \u03c6n and now add \u03c61. We\u2019ll take x0 1 > x0 k for all k = 2,..., n. Then we can pull \u03c61 out to the left of the time ordered product, writing T (\u03c61\u03c62... \u03c6n) = (\u03c6+ 1 + \u03c6\u2212 1 ) (: \u03c62... \u03c6n : + : contractions :) (3.44) The \u03c6\u2212 the right-hand side as a normal ordered product, the \u03c6+ past the crowd of \u03c6\u2212 \u03c61\u03c6k = \u2206F (x1 \u2212 xk) from the commutator. (Try it!) 1 term stays where it is since it is already normal ordered. But in order to write 1 term has to make its way k, we pick up a factor of k operators. Each time it moves past \u03c6\u2212 3.3.3 An Example: Nucleon Scattering Let\u2019s look at \u03c8\u03c8 \u2192 \u03c8\u03c8 scattering. We have the initial and \ufb01nal states |i = 2Ep1 |f = 2Ep 1 2Ep2 b\u2020 p1 b\u2020 p 1 2Ep 2 b\u2020 p2 b\u2020 p 2 |0 \u2261 |p1, p2 |0 \u2261 |p 1, p 2 (3.45) We can then look at the expansion of f | S |i. In fact, we really want to calculate f | S \u2212 1 |i since we\u2019re not interested in situations where no scattering occurs. At order g2 we have the term (\u2212ig)2 2 d4x1d4x2 T \u03c8\u2020(x1)\u03c8(x1)\u03c6(x1)\u03c8\u2020(x2)\u03c8(x2)\u03c6(x2) (3.46) \u2013 58 \u2013 Now, using Wick\u2019s theorem we see there is a piece in the string of operators which looks like : \u03c8\u2020(x1)\u03c8(x1)\u03c8\u2020(x2)\u03c8(x2) : \u03c6(x1)\u03c6(x2) (3.47) which will contribute to the scattering because the two \u03c8 \ufb01elds annihilate the \u03c8 particles, while the two \u03c8\u2020 \ufb01elds create \u03c8 particles. Any other way of ordering the \u03c8 and \u03c8\u2020 \ufb01elds will give zero contribution. This means that we have p = p 1, p 1, p eip = 2| : \u03c8\u2020(x1)\u03c8(x1)\u03c8\u2020(x2)\u03c8(x2) : |p1, p2 2| \u03c8\u2020(x1", ")\u03c8\u2020(x2) |0 0| \u03c8(x1)\u03c8(x2) |p1, p2 1\u00b7x1+ip 2\u00b7x1 1\u00b7x2+ip 2\u00b7x2 + eip = eix1\u00b7(p 1\u2212p1)+ix2\u00b7(p 2\u2212p2) + eix1\u00b7(p e\u2212ip1\u00b7x1\u2212ip2\u00b7x2 + e\u2212ip1\u00b7x2\u2212ip2\u00b7x1 2\u2212p1)+ix2\u00b7(p 1\u2212p2) + (x1 \u2194 x2) (3.48) where, in going to the third line, we\u2019ve used the fact that for relativistically normalized states, 0| \u03c8(x) |p = e\u2212ip\u00b7x (3.49) Now let\u2019s insert this into (3.46), to get the expression for f | S |i at order g2, (\u2212ig)2 2 d4x1d4x2 ei... + ei... + (x1 \u2194 x2) d4k (2\u03c0)4 ieik\u00b7(x1\u2212x2) k2 \u2212 m2 + i (3.50) where the expression in square brackets is (3.48), while the \ufb01nal integral is the \u03c6 propagator which comes from the contraction in (3.47). Now the (x1 \u2194 x2) terms double up with the others to cancel the factor of 1/2 out front. Meanwhile, the x1 and x2 integrals give delta-functions. We\u2019re left with the expression (\u2212ig)2 d4k (2\u03c0)4 i(2\u03c0)8 k2 \u2212 m2 + i \u03b4(4)(p 1 \u2212 p1 + k) \u03b4(4)(p 2 \u2212 p2 \u2212 k) + \u03b4(4)(p 2 \u2212 p1 + k) \u03b4(4)(p 1 \u2212 p2 \u2212 k) (3.51) Finally, we can trivially do the d4k integral using the delta-functions to get i(\u2212ig)2 1 (p1 \u2212 p 1)2 \u2212 m2 + i + 1 (p1 \u2212 p 2)2 \u2212 m2 + i (2\u03c0)4 \u03b4(4)(p1 + p2 \u2212 p 1 \u2212 p 2) In fact, for this process we may drop the +i terms since the denominator is never zero. To see this, we can go to the center of mass frame, where p1 = \u2212p2 and, by \u2013 59 \u2013 momentum conservation, |p1| = |p is k = (0, p \u2212 p ), so k2 < 0. We therefore have the end result, 1 |. This ensures that the 4-momentum of the meson i(\u2212ig)2 1 1)2 \u2212 m2 + (p1 \u2212 p 1 (p1 \u2212 p 2)2 \u2212 m2 (2\u03c0)4 \u03b4(4)(p1 + p2 \u2212 p 1 \u2212 p 2) (3.52) We will see another, much simpler way to reproduce this result shortly using Feynman diagrams. This will also shed light on the physical interpretation. This calculation is also relevant for other scattering processes, such as \u00af\u03c8 \u00af\u03c8 \u2192 \u00af\u03c8 \u00af\u03c8, \u03c8 \u00af\u03c8 \u2192 \u03c8 \u00af\u03c8. Each of these comes from the term (3.48) in Wick\u2019s theorem. However, we will never \ufb01nd a term that contributes to scattering \u03c8\u03c8 \u2192 \u00af\u03c8 \u00af\u03c8, for this would violate the conservation of Q charge. Another Example: Meson-Nucleon Scattering If we want to compute \u03c8\u03c6 \u2192 \u03c8\u03c6 scattering at order g2, we would need to pick out the term : \u03c8\u2020(x1)\u03c6(x1)\u03c8(x2)\u03c6(x2) : \u03c8(x1)\u03c8\u2020(x2) (3.53) and a similar term with \u03c8 and \u03c8\u2020 exchanged. Once more, this term also contributes to similar scattering processes, including \u00af\u03c8\u03c6 \u2192 \u00af\u03c8\u03c6 and \u03c6\u03c6 \u2192 \u03c8 \u00af\u03c8. 3.4 Feynman Diagrams \u201cLike the silicon chips of more recent years, the Feynman diagram was bringing computation to the masses.\u201d Julian Schwinger As the above example demonstrates, to actually compute scattering amplitudes using Wick\u2019s theorem is rather tedious. There\u2019s a much better way. It requires drawing pretty pictures. These pictures represent the expansion of f | S |i and we will learn how to associate numbers (or", " at least integrals) to them. These pictures are called Feynman diagrams. The object that we really want to compute is f | S \u22121 |i, since we\u2019re not interested in processes where no scattering occurs. The various terms in the perturbative expansion can be represented pictorially as follows \u2022 Draw an external line for each particle in the initial state |i and each particle in the \ufb01nal state |f. We\u2019ll choose dotted lines for mesons, and solid lines for nucleons. Assign a directed momentum p to each line. Further, add an arrow to \u2013 60 \u2013 solid lines to denote its charge; we\u2019ll choose an incoming (outgoing) arrow in the initial state for \u03c8 ( \u00af\u03c8). We choose the reverse convention for the \ufb01nal state, where an outgoing arrow denotes \u03c8. \u2022 Join the external lines together with trivalent vertices Each such diagram you can draw is in 1-1 correspondence with the terms in the expansion of f | S \u2212 1 |i. 3.4.1 Feynman Rules To each diagram we associate a number, using the Feynman rules \u2022 Add a momentum k to each internal line \u2022 To each vertex, write down a factor of (\u2212ig) (2\u03c0)4 \u03b4(4)( ki) i (3.54) where ki is the sum of all momenta \ufb02owing into the vertex. \u2022 For each internal dotted line, corresponding to a \u03c6 particle with momentum k, we write down a factor of d4k (2\u03c0)4 i k2 \u2212 m2 + i (3.55) We include the same factor for solid internal \u03c8 lines, with m replaced by the nucleon mass M. \u2013 61 \u2013 \u03c8\u03c8+\u03c6 3.5 Examples of Scattering Amplitudes Let\u2019s apply the Feynman rules to compute the amplitudes for various processes. We start with something familiar: Nucleon Scattering Revisited Let\u2019s look at how this works for the \u03c8\u03c8 \u2192 \u03c8\u03c8 scattering at order g2. We can write down the two simplest diagrams contributing to this process. They are shown in Figure 9. Figure 9: The two lowest order Feynman diagrams for nucleon scattering. Applying the Feynman rules to these diagrams, we get i(\u2212ig)2 1 1)2 \u2212 m2 + (p1 \u2212 p 1 (p1 \u2212 p 2)2 \u2212 m2 (2\u03c0)4 \u03b4(4)(p1 + p2 \u2212 p 1 \u2212 p 2) (3.56) which agrees with the calculation (3.51) that we performed earlier. There is a nice physical interpretation of these diagrams. We talk, rather loosely, of the nucleons exchanging a meson which, in the \ufb01rst diagram, has momentum k = (p1\u2212p 2\u2212p2). This meson doesn\u2019t satisfy the usual energy dispersion relation, because k2 = m2: the meson is called a virtual particle and is said to be o\ufb00-shell (or, sometimes, o\ufb00 massshell). Heuristically, it can\u2019t live long enough for its energy to be measured to great accuracy. In contrast, the momentum on the external, nucleon legs all satisfy p2 = M 2, the mass of the nucleon. They are on-shell. One \ufb01nal note: the addition of the two diagrams above ensures that the particles satisfy Bose statistics. 1) = (p There are also more complicated diagrams which will contribute to the scattering process at higher orders. For example, we have the two diagrams shown in Figures 1 and p 10 and 11, and similar diagrams with p 2 exchanged. Using the Feynman rules, each of these diagrams translates into an integral that we will not attempt to calculate here. And so we go on, with increasingly complicated diagrams, all appearing at higher order in the coupling constant g. \u2013 62 \u2013 p2p1p1/p2/p2p1/p/21+pkk Figure 10: A contribution at O(g4). Figure 11: A contribution at O(g6) Amplitudes Our \ufb01nal result for the nucleon scattering amplitude f | S \u2212 1 |i at order g2 was i(\u2212ig)2 1 1)2 \u2212 m2 + (p1 \u2212 p 1 (p1 \u2212 p 2)2 \u2212 m2 (2\u03c0)4 \u03b4(4)(p1 + p2 \u2212 p 1 \u2212 p 2) The \u03b4-function follows from the conservation of 4-momentum which, in turn, follows from spacetime translational invariance. It is common to all S-matrix elements. We will de\ufb01ne the amplitude Af i by stripping", " o\ufb00 this momentum-conserving delta-function, f | S \u2212 1 |i = i Af i (2\u03c0)4\u03b4(4)(pF \u2212 pI) (3.57) where pI (pF ) is the sum of the initial (\ufb01nal) 4-momenta, and the factor of i out front is a convention which is there to match non-relativistic quantum mechanics. We can now re\ufb01ne our Feynman rules to compute the amplitude iAf i itself: \u2022 Draw all possible diagrams with appropriate external legs and impose 4-momentum conservation at each vertex. \u2022 Write down a factor of (\u2212ig) at each vertex. \u2022 For each internal line, write down the propagator \u2022 Integrate over momentum k \ufb02owing through each loop d4k/(2\u03c0)4. This last step deserves a short explanation. The diagrams we\u2019ve computed so far have It\u2019s not hard to convince yourself that in no loops. They are tree level diagrams. tree diagrams, momentum conservation at each vertex is su\ufb03cient to determine the momentum \ufb02owing through each internal line. For diagrams with loops, such as those shown in Figures 10 and 11, this is no longer the case. \u2013 63 \u2013 p1p1/p2p2/p1p1/p2p2/ Figure 12: The two lowest order Feynman diagrams for nucleon to meson scattering. Nucleon to Meson Scattering Let\u2019s now look at the amplitude for a nucleon-anti-nucleon pair to annihilate into a pair of mesons: \u03c8 \u00af\u03c8 \u2192 \u03c6\u03c6. The simplest Feynman diagrams for this process are shown in Figure 12 where the virtual particle in these diagrams is now the nucleon \u03c8 rather than the meson \u03c6. This fact is re\ufb02ected in the denominator of the amplitudes which are given by iA = (\u2212ig)2 i 1)2 \u2212 M 2 + (p1 \u2212 p As in (3.52), we\u2019ve dropped the i from the propagators as the denominator never vanishes. i (p1 \u2212 p 2)2 \u2212 M 2 (3.58) Nucleon-Anti-Nucleon Scattering Figure 13: The two lowest order Feynman diagrams for nucleon-anti-nucleon scattering. For the scattering of a nucleon and an anti-nucleon, \u03c8 \u00af\u03c8 \u2192 \u03c8 \u00af\u03c8, the Feynman diagrams are a little di\ufb00erent. At lowest order, they are given by the diagrams of Figure 13. It is a simple matter to write down the amplitude using the Feynman rules, iA = (\u2212ig)2 i 1)2 \u2212 m2 + (p1 \u2212 p i (p1 + p2)2 \u2212 m2 + i (3.59) \u2013 64 \u2013 p2p1p2p1p1/p2/p/p/21+p2p1p1/p2/p2p1p1/p2/+ Notice that the momentum dependence in the second term is di\ufb00erent from that of nucleon-nucleon scattering (3.56), re\ufb02ecting the di\ufb00erent Feynman diagram that contributes to the process. In the center of mass frame, p1 = \u2212p2, the denominator of the second term is 4(M 2 + p 2 1 ) \u2212 m2. If m < 2M, then this term never vanishes and we may drop the i. In contrast, if m > 2M, then the amplitude corresponding to the second diagram diverges at some value of p. In this case it turns out that we may also neglect the i term, although for a di\ufb00erent reason: the meson is unstable when m > 2M, a result we derived in (3.30). When correctly treated, this instability adds a \ufb01nite imaginary piece to the denominator which overwhelms the i. Nonetheless, the increase in the scattering amplitude which we see in the second diagram when 4(M 2 + p 2) = m2 is what allows us to discover new particles: they appear as a resonance in the cross section. For example, the Figure 14 shows the cross-section (roughly the amplitude squared) plotted vertically for e+e\u2212 \u2192 \u00b5+\u00b5\u2212 scattering from the ALEPH experiment in CERN. The horizontal axis shows the center of mass energy. The curve rises sharply around 91 GeV, the mass of the Z-boson. Figure 14: Meson Scattering For \u03c6\u03c6 \u2192 \u03c6\u03c6, the simplest diagram we can write down has a single loop, and momentum conservation at each vertex is no longer", " su\ufb03cient to determine every momentum passing through the diagram. We choose to assign the single undetermined momentum k to the right-hand propagator. All other momenta are then determined. The amplitude corresponding to the diagram shown in the \ufb01gure is (\u2212ig)4 d4k (2\u03c0)4 1 (k2 \u2212 M 2 + i)((k + p 1)2 \u2212 M 2 + i) Figure 15: \u00d7 ((k + p 1 \u2212 p1)2 \u2212 M 2 + i)((k \u2212 p 2)2 \u2212 M 2 + i) 1 These integrals can be tricky. For large k, this integral goes as d4k/k8, which is at least convergent as k \u2192 \u221e. But this won\u2019t always be the case! \u2013 65 \u2013 p2p1/p/21pp1/p1/1pp/2kk+\u2212+k\u2212k 3.5.1 Mandelstam Variables We see that in many of the amplitudes above \u2014 in particular those that include the exchange of just a single particle \u2014 the same combinations of momenta are appearing frequently in the denominators. There are standard names for various sums and di\ufb00erences of momenta: they are known as Mandelstam variables. They are s = (p1 + p2)2 = (p t = (p1 \u2212 p u = (p1 \u2212 p 1 + p 1)2 = (p2 \u2212 p 2)2 = (p2 \u2212 p 2)2 2)2 1)2 (3.60) 1 and p where, as in the examples above, p1 and p2 are the momenta of the two initial particles, and p 2 are the momenta of the \ufb01nal two particles. We can de\ufb01ne these variables whether the particles involved in the scattering are the same or di\ufb00erent. To get a feel for what these variables mean, let\u2019s assume all four particles are the same. We sit in the center of mass frame, so that the initial two particles have four-momenta p1 = (E, 0, 0, p) and p2 = (E, 0, 0, \u2212p) (3.61) The particles then scatter at some angle \u03b8 and leave with momenta 1 = (E, 0, p sin \u03b8, p cos \u03b8) and p p 2 = (E, 0, \u2212p sin \u03b8, \u2212p cos \u03b8) (3.62) Then from the above de\ufb01nitions, we have that s = 4E2 and t = \u22122p2(1 \u2212 cos \u03b8) and u = \u22122p2(1 + cos \u03b8) (3.63) The variable s measures the total center of mass energy of the collision, while the variables t and u are measures of the momentum exchanged between particles. (They are basically equivalent, just with the outgoing particles swapped around). Now the amplitudes that involve exchange of a single particle can be written simply in terms of the Mandelstam variables. For example, for nucleon-nucleon scattering, the amplitude (3.56) is schematically A \u223c (t \u2212 m2)\u22121 + (u \u2212 m2)\u22121. For the nucleon-anti-nucleon scattering, the amplitude (3.59) is A \u223c (t \u2212 m2)\u22121 + (s \u2212 m2)\u22121. We say that the \ufb01rst case involves \u201ct-channel\u201d and \u201cu-channel\u201d diagrams. Meanwhile the nucleon-antinucleon scattering is said to involve \u201ct-channel\u201d and \u201cs-channel\u201d diagrams. (The \ufb01rst diagram indeed includes a vertex that looks like the letter \u201cT\u201d). Note that there is a relationship between the Mandelstam variables. When all the masses are the same we have s + t + u = 4M 2. When the masses of all 4 particles di\ufb00er, this becomes. \u2013 66 \u2013 3.5.2 The Yukawa Potential So far we\u2019ve computed the quantum amplitudes for various scattering processes. But these quantities are a little abstract. In Section 3.6 below (and again in next term\u2019s \u201cStandard Model\u201d course) we\u2019ll see how to turn amplitudes into measurable quantities such as cross-sections, or the lifetimes of unstable particles. Here we\u2019ll instead show how to translate the amplitude (3.52) for nucleon scattering into something familiar from Newtonian mechanics: a potential, or force, between the particles. Let\u2019s start by asking a simple question in classical \ufb01eld theory that will turn out to be relevant", ". Suppose that we have a \ufb01xed \u03b4-function source for a real scalar \ufb01eld \u03c6, that persists for all time. What is the pro\ufb01le of \u03c6(x)? To answer this, we must solve the static Klein-Gordon equation, \u2212\u22072\u03c6 + m2\u03c6 = \u03b4(3)(x) We can solve this using the Fourier transform, \u03c6(x) = d3k (2\u03c0)3 eik\u00b7x \u02dc\u03c6(k) Plugging this into (3.64) tells us that (k 2 + m2) \u02dc\u03c6(k) = 1, giving us the solution \u03c6(x) = d3k (2\u03c0)3 eik\u00b7x k 2 + m2 (3.64) (3.65) (3.66) Let\u2019s now do this integral. Changing to polar coordinates, and writing k \u00b7 x = kr cos \u03b8, we have \u03c6(x) = = = 1 (2\u03c0)2 1 (2\u03c0)2r 1 2\u03c0r Re \u221e dk 0 +\u221e k2 k2 + m2 2 sin kr kr dk k sin kr k2 + m2 keikr k2 + m2 dk 2\u03c0i \u2212\u221e +\u221e \u2212\u221e (3.67) We compute this last integral by closing the contour in the upper half plane k \u2192 +i\u221e, picking up the pole at k = +im. This gives \u03c6(x) = 1 4\u03c0r e\u2212mr (3.68) The \ufb01eld dies o\ufb00 exponentially quickly at distances 1/m, the Compton wavelength of the meson. \u2013 67 \u2013 Now we understand the pro\ufb01le of the \u03c6 \ufb01eld, what does this have to do with the force between \u03c8 particles? We do very similar calculations to that above in electrostatics where a charged particle acts as a \u03b4-function source for the gauge potential: \u2212\u22072A0 = \u03b4(3)(x), which is solved by A0 = 1/4\u03c0r. The pro\ufb01le for A0 then acts as the potential energy for another charged (test) particle moving in this background. Can we give the same interpretation to our scalar \ufb01eld? In other words, is there a classical limit of the scalar Yukawa theory where the \u03c8 particles act as \u03b4-function sources for \u03c6, creating the pro\ufb01le (3.68)? And, if so, is this pro\ufb01le then felt as a static potential? The answer is essentially yes, at least in the limit M m. But the correct way to describe the potential felt by the \u03c8 particles is not to talk about classical \ufb01elds at all, but instead work directly with the quantum amplitudes. Our strategy is to compare the nucleon scattering amplitude (3.52) to the corresponding amplitude in non-relativistic quantum mechanics for two particles interacting through a potential. To make this comparison, we should \ufb01rst take the non-relativistic limit of (3.52). Let\u2019s work in the center of mass frame, with p \u2261 p1 = \u2212p2 and p \u2261 p 2. The non-relativistic limit means |p| M which, by momentum conservation, ensures that |p | M. In fact one can check that, for this particular example, this limit doesn\u2019t change the scattering amplitude (3.52): it\u2019s given by 1 = \u2212p iA = +ig2 1 (p \u2212 p )2 + m2 + 1 (p + p )2 + m2 (3.69) How do we compare this to scattering in quantum mechanics? Consider two particles, separated by a distance r, interacting through a potential U (r). In non-relativistic quantum mechanics, the amplitude for the particles to scatter from momentum states \u00b1p into momentum states \u00b1p can be computed in perturbation theory, using the techniques described in Section 3.1. To leading order, known in this context as the Born approximation, the amplitude is given by p | U (r) |p = \u2212i d3r U (r)e\u2212i(p\u2212p)\u00b7r (3.70) There\u2019s a relative factor of (2M )2 that arises in comparing the quantum \ufb01eld theory amplitude A to p | U (r) |p, that can be traced to the relativistic normalization of the states |p1, p2. (It is also necessary to get the dimensions of the potential to work out correctly). Including this factor, and equ", "ating the expressions for the two amplitudes, we get d3r U (r) e\u2212i(p\u2212p)\u00b7r = \u2212\u03bb2 (p \u2212 p )2 + m2 (3.71) \u2013 68 \u2013 where we\u2019ve introduced the dimensionless parameter \u03bb = g/2M. We can trivially invert this to \ufb01nd, U (r) = \u2212\u03bb2 d3p (2\u03c0)3 eip\u00b7r p 2 + m2 (3.72) But this is exactly the integral (3.66) we just did in the classical theory. We have U (r) = \u2212\u03bb2 4\u03c0r e\u2212mr (3.73) This is the Yukawa potential. The force has a range 1/m, the Compton wavelength of the exchanged particle. The minus sign tells us that the potential is attractive. Notice that quantum \ufb01eld theory has given us an entirely new perspective on the nature of forces between particles. Rather than being a fundamental concept, the force arises from the virtual exchange of other particles, in this case the meson. In Section 6 of these lectures, we will see how the Coulomb force arises from quantum \ufb01eld theory due to the exchange of virtual photons. We could repeat the calculation for nucleon-anti-nucleon scattering. The amplitude from \ufb01eld theory is given in (3.59). The \ufb01rst term in this expression gives the same result as for nucleon-nucleon scattering with the same sign. The second term vanishes in the non-relativisitic limit (it is an example of an interaction that doesn\u2019t have a simple Newtonian interpretation). There is no longer a factor of 1/2 in (3.70), because the incoming/outgoing particles are not identical, so we learn that the potential between a nucleon and anti-nucleon is again given by (3.73). This reveals a key feature of forces arising due to the exchange of scalars: they are universally attractive. Notice that this is di\ufb00erent from forces due to the exchange of a spin 1 particle \u2014 such as electromagnetism \u2014 where the sign \ufb02ips when we change the charge. However, for forces due to the exchange of a spin 2 particle \u2014 i.e. gravity \u2014 the force is again universally attractive. 3.5.3 \u03c64 Theory Let\u2019s brie\ufb02y look at the Feynman rules and scattering amplitudes for the interaction Hamiltonian Hint = \u03bb 4! \u03c64 (3.74) The theory now has a single interaction vertex, which comes with a factor of (\u2212i\u03bb), while the other Feynman rules remain the same. Note that we assign (\u2212i\u03bb) to the \u2013 69 \u2013 vertex rather than (\u2212i\u03bb/4!). To see why this is, we can look at \u03c6\u03c6 \u2192 \u03c6\u03c6 scattering, which has its lowest contribution at order \u03bb, with the term \u2212i\u03bb 4! p 1, p 2| : \u03c6(x)\u03c6(x)\u03c6(x)\u03c6(x) : |p1, p2 (3.75) Any one of the \ufb01elds can do the job of annihilation or creation. This gives 4! di\ufb00erent contractions, which cancels the 1/4! sitting out front. Feynman diagrams in the \u03c64 theory sometimes come with extra combinatoric factors (typically 2 or 4) which are known as symmetry factors that one must take into account. For more details, see the book by Peskin and Schroeder. Using the Feynman rules, the scattering amplitude for \u03c6\u03c6 \u2192 \u03c6\u03c6 is simply iA = \u2212i\u03bb. Note that it doesn\u2019t depend on the angle at which in \u03c64 theory the leading order two-particle scattering the outgoing particles emerge: occurs with equal probability in all directions. Translating this into a potential between two mesons, we have Figure 16: U (r) = \u03bb (2m)2 d3p (2\u03c0)3 e+ip\u00b7r = \u03bb (2m)2 \u03b4(3)(r) (3.76) So scattering in \u03c64 theory is due to a \u03b4-function potential. The particles don\u2019t know what hit them until it\u2019s over. 3.5.4 Connected Diagrams and Amputated Diagrams We\u2019ve seen how one can compute scattering amplitudes by writing down all Feynman diagrams and assigning integrals to them using the Feynman rules. In fact, there are a couple of caveats about what Feynman diagrams you should write down. Both of these caveats are related to the assumption we made earlier", " that \u201cinitial and \ufb01nal states are eigenstates of the free theory\u201d which, as we mentioned at the time, is not strictly accurate. The two caveats which go some way towards ameliorating the problem are the following \u2022 We consider only connected Feynman diagrams, where every part of the diagram is connected to at least one external line. As we shall see shortly, this will be related to the fact that the vacuum |0 of the free theory is not the true vacuum |\u2126 of the interacting theory. An example of a diagram that is not connected is shown in Figure 17. \u2013 70 \u2013 \u2212i\u03bb \u2022 We do not consider diagrams with loops on external lines, for example the diagram shown in the Figure 18. We will not explain how to take these into account in this course, but you will discuss them next term. They are related to the fact that the one-particle states of the free theory are not the same as the one-particle states of the interacting theory. In particular, correctly dealing with these diagrams will account for the fact that particles in interacting quantum \ufb01eld theories are never alone, but surrounded by a cloud of virtual particles. We will refer to diagrams in which all loops on external legs have been cut-o\ufb00 as \u201camputated\u201d. Figure 17: A disconnected diagram. Figure 18: An un-amputated diagram 3.6 What We Measure: Cross Sections and Decay Rates So far we\u2019ve learnt to compute the quantum amplitudes for particles decaying or scattering. As usual in quantum theory, the probabilities for things to happen are the (modulus) square of the quantum amplitudes. In this section we will compute these probabilities, known as decay rates and cross sections. One small subtlety here is that the S-matrix elements f | S \u2212 1 |i all come with a factor of (2\u03c0)4\u03b4(4)(pF \u2212 pI), so we end up with the square of a delta-function. As we will now see, this comes from the fact that we\u2019re working in an in\ufb01nite space. 3.6.1 Fermi\u2019s Golden Rule Let\u2019s start with something familiar and recall how to derive Fermi\u2019s golden rule from Dyson\u2019s formula. For two energy eigenstates |m and |n, with Em = En, we have to leading order in the interaction, m| U (t) |n = \u2212i m| t 0 dt HI(t) |n t dt ei\u03c9t 0 ei\u03c9t \u2212 1 \u03c9 (3.77) = \u2212i m| Hint |n = \u2212 m| Hint |n \u2013 71 \u2013 where \u03c9 = Em \u2212 En. This gives us the probability for the transition from |n to |m in time t, as Pn\u2192m(t) = | m| U (t) |n | 2 = 2| m| Hint |n | 2 1 \u2212 cos \u03c9t \u03c92 (3.78) The function in brackets is plotted in Figure 19 for \ufb01xed t. We see that in time t, most transitions happen in a region between energy eigenstates separated by \u2206E = 2\u03c0/t. As t \u2192 \u221e, the function in the \ufb01gure starts to approach a deltafunction. To \ufb01nd the normalization, we can calculate +\u221e d\u03c9 1 \u2212 cos \u03c9t \u03c92 = \u03c0t \u2212\u221e 1 \u2212 cos \u03c9t \u03c92 \u21d2 \u2192 \u03c0t\u03b4(\u03c9) as t \u2192 \u221e Figure 19: Consider now a transition to a cluster of states with density \u03c1(E). In the limit t \u2192 \u221e, we get the transition probability Pn\u2192m = dEm \u03c1(Em) 2| m| Hint |n | 2 1 \u2212 cos \u03c9t \u03c92 \u2192 2\u03c0 | m| Hint |n | 2 \u03c1(En)t (3.79) which gives a constant probability for the transition per unit time for states around the same energy En \u223c Em = E. \u02d9Pn\u2192m = 2\u03c0| m| Hint |n |2 \u03c1(E) (3.80) This is Fermi\u2019s Golden Rule. In the above derivation, we were fairly careful with taking the limit as t \u2192 \u221e. Suppose we were a little sloppier, and \ufb01rst chose to compute the amplitude for the state |n at t \u2192 \u2212\u221e to transition to the state |m at t \u2192 +\u221e. Then we get \u2212i m| t=+\u221e t=\u2212\u221e HI(t) |n = \u2212i m| H", "int |n 2\u03c0\u03b4(\u03c9) (3.81) Now when squaring the amplitude to get the probability, we run into the problem of the square of the delta-function: Pn\u2192m = | m| Hint |n |2(2\u03c0)2\u03b4(\u03c9)2. Tracking through the previous computations, we realize that the extra in\ufb01nity is coming because Pm\u2192n \u2013 72 \u2013 is the probability for the transition to happen in in\ufb01nite time t \u2192 \u221e. We can write the delta-functions as (2\u03c0)2\u03b4(\u03c9)2 = (2\u03c0)\u03b4(\u03c9) T (3.82) where T is shorthand for t \u2192 \u221e (we used a very similar trick when looking at the vacuum energy in (2.25)). We now divide out by this power of T to get the transition probability per unit time, \u02d9Pn\u2192m = 2\u03c0| m| Hint |n |2 \u03b4(\u03c9) (3.83) which, after integrating over the density of \ufb01nal states, gives us back Fermi\u2019s Golden rule. The reason that we\u2019ve stressed this point is because, in our \ufb01eld theory calculations, we\u2019ve computed the amplitudes in the same way as (3.81), and the square of the \u03b4(4)-functions will just be re-interpreted as spacetime volume factors. 3.6.2 Decay Rates Let\u2019s now look at the probability for a single particle |i of momentum pI (I=initial) to decay into some number of particles |f with momentum pi and total momentum pF = i pi. This is given by P = | f | S |i |2 f | f i| i Our states obey the relativistic normalization formula (2.65), i| i = (2\u03c0)3 2EpI \u03b4(3)(0) = 2E pI V where we have replaced \u03b4(3)(0) by the volume of 3-space. Similarly, f | f = 2EpiV \ufb01nal states (3.84) (3.85) (3.86) If we place our initial particle at rest, so pI = 0 and EpI = m, we get the probability for decay P = |Af i|2 2mV (2\u03c0)4\u03b4(4)(pI \u2212 pF ) V T \ufb01nal states 1 2EpiV (3.87) where, as in the second derivation of Fermi\u2019s Golden Rule, we\u2019ve exchanged one of the delta-functions for the volume of spacetime: (2\u03c0)4\u03b4(4)(0) = V T. The amplitudes Af i are, of course, exactly what we\u2019ve been computing. (For example, in (3.30), we saw \u2013 73 \u2013 that A = \u2212g for a single meson decaying into two nucleons). We can now divide out by T to get the transition function per unit time. But we still have to worry about summing over all \ufb01nal states. There are two steps: the \ufb01rst is to integrate over all possible momenta of the \ufb01nal particles: V d3pi/(2\u03c0)3. The factors of spatial volume V in this measure cancel those in (3.87), while the factors of 1/2Epi in (3.87) conspire to produce the Lorentz invariant measure for 3-momentum integrals. The result is an expression for the density of \ufb01nal states given by the Lorentz invariant measure d\u03a0 = (2\u03c0)4\u03b4(4)(pF \u2212 pI) \ufb01nal states d3pi (2\u03c0)3 1 2Epi (3.88) The second step is to sum over all \ufb01nal states with di\ufb00erent numbers (and possibly types) of particles. This gives us our \ufb01nal expression for the decay probability per unit time, \u0393 = \u02d9P. \u0393 = 1 2m \ufb01nal states |Af i|2 d\u03a0 (3.89) \u0393 is called the width of the particle. It is equal to the reciprocal of the half-life \u03c4 = 1/\u0393. 3.6.3 Cross Sections Collide two beams of particles. Sometimes the particles will hit and bounce o\ufb00 each other; sometimes they will pass right through. The fraction of the time that they collide is called the cross section and is denoted by \u03c3. If the incoming \ufb02ux F is de\ufb01ned", " to be the number of incoming particles per area per unit time, then the total number of scattering events N per unit time is given by, N = F \u03c3 (3.90) We would like to calculate \u03c3 from quantum \ufb01eld theory. In fact, we can calculate a more sensitive quantity d\u03c3 known as the di\ufb00erential cross section which is the probability for a given scattering process to occur in the solid angle (\u03b8, \u03c6). More precisely d\u03c3 = Di\ufb00erential Probability Unit Time \u00d7 Unit Flux = 1 4E1E2V 1 F |Af i|2 d\u03a0 (3.91) where we\u2019ve used the expression for probability per unit time that we computed in the previous subsection. E1 and E2 are the energies of the incoming particles. We now need an expression for the unit \ufb02ux. For simplicity, let\u2019s sit in the center of mass frame of the collision. We\u2019ve been considering just a single particle per spatial volume V, \u2013 74 \u2013 meaning that the \ufb02ux is given in terms of the 3-velocities vi as F = |v1 \u2212 v2|/V. This then gives, d\u03c3 = 1 4E1E2 1 |v1 \u2212 v2| |Af i|2 d\u03a0 (3.92) If you want to write this in terms of momentum, then recall from your course on special 1 \u2212 v2 = relativity that the 3-velocities vi are related to the momenta by v = p/m p/p 0. \u221a Equation (3.92) is our \ufb01nal expression relating the S-matrix to the di\ufb00erential cross section. You may now take your favorite scattering amplitude, and compute the probability for particles to \ufb02y out at your favorite angles. This will involve doing the integral over the phase space of \ufb01nal states, with measure d\u03a0. Notice that di\ufb00erent scattering amplitudes have di\ufb00erent momentum dependence and will result in di\ufb00erent angular dependence in scattering amplitudes. For example, in \u03c64 theory the amplitude for tree level scattering was simply A = \u2212\u03bb. This results in isotropic scattering. In contrast, for nucleon-nucleon scattering we have schematically A \u223c (t \u2212 m2)\u22121 + (u \u2212 m2)\u22121. This gives rise to angular dependence in the di\ufb00erential cross-section, which follows from the fact that, for example, t = \u22122|p|2(1 \u2212 cos \u03b8), where \u03b8 is the angle between the incoming and outgoing particles. 3.7 Green\u2019s Functions So far we\u2019ve learnt to compute scattering amplitudes. These are nice and physical (well \u2013 they\u2019re directly related to cross-sections and decay rates which are physical) but there are many questions we want to ask in quantum \ufb01eld theory that aren\u2019t directly related to scattering experiments. For example, we might want to compute the viscosity of the quark gluon plasma, or the optical conductivity in a tentative model of strange metals, or \ufb01gure out the non-Gaussianity of density perturbations arising in the CMB from novel models of in\ufb02ation. All of these questions are answered in the framework of quantum \ufb01eld theory by computing elementary objects known as correlation functions. In this section we will brie\ufb02y de\ufb01ne correlation functions, explain how to compute them using Feynman diagrams, and then relate them back to scattering amplitudes. We\u2019ll leave the relationship to other physical phenomena to other courses. We\u2019ll denote the true vacuum of the interacting theory as |\u2126. We\u2019ll normalize H such that H |\u2126 = 0 (3.93) \u2013 75 \u2013 and \u2126| \u2126 = 1. Note that this is di\ufb00erent from the state we\u2019ve called |0 which is the vacuum of the free theory and satis\ufb01es H0 |0 = 0. De\ufb01ne G(n)(x1,..., xn) = \u2126| T \u03c6H(x1)... \u03c6H(xn) |\u2126 (3.94) where \u03c6H is \u03c6 in the Heisenberg picture of the full theory, rather than the interaction picture that we\u2019ve been dealing with so far. The G(n) are called correlation functions, or Green\u2019s functions. There are a number of di\ufb00erent ways of looking at these objects which tie together nicely. Let\u2019s start by asking how to compute G", "(n) using Feynman diagrams. We prove the following result Claim: We use the notation \u03c61 = \u03c6(x1), and write \u03c61H to denote the \ufb01eld in the Heisenberg picture, and \u03c61I to denote the \ufb01eld in the interaction picture. Then G(n)(x1,..., xn) = \u2126| T \u03c61H... \u03c6nH |\u2126 = 0| T \u03c61I... \u03c6nI S |0 0| S |0 (3.95) where the operators on the right-hand side are evaluated on |0, the vacuum of the free theory. Proof: Take t1 > t2 >... > tn. Then we can drop the T and write the numerator of the right-hand side as 0| UI(+\u221e, t1)\u03c61I U (t1, t2) \u03c62I... \u03c6nI UI(tn, \u2212\u221e) |0 We\u2019ll use the factors of UI(tk, tk+1) = T exp(\u2212i tk+1 HI) to convert each of the \u03c6I into \u03c6H and we choose operators in the two pictures to be equal at some arbitrary time t0. Then we can write tk 0| UI(+\u221e, t1)\u03c61I U (t1, t2) \u03c62I... \u03c6nI UI(tn, \u2212\u221e) |0 = 0| UI(+\u221e, t0)\u03c61H... \u03c6nH UI(t0, \u2212\u221e) |0 Now let\u2019s deal with the two remaining U (t0, \u00b1\u221e) at either end of the string of operators. Consider an arbitrary state |\u03a8 and look at \u03a8| UI(t, \u2212\u221e) |0 = \u03a8| U (t, \u2212\u221e) |0 (3.96) where U (t, \u2212\u221e) is the Schr\u00a8odinger evolution operator, and the equality above follows because H0 |0 = 0. Now insert a complete set of states, which we take to be energy eigenstates of H = H0 + Hint, \u03a8| U (t, \u2212\u221e) |0 = \u03a8| U (t, \u2212\u221e) |\u2126 \u2126| + = \u03a8| \u2126 \u2126| 0 + lim t\u2192\u2212\u221e n=0 |n n| |0 n=0 eiEn(t\u2212t) \u03a8| n n| 0 (3.97) \u2013 76 \u2013 But the last term vanishes. This follows from the Riemann-Lebesgue lemma which says that for any well-behaved function lim \u00b5\u2192\u221e b a dx f (x)ei\u00b5x = 0 (3.98) n in (3.97) is really an integral dn, Why is this relevant? The point is that the because all states are part of a continuum due to the momentum. (There is a caveat here: we want the vacuum |\u2126 to be special, so that it sits on its own, away from the continuum of the integral. This means that we must be working in a theory with a mass gap \u2013 i.e. with no massless particles). So the Riemann-Lebesgue lemma gives us lim t\u2192\u2212\u221e \u03a8| U (t, t) |0 = \u03a8| \u2126 \u2126| 0 (3.99) (Notice that to derive this result, Peskin and Schroeder instead send t \u2192 \u2212\u221e in a slightly imaginary direction, which also does the job). We now apply the formula (3.99), to the top and bottom of the right-hand side of (3.95) to \ufb01nd 0| \u2126 \u2126| T \u03c61H... \u03c6nH |\u2126 \u2126| 0 0| \u2126 \u2126| \u2126 \u2126| 0 (3.100) which, using the normalization \u2126| \u2126 = 1, gives us the left-hand side, completing the. proof. 3.7.1 Connected Diagrams and Vacuum Bubbles We\u2019re getting closer to our goal of computing the Green\u2019s functions G(n) since we can compute both 0| T \u03c6I(x1)... \u03c6I(xn) S |0 and 0| S |0 using the same methods we developed for S-matrix elements; namely Dyson", "\u2019s formula and Wick\u2019s theorem or, alternatively, Feynman diagrams. But what about dividing one by the other? What\u2019s that all about? In fact, it has a simple interpretation. For the following discussion, we will work in \u03c64 theory. Since there is no ambiguity in the di\ufb00erent types of line in Feynman diagrams, we will represent the \u03c6 particles as solid lines, rather than the dashed lines that we used previously. Then we have the diagramatic expansion for 0| S |0. 0| S |3.101) These diagrams are called vacuum bubbles. The combinatoric factors (as well as the symmetry factors) associated with each diagram are such that the whole series sums \u2013 77 \u2013 (++) to an exponential, 0| S |0 = exp (3.102) So the amplitude for the vacuum of the free theory to evolve into itself is 0| S |0 = exp(all distinct vacuum bubbles). A similar combinatoric simpli\ufb01cation occurs for generic correlation functions. Remarkably, the vacuum diagrams all add up to give the same exponential. With a little thought one can show that 0| T \u03c61... \u03c6nS |0 = connected diagrams 0| S |0 (3.103) where \u201cconnected\u201d means that every part of the diagram is connected to at least one of the external legs. The upshot of all this is that dividing by 0| S |0 has a very nice interpretation in terms of Feynman diagrams: we need only consider the connected Feynman diagrams, and don\u2019t have to worry about the vacuum bubbles. Combining this with (3.95), we learn that the Green\u2019s functions G(n)(x1..., xn) can be calculated by summing over all connected Feynman diagrams, \u2126| T \u03c6H(x1)... \u03c6H(xn) |\u2126 = Connected Feynman Graphs (3.104) An Example: The Four-Point Correlator: \u2126| T \u03c6H(x1)... \u03c6H(x4) |\u2126 As a simple example, let\u2019s look at the four-point correlation function in \u03c64 theory. The sum of connected Feynman diagrams is given by, All of these are connected diagrams, even though they don\u2019t look that connected! The point is that a connected diagram is de\ufb01ned by the requirement that every line is joined to an external leg. An example of a diagram that is not connected is shown in the \ufb01gure. As we have seen, such diagrams are taken care of in shifting the vacuum from |0 to |\u2126. Figure 20: Feynman Rules The Feynman diagrams that we need to calculate for the Green\u2019s functions depend on x1,..., xn. This is rather di\ufb00erent than the Feynman diagrams that we calculated for \u2013 78 \u2013 (+++...)x1x2x3x4x1x2x3x4x1x2x3x4++++5 Similar+...2 Similarx1x2x3x4 the S-matrix elements, where we were working primarily with momentum eigenstates, and ended up integrating over all of space. However, it\u2019s rather simple to adapt the Feynman rules that we had earlier in momentum space to compute G(n)(x1..., xn). For \u03c64 theory, we have \u2022 Draw n external points x1,..., xn, connected by the usual propagators and ver- tices. Assign a spacetime position y to the end of each line. \u2022 For each line \u2206F (x \u2212 y). from x to y write down a factor of the Feynman propagator \u2022 For each vertex at position y, write down a factor of \u2212i\u03bb d4y. 3.7.2 From Green\u2019s Functions to S-Matrices Having described how to compute correlation functions using Feynman diagrams, let\u2019s now relate them back to the S-matrix elements that we already calculated. The \ufb01rst step is to perform the Fourier transform, n d4xie\u2212ipi\u00b7xi G(n)(x1,..., xn) (3.105) \u02dcG(n)(p1,..., pn) = i=1 These are very closely related to the S-matrix elements that we\u2019ve computed above. The di\ufb00erence is that the Feynman rules for G(n)(x1,..., xn), e\ufb00ectively include propagators \u2206F", " for the external legs, as well as the internal legs. A related fact is that the 4momenta assigned to the external legs is arbitrary: they are not on-shell. Both of these problems are easily remedied to allow us to return to the S-matrix elements: we need to simply cancel o\ufb00 the propagators on the external legs, and place their momentum back on shell. We have p 1,..., p n| S \u2212 1 |p1..., pn = (\u2212i)n+n n (p 2 i \u2212 m2) n (p2 j \u2212 m2) (3.106) i=1 \u00d7 \u02dcG(n+n)(\u2212p j=1 1,..., \u2212p n, p1,..., pn) Each of the factors (p2\u2212m2) vanishes once the momenta are placed on-shell. This means that we only get a non-zero answer for diagrams contributing to G(n)(x1,..., xn) which have propagators for each external leg. So what\u2019s the point of all of this? We\u2019ve understood that ignoring the unconnected diagrams is related to shifting to the true vacuum |\u2126. But other than that, introducing the Green\u2019s functions seems like a lot of bother for little reward. The important point \u2013 79 \u2013 xyy is that this provides a framework in which to deal with the true particle states in the interacting theory through renormalization. Indeed, the formula (3.106), suitably interpreted, remains true even in the interacting theory, taking into account the swarm of virtual particles surrounding asymptotic states. This is the correct way to consider scattering. In this context, (3.106) is known as the LSZ reduction formula. You will derive it properly next term. \u2013 80 \u2013 4. The Dirac Equation \u201cA great deal more was hidden in the Dirac equation than the author had expected when he wrote it down in 1928. Dirac himself remarked in one of his talks that his equation was more intelligent than its author. It should be added, however, that it was Dirac who found most of the additional insights.\u201d Weisskopf on Dirac So far we\u2019ve only discussed scalar \ufb01elds such that under a Lorentz transformation x\u00b5 \u2192 (x)\u00b5 = \u039b\u00b5 \u03bdx\u03bd, the \ufb01eld transforms as \u03c6(x) \u2192 \u03c6(x) = \u03c6(\u039b\u22121x) (4.1) We have seen that quantization of such \ufb01elds gives rise to spin 0 particles. But most particles in Nature have an intrinsic angular momentum, or spin. These arise naturally in \ufb01eld theory by considering \ufb01elds which themselves transform non-trivially under the Lorentz group. In this section we will describe the Dirac equation, whose quantization gives rise to fermionic spin 1/2 particles. To motivate the Dirac equation, we will start by studying the appropriate representation of the Lorentz group. A familiar example of a \ufb01eld which transforms non-trivially under the Lorentz group is the vector \ufb01eld A\u00b5(x) of electromagnetism, A\u00b5(x) \u2192 \u039b\u00b5 \u03bdA\u03bd(\u039b\u22121x) (4.2) We\u2019ll deal with this in Section 6. (It comes with its own problems!). In general, a \ufb01eld can transform as \u03c6a(x) \u2192 D[\u039b]a b \u03c6b(\u039b\u22121x) (4.3) where the matrices D[\u039b] form a representation of the Lorentz group, meaning that D[\u039b1]D[\u039b2] = D[\u039b1\u039b2] (4.4) and D[\u039b\u22121] = D[\u039b]\u22121 and D[1] = 1. How do we \ufb01nd the di\ufb00erent representations? Typically, we look at in\ufb01nitesimal transformations of the Lorentz group and study the resulting Lie algebra. If we write, \u039b\u00b5 \u03bd = \u03b4\u00b5 \u03bd + \u03c9\u00b5 \u03bd (4.5) for in\ufb01nitesimal \u03c9, then the condition for a Lorentz transformation \u039b\u00b5 becomes the requirement that \u03c9 is anti-symmetric: \u03c3\u039b\u03bd \u03c1\u03b7\u03c3\u03c1 = \u03b7\u00b5\u03bd \u03c9\u00b5\u03bd + \u03c9\u03bd\u00b5 = 0 (4.6) \u2013 81 \u2013 Note that an ant", "isymmetric 4 \u00d7 4 matrix has 4 \u00d7 3/2 = 6 independent components, which agrees with the 6 transformations of the Lorentz group: 3 rotations and 3 boosts. It\u2019s going to be useful to introduce a basis of these six 4 \u00d7 4 anti-symmetric matrices. We could call them (MA)\u00b5\u03bd, with A = 1,..., 6. But in fact it\u2019s better for us (although initially a little confusing) to replace the single index A with a pair of antisymmetric indices [\u03c1\u03c3], where \u03c1, \u03c3 = 0,..., 3, so we call our matrices (M\u03c1\u03c3)\u00b5 \u03bd. The antisymmetry on the \u03c1 and \u03c3 indices means that, for example, M01 = \u2212M10, etc, so that \u03c1 and \u03c3 again label six di\ufb00erent matrices. Of course, the matrices are also antisymmetric on the \u00b5\u03bd indices because they are, after all, antisymmetric matrices. With this notation in place, we can write a basis of six 4 \u00d7 4 antisymmetric matrices as (M\u03c1\u03c3)\u00b5\u03bd = \u03b7\u03c1\u00b5 \u03b7\u03c3\u03bd \u2212 \u03b7\u03c3\u00b5\u03b7\u03c1\u03bd (4.7) where the indices \u00b5 and \u03bd are those of the 4 \u00d7 4 matrix, while \u03c1 and \u03c3 denote which basis element we\u2019re dealing with. If we use these matrices for anything practical (for example, if we want to multiply them together, or act on some \ufb01eld) we will typically need to lower one index, so we have (M\u03c1\u03c3)\u00b5 \u03bd = \u03b7\u03c1\u00b5 \u03b4\u03c3 \u03bd \u2212 \u03b7\u03c3\u00b5\u03b4\u03c1 \u03bd (4.8) Since we lowered the index with the Minkowski metric, we pick up various minus signs which means that when written in this form, the matrices are no longer necessarily antisymmetric. Two examples of these basis matrices are, (M01) and (M12)4.9) The \ufb01rst, M01, generates boosts in the x1 direction. It is real and symmetric. The second, M12, generates rotations in the (x1, x2)-plane. It is real and antisymmetric. \u03bd as a linear combination of the M\u03c1\u03c3, We can now write any \u03c9\u00b5 \u03c9\u00b5 \u03bd = 1 2 \u2126\u03c1\u03c3 (M\u03c1\u03c3)\u00b5 \u03bd (4.10) where \u2126\u03c1\u03c3 are just six numbers (again antisymmetric in the indices) that tell us what Lorentz transformation we\u2019re doing. The six basis matrices M\u03c1\u03c3 are called the generators of the Lorentz transformations. The generators obey the Lorentz Lie algebra relations, [M\u03c1\u03c3, M\u03c4 \u03bd] = \u03b7\u03c3\u03c4M\u03c1\u03bd \u2212 \u03b7\u03c1\u03c4M\u03c3\u03bd + \u03b7\u03c1\u03bdM\u03c3\u03c4 \u2212 \u03b7\u03c3\u03bdM\u03c1\u03c4 (4.11) \u2013 82 \u2013 where we have suppressed the matrix indices. A \ufb01nite Lorentz transformation can then be expressed as the exponential \u039b = exp 1 2 \u2126\u03c1\u03c3M\u03c1\u03c3 (4.12) Let me stress again what each of these objects are: the M\u03c1\u03c3 are six 4\u00d74 basis elements of the Lorentz Lie algebra; the \u2126\u03c1\u03c3 are six numbers telling us what kind of Lorentz transformation we\u2019re doing (for example, they say things like rotate by \u03b8 = \u03c0/7 about the x3-direction and run at speed v = 0.2 in the x1 direction). 4.1 The Spinor Representation We\u2019re interested in \ufb01nding other matrices which satisfy the Lorentz algebra commutation relations (4.11). We will construct the spinor representation. To do this, we start by de\ufb01ning something which, at \ufb01rst sight, has nothing to do with the Lorentz group. It is the Cli\ufb00ord algebra, {\u03b3\u00b5, \u03b3\u03bd} \u2261 \u03b3\u00b5\u03b3\u03bd + \u03b3\u03bd\u03b3\u00b5 = 2\u03b7\u00b5\u03bd1 (4.13) where \u03b3\u00b5, with \u00b5 = 0, 1, 2, 3, are a set of four matrices and the 1 on the right-hand side denotes the unit matrix. This means that we must \ufb01nd four matrices such that and \u03b3\u00b5\u03b3\u03bd = \u2212\u03b3\u03bd\u03b3\u00b5 when \u00b5 = \u03bd (\u03b30)2 = 1, (\u03b3i)2 = \u2212", "1 i = 1, 2, 3 (4.14) (4.15) It\u2019s not hard to convince yourself that there are no representations of the Cli\ufb00ord algebra using 2 \u00d7 2 or 3 \u00d7 3 matrices. The simplest representation of the Cli\ufb00ord algebra is in terms of 4 \u00d7 4 matrices. There are many such examples of 4 \u00d7 4 matrices which obey (4.13). For example, we may take \u03b30 = 0 1 1 0, \u03b3i = 0 \u03c3i \u2212\u03c3i 0 where each element is itself a 2 \u00d7 2 matrix, with the \u03c3i the Pauli matrices \u03c31 = 0 1 1 0, \u03c32 = 0 \u2212i i 0, \u03c33 = 1 0 0 \u22121 which themselves satisfy {\u03c3i, \u03c3j} = 2\u03b4ij. (4.16) (4.17) \u2013 83 \u2013 One can construct many other representations of the Cli\ufb00ord algebra by taking V \u03b3\u00b5V \u22121 for any invertible matrix V. However, up to this equivalence, it turns out that there is a unique irreducible representation of the Cli\ufb00ord algebra. The matrices (4.16) provide one example, known as the Weyl or chiral representation (for reasons that will soon become clear). We will soon restrict ourselves further, and consider only representations of the Cli\ufb00ord algebra that are related to the chiral representation by a unitary transformation V. So what does the Cli\ufb00ord algebra have to do with the Lorentz group? Consider the commutator of two \u03b3\u00b5, S\u03c1\u03c3 = 1 4 [\u03b3\u03c1, \u03b3\u03c3] = 0 2 \u03b3\u03c1\u03b3\u03c3 1 \u03c1 = \u03c3 \u03c1 = \u03c3 = 1 2\u03b3\u03c1\u03b3\u03c3 \u2212 1 2 \u03b7\u03c1\u03c3 (4.18) Let\u2019s see what properties these matrices have: Claim 4.1: [S\u00b5\u03bd, \u03b3\u03c1] = \u03b3\u00b5\u03b7\u03bd\u03c1 \u2212 \u03b3\u03bd\u03b7\u03c1\u00b5 Proof: When \u00b5 = \u03bd we have [S\u00b5\u03bd, \u03b3\u03c1] = 1 = 1 = 1 = \u03b3\u00b5\u03b7\u03bd\u03c1 \u2212 \u03b3\u03bd\u03b7\u03c1\u00b5 2[\u03b3\u00b5\u03b3\u03bd, \u03b3\u03c1] 2\u03b3\u00b5\u03b3\u03bd\u03b3\u03c1 \u2212 1 2\u03b3\u00b5{\u03b3\u03bd, \u03b3\u03c1} \u2212 1 2\u03b3\u03c1\u03b3\u00b5\u03b3\u03bd 2\u03b3\u00b5\u03b3\u03c1\u03b3\u03bd \u2212 1 2{\u03b3\u03c1, \u03b3\u00b5 }\u03b3\u03bd + 1 2\u03b3\u00b5\u03b3\u03c1\u03b3\u03bd Claim 4.2: The matrices S\u00b5\u03bd form a representation of the Lorentz algebra (4.11), meaning [S\u00b5\u03bd, S\u03c1\u03c3] = \u03b7\u03bd\u03c1S\u00b5\u03c3 \u2212 \u03b7\u00b5\u03c1S\u03bd\u03c3 + \u03b7\u00b5\u03c3S\u03bd\u03c1 \u2212 \u03b7\u03bd\u03c3S\u00b5\u03c1 (4.19) Proof: Taking \u03c1 = \u03c3, and using Claim 4.1 above, we have [S\u00b5\u03bd, S\u03c1\u03c3] = 1 = 1 = 1 2[S\u00b5\u03bd, \u03b3\u03c1\u03b3\u03c3] 2[S\u00b5\u03bd, \u03b3\u03c1]\u03b3\u03c3 + 1 2\u03b3\u00b5\u03b3\u03c3\u03b7\u03bd\u03c1 \u2212 1 2\u03b3\u03c1[S\u00b5\u03bd, \u03b3\u03c3] 2\u03b3\u03bd\u03b3\u03c3\u03b7\u03c1\u00b5 + 1 2\u03b3\u03c1\u03b3\u00b5\u03b7\u03bd\u03c3 \u2212 1 2\u03b3\u03c1\u03b3\u03bd\u03b7\u03c3\u00b5 Now using the expression (4.18) to write \u03b3\u00b5\u03b3\u03c3 = 2S\u00b5\u03c3 + \u03b7\u00b5\u03c3, we have [S\u00b5\u03bd, S\u03c1\u03c3] = S\u00b5\u03c3\u03b7\u03bd\u03c1 \u2212 S\u03bd\u03c3\u03b7\u03c1\u00b5 + S\u03c1\u00b5\u03b7\u03bd\u03c3 \u2212 S\u03c1\u03bd\u03b7\u03c3\u00b5 which is our desired expression. \u2013 84 \u2013 (4.20) (4.21) 4.1.1 Spinors The S\u00b5\u03bd are 4 \u00d7 4 matrices, because the \u03b3\u00b5 are 4 \u00d7 4 matrices. So far we haven\u2019t given an index name to the rows and columns of these matrices: we\u2019re going to call them \u03b1, \u03b2 = 1, 2, 3, 4. We need a \ufb01eld for the matrices (S\u00b5\u03bd)\u03b1 \u03b2 to act upon. We introduce the Dirac spinor \ufb01eld \u03c8\u03b1(x), an object with four complex components labelled by \u03b1 = 1, 2, 3, 4. Under Lorentz transformations, we have where \u03c8\u03b1(x) \u2192", " S[\u039b]\u03b1 \u03b2 \u03c8\u03b2(\u039b\u22121x) \u039b = exp 1 S[\u039b] = exp 1 2 \u2126\u03c1\u03c3M\u03c1\u03c3 2 \u2126\u03c1\u03c3S\u03c1\u03c3 (4.22) (4.23) (4.24) Although the basis of generators M\u03c1\u03c3 and S\u03c1\u03c3 are di\ufb00erent, we use the same six numbers \u2126\u03c1\u03c3 in both \u039b and S[\u039b]: this ensures that we\u2019re doing the same Lorentz transformation on x and \u03c8. Note that we denote both the generator S\u03c1\u03c3 and the full Lorentz transformation S[\u039b] as \u201cS\u201d. To avoid confusion, the latter will always come with the square brackets [\u039b]. Both \u039b and S[\u039b] are 4 \u00d7 4 matrices. So how can we be sure that the spinor representation is something new, and isn\u2019t equivalent to the familiar representation \u039b\u00b5 \u03bd? To see that the two representations are truly di\ufb00erent, let\u2019s look at some speci\ufb01c transformations. Rotations Sij = 1 2 0 \u03c3i \u2212\u03c3i 0 0 \u03c3j \u2212\u03c3j 0 = \u2212 i 2 ijk \u03c3k 0 0 \u03c3k (for i = j) (4.25) If we write the rotation parameters as \u2126ij = \u2212ijk\u03d5k (meaning \u212612 = \u2212\u03d53, etc) then the rotation matrix becomes S[\u039b] = exp 1 2 \u2126\u03c1\u03c3S\u03c1\u03c3 = e+i \u03d5\u00b7\u03c3/2 0 0 e+i\u03d5\u00b7\u03c3/2 (4.26) where we need to remember that \u212612 = \u2212\u212621 = \u2212\u03d53 when following factors of 2. Consider now a rotation by 2\u03c0 about, say, the x3-axis. This is achieved by \u03d5 = (0, 0, 2\u03c0), \u2013 85 \u2013 and the spinor rotation matrix becomes, S[\u039b] = e+i\u03c0\u03c33 0 0 e+i\u03c0\u03c33 = \u22121 Therefore under a 2\u03c0 rotation \u03c8\u03b1(x) \u2192 \u2212\u03c8\u03b1(x) (4.27) (4.28) which is de\ufb01nitely not what happens to a vector! To check that we haven\u2019t been cheating with factors of 2, let\u2019s see how a vector would transform under a rotation by \u03d5 = (0, 0, \u03d53). We have \u039b = exp 1 2 \u2126\u03c1\u03c3M\u03c1\u03c3 = exp 0 0 0 0 0 \u03d53 0 0 0 \u2212\u03d53 0 0 0 0 0 0 (4.29) So when we rotate a vector by \u03d53 = 2\u03c0, we learn that \u039b = 1 as you would expect. So S[\u039b] is de\ufb01nitely a di\ufb00erent representation from the familiar vector representation \u039b\u00b5 \u03bd. Boosts S0i = 1 2 0 1 1 0 0 \u03c3i \u2212\u03c3i 0 = 1 2 \u2212\u03c3i 0 0 \u03c3i Writing the boost parameter as \u2126i0 = \u2212\u21260i = \u03c7i, we have S[\u039b] = e+\u03c7\u00b7\u03c3/2 0 0 e\u2212\u03c7\u00b7\u03c3/2 (4.30) (4.31) Representations of the Lorentz Group are not Unitary Note that for rotations given in (4.26), S[\u039b] is unitary, satisfying S[\u039b]\u2020S[\u039b] = 1. But for boosts given in (4.31), S[\u039b] is not unitary. In fact, there are no \ufb01nite dimensional unitary representations of the Lorentz group. We have demonstrated this explicitly for the spinor representation using the chiral representation (4.16) of the Cli\ufb00ord algebra. We can get a feel for why it is true for a spinor representation constructed from any representation of the Cli\ufb00ord algebra. Recall that S[\u039b] = exp 1 2 \u2126\u03c1\u03c3S\u03c1\u03c3 (4.32) so the representation is unitary if S\u00b5\u03bd are anti-hermitian, i.e. (S\u00b5\u03bd)\u2020 = \u2212S\u00b5\u03bd. But we have (S\u00b5\u03bd)\u2020 = \u2212 1 4 [(\u03b3\u00b5)\u2020, (\u03b3\u03bd)\u2020] (4.33) \u2013 86 \u2013 which can be anti-hermitian if all \u03b3\u00b5 are hermitian or all are anti-hermitian. However, we can never arrange for this", " to happen since (\u03b30)2 = 1 \u21d2 Real Eigenvalues (\u03b3i)2 = \u22121 \u21d2 Imaginary Eigenvalues (4.34) So we could pick \u03b30 to be hermitian, but we can only pick \u03b3i to be anti-hermitian. Indeed, in the chiral representation (4.16), the matrices have this property: (\u03b30)\u2020 = \u03b30 and (\u03b3i)\u2020 = \u2212\u03b3i. In general there is no way to pick \u03b3\u00b5 such that S\u00b5\u03bd are anti-hermitian. 4.2 Constructing an Action We now have a new \ufb01eld to work with, the Dirac spinor \u03c8. We would like to construct a Lorentz invariant equation of motion. We do this by constructing a Lorentz invariant action. We will start in a naive way which won\u2019t work, but will give us a clue how to proceed. De\ufb01ne \u03c8\u2020(x) = (\u03c8)T (x) (4.35) which is the usual adjoint of a multi-component object. We could then try to form a Lorentz scalar by taking the product \u03c8\u2020\u03c8, with the spinor indices summed over. Let\u2019s see how this transforms under Lorentz transformations, \u03c8(x) \u2192 S[\u039b] \u03c8(\u039b\u22121x) \u03c8\u2020(x) \u2192 \u03c8\u2020(\u039b\u22121x) S[\u039b]\u2020 (4.36) So \u03c8\u2020(x)\u03c8(x) \u2192 \u03c8\u2020(\u039b\u22121x)S[\u039b]\u2020S[\u039b]\u03c8(\u039b\u22121x). But, as we have seen, for some Lorentz transformation S[\u039b]\u2020S[\u039b] = 1 since the representation is not unitary. This means that \u03c8\u2020\u03c8 isn\u2019t going to do it for us: it doesn\u2019t have any nice transformation under the Lorentz group, and certainly isn\u2019t a scalar. But now we see why it fails, we can also see how to proceed. Let\u2019s pick a representation of the Cli\ufb00ord algebra which, like the chiral representation (4.16), satis\ufb01es (\u03b30)\u2020 = \u03b30 and (\u03b3i)\u2020 = \u2212\u03b3i. Then for all \u00b5 = 0, 1, 2, 3 we have which, in turn, means that \u03b30\u03b3\u00b5\u03b30 = (\u03b3\u00b5)\u2020 (S\u00b5\u03bd)\u2020 = 1 4 [(\u03b3\u03bd)\u2020, (\u03b3\u00b5)\u2020] = \u2212\u03b30S\u00b5\u03bd\u03b30 (4.37) (4.38) \u2013 87 \u2013 so that S[\u039b]\u2020 = exp 1 2 \u2126\u03c1\u03c3(S\u03c1\u03c3)\u2020 = \u03b30 S[\u039b]\u22121\u03b30 With this in mind, we now de\ufb01ne the Dirac adjoint \u00af\u03c8(x) = \u03c8\u2020(x) \u03b30 (4.39) (4.40) Let\u2019s now see what Lorentz covariant objects we can form out of a Dirac spinor \u03c8 and its adjoint \u00af\u03c8. Claim 4.3: \u00af\u03c8\u03c8 is a Lorentz scalar. Proof: Under a Lorentz transformation, \u00af\u03c8(x) \u03c8(x) = \u03c8\u2020(x) \u03b30 \u03c8(x) \u2192 \u03c8\u2020(\u039b\u22121x) S[\u039b]\u2020\u03b30 S[\u039b]\u03c8(\u039b\u22121x) = \u03c8\u2020(\u039b\u22121x) \u03b30 \u03c8(\u039b\u22121x) = \u00af\u03c8(\u039b\u22121x) \u03c8(\u039b\u22121x) which is indeed the transformation law for a Lorentz scalar. Claim 4.4: \u00af\u03c8 \u03b3\u00b5 \u03c8 is a Lorentz vector, which means that \u00af\u03c8(x) \u03b3\u00b5 \u03c8(x) \u2192 \u039b\u00b5 \u03bd \u00af\u03c8(\u039b\u22121x) \u03b3\u03bd \u03c8(\u039b\u22121x) (4.41) (4.42) This equation means that we can treat the \u00b5 = 0, 1, 2, 3 index on the \u03b3\u00b5 matrices as a true vector index. In particular we can form Lorentz scalars by contracting it with other Lorentz indices. Proof: Suppressing the x argument, under a Lorentz transformation we have, \u00af\u03c8 \u03b3\u00b5", " \u03c8 \u2192 \u00af\u03c8 S[\u039b]\u22121 \u03b3\u00b5 S[\u039b]\u03c8 If \u00af\u03c8 \u03b3\u00b5 \u03c8 is to transform as a vector, we must have S[\u039b]\u22121 \u03b3\u00b5 S[\u039b] = \u039b\u00b5 \u03bd\u03b3\u03bd We\u2019ll now show this. We work in\ufb01nitesimally, so that 2 \u2126\u03c1\u03c3M\u03c1\u03c3 \u2248 1 + 1 2 \u2126\u03c1\u03c3S\u03c1\u03c3 \u2248 1 + 1 \u039b = exp 1 S[\u039b] = exp 1 2\u2126\u03c1\u03c3M\u03c1\u03c3 +... 2\u2126\u03c1\u03c3S\u03c1\u03c3 +... (4.43) (4.44) (4.45) (4.46) \u2013 88 \u2013 so the requirement (4.44) becomes \u2212[S\u03c1\u03c3, \u03b3\u00b5] = (M\u03c1\u03c3)\u00b5 \u03bd\u03b3\u03bd (4.47) where we\u2019ve suppressed the \u03b1, \u03b2 indices on \u03b3\u00b5 and S\u00b5\u03bd, but otherwise left all other indices explicit. In fact equation (4.47) follows from Claim 4.1 where we showed that [S\u03c1\u03c3, \u03b3\u00b5] = \u03b3\u03c1\u03b7\u03c3\u00b5 \u2212 \u03b3\u03c3\u03b7\u00b5\u03c1. To see this, we write the right-hand side of (4.47) by expanding out M, (M\u03c1\u03c3)\u00b5 \u03bd\u03b3\u03bd = (\u03b7\u03c1\u00b5\u03b4\u03c3 \u03bd \u2212 \u03b7\u03c3\u00b5\u03b4\u03c1 = \u03b7\u03c1\u00b5\u03b3\u03c3 \u2212 \u03b7\u03c3\u00b5\u03b3\u03c1 \u03bd)\u03b3\u03bd which means that the proof follows if we can show \u2212[S\u03c1\u03c3, \u03b3\u00b5] = \u03b7\u03c1\u00b5\u03b3\u03c3 \u2212 \u03b7\u03c3\u00b5\u03b3\u03c1 which is exactly what we proved in Claim 4.1. (4.48) (4.49) Claim 4.5: \u00af\u03c8\u03b3\u00b5\u03b3\u03bd\u03c8 transforms as a Lorentz tensor. More precisely, the symmetric part is a Lorentz scalar, proportional to \u03b7\u00b5\u03bd \u00af\u03c8\u03c8, while the antisymmetric part is a Lorentz tensor, proportional to \u00af\u03c8S\u00b5\u03bd\u03c8. Proof: As above. We are now armed with three bilinears of the Dirac \ufb01eld, \u00af\u03c8\u03c8, \u00af\u03c8\u03b3\u00b5\u03c8 and \u00af\u03c8\u03b3\u00b5\u03b3\u03bd\u03c8, each of which transforms covariantly under the Lorentz group. We can try to build a Lorentz invariant action from these. In fact, we need only the \ufb01rst two. We choose S = d4x \u00af\u03c8(x) (i\u03b3\u00b5\u2202\u00b5 \u2212 m) \u03c8(x) (4.50) This is the Dirac action. The factor of \u201ci\u201d is there to make the action real; upon complex conjugation, it cancels a minus sign that comes from integration by parts. (Said another way, it\u2019s there for the same reason that the Hermitian momentum operator \u2212i\u2207 in quantum mechanics has a factor i). As we will see in the next section, after quantization this theory describes particles and anti-particles of mass |m| and spin 1/2. Notice that the Lagrangian is \ufb01rst order, rather than the second order Lagrangians we were working with for scalar \ufb01elds. Also, the mass appears in the Lagrangian as m, which can be positive or negative. \u2013 89 \u2013 4.3 The Dirac Equation The equation of motion follows from the action (4.50) by varying with respect to \u03c8 and \u00af\u03c8 independently. Varying with respect to \u00af\u03c8, we have (i\u03b3\u00b5\u2202\u00b5 \u2212 m) \u03c8 = 0 (4.51) This is the Dirac equation. It\u2019s completely gorgeous. Varying with respect to \u03c8 gives the conjugate equation i\u2202\u00b5 \u00af\u03c8 \u03b3\u00b5 + m \u00af\u03c8 = 0 (4.52) The Dirac equation is \ufb01rst order in derivatives, yet miraculously Lorentz invariant. If we tried to write down a \ufb01rst order equation of motion for a scalar \ufb01eld, it would look like v\u00b5\u2202\u00b5\u03c6 =..., which necessarily includes a privileged vector in spacetime v\u00b5 and is not Lorentz invariant. However, for spinor \ufb01elds, the magic of the", " \u03b3\u00b5 matrices means that the Dirac Lagrangian is Lorentz invariant. The Dirac equation mixes up di\ufb00erent components of \u03c8 through the matrices \u03b3\u00b5. However, each individual component itself solves the Klein-Gordon equation. To see this, write (i\u03b3\u03bd\u2202\u03bd + m)(i\u03b3\u00b5\u2202\u00b5 \u2212 m) \u03c8 = \u2212 \u03b3\u00b5\u03b3\u03bd\u2202\u00b5\u2202\u03bd + m2 \u03c8 = 0 (4.53) But \u03b3\u00b5\u03b3\u03bd\u2202\u00b5\u2202\u03bd = 1 2{\u03b3\u00b5, \u03b3\u03bd}\u2202\u00b5\u2202\u03bd = \u2202\u00b5\u2202\u00b5, so we get \u2212(\u2202\u00b5\u2202\u00b5 + m2)\u03c8 = 0 (4.54) where this last equation has no \u03b3\u00b5 matrices, and so applies to each component \u03c8\u03b1, with \u03b1 = 1, 2, 3, 4. The Slash Let\u2019s introduce some useful notation. We will often come across 4-vectors contracted with \u03b3\u00b5 matrices. We write so the Dirac equation reads A\u00b5\u03b3\u00b5 \u2261 /A (i /\u2202 \u2212 m)\u03c8 = 0 \u2013 90 \u2013 (4.55) (4.56) 4.4 Chiral Spinors When we\u2019ve needed an explicit form of the \u03b3\u00b5 matrices, we\u2019ve used the chiral representation \u03b30 = 0 1 1 0, \u03b3i = 0 \u03c3i \u2212\u03c3i 0 (4.57) In this representation, the spinor rotation transformation S[\u039brot] and boost transformation S[\u039bboost] were computed in (4.26) and (4.31). Both are block diagonal, S[\u039brot] = e+i \u03d5\u00b7\u03c3/2 0 0 e+i\u03d5\u00b7\u03c3/2 and S[\u039bboost] = e+\u03c7\u00b7\u03c3/2 0 0 e\u2212\u03c7\u00b7\u03c3/2 (4.58) This means that the Dirac spinor representation of the Lorentz group is reducible. It decomposes into two irreducible representations, acting only on two-component spinors u\u00b1 which, in the chiral representation, are de\ufb01ned by \u03c8 = u+ u\u2212 (4.59) The two-component objects u\u00b1 are called Weyl spinors or chiral spinors. They transform in the same way under rotations, but oppositely under boosts, u\u00b1 \u2192 ei\u03d5\u00b7\u03c3/2 u\u00b1 u\u00b1 \u2192 e\u00b1\u03c7\u00b7\u03c3/2u\u00b1 (4.60) (4.61) In group theory language, u+ is in the ( 1 2, 0) representation of the Lorentz group, while u\u2212 is in the (0, 1 2, 0) \u2295 (0, 1 2) representation. (Strictly speaking, the spinor is a representation of the double cover of the Lorentz group SL(2, C)). 2) representation. The Dirac spinor \u03c8 lies in the ( 1 4.4.1 The Weyl Equation Let\u2019s see what becomes of the Dirac Lagrangian under the decomposition (4.59) into Weyl spinors. We have L = \u00af\u03c8(i /\u2202 \u2212 m)\u03c8 = iu\u2020 \u2212\u03c3\u00b5\u2202\u00b5u\u2212 + iu\u2020 +\u00af\u03c3\u00b5\u2202\u00b5u+ \u2212 m(u\u2020 +u\u2212 + u\u2020 \u2212u+) = 0 (4.62) \u2013 91 \u2013 where we have introduced some new notation for the Pauli matrices with a \u00b5 = 0, 1, 2, 3 index, \u03c3\u00b5 = (1, \u03c3i) and \u00af\u03c3\u00b5 = (1, \u2212\u03c3i) (4.63) From (4.62), we see that a massive fermion requires both u+ and u\u2212, since they couple through the mass term. However, a massless fermion can be described by u+ (or u\u2212) alone, with the equation of motion i\u00af\u03c3\u00b5\u2202\u00b5u+ = 0 i\u03c3\u00b5\u2202\u00b5u\u2212 = 0 or (4.64) These are the Weyl equations. Degrees of Freedom Let me comment here on the degrees of freedom in a spinor. The Dirac fermion has 4 complex components = 8 real components. How do we count degrees of freedom? In classical mechanics, the number of degrees of freedom of a system is equal to the dimension of the con\ufb01guration space or, equivalently, half the dimension of the phase space. In \ufffd", "\ufffdeld theory we have an in\ufb01nite number of degrees of freedom, but it makes sense to count the number of degrees of freedom per spatial point: this should at least be \ufb01nite. For example, in this sense a real scalar \ufb01eld \u03c6 has a single degree of freedom. At the quantum level, this translates to the fact that it gives rise to a single type of particle. A classical complex scalar \ufb01eld has two degrees of freedom, corresponding to the particle and the anti-particle in the quantum theory. But what about a Dirac spinor? One might think that there are 8 degrees of freedom. But this isn\u2019t right. Crucially, and in contrast to the scalar \ufb01eld, the equation of motion is \ufb01rst order rather than second order. In particular, for the Dirac Lagrangian, the momentum conjugate to the spinor \u03c8 is given by \u03c0\u03c8 = \u2202L/\u2202 \u02d9\u03c8 = i\u03c8\u2020 (4.65) It is not proportional to the time derivative of \u03c8. This means that the phase space for a spinor is therefore parameterized by \u03c8 and \u03c8\u2020, while for a scalar it is parameterized by \u03c6 and \u03c0 = \u02d9\u03c6. So the phase space of the Dirac spinor \u03c8 has 8 real dimensions and correspondingly the number of real degrees of freedom is 4. We will see in the next section that, in the quantum theory, this counting manifests itself as two degrees of freedom (spin up and down) for the particle, and a further two for the anti-particle. A similar counting for the Weyl fermion tells us that it has two degrees of freedom. \u2013 92 \u2013 4.4.2 \u03b35 The Lorentz group matrices S[\u039b] came out to be block diagonal in (4.58) because we chose the speci\ufb01c representation (4.57). In fact, this is why the representation (4.57) is called the chiral representation: it\u2019s because the decomposition of the Dirac spinor \u03c8 is simply given by (4.59). But what happens if we choose a di\ufb00erent representation \u03b3\u00b5 of the Cli\ufb00ord algebra, so that \u03b3\u00b5 \u2192 U \u03b3\u00b5U \u22121 and \u03c8 \u2192 U \u03c8? (4.66) Now S[\u039b] will not be block diagonal. Is there an invariant way to de\ufb01ne chiral spinors? We can do this by introducing the \u201c\ufb01fth\u201d gamma-matrix \u03b35 = \u2212i\u03b30\u03b31\u03b32\u03b33 You can check that this matrix satis\ufb01es {\u03b35, \u03b3\u00b5} = 0 and (\u03b35)2 = +1 (4.67) (4.68) The reason that this is called \u03b35 is because the set of matrices \u02dc\u03b3A = (\u03b3\u00b5, i\u03b35), with A = 0, 1, 2, 3, 4 satisfy the \ufb01ve-dimensional Cli\ufb00ord algebra {\u02dc\u03b3A, \u02dc\u03b3B} = 2\u03b7AB. (You might think that \u03b34 would be a better name! But \u03b35 is the one everyone chooses - it\u2019s a more sensible name in Euclidean space, where A = 1, 2, 3, 4, 5). You can also check that [S\u00b5\u03bd, \u03b35] = 0, which means that \u03b35 is a scalar under rotations and boosts. Since (\u03b35)2 = 1, this means we may form the Lorentz invariant projection operators P\u00b1 = 1 2 1 \u00b1 \u03b35 (4.69) such that P 2 representation (4.57), + = P+ and P 2 \u2212 = P\u2212 and P+P\u2212 = 0. One can check that for the chiral \u03b35 = 1 0 0 \u22121 (4.70) from which we see that the operators P\u00b1 project onto the Weyl spinors u\u00b1. However, for an arbitrary representation of the Cli\ufb00ord algebra, we may use \u03b35 to de\ufb01ne the chiral spinors, \u03c8\u00b1 = P\u00b1\u03c8 (4.71) which form the irreducible representations of the Lorentz group. \u03c8+ is often called a \u201cleft-handed\u201d spinor, while \u03c8\u2212 is \u201cright-handed\u201d. The name comes from the way the spin precesses as a massless fermion moves: we\u2019ll see this in Section", " 4.7.2. \u2013 93 \u2013 4.4.3 Parity The spinors \u03c8\u00b1 are related to each other by parity. Let\u2019s pause to de\ufb01ne this concept. The Lorentz group is de\ufb01ned by x\u00b5 \u2192 \u039b\u00b5 \u03bdx\u03bd such that \u039b\u00b5 \u03bd\u039b\u03c1 \u03c3\u03b7\u03bd\u03c3 = \u03b7\u00b5\u03c1 (4.72) So far we have only considered transformations \u039b which are continuously connected to the identity; these are the ones which have an in\ufb01nitesimal form. However there are also two discrete symmetries which are part of the Lorentz group. They are Time Reversal T : x0 \u2192 \u2212x0 ; xi \u2192 xi Parity P : x0 \u2192 x0 ; xi \u2192 \u2212xi (4.73) We won\u2019t discuss time reversal too much in this course. (It turns out to be represented by an anti-unitary transformation on states. See, for example the book by Peskin and Schroeder). But parity has an important role to play in the standard model and, in particular, the theory of the weak interaction. Under parity, the left and right-handed spinors are exchanged. This follows from the transformation of the spinors under the Lorentz group. In the chiral representation, we saw that the rotation (4.60) and boost (4.61) transformations for the Weyl spinors u\u00b1 are u\u00b1 rot\u2212\u2192 ei\u03d5\u00b7\u03c3/2 u\u00b1 and u\u00b1 boost\u2212\u2192 e\u00b1\u03c7\u00b7\u03c3/2u\u00b1 (4.74) Under parity, rotations don\u2019t change sign. But boosts do \ufb02ip sign. This con\ufb01rms that parity exchanges right-handed and left-handed spinors, P : u\u00b1 \u2192 u\u2213, or in the notation \u03c8\u00b1 = 1 2(1 \u00b1 \u03b35)\u03c8, we have P : \u03c8\u00b1(x, t) \u2192 \u03c8\u2213(\u2212x, t) (4.75) Using this knowledge of how chiral spinors transform, and the fact that P 2 = 1, we see that the action of parity on the Dirac spinor itself can be written as P : \u03c8(x, t) \u2192 \u03b30\u03c8(\u2212x, t) (4.76) Notice that if \u03c8(x, t) satis\ufb01es the Dirac equation, then the parity transformed spinor \u03b30\u03c8(\u2212x, t) also satis\ufb01es the Dirac equation, meaning (i\u03b30\u2202t + i\u03b3i\u2202i \u2212 m)\u03b30\u03c8(\u2212x, t) = \u03b30(i\u03b30\u2202t \u2212 i\u03b3i\u2202i \u2212 m)\u03c8(\u2212x, t) = 0 (4.77) where the extra minus sign from passing \u03b30 through \u03b3i is compensated by the derivative acting on \u2212x instead of +x. \u2013 94 \u2013 4.4.4 Chiral Interactions Let\u2019s now look at how our interaction terms change under parity. We can look at each of our spinor bilinears from which we built the action, P : \u00af\u03c8\u03c8(x, t) \u2192 \u00af\u03c8\u03c8(\u2212x, t) (4.78) which is the transformation of a scalar. For the vector \u00af\u03c8\u03b3\u00b5\u03c8, we can look at the temporal and spatial components separately, P : \u00af\u03c8\u03b30\u03c8(x, t) \u2192 \u00af\u03c8\u03b30\u03c8(\u2212x, t) P : \u00af\u03c8\u03b3i\u03c8(x, t) \u2192 \u00af\u03c8\u03b30\u03b3i\u03b30\u03c8(\u2212x, t) = \u2212 \u00af\u03c8\u03b3i\u03c8(\u2212x, t) (4.79) which tells us that \u00af\u03c8\u03b3\u00b5\u03c8 transforms as a vector, with the spatial part changing sign. You can also check that \u00af\u03c8S\u00b5\u03bd\u03c8 transforms as a suitable tensor. However, now we\u2019ve discovered the existence of \u03b35, we can form another Lorentz scalar and another Lorentz vector, \u00af\u03c8\u03b35\u03c8 and \u00af\u03c8\u03b35\u03b3\u00b5\u03c8 (4.80) How do these transform under parity? We can check: P : \u00af\u03c8\u03b35\u03c8(x, t) \u2192 \u00af\u03c8\u03b30\u03b35\u03b30 \u03c8(\u2212x, t) = \u2212 \u00af\u03c8\u03b35\u03c8(\u2212x, t) (4.81) P : \u00af\u03c8\u03b35\u03b3\u00b5\u03c8(x,", " t) \u2192 \u00af\u03c8\u03b30\u03b35\u03b3\u00b5\u03b30\u03c8(\u2212x, t) = \u2212 \u00af\u03c8\u03b35\u03b30\u03c8(\u2212x, t) + \u00af\u03c8\u03b35\u03b3i\u03c8(\u2212x, t) \u00b5 = 0 \u00b5 = i which means that \u00af\u03c8\u03b35\u03c8 transforms as a pseudoscalar, while \u00af\u03c8\u03b35\u03b3\u00b5\u03c8 transforms as an axial vector. To summarize, we have the following spinor bilinears, \u00af\u03c8\u03c8 : \u00af\u03c8\u03b3\u00b5\u03c8 : \u00af\u03c8S\u00b5\u03bd\u03c8 : \u00af\u03c8\u03b35\u03c8 : \u00af\u03c8\u03b35\u03b3\u00b5\u03c8 : scalar vector tensor pseudoscalar axial vector (4.82) The total number of bilinears is 1 + 4 + (4 \u00d7 3/2) + 4 + 1 = 16 which is all we could hope for from a 4-component object. \u2013 95 \u2013 We\u2019re now armed with new terms involving \u03b35 that we can start to add to our Lagrangian to construct new theories. Typically such terms will break parity invariance of the theory, although this is not always true. (For example, the term \u03c6 \u00af\u03c8\u03b35\u03c8 doesn\u2019t break parity if \u03c6 is itself a pseudoscalar). Nature makes use of these parity violating interactions by using \u03b35 in the weak force. A theory which treats \u03c8\u00b1 on an equal footing is called a vector-like theory. A theory in which \u03c8+ and \u03c8\u2212 appear di\ufb00erently is called a chiral theory. 4.5 Majorana Fermions Our spinor \u03c8\u03b1 is a complex object. It has to be because the representation S[\u039b] is typically also complex. This means that if we were to try to make \u03c8 real, for example by imposing \u03c8 = \u03c8, then it wouldn\u2019t stay that way once we make a Lorentz transformation. However, there is a way to impose a reality condition on the Dirac spinor \u03c8. To motivate this possibility, it\u2019s simplest to look at a novel basis for the Cli\ufb00ord algebra, known as the Majorana basis. \u03b30 = 0 \u03c32 \u03c32 0, \u03b31 = i\u03c33 0 0 i\u03c33, \u03b32 = 0 \u2212\u03c32 \u03c32 0, \u03b33 = \u2212i\u03c31 0 0 \u2212i\u03c31 These matrices satisfy the Cli\ufb00ord algebra. What is special about them is that they are all pure imaginary (\u03b3\u00b5) = \u2212\u03b3\u00b5. This means that the generators of the Lorentz group S\u00b5\u03bd = 1 4[\u03b3\u00b5, \u03b3\u03bd], and hence the matrices S[\u039b] are real. So with this basis of the Cli\ufb00ord algebra, we can work with a real spinor simply by imposing the condition, \u03c8 = \u03c8 (4.83) which is preserved under Lorentz transformation. Such spinors are called Majorana spinors. So what\u2019s the story if we use a general basis for the Cli\ufb00ord algebra? We\u2019ll ask only that the basis satis\ufb01es (\u03b30)\u2020 = \u03b30 and (\u03b3i)\u2020 = \u2212\u03b3i. We then de\ufb01ne the charge conjugate of a Dirac spinor \u03c8 as Here C is a 4 \u00d7 4 matrix satisfying \u03c8(c) = C\u03c8 C \u2020C = 1 and C \u2020\u03b3\u00b5C = \u2212(\u03b3\u00b5) (4.84) (4.85) Let\u2019s \ufb01rstly check that (4.84) is a good de\ufb01nition, meaning that \u03c8(c) transforms nicely under a Lorentz transformation. We have \u03c8(c) \u2192 CS[\u039b]\u03c8 = S[\u039b]C\u03c8 = S[\u039b]\u03c8(c) (4.86) \u2013 96 \u2013 where we\u2019ve made use of the properties (4.85) in taking the matrix C through S[\u039b]. In fact, not only does \u03c8(c) transform nicely under the Lorentz group, but if \u03c8 satis\ufb01es the Dirac equation, then \u03c8(c) does too. This follows from, (i /\u2202 \u2212 m)\u03c8 = 0 \u21d2 (\u2212i /\u2202 \u2212 m)\u03c8 = 0 \u21d2 C(\u2212i /\u2202 \u2212 m)\u03c8 = (+i /\u2202 \u2212 m)\u03c8(c) = 0 Finally, we can now", " impose the Lorentz invariant reality condition on the Dirac spinor, to yield a Majorana spinor, \u03c8(c) = \u03c8 (4.87) After quantization, the Majorana spinor gives rise to a fermion that is its own antiparticle. This is exactly the same as in the case of scalar \ufb01elds, where we\u2019ve seen that a real scalar \ufb01eld gives rise to a spin 0 boson that is its own anti-particle. (Be aware: In many texts an extra factor of \u03b30 is absorbed into the de\ufb01nition of C). So what is this matrix C? Well, for a given representation of the Cli\ufb00ord algebra, it is something that we can \ufb01nd fairly easily. In the Majorana basis, where the gamma matrices are pure imaginary, we have simply CMaj = 1 and the Majorana condition \u03c8 = \u03c8(c) becomes \u03c8 = \u03c8. In the chiral basis (4.16), only \u03b32 is imaginary, and we. (The matrix i\u03c32 that appears here is simply the may take Cchiral = i\u03b32 = anti-symmetric matrix \u03b1\u03b2). It is interesting to see how the Majorana condition (4.87) looks in terms of the decomposition into left and right handed Weyl spinors (4.59). Plugging in the various de\ufb01nitions, we \ufb01nd that u+ = i\u03c32u +. In other words, a Majorana spinor can be written in terms of Weyl spinors as \u2212 and u\u2212 = \u2212i\u03c32u 0 \u2212i\u03c32 i\u03c32 0 \u03c8 = u+ \u2212i\u03c32u + (4.88) Notice that it\u2019s not possible to impose the Majorana condition \u03c8 = \u03c8(c) at the same time as the Weyl condition (u\u2212 = 0 or u+ = 0). Instead the Majorana condition relates u\u2212 and u+. An Aside: Spinors in Di\ufb00erent Dimensions: The ability to impose Majorana or Weyl conditions on Dirac spinors depends on both the dimension and the signature of spacetime. One can always impose the Weyl condition on a spinor in even dimensional Minkowski space, basically because you can always build a suitable \u201c\u03b35\u201d projection matrix by multiplying together all the other \u03b3-matrices. The pattern for when the Majorana condition can be imposed is a little more sporadic. Interestingly, although the Majorana condition and Weyl condition cannot be imposed simultaneously in four dimensions, you can do this in Minowski spacetimes of dimension 2, 10, 18,.... \u2013 97 \u2013 4.6 Symmetries and Conserved Currents The Dirac Lagrangian enjoys a number of symmetries. Here we list them and compute the associated conserved currents. Spacetime Translations Under spacetime translations the spinor transforms as \u03b4\u03c8 = \u00b5\u2202\u00b5\u03c8 (4.89) The Lagrangian depends on \u2202\u00b5\u03c8, but not \u2202\u00b5 the energy-momentum tensor \u00af\u03c8, so the standard formula (1.41) gives us T \u00b5\u03bd = i \u00af\u03c8\u03b3\u00b5\u2202\u03bd\u03c8 \u2212 \u03b7\u00b5\u03bdL (4.90) Since a current is conserved only when the equations of motion are obeyed, we don\u2019t lose anything by imposing the equations of motion already on T \u00b5\u03bd. In the case of a scalar \ufb01eld this didn\u2019t really buy us anything because the equations of motion are second order in derivatives, while the energy-momentum is typically \ufb01rst order. However, for a spinor \ufb01eld the equations of motion are \ufb01rst order: (i /\u2202 \u2212 m)\u03c8 = 0. This means we can set L = 0 in T \u00b5\u03bd, leaving T \u00b5\u03bd = i \u00af\u03c8\u03b3\u00b5\u2202\u03bd\u03c8 (4.91) In particular, we have the total energy E = d3x T 00 = d3x i \u00af\u03c8\u03b30 \u02d9\u03c8 = d3x \u03c8\u2020\u03b30(\u2212i\u03b3i\u2202i + m)\u03c8 (4.92) where, in the last equality, we have again used the equations of motion. Lorentz Transformations Under an in\ufb01nitesimal Lorentz transformation, the Dirac spinor transforms as (4.22) which, in in\ufb01nitesimal form, reads \u03b4\u03c8\u03b1 = \u2212\u03c9\u00b5 \u03bdx\u03bd\u2202\u00b5\u03c8\u03b1 + 1 2\u2126", "\u03c1\u03c3(S\u03c1\u03c3)\u03b1 \u03b2\u03c8\u03b2 (4.93) where, following (4.10), we have \u03c9\u00b5 the Lorentz algebra given by (4.8) \u03bd = 1 2\u2126\u03c1\u03c3(M\u03c1\u03c3)\u00b5 \u03bd, and M\u03c1\u03c3 are the generators of (M\u03c1\u03c3)\u00b5 \u03bd = \u03b7\u03c1\u00b5 \u03b4\u03c3 \u03bd \u2212 \u03b7\u03c3\u00b5\u03b4\u03c1 \u03bd (4.94) \u2013 98 \u2013 which, after direct substitution, tells us that \u03c9\u00b5\u03bd = \u2126\u00b5\u03bd. So we get \u03b4\u03c8\u03b1 = \u2212\u03c9\u00b5\u03bd x\u03bd\u2202\u00b5\u03c8\u03b1 \u2212 1 2(S\u00b5\u03bd)\u03b1 \u03b2\u03c8\u03b2 (4.95) The conserved current arising from Lorentz transformations now follows from the same calculation we saw for the scalar \ufb01eld (1.54) with two di\ufb00erences: \ufb01rstly, as we saw above, the spinor equations of motion set L = 0; secondly, we pick up an extra piece in the current from the second term in (4.95). We have (J \u00b5)\u03c1\u03c3 = x\u03c1T \u00b5\u03c3 \u2212 x\u03c3T \u00b5\u03c1 \u2212 i \u00af\u03c8\u03b3\u00b5S\u03c1\u03c3\u03c8 (4.96) After quantization, when (J \u00b5)\u03c1\u03c3 is turned into an operator, this extra term will be responsible for providing the single particle states with internal angular momentum, telling us that the quantization of a Dirac spinor gives rise to a particle carrying spin 1/2. Internal Vector Symmetry The Dirac Lagrangian is invariant under rotating the phase of the spinor, \u03c8 \u2192 e\u2212i\u03b1\u03c8. This gives rise to the current V = \u00af\u03c8\u03b3\u00b5\u03c8 j\u00b5 (4.97) where \u201cV \u201d stands for vector, re\ufb02ecting the fact that the left and right-handed components \u03c8\u00b1 transform in the same way under this symmetry. We can easily check that j\u00b5 V is conserved under the equations of motion, \u2202\u00b5j\u00b5 V = (\u2202\u00b5 \u00af\u03c8)\u03b3\u00b5\u03c8 + \u00af\u03c8\u03b3\u00b5(\u2202\u00b5\u03c8) = im \u00af\u03c8\u03c8 \u2212 im \u00af\u03c8\u03c8 = 0 (4.98) where, in the last equality, we have used the equations of motion i /\u2202\u03c8 = m\u03c8 and i\u2202\u00b5 \u00af\u03c8\u03b3\u00b5 = \u2212m \u00af\u03c8. The conserved quantity arising from this symmetry is Q = d3x \u00af\u03c8\u03b30\u03c8 = d3x \u03c8\u2020\u03c8 (4.99) We will see shortly that this has the interpretation of electric charge, or particle number, for fermions. Axial Symmetry When m = 0, the Dirac Lagrangian admits an extra internal symmetry which rotates left and right-handed fermions in opposite directions, \u03c8 \u2192 ei\u03b1\u03b35\u03c8 and \u00af\u03c8 \u2192 \u00af\u03c8ei\u03b1\u03b35 (4.100) \u2013 99 \u2013 Here the second transformation follows from the \ufb01rst after noting that e\u2212i\u03b1\u03b35\u03b30 = \u03b30e+i\u03b1\u03b35. This gives the conserved current, A = \u00af\u03c8\u03b3\u00b5\u03b35\u03c8 j\u00b5 (4.101) where A is for \u201caxial\u201d since j\u00b5 Indeed, with the full Dirac Lagrangian we may compute A is an axial vector. This is conserved only when m = 0. \u2202\u00b5j\u00b5 A = (\u2202\u00b5 \u00af\u03c8)\u03b3\u00b5\u03b35\u03c8 + \u00af\u03c8\u03b3\u00b5\u03b35\u2202\u00b5\u03c8 = 2im \u00af\u03c8\u03b35\u03c8 (4.102) which vanishes only for m = 0. However, in the quantum theory things become more interesting for the axial current. When the theory is coupled to gauge \ufb01elds (in a manner we will discuss in Section 6), the axial transformation remains a symmetry of the classical Lagrangian. But it doesn\u2019t survive the quantization process. It is the archetypal example of an anomaly: a symmetry of the classical theory that is not preserved in the quantum theory. 4.7 Plane Wave Solutions Let\u2019s now study the solutions to the Dirac equation (i\u03b3\u00b5\u2202\u00b5 \u2212 m)\u03c8 = 0 We start by making a simple ansatz: \u03c8 = u(p) e\u2212ip\u00b7x (4.103) (4.104) where u(p)", " is a four-component spinor, independent of spacetime x which, as the notation suggests, can depend on the 3-momentum p. The Dirac equation then becomes (\u03b3\u00b5p\u00b5 \u2212 m)u(p) = \u2212m p\u00b5\u03c3\u00b5 p\u00b5\u00af\u03c3\u00b5 \u2212m u(p) = 0 (4.105) where we\u2019re again using the de\ufb01nition, \u03c3\u00b5 = (1, \u03c3i) and \u00af\u03c3\u00b5 = (1, \u2212\u03c3i) (4.106) Claim: The solution to (4.105) is u(p\u03c3 \u03be \u2013 100 \u2013 (4.107) for any 2-component spinor \u03be which we will normalize to \u03be\u2020\u03be = 1. Proof: Let\u2019s write u(p)T = (u1, u2). Then equation (4.105) reads (p \u00b7 \u03c3) u2 = mu1 and (p \u00b7 \u00af\u03c3)u1 = mu2 (4.108) Either one of these equations implies the other, a fact which follows from the identity 0 \u2212 pipj\u03b4ij = p\u00b5p\u00b5 = m2. To start with, let\u2019s try (p \u00b7 \u03c3)(p \u00b7 \u00af\u03c3) = p2 the ansatz u1 = (p \u00b7 \u03c3)\u03be for some spinor \u03be. Then the second equation in (4.108) immediately tells us that u2 = m\u03be. So we learn that any spinor of the form 0 \u2212 pipj\u03c3i\u03c3j = p2 u(p) = A (p \u00b7 \u03c3) \u03be m\u03be (4.109) with constant A is a solution to (4.105). To make this more symmetric, we choose A = 1/m and \u03be = p \u00b7 \u03c3 \u03be. So we get the promised result (4.107) p \u00b7 \u00af\u03c3 \u03be with constant \u03be. Then u1 = (p \u00b7 \u03c3) p \u00b7 \u00af\u03c3 \u03be = m \u221a \u221a \u221a Negative Frequency Solutions We get further solutions to the Dirac equation from the ansatz \u03c8 = v(p) e+ip\u00b7x (4.110) Solutions of the form (4.104), which oscillate in time as \u03c8 \u223c e\u2212iEt, are called positive frequency solutions. If we compute the energy of these solutions using (4.92), we \ufb01nd that it is positive. Those of the form (4.110), which oscillate as \u03c8 \u223c e+iEt, are negative frequency solutions. Now if we compute the energy using (4.92), it is negative. The Dirac equation requires that the 4-component spinor v(p) satis\ufb01es (\u03b3\u00b5p\u00b5 + m)v(p) = m p\u00b5\u03c3\u00b5 p\u00b5\u00af\u03c3\u00b5 m v(p) = 0 (4.111) which is solved by v(p\u03c3 \u03b7 (4.112) for some 2-component spinor \u03b7 which we take to be constant and normalized to \u03b7\u2020\u03b7 = 1. \u2013 101 \u2013 4.7.1 Some Examples Consider the positive frequency solution with mass m and 3-momentum p = 0, u(p) = \u221a m \u03be \u03be (4.113) where \u03be is any 2-component spinor. Spatial rotations of the \ufb01eld act on \u03be by (4.26), \u03be \u2192 e+i\u03d5\u00b7\u03c3/2\u03be (4.114) The 2-component spinor \u03be de\ufb01nes the spin of the \ufb01eld. This should be familiar from quantum mechanics. A \ufb01eld with spin up (down) along a given direction is described by the eigenvector of the corresponding Pauli matrix with eigenvalue +1 (-1 respectively). For example, \u03beT = (1, 0) describes a \ufb01eld with spin up along the z-axis. After quantization, this will become the spin of the associated particle. In the rest of this section, we\u2019ll indulge in an abuse of terminology and refer to the classical solutions to the Dirac equations as \u201cparticles\u201d, even though they have no such interpretation before quantization. Consider now boosting the particle with spin \u03beT = (1, 0) along the x3 direction, with p\u00b5 = (E, 0, 0, p). The solution to the Dirac equation becomes u(p \u2212 p3 E + p3 1 \uf8f6 0 1 \uf8f8 0 (4.115) In fact, this expression also makes sense for", " a massless \ufb01eld, for which E = p3. (We picked the normalization (4.107) for the solutions so that this would be the case). For a massless particle we have u(p) = \u221a 2E 0 0 1 0 (4.116) Similarly, for a boosted solution of the spin down \u03beT = (0, 1) \ufb01eld, we have u(p + p3 E \u2212 p3 0 \uf8f6 1 0 \uf8f8 1 m\u21920\u2212\u2192 \u221a 2E 0 1 0 0 (4.117) \u2013 102 \u2013 4.7.2 Helicity The helicity operator is the projection of the angular momentum along the direction of momentum, h = i 2ijk \u02c6pi Sjk = 1 2 \u02c6pi \u03c3i 0 0 \u03c3i (4.118) where Sij is the rotation generator given in (4.25). The massless \ufb01eld with spin \u03beT = (1, 0) in (4.116) has helicity h = 1/2: we say that it is right-handed. Meanwhile, the \ufb01eld (4.117) has helicity h = \u22121/2: it is left-handed. 4.7.3 Some Useful Formulae: Inner and Outer Products There are a number of identities that will be very useful in the following section, It\u2019s \ufb01rstly regarding the inner (and outer) products of the spinors u(p) and v(p). convenient to introduce a basis \u03bes and \u03b7s, s = 1, 2 for the two-component spinors such that for example, \u03ber \u2020\u03bes = \u03b4rs and \u03b7r \u2020\u03b7s = \u03b4rs \u03be1 = 1 0 and \u03be2 = 0 1 (4.119) (4.120) and similarly for \u03b7s. Let\u2019s deal \ufb01rst with the positive frequency plane waves. The two independent solutions are now written as us(p) = \u221a \u221a p \u00b7 \u03c3 \u03bes p \u00b7 \u00af\u03c3 \u03bes (4.121) We can take the inner product of four-component spinors in two di\ufb00erent ways: either as u\u2020 \u00b7 u, or as \u00afu \u00b7 u. Of course, only the latter will be Lorentz invariant, but it turns out that the former is needed when we come to quantize the theory. Here we state both: ur \u2020(p) \u00b7 us(p) = \u03ber \u2020 \u221a p \u00b7 \u03c3, \u03ber \u2020\u221a p \u00b7 \u00af\u03c3 \u221a \u221a p \u00b7 \u03c3 \u03bes p \u00b7 \u00af\u03c3 \u03bes = \u03ber \u2020p \u00b7 \u03c3\u03bes + \u03ber \u2020p \u00b7 \u00af\u03c3\u03bes = 2\u03ber \u2020p0\u03bes = 2p0\u03b4rs (4.122) while the Lorentz invariant inner product is \u00afur(p) \u00b7 us(p) = \u03ber \u2020 \u221a p \u00b7 \u03c3, \u03ber \u2020\u221a p \u00b7 \u00af \u03bes p \u00b7 \u00af\u03c3 \u03bes = 2m\u03b4rs (4.123) \u2013 103 \u2013 We have analogous results for the negative frequency solutions, which we may write as vs(p) = \u221a p \u00b7 \u03c3 \u03b7s \u221a p \u00b7 \u00af\u03c3 \u03b7s \u2212 with and vr \u2020(p) \u00b7 vs(p) = 2p0\u03b4rs \u00afvr(p) \u00b7 vs(p) = \u22122m\u03b4rs (4.124) We can also compute the inner product between u and v. We have \u00afur(p) \u00b7 vs(p) = \u03ber \u2020 \u221a p \u00b7 \u03c3, \u03ber \u2020\u221a p \u00b7 \u00af\u03c3 \u03b30 \u221a p \u00b7 \u03c3 \u03b7s \u221a p \u00b7 \u00af\u03c3 \u03b7s \u2212 = \u03ber\u2020(p \u00b7 \u00af\u03c3)(p \u00b7 \u03c3)\u03b7s \u2212 \u03ber \u2020(p \u00b7 \u00af\u03c3)(p \u00b7 \u03c3)\u03b7s = 0 (4.125) and similarly, \u00afvr(p) \u00b7 us(p) = 0. However, when we come to u\u2020 \u00b7 v, it is a slightly di\ufb00erent combination that has nice properties (and this same combination appears when we quantize the theory). We look at ur \u2020(p) \u00b7 vs(\u2212p), with the 3-momentum in the spinor v taking the opposite sign. De\ufb01ning the 4-momentum (p)\u00b5 = (p0, \u2212p),", " we have ur \u2020(p) \u00b7 vs(\u2212p) = \u03ber \u2020 \u221a p \u00b7 \u03c3, \u03ber \u2020\u221a p \u00b7 \u00af\u03c3 \u221a p \u00b7 \u03c3 \u03b7s \u221a p \u00b7 \u00af\u03c3 \u03b7s \u2212 = \u03ber\u2020(p \u00b7 \u03c3)(p \u00b7 \u03c3)\u03b7s \u2212 \u03ber \u2020(p \u00b7 \u00af\u03c3)(p \u00b7 \u00af\u03c3)\u03b7s (4.126) Now the terms under the square-root are given by (p\u00b7\u03c3)(p \u00b7\u03c3) = (p0 +pi\u03c3i)(p0 \u2212pi\u03c3i) = 0 \u2212 p 2 = m2. The same expression holds for (p \u00b7 \u00af\u03c3)(p \u00b7 \u00af\u03c3), and the two terms cancel. p2 We learn ur \u2020(p) \u00b7 vs(\u2212p) = vr \u2020(p) \u00b7 us(\u2212p) = 0 (4.127) Outer Products There\u2019s one last spinor identity that we need before we turn to the quantum theory. It is: Claim: 2 s=1 us(p) \u00afus(p) = /p + m (4.128) where the two spinors are not now contracted, but instead placed back to back to give a 4 \u00d7 4 matrix. Also, 2 s=1 vs(p) \u00afvs(p) = /p \u2212 m (4.129) \u2013 104 \u2013 Proof: 2 s=1 us(p) \u00afus(p) = \u221a \u221a p \u00b7 \u03c3 \u03bes p \u00b7 \u00af\u03c3 \u03bes 2 s=1 \u03bes \u2020 \u221a p \u00b7 \u00af\u03c3, \u03bes \u2020\u221a p \u00b7 \u03c3 (4.130) But s \u03bes \u03bes \u2020 = 1, the 2 \u00d7 2 unit matrix, which then gives us 2 s=1 us(p)\u00afus(p) = m p \u00b7 \u03c3 p \u00b7 \u00af\u03c3 m which is the desired result. A similar proof works for s vs(p)\u00afvs(p). (4.131) \u2013 105 \u2013 5. Quantizing the Dirac Field We would now like to quantize the Dirac Lagrangian, L = \u00af\u03c8(x) i /\u2202 \u2212 m \u03c8(x) (5.1) We will proceed naively and treat \u03c8 as we did the scalar \ufb01eld. But we\u2019ll see that things go wrong and we will have to reconsider how to quantize this theory. 5.1 A Glimpse at the Spin-Statistics Theorem We start in the usual way and de\ufb01ne the momentum, \u03c0 = \u2202L \u2202 \u02d9\u03c8 = i \u00af\u03c8\u03b30 = i\u03c8\u2020 (5.2) For the Dirac Lagrangian, the momentum conjugate to \u03c8 is i\u03c8\u2020. It does not involve the time derivative of \u03c8. This is as it should be for an equation of motion that is \ufb01rst order in time, rather than second order. This is because we need only specify \u03c8 and \u03c8\u2020 on an initial time slice to determine the full evolution. To quantize the theory, we promote the \ufb01eld \u03c8 and its momentum \u03c8\u2020 to operators, satisfying the canonical commutation relations, which read [\u03c8\u03b1(x), \u03c8\u03b2(y)] = [\u03c8\u2020 \u03b1(x), \u03c8\u2020 \u03b2(y)] = \u03b4\u03b1\u03b2 \u03b4(3)(x \u2212 y) \u03b2(y)] = 0 [\u03c8\u03b1(x), \u03c8\u2020 (5.3) It\u2019s this step that we\u2019ll soon have to reconsider. Since we\u2019re dealing with a free theory, where any classical solution is a sum of plane waves, we may write the quantum operators as 2 \u03c8(x) = \u03c8\u2020(x) = s=1 2 s=1 d3p (2\u03c0)3 1 2Ep p us(p)e+ip\u00b7x + cs \u2020 bs p vs(p)e\u2212ip\u00b7x d3p (2\u03c0)3 1 2Ep bs \u2020 p us(p)\u2020e\u2212ip\u00b7x + cs p vs(p)\u2020e+ip\u00b7x (5.4) where the operators bs \u2020 p create particles associated to vs(p). As with the scalars, the commutation relations of the \ufb01elds imply commutation relations for the annihilation and creation operators p create particles associated to the spinors us(p), while cs \u2020 \u2013 106 \u2013 Claim: The \ufb01eld commutation relations (5.3) are equivalent to p, bs \u2020 [br p, cs \u2020", " [cr q ] = (2\u03c0)3\u03b4rs \u03b4(3)(p \u2212 q) q ] = \u2212(2\u03c0)3\u03b4rs \u03b4(3)(p \u2212 q) (5.5) with all other commutators vanishing. Note the strange minus sign in the [c, c\u2020] term. This means that we can\u2019t de\ufb01ne the ground state |0 as something annihilated by cr p |0 = 0, because then the excited states cs \u2020 p |0 would have negative norm. To avoid this, we will have to \ufb02ip the interpretation of c and c\u2020, with the vacuum de\ufb01ned by cs \u2020 p |0. This, as we will see, will be our undoing. p |0 = 0 and the excited states by cr Proof: Let\u2019s show that the [b, b\u2020] and [c, c\u2020] commutators reproduce the \ufb01eld commutators (5.3), [\u03c8(x), \u03c8\u2020(y)] = d3p d3q (2\u03c0)6 r,s 1 4EpEq p, bs \u2020 [br q ]ur(p)us(q)\u2020ei(x\u00b7p\u2212y\u00b7q) + [cr \u2020 p, cs q]vr(p)vs(q)\u2020e\u2212i(x\u00b7p\u2212y\u00b7q) = s d3p (2\u03c0)3 1 2Ep us(p)\u00afus(p)\u03b30eip\u00b7(x\u2212y) + vs(p)\u00afvs(p)\u03b30e\u2212ip\u00b7(x\u2212y)(5.6) At this stage we use the outer product formulae (4.128) and (4.129) which tell us s vs(p)\u00afvs(p) = /p \u2212 m, so that s us(p)\u00afus(p) = /p + m and [\u03c8(x), \u03c8\u2020(y)] = d3p (2\u03c0)3 d3p (2\u03c0)3 1 2Ep 1 2Ep = ( /p + m)\u03b30eip\u00b7(x\u2212y) + ( /p \u2212 m)\u03b30e\u2212ip\u00b7(x\u2212y) (p0\u03b30 + pi\u03b3i + m)\u03b30 + (p0\u03b30 \u2212 pi\u03b3i \u2212 m)\u03b30 e+ip\u00b7(x\u2212y) where, in the second term, we\u2019ve changed p \u2192 \u2212p under the integration sign. Now, using p0 = Ep we have [\u03c8(x), \u03c8\u2020(y)] = d3p (2\u03c0)3 e+ip\u00b7(x\u2212y) = \u03b4(3)(x \u2212 y) (5.7) as promised. Notice that it\u2019s a little tricky in the middle there, making sure that the pi\u03b3i terms cancel. This was the reason we needed the minus sign in the [c, c\u2020] commutator terms in (5.5). \u2013 107 \u2013 5.1.1 The Hamiltonian To proceed, let\u2019s construct the Hamiltonian for the theory. Using the momentum \u03c0 = i\u03c8\u2020, we have H = \u03c0 \u02d9\u03c8 \u2212 L = \u00af\u03c8(\u2212i\u03b3i\u2202i + m)\u03c8 (5.8) which means that H = d3x H agrees with the conserved energy computed using Noether\u2019s theorem (4.92). We now wish to turn the Hamiltonian into an operator. Let\u2019s \ufb01rstly look at (\u2212i\u03b3i\u2202i + m)\u03c8 = d3p (2\u03c0)3 1 2Ep p (\u2212\u03b3ipi + m)us(p) e+ip\u00b7x + cs \u2020 bs p (\u03b3ipi + m)vs(p) e\u2212ip\u00b7x where, for once we\u2019ve left the sum over s = 1, 2 implicit. There\u2019s a small subtlety with the minus signs in deriving this equation that arises from the use of the Minkowski metric in contracting indices, so that p \u00b7 x \u2261 i xipi = \u2212xipi. Now we use the de\ufb01ning equations for the spinors us(p) and vs(p) given in (4.105) and (4.111), to replace (\u2212\u03b3ipi + m)us(p) = \u03b30p0us(p) and (\u03b3ipi + m)vs(p) = \u2212\u03b30p0vs(p) (5.9) so we can write (\u2212i\u03b3i\u2202i + m)\u03c8 = d3", "p (2\u03c0)3 Ep 2 \u03b30 p us(p) e+ip\u00b7x \u2212 cs \u2020 bs p vs(p) e\u2212ip\u00b7x (5.10) We now use this to write the operator Hamiltonian H = d3x \u03c8\u2020\u03b30(\u2212i\u03b3i\u2202i + m)\u03c8 = d3x d3p d3q (2\u03c0)6 Ep 4Eq br \u2020 q ur(q)\u2020e\u2212iq\u00b7x + cr q vr(q)\u2020 e+iq\u00b7x \u00b7 p us(p)e+ip\u00b7x \u2212 cs \u2020 bs p vs(p) e\u2212ip\u00b7x = d3p (2\u03c0)3 1 2 br \u2020 p bs p[ur(p)\u2020 \u00b7 us(p)] \u2212 cr pcs \u2020 p [vr(p)\u2020 \u00b7 vs(p)] \u2212br \u2020 p cs \u2020 \u2212p[ur(p)\u2020 \u00b7 vs(\u2212p)] + cr p bs \u2212p[vr(p)\u2020 \u00b7 us(\u2212p)] where, in the last two terms we have relabelled p \u2192 \u2212p. We now use our inner product formulae (4.122), (4.124) and (4.127) which read ur(p)\u2020 \u00b7 us(p) = vr(p)\u2020 \u00b7 vs(p) = 2p0\u03b4rs and ur(p)\u2020 \u00b7 vs(\u2212p) = vr(p)\u2020 \u00b7 us(\u2212p) = 0 \u2013 108 \u2013 giving us H = = d3p (2\u03c0)3 Ep d3p (2\u03c0)3 Ep bs \u2020 p bs p \u2212 cs pcs \u2020 p bs \u2020 p bs p \u2212 cs \u2020 p cs p + (2\u03c0)3\u03b4(3)(0) (5.11) (5.12) The \u03b4(3) term is familiar and easily dealt with by normal ordering. The b\u2020b term is familiar and we can check that b\u2020 create positive energy states as expected, [H, bs \u2020 p ] = Ep bs \u2020 p The minus sign in front of the c\u2020c term should make us nervous. If we think of c\u2020 as creation operators then there\u2019s no problem since, using the commutation relation (5.5), we still \ufb01nd that c\u2020 creates positive energy states, [H, cs \u2020 p ] = Ep cs \u2020 p However, as we noted after (5.5), these states have negative norm. To have a sensible Hilbert space, we need to interpret c as the creation operator. But then the Hamiltonian is not bounded below because [H, cs p] = \u2212Ep cs p This is a disaster. Taken seriously it would tell us that we could tumble to states of lower and lower energy by continually producing c particles. As the English would say, it\u2019s all gone a bit Pete Tong. (No relation). Since the above calculation was a little tricky, you might think that it\u2019s possible to rescue the theory to get the minus signs to work out right. You can play around with di\ufb00erent things, but you\u2019ll always \ufb01nd this minus sign cropping up somewhere. And, in fact, it\u2019s telling us something important that we missed. 5.2 Fermionic Quantization The key piece of physics that we missed is that spin 1/2 particles are fermions, meaning that they obey Fermi-Dirac statistics with the quantum state picking up a minus sign upon the interchange of any two particles. This fact is embedded into the structure of relativistic quantum \ufb01eld theory: the spin-statistics theorem says that integer spin \ufb01elds must be quantized as bosons, while half-integer spin \ufb01elds must be quantized as fermions. Any attempt to do otherwise will lead to an inconsistency, such as the unbounded Hamiltonian we saw in (5.12). \u2013 109 \u2013 So how do we go about quantizing a \ufb01eld as a fermion? Recall that when we quantized the scalar \ufb01eld, the resulting particles obeyed bosonic statistics because the creation and annihilation operators satis\ufb01ed the commutation relations, [a\u2020 p, a\u2020 q] = 0 \u21d2 a\u2020 pa\u2020 q |0 \u2261 |p, q = |q, p (5.13) To have states obeying fermionic statistics, we need anti-commutation relations, {A, B} \u2261 AB + BA. Rather than (5.3), we will ask that the spinor \ufb01elds satisfy {\u03c8\u03b1(x", "), \u03c8\u03b2(y)} = {\u03c8\u2020 \u03b1(x), \u03c8\u2020 \u03b2(y)} = \u03b4\u03b1\u03b2 \u03b4(3)(x \u2212 y) \u03b2(y)} = 0 {\u03c8\u03b1(x), \u03c8\u2020 (5.14) We still have the expansion (5.4) of \u03c8 and \u03c8\u2020 in terms of b, b\u2020, c and c\u2020. But now the same proof that led us to (5.5) tells us that {br {cr p, bs \u2020 p, cs \u2020 q } = (2\u03c0)3\u03b4rs \u03b4(3)(p \u2212 q) q } = (2\u03c0)3\u03b4rs \u03b4(3)(p \u2212 q) with all other anti-commutators vanishing, {br p, bs q} = {cr p, cs q} = {br p, cs \u2020 q } = {br p, cs q} =... = 0 (5.15) (5.16) The calculation of the Hamiltonian proceeds as before, all the way through to the penultimate line (5.11). At that stage, we get H = = d3p (2\u03c0)3 Ep d3p (2\u03c0)3 Ep bs \u2020 p bs p \u2212 cs pcs \u2020 p bs \u2020 p bs p + cs \u2020 p cs p \u2212 (2\u03c0)3\u03b4(3)(0) (5.17) The anti-commutators have saved us from the indignity of an unbounded Hamiltonian. Note that when normal ordering the Hamiltonian we now throw away a negative contribution \u2212(2\u03c0)3\u03b4(3)(0). In principle, this could partially cancel the positive contribution from bosonic \ufb01elds. Cosmological constant problem anyone?! 5.2.1 Fermi-Dirac Statistics Just as in the bosonic case, we de\ufb01ne the vacuum |0 to satisfy, p |0 = cs bs p |0 = 0 (5.18) \u2013 110 \u2013 Although b and c obey anti-commutation relations, the Hamiltonian (5.17) has nice commutation relations with them. You can check that [H, br [H, cr p] = \u2212Ep br p p] = \u2212Ep cr p and and [H, br \u2020 [H, cr \u2020 p ] = Ep br \u2020 p ] = Ep cr \u2020 p p (5.19) This means that we can again construct a tower of energy eigenstates by acting on the vacuum by br \u2020 to create particles and antiparticles, just as in the bosonic case. For example, we have the one-particle states p and cr,\u2020 p The two particle states now satisfy |p, r = br \u2020 p |0 |p1, r1; p2, r2 \u2261 br1 \u2020 p1 br2 \u2020 p2 |0 = \u2212 |p2, r2; p1, r1 (5.20) (5.21) con\ufb01rming that the particles do indeed obey Fermi-Dirac statistics. In particular, we have the Pauli-Exclusion principle |p, r; p, r = 0. Finally, if we wanted to be sure about the spin of the particle, we could act with the angular momentum operator (4.96) to con\ufb01rm that a stationary particle |p = 0, r does indeed carry intrinsic angular momentum 1/2 as expected. 5.3 Dirac\u2019s Hole Interpretation \u201cIn this attempt, the success seems to have been on the side of Dirac rather than logic\u201d Pauli on Dirac Let\u2019s pause our discussion to make a small historical detour. Dirac originally viewed his equation as a relativistic version of the Schr\u00a8odinger equation, with \u03c8 interpreted as the wavefunction for a single particle with spin. To reinforce this interpretation, he wrote (i /\u2202 \u2212 m)\u03c8 = 0 as i \u2202\u03c8 \u2202t = \u2212i\u03b1 \u00b7 \u2207\u03c8 + m\u03b2\u03c8 \u2261 \u02c6H\u03c8 (5.22) where \u03b1 = \u2212\u03b30\u03b3 and \u03b2 = \u03b30. Here the operator \u02c6H is interpreted as the one-particle Hamiltonian. This is a very di\ufb00erent viewpoint from the one we now have, where \u03c8 is a classical \ufb01eld that should be quantized. In Dirac\u2019s view, the Hamiltonian of the system is \u02c6H de\ufb01ned above, while for us the Hamiltonian is the \ufb01eld operator (5.17). Let\u2019s see where Dirac\u2019s viewpoint", " leads. \u2013 111 \u2013 With the interpretation of \u03c8 as a single-particle wavefunction, the plane-wave solutions (4.104) and (4.110) to the Dirac equation are thought of as energy eigenstates, with \u03c8 = u(p) e\u2212ip\u00b7x \u21d2 i \u03c8 = v(p) e+ip\u00b7x \u21d2 i \u2202\u03c8 \u2202t \u2202\u03c8 \u2202t = Ep \u03c8 = \u2212Ep \u03c8 (5.23) which look like positive and negative energy solutions. The spectrum is once again unbounded below; there are states v(p) with arbitrarily low energy \u2212Ep. At \ufb01rst glance this is disastrous, just like the unbounded \ufb01eld theory Hamiltonian (5.12). Dirac postulated an ingenious solution to this problem: since the electrons are fermions (a fact which is put in by hand to Dirac\u2019s theory) they obey the Pauli-exclusion principle. So we could simply stipulate that in the true vacuum of the universe, all the negative energy states are \ufb01lled. Only the positive energy states are accessible. These \ufb01lled negative energy states are referred to as the Dirac sea. Although you might worry about the in\ufb01nite negative charge of the vacuum, Dirac argued that only charge di\ufb00erences would be observable (a trick reminiscent of the normal ordering prescription we used for \ufb01eld operators). Having avoided disaster by \ufb02oating on an in\ufb01nite sea comprised of occupied negative energy states, Dirac realized that his theory made a shocking prediction. Suppose that a negative energy state is excited to a positive energy state, leaving behind a hole. The hole would have all the properties of the electron, except it would carry positive charge. After \ufb02irting with the idea that it may be the proton, Dirac \ufb01nally concluded that the hole is a new particle: the positron. Moreover, when a positron comes across an electron, the two can annihilate. Dirac had predicted anti-matter, one of the greatest achievements of theoretical physics. It took only a couple of years before the positron was discovered experimentally in 1932. Although Dirac\u2019s physical insight led him to the right answer, we now understand that the interpretation of the Dirac spinor as a single-particle wavefunction is not really correct. For example, Dirac\u2019s argument for anti-matter relies crucially on the particles being fermions while, as we have seen already in this course, anti-particles exist for both fermions and bosons. What we really learn from Dirac\u2019s analysis is that there is no consistent way to interpret the Dirac equation as describing a single particle. It is instead to be thought of as a classical \ufb01eld which has only positive energy solutions because the Hamiltonian (4.92) is positive de\ufb01nite. Quantization of this \ufb01eld then gives rise to both particle and anti-particle excitations. \u2013 112 \u2013 This from Julian Schwinger: \u201cUntil now, everyone thought that the Dirac equation referred directly to physical particles. Now, in \ufb01eld theory, we recognize that the equations refer to a sublevel. Experimentally we are concerned with particles, yet the old equations describe \ufb01elds.... When you begin with \ufb01eld equations, you It is operate on a level where the particles are not there from the start. when you solve the \ufb01eld equations that you see the emergence of particles.\u201d 5.4 Propagators Let\u2019s now move to the Heisenberg picture. We de\ufb01ne the spinors \u03c8(x, t) at every point in spacetime such that they satisfy the operator equation \u2202\u03c8 \u2202t = i[H, \u03c8] (5.24) We solve this by the expansion \u03c8(x) = \u03c8\u2020(x) = 2 s=1 2 s=1 d3p (2\u03c0)3 1 2Ep d3p (2\u03c0)3 1 2Ep pus(p)e\u2212ip\u00b7x + cs \u2020 bs p vs(p)e+ip\u00b7x bs \u2020 p us(p)\u2020e+ip\u00b7x + cs pvs(p)\u2020e\u2212ip\u00b7x (5.25) Let\u2019s now look at the anti-commutators of these \ufb01elds. We de\ufb01ne the fermionic propagator to be iS\u03b1\u03b2 = {\u03c8\u03b1(x), \u00af\u03c8\u03b2(y)} (5.26) In what follows we will often drop the indices and simply write iS", "(x\u2212y) = {\u03c8(x), \u00af\u03c8(y)}, but you should remember that S(x\u2212y) is a 4\u00d74 matrix. Inserting the expansion (5.25), we have iS(x \u2212 y) = d3p d3q (2\u03c0)6 1 4EpEq {bs p, br \u2020 q }us(p)\u00afur(q)e\u2212i(p\u00b7x\u2212q\u00b7y) = = d3p (2\u03c0)3 d3p (2\u03c0)3 1 2Ep 1 2Ep +{cs \u2020 q}vs(p)\u00afvr(q)e+i(p\u00b7x\u2212q\u00b7y) us(p)\u00afus(p)e\u2212ip\u00b7(x\u2212y) + vs(p)\u00afvs(p)e+ip\u00b7(x\u2212y) p, cr ( /p + m)e\u2212ip\u00b7(x\u2212y) + ( /p \u2212 m)e+ip\u00b7(x\u2212y) (5.27) \u2013 113 \u2013 where to reach the \ufb01nal line we have used the outer product formulae (4.128) and (4.129). We can then write iS(x \u2212 y) = (i /\u2202x + m)(D(x \u2212 y) \u2212 D(y \u2212 x)) (5.28) in terms of the propagator for a real scalar \ufb01eld D(x \u2212 y) which, recall, can be written as (2.90) D(x \u2212 y) = d3p (2\u03c0)3 1 2Ep e\u2212ip\u00b7(x\u2212y) (5.29) Some comments: \u2022 For spacelike separated points (x \u2212 y)2 < 0, we have already seen that D(x \u2212 y) \u2212 D(y \u2212 x) = 0. In the bosonic theory, we made a big deal of this since it ensured that [\u03c6(x), \u03c6(y)] = 0 (x \u2212 y)2 < 0 (5.30) outside the lightcone, which we trumpeted as proof that our theory was causal. However, for fermions we now have {\u03c8\u03b1(x), \u03c8\u03b2(y)} = 0 (x \u2212 y)2 < 0 (5.31) outside the lightcone. What happened to our precious causality? The best that we can say is that all our observables are bilinear in fermions, for example the Hamiltonian (5.17). These still commute outside the lightcone. The theory remains causal as long as fermionic operators are not observable. If you think this is a little weak, remember that no one has ever seen a physical measuring apparatus come back to minus itself when you rotate by 360 degrees! \u2022 At least away from singularities, the propagator satis\ufb01es (i /\u2202x \u2212 m)S(x \u2212 y) = 0 (5.32) which follows from the fact that ( /\u22022 condition p2 = m2. x + m2)D(x \u2212 y) = 0 using the mass shell 5.5 The Feynman Propagator By a similar calculation to that above, we can determine the vacuum expectation value, 0| \u03c8\u03b1(x) \u00af\u03c8\u03b2(y) |0 = 0| \u00af\u03c8\u03b2(y)\u03c8\u03b1(x) |0 = d3p (2\u03c0)3 d3p (2\u03c0)3 1 2Ep 1 2Ep ( /p + m)\u03b1\u03b2 e\u2212ip\u00b7(x\u2212y) ( /p \u2212 m)\u03b1\u03b2 e+ip\u00b7(x\u2212y) (5.33) \u2013 114 \u2013 We now de\ufb01ne the Feynman propagator SF (x \u2212 y), which is again a 4 \u00d7 4 matrix, as the time ordered product, SF (x \u2212 y) = 0| T \u03c8(x) \u00af\u03c8(y) |0 \u2261 0| \u03c8(x) \u00af\u03c8(y) |0 0| \u2212 \u00af\u03c8(y)\u03c8(x) |0 x0 > y0 y0 > x0 (5.34) Notice the minus sign! It is necessary for Lorentz invariance. When (x\u2212y)2 < 0, there is no invariant way to determine whether x0 > y0 or y0 > x0. In this case the minus sign is necessary to make the two de\ufb01nitions agree since {\u03c8(x), \u00af\u03c8(y)} = 0 outside the lightcone. We have the 4-momentum integral representation for the Feynman propagator, d4p (2\u03c0)4 e\u2212ip", "\u00b7(x\u2212y) which satis\ufb01es (i /\u2202x \u2212 m)SF (x \u2212 y) = i\u03b4(4)(x \u2212 y), so that SF is a Green\u2019s function for the Dirac operator. \u03b3 \u00b7 p + m p2 \u2212 m2 + i SF (x \u2212 y) = i (5.35) The minus sign that we see in (5.34) also occurs for any string of operators inside a time ordered product T (...). While bosonic operators commute inside T, fermionic operators anti-commute. We have this same behaviour for normal ordered products as well, with fermionic operators obeying : \u03c81\u03c82 := \u2212 : \u03c82\u03c81 :. With the understanding that all fermionic operators anti-commute inside T and ::, Wick\u2019s theorem proceeds just as in the bosonic case. We de\ufb01ne the contraction \u03c8(x) \u00af\u03c8(y) = T (\u03c8(x) \u00af\u03c8(y)) \u2212 : \u03c8(x) \u00af\u03c8(y) : = SF (x \u2212 y) (5.36) 5.6 Yukawa Theory The interaction between a Dirac fermion of mass m and a real scalar \ufb01eld of mass \u00b5 is governed by the Yukawa theory, L = 1 2\u2202\u00b5\u03c6\u2202\u00b5\u03c6 \u2212 1 2\u00b52\u03c62 + \u00af\u03c8(i\u03b3\u00b5\u2202\u00b5 \u2212 m)\u03c8 \u2212 \u03bb\u03c6 \u00af\u03c8\u03c8 (5.37) which is the proper version of the baby scalar Yukawa theory we looked at in Section 3. Couplings of this type appear in the standard model, between fermions and the Higgs boson. In that context, the fermions can be leptons (such as the electron) or quarks. Yukawa originally proposed an interaction of this type as an e\ufb00ective theory of nuclear forces. With an eye to this, we will again refer to the \u03c6 particles as mesons, and the \u03c8 particles as nucleons. Except, this time, the nucleons have spin. (This is still not a particularly realistic theory of nucleon interactions, not least because we\u2019re omitting isospin. Moreover, in Nature the relevant mesons are pions which are pseudoscalars, so a coupling of the form \u03c6 \u00af\u03c8\u03b35\u03c8 would be more appropriate. We\u2019ll turn to this brie\ufb02y in Section 5.7.3). \u2013 115 \u2013 Note the dimensions of the various \ufb01elds. We still have [\u03c6] = 1, but the kinetic terms require that [\u03c8] = 3/2. Thus, unlike in the case with only scalars, the coupling is dimensionless: [\u03bb] = 0. We\u2019ll proceed as we did in Section 3, \ufb01rstly computing the amplitude of a particular scattering process then, with that calculation as a guide, writing down the Feynman rules for the theory. We start with: 5.6.1 An Example: Putting Spin on Nucleon Scattering Let\u2019s study \u03c8\u03c8 \u2192 \u03c8\u03c8 scattering. This is the same calculation we performed in Section (3.3.3) except now the fermions have spin. Our initial and \ufb01nal states are |i = 4EpEq bs \u2020 p br \u2020 |f = 4EpEq bs \u2020 p br \u2020 q q |0 \u2261 |p, s; q, r |0 \u2261 |p, s; q, r (5.38) We need to be a little cautious about minus signs, because the b\u2020\u2019s now anti-commute. In particular, we should be careful when we take the adjoint. We have f | = 4EpEq 0| br q bs p (5.39) We want to calculate the order \u03bb2 terms from the S-matrix element f | S \u2212 1 |i. (\u2212i\u03bb)2 2 d4x1d4x2 T \u00af\u03c8(x1)\u03c8(x1)\u03c6(x1) \u00af\u03c8(x2)\u03c8(x2)\u03c6(x2) (5.40) where, as usual, all \ufb01elds are in the interaction picture. Just as in the bosonic calculation, the contribution to nucleon scattering comes from the contraction : \u00af\u03c8(x1)\u03c8(x1) \u00af\u03c8(x2)\u03c8(x2) : \u03c6(x1)\u03c6(x2)", " (5.41) We just have to be careful about how the spinor indices are contracted. Let\u2019s start by looking at how the fermionic operators act on |i. We expand out the \u03c8 \ufb01elds, leaving the \u00af\u03c8 \ufb01elds alone for now. We may ignore the c\u2020 pieces in \u03c8 since they give no contribution at order \u03bb2. We have : \u00af\u03c8(x1)\u03c8(x1) \u00af\u03c8(x2)\u03c8(x2) : bs \u2020 p br \u2020 q |0 = \u2212 d3k1 d3k2 (2\u03c0)6 [ \u00af\u03c8(x1) \u00b7 um(k1)] [ \u00af\u03c8(x2) \u00b7 un(k2)] e\u2212ik1\u00b7x1\u2212ik2\u00b7x2 4Ek1 Ek2 bm k1 bn k2 p br \u2020 bs \u2020 q |0 (5.42) \u2013 116 \u2013 where we\u2019ve used square brackets [\u00b7] to show how the spinor indices are contracted. The minus sign that sits out front came from moving \u03c8(x1) past \u00af\u03c8(x2). Now anti-commuting the b\u2019s past the b\u2020\u2019s, we get = \u22121 2EpEq [ \u00af\u03c8(x1) \u00b7 ur(q)] [ \u00af\u03c8(x2) \u00b7 us(p)]e\u2212ip\u00b7x2\u2212iq\u00b7x1 \u2212 [ \u00af\u03c8(x1) \u00b7 us(p)] [ \u00af\u03c8(x2) \u00b7 ur(q)]e\u2212ip\u00b7x1\u2212iq\u00b7x2 |0 (5.43) Note, in particular, the relative minus sign that appears between these two terms. Now let\u2019s see what happens when we hit this with f |. We look at 0| br q bs p [ \u00af\u03c8(x1) \u00b7 ur(q)] [ \u00af\u03c8(x2) \u00b7 us(p)] |0 = [\u00afus(p ) \u00b7 ur(q)] [\u00afur(q ) \u00b7 us(p)] e+ip\u00b7x1+iq\u00b7x2 2EpEq e+ip\u00b7x2+iq\u00b7x1 2EpEq \u2212 [\u00afur(q ) \u00b7 ur(q)] [\u00afus(p ) \u00b7 us(p)] The [ \u00af\u03c8(x1) \u00b7 us(p)] [ \u00af\u03c8(x2) \u00b7 ur(q)] term in (5.43) doubles up with this, cancelling the factor of 1/2 in front of (5.40). Meanwhile, the 1/ E terms cancel the relativistic state normalization. Putting everything together, we have the following expression for f | S \u2212 1 |i \u221a (\u2212i\u03bb)2 d4x1d4x2d4k (2\u03c0)4 ieik\u00b7(x1\u2212x2) k2 \u2212 \u00b52 + i [\u00afus(p ) \u00b7 us(p)] [\u00afur(q ) \u00b7 ur(q)]e+ix1\u00b7(q\u2212q)+ix2\u00b7(p\u2212p) \u2212 [\u00afus(p ) \u00b7 ur(q)] [\u00afur(q ) \u00b7 us(p)]eix1\u00b7(p\u2212q)+ix2\u00b7(q\u2212p) where we\u2019ve put the \u03c6 propagator back in. Performing the integrals over x1 and x2, this becomes, d4k (2\u03c0)4i(\u2212i\u03bb)2 k2 \u2212 \u00b52 + i [\u00afus(p ) \u00b7 us(p)] [\u00afur(q ) \u00b7 ur(q)]\u03b4(4)(q \u2212 q + k)\u03b4(4)(p \u2212 p \u2212 k) \u2212 [\u00afus(p ) \u00b7 ur(q)] [\u00afur(q ) \u00b7 us(p)]\u03b4(4)(p \u2212 q + k)\u03b4(4)(q \u2212 p \u2212 k) And we\u2019re almost there! Finally, writing the S-matrix element in terms of the amplitude in the usual way, f | S \u2212 1 |i = iA(2\u03c0)4\u03b4(4)(p + q \u2212 p \u2212 q), we have A = (\u2212i\u03bb)2 [\u00afus(p ) \u00b7 us(p)] [\u00afur(q ) \u00b7 ur(q)] (p \u2212 p)2 \u2212 \u00b52 + i \u2212 [\u00afus(p ) \u00b7 ur(q)] [\u00afur(q ) \u00b7 us(p)] (q \u2212 p)2 \u2212 \u00b52 + i which is our \ufb01nal answer for the amplitude. \u2013 117 \u2013 5.7 Feynman Rules for Ferm", "ions It\u2019s important to bear in mind that the calculation we just did kind of blows. Thankfully the Feynman rules will once again encapsulate the combinatoric complexities and make life easier for us. The rules to compute amplitudes are the following \u2022 To each incoming fermion with momentum p and spin r, we associate a spinor ur(p). For outgoing fermions we associate \u00afur(p). Figure 21: An incoming fermion Figure 22: An outgoing fermion \u2022 To each incoming anti-fermion with momentum p and spin r, we associate a spinor \u00afvr(p). For outgoing anti-fermions we associate vr(p). Figure 23: An incoming anti-fermion Figure 24: An outgoing anti-fermion \u2022 Each vertex gets a factor of \u2212i\u03bb. \u2022 Each internal line gets a factor of the relevant propagator. i p2 \u2212 \u00b52 + i i( /p + m) p2 \u2212 m2 + i for scalars for fermions (5.44) The arrows on the fermion lines must \ufb02ow consistently through the diagram (this ensures fermion number conservation). Note that the fermionic propagator is a 4\u00d74 matrix. The matrix indices are contracted at each vertex, either with further propagators, or with external spinors u, \u00afu, v or \u00afv. \u2022 Impose momentum conservation at each vertex, and integrate over undetermined loop momenta. \u2022 Add extra minus signs for statistics. Some examples will be given below. \u2013 118 \u2013 u (p)rpu (p)rpv (p)rprv (p)ppp 5.7.1 Examples Let\u2019s run through the same examples we did for the scalar Yukawa theory. Firstly, we have Nucleon Scattering For the example we worked out previously, the two lowest order Feynman diagrams are shown in Figure 25. We\u2019ve drawn the second Feynman diagram with the legs crossed Figure 25: The two Feynman diagrams for nucleon scattering to emphasize the fact that it picks up a minus sign due to statistics. (Note that the way the legs point in the Feynman diagram doesn\u2019t tell us the direction in which the particles leave the scattering event: the momentum label does that. The two diagrams above are di\ufb00erent because the incoming legs are attached to di\ufb00erent outgoing legs). Using the Feynman rules we can read o\ufb00 the amplitude. [\u00afus(p ) \u00b7 us(p)] [\u00afur(q ) \u00b7 ur(q)] (p \u2212 p)2 \u2212 \u00b52 [\u00afus(p ) \u00b7 ur(q)] [\u00afur(q ) \u00b7 us(p)] (p \u2212 q)2 \u2212 \u00b52 A = (\u2212i\u03bb)2 (5.45) \u2212 The denominators in each term are due to the meson propagator, with the momentum determined by conservation at each vertex. This agrees with the amplitude we computed earlier using Wick\u2019s theorem. Nucleon to Meson Scattering Let\u2019s now look at \u03c8 \u00af\u03c8 \u2192 \u03c6\u03c6. The two lowest order Feynman diagrams are shown in Figure 26. Applying the Feynman rules, we have A = (\u2212i\u03bb)2 \u00afvr(q)[\u03b3\u00b5(p\u00b5 \u2212 p \u00b5) + m]us(p) (p \u2212 p)2 \u2212 m2 + \u00afvr(q)[\u03b3\u00b5(p\u00b5 \u2212 q \u00b5) + m]us(p) (p \u2212 q)2 \u2212 m2 Since the internal line is now a fermion, the propagator contains \u03b3\u00b5(p\u00b5 \u2212p \u00b5)+m factors. This is a 4 \u00d7 4 matrix which sits on the top, sandwiched between the two external spinors. Now the exchange statistics applies to the \ufb01nal meson states. These are bosons and, correspondingly, there is no relative minus sign between the two diagrams. \u2013 119 \u2013 /p,s/q,r///p,s/q,r//+q,rp,sq,rp,s Figure 26: The two Feynman diagrams for nucleon to meson scattering Nucleon-Anti-Nucleon Scattering For \u03c8 \u00af\u03c8 \u2192 \u03c8 \u00af\u03c8, the two lowest order Feynman diagrams are of two distinct types, just like in the bosonic case. They are shown in Figure 27. The corresponding amplitude is given by, A = (\u2212i\u03bb)2 \u2212 [\u00afus(p ) \u00b7 us(p)] [\u00afvr(q) \u00b7 vr(q )] (p \u2212 p)2 \u2212 \u00b52 + [\u00afvr(q) \u00b7 us(p)] [\u00afus(p ) \u00b7 vr(", "q )] (p + q)2 \u2212 \u00b52 + i (5.46) As in the bosonic diagrams, there is again the di\ufb00erence in the momentum dependence in the denominator. But now the di\ufb00erence in the diagrams is also re\ufb02ected in the spinor contractions in the numerator. More subtle are the minus signs. The fermionic statistics mean that the \ufb01rst diagram has an extra minus sign relative to the \u03c8\u03c8 scattering of Figure 25. Since this minus sign will be important when we come to \ufb01gure out whether the Yukawa force is attractive or repulsive, let\u2019s go back to basics and see where it comes from. The initial and \ufb01nal states for this scattering process are |i = 4EpEq bs \u2020 p cr \u2020 |f = 4EpEq bs \u2020 p cr \u2020 q The ordering of b\u2020 and c\u2020 in these states is crucial and re\ufb02ects the scattering \u03c8 \u00af\u03c8 \u2192 \u03c8 \u00af\u03c8, as opposed to \u03c8 \u00af\u03c8 \u2192 \u00af\u03c8\u03c8 which would di\ufb00er by a minus sign. The \ufb01rst diagram in Figure 27 comes from the term in the perturbative expansion, |0 \u2261 |p, s; q, r |0 \u2261 |p, s; q, r (5.47) q f | : \u00af\u03c8(x1)\u03c8(x1) \u00af\u03c8(x2)\u03c8(x2) : bs \u2020 p cr \u2020 bs \u2020 where we\u2019ve neglected a bunch of objects in this equation like d4ki and exponential factors because we only want to keep track of the minus signs. Moving the annihilation operators past the creation operators, we have |0 \u223c f | [\u00afvm(k1) \u00b7 \u03c8(x1)] [ \u00af\u03c8(x2) \u00b7 un(k2)]cm k1 p cr \u2020 bn k2 q q |0 + f | [\u00afvr(q) \u00b7 \u03c8(x1)] [ \u00af\u03c8(x2) \u00b7 us(p)] |0 (5.48) \u2013 120 \u2013 /p,s/q,r///p,s/q,r//+q,rp,sq,rp,s Figure 27: The two Feynman diagrams for nucleon-anti-nucleon scattering Repeating the process by expanding out the \u03c8(x1) and \u00af\u03c8(x2) \ufb01elds and moving them to the left to annihilate f |, we have 0| cr q bs p cm \u2020 l1 bn \u2020 l2 [\u00afvr(q) \u00b7 vm(l1)] [\u00afun(l2) \u00b7 us(p)] |0 \u223c \u2212[\u00afvr(q) \u00b7 vr(q )] [\u00afus(p ) \u00b7 us(p)] where the minus sign has appeared from anti-commuting cm \u2020 p. This is the l1 overall minus sign found in (5.46). One can also follow similar contractions to compute the second diagram in Figure 27. past bs Meson Scattering Finally, we can also compute the scattering of \u03c6\u03c6 \u2192 \u03c6\u03c6 which, as in the bosonic case, picks up its leading contribution at one-loop. The amplitude for the diagram shown in the \ufb01gure is iA = \u2212(\u2212i\u03bb)4 d4k (2\u03c0)4 Tr /k + m (k2 \u2212 m2 + i) /k + /p 1 + m 1)2 \u2212 m2 + i) /k + /p ((k + p 1 \u2212 /p1 + m 1 \u2212 p1)2 \u2212 m2 + i) \u00d7 ((k + p Figure 28: /k \u2212 /p 2 + m 2)2 \u2212 m2 + i) ((k \u2212 p Notice that the high momentum limit of the integral is d4k/k4, which is no longer \ufb01nite, but diverges logarithmically. You will have to wait until next term to make sense of this integral. There\u2019s an overall minus sign sitting in front of this amplitude. This is a generic feature of diagrams with fermions running in loops: each fermionic loop in a diagram gives rise to an extra minus sign. We can see this rather simply in the diagram \u2013 121 \u2013 /p,s/q,r///p,s/q,r//+p,sq,rq,rp,sp2p1/p/21p which involves the expression \u00af\u03c8\u03b1(x) \u03c8\u03b1(x) \u00af\u03c8\u03b2(y) \u03c8\u03b2", "(y) = \u2212 \u03c8\u03b2(y) \u00af\u03c8\u03b1(x) \u03c8\u03b1(x) \u00af\u03c8\u03b2(y) = \u2212Tr (SF (y \u2212 x)SF (x \u2212 y)) After passing the fermion \ufb01elds through each other, a minus sign appears, sitting in front of the two propagators. 5.7.2 The Yukawa Potential Revisited We saw in Section 3.5.2, that the exchange of a real scalar particle gives rise to a universally attractive Yukawa potential between two spin zero particles. Does the same hold for the spin 1/2 particles? Recall that the strategy to compute the potential is to take the non-relativistic limit of the scattering amplitude, and compare with the analogous result from quantum mechanics. Our new amplitude now also includes the spinor degrees of freedom u(p) and v(p). In the non-relativistic limit, p \u2192 (m, p), and u(p) = \u221a \u221a p \u00b7 \u03c3\u03be p \u00b7 \u00af\u03c3\u03be v(p) = \u221a p \u00b7 \u03c3\u03be \u221a \u2212 p \u00b7 \u00af\u03c3\u03be \u221a m \u2192 \u221a m \u2192 \u03be \u03be \u03be \u2212\u03be (5.49) In this limit, the spinor contractions in the amplitude for \u03c8\u03c8 \u2192 \u03c8\u03c8 scattering (5.45) become \u00afus \u00b7 us = 2m\u03b4ss and the amplitude is = \u2212i(\u2212i\u03bb)2 (2m) \u03b4ss\u03b4rr (p \u2212 p) + \u00b52 \u2212 \u03b4sr\u03b4rs (p \u2212 q) + \u00b52 (5.50) The \u03b4 symbols tell us that spin is conserved in the non-relativistic limit, while the momentum dependence is the same as in the bosonic case, telling us that once again the particles feel an attractive Yukawa potential, U (r) = \u2212 \u03bb2e\u2212\u00b5r 4\u03c0r (5.51) Repeating the calculation for \u03c8 \u00af\u03c8 \u2192 \u03c8 \u00af\u03c8, there are two minus signs which cancel each other. The \ufb01rst is the extra overall minus sign in the scattering amplitude (5.46), \u2013 122 \u2013 /p,s/q,r//q,rp,s due to the fermionic nature of the particles. The second minus sign comes from the non-relativistic limit of the spinor contraction for anti-particles in (5.46), which is \u00afvs \u00b7 vs = \u22122m\u03b4ss. These two signs cancel, giving us once again an attractive Yukawa potential (5.51). 5.7.3 Pseudo-Scalar Coupling Rather than the standard Yukawa coupling, we could instead consider LYuk = \u2212\u03bb\u03c6 \u00af\u03c8\u03b35\u03c8 This still preserves parity if \u03c6 is a pseudoscalar, i.e. P : \u03c6(x, t) \u2192 \u2212\u03c6(\u2212x, t) (5.52) (5.53) We can compute in this theory very simply: the Feynman rule for the interaction vertex is now changed to a factor of \u2212i\u03bb\u03b35. For example, the Feynman diagrams for \u03c8\u03c8 \u2192 \u03c8\u03c8 scattering are again given by Figure 25, with the amplitude now A = (\u2212i\u03bb)2 [\u00afus(p )\u03b35us(p)] [\u00afur(q )\u03b35ur(q)] (p \u2212 p)2 \u2212 \u00b52 \u2212 [\u00afus(p )\u03b35ur(q)] [\u00afur(q )\u03b35us(p)] (p \u2212 q)2 \u2212 \u00b52 We could again try to take the non-relativistic limit for this amplitude. But this time, things work a little di\ufb00erently. Using the expressions for the spinors (5.49), we have \u00afus\u03b35us \u2192 0 in the non-relativistic limit. To \ufb01nd the non-relativistic amplitude, we must go to next to leading order. One can easily check that \u00afus(p )\u03b35us(p) \u2192 m \u03bes T (p \u2212 p ) \u00b7\u03c3\u03bes. So, in the non-relativistic limit, the leading order amplitude arising from pseudoscalar exchange is given by a spin-spin coupling, \u2192 +im(\u2212i\u03bb)2 [\u03bes T (p \u2212 p ) \u00b7 \u03c3\u03bes] [\u03ber T (p \u2212 p ) \u00b7 \u03c3\u03ber] (p \u2212 p )2 + \u00b52 (5.54) \u2013 123 \u2013 /p,s/q,r//q,rp,s 6", ". Quantum Electrodynamics In this section we \ufb01nally get to quantum electrodynamics (QED), the theory of light interacting with charged matter. Our path to quantization will be as before: we start with the free theory of the electromagnetic \ufb01eld and see how the quantum theory gives rise to a photon with two polarization states. We then describe how to couple the photon to fermions and to bosons. 6.1 Maxwell\u2019s Equations The Lagrangian for Maxwell\u2019s equations in the absence of any sources is simply L = \u2212 1 4 F\u00b5\u03bdF \u00b5\u03bd where the \ufb01eld strength is de\ufb01ned by F\u00b5\u03bd = \u2202\u00b5A\u03bd \u2212 \u2202\u03bdA\u00b5 The equations of motion which follow from this Lagrangian are \u2202\u00b5 \u2202L \u2202(\u2202\u00b5A\u03bd) = \u2212\u2202\u00b5F \u00b5\u03bd = 0 (6.1) (6.2) (6.3) Meanwhile, from the de\ufb01nition of F\u00b5\u03bd, the \ufb01eld strength also satis\ufb01es the Bianchi identity \u2202\u03bbF\u00b5\u03bd + \u2202\u00b5F\u03bd\u03bb + \u2202\u03bdF\u03bb\u00b5 = 0 (6.4) To make contact with the form of Maxwell\u2019s equations you learn about in high school, we need some 3-vector notation. If we de\ufb01ne A\u00b5 = (\u03c6, A), then the electric \ufb01eld E and magnetic \ufb01eld B are de\ufb01ned by E = \u2212\u2207\u03c6 \u2212 \u2202 A \u2202t and B = \u2207 \u00d7 A which, in terms of F\u00b5\u03bd, becomes F\u00b5\u03bd = 0 \u2212Ex \u2212Ey Bz \u2212Ez \u2212By Bx Ex Ey Ez 0 \u2212Bz By 0 \u2212Bx 0 The Bianchi identity (6.4) then gives two of Maxwell\u2019s equations, \u2207 \u00b7 B = 0 and \u2202 B \u2202t = \u2212\u2207 \u00d7 E \u2013 124 \u2013 (6.5) (6.6) (6.7) These remain true even in the presence of electric sources. Meanwhile, the equations of motion give the remaining two Maxwell equations, \u2207 \u00b7 E = 0 and \u2202 E \u2202t = \u2207 \u00d7 B (6.8) As we will see shortly, in the presence of charged matter these equations pick up extra terms on the right-hand side. 6.1.1 Gauge Symmetry The massless vector \ufb01eld A\u00b5 has 4 components, which would naively seem to tell us that the gauge \ufb01eld has 4 degrees of freedom. Yet we know that the photon has only two degrees of freedom which we call its polarization states. How are we going to resolve this discrepancy? There are two related comments which will ensure that quantizing the gauge \ufb01eld A\u00b5 gives rise to 2 degrees of freedom, rather than 4. \u2022 The \ufb01eld A0 has no kinetic term \u02d9A0 in the Lagrangian: it is not dynamical. This means that if we are given some initial data Ai and \u02d9Ai at a time t0, then the \ufb01eld A0 is fully determined by the equation of motion \u2207 \u00b7 E = 0 which, expanding out, reads This has the solution \u22072A0 + \u2207 \u00b7 \u2202 A \u2202t = 0 A0(x) = d3x (\u2207 \u00b7 \u2202 A/\u2202t)(x ) 4\u03c0|x \u2212 x | (6.9) (6.10) So A0 is not independent: we don\u2019t get to specify A0 on the initial time slice. It looks like we have only 3 degrees of freedom in A\u00b5 rather than 4. But this is still one too many. \u2022 The Lagrangian (6.3) has a very large symmetry group, acting on the vector potential as A\u00b5(x) \u2192 A\u00b5(x) + \u2202\u00b5\u03bb(x) (6.11) for any function \u03bb(x). We\u2019ll ask only that \u03bb(x) dies o\ufb00 suitably quickly at spatial x \u2192 \u221e. We call this a gauge symmetry. The \ufb01eld strength is invariant under the gauge symmetry: F\u00b5\u03bd \u2192 \u2202\u00b5(A\u03bd + \u2202\u03bd\u03bb) \u2212 \u2202\u03bd(A\u00b5 + \u2202\u00b5\u03bb) = F\u00b5\u03bd (6.12) \u2013 125 \u2013 So what are we to make of this? We have a theory with an in\ufb01nite number of symmetries", ", one for each function \u03bb(x). Previously we only encountered symmetries which act the same at all points in spacetime, for example \u03c8 \u2192 ei\u03b1\u03c8 for a complex scalar \ufb01eld. Noether\u2019s theorem told us that these symmetries give rise to conservation laws. Do we now have an in\ufb01nite number of conservation laws? The answer is no! Gauge symmetries have a very di\ufb00erent interpretation than the global symmetries that we make use of in Noether\u2019s theorem. While the latter take a physical state to another physical state with the same properties, the gauge symmetry is to be viewed as a redundancy in our description. That is, two states related by a gauge symmetry are to be identi\ufb01ed: they are the same physical state. (There is a small caveat to this statement which is explained in Section 6.3.1). One way to see that this interpretation is necessary is to notice that Maxwell\u2019s equations are not su\ufb03cient to specify the evolution of A\u00b5. The equations read, [\u03b7\u00b5\u03bd(\u2202\u03c1\u2202\u03c1) \u2212 \u2202\u00b5\u2202\u03bd] A\u03bd = 0 (6.13) But the operator [\u03b7\u00b5\u03bd(\u2202\u03c1\u2202\u03c1)\u2212\u2202\u00b5\u2202\u03bd] is not invertible: it annihilates any function of the form \u2202\u00b5\u03bb. This means that given any initial data, we have no way to uniquely determine A\u00b5 at a later time since we can\u2019t distinguish between A\u00b5 and A\u00b5 + \u2202\u00b5\u03bb. This would be problematic if we thought that A\u00b5 is a physical object. However, if we\u2019re happy to identify A\u00b5 and A\u00b5 + \u2202\u00b5\u03bb as corresponding to the same physical state, then our problems disappear. Since gauge invariance is a redundancy of the system, we might try to formulate the theory purely in terms of the local, physical, gauge invariant objects E and B. This is \ufb01ne for the free classical theory: Maxwell\u2019s equations were, after all, \ufb01rst written in terms of E and B. But it is not possible to describe certain quantum phenomena, such as the Aharonov-Bohm e\ufb00ect, without using the gauge potential A\u00b5. We will see shortly that we also require the gauge potential to describe classically charged \ufb01elds. To describe Nature, it appears that we have to introduce quantities A\u00b5 that we can never measure. Figure 29: The picture that emerges for the theory of electromagnetism is of an enlarged phase space, foliated by gauge orbits as shown in the \ufb01gure. All states that lie along a given \u2013 126 \u2013 Gauge OrbitsGaugeFixing line can be reached by a gauge transformation and are identi\ufb01ed. To make progress, we pick a representative from each gauge orbit. It doesn\u2019t matter which representative we pick \u2014 after all, they\u2019re all physically equivalent. But we should make sure that we pick a \u201cgood\u201d gauge, in which we cut the orbits. Di\ufb00erent representative con\ufb01gurations of a physical state are called di\ufb00erent gauges. There are many possibilities, some of which will be more useful in di\ufb00erent situations. Picking a gauge is rather like picking coordinates that are adapted to a particular problem. Moreover, di\ufb00erent gauges often reveal slightly di\ufb00erent aspects of a problem. Here we\u2019ll look at two di\ufb00erent gauges: \u2022 Lorentz Gauge: \u2202\u00b5A\u00b5 = 0 To see that we can always pick a representative con\ufb01guration satisfying \u2202\u00b5A\u00b5 = 0, \u00b5 satisfying \u2202\u00b5(A)\u00b5 = f (x). Then we suppose that we\u2019re handed a gauge \ufb01eld A choose A\u00b5 = A \u00b5 + \u2202\u00b5\u03bb, where \u2202\u00b5\u2202\u00b5\u03bb = \u2212f (6.14) This equation always has a solution. In fact this condition doesn\u2019t pick a unique representative from the gauge orbit. We\u2019re always free to make further gauge transformations with \u2202\u00b5\u2202\u00b5\u03bb = 0, which also has non-trivial solutions. As the name suggests, the Lorentz gauge3 has the advantage that it is Lorentz invariant. \u2022 Coulomb Gauge: \u2207 \u00b7 A = 0 We can make use of the residual gauge transformations in Lorentz gauge to pick \u2207 \u00b7 A = 0. (The argument is the same as before", "). Since A0 is \ufb01xed by (6.10), we have as a consequence A0 = 0 (6.15) (This equation will no longer hold in Coulomb gauge in the presence of charged matter). Coulomb gauge breaks Lorentz invariance, so may not be ideal for some purposes. However, it is very useful to exhibit the physical degrees of freedom: the 3 components of A satisfy a single constraint: \u2207 \u00b7 A = 0, leaving behind just 2 degrees of freedom. These will be identi\ufb01ed with the two polarization states of the photon. Coulomb gauge is sometimes called radiation gauge. 3Named after Lorenz who had the misfortune to be one letter away from greatness. \u2013 127 \u2013 6.2 The Quantization of the Electromagnetic Field In the following we shall quantize free Maxwell theory twice: once in Coulomb gauge, and again in Lorentz gauge. We\u2019ll ultimately get the same answers and, along the way, see that each method comes with its own subtleties. The \ufb01rst of these subtleties is common to both methods and comes when computing the momentum \u03c0\u00b5 conjugate to A\u00b5, \u03c00 = \u03c0i = \u2202L \u2202 \u02d9A0 \u2202L \u2202 \u02d9Ai = 0 = \u2212F 0i \u2261 Ei (6.16) so the momentum \u03c00 conjugate to A0 vanishes. This is the mathematical consequence of the statement we made above: A0 is not a dynamical \ufb01eld. Meanwhile, the momentum conjugate to Ai is our old friend, the electric \ufb01eld. We can compute the Hamiltonian, H = = d3x \u03c0i \u02d9Ai \u2212 L d3x \u2212 A0(\u2207 \u00b7 E) So A0 acts as a Lagrange multiplier which imposes Gauss\u2019 law \u2207 \u00b7 E = 0 (6.17) (6.18) which is now a constraint on the system in which A are the physical degrees of freedom. Let\u2019s now see how to treat this system using di\ufb00erent gauge \ufb01xing conditions. 6.2.1 Coulomb Gauge In Coulomb gauge, the equation of motion for A is \u2202\u00b5\u2202\u00b5 A = 0 which we can solve in the usual way, A = d3p (2\u03c0)3 \u03be(p) eip\u00b7x with p2 0 = |p|2. The constraint \u2207 \u00b7 A = 0 tells us that \u03be must satisfy \u03be \u00b7 p = 0 \u2013 128 \u2013 (6.19) (6.20) (6.21) which means that \u03be is perpendicular to the direction of motion p. We can pick \u03be(p) to be a linear combination of two orthonormal vectors r, r = 1, 2, each of which satis\ufb01es r(p) \u00b7 p = 0 and r(p) \u00b7 s(p) = \u03b4rs r, s = 1, 2 (6.22) These two vectors correspond to the two polarization states of the photon. It\u2019s worth pointing out that you can\u2019t consistently pick a continuous basis of polarization vectors for every value of p because you can\u2019t comb the hair on a sphere. But this topological fact doesn\u2019t cause any complications in computing QED scattering processes. To quantize we turn the Poisson brackets into commutators. Naively we would write [Ai(x), Aj(y)] = [Ei(x), Ej(y)] = 0 [Ai(x), Ej(y)] = i\u03b4j i \u03b4(3)(x \u2212 y) (6.23) But this can\u2019t quite be right, because it\u2019s not consistent with the constraints. We still want to have \u2207 \u00b7 A = \u2207 \u00b7 E = 0, now imposed on the operators. But from the commutator relations above, we see [\u2207 \u00b7 A(x), \u2207 \u00b7 E(y)] = i\u22072 \u03b4(3)(x \u2212 y) = 0 (6.24) What\u2019s going on? In imposing the commutator relations (6.23) we haven\u2019t correctly taken into account the constraints. In fact, this is a problem already in the classical theory, where the Poisson bracket structure is already altered4. The correct Poisson bracket structure leads to an alteration of the last commutation relation, [Ai(x), Ej(y)] = i \u03b4ij \u2212 \u2202i\u2202j \u22072 \u03b4(3)(x \u2212 y) (6.25) To see that this is now consistent with the constraints, we can rewrite", " the right-hand side of the commutator in momentum space, [Ai(x), Ej(y)] = i d3p (2\u03c0)3 \u03b4ij \u2212 pipj |p| 2 eip\u00b7(x\u2212y) (6.26) which is now consistent with the constraints, for example [\u2202iAi(x), Ej(y)] = i d3p (2\u03c0)3 \u03b4ij \u2212 pipj |p| 2 ipi eip\u00b7(x\u2212y) = 0 (6.27) 4For a nice discussion of the classical and quantum dynamics of constrained systems, see the small book by Paul Dirac, \u201cLectures on Quantum Mechanics\u201d \u2013 129 \u2013 We now write A in the usual mode expansion, A(x) = E(x) = d3p (2\u03c0)3 1 2|p| d3p (2\u03c0)3 (\u2212i) r(p) p eip\u00b7x + ar \u2020 ar p e\u2212ip\u00b7x 2 r=1 |p| 2 2 r=1 r(p) p eip\u00b7x \u2212 ar \u2020 ar p e\u2212ip\u00b7x (6.28) where, as before, the polarization vectors satisfy r(p) \u00b7 p = 0 and r(p) \u00b7 s(p) = \u03b4rs (6.29) It is not hard to show that the commutation relations (6.25) are equivalent to the usual commutation relations for the creation and annihilation operators, [ar p, as \u2020 q] = [ar \u2020 p, as \u2020 p, as q ] = (2\u03c0)3\u03b4rs \u03b4(3)(p \u2212 q) q ] = 0 [ar (6.30) where, in deriving this, we need the completeness relation for the polarization vectors, 2 r=1 r(p)j i r(p) = \u03b4ij \u2212 pipj |p| 2 (6.31) You can easily check that this equation is true by acting on both sides with a basis of vectors (1(p),2(p), p). We derive the Hamiltonian by substituting (6.28) into (6.17). The last term vanishes in Coulomb gauge. After normal ordering, and playing around with r polarization vectors, we get the simple expression H = d3p (2\u03c0)3 |p| 2 r=1 ar \u2020 p ar p (6.32) The Coulomb gauge has the advantage that the physical degrees of freedom are manifest. However, we\u2019ve lost all semblance of Lorentz invariance. One place where this manifests itself is in the propagator for the \ufb01elds Ai(x) (in the Heisenberg picture). In Coulomb gauge the propagator reads Dtr ij(x \u2212 y) \u2261 0| T Ai(x)Aj(y) |0 = d4p (2\u03c0)4 i p2 + i \u03b4ij \u2212 pipj |p|2 e\u2212ip\u00b7(x\u2212y) (6.33) The tr superscript on the propagator refers to the \u201ctransverse\u201d part of the photon. When we turn to the interacting theory, we will have to \ufb01ght to massage this propagator into something a little nicer. \u2013 130 \u2013 6.2.2 Lorentz Gauge We could try to work in a Lorentz invariant fashion by imposing the Lorentz gauge condition \u2202\u00b5A\u00b5 = 0. The equations of motion that follow from the action are then \u2202\u00b5\u2202\u00b5A\u03bd = 0 (6.34) Our approach to implementing Lorentz gauge will be a little di\ufb00erent from the method we used in Coulomb gauge. We choose to change the theory so that (6.34) arises directly through the equations of motion. We can achieve this by taking the Lagrangian L = \u2212 1 4 F\u00b5\u03bdF \u00b5\u03bd \u2212 1 2 (\u2202\u00b5A\u00b5)2 The equations of motion coming from this action are \u2202\u00b5F \u00b5\u03bd + \u2202\u03bd(\u2202\u00b5A\u00b5) = \u2202\u00b5\u2202\u00b5A\u03bd = 0 (6.35) (6.36) (In fact, we could be a little more general than this, and consider the Lagrangian L = \u2212 1 4F\u00b5\u03bdF \u00b5\u03bd \u2212 1 2\u03b1 (\u2202\u00b5A\u00b5)2 (6.37) with arbitrary \u03b1 and reach similar conclusions. The quantization of the theory is independent of \u03b1 and, rather confusingly, di\ufb00erent choices of \u03b1 are sometimes also referred to as di\ufb00erent \u201cgauges\u201d. We will", " use \u03b1 = 1, which is called \u201cFeynman gauge\u201d. The other common choice, \u03b1 = 0, is called \u201cLandau gauge\u201d.) Our plan will be to quantize the theory (6.36), and only later impose the constraint \u2202\u00b5A\u00b5 = 0 in a suitable manner on the Hilbert space of the theory. As we\u2019ll see, we will also have to deal with the residual gauge symmetry of this theory which will prove a little tricky. At \ufb01rst, we can proceed very easily, because both \u03c00 and \u03c0i are dynamical: \u03c00 = \u03c0i = \u2202L \u2202 \u02d9A0 \u2202L \u2202 \u02d9Ai = \u2212\u2202\u00b5A\u00b5 = \u2202iA0 \u2212 \u02d9Ai (6.38) Turning these classical \ufb01elds into operators, we can simply impose the usual commutation relations, [A\u00b5(x), A\u03bd(y)] = [\u03c0\u00b5(x), \u03c0\u03bd(y)] = 0 [A\u00b5(x), \u03c0\u03bd(y)] = i\u03b7\u00b5\u03bd \u03b4(3)(x \u2212 y) (6.39) \u2013 131 \u2013 and we can make the usual expansion in terms of creation and annihilation operators and 4 polarization vectors (\u00b5)\u03bb, with \u03bb = 0, 1, 2, 3. A\u00b5(x) = \u03c0\u00b5(x) = d3p (2\u03c0)3 d3p (2\u03c0)3 1 2|p| |p| 2 \u03bb \u00b5(p) p eip\u00b7x + a\u03bb \u2020 a\u03bb p e\u2212ip\u00b7x 3 \u03bb=0 (+i) 3 (\u00b5)\u03bb(p) p eip\u00b7x \u2212 a\u03bb \u2020 a\u03bb p e\u2212ip\u00b7x (6.40) \u03bb=0 Note that the momentum \u03c0\u00b5 comes with a factor of (+i), rather than the familiar (\u2212i) that we\u2019ve seen so far. This can be traced to the fact that the momentum (6.38) for the classical \ufb01elds takes the form \u03c0\u00b5 = \u2212 \u02d9A\u00b5 +.... In the Heisenberg picture, it becomes clear that this descends to (+i) in the de\ufb01nition of momentum. There are now four polarization 4-vectors \u03bb(p), instead of the two polarization 3vectors that we met in the Coulomb gauge. Of these four 4-vectors, we pick 0 to be timelike, while 1,2,3 are spacelike. We pick the normalization which also means that \u03bb \u00b7 \u03bb = \u03b7\u03bb\u03bb (\u00b5)\u03bb (\u03bd)\u03bb \u03b7\u03bb\u03bb = \u03b7\u00b5\u03bd (6.41) (6.42) The polarization vectors depend on the photon 4-momentum p = (|p|, p). Of the two spacelike polarizations, we will choose 1 and 2 to lie transverse to the momentum6.43) The third vector 3 is the longitudinal polarization. For example, if the momentum lies along the x3 direction, so p \u223c (1, 0, 0, 1), then 6.44) For other 4-momenta, the polarization vectors are the appropriate Lorentz transformations of these vectors, since (6.43) are Lorentz invariant. We do our usual trick, and translate the \ufb01eld commutation relations (6.39) into those for creation and annihilation operators. We \ufb01nd [a\u03bb p, a\u03bb q ] = [a\u03bb \u2020 p, a\u03bb \u2020 q ] = 0 and [a\u03bb p, a\u03bb \u2020 q ] = \u2212\u03b7\u03bb\u03bb (2\u03c0)3 \u03b4(3)(p \u2212 q) (6.45) \u2013 132 \u2013 The minus signs here are odd to say the least! For spacelike \u03bb = 1, 2, 3, everything looks \ufb01ne, [a\u03bb p, a\u03bb \u2020 q ] = \u03b4\u03bb\u03bb (2\u03c0)3 \u03b4(3)(p \u2212 q) \u03bb, \u03bb = 1, 2, 3 (6.46) But for the timelike annihilation and creation operators, we have [a0 p, a0 \u2020 q ] = \u2212(2\u03c0)3 \u03b4(3)(p \u2212 q) (6.47) This is very odd! To see just how strange this is, we take the Lorentz invariant vacuum |0 de\ufb01ned by a\u03bb p |0 = 0 Then we can create one-particle states in the usual way, |p,", " \u03bb = a\u03bb \u2020 p |0 (6.48) (6.49) For spacelike polarization states, \u03bb = 1, 2, 3, all seems well. But for the timelike polarization \u03bb = 0, the state |p, 0 has negative norm, p, 0| q, 0 = 0| a0 p a0 \u2020 q |0 = \u2212(2\u03c0)3 \u03b4(3)(p \u2212 q) (6.50) Wtf? That\u2019s very very strange. A Hilbert space with negative norm means negative probabilities which makes no sense at all. We can trace this negative norm back to the wrong sign of the kinetic term for A0 in our original Lagrangian: L = + 1 0 +.... 2 \u02d9A2 \u2212 1 \u02d9A2 2 At this point we should remember our constraint equation, \u2202\u00b5A\u00b5 = 0, which, until now, we\u2019ve not imposed on our theory. This is going to come to our rescue. We will see that it will remove the timelike, negative norm states, and cut the physical polarizations down to two. We work in the Heisenberg picture, so that \u2202\u00b5A\u00b5 = 0 (6.51) makes sense as an operator equation. Then we could try implementing the constraint in the quantum theory in a number of di\ufb00erent ways. Let\u2019s look at a number of increasingly weak ways to do this \u2022 We could ask that \u2202\u00b5A\u00b5 = 0 is imposed as an equation on operators. But this can\u2019t possibly work because the commutation relations (6.39) won\u2019t be obeyed for \u03c00 = \u2212\u2202\u00b5A\u00b5. We need some weaker condition. \u2013 133 \u2013 \u2022 We could try to impose the condition on the Hilbert space instead of directly on the operators. After all, that\u2019s where the trouble lies! We could imagine that there\u2019s some way to split the Hilbert space up into good states |\u03a8 and bad states that somehow decouple from the system. With luck, our bad states will include the weird negative norm states that we\u2019re so disgusted by. But how can we de\ufb01ne the good states? One idea is to impose \u2202\u00b5A\u00b5 |\u03a8 = 0 (6.52) on all good, physical states |\u03a8. But this can\u2019t work either! Again, the condition is too strong. For example, suppose we decompose A\u00b5(x) = A+ \u00b5 (x) with \u00b5 (x) + A\u2212 A+ \u00b5 (x) = A\u2212 \u00b5 (x) = d3p (2\u03c0)3 1 2|p| d3p (2\u03c0)3 1 2|p| 3 \u03bb \u00b5a\u03bb p e\u2212ip\u00b7x \u03bb=0 3 \u00b5a\u03bb \u2020 \u03bb p e+ip\u00b7x \u03bb=0 (6.53) Then, on the vacuum A+ the vacuum is a physical state if we use (6.52) as our constraint \u00b5 |0 = 0 automatically, but \u2202\u00b5A\u2212 \u00b5 |0 = 0. So not even \u2022 Our \ufb01nal attempt will be the correct one. In order to keep the vacuum as a good physical state, we can ask that physical states |\u03a8 are de\ufb01ned by This ensures that \u2202\u00b5A+ \u00b5 |\u03a8 = 0 \u03a8| \u2202\u00b5A\u00b5 |\u03a8 = 0 (6.54) (6.55) so that the operator \u2202\u00b5A\u00b5 has vanishing matrix elements between physical states. Equation (6.54) is known as the Gupta-Bleuler condition. The linearity of the constraint means that the physical states |\u03a8 span a physical Hilbert space Hphys. So what does the physical Hilbert space Hphys look like? And, in particular, have we rid ourselves of those nasty negative norm states so that Hphys has a positive de\ufb01nite inner product de\ufb01ned on it? The answer is actually no, but almost! Let\u2019s consider a basis of states for the Fock space. We can decompose any element of this basis as |\u03a8 = |\u03c8T |\u03c6, where |\u03c8T contains only transverse photons, created by \u2013 134 \u2013 a1,2 \u2020 p created by a3 \u2020, while |\u03c6 contains the timelike photons created by a0 \u2020 p. The Gupta-Bleuler condition (6.54) requires p and longitudinal photons (a3 p \u2212 a0 p) |\u03c6 = 0 (6.56) This means that the physical states must contain combinations of timelike", " and longitudinal photons. Whenever the state contains a timelike photon of momentum p, it must also contain a longitudinal photon with the same momentum. In general |\u03c6 will be a linear combination of states |\u03c6n containing n pairs of timelike and longitudinal photons, which we can write as |\u03c6 = \u221e n=0 Cn |\u03c6n (6.57) where |\u03c60 = |0 is simply the vacuum. It\u2019s not hard to show that although the condition (6.56) does indeed decouple the negative norm states, all the remaining states involving timelike and longitudinal photons have zero norm \u03c6m| \u03c6n = \u03b4n0\u03b4m0 (6.58) This means that the inner product on Hphys is positive semi-de\ufb01nite. Which is an improvement. But we still need to deal with all these zero norm states. The way we cope with the zero norm states is to treat them as gauge equivalent to the vacuum. Two states that di\ufb00er only in their timelike and longitudinal photon content, |\u03c6n with n \u2265 1 are said to be physically equivalent. We can think of the gauge symmetry of the classical theory as descending to the Hilbert space of the quantum theory. Of course, we can\u2019t just stipulate that two states are physically identical unless they give the same expectation value for all physical observables. We can check that this is true for the Hamiltonian, which can be easily computed to be H = d3p (2\u03c0)3 |p| 3 i=1 ai \u2020 p ai p \u2212 a0 \u2020 p a0 p (6.59) But the condition (6.56) ensures that \u03a8| a3 \u2020 p |\u03a8 so that the contributions from the timelike and longitudinal photons cancel amongst themselves in the Hamiltonian. This also renders the Hamiltonian positive de\ufb01nite, leaving us just with the contribution from the transverse photons as we would expect. p |\u03a8 = \u03a8| a0 \u2020 p a3 p a0 In general, one can show that the expectation values of all gauge invariant operators evaluated on physical states are independent of the coe\ufb03cients Cn in (6.57). \u2013 135 \u2013 Propagators Finally, it\u2019s a simple matter to compute the propagator in Lorentz gauge. It is given by 0| T A\u00b5(x)A\u03bd(y) |0 = d4p (2\u03c0)4 \u2212i\u03b7\u00b5\u03bd p2 + i e\u2212ip\u00b7(x\u2212y) (6.60) This is a lot nicer than the propagator we found in Coulomb gauge: in particular, it\u2019s Lorentz invariant. We could also return to the Lagrangian (6.37). Had we pushed through the calculation with arbitrary coe\ufb03cient \u03b1, we would \ufb01nd the propagator, 0| T A\u00b5(x)A\u03bd(y) |0 = d4p (2\u03c0)4 \u2212i p2 + i \u03b7\u00b5\u03bd + (\u03b1 \u2212 1) p\u00b5p\u03bd p2 e\u2212ip\u00b7(x\u2212y) (6.61) 6.3 Coupling to Matter Let\u2019s now build an interacting theory of light and matter. We want to write down a Lagrangian which couples A\u00b5 to some matter \ufb01elds, either scalars or spinors. For example, we could write something like where j\u00b5 is some function of the matter \ufb01elds. The equations of motion read L = \u2212 1 4F\u00b5\u03bdF \u00b5\u03bd \u2212 j\u00b5A\u00b5 so, for consistency, we require \u2202\u00b5F \u00b5\u03bd = j\u03bd \u2202\u00b5j\u00b5 = 0 (6.62) (6.63) (6.64) In other words, j\u00b5 must be a conserved current. But we\u2019ve got lots of those! Let\u2019s look at how we can couple two of them to electromagnetism. 6.3.1 Coupling to Fermions The Dirac Lagrangian L = \u00af\u03c8(i /\u2202 \u2212 m)\u03c8 (6.65) has an internal symmetry \u03c8 \u2192 e\u2212i\u03b1\u03c8 and \u00af\u03c8 \u2192 e+i\u03b1 \u00af\u03c8, with \u03b1 \u2208 R. This gives rise to V = \u00af\u03c8\u03b3\u00b5\u03c8. So we could look at the theory of electromagnetism the conserved current j\u00b5 coupled to fermions, with the Lagrangian, L = \u2212 1 4F\u00b5\u03bdF \u00b5\u03bd + \u00af\u03c8(i /\u2202 \u2212 m)\u03c8 \u2212", " e \u00af\u03c8\u03b3\u00b5A\u00b5\u03c8 (6.66) \u2013 136 \u2013 where we\u2019ve introduced a coupling constant e. For the free Maxwell theory, we have seen that the existence of a gauge symmetry was crucial in order to cut down the physical degrees of freedom to the requisite 2. Does our interacting theory above still have a gauge symmetry? The answer is yes. To see this, let\u2019s rewrite the Lagrangian as L = \u2212 1 4F\u00b5\u03bdF \u00b5\u03bd + \u00af\u03c8(i /D \u2212 m)\u03c8 (6.67) where D\u00b5\u03c8 = \u2202\u00b5\u03c8 + ieA\u00b5\u03c8 is called the covariant derivative. This Lagrangian is invariant under gauge transformations which act as A\u00b5 \u2192 A\u00b5 + \u2202\u00b5\u03bb and \u03c8 \u2192 e\u2212ie\u03bb\u03c8 (6.68) for an arbitrary function \u03bb(x). The tricky term is the derivative acting on \u03c8, since this will also hit the e\u2212ie\u03bb piece after the transformation. To see that all is well, let\u2019s look at how the covariant derivative transforms. We have D\u00b5\u03c8 = \u2202\u00b5\u03c8 + ieA\u00b5\u03c8 \u2192 \u2202\u00b5(e\u2212ie\u03bb\u03c8) + ie(A\u00b5 + \u2202\u00b5\u03bb)(e\u2212ie\u03bb\u03c8) = e\u2212ie\u03bbD\u00b5\u03c8 (6.69) so the covariant derivative has the nice property that it merely picks up a phase under the gauge transformation, with the derivative of e\u2212ie\u03bb cancelling the transformation of the gauge \ufb01eld. This ensures that the whole Lagrangian is invariant, since \u00af\u03c8 \u2192 e+ie\u03bb(x) \u00af\u03c8. Electric Charge The coupling e has the interpretation of the electric charge of the \u03c8 particle. This follows from the equations of motion of classical electromagnetism \u2202\u00b5F \u00b5\u03bd = j\u03bd: we know that the j0 component is the charge density. We therefore have the total charge Q given by Q = e d3x \u00af\u03c8(x)\u03b30\u03c8(x) After treating this as a quantum equation, we have Q = e d3p (2\u03c0)3 2 s=1 (bs \u2020 p bs p \u2212 cs \u2020 p cs p) (6.70) (6.71) which is the number of particles, minus the number of antiparticles. Note that the particle and the anti-particle are required by the formalism to have opposite electric \u2013 137 \u2013 charge. For QED, the theory of light interacting with electrons, the electric charge is usually written in terms of the dimensionless ratio \u03b1, known as the \ufb01ne structure constant \u03b1 = e2 4\u03c0c \u2248 1 137 (6.72) Setting = c = 1, we have e = \u221a 4\u03c0\u03b1 \u2248 0.3. There\u2019s a small subtlety here that\u2019s worth elaborating on. I stressed that there\u2019s a radical di\ufb00erence between the interpretation of a global symmetry and a gauge symmetry. The former takes you from one physical state to another with the same properties and results in a conserved current through Noether\u2019s theorem. The latter is a redundancy in our description of the system. Yet in electromagnetism, the gauge symmetry \u03c8 \u2192 e+ie\u03bb(x)\u03c8 seems to lead to a conservation law, namely the conservation of electric charge. This is because among the in\ufb01nite number of gauge symmetries parameterized by a function \u03bb(x), there is also a single global symmetry: that with \u03bb(x) = constant. This is a true symmetry of the system, meaning that it takes us to another physical state. More generally, the subset of global symmetries from among the gauge symmetries are those for which \u03bb(x) \u2192 \u03b1 = constant as x \u2192 \u221e. These take us from one physical state to another. Finally, let\u2019s check that the 4 \u00d7 4 matrix C that we introduced in Section 4.5 really deserves the name \u201ccharge conjugation matrix\u201d. If we take the complex conjugation of the Dirac equation, we have (i\u03b3\u00b5\u2202\u00b5 \u2212 e\u03b3\u00b5A\u00b5 \u2212 m)\u03c8 = 0 \u21d2 (\u2212i(\u03b3\u00b5)\u2202\u00b5 \u2212 e(\u03b3\u00b5)A\u00b5 \u2212 m)\u03c8 = 0 Now using the de\ufb01ning equation C \u2020\u03b3\u00b5C = \u2212(\u03b3\u00b5), and the de\ufb01nition \u03c8(c) = C\u03c8, we see that the charge conjugate spinor \u03c8(", "c) satis\ufb01es (i\u03b3\u00b5\u2202\u00b5 + e\u03b3\u00b5A\u00b5 \u2212 m)\u03c8(c) = 0 (6.73) So we see that the charge conjugate spinor \u03c8(c) satis\ufb01es the Dirac equation, but with charge \u2212e instead of +e. 6.3.2 Coupling to Scalars For a real scalar \ufb01eld, we have no suitable conserved current. This means that we can\u2019t couple a real scalar \ufb01eld to a gauge \ufb01eld. \u2013 138 \u2013 Let\u2019s now consider a complex scalar \ufb01eld \u03d5. (For this section, I\u2019ll depart from our previous notation and call the scalar \ufb01eld \u03d5 to avoid confusing it with the spinor). We have a symmetry \u03d5 \u2192 e\u2212i\u03b1\u03d5. We could try to couple the associated current to the gauge \ufb01eld, Lint = \u2212i((\u2202\u00b5\u03d5)\u03d5 \u2212 \u03d5\u2202\u00b5\u03d5)A\u00b5 (6.74) But this doesn\u2019t work because \u2022 The theory is no longer gauge invariant \u2022 The current j\u00b5 that we coupled to A\u00b5 depends on \u2202\u00b5\u03d5. This means that if we try to compute the current associated to the symmetry, it will now pick up a contribution from the j\u00b5A\u00b5 term. So the whole procedure wasn\u2019t consistent. We solve both of these problems simultaneously by remembering the covariant derivative. In this scalar theory, the combination D\u00b5\u03d5 = \u2202\u00b5\u03d5 + ieA\u00b5\u03d5 (6.75) again transforms as D\u00b5\u03d5 \u2192 e\u2212ie\u03bbD\u00b5\u03d5 under a gauge transformation A\u00b5 \u2192 A\u00b5 + \u2202\u00b5\u03bb and \u03d5 \u2192 e\u2212ie\u03bb\u03d5. This means that we can construct a gauge invariant action for a charged scalar \ufb01eld coupled to a photon simply by promoting all derivatives to covariant derivatives L = \u2212 1 4 F\u00b5\u03bdF \u00b5\u03bd + D\u00b5\u03d5D\u00b5\u03d5 \u2212 m2|\u03d5|2 (6.76) In general, this trick works for any theory. If we have a U (1) symmetry that we wish to couple to a gauge \ufb01eld, we may do so by replacing all derivatives by suitable covariant derivatives. This procedure is known as minimal coupling. 6.4 QED Let\u2019s now work out the Feynman rules for the full theory of quantum electrodynamics (QED) \u2013 the theory of electrons interacting with light. The Lagrangian is L = \u2212 1 4 F\u00b5\u03bdF \u00b5\u03bd + \u00af\u03c8(i /D \u2212 m)\u03c8 (6.77) where D\u00b5 = \u2202\u00b5 + ieA\u00b5. The route we take now depends on the gauge choice. If we worked in Lorentz gauge previously, then we can jump straight to Section 6.5 where the Feynman rules for QED are written down. If, however, we worked in Coulomb gauge, then we still have a bit of work in front of us in order to massage the photon propagator into something Lorentz invariant. We will now do that. \u2013 139 \u2013 In Coulomb gauge \u2207 \u00b7 A = 0, the equation of motion arising from varying A0 is now which has the solution \u2212\u22072A0 = e\u03c8\u2020\u03c8 \u2261 ej0 A0(x, t) = e d3x j0(x, t) 4\u03c0|x \u2212 x | (6.78) (6.79) In Coulomb gauge we can rewrite the Maxwell part of the Lagrangian as LMaxwell = = = d3x 1 2 E2 \u2212 1 2 B2 \u02d9A + \u2207A0)2 \u2212 1 B2 2 d3x 1 2( \u02d9A 2 + 1 d3x 1 2 2(\u2207A0)2 \u2212 1 2 B2 (6.80) where the cross-term has vanished using \u2207 \u00b7 A = 0. After integrating the second term by parts and inserting the equation for A0, we have LMaxwell = d3x \u02d9A 2 \u2212 1 2 1 2 B 2 + e2 2 d3x j0(x)j0(x ) 4\u03c0|x \u2212 x | (6.81) We \ufb01nd ourselves with a nonlocal term in the action. This is exactly the type of interaction that we boasted in Section 1.1.4 never arises in Nature! It appears here as an artifact of", " working in Coulomb gauge: it does not mean that the theory of QED is nonlocal. For example, it wouldn\u2019t appear if we worked in Lorentz gauge. We now compute the Hamiltonian. Changing notation slightly from previous chap- ters, we have the conjugate momenta, \u03a0 = \u03c0\u03c8 = \u02d9A = = i\u03c8\u2020 \u2202L \u02d9A \u2202 \u2202L \u2202 \u02d9\u03c8 (6.82) which gives us the Hamiltonian H = d3x \u02d9A 2 + 1 2 1 2 B2 + \u00af\u03c8(\u2212i\u03b3i\u2202i + m)\u03c8 \u2212 ej \u00b7 A + e2 2 d3x j0(x)j0(x ) 4\u03c0|x \u2212 x | where j = \u00af\u03c8\u03b3\u03c8 and j0 = \u00af\u03c8\u03b30\u03c8. \u2013 140 \u2013 6.4.1 Naive Feynman Rules We want to determine the Feynman rules for this theory. For fermions, the rules are the same as those given in Section 5. The new pieces are: \u2022 We denote the photon by a wavy line. Each end of the line comes with an i, j = 1, 2, 3 index telling us the component of A. We calculated the transverse photon propagator in (6.33): it is and contributes Dtr ij = i p2 + i \u03b4ij \u2212 pipj |p|2 \u2022 The vertex contributes \u2212ie\u03b3i. The index on \u03b3i contracts with the index on the photon line. \u2022 The non-local interaction which, in position space, is given by contributes a factor of i(e\u03b30)2\u03b4(x0 \u2212 y0) 4\u03c0|x \u2212 y| These Feynman rules are rather messy. This is the price we\u2019ve paid for working in Coulomb gauge. We\u2019ll now show that we can massage these expressions into something much more simple and Lorentz invariant. Let\u2019s start with the o\ufb00ending instantaneous interaction. Since it comes from the A0 component of the gauge \ufb01eld, we could try to rede\ufb01ne the propagator to include a D00 piece which will capture this term. In fact, it \ufb01ts quite nicely in this form: if we look in momentum space, we have \u03b4(x0 \u2212 y0) 4\u03c0|x \u2212 y| = d4p (2\u03c0)4 eip\u00b7(x\u2212y) |p|2 (6.83) so we can combine the non-local interaction with the transverse photon propagator by de\ufb01ning a new photon propagator D\u00b5\u03bd(p) = \uf8f1 \uf8f4\uf8f4\uf8f4\uf8f4\uf8f2 \uf8f4\uf8f4\uf8f4\uf8f4\uf8f3 i p2 + i i |p|2 + \u03b4ij \u2212 0 pipj |p|2 \u00b5, \u03bd = 0 \u00b5 = i = 0, \u03bd = j = 0 (6.84) otherwise With this propagator, the wavy photon line now carries a \u00b5, \u03bd = 0, 1, 2, 3 index, with the extra \u00b5 = 0 component taking care of the instantaneous interaction. We now need to change our vertex slightly: the \u2212ie\u03b3i above gets replaced by \u2212ie\u03b3\u00b5 which correctly accounts for the (e\u03b30)2 piece in the instantaneous interaction. \u2013 141 \u2013 xy The D00 piece of the propagator doesn\u2019t look a whole lot di\ufb00erent from the transverse photon propagator. But wouldn\u2019t it be nice if they were both part of something more symmetric! In fact, they are. We have the following: Claim: We can replace the propagator D\u00b5\u03bd(p) with the simpler, Lorentz invariant propagator D\u00b5\u03bd(p) = \u2212i \u03b7\u00b5\u03bd p2 (6.85) Proof: There is a general proof using current conservation. Here we\u2019ll be more pedestrian and show that we can do this for certain Feynman diagrams. In particular, we focus on a particular tree-level diagram that contributes to e\u2212e\u2212 \u2192 e\u2212e\u2212 scattering, \u223c e2[\u00afu(p)\u03b3\u00b5u(p)] D\u00b5\u03bd(k) [\u00afu(q)\u03b3\u03bdu(q)] (6.86) where k = p \u2212 p = q \u2212 q. Recall that u(p) satis\ufb01es the equation ( /p \u2212 m)u(p) = 0 (6.87) Let\u2019s de\ufb01ne the spinor contractions \u03b1\u00b5 = \u00afu", "(p )\u03b3\u00b5u(p) and \u03b2\u03bd = \u00afu(q )\u03b3\u03bdu(q). Then since k = p \u2212 p = q \u2212 q, we have k\u00b5\u03b1\u00b5 = \u00afu(p )( /p \u2212 /p )u(p) = \u00afu(p )(m \u2212 m)u(p) = 0 (6.88) and, similarly, k\u03bd\u03b2\u03bd = 0. Using this fact, the diagram can be written as \u03b1\u00b5D\u00b5\u03bd\u03b2\u03bd = i = i = i \u03b1 \u00b7 \u03b2 k2 \u2212 \u03b1 \u00b7 \u03b2 k2 \u2212 \u03b1 \u00b7 \u03b2 k2 \u2212 \u03b10\u03b20 |k|2 (\u03b1 \u00b7 k)(\u03b2 \u00b7 k) k2|k|2 + k2 0\u03b10\u03b20 k2|k|2 + \u03b10\u03b20 |k|2 (k2 0 \u2212 k2) \u03b10\u03b20 1 k2|k|2 = \u2212 i k2 \u03b1 \u00b7 \u03b2 = \u03b1\u00b5 \u03b2\u03bd i\u03b7\u00b5\u03bd k2 (6.89) \u2212 \u2013 142 \u2013 /qp/pq which is the claimed result. You can similarly check that the same substitution is legal in the diagram \u223c e2[\u00afv(q)\u03b3\u00b5u(p)]D\u00b5\u03bd(k)[\u00afu(p )\u03b3\u03bdv(q )] (6.90) In fact, although we won\u2019t show it here, it\u2019s a general fact that in every Feynman dia- gram we may use the very nice, Lorentz invariant propagator D\u00b5\u03bd = \u2212i\u03b7\u00b5\u03bd/p2. Note: This is the propagator we found when quantizing in Lorentz gauge (using the Feynman gauge parameter). In general, quantizing the Lagrangian (6.37) in Lorentz gauge, we have the propagator D\u00b5\u03bd = \u2212 i p2 \u03b7\u00b5\u03bd + (\u03b1 \u2212 1) p\u00b5p\u03bd p2 (6.91) Using similar arguments to those given above, you can show that the p\u00b5p\u03bd/p2 term cancels in all diagrams. For example, in the following diagrams the p\u00b5p\u03bd piece of the propagator contributes as \u223c \u00afu(p)\u03b3\u00b5u(p) k\u00b5 = \u00afu(p)( /p \u2212 /p )u(p) = 0 \u223c \u00afv(p)\u03b3\u00b5u(q) k\u00b5 = \u00afu(p)( /p + /q )u(q) = 0 (6.92) 6.5 Feynman Rules Finally, we have the Feynman rules for QED. For vertices and internal lines, we write \u2022 Vertex: \u2212ie\u03b3\u00b5 \u2022 Photon Propagator: \u2022 Fermion Propagator: \u2212 i\u03b7\u00b5\u03bd p2 + i i( /p + m) p2 \u2212 m2 + i For external lines in the diagram, we attach \u2022 Photons: We add a polarization vector \u00b5 In Coulomb gauge, 0 = 0 and \u00b7 p = 0. in/\u00b5 out for incoming/outgoing photons. \u2022 Fermions: We add a spinor ur(p)/\u00afur(p) for incoming/outgoing fermions. We add a spinor \u00afvr(p)/vr(p) for incoming/outgoing anti-fermions. \u2013 143 \u2013 pqpq// 6.5.1 Charged Scalars \u201cPauli asked me to calculate the cross section for pair creation of scalar It was only a short time after Bethe and Heitler particles by photons. had solved the same problem for electrons and positrons. I met Bethe in Copenhagen at a conference and asked him to tell me how he did the calculations. I also inquired how long it would take to perform this task; he answered, \u201cIt would take me three days, but you will need about three weeks.\u201d He was right, as usual; furthermore, the published cross sections were wrong by a factor of four.\u201d Viki Weisskopf The interaction terms in the Lagrangian for charged scalars come from the covariant derivative terms, L = D\u00b5\u03c8\u2020 D\u00b5\u03c8 = \u2202\u00b5\u03c8\u2020\u2202\u00b5\u03c8 \u2212 ieA\u00b5(\u03c8\u2020\u2202\u00b5\u03c8 \u2212 \u03c8\u2202\u00b5\u03c8\u2020) + e2A\u00b5A\u00b5\u03c8\u2020\u03c8 (6.93) This gives rise to two interaction vertices. But the cubic vertex is something we haven\u2019t seen before: it contains kinetic terms. How do these appear in the Feynman rules? After a Fourier transform, the", " derivative term means that the interaction is stronger for fermions with higher momentum, so we include a momentum factor in the Feynman rule. There is also a second, \u201cseagull\u201d graph. The two Feynman rules are \u2212 ie(p + q)\u00b5 and + 2ie2\u03b7\u00b5\u03bd The factor of two in the seagull diagram arises because of the two identical particles appearing in the vertex. (It\u2019s the same reason that the 1/4! didn\u2019t appear in the Feynman rules for \u03c64 theory). 6.6 Scattering in QED Let\u2019s now calculate some amplitudes for various processes in quantum electrodynamics, with a photon coupled to a single fermion. We will consider the analogous set of processes that we saw in Section 3 and Section 5. We have \u2013 144 \u2013 pq Electron Scattering Electron scattering e\u2212e\u2212 \u2192 e\u2212e\u2212 is described by the two leading order Feynman diagrams, given by + = \u2212i(\u2212ie)2 [\u00afus(p )\u03b3\u00b5us(p)] [\u00afur(q )\u03b3\u00b5ur(q)] (p \u2212 p)2 [\u00afus(p )\u03b3\u00b5ur(q)] [\u00afur(q )\u03b3\u00b5us(p)] (p \u2212 q)2 \u2212 The overall \u2212i comes from the \u2212i\u03b7\u00b5\u03bd in the propagator, which contract the indices on the \u03b3-matrices (remember that it\u2019s really positive for \u00b5, \u03bd = 1, 2, 3). Electron Positron Annihilation Let\u2019s now look at e\u2212e+ \u2192 2\u03b3, two gamma rays. The two lowest order Feynman diagrams are, + = i(\u2212ie)2 \u00afvr(q) \u03b3\u00b5( /p \u2212 /p + m)\u03b3\u03bd (p \u2212 p)2 \u2212 m2 + \u03b3\u03bd( /p \u2212 /q + m)\u03b3\u00b5 (p \u2212 q)2 \u2212 m2 us(p)\u03bd 1(p )\u00b5 2 (q ) Electron Positron Scattering For e\u2212e+ \u2192 e\u2212e+ scattering (sometimes known as Bhabha scattering) the two lowest order Feynman diagrams are + = \u2212i(\u2212ie)2 \u2212 + [\u00afus(p )\u03b3\u00b5us(p)] [\u00afvr(q)\u03b3\u00b5vr(q )] (p \u2212 p)2 [\u00afvr(q)\u03b3\u00b5us(p)] [\u00afus(p )\u03b3\u00b5vr(q )] (p + q)2 Compton Scattering The scattering of photons (in particular x-rays) o\ufb00 electrons e\u2212\u03b3 \u2192 e\u2212\u03b3 is known as Compton scattering. Historically, the change in wavelength of the photon in the \u2013 145 \u2013 //q,rp,sp,s/q,r/p,s////q,rq,rp,sp//qq,rp,s\u03b5\u03b5\u00b52\u03bd1\u03b51\u03bdp/\u03b5\u00b52/qq,rp,s//q,rp,sp,s/q,r/p,s////q,rp,sq,r scattering process was one of the conclusive pieces of evidence that light could behave as a particle. The amplitude is given by, + = i(\u2212ie)2 \u00afur(p ) \u03b3\u00b5( /p + /q + m)\u03b3\u03bd (p + q)2 \u2212 m2 + \u03b3\u03bd( /p \u2212 /q + m)\u03b3\u00b5 (p \u2212 q)2 \u2212 m2 us(p) \u03bd in \u00b5 out This amplitude vanishes for longitudinal photons. For example, suppose in \u223c q. Then, using momentum conservation p + q = p + q, we may write the amplitude as iA = i(\u2212ie)2 \u00afur(p ) /out( /p + /q + m) /q (p + q)2 \u2212 m2 + /q( /p \u2212 /q + m) /out (p \u2212 q)2 \u2212 m2 us(p) = i(\u2212ie)2 \u00afur(p ) /outus(p) 2p \u00b7 q (p + q)2 \u2212 m2 + 2p \u00b7 q (p \u2212 q)2 \u2212 m2 (6.94) where, in going to the second line, we\u2019ve performed some \u03b3-matrix manipulations, and invoked the fact that q is null, so /q /q = 0, together with the spinor equations ( /p \u2212 m)u(p) and \u00afu(p )( /p \u2212 m) = 0. We now recall the fact that q is a null vector, while p", "2 = (p)2 = m2 since the external legs are on mass-shell. This means that the two denominators in the amplitude read (p+q)2\u2212m2 = 2p\u00b7q and (p\u2212q)\u2212m2 = \u22122p\u00b7q. This ensures that the combined amplitude vanishes for longitudinal photons as promised. A similar result holds when /out \u223c q. Photon Scattering In QED, photons no longer pass through each other unimpeded. At one-loop, there is a diagram which leads to photon scattering. Although naively logarithmically divergent, the diagram is actually rendered \ufb01nite by gauge invariance. Adding Muons Figure 30: Adding a second fermion into the mix, which we could identify as a muon, new processes become possible. For example, we can now have processes such as e\u2212\u00b5\u2212 \u2192 e\u2212\u00b5\u2212 scattering, and e+e\u2212 annihilation into a muon anti-muon pair. Using our standard notation of p and q for incoming momenta, and p and q for outgoing \u2013 146 \u2013 out\u03b5/q\u03b5inq()()u(p)u(p)/\u2212\u03b5inq)(out\u03b5/q()u(p)u(p)/\u2212 momenta, we have the amplitudes given by \u223c 1 (p \u2212 p)2 and \u223c 1 (p + q)2 (6.95) 6.6.1 The Coulomb Potential We\u2019ve come a long way. We\u2019ve understood how to compute quantum amplitudes in a large array of \ufb01eld theories. To end this course, we use our newfound knowledge to rederive a result you learnt in kindergarten: Coulomb\u2019s law. To do this, we repeat our calculation that led us to the Yukawa force in Sections 3.5.2 and 5.7.2. We start by looking at e\u2212e\u2212 \u2192 e\u2212e\u2212 scattering. We have = \u2212i(\u2212ie)2 [\u00afu(p )\u03b3\u00b5u(p)] [\u00afu(q )\u03b3\u00b5u(q)] (p \u2212 p)2 (6.96) Following (5.49), the non-relativistic limit of the spinor is u(p) \u2192 \u221a m \u03be \u03be. This means that the \u03b30 piece of the interaction gives a term \u00afus(p)\u03b30ur(q) \u2192 2m\u03b4rs, while the spatial \u03b3i, i = 1, 2, 3 pieces vanish in the non-relativistic limit: \u00afus(p)\u03b3iur(q) \u2192 0. Comparing the scattering amplitude in this limit to that of non-relativistic quantum mechanics, we have the e\ufb00ective potential between two electrons given by, U (r) = +e2 d3p (2\u03c0)3 eip\u00b7r |p|2 = + e2 4\u03c0r (6.97) We \ufb01nd the familiar repulsive Coulomb potential. We can trace the minus sign that gives a repulsive potential to the fact that only the A0 component of the intermediate propagator \u223c \u2212i\u03b7\u00b5\u03bd contributes in the non-relativistic limit. For e\u2212e+ \u2192 e\u2212e+ scattering, the amplitude is = +i(\u2212ie)2 [\u00afu(p )\u03b3\u00b5u(p)] [\u00afv(q)\u03b3\u00b5v(q )] (p \u2212 p)2 (6.98) \u2013 147 \u2013 \u00b5\u2212\u00b5\u2212e\u2212e\u2212\u00b5\u00b5ee\u2212++\u2212//q,rp,sp,s/q,r///q,rp,sp,s/q,r/ The overall + sign comes from treating the fermions correctly: we saw the same minus sign when studying scattering in Yukawa theory. The di\ufb00erence now comes from looking at the non-relativistic limit. We have \u00afv\u03b30v \u2192 2m, giving us the potential between opposite charges, U (r) = \u2212e2 d3p (2\u03c0)3 eip\u00b7r |p|2 = \u2212 e2 4\u03c0r (6.99) Reassuringly, we \ufb01nd an attractive force between an electron and positron. The di\ufb00erence from the calculation of the Yukawa force comes again from the zeroth component of the gauge \ufb01eld, this time in the guise of the \u03b30 sandwiched between \u00afv\u03b30v \u2192 2m, rather than the \u00afvv \u2192 \u22122m that we saw in the Yukawa case. The Coulomb Potential for Scalars There are many minus signs in the above calculation which somewhat obscure the crucial one which gives rise", " to the repulsive force. A careful study reveals the o\ufb00ending sign to be that which sits in front of the A0 piece of the photon propagator \u2212i\u03b7\u00b5\u03bd/p2. Note that with our signature (+ \u2212 \u2212\u2212), the propagating Ai have the correct sign, while A0 comes with the wrong sign. This is simpler to see in the case of scalar QED, where we don\u2019t have to worry about the gamma matrices. From the Feynman rules of Section 6.5.1, we have the non-relativistic limit of scalar e\u2212e\u2212 scattering, = \u2212i\u03b7\u00b5\u03bd(\u2212ie)2 (p + p)\u00b5(q + q)\u03bd (p \u2212 p)2 \u2192 \u2212i(\u2212ie)2 (2m)2 \u2212(p \u2212 p )2 where the non-relativistic limit in the numerator involves (p+p)\u00b7(q +q) \u2248 (p+p)0(q + q)0 \u2248 (2m)2 and is responsible for selecting the A0 part of the photon propagator rather than the Ai piece. This shows that the Coulomb potential for spin 0 particles of the same charge is again repulsive, just as it is for fermions. For e\u2212e+ scattering, the amplitude picks up an extra minus sign because the arrows on the legs of the Feynman rules in Section 6.5.1 are correlated with the momentum arrows. Flipping the arrows on one pair of legs in the amplitude introduces the relevant minus sign to ensure that the non-relativistic potential between e\u2212e+ is attractive as expected. \u2013 148 \u2013 //q,rp,sp,s/q,r/ 6.7 Afterword In this course, we have laid the foundational framework for quantum \ufb01eld theory. Most of the developments that we\u2019ve seen were already in place by the middle of the 1930s, pioneered by people such as Jordan, Dirac, Heisenberg, Pauli and Weisskopf 5. Yet by the end of the 1930s, physicists were ready to give up on quantum \ufb01eld theory. The di\ufb03culty lies in the next terms in perturbation theory. These are the terms that correspond to Feynamn diagrams with loops in them, which we have scrupulously avoided computing in this course. The reason we\u2019ve avoided them is because they typically give in\ufb01nity! And, after ten years of trying, and failing, to make sense of this, the general feeling was that one should do something else. This from Dirac in 1937, Because of its extreme complexity, most physicists will be glad to see the end of QED But the leading \ufb01gures of the day gave up too soon. It took a new generation of postwar physicists \u2014 people like Schwinger, Feynman, Tomonaga and Dyson \u2014 to return to quantum \ufb01eld theory and tame the in\ufb01nities. The story of how they did that will be told in next term\u2019s course. 5For more details on the history of quantum \ufb01eld theory, see the excellent book \u201cQED and the Men who Made it\u201d by Sam Schweber. \u2013 149 \u2013bout 0.2%. Because the interaction energy of the magnetic moment with the magnetic \ufb01eld is just \u2212\u03bc \u00b7 B, the z-component of the force experienced by the atom is given by Fz = \u2202 \u2202z (\u03bc \u00b7 B) \u03bcz \u2202Bz \u2202z, (1.2) where we have ignored the components of B in directions other than the z-direction. Because the atom as a whole is very heavy, we expect that the classical concept of trajectory can be legitimately applied, a point which can be justi\ufb01ed using the Heisenberg uncertainty principle to be derived later. With the arrangement of Figure 1.1, the \u03bcz > 0 (Sz < 0) atom experiences an upward force, while the \u03bcz < 0 (Sz > 0) atom experiences a downward force. The beam is then expected to be split according to the values of \u03bcz. In other words, the SG (Stern\u2013Gerlach) apparatus \u201cmeasures\u201d the z-component of \u03bc or, equivalently, the z-component of S up to a proportionality factor. The atoms in the oven are randomly oriented; there is no preferred direction for the orientation of \u03bc. If the electron were like a classical spinning object, we would expect all values of \u03bcz to be realized between |\u03bc| and \u2212|\u03bc|. This would lead us to expect a continuous bundle of beams coming out of the SG apparatus, as indicated in Figure 1.1", ", spread more or less evenly over the expected range. Instead, what we experimentally observe is more like the situation also shown in Figure 1.1, where two \u201cspots\u201d are observed, corresponding to one \u201cup\u201d and one \u201cdown\u201d orientation. In other words, the SG apparatus splits the original silver beam from the oven into two distinct components, a phenomenon referred to in the early days of quantum theory as \u201cspace quantization.\u201d To the extent that \u03bc can be identi\ufb01ed within a proportionality factor with the electron spin S, only two possible values of the zcomponent of S are observed to be possible, Sz up and Sz down, which we call Sz+ and Sz\u2212. The two possible values of Sz are multiples of some fundamental unit of angular momentum; numerically it turns out that Sz = \u00afh/2 and \u2212\u00afh/2, where \u00afh = 1.0546 \u00d7 10\u221227 erg-s = 6.5822 \u00d7 10\u221216 eV-s. (1.3) This \u201cquantization\u201d of the electron spin angular momentum3 is the \ufb01rst important feature we deduce from the Stern\u2013Gerlach experiment. Figure 1.2a shows the result one would have expected from the experiment. According to classical physics, the beam should have spread itself over a vertical distance corresponding 3 An understanding of the roots of this quantization lies in the application of relativity to quantum mechanics. See Section 8.2 of this book for a discussion. 4 Fundamental Concepts (a) (b) Fig. 1.2 (a) Classical physics prediction for results from the Stern\u2013Gerlach experiment. The beam should have been spread out vertically, over a distance corresponding to the range of values of the magnetic moment times the cosine of the orientation angle. Stern and Gerlach, however, observed the result in (b), namely that only two orientations of the magnetic moment manifested themselves. These two orientations did not span the entire expected range. to the (continuous) range of orientation of the magnetic moment. Instead, one observes Figure 1.2b which is completely at odds with classical physics. The beam mysteriously splits itself into two parts, one corresponding to spin \u201cup\u201d and the other to spin \u201cdown.\u201d Of course, there is nothing sacred about the up-down direction or the z-axis. We could just as well have applied an inhomogeneous \ufb01eld in a horizontal direction, say in the x-direction, with the beam proceeding in the y-direction. In this manner we could have separated the beam from the oven into an Sx+ component and an Sx\u2212 component. 1.1.2 Sequential Stern\u2013Gerlach Experiments Let us now consider a sequential Stern\u2013Gerlach experiment. By this we mean that the atomic beam goes through two or more SG apparatuses in sequence. The \ufb01rst arrangement we consider is relatively straightforward. We subject the beam coming out of the oven to the arrangement shown in Figure 1.3a, where SG\u02c6z stands for an apparatus with the inhomogeneous magnetic \ufb01eld in the z-direction, as usual. We then block the Sz\u2212 component coming out of the \ufb01rst SG\u02c6z apparatus and let the remaining Sz+ component be subjected to another SG\u02c6z apparatus. This time there is only one beam component coming out of the second apparatus, just the Sz+ component. This is perhaps not so surprising; after all if the atom spins are up, they are expected to remain so, short of any external \ufb01eld that rotates the spins between the \ufb01rst and the second SG\u02c6z apparatuses. A little more interesting is the arrangement shown in Figure 1.3b. Here the \ufb01rst SG apparatus is the same as before but the second one (SG\u02c6x) has an inhomogeneous magnetic \ufb01eld in the x-direction. The Sz+ beam that enters the second apparatus (SG\u02c6x) is now split into two components, an Sx+ component and an Sx\u2212 component, with equal intensities. How can we explain this? Does it mean that 50% of the atoms in the Sz+ beam coming out of the \ufb01rst apparatus (SG\u02c6z) are made up of atoms characterized by both Sz+ and Sx+, while the remaining 50% have both Sz+ and Sx\u2212? It turns out that such a picture runs into dif\ufb01culty, as will be shown below. 5 1.1 The Stern\u2013Gerlach Experiment Oven SGz\u02c6 SGz\u02c6 Sz+ comp. Sz\u2013 comp. Oven SGz\u02c6 Oven SGz\u02c6 Sz+ beam Sz\u2013 beam Sz+ beam", " Sz\u2013 beam (a) (b) (c) SGx\u02c6 Sx+ beam SGx\u02c6 SGz\u02c6 Sx\u2013 beam Sz+ comp. No Sz\u2013 comp. Sx+ beam Sx\u2013 beam Sz+ beam Sz\u2013 beam Fig. 1.3 Sequential Stern\u2013Gerlach experiments. We now consider a third step, the arrangement shown in Figure 1.3c, which most dramatically illustrates the peculiarities of quantum-mechanical systems. This time we add to the arrangement of Figure 1.3b yet a third apparatus, of the SG\u02c6z type. It is observed experimentally that two components emerge from the third apparatus, not one; the emerging beams are seen to have both an Sz+ component and an Sz\u2212 component. This is a complete surprise because after the atoms emerged from the \ufb01rst apparatus, we made sure that the Sz\u2212 component was completely blocked. How is it possible that the Sz\u2212 component which, we thought, we eliminated earlier reappears? The model in which the atoms entering the third apparatus are visualized to have both Sz+ and Sx+ is clearly unsatisfactory. This example is often used to illustrate that in quantum mechanics we cannot determine both Sz and Sx simultaneously. More precisely, we can say that the selection of the Sx+ beam by the second apparatus (SG\u02c6x) completely destroys any previous information about Sz. It is amusing to compare this situation with that of a spinning top in classical mechanics, where the angular momentum L = I\u03c9 (1.4) can be measured by determining the components of the angular velocity vector \u03c9. By observing how fast the object is spinning in which direction we can determine \u03c9x, \u03c9y, and \u03c9z simultaneously. The moment of inertia I is computable if we know the mass density and the geometric shape of the spinning top, so there is no dif\ufb01culty in specifying both Lz and Lx in this classical situation. It is to be clearly understood that the limitation we have encountered in determining Sz and Sx is not due to the incompetence of the experimentalist. By improving the 6 Fundamental Concepts experimental techniques we cannot make the Sz\u2212 component out of the third apparatus in Figure 1.3c disappear. The peculiarities of quantum mechanics are imposed upon us by the experiment itself. The limitation is, in fact, inherent in microscopic phenomena. 1.1.3 Analogy with Polarization of Light Because this situation looks so novel, some analogy with a familiar classical situation may be helpful here. To this end we now digress to consider the polarization of light waves. This analogy will help us develop a mathematical framework for formulating the postulates of quantum mechanics. Consider a monochromatic light wave propagating in the z-direction. A linearly polarized (or plane polarized) light with a polarization vector in the x-direction, which we call for short an x-polarized light, has a space-time dependent electric \ufb01eld oscillating in the x-direction E = E0 \u02c6x cos(kz \u2212 \u03c9t). Likewise, we may consider a y-polarized light, also propagating in the z-direction, E = E0 \u02c6y cos(kz \u2212 \u03c9t). (1.5) (1.6) Polarized light beams of type (1.5) or (1.6) can be obtained by letting an unpolarized light beam go through a Polaroid \ufb01lter. We call a \ufb01lter that selects only beams polarized in the x-direction an x-\ufb01lter. Anx -\ufb01lter, of course, becomes a y-\ufb01lter when rotated by 90\u25e6 about the propagation (z) direction. It is well known that when we let a light beam go through an x-\ufb01lter and subsequently let it impinge on a y-\ufb01lter, no light beam comes out provided, of course, we are dealing with 100% ef\ufb01cient Polaroids; see Figure 1.4a. The situation is even more interesting if we insert between the x-\ufb01lter and the y-\ufb01lter yet another Polaroid that selects only a beam polarized in the direction \u2013 which we call the xdirection \u2013 that makes an angle of 45\u25e6 with the x-direction in the xy plane; see Figure 1.4b. x-filter y-filter No beam (a) No light 100% x-filter x -filter y-filter (45\u00b0 diagonal) (b) Fig. 1.4 Light beams subjected to Polaroid filters. 7 1.1 The Stern\u2013Gerlach Experiment y y\u02c6 y y\u02c6 x x\u02c6 x\u02c6 x Fig. 1.5 Orientations of the x", "- andy -axes. This time, there is a light beam coming out of the y-\ufb01lter despite the fact that right after the beam went through the x-\ufb01lter it did not have any polarization component in the ydirection. In other words, once the x-\ufb01lter intervenes and selects the x-polarized beam, it is immaterial whether the beam was previously x-polarized. The selection of the x-polarized beam by the second Polaroid destroys any previous information on light polarization. Notice that this situation is quite analogous to the situation that we encountered earlier with the SG arrangement of Figure 1.3b, provided that the following correspondence is made: Sz \u00b1 atoms \u2194 x-, y-polarized light Sx \u00b1 atoms \u2194 x-, y-polarized light, (1.7) where the x- and the y-axes are de\ufb01ned as in Figure 1.5. Let us examine how we can quantitatively describe the behavior of 45\u25e6-polarized beams (x- andy -polarized beams) within the framework of classical electrodynamics. Using Figure 1.5 we obtain E0 \u02c6x cos(kz \u2212 \u03c9t) = E0 E0 \u02c6y cos(kz \u2212 \u03c9t) = E0 1\u221a 2 \u02c6x cos(kz \u2212 \u03c9t) + 1\u221a 2 \u02c6y cos(kz \u2212 \u03c9t), \u2212 1\u221a 2 \u02c6x cos(kz \u2212 \u03c9t) + 1\u221a 2 (1.8) \u02c6y cos(kz \u2212 \u03c9t). In the triple-\ufb01lter arrangement of Figure 1.4b the beam coming out of the \ufb01rst Polaroid is an \u02c6x-polarized beam, which can be regarded as a linear combination of an x-polarized beam and a y-polarized beam. The second Polaroid selects the x-polarized beam, which can in turn be regarded as a linear combination of an x-polarized and a y-polarized beam. And \ufb01nally, the third Polaroid selects the y-polarized component. 8 Fundamental Concepts Applying correspondence (1.7) from the sequential Stern\u2013Gerlach experiment of Figure 1.3c, to the triple-\ufb01lter experiment of Figure 1.4b suggests that we might be able to represent the spin state of a silver atom by some kind of vector in a new kind of two-dimensional vector space, an abstract vector space not to be confused with the usual two-dimensional (xy) space. Just as \u02c6x and \u02c6y in (1.8) are the base vectors used to decompose the polarization vector \u02c6x of the \u02c6x-polarized light, it is reasonable to represent the Sx+ state by a vector, which we call a ket in the Dirac notation to be developed fully in the next section. We denote this vector by |Sx; + and write it as a linear combination of two base vectors, |Sz; + and |Sz; \u2212, which correspond to the Sz+ and the Sz\u2212 states, respectively. So we may conjecture |Sx; +?= 1\u221a 2 |Sz; + + 1\u221a 2 |Sz; \u2212 |Sx; \u2212?= \u2212 1\u221a 2 |Sz; + + 1\u221a 2 |Sz; \u2212 (1.9a) (1.9b) in analogy with (1.8). Later we will show how to obtain these expressions using the general formalism of quantum mechanics. Thus the unblocked component coming out of the second (SG\u02c6x) apparatus of Figure 1.3c is to be regarded as a superposition of Sz+ and Sz\u2212 in the sense of (1.9a). It is for this reason that two components emerge from the third (SG\u02c6z) apparatus. The next question of immediate concern is: How are we going to represent the Sy\u00b1 states? Symmetry arguments suggest that if we observe an Sz\u00b1 beam going in the x-direction and subject it to an SG\u02c6y apparatus, the resulting situation will be very similar to the case where an Sz\u00b1 beam going in the y-direction is subjected to an SG\u02c6x apparatus. The kets for Sy\u00b1 should then be regarded as a linear combination of |Sz; \u00b1, but it appears from (1.9) that we have already used up the available possibilities in writing |Sx; \u00b1. How can our vector space formalism distinguish Sy\u00b1 states from Sx\u00b1 states? An analogy with", " polarized light again rescues us here. This time we consider a circularly polarized beam of light, which can be obtained by letting a linearly polarized light pass through a quarter-wave plate. When we pass such a circularly polarized light through an x-\ufb01lter or a y-\ufb01lter, we again obtain either an x-polarized beam or a y-polarized beam of equal intensity. Yet everybody knows that the circularly polarized light is totally different from the 45\u25e6-linearly polarized (x-polarized or y-polarized) light. Mathematically, how do we represent a circularly polarized light? A right circularly polarized light is nothing more than a linear combination of an x-polarized light and a y-polarized light, where the oscillation of the electric \ufb01eld for the y-polarized component is 90\u25e6 out of phase with that of the x-polarized component:4 E = E0 1\u221a 2 \u02c6x cos(kz \u2212 \u03c9t) + \u02c6y cos kz \u2212 \u03c9t + \u03c0 2 1\u221a 2. (1.10) 4 Unfortunately, there is no unanimity in the de\ufb01nition of right versus left circularly polarized light in the literature. 9 1.1 The Stern\u2013Gerlach Experiment It is more elegant to use complex notation by introducing \u03b5 as follows: Re(\u03b5) =E/ E0. For a right circularly polarized light, we can then write \u03b5 = where we have usedi = ei\u03c0/2. 1\u221a 2 \u02c6xei(kz\u2212\u03c9t) + i\u221a 2 \u02c6yei(kz\u2212\u03c9t), We can make the following analogy with the spin states of silver atoms: Sy + atom \u2194 right circularly polarized beam, Sy \u2212 atom \u2194 left circularly polarized beam. (1.11) (1.12) (1.13) Applying this analogy to (1.12), we see that if we are allowed to make the coef\ufb01cients preceding base kets complex, there is no dif\ufb01culty in accommodating the Sy\u00b1 atoms in our vector space formalism: |Sy; \u00b1?= 1\u221a 2 |Sz; + \u00b1 i\u221a 2 |Sz; \u2212, (1.14) which are obviously different from (1.9). We thus see that the two-dimensional vector space needed to describe the spin states of silver atoms must be a complex vector space; an arbitrary vector in the vector space is written as a linear combination of the base vectors |Sz; \u00b1 with, in general, complex coef\ufb01cients. The fact that the necessity of complex numbers is already apparent in such an elementary example is rather remarkable. The reader must have noted by this time that we have deliberately avoided talking about photons. In other words, we have completely ignored the quantum aspect of light; nowhere did we mention the polarization states of individual photons. The analogy we worked out is between kets in an abstract vector space that describes the spin states of individual atoms with the polarization vectors of the classical electromagnetic \ufb01eld. Actually we could have made the analogy even more vivid by introducing the photon concept and talking about the probability of \ufb01nding a circularly polarized photon in a linearly polarized state, and so forth; however, that is not needed here. Without doing so, we have already accomplished the main goal of this section: to introduce the idea that quantum-mechanical states are to be represented by vectors in an abstract complex vector space.5 Finally, before outlining the mathematical formalism of quantum mechanics, we remark that the physics of a Stern\u2013Gerlach apparatus is of far more than simply academic interest. The ability to separate spin states of atoms has tremendous practical interest as well. Figure 1.6 shows the use of the Stern\u2013Gerlach technique to analyze the result of spin manipulation in an atomic beam of cesium atoms. The only stable isotope, 133Cs, of this alkali atom has a nuclear spin I = 7/2, and the experiment sorts out the F = 4 hyper\ufb01ne magnetic substate, giving nine spin orientations. This is only one of many examples where this once mysterious effect is used for practical devices. Of course, all of these uses only go 5 The reader who is interested in grasping the basic concepts of quantum mechanics through a careful study of photon polarization may \ufb01nd Chapter 1 of Baym (1969) extremely illuminating. 10 Fundamental Concepts CCD camera image Detection laser Permanent magnet (movable.0 0.8 0.6 0.4 0.2 0.0 1.0 0.8 0.6 0.4 0.2 0.0 \u22122 \u22121 \u2212", "3 (a) 0 +1 +3 +2 +4 (b) +4 0 5 10 15 Position [mm] \u22124 \u22124 20 25 m c 0 6 Cesium atomic beam Fig. 1.6 A modern Stern\u2013Gerlach apparatus, used to separate spin states of atomic cesium, taken from Lison et al., Phys. Rev. A, 61 (1999) 013405. The apparatus is shown on the left, while the data show the nine different projections for the spin-four atom, (a) before and (b) after optical pumping is used to populate only extreme spin projections. The spin quantum number F = 4 is a coupling between the outermost electron in the atom and the nuclear spin I = 7/2. to \ufb01rmly establish this effect, and the quantum-mechanical principles which we will now present and further develop. 1.2 Kets, Bras, and Operators In the preceding section we showed how analyses of the Stern\u2013Gerlach experiment led us to consider a complex vector space. In this and the following section we formulate the basic mathematics of vector spaces as used in quantum mechanics. Our notation throughout this book is the bra and ket notation developed by P. A. M. Dirac. The theory of linear vector spaces had, of course, been known to mathematicians prior to the birth of quantum mechanics, but Dirac\u2019s way of introducing vector spaces has many advantages, especially from the physicist\u2019s point of view. 1.2.1 Ket Space We consider a complex vector space whose dimensionality is speci\ufb01ed according to the nature of a physical system under consideration. In Stern\u2013Gerlach type experiments where the only quantum-mechanical degree of freedom is the spin of an atom, the dimensionality is determined by the number of alternative paths the atoms can follow when subjected to an SG apparatus; in the case of the silver atoms of the previous section, the dimensionality is just two, corresponding to the two possible values Sz can assume.6 Later, in Section 1.6, 6 For many physical systems the dimension of the state space is denumerably in\ufb01nite. While we will usually indicate a \ufb01nite number of dimensions, N, of the ket space, the results also hold for denumerably in\ufb01nite dimensions. 11 1.2 Kets, Bras, and Operators we consider the case of continuous spectra, for example, the position (coordinate) or momentum of a particle, where the number of alternatives is nondenumerably in\ufb01nite, in which case the vector space in question is known as a Hilbert space after D. Hilbert, who studied vector spaces in in\ufb01nite dimensions. In quantum mechanics a physical state, for example, a silver atom with a de\ufb01nite spin orientation, is represented by a state vector in a complex vector space. Following Dirac, we call such a vector a ket and denote it by| \u03b1. This state ket is postulated to contain complete information about the physical state; everything we are allowed to ask about the state is contained in the ket. Two kets can be added: |\u03b1 + |\u03b2 = |\u03b3. (1.15) The sum |\u03b3 is just another ket. If we multiply |\u03b1 by a complex number c, the resulting product c|\u03b1 is another ket. The number c can stand on the left or on the right of a ket; it makes no difference: c|\u03b1 = |\u03b1c. (1.16) In the particular case where c is zero, the resulting ket is said to be a null ket. One of the physics postulates is that |\u03b1 and c|\u03b1, with c = 0, represent the same physical state. In other words, only the \u201cdirection\u201d in vector space is of signi\ufb01cance. Mathematicians may prefer to say that we are here dealing with rays rather than vectors. An observable, such as momentum and spin components, can be represented by an operator, such as A, in the vector space in question. Quite generally, an operator acts on a ket from the left, A \u00b7 (|\u03b1) = A|\u03b1, (1.17) which is yet another ket. There will be more on multiplication operations later. In general, A|\u03b1 is not a constant times |\u03b1. However, there are particular kets of importance, known as eigenkets of operator A, denoted by with the property |a, |a, |a,... A|a = a|a, A|a = a|a,... (1.18) (1.19) where a, a,... are just numbers. Notice that applying A to an eigenket just reproduces the same ket apart from a multiplicative number. The set of numbers", " {a, a, a,...}, more compactly denoted by {a}, is called the set of eigenvalues of operator A. When it becomes necessary to order eigenvalues in a speci\ufb01c manner, {a(1), a(2), a(3),...} may be used in place of {a, a, a,...}. The physical state corresponding to an eigenket is called an eigenstate. In the simplest 2 systems, the eigenvalue-eigenket relation (1.19) is expressed as case of spin 1 Sz|Sz; + = \u00afh 2 |Sz; +, Sz|Sz; \u2212 = \u2212 \u00afh 2 |Sz; \u2212, (1.20) where |Sz; \u00b1 are eigenkets of operator Sz with eigenvalues \u00b1\u00afh/2. Here we could have used just |\u00afh/2 for |Sz; + in conformity with the notation |a, where an eigenket is labeled 12 Fundamental Concepts by its eigenvalue, but the notation |Sz; \u00b1, already used in the previous section, is more convenient here because we also consider eigenkets of Sx: Sx|Sx; \u00b1 = \u00b1 \u00afh 2 |Sx; \u00b1. (1.21) We remarked earlier that the dimensionality of the vector space is determined by the number of alternatives in Stern\u2013Gerlach type experiments. More formally, we are concerned with an N-dimensional vector space spanned by the N eigenkets of observable A. Any arbitrary ket |\u03b1 can be written as |\u03b1 = \u2211 a with a, a,... up to a( N ), where ca is a complex coef\ufb01cient. The question of the uniqueness of such an expansion will be postponed until we prove the orthogonality of eigenkets. ca |a, (1.22) 1.2.2 Bra Space and Inner Products The vector space we have been dealing with is a ket space. We now introduce the notion of a bra space, a vector space \u201cdual to\u201d the ket space. We postulate that corresponding to every ket |\u03b1 there exists a bra, denoted by \u03b1|, in this dual, or bra, space. The bra space is spanned by eigenbras {a|} which correspond to the eigenkets {|a}. There is a one-to-one correspondence between a ket space and a bra space: |\u03b1DC\u2194\u03b1| |a, |a,... DC\u2194a|, a|,... |\u03b1 + |\u03b2DC\u2194\u03b1| + \u03b2| (1.23) where DC stands for dual correspondence. Roughly speaking, we can regard the bra space as some kind of mirror image of the ket space. The bra dual to c|\u03b1 is postulated to be c\u2217\u03b1|, not c\u03b1|, which is a very important point. More generally, we have c\u03b1|\u03b1 + c\u03b2|\u03b2DC\u2194c\u2217 \u03b1\u03b1| + c\u2217 \u03b2\u03b2|. (1.24) We now de\ufb01ne the inner product of a bra and a ket.7 The product is written as a bra standing on the left and a ket standing on the right, for example, \u03b2|\u03b1 = (\u03b2|) \u00b7 (|\u03b1). bra (c) ket (1.25) This product is, in general, a complex number. Notice that in forming an inner product we always take one vector from the bra space and one vector from the ket space. We postulate two fundamental properties of inner products. First, \u03b2|\u03b1 = \u03b1|\u03b2\u2217. (1.26) 7 In the literature an inner product is often referred to as a scalar product because it is analogous to a \u00b7 b in Euclidean space; in this book, however, we reserve the term scalar for a quantity invariant under rotations in the usual three-dimensional space. 13 1.2 Kets, Bras, and Operators In other words, \u03b2|\u03b1 and \u03b1|\u03b2 are complex conjugates of each other. Notice that even though the inner product is, in some sense, analogous to the familiar scalar product a \u00b7 b, \u03b2|\u03b1 must be clearly distinguished from \u03b1|\u03b2; the analogous distinction is not needed in real vector space because a \u00b7 b is equal to b \u00b7 a. Using (1.26) we can immediately deduce that \u03b1|\u03b1 must be a real number. To prove this just let \u03b2| \u2192 \u03b1|. The second postulate on inner products is \u03b1|\u03b1 \u2265 0, (1.27) where the equality sign holds only if |\u03b1 is a null ket. This is sometimes known as the postulate of positive de\ufffd", "\ufffdnite metric. From a physicist\u2019s point of view, this postulate is essential for the probabilistic interpretation of quantum mechanics, as will become apparent later.8 Two kets |\u03b1 and |\u03b2 are said to be orthogonal if even though in the de\ufb01nition of the inner product the bra \u03b1| appears. The orthogonality relation (1.28) also implies, via (1.26), \u03b1|\u03b2 = 0, (1.28) Given a ket which is not a null ket, we can form a normalized ket | \u02dc\u03b1, where \u03b2|\u03b1 = 0. with the property | \u02dc\u03b1 = 1 \u03b1|\u03b1 |\u03b1, \u02dc\u03b1| \u02dc\u03b1 = 1. \u03b1|\u03b1 is known as the norm of |\u03b1, analogous to the magnitude of Quite generally, \u221a a \u00b7 a = |a| in Euclidean vector space. Because |\u03b1 and c|\u03b1 represent the same vector physical state, we might as well require that the kets we use for physical states be normalized in the sense of (1.31).9 1.2.3 Operators As we remarked earlier, observables like momentum and spin components are to be represented by operators that can act on kets. We can consider a more general class of operators that act on kets; they will be denoted by X, Y, and so forth, while A, B, and so on will be used for a restrictive class of operators that correspond to observables. An operator acts on a ket from the left side, X \u00b7 (|\u03b1) = X |\u03b1, (1.32) 8 Attempts to abandon this postulate led to physical theories with \u201cinde\ufb01nite metric.\u201d We shall not be concerned with such theories in this book. 9 For eigenkets of observables with continuous spectra, different normalization conventions will be used; see Section 1.6. (1.29) (1.30) (1.31) 14 Fundamental Concepts and the resulting product is another ket. Operators X and Y are said to be equal, if X = Y, X |\u03b1 = Y |\u03b1 (1.33) (1.34) for an arbitrary ket in the ket space in question. Operator X is said to be the null operator if, for any arbitrary ket |\u03b1, we have Operators can be added; addition operations are commutative and associative: X |\u03b1 = 0. X + Y = Y + X, X + (Y + Z) = (X + Y ) +Z. (1.35) (1.36a) (1.36b) With the single exception of the time-reversal operator to be considered in Chapter 4, the operators that appear in this book are all linear, that is, X(c\u03b1|\u03b1 + c\u03b2|\u03b2) =c \u03b1X |\u03b1 + c\u03b2X |\u03b2. An operator X always acts on a bra from the right side (\u03b1|) \u00b7 X = \u03b1|X, (1.37) (1.38) and the resulting product is another bra. The ket X |\u03b1 and the bra \u03b1|X are, in general, not dual to each other. We de\ufb01ne the symbol X \u2020 as X |\u03b1DC\u2194\u03b1|X \u2020. (1.39) The operator X \u2020 is called the Hermitian adjoint, or simply the adjoint, of X. An operator X is said to be Hermitian if X = X \u2020. (1.40) 1.2.4 Multiplication Operators X and Y can be multiplied. Multiplication operations are, in general, noncommutative, that is, XY = YX. Multiplication operations are, however, associative: X(YZ) = (XY )Z = XYZ. We also have (1.41) (1.42) X(Y |\u03b1) = (XY )|\u03b1 = XY |\u03b1, (\u03b2|X )Y = \u03b2|(XY ) = \u03b2|XY. (1.43) 15 1.2 Kets, Bras, and Operators Notice that because (XY )\u2020 = Y \u2020X \u2020 (1.44) XY |\u03b1 = X(Y |\u03b1)DC\u2194(\u03b1|Y \u2020)X \u2020 = \u03b1|Y \u2020X \u2020. So far, we have considered the following products: \u03b2|\u03b1, X |\u03b1, \u03b1|X, and XY. Are there other products we are allowed to form? Let us multiply |\u03b2 and \u03b1|, in that order. The resulting product (1.45) (|\u03b2) \u00b7 (\u03b1|) = |\u03b2\u03b1| (1.46) is known as the outer product of |\u03b2 and", " \u03b1|. We will emphasize in a moment that |\u03b2\u03b1| is to be regarded as an operator; hence it is fundamentally different from the inner product \u03b2|\u03b1, which is just a number. There are also \u201cillegal products.\u201d We have already mentioned that an operator must stand on the left of a ket or on the right of a bra. In other words, |\u03b1X and X\u03b1| are examples of illegal products. They are neither kets, nor bras, nor operators; they are simply nonsensical. Products like |\u03b1|\u03b2 and \u03b1|\u03b2| are also illegal when |\u03b1 and |\u03b2 (\u03b1| and \u03b2|) are ket (bra) vectors belonging to the same ket (bra) space.10 1.2.5 The Associative Axiom As is clear from (1.42), multiplication operations among operators are associative. Actually the associative property is postulated to hold quite generally as long as we are dealing with \u201clegal\u201d multiplications among kets, bras, and operators. Dirac calls this important postulate the associative axiom of multiplication. To illustrate the power of this axiom let us \ufb01rst consider an outer product acting on a ket: (|\u03b2\u03b1|) \u00b7 |\u03b3. Because of the associative axiom, we can regard this equally well as |\u03b2 \u00b7( \u03b1|\u03b3), (1.47) (1.48) where \u03b1|\u03b3 is just a number. So the outer product acting on a ket is just another ket; in other words, |\u03b2\u03b1| can be regarded as an operator. Because (1.47) and (1.48) are equal, we may as well omit the dots and let |\u03b2\u03b1|\u03b3 stand for the operator |\u03b2\u03b1| acting on |\u03b3 or, equivalently, the number \u03b1|\u03b3 multiplying |\u03b2. (On the other hand, if (1.48) is written as (\u03b1|\u03b3)\u00b7|\u03b2, we cannot afford to omit the dot and brackets because the resulting expression would look illegal.) Notice that the operator |\u03b2\u03b1| rotates |\u03b3 into the direction of |\u03b2. It is easy to see that if X = |\u03b2\u03b1|, (1.49) 10 Later in the book we will encounter products like |\u03b1|\u03b2, which are more appropriately written as |\u03b1\u2297|\u03b2, but in such cases |\u03b1 and |\u03b2 always refer to kets from different vector spaces. For instance, the \ufb01rst ket belongs to the vector space for electron spin, the second ket to the vector space for electron orbital angular momentum; or the \ufb01rst ket lies in the vector space of particle 1, the second ket in the vector space of particle 2, and so forth. 16 Fundamental Concepts then X \u2020 = |\u03b1\u03b2|, which is left as an exercise. In a second important illustration of the associative axiom, we note that (\u03b2|) bra \u00b7 (X |\u03b1) ket = (\u03b2|X ) bra \u00b7 (|\u03b1) ket. Because the two sides are equal, we might as well use the more compact notation \u03b2|X |\u03b1 (1.50) (1.51) (1.52) to stand for either side of (1.51). Recall now that \u03b1|X \u2020 is the bra that is dual to X |\u03b1, so \u03b2|X |\u03b1 = \u03b2| \u00b7(X |\u03b1) = {(\u03b1|X \u2020) \u00b7 |\u03b2}\u2217 = \u03b1|X \u2020|\u03b2\u2217, (1.53) where, in addition to the associative axiom, we used the fundamental property of the inner product (1.26). For a Hermitian X we have \u03b2|X |\u03b1 = \u03b1|X |\u03b2\u2217. (1.54) 1.3 Base Kets and Matrix Representations 1.3.1 Eigenkets of an Observable Let us consider the eigenkets and eigenvalues of a Hermitian operator A. We use the symbol A, reserved earlier for an observable, because in quantum mechanics Hermitian operators of interest quite often turn out to be the operators representing some physical observables. We begin by stating an important theorem. Theorem 1 corresponding to different eigenvalues are orthogonal. The eigenvalues of a Hermitian operator A are real; the eigenkets of A Proof First, recall that A|a = a|a. Because A is Hermitian, we also have a|A = a\u2217a|, (1.55) (1.56) where, a, a,... are eigenvalues of A. If we multiply both sides of (1.55) bya | on the left, both sides of (1.56) by|a on the right, and subtract, we obtain (", "a \u2212 a\u2217)a|a = 0. (1.57) 17 1.3 Base Kets and Matrix Representations Now a and a can be taken to be either the same or different. Let us \ufb01rst choose them to be the same; we then deduce the reality condition (the \ufb01rst half of the theorem) a = a\u2217, (1.58) where we have used the fact that |a is not a null ket. Let us now assume a and a to be different. Because of the just proved reality condition, the difference a \u2212 a\u2217 that appears in (1.57) is equal toa \u2212 a, which cannot vanish, by assumption. The inner product a|a must then vanish: which proves the orthogonality property (the second half of the theorem). a|a = 0 (a = a), (1.59) We expect on physical grounds that an observable has real eigenvalues, a point that will become clearer in the next section, where measurements in quantum mechanics will be discussed. The theorem just proved guarantees the reality of eigenvalues whenever the operator is Hermitian. That is why we talk about Hermitian observables in quantum mechanics. It is conventional to normalize |a so the {|a} form an orthonormal set: a|a = \u03b4aa. (1.60) We may logically ask: Is this set of eigenkets complete? Since we started our discussion by asserting that the whole ket space is spanned by the eigenkets of A, the eigenkets of A must therefore form a complete set by construction of our ket space.11 1.3.2 Eigenkets as Base Kets We have seen that the normalized eigenkets of A form a complete orthonormal set. An arbitrary ket in the ket space can be expanded in terms of the eigenkets of A. In other words, the eigenkets of A are to be used as base kets in much the same way as a set of mutually orthogonal unit vectors is used as base vectors in Euclidean space. Given an arbitrary ket| \u03b1 in the ket space spanned by the eigenkets of A, let us attempt to expand it as follows: |\u03b1 = \u2211 a Multiplying a| on the left and using the orthonormality property (1.60), we can immediately \ufb01nd the expansion coef\ufb01cient, ca |a. (1.61) In other words, we have ca = a|\u03b1. |\u03b1 = \u2211 a |aa|\u03b1, (1.62) (1.63) 11 The astute reader, already familiar with wave mechanics, may point out that the completeness of eigenfunctions we use can be proved by applying the Sturm\u2013Liouville theory to the Schr\u00a8odinger wave equation. But to \u201cderive\u201d the Schr\u00a8odinger wave equation from our fundamental postulates, the completeness of the position eigenkets must be assumed. 18 Fundamental Concepts which is analogous to an expansion of a vector V in (real) Euclidean space: V = \u2211 \u02c6ei(\u02c6ei \u00b7 V), i (1.64) where {\u02c6ei} form an orthogonal set of unit vectors. We now recall the associative axiom of multiplication: |aa|\u03b1 can be regarded either as the number a|\u03b1 multiplying |a or, equivalently, as the operator |aa| acting on |\u03b1. Because |\u03b1 in (1.63) is an arbitrary ket, we must have |aa| = 1, \u2211 a (1.65) where the 1 on the right-hand side is to be understood as the identity operator. Equation (1.65) is known as the completeness relation or closure. It is dif\ufb01cult to overestimate the usefulness of (1.65). Given a chain of kets, operators, or bras multiplied in legal orders, we can insert, in any place at our convenience, the identity operator written in form (1.65). Consider, for example \u03b1|\u03b1; by inserting the identity operator between \u03b1| and |\u03b1, we obtain \u03b1|\u03b1 = \u03b1| \u00b7 |aa| \u2211 a |a|\u03b1|2. = \u2211 a \u00b7 |\u03b1 (1.66) This, incidentally, shows that if |\u03b1 is normalized, then the expansion coef\ufb01cients in (1.61) must satisfy \u2211 a |ca |2 = \u2211 a |a|\u03b1|2 = 1. (1.67) Let us now look at |aa| that appears in (1.65). Since this is an outer product, it must be an operator. Let it operate on |\u03b1: (|aa", "|) \u00b7 |\u03b1 = |aa|\u03b1 = ca |a. We see that |aa| selects that portion of the ket |\u03b1 parallel to |a, so|a a| is known as the projection operator along the base ket |a and is denoted by \u039ba : (1.68) The completeness relation (1.65) can now be written as \u039ba \u2261 |aa|. \u039ba = 1. \u2211 a 1.3.3 Matrix Representations (1.69) (1.70) Having speci\ufb01ed the base kets, we now show how to represent an operator, say X, by a square matrix. First, using (1.65) twice, we write the operator X as X = \u2211 a |aa|X |aa|. \u2211 a (1.71) 19 1.3 Base Kets and Matrix Representations There are altogether N2 numbers of form a|X |a, where N is the dimensionality of the ket space. We may arrange them into an N \u00d7 N square matrix such that the column and row indices appear as follows: a| row X |a column. Explicitly we may write the matrix as \u239b X. = \u239c \u239c \u239d a(1)|X |a(1) a(1)|X |a(2) a(2)|X |a(1) a(2)|X |a(2)....... = stands for \u201cis represented by.\u201d12 where the symbol Using (1.53), we can write a|X |a = a|X \u2020|a\u2217. \u239e \u239f \u239f \u23a0, \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7... (1.72) (1.73) (1.74) At last, the Hermitian adjoint operation, originally de\ufb01ned by (1.39), has been related to the (perhaps more familiar) concept of complex conjugate transposed. If an operator B is Hermitian, we have a|B|a = a|B|a\u2217. (1.75) The way we arranged a|X |a into a square matrix is in conformity with the usual rule of matrix multiplication. To see this just note that the matrix representation of the operator relation reads Z = XY a|Z|a = a|XY |a = \u2211 a a|X |aa|Y |a. (1.76) (1.77) Again, all we have done is to insert the identity operator, written in form (1.65), between X and Y! Let us now examine how the ket relation can be represented using our base kets. The expansion coef\ufb01cients of |\u03b3 can be obtained by multiplying a| on the left: |\u03b3 = X |\u03b1 (1.78) a|\u03b3 = a|X |\u03b1 = \u2211 a a|X |aa|\u03b1. (1.79) 12 We do not use the equality sign here because the particular form of a matrix representation depends on the particular choice of base kets used. The operator is different from a representation of the operator just as the actress is different from a poster of the actress. 20 Fundamental Concepts But this can be seen as an application of the rule for multiplying a square matrix with a column matrix, once the expansion coef\ufb01cients of |\u03b1 and |\u03b3 are themselves arranged to form column matrices as follows: \u239b \u239e \u239b \u239e |\u03b1. = \u239c \u239c \u239c \u239c \u239d a(1)|\u03b1 a(2)|\u03b1 a(3)|\u03b1... \u239f \u239f \u239f \u239f \u23a0, |\u03b3. = \u239c \u239c \u239c \u239c \u239d a(1)|\u03b3 a(2)|\u03b3 a(3)|\u03b3... Likewise, given we can regard \u03b3| = \u03b1|X, \u03b3|a = \u2211 a \u03b1|aa|X |a. \u239f \u239f \u239f \u239f \u23a0. (1.80) (1.81) (1.82) So a bra is represented by a row matrix as follows: \u03b3|. = (\u03b3|a(1), \u03b3|a(2), \u03b3|a(3),...) = (a(1)|\u03b3\u2217, a(2)|\u03b3\u2217, a(3)|\u03b3\u2217,...). (1.83) Note the appearance of complex conjugation when the elements of the column matrix are written as in (1.83). The inner product \u03b2|\u03b1 can be written as the product of the row matrix representing \u03b2| with the column matrix representing |", "\u03b1: \u03b2|\u03b1 = \u2211 a \u03b2|aa|\u03b1 = (a(1)|\u03b2\u2217, a(2)|\u03b2\u2217,...) \u239b \u239c \u239c \u239d a(1)|\u03b1 a(2)|\u03b1... \u239e \u239f \u239f \u23a0. (1.84) If we multiply the row matrix representing \u03b1| with the column matrix representing |\u03b2, then we obtain just the complex conjugate of the preceding expression, which is consistent with the fundamental property of the inner product (1.26). Finally, the matrix representation of the outer product |\u03b2\u03b1| is easily seen to be \u239b |\u03b2\u03b1|. = \u239c \u239c \u239d a(1)|\u03b2a(1)|\u03b1\u2217 a(2)|\u03b2a(1)|\u03b1\u2217... a(1)|\u03b2a(2)|\u03b1\u2217 a(2)|\u03b2a(2)|\u03b1\u2217... \u239f \u239f \u23a0.......... (1.85) \u239e The matrix representation of an observable A becomes particularly simple if the eigenkets of A themselves are used as the base kets. First, we have A = \u2211 a \u2211 a |aa|A|aa|. But the square matrix a|A|a is obviously diagonal, a|A|a = a|A|a\u03b4aa = a\u03b4aa, (1.86) (1.87) 21 1.3 Base Kets and Matrix Representations so A = \u2211 a = \u2211 a a|aa| a\u039ba. (1.88) 1.3.4 Spin 1 2 Systems It is here instructive to consider the special case of spin 1 2 systems. The base kets used are |Sz; \u00b1, which we denote, for brevity, as |\u00b1. The simplest operator in the ket space spanned by |\u00b1 is the identity operator, which, according to (1.65), can be written as 1 = |++| + |\u2212\u2212|. According to (1.88), we must be able to write Sz as Sz = (\u00afh/2)[(|++|) \u2212 (|\u2212\u2212|)]. The eigenket-eigenvalue relation Sz|\u00b1 = \u00b1(\u00afh/2)|\u00b1 immediately follows from the orthonormality property of |\u00b1. It is also instructive to look at two other operators, S+ \u2261 \u00afh|+\u2212|, S\u2212 \u2261 \u00afh|\u2212+|, (1.89) (1.90) (1.91) (1.92) which are both seen to be non-Hermitian. The operator S+, acting on the spin-down ket |\u2212, turns |\u2212 into the spin-up ket |+ multiplied by \u00afh. On the other hand, the spin-up ket |+, when acted upon by S+, becomes a null ket. So the physical interpretation of S+ is that it raises the spin component by one unit of \u00afh; if the spin component cannot be raised any further, we automatically get a null state. Likewise, S\u2212 can be interpreted as an operator that lowers the spin component by one unit of \u00afh. Later we will show that S\u00b1 can be written as Sx \u00b1 iSy. In constructing the matrix representations of the angular-momentum operators, it is customary to label the column (row) indices in descending order of angular-momentum components, that is, the \ufb01rst entry corresponds to the maximum angular-momentum component, the second, the next highest, and so forth. In our particular case of spin 1 2 systems, we have |+. = 1 0 |\u2212. =, 1 0, 0 0 0 1, (1.93a) (1.93b). 0 0 1 0. = \u00afh S\u2212. = Sz \u00afh 2 0 1 0 \u22121,. = \u00afh S+ We will come back to these explicit expressions when we discuss the Pauli two-component formalism in Chapter 3. 22 Fundamental Concepts 1.4 Measurements, Observables, and the Uncertainty Relations 1.4.1 Measurements Having developed the mathematics of ket spaces, we are now in a position to discuss the quantum theory of measurement processes. This is not a particularly easy subject for beginners, so we \ufb01rst turn to the words of the great master, P. A. M. Dirac, for guidance (Dirac (1958), p. 36): \u201cA measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured.\u201d What does all this mean? We interpret Dirac\u2019s words as follows: Before a measurement of observable", " A is made, the system is assumed to be represented by some linear combination |\u03b1 = \u2211 a ca |a = \u2211 a |aa|\u03b1. (1.94) When the measurement is performed, the system is \u201cthrown into\u201d one of the eigenstates, say |a of observable A. In other words, |\u03b1 A measurement \u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2192 |a. (1.95) For example, a silver atom with an arbitrary spin orientation will change into either |Sz; + or |Sz; \u2212 when subjected to an SG apparatus of type SG\u02c6z. Thus a measurement usually changes the state. The only exception is when the state is already in one of the eigenstates of the observable being measured, in which case |a A measurement \u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2192 |a (1.96) with certainty, as will be discussed further. When the measurement causes |\u03b1 to change into |a, it is said that A is measured to be a. It is in this sense that the result of a measurement yields one of the eigenvalues of the observable being measured. Given (1.94), which is the state ket of a physical system before the measurement, we do not know in advance into which of the various |a the system will be thrown as the result of the measurement. We do postulate, however, that the probability for jumping into some particular |a is given by Probability for a = |a|\u03b1|2, (1.97) provided that |\u03b1 is normalized. Although we have been talking about a single physical system, to determine probability (1.97) empirically, we must consider a great number of measurements performed on an ensemble, that is, a collection, of identically prepared physical systems, all characterized by the same ket |\u03b1. Such an ensemble is known as a pure ensemble. (We will say more about ensembles in Chapter 3.) As an example, a beam of silver atoms which survive the \ufb01rst SG\u02c6z apparatus of Figure 1.3 with the Sz\u2212 component blocked is an example of a pure ensemble because every member atom of the ensemble is characterized by |Sz; +. The probabilistic interpretation (1.97) for the squared inner product |a|\u03b1|2 is one of the fundamental postulates of quantum mechanics, so it cannot be proven. Let us note, 23 1.4 Measurements, Observables, and the Uncertainty Relations however, that it makes good sense in extreme cases. Suppose the state ket is |a itself even before a measurement is made; then according to (1.97), the probability for getting a, or, more precisely, for being thrown into |a, as the result of the measurement is predicted to be 1, which is just what we expect. By measuring A once again, we, of course, get |a only; quite generally, repeated measurements of the same observable in succession yield the same result.13 If, on the other hand, we are interested in the probability for the system initially characterized by |a to be thrown into some other eigenket |a with a = a, then (1.97) gives zero because of the orthogonality between |a and |a. From the point of view of measurement theory, orthogonal kets correspond to mutually exclusive alternatives; for 2 system is in |Sz; +, it is not in|S z; \u2212 with certainty. example, if a spin 1 Quite generally, the probability for anything must be nonnegative. Furthermore, the probabilities for the various alternative possibilities must add up to unity. Both of these expectations are met by our probability postulate (1.97). We de\ufb01ne the expectation value of A taken with respect to state |\u03b1 as A \u2261 \u03b1|A|\u03b1. (1.98) To make sure that we are referring to state |\u03b1, the notation A\u03b1 is sometimes used. Equation (1.98) is a de\ufb01nition; however, it agrees with our intuitive notion of average measured value because it can be written as \u03b1|aa|A|aa| \u2191 measured value a |a|\u03b1|2. probability for obtaining a (1.99) It is very important not to confuse eigenvalues with expectation values. For example, the 2 systems can assume any real value between \u2212\u00afh/2 and expectation value of Sz for spin 1 +\u00afh/2, say 0.273\u00afh; in contrast, the eigenvalue of Sz assumes only two values, \u00afh/2 and \u2212\u00afh/2. To clarify further the meaning of measurements in quantum mechanics, we introduce the notion of a selective measurement, or \ufb01ltration. In Section 1.1 we considered a Stern\u2013Gerlach arrangement where we let only one of the spin components pass out of the apparatus while", " we completely blocked the other component. More generally, we imagine a measurement process with a device that selects only one of the eigenkets of A, say |a, and rejects all others; see Figure 1.7. This is what we mean by a selective measurement; it is also called \ufb01ltration because only one of the A eigenkets \ufb01lters through the ordeal. Mathematically we can say that such a selective measurement amounts to applying the projection operator \u039ba to |\u03b1: \u039ba|\u03b1 = |aa|\u03b1. (1.100) J. Schwinger has developed a formalism of quantum mechanics based on a thorough examination of selective measurements. He introduces a measurement symbol M(a) in the beginning, which is identical to \u039ba or |aa| in our notation, and deduces a number 13 Here successive measurements must be carried out immediately afterward. This point will become clear when we discuss the time evolution of a state ket in Chapter 2. 24 Fundamental Concepts a A Measurement a a with a a Fig. 1.7 Selective measurement. of properties of M(a) (and also of M(b, a) which amount to |ba|) by studying the outcome of various Stern\u2013Gerlach type experiments. In this way he motivates the entire mathematics of kets, bras, and operators. In this book we do not follow Schwinger\u2019s path; the interested reader may consult Gottfried (1966). 1.4.2 Spin 1 2 Systems, Once Again Before proceeding with a general discussion of observables, we once again consider spin 1 2 systems. This time we show that the results of sequential Stern\u2013Gerlach experiments, when combined with the postulates of quantum mechanics discussed so far, are suf\ufb01cient to determine not only the Sx,y eigenkets, |Sx; \u00b1 and |Sy; \u00b1, but also the operators Sx and Sy themselves. First, we recall that when the Sx+ beam is subjected to an apparatus of type SG\u02c6z, the beam splits into two components with equal intensities. This means that the probability for the Sx+ state to be thrown into |Sz; \u00b1, simply denoted as |\u00b1, is 1 2 each; hence, |+|Sx; +| = |\u2212|Sx; +| = 1\u221a 2. We can therefore construct the Sx+ ket as follows: 1\u221a 2 |Sx; + = 1\u221a 2 |+ + ei\u03b41 |\u2212, (1.101) (1.102) with \u03b41 real. In writing (1.102) we have used the fact that the overall phase (common to both |+ and |\u2212) of a state ket is immaterial; the coef\ufb01cient of |+ can be chosen to be real and positive by convention. The Sx\u2212 ket must be orthogonal to the Sx+ ket because the Sx+ alternative and Sx\u2212 alternative are mutually exclusive. This orthogonality requirement leads to |Sx; \u2212 = 1\u221a 2 |+ \u2212 1\u221a 2 ei\u03b41 |\u2212, (1.103) where we have, again, chosen the coef\ufb01cient of |+ to be real and positive by convention. We can now construct the operator Sx using (1.88) as follows: Sx = = \u00afh 2 \u00afh 2 [(|Sx; +Sx; +|) \u2212 (|Sx; \u2212Sx; \u2212|)] [e\u2212i\u03b41(|+\u2212|) +e i\u03b41(|\u2212+|)]. (1.104) 25 1.4 Measurements, Observables, and the Uncertainty Relations Notice that the Sx we have constructed is Hermitian, just as it must be. A similar argument with Sx replaced by Sy leads to |Sy; \u00b1 = 1\u221a 2 |+ \u00b1 1\u221a 2 ei\u03b42 |\u2212, Sy = \u00afh 2 [e\u2212i\u03b42(|+\u2212|) +e i\u03b42(|\u2212+|)]. (1.105) (1.106) Is there any way of determining \u03b41 and \u03b42? Actually there is one piece of information we have not yet used. Suppose we have a beam of spin 1 2 atoms moving in the z-direction. We can consider a sequential Stern\u2013Gerlach experiment with SG\u02c6x followed by SG\u02c6y. The results of such an experiment are completely analogous to the earlier case leading to (1.101): |Sy; \u00b1|Sx; +| = |Sy; \u00b1|Sx; \u2212| = 1\u221a 2, (1.107) which is not surprising in view of the", " invariance of physical systems under rotations. Inserting (1.103) and (1.105) into (1.107), we obtain |1 \u00b1 ei(\u03b41\u2212\u03b42)| = 1 2 1\u221a 2, (1.108) which is satis\ufb01ed only if \u03b42 \u2212 \u03b41 = \u03c0 or \u2212 \u03c0. (1.109) 2 2 We thus see that the matrix elements of Sx and Sy cannot all be real. If the Sx matrix elements are real, the Sy matrix elements must be purely imaginary (and vice versa). Just from this extremely simple example, the introduction of complex numbers is seen to be an essential feature in quantum mechanics. It is convenient to take the Sx matrix elements to be real14 and set \u03b41 = 0; if we were to choose \u03b41 = \u03c0, the positive x-axis would be oriented in the opposite direction. The second phase angle \u03b42 must then be \u2212\u03c0/2 or \u03c0/2. The fact that there is still an ambiguity of this kind is not surprising. We have not yet speci\ufb01ed whether the coordinate system we are using is right-handed or left-handed; given the x- and the z-axes there is still a twofold ambiguity in the choice of the positive y-axis. Later we will discuss angular momentum as a generator of rotations using the right-handed coordinate system; it can then be shown that \u03b42 = \u03c0/2 is the correct choice. To summarize, we have |Sx; \u00b1 = 1\u221a 2 |+ \u00b1 1\u221a 2 |\u2212, |Sy; \u00b1 = 1\u221a 2 |+ \u00b1 i\u221a 2 |\u2212, (1.110a) (1.110b) 14 This can always be done by adjusting arbitrary phase factors in the de\ufb01nition of |+ and |\u2212. This point will become clearer in Chapter 3, where the behavior of |\u00b1 under rotations will be discussed. 26 Fundamental Concepts and Sx = \u00afh 2 [(|+\u2212|) + (|\u2212+|)], Sy = \u00afh 2 [\u2212i(|+\u2212|) +i (|\u2212+|)]. (1.111a) (1.111b) The Sx\u00b1 and Sy\u00b1 eigenkets given here are seen to be in agreement with our earlier guesses (1.9) and (1.14) based on an analogy with linearly and circularly polarized light. (Note, in this comparison, that only the relative phase between the |+ and \u2212| components is of physical signi\ufb01cance.) Furthermore, the non-Hermitian S\u00b1 operators de\ufb01ned by (1.92) can now be written as S\u00b1 = Sx \u00b1 iSy. (1.112) The operators Sx and Sy, together with Sz given earlier, can be readily shown to satisfy the commutation relations and the anticommutation relations [Si, Sj] =i \u03b5ijk \u00afhSk, {Si, Sj} = 1 2 \u00afh2\u03b4ij, where the commutator [, ] and the anticommutator {, } are de\ufb01ned by [A, B] \u2261 AB \u2212 BA, (1.113) (1.114) (1.115a) {A, B} \u2261 AB + BA. (1.115b) (We make use of the totally antisymmetric symbol \u03b5ijk which has the value +1 for \u03b5123 and any cyclic permutation of indices, the value \u22121 for \u03b5213 and any cyclic permutation of indices, and the value 0 when any two indices are the same. We also make use of the implied summation convention, that is the assumption that we perform a summation over any pair of repeated indices.) The commutation relations in (1.113) will be recognized as the simplest realization of the angular-momentum commutation relations, whose signi\ufb01cance will be discussed in detail in Chapter 3. In contrast, the anticommutation relations in (1.114) turn out to be a special property of spin 1 2 systems. We can also de\ufb01ne the operator S \u00b7 S, orS 2 for short, as follows: x + S2 Because of (1.114), this operator turns out to be just a constant multiple of the identity operator y + S2 z. (1.116) S2 \u2261 S2 3 4 S2 = \u00afh2. We obviously have [S2, Si] =0. (1.117) (1.118) As will be shown in Chapter 3, for spins higher than 1 identity operator; however, (1.118) still holds. 2, S2 is", " no longer a multiple of the 27 1.4 Measurements, Observables, and the Uncertainty Relations 1.4.3 Compatible Observables Returning now to the general formalism, we will discuss compatible versus incompatible observables. Observables A and B are de\ufb01ned to be compatible when the corresponding operators commute, and incompatible when [A, B] =0, [A, B] = 0. (1.119) (1.120) For example, S2 and Sz are compatible observables, while Sx and Sz are incompatible observables. Let us \ufb01rst consider the case of compatible observables A and B. As usual, we assume that the ket space is spanned by the eigenkets of A. We may also regard the same ket space as being spanned by the eigenkets of B. We now ask: How are the A eigenkets related to the B eigenkets when A and B are compatible observables? Before answering this question we must touch upon a very important point we have bypassed earlier, the concept of degeneracy. Suppose there are two (or more) linearly independent eigenkets of A having the same eigenvalue; then the eigenvalues of the two eigenkets are said to be degenerate. In such a case the notation |a that labels the eigenket by its eigenvalue alone does not give a complete description; furthermore, we may recall that our earlier theorem on the orthogonality of different eigenkets was proved under the assumption of no degeneracy. Even worse, the whole concept that the ket space is spanned by {|a} appears to run into dif\ufb01culty when the dimensionality of the ket space is larger than the number of distinct eigenvalues of A. Fortunately, in practical applications in quantum mechanics, it is usually the case that in such a situation the eigenvalues of some other commuting observable, say B, can be used to label the degenerate eigenkets. Now we are ready to state an important theorem. Theorem 2 Suppose that A and B are compatible observables, and the eigenvalues of A are nondegenerate. Then the matrix elements a|B|a are all diagonal. (Recall here that the matrix elements of A are already diagonal if {|a} are used as the base kets.) Proof The proof of this important theorem is extremely simple. Using the de\ufb01nition (1.119) of compatible observables, we observe that a|[A, B]|a = (a \u2212 a)a|B|a = 0. So a|B|a must vanish unless a = a, which proves our assertion. We can write the matrix elements of B as a|B|a = \u03b4aa a|B|a. (1.121) (1.122) So both A and B can be represented by diagonal matrices with the same set of base kets. Using (1.71) and (1.122) we can write B as B = \u2211 a |aa|B|aa|. (1.123) 28 Fundamental Concepts Suppose that this operator acts on an eigenket of A: B|a = \u2211 a |aa|B|aa|a = (a|B|a)|a. (1.124) But this is nothing other than the eigenvalue equation for the operator B with eigenvalue b \u2261 a|B|a. (1.125) The ket |a is therefore a simultaneous eigenket of A and B. Just to be impartial to both operators, we may use |a, b to characterize this simultaneous eigenket. We have seen that compatible observables have simultaneous eigenkets. Even though the proof given is for the case where the A eigenkets are nondegenerate, the statement holds even if there is an n-fold degeneracy, that is, A|a(i) = a|a(i) (1.126) where |a(i) are n mutually orthonormal eigenkets of A, all with the same eigenvalue a. To see this, all we need to do is construct appropriate linear combinations of |a(i) that diagonalize the B operator by following the diagonalization procedure to be discussed in Section 1.5. i = 1, 2,..., n for A simultaneous eigenket of A and B, denoted by |a, b, has the property A|a, b = a|a, b, B|a, b = b|a, b. (1.127a) (1.127b) When there is no degeneracy, this notation is somewhat super\ufb02uous because it is clear from (1.125) that if we specify a,", " we necessarily know the b that appears in |a, b. The notation |a, b is much more powerful when there are degeneracies. A simple example may be used to illustrate this point. Even though a complete discussion of orbital angular momentum will not appear in this book until Chapter 3, the reader may be familiar from his or her earlier training in elementary wave mechanics that the eigenvalues of L2 (orbital angular momentum squared) and Lz (the z-component of orbital angular momentum) are \u00afh2l(l + 1) and ml \u00afh, respectively, with l an integer and ml = \u2212l, \u2212l+1,..., +l. To characterize an orbital angular momentum state completely, it is necessary to specify both l and ml. For example, if we just say l = 1, the ml value can still be 0, +1, or \u22121; if we just say ml = 1, l can be 1, 2, 3, 4, and so on. Only by specifying both l and ml do we succeed in uniquely characterizing the orbital angular momentum state in question. Quite often a collective index K is used to stand for (a, b), so that |K = |a, b. (1.128) We can obviously generalize our considerations to a situation where there are several (more than two) mutually compatible observables, namely, [A, B] = [B, C] = [A, C] =\u00b7 \u00b7 \u00b7 = 0. (1.129) Assume that we have found a maximal set of commuting observables; that is, we cannot add any more observables to our list without violating (1.129). The eigenvalues 29 1.4 Measurements, Observables, and the Uncertainty Relations of individual operators A, B, C,... may have degeneracies, but if we specify a combination (a, b, c,...), then the corresponding simultaneous eigenket of A, B, C,... is uniquely speci\ufb01ed. We can again use a collective index K to stand for (a, b, c,...). The orthonormality relation for |K = |a, b, c,... reads K|K = \u03b4KK = \u03b4aa \u03b4bb \u03b4cc..., while the completeness relation, or closure, can be written as (1.130) (1.131) \u2211 K |KKa, b, c,...a, b, c,... | = 1. (1.132) We now consider measurements of A and B when they are compatible observables. Suppose we measure A \ufb01rst and obtain result a. Subsequently, we may measure B and get result b. Finally we measure A again. It follows from our measurement formalism that the third measurement always gives a with certainty, that is, the second (B) measurement does not destroy the previous information obtained in the \ufb01rst (A) measurement. This is rather obvious when the eigenvalues of A are nondegenerate: |\u03b1 A measurement \u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2192 |a, b B measurement \u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2192 |a, b A measurement \u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2192 |a, b. (1.133) When there is degeneracy, the argument goes as follows: After the \ufb01rst (A) measurement, which yields a, the system is thrown into some linear combination n \u2211 i a |a, b(i), c(i) (1.134) where n is the degree of degeneracy and the kets |a, b(i) all have the same eigenvalue a as far as operator A is concerned. The second (B) measurement may select just one of the terms in the linear combination (1.134), say, |a, b( j), but the third (A) measurement applied to it still yields a. Whether or not there is degeneracy, A measurements and B measurements do not interfere. The term compatible is indeed deemed appropriate. 1.4.4 Incompatible Observables We now turn to incompatible observables, which are more nontrivial. The \ufb01rst point to be emphasized is that incompatible observables do not have a complete set of simultaneous eigenkets. To show this let us assume the converse to be true. There would then exist a set of simultaneous eigenkets with property (1.127a) and (1.127b). Clearly, Likewise, AB|a, b = Ab|a, b = ab|a, b. BA|a, b = Ba|a, b = ab|a, b; (1.135) (1.136) 30 Fundamental Concepts a b A B A (a) a = b b a (b", "s all others, and the third (C) \ufb01lter selects some particular |c and rejects all others. We are interested in the probability of obtaining |c when the beam coming out of the \ufb01rst \ufb01lter is normalized to unity. Because the probabilities are multiplicative, we obviously have |c|b|2|b|a|2. (1.138) Now let us sum over b to consider the total probability for going through all possible b routes. Operationally this means that we \ufb01rst record the probability of obtaining c with all but the \ufb01rst b route blocked, then we repeat the procedure with all but the second b blocked, and so on; then we sum the probabilities at the end and obtain \u2211 b |c|b|2|b|a|2 = \u2211 b c|bb|aa|bb|c. (1.139) 31 1.4 Measurements, Observables, and the Uncertainty Relations We now compare this with a different arrangement, where the B \ufb01lter is absent (or not operative); see Figure 1.8b. Clearly, the probability is just |c|a|2, which can also be written as follows: |c|a|2 = |\u2211 b c|bb|a|2 = \u2211 b c|bb|aa|bb|c. \u2211 b (1.140) Notice that expressions (1.139) and (1.140) are different! This is remarkable because in both cases the pure |a beam coming out of the \ufb01rst (A) \ufb01lter can be regarded as being made up of the B eigenkets |a = \u2211 b |bb|a, (1.141) where the sum is over all possible values of b. The crucial point to be noted is that the result coming out of the C \ufb01lter depends on whether or not B measurements have actually been carried out. In the \ufb01rst case we experimentally ascertain which of the B eigenvalues are actually realized; in the second case, we merely imagine |a to be built up of the various |b in the sense of (1.141). Put in another way, actually recording the probabilities of going through the various b routes makes all the difference even though we sum over b afterwards. Here lies the heart of quantum mechanics. Under what conditions do the two expressions become equal? It is left as an exercise for the reader to show that for this to happen, in the absence of degeneracy, it is suf\ufb01cient that [A, B] =0 or [B, C] =0. (1.142) In other words, the peculiarity we have illustrated is characteristic of incompatible observables. 1.4.5 The Uncertainty Relation The last topic to be discussed in this section is the uncertainty relation. Given an observable A, we de\ufb01ne an operator \u0394A \u2261 A \u2212 A, (1.143) where the expectation value is to be taken for a certain physical state under consideration. The expectation value of (\u0394A)2 is known as the dispersion of A. Because we have (\u0394A)2 = (A2 \u2212 2AA + A2) = A2 \u2212 A2, (1.144) the last equality of (1.144) may be taken as an alternative de\ufb01nition of dispersion. Sometimes the terms variance and mean square deviation are used for the same quantity. Clearly, the dispersion vanishes when the state in question is an eigenstate of A. Roughly speaking, the dispersion of an observable characterizes \u201cfuzziness.\u201d For example, for the Sz+ state of a spin 1 2 system, the dispersion of Sx can be computed to be \u2212 Sx2 = \u00afh2/4. In contrast the dispersion (\u0394Sz)2 obviously vanishes for the Sz+ state. So, for the Sz+ state, Sz is \u201csharp,\u201d a vanishing dispersion for Sz, while Sx is fuzzy. (1.145) S2 x 32 Fundamental Concepts We now state the uncertainty relation, which is the generalization of the well-known x-p uncertainty relation to be discussed in Section 1.6. Let A and B be observables. Then for any state we must have the following inequality: (\u0394A)2(\u0394B)2 \u2265 1 4 |[A, B]|2. (1.146) To prove this we \ufb01rst state three lemmas. Lemma 1 The Schwarz inequality \u03b1|\u03b1\u03b2|\u03b2 \u2265 |\u03b1|\u03b2|2, |a|2|b|2 \u2265 |a \u00b7 b|2 (1.147) (1", ".148) which is analogous to in real Euclidian space. Proof First note (\u03b1| + \u03bb\u2217\u03b2|) \u00b7 (|\u03b1 + \u03bb|\u03b2) \u2265 0, (1.149) where \u03bb can be any complex number. This inequality must hold when \u03bb is set equal to \u2212\u03b2|\u03b1/\u03b2|\u03b2: \u03b1|\u03b1\u03b2|\u03b2 \u2212 |\u03b1|\u03b2|2 \u2265 0, (1.150) which is the same as (1.147). Lemma 2 The expectation value of a Hermitian operator is purely real. Proof The proof is trivial, just use (1.75). Lemma 3 The expectation value of an anti-Hermitian operator, de\ufb01ned by C = \u2212C\u2020, is purely imaginary. Proof The proof is also trivial. Armed with these lemmas, we are in a position to prove the uncertainty relation (1.146). Using Lemma 1 with |\u03b1 = \u0394A|, |\u03b2 = \u0394B|, (1.151) where the blank ket | emphasizes the fact that our consideration may be applied to any ket, we obtain (\u0394A)2(\u0394B)2 \u2265 |\u0394A\u0394B|2, (1.152) 33 1.5 Change of Basis where the Hermiticity of \u0394A and \u0394B has been used. To evaluate the right-hand side of (1.152), we note \u0394A\u0394B = 1 2 [\u0394A, \u0394B] + {\u0394A, \u0394B}, 1 2 (1.153) where the commutator [\u0394A, \u0394B], which is equal to [A, B], is clearly anti-Hermitian ([A, B])\u2020 = (AB \u2212 BA)\u2020 = BA \u2212 AB = \u2212 [A, B]. (1.154) In contrast, the anticommutator {\u0394A, \u0394B} is obviously Hermitian, so \u0394A\u0394B = 1 2 [A, B] purely imaginary + 1 2 {\u0394A, \u0394B}, purely real (1.155) where Lemmas 2 and 3 have been used. The right-hand side of (1.152) now becomes |\u0394A\u0394B|2 = |[A, B]|2 + 1 4 1 4 |{\u0394A, \u0394B}|2. (1.156) The proof of (1.146) is now complete because the omission of the second (the anticommutator) term of (1.156) can only make the inequality relation stronger.15 Applications of the uncertainty relation to spin 1 2 systems will be left as exercises. We come back to this topic when we discuss the fundamental x-p commutation relation, that is, the Heisenberg uncertainty principle, in Section 1.6. 1.5 Change of Basis 1.5.1 Transformation Operator Suppose we have two incompatible observables A and B. The ket space in question can be viewed as being spanned either by the set {|a} or by the set {|b}. For example, for 2 systems |S\u02c6x\u00b1 may be used as our base kets; alternatively, |Sz\u00b1 may be used as spin 1 our base kets. The two different sets of base kets, of course, span the same ket space. We are interested in \ufb01nding out how the two descriptions are related. Changing the set of base kets is referred to as a change of basis or a change of representation. The basis in which the base eigenkets are given by {|a} is called the A representation or, sometimes, the A diagonal representation because the square matrix corresponding to A is diagonal in this basis. Our basic task is to construct a transformation operator that connects the old orthonormal set {|a} and the new orthonormal set {|b}. To this end, we \ufb01rst show the following. Theorem 3 Given two sets of base kets, both satisfying orthonormality and completeness, there exists a unitary operator U such that |b(1) = U|a(1), |b(2) = U|a(2),..., |b( N ) = U|a( N ). (1.157) 15 In the literature most authors use \u0394A for our (\u0394A)2 so the uncertainty relation is written as \u0394A\u0394B \u2265 |[A, B]|. In this book, however, \u0394A and \u0394B are to be understood as operators [see (1.143)], not numbers. 1 2 34 Fundamental Concepts By a unitary operator we mean an operator ful\ufb01lling the conditions as well as U \u2020U = 1 UU \u2020 = 1. (1.158", ") (1.159) Proof We prove this theorem by explicit construction. We assert that the operator U = \u2211 |b(k)a(k)| (1.160) will do the job and we apply this U to |a( l ). Clearly, k U|a( l ) = |b( l ) (1.161) is guaranteed by the orthonormality of {|a}. Furthermore, U is unitary: U \u2020U = \u2211 \u2211 |a( l )b( l )|b(k)a(k)| = \u2211 |a(k)a(k)| = 1, (1.162) k l k where we have used the orthonormality of {|b} and the completeness of {|a}. We obtain relation (1.159) in an analogous manner. 1.5.2 Transformation Matrix It is instructive to study the matrix representation of the U operator in the old {|a} basis. We have a(k)|U|a( l ) = a(k)|b( l ), (1.163) which is obvious from (1.161). In other words, the matrix elements of the U operator are built up of the inner products of old base bras and new base kets. We recall that the rotation matrix in three dimensions that changes one set of unit base vectors (\u02c6x, \u02c6y, \u02c6z) into another set (\u02c6x, \u02c6y, \u02c6z) can be written as (Goldstein et al. (2002), pp. 134\u2013144 for example) \u02c6x \u00b7 \u02c6x \u02c6y \u00b7 \u02c6x \u02c6z \u00b7 \u02c6x \u02c6x \u00b7 \u02c6y \u02c6y \u00b7 \u02c6y \u02c6z \u00b7 \u02c6y \u02c6x \u00b7 \u02c6z \u02c6y \u00b7 \u02c6z \u02c6z \u00b7 \u02c6z (1.164) \u23a0. R = \u239e \u239d \u239b The square matrix made up of a(k)|U|a( l ) is referred to as the transformation matrix from the {|a} basis to the {|b} basis. Given an arbitrary ket |\u03b1 whose expansion coef\ufb01cients a|\u03b1 are known in the old basis, |\u03b1 = \u2211 a |aa|\u03b1, (1.165) how can we obtain b|\u03b1, the expansion coef\ufb01cients in the new basis? The answer is very simple: Just multiply (1.165) (with a replaced by a( l ) to avoid confusion) by b(k)| b(k)|a( l )a( l )|\u03b1 = \u2211 a(k)|U \u2020|a( l )a( l )|\u03b1. b(k)|\u03b1 = \u2211 (1.166) l l 35 1.5 Change of Basis In matrix notation, (1.166) states that the column matrix for |\u03b1 in the new basis can be obtained just by applying the square matrix U \u2020 to the column matrix in the old basis: (new) = (U \u2020)(old). (1.167) The relationships between the old matrix elements and the new matrix elements are also easy to obtain: b(k)|X |b( l ) = \u2211 \u2211 b(k)|a(m)a(m)|X |a(n)a(n)|b( l ) m n = \u2211 \u2211 a(k)|U \u2020|a(m)a(m)|X |a(n)a(n)|U|a( l ). (1.168) m n This is simply the well-known formula for a similarity transformation in matrix algebra, X = U \u2020XU. (1.169) The trace of an operator X is de\ufb01ned as the sum of diagonal elements: tr(X ) = \u2211 a a|X |a. (1.170) Even though a particular set of base kets is used in the de\ufb01nition, tr(X) turns out to be independent of representation, as shown: \u2211 a|X ||X |b. b \u2211 b b|bb|X |b a|bb|X |bb|a (1.171) \u2211 a We can also prove tr(XY ) =tr(YX ), tr(U \u2020XU) =tr(X ), tr(|aa|) = \u03b4a a, tr(|ba|) =a |b. 1.5.3 Diagonalization (1.172a) (1.172b) (1.172c) (1.172d) So far we have not discussed", " how to \ufb01nd the eigenvalues and eigenkets of an operator B whose matrix elements in the old {|a} basis are assumed to be known. This problem turns out to be equivalent to that of \ufb01nding the unitary matrix that diagonalizes B. Even though the reader may already be familiar with the diagonalization procedure in matrix algebra, it is worth working out this problem using the Dirac bra-ket notation. We are interested in obtaining the eigenvalue b and the eigenket |b with the property B|b = b|b. (1.173) 36 Fundamental Concepts First, we rewrite this as a|B|aa|b = ba|b. \u2211 a (1.174) When |b in (1.173) stands for the lth eigenket of operator B, we can write (1.174) in matrix notation as follows: \u239b \u239c \u239d B11 B12 B13 B21 B22 B23.................. \u239e \u239f \u23a0 \u239b \u239c \u239c \u239d C( l ) 1 C( l ) 2... \u239e \u239b \u239e \u239f \u239f \u23a0 = b( l ) \u239c \u239c \u239d C( l ) 1 C( l ) 2... \u239f \u239f \u23a0, (1.175) (1.176a) (1.176b) with and Bij = a(i)|B|a( j), C( l ) k = a(k)|b( l ), where i, j, k run up to N, the dimensionality of the ket space. As we know from linear algebra, nontrivial solutions for C( l ) k are possible only if the characteristic equation det(B \u2212 \u03bb1) =0 (1.177) is satis\ufb01ed. This is an Nth order algebraic equation for \u03bb, and the N roots obtained are to be identi\ufb01ed with the various b( l ) we are trying to determine. Knowing b( l ) we can solve for the corresponding C( l ) k up to an overall constant to be determined from the normalization condition. Comparing (1.176b) with (1.163), we see that the C( l ) are just the elements of k the unitary matrix involved in the change of basis {|a} \u2192 {|b}. For this procedure the Hermiticity of B is important. For example, consider S+ de\ufb01ned by (1.92) or (1.112). This operator is obviously non-Hermitian. The corresponding matrix, which reads in the Sz basis as. = \u00afh S+ 0 1 0 0, (1.178) cannot be diagonalized by any unitary matrix. In Chapter 2 we will encounter eigenkets of a non-Hermitian operator in connection with a coherent state of a simple harmonic oscillator. Such eigenkets, however, are known not to form a complete orthonormal set, and the formalism we have developed in this section cannot be immediately applied. 1.5.4 Unitary Equivalent Observables We conclude this section by discussing a remarkable theorem on the unitary transform of an observable. Theorem 4 Consider again two sets of orthonormal basis {|a} and {|b} connected by the U operator (1.160). Knowing U, we may construct a unitary transform of A, 37 1.6 Position, Momentum, and Translation UAU\u22121; then A and UAU\u22121 are said to be unitary equivalent observables. The eigenvalue equation for A, clearly implies that But this can be rewritten as A|a( l ) = a( l )|a( l ), UAU\u22121U|a( l ) = a( l )U|a( l ). (UAU\u22121)|b( l ) = a( l )|b( l ). (1.179) (1.180) (1.181) This deceptively simple result is quite profound. It tells us that the |b are eigenkets of UAU\u22121 with exactly the same eigenvalues as the A eigenvalues. In other words, unitary equivalent observables have identical spectra. The eigenket |b( l ), by de\ufb01nition, satis\ufb01es the relationship B|b( l ) = b( l )|b( l ). (1.182) Comparing (1.181) and (1.182), we infer that B and UAU\u22121 are simultaneously diagonalizable. A natural question is, is UAU\u22121 the same as B itself", "? The answer quite often is yes in cases of physical interest. Take, for example, Sx and Sz. They are related by a unitary operator, which, as we will discuss in Chapter 3, is actually the rotation operator around the y-axis by angle \u03c0/2. In this case Sx itself is the unitary transform of Sz. Because we know that Sx and Sz exhibit the same set of eigenvalues, namely, +\u00afh/2 and \u2212\u00afh/2, we see that our theorem holds in this particular example. 1.6 Position, Momentum, and Translation 1.6.1 Continuous Spectra The observables considered so far have all been assumed to exhibit discrete eigenvalue spectra. In quantum mechanics, however, there are observables with continuous eigenvalues. Take, for instance, pz, the z-component of momentum. In quantum mechanics this is again represented by a Hermitian operator. In contrast to Sz, however, the eigenvalues of pz (in appropriate units) can assume any real value between \u2212\u221e and \u221e. The rigorous mathematics of a vector space spanned by eigenkets that exhibit a continuous spectrum is rather treacherous. The dimensionality of such a space is obviously in\ufb01nite. Fortunately, many of the results we worked out for a \ufb01nite-dimensional vector space with discrete eigenvalues can immediately be generalized. In places where straightforward generalizations do not hold, we indicate danger signals. We start with the analogue of eigenvalue equation (1.19), which, in the continuous spectrum case, is written as \u03be|\u03be = \u03be|\u03be, (1.183) 38 Fundamental Concepts where \u03be is an operator and \u03be is simply a number. The ket |\u03be is, in other words, an eigenket of operator \u03be with eigenvalue \u03be, just as |a is an eigenket of operator A with eigenvalue a. In pursuing this analogy we replace the Kronecker symbol by Dirac\u2019s \u03b4-function, a discrete sum over the eigenvalues {a} by an integral over the continuous variable \u03be, so a|a = \u03b4a a \u2192 \u03be|\u03be = \u03b4(\u03be \u2212 \u03be), (1.184a) |aa| = 1 \u2192 \u2211 a |\u03b1 = \u2211 a |a|\u03b1|2 = 1 \u2192 |aa|\u03b1 \u2192 |\u03b1 = d\u03be|\u03be\u03be| = 1, d\u03be|\u03be\u03be|\u03b1, d\u03be|\u03be|\u03b1|2 = 1, \u2211 a \u03b2|\u03b1 = \u2211 a \u03b2|aa|\u03b1 \u2192 \u03b2|\u03b1 = d\u03be\u03b2|\u03be\u03be|\u03b1, (1.184e) (1.184b) (1.184c) (1.184d) a|A|a = a\u03b4a a \u2192 \u03be|\u03be|\u03be = \u03be\u03b4(\u03be \u2212 \u03be). (1.184f) Notice in particular how the completeness relation (1.184b) is used to obtain (1.184c) and (1.184e). 1.6.2 Position Eigenkets and Position Measurements In Section 1.4 we emphasized that a measurement in quantum mechanics is essentially a \ufb01ltering process. To extend this idea to measurements of observables exhibiting continuous spectra it is best to work with a speci\ufb01c example. To this end we consider the position (or coordinate) operator in one dimension. The eigenkets |x of the position operator x satisfying x|x = x|x (1.185) are postulated to form a complete set. Here x is just a number with the dimension of length 0.23 cm, for example, while x is an operator. The state ket for an arbitrary physical state can be expanded in terms of {|x}: |\u03b1 = dx|xx|\u03b1. (1.186) \u221e \u2212\u221e We now consider a highly idealized selective measurement of the position observable. Suppose we place a very tiny detector that clicks only when the particle is precisely at x and nowhere else. Immediately after the detector clicks, we can say that the state in question is represented by |x. In other words, when the detector clicks, |\u03b1 abruptly \u201cjumps into\u201d |x in much the same way as an arbitrary spin state jumps into the Sz+ (or Sz\u2212) state when subjected to an SG apparatus of the Sz type. In practice the best the detector can do is to locate the particle within a narrow interval around x. A realistic detector clicks when a particle is observed to be located within some 39 1.6 Position, Momentum, and Translation narrow range (x \u2212 \u0394/2,", " x + \u0394/2). When a count is registered in such a detector, the state ket changes abruptly as follows: |\u03b1 = \u221e \u2212\u221e dx|xx|\u03b1 measurement \u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2192 x+\u0394/2 x\u2212\u0394/2 dx|xx|\u03b1. (1.187) Assuming that x|\u03b1 does not change appreciably within the narrow interval, probability for the detector to click is given by the |x|\u03b1|2dx, (1.188) where we have written dx for \u0394. This is analogous to |a|\u03b1|2 for the probability for |\u03b1 to be thrown into |a when A is measured. The probability of recording the particle somewhere between \u2212\u221e and \u221e is given by \u221e \u2212\u221e dx|x|\u03b1|2, (1.189) which is normalized to unity if |\u03b1 is normalized: \u03b1|\u03b1 = 1 \u21d2 \u221e \u2212\u221e dx\u03b1|xx|\u03b1 = 1. (1.190) The reader familiar with wave mechanics may have recognized by this time that x|\u03b1 is the wave function for the physical state represented by |\u03b1. We will say more about this identi\ufb01cation of the expansion coef\ufb01cient with the x-representation of the wave function in Section 1.7. The notion of a position eigenket can be extended to three dimensions. It is assumed in nonrelativistic quantum mechanics that the position eigenkets |x are complete. The state ket for a particle with internal degrees of freedom, such as spin, ignored can therefore be expanded in terms of {|x} as follows: |\u03b1 = d3x|xx|\u03b1, (1.191) where x stands for x, y, andz ; in other words, |x is a simultaneous eigenket of the observables x, y, andz in the sense of Section 1.4: |x \u2261 |x, y, z, x|x = x|x, y|x = y|x, z|x = z|x. (1.192a) (1.192b) To be able to consider such a simultaneous eigenket at all, we are implicitly assuming that the three components of the position vector can be measured simultaneously to arbitrary degrees of accuracy; hence, we must have [xi, xj] =0, (1.193) where x1, x2, and x3 stand for x, y, andz, respectively. 40 Fundamental Concepts 1.6.3 Translation We now introduce the very important concept of translation, or spatial displacement. Suppose we start with a state that is well localized around x. Let us consider an operation that changes this state into another well-localized state, this time around x + dx with everything else (for example, the spin direction) unchanged. Such an operation is de\ufb01ned to be an in\ufb01nitesimal translation by dx, and the operator that does the job is denoted by J (dx): J (dx)|x = |x + dx, (1.194) where a possible arbitrary phase factor is set to unity by convention. Notice that the righthand side of (1.194) is again a position eigenket, but this time with eigenvalue x + dx. Obviously |x is not an eigenket of the in\ufb01nitesimal translation operator. By expanding an arbitrary state ket |\u03b1 in terms of the position eigenkets we can examine the effect of in\ufb01nitesimal translation on |\u03b1: |\u03b1 \u2192J (dx)|\u03b1 = J (dx) d3x|xx|\u03b1 = d3x|x + dxx|\u03b1. (1.195) We also write the right-hand side of (1.195) as d3x|x + dxx|\u03b1 = d3x|xx \u2212 dx|\u03b1 (1.196) because the integration is over all space and x is just an integration variable. This shows that the wave function of the translated state J (dx)|\u03b1 is obtained by substituting x \u2212 dx for x in x|\u03b1. There is an equivalent approach to translation that is often treated in the literature. Instead of considering an in\ufb01nitesimal translation of the physical system itself, we consider a change in the coordinate system being used such that the origin is shifted in the opposite direction, \u2212dx. Physically, in this alternative approach we are asking how the same state ket would look to another observer whose coordinate system is shifted by \u2212dx. In this book we try not to use this approach. Obviously it is important that we do not mix the two approaches! We now list the properties of the in\ufb01nitesimal translation operator J (\u2212dx). The \ufb01rst property we demand is the unitarity property imposed", " by probability conservation. It is reasonable to require that if the ket |\u03b1 is normalized to unity, the translated ket J (dx)|\u03b1 also be normalized to unity, so \u03b1|\u03b1 = \u03b1|J \u2020(dx)J (dx)|\u03b1. (1.197) This condition is guaranteed by demanding that the in\ufb01nitesimal translation be unitary: J \u2020(dx)J (dx) =1. (1.198) Quite generally, the norm of a ket is preserved under unitary transformations. For the second property, suppose we consider two successive in\ufb01nitesimal translations, \ufb01rst by dx and subsequently by dx, wheredx and dx need not be in the same direction. We expect 41 1.6 Position, Momentum, and Translation the net result to be just a single translation operation by the vector sum dx + dx, so we demand that J (dx)J (dx) =J (dx + dx). (1.199) For the third property, suppose we consider a translation in the opposite direction; we expect the opposite-direction translation to be the same as the inverse of the original translation: J (\u2212dx) =J \u22121(dx). (1.200) For the fourth property, we demand that as dx \u2192 0, the translation operation reduce to the identity operation J (dx) =1 lim dx\u21920 (1.201) and that the difference between J (dx) and the identity operator be of \ufb01rst order in dx. We now demonstrate that if we take the in\ufb01nitesimal translation operator to be J (dx) =1 \u2212 iK \u00b7 dx, (1.202) where the components of K, Kx, Ky, and Kz, are Hermitian operators, then all the properties listed are satis\ufb01ed. The \ufb01rst property, the unitarity of J (dx), is checked as follows: J \u2020(dx)J (dx) = (1 + iK\u2020 \u00b7 dx)(1 \u2212 iK \u00b7 dx) = 1 \u2212 i(K \u2212 K\u2020) \u00b7 dx + 0[(dx)2] 1, (1.203) where terms of second order in dx have been ignored for an in\ufb01nitesimal translation. The second property (1.199) can also be proved as follows: J (dx)J (dx) = (1 \u2212 iK \u00b7 dx)(1 \u2212 iK \u00b7 dx) 1 \u2212 iK \u00b7(dx + dx) = J (dx + dx). (1.204) The third and fourth properties are obviously satis\ufb01ed by (1.202). Accepting (1.202) to be the correct form for J (dx), we are in a position to derive an extremely fundamental relation between the K operator and the x operator. First, note that xJ (dx)|x = x|x + dx = (x + dx)|x + dx (1.205a) and hence, J (dx)x|x = xJ (dx)|x = x|x + dx; (1.205b) [x, J (dx)] |x = dx|x + dx dx |x, (1.206) 42 Fundamental Concepts where the error made in writing the last part of (1.206) is of second order in dx. Now|x can be any position eigenket, and the position eigenkets are known to form a complete set. We must therefore have an operator identity or [x, J (dx)] = dx, \u2212ixK \u00b7 dx + iK \u00b7 dxx = dx, (1.207) (1.208) where on the right-hand sides of (1.207) and (1.208) dx is understood to be the number dx multiplied by the identity operator in the ket space spanned by |x. By choosing dx in the direction of \u02c6xj and forming the scalar product with \u02c6xi, we obtain [xi, Kj] =i \u03b4ij, (1.209) where again \u03b4ij is understood to be multiplied by the identity operator. 1.6.4 Momentum as a Generator of Translation Equation (1.209) is the fundamental commutation relation between the position operators x, y, z and the K operators Kx, Ky, Kz. Remember that so far the K operator is de\ufb01ned in terms of the in\ufb01nitesimal translation operator by (1.202). What is the physical signi\ufb01cance we can attach to K? J. Schwinger, lecturing on quantum mechanics, once remarked, \u201c... for fundamental properties we will borrow only names from classical physics.\u201d In the present case we would like to borrow from classical mechanics the notion", " that momentum is the generator of an in\ufb01nitesimal translation. An in\ufb01nitesimal translation in classical mechanics can be regarded as a canonical transformation, xnew \u2261 X = x + dx, pnew \u2261 P = p, obtainable from the generating function (Goldstein et al. (2002), pp. 386 and 403) F(x, P) =x \u00b7 P + p \u00b7 dx, (1.210) (1.211) where p and P refer to the corresponding momenta. This equation has a striking similarity to the in\ufb01nitesimal translation operator (1.202) in quantum mechanics, particularly if we recall that x \u00b7 P in (1.211) is the generating function for the identity transformation (X = x, P = p). We are therefore led to speculate that the operator K is in some sense related to the momentum operator in quantum mechanics. Can the K operator be identi\ufb01ed with the momentum operator itself? Unfortunately the dimension is all wrong; the K operator has the dimension of 1/length because K \u00b7 dx must be dimensionless. But it appears legitimate to set K = p universal constant with the dimension of action. (1.212) From the fundamental postulates of quantum mechanics there is no way to determine the actual numerical value of the universal constant. Rather, this constant is needed here 43 1.6 Position, Momentum, and Translation because, historically, classical physics was developed before quantum mechanics using units convenient for describing macroscopic quantities \u2013 the circumference of the Earth, the mass of 1 cm3 of water, the duration of a mean solar day, and so forth. Had microscopic physics been formulated before macroscopic physics, the physicists would have almost certainly chosen the basic units in such a way that the universal constant appearing in (1.212) would be unity. An analogy from electrostatics may be helpful here. The interaction energy between two particles of charge e separated at a distance r is proportional to e2/r; in unrationalized Gaussian units, the proportionality factor is just 1, but in rationalized mks units, which may be more convenient for electrical engineers, the proportionality factor is 1/4\u03c0\u03b50. (See Appendix A.) The universal constant that appears in (1.212) turns out to be the same as the constant \u00afh that appears in L. de Broglie\u2019s relation, written in 1924, 2\u03c0 \u03bb = where \u03bb is the wavelength of a \u201cparticle wave.\u201d In other words, the K operator is the quantum-mechanical operator that corresponds to the wave number, that is, 2\u03c0 times the reciprocal wavelength, usually denoted by k. With this identi\ufb01cation the in\ufb01nitesimal translation operator J (dx) reads (1.213) p \u00afh, J (dx) =1 \u2212 ip \u00b7 dx/\u00afh, (1.214) where p is the momentum operator. The commutation relation (1.209) now becomes [xi, pj] =i \u00afh\u03b4ij. (1.215) The commutation relations (1.215) imply, for example, that x and px (but not x and py) are incompatible observables. It is therefore impossible to \ufb01nd simultaneous eigenkets of x and px. The general formalism of Section 1.4 can be applied here to obtain the positionmomentum uncertainty relation of W. Heisenberg: (\u0394x)2(\u0394px)2 \u2265 \u00afh2/4. (1.216) Some applications of (1.216) will appear in Section 1.7. So far we have concerned ourselves with in\ufb01nitesimal translations. A \ufb01nite translation, that is, a spatial displacement by a \ufb01nite amount, can be obtained by successively compounding in\ufb01nitesimal translations. Let us consider a \ufb01nite translation in the xdirection by an amount \u0394x: J (\u0394x \u02c6x)|x = |x + \u0394x \u02c6x. (1.217) By compounding N in\ufb01nitesimal translations, each of which is characterized by a spatial displacement \u0394x/N in the x-direction, and letting N \u2192 \u221e, we obtain J (\u0394x \u02c6x) = lim N\u2192\u221e N 1 \u2212 ipx\u0394x N\u00afh \u2212 ipx\u0394x \u00afh = exp. (1.218) 44 Fundamental Concepts D A B \u0394y y\u02c6 C \u0394 x x\u02c6 Fig. 1.9 Successive translations in different directions. Here exp(\u2212ipx\u0394x/\u00afh) is understood to be a function of the operator px; generally, for any operator X we have exp", "(X ) \u2261 1 + X + X2 2! + \u00b7 \u00b7 \u00b7. (1.219) A fundamental property of translations is that successive translations in different directions, say in the x- andy -directions, commute. We see this clearly in Figure 1.9; in shifting from A and B it does not matter whether we go via C or via D. Mathematically, J (\u0394y \u02c6y)J (\u0394x \u02c6x) =J (\u0394x \u02c6x + \u0394y \u02c6y), J (\u0394x \u02c6x)J (\u0394y \u02c6y) =J (\u0394x \u02c6x + \u0394y \u02c6y). (1.220) This point is not so trivial as it may appear; we will show in Chapter 3 that rotations about different axes do not commute. Treating \u0394x and \u0394y up to second order, we obtain [J (\u0394y \u02c6y), J (\u0394x \u02c6x)] = 1 \u2212 ipy\u0394y \u00afh \u2212 y(\u0394y)2 p2 2\u00afh2 + \u00b7 \u00b7 \u00b7 1 \u2212 ipx\u0394x \u00afh \u2212 p2 x(\u0394x)2 2\u00afh2 + \u00b7 \u00b7 \u00b7 \u2212 (\u0394x)(\u0394y)[ py, px] \u00afh2. Because \u0394x and \u0394y are arbitrary, requirement (1.220), or [J (\u0394y \u02c6y), J (\u0394x \u02c6x)] = 0, immediately leads to or, more generally, [ px, py] =0, [ pi, pj] =0., (1.221) (1.222) (1.223) (1.224) 45 1.6 Position, Momentum, and Translation This commutation relation is a direct consequence of the fact that translations in different directions commute. Whenever the generators of transformations commute, the corresponding group is said to be Abelian. The translation group in three dimensions is Abelian. Equation (1.224) implies that px, py, andp z are mutually compatible observables. We can therefore conceive of a simultaneous eigenket of px, py, pz, namely, |p \u2261 |p x, p y, p z, x |p, px|p = p py|p = p It is instructive to work out the effect of J (dx) on such a momentum eigenket: 1 \u2212 ip \u00b7 dx \u00afh 1 \u2212 ip \u00b7 dx \u00afh J (dx)|p = pz|p = p |p = |p. |p, |p. y z (1.225a) (1.225b) (1.226) We see that the momentum eigenket remains the same even though it suffers a slight phase change, so unlike |x, |p is an eigenket of J (dx), which we anticipated because [p, J (dx)] = 0. (1.227) Notice, however, that the eigenvalue of J (dx) is complex; we do not expect a real eigenvalue here because J (dx), though unitary, is not Hermitian. 1.6.5 The Canonical Commutation Relations We summarize the commutator relations we inferred by studying the properties of translation: [xi, xj] =0, [ pi, pj] =0, [xi, pj] = i\u00afh\u03b4ij. (1.228) These relations form the cornerstone of quantum mechanics; in his book, P. A. M. Dirac (1958) calls them the \u201cfundamental quantum conditions.\u201d More often they are known as the canonical commutation relations, or the fundamental commutation relations. Historically it was W. Heisenberg who, in 1925, showed that the combination rule for atomic transition lines known at that time could best be understood if one associated arrays of numbers obeying certain multiplication rules with these frequencies. Immediately afterward M. Born and P. Jordan pointed out that Heisenberg\u2019s multiplication rules are essentially those of matrix algebra, and a theory was developed based on the matrix analogues of (1.228), which is now known as matrix mechanics.16 Also in 1925, P. A. M. Dirac observed that the various quantum-mechanical relations can be obtained from the corresponding classical relations just by replacing classical Poisson brackets by commutators, as follows: [, ]classical \u2192 [ ],, i\u00afh (1.229) 16 Appropriately, pq \u2212 qp = h/2\u03c0i is inscribed on the gravestone of M. Born in G\u00a8ottingen. 46 Fundamental Concepts where we may recall that the classical Poisson brackets are de\ufb01ned for functions", " of q and p as [A(q, p), B(q, p)]classical \u2261 \u2211 s For example, in classical mechanics, we have \u2202A \u2202qs \u2202B \u2202ps \u2212 \u2202A \u2202ps \u2202B \u2202qs. (1.230) [xi, pj]classical = \u03b4ij, (1.231) which in quantum mechanics turns into (1.215). Dirac\u2019s rule (1.229) is plausible because the classical Poisson brackets and quantummechanical commutators satisfy similar algebraic properties. In particular, the following relations can be proved regardless of whether [, ] is understood as a classical Poisson bracket or as a quantum-mechanical commutator: [A, A] =0 [A, B] =\u2212 [B, A] [A, c] =0 (c is just a number) [A + B, C] = [A, C] + [B, C] [A, BC] = [A, B] C + B [A, C] [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0, (1.232a) (1.232b) (1.232c) (1.232d) (1.232e) (1.232f) where the last relation is known as the Jacobi identity.17 However, there are important differences. First, the dimension of the classical Poisson bracket differs from that of the quantum-mechanical commutator because of the differentiations with respect to q and p appearing in (1.230). Second, the Poisson bracket of real functions of q and p is purely real, while the commutator of two Hermitian operators is anti-Hermitian (see Lemma 3 of Section 1.4). To take care of these differences the factor i\u00afh is inserted in (1.229). We have deliberately avoided exploiting Dirac\u2019s analogy in obtaining the canonical commutation relations. Our approach to the commutation relations is based solely on (1) the properties of translations and (2) the identi\ufb01cation of the generator of translation with the momentum operator modulo a universal constant with the dimension of action. We believe that this approach is more powerful because it can be generalized to situations where observables have no classical analogues. For example, the spin angular-momentum components we encountered in Section 1.4 have nothing to do with the p and q of classical mechanics; yet, as we will show in Chapter 3, the spin angular-momentum commutation relations can be derived using the properties of rotations just as we derived the canonical commutation relations using the properties of translations. 17 It is amusing that the Jacobi identity in quantum mechanics is much easier to prove than its classical analogue. 47 1.7 Wave Functions in Position and Momentum Space 1.7 Wave Functions in Position and Momentum Space 1.7.1 Position-Space Wave Function In this section we present a systematic study of the properties of wave functions in both position and momentum space. For simplicity let us return to the one-dimensional case. The base kets used are the position kets satisfying x|x = x|x, normalized in such a way that the orthogonality condition reads x|x = \u03b4(x \u2212 x). (1.233) (1.234) We have already remarked that the ket representing a physical state can be expanded in terms of |x, |\u03b1 = dx|xx|\u03b1, and that the expansion coef\ufb01cient x|\u03b1 is interpreted in such a way that |x|\u03b1|2 dx (1.235) (1.236) is the probability for the particle to be found in a narrow interval dx around x. In our formalism the inner product x|\u03b1 is what is usually referred to as the wave function \u03c8\u03b1(x) for state |\u03b1: x|\u03b1 = \u03c8\u03b1(x). (1.237) In elementary wave mechanics the probabilistic interpretations for the expansion coef\ufb01cient ca ( = a|\u03b1) and for the wave function \u03c8\u03b1(x)( = x|\u03b1) are often presented as separate postulates. One of the major advantages of our formalism, originally due to Dirac, is that the two kinds of probabilistic interpretations are uni\ufb01ed; \u03c8\u03b1(x) is an expansion coef\ufb01cient [see (1.235)] in much the same way as ca is. By following the footsteps of Dirac we come to appreciate the unity of quantum mechanics. Consider the inner product \u03b2|\u03b1. Using the completeness of |x, we have \u03b2|\u03b1 = dx\u03b2|xx|\u03b1 = dx\u03c8\ufffd", "\ufffd \u03b2(x)\u03c8\u03b1(x), (1.238) so \u03b2|\u03b1 characterizes the overlap between the two wave functions. Note that we are not de\ufb01ning \u03b2|\u03b1 as the overlap integral; the identi\ufb01cation of \u03b2|\u03b1 with the overlap integral follows from our completeness postulate for |x. The more general interpretation of \u03b2|\u03b1, independent of representations, is that it represents the probability amplitude for state |\u03b1 to be found in state |\u03b2. 48 Fundamental Concepts This time let us interpret the expansion |\u03b1 = \u2211 a |aa|\u03b1 (1.239) using the language of wave functions. We just multiply both sides of (1.239) by the position eigenbra x| on the left. Thus x|\u03b1 = \u2211 a x|aa|\u03b1. (1.240) In the usual notation of wave mechanics this is recognized as \u03c8\u03b1(x) = \u2211 a ca ua (x), where we have introduced an eigenfunction of operator A with eigenvalue a: ua (x) =x |a. Let us now examine how \u03b2|A|\u03b1 can be written using the wave functions for |\u03b1 and (1.241) |\u03b2. Clearly, we have \u03b2|A|\u03b1 = = dx dx dx\u03b2|xx|A|xx|\u03b1 dx\u03c8\u2217 \u03b2(x)x|A|x\u03c8\u03b1(x). (1.242) So to be able to evaluate \u03b2|A|\u03b1, we must know the matrix element x|A|x, which is, in general, a function of the two variables x and x. An enormous simpli\ufb01cation takes place if observable A is a function of the position operator x. In particular, consider A = x2, (1.243) which actually appears in the Hamiltonian for the simple harmonic oscillator problem to be discussed in Chapter 2. We have x|x2|x = (x|) \u00b7 (x2|x) =x 2\u03b4(x \u2212 x), (1.244) where we have used (1.233) and (1.234). The double integral (1.242) is now reduced to a single integral: In general, \u03b2|x2|\u03b1 = dx\u03b2|xx2x|\u03b1 = dx\u03c8\u2217 \u03b2(x)x2\u03c8\u03b1(x). \u03b2| f (x)|\u03b1 = dx\u03c8\u2217 \u03b2(x)f (x)\u03c8\u03b1(x). (1.245) (1.246) Note that the f (x) on the left-hand side of (1.246) is an operator, while the f (x) on the right-hand side is not an operator. 49 1.7 Wave Functions in Position and Momentum Space 1.7.2 Momentum Operator in the Position Basis We now examine how the momentum operator may look in the x-basis, that is, in the representation where the position eigenkets are used as base kets. Our starting point is the de\ufb01nition of momentum as the generator of in\ufb01nitesimal translations: 1 \u2212 ip\u0394x \u00afh |\u03b1 = dxJ (\u0394x)|xx|\u03b1 dx|x + \u0394xx|\u03b1 dx|xx \u2212 \u0394x|\u03b1 x|\u03b1 \u2212\u0394x \u2202 \u2202x dx|x x|\u03b1. = = = (1.247) (1.248) (1.249) Comparison of both sides yields p|\u03b1 = dx|x \u2212i\u00afh or \u2202 \u2202x x|\u03b1 x|p|\u03b1 = \u2212i\u00afh \u2202 \u2202x x|\u03b1, where we have used the orthogonality property (1.234). For the matrix element p in the x-representation, we obtain x|p|x = \u2212i\u00afh \u2202 \u2202x \u03b4(x \u2212 x). (1.250) From (1.248) we get a very important identity: \u03b2|p|\u03b1 = dx\u03b2|x = dx\u03c8\u2217 \u03b2(x) \u2212i\u00afh \u2212i\u00afh x|\u03b1 \u2202 \u2202x \u2202 \u2202x \u03c8\u03b1(x). (1.251) In our formalism (1.251) is not a postulate; rather, it has been derived using the basic properties of momentum. By repeatedly applying (1.249), we can also obtain x|pn|\u03b1 = (\u2212i\u00afh)n \u2202n \u2202xn x|\u03b1, \u03b2|pn|\u03b1 = dx\u03c8\u2217 \u03b2(x)(\u2212i\u00afh)n \u2202n \u2202xn \u03c8\u03b1(x). (1.252) (1.253) 1.7", ".3 Momentum-Space Wave Function So far we have worked exclusively in the x-basis. There is actually a complete symmetry between x and p, apart from occasional minus signs, which we can infer from the 50 Fundamental Concepts canonical commutation relations. Let us now work in the p-basis, that is, in the momentum representation. For simplicity we continue working in one-space. The base eigenkets in the p-basis specify and p|p = p|p p|p = \u03b4(p \u2212 p). (1.254) (1.255) The momentum eigenkets {|p} span the ket space in much the same way as the position eigenkets {|x}. An arbitrary state ket |\u03b1 can therefore be expanded as follows: |\u03b1 = dp|p p|\u03b1. (1.256) We can give a probabilistic interpretation for the expansion coef\ufb01cient p|\u03b1; the probability that a measurement of p gives eigenvalue p within a narrow interval dp is | p|\u03b1|2dp. It is customary to call p|\u03b1 the momentum-space wave function; the notation \u03c6\u03b1( p) is often used: p|\u03b1 = \u03c6\u03b1(p). If |\u03b1 is normalized, we obtain dp\u03b1|p p|\u03b1 = dp|\u03c6\u03b1(p)|2 = 1. (1.257) (1.258) Let us now establish the connection between the x-representation and the prepresentation. We recall that in the case of the discrete spectra, the change of basis from the old set {|a} to the new set {|b} is characterized by the transformation matrix (1.163). Likewise, we expect that the desired information is contained in x|p, which is a function of x and p, usually called the transformation function from the x-representation to the p-representation. To derive the explicit form of x|p, \ufb01rst recall (1.249); letting |\u03b1 be the momentum eigenket |p, we obtain or x|p|p = \u2212i\u00afh \u2202 \u2202x x|p px|p = \u2212i\u00afh \u2202 \u2202x x|p. The solution to this differential equation for x|p is ipx \u00afh x|p = N exp, (1.259) (1.260) (1.261) where N is the normalization constant to be determined in a moment. Even though the transformation function x|p is a function of two variables, x and p, we can temporarily regard it as a function of x with p \ufb01xed. It can then be viewed as the probability amplitude for the momentum eigenstate speci\ufb01ed by p to be found at position x; in other words, it is 51 1.7 Wave Functions in Position and Momentum Space just the wave function for the momentum eigenstate |p, often referred to as the momentum eigenfunction (still in the x-space). So (1.261) simply says that the wave function of a momentum eigenstate is a plane wave. It is amusing that we have obtained this plane wave solution without solving the Schr\u00a8odinger equation (which we have not yet written down). To get the normalization constant N let us \ufb01rst consider x|x = dpx|p p|x. (1.262) The left-hand side is just \u03b4(x \u2212 x); the right-hand side can be evaluated using the explicit form of x|p: \u03b4(x \u2212 x) =|N |2 dp exp ip(x \u2212 x) \u00afh = 2\u03c0\u00afh|N|2\u03b4(x \u2212 x). (1.263) Choosing N to be purely real and positive by convention, we \ufb01nally have x|p = 1\u221a 2\u03c0\u00afh exp ipx \u00afh. (1.264) We can now demonstrate how the position-space wave function is related to the momentum-space wave function. All we have to do is rewrite x|\u03b1 = dpx|p p|\u03b1 and as and p|\u03b1 = dx p|xx|\u03b1 \u03c8\u03b1(x) = 1\u221a 2\u03c0\u00afh dp exp ipx \u00afh \u03c6\u03b1(p) \u03c6\u03b1(p) = 1\u221a 2\u03c0\u00afh dx exp \u2212ipx \u00afh \u03c8\u03b1(x). (1.265a) (1.265b) (1.266a) (1.266b) The pair of equations is just what one expects from Fourier\u2019s inversion theorem. Apparently the mathematics we have developed somehow \u201cknows\u201d Fourier\u2019s work on integral transforms. 1.7", ".4 Gaussian Wave Packets It is instructive to look at a physical example to illustrate our basic formalism. We consider what is known as a Gaussian wave packet, whose x-space wave function is given by x|\u03b1 = exp ikx \u2212 x2 2d2 1 \u221a \u03c01/4 d. (1.267) 52 Fundamental Concepts This is a plane wave with wave number k modulated by a Gaussian pro\ufb01le centered on the origin. The probability of observing the particle vanishes very rapidly for |x| > d; more quantitatively, the probability density |x|\u03b1|2 has a Gaussian shape with width d. We now compute the expectation values of x, x2, p, and p2. The expectation value of x is clearly zero by symmetry: \u221e x = dx\u03b1|xxx|\u03b1 = \u221e \u2212\u221e dx|x|\u03b1|2x = 0. (1.268) \u2212\u221e For x2 we obtain x2 = = = dxx2|x|\u03b1|2 \u221e 1\u221a \u03c0d dx x2 exp \u2212\u221e \u221e \u2212\u221e d2 2, \u2212x2 d2 which leads to (\u0394x)2 = x2 \u2212 x2 = d2 2 (1.269) (1.270) for the dispersion of the position operator. The expectation values of p and p2 can also be computed as follows: p = \u00afhk p2 = \u00afh2 2d2 + \u00afh2k2, which is left as an exercise. The momentum dispersion is therefore given by (\u0394p)2 = p2 \u2212 p2 = \u00afh2 2d2. (1.271a) (1.271b) (1.272) Armed with (1.270) and (1.272), we can check the Heisenberg uncertainty relation (1.216); in this case the uncertainty product is given by = (\u0394p)2 (\u0394x)2 \u00afh2 4, (1.273) independent of d, so for a Gaussian wave packet we actually have an equality relation rather than the more general inequality relation (1.216). For this reason a Gaussian wave packet is often called a minimum uncertainty wave packet. We now go to momentum space. By a straightforward integration, just completing the square in the exponent, we obtain p|\u03b1 = 1\u221a 2\u03c0\u00afh = d \u221a \u00afh \u03c0 exp \u221e 1 \u221a \u03c01/4 \u2212\u221e d \u2212(p \u2212 \u00afhk)2d2 2\u00afh2 dx exp \u2212ipx \u00afh + ikx \u2212 x2 2d2. (1.274) 53 1.7 Wave Functions in Position and Momentum Space This momentum-space wave function provides an alternative method for obtaining p and p2, which is also left as an exercise. The probability of \ufb01nding the particle with momentum p is Gaussian (in momentum space) centered on \u00afhk, just as the probability of \ufb01nding the particle at x is Gaussian (in position space) centered on zero. Furthermore, the widths of the two Gaussians are inversely proportional to each other, which is just another way of expressing the constancy of the uncertainty product (\u0394x)2\u0394p)2 explicitly computed in (1.273). The wider the spread in the p-space, the narrower the spread in the x-space, and vice versa. As an extreme example, suppose we let d \u2192 \u221e. The position-space wave function (1.267) then becomes a plane wave extending over all space; the probability of \ufb01nding the particle is just constant, independent of x. In contrast, the momentum-space wave function is \u03b4-function-like and is sharply peaked at \u00afhk. In the opposite extreme, by letting d \u2192 0, we obtain a position-space wave function localized like the \u03b4-function, but the momentum-space wave function (1.274) is just constant, independent of p. We have seen that an extremely well-localized (in the x-space) state is to be regarded as a superposition of momentum eigenstates with all possible values of momenta. Even those momentum eigenstates whose momenta are comparable to or exceed mc must be included in the superposition. However, at such high values of momentum, a description based on nonrelativistic quantum mechanics is bound to break down.18 Despite this limitation our formalism, based on the existence of the position eigenket |x, has a wide domain of applicability. 1.7.5 Generalization to Three Dimensions So far in this section we have worked exclusively in one-space for simplicity, but everything we have done can be generalized to three-space", ", if the necessary changes are made. The base kets to be used can be taken as either the position eigenkets satisfying x|x = x|x or the momentum eigenkets satisfying p|p = p|p. They obey the normalization conditions and x|x = \u03b43(x \u2212 x) p|p = \u03b43(p \u2212 p), (1.275) (1.276) (1.277a) (1.277b) where \u03b43 stands for the three-dimensional \u03b4-function \u03b43(x \u2212 x) = \u03b4(x \u2212 x)\u03b4(y \u2212 y)\u03b4(z \u2212 z). (1.278) 18 It turns out that the concept of a localized state in relativistic quantum mechanics is far more intricate because of the possibility of \u201cnegative energy states,\u201d or pair creation. See Chapter 8 of this textbook. 54 Fundamental Concepts The completeness relations read and d3x|xx| = 1 d3p|pp| = 1, which can be used to expand an arbitrary state ket: |\u03b1 = d3x|xx|\u03b1, |\u03b1 = d3p|pp|\u03b1. (1.279a) (1.279b) (1.280a) (1.280b) The expansion coef\ufb01cients x|\u03b1 and p|\u03b1 are identi\ufb01ed with the wave functions \u03c8\u03b1(x) and \u03c6\u03b1(p) in position and momentum space, respectively. The momentum operator, when taken between |\u03b2 and |\u03b1, becomes \u03b2|p|\u03b1 = d3x\u03c8\u2217 \u03b2(x)(\u2212i\u00afh\u2207)\u03c8\u03b1(x). The transformation function analogous to (1.264) is x|p = 1 (2\u03c0\u00afh)3/2 exp, ip \u00b7 x \u00afh so that and \u03c8\u03b1(x) = 1 (2\u03c0\u00afh)3/2 d3p exp ip \u00b7 x \u00afh \u03c6\u03b1(p) \u03c6\u03b1(p) = 1 (2\u03c0\u00afh)3/2 d3x exp \u2212ip \u00b7 x \u00afh \u03c8\u03b1(x). (1.281) (1.282) (1.283a) (1.283b) It is interesting to check the dimension of the wave functions. In one-dimensional problems the normalization requirement (1.190) implies that |x|\u03b1|2 has the dimension of inverse length, so the wave function itself must have the dimension of (length)\u22121/2. In contrast, the wave function in three-dimensional problems must have the dimension of (length)\u22123/2 because |x|\u03b1|2 integrated over all spatial volume must be unity (dimensionless). Problems 1.1 A beam of silver atoms is created by heating a vapor in an oven to 1000\u25e6C, and selecting atoms with a velocity close to the mean of the thermal distribution. The beam moves through a one-meter long magnetic \ufb01eld with a vertical gradient 10 T/m, and impinges a screen one meter downstream of the end of the magnet. Assuming the silver atom has spin 1 2 with a magnetic moment of one Bohr magneton, \ufb01nd the separation distance in millimeters of the two states on the screen. 55 Problems 1.2 Prove [AB, CD] =\u2212 AC{D, B} + A{C, B}D \u2212 C{D, A}B + {C, A}DB. 1.3 For the spin 1 2 state |Sx; +, evaluate both sides of the inequality (1.146), that is \u2265 1 4 |[A, B]|2 (\u0394B)2 (\u0394A)2 for the operators A = Sx and B = Sy, and show that the inequality is satis\ufb01ed. Repeat for the operators A = Sz and B = Sy. 1.4 Suppose a 2 \u00d7 2 matrixX (not necessarily Hermitian, nor unitary) is written as X = a0 + \u03c3 \u00b7 a, where the matrices \u03c3 are given in (3.50) and a0 and a1,2,3 are numbers. a. How are a0 and ak(k = 1, 2, 3) related to tr(X ) and tr(\u03c3kX )? b. Obtain a0 and ak in terms of the matrix elements Xij. 1.5 Show that the determinant of a 2 \u00d7 2 matrix \u03c3 \u00b7 a is invariant under \u2212i\u03c3 \u00b7 \u02c6n\u03c6 2 \u03c3 \u00b7 a \u2192 \u03c3 \u00b7 a \u2261 exp i\u03c3 \u00b7 \u02c6n\u03c6 2 \u03c3 \u00b7 a exp, where the matrices \u03c3 are given in (3.50", "). Find a positive z-direction and interpret your result. k in terms of ak when \u02c6n is in the 1.6 Using the rules of bra-ket algebra, prove or evaluate the following: a. tr(XY ) =tr(YX ), whereX and Y are operators; b. (XY )\u2020 = Y \u2020X \u2020, whereX and Y are operators; c. exp[if (A)] =? in ket-bra form, where A is a Hermitian operator whose eigenvalues are known; 1.7 a (x)\u03c8a (x), where \u03c8a (x) =x |a. d. \u2211a \u03c8\u2217 a. Consider two kets |\u03b1 and |\u03b2. Suppose a|\u03b1, a|\u03b1,... and a|\u03b2, a|\u03b2,... are all known, where |a, |a,... form a complete set of base kets. Find the matrix representation of the operator |\u03b1\u03b2| in that basis. b. We now consider a spin 1 2 system and let |\u03b1 and |\u03b2 be |Sz; + and |Sx; +, respectively. Write down explicitly the square matrix that corresponds to |\u03b1\u03b2| in the usual (sz diagonal) basis. 1.8 Suppose |i and | j are eigenkets of some Hermitian operator A. Under what condition can we conclude that |i + | j is also an eigenket of A? Justify your answer. 1.9 Consider a ket space spanned by the eigenkets {|a} of a Hermitian operator A. There is no degeneracy. a. Prove that is the null operator. (A \u2212 a) \u220f a 56 Fundamental Concepts b. What is the signi\ufb01cance of (A \u2212 a) (a \u2212 a)? \u220f a= a c. Illustrate (a) and (b) using A set equal to Sz of a spin 1 2 system. 1.10 Using the orthonormality of |+ and |\u2212, prove [Si, Sj] = i\u03b5ijk \u00afhSk, {Si, Sj} = \u00afh2 2 \u03b4ij, where Sx = Sz = \u00afh 2 \u00afh 2 (|+\u2212| + |\u2212+|), Sy = i\u00afh 2 (\u2212|+\u2212| + |\u2212+|), (|++| \u2212 |\u2212\u2212|). 1.11 Construct |S \u00b7 \u02c6n; + such that S \u00b7 \u02c6n|S \u00b7 \u02c6n; + = \u00afh 2 |S \u00b7 \u02c6n; + where \u02c6n is characterized by the angles shown in the \ufb01gure. Express your answer as a linear combination of |+ and |\u2212. [Note: The answer is cos |+ + sin ei\u03b1|\u2212. \u03b2 2 \u03b2 2 But do not just verify that this answer satis\ufb01es the above eigenvalue equation. Rather, treat the problem as a straightforward eigenvalue problem. Also do not use rotation operators, which we will introduce later in this book.] n\u02c6 z \u03b1 \u03b2 y x 1.12 The Hamiltonian operator for a two-state system is given by H = a(|11| \u2212 |22| + |12| + |21|), where a is a number with the dimension of energy. Find the energy eigenvalues and the corresponding energy eigenkets (as linear combinations of |1 and |2). 57 Problems 1.13 A two-state system is characterized by the Hamiltonian H = H11|11| + H22|22| + H12[|12| + |21|] where H11, H22, andH 12 are real numbers with the dimension of energy, and |1 and |2 are eigenkets of some observable ( = H). Find the energy eigenkets and corresponding energy eigenvalues. Make sure that your answer makes good sense for H12 = 0. 1.14 A spin 1 2 system is known to be in an eigenstate of S \u00b7 \u02c6n with eigenvalue \u00afh/2, where \u02c6n is a unit vector lying in the xz-plane that makes an angle \u03b3 with the positive z-axis. a. Suppose Sx is measured. What is the probability of getting + \u00afh/2? b. Evaluate the dispersion in Sx, that is, (Sx \u2212 Sx)2. (For your own peace of mind check your answers for the special cases \u03b3 = 0, \u03c0/2, and \u03c0.) 2 atoms goes through a series of Stern\u2013Gerlach type measurements 1.15 A beam of spin 1 as follows. a. The \ufb01r", "st measurement accepts sz = \u00afh/2 atoms and rejects sz = \u2212\u00afh/2 atoms. b. The second measurement accepts sn = \u00afh/2 atoms and rejects sn = \u2212\u00afh/2 atoms, where sn is the eigenvalue of the operator S \u00b7 \u02c6n, with \u02c6n making an angle \u03b2 in the xz-plane with respect to the z-axis. c. The third measurement accepts sz = \u2212\u00afh/2 atoms and rejects sz = \u00afh/2 atoms. What is the intensity of the \ufb01nal sz = \u2212\u00afh/2 beam when the sz = \u00afh/2 beam surviving the \ufb01rst measurement is normalized to unity? How must we orient the second measuring apparatus if we are to maximize the intensity of the \ufb01nal sz = \u2212\u00afh/2 beam? 1.16 A certain observable in quantum mechanics has a 3 \u00d7 3 matrix representation as follows: \u239b \u239d 1. Find the normalized eigenvectors of this observable and the corresponding eigenvalues. Is there any degeneracy? b. Give a physical example where all this is relevant. 1.17 Let A and B be observables. Suppose the simultaneous eigenkets of A and B {|a, b} form a complete orthonormal set of base kets. Can we always conclude that [A, B] =0? If your answer counterexample. is yes, prove the assertion. If your answer is no, give a 58 Fundamental Concepts 1.18 Two Hermitian operators anticommute: {A, B} = AB + BA = 0. Is it possible to have a simultaneous (that is, common) eigenket of A and B? Prove or illustrate your assertion. 1.19 Two observables A1 and A2, which do not involve time explicitly, are known not to commute, [A1, A2] = 0, yet we also know that A1 and A2 both commute with the Hamiltonian: [A1, H] = 0, [A2, H] =0. Prove that the energy eigenstates are, in general, degenerate. Are there exceptions? As an example, you may think of the central-force problem H = p2/2m + V(r), with A1 \u2192 Lz, A2 \u2192 Lx. 1.20 a. The simplest way to derive the Schwarz inequality goes as follows. First, observe (\u03b1| + \u03bb\u2217\u03b2|) \u00b7 (|\u03b1 + \u03bb|\u03b2) \u2265 0 for any complex number \u03bb; then choose \u03bb in such a way that the preceding inequality reduces to the Schwarz inequality. b. Show that the equality sign in the generalized uncertainty relation holds if the state in question satis\ufb01es with \u03bb purely imaginary. \u0394A|\u03b1 = \u03bb\u0394B|\u03b1 c. Explicit calculations using the usual rules of wave mechanics show that the wave function for a Gaussian wave packet given by x|\u03b1 = (2\u03c0d2)\u22121/4 exp i px \u00afh \u2212 (x \u2212 x)2 4d2 satis\ufb01es the minimum uncertainty relation (\u0394x)2 (\u0394p)2 = \u00afh 2. Prove that the requirement x|\u0394x|\u03b1 = (imaginary number)x|\u0394p|\u03b1 is indeed satis\ufb01ed for such a Gaussian wave packet, in agreement with (b). 1.21 a. Compute (\u0394Sx)2 \u2261 S2 x \u2212 Sx2, 59 Problems where the expectation value is taken for the Sz+ state. Using your result, check the generalized uncertainty relation (\u0394A)2(\u0394B)2 \u2265 1 4 |[A, B]|2, with A \u2192 Sx, B \u2192 Sy. b. Check the uncertainty relation with A \u2192 Sx, B \u2192 Sy for the Sx+ state. 1.22 Find the linear combination of |+ and |\u2212 kets that maximizes the uncertainty product (\u0394Sx)2(\u0394Sy)2. Verify explicitly that for the linear combination you found, the uncertainty relation for Sx and Sy is not violated. 1.23 Evaluate the x-p uncertainty product (\u0394x)2(\u0394p)2 for a one-dimensional particle con\ufb01ned between two rigid walls V = for 0 < x < a, 0 \u221e otherwise. Do this for both the ground and excited states. 1.24 Estimate the rough order of magnitude of the length of time that an ice pick can be balanced on its point if the only limitation is that set by the Heisenberg uncertainty principle. Assume that the point is sharp and that the point and the surface", " on which it rests are hard. You may make approximations which do not alter the general order of magnitude of the result. Assume reasonable values for the dimensions and weight of the ice pick. Obtain an approximate numerical result and express it in seconds. 1.25 Consider a three-dimensional ket space. If a certain set of orthonormal kets, say, |1, |2, and |3, are used as the base kets, the operators A and B are represented by \u239e \u239e \u239b \u239b. = A \u239d a 0 0 \u2212a 0 0 0 0 \u2212ib 0 ib \u23a0 with a and b both real. a. Obviously A exhibits a degenerate spectrum. Does B also exhibit a degenerate spectrum? b. Show that A and B commute. c. Find a new set of orthonormal kets which are simultaneous eigenkets of both A and B. Specify the eigenvalues of A and B for each of the three eigenkets. Does your speci\ufb01cation of eigenvalues completely characterize each eigenket? \u221a 1.26 a. Prove that (1/ 2)(1 + i\u03c3x), where the matrix \u03c3x is given in (3.50), acting on a two-component spinor can be regarded as the matrix representation of the rotation operator about the x-axis by angle \u2212\u03c0/2. (The minus sign signi\ufb01es that the rotation is clockwise.) 60 Fundamental Concepts b. Construct the matrix representation of Sz when the eigenkets of Sy are used as base vectors. 1.27 Some authors de\ufb01ne an operator to be real when every member of its matrix elements b|A|b is real in some representation ({|b} basis in this case). Is this concept representation independent, that is, do the matrix elements remain real even if some basis other than {|b} is used? Check your assertion using familiar operators such as Sy and Sz (see Problem 1.26) or x and px. 1.28 Construct the transformation matrix that connects the Sz diagonal basis to the Sx diagonal basis. Show that your result is consistent with the general relation U = \u2211 |b(r)a(r)|. r 1.29 a. Suppose that f (A) is a function of a Hermitian operator A with the property A|a = a|a. Evaluateb | f (A)|b when the transformation matrix from the a basis to the b basis is known. b. Using the continuum analogue of the result obtained in (a), evaluate p|F(r)|p. Simplify your expression as far as you can. Note that r is y, andz are operators. x2 + y2 + z2, wherex, 1.30 a. Let x and px be the coordinate and linear momentum in one dimension. Evaluate the classical Poisson bracket [x, F(px)]classical. b. Let x and px be the corresponding quantum-mechanical operators this time. Evaluate the commutator x, exp ipxa \u00afh. c. Using the result obtained in (b), prove that ipxa \u00afh exp |x (x|x = x|x) is an eigenstate of the coordinate operator x. What is the corresponding eigenvalue? 1.31 a. On p. 247, Gottfried (1966) states that \u2202G \u2202pi [xi, G(p)] = i\u00afh, [ pi, F(x)] = \u2212i\u00afh \u2202F \u2202xi can be \u201ceasily derived\u201d from the fundamental commutation relations for all functions of F and G that can be expressed as power series in their arguments. Verify this statement. b. Evaluate [x2, p2]. Compare your result with the classical Poisson bracket [x2, p2]classical. 61 Problems 1.32 The translation operator for a \ufb01nite (spatial) displacement is given by J (l) =exp where p is the momentum operator. a. Evaluate, \u2212ip \u00b7 l \u00afh [xi, J (l)]. b. Using (a) (or otherwise), demonstrate how the expectation value x changes under translation. 1.33 In the main text we discussed the effect of J (dx) on the position and momentum eigenkets and on a more general state ket |\u03b1. We can also study the behavior of expectation values x and p under in\ufb01nitesimal translation. Using (1.207), (1.227), and |\u03b1 \u2192J (dx)|\u03b1 only, prove x \u2192 x + dx, p \u2192 p under in\ufb01nitesimal translation. 1.34 Starting with a momentum operator p having eigenstates |p, de\ufb01", "ne an in\ufb01nitesimal boost operator B(dp) that changes one momentum eigenstate into another, that is B(dp)|p = |p + dp. Show that the form B(dp) =1 + iW \u00b7 dp, whereW is Hermitian, satis\ufb01es the unitary, associative, and inverse properties that are appropriate for B(dp). Use dimensional analysis to express W in terms of the position operator x, and show that the result satis\ufb01es the canonical commutation relations [xi, pj] =i \u00afh\u03b4ij. Derive an expression for the matrix element p|x|\u03b1 in terms of a derivative with respect to p of p|\u03b1. 1.35 a. Verify (1.271a) and (1.271b) for the expectation value of p and p2 from the Gaussian wave packet (1.267). b. Evaluate the expectation value of p and p2 using the momentum-space wave function (1.274). 1.36 a. Prove the following: \u2202 \u2202p dp\u03c6\u2217 (i) p|x|\u03b1 = i\u00afh p|\u03b1, (ii) \u03b2|x|\u03b1 = where \u03c6\u03b1(p) = p|\u03b1 and \u03c6\u03b2(p) = p|\u03b2 are momentum-space wave functions. \u03c6\u03b1(p), \u03b2(p)i\u00afh \u2202 \u2202p b. What is the physical signi\ufb01cance of ix\u039e \u00afh where x is the position operator and \u039e is some number with the dimension of momentum? Justify your answer. exp, 2 Quantum Dynamics So far we have not discussed how physical systems change with time. This chapter is devoted exclusively to the dynamic development of state kets and/or observables. In other words, we are concerned here with the quantum-mechanical analogue of Newton\u2019s (or Lagrange\u2019s or Hamilton\u2019s) equations of motion. 2.1 Time Evolution and the Schr\u00f6dinger Equation The \ufb01rst important point we should keep in mind is that time is just a parameter in quantum mechanics, not an operator. In particular, time is not an observable in the language of the previous chapter. It is nonsensical to talk about the time operator in the same sense as we talk about the position operator. Ironically, in the historical development of wave mechanics both L. de Broglie and E. Schr\u00a8odinger were guided by a kind of covariant analogy between energy and time on the one hand and momentum and position (spatial coordinate) on the other. Yet when we now look at quantum mechanics in its \ufb01nished form, there is no trace of a symmetrical treatment between time and space. The relativistic quantum theory of \ufb01elds does treat the time and space coordinates on the same footing, but it does so only at the expense of demoting position from the status of being an observable to that of being just a parameter. 2.1.1 Time-Evolution Operator Our basic concern in this section is, How does a state ket change with time? Suppose we have a physical system whose state ket at t0 is represented by |\u03b1. At later times, we do not, in general, expect the system to remain in the same state |\u03b1. Let us denote the ket corresponding to the state at some later time by |\u03b1, t0; t (t > t0), (2.1) where we have written \u03b1, t0 to remind ourselves that the system used to be in state |\u03b1 at some earlier reference time t0. Because time is assumed to be a continuous parameter, we expect |\u03b1, t0; t = |\u03b1 lim t\u2192t0 (2.2) 62 63 2.1 Time Evolution and the Schr\u00f6dinger Equation and we may as well use a shorthand notation, |\u03b1, t0; t0 = |\u03b1, t0, for this. Our basic task is to study the time evolution of a state ket: |\u03b1, t0 = |\u03b1 time evolution \u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2192 |\u03b1, t0; t. (2.3) (2.4) Put in another way, we are interested in asking how the state ket changes under a time displacement t0 \u2192 t. As in the case of translation, the two kets are related by an operator which we call the time-evolution operator U (t, t0): |\u03b1, t0; t = U (t, t0)|\u03b1, t0. (2.5) What are some of the properties we would like to ascribe to the time-evolution operator? The \ufb01rst important property is the unitary requirement for U (t, t0)", " that follows from probability conservation. Suppose that at t0 the state ket is expanded in terms of the eigenkets of some observable A: |\u03b1, t0 = \u2211 a ca (t0)|a. Likewise, at some later time, we have |\u03b1, t0; t = \u2211 a ca(t)|a. (2.6) (2.7) In general, we do not expect the modulus of the individual expansion coef\ufb01cient to remain the same:1 |ca (t)| = |ca(t0)|. (2.8) For instance, consider a spin 1 2 system with its spin magnetic moment subjected to a uniform magnetic \ufb01eld in the z-direction. To be speci\ufb01c, suppose that at t0 the spin is in the positive x-direction; that is, the system is prepared in an eigenstate of Sx with eigenvalue \u00afh/2. As time goes on, the spin precesses in the xy-plane, as will be quantitatively demonstrated later in this section. This means that the probability for observing Sx + is no longer unity at t > t0; there is a \ufb01nite probability for observing Sx \u2212 as well. Yet the sum of the probabilities for Sx + and Sx \u2212 remains unity at all times. Generally, in the notation of (2.6) and (2.7), we must have \u2211 a |ca(t0)|2 = \u2211 a |ca(t)|2 (2.9) despite (2.8) for the individual expansion coef\ufb01cients. Stated another way, if the state ket is initially normalized to unity, it must remain normalized to unity at all later times: \u03b1, t0|\u03b1, t0 = 1 \u21d2 \u03b1, t0; t|\u03b1, t0; t = 1. (2.10) 1 We later show, however, that if the Hamiltonian commutes with A, then| ca (t)| is indeed equal to |ca (t0)|. 64 Quantum Dynamics As in the translation case, this property is guaranteed if the time-evolution operator is taken to be unitary. For this reason we take unitarity, U \u2020(t, t0)U (t, t0) =1, to be one of the fundamental properties of the U operator. It is no coincidence that many authors regard unitarity as being synonymous with probability conservation. Another feature we require of the U operator is the composition property: (2.11) U (t2, t0) =U (t2, t1)U (t1, t0) (t2 > t1 > t0). (2.12) This equation says that if we are interested in obtaining time evolution from t0 to t2, then we can obtain the same result by \ufb01rst considering time evolution from t0 to t1, then from t1 to t2, a reasonable requirement. Note that we read (2.12) from right to left! It also turns out to be advantageous to consider an in\ufb01nitesimal time-evolution operator U (t0 + dt, t0): |\u03b1, t0; t0 + dt = U (t0 + dt, t0)|\u03b1, t0. (2.13) Because of continuity [see (2.2)], the in\ufb01nitesimal time-evolution operator must reduce to the identity operator as dt goes to zero, U (t0 + dt, t0) = 1, lim dt\u21920 (2.14) and as in the translation case, we expect the difference between U (t0 + dt, t0) and 1 to be of \ufb01rst order in dt. We assert that all these requirements are satis\ufb01ed by U (t0 + dt, t0) =1 \u2212 i\u03a9 dt, where \u03a9 is a Hermitian operator,2 \u03a9\u2020 = \u03a9. (2.15) (2.16) With (2.15) the in\ufb01nitesimal time-displacement operator satis\ufb01es the composition property U (t0 + dt1 + dt2, t0) =U (t0 + dt1 + dt2, t0 + dt1)U (t0 + dt1, t0); (2.17) it differs from the identity operator by a term of order dt. The unitarity property can also be checked as follows: U \u2020(t0 + dt, t0)U (t0 + dt, t0) = (1 + i\u03a9\u2020dt)(1", " \u2212 i\u03a9dt) 1, (2.18) to the extent that terms of order (dt)2 or higher can be ignored. The operator \u03a9 has the dimension of frequency or inverse time. Is there any familiar observable with the dimension of frequency? We recall that in the old quantum theory, angular frequency \u03c9 is postulated to be related to energy by the Planck\u2013Einstein relation E = \u00afh\u03c9. (2.19) 2 If the \u03a9 operator depends on time explicitly, it must be evaluated at t0. 65 2.1 Time Evolution and the Schr\u00f6dinger Equation Let us now borrow from classical mechanics the idea that the Hamiltonian is the generator of time evolution (Goldstein et al. (2002), pp. 401\u2013402). It is then natural to relate \u03a9 to the Hamiltonian operator H: \u03a9 = H \u00afh. To sum up, the in\ufb01nitesimal time-evolution operator is written as U (t0 + dt, t0) =1 \u2212 iH dt \u00afh, (2.20) (2.21) where H, the Hamiltonian operator, is assumed to be Hermitian. The reader may ask whether the \u00afh introduced here is the same as the \u00afh that appears in the expression for the translation operator (1.214). This question can be answered by comparing the quantummechanical equation of motion we derive later with the classical equation of motion. It turns out that unless the two \u00afh are taken to be the same, we are unable to obtain a relation like dx dt = p m (2.22) as the classical limit of the corresponding quantum-mechanical relation. 2.1.2 The Schr\u00f6dinger Equation We are now in a position to derive the fundamental differential equation for the timeevolution operator U (t, t0). We exploit the composition property of the time-evolution operator by letting t1 \u2192 t, t2 \u2192 t + dt in (2.12): U (t + dt, t0) =U (t + dt, t)U (t, t0) = U (t, t0), (2.23) where the time difference t \u2212 t0 need not be in\ufb01nitesimal. We have U (t + dt, t0) \u2212 U (t, t0) =\u2212 i which can be written in differential equation form: 1 \u2212 iH dt \u00afh dtU (t, t0), H \u00afh (2.24) (2.25) i\u00afh \u2202 \u2202t U (t, t0) =HU (t, t0). This is the Schr\u00a8odinger equation for the time-evolution operator. Everything that has to do with time development follows from this fundamental equation. Equation (2.25) immediately leads to the Schr\u00a8odinger equation for a state ket. Multiply- ing both sides of (2.25) by| \u03b1, t0 on the right, we obtain \u2202 \u2202t U (t, t0)|\u03b1, t0 = HU (t, t0)|\u03b1, t0. i\u00afh But |\u03b1, t0 does not depend on t, so this is the same as i\u00afh \u2202 \u2202t |\u03b1, t0; t = H|\u03b1, t0; t, where (2.5) has been used. (2.26) (2.27) 66 Quantum Dynamics If we are given U (t, t0) and, in addition, know how U (t, t0) acts on the initial state ket |\u03b1, t0, it is not necessary to bother with the Schr\u00a8odinger equation for the state ket (2.27). All we have to do is applyU (t, t0) to |\u03b1, t0; in this manner we can obtain a state ket at any t. Our \ufb01rst task is therefore to derive formal solutions to the Schr\u00a8odinger equation for the time-evolution operator (2.25). There are three cases to be treated separately. Case 1. The Hamiltonian operator is independent of time. By this we mean that even when the parameter t is changed, the H operator remains unchanged. The Hamiltonian for a spin-magnetic moment interacting with a time-independent magnetic \ufb01eld is an example of this. The solution to (2.25) in such a case is given by U (t, t0) =exp. (2.28) \u2212iH(t \u2212 t0) \u00afh To prove this let us expand the exponential as follows: exp \u2212iH(t \u2212 t0) \u00afh = 1 + \u2212iH(t \u2212 t0) \u00afh + (\u2212i)2 2 H(", "t \u2212 t0) \u00afh 2 + \u00b7 \u00b7 \u00b7. (2.29) Because the time derivative of this expansion is given by \u2202 \u2202t exp \u2212iH(t \u2212 t0) \u00afh = \u2212iH \u00afh + (\u2212i)2 2 2 2 H \u00afh (t \u2212 t0) +\u00b7 \u00b7 \u00b7, (2.30) expression (2.28) obviously satis\ufb01es differential equation (2.25). The boundary condition is also satis\ufb01ed because as t \u2192 t0, (2.28) reduces to the identity operator. An alternative way to obtain (2.28) is to compound successively in\ufb01nitesimal time-evolution operators just as we did to obtain (1.218) for \ufb01nite translation: 1 \u2212 (iH/\u00afh)(t \u2212 t0) N lim N\u2192\u221e = exp \u2212iH(t \u2212 t0) \u00afh N. (2.31) Case 2. The Hamiltonian operator H is time dependent but the H at different times commute. As an example, let us consider the spin-magnetic moment subjected to a magnetic \ufb01eld whose strength varies with time but whose direction is always unchanged. The formal solution to (2.25) in this case is U (t, t0) =exp \u2212 dtH(t). (2.32) t t0 i \u00afh This can be proved in a similar way. We simply replace H(t \u2212 t0) in (2.29) and (2.30) by t t0 dtH(t). Case 3. The H at different times do not commute. Continuing with the example involving spin-magnetic moment, we suppose, this time, that the magnetic \ufb01eld direction also changes with time: at t = t1 in the x-direction, at t = t2 in the y-direction, and so forth. Because Sx and Sy do not commute, H(t1) and H(t2), which go like S \u00b7 B, do not commute either. The formal solution in such a situation is given by U (t, t0) =1 + \u221e \u2211 n=1 \u2212i \u00afh n t dt1 t0 t1 t0 dt2 \u00b7 \u00b7 \u00b7 tn\u22121 t0 dtn H(t1)H(t2) \u00b7 \u00b7 \u00b7 H(tn), (2.33) 67 2.1 Time Evolution and the Schr\u00f6dinger Equation which is sometimes known as the Dyson series, after F. J. Dyson, who developed a perturbation expansion of this form in quantum \ufb01eld theory. We do not prove (2.33) now because the proof is very similar to the one presented in Chapter 5 for the time-evolution operator in the interaction picture. In elementary applications, only case 1 is of practical interest. In the remaining part of this chapter we assume that the H operator is time independent. We will encounter timedependent Hamiltonians in Chapter 5. 2.1.3 Energy Eigenkets To be able to evaluate the effect of the time-evolution operator (2.28) on a general initial ket |\u03b1, we must \ufb01rst know how it acts on the base kets used in expanding |\u03b1. This is particularly straightforward if the base kets used are eigenkets of A such that [A, H] =0; (2.34) then the eigenkets of A are also eigenkets of H, called energy eigenkets, whose eigenvalues are denoted by Ea : H|a = Ea |a. (2.35) We can now expand the time-evolution operator in terms of |aa|. Takingt 0 = 0 for simplicity, we obtain \u2212iHt \u00afh exp \u2212iHt \u00afh |aaaa| exp |a exp \u2212iEat \u00afh a|. (2.36) The time-evolution operator written in this form enables us to solve any initial-value problem once the expansion of the initial ket in terms of {|a} is known. As an example, suppose that the initial ket expansion reads |\u03b1, t0 = 0 = \u2211 a |aa|\u03b1 = \u2211 a ca |a. (2.37) We then have |\u03b1, t0 = 0; t = exp \u2212iHt \u00afh |\u03b1, t0 = 0 = \u2211 a |aa|\u03b1 exp. \u2212iEat \u00afh In other words, the expansion coef\ufb01cient changes with time as ca (t = 0) \u2192 ca (t) = ca (t = 0) exp \u2212iEat \u00afh (2.38) (2.39) with its mod", "ulus unchanged. Notice that the relative phases among various components do vary with time because the oscillation frequencies are different. A special case of interest is where the initial state happens to be one of {|a} itself. We have |\u03b1, t0 = 0 = |a (2.40) 68 Quantum Dynamics initially, and at a later time |a, t0 = 0; t = |a exp, \u2212iEat \u00afh (2.41) so if the system is initially a simultaneous eigenstate of A and H, it remains so at all times. The most that can happen is the phase modulation, exp(\u2212iEat/\u00afh). It is in this sense that an observable compatible with H [see (2.34)] is a constant of the motion. We will encounter this connection once again in a different form when we discuss the Heisenberg equation of motion. In the foregoing discussion the basic task in quantum dynamics is reduced to \ufb01nding an observable that commutes with H and evaluating its eigenvalues. Once that is done, we expand the initial ket in terms of the eigenkets of that observable and just apply the timeevolution operator. This last step merely amounts to changing the phase of each expansion coef\ufb01cient, as indicated by (2.39). Even though we worked out the case where there is just one observable A that commutes with H, our considerations can easily be generalized when there are several mutually compatible observables all also commuting with H: [A, B] = [B, C] = [A, C] =\u00b7 \u00b7 \u00b7 = 0, [A, H] = [B, H] = [C, H] =\u00b7 \u00b7 \u00b7 = 0. Using the collective index notation of Section 1.4 [see (1.130)], we have exp \u2212iHt \u00afh |K exp = \u2211 K \u2212iEKt \u00afh K|, (2.42) (2.43) where EK is uniquely speci\ufb01ed once a, b, c,... are speci\ufb01ed. It is therefore of fundamental importance to \ufb01nd a complete set of mutually compatible observables that also commute with H. Once such a set is found, we express the initial ket as a superposition of the simultaneous eigenkets of A, B, C,... and H. The \ufb01nal step is just to apply the timeevolution operator, written as (2.43). In this manner we can solve the most general initialvalue problem with a time-independent H. 2.1.4 Time Dependence of Expectation Values It is instructive to study how the expectation value of an observable changes as a function of time. Suppose that at t = 0 the initial state is one of the eigenstates of an observable A that commutes with H, as in (2.40). We now look at the expectation value of some other observable B, which need not commute with A nor with H. Because at a later time we have |a, t0 = 0; t = U (t, 0)|a (2.44) for the state ket, B is given by B = (a|U \u2020(t, 0)) \u00b7 B \u00b7 (U (t, 0)|a) \u2212iEat iEat \u00afh \u00afh = a| exp B exp |a = a|B|a, (2.45) 69 2.1 Time Evolution and the Schr\u00f6dinger Equation which is independent of t. So the expectation value of an observable taken with respect to an energy eigenstate does not change with time. For this reason an energy eigenstate is often referred to as a stationary state. The situation is more interesting when the expectation value is taken with respect to a superposition of energy eigenstates, or a nonstationary state. Suppose that initially we have |\u03b1, t0 = 0 = \u2211 a ca |a. We easily compute the expectation value of B to be \u2211 a B = a a| exp c\u2217 iEat \u00afh = \u2211 a \u2211 a a ca a|B|a exp c\u2217 \u00b7 B \u00b7 ca exp \u2211 a \u2212i(Ea \u2212 Ea)t \u00afh (2.46) (2.47) |a \u2212iEat \u00afh. So this time the expectation value consists of oscillating terms whose angular frequencies are determined by N. Bohr\u2019s frequency condition \u03c9aa = (Ea \u2212 Ea) \u00afh. 2.1.5 Spin Precession (2.48) It is appropriate to treat an example here. We consider an extremely simple system which, however, illustrates the basic formalism we have developed. We start with a Hamiltonian of a spin 1 2 system with magnetic moment e\u00afh/2mec subjected to", " an external magnetic \ufb01eld B: H = \u2212 e mec S \u00b7 B (2.49) (e < 0 for the electron). Furthermore, we take B to be a static, uniform magnetic \ufb01eld in the z-direction. We can then write H as H = \u2212 eB mec Sz. (2.50) Because Sz and H differ just by a multiplicative constant, they obviously commute. The Sz eigenstates are also energy eigenstates, and the corresponding energy eigenvalues are E\u00b1 = \u2213 e\u00afhB 2mec, for Sz \u00b1. (2.51) It is convenient to de\ufb01ne \u03c9 in such a way that the difference in the two energy eigenvalues is \u00afh\u03c9: \u03c9 \u2261 |e|B mec. (2.52) 70 Quantum Dynamics We can then rewrite the H operator simply as H = \u03c9Sz. (2.53) All the information on time development is contained in the time-evolution operator U (t, 0) =exp. (2.54) \u2212i\u03c9Szt \u00afh We apply this to the initial state. The base kets we must use in expanding the initial ket are obviously the Sz eigenkets, |+ and |\u2212, which are also energy eigenkets. Suppose that at t = 0 the system is characterized by |\u03b1 = c+|+ + c\u2212|\u2212. Upon applying (2.54), we see that the state ket at some later time is +i\u03c9t 2 |\u03b1, t0 = 0; t = c+ exp |+ + c\u2212 exp \u2212i\u03c9t 2 (2.55) (2.56) |\u2212, where we have used \u00b1\u00afh\u03c9 2 Speci\ufb01cally, let us suppose that the initial ket |\u03b1 represents the spin-up (or, more H|\u00b1 = (2.57) |\u00b1. precisely, Sz+) state |+, which means that c+ = 1, c\u2212 = 0. (2.58) At a later time, (2.56) tells us that it is still in the spin-up state, which is no surprise because this is a stationary state. Next, let us suppose that initially the system is in the Sx+ state. Comparing (1.110a) with (2.55), we see that c+ = c\u2212 = 1\u221a 2. (2.59) It is straightforward to work out the probabilities for the system to be found in the Sx\u00b1 state at some later time t: |Sx; \u00b1|\u03b1, t0 = 0; t|2 = 1\u221a 2 +| \u00b1 1\u221a 2 \u2212| \u00b7 + \u2212i\u03c9t 2 +i\u03c9t 2 |+ exp exp 1\u221a 2 1\u221a 2 2 = 1 2 exp \u2212i\u03c9t 2 \u00b1 1 2 exp +i\u03c9t 2 = cos2 = sin2 \u03c9t 2 \u03c9t 2 for for Sx+, Sx \u2212. 2 |\u2212 (2.60a) (2.60b) Even though the spin is initially in the positive x-direction, the magnetic \ufb01eld in the z-direction causes it to rotate; as a result, we obtain a \ufb01nite probability for \ufb01nding 71 2.1 Time Evolution and the Schr\u00f6dinger Equation Sx\u2212 at some later time. The sum of the two probabilities is seen to be unity at all times, in agreement with the unitarity property of the time-evolution operator. Using (1.99), we can write the expectation value of Sx as \u03c9t 2 + \u2212\u00afh 2 sin2 \u03c9t 2 cos2 Sx = = \u00afh 2 \u00afh 2 cos \u03c9t, (2.61) so this quantity oscillates with an angular frequency corresponding to the difference of the two energy eigenvalues divided by \u00afh, in agreement with our general formula (2.47). Similar exercises with Sy and Sz show that Sy = sin \u03c9t \u00afh 2 and Sz = 0. (2.62a) (2.62b) Physically this means that the spin precesses in the xy-plane. We will comment further on spin precession when we discuss rotation operators in Chapter 3. Experimentally, spin precession is well established. In fact, it is used as a tool for other investigations of fundamental quantum-mechanical phenomena. For example, the form of the Hamiltonian (2.49) can be derived for pointlike particles, such as electrons or muons, which obey the Dirac equation, for which the gyromagnetic ratio g = 2. (See Section 8.2.) However, higher-order corrections from quantum \ufb01eld theory predict a small but precisely calcul", "able deviation from this, and it is a high priority to produce competitively precise measurements of g \u2212 2. Such an experiment has been recently completed. See Bennett et al., Phys. Rev. D, 73 (2006) 072003. Muons are injected into a \u201cstorage ring\u201d designed so that their spins would precess in lock step with their momentum vector only if g \u2261 2. Consequently, observation of their precession measures g \u2212 2 directly, facilitating a very precise result. Figure 2.1 shows the experimenters\u2019 observation of the muon spin rotation over more than one hundred periods. They determine a value for g \u2212 2 to a precision smaller than one part per million, which agrees reasonably well with the theoretical value. 2.1.6 Neutrino Oscillations A lovely example of quantum-mechanical dynamics leading to interference in a two-state system, based on current physics research, is provided by the phenomenon known as neutrino oscillations.3 Neutrinos are elementary particles with no charge, and very small mass, much smaller than that of an electron. They are known to occur in nature in three distinct \u201c\ufb02avors,\u201d 3 The treatment here is the straightforward approach usually covered in the literature, but it has shortcomings. See, for example, Cohen et al., Phys. Lett. B, 678 (2009) 191 and Akhmedov (2018) arXiv:1901.05232 [hep-ph]. 72 Quantum Dynamics 10 1 10\u20131 10\u2013 10\u20133 0 20 40 60 80 100 Time modulo 100 \u03bcs Fig. 2.1 Observations of the precession of muon spin by Bennett et al., Phys. Rev. D, 73 (2006) 072003. Data points are wrapped around every 100 \u03bcs. The size of the signal decreases with time because the muons decay. although for this discussion it suf\ufb01ces to only consider two of them. These two \ufb02avors are identi\ufb01ed by their interactions which may be either with electrons, in which case we write \u03bde, or with muons, that is \u03bd\u03bc. These are in fact eigenstates of a Hamiltonian which controls those interactions. On the other hand, it is possible (and, in fact, now known to be true) that neutrinos may have some other interactions, in which case their energy eigenvalues correspond to states that have a well-de\ufb01ned mass. These \u201cmass eigenstates\u201d would have eigenvalues E1 and E2, say, corresponding to masses m1 and m2, and might be denoted as |\u03bd1 and |\u03bd2. The \u201c\ufb02avor eigenstates\u201d are related to these through a simple unitary transformation, speci\ufb01ed by some mixing angle \u03b8, as follows: |\u03bde = cos \u03b8|\u03bd1 \u2212sin \u03b8|\u03bd2 |\u03bd\u03bc = sin \u03b8|\u03bd1 + cos \u03b8|\u03bd2. (2.63a) (2.63b) If the mixing angle were zero, then |\u03bde and |\u03bd\u03bc would respectively be the same as |\u03bd1 and |\u03bd2. However, we know of no reason why this should be the case. Indeed, there is no strong theoretical bias for any particular value of \u03b8, and it is a free parameter which, today, can only be determined through experiment. Neutrino oscillation is the phenomenon by which we can measure the mixing angle. Suppose we prepare, at time t = 0, a momentum eigenstate of one \ufb02avor of neutrino, say |\u03bde. Then according to (2.63a) the two different mass eigenstate components will evolve with different frequencies, and therefore develop a relative phase difference. If the difference in the masses is small enough, then this phase difference can build up over a macroscopic distance. In fact, by measuring the interference as a function of difference, 73 2.1 Time Evolution and the Schr\u00f6dinger Equation one can observe oscillations with a period that depends on the difference of masses, and an amplitude that depends on the mixing angle. It is straightforward (see Problem 2.4 at the end of this chapter) to use (2.63) along with (2.28) and our quantum-mechanical postulates, and \ufb01nd a measurable quantity that exhibits neutrino oscillations. In this case, the Hamiltonian is just that for a free particle, but we need to take some care. Neutrinos are very low mass, so they are highly relativistic for any practical experimental conditions. Therefore, for a \ufb01xed momentum p, the energy eigenvalue for a neutrino of mass m is given to an extremely good approximation", " as! E = p2c2 + m2c4 \"1/2 \u2248 pc 1 + m2c2 2p2. (2.64) If we then allow our state |\u03bde to evolve, and then at some later time t ask what is the probability that it still appears as a |\u03bde (as opposed to a |\u03bd\u03bc), we \ufb01nd P(\u03bde \u2192 \u03bde) =1 \u2212 sin2 2\u03b8sin2 (2.65) \u0394m2c4 L 4E\u00afhc where \u0394m2 \u2261 m2 1 nominal neutrino energy. \u2212 m2 2, L = ct is the \ufb02ight distance of the neutrino, and E = pc is the The oscillations predicted by (2.65) have been dramatically observed by the KamLAND experiment. See Figure 2.2. Neutrinos from a series of nuclear reactors are detected at a distance of \u223c150km, and the rate is compared to that expected from reactor power and properties. The curve is not a perfect sine wave because the reactors are not all at the same distance from the detector.0 0.8 0.6 0.4 0.2 0 Data - BG - Geo ve Expectation based on osci. parameters determined by KamLAND 20 30 40 50 60 70 (km/MeV) L0 /E ve 80 90 100 Fig. 2.2 Neutrino oscillations as observed by the KamLAND experiment, taken from Abe et al., Phys. Rev. Lett., 100 (2008) 221803. The oscillations as a function of L/E demonstrate interference between different mass eigenstates of neutrinos. 74 Quantum Dynamics 2.1.7 Correlation Amplitude and the Energy-Time Uncertainty Relation We conclude this section by asking how state kets at different times are correlated with each other. Suppose the initial state ket at t = 0 of a physical system is given by |\u03b1. With time it changes into |\u03b1, t0 = 0; t, which we obtain by applying the time-evolution operator. We are concerned with the extent to which the state ket at a later time t is similar to the state ket at t = 0; we therefore construct the inner product between the two state kets at different times: C(t) \u2261 \u03b1|\u03b1, t0 = 0; t = \u03b1|U (t, 0)|\u03b1, (2.66) which is known as the correlation amplitude. The modulus of C(t) provides a quantitative measure of the \u201cresemblance\u201d between the state kets at different times. As an extreme example, consider the very special case where the initial ket |\u03b1 is an eigenket of H; we then have C(t) =a |a, t0 = 0; t = exp, \u2212iEat \u00afh (2.67) so the modulus of the correlation amplitude is unity at all times, which is not surprising for a stationary state. In the more general situation where the initial ket is represented by a superposition of {|a}, as in (2.37), we have C(t) = a a| c\u2217 \u2211 a = \u2211 a |ca|2 exp \u2211 a \u2212iEat \u00afh ca exp \u2212iEat \u00afh |a. (2.68) As we sum over many terms with oscillating time dependence of different frequencies, a strong cancellation is possible for moderately large values of t. We expect the correlation amplitude that starts with unity at t = 0 to decrease in magnitude with time. To estimate (2.68) in a more concrete manner, let us suppose that the state ket can be regarded as a superposition of so many energy eigenkets with similar energies that we can regard them as exhibiting essentially a quasi-continuous spectrum. It is then legitimate to replace the sum by the integral \u2192 \u2211 a dE\u03c1(E), ca \u2192 g(E), EEa (2.69) where \u03c1(E) characterizes the density of energy eigenstates. Expression (2.68) now becomes C(t) = dE|g(E)|2\u03c1(E) exp \u2212iEt \u00afh, (2.70) subject to the normalization condition dE|g(E)|2\u03c1(E) =1. (2.71) 75 2.2 The Schr\u00f6dinger Versus the Heisenberg Picture In a realistic physical situation |g(E)|2\u03c1(E) may be peaked around E = E0 with width \u0394E. Writing (2.70) as C(t) = exp dE|g(E)|2\u03c1(E) exp, (2.72) \u2212iE0t \u00afh \u2212i(E \u2212 E0)t \u00afh we see that as t becomes", " large, the integrand oscillates very rapidly unless the energy interval |E \u2212 E0| is small compared with \u00afh/t. If the interval for which |E \u2212 E0| \u00afh/t holds is much narrower than \u0394E, the width of |g(E)|2\u03c1(E), we get essentially no contribution to C(t) because of strong cancellations. The characteristic time at which the modulus of the correlation amplitude starts becoming appreciably different from 1 is given by t \u00afh \u0394E. (2.73) Even though this equation is obtained for a superposition state with a quasi-continuous energy spectrum, it also makes sense for a two-level system; in the spin-precession problem considered earlier, the state ket, which is initially |Sx+, starts losing its identity after \u223c 1/\u03c9 = \u00afh/(E+ \u2212 E\u2212), as is evident from (2.60). To summarize, as a result of time evolution the state ket of a physical system ceases to retain its original form after a time interval of order \u00afh/\u0394E. In the literature this point is often said to illustrate the energy-time uncertainty relation \u0394t\u0394E \u00afh. (2.74) However, it is to be clearly understood that this energy-time uncertainty relation is of a very different nature from the uncertainty relation between two incompatible observables discussed in Section 1.4. InChapter 5 we will come back to (2.74) in connection with time-dependent perturbation theory. 2.2 The Schr\u00f6dinger Versus the Heisenberg Picture 2.2.1 Unitary Operators In the previous section we introduced the concept of time development by considering the time-evolution operator that affects state kets; that approach to quantum dynamics is known as the Schr\u00a8odinger picture. There is another formulation of quantum dynamics where observables, rather than state kets, vary with time; this second approach is known as the Heisenberg picture. Before discussing the differences between the two approaches in detail, we digress to make some general comments on unitary operators. Unitary operators are used for many different purposes in quantum mechanics. In this book we introduced (Section 1.5) an operator satisfying the unitarity property. In that section we were concerned with the question of how the base kets in one representation are related to those in some other representations. The state kets themselves are assumed not to change as we switch to a different set of base kets even though the numerical values of the expansion coef\ufb01cients for |\u03b1 are, of course, different in different representations. 76 Quantum Dynamics Subsequently we introduced two unitary operators that actually change the state kets, the translation operator of Section 1.6 and the time-evolution operator of Section 2.1. We have |\u03b1 \u2192U| \u03b1, (2.75) where U may stand for T (dx) or U (t, t0). HereU| \u03b1 is the state ket corresponding to a physical system that actually has undergone translation or time evolution. It is important to keep in mind that under a unitary transformation that changes the state kets, the inner product of a state bra and a state ket remains unchanged: \u03b2|\u03b1 \u2192 \u03b2|U\u2020U|\u03b1 = \u03b2|\u03b1. (2.76) Using the fact that these transformations affect the state kets but not operators, we can infer how \u03b2|X |\u03b1 must change: \u03b2|X |\u03b1 \u2192( \u03b2|U\u2020) \u00b7 X \u00b7 (U|\u03b1) = \u03b2|U\u2020XU|\u03b1. (2.77) We now make a very simple mathematical observation that follows from the associative axiom of multiplication. (\u03b2|U\u2020) \u00b7 X \u00b7 (U|\u03b1) = \u03b2| \u00b7(U \u2020XU) \u00b7 |\u03b1. (2.78) Is there any physics in this observation? This mathematical approaches to unitary transformations. identity suggests two Approach 1: Approach 2: |\u03b1 \u2192U| \u03b1, with operators unchanged. X \u2192 U\u2020XU, with state kets unchanged. (2.79a) (2.79b) In classical physics we do not introduce state kets, yet we talk about translation, time evolution, and the like. This is possible because these operations actually change quantities such as x and L, which are observables of classical mechanics. We therefore conjecture that a closer connection with classical physics may be established if we follow approach 2. A simple example may be helpful here. We go back to the in\ufb01nitesimal translation operator T (dx). The formalism presented in Section 1.6 is based on approach 1; T (dx) affects the state kets, not the position operator: 1 \u2212 ip \u00b7 dx \u00afh |\u03b1 \u2192 x \u2192 x. |\u03b1. (2.80) In contrast, if we follow approach 2, we", " obtain |\u03b1 \u2192 |\u03b1, x \u2192 ip \u00b7 dx \u00afh 1 + x 1 \u2212 ip \u00b7 dx \u00afh = x + i \u00afh = x + dx. [p \u00b7 dx, x] (2.81) 77 2.2 The Schr\u00f6dinger Versus the Heisenberg Picture We leave it as an exercise for the reader to show that both approaches lead to the same result for the expectation value of x: x \u2192 x + dx. (2.82) 2.2.2 State Kets and Observables in the Schr\u00f6dinger and the Heisenberg Pictures We now return to the time-evolution operator U (t, t0). In the previous section we examined how state kets evolve with time. This means that we were following approach 1, known as the Schr\u00a8odinger picture when applied to time evolution. Alternatively we may follow approach 2, known as the Heisenberg picture when applied to time evolution. In the Schr\u00a8odinger picture the operators corresponding to observables like x, py, andS z are \ufb01xed in time, while state kets vary with time, as indicated in the previous section. In contrast, in the Heisenberg picture the operators corresponding to observables vary with time; the state kets are \ufb01xed, frozen so to speak, at what they were at t0. It is convenient to set t0 in U (t, t0) to zero for simplicity and work with U (t), which is de\ufb01ned by U (t, t0 = 0) \u2261 U (t) =exp. \u2212iHt \u00afh (2.83) Motivated by (2.79b) of approach 2, we de\ufb01ne the Heisenberg picture observable by A(H)(t) \u2261 U \u2020(t)A(S)U (t), (2.84) where the superscripts H and S stand for Heisenberg and Schr\u00a8odinger, respectively. At t = 0, the Heisenberg picture observable and the corresponding Schr\u00a8odinger picture observable coincide: A(H)(0) = A(S). (2.85) The state kets also coincide between the two pictures at t = 0; at later t the Heisenberg picture state ket is frozen to what it was at t = 0: |\u03b1, t0 = 0; tH = |\u03b1, t0 = 0, independent of t. This is in dramatic contrast with the Schr\u00a8odinger picture state ket, |\u03b1, t0 = 0; tS = U (t)|\u03b1, t0 = 0. The expectation value A is obviously the same in both pictures: (2.86) (2.87) S\u03b1, t0 = 0; t|A(S)|\u03b1, t0 = 0; tS = \u03b1, t0 = 0|U \u2020A(S)U |\u03b1, t0 = 0 = H\u03b1, t0 = 0; t|A(H)(t)|\u03b1, t0 = 0; tH. (2.88) 78 Quantum Dynamics 2.2.3 The Heisenberg Equation of Motion We now derive the fundamental equation of motion in the Heisenberg picture. Assuming that A(S) does not depend explicitly on time, which is the case in most physical situations of interest, we obtain [by differentiating (2.84)] dA(H) dt A(S)U + U \u2020A(S) \u2202U \u2202t U \u2020HU U \u2020A(S)U + \u2202U \u2020 = \u2202t = \u2212 1 i\u00afh 1 [A(H), U \u2020HU ], i\u00afh = 1 i\u00afh U \u2020A(S)U U \u2020HU where we have used [see (2.25)] \u2202U \u2202t = 1 i\u00afh HU, \u2202U \u2020 \u2202t = \u2212 1 i\u00afh U \u2020H. (2.89) (2.90a) (2.90b) Because H was originally introduced in the Schr\u00a8odinger picture, we may be tempted to de\ufb01ne H(H) = U \u2020HU (2.91) in accordance with (2.84). But in elementary applications where U is given by (2.83), U and H obviously commute; as a result, so it is all right to write (2.89) as U \u2020HU = H, dA(H) dt = 1 i\u00afh # $ A(H), H. (2.92) (2.93) This equation is known as the Heisenberg equation of motion. Notice that we have derived it using the properties of the time-evolution operator and the de\ufffd", "\ufffd\ufffdning equation for A(H). It is instructive to compare (2.93) with the classical equation of motion in Poisson bracket form. In classical physics, for a function A of q and p that does not involve time explicitly, we have (Goldstein et al. (2002), pp. 396\u2013397) dA dt = [A, H]classical. (2.94) Again, we see that Dirac\u2019s quantization rule (1.6.47) leads to the correct equation in quantum mechanics. Indeed, historically (2.93) was \ufb01rst written by P. A. M. Dirac, who, with his characteristic modesty, called it the Heisenberg equation of motion. It is worth noting, however, that (2.93) makes sense whether or not A(H) has a classical analogue. For example, the spin operator in the Heisenberg picture satis\ufb01es 79 2.2 The Schr\u00f6dinger Versus the Heisenberg Picture # $ dS(H) i dt = S(H) i 1 i\u00afh which can be used to discuss spin precession, but this equation has no classical counterpart because Sz cannot be written as a function of q and p. Rather than insisting on Dirac\u2019s rule, (1.229), we may argue that for quantities possessing classical counterparts, the correct classical equation can be obtained from the corresponding quantum-mechanical equation via the ansatz, (2.95), H, Classical mechanics can be derived from quantum mechanics, but the opposite is not true.4 [, ] i\u00afh \u2192 [, ]classical. (2.96) 2.2.4 Free Particles: Ehrenfest\u2019s Theorem Whether we work in the Schr\u00a8odinger picture or in the Heisenberg picture, to be able to use the equations of motion we must \ufb01rst learn how to construct the appropriate Hamiltonian operator. For a physical system with classical analogues, we assume the Hamiltonian to be of the same form as in classical physics; we merely replace the classical xi and pi by the corresponding operators in quantum mechanics. With this assumption we can reproduce the correct classical equations in the classical limit. Whenever an ambiguity arises because of noncommuting observables, we attempt to resolve it by requiring H to be Hermitian; for instance, we write the quantum-mechanical analogue of the classical product xp as 1 2 (xp + px). When the physical system in question has no classical analogues, we can only guess the structure of the Hamiltonian operator. We try various forms until we get the Hamiltonian that leads to results agreeing with empirical observation. In practical applications it is often necessary to evaluate the commutator of xi (or pi) with functions of xj and pj. To this end the following formulas are found to be useful: and [xi, F(p)] = i\u00afh \u2202F \u2202pi [ pi, G(x)] = \u2212i\u00afh \u2202G \u2202xi, (2.97a) (2.97b) where F and G are functions that can be expanded in powers of pj and xj, respectively. We can easily prove both formulas by repeatedly applying (1.232e). We are now in a position to apply the Heisenberg equation of motion to a free particle of mass m. The Hamiltonian is taken to be of the same form as in classical mechanics: % & H = p2 2m = y + p2 z x + p2 p2 2m. (2.98) 4 In this book we follow the order: the Schr\u00a8odinger picture \u2192 the Heisenberg picture \u2192 classical. For an enlightening treatment of the same subject in opposite order, classical \u2192 the Heisenberg picture \u2192 the Schr\u00a8odinger picture, see Finkelstein (1973), pp. 68\u201370 and 109. 80 Quantum Dynamics We look at the observables pi and xi, which are understood to be the momentum and the position operator in the Heisenberg picture even though we omit the superscript (H). Because pi commutes with any function of pj, we have dpi dt = 1 i\u00afh [ pi, H] =0. (2.99) Thus for a free particle, the momentum operator is a constant of the motion, which means that pi(t) is the same as pi(0) at all times. Quite generally, it is evident from the Heisenberg equation of motion (2.93) that whenever A(H) commutes with the Hamiltonian, A(H) is a constant of the motion. Next, dxi dt = 1 i\u00afh [xi, H] = 1 i\u00afh 1 2m i\u00afh \u2202 \u2202pi 3 \u2211 j=1 p", "2 j pi m where we have taken advantage of (2.97a), so we have the solution pi(0) m = =, (2.100) xi(t) =x i(0) + t, (2.101) pi(0) m which is reminiscent of the classical trajectory equation for a uniform rectilinear motion. It is important to note that even though we have [xi(0), xj(0)] = 0 (2.102) at equal times, the commutator of the xi at different times does not vanish; speci\ufb01cally, [xi(t), xi(0)] =, xi(0) = pi(0)t m \u2212i\u00afht m. (2.103) (2.104) Applying the uncertainty relation (1.146) to this commutator, we obtain (\u0394xi)2t(\u0394xi)2t=0 \u2265 \u00afh2t2 4m2. Among other things, this relation implies that even if the particle is well localized at t = 0, its position becomes more and more uncertain with time, a conclusion which can also be obtained by studying the time-evolution behavior of free-particle wave packets in wave mechanics. We now add a potential V(x) to our earlier free-particle Hamiltonian: H = p2 2m + V(x). (2.105) Here V(x) is to be understood as a function of the x-, y-, and z-operators. Using (2.97b) this time, we obtain dpi dt = 1 i\u00afh [ pi, V(x)] = \u2212 \u2202 \u2202xi V(x). On the other hand, we see that dxi dt = pi m (2.106) (2.107) 81 2.2 The Schr\u00f6dinger Versus the Heisenberg Picture still holds because xi commutes with the newly added term V(x). We can use the Heisenberg equation of motion once again to deduce, H = $ # pi m 1 i\u00afh d2xi dt2 = = 1 i\u00afh 1 m dxi dt dpi dt., H (2.108) Combining this with (2.32), we \ufb01nally obtain in vectorial form d2x dt2 = \u2212\u2207V(x). This is the quantum-mechanical analogue of Newton\u2019s second law. By taking the expectation values of both sides with respect to a Heisenberg state ket that does not move with time, we obtain (2.109) m m d2 dt2 x = dp dt = \u2212\u2207V(x). (2.110) This is known as the Ehrenfest theorem after P. Ehrenfest, who derived it in 1927 using the formalism of wave mechanics. When written in this expectation form, its validity is independent of whether we are using the Heisenberg or the Schr\u00a8odinger picture; after all, the expectation values are the same in the two pictures. In contrast, the operator form (2.109) is meaningful only if we understand x and p to be Heisenberg picture operators. We note that in (2.110) the \u00afh have completely disappeared. It is therefore not surprising that the center of a wave packet moves like a classical particle subjected to V(x). 2.2.5 Base Kets and Transition Amplitudes So far we have avoided asking how the base kets evolve in time. A common misconception is that as time goes on, all kets move in the Schr\u00a8odinger picture and are stationary in the Heisenberg picture. This is not the case, as we will make clear shortly. The important point is to distinguish the behavior of state kets from that of base kets. We started our discussion of ket spaces in Section 1.2 by remarking that the eigenkets of observables are to be used as base kets. What happens to the de\ufb01ning eigenvalue equation A|a = a|a (2.111) with time? In the Schr\u00a8odinger picture, A does not change, so the base kets, obtained as the solutions to this eigenvalue equation at t = 0, for instance, must remain unchanged. Unlike state kets, the base kets do not change in the Schr\u00a8odinger picture. The whole situation is very different in the Heisenberg picture, where the eigenvalue equation we must study is for the time-dependent operator A(H)(t) = U \u2020A(0)U. From (2.111) evaluated at t = 0, when the two pictures coincide, we deduce U \u2020A(0)U U \u2020|a = aU \u2020|a", ", (2.112) (2.113) 82 Quantum Dynamics which implies an eigenvalue equation for A(H): A(H)(U \u2020|a) =a (U \u2020|a). (2.114) If we continue to maintain the view that the eigenkets of observables form the base kets, then {U \u2020|a} must be used as the base kets in the Heisenberg picture. As time goes on, the Heisenberg picture base kets, denoted by |a, tH, move as follows: |a, tH = U \u2020|a. Because of the appearance of U \u2020 rather than U in (2.115), the Heisenberg picture base kets are seen to rotate oppositely when compared with the Schr\u00a8odinger picture state kets; speci\ufb01cally, |a, tH satis\ufb01es the \u201cwrong-sign Schr\u00a8odinger equation\u201d (2.115) i\u00afh \u2202 \u2202t |a, tH = \u2212H|a, tH. (2.116) As for the eigenvalues themselves, we see from (2.114) that they are unchanged with time. This is consistent with the theorem on unitary equivalent observables discussed in Section 1.5. Notice also the following expansion for A(H)(t) in terms of the base kets and bras of the Heisenberg picture: A(H)(t) = \u2211 a = \u2211 a |a, tHa Ha, t| U \u2020|aaa|U = U \u2020A(S)U, (2.117) which shows that everything is quite consistent provided that the Heisenberg base kets change as in (2.115). We see that the expansion coef\ufb01cients of a state ket in terms of base kets are the same in both pictures: ca (t) = a| base bra \u00b7 (U |\u03b1, t0 = 0) state ket \u00b7 |\u03b1, t0 = 0 state ket ca (t) = (a|U ) base bra (the Schr\u00a8odinger picture) (2.118a) (the Heisenberg picture). (2.118b) Pictorially, we may say that the cosine of the angle between the state ket and the base ket is the same whether we rotate the state ket counterclockwise or the base ket clockwise. These considerations apply equally well to base kets that exhibit a continuous spectrum; in particular, the wave function x|\u03b1 can be regarded either as (1) the inner product of the stationary position eigenbra with the moving state ket (the Schr\u00a8odinger picture) or as (2) the inner product of the moving position eigenbra with the stationary state ket (the Heisenberg picture). We will discuss the time dependence of the wave function in Section 2.4, where we will derive the celebrated wave equation of Schr\u00a8odinger. To illustrate further the equivalence between the two pictures, we study transition amplitudes, which will play a fundamental role in Section 2.6. Suppose there is a physical system prepared at t = 0 to be in an eigenstate of observableA with eigenvalue a. At some later time t we may ask: What is the probability amplitude, known as the transition 83 2.3 Simple Harmonic Oscillator Table 2.1 The Schr\u00f6dinger Picture Versus the Heisenberg Picture Schr\u00a8odinger picture Heisenberg picture Moving: (2.5), (2.27) Stationary State ket Observable Stationary Stationary Base ket Moving: (2.84), (2.93) Moving oppositely: (2.115), (2.116) amplitude, for the system to be found in an eigenstate of observable B with eigenvalue b? Here A and B can be the same or different. In the Schr\u00a8odinger picture the state ket at t is given by U |a, while the base kets |a and |b do not vary with time; so we have b| base bra \u00b7 (U |a) state ket (2.119) for this transition amplitude. In contrast, in the Heisenberg picture the state ket is stationary, that is, it remains as |a at all times, but the base kets evolve oppositely. So the transition amplitude is \u00b7 (b|U ) base bra. |a state ket (2.120) Obviously (2.119) and (2.120) are the same. They can both be written as b|U (t, 0)|a. In some loose sense this is the transition amplitude for \u201cgoing\u201d from state |a to state |b. To conclude this section let us summarize the differences between the Schr\u00a8odinger (2.121)", " picture and the Heisenberg picture; see Table 2.1. 2.3 Simple Harmonic Oscillator The simple harmonic oscillator is one of the most important problems in quantum mechanics. It not only illustrates many of the basic concepts and methods of quantum mechanics, it also has much practical value. Essentially any potential well can be approximated by a simple harmonic oscillator, so it describes phenomena from molecular vibrations to nuclear structure. Furthermore, since the Hamiltonian is basically the sum of squares of two canonically conjugate variables, it is also an important starting point for much of quantum \ufb01eld theory. 2.3.1 Energy Eigenkets and Energy Eigenvalues We begin our discussion with Dirac\u2019s elegant operator method, which is based on the earlier work of M. Born and N. Wiener, to obtain the energy eigenkets and energy eigenvalues of the simple harmonic oscillator. The basic Hamiltonian is 84 Quantum Dynamics H = p2 2m + m\u03c92x2 2, (2.122) where \u03c9 is the angular frequency of the classical oscillator related to the spring constant k in Hooke\u2019s law via \u03c9 = k/m. The operators x and p are, of course, Hermitian. It is convenient to de\ufb01ne two non-Hermitian operators, \u2019 \u2019 \u221a a = m\u03c9 2\u00afh x + ip m\u03c9, a\u2020 = m\u03c9 2\u00afh x \u2212 ip m\u03c9, (2.123) known as the annihilation operator and the creation operator, respectively, for reasons that will become evident shortly. Using the canonical commutation relations, we readily obtain We also de\ufb01ne the number operator! \" = a, a\u2020 1 2\u00afh (\u2212i[x, p] +i [ p, x]) = 1. (2.124) N = a\u2020a, (2.125) which is obviously Hermitian. It is straightforward to show that p2 m2\u03c92 + i 2\u00afh a\u2020a = = m\u03c9 2\u00afh \u2212 1 2 H \u00afh\u03c9 x2 +, [x, p] (2.126) so we have an important relation between the number operator and the Hamiltonian operator: H = \u00afh\u03c9(N + 1 2 ). (2.127) Because H is just a linear function of N, N can be diagonalized simultaneously with H. We denote an energy eigenket of N by its eigenvalue n, so N|n = n|n. (2.128) We will later show that n must be a nonnegative integer. Because of (2.127) we also have which means that the energy eigenvalues are given by H|n = (n + 1 2 )\u00afh\u03c9|n, En = (n + 1 2 )\u00afh\u03c9. To appreciate the physical signi\ufb01cance of a, a\u2020, andN, let us \ufb01rst note that \" \"!! [N, a] = a\u2020a, a = a\u2020 [a, a] + a\u2020, a a = \u2212a, where we have used (2.124). Likewise, we can derive \"! N, a\u2020 = a\u2020. (2.129) (2.130) (2.131) (2.132) 85 2.3 Simple Harmonic Oscillator As a result, we have and Na\u2020|n = ([N, a\u2020] +a \u2020N)|n = (n + 1)a\u2020|n (2.133a) Na|n = ([N, a] +aN )|n = (n \u2212 1)a|n. (2.133b) These relations imply that a\u2020|n(a|n) is also an eigenket of N with eigenvalue increased (decreased) by one. Because the increase (decrease) of n by one amounts to the creation (annihilation) of one quantum unit of energy \u00afh\u03c9, the term creation operator (annihilation operator) for a\u2020 (a) is deemed appropriate. Equation (2.133b) implies that a|n and |n \u2212 1 are the same up to a multiplicative constant. We write a|n = c|n \u2212 1, (2.134) where c is a numerical constant to be determined from the requirement that both |n and |n \u2212 1 be normalized. First, note that n|a\u2020a|n = |c| 2. (2.135) We can evaluate the left-hand side of (2.135) by noting that a\u2020a is just the number operator, so Taking c to be real and positive by convention, we \ufb01nally obtain n = |c| 2. Similarly, it is easy to show", " that \u221a a|n = n|n \u2212 1. a\u2020|n = \u221a n + 1|n + 1. (2.136) (2.137) (2.138) Suppose that we keep on applying the annihilation operator a to both sides of (2.137): a2|n = a3|n =... n(n \u2212 1)|n \u2212 2, n(n \u2212 1)(n \u2212 2)|n \u2212 3, (2.139) We can obtain numerical operator eigenkets with smaller and smaller n until the sequence terminates, which is bound to happen whenever we start with a positive integer n. One may argue that if we start with a noninteger n, the sequence will not terminate, leading to eigenkets with a negative value of n. But we also have the positivity requirement for the norm of a|n: n = n|N|n = (n|a\u2020) \u00b7 (a|n) \u2265 0, (2.140) which implies that n can never be negative! So we conclude that the sequence must terminate with n = 0 and that the allowed values of n are nonnegative integers. 86 Quantum Dynamics Because the smallest possible value of n is zero, the ground state of the harmonic oscillator has 1 2 We can now successively apply the creation operator a\u2020 to the ground state |0. Using (2.138), we obtain (2.141) E0 = \u00afh\u03c9. a\u2020 \u221a 2 a\u2020 \u221a 3 |1 = |2 = (a\u2020)2 \u221a 2 (a\u2020)3 \u221a 3! |0, |0, (2.142) |1 = a\u2020|0, |2 = |3 =... |n = (a\u2020)n \u221a n! |0. In this way we have succeeded in constructing simultaneous eigenkets of N and H with energy eigenvalues & % n + 1 2 En = \u00afh\u03c9 (n = 0, 1, 2, 3,...). (2.143) From (2.137), (2.138), and the orthonormality requirement for {|n}, we obtain the matrix elements n|a|n = \u221a n\u03b4n,n\u22121, n|a\u2020|n = \u221a n + 1\u03b4n,n+1. (2.144) Using these together with \u2019 \u2019 m\u00afh\u03c9 2 we derive the matrix elements of the x and p operators: \u00afh 2m\u03c9 (a + a\u2020), p = i x = (\u2212a + a\u2020), (2.145) \u2019 n|x|n = \u2019 n|p|n = i \u221a n\u03b4n,n\u22121 + \u00afh 2m\u03c9 ( \u221a n + 1\u03b4n,n+1), m\u00afh\u03c9 2 (\u2212 \u221a n\u03b4n,n\u22121 + \u221a n + 1\u03b4n,n+1). (2.146a) (2.146b) Notice that neither x nor p is diagonal in the N-representation we are using. This is not surprising because x and p, like a and a\u2020, do not commute with N. The operator method can also be used to obtain the energy eigenfunctions in position space. Let us start with the ground state de\ufb01ned by a|0 = 0, which, in the x-representation, reads \u2019 x|a|0 = x| x + ip m\u03c9 m\u03c9 2\u00afh |0 = 0. (2.147) (2.148) 87 2.3 Simple Harmonic Oscillator Recalling (1.249), we can regard this as a differential equation for the ground-state wave function x|0: x + x2 0 d dx x|0 = 0, where we have introduced \u2019 \u00afh m\u03c9, x0 \u2261 (2.149) (2.150) which sets the length scale of the oscillator. We see that the normalized solution to (2.149) is x|0 = 1 \u221a \u03c01/4 x0 exp 2 x x0 \u2212 1 2. (2.151) We can also obtain the energy eigenfunctions for excited states by evaluating x|1 = x|a\u2020|0 = 1\u221a 2x0 d x \u2212 x2 dx 0 x|2 = x|(a\u2020)2|0 = 1\u221a 2 1\u221a 2! x|0, 2 1\u221a 2x0 2 x \u2212 x2 0 d dx x|0,.... (2.152) In general, we obtain x|n = 1 \u221a \u03c01/4 2nn! x \u2212 x2 0 d dx n exp 2 x x0 \u2212 1 2 1 xn+", "1/2 0. (2.153) It is instructive to look at the expectation values of x2 and p2 for the ground state. First, note that \u00afh 2m\u03c9 x2 = (a2 + a\u20202 + a\u2020a + aa\u2020). (2.154) When we take the expectation value of x2, only the last term in (2.154) yields a nonvanishing contribution: Likewise, x2 = \u00afh 2m\u03c9 = x2 0 2. p2 = \u00afhm\u03c9 2. (2.155) (2.156) It follows that the expectation values of the kinetic and the potential energies are, respectively, ( ) ( ) p2 2m = \u00afh\u03c9 4 = H 2 and m\u03c92x2 2 = \u00afh\u03c9 4 = H 2, as expected from the virial theorem. From (2.146a) and (2.146b), it follows that x = p = 0, (2.157) (2.158) 88 Quantum Dynamics which also holds for the excited states. We therefore have (\u0394x)2 = x2 = \u00afh 2m\u03c9 and (\u0394p)2 = p2 = \u00afhm\u03c9 2, (2.159) and we see that the uncertainty relation is satis\ufb01ed in the minimum uncertainty product form: (\u0394x)2(\u0394p)2 = \u00afh2 4. (2.160) This is not surprising because the ground-state wave function has a Gaussian shape. In contrast, the uncertainty products for the excited states are larger: (\u0394x)2(\u0394p)2 = n + 2 1 2 \u00afh2, (2.161) as the reader may easily verify. 2.3.2 Time Development of the Oscillator So far we have not discussed the time evolution of oscillator state kets nor of observables like x and p. Everything we have done is supposed to hold at some instant of time, say at t = 0; the operators x, p, a, anda \u2020 are to be regarded either as Schr\u00a8odinger picture operators (at all t) or as Heisenberg picture operators at t = 0. In the remaining part of this section, we work exclusively in the Heisenberg picture, which means that x, p, a, anda \u2020 are all time dependent even though we do not explicitly write x(H)(t), and so forth. The Heisenberg equations of motion for p and x are, from (2.106) and (2.107), and dp dt = \u2212m\u03c92x dx dt = p m. (2.162a) (2.162b) This pair of coupled differential equations is equivalent to two uncoupled differential equations for a and a\u2020, namely, da dt = and whose solutions are \u2019 m\u03c9 2\u00afh p m \u2212 i\u03c9x = \u2212i \u03c9 a da\u2020 dt = i \u03c9 a\u2020, (2.163a) (2.163b) a(t) =a(0) exp(\u2212i\u03c9t) and a\u2020(t) =a \u2020(0) exp(i\u03c9t). (2.164) Incidentally, these relations explicitly show that N and H are time-independent operators even in the Heisenberg picture, as they must be. In terms of x and p, we can rewrite (2.164) as 89 2.3 Simple Harmonic Oscillator x(t) + ip(t) m\u03c9 = x(0) exp(\u2212i \u03c9 t) +i p(0) m\u03c9 x(t) \u2212 ip(t) m\u03c9 = x(0) exp(i \u03c9 t) \u2212 i p(0) m\u03c9 exp(\u2212i \u03c9 t), exp(i \u03c9 t). (2.165) Equating the Hermitian and anti-Hermitian parts of both sides separately, we deduce x(t) =x (0) cos \u03c9t + p(0) m\u03c9 sin \u03c9 t (2.166a) and p(t) =\u2212 m\u03c9x(0) sin \u03c9t + p(0) cos \u03c9t. (2.166b) These look the same as the classical equations of motion. We see that the x and p operators \u201coscillate\u201d just like their classical analogues. For pedagogical reasons we now present an alternative derivation of (2.166a). Instead of solving the Heisenberg equation of motion, we attempt to evaluate x(t) =exp iHt \u00afh x(0) exp. \u2212iHt \u00afh (2.167) To this end we record a very useful formula: exp(iG\u03bb", ")A exp(\u2212iG\u03bb) = A + i\u03bb [G, A] + i2\u03bb2 2! [G, [G, A]] + \u00b7 \u00b7 \u00b7 + [G, [G, [G,... [G, A]]]...] +\u00b7 \u00b7 \u00b7, (2.168) in\u03bbn n! where G is a Hermitian operator and \u03bb is a real parameter. We leave the proof of this formula, known as the Baker\u2013Hausdorff lemma as an exercise. Applying this formula to (2.167), we obtain iHt \u00afh exp x(0) exp \u2212iHt \u00afh = x(0) + it \u00afh [H, x(0)] + i2t2 2! \u00afh2 [H, [H, x(0)]] + \u00b7 \u00b7 \u00b7. (2.169) Each term on the right-hand side can be reduced to either x or p by repeatedly using and [H, x(0)] = \u2212i\u00afhp(0) m (2.170a) [H, p(0)] = i\u00afhm\u03c92x(0). (2.170b) 90 Quantum Dynamics Thus iHt \u00afh exp x(0) exp \u2212iHt \u00afh = x(0) + \u2212 1 3! t \u2212 p(0) m t3\u03c92p(0) m p(0) m\u03c9 = x(0) cos \u03c9t + 1 2! t2\u03c92x(0) + \u00b7 \u00b7 \u00b7 sin \u03c9t, (2.171) in agreement with (2.166a). From (2.166a) and (2.166b), one may be tempted to conclude that x and p always oscillate with angular frequency \u03c9. However, this inference is not correct. Take any energy eigenstate characterized by a de\ufb01nite value of n; the expectation value n|x(t)|n vanishes because the operators x(0) and p(0) change n by \u00b1 1 and |n and |n \u00b1 1 are orthogonal. This point is also obvious from our earlier conclusion (see Section 2.1) that the expectation value of an observable taken with respect to a stationary state does not vary with time. To observe oscillations reminiscent of the classical oscillator, we must look at a superposition of energy eigenstates such as |\u03b1 = c0|0 + c1|1. (2.172) The expectation value of x(t) taken with respect to (2.172) does oscillate, as the reader may readily verify. We have seen that an energy eigenstate does not behave like the classical oscillator \u2013 in the sense of oscillating expectation values for x and p \u2013 no matter how large n may be. We may logically ask: How can we construct a superposition of energy eigenstates that most closely imitates the classical oscillator? In wave function language, we want a wave packet that bounces back and forth without spreading in shape. It turns out that a coherent state de\ufb01ned by the eigenvalue equation for the non-Hermitian annihilation operator a, a|\u03bb = \u03bb|\u03bb, (2.173) with, in general, a complex eigenvalue \u03bb does the desired job. The coherent state has many other remarkable properties. 1. When expressed as a superposition of energy (or N) eigenstates, |\u03bb = \u221e \u2211 n = 0 f (n)|n, (2.174) the distribution of | f (n)|2 with respect to n is of the Poisson type about some mean value \u00afn: | f (n)|2 = exp(\u2212\u00afn). (2.175) \u00afnn n! 2. It can be obtained by translating the oscillator ground state by some \ufb01nite distance. 3. It satis\ufb01es the minimum uncertainty product relation at all times. 91 2.4 Schr\u00f6dinger\u2019s Wave Equation A systematic study of coherent states, pioneered by R. Glauber, is very rewarding; the reader is urged to work out Exercise 2.21 on this subject at the end of this chapter.5 2.4 Schr\u00f6dinger\u2019s Wave Equation 2.4.1 Time-Dependent Wave Equation We now turn to the Schr\u00a8odinger picture and examine the time evolution of |\u03b1, t0; t in the x-representation. In other words, our task is to study the behavior of the wave function \u03c8(x, t) =x |\u03b1, t0; t as a function of time, where |\u03b1, t0; t is a state ket in the Schr\u00a8odinger picture at time t, and x| is a time-independent position eigen", "bra with eigenvalue x. The Hamiltonian operator is taken to be (2.176) H = p2 2m + V(x). (2.177) The potential V(x) is a Hermitian operator; it is also local in the sense that in the xrepresentation we have x|V(x)|x = V(x)\u03b43(x \u2212 x), (2.178) where V(x) is a real function of x. Later in this book we will consider more complicated Hamiltonians: a time-dependent potential V(x, t); a nonlocal but separable potential where the right-hand side of (2.178) is replaced by v1(x)v2(x); a momentum-dependent interaction of the form p \u00b7 A + A \u00b7 p, whereA is the vector potential in electrodynamics, and so on. We now derive Schr\u00a8odinger\u2019s time-dependent wave equation. We \ufb01rst write the Schr\u00a8odinger equation for a state ket (2.27) in thex-representation: \u2202 \u2202t x|\u03b1, t0; t = x|H|\u03b1, t0; t, i\u00afh (2.179) where we have used the fact that the position eigenbras in the Schr\u00a8odinger picture do not change with time. Using (1.252), we can write the kinetic-energy contribution to the righthand side of (2.179) as ( x p2 2m \u03b1, t0; t = \u2212 \u2207 2x|\u03b1, t0; t. ) \u00afh2 2m As for V(x), we simply use x|V(x) =x |V(x), (2.180) (2.181) 5 For applications to laser physics, see Sargent, et al. (1974) and Loudon (2000). See also the discussion on squeezed light at the end of Section 7.8 of this book. 92 Quantum Dynamics where V(x) is no longer an operator. Combining everything, we deduce i\u00afh \u2202 \u2202t x|\u03b1, t0; t = \u2212 \u00afh2 2m \u2207 2x|\u03b1, t0; t + V(x)x|\u03b1, t0; t, (2.182) which we recognize to be the celebrated time-dependent wave equation of E. Schr\u00a8odinger, usually written as i\u00afh \u2202 \u2202t \u03c8(x, t) =\u2212 \u00afh2 2m \u2207 2\u03c8(x, t) +V (x)\u03c8(x, t). (2.183) The quantum mechanics based on wave equation (2.183) is known as wave mechanics. This equation is, in fact, the starting point of many textbooks on quantum mechanics. In our formalism, however, this is just the Schr\u00a8odinger equation for a state ket written explicitly in the x-basis when the Hamiltonian operator is taken to be (2.177). 2.4.2 The Time-Independent Wave Equation We now derive the partial differential equation satis\ufb01ed by energy eigenfunctions. We showed in Section 2.1 that the time dependence of a stationary state is given by exp(\u2212iEat/\u00afh). This enables us to write its wave function as x|a, t0; t = x|a exp, \u2212iEat \u00afh (2.184) where it is understood that initially the system is prepared in a simultaneous eigenstate of A and H with eigenvalues a and Ea, respectively. Let us now substitute (2.184) into the time-dependent Schr\u00a8odinger equation (2.182). We are then led to \u2212 \u00afh2 2m \u2207 2x|a + V(x)x|a = Ea x|a. (2.185) This partial differential equation is satis\ufb01ed by the energy eigenfunction x|a with energy eigenvalue Ea. Actually, in wave mechanics where the Hamiltonian operator is given as a function of x and p, as in (2.177), it is not necessary to refer explicitly to observable A that commutes with H because we can always choose A to be that function of the observables x and p which coincides with H itself. We may therefore omit reference to a and simply write (2.185) as the partial differential equation to be satis\ufb01ed by the energy eigenfunction uE(x): \u2212 \u00afh2 2m \u2207 2uE(x) +V (x)uE(x) =Eu E(x). (2.186) This is the time-independent wave equation of E. Schr\u00a8odinger, announced in the \ufb01", "rst of four monumental papers, all written in the \ufb01rst half of 1926, that laid the foundations of wave mechanics. In the same paper, Schr\u00a8odinger immediately applied (2.186) to derive the energy spectrum of the hydrogen atom. 93 2.4 Schr\u00f6dinger\u2019s Wave Equation To solve (2.186) some boundary condition has to be imposed. Suppose we seek a solution to (2.186) with E < lim |x|\u2192\u221e where the inequality relation is to hold for |x| \u2192\u221e in any direction. The appropriate boundary condition to be used in this case is (2.187) V(x), uE(x) \u2192 0 as |x| \u2192\u221e. (2.188) Physically this means that the particle is bound or con\ufb01ned within a \ufb01nite region of space. We know from the theory of partial differential equations that (2.186) subject to boundary condition (2.188) allows nontrivial solutions only for a discrete set of values of E. It is in this sense that the time-independent Schr\u00a8odinger equation (2.186) yields the quantization of energy levels.6 Once the partial differential equation (2.186) is written, the problem of \ufb01nding the energy levels of microscopic physical systems is as straightforward as that of \ufb01nding the characteristic frequencies of vibrating strings or membranes. In both cases we solve boundary-value problems in mathematical physics. A short digression on the history of quantum mechanics is in order here. The fact that exactly soluble eigenvalue problems in the theory of partial differential equations can also be treated using matrix methods was already known to mathematicians in the \ufb01rst quarter of the twentieth century. Furthermore, theoretical physicists like M. Born frequently consulted great mathematicians of the day \u2013 D. Hilbert and H. Weyl, in particular. Yet when matrix mechanics was born in the summer of 1925, it did not immediately occur to the theoretical physicists or to the mathematicians to reformulate it using the language of partial differential equations. Six months after Heisenberg\u2019s pioneering paper, wave mechanics was proposed by Schr\u00a8odinger. However, a close inspection of his papers shows that he was not at all in\ufb02uenced by the earlier works of Heisenberg, Born, and Jordan. Instead, the train of reasoning that led Schr\u00a8odinger to formulate wave mechanics has its roots in W. R. Hamilton\u2019s analogy between optics and mechanics, on which we will comment later, and the particle-wave hypothesis of L. de Broglie. Once wave mechanics was formulated, many people, including Schr\u00a8odinger himself, showed the equivalence between wave mechanics and matrix mechanics. It is assumed that the reader of this book has some experience in solving the timedependent and time-independent wave equations. He or she should be familiar with the time evolution of a Gaussian wave packet in a force-free region; should be able to solve one-dimensional transmission-re\ufb02ection problems involving a rectangular potential barrier, and the like; should have seen derived some simple solutions of the time-independent wave equation \u2013 a particle in a box, a particle in a square well, the simple harmonic oscillator, the hydrogen atom, and so on \u2013 and should also be familiar with some general properties of the energy eigenfunctions and energy eigenvalues, such as (1) the fact that the energy levels exhibit a discrete or continuous spectrum depending on whether or not (2.187) is 6 Schr\u00a8odinger\u2019s paper that announced (2.186) is appropriately entitled Quantisierung als Eigenwertproblem (Quantization as an Eigenvalue Problem). 94 Quantum Dynamics satis\ufb01ed and (2) the property that the energy eigenfunction in one dimension is sinusoidal or damped depending on whether E \u2212 V(x) is positive or negative. In this book, we do not thoroughly cover these more elementary topics and solutions. Some of these are pursued, for example the harmonic oscillator and hydrogen atom, but at a mathematical level somewhat higher than what is usually seen in undergraduate courses. In any case, a brief summary of elementary solutions to Schr\u00a8odinger\u2019s equations is presented in Appendix B. 2.4.3 Interpretations of the Wave Function We now turn to discussions of the physical interpretations of the wave function. In Section 1.7 we commented on the probabilistic interpretation of |\u03c8|2 that follows from the fact that x|\u03b1, t0; t is to be regarded as an expansion coef\ufb01cient of |\u03b1, t0; t in terms of the position eigenkets {|x}. The quantity \u03c1(x, t) de\ufb01ned by \u03c1(x, t) =| \u03c8", "(x, t)| 2 = |x|\u03b1, t0; t| 2 (2.189) is therefore regarded as the probability density in wave mechanics. Speci\ufb01cally, when we use a detector that ascertains the presence of the particle within a small volume element d3x around x, the probability of recording a positive result at time t is given by \u03c1(x, t)d3x. In the remainder of this section we use x for x because the position operator will not appear. Using Schr\u00a8odinger\u2019s time-dependent wave equation, it is straightforward to derive the continuity equation \u2202\u03c1 \u2202t + \u2207 \u00b7 j = 0, (2.190) where \u03c1(x, t) stands for |\u03c8| 2 as before, and j(x, t), known as the probability \ufb02ux, is given by i\u00afh 2m j(x, t) =\u2212 = \u00afh m [\u03c8\u2217\u2207\u03c8 \u2212 (\u2207\u03c8\u2217)\u03c8] Im(\u03c8\u2217\u2207\u03c8). (2.191) The reality of the potential V (or the Hermiticity of the V operator) has played a crucial role in our obtaining this result. Conversely, a complex potential can phenomenologically account for the disappearance of a particle; such a potential is often used for nuclear reactions where incident particles get absorbed by nuclei. We may intuitively expect that the probability \ufb02ux j is related to momentum. This is indeed the case for j integrated over all space. From (2.191) we obtain d3xj(x, t) =, (2.192) pt m where pt is the expectation value of the momentum operator at time t. Equation (2.190) is reminiscent of the continuity equation in \ufb02uid dynamics that characterizes a hydrodynamic \ufb02ow of a \ufb02uid in a source-free, sink-free region. Indeed, 95 2.4 Schr\u00f6dinger\u2019s Wave Equation historically Schr\u00a8odinger was \ufb01rst led to interpret |\u03c8| 2 as the actual matter density, or e|\u03c8| 2 as the actual electric charge density. If we adopt such a view, we are led to face some bizarre consequences. A typical argument for a position measurement might go as follows. An atomic electron is to be regarded as a continuous distribution of matter \ufb01lling up a \ufb01nite region of space around the nucleus; yet, when a measurement is made to make sure that the electron is at some particular point, this continuous distribution of matter suddenly shrinks to a pointlike particle with no spatial extension. The more satisfactory statistical interpretation of |\u03c8| 2 as the probability density was \ufb01rst given by M. Born. To understand the physical signi\ufb01cance of the wave function, let us write it as \u03c8(x, t) = \u03c1(x, t) exp iS(x, t) \u00afh, (2.193) with S real and \u03c1 > 0, which can always be done for any complex function of x and t. The meaning of \u03c1 has already been given. What is the physical interpretation of S? Noting \u03c8\u2217\u2207\u03c8 = \u221a \u221a \u03c1 \u2207( \u03c1) + we can write the probability \ufb02ux as [see (2.191)] \u03c1\u2207S m j =. i \u00afh \u03c1\u2207S, (2.194) (2.195) We now see that there is more to the wave function than the fact that |\u03c8|2 is the probability density; the gradient of the phase S contains a vital piece of information. From (2.195) we see that the spatial variation of the phase of the wave function characterizes the probability \ufb02ux; the stronger the phase variation, the more intense the \ufb02ux. The direction of j at some point x is seen to be normal to the surface of a constant phase that goes through that point. In the particularly simple example of a plane wave (a momentum eigenfunction) \u03c8(x, t) \u221d exp, (2.196) ip \u00b7 x \u00afh \u2212 iEt \u00afh where p stands for the eigenvalue of the momentum operator. All this is evident because More generally, it is tempting to regard \u2207S/m as some kind of \u201cvelocity,\u201d \u2207S = p. \u201cv\u201d = \u2207S m, and to write the continuity equation (2.190) as \u2202\u03c1 \u2202t + \u2207 \u00b7(\u03c1 \u201cv\u201d) = 0, (2.197) (2.198) (2.199) just as in \ufb02", "uid dynamics. However, we would like to caution the reader against a too literal interpretation of j as \u03c1 times the velocity de\ufb01ned at every point in space, because a simultaneous precision measurement of position and velocity would necessarily violate the uncertainty principle. 96 Quantum Dynamics 2.4.4 The Classical Limit We now discuss the classical limit of wave mechanics. First, we substitute \u03c8 written in form (2.193) into both sides of the time-dependent wave equation. Straightforward differentiations lead to \u221a (\u2207 \u03c1) \u00b7 (\u2207S) \u2212 \u221a \u03c1|\u2207S| 2 + \u221a \u221a \u03c1 \u22072S + \u03c1V 1 \u00afh2 i \u00afh \u2212 \u00afh2 2m = i\u00afh \u22072 \u221a \u2202 \u03c1 \u2202t \u221a \u03c1 + + i \u00afh 2i \u00afh \u221a. \u03c1 \u2202S \u2202t (2.200) So far everything has been exact. Let us suppose now that \u00afh can, in some sense, be regarded as a small quantity. The precise physical meaning of this approximation, to which we will come back later, is not evident now, but let us assume \u00afh|\u2207 2S| |\u2207S| 2, (2.201) and so forth. We can then collect terms in (2.200) that do not explicitly contain \u00afh to obtain a nonlinear partial differential equation for S: 1 2m |\u2207S(x, t)| 2 + V(x) + \u2202S(x, t) \u2202t = 0. (2.202) We recognize this to be the Hamilton\u2013Jacobi equation in classical mechanics, \ufb01rst written in 1836, where S(x, t) stands for Hamilton\u2019s principal function. So, not surprisingly, in the \u00afh \u2192 0 limit, classical mechanics is contained in Schr\u00a8odinger\u2019s wave mechanics. We have a semiclassical interpretation of the phase of the wave function: \u00afh times the phase is equal to Hamilton\u2019s principal function provided that \u00afh can be regarded as a small quantity. Let us now look at a stationary state with time dependence exp(\u2212iEt/\u00afh). This time dependence is anticipated from the fact that for a classical system with a constant Hamiltonian, Hamilton\u2019s principal function S is separable: S(x, t) = W(x) \u2212 Et, (2.203) where W(x) is called Hamilton\u2019s characteristic function (Goldstein et al. (2002), pp. 440\u2013444). As time goes on, a surface of a constant S advances in much the same way as a surface of a constant phase in wave optics, a \u201cwave front,\u201d advances. The momentum in the classical Hamilton\u2013Jacobi theory is given by Pclass = \u2207S = \u2207W, (2.204) which is consistent with our earlier identi\ufb01cation of \u2207S/m with some kind of velocity. In classical mechanics the velocity vector is tangential to the particle trajectory, and as a result we can trace the trajectory by following continuously the direction of the velocity vector. The particle trajectory is like a ray in geometric optics because the \u2207S that traces the trajectory is normal to the wave front de\ufb01ned by a constant S. In this sense geometrical optics is to wave optics what classical mechanics is to wave mechanics. One might wonder, in hindsight, why this optical-mechanical analogy was not fully exploited in the nineteenth century. The reason is that there was no motivation for regarding 97 2.5 Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation Hamilton\u2019s principal function as the phase of some traveling wave; the wave nature of a material particle did not become apparent until the 1920s. Besides, the basic unit of action \u00afh, which must enter into (2.193) for dimensional reasons, was missing in the physics of the nineteenth century. 2.5 Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation It is both instructive and useful to look at some relatively elementary solutions to (2.186) for particular choices of the potential energy function V(x). In this section we choose some particular examples to illustrate contemporary physics and/or which will be useful in later chapters of this textbook. 2.5.1 Free Particle in Three Dimensions The case V(x) =0 has fundamental signi\ufb01cance. We will consider the solution to Schr\u00a8odinger\u2019s equation here in three dimensions using Cartesian coordinates. The solution in spherical coordinates will be left until our treatment of angular momentum is presented in the next chapter. Equation (2.186) becomes \u22072uE(x) =\u2212 2mE \u00afh2 uE(x). De\ufffd", "\ufffd\ufffdne a vector k where k2 = k2 x + k2 y + k2 z \u2261 2mE \u00afh2 = p2 \u00afh2 (2.205) (2.206) that is, p = \u00afhk. Differential equation (2.205) is easily solved using the technique known as \u201cseparation of variables.\u201d Writing uE(x) =u x(x)uy(y)uz(z) (2.207) we arrive at 1 ux d2ux dx2 + k2 d2uy dy2 + k2 This leads to individual plane wave solutions uw(w) =c weikww for w = x, y, z. Note that one gets the same energy E for values \u00b1kw. d2uz dz2 + k2 (2.208) 1 uy 1 uz = 0. + + x y z Collecting these solutions, and combining the normalization constants, we obtain uE(x) =c xcyczeikxx+ikyy+ikzz = Ceik\u00b7x. (2.209) The normalization constant C presents the usual dif\ufb01culties, which are generally handled by using a \u03b4-function normalization condition. It is convenient in many case, however, to use a \u201cbig box\u201d normalization, where all space is contained within a cube of side length L. We impose periodic boundary conditions on the box, and thereby obtain a \ufb01nite 98 Quantum Dynamics normalization constant C. For any real calculation, we simply let the size L \u2192 \u221e at the end of the calculation. Imposing the condition ux(x + L) = ux(x) we have kxL = 2\u03c0nx where nx is an integer. That is kx = 2\u03c0 L nx ky = 2\u03c0 L ny kz = 2\u03c0 L nz and the normalization criterion becomes L L 1 = dx dy 0 0 in which case C = 1/L3/2 and L 0 dz u\u2217 E(x)uE(x) =L 3|C|2 uE(x) = 1 L3/2 eik\u00b7x. The energy eigenvalue is E = p2 2m = \u00afh2k2 2m = \u00afh2 2m 2 % 2\u03c0 L x + n2 n2 y + n2 z &. (2.210) (2.211) (2.212) (2.213) The sixfold degeneracy we mentioned earlier corresponds to the six combinations of (\u00b1nx, \u00b1ny, \u00b1nz), but the degeneracy can actually be much larger since, in some cases, there are various combinations of nx, ny, andn z which can give the same E. In fact, in the (realistic) limit where L is very large, there can be a large number of states N which have an energy between E and E + dE. This \u201cdensity of states\u201d dN/dE is an important quantity for calculations of processes which include free particles. See, for example, the discussion of the photoelectric effect in Section 5.8. To calculate the density of states, imagine a spherical shell in k space with radius |k| = 2\u03c0|n|/L and thickness d|k| = 2\u03c0d|n|/L. All states within this shell have energy E = \u00afh2k2/2m. The number of states dN within this shell is 4\u03c0n2d|n|. Therefore = 4\u03c0 \u00afh2 m 2 L 2\u03c0 |k| L 2\u03c0 dN dE = = 4\u03c0n2d|n| \u00afh2|k|d|k|/m m3/2E1/2L3 \u221a 2\u03c02 \u00afh3. (2.214) In a typical \u201creal\u201d calculation, the density of states will be multiplied by some probability that involves u\u2217 E(x)uE(x). In this case, the factors of L3 will cancel explicitly, so the limit L \u2192 \u221e is trivial. This \u201cbig box\u201d normalization also yields the correct answer for the probability \ufb02ux. Rewriting (2.196) with this normalization, we have \u03c8(x, t) = 1 L3/2 exp ip \u00b7 x \u00afh \u2212 iEt \u00afh in which case we \ufb01nd j(x, t) = \u00afh m Im(\u03c8\u2217\u2207\u03c8) = \u00afhk m 1 L3 = v\u03c1 where \u03c1 = 1/L3 is indeed the probability density. (2.215) (2.216) 99 2.5 Elementary Solutions to Schr\u00f6dinger", "\u2019s Wave Equation 2.5.2 The Simple Harmonic Oscillator We saw an elegant solution for the case V(x) = m\u03c92x2/2 in Section 2.3, which yielded the energy eigenvalues, eigenstates, and wave functions. Here, we demonstrate a different approach which solves the differential equation \u2212 \u00afh2 2m d2 dx2 uE(x) + 1 2 m\u03c92x2uE(x) =Eu E(x). (2.217) Our approach will introduce the concept of generating functions, a generally useful technique which arises in many treatments of differential eigenvalue problems. First, transform (2.217) using the dimensionless position y \u2261 x/x0 where x0 \u2261 \u00afh/m\u03c9. Also introduce a dimensionless energy variable \u03b5 \u2261 2E/\u00afh\u03c9. The differential equation we need to solve becomes therefore \u221a d2 dy2 u(y) + (\u03b5 \u2212 y2)u(y) =0. For y \u2192 \u00b1\u221e, the solution must tend to zero, otherwise the wave function will not be normalizable and hence unphysical. The differential equation w(y) \u2212 y2w(y) = 0 has solutions w(y) \u221d exp(\u00b1y2/2), so we would have to choose the minus sign. We then \u201cremove\u201d the asymptotic behavior of the wave function by writing (2.218) where the function h(y) satis\ufb01es the differential equation u(y) =h(y )e\u2212y2/2 d2h dy2 \u2212 2y dh dy + (\u03b5 \u2212 1)h(y) = 0. (2.219) (2.220) To this point, we have followed the traditional solution of the simple harmonic oscillator as found in many textbooks. Typically, one would now look for a series solution for h(y) and discover that a normalizable solution is only possible if the series terminates. (In fact, we use this approach for the three-dimensional isotropic harmonic oscillator in this book. See Section 3.7.) One forces this termination by imposing the condition that \u03b5 \u2212 1 be an even, nonnegative integer 2n, n = 0, 1, 2,.... The solutions are then written using the resulting polynomials hn(y). Of course, \u03b5 \u2212 1 = 2n is equivalent to E = (n + 1 2 )\u00afh\u03c9, the quantization relation (2.143). Let us take a different approach. Consider the \u201cHermite polynomials\u201d Hn(x) de\ufb01ned by the \u201cgenerating function\u201d g(x, t) through g(x, t) \u2261 e\u2212t2+2tx \u2261 \u221e \u2211 n=0 Hn(x) tn n!. (2.221a) (2.221b) Some properties of the Hn(x) are immediately obvious. For example, H0(x) =1. Also, since g(0, t) =e \u2212t2 = \u221e \u2211 n=0 (\u22121)n n! t2n (2.222) 100 Quantum Dynamics it is clear that Hn(0) = 0 if n is odd, since this series only involves even powers of t. On the other hand, if we restrict ourselves to even values of n, we have g(0, t) =e \u2212t2 = \u221e \u2211 n=0 (\u22121)(n/2) (n/2)! tn = \u221e \u2211 n=0 (\u22121)(n/2) (n/2)! n! n! tn (2.223) and so Hn(0) = (\u22121)n/2n! /(n/2)!. Also, since g(\u2212x, t) reverses the sign only on terms with odd powers of t, Hn(\u2212x) = (\u22121)nHn(x). We can take derivatives of g(x, t) to build the Hermite polynomials using recursion relations between them and their derivatives. The trick is that we can differentiate the analytic form of the generating function (2.221a) or the series form (2.221b) and then compare results. For example, if we take the derivative using (2.221a), then \u2202g \u2202x = 2tg(x, t) = \u221e \u2211 n=0 2Hn(x) tn+1 n! = \u221e \u2211 n=0 2(n + 1)Hn(x) tn+1 (n + 1)! (2.224) where we insert the series de\ufb01nition of the", " generating function after taking the derivative. On the other hand, we can take the derivative of (2.221b) directly, in which case \u2202g \u2202x = \u221e \u2211 n=0 H n(x) tn n!. Comparing (2.224) and (2.225) shows that H n(x) =2nH n\u22121(x) This is enough information for us build the Hermite polynomials: H0(x) =1 (2.225) (2.226) so H so H therefore H1(x) =2x 1(x) =2, 2(x) =8x, therefore H2(x) =4 x 2 \u2212 2 so H 3(x) =24 x 2 \u2212 12, therefore H3(x) =8x 3 \u2212 12x... So far, this is just a curious mathematical exercise. To see why it is relevant to the simple harmonic oscillator, consider the derivative of the generating function with respect to t. If we start with (2.221a) then \u2202g \u2202t = \u22122tg(x, t) +2xg(x, t) \u221e \u2211 + 2xHn(x) = \u2212 = \u2212 \u221e \u2211 n=0 \u221e \u2211 n=0 2Hn(x) tn+1 n! 2nHn\u22121(x) tn n! n=0 \u221e + \u2211 n=0 tn n! tn n! 2xHn(x). (2.227) Or, if we differentiate (2.221b) then we have \u2202g \u2202t = \u221e \u2211 n=0 nHn(x) tn\u22121 n! = \u221e \u2211 n=0 Hn+1(x) tn n!. (2.228) 101 2.5 Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation Comparing (2.227) and (2.228) gives us the recursion relation Hn+1(x) =2xH n(x) \u2212 2nHn\u22121(x) (2.229) which we combine with (2.226) to \ufb01nd H n(x) =2n \u00b7 2(n \u2212 1)Hn\u22122(x) = 2n [2xHn\u22121(x) \u2212 Hn(x)] = 2xH n(x) \u2212 2nHn(x). In other words, the Hermite polynomials satisfy the differential equation n(x) \u2212 2xH H n(x) +2nH n(x) =0 (2.230) (2.231) where n is a nonnegative integer. This, however, is the same as the Schr\u00a8odinger equation written as (2.220) since \u03b5 \u2212 1 = 2n. That is, the wave functions for the simple harmonic oscillator are given by un(x) =c nHn x e\u2212m\u03c9x2/2\u00afh (2.232) \u2019 m\u03c9 \u00afh up to some normalization constant cn. This constant can be determined from the orthogonality relationship \u221e \u2212\u221e Hn(x)Hm(x)e\u2212x2 dx = \u03c01/22nn! \u03b4nm (2.233) which is easily proved using the generating function. See Problem 2.25 at the end of this chapter. Generating functions have a usefulness that far outreaches our limited application here. Among other things, many of the orthogonal polynomials which arise from solving the Schr\u00a8odinger equation for different potentials, can be derived from generating functions. See, for example, Problem 3.30 in Chapter 3. The interested reader is encouraged to pursue this further, probably best from any one of the many excellent texts on mathematical physics. 2.5.3 The Linear Potential Perhaps the \ufb01rst potential energy function, with bound states, to come to mind is the linear potential, namely V(x) =k |x| (2.234) where k is an arbitrary positive constant. Given a total energy E, this potential has a classical turning point at a value x = a where E = ka. This point will be important for understanding the quantum behavior of a particle of mass m bound by this potential. The Schr\u00a8odinger equation becomes \u2212 \u00afh2 2m d2uE dx2 + k|x|uE(x) =Eu E(x). (2.235) 102 Quantum Dynamics It is easiest to deal with the absolute value by restricting our attention to x \u2265 0. We can do this because V(\u2212x) = V(x), so there are", " two types of solutions, namely uE(\u2212x) =\u00b1 uE(x). In either case, we need uE(x) to tend towards zero as x \u2192 \u221e. Ifu E(\u2212x) =\u2212 uE(x), then we need uE(0) = 0. On the other hand, if uE(\u2212x) = +uE(x), then we have u E(0) = 0, since uE(\u03b5) \u2212 uE(\u2212\u03b5) \u2261 0, even for \u03b5 \u2192 0. (As we will discuss in Chapter 4, we refer to these solutions as \u201codd\u201d and \u201ceven\u201d parity.) Once again, we write the differential equation in terms of dimensionless variables, based on appropriate scales for length and energy. In this case, the dimensionless length scale is x0 = (\u00afh2/mk)1/3 and the dimensionless enery scale is E0 = kx0 = (\u00afh2k2/m)1/3. De\ufb01ning y \u2261 x/x0 and \u03b5 \u2261 E/E0 allows us to rewrite (2.235) as d2uE dy2 \u2212 2(y \u2212 \u03b5)uE(y) =0, y \u2265 0. (2.236) Notice that y = \u03b5 when x = E/k, i.e. the classical turning point x = a. In fact, de\ufb01ning a translated position variable z \u2261 21/3(y \u2212 \u03b5), (2.236) becomes d2uE dz2 \u2212 zuE(z) =0. (2.237) This is the Airy equation, and the solution is the Airy function Ai(z), plotted in Figure 2.3. The Airy function has a peculiar behavior, oscillatory for negative values of the argument, and decreasing rapidly towards zero for positive values. Of course, this is exactly the behavior we expect for the wave function, since z = 0 is the classical turning point. Note that the boundary conditions at x = 0 translate into zeros for either Ai(z) or Ai(z) where z = \u221221/3\u03b5. In other words, the zeros of the Airy function or its derivative determine the quantized energies. One \ufb01nds that Ai(z) = 0 Ai(z) = 0 for z = \u22121.019, \u2212 3.249, \u2212 4.820,... for z = \u22122.338, \u2212 4.088, \u2212 5.521,... (even) (odd). (2.238) (2.239) For example, the ground-state energy is E = (1.019/21/3)(\u00afh2k2/m)1/3. 0.5 ) z ( i A 0 \u22120.5 \u221210 Fig. 2.3 The Airy function. 0 z 10 103 2.5 Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation 0.1 1 \u2212 s, N 0.01 1 \u00d7 10\u22123 0 10 20 30 40 \u0394h, \u03bcm Fig. 2.4 Experimental observation of the quantum-mechanical states of a bouncing neutron, from Nesvizhevsky et al., Phys. Rev. D, 67 (2003) 102002. The solid curve is a fit to the data based on classical physics. Note that the vertical scale is logarithmic. The quantum-theoretical treatment of the linear potential may appear to have little to do with the real world. It turns out, however, that a potential of type (2.234) is actually of practical interest in studying the energy spectrum of a quark-antiquark bound system, called quarkonium. In this case, the x in (2.234) is replaced by the quark-antiquark separation distance r. This constant k is empirically estimated to be in the neighborhood of 1 GeV/fm 1.6 \u00d7 105 N, (2.240) which corresponds to a gravitational force of about 16 tons. Indeed, another real world example of the linear potential is the \u201cbouncing ball.\u201d One interprets (2.234) as the potential energy of a ball of mass m at a height x above the \ufb02oor, and k = mg where g is the local acceleration due to gravity. Of course, this is the potential energy only for x \u2265 0 as there is an in\ufb01nite potential barrier which causes the ball to \u201cbounce.\u201d Quantum mechanically, this means that only the odd parity solutions (2.239) are allowed. The bouncing ball happens to be one of those rare cases where quantum-mechanical effects can be observed macroscopically. The trick is to have a very", " low mass \u201cball,\u201d which has been achieved with neutrons by a group7 working at the Institut Laue-Langevin (ILL) in Grenoble, France. For neutrons with m = 1.68 \u00d7 10\u221227 kg, the characteristic length scale is x0 = (\u00afh2/m2g)1/3 = 7.40 \u03bcm. The \u201callowed heights\u201d to which a neutron can bounce are (2.338/21/3)x0 = 14 \u03bcm, (4.088/21/3)x0 = 24 \u03bcm, (5.521/21/3)x0 = 32 \u03bcm, and so on. These are small, but measurable with precision mechanical devices and very low energy, aka \u201cultracold,\u201d neutrons. The experimenters\u2019 results are shown in Figure 2.4. Plotted is the detected neutron rate as a function of the height of a slit which only allows neutrons 7 See Nesvizhevsky et al., Phys. Rev. D, 67 (2003) 102002, and Eur. Phys. J., C40 (2005) 479. See also Jenke et al., Nature Phys., 7 (2011) 468. 104 Quantum Dynamics to pass if they exceed this height. No neutrons are observed unless the height is at least \u2248 14 \u03bcm, and clear breaks are observed at \u2248 24 \u03bcm and \u2248 32 \u03bcm, in excellent agreement with the predictions of quantum mechanics. 2.5.4 The WKB (Semiclassical) Approximation Having solved the problem of a linear potential, it is worthwhile to introduce an important approximation technique known as the WKB solution, after G. Wentzel, A. Kramers, and L. Brillouin.8 This technique is based on making use of regions where the wavelength is much shorter than the typical distance over which the potential energy varies. Such is never the case near classical turning points, but this is where the linear potential solution can be used to join the solutions on either side of them. Again restricting ourselves to one dimension, we write Schr\u00a8odinger\u2019s wave equation as d2uE dx2 + 2m \u00afh2 (E \u2212 V(x))uE(x) =0. (2.241) De\ufb01ne the quantities k(x) \u2261 k(x) \u2261 \u2212i\u03ba(x) \u2261 \u2212i and so (2.241) becomes 1/2 2m \u00afh2 (E \u2212 V(x)) 2m \u00afh2 (V(x) \u2212 E) for E > V(x) (2.242a) 1/2 for E < V(x) (2.242b) d2uE dx2 + [k(x)]2 uE(x) =0. Now, if V(x) were not changing with x, then k(x) would be a constant, and u(x) \u221d exp(\u00b1ikx) would solve (2.243). Consequently, if we assume that V(x) varies only \u201cslowly\u201d with x, then we are tempted to try a solution of the form (2.243) uE(x) \u2261 exp [iW(x)/\u00afh]. (2.244) (The reason for including the \u00afh will become apparent at the end of this section, when we discuss the physical interpretation of the WKB approximation.) In this case, (2.243) becomes 2 i\u00afh d2W dx2 \u2212 dW dx + \u00afh2 [k(x)]2 = 0 (2.245) which is completely equivalent to Schr\u00a8odinger\u2019s equation, although rewritten in what appears to be a nasty form. However, we consider a solution to this equation under the condition that \u00afh d2W dx2 2. dW dx (2.246) 8 A similar technique was used earlier by H. Jeffreys; this solution is referred to as the JWKB solution in some English books. 105 2.5 Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation This quanti\ufb01es our notion of a \u201cslowly varying\u201d potential V(x), and we will return soon to the physical signi\ufb01cance of this condition. Forging ahead for now, we use the condition (2.246) with our differential equation (2.245) to write a lowest-order approximation for W(x), namely W 0(x) =\u00b1 \u00afhk(x) leading to a \ufb01rst-order approximation for W(x), based on 2 dW1 dx = \u00afh2 [k(x)]2 + i\u00afhW 0(x) = \u00afh2 [k(x)]2", " \u00b1 i\u00afh2k(x) (2.247) (2.248) where the second term in (2.248) is much smaller than the \ufb01rst, so that W(x) \u2248 W1(x) =\u00b1 \u00afh \u2248 \u00b1\u00afh = \u00b1\u00afh x x x! dx k2(x) \u00b1 ik(x) dxk(x) 1 \u00b1 i 2 k(x) k2(x) \"1/2 dxk(x) + i 2 \u00afh ln [k(x)]. (2.249) The WKB approximation for the wave function is given by (2.244) and the \ufb01rst-order approximation for (2.249) forW(x ), namely uE(x) \u2248 exp [iW(x)/\u00afh] = 1 [k(x)]1/2 exp x \u00b1i dxk(x). (2.250) Note that this speci\ufb01es a choice of two solutions (\u00b1) either in the region where E > V(x), with k(x) from (2.242a), or in the region where E < V(x), with k(x) from (2.242b). Joining these two solutions across the classical turning point is the next task. We do not discuss this joining procedure in detail, as it is discussed in many places (Schiff (1968), pp. 268\u2013276, or Merzbacher (1998), Chapter 7, for example). Instead, we content ourselves with presenting the results of such an analysis for a potential well, schematically shown in Figure 2.5, with two turning points, x1 and x2. The wave function must behave like (2.250), with k(x) given by (2.242a) in region II, and by (2.242b) in regions I and III. The solutions in the neighborhood of the turning points, shown as a dashed line in Figure 2.5, are given by Airy functions, since we assume a linear approximation to the potential in these regions. Note that the asymptotic dependences of the Airy function9 are Ai(z) \u2192 1 \u221a 2 Ai(z) \u2192 1\u221a \u03c0 \u03c0 z\u22121/4 exp |z|\u22121/4 cos \u2212 2 3 z3/2 |z|3/2 \u2212 2 3 z \u2192 +\u221e (2.251a) z \u2192 \u2212\u221e. (2.251b) \u03c0 4 9 There is actually a second Airy function, Bi(z), which is very similar to Ai(z) but which is singular at the origin. It is relevant for this discussion, but we are glossing over the details. 106 Quantum Dynamics Fig. 2.5 Schematic diagram for behavior of the wave function uE(x) in a potential well V(x) with turning points x1 and x2. Note the similarity with Figure 2.3 near the turning points. For connecting regions I and II, the correct linear combination of the two solutions (2.250) is determined by choosing the integration constants in such a way that * + 2m [V(x) \u2212 E] x1 dx 1 [V(x) \u2212 E]1/4 * exp \u2212 + \u2192 2 [E \u2212 V(x)]1/4 cos 1 \u00afh x 1 \u00afh x dx x1 2m [E \u2212 V(x)] \u2212. \u03c0 4 (2.252) Likewise, from region III into region II we have + * 2m [V(x) \u2212 E] x dx 1 [V(x) \u2212 E]1/4 * exp \u2212 + 1 \u00afh \u2192 2 [E \u2212 V(x)]1/4 cos \u2212 x2 1 \u00afh x x2 dx 2m [E \u2212 V(x)] +. \u03c0 4 (2.253) Of course, we must obtain the same form for the wave function in region II, regardless of which turning point is analyzed. This implies that the arguments of the cosine in (2.252) and (2.253) must differ at most by an integer multiple of \u03c0 [not of 2\u03c0, because the signs of both sides of (2.253) can be reversed]. In this way we obtain a very interesting consistency condition, x2 dx x1 2m [E \u2212 V(x)] = (n + 1 2 )\u03c0\u00afh (n = 0, 1, 2, 3,...). (2.254) Apart from the difference between n + 1 2 and n, this equation is simply the quantization condition of the old quantum theory due to A. Sommerfeld and W. Wilson, originally written in 1915 as, p dq = nh, (2.255) where h is Plan", "ck\u2019s h, not Dirac\u2019s \u00afh, and the integral is evaluated over one whole period of classical motion, from x1 to x2 and back. 107 2.5 Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation Equation (2.254) can be used to obtain approximate expressions for the energy levels of a particle con\ufb01ned in a potential well. As an example, we consider the energy spectrum of a ball bouncing up and down over a hard surface, that is the \u201cbouncing neutrons\u201d discussed earlier in this section, namely V = mgx \u221e for x > 0 for x < 0, (2.256) where x stands for the height of the ball measured from the hard surface. One might be tempted to use (2.254) directly with x1 = 0, x2 = E mg, (2.257) which are the classical turning points of this problem. We note, however, that (2.254) was derived under the assumption that the WKB wave function \u201cleaks into\u201d the x < x1 region, while in our problem the wave function must strictly vanish for x \u2264 x1 = 0. A much more satisfactory approach to this problem is to consider the odd-parity solutions, guaranteed to vanish at x = 0, of a modi\ufb01ed problem de\ufb01ned by whose turning points are V(x) =mg|x | (\u2212 \u221e < x < \u221e) x1 = \u2212 E mg, x2 = E mg. (2.258) (2.259) The energy spectrum of the odd-parity states for this modi\ufb01ed problem must clearly be the same as that of the original problem. The quantization condition then becomes E/mg dx \u2212E/mg 2m(E \u2212 mg|x|) = (nodd + 1 2 )\u03c0\u00afh (nodd = 1, 3, 5,...) (2.260) or, equivalently, E/mg dx 0 2m(E \u2212 mgx) = (n \u2212 1 4 )\u03c0\u00afh (n = 1, 2, 3, 4,...). (2.261) This integral is elementary, and we obtain *! % 3 n \u2212 1 4 2 En = + \" 2/3 & \u03c0 (mg2 \u00afh2)1/3 (2.262) for the quantized energy levels of the bouncing ball. Table 2.2 compares the WKB approximation to the exact solution, using zeros of the Airy function, for the \ufb01rst 10 energy levels. We see that agreement is excellent even for small values of n and essentially exact for n 10. Before concluding, let us return to the interpretation of the condition (2.246). It is exact in the case \u00afh \u2192 0, suggesting a connection between the WKB approximation and the classical limit. In fact, using (2.244) the time-dependent wave function becomes \u03c8(x, t) \u221d uE(x) exp(\u2212iEt/\u00afh) =exp (iW(x)/\u00afh \u2212 iEt/\u00afh). (2.263) 108 Quantum Dynamics Table 2.2 The Quantized Energies of a Bouncing Ball in Units of (mg2 \u00afh2/2)1/ 10 WKB 2.320 4.082 5.517 6.784 7.942 9.021 10.039 11.008 11.935 12.828 Exact 2.338 4.088 5.521 6.787 7.944 9.023 10.040 11.009 11.936 12.829 Comparing this to (2.193) and (2.203) we see that W(x) corresponds directly to Hamilton\u2019s characteristic function. Indeed, condition (2.246) is the same as (2.201), the condition for reaching the classical limit. For these reasons, the WKB approximation is frequently referred to as a \u201csemiclassical\u201d approximation. We also note that condition (2.246) is equivalent to |k(x)| |k2(x)|. In terms of the de Broglie wavelength divided by 2\u03c0, this condition amounts to 2 [E \u2212 V(x)] |dV/dx| \u00afh 2m [E \u2212 V(x)] \u03bb =. (2.264) In other words, \u03bb must be small compared with the characteristic distance over which the potential varies appreciably. Roughly speaking, the potential must be essentially constant over many wavelengths. Thus we see that the semiclassical picture is reliable in the shortwavelength limit. 2.6 Propagators and Feynman Path Integrals 2.6.1 Propagators in Wave", " Mechanics In Section 2.1 we showed how the most general time-evolution problem with a timeindependent Hamiltonian can be solved once we expand the initial ket in terms of the eigenkets of an observable that commutes with H. Let us translate this statement into the language of wave mechanics. We start with |\u03b1, t0; t = exp \u2212iH(t \u2212 t0) \u00afh |\u03b1, t0 |aa|\u03b1, t0 exp = \u2211 a \u2212iEa(t \u2212 t0) \u00afh. (2.265) 109 2.6 Propagators and Feynman Path Integrals Multiplying both sides by x| on the left, we have x|\u03b1, t0; t = \u2211 a x|aa|\u03b1, t0 exp \u2212iEa(t \u2212 t0) \u00afh, which is of the form with \u03c8(x, t) = \u2211 a ca (t0)ua(x) exp \u2212iEa(t \u2212 t0) \u00afh, ua (x) =x |a standing for the eigenfunction of operator A with eigenvalue a. Note also that a|\u03b1, t0 = d3xa|xx|\u03b1, t0, (2.266) (2.267) (2.268) (2.269) which we recognize as the usual rule in wave mechanics for getting the expansion coef\ufb01cients of the initial state: ca (t0) = d3xu\u2217 a (x)\u03c8(x, t0). (2.270) All this should be straightforward and familiar. Now (2.266) together with (2.269) can also be visualized as some kind of integral operator acting on the initial wave function to yield the \ufb01nal wave function: \u03c8(x, t) = d3xK(x, t; x, t0)\u03c8(x, t0). (2.271) Here the kernel of the integral operator, known as the propagator in wave mechanics, is given by K(x, t; x, t0) = \u2211 a x|aa|x exp \u2212iEa(t \u2212 t0) \u00afh. (2.272) In any given problem the propagator depends only on the potential and is independent of the initial wave function. It can be constructed once the energy eigenfunctions and their eigenvalues are given. Clearly, the time evolution of the wave function is completely predicted if K(x, t; x, t0) is known and \u03c8(x, t0) is given initially. In this sense Schr\u00a8odinger\u2019s wave mechanics is a perfectly causal theory. The time development of a wave function subjected to some potential is as \u201cdeterministic\u201d as anything else in classical mechanics provided that the system is left undisturbed. The only peculiar feature, if any, is that when a measurement intervenes, the wave function changes abruptly, in an uncontrollable way, into one of the eigenfunctions of the observable being measured. There are two properties of the propagator worth recording here. First, for t > t0, K(x, t; x, t0) satis\ufb01es Schr\u00a8odinger\u2019s time-dependent wave equation in the variables x and t, with x and t0 \ufb01xed. This is evident from (2.272) because x|a exp[\u2212iEa(t \u2212 t0)/\u00afh], being the wave function corresponding to U (t, t0)|a, satis\ufb01es the wave equation. Second, K(x, t; x, t0) = \u03b43(x \u2212 x), lim t\u2192t0 (2.273) 110 Quantum Dynamics which is also obvious; as t \u2192 t0, because of the completeness of {|a}, sum (2.272) just reduces to x|x. Because of these two properties, the propagator (2.272), regarded as a function of x, is simply the wave function at t of a particle which was localized precisely at x at some earlier time t0. Indeed, this interpretation follows, perhaps more elegantly, from noting that (2.272) can also be written as K(x, t; x, t0) =x | exp \u2212iH(t \u2212 t0) \u00afh |x, (2.274) where the time-evolution operator acting on |x is just the state ket at t of a system that was localized precisely at x at time t0 (< t). If we wish to solve a more general problem where the initial wave function extends over a \ufb01nite region of space, all we have to do is multiply \u03c8(x, t0", ") by the propagator K(x, t; x, t0) and integrate over all space (that is, over x). In this manner we can add the various contributions from different positions (x). This situation is analogous to one in electrostatics; if we wish to \ufb01nd the electrostatic potential due to a general charge distribution \u03c1(x), we \ufb01rst solve the point-charge problem, multiply the point-charge solution with the charge distribution, and integrate: \u03c6(x) = d3x \u03c1(x) |x \u2212 x|. (2.275) The reader familiar with the theory of the Green functions must have recognized by this time that the propagator is simply the Green function for the time-dependent wave equation satisfying \u22072 + V(x) \u2212 i\u00afh K(x, t; x, t0) =\u2212 i\u00afh\u03b43(x \u2212 x)\u03b4(t \u2212 t0) (2.276) \u2212 \u00afh2 2m \u2202 \u2202t with the boundary condition K(x, t; x, t0) =0, for t < t0. (2.277) The delta function \u03b4(t \u2212 t0) is needed on the right-hand side of (2.276) because K varies discontinuously at t = t0. The particular form of the propagator is, of course, dependent on the particular potential to which the particle is subjected. Consider, as an example, a free particle in one dimension. The obvious observable that commutes with H is momentum; |p is a simultaneous eigenket of the operators p and H: p|p = p|p H|p = p2 2m |p. (2.278) The momentum eigenfunction is just the transformation function of Section 1.7 [see (1.264)] which is of the plane wave form. Combining everything, we have K(x, t; x, t0) = \u221e \u2212\u221e 1 2\u03c0\u00afh dp exp ip(x \u2212 x) \u00afh \u2212 ip2(t \u2212 t0) 2m\u00afh. (2.279) 111 2.6 Propagators and Feynman Path Integrals The integral can be evaluated by completing the square in the exponent. Here we simply record the result: \u2019 K(x, t; x, t0) = m 2\u03c0i\u00afh(t \u2212 t0) This expression may be used, for example, to study how a Gaussian wave packet spreads out as a function of time. (2.280) im(x \u2212 x)2 2\u00afh(t \u2212 t0) exp. For the simple harmonic oscillator, where the wave function of an energy eigenstate is given by un(x) exp \u2212iEnt \u00afh = 1 \u221a 2n/2 n! \u2019 \u00d7 Hn m\u03c9 \u00afh m\u03c9 \u03c0\u00afh 1/4 exp x exp \u2212i\u03c9 \u2212m\u03c9x2 2\u00afh 1 2 n + t, (2.281) the propagator is given by \u2019 K(x, t; x, t0) = m\u03c9 2\u03c0i\u00afh sin[\u03c9(t \u2212 t0)] exp - im\u03c9 2\u00afh sin[\u03c9(t \u2212 t0)] \u00d7 {(x2 + x2) cos[\u03c9(t \u2212 t0)] \u2212 2xx}. (2.282) One way to prove this is to use 1 1 \u2212 \u03b62 exp \u2212(\u03be2 + \u03b72 \u2212 2\u03be\u03b7\u03b6) (1 \u2212 \u03b62) \" \u03b6n \u2211 n=0 2nn!! = exp \u2212(\u03be2 + \u03b72) Hn(\u03be)Hn(\u03b7), (2.283) which is found in books on special functions (Morse and Feshbach (1953), p. 786). It can also be obtained using the a, a\u2020 operator method (Saxon (1968), pp. 144\u2013145) or, alternatively, the path-integral method to be described later. Notice that (2.282) is a periodic function of t with angular frequency \u03c9, the classical oscillator frequency. This means, among other things, that a particle initially localized precisely at x will return to its original position with certainty at 2\u03c0/\u03c9 (4\u03c0/\u03c9, and so forth) later. Certain space and time integrals derivable from K(x, t; x, t0) are of considerable interest. Without loss of generality we set t0 = 0 in the following. The \ufb01rst integral we consider is obtained by setting x = x and integrating over all space. We have G(t) \u2261 d3xK(x, t; x,", " 0) = d3x \u2211 a |x|a|2 exp \u2212iEat \u00afh \u2212iEat \u00afh This result is anticipated; recalling (2.274), we observe that setting x = x and integrating are equivalent to taking the trace of the time-evolution operator in the x-representation. But the trace is independent of representations; it can be evaluated more readily using the = \u2211 a (2.284) exp. 112 Quantum Dynamics {|a} basis where the time-evolution operator is diagonal, which immediately leads to the last line of (2.284). Now we see that (2.284) is just the \u201csum over states,\u201d reminiscent of the partition function in statistical mechanics. In fact, if we analytically continue in the t variable and make t purely imaginary, with \u03b2 de\ufb01ned by \u03b2 = it \u00afh real and positive, we can identify (2.284) with the partition function itself: Z = \u2211 a exp(\u2212\u03b2Ea). (2.285) (2.286) For this reason some of the techniques encountered in studying propagators in quantum mechanics are also useful in statistical mechanics. Next, let us consider the Laplace\u2013Fourier transform of G(t): \u221e \u02dcG(E) \u2261 \u2212i = \u2212i 0 \u221e 0 dtG(t) exp(iEt/\u00afh)/\u00afh dt\u2211 a exp(\u2212iEat/\u00afh) exp(iEt/\u00afh)/\u00afh. (2.287) The integrand here oscillates inde\ufb01nitely. But we can make the integral meaningful by letting E acquire a small positive imaginary part: We then obtain, in the limit \u03b5 \u2192 0, E \u2192 E + i\u03b5. \u02dcG(E) = \u2211 a 1 E \u2212 Ea. (2.288) (2.289) \u02dcG(E) in Observe now that the complete energy spectrum is exhibited as simple poles of the complex E-plane. If we wish to know the energy spectrum of a physical system, it is suf\ufb01cient to study the analytic properties of \u02dcG(E). 2.6.2 Propagator as a Transition Amplitude To gain further insight into the physical meaning of the propagator, we wish to relate it to the concept of transition amplitudes introduced in Section 2.2. But \ufb01rst, recall that the wave function which is the inner product of the \ufb01xed position bra x| with the moving state ket |\u03b1, t0; t can also be regarded as the inner product of the Heisenberg picture position bra x, t|, which moves \u201coppositely\u201d with time, with the Heisenberg picture state ket |\u03b1, t0, which is \ufb01xed in time. Likewise, the propagator can also be written as K(x, t; x, t0) = \u2211 a = \u2211 a x|aa|x exp x| exp \u2212iHt \u00afh \u2212iEa(t \u2212 t0) \u00afh |aa| exp |x iHt0 \u00afh = x, t|x, t0, (2.290) 113 2.6 Propagators and Feynman Path Integrals where |x, t0 and x, t| are to be understood as an eigenket and an eigenbra of the position operator in the Heisenberg picture. In Section 2.2 we showed that b, t|a, in the Heisenberg picture notation, is the probability amplitude for a system originally prepared to be an eigenstate of A with eigenvalue a at some initial time t0 = 0 to be found at a later time t in an eigenstate of B with eigenvalue b, and we called it the transition amplitude for going from state |a to state |b. Because there is nothing special about the choice of t0, only the time difference t \u2212 t0 is relevant, we can identify x, t|x, t0 as the probability amplitude for the particle prepared at t0 with position eigenvalue x to be found at a later time t at x. Roughly speaking, x, t|x, t0 is the amplitude for the particle to go from a space-time point (x, t0) to another space-time point (x, t), so the term transition amplitude for this expression is quite appropriate. This interpretation is, of course, in complete accord with the interpretation we gave earlier for K(x, t; x, t0). Yet another way to interpret x, t|x, t0 is as follows. As we emphasized earlier, |x, t0 is the position eigenket at t0 with the eigenvalue x in the He", "isenberg picture. Because at any given time the Heisenberg picture eigenkets of an observable can be chosen as base kets, we can regard x, t|x, t0 as the transformation function that connects the two sets of base kets at different times. So in the Heisenberg picture, time evolution can be viewed as a unitary transformation, in the sense of changing bases, that connects one set of base kets formed by {|x, t0} to another formed by {|x, t}. This is reminiscent of classical physics, in which the time development of a classical dynamic variable such as x(t) is viewed as a canonical (or contact) transformation generated by the classical Hamiltonian (Goldstein et al. (2002), pp. 401\u2013402). It turns out to be convenient to use a notation that treats the space and time coordinates more symmetrically. To this end we write x, t|x, t in place of x, t|x, t0. Because at any given time the position kets in the Heisenberg picture form a complete set, it is legitimate to insert the identity operator written as d3x|x, tx, t| = 1 (2.291) at any place we desire. For example, consider the time evolution from t to t; by dividing the time interval (t, t) into two parts, (t, t) and (t, t), we have x, t|x, t = d3xx, t|x, tx, t|x, t (t > t > t). (2.292) We call this the composition property of the transition amplitude.10 Clearly, we can divide the time interval into as many smaller subintervals as we wish. We have x, t|x, t = d3x d3xx, t|x, tx, t|x, t (2.293) and so on. If we somehow guess the form of x, t|x, t for an in\ufb01nitesimal time interval (between t and t = t + dt), we should be able to obtain the amplitude x, t|x, t for a \u00d7 x, t|x, t (t > t > t > t), 10 The analogue of (2.292) in probability theory is known as the Chapman\u2013Kolmogoroff equation, and in diffusion theory, the Smoluchowsky equation. 114 Quantum Dynamics \ufb01nite time interval by compounding the appropriate transition amplitudes for in\ufb01nitesimal time intervals in a manner analogous to (2.293). This kind of reasoning leads to an independent formulation of quantum mechanics due to R. P. Feynman, published in 1948, to which we now turn our attention Feynman, Rev. Mod. Phys., 20 (1948) 367. 2.6.3 Path Integrals as the Sum over Paths Without loss of generality we restrict ourselves to one-dimensional problems. Also, we avoid awkward expressions like x\u00b7\u00b7\u00b7 (N times) by using notation such as xN. With this notation we consider the transition amplitude for a particle going from the initial space-time point (x1, t1) to the \ufb01nal space-time point (xN, tN). The entire time interval between t1 and tN is divided into N \u2212 1 equal parts: tj \u2212 tj\u22121 = \u0394t = (tN \u2212 t1) (N \u2212 1). (2.294) Exploiting the composition property, we obtain xN, tN|x1, t1 = dxN\u22121 dxN\u22122 \u00b7 \u00b7 \u00b7 dx2xN, tN|xN\u22121, tN\u22121 \u00d7 xN\u22121, tN\u22121|xN\u22122, tN\u22122 \u00b7 \u00b7 \u00b7 x2, t2|x1, t1. (2.295) To visualize this pictorially, we consider a space-time plane, as shown in Figure 2.6. The initial and \ufb01nal space-time points are \ufb01xed to be (x1, t1) and (xN, tN), respectively. For each time segment, say between tn\u22121 and tn, we are instructed to consider the transition amplitude to go from (xn\u22121, tn\u22121) to (xn, tn); we then integrate over x2, x3,..., xN\u22121. This means that we must sum over all possible paths in the space-time plane with the end points \ufb01xed. Before proceeding further, it is pro\ufb01table to review here how paths appear in classical mechanics. Suppose we have a particle subjected to a force \ufb01eld derivable from a potential V(", "x). The classical Lagrangian is written as t (xN, tN) Fig. 2.6 Paths in the xt-plane. (x1, t1) tN tN\u22121 tr t3 t2 t1 x 115 2.6 Propagators and Feynman Path Integrals Lclassical(x, \u02d9x) = m\u02d9x2 2 \u2212 V(x). (2.296) Given this Lagrangian with the end points (x1, t1) and (xN, tN) speci\ufb01ed, we do not consider just any path joining (x1, t1) and (xN, tN) in classical mechanics. On the contrary, there exists a unique path that corresponds to the actual motion of the classical particle. For example, given V(x) = mgx, (x1, t1) = (h, 0), (xN, tN) = 0, 2h g, (2.297) where h may stand for the height of the Leaning Tower of Pisa, the classical path in the xt-plane can only be x = h \u2212 gt2 2. (2.298) More generally, according to Hamilton\u2019s principle, the unique path is that which minimizes the action, de\ufb01ned as the time integral of the classical Lagrangian: \u03b4 t2 t1 dtLclassical(x, \u02d9x) =0, (2.299) from which Lagrange\u2019s equation of motion can be obtained. 2.6.4 Feynman\u2019s Formulation The basic difference between classical mechanics and quantum mechanics should now be apparent. In classical mechanics a de\ufb01nite path in the xt-plane is associated with the particle\u2019s motion; in contrast, in quantum mechanics all possible paths must play roles including those which do not bear any resemblance to the classical path. Yet we must somehow be able to reproduce classical mechanics in a smooth manner in the limit \u00afh \u2192 0. How are we to accomplish this? As a young graduate student at Princeton University, R. P. Feynman tried to attack this problem. In looking for a possible clue, he was said to be intrigued by a mysterious remark in Dirac\u2019s book which, in our notation, amounts to the following statement: exp i t2 t1 dt Lclassical(x, \u02d9x) \u00afh corresponds to x2, t2|x1, t1. Feynman attempted to make sense out of this remark. Is \u201ccorresponds to\u201d the same thing as \u201cis equal to\u201d or \u201cis proportional to\u201d? In so doing he was led to formulate a space-time approach to quantum mechanics based on path integrals. In Feynman\u2019s formulation the classical action plays a very important role. For compact- ness, we introduce a new notation: S(n, n \u2212 1) \u2261 tn tn\u22121 Because Lclassical is a function of x and \u02d9x, S(n, n \u2212 1) is de\ufb01ned only after a de\ufb01nite path is speci\ufb01ed along which the integration is to be carried out. So even though the dt Lclassical(x, \u02d9x). (2.300) 116 Quantum Dynamics path dependence is not explicit in this notation, it is understood that we are considering a particular path in evaluating the integral. Imagine now that we are following some prescribed path. We concentrate our attention on a small segment along that path, say between (xn\u22121, tn\u22121) and (xn, tn). According to Dirac, we are instructed to associate exp[iS(n, n \u2212 1)/\u00afh] with that segment. Going along the de\ufb01nite path we are set to follow, we successively multiply expressions of this type to obtain N \u220f exp iS(n, n \u2212 1) \u00afh = exp i \u00afh N \u2211 S(n, n \u2212 1) = exp. (2.301) iS(N, 1) \u00afh n=2 n=2 This does not yet give xN, tN|x1, t1; rather, this equation is the contribution to xN, tN|x1, t1 arising from the particular path we have considered. We must still integrate over x2, x3,..., xN\u22121. At the same time, exploiting the composition property, we let the time interval between tn\u22121 and tn be in\ufb01nitesimally small. Thus our candidate expression for xN, tN|x1, t1 may be written, in some loose", " sense, as xN, tN|x1, t1 \u223c \u2211 exp all paths iS(N, 1) \u00afh, (2.302) where the sum is to be taken over an innumerably in\ufb01nite set of paths! Before presenting a more precise formulation, let us see whether considerations along this line make sense in the classical limit. As \u00afh \u2192 0, the exponential in (2.302) oscillates very violently, so there is a tendency for cancellation among various contributions from neighboring paths. This is because exp[iS/\u00afh] for some de\ufb01nite path and exp[iS/\u00afh] for a slightly different path have very different phases because of the smallness of \u00afh. So most paths do not contribute when \u00afh is regarded as a small quantity. However, there is an important exception. Suppose that we consider a path that satis\ufb01es \u03b4S(N, 1) = 0, (2.303) where the change in S is due to a slight deformation of the path with the end points \ufb01xed. This is precisely the classical path by virtue of Hamilton\u2019s principle. We denote the S that satis\ufb01es (2.303) by Smin. We now attempt to deform the path a little bit from the classical path. The resulting S is still equal to Smin to \ufb01rst order in deformation. This means that the phase of exp[iS/\u00afh] does not vary very much as we deviate slightly from the classical path even if \u00afh is small. As a result, as long as we stay near the classical path, constructive interference between neighboring paths is possible. In the \u00afh \u2192 0 limit, the major contributions must then arise from a very narrow strip (or a tube in higher dimensions) containing the classical path, as shown in Figure 2.7. Our (or Feynman\u2019s) guess based on Dirac\u2019s mysterious remark makes good sense because the classical path is singled out in the \u00afh \u2192 0 limit. To formulate Feynman\u2019s conjecture more precisely, let us go back to xn, tn|xn\u22121, tn\u22121, where the time difference tn \u2212 tn\u22121 is assumed to be in\ufb01nitesimally small. We write xn, tn|xn\u22121, tn\u22121 = 1 w(\u0394t) exp iS(n, n \u2212 1) \u00afh, (2.304) 117 2.6 Propagators and Feynman Path Integrals (xN, tN) Fig. 2.7 Paths important in the \u00afh \u2192 0 limit. (x1, t1) where we evaluate S(n, n\u2212 1) in a moment in the \u0394t \u2192 0 limit. Notice that we have inserted a weight factor, 1/w(\u0394t), which is assumed to depend only on the time interval tn \u2212 tn\u22121 and not on V(x). That such a factor is needed is clear from dimensional considerations; according to the way we normalized our position eigenkets, xn, tn|xn\u22121, tn\u22121 must have the dimension of 1/length. We now look at the exponential in (2.304). Our task is to evaluate the \u0394t \u2192 0 limit of S(n, n \u2212 1). Because the time interval is so small, it is legitimate to make a straight-line approximation to the path joining (xn\u22121, tn\u22121) and (xn, tn) as follows: S(n, n \u2212 1) = tn dt \u2212 V(x) m\u02d9x2 2 tn\u22121 * = \u0394t m 2 (xn \u2212 xn\u22121) \u0394t 2 \u2212 V (xn + xn\u22121) 2 +. (2.305) As an example, we consider speci\ufb01cally the free-particle case, V = 0. Equation (2.304) now becomes xn, tn|xn\u22121, tn\u22121 = 1 w(\u0394t) exp im(xn \u2212 xn\u22121)2 2\u00afh\u0394t. (2.306) We see that the exponent appearing here is completely identical to the one in the expression for the free-particle propagator (2.280). The reader may work out a similar comparison for the simple harmonic oscillator. We remarked earlier that the weight factor 1/w(\u0394t) appearing in (2.304) is assumed to be independent of V(x), so we may as well evaluate it for the free particle. Noting the orthonormality, in the sense of \u03b4-", "function, of Heisenberg picture position eigenkets at equal times, we obtain xn, tn|xn\u22121, tn\u22121|tn=tn\u22121 = \u03b4(xn \u2212 xn\u22121), 1 w(\u0394t) = \u2019 m 2\u03c0i\u00afh\u0394t, (2.307) (2.308) 118 Quantum Dynamics where we have used and \u221e \u2212\u221e d\u03be exp im\u03be2 2\u00afh\u0394t \u2019 = 2\u03c0i\u00afh\u0394t m \u2019 m 2\u03c0i\u00afh\u0394t exp im\u03be2 2\u00afh\u0394t lim \u0394t\u21920 = \u03b4(\u03be). (2.309a) (2.309b) This weight factor is, of course, anticipated from the expression for the free-particle propagator (2.280). To summarize, as \u0394t \u2192 0, we are led to \u2019 xn, tn|xn\u22121, tn\u22121 = m 2\u03c0i\u00afh\u0394t exp iS(n, n \u2212 1) \u00afh. (2.310) The \ufb01nal expression for the transition amplitude with tN \u2212 t1 \ufb01nite is xN, tN|x1, t1 = lim N\u2192\u221e (N\u22121)/2 m 2\u03c0i\u00afh\u0394t \u00d7 dxN\u22121 dxN\u22122 \u00b7 \u00b7 \u00b7 dx2 N \u220f n=2 exp iS(n, n \u2212 1) \u00afh, (2.311) where the N \u2192 \u221e limit is taken with xN and tN \ufb01xed. It is customary here to de\ufb01ne a new kind of multidimensional (in fact, in\ufb01nite-dimensional) integral operator xN x1 D[x(t)] \u2261 lim N\u2192\u221e (N\u22121)/2 m 2\u03c0i\u00afh\u0394t dxN\u22121 dxN\u22122 \u00b7 \u00b7 \u00b7 dx2 (2.312) and write (2.311) as xN, tN|x1, t1 = xN x1 D[x(t)] exp i tN t1 dt Lclassical(x, \u02d9x) \u00afh. (2.313) This expression is known as Feynman\u2019s path integral. Its meaning as the sum over all possible paths should be apparent from (2.311). Our steps leading to (2.313) are not meant to be a derivation. Rather, we (or Feynman) have attempted a new formulation of quantum mechanics based on the concept of paths, motivated by Dirac\u2019s mysterious remark. The only ideas we borrowed from the conventional form of quantum mechanics are (1) the superposition principle (used in summing the contributions from various alternate paths), (2) the composition property of the transition amplitude, and (3) classical correspondence in the \u00afh \u2192 0 limit. Even though we obtained the same result as the conventional theory for the free-particle case, it is now obvious, from what we have done so far, that Feynman\u2019s formulation is completely equivalent to Schr\u00a8odinger\u2019s wave mechanics. We conclude this section by proving that Feynman\u2019s expression for xN, tN|x1, t1 indeed satis\ufb01es Schr\u00a8odinger\u2019s timedependent wave equation in the variables xN, tN, just as the propagator de\ufb01ned by (2.272). 119 2.6 Propagators and Feynman Path Integrals We start with xN, tN|x1, t1 = dxN\u22121xN, tN|xN\u22121, tN\u22121xN\u22121, tN\u22121|x1, t1 \u2019 \u221e = \u2212\u221e dxN\u22121 m 2\u03c0i\u00afh\u0394t \u00d7 xN\u22121, tN\u22121|x1, t1, im 2\u00afh exp (xN \u2212 xN\u22121)2 \u0394t \u2212 iV\u0394t \u00afh where we have assumedt N \u2212 tN\u22121 to be in\ufb01nitesimal. Introducing \u03be = xN \u2212 xN\u22121 (2.314) (2.315) and letting xN \u2192 x and tN \u2192 t + \u0394t, we obtain \u2019 x, t + \u0394t|x1, t1 = m 2\u03c0i\u00afh\u0394t (\u221e) \u2212\u221e d\u03be exp im\u03be2 2\u00afh\u0394t \u2212 iV\u0394t \u00afh x \u2212 \u03be, t|x1, t1. (2.316) As", " is evident from (2.309b), in the limit \u0394t \u2192 0, the major contribution to this integral comes from the \u03be 0 region. It is therefore legitimate to expand x \u2212 \u03be, t|x1, t1 in powers of \u03be. We also expand x, t + \u0394t|x1, t1 and exp(\u2212iV\u0394t/\u00afh) in powers of \u0394t, so 1 \u2212 iV\u0394t \u00afh x, t|x1, t1 + \u0394t x, t|x1, t1 = \u2202 \u2202t \u221e \u2019 + \u00b7 \u00b7 \u00b7 m 2\u03c0i\u00afh\u0394t x, t|x1, t1 + \u2212\u221e \u00d7 d\u03be exp \u03be2 2 im\u03be2 2\u00afh\u0394t \u22022 \u2202x2 x, t|x1, t1 + \u00b7 \u00b7 \u00b7, (2.317) where we have dropped a term linear in \u03be because it vanishes when integrated with respect to \u03be. The x, t|x1, t1 term on the left-hand side just matches the leading term on the righthand side because of (2.309a). Collecting terms \ufb01rst order in \u0394t, we obtain \u0394t \u2202 \u2202t x, t|x1, t1 = ( \u221a 2\u03c0) i\u00afh\u0394t m 3/2 1 2 \u22022 \u2202x2 x, t|x1, t1 \u2019 \u2212 m 2\u03c0i\u00afh\u0394t i \u00afh \u0394tVx, t|x1, t1, where we have used \u221e \u2212\u221e d\u03be\u03be2 exp \u221a 2\u03c0 = 3/2, i\u00afh\u0394t m im\u03be2 2\u00afh\u0394t (2.318) (2.319) obtained by differentiating (2.309a) with respect to \u0394t. In this manner we see that x, t|x1, t1 satis\ufb01es Schr\u00a8odinger\u2019s time-dependent wave equation: i\u00afh \u2202 \u2202t x, t|x1, t1 = \u2212 \u22022 \u2202x2 Thus we can conclude that x, t|x1, t1 constructed according to Feynman\u2019s prescription is the same as the propagator in Schr\u00a8odinger\u2019s wave mechanics. x, t|x1, t1 + Vx, t|x1, t1. \u00afh2 2m (2.320) Feynman\u2019s space-time approach based on path integrals is not too convenient for attacking practical problems in nonrelativistic quantum mechanics. Even for the simple 120 Quantum Dynamics harmonic oscillator it is rather cumbersome to evaluate explicitly the relevant path integral.11 However, his approach is extremely gratifying from a conceptual point of view. By imposing a certain set of sensible requirements on a physical theory, we are inevitably led to a formalism equivalent to the usual formulation of quantum mechanics. It makes us wonder whether it is at all possible to construct a sensible alternative theory that is equally successful in accounting for microscopic phenomena. Methods based on path integrals have been found to be very powerful in other branches of modern physics, such as quantum \ufb01eld theory and statistical mechanics. In this book the path-integral method will appear again when we discuss the Aharonov\u2013Bohm effect.12 2.7 Potentials and Gauge Transformations 2.7.1 Constant Potentials In classical mechanics it is well known that the zero point of the potential energy is of no physical signi\ufb01cance. The time development of dynamic variables such as x(t) and L(t) is independent of whether we use V(x) or V(x) +V 0 with V0 constant in both space and time. The force that appears in Newton\u2019s second law depends only on the gradient of the potential; an additive constant is clearly irrelevant. What is the analogous situation in quantum mechanics? We look at the time evolution of a Schr\u00a8odinger picture state ket subject to some potential..|\u03b1, t0; t, the corresponding state ket Let |\u03b1, t0; t be a state ket in the presence of V(x), and appropriate for \u02dcV(x) = V(x) +V 0. (2.321) To be precise, let us agree that the initial conditions are such that both kets coincide with |\u03b1 at t = t0. If they represent the same physical situation, this can always be done by a suitable choice of the phase. Recalling that the state ket at t can be obtained by applying the time-evolution operator U (t, t0) to the state ket at t0, we", " obtain (t \u2212 t0) \u00afh |.\u03b1, t0; t = exp + V(x) +V 0 p2 2m |\u03b1 \u2212i = exp \u2212iV0(t \u2212 t0) \u00afh |\u03b1, t0; t. (2.322) In other words, the ket computed under the in\ufb02uence of \u02dcV has a time dependence different only by a phase factor exp[\u2212iV0(t \u2212 t0)/\u00afh]. For stationary states, this means that if the time dependence computed with V(x) is exp[\u2212iE(t \u2212 t0)/\u00afh], then the corresponding time 11 The reader is challenged to solve the simple harmonic oscillator problem using the Feynman path-integral method in Problem 2.44 of this chapter. 12 The reader who is interested in the fundamentals and applications of path integrals may consult Feynman and Hibbs (1965) and also Zee (2010). 121 2.7 Potentials and Gauge Transformations dependence computed with V(x) +V 0 is exp[\u2212i(E + V0)(t \u2212 t0)/\u00afh]. In other words, the use of \u02dcV in place of V just amounts to the following change: E \u2192 E + V0, (2.323) which the reader probably guessed immediately. Observable effects such as the time evolution of expectation values of x and S always depend on energy differences [see (2.47)]; the Bohr frequencies that characterize the sinusoidal time dependence of expectation values are the same whether we use V(x) or V(x) +V 0. In general, there can be no difference in the expectation values of observables if every state ket in the world is multiplied by a common factor exp[\u2212iV0(t \u2212 t0)/\u00afh]. Trivial as it may seem, we see here the \ufb01rst example of a class of transformations known as gauge transformations. The change in our convention for the zero point energy of the potential (2.324) (2.325) (2.326) (2.327) must be accompanied by a change in the state ket V(x) \u2192 V(x) +V 0 |\u03b1, t0; t \u2192exp \u03c8(x, t) \u2192 exp \u2212iV0(t \u2212 t0) \u00afh |\u03b1, t0; t. \u2212iV0(t \u2212 t0) \u00afh \u03c8(x, t). Of course, this change implies the following change in the wave function: Next we consider V0 that is spatially uniform but dependent on time. We then easily see that the analogue of (2.325) is |\u03b1, t0; t \u2192exp \u2212i t dt V0(t) \u00afh |\u03b1, t0; t. t0 Physically, the use of V(x) +V 0(t) in place of V(x) simply means that we are choosing a new zero point of the energy scale at each instant of time. Even though the choice of the absolute scale of the potential is arbitrary, potential differences are of nontrivial physical signi\ufb01cance and, in fact, can be detected in a very striking way. To illustrate this point, let us consider the arrangement shown in Figure 2.8. A beam of charged particles is split into two parts, each of which enters a metallic cage. If we so desire, we can maintain a \ufb01nite potential difference between the two cages by turning on a switch, as shown. A particle in the beam can be visualized as a wave packet whose dimension is much smaller than the dimension of the cage. Suppose we switch on the potential difference only after the wave packets enter the cages and switch it off before the wave packets leave the cages. The particle in the cage experiences no force because inside the cage the potential is spatially uniform; hence no electric \ufb01eld is present. Now let us recombine the two beam components in such a way that they meet in the interference region of Figure 2.8. Because of the existence of the potential, each beam component suffers a phase change, as indicated by (2.327). As a result, there is an observable interference term in the beam intensity in the interference region, namely, cos(\u03c61 \u2212 \u03c62), sin(\u03c61 \u2212 \u03c62), (2.328) 122 Quantum Dynamics Interference region Fig. 2.8 Quantum-mechanical interference to detect a potential difference. where \u03c61 \u2212 \u03c62 = 1 \u00afh tf ti dt[V2(t) \u2212 V1(t)]. (2.329) So despite the fact that the particle experiences no force, there is an observable effect that depends on whether V2(t) \u2212 V1(t)", " has been applied. Notice that this effect is purely quantum mechanical; in the limit \u00afh \u2192 0, the interesting interference effect is washed out because the oscillation of the cosine becomes in\ufb01nitely rapid.13 2.7.2 Gravity in Quantum Mechanics There is an experiment that exhibits in a striking manner how a gravitational effect appears in quantum mechanics. Before describing it, we \ufb01rst comment on the role of gravity in both classical and quantum mechanics. Consider the classical equation of motion for a purely falling body: m\u00a8x = \u2212m\u2207\u03a6grav = \u2212mg\u02c6z. (2.330) The mass term drops out; so in the absence of air resistance, a feather and a stone would behave in the same way \u2013 `a la Galileo \u2013 under the in\ufb02uence of gravity. This is, of course, a direct consequence of the equality of the gravitational and the inertial masses. Because the mass does not appear in the equation of a particle trajectory, gravity in classical mechanics is often said to be a purely geometric theory. The situation is rather different in quantum mechanics. In the wave-mechanical formu- lation, the analogue of (2.330) is \u2212 \u22072 + m\u03a6grav \u03c8 = i\u00afh \u2202\u03c8 \u2202t. (2.331) \u00afh2 2m The mass no longer cancels; instead it appears in the combination \u00afh/m, so in a problem where \u00afh appears, m is also expected to appear. We can see this point also using the Feynman path-integral formulation of a falling body based on 13 This gedanken experiment is the Minkowski-rotated form of the Aharonov\u2013Bohm experiment to be discussed later in this section. 123 2.7 Potentials and Gauge Transformations xn, tn|xn\u22121, tn\u22121 = \u2019 m 2\u03c0i\u00afh\u0394t exp i tn tn\u22121 dt 2 m\u02d9x2 \u2212 mgz) ( 1 \u00afh (tn \u2212 tn\u22121 = \u0394t \u2192 0). (2.332) Here again we see that m appears in the combination m/\u00afh. This is in sharp contrast with Hamilton\u2019s classical approach, based on t2 \u03b4 t1 dt m\u02d9x2 2 \u2212 mgz = 0, (2.333) where m can be eliminated in the very beginning. Starting with the Schr\u00a8odinger equation (2.331), we may derive the Ehrenfest theorem d2 dt2 x = \u2212g\u02c6z. (2.334) However, \u00afh does not appear here, nor does m. To see anontrivial quantum-mechanical effect of gravity, we must study effects in which \u00afh appears explicitly \u2013 and consequently where we expect the mass to appear \u2013 in contrast with purely gravitational phenomena in classical mechanics. Until 1975, there had been no direct experiment that established the presence of the m\u03a6grav term in (2.331). To be sure, a free fall of an elementary particle had been observed, but the classical equation of motion, or the Ehrenfest theorem (2.334), where \u00afh does not appear, suf\ufb01ced to account for this. The famous \u201cweight of photon\u201d experiment of V. Pound and collaborators did not test gravity in the quantum domain either because they measured a frequency shift where \u00afh does not explicitly appear. On the microscopic scale, gravitational forces are too weak to be readily observable. To appreciate the dif\ufb01culty involved in seeing gravity in bound-state problems, let us consider the ground state of an electron and a neutron bound by gravitational forces. This is the gravitational analogue of the hydrogen atom, where an electron and a proton are bound by Coulomb forces. At the same distance, the gravitational force between the electron and the neutron is weaker than the Coulomb force between the electron and the proton by a factor of \u223c 2 \u00d7 1039. The Bohr radius involved here can be obtained simply: a0 = \u00afh2 e2me \u2192 \u00afh2 GNm2 emn, (2.335) where GN is Newton\u2019s gravitational constant. If we substitute numbers in the equation, the Bohr radius of this gravitationally bound system turns out to be \u223c 1031, or\u223c 1013 light years, which is larger than the estimated radius of the universe by a few orders of magnitude! We now discuss a remarkable phenomenon known as gravity-induced quantum interference. A nearly monoenergetic beam of particles, in practice, thermal neutrons, is split into two parts and then brought together as shown in Figure 2.9. In actual experiments the neutron beam is split and bent by silicon crystals, but the details of this beautiful art of neutron interferometry do", " not concern us here. Because the size of the wave packet can be assumed to be much smaller than the macroscopic dimension of the loop formed by the two 124 Quantum Dynamics B A Interference region D l2 l1 C Fig. 2.9 Experiment to detect gravity-induced quantum interference. alternate paths, we can apply the concept of a classical trajectory. Let us \ufb01rst suppose that path A \u2192 B \u2192 D and path A \u2192 C \u2192 D lie in a horizontal plane. Because the absolute zero of the potential due to gravity is of no signi\ufb01cance, we can set V = 0 for any phenomenon that takes place in this plane; in other words, it is legitimate to ignore gravity altogether. The situation is very different if the plane formed by the two alternate paths is rotated around segment AC by \u03b4. This time the potential at level BD is higher than that at level AC by mgl2 sin \u03b4, which means that the state ket associated with path BD \u201crotates faster.\u201d This leads to a gravity-induced phase difference between the amplitudes for the two wave packets arriving at D. Actually there is also a gravity-induced phase change associated with AB and also with CD, but the effects cancel as we compare the two alternate paths. The net result is that the wave packet arriving at D via path ABD suffers a phase change exp \u2212imngl2 (sin \u03b4)T \u00afh (2.336) relative to that of the wave packet arriving at D via path ACD, where T is the time spent for the wave packet to go from B to D (or from A to C) and mn is the neutron mass. We can control this phase difference by rotating the plane of Figure 2.9; \u03b4 can change from 0 to \u03c0/2, or from 0 to \u2212\u03c0/2. Expressing the time spent T, or l1/\u03bdwavepacket, in terms of \u03bb, the de Broglie wavelength of the neutron, we obtain the following expression for the phase difference: \u03c6ABD \u2212 \u03c6ACD = \u2212 (m2 ngl1l2\u03bb sin \u03b4) \u00afh2. (2.337) In this manner we predict an observable interference effect that depends on angle \u03b4, which is reminiscent of fringes in Michelson-type interferometers in optics. An alternative, more wave-mechanical way to understand (2.337) follows. Because we are concerned with a time-independent potential, the sum of the kinetic energy and the potential energy is constant: p2 2m + mgz = E. (2.338) 125 2.7 Potentials and Gauge Transformations 1200 1000 800 600 \u221230 \u221220 \u221210 0 \u03b4 10 20 30 Fig. 2.10 Dependence of gravity-induced phase on the angle of rotation \u03b4. From Colella et al. Phys. Rev. Lett., 34 (1975) 1472. The difference in height between level BD and level AC implies a slight difference in p, or \u03bb. As a result, there is an accumulation of phase differences due to the \u03bb difference. It is left as an exercise to show that this wave-mechanical approach also leads to result (2.337). What is interesting about expression (2.337) is that its magnitude is neither too small nor too large; it is just right for this interesting effect to be detected with thermal neutrons traveling through paths of \u201ctable-top\u201d dimensions. For \u03bb = 1.42 \u02daA (comparable ngl1l2\u03bb/\u00afh2. to interatomic spacing in silicon) and l1l2 = 10 cm2, we obtain 55.6 for m2 As we rotate the loop plane gradually by 90\u25e6, we predict the intensity in the interference region to exhibit a series of maxima and minima; quantitatively we should see 55.6/2\u03c0 9 oscillations. It is extraordinary that such an effect has indeed been observed experimentally; see Figure 2.10 taken from a 1975 experiment of R. Colella, A. Overhauser, and S. A. Werner. The phase shift due to gravity is seen to be veri\ufb01ed to well within 1%. We emphasize that this effect is purely quantum mechanical because as \u00afh \u2192 0, the interference pattern gets washed out. The gravitational potential has been shown to enter into the Schr\u00a8odinger equation just as expected. This experiment also shows that gravity is not purely geometric at the quantum level because the effect depends on (m/\u00afh)2.14 14 However, this does not imply that the equivalence principle is unimportant in understanding an effect of this sort. If the gravitational mass (mgrav) and inertial mass (minert) were unequal, (m/\u00afh)2 would have to be replaced by mgravminert/", "\u00afh2. The fact that we could correctly predict the interference pattern without making a distinction between mgrav and minert shows some support for the equivalence principle at the quantum level. 126 Quantum Dynamics 2.7.3 Gauge Transformations in Electromagnetism Let us now turn to potentials that appear in electromagnetism. We consider an electric and a magnetic \ufb01eld derivable from the time-independent scalar and vector potential, \u03c6(x) and A(x): E = \u2212\u2207\u03c6, B = \u2207 \u00d7 A. (2.339) The Hamiltonian for a particle of electric charge e (e < 0 for the electron) subjected to the electromagnetic \ufb01eld is taken from classical physics to be H = 1 2m 2 p \u2212 eA c + e\u03c6. (2.340) In quantum mechanics \u03c6 and A are understood to be functions of the position operator x of the charged particle. Because p and A do not commute, some care is needed in interpreting (2.340). The safest procedure is to write 2 p \u2212 eA c e c 2 e c \u2192 p2 \u2212 (p \u00b7 A + A \u00b7 p) + A2. (2.341) In this form the Hamiltonian is obviously Hermitian. To study the dynamics of a charged particle subjected to \u03c6 and A, let us \ufb01rst proceed in the Heisenberg picture. We can evaluate the time derivative of x in a straightforward manner as dxi dt = [xi, H] i\u00afh = (pi \u2212 eAi/c) m, (2.342) which shows that the operator p, de\ufb01ned in this book to be the generator of translation, is not the same as m dx/ dt. Quite often p is called canonical momentum, as distinguished from kinematical (or mechanical) momentum, denoted by \u03a0: Even though we have \u03a0 \u2261 m dx dt = p \u2212 eA c. [ pi, pj] =0 (2.343) (2.344) for canonical momentum, the analogous commutator does not vanish for mechanical momentum. Instead we have [\u03a0i, \u03a0j] = \u03b5ijkBk, (2.345) as the reader may easily verify. Rewriting the Hamiltonian as i\u00afhe c \u03a02 H = + e\u03c6 2m and using the fundamental commutation relation, we can derive the quantum-mechanical version of the Lorentz force, namely, m d2x dt2 = d\u03a0 dt = e E + 1 2c dx dt \u00d7 B \u2212 B \u00d7 dx dt. (2.347) (2.346) 127 2.7 Potentials and Gauge Transformations This then is Ehrenfest\u2019s theorem, written in the Heisenberg picture, for the charged particle in the presence of E and B. We now study Schr\u00a8odinger\u2019s wave equation with \u03c6 and A. Our \ufb01rst task is to sandwich H between x| and |\u03b1, t0; t. The only term with which we have to be careful is x| p \u2212 eA(x) c 2 |\u03b1, t0; t = = \u2212i\u00afh\u2207 \u2212 eA(x) c \u2212i\u00afh\u2207 \u2212 eA(x) c x| p \u2212 eA(x) c \u2212i\u00afh\u2207 \u2212 eA(x) \u00b7 c |\u03b1, t0; t x|\u03b1, t0; t. (2.348) It is important to emphasize that the \ufb01rst \u2207 in the last line can differentiate both x|\u03b1, t0; t and A(x). Combining everything, we have 1 2m \u2212i\u00afh\u2207 \u2212 eA(x) \u00b7 c x|\u03b1, t0; t \u2212i\u00afh\u2207 \u2212 eA(x) \u2202 \u2202t c + e\u03c6(x)x|\u03b1, t0; t = i\u00afh x|\u03b1, t0; t. From this expression we readily obtain the continuity equation \u2202\u03c1 \u2202t + \u2207\u00b7 j = 0, where \u03c1 is |\u03c8|2 as before, with x|\u03b1, t0; t written as \u03c8, but for the probability \ufb02ux j we have \u00afh m j = e mc Im(\u03c8\u2217\u2207\u03c8) \u2212 A|\u03c8|2, (2.351) which is just what we expect from the substitution (2.349) (2.350) Writing the wave function of j, namely, \u221a \u2207 \u2192 \u2207 \u2212 ie \u00afhc A. (2.352) \u03c1 exp(iS/\u00afh)", " [see (2.193)], we obtain an alternative form for \u03c1 m j = \u2207S \u2212 eA c, (2.353) which is to be compared with (2.195). We will \ufb01nd this form to be convenient in discussing superconductivity, \ufb02ux quantization, and so on. We also note that the space integral of j is the expectation value of kinematical momentum (not canonical momentum) apart from 1/m: d3xj = p \u2212 eA/c m = \u03a0/m. (2.354) We are now in a position to discuss the subject of gauge transformations in electro- magnetism. First, consider \u03c6 \u2192 \u03c6 + \u03bb, A \u2192 A, (2.355) 128 Quantum Dynamics with \u03bb constant, that is, independent of x and t. BothE and B obviously remain unchanged. This transformation just amounts to a change in the zero point of the energy scale, a possibility treated in the beginning of this section; we just replace V by e\u03c6. We have already discussed the accompanying change needed for the state ket [see (2.325)], so we do not dwell on this transformation any further. Much more interesting is the transformation \u03c6 \u2192 \u03c6, A \u2192 A + \u2207\u039b, (2.356) where \u039b is a function of x. The static electromagnetic \ufb01elds E and B are unchanged under (2.356). Both (2.355) and (2.356) are special cases of \u03c6 \u2192 \u03c6 \u2212 1 c \u2202\u039b \u2202t, A \u2192 A + \u2207\u039b, (2.357) which leave E and B, given by \u2202A \u2202t unchanged, but in the following we do not consider time-dependent \ufb01elds and potentials. In the remaining part of this section the term gauge transformation refers to (2.356). E = \u2212\u2207\u03c6 \u2212 1 c B = \u2207 \u00d7 A, (2.358), In classical physics observable effects such as the trajectory of a charged particle are independent of the gauge used, that is, of the particular choice of \u039b we happen to adopt. Consider a charged particle in a uniform magnetic \ufb01eld in the z-direction B = B\u02c6z. (2.359) This magnetic \ufb01eld may be derived from Ax = \u2212By 2, Ay = Bx 2, Az = 0 (2.360) or also from Ax = \u2212By, Ay = 0, Az = 0. The second form is obtained from the \ufb01rst by A \u2192 A \u2212 \u2207, Bxy 2 (2.361) (2.362) which is indeed of the form of (2.356). Regardless of which A we may use, the trajectory of the charged particle with a given set of initial conditions is the same; it is just a helix \u2013 a uniform circular motion when projected in the xy-plane, superposed with a uniform rectilinear motion in the z-direction. Yet if we look at px and py, the results are very different. For one thing, px is a constant of the motion when (2.361) is used but not when (2.360) is used. Recall Hamilton\u2019s equations of motion: dpx dt = \u2212 \u2202H \u2202x, dpy dt = \u2212 \u2202H \u2202y,.... (2.363) In general, the canonical momentum p is not a gauge-invariant quantity; its numerical value depends on the particular gauge used even when we are referring to the same physical 129 2.7 Potentials and Gauge Transformations situation. In contrast, the kinematic momentum \u03a0, ormdx/ dt, that traces the trajectory of the particle is a gauge-invariant quantity, as one may explicitly verify. Because p and mdx/dt are related via (2.343), p must change to compensate for the change in A given by (2.362). We now return to quantum mechanics. We believe that it is reasonable to demand that the expectation values in quantum mechanics behave in a manner similar to the corresponding classical quantities under gauge transformations, so x and \u03a0 are not to change under gauge transformations, while p is expected to change. Let us denote by |\u03b1 the state ket in the presence of A; the state ket for the same physical situation when \u02dcA = A + \u2207\u039b (2.364) is used in place of A is denoted by | \u02dc\u03b1. Here\u039b, as well asA, is a function of the position operator x. Our basic requirements are and \u03b1|x|\u03b1 = \u02dc\u03b1|x| \u02dc\u03b1 p \u2212 eA c \u03b1| |\u03b1 = \u02dc\u03b1", "| p \u2212 e \u02dcA c | \u02dc\u03b1. In addition, we require, as usual, the norm of the state ket to be preserved: We must construct an operator G that relates | \u02dc\u03b1 to |\u03b1: \u03b1|\u03b1 = \u02dc\u03b1| \u02dc\u03b1. | \u02dc\u03b1 = G |\u03b1. Invariance properties (2.365a) and (2.365b) are guaranteed if and We assert that G \u2020xG = x p \u2212 eA c \u2212 e\u2207\u039b c G \u2020 G = p \u2212 eA c. G = exp ie\u039b(x) \u00afhc (2.365a) (2.365b) (2.366) (2.367) (2.368a) (2.368b) (2.369) will do the job. First, G is unitary, so (2.366) is all right. Second, (2.368a) is obviously satis\ufb01ed because x commutes with any function of x. As for (2.368b), just note that exp p exp \u2212ie\u039b \u00afhc ie\u039b \u00afhc = exp = \u2212 exp \u2212ie\u039b \u00afhc \u2212ie\u039b \u00afhc p, exp ie\u039b \u00afhc i\u00afh\u2207 exp ie\u039b \u00afhc + p + p (2.370) where we have used (2.97b). = p + e\u2207\u039b c, 130 Quantum Dynamics The invariance of quantum mechanics under gauge transformations can also be demonstrated by looking directly at the Schr\u00a8odinger equation. Let |\u03b1, t0; t be a solution to the Schr\u00a8odinger equation in the presence of A: + e\u03c6 (p \u2212 eA/c)2 2m The corresponding solution in the presence of \u02dcA must satisfy (p \u2212 eA/c \u2212 e\u2207\u039b/c)2 2m |.\u03b1, t0; t = i\u00afh |\u03b1, t0; t = i\u00afh + e\u03c6 \u2202 \u2202t We see that if the new ket is taken to be |.\u03b1, t0; t = exp ie\u039b \u00afhc |\u03b1, t0; t |\u03b1, t0; t. \u2202 \u2202t |.\u03b1, t0; t. (2.371) (2.372) (2.373) in accordance with (2.369), then the new Schr\u00a8odinger equation (2.372) will be satis\ufb01ed; all we have to note is that \u2212ie\u039b \u00afhc \u2212 e\u2207\u039b c p \u2212 eA c p \u2212 eA c ie\u039b \u00afhc (2.374) 2 exp exp =, 2 which follows from applying (2.370) twice. Equation (2.373) also implies that the corresponding wave equations are related via \u02dc\u03c8(x, t) =exp \u03c8(x, t), (2.375) ie\u039b(x) \u00afhc where \u039b(x) is now a real function of the position vector eigenvalue x. This can, of course, be veri\ufb01ed also by directly substituting (2.375) into Schr\u00a8odinger\u2019s wave equation with A replaced by A + \u2207\u039b. In terms of \u03c1 and S, we see that \u03c1 is unchanged but S is modi\ufb01ed as follows: S \u2192 S + e\u039b c. (2.376) This is highly satisfactory because we see that the probability \ufb02ux given by (2.353) is then gauge invariant. To summarize, when vector potentials in different gauges are used for the same physical situation, the corresponding state kets (or wave functions) must necessarily be different. However, only a simple change is needed; we can go from a gauge speci\ufb01ed by A to another speci\ufb01ed by A + \u2207\u039b by merely multiplying the old ket (the old wave function) by exp[ie\u039b(x)/\u00afhc] (exp[ie\u039b(x)/\u00afhc]). Thecanonical momentum, de\ufb01ned as the generator of translation, is manifestly gauge dependent in the sense that its expectation value depends on the particular gauge chosen, while the kinematic momentum and the probability \ufb02ux are gauge invariant. The reader may wonder why invariance under (2.369) is called gauge invariance. This word is the translation of the German Eichinvarianz, whereEich means gauge. There is a historical anecdote that goes with the origin of this term. Consider some function of position at x: F(x). At a neighboring point we obviously have", " F(x + dx) F(x) + (\u2207F) \u00b7 dx. (2.377) 131 2.7 Potentials and Gauge Transformations But suppose we apply a scale change as we go from x to x + dx as follows: 1|at x \u2192 [1 + \u03a3(x) \u00b7 dx]|at x+dx. We must then rescale F(x) as follows: F(x + dx)|rescaled F(x) + [(\u2207 + \u03a3)F] \u00b7 dx (2.378) (2.379) instead of (2.377). The combination \u2207 + \u03a3 is similar to the gauge-invariant combination \u2207 \u2212 ie \u00afhc A (2.380) encountered in (2.352) except for the absence of i. Historically, H. Weyl unsuccessfully attempted to construct a geometric theory of electromagnetism based on Eichinvarianz by identifying the scale function \u03a3(x) in (2.378) and (2.379) with the vector potential A itself. With the birth of quantum mechanics, V. Fock and F. London realized the importance of the gauge-invariant combination (2.380), and they recalled Weyl\u2019s earlier work by comparing \u03a3 with i times A. We are stuck with the term gauge invariance even though the quantummechanical analogue of (2.378) 1 at x \u2192 1 \u2212 ie \u00afhc A \u00b7 dx at x+dx (2.381) would actually correspond to \u201cphase change\u201d rather than to \u201cscale change.\u201d 2.7.4 The Aharonov\u2013Bohm Effect The use of vector potential in quantum mechanics has many far-reaching consequences, some of which we are now ready to discuss. We start with a relatively innocuous-looking problem. For \ufb01lling in the details, plus an extension into a discussion of quantized \ufb02ux, see Problem 2.37 at the end of this chapter. Consider a hollow cylindrical shell, as shown in Figure 2.11a. We assume that a particle of charge e can be completely con\ufb01ned to the interior of the shell with rigid walls. The wave function is required to vanish on the inner (\u03c1 = \u03c1a) and outer (\u03c1 = \u03c1b) walls as well as at the top and the bottom. It is a straightforward boundary-value problem in mathematical physics to obtain the energy eigenvalues. Let us now consider a modi\ufb01ed arrangement where the cylindrical shell encloses a uniform magnetic \ufb01eld, as shown in Figure 2.11b. Speci\ufb01cally, you may imagine \ufb01tting a very long solenoid into the hole in the middle in such a way that no magnetic \ufb01eld leaks into the region \u03c1 \u2265 \u03c1a. The boundary conditions for the wave function are taken to be the same as before; the walls are assumed to be just as rigid. Intuitively we may conjecture that the energy spectrum is unchanged because the region with B = 0 is completely inaccessible to the charged particle trapped inside the shell. However, quantum mechanics tells us that this conjecture is not correct. Even though the magnetic \ufb01eld vanishes in the interior, the vector potential A is nonvanishing. We need A = 0 for \u03c1 < \u03c1a, but we cannot set A = 0 for \u03c1 > \u03c1a because the discontinuity leads to an unphysical magnetic \ufb01eld. However, we can use a gauge 132 Quantum Dynamics \u03c1b \u03c1a L Fig. 2.11 Hollow cylindrical shell (a) without a magnetic field, (b) with a uniform magnetic field. (a) (b) transformation to set B = 0, and then apply (2.375) so that the wave function gains the appropriate phase factor. The correspondingly modi\ufb01ed Schr\u00a8odinger equation is then solved for the energy eigenvalues in the region \u03c1a \u2264 \u03c1 \u2264 \u03c1b. Using Stokes\u2019s theorem we can infer that the vector potential needed to produce the magnetic \ufb01eld B (= B\u02c6z) is \u02c6\u03c6, B\u03c12 a 2\u03c1 A = (2.382) where \u02c6\u03c6 is the unit vector in the direction of increasing azimuthal angle. In attempting to solve the Schr\u00a8odinger equation to \ufb01nd the energy eigenvalues for this new problem, we need only to replace the gradient \u2207 by \u2207 \u2212 (ie/\u00afhc)A; we can accomplish this in cylindrical coordinates by replacing the partial derivative with respect to \u03c6 as follows: \u2202 \u2202\u03c6 \u2192 \u2202 \u2202\u03c6 \u2212 ie \u00afh", "c B\u03c12 a 2 ; recall the expression for gradient in cylindrical coordinates: \u2207 = \u02c6\u03c1 \u2202 \u2202\u03c1 + \u02c6z \u2202 \u2202z + \u02c6\u03c6 1 \u03c1 \u2202 \u2202\u03c6. (2.383) (2.384) The replacement (2.383) results in an observable change in the energy spectrum, as the reader may verify explicitly. This is quite remarkable because the particle never \u201ctouches\u201d the magnetic \ufb01eld; the Lorentz force the particle experiences is identically zero in this problem, yet the energy levels depend on whether or not the magnetic \ufb01eld is \ufb01nite in the hole region inaccessible to the particle. 133 2.7 Potentials and Gauge Transformations A Source region B 0 Impenetrable cylinder B Interference region Fig. 2.12 The Aharonov\u2013Bohm effect. The problem we have just treated is the bound-state version of what is commonly referred to as the Aharonov\u2013Bohm effect.15 We are now in a position to discuss the original form of the Aharonov\u2013Bohm effect itself. Consider a particle of charge e going above or below a very long impenetrable cylinder, as shown in Figure 2.12. Inside the cylinder is a magnetic \ufb01eld parallel to the cylinder axis, taken to be normal to the plane of Figure 2.12. So the particle paths above and below enclose a magnetic \ufb02ux. Our object is to study how the probability of \ufb01nding the particle in the interference region B depends on the magnetic \ufb02ux. Even though this problem can be attacked by comparing the solutions to the Schr\u00a8odinger equation in the presence and absence of B, for pedagogical reasons we prefer to use the Feynman path-integral method. Let x1 and xN be typical points in source region A and interference region B, respectively. We recall from classical mechanics that the Lagrangian in the presence of the magnetic \ufb01eld can be obtained from that in the absence of the magnetic \ufb01eld, denoted by L(0) classical, as follows: 2 L(0) classical = m 2 dx dt \u2192 L(0) classical + e c dx dt \u00b7 A. (2.385) The corresponding change in the action for some de\ufb01nite path segment going from (xn\u22121, tn\u22121) to (xn, tn) is then given by S(0)(n, n \u2212 1) \u2192 S(0)(n, n \u2212 1) + e c tn tn\u22121 dt dx dt \u00b7 A. (2.386) But this last integral can be written as tn dt tn\u22121 e c dx dt \u00b7 A = xn xn\u22121 e c A \u00b7 ds, (2.387) where ds is the differential line element along the path segment, so when we consider the entire contribution from x1 to xN, we have the following change: - \u220f exp iS(0)(n, n \u2212 1) \u00afh \u2192 \u220f exp iS(0)(n, n \u2212 1) \u00afh exp A \u00b7 ds. (2.388) xN x1 ie \u00afhc All this is for a particular path, such as going above the cylinder. We must still sum over all possible paths, which may appear to be a formidable task. Fortunately, we know from the 15 After a 1959 paper by Y. Aharonov and D. Bohm (Phys. Rev., 115 (1959) 485). Essentially the same effect was discussed 10 years earlier by W. Ehrenberg and R. E. Siday. 134 Quantum Dynamics A \u00b7 ds is independent of paths, that is, it theory of electromagnetism that the line integral is dependent only on the end points, as long as the loop formed by a pair of different paths does not enclose a magnetic \ufb02ux. As a result, the contributions due to A = 0 to all paths going above the cylinder are given by a common phase factor; similarly, the contributions from all paths going below the cylinder are multiplied by another common phase factor. In the path-integral notation we have, for the entire transition amplitude, D[x(t)] exp iS(0)(N, 1) \u00afh + D[x(t)] exp below iS(0)(N, 1) \u00afh above \u2192 above D[x(t)] exp + D[x(t)] exp below iS(0)(N, 1) \u00afh iS(0)(N, 1) \u00afh ie \u00afhc exp exp - xN A\u00b7ds x1 xN x1 above A\u00b7ds below -", ". (2.389) ie \u00afhc The probability for \ufb01nding the particle in the interference region B depends on the modulus squared of the entire transition amplitude and hence on the phase difference between the contribution from the paths going above and below. The phase difference due to the presence of B is just, e \u00afhc xN A \u00b7 ds \u2212 x1 above e \u00afhc xN A \u00b7 ds x1 below = = e \u00afhc e \u00afhc A \u00b7 ds \u03a6B, (2.390) where \u03a6B stands for the magnetic \ufb02ux inside the impenetrable cylinder. This means that as we change the magnetic \ufb01eld strength, there is a sinusoidal component in the probability for observing the particle in region B with a period given by a fundamental unit of magnetic \ufb02ux, namely, 2\u03c0\u00afhc |e| = 4.135 \u00d7 10\u22127 Gauss-cm2. (2.391) We emphasize that the interference effect discussed here is purely quantum mechanical. Classically, the motion of a charged particle is determined solely by Newton\u2019s second law supplemented by the force law of Lorentz. Here, as in the previous bound-state problem, the particle can never enter the region in which B is \ufb01nite; the Lorentz force is identically zero in all regions where the particle wave function is \ufb01nite. Yet there is a striking interference pattern that depends on the presence or absence of a magnetic \ufb01eld inside the impenetrable cylinder. This point has led some people to conclude that in quantum mechanics it is A rather than B that is fundamental. It is to be noted, however, that the observable effects in both examples depend only on \u03a6B, which is directly expressible in terms of B. Experiments to verify the Aharonov\u2013Bohm effect have been performed using a thin magnetized iron \ufb01lament called a whisker.16 16 One such recent experiment is that of Tonomura et al., Phys. Rev. Lett., 48 (1982) 1443. 135 2.7 Potentials and Gauge Transformations 2.7.5 Magnetic Monopole We conclude this section with one of the most remarkable predictions of quantum physics, which has yet to be veri\ufb01ed experimentally. An astute student of classical electrodynamics may be struck by the fact that there is a strong symmetry between E and B, yet a magnetic charge, commonly referred to as a magnetic monopole, analogous to electric charge is peculiarly absent in Maxwell\u2019s equations. The source of a magnetic \ufb01eld observed in nature is either a moving electric charge or a static magnetic dipole, never a static magnetic charge. Instead of analogous to \u2207 \u00b7 B = 4\u03c0\u03c1M \u2207 \u00b7 E = 4\u03c0\u03c1, (2.392) (2.393) \u2207 \u00b7 B actually vanishes in the usual way of writing Maxwell\u2019s equations. Quantum mechanics does not predict that a magnetic monopole must exist. However, it unambiguously requires that if a magnetic monopole is ever found in nature, the magnitude of magnetic charge must be quantized in terms of e, \u00afh, andc, as we now demonstrate. Suppose there is a point magnetic monopole, situated at the origin, of strength eM analogous to a point electric charge. The static magnetic \ufb01eld is then given by eM r2 \u02c6r. B = At \ufb01rst sight it may appear that the magnetic \ufb01eld (2.394) can be derived from A = eM(1 \u2212 cos \u03b8) r sin \u03b8 \u02c6\u03c6. (2.394) (2.395) Recall the expression for curl in spherical coordinates: \u2207 \u00d7 A = \u02c6r 1 r sin \u03b8 \u2202 \u2202\u03b8(A\u03c6 sin \u03b8) \u2212 \u2202A\u03b8 \u2202\u03c6 + \u02c6\u03b8 1 r 1 sin \u03b8 \u2202Ar \u2202\u03c6 \u2212 \u2202 \u2202r (rA\u03c6) \u2202 \u2202r (rA\u03b8) \u2212. \u2202Ar \u2202\u03b8 (2.396) + \u02c6\u03c6 1 r But vector potential (2.395) has one dif\ufb01culty \u2013 it is singular on the negative z-axis (\u03b8 = \u03c0). In fact, it turns out to be impossible to construct a singularity-free potential valid everywhere for this problem. To see this we \ufb01rst note \u201cGauss\u2019s law\u201d B \u00b7 d\u03c3 = 4\u03c0eM closed surface (2.397) for any surface boundary enclosing the origin at which the magnetic monopole is located. On the", " other hand, if A were nonsingular, we would have \u2207 \u00b7(\u2207 \u00d7 A) = 0 (2.398) 136 Quantum Dynamics \u03b5 \u03b5 \u03b5 \u03b5 I II Fig. 2.13 Regions of validity for the potentials A(I) and A(II). everywhere; hence, B \u00b7 d\u03c3 = closed surface volume inside in contradiction with (2.397). \u2207 \u00b7(\u2207 \u00d7 A)d3x = 0, (2.399) However, one might argue that because the vector potential is just a device for obtaining B, we need not insist on having a single expression for A valid everywhere. Suppose we construct a pair of potentials, A(I) = A(II) = \u2212 eM(1 \u2212 cos \u03b8) r sin \u03b8 eM(1 + cos \u03b8) r sin \u03b8 \u02c6\u03c6 (\u03b8 < \u03c0 \u2212 \u03b5) \u02c6\u03c6 (\u03b8 > \u03b5), (2.400a) (2.400b) such that the potential A(I) can be used everywhere except inside the cone de\ufb01ned by \u03b8 = \u03c0 \u2212 \u03b5 around the negative z-axis; likewise, the potential A(II) can be used everywhere except inside the cone \u03b8 = \u03b5 around the positive z-axis; see Figure 2.13. Together they lead to the correct expression for B everywhere.17 Consider now what happens in the overlap region \u03b5 < \u03b8 < \u03c0 \u2212 \u03b5, where we may use either A(I) or A(II). Because the two potentials lead to the same magnetic \ufb01eld, they must be related to each other by a gauge transformation. To \ufb01nd \u039b appropriate for this problem we \ufb01rst note that A(II) \u2212 A(I) = \u2212 \u02c6\u03c6. (2.401) 2eM r sin \u03b8 17 An alternative approach to this problem uses A(I) everywhere, but taking special care of the string of singularities, known as a Dirac string, along the negative z-axis. 137 2.7 Potentials and Gauge Transformations Recalling the expression for gradient in spherical coordinates, \u2207\u039b = \u02c6r \u2202\u039b \u2202r + \u02c6\u03b8 1 r \u2202\u039b \u2202\u03b8 + \u02c6\u03c6 1 r sin \u03b8 \u2202\u039b \u2202\u03c6, we deduce that will do the job. \u039b = \u22122eM\u03c6 (2.402) (2.403) Next, we consider the wave function of an electrically charged particle of charge e subjected to magnetic \ufb01eld (2.394). As we emphasized earlier, the particular form of the wave function depends on the particular gauge used. In the overlap region where we may use either A(I) or A(II), the corresponding wave functions are, according to (2.375), related to each other by \u03c8(II) = exp \u03c8(I). (2.404) \u22122ieeM\u03c6 \u00afhc Wave functions \u03c8(I) and \u03c8(II) must each be single valued because once we choose a particular gauge, the expansion of the state ket in terms of the position eigenkets must be unique. After all, as we have repeatedly emphasized, the wave function is simply an expansion coef\ufb01cient for the state ket in terms of the position eigenkets. Let us now examine the behavior of wave function \u03c8(II) on the equator \u03b8 = \u03c0/2 with some de\ufb01nite radius r, which is a constant. When we increase the azimuthal angle \u03c6 along the equator and go around once, say from \u03c6 = 0 to \u03c6 = 2\u03c0, \u03c8(II), as well as \u03c8(I), must return to its original value because each is single valued. According to (2.404), this is possible only if 2eeM \u00afhc = \u00b1N, N = 0, \u00b11, \u00b12,.... (2.405) So we reach a very far-reaching conclusion: The magnetic charges must be quantized in units of \u00afhc 2|e| 137 2 The smallest magnetic charge possible is \u00afhc/2|e|, where e is the electronic charge. It is amusing that once a magnetic monopole is assumed to exist, we can use (2.405) backward, so to speak, to explain why the electric charges are quantized, for example, why the proton charge cannot be 0.999972 times |e|.18 (2.406) |e|. We repeat once again that quantum mechanics does not require magnetic monopoles to exist. However, it", " unambiguously predicts that a magnetic charge, if it is ever found in nature, must be quantized in units of \u00afhc/2|e|. The quantization of magnetic charges in quantum mechanics was \ufb01rst shown in 1931 by P. A. M. Dirac. The derivation given here is due to T. T. Wu and C. N. Yang. A different solution, which connects 18 Empirically the equality in magnitude between the electron charge and the proton charge is established to an accuracy of four parts in 1019. 138 Quantum Dynamics the Dirac quantization condition to the quantization of angular momentum, is discussed by Lipkin et al., Ann. Phys., 53 (1969) 203. Finally, we will revisit this subject again in Section 5.6 when we discuss Berry\u2019s phase in conjunction with the adiabatic approximation. Problems 2.1 Consider the spin-precession problem discussed in the text. It can also be solved in the Heisenberg picture. Using the Hamiltonian H = \u2212 Sz = \u03c9Sz, eB mc write the Heisenberg equations of motion for the time-dependent operators Sx(t), Sy(t), and Sz(t). Solve them to obtain Sx,y,z as functions of time. 2.2 Look again at the Hamiltonian of Chapter 1, Problem 1.13. Suppose the typist made an error and wrote H as H = H11|11| + H22|22| + H12|12|. What principle is now violated? Illustrate your point explicitly by attempting to solve the most general time-dependent problem using an illegal Hamiltonian of this kind. (You may assume H11 = H22 = 0 for simplicity.) 2.3 An electron is subject to a uniform, time-independent magnetic \ufb01eld of strength B in the positive z-direction. At t = 0 the electron is known to be in an eigenstate of S \u00b7 \u02c6n with eigenvalue \u00afh/2, where \u02c6n is a unit vector, lying in the xz-plane, that makes an angle \u03b2 with the z-axis. a. Obtain the probability for \ufb01nding the electron in the Sx = \u00afh/2 state as a function of time. b. Find the expectation value of Sx as a function of time. c. For your own peace of mind show that your answers make good sense in the extreme cases (i) \u03b2 \u2192 0 and (ii) \u03b2 \u2192 \u03c0/2. 2.4 Derive the neutrino oscillation probability (2.65) and use it, along with the data in Figure 2.2, to estimate the values of \u0394m2c4 (in units of eV2) and \u03b8. 2.5 Let x(t) be the coordinate operator for a free particle in one dimension in the Heisenberg picture. Evaluate [x(t), x(0)]. 2.6 Consider a particle in one dimension whose Hamiltonian is given by H = p2 2m + V(x). 139 Problems By calculating [[H, x], x] prove \u2211 a where |a is an energy eigenket with eigenvalue Ea. |a|x|a|2(Ea \u2212 Ea) = \u00afh2 2m, 2.7 Consider a particle in three dimensions whose Hamiltonian is given by H = p2 2m + V(x). By calculating [x \u00b7 p, H] obtain d dt x \u00b7 p = ) ( p2 m \u2212 x \u00b7 \u2207V. To identify the preceding relation with the quantum-mechanical analogue of the virial theorem it is essential that the left-hand side vanish. Under what condition would this happen? 2.8 Consider a free-particle wave packet in one dimension. At t = 0 it satis\ufb01es the minimum uncertainty relation In addition, we know (\u0394x)2(\u0394p)2 = \u00afh2 4 (t = 0). x = p = 0 (t = 0). Using the Heisenberg picture, obtain (\u0394x)2t as a function of t (t \u2265 0) when (\u0394x)2t=0 is given. (Hint: Take advantage of the property of the minimumuncertainty wave packet you worked out in Chapter 1, Problem 1.20.) 2.9 For a wave function x|\u03b1 = A(x \u2212 a)2(x + a)2eikx for \u2212a \u2264 x \u2264 a and zero otherwise, carry out the following. a. Find the constant A. b. Find the expectation values x, p, x2, and p2. c. Find the expectation values (\u0394x)2 and (\u0394p)2, and compare their product to that", " for a Gaussian wave packet. 2.10 Let |a and |a be eigenstates of a Hermitian operator A with eigenvalues a and a, respectively (a = a). The Hamiltonian operator is given by H = |a\u03b4a| + |a\u03b4a|, where \u03b4 is just a real number. a. Clearly, |a and |a are not eigenstates of the Hamiltonian. Write down the eigenstates of the Hamiltonian. What are their energy eigenvalues? b. Suppose the system is known to be in state |a at t = 0. Write down the state vector in the Schr\u00a8odinger picture for t > 0. 140 Quantum Dynamics c. What is the probability for \ufb01nding the system in |a for t > 0 if the system is known to be in state |a at t = 0? d. Can you think of a physical situation corresponding to this problem? 2.11 A box containing a particle is divided into a right and a left compartment by a thin partition. If the particle is known to be on the right (left) side with certainty, the state is represented by the position eigenket |R (|L), where we have neglected spatial variations within each half of the box. The most general state vector can then be written as |\u03b1 = |RR|\u03b1 + |LL|\u03b1, where R|\u03b1 and L|\u03b1 can be regarded as \u201cwave functions.\u201d The particle can tunnel through the partition; this tunneling effect is characterized by the Hamiltonian H = \u0394(|LR| + |RL|), where \u0394 is a real number with the dimension of energy. a. Find the normalized energy eigenkets. What are the corresponding energy eigenvalues? b. In the Schr\u00a8odinger picture the base kets |R and |L are \ufb01xed, and the state vector moves with time. Suppose the system is represented by |\u03b1 as given above at t = 0. Find the state vector |\u03b1, t0 = 0; t for t > 0 by applying the appropriate time-evolution operator to |\u03b1. c. Suppose at t = 0 the particle is on the right side with certainty. What is the probability for observing the particle on the left side as a function of time? d. Write down the coupled Schr\u00a8odinger equations for the wave functions R|\u03b1, t0 = 0; t and L|\u03b1, t0 = 0; t. Show that the solutions to the coupled Schr\u00a8odinger equations are just what you expect from (b). e. Suppose the printer made an error and wrote H as H = \u0394|LR|. By explicitly solving the most general time-evolution problem with this Hamiltonian, show that probability conservation is violated. 2.12 A one-dimensional simple harmonic oscillator with natural frequency \u03c9 is in initial state |\u03b1 = 1\u221a 2 |0 + ei\u03b4 \u221a 2 |1 where \u03b4 is a real number. a. Find the time-dependent wave function x|\u03b1; t and evaluate the (time-dependent) expectation values x and p in the state |\u03b1; t, i.e. in the Schr\u00a8odinger picture. b. Now calculate x and p in the Heisenberg picture and compare the results. 2.13 A particle with mass m moves in one dimension and is acted on by a constant force F. Find the operators x(t) and p(t) in the Heisenberg picture, and \ufb01nd their expectation values for an arbitrary state |\u03b1. Usex (0) = x0 and p(0) = p0. The result should 141 Problems be obvious. Comment on how to do this problem in the Schr\u00a8odinger picture, but do not try to work it through. 2.14 Consider a particle subject to a one-dimensional simple harmonic oscillator potential. Suppose at t = 0 the state vector is given by |0, \u2212ipa \u00afh exp where p is the momentum operator, a is some number with dimension of length, and the state |0 is the one for which x = 0 = p. Using the Heisenberg picture, evaluate the expectation value x for t \u2265 0. 2.15 a. Write down the wave function (in coordinate space) for the state speci\ufb01ed in Problem 2.14 at t = 0. You may use x|0 = \u03c0\u22121/4x\u22121/2 0 exp 2 x x0 \u2212 1 2, x0 \u2261 1/2. \u00afh m\u03c9 b. Obtain a simple expression for the probability that the state is found in the ground state at t = 0. Does this probability change for t > 0? 2.16 Consider a one-dimensional simple harmonic oscillator. a. Using - \u2019 = m\u03c9 2\u00afh x \u00b1", " ip m\u03c9, a a\u2020 - a|n a\u2020|n = \u221a \u221a n|n \u2212 1 n + 1|n + 1, evaluate m|x|n, m|p|n, m|{x, p}|n, m|x2|n, and m|p2|n. b. Translated from classical physics, the virial theorem states that ) ( p2 m = x \u00b7 \u2207V (3D) or ) ( p2 m = x dV dx (1D) Check that the virial theorem holds for the expectation values of the kinetic and the potential energy taken with respect to an energy eigenstate. 2.17 a. Using prove x|p = (2\u03c0\u00afh)\u22121/2eip x/\u00afh (one dimension) p|x|\u03b1 = i\u00afh \u2202 \u2202p p|\u03b1. b. Consider a one-dimensional simple harmonic oscillator. Starting with the Schr\u00a8odinger equation for the state vector, derive the Schr\u00a8odinger equation for the momentum-space wave function. (Make sure to distinguish the operator p from the eigenvalue p.) Can you guess the energy eigenfunctions in momentum space? 142 Quantum Dynamics 2.18 Consider a function, known as the correlation function, de\ufb01ned by C(t) =x (t)x(0), where x(t) is the position operator in the Heisenberg picture. Evaluate the correlation function explicitly for the ground state of a one-dimensional simple harmonic oscillator. 2.19 Consider again a one-dimensional simple harmonic oscillator. Do the following algebraically, that is, without using wave functions. a. Construct a linear combination of |0 and |1 such that x is as large as possible. b. Suppose the oscillator is in the state constructed in (a) at t = 0. What is the state vector for t > 0 in the Schr\u00a8odinger picture? Evaluate the expectation value x as a function of time for t > 0 using (i) the Schr\u00a8odinger picture and (ii) the Heisenberg picture. c. Evaluate (\u0394x)2 as a function of time using either picture. 2.20 Show for the one-dimensional simple harmonic oscillator 0|eikx|0 = exp[\u2212k20|x2|0/2], where x is the position operator. 2.21 The problem covers some fundamental concepts in quantum optics. See Glauber, Phys. Rev., 84 (1951) 395 and his Nobel lecture, Rev. Mod. Phys., 78 (2006) 1267; Gottfried (1966), Section 31; Merzbacher (1998), Section 10.7; and Gottfried and Yan (2003), Section 4.2. A coherent state of a one-dimensional simple harmonic oscillator is de\ufb01ned to be an eigenstate of the (non-Hermitian) annihilation operator a: a|\u03bb = \u03bb|\u03bb, where \u03bb is, in general, a complex number. a. Prove that |\u03bb = e\u2212|\u03bb|2/2e\u03bb a\u2020 |0 is a normalized coherent state. b. Prove the minimum uncertainty relation for such a state. c. Write |\u03bb as |\u03bb = \u221e \u2211 n=0 f (n)|n. Show that the distribution of | f (n)|2 with respect to n is in the form of a Poisson distribution, that is Pn(\u03bc) = e\u2212\u03bc\u03bcn/n! where \u03bc is the mean of the distribution. Find the most probable (integer) value of n, hence of E. d. Show that a coherent state can also be obtained by applying the translation (\ufb01nite displacement) operator e\u2212ipl/\u00afh (where p is the momentum operator, and l is the 143 Problems displacement distance) to the ground state. (See also Gottfried (1966), pp. 262\u2013 264; Problem 2.13 in Gottfried and Yan (2003), and Eq. 39 in Glauber\u2019s 1951 paper.) 2.22 Write a computer program to animate the time dependence of an arbitrary linear combination of stationary states for the simple harmonic oscillator in one dimension. The animation should display the time dependence of both the wave function and the probability density distribution for \ufb01nding the particle, both as a function of position. Check your animation by using pure eigenstates as input, and consider combinations that would approximate classical motion. Also animate the coherent state |\u03bb in Problem 2.21 above, for different mean values of the Poisson distribution. 2.23 Make the de\ufb01nitions J\u00b1 \u2261 \u00afha\u2020 \u00b1a\u2213, Jz \u2261 \u00afh 2 (a\u2020 +a+ \u2212 a\u2020 \u2212a\u2212), N \u2261 a\u2020 +", "a+ + a\u2020 \u2212a\u2212 where a\u00b1 and a\u2020 \u00b1 are the annihilation and creation operators of two independent simple harmonic oscillators satisfying the usual simple harmonic oscillator commutation relations. Also make the de\ufb01nition J2 \u2261 J2 z + 1 2 (J+J\u2212 + J\u2212J+). Prove [Jz, J\u00b1] =\u00b1 \u00afhJ\u00b1, [J2, Jz] =0, J2 = \u00afh2 2 N + 1. N 2 2.24 This exercise has to do with proving conservation laws in classical and quantum physics. a. Show that if a quantity Q with a density \u03c1(x, t) in some region R can only be changed by a \ufb02ux density j(x, t) through the surface bordering R, then \u2202\u03c1 \u2202t b. Prove that Maxwell\u2019s equations imply that electric charge is conserved c. Prove that Schr\u00a8odinger\u2019s equation implies that probability is conserved, i.e. + \u2207 \u00b7 j = 0. (2.190) 2.25 Derive the normalization constant cn in (2.232) by deriving the orthogonality relationship (2.233) using generating functions. Start by working out the integral \u221e I = \u2212\u221e g(x, t)g(x, s)e\u2212x2 dx and then consider the integral again with the generating functions in terms of series with Hermite polynomials. 2.26 Derive an expression for the action of the position operator on an arbitrary state in the momentum representation. That is, \ufb01nd p|x|\u03b1 in terms of p|\u03b1. Use this to solve the linear potential (2.234) problem in momentum space, and show that the 144 Quantum Dynamics Fourier transform of your solution is indeed the appropriate Airy function. Note that an alternative form of the Airy function (see http://dlmf.nist.gov/9.5) is \u221e Ai(x) = 1 \u03c0 cos 0 1 3 t3 + xt dt. You need not be concerned with normalizing the wave function. 2.27 Consider a particle of mass m subject to a one-dimensional potential of the following form: V = \u23a7 \u23a8 \u23a9 1 2 \u221e kx2 for for x > 0 x < 0. a. What is the ground-state energy? b. What is the expectation value x2 for the ground state? 2.28 A particle in one dimension is trapped between two rigid walls: V(x) = 0 \u221e for for 0 < x < L x < 0, x > L. At t = 0 it is known to be exactly at x = L/2 with certainty. What are the relative probabilities for the particle to be found in various energy eigenstates? Write down the wave function for t \u2265 0. (You need not worry about absolute normalization, convergence, and other mathematical subtleties.) 2.29 Consider a particle in one dimension bound to a \ufb01xed center by a \u03b4-function potential of the form V(x) =\u2212 \u03bd0\u03b4(x) where \u03bd0 is real and positive. Find the wave function and the binding energy of the ground state. Are there excited bound states? 2.30 A particle of mass m in one dimension is bound to a \ufb01xed center by an attractive \u03b4-function potential: V(x) =\u2212 \u03bb\u03b4(x) (\u03bb > 0). At t = 0, the potential is suddenly switched off (that is, V = 0 for t > 0). Find the wave function for t > 0. (Be quantitative! But you need not attempt to evaluate an integral that may appear.) 2.31 A particle in one dimension (\u2212\u221e < x < \u221e) is subjected to a constant force derivable from V = \u03bbx (\u03bb > 0). a. Is the energy spectrum continuous or discrete? Write down an approximate expression for the energy eigenfunction speci\ufb01ed by E. Also sketch it crudely. b. Discuss brie\ufb02y what changes are needed if V is replaced by V = \u03bb|x|. 145 Problems 2.32 A particle of mass m is con\ufb01ned to a one-dimensional square well with \ufb01nite walls. That is, a potential V(x) =0 for\u2212 a \u2264 x \u2264 +a, and V(x) =V 0 = \u03b7(\u00afh2/2ma2) otherwise. You are to \ufb01nd the bound-state energy eigenvalues as E = \u03b5V0 along with their wave functions. a. Set the problem up with a wave function Ae\u03b1x for x \u2264 \u2212a, De\u2212", "\u03b1x for x \u2265 +a, and Beikx + Ce\u2212ikx inside the well. Match the boundary conditions at x = \u00b1a and show that k and \u03b1 must satisfy z = \u00b1z\u2217 where z \u2261 eiak(k \u2212 i\u03b1). Proceed to \ufb01nd a purely real or purely imaginary expression for z in terms of k and \u03b1. b. Find the wave functions for the two choices of z and show that the purely real (imaginary) choice leads to a wave function that is even (odd) under the exchange x \u2192 \u2212x. c. Find a transcendental equation for each of the two wave functions relating \u03b7 and \u03b5. Show that even a very shallow well (\u03b7 \u2192 0) has at least one solution for the even wave function, but you are not guaranteed any solution for an odd wave function. d. For \u03b7 = 10, \ufb01nd all the energy eigenvalues and plot their normalized wave functions. 2.33 Consider a particle of mass m moving in one dimension x under the in\ufb02uence of a potential energy function V(x). a. Show that if V(\u2212x) = V(x), then a solution u(x) to the time-independent Schr\u00a8odinger equation must have the property u(\u2212x) =\u00b1 u(x). (This is a simple example of parity symmetry, which will be discussed in more detail in Section 4.2.) b. Consider a potential that is an in\ufb01nite square well but with a rectangular barrier in the middle. That is V(x) =V 0 > 0 for \u2212b \u2264 x \u2264 b, in\ufb01nity for |x| > a, and zero for b \u2264 |x| \u2264a. For b/a = 1/3 and V0 = 20(\u00afh2/2ma2), \ufb01nd the two energy eigenvalues along with their normalized eigenfunctions. (Plots of the eigenfunctions are shown in Figure 4.3.) c. For V0 = 10(\u00afh2/2ma2) show that there is only one energy eigenstate, and \ufb01nd the eigenvalue. You will need to write a computer program to calculate the eigenvalues and plot the wave functions. It is easiest to scale the eigenvalues by \u00afh2/2ma2 and \ufb01nd transcendental equations to solve (numerically) for the eigenvalues in the positive and negative parity cases. 2.34 A particle of mass m moves in one dimension x under a potential energy V(x). a. For V(x) =\u2212 V0 b \u03b4(x), V0 > 0, b > 0, \ufb01nd the bound-state energy eigenvalue E. b. Generalize this to the \u201cdouble delta function\u201d potential V(x) =\u2212 V0 # and \ufb01nd the bound-state energy eigenvalues and plot the corresponding eigenfunctions. Also show that you get the expected results as a \u2192 0. 146 Quantum Dynamics 2.35 Derive an expression for the density of free-particle states in two dimensions, normalized with periodic boundary conditions inside a box of side length L. 2.36 Use the WKB method to \ufb01nd the (approximate) energy eigenvalues for the one- dimensional simple harmonic oscillator potential V(x) =m \u03c92x2/2. 2.37 Consider an electron con\ufb01ned to the interior of a hollow cylindrical shell whose axis coincides with the z-axis. The wave function is required to vanish on the inner and outer walls, \u03c1 = \u03c1a and \u03c1b, and also at the top and bottom, z = 0 and L. a. Find the energy eigenfunctions. (Do not bother with normalization.) Show that the energy eigenvalues are given by Elmn = \u00afh2 2me l\u03c0 L k2 mn + 2 (l = 1, 2, 3,..., m = 0, 1, 2,...), where kmn is the nth root of the transcendental equation Jm(kmn\u03c1b)Nm(kmn\u03c1a) \u2212 Nm(kmn\u03c1b)Jm(kmn\u03c1a) =0. b. Repeat the same problem when there is a uniform magnetic \ufb01eld B = B\u02c6z for 0 < \u03c1 < \u03c1a. Note that the energy eigenvalues are in\ufb02uenced by the magnetic \ufb01eld even though the electron never \u201ctouches\u201d the magnetic \ufb01eld. c. Compare, in particular, the ground state of the B = 0 problem with that of the B = 0 problem. Show that", " if we require the ground-state energy to be unchanged in the presence of B, we obtain \u201c\ufb02ux quantization\u201d \u03c0\u03c12 aB = 2\u03c0N\u00afhc e (N = 0, \u00b11, \u00b12,...). 2.38 Consider a particle moving in one dimension under the in\ufb02uence of a potential V(x). Suppose its wave function can be written as exp[iS(x, t)/\u00afh]. Prove that S(x, t) satis\ufb01es the classical Hamilton\u2013Jacobi equation to the extent that \u00afh can be regarded as small in some sense. Show how one may obtain the correct wave function for a plane wave by starting with the solution of the classical Hamilton\u2013Jacobi equation with V(x) set equal to zero. Why do we get the exact wave function in this particular case? 2.39 Using spherical coordinates, obtain an expression for j for the ground and excited states of the hydrogen atom. Show, in particular, that for ml = 0 states, there is a circulating \ufb02ux in the sense that j is in the direction of increasing or decreasing \u03c6, depending on whether ml is positive or negative. 2.40 Derive (2.280) and obtain the three-dimensional generalization of (2.280). 2.41 A particle of mass m moves along one of two \u201cpaths\u201d through space and time connecting the points (x, t) = (0, 0) and (x, t) = (D, T). One path is quadratic in time, i.e. x1(t) = 1 2 at2 where a is a constant. The second path is linear in time, i.e. x2(t) =vt where v is a constant. The correct classical path is the quadratic path, that is x1(t). a. Find the acceleration a for the correct classical path. Use freshman physics to \ufb01nd the force F = ma = \u2212dV/dx and then the potential energy function V(x) in terms of m, D, andT. Also \ufb01nd the velocity v for the linear (i.e. incorrect classical) path. 147 Problems! \" b. Calculate the classical action S[x(t)] = T dt for each of the two 0 paths x1(t) and x2(t). Con\ufb01rm that S1 \u2261 S[x1(t)] < S2 \u2261 S[x2(t)], and \ufb01nd \u0394S = S2 \u2212 S1. 2 m\u02d9x2 \u2212 V(x) 1 c. Calculate \u0394S/\u00afh for a particle which moves 1 mm in 1 ms for two cases. The particle is a nanoparticle made up of 100 carbon atoms in one case. The other case is an electron. For which of these would you consider the motion \u201cquantum mechanical\u201d and why? 2.42 De\ufb01ne the partition function as Z = d3xK(x, t; x, 0)|\u03b2=it/\u00afh, as in (2.284)\u2013(2.286). Show that the ground-state energy is obtained by taking \u2202Z \u2202\u03b2 (\u03b2 \u2192 \u221e). \u2212 1 Z Illustrate this for a particle in a one-dimensional box. 2.43 The propagator in momentum space analogous to (2.290) is given by p, t|p, t0. Derive an explicit expression for p, t|p, t0 for the free-particle case. 2.44 a. Write down an expression for the classical action for a simple harmonic oscillator for a \ufb01nite time interval. b. Construct xn, tn|xn\u22121, tn\u22121 for a simple harmonic oscillator using Feynman\u2019s prescription for tn \u2212 tn\u22121 = \u0394t small. Keeping only terms up to order (\u0394t)2, show that it is in complete agreement with the t \u2212 t0 \u2192 0 limit of the propagator given by (2.290). 2.45 State the Schwinger action principle (see Finkelstein (1973), p. 155). Obtain the solution for x2t2|x1t1 by integrating the Schwinger principle and compare it with the corresponding Feynman expression for x2t2|x1t1. Describe the classical limits of these two expressions. 2.46 Show that the wave-mechanical approach to the gravity-induced problem discussed in Section 2.7 also leads to phase-difference expression (2.337). 2.47 a. Verify (2.345) and (2.347). b. Verify continuity equation (2.350) with j given by (2", ".351). 2.48 Consider the Hamiltonian of a spinless particle of charge e. In the presence of a static magnetic \ufb01eld, the interaction terms can be generated by poperator \u2192 poperator \u2212 eA c, where A is the appropriate vector potential. Suppose, for simplicity, that the magnetic \ufb01eld B is uniform in the positive z-direction. Prove that the above prescription indeed leads to the correct expression for the interaction of the orbital magnetic moment (e/2mc)L with the magnetic \ufb01eld B. Show that there is also an extra term proportional to B2(x2 + y2), and comment brie\ufb02y on its physical signi\ufb01cance. 148 Quantum Dynamics 2.49 An electron moves in the presence of a uniform magnetic \ufb01eld in the z-direction (B = B\u02c6z). a. Evaluate where [\u03a0x, \u03a0y], \u03a0x \u2261 px \u2212 eAx c, \u03a0y \u2261 py \u2212 eAy c. b. By comparing the Hamiltonian and the commutation relation obtained in (a) with those of the one-dimensional oscillator problem, show how we can immediately write the energy eigenvalues as Ek,n = \u00afh2k2 2m + |eB|\u00afh mc, n + 1 2 where \u00afhk is the continuous eigenvalue of the pz operator and n is a nonnegative integer including zero. 2.50 Consider the neutron interferometer. p = h /\u03bb B l Interference region Prove that the difference in the magnetic \ufb01elds that produce two successive maxima in the counting rates is given by B = 4\u03c0\u00afhc |e|gn\u03bbl, where gn(= \u22121.91) is the neutron magnetic moment in units of \u2212e\u00afh/2mnc, and \u03bb \u2261 \u03bb/2\u03c0. This problem was in fact analyzed in the paper by Bernstein, Phys. Rev. Lett., 18 (1967) 1102. 3 Theory of Angular Momentum This chapter is concerned with a systematic treatment of angular momentum and related topics. The importance of angular momentum in modern physics can hardly be overemphasized. A thorough understanding of angular momentum is essential in molecular, atomic, and nuclear spectroscopy; angular momentum considerations play an important role in scattering and collision problems as well as in bound-state problems. Furthermore, angularmomentum concepts have important generalizations \u2013 isospin in nuclear physics, SU(3), SU(2)\u2297U(1) in particle physics, and so forth. 3.1 Rotations and Angular Momentum Commutation Relations 3.1.1 Finite Versus Infinitesimal Rotations We recall from elementary physics that rotations about the same axis commute, whereas rotations about different axes do not. For instance, a 30\u25e6 rotation about the z-axis followed by a 60\u25e6 rotation about the same z-axis is obviously equivalent to a 60\u25e6 rotation followed by a 30\u25e6 rotation, both about the same axis. However, let us consider a 90\u25e6 rotation about the z-axis, denoted by Rz(\u03c0/2), followed by a 90\u25e6 rotation about the x-axis, denoted by Rx(\u03c0/2); compare this with a 90\u25e6 rotation about the x-axis followed by a 90\u25e6 rotation about the z-axis. The net results are different, as we can see from Figure 3.1. Our \ufb01rst basic task is to work out quantitatively the manner in which rotations about different axes fail to commute. To this end, we \ufb01rst recall how to represent rotations in three dimensions by 3 \u00d7 3 real, orthogonal matrices. Consider a vector V with components Vx, Vy, and Vz. When we rotate, the three components become some other set of numbers, x, V V z. The old and new components are related via a 3 \u00d7 3 orthogonal matrix R: \u239b y, and R \u23a0 = V Vx x V Vy y V Vz z RRT = RTR = 1, \u23a0, (3.1a) (3.1b) where the superscript T stands for a transpose of a matrix. It is a property of orthogonal matrices that is automatically satis\ufb01ed. 149 3.2) 150 Theory of Angular Momentum z z z z x x z z Rx(\u03c0/2) x Rz(\u03c0/2) x Rz(\u03c0/2) x Rx(\u03c0/2) x Fig. 3.1 Example to illustrate the noncommutativity of finite rotations. To be de\ufb01nite, we consider a rotation about the z-axis", " by angle \u03c6. The convention we follow throughout this book is that a rotation operation affects a physical system itself, as in Figure 3.1, while the coordinate axes remain unchanged. The angle \u03c6 is taken to be positive when the rotation in question is counterclockwise in the xy-plane, as viewed from the positive z-side. If we associate a right-handed screw with such a rotation, a positive \u03c6 rotation around the z-axis means that the screw is advancing in the positive z-direction. With this convention, we easily verify that Rz(\u03c6) = \u239b \u239d cos \u03c6 \u2212 sin \u03c6 0 sin \u03c6 0 1 0 cos \u03c6 0 \u239e \u23a0. (3.3) Had we adopted a different convention, in which a physical system remained \ufb01xed but the coordinate axes rotated, this same matrix with a positive \u03c6 would have represented a clockwise rotation of the x- and y-axes, when viewed from the positive z-side. It is obviously important not to mix the two conventions! Some authors distinguish the two approaches by using \u201cactive rotations\u201d for physical systems rotated and \u201cpassive rotations\u201d for the coordinate axes rotated. We are particularly interested in an in\ufb01nitesimal form of Rz: \u239e \u239b \u03b52 2 Rz(\u03b5\u03b5 \u03b52 3.4) 151 3.1 Rotations and Angular Momentum Commutation Relations where terms of order \u03b53 and higher are ignored. Likewise, we have \u239b Rx(\u03b5 \u2212 \u03b52 2 \u03b5 0 \u2212\u03b5 1 \u2212 \u03b52 2 and Ry(\u03b5) = \u239b \u239c \u239c \u239c \u239d \u03b52 2 1 \u2212 0 \u2212 \u2212 \u03b52 2 (3.5a) (3.5b) which may be read from (3.4) by cyclic permutations of x, y, z, that is, x \u2192 y, y \u2192 z, z \u2192 x. Compare now the effect of a y-axis rotation followed by an x-axis rotation with that of an x-axis rotation followed by a y-axis rotation. Elementary matrix manipulations lead to Rx(\u03b5)Ry(\u03b5) = \u239c \u239c \u239c \u239d \u239b 1 \u2212 \u03b52 2 \u03b52 \u2212\u03b5 \u239e 0 \u03b52 \u03b5 1 \u2212 \u03b52 and Ry(\u03b5)Rx(\u03b5) = \u239b \u239c \u239c \u239c \u239d 1 \u2212 \u03b52 2 0 \u2212\u03b5 \u239e \u239f \u239f \u239f \u23a0. \u03b52 \u03b5 \u03b52 2 1 \u2212 \u03b5 \u2212\u03b5 1 \u2212 \u03b52 (3.6a) (3.6b) From (3.6a) and (3.6b) we have the \ufb01rst important result: In\ufb01nitesimal rotations about different axes do commute if terms of order \u03b52 and higher are ignored.1 The second and even more important result concerns the manner in which rotations about different axes fail to commute when terms of order \u03b52 are kept: Rx(\u03b5)Ry(\u03b5) \u2212 Ry(\u03b5)Rx(\u03b5) = \u239d \u239b \u239e \u23a0 0 \u2212\u03b52 \u03b52 0 0 0 0 0 0 where all terms of order higher than \u03b52 have been ignored throughout this derivation. We also have = Rz(\u03b52) \u2212 1, (3.7) 1 = Rany(0) (3.8) 1 Actually there is a familiar example of this in elementary mechanics. The angular velocity vector \u03c9 that characterizes an in\ufb01nitesimal change in rotation angle during an in\ufb01nitesimal time interval follows the usual rule of vector addition, including commutativity of vector addition. However, we cannot ascribe a vectorial property to a \ufb01nite angular change. 152 Theory of Angular Momentum where any stands for any rotation axis. Thus the \ufb01nal result can be written as Rx(\u03b5)Ry(\u03b5) \u2212 Ry(\u03b5)Rx(\u03b5) =R z(\u03b52) \u2212 Rany(0). (3.9) This is an example of the commutation relations between rotation operations about different axes, which we will use later in deducing the angular-momentum commutation relations in quantum mechanics. 3.1.2 Infinitesimal Rotations in Quantum Mechanics So far we have not used quantum-mechanical concepts. The matrix R is just a 3 \u00d7 3 orthogonal matrix acting on a vector V written in column matrix form. We must now understand how to characterize rotations in", " quantum mechanics. Because rotations affect physical systems, the state ket corresponding to a rotated system is expected to look different from the state ket corresponding to the original unrotated system. Given a rotation operation R, characterized by a 3 \u00d7 3 orthogonal matrix R, we associate an operator D(R) in the appropriate ket space such that |\u03b1R = D(R)|\u03b1, where |\u03b1R and |\u03b1 stand for the kets of the rotated and original system, respectively.2 Note that the 3 \u00d7 3 orthogonal matrix R acts on a column matrix made up of the three components of a classical vector, while the operator D(R) acts on state vectors in ket space. The matrix representation of D(R), which we will study in great detail in the subsequent sections, depends on the dimensionality N of the particular ket space in question. For N = 2, 2 system with no other degrees of freedom, D(R) is appropriate for describing a spin 1 represented by a 2 \u00d7 2 matrix; for a spin 1 system, the appropriate representation is a 3 \u00d7 3 unitary matrix, and so on. (3.10) To construct the rotation operator D(R), it is again fruitful to examine \ufb01rst its properties under an in\ufb01nitesimal rotation. We can almost guess how we must proceed by analogy. In both translations and time evolution, which we studied in Sections 1.6 and 2.1, respectively, the appropriate in\ufb01nitesimal operators could be written as U\u03b5 = 1 \u2212 iG\u03b5 with a Hermitian operator G. Speci\ufb01cally, G \u2192 px \u00afh, \u03b5 \u2192 dx for an in\ufb01nitesimal translation by a displacement dx in the x-direction, and G \u2192 H \u00afh, \u03b5 \u2192 dt (3.11) (3.12) (3.13) for an in\ufb01nitesimal time evolution with time displacement dt. We know from classical mechanics that angular momentum is the generator of rotation in much the same way as momentum and Hamiltonian are the generators of translation and time evolution, 2 The symbol D stems from the German word Drehung, meaning rotation. 153 3.1 Rotations and Angular Momentum Commutation Relations respectively. We therefore de\ufb01ne the angular-momentum operator Jk in such a way that the operator for an in\ufb01nitesimal rotation around the kth axis by angle d\u03c6 can be obtained by letting G \u2192 Jk \u00afh, \u03b5 \u2192 d\u03c6 (3.14) in (3.11). With Jk taken to be Hermitian, the in\ufb01nitesimal rotation operator is guaranteed to be unitary and reduces to the identity operator in the limit d\u03c6 \u2192 0. More generally, we have D( \u02c6n, d\u03c6) =1 \u2212 i d\u03c6 (3.15) J\u00b7 \u02c6n \u00afh for a rotation about the direction characterized by a unit vector \u02c6n by an in\ufb01nitesimal angle d\u03c6. We stress that in this book we do not de\ufb01ne the angular-momentum operator to be x \u00d7 p. This is important because spin angular momentum, to which our general formalism also applies, has nothing to do with xi and pj. Put in another way, in classical mechanics one can prove that the angular momentum de\ufb01ned to be x \u00d7 p is the generator of a rotation; in contrast, in quantum mechanics we de\ufb01ne J so that the operator for an in\ufb01nitesimal rotation takes the form (3.15). 3.1.3 Finite Rotations in Quantum Mechanics A \ufb01nite rotation can be obtained by compounding successively in\ufb01nitesimal rotations about the same axis. For instance, if we are interested in a \ufb01nite rotation about the z-axis by angle \u03c6, we consider Dz(\u03c6) = lim N\u2192\u221e 1 \u2212 i \u2212iJz\u03c6 \u00afh = exp N \u03c6 N Jz \u00afh = 1 \u2212 iJz\u03c6 \u00afh \u03c62 \u2212 J 2 z 2\u00afh2 + \u00b7 \u00b7 \u00b7. (3.16) In order to obtain the angular-momentum commutation relations, we need one more concept. As we remarked earlier, for every rotation R represented by a 3 \u00d7 3 orthogonal matrix R there exists a rotation operator D(R) in the appropriate ket space. We further postulate that D(R) has the same group properties as R: Identity: R \u00b7 1 = R \u21d2 D(R) \u00b7 1 = D(R) Closure: R1R2 = R3 \u21d2 D(R1)D(R2", ") =D (R3) Inverses: RR\u22121 = 1 \u21d2 D(R)D \u22121(R) =1 R\u22121R = 1 \u21d2 D \u22121(R)D(R) = 1 (3.17a) (3.17b) (3.17c) 154 Theory of Angular Momentum Associativity: R1(R2R3) = (R1R2)R3 = R1R2R3 \u21d2 D(R1)[D(R2)D(R3)] = [D(R1)D(R2)]D(R3) = D(R1)D(R2)D(R3). (3.17d) 3.1.4 Commutation Relations for Angular Momentum Let us now return to the fundamental commutation relations for rotation operations (3.9) written in terms of the R matrices. Its rotation operator analogue would read \u03b52 \u2212 J2 x 2\u00afh2 1 \u2212 iJx\u03b5 \u00afh 1 \u2212 iJy\u03b5 \u00afh \u2212 \u2212 \u03b52 J 2 y 2\u00afh2 1 \u2212 iJy\u03b5 \u00afh \u2212 \u03b52 J 2 y 2\u00afh2 1 \u2212 iJx\u03b5 \u00afh \u03b52 \u2212 J 2 x 2\u00afh2 = 1 \u2212 iJz\u03b52 \u00afh \u2212 1. (3.18) Terms of order \u03b5 automatically drop out. Equating terms of order \u03b52 on both sides of (3.18), we obtain [Jx, Jy] =i \u00afhJz. Repeating this kind of argument with rotations about other axes, we obtain [Ji, Jj] =i \u00afh\u03b5ijkJk, (3.19) (3.20) known as the fundamental commutation relations of angular momentum. In general, when the generators of in\ufb01nitesimal transformations do not commute, the corresponding group of operations is said to be non-Abelian. Because of (3.20), the rotation group in three dimensions is non-Abelian. In contrast, the translation group in three dimensions is Abelian because pi and pj commute even with i = j. We emphasize that in obtaining the commutation relations (3.20) we have used the following two concepts. 1. Jk is the generator of rotation about the kth axis. 2. Rotations about different axes fail to commute. It is no exaggeration to say that commutation relations (3.20) summarize in a compact manner all the basic properties of rotations in three dimensions. The fundamental commutation relations make it possible to show that the angularmomentum operators themselves transform as expected under rotations. Consider a rotation by a \ufb01nite angle \u03c6 about the z-axis. If the ket of the system before rotation is given by |\u03b1, the ket after rotation is given by |\u03b1R = Dz(\u03c6)|\u03b1 (3.21) 155 3.2 Spin 1 2 Systems and Finite Rotations where Dz(\u03c6) is given by (3.16). To see that this operator really rotates the physical system, let us look at its effect on Jx. Under rotation this expectation value changes as follows: R = \u03b1|D \u2020 z (\u03c6)JxDz(\u03c6)|\u03b1. Jx \u2192 R\u03b1|Jx|\u03b1 (3.22) We must therefore compute iJz\u03c6 \u00afh exp Jx exp. \u2212iJz\u03c6 \u00afh (3.23) Using (2.168) to evaluate (3.23), we have iJz\u03c6 \u00afh \u2212iJz\u03c6 \u00afh Jx exp exp = Jx + i\u03c6 \u00afh + 1 2! 2 i\u03c6 \u00afh [Jz, [Jz, Jx]] i\u00afhJy \u00afh2Jx + 1 3! i\u03c6 \u00afh \u03c63 3! + \u00b7 \u00b7 \u00b7 \u03c62 = Jx 1 \u2212 + \u00b7 \u00b7 \u00b7 \u2212 Jy \u03c6 \u2212 2! = Jx cos \u03c6 \u2212 Jy sin \u03c6. [Jz, Jx] i\u00afhJy 3 [Jz, [Jz, [Jz, Jx]]] + \u00b7 \u00b7 \u00b7 \u00afh2Jx i\u00afh3Jy (3.24) We will return to transformations of this sort when we discuss symmetry operations in Chapter 4. 3.2 Spin 1 2 Systems and Finite Rotations 3.2.1 Rotation Operator for Spin 1 2 The lowest number, N, of dimensions in which the angular-momentum commutation relations (3.20) are realized is N = 2. The reader has already checked in Problem 1.10 of Chapter 1 that the operators de\ufb01ned by S", "x = Sy = Sz = \u00afh 2 i\u00afh 2 \u00afh 2 {(|+\u2212|) + (|\u2212+|)}, {\u2212(|+\u2212|) + (|\u2212+|)}, (3.25) {(|++|) \u2212 (|\u2212\u2212|)}, satisfy commutation relations (3.20) with Jk replaced by Sk. It is not a priori obvious that nature takes advantage of the lowest dimensional realization of (3.20), but numerous experiments \u2013 from atomic spectroscopy to nuclear magnetic resonance \u2013 suf\ufb01ce to convince us that this is in fact the case. 156 Theory of Angular Momentum We showed in (3.24) that components of the general angular-momentum operator behave as expected under \ufb01nite rotations. This is of course also true for spin 1 2, that is Sx \u2192 R\u03b1|Sx|\u03b1 R = Sx cos \u03c6 \u2212 Sy sin \u03c6, (3.26) where the expectation value without subscripts is understood to be taken with respect to the (old) unrotated system. Similarly, Sy \u2192 Sy cos \u03c6 + Sx sin \u03c6. (3.27) As for the expectation value of Sz, there is no change because Sz commutes with Dz(\u03c6): Sz \u2192 Sz. (3.28) Relations (3.26), (3.27), and (3.28) are quite reasonable. They show that rotation operator (3.16), when applied to the state ket, does rotate the expectation value of S around the z-axis by angle \u03c6. In other words, the expectation value of the spin operator behaves as though it were a classical vector under rotation: Sk \u2192 \u2211 RklSl, (3.29) l where Rkl are the elements of the 3 \u00d7 3 orthogonal matrix R that speci\ufb01es the rotation in question. It should be clear from our derivation that this property is not restricted to the spin operator of spin 1 2 systems. In general, we have RklJl Jk \u2192 \u2211 (3.30) l under rotation, where Jk are the generators of rotations satisfying the angular-momentum commutation relations (3.20). Later we will show that relations of this kind can be further generalized to any vector operator. So far everything has been as expected. But now, be prepared for a surprise! We examine the effect of rotation operator (3.16) on a general ket, |\u03b1 = |++|\u03b1 + |\u2212\u2212|\u03b1, (3.31) a little more closely. We see that exp \u2212iSz\u03c6 \u00afh |\u03b1 = e\u2212i\u03c6/2|++|\u03b1 + ei\u03c6/2|\u2212\u2212|\u03b1. (3.32) The appearance of the half-angle \u03c6/2 here has an extremely interesting consequence. Let us consider a rotation by 2\u03c0. We then have |\u03b1Rz(2\u03c0) \u2192 \u2212|\u03b1. So the ket for the 360\u25e6 rotated state differs from the original ket by a minus sign. We would need a 720\u25e6 (\u03c6 = 4\u03c0) rotation to get back to the same ket with a plus sign. Notice that this minus sign disappears for the expectation value of S because S is sandwiched by |\u03b1 and \u03b1|, both of which change sign. Will this minus sign ever be observable? We will give the answer to this interesting question after we discuss spin precession once again. (3.33) 157 3.2 Spin 1 2 Systems and Finite Rotations 3.2.2 Spin Precession Revisited We now treat the problem of spin precession, already discussed in Section 2.1, from a new point of view. We recall that the basic Hamiltonian of the problem is given by H = \u2212 where e mec S\u00b7B = \u03c9Sz, (3.34) (3.35) \u03c9 \u2261 |e|B mec. The time-evolution operator based on this Hamiltonian is given by U (t, 0) =exp = exp. (3.36) \u2212iHt \u00afh \u2212iSz\u03c9t \u00afh Comparing this equation with (3.16), we see that the time-evolution operator here is precisely the same as the rotation operator in (3.16) with \u03c6 set equal to \u03c9t. In this manner we see immediately why this Hamiltonian causes spin precession. Paraphrasing (3.26), (3.27), and (3.28), we obtain Sxt = Sxt=0 cos \u03c9t \u2212 Syt=0 sin \u03c9t, Syt = Syt=0 cos \u03c9t + Sxt=0 sin \u03c9t, Szt = Szt=0. (", "3.37a) (3.37b) (3.37c) After t = 2\u03c0/\u03c9, the spin returns to its original direction. This set of equations can be used to discuss the spin precession of a muon, an electronlike particle which, however, is 210 times as heavy. The muon magnetic moment can be determined from other experiments \u2013 for example, the hyper\ufb01ne splitting in muonium, a bound state of a positive muon and an electron \u2013 to be e\u00afh/2m\u03bcc, just as expected from Dirac\u2019s relativistic theory of spin 1 2 particles. (We will here neglect very small corrections that arise from quantum \ufb01eld theory effects.) Knowing the magnetic moment we can predict the angular frequency of precession. So (3.37) can be and, in fact, has been checked experimentally (see Figure 2.1). In practice, as the external magnetic \ufb01eld causes spin precession, the spin direction is analyzed by taking advantage of the fact that electrons from muon decay tend to be emitted preferentially in the direction opposite to the muon spin. Let us now look at the time evolution of the state ket itself. Assuming that the initial (t = 0) ket is given by (3.31), we obtain after time t |\u03b1, t0 = 0; t = e\u2212i\u03c9t/2|++|\u03b1 + e+i\u03c9t/2|\u2212\u2212|\u03b1. (3.38) Expression (3.38) acquires a minus sign at t = 2\u03c0/\u03c9, and we must wait until t = 4\u03c0/\u03c9 to get back to the original state ket with the same sign. To sum up, the period for the state ket is twice as long as the period for spin precession 158 Theory of Angular Momentum \u03c4precession = \u03c4state ket = 2\u03c0 \u03c9, 4\u03c0 \u03c9. (3.39a) (3.39b) 3.2.3 Neutron Interferometry Experiment to Study 2\u03c0 Rotations We now describe an experiment performed to detect the minus sign in (3.33). Quite clearly, if every state ket in the universe is multiplied by a minus sign, there will be no observable effect. The only way to detect the predicted minus sign is to make a comparison between an unrotated state and a rotated state. As in gravity-induced quantum interference, discussed in Section 2.7, we rely on the art of neutron interferometry to verify this extraordinary prediction of quantum mechanics. A nearly monoenergetic beam of thermal neutrons is split into two parts, path A and path B; see Figure 3.2. Path A always goes through a magnetic-\ufb01eld-free region; in contrast, path B enters a small region where a static magnetic \ufb01eld is present. As a result, the neutron state ket going via path B suffers a phase change e\u2213 i \u03c9T/2, where T is the time spent in the B = 0 region and \u03c9 is the spin-precession frequency \u03c9 = gneB mpc (gn \u22121.91) (3.40) for the neutron with a magnetic moment of gne\u00afh/2mpc, as we can see if we compare this with (3.35), which is appropriate for the electron with magnetic moment e\u00afh/2mec. When path A and path B meet again in the interference region of Figure 3.2, the amplitude of the neutron arriving via path B is c2 = c2(B = 0)e\u2213 i \u03c9T/2, (3.41) while the amplitude of the neutron arriving via path A is c1, independent of B. So the intensity observable in the interference region must exhibit a sinusoidal variation \u2213 \u03c9T 2 where \u03b4 is the phase difference between c1 and c2 (B = 0). In practice, T, the time spent in the B = 0 region, is \ufb01xed but the precession frequency \u03c9 is varied by changing the (3.42) + \u03b4 cos, A B B A B Interference region Fig. 3.2 Experiment to study the predicted minus sign under a 2\u03c0 rotation. 159 3.2 Spin 1 2 Systems and Finite Rotations strength of the magnetic \ufb01eld. The intensity in the interference region as a function of B is predicted to have a sinusoidal variation. If we call \u0394B the difference in B needed to produce successive maxima, we can easily show that \u0394B = 4\u03c0\u00afhc egn\u03bbl, (3.43) where l is the path length. In deriving this formula we used the fact that a 4\u03c0 rotation is needed for the state ket to return to the original ket with the same sign, as required", " by our formalism. If, on the other hand, our description of spin 1 2 systems were incorrect and the ket were to return to its original ket with the same sign under a 2\u03c0 rotation, the predicted value for \u0394B would be just one-half of (3.43). Two different groups have conclusively demonstrated experimentally that prediction (3.43) is correct to an accuracy of a fraction of a percent.3 This is another triumph of quantum mechanics. The nontrivial prediction (3.33) has been experimentally established in a direct manner. 3.2.4 Pauli Two-Component Formalism Manipulations with the state kets of spin 1 2 systems can be conveniently carried out using the two-component spinor formalism introduced by W. Pauli in 1926. In Section 1.3 we learned how a ket (bra) can be represented by a column (row) matrix; all we have to do is arrange the expansion coef\ufb01cients in terms of a certain speci\ufb01ed set of base kets into a column (row) matrix. In the spin 1 2 case we have \u2261 \u03c7+ |\u2212 \u02d9= |+ \u02d9= 1 0 +| \u02d9= (1, 0) = \u03c7\u2020 + 0 1 \u2212| \u02d9= (0, 1) = \u03c7\u2020 \u2212 \u2261 \u03c7\u2212 for the base kets and bras and |\u03b1 = |++|\u03b1 + |\u2212\u2212|\u03b1 \u02d9= and +|\u03b1 \u2212|\u03b1 (3.44) (3.45a) \u03b1| = \u03b1|++| + \u03b1|\u2212\u2212| \u02d9= (\u03b1|+, \u03b1|\u2212) (3.45b) for an arbitrary state ket and the corresponding state bra. Column matrix (3.45a) is referred to as a two-component spinor and is written as \u03c7 = +|\u03b1 \u2212|\u03b1 \u2261 c+ c\u2212 = c+\u03c7+ + c\u2212\u03c7\u2212, (3.46) 3 Rauch et al., Phys. Lett. A, 54 (1975) 425; Werner et al., Phys. Rev. Lett., 35 (1975) 1053. 160 Theory of Angular Momentum where c+ and c\u2212 are, in general, complex numbers. For \u03c7\u2020 we have \u03c7\u2020 = (\u03b1|+, \u03b1|\u2212) = (c\u2217 +, c\u2217 \u2212). (3.47) The matrix elements \u00b1|Sk|+ and \u00b1|Sk|\u2212, apart from \u00afh/2, are to be set equal to those of 2 \u00d7 2 matrices \u03c3k, known as the Pauli matrices. We identify \u00b1|Sk|+ \u2261 (\u03c3k)\u00b1, +, \u00b1|Sk|\u2212 \u2261 (\u03c3k)\u00b1, \u2212. (3.48) \u00afh 2 We can now write the expectation value Sk in terms of \u03c7 and \u03c3k: Sk = \u03b1|Sk|\u03b1 = \u2211 a=+,\u2212 \u2211 a=+,\u2212 \u03b1|aa|Sk|aa|\u03b1 = \u03c7\u2020\u03c3k\u03c7, (3.49) \u00afh 2 \u00afh 2 where the usual rule of matrix multiplication is used in the last line. Explicitly, we see from (3.25) together with (3.48) that \u03c31 = 0 1 1 0, \u03c32 = 0 \u2212i 0 i, \u03c33 = 0 1 0 \u22121, (3.50) where the subscripts 1, 2, and 3 refer to x, y, andz, respectively. We record some properties of the Pauli matrices. First, \u03c32 i = 1 \u03c3i\u03c3j + \u03c3j\u03c3i = 0, for i = j, (3.51a) (3.51b) where the right-hand side of (3.51a) is to be understood as the 2 \u00d7 2 identity matrix. These two relations are, of course, equivalent to the anticommutation relations {\u03c3i, \u03c3j} = 2\u03b4ij. We also have the commutation relations [\u03c3i, \u03c3j] =2i \u03b5ijk\u03c3k, (3.52) (3.53) which we see to be the explicit 2 \u00d7 2 matrix realizations of the angular-momentum commutation relations (3.20). Combining (3.52) and (3.53), we can obtain Notice also that \u03c31\u03c32 = \u2212\u03c32\u03c31 = i\u03c33.... i = \u03c3i, \u03c3\u2020 det(\u03c3i) =\u2212 1, Tr(\u03c3i) =0. (3.54) (3.55a) (3.55b) (3.55c) 161 3.2 Spin 1", " 2 Systems and Finite Rotations We now consider \u03c3\u00b7a, wherea is a vector in three dimensions. This is actually to be understood as a 2 \u00d7 2 matrix. Thus \u03c3\u00b7a \u2261 \u2211 k = ak\u03c3k + a3 a1 + ia2 a1 \u2212 ia2 \u2212a3. There is also a very important identity, (\u03c3\u00b7a)(\u03c3\u00b7b) =a\u00b7b + i\u03c3\u00b7(a \u00d7 b). (3.56) (3.57) To prove this all we need are the anticommutation and commutation relations, (3.52) and (3.53), respectively: \u2211 \u03c3jaj \u2211 \u03c3kbk = \u2211 \u2211 \u03c3j, \u03c3k} + 1 2 [\u03c3j, \u03c3k] ajbk (\u03b4jk + i\u03b5jkl\u03c3l) ajbk = a\u00b7b + i\u03c3\u00b7(a \u00d7 b). If the components of a are real, we have (\u03c3\u00b7a)2 = |a|2, where |a| is the magnitude of the vector a. (3.58) (3.59) 3.2.5 Rotations in the Two-Component Formalism Let us now study the 2 \u00d7 2 matrix representation of the rotation operator D( \u02c6n, \u03c6). We have Using exp \u2212iS\u00b7 \u02c6n\u03c6 \u00afh \u02d9= exp. \u2212i\u03c3\u00b7 \u02c6n\u03c6 2 (\u03c3\u00b7 \u02c6n)n = for n even 1 \u03c3\u00b7 \u02c6n for n odd, which follows from (3.59), we can write exp \u2212i\u03c3\u00b7 \u02c6n\u03c6 2 = 1 \u2212 (\u03c3\u00b7 \u02c6n)2 2! 2 \u03c6 + 2 \u2212 (\u03c3\u00b7 \u02c6n)3 3! (\u03c3\u00b7 \u02c6n)4 4\u03c3\u00b7 \u02c6n) 2 \u03c6 2 = 1 cos \u2212 i\u03c3\u00b7 \u02c6n sin \u2212 \u00b7 \u00b7 \u00b7 \u03c6 2 4 + \u00b7 \u00b7 \u00b7 (3.60) (3.61) (3.62) 162 Theory of Angular Momentum 2 \u2212i\u03c3\u00b7 \u02c6n\u03c6 2 Explicitly, in 2 \u00d7 2 form we have exp \u2212i\u03c3\u00b7 \u02c6n\u03c6 2 \u239c \u239c \u239d = \u239b \u03c6 cos \u2212 inz sin \u03c6 2 (\u2212inx + ny) sin \u03c6 2 \u03c6 2 (\u2212inx \u2212 ny) sin \u03c6 cos + inz sin 2 \u03c6 2 \u239e \u239f \u239f \u23a0. (3.63) Just as the operator exp(\u2212iS\u00b7 \u02c6n\u03c6/\u00afh) acts on a state ket |\u03b1, the 2 \u00d7 2 matrixexp (\u2212i\u03c3\u00b7 \u02c6n\u03c6/2) acts on a two-component spinor \u03c7. Under rotations we change \u03c7 as follows: \u03c7 \u2192 exp \u03c7. (3.64) On the other hand, the \u03c3k themselves are to remain unchanged under rotations. So strictly speaking, despite its appearance, \u03c3 is not to be regarded as a vector; rather, it is \u03c7\u2020\u03c3\u03c7 which obeys the transformation property of a vector: \u03c7\u2020\u03c3k\u03c7 \u2192 \u2211 Rkl\u03c7\u2020\u03c3l\u03c7. l (3.65) An explicit proof of this may be given using exp i\u03c33\u03c6 2 \u03c31 exp \u2212i\u03c33\u03c6 2 = \u03c31 cos \u03c6 \u2212 \u03c32 sin \u03c6 (3.66) and so on, which is the 2 \u00d7 2 matrix analogue of (3.16). In discussing a 2\u03c0 rotation using the ket formalism, we have seen that a spin 1 2 ket |\u03b1 goes into \u2212|\u03b1. The 2 \u00d7 2 analogue of this statement is \u2212i\u03c3\u00b7 \u02c6n\u03c6 2 = \u22121, for any \u02c6n, (3.67) \u03c6 = 2 \u03c0 exp which is evident from (3.62). As an instructive application of rotation matrix (3.63), let us see how we can construct an eigenspinor of \u03c3\u00b7 \u02c6n with eigenvalue + 1, where \u02c6n is a unit vector in some speci\ufb01ed direction. Our object is to construct \u03c7 satisfying \u03c3\u00b7 \u02c6n\u03c7 = \u03c7. (3.68) In other words, we look for the two-component column matrix representation of |S\u00b7 \u02c6n; + de\ufb01ned by S\u00b7 \u02c6n|S\u00b7 \u02c6n; + = |S\u00b7 \u02c6n;", " +. (3.69) \u00afh 2 Actually this can be solved as a straightforward eigenvalue problem (see Problem 1.11 in Chapter 1), but here we present an alternative method based on rotation matrix (3.63). Let the polar and the azimuthal angles that characterize \u02c6n be \u03b2 and \u03b1, respectively. We start with, the two-component spinor that represents the spin-up state. Given this, we \ufb01rst rotate about the y-axis by angle \u03b2; we subsequently rotate by angle \u03b1 about the z-axis. We see that the desired spin state is then obtained; see Figure 3.3. In the Pauli 1 0 163 3.3 SO(3), SU(2), and Euler Rotations Second rotation \u03b2 \u03b1 First rotation Fig. 3.3 Construction of the \u03c3\u00b7\u02c6n eigenspinor. spinor language this sequence of operations is equivalent to applying exp(\u2212i\u03c32\u03b2/2) to 1 0 followed by an application of exp(\u2212i\u03c33\u03b1/2). The net result is $ # \u03c7 = cos \u03b1 2 \u03b1 2 \u2212 i\u03c33 sin \u2212 i sin \u03b1 2 \u03b1 2 cos = = cos sin \u03b2 2 \u03b2 2 0 e\u2212i\u03b1/2, ei\u03b1/2 \u03b2 2 0 cos cos \u03b1 2 \u2212 i\u03c32 sin 1 0 \u03b2 2 cos \u03b2 2 + i sin \u03b1 2 sin \u03b2 2 \u2212 sin \u03b2 2 \u03b2 2 cos 1 0 (3.70) in complete agreement with Problem 1.11 of Chapter 1 if we realize that a phase common to both the upper and lower components is devoid of physical signi\ufb01cance. 3.3 SO(3), SU(2), and Euler Rotations 3.3.1 Orthogonal Group We will now study a little more systematically the group properties of the operations with which we have been concerned in the previous two sections. The most elementary approach to rotations is based on specifying the axis of rotation and the angle of rotation. It is clear that we need three real numbers to characterize a general rotation: the polar and the azimuthal angles of the unit vector \u02c6n taken in the direction of 164 Theory of Angular Momentum the rotation axis and the rotation angle \u03c6 itself. Equivalently, the same rotation can be speci\ufb01ed by the three Cartesian components of the vector \u02c6n\u03c6. However, these ways of characterizing rotation are not so convenient from the point of view of studying the group properties of rotations. For one thing, unless \u03c6 is in\ufb01nitesimal or \u02c6n is always in the same direction, we cannot add vectors of the form \u02c6n\u03c6 to characterize a succession of rotations. It is much easier to work with a 3 \u00d7 3 orthogonal matrix R because the effect of successive rotations can be obtained just by multiplying the appropriate orthogonal matrices. How many independent parameters are there in a 3 \u00d7 3 orthogonal matrix? A real 3 \u00d7 3 matrix has 9 entries, but we have the orthogonality constraint RRT = 1 (3.71) which corresponds to 6 independent equations because the product RRT, being the same as RTR, is a symmetrical matrix with 6 independent entries. As a result, there are 3 (that is, the same number previously obtained by a more 9\u20136) independent numbers in R, elementary method. The set of all multiplication operations with orthogonal matrices forms a group. By this we mean that the following four requirements are satis\ufb01ed. 1. The product of any two orthogonal matrices is another orthogonal matrix, which is satis\ufb01ed because 2. The associative law holds: (R1R2)(R1R2)T = R1R2RT 2RT 1 = 1. 3. The identity matrix 1, physically corresponding to no rotation, de\ufb01ned by R1(R2R3) = (R1R2)R3. R1 = 1R = R (3.72) (3.73) (3.74) is a member of the class of all orthogonal matrices. 4. The inverse matrix R\u22121, physically corresponding to rotation in the opposite sense, de\ufb01ned by is also a member. RR\u22121 = R\u22121R = 1 (3.75) This group has the name SO(3), where S stands for special, O stands for orthogonal, 3 for three dimensions. Note only rotational operations are considered here, hence we have SO(3) rather than O(3) (which can include the inversion operation of Chapter 4 later). 3.3.2 Unitary Unimodular Group In the previous section we learned yet another way to characterize an arbitrary rotation", ", that is, to look at the 2 \u00d7 2 matrix (3.63) that acts on the two-component spinor \u03c7. Clearly, (3.63) is unitary. As a result, for the c+ and c\u2212, de\ufb01ned in (3.46), |c+| 2 + |c\u2212| 2 = 1 (3.76) 165 3.3 SO(3), SU(2), and Euler Rotations is left invariant. Furthermore, matrix (3.63) is unimodular; that is, its determinant is 1, as will be shown explicitly below. We can write the most general unitary unimodular matrix as U(a, b) = a \u2212b\u2217, b a\u2217 where a and b are complex numbers satisfying the unimodular condition |a| 2 + |b| 2 = 1. (3.77) (3.78) We can easily establish the unitary property of (3.77) as follows: U(a, b)\u2020U(a, b) = a\u2217 \u2212b b\u2217 a a \u2212b\u2217 b a\u2217 = 1. (3.79) We can readily see that the 2 \u00d7 2 matrix (3.63) that characterizes a rotation of a spin 1 2 system can be written as U(a, b). Comparing (3.63) with (3.77), we identify \u03c6 Re(a) =cos 2 Re(b) =\u2212 nysin Im(a) =\u2212 nzsin, Im(b) =\u2212 nxsin, \u03c6 2 \u03c6 2 \u03c6 2,, (3.80) from which the unimodular property of (3.78) is immediate. Conversely, it is clear that the most general unitary unimodular matrix of form (3.77) can be interpreted as representing a rotation. The two complex numbers a and b are known as Cayley\u2013Klein parameters. Historically the connection between a unitary unimodular matrix and a rotation was known long before the birth of quantum mechanics. In fact, the Cayley\u2013Klein parameters were used to characterize complicated motions of gyroscopes in rigid-body kinematics. Without appealing to the interpretations of unitary unimodular matrices in terms of rotations, we can directly check the group properties of multiplication operations with unitary unimodular matrices. Note in particular that U(a1, b1)U(a2, b2) =U(a 1a2 \u2212 b1b\u2217 2, a1b2 + a\u2217 2b1), where the unimodular condition for the product matrix is |a1a2 \u2212 b1b\u2217 2 |2 + |a1b2 + a\u2217 2b1|2 = 1. For the inverse of U we have U\u22121(a, b) =U(a \u2217, \u2212b). (3.81) (3.82) (3.83) This group is known as SU(2), where S stands for special, U for unitary, and 2 for dimensionality 2. In contrast, the group de\ufb01ned by multiplication operations with general 2 \u00d7 2 unitary matrices (not necessarily constrained to be unimodular) is known as U(2). The most general unitary matrix in two dimensions has four independent parameters and can be written as ei\u03b3 (with \u03b3 real) times a unitary unimodular matrix:, |a|2 + |b|2 = 1, \u03b3\u2217 = \u03b3. (3.84) U = ei\u03b3 a \u2212b\u2217 b a\u2217 SU(2) is called a subgroup of U(2). 166 Theory of Angular Momentum Because we can characterize rotations using both the SO(3) language and the SU(2) language, we may be tempted to conclude that the groups SO(3) and SU(2) are isomorphic, that is, that there is a one-to-one correspondence between an element of SO(3) and an element of SU(2). This inference is not correct. Consider a rotation by 2\u03c0 and another one by 4\u03c0. In the SO(3) language, the matrices representing a 2\u03c0 rotation and a 4\u03c0 rotation are both 3 \u00d7 3 identity matrices; however, in the SU(2) language the corresponding matrices are \u22121 times the 2 \u00d7 2 identity matrix and the identity matrix itself, respectively. More generally, U(a, b) and U(\u2212a, \u2212b) both correspond to a single 3 \u00d7 3 matrix in the SO(3) language. The correspondence therefore is two-to-one; for a given R, the corresponding U is double valued. One can say, however, that the two", " groups are locally isomorphic. 3.3.3 Euler Rotations From classical mechanics the reader may be familiar with the fact that an arbitrary rotation of a rigid body can be accomplished in three steps, known as Euler rotations. The Euler rotation language, speci\ufb01ed by three Euler angles, provides yet another way to characterize the most general rotation in three dimensions. The three steps of Euler rotations are as follows. First, rotate the rigid body counterclockwise (as seen from the positive z-side) about the z-axis by angle \u03b1. Imagine now that there is a body y-axis embedded, so to speak, in the rigid body such that before the z-axis rotation is carried out, the body y-axis coincides with the usual y-axis, referred to as the space-\ufb01xed y-axis. Obviously, after the rotation about the z-axis, the body y-axis no longer coincides with the space-\ufb01xed y-axis; let us call the former the y-axis. To see how all this may appear for a thin disk, refer to Figure 3.4a. We now perform a second rotation, this time about the y-axis by angle \u03b2. As a result, the body z-axis no longer points in the space\ufb01xed z-axis direction. We call the body-\ufb01xed z-axis after the second rotation the z-axis; see Figure 3.4b. The third and \ufb01nal rotation is about the z-axis by angle \u03b3. The body y-axis now becomes the y-axis of Figure 3.4c. In terms of 3 \u00d7 3 orthogonal matrices the product of the three operations can be written as R(\u03b1, \u03b2, \u03b3) \u2261 Rz (\u03b3)Ry(\u03b2)Rz(\u03b1). (3.85) A cautionary remark is in order here. Most textbooks in classical mechanics prefer to perform the second rotation (the middle rotation) about the body x-axis rather than about the body y-axis (see, for example, Goldstein et al. (2002)). This convention is to be avoided in quantum mechanics for a reason that will become apparent in a moment. In (3.85) there appear Ry and Rz, which are matrices for rotations about body axes. This approach to Euler rotations is rather inconvenient in quantum mechanics because we earlier obtained simple expressions for the space-\ufb01xed (unprimed) axis components of the S operator, but not for the body-axis components. It is therefore desirable to express the body-axis rotations we considered in terms of space-\ufb01xed axis rotations. Fortunately there is a very simple relation, namely, Ry(\u03b2) =R z(\u03b1)Ry(\u03b2)R\u22121 z (\u03b1). (3.86) 167 3.3 SO(3), SU(2), and Euler Rotations z (a Fig. 3.4 Euler rotations. (b) (c) The meaning of the right-hand side is as follows. First, bring the body y-axis of Figure 3.4a (that is, the y-axis) back to the original \ufb01xed-space y-direction by rotating clockwise (as seen from the positive z-side) about the z-axis by angle \u03b1 then rotate about the y-axis by angle \u03b2. Finally, return the body y-axis to the direction of the y-axis by rotating about the \ufb01xed-space z-axis (not about the z-axis!) by angle \u03b1. Equation (3.86) tells us that the net effect of these rotations is a single rotation about the y-axis by angle \u03b2. To prove this assertion, let us look more closely at the effect of both sides of (3.86) on the circular disk of Figure 3.4a. Clearly, the orientation of the body y-axis is unchanged in both cases, namely, in the y-direction. Furthermore, the orientation of the \ufb01nal body z-axis is the same whether we apply Ry(\u03b2) or Rz(\u03b1)Ry(\u03b2)R\u22121 z (\u03b1). In both cases the \ufb01nal body z-axis makes a polar angle \u03b2 with the \ufb01xed z-axis (the same as the initial z-axis), and its azimuthal angle, as measured in the \ufb01xed-coordinate system, is just \u03b1. In other words, the \ufb01nal body z-axis is the same as the z-axis of Figure 3.4b. Similarly, we can prove Rz(\u03b3) =R y (\u03b2)Rz(\u03b3)R\u22121 y (\u03b2).", " (3.87) 168 Theory of Angular Momentum Using (3.86) and (3.87), we can now rewrite (3.85). We obtain Rz (\u03b3)Ry(\u03b2)Rz(\u03b1) =R y (\u03b2)Rz(\u03b3)R\u22121 = Rz(\u03b1)Ry(\u03b2)R\u22121 = Rz(\u03b1)Ry(\u03b2)Rz(\u03b3), y (\u03b2)Ry(\u03b2)Rz(\u03b1) z (\u03b1)Rz(\u03b3)Rz(\u03b1) (3.88) where in the \ufb01nal step we used the fact that Rz(\u03b3) and Rz(\u03b1) commute. To summarize, R(\u03b1, \u03b2, \u03b3) =R z(\u03b1)Ry(\u03b2)Rz(\u03b3), (3.89) where all three matrices on the right-hand side refer to \ufb01xed-axis rotations. Now let us apply this set of operations to spin 1 2 systems in quantum mechanics. Corresponding to the product of orthogonal matrices in (3.89) there exists a product of rotation operators in the ket space of the spin 1 2 system under consideration: D(\u03b1, \u03b2, \u03b3) =D z(\u03b1)Dy(\u03b2)Dz(\u03b3). (3.90) The 2 \u00d7 2 matrix representation of this product is exp exp exp \u2212i\u03c33\u03b1 2 \u2212i\u03c32\u03b2 2 \u2212i\u03c33\u03b3 2 = = 0 ei\u03b1/2 cos(\u03b2/2) \u2212 sin(\u03b2/2) e\u2212i\u03b1/2 sin(\u03b2/2) 0 e\u2212i(\u03b1+\u03b3)/2 cos(\u03b2/2) \u2212e\u2212i(\u03b1\u2212\u03b3)/2 sin(\u03b2/2) ei(\u03b1\u2212\u03b3)/2 sin(\u03b2/2) ei(\u03b1+\u03b3)/2 cos(\u03b2/2) cos(\u03b2/2) e\u2212i\u03b3/2 0 0 ei\u03b3/2, (3.91) where (3.62) was used. This matrix is clearly of the unitary unimodular form. Conversely, the most general 2\u00d72 unitary unimodular matrix can be written in this Euler angle form. Notice that the matrix elements of the second (middle) rotation exp(\u2212i\u03c3y\u03c6/2) are purely real. This would not have been the case had we chosen to rotate about the x-axis rather than the y-axis, as done in most textbooks in classical mechanics. In quantum mechanics it pays to stick to our convention because we prefer the matrix elements of the second rotation, which is the only rotation matrix containing off-diagonal elements, to be purely real.4 The 2 \u00d7 2 matrix in (3.91) is called the j = 1 operator D(\u03b1, \u03b2, \u03b3). Its matrix elements are denoted by D (1/2) angular-momentum operators we have ( D (1/2) mm (\u03b1, \u03b2, \u03b3) = j =, m exp 1 2 \u2212iJz\u03b1 \u00afh \u00d7 exp \u2212iJy\u03b2 \u00afh exp 2 irreducible representation of the rotation mm (\u03b1, \u03b2, \u03b3). In terms of the \u2212iJz\u03b3 \u00af3.92) In Section 3.5 we will extensively study higher j-analogues of (3.91). 4 This, of course, depends on our convention that the matrix elements of Sy (or, more generally, Jy) are taken to be purely imaginary. 169 3.4 Density Operators and Pure Versus Mixed Ensembles 3.4 Density Operators and Pure Versus Mixed Ensembles 3.4.1 Polarized Versus Unpolarized Beams The formalism of quantum mechanics developed so far makes statistical predictions on an ensemble, that is, a collection, of identically prepared physical systems. More precisely, in such an ensemble all members are supposed to be characterized by the same state ket |\u03b1. A good example of this is a beam of silver atoms coming out of an SG \ufb01ltering apparatus. Every atom in the beam has its spin pointing in the same direction, namely, the direction determined by the inhomogeneity of the magnetic \ufb01eld of the \ufb01ltering apparatus. We have not yet discussed how to describe quantum mechanically an ensemble of physical systems for which some, say 60%, are characterized by |\u03b1, and the remaining 40% are characterized by some other ket |\u03b2. To illustrate vividly the incompleteness of the formalism developed so far, let us consider silver atoms coming directly out of a hot oven, yet to be subjected to a \ufb01ltering apparatus of the Stern\u2013Gerlach type. On symmetry grounds we expect that such atoms have random spin orientations; in other words", ", there should be no preferred direction associated with such an ensemble of atoms. According to the formalism developed so far, the most general state ket of a spin 1 2 system is given by |\u03b1 = c+|+ + c\u2212|\u2212. (3.93) Is this equation capable of describing a collection of atoms with random spin orientations? The answer is clearly no; (3.93) characterizes a state ket whose spin is pointing in some de\ufb01nite direction, namely, in the direction of \u02c6n, whose polar and azimuthal angles, \u03b2 and \u03b1, respectively, are obtained by solving c+ c\u2212 = cos(\u03b2/2) ei\u03b1 sin(\u03b2/2) ; (3.94) see (3.70). To cope with a situation of this kind we introduce the concept of fractional population, or probability weight. An ensemble of silver atoms with completely random spin orientation can be viewed as a collection of silver atoms in which 50% of the members of the ensemble are characterized by |+ and the remaining 50% by |\u2212. We specify such an ensemble by assigning w+ = 0.5, w\u2212 = 0.5, (3.95) where w+ and w\u2212 are the fractional population for spin-up and spin-down, respectively. Because there is no preferred direction for such a beam, it is reasonable to expect that this same ensemble can be regarded equally well as a 50:50 mixture of |Sx; + and |Sx; \u2212. The mathematical formalism needed to accomplish this will appear shortly. It is very important to note that we are simply introducing two real numbers w+ and w\u2212. There is no information on the relative phase between the spin-up and the spin-down 170 Theory of Angular Momentum ket. Quite often we refer to such a situation as an incoherent mixture of spin-up and spindown states. What we are doing here is to be clearly distinguished from what we did with a coherent linear superposition, for example, 1\u221a 2 1\u221a 2 |+ + |\u2212, (3.96) where the phase relation between |+ and |\u2212 contains vital information on the spin orientation in the xy-plane, in this case in the positive x-direction. In general, we should not confuse w+ and w\u2212 with |c+|2 and |c\u2212|2. The probability concept associated with w+ and w\u2212 is much closer to that encountered in classical probability theory. The situation encountered in dealing with silver atoms directly from the hot oven may be compared with that of a graduating class in which 50% of the graduating seniors are male, the remaining 50% female. When we pick a student at random, the probability that the particular student is male (or female) is 0.5. Whoever heard of a student referred to as a coherent linear superposition of male and female with a particular phase relation? The beam of silver atoms coming directly out of the oven is an example of a completely random ensemble; the beam is said to be unpolarized because there is no preferred direction for spin orientation. In contrast, the beam that has gone through a selective Stern\u2013 Gerlach type measurement is an example of a pure ensemble; the beam is said to be polarized because all members of the ensemble are characterized by a single common ket that describes a state with spin pointing in some de\ufb01nite direction. To appreciate the difference between a completely random ensemble and a pure ensemble, let us consider a rotatable SG apparatus where we can vary the direction of the inhomogeneous B just by rotating the apparatus. When a completely unpolarized beam directly out of the oven is subjected to such an apparatus, we always obtain two emerging beams of equal intensity no matter what the orientation of the apparatus may be. In contrast, if a polarized beam is subjected to such an apparatus, the relative intensities of the two emerging beams vary as the apparatus is rotated. For some particular orientation the ratio of the intensities actually becomes one to zero. In fact, the formalism we developed in Chapter 1 tells us that the relative intensities are simply cos2(\u03b2/2) and sin2(\u03b2/2), where \u03b2 is the angle between the spin direction of the atoms and the direction of the inhomogeneous magnetic \ufb01eld in the SG apparatus. A complete random ensemble and a pure ensemble can be regarded as the extremes of what is known as a mixed ensemble. In a mixed ensemble a certain fraction, for example, 70%, of the members are characterized by a state ket |\u03b1, the remaining 30% by |\u03b2. In such a case the beam is said to be partially polarized. Here |\u03b1 and |\u03b2 need not even be orthogonal; we can, for example, have 70% with spin in the positive x-direction and 30% with spin in the negative z-direction.5 3.4.2 En", "semble Averages and Density Operator We now present the density operator formalism, pioneered by J. von Neumann in 1927, that quantitatively describes physical situations with mixed as well as pure ensembles. Our 5 In the literature what we call pure and mixed ensembles are often referred to as pure and mixed states. In this book, however, we use state to mean a physical system described by a de\ufb01nite state ket |\u03b1. 171 3.4 Density Operators and Pure Versus Mixed Ensembles general discussion here is not restricted to spin 1 return repeatedly to spin 1 2 systems. 2 systems, but for illustrative purposes we A pure ensemble by de\ufb01nition is a collection of physical systems such that every member is characterized by the same ket |\u03b1. In contrast, in a mixed ensemble, a fraction of the members with relative population w1 is characterized by |\u03b1(1), some other fraction with relative population w2, by |\u03b1(2), and so on. Roughly speaking, a mixed ensemble can be viewed as a mixture of pure ensembles, just as the name suggests. The fractional populations are constrained to satisfy the normalization condition \u2211 wi = 1. i (3.97) As we mentioned previously, |\u03b1(1) and |\u03b1(2) need not be orthogonal. Furthermore, the number of terms in the i sum of (3.97) need not coincide with the dimensionality N of the ket space; it can easily exceed N. For example, for spin 1 2 systems with N = 2, we may consider 40% with spin in the positive z-direction, 30% with spin in the positive xdirection, and the remaining 30% with spin in the negative y-direction. Suppose we make a measurement on a mixed ensemble of some observable A. We may ask what is the average measured value of A when a large number of measurements are carried out. The answer is given by the ensemble average of A, which is de\ufb01ned by [A] \u2261 \u2211 wi\u03b1(i)|A|\u03b1(i) i = \u2211 \u2211 a i wi|a|\u03b1(i)|2a, (3.98) where |a is an eigenket of A. Recall that \u03b1(i)|A|\u03b1(i) is the usual quantum-mechanical expectation value of A taken with respect to state |\u03b1(i). Equation (3.98) tells us that these expectation values must further be weighted by the corresponding fractional populations wi. Notice how probabilistic concepts enter twice; \ufb01rst in |a|\u03b1(i)|2 for the quantummechanical probability for state |\u03b1(i) to be found in an A eigenstate |a; second, in the probability factor wi for \ufb01nding in the ensemble a quantum-mechanical state characterized by |\u03b1(i).6 We can now rewrite ensemble average (3.98) using a more general basis, {|b}: [A] = \u2211 wi\u2211 b \u2211 b i \u03b1(i)|bb|A|bb|\u03b1(i) = \u2211 b \u2211 b \u2211 wib|\u03b1(i)\u03b1(i)|b b|A|b. (3.99) i The number of terms in the sum of the b (b) is just the dimensionality of the ket space, while the number of terms in the sum of the i depends on how the mixed ensemble is viewed as a mixture of pure ensembles. Notice that in this form the basic property of the ensemble 6 Quite often in the literature the ensemble average is also called the expectation value. However, in this book, the term expectation value is reserved for the average measured value when measurements are carried out on a pure ensemble. 172 Theory of Angular Momentum which does not depend on the particular observable A is factored out. This motivates us to de\ufb01ne the density operator \u03c1 as follows: \u03c1 \u2261 \u2211 wi|\u03b1(i)\u03b1(i)|. (3.100) The elements of the corresponding density matrix have the following form: i b|\u03c1|b = \u2211 wib|\u03b1(i)\u03b1(i)|b. i (3.101) The density operator contains all the physically signi\ufb01cant information we can possibly obtain about the ensemble in question. Returning to (3.99), we see that the ensemble average can be written as b|\u03c1|bb|A|b [A] = \u2211 \u2211 b b = tr(\u03c1A). (3.102) Because the trace is independent of representations, tr(\u03c1A) can be evaluated using any convenient basis. As a result, (3.102) is an extremely powerful relation. There are two properties of the density operator worth recording. First, the density operator", " is Hermitian, as is evident from (3.100). Second, the density operator satis\ufb01es the normalization condition tr(\u03c1) = \u2211 wib|\u03b1(i)\u03b1(i)|b \u2211 b wi\u03b1(i)|\u03b1(i) i = \u2211 i = 1. (3.103) Because of the Hermiticity and the normalization condition, for spin 1 2 systems with dimensionality 2, the density operator, or the corresponding density matrix, is characterized by three independent real parameters. Four real numbers characterize a 2 \u00d7 2 Hermitian matrix. However, only three are independent because of the normalization condition. The three numbers needed are [Sx], [Sy], and[S z]; the reader may verify that knowledge of these three ensemble averages is suf\ufb01cient to reconstruct the density operator. The manner in which a mixed ensemble is formed can be rather involved. We may mix pure ensembles characterized by all kinds of |\u03b1(i) with appropriate wi; yet for spin 1 2 systems three real numbers completely characterize the ensemble in question. This strongly suggests that a mixed ensemble can be decomposed into pure ensembles in many different ways. A problem to illustrate this point appears at the end of this chapter. A pure ensemble is speci\ufb01ed by wi = 1 for some |\u03b1(i), with i = n, for instance, and wi = 0 for all other conceivable state kets, so the corresponding density operator is written as \u03c1 = |\u03b1(n)\u03b1(n)| (3.104) with no summation. Clearly, the density operator for a pure ensemble is idempotent, that is, \u03c12 = \u03c1 (3.105) 173 3.4 Density Operators and Pure Versus Mixed Ensembles or, equivalently, \u03c1(\u03c1 \u2212 1) =0. Thus, for a pure ensemble only, we have tr(\u03c12) =1 (3.106) (3.107) in addition to (3.103). The eigenvalues of the density operator for a pure ensemble are zero or one, as can be seen by inserting a complete set of base kets that diagonalize the Hermitian operator \u03c1 between \u03c1 and (\u03c1 \u2212 1) of (3.106). When diagonalized, the density matrix for a pure ensemble must therefore look like \u239b \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239c \u239d diagonal form). (3.108) It can be shown that tr(\u03c12) is maximal when the ensemble is pure; for a mixed ensemble tr(\u03c12) is a positive number less than one. Given a density operator, let us see how we can construct the corresponding density matrix in some speci\ufb01ed basis. To this end we \ufb01rst recall that |\u03b1\u03b1| = \u2211 b \u2211 b |bb|\u03b1\u03b1|bb|. (3.109) This shows that we can form the square matrix corresponding to |\u03b1(i)\u03b1(i)| by combining, in the sense of outer product, the column matrix formed by b|\u03b1(i) with the row matrix formed by \u03b1(i)|b, which, of course, is equal to b|\u03b1(i)\u2217. The \ufb01nal step is to sum such square matrices with weighting factors wi, as indicated in (3.100). The \ufb01nal form agrees with (3.101), as expected. It is instructive to study several examples, all referring to spin 1 2 systems. Example 1 A completely polarized beam with Sz+: (1, 0) \u03c1 = |++|. = 1 0 = 1 0 0 0. (3.110) 174 Theory of Angular Momentum Example 2 A completely polarized beam with Sx\u00b1: \u03c1 = |Sx; \u00b1Sx; \u00b1| = \u239e \u239b 1\u221a 2 (|+ \u00b1 |\u2212). = \u239c \u239c \u239d 1 2 \u00b1 1 2 The ensembles of Examples 1 and 2 are both pure\u221a 2 (+| \u00b1 \u2212|) (3.111) Example 3 An unpolarized beam. This can be regarded as an incoherent mixture of a spinup ensemble and a spin-down ensemble with equal weights (50% each): \u03c1 = ( 1 2 )|++| + ( 1 2 )|\u2212\u2212|. = 1 2 0 0 1 2, (3.112) which is just the identity matrix divided by 2. As we remarked earlier, the same ensemble can also be regarded as an incoherent mixture of an S", "x+ ensemble and an Sx\u2212 ensemble with equal weights. It is gratifying that our formalism automatically satis\ufb01es the expectation 3.113) where we see from Example 2 that the two terms on the right-hand side are the density matrices for pure ensemble with Sx+ and Sx\u2212. Because \u03c1 in this case is just the identity operator divided by 2 (the dimensionality), we have tr(\u03c1Sx) =tr (\u03c1Sy) =tr( \u03c1Sz) =0, (3.114) where we used the fact that Sk is traceless. Thus for the ensemble average of S we have [S] =0. (3.115) This is reasonable because there should be no preferred spin direction in a completely random ensemble of spin 1 2 systems. Example 4 As an example of a partially polarized beam, let us consider a 75:25 mixture of two pure ensembles, one with Sz+ and the other with Sx+: w(Sz+) = 0.75, w(Sx+) = 0.25. (3.116) The corresponding \u03c1 can be represented by 3.117) 175 3.4 Density Operators and Pure Versus Mixed Ensembles from which follows [Sx] = \u00afh 8, [Sy] =0, [Sz] = 3\u00afh 8. (3.118) We leave as an exercise for the reader the task of showing that this ensemble can be decomposed in ways other than (3.116). 3.4.3 Time Evolution of Ensembles How does the density operator \u03c1 change as a function of time? Let us suppose that at some time t0 the density operator is given by \u03c1(t0) = \u2211 wi|\u03b1(i)\u03b1(i)|. i (3.119) If the ensemble is to be left undisturbed, we cannot change the fractional population wi. So the change in \u03c1 is governed solely by the time evolution of state ket |\u03b1(i): |\u03b1(i) at t0 \u2192 |\u03b1(i), t0; t. (3.120) From the fact that |\u03b1(i), t0; t satis\ufb01es the Schr\u00a8odinger equation we obtain i\u00afh \u2202\u03c1 \u2202t = \u2211 wi(H|\u03b1(i), t0; t\u03b1(i), t0; t| \u2212 |\u03b1(i), t0; t\u03b1(i), t0; t|H) i = \u2212[\u03c1, H]. (3.121) This looks like the Heisenberg equation of motion except that the sign is wrong! This is not disturbing because \u03c1 is not a dynamic observable in the Heisenberg picture. On the contrary, \u03c1 is built up of Schr\u00a8odinger picture state kets and state bras which evolve in time according to the Schr\u00a8odinger equation. It is amusing that (3.121) can be regarded as the quantum-mechanical analogue of Liouville\u2019s theorem in classical statistical mechanics, \u2202\u03c1classical \u2202t = \u2212[\u03c1classical, H]classical, (3.122) where \u03c1classical stands for the density of representative points in phase space.7 Thus the name density operator for the \u03c1 appearing in (3.121) is indeed appropriate. The classical analogue of (3.102) for the ensemble average of some observable A is given by \u03c1classicalA(q, p)d \u0393q,p \u03c1classicald \u0393q,p where d \u0393q,p stands for a volume element in phase space. Aaverage =, (3.123) 7 Remember, a pure classical state is one represented by a single moving point in phase space (q1,..., qf, p1,..., pf) at each instant of time. A classical statistical state, on the other hand, is described by our nonnegative density function \u03c1classical(q1,..., qf, p1,..., pf, t) such that the probability that a system is found in the interval dq1,..., dpf at time t is \u03c1classicaldq1,..., dpf. 176 Theory of Angular Momentum 3.4.4 Continuum Generalizations So far we have considered density operators in ket space where the base kets are labeled by the discrete eigenvalues of some observable. The concept of density matrix can be generalized to cases where the base kets used are labeled by continuous eigenvalues. In particular, let us consider the ket space spanned by the position eigenkets |x. The analogue of (3.102) is given by [A] = d3x d3xx|\u03c1|xx|A|", "x. (3.124) The density matrix here is actually a function of x and x, namely, x|\u03c1|x = x| wi|\u03b1(i)\u03b1(i)| |x \u2211 i = \u2211 wi\u03c8i(x)\u03c8\u2217 i (x), i (3.125) where \u03c8i is the wave function corresponding to the state ket |\u03b1(i). Notice that the diagonal element (that is, x = x) of this is just the weighted sum of the probability densities. Once again, the term density matrix is indeed appropriate. In continuum cases, too, it is important to keep in mind that the same mixed ensemble can be decomposed in different ways into pure ensembles. For instance, it is possible to regard a \u201crealistic\u201d beam of particles either as a mixture of plane wave states (monoenergetic free-particle states) or as a mixture of wave packet states. 3.4.5 Quantum Statistical Mechanics We conclude this section with a brief discussion on the connection between the density operator formalism and statistical mechanics. Let us \ufb01rst record some properties of completely random and of pure ensembles. The density matrix of a completely random ensemble looks like \u239e \u239b 3.126) in any representation [compare Example 3 with (3.112)]. This follows from the fact that all states corresponding to the base kets with respect to which the density matrix is written are equally populated. In contrast, in the basis where \u03c1 is diagonalized, we have (3.108) for the matrix representation of the density operator for a pure ensemble. The two diagonal matrices (3.126) and (3.108), both satisfying the normalization requirement (3.103), cannot 177 3.4 Density Operators and Pure Versus Mixed Ensembles look more different. It would be desirable if we could somehow construct a quantity that characterizes this dramatic difference. Thus we de\ufb01ne a quantity called \u03c3 by \u03c3 = \u2212tr(\u03c1 ln \u03c1). (3.127) The logarithm of the operator \u03c1 may appear rather formidable, but the meaning of (3.127) is quite unambiguous if we use the basis in which \u03c1 is diagonal: \u03c3 = \u2212\u2211 \u03c1(diag) kk ln \u03c1(diag) kk. k (3.128) Because each element \u03c1(diag) semide\ufb01nite. For a completely random ensemble (3.126), we have is a real number between 0 and 1, \u03c3 is necessarily positive kk \u03c3 = \u2212 N \u2211 k=1 1 N ln 1 N = ln N. In contrast, for a pure ensemble (3.108) we have \u03c3 = 0 where we have used \u03c1(diag) kk = 0 or ln \u03c1(diag) kk = 0 (3.129) (3.130) (3.131) for each term in (3.128). We now argue that physically \u03c3 can be regarded as a quantitative measure of disorder. A pure ensemble is an ensemble with a maximum amount of order because all members are characterized by the same quantum-mechanical state ket; it may be likened to marching soldiers in a well-regimented army. According to (3.130), \u03c3 vanishes for such an ensemble. On the other extreme, a completely random ensemble, in which all quantum-mechanical states are equally likely, may be likened to drunken soldiers wandering around in random directions. According to (3.129), \u03c3 is large; indeed, we will show later that ln N is the maximum possible value for \u03c3 subject to the normalization condition \u03c1kk = 1. \u2211 k (3.132) In thermodynamics we learn that a quantity called entropy measures disorder. It turns out that our \u03c3 is related to the entropy per constituent member, denoted by S, of the ensemble via S = k \u03c3, (3.133) where k is a universal constant identi\ufb01able with the Boltzmann constant. In fact, (3.133) may be taken as the de\ufb01nition of entropy in quantum statistical mechanics. We now show how the density operator \u03c1 can be obtained for an ensemble in thermal equilibrium. The basic assumption we make is that nature tends to maximize \u03c3 subject to the constraint that the ensemble average of the Hamiltonian has a certain prescribed value. To justify this assumption would involve us in a delicate discussion of how equilibrium is 178 Theory of Angular Momentum established as a result of interactions with the environment, which is beyond the scope of this book. In any case, once thermal equilibrium is established, we expect \u2202\u03c1 \u2202t = 0. (3.134) Because of (3", ".121), this means that \u03c1 and H can be simultaneously diagonalized. So the kets used in writing (3.128) may be taken to be energy eigenkets. With this choice \u03c1kk stands for the fractional population for an energy eigenstate with energy eigenvalue Ek. Let us maximize \u03c3 by requiring that \u03b4\u03c3 = 0. (3.135) However, we must take into account the constraint that the ensemble average of H has a certain prescribed value. In the language of statistical mechanics, [H] is identi\ufb01ed with the internal energy per constituent denoted by U: [H] =tr( \u03c1H) =U. (3.136) In addition, we should not forget the normalization constraint (3.132). So our basic task is to require (3.135) subject to the constraints \u03b4[H] = \u2211 (3.137a) \u03b4\u03c1kkEk = 0 and k \u03b4(tr\u03c1) = \u2211 \u03b4\u03c1kk = 0. k (3.137b) We can most readily accomplish this by using Lagrange multipliers. We obtain \u03b4\u03c1kk[(ln \u03c1kk + 1) + \u03b2Ek + \u03b3] =0, (3.138) \u2211 k which for an arbitrary variation is possibly only if \u03c1kk = exp(\u2212\u03b2Ek \u2212 \u03b3 \u2212 1). The constant \u03b3 can be eliminated using the normalization condition (3.132), and our \ufb01nal result is (3.139) \u03c1kk = exp(\u2212\u03b2Ek) N \u2211 exp(\u2212\u03b2El), l (3.140) which directly gives the fractional population for an energy eigenstate with eigenvalue Ek. It is to be understood throughout that the sum is over distinct energy eigenstates; if there is degeneracy we must sum over states with the same energy eigenvalue. The density matrix element (3.140) is appropriate for what is known in statistical mechanics as a canonical ensemble. Had we attempted to maximize \u03c3 without the internalenergy constraint (3.137a), we would have obtained instead \u03c1kk = 1 N (independent of k), (3.141) 179 3.4 Density Operators and Pure Versus Mixed Ensembles which is the density matrix element appropriate for a completely random ensemble. Comparing (3.140) with (3.141), we infer that a completely random ensemble can be regarded as the \u03b2 \u2192 0 limit (physically the high-temperature limit) of a canonical ensemble. We recognize the denominator of (3.140) as the partition function Z = N \u2211 k exp(\u2212\u03b2Ek) in statistical mechanics. It can also be written as Z = tr(e\u2212\u03b2H). Knowing \u03c1kk given in the energy basis, we can write the density operator as \u03c1 = e\u2212\u03b2H Z. (3.142) (3.143) (3.144) This is the most basic equation from which everything follows. We can immediately evaluate the ensemble average of any observable A: [A] = tr(e\u2212\u03b2HA) Z \u2211 N Ak exp(\u2212\u03b2Ek) k = N \u2211 k exp(\u2212\u03b2Ek). \u2211 N Ek exp(\u2212\u03b2Ek) In particular, for the internal energy per constituent we obtain U = k N \u2211 exp(\u2212\u03b2Ek) k \u2202 \u2202\u03b2 (ln Z), = \u2212 a formula well known to every student of statistical mechanics. The parameter \u03b2 is related to the temperature T as follows: \u03b2 = 1 kT (3.145) (3.146) (3.147) where k is the Boltzmann constant. It is instructive to convince ourselves of this identi\ufb01cation by comparing the ensemble average [H] of simple harmonic oscillators with the kT expected for the internal energy in the classical limit, which is left as an exercise. We have already commented that in the high-temperature limit, a canonical ensemble becomes a completely random ensemble in which all energy eigenstates are equally populated. In the opposite low-temperature limit (\u03b2 \u2192 \u221e), (3.140) tells us that a canonical ensemble becomes a pure ensemble where only the ground state is populated. 180 Theory of Angular Momentum As a simple illustrative example, consider a canonical ensemble made up of spin 1 2 systems, each with a magnetic moment e\u00afh/2mec subjected to a uniform magnetic \ufb01eld in the z-direction. The Hamiltonian relevant to this problem has already been given [see (3.34)]. Because H and Sz commute, the density matrix for this canonical ensemble is diagonal in the Sz basis. Thus e\u2212\u03b2\u00afh\u03c9/2 0 0 e\u03b2\u00afh\u03c9/2 Z,", " (3.148) \u03c1. = where the partition function is just Z = e\u2212\u03b2\u00afh\u03c9/2 + e\u03b2\u00afh\u03c9/2. From this we compute [Sx] = [Sy] = 0, [Sz] =\u2212 tanh \u00afh 2. \u03b2\u00afh\u03c9 2 (3.149) (3.150) The ensemble average of the magnetic moment component is just e/mec times [Sz]. The paramagnetic susceptibility \u03c7 may be computed from e mec [Sz] = \u03c7B. In this way we arrive at Brillouin\u2019s formula for \u03c7: |e|\u00afh 2mecB \u03c7 =. \u03b2\u00afh\u03c9 2 tanh (3.151) (3.152) 3.5 Eigenvalues and Eigenstates of Angular Momentum Up to now our discussion of angular momentum has been con\ufb01ned exclusively to spin 1 2 systems with dimensionality N = 2. In this and subsequent sections we study more general angular-momentum states. To this end we \ufb01rst work out the eigenvalues and eigenkets of J2 and Jz and derive the expressions for matrix elements of angular-momentum operators, \ufb01rst obtained in a 1926 paper by M. Born, W. Heisenberg, and P. Jordan. 3.5.1 Commutation Relations and the Ladder Operators Everything we will do follows from the angular-momentum commutation relations (3.20), where we may recall that Ji is de\ufb01ned as the generator of in\ufb01nitesimal rotation. The \ufb01rst important property we derive from the basic commutation relations is the existence of a new operator J2, de\ufb01ned by J2 \u2261 JxJx + JyJy + JzJz, (3.153) 181 3.5 Eigenvalues and Eigenstates of Angular Momentum that commutes with every one of Jk: [J2, Jk] = 0 (k = 1, 2, 3). (3.154) To prove this let us look at the k = 3 case: [JxJx + JyJy + JzJz, Jz] =J x[Jx, Jz] + [Jx, Jz]Jx + Jy[Jy, Jz] + [Jy, Jz]Jy = Jx(\u2212i\u00afhJy) + (\u2212i\u00afhJy)Jx + Jy(i\u00afhJx) + (i\u00afhJx)Jy = 0. (3.155) The proofs for the cases where k = 1 and 2 follow by cyclic permutation (1 \u2192 2 \u2192 3 \u2192 1) of the indices. Because Jx, Jy, and Jz do not commute with each other, we can choose only one of them to be the observable to be diagonalized simultaneously with J2. By convention we choose Jz for this purpose. We now look for the simultaneous eigenkets of J2 and Jz. We denote the eigenvalues of J2 and Jz by a and b, respectively: J2|a, b = a|a, b Jz|a, b = b|a, b. (3.156a) (3.156b) To determine the allowed values for a and b, it is convenient to work with the nonHermitian operators J\u00b1 \u2261 Jx \u00b1 iJy, (3.157) called the ladder operators, rather than with Jx and Jy. They satisfy the commutation relations and [J+, J\u2212] =2 \u00afhJz [Jz, J\u00b1] = \u00b1\u00afhJ\u00b1, which can easily be obtained from (3.20). Note also that [J2, J\u00b1] =0, (3.158a) (3.158b) (3.159) which is an obvious consequence of (3.154). What is the physical meaning of J\u00b1? To answer this we examine how Jz acts on J\u00b1|a, b: Jz(J\u00b1|a, b) = ([Jz, J\u00b1] +J \u00b1Jz)|a, b = (b \u00b1 \u00afh)(J\u00b1|a, b) (3.160) where we have used (3.158b). In other words, if we apply J+(J\u2212) to a Jz eigenket, the resulting ket is still a Jz eigenket except that its eigenvalue is now increased (decreased) by one unit of \u00afh. So now we see why J\u00b1, which step one step up (down) on the \u201cladder\u201d of Jz eigenvalues,", " are known as the ladder operators. 182 Theory of Angular Momentum We now digress to recall that the commutation relations in (3.158b) are reminiscent of some commutation relations we encountered in the earlier chapters. In discussing the translation operator T (l) we had [xi, T (l)] = liT (l); also, in discussing the simple harmonic oscillator we had [N, a\u2020] =a \u2020, [N, a] = \u2212a. (3.161) (3.162) We see that both (3.161) and (3.162) have a structure similar to (3.158b). The physical interpretation of the translation operator is that it changes the eigenvalue of the position operator x by l in much the same way as the ladder operator J+ changes the eigenvalue of Jz by one unit of \u00afh. Likewise, the oscillator creation operator a\u2020 increases the eigenvalue of the number operator N by unity. Even though J\u00b1 changes the eigenvalue of Jz by one unit of \u00afh, it does not change the eigenvalue of J2: J2(J\u00b1|a, b) =J \u00b1J2|a, b = a(J\u00b1|a, b), (3.163) where we have used (3.159). To summarize, J\u00b1|a, b are simultaneous eigenkets of J2 and Jz with eigenvalues a and b \u00b1 \u00afh. We may write J\u00b1|a, b = c\u00b1|a, b \u00b1 \u00afh, (3.164) where the proportionality constant c\u00b1 will be determined later from the normalization requirement of the angular-momentum eigenkets. 3.5.2 Eigenvalues of J2 and Jz We now have the machinery needed to construct angular-momentum eigenkets and to study their eigenvalue spectrum. Suppose we apply J+ successively, say n times, to a simultaneous eigenket of J2 and Jz. We then obtain another eigenket of J2 and Jz with the Jz eigenvalue increased by n\u00afh, while its J2 eigenvalue is unchanged. However, this process cannot go on inde\ufb01nitely. It turns out that there exists an upper limit to b (the Jz eigenvalue) for a given a (the J2 eigenvalue): To prove this assertion we \ufb01rst note that a \u2265 b2. J2 \u2212 J 2 z = 1 = 1 2 (J+J\u2212 + J\u2212J+) + + J \u2020 2 (J+J \u2020 +J+). Now J+J \u2020 + and J \u2020 +J+ must have nonnegative expectation values because J \u2020 + |a, bDC\u2194a, b|J+, J+|a, bDC\u2194a, b|J \u2020 +; (3.165) (3.166) (3.167) 183 3.5 Eigenvalues and Eigenstates of Angular Momentum thus a, b|(J2 \u2212 J 2 z )|a, b \u22650, (3.168) which, in turn, implies (3.165). It therefore follows that there must be a bmax such that J+|a, bmax = 0. (3.169) Stated another way, the eigenvalue of b cannot be increased beyond bmax. Now (3.169) also implies J\u2212J+|a, bmax = 0. J\u2212J+ = J 2 x + J 2 y = J2 \u2212 J 2 z \u2212 i(JyJx \u2212 JxJy) \u2212 \u00afhJz. But So (J2 \u2212 J 2 z Because |a, bmax itself is not a null ket, this relationship is possible only if \u2212 \u00afhJz)|a, bmax = 0. or a \u2212 b2 max \u2212 bmax \u00afh = 0 a = bmax(bmax + \u00afh). (3.170) (3.171) (3.172) (3.173) (3.174) In a similar manner we argue from (3.165) that there must also exist a bmin such that By writing J+J\u2212 as J\u2212|a, bmin = 0. J+J\u2212 = J2 \u2212 J 2 z + \u00afhJz in analogy with (3.171), we conclude that a = bmin(bmin \u2212 \u00afh). By comparing (3.174) with (3.177) we infer that bmax = \u2212bmin, with bmax positive, and that the allowed values of b lie within \u2212bmax \u2264 b \u2264 bmax. (3.175) (3.176) (3.177) (3.", "178) (3.179) Clearly, we must be able to reach |a, bmax by applying J+ successively to |a, bmin a \ufb01nite number of times. We must therefore have bmax = bmin + n\u00afh, (3.180) 184 Theory of Angular Momentum where n is some integer. As a result, we get bmax = n\u00afh 2. (3.181) It is more conventional to work with j, de\ufb01ned to be bmax/\u00afh, instead of with bmax so that j = n 2. (3.182) The maximum value of the Jz eigenvalue is j\u00afh, wherej is either an integer or a half-integer. Equation (3.174) implies that the eigenvalue of J2 is given by Let us also de\ufb01ne m so that a = \u00afh2j( j + 1). b \u2261 m\u00afh. (3.183) (3.184) If j is an integer, all m values are integers; if j is a half-integer, all m values are half-integers. The allowed m-values for a given j are 2j+1 states Instead of |a, b it is more convenient to denote a simultaneous eigenket of J2 and Jz by m = \u2212j, \u2212j + 1,..., j \u2212 1, j. (3.185) | j, m. The basic eigenvalue equations now read and J2| j, m = j( j + 1)\u00afh2| j, m Jz| j, m = m\u00afh| j, m, (3.186a) (3.186b) with j either an integer or a half-integer and m given by (3.185). It is very important to recall here that we have used only the commutation relations (3.20) to obtain these results. The quantization of angular momentum, manifested in (3.186), is a direct consequence of the angular-momentum commutation relations, which, in turn, follow from the properties of rotations together with the de\ufb01nition of Jk as the generator of rotation. 3.5.3 Matrix Elements of Angular-Momentum Operators Let us work out Assuming | j, m to be normalized, we obviously have from (3.186) the matrix elements of the various angular-momentum operators. and j, m|J2| j, m = j( j + 1)\u00afh2\u03b4jj\u03b4mm j, m|Jz| j, m = m\u00afh\u03b4jj\u03b4mm. (3.187a) (3.187b) To obtain the matrix elements of J\u00b1, we \ufb01rst consider +J+| j, m = j, m|(J2 \u2212 J 2 z j, m|J\u2020 \u2212 \u00afhJz)| j, m = \u00afh2[ j( j + 1) \u2212 m2 \u2212 m]. (3.188) 185 3.5 Eigenvalues and Eigenstates of Angular Momentum Now J+| j, m must be the same as | j, m + 1 (normalized) up to a multiplicative constant [see (3.164)]. Thus Comparison with (3.188) leads to J+| j, m = c+ jm | j, m + 1. |c+ jm |2 = \u00afh2[ j( j + 1) \u2212 m(m + 1)] = \u00afh2( j \u2212 m)( j + m + 1). (3.189) (3.190) So we have determined c+ be real and positive by convention. So jm up to an arbitrary phase factor. It is customary to choose c+ jm to J+| j, m = ( j \u2212 m)( j + m + 1)\u00afh| j, m + 1. Similarly, we can derive J\u2212| j, m = ( j + m)( j \u2212 m + 1)\u00afh| j, m \u2212 1. Finally, we determine the matrix elements of J\u00b1 to be j, m|J\u00b1| j, m = ( j \u2213 m)( j \u00b1 m + 1)\u00afh\u03b4jj\u03b4m,m\u00b11. 3.5.4 Representations of the Rotation Operator (3.191) (3.192) (3.193) Having obtained the matrix elements of Jz and J\u00b1, we are now in a position to study the matrix elements of the rotation operator D(R). If a rotation R is speci\ufb01ed by \u02c6n and \u03c6, we can de\ufb01ne its matrix elements by mm(R) = j, m| exp D (", " j) | j, m. (3.194) \u2212iJ\u00b7 \u02c6n\u03c6 \u00afh These matrix elements are sometimes called Wigner functions after E. P. Wigner, who made pioneering contributions to the group-theoretical properties of rotations in quantum mechanics. Notice here that the same j-value appears in the ket and bra of (3.194); we need not consider matrix elements of D(R) between states with different j-values because they all vanish trivially. This is because D(R)| j, m is still an eigenket of J2 with the same eigenvalue j( j + 1)\u00afh2: J2D(R)| j, m = D(R)J2| j, m = j( j + 1)\u00afh2[D(R)| j, m], (3.195) which follows directly from the fact that J2 commutes with Jk (hence with any function of Jk). Simply stated, rotations cannot change the j-value, which is an eminently sensible result. Often in the literature the (2j + 1) \u00d7 (2j + 1) matrix formed by D ( j) mm(R) is referred to as the (2j + 1)-dimensional irreducible representation of the rotation operator D(R). This means that the matrix which corresponds to an arbitrary rotation operator in ket space not 186 Theory of Angular Momentum necessarily characterized by a single j-value can, with a suitable choice of basis, be brought to block-diagonal form: (3.196) where each shaded square is a (2j + 1) \u00d7 (2j + 1) square matrix formed by D ( j) mm with some de\ufb01nite value of j. Furthermore, each square matrix itself cannot be broken into smaller blocks 2j + 1 2j + 1 k k 2j + 1 \u2212 k 2j + 1 \u2212 k (3.197) with any choice of basis. The rotation matrices characterized by de\ufb01nite j form a group. First, the identity is a member because the rotation matrix corresponding to no rotation (\u03c6 = 0) is the (2j + 1) \u00d7 (2j + 1) identity matrix. Second, the inverse is also a member; we simply reverse the rotation angle \u03c6 \u2192 \u2212\u03c6 without changing the rotation axis \u02c6n. Third, the product of any two members is also a member; explicitly we have D ( j) mm(R1)D ( j) mm(R2) = D ( j) mm(R1R2), (3.198) \u2211 m where the product R1R2 represents a single rotation. We also note that the rotation matrix is unitary because the corresponding rotation operator is unitary; explicitly we have Dmm(R\u22121) = D \u2217 mm (R). (3.199) 187 3.5 Eigenvalues and Eigenstates of Angular Momentum To appreciate the physical signi\ufb01cance of the rotation matrix let us start with a state represented by | j, m. We now rotate it: | j, m \u2192 D(R)| j, m. (3.200) Even though this rotation operation does not change j, we generally obtain states with mvalues other than the original m. To \ufb01nd the amplitude for being found in | j, m, we simply expand the rotated state as follows: D(R)| j, m j, m|D(R)| j, m | j, mD ( j) mm(R), (3.201) where, in using the completeness relation, we took advantage of the fact that D(R) connects only states with the same j. So the matrix element D ( j) mm(R) is simply the amplitude for the rotated state to be found in | j, m when the original unrotated state is given by | j, m. In Section 3.3 we saw how Euler angles may be used to characterize the most general rotation. We now consider the matrix realization of (3.90) for an arbitrary j (not necessarily 1 2 ): mm(\u03b1, \u03b2, \u03b3) = j, m|exp D ( j) \u2212iJz\u03b1 \u00afh exp \u2212iJy\u03b2 \u00afh \u2212iJy\u03b2 \u00afh exp \u2212iJz\u03b3 \u00afh | j, m = e\u2212i(m \u03b1+m\u03b3) j, m|exp | j, m. (3.202) Notice that the only nontrivial part is the middle rotation about the y-axis, which mixes different m-values. It is convenient to de\ufb01ne a new matrix d ( j)(\u03b2) as mm(\u03b2) \u2261 j, m| exp d ( j) \u2212iJy\u03b2", " \u00afh | j, m. (3.203) Finally, let us turn to some examples. The j = 1 2 case has already been worked out in Section 3.3. See the middle matrix of (3.91), d 1/2 = \u239b \u239c \u239c \u239d cos sin \u2212 sin cos (3.204) The next simplest case is j = 1, which we consider in some detail. Clearly, we must \ufb01rst obtain the 3 \u00d7 3 matrix representation of Jy. Because Jy = (J+ \u2212 J\u2212) 2i (3.205) from the de\ufb01ning equation (3.157) for J\u00b1, we can use (3.193) to obtain m = 1, m = 0, m = \u22121, 188 Theory of Angular Momentum J ( j=1) y = \u00afh 2 \u239b \u239d 0 \u221a 2i 0 \u221a 2i \u2212 0 \u221a 2i \u2212 0 \u221a 0 \u239e \u23a0 2i m = 1 m = 0 m = \u22121. (3.206) Our next task is to work out the Taylor expansion of exp(\u2212iJy\u03b2/\u00afh). Unlike the case j = 2, [J ( j=1) 1 ]2 is independent of 1 and J ( j=1). However, it is easy to work out y y 3 J ( j=1) y \u00afh = J ( j=1) y \u00afh. Consequently, for j = 1 only, it is legitimate to replace exp \u2212iJy\u03b2 \u00afh \u2192 1 \u2212 2 Jy \u00afh (1 \u2212 cos \u03b2) \u2212 i Jy \u00afh sin \u03b2, as the reader may verify in detail. Explicitly we have d(1)(\u03b21 + cos \u03b2) \u2212 sin \u03b2 1\u221a 2 (1 \u2212 cos \u03b2) 1\u221a 2 1\u221a 2 sin \u03b2 1 2 cos \u03b2 \u2212 & sin \u03b2 % 1 2 & (1 \u2212 cos \u03b2) sin \u03b2 (1 + cos \u03b2) 1\u221a 2 \u239e \u239f \u239f \u239f \u23a0. (3.207) (3.208) (3.209) Clearly, this method becomes time consuming for large j. Other, much easier methods are possible, but we will not pursue them in this book. 3.6 Orbital Angular Momentum We introduced the concept of angular momentum by de\ufb01ning it to be the generator of an in\ufb01nitesimal rotation. There is another way to approach the subject of angular momentum when spin-angular momentum is zero or can be ignored. The angular momentum J for a single particle is then the same as orbital angular momentum, which is de\ufb01ned as L = x \u00d7 p. (3.210) In this section we explore the connection between the two approaches. 3.6.1 Orbital Angular Momentum as Rotation Generator We \ufb01rst note that the orbital angular-momentum operator de\ufb01ned as (3.210) satis\ufb01es the angular-momentum commutation relations [Li, Lj] = i\u03b5ijk \u00afhLk (3.211) 189 3.6 Orbital Angular Momentum by virtue of the commutation relations among the components of x and p. This can easily be proved as follows: [Lx, Ly] = [ypz \u2212 zpy, zpx \u2212 xpz] = [ypz, zpx] + [zpy, xpz] = ypx[ pz, z] +p yx[z, pz] = i\u00afh(xpy \u2212 ypx) = i\u00afhLz... Next we let \u03b4\u03c6 \u00afh 1 \u2212 i Lz = 1 \u2212 i \u03b4\u03c6 \u00afh (xpy \u2212 ypx) (3.212) (3.213) act on an arbitrary position eigenket |x, y, z to examine whether it can be interpreted as the in\ufb01nitesimal rotation operator about the z-axis by angle \u03b4\u03c6. Using the fact that momentum is the generator of translation, we obtain [see (1.214)] $ # 1 \u2212 i Lz |x, y, z = 1 \u2212 i (\u03b4\u03c6x) +i (\u03b4\u03c6y) |x, y, z py \u00afh px \u00afh \u03b4\u03c6 \u00afh = |x \u2212 y\u03b4\u03c6, y + x\u03b4\u03c6, z. (3.214) This is precisely what we expect if Lz generates an in\ufb01nitesimal rotation about the z-axis. So we have demonstrated that if p generates translation, then L generates rotation. Suppose the wave function for an arbitrary physical state of a spinless particle is given by x", ", y, z|\u03b1. After an in\ufb01nitesimal rotation about the z-axis is performed, the wave function for the rotated state is x, y, z| 1 \u2212 i Lz |\u03b1 = x + y\u03b4\u03c6, y \u2212 x\u03b4\u03c6, z|\u03b1. (3.215) \u03b4\u03c6 \u00afh It is actually more transparent to change the coordinate basis x, y, z|\u03b1 \u2192 r, \u03b8, \u03c6|\u03b1. (3.216) For the rotated state we have, according to (3.215), 1 \u2212 i r, \u03b8, \u03c6| \u03b4\u03c6 \u00afh Lz |\u03b1 = r, \u03b8, \u03c6 \u2212 \u03b4\u03c6|\u03b1 = r, \u03b8, \u03c6|\u03b1 \u2212 \u03b4\u03c6 \u2202 \u2202\u03c6 r, \u03b8, \u03c6|\u03b1. (3.217) Because r, \u03b8, \u03c6| is an arbitrary position eigenket, we can identify x|Lz|\u03b1 = \u2212i\u00afh \u2202 \u2202\u03c6 x|\u03b1, (3.218) which is a well-known result from wave mechanics. Even though this relation can also be obtained just as easily using the position representation of the momentum operator, the derivation given here emphasizes the role of Lz as the generator of rotation. 190 Theory of Angular Momentum We next consider a rotation about the x-axis by angle \u03b4\u03c6x. In analogy with (3.215) we have x, y, z| 1 \u2212 i \u03b4\u03c6x \u00afh Lx |\u03b1 = x, y + z\u03b4\u03c6x, z \u2212 y\u03b4\u03c6x|\u03b1. (3.219) By expressing x, y, andz in spherical coordinates, we can show that x|Lx|\u03b1 = \u2212i\u00afh \u2212 sin \u03c6 \u2212 cot \u03b8cos \u03c6 x|\u03b1. (3.220) \u2202 \u2202\u03b8 \u2202 \u2202\u03c6 Likewise, x|Ly|\u03b1 = \u2212i\u00afh cos \u03c6 \u2202 \u2202\u03b8 \u2212 cot \u03b8sin \u03c6 \u2202 \u2202\u03c6 x|\u03b1. (3.221) Using (3.220) and (3.221), for the ladder operator L\u00b1 de\ufb01ned as in (3.157), we have x|L\u00b1|\u03b1 = \u2212i\u00afhe\u00b1i\u03c6 \u00b1i \u2212 cot \u03b8 x|\u03b1. (3.222) \u2202 \u2202\u03b8 \u2202 \u2202\u03c6 Finally, it is possible to write x|L2|\u03b1 using L2 = L2 z + 1 2 (L+L\u2212 + L\u2212L+), (3.223) (3.218), and (3.222), as follows: x|L2|\u03b1 = \u2212\u00afh2 1 sin2 \u03b8 \u22022 \u2202\u03c62 + 1 sin \u03b8 \u2202 \u2202\u03b8 sin \u03b8 \u2202 \u2202\u03b8 x|\u03b1. (3.224) Apart from 1/r2, we recognize the differential operator that appears here to be just the angular part of the Laplacian in spherical coordinates. It is instructive to establish this connection between the L2 operator and the angular part of the Laplacian in another way by looking directly at the kinetic-energy operator. We \ufb01rst record an important operator identity, L2 = x2p2 \u2212 (x\u00b7p)2 + i\u00afhx\u00b7p, (3.225) where x2 is understood to be the operator x\u00b7x, just asp 2 stands for the operator p\u00b7p. The proof of this is straightforward: L2 = \u2211 \u03b5ijkxipj\u03b5lmkxlpm ijlmk = \u2211 (\u03b4il\u03b4jm \u2212 \u03b4im\u03b4jl)xipjxlpm ijlm = \u2211 ijlm [\u03b4il\u03b4jmxi(xlpj \u2212 i\u00afh\u03b4jl)pm \u2212 \u03b4im\u03b4jlxipj(pmxl + i\u00afh\u03b4lm)] = x2p2 \u2212 i\u00afhx \u00b7 p \u2212 \u2211 \u03b4im\u03b4jl[xipm(xlpj \u2212 i\u00afh\u03b4jl) +i \u00afh\u03b4lmxipj] ijlm = x2p2 \u2212 (x\u00b7p)2 + i\u00afhx\u00b7p. (3.226) 191 3.6 Orbital Angular Momentum Before taking the preceding expression between x| and |\u03b1, \ufb01rst note that x|x\u00b7p|\u03b1 = x\u00b7(\u2212i\u00afh\u2207x|\u03b1) = \u2212i\u00af", "hr \u2202 \u2202r x|\u03b1. Likewise, Thus \u2202 x|(x\u00b7p)2|\u03b1 = \u2212\u00afh2r \u2202r = \u2212\u00afh2 r2 \u2202 \u2202r r x|\u03b1 \u22022 \u2202r2 x|\u03b1 + r x|\u03b1. \u2202 \u2202r x|L2|\u03b1 = r2x|p2|\u03b1 + \u00afh2 r2 \u22022 \u2202r2 x|\u03b1 + 2r x|\u03b1. \u2202 \u2202r (3.227) (3.228) (3.229) In terms of the kinetic energy p2/2m, we have 1 2m x|p2|\u03b1 = \u2212 = \u2212 \u00afh2 2m \u00afh2 2m \u22072x|\u03b1 \u22022 \u2202r2 x|\u03b1 + 2 r \u2202 \u2202r x|\u03b1 \u2212 1 \u00afh2r2 x|L2|\u03b1. (3.230) The \ufb01rst two terms in the last line are just the radial part of the Laplacian acting on x|\u03b1. The last term must then be the angular part of the Laplacian acting on x|\u03b1, in complete agreement with (3.224). 3.6.2 Spherical Harmonics Consider a spinless particle subjected to a spherical symmetrical potential. The wave equation is known to be separable in spherical coordinates and the energy eigenfunctions can be written as l (\u03b8, \u03c6), x|n, l, m = Rnl(r)Ym (3.231) where the position vector x is speci\ufb01ed by the spherical coordinates r, \u03b8, and \u03c6, andn stands for some quantum number other than l and m, for example, the radial quantum number for bound-state problems or the energy for a free-particle spherical wave. As will be made clearer in Section 3.11, this form can be regarded as a direct consequence of the rotational invariance of the problem. When the Hamiltonian is spherically symmetrical, H commutes with Lz and L2, and the energy eigenkets are expected to be eigenkets of L2 and Lz also. Because Lk with k = 1, 2, 3 satisfy the angular-momentum commutation relations, the eigenvalues of L2 and Lz are expected to be l(l+1)\u00afh2, and m\u00afh = [\u2212l\u00afh, (\u2212l+1) \u00afh,..., (l \u2212 1)\u00afh, l\u00afh]. Because the angular dependence is common to all problems with spherical symmetry, we can isolate it and consider \u02c6n|l, m = Ym l (\u03b8, \u03c6) =Y m l ( \u02c6n), (3.232) 192 Theory of Angular Momentum where we have de\ufb01ned a direction eigenket | \u02c6n. From this point of view, Ym l (\u03b8, \u03c6) is the amplitude for a state characterized by l, m to be found in the direction \u02c6n speci\ufb01ed by \u03b8 and \u03c6. Suppose we have relations involving orbital angular-momentum eigenkets. We can immediately write the corresponding relations involving the spherical harmonics. For example, take the eigenvalue equation Lz|l, m = m\u00afh|l, m. Multiplying \u02c6n| on the left and using (3.218), we obtain \u2212i\u00afh \u2202 \u2202\u03c6 \u02c6n|l, m = m\u00afh \u02c6n|l, m. We recognize this equation to be \u2212i\u00afh \u2202 \u2202\u03c6 Ym l (\u03b8, \u03c6) =m \u00afhYm l (\u03b8, \u03c6), (3.233) (3.234) (3.235) which implies that the \u03c6-dependence Ym sponding to l (\u03b8, \u03c6) must behave like eim\u03c6. Likewise, corre- we have [see (3.224)] 1 sin \u03b8 \u2202 \u2202\u03b8 L2|l, m = l(l + 1)\u00afh2|l, m, sin \u03b8 \u2202 \u2202\u03b8 + 1 sin2 \u03b8 \u22022 \u2202\u03c62 + l(l + 1) Ym l = 0, (3.236) (3.237) which is simply the partial differential equation satis\ufb01ed by Ym l relation itself. The orthogonality l, m|l, m = \u03b4ll \u03b4mm leads to 2\u03c0 0 d\u03c6 1 \u22121 d(cos \u03b8", ")Ym\u2217 l (\u03b8, \u03c6)Ym l (\u03b8, \u03c6) = \u03b4ll \u03b4mm, where we have used the completeness relation for the direction eigenkets, d\u03a9 \u02c6n| \u02c6n \u02c6n| = 1. To obtain the Ym l themselves, we may start with the m = l case. We have L+|l, l = 0, which, because of (3.222), leads to \u2212i\u00afhei\u03c6 \u2202 \u2202\u03b8 i \u2212 cot \u03b8 \u2202 \u2202\u03c6 \u02c6n|l, l = 0. (3.238) (3.239) (3.240) (3.241) (3.242) Remembering that the \u03c6-dependence must behave like eil\u03c6, we can easily show that this partial differential equation is satis\ufb01ed by \u02c6n|l, l = Yl l(\u03b8, \u03c6) =c leil\u03c6 sinl \u03b8, (3.243) 193 3.6 Orbital Angular Momentum where cl is the normalization constant determined from (3.239) to be8 \u2019 cl = (\u22121)l 2ll! [(2l + 1)(2l)! ] 4\u03c0. (3.244) Starting with (3.243) we can use \u02c6n|l, m \u2212 1 = \u02c6n|L\u2212|l, m (l + m)(l \u2212 m + 1)\u00afh = 1 (l + m)(l \u2212 m + 1) e\u2212i\u03c6 \u2212 \u2202 \u2202\u03b8 + i cot \u03b8 \u2202 \u2202\u03c6 \u02c6n|l, m (3.245) successively to obtain all Ym l with l \ufb01xed. Because this is done in many textbooks on elementary quantum mechanics, for example Townsend (2000), we will not work out the details here. The result for m \u2265 0 is l (\u03b8, \u03c6) = Ym (\u22121)l 2ll! (2l + 1) 4\u03c0 (l + m)! (l \u2212 m)! eim\u03c6 1 sinm \u03b8 dl\u2212m d(cos \u03b8)l\u2212m (sin \u03b8)2l, and we de\ufb01ne Y\u2212m l by Y\u2212m l (\u03b8, \u03c6) = (\u22121)m[Ym l (\u03b8, \u03c6)]\u2217. (3.246) (3.247) Regardless of whether m is positive or negative, the \u03b8-dependent part of Ym l (\u03b8, \u03c6) is [sin \u03b8]|m| times a polynomial in cos \u03b8 with a highest power of l \u2212 |m|. For m = 0, we obtain \u2019 l (\u03b8, \u03c6) = Y0 2 l + 1 4\u03c0 Pl(cos \u03b8). (3.248) From the point of view of the angular-momentum commutation relations alone, it might not appear obvious why l cannot be a half-integer. It turns out that several arguments can be advanced against half-integer l-values. First, for half-integer l, and hence for half-integer m, the wave function would acquire a minus sign, eim(2\u03c0) = \u22121, (3.249) under a 2\u03c0 rotation. As a result, the wave function would not be single valued; we pointed out in Section 2.4 that the wave function must be single valued because of the requirement that the expansion of a state ket in terms of position eigenkets be unique. We can prove that if L, de\ufb01ned to be x \u00d7 p, is to be identi\ufb01ed as the generator of rotation, then the wave function must acquire a plus sign under a 2\u03c0 rotation. This follows from the fact that the wave function for a 2\u03c0-rotated state is the original wave function itself with no sign change: x| exp \u2212iLz2\u03c0 \u00afh |\u03b1 = x cos 2\u03c0 + y sin 2\u03c0, y cos 2\u03c0 \u2212 x sin 2\u03c0, z|\u03b1 = x|\u03b1, (3.250) 8 Normalization condition (3.239), of course, does not determine the phase of cl. The factor (\u22121)l is inserted so l with the same sign as the that when we use the L\u2212 operator successively to reach the state m = 0, we obtain Y0 Legendre polynomial Pl(cos \u03b8) whose phase is \ufb01xed by Pl(1) = 1 [see( 3.248)]. 194 Theory of Angular Momentum (3.253) (3.254) l (\ufffd", "\ufffd, \u03c6) with where we have used the \ufb01nite-angle version of (3.215). Next, let us suppose Ym a half-integer l were possible. To be speci\ufb01c, we choose the simplest case, l = m = 1 2. According to (3.243) we would have 1/2(\u03b8, \u03c6) =c 1/2ei\u03c6/2 Y1/2 sin \u03b8. \u221a From the property of L\u2212 [see (3.245)] we would then obtain 1/2 (\u03b8, \u03c6) =e \u2212i\u03c6 Y\u22121/2 \u2212 \u2202 \u2202\u03b8 + i cot \u03b8 \u2202 \u2202\u03c6 \u221a c1/2ei\u03c6/2 = \u2212c1/2e\u2212i\u03c6/2 cot \u03b8 sin \u03b8. (3.252) This expression is not permissible because it is singular at \u03b8 = 0, \u03c0. What is worse, from the partial differential equation ( ( ) ) (3.251) \u221a sin \u03b8 \u02c6n|L\u2212| 1 2, \u2212 1 2 = \u2212i\u00afhe\u2212i\u03c6 \u2212i = 0 \u2202 \u2202\u03b8 \u2212 cot \u03b8 \u2202 \u2202\u03c6 \u02c6n| 1 2, \u2212 1 2 we directly obtain 1/2 = c Y\u22121/2 1/2e\u2212i\u03c6/2 \u221a sin \u03b8, in sharp contradiction with (3.252). Finally, we know from the Sturm\u2013Liouville theory of differential equations that the solutions of (3.237) with l integer form a complete set. An arbitrary function of \u03b8 and \u03c6 can be expanded in terms of Ym l with integer l and m only. For all these reasons it is futile to contemplate orbital angular momentum with half-integer l-values. 3.6.3 Spherical Harmonics as Rotation Matrices We conclude this section on orbital angular momentum by discussing the spherical harmonics from the point of view of the rotation matrices introduced in the last section. We can readily establish the desired connection between the two approaches by constructing the most general direction eigenket | \u02c6n by applying appropriate rotation operators to |\u02c6z, the direction eigenket in the positive z-direction. We wish to \ufb01nd D(R) such that | \u02c6n = D(R)|\u02c6z. (3.255) We can rely on the technique used in constructing the eigenspinor of \u03c3\u00b7 \u02c6n in Section 3.2. We \ufb01rst rotate about the y-axis by angle \u03b8, then around the z-axis by angle \u03c6; see Figure 3.3 with \u03b2 \u2192 \u03b8, \u03b1 \u2192 \u03c6. In the notation of Euler angles we have Writing (3.255) as D(R) = D(\u03b1 = \u03c6, \u03b2 = \u03b8, \u03b3 = 0). | \u02c6n = \u2211 \u2211 D(R)|l, ml, m|\u02c6z, l m (3.256) (3.257) 195 3.7 Schr\u00f6dinger\u2019s Equation for Central Potentials we see that | \u02c6n, when expanded in terms of |l, m, contains all possible l-values. However, when this equation is multiplied by l, m| on the left, only one term in the l-sum contributes, namely, l, m| \u02c6n = \u2211 m D (l) mm(\u03b1 = \u03c6, \u03b2 = \u03b8, \u03b3 = 0)l, m|\u02c6z. (3.258) Now l, m|\u02c6z is just a number; in fact, it is precisely Ym\u2217 l (\u03b8, \u03c6) evaluated at \u03b8 = 0 with \u03c6 undetermined. At \u03b8 = 0, Ym is known to vanish for m = 0, which can also be seen directly l from the fact that |\u02c6z is an eigenket of Lz (which equals xpy \u2212 ypx) with eigenvalue zero. So we can write l, m|\u02c6z = Ym\u2217 l (\u03b8 = 0, \u03c6 undetermined)\u03b4m 0 \u2019 (2 l + 1) = 4\u03c0 Pl(cos \u03b8) \u03b4m 0 cos \u03b8= 1 Returning to (3.258), we have \u2019 = \u2019 (2 l + 1) 4\u03c0 \u03b4m 0. (3.259) or Ym\u2217 l (\u03b8, \u03c6) = (2 l + 1) 4\u03c0 D (l) m0(\u03b1 = \u03c6, \u03b2 = \u03b8", ", \u03b3 = 0) (3.260) D (l) m 0(\u03b1, \u03b2, \u03b3 = 0) = 4\u03c0 (2 l + 1) Ym\u2217 l (\u03b8, \u03c6). (3.261) \u03b8= \u03b2,\u03c6 = \u03b1 Notice the m = 0 case, which is of particular importance: d( l ) 00 (\u03b2) \u03b2 = \u03b8 = P l(cos \u03b8). (3.262) 3.7 Schr\u00f6dinger\u2019s Equation for Central Potentials Problems described by Hamiltonians of the form H = p2 2m + V(r) r2 = x2 (3.263) form the basis for very many situations in the physical world. The fundamental importance of this Hamiltonian lies in the fact that it is spherically symmetric. Classically, we expect orbital angular momentum to be conserved in such a system. This is also true quantum mechanically, since it is easy to show that and therefore [L, p2] = [L, x2] = 0 [L, H] = [L2, H] =0 (3.264) (3.265) 196 Theory of Angular Momentum if H is given by (3.263). We refer to such problems as \u201ccentral potential\u201d or \u201ccentral force\u201d problems. Even if the Hamiltonian is not strictly of this form, it is often the case that this is a good starting point when we consider approximation schemes that build on \u201csmall\u201d corrections to central potential problems. In this section we will discuss some general properties of bound-state eigenfunctions generated by (3.263), and a few representative central potential problems. (Scattering solutions will be discussed in Chapter 6.) For more detail, the reader is referred to any number of excellent texts which explore such problems in greater depth. 3.7.1 The Radial Equation Equation (3.265) makes it clear that we should search for energy eigenstates |\u03b1 = |Elm where H|Elm = E|Elm L2|Elm = l(l + 1)\u00afh2|Elm Lz|Elm = m\u00afh|Elm. (3.266) (3.267) (3.268) It is easiest to work in the coordinate representation, and solve the appropriate differential equation for eigenfunctions in terms of a radial function REl(r) and spherical harmonics, as shown in (3.231). Combining (3.263), (3.266), and (3.267) with (3.230) and (3.231) we arrive at the radial equation9 \u2212 \u00afh2 2mr2 d dr r2 d dr l(l + 1)\u00afh2 + 2mr2 + V(r) REl(r) =ER El(r). (3.269) Depending on the speci\ufb01c form of V(r), we may work with this equation or some variant of it, to identify the radial part REl(r) of the eigenfunction and/or the energy eigenvalues E. In fact, we can immediately gain some insight to the effects of angular momentum on the eigenfunctions by making the substitution REl(r) = uEl(r) r (3.270) which reduces (3.269) to \u2212 \u00afh2 2m d2uEl dr2 + l(l + 1)\u00afh2 2mr2 + V(r) Coupled with the fact that the spherical harmonics are separately normalized, so that the overall normalization condition becomes uEl(r) = EuEl(r). (3.271) 1 = r2dr R\u2217 El(r)REl(r) = dr u\u2217 El(r)uEl(r), (3.272) we see that uEl(r) can be interpreted as a wave function in one dimension for a particle moving in an \u201ceffective potential\u201d 9 We apologize for using m for both \u201cmass\u201d and the quantum number for angular momentum. However, in this section, it should be clear from the context which is which. 197 3.7 Schr\u00f6dinger\u2019s Equation for Central Potentials ) r ( f f e V 2 0 \u22122 \u22124 0 V(r) 2 r 4 Fig. 3.5 The \u201ceffective potential\u201dwhich governs the behavior of the \u201cradial wave function\u201duEl(r). If the potential energy V(r) (shown as a dashed line) is not too singular at the origin, then there is an angular momentum barrier for all states with l = 0, which makes it very improbable for a particle to located near the origin. Veff(r) =V(r) + l", "(l + 1)\u00afh2 2mr2. (3.273) Equation (3.273) demonstrates the existence of an \u201cangular momentum barrier\u201d if l = 0, as shown in Figure 3.5. Quantum mechanically, this means that the amplitude (and therefore the probability) is small for locating the particle near the origin, except for s-states. As we will see later on, this fact has important physical consequences in atoms, for example. We can be more quantitative about this interpretation. Let us assume that the potential energy function V(r) is not so singular so that limr\u21920 r2V(r) = 0. Then, for small values of r, (3.271) becomes d2uEl dr2 = l(l + 1) r2 uEl(r) (r \u2192 0) which has the general solution uEl(r) = Arl+1 + B rl. (3.274) (3.275) It is tempting to set B = 0 out of hand, because 1/rl produces severe singularities as r \u2192 0, especially for large l. However, there are better reasons for setting B = 0, that are rooted in the foundations of quantum mechanics. For example, under certain conditions, one can show that the interference between the two terms leads to nonconservation of probability. (See Problem 3.27 at the end of this chapter.) Other rather sophisticated explanations are given in some textbooks, but we give a relatively simple approach here. If B = 0 in (3.275), then the radial wave function REl(r) \u2192 1/rl+1 as r \u2192 0, and this is not normalizable for l \u2265 1. The wave function can be normalized for l = 0, but in this case the Schr\u00a8odinger equation involves \u22072(1/r) = \u22124\u03c0\u03b4(x) and solutions cannot be found (unless the potential is a \u03b4-function). Consequently, we are left with REl(r) \u2192 rl as r \u2192 0. (3.276) 198 Theory of Angular Momentum This relation has profound consequences. Firstly, it embodies the \u201cangular-momentum barrier\u201d shown in Figure 3.5, since the wave function goes to zero except for s-states. More practically, it means that the probability of \ufb01nding, say, an electron in an atom in the region of the nucleus, goes like (R/a0)2l where R a0 is the size of the nucleus and a0 is the Bohr radius. These concepts will become explicit when we come to the study of atomic structure. When considering bound states of potential energy functions V(r) which tend to zero at large r, there is another form of the radial equation which we can consider. For r \u2192 \u221e, (3.271) becomes d2uE dr2 = \u03ba2u \u03ba2 \u2261 \u22122mE/\u00afh2 > 0 r \u2192 \u221e (3.277) since E < 0 for bound states. The solution to this equation is simply uE(r) \u221d e\u2212\u03bar. Also, it makes it clear that the dimensionless variable \u03c1 \u2261 \u03bar would be useful to recast the radial equation. Consequently, we remove both the short and long distance behavior of the wave function and write10 uEl(\u03c1) = \u03c1l+1e\u2212\u03c1w(\u03c1) where the function w(\u03c1) is \u201cwell behaved\u201d and satis\ufb01es d2w d\u03c12 + 2 dw d\u03c1 + (The manipulations which lead to this equation are left to the reader.) One then attacks the solution w(\u03c1) of (3.279) for the particular function V(r = \u03c1/\u03ba). \u2212 2(l + 1) \u03c1 l + 1 \u03c1 (3.279) w = 0. \u2212 1 V E (3.278) 3.7.2 The Free Particle and Infinite Spherical Well In Section 2.5.1 we saw the solution to the free-particle problem in three dimensions, using Cartesian coordinates. We can of course approach the same problem by exploiting spherical symmetry and angular momentum. Starting from (3.269) we write E \u2261 \u00afh2k2 2m \u03c1 \u2261 kr (3.280) and and arrive at the modi\ufb01ed radial equation d2R d\u03c12 + 2 \u03c1 dR d\u03c1 + 1 \u2212 l(l + 1) \u03c12 R = 0. (3.281) This is a well-known differential equation whose solutions are called spherical Bessel functions jl(\u03c1) and nl(\u03c1), where 10 We are being a bit careless with notation when switching arguments from r to \u03c1, as is typical in these kinds of treatments. For", " example, going from (3.277) to( 3.278) and ignoring subscripts, we should de\ufb01ne the \u221e dimensionless function \u02dcu(\u03c1) \u2261 \u03ba\u22121/2u(\u03c1/\u03ba) in which case 0 \u02dcu2(\u03c1)d\u03c1 = 1. We can otherwise ignore the factor \u03ba\u22121/2 which gets absorbed into the normalization, and then drop the \u201ctilde,\u201d that is \u02dcu(\u03c1) \u2192 u(\u03c1). 199 3.7 Schr\u00f6dinger\u2019s Equation for Central Potentials jl(\u03c1) = (\u2212\u03c1)l 1 \u03c1 nl(\u03c1) =\u2212 (\u2212\u03c1)l l sin \u03c1 \u03c1 d d\u03c1 l. cos \u03c1 \u03c1 1 \u03c1 d d\u03c1 (3.282a) (3.282b) It is easy to show that as \u03c1 \u2192 0, jl(\u03c1) \u2192 \u03c1l and nl(\u03c1) \u2192 \u03c1\u2212l\u22121. Hence, jl(\u03c1) corresponds to (3.276) and these are the only solutions we consider here.11 It is also useful to point out that the spherical Bessel functions are de\ufb01ned over the entire complex plane, and it can be shown that jl(z) = 1 2il 1 \u22121 ds eizsPl(s). The \ufb01rst few spherical Bessel functions are j0(\u03c1) = j1(\u03c1) = j2(\u03c1) = sin \u03c1 \u03c1 sin \u03c1 \u03c12 3 \u03c13 \u2212 cos \u03c1 \u03c1 sin \u03c1 \u2212 3 cos \u03c1 \u03c12 \u2212 1 \u03c1 (3.283) (3.284) (3.285) (3.286). This result can be immediately applied to the case of a particle con\ufb01ned to an in\ufb01nite spherical well, that is, a potential energy function V(r) =0 within r < a, but with the wave function constrained to be zero at r = a. For any given value of l, this leads to the \u201cquantization condition\u201d jl(ka) =0, that is, ka equals the set of zeros of the spherical Bessel function. For l = 0 these are obviously ka = \u03c0, 2\u03c0, 3\u03c0,.... For other values of l, computer programs are readily available which can compute the zeros. We \ufb01nd that El=0 = El=1 = El=2 =!!! \u00afh2 2ma2 \u00afh2 2ma2 \u00afh2 2ma2 \u03c02, (2\u03c0)2, (3\u03c0)2,... \" 4.492, 7.732, 10.902,... 5.762, 9.102, 12.322,... \" \". (3.287) (3.288) (3.289) It should be noted that this series of energy levels shows no degeneracies in l. Indeed, such degenerate energy levels are impossible, except for any accidental equality between zeros of spherical Bessel functions of different orders. 3.7.3 The Isotropic Harmonic Oscillator Energy eigenvalues for the Hamiltonian H = p2 2m + 1 2 m\u03c92r2 (3.290) 11 In a treatment of \u201chard sphere scattering\u201d problems, the origin is explicitly excluded, and the solutions nl(\u03c1) are also kept. The relative phase between the two solutions for a given l is called the \u201cphase shift.\u201d 200 Theory of Angular Momentum are straightforward to determine. Introducing dimensionless energy \u03bb and radial coordinate \u03c1 through E = 1 2 \u00afh\u03c9\u03bb and r = 1/2 \u03c1 \u00afh m\u03c9 (3.291) we transform (3.271) into d 2u d\u03c12 \u2212 l(l + 1) \u03c12 u(\u03c1) + (\u03bb \u2212 \u03c12)u(\u03c1) =0. (3.292) It is again worthwhile to explicitly remove the behavior for large (and small) \u03c1, although we cannot use (3.278) because V(r) does not tend to zero for large r. Instead, we write u(\u03c1) = \u03c1l+1e\u2212\u03c12/2f (\u03c1). This yields the following differential equation for the function f (\u03c1): \u03c1 d 2f d\u03c12 + 2[(l + 1) \u2212 \u03c12] We solve (3.294) by writing f (\u03c1) as an in\ufb01nite series, namely df d\u03c1 + [\u03bb \u2212 (2l + 3)]\u03c1f (\u03c1) = 0. f (\u03c1) = \u221e \u2211 an\u03c1n", ". (3.293) (3.294) (3.295) n=0 We insert this into the differential equation, and set each term to zero by powers of \u03c1. The only surviving term in \u03c10 is 2(l + 1)a1, so The terms proportional to \u03c11 allow us to relate a2 to a0, which in turn can be set through the normalization condition. Continuing, (3.294) becomes a1 = 0. (3.296) \u221e \u2211 n=2 {(n + 2)(n + 1)an+2 + 2(l + 1)(n + 2)an+2 \u2212 2nan + [\u03bb \u2212 (2l + 3]an} \u03c1n+1 = 0 which leads, \ufb01nally, to the recursion relation an+2 = 2n + 2l + 3 \u2212 \u03bb (n + 2)(n + 2l + 3) an. (3.297) (3.298) Immediately we see that f (\u03c1) involves only even powers of \u03c1, since (3.296) and (3.298) imply that an = 0 for odd n. Also, asn \u2192 \u221e, we have 1 q \u2192 2 n (3.299) an+2 an = where q = n/2 includes both odd and even integers. Therefore, for large values of \u03c1, (3.295) becomes f (\u03c1) \u2192 constant \u00d7\u2211 q % &q \u221d e\u03c12. \u03c12 1 q! (3.300) 201 3.7 Schr\u00f6dinger\u2019s Equation for Central Potentials In other words, u(\u03c1) from (3.293) would grow exponentially for large \u03c1 (and therefore be unable to meet the normalization condition) unless the series terminates. Therefore 2n + 2l + 3 \u2212 \u03bb = 0 for some even value of n = 2q, and the energy eigenvalues are Eql = 2q + l + \u00afh\u03c9 \u2261 N + \u00afh\u03c9 3 2 3 2 (3.301) (3.302) for q = 0, 1, 2,... and l = 0, 1, 2,..., and N \u2261 2q + l. One frequently refers to N as the \u201cprincipal\u201d quantum number. It can be shown that q counts the number of nodes in the radial function. Quite unlike the square well, the three-dimensional isotropic harmonic oscillator has degenerate energy eigenvalues in the l quantum number. There are three states (all l = 1) for N = 1. For N = 2 there are \ufb01ve states with l = 2, plus one state with q = 1 andl = 0, giving a total of six. Notice that for even (odd) values of N, only even (odd) values of l are allowed. Therefore, the parity of the wave function is even or odd with the value of N. These wave functions are popular basis states for calculations of various natural phenomena, when the potential energy function is a \u201cwell\u201d of some \ufb01nite size. One of the greatest successes of such an approach is the nuclear shell model, where individual protons and neutrons are pictured to move independently in a potential energy function generated by the cumulative effect of all nucleons in the nucleus. Figure 3.6 compares the energy levels observed in nuclei with those obtained for the isotropic harmonic oscillator and for the in\ufb01nite spherical well. It is natural to label the eigenstates of the Hamiltonian (3.290) as |qlm or |Nlm. However, this Hamiltonian may also be written as H = Hx + Hy + Hz (3.303) where Hi = a\u2020 2 is an independent one-dimensional harmonic oscillator in direction i = x, y, z. In this way, we would label the eigenstates |nx, ny, nz and the energy eigenvalues are i ai + 1 E = nx + = N + + nx + 1 2 1 2 1 + nx + 2 3 2 \u00afh\u03c9 \u00afh\u03c9 (3.304) where, now, N = nx + ny + nz. It is simple to show numerically for the \ufb01rst few energy levels, that the degeneracy is the same regardless of which basis is used. It is an interesting exercise to show this in general, and also to derive the unitary transformation matrix nx, ny, nz|qlm which changes from one basis to the other. (See Problem 3.29 at the end of this chapter.) 3.7.4 The Coulomb Potential Perhaps the most important potential energy function in physics is V(x) =\u2212 Ze2 r (3.305)", " 202 Theory of Angular Momentum 6 5 4 3 2 1 0 4s 3d 2g 3p 1i 2f 3s 1h 2d 1g 2p 1f 2s 1d 1p 4s 3d 2g 1i 3p 2f 1h 3s 2d 1g 2p 1f 2s 1d 1p Harmonic oscillator 1s Infinite square well Finite square well 1s Square well with rounded edges Plus spin-orbit coupling 4s1/2 3d3/2 2g7/2 3d5/2 1i11/2 2g9/2 2f5/2 3p1/2 1i13/2 3p3/2 2f7/2 1h 9/2 3s1/2 1h11/2 2d3/2 2d5/2 1g7/2 1g9/2 2p1/2 2p3/2 1f5/2 1f7/2 2s1/2 1d3/2 1d5/2 1p1/2 1p3/2 1s1/2 126 82 50 28 20 8 2 Fig. 3.6 Energy levels in the nuclear shell model, adapted from Haxel et al., Z. Phys., 128 (1950) 295. Energy levels of the three-dimensional isotropic harmonic oscillator are on the left, followed by the infinite spherical well. Modifications of the infinite square well, for finite walls and then for \u201crounded corners\u201dfollow. The rightmost plot of energy levels shows those obtained by including the interaction between the nucleon spin and the orbital angular momentum. The final column indicates the total angular-momentum quantum number. 203 3.7 Schr\u00f6dinger\u2019s Equation for Central Potentials where the constant Ze2 is obviously chosen so that (3.305) represents the potential for a one-electron atom with atomic number Z. In addition to Coulomb forces, and classical gravity, it is widely used in models applied to very many physical systems.12 We consider here the radial equation based on such a function and the resulting energy eigenvalues. The 1/r potential satis\ufb01es all the requirements that led us to (3.279). We therefore search for solutions of the form (3.278) by determining the function w(\u03c1). Making the de\ufb01nition 2m \u2212E 1/2 Ze2 \u00afh = 2mc2 \u2212E 1/2 Z\u03b1 \u03c10 = where \u03b1 \u2261 e2/\u00afhc \u2248 1/137 is the \ufb01ne-structure constant, (3.279) becomes \u03c1 d2w d\u03c12 + 2(l + 1 \u2212 \u03c1) dw d\u03c1 + [\u03c10 \u2212 2(l + 1)]w(\u03c1) = 0. (3.306) (3.307) We could of course proceed to solve (3.307) using a series approach and derive a recursion relation for the coef\ufb01cients, just as we did with (3.294). However, it turns out that the solution is in fact already well known. Equation (3.307) can be written as Kummer\u2019s equation where x d2F dx2 + (c \u2212 x) dF dx \u2212 aF = 0 x = 2\u03c1 c = 2(l + 1) 2a = 2(l + 1) \u2212 \u03c10. (3.308) (3.309) The solution to (3.308) is called the con\ufb02uent hypergeometric function, written as the series and so F(a; c; x) =1 + a c x 1! + a(a + 1) c(c + 1) x2 2! + \u00b7 \u00b7 \u00b7 w(\u03c1) =F l + 1 \u2212 ; 2(l + 1) ; 2\u03c1. \u03c10 2 (3.310) (3.311) Note that for large \u03c1, we have w(\u03c1) \u2248 \u2211 Large N a(a + 1) \u00b7 \u00b7 \u00b7 c(c + 1) \u00b7 \u00b7 \u00b7 (2\u03c1)N N! \u2248 \u2211 Large N (N/2)N NN (2\u03c1)N N! \u2248 \u2211 Large N (\u03c1)N N! \u2248 e\u03c1. 12 Indeed, 1/r potential energy functions result from any quantum \ufb01eld theory in three spatial dimensions with massless intermediate exchange particles. See Chapter I.6 in Zee (2010). 204 Theory of Angular Momentum Therefore, once again, (3.278) gives a radial wave function which would grow without bound unless the series (3.310) terminates. So, for some integer N we must have a + N = 0 which leads to \u03c10 = 2", "(N + l + 1) and N = 0, 1, 2..., where l = 0, 1, 2,.... (3.312) It is customary (and, as we shall soon see, instructive) to de\ufb01ne the principal quantum number n as n \u2261 N + l + 1 = 1, 2, 3,... where l = 0, 1,..., n \u2212 1. (3.313) We point out that it is possible to solve the radial equation for the Coulomb problem using the generating function techniques described in Section 2.5. SeeProblem 3.30 at the end of this chapter. Energy eigenvalues arise by combining (3.306) and (3.312) in terms of the principal quantum number, that is which leads to 1/2 2mc2 \u2212E \u03c10 = Z\u03b1 = 2n E = \u2212 1 2 mc2 Z2\u03b12 n2 = \u221213.6 eV Z2 n2 (3.314) (3.315) where the numerical result is for a one-electron atom, i.e. mc2 = 511 keV. Equation (3.315) is of course the familiar Balmer formula. It is time to make various points. First, there is a stark disagreement between the energy level properties predicted by modern quantum mechanics, and those of the old Bohr model of the atom. The Bohr model had a one-to-one correspondence between angular-momentum eigenvalues l and principal quantum number n. In fact the ground state corresponded to n = l = 1. We see instead that only l = 0 is allowed for n = 1, and that different values of l are allowed for higher energy levels. Second, a natural length scale a0 has emerged. Since \u03c1 = \u03bar where \u03ba = \u22122mE/\u00afh2 (see (3.277) we have where 1 \u03ba = \u00afh mc\u03b1 n Z \u2261 a0 n Z (3.316) \u00afh mc\u03b1 = is called the Bohr radius. For an electron, a0 = 0.53 \u00d7 10\u22128 cm = 0.53 \u02daA. This is indeed the typical size of an atom. (3.317) a0 = \u00afh2 me2 Finally, the energy eigenvalues (3.315) demonstrate an interesting degeneracy. The eigenvalues depend only on n, and not on l or m. The level of degeneracy for a state |nlm is therefore given by 205 3.8 Addition of Angular Momenta 2.0 1.5 1.0 0. Zr/a0 ).0 0. Zr/a0 Fig. 3.7 Radial wave functions for the Coulomb potential and principal quantum numbers n = 1 (left) and n = 2 (right). Degeneracy = n\u22121 \u2211 (2l + 1) = n2. l=0 (3.318) This degeneracy is in fact not accidental, but re\ufb02ects a subtle symmetry of the Coulomb potential. We will return to this question in Chapter 4. We can now write down the hydrogen atom wave functions explicitly. Going back to (3.231) and putting in the appropriate normalization (see Problem 3.30), we have \u03c8nlm(x) = x|nlm = Rnl(r)Ym l (\u03b8, \u03c6) (3.319) where Rnl(r) = l 2Zr na0 1 (2l + 1)! 2Z na0 \u00d7 F(\u2212n + l + 1; 2l + 2; 2Zr/na0). e\u2212Zr/na0 1/2 3 (n + l)! 2n(n \u2212 l \u2212 1)! (3.320) Figure 3.7 plots these radial wave functions for n = 1 and n = 2. As we have discussed, only the l = 0 wave functions are nonzero at the origin. Also note that there are n \u2212 1 nodes in the wave function for l = 0, and no nodes for the wave function with l = n \u2212 1. 3.8 Addition of Angular Momenta Angular-momentum addition has important applications in all areas of modern physics \u2013 from atomic spectroscopy to nuclear and particle collisions. Furthermore, a study of angular-momentum addition provides an excellent opportunity to illustrate the concept of change of basis, which we discussed extensively in Chapter 1. 3.8.1 Simple Examples of Angular-Momentum Addition Before studying a formal theory of angular-momentum addition, it is worth looking at two simple examples with which the reader may be familiar: (1) how to add orbital angular momentum and spin-angular momentum and (2) how to add the spin-angular momenta of two", " spin 1 2 particles. 206 Theory of Angular Momentum Previously we studied both spin 1 2 systems with all quantum-mechanical degrees of freedom other than spin, such as position and momentum, ignored and quantummechanical particles with the space degrees of freedom (such as position or momentum) taken into account but the internal degrees of freedom (such as spin) ignored. A realistic description of a particle with spin must of course take into account both the space degree of freedom and the internal degrees of freedom. The base ket for a spin 1 2 particle may be visualized to be in the direct-product space of the in\ufb01nite-dimensional ket space spanned by the position eigenkets {|x} and the two-dimensional spin space spanned by |+ and |\u2212. Explicitly, we have for the base ket, \u00b1 = |x \u2297 |\u00b1, where any operator in the space spanned by {|x} commutes with any operator in the two-dimensional space spanned by |\u00b1. (3.321) |x The rotation operator still takes the form exp(\u2212iJ\u00b7 \u02c6n\u03c6/\u00afh) but J, the generator of rotations, is now made up of two parts, namely, J = L + S. It is actually more obvious to write (3.322) as J = L \u2297 1 + 1 \u2297 S, (3.322) (3.323) where the 1 inL \u2297 1 stands for the identity operator in the spin space, while the 1 in 1 \u2297 S stands for the identity operator in the in\ufb01nite-dimensional ket space spanned by the position eigenkets. Because L and S commute, we can write D(R) =D (orb)(R) \u2297 D (spin)(R) =exp \u2212iL\u00b7 \u02c6n\u03c6 \u00afh \u2297 exp \u2212iS\u00b7 \u02c6n\u03c6 \u00afh. (3.324) (3.325) The wave function for a particle with spin is written as x, \u00b1|\u03b1 = \u03c8 \u00b1(x)., The two components \u03c8 \u00b1 are often arranged in column matrix form as follows: (3.326) \u03c8 +(x) \u03c8 \u2212(x) where |\u03c8 \u00b1(x)| 2 stands for the probability density for the particle to be found at x with spin up and spin down, respectively. Instead of |x as the base kets for the space part, we may use |n, l, m, which are eigenkets of L 2 and Lz with eigenvalues \u00afh2l(l + 1) and ml \u00afh, respectively. For the spin part, |\u00b1 are eigenkets of S2 and Sz with eigenvalues 3\u00afh2/4 and \u00b1\u00afh/2, respectively. However, as we will show later, we can also use base kets which are eigenkets of J2, Jz, L2, andS 2. In other words, we can expand a state ket of a particle with spin in terms of simultaneous eigenkets of L2, S2, Lz, and Sz or in terms of simultaneous eigenkets of J2, Jz, L2, andS 2. We will study in detail how the two descriptions are related. 2 particles, say two electrons, with the orbital As a second example, we study two spin 1 degree of freedom suppressed. The total spin operator is usually written as S = S1 + S2, (3.327) 207 3.8 Addition of Angular Momenta but again it is to be understood as S1 \u2297 1 + 1 \u2297 S2, (3.328) where the 1 in the \ufb01rst (second) term stands for the identity operator in the spin space of electron 2 (1). We, of course, have [S1x, S2y] =0 (3.329) and so forth. Within the space of electron 1(2) we have the usual commutation relations [S1x, S1y] = i\u00afhS1z, [S2x, S2y] =i \u00afhS2z,.... (3.330) As a direct consequence of (3.329) and (3.330), we have [Sx, Sy] =i \u00afhSz (3.331) and so on for the total spin operator. The eigenvalues of the various spin operators are denoted as follows: S2 = (S1 + S2)2 : s(s + 1)\u00afh2 Sz = S1z + S2z S1z S2z : m\u00afh : m1 \u00afh : m2 \u00afh. (", "3.332) Again, we can expand the ket corresponding to an arbitrary spin state of two electrons in terms of either the eigenkets of S2 and Sz or the eigenkets of S1z and S2z. The two possibilities are as follows. 1. The {m1, m2} representation based on the eigenkets of S1z and S2z: | + +, | + \u2212, | \u2212+, and | \u2212 \u2212, (3.333) where | + \u2212 stands for m1 = 1 2, m2 = \u2212 1 2, and so forth. 2. The {s, m} representation (or the triplet-singlet representation) based on the eigenkets of S2 and Sz: |s = 1, m = \u00b11, 0, |s = 0, m = 0, (3.334) where s = 1 (s = 0) is referred to as the spin triplet (spin singlet). Notice that in each set there are four base kets. The relationship between the two sets of base kets is as follows: |s = 1, m = 1 = | + +, |s = 1, m = 0 = (| + \u2212 + | \u2212+ ), |s = 1, m = \u22121 = | \u2212 \u2212, |s = 0, m = 0 = (| + \u2212 \u2212 | \u2212 +). 1\u221a 2 1\u221a 2 (3.335a) (3.335b) (3.335c) (3.335d) 208 Theory of Angular Momentum The right-hand side of (3.335a) tells us that we have both electrons with spin up; this situation can correspond only to s = 1, m = 1. We can obtain (3.335b) from (3.335a) by applying the ladder operator S\u2212 \u2261 S1\u2212 + S2\u2212 = (S1x \u2212 iS1y) + (S2x \u2212 iS2y) (3.336) to both sides of (3.335a). In doing so we must remember that an electron 1 operator like S1\u2212 affects just the \ufb01rst entry of | + +, and so on. We can write S\u2212|s = 1, m = 1 = (S1\u2212 + S2\u2212)| + + (3.337) as (1 + 1)(1 \u2212 1 + 1)|s = 1 \u00d7 | \u2212+ & | + \u2212, (3.338) which immediately leads to (3.335b). Likewise, we can obtain |s = 1, m = \u22121 by applying (3.336) once again to (3.335b). Finally, we can obtain (3.335d) by requiring it to be orthogonal to the other three kets, in particular to (3.335b). The coef\ufb01cients that appear on the right-hand side of (3.335) are the simplest example of Clebsch\u2013Gordan coef\ufb01cients to be discussed further at a later time. They are simply the elements of the transformation matrix that connects the {m1, m2} basis to the {s, m} basis. It is instructive to derive these coef\ufb01cients in another way. Suppose we write the 4 \u00d7 4 matrix corresponding to S2 = S2 = S2 1 + S2 1 + S2 2 + 2S1\u00b7S2 2 + 2S1zS2z + S1+S2\u2212 + S1\u2212S2+ (3.339) using the (m1, m2) basis. The square matrix is obviously not diagonal because an operator like S1+ connects | \u2212+ with | + +. The unitary matrix that diagonalizes this matrix carries the |m1, m2 base kets into the |s, m base kets. The elements of this unitary matrix are precisely the Clebsch\u2013Gordan coef\ufb01cients for this problem. The reader is encouraged to work out all this in detail. See Problem 3.33 at the end of this chapter. 3.8.2 Formal Theory of Angular-Momentum Addition Having gained some physical insight by considering simple examples, we are now in a position to study more systematically the formal theory of angular-momentum addition. Consider two angular-momentum operators J1 and J2 in different subspaces. The components of J1(J2) satisfy the usual angular-momentum commutation relations: and [J1i, J1j] =i \u00afh\u03b5ijkJ1k [J2i, J2j] =i \u00afh\u03b5ijkJ2k. (3.340a) (3.340b) 209 3.8 Addition of Angular Momenta", " However, we have [J1k, J2l] = 0 (3.341) between any pair of operators from different subspaces. The in\ufb01nitesimal rotation operator that affects both subspace 1 and subspace 2 is written as 1 \u2212 iJ1\u00b7 \u02c6n\u03b4\u03c6 \u00afh 1 \u2212 iJ2\u00b7 \u02c6n\u03b4\u03c6 \u00afh \u2297 = 1 \u2212 i(J1 \u2297 1 + 1 \u2297 J2) \u00b7 \u02c6n\u03b4\u03c6 \u00afh. (3.342) We de\ufb01ne the total angular momentum by J \u2261 J1 \u2297 1 + 1 \u2297 J2, which is more commonly written as J = J1 + J2. The \ufb01nite-angle version of (3.342) is D1(R) \u2297 D2(R) =exp \u2212iJ1\u00b7 \u02c6n\u03c6 \u00afh \u2297 exp. \u2212iJ2\u00b7 \u02c6n\u03c6 \u00afh (3.343) (3.344) (3.345) Notice the appearance of the same axis of rotation and the same angle of rotation. It is very important to note that the total J satis\ufb01es the angular-momentum commutation relations [Ji, Jj] =i \u00afh\u03b5ijkJk (3.346) as a direct consequence of (3.340) and (3.341). In other words, J is an angular momentum in the sense of Section 3.1. Physically this is reasonable because J is the generator for the entire system. Everything we learned in Section 3.5, for example, the eigenvalue spectrum of J2 and Jz and the matrix elements of the ladder operators, also holds for the total J. As for the choice of base kets we have two options. Option A: Simultaneous eigenkets of J2 1, J2 2, J1z, andJ 2z, denoted by | j1 j2; m1m2. Obviously the four operators commute with each other. The de\ufb01ning equations are J2 1 | j1 j2; m1m2 = j1( j1 + 1)\u00afh2| j1 j2; m1m2, J1z| j1 j2; m1m2 = m1 \u00afh| j1 j2; m1m2, J2 2 | j1 j2; m1m2 = j2( j2 + 1)\u00afh2| j1 j2; m1m2, J2z| j1 j2; m1m2 = m2 \u00afh| j1 j2; m1m2. (3.347a) (3.347b) (3.347c) (3.347d) Option B: Simultaneous eigenkets of J2, J2 1, J2 operators mutually commute. In particular, we have 2, and Jz. First, note that this set of which can readily be seen by writing J2 as [J2, J2 1] =0, J2 = J2 1 + J2 2 + 2J1zJ2z + J1+J2\u2212 + J1\u2212J2+. (3.348) (3.349) 210 Theory of Angular Momentum We use | j1, j2; jm to denote the base kets of option B: J2 1 J2 2 | j1 j2; jm = j1( j1 + 1)\u00afh2| j1 j2; jm, | j1 j2; jm = j2( j2 + 1)\u00afh2| j1 j2; jm, J2| j1 j2; jm = j( j + 1)\u00afh2| j1 j2; jm, Jz| j1 j2; jm = m\u00afh| j1 j2; jm. Quite often j1, j2 are understood, and the base kets are written simply as | j, m. It is very important to note that even though we have [J2, Jz] =0, [J2, J1z] = 0, [J2, J2z] = 0, (3.350a) (3.350b) (3.350c) (3.350d) (3.351) (3.352) as the reader may easily verify using (3.349). This means that we cannot add J2 to the set of operators of option A. Likewise, we cannot add J1z and/or J2z to the set of operators of option B. We have two possible", " sets of base kets corresponding to the two maximal sets of mutually compatible observables we have constructed. Let us consider the unitary transformation in the sense of Section 1.5 that connects the two bases: | j1 j2; jm = \u2211 m1 \u2211 m2 | j1 j2; m1m2 j1 j2; m1m2| j1 j2; jm, (3.353) where we have used \u2211 m1 \u2211 m2 | j1 j2; m1m2 j1 j2; m1m2| = 1 (3.354) and where the right-hand side is the identity operator in the ket space of given j1 and j2. The elements of this transformation matrix j1 j2; m1m2| j1 j2; jm are Clebsch\u2013Gordan coef\ufb01cients. There are many important properties of Clebsch\u2013Gordan coef\ufb01cients that we are now ready to study. First, the coef\ufb01cients vanish unless To prove this, \ufb01rst note that m = m1 + m2. (Jz \u2212 J1z \u2212 J2z)| j1 j2; jm = 0. Multiplying j1 j2; m1m2| on the left, we obtain (m \u2212 m1 \u2212 m2) j1 j2; m1m2| j1 j2; jm = 0, (3.355) (3.356) (3.357) which proves our assertion. Admire the power of the Dirac notation! It really pays to write the Clebsch\u2013Gordan coef\ufb01cients in Dirac\u2019s bracket form, as we have done. 211 3.8 Addition of Angular Momenta Second, the coef\ufb01cients vanish unless | j1 \u2212 j2| \u2264j \u2264 j1 + j2. (3.358) This property may appear obvious from the vector model of angular-momentum addition, where we visualize J to be the vectorial sum of J1 and J2. However, it is worth checking this point by showing that if (3.358) holds, then the dimensionality of the space spanned by {| j1 j2; m1m2} is the same as that of the space spanned by {| j1 j2; jm}. For the (m1, m2) way of counting we obtain N = (2j1 + 1)(2j2 + 1) (3.359) because for given j1 there are 2j1 + 1 possible values of m1; a similar statement is true for the other angular momentum j2. As for the ( j, m) way of counting, we note that for each j, there are 2j + 1 states, and according to (3.358), j itself runs from j1 \u2212 j2 to j1 + j2, where we have assumed, without loss of generality, that j1 \u2265 j2. We therefore obtain N = (2j + 1) j1+j2\u2211 j=j1\u2212j2 2 [{2( j1 \u2212 j2) +1} + {2( j1 + j2) +1}](2j 2 + 1) = 1 = (2j1 + 1)(2j2 + 1). (3.360) Because both ways of counting give the same N-value, we see that (3.358) is quite consistent.13 The Clebsch\u2013Gordan coef\ufb01cients form a unitary matrix. Furthermore, the matrix elements are taken to be real by convention. An immediate consequence of this is that the inverse coef\ufb01cient j1 j2; jm| j1 j2; m1m2 is the same as j1 j2; m1m2| j1 j2; jm itself. A real unitary matrix is orthogonal, so we have the orthogonality condition \u2211 \u2211 j1 j2; m1m2| j1 j2; jm j1 j2; m 1m 2 j m | j1 j2; jm = \u03b4 m1m 1 \u03b4 m2m 2, (3.361) which is obvious from the orthonormality of {| j1 j2; m1m2} together with the reality of the Clebsch\u2013Gordan coef\ufb01cients. Likewise, we also have \u2211 m1 j1 j2; m1m2| j1 j2; jm j1 j2; m1m2| j1 j2; j m = \u03b4jj \u03b4mm. \u2211 m2 (3.362) As a", " special case of this we may set j = j, m = m = m1 + m2. We then obtain \u2211 m1,m2=m\u2212m1 | j1 j2; m1m2| j1 j2; jm| 2 = 1, (3.363) which is just the normalization condition for | j1 j2; jm. Some authors use somewhat different notations for the Clebsch\u2013Gordan coef\ufb01cients. Instead of j1 j2; m1m2| j1 j2; jm we sometimes see j1m1 j2m2| j1 j2 jm, C( j1 j2 j; m1m2m), 13 A complete proof of (3.358) is given in Gottfried (1966), p. 215, and also in Appendix D of this book. 212 Theory of Angular Momentum Cj1 j2( jm; m1m2), and so on. They can also be written in terms of Wigner\u2019s 3-j symbol, which is occasionally found in the literature: j1 j2; m1m2| j1 j2; jm = (\u22121)j1\u2212j2+m 2j + 1 j2 j j1 m1 m2 \u2212m. (3.364) 3.8.3 Recursion Relations for the Clebsch\u2013Gordan Coefficients14 With j1, j2, and j \ufb01xed, the coef\ufb01cients with different m1 and m2 are related to each other by recursion relations. We start with J\u00b1| j1 j2; jm = (J1\u00b1 + J2\u00b1)\u2211 m1 | j1 j2; m1m2 j1 j2; m1m2| j1 j2; jm. (3.365) \u2211 m2 1, m2 \u2192 m 2) Using (3.191) and (3.192) we obtain (with m1 \u2192 m ( j \u2213 m)( j \u00b1 m + 1)| j1 j2; j, m \u00b1 1 = \u2211 m 1 1)( j1 \u00b1 m ( j1 \u2213 m \u2211 m 2 1 + 1) | j1 j2; m 1 \u00b1 1, m 2 + ( j2 \u2213 m 2)( j2 \u00b1 m 2 + 1)| j1 j2; m 1, m 2 \u00b1 1 \u00d7 j1 j2; m 1m 2 | j1 j2; jm. (3.366) Our next step is to multiply by j1 j2; m1m2| on the left and use orthonormality, which means that nonvanishing contributions from the right-hand side are possible only with for the \ufb01rst term and m1 = m 1 \u00b1 1, m2 = m 2 m1 = m 1, m2 = m 2 \u00b1 1 for the second term. In this manner we obtain the desired recursion relations: ( j \u2213 m)( j \u00b1 m + 1) j1 j2; m1m2| j1 j2; j, m \u00b1 1 = + ( j1 \u2213 m1 + 1) ( j1 \u00b1 m1) j1 j2; m1 \u2213 1, m2| j1 j2; jm ( j2 \u2213 m2 + 1) (j 2 \u00b1 m2) j1 j2; m1, m2 \u2213 1| j1 j2; jm. (3.367) (3.368) (3.369) It is important to note that because the J\u00b1 operators have shifted the m-values, the nonvanishing condition (3.355) for the Clebsch\u2013Gordan coef\ufb01cients has now become [when applied to (3.369)] m1 + m2 = m \u00b1 1. (3.370) We can appreciate the signi\ufb01cance of the recursion relations by looking at (3.369) in an m1m2-plane. The J+ recursion relation (upper sign) tells us that the coef\ufb01cient at (m1, m2) 14 More detailed discussion of Clebsch\u2013Gordan and Racah coef\ufb01cients, recoupling, and the like is given in Edmonds (1974), for instance. 213 3.8 Addition of Angular Momenta (m1 \u2212 1, m2) RHS LHS (m1, m2) (m1, m2 + 1) RHS J+ J\u2212 RHS (m1, m2 \u2212 1) LHS (m1, m2) RHS (m1 + 1, m2) (a) J+ relation (b", ") J\u2212 relation Fig. 3.8 m1m2-plane showing the Clebsch\u2013Gordan coefficients related by the recursion relations ( 3.369). m2 = j2 m1 + m2 = j A m1 = \u2212j1 m1 = j1 m1 + m2 = \u2212j m2 = \u2212j2 (a) Forbidden!! x D E F j+ j+ j\u2212 j\u2212 A j\u2212 B C (b) Fig. 3.9 Use of the recursion relations to obtain the Clebsch\u2013Gordan coefficients. is related to the coef\ufb01cients at (m1 \u2212 1, m2) and (m1, m2 \u2212 1), as shown in Figure 3.8a. Likewise, the J\u2212 recursion relation (lower sign) relates the three coef\ufb01cients whose m1, m2 values are given in Figure 3.8b. Recursion relations (3.369), together with normalization condition (3.363), almost uniquely determine all Clebsch\u2013Gordan coef\ufb01cients. (We say \u201calmost uniquely\u201d because certain sign conventions have yet to be speci\ufb01ed.) Our strategy is as follows. We go back to the m1m2-plane, again for \ufb01xed j1, j2, and j, and plot the boundary of the allowed region determined by |m1| \u2264j 1, |m2| \u2264j 2, \u2212j \u2264 m1 + m2 \u2264 j; (3.371) see Figure 3.9a. We may start with the upper right-hand corner, denoted by A. Because we work near A at the start, a more detailed \u201cmap\u201d is in order; see Figure 3.6b. We apply the J\u2212 recursion relation (3.369) (lower sign), with (m1, m2 + 1) corresponding to A. Observe now that the recursion relation connects A with only B because the site corresponding to (m1 + 1, m2) is forbidden by m1 \u2264 j1. As a result, we can obtain the Clebsch\u2013Gordan coef\ufb01cient of B in terms of the coef\ufb01cient of A. Next, we form a J+ triangle made up of A, 214 Theory of Angular Momentum x x x ms j\u2212 j\u2212 j\u2212 ml Fig. 3.10 Recursion relations used to obtain the Clebsch\u2013Gordan coefficients for ji = l and j2 = s = 1 2. B, andD. This enables us to obtain the coef\ufb01cient of D once the coef\ufb01cient of A is speci\ufb01ed. We can continue in this fashion: Knowing B and D, we can get to E; knowing B and E we can get to C, and so on. With enough patience we can obtain the Clebsch-Gordan coef\ufb01cient of every site in terms of the coef\ufb01cient of starting site, A. For overall normalization we use (3.363). The \ufb01nal overall sign is \ufb01xed by convention. (See the following example.) As an important practical example we consider the problem of adding the orbital and spin-angular momenta of a single spin 1 2 particle. We have (integer), m1 = ml, j1 = l j2 = s = 1 2, The allowed values of j are given by m2 = ms = \u00b1 ; j = 1 2, l = 0. (3.372) (3.373) So for each l there are two possible j-values; for example, for l = 1 (p state) we get, in spectroscopic notation, p3/2 and p1/2, where the subscript refers to j. The m1m2-plane, or better the mlms-plane, of this problem is particularly simple. The allowed sites form only two rows: the upper row for ms = 1 2 (see Figure 3.10). Speci\ufb01cally, we work out the case j = l + 1 2, we can use the J\u2212 recursion in such a way that we always stay in the upper row (m2 = ms = 1 2 ), while the ml-value changes by one unit each time we consider a new J\u2212 triangle. Suppressing j1 = l, j2 = 1 2, in writing the Clebsch\u2013Gordan coef\ufb01cient, we obtain from (3.369) (lower sign) 2 and the lower row for ms = \u2212 1 2. Because ms cannot exceed l + 1 2, m + 1 = where we have used m1 = ml = m \u2212 1 2, m2 = ms = 1 2. In", " this way we can move horizontally by one unit3.374) (3.375), ), m + 1. (3.376) 215 3.8 Addition of Angular Momenta We can in turn express m + 1 2, m + 2, and so forth. Clearly, this procedure can be continued until ml reaches l, the maximum possible value: 2, m + 1 in terms of m + 3 |l + 1 | )... = l + m + 1 2 2l + 1 ( l3.377) Consider the angular-momentum con\ufb01guration in which ml and ms are both maximal, that is, l and 1 2, which is possible only for, j = l + 1 up to a phase factor. We take this phase factor to be real and positive by convention. With this choice we have 2, respectively. The total m = ml + ms is l + 1 2. So |ml = l, ms = 1 must be equal to | j = l + 1 2 and not for. (3.378) Returning to (3.377), we \ufb01nally obtain ( 2l + 1. (3.379) But this is only about one-fourth of the story. We must still determine the value of the question marks that appear in the following 2l + 1 ml = m \u2212 1 2, ms = ) 1 2 +? ml = m + ), m =? j = l \u2212 1 2 ml = m \u2212 1 2, ms = +? | ml = m + 1 2 ). 1 2, ms = \u2212 1 2 (3.380) ), 1 2, ms = \u2212 1 2 ) We note that the transformation matrix with \ufb01xed m from the (ml, ms) basis to the ( j, m) basis is, because of orthogonality, expected to have the form cos \u03b1 sin \u03b1 \u2212 sin \u03b1 cos \u03b1. (3.381) 216 Theory of Angular Momentum Comparison with (3.380) shows that cos \u03b1 is (3.379) itself; so we can readily determine sin \u03b1 up to a sign ambiguity: sin2 2l + 1) = l \u2212 m + 1 2 (2l + 1). (3.382) 2, ms = \u2212 1 2, m must be positive because all j = l + 1 We claim that ml = m + 1 2 2 states are reachable by applying the J\u2212 operator successively to | j = l + 1, and the matrix elements of J\u2212 are always positive by convention. So the 2 \u00d7 2 transformation matrix (3.381) can be only 2l + 1 l \u2212 m + 1 2 2l + 1 l \u2212 m + 1 2 2l + 1 l + m + 1 2 2l + 1 \u239f \u239f \u239f \u239f \u23a0. (3.383) We de\ufb01ne spin-angular functions in two-component form as follows: Y j=l\u00b11/2, 2l + 1 Ym\u22121/2 l (\u03b8, \u03c6)\u03c7+ + l \u2213 m + 1 2 2l + 1 \u239b Ym+1/2 l (\u03b8, \u03c6)\u03c7\u2212 = 1\u221a 2l + \u03b8, \u03c6) l 2 Ym\u22121/2 2 Ym+1/2 l (\u03b8, \u03c6) \u239e \u23a0. (3.384) They are, by construction, simultaneous eigenfunctions of L2, S2, J2, and Jz. They are also eigenfunctions of L\u00b7S but L\u00b7S, being just L\u00b7S = 1 2 % J2 \u2212 L2 \u2212 S2 &, (3.385) (3.386) is not independent. Indeed, its eigenvalue can easily be computed as follows: \u00afh2 2 j( j + 1) \u2212 l(l + 1) \u2212 3 4 = \u23a7 \u23aa\u23aa\u23aa\u23a8 l\u00afh2 2 \u23aa\u23aa\u23aa\u23a9 \u2212 (l + 1)\u00afh2 2 for j = l + 1 2, for j = l \u2212 1 2. 3.8.4 Clebsch\u2013Gordan Coefficients and Rotation Matrices Angular-momentum addition may be discussed from the point of view of rotation matrices. Consider the rotation operator D ( j1)(R) in the ket space spanned by the angular-momentum eigenkets with eigenvalue j1. Likewise, consider D ( j2)(R). The product D ( j1) \u2297 D ( j2) is reducible in the sense that after suitable choice of base kets, its matrix representation can take the following form: 217 3.8", " Addition of Angular Momenta ( j1 + j2) ( j1 + j2 \u2212 1) ( j1 + j2 \u2212 2). (| j1 \u2212 j2|) (3.387) In the notation of group theory this is written as D ( j1) \u2297 D ( j2) = D ( j1+j2) \u2295 D ( j1+j2\u22121) \u2295 \u00b7 \u00b7 \u00b7 \u2295 D (| j1\u2212j2|). (3.388) In terms of the elements of rotation matrices, we have an important expansion known as the Clebsch\u2013Gordan series: D ( j1) m1m 1 (R)D ( j2) m2m 2 (R) = \u2211 \u2211 j m j1 j2; m1m2| j1 j2; jm \u2211 m \u00d7 j1 j2; m where the j-sum runs from | j1 \u2212 j2| to j1 + j2. The proof of this equation follows. First, note that the left-hand side of (3.389) is the same as j1j2; m1m2|D(R)| j1j2; m j2m2|D(R)| j2m | j1 j2; jmD ( j) mm (R), (3.389) 1m 2 1m 2 2 = j1m1|D(R)| j1m = D ( j1) m1m 1 (R)D ( j2) m2m 2 1 (R). (3.390) But the same matrix element is also computable by inserting a complete set of states in the ( j, m) basis. Thus j1 j2; m1m2| j1 j2; jm j1 j2; jm|D(R)| j1 j2; j m j1 j2; m1m2|D(R)| j1 j2; m \u2211 = \u2211 1m 2 j m \u2211 j \u2211 m \u00d7 j1 j2; j m| j1 j2; m \u2211 \u2211 m j 1m \u00d7 j1 j2; m \u2211 m 2 j = \u2211 | j1 j2; j m, 1m 2 j1 j2; m1m2| j1 j2; jmD ( j) mm (R)\u03b4jj (3.391) which is just the right-hand side of (3.389). As an interesting application of (3.389), we derive an important formula for an integral m0 and Ym\u2217 involving three spherical harmonics. First, recall the connection between D ( l ) l 218 Theory of Angular Momentum given by (3.261). Letting j1 \u2192 l1, j2 \u2192 l2, m 1 obtain, after complex conjugation, \u2192 0, m 2 \u2192 0 (hence m \u2192 0) in (3.389), we Ym1 l1 (\u03b8, \u03c6)Ym2 l2 (\u03b8, \u03c6) = \u2211 l (2l1 + 1)(2l2 + 1) 4\u03c0 l1l2; m1m2|l1l2; lm \u2211 m 4\u03c0 2l + 1 l (\u03b8, \u03c6) and integrate over solid angles. The summations drop \u00d7 l1l2; 00|l1l2; l0 Ym l (\u03b8, \u03c6). (3.392) \u2019 We multiply both sides by Ym\u2217 out because of the orthogonality of spherical harmonics, and we are left with d\u03a9 Ym\u2217 l (\u03b8, \u03c6)Ym1 l1 (\u03b8, \u03c6)Ym2 l2 (\u03b8, \u03c6) = (2l1 + 1)(2l2 + 1) 4\u03c0(2l + 1) l1l2; 00|l1l2; l0l1l2; m1m2|l1l2; lm. (3.393) The square root factor times the \ufb01rst Clebsch\u2013Gordan coef\ufb01cient is independent of orientations, that is, of m1 and m2. The second Clebsch\u2013Gordan coef\ufb01cient is the one appropriate for adding l1 and l2 to obtain total l. Equation (3.393) turns out to be a special case of the Wigner\u2013Eckart theorem to be derived in Section 3.11. This formula is extremely useful in evaluating multipole matrix elements in atomic and nuclear spectroscopy. 3.9 Schwinger\u2019s Oscillator Model of", " Angular Momentum 3.9.1 Angular Momentum and Uncoupled Oscillators There exists a very interesting connection between the algebra of angular momentum and the algebra of two independent (that is, uncoupled) oscillators, which was worked out in J. Schwinger\u2019s notes. See Biedenharn and Van Dam (1965), p. 229. Let us consider two simple harmonic oscillators, which we call the plus type and the minus type. We have the annihilation and creation operators, denoted by a+ and a\u2020 + for the plus-type oscillator; likewise, we have a\u2212 and a\u2020 \u2212 for the minus-type oscillator. We also de\ufb01ne the number operators N+ and N\u2212 as follows: N+ \u2261 a\u2020 +a+, N\u2212 \u2261 a\u2020 \u2212a\u2212. (3.394) We assume that the usual commutation relations among a, a\u2020, andN hold for oscillators of the same type (see Section 2.3): [a+, a\u2020 +] =1, [a\u2212, a\u2020 \u2212] =1, [N+, a+] = \u2212a+, [N\u2212, a\u2212] =\u2212 a\u2212, [N+, a\u2020 +] =a \u2020 +, [N\u2212, a\u2020 \u2212] =a \u2020 \u2212. (3.395a) (3.395b) (3.395c) 219 3.9 Schwinger\u2019s Oscillator Model of Angular Momentum However, we assume that any pair of operators between different oscillators commute: [a+, a\u2020 \u2212] = [a\u2212, a\u2020 +] =0 (3.396) and so forth. So it is in this sense that we say the two oscillators are uncoupled. Because N+ and N\u2212 commute by virtue of (3.396), we can build up simultaneous eigenkets of N+ and N\u2212 with eigenvalues n+ and n\u2212, respectively. So we have the following eigenvalue equations for N\u00b1: N+|n+, n\u2212 = n+|n+, n\u2212, N\u2212|n+, n\u2212 = n\u2212|n+, n\u2212. (3.397) In complete analogy with (2.137) and (2.138), the creation and annihilation operators, a\u2020 \u00b1 and a\u00b1, act on |n+, n\u2212 as follows: a\u2020 + |n+, n\u2212 = n+ + 1|n+ + 1, n\u2212, \u2212|n+, n\u2212 = a\u2020 n\u2212 + 1|n+, n\u2212 + 1, a+|n+, n\u2212 = \u221a n+|n+ \u2212 1, n\u2212, a\u2212|n+, n\u2212 = \u221a n\u2212|n+, n\u2212 \u2212 1. (3.398a) (3.398b) We can obtain the most general eigenkets of N+ and N\u2212 by applying a\u2020 successively to the vacuum ket de\ufb01ned by + and a\u2020 \u2212 In this way we obtain a+|0, 0 = 0, a\u2212|0, 0 = 0. |n+, n\u2212 = n\u2212 n+ (a\u2020 (a\u2020 \u2212) +) \u221a \u221a n\u2212! n+! |0, 0. (3.399) (3.400) Next, we de\ufb01ne and J+ \u2261 \u00afha\u2020 +a\u2212, J\u2212 \u2261 \u00afha\u2020 \u2212a+, (3.401a) \u00afh 2 Jz \u2261 +a+ \u2212 a\u2020 a\u2020 \u2212a\u2212 = \u00afh 2 (N+ \u2212 N\u2212). (3.401b) We can readily prove that these operators satisfy the angular-momentum commutation relations of the usual form [Jz, J\u00b1] = \u00b1\u00afhJ\u00b1, [J+, J\u2212] = 2\u00afhJz. For example, we prove (3.402) as follows: \u00afh2[a\u2020 +a\u2212, a\u2020 \u2212a+] = \u00afh2a\u2020 = \u00afh2a\u2020 = \u00afh2(a\u2020 \u2212a+ \u2212 \u00afh2a\u2020 +a\u2212a\u2020 +(a\u2020 +a+ \u2212 a\u2020 \u2212a+a\u2020 \u2212a\u2212 + 1)a+ \u2212 \u00afh2a\u2020 \u2212a\u2212) =2 \u00afhJz. +a\u2212 \u2212(a\u2020 +a+ + 1)a\u2212 De\ufb01ning the total N to be N \u2261 N+ + N\u2212 = a\u2020 +a+ + a\u2020 \u2212a\u2212, (3.402a) (3.402b) (3.403) (3.404) 220 Theory of Angular Momentum we can also prove = which is left as an exercise. J2 \u2261 J 2 z + (J+J\u2212 +", " J\u2212J+) 1 2 N N 2 \u00afh2 2 + 1, (3.405) What are the physical interpretations of all this? We associate spin up (m = 1 2 ) with one quantum unit of the plus-type oscillator and spin down (m = \u2212 1 2 ) with one quantum unit of the minus-type oscillator. If you like, you may imagine one spin 1 2 \u201cparticle\u201d with spin up (down) with each quantum unit of the plus- (minus-) type oscillator. The eigenvalues n+ and n\u2212 are just the number of spins up and spins down, respectively. The meaning of J+ is that it destroys one unit of spin down with the z-component of spin-angular momentum \u2212\u00afh/2 and creates one unit of spin up with the z-component of spin-angular momentum +\u00afh/2; the z-component of angular momentum is therefore increased by \u00afh. Likewise J\u2212 destroys one unit of spin up and creates one unit of spin down; the z-component of angular momentum is therefore decreased by \u00afh. As for theJ z operator, it simply counts \u00afh/2 times the difference of n+ and n\u2212, just the z-component of the total angular momentum. With (3.398) at our disposal we can easily examine how J\u00b1 and Jz act on |n+, n\u2212 as follows: J+|n+, n\u2212 = \u00afha\u2020 +a\u2212|n+, n\u2212 = n\u2212(n+ + 1)\u00afh|n+ + 1, n\u2212 \u2212 1, n+(n\u2212 + 1)\u00afh|n+ \u2212 1, n\u2212 + 1, \u2212a+|n+, n\u2212 = J\u2212|n+, n\u2212 = \u00afha\u2020 Jz|n+, n\u2212 = (N+ \u2212 N\u2212)|n+, n\u2212 = \u00afh 2 1 2 (n+ \u2212 n\u2212)\u00afh|n+, n\u2212. (3.406c) (3.406a) (3.406b) Notice that in all these operations, the sum n+ + n\u2212, which corresponds to the total number of spin 1 2 particles, remains unchanged. Observe now that (3.406a), (3.406b), and (3.406c) reduce to the familiar expressions for the J\u00b1 and Jz operators we derived in Section 3.5, provided we substitute n+ \u2192 j + m, n\u2212 \u2192 j \u2212 m. The square root factors in (3.406a) and (3.406b) change to n\u2212(n+ + 1) \u2192 n+(n\u2212 + 1) \u2192 ( j \u2212 m)( j + m + 1), ( j + m)( j \u2212 m + 1), which are exactly the square root factors appearing in (3.191) and (3.193). (3.407) (3.408) Notice also that the eigenvalue of the J2 operator de\ufb01ned by (3.405) changes as follows: \u00afh2 2 (n+ + n\u2212) 2 (n+ + n\u2212) + 1 \u2192 \u00afh2j( j + 1). (3.409) All this may not be too surprising because we have already proved that J\u00b1 and J2 operators we constructed out of the oscillator operators satisfy the usual angularmomentum commutation relations. But it is instructive to see in an explicit manner 221 3.9 Schwinger\u2019s Oscillator Model of Angular Momentum the connection between the oscillator matrix elements and angular-momentum matrix elements. In any case, it is now natural to use j \u2261 (n+ + n\u2212) 2, m \u2261 (n+ \u2212 n\u2212) 2 (3.410) in place of n+ and n\u2212 to characterize simultaneous eigenkets of J2 and Jz. According to (3.406a) the action of J+ changes n+ into n+ + 1, n\u2212 into n\u2212 \u2212 1, which means that j is unchanged and m goes into m + 1. Likewise, we see that the J\u2212 operator that changes n+ into n+ \u2212 1 and n\u2212 into n+ \u2212 1 lowers m by one unit without changing j. We can now write as (3.400) for the most general N+, N\u2212 eigenket | j, m = j\u2212m j+m (a\u2020 (a\u2020 +) ( j + m)! ( j \u2212 m)! \u2212) |0, (3.411) where we have used|0 for the vacuum ket, earlier denoted by |0, 0. A special case of (3.411) is of interest. Let us set m = j, which physically means that the eigenvalue of Jz is as large as possible for a given j. We have | j, j", " = (a\u2020 +)2j (2j)! |0. (3.412) We can imagine this state to be built up of 2j spin 1 in the positive z-direction. 2 particles with their spins all pointing In general, we note that a complicated object of high j can be visualized as being made 2 particles, j + m of them with spin up and the remaining j \u2212 m of them up of primitive spin 1 with spin down. This picture is extremely convenient even though we obviously cannot always regard an object of angular momentum j literally as a composite system of spin 1 2 particles. All we are saying is that, as far as the transformation properties under rotations are concerned, we can visualize any object of angular momentum j as a composite system of 2j spin 1 2 particles formed in the manner indicated by (3.411). From the point of view of angular-momentum addition developed in the previous section, we can add the spins of 2j spin 1 2 particles to obtain states with angular momentum j, j \u2212 1, j \u2212 2,.... As a simple example, we can add the spin-angular momenta of two spin 1 2 particles to obtain a total angular momentum of zero as well as one. In Schwinger\u2019s oscillator scheme, however, we obtain only states with angular momentum j when we start with 2j spin 1 2 particles. In the language of permutation symmetry to be developed in Chapter 7, only totally symmetrical states are constructed by this method. The primitive spin 1 2 particles appearing here are actually bosons! This method is quite adequate if our purpose is to examine the properties under rotations of states characterized by j and m without asking how such states are built up initially. The reader who is familiar with isospin in nuclear and particle physics may note that what we are doing here provides a new insight into the isospin (or isotopic spin) formalism. The operator J+ that destroys one unit of the minus type and creates one unit of the plus type is completely analogous to the isospin ladder operator T+ (sometimes denoted by I+) that annihilates a neutron (isospin down) and creates a proton (isospin up), thus raising the 222 Theory of Angular Momentum z-component of isospin by one unit. In contrast, Jz is analogous to Tz, which simply counts the difference between the number of protons and neutrons in nuclei. 3.9.2 Explicit Formula for Rotation Matrices Schwinger\u2019s scheme can be used to derive, in a very simple way, a closed formula for rotation matrices, \ufb01rst obtained by E. P. Wigner using a similar (but not identical) method. We apply the rotation operator D(R ) to | j, m, written as (3.411). In the Euler angle notation the only nontrivial rotation is the second one about the y -axis, so we direct our attention to D(R) =D (\u03b1, \u03b2, \u03b3)|\u03b1=\u03b3=0 = exp \u2212iJy\u03b2 \u00afh We have D(R)| j, m = [D(R)a\u2020 + j+m D \u22121(R)] [D(R)a\u2020 ( j + m)! ( j \u2212 m)! \u2212D \u22121(R)] j\u2212m. (3.413) D(R)|0. (3.414) Now, D(R) acting on |0 just reproduces |0 because, by virtue of (3.399), only the leading term, 1, in the expansion of exponential (3.413) contributes. So \u2212iJy\u03b2 \u00afh a\u2020 \u00b1exp iJy\u03b2 \u00afh. (3.415) D(R)a\u2020 \u00b1D \u22121(R) =exp Thus we may use formula (2.168). Letting G \u2192 \u2212Jy \u00afh, \u03bb \u2192 \u03b2 in (2.168), we realize that we must look at various commutators, namely,, a\u2020 + \u2212Jy \u00afh \u2212Jy \u00afh, a\u2020 + = = 1 2i \u2212Jy \u00afh 1 2i [a\u2020 \u2212a+, a\u2020 +] = a\u2020 \u2212 2i, = 1 4 a\u2020 + \u2212Jy \u00afh, a\u2020 \u2212, + or a\u2020 and so forth. Clearly, we always obtain either a\u2020 \u03b2 \u2212. Collecting terms, we get D(R)a\u2020 + D \u22121(R) =a \u2020 + cos + a\u2020 \u2212 sin Likewise, D(R)a\u2020 \u2212D \u22121(R) =a \u2020 \u2212 cos \u2212 a\u2020 + sin 3.416) (3.417) (3.418) (3.419) Actually this result is not surprising. After all, the basic spin-up state is supposed", " to transform as a\u2020 + |0 \u2192cos a\u2020 + |0 + sin \u2212|0 a\u2020 (3.420) \u03b2 2 \u03b2 2 223 3.9 Schwinger\u2019s Oscillator Model of Angular Momentum under a rotation about the y-axis. Substituting (3.418) and (3.419) into (3.414) and recalling the binomial theorem (x + y)N = \u2211 k N! xN\u2212kyk (N \u2212 k)! k!, we obtain D(\u03b1 = 0, \u03b2, \u03b3 = 0| j, m = \u2211 \u2211 ( j + m)! ( j \u2212 m)! ( j + m \u2212 k)! k! ( j \u2212 m \u2212 l)! l! + cos(\u03b2/2)]j+m\u2212k[a\u2020 \u2212 sin(\u03b2/2)]k k l \u00d7 [a\u2020 ( j + m)! ( j \u2212 m)! \u00d7 [\u2212a\u2020 + sin(\u03b2/2)]j\u2212m\u2212l[a\u2020 \u2212 cos(\u03b2/2)]l|0. (3.421) (3.422) We may compare (3.422) with D(\u03b1 = 0, \u03b2, \u03b3 = 0)| j, m = \u2211 m | j, md( j) mm(\u03b2) = \u2211 m d( j) mm(\u03b2) \u2212)j\u2212m +)j+m (a\u2020 (a\u2020 ( j + m)! ( j \u2212 m)! |0. (3.423) We can obtain an explicit form for d( j) (3.422) and (3.423). Speci\ufb01cally, we want to compare a\u2020 + raised to 2j \u2212 k \u2212 l, so we identify a\u2020 mm(\u03b2) by equating the coef\ufb01cients of powers of a\u2020 + in + raised to j + m in (3.423) with l = j \u2212 k \u2212 m. (3.424) We are seeking dmm(\u03b2) with m \ufb01xed. The k-sum and the l-sum in (3.422) are not independent of each other; we eliminate l in favor of k by taking advantage of (3.424). \u2212 raised to j \u2212 m in (3.423) automatically matches As for the powers of a\u2020 with a\u2020 \u2212 raised to k + l in (3.422) when (3.424) is imposed. The last step is to identify the exponents of cos(\u03b2/2), sin(\u03b2/2), and (\u22121), which are, respectively, \u2212, we note that a\u2020 j + m \u2212 k + l = 2j \u2212 2k + m \u2212 m, k + j \u2212 m \u2212 l = 2k \u2212 m + m, (3.425a) (3.425b) (3.425c) where we have used (3.424) to eliminate l. In this way we obtain Wigner\u2019s formula for d( j) mm(\u03b2): mm(\u03b2) = \u2211 (\u22121)k\u2212m+m d( j) k ( j + m)! ( j \u2212 m)! ( j + m)! ( j \u2212 m)! ( j + m \u2212 k)! k! ( j \u2212 k \u2212 m)! (k \u2212 m + m)! 2k\u2212m+m 2j\u22122k+m\u2212m \u00d7 cos \u03b2 2 sin \u03b2 2, (3.426) where we take the sum over k whenever none of the arguments of factorials in the denominator is negative. 224 Theory of Angular Momentum 3.10 Spin Correlation Measurements and Bell\u2019s Inequality 3.10.1 Correlations in Spin-Singlet States The simplest example of angular-momentum addition we encountered in Section 3.8 was concerned with a composite system made up of spin 1 2 particles. In this section we use such a system to illustrate one of the most astonishing consequences of quantum mechanics. Consider a two-electron system in a spin-singlet state, that is, with a total spin of zero. We have already seen that the state ket can be written as [see (3.335d)] |spin singlet = (|\u02c6z+; \u02c6z\u2212 \u2212 |\u02c6z\u2212; \u02c6z+), (3.427) 1\u221a 2 where we have explicitly indicated the quantization direction. Recall that |\u02c6z+; \u02c6z\u2212 means that electron 1 is in the spin-up state and electron 2 is in the spin-down state. The same is true for |\u02c6z\u2212; \u02c6z+. Suppose we make a measurement on the spin component of one of the electrons. Clearly, there is a 50% chance of getting either up or down because the composite system may", " be in |\u02c6z+; \u02c6z\u2212 or |\u02c6z\u2212; \u02c6z+ with equal probabilities. But if one of the components is shown to be in the spin-up state, the other is necessarily in the spin-down state, and vice versa. When the spin component of electron 1 is shown to be up, the measurement apparatus has selected the \ufb01rst term, |\u02c6z+; \u02c6z\u2212 of (3.427); a subsequent measurement of the spin component of electron 2 must ascertain that the state ket of the composite system is given by |\u02c6z+; \u02c6z\u2212. It is remarkable that this kind of correlation can persist even if the two particles are well separated and have ceased to interact provided that as they \ufb02y apart, there is no change in their spin states. This is certainly the case for a J = 0 system disintegrating spontaneously into two spin 1 2 particles with no relative orbital angular momentum, because angularmomentum conservation must hold in the disintegration process. An example of this would be a rare decay of the \u03b7 meson (mass 549 MeV/c2) into a muon pair \u03b7 \u2192 \u03bc+ + \u03bc\u2212 (3.428) which, unfortunately, has a branching ratio of only approximately 6 \u00d7 10\u22126. More realistically, in proton-proton scattering at low kinetic energies, the Pauli principle to be discussed in Chapter 7 forces the interacting protons to be in 1S0 (orbital angular momentum 0, spinsinglet state), and the spin states of the scattered protons must be correlated in the manner indicated by (3.427) even after they become separated by a macroscopic distance. To be more pictorial we consider a system of two spin 1 2 particles moving in opposite directions, as in Figure 3.11. Observer A specializes in measuring Sz of particle 1 (\ufb02ying B Particle 2 Particle 1 A Fig. 3.11 Spin correlation in a spin-singlet state. 225 3.10 Spin Correlation Measurements and Bell\u2019s Inequality to the right), while observer B specializes in measuring Sz of particle 2 (\ufb02ying to the left). To be speci\ufb01c, let us assume that observer A \ufb01nds Sz to be positive for particle 1. Then he or she can predict, even before B performs any measurement, the outcome of B\u2019s measurement with certainty: B must \ufb01nd Sz to be negative for particle 2. On the other hand, if A makes no measurement, B has a 50% chance of getting Sz+ or Sz\u2212. This by itself might not be so peculiar. One may say, \u201cIt is just like an urn known to contain one black ball and one white ball. When we blindly pick one of them, there is a 50% chance of getting black or white. But if the \ufb01rst ball we pick is black, then we can predict with certainty that the second ball will be white.\u201d It turns out that this analogy is too simple. The actual quantum-mechanical situation is far more sophisticated than that! This is because observers may choose to measure Sx in place of Sz. The same pair of \u201cquantum-mechanical balls\u201d can be analyzed either in terms of black and white or in terms of blue and red! Recall now that for a single spin 1 2 system the Sx eigenkets and Sz eigenkets are related as follows: |\u02c6x\u00b1 = 1\u221a 2 (|\u02c6z+ \u00b1 |\u02c6z\u2212), |\u02c6z\u00b1 = 1\u221a 2 (|\u02c6x+ \u00b1 |\u02c6x\u2212). (3.429) Returning now to our composite system, we can rewrite spin-singlet ket (3.427) by choosing the x-direction as the axis of quantization: |spin singlet = 1\u221a 2 (|\u02c6x\u2212; \u02c6x+ \u2212 |\u02c6x+; \u02c6x\u2212). (3.430) Apart from the overall sign, which in any case is a matter of convention, we could have guessed this form directly from (3.427) because spin-singlet states have no preferred direction in space. Let us now suppose that observer A can choose to measure Sz or Sx of particle 1 by changing the orientation of his or her spin analyzer, while observer B always specializes in measuring Sx of particle 2. If A determines Sz of particle 1 to be positive, B clearly has a 50% chance for getting Sx+ or Sx\u2212; even though Sz of particle 2 is known to be negative with certainty, its Sx is completely undetermined. On the other hand, let us suppose that A also chooses to measure Sx; if observer A determines S", "x of particle 1 to be positive, then without fail, observer B will measure Sx of particle 2 to be negative. Finally, if A chooses to make no measurement, B, of course, will have a 50% chance of getting Sx+ or Sx\u2212. To sum up, we have the following. 1. If A measures Sz and B measures Sx, there is a completely random correlation between the two measurements. 2. If A measures Sx and B measures Sx, there is a 100% (opposite sign) correlation between the two measurements. 3. If A makes no measurement, B\u2019s measurements show random results. Table 3.1 shows all possible results of such measurements when B and A are allowed to choose to measure Sx or Sz. These considerations show that the outcome of B\u2019s measurement appears to depend on what kind of measurement A decides to perform: an Sx measurement, an Sz measurement, or no measurement. Notice again that A and B can be 226 Theory of Angular Momentum Table 3.1 Spin-Correlation Measurements Spin component measured by A A\u2019s result Spin component measured by B B\u2019s result \u2212 + \u2212 + \u2212 \u2212 + + + \u2212 + \u2212 miles apart with no possibility of communications or mutual interactions. Observer A can decide how to orient his or her spin-analyzer apparatus long after the two particles have separated. It is as though particle 2 \u201cknows\u201d which spin component of particle 1 is being measured. The orthodox quantum-mechanical interpretation of this situation is as follows. A measurement is a selection (or \ufb01ltration) process. When Sz of particle 1 is measured to be positive, then component |\u02c6z+; \u02c6z\u2212 is selected. A subsequent measurement of the other particle\u2019s Sz merely ascertains that the system is still in |\u02c6z+; \u02c6z\u2212. We must accept that a measurement on what appears to be a part of the system is to be regarded as a measurement on the whole system. 3.10.2 Einstein\u2019s Locality Principle and Bell\u2019s Inequality Many physicists have felt uncomfortable with the preceding orthodox interpretation of spin-correlation measurements. Their feelings are typi\ufb01ed in the following frequently quoted remarks by A. Einstein, which we call Einstein\u2019s locality principle: \u201cBut on one supposition we should, in my opinion, absolutely hold fast: The real factual situation of the system S2 is independent of what is done with the system S1, which is spatially separated from the former.\u201d Because this problem was \ufb01rst discussed in a 1935 paper of A. Einstein, B. Podolsky, and N. Rosen, it is sometimes known as the Einstein\u2013Podolsky\u2013 Rosen paradox.15 Some have argued that the dif\ufb01culties encountered here are inherent in the probabilistic interpretations of quantum mechanics and that the dynamic behavior at the microscopic level appears probabilistic only because some yet unknown parameters, so-called hidden 15 To be historically accurate, the original Einstein\u2013Podolsky\u2013Rosen paper dealt with measurements of x and p. 2 systems to illustrate the Einstein\u2013Podolsky\u2013Rosen paradox started with D. Bohm. The use of composite spin 1 227 3.10 Spin Correlation Measurements and Bell\u2019s Inequality variables, have not been speci\ufb01ed. It is not our purpose here to discuss various alternatives to quantum mechanics based on hidden-variable or other considerations. Rather, let us ask: Do such theories make predictions different from those of quantum mechanics? Until 1964, it could be thought that the alternative theories could be concocted in such a way that they would give no predictions, other than the usual quantum-mechanical predictions, that could be veri\ufb01ed experimentally. The whole debate would have belonged to the realm of metaphysics rather than physics. It was then pointed out by J. S. Bell that the alternative theories based on Einstein\u2019s locality principle actually predict a testable inequality relation among the observables of spin-correlation experiments that disagrees with the predictions of quantum mechanics. We derive Bell\u2019s inequality within the framework of a simple model, conceived by E. P. Wigner, that incorporates the essential features of the various alternative theories. Proponents of this model agree that it is impossible to determine Sx and Sz simultaneously. However, when we have a large number of spin 1 2 particles, we assign a certain fraction of them to have the following property. If Sz is measured, we obtain a plus sign with certainty. If Sx is measured, we obtain a minus sign with certainty. A particle satisfying this property is said to belong to type (\u02c6z+, \u02c6x\u2212). Notice that we are not asserting that we can simultaneously measure Sz and Sx to be + and", " \u2212, respectively. When we measure Sz, we do not measure Sx, and vice versa. We are assigning de\ufb01nite values of spin components in more than one direction with the understanding that only one or the other of the components can actually be measured. Even though this approach is fundamentally different from that of quantum mechanics, the quantummechanical predictions for Sz and Sx measurements performed on the spin-up (Sz+) state are reproduced provided there are as many particles belonging to type (\u02c6z+, \u02c6x+) as to type (\u02c6z+, \u02c6x\u2212). Let us now examine how this model can account for the results of spin-correlation measurements made on composite spin-singlet systems. Clearly, for a particular pair, there must be a perfect matching between particle 1 and particle 2 to ensure zero total angular momentum: If particle 1 is of type (\u02c6z+, \u02c6x\u2212), then particle 2 must belong to type (\u02c6z\u2212, \u02c6x+), and so forth. The results of correlation measurements, such as in Table 3.1, can be reproduced if particle 1 and particle 2 are matched as follows: particle 1 particle 2 (\u02c6z+, \u02c6x\u2212) \u2194 (\u02c6z\u2212, \u02c6x+), (\u02c6z+, \u02c6x+) \u2194 (\u02c6z\u2212, \u02c6x\u2212), (\u02c6z\u2212, \u02c6x+) \u2194 (\u02c6z+, \u02c6x\u2212), (\u02c6z\u2212, \u02c6x\u2212) \u2194 (\u02c6z+, \u02c6x+), (3.431a) (3.431b) (3.431c) (3.431d) with equal populations, that is, 25% each. A very important assumption is implied here. Suppose a particular pair belongs to type (3.431a) and observer A decides to measure Sz of 228 Theory of Angular Momentum Table 3.2 Spin-Component Matching in the Alternative Theories Population Particle 1 Particle 2 N1 N2 N3 N4 N5 N6 N7 N8 (\u02c6a+, \u02c6b+, \u02c6c+) (\u02c6a+, \u02c6b+, \u02c6c\u2212) (\u02c6a+, \u02c6b\u2212, \u02c6c+) (\u02c6a+, \u02c6b\u2212, \u02c6c\u2212) (\u02c6a\u2212, \u02c6b+, \u02c6c+) (\u02c6a\u2212, \u02c6b+, \u02c6c\u2212) (\u02c6a\u2212, \u02c6b\u2212, \u02c6c+) (\u02c6a\u2212, \u02c6b\u2212, \u02c6c\u2212) (\u02c6a\u2212, \u02c6b\u2212, \u02c6c\u2212) (\u02c6a\u2212, \u02c6b\u2212, \u02c6c+) (\u02c6a\u2212, \u02c6b+, \u02c6c\u2212) (\u02c6a\u2212, \u02c6b+, \u02c6c+) (\u02c6a+, \u02c6b\u2212, \u02c6c\u2212) (\u02c6a+, \u02c6b\u2212, \u02c6c+) (\u02c6a+, \u02c6b+, \u02c6c\u2212) (\u02c6a+, \u02c6b+, \u02c6c+) particle 1; then he or she necessarily obtains a plus sign regardless of whether B decides to measure Sz or Sx. It is in this sense that Einstein\u2019s locality principle is incorporated in this model: A\u2019s result is predetermined independently of B\u2019s choice as to what to measure. In the examples considered so far, this model has been successful in reproducing the predictions of quantum mechanics. We now consider more complicated situations where the model leads to predictions different from the usual quantum-mechanical predictions. This time we start with three unit vectors \u02c6a, \u02c6b, and \u02c6c, which are, in general, not mutually orthogonal. We imagine that one of the particles belongs to some de\ufb01nite type, say (\u02c6a\u2212, \u02c6b+, \u02c6c+), which means that if S \u00b7 \u02c6a is measured, we obtain a minus sign with certainty; if S \u00b7 \u02c6b is measured, we obtain a plus sign with certainty; if S \u00b7 \u02c6c is measured, we obtain a plus sign with certainty. Again there must be a perfect matching in the sense that the other particle necessarily belongs to type (\u02c6a+, \u02c6b\u2212, \u02c6c\u2212) to ensure zero total angular momentum. In any given event, the particle pair in question must be a member of one of the eight types shown in Table 3.", "2. These eight possibilities are mutually exclusive and disjoint. The population of each type is indicated in the \ufb01rst column. Let us suppose that observer A \ufb01nds S1 \u00b7 \u02c6a to be plus and observer B \ufb01nds S2 \u00b7 \u02c6b to be plus also. It is clear from Table 3.2 that the pair belong to either type 3 or type 4, so the number of particle pairs for which this situation is realized is N3 + N4. Because Ni is positive semide\ufb01nite, we must have inequality relations like N3 + N4 \u2264 (N2 + N4) + (N3 + N7). (3.432) Let P(\u02c6a+; \u02c6b+) be the probability that, in a random selection, observer A measures S1 \u00b7 \u02c6a to be + and observer B measures S2 \u00b7 \u02c6b to be +, and so on. Clearly, we have P(\u02c6a+; \u02c6b+) = (N3 + N4) \u22118 i Ni. In a similar manner, we obtain P(\u02c6a+; \u02c6c+) = (N2 + N4) \u22118 i Ni and P(\u02c6c+; \u02c6b+) = (N3 + N7) \u22118 i Ni. (3.433) (3.434) 229 3.10 Spin Correlation Measurements and Bell\u2019s Inequality The positivity condition (3.432) now becomes P(\u02c6a+; \u02c6b+) \u2264 P(\u02c6a+; \u02c6c+) + P(\u02c6c+; \u02c6b+). (3.435) This is Bell\u2019s inequality, which follows from Einstein\u2019s locality principle. 3.10.3 Quantum Mechanics and Bell\u2019s Inequality We now return to the world of quantum mechanics. In quantum mechanics we do not talk about a certain fraction of particle pairs, say N3/ \u22118 i Ni, belonging to type 3. Instead, we characterize all spin-singlet systems by the same ket (3.427); in the language of Section 3.4 we are concerned here with a pure ensemble. Using this ket and the rules of quantum mechanics we have developed, we can unambiguously calculate each of the three terms in inequality (3.435). We \ufb01rst evaluate P(\u02c6a+; \u02c6b+). Suppose observer A \ufb01nds S1 \u00b7 \u02c6a to be positive; because of the 100% (opposite sign) correlation we discussed earlier, B\u2019s measurement of S2 \u00b7 \u02c6a will yield a minus sign with certainty. But to calculate P(\u02c6a+; \u02c6b+) we must consider a new quantization axis \u02c6b that makes an angle \u03b8ab with \u02c6a; seeFigure 3.12. According to the formalism of Section 3.2, the probability that the S2 \u00b7 \u02c6b measurement yields + when particle 2 is known to be in an eigenket of S2 \u00b7 \u02c6a with negative eigenvalue is given by cos2 (\u03c0 \u2212 \u03b8ab) 2 = sin2 \u03b8ab 2 As a result, we obtain P(\u02c6a+; \u02c6b+) = sin2 1 2 \u03b8ab 2., (3.436) (3.437) 2 arises from the probability of initially obtaining S1 \u00b7 \u02c6a with +. Using where the factor 1 (3.437) and its generalization to the other two terms of (3.435), we can write Bell\u2019s inequality as \u02c6 b-direction \u02c6 a-direction S1 \u03b8ab S2 Fig. 3.12 Evaluation of P(\u02c6a+; \u02c6b+). 230 Theory of Angular Momentum \u03b8ab 2 sin2 \u2264 sin2 \u03b8ac 2 + sin2. \u03b8cb 2 (3.438) We now show that inequality (3.438) is not always possible from a geometric point of view. For simplicity let us choose \u02c6a, \u02c6b, and \u02c6c to lie in a plane, and let \u02c6c bisect the two directions de\ufb01ned by \u02c6a and \u02c6b: Inequality (3.438) is then violated for \u03b8ab = 2\u03b8, \u03b8ac = \u03b8cb = \u03b8. 0 < \u03b8 < \u03c0 2. For example, take \u03b8 = \u03c0/4; we then obtain 0.500 \u2264 0.292?? (3.439) (3.440) (3.441) So the", " quantum-mechanical predictions are not compatible with Bell\u2019s inequality. There is a real observable, in the sense of being experimentally veri\ufb01able, difference between quantum mechanics and the alternative theories satisfying Einstein\u2019s locality principle. Several experiments have been performed to test Bell\u2019s inequality. For a review, see \u201cBell\u2019s inequality test: more ideal than ever\u201d by Aspect, Nature, 398 (1999) 189. In one of the experiments spin correlations between the \ufb01nal protons in low-energy protonproton scattering were measured. In all other experiments photon polarization correlations between a pair of photons in a cascade transition of an excited atom (Ca, Hg,...), ( j = 0) \u03b3 \u2192( j = 1) \u03b3 \u2192( j = 0), (3.442) or in the decay of a positronium (an e+e\u2212 bound state in 1S0) were measured; studying photon polarization correlations should be just as good in view of the analogy developed in Section 1.1 Sz+ \u2192 \u02c6\u03b5 Sz\u2212 \u2192 \u02c6\u03b5 Sx+ \u2192 \u02c6\u03b5 Sx\u2212 \u2192 \u02c6\u03b5 in x-direction, in y-direction, in 45\u25e6 diagonal direction, in 135\u25e6 diagonal direction. (3.443a) (3.443b) (3.443c) (3.443d) The results of all recent precision experiments have conclusively established that Bell\u2019s inequality was violated, in one case by more than nine standard deviations. Furthermore, in all these experiments the inequality relation was violated in such a way that the quantummechanical predictions were ful\ufb01lled within error limits. In this controversy, quantum mechanics has triumphed with \ufb02ying colors. The fact that the quantum-mechanical predictions have been veri\ufb01ed does not mean that the whole subject is now a triviality. Despite the experimental verdict we may still feel psychologically uncomfortable about many aspects of measurements of this kind. Consider in particular the following point: Right after observer A performs a measurement on particle 1, how does particle 2 \u2013 which may, in principle, be many light years away from particle 1 \u2013 get to \u201cknow\u201d how to orient its spin so that the remarkable correlations apparent 231 3.11 Tensor Operators in Table 3.1 are realized? In one of the experiments to test Bell\u2019s inequality (performed by A. Aspect and collaborators) the analyzer settings were changed so rapidly that A\u2019s decision as to what to measure could not be made until it was too late for any kind of in\ufb02uence, traveling slower than light, to reach B. We conclude this section by showing that despite these peculiarities, we cannot use spin-correlation measurements to transmit any useful information between two macroscopically separated points. In particular, superluminal (faster than light) communications are impossible. Suppose A and B both agree in advance to measure Sz; then, without asking A, B knows precisely what A is getting. But this does not mean that A and B are communicating; B just observes a random sequence of positive and negative signs. There is obviously no useful information contained in it. B veri\ufb01es the remarkable correlations predicted by quantum mechanics only after he or she gets together with A and compares the notes (or computer sheets). It might be thought that A and B can communicate if one of them suddenly changes the orientation of his or her analyzing apparatus. Let us suppose that A agrees initially to measure Sz, and B, Sx. The results of A\u2019s measurements are completely uncorrelated with the results of B\u2019s measurements, so there is no information transferred. But then, suppose A suddenly breaks his or her promise and without telling B starts measuring Sx. There are now complete correlations between A\u2019s results and B\u2019s results. However, B has no way of inferring that A has changed the orientation of his or her analyzer. B continues to see just a random sequence of + and \u2212 by looking at his or her own notebook only. So again there is no information transferred. 3.11 Tensor Operators 3.11.1 Vector Operator We have been using notations such as x, p, S, andL, but as yet we have not systematically discussed their rotational properties. They are vector operators, but what are their properties under rotations? In this section we give a precise quantum-mechanical de\ufb01nition of vector operators based on their commutation relations with the angular-momentum operator. We then generalize to tensor operators with more complicated transformation properties and derive an important theorem on the matrix elements of vector and tensor operators. We all know that a vector in classical physics is a quantity with three components that transforms by de\ufffd", "\ufffd\ufffdnition like Vi \u2192 \u03a3jRijVj under a rotation. It is reasonable to demand that the expectation value of a vector operator V in quantum mechanics be transformed like a classical vector under rotation. Speci\ufb01cally, as the state ket is changed under rotation according to |\u03b1 \u2192D (R)|\u03b1, (3.444) 232 Theory of Angular Momentum the expectation value of V is assumed to change as follows: \u03b1|Vi|\u03b1 \u2192 \u03b1|D \u2020(R)ViD(R)|\u03b1 = \u2211 Rij\u03b1|Vj|\u03b1. j This must be true for an arbitrary ket |\u03b1. Therefore, D \u2020(R)ViD(R) = \u2211 RijVj j (3.445) (3.446) must hold as an operator equation, where Rij is the 3 \u00d7 3 matrix that corresponds to rotation R. Let us now consider a speci\ufb01c case, an in\ufb01nitesimal rotation. When the rotation is in\ufb01nitesimal, we have D(R) =1 \u2212 i\u03b5J \u00b7 \u02c6n \u00afh. We can now write (3.446) as Vi + \u03b5 i\u00afh [Vi, J \u00b7 \u02c6n] = \u2211 Rij( \u02c6n; \u03b5)Vj. j In particular, for \u02c6n along the z-axis, we have \u239b R(\u02c6z; \u03b5) = \u239d \u239e \u23a0, 1 \u2212\u03b5 \u03b5 0 1 0 0 1 so 0 i = 1, Vx + i = 2, Vy + i = 3, Vz + \u03b5 i\u00afh \u03b5 i\u00afh \u03b5 i\u00afh [Vx, Jz] =V x \u2212 \u03b5Vy [Vy, Jz] = \u03b5Vx + Vy [Vz, Jz] =V z. This means that V must satisfy the commutation relations [Vi, Jj] =i \u03b5ijk \u00afhVk. (3.447) (3.448) (3.449) (3.450a) (3.450b) (3.450c) (3.451) Clearly, the behavior of V under a \ufb01nite rotation is completely determined by the preceding commutation relations; we just apply the by now familiar formula (2.168) to iJj\u03c6 \u00afh \u2212iJj\u03c6 \u00afh Vi exp. (3.452) exp We simply need to calculate [Jj, [Jj, [\u00b7 \u00b7 \u00b7 [Jj, Vi] \u00b7 \u00b7 \u00b7 ]]]. Multiple commutators keep on giving back to us Vi or Vk (k = i, j) as in spin case (3.24). (3.453) We can use (3.451) as the de\ufb01ning property of a vector operator. Notice that the angularmomentum commutation relations are a special case of (3.451) in which we letV i \u2192 Ji, Vk \u2192 Jk. Other special cases are [y, Lz] =i \u00afhx, [x, Lz] =\u2212 i\u00afhy, [ px, Lz] =\u2212 i\u00afhpy, [ py, Lz] = i\u00afhpx; these can be proved explicitly. 233 3.11 Tensor Operators 3.11.2 Cartesian Tensors Versus Irreducible Tensors In classical physics it is customary to de\ufb01ne a tensor Tijk\u00b7\u00b7\u00b7 by generalizing Vi \u2192 \u03a3jRijVj as follows: Tijk\u00b7\u00b7\u00b7 \u2192 \u2211 i \u2211 j \u2211 k \u00b7 \u00b7 \u00b7 RiiRjj\u00b7\u00b7\u00b7Tijk\u00b7\u00b7\u00b7 (3.454) under a rotation speci\ufb01ed by the 3 \u00d7 3 orthogonal matrix R. The number of indices is called the rank of a tensor. Such a tensor is known as a Cartesian tensor. The simplest example of a Cartesian tensor of rank 2 is a dyadic formed out of two vectors U and V. One simply takes a Cartesian component of U and a Cartesian component of V and puts them together: Tij \u2261 UiVj. (3.455) Notice that we have nine components altogether. They obviously transform like (3.454) under rotation. The trouble with a Cartesian tensor like (3.455) is that it is reducible, that is, it can be decomposed into objects that transform differently under rotations. Speci\ufb01cally, for the dyadic in (3.455) we have UiVj = U \u00b7 V 3 \u03b4ij + (UiVj \u2212 UjVi) 2 +", " UiVj + UjVi 2 \u2212 U \u00b7 V 3 \u03b4ij. (3.456) The \ufb01rst term on the right-hand side, U \u00b7 V, is a scalar product invariant under rotation. The second is an antisymmetric tensor which can be written as a vector product \u03b5ijk(U \u00d7 V)k. There are altogether 3 independent components. The last is a 3 \u00d7 3 symmetric traceless tensor with 5 (= 6 \u2212 1, where 1 comes from the traceless condition) independent components. The number of independent components checks. (3.457) We note that the numbers appearing on the right-hand side of (3.457) are precisely the multiplicities of objects with angular momentum l = 0, l = 1, and l = 2, respectively. This suggests that the dyadic has been decomposed into tensors that can transform like spherical harmonics with l = 0, 1, and 2. In fact, (3.456) is the simplest nontrivial example to illustrate the reduction of a Cartesian tensor into irreducible spherical tensors. Before presenting the precise de\ufb01nition of a spherical tensor, we \ufb01rst give an example l (\u03b8, \u03c6). We have of a spherical tensor of rank k. Suppose we take a spherical harmonic Ym already seen that it can be written as Ym l ( \u02c6n), where the orientation of \u02c6n is characterized by \u03b8 and \u03c6. We now replace \u02c6n by some vector V. The result is that we have a spherical tensor of rank k (in place of l) with magnetic quantum number q (in place of m), namely q = Ym=q T(k) l=k (V). (3.458) 234 Theory of Angular Momentum Speci\ufb01cally, in the case k = 1 we take spherical harmonics with l = 1 and replace (z/r) = ( \u02c6n)z by Vz, and so on1 1 = \u2213 3 4\u03c0 cos \u03b8 = \u2019 x \u00b1 iy\u221a 3 4\u03c0 2r 3 4\u03c0 z r \u2192 T (1) \u2019 3 0 = 4\u03c0 Vz, \u2213 Vx \u00b1 iVy\u221a \u2192 T (1) \u00b11 = 2. (3.459) 3 4\u03c0 \u2019 Obviously this can be generalized for higher k, for example, \u2019 Y \u00b12 2 = 15 32\u03c0 (x \u00b1 iy)2 r2 \u2192 T (2) \u00b12 = 15 32\u03c0 (Vx \u00b1 iVy)2 (3.460) T (k) q are irreducible, just as Y (m) satisfactory than working with Cartesian tensors. l are. For this reason, working with spherical tensors is more To see the transformation of spherical tensors constructed in this manner, let us \ufb01rst review how Y m l transform under rotations. First, we have for the direction eigenket; | \u02c6n \u2192D (R)| \u02c6n \u2261 | \u02c6n, (3.461) which de\ufb01nes the rotated eigenket | \u02c6n. We wish to examine how Ym look in terms of Ym l ( \u02c6n) = \u02c6n|l, m would l ( \u02c6n). We can easily see this by starting with mm(R\u22121) |l, mD ( l ) D(R\u22121)|l, m = \u2211 m and contracting with \u02c6n| on the left, using (3.461): l ( \u02c6n) = \u2211 Ym m Ym l ( \u02c6n)D ( l ) mm(R\u22121). If there is an operator that acts like Ym D \u2020(R)Ym l (V), it is then reasonable to expect l (V)D (l)\u2217 Ym mm(R), l (V)D(R) = \u2211 m (3.462) (3.463) (3.464) where we have used the unitarity of the rotation operator to rewrite D ( l ) mm(R\u22121). All this work is just to motivate the de\ufb01nition of a spherical tensor. We now consider spherical tensors in quantum mechanics. Motivated by (3.464) we de\ufb01ne a spherical tensor operator of rank k with (2k + 1) components as or, equivalently, D \u2020(R)T(k) q D(R) = k \u2211 q=\u2212k D (k)\u2217 qq (R)T(k) q D(R)T(", "k) q D \u2020(R) = k \u2211 q=\u2212k D (k) qq(R)T(k) q. (3.465a) (3.465b) This de\ufb01nition holds regardless of whether T(k) l=k (V); for example, (Ux + iUy)(Vx + iVy) is the q = +2 component of a spherical tensor of rank 2 even though, unlike (Vx + iVy)2, it cannot be written as Yq k(V). q can be written as Ym=q 235 3.11 Tensor Operators A more convenient de\ufb01nition of a spherical tensor is obtained by considering the in\ufb01nitesimal form of (3.465b), namely, 1 \u2212 iJ \u00b7 \u02c6n\u03b5 \u00afh T(k) q 1 + iJ \u00b7 \u02c6n\u03b5 \u00afh = k \u2211 q=\u2212k q kq| T(k) 1 \u2212 iJ \u00b7 \u02c6n\u03b5 \u00afh |kq or [J \u00b7 \u02c6n, T(k) q ] = \u2211 q q kq|J \u00b7 \u02c6n|kq. T(k) (3.466) (3.467) By taking \u02c6n in the \u02c6z- and in the (\u02c6x \u00b1 i\u02c6y) directions and using the nonvanishing matrix elements of Jz and J\u00b1 [see (3.187b) and (3.193)], we obtain and [Jz, T(k) q ] = \u00afhqT(k) q [J\u00b1, T(k) q ] = \u00afh (k \u2213 q)(k \u00b1 q + 1)T(k) q\u00b11. (3.468a) (3.468b) These commutation relations can be considered as a de\ufb01nition of spherical tensors in place of (3.465). 3.11.3 Product of Tensors We have seen how to form a scalar, vector (or antisymmetric tensor), and a traceless symmetric tensor out of two vectors using the Cartesian tensor language. Of course, spherical tensor language can also be used (Baym (1969), Chapter 17), for example, (U+1V\u22121 + U\u22121V+1 \u2212 U0V0) 3, T (0) 0 = T (1) q = = \u2212U \u00b7 V 3 (U \u00d7 V)q \u221a 2 i, T (2) \u00b12 = U\u00b11V\u00b11, T (2) \u00b11 = T (2) 0 = U\u00b11V0 + U0V\u00b11 \u221a 2, U+1V\u22121 + 2U0V0 + U\u22121V+1 \u221a 6, (3.469) where Uq(Vq) is the qth component of a spherical tensor of rank 1, corresponding to vector U(V). The preceding transformation properties can be checked by comparing with Ym l and remembering that U+1 = \u2212(Ux + iUy)/ 2, U0 = Uz. A similar check can be made for V\u00b11,0. For instance, 2, U\u22121 = (Ux \u2212 iUy)/ \u221a \u221a \u2019 Y0 2 = 5 16\u03c0 3z2 \u2212 r2 r2, where 3z2 \u2212 r2 can be written as 2z2 + 2 hence, Y0 2 is just a special case of T (2) 0 \u2212 (x + iy)\u221a 2 ; (x \u2212 iy)\u221a 2 for U = V = r. 236 Theory of Angular Momentum A more systematic way of forming tensor products goes as follows. We start by stating a theorem. Theorem 5 respectively. Then Let X (k1) q1 and Z (k2) q2 be irreducible spherical tensors of rank k1 and k2, q = \u2211 T (k) q1 k1k2; q1q2|k1k2; kqX(k1) q1 \u2211 q2 Z(k2) q2 (3.470) is a spherical (irreducible) tensor of rank k. Proof We must show that under rotation T(k) q must transform according to (3.465) D \u2020(R)T (k) \u2211 q2 D(R) k1k2; q1q2|k1k2; kq q D(R) = \u2211 q1 \u00d7D \u2020(R)X (k1) D(R)D \u2020(R)Z (k2", ") q1 q2 = \u2211 \u2211 \u2211 \u2211 k1k2; q1q2|k1k2; kq q q q1 q2 2 1 D (k1) \u00d7X (k1) (R\u22121)Z (k2) q q q 1q1 1 2 = \u2211 \u2211 \u2211 \u2211 \u2211 \u2211 k1k2; q1q2|k1k2; kq q q k q q1 q2 1 2 \u00d7k1k2; q |k1k2; kq 1q \u00d7k1k2; q1q2|k1k2; kqD (k) D (k2) q 2q2 (R\u22121) \u2211 q qq (R\u22121)X(k1) q 1 2 Z(k2) q 2, where we have used the Clebsch\u2013Gordan series formula (3.389). The preceding expression becomes = \u2211 \u03b4kk \u03b4qq k1k2; q 1q 2 |k1k2; kqD (k) qq (R\u22121)X(k1) q 1 Z(k2) q 2, where we have used the orthogonality of Clebsch\u2013Gordan coef\ufb01cients (3.362). Finally, this expression reduces to \u239b \u239e |k1k2; kqX(k1) q 1 Z(k2) q 2 \u23a0 D (k) qq(R\u22121) = \u2211 q = \u2211 q 2 1q k1k2; q \u239d\u2211 \u2211 q q 1 2 qq(R\u22121) = \u2211 q D (k) T(k) q D (k)\u2217 qq (R)T(k) q. The foregoing shows how we can construct tensor operators of higher or lower ranks by multiplying two tensor operators. Furthermore, the manner in which we construct tensor products out of two tensors is completely analogous to the manner in which we construct an angular-momentum eigenstate by adding two angular momenta; exactly the same Clebsch\u2013 Gordan coef\ufb01cients appear if we let k1,2 \u2192 j1,2 and q1,2 \u2192 m1,2. 3.11.4 Matrix Elements of Tensor Operators; the Wigner\u2013Eckart Theorem In considering the interactions of an electromagnetic \ufb01eld with atoms and nuclei, it to is often necessary to evaluate matrix elements of tensor operators with respect 237 3.11 Tensor Operators angular-momentum eigenstates. Examples of this will be given in Chapter 5. In general, it is a formidable dynamic task to calculate such matrix elements. However, there are certain properties of these matrix elements that follow purely from kinematic or geometric considerations, which we now discuss. First, there is a very simple m-selection rule: \u03b1, jm|T(k) q |\u03b1, jm = 0, unless m = q + m. (3.471) Proof. Using (3.468a), we have $ # \u03b1, jm| Jz, T(k) q \u2212 \u00afhqT(k) q |\u03b1, jm = [(m \u2212 m)\u00afh \u2212 \u00afhq] \u00d7 \u03b1, jm|T(k) q |\u03b1, jm = 0; hence, \u03b1, jm|T(k) q |\u03b1, jm = 0, unless m = q + m. Another way to see this is to note the transformation property of T(k) namely, q |\u03b1, jm under rotation, DT(k) q |\u03b1, jm = DT(k) q D \u2020D|\u03b1, jm. (3.472) If we now let D stand for a rotation operator around the z-axis, we get [see (3.465b) and (3.16)] D(\u02c6z, \u03c6)T(k) q |\u03b1, jm = e\u2212iq\u03c6e\u2212im\u03c6T(k) q |\u03b1, jm, (3.473) which is orthogonal to |\u03b1, jm unless q + m = m. We are going to prove one of the most important theorems in quantum mechanics, the Wigner\u2013Eckart theorem. The Wigner\u2013Eckart Theorem The matrix elements of tensor operators with respect to angular-momentum eigenstates satisfy \u03b1, jm|T(k) q |\u03b1, jm = jk; mq| jk; jm \u03b1j||T(k)||\u03b1j 2j + 1 \u221a,", " (3.474) where the double-bar matrix element is independent of m and m, and q. Before we present a proof of this theorem, let us look at its signi\ufb01cance. First, we see that the matrix element is written as the product of two factors. The \ufb01rst factor is a Clebsch\u2013Gordan coef\ufb01cient for adding j and k to get j. It depends only on the geometry, that is, the way the system is oriented with respect to the z-axis. There is no reference whatsoever to the particular nature of the tensor operator. The second factor does depend on the dynamics, for instance, \u03b1 may stand for the radial quantum number and its evaluation may evolve, for example, evaluation of radial integrals. On the other hand, it is completely independent of the magnetic quantum numbers m, m, and q, which specify the orientation of the physical system. To evaluate \u03b1, jm|T(k) q |\u03b1, jm with various combinations of m, m, andq it is suf\ufb01cient to know just one of them; all others can be related geometrically because they are proportional to Clebsch\u2013Gordan coef\ufb01cients, which 238 Theory of Angular Momentum are known. The common proportionality factor is \u03b1j||T(k)||\u03b1j, which makes no reference whatsoever to the geometric features. The selection rules for the tensor operator matrix element can be immediately read off from the selection rules for adding angular momentum. Indeed, from the requirement that the Clebsch\u2013Gordan coef\ufb01cient be nonvanishing, we immediately obtain the m-selection rule (3.471) derived before and also the triangular relation | j \u2212 k| \u2264 j \u2264 j + k. (3.475) Now we prove the theorem. Proof Using (3.468b) we have \u03b1, jm|[J\u00b1, T(k) q ]|\u03b1, jm = \u00afh or using (3.191) and (3.192) we have (k \u2213 q)(k \u00b1 q + 1)\u03b1, jm|T(k) q\u00b11 |\u03b1, jm, (3.476) = ( j \u00b1 m)( j \u2213 m + 1)\u03b1, j, m \u2213 1|T(k) q ( j \u2213 m)( j \u00b1 m + 1)\u03b1, jm|T(k) q + (k \u2213 q)(k \u00b1 q + 1)\u03b1, jm|T(k) q\u00b11 |\u03b1, jm |\u03b1, j, m \u00b1 1 |\u03b1, jm. (3.477) Compare this with the recursion relation for the Clebsch\u2013Gordan coef\ufb01cient (3.369). Note the striking similarity if we substitute j \u2192 j, m \u2192 m, j \u2192 j1, m \u2192 m1, k \u2192 j2, andq \u2192 m2. Both recursion relations are of the form \u2211j aijxj = 0, that is, \ufb01rst-order linear homogeneous equations with the same coef\ufb01cients aij. Whenever we have \u2211 aijxj = 0, \u2211 aijyj = 0, j j we cannot solve for the xj (or yj) individually but we can solve for the ratios; so xj xk = yj yk or xj = cyj, (3.478) (3.479) where c is a universal proportionality factor. Noting that j1 j2; m1, m2 \u00b1 1| j1 j2; jm in the Clebsch\u2013Gordan recursion relation (3.369) corresponds to \u03b1, jm|T(k) |\u03b1, jm, we see that q\u00b11 \u03b1, jm|T(k) q\u00b11 |\u03b1, jm = (universal proportionality constant independent of m, q, andm ) jk; m q \u00b1 1| jk; jm, which proves the theorem. (3.480) There are several conventions for the \u201cuniversal proportionality constant.\u201d Our choice (3.474) is consistent with Gottfried and Yan (2003). Edmonds (1974) uses a different, popular convention, and also includes a guide to other choices. Let us now look at two simple examples of the Wigner\u2013Eckart theorem. 239 3.11 Tensor Operators Example 6 Tensor of rank 0, that is, scalar T(0) satis\ufb01es 0 = S. The matrix element of a scalar operator \u03b1, jm|S|\u03b1, jm = \u03b4jj \u03b4mm \u03b1j||S||\u03b1j \u221a 2j + 1", " (3.481) because S acting on |\u03b1, jm is like adding an angular momentum of zero. Thus the scalar operator cannot change j, m values. Example 7 Vector operator which in the spherical tensor language is a rank 1 tensor. The spherical component of V can be written as Vq=\u00b11,0, so we have the selection rule \u0394m \u2261 m \u2212 m = \u00b11, 0, \u0394j \u2261 j \u2212 j = \u00b11 0. (3.482) In addition, the 0 \u2192 0 transition is forbidden. This selection rule is of fundamental importance in the theory of radiation; it is the dipole selection rule obtained in the longwavelength limit of emitted photons. For j = j the Wigner\u2013Eckart theorem, when applied to the vector operator, takes a particularly simple form, often known as the projection theorem for obvious reasons. The Projection Theorem This theorem states that \u03b1, jm|Vq|\u03b1, jm = \u03b1, jm|J \u00b7 V|\u03b1, jm \u00afh2j( j + 1) jm|Jq| jm, (3.483) where analogous to our discussion after (3.469) we choose J\u00b11 = \u2213 1\u221a 2 (Jx \u00b1 iJy) =\u2213 1\u221a 2 J\u00b1, J0 = Jz. (3.484) Proof Noting (3.469) we have \u03b1, jm|J \u00b7 V|\u03b1, jm = \u03b1, jm|(J0V0 \u2212 J+1V\u22121 \u2212 J\u22121V+1)|\u03b1, jm = m\u00afh\u03b1, jm|V0|\u03b1, jm + ( j + m)( j \u2212 m + 1) \u00afh\u221a 2 \u00d7 \u03b1, jm \u2212 1|V\u22121|\u03b1, jm ( j \u2212 m)( j + m + 1)\u03b1, jm + 1|V+1|\u03b1, jm \u2212 \u00afh\u221a 2 = cjm\u03b1j||V||\u03b1j (3.485) by the Wigner\u2013Eckart theorem (3.474), where cjm is independent of \u03b1, \u03b1, and V, and the matrix elements of V0,\u00b11 are all proportional to the double-bar matrix element (sometimes also called the reduced matrix element). Furthermore, cjm is independent of m because J \u00b7 V is a scalar operator, so we may as well write it as cj. Because cj does not depend on V, (3.485) holds even if we let V \u2192 J and \u03b1 \u2192 \u03b1, that is, \u03b1, jm|J2|\u03b1, jm = cj\u03b1j||J||\u03b1j. (3.486) 240 Theory of Angular Momentum Returning to the Wigner\u2013Eckart theorem applied to Vq and Jq, we have \u03b1, jm|Vq|\u03b1, jm \u03b1, jm|Jq|\u03b1, jm = \u03b1j||V||\u03b1j \u03b1j||J||\u03b1j. (3.487) But the right-hand side of (3.487) is the same as \u03b1, jm|J \u00b7 V|\u03b1, jm/\u03b1, jm|J2|\u03b1, jm by (3.485) and (3.486). Moreover, the left-hand side of (3.486) is just j( j + 1)\u00afh2. So \u03b1, jm|Vq|\u03b1, jm = \u03b1, jm|J \u00b7 V|\u03b1, jm \u00afh2j( j + 1) jm|Jq| jm, which proves the projection theorem. We will give applications of the theorem in subsequent sections. (3.488) Problems 3.2 3.3 3.1 Use the speci\ufb01c form of Sx given by (3.25) to evaluate (3.23) and show that Sx rotates as expected through an angle \u03c6 about the z-axis. Find the eigenvalues and eigenvectors of \u03c3y = 0 \u2212i 0 i. Suppose an electron is in the spin state. Ifs y is measured, what is the probability of the result \u00afh/2? \u03b1 \u03b2 Find, by explicit construction using Pauli matrices, Hamiltonian the eigenvalues for the H = \u2212 2\u03bc \u00afh 2 particle in the presence of a magnetic \ufb01eld B = Bx \u02c6x + By \u02c6y + Bz \u02c6z. S \u00b7 B for a spin 1 3.4 Consider the 2 \u00d7 2 matrix de\ufb01ned by U = a0 + i\u03c3 \u00b7 a a0 \u2212 i\u03c3 \u00b7 a, where a0 is a real number and a", " is a three-dimensional vector with real components. a. Prove that U is unitary and unimodular. b. In general, a 2 \u00d7 2 unitary unimodular matrix represents a rotation in three dimensions. Find the axis and angle of rotation appropriate for U in terms of a0, a1, a2, anda 3. 3.5 The spin-dependent Hamiltonian of an electron-positron system in the presence of a uniform magnetic \ufb01eld in the z-direction can be written as H = AS(e\u2212) \u00b7 S(e+) + eB mc S(e\u2212) z \u2212 S(e+) z. Suppose the spin function of the system is given by \u03c7(e\u2212) + \u03c7(e+) \u2212. 241 Problems a. Is this an eigenfunction of H in the limit A \u2192 0, eB/mc = 0? If it is, what is the energy eigenvalue? If it is not, what is the expectation value of H? b. Same problem when eB/mc \u2192 0, A = 0. 3.6 Consider a spin 1 particle. Evaluate the matrix elements of Sz(Sz + \u00afh)(Sz \u2212 \u00afh) and Sx(Sx + \u00afh)(Sx \u2212 \u00afh). 3.7 Let the Hamiltonian of a rigid body be H = 1 2, + K2 2 I2 + K2 3 I3 K2 1 I1 where K is the angular momentum in the body frame. From this expression obtain the Heisenberg equation of motion for K and then \ufb01nd Euler\u2019s equation of motion in the correspondence limit. 3.8 Let U = eiG3 \u03b1eiG2 \u03b2eiG3 \u03b3, where (\u03b1, \u03b2, \u03b3) are the Eulerian angles. In order that U represent a rotation (\u03b1, \u03b2, \u03b3), what are the commutation rules satis\ufb01ed by the Gk? Relate G to the angular-momentum operators. 3.9 What is the meaning of the following equation: U\u22121AkU = \u2211 RklAl, where the three components of A are matrices? From this equation show that matrix elements m|Ak|n transform like vectors. 3.10 Consider a sequence of Euler rotations represented by D (1/2)(\u03b1, \u03b2, \u03b3) = exp \u2212i\u03c33\u03b1 2 exp \u2212i\u03c32\u03b2 2 exp \u03b2 e\u2212i(\u03b1+\u03b3)/2 cos 2 \u03b2 ei(\u03b1\u2212\u03b3)/2 sin 2 \u2212e\u2212i(\u03b1\u2212\u03b3)/2 sin \u03b2 ei(\u03b1+\u03b3)/2 cos 2 \u2212i\u03c33\u03b3 2 \u239e \u03b2 2 \u239f \u23a0. \u239b \u239c \u239d = Because of the group properties of rotations, we expect that this sequence of operations is equivalent to a single rotation about some axis by an angle \u03b8. Find \u03b8. 3.11 Use the triangle inequality (1.147) and the de\ufb01nition (3.100) of the density operator \u03c1 to prove that 0 \u2264 Tr(\u03c12) \u2264 1. 3.12 A large collection of spin 1 2 particles is in a mixture of the two states |Sz; + and |Sy; \u2212. The fraction of particles in the state |Sz; + is a. Find the ensemble averages [Sx], [Sy], and [Sz] in terms of a. Con\ufb01rm that your expression gives the answers you expect for a = 0 and a = 1. 3.13 a. Consider a pure ensemble of identically prepared spin 1 2 systems. Suppose the expectation values Sx and Sz and the sign of Sy are known. Show how we may determine the state vector. Why is it unnecessary to know the magnitude of Sy? 242 Theory of Angular Momentum b. Consider a mixed ensemble of spin 1 2 systems. Suppose the ensemble averages [Sx], [Sy], and [Sz] are all known. Show how we may construct the 2 \u00d7 2 density matrix that characterizes the ensemble. 3.14 Consider a one-dimensional simple harmonic oscillator with frequency \u03c9 and eigenstates |0, |1, |2,.... A mixed ensemble is formed with equal parts of each of the three states |\u03b1 \u2261 1\u221a 2 |\u03b2 \u2261 1\u221a 2 [|0 + |1], [|1 + |2], |2. and Find the density operator \u03c1 and calculate the ensemble average of the energy. 3.15 a. Prove that the time evolution of the density operator \u03c1 (in the Schr\u00a8odinger picture) is given by \u03c1(t) =U (t, t0)\u03c1(t0)U \u2020(t, t0).", " b. Suppose we have a pure ensemble at t = 0. Prove that it cannot evolve into a mixed ensemble as long as the time evolution is governed by the Schr\u00a8odinger equation. 3.16 Consider an ensemble of spin 1 systems. The density matrix is now a 3 \u00d7 3 matrix. How many independent (real) parameters are needed to characterize the density matrix? What must we know in addition to [Sx], [Sy], and [Sz] to characterize the ensemble completely? 3.17 An angular-momentum eigenstate | j, m = mmax = j is rotated by an in\ufb01nitesimal angle \u03b5 about the y-axis. Without using the explicit form of the d( j) mm function, obtain an expression for the probability for the new rotated state to be found in the original state up to terms of order \u03b52. 3.18 Show that the 3 \u00d7 3 matrices Gi(i = 1, 2, 3) whose elements are given by (Gi)jk = \u2212i\u00afh\u03b5ijk, where j and k are the row and column indices, satisfy the angular-momentum commutation relations. What is the physical (or geometric) signi\ufb01cance of the transformation matrix that connects Gi to the more usual 3 \u00d7 3 representations of the angular-momentum operator Ji with J3 taken to be diagonal? Relate your result to V \u2192 V + \u02c6n\u03b4\u03c6 \u00d7 V under in\ufb01nitesimal rotations. (Note: This problem may be helpful in understanding the photon spin.) 3.19 a. Using the fact that Jx, Jy, Jz, and J\u00b1 \u2261 Jx \u00b1iJy satisfy the usual angular-momentum commutation relations, prove that J2 = J 2 z + J+J\u2212 \u2212 \u00afhJz. 243 Problems b. Using this result, or otherwise, derive the coef\ufb01cient c\u2212 that appears in J\u2212| jm = c\u2212| j, m \u2212 1. 3.20 Construct the matrix representations of the operators Jx and Jy for a spin 1 system, in the Jz basis, spanned by the kets |+ \u2261 |1, 1, |0 \u2261 |1, 0, and|\u2212 \u2261 |1, \u22121. Use these matrices to \ufb01nd the three analogous eigenstates for each of the two operators Jx and Jy in terms of |+, |0, and |\u2212. 3.21 Show that the orbital angular-momentum operator L commutes both with the operators p2 and x2, that is prove (3.264). 3.22 For the orbital angular-momentum operator L = x \u00d7 p, derive (3.218), that is x|Lz|\u03b1 = \u2212i\u00afh \u2202 \u2202\u03c6 x|\u03b1 by using the standard spherical coordinate transformation x = r cos \u03c6 sin \u03b8, y = r sin \u03c6 sin \u03b8, z = r cos \u03b8. 3.23 The wave function of a particle subjected to a spherically symmetrical potential V(r) is given by \u03c8(x) = (x + y + 3z)f (r). a. Is \u03c8 an eigenfunction of L2? If so, what is the l-value? If not, what are the possible values of l we may obtain when L2 is measured? b. What are the probabilities for the particle to be found in various ml states? c. Suppose it is known somehow that \u03c8(x) is an energy eigenfunction with eigenvalue E. Indicate how we may \ufb01nd V(r). 3.24 A particle in a spherically symmetrical potential is known to be in an eigenstate of L2 and Lz with eigenvalues \u00afh2l(l + 1) and m\u00afh, respectively. Prove that the expectation values between |lm states satisfy Lx = Ly = 0, L2 x = L2 y = [l(l + 1)\u00afh2 \u2212 m2 \u00afh2] 2. Interpret this result semiclassically. 3.25 Suppose a half-integer l-value, say 1 2, were allowed for orbital angular momentum. From we may deduce, as usual, L+Y1/2 1/2(\u03b8, \u03c6) =0, 1/2(\u03b8, \u03c6) \u221d ei\u03c6/2 Y1/2 \u221a sin \u03b8. 1/2(\u03b8, \u03c6) and (b) using 1/2 (\u03b8, \u03c6) (a) by applying L\u2212 to Y1/2 1/2 (\u03b8, \u03c6) = 0. Show that the two procedures lead to contradictory results", ". (This Now try to construct Y\u22121/2 L\u2212Y\u22121/2 gives an argument against half-integer l-values for orbital angular momentum.) 244 Theory of Angular Momentum 3.26 Consider an orbital angular-momentum eigenstate |l = 2, m = 0. Suppose this state is rotated by an angle \u03b2 about the y-axis. Find the probability for the new state to be found in m = 0, \u00b11, and \u00b12. (The spherical harmonics for l = 0, 1, and 2 given in Section B.5 in Appendix B may be useful.) 3.27 Show that by keeping both terms in (3.275), the origin becomes a source of probability, provided that there is a relative phase between the constants A and B. 3.28 Consider the energy eigenvalues for a spherically symmetric \u201cbox\u201d of radius a. a. For the box with in\ufb01nite walls, check the eigenvalues for the l = 0, l = 1, and l = 2 states, given in (3.287), (3.288), and (3.289). b. Find the lowest energy eigenvalues with l = 0 for a \ufb01nite spherical box with potential wall height V0 = \u00afh2\u03b22/2ma2 where \u03b2 = 4, 10, 25, and 100, and show that your numerical results approach the appropriate value given in (a). 3.29 The goal of this problem is to determine degenerate eigenstates of the threedimensional isotropic harmonic oscillator written as eigenstates of L2 and Lz, in terms of the Cartesian eigenstates |nxnynz. a. Show that the angular-momentum operators are given by Li = i\u00afh\u03b5ijkaja\u2020 k # $ L2 = \u00afh2 N(N + 1) \u2212 a\u2020 ka\u2020 kajaj where summation is implied over repeated indices, \u03b5ijk is the totally antisymmetric symbol, and N \u2261 a\u2020 j aj counts the total number of quanta. b. Use these relations to express the states |qlm = |01m, m = 0, \u00b11, in terms of the three eigenstates |nxnynz that are degenerate in energy. Write down the representation of your answer in coordinate space, and check that the angular and radial dependences are correct. c. Repeat for |qlm = |100. d. Repeat for |qlm = |02m, with m = 0, 1, and 2. 3.30 By considering the associated Laguerre polynomials Lq p(x), which are also solutions to Kummer\u2019s equation (3.308), we can derive the normalization constant in (3.320). a. First consider the Laguerre polynomials Lp(x), which are de\ufb01ned according to a generating function as g(x, t) = e\u2212xt/(1\u2212t) 1 \u2212 t = \u221e \u2211 p=0 Lp(x) t p p! where 0 < t < 1. Prove that Ln(0) =n! andL 0(x) = 1. Differentiate g(x, t) with respect to x to show that p(x) \u2212 pL L p\u22121(x) =\u2212 pLp\u22121(x). Then differentiate g(x, t) with respect to t to show that Lp+1(x) \u2212 (2p + 1 \u2212 x)Lp(x) +p 2Lp\u22121(x) =0. 245 Problems Combine these equations to derive the differential equation for the Lp(x), namely p(x) + (1 \u2212 x)L b. The associated laguerre Polynomials Lq xL p(x) +pL p(x) =0. p(x) are de\ufb01ned from the Lp(x) as p(x) = (\u22121)q dq Lq dxq [Lp+q(x)]. Use this to show that the Lq p(x) satisfy the differential equation xLq p (x) + (q + 1 \u2212 x)Lq p (x) +pL q p(x) =0 which is the same as (3.308) where q = c \u2212 1 and p = \u2212a. Also show that h(x, t) = (\u22121)q (1 \u2212 t)q+1 e\u2212xt/(1\u2212t) = \u221e \u2211 p=0 Lq p(x) tp (p + q)! which gives Lq p(0) = (\u22121)q[(p + q)! ]2/p! q!", " through a generating function h(x, t). c. Now \ufb01nd the normalization constant in (3.320). Start with the relationship \u221e xq+1e\u2212xh(x, t)h(x, s) dx = \u221e \u2211 p=0 \u221e \u2211 p=0 tp (p + q)! sp (p + q)! Iq pp 0 \u221e 0 xq+1e\u2212xLq where Iq pp by isolating the terms with the same powers of s and t. Make use of the generalized binomial expansion p(x) dx, and determine Iq p(x)Lq pp \u2261 1 (1 \u2212 x)n = \u221e \u2211 p=0 n + p \u2212 1 p xp = \u221e \u2211 p=0 (n + p \u2212 1)! p! (n \u2212 1)! xp. 3.31 Consider the Coulomb potential V(x) =\u2212 Ze2/r and de\ufb01ne the (quantum-mechanical operator analogue of the) Runge\u2013Lenz vector M = 1 2m (p \u00d7 L \u2212 L \u00d7 p) \u2212 Ze2 r x. Prove that M is Hermitian and that it commutes with the Hamiltonian. We will return to M when we go through Pauli\u2019s algebraic solution for this Hamiltonian in Section 4.1.4. 3.32 What is the physical signi\ufb01cance of the operators K+ \u2261 a\u2020 +a\u2020 \u2212 and K\u2212 \u2261 a+a\u2212 in Schwinger\u2019s scheme for angular momentum? Give the nonvanishing matrix elements of K\u00b1. 3.33 Carry through the argument outlined on p. 208 for adding two spin 1 2 particles by diagonalizing the 4 \u00d7 4 matrix corresponding to the operator S2 given in (3.339). That is, construct the matrix representation of S2 in the | \u00b1 \u00b1 basis, and \ufb01nd the eigenvalues and eigenvectors. Your result should agree with (3.335). 246 Theory of Angular Momentum 3.34 Find all nine states | j, m for j = 2, 1, and 0 formed by adding j1 = 1 and j2 = 1. Use a simpli\ufb01ed notation, where | j, m is explicit and \u00b1, 0 stand for m1,2 = \u00b11, 0, respectively, for example |1, 1 = 1\u221a 2 | + 0 \u2212 1\u221a 2 |0+. You may also want to make use of the ladder operators J\u00b1, or recursion relations, as well as orthonormality. Check your answers by \ufb01nding a table of Clebsch\u2013Gordan coef\ufb01cients for comparison; see Appendix E. 3.35 A spin 1 2 particle is in an orbital angular momentum l = 1 state. a. Starting with the total angular momentum state | j, m = | j, j for j = 3/2, use the operator J = L + S to construct all the states | j, mj in terms of the eigenstates |l, ml for L2 and Lz, and the eigenstates |1/2, \u00b11/2 for S2 and Sz. Check your answers against any available table of Clebsch\u2013Gordan coef\ufb01cients; see Appendix E. b. If the particle is in a total angular-momentum eigenstate with z-component +\u00afh/2, calculate the probability of \ufb01nding the z-component of the spin of the particle to have the value ms = +1/2. 3.36 The \u201cspin-angular functions\u201d (aka \u201cspinor spherical harmonics\u201d) are de\ufb01ned as \u23a1 \u23a4 Y j=l\u00b11/2,m l = 1\u221a 2l + Ym\u22121/2 2 Ym+1/2 l (\u03b8, \u03c6) (\u03b8, \u03c6) \u23a6. See (3.384). These were constructed to be properly normalized eigenfunctions of L2, S2, J2, andJ z, whereJ \u2261 L + S. Use explicit calculations to prove the following: a. The Y j=l\u00b11/2,m l are normalized, that is \u03c0 2\u03c0 d\u03c6 sin \u03b8d\u03b8 0 0 Y j=l\u00b11/2,m l \u2020 Y j=l\u00b11/2,m l = 1. b. The Y j=l\u00b11/2,m l have the correct eigenvalues for J2 = L2 + S2 + 2LzSz + L+S\u2212 + L\u2212S+, that is see (3.349), and Jz. 3", ".37 a. Evaluate j \u2211 m=\u2212j |d( j) mm (\u03b2)|2m for any j (integer or half-integer); then check your answer for j = 1 2. b. Prove, for any j, j \u2211 m=\u2212j m2|d( j) mm(\u03b2)|2 = 1 2 j( j + 1) sin2 \u03b2 + m2 1 2 (3 cos2 \u03b2 \u2212 1). [Hint: This can be proved in many ways. You may, for instance, examine the rotational properties of J 2 z using the spherical (irreducible) tensor language.] 247 Problems 3.38 a. Consider a system with j = 1. Explicitly write j = 1, m|Jy| j = 1, m in 3 \u00d7 3 matrix form. b. Show that for j = 1 only, it is legitimate to replace e\u2212iJy\u03b2/\u00afh by 1 \u2212 i sin \u03b2 \u2212 (1 \u2212 cos \u03b2). Jy \u00afh 2 Jy \u00afh c. Using (b), prove d( j=1)(\u03b21 + cos \u03b2) \u2212 sin \u03b2 1\u221a 2 (1 \u2212 cos \u03b2) sin \u03b2 cos \u03b2 sin \u03b2 1\u221a 2 11 \u2212 cos \u03b2) sin \u03b2 (1 + cos \u03b2) 1\u221a 2 3.39 Express the matrix element \u03b12\u03b22\u03b32|J 2 3 |\u03b11\u03b21\u03b31 in terms of a series in D j mn(\u03b1\u03b2\u03b3) = \u03b1\u03b2\u03b3| jmn. 3.40 Consider a system made up of two spin 1 2 particles. Observer A specializes in measuring the spin components of one of the particles (s1z, s1x and so on), while observer B measures the spin components of the other particle. Suppose the system is known to be in a spin-singlet state, that is, Stotal = 0. a. What is the probability for observer A to obtain s1z = \u00afh/2 when observer B makes no measurement? Same problem for s1x = \u00afh/2. b. Observer B determines the spin of particle 2 to be in the s2z = \u00afh/2 state with certainty. What can we then conclude about the outcome of observer A\u2019s measurement if (i) A measures s1z and (ii) A measures s1x? Justify your answer. 3.41 Using the de\ufb01ning property (3.451) for a vector operator, prove that the momentum operator p is a vector, based on its commutation relations with angular momentum L. 3.42 Consider a spherical tensor of rank 1 (that is, a vector) \u00b11 = \u2213 Vx \u00b1 iVy\u221a V(1) 2, V(1) 0 = Vz. Using the expression for d( j=1) given in Problem 3.38, evaluate qq (\u03b2)V(1) d(1) q \u2211 q and show that your results are just what you expect from the transformation properties of Vx,y,z under rotations about the y-axis. 3.43 a. Construct a spherical tensor of rank 1 out of two different vectors U = (Ux, Uy, Uz) and V = (Vx, Vy, Vz). Explicitly write T(1) \u00b11,0 in terms of Ux,y,z and Vx,y,z. b. Construct a spherical tensor of rank 2 out of two different vectors U and V. Write down explicitly T(2) \u00b12,\u00b11,0 in terms of Ux,y,z and Vx,y,z. 248 Theory of Angular Momentum 3.44 Consider a spinless particle bound to a \ufb01xed center by a central force potential. a. Relate, as much as possible, the matrix elements n, l, m| \u2213 1\u221a 2 (x \u00b1 iy)|n, l, m and n, l, m|z|n, l, m using only the Wigner\u2013Eckart theorem. Make sure to state under what conditions the matrix elements are nonvanishing. b. Do the same problem using wave functions \u03c8(x) = Rnl(r)Ym l (\u03b8, \u03c6). 3.45 a. Write xy, xz, and(x 2 \u2212 y2) as components of the components of a spherical (irreducible) tensor of rank 2. b. The expectation value Q \u2261 e\u03b1, j, m = j|(3z2 \u2212 r2)|\u03b1, j, m = j is known as the quadrupole moment. Evaluate e\u03b1, j, m|(x2 \u2212 y2)|\u03b1, j, m = j, (where m =", " j, j \u2212 1, j \u2212 2,...) in terms ofQ and appropriate Clebsch\u2013Gordan coef\ufb01cients. 3.46 A spin 3 2 nucleus situated at the origin is subjected to an external inhomogeneous electric \ufb01eld. The basic electric quadrupole interaction may by taken to be Hint = eQ 2s(s \u2212 1)\u00afh2 \u22022\u03c6 \u2202x2 S2 x + 0 \u22022\u03c6 \u2202y2 S2 y + 0 \u22022\u03c6 \u2202z2 S2 z, 0 where \u03c6 is the electrostatic potential satisfying Laplace\u2019s equation and the coordinate axes are so chosen that \u22022\u03c6 \u2202x\u2202y = 0 \u22022\u03c6 \u2202y\u2202z = 0 \u22022\u03c6 \u2202x\u2202z = 0. 0 Show that the interaction energy can be written as A(3S2 z \u2212 S2) +B (S2 + + S2 \u2212), and express A and B in terms of (\u22022\u03c6/\u2202x2)0 and so on. Determine the energy eigenkets (in terms of |m, where m = \u00b1 3 2 ) and the corresponding energy eigenvalues. Is there any degeneracy? 2, \u00b1 1 4 Symmetry in Quantum Mechanics Having studied the theory of rotation in detail, we are in a position to discuss, in more general terms, the connection between symmetries, degeneracies, and conservation laws. We have deliberately postponed this very important topic until now so that we can discuss it using the rotation symmetry of Chapter 3 as an example. 4.1 Symmetries, Conservation Laws, and Degeneracies 4.1.1 Symmetries in Classical Physics We begin with an elementary review of the concepts of symmetry and conservation law in classical physics. In the Lagrangian formulation of quantum mechanics, we start with the Lagrangian L, which is a function of a generalized coordinate qi and the corresponding generalized velocity \u02d9qi. IfL is unchanged under displacement, then we must have qi \u2192 qi + \u03b4qi, \u2202L \u2202qi = 0. It then follows, by virtue of the Lagrange equation, d/dt(\u2202L/\u2202\u02d9qi) \u2212 \u2202L/\u2202qi = 0, that where the canonical momentum is de\ufb01ned as dpi dt = 0, pi = \u2202L \u2202\u02d9qi. (4.1) (4.2) (4.3) (4.4) So if L is unchanged under displacement (4.1), then we have a conserved quantity, the canonical momentum conjugate to qi. Likewise, in the Hamiltonian formulation based on H regarded as a function of qi and pi, we have 249 dpi dt = 0 (4.5) 250 Symmetry in Quantum Mechanics whenever \u2202H \u2202qi = 0. (4.6) So if the Hamiltonian does not explicitly depend on qi, which is another way of saying H has a symmetry under qi \u2192 qi + \u03b4qi, we have a conserved quantity. 4.1.2 Symmetry in Quantum Mechanics In quantum mechanics we have learned to associate a unitary operator, say S, with an operation like translation or rotation. It has become customary to call S a symmetry operator regardless of whether the physical system itself possesses the symmetry corresponding to S. Further, we have learned that for symmetry operations that differ in\ufb01nitesimally from the identity transformation, we can write S = 1 \u2212 i\u03b5 \u00afh G, (4.7) where G is the Hermitian generator of the symmetry operator in question. Let us now suppose that H is invariant under S. We then have But this is equivalent to S \u2020HS = H. [G, H] =0. By virtue of the Heisenberg equation of motion, we have dG dt = 0; (4.8) (4.9) (4.10) hence, G is a constant of the motion. For instance, if H is invariant under translation, then momentum is a constant of the motion; if H is invariant under rotation, then angular momentum is a constant of the motion. It is instructive to look at the connection between (4.9) and conservation of G from the point of view of an eigenket of G when G commutes with H. Suppose at t0 the system is in an eigenstate of G. Then the ket at a later time obtained by applying the time-evolution operator |g, t0; t = U(t, t0)|g (4.11) is also an eigenket of G with the same eigenvalue g", ". In other words, once a ket is a G eigenket, it is always a G eigenket with the same eigenvalue. The proof of this is extremely simple once we realize that (4.9) and (4.10) also imply that G commutes with the timeevolution operator, namely, G[U(t, t0)|g] =U(t, t0)G|g = g[U(t, t0)|g]. (4.12) 251 4.1 Symmetries, Conservation Laws, and Degeneracies 4.1.3 Degeneracies Let us now turn to the concept of degeneracies. Even though degeneracies may be discussed at the level of classical mechanics \u2013 for instance in discussing closed (nonprecessing) orbits in the Kepler problem (Goldstein et al. (2002)) \u2013 this concept plays a far more important role in quantum mechanics. Let us suppose that [H, S ] =0 (4.13) for some symmetry operator and |n is an energy eigenket with eigenvalue En. ThenS |n is also an energy eigenket with the same energy, because H(S |n) =S H|n = En(S |n). Suppose |n and S |n represent different states. Then these are two states with the same energy, that is, they are degenerate. Quite often S is characterized by continuous parameters, say \u03bb, in which case all states of the form S (\u03bb)|n have the same energy. (4.14) We now consider rotation speci\ufb01cally. Suppose the Hamiltonian is rotationally invari- ant, so which necessarily implies that [D(R), H] = 0, [J, H] =0, [J2, H] =0. (4.15) (4.16) We can then form simultaneous eigenkets of H, J2, and Jz, denoted by |n; j, m. The argument just given implies that all states of the form D(R)|n; j, m (4.17) have the same energy. We saw in Chapter 3 that under rotation different m-values are mixed up. In general, D(R)|n; j, m is a linear combination of 2j + 1 independent states. Explicitly, mm(R), |n; j, mD ( j) D(R)|n; j, m = \u2211 m and by changing the continuous parameter that characterizes the rotation operator D(R), we can get different linear combinations of |n; j, m. If all states of form D(R)|n; j, m with arbitrary D(R) are to have the same energy, it is then essential that each of |n; j, m with different m must have the same energy. So the degeneracy here is (2j + 1)-fold, just equal to the number of possible m-values. This point is also evident from the fact that all states obtained by successively applying J\u00b1, which commutes with H, to |n; jm have the same energy. (4.18) As an application, consider an atomic electron whose potential is written as V(r) + VLS(r)L \u00b7 S. Because r and L \u00b7 S are both rotationally invariant, we expect a (2j + 1)-fold degeneracy for each atomic level. On the other hand, suppose there is an external electric or magnetic \ufb01eld, say in the z-direction. The rotational symmetry is now manifestly broken; as a result, the (2j + 1)-fold degeneracy is no longer expected and states characterized by different m-values no longer have the same energy. We will examine how this splitting arises in Chapter 5. 252 Symmetry in Quantum Mechanics 4.1.4 SO(4) Symmetry in the Coulomb Potential A \ufb01ne example of continuous symmetry in quantum mechanics is afforded by the hydrogen atom problem and the solution for the Coulomb potential. We carried out the solution to this problem in Section 3.7 where we discovered that the energy eigenvalues in (3.315) show the striking degeneracy summarized in (3.318). It would be even more striking if this degeneracy were just an accident, but indeed, it is the result of an additional symmetry that is particular to the problem of bound states of 1/r potentials. The classical problem of orbits in such potentials, the Kepler problem, was of course well studied long prior to quantum mechanics. The fact that the solution leads to elliptical orbits that are closed means that there should be some (vector) constant of the motion which maintains the orientation of the major axis of the ellipse. We know that even a small deviation from a 1/r potential leads to precession of this", " axis, so we expect that the constant of the motion we seek is in fact particular to 1/r potentials. Classically, this new constant of the motion is M = p \u00d7 L m \u2212 Ze2 r x (4.19) where we refer back to the notation used in Section 3.7. This quantity is generally known as the Lenz vector or sometimes the Runge\u2013Lenz vector. Rather than belabor the classical treatment here, we will move on to the quantum-mechanical treatment in terms of the symmetry responsible for this constant of the motion. This new symmetry is called SO(4), completely analogous to the symmetry SO(3) studied in Section 3.3. That is, SO(4) is the group of rotation operators in four spatial dimensions. Equivalently, it is the group of orthogonal 4\u00d74 matrices with unit determinant. Let us build up the properties of the symmetry that leads to the Lenz vector as a constant of the motion, and then we will see that these properties of those we expect from SO(4). Our approach1 closely follows that given by Schiff (1968), pp. 235\u2013239. We \ufb01rst need to modify (4.19) to construct a Hermitian operator. For two Hermitian vector operators A and B, it is easy to show that (A \u00d7 B)\u2020 = \u2212B \u00d7 A. Therefore a Hermitian version of the Lenz vector is M = 1 2m (p \u00d7 L \u2212 L \u00d7 p) \u2212 Ze2 r x. It can be shown that M commutes with the Hamiltonian that is, H = p2 2m \u2212 Ze2 r [M, H] =0, (4.20) (4.21) (4.22) 1 A classical treatment of the Runge\u2013Lenz vector is given in Section 3.9 of Goldstein et al. (2002). The quantum-mechanical treatment was \ufb01rst done by Pauli, Z. Phys., 33 (1925) 879, and an English translation \u201cOn the hydrogen spectrum from the standpoint of the new quantum mechanics\u201d is published in Van der Waerden (1967). 253 4.1 Symmetries, Conservation Laws, and Degeneracies so that indeed M is a (quantum mechanical) constant of the motion. (See Problem 3.31 in Chapter 3.) Other useful relations can be proven, namely L \u00b7 M = 0 = M \u00b7 L M2 = % 2 m H & L2 + \u00afh2 + Z2e4. (4.23) (4.24) In order to identify the symmetry responsible for this constant of the motion, it is instructive to review the algebra of the generators of this symmetry. We already know part of this algebra, namely [Li, Lj] =i \u00afh\u03b5ijkLk (4.25) which we wrote earlier as (3.211) in a notation where repeated indices (k in this case) are automatically summed over components. One can also show that [Mi, Lj] =i \u00afh\u03b5ijkMk (4.26) which in fact establish M as a vector operator in the sense of (3.451). Finally, it is possible to derive [Mi, Mj] =\u2212 i\u00afh\u03b5ijk 2 m HLk. (4.27) To be sure, (4.25), (4.26), and (4.27) do not form a closed algebra, due to the presence of H in (4.27), and that makes it dif\ufb01cult to identify these operators as generators of a continuous symmetry. However, we can consider the problem of speci\ufb01c bound states. In this case, the vector space is truncated only to those that are eigenstates of H, with eigenvalue E < 0. In that case, we replace H with E in (4.27) and the algebra is closed. It is instructive to replace M with the scaled vector operator N \u2261 1/2 \u2212 m 2E M. In this case we have the closed algebra [Li, Lj] =i \u00afh\u03b5ijkLk [Ni, Lj] =i \u00afh\u03b5ijkNk [Ni, Nj] =i \u00afh\u03b5ijkLk. (4.28) (4.29a) (4.29b) (4.29c) So what is the symmetry operation generated by the operators L and N in (4.29)? Although it is far from obvious, the answer is \u201crotation in four spatial dimensions.\u201d The \ufb01rst clue is in the number of generators, namely six, each of which should correspond to rotation about some axis. Think of a rotation as an operation which mixes two orthogonal axes. Then, the number of generators for rotations in", " n spatial dimensions should be the number of combinations of n things taken two at a time, namely n(n \u2212 1)/2. Consequently, rotations in two dimensions require one generator, that is Lz. Rotations in three dimensions require three, namely L, and four-dimensional rotations require six generators. It is harder to see that (4.29) is the appropriate algebra for this kind of rotation, but we proceed as follows. In three spatial dimensions, the orbital angular-momentum operator 254 Symmetry in Quantum Mechanics (3.210) generates rotations. We saw this clearly in (3.215) where an in\ufb01nitesimal z-axis rotation on a state |\u03b1 is represented in a rotated version of the |x, y, z basis. This was just a consequence of the momentum operator being the generator of translations in space. In fact, a combination like Lz = xpy \u2212 ypx indeed mixes the x-axis and y-axis, just as one would expect from the generator of rotations about the z-axis. To generalize this to four spatial dimensions, we \ufb01rst associate (x, y, z) and (px, py, pz) with (x1, x2, x3) and (p1, p2, p3). We are led to rewrite the generators as L3 = \u02dcL12 = x1p2 \u2212 x2p1, L1 = \u02dcL23, andL 2 = \u02dcL31. If we then invent a new spatial dimension x4 and its conjugate momentum p4 (with the usual commutation relations) we can de\ufb01ne \u02dcL14 = x1p4 \u2212 x4p1 \u2261 N1 \u02dcL24 = x2p4 \u2212 x4p2 \u2261 N2 \u02dcL34 = x3p4 \u2212 x4p3 \u2261 N3. It is easy to show that these operators Ni obey the algebra (4.29). For example [N1, L2] = [x1p4 \u2212 x4p1, x3p1 \u2212 x1p3] = p4[x1, p1]x3 + x4[ p1, x1]p3] = i\u00afh(x3p4 \u2212 x4p3) = i\u00afhN3. (4.30a) (4.30b) (4.30c) (4.31) In other words, this is the algebra of four spatial dimensions. We will return to this notion in a moment, but for now we will press on with the degeneracies in the Coulomb potential that are implied by (4.14). De\ufb01ning the operators I \u2261 (L + N)/2 K \u2261 (L \u2212 N)/2 we easily can prove the following algebra: [Ii, Ij] =i \u00afh\u03b5ijkIk [Ki, Kj] =i \u00afh\u03b5ijkKk [Ii, Kj] =0. (4.32) (4.33) (4.34a) (4.34b) (4.34c) Therefore, these operators obey independent angular momentum algebras. It is also evident that [I, H] = [K, H] =0. Thus, these \u201cangular momenta\u201d are conserved quantities, and we denote the eigenvalues of the operators I2 and K2 by i(i + 1)\u00afh2 and k(k + 1)\u00afh2 respectively, with i, k = 0, 1 Since I2 \u2212 K2 = L \u00b7 N = 0 by (4.23) and (4.28), we must have i = k. On the other hand, 2, 1, 3 2,.... the operator I2 + K2 = % 1 2 L2 + N2 & = 1 2 L2 \u2212 m 2E M2 leads, with (4.24), to the numerical relation 2k(k + 1)\u00afh2 = 1 2 \u2212\u00afh2 \u2212 m 2E Z2e4. (4.35) (4.36) 255 4.1 Symmetries, Conservation Laws, and Degeneracies Solving for E we \ufb01nd E = \u2212 mZ2e4 2\u00afh2 1 (2k + 1)2. (4.37) This is the same as (3.315) with the principal quantum number n replaced by 2k + 1. We now see that the degeneracy in the Coulomb problem arises from the two \u201crotational\u201d symmetries represented by the operators I and K. The degree of degeneracy, in fact, is (2i + 1)(2k + 1) = (2k + 1)2 = n2", ". This is exactly what we arrived at in (3.318) except it is now clear that the degeneracy is no accident. It is worth noting that we have just solved for the eigenvalues of the hydrogen atom without ever resorting to solving the Schr\u00a8odinger equation. Instead, we exploited the inherent symmetries to arrive at the same answer. This solution was apparently \ufb01rst carried out by Pauli. In the language of the theory of continuous groups, which we started to develop in Section 3.3, we see that the algebra (4.29) corresponds to the group SO(4). Furthermore, rewriting this algebra as (4.34) shows that this can also be thought of as two independent groups SU(2), that is SU(2)\u00d7SU(2). Although it is not the purpose of this book to include an introduction to group theory, we will carry this a little further to show how one formally carries out rotations in n spatial dimensions, that is, the group SO(n). Generalizing the discussion in Section 3.3, consider the group of n \u00d7 n orthogonal matrices R which carry out rotations in n dimensions. They can be parameterized as R = exp i \u03c6q\u03c4q (4.38) n(n\u22121)/2 \u2211 q=1 where the \u03c4q are purely imaginary, antisymmetric n \u00d7 n matrices, that is (\u03c4q)T = \u2212\u03c4q and the \u03c6q are generalized rotation angles. The antisymmetry condition ensures that R is orthogonal. The overall factor of i implies that the imaginary matrices \u03c4q are also Hermitian. The \u03c4q are obviously related to the generators of the rotation operator. In fact, it is their commutation relations which should be parroted by the commutation relations of these generators. Following along as in Section 3.1, we compare the action of performing an in\ufb01nitesimal rotation \ufb01rst about axis q and then about axis p with the rotation carried out in reverse order. Then, (1 + i\u03c6p\u03c4p) (1 + i\u03c6q\u03c4q) \u2212 (1 + i\u03c6q\u03c4q) (1 + i\u03c6p\u03c4p) = \u2212\u03c6p\u03c6q [\u03c4p, \u03c4q] = 1 \u2212 1 + i\u03c6p\u03c6q \u2211 f pq r \u03c4r r (4.39) where the last line of (4.39) recognizes that the result must be a second-order rotation about the two axes with some linear combination of generators. The f pq are called structure constants for this group of rotations. This gives us the commutation relations r [\u03c4p, \u03c4q] =i \u2211 f pq r \u03c4r. r (4.40) 256 Symmetry in Quantum Mechanics RH z LH New x New y y x New z Fig. 4.1 Right-handed (RH) and left-handed (LH) systems. To go further, one would need to determine the structure constants f pq, and we leave these details to textbooks devoted to group theory. It is not hard to show, however, that in three dimensions, f pq r = \u03b5pqr as expected. r 4.2 Discrete Symmetries, Parity, or Space Inversion So far we have considered continuous symmetry operators, that is, operations that can be obtained by applying successively in\ufb01nitesimal symmetry operations. All symmetry operations useful in quantum mechanics are not necessarily of this form. In this chapter we consider three symmetry operations that can be considered to be discrete, as opposed to continuous \u2013 parity, lattice translation, and time reversal. The \ufb01rst operation we consider is parity, or space inversion. The parity operation, as applied to transformation on the coordinate system, changes a right-handed (RH) system into a left-handed (LH) system, as shown in Figure 4.1. However, in this book we consider a transformation on state kets rather than on the coordinate system. Given |\u03b1, we consider a space-inverted state, assumed to be obtained by applying a unitary operator \u03c0 known as the parity operator, as follows: |\u03b1 \u2192 \u03c0|\u03b1. (4.41) We require the expectation value of x taken with respect to the space-inverted state to be opposite in sign: a very reasonable requirement. This is accomplished if \u03b1|\u03c0\u2020x\u03c0|\u03b1 = \u2212\u03b1|x|\u03b1, or \u03c0\u2020x\u03c0 = \u2212x x\u03c0 = \u2212\u03c0x, (4.42) (4.43) (4.44) where we have used the fact that \u03c0 is unitary. In other words, x and \u03c0 must anticommute. 257 4.2 Discrete Sy", "mmetries, Parity, or Space Inversion How does an eigenket of the position operator transform under parity? We claim that \u03c0|x = ei\u03b4| \u2212x where ei\u03b4 is a phase factor (\u03b4 real). To prove this assertion let us note that x\u03c0|x = \u2212\u03c0x|x = (\u2212x)\u03c0|x. (4.45) (4.46) This equation says that \u03c0|x is an eigenket of x with eigenvalue \u2212x, so it must be the same as a position eigenket | \u2212x up to a phase factor. It is customary to take ei\u03b4 = 1 by convention. Substituting this in (4.45), we have \u03c02|x = |x; hence, \u03c02 = 1. That is, we come back to the same state by applying \u03c0 twice. We easily see from (4.45) that \u03c0 is now not only unitary but also Hermitian: \u03c0\u22121 = \u03c0\u2020 = \u03c0. (4.47) Its eigenvalue can be only +1 or \u22121. What about the momentum operator? The momentum p is like mdx/dt, so it is natural to expect it to be odd under parity, like x. A more satisfactory argument considers the momentum operator as the generator of translation. Since translation followed by parity is equivalent to parity followed by translation in the opposite direction, as can be seen from Figure 4.2, then \u03c0T (dx) =T (\u2212dx)\u03c0 \u03c0 1 \u2212 ip \u00b7 dx \u00afh \u03c0\u2020 = 1 + ip \u00b7 dx \u00afh, {\u03c0, p} = 0 or \u03c0\u2020p\u03c0 = \u2212p. from which follows (4.48) (4.49) (4.50) dx \u2212dx Fig. 4.2 Translation followed by parity, and vice versa. 258 Symmetry in Quantum Mechanics We can now discuss the behavior of J under parity. First, for orbital angular momentum we clearly have because [\u03c0, L] =0 L = x \u00d7 p, (4.51) (4.52) and both x and p are odd under parity. However, to show that this property also holds for spin, it is best to use the fact that J is the generator of rotation. For 3 \u00d7 3 orthogonal matrices, we have where explicitly R(parity)R(rotation) = R(rotation)R(parity), R(parity) = \u239b \u239d \u22121 0 \u22121 \u239e \u23a0 ; 0 \u22121 (4.53) (4.54) that is, the parity and rotation operations commute. In quantum mechanics, it is natural to postulate the corresponding relation for the unitary operators, so \u03c0D(R) = D(R)\u03c0, where D(R) = 1 \u2212 iJ \u00b7 \u02c6n\u03b5/\u00afh. From (4.55) it follows that [\u03c0, J] =0 or \u03c0\u2020J\u03c0 = J. (4.55) (4.56) This together with (4.51) means that the spin operator S (leading to the total angular momentum J = L + S) also transforms in the same way as L. Under rotations, x and J transform in the same way, so they are both vectors, or spherical tensors, of rank 1. However, x (or p) is odd under parity [see (4.43) and (4.50)], while J is even under parity [see (4.56)]. Vectors that are odd under parity are called polar vectors, while vectors that are even under parity are called axial vectors, orpseudovectors. Let us now consider operators like S \u00b7 x. Under rotations they transform like ordinary scalars, such as S \u00b7 L or x \u00b7 p. Yet under space inversion we have while for ordinary scalars we have \u03c0\u22121S \u00b7 x\u03c0 = \u2212S \u00b7 x, \u03c0\u22121L \u00b7 S\u03c0 = L \u00b7 S and so on. The operator S \u00b7 x is an example of a pseudoscalar. 4.2.1 Wave Functions under Parity (4.57) (4.58) Let us now look at the parity property of wave functions. First, let \u03c8 be the wave function of a spinless particle whose state ket is |\u03b1: \u03c8(x) =x |\u03b1. (4.59) 259 4.2 Discrete Symmetries, Parity, or Space Inversion The wave function of the space-inverted state, represented by the state ket \u03c0|\u03b1, is x|\u03c0|\u03b1 = \u2212x|\u03b1 = \u03c8(\u2212x). (4.60) Suppose |\u03b1 is an eigen", "ket of parity. We have already seen that the eigenvalue of parity must be \u00b11, so Let us look at its corresponding wave function, \u03c0|\u03b1 = \u00b1|\u03b1. But we also have x|\u03c0|\u03b1 = \u00b1x|\u03b1. x|\u03c0|\u03b1 = \u2212x|\u03b1, (4.61) (4.62) (4.63) so the state |\u03b1 is even or odd under parity depending on whether the corresponding wave function satis\ufb01es \u03c8(\u2212x) =\u00b1 \u03c8(x) even parity odd parity. (4.64) Not all wave functions of physical interest have de\ufb01nite parities in the sense of (4.64). Consider, for instance, the momentum eigenket. The momentum operator anticommutes with the parity operator, so the momentum eigenket is not expected to be a parity eigenket. Indeed, it is easy to see that the plane wave, which is the wave function for a momentum eigenket, does not satisfy (4.64). An eigenket of orbital angular momentum is expected to be a parity eigenket because L and \u03c0 commute [see (4.51)]. To see how an eigenket of L2 and Lz behaves under parity, let us examine the properties of its wave function under space inversion, The transformation x \u2192 \u2212x is accomplished by letting x|\u03b1, lm = R\u03b1(r)Ym l (\u03b8, \u03c6). cos \u03b8 \u2192 \u2212 cos \u03b8) (eim\u03c6 \u2192 (\u22121)meim\u03c6). Using the explicit form of l = (\u22121)m Ym (2l + 1)(l \u2212 m)! 4\u03c0(l + m)! l (cos \u03b8)eim\u03c6 Pm (4.65) (4.66) (4.67) for positive m, with (3.247), where |m| l (cos \u03b8) = P (\u22121)m+l 2ll! (l + |m|)! (l \u2212 |m|)! sin\u2212|m| \u03b8 l\u2212|m| d d(cos \u03b8) sin2l \u03b8, (4.68) we can readily show that Ym l \u2192 (\u22121)lYm l (4.69) 260 Symmetry in Quantum Mechanics as \u03b8 and \u03c6 are changed, as in (4.66). Therefore, we can conclude that \u03c0|\u03b1, lm = (\u22121)l|\u03b1, lm. (4.70) It is actually not necessary to look at Ym with m = 0 and note that Lr and (L\u00b1)r commute. l ; an easier way to obtain the same result is to work \u00b1|l, m = 0(r = 0, 1,..., l) must have the same parity because \u03c0 Let us now look at the parity properties of energy eigenstates. We begin by stating a very important theorem. Theorem 6 Suppose [H, \u03c0] = 0 and |nis a nondegenerate eigenket of H with eigenvalue En: H|n = En|n; then |n is also a parity eigenket. Proof We prove this theorem by \ufb01rst considering the state |\u03b1 \u2261 1 2 (1 \u00b1 \u03c0)|n. (4.71) (4.72) (4.73) Since \u03c02 = 1, it is simple to show that \u03c0|\u03b1 = \u00b1|\u03b1, so| \u03b1 is a parity eigenket. Also, since H\u03c0 = \u03c0H, it is simple to show that H|\u03b1 = En|\u03b1, so| \u03b1 is also an energy eigenket with eigenvalue En. However, we assumed that |n was nondegenerate. Therefore |n and |\u03b1 must be the same, to within a multiplicative constant, and |n must be a parity eigenket with parity \u00b11, that is \u03c0|n = \u00b1|n. (It is worth noting that |\u03b1 = 0 if we choose the \u201cwrong\u201d sign in (4.73). That is, despite the \u00b1, (4.73) de\ufb01nes just one state.) As an example, let us look at the simple harmonic oscillator. The ground state |0 has even parity because its wave function, being Gaussian, is even under x \u2192 \u2212x. The \ufb01rst excited state, |1 = a\u2020|0, (4.74) must have odd parity because a\u2020 is linear in x and p, which are both odd [see (2.123)]. In general, the parity of the nth excited state of the simple harmonic operator is given by (\u22121)n. It is important to note that the nondegenerate", " assumption is essential here. For instance, consider the hydrogen atom in nonrelativistic quantum mechanics. As is well known, the energy eigenvalues depend only on the principal quantum number n (for example, 2p and 2s states are degenerate) \u2013 the Coulomb potential is obviously invariant under parity \u2013 yet an energy eigenket cp|2p + cs|2s (4.75) is obviously not a parity eigenket. 261 4.2 Discrete Symmetries, Parity, or Space Inversion As another example, consider a momentum eigenket. Momentum anticommutes with parity, so \u2013 even though the free-particle Hamiltonian H is invariant under parity \u2013 the momentum eigenket (though obviously an energy eigenket) is not a parity eigenket. Our theorem remains intact because we have here a degeneracy between |p and | \u2212p, which 2)(|p \u00b1 have the same energy. In fact, we can easily construct linear combinations (1/ | \u2212 p), which are parity eigenkets with eigenvalues \u00b11. In terms of wave function language, eip\u00b7x/\u00afh does not have a de\ufb01nite parity, but cos p \u00b7 x/\u00afh and sin p \u00b7 x/\u00afh do. \u221a 4.2.2 Symmetrical Double-Well Potential As an elementary but instructive example, we consider a symmetrical double-well potential; see Figure 4.3. The Hamiltonian is obviously invariant under parity. In fact, the two lowest-lying states are as shown in Figure 4.3, as we can see by working out the explicit solutions involving sine and cosine in classically allowed regions and sinh and cosh in the classically forbidden region. The solutions are matched where the potential is discontinuous; we call them the symmetrical state |S and the antisymmetrical state |A. Of course, they are simultaneous eigenkets of H and \u03c0. Calculation also shows that EA > ES, (4.76) which we can infer from Figure 4.3 by noting that the wave function of the antisymmetrical state has a greater curvature. The energy difference is very tiny if the middle barrier is high, a point which we will discuss later. We can form and |R = 1\u221a 2 (|S + |A) |L = 1\u221a 2 (|S \u2212 |A). (4.77a) (4.77b) The wave functions of (4.77a) and (4.77b) are largely concentrated in the right-hand side and the left-hand side, respectively. They are obviously not parity eigenstates; in fact, under parity |R and |L are interchanged. Note that they are not energy eigenstates either. Indeed, they are typical examples of nonstationary states. To be precise, let us assume that the system is represented by |R at t = 0. At a later time, we have Fig. 4.3 The symmetrical double well with the two lowest-lying states |S (symmetrical) and |A (antisymmetrical) shown. Symmetrical S Antisymmetrical A 262 Symmetry in Quantum Mechanics Fig. 4.4 The symmetrical double well with an infinitely high middle barrier. |R, t0 = 0; t = = % e\u2212iESt/\u00afh|S + eiEAt/\u00afh|A & e\u2212iESt/\u00afh |S + ei(EA\u2212ES)t/\u00afh|A. 1\u221a 2 1\u221a 2 (4.78) At time t = T/2 \u2261 2\u03c0\u00afh/2(EA \u2212 ES), the system is found in pure |L. At t = T, we are back to pure |R, and so forth. Thus, in general, we have an oscillation between |R and |L with angular frequency \u03c9 = (EA \u2212 ES) \u00afh. (4.79) This oscillatory behavior can also be considered from the viewpoint of tunneling in quantum mechanics. A particle initially con\ufb01ned to the right-hand side can tunnel through the classically forbidden region (the middle barrier) into the left-hand side, then back to the right-hand side, and so on. But now let the middle barrier become in\ufb01nitely high; see Figure 4.4. The |S and |A states are now degenerate, so (4.77a) and (4.77b) are also energy eigenkets even though they are not parity eigenkets. Once the system is found in |R, it remains so forever (oscillation time between |S and |A is now \u221e). Because the middle barrier is in\ufb01", "nitely high, there is no possibility for tunneling. Thus when there is degeneracy, the physically realizable energy eigenkets need not be parity eigenkets. We have a ground state which is asymmetrical despite the fact that the Hamiltonian itself is symmetrical under space inversion, so with degeneracy the symmetry of H is not necessarily obeyed by energy eigenstates |S and |A. This is a very simple example of broken symmetry and degeneracy. Nature is full of situations analogous to this. Consider a ferromagnet. The basic Hamiltonian for iron atoms is rotationally invariant, but the ferromagnet clearly has a de\ufb01nite direction in space; hence, the (in\ufb01nite) number of ground states is not rotationally invariant, since the spins are all aligned along some de\ufb01nite (but arbitrary) direction. A textbook example of a system that illustrates the actual importance of the symmetrical double well is an ammonia molecule, NH3; seeFigure 4.5. We imagine that the three H atoms form the three corners of an equilateral triangle. The N atom can be up or down, where the directions up and down are de\ufb01ned because the molecule is rotating around the axis as shown in Figure 4.5. The up and down positions for the N atom are analogous to 263 4.2 Discrete Symmetries, Parity, or Space Inversion H H H N H (a) H H N (b) Fig. 4.5 An ammonia molecule, NH3, where the three H atoms form the three corners of an equilateral triangle. R and L of the double-well potential. The parity and energy eigenstates are superpositions of Figure 4.5a and Figure 4.5b in the sense of (4.77a) and (4.77b), respectively, and the energy difference between the simultaneous eigenstates of energy and parity corresponds to an oscillation frequency of 24,000 MHz, a wavelength of about 1 cm, which is in the microwave region. In fact, NH3 is of fundamental importance in maser physics. There are naturally occurring organic molecules, such as sugars or amino acids, which are of the R-type (or L-type) only. Such molecules which have de\ufb01nite handedness are called optical isomers. In many cases the oscillation time is practically in\ufb01nite \u2013 on the order of 104\u2013106 years \u2013 so R-type molecules remain right-handed for all practical purposes. It is amusing that if we attempt to synthesize such organic molecules in the laboratory, we \ufb01nd equal mixtures of R and L. Why we have a preponderance of one type is nature\u2019s deepest mystery. Is it due to a genetic accident, like the spiral shell of a snail or the fact that our hearts are on the left-hand side?2 4.2.3 Parity-Selection Rule Suppose |\u03b1 and |\u03b2 are parity eigenstates: and \u03c0|\u03b1 = \u03b5\u03b1|\u03b1 \u03c0|\u03b2 = \u03b5\u03b2|\u03b2, (4.80a) (4.80b) 2 It has been suggested that parity violation in nuclear processes active during the formation of life, may have contributed to this handedness. See Bonner, Chirality, 12 (2000) 114. 264 Symmetry in Quantum Mechanics where \u03b5\u03b1, \u03b5\u03b2 are the parity eigenvalues (\u00b11). We can show that \u03b2|x|\u03b1 = 0 (4.81) unless \u03b5\u03b1 = \u2212\u03b5\u03b2. In other words, the parity-odd operator x connects states of opposite parity. The proof of this follows: \u03b2|x|\u03b1 = \u03b2|\u03c0\u22121\u03c0x\u03c0\u22121\u03c0|\u03b1 = \u03b5\u03b1 \u03b5\u03b2(\u2212\u03b2|x|\u03b1), (4.82) which is impossible for a \ufb01nite nonzero \u03b2|x|\u03b1 unless \u03b5\u03b1 and \u03b5\u03b2 are opposite in sign. Perhaps the reader is familiar with this argument from \u03c8\u2217 \u03b2x\u03c8\u03b1d\u03c4 = 0 (4.83) if \u03c8\u03b2 and \u03c8\u03b1 have the same parity. This selection rule, due to Wigner, is of importance in discussing radiative transitions between atomic states. As we will discuss in greater detail later, radiative transitions take place between states of opposite parity as a consequence of multipole expansion formalism. This rule was known phenomenologically from analysis of spectral lines, before the birth of quantum mechanics, as Laporte\u2019s rule. It was Wigner who showed that Laporte\u2019s rule is a consequence of the parity-selection rule. If the basic Hamiltonian H is invariant under parity, nondegenerate energy eigenstates [as a corollary", " of (4.83)] cannot possess a permanent electric dipole moment: n|x|n = 0. (4.84) This follows trivially from (4.83), because with the nondegenerate assumption, energy eigenstates are also parity eigenstates [see (4.72) and (4.73)]. For a degenerate state, it is perfectly all right to have an electric dipole moment. We will see an example of this when we discuss the linear Stark effect in Chapter 5. Our considerations can be generalized: Operators that are odd under parity, like p or S\u00b7x, have nonvanishing matrix elements only between states of opposite parity. In contrast, operators that are even under parity connect states of the same parity. 4.2.4 Parity Nonconservation The basic Hamiltonian responsible for the so-called weak interaction of elementary particles is not invariant under parity. In decay processes we can have \ufb01nal states which are superpositions of opposite parity states. Observable quantities like the angular distribution of decay products can depend on pseudoscalars such as S \u00b7 p. It is remarkable that parity conservation was believed to be a sacred principle until 1956, when Lee and Yang speculated that parity is not conserved in weak interactions and proposed crucial experiments to test the validity of parity conservation. Subsequent experiments indeed showed that observable effects do depend on pseudoscalar quantities such as correlation between S and p. To this day, one of the clearest demonstrations of parity nonconservation is the experiment which \ufb01rst discovered it. This result, from Wu et al., Phys. Rev., 105 (1957) 1413, shows a decay rate that depends on S \u00b7 p. The decay observed is 60Co \u219260 Ni + e\u2212 + \u00af\u03bde 265 4.3 Lattice Translation as a Discrete Symmetry.3 1.2 1.1 1.0 0.9 0.8 0.7 0.3 \u03b5\u03b3 0.2 0.1 0 1.20 m r a w 1.10.00 0.9 0. Gamma-anisotropy a) Equatorial counter b) Polar counter a) b) Gamma-anisotropy calculated from (a) & (b) \u03b5 = W(\u03c0/2) \u2212 W(O) W(\u03c0/2) for both polarizing field up & down (At pulse height IOV) Exchange gas in \u03b2 asymmetry H H Beta rays Spinning cobalt nuclei Beta rays (electrons) Mirror world This world 0.7 0 2 4 6 8 10 12 Time in minutes 14 16 18 Fig. 4.6 Experimental demonstration of parity nonconservation. The key observation, shown on the left, is that radioactive cobalt nuclei, oriented according to their nuclear spin, emit \u201cbeta rays\u201d(i.e. electrons) preferentially in the the opposite direction. The experiment data, shown on the right, show how the \u201cup/down\u201dbeta decay asymmetry (bottom panel) correlates perfectly with the signal which indicates the degree of nuclear polarization (upper panel). As time goes on, the sample warms up and the cobalt nuclei depolarize. where S is the spin of the 60Co nucleus and the momentum of the emitted e\u2212 is p. A sample of spin-polarized radioactive 60Co nuclei is prepared at low temperature, and the decay e\u2212 are detected in the direction parallel or antiparallel to the spin, depending on the sign of the polarizing magnetic \ufb01eld. The polarization of the sample is monitored by observing the anisotropy of the \u03b3-rays in the decay of the excited 60Ni daughter nuclei, a parity conserving effect. The results are shown in Figure 4.6. Over a period of several minutes, the sample warms up, and the \u03b2-decay asymmetry disappears at exactly the same rate as the \u03b3-ray anisotropy. Because parity is not conserved in weak interactions, previously thought \u201cpure\u201d nuclear and atomic states are, in fact, parity mixtures. These subtle effects have also been found experimentally. 4.3 Lattice Translation as a Discrete Symmetry We now consider another kind of discrete symmetry operation, namely, lattice translation. This subject has extremely important applications in solid-state physics. 266 Symmetry in Quantum Mechanics a a a a (a) a a (b) a a Fig. 4.7 (a) Periodic potential in one dimension with periodicity a. (b) The periodic potential when the barrier height between two adjacent lattice sites becomes infinite. Consider a periodic potential in one dimension, where V(x \u00b1 a) = V(x), as depicted in Figure 4.7. Realistically, we may consider the motion of an electron in a chain of regularly spaced positive ions", ". In general, the Hamiltonian is not invariant under a translation represented by \u03c4(l) with l arbitrary, where \u03c4(l) has the property (see Section 1.6) \u03c4\u2020(l)x\u03c4(l) = x + l, \u03c4(l)|x = |x + l. However, when l coincides with the lattice spacing a, we do have \u03c4\u2020(a)V(x)\u03c4(a) = V(x + a) =V (x). (4.85) (4.86) Because the kinetic-energy part of the Hamiltonian H is invariant under the translation with any displacement, the entire Hamiltonian satis\ufb01es \u03c4\u2020(a)H\u03c4(a) = H. Because \u03c4(a) is unitary, we have [from (4.87)] [H, \u03c4(a)] = 0, (4.87) (4.88) so the Hamiltonian and \u03c4(a) can be simultaneously diagonalized. Although \u03c4(a) is unitary, it is not Hermitian, so we expect the eigenvalue to be a complex number of modulus 1. Before we determine the eigenkets and eigenvalues of \u03c4(a) and examine their physical signi\ufb01cance, it is instructive to look at a special case of periodic potential when the barrier height between two adjacent lattice sites is made to go to in\ufb01nity, as in Figure 4.7b. What is the ground state for the potential of Figure 4.7b? Clearly, a state in which the particle is completely localized in one of the lattice sites can be a candidate for the ground state. To be speci\ufb01c let us assume that the particle is localized at the nth site and denote the corresponding ket by |n. This is an energy eigenket with energy eigenvalue E0, namely, H|n = E0|n. Its wave function x|n is \ufb01nite only in the nth site. However, we note that a 267 4.3 Lattice Translation as a Discrete Symmetry similar state localized at some other site also has the same energy E0, so actually there are denumerably in\ufb01nite ground states n, where n runs from \u2212\u221e to +\u221e. Now |n is obviously not an eigenket of the lattice translation operator, because when the lattice translation operator is applied to it, we obtain |n + 1: \u03c4(a)|n = |n + 1. (4.89) So despite the fact that \u03c4(a) commutes with H, |n \u2013 which is an eigenket of H \u2013 is not an eigenket of \u03c4(a). This is quite consistent with our earlier theorem on symmetry because we have an in\ufb01nitefold degeneracy. When there is such degeneracy, the symmetry of the world need not be the symmetry of energy eigenkets. Our task is to \ufb01nd a simultaneous eigenket of H and \u03c4(a). Here we may recall how we handled a somewhat similar situation with the symmetrical double-well potential of the previous section. We noted that even though neither |R nor |L is an eigenket of \u03c0, we could easily form a symmetrical and an antisymmetrical combination of |R and |L that are parity eigenkets. The case is analogous here. Let us speci\ufb01cally form a linear combination3 \u221e |\u03b8 \u2261 ein\u03b8|n, \u2211 n=\u2212\u221e where \u03b8 is a real parameter with \u2212\u03c0 \u2264 \u03b8 \u2264 \u03c0. We assert that |\u03b8 is a simultaneous eigenket of H and \u03c4(a). That it is an H eigenket is obvious because |n is an energy eigenket with eigenvalue E0, independent of n. To show that it is also an eigenket of the lattice translation operator we apply \u03c4(a) as follows: (4.90) \u03c4(a)|\u03b8 = ein\u03b8|n + 1 = \u221e \u2211 n=\u2212\u221e = e\u2212i\u03b8|\u03b8. \u221e \u2211 n=\u2212\u221e ei(n\u22121)\u03b8|n (4.91) Note that this simultaneous eigenket of H and \u03c4(a) is parameterized by a continuous parameter \u03b8. Furthermore, the energy eigenvalue E0 is independent of \u03b8. Let us now return to the more realistic situation of Figure 4.7a, where the barrier between two adjacent lattice sites is not in\ufb01nitely high. We can construct a localized ket |n just as before with the property \u03c4(a)|n = |n + 1.", " However, this time we expect that there is some leakage possible into neighboring lattice sites due to quantum-mechanical tunneling. In other words, the wave function x|n has a tail extending to sites other than the nth site. The diagonal elements of H in the {|n} basis are all equal because of translation invariance, that is, n|H|n = E0, (4.92) independent of n, as before. However we suspect that H is not completely diagonal in the {|n} basis due to leakage. Now, suppose the barriers between adjacent sites are high (but not in\ufb01nite). We then expect matrix elements of H between distant sites to be 3 Beware that the states |\u03b8 are not orthonormal. That is, the inner product \u03b8|\u03b8 is in\ufb01nite when \u03b8 = \u03b8, but is not necessarily zero otherwise. 268 Symmetry in Quantum Mechanics completely negligible. Let us assume that the only nondiagonal elements of importance connect immediate neighbors. That is, n|H|n = 0 only if n = n or n = n \u00b1 1. (4.93) In solid-state physics this assumption is known as the tight-binding approximation. Let us de\ufb01ne n \u00b1 1|H|n = \u2212\u0394. (4.94) Clearly, \u0394 is again independent of n due to translation invariance of the Hamiltonian. To the extent that |n and |n are orthogonal when n = n, we obtain H|n = E0|n \u2212\u0394|n + 1 \u2212\u0394|n \u2212 1. (4.95) Note that |n is no longer an energy eigenket. As we have done with the potential of Figure 4.7b, let us form a linear combination |\u03b8 = \u221e \u2211 n=\u2212\u221e ein\u03b8|n. (4.96) Clearly, |\u03b8 is an eigenket of translation operator \u03c4(a) because the steps in (4.91) still hold. A natural question is, is |\u03b8 an energy eigenket? To answer this question, we apply H: H\u2211 ein\u03b8|n = E0 \u2211 ein\u03b8|n \u2212\u0394 \u2211 ein\u03b8|n + 1 \u2212\u0394 \u2211 ein\u03b8|n \u2212 1 = E0 \u2211 ein\u03b8|n \u2212\u0394 \u2211(ein\u03b8\u2212i\u03b8 + ein\u03b8+i\u03b8)|n = (E0 \u2212 2\u0394 cos \u03b8)\u2211 ein\u03b8|n. (4.97) The big difference between this and the previous situation is that the energy eigenvalue now depends on the continuous real parameter \u03b8. The degeneracy is lifted as \u0394 becomes \ufb01nite, and we have a continuous distribution of energy eigenvalues between E0 \u22122\u0394 and E0 +2\u0394. See Figure 4.8, where we visualize how the energy levels start forming a continuous energy band as \u0394 is increased from zero. To see the physical meaning of the parameter \u03b8 let us study the wave function x|\u03b8. For the wave function of the lattice translated state \u03c4(a)|\u03b8, we obtain x|\u03c4(a)|\u03b8 = x \u2212 a|\u03b8 (4.98) by letting \u03c4(a) act on x|. But we can also let \u03c4(a) operate on |\u03b8 and use (4.91). Thus x|\u03c4(a)|\u03b8 = e\u2212i\u03b8x|\u03b8, so We solve this equation by setting x \u2212 a|\u03b8 = x|\u03b8e\u2212i\u03b8. x|\u03b8 = eikx uk(x), (4.99) (4.100) (4.101) 269 4.3 Lattice Translation as a Discrete Symmetry E 0 0 Fig. 4.8 Energy levels forming a continuous energy band as \u0394 is increased from zero. with \u03b8 = ka, where uk(x) is a periodic function with period a, as we can easily verify by explicit substitutions, namely, eik(x\u2212a)uk(x \u2212 a) =e ikx uk(x)e\u2212ika. (4.102) Thus we get the important condition known as Bloch\u2019s theorem: The wave function of |\u03b8, which is an eigenket of \u03c4(a), can be written as a plane wave eikx times a periodic function with periodicity a. Notice that the only fact we used was that |\u03b8 is an eigenket of \u03c4(a) with eigenvalue e\u2212i\u03b8 [see (4.91)]. In particular, the", " theorem holds even if the tight-binding approximation (4.93) breaks down. We are now in a position to interpret our earlier result (4.97) for |\u03b8 given by (4.96). We know that the wave function is a plane wave characterized by the propagation wave vector k modulated by a periodic function uk(x) [see (4.101)]. As \u03b8 varies from \u2212\u03c0 to \u03c0, the wave vector k varies from \u2212\u03c0/a to \u03c0/a. The energy eigenvalue E now depends on k as follows: E(k) =E 0 \u2212 2\u0394 cos ka. (4.103) Notice that this energy eigenvalue equation is independent of the detailed shape of the potential as long as the tight-binding approximation is valid. Note also that there is a cutoff in the wave vector k of the Bloch wave function (4.101) given by|k| = \u03c0/a. Equation (4.103) de\ufb01nes a dispersion curve, as shown in Figure 4.9. As a result of tunneling, the denumerably in\ufb01nitefold degeneracy is now completely lifted, and the allowed energy values form a continuous band between E0 \u2212 2\u0394 and E0 + 2\u0394, known as the Brillouin zone. 270 Symmetry in Quantum Mechanics E(k) E 0 + 2 E0 E 0 \u2212 2 \u2013\u03c0/a 0 +\u03c0/a Fig. 4.9 Dispersion curve for E(k) versusk in the Brillouin zone |k| \u2264 \u03c0/a. So far we have considered only one particle moving in a periodic potential. In a more realistic situation we must look at many electrons moving in such a potential. Actually the electrons satisfy the Pauli exclusion principle, as we will discuss more systematically in Chapter 7, and they start \ufb01lling the band. In this way, the main qualitative features of metals, semiconductors, and the like can be understood as a consequence of translation invariance supplemented by the exclusion principle. The reader may have noted the similarity between the symmetrical double-well problem of Section 4.2 and the periodic potential of this section. Comparing Figures 4.3 and 4.7, we note that they can be regarded as opposite extremes (two versus in\ufb01nite) of potentials with a \ufb01nite number of troughs. 4.4 The Time-Reversal Discrete Symmetry In this section we study another discrete symmetry operator, called time reversal. This is a dif\ufb01cult topic for the novice, partly because the term time reversal is a misnomer; it reminds us of science \ufb01ction. Actually what we do in this section can be more appropriately characterized by the term reversal of motion. Indeed, that is the terminology used by E. Wigner, who formulated time reversal in a very fundamental paper written in 1932. For orientation purposes let us look at classical mechanics. Suppose there is a trajectory of a particle subject to a certain force \ufb01eld; see Figure 4.10. At t = 0, let the particle stop and reverse its motion: p|t=0 \u2192 \u2212p|t=0. The particle traverses backward along the same trajectory. If you run the motion picture of trajectory (a) backward as in (b), you may have a hard time telling whether this is the correct sequence. More formally, if x(t) is a solution to m\u00a8x = \u2212\u2207V(x), (4.104) then x(\u2212t) is also a possible solution in the same force \ufb01eld derivable from V. It is, of course, important to note that we do not have a dissipative force here. A block sliding on a 271 4.4 The Time-Reversal Discrete Symmetry At t = 0 Stop Reverse p t=0 \u2192 \u2212p t=0 (a) (b) Fig. 4.10 Classical trajectory which (a) stops at t = 0 and (b) reverses its motion p|t=0 \u2192 \u2212p|t=0. table decelerates (due to friction) and eventually stops. But have you ever seen a block on a table spontaneously start to move and accelerate? With a magnetic \ufb01eld you may be able to tell the difference. Imagine that you are taking the motion picture of a spiraling electron trajectory in a magnetic \ufb01eld. You may be able to tell whether the motion picture is run forward or backward by comparing the sense of rotation with the magnetic pole labeling N and S. However, from a microscopic point of view, B is produced by moving charges via an electric current; if you could reverse the current that causes B, then the situation would be quite symmetrical. In terms of the picture shown in Figure", " 4.11, you may have \ufb01gured out that N and S are mislabeled! Another more formal way of saying all this is that the Maxwell equations, for example, \u2207 \u00b7 E = 4\u03c0\u03c1, \u2207 \u00d7 B \u2212 1 c \u2202E \u2202t and the Lorentz force equation F = e[E + (1/c)(v \u00d7 B)] are invariant under t \u2192 \u2212t provided we also let, \u2207 \u00d7 E = \u2212 1 c 4\u03c0j c \u2202B \u2202t (4.105) =, E \u2192 E, B \u2192 \u2212B, \u03c1 \u2192 \u03c1, j \u2192 \u2212j, v \u2192 \u2212v. (4.106) Let us now look at wave mechanics, where the basic equation of the Schr\u00a8odinger wave equation is i\u00afh \u2202\u03c8 \u2202t = \u2212 \u00afh2 2m \u22072 + V \u03c8. (4.107) Suppose \u03c8(x, t) is a solution. We can easily verify that \u03c8(x, \u2212t) is not a solution, because of the appearance of the \ufb01rst-order time derivative. However, \u03c8\u2217(x, \u2212t) is a solution, as you may verify by complex conjugation of (4.107). It is instructive to convince ourselves of this point for an energy eigenstate, that is, by substituting \u03c8(x, t) =u n(x)e\u2212iEnt/\u00afh, \u03c8\u2217(x, \u2212t) = u\u2217 n(x)e\u2212iEnt/\u00afh (4.108) into the Schr\u00a8odinger equation (4.107). Thus we conjecture that time reversal must have something to do with complex conjugation. If at t = 0 the wave function is given by \u03c8 = x|\u03b1, (4.109) 272 Symmetry in Quantum Mechanics S B N Electron trajectory Fig. 4.11 Electron trajectory between the north and south poles of a magnet. then the wave function for the corresponding time-reversed state is given by x|\u03b1\u2217. We will later show that this is indeed the case for the wave function of a spinless system. As an example, you may easily check this point for the wave function of a plane wave; see Problem 4.8 of this chapter. 4.4.1 Digression on Symmetry Operations Before we begin a systematic treatment of the time-reversal operator, some general remarks on symmetry operations are in order. Consider a symmetry operation |\u03b1 \u2192 | \u02dc\u03b1, |\u03b2 \u2192 | \u02dc\u03b2. (4.110) One may argue that it is natural to require the inner product \u03b2|\u03b1 to be preserved, that is, \u02dc\u03b2| \u02dc\u03b1 = \u03b2|\u03b1. (4.111) Indeed, for symmetry operations such as rotations, translations, and even parity, this is the case. If |\u03b1 is rotated and |\u03b2 is also rotated in the same manner, \u03b2|\u03b1 is unchanged. Formally this arises from the fact that, for the symmetry operations considered in the previous sections, the corresponding symmetry operator is unitary, so 273 4.4 The Time-Reversal Discrete Symmetry \u03b2|\u03b1 \u2192 \u03b2|U \u2020U|\u03b1 = \u03b2|\u03b1. (4.112) However, in discussing time reversal, we see that requirement (4.111) turns out to be too restrictive. Instead, we merely impose the weaker requirement that | \u02dc\u03b2| \u02dc\u03b1| = |\u03b2|\u03b1|. Requirement (4.111) obviously satis\ufb01es (4.113). But this is not the only way; \u02dc\u03b2| \u02dc\u03b1 = \u03b2|\u03b1\u2217 = \u03b1|\u03b2 (4.113) (4.114) works equally well. We pursue the latter possibility in this section because from our earlier discussion based on the Schr\u00a8odinger equation we inferred that time reversal has something to do with complex conjugation. Definition 1 The transformation |\u03b1 \u2192 | \u02dc\u03b1 = \u03b8|\u03b1, |\u03b2 \u2192 | \u02dc\u03b2 = \u03b8|\u03b2 (4.115) is said to be antiunitary if \u02dc\u03b2| \u02dc\u03b1 = \u03b2|\u03b1\u2217, \u03b8(c1|\u03b1 + c2|\u03b2) =c \u2217 1 \u03b8|\u03b1 + c\u2217 2 \u03b8|\u03b2. (4.116a) (4.116b) In such a case the operator \u03b8 is an antiunitary operator. Relation (4.116b) alone de\ufb01nes an antilinear operator. We now claim that an antiunitary operator can be written as \u03b8 = UK, (4.117) where U is a unitary operator", " and K is the complex conjugate operator that forms the complex conjugate of any coef\ufb01cient that multiplies a ket (and stands on the right of K). Before checking (4.116) let us examine the property of the K operator. Suppose we have a ket multiplied by a complex number c. We then have Kc|\u03b1 = c\u2217K|\u03b1. (4.118) One may further ask: What happens if |\u03b1 is expanded in terms of base kets {|a}? Under the action K we have |\u03b1 = \u2211 a = \u2211 a |aa|\u03b1 K\u2192| \u02dc\u03b1 = \u2211 a a|\u03b1\u2217K|a a|\u03b1\u2217|a. (4.119) Notice that K acting on the base ket does not change the base ket. The explicit representation of |a is 274 Symmetry in Quantum Mechanics |... 0 1 0... 0 (4.120) and there is nothing to be changed by K. The reader may wonder, for instance, whether the Sy eigenkets for a spin 1 2 system change under K. The answer is that if the Sz eigenkets are used as base kets, we must change the Sy eigenkets because the Sy eigenkets (1.14) undergo under K 1\u221a 2 |+ \u00b1 i\u221a 2 |\u2212 \u2192 1\u221a 2 |+ \u2213 i\u221a |\u2212. 2 (4.121) K On the other hand, if the Sy eigenkets themselves are used as the base kets, we do not change the Sy eigenkets under the action of K. Thus the effect of K changes with the basis. As a result, the form of U in (4.117) also depends on the particular representation (that is, the choice of base kets) used. Returning to \u03b8 = UK and (4.116), let us \ufb01rst check property (4.116b). We have \u03b8(c1|\u03b1 + c2|\u03b2) =UK (c1|\u03b1 + c2|\u03b2) = c\u2217 2UK|\u03b2 = c\u2217 \u03b8|\u03b2, 1 1UK|\u03b1 + c\u2217 \u03b8|\u03b1 + c\u2217 2 (4.122) so (4.116b) indeed holds. Before checking (4.116a), we assert that it is always safer to work with the action of \u03b8 on kets only. We can \ufb01gure out how the bras change just by looking at the corresponding kets. In particular, it is not necessary to consider \u03b8 acting on bras from the right, nor is it necessary to de\ufb01ne \u03b8\u2020. We have |\u03b1 \u03b8\u2192| \u02dc|\u03b1\u2217UK|a a|\u03b1\u2217U|a \u03b1|aU|a. As for |\u03b2, we have a|\u03b2a|U \u2020 a|\u03b2\u2217U|aDC\u2194 \u02dc\u03b2| = \u2211 a a|\u03b2a|U \u2020U|a\u03b1|a \u2211 a \u03b1|aa|\u03b2 = \u03b1|\u03b2 | \u02dc\u03b2 = \u2211 a \u02dc\u03b2| \u02dc\u03b1 = \u2211 a = \u2211 a = \u03b2|\u03b1\u2217, (4.123) (4.124) so this checks. (Recall the notion of dual correspondence, or DC, from Section 1.2.) 275 4.4 The Time-Reversal Discrete Symmetry In order for (4.113) to be satis\ufb01ed, it is of physical interest to consider just two types of transformation, unitary and antiunitary. Other possibilities are related to either of the preceding via trivial phase changes. The proof of this assertion is actually very dif\ufb01cult and will not be discussed further here. See, however, Gottfried and Yan (2003), Section 7.1. 4.4.2 Time-Reversal Operator We are \ufb01nally in a position to present a formal theory of time reversal. Let us denote the time-reversal operator by \u0398, to be distinguished from \u03b8, a general antiunitary operator. Consider |\u03b1 \u2192\u0398| \u03b1, (4.125) where \u0398|\u03b1 is the time-reversed state. More appropriately, \u0398|\u03b1 should be called the motion-reversed state. If |\u03b1 is a momentum eigenstate |p, we expect \u0398|\u03b1 to be | \u2212p up to a possible phase. Likewise, J is to be reversed under time reversal. We now deduce the fundamental property of the time-reversal operator by looking at the time evolution of the time-reversed state. Consider a physical system represented by a ket |\u03b1,", " say at t = 0. Then at a slightly later time t = \u03b4t, the system is found in |\u03b1, t0 = 0; t = \u03b4t = \u03b4t |\u03b1, (4.126) 1 \u2212 iH \u00afh where H is the Hamiltonian that characterizes the time evolution. Instead of the preceding equation, suppose we \ufb01rst apply \u0398, say at t = 0, and then let the system evolve under the in\ufb02uence of the Hamiltonian H. We then have at \u03b4t 1 \u2212 iH\u03b4t \u00afh \u0398|\u03b1. (4.127a) If motion obeys symmetry under time reversal, we expect the preceding state ket to be the same as \u0398|\u03b1, t0 = 0; t = \u2212\u03b4t (4.127b) that is, \ufb01rst consider a state ket at earlier time t = \u2212\u03b4t, and then reverse p and J; see Figure 4.12. Mathematically, 1 \u2212 iH \u00afh 1 \u2212 iH \u00afh \u03b4t \u0398|\u03b1 = \u0398 (\u2212\u03b4t) |\u03b1. (4.128) If the preceding relation is to be true for any ket, we must have \u2212iH\u0398| = \u0398iH|, (4.129) where the blank ket | emphasizes that (4.129) is to be true for any ket. We now argue that \u0398 cannot be unitary if the motion of time reversal is to make sense. Suppose \u0398 were unitary. It would then be legitimate to cancel the i in (4.129), and we would have the operator equation \u2212H\u0398 = \u0398H. (4.130) 276 Symmetry in Quantum Mechanics Momentum before reversal Momentum after reversal Momentum before reversal t = 0 t = 0 At t = +\u03b4t At t = \u2212\u03b4t Momentum after reversal (a) (b) Fig. 4.12 Momentum before and after time reversal at time t = 0 and t = \u00b1\u03b4t. Consider an energy eigenket |n with energy eigenvalue En. The corresponding timereversed state would be \u0398|n, and we would have, because of (4.130), H\u0398|n = \u2212\u0398H|n = (\u2212En)\u0398|n. (4.131) This equation says that \u0398|n is an eigenket of the Hamiltonian with energy eigenvalues \u2212En. But this is nonsensical even in the very elementary case of a free particle. We know that the energy spectrum of the free particle is positive semide\ufb01nite, from 0 to +\u221e. There is no state lower than a particle at rest (momentum eigenstate with momentum eigenvalue zero); the energy spectrum ranging from \u2212\u221e to 0 would be completely unacceptable. We can also see this by looking at the structure of the free-particle Hamiltonian. We expect p to change sign but not p2; yet (4.130) would imply that \u0398\u22121 p2 2m \u0398 = \u2212p2 2m. (4.132) All these arguments strongly suggest that if time reversal is to be a useful symmetry at all, we are not allowed to cancel the i in (4.129); hence, \u0398 had better be antiunitary. In this case the right-hand side of (4.129) becomes \u0398iH| = \u2212i\u0398H| (4.133) by antilinear property (4.116b). Now at last we can cancel the i in (4.129) leading, \ufb01nally, via (4.133) to \u0398H = H\u0398. (4.134) Equation (4.134) expresses the fundamental property of the Hamiltonian under time reversal. With this equation the dif\ufb01culties mentioned earlier [see (4.130) to (4.132)] are absent, and we obtain physically sensible results. From now on we will always take \u0398 to be antiunitary. 277 4.4 The Time-Reversal Discrete Symmetry We mentioned earlier that it is best to avoid an antiunitary operator acting on bras from the right. Nevertheless, we may use which is always to be understood as and never as \u03b2|\u0398|\u03b1, (\u03b2|) \u00b7 (\u0398|\u03b1) (\u03b2|\u0398) \u00b7 |\u03b1. (4.135) (4.136) (4.137) In fact, we do not even attempt to de\ufb01ne \u03b2|\u0398. This is one place where the Dirac bra-ket notation is a little confusing. After all, that notation was invented to handle linear operators, not antilinear", " operators. With this cautionary remark, we are in a position to discuss the behavior of operators under time reversal. We continue to take the point of view that the \u0398 operator is to act on kets | \u02dc\u03b1 = \u0398|\u03b1, | \u02dc\u03b2 = \u0398|\u03b2, (4.138) yet it is often convenient to talk about operators \u2013 in particular, observables \u2013 which are odd or even under time reversal. We start with an important identity, namely, \u03b2|X |\u03b1 = \u02dc\u03b1|\u0398X \u2020\u0398\u22121| \u02dc\u03b2, (4.139) where X is a linear operator. This identity follows solely from the antiunitary nature of \u0398. To prove this let us de\ufb01ne By dual correspondence we have Hence, |\u03b3 \u2261 X \u2020|\u03b2. |\u03b3DC\u2194\u03b2|X = \u03b3|. \u03b2|X |\u03b1 = \u03b3|\u03b1 = \u02dc\u03b1| \u02dc\u03b3 = \u02dc\u03b1|\u0398X \u2020|\u03b2 = \u02dc\u03b1|\u0398X \u2020\u0398\u22121\u0398|\u03b2 = \u02dc\u03b1|\u0398X \u2020\u0398\u22121| \u02dc\u03b2, which proves the identity. In particular, for Hermitian observables A we get \u03b2|A|\u03b1 = \u02dc\u03b1|\u0398A\u0398\u22121| \u02dc\u03b2. (4.140) (4.141) (4.142) (4.143) We say that observables are even or odd under time reversal according to whether we have the upper or lower sign in \u0398A\u0398\u22121 = \u00b1A. (4.144) 278 Symmetry in Quantum Mechanics Note that this equation, together with (4.143), gives a phase restriction on the matrix elements of A taken with respect to time reversed states as follows: \u03b2|A|\u03b1 = \u00b1 \u02dc\u03b2|A| \u02dc\u03b1\u2217. If |\u03b2 is identical to |\u03b1, so that we are talking about expectation values, we have \u03b1|A|\u03b1 = \u00b1 \u02dc\u03b1|A| \u02dc\u03b1, (4.145) (4.146) where \u02dc\u03b1|A| \u02dc\u03b1 is the expectation value taken with respect to the time-reversed state. As an example, let us look at the expectation value of p. It is reasonable to expect that the expectation value of p taken with respect to the time-reversed state be of opposite sign. Thus so we take p to be an odd operator, namely, \u03b1|p|\u03b1 = \u2212 \u02dc\u03b1|p| \u02dc\u03b1, This implies that \u0398p\u0398\u22121 = \u2212p. p\u0398|p = \u2212\u0398p\u0398\u22121\u0398|p = (\u2212p)\u0398|p. (4.147) (4.148) (4.149) Equation (4.149) agrees with our earlier assertion that \u0398|p is a momentum eigenket with eigenvalue \u2212p. It can be identi\ufb01ed with | \u2212p itself with a suitable choice of phase. Likewise, we obtain \u0398x\u0398\u22121 = x \u0398|x = |x (up to a phase) from the (eminently reasonable) requirement \u03b1|x|\u03b1 = \u02dc\u03b1|x| \u02dc\u03b1. We can now check the invariance of the fundamental commutation relation [xi, pj]| = i\u00afh\u03b4ij|, (4.150) (4.151) (4.152) where the blank ket | stands for any ket. Applying \u0398 to both sides of (4.152), we have \u0398 [xi, pj]\u0398\u22121\u0398| = \u0398i\u00afh\u03b4ij|, (4.153) which leads to, after passing \u0398 through i\u00afh, [xi, (\u2212pj)]\u0398| = \u2212i\u00afh\u03b4ij\u0398|. (4.154) Note that the fundamental commutation relation [xi, pj] = i\u00afh\u03b4ij is preserved by virtue of the fact that \u0398 is antiunitary. This can be given as yet another reason for taking \u0398 to be antiunitary; otherwise, we will be forced to abandon either (4.148) or (4.150)! Similarly, to preserve [Ji, Jj] =i \u00afh\u03b5ijkJk, (4.155) 279 4.4 The Time-Reversal Discrete Symmetry the angular-momentum operator must be odd under time reversal, that is, \u0398J\u0398\u22121 = \u2212J. (4.156) This is consistent for a spinless system where J is just x \u00d7 p. Alternatively, we could have deduced", " this relation by noting that the rotational operator and the time-reversal operator commute (note the extra i!). 4.4.3 Wave Function Suppose at some given time, say at t = 0, a spinless single-particle system is found in a state represented by |\u03b1. Its wave function x|\u03b1 appears as the expansion coef\ufb01cient in the position representation |\u03b1 = d3x|xx|\u03b1. (4.157) Applying the time-reversal operator \u0398|\u03b1 = d3x\u0398|xx|\u03b1\u2217 = d3x|xx|\u03b1\u2217, (4.158) where we have chosen the phase convention so that \u0398|x is |x itself. We then recover the rule \u03c8 (x) \u2192 \u03c8\u2217(x) (4.159) inferred earlier by looking at the Schr\u00a8odinger wave equation [see (4.108)]. The angular part of the wave function is given by a spherical harmonic Ym l. With the usual phase convention we have l (\u03b8, \u03c6) = (\u22121) mY\u2212m l (\u03b8, \u03c6) is the wave function for |l, m [see (3.232)]; therefore, from (4.159) we l (\u03b8, \u03c6) \u2192 Ym\u2217 Ym (\u03b8, \u03c6). (4.160) l Now Ym deduce \u0398|l, m = (\u22121)m|l, \u2212m. (4.161) If we study the probability current density (2.191) for a wave function of type (3.231) going like R(r)Ym l, we shall conclude that for m > 0 the current \ufb02ows in the counterclockwise direction, as seen from the positive z-axis. The wave function for the corresponding timereversed state has its probability current \ufb02owing in the opposite direction because the sign of m is reversed. All this is very reasonable. As a nontrivial consequence of time-reversal invariance, we state an important theorem on the reality of the energy eigenfunction of a spinless particle. Theorem 7 Suppose the Hamiltonian is invariant under time reversal and the energy eigenket |n is nondegenerate; then the corresponding energy eigenfunction is real (or, more generally, a real function times a phase factor independent of x). 280 Symmetry in Quantum Mechanics Proof To prove this, \ufb01rst note that H\u0398|n = \u0398H|n = En\u0398|n, so |n and \u0398|n have the same energy. The nondegeneracy assumption prompts us to conclude that |n and \u0398|n must represent the same state; otherwise there would be two different states with the same energy En, an obvious contradiction! Let us recall that the wave functions for |n and \u0398|n are x|n and x|n\u2217, respectively. They must be the same, that is, (4.162) x|n = x|n\u2217 (4.163) for all practical purposes, or, more precisely, they can differ at most by a phase factor independent of x. Thus if we have, for instance, a nondegenerate bound state, its wave function is always real. On the other hand, in the hydrogen atom with l = 0, m = 0, the energy eigenfunction characterized by de\ufb01nite (n, l, m) quantum numbers is complex because Ym is complex; this l does not contradict the theorem because |n, l, m and |n, l, \u2212m are degenerate. Similarly, the wave function of a plane wave eip\u00b7x/\u00afh is complex, but it is degenerate with e\u2212ip\u00b7x/\u00afh. We see that for a spinless system, the wave function for the time-reversed state, say at t = 0, is simply obtained by complex conjugation. In terms of ket |\u03b1 written as in (4.119) or in (4.157), the \u0398 operator is the complex conjugate operator K itself because K and \u0398 have the same effect when acting on the base ket |a (or |x). We may note, however, that the situation is quite different when the ket |\u03b1 is expanded in terms of the momentum eigenket because \u0398 must change |p into | \u2212p as follows: \u0398|\u03b1 = d3p| \u2212p p|\u03b1\u2217 = d3p|p\u2212p|\u03b1\u2217. (4.164) It is apparent that the momentum-space wave function of the time-reversed state is not just the complex conjugate of the original momentum-space wave function; rather, we must identify", " \u03c6\u2217(\u2212p) as the momentum-space wave function for the time-reversed state. This situation once again illustrates the basic point that the particular form of \u0398 depends on the particular representation used. 4.4.4 Time Reversal for a Spin 1 2 System The situation is even more interesting for a particle with spin, and for spin 1 2, in particular. We recall from Section 3.2 that the eigenket of S\u00b7 \u02c6n with eigenvalue \u00afh/2 can be written as | \u02c6n; + = e\u2212iSz \u03b1/\u00afhe\u2212iSy\u03b2/\u00afh|+, where \u02c6n is characterized by the polar and azimuthal angles \u03b2 and \u03b1, respectively. Noting (4.156) we have (4.165) \u0398| \u02c6n; + = e\u2212iSz\u03b1/\u00afhe\u2212iSy \u03b2/\u00afh\u0398|+ = \u03b7| \u02c6n; \u2212. On the other hand, we can easily verify that | \u02c6n; \u2212 = e\u2212i\u03b1Sz/\u00afhe\u2212i(\u03c0+\u03b2)Sy/\u00afh|+. (4.166) (4.167) 281 4.4 The Time-Reversal Discrete Symmetry In general, we saw earlier that the product UK is an antiunitary operator. Comparing (4.166) and (4.167) with \u0398 set equal to UK, and noting that K acting on the base ket |+ gives just |+, we see that \u0398 = \u03b7e\u2212i\u03c0Sy/\u00afhK = \u2212i\u03b7 K, (4.168) 2Sy \u00afh where \u03b7 stands for an arbitrary phase (a complex number of modulus unity). Another way to be convinced of (4.168) is to verify that if \u03c7( \u02c6n; +) is the two-component eigenspinor corresponding to | \u02c6n; + [in the sense that \u03c3\u00b7 \u02c6n\u03c7( \u02c6n; +) = \u03c7( \u02c6n; +)], then \u2212i\u03c3y\u03c7\u2217( \u02c6n; +) (4.169) (note the complex conjugation!) is the eigenspinor corresponding to | \u02c6n; \u2212, again up to an arbitrary phase, see Problem 4.7 of this chapter. The appearance of Sy or \u03c3y can be traced to the fact that we are using the representation in which Sz is diagonal and the nonvanishing matrix elements of Sy are purely imaginary. Let us now note e\u2212i\u03c0Sy/\u00afh|+ = +|\u2212, e\u2212i\u03c0Sy/\u00afh|\u2212 = \u2212|+. (4.170) Using (4.170), we are in a position to work out the effect of \u0398, written as (4.168), on the most general spin 1 2 ket: \u0398(c+|+ + c\u2212|\u2212) = +\u03b7c\u2217 + |\u2212 \u2212 \u03b7c\u2217 \u2212|+. (4.171) Let us apply \u0398 once again: \u03982(c+|+ + c\u2212|\u2212) = \u2212|\u03b7|2c+|+ \u2212 |\u03b7|2c\u2212|\u2212 = \u2212(c+|+ + c\u2212|\u2212) or \u03982 = \u22121 (4.172) (4.173) (where \u22121 is to be understood as \u22121 times the identity operator), for any spin orientation. This is an extraordinary result. It is crucial to note here that our conclusion is completely independent of the choice of phase; (4.173) holds no matter what phase convention we may use for \u03b7. In contrast, we may note that two successive applications of \u0398 to a spinless state give \u03982 = +1 as is evident from, say, (4.161). More generally, we now prove \u03982| j half-integer = \u2212| j half-integer \u03982| j integer = +| j integer. (4.174) (4.175a) (4.175b) 282 Symmetry in Quantum Mechanics Thus the eigenvalue of \u03982 is given by (\u22121)2j. We \ufb01rst note that (4.168) generalizes for arbitrary j to \u0398 = \u03b7e\u2212i\u03c0Jy/\u00afhK. For a ket |\u03b1 expanded in terms of | j, m base eigenkets, we have \u0398 (\u0398 \u2211 | jmjm|\u03b1) = \u0398 % \u03b7 \u2211 e\u2212i\u03c0Jy/\u00afh| jmjm|\u03b1\u2217 = |", "\u03b7|2e\u22122i\u03c0Jy/\u00afh \u2211 | jmjm|\u03b1. But e\u22122i\u03c0Jy/\u00afh|jm = (\u22121)2j| jm, & (4.176) (4.177) (4.178) as is evident from the properties of angular-momentum eigenstates under rotation by 2\u03c0. In (4.175b), | j integer may stand for the spin state 1\u221a 2 (| + \u2212 \u00b1 | \u2212 +) (4.179) of a two-electron system or the orbital state |l, m of a spinless particle. It is important only that j is an integer. Likewise, | j half-integer may stand, for example, for a three-electron system in any con\ufb01guration. Actually, for a system made up exclusively of electrons, any system with an odd (even) number of electrons \u2013 regardless of their spatial orientation (for example, relative orbital angular momentum) \u2013 is odd (even) under \u03982; they need not even be J2 eigenstates! We make a parenthetical remark on the phase convention. In our earlier discussion based on the position representation, we saw that with the usual convention for spherical harmonics it is natural to choose the arbitrary phase for |l, m under time reversal so that \u0398|l, m = (\u22121)m|l, \u2212m. Some authors \ufb01nd it attractive to generalize this to obtain \u0398|j, m = (\u22121)m| j, \u2212m ( j an integer) (4.180) (4.181) regardless of whether j refers to l or s (for an integer spin system). We may naturally ask 2 system when we visualize | j, m as whether this is compatible with (4.175a) for a spin 1 being built up of \u201cprimitive\u201d spin 1 2 objects according to Wigner and Schwinger. It is easy to see that (4.175a) is indeed consistent provided we choose \u03b7 in (4.176) to be+ i. In fact, in general, we can take \u0398| j, m = i2m| j, \u2212m (4.182) for any j, either a half-integer j or an integer j; seeProblem 4.10 of this chapter. The reader should be warned, however, that this is not the only convention found in the literature. See, for instance, Henley and Garcia (2007). For some physical applications, it is more convenient to use other choices; for instance, the phase convention that makes the J\u00b1 operator matrix elements simple is not the phase convention that makes the timereversal operator properties simple. We emphasize once again that (4.175) is completely independent of phase convention. 283 4.4 The Time-Reversal Discrete Symmetry Having worked out the behavior of angular-momentum eigenstates under time reversal, we are in a position to study once again the expectation values of a Hermitian operator. Recalling (4.146), we obtain under time reversal (canceling the i2m factors) \u03b1, j, m|A|\u03b1, j, m = \u00b1\u03b1, j, \u2212m|A|\u03b1, j, \u2212m. (4.183) Now suppose A is a component of a spherical tensor T(k) q. Because of the Wigner\u2013Eckart theorem, it is suf\ufb01cient to examine just the matrix element of the q = 0 component. In general, T(k) (assumed to be Hermitian) is said to be even or odd under time reversal depending on how its q = 0 component satis\ufb01es the upper or lower sign in \u0398T(k) q=0 \u0398\u22121 = \u00b1T(k) q=0. Equation (4.183) for A = T(k) 0 becomes \u03b1, j, m|T(k) 0 |\u03b1, j, m = \u00b1\u03b1, j, \u2212m|T(k) 0 |\u03b1, j, \u2212m. (4.184) (4.185) Due to (3.255)\u2013(3.258), we expect |\u03b1, j, \u2212m = D(0, \u03c0, 0)|\u03b1, j, m up to a phase. We next use (3.465) forT (k) 0, which leads to D \u2020(0, \u03c0, 0)T(k) 0 D(0, \u03c0, 0) = (\u22121)kT(k) 0 + (q = 0 components), (4.186) where we have used D (k) 00 (0, \u03c0, 0) =P k(cos \u03c0) =", " (\u22121)k, and the q = 0 components give vanishing contributions when sandwiched between \u03b1, j, m| and |\u03b1, j, m. The net result is \u03b1, j, m|T(k) 0 |\u03b1, j, m = \u00b1(\u22121)k\u03b1, j, m|T(k) 0 |\u03b1, j, m. (4.187) As an example, taking k = 1, the expectation value x taken with respect to eigenstates of j, m vanishes. We may argue that we already know x = 0 from parity inversion if the expectation value is taken with respect to parity eigenstates [see (4.81)]. But note that here |\u03b1, j, m need not be parity eigenkets! For example, the | j, m for spin 1 2 particles could be cs|s1/2 + cp|p1/2. 4.4.5 Interactions with Electric and Magnetic Fields; Kramers Degeneracy Consider charged particles in an external electric or magnetic \ufb01eld. If we have only a static electric \ufb01eld interacting with the electric charge, the interaction part of the Hamiltonian is just V(x) =e \u03c6(x), (4.188) where \u03c6(x) is the electrostatic potential. Because \u03c6(x) is a real function of the timereversal even operator x, we have [\u0398, H] =0. (4.189) Unlike the parity case, (4.189) does not lead to an interesting conservation law. The reason is that \u0398U(t, t0) = U(t, t0)\u0398 (4.190) 284 Symmetry in Quantum Mechanics even if (4.189) holds, so our discussion following (4.9) ofSection 4.1 breaks down. As a result, there is no such thing as the \u201cconservation of time-reversal quantum number.\u201d As we already mentioned, requirement (4.189) does, however, lead to a nontrivial phase restriction \u2013 the reality of a nondegenerate wave function for a spinless system [see (4.162) and (4.163)]. Another far-reaching consequence of time-reversal invariance is the Kramers degeneracy. Suppose H and \u0398 commute, and let |n and \u0398|n be the energy eigenket and its time-reversed state, respectively. It is evident from (4.189) that |n and \u0398|n belong to the same energy eigenvalue En(H\u0398|n = \u0398H|n = En\u0398|n). The question is, does |n represent the same state as \u0398|n? If it does, |n and \u0398|n can differ at most by a phase factor. Hence, \u0398|n = ei\u03b4|n. (4.191) Applying \u0398 again to (4.191), we have \u03982|n = \u0398ei\u03b4|n = e\u2212i\u03b4\u0398|n = e\u2212i\u03b4e+i\u03b4|n; hence, \u03982|n = +|n. (4.192) But this relation is impossible for half-integer j systems, for which \u03982 is always \u22121, so we are led to conclude that |n and \u0398|n, which have the same energy, must correspond to distinct states, that is, there must be a degeneracy. This means, for instance, that for a system composed of an odd number of electrons in an external electric \ufb01eld E, each energy level must be at least twofold degenerate no matter how complicated E may be. Considerations along this line have interesting applications to electrons in crystals where odd-electron and even-electron systems exhibit very different behaviors. Historically, Kramers inferred degeneracy of this kind by looking at explicit solutions of the Schr\u00a8odinger equation; subsequently, Wigner pointed out that Kramers degeneracy is a consequence of time-reversal invariance. Let us now turn to interactions with an external magnetic \ufb01eld. The Hamiltonian H may then contain terms like S\u00b7B, p\u00b7A + A\u00b7p (B = \u2207 \u00d7 A), (4.193) where the magnetic \ufb01eld is to be regarded as external. The operators S and p are odd under time reversal; these interaction terms therefore do lead to \u0398H = H\u0398. (4.194) 2 system the spin-up state |+ and its time-reversed state As a trivial example, for a spin 1 |\u2212 no longer have the same energy in the presence of an external magnetic \ufb01eld. In", " general, Kramers degeneracy in a system containing an odd number of electrons can be lifted by applying an external magnetic \ufb01eld. Notice that when we treat B as external, we do not change B under time reversal; this is because the atomic electron is viewed as a closed quantum-mechanical system to which we apply the time-reversal operator. This should not be confused with our earlier remarks concerning the invariance of the Maxwell equations (4.105) and the Lorentz force equation under t \u2192 \u2212t and (4.106). There we were to apply time reversal to the whole world, for example, even to the currents in the wire that produces the B \ufb01eld! 285 Problems Problems 4.1 Calculate the three lowest energy levels, together with their degeneracies, for the following systems (assume equal mass distinguishable particles). a. Three noninteracting spin 1 b. Four noninteracting spin 1 2 particles in a cubic box of side length L. 2 particles in a cubic box of side length L. 4.2 Let Td denote the translation operator (displacement vector d), D( \u02c6n, \u03c6) the rotation operator ( \u02c6n and \u03c6 are the axis and angle of rotation, respectively), and \u03c0 the parity operator. Which, if any, of the following pairs commute? Why? a. Td and Td (d and d in different directions). b. D( \u02c6n, \u03c6) and D( \u02c6n, \u03c6)( \u02c6n and \u02c6n in different directions). c. Td and \u03c0. d. D( \u02c6n, \u03c6) and \u03c0. 4.3 A quantum-mechanical state |\u03a8 is known to be a simultaneous eigenstate of two Hermitian operators A and B which anticommute, AB + BA = 0. What can you say about the eigenvalues of A and B for state |\u03a8? Illustrate your point using the parity operator (which can be chosen to satisfy \u03c0 = \u03c0\u22121 = \u03c0\u2020) and the momentum operator. 4.4 A spin 1 2 particle is bound to a \ufb01xed center by a spherically symmetrical potential. a. Write down the spin angular function Y j=1/2,m=1/2 b. Express (\u03c3\u00b7x) Y j=1/2,m=1/2 c. Show that your result in (b) is understandable in view of the transformation properties of the operator S\u00b7x under rotations and under space inversion (parity). l=0 in terms of some other Y j,m l=0.. l 4.5 Because of weak (neutral-current) interactions there is a parity-violating potential between the atomic electron and the nucleus as follows: V = \u03bb[\u03b4(3)(x)S\u00b7p + S\u00b7p\u03b4(3)(x)], where S and p are the spin and momentum operators of the electron, and the nucleus is assumed to be situated at the origin. As a result, the ground state of an alkali atom, usually characterized by |n, l, j, m, actually contains very tiny contributions from other eigenstates as follows: |n, l, j, m \u2192 |n, l, j, m + \u2211 nljm Cnljm |n, l, j, m. On the basis of symmetry considerations alone, what can you say about (n, l, j, m), which give rise to nonvanishing contributions? Suppose the radial wave functions and the energy levels are all known. Indicate how you may calculate Cnljm. Do we get further restrictions on (n, l, j, m)? 286 Symmetry in Quantum Mechanics 4.6 Consider a symmetric rectangular double-well potential: \u23a7 \u23a8 \u221e V = \u23a9 0 V0 > 0 for |x| > a + b, for a < |x| < a + b, for |x| < a. Assuming that V0 is very high compared to the quantized energies of low-lying states, obtain an approximate expression for the energy splitting between the two lowestlying states. 4.7 a. Let \u03c8(x, t) be the wave function of a spinless particle corresponding to a plane wave in three dimensions. Show that \u03c8\u2217(x, \u2212t) is the wave function for the plane wave with the momentum direction reversed. b. Let \u03c7( \u02c6n) be the two-component eigenspinor of \u03c3 \u00b7 \u02c6n with eigenvalue +1. Using the explicit form of \u03c7( \u02c6n) (in terms of the polar and azimuthal angles \u03b2", " and \u03b3 that characterize \u02c6n) verify that\u2212 i\u03c32\u03c7\u2217( \u02c6n) is the two-component eigenspinor with the spin direction reversed. 4.8 a. Assuming that the Hamiltonian is invariant under time reversal, prove that the wave function for a spinless nondegenerate system at any given instant of time can always be chosen to be real. b. The wave function for a plane wave state at t = 0 is given by a complex function eip\u00b7x/\u00afh. Why does this not violate time-reversal invariance? 4.9 Let \u03c6(p) be the momentum-space wave function for state |\u03b1, that is, \u03c6(p) = p|\u03b1. Is the momentum-space wave function for the time-reversed state \u0398|\u03b1 given by \u03c6(p), \u03c6(\u2212p), \u03c6\u2217(p), or \u03c6\u2217(\u2212p)? Justify your answer. 4.10 a. For the time-reversal operator \u0398, use (4.156) to show that \u0398| jm equals | j, \u2212m up to some phase that includes the factor (\u22121)m. In other words, show that \u0398| jm = ei\u03b4(\u22121)m| j, \u2212m, where \u03b4 is independent of m. b. Using the same phase convention, \ufb01nd the time-reversed state corresponding to D(R)| jm. Consider using the in\ufb01nitesimal form D( \u02c6n, d\u03c6) and then generalize to \ufb01nite rotations. c. From these results, prove that, independent of \u03b4, one \ufb01nds D ( j)\u2217 mm(R) = (\u22121)m\u2212mD ( j) \u2212m,\u2212m(R). d. Conclude that we are free to choose \u03b4 = 0, and \u0398| jm = (\u22121)m| j, \u2212m = i2m| j, \u2212m. 4.11 Suppose a spinless particle is bound to a \ufb01xed center by a potential V(x) so asymmetrical that no energy level is degenerate. Using time-reversal invariance prove L = 0 287 Problems for any energy eigenstate. (This is known as quenching of orbital angular momentum.) If the wave function of such a nondegenerate eigenstate is expanded as \u2211 \u2211 l m Flm(r)Ym l (\u03b8, \u03c6), what kind of phase restrictions do we obtain on Flm(r)? It is useful to review Theorem 7 on p. 279. 4.12 The Hamiltonian for a spin 1 system is given by H = AS2 z + B(S2 x \u2212 S2 y). Solve this problem exactly to \ufb01nd the normalized energy eigenstates and eigenvalues. (A spin-dependent Hamiltonian of this kind actually appears in crystal physics.) Is this Hamiltonian invariant under time reversal? How do the normalized eigenstates you obtained transform under time reversal? 5 Approximation Methods Few problems in quantum mechanics \u2013 with either time-independent or time-dependent Hamiltonians \u2013 can be solved exactly. Inevitably we are forced to resort to some form of approximation method. One may argue that with the advent of high-speed computers it is always possible to obtain the desired solution numerically to the requisite degree of accuracy; nevertheless, it remains important to understand the basic physics of the approximate solutions even before we embark on ambitious computer calculations. This chapter is devoted to a fairly systematic discussion of approximate solutions to bound-state problems. 5.1 Time-Independent Perturbation Theory: Nondegenerate Case 5.1.1 Statement of the Problem The approximation method we consider here is time-independent perturbation theory, sometimes known as the Rayleigh\u2013Schr\u00a8odinger perturbation theory. We consider a timeindependent Hamiltonian H such that it can be split into two parts, namely, H = H0 + V, (5.1) where the V = 0 problem is assumed to have been solved in the sense that both the exact energy eigenkets |n(0) and the exact energy eigenvalues E(0) n are known: H0|n(0) = E(0) n |n(0). (5.2) We are required to \ufb01nd approximate eigenkets and eigenvalues for the full Hamiltonian problem (H0 + V)|n = En|n, (5.3) where V is known as the perturbation; it is not, in general, the full-potential operator. For example, suppose we consider the hydrogen atom", " in an external electric or magnetic \ufb01eld. The unperturbed Hamiltonian H0 is taken to be the kinetic energy p2/2m and the Coulomb potential due to the presence of the proton nucleus \u2212e2/r. Only that part of the potential due to the interaction with the external E or B \ufb01eld is represented by the perturbation V. 288 289 5.1 Time-Independent Perturbation Theory: Nondegenerate Case Instead of (5.3) it is customary to solve (H0 + \u03bbV)|n = En|n, (5.4) where \u03bb is a continuous real parameter. This parameter is introduced to keep track of the number of times the perturbation enters. At the end of the calculation we may set \u03bb \u2192 1 to get back to the full-strength case. In other words, we assume that the strength of the perturbation can be controlled. The parameter \u03bb can be visualized to vary continuously from 0 to 1, the \u03bb = 0 case corresponding to the unperturbed problem and \u03bb = 1 corresponding to the full-strength problem of (5.3). In physical situations where this approximation method is applicable, we expect to see a smooth transition of |n0 into |n and E(0) n into En as \u03bb is \u201cdialed\u201d from 0 to 1. The method rests on the expansion of the energy eigenvalues and energy eigenkets in powers of \u03bb. This means that we implicitly assume the analyticity of the energy eigenvalues and eigenkets in a complex \u03bb-plane around \u03bb = 0. Of course, if our method is to be of practical interest, good approximations can better be obtained by taking only one or two terms in the expansion. 5.1.2 The Two-State Problem Before we embark on a systematic presentation of the basic method, let us see how the expansion in \u03bb might indeed be valid in the exactly soluble two-state problem we have encountered many times already. Suppose we have a Hamiltonian that can be written as H = E(0) 1 |1(0)1(0)| + E(0) 2 |2(0)2(0)| + \u03bbV12|1(0)2(0)| + \u03bbV21|2(0)1(0)|, (5.5) where |1(0) and |2(0) are the energy eigenkets for the \u03bb = 0 problem, and we consider the case V11 = V22 = 0. In this representation the H may be represented by a square matrix as follows: E(0) \u03bbV12 1 \u03bbV21 E(0) 2 H =, (5.6) where we have used the basis formed by the unperturbed energy eigenkets. The V matrix must, of course, be Hermitian; let us solve the case when V12 and V21 are real: V12 = V\u2217 12, V21 = V\u2217 21; (5.7) hence, by Hermiticity V12 = V21. (5.8) This can always be done by adjusting the phase of |2(0) relative to that of |1(0). The problem of obtaining the energy eigenvalues here is completely analogous to that of solving the spin-orientation problem, where the analogue of (5.6) is H = a0 + \u03c3\u00b7a = a0 + a3 a1 a1 a0 \u2212 a3, (5.9) 290 Approximation Methods where we assume a = (a1, 0,a 3) is small and a0, a1, a3 are all real. The eigenvalues for this problem are known to be just E = a0 \u00b1 1 + a2 a2 3. By analogy the corresponding eigenvalues for (5.6) are - E1 E2 = 1 + E(0) E(0(0) 1 \u2212 E(0) 2 2 4 \u23a4 \u23a5 \u23a6. + \u03bb2|V12|2 (5.10) (5.11) Let us suppose \u03bb|V12| is small compared with the relevant energy scale, the difference of the energy eigenvalues of the unperturbed problem: \u03bb|V12| |E(0) 1 \u2212 E(0) 2 |. (5.12) We can then use \u221a 1 2 to obtain the expansion of the energy eigenvalues in the presence of perturbation \u03bb|V12|, namely, 1 + \u03b5 = 1 + (5.13) + \u00b7 \u00b7 \u00b7 \u03b5 \u2212 8 \u03b52 E1 = E(0) 1 + E2 = E(0)", " 2 + \u03bb2|V12|2 \u2212 E(0) E(0) 1 2 \u03bb2|V12|2 \u2212 E(0) E(05.14) These are expressions that we can readily obtain using the general formalism to be developed shortly. It is also possible to write down the energy eigenkets in analogy with the spin-orientation problem. The reader might be led to believe that a perturbation expansion always exists for a suf\ufb01ciently weak perturbation. Unfortunately this is not necessarily the case. As an elementary example, consider a one-dimensional problem involving a particle of mass m in a very weak square-well potential of depth V0 (i.e. V = \u2212V0 for \u2212a < x < a and V = 0 for |x| > a). This problem admits one bound state of energy, E = \u2212(2ma2/\u00afh2)|\u03bbV|2, \u03bb > 0 for attraction. (5.15) We might regard the square well as a very weak perturbation to be added to the free-particle Hamiltonian and interpret result (5.15) as the energy shift in the ground state from zero to |\u03bbV|2. Speci\ufb01cally, because (5.15) is quadratic in V, we might be tempted to associate this as the energy shift of the ground state computed according to second-order perturbation theory. However, this view is false because if this were the case, the system would also admit an E < 0 state for a repulsive potential case with \u03bb negative, which would be sheer nonsense. 291 5.1 Time-Independent Perturbation Theory: Nondegenerate Case Let us now examine the radius of convergence of series expansion (5.14). If we go back to the exact expression of (5.11) and regard it as a function of a complex variable \u03bb, we see that as |\u03bb| is increased from zero, branch points are encountered at The condition for the convergence of the series expansion for the \u03bb = 1 full-strength case is \u03bb|V12| = \u00b1i(E(0) \u2212 E(0) 2 ) 1 2. (5.16) |V12| < |E(0) 1 | \u2212 E(0) 2 2. (5.17) If this condition is not met, perturbation expansion (5.14) is meaningless.1 5.1.3 Formal Development of Perturbation Expansion We now state in more precise terms the basic problem we wish to solve. Suppose we know completely and exactly the energy eigenkets and energy eigenvalues of the unperturbed Hamiltonian H0, that is H0|n(0) = E(0) n |n(0). (5.18) The set {|n(0)} is complete in the sense that the closure relation 1 = \u2211n |n(0)n(0)| holds. Furthermore, we assume here that the energy spectrum is nondegenerate; in the next section we will relax this assumption. We are interested in obtaining the energy eigenvalues and eigenkets for the problem de\ufb01ned by (5.4). To be consistent with (5.18) we should write (5.4) as (H0 + \u03bbV)|n\u03bb = E(\u03bb) n |n\u03bb (5.19) to denote the fact that the energy eigenvalues E(\u03bb) and energy eigenkets |n\u03bb are functions of the continuous parameter \u03bb; however, we will usually dispense with this correct but more cumbersome notation. n As the continuous parameter \u03bb is increased from zero, we expect the energy eigenvalue n, so we de\ufb01ne the energy En for the nth eigenket to depart from its unperturbed value E(0) shift for the nth level as follows: \u0394n \u2261 En \u2212 E(0) n. The basic Schr\u00a8odinger equation to be solved (approximately) is (E(0) n \u2212 H0)|n = (\u03bbV \u2212 \u0394n)|n. (5.20) (5.21) We may be tempted to invert the operator E(0) n \u2212 H0; however, in general, the inverse operator 1/(E(0) n \u2212 H0) is ill de\ufb01ned because it may act on |n(0). Fortunately in our case (\u03bbV \u2212 \u0394n)|n has no component along |n(0), as can easily be seen by multiplying both sides of (5.21) by n(0)| on the left: n(0)|(\u03bbV \u2212 \u0394n)|n = 0. (5.22) 1 See the discussion on convergence following (5.", "44), under general remarks. 292 Approximation Methods Suppose we de\ufb01ne the complementary projection operator \u03c6n \u2261 1 \u2212 |n(0)n(0)| = \u2211 k=n |k(0)k(0)|. (5.23) The inverse operator 1/(E(0) Explicitly, n \u2212 H0) is well de\ufb01ned when it multiplies \u03c6n on the right. 1 E(0) n \u2212 H0 \u03c6n = \u2211 k=n 1 n \u2212 E(0) E(0) k |k(0)k(0)|. Also from (5.22) and (5.23), it is evident that (\u03bbV \u2212 \u0394n)|n = \u03c6n(\u03bbV \u2212 \u0394n)|n. We may therefore be tempted to rewrite (5.21) as |n?= 1 E(0) n \u2212 H0 \u03c6n(\u03bbV \u2212 \u0394n)|n. (5.24) (5.25) (5.26) However, this cannot be correct because as \u03bb \u2192 0, we must have |n \u2192 |n(0) and \u0394n \u2192 0. Nevertheless, even for \u03bb = 0, we can always add to |n a solution to the homogeneous equation (5.18), namely, cn|n(0), so a suitable \ufb01nal form is |n = cn(\u03bb)|n(0) + 1 E(0) n \u2212 H0 \u03c6n(\u03bbV \u2212 \u0394n)|n, where Note that cn(\u03bb) =1. lim \u03bb\u21920 cn(\u03bb) =n (0)|n. (5.27) (5.28) (5.29) For reasons we will see later, it is convenient to depart from the usual normalization convention Rather, we set n|n = 1. (5.30) n(0)|n = cn(\u03bb) =1, even for \u03bb = 0. We can always do this if we are not worried about the overall normalization because the only effect of setting cn = 1 is to introduce a common multiplicative factor. Thus, if desired, we can always normalize the ket at the very end of the calculation. It is also customary to write (5.31) 1 E(0) n \u2212 H0 \u03c6n \u2192 \u03c6n E(0) n \u2212 H0 and similarly 1 E(0) n \u2212 H0 \u03c6n = \u03c6n 1 E(0) n \u2212 H0 = \u03c6n 1 E(0) n \u2212 H0 \u03c6n, (5.32) (5.33) 293 5.1 Time-Independent Perturbation Theory: Nondegenerate Case so we have |n = |n(0) + \u03c6n E(0) n \u2212 H0 (\u03bbV \u2212 \u0394n)|n. We also note from (5.22) and (5.31) that \u0394n = \u03bbn(0)|V|n. (5.34) (5.35) Everything depends on the two equations in (5.34) and (5.35). Our basic strategy is to expand |n and \u0394n in the powers of \u03bb and then match the appropriate coef\ufb01cients. This is justi\ufb01ed because (5.34) and (5.35) are identities which hold for all values of \u03bb between 0 and 1. We begin by writing |n = |n(0) + \u03bb|n(1) + \u03bb2|n(2) + \u00b7 \u00b7 \u00b7 \u0394n = \u03bb\u0394(1) n + \u03bb2\u0394(2) n + \u00b7 \u00b7 \u00b7. (5.36) Substituting (5.36) into (5.35) and equating the coef\ufb01cient of various powers of \u03bb, we obtain \u0394(1) n \u0394(2) n O(\u03bb1) : O(\u03bb2) :... O(\u03bbN) : \u0394( N ) n... = n(0)|V|n(0) = n(0)|V|n(1)... = n(0)|V|n(N\u22121)... (5.37) so to evaluate the energy shift up to order \u03bbN it is suf\ufb01cient to know |n only up to order \u03bbN\u22121. We now look at (5.34); when it is expanded using (5.36), we get |n(0) + \u03bb|n(1) + \u03bb2|n(2) + \u00b7 \u00b7 \u00b7 = |n(0) + \u03c6n E(0) n \u2212 H0 (\u03bbV", " \u2212 \u03bb\u0394(1) n \u2212 \u03bb2\u03942 n \u2212 \u00b7 \u00b7 \u00b7 ) \u00d7 (|n(0) + \u03bb|n(1) + \u00b7 \u00b7 \u00b7 ). Equating the coef\ufb01cient of powers of \u03bb, we have O(\u03bb) : |n(1) = \u03c6n E(0) n \u2212 H0 V|n(0), (5.38) (5.39) where we have used \u03c6n\u0394(1) back to our earlier expression for \u0394(2) n [see (5.37)]: n |n(0) = 0. Armed with |n(1), it is now pro\ufb01table for us to go n = n(0)|V \u0394(2) \u03c6n E(0) n \u2212 H0 V|n(0). (5.40) 294 Approximation Methods Knowing \u0394(2) follows: n, we can work out the \u03bb2-term in ket equation (5.38) also using (5.39) as O(\u03bb2) : |n(2) = \u03c6n E(0) n \u2212 H0 V \u03c6n E(0) n \u2212 H0 V|n(0) \u2212 \u03c6n E(0) n \u2212 H0 n(0)|V|n(0) \u03c6n E(0) n \u2212 H0 V|n(0). (5.41) Clearly, we can continue in this fashion as long as we wish. Our operator method is very compact; it is not necessary to write down the indices each time. Of course, to do practical calculations we must use at the end the explicit form of \u03c6n as given by (5.23). To see how all this works, we write down the explicit expansion for the energy shift where \u0394n \u2261 En \u2212 E(0) n = \u03bbVnn + \u03bb2 \u2211 k=n |Vnk|2 n \u2212 E(0) E(0) k + \u00b7 \u00b7 \u00b7, Vnk \u2261 n(0)|V|k(0) = n|V|k, (5.42) (5.43) that is, the matrix elements are taken with respect to unperturbed kets. Notice that when we apply the expansion to the two-state problem we recover the earlier expression (5.14). The expansion for the perturbed ket goes as follows: |n = |n(0) + \u03bb\u2211 k=n |k(0) Vkn n \u2212 E(0) E(0) k |k(0)VklVln k )(E(0) (E(0) n \u2212 E(0) n \u2212 E(0) l ) \u2211 k=n \u2211 l=n + \u03bb2 + \u00b7 \u00b7 \u00b7. \u2212 \u2211 k=n |k(0)VnnVkn n \u2212 E(0) (E(0) k )2 (5.44) Equation (5.44) says that the nth level is no longer proportional to the unperturbed ket |n(0) but acquires components along other unperturbed energy kets; stated another way, the perturbation V mixes various unperturbed energy eigenkets. A few general remarks are in order. First, to obtain the \ufb01rst-order energy shift it is suf\ufb01cient to evaluate the expectation value of V with respect to the unperturbed kets. Second, it is evident from the expression of the second-order energy shift (5.42) that two energy levels, say the ith level and the jth level, when connected by Vij tend to repel each other; the lower one, say the ith level, tends to get depressed as a result of mixing with the higher jth level by |Vij|2/(E(0) i ), while the energy of the jth level goes up by the same amount. This is a special case of the no-level crossing theorem, which states that a pair of energy levels connected by perturbation do not cross as the strength of the perturbation is varied. \u2212 E(0) j Suppose there is more than one pair of levels with appreciable matrix elements but the ket |n, whose energy we are concerned with, refers to the ground state; then each term in (5.42) for the second-order energy shift is negative. This means that the second-order 295 5.1 Time-Independent Perturbation Theory: Nondegenerate Case energy shift is always negative for the ground state; the lowest state tends to get even lower as a result of mixing. It is clear that perturbation expansions (5.42) and (5.44) will converge if |V", "il/(E(0) \u2212 E(0) l )| is suf\ufb01ciently \u201csmall.\u201d A more speci\ufb01c criterion can be given for the case in which H0 is simply the kinetic-energy operator (then this Rayleigh\u2013Schr\u00a8odinger perturbation expansion is just the Born series): At an energy E0 < 0, the Born series converges if and only if neither H0 + V nor H0 \u2212 V has bound states of energy E \u2264 E0. See Newton (1982), p. 233. i 5.1.4 Wave Function Renormalization We are in a position to look at the normalization of the perturbed ket. Recalling the normalization convention we use, (5.31), we see that the perturbed ket |n is not normalized in the usual manner. We can renormalize the perturbed ket by de\ufb01ning |nN = Z1/2 n |n, (5.45) where Zn is simply a constant with Nn|nN = 1. Multiplying n(0)| on the left we obtain [because of (5.31)] n = n0|nN. Z1/2 (5.46) What is the physical meaning of Zn? Because |nN satis\ufb01es the usual normalization requirement (5.30), Zn can be regarded as the probability for the perturbed energy eigenstate to be found in the corresponding unperturbed energy eigenstate. Noting Nn|nN = Znn|n = 1, (5.47) we have Z\u22121 n = n|n = (n(0)| + \u03bbn(1)| + \u03bb2n(2)| + \u00b7 \u00b7 \u00b7 ) \u00d7 (|n(0) + \u03bb|n(1) + \u03bb2|n(2) + \u00b7 \u00b7 \u00b7 ) = 1 + \u03bb2n(1)|n(1) + 0(\u03bb3) = 1 + \u03bb2\u2211 k=n |Vkn|2 n \u2212 E(0) (E(0) k )2 + 0(\u03bb3), (5.48a) so up to order \u03bb2, we get for the probability of the perturbed state to be found in the corresponding unperturbed state Zn 1 \u2212 \u03bb2\u2211 k=n |Vkn|2 \u2212 E0 k)2 (E0 n. (5.48b) The second term in (5.48b) is to be understood as the probability for \u201cleakage\u201d to states other than |n(0). Notice that Zn is less than 1, as expected on the basis of the probability interpretation for Z. 296 Approximation Methods It is also amusing to note from (5.42) that to order \u03bb2, Z is related to the derivative of En with respect to E(0) n as follows: Zn = \u2202En \u2202E(0) n. (5.49) We understand, of course, that in taking the partial derivative of En with respect to E(0) n, we must regard the matrix elements of V as \ufb01xed quantities. Result (5.49) is actually quite general and not restricted to second-order perturbation theory. 5.1.5 Elementary Examples To illustrate the perturbation method we have developed, let us look at two examples. The \ufb01rst one concerns a simple harmonic oscillator whose unperturbed Hamiltonian is the usual one: H0 = p2 2m + 1 2 m\u03c92x2. (5.50) Suppose the spring constant k = m\u03c92 is changed slightly. We may represent the modi\ufb01cation by adding an extra potential V = 1 2 \u03b5m\u03c92x2, (5.51) where \u03b5 is a dimensionless parameter such that \u03b5 1. From a certain point of view this is the silliest problem in the world to which to apply perturbation theory; the exact solution is immediately obtained just by changing \u03c9 as follows \u221a \u03c9 \u2192 1 + \u03b5 \u03c9, (5.52) yet this is an instructive example because it affords a comparison between the perturbation approximation and the exact approach. We are concerned here with the new ground-state ket |0 in the presence of V and the ground-state energy shift \u03940: and |0 = |0(0) + \u2211 k=0 |k(0) Vk0 \u2212 E(0) k E(0) 0 + \u00b7 \u00b7 \u00b7 \u03940 = V00 + \u2211 k=0 |Vk0|2 E(0) 0 \u2212 E(0) k + \u00b7 \u00b7 \u00b7. The relevant matrix elements are (", "see Problem 5.8 in this chapter) V00 = V20 = \u03b5m\u03c92 2 \u03b5m\u03c92 2 0(0)|x2|0(0) = 2(0)|x2|0(0) = \u03b5\u00afh\u03c9 4 \u03b5\u00afh\u03c9 \u221a 2 2. (5.53a) (5.53b) (5.54) 297 5.1 Time-Independent Perturbation Theory: Nondegenerate Case All other matrix elements of form Vk0 vanish. Noting that the nonvanishing energy denominators in (5.53a) and (5.53b) are \u22122\u00afh\u03c9, we can combine everything to obtain and |0 = |0(0) \u2212 \u03b5 \u221a 2 4 \u03940 = E0 \u2212 E(0) 0 = \u00afh\u03c9 |2(0) + 0(\u03b52) + 0(\u03b53). \u03b5 4 \u2212 \u03b52 16 (5.55a) (5.55b) Notice that as a result of perturbation, the ground-state ket, when expanded in terms of original unperturbed energy eigenkets {|n(0)}, acquires a component along the second excited state. The absence of a component along the \ufb01rst excited state is not surprising because our total H is invariant under parity; hence, an energy eigenstate is expected to be a parity eigenstate. A comparison with the exact method can easily be made for the energy shift as follows: \u00afh\u03c9 2 1 + \u03b5 = \u00afh\u03c9 2 \u00afh\u03c9 2 (5.56) + \u00b7 \u00b7 \u00b7 1 + \u2192 \u03b52 \u221a \u2212 8 2 \u03b5, in complete agreement with (5.55b). As for the perturbed ket, we look at the change in the wave function. In the absence of V the ground-state wave function is where Substitution (5.52) leads to x|0(0) = 1 \u03c01/4 1\u221a x0 e\u2212x2/2x2 0, \u2019 \u00afh m\u03c9. x0 \u2261 x0 \u2192 x0 (1 + \u03b5)1/4 ; hence, x|0(0) \u2192 1 \u221a \u03c01/4 x0 (1 + \u03b5)1/8 exp \u2212 x2 2x2 0 1 \u03c01/4 1\u221a x0 e\u2212x2/2x2 0 + \u03b5 \u221a \u03c01/4 x0 = x|0(0) \u2212 \u03b5 \u221a 4 2 x|2(0), e\u2212x2/2x2 0 (1 + \u03b5)1/2 1 8 \u2212 1 4 x2 x2 0 where we have used x|2(0) = 1 \u221a 2 2 x|0(0)H2 x x0 = 1 \u221a 2 2 1 \u03c01/4 1\u221a x0 e\u2212x2/2x2 0 \u22122 + 4 2, x x0 and H2(x/x0) is a Hermite polynomial of order 2. (5.57) (5.58) (5.59) (5.60) (5.61) 298 Approximation Methods As another illustration of nondegenerate perturbation theory, we discuss the quadratic Stark effect. A one-electron atom \u2013 the hydrogen atom or a hydrogenlike atom with one valence electron outside the closed (spherically symmetrical) shell \u2013 is subjected to a uniform electric \ufb01eld in the positive z-direction. The Hamiltonian H is split into two parts, H0 = p2 2m + V0(r) and V = \u2212e|E|z (e < 0 for the electron). (5.62) [Editor\u2019s Note: Since the perturbation V \u2192 \u2212\u221e as z \u2192 \u2212\u221e, particles bound by H0 can, of course, escape now, and all formerly bound states acquire a \ufb01nite lifetime. However, we can still formally use perturbation theory to calculate the shift in the energy. (The imaginary part of this shift, which we shall ignore here, would give us the lifetime of the state or the width of the corresponding resonance.)] It is assumed that the energy eigenkets and the energy spectrum for the unperturbed problem (H0 only) are completely known. The electron spin turns out to be irrelevant in this problem, and we assume that with spin degrees of freedom ignored, no energy level is degenerate. This assumption does not hold for n = 1 levels of the hydrogen atoms, where V0 is the pure Coulomb potential; we will treat such cases later. The energy shift is given by \u0394k = \u2212e|", "E|zkk + e2|E|2\u2211 j=k |zkj|2 \u2212 E(0) j E(0) k + \u00b7 \u00b7 \u00b7, (5.63) where we have used k rather than n to avoid confusion with the principal quantum number n. With no degeneracy, |k(0) is expected to be a parity eigenstate; hence, zkk = 0, (5.64) as we saw in Section 4.2. Physically speaking, there can be no linear Stark effect, that is, there is no term in the energy shift proportional to |E| because the atom possesses a vanishing permanent electric dipole, so the energy shift is quadratic in |E| if terms of order e3|E|3 or higher are ignored. Let us now look at zkj, which appears in (5.63), where k (or j) is thecollective index that stands for (n, l, m) and (n, l, m). First, we recall the selection rule [see (3.482)] n, lm|z|n, lm = 0, unless l = l \u00b1 1 m = m (5.65) that follows from angular momentum (the Wigner\u2013Eckart theorem with T(1) considerations. q=0) and parity There is another way to look at the m-selection rule. In the presence of V, the full spherical symmetry of the Hamiltonian is destroyed by the external electric \ufb01eld that selects the positive z-direction, but V (hence the total H) is still invariant under rotation around the z-axis; in other words, we still have a cylindrical symmetry. Formally this is re\ufb02ected by the fact that [V, Lz] =0. (5.66) 299 5.1 Time-Independent Perturbation Theory: Nondegenerate Case This means that Lz is still a good quantum number even in the presence of V. As a result, the perturbation can be written as a superposition of eigenkets of Lz with the same m, where m = 0 in our case. This statement is true for all orders, in particular, for the \ufb01rst-order ket. Also, because the second-order energy shift is obtained from the \ufb01rst-order ket [see (5.40)], we can understand why only the m = 0 terms contribute to the sum. The polarizability \u03b1 of an atom is de\ufb01ned in terms of the energy shift of the atomic state as follows: \u0394 = \u2212 1 2 Let us consider the special case of the ground state of the hydrogen atom. Even though the spectrum of the hydrogen atom is degenerate for excited states, the ground state (with spin ignored) is nondegenerate, so the formalism of nondegenerate perturbation theory can be applied. The ground state |0(0) is denoted in the (n, l, m) notation by (1, 0, 0), so \u03b1|E|2. (5.67) \u03b1 = \u22122e2 \u221e \u2211 k=0 |k(0)|z|1, 0, 0|2 $, # E(0) 0 \u2212 E(0) k (5.68) where the sum over k includes not only all bound states |n, l, m (for n > 1) but also the positive-energy continuum states of hydrogen. There are many ways to estimate approximately or evaluate exactly the sum in (5.68) with various degrees of sophistication. We present here the simplest of all the approaches. Suppose the denominator in (5.68) were constant. Then we could obtain the sum by considering |k(0)|z|1, 0, 0|2 = \u2211 |k(0)|z|1, 0, 0|2 \u2211 k=0 all k = 1, 0, 0|z2|1, 0, 0, (5.69) where we have used the completeness relation in the last step. But we can easily evaluate z2 for the ground state as follows: z2 = x2 = y2 = r2, 1 3 (5.70) and using the explicit form for the wave function we obtain r2 = 3a2 0, where a0 stands for the Bohr radius. Unfortunately the expression for polarizability \u03b1 involves the energy denominator that depends on E(0) k, but we know that the inequality \u2212E(0) 0 + E(0) k \u2265 \u2212E(0) 0 + E(0) 1 = e2 2a0 1 \u2212 1 4 (5.71) holds for every energy denominator in (5.68). As a result, we can obtain an upper limit for the", " polarizability of the ground state of the hydrogen atom, namely, \u03b1 < 16a3 0 3 5.3a3 0. (5.72) 300 Approximation Methods It turns out that we can evaluate exactly the sum in (5.68) using a method due to A. Dalgarno and J. T. Lewis (Merzbacher (1970), p. 424, for example), which also agrees with the experimentally measured value. This gives \u03b1 = 9a3 0 2 = 4.5a3 0. (5.73) We obtain the same result (without using perturbation theory) by solving the Schr\u00a8odinger equation exactly using parabolic coordinates. 5.2 Time-Independent Perturbation Theory: The Degenerate Case The perturbation method we developed in the previous section fails when the unperturbed energy eigenkets are degenerate. The method of the previous section assumes that there is a unique and well-de\ufb01ned unperturbed ket of energy E(0) n which the perturbed ket approaches as \u03bb \u2192 0. With degeneracy present, however, any linear combination of unperturbed kets has the same unperturbed energy; in such a case it is not a priori obvious to what linear combination of the unperturbed kets the perturbed ket is reduced in the limit \u03bb \u2192 0. Here specifying just the energy eigenvalue is not enough; some other observable is needed to complete the picture. To be more speci\ufb01c, with degeneracy we can take as our base kets simultaneous eigenkets of H0 and some other observable A, and we can continue labeling the unperturbed energy eigenket by |k(0), wherek now symbolizes a collective index that stands for both the energy eigenvalue and the A eigenvalue. When the perturbation operator V does not commute with A, the zeroth-order eigenkets for H (including the perturbation) are in fact not A eigenkets. From a more practical point of view, a blind application of formulas like (5.42) and (5.44) obviously runs into dif\ufb01culty because Vnk n \u2212 E(0) E(0) k becomes singular if Vnk is nonvanishing and E(0) n and E(0) k method of the previous section to accommodate such a situation. (5.74) are equal. We must modify the Whenever there is degeneracy we are free to choose our base set of unperturbed kets. We should, by all means, exploit this freedom. Intuitively we suspect that the catastrophe of vanishing denominators may be avoided by choosing our base kets in such a way that V has no off-diagonal matrix elements [such as Vnk = 0 in (5.74)]. In other words, we should use the linear combinations of the degenerate unperturbed kets that diagonalize H in the subspace spanned by the degenerate unperturbed kets. This is indeed the correct procedure to use. Suppose there is a g-fold degeneracy before the perturbation V is switched on. This means that there are g different eigenkets all with the same unperturbed energy E(0) D. Let us denote these kets by {|m(0)}. In general, the perturbation removes the degeneracy in the sense that there will be g perturbed eigenkets all with different energies. Let them form a 301 5.2 Time-Independent Perturbation Theory: The Degenerate Case set {|l}. As \u03bb goes to zero |l \u2192 |l(0), and various |l(0) are eigenkets of H0 all with the same energy E(0) m. However, the set |l(0) need not coincide with {|m(0)} even though the two sets of unperturbed eigenkets span the same degenerate subspace, which we call D. We can write |l(0) = \u2211 m\u2208D m(0)|l(0)|m(0), where the sum is over the energy eigenkets in the degenerate subspace. Before expanding in \u03bb, there is a rearrangement of the Schr\u00a8odinger equation that will make it much easier to carry out the expansion. Let P0 be a projection operator onto the space de\ufb01ned by {|m(0)}. (Recall the discussion of projection operators in Section 1.3.) We de\ufb01ne P1 = 1 \u2212 P0 to be the projection onto the remaining states. We shall then write the Schr\u00a8odinger equation for the states |l as 0 = (E \u2212 H0 \u2212 \u03bbV)|l = (E \u2212", " E(0) D \u2212 \u03bbV)P0|l + (E \u2212 H0 \u2212 \u03bbV)P1|l. (5.75) We next separate (5.75) into two equations by projecting from the left on (5.75) with P0 and P1, (E \u2212 E(0) \u2212 \u03bbP0V)P0|l \u2212 \u03bbP0VP1|l = 0 D \u2212\u03bbP1VP0|l + (E \u2212 H0 \u2212 \u03bbP1V)P1|l = 0. We can solve (5.77) in theP 1 subspace because P1(E \u2212 H0 \u2212 \u03bbP1VP1) is not singular in this subspace since E is close to E(0) D and the eigenvalues of P1H0P1 are all different from E(0) D. Hence we can write (5.76) (5.77) P1|l = P1 \u03bb E \u2212 H0 \u2212 \u03bbP1VP1 P1VP0|l or written out explicitly to order \u03bb when |l is expanded as |l = |l(0) + \u03bb|l(1) + \u00b7 \u00b7 \u00b7 P1|l(1) = \u2211 k /\u2208D |k(0)Vkl \u2212 E(0) E(0) D k. (5.78) (5.79) To calculate P0|l, we substitute (5.78) into (5.76) to obtain E \u2212 E(0) D \u2212 \u03bbP0VP0 \u2212 \u03bb2P0VP1 1 E \u2212 H0 \u2212 \u03bbV P1VP0 P0|l = 0. (5.80) Although there is a term of order \u03bb2 in (5.80) that results from the substitution, we shall \ufb01nd that it produces a term of order \u03bb in the state P0|l. So we obtain the equation for the energies to order \u03bb and eigenfunctions to order zero, (E \u2212 E(0) D \u2212 \u03bbP0VP0)(P0|l(0)) =0. (5.81) This is an equation in the g-dimensional degenerate subspace and clearly means that the eigenvectors are just the eigenvectors of the g \u00d7 g matrix P0VP0 and the eigenvalues E(1) are just the roots of the secular equation det[V \u2212 (E \u2212 E(0) D )] = 0 (5.82) 302 Approximation Methods where V = matrix of P0VP0 with matrix elements m(0)|V|m(0). Explicitly in matrix form we have \u239b \u239c \u239d V11 V12 V21 V22...... \u239e \u239b \u239f \u23a0 \u239c \u239d \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7... 1(0)|l(0) 2(0)|l(0)... \u239e \u239b \u239f \u23a0 = \u0394(1) l \u239c \u239d \u239e 1(0)|l(0) 2(0)|l(0)... \u239f \u23a0. (5.83) l The roots determine the eigenvalues \u0394(1), there are g altogether, and by substituting them into (5.83), we can solve for m(0)|l(0) for each l up to an overall normalization constant. Thus by solving the eigenvalue problem, we obtain in one stroke both the \ufb01rst-order energy shifts and the correct zeroth-order eigenkets. Notice that the zeroth-order kets we obtain as \u03bb \u2192 0 are just the linear combinations of the various |m(0) that diagonalize the perturbation V, the diagonal elements immediately giving the \ufb01rst-order shift \u0394(1) l = l(0)|V|l(0). (5.84) Note also that if the degenerate subspace were the whole space, we would have solved the problem exactly in this manner. The presence of unperturbed \u201cdistant\u201d eigenkets not belonging to the degenerate subspace will show up only in higher orders \u2013 \ufb01rst order and higher for the energy eigenkets and second order and higher for the energy eigenvalues. Expression (5.84) looks just like the \ufb01rst-order energy shift [see (5.37)] in the nondegenerate case except that here we have to make sure that the base kets used are such that V does not have nonvanishing off-diagonal matrix elements in the subspace spanned by", " the degenerate unperturbed eigenkets. If the V operator is already diagonal in the base ket representation we are using, we can immediately write down the \ufb01rst-order shift by taking the expectation value of V, just as in the nondegenerate case. Let us now look at (5.80). To be safe we keep all terms in the g \u00d7 g effective Hamiltonian that appears in (5.80) to order \u03bb2 although we want P0|l only to order \u03bb. We \ufb01nd E \u2212 E(0) D \u2212 \u03bbP0VP0 \u2212 \u03bb2P0VP1 1 \u2212 H0 E(0) D P1VP0 P0|l = 0. (5.85) For the g \u00d7 g matrix P0VP0, let us call the eigenvalues vi and the eigenvectors P0|l(0). i = E(0) The eigen energies to \ufb01rst order are E(1) D + \u03bbvi. We assume that the degeneracy is completely resolved so that E(1) \u2212 E(1) j = \u03bb(vi \u2212 vj) are all nonzero. We can now apply nondegenerate perturbation theory (5.39) to theg \u00d7g dimensional Hamiltonian that appears in (5.85). The resulting correction to the eigenvectors P0|l(0) is i i i or more explicitly P0|l(1) i = \u2211 j=i P0|l(0) j vi \u2212 vj l(0) j |VP1 1 \u2212 H0 E(0) D P1V|l(0) i P0|l(1) i = \u2211 j=i P0|l(0) j vi \u2212 vj l(0) j \u2211 k\u2208D |V|k 1 \u2212 E(0) k E(0) D k|V|l(0) i. (5.86) (5.87) 303 5.2 Time-Independent Perturbation Theory: The Degenerate Case Thus, although the third term in the effective Hamiltonian that appears in (5.85) is of order \u03bb2, it is divided by energy denominators of order \u03bb in forming the correction to the eigenvector, which then gives terms of order \u03bb in the vector. If we add together (5.79) and (5.87), we get the eigenvector accurate to order \u03bb. As in the nondegenerate case, it is convenient to adopt the normalization convention l + \u03bb2\u0394(2) l(0)|l = 1. We then have, from (5.76) and (5.77), \u03bbl(0)|V|l = \u0394l = \u03bb\u0394(1) l + \u00b7 \u00b7 \u00b7. The \u03bb-term just reproduces (5.84). As for the \u03bb2-term, we obtain \u0394(2) l = l(0)|V|l(1) = l(0)|V|P1l(1) + l(0)|V|P0l(1) are eigenvectors of V, the correction to the vector, (5.87), gives no contribution to the second-order energy shift, so we \ufb01nd using (5.79). Since the vectors P0|l(0) i j l = \u2211 \u0394(2) k /\u2208D |Vkl|2 E(0) D \u2212 E(0) k. (5.88) Our procedure works provided that there is no degeneracy in the roots of secular equation (5.82). Otherwise we still have an ambiguity as to which linear combination of the degenerate unperturbed kets the perturbed kets are reduced in the limit \u03bb \u2192 0. Put in another way, if our method is to work, the degeneracy should be removed completely in \ufb01rst order. A challenge for the experts: How must we proceed if the degeneracy is not removed in \ufb01rst order, that is, if some of the roots of the secular equation are equal? (See Problem 5.15 of this chapter.) Let us now summarize the basic procedure of degenerate perturbation theory. 1. Identify degenerate unperturbed eigenkets and construct the perturbation matrix V, a g \u00d7 g matrix if the degeneracy is g-fold. 2. Diagonalize the perturbation matrix by solving, as usual, the appropriate secular equation. 3. Identify the roots of the secular equation with the \ufb01rst-order energy shifts; the base kets that diagonalize the V matrix are the correct zeroth-order k", "ets to which the perturbed kets approach in the limit \u03bb \u2192 0. 4. For higher orders use the formulas of the corresponding nondegenerate perturbation theory except in the summations, where we exclude all contributions from the unperturbed kets in the degenerate subspace D. 5.2.1 Linear Stark Effect As an example of degenerate perturbation theory, let us study the effect of a uniform electric \ufb01eld on excited states of the hydrogen atom. As is well known, in the Schr\u00a8odinger theory with a pure Coulomb potential with no spin dependence, the bound-state energy of the hydrogen atom depends only on the principal quantum number n. This leads to degeneracy for all but the ground state because the allowed values of l for a given n satisfy 0 \u2264 l < n. (5.89) 304 Approximation Methods To be speci\ufb01c, for the n = 2 level, there is an l = 0 state called 2s and three l = 1 (m = \u00b11, 0) states called 2p, all with the same energy, \u2212e2/8a0. As we apply a uniform electric \ufb01eld in the z-direction, the appropriate perturbation operator is given by V = \u2212ez|E|, (5.90) which we must now diagonalize. Before we evaluate the matrix elements in detail using the usual (nlm) basis, let us note that the perturbation (5.90) has nonvanishing matrix elements only between states of opposite parity, that is, between l = 1 and l = 0 in our case. Furthermore, in order for the matrix element to be nonvanishing, the m-values must be the same because z behaves like a spherical tensor of rank one with spherical component (magnetic quantum number) zero. So the only nonvanishing matrix elements are between 2s (m = 0 necessarily) and 2p with m = 0. Thus V = \u239b \u239c \u239c \u239d 2s 0 2p, m = 0|V|2s 0 0 2p m = 0 2s|V|2p, m = 0 0 0 0 Explicitly, 2p m = 1 2p m = \u22125.91) 2s|V|2p, m = 0 = 2p, m = 0|V|2s = 3ea0|E|. (5.92) It is suf\ufb01cient to concentrate our attention on the upper left-hand corner of the square matrix. It then looks very much like the \u03c3x matrix, and we can immediately write down the answer. For the energy shifts we get \u0394(1) \u00b1 = \u00b13ea0|E|, where the subscripts \u00b1 refer to the zeroth-order kets that diagonalize V: 1\u221a 2 (|2s, m = 0 \u00b1 |2p, m = 0). |\u00b1 = (5.93) (5.94) Schematically the energy levels are as shown in Figure 5.1. 3ea 0 E 3ea 0 E ( 2s, m = 0 \u2212 2p, m = 0 ) 2 No change for 2p, m = \u00b11 ( 2s, m = 0 + 2p, m = 0 ) 2 Fig. 5.1 Schematic energy-level diagram for the linear Stark effect as an example of degenerate perturbation theory. 305 5.3 Hydrogenlike Atoms: Fine Structure and the Zeeman Effect Notice that the shift is linear in the applied electric \ufb01eld strength, hence the term the linear Stark effect. One way we can visualize the existence of this effect is to note that the energy eigenkets (5.94) are not parity eigenstates and are therefore allowed to have nonvanishing electric permanent dipole moments, as we can easily see by explicitly evaluating z. Quite generally, for an energy state that we can write as a superposition of opposite parity states, it is permissible to have a nonvanishing permanent electric dipole moment, which gives rise to the linear Stark effect. An interesting question can now be asked. If we look at the \u201creal\u201d hydrogen atom, the 2s level and 2p level are not really degenerate. Due to the spin-orbit force, 2p3/2 is separated from 2p1/2, as we will show in the next section, and even the degeneracy between the 2s1/2 and 2p1/2 levels that persists in the single-particle Dirac theory is removed by quantum electrodynamics effects (the Lamb shift). We might therefore ask: Is it realistic to apply degenerate perturbation theory to this problem? A comparison with the exact result shows that if the perturbation matrix elements are much larger when compared to the Lamb shift", " splitting, then the energy shift is linear in |E| for all practical purposes and the formalism of degenerate perturbation theory is applicable. On the opposite extreme, if the perturbation matrix elements are small compared to the Lamb shift splitting, then the energy shift is quadratic and we can apply nondegenerate perturbation theory; see Problem 5.19 of this chapter. This incidentally shows that the formalism of degenerate perturbation theory is still useful when the energy levels are almost degenerate compared to the energy scale de\ufb01ned by the perturbation matrix element. In intermediate cases we must work harder; it is safer to attempt to diagonalize the Hamiltonian exactly in the space spanned by all the nearby levels. 5.3 Hydrogenlike Atoms: Fine Structure and the Zeeman Effect 5.3.1 The Relativistic Correction to the Kinetic Energy A hydrogen like atom with a single electron has the potential energy function (3.305) giving the Hamiltonian H0 = p2 2me \u2212 Ze2 r (5.95) where the \ufb01rst term is the nonrelativistic kinetic-energy operator. However, the relativistically correct kinetic energy is K = p2c2 + m2 ec4 \u2212 mec2 \u2248 p2 2me \u2212 (p2)2 ec2. 8m3 (5.96) 306 Approximation Methods Therefore, following the notation in (5.1), we can treat this problem in perturbation theory where H0 is given by (5.95) and the perturbation is V = \u2212 (p2)2 ec2. 8m3 (5.97) Now in principle, this is a complicated problem because of the highly degenerate eigenstates |nlm of the hydrogen atom. However, since L commutes with p2 as we noted in (3.264), we also have [L, V] =0. (5.98) In other words, V is rotationally symmetric, and is therefore already diagonal in the |nlm basis. Therefore the \ufb01rst-order energy shifts due to V are just equal to the expectation values in these basis states. Following (5.37) we write nl = nlm|V|nlm = \u2212nlm| (p2)2 \u0394(1) ec2 8m3 |nlm (5.99) where the rotational symmetry assures us that the \ufb01rst-order energy shifts cannot depend on m. In principle, (5.99) could be evaluated by brute force, but there is a more elegant way. Since (p2)2 8m3 ec2 = 1 2mec2 2 p2 2me = 1 2mec2 H0 + 2 Ze2 r (5.100) we immediately see that nl = \u2212 1 \u0394(1) 2mec2 2 E(0) n + 2E(0) n nlm| Ze2 r |nlm + nlm| (Ze2)2 r2 |nlm. (5.101) The problem is therefore reduced to calculating the expectation values for Ze2/r and (Ze2)2/r2. In fact, both of these expectation values can be evaluated using some clever tricks. We simply outline the approach here, but the interested reader is referred to Shankar (1994) or Townsend (2000) for more details. If one imagines a hydrogen atom with a \u201cperturbation\u201d V\u03b3 = \u03b3/r, then the expectation value in the second term of (5.101) is simply the \ufb01rst-order correction to the energy with \u03b3 = Ze2. On the other hand, it is simple to solve this problem exactly since it corresponds to the hydrogen atom with Ze2 \u2192 Ze2 \u2212 \u03b3, and it is straightforward to \ufb01nd the \ufb01rst-order correction from the exact solution. One \ufb01nds that nlm| Ze2 r |nlm = \u22122E(0) n. (5.102) Indeed, this is actually a statement of the virial theorem for the Coulomb potential. A similar approach can be taken for the third term in (5.101). In this case, imagine a perturbation V\u03b3 = \u03b3/r2 which modi\ufb01es the centrifugal barrier term in the effective potential. That is, it takes l to a form that includes \u03b3, which can again be used to write down the \ufb01rst-order correction. One \ufb01nds that nlm| (Ze2)2 r2 |nlm = 4n l + 1 2 2 E(0) n. (5.103) 307 5.3 Hydrogenlike Atoms: Fine Structure and the", " Zeeman Effect So, using (5.102) and (5.103), along with E(0) n from (3.315), we rewrite (5.101) as \u0394(1) nl = E(0) n Z2\u03b12 n2 + mec2Z4\u03b14 \u2212 3 4n4 + & % 1 l + 1 2 n3. (5.104a) (5.104b) Not unexpectedly, the relative size of the \ufb01rst-order correction is proportional to Z2\u03b12, the square of the classical electron orbital velocity (in units of c). 5.3.2 Spin-Orbit Interaction and Fine Structure Now let us move on to the study of the atomic levels of general hydrogenlike atoms, that is, atoms with one valence electron outside the closed shell. Alkali atoms such as sodium (Na) and potassium (K) belong to this category. The central (spin-independent) potential Vc(r) appropriate for the valence electron is no longer of the pure Coulomb form. This is because the electrostatic potential \u03c6(r) that appears in Vc(r) = e\u03c6(r) (5.105) is no longer due just to the nucleus of electric charge |e|Z; we must take into account the cloud of negatively charged electrons in the inner shells. A precise form of \u03c6(r) does not concern us here. We simply remark that the degeneracy characteristics of the pure Coulomb potential are now removed in such a way that the higher l states lie higher for a given n. Physically this arises from the fact that the higher l states are more susceptible to the repulsion due to the electron cloud. Instead of studying the details of Vc(r), which determines the gross structure of hydrogenlike atoms, we discuss the effect of the spin-orbit (L \u00b7 S) interaction that gives rise to \ufb01ne structure. We can understand the existence of this interaction in a qualitative fashion as follows. Because of the central force part (5.105), the valence electron experiences the electric \ufb01eld E = \u2212 \u2207Vc(r). (5.106) 1 e But whenever a moving charge is subjected to an electric \ufb01eld, it \u201cfeels\u201d an effective magnetic \ufb01eld given by Beff = \u2212 v c \u00d7 E. Because the electron has a magnetic moment \u03bc given by \u03bc = eS mec, (5.107) (5.108) 308 Approximation Methods we suspect a spin-orbit potential VLS contribution to H as follows: HLS?= \u2212 \u03bc \u00b7 Beff \u00d7 E = \u03bc \u00b7 v c eS mec \u00b7 p mec \u00d7 x r 1 (\u2212e) dVc dr = = 1 ec2 m2 1 r dVc dr (L \u00b7 S). (5.109) When this expression is compared with the observed spin-orbit interaction, it is seen to have the correct sign, but the magnitude turns out to be too large by a factor of two.2 There is a classical explanation for this due to spin precession (Thomas precession after L. H. Thomas), but we shall not bother with that. See Jackson (1975), for example. We simply treat the spin-orbit interaction phenomenologically and take VLS to be one-half of (5.109). The correct quantum-mechanical explanation for this discrepancy must await the Dirac (relativistic) theory of the electron discussed in the last chapter of this textbook. We are now in a position to apply perturbation theory to hydrogen like atoms using VLS as the perturbation (V of Sections 5.1 and 5.2). The unperturbed Hamiltonian H0 is taken to be H0 = p2 2m + Vc(r), (5.110) where the central potential Vc is no longer of the pure Coulomb form for alkali atoms. With just H0 we have freedom in choosing the base kets. Set 1: The eigenkets of L2, Lz, S2, Sz. Set 2: The eigenkets of L2, S2, J2, Jz. (5.111) Without VLS (or HLS) either set is satisfactory in the sense that the base kets are also energy eigenkets. With HLS added it is far superior to use set 2 of (5.111) because L \u00b7 S does not commute with Lz and Sz, while it does commute with J2 and Jz. Remember the cardinal rule: choose unperturbed kets that diagonalize the perturbation. You have to be either a fool or a masochist to use the Lz, Sz eigenkets [set 1 of (5.111)] as the base kets for this problem;", " if we proceeded to apply blindly the method of degenerate perturbation theory starting with set 1 as our base kets, we would be forced to diagonalize the VLS(HLS) matrix written in the Lz, Sz representation. The results of this, after a lot of hard algebra, give us just the J2, Jz eigenkets as the zeroth-order unperturbed kets to be used! In degenerate perturbation theory, if the perturbation is already diagonal in the representation we are using, all we need to do for the \ufb01rst-order energy shift is to take the expectation value. The wave function in the two-component form is explicitly written as \u03c8nlm = Rnl(r)Y j=l\u00b11/2,m (5.112) l 2 Indeed, this is a misleading statement, although it appears often in physics textbooks. See Haar and Curtis, Am. J. Phys., 55 (1987) 1044. 309 5.3 Hydrogenlike Atoms: Fine Structure and the Zeeman Effect Doublet or \u201cfine\u201d structure 3p3/2 3p1/2 \u03bb = 5,890 \u00c5 \u03bb = 5,896 \u00c5 3s1/2 Fig. 5.2 Schematic diagram of 3s and 3p lines. The 3s and 3p degeneracy is lifted because Vc(r) is now the screened Coulomb potential due to core electrons rather than pure Coulombic potential; VLS then removes the 3p1/2 and3p3/2 degeneracy. where Y j=l\u00b11/2,m order shift, we obtain l is the spin-angular function of Section 3.8 [see (3.384)]. For the \ufb01rst- \u0394nlj = ) \u2261 1 2m2 ec2 \u221e Rnl nl 0 ( 1 r dVc dr ( 1 r dVc dr ) nl \u00afh2 2 * + l \u2212(l + 1) 1 r dVc dr Rnlr2dr where we have used the m-independent identity [see (3.386)] Y \u2020 S \u00b7 LY d\u03a9 = 1 2 j( j + 1) \u2212 l(l + 1) \u2212 3 4 \u00afh2 = \u00afh2 2 l \u2212(l + 15.113) (5.114). Equation (5.113) is known as Lande\u2019s interval rule. To be speci\ufb01c, consider a sodium atom. From standard atomic spectroscopy notation, the ground-state con\ufb01guration is (1s)2(2s)2(2p)6(3s). (5.115) The inner 10 electrons can be visualized to form a spherically symmetrical electron cloud. We are interested in the excitation of the eleventh electron from 3s to a possible higher state. The nearest possibility is excitation to 3p. Because the central potential is no longer of the pure Coulomb form, 3s and 3p are now split. The \ufb01ne structure brought about by VLS refers to an even \ufb01ner split within 3p, between 3p1/2 and 3p3/2, where the subscript refers to j. Experimentally, we observe two closely separated yellow lines, known as the sodium D lines, one at 5896 \u02daA and the other at 5890 \u02daA; see Figure 5.2. Notice that 3p3/2 lies higher because the radial integral in (5.113) is positive. 310 Approximation Methods To appreciate the order of magnitude of the \ufb01ne-structure splitting, let us note that for Z 1, ( 1 r dVc dr ) nl \u223c e2 a3 0 (5.116) just on the basis of dimensional considerations. So the \ufb01ne-structure splitting is of order (e2/a3 0)(\u00afh/mec)2, which is to be compared with Balmer splittings of order e2/a0. It is useful to recall here that the classical radius of the electron, the Compton wavelength of the electron, and the Bohr radius are related in the following way: e2 mec2 : \u00afh mec : a0 : : 1 : 137 : (137)2, (5.117) where we have used e2 \u00afhc = 1 137. Typically, \ufb01ne-structure splittings are then related to typical Balmer splittings via e2 a3 0 \u00afh2 ec2 m2 : e2 a0 : : 1 137 2 : 1, (5.118) (5.119) which explains the origin of the term \ufb01ne structure. There are other", " effects of similar orders of magnitude, for example, the relativistic correction to kinetic energy discussed earlier in this section. Before leaving this discussion, let us calculate out (5.113) for the case of the Coulomb potential, that is, a hydrogen atom or one-electron ion with Z protons. In this case ) ) ( ( 1 r dVc dr = nl Ze2 r3. nl (5.120) We can evaluate this expectation value with the help of yet another trick. First we note that with H0 given by (5.95) we have nlm|[H0, A]|nlm = 0 (5.121) for any operator A, since H0 acting to the right or left just gives E(0) n. If we letA = pr, the radial momentum operator, then it obviously commutes with the radial part of the kineticenergy term in H0. Hence we are left with nlm| l(l + 1)\u00afh2 2mer2 \u2212 Ze2 r, pr |nlm = 0. (5.122) Now in coordinate space, pr does not commute with functions of the coordinate r because of the presence of the derivative \u2202/\u2202r. Therefore we can explicitly carry out the commutator in (5.122) to arrive at nlm| \u2212 l(l + 1)\u00afh2 mer3 + Ze2 r2 |nlm = 0. (5.123) 311 5.3 Hydrogenlike Atoms: Fine Structure and the Zeeman Effect Finally, then, we make use of (5.103) and (3.315) to write ( ) Ze2 r3 me l(l + 1)\u00afh2 = nl ( ) (Ze2)2 r2 nl = \u2212 2m2 ec2Z2\u03b12 nl(l + 1)(l + 1/2)\u00afh2 E(0) n. (5.124) We therefore have the spin-orbit correction to the energy eigenstates of the hydrogen atom from (5.113) as \u0394nlj = \u2212 Z2\u03b12 2nl(l + 1)(l + 1/2) E(0) n * + l \u2212(l + 15.125) Interestingly, this expression is nonzero for l = 0. Nevertheless, it gives the correct answer for the energy eigenvalues of the Dirac equation, as we shall see later in this textbook. The origin of this shift, attributed to something called the Darwin term, is discussed elsewhere. See, for example, Townsend (2000). 5.3.3 The Zeeman Effect We now discuss hydrogen or hydrogenlike (one-electron) atoms in a uniform magnetic \ufb01eld \u2013 the Zeeman effect, sometimes called the anomalous Zeeman effect with the electron spin taken into account. Recall that a uniform magnetic \ufb01eld B is derivable from a vector potential A = 1 2 (B \u00d7 x). For B in the positive z-direction, that is B = B\u02c6z, this becomes A = \u2212 1 2 (By\u02c6x \u2212 Bx\u02c6y). (5.126) (5.127) Apart from the spin term, the interaction Hamiltonian is generated by the substitution p \u2192 p \u2212 eA c. (5.128) We therefore have Because H = p2 2me + Vc(r) \u2212 e 2mec (p \u00b7 A + A \u00b7 p) + e2A2 2mec2. (5.129) x|p \u00b7 A(x)| = \u2212i\u00afh\u2207 \u00b7 [A(x)x|] = x|A(x) \u00b7 p| + x|[\u2212i\u00afh\u2207 \u00b7 A(x)], (5.130) it is legitimate to replace p \u00b7 A by A \u00b7 p whenever \u2207 \u00b7 A(x) =0, (5.131) 312 Approximation Methods which is the case for the vector potential of (5.127). Noting A \u00b7 p = B(\u2212 1 2 ypx + 1 2 xpy) and we obtain for (5.129) = 1 2 BLz A2 = 1 4 B2(x2 + y2), H = p2 2me + Vc(r) \u2212 e 2mec BLz + e2 8mec2 B2(x2 + y2). To this we may add the spin magnetic-moment interaction \u2212\u03bc \u00b7 B = \u2212e mec S \u00b7 B = \u2212e mec BSz. (5.132) (5.133) (5.134) (5.135) The quadratic B2(x2 + y2) is unimportant for a one-electron atom; the analogous term is important for the ground state of the helium atom", " where L(tot) both vanish. The reader may come back to this problem when he or she computes diamagnetic susceptibilities in Problems 5.25 and 5.26 of this chapter. and S(tot) z z To summarize, omitting the quadratic term, the total Hamiltonian is made up of the following three terms: H0 = p2 2me + Vc(r) HLS = 1 ec2 2m2 1 r dVc(r) dr L \u00b7 S HB = \u2212eB 2mec (Lz + 2Sz). (5.136a) (5.136b) (5.136c) Notice the factor 2 in front of Sz; this re\ufb02ects the fact that the g-factor of the electron is 2. Suppose HB is treated as a small perturbation. We can study the effect of HB using the eigenkets of H0 + HLS, that is, the J2, Jz eigenkets, as our base kets. Noting the \ufb01rst-order shift can be written as Lz + 2Sz = Jz + Sz, \u2212eB 2mec Jz + Szj=l\u00b11/2,m. (5.137) (5.138) The expectation value of Jz immediately gives m\u00afh. As forS z, we \ufb01rst recall ), 2l + 1 \u00d7 ml = m \u2212 1 2, ms = 2l + 1 ml = m + ). 1 2, ms = \u2212 1 2 (5.139) 313 5.3 Hydrogenlike Atoms: Fine Structure and the Zeeman Effect The expectation value of Sz can then easily be computed: Szj=l\u00b11/2,m = \u00afh 2 (|c+|2 \u2212 |c\u2212|2) = \u00afh 2 1 (2l + 1\u00afh (2l + 1). In this manner we obtain Lande\u2019s formula for the energy shift (due to B \ufb01eld), \u2212e\u00afhB 2mec 1 (2l + 1) \u0394EB = 1 \u00b1 m. (5.140) (5.141) We see that the energy shift of (5.141) is proportional to m. To understand the physical origin for this, we present another method for deriving (5.140). We recall that the expectation value of Sz can also be obtained using the projection theorem of Section 3.11. We get [see (3.488)] Szj=l\u00b11/2,m = [S \u00b7 Jj=l\u00b11/2] m\u00afh \u00afh2j( j + 1) = mJ2 + S2 \u2212 L2j=l\u00b11/2 2\u00afhj( j + 1\u00afh \u2212 l(l + 1\u00afh (2l + 1), (5.142) which is in complete agreement with (5.140). In the foregoing discussion the magnetic \ufb01eld is treated as a small perturbation. We now consider the opposite extreme \u2013 the Paschen\u2013Back limit \u2013 with a magnetic \ufb01eld so intense that the effect of HB is far more important than that of HLS, which we later add as a small perturbation. With H0 + HB only, the good quantum numbers are Lz and Sz. EvenJ 2 is no good because spherical symmetry is completely destroyed by the strong B \ufb01eld that selects a particular direction in space, the z-direction. We are left with cylindrical symmetry only, that is, invariance under rotation around the z-axis. So the Lz, Sz eigenkets |l, s = 1 2, ml, ms are to be used as our base kets. The effect of the main term HB can easily be computed: HBmlms = \u2212eB\u00afh 2mec The 2(2l + 1) degeneracy in ml and ms we originally had with H0 [see (5.136a)] is now reduced by HB to states with the same (ml)+(2ms), namely, (ml)+(1) and (ml +2)+(\u22121). Clearly we must evaluate the expectation value of L \u00b7 S with respect to |ml, ms: (ml + 2ms). (5.143) L \u00b7 S = LzSz + 1 2 (L+S\u2212 + L\u2212S+)mlms = \u00afh2mlms, (5.144) 314 Approximation Methods Table 5.1 Quantum Numbers in Weak and Strong Magnetic Fields Dominant interaction Almost good No good Always good Weak B HLS Strong B HB J2 (or L\u00b7S) \u2217 Lz, Sz Lz, Sz J2 (or L\u00b7S) L2, S", "2, Jz \u2217The exception is the stretched con\ufb01guration, e.g., p3/2 with m = \u00b1 3 2. HereL z and Sz are both good; this is because the magnetic quantum number of Jz, m = ml + ms can be satis\ufb01ed in only one way. where we have used Hence, L\u00b1ml = 0, S\u00b1ms = 0. HLSmlms = \u00afh2mlms ec2 2m2 ( ). 1 r dVc dr (5.145) (5.146) We summarize our results in Table 5.1, where weak and strong B \ufb01elds are \u201ccalibrated\u201d by comparing their magnitudes e\u00afhB/2mec with (1/137)2e2/a0. In this tablealmost good simply means good to the extent that the less dominant interaction could be ignored. Speci\ufb01cally, let us look at the level scheme of a p electron l = 1(p3/2, p1/2). In the weak B case the energy shifts are linear in B, with slopes determined by m 1 \u00b1. 1 2l + 1 As we now increase B, mixing becomes possible between states with the same m-value, for example, p3/2 with m = \u00b1 1 2 ; in this connection note the operator Lz + 2Sz that appears in HB [(5.136c)] is a rank 1 tensor operator T(k=1) q=0 with spherical component q = 0. 2 and p1/2 with m = \u00b1 1 It is instructive to solve the full problem, including \ufb01ne structure along with the Zeeman effect for arbitrary \ufb01eld strength, and see how the behavior transitions from weak to strong magnetic \ufb01elds. For the n = 2 states, one diagonalizes the 8 \u00d7 8 matrix of the perturbations in the degenerate subspace. This is not as dif\ufb01cult as it sounds, and is the object of Problem 5.24 at the end of this chapter. The resulting behavior is plotted in Figure 5.3. 5.3.4 Van der Waals\u2019 Interaction An important, nice application of the Rayleigh\u2013Schr\u00a8odinger perturbation theory is to calculate the long range interaction, or van der Waals\u2019 force, between two hydrogen atoms in their ground states. It is easy to show that the energy between the two atoms for large separation r is attractive and varies as r\u22126. 315 5.3 Hydrogenlike Atoms: Fine Structure and the Zeeman Effect 2 \u20134 0.0 0.5 1.0 1.5 2.0 Magnetic Field Strength Fig. 5.3 First-order energy shifts for the n = 2 states in hydrogen, as a function of the applied magnetic field strength. See Problem 5.24 at the end of this chapter. Note that there is a horizontal line at zero, in addition to the eight curves for the 2s and 2p states. The evolution from eight states at low field to five states at high field is evident. \u2212 r1 + \u2212 r2 r + z-axis Fig. 5.4 Two hydrogen atoms with their protons (+) separated by a fixed distance r and their electrons (\u2212) at displacements ri from them. Consider the two protons of the hydrogen atoms to be \ufb01xed at a distance r (along the z-axis) with r1 the vector from the \ufb01rst proton to its electron and r2 the vector from the second proton to its electron; see Figure 5.4. Then the Hamiltonian H can be written as H = H0 + V H0 = \u2212 \u00afh2 2m (\u22072 1 + \u22072 2) \u2212 e2 \u2212 e2 r1 r2 \u2212 e2 V = e2 r + e2 |r + r2 \u2212 r1| \u2212 e2 |r \u2212 r1|. (5.147) |r + r2| The lowest-energy solution of H0 is simply the product of the ground-state wave functions of the noninteracting hydrogen atoms 0 = U(0) U(0) 100(r1)U(0) 100(r2). (5.148) 316 Approximation Methods Now for large r ( the Bohr radius a0) expand the perturbation V in powers of ri/r to obtain V = e2 r3 (x1x2 + y1y2 \u2212 2z1z2) +0 1 r4 + \u00b7 \u00b7 \u00b7. (5.149) The lowest-order r\u22123 term in (5.149) corresponds to the interaction of two electric dipoles er1 and er2 separated by r. The higher-order terms", " represent higher-order multipole interactions, and thus every term in V involves spherical harmonics Ym l with li > 0 for each hydrogen atom. Hence, for each term in (5.149) the \ufb01rst-order perturbation energy matrix element V00 0, since the ground-state U(0) 0 wave function (5.148) has li = 0 and d\u03a9Ym l (\u03a9) =0 forl and m = 0. The second-order perturbation e4 r6 \u2211 |k(0)|x1x2 + y1y2 \u2212 2z1z2|0(0)|2 E(2)(r) = k=0 E(0) 0 \u2212 E(0) k (5.150) will be nonvanishing. We immediately see that this interaction varies as 1/r6; since E(0) k > E(0) 0, it is negative. This 1/r6 long range attractive van der Waals\u2019 potential is a general property of the interaction between two atoms in their ground state.3 5.4 Variational Methods The perturbation theory developed in the previous sections is, of course, of no help unless we already know exact solutions to a problem whose Hamiltonian is suf\ufb01ciently similar. The variational method we now discuss is very useful for estimating the ground-state energy E0 when such exact solutions are not available. We attempt to guess the ground-state energy E0 by considering a \u201ctrial ket\u201d |\u02dc0, which tries to imitate the true ground-state ket |0. To this end we \ufb01rst obtain a theorem of great practical importance. We de\ufb01ne H such that H \u2261 \u02dc0|H|\u02dc0 \u02dc0|\u02dc0, (5.151) where we have accommodated the possibility that |\u02dc0 might not be normalized. We can then prove the following. Theorem 8 This means that we can obtain an upper bound to E0 by considering various kinds of |\u02dc0. The proof of this is very straightforward. H \u2265 E0. (5.152) 3 See the treatment in Schiff (1968), pp. 261\u2013263, which gives a lower and upper bound on the magnitude of the van der Waals\u2019 potential from (5.150) and from a variational calculation. Also note the \ufb01rst footnote on p. 263 of Schiff concerning retardation effects. 317 5.4 Variational Methods Proof Even though we do not know the energy eigenket of the Hamiltonian H, we can imagine that |\u02dc0 can be expanded as |\u02dc0 = \u221e \u2211 k=0 |kk|\u02dc0 (5.153) where |k is an exact energy eigenket of H: H|k = Ek|k. The theorem (5.152) follows when we use Ek = Ek \u2212 E0 + E0 to evaluate H in (5.151). We have (5.154) \u2211 |k|\u02dc0|2Ek H = k=0 \u2211 |k|\u02dc0|2 k=0 \u221e \u2211 k=1 = \u2265 E0, |k|\u02dc0|2(Ek \u2212 E0) |k|\u02dc0|2 \u2211 k=0 + E0 (5.155a) (5.155b) (5.155c) where we have used the fact that Ek \u2212 E0 in the \ufb01rst sum of (5.155b) is necessarily positive. It is also obvious from this proof that the equality sign in (5.152) holds only if |\u02dc0 coincides exactly with |0, that is, if the coef\ufb01cients k|\u02dc0 all vanish for k = 0. The theorem (5.152) is quite powerful because H provides an upper bound to the true ground-state energy. Furthermore, a relatively poor trial ket can give a fairly good energy estimate for the ground state because if then from (5.155) we have k|\u02dc0 \u223c 0(\u03b5) for k = 0, H \u2212 E0 \u223c 0(\u03b52). (5.156) (5.157) We see an example of this in a moment. Of course, the method does not say anything about the discrepancy between H and E0; all we know is that H is larger than (or equal to) E0. Another way to state the theorem is to assert that H is stationary with respect to the variation |\u02dc0 \u2192 |\u02dc0 + \u03b4|\u02dc0; (5.158) that is, \u03b4H = 0 when |\u02dc0 coincides with |0. By this we mean that if |0 + \u03b4|\u02dc0 is used in place of |\u02dc0 in (5.155) and we calculate H", ", then the error we commit in estimating the true ground-state energy involves |\u02dc0 to order (\u03b4|\u02dc0)2. The variational method per se does not tell us what kind of trial kets are to be used to estimate the ground-state energy. Quite often we must appeal to physical intuition, for example, the asymptotic behavior of a wave function at large distances. What we do in practice is to characterize trial kets by one or more parameters \u03bb1, \u03bb2,..., and compute H 318 Approximation Methods as a function of \u03bb1, \u03bb2,.... We then minimize H by (1) setting the derivative with respect to the parameters all zero, namely, \u2202H \u2202\u03bb1 = 0, \u2202H \u2202\u03bb2 = 0,..., (5.159) (2) determining the optimum values of \u03bb1, \u03bb2,..., and (3) substituting them back to the expression for H. If the wave function for the trial ket already has a functional form of the exact groundstate energy eigenfunction, we of course obtain the true ground-state energy function by this method. For example, suppose somebody has the foresight to guess that the wave function for the ground state of the hydrogen atom must be of the form x|0 \u221d e\u2212r/a, (5.160) where a is regarded as a parameter to be varied. We then \ufb01nd, upon minimizing H with (5.160), the correct ground-state energy \u2212e2/2a0. Not surprisingly, the minimum is achieved when a coincides with the Bohr radius a0. As a second example, we attempt to estimate the ground state of the in\ufb01nite-well (one- dimensional box) problem de\ufb01ned by V = for |x| < a 0 \u221e for |x| > a. The exact solutions are, of course, well known: (5.161) x|0 = cos, \u03c0x 2a 1\u221a a \u00afh2 2m But suppose we did not know these. Evidently the wave function must vanish at x = \u00b1a; furthermore, for the ground state the wave function cannot have any wiggles. The simplest analytic function that satis\ufb01es both requirements is just a parabola going through x = \u00b1a: E0 = 4a2. \u03c02 (5.162) x|\u02dc0 = a2 \u2212 x2, (5.163) where we have not bothered to normalize |\u02dc0. Here there is no variational parameter. We can compute H as follows: \u2212\u00afh2 2m a \u2212a H = (a2 \u2212 x2) d2 dx2 (a2 \u2212 x2)dx a \u2212a (a2 \u2212 x2)2 dx 10 \u03c02 = \u03c02 \u00afh2 8a2m 1.0132 E0. (5.164) It is remarkable that with such a simple trial function we can come within 1.3% of the true ground-state energy. 319 5.4 Variational Methods A much better result can be obtained if we use a more sophisticated trial function. We try x|\u02dc0 = |a|\u03bb \u2212 |x|\u03bb, (5.165) where \u03bb is now regarded as a variational parameter. Straightforward algebra gives H = which has a minimum at (\u03bb + 1)(2\u03bb + 1) (2\u03bb \u2212 1) \u00afh2 4ma2, (5.166) \u221a \u03bb = (1 + 2 6) 1.72, (5.167) not far from \u03bb = 2 (a parabola) considered earlier. This gives \u221a 6 5 + 2 \u03c02 Hmin = E0 1.00298E0. (5.168) So the variational method with (5.165) gives the correct ground-state energy within 0.3% \u2013 a fantastic result considering the simplicity of the trial function used. How well does this trial function imitate the true ground-state wave function? It is amusing that we can answer this question without explicitly evaluating the overlap integral 0|\u02dc0. Assuming that |\u02dc0 is normalized, we have [from (5.151)\u2013(5.154)] \u221e Hmin = \u2211 |k|\u02dc0|2Ek k=0 \u2265 |0|\u02dc0|2E0 + 9E0(1 \u2212 |0|\u02dc0|2) (5.169) where 9E0 is the energy of the second excited state; the \ufb01rst excited state (k = 1) gives no contribution by parity conservation. Solving for |0|\u02dc0| and using (5.168), we have |0|\u02dc0|2 \u2265 9", "E0 \u2212 Hmin = 0.99963. (5.170) 8E0 Departure from unity characterizes a component of |\u02dc0 in a direction orthogonal to |0. If we are talking about the \u201cangle\u201d \u03b8 de\ufb01ned by then (5.170) corresponds to 0|\u02dc0 = cos \u03b8, \u03b8< \u223c 1.1\u25e6, (5.171) (5.172) so |0 and |\u02dc0 are nearly \u201cparallel.\u201d One of the earliest applications of the variational method involved the ground-state energy of the helium atom, which we will discuss in Section 7.4. We can also use the variational method to estimate the energies of \ufb01rst excited states; all we need to do is work with a trial ket orthogonal to the ground-state wave function, either exact, if known, or an approximate one obtained by the variational method. 320 Approximation Methods 5.5 Time-Dependent Potentials: The Interaction Picture 5.5.1 Statement of the Problem So far in this book we have been concerned with Hamiltonians that do not contain time explicitly. In nature, however, there are many quantum-mechanical systems of importance with time dependence. In the remaining part of this chapter we show how to deal with situations with time-dependent potentials. We consider a Hamiltonian H such that it can be split into two parts, H = H0 + V(t), (5.173) where H0 does not contain time explicitly. The problem V(t) =0 is assumed to be solved in the sense that the energy eigenkets |n and the energy eigenvalues En de\ufb01ned by H0|n = En|n (5.174) are completely known.4 We may be interested in situations where initially only one of the energy eigenstates of H0, for example, |i is populated. As time goes on, however, states other than |i are populated because with V(t) = 0 we are no longer dealing with \u201cstationary\u201d problems; the time-evolution operator is no longer as simple as e\u2212iHt/\u00afh when H itself involves time. Quite generally the time-dependent potential V(t) can cause transitions to states other than |i. The basic question we address is, what is the probability as a function of time for the system to be found in |n, with n = i? More generally, we may be interested in how an arbitrary state ket changes as time goes on, where the total Hamiltonian is the sum of H0 and V(t). Suppose at t = 0, the state ket of a physical system is given by |\u03b1 = \u2211 cn(0)|n. n We wish to \ufb01nd cn(t) for t > 0 such that |\u03b1, t0 = 0; t = \u2211 cn(t)e\u2212iEnt/\u00afh|n n (5.175) (5.176) where the ket on the left side stands for the state ket in the Schr\u00a8odinger picture at t of a physical system whose state ket at t = 0 was found to be |\u03b1. The astute reader may have noticed the manner in which we have separated the time dependence of the coef\ufb01cient of |n in (5.176). The factor e\u2212iEnt/\u00afh is present even if V is absent. This way of writing the time dependence makes it clear that the time evolution of cn(t) is due solely to the presence of V(t); cn(t) would be identically equal to cn(0) and hence independent of t if V were zero. As we shall see in a moment, this separation is convenient because cn(t) satis\ufb01es a relatively simple differential equation. The probability of \ufb01nding |n is found by evaluating |cn(t)|2. 4 In (5.174) we no longer use the notation |n(0), E(0) n. 321 5.5 Time-Dependent Potentials: The Interaction Picture 5.5.2 The Interaction Picture Before we discuss the differential equation for cn(t), we discuss the interaction picture. Suppose we have a physical system such that its state ket coincides with |\u03b1 at t = t0, where t0 is often taken to be zero. At a later time, we denote the state ket in the Schr\u00a8odinger picture by |\u03b1, t0; tS, where the subscript S reminds us that we are dealing with the state ket of the Schr\u00a8odinger picture. We now de\ufb01ne |\u03b1, t0; tI = eiH0t/\u00afh|\u03b1, t0", "; tS, (5.177) where | I stands for a state ket that represents the same physical situation in the interaction picture. At t = 0, | I evidently coincides with | S. For operators (representing observables) we de\ufb01ne observables in the interaction picture as AI \u2261 eiH0 t / \u00afhAS e\u2212iH0t / \u00afh. (5.178) In particular, VI = eiH0t / \u00afhV e\u2212iH0t / \u00afh where V without a subscript is understood to be the time-dependent potential in the Schr\u00a8odinger picture. The reader may recall here the connection between the Schr\u00a8odinger picture and the Heisenberg picture: (5.179) |\u03b1H = e+iHt/\u00afh|\u03b1, t0 = 0; tS AH = eiHt/\u00afhASe\u2212iHt/\u00afh. (5.180) (5.181) The basic difference between (5.180) and (5.181) on the one hand and (5.177) and (5.178) on the other is that H rather than H0 appears in the exponential. We now derive the fundamental differential equation that characterizes the time evolution of a state ket in the interaction picture. Let us take the time derivative of (5.177) with the full H given by (5.173): i\u00afh \u2202 \u2202t |\u03b1, t0; tI = i\u00afh \u2202 \u2202t (eiH0t/\u00afh|\u03b1, t0; tS) = \u2212H0eiH0t/\u00afh|\u03b1, t0; tS + eiH0t/\u00afh(H0 + V)|\u03b1, t0; tS = eiH0t/\u00afhVe\u2212iH0t/\u00afheiH0t/\u00afh|\u03b1, t0; tS. We thus see i\u00afh \u2202 \u2202t |\u03b1, t0; tI = VI|\u03b1, t0; tI, (5.182) (5.183) which is a Schr\u00a8odinger-like equation with the total H replaced by VI. In other words |\u03b1, t0; tI would be a ket \ufb01xed in time if VI were absent. We can also show for an observable A (that does not contain time t explicitly in the Schr\u00a8odinger picture) that dAI dt = 1 i\u00afh [AI, H0], (5.184) which is a Heisenberg-like equation with H replaced by H0. 322 Approximation Methods Table 5.2 Heisenberg, Interaction, and Schr\u00f6dinger Pictures Heisenberg picture Interaction picture Schr\u00a8odinger picture State ket No change Evolution determined Evolution determined by VI by H Observable Evolution determined Evolution determined No change by H by H0 In many respects, the interaction picture, or Dirac picture, is intermediate between the Schr\u00a8odinger picture and the Heisenberg picture. This should be evident from Table 5.2. In the interaction picture we continue using |n as our base ket. Thus we expand | I as follows: |\u03b1, t0; tI = \u2211 cn(t)|n. n (5.185) With t0 set equal to 0, we see that the cn(t) appearing here are the same as the cn(t) introduced earlier in (5.176), as can easily be veri\ufb01ed by multiplying both sides of (5.176) by eiH0t/\u00afh using (5.174). We are \ufb01nally in a position to write the differential equation for cn(t). Multiplying both sides of (5.183) byn| from the left, we obtain i\u00afh \u2202 \u2202t n|\u03b1, t0; tI = \u2211 n|VI|mm|\u03b1, t0; tI. m (5.186) This can also be written using n|eiH0t/\u00afhV(t)e\u2212iH0t/\u00afh|m = Vnm(t)ei(En\u2212Em)t/\u00afh cn(t) = n|\u03b1, t0; tI i\u00afh d dt cn(t) = \u2211 Vnmei\u03c9nmtcm(t), m \u03c9nm \u2261 (En \u2212 Em) \u00afh = \u2212\u03c9mn. and [from (5.185)] as where Explicitly, i\u00afc1 \u02d9c2 \u02d9c3... V11 V21ei\u03c921t V12ei\u03c912t V22 V33 \u239e \u239b \u239e \u239f \ufffd", "\ufffd\ufffd \u239f \u23a0 \u239c \u239c \u239c \u239d \u239f \u239f \u239f \u23a0. c1 c2 c3... \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7...... This is the basic coupled differential equation that must be solved to obtain the probability of \ufb01nding |n as a function of t.... (5.187) (5.188) (5.189) 323 5.5 Time-Dependent Potentials: The Interaction Picture 5.5.3 Time-Dependent Two-State Problems: Nuclear Magnetic Resonance, Masers, and So Forth Exact soluble problems with time-dependent potentials are rather rare. In most cases we have to resort to perturbation expansion to solve the coupled differential equations (5.189), as we will discuss in the next section. There is, however, a problem of enormous practical importance, which can be solved exactly \u2013 a two-state problem with a sinusoidal oscillating potential. The problem is de\ufb01ned by H0 = E1|11| + E2|22| V(t) = \u03b3ei\u03c9t|12| + \u03b3e\u2212i\u03c9t|21|, (E2 > E1) (5.190) where \u03b3 and \u03c9 are real and positive. In the language of (5.186) and (5.187), we have 21 = \u03b3ei\u03c9t V12 = V\u2217 V11 = V22 = 0. (5.191) We thus have a time-dependent potential that connects the two energy eigenstates of H0. In other words, we can have a transition between the two states |1 \u2192 \u2190 |2. An exact solution to this problem is available. If initially, at t = 0, only the lower level is populated so that [see (5.175)] c1(0) =1, c2(0) =0, (5.192) then the probability for being found in each of the two states is given by Rabi\u2019s formula (after I. I. Rabi, who is the father of molecular beam techniques) 1/2 * + |c2(t)|2 = \u03b32/\u00afh2 \u03b32/\u00afh2 + (\u03c9 \u2212 \u03c921)2/4 sin2 \u03b32 \u00afh2 + (\u03c9 \u2212 \u03c921)2 4 |c1(t)|2 = 1 \u2212 |c2(t)|2, where \u03c921 \u2261 (E2 \u2212 E1) \u00afh, t (5.193a) (5.193b) (5.194) as the reader may verify by working out Problem 5.37 of this chapter. Let us now look at |c2|2 a little more closely. We see that the probability for \ufb01nding the upper state E2 exhibits an oscillatory time dependence with angular frequency, two times that of \u03a9 = \u03b32 \u00afh2 + (\u03c9 \u2212 \u03c921)2 4. The amplitude of oscillation is very large when \u03c9 \u03c921 = (E2 \u2212 E1) \u00afh, (5.195) (5.196) 324 Approximation Methods |c1(t)|2 |c2(t)|2 1 0 \u03c0h/2\u03b3 \u03c0h/\u03b3 3\u03c0h/2\u03b3 t Absorption Emission Absorption Fig. 5.5 Plot of |c1(t)|2 and |c2(t)|2 against time t exactly at resonance \u03c9 = \u03c921 and \u03a9 = \u03b3/\u00afh. The graph also illustrates the back-and-forth behavior between |1 and |2. |c2(t)|2 max 1 1/2 Full width at half maximum = 4\u03b3 /h Fig. 5.6 Graph of |c2(t)|2 max as a function of \u03c9, where \u03c9 = \u03c921 corresponds to the resonant frequency. \u03c9 21 \u03c9 that is, when the angular frequency of the potential, usually due to an externally applied electric or magnetic \ufb01eld, is nearly equal to the angular frequency characteristic of the two-state system. Equation (5.196) is therefore known as the resonance condition. It is instructive to look at (5.193a) and (5.193b) a little closely exactly at resonance: \u03c9 = \u03c921, \u03a9 = \u03b3 \u00afh. (5.197) We can plot |c1(t)|2 and |c2(t)|2 as a function of t; seeFigure 5.5. From t = 0 tot = \u03c0\u00afh/2\u03b3, the two-level system absorbs energy from the time-dependent potential V(t); |", "c1(t)|2 decreases from unity as |c2(t)|2 grows. At t = \u03c0\u00afh/2\u03b3, only the upper state is populated. From t = \u03c0\u00afh/2\u03b3 to t = \u03c0\u00afh/\u03b3, the system gives up its excess energy [of the excited (upper) state] to V(t); |c2|2 decreases and |c1|2 increases. This absorption-emission cycle is repeated inde\ufb01nitely, as is also shown in Figure 5.5, so V(t) can be regarded as a source or sink of energy; put in another way, V(t) can cause a transition from |1 to |2 (absorption) or from |2 to |1 (emission). We will come back to this point of view when we discuss emission and absorption of radiation. The absorption-emission cycle takes place even away from resonance. However, the max is no longer 1 and |c1(t)|2 does amplitude of oscillation for |2 is now reduced; |c2(t)|2 not go down all the way to 0. In Figure 5.6 we plot |c2(t)|2 max as a function of \u03c9. This curve has a resonance peak centered around \u03c9 = \u03c921, and the full width at half maximum is 325 5.5 Time-Dependent Potentials: The Interaction Picture given by 4\u03b3/\u00afh. It is worth noting that the weaker the time-dependent potential (\u03b3 small), the narrower the resonance peak. 5.5.4 Spin Magnetic Resonance The two-state problem de\ufb01ned by (5.190) has many physical applications. As a \ufb01rst example, consider a spin 1 2 system, say a bound electron, subjected to a t-independent uniform magnetic \ufb01eld in the z-direction and, in addition, a t-dependent magnetic \ufb01eld rotating in the xy-plane: B = B0 \u02c6z + B1(\u02c6x cos \u03c9t + \u02c6y sin \u03c9t) (5.198) with B0 and B1 constant. We can treat the effect of the uniform t-independent \ufb01eld as H0 and the effect of the rotating \ufb01eld as V. For we have \u03bc = e mec S (5.199) H0 = \u2212 V(t) = \u2212 e\u00afhB0 2mec e\u00afhB1 2mec (|++| \u2212 |\u2212\u2212|) [cos \u03c9t(|+\u2212| + |\u2212+|) (5.200) + sin \u03c9t(\u2212i|+\u2212| + i|\u2212+|)], where we have used the ket-bra forms of 2Sj/\u00afh [see (3.25)]. With e < 0, E+ has a higher energy than E\u2212, and we can identify |+ \u2192 |2 |\u2212 \u2192 |1 (upper level) (lower level) (5.201) to make correspondence with the notation of (5.190). The angular frequency characteristic of the two-state system is \u03c921 = |e|B0 mec, (5.202) which is just the spin-precession frequency for the B0 = 0, B1 = 0 problem already treated in Section 2.1. Even though the expectation values of Sx,y change due to spin precession in the counterclockwise direction (seen from the positive z-side), |c+|2 and |c\u2212|2 remain unchanged in the absence of the rotating \ufb01eld. We now add a new feature as a result of the rotating \ufb01eld: |c+|2 and |c\u2212|2 do change as a function of time. This can be seen by identifying \u2212e\u00afhB1 2mec \u2192 \u03b3, \u03c9 \u2192 \u03c9 (5.203) to make correspondence with the notation of (5.190); our time-dependent interaction (5.200) is precisely of form (5.190). The fact that |c+(t)|2 and |c\u2212(t)|2 vary in the manner indicated by Figure 5.5 for \u03c9 = \u03c921 and the correspondence (5.201), for example, implies 326 Approximation Methods 2 system undergoes a succession of spin-\ufb02ips, |+ \u2192 \u2190 |\u2212, in addition to spin that the spin 1 precession. Semiclassically, spin-\ufb02ips of this kind can be interpreted as being due to the driving torque exerted by rotating B. The resonance condition is satis\ufb01ed whenever the frequency of the rotating magnetic \ufb01eld coincides with the frequency of spin precession determined by", " the strength of the uniform magnetic \ufb01eld. We see that the probability of spin-\ufb02ips is particularly large. In practice, a rotating magnetic \ufb01eld may be dif\ufb01cult to produce experimentally. Fortunately, a horizontally oscillating magnetic \ufb01eld, for instance, in the x-direction, is just as good. To see this, we \ufb01rst note that such an oscillating \ufb01eld can be decomposed into a counterclockwise component and a clockwise component as follows: 2B1 \u02c6x cos \u03c9t = B1(\u02c6x cos \u03c9t + \u02c6y sin \u03c9t) +B 1(\u02c6x cos \u03c9t \u2212 \u02c6y sin \u03c9t). (5.204) We can obtain the effect of the counterclockwise component simply by reversing the sign of \u03c9. Suppose the resonance condition is met for the counterclockwise component \u03c9 \u03c921. Under a typical experimental condition, B1 B0 which implies from (5.202) and (5.203) that 1, \u03b3 \u00afh \u03c921; (5.205) (5.206) (5.207) as a result, whenever the resonance condition is met for the counterclockwise component, the effect of the clockwise component becomes completely negligible, since it amounts to \u03c9 \u2192 \u2212\u03c9, and the amplitude becomes small in magnitude as well as very rapidly oscillating. The resonance problem we have solved is of fundamental importance in interpreting atomic molecular beam and nuclear magnetic resonance experiments. By varying the frequency of oscillating \ufb01eld, it is possible to make a very precise measurement of magnetic moment. We have based our discussion on the solution to differential equations (5.189); this problem can also be solved, perhaps more elegantly, by introducing the rotating axis representation of Rabi, Schwinger, and Van Vleck. 5.5.5 Maser As another application of the time-dependent two-state problem, let us consider a maser. Speci\ufb01cally, we consider an ammonia molecule NH3, which \u2013 as we may recall from Section 4.2 \u2013 has two parity eigenstates |S and |A lying close together such that |A is slightly higher. Let \u03bcel be the electric dipole operator of the molecule. From symmetry considerations we expect that \u03bcel is proportional to x, the position operator for the N atom. The basic interaction is like \u2212\u03bcel \u00b7 E, where for a maser E is a time-dependent electric \ufb01eld in a microwave cavity: E = |E|max \u02c6z cos \u03c9t. (5.208) 327 5.6 Hamiltonians with Extreme Time Dependence It is legitimate to ignore the spatial variation of E because the wavelength in the microwave region is far larger than the molecular dimension. The frequency \u03c9 is tuned to the energy difference between |A and |S: \u03c9 (EA \u2212 ES) \u00afh. The diagonal matrix elements of the dipole operator vanish by parity, A|\u03bcel|A = S|\u03bcel|S = 0, but the off-diagonal elements are, in general, nonvanishing: S|x|A = A|x|S = 0. (5.209) (5.210) (5.211) This means that there is a time-dependent potential that connects |S and |A, and the general two-state problem we discussed earlier is now applicable. We are now in a position to discuss how masers work. Given a molecular beam of NH3 containing both |S and |A, we \ufb01rst eliminate the |S component by letting the beam go through a region of time-independent inhomogeneous electric \ufb01eld. Such an electric \ufb01eld separates |S from |A in much the same way as the inhomogeneous magnetic \ufb01eld in the Stern\u2013Gerlach experiment separates |+ from |\u2212. A pure beam of |A then enters a microwave cavity tuned to the energy difference EA \u2212 ES. The dimension of the cavity is such that the time spent by the molecule is just (\u03c0/2)\u00afh/\u03b3. As a result we stay in the \ufb01rst emission phase of Figure 5.5; we have|A in and |S out. The excess energy of |A is given up to the time-dependent potential as |A turns into |S and the radiation (microwave) \ufb01eld gains energy. In this way we obtain Microwave Ampli\ufb01cation by Stimulated Emission of Radiation, or MASER. There are many other applications of the general time-dependent two-state problem, such as the atomic clock and optical pumping. In fact, it is amusing to see that", " as many as four Nobel Prizes in physics have been awarded to those who exploited time-dependent two-state systems of some form.5 5.6 Hamiltonians with Extreme Time Dependence This section is devoted to time-dependent Hamiltonians, with some \u201cobvious\u201d approximations in the case of very fast or very slow time dependence. A careful look, though, points out some interesting phenomena, one of which did not come to light until very late in the twentieth century. Our treatment here is limited to a discussion of the basics, followed by some speci\ufb01c examples. An instructive example which we do not discuss is that of the square well with 5 Nobel Prize winners who took advantage of resonance in the two-level systems are Rabi (1944) on molecular beams and nuclear magnetic resonance; Bloch and Purecell (1952) on B \ufb01eld in atomic nuclei and nuclear magnetic moments; Townes, Basov, and Prochorov (1964) on masers, lasers, and quantum optics; and Kastler (1966) on optical pumping. 328 Approximation Methods contracting or expanding walls, where the walls may move quickly or slowly. For these cases, we refer the interested reader to Pinder, Am. J. Phys., 58 (1990) 54, and D. W. Schlitt and Stutz, Am. J. Phys., 38 (1970) 70. 5.6.1 Sudden Approximation If a Hamiltonian changes very quickly, then the system \u201cdoesn\u2019t have time\u201d to adjust to the change. This leaves the system in the same state it was before the change, and is the essence of the so-called \u201csudden approximation.\u201d Of course, even though it may have been in an eigenstate beforehand, there is no reason to believe that this is an eigenstate of the transformed Hamiltonian. Therein lie opportunities for interesting physics. One classic example is the calculation of the population of electronic \ufb01nal states in the 3He+ ion following beta decay of the tritium atom.6 See Problem 5.42 at the end of this chapter. Let us consider a more precise statement of the sudden approximation, and work through some of the consequences. Rewrite the Schr\u00a8odinger equation for the time-evolution operator (2.25) as \u2202 \u2202s i U (t, t0) = H \u00afh/T U (t, t0) = H \u00afh\u03a9 U (t, t0), (5.212) where we have written time t = sT in terms of a dimensionless parameter s and a time scale T, and de\ufb01ned \u03a9 \u2261 1/T. In the sudden approximation, the time scale T \u2192 0 which means that \u00afh\u03a9 will be much larger than the energy scale represented by H. Assuming we can rede\ufb01ne H by adding or subtracting an arbitrary constant, introducing some overall phase factor in the state vectors, we see that U (t, t0) \u2192 1 as T \u2192 0. (5.213) This proves the validity of the sudden approximation. It should be appropriate if T is small compared to times 2\u03c0/\u03c9ab where Eab = \u00afh\u03c9ab is the difference between two relevant eigenvalues of the Hamiltonian H. 5.6.2 Adiabatic Approximation We tend to take the adiabatic approximation for granted. Given a Hamiltonian which depends on some set of parameters, we will \ufb01nd energy eigenvalues which depend on the values of those parameters. If the parameters vary \u201cslowly\u201d with time, then the energy eigenvalues should just follow the values one gets as the parameters themselves change. The key is what we mean by \u201cslowly.\u201d Quantum mechanically or otherwise, presumably we mean that the parameters change on a time scale T that is much larger than 2\u03c0/\u03c9ab = 2\u03c0\u00afh/Eab for some difference Eab in energy eigenvalues. 6 This has important implications for modern experiments which try to infer a nonzero neutrino mass from beta decay measurements. The Karlsruhe Tritium Neutrino Experiment (KATRIN), for example, has just published \ufb01rst results at the time of this writing. See Aker et al., Phys. Rev. Lett., 123 (2019) 221802. 329 5.6 Hamiltonians with Extreme Time Dependence An obvious classical example is a pendulum which is transported around near the surface of the Earth. The pendulum will behave normally as you climb a mountain, with only the period slowly lengthening as the force of gravity decreases, so long as the time over which the height is changed is long compared to pendulum period. If one slowly changes the electric \ufb01eld which permeates a", " hydrogen atom, the energy levels will change in pace according to the Stark effect calculation in Section 5.2. Let us consider the mathematics of adiabatic change from a quantum-mechanical point of view. We follow the treatment given in Grif\ufb01ths (2005), and pay particular attention to the phase change as a function of time. We number the states in order using the index n and assume no degeneracy7 so there is no confusion with the ordering of states crossing as time changes. Our starting point is essentially (2.27) but we will take t0 = 0 and suppress the initial time in our notation. Begin with the eigenvalue equation using the notation H(t)|n; t = En(t)|n; t, (5.214) simply noting that at any particular time t, the states and eigenvalues may change. If we now look for general solutions to the Schr\u00a8odinger equation of the form i\u00afh \u2202 \u2202t |\u03b1; t = H(t)|\u03b1; t |\u03b1; t = \u2211 cn(t)ei\u03b8n(t)|n; t n (5.215) (5.216) then we can write where \u03b8n(t) \u2261 \u2212 1 \u00afh The separation of the expansion coef\ufb01cient into the factors cn(t) and exp(i\u03b8n(t)) will prove useful in a moment. Substituting (5.216) into (5.215) and using (5.214) we \ufb01nd En(t)dt. (5.217) 0 t \u2211 ei\u03b8n(t) n \u02d9cn(t)|n; t + cn(t) \u2202 \u2202t |n; t = 0. (5.218) Now, taking the inner product with m; t| and invoking orthonormality of the eigenstates at equal times, we arrive at a differential equation for the cn(t), namely \u02d9cm(t) =\u2212 \u2211 cn(t)ei[\u03b8n(t)\u2212\u03b8m(t)]m; t| n |n; t. (5.219) \u2202 \u2202t The inner product m; t|(\u2202/\u2202t)|n; t is a new feature. If H were not time dependent, then the |n; t would be stationary states, and the usual exponential time dependence would emerge. In order to treat this in the general case, we can go back to (5.214) and take the time derivative of both sides. For the case where m = n we \ufb01nd m; t| \u02d9H|n; t = [En(t) \u2212 Em(t)] m; t| |n; t. (5.220) \u2202 \u2202t 7 This is not a signi\ufb01cant constraint. If the degeneracy is broken by H(t) after some time, we can just \u201cstart\u201d there. If the degeneracy is never broken by H(t) then it is irrelevant. 330 Approximation Methods This \ufb01nally allows us to rewrite (5.219) as \u02d9cm(t) =\u2212 cm(t)m; t| \u2202 \u2202t |m; t \u2212\u2211 cn(t)ei(\u03b8n\u2212\u03b8m) n m; t| \u02d9H|n; t En \u2212 Em (5.221) which is a formal solution to the general time-dependent problem. Equation (5.221) demonstrates that as time goes on, states with n = m will mix with |m; t due to the time dependence of the Hamiltonian H, by virtue of the second term. Now we can apply the adiabatic approximation, which amounts to neglecting the second term in (5.221). Roughly, this means that |m; t| \u02d9H|n; t| Enm \u2261 1 \u03c4 m; t| |m; t \u2202 \u2202t \u223c Em \u00afh. (5.222) In other words, the time scale \u03c4 for changes in the Hamiltonian must be very large compared to the inverse natural frequency of the state phase factor. That is, just as for the pendulum being carried around the Earth, the Hamiltonian changes much more slowly than the oscillation frequency of the system. Consequently, we have where cn(t) =e i\u03b3n(t)cn(0) \u03b3n(t) \u2261 i t 0 n; t| \u2202 \u2202t |n; t dt. Note that by this de\ufb01nition, \u03b3n(t) is real, since 0 = \u2202 \u2202t n; t|n", "; t = \u2202 \u2202t n; t| |n; t + n; t| |n; t \u2202 \u2202t or, in other words, n; t| \u2202 \u2202t \u2217 |n; t = \u2212n; t| |n; t \u2202 \u2202t (5.223) (5.224) (5.225) (5.226) in which case the integrand in (5.224) is purely imaginary. Therefore, in the adiabatic approximation, if a system starts out in an eigenstate |n of H(0), then it remains in the eigenstate |n; t of H(t), since ci(0) = 0 unless i = n in which case cn(0) =1. Using (5.216) with ( 5.223) we have, in an obvious notation, |\u03b1(n); t = ei\u03b3n(t)ei\u03b8n(t)|n; t. (5.227) It would appear that (5.227) is dif\ufb01cult to use, since the de\ufb01nition (5.224) assumes the time dependence of the state is given, but we will \ufb01nd ways to make good use of this result. In any case, it is easy to see that the result is self consistent. We know that for the case when H is not time dependent, we expect and so |n; t = e\u2212iEnt/\u00afh|n \u2202 \u2202t n; t| |n; t = \u2212i En \u00afh (5.228) (5.229) 331 5.6 Hamiltonians with Extreme Time Dependence which, by (5.224) gives \u03b3n(t) = +Ent/\u00afh. On the other hand, (5.217) says that \u03b8n(t) = \u2212Ent/\u00afh. So, the two exponential factors in (5.227) cancel each other, and we \ufb01nd |\u03b1(n); t = |n; t for H = H(t) (5.230) as we should expect. The addition of this new phase \u03b3n(t) is the only result of the adiabatic approximation that is less than obvious. It was not considered worth pursuing for many years, until it was discovered that it is in fact measurable. Indeed, it turns out to be the quantum-mechanical manifestation of very many physical phenomena that involve systems that are cyclic in time. 5.6.3 Berry\u2019s Phase Excitement in the implications of (5.224) grew dramatically with the publication of \u201cQuantal phase factors accompanying adiabatic changes\u201d, by M. V. Berry (Proc. R. Soc. London, Ser. A, 392 (1984) 45). Indeed, the accumulated phase for systems which travel in a closed loop is generally called \u201cBerry\u2019s phase,\u201d although Berry himself refers to this as a \u201cgeometric phase.\u201d Berry\u2019s paper is widely cited, and the interested reader will have no dif\ufb01culty \ufb01nding many references. One particular paper that provides a succinct summary and interesting implications is \u201cThe adiabatic theorem and Berry\u2019s phase\u201d by Holstein (Am. J. Phys., 57 (1989) 1079). Berry, in fact, gives a lovely history of work prior to his own publication, in \u201cAnticipations of the geometric phase\u201d (Phys. Today, 43 (1990) 34). Assume that the time dependence of the Hamiltonian is represented by a \u201cvector of parameters\u201d R(t). That is, there exists some space in which the components of a vector R(t) specify the Hamiltonian, and change as a function of time. (In an example below, R(t) will be the magnetic \ufb01eld.) Therefore, we have En(t) = En(R(t)) and |n; t = |n(R(t)), and also \u2202 \u2202t |n; t n; t| = n; t| [\u2207R|n; t] \u00b7 dR dt (5.231) where \u2207R is simply a gradient operator in the space and direction of R. The geometric phase (5.224) then becomes \u03b3n(T) =i n; t| [\u2207R|n; t] \u00b7 dR dt dt T 0 R(T) = i R(0) n; t| [\u2207R|n; t] \u00b7 dR. (5.232) In the case where T represents the period for one full cycle, so that R(T) =R(0)", " where the vector R traces a curve C, we have, \u03b3n(C) =i n; t| [\u2207R|n; t] \u00b7 dR. (5.233) 332 Approximation Methods With a notation that shows a bias of how we can proceed, de\ufb01ne An(R) \u2261 in; t| [\u2207R|n; t] in which case \u03b3n(C) =, C An(R) \u00b7 dR = [\u2207R \u00d7 An(R)] \u00b7 da (5.234) (5.235) using Stokes\u2019s theorem, generalized8 for the dimensionality of R. (The measure da is a small area element on some surface bounded by the closed path.) Thus, Berry\u2019s phase is determined by the \u201c\ufb02ux\u201d of a generalized \ufb01eld Bn(R) \u2261 \u2207R \u00d7 An(R) (5.236) through a surface S bounded by the circuit followed by R(t) over one complete cycle. One obtains the same phase \u03b3n so long as one encounters the same total \ufb02ux, regardless of the actual path followed by R(t). Note that, quite similarly to our derivation of the result (5.226), both An(R) and Bn(R) are purely real quantities. Soon we will be concerning ourselves with sources of the \ufb01eld Bn(R). Equation (5.235) has a remarkable property that betrays the notation we have chosen using An(R). Suppose that we multiply |n; t by an arbitrary phase factor that changes through R-space. That is |n; t \u2212\u2192e i\u03b4(R)|n; t. (5.237) Then by (5.234) we have An(R) \u2212\u2192 An(R) \u2212 \u2207R\u03b4(R) which leaves (5.235) unchanged. In other words, the value of \u03b3n(C) does not depend on the details of the phase behavior along the path, despite our starting point (5.227). Indeed, \u03b3n(C) depends only on the geometry of the path traced out by R(t), hence the name \u201cgeometric\u201d phase. Of course, it remains for us to show that \u03b3n(C) is nonzero, at least under certain conditions. Note also, that (5.237) and (5.238) have exactly the same form as that for gauge transformations in electromagnetism. See (2.356) and (2.369). This analogy will be exploited more fully before we conclude this section. (5.238) We now turn to an evaluation of \u03b3n(C). Noting \ufb01rst that since the curl of a gradient vanishes, we can combine (5.234) and (5.236) to get Bn(R) =i [\u2207Rn; t|] \u00d7 [\u2207R|n; t]. (5.239) Now insert a complete set of states |m; t to \ufb01nd Bn(R) = i \u2211 m=n [\u2207Rn; t|] |m; t \u00d7 m; t| [\u2207R|n; t]. (5.240) 8 To be sure, generalizing Stokes\u2019s theorem for higher dimensionality, is not trivial. See a discussion of this in Berry\u2019s original paper. In our case, however, all of our examples will involve only three-dimensional parameter vectors R. 333 5.6 Hamiltonians with Extreme Time Dependence We explicitly discard the term with m = n, but it is easily shown to be zero, since n; t|n; t = 1 implies that [\u2207Rn; t|] |n; t = \u2212n; t| [\u2207R|n; t] and so the cross product in (5.240) must be zero. Now, by taking the R-gradient of (5.214) and taking the inner product with m; t| we determine m; t| [\u2207R|n; t] = m; t| [\u2207RH] |n; t En \u2212 Em m = n. (5.241) This allows us to write, \ufb01nally, \u03b3n(C) = Bn(R) \u00b7 da where Bn(R) =i \u2211 m=n n; t| [\u2207RH] |m; t \u00d7 m; t| [\u2207RH] |n; t (Em \u2212 En)2. (5.242) (5.243) As Berry states in his original paper, these last two equations \u201cembody the central results\u201d of his work", ". Points in R-space where Em(R) =E n(R) will contribute to the surface integral (5.242) even though the path enclosing that surface does not include those points. It was realized early on that Berry\u2019s phase could be observed using photons moving through a twisted optical \ufb01ber, and that the geometric character of this phase could be tested experimentally. See Tomita and Chiao, Phys. Rev. Lett., 57 (1986) 937. Indeed, this experiment can be carried out in the student laboratory. A description of the setup can be found in Melissinos and Napolitano (2003). The next section carries through an example strongly connected to concepts already developed in this textbook, and compared to measurement. However, there are many examples that can be experimentally veri\ufb01ed. One is the precise demonstration of Berry\u2019s phase in chemical kinetics, described in Yuan et al., Science, 362 (2018) 1289. 5.6.4 Example: Berry\u2019s Phase for Spin 1 2 Let us now turn to a speci\ufb01c example, and carry through a calculation of \u03b3n(C) from (5.242). We will study the phase motion for a spin 1 2 particle manipulated slowly through a time varying magnetic \ufb01eld. This particular example has in fact been studied experimentally. We return to (2.49), our familiar Hamiltonian for a spin 1 2 particle in a magnetic \ufb01eld, but with some modi\ufb01cation for a particle with arbitrary magnetic moment. Since in this case, it is the magnetic \ufb01eld which changes slowly in time, let the magnetic \ufb01eld be given by the three-dimensional vector9 R(t). That is,R(t ) is the vector of parameters that we will change slowly. For a magnetic moment \u03bc, our Hamiltonian is written as H(t) =H(R(t)) = \u2212 2\u03bc \u00afh S \u00b7 R(t) (5.244) where S is the spin 1 value for the magnetic moment in the spin-up state is simply \u03bc. 2 angular-momentum operator. Written in this way, the expectation 9 We do not use B to represent the magnetic \ufb01eld, so that we avoid confusions with (5.243). 334 Approximation Methods Now on to the evaluation of B(R) using (5.243). First, it is simple enough to show, either explicitly (see Problem 3.3 in Chapter 3) or using rotational symmetry to \ufb01x R in the \u02c6z-direction, that the two energy eigenvalues for (5.244) are E\u00b1(t) = \u2213\u03bcR(t) (5.245) where R(t) is the magnitude of the magnetic \ufb01eld vector, and the spin-up(down) eigenstates (with respect to the direction of R(t)) are |\u00b1; t. The summation in (5.243) consists of only one term, with denominator It is also clear that (E\u00b1 \u2212 E\u2213)2 = 4\u03bc2R2. \u2207RH = \u2212 2\u03bc \u00afh S (5.246) (5.247) leaving us with the need to evaluate the cross product \u00b1; t|S|\u2213; t \u00d7 \u2213; t|S|\u00b1; t = \u00b1; t|S|\u2213; t \u00d7 \u00b1; t|S|\u2213; t\u2217. (5.248) Evaluating this matrix element would be tedious, except that we can invoke rotational symmetry and de\ufb01ne the components of S relative to the direction of R. That is,|\u00b1; t can be taken to be eigenstates of Sz. So, using (3.157) to write S = 1 2 (S+ + S\u2212) \u02c6x + 1 2i (S+ \u2212 S\u2212) \u02c6y + Sz \u02c6z, we invoke (3.191) and (3.192) to \ufb01nd \u00afh 2 Combining (5.246), (5.248), and (5.250), we have B\u00b1(R) =\u2213 1 \u00b1; t|S|\u2213; t = (\u02c6x \u2213 i\u02c6y). \u02c6z. 2R2(t) (5.249) (5.250) (5.251) Of course, this result was derived by taking |\u00b1; t to be eigenstates of Sz, when in fact they are in the direction of R. Therefore, we actually have B\u00b1(R) =\u2213 1 2R2(t) \u02c6R. Finally, we calculate Berry\u2019s phase (", "5.242) to be \u03b3\u00b1(C) =\u2213 1 2 \u02c6R \u00b7 da R2 = \u2213 1 2 \u03a9 (5.252) (5.253) where \u03a9 is the \u201csolid angle\u201d subtended by the path through which the parameter vector R(t) travels, relative to an origin R = 0 which is the source point for the \ufb01eld B. This emphasizes the \u201cgeometric\u201d character of Berry\u2019s phase. Speci\ufb01cs of the path do not matter, so long as the solid angle subtended by the path is the same. The result is also independent of the magnetic moment \u03bc. Soon after Berry\u2019s prediction for this effect in spin 1 2 systems, two groups carried out measurements using neutrons at the Institut Laue-Langevin in Grenoble, France. One of 335 5.6 Hamiltonians with Extreme Time Dependence 1 0 \u03c0 2 / \u03b3 \u22121 0 \u03b3\u2212 \u03b3+ 1 2 /2\u03c0 3 Fig. 5.7 Observation of Berry\u2019s phase for spin 1 2 particles using ultra cold neutrons, from Richardson et al., Phys. Rev. Lett., 61 (1988) 2030. Data are taken from Table 1 of their paper, and show Berry\u2019s phase as a function of \u201csolid angle\u201dfor the rotating magnetic field. Both spin-up and spin-down phases were measured. Uncertainties on the data points are about the same size as, or smaller than, the points themselves. The solid lines are taken from( 5.253). these, Bitter and Dubbers, Phys. Rev. Lett., 59 (1987) 251, used a slow (500 m/sec) neutron beam passing through a twisted magnetic \ufb01eld. The second, Richardson et al., Phys. Rev. Lett., 61 (1988) 2030, made use of ultra cold neutrons (UCN) and is more precise. UCN can be stored for long periods of time, so a cycle period T = 7.387 sec was used, ensuring the validity of the adiabatic theorem. Initially polarized in the z-direction, the neutrons are subjected to a rotating magnetic \ufb01eld component that is switched on at t = 0 and switched off at t = T. The magnetic \ufb01eld vector traces out a circle (or ellipse, depending on adjustable parameters) in the yz-plane, depolarizing the neutrons by an amount depending on the integrated phase. Measuring the \ufb01nal polarization determines the integrated phase, and the dynamical phase (5.217) is subtracted off, leaving Berry\u2019s phase. Figure 5.7 shows the results obtained by Richardson et al. Both spin-up and spin-down phases are measured, and both agree extremely well with Berry\u2019s analysis. Even though the value of the magnetic moment does not enter the calculation, its sign determines the direction of rotation, and this experiment con\ufb01rms that the neutron magnetic moment is indeed negative. 5.6.5 Aharonov\u2013Bohm and Magnetic Monopoles Revisited We have seen that Berry\u2019s phase uses a formalism that is closely tied to the formalism of gauge transformations. See (5.237) and (5.238). Let us now makes this connection closer to some physics we have already seen in our study of gauge transformation in Section 2.7. First we can show that the Aharonov\u2013Bohm effect due to the magnetic \ufb01eld can be shown to be just a consequence of a geometrical phase factor. Let a small box con\ufb01ning an 336 Approximation Methods Magnetic flux line r R C Fig. 5.8 The Aharonov\u2013Bohm effect as a manifestation of Berry\u2019s phase. An electron in a box takes one turn around a magnetic flux line. electron (charge e < 0) make one turn along a closed loop C, which surrounds a magnetic \ufb02ux line \u03a6B, as shown in Figure 5.8. Let R be the vector connecting the origin \ufb01xed in the space and a reference point in the box. In this case the vector R is an external parameter in the real space itself. By using the vector-potential A to describe the magnetic \ufb01eld B, the nth wave function of the electron in the box (with position vector r) is written as - r A(r) \u00b7 dr ie \u00afhc r|n(R) = exp \u03c8n(r \u2212 R) (5.254) R where \u03c8n(r) is the wave function of the electron at the r position coordinates of the box in the absence of magnetic \ufb01eld. Now let R travel around the loop", " C and calculate Berry\u2019s phase. We can easily calculate the derivative of the wave function with respect to the external parameter to obtain n(R) | [\u2207R|n(R)] = \u00d7 d3x\u03c8\u2217 \u2212 ie \u00afhc n(r \u2212 R) A(R)\u03c8n(r \u2212 R) + \u2207R\u03c8n(r \u2212 R) - = \u2212 ieA(R) \u00afhc. (5.255) The second term under the integral vanishes for the electron in the box. From (5.232) and (5.255) we see that the geometrical phase is given by, \u03b3n(C) = e \u00afhc A \u00b7 dR = C e \u00afhc B(R) \u00b7 dS = S(C) e \u00afhc \u03a6B. (5.256) This result is just the expression (2.390) of the Aharonov\u2013Bohm effect obtained earlier in Section 2.7. 337 5.7 Time-Dependent Perturbation Theory A second example10 of the physical connection between gauge invariance and Berry\u2019s phase, is the Dirac quantization condition (2.405) for magnetic monopoles. Consider two surfaces a1 and a2 in R-space, each bounded by the same curve C. Since the Berry\u2019s phase that results from following C is physically measurable, the surface integral (5.242) must be the same for a1 and a2 to within a multiple of 2\u03c0. That is B \u00b7 da = B \u00b7 da + 2N\u03c0 N = 0, \u00b11, \u00b12,.... (5.257) a1 a2 Now construct a closed surface by putting a1 \u201cabove\u201d C and a2 below. Assuming that we use something like the right-hand rule to consistently determine the orientation of da for each of the two integrals in (5.257), then da points inward for one of them. So, reverse the sign of da for that integral, and rewrite (5.257) as, B \u00b7 da = 2N\u03c0. (5.258) Of course, inserting (5.252) into (5.258) yields \u00b12\u03c0, i.e. N = \u00b11, but that is an artifact of our choosing a spin 1 2 system for deriving (5.252). If we carried through that example for a system with arbitrary spin, then n works out to be equal to twice the spin projection quantum number. Now (5.252) looks suspiciously like the magnetic \ufb01eld from a monopole with charge 1/2, but recall that it is in fact a Berry \ufb01eld. How can we relate this to an actual magnetic \ufb01eld for a monopole? It is easiest to see this when we express Berry\u2019s phase in terms of the line integral in (5.235) of the vector potential around the curve C. The gauge transformation (2.403) gives the form of the vector potential, and the line integral is just an integral over \u03c6 leading to a factor of 2\u03c0. We can then use this to evaluate the left side of (5.258), including the interaction with an single electric charge e to complete the phase, leading to the factor e/\u00afhc as in (2.404). Thus, (5.258) becomes or e \u00afhc (2eM)2\u03c0 = 2N\u03c0 2eeM \u00afhc = N (5.259) which is the same result (2.405) as we obtained earlier in Section 2.7. 5.7 Time-Dependent Perturbation Theory 5.7.1 Dyson Series With the exception of a few problems like the two-level time-dependent problem of the previous section, exact solutions to the differential equation for cn(t) are usually 10 Our discussion here closely follows that given by Holstein, Am. J. Phys., 57 (1989) 1079. 338 Approximation Methods not available. We must be content with approximate solutions to (5.189) obtained by perturbation expansion: cn(t) =c (0) n + c(1) n + c(2) n + \u00b7 \u00b7 \u00b7, (5.260) n, c(2) where c(1) n,... signify amplitudes of \ufb01rst order, second order, and so on in the strength parameter of the time-dependent potential. The iteration method used to solve this problem is similar to what we did in time-independent perturbation theory. If initially only the state i is populated, we approximate cn on the right-hand side of differential equation (5.189) by c(0) n, integrate the differential equation to obtain c(1) into the right-", "hand side [of (5.189)] again to obtain the differential equation for c(2) n, and so on. This is how Dirac developed timedependent perturbation theory in 1927. n = \u03b4ni (independent of t) and relate it to the time derivative of c(1) n, plugc (1) n Instead of working with cn(t), we propose to look at the time-evolution operator UI(t, t0) in the interaction picture, which we will de\ufb01ne later. We obtain a perturbation expansion for UI(t, t0), and at the very end we relate the matrix elements of UI to cn(t). If we are interested only in solving simple problems in nonrelativistic quantum mechanics, all this might look super\ufb02uous; however, the operator formalism we develop is very powerful because it can immediately be applied to more advanced problems, such as relativistic quantum \ufb01eld theory and many-body theory. The time-evolution operator in the interaction picture is de\ufb01ned by |\u03b1, t0; tI = UI(t, t0)|\u03b1, t0; t0I. (5.261) Differential equation (5.183) for the state ket of the interaction picture is equivalent to i\u00afh d dt UI(t, t0) =V I(t)UI(t, t0). (5.262) We must solve this operator differential equation subject to the initial condition UI(t, t0)|t=t0 = 1. (5.263) First, let us note that the differential equation together with the initial condition is equivalent to the following integral equation: UI(t, t0) =1 \u2212 i \u00afh t t0 VI(t)UI(t, t0)dt. (5.264) We can obtain an approximate solution to this equation by iteration: UI(t, t0) =1 \u2212 i \u00afh =1 \u2212 i \u00afh VI(t) 1 \u2212 i \u00afh dtVI(t) + t t0 t t0 t t0 \u2212i \u00afh VI(t)UI(t, t0)dt 2 t t dt t0 t0 dt dtVI(t)VI(t) 339 5.7 Time-Dependent Perturbation Theory + \u00b7 \u00b7 \u00b7 + n \u2212i \u00afh t t dt t0 t0 dt \u00b7 \u00b7 \u00b7 dt(n)VI(t)VI(t) \u00b7 \u00b7 \u00b7 VI(t(n)) t(n\u22121) \u00d7 t0 + \u00b7 \u00b7 \u00b7. (5.265) This series is known as the Dyson series after Freeman J. Dyson, who applied this method to covariant quantum electrodynamics (QED).11 Setting aside the dif\ufb01cult question of convergence, we can compute UI(t, t0) to any \ufb01nite order of perturbation theory. 5.7.2 Transition Probability Once UI(t, t0) is given, we can predict the time development of any state ket. For example, if the initial state at t = 0 is one of the energy eigenstates of H0, then to obtain the initial state ket at a later time, all we need to do is multiply by UI(t, 0): |i, t0 = 0; tI = UI(t, 0)|i = \u2211 |nn|UI(t, 0)|i. n (5.266) In fact, n|UI(t, 0)|i is nothing more than what we called cn(t) earlier [see (5.185)]. We will say more about this later. We earlier introduced the time-evolution operator U(t, t0) in the Schr\u00a8odinger picture (see Section 2.2). Let us now explore the connection between U(t, t0) and UI(t, t0). We note from (2.87) and (5.177) that |\u03b1, t0; tI = eiH0t/\u00afh|\u03b1, t0; tS = eiH0t/\u00afhU(t, t0)|\u03b1, t0; t0S = eiH0t/\u00afhU(t, t0)e\u2212iH0t0/\u00afh|\u03b1, t0; t0I. UI(t, t0) =e iH0t/\u00afhU(t, t0)e\u2212iH0t0/\u00afh. (5.267) (5.268) So we have Let us now look at the matrix element of UI(t, t0) between energy eigenstates of H0: n|UI(", "t, t0)|i = ei(Ent\u2212Eit0)/\u00afhn|U(t, t0)|i. We recall from Section 2.2 that n|U(t, t0)|i is de\ufb01ned to be the transition amplitude. Hence our n|UI(t, t0)|i here is not quite the same as the transition amplitude de\ufb01ned earlier. However, the transition probability de\ufb01ned as the square of the modulus of n|U(t, t0)|i is the same as the analogous quantity in the interaction picture (5.269) |n|UI(t, t0)|i|2 = |n|U(t, t0)|i|2. (5.270) 11 Note that in QED, the time-ordered product (t > t > \u00b7 \u00b7 \u00b7 ) is introduced, and then this perturbation series can be summed into an exponential form. This exponential form immediately gives U(t, t0) = U(t, t1)U(t1, t0) (Bjorken and Drell (1965), pp. 175\u2013178). 340 Approximation Methods Parenthetically, we may remark that if the matrix elements of UI are taken between initial and \ufb01nal states that are not energy eigenstates, for example, between |a and |b (eigenkets of A and B, respectively), where [H0, A] = 0 and/or [H0, B] = 0, then we have, in general, |b|UI(t, t0)|a| = |b|U(t, t0)|a|, as the reader may easily verify. Fortunately, in problems where the interaction picture is found to be useful, the initial and \ufb01nal states are usually taken to be H0 eigenstates. Otherwise, all that is needed is to expand |a, |b, and so on in terms of the energy eigenkets of H0. Coming back to n|UI(t, t0)|i, we illustrate by considering a physical situation where at t = t0, the system is known to be in state |i. The state ket in the Schr\u00a8odinger picture |i, t0; tS is then equal to |i up to a phase factor. In applying the interaction picture, it is convenient to choose the phase factor at t = t0 so that which means that in the interaction picture we have the simple equation |i, t0; t0S = e\u2212iEit0/\u00afh|i, At a later time we have |i, t0; t0I = |i. |i, t0; tI = UI(t, t0)|i. Comparing this with the expansion we see |i, t0; tI = \u2211 cn(t)|n, n cn(t) =n|U I(t, t0)|i. (5.271) (5.272) (5.273) (5.274) (5.275) We now go back to the perturbation expansion for UI(t, t0) [see (5.265)]. We can also is second order in VI(t), expand cn(t) as in (5.260), where c(1) n and so on. Comparing the expansion of both sides of (5.275), we obtain [using (5.179)] is \ufb01rst order in VI(t), c(2) n c(1) n (t) = c(0) n (t) = \u03b4ni \u2212i \u00afh \u2212i \u00afh = (independent of t) n|VI(t)|idt t t0 Vni(t)dt t ei\u03c9nit t0 \u2212i \u00afh \u2211 2 m c(2) n (t) = where we have used t t dt t0 t0 dtei\u03c9nmt Vnm(t)ei\u03c9mit Vmi(t), ei(En\u2212Ei)t/\u00afh = ei\u03c9nit. The transition probability for |i \u2192 |n with n = i is obtained by n (t) +\u00b7 \u00b7 \u00b7 | 2. P(i \u2192 n) = |c(1) n (t) +c (2) (5.276) (5.277) (5.278) 341 5.7 Time-Dependent Perturbation Theory 5.7.3 Constant Perturbation As an application of (5.276), let us consider a constant perturbation turned on at t = 0: V(t) = 0 V (independent of t) for for t < 0 t \u2265 0. (5.279) Even", " though the operator V has no explicit dependence on time, it is, in general, made up t = 0, we have only |i. With t0 taken to be of operators like x, p, ands. Now suppose at zero, we obtain n = c(0) c(0) n (0) = \u03b4ni, \u2212i \u00afh c(1) n = Vni ei\u03c9nit 0 t dt (5.280) or = Vni En \u2212 Ei (1 \u2212 ei\u03c9nit), |c(1) n |2 = = |Vni|2 |En \u2212 Ei|2 (2 \u2212 2 cos \u03c9nit) 4|Vni|2 (En \u2212 Ei)t |En \u2212 Ei|2 sin2 2\u00afh. (5.281) The probability of \ufb01nding |n depends not only on |Vni|2 but also on the energy difference En \u2212 Ei, so let us try to see how (5.281) looks as a function of En. In practice, we are interested in this way of looking at (5.281) when there are many states with E \u223c En so that we can talk about a continuum of \ufb01nal states with nearly the same energy. To this end we de\ufb01ne \u03c9 \u2261 En \u2212 Ei \u00afh (5.282) and plot 4 sin2(\u03c9t/2)/\u03c92 as a function of \u03c9 for \ufb01xed t, the time interval during which the perturbation has been on; see Figure 5.9. We see that the height of the middle peak, centered around \u03c9 = 0, is t2 and the width is proportional to 1/t. Ast becomes large, |c(1) n (t)|2 is appreciable only for those \ufb01nal states that satisfy t \u223c 2\u03c0 |\u03c9| = 2\u03c0\u00afh |En \u2212 Ei|. (5.283) If we call \u0394t the time interval during which the perturbation has been turned on, a transition with appreciable probability is possible only if \u0394t\u0394E \u223c \u00afh, (5.284) where by \u0394E we mean the energy change involved in a transition with appreciable probability. If \u0394t is small, we have a broader peak in Figure 5.9, and as a result we can tolerate a fair amount of energy nonconservation. On the other hand, if the perturbation has been on for a very long time, we have a very narrow peak, and approximate energy 342 Approximation Methods 4 sin2(\u03c9t/2)/\u03c92 t2 Fig. 5.9 \u22124\u03c0/t \u03c9 = 0 Plot of 4 sin2(\u03c9t/2)/\u03c92 versus \u03c9 for a fixed t, where in \u03c9 = (En \u2212 Ei)/\u00afh we have regarded En as a continuous variable. \u22122\u03c0/t 2\u03c0/t 4\u03c0/t \u03c9 Nearly the same energy + z-direction Fig. 5.10 Elastic scattering of plane wave by some finite range potential. conservation is required for a transition with appreciable probability. Note that this \u201cuncertainty relation\u201d is fundamentally different from the x \u2212 p uncertainty relation of Section 1.6. There x and p are both observables. In contrast, time in nonrelativistic quantum mechanics is a parameter, not an observable. For those transitions with exact energy conservation En = Ei, we have |c(1) n (t)|2 = 1 \u00afh2 |Vni|2t2. (5.285) The probability of \ufb01nding |n after a time interval t is quadratic, not linear, in the time interval during which V has been on. This may appear intuitively unreasonable. There is no cause for alarm, however. In a realistic situation where our formalism is applicable, there is usually a group of \ufb01nal states, all with nearly the same energy as the energy of the initial state |i. In other words, a \ufb01nal state forms a continuous energy spectrum in the neighborhood of Ei. We give two examples along this line. Consider for instance, elastic scattering by some \ufb01nite range potential (see Figure 5.10), which we will consider in detail in Chapter 6. The initial state is taken to be a plane wave state with its propagation direction oriented in the positive z-direction; the \ufb01nal state may also be a plane wave state of the same energy but with its propagation direction, in general, in a direction other than the positive 343 5.7 Time-Dependent Perturbation Theory One of the electrons is in 1s state and the other is in free state (2s)2 (1s)(2s) (1s)2", " 1s Fig. 5.11 Schematic diagram of two electron energy levels of a helium atom. z-direction. Another example of interest is the de-excitation of an excited atomic state via the emission of an Auger electron. The simplest example is a helium atom. The initial state may be (2s)2, where both the electrons are excited; the \ufb01nal state may be (1s) (that is, one of the electrons is still bound) of the He+ ion, while the second electron escapes with a positive energy E; see Figure 5.11. In such a case we are interested in the total probability, that is, the transition probabilities summed over \ufb01nal states with En Ei: \u2211 n,EnEi |c(1) n |2. (5.286) It is customary to de\ufb01ne the density of \ufb01nal states as the number of states within energy interval (E, E + dE) as \u03c1(E) dE. (5.287) We can then write (5.286) as \u2211 n,EnEi |c(1) n |2 \u21d2 dEn\u03c1(En)|c(1) n |2 = 4 sin2 (En \u2212 Ei)t 2\u00afh |Vni|2 |En \u2212 Ei|2 \u03c1(En) dEn. (5.288) As t \u2192 \u221e, we can take advantage of lim t\u2192\u221e 1 |En \u2212 Ei|2 sin2 which follows from (En \u2212 Ei)t 2\u00afh = \u03c0t 2\u00afh \u03b4(En \u2212 Ei), (5.289) lim \u03b1\u2192\u221e 1 \u03c0 sin2 \u03b1x \u03b1x2 = \u03b4(x). (5.290) It is now possible to take the average of |Vni|2 outside the integral sign and perform the integration with the \u03b4-function: lim t\u2192\u221e dEn\u03c1(En)|c(1) n (t)|2 = 2\u03c0 \u00afh |Vni|2\u03c1(En)t. EnEi (5.291) 344 Approximation Methods Thus the total transition probability is proportional to t for large values of t, which is quite reasonable. Notice that this linearity in t is a consequence of the fact that the total transition probability is proportional to the area under the peak of Figure 5.9, where the height varies as t2 and the width varies as 1/t. It is conventional to consider the transition rate, that is, the transition probability per unit time. Expression (5.291) tells us that the total transition rate, de\ufb01ned by d dt \u2211 |c(1) n |2, n (5.292) is constant in t for large t. Calling (5.292) wi\u2192[n], where [n] stands for a group of \ufb01nal states with energy similar to i, we obtain wi\u2192[n] = 2\u03c0 \u00afh |Vni|2\u03c1(En)EnEi (5.293) independent of t, provided the \ufb01rst-order time-dependent perturbation theory is valid. This formula is of great practical importance; it is called Fermi\u2019s golden rule even though the basic formalism of t-dependent perturbation theory is due to Dirac.12 We sometimes write (5.293) as 2\u03c0 \u00afh wi\u2192n = |Vni|2\u03b4(En \u2212 Ei), (5.294) dEn\u03c1(En). where it must be understood that this expression is integrated with We should also understand what is meant by |Vni|2. If the \ufb01nal states |n form a quasicontinuum, the matrix elements Vni are often similar if |n are similar. However, it may happen that all energy eigenstates with the same En do not necessarily have similar matrix elements. Consider, for example, elastic scattering. The |Vni|2 that determines the scattering cross section may depend on the \ufb01nal momentum direction. In such a case the group of \ufb01nal states we should consider must not only have approximately the same energy, but they must also have approximately the same momentum direction. This point becomes clearer when we discuss the photoelectric effect. Let us now look at the second-order term, still with the constant perturbation of (5.279). From (5.276) we have c(2) n = 2 \u2212i \u00afh \u2211 m VnmVmi t 0 dtei\u03c9nmt t 0 dtei\u03c9mit = i \u00afh \u2211 m VnmVmi Em \u2212 Ei t 0 (ei\u03c9nit \u2212 ei\u03c9nmt )dt. (5.", "295) The \ufb01rst term on the right-hand side has the same t-dependence as c(1) [see (5.280)]. If this n were the only term, we could then repeat the same argument as before and conclude that as t \u2192 \u221e, the only important contribution arises from En Ei. Indeed, when Em differs from En and Ei, the second contribution gives rise to a rapid oscillation, which does not give a contribution to the transition probability that grows with t. 12 For an interesting discussion on the history of Fermi\u2019s golden rule, see Visser, \u201cWhose golden rule is it anyway?\u201d, Am. J. Phys., 77 (2009) 487. 345 5.7 Time-Dependent Perturbation Theory With c(1) and c(2) together, we have wi\u2192[n] = 2\u03c0 \u00afh Vni +\u2211 m VnmVmi Ei \u2212 Em 2 \u03c1(En). EnEi (5.296) The formula has the following physical interpretation. We visualize that the transition due to the second-order term takes place in two steps. First, |i makes an energy nonconserving transition to |m; subsequently, |m makes an energy nonconserving transition to |n, where between |n and |i there is overall energy conservation. Such energy nonconserving transitions are often called virtual transitions. Energy need not be conserved for those virtual transitions into (or from) virtual intermediate states. In contrast, the \ufb01rst-order term Vni is often said to represent a direct energy-conserving \u201creal\u201d transition. A special treatment is needed if VnmVmi = 0 with Em Ei. The best way to treat this is to use the slow-turn-on method V \u2192 e\u03b7tV, which we will discuss in Section 5.9 and Problem 5.38 of this chapter. The net result is to change the energy denominator in (5.296) as follows: Ei \u2212 Em \u2192 Ei \u2212 Em + i\u03b5. (5.297) 5.7.4 Harmonic Perturbation We now consider a sinusoidally varying time-dependent potential, commonly referred to as harmonic perturbation: V(t) =V ei\u03c9t + V \u2020e\u2212i\u03c9t, (5.298) where V may still depend on x, p, s, and so on. Actually, we have already encountered a time-dependent potential of this kind in Section 5.5 in discussing t-dependent two-level problems. Again assume that only one of the eigenstates of H0 is populated initially. Perturbation (5.298) is assumed to be turned on at t = 0, so (Vniei\u03c9t + V \u2020 ni e\u2212i\u03c9t )ei\u03c9nit dt t c(1) n = = \u2212i \u00afh 1 \u00afh 0 1 \u2212 ei(\u03c9+\u03c9ni)t \u03c9 + \u03c9ni Vni + 1 \u2212 ei(\u03c9ni\u2212\u03c9)t \u2212\u03c9 + \u03c9ni V \u2020 ni (5.299) ni actually stands for (V \u2020)ni. We see that this formula is similar to the constant where V \u2020 perturbation case. The only change needed is En \u2212 Ei \u00afh \u03c9ni = \u2192 \u03c9ni \u00b1 \u03c9. So as t \u2192 \u221e, |c(1) n |2 is appreciable only if \u03c9ni + \u03c9 0 \u03c9ni \u2212 \u03c9 0 orE n Ei \u2212 \u00afh\u03c9 orE n Ei + \u00afh\u03c9. (5.300) (5.301a) (5.301b) Clearly, whenever the \ufb01rst term is important because of (5.301a), the second term is unimportant, and vice versa. We see that we have no energy-conservation condition 346 Approximation Methods (a) Ei En h \u03c9 (b) En Ei h \u03c9 Fig. 5.12 (a) Stimulated emission: Quantum-mechanical system gives up \u00afh\u03c9 to V (possible only if the initial state is excited). (b) Absorption: Quantum-mechanical system receives \u00afh\u03c9 from V and ends up as an excited state. satis\ufb01ed by the quantum-mechanical system alone; rather the apparent lack of energy conservation is compensated by the energy given out to, or energy taken away from, the \u201cexternal\u201d potential V(t). Pictorially, we have Figure 5.12. In the \ufb01rst case (stimulated emission), the quantum-mechanical system gives up energy \u00afh\u03c9 to V; this is clearly possible only if the", " initial state is excited. In the second case (absorption), the quantum-mechanical system receives energy \u00afh\u03c9 from V and ends up as an excited state. Thus a time-dependent perturbation can be regarded as an inexhaustible source or sink of energy. In complete analogy with (5.293), we have wi\u2192[n] = wi\u2192[n] = 2\u03c0 \u00afh 2\u03c0 \u00afh |Vni|2\u03c1(En) \u223c =Ei\u2212\u00afh\u03c9 En |V \u2020 ni |2\u03c1(En) \u223c =Ei+\u00afh\u03c9 En (5.302a) (5.302b) wi\u2192n = - |Vni|2 |2 |V \u2020 ni 2\u03c0 \u00afh \u03b4(En \u2212 Ei \u00b1 \u00afh\u03c9). (5.303) or, more commonly, Note also that which is a consequence of |Vni|2 = |V \u2020 in |2, i|V \u2020|n = n|V |i\u2217 (remember V \u2020|nDC\u2194n|V ). Combining (5.302) and (5.304), we have emission rate for i \u2192 [n] density of \ufb01nal states for [n] = absorption rate for n \u2192 [i] density of \ufb01nal states for [i], (5.304) (5.305) (5.306) where in the absorption case we let i stand for \ufb01nal states. Equation (5.306), which expresses symmetry between emission and absorption, is known as detailed balancing. To summarize, for constant perturbation, we obtain appreciable transition probability for |i \u2192 |n only if En Ei. In contrast, for harmonic perturbation we have appreciable transition probability only if En Ei \u2212 \u00afh\u03c9 (stimulated emission) or En Ei + \u00afh\u03c9 (absorption). 347 5.8 Applications to Interactions with the Classical Radiation Field 5.8 Applications to Interactions with the Classical Radiation Field 5.8.1 Absorption and Stimulated Emission We apply the formalism of time-dependent perturbation theory to the interactions of atomic electrons with the classical radiation \ufb01eld. By a classical radiation \ufb01eld we mean the electric or magnetic \ufb01eld derivable from a classical (as opposed to quantized) radiation \ufb01eld. For an electron with charge e in an electromagnetic \ufb01eld speci\ufb01ed by a vector potential A(x) and scalar potential \u03c6(x), the basic Hamiltonian, with |A|2 omitted, is which is justi\ufb01ed if H = p2 2me + e\u03c6(x) \u2212 e mec A \u00b7 p, \u2207 \u00b7 A = 0; speci\ufb01cally, we work with a monochromatic \ufb01eld of the plane wave for A = 2A0 \u02c6\u03b5 cos \u02c6n \u00b7 x \u2212 \u03c9t \u03c9 c (5.307) (5.308) (5.309) where \u02c6\u03b5 and \u02c6n are the (linear) polarization and propagation direction. Equation (5.309) obviously satis\ufb01es (5.308) because \u02c6\u03b5 is perpendicular to the propagation direction \u02c6n. We write \u03c9 c cos \u02c6n \u00b7 x \u2212 \u03c9t = 1 2 [ei(\u03c9/c) \u02c6n\u00b7x\u2212i\u03c9t + e\u2212i(\u03c9/c) \u02c6n\u00b7x+i\u03c9t] (5.310) and treat \u2212(e/mec)A \u00b7 p as a time-dependent potential, where we express A in (5.309) as A = A0 \u02c6\u03b5[ei(\u03c9/c) \u02c6n\u00b7x\u2212i\u03c9t + e\u2212i(\u03c9/c) \u02c6n\u00b7x+i\u03c9t]. Comparing this result with (5.298), we see that the e\u2212i\u03c9t-term in (5.311) \u2212 e mec e mec A \u00b7 p = \u2212 A0 \u02c6\u03b5 \u00b7 p[ei(\u03c9/c) \u02c6n\u00b7x\u2212i\u03c9t + e\u2212i(\u03c9/c) \u02c6n \u00b7 x+i\u03c9t] (5.312) is responsible for absorption, while the e+i\u03c9t-term is responsible for stimulated emission. Let us now treat the absorption case in detail. We have ni = \u2212 eA0 V \u2020 mec (ei(\u03c9/c)( \u02c6n\u00b7x) \u02c6\u03b5 \u00b7 p)ni (5.313) and wi\u2192n = 2\u03c0", " \u00afh e2 ec2 m2 |A0|2|n|ei(\u03c9/c)( \u02c6n\u00b7x) \u02c6\u03b5 \u00b7 p|i|2\u03b4(En \u2212 Ei \u2212 \u00afh\u03c9). (5.314) The meaning of the \u03b4-function is clear. If |n forms a continuum, we simply integrate with \u03c1(En). But even if |n is discrete, because |n cannot be a ground state (albeit a bound-state energy level), its energy is not in\ufb01nitely sharp; there may be a natural broadening due to 348 Approximation Methods a \ufb01nite lifetime (see Section 5.9); there can also be a mechanism for broadening due to collisions. In such cases, we regard \u03b4(\u03c9 \u2212 \u03c9ni) as \u03b4(\u03c9 \u2212 \u03c9ni) = lim \u03b3\u21920 \u03b3 2\u03c0 1 ([\u03c9 \u2212 \u03c9ni)2 + \u03b32/4]. (5.315) Finally, the incident electromagnetic wave itself is not perfectly monochromatic; in fact, there is always a \ufb01nite frequency width. We derive an absorption cross section as (Energy/unit time) absorbed by the atom (i \u2192 n) Energy \ufb02ux of the radiation \ufb01eld. (5.316) For the energy \ufb02ux (energy per area per unit time), classical electromagnetic theory gives us cU = 1 2\u03c0 \u03c92 c |A0|2, where we have used E2 8\u03c0 + for energy density (energy per unit volume) with U = 1 2 max B2 max 8\u03c0 E = \u2212 1 c \u2202 \u2202t A, B = \u2207 \u00d7 A. (5.317) (5.318) (5.319) Putting everything together, we get (remembering that \u00afh\u03c9 is the energy absorbed by the atom for each absorption process) \u00afh\u03c9(2\u03c0/\u00afh)(e2/m2 ec2)|A0|2|n|ei(\u03c9/c)( \u02c6n\u00b7x) \u02c6\u03b5 \u00b7 p|i|2\u03b4(En \u2212 Ei \u2212 \u00afh\u03c9) \u03c3abs = (1/2\u03c0)(\u03c92/c)|A0|2 = 4\u03c02 \u00afh \u03c9 m2 e e2 \u00afhc |n|ei(\u03c9/c)( \u02c6n\u00b7x) \u02c6\u03b5 \u00b7 p|i|2\u03b4(En \u2212 Ei \u2212 \u00afh\u03c9). (5.320) Equation (5.320) has the correct dimension [1/(M2/T)](M2L2/T2)T = L2 if we recognize that \u03b1 = e2/\u00afhc 1/137 (dimensionless) and \u03b4(En \u2212 Ei \u2212 \u00afh\u03c9) = (1/\u00afh)\u03b4(\u03c9ni \u2212 \u03c9), where \u03b4(\u03c9ni \u2212 \u03c9) has time dimension T. 5.8.2 Electric Dipole Approximation The electric dipole approximation (E1 approximation) is based on the fact that the wavelength of the radiation \ufb01eld is far longer than the atomic dimension, so that the series (remember \u03c9/c = 1/\u03bb) ei(\u03c9/c) \u02c6n\u00b7x = 1 + i \u03c9 c \u02c6n \u00b7 x + \u00b7 \u00b7 \u00b7 (5.321) can be approximated by its leading term, 1. The validity of this for a light atom is as follows. First, the \u00afh\u03c9 of the radiation \ufb01eld must be of order of atomic level spacing, so \u00afh\u03c9 \u223c Ze2 (a0/Z) Ze2 Ratom. (5.322) 349 5.8 Applications to Interactions with the Classical Radiation Field This leads to In other words, c \u03c9 = \u03bb \u223c c\u00afhRatom Ze2 137Ratom Z. 1 \u03bb Ratom \u223c Z 137 1 (5.323) (5.324) for light atoms (small Z). Because the matrix element of x is of order Ratom, that ofx 2, of order R2 atom, and so on, we see that the approximation of replacing (5.321) by its leading term is an excellent one. Now we have n|ei(\u03c9/c)( \u02c6n\u00b7x) \u02c6\u03b5 \u00b7 p|i \u2192 \u02c6\u03b5 \u00b7 n|p|i. (5.325) In particular, we take \u02c6\u03b5 along the x-axis (and \u02c6n along the z-axis). We must calculate n|px|i. Using we have [x, H0] = i\u00afhpx m, n|px|i =", " m i\u00afh n|[x, H0]|i = im\u03c9nin|x|i. (5.326) (5.327) Because of the approximation of the dipole operator, this approximation scheme is called the electric dipole approximation. We may here recall [see (3.482)] the selection rule for the dipole matrix element. Since x is a spherical tensor of rank 1 with q = \u00b11, we must have m \u2212 m = \u00b11, | j \u2212 j| = 0, 1 (no 0 \u2192 0 transition). If \u02c6\u03b5 is along the y-axis, the same selection rule applies. On the other hand, if \u02c6\u03b5 is in the z-direction, q = 0; hence, m = m. With the electric dipole approximation, the absorption cross section (5.320) now takes a simpler form upon using (5.325) and (5.327) as \u03c3abs = 4\u03c02\u03b1\u03c9ni|n|x|i|2\u03b4(\u03c9 \u2212 \u03c9ni). In other words, \u03c3abs treated as a function of \u03c9 exhibits a sharp \u03b4-function-like peak whenever \u00afh\u03c9 corresponds to the energy-level spacing at \u03c9 (En \u2212 Ei)/\u00afh. Suppose |i is the ground state, then \u03c9ni is necessarily positive; integrating (5.328), we get (5.328) \u03c3abs(\u03c9) d\u03c9 = \u2211 4\u03c02\u03b1\u03c9ni|n|x|i|2. n In atomic physics we de\ufb01ne oscillator strength, fni, as fni \u2261 2m\u03c9ni \u00afh |n|x|i|2. (5.329) (5.330) It is then straightforward (consider [x, [x, H0]]) to establish theThomas\u2013Reiche\u2013Kuhn sum rule, fni = 1. \u2211 n (5.331) 350 Approximation Methods In terms of the integration over the absorption cross section, we have \u03c3abs(\u03c9) d\u03c9 = 4\u03c02\u03b1\u00afh 2me = 2\u03c02c e2 mec2. (5.332) Notice how \u00afh has disappeared. Indeed, this is just the oscillation sum rule already known in classical electrodynamics (Jackson (1975), for instance). Historically, this was one of the \ufb01rst examples of how \u201cnew quantum mechanics\u201d led to the correct classical result. This sum rule is quite remarkable because we did not specify in detail the form of the Hamiltonian. 5.8.3 Photoelectric Effect We now consider the photoelectric effect, that is, the ejection of an electron when an atom is placed in the radiation \ufb01eld. The basic process is considered to be the transition from an atomic (bound) state to a continuum state E > 0. Therefore, |i is the ket for an atomic state, while |n is the ket for a continuum state, which can be taken to be a plane wave state |kf, an approximation that is valid if the \ufb01nal electron is not too slow. Our earlier formula for \u03c3abs(\u03c9) can still be used, except that we must now integrate \u03b4(\u03c9ni \u2212 \u03c9) together with the density of \ufb01nal states \u03c1(En). In fact, we calculated the density of states for a free particle in Section 2.5.1. To review, our basic task is to calculate the number of \ufb01nal states per unit energy interval. As we will see in a moment, this is an example where the matrix element depends not only on the \ufb01nal state energy but also on the momentum direction. We must therefore consider a group of \ufb01nal states with both similar momentum directions and similar energies. To count the number of states it is convenient to use the box normalization convention for plane wave states. We consider a plane wave state to be normalized if, when we integrate the square modulus of its wave function for a cubic box of side L, we obtain unity. Furthermore, the state is assumed to satisfy the periodic boundary condition with periodicity of the side of the box. The wave function must then be of form x|kf = eikf \u00b7 x L3/2, (5.333) where the allowed values of kx must satisfy 2\u03c0nx L with nx a positive or negative integer. Similar restrictions hold for ky and kz. Notice that as L \u2192 \u221e, kx, ky, and kz become continuous variables. (5.334) kx =,..., The problem of counting the number of states is reduced to that of counting the number of dots in three-dimensional lattice space. We de\ufb01ne n such that x +", " n2 As L \u2192 \u221e, it is a good approximation to treat n as a continuous variable; in fact it is just the magnitude of the radial vector in the lattice space. Let us consider a small-volume element such that the radial vector falls within n and n + dn and the solid angle element d\u03a9; clearly, y + n2 z. n2 = n2 (5.335) 351 5.8 Applications to Interactions with the Classical Radiation Field it is of volume n2 dn d\u03a9. The energy of the \ufb01nal state plane wave is related to kf and hence to n; we have E = \u00afh2k2 f 2me = \u00afh2 2me n2(2\u03c0)2 L2. (5.336) Furthermore, the direction of the radial vector in the lattice space is just the momentum direction of the \ufb01nal state, so the number of states in the interval between E and E + dE with direction into d\u03a9 being kf is (remember dE = (\u00afh2kf /me)dkf) given by13 n2 d\u03a9 dn dE dE = (k2 f ) dkf dE d\u03a9 dE 3 L 2\u03c0 L 2\u03c0 = 3 me \u00afh2 kf dE d\u03a9. (5.337) We can now put everything together to obtain an expression for the differential cross section for the photoelectric effect: 4\u03c02\u03b1\u00afh \u03c9 m2 e |kf|ei(\u03c9/c)( \u02c6n\u00b7x) \u02c6\u03b5 \u00b7 p|i|2 mekfL3 \u00afh2(2\u03c0)3 d\u03c3 d\u03a9 = (5.338). To be speci\ufb01c, let us consider the ejection of a K shell (the innermost shell) electron caused by absorption of light. The initial state wave function is essentially the same as the ground-state hydrogen atom wave function except that the Bohr radius a0 is replaced by a0/Z. Thus kf|ei(\u03c9/c)( \u02c6n\u00b7x) \u02c6\u03b5 \u00b7 p|i = \u02c6\u03b5 \u00b7 d3x e\u2212ikf\u00b7x L3/2 ei(\u03c9/c)( \u02c6n\u00b7x) 1\u221a \u03c0 e\u2212Zr/a0 \u00d7 (\u2212i\u00afh\u2207) 3/2 Z a0. (5.339) Integrating by parts, we can pass \u2207 to the left side. Furthermore, \u02c6\u03b5 \u00b7 [\u2207ei(\u03c9/c)( \u02c6n\u00b7x)] =0 (5.340) because \u02c6\u03b5 is perpendicular to \u02c6n. On the other hand, \u2207 acting on e\u2212ikf \u00b7 x brings down \u2212ikf, which can be taken outside the integral. Thus to evaluate (5.339), all we need to do is take the Fourier transform of the atomic wave function with respect to q \u2261 kf \u2212 \u03c9 c \u02c6n. (5.341) The \ufb01nal answer is (see Problem 5.48 of this chapter for the Fourier transform of the hydrogen atom wave function) d\u03c3 d\u03a9 = 32e2kf (\u02c6\u03b5 \u00b7 kf)2 mec\u03c9 Z5 a5 0 1 [(Z2/a2 0) +q 2]4. (5.342) 13 This is equivalent to taking one state per cube d3x d3p/(2\u03c0\u00afh)3 in phase space. 352 Approximation Methods k f \u03b8 z n\u02c6 0 \u03c6 y e \u02c6 x Fig. 5.13 Polar coordinate system with \u02c6\u03b5 and \u02c6n along the x- andz -axes, respectively, and kf = (kf sin \u03b8 cos \u03c6, kf sin \u03b8 sin \u03c6, kf cos \u03b8). If we introduce the coordinate system shown in Figure 5.13, we can write the differential cross section in terms of \u03b8 and \u03c6 using (\u02c6\u03b5 \u00b7 kf)2 = k2 f sin2 \u03b8cos2 \u03c6 \u03c9 q2 = k2 f \u2212 2kf cos \u03b8+ c 2. \u03c9 c (5.343) 5.8.4 Spontaneous Emission An atom in an excited state will spontaneously emit a photon, making a transition to a lower energy state. We now understand the underlying source of these photons to be electromagnetic energy in the so-called vacuum, but it is possible to accurately calculate this phenomenon using the classical electromagnetic \ufb01eld and the techniques discussed in this section. Most of the formalism has already been worked out. The Hamiltonian H = H0 + V(", "t) e/2me + e\u03c6(x) describes the atomic system. We is taken to be (5.307) where H0 = p2 are considering transitions between eigenstates of H0, and the perturbation V(t) = \u2212(e/mec)A \u00b7 p where A(x, t) is a plane wave. Thus we once again have a harmonic perturbation, and the transition rate for emission is given by (5.302a). As in (5.309) we write A(x, t) as A = 2A0 \u02c6\u03b5 cos (k \u00b7 x \u2212 \u03c9t) (5.344) where k \u2261 (\u03c9/c) \u02c6n and \u02c6\u03b5 is the polarization unit vector. We contain this \ufb01eld inside a \u201cbig box\u201d of side length L, large enough to include the frequency \u03c9 that corresponds to the 353 5.8 Applications to Interactions with the Classical Radiation Field atomic transition, applying periodic boundary conditions, similar to the way we describe the ejected photoelectron in (5.333). This means that the wave vector k can be written as k = 2\u03c0 L (nx \u02c6x + ny \u02c6y + nz \u02c6z) (5.345) where nx, ny, and nz are integers. In order to determine the amplitude A0, we require that the time-averaged electromagnetic energy in the box equal the transition energy \u00afh\u03c9. From Maxwell\u2019s equations in free space, the electric and magnetic \ufb01elds are \u2202A \u2202 = 2A0(\u02c6\u03b5 \u00d7 k) sin (k \u00b7 x \u2212 \u03c9t). = 2A0k\u02c6\u03b5 sin (k \u00b7 x \u2212 \u03c9t) The energy density is u = (E2 + B2)/8\u03c0, so the time-averaged energy is L3 d3x u = \u03c92 2\u03c0c2 A2 0L3 = \u00afh\u03c9. Therefore, the amplitude of the classical14 electromagnetic wave is \u2019 2\u03c0\u00afhc2 \u03c9L3. A0 = (5.346a) (5.346b) (5.347) (5.348) Referring back to (5.302a), we need to \ufb01nd the density of states \u03c1(En) = \u03c1(E ) where y + E = \u00afh\u03c9 is the energy of the emitted radiation. Writing E = 2\u03c0\u00afhcn/L where n \u2261 (n2 z )1/2, we integrate over all directions in n-space while keeping n constant to \ufb01nd n2 \u03c9L 2\u03c0c \u03c92 2\u03c02 \u00afhc3 L3dE. \u03c1(E )dE = 4\u03c0n2dn = 4\u03c0 2 L 2\u03c0\u00afhc x + n2 (5.349) dE = We also need to calculate the transition matrix element Vni = \u2212 eA0 mec n|\u02c6\u03b5 \u00b7 e\u2212ik\u00b7xp|i. (5.350) The wavelength of emitted radiation in atomic transitions is on the order of 100 nm or longer, three orders of magnitude larger than a typical atom. Therefore the long wavelength approximation (5.321) is valid. In fact, let us go further, and assume the transition is between states that satisfy the electric dipole transition rules (see p. 349). Therefore Vni = \u2212 eA0 mec n|\u02c6\u03b5 \u00b7 p|i = \u2212i e\u03c9A0 c n|\u02c6\u03b5 \u00b7 x|i. (5.351) The rate w of a spontaneous transition is usually expressed terms of its inverse \u03c4, called the \u201clifetime\u201d of the initial state. Then (5.302a) becomes 2 1 \u03c4i\u2192n = 2\u03c0 \u00afh e\u03c9A0 c |n|\u02c6\u03b5 \u00b7 x|i|2 \u03c92 2\u03c02 \u00afhc3 L3 = 2\u03b1 \u03c93 c2 |n|\u02c6\u03b5 \u00b7 x|i|2 (5.352) 14 See Section 7.8 for a quantum-mechanical treatment of the free electromagnetic \ufb01eld. Note that (5.348) is the same result as (7.180). 354 Approximation Methods where again we make use of \u03b1 \u2261 e2/\u00afhc \u2248 1/137. Note that this result is free of the box dimension L. Given initial and \ufb01nal state atomic wave functions, we could calculate the rate, perhaps numerically for a complicated system. The rate will depend on the polarization direction \u02c6\u03b5 if the atom itself is somehow polarized. It is useful to write the dipole operator in terms of circular and linear", " polarization directions as \u03b5x \u2212 i\u03b5y\u221a 2 x + iy\u221a 2 + \u03b5x + i\u03b5y\u221a 2 x \u2212 iy\u221a 2 + \u03b5zz % 4\u03c0 3 \u2212\u03b5\u2212Y+1 1 r + \u03b5\u2212Y\u22121 1 r + \u03b50Y0 1r & (5.353) \u02c6\u03b5 \u00b7 x = = \u2019 \u221a where \u03b5\u00b1 \u2261 (\u03b5x \u2212 i\u03b5y) matrix elements are subject to the Wigner\u2013Eckart theorem (3.474). 2 and \u03b50 \u2261 \u03b5z. Then recall that the Ym l are spherical tensors, whose Let us illustrate using a p \u2192 s transition, where the initial state is unpolarized. That is, the initial atomic p-state is an equal mixture of the m = 0, \u00b11 states. Now, from (3.471), the matrix element s|Y q |p, m = 0 unless m = \u2212q. Therefore, each of the three terms in 1 (5.353) gives a nonzero transition matrix element for exactly one m value for the initial p-state. Furthermore, by (3.474), \u03b1; s|Y q 1 r|\u03b1; p, m = 11; \u2212q, q|11; 00 \u03b1; s||Y1r||\u03b1; p \u221a 3 (5.354) where \u03b1 and \u03b1 represent other information needed to specify the state (perhaps a radial quantum number). The reduced matrix element \u03b1; s||Y1r||\u03b1; p is the same for all three terms in (5.353), so we just need to calculate the matrix element \u03b1; s|Yq 1r|\u03b1; p, m for one of them. Choosing the m = 0 term for this calculation, we \ufb01nd where and so \u03b1; s|Y 0 1 r|\u03b1; p, m = 0 = 1\u221a 4\u03c0 R\u03b1 \u03b1 R\u03b1 \u03b1 \u2261 \u221e 0 r2dr R\u2217 \u03b1 (r)rR\u03b1(r) \u03b1; s||Y1r||\u03b1; p = \u2212 3\u221a 4\u03c0 R\u03b1 \u03b1. (5.355) (5.356) (5.357) We can then write down the matrix element of (5.353) as \u03b1; s|\u02c6\u03b5 \u00b7 x|\u03b1; p, m = \u2212 1\u221a 3 (\u2212\u03b5\u2212\u03b4m,\u22121 + \u03b5\u2212\u03b4m,1 \u2212 \u03b50\u03b4m,0) R\u03b1 \u03b1. (5.358) We do not know the value of m for the initial state. However, (5.358) shows that each of the three m states gives the same result in (5.352). The initial state is unpolarized, and we are integrating over all directions of the emitted radiation, so we average over the contributions from the three initial states. Since |\u03b5+|2 + |\u03b5\u2212|2 + |\u03b50|2 = 1, this average 355 5.9 Energy Shift and Decay Width introduces an additional factor of 1/3. Realizing that we must also multiply (5.352) by two to account for the two orthogonal polarization directions, we have, \ufb01nally, 1 \u03c4p\u2192s = \u03b1 4 9 \u03c93 c2 |R\u03b1 \u03b1|2. (5.359) Before concluding, we note that if the electric dipole selection rules are not satis\ufb01ed by the initial and \ufb01nal atomic states, we can still use the long wavelength approximation to evaluate (5.350). However, the matrix element will contain higher powers of x and is therefore smaller by additional powers of atomic size divided by the wavelength. That is, the lifetimes will be much longer, so if there are \ufb01nal states available which do satisfy the selection rules, these will dominate. 5.9 Energy Shift and Decay Width Our considerations so far have been restricted to the question of how states other than the initial state become populated. In other words, we have been concerned with the time development of the coef\ufb01cient cn(t) with n = i. The question naturally arises, What happens to ci(t) itself? To avoid the effect of a sudden change in the Hamiltonian, we propose to increase the perturbation very slowly. In the remote past (t \u2192 \u2212\u221e) the time-dependent potential is assumed to be zero. We then gradually turn on the perturbation to its full value; speci\ufb01cally, V(t) =e \u03b7tV (5.360) where V is assumed to be constant and \u03b7 is small and positive. At the end of the calculation, we let \u03b7 \u2192 0 (see Figure 5.14), and the potential", " then becomes constant at all times. In the remote past, we take this time to be \u2212\u221e, so the state ket in the interaction picture is assumed to be |i. Our basic aim is to evaluate ci(t). However, before we do that, let us make sure that the old formula of the golden rule (see Section 5.7) can be reproduced using this slow-turn-on method. For cn(t) with n = i, we have [using (5.276)] V(t) \u03b7 > 0 V(t) = V as \u03b7 \u2192 0 t Fig. 5.14 Plot of V(t) versust in the adiabatic (slow-turn-on) picture. 356 Approximation Methods c(0) n (t) =0 c(1) n (t) = t Vni lim t0\u2192\u2212\u221e t0 e\u03b7t+i\u03c9nit \u03b7 + i\u03c9ni Vni e\u03b7t ei\u03c9nit dt. \u2212i \u00afh \u2212i \u00afh = To lowest nonvanishing order, the transition probability is therefore given by or |cn(t)|2 |Vni|2 \u00afh2 e2\u03b7t \u03b72 + \u03c92 ni, d dt |cn(t)|2 2|Vni|2 \u00afh2. \u03b7e2\u03b7t \u03b72 + \u03c92 ni (5.361) (5.362) (5.363) We now let \u03b7 \u2192 0. Clearly, it is all right to replace e\u03b7t by unity, but note \u03b7 \u03b72 + \u03c92 ni lim \u03b7\u21920 = \u03c0\u03b4(\u03c9ni) = \u03c0\u00afh\u03b4(En \u2212 Ei). (5.364) This leads to the golden rule, wi\u2192n 2\u03c0 \u00afh |Vni|2\u03b4(En \u2212 Ei). (5.365) Encouraged by this result, let us calculate c(0) i, c(1) i, and c(2) i, again using (5.276). We have c(0) i = 1 c(1) i = \u2212i \u00afh c(2) i = = \u2212i \u00afh \u2212i \u00afh t t0 e\u03b7t dt = Vii lim t0\u2192\u2212\u221e 2 2 \u2211 m |Vmi|2 lim t0\u2192\u2212\u221e |Vii|2 e2\u03b7t 2\u03b72 + t \u2212i \u00afh\u03b7 Viie\u03b7t dtei\u03c9imt+\u03b7t ei\u03c9mit+\u03b7t i(\u03c9mi \u2212 i\u03b7) |Vmi|2e2\u03b7t 2\u03b7(Ei \u2212 Em + i\u00afh\u03b7) \u2211 m=i t0 \u2212i \u00afh (5.366). Thus up to second order we have ci(t) 1 \u2212 i \u00afh\u03b7 Viie\u03b7t + 2 \u2212i \u00afh |Vii|2 e2\u03b7t 2\u03b72 + \u2212i \u00afh |Vmi|2e2\u03b7t 2\u03b7(Ei \u2212 Em + i\u00afh\u03b7). \u2211 m=i (5.367) Now consider the time derivative of ci[dci(t)/dt \u2261 \u02d9ci], which we have from (5.367). Upon dividing by ci and letting \u03b7 \u2192 0 (thus replacing e\u03b7t and e2\u03b7t by unity), we get \u2212i \u00afh Vii + \u2212i \u00afh \u02d9ci ci 2 |Vii|2 \u03b7 + \u2212i \u00afh |Vmi|2 (Ei \u2212 Em + i\u00afh\u03b7) \u2211 m=i 1 \u2212 i \u00afh Vii \u03b7 \u2212i \u00afh Vii + \u2212i \u00afh |Vmi|2 Ei \u2212 Em + i\u00afh\u03b7. \u2211 m=i (5.368) 357 5.9 Energy Shift and Decay Width Expansion (5.368) is formally correct up to second order in V. Note here that \u02d9ci(t)/ci(t) is now independent of t. Equation (5.368) is a differential equation that is to hold at all times. Having obtained this, it is convenient to renormalize ci so that ci(0) =1. We now try the ansatz ci(t) = e\u2212i\u0394it/\u00afh, \u02d9ci(t) ci(t) = \u2212i \u00afh \u0394i (5.369) with \u0394i constant (in time) but not necessarily real. Clearly (5.369) is consistent with (5.368) because the", " right-hand side of (5.369) is constant. We can see the physical meaning of \u0394i by noting that e\u2212i\u0394it/\u00afh|i in the interaction picture implies e\u2212i\u0394it/\u00afh\u2212iEit/\u00afh|i in the Schr\u00a8odinger picture. In other words, Ei \u2192 Ei + \u0394i (5.370) as a result of perturbation. That is, we have calculated the level shift using time-dependent perturbation theory. Now expand, as usual, \u0394i = \u0394(1) (5.371) i + \u0394(2) i + \u00b7 \u00b7 \u00b7, and compare (5.369) with (5.368); we get to \ufb01rst order \u0394(1) i = Vii. (5.372) But this is just what we expect from t-independent perturbation theory. Before we look at \u0394(2) i, recall 1 x + i\u03b5 = Pr 1 x lim \u03b5\u21920 \u2212 i\u03c0\u03b4(x). Thus |Vmi|2 Ei \u2212 Em Re(\u0394(2) Im(\u0394(2) i ) =Pr \u2211 m=i i ) =\u2212 \u03c0\u2211 m=i |Vmi|2\u03b4(Ei \u2212 Em). (5.373) (5.374a) (5.374b) But the right-hand side of (5.374b) is familiar from the golden rule, so we can identify wi\u2192m = \u2211 m=i 2\u03c0 \u00afh |Vmi|2\u03b4(Ei \u2212 Em) =\u2212 2 \u00afh \u2211 m=i Im[\u0394(2) i ]. Coming back to ci(t), we can write (5.369) as ci(t) =e \u2212(i/\u00afh)[Re(\u0394i)t]+(1/\u00afh)[Im(\u0394i)t]. If we de\ufb01ne then \u0393i \u00afh \u2261 \u2212 2 \u00afh Im(\u0394i), |ci|2 = e2Im(\u0394i)t/\u00afh = e\u2212\u0393it/\u00afh. Therefore, \u0393i characterizes the rate at which state |i disappears. (5.375) (5.376) (5.377) (5.378) 358 Approximation Methods It is worth checking the probability conservation up to second order in V for small t: |cm|2 = (1 \u2212 \u0393it/\u00afh) + \u2211 m=i |ci|2 + \u2211 m=i wi\u2192mt = 1, (5.379) where (5.375) has been used. Thus the probabilities for \ufb01nding the initial state and all other states add up to 1. Put in another way, the depletion of state |i is compensated by the growth of states other than |i. To summarize, the real part of the energy shift is what we usually associate with the level shift. The imaginary part of the energy shift is, apart from \u22122 [see (5.377)], the decay width. Note also \u00afh \u0393i = \u03c4i where \u03c4i is the mean lifetime of state |i because |ci|2 = e\u2212t/\u03c4i. To see why \u0393i is called width, we look at the Fourier decomposition f (E)e\u2212iEt/\u00afhdE = e\u2212i[Ei+Re(\u0394i)]t/\u00afh\u2212\u0393it/2\u00afh. Using the Fourier inversion formula, we get | f (E)|2 \u221d 1 {E \u2212 [Ei + Re(\u0394i)]}2 + \u03932 i /4. (5.380) (5.381) (5.382) (5.383) Therefore, \u0393i has the usual meaning of full width at half maximum. Notice that we get the energy-time uncertainty relation from (5.380) \u0394t\u0394E \u223c \u00afh, (5.384) where we identify the uncertainty in the energy with \u0393i and the mean lifetime with \u0394t. Even though we discussed the subject of energy shift and decay width using the constant perturbation V obtained as the limit of (5.360) when \u03b7 \u2192 0, we can easily generalize our considerations to the harmonic perturbation case discussed in Section 5.7. All we must do is to let En(m) \u2212 Ei \u2192 En(m) \u2212 Ei \u00b1 \u00afh\u03c9 (5.385) in (5.361), (5.367), and (5.374), and so on. The quantum-mechanical description of unstable states we have developed here is originally due to Wigner and Weissk", "opf in 1930. Problems 5.1 A simple harmonic oscillator (in one dimension) is subjected to a perturbation H1 = bx where b is a real constant. a. Calculate the energy shift of the ground state to lowest nonvanishing order. b. Solve this problem exactly and compare with your result obtained in (a). 359 Problems 5.2 A one-dimensional potential well has in\ufb01nite walls at x = 0 and x = L. The bottom of the well is not \ufb02at, but rather increases linearly from 0 at x = 0 to V at x = L. Find the \ufb01rst-order shift in the energy levels as a function of principal quantum number n. 5.3 A particle of mass m moves in a potential well V(x) =m \u03c92x2/2. Treating relativistic effects to order \u03b22 = (p/mc)2, \ufb01nd the ground-state energy shift. 5.4 A diatomic molecule can be modeled as a rigid rotor with moment of inertia I and an electric dipole moment d along the axis of the rotor. The rotor is constrained to rotate in a plane, and a weak uniform electric \ufb01eld E lies in the plane. Write the classical Hamiltonian for the rotor, and \ufb01nd the unperturbed energy levels by quantizing the angular-momentum operator. Then treat the electric \ufb01eld as a perturbation, and \ufb01nd the \ufb01rst nonvanishing corrections to the energy levels. 5.5 In nondegenerate time-independent perturbation theory, what is the probability of \ufb01nding in a perturbed energy eigenstate (|k) the corresponding unperturbed eigenstate (|k(0))? Solve this up to terms of order \u03bb2. 5.6 Consider a particle in a two-dimensional potential V0 = for 0 \u2264 x \u2264 L, 0 \u2264 y \u2264 L 0 \u221e otherwise. Write the energy eigenfunctions for the ground and \ufb01rst excited states. We now add a time-independent perturbation of the form V1 = \u03bbxy 0 for 0 \u2264 x \u2264 L, 0 \u2264 y \u2264 L otherwise. Obtain the zeroth-order energy eigenfunctions and the \ufb01rst-order energy shifts for the ground and \ufb01rst excited states. 5.7 Consider an isotropic harmonic oscillator in two dimensions. The Hamiltonian is H0 = p2 x 2m + p2 y 2m + m\u03c92 2 (x2 + y2). a. What are the energies of the three lowest-lying states? Is there any degeneracy? b. We now apply a perturbation V = \u03b4m\u03c92xy, where \u03b4 is a dimensionless real number much smaller than unity. Find the zerothorder energy eigenket and the corresponding energy to \ufb01rst order [that is, the unperturbed energy obtained in (a) plus the \ufb01rst-order energy shift] for each of the three lowest-lying states. c. Solve the H0 + V problem exactly. Compare with the perturbation results obtained in (b). 360 Approximation Methods 5.8 Establish (5.54) for the one-dimensional harmonic oscillator given by (5.50) with \u03b5m\u03c92x2. Show that all other matrix elements Vk0 an additional perturbation V = 1 2 vanish. 5.9 A slightly anisotropic three-dimensional harmonic oscillator has \u03c9x = \u03c9y \u2261 \u03c9 and \u03c9z = (1 + \u03b5)\u03c9 where \u03b5 1. (See Section 3.7.3 for nomenclature and wave functions.) A charged particle moves in the \ufb01eld of this oscillator and is at the same time exposed to a uniform magnetic \ufb01eld in the x-direction. Assuming that the Zeeman splitting is comparable to the splitting produced by the anisotropy, but small compared to \u00afh\u03c9, calculate to \ufb01rst order the energies of the components of the \ufb01rst excited state. Discuss various limiting cases. (This is taken from Problem 17.7 in Merzbacher (1970). You might \ufb01nd it useful to consult Problem 2.16 in Chapter 2 and Problem 3.29 in Chapter 3.) 5.10 A one-electron atom whose ground state is nondegenerate is placed in a uniform electric \ufb01eld in the z-direction. Obtain an approximate expression for the induced electric dipole moment of the ground state by considering the expectation value of ez with respect to the perturbed state vector computed to \ufb01r", "st order. Show that the same expression can also be obtained from the energy shift \u0394 = \u2212\u03b1|E|2/2 of the ground state computed to second order. (Note: \u03b1 stands for the polarizability.) Ignore spin. 5.11 Evaluate the matrix elements (or expectation values) given below. If any vanishes, explain why it vanishes using simple symmetry (or other) arguments. a. n = 2, l = 1, m = 0|x|n = 2, l = 0, m = 0. b. n = 2, l = 1, m = 0|pz|n = 2, l = 0, m = 0. [In (a) and (b), |nlm stands for the energy eigenket of a nonrelativistic hydrogen atom with spin ignored.] c. Lz for an electron in a central \ufb01eld with j = 9 d. singlet, ms = 0|S(e\u2212) z \u2212 S(e+) z e. S(1) \u00b7 S(2) for the ground state of a hydrogen molecule. 2, m = 7 |triplet, ms = 0 for an s-state positronium. 2, l = 4. 5.12 A p-orbital electron characterized by |n, l = 1, m = \u00b11, 0 (ignore spin) is subjected to a potential V = \u03bb(x2 \u2212 y2) (\u03bb = constant). a. Obtain the \u201ccorrect\u201d zeroth-order energy eigenstates that diagonalize the perturbation. You need not evaluate the energy shifts in detail, but show that the original threefold degeneracy is now completely removed. b. Because V is invariant under time reversal and because there is no longer any degeneracy, we expect each of the energy eigenstates obtained in (a) to go into itself (up to a phase factor or sign) under time reversal. Check this point explicitly. 5.13 Consider a spinless particle in a two-dimensional in\ufb01nite square well: V = for 0 \u2264 x \u2264 a, 0 \u2264 y \u2264 a 0 \u221e otherwise. 361 Problems a. What are the energy eigenvalues for the three lowest states? Is there any degeneracy? b. We now add a potential V1 = \u03bbxy, 0 \u2264 x \u2264 a, 0 \u2264 y \u2264 a. Taking this as a weak perturbation, answer the following. (i) Is the energy shift due to the perturbation linear or quadratic in \u03bb for each of the three states? (ii) Obtain expressions for the energy shifts of the three lowest states accurate to order \u03bb. (You need not evaluate integrals that may appear.) (iii) Draw an energy diagram with and without the perturbation for the three energy states. Make sure to specify which unperturbed state is connected to which perturbed state. 5.14 The Hamiltonian matrix for a two-state system can be written as H =. \u03bb\u0394 E0 1 \u03bb\u0394 E0 2 Clearly the energy eigenfunctions for the unperturbed problems (\u03bb = 0) are given by \u03c6(0) 1 =, \u03c6(0) 2 = 1 0 0 1. a. Solve this problem exactly to \ufb01nd the energy eigenfunctions \u03c81 and \u03c82 and the energy eigenvalues E1 and E2. \u2212 E0 2 b. Assuming that \u03bb|\u0394| |E0 1 |, solve the same problem using time-independent perturbation theory up to \ufb01rst order in the energy eigenfunctions and up to second order in the energy eigenvalues. Compare with the exact results obtained in (a). c. Suppose the two unperturbed energies are \u201calmost degenerate,\u201d that is, |E0 1 \u2212 E0 2 | \u03bb|\u0394|. Show that the exact results obtained in (a) closely resemble what you would expect by applying degenerate perturbation theory to this problem with E0 1 set exactly equal to E0 2. 5.15 (This is a tricky problem because the degeneracy between the \ufb01rst and the second state is not removed in \ufb01rst order. See also Gottfried (1966), p. 397, Problem 1.) This problem is from Schiff (1968), p. 295, Problem 4. A system that has three unperturbed states can be represented by the perturbed Hamiltonian matrix \u239b \u239d E1 0 a\u2217 a 0 b E1 b\u2217 E2 \u239e \u23a0 where E2 > E1. The quantities a and b are to be regarded as perturbations that are of the same order and are small compared with E2", " \u2212 E1. Use the second-order nondegenerate perturbation theory to calculate the perturbed eigenvalues. (Is this 362 Approximation Methods procedure correct?) Then diagonalize the matrix to \ufb01nd the exact eigenvalues. Finally, use the second-order degenerate perturbation theory. Compare the three results obtained. 5.16 Use perturbation theory to calculate the effect of the proton\u2019s \ufb01nite size on the n = 1 and n = 2 energy levels of the hydrogen atom. Assume the proton is a uniformly charged sphere of radius R. Give a physical explanation for why the = 1 shifts are so much smaller than for = 0. 5.17 This chapter derived two of the three relativistic corrections to the one-electron atom, namely \u0394(1) K from \u201crelativistic kinetic energy,\u201d and \u0394(1) LS from the spin-orbit interaction. A third term comes from the spread of the electron wave function in the region of changing electric \ufb01eld. The perturbation for this \u201cDarwin term\u201d is VD = \u2212 1 8m2c2 3 \u2211 [ pi, [ pi, e\u03c6(r)]] i=1 where \u03c6(r) is the Coulomb potential. Find \u0394(1) \u0394(1) nj \u2261 \u0394(1) K + \u0394(1) LS + \u0394(1) D = D and show that mc2(Z\u03b1)4 2n3 3 4n \u2212 1 j + 1/2. In Section 8.4 we will compare this expression to the result of solving the Dirac equation in the presence of the Coulomb potential. 5.18 These questions are meant to associate numbers with atomic hydrogen phenomena. a. The red n = 3 \u2192 2 Balmer transition has a wavelength \u03bb \u2248 656 nm. Calculate the wavelength difference \u0394\u03bb (in nm) between the 3p3/2 \u2192 2s1/2 and 3p1/2 \u2192 2s1/2 transitions due to the spin-orbit interaction. Comment on how you might measure this splitting. b. How large an electric \ufb01eld E is needed so that the Stark splitting in the n = 2 level is the same as the correction from relativistic kinetic energy between the 2s and 2p levels? How easy or dif\ufb01cult is it to achieve an electric \ufb01eld of this magnitude in the laboratory? c. The Zeeman effect can be calculated with a \u201cweak\u201d or \u201cstrong\u201d magnetic \ufb01eld, depending on the size of the energy shift relative to the spin-orbit splitting. Give examples of a weak and a strong \ufb01eld. How easy or dif\ufb01cult is it to achieve such a magnetic \ufb01eld? 5.19 Compute the Stark effect for the 2s1/2 and 2p1/2 levels of hydrogen for a \ufb01eld E suf\ufb01ciently weak so that eE a0 is small compared to the \ufb01ne structure, but take the Lamb shift \u03b4 (\u03b4 = 1057 MHz) into account (that is, ignore 2p3/2 in this calculation). Show that for eE a0 \u03b4, the energy shifts are quadratic in E, whereas for eE a0 \u03b4 they are linear in E. Brie\ufb02y discuss the consequences (if any) of time reversal for this problem. This problem is from Gottfried (1966), Problem 7-3. 5.20 Work out the Stark effect to lowest nonvanishing order for the n = 3 level of the hydrogen atom. Ignoring the spin-orbit force and relativistic correction (Lamb 363 Problems shift), obtain not only the energy shifts to lowest nonvanishing order but also the corresponding zeroth-order eigenket. el the spin magnetic moment (that is, \u03bc \u2212\u03bc 5.21 Suppose the electron had a very small intrinsic electric dipole moment analogous to el proportional to \u03c3). Treating the hypothetical \u00b7 E interaction as a small perturbation, discuss qualitatively how the energy levels of the Na atom (Z = 11) would be altered in the absence of any external electromagnetic \ufb01eld. Are the level shifts \ufb01rst order or second order? State explicitly which states get mixed with each other. Obtain an expression for the energy shift of the lowest level that is affected by the perturbation. Assume throughout that only the valence electron is subjected to the hypothetical interaction. 5.22 Consider a particle bound to a \ufb01xed center by a spherically symmetric potential V(r). a. Prove |\u03c8(0)|2 = ( ) dV dr m 2\u03c0", "\u00afh2 for all s states, ground and excited. b. Check this relation for the ground state of a three-dimensional isotropic oscillator, the hydrogen atom, and so on. (Note: This relation has actually been found to be useful in guessing the form of the potential between a quark and an antiquark. See Moxhay and Rosner, J. Math. Phys., 21 (1980) 1688.) 5.23 a. Suppose the Hamiltonian of a rigid rotator in a magnetic \ufb01eld perpendicular to the axis is of the form (Merzbacher 1970, Problem 17-1) AL2 + BLz + CLy if terms quadratic in the \ufb01eld are neglected. Assuming B C, use perturbation theory to lowest nonvanishing order to get approximate energy eigenvalues. b. Consider the matrix elements nlm lm s nlm |(3z2 \u2212 r2)|nlmlms, lm |xy|nlmlms s of a one-electron (for example, alkali) atom. Write the selection rules for \u0394l, \u0394ml, and \u0394ms. Justify your answer. 5.24 The n = 2 state of hydrogen is eightfold degenerate, accounting for both spin and orbital angular momentum. This degeneracy is broken by a perturbation V = A \u00afh2 L \u00b7 S + B \u00afh (Lz + 2Sz) where L and S are the orbital and spin angular-momentum operators, A is a constant, and B is proportional (but not equal) to an applied magnetic \ufb01eld in the z-direction. a. Write V in terms of J2, L2, S2, Jz, andS z, whereJ = L + S. b. Find all nonzero matrix elements of V in the basis |l, s = 1/2, j = l \u00b1 1/2, m for the eight n = 2 states. Hint: Show that the 8 \u00d7 8 matrix decouples into four 2 \u00d7 2 matrices, two of which are diagonal. 364 Approximation Methods c. Use degenerate perturbation theory to \ufb01nd the \ufb01rst-order energy shifts \u0394. For all eight states, plot \u0394/A as a function of B/A. See Figure 5.3. Explain why the resulting spectrum looks qualitatively different for B/A 1 and B/A 1. 5.25 Work out the quadratic Zeeman effect for the ground-state hydrogen atom due to the usually neglected e2A2/2mec2-term in the Hamiltonian taken to \ufb01rst order. Write the energy shift as and obtain an expression for diamagnetic susceptibility, \u03c7. \u0394 = \u2212 1 2 \u03c7B2 5.26 (Merzbacher (1970), p. 448, Problem 11.) For the He wave function, use \u03c8(x1, x2) = (Z3 eff/\u03c0a3 0)exp \u2212Zeff(r1 + r2) a0 16, as obtained by the variational method. The measured value of the with Zeff = 2 \u2212 5 diamagnetic susceptibility is 1.88 \u00d7 10\u22126 cm3/mole. a. Using the Hamiltonian for an atomic electron in a magnetic \ufb01eld, determine, for a state of zero angular momentum, the energy change to order B2 if the system is in a uniform magnetic \ufb01eld represented by the vector potential A = 1 2 B \u00d7 x. \u03c7B2, calculate \u03c7 for a helium atom in the ground state and compare the result with the measured value. b. De\ufb01ning the atomic diamagnetic susceptibility \u03c7 by E = \u2212 1 2 5.27 Estimate the ground-state energy of a one-dimensional simple harmonic oscillator using x|\u02dc0 = e\u2212\u03b2|x| as a trial function with \u03b2 to be varied. 5.28 Estimate the lowest eigenvalue (\u03bb) of the differential equation d2\u03c8 dx2 + (\u03bb \u2212 |x|)\u03c8 = 0 where \u03c8 \u2192 0 as |x| \u2192\u221e using the variational method with \u03c8 = for |x| < \u03b1 for |x| > \u03b1 as a trial function. (Caution: d\u03c8/dx is discontinuous at x = 0.) The exact value of the lowest eigenvalue can be shown to be 1.019. c(\u03b1 \u2212 |x|) 0 (\u03b1 to be varied) 5.29 Consider a one-dimensional simple harmonic oscillator whose classical angular frequency is \u03c90. For t < 0 it is known to be in the ground state. For t > 0 there is also a time-dependent potential V(t) =F 0x cos \u03c9t where F0", " is constant in both space and time. Obtain an expression for the expectation value x as a function of time using time-dependent perturbation theory to lowest nonvanishing order. Is this procedure valid for \u03c9 \u03c90? 365 Problems 5.30 A one-dimensional harmonic oscillator is in its ground state for t < 0. For t \u2265 0 it is subjected to a time-dependent but spatially uniform force (not potential!) in the x-direction, F(t) =F 0e\u2212t/\u03c4. a. Using time-dependent perturbation theory to \ufb01rst order, obtain the probability of \ufb01nding the oscillator in its \ufb01rst excited state for t > 0. Show that the t \u2192 \u221e (\u03c4 \ufb01nite) limit of your expression is independent of time. Is this reasonable or surprising? b. Can we \ufb01nd higher excited states? 5.31 Consider a particle bound in a simple harmonic oscillator potential. Initially (t < 0), it is in the ground state. At t = 0 a perturbation of the form H(x, t) =Ax 2e\u2212t/\u03c4 is switched on. Using time-dependent perturbation theory, calculate the probability that, after a suf\ufb01ciently long time (t \u03c4), the system will have made a transition to a given excited state. Consider all \ufb01nal states. 5.32 The unperturbed Hamiltonian of a two-state system is represented by E0 1 0. 0 E0 2 H0 = There is, in addition, a time-dependent perturbation V(t) = 0 \u03bb cos \u03c9t \u03bb cos \u03c9t 0 (\u03bb real). a. At t = 0 the system is known to be in the \ufb01rst state, represented by 1 0. Using time-dependent perturbation theory and assuming that E0 2 is not close 1 to \u00b1\u00afh\u03c9, derive an expression for the probability that the system be found in the second state represented by \u2212 E0 0 1 as a function of t(t > 0). b. Why is this procedure not valid when E0 1 \u2212 E0 2 is close to \u00b1\u00afh\u03c9? 5.33 A one-dimensional simple harmonic oscillator of angular frequency \u03c9 is acted upon by a spatially uniform but time-dependent force (not potential) F(t) = (F0\u03c4/\u03c9) (\u03c42 + t2), \u2212\u221e < t < \u221e. 366 Approximation Methods At t = \u2212\u221e, the oscillator is known to be in the ground state. Using the time-dependent perturbation theory to \ufb01rst order, calculate the probability that the oscillator is found in the \ufb01rst excited state at t = +\u221e. Challenge for experts: F(t) is so normalized that the impulse F(t)dt imparted to the oscillator is always the same, that is, independent of \u03c4; yet for \u03c4 1/\u03c9, the probability for excitation is essentially negligible. Is this reasonable? 5.34 Consider a particle in one dimension moving under the in\ufb02uence of some timeindependent potential. The energy levels and the corresponding eigenfunctions for this problem are assumed to be known. We now subject the particle to a traveling pulse represented by a time-dependent potential, V(t) =A \u03b4(x \u2212 ct). a. Suppose at t = \u2212\u221e the particle is known to be in the ground state whose energy eigenfunction is x|i = ui(x). Obtain the probability for \ufb01nding the system in some excited state with energy eigenfunction x| f = uf(x) at t = +\u221e. b. Interpret your result in (a) physically by regarding the \u03b4-function pulse as a superposition of harmonic perturbations; recall \u03b4(x \u2212 ct) = 1 2\u03c0c \u221e \u2212\u221e d\u03c9ei\u03c9[(x/c)\u2212t]. Emphasize the role played by energy conservation, which holds even quantum mechanically as long as the perturbation has been on for a very long time. 5.35 A hydrogen atom in its ground state [(n, l, m) = (1, 0, 0)] is placed between the plates of a capacitor. A time-dependent but spatial uniform electric \ufb01eld (not potential!) is applied as follows: E = 0 E0e\u2212t/\u03c4 for t < 0 for t > 0 (E0 in the positive z-direction). Using \ufb01rst-order time-dependent perturbation theory, compute the probability for the atom to be found at t \u03c4 in each of the three 2p states: (n,", " l, m) = (2, 1, \u00b11 or 0). Repeat the problem for the 2s state: (n, l, m) = (2, 0, 0). Consider the limit \u03c4 \u2192 \u221e. 5.36 Consider a composite system made up of two spin 1 2 objects. For t < 0, the Hamiltonian does not depend on spin and can be taken to be zero by suitably adjusting the energy scale. For t > 0, the Hamiltonian is given by 4\u0394 \u00afh2 H = S1 \u00b7 S2. Suppose the system is in | + \u2212 for t \u2264 0. Find, as a function of time, the probability for being found in each of the following states | + +, | + \u2212, | \u2212+, and | \u2212 \u2212. 367 Problems a. By solving the problem exactly. b. By solving the problem assuming the validity of \ufb01rst-order time-dependent perturbation theory with H as a perturbation switched on at t = 0. Under what condition does (b) give the correct results? 5.37 Consider a two-level system with E1 < E2. There is a time-dependent potential that connects the two levels as follows: V11 = V22 = 0, V12 = \u03b3ei\u03c9t, V21 = \u03b3e\u2212i\u03c9t (\u03b3 real). At t = 0, it is known that only the lower level is populated, that is, c1(0) =1, c2(0) =0. a. Find |c1(t)|2 and |c2(t)|2 for t > 0 by exactly solving the coupled differential equation i\u00afh\u02d9ck = 2 \u2211 Vkn(t)ei\u03c9kntcn (k = 1, 2). n=1 b. Do the same problem using time-dependent perturbation theory to lowest nonvanishing order. Compare the two approaches for small values of \u03b3. Treat the following two cases separately: (i) \u03c9 very different from \u03c921 and (ii) \u03c9 close to \u03c921. Answer for (a): (Rabi\u2019s formula) |c2(t)|2 = \u03b32/\u00afh2 \u03b32/\u00afh2 + (\u03c9 \u2212 \u03c921)2/4 sin2 |c1(t)|2 = 1 \u2212 |c2(t)|2. * 1/2 \u03b32 \u00afh2 + (\u03c9 \u2212 \u03c921)2 4 + t, 5.38 Show that the slow-turn-on of perturbation V \u2192 Ve\u03b7t (see Baym (1969), p. 257) can generate contribution from the second term in (5.295). 5.39 a. Consider the positronium problem solved in Chapter 3, Problem 3.5. In the presence of a uniform and static magnetic \ufb01eld B along the z-axis, the Hamiltonian is given by H = AS1 \u00b7 S2 + (S1z \u2212 S2z). eB mec Solve this problem to obtain the energy levels of all four states using degenerate time-independent perturbation theory (instead of diagonalizing the Hamiltonian matrix). Regard the \ufb01rst and the second terms in the expression for H as H0 and V, respectively. Compare your results with the exact expressions E = \u2212 \u00afh2A 4 E = \u00afh2A 4 \u23a1 \u23a31 \u00b1 2 1 + 4 2 eB mec\u00afhA \u23a4 * \u23a6 for singlet m = 0 triplet m = 0 for triplet m = \u00b11, where triplet (singlet) m = 0 stands for the state that becomes a pure triplet (singlet) with m = 0 as B \u2192 0. 368 Approximation Methods b. We now attempt to cause transitions (via stimulated emission and absorption) between the two m = 0 states by introducing an oscillating magnetic \ufb01eld of the \u201cright\u201d frequency. Should we orient the magnetic \ufb01eld along the z-axis or along the x- (ory -) axis? Justify your choice. (The original static \ufb01eld is assumed to be along the z-axis throughout.) c. Calculate the eigenvectors to \ufb01rst order. 5.40 Repeat Problem 5.39 above, but with the atomic hydrogen Hamiltonian H = AS1 \u00b7 S2 + S1 \u00b7 B eB mec where in the hyper\ufb01ne term AS1 \u00b7 S2, S1 is the electron spin, while S2 is the proton spin. (Note the problem here has less symmetry than that of the positronium case.) 5.41 Consider the spontaneous emission of a photon by an excited atom. The process is known to be an", " E1 transition. Suppose the magnetic quantum number of the atom decreases by one unit. What is the angular distribution of the emitted photon? Also discuss the polarization of the photon with attention to angular-momentum conservation for the whole (atom plus photon) system. 5.42 Consider an atom made up of an electron and a singly charged (Z = 1) triton (3H). Initially the system is in its ground state (n = 1, l = 0). Suppose the system undergoes beta decay, in which the nuclear charge suddenly increases by one unit (realistically by emitting an electron and an antineutrino). This means that the tritium nucleus (called a \u201ctriton\u201d) turns into a helium (Z = 2) nucleus of mass 3 (3He). a. Obtain the probability for the system to be found in the ground state of the resulting helium ion. b. The available energy in tritium beta decay is about 18 keV and the size of the 3He atom is about 1 \u02daA. Check that the time scale T for the transformation satis\ufb01es the criterion of validity for the sudden approximation. 5.43 Show that An(R) de\ufb01ned in (5.234) is a purely real quantity. 5.44 Consider a neutron in a magnetic \ufb01eld, \ufb01xed at an angle \u03b8 with respect to the z-axis, but rotating slowly in the \u03c6 direction. That is, the tip of the magnetic \ufb01eld traces out a circle on the surface of the sphere, at \u201clatitude\u201d \u03c0 \u2212 \u03b8. Explicitly calculate the Berry potential A for the spin-up state from (5.234), take its curl, and determine Berry\u2019s phase \u03b3+. Thus, verify (5.253) for this particular example of a curve C. (For hints, see \u201cThe adiabatic theorem and Berry\u2019s phase\u201d by Holstein, Am. J. Phys., 57 (1989) 1079.) 5.45 The ground state of a hydrogen atom (n = 1, l = 0) is subjected to a time-dependent potential as follows: V(x, t) =V 0 cos(kz \u2212 \u03c9t). Using time-dependent perturbation theory, obtain an expression for the transition rate at which the electron is emitted with momentum p. Show, in particular, how you 369 Problems may compute the angular distribution of the ejected electron (in terms of \u03b8 and \u03c6 de\ufb01ned with respect to the z-axis). Discuss brie\ufb02y the similarities and the differences between this problem and the (more realistic) photoelectric effect. (Note: For the initial wave function see Problem 5.42. If you have a normalization problem, the \ufb01nal wave function may be taken to be \u03c8f(x) = 1 L3/2 eip \u00b7 x/\u00afh with L very large, but you should be able to show that the observable effects are independent of L.) 5.46 A particle of mass m constrained to move in one dimension is con\ufb01ned within 0 < x < L by an in\ufb01nite-wall potential V = \u221e for x < 0, x > L, V = 0 for 0 \u2264 x \u2264 L. Obtain an expression for the density of states (that is, the number of states per unit energy interval) for high energies as a function of E. (Check your dimension!) 5.47 Linearly polarized light of angular frequency \u03c9 is incident on a one-electron \u201catom\u201d whose wave function can be approximated by the ground state of a three-dimensional isotropic harmonic oscillator of angular frequency \u03c90. Show that the differential cross section for the ejection of a photoelectron is given by d\u03c3 d\u03a9 = 4\u03b1\u00afh2k3 f m2\u03c9\u03c90 \u03c0\u00afh m\u03c90 exp \u00d7 sin2 \u03b8cos2 \u03c6 exp - 2 k2 f + \u03c9 c cos \u03b8 \u2212 \u00afh m\u03c90 2\u00afhkf\u03c9 m\u03c90c provided the ejected electron of momentum \u00afhkf can be regarded as being in a plane wave state. (The coordinate system used is shown in Figure 5.13.) 5.48 Find the probability |\u03c6(p)|2d3p of the particular momentum p for the ground-state hydrogen atom. (This is a nice exercise in three-dimensional Fourier transforms. To perform the angular integration choose the z-axis in the direction of p.) 5.49 Calculate the lifetimes for the 2p \u2192 1s and 3p \u2192 1s transitions in the hydrogen atom. You can \ufb01nd measurements of these lifetimes in Bickel and Goodman, Phys", ". Rev., 148 (1966) 1. 5.50 A hydrogen atom is prepared in the 2p, m = +1 state. Find the rate for the 2p \u2192 1s transition as a function of polar angle \u03b8 with respect to the z-axis. Describe the polarization of the emitted radiation as a function of \u03b8 and use this to qualitatively explain the intensity pattern. 5.51 This problem highlights anomalies in the \u201cexponential\u201d decay of a state. It is inspired by Winter, Phys. Rev., 123 (1961) 1503, but modern computer applications make it straightforward to directly evaluate the integrals numerically. 370 Approximation Methods Consider a particle of mass m that is initially inside a \u201cwell\u201d bounded by an in\ufb01nite wall to the left and a \u03b4-function potential on the right: U d (x) (x, 0) \u2013a 0 x The in\ufb01nite wall is located at x = \u2212a, and the potential at x = 0 isU \u03b4(x) where U is a positive constant. The \ufb01gure also shows a plausible \u201cground state\u201d initial wave function \u03a8(x, t = 0) = (2/a)1/2 sin(n\u03c0x/a) with n = 1. a. Show that the wave function at all times can be written as 1/2 \u221e 2 a e\u2212iTq2q sin q [q sin(l + 1)q + f ] (q2 \u2212 n2\u03c02)(q2 + Gq sin 2q + G2 sin2 q) dq 0 \u03a8(x, t) =2n where q \u2261 [a(2mE)1/2]/\u00afh for a particle with energy E, T \u2261 \u00afht/2ma2, l \u2261 x/a, and G \u2261 2maU/\u00afh2 are all dimensionless quantities, and f = 0 for \u2212a \u2264 x \u2264 0 and f = G sin q sin lq for x \u2265 0. This is most easily done by expanding the wave function in energy eigenstates |E, as \u03a8(x, t) = dE\u03c6E(x)e\u2212iEt/\u00afh \u221e 0 where \u03c6E(x) is an energy eigenfunction and E|E = \u03b4(E \u2212 E). b. Write a computer program to (numerically) integrate the probability of \ufb01nding the particle inside the well. Carry out the integration for a series of values of T between zero and 12, using the same parameters as Winter, namely n = 1 and G = 6. Plotting these probabilities as a function of T should resemble Figure 2 of Winter\u2019s paper. Fit the points for 2 \u2264 T \u2264 8 to an exponential, and compare the decay time to Winter\u2019s value of 0.644. c. Examine the behavior for T \u2265 8, and compare to the behavior Winter found for the current at x = 0. This suggests an experimental measurement. See Norman et al., Phys. Rev. Lett., 60 (1988) 2246. 6 Scattering Theory This chapter is devoted to the theory of scattering processes. These are processes in which a continuum initial state is transformed into a continuum \ufb01nal state, through the action of some potential which we will treat as a time-dependent perturbation. Such processes are of enormous signi\ufb01cance. They are the primary way in which we learn experimentally about distributions in mass, charge, and, in general, potential energy for molecular, atomic, and subatomic systems. 6.1 Scattering as a Time-Dependent Perturbation We assume that the Hamiltonian can be written as where H = H0 + V(x) H0 = p2 2m stands for the kinetic-energy operator, with eigenvalues Ek = \u00afh2k2 2m. (6.1) (6.2) (6.3) We denote the plane wave eigenvectors of H0 by |k and we assume that the scattering potential V(r) is time independent. Our treatment realizes that an incoming particle will \u201csee\u201d the scattering potential as a perturbation which is \u201cturned on\u201d only during the time that the particle is in the vicinity of the scatterer. Therefore, we can analyze the problem in terms of time-dependent perturbation theory in the interaction picture. To review (see Section 5.7), the state |\u03b1, t0; t0I evolves into the state |\u03b1, t0; tI according to |\u03b1, t0; tI = UI(t, t0)|\u03b1, t0; t0I where UI(t, t0) satis\ufb01es the", " equation i\u00afh \u2202 \u2202t UI(t, t0) =V I(t)UI(t, t0) 371 (6.4) (6.5) 372 Scattering Theory with UI(t0, t0) =1 and VI(t) = exp(iH0t/\u00afh)V exp(\u2212iH0t/\u00afh). The solution of this equation can be formally written as t UI(t, t0) =1 \u2212 i \u00afh VI(t)UI(t, t0)dt. (6.6) t0 Therefore, the \u201ctransition amplitude\u201d for an initial state |i to transform into a \ufb01nal state |n, where both are eigenstates of H0, is given by n|UI(t, t0)|i = \u03b4ni \u2212 i \u00afh ei\u03c9nmtm|UI(t, t0)|idt n|V|m (6.7) \u2211 t t0 m where n|i = \u03b4ni and \u00afh\u03c9nm = En \u2212 Em. To apply this formalism to scattering theory, we need to make some adjustments. First, there is the normalization of the initial and \ufb01nal states. Equation (6.7) assumes discrete states, but our scattering states are in the continuum. We deal with this by quantizing our scattering states in a \u201cbig box,\u201d i.e. a cube of side L. In the coordinate representation, this gives x|k = 1 L3/2 eik\u00b7x (6.8) in which case k|k = \u03b4kk where the k take on discrete values. We will take L \u2192 \u221e at the end of any calculation. We also need to deal with the fact that both the initial and \ufb01nal states exist only asymptotically. That is, we need to work with both t \u2192 \u221e and t0 \u2192 \u2212\u221e. We can take a hint from a \ufb01rst-order treatment of (6.7) in which case we set m|UI(t, t0)|i = \u03b4mi inside the integral, that is t n|UI(t, t0)|i = \u03b4ni \u2212 i \u00afh In this case, as t \u2192 \u221e we saw a \u201ctransition rate\u201d emerge as Fermi\u2019s golden rule. So, in order to also accommodate t0 \u2192 \u2212\u221e, we de\ufb01ne a matrix T as follows: n|V|i ei\u03c9nit (6.9) dt. t0 n|UI(t, t0)|i = \u03b4ni \u2212 i \u00afh Tni t t0 ei\u03c9nit +\u03b5t dt (6.10) where \u03b5 > 0 and t (1/\u03b5). These conditions ensure that e\u03b5t is close to unity as t \u2192 \u221e, and that the integrand goes to zero as t0 \u2192 \u2212\u221e. We just need to make sure that we take the limit \u03b5 \u2192 0 \ufb01rst, before we take t \u2192 +\u221e. We can now de\ufb01ne the scattering (or S) matrix in terms of theT matrix: Sni \u2261 lim t\u2192\u221e n|UI(t, \u2212\u221e)|i lim \u03b5\u21920 = \u03b4ni \u2212 i \u00afh Tni \u221e \u2212\u221e ei\u03c9nit dt = \u03b4ni \u2212 2\u03c0i\u03b4(En \u2212 Ei)Tni. (6.11) Clearly, the S matrix consists of two parts. One part is that in which the \ufb01nal state is the same as the initial state. The second part, governed by the T matrix, is one in which some sort of scattering occurs. 373 6.1 Scattering as a Time-Dependent Perturbation 6.1.1 Transition Rates and Cross Sections Proceeding as in Section 5.7, we de\ufb01ne the transition rate as w(i \u2192 n) = d dt |n|UI(t, \u2212\u221e)|i|2 where for |i = |n we have and therefore n|UI(t, \u2212\u221e)|i = \u2212 i \u00afh Tni t \u2212\u221e ei\u03c9nit +\u03b5t dt = \u2212 i \u00afh Tni ei\u03c9nit+\u03b5t i\u03c9ni + \u03b5 w(i \u2192 n) = 1 \u00afh2 d dt |Tni|2 e2\u03b5t ni + \u03b52 \u03c92 = 1 \u00afh2 |Tni|2 2\u03b5 e2\u03b5t ni + \u03b52 \u03c92. (6.12) (6.13) We need to take \u03b5 \u2192", " 0 for \ufb01nite values of t, and thent \u2192 \u221e. Clearly, this will send w \u2192 0 if \u03c9ni = 0 so we see something like \u03b4(\u03c9ni) emerging, which is not unexpected based on (6.11). In fact, since \u221e 1 \u03c92 + \u03b52 d\u03c9 = \u03c0 \u03b5 for \u03b5 > 0, we have, for \ufb01nite t, \u2212\u221e \u03b5 e2\u03b5t ni + \u03b52 \u03c92 lim \u03b5\u21920 = \u03c0\u03b4(\u03c9ni) = \u03c0\u00afh\u03b4(En \u2212 Ei). Therefore, the transition rate is w(i \u2192 n) = 2\u03c0 \u00afh |Tni|2 \u03b4(En \u2212 Ei) (6.14) (6.15) (6.16) which is independent of time, so the limit as t \u2192 \u221e is trivial. This expression is strikingly similar to Fermi\u2019s golden rule (5.294), except that Vni has been replaced by the more general Tni. We will see below how to determine the matrix elements Tni in general. First, however, let us continue with this discussion and use the transition rate to express the scattering cross section. As with Fermi\u2019s golden rule, in order to integrate over the \ufb01nal state energy En, we need to determine the density of \ufb01nal states \u03c1(En) = \u0394n/\u0394En. We will determine the density of states for elastic scattering, where |i = |k and |n = |k and |k| = |k | \u2261k. (Recall our discussion of \u201cThe Free Particle in Three Dimensions\u201d in Section 2.5.1.) For our \u201cbig box\u201d normalization, we write 2 En = \u00afh2k 2 2m = \u00afh2 2m 2\u03c0 L |n|2 so \u0394En = \u00afh2 m 2\u03c0 L 2 |n| \u0394|n| (6.17) where n = nx\u02c6i + ny\u02c6j + nz \u02c6k and nx,y,z are integers. Since n = (L/2\u03c0)|k | = (L/2\u03c0)k and L is large, we can think of |n| as nearly continuous, and the number of states within a spherical shell of radius |n| and thickness \u0394|n| is \u0394n = 4\u03c0|n|2\u0394|n| \u00d7 d\u03a9 4\u03c0 (6.18) 374 Scattering Theory taking into account the fraction of solid angle represented by the \ufb01nal state wave vector k. Therefore 3 \u03c1(En) = \u0394n \u0394En = m \u00afh2 L 2\u03c0 2 |n|d\u03a9 = mk \u00afh2 L 2\u03c0 d\u03a9 (6.19) and after integrating over \ufb01nal states, the transition rate is given by w(i \u2192 n) = mkL3 (2\u03c0)2 \u00afh3 |Tni|2 d\u03a9. (6.20) We use the concept of cross section to interpret the transition rate in scattering experiments. That is, we determine the rate at which particles are scattered into a solid angle d\u03a9 from a \u201cbeam\u201d of particles with momentum \u00afhk. The speed of these particles is v = \u00afhk/m so the time it takes for a particle to cross the \u201cbig box\u201d is L/v. Therefore the \ufb02ux in the particle beam is (1/L2) \u00f7 (L/v) = v/L3. Indeed, the probability \ufb02ux (2.191) for the wave function (6.8) becomes \u00afh m The cross section d\u03c3 is simply de\ufb01ned as the transition rate divided by the \ufb02ux. Putting this all together, we have k L3 = j(x, t) = v L3. (6.21) 2 d\u03c3 d\u03a9 = mL3 2\u03c0\u00afh2 |Tni|2. (6.22) The job now before us is to relate the matrix elements Tni to the scattering potential distribution V(r). 6.1.2 Solving for the T Matrix We return to the de\ufb01nition of the T matrix. From (6.10) and (6.13) we have n|UI(t, \u2212\u221e)|i = \u03b4ni + 1 \u00afh Tni ei\u03c9nit+\u03b5t \u2212\u03c9ni + i\u03b5. (6.23) We can also return to (6.7). Writing Vnm = n|V|m, we have n|UI(t, \u2212\u221e)|i =", " \u03b4ni \u2212 i \u00afh \u2211 m Vnm t \u2212\u221e ei\u03c9nmt m|UI(t, \u2212\u221e)|idt. (6.24) Now insert (6.23) into the integrand of (6.24). This results in three terms, the \ufb01rst of which is \u03b4ni and the second looks just like (6.23) but with Tni replaced with Vni. The third term is \u2212 i \u00afh 1 \u00afh \u2211 m Vnm Tmi \u2212\u03c9mi + i\u03b5 t \u2212\u221e ei\u03c9nmt +i\u03c9mit +\u03b5t dt. (6.25) The integral is then carried out, and since \u03c9nm + \u03c9mi = \u03c9ni, the result can be taken outside the summation. Gathering terms and comparing the result to (6.23), we discover the following relation: 375 6.1 Scattering as a Time-Dependent Perturbation Tni = Vni + 1 \u00afh \u2211 m Vnm Tmi \u2212\u03c9mi + i\u03b5 = Vni +\u2211 m Vnm Tmi Ei \u2212 Em + i\u00afh\u03b5. (6.26) This is an inhomogeneous system of linear equations which can be solved for the values Tni, in terms of the known matrix elements Vnm. It is convenient to de\ufb01ne a set of vectors |\u03c8(+) in terms of components in some basis | j, so that Tni = \u2211 n|V| jj|\u03c8(+) = n|V|\u03c8(+). j (The choice of notation will be become apparent shortly.) Therefore (6.26) becomes n|V|\u03c8(+) = n|V|i +\u2211 n|V|m m m|V|\u03c8(+) Ei \u2212 Em + i\u00afh\u03b5. Since this must be true for all |n, we have an expression for the |\u03c8(+), namely (6.27) (6.28) |\u03c8(+) = |i +\u2211 |m m m|V|\u03c8(+) Ei \u2212 Em + i\u00afh\u03b5 = |i +\u2211 m 1 Ei \u2212 H0 + i\u00afh\u03b5 |mm|V|\u03c8(+) or |\u03c8(+) = |i + 1 Ei \u2212 H0 + i\u00afh\u03b5 V|\u03c8(+). (6.29) This is known as the Lippmann\u2013Schwinger equation. The physical meaning of (+) is to be discussed in a moment by looking at x|\u03c8(+) at large distances. Clearly, the states |\u03c8(+) have a fundamental importance, allowing us to rewrite (6.22) as d\u03c3 d\u03a9 = mL3 2\u03c0\u00afh2 2 n|V|\u03c8(+) 2. (6.30) We have introduced the matrix elements Tni simply as complex numbers, de\ufb01ned by (6.10). However, we can also de\ufb01ne an operator T with matrix elements n|T|i = Tni by writing T|i = V|\u03c8(+). We can then operate on (6.29) from the left with V which leads to the succinct operator equation T = V + V 1 Ei \u2212 H0 + i\u00afh\u03b5 T. (6.31) To the extent that the scattering potential V is \u201cweak\u201d, an order-by-order approximation scheme presents itself for the transition operator T, namely T = V + V 1 Ei \u2212 H0 + i\u00afh\u03b5 V + V 1 Ei \u2212 H0 + i\u00afh\u03b5 V 1 Ei \u2212 H0 + i\u00afh\u03b5 V + \u00b7 \u00b7 \u00b7. (6.32) We will return to this approximation scheme in Section 6.3. 376 Scattering Theory 6.1.3 Scattering from the Future to the Past We can also picture the scattering process as evolving backwards in time from a plane wave state |i in the far future, to a state |n in the distant past. In this case, we would write the formal solution (6.6) as t0 i \u00afh UI(t, t0) =1 + VI(t )UI(t, t0)dt (6.33) t which is a form suitable for taking t0 \u2192 +\u221e. Our T matrix is then de\ufb01ned by regularizing the integral with the opposite sign exponential, namely t0 n|UI(t, t0)|i = \u03b4ni + i \u00afh In this case, the T operator is de\ufb01ned through a different set of states |\u03c8(\u2212) through T|i = V|\u03c8(\u2212). We are now prepared to study practical", " solutions to the scattering problem, and gain insight as to the different scattering states |\u03c8(+) and |\u03c8(\u2212). ei\u03c9nit \u2212\u03b5t (6.34) dt. Tni t 6.2 The Scattering Amplitude Let us replace \u00afh\u03b5 in the Lippman\u2013Schwinger equation with \u03b5; this will be handy and presents no dif\ufb01culties since the only constraints on \u03b5 are that it be positive and arbitrarily small. We will also continue to anticipate application to elastic scattering, and use E for the initial (and \ufb01nal) energy. We therefore rewrite (6.29) as |\u03c8(\u00b1) = |i + 1 E \u2212 H0 \u00b1 i\u03b5 V|\u03c8(\u00b1). (6.35) We now con\ufb01ne ourselves to the position basis by multiplying x| from the left, and inserting a complete set of position basis states. Thus x|\u03c8(\u00b1) = x|i + d3x x x |V|\u03c8(\u00b1). (6.36) ( ) 1 E \u2212 H0 \u00b1 i\u03b5 x This is an integral equation for scattering because the unknown ket |\u03c8(\u00b1) appears under an integral sign. To make progress we must \ufb01rst evaluate the function G\u00b1(x, x ) \u2261 \u00afh2 2m 1 E \u2212 H0 \u00b1 i\u03b5 (6.37) x ) ( x. Since the eigenstates of H0 are most easily evaluated in the momentum basis, we proceed by inserting complete sets of states |k. (Recall that these are discrete states in our normalization scheme.) We then write G\u00b1(x, x ) = \u00afh2 2m \u2211 k x|k \u2211 k ( k 1 E \u2212 H0 \u00b1 i\u03b5 k ) k|x. (6.38) 377 6.2 The Scattering Amplitude Now let H0 act on k |, and use ( k 1 E \u2212 (\u00afh2k 2/2m) \u00b1 i\u03b5 k ) = x|k = k|x = and put E = \u00afh2k2/2m. Equation (6.37) then becomes \u03b4k k E \u2212 (\u00afh2k2/2m) \u00b1 i\u03b5 eik \u00b7x L3/2 e\u2212ik\u00b7x L3/2 G\u00b1(x, x ) = 1 L3 \u2211 k eik \u00b7(x\u2212x ) k2 \u2212 k2 \u00b1 i\u03b5 (6.39) (6.40) (6.41) (6.42) where we have once again rede\ufb01ned \u03b5. This sum is actually easiest to do if we take L \u2192 \u221e and convert it to an integral. Since ki = 2\u03c0ni/L (i = x, y, z), the integral measure becomes d3k = (2\u03c0)3/L3 and we have G\u00b1(x, x ) = 1 (2\u03c0)3 d3k eik \u00b7(x\u2212x ) k2 \u2212 k2 \u00b1 i\u03b5. (6.43) This integral can be done analytically. First, convert to spherical coordinates, with angles measured relative to the k direction. The integral over \u03c6k just gives a factor of 2\u03c0. For the integral over sin \u03b8k d\u03b8k, put \u03bc \u2261 cos \u03b8k to \ufb01nd G\u00b1(x, x ) = 1 (2\u03c0)2 \u221e 0 k2dk +1 \u22121 = 1 8\u03c02 1 i|x \u2212 x | \u221e \u2212\u221e k dk d\u03bc eik |x\u2212x |\u03bc k2 \u2212 k2 \u00b1 i\u03b5 eik |x\u2212x | \u2212 e\u2212ik |x\u2212x | k2 \u2212 k2 \u00b1 i\u03b5 (6.44) where we made use of the fact that the integrand is even in k, allowing us to extend the integration lower limit to \u2212\u221e while dividing by two. We complete the calculation using complex contour integration.1 This procedure will in fact demonstrate the importance of \u03b5 and its sign. Treating k as a complex variable, imagine an integration contour running along the Re(k ) axis, and then closed with a semicircle in either the upper or lower plane. See Figure 6.1. Now make use of the Cauchy integral formula (F.5), which we rewrite as, f (z) z \u2212 z0 C dz = 2\u03c0if (z0) (6.45) where the contour C is followed counter-clockwise. The denominator of the integrand of (6.44) can be written as \u2212(k \u2212 k0)(k + k0) where k0 \u2261 k \u00b1 i\u03b5. (Once again, \u03b5 is rede\ufffd", "\ufffd\ufffdned, keeping its sign intact, since k is real and positive.) Therefore, the integral over the complex k plane has poles at \u00b1k0, and we can apply (6.45). 1 Integration over the complex plane arises naturally in many subjects, including scattering theory. For a brief summary of the mathematics, see Appendix F of this book. Complex analysis is also covered in just about any textbook on mathematical physics, for example Arfken et al. (2013) or Byron and Fuller (1992). 378 Scattering Theory Im ( )k Re( k ) Fig. 6.1 Integrating the two terms in (6.44) using complex contours. The dots (crosses) mark the positions of the two poles for the + (\u2212) form of G\u00b1(x, x ). We replace the integral over a real-valued k in (6.44) with one of the two contours in the figure, choosing the one on which the factor e\u00b1ik |x\u2212x | tends to zero along the semicircle at large Im(k ). Thus, the only contribution to the contour integral is along the real axis. Consider separately the two terms in the integrand of (6.44). For the \ufb01rst term, close the contour in the upper plane. In this case, the contribution to the integrand along the semicircle goes to zero exponentially with eik |x\u2212x | as Im(k ) \u2192 +\u221e. Closing in the upper plane encloses the pole at k = +k + i\u03b5 (k = \u2212k + i\u03b5) in the calculation of G+ (G\u2212). Applying (6.45), the integral of the \ufb01rst term in brackets of (6.44) becomes 2\u03c0i(\u00b1k) ei(\u00b1k) |x\u2212x | (\u2212)(\u00b12k) = \u2212\u03c0ie\u00b1ik |x\u2212x | (6.46) where we have let \u03b5 \u2192 0. The second term is treated the same way, except that the contour is closed in the lower plane, with another overall minus sign because the contour is traced clockwise. The contribution to the integral turns out to be the same as the \ufb01rst-term. We therefore get our \ufb01nal result, namely G\u00b1(x, x ) =\u2212 1 4\u03c0 e\u00b1ik|x\u2212x | |x \u2212 x |. (6.47) 379 6.2 The Scattering Amplitude The reader may recognize that G \u00b1 is nothing more than Green\u2019s function for the Helmholtz equation, (\u22072 + k2)G \u00b1 (x, x ) = \u03b4(3)(x \u2212 x ). (6.48) That is, for x = x, G\u00b1(x, x ) solves the eigenvalue equation H0G\u00b1 = EG\u00b1. We can now rewrite (6.36) in a more explicit form, using (6.47), namely x|\u03c8(\u00b1) = x|i \u2212 2m \u00afh2 d3x e \u00b1 ik |x\u2212x | 4\u03c0|x \u2212 x | x |V|\u03c8( \u00b1). (6.49) Notice that the wave function x|\u03c8( \u00b1 ) in the presence of the scatterer is written as the sum of the wave function for the incident wave x|i and a term that represents the effect of scattering. As we will see explicitly later, at suf\ufb01ciently large distances r, the spatial dependence of the second term is e \u00b1 ik r/r provided that the potential is of \ufb01nite range. This means that the positive solution (negative solution) corresponds to the plane wave plus an outgoing (incoming) spherical wave. This is in keeping with the origin of the sign in terms of scattering forward (backward) in time. In most physical problems we are interested in the positive solution because it is dif\ufb01cult to prepare a system satisfying the boundary condition appropriate for the negative solution. To see the behavior of x|\u03c8( \u00b1 ) more explicitly, let us consider the speci\ufb01c case where V is a local potential, that is, a potential diagonal in the x-representation. Potentials that are functions only of the position operator x belong to this category. In precise terms V is said to be local if it can be written as x |V|x = V(x )\u03b4(3)(x \u2212 x). (6.50) As a result, we obtain x |V|\u03c8( \u00b1 ) = d3xx |V|xx|\u03c8( \u00b1 ) = V(x )x |\u03c8( \u00b1 ). (6.51) The integral equation (6.49) now simpli\ufb01es as x|\u03c8( \u00b1 )", " = x|i \u2212 2m \u00afh2 d3x e \u00b1 ik |x\u2212x | 4\u03c0|x \u2212 x | V(x )x |\u03c8( \u00b1 ). (6.52) Let us attempt to understand the physics contained in this equation. The vector x is understood to be directed towards the observation point at which the wave function is evaluated. For a \ufb01nite range potential, the region that gives rise to a nonvanishing contribution is limited in space. In scattering processes we are interested in studying the effect of the scatterer (that is, the \ufb01nite range potential) at a point far outside the range of the potential. This is quite relevant from a practical point of view because we cannot put a detector at short distance near the scattering center. Observation is always made by a detector placed very far away from the scatterer at r greatly larger than the range of the potential. In other words, we can safely set |x| |x |, (6.53) as depicted in Figure 6.2. 380 Scattering Theory P Observation point x x \u2212 x k = pl /h 0 \u03b1 x Fig. 6.2 Finite-range scattering potential. The observation point P is where the wave function x|\u03c8 (\u00b1) is to be evaluated, while the contribution to the integral in (6.52) is for|x | less than the range of the potential, as depicted by the shaded region of the figure. Introducing r = |x|, r = |x |, and \u03b1 = (x, x ), we have forr r, |x \u2212 x | = r2 \u2212 2rr cos \u03b1 + r 2 r 2 1 \u2212 2r r2 r cos \u03b1 + = r 1/2 where \u2248 r \u2212 \u02c6r\u00b7x \u02c6r \u2261 x |x| in which case k \u2261 k\u02c6r. We then obtain e\u00b1 ik |x\u2212x | \u2248 e\u00b1 ikre\u2213 ik \u00b7x (6.54) (6.55) (6.56) for large r. It is also legitimate to replace 1/|x \u2212 x | by just 1/r. At this point, we specify the initial state as an eigenstate of the free-particle Hamiltonian H0, that is|i = |k. Putting this all together, we have, \ufb01nally, x|\u03c8(+) r large\u2212\u2192 x|k \u2212 1 4\u03c0 eikr r d3x e\u2212ik \u00b7x V(x )x |\u03c8(+) 2m \u00afh2 eikr r = 1 L3/2 eik\u00b7x + f (k, k). (6.57) This form makes it very clear that we have the original plane wave in propagation direction k plus an outgoing spherical wave with amplitude f (k, k) given by d3x e\u2212ik \u00b7x L3/2 V(x )x |\u03c8(+) f (k, k) \u2261 \u2212 1 4\u03c0 = \u2212 mL3 2\u03c0\u00afh2 2m \u00afh2 L3 k |V|\u03c8(+). (6.58) We can also show from (6.52) and (6.56) thatx| \u03c8(\u2212) corresponds to the original plane wave in propagation direction k plus an incoming spherical wave with spatial dependence e\u2212ikr/r and amplitude \u2212(mL3/2\u03c0\u00afh2)\u2212k |V|\u03c8(\u2212). 381 6.2 The Scattering Amplitude (a) (b) Fig. 6.3 (a) Incident wave packet approaching scattering center initially. (b) Incident wave packet continuing to move in the original direction plus spherical outgoing wave front (after a long time duration). We refer to f (k, k) as the scattering amplitude. By comparing (6.58) with (6.30), we see that the differential cross section can be written as d\u03c3 d\u03a9 = | f (k, k)| 2. 6.2.1 Wave Packet Description (6.59) The reader may wonder here whether our formulation of scattering has anything to do with the motion of a particle being bounced by a scattering center. The incident plane wave we have used is in\ufb01nite in extent in both space and time. In a more realistic situation, we consider a wave packet (a dif\ufb01cult subject!) that approaches the scattering center.2 After a long time we have both the original wave packet moving in the original direction plus a spherical wave front that moves outward, as in Figure 6.3. Actually the use of a plane wave is satisfactory as long as the dimension of the wave packet is much larger than the size of the scatterer (or range of V). 6.2.2", " The Optical Theorem There is a fundamental and useful relationship popularly attributed to Bohr, Peierls, and Placzek3 called the optical theorem, which relates the imaginary part of the forward scattering amplitude f (\u03b8 = 0) \u2261 f (k, k) to the total cross section \u03c3tot \u2261 d\u03a9 (d\u03c3/d\u03a9), as follows: Im f (\u03b8 = 0) = k\u03c3tot 4\u03c0. (6.60) 2 For a fuller account of the wave packet approach, see Chapter 3 in Goldberger and Watson (1964), and Chapter 6 in Newton (1982). 3 This relationship is in fact due to Eugene Feenberg, Phys. Rev., 40 (1932) 40. See Newton, Am. J. Phys., 44 (1976) 639 for the historical background. 382 Scattering Theory y Fig. 6.4 Contour used to integrate around a singularity located at z0 = x0 + i\u03b5. x 0 x To prove this, start with the Lippman\u2013Schwinger Equation (6.35) with |i = |k to write k|V|\u03c8(+) = \u03c8(+)| \u2212 \u03c8(+)|V V|\u03c8(+) 1 E \u2212 H0 \u2212 i\u03b5 = \u03c8(+)|V|\u03c8(+) \u2212 \u03c8(+)|V 1 E \u2212 H0 \u2212 i\u03b5 V|\u03c8(+). (6.61) Comparing (6.58) and (6.60) we see that we want to take the imaginary part of both sides of (6.61). The \ufb01rst term on the right side of (6.61) is a real number, since it is the expectation value of a Hermitian operator. Finding the imaginary part of the second term is more dif\ufb01cult, because of singularities along the real axis as \u03b5 \u2192 0. To do this, we use a trick borrowed from the concept of the Cauchy principal value in complex integration. Figure 6.4 shows a complex integration contour which follows the real axis except for a small semicircle which jumps over a singularity near the real axis. The singularity is located at z0 = x0 + i\u03b5, with \u03b5 > 0, which is always above the x-axis. So, let the semicircle be centered on the real axis at x0, and extend into the lower half of the complex plane with a radius \u03b4. The semicircle is described by z \u2212 x0 = \u03b4ei\u03c6 with \u03c6 running from \u2212\u03c0 to zero. Next consider a complex function f (z), with z = x + iy. We can write \u221e \u2212\u221e f (x) x \u2212 x0 dx = x0\u2212\u03b4 f (x) x \u2212 x0 +\u221e \u2212\u221e dx + f (z) z \u2212 z0 c = P f (x) x \u2212 x0 dx + c f (z) z \u2212 z0 \u2212\u221e dz + +\u221e x0+\u03b4 f (x) x \u2212 x0 dx = 0 dz (6.62) where c denotes the small semicircular contour around the singularity. The Cauchy principal value is de\ufb01ned as P +\u221e \u2212\u221e f (x) x \u2212 x0 dx = lim \u03b4\u21920 x0\u2212\u03b4 \u2212\u221e f (x) x \u2212 x0 dx + +\u221e x0+\u03b4 f (x) x \u2212 x0 - We can evaluate the second term in (6.62) as c f (z) z \u2212 z0 dz = 0 f (x0) \u03b4ei\u03c6 \u2212\u03c0 \u2192 i\u03c0f (x0) % i\u03c6\u03b4ei\u03c6d\u03c6 & as \u03b4 \u2192 0. Consequently, we rewrite (6.62) as \u221e \u2212\u221e f (x) x \u2212 x0 dx = P +\u221e \u2212\u221e f (x) x \u2212 x0 dx + i\u03c0f (x0). dx. (6.63) (6.64) (6.65) 383 6.2 The Scattering Amplitude Now we can return to \ufb01nding the imaginary part of the right side of (6.61). We have lim \u03b5\u21920 1 E \u2212 H0 \u2212 i\u03b5 +\u221e \u2212\u221e \u03b4(E \u2212 E ) E \u2212 H0 \u2212 i\u03b5 dE = lim \u03b5\u21920 = i\u03c0\u03b4(E \u2212 H0) (6.66) where we have made use of (6.65). Therefore Imk|V|\u03c8(+) = \u2212\u03c0\u03c8(+)|V\u03b4(E \u2212 H0)V|\u03c8(", "+) = \u2212\u03c0k|T\u2020\u03b4(E \u2212 H0)T|k (6.67) where we recall that T is de\ufb01ned through T|k = V|\u03c8(+). Consequently, using (6.58), Im f (k, k) =\u2212 mL3 k|T\u2020\u03b4(E \u2212 H0)T|k 2\u03c0\u00afh2 Imk|V|\u03c8(+) mL3 2\u00afh2 mL3 2\u00afh2 \u2211 2\u00afh2 \u2211 |k |T|k|2 \u03b4 mL3 k k = = = k|T\u2020\u03b4(E \u2212 H0)|k k |T|k E,\u00afh2k 2/2m (6.68) where E = \u00afh2k2/2m. The optical theorem (6.60) now begins to appear. The factor |k |T|k|2 is proportional to the differential cross section (6.59). The sum, including the \u03b4 function, is over all scattered momenta which conserve energy; in other words it is over all directions in space. Therefore, the right-hand side of (6.68) is an integral of the differential cross section over all directions and so is proportional to the total cross section. To carry (6.68) through to the end, we make use of k |T|k = k |V|\u03c8(+) with (6.58), and converting the sum to an integral as we did to go from (6.42) to (6.44). This gives 2 Im f (k, k) = 2\u03c0\u00afh2 mL3 mL3 2\u00afh2 \u2212\u2192 2\u03c02 \u00afh2 m(2\u03c0)3 \u2211 k | f (k, k)|2 \u03b4 E,\u00afh2k 2/2m d3k | f (k, k)|2 \u03b4 E \u2212 \u00afh2k 2 2m 1 \u00afh2k/m k2 d\u03a9k d\u03c3 d\u03a9k \u00afh2 4\u03c0m k 4\u03c0 \u03c3tot = = (6.69) thus proving (6.60). Section 6.5 will provide some insights as to the physical signi\ufb01cance of the optical theorem. 384 Scattering Theory 6.3 The Born Approximation Our task now is to calculate the scattering amplitude f (k, k) for some given potential energy function V(x). This amounts to calculating the matrix element k |V|\u03c8(+) = k |T|k. (6.70) This task is not straightforward, however, since we do not have closed analytic expressions for either x |\u03c8(+) or T. Consequently, one typically resorts to approximations at this point. We have already alluded to a useful approximation scheme in (6.32). Again replacing \u00afh\u03b5 with \u03b5, this is T = V + V 1 E \u2212 H0 + i\u03b5 V + V 1 E \u2212 H0 + i\u03b5 V 1 E \u2212 H0 + i\u03b5 V + \u00b7 \u00b7 \u00b7 (6.71) which is an expansion in powers of V. We will shortly examine the conditions under which truncations of this expansion should be valid. First, however, we will make use of this scheme and see where it leads us. Taking the \ufb01rst term in the expansion, i.e. T = V or, equivalently, |\u03c8(+) = |k, is called the \ufb01rst-order Born approximation. In this case, the scattering amplitude is denoted by f (1), where f (1)(k, k) = \u2212 m 2\u03c0\u00afh2 d3x ei(k\u2212k )\u00b7x V(x ) (6.72) after inserting a complete set of states |x into (6.58). In other words, apart from an overall factor, the \ufb01rst-order amplitude is just the three-dimensional Fourier transform of the potential V with respect to q \u2261 k \u2212 k. An important special case is when V is a spherically symmetric potential. This implies that f (1)(k, k) is a function of q \u2261 |q|, which is simply related to kinematic variables easily accessible by experiment. See Figure 6.5. Since |k | = k by energy conservation, we have q k \u03b8 k Fig. 6.5 Scattering through angle \u03b8, whereq = k \u2212 k. 385 6.3 The Born Approximation q = |k \u2212 k | = 2 k sin \u03b8 2. We can perform the angular integration in (6.72) explicitly to obtain 2m f (1)(\u03b8) =\u2212 1 \u00afh2 2 = \u2212 2m 1 \u00afh2 q \u221e 1 iq", " \u221e 0 r2 r V(r)(eiqr \u2212 e\u2212iqr) dr rV(r) sin qr dr. 0 A simple but important example is scattering by a \ufb01nite square well, that is V(r) = V0 0 r \u2264 a r > a. The integral in (6.74) is readily done, and yields f (1)(\u03b8) =\u2212 2m \u00afh2 V0a3 (qa)2 \u2212 cos qa. sin qa qa (6.73) (6.74) (6.75) (6.76) This function has zeros at qa = 4.49, 7.73, 10.9... and the position of these zeros, along with (6.73), can be used to determine the well radius a. Figure 6.6 shows elastic proton scattering from several nuclei, each of which is an isotope of calcium. The nuclear potential is approximated rather nicely by a \ufb01nite square well, and the differential cross section shows the characteristic minima predicted by (6.76). Furthermore, the data indicate that as neutrons are added to the calcium nucleus, the minima appear at smaller angles, showing that the nuclear radius in fact increases. Proton elastic scattering 800 MeV 48Ca 44Ca \u00d7 0.1 42Ca \u00d7 0.01 40Ca \u00d7 0.001 ) 104 103 102 101 100 10\u20131 10\u20132 10\u20133 10\u20134 2 4 6 8 10 12 14 16 18 20 22 24 \u03b8c.m. (deg) Fig. 6.6 Data on elastic scattering of protons from the nuclei of four different isotopes of calcium. The angles at which the cross sections show minima, decrease consistently with increasing neutron number. Therefore, the radius of the calcium nucleus increases as more neutrons are added, as one expects. From Ray et al.,Phys. Rev. C, 23 (1981) 828. 386 Scattering Theory Another important example is scattering by a Yukawa potential V(r) = V0e\u2212\u03bcr \u03bcr (6.77) where V0 is independent of r, and 1/\u03bc corresponds, in a certain sense, to the range of the potential. Notice that V goes to zero very rapidly for r 1/\u03bc. For this potential we obtain, from (6.74), (6.78) (6.79) (6.80) (6.82) (6.83) f (1)(\u03b8) =\u2212 2mV0 \u03bc\u00afh2 1 q2 + \u03bc2, where we note that sin qr = Im(eiqr) and have used \u221e e\u2212\u03bcrei q rdr = \u2212Im Im 0 Notice also that 1 \u2212\u03bc + iq = q \u03bc2 + q2. q2 = 4k2 sin2 \u03b8 2 = 2k2(1 \u2212 cos \u03b8). So, in the \ufb01rst Born approximation, the differential cross section for scattering by a Yukawa potential is given by d\u03c3 d\u03a9 2mV0 \u03bc\u00afh2 2 1 [2k2(1 \u2212 cos \u03b8) + \u03bc2]2. (6.81) It is amusing to observe here that as \u03bc \u2192 0, the Yukawa potential is reduced to the Coulomb potential, provided the ratio V0/\u03bc is \ufb01xed, for example, to be ZZ e2, in the limiting process. We see that the \ufb01rst Born differential cross section obtained in this manner becomes d\u03c3 d\u03a9 (2m)2(ZZ e2)2 \u00afh4 1 16k4 sin4(\u03b8/2). Even the \u00afh disappears if \u00afhk is identi\ufb01ed as |p|, so d\u03c3 d\u03a9 = 1 16 ZZ e2 EKE 2 1 sin4(\u03b8/2), where EKE = |p| 2/2m; this is precisely the Rutherford scattering cross section that can be obtained classically. Coming back to (6.74), the Born amplitude with a spherically symmetric potential, there are several general remarks we can make if f (k, k) can be approximated by the corresponding \ufb01rst Born amplitude, f (1). 1. d\u03c3/d\u03a9, or f (\u03b8), is a function of q only; that is, f (\u03b8) depends on the energy (\u00afh2k2/2m) and \u03b8 only through the combination 2k2(1 \u2212 cos \u03b8). 2. f (\u03b8) is always real. 3. d\u03c3/d\u03a9 is independent of the sign of V. 387 6.3 The Born Approximation 4. For small k (q necessarily small), 2", "m \u00afh2 involving a volume integral independent of \u03b8. f (1)(\u03b8) =\u2212 1 4\u03c0 V(r)d3x 5. f(\u03b8) is small for large q due to rapid oscillation of the integrand. In order to study the conditions under which the Born approximation should be valid, let us return to (6.52), slightly rewritten as x|\u03c8(+) = x|k \u2212 2m \u00afh2 d3x eik |x\u2212x | 4\u03c0|x \u2212 x | V(x )x |\u03c8(+). The approximation is that T \u2248 V, which means that |\u03c8(+) can be replaced by |k. Therefore, the second term on the right-hand side in this equation must be much smaller than the \ufb01rst. Let us assume that a \u201ctypical\u201d value for the potential energy V(x) is V0, and that it acts within some \u201crange\u201d a. Writing r = |x \u2212 x | and carrying out a rough approximation on the integral, our validity condition becomes 2m \u00afh2 4\u03c0 3 a3 eikr 4\u03c0a eik\u00b7x L3/2 V0 eik\u00b7x L3/2. Now for low energies (i.e. ka 1), the exponential factors can be replaced by unity. Then, ignoring numerical factors of order unity, the following succinct criterion emerges: m|V0|a2 \u00afh2 1. (6.84) Consider the special case of the Yukawa potential in (6.77), in which the range a = 1/\u03bc. The validity criterion becomes m|V0|/\u00afh2\u03bc2 1. This requirement may be compared with the condition for the Yukawa potential to develop a bound state, which we can show to be 2m|V0|/\u00afh2\u03bc2 \u2265 2.7, with V0 negative. In other words, if the potential is strong enough to develop a bound state, the Born approximation will probably give a misleading result. At high energies (i.e. ka 1), the factors eikr and eik\u00b7x oscillate strongly over the region of integration, so they cannot be set equal to unity. Instead, it can be shown that |V0|a k ln(ka) 1. 2m \u00afh2 (6.85) As k becomes larger, this inequality is more easily satis\ufb01ed. Quite generally, the Born approximation tends to get better at higher energies. 6.3.1 The Higher-Order Born Approximation Now, write T to second order in V, using (6.71), namely T = V + V 1 E \u2212 H0 + i\u03b5 V. It is natural to continue our Born approximation approach and write f (k, k) \u2248 f (1)(k, k) +f (2)(k, k) 388 Scattering Theory k k x x Fig. 6.7 Physical interpretation of the higher-order Born term f (2)(k, k). where f (1)(k, k) is given by (6.72) and d3x d3xk |x V(x ) ) x V(x)(x|k) f (2) = \u2212 1 4\u03c0 ( 2m \u00afh2 (2\u03c0)3 \u00d7 x 1 E \u2212 H0 + i\u03b5 d3x = \u2212 1 2m 4\u03c0 \u00afh2 \u00afh2 G+(x, x) 2m \u00d7 d3xe\u2212ik \u00b7x V(x ) V(x)eik\u00b7x. (6.86) This scheme can obviously be continued to higher orders A physical interpretation of (6.86) is given inFigure 6.7, where the incident wave interacts at x, which explains the appearance of V(x), and then propagates from x to x via Green\u2019s function for the Helmholtz equation (6.48). Subsequently, a second interaction occurs at x, thus the appearance of V(x ), and, \ufb01nally, the wave is scattered into the direction k. In other words,f (2) corresponds to scattering viewed as a two-step process. Likewise, f (3) can be viewed as a three-step process, and so on. 6.4 Phase Shifts and Partial Waves In considering scattering by a spherically symmetric potential, we often examine how states with de\ufb01nite angular momenta are affected by the scatterer. Such considerations lead to the method of partial waves, which we will discuss shortly. However, before discussing the angular momentum decomposition of scattering states, let us \ufb01rst talk about free-particle states, which are also eigenstates of angular momentum. 6.4.1 Free-Particle States For a free particle", " the Hamiltonian is just the kinetic-energy operator, which obviously commutes with the momentum operator. We note, however, that the free-particle 389 6.4 Phase Shifts and Partial Waves Hamiltonian also commutes with L2 and Lz. Thus it is possible to consider a simultaneous eigenket of H0, L2, and Lz. Ignoring spin, such a state is denoted by |E, l, m, often called a spherical wave state. More generally, the most general free-particle state can be regarded as a superposition of |E, l, m with various E, l, and m in much the same way as the most general free-particle state can be regarded as a superposition of |k with different k, different in both magnitude and direction. Put in another way, a free-particle state can be analyzed using either the plane wave basis {|k} or the spherical wave basis {|E, l, m}. We now derive the transformation function k|E, l, m that connects the plane wave basis with the spherical wave basis. We can also regard this quantity as the momentumspace wave function for the spherical wave characterized by E, l, andm. We adopt the normalization convention for the spherical wave eigenket as follows: E, l, m|E, l, m = \u03b4ll \u03b4mm \u03b4(E \u2212 E). (6.87) In analogy with the position-space wave function, we may guess the angular dependence to be k|E, l, m = glE(k)Ym l ( \u02c6k), (6.88) where the function glE(k) will be considered later. To prove this rigorously, we proceed as follows. First, consider the momentum eigenket |k\u02c6z, that is, a plane wave state whose propagation direction is along the positive z-axis. An important property of this state is that it has no orbital angular-momentum component in the z-direction: Lz|k\u02c6z = (xpy \u2212 ypx)|kx = 0, ky = 0, kz = k = 0. (6.89) Actually this is plausible from classical considerations: The angular-momentum component must vanish in the direction of propagation because L \u00b7 p = (x \u00d7 p) \u00b7 p = 0. Because of (6.89), and since E, l, m|k\u02c6z = 0 for m = 0, we must be able to expand |k\u02c6z as follows: |k\u02c6z = \u2211 l dE|E, l, m = 0E, l, m = 0|k\u02c6z. (6.90) Notice that there is no m sum; m is always zero. We can obtain the most general momentum eigenket, with the direction of k speci\ufb01ed by \u03b8 and \u03c6, from |k\u02c6z by just applying the appropriate rotation operator as follows [see Figure 3.3 and (3.256)]: |k = D(\u03b1 = \u03c6, \u03b2 = \u03b8, \u03b3 = 0)|k\u02c6z. (6.91) Multiplying this equation by E, l, m| on the left, we obtain E, l, m|k = \u2211 l \u00d7 E, l, m = 0|k\u02c6z dEE, l, m|D(\u03b1 = \u03c6, \u03b2 = \u03b8, \u03b3 = 0)|E, l, m = 0 dED (l) m0 (\u03b1 = \u03c6, \u03b2 = \u03b8, \u03b3 = 0) = \u2211 l \u00d7 \u03b4ll \u03b4(E \u2212 E)E, l, m = 0|k\u02c6z = D ( l ) m0 (\u03b1 = \u03c6, \u03b2 = \u03b8, \u03b3 = 0)E, l, m = 0|k\u02c6z. (6.92) 390 Scattering Theory (6.93) (6.94) (6.95) (6.96) Now E, l, m = 0|k\u02c6z is independent of the orientation of k, that is, independent of \u03b8 and \u03c6, and we may as well call it lE(k). So we can write, using (3.260), 2l+1 4\u03c0 g\u2217 k|E, l, m = glE (k) Ym l ( \u02c6k). Let us determine glE(k). First, we note that (H0 \u2212 E)|E, l, m = 0. But we also let H0 \u2212 E operate on a momentum eigenbra k| as follows: k|(H0 \u2212 E) = \u2212 E k|. \u00afh2k2 2m Multiplying (", "6.95) with |E, l, m on the right, we obtain \u00afh2k2 2m \u2212 E k|E, l, m = 0. This means that k|E, l, m can be nonvanishing only if E = \u00afh2k2/2m; so we must be able to write glE(k) as glE(k) =N \u03b4 \u2212 E. (6.97) \u00afh2k2 2m To determine N we go back to our normalization convention (6.87). We obtain E, lm|E, l, m = d3kE, l, m|kk|E, l, m = k2dk d\u03a9k |N|2\u03b4 \u00afh2k2 2m \u2212 E \u00d7 \u03b4 \u00afh2k2 2m \u2212 E Ym\u2217 l ( \u02c6k)Ym l ( \u02c6k) d\u03a9k |N|2\u03b4 \u00afh2k2 2m \u2212 E \u03b4 \u2212 E \u00afh2k2 2m = k2dE dE/dk l ( \u02c6k)Ym l ( \u02c6k) \u00d7 Ym\u2217 = |N|2 mk \u00afh2 \u03b4(E \u2212 E)\u03b4ll \u03b4mm, (6.98) where we have de\ufb01ned E = \u00afh2k2/2m to change k-integration into E-integration. Comparing this with (6.87), we see that N = \u00afh/ mk will suf\ufb01ce. Therefore, we can \ufb01nally write \u221a glE(k) = \u00afh\u221a \u03b4 mk \u00afh2k2 2m \u2212 E ; hence k|E, l, m = \u00afh2k2 2m \u00afh\u221a \u03b4 mk \u2212 E l ( \u02c6k). Ym (6.99) (6.100) 391 6.4 Phase Shifts and Partial Waves From (6.100) we infer that the plane wave state |k can be expressed as a superposition of free spherical wave states with all possible l-values; in particular, |k = \u2211 \u2211 dE|E, l, mE, l, m|k l \u221e \u2211 l=0 m l \u2211 m=\u22121 = |E, l, m \u00afh\u221a mk Ym\u2217 l ( \u02c6k). E=\u00afh2k2/2m (6.101) Because the transverse dimension of the plane wave is in\ufb01nite, we expect that the plane wave must contain all possible values of impact parameter b (semiclassically, the impact parameter b l\u00afh/p). From this point of view it is no surprise that the momentum eigenstates |k, when analyzed in terms of spherical wave states, contain all possible values of l. We have derived the wave function for |E, l, m in momentum space. Next, we consider the corresponding wave function in position space. From wave mechanics, the reader should be familiar with the fact that the wave function for a free spherical wave isjl(kr)Ym l (\u02c6r), wherej l(kr) is the spherical Bessel function of order l (see (3.282a) and also Appendix B). The second solution nl(kr), although it satis\ufb01es the appropriate differential equation, is inadmissible because it is singular at the origin. So we can write x|E, l, m = cljl(kr)Ym l (\u02c6r). (6.102) To determine cl, all we have to do is compare x|k = eik\u00b7x (2\u03c0)3/2 = \u2211 l \u2211 m dEx|E, l, mE, l, m|k = \u2211 \u2211 l m dE cljl(kr)Ym l (\u02c6r) E \u2212 \u00afh2k2 2m \u00afh\u221a \u03b4 mk Ym\u2217 l ( \u02c6k) = \u2211 l (2l + 1) 4\u03c0 Pl( \u02c6k\u00b7\u02c6r) \u00afh\u221a mk cljl(kr), where we have used the addition theorem l ( \u02c6k) = [(2l + 1)/4\u03c0]Pl( \u02c6k\u00b7\u02c6r) in the last step. Now x|k = eik\u00b7x/(2\u03c0)3/2 can also be written as \u03a3mYm l (\u02c6r)Ym\u2217 eik\u00b7x (2\u03c0)3/2 = 1 (2\u03c0)3/2 \u2211 l (2l + 1)iljl(", "kr)Pl( \u02c6k\u00b7\u02c6r), (6.103) (6.104) which can be proved by using the following integral representation for jl(kr): jl(kr) = +1 \u22121 1 2il eikr cos \u03b8Pl(cos \u03b8)d(cos \u03b8). (6.105) Comparing (6.103) with (6.104), we have \u2019 cl = il \u00afh 2mk \u03c0. (6.106) 392 Scattering Theory To summarize, we have k|E, l, m = E \u2212 \u00afh2k2 2m l ( \u02c6k) Ym \u00afh\u221a \u03b4 mk \u2019 x|E, l, m = il \u00afh 2mk \u03c0 jl(kr)Ym l (\u02c6r). (6.107a) (6.107b) These expressions are extremely useful in developing the partial-wave expansion. We conclude this section by applying (6.107a) to a decay process. Suppose a parent particle of spin j disintegrates into two spin zero particles A (spin j) \u2192 B (spin 0) + C (spin 0). The basic Hamiltonian responsible for such a decay process is, in general, very complicated. However, we do know that angular momentum is conserved because the basic Hamiltonian must be rotationally invariant. So the momentum-space wave function for the \ufb01nal state must be of the form (6.107a), with l identi\ufb01ed with the spin of the parent particle. This immediately enables us to compute the angular distribution of the decay product because the momentum-space wave function is nothing more than the probability amplitude for \ufb01nding the decay product with relative momentum direction k. As a concrete example from nuclear physics, let us consider the decay of an excited nucleus, Ne20\u2217: Ne20\u2217 \u2192 O16 + He4. Both O16 and He4 are known to be spinless particles. Suppose the magnetic quantum number of the parent nucleus is \u00b11, relative to some direction z. Then the angular distribution of the decay product is proportional to |Y\u00b11 1 (\u03b8, \u03c6)|2 = (3/8\u03c0) sin2 \u03b8, where (\u03b8, \u03c6) are the polar angles de\ufb01ning the relative direction k of the decay product. On the other hand, if the magnetic quantum number is 0 for a parent nucleus with spin 1, the decay 1(\u03b8, \u03c6)|2 = (3/4\u03c0) cos2 \u03b8. angular distribution varies as |Y0 (6.108) For a general spin orientation we obtain 1 \u2211 m=\u2212l w(m)|Ym l=1 |2. (6.109) For an unpolarized nucleus the various w(m) are all equal, and we obtain an isotropic distribution; this is not surprising because there is no preferred direction if the parent particle is unpolarized. For a higher spin object, the angular distribution of the decay is more involved; the higher the spin of the parent decaying system, the greater the complexity of the angular distribution of the decay products. Quite generally, through a study of the angular distribution of the decay products, it is possible to determine the spin of the parent nucleus. 6.4.2 Partial-Wave Expansion Let us now come back to the case V = 0. We assume that the potential is spherically symmetric, that is, invariant under rotations in three dimensions. It then follows that the transition operator T, which is given by (6.71), commutes with L2 and L. In other words, T is a scalar operator. 393 6.4 Phase Shifts and Partial Waves It is now useful to use the spherical wave basis because the Wigner\u2013Eckart theorem [see (3.481)], applied to a scalar operator, immediately gives E, l, m|T|E, l, m = Tl(E)\u03b4ll \u03b4mm. (6.110) In other words, T is diagonal both in l and in m; furthermore, the (nonvanishing) diagonal element depends on E and l but not on m. This leads to an enormous simpli\ufb01cation, as we will see shortly. Let us now look at the scattering amplitude (6.58): f (k, k) = \u2212 1 4\u03c0 2m \u00afh2 L3k|T|k \u2212\u2192 \u2212 1 4\u03c0 2m \u00afh2 (2\u03c0)3 \u2211 l dE dEk|Elm \u2211 m \u2211 l \u2211 m \u00d7 Elm|T|ElmElm|k = \u2212 1 4\u03c0 2m \u00afh2 (2\u03c0)3 \u00afh2 mk \u2211 \u2211", " l m Tl(E) E=\u00afh2k2/2m l ( \u02c6k)Ym\u2217 Ym l ( \u02c6k) = \u2212 4\u03c02 k \u2211 \u2211 l m Tl(E) E=\u00afh2k2/2m l ( \u02c6k)Ym\u2217 Ym l ( \u02c6k). (6.111) To obtain the angular dependence of the scattering amplitude, let us choose the coordinate system in such a way that k, as usual, is in the positive z-direction. We then have [see (3.259)] \u2019 l ( \u02c6k) = Ym 2l + 1 4\u03c0 \u03b4m0, (6.112) where we have used Pl(1) =1; hence only the terms m = 0 contribute. Taking \u03b8 to be the angle between k and k, we can write \u2019 l ( \u02c6k) = Y0 2l + 1 4\u03c0 Pl(cos \u03b8). It is customary here to de\ufb01ne the partial-wave amplitude fl(k) as follows: fl(k) \u2261 \u2212 \u03c0Tl(E) k. (6.113) (6.114) For (6.111) we then have f (k, k) =f (\u03b8) = \u221e \u2211 l=0 (2l + 1)fl(k)Pl(cos \u03b8), (6.115) where f (\u03b8) still depends on k (or the incident energy) even though k is suppressed. 394 Scattering Theory To appreciate the physical signi\ufb01cance of fl(k), let us study the large-distance behavior of the wave function x|\u03c8(+) given by (6.57). Using the expansion of a plane wave in terms of spherical waves [(6.104)] and noting that (Appendix B) jl(kr) large r\u2212\u2212\u2212\u2192 ei(kr\u2212(l\u03c0/2)) \u2212 e\u2212i(kr\u2212(l\u03c0/2)) 2ikr (il = ei(\u03c0/2)l) (6.116) and that f (\u03b8) is given by (6.115), we have x|\u03c8(+) large r\u2212\u2212\u2212\u2192 1 (2\u03c0)3/2 \u2211 = 1 (2\u03c0)3/2 l eikz + f (\u03b8) eikr r (2l + 1)Pl(cos \u03b8) eikr \u2212 e\u2212i(kr\u2212l\u03c0) 2ikr + \u2211 (2l + 1)fl(k)Pl(cos \u03b8) l eikr r = 1 (2\u03c0)3/2 \u2211 l (2l + 1) Pl 2ik [1 + 2ikfl(k)] eikr r \u2212 e\u2212i(kr\u2212l\u03c0) r. (6.117) The physics of scattering is now clear. When the scatterer is absent, we can analyze the plane wave as the sum of a spherically outgoing wave behaving like eikr/r and a spherically incoming wave behaving like \u2212e\u2212i(kr\u2212l\u03c0)/r for each l. The presence of the scatterer changes only the coef\ufb01cient of the outgoing wave, as follows: 1 \u2192 1 + 2ikfl(k). (6.118) The incoming wave is completely unaffected. 6.4.3 Unitarity and Phase Shifts We now examine the consequences of probability conservation, or unitarity. In a time-independent formulation, the \ufb02ux current density j must satisfy \u2207\u00b7j = \u2212 \u2202|\u03c8|2 \u2202t = 0. (6.119) Let us now consider a spherical surface of very large radius. By Gauss\u2019s theorem, we must have j\u00b7dS = 0. spherical surface (6.120) Physically (6.119) and (6.120) mean that there is no source or sink of particles. The outgoing \ufb02ux must equal the incoming \ufb02ux. Furthermore, because of angular-momentum conservation, this must hold for each partial wave separately. In other words, the coef\ufb01cient of eikr/r must be the same in magnitude as the coef\ufb01cient of e\u2212ikr/r. De\ufb01ning Sl(k) to be Sl(k) \u2261 1 + 2ikfl(k), (6.121) 395 6.4 Phase Shifts and Partial Waves this means [from (6.118)] that |Sl(k)| = 1, (6.122) that is, the most that can happen is a change in the phase of the outgoing wave. Equation (", "6.122) is known as the unitarity relation for the lth partial wave. In a more advanced treatment of scattering, Sl(k) can be regarded as the lth diagonal element of the S operator, which is required to be unitary as a consequence of probability conservation. We thus see that the only change in the wave function at a large distance as a result of scattering is to change the phase of the outgoing wave. Calling this phase 2\u03b4l (the factor of 2 here is conventional), we can write Sl = e2i\u03b4l, with \u03b4l real. It is understood here that \u03b4l is a function of k even though we do not explicitly write \u03b4l as \u03b4l(k). Returning to fl, we can write [from (6.121)] (6.123) or, explicitly in terms of \u03b4l, fl = (Sl \u2212 1) 2ik fl = e2i\u03b4l \u2212 1 2ik = ei\u03b4l sin \u03b4l k = 1 k cot \u03b4l \u2212 ik, (6.124) (6.125) whichever is convenient. For the full scattering amplitude we have f (\u03b8) = \u2211 (2l + 1) e2i\u03b4l \u2212 1 2ik Pl(cos \u03b8) l=0 1 k \u2211 l=0 = (2l + 1)ei\u03b4l sin \u03b4lPl(cos \u03b8) (6.126) with \u03b4l real. This expression for f (\u03b8) rests on the twin principles of rotational invariance and probability conservation. In many books on wave mechanics, (6.126) is obtained by explicitly solving the Schr\u00a8odinger equation with a real, spherically symmetric potential; our derivation of (6.126) may be of interest because it can be generalized to situations when the potential described in the context of nonrelativistic quantum mechanics may fail. The differential cross section d\u03c3/d\u03a9 can be obtained by just taking the modulus squared of (6.126). To obtain the total cross section we have \u03c3tot = | f (\u03b8)|2d\u03a9 = 1 k2 2\u03c0 0 +1 \u22121 d\u03c6 d(cos \u03b8)\u2211 (2l + 1)(2l + 1) \u2211 l l l sin \u03b4l PlPl \u00d7 ei\u03b4l sin \u03b4le\u2212i\u03b4 4\u03c0 k2 \u2211 l = (2l + 1) sin2 \u03b4l. (6.127) We can check the optical theorem (6.60), which we obtained earlier using a more general argument. All we need to do is note from (6.126) that 396 Scattering Theory Imf (\u03b8 = 0) = \u2211 l (2l + 1)Im[ei\u03b4l sin \u03b4l] k Pl(cos \u03b8) \u03b8=0 = \u2211 l (2l + 1) k sin2 \u03b4l, (6.128) which is the same as (6.127) except for 4\u03c0/k. As a function of energy, \u03b4l changes; hence fl(k) changes also. The unitarity relation of (6.122) is a restriction on the manner in which fl can vary. This can be most conveniently seen by drawing an Argand diagram for kfl. We plotkf l in a complex plane, as shown in Figure 6.8, which is self-explanatory if we note from (6.125) that kfl = i 2 + 1 2 e\u2212(i\u03c0/2)+2i\u03b4l. (6.129) Notice that there is a circle of radius 1 2, known as the unitary circle, on whichkf l must lie. We can see many important features from Figure 6.8. Suppose \u03b4l is small. Then fl must stay near the bottom of the circle. It may be positive or negative, but fl is almost purely real: fl = ei\u03b4l sin \u03b4l k (1 + i\u03b4l)\u03b4l k \u03b4l k. (6.130) Im(k f l) 2\u03b4l C 1 2 O P Re(k f l) Fig. 6.8 Argand diagram for kfl. OP is the magnitude of kfl, whileCO and CP are each radii of length 1 angle OCP = 2\u03b4l. 2 on the unitary circle; 397 6.4 Phase Shifts and Partial Waves On the other hand, if \u03b4l is near \u03c0/2, kfl is almost purely imaginary, and the magnitude of kfl is maximal. Under such a condition the lth partial wave may be in resonance, a concept to be discussed in some detail in Section 6.7", ". Note that the maximum partial cross section \u03c3( l ) max = 4\u03c0 k2 (2l + 1) (6.131) is achieved [see (6.127)] when sin2 \u03b4l = 1. 6.4.4 Determination of Phase Shifts Let us now consider how we may actually determine the phase shifts given a potential V. We assume that V vanishes for r > R, R being the range of the potential. Outside (that is, for r > R) the wave function must be that of a free spherical wave. This time, however, there is no reason to exclude nl(r) because the origin is excluded from our consideration. The wave function is therefore a linear combination of jl(kr)Pl(cos \u03b8) and nl(kr)Pl(cos \u03b8) or, equivalently, h(1) are the spherical Hankel functions de\ufb01ned by l Pl, whereh (1) l Pl and h(2) and h(2) l l h(1) l = jl + inl, h(2) l = jl \u2212 inl; (6.132) these have the asymptotic behavior (see Appendix B) h(1) l r large\u2212\u2192 ei(kr\u2212(l\u03c0/2)) ikr, h(2) l r large\u2212\u2192 \u2212 e\u2212i(kr\u2212(l\u03c0/2)) ikr. (6.133) The full-wave function at any r can then be written as: x|\u03c8(+) = 1 (2\u03c0)3/2 \u2211 il(2l + 1)Al(r)Pl(cos \u03b8) (r > R). (6.134) For r > R we have (for the radial wave function) Al = c(1) l h(1) l (kr) +c (2) l h(2) l (kr), (6.135) where the coef\ufb01cient that multiplies Al in (6.134) is chosen so that, for V = 0, Al(r) coincides with jl(kr) everywhere [see (6.104)]. Using (6.133), we can compare the behavior of the wave function for large r given by (6.134) and (6.135) with \u2212 e\u2212i(kr\u2212l\u03c0) 2ikr (2\u03c0)3/2 \u2211 e2i\u03b4l eikr 2ikr (2l + 1)Pl (6.136) 1. l Clearly, we must have c(1) l = 1 2 e2i\u03b4l, c(2) l = 1 2. So the radial wave function for r > R is now written as Al(r) =e i\u03b4l [cos \u03b4ljl(kr) \u2212 sin \u03b4lnl(kr)]. (6.137) (6.138) 398 Scattering Theory Using this, we can evaluate the logarithmic derivative at r = R, that is, just outside the range of the potential, as follows: \u03b2l \u2261 r Al r=R dAl dr l(kR) cos \u03b4l \u2212 n j l(kR) sin \u03b4l jl(kR) cos \u03b4l \u2212 nl(kR) sin \u03b4l = kR, (6.139) where j l(kR) stands for the derivative of jl with respect to kr evaluated at kr = kR. Conversely, knowing the logarithmic derivative at R, we can obtain the phase shift as follows: kRj kRn The problem of determining the phase shift is thus reduced to that of obtaining \u03b2l. l(kR) \u2212 \u03b2ljl(kR) l(kR) \u2212 \u03b2lnl(kR) tan \u03b4l =. (6.140) We now look at the solution to the Schr\u00a8odinger equation for r < R, that is, inside the range of the potential. For a spherically symmetric potential, we can solve the Schr\u00a8odinger equation in three dimensions by looking at the equivalent one-dimensional equation d2ul dr2 + where k2 \u2212 2m \u00afh2 V \u2212 l(l + 1) r2 ul = 0, (6.141) subject to the boundary condition ul = rAl(r) ul|r=0 = 0. (6.142) (6.143) We integrate this one-dimensional Schr\u00a8odinger equation, if necessary numerically, up to r = R, starting at r = 0. In this way we obtain the logarithmic derivative at R. By continuity we must be able to match the logarithmic derivative for the inside and outside solutions at r = R", ": \u03b2l|inside solution = \u03b2l|outside solution, (6.144) where the left-hand side is obtained by integrating the Schr\u00a8odinger equation up to r = R, while the right-hand side is expressible in terms of the phase shifts that characterize the large-distance behavior of the wave function. This means that the phase shifts are obtained simply by substituting \u03b2l for the inside solution into tan \u03b4l [(6.140)]. For an alternative approach it is possible to derive an integral equation for Al(r), from which we can obtain phase shifts (see Problem 6.9 of this chapter). 6.4.5 Hard-Sphere Scattering Let us work out a speci\ufb01c example. We consider scattering by a hard, or rigid, sphere \u221e for for 0 V = r < R r > R. (6.145) 399 6.4 Phase Shifts and Partial Waves In this problem we need not even evaluate \u03b2l (which is actually \u221e). All we need to know is that the wave function must vanish at r = R because the sphere is impenetrable. Therefore, or, from (6.138), or Al(r)|r=R = 0 jl(kR) cos \u03b4l \u2212 nl(kR) sin \u03b4l = 0 tan \u03b4l = jl(kR) nl(kR). (6.146) (6.147) (6.148) Thus the phase shifts are now known for any l. Notice that no approximations have been made so far. To appreciate the physical signi\ufb01cance of the phase shifts, let us consider the l = 0 case (S-wave scattering) speci\ufb01cally. Equation (6.148) becomes, for l = 0 tan \u03b40 = sin kR/kR \u2212 cos kR/kR = \u2212 tan kR, (6.149) or \u03b40 = \u2212kR. The radial wave function (6.138) with ei\u03b40 omitted varies as Al=0(r) \u221d sin kr kr cos \u03b40 + cos kr kr sin \u03b40 = 1 kr sin(kr + \u03b40). (6.150) Therefore, if we plot rAl=0(r) as a function of distance r, we obtain a sinusoidal wave, which is shifted when compared to the free sinusoidal wave by amount R; see Figure 6.9. rAl = 0(r) R = \u2013\u03b40/k r Fig. 6.9 Plot of rAl=0(r) versusr (with the ei\u03b40 factor removed). The dashed curve for V = 0 behaves like sin kr, while the solid curve is for S-wave hard-sphere scattering, shifted by R = \u2212\u03b40/k from the case V = 0. 400 Scattering Theory Let us now study the low and high-energy limits of tan \u03b4l. Low energy means kR small, kR 1. We can then use4 to obtain jl(kr) (kr)l (2l + 1)!! nl(kr) \u2212 (2l \u2212 1)!! (kr)l+1 tan \u03b4l = \u2212(kR)2l+1 {(2l + 1)[(2l \u2212 1)!!] 2}. (6.151) (6.152) It is therefore all right to ignore \u03b4l with l = 0. In other words, we have S-wave scattering only, which is actually expected for almost any \ufb01nite-range potential at low energy. Because \u03b40 = \u2212kR regardless of whether k is large or small, we obtain d\u03c3 d\u03a9 = sin2 \u03b40 k2 R2 for kR 1. It is interesting that the total cross section, given by \u03c3tot = d\u03c3 d\u03a9 d\u03a9 = 4\u03c0R2, (6.153) (6.154) is four times the geometric cross section \u03c0R2. By geometric cross section we mean the area of the disc of radius R that blocks the propagation of the plane wave (and has the same cross section area as that of a hard sphere). Low-energy scattering, of course, means a very large wavelength scattering, and we do not necessarily expect a classically reasonable result. We will consider what happens in the high-energy limit when we discuss the eikonal approximation in the next section. 6.5 Eikonal Approximation This approximation covers a situation in which V(x) varies very little over a distance of order of wavelength \u00af\u03bb (which can be regarded as \u201csmall\u201d). Note that V itself need not be weak as long as E |V|; hence", " the domain of validity here is different from the Born approximation. Under these conditions, the semiclassical path concept becomes applicable, and we replace the exact wave function \u03c8(+) by the semiclassical wave function [see (2.193) and (2.197)], namely, This leads to the Hamilton\u2013Jacobi equation for S, \u03c8(+) \u223c eiS(x)/\u00afh. (\u2207S)2 2m + V = E = \u00afh2k2 2m, 4 Note that (2n + 1)!!\u2261 (2n + 1)(2n \u2212 1)(2n \u2212 3) \u00b7 \u00b7 \u00b7 1. (6.155) (6.156) 401 6.5 Eikonal Approximation z b b x z-direction Scattering region Fig. 6.10 Schematic diagram of eikonal approximation scattering where the classical straight-line trajectory is along the z-direction, |x| = r, andb = |b| is the impact parameter. as discussed in Section 2.4. We propose to compute S from (6.156) by making the further approximation that the classical trajectory is a straight-line path, which should be satisfactory for small de\ufb02ection at high energy.5 Consider the situation depicted in Figure 6.10, where the straight-line trajectory is along the z-direction. Integrating (6.156) we have z \u2212\u221e S \u00afh = b2 + z 2 k2 \u2212 2m \u00afh2 V 1/2 dz + constant. (6.157) The additive constant is to be chosen in such a way that S \u00afh \u2192 kz as V \u2192 0 (6.158) so that the plane wave form for (6.155) is reproduced in this zero-potential limit. We can then write (6.157) as \u2019 S \u00afh = kz + z \u2212\u221e k2 \u2212 2m \u00afh2 V b2 + z 2 \u2212 k = kz \u2212 m \u223c \u00afh2k z V \u2212\u221e dz b2 + z 2 where for E V, we have used \u2019 k2 \u2212 2m \u00afh2 V b2 + z 2 \u223c k \u2212 mV \u00afh2k dz (6.159) at high E = \u00afh2k2/2m. So \u03c8(+)(x) = \u03c8(+)(b + z\u02c6z) 1 (2\u03c0)3/2 ei k z exp \u2212im \u00afh2k z V \u2212\u221e dz. b2 + z 2 (6.160) Though (6.160) does not have the correct asymptotic form appropriate for an incident plus spherical outgoing wave (that is, it is not of form eik\u00b7x + f (\u03b8)(eikr/r) and indeed refers only 5 Needless to say, solution of (6.156) to determine the classical trajectory would be a forbidding task in general. 402 Scattering Theory to motion along the original direction), it can nevertheless still be used in (6.58) to obtain an approximate expression for f (k, k), to wit,6 f (k, k) = \u2212 1 4\u03c0 \u00d7 exp 2m \u00afh2 \u2212 im \u00afh2k d3x e\u2212ik \u00b7x V b2 + z 2 z V \u2212\u221e b2 + z 2 eik\u00b7x dz. (6.161) Note that without the last factor, exp [...], (6.161) is just like the \ufb01rst-order Born amplitude in (6.72). We perform the three-dimensional (d3x ) integration in (6.161) by introducing cylindrical coordinates d3x = b db d \u03c6b dz (see Figure 6.10) and noting that (k \u2212 k)\u00b7x = (k \u2212 k)\u00b7(b + z \u02c6z) \u2212k\u00b7b, (6.162) where we have used k \u22a5 b and (k \u2212 k) \u00b7 \u02c6z \u223c 0(\u03b82), which can be ignored for small de\ufb02ection \u03b8. Without loss of generality we choose scattering to be in the xz-plane and write k\u00b7b = (k sin \u03b8\u02c6x + k cos \u03b8\u02c6z) \u00b7 (b cos \u03c6b \u02c6x + b sin \u03c6b \u02c6y) kb\u03b8cos \u03c6b. (6.163) The expression for f (k, k) becomes f (k, k) = \u2212 1 4\u03c0 \u221e 0 b db 2m \u00afh2 +\u221e 2\u03c0 0 \u2212im \u00afh2k z \u2212\u221e \u00d7 dz V exp \u2212\u221e d\u03c6be\u2212ik", "b\u03b8cos \u03c6b V dz. (6.164) We next use the following identities: 2\u03c0 0 d\u03c6be\u2212ikb\u03b8cos \u03c6b = 2\u03c0J0(kb\u03b8) (6.165) and z=+\u221e z \u2212\u221e +\u221e dz V exp i\u00afh2k m where, of course, the contribution from z = \u2212\u221e on the right-hand side of (6.166) vanishes in the exponent. So, \ufb01nally \u2212im \u00afh2k \u2212im \u00afh2k (6.166) V dz V dz z=\u2212\u221e exp = \u2212\u221e \u2212\u221e z f (k, k) = \u2212ik db bJ0(kb\u03b8)[e2i\u0394(b) \u2212 1], \u221e 0 where \u0394(b) \u2261 +\u221e V \u2212\u221e \u2212m 2k\u00afh2 b2 + z2 dz. (6.167) (6.168) In (6.168) we \ufb01x the impact parameter b and integrate along the straight-line path z, shown in Figure 6.10. There is no contribution from [e2i\u0394(b) \u2212 1] in (6.167) if b is greater than the range of V. 6 We leave behind the \u201cbig box,\u201d and write f (k, k) assuming a continuum normalization. 403 6.5 Eikonal Approximation It can be shown in a straightforward manner that the eikonal approximation satis\ufb01es the optical theorem (6.60). This proof plus some interesting applications, for example, when V is a Gaussian potential \u0394(b) becomes Gaussian in b-space, are discussed in the literature (Gottfried, 1966). See also Problem 6.8 at the end of this chapter, which includes the case where V is a Yukawa potential. 6.5.1 Partial Waves and the Eikonal Approximation The eikonal approximation is valid at high energies (\u03bb range R); hence many partial waves contribute. We may regard l as a continuous variable. As an aside we note the semiclassical argument that l = bk (because angular momentum l\u00afh = bp, where b is the impact parameter and momentum p = \u00afhk). We take lmax = kR; (6.169) then we make the following substitutions in expression (6.126): lmax=kR \u2211 \u2192 k l db, Pl(cos \u03b8) large l small \u03b8 J0(l\u03b8) =J 0(kb\u03b8), \u03b4l \u2192 \u0394(b)|b=l/k, where lmax = kR implies that We have e2i\u03b4l \u2212 1 = e2i\u0394(b) \u2212 1 = 0 for l > lmax. f (\u03b8) \u2192 k 2kb 2ik db (e2i\u0394(b) \u2212 1)J0(kb\u03b8) (6.170) (6.171) = \u2212ik dbbJ0(kb\u03b8)[e2i\u0394(b) \u2212 1]. (6.172) The computation of \u03b4l can be done by using the explicit form for \u0394(b) given by (6.168) (see Problem 6.8 in this chapter). Recall now our discussion of partial waves and the \u201chard sphere\u201d example, from the last section. There, we found that the total cross section was four times the geometric cross section in the low-energy (long wavelength) limit. However, one might conjecture that the geometric cross section is reasonable to expect for high-energy scattering because at high energies the situation might look similar to the semiclassical situation. At high energies many l-values contribute, up to lmax kR, a reasonable assumption. The total cross section is therefore given by \u03c3tot = 4\u03c0 k2 lkR \u2211 (2l + 1) sin2 \u03b4l. l=0 But using (6.148), we have tan2 \u03b4l 1 + tan2 \u03b4l sin2 \u03b4l = = [jl(kR)]2 [jl(kR)]2 + [nl(kR)]2 sin2 kR \u2212, \u03c0l 2 (6.173) (6.174) 404 Scattering Theory where we have used sin jl(kr) \u223c 1 kr nl(kr) \u223c \u2212 1 kr cos kr \u2212 l\u03c0 2 kr \u2212 l\u03c0 2. (6.175) We see that \u03b4l decreases by 90\u25e6 each time l increases by one unit. Thus, for an adjacent pair of partial waves, sin2 \u03b4l + sin2 \u03b4l+1", " = sin2 \u03b4l + sin2(\u03b4l \u2212 \u03c0/2) =sin 2 \u03b4l + cos2 \u03b4l = 1, and with so many l-values contributing to (6.173), it is legitimate to replace sin2 \u03b4l by its average value, 1 2. The number of terms in the l-sum is roughly kR, as is the average of 2l + 1. Putting all the ingredients together, (6.173) becomes 4\u03c0 k2 (kR)2 1 which is not the geometric cross section \u03c0R2 either! To see the origin of the factor of 2, we may split (6.126) into two parts: = 2\u03c0R2, \u03c3tot = (6.176) 2 f (\u03b8) = 1 2ik kR \u2211 (2l + 1)e2i\u03b4lPl(cos \u03b8) + l=0 i 2k kR \u2211 (2l + 1)Pl(cos \u03b8) l=0 = fre\ufb02ection + fshadow. (6.177) | fre\ufb02|2d\u03a9, the orthogonality of the Pl(cos \u03b8) ensures that there is no In evaluating interference amongst contributions from different l, and we obtain the sum of the square of partial-wave contributions: lmax\u2211 (2l + 1)2[Pl(cos \u03b8)]2d(cos \u03b8) = | fre\ufb02|2d\u03a9 = (6.178) max +1 \u03c0l2 k2 = \u03c0R2. 2\u03c0 4k2 \u22121 l=0 Turning our attention to fshad, we note that it is pure imaginary. It is particularly strong in the forward direction because Pl(cos \u03b8) =1 for \u03b8 = 0, and the contributions from various l-values all add up coherently, that is, with the same phase, pure imaginary and positive in our case. We can use the small-angle approximation for Pl to obtain fshad i 2k ik \u2211(2l + 1)J0(l\u03b8) bdbJ0(kb\u03b8) R 0 iRJ1(kR\u03b8) \u03b8. = (6.179) But this is just the formula for Fraunhofer diffraction in optics with a strong peaking near \u03b8 0. Letting \u03be = kR\u03b8 and d\u03be/\u03be = d\u03b8/\u03b8, we can evaluate R2[J1(kR\u03b8)]2 \u03b82 [J1(\u03be)]2 \u03be | fshad|2d\u03a9 = 2\u03c0 d(cos \u03b8) 2\u03c0R2 \u22121 \u221e d\u03be +1 0 \u03c0R2. (6.180) 405 6.6 Low-Energy Scattering and Bound States Finally, the interference between fshad and fre\ufb02 vanishes: Re( f \u2217 shad fre\ufb02) 0 (6.181) because the phase of fre\ufb02 oscillates (2\u03b4l+1 = 2\u03b4l \u2212 \u03c0), approximately averaging to zero, while fshad is pure imaginary. Thus \u03c3tot = \u03c0R2 \u2191 \u03c3re\ufb02 + \u03c0R2. \u2191 \u03c3shad (6.182) The second term (coherent contribution in the forward direction) is called a shadow because for hard-sphere scattering at high energies, waves with impact parameter less than R must be de\ufb02ected. So, just behind the scatterer there must be zero probability for \ufb01nding the particle and a shadow must be created. In terms of wave mechanics, this shadow is due to destructive interference between the original wave (which would be there even if the scatterer were absent) and the newly scattered wave. Thus we need scattering in order to create a shadow. That this shadow amplitude must be pure imaginary may be seen by recalling from (6.117) that the coef\ufb01cient of eikr/2ikr for the lth partial wave behaves like 1 + 2ikfl(k), where the 1 would be present even without the scatterer; hence there must be a positive imaginary term in fl to get cancellation. In fact, this gives a physical interpretation of the optical theorem, which can be checked explicitly. First note that 4\u03c0 k Imf (0) 4\u03c0 k Im[fshad(0)] (6.183) because Im[fre\ufb02(0)] averages to zero due to oscillating phase. Using (6.177), we obtain 4\u03c0 k Imfshad(0) = 4\u03c0 k Im i 2k kR \u2211 (2l", " + 1)Pl(1) = 2\u03c0R2 l=0 (6.184) which is indeed equal to \u03c3tot. 6.6 Low-Energy Scattering and Bound States At low energies \u2013 or, more precisely, when \u03bb = 1/k is comparable to or larger than the range R \u2013 partial waves for higher l are, in general, unimportant. This point may be obvious classically because the particle cannot penetrate the centrifugal barrier; as a result the potential inside has no effect. In terms of quantum mechanics, the effective potential for the lth partial wave is given by Veff = V(r) + \u00afh2 2m l(l + 1) r2 ; (6.185) unless the potential is strong enough to accommodate l = 0 bound states near E 0, the behavior of the radial wave function is largely determined by the centrifugal barrier term, which means that it must resemble jl(kr). More quantitatively, it is possible to estimate the 406 Scattering Theory behavior of the phase shift using the integral equation for the partial wave (see Problem 6.9 of this chapter): ei\u03b4l sin\u03b4l k = \u2212 2m \u00afh2 \u221e 0 jl(kr)V(r)Al(r)r2dr. (6.186) If Al(r) is not too different from jl(kr) and 1/k is much larger than the range of the potential, the right-hand side would vary as k2l; for small \u03b4l, the left-hand side must vary as \u03b4l/k. Hence, the phase shift k goes to zero as \u03b4l \u223c k2l+1 (6.187) for small k. This is known as threshold behavior. It is therefore clear that at low energies with a \ufb01nite range potential, S-wave scattering is important. 6.6.1 Rectangular Well or Barrier To be speci\ufb01c let us consider S-wave scattering by V = V0 = constant 0 for r < R otherwise V0 > 0 V0 < 0 repulsive attractive. (6.188) Many of the features we obtain here are common to more complicated \ufb01nite range potentials. We have already seen that the outside wave function [see (6.138) and (6.150)] must behave like ei\u03b40 [j0(kr) cos \u03b40 \u2212 n0(kr) sin \u03b40] ei\u03b40 sin(kr + \u03b40) kr. (6.189) The inside solution can also easily be obtained for V0 a constant: with k determined by u \u2261 rAl=0(r) \u221d sin kr, E \u2212 V0 = \u00afh2k2 2m, (6.190) (6.191) where we have used the boundary condition u = 0 at r = 0. In other words, the inside wave function is also sinusoidal as long as E > V0. The curvature of the sinusoidal wave is different than in the free-particle case; as a result the wave function can be pushed in (\u03b40 > 0) or pulled out (\u03b40 < 0) depending on whether V0 < 0 (attractive) or V0 > 0 (repulsive), as shown in Figure 6.11. Notice also that (6.190) and (6.191) hold even if V0 > E, provided we understand sin to mean sinh, that is, the wave function behaves like (6.190) u(r) \u221d sinh[\u03bar], where \u00afh2\u03ba2 2m = (V0 \u2212E). (6.191) 407 6.6 Low-Energy Scattering and Bound States No potential u(r) 0 u(r) Larger curvature \u03b40/k 0 R V0 < 0 u(r) V0 > 0 0 R sinh function if V0 > E \u03b40/k r (a) r (b) r (c) Fig. 6.11 Plot of u(r) versusr. (a) ForV = 0 (dashed line). (b) For V0 < 0, \u03b40 > 0 with the wave function (solid line) pushed in. (c) For V0 > 0, \u03b40 < 0 with the wave function (solid line) pulled out. We now concentrate on the attractive case and imagine that the magnitude of V0 is increased. Increased attraction will result in a wave function with a larger curvature. Suppose the attraction is such that the interval [0, R] just accommodates one-fourth cycle of the sinusoidal wave. Working in the low-energy kR 1 limit, the phase shift is now \u03b40 = \u03c0/2, and this", " results in a maximal S-wave cross section for a given k because sin2 \u03b40 is unity. Now increase the well depth V0 even further. Eventually the attraction is so strong that one-half cycle of the sinusoidal wave can be \ufb01tted within the range of the potential. The phase shift \u03b40 is now \u03c0; in other words, the wave function outside R is 180\u25e6 out of phase compared to the free-particle wave function. What is remarkable is that the partial cross section vanishes (sin2 \u03b40 = 0), \u03c3l=0 = 0, despite the very strong attraction of the potential. In addition, if the energy is low enough for l = 0 waves still to be unimportant, we then have an almost perfect transmission of the incident wave. This kind of situation, known as the Ramsauer\u2013Townsend effect, is actually observed experimentally for scattering of electrons by such rare gases as argon, krypton, and xenon. This effect was \ufb01rst observed in 1923 prior to the birth of wave mechanics and was considered to be a great mystery. Note the typical parameters here are R \u223c 2 \u00d7 10\u22128 cm for electron kinetic energy of order 0.1 eV, leading to kR \u223c 0.324. (6.192) 408 Scattering Theory 6.6.2 Zero-Energy Scattering and Bound States Let us consider scattering at extremely low energies (k 0). For r > R and for l = 0, the outside radial wave function satis\ufb01es d2u dr2 = 0. The obvious solution to this equation is u(r) =constant (r \u2212 a), (6.193) (6.194) just a straight line! This can be understood as an in\ufb01nitely long wavelength limit of the usual expression for the outside wave function [see (6.142) and (6.150)], lim k\u21920 sin(kr + \u03b40) = lim k\u21920 sin k r + \u03b40 k, (6.195) which looks like (6.194). We have = k cot k r + \u03b40 k k\u21920\u2192 1 r \u2212 a. (6.196) u u Setting r = 0 [even though at r = 0, (6.194) is not the true wave function], we obtain k cot \u03b40 lim k\u21920 k\u21920\u2192 \u2212 1 a. (6.197) The quantity a is known as the scattering length. The limit of the total cross section as k \u2192 0 is given by [see (6.125)] \u03c3tot = \u03c3l=0 = 4\u03c0 lim k\u21920 2 1 k cot \u03b40 \u2212 ik = 4\u03c0a2. (6.198) Even though a has the same dimension as the range of the potential R, a and R can differ by orders of magnitude. In particular, for an attractive potential, it is possible for the magnitude of the scattering length to be far greater than the range of the potential. To see the physical meaning of a, we note that a is nothing more than the intercept of the outside wave function. For a repulsive potential, a > 0 and is roughly of order of R, as seen in Figure 6.12a. However, for an attractive potential, the intercept is on the negative side (Figure 6.12b). If we increase the attraction, the outside wave function can again cross the r-axis on the positive side (Figure 6.12c). The sign change resulting from increased attraction is related to the development of a bound state. To see this point quantitatively, we note from Figure 6.12c that for a very large and positive, the wave function is essentially \ufb02at for r > R. But (6.194) with a very large is not too different from e\u2212\u03bar with \u03ba essentially zero. Now e\u2212\u03bar with \u03ba 0 is just a bound-state wave function for r > R with energy E in\ufb01nitesimally negative. The inside wave functions (r < R) for the E = 0+ case (scattering with zero kinetic energy) and the E = 0\u2212 case (bound state with in\ufb01nitesimally small binding energy) are essentially the same because in both cases k in sin kr [(6.190)] is determined by \u00afh2k2 2m = E \u2212 V0 |V0| (6.199) with E in\ufb01nitesimal (positive or negative). 409 6.6 Low-Energy Scattering and Bound States u(r) u(r) R 0 r 0 |aa) u(r) 0 R a a > 0 (c) (b) r Fig. 6.12 Plot of u(r) versusr for (a) rep", "ulsive potential, (b) attractive potential, and (c) deeper attraction. The intercept a of the zero-energy outside wave function with the r-axis is shown for each of three cases. Because the inside wave functions are the same for the two physical situations (E = 0+ and E = 0\u2212), we can equate the logarithmic derivative of the bound-state wave function with that of the solution involving zero kinetic-energy scattering, \u2212 \u03bae\u2212\u03bar e\u2212\u03bar = r=R 1 r \u2212 a, r=R or, if R a, The binding energy satis\ufb01es \u03ba 1 a. EBE = \u2212Ebound state = \u00afh2\u03ba2 2m \u00afh2 2ma2, (6.200) (6.201) (6.202) 410 Scattering Theory and we have a relation between scattering length and bound-state energy. This is a remarkable result. To wit, if there is a loosely bound state, we can infer its binding energy by performing scattering experiments near zero kinetic energy, provided a is measured to be large compared with the range R of the potential. This connection between the scattering length and the bound-state energy was \ufb01rst pointed out by Wigner, who attempted to apply (6.202) to np-scattering. Experimentally, the 3S1-state of the np-system has a bound state, that is, the deuteron with EBE = 2.22 MeV. The scattering length is measured to be atriplet = 5.4 \u00d7 10\u221213 cm, (6.203) (6.204) leading to the binding-energy prediction \u00afh2 2\u03bca2 = \u00afh2 mNa2 = mNc2 = (938 MeV) 2 \u00afh mNca 2.1 \u00d7 10\u221214 cm 5.4 \u00d7 10\u221213 cm 2 = 1.4 MeV (6.205) where \u03bc is the reduced mass approximated by mn,p/2. The agreement between experiment and prediction is not too satisfactory. The discrepancy is due to the fact that the inside wave functions are not exactly the same and that atriplet R is not really such a good approximation for the deuteron. A better result can be obtained by keeping the next term in the expansion of k cot \u03b4 as a function of k, where r0 is known as the effective range (see, for example, Preston (1962), p. 23). k cot \u03b40 = \u2212 1 a + 1 2 r0k2, (6.206) 6.6.3 Bound States as Poles of Sl(k) We conclude this section by studying the analytic properties of the amplitude Sl(k) for l = 0. Let us go back to (6.117) and (6.121), where the radial wave function for l = 0 at large distance was found to be proportional to Sl=0(k) eikr r \u2212 e\u2212ikr r. (6.207) Compare this with the wave function for a bound state at large distance, e\u2212\u03bar r The existence of a bound state implies that a nontrivial solution to the Schr\u00a8odinger equation with E < 0 exists only for a particular (discrete) value of \u03ba. We may argue that e\u2212\u03bar/r is like eikr/r, except that k is now purely imaginary. Apart from k being imaginary, the important difference between (6.207) and (6.208) is that in the bound-state case, e\u2212\u03bar/r is present even (6.208). 411 6.6 Low-Energy Scattering and Bound States Im k Pole k > 0 (real) \u03ba Region of physical scattering Fig. 6.13 The complex k-plane with bound-state pole at k = +i\u03ba. Re k without the analogue of the incident wave. Quite generally only the ratio of the coef\ufb01cient of eikr/r to that of e\u2212ikr/r is of physical interest, and this is given by Sl(k). In the bound-state case we can sustain the outgoing wave (with imaginary k) even without an incident wave. So the ratio is \u221e, which means that Sl=0(k), regarded as a function of a complex variable k, has a pole at k = i\u03ba. Thus a bound state implies a pole (which can be shown to be a simple pole) on the positive imaginary axis of the complex k-plane; see Figure 6.13. For k real and positive, we have the region of physical scattering. Here we must require [compare with (6.123)] Sl=0 = e2i\u03b40 with \u03b40 real. Furthermore, as k \u2192 0, k cot \u03b40 has a limiting value \u2212", "1/a (6.197), which is \ufb01nite, so \u03b40 must behave as follows: (6.209) \u03b40 \u2192 0, \u00b1 \u03c0,.... Hence Sl=0 = e2i\u03b40 \u2192 1 as k \u2192 0. Now let us attempt to construct a simple function satisfying: 1. Pole at k = i\u03ba (existence of bound state) |Sl=0| = 1 for k > 0 real (unitarity) 2. Sl=0 = 1 at k = 0 (threshold behavior). 3. The simplest function that satis\ufb01es all three conditions of (6.211) is Sl=0(k) = \u2212k \u2212 i\u03ba k \u2212 i\u03ba. (6.210) (6.211) (6.212) [Editor\u2019s Note: Equation (6.212) is chosen for simplicity rather than as a physically realistic example. For reasonable potentials (not hard spheres!) the phase shift vanishes as k \u2192 \u221e.] An assumption implicit in choosing this form is that there is no other singularity that is important apart from the bound-state pole. We can then use (6.124) to obtain, for fl=0(k), Comparing this with (6.125), fl=0 = Sl=0 \u2212 1 2ik = 1 \u2212\u03ba \u2212 ik. fl=0 = 1 k cot \u03b40 \u2212 ik, (6.213) (6.214) 412 Scattering Theory we see that k cot \u03b40 = \u2212 1 a lim k\u21920 = \u2212\u03ba, (6.215) precisely the relation between bound state and scattering length (6.201). It thus appears that by exploiting unitarity and analyticity of Sl(k) in the k-plane, we may obtain the kind of information that can be secured by solving the Schr\u00a8odinger equation explicitly. This kind of technique can be very useful in problems where the details of the potential are not known. 6.7 Resonance Scattering In atomic, nuclear, and particle physics, we often encounter a situation where the scattering cross section for a given partial wave exhibits a pronounced peak. This section is concerned with the dynamics of such a resonance. We continue to consider a \ufb01nite-ranged potential V(r). The effective potential appropriate for the radial wave function of the lth partial wave is V(r) plus the centrifugal barrier term as given by (6.185). Suppose V(r) itself is attractive. Because the second term, \u00afh2 2m l(l + 1) r2 is repulsive, we have a situation where the effective potential has an attractive well followed by a repulsive barrier at larger distances, as shown in Figure 6.14. Suppose the barrier were in\ufb01nitely high. It would then be possible for particles to be trapped inside, which is another way of saying that we expect bound states, with energy E > 0. They are genuine bound states in the sense that they are eigenstates of the Hamiltonian with de\ufb01nite values of E. In other words, they are stationary states with in\ufb01nite lifetime. Veff (or V for l = 0) Quasi-bound state Barrier r Fig. 6.14 Veff = V(r) + (\u00afh2/2m)[l(l + 1)/r2] versusr. Forl barrier must be due to V itself. = 0 the barrier can be due to (\u00afh2/2m)[l(l + 1)/r2]; for l = 0 413 6.7 Resonance Scattering In the more realistic case of a \ufb01nite barrier, the particle can be trapped inside, but it cannot be trapped forever. Such a trapped state has a \ufb01nite lifetime due to quantummechanical tunneling. In other words, a particle leaks through the barrier to the outside region. Let us call such a state a quasi-bound state because it would be an honest bound state if the barrier were in\ufb01nitely high. The corresponding scattering phase shift \u03b4l rises through the value \u03c0/2 as the incident energy rises through that of the quasi-bound state, and at the same time the corresponding partial-wave cross section passes through its maximum possible value (6.131) 4\u03c0(2l + 1)/k2. [Editor\u2019s Note: Such a sharp rise in the phase shift is, in the time-dependent Schr\u00a8odinger equation, associated with a delay of the emergence of the trapped particles, rather than an unphysical advance, as would be the case for a sharp decrease through \u03c0/2.] It is instructive to verify this point with explicit calculations for some known potential. The result of a numerical calculation shows that a resonance behavior is in fact possible for", " l = 0 with a spherical-well potential. To be speci\ufb01c we show the results for a spherical well with 2mV0R2/\u00afh2 = 5.52 and l = 3 inFigure 6.15. The phase shift ( Figure 6.15b), which is small at extremely low energies, starts increasing rapidly past k \u2248 1/R, and goes through \u03c0/2 at k = 1.41/R. Another very instructive example is provided by a repulsive \u03b4-shell potential that is exactly soluble (see Problem 6.10 in this chapter): 2m \u00afh2 V(r) = \u03b3\u03b4(r \u2212 R). Here resonances are possible for l = 0 because the \u03b4-shell potential itself can trap the particle in the region 0 < r < R. For the case \u03b3 = \u221e, we expect a series of bound states in the region r < R with (6.216) kR = \u03c0, 2\u03c0,... ; (6.217) this is because the radial wave function for l = 0 must vanish not only at r = 0 but also at r = R\u2212 in this case. For the region r > R, we simply have hard-sphere scattering with the S-wave phase shift, given by \u03b40 = \u2212kR. (6.218) With \u03b3 = \u221e, there is no connection between the two problems because the wall at r = R cannot be penetrated. The situation is more interesting with a \ufb01nite barrier, as we can show explicitly. The scattering phase shift exhibits a resonance behavior whenever Eincident Equasi-bound state. (6.219) Moreover, the larger the \u03b3, the sharper the resonance peak. However, away from the resonance \u03b40 looks very much like the hard-sphere phase shift. Thus we have a situation in which a resonance behavior is superimposed on a smoothly behaving background scattering. This serves as a model for neutron-nucleus scattering, where a series of sharp resonance peaks are observed on top of a smoothly varying cross section. 414 Scattering Theory 3 (k) 50 40 30 20 10 0 3 (k) 2.5 2.0 1.5 1.0 0.5 0.0 28 k2 k res R 1.35 1.40 1.45 1.50 /2 k res R (a) (b) kR kR 1.35 1.40 1.45 1.50 Fig. 6.15 Plots of (a) \u03c3l=3 versus k, where at resonance \u03b43(kres) = \u03c0/2, along with the unitarity limit 4\u03c0(7)/k2 from (6.131); and (b) \u03b43(k) versusk. The curves are for a spherical well with 2 mV0R2/\u00afh2 = 5.52. Coming back to our general discussion of resonance scattering, we ask how the scattering amplitudes vary in the vicinity of the resonance energy. If we are to have any connection between \u03c3l being large and the quasi-bound states, \u03b4l must go through \u03c0/2 (or 3\u03c0/2,...) from below, as discussed above. In other words \u03b4l must go through zero from above. Assuming that cot \u03b4l is smoothly varying near the vicinity of resonance, that is, we may attempt to expand \u03b4l as follows: E Er, cot \u03b4l = cot \u03b4l|E=Er 0 \u2212c(E \u2212 Er) +O! (E \u2212 Er)2 \". (6.220) (6.221) 415 6.7 Resonance Scattering This leads to fl(k) = 1 k cot \u03b4l \u2212 ik \u0393/2 = \u2212 k (E \u2212 Er) + where we have de\ufb01ned the width \u0393 by, i\u0393 2 = 1 k 1 [\u2212c(E \u2212 Er) \u2212 i] (6.222) (6.223) d(cot \u03b4l) dE = \u2212c \u2261 \u2212 2 \u0393. E=Er Notice that \u0393 is very small if cot \u03b4l varies rapidly. If a simple resonance dominates the lth partial-wave cross section, we obtain a one-level resonance formula (the Breit\u2013Wigner formula): \u03c3l = 4\u03c0 k2 (2l + 1)(\u0393/2)2 (E \u2212 Er)2 + \u03932/4. (6.224) So it is legitimate to regard \u0393 as the full width at half maximum, provided the resonance is reasonably narrow so that variation in 1/k2 can be ignored. It is worthwhile to demonstrate these concepts with actual scattering measurements, but that is typically dif\ufb01cult. For", " one thing, real world interaction potentials are complicated and not well modeled by something as simple as a \u03b4-function shell or a hard wall spherical well. Furthermore, it usually happens that many different partial waves overlap, so that a sophisticated data analysis is required to separate them. One good example, however, of an isolated partial wave giving a narrow resonance for an interaction that is reasonably well modeled by a spherical well, is the scattering of positive \u03c0-mesons from protons. Pions and protons are strongly interacting elementary particles, with \u201csizes\u201d on the order of 1 fm \u2261 10\u221215 m. The force between them is attractive, and roughly constant when the particles overlap, but falls off rapidly outside this range. It is therefore perhaps not unreasonable that one might observe a relatively narrow resonance in certain partial waves. Figure 6.16 plots the data7 with \u03c0+p elastic scattering, for \u03c0+ center-of-mass momenta in the region of a few hundred MeV/c. There is a clear peak in the cross section, wider than, but otherwise similar to, Figure 6.15. One indication that this is in fact a resonance, is the comparison to the unitary limit, but \ufb01rst we need to reconsider the case when the scattering particles are not spinless. The factor 2l + 1 in (6.131) comes from summing over the \ufb01nal states with the same l but different m. If the scattering particles have spin, then we would replace this factor with 2j + 1, where j is the total angular-momentum quantum number of the appropriate partial wave. However, we also need to consider the spins s1 and s2 of the scattering particles 7 The data are available from http://pdg.lbl.gov. In fact, this \ufb01gure plots both the elastic and total cross sections on the same plot. There are many more total cross section data points, but as there are no reactions other than \u03c0+p \u2192 \u03c0+p in this energy range, the cross sections agree. 416 Scattering Theory ) 250 200 150 100 50 0 j = 3/2 j = 1/2 j = 5/2 0.15 0.20 0.25 0.30 0.35 0.40 Pion CM Momentum (GeV/c) Fig. 6.16 Elastic cross section data for the elastic scattering reaction \u03c0+p \u2192 \u03c0+p, as a function of momentum in the center-of-mass. Also plotted are the unitary limits for different possibilities of the dominant partial wave. Compare to the calculated cross section for a hard walled square well in Figure 6.15. There is a clear suggestion that this is a resonance with total angular momentum j = 3/2. when considering the initial scattering state. Averaging over all the possibilities gives an additional factor of 1/[(2s1 + 1)(2s2 + 1)]. Therefore (6.131) becomes \u03c3( l ) max = 4\u03c0 k2 g where g \u2261 2j + 1 (2s1 + 1)(2s2 + 1) (6.225) is usually called the \u201cstatistical factor.\u201d More detailed treatments can be found in nearly any textbook on nuclear physics, for example Section VIII.10 in the classic text Blatt and Weisskopf (1952). For \u03c0+p scattering, the \u03c0+ is spinless and the proton has spin 1 2, sog = (2j + 1)/2 and j can take on any positive half-integer value. Figure 6.16 includes curves using (6.225) for different values of j. The agreement with the maximum cross section for j = 3/2 strongly suggests that this is a spin 3 2 resonance. Indeed, a detailed analysis of the data including relative l = 0 and l = 1 phase differences shows that j = 3/2. This resonances is known as the \u0394(1232), where 1232 MeV/c2 is the invariant mass of the \u03c0+p system at the peak of the cross section. 6.8 Symmetry Considerations in Scattering Let us consider the scattering of two identical spinless charged particles via some central potential, such as the Coulomb potential.8 The spatial part of the wave function must now be symmetric, so the asymptotic wave function must look like 8 For the student unfamiliar with the elements of permutation symmetry with identical particles, see Chapter 7 of this textbook. 417 6.8 Symmetry Considerations in Scattering eik\u00b7x + e\u2212ik\u00b7x + [ f (\u03b8) +f (\u03c0 \u2212 \u03b8)] eikr r, (6.226) where x = x1 \u2212 x2 is the relative position vector between the two particles 1 and 2. This results in a differential cross section, d", "\u03c3 d\u03a9 = | f (\u03b8) +f (\u03c0 \u2212 \u03b8)|2 = | f (\u03b8)|2 + | f (\u03c0 \u2212 \u03b8)|2 + 2Re[ f (\u03b8)f\u2217(\u03c0 \u2212 \u03b8)]. (6.227) The cross section is enhanced through constructive interference at \u03b8 \u03c0/2. \u2212 spin 1 In contrast, for spin 1 2 2 scattering with unpolarized beam and V independent of spin, we have the spin-singlet scattering going with space-symmetrical function and the spin-triplet scattering going with space-antisymmetrical wave function (see Section 7.3). If the initial beam is unpolarized, we have the statistical contribution 1 4 for spin singlet and 3 4 for spin triplet; hence d\u03c3 d\u03a9 = | f (\u03b8) +f (\u03c0 \u2212 \u03b8)|2 + | f (\u03b8) \u2212 f (\u03c0 \u2212 \u03b8)|2 3 4 1 4 = | f (\u03b8)|2 + | f (\u03c0 \u2212 \u03b8)|2 \u2212 Re[f (\u03b8)f\u2217(\u03c0 \u2212 \u03b8)]. (6.228) In other words, we expect destructive interference at \u03b8 \u03c0/2. This has, in fact, been observed. Now consider symmetries other than exchange symmetry. Suppose V and H0 are both invariant under some symmetry operation. We may ask what this implies for the matrix element of T or for the scattering amplitude f (k, k). If the symmetry operator is unitary (for example, rotation and parity), everything is quite straightforward. Using the explicit form of T as given by (6.32), we see that UH0U \u2020 = H0, UVU \u2020 = V implies that T is also invariant under U, that is, UTU \u2020 = T. We de\ufb01ne Then | \u02dck \u2261U|k, | \u02dck \u2261U|k. \u02dck|T| \u02dck = k|U \u2020UTU \u2020U|k = k|T|k. (6.229) (6.230) (6.231) (6.232) As an example, we consider the speci\ufb01c case where U stands for the parity operator \u03c0|k = | \u2212k, \u03c0| \u2212k = |k. Thus invariance of H0 and V under parity would mean \u2212k|T| \u2212k = k|T|k. Pictorially, we have the situation illustrated in Figure 6.17a. (6.233) (6.234) 418 Scattering Theory k \u2212k k \u2212k (a) k k Rotated k k (b) Fig. 6.17 (a) Equality of T matrix elements between k \u2192 k and \u2212k \u2192 \u2212k. (b) Equality of T matrix elements under rotation. We exploited the consequence of angular-momentum conservation when we developed the method of partial waves. The fact that T is diagonal in the |Elm representation is a direct consequence of T being invariant under rotation. Notice also that k|T|k depends only on the relative orientation of k and k, as depicted in Figure 6.17b. When the symmetry operation is antiunitary (as in time reversal), we must be more careful. First, we note that the requirement that V as well as H0 be invariant under time reversal invariance requires that This is because the antiunitary operator changes \u0398T\u0398\u22121 = T\u2020. 1 E \u2212 H0 + i\u03b5 into 1 E \u2212 H0 \u2212 i\u03b5 in (6.32). We also recall that for an antiunitary operator [see (4.114)], \u03b2|\u03b1 = \u02dc\u03b1| \u02dc\u03b2, | \u02dc\u03b1 \u2261\u0398| \u03b1 and | \u02dc\u03b2 \u2261\u0398| \u03b2. |\u03b1 = T|k, \u03b2| = k|; where Let us consider then | \u02dc\u03b1 = \u0398T|k = \u0398T\u0398\u22121\u0398|k = T \u2020| \u2212k | \u02dc\u03b2 = \u0398|k = | \u2212k. As a result (6.237) becomes k|T|k = \u2212k|T| \u2212k. (6.235) (6.236) (6.237) (6.238) (6.239) (6.240) (6.241) Notice that the initial and \ufb01nal momenta are interchanged, in addition to the fact that the directions of the momenta have been reversed. 419 6.9 Inelastic Electron-Atom Scattering It is also interesting to combine the requirements of time reversal [(6.241)] and parity [(6.234)]: k|T|k under \u0398", " = \u2212k|T| \u2212k under \u03c0 = k|T|k; that is, from (6.58) and (6.70) we have which results in f (k, k) =f (k, k), d\u03c3 d\u03a9 (k \u2192 k) = d\u03c3 d\u03a9 (k \u2192 k). (6.242) (6.243) (6.244) Equation (6.244) is known as detailed balance. It is more interesting to look at the analogue of (6.242) when we have spin. Here we may characterize the initial free-particle ket by |k, ms, and we exploit (4.182) for the time reversal portion: k, m s |T|k, ms = i\u22122ms+2ms = i\u22122ms+2ms \u2212k, \u2212ms|T| \u2212k, \u2212m k, \u2212ms|T|k, \u2212m. s s (6.245) For unpolarized initial states, we sum over the initial spin states and divide by (2s + 1); if the \ufb01nal polarization is not observed, we must sum over \ufb01nal states. We then obtain detailed balance in the form d\u03c3 d\u03a9 (k \u2192 k) = d\u03c3 d\u03a9 (k \u2192 k), (6.246) where we understand the bar on the top of d\u03c3/d\u03a9 in (6.246) to mean that we average over the initial spin states and sum over the \ufb01nal spin states. 6.9 Inelastic Electron-Atom Scattering Let us consider the interactions of electron beams with atoms assumed to be in their ground states. The incident electron may be scattered elastically with \ufb01nal atoms unexcited: e\u2212 + atom (ground state) \u2192 e\u2212 + atom (ground state). (6.247) This is an example of elastic scattering. To the extent that the atom can be regarded as in\ufb01nitely heavy, the kinetic energy of the electron does not change. It is also possible for the target atom to be excited: e\u2212 + atom (ground state) \u2192 e\u2212 + atom (excited state). (6.248) In this case we talk about inelastic scattering because the kinetic energy of the \ufb01nal outgoing electron is now less than that of the initial incoming electron, the difference being used to excite the target atom. 420 Scattering Theory The initial ket of the electron plus the atomic system is written as |k, 0 (6.249) where k refers to the wave vector of the incident electron and 0 stands for the atomic ground state. Strictly speaking (6.249) should be understood as the direct product of the incident electron ket |k and the ground-state atomic ket |0. The corresponding wave function is 1 L3/2 eik\u00b7x\u03c80(x1, x2,..., xz) (6.250) where we use the box normalization for the plane wave. We may be interested in a \ufb01nal state electron with a de\ufb01nite wave vector k. The \ufb01nal state ket and the corresponding wave function are |k, n and 1 L3/2 eik\u00b7x\u03c8n(x1,..., xz), (6.251) where n = 0 for elastic scattering and n = 0 for inelastic scattering. Assuming that time-dependent perturbation theory is applicable, we can immediately write the differential cross section, as in the previous section: d\u03c3 d\u03a9 (0 \u2192 n) = = 1 (\u00afhk/meL3) 2\u03c0 \u00afh |kn|V|k0|2 3 L 2\u03c0 kme \u00afh2 k k L6 1 4\u03c0 k, n|V|k, 0 2. 2me \u00afh2 (6.252) Everything is similar, including the cancellation of terms such as L3, with one important exception: k \u2261 |k| is not, in general, equal to k \u2261 |k| for inelastic scattering. The next question is, what V is appropriate for this problem? The incident electron can interact with the nucleus, assumed to be situated at the origin; it can also interact with each of the atomic electrons. So V is to be written as V = \u2212 Ze2 r +\u2211 i e2 |x \u2212 xi|. (6.253) Here complications may arise because of the identity of the incident electron with one of the atomic electrons; to treat this rigorously is a nontrivial task. Fortunately, for a relatively fast electron we can legitimately ignore the question of identity; this is because there is little overlap between the bound-state electron and the incident electron in momentum space. We must evaluate the matrix element k", ", n|V|k0, which, when explicitly written, is kn|V|k0 = = 1 L3 1 L3 d3xeiq\u00b7xn| \u2212 Ze2 r +\u2211 i e2 |x \u2212 xi| |0 d3xeiq\u00b7x z \u220f i d3xi\u03c8\u2217 n(x1,..., xz) \u2212 Ze2 r +\u2211 i e2 |x \u2212 xi| \u00d7 \u03c80(x1,..., xz) with q \u2261 k \u2212 k. (6.254) 421 6.9 Inelastic Electron-Atom Scattering Let us see how to evaluate the matrix element of the \ufb01rst term, \u2212Ze2/r, where r actually means |x|. First we note that this is a potential between the incident electron and the nucleus, which is independent of the atomic electron coordinates. So it can be taken outside the integration in (6.254); we simply obtain z \u220f i d3xi n|0 = \u03b4n0 (6.255) for the remainder. In other words, this term contributes only to the elastic scattering case, where the target atom remains unexcited. In the elastic case we must still integrate eiq\u00b7x/r with respect to x, which amounts to taking the Fourier transform of the Coulomb potential. This can readily be done because we already evaluated the Fourier transform of the Yukawa potential; see (6.78). Hence d3x eiq\u00b7x r = lim \u03bc\u21920 d3xeiq\u00b7x\u2212\u03bcr r = 4\u03c0 q2. (6.256) As for the second term in (6.254), we can evaluate the Fourier transform of 1/|x \u2212 xi|. We can accomplish this by shifting the coordinate variables x \u2192 x + xi: \u2211 i d3xeiq\u00b7x |x \u2212 xi| = \u2211 i d3xeiq\u00b7(x+xi) |x| = 4\u03c0 q2 \u2211 i eiq\u00b7xi. (6.257) Notice that this is just the Fourier transform of the Coulomb potential multiplied by the Fourier transform of the electron density due to the atomic electrons situated at xi: \u03c1atom(x) = \u2211 \u03b4(3)(x \u2212 xi). i (6.258) We customarily de\ufb01ne the form factor Fn(q) for excitation |0 to |n as follows: ZFn(q) \u2261 n|\u2211 eiq\u00b7xi |0, (6.259) i which is made of coherent, tions from the various electrons. Notice that as q \u2192 0, we have in the sense of de\ufb01nite phase relationships, contribu- 1 Z n|\u2211 eiq\u00b7xi|0 \u21921 i for n = 0; hence the form factor approaches unity in the elastic scattering case. For n = 0 (inelastic scattering), Fn(q) \u2192 0 asq \u2192 0 by orthogonality between |n and |0. We can then write the matrix element in (6.254) as d3xeiq\u00b7xn| \u2212 Ze2 r +\u2211 i e2 |x \u2212 xi| |0 = 4\u03c0Ze2 q2 [\u2212\u03b4n0 + Fn(q)]. (6.260) 422 Scattering Theory We are \ufb01nally in a position to write the differential cross section for inelastic (or elastic) scattering of electrons by atoms: d\u03c3 d\u03a9 (0 \u2192 n) = k k 1 4\u03c0 2 [\u2212\u03b4n0 + Fn(q)] 2me \u00afh2 4\u03c0Ze2 q2 = 4m2 e \u00afh4 (Ze2)2 q4 k k | \u2212 \u03b4n0 + Fn(q)|2. (6.261) For inelastic scattering the \u03b4n0-term does not contribute, and it is customary to write the differential cross section in terms of the Bohr radius, as follows: a0 = d\u03c3 d\u03a9 (0 \u2192 n) = 4Z2a2 0 \u00afh2 e2me, k k 1 (qa0)4 |Fn(q)|2. Quite often d\u03c3/dq is used in place of d\u03c3/d\u03a9; using q2 = |k \u2212 k|2 = k2 + k 2 \u2212 2kk cos \u03b8 (6.262) (6.263) (6.264) = d\u03c3 d\u03a9. and dq = \u2212d(cos \u03b8)kk/q, we can write d\u03c3 dq 2\u03c0q kk The inelastic cross section we have obtained can be used to discuss stopping power \u2013 the energy", " loss of a charged particle as it goes through matter. A number of people, including H. A. Bethe and F. Bloch, have discussed the quantum-mechanical derivation of stopping power from the point of view of the inelastic scattering cross section. We are interested in the energy loss of a charged particle per unit length traversed by the incident charged particle. The collision rate per unit length is N\u03c3, where N is the number of atoms per unit volume; at each collision process the energy lost by the charged particle is En \u2212 E0. So dE/dx is written as (6.265) (0 \u2192 n) dq dE dx = N\u2211 (En \u2212 E0) n = N\u2211 (En \u2212 E0) n d\u03c3 dq 4Z2 a2 0 = 8\u03c0N k2a2 0 \u2211 (En \u2212 E0) n qmax qmin qmax qmin k k 1 q4 2\u03c0q kk |Fn(q)|2dq n| z \u2211 i=1 eiq\u00b7xi|0 2 dq q3. (6.266) There are many papers written on how to evaluate the sum in (6.266).9 The upshot of all this is to justify quantum mechanically Bohr\u2019s 1913 formula for stopping power, 4\u03c0NZe4 mev2 2mev2 I (6.267) dE dx ln =, 9 For a relatively elementary discussion, see Gottfried (1966) and Bethe and Jackiw (1968). 423 6.9 Inelastic Electron-Atom Scattering where I is a semiempirical parameter related to the average excitation energy En \u2212 E0. If the charged particle has electric charge \u00b1 ze, we just replace Ze4 by z2Ze4. It is also important to note that even if the projectile is not an electron, the me that appears in (6.267) is still the electron mass, not the mass of the charged particle. So the energy loss is dependent on the charge and the velocity of the projectile but is independent of the mass of the projectile. This has an important application to the detection of charged particles. Quantum mechanically, we view the energy loss of a charged particle as a series of inelastic scattering processes. At each interaction between the charged particle and an atom, we may imagine that a \u201cmeasurement\u201d of the position of the charged particle is made. We may wonder why particle tracks in such media as cloud chambers and nuclear emulsions are nearly straight. The reason is that the differential cross section (6.263) is sharply peaked at small q; in an overwhelming number of collisions, the \ufb01nal direction of momentum is nearly the same as the incident electron due to the rapid falloff of q\u22124 and Fn(q) for large q. 6.9.1 Nuclear Form Factor The excitation of atoms due to inelastic scattering is important for q \u223c 109 cm\u22121 to 1010 cm\u22121. Ifq is too large, the contributions due to F0(q) or Fn(q) drop off very rapidly. At extremely high q, where q is now of order 1/Rnucleus \u223c 1012 cm\u22121, the structure of the nucleus becomes important. The Coulomb potential due to the point nucleus must now be replaced by a Coulomb potential due to an extended object, \u2212 Ze2 r \u2192 \u2212Ze2 d3xN(r) |x \u2212 x|, where N(r) is a nuclear charge distribution, normalized so that d3xN(r) =1. The pointlike nucleus can now be regarded as a special case, with N(r) = \u03b4(3)(r). (6.268) (6.269) (6.270) We can evaluate the Fourier transform of the right-hand side of (6.268) in analogy with (6.256) as follows: Ze2 d3x d3xeiq\u00b7xN(r) |x \u2212 x| = Ze2 d3xeiq\u00b7x N(r) d3xeiq\u00b7x r = Ze2 4\u03c0 q2 Fnucleus(q) where we have shifted the coordinates x \u2192 x + x in the \ufb01rst step and Fnucleus \u2261 d3xeiq\u00b7xN(r). (6.271) (6.272) 424 Scattering Theory We thus obtain the deviation from the Rutherford formula due to the \ufb01nite size of the nucleus, d\u03c3 d\u03a9 = d\u03c3 d\u03a9 |F(q)|2, Rutherford (6.273) where (d\u03c3/d\u03a9)Rutherford is the differential cross section for the electric scattering of electrons by a pointlike nucleus of charge Z|e|. For small q we have Fnucle", "us(q) = d3x q2r2( \u02c6q\u00b7\u02c6r)2 + \u00b7 \u00b7 \u00b7 N(r) 1 + iq\u00b7x \u2212 1 2 = 1 \u2212 1 6 q2r2nucleus + \u00b7 \u00b7 \u00b7. (6.274) The q \u00b7 x-term vanishes because of spherical symmetry, and in the q2-term we have used the fact that the angular average of cos2 \u03b8 (where \u03b8 is the angle between \u02c6q and \u02c6r) is just 1 3 : +1 \u22121 1 2 d(cos \u03b8) cos2 \u03b8 = 1 3. (6.275) The quantity r2nucleus is known as the mean square radius of the nucleus. In this way it is possible to \u201cmeasure\u201d the size of the nucleus and also of the proton, as done by R. Hofstadter and coworkers. In the proton case the spin (magnetic moment) effect is also important. Problems 6.1 Consider scattering in one dimension x from a potential V(x) localized near x = 0. 2\u03c0. The initial state is a plane wave coming from the left, that is \u03c6(x) \u2261 x|i = eikx/ a. Find the scattering Green\u2019s function G(x, x ), de\ufb01ned in one dimension analo- \u221a gously with (6.37), for G+(x, x ). b. For the case of an attractive \u03b4-function potential V(x) =\u2212 \u03b3\u00afh2\u03b4(x)/2m, with \u03b3 > 0, use the Lippman\u2013Schwinger equation to \ufb01nd the outgoing wave function \u03c8(x) \u2261 x|\u03c8(+). c. Determine the transmission and re\ufb02ection coef\ufb01cients T(k) and R(k), de\ufb01ned as \u03c8(x) =T(k) \u03c6(x) for x > 0 and \u03c8(x) = \u03c6(x) +R (k) e\u2212ikx \u221a 2\u03c0 for x < 0. Show that |T|2 + |R|2 = 1, as must be the case. d. Con\ufb01rm that you get the same result by matching right and left going waves on the left with a right going wave on the right at x = 0. e. We showed in Problem 2.29 that this potential has one, and only one, bound state. Show that your results for T(k) and R(k) have bound-state poles at the expected positions when k is treated as a complex variable. 425 Problems 6.2 Prove \u03c3tot m2 \u03c0\u00afh4 d3x d3xV(r)V(r) sin2 k|x \u2212 x| k2|x \u2212 x|2 in each of the following ways. a. By integrating the differential cross section computed using the \ufb01rst-order Born approximation. b. By applying the optical theorem to the forward-scattering amplitude in the second-order Born approximation. [Note that f (0) is real if the \ufb01rst-order Born approximation is used.] 6.3 Estimate the radius of the 40Ca nucleus from the data in Figure 6.6 and compare to that expected from the empirical value \u22481.4A1/3 fm, where A is the nuclear mass number. Check the validity of using the \ufb01rst-order Born approximation for these data. 6.4 Consider a potential V = 0 for r > R, V = V0 = constant for r < R, where V0 may be positive or negative. Using the method of partial waves, show that for |V0| E = \u00afh2k2/2m and kR 1 the differential cross section is isotropic and that the total cross section is given by \u03c3tot = 16\u03c0 9 m2V2 0R6 \u00afh4. Suppose the energy is raised slightly. Show that the angular distribution can then be written as d\u03c3 d\u03a9 = A + B cos \u03b8. Obtain an approximate expression for B/A. 6.5 A spinless particle is scattered by a weak Yukawa potential V = V0e\u2212\u03bcr \u03bcr where \u03bc > 0 but V0 can be positive or negative. It was shown in the text that the \ufb01rst-order Born amplitude is given by f (1)(\u03b8) =\u2212 2mV0 \u00afh2\u03bc 1 [2k2(1 \u2212 cos \u03b8) + \u03bc2]. a. Using f (1)(\u03b8) and assuming |\u03b4l| 1, obtain an expression for \u03b4l in terms of", " a Legendre function of the second kind, Ql(\u03b6) = 1 \u22121 1 2 Pl(\u03b6) \u03b6 \u2212 \u03b6 d\u03b6. 426 Scattering Theory b. Use the expansion formula Ql(\u03b6) = l2l + 1) \u00d7 1 \u03b6l+1 + (l + 1)(l + 2) 2(2l + 3) 1 \u03b6l+3 + (l + 1)(l + 2)(l + 3)(l + 4) 2 \u00b7 4 \u00b7 (2l + 3)(2l + 5) 1 \u03b6l+5 + \u00b7 \u00b7 \u00b7 - (|\u03b6| > 1) to prove each assertion. (i) \u03b4l is negative (positive) when the potential is repulsive (attractive). (ii) When the de Broglie wavelength is much longer than the range of the potential, \u03b4l is proportional to k2l+1. Find the proportionality constant. 6.6 Check explicitly the x \u2212 px uncertainty relation for the ground state of a particle con\ufb01ned inside a hard sphere: V = \u221e for r > a, V = 0 forr < a. (Hint: Take advantage of spherical symmetry.) 6.7 Consider the scattering of a particle by an impenetrable sphere V(r) = for r > a 0 \u221e for r < a. a. Derive an expression for the s-wave (l = 0) phase shift. (You need not know the detailed properties of the spherical Bessel functions to be able to do this simple problem!) b. What is the total cross section \u03c3[\u03c3 = (d\u03c3/d\u03a9)d\u03a9] in the extreme low-energy limit k \u2192 0? Compare your answer with the geometric cross section \u03c0a2. You may assume without proof: d\u03c3 d\u03a9 = | f (\u03b8)|2, \u221e \u2211 f (\u03b8) = 1 k l=0 (2l + 1)ei\u03b4l sin \u03b4lPl(cos \u03b8). 6.8 Use \u03b4l = \u0394(b)|b=l/k to obtain the phase shift \u03b4l for scattering at high energies by (a) the Gaussian potential, V = V0 exp(\u2212r2/a2), and (b) the Yukawa potential, V = V0 exp(\u2212\u03bcr)/\u03bcr. Verify the assertion that \u03b4l goes to zero very rapidly with increasing l (k \ufb01xed) for l kR, where R is the \u201crange\u201d of the potential. [The formula for \u0394(b) is given in (6.168).] It is useful to realize that \u221e 1 e\u2212ax (x2 \u2212 1)1/2 dx = K0(x) where K0(x) is the modi\ufb01ed Bessel function of the second kind, of order zero. 6.9 a. Prove \u00afh2 2m x| 1 E \u2212 H0 + i\u03b5 |x = \u2212ik\u2211 \u2211 l m l (\u02c6r)Ym\u2217 Ym l (\u02c6r)jl(kr<)h(1) l (kr>) where r< (r>) stands for the smaller (larger) of r and r. 427 Problems b. For spherically symmetric potentials, the Lippmann\u2013Schwinger equation can be written for spherical waves: |Elm(+) = |Elm + 1 E \u2212 H0 + i\u03b5 V|Elm(+). Using (a), show that this equation, written in the x-representation, leads to an equation for the radial function, Al(k; r), as follows: Al(k; r) =j l(kr) \u2212 2mik \u00afh2 \u221e \u00d7 0 jl(kr<)h(1) l (kr>)V(r)Al(k; r)r 2dr. By taking r very large, also obtain sin \u03b4l k 2m \u00afh2 fl(k) =e i\u03b4l = \u2212 \u221e jl(kr)Al(k; r)V(r)r2dr. 0 6.10 Consider scattering by a repulsive \u03b4-shell potential: 2m \u00afh2 V(r) = \u03b3\u03b4(r \u2212 R) (\u03b3 > 0). a. Set up an equation that determines the s-wave phase shift \u03b40 as a function of k (E = \u00afh2k2/2m). b. Assume now that \u03b3 is very large, \u03b3 1 R, k. Show that if tan kR is not close to zero, the s-wave phase shift resembles the hardsphere result discussed in the text.", " Show also that for tan kR close to (but not exactly equal to) zero, resonance behavior is possible; that is, cot \u03b40 goes through zero from the positive side as k increases. Determine approximately the positions of the resonances keeping terms of order 1/\u03b3; compare them with the bound-state energies for a particle con\ufb01ned inside a spherical wall of the same radius, V = 0, r < R; V = \u221e, r > R. Also obtain an approximate expression for the resonance width \u0393 de\ufb01ned by \u0393 = \u22122 [d(cot \u03b40)/dE]|E=Er and notice, in particular, that the resonances become extremely sharp as \u03b3 becomes large. (Note: For a different, more sophisticated approach to this problem see Gottfried (1966), pp. 131\u2013141, who discusses the analytic properties of the Dl-function de\ufb01ned by Al = jl/Dl.) 428 Scattering Theory 6.11 A spinless particle is scattered by a time-dependent potential V (r, t) =V (r) cos \u03c9t. Show that if the potential is treated to \ufb01rst order in the transition amplitude, the energy of the scattered particle is increased or decreased by \u00afh\u03c9. Obtaind \u03c3/d\u03a9. Discuss qualitatively what happens if the higher-order terms are taken into account. 6.12 Show that the differential cross section for the elastic scattering of a fast positron by the ground state of the hydrogen atom is given by * d\u03c3 d\u03a9 = 4m2e4 \u00afh4q4 1 \u2212 16 [4 + (qa0)2]2 + 2. 6.13 Write a computer program to reproduce Figure 6.15. Also include a plot showing the relative sizes of the energy, well depth, and effective potential, that is, the analogue of Figure 6.14 for this problem. 6.14 Let the energy of a particle moving in a central \ufb01eld be E(J1J2J3), where (J1, J2, J3) are the three action variables. How does the functional form of E specialize for the Coulomb potential? Using the recipe of the action-angle method, compare the degeneracy of the central \ufb01eld and the Coulomb problems and relate it to the vector A. If the Hamiltonian is H = p2 2\u03bc + V(r) +F(A 2), how are these statements changed? Describe the corresponding degeneracies of the central \ufb01eld and Coulomb problems in quantum theory in terms of the usual quantum numbers (n, l, m) and also in terms of the quantum numbers (k, m, n). Here the second set, (k, m, n), labels the wave functions D k mn(\u03b1\u03b2\u03b3). How are the wave functions D k mn(\u03b1\u03b2\u03b3) related to Laguerre times spherical harmonics? 7 Identical Particles This chapter is devoted to a discussion of some striking quantum-mechanical effects arising from the identity of particles. First we present a suitable formalism and the way that nature deals with what appears to be an arbitrary choice. We then consider some applications to atoms more complex than hydrogen like atoms. Next we generalize to systems with many identical particles, and discuss two different formalisms for calculations. Lastly, we will cover one concrete example of a many-particle quantum-mechanical \ufb01eld theory, namely quantizing the electromagnetic \ufb01eld. 7.1 Permutation Symmetry In classical physics it is possible to keep track of individual particles even though they may look alike. When we have particle 1 and particle 2 considered as a system, we can, in principle, follow the trajectory of particle 1 and that of particle 2 separately at each instant of time. For bookkeeping purposes, you may color one of them blue and the other red and then examine how the red particle moves and how the blue particle moves as time passes. In quantum mechanics, however, identical particles are truly indistinguishable. This is because we cannot specify more than a complete set of commuting observables for each of the particles; in particular, we cannot label the particle by coloring it blue. Nor can we follow the trajectory because that would entail a position measurement at each instant of time, which necessarily disturbs the system; in particular the two situations (a) and (b) shown in Figure 7.1 cannot be distinguished \u2013 not even in principle. For simplicity consider just two particles. Suppose one of the particles, which we call particle 1, is characterized by |k, where k is a collective index for a complete set of observables. Likewise, we call the ket of the remaining particle |k. The state ket for the two particles can be written in product form, |k|k,", " (7.1) where it is understood that the \ufb01rst ket refers to particle 1 and the second ket to particle 2. We can also consider |k|k, where particle 1 is characterized by |k and particle 2 by |k. Even though the two particles are indistinguishable, it is worth noting that mathematically (7.1) and (7.2) are distinct kets for k = k. In fact, withk = k, they are orthogonal to each other. (7.2) 429 430 Identical Particles Fig. 7.1 Two different paths, (a) and (b), of a two-electron system, for example, in which we cannot assert even in principle through which of the paths the electrons pass. (a) (b) Suppose we make a measurement on the two-particle system. We may obtain k for one particle and k for the other. However, we do not know a priori whether the state ket is |k|k, |k|k, or, for that matter, any linear combination of the two. Put in another way, all kets of form c1|k|k + c 2|k|k (7.3) lead to an identical set of eigenvalues when measurement is performed. This is known as exchange degeneracy. Exchange degeneracy presents a dif\ufb01culty because, unlike the single-particle case, a speci\ufb01cation of the eigenvalue of a complete set of observables does not completely determine the state ket. The way nature avoids this dif\ufb01culty is quite ingenious. But before proceeding further, let us develop the mathematics of permutation symmetry. We de\ufb01ne the permutation operator P12 by Clearly, P12|k|k = |k|k. P21 = P12 and P 2 12 = 1. (7.4) (7.5) Under P12, particle 1 having k becomes particle 1 having k; particle 2 having k becomes particle 2 having k. In other words, it has the effect of interchanging 1 and 2. In practice we often encounter an observable that has particle labels. For example in S1 \u00b7 S2 for a two-electron system, S1 (S2) stands for the spin operator of particle 1 (2). For simplicity we consider a speci\ufb01c case where the two-particle state ket is completely speci\ufb01ed by the eigenvalues of a single observable A for each of the particles: and A1|a|a = a|a|a A2|a|a = a|a|a, (7.6a) (7.6b) where the subscripts on A denote the particle labels, and A1 and A2 are thus the observables A for particles 1 and 2, respectively. Applying P12 to both sides of (7.6a), and inserting 1 = P\u22121 12 P12, we have 431 7.1 Permutation Symmetry P12A1P\u22121 12 P12|a|a = aP12|a|a |a|a = a|a|a. P12A1P\u22121 12 This is consistent with (7.6b) only if P12A1P\u22121 12 = A2. (7.7) (7.8) It follows that P12 must change the particle labels of observables. Let us now consider the Hamiltonian of a system of two identical particles. The observables, such as momentum and position operators, must necessarily appear symmetrically in the Hamiltonian, for example, H = p2 1 2m + p2 2 2m + Vpair (|x1 \u2212 x2|) + Vext(x1) +V ext(x2). (7.9) Here we have separated the mutual interaction between the two particles from their interaction with some other external potential. Clearly, we have P12HP\u22121 12 = H (7.10) for H made up of observables for two identical particles. Because P12 commutes with H, we can say that P12 is a constant of the motion. The eigenvalues of P12 allowed are +1 and \u22121 because of (7.5). It therefore follows that if the two-particle state ket is symmetric (antisymmetric) to start with, it remains so at all times. If we insist on eigenkets of P12, two particular linear combinations are selected: and |kk+ \u2261 1\u221a 2 (|k|k + |k|k), |kk\u2212 \u2261 1\u221a 2 (|k|k \u2212 |k|k). We can de\ufb01ne the symmetrizer and antisymmetrizer as follows: 2 (", "1 \u2212 P12). 2 (1 + P12), A12 \u2261 1 S12 \u2261 1 (7.11a) (7.11b) (7.12) We can extend this formalism to include states with more than two identical particles. From (7.12), if we apply S12(A12) to an arbitrary linear combination of |k|k and |k|k, the resulting ket is necessarily symmetric (antisymmetric). This can easily be seen as follows: - S12 A12 [c1|k|k + c 2|k|k] = 1 = 2 (c1|k|k + c 2|k|k) \u00b1 1 c1 \u00b1 c2 (|k|k \u00b1 |k|k). 2 2 (c1|k|k + c 2|k|k) (7.13) In Section 7.5 we will build on this approach. Before closing this section, we pause to point out that the consequences can be dramatic when permutation symmetry is ignored. Figure 7.2 shows a result which compares two experiments, before and after an error was corrected which ignored permutation symmetry in the analysis. 432 Identical Particles ) E n / M ( \u2212 \u03b4 16 14 12 10 8 6 4 2 0 0 CVC prediction 4 8 E (MeV 12 10 8 6 4 2 0 0 CVC prediction 4 8 E (MeV) 12 Fig. 7.2 Dramatic consequences arise when permutation symmetry is neglected. The data points are from McKeown et al., Phys. Rev. C, 22 (1980) 738, who tested a prediction of the CVC hypothesis. The \u03b2\u00b1 \u2212 \u03b1 correlation \u03b4\u2212 is plotted against the \u03b2\u00b1 energy. The prediction comes from a different, previous experiment, which at first neglected permutation symmetry. The corrected plot is on the right, from McKeown et al., Phys. Rev. C, 26 (1982) 2336, where the CVC prediction is smaller by a factor of \u221a 2. The object of this set of experiments was to test something called the Conserved Vector Current (CVC) hypothesis, based on the assumption of an intimate connection1 between the electromagnetic and weak interactions. Con\ufb01rmation, or refutation, of the CVC hypothesis was a high priority, and this experiment was one of the most precise tests. The initial result, shown on the left in Figure 7.2, was less than clear. The corrected result, on the right, was \ufb01nally a decisive con\ufb01rmation of CVC. The data points in Figure 7.2, which are identical for the left-hand and right-hand plots, are from a measurement of the beta decays of 8Li and 8B, each of which leads to a \ufb01nal state with two (identical) \u03b1 particles, through an excited state of 8Be. That is followed by 8Li \u21928 Be\u2217 + e\u2212 + \u00af\u03bde 8B \u21928 Be\u2217 + e+ + \u03bde 8Be\u2217 \u2192 \u03b1 + \u03b1. (7.14a) (7.14b) (7.14c) The experiment determines \u03b4\u2212, the correlation in direction between the e\u00b1 and \u03b1 directions for the two beta decays, as a function of e\u00b1 energy. The result of this measurement is published as McKeown et al., Phys. Rev. C, 22 (1980) 738. The hatched area shows the CVC prediction derived from an earlier experiment, published as Bowles and Garvey Phys. Rev. C, 18 (1978) 1447. This work measured the reaction followed by \u03b1 + \u03b1 \u21928 Be\u2217 8Be\u2217 \u21928 Be + \u03b3. (7.15a) (7.15b) 1 The CVC hypothesis predates the uni\ufb01cation of electromagnetic and weak interactions in what today is referred to as the \u201cStandard Model.\u201d 433 7.2 Symmetrization Postulate That is, a process which is rather the inverse of that in (7.14) and which proceeds through an electromagnetic interaction instead of the weak interaction. Deriving the CVC prediction from this result requires that the \u03b1\u03b1 wave function be symmetrized, but this was neglected at \ufb01rst, for the plot shown on the left of Figure 7.2. Some time later, this \u221a 2, and the plot on the right was published, error was corrected for the missing factor of showing much better agreement between prediction and measurement. 7.2 Symmetrization Postulate So far we have not discussed whether nature takes advantage of totally symmetrical or totally antisymmetrical states. It turns out2 that systems containing N identical particles are either totally symmetrical under the interchange of any pair, in which case the particles are said to", " satisfy Bose\u2013Einstein (B-E) statistics, hence known as bosons, or totally antisymmetrical, in which case the particles are said to satisfy Fermi\u2013Dirac (F-D) statistics, hence known as fermions. Thus Pij|N identical bosons = +|N identical bosons Pij|N identical fermions = \u2212|N identical fermions, (7.16a) (7.16b) where Pij is the permutation operator that interchanges the ith and the jth particle, with i and j arbitrary. It is an empirical fact that a mixed symmetry does not occur. Even more remarkable is that there is a connection between the spin of a particle and the statistics obeyed by it: Half-integer spin particles are fermions; Integer spin particles are bosons. (7.17a) (7.17b) Here particles can be composite; for example, a 3He nucleus is a fermion just as the e\u2212 or the proton; a 4He nucleus is a boson just as the \u03c0 meson or Z0 gauge boson. This spin-statistics connection is, as far as we can tell, an exact law of nature with no known exceptions. In the framework of nonrelativistic quantum mechanics, this principle must be accepted as an empirical postulate. In the relativistic quantum theory, however, it can be proved that half-integer spin particles cannot be bosons and integer spin particles cannot be fermions. An immediate consequence of the electron being a fermion is that the electron must satisfy the Pauli exclusion principle, which states that no two electrons can occupy the same state. This follows because a state like |k|k is necessarily symmetrical, which is not 2 To be sure, there is an important subtlety that relies on our living in three-dimensional space. It is possible to have objects, called anyons, which have a continuum of statistical properties spanning the range between fermions and bosons, if they are constrained to two spatial dimensions. The literature on this subject is fascinating, but scattered. The reader is referred to two early papers, namely \u201cQuantum mechanics of fractionalspin particles,\u201d Wilczek, Phys. Rev. Lett., 49 (1982) 957 and \u201cQuantum spectrum of three anyons in an oscillator potential,\u201d Murthy et al., Phys. Rev. Lett., 67 (1991) 817. 434 Identical Particles possible for a fermion. As is well known, the Pauli exclusion principle is the cornerstone of atomic and molecular physics as well as the whole of chemistry. To illustrate the dramatic differences between fermions and bosons, let us consider two particles, each of which can occupy only two states, characterized by k and k. For a system of two fermions, we have no choice; there is only one possibility: 1\u221a 2 (|k|k \u2212 |k|k). (7.18) For bosons there are three states possible: |k|k, |k|k, 1\u221a 2 (|k|k + |k|k). (7.19) In contrast, for \u201cclassical\u201d particles satisfying Maxwell\u2013Boltzmann (M-B) statistics with no restriction on symmetry, we have altogether four independent states: |k|k, |k|k, |k|k, |k|k. (7.20) We see that in the fermion case it is impossible for both particles to occupy the same state. In the boson case, for two out of the three allowed kets, both particles occupy the same state. In the classical (M-B) statistics case, both particles occupy the same state for two out of the four allowed kets. In this sense fermions are the least sociable; they avoid each other to make sure that they are not in the same state; in contrast, bosons are the most sociable, they really love to be in the same state, even more so than classical particles obeying M-B statistics. The difference between fermions and bosons shows up most dramatically at low temperatures; a system made up of bosons, such as liquid 4He, exhibits a tendency for all particles to get down to the same ground state at extremely low temperatures.3 This is known as Bose\u2013Einstein condensation, a feature not shared by a system made up of fermions. 7.3 Two-Electron System Let us now consider speci\ufb01cally a two-electron system. The eigenvalue of the permutation operator is necessarily \u22121. Suppose the base kets we use may be speci\ufb01ed by x1, x2, ms1, and ms2, wherem s1 and", " ms2 stand for the spin-magnetic quantum numbers of electron 1 and electron 2, respectively. We can express the wave function for a two-electron system as a linear combination of the state ket with eigenbras of x1, x2, ms1, and ms2 as follows: \u03c8 = \u2211 ms1 \u2211 ms2 C (ms1, ms2) x1, ms1; x2, ms2|\u03b1. (7.21) 3 The visual behavior of liquid helium, as it is cooled past the critical temperature, is striking. Various video examples can be seen at www.youtube.com/ including a classic physics demonstration movie, \u201cLiquid Helium II: The Super\ufb02uid\u201d by A. Leitner, from 1963. See also the site http://alfredleitner.com/. 435 7.3 Two-Electron System If the Hamiltonian commutes with S2 tot,! \" S2 tot, H = 0, then the energy eigenfunction is expected to be an eigenfunction of S2 written as then the spin function \u03c7 is expected to be one of the following: \u03c8 = \u03c6(x1, x2)\u03c7, \u03c7(ms1, ms2) = \u23ab \u23aa\u23aa\u23ac \u23aa\u23aa\u23ad \u23a7 \u23aa\u23aa\u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23aa\u23aa\u23a9 \u03c7++ (\u03c7+\u2212 + \u03c7\u2212+) 1\u221a 2 \u03c7\u2212\u2212 (\u03c7+\u2212 \u2212 \u03c7\u2212+) 1\u221a 2 triplet (symmetrical) singlet (antisymmetrical), (7.22) tot, and if \u03c8 is (7.23) (7.24) where \u03c7+\u2212 corresponds to \u03c7(ms1 = 1 2 ). Notice that the triplet spin functions are all symmetrical; this is reasonable because the ladder operator S1\u2212 + S2\u2212 commutes with P12 and the |+|+ state is even under P12. 2, ms2 = \u2212 1 We note x1, ms1; x2, ms2|P12|\u03b1 = x2, ms2; x1, ms1|\u03b1. Fermi\u2013Dirac statistics thus requires x1, ms1; x2, ms2|\u03b1 = \u2212x2, ms2; x1, ms1|\u03b1. Clearly, P12 can be written as P12 = P (space) 12 P (spin) 12 where P (space) spin states. It is amusing that we can express P (spin) just interchanges the position coordinate, while P (spin) as 12 12 P (spin) 12 = 1 2 12 4 1 + \u00afh2 S1 \u00b7 S2, which follows because S1 \u00b7 S2 = \u23a7 \u23aa\u23aa\u23aa\u23a8 \u23aa\u23aa\u23aa\u23a9 \u00afh2 4 \u22123\u00afh2 4 (triplet) (singlet). It follows from (7.23) that letting |\u03b1 \u2192P 12|\u03b1 amounts to \u03c6(x1, x2) \u2192 \u03c6(x2, x1), \u03c7(ms1, ms2) \u2192 \u03c7(ms2, ms1). (7.25) (7.26) (7.27) just interchanges the (7.28) (7.29) (7.30) (7.31) This together with (7.26) implies that if the space part of the wave function is symmetrical (antisymmetrical) the spin part must be antisymmetrical (symmetrical). As a result, the spin 436 Identical Particles triplet state has to be combined with an antisymmetrical space function and the spin-singlet state has to be combined with a space symmetrical function. The space part of the wave function \u03c6(x1, x2) provides the usual probabilistic interpretation. The probability for \ufb01nding electron 1 in a volume element d3x1 centered around x1 and electron 2 in a volume element d3x2 is |\u03c6(x1, x2)|2d3x1d3x2. (7.32) To see the meaning of this more closely, let us consider the speci\ufb01c case where the mutual interaction between the two electrons [for example, Vpair(|x1 \u2212 x2|), S1\u00b7 S2] can be ignored. If there is no spin dependence, the wave equation for the energy eigenfunction \u03c8 [see (7.9)], \u2212\u00afh2 2m \u22072 1 \u2212", " \u00afh2 2m \u22072 2 + Vext(x1) +V ext(x2) \u03c8 = E\u03c8, (7.33) is now separable. We have a solution of the form \u03c9A(x1)\u03c9B(x2) times the spin function. With no spin dependence S2 tot necessarily (and trivially) commutes with H, so the spin part must be a triplet or a singlet, which have de\ufb01nite symmetry properties under P (spin). The space part must then be written as a symmetrical and antisymmetrical combination of \u03c9A(x1)\u03c9B(x2) and \u03c9A(x2)\u03c9B(x1): 1\u221a 2 [\u03c9A(x1)\u03c9B(x2) \u00b1 \u03c9A(x2)\u03c9B(x1)] \u03c6(x1, x2) = (7.34) 12 where the upper sign is for a spin singlet and the lower is for a spin triplet. The probability of observing electron 1 in d3x1 around x1 and electron 2 in d3x2 around x2 is given by |\u03c9A(x1)| 2|\u03c9B(x2)| 2 + |\u03c9A(x2)| 2|\u03c9B(x1)|2? 1 2 (7.35) \u00b1 2 Re [\u03c9A(x1)\u03c9B(x2)\u03c9\u2217 A(x2)\u03c9\u2217 B(x1)]} d3x1d3x2. The last term in the curly bracket is known as the exchange density. We immediately see that when the electrons are in a spin-triplet state, the probability of \ufb01nding the second electron at the same point in space vanishes. Put another way, the electrons tend to avoid each other when their spins are in a triplet state. In contrast, when their spins are in a singlet state, there is enhanced probability of \ufb01nding them at the same point in space because of the presence of the exchange density. Clearly, the question of identity is important only when the exchange density is nonnegligible or when there is substantial overlap between function \u03c9A and function \u03c9B. To see this point clearly, let us take the extreme case where |\u03c9A(x)| 2 (where x may refer to x1 or x2) is big only in region A and |\u03c9B(x)| 2 is big only in region B such that the two regions are widely separated. Now choose d3x1 in region A and d3x2 in region B; see Figure 7.3. The only important term then is just the \ufb01rst term in (7.35), |\u03c9A(x1)| 2|\u03c9B(x2)| 2, (7.36) which is nothing more than the joint probability density expected for classical particles. In this connection, recall that classical particles are necessarily well localized and the question of identity simply does not arise. Thus the exchange density term is unimportant 437 7.4 The Helium Atom d3x1 d3x2 A B Fig. 7.3 Two widely separated regions A and B; |\u03c9A(x)| 2 is large in region A while |\u03c9B(x)| 2 is large in region B. if regions A and B do not overlap. There is no need to antisymmetrize if the electrons are far apart and the overlap is negligible. This is quite gratifying. We never have to worry about the question of antisymmetrization with 10 billion electrons, nor is it necessary to take into account the antisymmetrization requirement between an electron in New York and an electron in Beijing. 7.4 The Helium Atom A study of the helium atom is rewarding for several reasons. First of all, it is the simplest realistic problem where the question of identity, which we encountered in Section 7.3, plays an important role. Second, even though it is a simple system, the two-particle Schr\u00a8odinger equation cannot be solved analytically; therefore, this is a nice place to illustrate the use of perturbation theory and also the use of the variational method. The basic Hamiltonian is given by H = p2 1 2m + p2 2 2m \u2212 2e2 r1 \u2212 2e2 r2 + e2 r12, (7.37) where r1 \u2261 |x1|, r2 \u2261 |x2|, and r12 \u2261 |x1 \u2212 x2|; seeFigure 7.4. Suppose the e2/r12-term were absent. Then, with the identity question ignored, the wave function would be just the product of two hydrogen atom wave functions with Z = 1 changed into Z = 2. The total spin", " is a constant of the motion, so the spin state is either singlet or triplet. The space part of the wave function for the important case where one of the electrons is in the ground state and the other in an excited state characterized by (nlm) is \u03c6(x1, x2) = 1\u221a 2 [\u03c8100(x1)\u03c8nlm(x2) \u00b1 \u03c8100(x2)\u03c8nlm(x1)] (7.38) where the upper (lower) sign is for the spin singlet (triplet). We will come back to this general form for an excited state later. 438 Identical Particles Electron 1 r12 x1 \u03b3 x2 Electron 2 Fig. 7.4 Schematic diagram of the helium atom. Nucleus of charge + 2 e For the ground state, we need a special treatment. Here the con\ufb01guration is characterized by (1s) 2, that is, both electrons in n = 1, l = 0. The space function must then necessarily be symmetric and only the spin-singlet function is allowed. So we have \u03c8100(x1)\u03c8100(x2)\u03c7singlet = Z3 \u03c0a3 0 e\u2212Z(r1 + r2)/a0 \u03c7 with Z = 2. Not surprisingly, this \u201cunperturbed\u201d wave function gives E = 2 \u00d7 4 \u2212 e2 2a0 = \u2212108.8 eV (7.39) (7.40) for the ground-state energy, which is about 30% larger than the experimental value. This is just the starting point of our investigation because in obtaining the above form (7.39), we have completely ignored the last term in (7.37) that describes the interaction between the two electrons. One way to approach the problem of obtaining a better energy value is to apply \ufb01rst-order perturbation theory using (7.39) as the unperturbed wave function and e2/r12 as the perturbation. We obtain ( ) \u0394 (1s)2 = e2 r12 = (1s)2 Z6 \u03c02a6 0 e2 r12 e\u22122Z(r1 + r2)/a0 d3x1d3x2. (7.41) To carry out the indicated integration we \ufb01rst note 1 r12 = 1 + r2 r2 2 1 \u2212 2r1r2 cos \u03b3 = \u221e \u2211 l = 0 rl < rl + 1 > Pl(cos \u03b3), (7.42) where r> (r<) is the larger (smaller) of r1 and r2 and \u03b3 is the angle between x1 and x2. The angular integration is easily performed by expressing P l(cos \u03b3) in terms of Ym l (\u03b81, \u03c61) and l (\u03b82, \u03c62) using the addition theorem of spherical harmonics. (See, for example, Section Ym 12.8 of Arfken and Weber (1995).) We have P l(cos \u03b3) = 4\u03c0 2 l + 1 l \u2211 m = \u22121 Ym\u2217 l (\u03b81, \u03c61)Ym l (\u03b82, \u03c62). (7.43) 439 7.4 The Helium Atom The angular integration is now trivial: l (\u03b8i, \u03c6i) d\u03a9i = Ym 1\u221a 4\u03c0 (4\u03c0)\u03b4l 0\u03b4m 0. The radial integration is elementary (but involves tedious algebra!); it leads to \u221e 0 0 r1 1 r1 e\u2212(2Z/a0)(r1 + r2)r2 2 dr2 + \u221e r1 1 r2 e\u2212(2Z/a0)(r1 +r2 )r2 2 dr2 r2 1 dr1 a5 0 Z5. Combining everything, we have (for Z = 2) 5 128 = \u0394 (1s)2 = Z6e2 \u03c02a6 0 \u221a 4\u03c0)2 4\u03c0( Adding this energy shift to (7.40), we have Ecal = \u22128 + 5 2 = 5 2 a5 0 Z5. e2 2a0 5 128 e2 2a0 \u221274.8 eV. Compare this with the experimental value, Eexp = \u221279.005151042(40) eV, as determined from the NIST Atomic Spectra Database. (7.44) (7.45) (7.46) (7.47) (7.48) This is not bad, but we can do better! We propose to use the variational method with Z", ", which we call Zeff, as a variational parameter. The physical reason for this choice is that the effective Z seen by one of the electrons is smaller than 2 because the positive charge of 2 units at the origin (see Figure 7.4) is \u201cscreened\u201d by the negatively charged cloud of the other electron; in other words, the other electron tends to neutralize the positive charge due to the helium nucleus at the center. For the normalized trial function we use x1, x2|\u02dc0 = From this we obtain ( H = \u02dc0 ) \u2212 p2 1 2m + p2 2 2m \u02dc0 Z3 eff \u03c0a3 0 ( \u02dc0 Ze2 r1 = 2 Z2 eff 2 \u2212 2ZZeff + 5 8 Zeff. e2 a0 e\u2212Zeff(r1 + r2)/a0. (7.49) + Ze2 r2 ) \u02dc0 + ( \u02dc0 ) e2 r12 \u02dc0 We easily see that the minimization of H is at Zeff = 2 \u2212 5 16 = 1.6875. This is smaller than 2, as anticipated. Using this value for Zeff we get Ecal = \u221277.5 eV, which is already very close considering the crudeness of the trial wave function. (7.50) (7.51) (7.52) 440 Identical Particles E100 + Enlm Fig. 7.5 Schematic diagram for the energy-level splittings of (1s)(nl) for the helium atom. I J Singlet J Triplet Historically, this achievement was considered to be one of the earliest signs that Schr\u00a8odinger\u2019s wave mechanics was on the right track. We cannot get this kind of number by the purely algebraic (operator) method. The helium calculation was \ufb01rst done by A. Uns\u00a8old in 1927.4 Let us brie\ufb02y consider excited states. This is more interesting from the point of view of illustrating quantum-mechanical effects due to identity. We consider just (1s)(nl). We write the energy of this state as E = E100 + Enlm + \u0394E. In \ufb01rst-order perturbation theory, \u0394E is obtained by evaluating the expectation value of e2/r12. We can write (7.53) ( ) e2 r12 = I \u00b1 J, (7.54) where I and J, known respectively as the direct integral and the exchange integral, are given by I = d3x1 J = d3x1 d3x2|\u03c8100(x1)| 2|\u03c8nlm(x2)| 2 e2 r12 100(x2)\u03c8\u2217 \u03c8\u2217 d3x2\u03c8100(x1)\u03c8nlm(x2), e2 r12 nlm(x1). (7.55a) (7.55b) The upper (lower) sign goes with the spin-singlet (spin-triplet) state. Obviously, I is positive; we can also show that J is positive. So the net result is such that for the same con\ufb01guration, the spin-singlet state lies higher, as shown in Figure 7.5. The physical interpretation for this is as follows: In the singlet case the space function is symmetric and the electrons have a tendency to come close to each other. Therefore, the effect of the electrostatic repulsion is more serious; hence, a higher energy results. In the triplet case, the space function is antisymmetric and the electrons tend to avoid each other. Helium in spin-singlet states is known as parahelium, while helium in spin-triplet states is known as orthohelium. Each con\ufb01guration splits into the para state and the ortho state, the para state lying higher. For the ground state only parahelium is possible. See Figure 7.6 for a schematic energy-level diagram of the helium atom. It is very important to recall that the original Hamiltonian is spin independent because the potential is made up of just three Coulomb terms. There was no S1\u00b7S2-term whatsoever. Yet there is a spin-dependent effect \u2013 the electrons with parallel spins have a lower energy \u2013 that arises from Fermi\u2013Dirac statistics. 4 Uns\u00a8old, Ann. Phys., 82 (1927) 355. 441 7.5 Multiparticle States (1s)(2p) (1s)(2s) Para 1P1 Ortho 3P2,1,0 Para 1S0 Ortho 3S1 (1s)2 1S0 Spin singlet, necessarily \u201cpara\u201d Fig. 7.6 Schematic", " energy-level diagram for low-lying configurations of the helium atom. This explanation of the apparent spin dependence of the helium atom energy levels is due to Heisenberg. The physical origin of ferromagnetism \u2013 alignment of the electron spins extended over microscopic distances \u2013 is also believed to be essentially the same, but the properties of ferromagnets are much harder to calculate quantitatively from \ufb01rst principles. 7.5 Multiparticle States It is not dif\ufb01cult to extend the symmetrization of Section 7.2 to a system made up of many identical particles. Recalling (7.13) we de\ufb01ne Pij|k|k \u00b7 \u00b7 \u00b7 |ki|ki + 1 \u00b7 \u00b7 \u00b7 |kj \u00b7 \u00b7 \u00b7 = |k|k \u00b7 \u00b7 \u00b7 |kj|ki + 1 \u00b7 \u00b7 \u00b7 |ki \u00b7 \u00b7 \u00b7. (7.56) Clearly, just as before, and the allowed eigenvalues of Pij are +1 and \u22121. It is important to note, however, that in general P2 ij = 1 (7.57) [Pij, Pkl] = 0. (7.58) It is worth explicitly working out a system of three identical particles. First, there are 3! = 6 possible kets of form |k|k|k (7.59) 442 Identical Particles where k, k, and k are all different. Thus there is sixfold exchange degeneracy. Yet if we insist that the state be totally symmetrical or totally antisymmetrical, we can form only one linear combination each. Explicitly, we have |kkk\u00b1 \u2261 1\u221a 6 {|k|k|k \u00b1 |k|k|k + |k|k|k \u00b1 |k|k|k + |k|k|k \u00b1 |k|k|k}. (7.60) These are both simultaneous eigenkets of P12, P23, and P13. We remarked that there are altogether six independent state kets. It therefore follows that there are four independent kets that are neither totally symmetrical nor totally antisymmetrical. We could also introduce the operator P123 by de\ufb01ning P123 (|k|k|k) =|k |k|k. (7.61) Note that P123 = P12P13 because P12P13 (|k|k|k) =P 12 (|k|k|k) = |k|k|k. In writing (7.60) we assumed that k, k, andk are all different. If two of the three indices coincide, it is impossible to have a totally antisymmetrical state. The totally symmetrical state is given by (7.62) |kkk+ = 1\u221a 3 (|k|k|k + |k|k|k + |k|k|k), (7.63) \u221a 2! /3!. For more general cases we have where the normalization factor is understood to be a normalization factor \u2019 N1! N2! \u00b7 \u00b7 \u00b7 Nn! N!, (7.64) where N is the total number of particles and Ni the number of times |k(i) occurs. For more general cases of the totally antisymmetrical state, it is helpful to write the antisymmetric case of (7.60) as |kkk\u2212 = 1\u221a 3! |k |k |k |k |k |k |k |k |k (7.65) where the ordering of kets in the determinant is set by the row from which they originate. This construction is known as a Slater determinant. Its generalization to the N-particle case is straightforward. In principle, we can follow along the lines of Section 7.4 to apply this formalism to N-particle systems, but it can quickly become unwieldy, even with modern computational resources. If N is on the order of Avogadro\u2019s number, as it is for problems in condensed matter physics or materials chemistry, then this approach is intractable. Other techniques of calculation needed to be invented in order to handle these cases. 443 7.6 Density Functional Theory The remainder of this chapter develops two such techniques. Section 7.6 covers the basics of density functional theory, a coordinate representation-based approach that reduces the problem to one with only three (spatial) degrees of freedom. Section 7.7 introduces the foundations of quantum \ufb01eld theory using the formalism known as \u201csecond quantization.\u201d Each of these techniques is both powerful and widely used. Finally, Section 7.8 applies second quantization to the electromagnetic \ufb01eld in free space. (Second quantization is applied to the relativistic Klein\u2013Gordon \ufffd", "\ufffdeld in Section 8.1, see p. 485.) 7.6 Density Functional Theory The problem of having many identical particles, interacting with each other and with some external potential, can be written down as an N-body Schr\u00a8odinger equation for a wave function in coordinate space. This equation cannot be solved exactly, however, so approximation schemes must be developed. One particular approach has emerged in the most recent decades as a popular choice, especially for the electronic structure of condensed matter systems. This approach, known as density functional theory (DFT), is based on two seminal papers, namely Hohenberg and Kohn, Phys. Rev., 136 (1964) B864 and Kohn and Sham, Phys. Rev., 140 (1965) A1133. DFT has found wide application in condensed matter and materials physics and chemistry. In fact, the 1998 Nobel Prize in Chemistry was awarded to physicist Walter Kohn and chemist John Pople for the development and application of this formalism. Our treatment here is strictly an introduction. See Jones and Gunnarsson, Rev. Mod. Phys., 61 (1989) 689 for a thorough review after DFT had matured. Today DFT is used widely in condensed matter physics, and described in modern textbooks on the subject, for example Cohen and Louie (2016). For interesting historical and contextual reading, see Walter Kohn\u2019s Nobel Prize lecture, Rev. Mod. Phys., 71 (1999) 1253 and also Perdew et al., J. Chem. Theory Comput., 5 (2009) 902. 7.6.1 The Energy Functional for a Single Particle In Section 5.4 we proved a theorem that was the basis of the variational approximation to the ground-state energy. Cast in terms of wave mechanics, we found that E[ \u02dc\u03c8] \u2261 \u02dc\u03c8\u2217H \u02dc\u03c8 d3x \u2212 \u00afh2 2m = \u02dc\u03c8\u2217 \u22072 + V(x) \u02dc\u03c8 d3x \u2265 E0 (7.66) for the Hamiltonian H and a (properly normalized) trial wave function \u02dc\u03c8(x), whereE 0 is the true ground-state energy. The notation E[ \u02dc\u03c8] means that the energy is a functional of the trial wave function. That is, given some function \u02dc\u03c8(x), we can calculate a value for E[ \u02dc\u03c8]. 444 Identical Particles We made use of this in the variational approximation by giving \u02dc\u03c8 free parameters, which we varied to minimize E[ \u02dc\u03c8], thereby providing an approximation to E0. The key idea behind DFT is that the energy can be considered a functional of the density instead of the wave function. Hohenberg, Kohn, and Sham proved rigorously that this converts the quantum many-body problem into an effective one-body problem, a tractable calculation. Before describing their work, though, let us see how this works for the singleparticle case. We will \ufb01nd it convenient to rewrite (7.66) using the divergence theorem, and the fact that the wave function tends to zero at in\ufb01nity for bound states. That is \u02dc\u03c8\u2217\u22072 \u02dc\u03c8 d3x =! \u2207 \u00b7 ( \u02dc\u03c8\u2217\u2207 \u02dc\u03c8) \u2212 \u2207 \u02dc\u03c8\u2217 \u00b7 \u2207 \u02dc\u03c8 \" d3x = \u2212 \u2207 \u02dc\u03c8\u2217 \u00b7 \u2207 \u02dc\u03c8 d3x (7.67) since the term with the divergence converts to a surface integral at in\ufb01nity. This transformation is sometimes called a \u201cchoice of gauge,\u201d for its similarity to the transformations discussed in Section 2.7. The energy functional (7.66) then becomes E[ \u02dc\u03c8] = \u2207 \u02dc\u03c8\u2217 \u00b7 \u2207 \u02dc\u03c8 + V(x) \u02dc\u03c8\u2217 \u02dc\u03c8 d3x. (7.68) \u00afh2 2m Now, the single-particle density for a wave function \u03c8(x) is \u03c1(x) = \u03c8\u2217\u03c8. We can take the ground-state trial wave function \u02dc\u03c8(x) to be real, and then write the energy as a functional of the density \u02dc\u03c1(x). Using \u02dc\u03c8 = E[ \u02dc\u03c1] = &2 \u2207 \u02dc\u03c11/2 % \u00afh2 2m \u02dc\u03c1 we \ufb01nd \u00afh2 8m (\u2207 \u02dc\u03c1)2 \u02dc\u03c1 + V(x) \u02dc\u03c1 + V(x) \u02dc\u03c1 d3x = d3x (7.69) for the ground-state energy as a functional of density", ". At this point, we could make a guess for the function \u02dc\u03c1(x), including one or more free parameters, calculate E[ \u02dc\u03c1], and minimize the result with respect to those parameters. This is directly analogous to the approach we took in Section 5.4. However, we can take a different approach. Equation (7.69) shows that the energy E[ \u02dc\u03c1] is a functional of \u02dc\u03c1 and its \ufb01rst derivatives. Minimizing E[ \u02dc\u03c1] is equivalent to \ufb01nding a stationary value of the energy, that is \u03b4E[ \u02dc\u03c1] = 0, a well-known problem in the calculus of variations.5 First we write (7.69) as E[ \u02dc\u03c1] = \u03b5( \u02dc\u03c1, \u2207 \u02dc\u03c1)d3x = \u03b5( \u02dc\u03c1, \u02dc\u03c1x, \u02dc\u03c1y, \u02dc\u03c1z)d3x (7.70) where \u02dc\u03c1x \u2261 \u2202 \u02dc\u03c1/\u2202x, \u02dc\u03c1y \u2261 \u2202 \u02dc\u03c1/\u2202y, and \u02dc\u03c1z \u2261 \u2202 \u02dc\u03c1/\u2202z, so that \u03b5( \u02dc\u03c1, \u2207 \u02dc\u03c1) = \u00afh2 8m (\u2207 \u02dc\u03c1)2 \u02dc\u03c1 + V(x) \u02dc\u03c1 = \u00afh2 8m x + \u02dc\u03c12 y + \u02dc\u03c12 \u02dc\u03c12 z \u02dc\u03c1 + V(x) \u02dc\u03c1. (7.71) 5 The calculus of variations is covered in every textbook on classical mechanics, treating the action as a functional of the spatial coordinates and their derivatives, with time as the independent variable. Here we make use of the case of several independent variables, typically covered in textbooks on mathematical physics. See, for example, Arfken, Weber, and Harris (2013), Chapter 22, or Byron and Fuller (1992), Chapter 2. 445 7.6 Density Functional Theory Before imposing \u03b4E[ \u02dc\u03c1] = 0, we must include the constraint that the density is normalized, that is \u02dc\u03c1 d3x = 1. (7.72) This is most easily done by multiplying (7.72) by a constant, which we take to be \u2212\u03bc, and adding it to (7.70), creating a new functional E(\u03bc)[ \u02dc\u03c1]. (This is formally referred to as the method of Lagrange multipliers.) Because of the constraint (7.72), minimizing E[ \u02dc\u03c1] is equivalent to minimizing E(\u03bc)[ \u02dc\u03c1], that is setting \u03b4E(\u03bc)[ \u02dc\u03c1] = 0. We have \u03b4E(\u03bc)[ \u02dc\u03c1] = \u03b4 \u03b5( \u02dc\u03c1, \u2207 \u02dc\u03c1)d3x \u2212 \u03bc \u02dc\u03c1 d3x = \u03b4 \u02dc\u03c1 + \u2202\u03b5 \u2202 \u02dc\u03c1 \u2202\u03b5 \u2202 \u02dc\u03c1x \u2202\u03b5 \u2202 \u02dc\u03c1y Recognizing that \u03b4 \u02dc\u03c1x = \u2202(\u03b4 \u02dc\u03c1)/\u2202x, and similarly for \u02dc\u03c1y and \u02dc\u03c1z, we integrate those terms by parts with \u03b4 \u02dc\u03c1 \ufb01xed to zero at in\ufb01nity. Then factor out \u03b4 \u02dc\u03c1 and set the integrand to zero. The result is \u03b4 \u02dc\u03c1z \u2212 \u03bc\u03b4 \u02dc\u03c1 \u2202\u03b5 \u2202 \u02dc\u03c1z d3x = 0. \u03b4 \u02dc\u03c1y + \u03b4 \u02dc\u03c1x + (7.73) Noting that \u2202\u03b5 \u2202 \u02dc\u03c1 \u2212 \u2202 \u2202x \u2202\u03b5 \u2202 \u02dc\u03c1x \u2212 \u2202 \u2202y \u2202\u03b5 \u2202 \u02dc\u03c1y \u2212 \u2202 \u2202z \u2202\u03b5 \u2202 \u02dc\u03c1z = \u03bc. \u2202 \u2202x \u2202\u03b5 \u2202 \u02dc\u03c1x = \u00afh2 4m \u2202 \u2202x \u02dc\u03c1x \u02dc\u03c1 = \u00afh2 4m \u02dc\u03c1xx \u02dc\u03c1 \u2212 \u02dc\u03c12 x \u02dc\u03c12 we insert (7.71) into (7.74) to \ufb01nd \u00afh2 8m (\u2207 \u02dc\u03c1)2 \u02dc\u03c1 + V(x) \u02dc\u03c1 \u2212 \u00afh2 4m \u22072 \u02dc\u03c1 = \u03bc \u02dc\u03c1. (7.74) (7.75) (7.76) This is a differential equation that could be solved for Schr\u00a8odinger equation could be solved for the wave function. \u02dc\u03c1(x), similar to the way the Note that integrating both sides of (7.76) proves that \u03bc equals the ground-state energy, for the exact", " solution \u02dc\u03c1. This is because the \ufb01rst two terms on the left are \u03b5( \u02dc\u03c1, \u2207 \u02dc\u03c1), we have the constraint (7.72), and (similar to the way we dealt with the kinetic-energy term (7.67)) the integral of \u22072 \u02dc\u03c1 becomes a surface integral at in\ufb01nity and vanishes. Problem 7.12 at the end of this chapter gives an example of applying these ideas to a single particle in a one-dimensional harmonic oscillator potential. In three dimensions, this could also be demonstrated with the nonisotropic harmonic oscillator or the hydrogen atom. 7.6.2 The Hohenberg\u2013Kohn Theorem Now consider the ground state of a system with N identical particles, with wave function \u03a8(x1, x2,..., xN), where the argument position identi\ufb01es the particle number. (We will abandon the notation \u02dc\u03a8.) The wave function must be appropriately symmetrized, that is \u03a8(x1, x2,..., xi,..., xj,..., xN) =\u00b1 \u03a8(x1, x2,..., xj,..., xi,..., xN). (7.77) 446 Identical Particles Other than acknowledging the symmetry of the wave function, we will not be considering spin degrees of freedom in our discussion. The number density n(x) in this case will be the expectation value of the operator that locates a particular particle at position x, summed over all the particles. That is n(x) = N \u2211 \u03a8|\u03b43(x \u2212 xi)|\u03a8 i=1 N \u2211 = \u03a8\u2217(x1, x2,..., xi = x,..., xN) i=1 \u00d7 \u03a8(x1, x2,..., xi = x,..., xN)d3x1 d3x2 \u00b7 \u00b7 \u00b7 d3xN (7.78) where the measure of the integral does not include the factor d3xi. Invoking the symmetry (7.77), we can exchange x1 for xi for each term in (7.78) to write n(x) =N \u03a8\u2217(x, x2,..., xN)\u03a8(x, x2,..., xN)d3x2 \u00b7 \u00b7 \u00b7 d3xN. (7.79) For a properly normalized wave function \u03a8, we clearly have n(x)d3x = N. (7.80) We consider a Hamiltonian that is a straightforward generalization of (7.9), namely p2 i 2m H = \u2211 + \u2211 \u2211 j=i \u2261 T + Upair + Uext. i i 1 2 Vpair(|xi \u2212 xj|) +\u2211 Vext(xi) i (7.81) Our goal is to write the ground-state energy E = \u03a8|H|\u03a8 as a functional of the density n(x). We will take the point of view that it is the external potential Vext(x) that de\ufb01nes our problem. For example, if we are trying to \ufb01nd the ground state of a many-electron system, then T + Upair is universal for all such problems. In other words, the wave function \u03a8 is determined by Vext(x). The main point of the Hohenberg\u2013Kohn theorem is that n(x) is also determined by Vext(x). This was trivial when we wrote \u03c1 = \u03c82 for the one-particle system, but it is more subtle for the many-particle case. First let us \ufb01nd the ground-state expectation value for the Uext term in (7.81). Returning for a moment to our formal distinction between an operator x and its eigenvalue x, we have \u03a8|Uext|\u03a8 = \u2211 d3x i \u03a8|Vext(xi)|x i x i |\u03a8 = \u2211 d3x i Vext(x i)\u03a8|x i i = \u2211 d3x 1d3x 2 i \u00b7 \u00b7 \u00b7 d3x i \u00b7 \u00b7 \u00b7 d3x N Vext(x i) i \u00d7 \u03a8\u2217(x 1, x 2,..., x i,..., x N)\u03a8(x 1, x 2,..., x i,..., x N). x i |\u03a8 (7.82) We then again change integration variables in each term of the sum, exchanging x i for x 1, and realizing that each term is the same because", " of (7.77). Renaming the dummy integration variable x 1 to x, we use (7.79) to write \u03a8|Uext|\u03a8 = Vext(x)n(x) d3x. (7.83) 447 7.6 Density Functional Theory This shows that this expectation value is a functional of n, and is of course completely analogous to the second term in (7.69). Now we need to show that \u03a8|(T + Upair)|\u03a8 is a functional of n(x). That is, we need to show that the external potential Vext uniquely determines n(x), just as it does the groundstate6 wave function \u03a8. The variational theorem of Section 5.4 provides the proof we need. Suppose there is another external potential V ext, differing by more than an additive constant from Vext, which gives a different wave function \u03a8 but the same density n(x). If we consider \u03a8 as a trial wave function, then \u03a8|H|\u03a8 > \u03a8|H|\u03a8. (7.84) However, since the density n(x) is the same, we invoke (7.83) and write \u03a8|(T + Upair)|\u03a8 > \u03a8|(T + Upair)|\u03a8. This result is absurd, of course. We could just have easily let \u03a8 be the trial wave function, and would have found the opposite result. The only way out is that \u03a8 is the same as \u03a8 (to within an overall phase), resulting in the same density n(x). (7.85) Using a standard notation, we write F[n] \u2261 \u03a8|(T + Upair)|\u03a8 (7.86) for the \u201cuniversal\u201d function of a problem involving a speci\ufb01c particle species and two-body interaction. We then have \u03a8|H|\u03a8 = Vext(x)n(x) d3x + F[n] \u2261 E[n] (7.87) which is known as the Hohenberg\u2013Kohn theorem. This result is, of course, a tremendous simpli\ufb01cation of the quantum-mechanical many-body problem, reducing it to \ufb01nding a function n(x) of a single position variable, instead of needing to \ufb01nd the multiparticle wave function \u03a8(x1, x2,..., xN). 7.6.3 The Kohn\u2013Sham Equations Even though (7.87) simpli\ufb01es the approach to the quantum-mechanical many-body problem in principle, it is dif\ufb01cult to directly apply it to practical problems. Hohenberg and Kohn used it to study two limiting cases of density inhomogeneities in an electron gas, but without a prescription for F[n], detailed calculations of the many-body ground state are not possible. Kohn and Sham found a way to derive a self-consistent approximation scheme, based on single-particle wave functions7 \u03c6j(x) that solve a particular, albeit \ufb01ctitious, one-body Schr\u00a8odinger equation. The \u03c6j(x) form the multiparticle wave function \u03a8(x1, x2,..., xN) through the N-body generalizations of (7.60), that is, 6 We are only considering nondegenerate ground states. In fact, the Hohenberg\u2013Kohn theorem can be proved in general, using a constrained search approach. 7 In much of the literature on DFT, particularly that aimed at chemistry, these wave functions are called \u201corbitals.\u201d 448 Identical Particles \u03a8(x1, x2,..., xN) = \u03c61(x1)\u03c62(x2) \u00b7 \u00b7 \u00b7 \u03c6N(xN) \u00b1 \u03c62(x1)\u03c61(x2) \u00b7 \u00b7 \u00b7 \u03c6N(xN) \u00b1.... (7.88) Note that although we labeled the \u03c6j(xi) with j = 1, 2,..., N, they do not necessarily have to be distinct functions. For bosons, for example, they could all be the same function, and for fermions, they could be pairwise the same, each with a different spin quantum number. Next, a constraint is applied so that the \u03c6j(x) yield the correct density n(x). Since the \u03c6j(x) solve a Schr\u00a8odinger equation, they are orthonormal, so inserting (7.88) into (7.78), this constraint takes the form n(x) = N \u2211 |\u03c6j(x)|2. (7.89) j", "=1 The \u03c6j(x) are used to construct F[n], and then the calculus of variations is applied to minimize the energy functional. The Kohn\u2013Sham wave functions \u03c6j(x) are solutions to Schr\u00a8odinger equations of the form x|HKS|\u03c6j = \u2212 \u00afh2 2m where the \u03b5j are the energy eigenvalues. To the extent that \u22072\u03c6j(x) +V KS(x)\u03c6j(x) = \u03b5j\u03c6j(x) \u03a8|HKS|\u03a8 = = d3x1 \u00b7 \u00b7 \u00b7 d3xN \u03a8| {|x1x1| \u00b7 \u00b7 \u00b7 |xNxN|HKS|} |\u03a8 d3xi \u03c6\u2217 j (xi) \u22072 i \u03c6j(xi) +V KS(xi)\u03c6j(xi) = \u2212 \u00afh2 2m N \u2211 j=1 (7.90) N \u2211 j=1 \u03b5j (7.91) approximates the true ground-state energy \u03a8|H|\u03a8, we can rely on the kinetic-energy term \u03a8|TKS|\u03a8 = N \u2211 j=1 d3xi \u03c6\u2217 j (xi) \u2212 \u00afh2 2m \u22072 i \u03c6j(xi) (7.92) to approximate \u03a8|T|\u03a8 in (7.86). In fact, an important underlying notion for the Kohn\u2013 Sham formalism is that the constraint (7.89) implies that (7.92) is an exceptionally good approximation for \u03a8|T|\u03a8. Unlike the single-particle case, it is not simple to explicitly write \u03a8|TKS|\u03a8 in terms of the density n(x), but we can make use of the Hohenberg\u2013Kohn theorem to write \u03a8|TKS|\u03a8 \u2261T KS[n]. We then write F[n] =T[n] +U pair[n] = TKS[n] +U lr[n] +U xc[n] (7.93a) (7.93b) where Ulr[n] is the long range component of the two-particle interaction, and Uxc[n], called the \u201cexchange-correlation energy,\u201d is what remains. As we believe that TKS[n] is a good approximation to the kinetic energy, and that in typical problems we expect the long range interaction to dominate, Uxc[n] turns out to be small in many problems of interest. 449 7.6 Density Functional Theory Therefore, good approximations to Uxc[n] should yield excellent approximations for E[n] in (7.86). We note that for the many-electron problem in atomic, molecular, or condensed matter systems, which is by far the most common application of DFT, the long range two-particle interaction is just the Coulomb potential between pairs of electrons. In this case, we can use a classical approach to writing down Ulr[n] =U ee[n]. The charge density at any point is just e n(x) so the Coulomb energy is Uee[n] = d3x e2 2 d3x n(x)n(x) |x \u2212 x| (7.94) where the factor of 1/2 just removes the effect of double-counting the electrons, as in (7.81). At this point, we will proceed as we would for a many-electron problem, that is with Ulr[n] = Uee[n]. Following (7.73) to minimize the energy functional, and using (7.87) with (7.93b) and applying the constraint (7.80), we have \u03b4 Vext(x)n(x) d3x + TKS[n] +U ee[n] +U xc[n] \u2212 \u03bc = 0 (7.95) n(x) d3x where the variation is in terms of the density n(x). The \ufb01rst and last terms are simple, [Vext \u2212 \u03bc]\u03b4n d3x. The long range electron-electron interaction is also easy, becoming namely \u03b4Uee[n] =e 2 d3x n(x) |x \u2212 x| \u03b4n(x) d3x. (7.96) This allows us to use (7.95) to write the chemical potential \u03bc = \u03b4E[n]/\u03b4n as \u03bc = \u03b4TKS[n] \u03b4n + Vext(x) +e 2 d3x n(x) |x \u2212 x", "| + \u03b4Uxc[n] \u03b4n. (7.97) Now imagine, instead, that the system is governed by a collection of noninteracting electrons with single-particle wave functions \u03c6j(x), determined by an effective \u201cexternal\u201d potential VKS(r). Using (7.90) we would obviously \ufb01nd \u03bc = \u03b4TKS[n] \u03b4n + VKS(x) which means we can write down the Kohn\u2013Sham potential of (7.90) as VKS(x) =V ext(x) +e 2 d3x n(x) |x \u2212 x| + \u03b4Uxc[n] \u03b4n. (7.98) (7.99) Given some external potential Vext(x) (perhaps a screened atomic or molecular potential, or even a periodic lattice of them) and exchange-correlation functional Uxc[n], we now see how we might carry through a self-consistent iterative approximation. First, choose a reasonable approximation for the density n(x). Then use (7.99) to calculate the Kohn\u2013 Sham single-particle potential VKS(x). Solve (7.90) using this potential to \ufb01nd the singleparticle wave functions \u03c6j(x), and redetermine n(x) using (7.89). Then repeat this process using the new density, comparing the function n(x) at the end with what you started out 450 Identical Particles with, until you are satis\ufb01ed that convergence is reached. When you have your \ufb01nal function n(x), you then calculate the ground-state energy from E[n] = Vext(x)n(x) d3x + TKS[n] +U ee[n] +U xc[n] (7.100) where we understand that the kinetic-energy functional TKS[n] is calculated from (7.92) using the wave functions determined by the iteration procedure that found n(x). This procedure also gives us a good approximation for the wave function (7.88), from which other ground-state properties can be calculated. One remaining issue is which eigenfunctions \u03c6j(x), obtained by solving (7.90), should be used to construct n(x). Presumably, our iterative procedure will converge most quickly if we use the lowest energy single-particle wave functions to construct the wave function (7.88). For a system of bosons, this would be all \u03c6j(x) = \u03c60(x), the ground-state eigenfunction. For fermions, we would use the lowest energy eigenfunctions consistent with total antisymmetry, including accounting for the spin degrees of freedom. Rarely is there an analytic solution to (7.90). However, it can be solved numerically, or the Hamiltonian can be represented in a (truncated) basis of known eigenfunctions, for example the hydrogen atom or harmonic oscillator, and its matrix diagonalized. It is important to realize that the Kohn\u2013Sham procedure yields much more than just the ground-state energy of the system. Having determined the eigenfunctions \u03c6j(x) that give the ground-state density, we then have an excellent approximation (7.88) to the groundstate wave function. This wave function can be used to calculate expectation values of any dynamical operator. For example, in condensed matter physics where DFT \ufb01nds many applications, results of calculations include electrical and thermal resistivity, and response to electromagnetic \ufb01elds, mechanical strain, and other perturbations. See Cohen and Louie (2016), Section 7.4 for a detailed discussion. 7.6.4 Models of the Exchange-Correlation Energy Density functional theory is widely used because it can give precise, accurate results for many-electron systems. This is because everything in (7.100) is known from \ufb01rst principles, except for the exchange-correlation energy Uxc[n], which can be approximated well. It contains the corrections to the Kohn\u2013Sham one-body effective potential, as well as the effects of the correlations between identical particles, for example as in (7.35). Several authors have derived a set of exact constraints that the exchange-correlation functional must satisfy. Model functionals can then be produced which satisfy as many of those constraints as possible. These models can be tested against simple systems like the uniform electron gas, or systems that are somewhat more complicated, but exactly calculable. In this way, different functionals have been derived for systems ranging from single atoms to complex materials. Functionals Uxc[n] are generally divided into two classes, namely the local density approximation (LDA) and the", " generalized gradient approximation (GGA). They take the forms ULDA xc [n] = d3x n(x) \u03b5(n) (7.101a) 451 7.6 Density Functional Theory UGGA xc [n] = d3x f(n, \u2207n) (7.101b) where \u03b5(n) and f (n, \u2207n) can be parameterized functions. A goal is to determine those parameters from fundamental \ufb01rst principles, but it is also possible to determine them by \ufb01tting to data. In the original Kohn\u2013Sham formalism, \u03b5(n) in (7.101) is the exchangecorrelation energy per electron for an electron gas of uniform density, and f (n, 0) =n \u03b5(n). A particularly popular form for f (n, \u2207n) which otherwise includes only fundamental constants is described in a paper8 by Perdew et al., Phys. Rev. Lett., 77 (1996) 3865. 7.6.5 Application to the Helium Atom We conclude this section on DFT with an application to the two-electron system that is the helium atom. It is also a good opportunity to compare to the results in Section 7.4. Our discussion here outlines the solution, with details left to Problem 7.14 at the end of this chapter. As always, we want to solve differential equations written in terms of dimensionless variables. It is common in DFT calculations to use \u201catomic units,\u201d where \u00afh = e = m = 1. This means that all distances are measured in units of the Bohr radius (3.317) a0 = \u00afh2/me2 = 0.53 \u02daA, and energies are measured in units of e2/a0, called a \u201cHartree.\u201d Recall from (3.315) that the magnitude of the ground-state energy of a hydrogen atom is E0 = e2/2a0 = 13.6 eV, so one Hartree equals 27.2 eV. The Kohn\u2013Sham Schr\u00a8odinger equation (7.90) becomes \u2212 1 2 \u22072\u03c6j(x) +V KS(x)\u03c6j(x) = \u03b5j\u03c6j(x). (7.102) We will take the external potential Vext(x) to be Coulomb\u2019s law, which in atomic units becomes Vext(x) =V ext(r) = \u2212 Z r, (7.103) where Z = 2 for helium. We need to choose an initial approximation n(0)(x) to the density. We learned in Section 7.4 that a good starting wave function would be proportional to e\u2212Zeffr where Zeff = 16. This suggests a starting density proportional to e\u22122Zeffr. However, it is frequently 2 \u2212 5 the case with DFT calculations, that we do not have such a good guess for n(0)(x). So, we illustrate the convergence process by making a poor guess, leaving off the factor of two and starting with n(0)(x) proportional to e\u2212Zeffr. With the proper normalization, we \ufb01nd n(0)(r) = Z3 4\u03c0 e\u2212Zeffr. eff (7.104) Note that for this, or any other spherically symmetric density n(r), the long range ee interaction term in (7.99) can be written as 8 With nearly 100,000 citations, this paper lays claim to the most highly cited article ever published in Physical Review Letters. 452 Identical Particles \u221e |x \u2212 x| = 8\u03c0 which is a useful expression for the numerical computations that follow. 0 r2 dr n(r) (r \u2212 r)2 + r + r d3x n(x) (7.105) For the exchange-correlation functional Uxc[n], we use a form of (7.101a) derived by Sun et al., J. Chem. Phys., 144 (2016) 191101. This functional was developed speci\ufb01cally for precise calculations in two-electron systems, and uses only fundamental constants. We have \u03b5xc(n) = \u03b5x(n) + \u03b5c(n) (7.106) \u03b5x(n) =\u2212 h0 x \u03b5c(n) =\u2212 % 3 4\u03c0 &1/3 3\u03c02n \u22120.0233504 1 + 0.1018r1/2 s + 0.102582rs x = 1.174 and rs = (4\u03c0n/3)\u22121/3. From here, it is straightforward to write down where h0 \u03b4Uxc[n]/", "\u03b4n for use in (7.99). Thus we have all the ingredients we need to build our Kohn\u2013Sham potential (7.99), in particular (7.103) and (7.106). This potential is used to solve the Schr\u00a8odinger equation (7.102). Note that for the two-electron system, the ground state is a spin-singlet. Therefore the two-particle wave function is symmetric, and we construct it from the ground-state wave function \u03c60(x) that solves (7.90). That is \u03a8(x1, x2) = \u03c60(x1)\u03c60(x2). (7.107) Therefore, our goal now is to solve (7.102) for the ground-state wave function. We do this \ufb01rst using n(r) =n (0)(r), using the result to \ufb01nd the next iteration of the density n(1)(r) from (7.89). This process is repeated until the result converges, and the ground-state energy is calculated from (7.100). In principle we could solve (7.102) numerically, but singular potentials like (7.103) will make this dif\ufb01cult. It is easier to diagonalize the Hamiltonian in some basis, and for this we will use the one-electron atomic wave functions based on (7.103). Since the problem has spherical symmetry, only the s-states will contribute. We proceed, therefore, using the 1s, 2s, and 3s wave functions for a one-electron atom with Z = Zeff, and check that the contribution from the 3s state is small. At this point, the calculation is numerical and done on a computer. (The results here were obtained using MATHEMATICA, but there are many other options available.) A 3 \u00d7 3 matrix is constructed using the Hamiltonian from (7.102), and diagonalized to \ufb01nd the ground-state eigenfunction. One \ufb01nds 1\u221a [0.99851R1s(r) +0.04831R 2s(r) +0.02519R 3s(r)]. (7.108) \u03c6(1) 0 (x) = 4\u03c0 As expected, the Kohn\u2013Sham wave function is dominated by the ground 1s state for Z = Zeff. We use this wave function, with (7.89), to get the next iteration of the density, namely # $ n(1)(r) = 2 \u03c6(1) 0 (x) 2. (7.109) The \ufb01rst iteration of the energy functional (7.100) is found using n(1)(r), and also \u03c6(1) for TKS[n], and this is our \ufb01rst estimate of the ground-state energy. 0 (x) 453 7.6 Density Functional Theory Table 7.1 Contributions to the Ground-State Energy (7.100) of the Helium Atom for Three Iterations of the Kohn\u2013Sham Equations. The values for the functionals are in Hartree, and their sum for the ground-state energy is given in eV. The numerical calculation is accurate for all significant figures shown, including the value used to convert Hartree to eV. Iteration Uext[n] 1 2 3 \u20136.90876 \u20136.48303 \u20136.56468 TKS[n] 2.9882 2.6350 2.6978 Uee[n] 2.17546 1.99548 2.03054 Uxc[n] \u20131.13109 \u20131.04472 \u20131.06127 E \u201378.2649 \u201378.8379 \u201378.8478 r 2n(r) 0.15 0.10 0.05 0.00 0 r 2n(0) r 2n(1) r 2n(2) r 2n(3) 1 2 3 4 r 5 Fig. 7.7 The initial guess for the two-electron density function for the helium atom, and the results for the first three iterations. Plotted is the density times r2 as a function ofr, withr in Bohr radii. The first iteration is already close to the final function, and convergence is rapid. The second iteration \ufb01nds a new Kohn\u2013Sham potential (7.99), and the process is repeated, yielding a new Kohn\u2013Sham wave function \u03c6(2) 0 (x), a new density n(2)(r), and a new calculation of the ground-state energy. This can be repeated as often as desired, to achieve the desired precision. Of course, the ultimate accuracy of the calculation will only be as", " good as the form used for the exchange-correlation energy functional. Table 7.1 shows the results for three iterations of this procedure. Recall from (7.48) that the experimental value is \u221279.0 eV, and the result (7.52) from the variational principle is \u221277.5 eV. The \ufb01rst iteration is better than (7.52), and iterations converge rapidly to the measured value. Note that for this problem, typical of small systems, the exchange-correlation energy is not a small contribution. Table 7.1 shows that here it is in fact relatively large, \u2248 40% of the total ground-state energy. It is instructive to examine the density functions for each iteration. See Figure 7.7. Even though our initial guess is rather different from the function on which the procedure converges, even one iteration gives something very close to the \ufb01nal answer. We note again that in addition to calculating the ground-state energy, we also now have a good approximation for the two-particle wave function (7.107) that can be used for other calculations of the properties of the helium atom. 454 Identical Particles 7.7 Quantum Fields A different approach to dealing with multiparticle states involves the introduction of quantum-mechanical \ufb01elds. One of the major attractions of quantum \ufb01eld theory is that it can deal with relativistic systems, where particles can be created or destroyed. The techniques are also very useful for nonrelativistic systems, however, so we will take a more general approach here. 7.7.1 Second Quantization This approach known as9 second quantization reexamines the way in which we de\ufb01ne the state vector. De\ufb01ne a multiparticle state vector as |n1, n2,..., ni,... (7.110) where the ni specify the number of particles with eigenvalue ki for some operator. Although we take it as a perfectly valid nomenclature for a state vector, it is a member of a new kind of vector space, called \u201cFock space,\u201d in which we need to build in the necessary permutation symmetry. A word of caution is in order. Our Fock space (or \u201coccupation number\u201d) notation for state vectors itself makes an important assumption, namely that there indeed exists a basis of noninteracting states. Interactions between particles can in fact, in principle, affect their very nature. Whether or not we can make a self-consistent theory starting on this assumption, and which in turn accurately describes nature, can only be tested by experiment. See the discussions in Landau (1996) and Merzbacher (1998). We will set this question aside, however, and move ahead at full steam. Let us now build a framework for a theory of many-particle systems using states in Fock space. We begin this task by recognizing two special cases of states in Fock space. The \ufb01rst of these is |0, 0,..., 0,... \u2261 |0 (7.111) for which there are no particles in any single-particle states. This state is called the \u201cvacuum\u201d and is, as usual, normalized to unity. The second special case is |0, 0,..., ni = 1,... \u2261 |ki (7.112) which is the state in which there is exactly one particle in the state with eigenvalue ki. Of course, this is just the single-particle state which has dominated our discussion of quantum mechanics, prior to this chapter. Now we need to learn how to build multiparticle states, and then make sure that this building process respects permutation symmetry. In an obvious nod to the creation and 9 The term \u201csecond quantization\u201d was apparently coined in the early days of trying to extend quantum mechanics into \ufb01eld theory. The idea was that wave functions were to be turned into operators, which in turn were subject to their own canonical quantization rules. Hence, quantization was enforced a \u201csecond\u201d time. See, for example, Section III.12.1 in Heitler (1954). 455 7.7 Quantum Fields annihilation operators that we \ufb01rst encountered in Section 2.3, we de\ufb01ne a \u201c\ufb01eld operator\u201d a\u2020 i which increases by one the number of particles in the state with eigenvalue ki, that is a\u2020 i |n1, n2,..., ni,... \u221d |n1, n2,..., ni + 1,... (7.113) where a normalization criterion will be used later to determine the proportionality constant. We postulate that the action of the particle creation operator a", "\u2020 i on the vacuum is to create a properly normalized single-particle state, namely This leads us to write which implies that a\u2020 i |0 = |ki. 1 = ki|ki = [0|ai] # # $ a\u2020 i |0 $ = 0| aia\u2020 i |0 = 0|ai|ki ai|ki = |0 (7.114) (7.115) (7.116) so that ai acts as a particle annihilation operator. We conclude with the following postulates for the particle annihilation operator, namely ai|n1, n2,..., ni,... \u221d |n1, n2,..., ni \u2212 1,... ai|0 = 0 ai|kj = 0 if i = j where an economy of notation lets us combine (7.116) and (7.119) into ai|kj = \u03b4ij|0. (7.117) (7.118) (7.119) (7.120) These are enough postulates to fully de\ufb01ne the \ufb01eld operators ai, short of actually incorporating permutation symmetry. The act of permuting two particles, one for the other, is most easily seen by putting the \u201c\ufb01rst\u201d particle in state |ki and then the \u201csecond\u201d particle in |kj, and comparing to what happens when we reverse the order in which these states are populated. That is, we expect that, for a two-particle state, i a\u2020 a\u2020 j |0 = \u00b1a\u2020 j a\u2020 i |0 (7.121) where the + (\u2212) sign is for bosons (fermions). Applying this same logic to particle exchange in multi-particle states, we are led to i, a\u2020 i, a\u2020 j where we make use of the \u201canticommutator\u201d {A, B} \u2261 AB + BA. Simply taking the adjoint of these equations tells us that \u2212 a\u2020 i = [a\u2020 j a\u2020 i a\u2020 a\u2020 i = {a\u2020 j a\u2020 j + a\u2020 i a\u2020 a\u2020 j ] =0 } = 0 Fermions (7.122b) (7.122a) Bosons j [ai, aj] =0 {ai, aj} = 0 Bosons Fermions. (7.123a) (7.123b) 456 Identical Particles Table 7.2 The Algebra for Identical Particles in Second Quantization Bosons Fermions j j a\u2020 \u2212 a\u2020 i, a\u2020 i = [a\u2020 a\u2020 i a\u2020 aiaj \u2212 ajai = [ai, aj] = 0 aia\u2020 j j ai = [ai, a\u2020 \u2212 a\u2020 j ] = \u03b4ij j ] = 0 } = 0 j a\u2020 i, a\u2020 j + a\u2020 i = {a\u2020 a\u2020 i a\u2020 aiaj + ajai = {ai, aj} = 0 aia\u2020 j ai = {ai, a\u2020 j + a\u2020 j j } = \u03b4ij i a\u2020 Note that the Pauli exclusion principle is automatically built into our formalism, since i = 0 for some single-particle state |ki. (7.122b) implies that a\u2020 Now what about the commutation rules for ai and a\u2020 j? We would like to de\ufb01ne a \u201cnumber operator\u201d Ni = a\u2020 i ai which would count the number of particles in the single-particle state |ki. Our experience from Section 2.3 argues shows that this is possible if we have [ai, a\u2020 i ] =1. In fact, a self-consistent picture of both bosons and fermions can be built in just this way, by replacing commutators with anticommutators. The complete algebra is summarized in Table 7.2. For both bosons and fermions, we can de\ufb01ne the operator N = \u2211 (7.124) a\u2020 i ai i which counts the total number of identical particles. (See Problem 7.9 at the end of this chapter.) We have taken a very ad hoc approach to coming up with the algebra in Table 7.2, rather contrary to the general tone of this book. It is in fact possible to do a somewhat better job in this respect, by postulating, for example, that certain quantities such as the total number of particles be unchanged under a basis change from single-particle states |ki to different states |lj which are connected by a unitary transformation.10 Nevertheless, it is not possible to do a fully self-consistent treatment minimizing ad hoc assumptions without developing relativistic quantum \ufb01eld theory, and", " that is not our mission here. 7.7.2 Dynamical Variables in Second Quantization How do we build operators in second quantization which do more than simply count the number of particles? The answer is straightforward, but once again it is necessary to make some ad hoc assumptions with our current approach. Suppose the single-particle states |ki are eigenstates of some \u201cadditive\u201d single-particle operator K. Examples might be momentum or kinetic energy. In some multiparticle state |\u03a8 = |n1, n2,..., ni,... we expect the eigenvalue of the multiparticle operator K to be \u2211i niki. This is easy to accomplish if we write (7.125) K = \u2211 kiNi = \u2211 kia\u2020 i ai. i i (7.126) 10 This approach, sometimes called the Principle of Unitary Symmetry, is exploited in Merzbacher (1998). 457 7.7 Quantum Fields Now suppose that the basis for which the single-particle states are speci\ufb01ed is different from the basis in which it is easiest to work. After all, we are used to working with the momentum operator in the coordinate basis, for example. If we use completeness to write then it makes sense to postulate that which implies that |ki = \u2211 |ljlj|ki j i = \u2211 a\u2020 b\u2020 j lj|ki j ai = \u2211 ki|ljbj j (7.127) (7.128a) (7.128b) j and bj create and annihilate particles in the single-particle states |lj. where the operators b\u2020 With these assignments, acting on the vacuum state (7.111) with (7.128a) yields (7.127). Equations (7.128) give us what we need to change the basis for our dynamical single particle operator. We have K = \u2211 ki \u2211 b\u2020 m lm|kiki|lnbn i mn = \u2211 mn = \u2211 mn mbn \u2211 b\u2020 lm|kikiki|ln i mbnlm| b\u2020 K\u2211 |kiki| |ln i or K = \u2211 mbnlm|K|ln. b\u2020 mn (7.129) This general form is suitable for writing down a second quantized version of any additive single-particle operator. Examples not only include momentum and kinetic energy, but also any \u201cexternal\u201d potential energy function which acts individually on each of the particles. All that matters is that the particles do not interact with each other. In the case of bosons, essentially all the particles may \ufb01nd themselves in the lowest energy level of such a potential well, so long as the temperature is low enough. (Experimentally, this phenomenon is referred to as a Bose\u2013Einstein condensate.) Fermions would behave differently, however. The Pauli exclusion principle will force the particles to populate increasingly higher energy levels in the well. For a system with a very large number of fermions, the total energy of the ground state could be enormous. The highest populated energy level (known as the \u201cFermi energy\u201d) might easily be much larger than the thermal energy \u223ckT. A classic example is a white dwarf star, a very dense object consisting basically of carbon atoms. The electrons in a white dwarf are, to a good approximation, bound in a potential well. The Fermi level is very high, much larger than the thermal energy for a temperature of tens of millions of degrees Kelvin. 458 Identical Particles Many-particle systems, however, present a new situation, namely the inevitable possibility that the particles actually interact among themselves. Once again, we postulate an additive operator, that is, one in which the individual two-particle interactions add up independently. Let the symmetric real matrix Vij specify the two-particle eigenvalue for an interaction between particles in single-particle states |ki and |kj. Then the second quantized version of this operator becomes V = 1 2 \u2211 i=j VijNiNj + 1 2 \u2211 i ViiNi(Ni \u2212 1). (7.130) The \ufb01rst term sums up all of the two-particle interactions, where the factor of 1/2 is necessary because this form double-counts pairs. The second term accounts for all \u201cself interactions\u201d for particles in the same state; there are n(n \u2212 1)/2 ways to take n things two at a time. The requirement that Vij be real ensures that V is Hermitian. The part of the self energy term in (7.130) containing N2 i exactly represents the parts of the sum in the \ufb01rst term removed by specifying i =", " j. Therefore, we can combine this more neatly as V = 1 2 \u2211 ij Vij (NiNj \u2212 Ni\u03b4ij) = 1 2 \u2211 ij Vij\u03a0ij (7.131) where \u03a0ij \u2261 NiNj \u2212 Ni\u03b4ij is called the pair distribution operator. Furthermore, we use Table 7.2 to write i ai\u03b4ij \u03a0ij = a\u2020 i aia\u2020 j aj \u2212 a\u2020 \u03b4ij \u00b1 a\u2020 i a\u2020 j aiaj j ai = a\u2020 i = \u00b1a\u2020 aj \u2212 a\u2020 i ai\u03b4ij or \u03a0ij = (\u00b1)(\u00b1)a\u2020 i a\u2020 j ajai (7.132) where we used (7.123) to reverse the order of the last two factors. This allows us to rewrite (7.130) as V = 1 2 \u2211 ij Vija\u2020 i a\u2020 j ajai. (7.133) This sequence of creation and annihilation operators, \ufb01rst one particle is annihilated, then another, and then creating them in reverse order, is called \u201cnormal ordering.\u201d Note that we see explicitly from (7.122b) or (7.123b) that there is no contribution from diagonal elements of V for fermions. We can use (7.128) to rewrite (7.133) in a different basis. We have V = 1 2 \u2211 mnpq mn|V|pqb\u2020 mb\u2020 nbqbp (7.134) 459 7.7 Quantum Fields where mn|V|pq \u2261 \u2211 Vijlm|kiki|lpln|kjkj|lq. (7.135) ij This result provides some insight into the physical meaning of our formalism. Suppose, for example, that the |ki are position basis states |x, and that the |li are momentum basis states |p = \u00afhk. Then Vij would represent an interaction between two particles, one located at x and the other at x. A natural example would be for a collection of particles each with charge q = \u2212e in which case we would write Vij \u2192 V(x, x) = e2 |x \u2212 x| \u2192 \u2211 ij d3x d3x (7.136) (7.137) but any mutual interaction between the particles would be treated in a similar way. The quantity mn|V|pq therefore represents a momentum-space version of the interaction, with m and p following one particle, and n and q following the other. (It is easy to show that mn|V|pq = nm|V|qp but interchanging one side and not the other will depend on whether the particles are bosons or fermions.) The four inner products in (7.135) lead to a factor ei(km\u2212kp)\u00b7x+i(kn\u2212kq)\u00b7x which, after the integrals (7.136) are carried out, results in a \u03b4-function which conserves momentum. One might diagrammatically represent the two-particle interaction as shown in Fig. 7.8. Clearly we are on our way towards developing a nonrelativistic version of quantum \ufb01eld theory. As a speci\ufb01c example, we will treat the quantum-mechanical version of the kp km V kq kn Fig. 7.8 Diagrammatic representation of the \u201cmomentum-space matrix element\u201dmn|V|pq. 460 Identical Particles noninteracting electromagnetic \ufb01eld shortly. However, we will not push the general case beyond this point, as it goes beyond the scope of this book and is treated well in any one of several other books. See, for example, Merzbacher (1998), Landau (1996), and Fetter and Walecka (2003). 7.7.3 Example: The Degenerate Electron Gas An excellent example of the principles discussed in this section, is the degenerate electron gas. This is a collection of electrons, interacting with each other through their mutual Coulomb repulsion, bound in some positively charged background medium. Physical examples might include a high temperature plasma, or even, to some approximation, bulk metals. This problem is treated thoroughly in Chapter One, Section 3 of Fetter and Walecka (2003). We present the problem and outline its solution here, but the interested reader is referred to the original reference to \ufb01ll in the details. Our task is to \ufb01nd the eigenvalues of the Hamiltonian for a system of N electrons. The electrons interact among themselves according to H = Hel + Hb + Hel-b Hel = \u2211 i p2 i 2m + 1 2 e2 \ufffd", "\ufffd i \u2211 j=i e\u2212\u03bc|xi\u2212xj| |xi \u2212 xj| (7.138) (7.139) where we employ a \u201cscreened\u201d Coulomb potential but will let \u03bc \u2192 0 before we \ufb01nish the calculation. The energy of the positive background is Hb = 1 2 e2 d3x d3x\u03c1(x)\u03c1(x) e\u2212\u03bc|x\u2212x| |x \u2212 x| (7.140) where \u03c1(x) is the number density of background particle sites. We will assume a uniform background, with \u03c1(x) = N/V for a system of volume V = L3. Then, translating to a variable x \u2261 x \u2212 x (7.140) becomes Hb = 1 2 e2 N V 2 d3x d3x e\u2212\u03bc|x| |x| = 1 2 e2 N2 V 4\u03c0 \u03bc2. (7.141) Thus Hb contributes simply an additive constant to the energy. The fact that this constant grows without bound as \u03bc \u2192 0 will not be a problem, as we shall see shortly. The interaction of the electrons with the constant background is Hel-b = \u2212e2 \u2211 i = \u2212e2 N V \u2211 i d3x\u03c1(x) d3x e\u2212\u03bc|x\u2212xi| |x \u2212 xi| |x \u2212 xi| = \u2212e2 N2 e\u2212\u03bc|x\u2212xi| V 4\u03c0 \u03bc2. Therefore (7.138) becomes H = \u2212 1 2 e2 N2 V 4\u03c0 \u03bc2 +\u2211 i p2 i 2m + 1 2 e2 \u2211 i \u2211 j=i e\u2212\u03bc|xi\u2212xj| |xi \u2212 xj|. (7.142) (7.143) 461 7.7 Quantum Fields The \ufb01rst term in this equation is just a number. The second is a one-body operator which we will express simply in terms of operators in second quantization in momentum space. The third term is a two-body operator which will involve a bit more work to write in second quantization. Writing the kinetic-energy term in (7.143) is just a matter of rewriting (7.129) forK being the momentum operator p and the |ln being momentum basis states. Single-particle states are denoted by i = {k, \u03bb} where \u03bb = \u00b1 indicates electron spin. We know that so we have k\u03bb|p|k\u03bb = \u00afhk\u03b4kk \u03b4\u03bb\u03bb p2 i 2m \u2211 i \u21d2 K = \u2211 k\u03bb \u00afh2k2 2m a\u2020 k\u03bbak\u03bb. (7.144) (7.145) Now we write the potential energy term in (7.143) in second quantization, using (7.134) and (7.135). Note (7.136) and (7.137). We have V = 1 2 \u2211 k1 \u03bb1 \u2211 k2 \u03bb2 \u2211 k3 \u03bb3 \u2211 k4 \u03bb4 where k1\u03bb1k2\u03bb2|V|k3\u03bb3k4\u03bb4a\u2020 k1 \u03bb1 a\u2020 k2 \u03bb2 ak4 \u03bb4ak3 \u03bb3 (7.146) k1\u03bb1k2\u03bb2|V|k4\u03bb4k3\u03bb3 = = = = = e2 V2 e2 V2 e2 V2 e2 V d3x d3xV(x, x)k1\u03bb1|xx|k3\u03bb3k2\u03bb2|xx|k4\u03bb4 d3x d3x \u03b4\u03bb1 \u03bb4 \u03b4\u03bb2 \u03bb3 \u03bb1 eik3\u00b7x \u03c7\u03bb3e\u2212ik2\u00b7x \u03c7\u2020 \u03bb2 d3x e\u2212\u03bc|x\u2212x| |x \u2212 x| e\u2212ik1\u00b7x \u03c7\u2020 e\u2212\u03bcy e\u2212ik2\u00b7x \u03b4\u03bb2 \u03bb4eik4\u00b7x e\u2212ik1\u00b7x \u03b4\u03bb1 \u03bb3 eik3\u00b7x y e\u2212\u03bcy y d3xe\u2212i(k1+k2\u2212k3\u2212k4)\u00b7x d3y d3y e\u2212i(k1\u2212k3)\u00b7y eik4\u00b7x \u03c7\u03bb4 \u03b4\u03bb1 \u03bb4 \u03b4\u03bb2 \u03bb3 \u03b4k1+k2,k3+k4 d3y e\u2212i(k1\u2212k3)\u00b7y (7.147) e\u2212\u03bcy y where the \u03c7\u03bb are spinor representations, and we use a change of variables x =", " x and y = x \u2212 x. Finally, we de\ufb01ne the momentum transfer q \u2261 k1 \u2212 k3 and \ufb01nd k1\u03bb1k2\u03bb2|V|k3\u03bb3k4\u03bb4 = e2 V \u03b4\u03bb1 \u03bb4 \u03b4\u03bb2 \u03bb3 \u03b4k1+k2,k3+k4 4\u03c0 q2 + \u03bc2. (7.148) The Kronecker deltas in the spin just insure that that no spins are \ufb02ipped by this interaction, which we expect since the interaction is spin independent. The Kronecker delta in wave number insures that momentum is conserved. Thus (7.146) becomes V = e2 2V \u2211 k1 \u03bb1 \u2211 k2 \u03bb2 \u2211 k3 \u2211 k4 \u03b4k1+k2,k3+k4 4\u03c0 q2 + \u03bc2 a\u2020 k1 \u03bb1 a\u2020 k2 \u03bb2 ak4 \u03bb2ak3 \u03bb1 (7.149) after reducing the summations using the spin-conserving Kronecker deltas. 462 Identical Particles An important feature of (7.149) becomes apparent if we \ufb01rst rede\ufb01ne k3 \u2261 k and k4 \u2261 p. Then the terms of (7.149) for whichq = 0 become e2 2V \u2211 kp \u2211 \u03bb1 \u03bb2 4\u03c0 \u03bc2 a\u2020 k\u03bb1 a\u2020 p\u03bb2 ap\u03bb2ak\u03bb1 = = e2 2V e2 2V a\u2020 k\u03bb1 k\u03bb1 \u2211 p\u03bb2 4\u03c0 \u03bc2 \u2211 4\u03c0 \u03bc2 (N2 \u2212 N) ak\u03bb1 a\u2020 p\u03bb2 ap\u03bb2 \u2212 \u03b4kp\u03b4\u03bb1 \u03bb2 (7.150) where we have made use of the fermion anticommutation relations, and the de\ufb01nition of the number operator. The \ufb01rst term in this relation just cancels the \ufb01rst term of (7.143). The second term represents an energy \u22122\u03c0e2/\u03bc2V per particle, but this will vanish in the limit where V = L3 \u2192 \u221e while always keeping \u03bc 1/L. Thus the terms with q = 0 do not contribute, and cancel the rapidly diverging terms in the Hamiltonian. Indeed, this \ufb01nally allows us to set the screening parameter \u03bc = 0 and write the second quantized Hamiltonian as H = H0 + H1 \u00afh2k2 2m H0 = \u2211 k\u03bb e2 2V H1 = a\u2020 k\u03bbak\u03bb \u2211 kpq \u2211 \u03bb1 \u03bb2 4\u03c0 q2 a\u2020 k+q,\u03bb1 a\u2020 p\u2212q,\u03bb2 ap\u03bb2ak\u03bb1 (7.151a) (7.151b) (7.151c) where the notation \u03a3 indicates that terms with q = 0 are to be omitted. Note that in the limit we have taken, a \ufb01nite density n = N/V is implicitly assumed. Finding the eigenvalues of (7.151) is a dif\ufb01cult problem, although solutions are possible. Our approach will be to \ufb01nd the ground-state energy by treating the second term as a perturbation on the \ufb01rst. Although reasonable arguments can be made why this should be a good approximation (see Fetter and Walecka), those arguments only hold in a particular range of densities. Fortunately, that range of densities is relevant to physical systems such as metals, so our approach indeed has practical interest. This is a good time to introduce some scaling variables. The density is determined by the interatomic spacing r0, that is n = N V = \u22121 4\u03c0 3 r3 0 (7.152) and a natural scale for r0 is the Bohr radius (3.317), i.e. a0 = \u00afh2/me2. We de\ufb01ne a dimensionless distance scale rs = r0/a0, called the Wigner\u2013Seitz radius. Our calculation of the ground-state energy will be as a function of rs. As an introduction to calculating the expectation value E(0) of the operator H0 for the ground state, we discuss the concept of Fermi energy. (Recall the discussion on p. 457.) Because of the Pauli exclusion principle, electrons will \ufb01ll the available energy levels up to some maximum wavenumber kF. We can relate kF to the total number of electrons by adding up all of the states with k \u2264 kF, that is 463 7.7 Quantum Fields \u03b8", "(kF \u2212 k) N = \u2211 k\u03bb \u2192 V V 3\u03c02 k3 where \u03b8(x) =0 forx < 0 and unity otherwise. This implies that d3k \u03b8(kF \u2212 k) = (2\u03c0)3 \u2211 \u03bb F 3\u03c02N V 1/3 = 9\u03c0 4 1/3 1 r0 kF = (7.153) (7.154) which shows that kF is about the same size as the inverse interparticle spacing. Now use the same approach to calculate the unperturbed energy E(0). Denoting the ground state as |F we have E(0) = F|H0|F = k2\u03b8(kF \u2212 k) \u00afh2 2m \u2211 k\u03bb d3k k2\u03b8(kF \u2212 k) = \u2192 \u00afh2 2m V (2\u03c0)3 \u2211 \u03bb e2 2a0 N 3 5 9\u03c0 4 2/3 1 r2 s. (7.155) Note that e2/2a0 \u2248 13.6 eV, the ground-state energy of the hydrogen atom. The \ufb01rst-order correction to the ground-state energy is E(1) = F|H1|F e2 \u2211 2V \u03bb1 \u03bb2 \u2211 kpq = 4\u03c0 q2 F|a\u2020 k+q,\u03bb1 a\u2020 p\u2212q,\u03bb2 ap\u03bb2ak\u03bb1 |F. (7.156) The summation is easy to reduce since |F is a collection of single-particle states with occupation numbers either zero or one. The only way for the matrix element in (7.156) to be nonzero is if the annihilation and creation operators pair up appropriately. Since q = 0 in the sum, the only way to pair up the operators is by setting {p \u2212 q, \u03bb2} = {k, \u03bb1} and {k + q, \u03bb1} = {p, \u03bb2}. Therefore E(1) = F|a\u2020 k+q,\u03bb1 a\u2020 k,\u03bb1 ak+q,\u03bb1ak\u03bb1 |F 4\u03c0 q2 4\u03c0 q2 \u2211 kq \u2211 e2 \u2211 2V \u03bb1 = \u2212 e2 \u2211 2V \u03bb1 = \u2212 e2 2V = \u2212e2 4\u03c0V (2\u03c0)6 2 kq V2 (2\u03c0)6 F| a\u2020 k+q,\u03bb1 ak+q,\u03bb1 a\u2020 k,\u03bb1 ak\u03bb1 |F d3k d3q 4\u03c0 q2 d3q 1 q2 \u03b8(kF \u2212 |k + q|)\u03b8(kF \u2212 k) d3P \u03b8 kF \u2212 |P + q| \u03b8 1 2 kF \u2212 |P \u2212 1 2 q|. (7.157) The integral over P is just the intersection between two spheres of radius kF but with centers separated by q, and is easy to evaluate. The result is E(1) = \u2212 e2 2a0 N 3 2\u03c0 9\u03c0 4 1/3 1 rs. (7.158) 464 Identical Particles E/N 0.10e2/2a0 Exact result as rs \u2192 0 2 4 6 8 10 rs rs = 4.83 \u201cWigner solid\u201d E/N = \u22120.095e2/2a0 \u22120.10e2/2a0 Fig. 7.9 The ground-state energy, to first order in perturbation theory, for a system of N electrons inside a uniform, positively charged background. The energy per electron is plotted as a function of the interparticle spacing in units of the Bohr radius. From Fetter and Walecka (2003). Therefore, the ground-state energy to \ufb01rst order is given by E N = e2 2a0 9\u03c0 4 2/3 3 5 1 r2 s \u2212 3 2\u03c0 1 rs. (7.159) This is plotted in Figure 7.9. The unperturbed energy decreases monotonically as rs \u2192 0, but the \ufb01rst-order correction is an attraction which falls more slowly. The result is a minimum at a value E/N = \u22120.095e2/2a0 = \u22121.29 eV where rs = 4.83. Our model is crude, and the solution only approximate, but the agreement with experiment is surprisingly good. For sodium metal, one \ufb01nds E/N = \u22121.13 eV where rs = 3.96. 7.8 Quantization of the Electromagnetic Field Maxwell\u2019s equations form a complete classical description of noninter", "acting electric and magnetic \ufb01elds in free space. It is tricky to apply quantum mechanics to that description, but it can be done in a number of ways. In this section, we will once again take a \u201cfollow our nose\u201d approach to the problem, based on the many-particle formalism developed in this chapter. The particles, of course, are photons,11 whose creation and annihilation operators obey Bose\u2013Einstein commutation relations. We start with a brief summary of Maxwell\u2019s equations to establish our notation, and their solution in terms of electromagnetic waves. Then we derive the energy, and associate it with the eigenvalues of a Hamiltonian constructed using bosonic creation and annihilation operators. 11 The concept of photons is not without controversy. See \u201cAnti-photon\u201d by Lamb Appl. Phys. B, 60 (1995) 77. The astute reader will note that we derived the cross section for the photoelectric effect in Section 5.8.3 using the classical electromagnetic \ufb01eld. 465 7.8 Quantization of the Electromagnetic Field Including interactions with electromagnetic \ufb01elds, through the inclusion of spin 1 2 charged electrons, is the subject of quantum electrodynamics. We do not pursue this subject in this book. (See Section 5.8 for a discussion of a more ad hoc way to apply electromagnetic \ufb01eld interactions to atomic systems.) However, there is a fascinating quantummechanical effect observable with free electromagnetic \ufb01elds, the Casimir effect, and we conclude this section with a description of the calculation and the experimental data. Our treatment here more or less follows Chapter Four in Loudon (2000), although the approach has become rather standard. See, for example, Chapter 23 in Merzbacher (1998). 7.8.1 Maxwell\u2019s Equations in Free Space In the absence of any charges or currents, Maxwell\u2019s equations (in Gaussian units; see Appendix A) take the form \u2207 \u00b7 E = 0 \u2207 \u00b7 B = 0 \u2202B 1 \u2202t c \u2202E \u2207 \u00d7 B \u2212 1 \u2202. Following standard procedure, we postulate a vector potential A(x, t) such that B = \u2207 \u00d7 A (7.160a) (7.160b) (7.160c) (7.160d) (7.161) which means that (7.160b) is immediately satis\ufb01ed. If we impose the further condition (which is known as \u201cchoosing the Coulomb gauge\u201d) then \u2207 \u00b7 A = 0 E = \u2212 1 c \u2202A \u2202t (7.162) (7.163) means that (7.160a) and (7.160c) are also satis\ufb01ed. Therefore, determining A(x, t) is equivalent to determining E(x, t) and B(x, t). A solution for A(x, t) is evident, though, by observing that (7.160d) leads directly to \u22072A \u2212 1 c2 \u22022A \u2202t2 = 0. (7.164) That is, A(x, t) satis\ufb01es the wave equation, with wave speed c, just as we might have guessed. The set of solutions to (7.164) are naturally written as A(x, t) =A(k)e \u00b1ik\u00b7xe\u00b1i\u03c9t (7.165) where \u03c9 = \u03c9k \u2261 |k|c = kc for the solution to be valid. The Coulomb gauge condition (7.162) implies that \u00b1ik \u00b7 A(x, t) = 0 or k \u00b7 A(k) =0. (7.166) 466 Identical Particles In other words, A(x, t) is perpendicular to the propagation direction k. For this reason, the Coulomb gauge is frequently referred to as the \u201ctransverse gauge.\u201d This allows us to write the general solution to (7.164) as A(x, t) = \u2211 k,\u03bb \u02c6ek\u03bbAk,\u03bb(x, t) (7.167) where \u02c6ek\u03bb are two unit vectors (corresponding to two values for \u03bb) perpendicular to k, and where Ak,\u03bb(x, t) =A k,\u03bbe\u2212i(\u03c9kt\u2212k\u00b7x) + A\u2217 k,\u03bbe+i(\u03c9kt\u2212k\u00b7x). (7.168) Note that in (7.168) the quantities written Ak,\u03bb on the right side of the equation are numerical coef\ufb01cients, not functions of either position or time. Note also that k and \u2212k represent different terms in the sum. We write the super", "position (7.167) as a sum, not an integral, because we envision quantizing the electromagnetic \ufb01eld inside a \u201cbig box\u201d whose dimensions may eventually be taken to grow without bound. We use the form (7.168) to insure that Ak,\u03bb(x, t) is real. When we quantize the electromagnetic \ufb01eld, Ak,\u03bb(x, t) will become a Hermitian operator. The coef\ufb01cients A\u2217 k,\u03bb and Ak,\u03bb will become creation and annihilation operators. As we shall see later, it is useful to take the unit vectors \u02c6ek\u03bb as directions of circular are the linear unit vectors polarization as opposed to linear. That is, if \u02c6e(1) k perpendicular to k, then and \u02c6e(2) k \u02c6ek\u00b1 = \u2213 1\u221a 2 \u02c6e(1) k \u00b1 i\u02c6e(2) k (7.169) where \u03bb = \u00b1 denotes the polarization state. With these de\ufb01nitions, it is easy to show that \u02c6e\u2217 k\u03bb \u00b7 \u02c6e\u00b1k\u03bb = \u00b1\u03b4\u03bb\u03bb \u02c6e\u2217 k\u03bb \u00d7 \u02c6e\u00b1k\u03bb = \u00b1i\u03bb\u03b4\u03bb\u03bb \u02c6k (7.170a) (7.170b) where \u02c6k is a unit vector in the direction of k. The electric \ufb01eld E(x, t) can now be written down from (7.163), and similarly for the magnetic \ufb01eld B(x, t) using (7.161). The energy E in the electromagnetic \ufb01eld is given by integrating the energy density over all space, that is E = 1 8\u03c0 # V |E(x, t)|2 + |B(x, t)|2 d3x (7.171) $ where, as discussed earlier, \u201call space\u201d is a \ufb01nite volume V = L3 with periodic boundary conditions. In other words, we are working inside an electromagnetic cavity with conducting walls. This means that k = (kx, ky, kz) = 2\u03c0 L (nx, ny, nz) (7.172) where nx, ny, and nz are integers. 467 7.8 Quantization of the Electromagnetic Field Consider \ufb01rst the term dependent on the electric \ufb01eld in (7.171). Using (7.163) with (7.167) and (7.168) we have E = i c \u2211 k,\u03bb \u03c9k # Ak,\u03bbe\u2212i(\u03c9kt\u2212k\u00b7x) \u2212 A\u2217 k,\u03bbe+i(\u03c9kt\u2212k\u00b7x) $ \u02c6ek\u03bb and # E\u2217 = \u2212 i c \u03c9k \u2211 k,\u03bb k,\u03bbe+i(\u03c9 A\u2217 t\u2212k\u00b7x) \u2212 Ak,\u03bbe\u2212i(\u03c9 k k t\u2212k\u00b7x) $ \u02c6e\u2217 k \u03bb. (7.173a) (7.173b) Since we have already suggested that the A\u2217 annihilation operators, we need to take care and keep their order intact. k,\u03bb and Ak,\u03bb will become creation and This all leads to an awkward expression for |E|2 = E\u2217 \u00b7 E, a summation over k, \u03bb, k, and \u03bb, with four terms inside the summation. However, an important simpli\ufb01cation follows from the integral over the spatial volume. Each term inside the sum packs all of its position dependence into an exponential so that the volume integral is of the form ei(k\u2213k)\u00b7xd3x = V\u03b4k,\u00b1k. V (7.174) Combining this with (7.170a) one \ufb01nds |E(x, t)|2 d3x = \u2211 k,\u03bb V! \u03c92 k c2 V k,\u03bbAk,\u03bb + Ak,\u03bbA\u2217 A\u2217 k,\u03bb + A\u2217 k,\u03bbA\u2217 \u2212k,\u03bbe2i\u03c9kt + Ak,\u03bbA\u2212k,\u03bbe\u22122i\u03c9kt \". Starting with (7.161), the calculation for |B|2 = B\u2217 \u00b7 B is very similar. The curl brings in factors like k \u00d7 \u02c6ek\u03bb instead of the \u03c9k/c in the calculation involving the electric \ufb01eld, but since k2 = \u03c92 k /c2 the result is nearly identical. The key difference, though, is that under the change k \u2192 \u2212", "k terms like k \u00d7 \u02c6ek\u03bb do not change sign. This means that the terms analogous to the third and fourth terms in (7.175) appear the same way but with opposite signs. Therefore, they cancel when evaluating (7.171). The result is (7.175) E = 1 4\u03c0 V\u2211 k,\u03bb! \u03c92 k c2 k,\u03bbAk,\u03bb + Ak,\u03bbA\u2217 A\u2217 k,\u03bb \". (7.176) 7.8.2 Photons and Energy Quantization Our goal now is to associate (7.176) with the eigenvalues of a Hamiltonian operator. We will do this by hypothesizing that the quantized electromagnetic \ufb01eld is made up of a collection of identical particles called photons. An operator a\u2020 \u03bb(k) creates a photon with polarization \u03bb and momentum \u00afhk, and a\u03bb(k) annihilates this photon. The energy of a photon is \u00afh\u03c9k = \u00afhck, so we will build our Hamiltonian operator based on (7.126) and write H = \u2211 k,\u03bb \u00afh\u03c9k a\u2020 \u03bb(k)a\u03bb(k) +E 0 (7.177) where we will \ufb01nd it convenient to allow an arbitrary constant E0 to be added to the Hamiltonian. We do not need to consider terms like (7.133) since, by our starting assumption, we are building a noninteracting electromagnetic \ufb01eld. 468 Identical Particles We are now faced with an important question. Are photons bosons or fermions? That is, what is the \u201cspin\u201d of the photon? We need to know whether it is integer or half-integer, in order to know which algebra is followed by the creation and annihilation operators. A fully relativistic treatment of the photon \ufb01eld demonstrates that the photon has spin one and is therefore a boson, but do we have enough preparation at this point to see that this should be the case? Yes, we do. We know from Chapter 3 that rotation through an angle \u03c6 about (say) the z-axis is carried out by the operator exp(\u2212iJz\u03c6/\u00afh). The possible eigenvalues m of Jz show up explicitly if we rotate a state which happens to be an eigenstate of Jz, introducing a phase factor exp(\u2212im\u03c6). (This is what gives rise to the \u201cfamous\u201d minus sign when a spin 2 state is rotated through 2\u03c0. Recall (3.33).) 1 So, consider what happens if we rotate about the photon direction k through an angle \u03c6 for a right- or left-handed circularly polarized electromagnetic wave. The polarization directions are the unit vectors \u02c6ek\u00b1 given by (7.169). The rotation is equivalent to the transformation (7.178a) \u02c6e(1) k \u02c6e(2) k \u2192 \u02c6e(1) \u2192 \u02c6e(2) k = cos \u03c6\u02c6e(1) k = sin \u03c6\u02c6e(1) \u2212 sin \u03c6\u02c6e(2) k k + cos \u03c6\u02c6e(2) k (7.178b) which means that the rotation introduces a phase change exp(\u2213i\u03c6) to the \u02c6ek\u00b1. Apparently, right- and left-handed circularly polarized photons correspond to eigenvalues \u00b11\u00afh of Jz. The photon seems to have spin one. k Consequently, we proceed under the assumption that photons are bosons. We therefore rewrite (7.177) slightly as H = \u2211 k,\u03bb = \u2211 k,\u03bb \u00afh\u03c9k \u00afh\u03c9k \u03bb(k)a\u03bb(k) +a \u2020 a\u2020 \u03bb(k)a\u03bb(k) + E0 $ $ \u03bb(k)a\u03bb(k) +a \u03bb(k)a\u2020 a\u2020 \u03bb(k) \u2212 1 + E0 (7.179) # # 1 2 1 2 and recover the classical energy (7.176) with the de\ufb01nition of the operator Ak,\u03bb = (4\u03c0\u00afhc2)1/2 1\u221a V along with the realization of a \u201czero point\u201d energy a\u03bb(k) 1\u221a 2\u03c9k E0 = 1 2 \u2211 k,\u03bb \u00afh\u03c9k = \u2211 \u00afh\u03c9k. k (7.180) (7.181) This is the energy in the electromagnetic \ufb01eld when there are zero photons present, and is sometimes called the vacuum energy. It is an in\ufb01nite number, but nevertheless a constant. More important, it has observable consequences. 7", ".8.3 The Casimir Effect The vacuum energy of the electromagnetic \ufb01eld has a number of physical consequences, but probably the most dramatic is its ability to exert a macroscopic force between conducting surfaces. This is called the Casimir effect, and it has been precisely measured 469 7.8 Quantization of the Electromagnetic Field and compared to calculations. Fine accounts by S. Lamoreaux have been published, including a popular article in Physics Today (60 (2007) 40), and a more technical review Rep. Prog. Phys., 68 (2005) 201. See also M. Fortun, \u201cFluctuating about zero, taking nothing\u2019s measure,\u201d in Marcus (2000). Casimir\u2019s calculation relies only on the assumption of the vacuum energy (7.181). We reproduce it here, following Lamoreaux\u2019s technical review article,12 above. Two large, parallel, conducting plates are separated by a distance d. De\ufb01ne a coordinate system where the (x, y) plane is parallel to the surface of the conducting plates, so z measures the distance perpendicularly away from one surface. This allows us to write down a potential energy function U(d) =E 0(d) \u2212 E0(\u221e) (7.182) which just gives the difference in the vacuum energy for plates with a \ufb01nite and in\ufb01nite separation. Combining (7.181) with (7.172) (and combining positive and negative integer values) we have E0(d) = \u00afh \u2211 kx,ky,n \u03c9k = \u00afhc \u2211 kx,ky,n \u2019 x + k2 k2 y + 2. n\u03c0 d (7.183) (This equation actually is missing a \u201clost\u201d factor of 1/2 on the n = 0 term. This is because only one polarization state should be counted in (7.181) for n = 0, since there is only one purely transverse mode when kz = 0. We will recover this factor below.) Now assume square plates with x and y lengths L d. SinceL is large, we can replace the summations over kx and ky with integrals and write 2 \u221e \u221e \u2019 E0(d) = \u00afhc L \u03c0 dkx dky \u2211 x + k2 k2 y +. (7.184) 2 n\u03c0 d 0 For the limit d \u2192 \u221e we can also replace the sum over n with an integral. This gives us all the necessary ingredients to evaluate (7.182). n 0 Unfortunately, however, (7.182) is the difference between two in\ufb01nite numbers. It is plausible that the difference is \ufb01nite, since for any particular value of d, terms with large enough n will give the same result for different values of d. That is, both terms in (7.182) should tend towards in\ufb01nity in the same way, and these parts will cancel when taking the difference. This suggests that we can handle the in\ufb01nities by multiplying the integrand in (7.184) by a function f (k) where f (k) \u2192 1 fork \u2192 0 andf (k) \u2192 0 fork \u2192 \u221e. This function \u201ccuts off\u201d the integrand before it gets too large, but does so in the same way to both terms in (7.182) so that the contributions from large k still cancel.13 It is also helpful to introduce the polar 12 We note that Lamoreaux\u2019s derivation closely follows that of Itzykson and Zuber (1980), Section 3-2-4. See also Holstein (1992) for a somewhat different approach, and attendant discussion, with a particularly physical perspective. 13 We can think of many physical reasons why there should be a cutoff at very high frequencies. In general, we expect the main contributions to come from values of k \u223c 1/d, but there are more speci\ufb01c examples such as the response of electrons in metals to very high-energy photons. It remains an interesting problem, in any case, to see whether the eventual result can be derived even if there is no cutoff frequency. 470 Identical Particles x + k2 k2 coordinate \u03c1 = in (7.184) correspond to 1/4 of the (kx, ky) plane. Then (7.182) becomes y in which case dkxdky = 2\u03c0\u03c1d\u03c1. Note that the integration limits \u2019 \u2019 \u221e U(d) =2 \u03c0\u00afhc \u03c1d\u03c1 \u2211 f \u03c12 + L \u03c0 2 1 4 2 n\u03c0 d \u03c12 + n\u03c0 d 2 dkzf \u03c12 + k2 z", " \u03c12 + k2 z. (7.185) 0 n \u2212 d \u03c0 \u221e 0 Now de\ufb01ne a function F(\u03ba) as \u221e F(\u03ba) = \u03c0 d \u03c0 d Putting \u03c12 = (\u03c0/d)2x and kz = (\u03c0/d)\u03ba allows us to write the potential energy more succinctly, reclaiming the lost factor of two, as x + \u03ba2 y (7.186b) (7.186a) x + \u03ba2 2y2f 0 \u221e dx f dy. = \u03ba U(d) = \u03c02 \u00afhc 4d3 L2 1 2 F(0) + \u221e \u2211 n=1 F(n) \u2212 \u221e 0 F(\u03ba)d\u03ba. (7.187) We are therefore left with evaluating the difference between an integral and a sum, both of which are reasonable approximations of each other. Indeed, if a function F(x), de\ufb01ned over range 0 \u2264 x \u2264 N, is evaluated at integer points x = i, then the approximation scheme known as the trapezoidal rule says that N 0 F(x)dx \u2248 F(0) +F ( N ) 2 + N \u2211 i=1 F(i). (7.188) In our case, N \u2192 \u221e with F( N ) \u2192 0 (thanks to the cutoff function f (k)), and our job is to \ufb01nd the difference between the left- and right-hand sides of (7.188). Fortunately, there is a theorem which evaluates this difference. It is called the Euler\u2013 Maclaurin summation formula and can be written as F(0) 2 + \u221e \u2211 i=1 F(i) \u2212 \u221e 0 F(x)dx = \u2212 1 12 F(0) + 1 720 F(0) +\u00b7 \u00b7 \u00b7. (7.189) The derivatives can be calculated using (7.186b). Since F(x) \u2192 0 as x \u2192 \u221e, we have F(y) =\u2212 2y2f y \u03c0 d which gives F(0) =0. If we make one further, but natural, assumption about the cutoff function f (k), namely that all of its derivatives go to zero as k \u2192 0, then we are only left with the third derivative term in (7.189). In fact, F(0) =\u2212 4 and U(d) = \u03c02 \u00afhc 4d3 L2 \u22124 720 = \u2212 \u03c02 \u00afhc 720d3 L2. So, \ufb01nally, we derive the Casimir force (per unit area) to be \u03c02 \u00afhc 240d4. \u2212 dU dd F (d) = = \u2212 1 L2 (7.190) (7.191) (7.192) 471 7.8 Quantization of the Electromagnetic Field Photodiodes B A 20 0 \u221220 \u221240 \u221260 \u221280 \u2212100 \u2212120 ) 100 200 Laser Sphere d Plate Piezo 400 300 700 500 Plate-sphere separation (nm) 600 800 900 Fig. 7.10 Experimental verification of the Casimir effect, from Mohideen and Anushree Phys. Rev. Lett., 81 (1998) 4549. For experimental reasons, the force is measured between a metallic sphere and a flat plate, rather than two flat plates. A laser precisely measures the small deflection, from which the force is deduced. The force (measured in 10\u221212 N) varies as a function of separation between the sphere and the plate, in excellent agreement with the prediction, shown by the line through the data points, based on a quantized electromagnetic field. Thus, there is an attractive force between the plates which varies as the inverse fourth power of the separation, due to the recon\ufb01guration of the vacuum energy in the quantized electromagnetic \ufb01eld. This is one of the examples in nature where a purely quantum-mechanical effect manifests itself in a macroscopic system. Indeed, the Casimir force between conductors has been precisely measured and the result is in excellent agreement with the theory. See Figure 7.10. This experiment makes use of the \u201catomic force microscope\u201d concept, which relies on the bending of a microscopic cantilever beam in response to a tiny force between nearby surfaces. For this reason, an apparatus is used which suspends a small conducting sphere from the cantilever, and measures the force between the sphere and a \ufb02at plate, given by \u2212(\u03c03R/360)(\u00afhc/d3) where R is the sphere radius. The force de\ufb02ects the cantilever, and this motion is detected using a laser which re\ufb02ects", " from the sphere surface. The measured force as a function of the distance d is shown in the \ufb01gure as data points, and is compared with the theoretical prediction. The Casimir effect has also been observed experimentally with parallel conducting surfaces. See, for example, Bressi et al., Phys. Rev. Lett., 88 (2002) 041804. If the Casimir Effect is due to the presence of electromagnetic \ufb01elds, and these \ufb01elds interact only with charges, then why does the electric charge e not appear anywhere in (7.192)? The answer lies in our starting point for the calculation, where we assumed the boundary conditions for conducting plates. These arise from the relative mobility of the electrons in the metal, with which the electromagnetic \ufb01eld interacts. In fact, we made use of a cutoff frequency whose physical basis can lie in the penetrability of electromagnetic radiation at short wavelengths. Indeed, if this penetrability existed for all wavelengths, there would be no Casimir effect. The Casimir effect has seen renewed interest potential application in nanomechanical devices, but also for in recent years, not only for its its calculation and 472 Identical Particles interpretation14 using fundamental quantum \ufb01eld theoretical principles. In a formulation in terms of path integrals, the Casimir energy can be written down in terms of the free \ufb01eld propagator with appropriate boundary conditions. The boundary conditions are simply de\ufb01ned by the objects under consideration. The result is an elegant expression for the Casimir energy in terms of the T-matrix scattering amplitudes for the free \ufb01eld from the objects, and transformation matrices that express each object\u2019s geometry in a natural way with respect to the other. This approach lends itself to a number of insights. For example, it allows one to calculate the Casimir energy for any \ufb01eld which can be expressed in terms of this constrained propagator, such as scalar or fermion \ufb01elds. It is also clearly amenable to any number of geometries, far beyond simple parallel plates. 7.8.4 Concluding Remarks Before leaving this chapter, we should point out that our treatment in this section only scratches the surface of applications of quantizing the electromagnetic \ufb01eld. Now that we have expression (7.180) (and its adjoint) which is an operator which destroys (or creates) photons of speci\ufb01c wavelength and polarization, we can incorporate it in any number of ways. For example, we have already seen in (2.343) how we can add the electromagnetic \ufb01eld into the conjugate momentum. This is built into the Hamiltonian in (2.346). Using the quantized version for A we have an ad hoc Hamiltonian operator which can create or destroy photons. Terms, then, proportional to A \u00b7 p can be treated as time-dependent perturbations. Thus we can let a photon be absorbed by an atom (the photoelectric effect) or let an excited state of an atom decay spontaneously and emit a photon. These applications, of course, can be brought to bear just as well in systems covered by nuclear physics or condensed matter physics. These topics are covered in a wide variety of books, some on quantum mechanics in general, but many in books which cover speci\ufb01c research areas. One particularly fascinating direction, which in fact involves noninteracting electromagnetic \ufb01elds, is quantum optics. This is a \ufb01eld that has come of age in the past few decades, spurred on partly by advances in laser technology and a growing interest in quantum computing and quantum information. A re\ufb02ective view of the \ufb01eld is given in Roy Glauber\u2019s Nobel Prize lecture, Rev. Mod. Phys., 78 (2006) 1267. In the remainder of this section, we give a very brief overview of this large subject. A hint to the richness of quantum optics is immediately apparent. By virtue of (7.180), the electric \ufb01eld vector (7.173a) becomes an operator which creates and destroys photons. The expectation value of this operator vanishes in any state |\u03a8 with a de\ufb01nite number of photons, that is |\u03a8 = |..., nk\u03bb,.... (7.193) This is simple to see, since (7.173a) changes the number of photons nk\u03bb in which case \u03a8|E|\u03a8 becomes the inner product between orthogonal states. Therefore any physical 14 There is quite a lot of recent literature. I recommend that the interested reader start with Emig and Jaffe, J. Phys. A, 41 (2008) 164001; Emig et al., Phys. Rev. Lett., 99 (", "2007) 170403; and Jaffe, Phys. Rev. D, 72 (2005) 021301. 473 7.8 Quantization of the Electromagnetic Field state needs to be a superposition of states with different number of photons. A wide variety of physical states with different properties can in principle be realized, if one can manipulate this superposition. It is the ability to carry out this manipulation which has given birth to quantum optics. Problem 2.21 of Chapter 2 suggests one possible manipulation leading to something known as a coherent state. Coherent states are eigenstates of the annihilation operator a, and therefore serve as eigenstates of positive or negative frequency parts of E. Let us explore one such type of manipulation of single-mode electric \ufb01eld operators, following Chapter Five of Loudon (2000). For a given direction of linear polarization, the electric \ufb01eld is given by E(\u03c7) = E+(\u03c7) +E \u2212(\u03c7) = ae\u2212i\u03c7 + 1 2 a\u2020ei\u03c7 1 2 (7.194) where \u03c7 \u2261 \u03c9t \u2212 kz \u2212 \u03c0/2. (We absorb a factor of \u2212(8\u03c0\u00afh\u03c9k/V)1/2 into the de\ufb01nition of the electric \ufb01eld.) The phase angle \u03c7 can be adjusted experimentally. Furthermore, \ufb01elds with different phase angles generally do not commute. From (2.124) it is easily shown that [E(\u03c71), E(\u03c72)] = \u2212 i 2 sin(\u03c71 \u2212 \u03c72). The uncertainty relation (1.146) therefore implies that \u0394E(\u03c71)\u0394E(\u03c72) \u2265 1 4 | sin(\u03c71 \u2212 \u03c72)| where the electric \ufb01eld variance (\u0394E(\u03c7))2 is de\ufb01ned in the usual way as (\u0394E(\u03c7))2 = (E(\u03c7))2 \u2212 E(\u03c7)2 = (E(\u03c7))2 since E(\u03c7) = 0 for a state with a single mode. A state |\u03b6 for which 0 \u2264 (\u0394E(\u03c7))2 < 1 4 (7.195) (7.196) (7.197) (7.198) is said to be quadrature squeezed. It is possible to write |\u03b6 as the action of a unitary operator on the vacuum, that is, |\u03b6 = exp 1 2 \u03b6\u2217a2 \u2212 1 2 \u03b6(a\u2020)2 (7.199) where \u03b6 = si\u03b8 is called the squeeze parameter. In this state, the electric \ufb01eld variance is - (\u0394E(\u03c7))2 = 1 4 \u03c7 \u2212 1 2 e2s sin2 \u03b8 + e\u22122s cos2 \u03b8. (7.200) \u03c7 \u2212 1 2 Thus one can achieve for \u0394E(\u03c7) a minimum \u0394Emin = e\u2212s 1 2 for \u03c7 = + m\u03c0 (7.201) where m is an integer, and a maximum \u0394Emax = 1 2 es for. (7.202) \u03b8 2 474 Identical Particles 10.0 Minimum uncertainty (1 + S+)(1 + S\u2212) = 1 ) + S + 1 ( 5.0 1. Vacuum state 0.5 1.0 1.5 (1 + S\u2212) Fig. 7.11 Observation of states of \u201csqueezed light,\u201dfrom Wu et al., J. Opt. Soc. Am. B, 4 (1987) 1465. (See also Chapter Five in Loudon (2000).) Data are obtained by measuring the electric field variance, that is the noise, for different scans of the phase angle \u03c7. The different points correspond to different squeezed states, formed by selecting different values of the magnitude s of the squeeze parameter \u03b6. The solid line through the points is given by (7.203). The resulting uncertainty relation is \u0394Emin\u0394Emax = 1 4 (7.203) which satis\ufb01es (7.196) as an equality. The observation of squeezed light is challenging, but such measurements have been carried out. See Figure 7.11. The squeezed states are prepared using an optical technique known as parametric down conversion which allows different magnitudes of \u03b6 to be selected. Each point is the result of sweeping over the phase \u03c7 and measuring the noise spectrum of the electric \ufb01eld. Problems 7.1 Liquid helium makes a transition to a macroscopic quantum \ufb02uid, called super\ufb02uid helium, when cooled below a phase transition temperature T = 2.17 K. Calculate the de Bro", "glie wavelength \u03bb = h/p for helium atoms with average energy at this temperature, and compare it to the size of the atom itself. Use this to predict the super\ufb02uid transition temperature for other noble gases, and explain why none of them can form super\ufb02uids. (You will need to look up some empirical data for these elements.) 475 Problems 7.2 Three identical particles are in a one-dimensional harmonic oscillator potential well with classical angular frequency \u03c9. a. Write the complete time-independent Hamiltonian for this system, and express it in coordinate space as a differential equation whose solution is the three-body wave function \u03a8(x1, x2, x3). b. Assume the particles have zero spin. Use the single-particle wave functions to construct the ground-state wave function \u03a80(x1, x2, x3), and show that it satis\ufb01es the differential equation in (a), and \ufb01nd the ground-state energy. c. Assume the particles have spin 1 2. Repeat (b), and also construct the ground state spin state from single-particle spin eigenstates. 7.3 a. N identical spin 1 2 particles are subjected to a one-dimensional simple harmonic oscillator potential. Ignore any mutual interactions between the particles. What is the ground-state energy? What is the Fermi energy? b. What are the ground-state and Fermi energies if we ignore the mutual interactions and assume N to be very large? 7.4 It is obvious that two nonidentical spin 1 particles with no orbital angular momenta (that is, s-states for both) can form j = 0, j = 1, and j = 2. Suppose, however, that the two particles are identical. What restrictions do we get? 7.5 Discuss what would happen to the energy levels of a helium atom if the electron were a spinless boson. Be as quantitative as you can. 7.6 Three spin 0 particles are situated at the corners of an equilateral triangle. Let us de\ufb01ne the z-axis to go through the center and in the direction normal to the plane of the triangle. The whole system is free to rotate about the z-axis. Using statistics considerations, obtain restrictions on the magnetic quantum numbers corresponding to Jz. 7.7 Consider three weakly interacting, identical spin 1 particles. a. Suppose the space part of the state vector is known to be symmetric under interchange of any pair. Using notation |+|0|+ for particle 1 in ms = +1, 476 Identical Particles particle 2 in ms = 0, particle 3 in ms = +1, and so on, construct the normalized spin states in the following three cases. (i) All three of them in |+. (ii) Two of them in |+, one in |0. (iii) All three in different spin states. What is the total spin in each case? b. Attempt to do the same problem when the space part is antisymmetric under interchange of any pair. 7.8 A porphyrin ring is a molecule which is present in chlorophyll, hemoglobin, and other biological compounds. It can be modeled as 18 electrons moving freely along a one-dimensional circular path of radius R = 0.4 nm. a. Using a polar angular coordinate \u03b8, write down the appropriately normalized single-particle wave functions \u03c8(\u03b8), including periodic boundary conditions. Find an expression for the single-particle energy eigenvalues. b. Find the electron con\ufb01gurations and energies for the ground state and \ufb01rst excited state of porphyrin. c. Find a numerical value for the wavelength of electromagnetic radiation that would excite the ground state into the \ufb01rst excited state. This is a very simple model, and porphyrin comes in many varieties, but compare your result to an experimental result. 7.9 Show that for an operator a which, with its adjoint, obeys the anticommutation relation {a, a\u2020} = aa\u2020 + a\u2020a = 1, then the operator N = a\u2020a has eigenstates with the eigenvalues 0 and 1. 7.10 Suppose the electron were a spin 3 2 particle obeying Fermi\u2013Dirac statistics. Write the con\ufb01guration of a hypothetical Ne (Z = 10) atom made up of such \u201celectrons\u201d [that is, the analogue of (1s)2(2s)2(2p)6]. Show that the con\ufb01guration is highly degenerate. What is the ground state (the lowest term) of the hypothetical Ne atom in spectroscopic notation (2 S+1LJ, where S, L, andJ stand for the total spin, the total", " orbital angular momentum, and the total angular momentum, respectively) when exchange splitting and spin-orbit splitting are taken into account? 7.11 Two identical spin 1 2 fermions move in one dimension under the in\ufb02uence of the in\ufb01nite-wall potential V = \u221e for x < 0, x > L, andV = 0 for 0 \u2264 x \u2264 L. a. Write the ground-state wave function and the ground-state energy when the two particles are constrained to a triplet spin state (ortho state). b. Repeat (a) when they are in a singlet spin state (para state). c. Let us now suppose that the two particles interact mutually via a very short range attractive potential that can be approximated by V = \u2212\u03bb\u03b4(x1 \u2212 x2) (\u03bb > 0). Assuming that perturbation theory is valid even with such a singular potential, discuss semiquantitatively what happens to the energy levels obtained in (a) and (b). 477 Problems 7.12 Consider the case of a single particle of mass m in a one-dimensional simple harmonic oscillator potential V(x) =m \u03c92x2/2. a. Using a form for \u02dc\u03c1(x) proportional to exp(\u2212ax2), calculate the energy functional (7.69) in terms ofa. Minimize the result with respect to a and show that you get the correct ground-state energy. b. Now show that a form proportional to exp(\u2212ax2) is a solution to the differential equation (7.76), for the appropriate value of a. Check that (7.69) gives you the correct ground-state energy, and show that it equals the value of \u03bc. 7.13 Show by explicit construction for N = 2 that the normalization condition (7.80) is met for n(x) as de\ufb01ned by (7.89) with the multiparticle wave function (7.88) for the cases when a. \u03c61(x) and \u03c62(x) are distinct orthonormal functions, b. \u03c61(x) = \u03c62(x). 7.14 Using whatever code or application you prefer, \ufb01ll in the details of the calculation on the helium atom in density functional theory. a. Con\ufb01rm the dimensionless forms (7.102) and (7.103). b. Show that (7.104) is properly normalized. c. Verify (7.105). d. Obtain (7.108), or something very close, based on your own numerical approach. e. Complete the calculation and reproduce Table 7.1 and Figure 7.7. f. Try repeating the calculation using an initial density n(0)(r) that is smarter than (7.104). In fact, (7.108) shows that the \ufb01rst Kohn\u2013Sham ground-state wave function is very close to the 1s state for Z = Zeff. What starting density does this imply? There are many variations to this problem you might consider, including trying different forms for the exchange-correlation energy, a larger or smaller set of basis states, or more iterations or different starting densities. 7.15 Prove the relations (7.170), and then carry through the calculation to derive (7.176). 7.16 A Hamiltonian for a system of bosons has the form H = \u2211 T(k)a\u2020 kak + \u03bb\u2211 \u2211 V(l + m)a\u2020 l a\u2020 \u2212lama\u2212m k l m where \u03bb is a constant. Prove that the number operator N = \u2211 a\u2020 kak k is a constant of the motion. 8 Relativistic Quantum Mechanics This \ufb01nal chapter provides a succinct look at how one constructs single-particle wave equations that are consistent with special relativity. To be sure, this effort is ultimately doomed to failure. Special relativity makes it possible to create particles out of energy, but much of our development of quantum mechanics was based on the conservation of probability, so we cannot expect to be entirely successful. The right way to attack this problem is by starting with the postulates of quantum mechanics and building a many-body theory of \ufb01elds that is relativistically consistent. Nevertheless, at energies low compared to the masses involved, we can expect single-particle quantum mechanics to be a very good approximation to nature. Furthermore, this is a natural way to develop the nomenclature and mathematics of a relativistic \ufb01eld theory. We will start with the general problem of forming a relativistic wave equation for a free particle. This leads more or less intuitively to the Klein\u2013Gordon equation, which we will discuss in some detail. Along the way, we introduce and continue to use the concepts of", " natural units, and of relativistically covariant notation. Then, we will go through Dirac\u2019s approach to \ufb01nding a relativistic wave equation that is linear in space-time derivatives, not quadratic. A study of the symmetries of the Dirac equation is presented. The chapter concludes with the solution of the one-electron atom problem and its comparison with data. This material is of course covered by many other authors, but one nice reference, written at the time when relativistic \ufb01eld theory was emerging out of decades of relativistic quantum mechanics, is \u201cElementary relativistic wave mechanics of spin 0 and spin 1/2 particles,\u201d Feshbach and Villars, Rev. Mod. Phys., 30 (1958) 24. 8.1 Paths to Relativistic Quantum Mechanics The early part of the twentieth century saw the more or less simultaneous developments of both relativity and quantum theory. Therefore, it is not surprising to learn that early attempts to develop wave mechanics produced relativistic wave equations.1 Although we now understand the many pitfalls which confounded these early pioneers, it took many decades to sort things out. We begin by focussing on the Hamiltonian operator, the (Hermitian) generator of time translations which led us to the Schr\u00a8odinger equation (2.25) for the time evolution of a state. That is, a state |\u03c8(t) evolves in time according to the equation 1 See Volume I, Section 1.1 of Weinberg (1995). 478 479 8.1 Paths to Relativistic Quantum Mechanics i\u00afh \u2202 \u2202t |\u03c8(t) = H|\u03c8(t). (8.1) We interpret the eigenvalues of the Hamiltonian, of course, as the allowed energies of the system. This is where we can start to incorporate special relativity. 8.1.1 Natural Units This is a good time to graduate to the use of so-called natural units, that is, units in which \u00afh = c = 1. Most people react to this with bewilderment when they \ufb01rst see it, but it is in fact very simple and useful. First consider the consequences of setting c = 1. Then, we measure time (= distance/c) in length units, like meters or centimeters. If you really need to know the value of time in seconds, just divide by c = 3 \u00d7 1010 cm/sec. Velocity becomes a dimensionless number, which we typically denote by \u03b2. Setting c = 1 means that we also measure both momentum and mass in units of energy, like eV or MeV. Frequently, one puts in the c explicitly and writes momentum units as MeV/c and mass as MeV/c2. Most physicists know that the electron mass, for example, is 0.511 MeV/c2, but very few know this value in kilograms without doing the conversion arithmetic! Just don\u2019t be surprised if someone tells you that the mass is 0.511 MeV, and leaves off the c2. Now consider what happens when we set \u00afh = 1 as well. This ties together units for length and units for energy. For example, the canonical commutation relation for the momentum and position operators says that their product has the same units as \u00afh. Therefore, we would measure position in units of MeV\u22121, or some other inverse energy unit. Remember that you can always put back the \u00afh and c in the right places if you need to go back to the old way of doing things. It is not uncommon to do this if you are trying to evaluate the result of some experiment, for example. It is handy to keep in mind that, to a very good approximation, \u00afhc = 200 MeV\u00b7fm for doing these conversions. As a \ufb01nal note, we point out that in a textbook on statistical mechanics, one would also \u201cnaturally\u201d set Boltzmann\u2019s constant k = 1. That is, temperature would be measured in units of energy as well. 8.1.2 The Energy of a Free Relativistic Particle Consider the energy of a free particle with momentum p = |p| and mass m, namely Ep = + p2 + m2. (8.2) We need to come up with a Hamiltonian which yields this energy eigenvalue for a state |p with momentum eigenvalue p. It is the square root, however, which plagued early efforts to come up with a relativistic wave equation, and which we must \ufb01gure out how to deal with here. We have faced transcendental functions of operators before, such as U(t) = exp(\u2212iHt), interpreting them in terms of their Taylor expansions. We could take the same approach here, and write 480 Relativ", "istic Quantum Mechanics H = p2 + m2 = m 1 + 1/2 p2 m2 p6 16m5 + \u00b7 \u00b7 \u00b7. \u2212 p4 p2 2m = m + 8m3 + In fact, this would be a viable way to proceed, but it has some serious shortcomings. For one, it would make it impossible to formulate a \u201ccovariant\u201d wave equation. That is, if we formed a coordinate space (or momentum space) representation of a state vector |\u03c8, the resulting wave equation would have one time derivative and an in\ufb01nite series of increasing spatial derivatives from the momentum operator. There would be no way to put time and space on an \u201cequal footing,\u201d so to speak. (8.3) This consideration actually leads to a more important problem. Let us go ahead and try to build this wave equation. From (8.1) we have \u2202 \u2202t i x|\u03c8(t) = d3px|pp|H|\u03c8(t) = = d3x d3px|pp|xx|Ep|\u03c8(t) d3x d3p eip\u00b7(x\u2212x) (2\u03c0)3 x|Ep|\u03c8(t) (8.4) and (8.3) means thatx |Ep|\u03c8(t) becomes an in\ufb01nite series of ever higher order derivatives; see (1.252). This renders this wave equation nonlocal since it must reach further and further away from the region near x in order to evaluate the time derivative. Eventually, causality will be violated for any spatially localized wave function x|\u03c8(t). The loss of covariance costs us a great deal indeed. We abandon this approach, and instead work with the square of the Hamiltonian, instead of the Hamiltonian itself. This removes the problem of the square root, and all of its attendant problems, but it will introduce a different problem. There will be solutions to the wave equation with negative energies, necessary to form a complete set of basis states, but which have no obvious physical meaning. Nevertheless, this approach is more useful than the one we are now leaving. 8.1.3 The Klein\u2013Gordon Equation Start with (8.1) and take the time derivative once more. That is \u2212 \u22022 \u2202t2 |\u03c8(t) = i \u2202 \u2202t H|\u03c8(t) = H2|\u03c8(t). (8.5) We can now write down a simple wave equation for \u03a8(x, t) \u2261 x|\u03c8(t). Taking H2 = p2 + m2 and using x|p2|\u03c8(t) = \u2212\u22072\u03a8(x, t), we obtain \u22022 \u2202t2 \u2212 \u22072 + m2 \u03a8(x, t) =0. (8.6) Equation (8.6) is known as the Klein\u2013Gordon equation. This looks very much like a classical wave equation, except for the m2 term. Putting back our \u00afh and c, we see that this term introduces a length scale \u00afh/mc, called the Compton wavelength. 481 8.1 Paths to Relativistic Quantum Mechanics The Klein\u2013Gordon equation has nearly all the desirable qualities of a relativistic wave equation. Firstly, it is relativistically covariant. You can see that because a Lorentz transformation leaves the square of the space-time interval ds2 = dt2 \u2212 dx2 invariant. Therefore, the combination of derivatives in (8.6) is the same if one changes frames from (x, t) to (x, t). In other words, \u03a8(x, t) solves the same equation as \u03a8(x, t). Relativistic covariance is easier to see if one uses relativistic covariant notation. We will use a notation which has become standard. That is, Greek indices run 0, 1, 2, 3 and Latin indices run 1, 2, 3. If an index is ever repeated in an expression, summation over that index is implied. A contravariant four-vector a\u03bc \u2261 (a0, a) has a dual covariant vector a\u03bc = \u03b7\u03bc\u03bda\u03bd, where \u03b700 = +1, \u03b711 = \u03b722 = \u03b733 = \u22121, and all other elements are zero. Thus a\u03bc = (a0, \u2212a). Inner products of four-vectors can only be taken between a contravariant vector and a covariant vector, e.g. a\u03bcb\u03bc = a0b0 \u2212 a \u00b7 b. In particular, a\u03bca\u03bc = 2 \u2212 a2. a0 % & A key point of", " Lorentz transformations is that inner products of four-vectors are invariant. That is a\u03bcb\u03bc will have the same value in any reference frame. This is the reason that covariant notation is very useful for demonstrating the covariance of a particular expression. The space-time position four-vector is x\u03bc = (t, x). This gives the four-gradient \u2202 \u2202x\u03bc = \u2202 \u2202t, \u2207 \u2261 \u2202\u03bc (8.7) which is a covariant vector operator, despite the positive sign in front of the spacelike part. Now, the covariance of (8.6) is absolutely clear. The Klein\u2013Gordon equation becomes! \" \u2202\u03bc\u2202\u03bc + m2 \u03a8(x, t) =0. (8.8) Sometimes, an even further economy of notation is used by writing \u22022 \u2261 \u2202\u03bc\u2202\u03bc. Another desirable property of the Klein\u2013Gordon equation is that it has solutions that are in fact what we expect for a free relativistic particle of mass m. We expect the time dependence to be like exp(\u2212iEt), where E is an eigenvalue of the Hamiltonian. We also expect the spatial dependence to be that of a plane wave, that is exp(+ip \u00b7 x) for momentum p. In other words, our solution should be \u03a8(x, t) = Ne\u2212i(Et\u2212p\u00b7x) = Ne\u2212ip\u03bcx\u03bc where p\u03bc = (E, p). Indeed (8.9) solves (8.8) so long as \u2212p\u03bcp\u03bc + m2 = \u2212E2 + p2 + m2 = 0 (8.9) (8.10) or E2 = E2 p. Thus the energy eigenvalues E = +Ep are included, as they should be. On the other hand, the negative energy eigenvalues E = \u2212Ep are also included. This was a serious stumbling block in the historical development of relativistic quantum mechanics, but we will take up a practical explanation of it shortly. Schr\u00a8odinger\u2019s nonrelativistic wave equation has a very important property, namely that it implies that probability is conserved. The probability density \u03c1(x, t) = \u03c8\u2217\u03c8 (2.189) is a positive de\ufb01nite quantity, and the probability \ufb02ux (2.191) obeys a continuity equation (2.190) which proves that the probability density can only be in\ufb02uenced by the \ufb02ux into or out of a particular region. 482 Relativistic Quantum Mechanics One would like to identify analogous expressions using the Klein\u2013Gordon equations, so that the wave function \u03a8(x, t) can be similarly interpreted. The form of the continuity equation strongly suggests that we construct a four-vector current j \u03bc with the property \u2202\u03bcj \u03bc = 0, with the probability density \u03c1 \u2261 j 0. In fact, if we follow (2.191) to write i 2m then it is easy to show that \u2202\u03bc j \u03bc = 0. Therefore we calculate a density [\u03a8\u2217\u2202\u03bc\u03a8 \u2212 (\u2202\u03bc\u03a8) j \u03bc = \u2217 \u03a8] (8.11) \u03c1(x, t) =j 0(x, t) = i 2m \u03a8\u2217 \u2202\u03a8 \u2202t \u2212 \u2202\u03a8 \u2202t \u2217 \u03a8. (8.12) Although this density is conserved, it is not positive de\ufb01nite! This was a tremendous problem in the development of relativistic quantum mechanics, since it rendered the standard probabilistic interpretation of the wave function impossible. Eventually a consistent physical interpretation was found. Before discussing this interpretation, though, we need to consider the effect of electromagnetic interactions within the context of our relativistic framework. The explicitly covariant nature of the Klein\u2013Gordon equation makes it straightforward to add electromagnetic interactions into the Hamiltonian. See Section 2.7, especially (2.343) and (2.346). As before, we assume that the particle has an electric charge e < 0. In a classical Hamiltonian, one simply makes the substitutions2 E \u2192 E \u2212 e \u03a6 and p \u2192 p \u2212 eA, where \u03a6 is the \u201cscalar\u201d electric potential and A is the vector potential. In covariant form, this becomes p\u03bc \u2192 p\u03bc \u2212 eA\u03bc where A\u03bc = (\u03a6, A), so A\u03bc = (\u03a6, \u2212A). This all amounts to rewriting (8.8) as D\u03bcD\u03bc + m2 \u03a8(x, t) =0! \" (8.13) (8.14) where D\u03bc \u2261 \u2202\u03bc + ieA\u03bc.", " We refer to D\u03bc as the covariant derivative. Unlike the nonrelativistic Schr\u00a8odinger wave equation, the Klein\u2013Gordon equation is second order in time derivatives, not \ufb01rst order. That implies that not only must one specify \u03a8(x, t)|t=0 for its solution, but also \u2202\u03a8(x, t)/\u2202t|t=0. Consequently, more information is necessary than one might have originally expected based on our experience from nonrelativistic quantum mechanics. In fact, this additional \u201cdegree of freedom\u201d shows up as the sign of the charge of the particle. This is clear by noting that if \u03a8(x, t) solves (8.14), then \u03a8\u2217(x, t) solves the same equation, but with e \u2192 \u2212e. More explicitly, we can reduce the second-order Klein\u2013Gordon equation to two \ufb01rstorder equations, and then interpret the result in terms of the sign of the electric charge. 2 It is worthwhile to take a moment and review the origin of these substitutions. A Lagrangian L is constructed which yields the Lorentz force law, F = e [E + v \u00d7 B/c]. For a coordinate xi, the canonical momentum is pi \u2261 \u2202L/\u2202\u02d9xi = m\u02d9xi +eAi. Hence, the kinetic energy uses the \u201ckinematic momentum\u201d m\u02d9xi = pi \u2212eAi. SeeAppendix C of this book, or Taylor (2005), Section 7.9 for more details. Extension to relativistic kinematics is relatively straightforward. The four-momentum p\u03bc is replaced by p\u03bc \u2212 eA\u03bc; see Jackson (1998), Section 12.1A. When working in coordinate space, the quantum-mechanical operator for the (covariant vector) p\u03bc = (E, \u2212p) is i\u2202\u03bc = (i\u2202t, i\u2207). Therefore, to incorporate electromagnetism, we replace i\u2202\u03bc with i\u2202\u03bc \u2212 eA\u03bc = i(\u2202\u03bc + ieA\u03bc) \u2261 iD\u03bc. 483 8.1 Paths to Relativistic Quantum Mechanics Using a rather obvious notation in which D\u03bcD\u03bc = D2 t two equations, each \ufb01rst order in time, de\ufb01ning two new functions \u2212 D2, we can easily write (8.14) as \u03c6(x, t) = \u03c7(x, t) = 1 2 1 2 \u03a8(x, t) + i m \u03a8(x, t) \u2212 i m Dt\u03a8(x, t) Dt\u03a8(x, t) (8.15a) (8.15b) so that instead of specifying \u03a8(x, t)|t=0 and \u2202\u03a8(x, t)/\u2202t|t=0, we can specify \u03c6(x, t)|t=0 and \u03c7(x, t)|t=0. Furthermore, \u03c6(x, t) and \u03c7(x, t) satisfy the coupled equations iDt\u03c6 = \u2212 1 2m 1 2m D2(\u03c6 + \u03c7) +m \u03c6 D2(\u03c6 + \u03c7) \u2212 m\u03c7 iDt\u03c7 = + (8.16b) (8.16a) which bear a striking resemblance to the nonrelativistic Schr\u00a8odinger equation. We can demonstrate this resemblance even more keenly be de\ufb01ning a two-component object \u03a5(x, t) in terms of the two functions \u03c6(x, t) and \u03c7(x, t), and using the Pauli matrices (3.50). That is, for the functions \u03c6(x, t) and \u03c7(x, t) which satisfy (8.16), we de\ufb01ne a column vector function We now write the Klein\u2013Gordon equation as \u03a5(x, t) \u2261 \u03c6(x, t) \u03c7(x, t). iDt\u03a5 = D2(\u03c43 + i\u03c42) +m \u03c43 \u03a5 \u2212 1 2m (8.17) (8.18) (Note that we use \u03c4 rather than \u03c3 to denote the Pauli matrices, to avoid any confusion with the concept of spin.) Equation (8.18) is completely equivalent to our formulation in (8.14), but it is a \ufb01rst-order differential equation in time. We have \u201chidden\u201d the additional degree of freedom in the two-component nature of \u03a5(x, t). Now let us return to the question of probability current density.", " Having rewritten the Klein\u2013Gordon equation using the covariant derivative as (8.14), the correct form of the conserved current is now j \u03bc = i 2m [\u03a8\u2217D\u03bc\u03a8 \u2212 (D\u03bc\u03a8) \u2217 \u03a8]. The \u201cprobability\u201d density (8.12) therefore becomes \u03c1 = j 0 = i 2m [\u03a8\u2217Dt\u03a8 \u2212 (Dt\u03a8) = \u03c6\u2217\u03c6 \u2212 \u03c7\u2217\u03c7 = \u03a5\u2020\u03c43\u03a5. \u2217 \u03a8] (8.19) (8.20) This is easy to see by using (8.15) to write \u03a8(x, t) and Dt\u03a8 in terms of \u03c6(x, t) and \u03c7(x, t). We are therefore led to interpret \u03c1 as a probability charge density, where \u03c6(x, t) is the wave function of a positive particle, and \u03c7(x, t) is the wave function of a negative particle. That is, the Klein\u2013Gordon equation has buried in it simultaneous degrees of freedom for a particle of a certain charge, as well as a particle that behaves identically but with the 484 Relativistic Quantum Mechanics opposite charge. Before going so far as to refer to these as \u201cparticle\u201d and \u201cantiparticle,\u201d we should go back and consider the interpretation of the negative energy solutions. 8.1.4 An Interpretation of Negative Energies free particles, in which case D\u03bc = \u2202\u03bc, and for which \u03a5(x, t) \u221d First consider exp [\u2212i(Et \u2212 p \u00b7 x)]. Inserting this into (8.18) yields the eigenvalues E = \u00b1Ep as it, of course, should. We \ufb01nd for the eigenfunctions (8.21a) \u03a5(x, t) = \u03a5(x, t) = e\u2212iEpt+ip\u00b7x for E = +Ep e+iEpt+ip\u00b7x for E = \u2212Ep 1 2(mEp)1/2 1 2(mEp)1/2 Ep + m m \u2212 Ep m \u2212 Ep Ep + m with a normalization that leads to a charge density \u03c1 = \u00b11 for E = \u00b1Ep. That is, we impose the condition that a free particle with negative charge is to be associated with a particle that has negative total energy. Also, for a particle at rest, Ep = m and the positive energy solution 8.21a has only an upper component (that is, \u03c7(x, t) = 0), while the negative energy solution 8.21b, has only a lower component (that is, \u03c6(x, t) = 0). This continues for the nonrelativistic case where p Ep and the positive energy solution is dominated by \u03c6(x, t) and the negative energy solution by \u03c7(x, t). (8.21b) More insight to the meaning of negative energies comes from considering the probability current density j. Making use of (3.52), (3.53), and (8.18) we have & % \u2202t\u03a5\u2020 \u03c43\u03a5 + \u03a5\u2020\u03c43 (\u2202t\u03a5) % &\" (1 + \u03c41)\u03a5 \u2212 \u03a5\u2020(1 + \u03c41) \u22072\u03a5!% \u2202t\u03c1 = \u2202t(\u03a5\u2020\u03c43\u03a5) = & 1 2im = \u2212\u2207 \u00b7 j \u22072\u03a5\u2020 = where j =! 1 2im \u03a5\u2020(1 + \u03c41) (\u2207\u03a5) \u2212 & % \u2207\u03a5\u2020 (1 + \u03c41)\u03a5 \". In the case of a free particle, for either positive or negative energies, this reduces to j = p m \u03a5\u2020(1 + \u03c41)\u03a5 = p Ep. (8.22) (8.23) Now this would appear to be quite peculiar. With a normalization that imposes positive and negative charges on positive and negative energy solutions, respectively, we end up with a charge current density that is the same regardless of the sign of the charge and energy. One way to \u201c\ufb01x\u201d this would be to recognize that the sign of the momentum vector p in (8.21b) is \u201cwrong\u201d since we want the exponent to have the form ip\u03bcx\u03bc in order to be relativistically invariant. We might reverse the sign of p, therefore, for negative energy solutions, in which case (8.23) would carry the \u201ccorrect\u201d sign to account for the charge of the particle. Another way of \u201c\ufb01xing\u201d this problem", " would be to say that the negative energy particles are moving \u201cbackwards in time.\u201d This not only reverses the sign of p, but also 485 8.1 Paths to Relativistic Quantum Mechanics lets the energy be positive in the exponent of (8.21b)! We have made some contact with the popular lore of particles and antiparticles. If we like, we can formally associate the positive energy solution \u03a8E>0(x, t) to the Klein\u2013Gordon equation with that for a \u201cparticle,\u201d and the complex conjugate of the negative energy solution \u03a8\u2217 E<0(x, t) with an \u201cantiparticle.\u201d In this case (8.14) yields us the two equations! (\u2202\u03bc \u2212 ieA\u03bc)(\u2202\u03bc \u2212 ieA\u03bc) +m 2! (\u2202\u03bc + ieA\u03bc)(\u2202\u03bc + ieA\u03bc) +m 2 \u03a8particle(x, t) =0 \u03a8antiparticle(x, t) =0. (8.24a) (8.24b) \" \" This makes it explicitly clear how to split the solutions to the Klein\u2013Gordon equation into two pieces that correspond individually to particles with charge \u00b1e. It is possible to continue along these lines and solve the Klein\u2013Gordon equation for an atomic system. In fact, the results compare well with experiment, so long as the orbiting charged particle has no spin. (See Problem 8.7 at the end of this chapter.) Nevertheless, the interpretation of the probability density and negative energies is strictly ad hoc, and not founded in the kind of \ufb01rst-principles derivations we espouse in this book. More importantly, this formalism does not allow for expected relativistic effects, in particular the creation and destruction of massive particles from energy. For this, we must resort to quantum \ufb01eld theory. Of course, Dirac did \ufb01nd a way to write down a singleparticle wave equation \u2013 with a different, and more palatable interpretation of negative energies \u2013 and we take this up in Section 8.2. First, though, we take a small detour to show how the conceptual problems with the Klein\u2013Gordon wave function are solved if we interpret the solution as a quantum \ufb01eld. 8.1.5 The Klein\u2013Gordon Field We now return to \ufb01nding a solution \u03a8(x, t) to the Klein\u2013Gordon equation (8.6). This time, however, we interpret \u03a8(x, t) as a second-quantized \ufb01eld, as discussed in Section 7.7.1. We will \ufb01nd it convenient to write \u03a8(x, t) as a complex linear combination of two Hermitian \ufb01elds, so we start by looking for a Hermitian solution \u03a6(x, t) of (8.6). We write 1 \u03a6(x, t) = L3/2 \u2211 where we interpret the qk(t) as operators in second quantization. Once again, we use a \u201cbig box\u201d normalization in which the k form a complete set of discrete wave vectors. We ensure that \u03a6(x, t) is Hermitian by requiring that qk(t)eik\u00b7x (8.25) k q\u2020 k(t) = q\u2212k(t). Inserting (8.25) into (8.6), we \ufb01nd that \u00a8qk + \u03c92 kqk = 0 (8.26) (8.27) where \u03c92 k \u2261 k2 + m2. (Recall that we are using natural units, in which \u00afh = c = 1.) Equation (8.27) suggests how to quantize the Klein\u2013Gordon \ufb01eld. If the qk(t) were classical quantities, then (8.27) would imply that they behave just like the position 486 Relativistic Quantum Mechanics coordinate in a simple harmonic oscillator, and we learned how to quantize the simple harmonic oscillator in Chapter 2. However, by virtue of (8.26), unlike the position coordinate, qk is not a Hermitian operator. Nevertheless, we can derive commutation relations for the qk from \ufb01rst principles of the quantum harmonic oscillator. Consider a two-dimensional isotropic harmonic oscillator with position coordinates x and y and natural frequency \u03c9. (The two degrees of freedom will correspond to qk and q\u2212k. Note that \u03c9 k = \u03c9\u2212k.) In terms of their respective creation and annihilation operators, in natural units, the position coordinates are x = % 1\u221a 2m\u03c9 & ax + a\u2020 x and y = % 1", "\u221a 2m\u03c9 &. ay + a\u2020 y Now de\ufb01ne the (non-Hermitian) operators q\u00b1 as \u2019 m 2 q\u00b1 = (x \u00b1 iy) + = q\u2212, which is similar to (8.26). Making the de\ufb01nitions a\u00b1 = (ax \u00b1 iay)/ so that q\u2020 write q+ = 1\u221a 2\u03c9 (a+ + a\u2020 \u2212). (8.28) (8.29) \u221a 2, we (8.30) Note that a\u00b1 each satisfy the appropriate commutation relations with their respective creation operator a\u2020 \u2212] =0. That is, the a+ and a\u2212 correspond to independent quanta. \u00b1, but also [a+, a\u2020 Therefore, following (8.30), the \ufb01eld \u03a6(x, t) is given by (8.25) with with qk(t) = # 1\u221a 2\u03c9k $ ake\u2212i\u03c9kt + a\u2020 \u2212kei\u03c9kt # $ ak, a\u2020 k = \u03b4k,k (8.31a) (8.31b) and [ak, ak ] =0 = [a\u2020 bosons. k, a\u2020 k]. Evidently, the Klein\u2013Gordon \ufb01eld speci\ufb01cally applies to Of course, (8.31) are contrived. When quantizing the simple harmonic oscillator in Chapter 2, we made use of the de\ufb01nition of momentum as the generator of translations, and this provided the physical basis for deriving the properties of the creation and annihilation operators. We simply used this as a model for building the qk, because (8.27) shows that they behave like a classical oscillator. Therefore, we should check that implications of (8.31) are consistent with the physical interpretation of momentum. The momentum operator, in second quantization, should be P = \u2211k Nkk = \u2211k ka\u2020 kak. Showing that it is indeed the generator of translation, as we de\ufb01ned in Section 1.6, is equivalent to showing that [\u03a6(x, t), P] = \u2212i\u2207\u03a6(x, t). Writing \u03a6(x, t) explicitly as \u03a6(x, t) = 1 L3/2 \u2211 k 1\u221a 2\u03c9 k ake\u2212i\u03c9kt + a\u2020 \u2212kei\u03c9kt eik\u00b7x (8.32) 487 8.1 Paths to Relativistic Quantum Mechanics we can evaluate the commutators # $ ak, a\u2020 kak $ # \u2212k, a\u2020 a\u2020 kak = aka\u2020 = a\u2020 \u2212ka\u2020 k ak \u2212 a\u2020 k ak \u2212 a\u2020 kak ak = kak a\u2020 # $ ak, a\u2020 k # ak = \u03b4k,kak $ \u2212k = a\u2020 k a\u2020 \u2212k, ak = \u2212\u03b4\u2212k,ka\u2020 k (8.33) (8.34) to calculate the commutator [\u03a6(x, t), P] = # $ ak, a\u2020 kak 2\u03c9 k akke\u2212i\u03c9 kt \u2212 a\u2020 # e\u2212i\u03c9 kt + \u2212k(\u2212k)ei\u03c9 kt k ake\u2212i\u03c9 kt + a\u2020 \u2212kei\u03c9 kt eik\u00b7x k ak \u2212k, a\u2020 a\u2020 eik\u00b7x 1\u221a 1 k 1 = \u2211 k 1\u221a L3/2 \u2211 L3/2 \u2211 L3/2 \u2211 = \u2212i\u2207\u03a6(x, t) 1\u221a = 1 k k 2\u03c9 k 2\u03c9 k $ ei\u03c9 kt keik\u00b7x (8.35) as it should be. This gives us con\ufb01dence in the physical interpretation of our construction of the Klein\u2013Gordon \ufb01eld (8.32). The Hamiltonian is simply given by the sum of the single oscillator energies, that is H = \u2211 k a\u2020 kak + 1 2 \u03c9k = \u2211 Nk\u03c9k +\u2211 k k 1 2 \u03c9k. (8.36) We interpret the energy eigenvalues as the sum over the numbers of quanta, times their respective energies, plus the zero point energy. This is similar to the situation with energy in the electromagnetic \ufb01eld, in terms of photons, discussed in Section 7.8. The energy eigenvalues are manifestly positive. There is no longer an issue with negative energies! We have solved one of the big problems of our relativistic wave equation without even trying. It remains, however, to see if we can resolve the problem of a probabilistic interpretation using our construction. We saw", " in (8.20) the suggestion that charge conservation might be associated with breaking up the wave function into two complex components. To that end, let us use the Hermitian \ufb01eld \u03a6(x, t) to form a complex (non-Hermitian) \ufb01eld \u03a8(x, t) as \u03a8(x, t) \u2261 1\u221a 2 [\u03a61(x, t) +i\u03a6 2(x, t)] (8.37) which now contains two independent sets of \ufb01eld operators ajk with j = 1, 2. That is \u03a6j(x, t) = # ajk, a\u2020 j k! ajk, aj k $ \" 1 L3/2 \u2211 k = \u03b4jj \u03b4k,k = 0. # 1\u221a 2\u03c9 k ajkei(k\u00b7x\u2212\u03c9kt) + a\u2020 jke\u2212i(k\u00b7x\u2212\u03c9kt) $ (8.38a) (8.38b) (8.38c) Note that in deriving (8.38a) from (8.32), we switched the dummy summation index in the second term from \u2212k to k. Now write \u03a8(x, t) in terms of its own \ufb01eld operators bk and ck 488 Relativistic Quantum Mechanics as \u03a8(x, t) = 1 L3/2 \u2211 k # 1\u221a 2\u03c9 k bkei(k\u00b7x\u2212\u03c9kt) + c\u2020 ke\u2212i(k\u00b7x\u2212\u03c9kt) $ bk \u2261 1\u221a 2 ck \u2261 1\u221a 2 [a1k + ia2k] [a1k \u2212 ia2k]. (8.39a) (8.39b) (8.39c) Note that bk and ck act as an alternative set of \ufb01eld operators, with respect to a1k and a2k. That is, for example, # # bk, b\u2020 k bk, c\u2020 k $ $ # # $ # + = = a1k, a\u2020 1k \u2212 ia\u2020 a1k + ia2k, a\u2020 1k 2k # $ 1 2 1 2 1 2 1 a1k, a\u2020 1k 2 1ka1k + a\u2020 kck = a\u2020 1 2 a1k + ia2k, a\u2020 1k + ia\u2020 # $ \u2212 1 2 2ka2k. = = a2k, a\u2020 2k $ a2k, a\u2020 2k 2k # kbk + c\u2020 b\u2020 = 1, (8.40a) $ $ = 0, (8.40b) (8.40c) (8.41a) (8.41b) In terms of bk and ck, the Hamiltonian and momentum operators become kbk + c\u2020 b\u2020 kck + 1 kbk + c\u2020 b\u2020 kck. H = \u2211 k P = \u2211 k \u03c9k k It begs the question, why convert to the \ufb01elds bk and ck from a1k and a2k if the energy and momentum are equivalently expressed in terms of either sets of quanta? The answer is that, as we will see in a moment, bk and ck will naturally take on the role of \u201cparticle\u201d and \u201cantiparticle\u201d operators. In fact, we are now ready to consider the conservation of charge3 with our free Klein\u2013 Gordon \ufb01eld. The \ufb01eld \u03a8(x, t) satis\ufb01es the Klein\u2013Gordon equation because \u03a61(x, t) and \u03a62(x, t) are solutions. Therefore, following (8.11), we know that \u2202\u03bcj \u03bc = 0 for the \ufb01eld j \u03bc(x, t) =C \u00d7! \u03a8(\u2202\u03bc\u03a8\u2020) \u2212 \u03a8\u2020(\u2202\u03bc\u03a8) \" where C is an arbitrary constant, so the quantity represented by the \ufb01eld Q = d3x j 0(x, t) =C d3x \u03a8 \u2212 \u03a8\u2020 \u2202\u03a8\u2020 \u2202t \u2202\u03a8 \u2202t V=L3 is conserved. (It is worth noting that (8.37) implies that\u03a8(x, t) and \u03a8\u2020(x, t) commute, since \u03a61(x, t) and \u03a62(x, t) are Hermitian and commute with each other.) Inserting (8.39a), we \ufb01nd (8.", "42) (8.43) 3 Since we are in a multiparticle \ufb01eld theory now, the concept of conservation of probability for a single particle is irrelevant. 489 8.1 Paths to Relativistic Quantum Mechanics Q = C 1 L3 # \u00d7 d3x\u2211 \u2211 k k V=L3 bkei(k\u00b7x\u2212\u03c9kt) + c\u2020 1\u221a 1\u221a 2\u03c9 k 2\u03c9 k ke\u2212i(k\u00b7x\u2212\u03c9kt) \u2212 ke\u2212i(k\u00b7x\u2212\u03c9kt) + ckei(k\u00b7x\u2212\u03c9kt) b\u2020 i\u03c9kb\u2020 k e\u2212i(k\u00b7x\u2212\u03c9 \u2212i\u03c9kbk ei(k\u00b7x\u2212\u03c9 k t) \u2212 i\u03c9kck ei(k\u00b7x\u2212\u03c9 t) k k t) + i\u03c9kc\u2020 k e\u2212i(k\u00b7x\u2212\u03c9 t) k $. (8.44) In the next step, we carry through the volume integrals. The procedure is reminiscent of our work with the electromagnetic \ufb01eld, and in fact invokes the same result as in (7.174). The two sums in (8.44) collapse to one, with k = \u00b1k, so \u03c9k = \u03c9k, and an overall factor of L3 appears. We \ufb01nd # bkb\u2020 k \u2212 bkc\u2212ke\u22122i\u03c9 k t + c\u2020 kb\u2020 \u2212ke2i\u03c9 k t \u2212 c\u2020 kck kbk \u2212 b\u2020 kc\u2020 \u2212ke2i\u03c9 +b\u2020 # $ k t + ckb\u2212ke\u22122i\u03c9 $ k t \u2212 ckc\u2020 k bkb\u2020 k \u2212 c\u2020 kck + b\u2020 kbk \u2212 ckc\u2020 k where the cross terms cancel because the bk and ck commute, and we switch dummy indices k for \u2212k in the second pair of terms. Finally, since bkb\u2020 kck + 1, we have kbk + 1 and ckc\u2020 k = c\u2020 k = b\u2020 $ # kbk \u2212 c\u2020 b\u2020 kck (8.45) k where \u02dcC \u2261 iC is still an arbitrary constant. Q = \u02dcC\u2211 Equation (8.45) is quite a triumph. It says that a conserved quantity Q, based on the conserved current (8.42), is proportional to the total number of \u201cb\u201d quanta, minus the total number of \u201cc\u201d quanta. Of course, the physical meaning of Q depends on the particular current in question, but electric charge is a good example if we incorporate Maxwell\u2019s equations. It is natural, therefore, to interpret the \u201cb\u201d quanta as \u201cparticles\u201d and the \u201cc\u201d quanta as \u201cantiparticles.\u201d 8.1.6 Summary: The Klein\u2013Gordon Equation and the Scalar Field A straightforward pursuit of a relativistic wave equation led us to (8.6), which gives the correct wave function for a free relativistic particle. One big problem, though, is that it implied a full complement of negative total energies, for which there is no good physical interpretation. Another big problem is that we were unable to come up with a conserved current that would provide a (necessarily positive) probability density. By adding electromagnetism in the appropriately covariant fashion, we developed some insight about these problems, but no good answers. It also allows a solution for oneelectron atoms (Problem 8.7 at the end of this chapter) but that solution gives \ufb01ne structure that does not agree with experiment. We then discovered that interpreting the solution to (8.6) as a \ufb01eld in second quantization, rather than a wave function, led to the implication that the energies of free particles are 490 Relativistic Quantum Mechanics all positive. We also found a perfectly consistent way to interpret the conserved current, in terms of \u201cparticles and antiparticles.\u201d The creation and annihilation operators are bosonic, and they have no additional degrees of freedom, so the Klein\u2013Gordon \ufb01eld is necessarily spin zero, that is, a scalar \ufb01eld. It would seem, then, that we have found our way out of the woods. The downside, though, is that we have traded a problem with a single degree of freedom in the oneparticle wave function, to one with an in\ufb01nite number of degrees of freedom. This presents a major hurdle to solving problems that include realistic interactions. Perturbation theory is generally a viable technique for arriving at a solution, covered in any textbook devoted to quantum \ufb01eld theory. The problem of getting the wrong \ufb01", "ne structure for one-electron atoms remains, however. For this, it took Dirac\u2019s insight to \ufb01nd a solution, and we cover that next. 8.2 The Dirac Equation Many of the dif\ufb01culties with interpreting the results from the Klein\u2013Gordon equation stem from the fact that it is a second-order differential equation in time. These include a nonpositive de\ufb01nite probability density, and additional degrees of freedom, although both of these can be identi\ufb01ed to some extent with particles and their oppositely charged antiparticles. Nevertheless, Dirac looked for a way to write a wave equation that is \ufb01rst order in time, and along the way discovered the need for j = 1/2 angular momentum states in nature. This also lent itself to a particularly useful interpretation of the negative energy states. The linear differential equation we seek can be written as (i\u03b3\u03bc\u2202\u03bc \u2212 m)\u03a8(x, t) = 0 (8.46) where the \u03b3\u03bc have yet to be determined. (Of course, the constant m is also yet to be determined, but it will turn out to be the mass.) We must still insist that the correct energy eigenvalues (8.10) are obtained for a free particle (8.9), as does (8.8). We can turn (8.46) into (8.8) simply by operating on it with \u2212i\u03b3\u03bd\u2202\u03bd \u2212 m to get (\u03b3\u03bd\u2202\u03bd \u03b3\u03bc\u2202\u03bc + m2)\u03a8(x, t) = 0 (8.47) and then imposing the condition that \u03b3\u03bd \u03b3\u03bc\u2202\u03bd\u2202\u03bc = \u2202\u03bc\u2202\u03bc = \u03b7\u03bc\u03bd\u2202\u03bd\u2202\u03bc. This condition can be written succinctly, by reversing dummy indices to symmetrize, as 1 2 (\u03b3\u03bc\u03b3\u03bd + \u03b3\u03bd \u03b3\u03bc) \u2261 1 2 {\u03b3\u03bc, \u03b3\u03bd} = \u03b7\u03bc\u03bd. (8.48) Thus the four quantities \u03b3\u03bc, \u03bc = 0, 1, 2, 3, are not simply complex numbers, but rather entities that obey something called a Clifford algebra. Clearly this algebra implies that % &2 & \u03b30 % = 1 2 = \u22121 \u03b3i \u03b3\u03bc\u03b3\u03bd = \u2212\u03b3\u03bd \u03b3\u03bc i = 1, 2, 3 if \u03bc = \u03bd. (8.49a) (8.49b) (8.49c) 491 8.2 The Dirac Equation Note that the anticommutation property of the \u03b3\u03bc means that each of these matrices (as well as \u03b1 and \u03b2, as de\ufb01ned below) are traceless. Now substitute the free-particle solution (8.9) into (8.46) to \ufb01nd \u03b3\u03bcp\u03bc \u2212 m = 0 (8.50) from which we can recover the free-particle energy eigenvalues E. Writing (8.50) out in terms of timelike and spacelike parts, and then multiplying through by \u03b30, we obtain E = \u03b30\u03b3 \u00b7 p + \u03b30m. (8.51) This leads to the Dirac Hamiltonian, written in a traditional form. Making the de\ufb01nitions we arrive at \u03b1i \u2261 \u03b30\u03b3i and \u03b2 \u2261 \u03b30 H = \u03b1 \u00b7 p + \u03b2m. (8.52) (8.53) Note that if we add electromagnetism by making the substitution (8.13) into (8.50), and then setting A = 0 and A0 = \u03a6, we have H = \u03b1 \u00b7 p + \u03b2m + e\u03a6 (8.54) which controls the motion of a charged particle in an electrostatic potential \u03a6. We will make use of this when we solve the relativistic one-electron atom in Section 8.4.1. Which form of the Dirac equation we use, whether it be (8.1) with (8.53) or (8.54), or the covariant form (8.46) perhaps with the substituion (8.13), depends on the speci\ufb01c problem at hand. For example, sometimes it is easiest to use (8.53) when solving problems involving dynamics and the Dirac equation, whereas it is easier to discuss symmetries of the Dirac equation in covariant forms using \u03b3\u03bc. The algebra (8.49) can be realized with square matrices, so long as they are at least 4 \u00d7 4. We know that 2 \u00d7 2 matrices are not large enough, for example, since the Pauli matrices \u03c3 form a complete set along with the identity matrix.", " However {\u03c3k, 1} = 2\u03c3k, so this set is not large enough to realize the Clifford algebra. Therefore \u03a8(x, t) in (8.46) would be a four-dimensional column vector. In order to keep a convention consistent with our matrix representation of states and operators, we insist that \u03b1 and \u03b2 are Hermitian matrices. Note that this implies that \u03b30 is Hermitian, while \u03b3 is anti-Hermitian. We choose to make use of the 2 \u00d7 2 Pauli spin matrices \u03c3 (3.50) and we write 0 \u03c3 \u03c3 0 \u03b1 = and \u03b2 =. (8.55) 1 0 0 \u22121 That is, we write these 4 \u00d7 4 matrices as 2 \u00d7 2 matrices of 2 \u00d7 2 matrices. 8.2.1 The Conserved Current The Dirac equation immediately solves the problem of the positive de\ufb01nite nature of the probability density. De\ufb01ning \u03a8\u2020 in the usual way, namely as the complex conjugate of the row vector corresponding to the column vector \u03a8, we can show that the quantity \u03c1 = \u03a8\u2020\u03a8 492 Relativistic Quantum Mechanics is in fact interpretable as a probability density. First, as the sum of the squared magnitudes of all four components of \u03a8(x, t), it is positive de\ufb01nite. Historically, the ability of the Dirac equation to provide a positive de\ufb01nite probability current was one of the main reasons it was adopted as the correct direction for relativistic quantum mechanics. An examination of its free-particle solutions led to an attractive interpretation of negative energies, and in fact to the discovery of the positron. Second, it satis\ufb01es the continuity equation \u2202\u03c1 \u2202t + \u2207 \u00b7 j = 0 (8.56) for j = \u03a8\u2020\u03b1\u03a8. (This is simple to prove. Just use the Schr\u00a8odinger equation and its adjoint. See Problem 8.10 at the end of this chapter.) This means that \u03c1 can only change on the basis of a \ufb02ow into or out of the immediate region of interest, and is therefore a conserved quantity. Instead of \u03a8\u2020 one usually uses \u03a8 \u2261 \u03a8\u2020\u03b2 = \u03a8\u2020\u03b30 in forming the probability density and current. In this case \u03c1 = \u03a8\u2020\u03a8 = \u03a8\u2020\u03b30\u03b30\u03a8 = \u03a8\u03b30\u03a8 and j = \u03a8\u2020\u03b30\u03b30\u03b1\u03a8 = \u03a8\u03b30\u03b1\u03a8. Since \u03b30\u03b1 = \u03b3 by (8.52) we have where % \u03a8\u03b30\u03a8 & % \u03a8\u03b3\u03a8 & + \u2207 \u00b7 = \u2202\u03bcj \u03bc = 0 \u2202 \u2202t j \u03bc = \u03a8\u03b3\u03bc\u03a8 is a four-vector current. Rewriting (8.46) in terms of four-momentum as (\u03b3\u03bcp\u03bc \u2212 m)\u03a8(x, t) = 0, and also taking the adjoint of this equation and using (8.49) to insert a factor of \u03b30, \u03a8(x, t)(\u03b3\u03bcp\u03bc \u2212 m) =0, (8.57) (8.58) (8.59) (8.60) we come to an insightful interpretation of the conserved current for a free particle. We write?! \" \u03a8\u03b3\u03bc?! \u03a8 + \u03a8 [\u03b3\u03bc\u03a8] \" @ \u03a8\u03b3\u03bc \u03b3\u03bdp\u03bd\u03a8 + \u03a8\u03b3\u03bdp\u03bd [\u03b3\u03bc\u03a8] @ \u03a8 [\u03b3\u03bc\u03b3\u03bd + \u03b3\u03bd \u03b3\u03bc] p\u03bd\u03a8 j \u03bc = = = = 1 2 1 2m 1 2m p\u03bc m \u03a8\u03a8. Thus, writing the usual Lorentz contraction factor as \u03b3, # $ \u03a8\u03a8 = \u03b3 \u03a8\u2020 up \u03a8up \u2212 \u03a8\u2020 down \u03a8down j 0 = E m # $ \u03a8\u03a8 = \u03b3v \u03a8\u2020 up \u03a8up \u2212 \u03a8\u2020 down \u03a8down. j = p m (8.61) (8.62) (8.63) 493 8.2 The Dirac Equation The factor of \u03b3 is expected because of the Lorentz contraction of the volume element d3x in the direction of motion. (See Holstein (1992) for an extended discussion.) The meaning of the relative negative sign of the upper and lower components becomes clearer after we study the speci\ufb01c solutions for the free particle. 8.2.2 Free-Particle Solutions We are now in a position to study solutions of the Dirac equation and their symmetry properties. Already we notice", ", however, that the wave function \u03a8(x, t) has four components, whereas the Klein\u2013Gordon wave function \u03a5(x, t) has two. We will see that the additional degree of freedom in the Dirac equation is the same quantity we called \u201cspin 1 2 \u201d at the very beginning of this book. The four-component object \u03a8(x, t) is called a \u201cspinor.\u201d We get immediate insight as to the nature of the solutions of the Dirac equation, just by considering free particles at rest (p = 0). In this case the Dirac equation is simply i\u2202t\u03a8 = \u03b2m\u03a8. Given the diagonal form of \u03b2 (8.55) we see that there are four independent solutions for \u03a8(x, t). These are \u23a4 \u23a4 \u23a4 \u23a1 \u23a1 \u23a4 \u23a1 \u23a1 \u03a81 = e\u2212imt \u23a2 \u23a2 \u23a3 \u23a5 \u23a5 \u23a6, \u03a82 = e\u2212imt \u23a2 \u23a2 \u23a3 \u23a5 \u23a5 \u23a6, \u03a83 = e+imt \u23a2 \u23a2 \u23a3 \u23a5 \u23a5 \u23a6, \u03a84 = e+imt \u23a2 \u23a2 \u23a3 8.64) Just as in the case of the Klein\u2013Gordon equation, the lower half of the wave function corresponds to negative energy, and we will need to deal with its interpretation later. Both the upper and lower halves of the Dirac wave function, however, have one component that it is tempting to call \u201cspin up\u201d and the other we would call \u201cspin down.\u201d This interpretation is in fact correct, but we need to be somewhat more ambitious before we can state this with con\ufb01dence. Let us go on and consider free-particle solutions with nonzero momentum p = p\u02c6z, that is, a particle moving freely in the z-direction. In this case, we want to solve the eigenvalue problem H\u03a8 = E\u03a8 for H = \u03b1zp + \u03b2m, which is no longer diagonal in spinor space. The eigenvalue equation becomes \u23a1 \u23a2 \u23a2 \u23a3 0 m 0 \u2212m m 0 p 0 \u2212p p 0 0 \u2212p 0 0 \u2212m \u23a4 \u23a1 \u23a5 \u23a5 \u23a6 \u23a2 \u23a2 \u23a3 u1 u2 u3 u4 \u23a4 \u23a1 \u23a5 \u23a5 \u23a6 = E \u23a2 \u23a2 \u23a3 \u23a4 \u23a5 \u23a5 \u23a6. u1 u2 u3 u4 (8.65) Notice that the equations for u1 and u3 are coupled together, as are the equations for components u2 and u4, but these components are otherwise independent of each other. This makes it simple to \ufb01nd the eigenvalues and eigenfunctions. Details are left as an exercise. (See Problem 8.11 at the end of this chapter.) From the two equations coupling u1 and u3, we \ufb01nd E = \u00b1Ep. We \ufb01nd the same for the two equations coupling u2 and u4. Once again, we \ufb01nd the expected \u201ccorrect\u201d positive energy eigenvalue, and also the \u201cspurious\u201d 494 Relativistic Quantum Mechanics negative energy solution. In the case of the Dirac equation, however, a relatively palatable interpretation is forthcoming, as we shall see shortly. First, however, let us return to the question of spin. Continue to construct the free-particle spinors. For E = +Ep we can either set u1 = 1 (andu 2 = u4 = 0) in which case u3 = +p/(Ep + m) or u2 = 1 (and u1 = u3 = 0) in which case u4 = \u2212p/(Ep + m). In both cases, as for the Klein\u2013Gordon equation, the upper components dominate in the nonrelativistic case. Similarly, for E = \u2212Ep the nonzero components are either u3 = 1 and u1 = \u2212p/(Ep + m) or u4 = 1 and u2 = p/(Ep + m) and the lower components dominate nonrelativistically. Now consider the behavior of the operator \u03a3 \u00b7 \u02c6p = \u03a3z where \u03a3 is the 4 \u00d7 4 matrix \u03a3 \u2261 \u03c3 0 0 \u03c3. (8.66) We expect this operator to project out components of spin in the direction of momentum. \u03a3 projects out positive (negative) Indeed, it is simple to see that", " the spin operator S = \u00afh 2 helicity for the positive energy solution with u1 = 0 (u2 = 0). We \ufb01nd the analogous results for the negative energy solutions. In other words, the free-particle solutions do indeed behave appropriately according to the spin-up/down assignment that we have conjectured. Putting this together, we write the positive energy solutions as \u23a1 \u23a4 \u23a1 \u23a4 u(+) R (p) = \u23a2 \u23a2 \u23a2 \u23a3 1 0 p Ep+m 0 \u23a5 \u23a5 \u23a5 \u23a6, u(+) L (pp Ep+m for E = +Ep (8.67a) where the subscript R (L) stands for right (left) handedness, i.e. positive (negative) helicity. For the negative energy solutions, we have u(\u2212) R (pp Ep+(\u2212) L (p Ep+m 0 1 for E = \u2212Ep. (8.67b) These spinors are normalized to the factor 2Ep/(Ep + m). The free-particle wave functions are formed by including the normalization, and also the factor exp(\u2212ip\u03bcx\u03bc). 8.2.3 Interpretation of Negative Energies Dirac made use of the Pauli exclusion principle in order to interpret the negative energy solutions. One conjectures a \u201cnegative energy sea\u201d that is \ufb01lled with electrons, as shown in Figure 8.1. (This represents a \u201cbackground\u201d of in\ufb01nite energy and in\ufb01nite charge, but it is possible to imagine that we would be insensitive to both of these.) Since this \ufb01lls all of the negative energy states, it is not possible for positive energy electrons to fall into negative energies. It would be possible, however, for a high-energy photon to promote electrons out of the sea into positive energies, where it would be observable. The \u201chole\u201d left in the sea would also be observable, as an object will all the properties of an electron, but with positive charge. 495 8.2 The Dirac Equation Energy m0c2 0 \u03b3 \u2212m0c2 Electron Hole (positron) Fig. 8.1 The figure on the left shows Dirac\u2019s interpretation of negative energy states, including the possibility that a negative energy electron can be promoted to positive energy, leaving a positively charge hole, or a \u201cpositron.\u201d The figure on the right shows the discovery of the positron, Anderson, Phys. Rev., 43 (1933) 491. This cloud chamber photograph shows a particle track moving upward, bending in a known magnetic field. The direction is determined because the curvature above the lead plate is larger than below, since the particle lost energy while traversing it. The amount of energy loss is not consistent with a proton of this momentum, but is consistent with a particle having the mass of an electron. Figure 8.1 also shows the original discovery of the positron, by Carl Anderson in 1933. After this discovery, the Dirac equation became the standard treatment of relativistic quantum mechanics, explaining the spin 1 2 electron and (as we will see) its electromagnetic interactions. 8.2.4 Electromagnetic Interactions We introduce electromagnetic interactions in the Dirac equation in the same way that we did for the Klein\u2013Gordon equation, that is through \u02dcp \u2261 p \u2212 eA and the Dirac equation becomes, in 2 \u00d7 2 matrix form, \u03c3 \u00b7 \u02dcp m \u03c3 \u00b7 \u02dcp \u2212m u v = E u v where \u03a8 = (8.68) (8.69). u v At nonrelativistic (positive) energies E = K + m, the kinetic energy K m and the lower equation becomes \u03c3 \u00b7 \u02dcpu = (E + m)v \u2248 2mv (8.70) which lets us write the upper equation as (\u03c3 \u00b7 \u02dcp) (\u03c3 \u00b7 \u02dcp) 2m u = \u00b7 ( \u02dcp \u00d7 \u02dcp) u = Ku (8.71) \u02dcp2 2m + i\u03c3 2m 496 Relativistic Quantum Mechanics where we have used (3.57). Now in the coordinate representation \u02dcp \u00d7 \u02dcpu = (i\u2207 + eA) \u00d7 (i\u2207u + eAu) = ie [\u2207 \u00d7 (Au) +A \u00d7 \u2207u] = ie(\u2207 \u00d7 A)u = ieBu (8.72) where B is the magnetic \ufb01eld associated with the vector potential A. Therefore (8.71) becomes where with and \u02dcp2 2m \u2212 \u03bc \u00b7 B u = Ku \u03bc = g e 2m S S =", " \u03c3 \u00afh 2 g = 2. (8.73) (8.74) (8.75) (8.76) In other words, the Dirac equation in the presence of an electromagnetic \ufb01eld, reduces nonrelativistically to (8.73), which is just the time-independent Schr\u00a8odinger equation (with energy eigenvalue K) for a particle with magnetic moment \u03bc in the presence of an external magnetic \ufb01eld. The magnetic moment is derived from the spin operator with a gyromagnetic ratio g = 2. This brings us full circle. We began this book by discussing the behavior of particles with magnetic moments in the presence of inhomogeneous magnetic \ufb01elds, where it appears that they behaved as if they had spin projections quantized in one of two states. We now see that this stems from a consideration of relativity and quantum mechanics, for particles that obey the Dirac equation. 8.3 Symmetries of the Dirac Equation Let us now examine some symmetries inherent in the Dirac equation. We will consider situations where a spin 1 2 particle sits in some external potential, that is, solutions to the equation where \u2202 \u2202t i \u03a8(x, t) =H\u03a8(x, t) =E \u03a8(x, t) H = \u03b1 \u00b7 p + \u03b2m + V(x) (8.77) (8.78) 497 8.3 Symmetries of the Dirac Equation for some potential energy function V(x). This form, of course, ruins our ability to write a covariant equation, but that is a necessary penalty if we want to talk about potential energy. We note, though, that in the case of electromagnetic interactions, we can end up with exactly this form from the covariant equation if we choose a frame in which the vector potential A = 0. 8.3.1 Angular Momentum Our discussion of rotational invariance for wave mechanics in three dimensions centered on the fact that the orbital angular momentum operator L = x \u00d7 p commuted with Hamiltonians with \u201ccentral potentials.\u201d This in turn hinges on the fact that L commutes with p2, and so the kinetic-energy operator, and also x2 (see (3.264)). Let us now consider the commutator [H, L] \ufb01rst for the free Dirac Hamiltonian (8.53). It is obvious that [\u03b2, L] =0 so, we need to consider the commutator [\u03b1 \u00b7 p, Li] = [\u03b1p, \u03b5ijkxjpk] = \u03b5ijk\u03b1[ p, xj]pk = \u2212i\u03b5ijk\u03b1jpk = 0. (8.79) (Recall our convention of summing over repeated indices.) In other words, the orbital angular momentum operator does not commute with the Dirac Hamiltonian! Therefore L will not be a conserved quantity for spin 1 2 particles that are either free or bound in central potentials. Consider, however, the spin operator (8.66) and its commutator with the Hamiltonian. It is simple to show that \u03b2\u03a3i = \u03a3i\u03b2. It is also easy to use (3.53) to show that [\u03b1i, \u03a3j] =2i \u03b5ijk\u03b1k. So, we need to evaluate [\u03b1 \u00b7 p, \u03a3j] = [\u03b1i, \u03a3j]pi = 2i\u03b5ijk\u03b1kpi. (8.80) Thus we see that (putting back in \u00afh momentarily) even though neither L nor \u03a3 commutes with the free Dirac Hamiltonian, the combined vector operator \u00afh 2 does commute. That is, the Dirac Hamiltonian conserves total angular momentum, but not orbital or spin angular momenta separately8.81) 8.3.2 Parity In the case where V(x) =V (|x|) we expect solutions to be parity symmetric. That is, we should have \u03a8(\u2212x) =\u00b1\u03a8(x). It does not appear that this is the case, however, since if x \u2192 \u2212x, thenp \u2192 \u2212p in (8.78) and the Hamiltonian changes form. However, this simple look does not take into account the parity transformation on spinors. Indeed, the parity transformation operator \u03c0, discussed in Section 4.2, only concerns coordinate re\ufb02ection. That is, \u03c0 is a unitary (and also Hermitian) operator with the properties 498 Relativistic Quantum Mechanics \u03c0\u2020x\u03c0 = \u2212x \u03c0\u2020p\u03c0 = \u2212p (8.82a) (8.82b) (See equations (4.43) and (4.50).) The full parity operator, let us call it P", ", needs to be augmented with a unitary operator UP, which is a 4 \u00d7 4 matrix in spinor space, and which renders the Hamiltonian (8.78) invariant under a parity transformation. That is where the matrix UP must have the properties P \u2261 \u03c0UP U\u2020 P U\u2020 P \u03b1UP = \u2212\u03b1 \u03b2UP = \u03b2 as well as (8.83) (8.84a) (8.84b) P = 1. Obviously UP = \u03b2 = \u03b2\u2020 is consistent with these requirements. Consequently, a parity transformation on a Dirac Hamiltonian consists of taking x \u2192 \u2212x and multiplying on both the left and the right by \u03b2. The transformation of a spinor \u03a8(x) yields \u03b2\u03a8(\u2212x). (8.84c) U2 Therefore, in cases where V(x) =V (|x|), we expect to \ufb01nd eigenstates of the Dirac Hamiltonian which are simultaneously eigenstates of parity, J2 and Jz. Luckily, we have already constructed angular and spinor parts of these eigenfunctions. These are the twocomponent spin-angular functions Y j=l\u00b11/2,m (\u03b8, \u03c6) de\ufb01ned in (3.384). We will make use of these as we proceed to solve the Dirac equation for a particular potential energy function of this form in Section 8.4. l 8.3.3 Charge Conjugation We saw in (8.24) that for the Klein\u2013Gordon equation, we could split the positive and negative energy solutions into \u201cparticle\u201d and \u201cantiparticle\u201d solutions on the basis of the association \u03a8particle(x, t) \u2261 \u03a8E>0(x, t) E<0(x, t). \u03a8antiparticle(x, t) \u2261 \u03a8\u2217 (8.85a) (8.85b) Let us work towards a similar association for the Dirac equation, and then explore a symmetry operation that connects the two solutions. For our purposes, an \u201cantiparticle\u201d is an object whose wave function behaves just as that for a \u201cparticle\u201d but with opposite electric charge. So, let us return to the covariant form (8.46) of the Dirac equation, and add an electromagnetic \ufb01eld according to our usual prescription (8.13). We have % & i\u03b3\u03bc\u2202\u03bc \u2212 e\u03b3\u03bcA\u03bc \u2212 m \u03a8(x, t) =0. (8.86) We seek an equation where e \u2192 \u2212e, and which relates the new wave function to \u03a8(x, t). The key is that the operator in (8.86) has three terms, only two of which contain \u03b3\u03bc, and only one of those two contains i. So, take the complex conjugate of this equation to \ufb01nd 499 8.3 Symmetries of the Dirac Equation! \u2212i (\u03b3\u03bc) \u2217 \u2217 \u2202\u03bc \u2212 e (\u03b3\u03bc) A\u03bc \u2212 m \" \u03a8\u2217(x, t) =0 (8.87) and the relative sign between the \ufb01rst two terms is reversed. If we can now identify a matrix \u02dcC such that \u02dcC (\u03b3\u03bc) \u2217 \u02dcC\u22121 = \u2212\u03b3\u03bc (8.88) we just insert 1 = \u02dcC\u22121 \u02dcC before the wave function in (8.87), and multiply on the left by \u02dcC. The result is! \" i\u03b3\u03bc\u2202\u03bc + e\u03b3\u03bcA\u03bc \u2212 m \u02dcC\u03a8\u2217(x, t) = 0. (8.89) Therefore, the wave function \u02dcC\u03a8\u2217(x, t) satis\ufb01es the \u201cpositron\u201d equation (8.89) where \u03a8(x, t) satis\ufb01es the \u201celectron\u201d equation (8.86). &\u2217 We need to identify the matrix \u02dcC. From (8.52) and (8.55) we see that \u03b30, \u03b31, and \u03b33 are % \u03b32 = \u2212\u03b32. This makes it possible to realize (8.88) with real matrices, but \u02dcC = i\u03b32. (8.90) Therefore, the \u201cpositron wave\u201d function corresponding to \u03a8(x, t) is i\u03b32\u03a8\u2217(x, t). It is more convenient, in turns out, to write this wave function in terms of \u03a8 = \u03a8\u2020\u03b30 = (\u03a8\u2217)T \u03b30. (The superscript T indicates \u201c", "transpose.\u201d) This means that the \u201cpositron\u201d wave function can be written as \u02dcC\u03a8\u2217(x, t) =i \u03b32 where % \u03a8\u03b30 &T &T % \u03a8 = UC UC \u2261 i\u03b32\u03b30. Therefore, the charge conjugation operator is C where &T % C \u03a8(x, t) = UC \u03a8. (8.91) (8.92) (8.93) Note that the change in the space-time part of the free-particle wave function \u03a8(x, t) \u221d exp(\u2212ip\u03bcx\u03bc) due to C is to, effectively, take x \u2192 \u2212x and t \u2192 \u2212t. 8.3.4 Time Reversal Let us now apply the ideas of Section 4.4 to the Dirac equation. First, a brief review. This discussion hinged on the de\ufb01nition (4.117) of an antiunitary operator \u03b8 \u03b8 = UK (8.94) where U is a unitary operator and K is an operator which takes the complex conjugate of any complex numbers which follow it. Clearly, K does not affect the kets to the right, and K2 = 1. Based on this, we de\ufb01ned an antiunitary operator \u0398 which takes an arbitrary state |\u03b1 to a time-reversed (or, more properly, a motion-reversed) state | \u02dc\u03b1, that is \u0398|\u03b1 = | \u02dc\u03b1. (8.95) 500 Relativistic Quantum Mechanics We imposed two reasonable requirements on \u0398, namely (4.148) and (4.150), that is and so \u0398p\u0398\u22121 = \u2212p \u0398x\u0398\u22121 = x \u0398J\u0398\u22121 = \u2212J. (8.96a) (8.96b) (8.96c) For Hamiltonians that commute with \u0398, we made the important observation (4.162) that an energy eigenstate |n has the same energy eigenvalue as its time-reversed counterpart state \u0398|n. We further learned that acting twice on a spin 1 2 state yields \u03982 = \u22121, since the two-component spinor form of (4.168) shows that \u0398 = \u2212i\u03c3yK. In other words, U = \u2212i\u03c3y in (8.94). Indeed, in this case \u03982 = \u2212i\u03c3yK(\u2212i\u03c3yK) = \u03c3y\u03c3\u2217 y K2 = \u22121. (8.97) (8.98) We will see something similar when we apply time reversal to the Dirac equation, which we now take up. Returning to the Schr\u00a8odinger equation (8.1) with the Dirac Hamiltonian (8.52), but using the \u03b3 matrices instead of \u03b1 and \u03b2, we have! i\u2202t\u03a8(x, t) = \u2212i\u03b30\u03b3 \u00b7 \u2207 + \u03b30m \" \u03a8(x, t). (8.99) We write our time-reversal operator, following the scheme earlier in this Section, as T instead of \u0398, where T = UTK (8.100) and UT is a unitary matrix which we need to identify. As before, insert T \u22121T before the wave function on the left and right sides, and multiply through from the left by T. The left side of (8.99) becomes T (i\u2202t)T \u22121T \u03a8(x, t) =U TK(i\u2202t)KU\u22121 T UT\u03a8\u2217(x, t) (8.101) reversing the sign of t in the derivative, as we need. In order for [UT\u03a8\u2217(x, t)] to satisfy the time-reversed form of (8.99), we must insist that = \u2212i\u2202tUT\u03a8\u2217(x, t) =i \u2202\u2212t [UT\u03a8\u2217(x, t)] T % i\u03b30\u03b3 % \u03b30 T & & T \u22121 = i\u03b30\u03b3 T \u22121 = \u03b30. (8.102a) (8.102b) These are easily converted to a more convenient form so that we can identify UT. First, do T \u22121 on the left and T on the right. Then, do K on the left and right. Finally, insert UTU\u22121 T in between the \u03b3 matrices in (8.102a) and then use the result from (8.102b). Both results then become U\u22121 T (\u03b3) UT = \u2212 (\u03b3) &\u2217 % U", "\u22121 UT = T \u03b30 \u03b30 & % \u2217. (8.103a) (8.103b) 501 8.3 Symmetries of the Dirac Equation We now specialize to our choice (8.55), with (8.52), for the \u03b3 matrices. Only \u03b32 is imaginary in this representation. Therefore, if we want to build UT out of \u03b3 matrices, then (8.103) says that we need a combination that does not change the sign when commuted with \u03b30 and \u03b32, but does change sign with \u03b31 and \u03b33. Rather obviously, this is accomplished by UT = \u03b31\u03b33 (8.104) up to some arbitrary phase factor. Indeed, this works out to be equivalent to the result U = i\u03c3y in (8.97). See Problem 8.13 at the end of this chapter. 8.3.5 CPT We conclude with a brief look of the operator combination C PT. Its action on a Dirac wave function \u03a8(x, t) is straightforward to work out, given the above discussion. That is C PT \u03a8(x, t) =i \u03b32 [PT \u03a8(x, t)] \u2217 \u2217 = i\u03b32\u03b30 [T \u03a8(\u2212x, t)] = i\u03b32\u03b30\u03b31\u03b33\u03a8(\u2212x, t) = i\u03b30\u03b31\u03b32\u03b33\u03a8(\u2212x, t). (8.105) This combination of \u03b3 matrices is well known, and in fact given a special name. We de\ufb01ne \u03b35 \u2261 i\u03b30\u03b31\u03b32\u03b33. (8.106) In our basis (8.55), again writing 4 \u00d7 4 matrices as 2 \u00d7 2 matrices of 2 \u00d7 2 matrices, we \ufb01nd that \u03b35 =. 0 1 1 0 (8.107) That is, \u03b35, and therefore also C PT, reverses the up and down two-component spinors in the Dirac wave function. The net effect of C PT on a free-particle electron wave function is in fact to convert it into the \u201cpositron\u201d wave function. See Problem 8.14 at the end of this chapter. This is a \u201ctip of the iceberg\u201d observation of a profound concept in relativistic quantum \ufb01eld theory. The notion of C PT invariance is equivalent to a total symmetry between matter and antimatter, so long as one integrates over all potential \ufb01nal states. For example, it predicts that the mass of any particle must equal the mass of the corresponding antiparticle. Indeed, one can show, although it is far from straightforward, that any Lorentz invariant quantum \ufb01eld theory is invariant under C PT. The implications are far reaching, particularly in this day and age when string theories offer the possibility that Lorentz invariance is broken at distances approaching the Planck mass. The reader is referred to any one of a number of advanced textbooks, and also the current literature, for more details. 502 Relativistic Quantum Mechanics 8.4 Solving with a Central Potential Our goal is to solve the eigenvalue problem where H\u03a8(x) =E \u03a8(x) H = \u03b1 \u00b7 p + \u03b2m + V(r) (8.108) (8.109) and we write the four-component wave function \u03a8(x) in terms of two two-component wave functions \u03c81(x) and \u03c82(x) as \u03a8(x) =. \u03c81(x) \u03c82(x) (8.110) Based on the symmetries of the Dirac equation that we have already discussed, we expect \u03a8(x) to be an eigenfunction of parity, J2 and Jz. Parity conservation implies that \u03b2\u03a8(\u2212x) = \u00b1\u03a8(x). Given the form (8.55) of \u03b2, this implies that \u03c81(\u2212x) \u2212\u03c82(\u2212x) = \u00b1. \u03c81(x) \u03c82(x) This leaves us with two choices, namely \u03c81(\u2212x) = +\u03c81(x) and \u03c82(\u2212x) = \u2212\u03c82(x) \u03c81(\u2212x) = \u2212\u03c81(x) and \u03c82(\u2212x) = +\u03c82(x). (8.111) (8.112a) (8.112b) These conditions are neatly realized by the spinor functions Y jm (\u03b8, \u03c6) de\ufb01ned in (3.384), where l = j \u00b1 (1/2).", " For a given value of j, one possible value of l is even and the other is just (\u22121)l, then we are presented with two is odd. Since the parity of any particular Ym l natural choices for the angular and spinor dependences for the conditions (8.112). We write l j\u22121/2(\u03b8, \u03c6) j+1/2(\u03b8, \u03c6) which is an even (odd) parity solution if j \u2212 1/2 is even (odd), or uA(r)Y jm \u2212ivA(r)Y jm \u03a8(x) =\u03a8 A(x) \u2261 \u03a8(x) =\u03a8 B(x) \u2261 uB(r)Y jm \u2212ivB(r)Y jm j+1/2(\u03b8, \u03c6) j\u22121/2(\u03b8, \u03c6) (8.113a) (8.113b) which is an odd (even) parity solution if j \u2212 1/2 is even (odd). (The factor of \u2212i on the lower spinors is included for later convenience.) Note that although both \u03a8A(x) and \u03a8B(x) have de\ufb01nite parity and quantum numbers j and m, they mix values of l. Orbital angular momentum is no longer a good quantum number when considering central potentials in the Dirac equation. 503 8.4 Solving with a Central Potential We are now ready to turn (8.108) into a differential equation in r for the functions uA(B)(r) and vA(B)(r). First, rewrite the Dirac equation as two coupled equations for the spinors \u03c81(x) and \u03c82(x). Thus [E \u2212 m \u2212 V(r)] \u03c81(x) \u2212 (\u03c3 \u00b7 p)\u03c82(x) =0 [E + m \u2212 V(r)] \u03c82(x) \u2212 (\u03c3 \u00b7 p)\u03c81(x) =0. (8.114a) (8.114b) Now make use of (3.57) and (3.59) to write \u03c3 \u00b7 p = = 1 r2 (\u03c3 \u00b7 x)(\u03c3 \u00b7 x)(\u03c3 \u00b7 p) 1 r2 (\u03c3 \u00b7 x)[x \u00b7 p + i\u03c3 \u00b7 (x \u00d7 p)] \u02c6r \u00b7 p + i\u03c3 \u00b7 L r. = (\u03c3 \u00b7 \u02c6r) Working in coordinate space, we have \u02c6r \u00b7 p \u2192 \u02c6r \u00b7 (\u2212i\u2207) = \u2212i \u2202 \u2202r which will act on the radial part of the wave function only. We also know that so that we can write where where \u03c3 \u00b7 L = 2S \u00b7 L = J2 \u2212 L2 \u2212 S2 (\u03c3 \u00b7 L)Y jm l = j( j + 1) \u2212 l(l + 1) \u2212 3 4 Y jm l \u2261 \u03ba( j, l)Y jm l \u03ba = \u2212 = \u2212(\u03bb + 1) for l = j + 2 = +(\u03bb \u2212 1) for. It is trickier to calculate the effect of the matrix factor \u03c3 \u00b7 \u02c6r = cos \u03b8 e\u2212i\u03c6 sin \u03b8 \u2212 cos \u03b8 ei\u03c6 sin \u03b8 (8.115) (8.116) (8.117) (8.118) (8.119a) (8.119b) (8.120) (8.121) on the spinor wave functions. In principle, we can carry out the multiplication on the Y jm as de\ufb01ned in (3.384) and then use the de\ufb01nition (3.246) to evaluate the result. There is an easier way, however. l We expect \u03c3 \u00b7 \u02c6r to behave as a (pseudo)-scalar under rotations, so if we evaluate its effect at one particular \u02c6r then it should behave this way for all \u02c6r. Choose \u02c6r = \u02c6z, that is, \u03b8 = 0. 504 Relativistic Quantum Mechanics (0, \u03c6) (0, \u03c6) (8.122) (8.123) Since the \u03b8-dependent part of any Ym write l (\u03b8, \u03c6) contains a factor [sin \u03b8]|m| we use (3.248) to l (\u03b8 = 0, \u03c6) = Ym \u2019 2l + 1 4\u03c0 \u03b4m0 in which case Y j=l\u00b11/2,m l (\u03b8 = 0, \u03c6) = = \u00b1 1/2 Ym\u22121/2 l \u2213 m + 1/2 Ym+1", "/2 l l \u00b1 m + 1/2 \u03b4m,1/2 l \u2213 m + 1/2 \u03b4m,\u22121/2 \u00b1 \u221a 2l + 1 1\u221a 4\u03c0 or Therefore Y j,m l=j\u22131/2(\u03b8 = 0, \u03c6) = \u2019 j + 1/2 4\u03c0 \u2019 j + 1/2 4\u03c0 \u2213\u03b4m,1/2 \u03b4m,\u22121/2. \u00b1\u03b4m,1/2 \u03b4m,\u22121/2 (\u03c3 \u00b7 \u02c6z)Y j,m l=j\u22131/2(\u03b8 = 0, \u03c6) = \u2212 = \u2212Y j,m l=j\u00b11/2(\u03b8 = 0, \u03c6) (8.124) and so, as we have argued that this result is independent of \u03b8 and \u03c6, we have (\u03c3 \u00b7 \u02c6r)Y j,m l=j\u00b11/2(\u03b8, \u03c6) =\u2212 Y j,m where we have used the fact that (\u03c3 \u00b7 \u02c6r)2 = 1. In other words, for a given j and m, Y j,m l=j\u00b11/2(\u03b8, \u03c6) is an eigenstate of \u03c3 \u00b7 \u02c6r with eigenvalue \u22121 (a consequence of the pseudoscalar nature of the operator) and changes l to the other allowed value, which naturally has opposite parity. l=j\u22131/2(\u03b8, \u03c6) (8.125) Now we return to the coupled equations (8.114) with solutions in the form (8.113). We have two choices for \u03c81(x) and \u03c82(x), namely \u201cChoice A\u201d \u03c81(x) =u A(r)Y jm j\u22121/2(\u03b8, \u03c6) and \u03c82(x) =\u2212 ivA(r)Y jm j+1/2(\u03b8, \u03c6) (8.126) or \u201cChoice B\u201d \u03c81(x) =u B(r)Y jm j+1/2(\u03b8, \u03c6) and \u03c82(x) =\u2212 ivB(r)Y jm j\u22121/2(\u03b8, \u03c6). (8.127) Note that for whichever of these two choices we pick, the effect of the factor (\u03c3 \u00b7 \u02c6r), which is part of (\u03c3 \u00b7 p) in (8.114) is to exchange l = j \u00b1 1/2 for l = j \u2213 1/2 in the angular spinor Y jm, that is, switching the angular spinor factor of the second term in each of (8.114) so l that it is the same as the \ufb01rst term. In other words, the angular factors drop out, and we are left with just the radial equations. Putting this all together, \ufb01nally, (8.114) becomes for \u201cChoice A\u201d [E \u2212 m \u2212 V(r)] uA(r) \u2212 [E + m \u2212 V(r)] vA(r) + d dr d dr + \u2212 \u03bb + 1 r \u03bb \u2212 1 r vA(r) = 0 uA(r) = 0 (8.128a) (8.128b) 505 8.4 Solving with a Central Potential \u2212 d dr \u03bb \u2212 1 r \u03bb + 1 r and, for \u201cChoice B\u201d [E \u2212 m \u2212 V(r)] uB(r) \u2212 vB(r) =0 (8.129a) [E + m \u2212 V(r)] vB(r) + d dr However, formally, equations (8.128) become (8.129) with the exchange \u03bb \u2194 \u2212\u03bb. Therefore, we can focus on the solution to (8.128) and drop the subscript A. uB(r) =0. (8.129b) + Equations (8.128) are coupled, \ufb01rst-order ordinary differential equations to be solved for the u(r) and v(r), subject to certain boundary conditions (i.e. normalizability) which will yield eigenvalues E. This solution can at least be carried out numerically, which is practical in many situations. We conclude this section, however, with one case that can be solved analytically. 8.4.1 The One-Electron Atom We can now consider atoms with one electron, with the potential energy function V(r) = \u2212 Ze2 r. (8.130) We expect that the \u201c\ufb01", "ne structure\u201d of the hydrogen atom, which we studied using perturbation theory in Section 5.3, should emerge naturally with our solution using the Dirac equation. Start by writing (8.128) in terms of scaled variables. That is \u03b5 \u2261 E m x \u2261 mr (8.131) (8.132) and recall that we write \u03b1 \u2261 e2(/\u00afhc) \u2248 1/137. This gives \u03bb + 1 x \u03bb \u2212 1 x Z\u03b1 x Z\u03b1 (x) \u2212 v(x) + d dx d dx \u2212 + v(x) =0 (8.133a) u(x) =0. (8.133b) Next consider the behavior of the solutions as x \u2192 \u221e. Equation (8.133a) becomes (\u03b5 \u2212 1)u \u2212 dv dx = 0 (8.134) and so (8.133b) implies that (\u03b5 + 1)v + du dx = (\u03b5 + 1)v + 1 \u03b5 \u2212 1 d2v dx2 = 0 which leads to d2v dx2 = (1 \u2212 \u03b52)v. (8.135) (8.136) 506 Relativistic Quantum Mechanics Note that, classically, bound states require that the kinetic energy E \u2212 m \u2212 V(r) = 0 at some distance r, and V(r) < 0 everywhere, so E \u2212 m < 0 and \u03b5 = E/m < 1. Therefore 1 \u2212 \u03b52 is guaranteed to be positive, and (8.136) implies that \" \u2212(1 \u2212 \u03b52)1/2x v(x) =exp x \u2192 \u221e (8.137) for! where we require that v(x) be normalizable as x \u2192 \u221e, but ignore the normalization constant for now. Similarly, (8.134) then implies that u(x) =exp! \" \u2212(1 \u2212 \u03b52)1/2x for x \u2192 \u221e (8.138) as well. Now write u(x) and v(x) as power series with their own expansion coef\ufb01cients, and look for relationships consistent with the differential equations. Using u(x) =e \u2212(1\u2212\u03b52)1/2xx\u03b3 v(x) =e \u2212(1\u2212\u03b52)1/2xx\u03b3 \u221e \u2211 i=0 \u221e \u2211 i=0 aixi bixi (8.139) (8.140) we tacitly assume that we can \ufb01nd series solutions making use of the same overall power \u03b3 for both u(x) and v(x). Indeed, inserting these expressions into (8.133) and \ufb01rst considering terms proportional to x\u03b3\u22121, we \ufb01nd, after a little rearrangement, (Z\u03b1)a0 \u2212 (\u03b3 + \u03bb + 1)b0 = 0 (\u03b3 \u2212 \u03bb + 1)a0 + (Z\u03b1)b0 = 0. (8.141a) (8.141b) Our approach will soon yield recursion relations for the coef\ufb01cients ai and bi. This means that we need to avoid a0 = 0 and b0 = 0, and the only way to do this is to require that the determinant of (8.141) vanish. That is or, solving for \u03b3, (Z\u03b1)2 + (\u03b3 + 1 + \u03bb)(\u03b3 + 1 \u2212 \u03bb) = 0 \u03b3 = \u22121 \u00b1! \u03bb2 \u2212 (Z\u03b1)2 \"1/2. (8.142) (8.143) Notice \ufb01rst \u03bb = j + 1/2 is of order unity, so that things will break down if Z\u03b1 \u223c 1. In the strong Coulomb \ufb01elds for Z \u2248 137, spontaneous e+e\u2212 production would occur, and the single-particle nature of the Dirac equation cannot be expected to prevail. Indeed, we have Z\u03b1 1 in cases of interest here. This also means that the expression within the brackets in (8.143) is of order unity. For the \u2212 sign, this gives \u03b3 \u223c \u22122 which would be too singular at the origin. As a result, we choose the + sign and have \u03b3 = \u22121 + j + 1/2 \u2212 (Z\u03b1)2. 2 1 2 (8.144) Note that for j = 1/2 we still have a singularity at the origin, since \u03b3 < 0, but this singularity is very weak, and integrable over space. 507 8.4 Solving with a Central Potential Starting with a value for a0 which is determined by normalization, and b0 = a0(Z\u03b1)/ (", "\u03b3 + \u03bb + 1), we can \ufb01nd the remaining ai and bi by going back to the result of inserting (8.139) and (8.140) into (8.133). Collecting powers of x\u03b3 and higher, we \ufb01nd (1 \u2212 \u03b5)ai\u22121 \u2212 Z\u03b1ai \u2212 (1 \u2212 \u03b52)1/2bi\u22121 + (\u03bb + 1 + \u03b3 + i)bi = 0 (1 + \u03b5)bi\u22121 + Z\u03b1bi \u2212 (1 \u2212 \u03b52)1/2ai\u22121 \u2212 (\u03bb \u2212 1 \u2212 \u03b3 \u2212 i)ai = 0. (8.145a) (8.145b) Multiply (8.145a) by (1 + \u03b5)1/2 and (8.145b) by (1 \u2212 \u03b5)1/2 and then add them. This leads to a relationship between the coef\ufb01cients ai and bi, namely bi ai = Z\u03b1(1 + \u03b5)1/2 + (\u03bb \u2212 1 \u2212 \u03b3 \u2212 i)(1 \u2212 \u03b5)1/2 Z\u03b1(1 \u2212 \u03b5)1/2 + (\u03bb + 1 + \u03b3 + i)(1 + \u03b5)1/2. (8.146) This relation shows that for large values of x, where terms with large i dominate, ai and bi are proportional to each other. Furthermore (8.145) also implies that ai/ai\u22121 \u223c 1/i for large i. (See Problem 8.15 at the end of this chapter.) In other words, the series (8.139) and (8.140) will grow exponentially, and not be normalizable, unless we force the series to terminate. If we then assume that ai = bi = 0 fori = n + 1 then (1 \u2212 \u03b5)an \u2212 (1 \u2212 \u03b52)1/2bn = 0 (1 + \u03b5)bn \u2212 (1 \u2212 \u03b52)1/2an = 0. (8.147a) (8.147b) Either of these leads to the same condition on the ratio of these terminating coef\ufb01cients, namely bn an = 1 \u2212 \u03b5 1 + \u03b5 1/2. (8.148) (8.149) Combining (8.146) and (8.147) we have (1 + \u03b3 + n)(1 \u2212 \u03b52)1/2 = Z\u03b1\u03b5 which, \ufb01nally, can be solved for \u03b5. Putting c back in, we determine the energy eigenvalues E = mc2 1/2. 1 + #\u221a (Z\u03b1)2 ( j+1/2)2\u2212(Z\u03b1)2+n $ 2 (8.150) We emphasize that, for any given quantum number n, the energy eigenvalues depend on the total angular momentum j. That is, for example, the energy will be the same for j = 1/2, regardless of whether or not it comes from coupling l = 0 or l = 1 with spin 1 2. To lowest order in Z\u03b1, (8.150) becomes E = mc2 \u2212 1 2 where n \u2261 j + 1/2 + n. Comparison to (3.315) shows that this is simply the familiar Balmer formula, with the addition of rest mass energy, with n being the principal quantum number. Including higher orders of Z\u03b1 leads to the well-known expressions for the relativistic correction to kinetic energy (5.104) and the spin-orbit interaction (5.125). (8.151) mc2(Z\u03b1)2 n2 508 Relativistic Quantum Mechanics Zero of energy (ionized hydrogen 10 S1/2 P1/2 P3/2 D3/2 D5/ Scale \u00d74 Visible near infrared Scale \u00d71 Ultraviolet P3/2 Fine structure ~3.01\u22126 hcR n = 2 S1/2 P1/2 n = 1 n = 2 Lamb shift aorbit) j (total) 0 1/2 1 1/2 1 3/2 2 3/2 2 5/2 (b) (c) Fig. 8.2 The energy levels of the hydrogen atom, with reference to the many high precision experiments which have measured electromagnetic transitions between them. Taken from Cagnac, Phys. Scr., T70 (1997) 24. (a) The full energy-level diagram, (b) with energy scale multiplied by four, and (c) fine structure of the n = 2 and n = 3 levels, detailing behavior on the quantum numbers l and j. Figure 8.2 shows the energy levels of the hydrogen atom, and the experiments which made these measurements. A number of ingenious techniques were devised", ", including twophoton absorption for connecting \u201cforbidden\u201d transitions, in order to obtain these results. The so-called \u201c\ufb01ne structure\u201d is clear, that is, the relativistic effects that lead to splitting between the S and P levels for the n = 2 states, and between the S, P, and D levels for the n = 3 states. Problem 8.16 at the end of this chapter shows that the full relativistic energy levels give the same splittings as obtained using perturbation theory. There is, of course, a profound discrepancy between the energy levels in Figure 8.2 and the result (8.150) we derived for the energy eigenvalues. According to (8.150) the energy can only depend on n and j. However, we see that there is a small splitting between, for 509 8.5 Relativistic Quantum Field Theory example, the 2P1/2 and 2S1/2 states. This splitting, called the Lamb shift after its discoverer, played a central role in recognizing the importance of relativistic quantum \ufb01eld theory in atomic structure. See Holstein (1992) for a discussion of the history, as well as both formal and \u201cphysically intuitive\u201d derivations of the size of the Lamb shift. Problem 8.17 at the end of this chapter compares energy levels predicted by (8.150) to available high precision data. 8.5 Relativistic Quantum Field Theory We now conclude our coverage of \u201cModern Quantum Mechanics.\u201d The framework outlined in this book continues to be the most fundamental basis by which we understand the physical world. Although the probabilistic interpretation of the concept of measurement is disturbing in some ways, it always prevails when confronted with experiment. The de\ufb01ciencies that remain, for example in the Lamb shift, are not the result of problems in the underlying axioms of quantum mechanics. Instead, they are the result of a necessary approximation we make when we try to develop quantum-mechanical wave equations that are consistent with relativity. The ability to create and destroy particles is inconsistent with our \u201csingle-particle\u201d approach to writing down dynamics in quantum mechanics. Instead, we would need to re-examine the Hamiltonian formalism, on which much of this book is based, in order address these issues. Quantum \ufb01eld theory is the correct framework for addressing relativistic quantum mechanics, and multiparticle quantum mechanics in general. There are essentially two ways to approach quantum \ufb01eld theory, neither of which is developed here. We only mention them for the reader interested in going on to the next steps. One approach is through the method of \u201csecond quantization\u201d where operators are introduced which create and destroy particles. These operators commute with each other if they have integer spin, and anticommute for half-integer spin. Work needs to be done in order to build in relativistic covariance, but it is relatively straightforward. It is also, however, not necessary if the problem does not warrant it. This is the case, for example, in a vast number of fascinating problems in condensed matter physics. Second quantization is discussed in Section 7.5 of this book. For other examples, see Quantum Mechanics, by Eugen Merzbacher (1998), and Quantum Theory of ManyParticle Systems, by Alexander L. Fetter and John Dirk Walecka (2003). The second approach is through the path-integral approach to quantum mechanics, famously pioneered by Richard Feynman in his Ph.D. thesis. This conceptually appealing formalism is straightforward to extend from particle quantum mechanics to quantum \ufb01elds. However, it is not straightforward to use this formalism for calculation of typical problems, until one makes the connection onto the \u201ccanonical\u201d formalism that eventually becomes second quantization. Nevertheless, it is a worthwhile subject for students who would like to have a better understanding of the principles that lead to the quantum many-body problem. Path integrals are not the basis for many books on quantum \ufb01eld theory, but they are beautifully exploited in Zee (2010). 510 Relativistic Quantum Mechanics Problems 8.1 These exercises are to give you some practice with natural units. a. Express the proton mass mp = 1.67262158 \u00d7 10\u221227 kg in units of GeV. b. Assume that a particle with negligible mass is con\ufb01ned to a box the size of the proton, around 1 fm = 10\u221215 m. Use the uncertainty principle to estimate the energy of the con\ufb01ned particle. You might be interested to know that the mass, in natural units, of the pion, the lightest strongly interacting particle, is m\u03c0 = 135 MeV. c. String theory concerns the physics at a scale which", " combines gravity, relativity, and quantum mechanics. Use dimensional analysis to \ufb01nd the \u201cPlanck mass\u201d MP which is formed from G, \u00afh, and c, and express the result in GeV. 8.2 Show that a matrix \u03b7\u03bc\u03bd with the same elements as the metric tensor \u03b7\u03bc\u03bd used in this chapter, has the property that \u03b7\u03bc\u03bb \u03b7\u03bb\u03bd = \u03b4\u03bc \u03bd, the identity matrix. Thus, show that the natural relationship \u03b7\u03bc\u03bd = \u03b7\u03bc\u03bb \u03b7\u03bd\u03c3 \u03b7\u03bb\u03c3 in fact holds with this de\ufb01nition. Show also that a\u03bcb\u03bc = a\u03bcb\u03bc for two four-vectors a\u03bc and b\u03bc. 8.3 Show that (8.11) is in fact a conserved current, when \u03a8(x, t) satis\ufb01es the Klein\u2013 Gordon equation. 8.4 Show that (8.14) follows from (8.8). 8.5 Derive (8.16a), (8.16b), and (8.18). 8.6 Show that the free-particle energy eigenvalues of (8.18) are E = \u00b1Ep and that the eigenfunctions are indeed given by (8.21), subject to the normalization that \u03a5\u2020\u03c43\u03a5 = \u00b11 for E = \u00b1Ep. 8.7 This problem is taken from Landau (1996). A spinless electron is bound by the Coulomb potential V(r) =\u2212 Ze2/r in a stationary state of total energy E \u2264 m. You can incorporate this interaction into the Klein\u2013Gordon equation by using the covariant derivative with V = \u2212e\u03a6 and A = 0. a. Assume that the radial and angular parts of the equation separate, and that the wave function can be written as e\u2212iEt[ul(r)/r]Ylm(\u03b8, \u03c6) and show that the radial equation becomes d2u d\u03c12 + 2EZ\u03b1 \u03b3\u03c1 \u2212 1 4 \u2212 l(l + 1) \u2212 (Z\u03b1)2 \u03c12 ul(\u03c1) =0 where \u03b1 = e2, \u03b32 = 4(m2 \u2212 E2), and \u03c1 = \u03b3r. b. Assume that this equation has a solution of the usual form of a power series times the \u03c1 \u2192 \u221e and \u03c1 \u2192 0 solutions, that is ul(\u03c1) = \u03c1k(1 + c1\u03c1 + c2\u03c12 + \u00b7 \u00b7 \u00b7 )e\u2212\u03c1/2 511 Problems and show that Z\u03b1)2 k = k\u00b1 = and that only for k+ is the expectation value of the kinetic energy \ufb01nite and that this solution has a nonrelativistic limit which agrees with the solution found for the Schr\u00a8odinger equation. c. Determine the recurrence relation among the ci for this to be a solution of the Klein\u2013Gordon equation, and show that unless the power series terminates, the wave function will have an incorrect asymptotic form. d. In the case where the series terminates, show that the energy eigenvalue for the k+ solution is E = 1 + (Z\u03b1)/2 \u22122 & l + 1 2 2 \u2212 (Z\u03b1)2 where n is the principal quantum number. e. Expand E in powers of (Z\u03b1)2 and show that the \ufb01rst-order term yields the Bohr formula. Connect the higher-order terms with relativistic corrections, and discuss the degree to which the degeneracy in l is removed. Jenkins and Kunselman, Phys. Rev. Lett., 17 (1966) 1148, report measurements of a large number of transition energies for \u03c0\u2212 atoms in large Z nuclei. Compare some of these to the calculated energies and discuss the accuracy of the prediction. (For example, consider the 3d \u2192 2p transition in 59Co which emits a photon with energy 384.6 \u00b1 1.0 keV.) You will probably either need a computer to carry out the energy differences with high enough precision, or else expand to higher powers of (Z\u03b1)2. 8.8 Prove that the traces of the \u03b3\u03bc, \u03b1, and \u03b2 are all zero. 8.9 a. Derive the matrices \u03b3\u03bc from (8.55) and show that they satisfy the Clifford algebra (8.49). b. Show that \u03b30 = \u03c3i I 0 0 \u2212I 0 \u2212\u03c3i = I \u2297 \u03c43 = \u03c3i \u2297 i\u03c42 \u03b3i = 0 where I is the 2 \u00d7 2 identity matrix, and \u03c3i and \u03c4i are the Pauli matrices. (The \u2297 notation is a formal way to write", " our 4 \u00d7 4 matrices as 2 \u00d7 2 matrices of 2 \u00d7 2 matrices.) 8.10 Prove the continuity equation (8.56) for the Dirac equation. 8.11 Find the eigenvalues for the free-particle Dirac equation (8.65). 8.12 Insert one of the four solutions u(\u00b1) current (8.58) and interpret the result. R,L (p) from (8.67) into the four-vector probability 512 Relativistic Quantum Mechanics 8.13 Make use of Problem 8.9 to show that UT as de\ufb01ned by (8.104) is just \u03c32 \u2297 I, up to a phase factor. 8.14 Write down the positive helicity, positive energy free-particle Dirac spinor wave function \u03a8(x, t). a. Construct the spinors P\u03a8, C \u03a8, T \u03a8. b. Construct the spinor C PT \u03a8 and interpret it using the discussion of negative energy solutions to the Dirac equation. 8.15 Show that (8.145) imply that u(x) and v(x) grow like exponentials if the series (8.139) and (8.140) do not terminate. 8.16 Expand the energy eigenvalues given by (8.150) in powers of Z\u03b1 and show that the result is equivalent to including the relativistic correction to kinetic energy (5.104) and the spin-orbit interaction (5.125) to the nonrelativistic energy eigenvalues for the one-electron atom (8.151). 8.17 The National Institute of Standards and Technology (NIST) maintains a website with up-to-date high precision data on the atomic energy levels of hydrogen and deuterium: http://physics.nist.gov/PhysRefData/HDEL/data.html Following is a table of data obtained from that website. These are the energies of transitions between the (n, l, j) = (1, 0, 1/2) energy level and the energy level indicated by the columns on the leftE(n, l, j) \u2212 E(1, 0, 1/2)]/hc (cm\u22121) 1/2 1/2 3/2 1/2 1/2 3/2 3/2 5/2 1/2 1/2 3/2 3/2 5/2 5/2 7/2 82 258.954 399 2832(15) 82 258.919 113 406(80) 82 259.285 001 249(80) 97 492.221 724 658(46) 97 492.211 221 463(24) 97 492.319 632 775(24) 97 492.319 454 928(23) 97 492.355 591 167(23) 102 823.853 020 867(68) 102 823.848 581 881(58) 102 823.894 317 849(58) 102 823.894 241 542(58) 102 823.909 486 535(58) 102 823.909 459 541(58) 102 823.917 081 991(58) 513 Problems (The number in parentheses is the numerical value of the standard uncertainty referred to the last \ufb01gures of the quoted value.) Compare these values to those predicted by (8.150) (you may want to make use of Problem 8.16, in particular the following). a. Compare \ufb01ne-structure splitting between the n = 2, j = 1/2 and n = 2, j = 3/2 states to (8.150). b. Compare \ufb01ne-structure splitting between the n = 4, j = 5/2 and n = 4, j = 7/2 states to (8.150). c. Compare the 1S \u2192 2S transition energy to the \ufb01rst line in the table. Use as many signi\ufb01cant \ufb01gures as necessary in the values of the fundamental constants, to compare the results within standard uncertainty. d. How many examples of the Lamb shift are demonstrated in this table? Identify one example near the top and the other near the bottom of the table, and compare their values. A Appendix A Electromagnetic Units Two divergent systems of units established themselves over the course of the twentieth century. One system, known as SI (from the French Le Syst`eme International d\u2019Unit\u00b4es), is rooted in the laboratory. It gained favor in the engineering community and forms the basis for most undergraduate curricula. The other system, called Gaussian, is aesthetically cleaner and is much favored in the theoretical physics community. We use", " the Gaussian system in this book, as do most graduate level physics texts on quantum mechanics and other subjects. The SI system is also known as MKSA (for meter, kilogram, second, Ampere), and the Gaussian system is sometimes called1 CGS (for centimeter, gram, second). For problems in mechanics, the difference is trivial, amounting only to some powers of ten. The dif\ufb01culty comes when incorporating electromagnetism. As discussed below, SI incorporates a fourth base unit, the Ampere, which implies that charge and other electromagnetic quantities are dimensionally distinct between the SI and Gaussian systems. This one point is the source of all the confusion. In other words, although we can write that a meter is equal to 100 centimeters, the SI unit of charge, called the Coulomb, is not equal to any number of electrostatic units (esu), the Gaussian unit of charge. The two kinds of charges literally have different physical meanings. They should probably have different names, but historically they are both called \u201ccharge.\u201d Similar comments apply to electric current, electric and magnetic \ufb01elds, the electric potential, and so on. This is why you see factors like \u03b50 and \u03bc0 appear in SI formulas, whereas factors of c are common in Gaussian formulas. This appendix gives an explanation for how these two systems of units diverge when it comes to electromagnetism. It also shows how physical quantities such as force and energy can be used to relate electromagnetic quantities between the two systems. For a more detailed discussion, see \u201cOn electric and magnetic units and dimensions,\u201d by Birge, Am. Phys. Teach., 2 (1934) 41. (This journal is now called the American Journal of Physics.) See also the articles in the American Journal of Physics, volume 3, 2005, pp. 90, 102, and 171. Electricity, Magnetism, and Electromagnetism: A Review Electricity and magnetism are not separate phenomena. They are different manifestations of the same phenomenon, called electromagnetism. You need to incorporate special 1 Some authors distinguish between Gaussian and CGS by including a factor of 4\u03c0 in Gauss\u2019s law. 514 515 Electromagnetic Units relativity to see how electricity and magnetism are united, and there were some decades between Maxwell and Einstein. Consequently, it was quite some time after they were separately established, that electricity and magnetism were realized to be just different ways that electromagnetism can exert a force. The starting place for an \u201celectric\u201d force is Coulomb\u2019s law. If some number of electrons is added to, or removed from, an object, then it acquires a \u201ccharge\u201d q. A force appears between two charged objects separated by some distance. This force is proportional to the product of the charges, and inversely proportional to the square of the distance between them, that is, F = kE q1q2 d2. (A.1) Here kE is an arbitrary constant of proportionality; without describing what we mean by \u201ccharge,\u201d we can say no more about it. A \u201cmagnetic\u201d force appears between two wires, each of which carries something called a \u201ccurrent.\u201d For two long, parallel wires, the force per unit length is proportional to the product of the currents and inversely proportional to the perpendicular separation of the wires, that is, F L = kM I1I2 d. (A.2) As with the electric force, kM is a generic constant of proportionality which depends on what we mean by \u201ccurrent.\u201d Today we understand that (A.1) and (A.2) are two different manifestations of \u201celectromagnetism.\u201d A \u201ccurrent\u201d is in fact a \ufb02ow of \u201ccharge,\u201d and the theory of electromagnetism tells us that 2kE = c2kM. (A.3) In other words, if we make a choice for kE, then (A.3) speci\ufb01es kM and vice versa. The essential point is that the SI and Gaussian systems make different choices for kE or kM. Other choices will lead to other systems of units, but we will not be discussing them here. There are also variations on whether or not to combine factors of 4\u03c0 into various quantities, but we are not going to distinguish between them in this appendix. The SI System: Inventing a Unit for Current The SI system is based on (A.2). People learned how to make current, well before we understood it in terms of charge. Perhaps for these reasons, a new base unit, the Ampere (A), was created. One Ampere is the amount of current \ufb02owing in each of two long", ", parallel wires, separated by one meter, such that the force between the wires is 2 \u00d7 10\u22127 N/m. We write kM = \u03bc0 2\u03c0 SI (A.4) 516 Appendix A where \u03bc0 \u2261 4\u03c0 \u00d7 10\u22127 N A2. (The factor of 4\u03c0 turns out to be handy to cancel out integrations over the unit sphere.) (A.5) This quantity \u03bc0 turns out to describe the magnetic properties of the vacuum. It shows up, for example, in the inductance of a loop of wire surrounding empty space. This is all forced upon us by the invention of the Ampere. Some books refer to \u03bc0 as the \u201cpermeability of free space.\u201d Equations (A.3) and (A.4) tell us how to write Coulomb\u2019s law (A.1) in the SI system. We have kE = c2 2 kM = \u03bc0c2 4\u03c0 SI = 8.99 \u00d7 109 N \u00b7 m2 (A \u00b7 s)2. (A.6) (A.7) This form of Coulomb\u2019s law shows that charge, in the SI system, has units Amperes\u00d7seconds (A \u00b7 s). This is de\ufb01ned to be the Coulomb (C). SI furthermore de\ufb01nes the quantity \u03b50 \u2261 1 \u03bc0c2 (A.8) called the \u201cpermittivity of free space.\u201d It is another property of the vacuum, showing up, for example, as the capacitance of parallel plates separated by empty space. It is, as with \u03bc0, forced upon us by choosing a new base unit for current. Thus, combining (A.1), (A.6), and (A.8), we write Coulomb\u2019s law in SI as which is the form presented in introductory physics textbooks that use the SI convention. F = 1 4\u03c0\u03b50 q1q2 d2 SI (A.9) The Gaussian System: No New Base Units In the Gaussian system, we take the point of view that no new base units are necessary. We write kE = 1 Gaussian that is, a dimensionless number. In other words, Coloumb\u2019s law (A.1) is simply q1q2 d2 Gaussian. F = (A.10) (A.11) The unit of charge in the Gaussian system is derived in terms of centimeters, grams, and seconds. It is called the electrostatic unit (esu), or sometimes the statcoulomb, and is simply2 esu \u2261 dyne \u00b7 cm2 = g1/2 \u00b7 cm3/2/s. (A.12) 2 Recall that the unit of force in CGS is called the dyne \u2261 g \u00b7 cm/s2 = 10\u22125 N. 517 Electromagnetic Units In this case, the magnetic force between wires is just (A.2) with (A.3), namely F L = 2 c2 I1I2 d CGS. (A.13) The Gaussian system is not without its sources of confusion. Some authors use (A.13) to de\ufb01ne a unit of current, the statampere, which gives 2 dyne/cm of force between two long parallel wires separated by 1 cm. Note that this is not the same as one esu/s, something called the \u201cabsolute ampere\u201d or \u201cabampere.\u201d The statampere and abampere differ by a factor of 2.998 \u00d7 1010, although they have the same dimensions, namely those of the esu/s = g1/2 \u00b7 cm3/2/s2. Converting between SI and CGS It should now be clear to you that the units of charge and current have different dimensions between SI and Gaussian, and this is why everyone encounters confusion when converting between one system and the other. Of course, all of this boils down to experiment. You make a measurement, and use some equations (whether they are Gaussian or SI) to interpret the result. We will take the point of view of Coulomb\u2019s law as a starting point, and the classic work by Millikan3 to measure the charge on a single electron, a (negative) quantity that we traditionally call \u2212e. The modern best value for his measurement is e = 4.8032042 \u00d7 10\u221210 esu. So let us start by giving ourselves the problem of expressing the charge on an electron in Coulombs. This is easy. Let eesu equal the dimensionless number 4.8032042 \u00d7 10\u221210. The force between two electrons separated by one meter is 10\u22124e2 esu N. So, in SI esu dyne = 10\u22129e2 10", "\u22129e2 esu N = 1 4\u03c0\u03b50 e2 (1 m)2 = \u03bc0 4\u03c0 c2e2 (1 m)2 = 10\u22127c2 SIe2 C N (A.14) where eC is the electron charge in Coulombs, and cSI \u2261 2.998 \u00d7 108, is yet another dimensionless number. Then eC = eesu/10cSI = 1.602 \u00d7 10\u221219. This procedure is obviously valid regardless of the charge on an electron. We therefore write qC = qesu/10cSI = qesu/2.998 \u00d7 109 (A.15) as a general conversion between charge in the CGS system to that in SI. That is, one Coulomb represents a much larger amount of charge (i.e. very many more electrons) than one esu, by a factor of 10cSI. The trick here was to recognize that the numerical difference between Coulombs and esu is absorbed by the factor c2 in (A.14). This is not to say that Coulombs and esu differ by the dimensions of velocity. One cannot equate Coulombs to esu without some conversion factor that explicitly cancels out the base unit Amperes. 3 This experiment was a tour de force, which Millikan carried out systematically and carefully over two decades. For a culmination of this work, see his paper \u201cThe most probable 1930 values of the electron and related constants,\u201d Phys. Rev., 35 (1930) 1231. He determined the value e = (4.770 \u00b1 0.005) \u00d7 10\u221210 esu. 518 Appendix A We can extend from here. Consider the units of electric potential, de\ufb01ned by 1 Joule = 1 Volt \u00b7 C SI 1 erg = 1 statvolt \u00b7 esu Gaussian therefore 1 Volt \u00b7 C = 107 statvolt \u00b7 esu (A.16) since one Joule equals 107 ergs. (I like this way of writing things because it means I can use the \u201c=\u201d sign. Energy is energy, whether Gaussian or SI.) Now thinking again in terms of number of electrons, we know that one Coulomb corresponds to 10cSI times as much charge as an esu. So we write, now having to abandon a strict equality, or 1 Volt \u00b7 10cSI \u21d0\u21d2 107 statvolt 1 statvolt \u21d0\u21d2 299.8 Volt. (A.17) (A.18) In other words, in practical terms, one statvolt is the same as 300 volts. Perhaps here is a reason that SI is more popular with electricians and engineers. We always prefer to use numbers on the order of unity when doing practical work. One volt is a reasonable potential difference from a human perspective, but 300 volts would give you a rather signi\ufb01cant shock. So, if we worked in Gaussian, practical electronics would be discussed in terms of \u201cmillistatvolts,\u201d a somewhat unwieldy term. \u221a The same thing works symbolically, of course. To go from Gaussian to SI we need to insert the factor \u03bc0c2/4\u03c0 = 1/4\u03c0\u03b50 in front of Coulomb\u2019s law, and rederive things. So, wherever we encounter a value of charge q in a Gaussian equation, we replace it with 4\u03c0\u03b50. (The trivial conversions from centimeters and grams to meters and kilograms q/ \u03bc0/4\u03c0. are irrelevant symbolically.) Similarly, any values of current I are replaced by cI You can easily check that these substitutions turn (A.11) into (A.9), and (A.13) into (A.2) w/(A.4). For a different example, the electric \ufb01eld E = limq0\u21920 F/q0 is multiplied by \u221a 4\u03c0\u03b50 leaving (of course) the force for a charge in an electric \ufb01eld F = qE unchanged. Now let us try going the other way, namely SI to Gaussian, with the Lorentz force law, namely F = qE + qv \u00d7 B SI. (A.19) 4\u03c0\u03b50 = q We know that the \ufb01rst term on the right is unchanged. In the second term, replace q with \u221a 4\u03c0/\u03bc0/c, but what about the B \ufb01eld? A study of electromagnetism leads to q Amp`ere\u2019s law, which relates magnetic \ufb01elds and currents. Indeed, steady currents give rise to static magnetic \ufb01elds according to \u2207 \u00d7 B = 4\u03c0 c j Gaussian \u2207 \u00d7 B = \u03bc0 j SI (A.20a) (A.20b) 519", " Electromagnetic Units where j is the current density. To go from CGS to SI, j is multiplied by c equations (A.20) tell us to multiply B by 4\u03c0/\u03bc0/c for q and the second term by to CGS.) We therefore arrive at the Lorentz force law \u03bc0/4\u03c0, so 4\u03c0/\u03bc0. So, getting back to (A.19) we multiply \u03bc0/4\u03c0 for B. (Remember, we are going from SI F = qE + q v c \u00d7 B Gaussian. (A.21) An important lesson in this last example is that in going from SI to Gaussian, the magnetic \ufb01eld changes its dimensions differently than does the electric \ufb01eld. Indeed, (A.19) shows that in SI, the dimensions of E are the dimensions of B multiplied by the dimensions of velocity. On the other hand, (A.21) shows that in the Gaussian system, E and B have the same dimensions. Pedagogically, this is an advantage that the Gaussian system has over SI. For a \ufb01nal example, let us express the magnetic moment in Gaussian units, which is the starting point for this book. In your introductory physics class, which most likely used SI units, you de\ufb01ned the magnetic moment \u03bc = IA for a current I moving in a closed loop that enclosed an area A. You also learned that the potential energy for such a current loop in a magnetic \ufb01eld B was \u2212\u03bc \u00b7 B, which of course must be the same expression as in the 4\u03c0/\u03bc0B. Therefore, in Gaussian system. We already know that I \u2192 c order to keep the same expression for potential energy, the de\ufb01nition of magnetic moment in Gaussian units must be \u03bc = IA /c. \u03bc0/4\u03c0I and B \u2192 B Appendix B Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation This appendix summarizes simple solutions of Schr\u00a8odinger\u2019s wave equation corresponding to a variety of soluble potential energy functions. Most of these are derived and discussed in the body of the textbook, but all are included in nearly any undergraduate textbook on quantum mechanics. For some solutions, MATHEMATICA code is given for calculating the wave functions. B.1 Free Particles (V = 0) The plane wave, or momentum, eigenfunction, from Section 1.7, is \u03c8k(x, t) = x|ke\u2212iEt/\u00afh = 1 (2\u03c0)3/2 eik\u00b7x\u2212i\u03c9t, where k = p \u00afh and \u03c9 = E \u00afh = p2 2m\u00afh = \u00afhk2 2m. The normalization, as described in Section 1.6, is k \u03c8k d3x = \u03b4(3)(k \u2212 k). \u03c8\u2217 (B.1) (B.2) (B.3) The superposition of plane waves leads to the wave packet description. In one dimension, \u03c8(x, t) = \u221e \u2212\u221e 1\u221a 2\u03c0 dk A(k) ei(kx\u2212\u03c9t) where \u03c9 = \u00afhk2 2m. For |A(k)| sharply peaked near k k0, the wave packet moves with a group velocity \u03c5g d\u03c9 dk = \u00afhk0 m. k0 (B.4) (B.5) The time evolution of a minimum wave packet can be described by \u03c8(x, t) = (\u0394x)2 0 2\u03c03 1/4 \u221e \u2212\u221e e\u2212(\u0394x)2 0(k\u2212k0)2+ikx\u2212i\u03c9(k)t dk where \u03c9(k) = \u00afhk2 2m, (B.6) 520 521 Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation and * |\u03c8(x, t)|2 =! 2\u03c0(\u0394x)2 0 * + 1/2 \" 1 1 + (\u00afh2t2/4m2)(\u0394x)\u22124 0 (x \u2212 \u00afhk0t/m)2 1 + (\u00afh2t2/4m2)(\u0394x)\u22124 0! 2(\u0394x)2 0 \u00d7 exp \u2212 + \". (B.7) So the width of the wave packet expands as (\u0394x)0 at t = 0 \u2192 (\u0394x)0 1 + 1/2 \u00afh2t2 4m2 (\u0394x)\u22124 0 at t > 0. (B.8) B.2 Piecewise Constant Potentials in One Dimension For a constant potential V = V0 and E > V, the solution", " is \u03c8E(x) =c +eikx + c\u2212e\u2212ikx where k = 2m(E \u2212 V0) \u00afh2. For E < V, that is the classically forbidden region, the solution is \u03c8E(x) =c +e\u03bax + c\u2212e\u2212\u03bax where \u03ba = 2m(V0 \u2212 E) \u00afh2. (B.9) (B.10) Note that c\u00b1 must be set equal to 0 if x = \u00b1 \u221e is included in the domain under discussion. B.2.1 Rigid-Wall Potential (One-Dimensional Box) For a well with in\ufb01nitely high walls at x = 0 and x = L, that is V = for 0 < x < L 0 \u221e otherwise, the wave functions and energy eigenstates are n\u03c0x L, n = 1, 2, 3..., \u03c8E(x) = E = \u2019 sin 2 L \u00afh2n2\u03c02 2mL2. B.2.2 Square-Well Potential For a well with \ufb01nite walls at x = \u00b1a, that is V = 0 \u2212V0 for |x| > a for |x| < a (V0 > 0), (B.11) (B.12) (B.13) 522 Appendix B the bound-state (E < 0) solutions are: \u23a7 \u23aa\u23aa\u23a8 \u23aa\u23aa\u23a9 e\u2212\u03ba|x| cos kx sin kx \u03c8E \u223c - (even parity) (odd parity) for |x| > a for |x| < a, where k = 2m(\u2212|E| + V0) \u00afh2 and \u03ba = 2m|E| \u00afh2. (B.14) (B.15) The allowed discrete values of energy E = \u2212\u00afh2\u03ba2/2m are to be determined by solving ka tan ka = \u03baa ka cot ka = \u2212\u03baa (even parity) (odd parity). Note also that \u03ba and k are related by 2mV0a2 \u00afh2 = (k2 + \u03ba2)a2. (B.16) (B.17) B.3 Transmission Reflection Problems In this discussion we de\ufb01ne the transmission coef\ufb01cient T to be the ratio of the \ufb02ux of the transmitted wave to that of the incident wave. We consider these simple examples. B.3.1 Square Well For a wave incident on a \ufb01nite square well de\ufb01ned by V = 0 for |x| > a and V = \u2212V0 for |x| < a, with V0 > 0, the transmission coef\ufb01cient is T =? 1! (k2 \u2212 k2)2/4k2k2 \" 1 + @ sin2 ka where =?! V2 0/4E(E + V0) \" 1 + 1 sin2 % 2a 2m(E + V0)/\u00afh2 &@ (B.18) \u2019 k = 2mE \u00afh2 and k = 2m(E + V0) \u00afh2. (B.19) Note that resonances occur whenever 2a 2m(E + V0) \u00afh2 = n\u03c0, n = 1, 2, 3,.... (B.20) 523 Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation B.3.2 Square Barrier This potential is the same as for the square well, but with V = V0 > 0 for |x| < a. One \ufb01nds for the case E < V0, T =? 1! (k2 + \u03ba2)2/4k2\u03ba2 \" 1 + @ sinh2 \u03baa =?! 0/4E(V0 \u2212 E) V2 \" 1 + 1 sinh2 % 2a 2m(V0 \u2212 E)/\u00afh2 &@. (B.21) For E > V0, the result is the same as the square-well case with V0 replaced by \u2212V0. B.3.3 Potential Step The potential is V = 0 forx < 0 and V = V0 for x > 0, and we take E > V0. One \ufb01nds for the transmission coef\ufb01cient T = 4kk (k + k)2 = 4 %\u221a (E \u2212 V0)E \u221a E \u2212 V0 & 2 E + with \u2019 k = 2mE \u00afh2 and k = 2m(E \u2212 V0) \u00afh2. B.3.4 General Potential Barrier For", " E < V(x) in the range a \u2264 x \u2264 b and E > V(x) elsewhere, the approximate JWKB1 solution for T is \u23a7 \u23a8 b a dx 2m[V(x) \u2212 E] \u00afh2 \u23ab \u23ac \u23ad, T exp \u22122 \u23a9 where a and b are the classical turning points. B.4 Simple Harmonic Oscillator See Section 2.5.2. The potential energy function is V(x) = m\u03c92x2. 1 2 1 JWKB stand for Jeffreys\u2013Wentzel\u2013Kramers\u2013Brillouin. (B.22) (B.23) (B.24) (B.25) 524 Appendix B We introduce a dimensionless variable \u03be = \u2019 m\u03c9 \u00afh x, and write the energy eigenfunctions as \u03c8E = (2nn! )\u22121/2 1/4 m\u03c9 \u03c0\u00afh e\u2212\u03be2/2Hn(\u03be) (B.26) (B.27) and the energy levels are E = \u00afh\u03c9 n + 1 2, n = 0, 1, 2,.... (B.28) The Hermite polynomials have the following properties: Hn(\u03be) = (\u22121)ne\u03be2 \u2202n \u2202\u03ben e\u2212\u03be2 Hn (\u03be) Hn (\u03be) e\u2212\u03be2 d\u03be = \u03c01/22n!\u03b4nn \u221e \u2212\u221e d2 d\u03be2 Hn \u2212 2\u03be dHn d\u03be + 2nHn = 0 H0(\u03be) =1, H1(\u03be) = 2\u03be, (B.29) H2(\u03be) = 4\u03be2 \u2212 2, H3(\u03be) = 8\u03be3 \u2212 12\u03be, H4(\u03be) = 16\u03be4 \u2212 48\u03be2 + 12. The Hermite polynomials are calculated in MATHEMATICA with HermiteH[n, \u03be]. The following code produces the \ufb01rst few wave functions as functions of x: \\[Psi]E = (2\u02c6{n} n!)\u02c6(-1/2) (m \\[Omega]/(Pi \\[HBar]))\u02c6(1/ 4) Exp[-\\[Xi]\u02c6{2}/2] HermiteH[n, \\[Xi]]; \\[Psi]0 = \\[Psi]E /. n -> 0 /. \\[Xi] -> x Sqrt[m \\[Omega]/\\[HBar]]; \\[Psi]1 = \\[Psi]E /. n -> 1 /. \\[Xi] -> x Sqrt[m \\[Omega]/\\[HBar]]; \\[Psi]2 = \\[Psi]E /. n -> 2 /. \\[Xi] -> x Sqrt[m \\[Omega]/\\[HBar]]; \\[Psi]3 = \\[Psi]E /. n -> 3 /. \\[Xi] -> x Sqrt[m \\[Omega]/\\[HBar]]; If you code these up into MATHEMATICA, or some other symbolic mathematics application, it is worth your while to check that your wave functions are orthonormal. B.5 The Central Force Problem See Section 3.7. The basic time-independent Schr\u00a8odinger equation is \u2212 \u00afh2 2m 1 r2 sin \u03b8 1 r2 sin2 \u03b8 \u2202\u03c8E \u2202r \u2202\u03c8E \u2202\u03b8 \u22022\u03c8E \u2202\u03c62 \u2202 \u2202\u03b8 \u2202 \u2202r sin \u03b8 1 r2 r2 + + + V(r)\u03c8E = E\u03c8E (B.30) 525 Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation where our spherically symmetrical potential V(r) satis\ufb01es r2V(r) \u2192 0. lim r\u21920 The method of separation of variables, \u03a8E(x) =R (r)Ym l (\u03b8, \u03c6), leads to the angular equation \u2212 1 sin \u03b8 \u2202 \u2202\u03b8 sin \u03b8 \u2202 \u2202\u03b8 + 1 sin2 \u03b8 \u22022 \u2202\u03c62 where the spherical harmonics (B.31) (B.32) l = l(l + 1)Ym Ym l, (B.33) l (\u03b8, \u03c6", "), Ym l = 0, 1, 2,..., m = \u2212l, \u2212l + 1,..., +l (B.34) satisfy \u2212i \u2202 \u2202\u03c6 Ym l = mYm l (B.35) and the Ym l (\u03b8, \u03c6) have the following properties: l (\u03b8, \u03c6) = (\u22121)m Ym 2l + 1 4\u03c0 (l \u2212 m)! (l + m)! l (cos \u03b8) eim\u03c6 Pm for m \u2265 0, for m \u2265 0, (B.36) l (\u03b8, \u03c6) = (\u22121)|m|Y Ym |m|\u2217 l (\u03b8, \u03c6) l (cos \u03b8) = (1 \u2212 cos2 \u03b8)m/2 Pm for m < 0, d(cos \u03b8)m Pl(cos \u03b8) dm Pl(cos \u03b8) = Y0 0 = 3 4\u03c0 cos \u03b8,, 3 1 =, Y0 dl(1 \u2212 cos2 \u03b8)l d(cos \u03b8)l \u2019 (\u22121)l 2ll! 1\u221a 4\u03c0 \u2019 8\u03c0 (sin \u03b8)e\u00b1i\u03c6, 5 16\u03c0 (3 cos2 \u03b8\u2212 1), \u2019 15 8\u03c0 (sin \u03b8cos \u03b8)e\u00b1i\u03c6, 32\u03c0 (sin2 \u03b8)e\u00b12i\u03c6 15 Y\u00b11 1 = \u2213 \u2019 Y0 2 = Y\u00b11 2 = \u2213 \u2019 Y\u00b12 2 = Ym\u2217 l (\u03b8, \u03c6)Ym l (\u03b8, \u03c6)d\u03a9 = \u03b4ll \u03b4mm d\u03a9 = For the radial piece of (B.32), we de\ufb01ne 2\u03c0 0 d\u03c6 +1 \u22121 d(cos \u03b8). uE(r) = rR(r) (B.37) 526 Appendix B and then the radial equation is reduced to an equivalent one-dimensional problem, namely, \u2212 \u00afh2 2m d2uE dr2 + subject to the boundary condition V(r) + l(l + 1)\u00afh2 2mr2 uE = EuE uE(r)|r=0 = 0. (B.38) For the case of free particles, that is V(r) =0, in our spherical coordinates: R(r) =c 1 jl(\u03c1) +c 2nl(\u03c1) (c2 = 0 if the origin is included) (B.39) where \u03c1 is a dimensionless variable \u03c1 \u2261 kr, with k = \u2019 2mE \u00afh2. (B.40) We need to list the commonly used properties of the Bessel functions and spherical Bessel and Hankel functions. The spherical Bessel functions are jl(\u03c1) = 1/2 \u03c0 2\u03c1 nl(\u03c1) = (\u22121)l+1 Jl+1/2(\u03c1), 1/2 \u03c0 2\u03c1 J\u2212l\u22121/2(\u03c1), n0(\u03c1) =\u2212 cos \u03c1 \u03c1, sin \u03c1 \u03c1, sin \u03c1 \u03c12 j0(\u03c1) = j1(\u03c1) = j2(\u03c1) = n2(\u03c1) = \u2212 \u2212 cos \u03c1 \u03c1, sin \u03c1 \u2212 3 3 \u03c13 \u2212 1 \u03c1 \u2212 1 \u03c1 3 \u03c13 \u03c12 cos \u03c1, cos \u03c1 \u2212 3 \u03c12 sin \u03c1. n1(\u03c1) =\u2212 cos \u03c1 \u03c12 \u2212 sin \u03c1 \u03c1, (B.41) For \u03c1 \u2192 0, the leading terms are where jl(\u03c1) \u2212\u2192 \u03c1\u21920 \u03c1l (2l + 1)!!, nl(\u03c1) \u2212\u2192 \u03c1\u21920 \u2212 (2l \u2212 1)!! \u03c1l+1, (2l + 1)!!\u2261 (2l + 1)(2l \u2212 1) \u00b7 \u00b7 \u00b7 5 \u00b7 3 \u00b7 1. In the large \u03c1-asymptotic limit, we have jl(\u03c1) \u2212\u2192 \u03c1\u2192\u221e 1 \u03c1 cos nl(\u03c1) \u2212\u2192 \u03c1\u2192\u221e 1 \u03c1 sin \u03c1 \u2212 (l + 1)\u03c0 2 \u03c1 \u2212 (l + 1)\u03c0 2,. (B.42) (B.43) (B.44) Because of constraints (B.37) and (B.38), R(r) must be \ufb01nite at r = 0; hence, from (B.39) and (B.42) we see that the nl(\u03c1)-term must be", " deleted because of its singular behavior as 527 Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation \u03c1 \u2192 0. Thus R(r) =c ljl(\u03c1) [or, in the notation of Section 6.4, Al(r) =R(r) =c ljl(\u03c1)]. For a three-dimensional square-well potential, V = \u2212V0 for r < R (with V0 > 0), the desired solution is R(r) =A l(r) =constant jl(\u03b1r), (B.45) where \u03b1 = 2m(V0 \u2212 |E|) \u00afh2 1/2 for r < R. (B.46) As discussed in (6.135), the exterior solution for r > R, whereV = 0, can be written as a linear combination of spherical Hankel functions. These are de\ufb01ned as follows: h(1) l (\u03c1) =j l(\u03c1) +in l(\u03c1) (\u03c1) =h (2) l (\u03c1) =j l(\u03c1) \u2212 inl(\u03c1) (B.48) (B.47) h(1)\u2217 l which, from (B.44), have the asymptotic forms for \u03c1 \u2192 \u221e as follows: h(1) l (\u03c1)\u2212\u2192 \u03c1\u2192\u221e h(1)\u2217 l (\u03c1) =h (2) l (\u03c1)\u2212\u2192 \u03c1\u2192\u221e 1 \u03c1 ei[\u03c1\u2212(l+1)\u03c0/2] 1 \u03c1 e\u2212i[\u03c1\u2212(l+1)\u03c0/2]. (B.49) (B.50) If we are interested in the bound-state energy levels of the three-dimensional square-well potential, where V(r) =0, r > R, we have ul(r) = rAl(r) =constant e\u2212\u03barf \u03ba = 1/2. 2m|E| \u00afh2 1 \u03bar (B.51) To the extent that the asymptotic expansions, of which (B.50) give the leading terms, do not contain terms with exponent of opposite sign to that given, we have, for r > R, the desired solution from (B.51): Al(r) =constant h(1) l (i\u03bar) =constant [jl (i\u03bar) +in l (i\u03bar)], (B.52) where the \ufb01rst three of these functions are 0 (i\u03bar) =\u2212 1 h(1) \u03bar e\u2212\u03bar h(1) 1 (i\u03bar) =i h(1) 2 (i\u03bar) = e\u2212\u03bar + 1 \u03ba2r2 1 \u03bar 1 \u03bar + 3 \u03ba2r2 + 3 \u03ba3r3 (B.53) e\u2212\u03bar. Finally, we note that in considering the shift from free particles, i.e. V(r) = 0 to the case of the constant potential V(r) =V 0, we need only to replace the E in the free-particle solution (B.39) and (B.40) by E \u2212 V0. Note, though, that if E < V0, then h(1,2) (i\u03bar) is to be used with \u03ba = 2m(V0 \u2212 E)/\u00afh2. l 528 Appendix B B.6 Hydrogen Atom See Section 3.7.4. The potential energy function for this problem is given by V(r) =\u2212 Ze2 r and for bound states (E < 0) we introduce the dimensionless variable 1/2 \u03c1 = \u2212 2meE \u00afh2 r = Zr na0 (B.54) (B.55) for electron mass me and Bohr radius a0 = \u00afh2/mee2. The energy eigenfunctions and eigenvalues (energy levels) are \u03c8nlm = Rnl(r)Ym l (\u03b8, \u03c6) R(r) = 1 (2l + 1)! l e\u2212Zr/na0 3 2Z na0 (n + l)! 2n(n \u2212 l \u2212 1)! 1/2 F (l + 1 \u2212 n; 2l + 2; 2\u03c1) 2Zr na0 En = \u2212 Z2e2 2a0 1 n2 (B.56) where n = 1, 2, 3,... with l = 0, 1,..., n \u2212 1, and F(a; c; x) is the con\ufb02uent hypergeometric function, which solves Kummer\u2019s equation and is normalized as F(", "a; c; 0) =1. One can instead write the radial wave function using associated Laguerre polynomials, but conventions and normalizations vary. Using the de\ufb01nition developed in Problem 3.30, l 2Zr na0 e\u2212Zr/na0 2Z na0 3 (n \u2212 l \u2212 1)! 2n[(n + l)! ]3 1/2 R(r) = n\u2212l\u22121(2\u03c1). L2l+1 (B.57) The radial functions for low n are: R10(r) = R20(r) = R21(r) = 3/2 Z a0 2e\u2212Zr/a0 3/2 (2 \u2212 Zr/a0)e\u2212Zr/2a0 3/2 Zr\u221a 3 a0 e\u2212Zr/2a0. Z 2a0 Z 2a0 (B.58) 529 Elementary Solutions to Schr\u00f6dinger\u2019s Wave Equation Some convenient radial integrals are \u221e [3n2 \u2212 l(l + 1)] [5n2 + 1 \u2212 3l(l + 1)] (B.59) dr r2+k[Rnl(r)]2, 0 a0 2Z 0n2 a2 2Z2 Z n2a0, rk \u2261 r = r2 = ) ( 1 r 1 r2 ) = = ( Z2 % l + 1 2 &. n3a2 0 The following MATHEMATICA code is useful for calculating these wave functions: \\[Psi]nlm[r_, \\[Theta]_, \\[Phi]_] = Nnl Rnl[r] Ylm[\\[Theta], \\[Phi]]; Ylm[\\[Theta]_, \\[Phi]_] = SphericalHarmonicY[l, m, \\[Theta], \\[Phi]]; Rnl[r_] = (2 Z r/(n a0))\u02c6{l} Exp[-Z r/(n a0)] Hypergeometric1F1[-n + l + 1, 2 l + 2, 2 Zr/(n a0)]; Nnl = 1/(2 l + 1)! Sqrt[(2 Z/(n a0))\u02c6{3} (n + l)!/(2 n (n - l - 1)!)]; Given these de\ufb01nitions, then, for example, the 1s and 2p, m = +1 wave functions are \\[Psi]100 = \\[Psi]nlm[r,\\[Theta],\\[Phi]] /. {n -> 1, l -> 0, m -> 0} \\[Psi]211 = \\[Psi]nlm[r,\\[Theta],\\[Phi]] /. {n -> 2, l -> 1, m -> 1} As always, it is wise to check that your code produces proper orthonormal wave functions. C Appendix C Hamiltonian for a Charge in an Electromagnetic Field We will often need to know the Hamiltonian for a particle with mass m and charge q moving while in the presence of a static electric \ufb01eld E(x) and static magnetic \ufb01eld B(x). We will take a very utilitarian approach, namely to write down the equation of motion and then \ufb01nd the Lagrangian which, in classical physics, yields this equation. We then construct the Hamiltonian from the Lagrangian. We know that the equation of motion is given by the Lorentz force law, namely m\u00a8x = qE + q\u02d9x \u00d7 B. 1 c (C.1) We can derive the Hamiltonian from the Lagrangian, but it is not obvious how to build the Lagrangian when there is no obvious \u201cpotential energy\u201d function for a charged particle in a magnetic \ufb01eld. Our approach1 will be to start with the conjecture that the correct Lagrangian is given by L (x, \u02d9x) = 1 2 m\u02d9x2 \u2212 q\u03c6(x) + \u02d9x \u00b7 A(x) q c where \u03c6 and A are the standard electrostatic and magnetic vector potentials, that is E(x) =\u2212 \u2207\u03c6(x) and B(x) = \u2207 \u00d7 A(x), and then show that Lagrange\u2019s equations lead us to (C.1). We have 0 = = \u2212 \u2202L \u2202\u02d9xi # m\u02d9xi + d dt d dt \u2202L \u2202xi q c Ai(x) $ = m(\u00a8x)i +", " q(\u2207\u03c6)i + \u2202\u03c6 \u2202xi d dt + q q c \u2212 q c \u02d9x \u00b7 Ai(x) \u2212 \u02d9x \u00b7 \u2202A \u2202xi \u2202A \u2202xi. (C.2) (C.3) (C.4) The \ufb01rst term on the right of (C.4) is just the left-hand side of (C.1), and the second term on the right of (C.4) is just the \ufb01rst term on the right-hand side of (C.1). It therefore remains to evaluate \u02d9x \u00b7 \u2202A \u2202xi \u2212 d dt Ai(x) = \u2211 \u02d9xj j \u2202Aj \u2202xi \u2212\u2211 j \u2202Ai \u2202xj \u02d9xj = \u2211 \u02d9xj j \u2202Aj \u2202xi \u2212 \u2202Ai \u2202xj (C.5) 1 One can derive this Lagrangian from \ufb01rst principles of relativity and electromagnetism, but we will leave that approach to an advanced course on classical or quantum \ufb01eld theory. 530 531 Hamiltonian for a Charge in an Electromagnetic Field and show that it equals (\u02d9x \u00d7 B)i = (\u02d9x \u00d7 (\u2207 \u00d7 A))i. We can use the totally antisymmetric \u03b5ijkakbk. Therefore symbol \u03b5ijk to write the cross product of two vectors as (a \u00d7 b)i = \u2211j \u2211k (\u02d9x \u00d7 B)i = \u2211 \u2211 \u03b5ijk \u02d9xj \u2211 \u2211 \u03b5klm j k l m \u2202Al \u2202xm = \u2211 \u02d9xj \u2211 \u2211 \u2211 \u03b5kij\u03b5klm j l m k \u2202Al \u2202xm (C.6) where \u03b5ijk = \u03b5kij because the indices are rearranged by an even number of exchanges. We then make use of the theorem \u2211k \u03b5kij\u03b5klm = \u03b4il\u03b4jm \u2212 \u03b4im\u03b4jl to write (\u02d9x \u00d7 B)i = \u2211 \u02d9xj \u2211 \u2211 (\u03b4il\u03b4jm \u2212 \u03b4im\u03b4jl) j l m \u2202Al \u2202xm = \u2211 \u02d9xj j \u2202Ai \u2202xj \u2212 \u2202Aj \u2202xi (C.7) which is the same as (C.5). This proves our conjecture that (C.2) is the correct Lagrangian. We can now use de\ufb01nitions from classical mechanics to derive the Hamiltonian H (p, x) from (C.2). First we determine the canonical momentum p from pi \u2261 \u2202L \u2202\u02d9xi = m\u02d9xi + q c Ai(x). Then we construct the Hamiltonian using the Legendre transformation H = \u2211 \u02d9xipi \u2212 2m = = m \u00b7 p \u2212 1 2 2 + q\u03c6. 1 m2 p \u2212 q c A 2 + q\u03c6 \u2212 It is important to note that the canonical momentum p is not equal to m\u02d9x in this case. (C.8) (C.9) D Appendix D Proof of the Angular-Momentum Rule (3.358) It will be instructive to discuss the angular-momentum addition rule from the quantummechanical point of view. Let us, for the moment, label our angular momenta, so that j1> = j2. This we can always do. From (3.355), the maximum value of m, mmax, is mmax = mmax 1 + mmax There is only one ket that corresponds to the eigenvalue mmax, whether the description is in terms of | j1 j2; m1m2 or | j1 j2; jm. In other words, choosing the phase factor to be 1, we have 2 = j1 + j2. (D.1) | j1 j2; j1 j2 = | j1 j2; j1 + j2, j1 + j2. In the | j1 j2; m1m2 basis, there are two kets that correspond to the m eigenvalue mmax \u2212 1, \u2212 1 and m2 = mmax namely, one ket with m1 = mmax and 2 \u2212 1. There is thus a twofold degeneracy in this basis; therefore, there must be a m2 = mmax twofold degeneracy in the | j1 j2; jm basis as well. From where could this come? Clearly", ", mmax \u2212 1 is a possible m-value for j = j1 + j2. It is also a possible m-value for j = j1 + j2 \u2212 1, in fact, the maximum m-value for this j. So j1, j2 can add to j of j1 + j2 and j1 + j2 \u2212 1. and one ket with m1 = mmax (D.2) 1 2 1 We can continue in this way, but it is clear that the degeneracy cannot increase inde\ufb01nitely. Indeed, for mmin = \u2212j1 \u2212 j2, there is once again a single ket. The maximum degeneracy is (2j2 + 1)-fold, as is apparent from Table D.1 constructed for two special examples: for j1 = 2, j2 = 1 and for j1 = 2, j2 = 1 2. This (2j2 + 1)-fold degeneracy must be associated with the 2j2 + 1 states j: j1 + j2, j1 + j2 \u2212 1,..., j1 \u2212 j2. (D.3) If we lift the restriction j1 \u2265 j2, we obtain (3.358). 532 533 Proof of the Angular-Momentum Rule (3.358) Table D.1 Special Examples of Values of m, m1, andm 2 for the Two Casesj 1 = 2,j 2 = 1 and j1 = 2, j2 = 1 2, Respectively j1 = 2, j2 = 1 m (m1, m2) Numbers of States j1 = 2, j2 = 1 2 m 3 (2,1) 1 5 2 2 (1,1) (2,0) 1 0 (0,1) (1,0) (2,\u22121) (\u22121,1) (0,0) (1,\u221211 (\u22122,1) (\u22121,0) (0,\u22121) 3 \u2212 3 2 \u22122 \u22123 (\u22122,0) (\u22121, \u22121) (\u22122, \u22121) 1 2 \u2212 5 2 (m1, m2) (2, 1 2 ) (1, 1 2 ) (2,\u2212 1 2 ) (0, 1 2 ) (1,\u2212 1 2 ) (\u22121, 1 2 ) (0,\u2212 1 2 ) (\u22122, 1 2 ) (\u22121,\u2212 1 2 ) (\u22122,\u2212 1 2 ) Numbers of States 1 2 2 2 2 1 E Appendix E Finding Clebsch\u2013Gordan Coefficients Section 3.8 includes a discussion of how one goes about calculating Clebsch\u2013Gordan coef\ufb01cients. However, in practice, we generally use tables to look them up, or computer applications that carry out the calculation for us. Of the many tables available, I am partial to the one available online from the Particle Data Group (PDG) (http://pdg.lbl.gov). A concise table for all of the Clebsch\u2013Gordan coef\ufb01cients is posted at http://pdg.lbl.gov/2018/reviews/rpp2018-rev-clebsch-gordan-coefs.pdf This page also lists the \ufb01rst several spherical harmonics and d-functions. Programming languages such as PYTHON include libraries for calculating Clebsch\u2013 Gordan coef\ufb01cients, as do symbolic manipulation applications such as MATHEMATICA. For example, in MATHEMATICA, the syntax is ClebschGordan[{j1, m1}, {j2, m2}, {j, m}] which returns the coef\ufb01cient for adding j1 and j2 to get j, with m1 +m2 = m. If an unphysical combination of quantum numbers is given, then MATHEMATICA returns the value zero, along with a warning that the request was not physical. Purely symbolic function calls are of course possible, and return the conditions under which a nonzero answer is returned. As a trivial example that compares the result from the PDG table with the output from MATHEMATICA, executing the cell ClebschGordan[{2, 1}, {3/2, -1/2}, {7/2, 1/2}] == Sqrt[12/35] returns the value True. 534 F Appendix F Notes on Complex Variables These notes are meant to accompany a graduate level physics course, to provide a basic introduction to the necessary concepts in complex analysis. They are not complete, nor are any of the proofs considered rigorous. The immediate goal is to carry through enough of the work needed to explain the Cauchy residue theorem. F.1 Complex Numbers and Complex Functions A complex number z can be", " written as z = x + iy or z = rei\u03c6 with r \u2265 0 \u221a where i = leading to a description in terms of the \u201ccomplex plane.\u201d The complex conjugate of z is \u22121, and x, y, r, and \u03c6 are real numbers. Clearly, x = r cos \u03c6 and y = r sin \u03c6, The \u201cmodulus\u201d of z is |z| \u2261 z\u2217 = x \u2212 iy \u221a z\u2217z = r = or z\u2217 = re\u2212i\u03c6. x2 + y2, and \u03c6 is often called the \u201cphase\u201d of z. A complex function f (z) typically returns a complex number. Generically, we write f (z) =u(x, y) +iv (x, y) (F.1) for purposes of proofs or illustrations. The behavior of the (real) functions u(x, y) and v(x, y) are critical for classifying complex functions, as seen when we consider taking derivatives. F.2 Differentiation and Analyticity We de\ufb01ne the derivative f (z) = df/dz of a complex function f (z) in the same way as we do for the derivatives of real functions. That is, for z0 \u2261 x0 + iy0, f (z) \u2212 f (z0) z \u2212 z0 f (z0) = = lim z\u2192z0 df dz. z=z0 535 536 Appendix F However, there is clearly an ambiguity, depending on whether we approach z0 along the line y = y0 or along x = x0. (Of course, we could also say the ambiguity is along any line of constant \u03c6 = \u03c60, but it is suf\ufb01cient to consider just two orthogonal directions.) That is, f (z0) = lim x\u2192x0 u(x, y0) \u2212 u(x0, y0) x \u2212 x0 + i lim x\u2192x0 v(x, y0) \u2212 v(x0, y0) x \u2212 x0 = \u2202u \u2202x + i \u2202v \u2202x or f (z0) = lim y\u2192y0 u(x0, y) \u2212 u(x0, y0) iy \u2212 iy0 + i lim y\u2192y0 v(x0, y) \u2212 v(x0, y0) iy \u2212 iy0 = \u2212i \u2202u \u2202y + \u2202v \u2202y. Therefore, in order to remove the ambiguity and have a consistent de\ufb01nition of the derivative, \u2202u \u2202x = \u2202v \u2202y and \u2202u \u2202y = \u2212 \u2202v \u2202x. (F.2) These are called the Cauchy\u2013Riemann conditions. A function f (z) which satis\ufb01es these rather restrictive conditions is called analytic. Indeed, analytic functions have very many applications in physics, and we will merely scratch the surface here. For example, the function f (z) = ez = ex(cos y + i sin y) is analytic. This is easy to prove. Putting u(x, y) =e x cos y and v(x, y) =e x sin y, \u2202u \u2202x = ex cos y = \u2202v \u2202y and \u2202u \u2202y = \u2212ex sin y = \u2212 \u2202v \u2202x so the Cauchy\u2013Riemann conditions (F.2) are satis\ufb01ed. It is simple to show that f (z) = az is analytic, where a is a complex constant. It is also not hard to show that the product of two analytic functions is analytic, so any function of the form f (z) = anzn, wheren is a nonnegative integer, is also analytic. Of course, any sum of analytic functions is analytic, so we see that any polynomial in z is analytic in the entire complex plane. These examples beg the question: If a function f (z) can be written explicitly in terms of z, is it analytic? The answer is \u201cYes.\u201d To see this, realize that instead of x and y, we could always write a complex function in terms of z and z\u2217 using x = (z+z\u2217)/2 and y = (z\u2212z\u2217)/2i. Now consider \u2202f \u2202z\u2217 = \u2202f \u2202x \u2202x \u2202z\u2217 + \u2202f \u2202y \u2202y \u2202z\u2217 = + i \u2202v \u2202x \u2202u \u2202x 1 2 \u2212 \u2202v", " \u2202y + i 2 \u2202u \u2202x = 1 2 \u2202u \u2202y + i \u2202v \u2202y + \u2202u \u2202y + \u2202v \u2202x = 0 \u2212 1 2i so long as the Cauchy\u2013Riemann conditions (F.2) are satis\ufb01ed. That is, if the expression for f (z) contains only z (and not z\u2217) then the function is analytic. There is some common terminology. A function f (z) need not be analytic in the entire complex plane. (If it is, we called the function \u201centire.\u201d) If it is analytic at a point z0 then we call that a \u201cregular point.\u201d Otherwise, z0 is called a \u201csingular point.\u201d Much of our discussion of complex integration will focus on the notion of singular points. 537 Notes on Complex Variables F.3 Integration and Series Expansion Similarly to differentiation, we approach integration of complex functions the same way as with real functions, but we need to be aware that there is now an arbitrariness of the \u201cpath\u201d of integration. With dz = dx + idy and using using (F.1), we have z2 z1 f (z)dz = z2 z1 (u dx \u2212 v dy) +i z2 z1 (v dx + u dy) = z2 z1 A \u00b7 dx + i z2 z1 B \u00b7 dx (F.3) where A = u\u02c6x \u2212 v\u02c6y and B = v\u02c6x + u\u02c6y. So, we can now think of the two integrals on the right as real integrals of vector functions over curves in the xy plane. However, if we invoke Stokes\u2019s theorem, these become integrals of the curls, and using (F.2), we \ufb01nd \u2207 \u00d7 A = \u2212 \u2202v \u2202x \u2212 \u2202u \u2202y = 0 and \u2207 \u00d7 B = \u2202u \u2202x \u2212 \u2202v \u2202y = 0 (F.4) and each of the two integrals on the right in (F.3) is path independent. Hence, the integral of an analytic complex function f (z) is path independent and can be unambiguously de\ufb01ned. From here on, we assume all functions to be analytic unless explicitly noted otherwise. It is obvious from (F.3) that, when integrating around a closed path C,, C f (z)dz = 0 which is known as the Cauchy\u2013Goursat theorem. We will be exploring circumstances where the integrand is explicitly singular at one or more points. For the \ufb01rst example, we prove the Cauchy integral formula, namely f (z0) =, 1 2\u03c0i f (z) z \u2212 z0 C dz (F.5) where C is a closed contour in the complex plane that contains the point z0 and traversed in the counterclockwise direction. We can break up a contour C into something that looks like Figure F.1. Notice that C0 is a tiny circular contour around the singular point, but in the clockwise direction. That is, we replace C with limC0\u21920(C + C0). However, for C0 = 0, the new contour C does not include the singular point, so by (F.4) we write (F.5) as f (z0) =\u2212 1 2\u03c0i, C0 f (z) z \u2212 z0 dz. (F.6) The shrinking contour C0 is parameterized as z \u2212 z0 = rei\u03c6 for r \u2192 0 and \u03c6 = 2\u03c0 \u2192 0, so \u2212 1 2\u03c0i, C0 f (z) z \u2212 z0 dz = \u2212 1 2\u03c0i f (z0) 0 2\u03c0 1 rei\u03c6 irei\u03c6d\u03c6 = \u2212 1 2\u03c0i f (z0)i(\u22122\u03c0) = f (z0) proving the Cauchy integral formula (F.5). A trivial, but suggestive, rewriting of (F.5) gives, f (z) = 1 2\u03c0i f (\u03be) \u03be \u2212 z C d\u03be (F.7) 538 Appendix F Fig. F.1 A contour in the complex plane for proving the Cauchy integral formula. which leads to a convenient way to write the derivatives of a complex function, namely f(n)(z) = dnf dzn = n! 2\u03c0i f (\u03be) (\u03be \u2212 z)n+1 d\u03be. (F.8) C, Now consider the series expansion of an analytic", " function f (z). We would naturally write where f (z) = f (z0) +f (z0)(z \u2212 z0) +\u00b7 \u00b7 \u00b7 = \u221e \u2211 n=0 an(z \u2212 z0)n an \u2261 1 n! f(n)(z0) =, 1 2\u03c0i f (\u03be) (\u03be \u2212 z0)n+1 d\u03be. C (F.9) (F.10) Such a Taylor series expansion works out as expected, but the curve C speci\ufb01es regions in which the series converges. This idea can be expanded to include \u2212\u221e \u2264 n \u2264 \u221e, still using the right side of (F.10) to de\ufb01ne an, and with modi\ufb01ed regions of convergence. Such an expansion is called a Laurent series. It clearly is not, in general, an analytic function because of poles that appear for n < 0. These, however, lead us to one of the most important theorems of complex analysis, so far as mathematical physics is concerned. F.4 The Cauchy Residue Theorem Let g(z) have an isolated singularity at z = z0. If the Laurent expansion can be written as g(z) = \u221e \u2211 n=\u2212\u221e an(z \u2212 z0)n = b1 z \u2212 z0 + \u221e \u2211 n=0 an(z \u2212 z0)n (F.11) 539 Notes on Complex Variables then we say that g(z) has a \u201csimple pole\u201d at z = z0. Higher-order poles are possible, but we are not going to consider them here. Consider a contour C within the radius of convergence of g(z). Separate the integral of g(z) around this contour into two terms, one for each of the two terms on the right in (F.11). The second term is a polynomial in z; therefore it is analytic and the integral is zero. Recall that we reduced the contour to a small circle around the pole in order to prove the Cauchy integral formula. We can do the same thing here, and, C g(z)dz = b1, C 1 z \u2212 z0 = 2\u03c0i b1. (F.12) We refer to b1 as the \u201cresidue\u201d of g(z0), sometimes written as Res[g(z0)]. We have Res[g(z0)] = lim z\u2192z0 (z \u2212 z0)g(z) for a simple pole at z0. If there is more than one simple pole within the contour C, this result is easy to generalize. Instead of redrawing the contour with a small loop about the single pole, do it for all N poles within the contour. The result is clearly, C g(z)dz = 2\u03c0i N \u2211 k=1 Res[g(zk)]. (F.13) We refer to this as the Cauchy residue theorem. It is widely used in mathematical physics. The usefulness of the residue theorem can be illustrated in many ways, but here is one important example. It is a warm-up to evaluating the integral in (6.44). The exercise is to evaluate the integral \u221e \u2212\u221e I = eika q2 \u2212 k2 k dk = lim \u03b5\u21920 \u221e \u2212\u221e eika q2 \u2212 k2 + i\u03b5 k dk (F.14) where k, a, q, and \u03b5 > 0 are all real variables. We use the second version above because this moves the singularities at k = \u00b1q off the real axis. To be sure, we could have moved off the real axis by using \u2212i\u03b5 instead of +i\u03b5, and in fact, this would give us a different answer. A physical rationale is needed to justify one sign or the other. Leave that for a physics course. We can evaluate (F.14) using contour integration by \ufb01rst allowing k to be complex and then noting that eikx \u2192 0 as Im(k) \u2192 +\u221e. Therefore (F.14) can be rewritten as an integral over a semicircular contour C that runs (counterclockwise) along the Re(k) axis and closes as a semicircle in the Im(k) > 0 plane. Then for \u03b5 \u2192 0, the integrand in (F.14) has poles at k = \u00b1 q2 \u2212 i\u03b5 = \u00b1q 1 \u2212 i \u21d2 \u00b1q \u2213 i\u03b5 \u03b5 q2 where we rede\ufb01ne \u03b5 (with q > 0) so that it is still small and has the same sign. 540 Appendix F The pole at k = k0", " \u2261 +q \u2212 i\u03b5 does not matter to us, since it is outside the integration contour. However, the pole at k = \u2212k0 = \u2212q + i\u03b5 is inside, so we use the residue theorem to write, I = lim \u03b5\u21920 C eika (k \u2212 k0)(k + k0) k dk = lim \u03b5\u21920 2\u03c0i eika k \u2212 k0 k k=\u2212k0 = \u03c0i lim \u03b5\u21920 e\u2212ik0a = \u03c0ie\u2212iqa. Bibliography The following list includes textbooks and monographs that are, in general, cited earlier in this book. Some entries are not cited, but are included here to indicate general upper level textbooks on quantum mechanics that students and instructors may \ufb01nd useful for alternative explanations and additional exercises. Arfken, G. B., and H. J. Weber (1995). Mathematical Methods for Physicists, 4th edn. Academic Press. Arfken, G. B., H. J. Weber, and F. E. Harris (2013). Mathematical Methods for Physicists: A Comprehensive Guide, 7th edn. Academic Press. Baym, G. (1969). Lectures on Quantum Mechanics. New York: W. A. Benjamin. Republished in paperback by CRC Press, 1974. Bethe, H. A., and R. W. Jackiw (1968). Intermediate Quantum Mechanics, 2nd edn. New York: W. A. Benjamin. Biedenharn, L. C., and H. Van Dam (editors) (1965). Quantum Theory of Angular Momentum. New York: Academic Press. Bjorken, J. D., and S. D. Drell (1965). Relativistic Quantum Fields. New York: McGraw-Hill. Blatt, J. M., and V. W. Weisskopf (1952). Theoretical Nuclear Physics. John Wiley & Sons. Bohm, D. (1951). Quantum Theory. Englewood Cliffs, NJ: Prentice-Hall. Bransden, B. H., and C. J. Joachain (2003). Physics of Atoms and Molecules. Pearson. Byron, F. W., and R. W. Fuller (1992). Mathematics of Classical and Quantum Physics. Dover. Cohen, M. L., and S. G. Louie (2016). Fundamentals of Condensed Matter Physics. Cambridge University Press. Dicke, R. H., and J. P. Wittke (1960). Introduction to Quantum Mechanics. Reading, MA: Addison-Wesley. Dirac, P. A. M. (1958). Quantum Mechanics, 4th edn. London: Oxford University Press. Edmonds, A. R. (1974). Angular Momentum in Quantum Mechanics, 3rd printing. Princeton University Press. Fermi, E. (1950). Nuclear Physics. Chicago, IL: University of Chicago Press. Fetter, A. L., and J. D. Walecka (2003). Quantum Theory of Many-Particle Systems. Dover. Feynman, R. P., and A. R. Hibbs (1964). Quantum Mechanics and Path Integrals. New York: McGraw-Hill. Finkelstein, R. J. (1973). Nonrelativistic Mechanics. Reading, MA: W. A. Benjamin. French, A. P., and E. F. Taylor (1978). An Introduction to Quantum Physics. New York: W. W. Norton. 541 542 Bibliography Goldberger, M. L., and K. M. Watson (1964). Collision Theory. New York: Wiley. Goldstein, H., C. Poole, and J. Safko (2002). Classical Mechanics, 3rd edn. Reading, MA: Addison-Wesley. Gottfried, K. (1966). Quantum Mechanics, vol. I. New York: W. A. Benjamin. Gottfried, K., and T.-M. Yan (2003). Quantum Mechanics: Fundamentals, 2nd edn. Springer-Verlag. Grif\ufb01ths, D. J. (2005). Introduction to Quantum Mechanics, 2nd edn. Pearson. Heitler, W. (1954). The Quantum Theory of Radiation, 3rd edn. Oxford: Oxford University Press. Henley, E. M., and A. Garcia (2007). Subatomic Physics, 3rd edn. World Scienti\ufb01c. Holstein, B. R. (1992). Topics in Advanced Quantum Mechanics. Addison-Wesley. Itzykson, C., and J.-B. Zuber (1980). Quantum Field Theory. McGraw-Hill. Jackson, J. D. (1975). Classical Electrodynamics, 2nd edn. New York", ": Wiley. Jackson, J. D. (1998). Classical Electrodynamics, 3rd edn. New York: Wiley. Landau, R. H. (1996). Quantum Mechanics II: A Second Course in Quantum Theory. Wiley. Loudon, R. (2000). The Quantum Theory of Light, 3rd edn. Oxford Science Publications. Mandl, F. (1957). Quantum Mechanics. London: Butterworths Scienti\ufb01c Publications. Marcus, G. E. (editor) (2000). Zeroing in on the Year 2000: The Final Edition. Chicago, IL: University of Chicago Press. Matthews, P. T. (1974). Introduction to Quantum Mechanics, 3rd edn. London: McGraw- Hill. Melissinos, A., and J. Napolitano (2003). Modern Physics. Academic Press. Merzbacher, E. (1970). Quantum Mechanics, 2nd edn. New York: Wiley. Merzbacher, E. (1998). Quantum Mechanics, 3rd edn. Wiley. Messiah, A. (1961). Quantum Mechanics. New York: Interscience. Morse, P. M., and H. Feshbach (1953). Methods of Theoretical Physics. New York: McGraw-Hill. Mott, N. F. (1952). Elements of Wave Mechanics. London: Cambridge University Press. Newton, R. G. (1982). Scattering Theory of Waves and Particles, 2nd. edn. New York: McGraw-Hill. Park, D. (1974). Introduction to the Quantum Theory, 2nd. edn. New York: McGraw-Hill. Pauling, L., and E. B. Wilson (1935). Introduction to Quantum Mechanics. New York: McGraw-Hill. Preston, M. (1962). Physics of the Nucleus. Reading, MA: Addison-Wesley. Rojansky, V. (1938). Introductory Quantum Mechanics. Englewood Cliffs, NJ: Prentice- Hall. Sakurai, J. J. (1964). Invariance Principles and Elementary Particles. Princeton University Press. Republished 2016. Sakurai, J. J. (1967). Advanced Quantum Mechanics. Reading, MA: Addison-Wesley. Sargent III, M., M. O. Scully, and W. E. Lamb, Jr. (1974). Laser Physics. Reading, MA: Addison-Wesley. Saxon, D. S. (1968). Elementary Quantum Mechanics. San Francisco, CA: Holden-Day. 543 Bibliography Schiff, L. (1968). Quantum Mechanics, 3rd. edn. New York: McGraw-Hill. Shankar, R. (1994). Principles of Quantum Mechanics, 2nd edn. Plenum. Taylor, J. R. (2005). Classical Mechanics. University Science Books. Tomanaga, S. (1962). Quantum Mechanics I. Amsterdam: North-Holland. Townsend, J. S. (2000). A Modern Approach to Quantum Mechanics. University Science Books. Van der Waerden, B. L. (editor) (1967). Sources of Quantum Mechanics. Dover. Weinberg, S. (1995). The Quantum Theory of Fields. Cambridge University Press. Zee, A. (2010). Quantum Field Theory in a Nutshell, 2nd edn. Princeton University Press. Index Born, M., 45, 83, 93, 95, 180 Bose\u2013Einstein condensation, 434, 457 bosons, 433, 455 bra, 12 representation as row matrix, 18 Brillouin zone, 269 Brillouin, L., 104 Burke, K., 451 canonical commutation relations, 45 Casimir effect, 468 Cayley\u2013Klein parameters, 165 charge conjugation symmetry, 498 classical correspondence, 76, 79, 81, 90, 96, 116, 326 Clebsch\u2013Gordan coef\ufb01cients, 208, 210 and tensor operators, 236 recursion relations, 212 rotation operator, 216 unitary matrix, 211 Clifford algebra, 490 commutator, 26 compatibility, 27, 68 completeness, 17, 18 Compton effect, 1 Compton wavelength, 480 con\ufb02uent hypergeometric function, 203 conservation laws, 249 and symmetry operators, 250 classical Hamiltonian, 250 constant of the motion, 250, 252 continuity equation, 94, 482, 492 Coulomb potential, 201 covariance, 481 CPT symmetry, 501 cross section photoelectric effect, 351 radiation absorption, 348 Rutherford, 386, 424 scattering, 373, 381 Dalgarno, A., 300 Darwin term, 311 Davisson\u2013Germer\u2013Thompson experiment, 1 de Broglie, L., 1, 43, 62, 93 degeneracy, 27, 98, 201, 205 and parity symmetry, 260 2, 206 Abelian group,", " 154 absorption of radiation, 347 adiabatic approximation, 328 Aharonov, Y., 133 Aharonov\u2013Bohm effect, 120, 131, 335 Airy function, 102, 107 ammonia maser, 326 ammonia molecule, 262 Anderson, C., 495 angular momentum addition, 205 addition of orbital plus spin 1 addition of two spin 1 barrier, 197 commutation relations, 26, 154, 209 Dirac equation, 497 eigenvalues, 180 orbital, 28, 188 partial wave decomposition, 388 quantization, 3 anticommutator, 26 anyons, 433 Argand diagram, 396 Aspect, A., 231 atomic \ufb01ne structure, 307, 309, 310 atomic force microscope, 471 atomic polarizability, 299 2, 206 Baker\u2013Hausdorff lemma, 89 Balmer formula, 204, 507 Bell\u2019s inequality, 227 Bell, J. S., 227 Berry\u2019s phase, 138, 331 Aharonov\u2013Bohm effect, 336 neutrons, 334 photons, 333 Berry, M. V.,331 Bethe, H. A., 422 Bloch\u2019s theorem, 269 Bloch, F., 422 Bohm, D., 133, 226 Bohr model, 204 Bohr radius, 204, 318, 462 Bohr, N., 1, 69 Born approximation, 375, 384 higher orders, 387 544 545 Index and perturbation theory, 300 connection to symmetry, 251 exchange, 430, 442 density matrix, 172 density of states, 98, 344, 350, 353, 373 density operator, 172 diagonalization, 28, 35 in degenerate subspace, 302 Dirac delta function, 38 Dirac equation, 478, 491 Dirac picture, 322 Dirac, P. A. M., 1, 10, 15, 22, 45, 47, 78, 83, 115, 137, 338, 344, 490 Dyson, F. J., 67, 339 effective potential, 196 Ehrenberg, W., 133 Ehrenfest theorem, 81 Ehrenfest, P., 81 eigenket, 11, 16, 81 energy, 67, 70 position or momentum, 38, 53 simultaneous, 28, 181 eigenvalue, 11, 16, 23 angular momentum, 180 continuous, 37 Einstein\u2019s locality principle, 226 Einstein, A., 226 Einstein\u2013Debye theory, 1 Einstein\u2013Podolsky\u2013Rosen paradox, 226 electric dipole approximation, 348 energy bands, 268 energy-time uncertainty relation, 74 ensemble, 22, 169 pure, 172, 177 time evolution, 175 ensemble average, 171 Ernzerhof, M., 451 Euler angles, 166, 187, 194, 222 exchange degeneracy, 430, 442 exchange density, 436 exchange-correlation energy, 448, 450 expectation value, 23, 68 under rotation, 155 Fermi energy, 457, 462 Fermi\u2019s golden rule, 344, 373 fermions, 433, 455 Fetter, A. L., 509 Feynman, R. P., 114, 115, 509 Fock space, 454 form factor, 421, 423, 428 Fortun, M., 469 Fourier transform, 51, 351 Franck\u2013Hertz experiment, 1 free particle, 79, 98, 110, 146, 198, 479, 484, 491\u2013493 gauge transformations, 120 electromagnetism, 126 gravitational effect, 122 generating function, 99, 101, 204 geometric phase, 331 Gerlach, W., 2 Glauber, R., 91, 142, 472 Green\u2019s function, 110 Helmholtz equation, 379, 388 scattering, 376, 424 Groups SO(3), 163 SO(4), 252 SU(2), 164 SU(2)\u00d7SU(2), 255 Hamilton, W. R., 93 Hamilton\u2013Jacobi equation, 96, 400 Hamiltonian, 63, 65 Heisenberg equation of motion, 68, 78 Heisenberg picture, 77, 321 Heisenberg uncertainty principle, 3, 33, 43, 88 Heisenberg, W., 1, 45, 180, 441 helium atom, 312, 319, 437, 451 Hermite polynomials, 99 hidden variables, 227 Hilbert space, 11 Hilbert, D., 11, 93 Hofstadter, R., 424 Hohenberg, P., 444, 447 Hohenberg\u2013Kohn theorem, 447 hydrogenlike atoms, 201 degeneracy, 205, 252, 255, 260 Dirac equation solution, 505 energy levels, 204 generating function, 204, 244 Klein\u2013Gordon equation solution, 485 linear Stark effect, 298, 303 perturbation, 288 principal quantum number, 204 quadratic Stark effect, 298 relativistic perturbation, 306 spin-orbit perturbation, 308 variational principle, 318 versus Bohr model, 204 wave function, 205, 528 inner product, 12 interaction picture, 321, 340", ", 355, 371 Jeffreys, H., 104 Jordan, P., 45, 180 ket, 8, 10 representation as column matrix, 18 Klein\u2013Gordon equation, 478, 480 Klein\u2013Gordon \ufb01eld, 443, 485 546 Index Kohn, W., 443, 444, 447 Kohn\u2013Sham potential, 449 Kramers degeneracy, 284 Kramers, A., 104 Kummer\u2019s equation, 203 Lamb shift, 305, 509 Lamoreaux, S., 469 Lande\u2019s formula, 313 Laporte\u2019s rule, 264 Lee, T. D., 264 Lewis, J. T., 300 Liouville\u2019s theorem, 175 Lipkin, H. J., 138 Lippmann\u2013Schwinger equation, 375, 382, 387, 424, 427 long wavelength approximation, 353, 355 Lorentz force, 126 Lorentz transformation, 481 magnetic moment, 3, 4, 55, 307, 326 from Dirac Equation, 496 time evolution in magnetic \ufb01eld, 63, 66, 69, 147, 157, 158, 180, 333 magnetic monopole, 135, 337 matrix mechanics, 45 matrix representation, 18 angular-momentum operators, 184 diagonal matrix, 27 eigenvalue problem, 35 rotation operator, 185, 216 unitary transformation, 34 Maxwell\u2019s equations, 464 Merzbacher, E., 509 muon g \u2212 2, 71 neutrinos, 71, 138 neutrons Berry\u2019s phase, 334, 368 bouncing energy eigenvalues, 103, 107 gravitational phase, 123 interferometry, 148, 158 np bound state, 410 normal ordering, 458 normalization, 13, 17, 22, 292 nuclear magnetic resonance, 326 observable, 11, 16, 22 compatible, 27, 28 incompatible, 29 unitary equivalent, 36 operator, 13 annihilation and creation, 84 antiunitary, 273 generator, 25, 42, 152, 189, 250 Hermitian, 14, 16, 19 ladder, 181 linear, 14 number, 84 pair distribution, 458 permutation, 430 projection, 18, 292, 301 representation as square matrix, 18 rotation, 152, 168, 216 symmetry, 250 tensor, 233 time evolution, 63, 65 trace of, 35 translation, 40 unitary, 34, 75, 210 vector, 232 optical theorem, 381 orthogonality, 13, 17 orthohelium, 440 outer product, 15 parabolic coordinates, 300 parahelium, 440 parity symmetry, 256 Dirac equation, 497 one-dimensional double well, 261 perturbation theory, 297 violation, 264 partition function, 112, 179 Paschen\u2013Back effect, 313 path-integral formulation, 114 Pauli exclusion principle, 433, 456, 494 Pauli matrices, 160 Pauli, W., 21, 159 Perdew, J., 443, 451 Peshkin, M., 138 photoelectric effect, 98, 350, 464 photon, 464, 467 pion-proton scattering, 415 Planck radiation law, 1 Poisson brackets, 46 polarization circular, 8 of light, 6 spin, 170 Pople, J., 443 Positron, 492, 495 potential energy function \u03b4-function well, 144 Coulomb potential, 201, 386, 505, 528 double \u03b4-function, 145 Gaussian potential, 403 linear, 101 local, 379 one-dimensional \ufb01nite well, 145, 521 one-dimensional in\ufb01nite well, 318, 521 periodic, 266 screened Coulomb, 309 simple harmonic oscillator, 83, 99, 523 547 Index spherical well, 198, 385 symmetric double well, 261 three-dimensional harmonic oscillator, 199 Yukawa potential, 386, 387, 403 precession, 69, 157 projection theorem, 239 propagator, 109 quantum optics, 472 quarkonium, 103 Rabi, I. I.,323 Ramsauer\u2013Townsend effect, 407 reduced matrix element, 239 representations, 18 resonance magnetic, 325 scattering, 397, 412 two-state system, 324 rotation, 149 \ufb01nite, 153 in\ufb01nitesimal, 150 operator, 152 orthogonal group, 163 orthogonal matrix, 149 two-component formalism, 161 unitary unimodular group, 164 vector, 231 rotational symmetry angular momentum conservation, 251 Runge\u2013Lenz vector, 245, 252 S matrix, 372 poles as bound states, 410 scattering phase shift, 394 scattering shadow amplitude, 405 Schr\u00a8odinger equation, 65, 82, 91 energy quantization, 93 probability density and \ufb02ux, 94 spherically symmetric potentials, 195 Schr\u00a8odinger picture, 77, 321, 339 Schr\u00a8odinger, E., 1, 62, 92 Schwinger, J., 23, 42, 218 second quantization, 443,", " 454, 485 selection rules parity, 263 Sham, L.J., 444, 447 Siday, R.E., 133 simple harmonic oscillator, 83 as solution to the Schr\u00a8odinger equation, 99 coherent states, 91, 142, 473 generating function, 99 ground state, 86 parity symmetry, 260 propagator, 111 quadratic perturbation, 296 three-dimensional, 199 Slater determinant, 442 sodium atom, 309 sodium metal, 464 solid angle, 374 Sommerfeld, A., 106 space quantization, 3 spherical Bessel function, 199 spherical harmonics, 191, 217 as spherical tensors, 233 spin-angular functions, 216, 246, 309, 502 spin magnetic resonance, 325 spin-orbit interaction, 307 spin-statistics theorem, 433 spinor, 493 spinor spherical harmonics, 216, 246, 309, 502 spontaneous emission of radiation, 352 squeezed states, 474 Stark effect, 264, 298, 329 stationary state, 69 statistical mechanics, 176 statistics Bose\u2013Einstein, 433 Fermi\u2013Dirac, 433 Maxwell\u2013Boltzmann, 434 Stern, O., 1 Stern\u2013Gerlach experiment, 1, 54 sequential, 4, 8, 24 stimulated emission of radiation, 347 sudden approximation, 328 super\ufb02uid helium, 434, 474 T matrix, 372, 374 tensor operator Cartesian, 233 irreducible spherical, 234 reducibility, 233 Thomas precession, 308 Thomas, L. H., 308 Thomas\u2013Reiche\u2013Kuhn sum rule, 349 time reversal, 14, 270 Dirac equation, 499 totally antisymmetric symbol \u03b5ijk, 26 transition amplitude, 339, 372 transition probability, 340 translation, 40 \ufb01nite, 43 tritium atom, 328 two-state systems, 2, 289, 323 unitarity limit, 413\u2013415 units, 43, 479, 510 Uns\u00a8old, A., 440 vacuum, 454, 468 van der Waals\u2019 force, 314 variational theorem, 316, 443, 447 548 Index vector, 149 axial, 258 operator de\ufb01nition, 232 polari, 258 rotations, 149, 152, 156 vector space, 10 bra space, 12, 15 dual space, 12 ket space, 10, 15 virial theorem, 87, 306 von Neumann, J., 170 Walecka, J. D., 509 wave function, 39, 47, 82, 94 orbital plus spin angular momentum, 206 plane wave, 51, 97, 520 renormalization, 295 spherical well, 198, 397 under parity, 258 under time reversal, 279 wave mechanics, 92 wave packet, 51, 81, 111 in scattering theory, 381 Weisberger, W. I., 138 Weisskopf, V. F., 358 Wentzel, G., 104 Weyl, H., 93, 131 Wiener, N., 83 Wigner, E. P., 185, 222, 227, 264, 270, 284, 358 Wigner\u2013Eckart theorem, 218, 237, 298, 393 examples, 238 projection theorem, 239 Wigner\u2013Seitz radius, 462 Wilson, W., 106 WKB approximation, 104 Wu, T. T., 137 Yang, C. N., 137, 264 Zee, A., 509 Zeeman effect, 311, 360, 362 quadratic, 312, 364 zero point energy, 468, 487ier, more technical, biography by Carlo Cercignani. \u2013 16 \u2013 It is often said that Boltzmann\u2019s great insight was the equation which is now engraved on his tombstone, S = kB log \u2126, which lead to the understanding of the second law of thermodynamics in terms of microscopic disorder. Yet these days it is di\ufb03cult to appreciate the boldness of this proposal simply because we rarely think of any other de\ufb01nition of entropy. We will, in fact, meet the older thermodynamic notion of entropy and the second law in Section 4. In the meantime, perhaps Boltzmann\u2019s genius is better re\ufb02ected in the surprising equation for temperature: 1/T = \u2202S/\u2202E. Boltzmann gained prominence during his lifetime, holding professorships at Graz, Vienna, Munich and Leipzig (not to mention a position in Berlin that he somehow failed to turn up for). Nonetheless, his work faced much criticism from those who would deny the existence of atoms, most notably Mach. It is not known whether these battles contributed to the depression Boltzmann su\ufb00ered later in life, but the true signi\ufb01cance of his work was only appreciated after his body was found hanging from the rafters of a guest house near Trieste. 1.3 The Canonical Ensemble The microcanonical ensemble describes systems that", " have a \ufb01xed energy E. From this, we deduce the equilibrium temperature T. However, very often this is not the best way to think about a system. For example, a glass of water sitting on a table has a well de\ufb01ned average energy. But the energy is constantly \ufb02uctuating as it interacts with the environment. For such systems, it is often more appropriate to think of them as sitting at \ufb01xed temperature T, from which we then deduce the average energy. To model this, we will consider a system \u2014 let\u2019s call it S \u2014 in contact with a second system which is a large heat reservoir \u2013 let\u2019s call it R. This reservoir is taken to be at some equilibrium temperature T. The term \u201creservoir\u201d means that the energy of S is negligible compared with that of R. In particular, S can happily absorb or donate energy from or to the reservoir without changing the ambient temperature T. How are the energy levels of S populated in such a situation? We label the states of S as |n, each of which has energy En. The number of microstates of the combined systems S and R is given by the sum over all states of S, \u2126(Etotal) = n \u2126R(Etotal \u2212 En) \u2261 exp n SR(Etotal \u2212 En) kB I stress again that the sum above is over all the states of S, rather than over the energy levels of S. (If we\u2019d written the latter, we would have to include a factor of \u2126S(En) in the sum to take into account the degeneracy of states with energy En). The fact that \u2013 17 \u2013 R is a reservoir means that En Etotal. This allows us to Taylor expand the entropy, keeping just the \ufb01rst two terms, \u2126(Etotal) \u2248 exp n SR(Etotal) kB \u2212 \u2202SR \u2202Etotal En kB But we know that \u2202SR/\u2202Etotal = 1/T, so we have \u2126(Etotal) = eSR(Etotal)/kB e\u2212En/kBT n We now apply the fundamental assumption of statistical mechanics \u2014 that all accessible energy states are equally likely \u2014 to the combined system + reservoir. This means that each of the \u2126(Etotal) states above is equally likely. The number of these states for which the system sits in |n is eSR/kB e\u2212En/kBT. So the probabilty that the system sits in state |n is just the ratio of this number of states to the total number of states, p(n) = e\u2212En/kBT m e\u2212Em/kBT (1.19) This is the Boltzmann distribution, also known as the canonical ensemble. Notice that the details of the reservoir have dropped out. We don\u2019t need to know SR(E) for the reservoir; all that remains of its in\ufb02uence is the temperature T. The exponential suppression in the Boltzmann distribution means that it is very unlikely that any of the states with En kBT are populated. However, all states with energy En \u2264 kBT have a decent chance of being occupied. Note that as T \u2192 0, the Boltzmann distribution forces the system into its ground state (i.e. the state with lowest energy); all higher energy states have vanishing probability at zero temperature. 1.3.1 The Partition Function Since we will be using various quantities a lot, it is standard practice to introduce new notation. Firstly, the inverse factor of the temperature is universally denoted, And the normalization factor that sits in the denominator of the probability is written, \u03b2 \u2261 1 kBT (1.20) Z = e\u2212\u03b2En n In this notation, the probability for the system to be found in state |n is p(n) = e\u2212\u03b2En Z \u2013 18 \u2013 (1.21) (1.22) Rather remarkably, it turns out that the most important quantity in statistical mechanics is Z. Although this was introduced as a fairly innocuous normalization factor, it actually contains all the information we need about the system. We should think of Z, as de\ufb01ned in (1.21), as a function of the (inverse) temperature \u03b2. When viewed in this way, Z is called the partition function. We\u2019ll see lots of properties of Z soon. But we\u2019ll start with a fairly basic, yet important, point: for independent systems, Z\u2019s multiply. This is easy to prove. Suppose that we have two systems which don\u2019t interact with each other. The energy of the combined system is then just the sum of the individual energies. The partition function for the combined system is (in, hopefully, obvious notation) Z = = =", " n,m n,m n e\u2212\u03b2(E(1) n +E(2) m ) e\u2212\u03b2E(1) n e\u2212\u03b2E(2) m e\u2212\u03b2E(1) n e\u2212\u03b2E(2) m = Z1Z2 m (1.23) A Density Matrix for the Canonical Ensemble In statistical mechanics, the inherent probabilities of the quantum world are joined with probabilities that arise from our ignorance of the underlying state. The correct way to describe this is in term of a density matrix, \u02c6\u03c1. The canonical ensemble is really a choice of density matrix, \u02c6\u03c1 = e\u2212\u03b2 \u02c6H Z (1.24) If we make a measurement described by an operator \u02c6O, then the probability that we \ufb01nd ourselves in the eigenstate |\u03c6 is given by p(\u03c6) = \u03c6|\u02c6\u03c1|\u03c6 For energy eigenstates, this coincides with our earlier result (1.22). We won\u2019t use the language of density matrices in this course, but it is an elegant and conceptually clear framework to describe more formal results. 1.3.2 Energy and Fluctuations Let\u2019s see what information is contained in the partition function. We\u2019ll start by thinking about the energy. In the microcanonical ensemble, the energy was \ufb01xed. In the \u2013 19 \u2013 canonical ensemble, that is no longer true. However, we can happily compute the average energy, E = n p(n)En = Ene\u2212\u03b2En Z n But this can be very nicely expressed in terms of the partition function by E = \u2212 \u2202 \u2202\u03b2 log Z (1.25) We can also look at the spread of energies about the mean \u2014 in other words, about \ufb02uctuations in the probability distribution. As usual, this spread is captured by the variance, \u2206E2 = (E \u2212 E)2 = E2 \u2212 E2 This too can be written neatly in terms of the partition function, \u2206E2 = \u22022 \u2202\u03b22 log Z = \u2212 \u2202E \u2202\u03b2 (1.26) There is another expression for the \ufb02uctuations that provides some insight. Recall our de\ufb01nition of the heat capacity (1.9) in the microcanonical ensemble. In the canonical ensemble, where the energy is not \ufb01xed, the corresponding de\ufb01nition is CV = \u2202E \u2202T V Then, since \u03b2 = 1/kBT, the spread of energies in (1.26) can be expressed in terms of the heat capacity as \u2206E2 = kBT 2CV (1.27) There are two important points hiding inside this small equation. The \ufb01rst is that the equation relates two rather di\ufb00erent quantities. On the left-hand side, \u2206E describes the probabilistic \ufb02uctuations in the energy of the system. On the right-hand side, the heat capacity CV describes the ability of the system to absorb energy. If CV is large, the system can take in a lot of energy without raising its temperature too much. The equation (1.27) tells us that the \ufb02uctuations of the systems are related to the ability of the system to dissipate, or absorb, energy. This is the \ufb01rst example of a more general result known as the \ufb02uctuation-dissipation theorem. \u2013 20 \u2013 The other point to take away from (1.27) is the size of the \ufb02uctuations as the number of particles N in the system increases. Typically E \u223c N and CV \u223c N. Which means that the relative size of the \ufb02uctuations scales as \u2206E E \u223c 1 \u221a N (1.28) The limit N \u2192 \u221e in known as the thermodynamic limit. The energy becomes peaked closer and closer to the mean value E and can be treated as essentially \ufb01xed. But this was our starting point for the microcanonical ensemble. In the thermodynamic limit, the microcanonical and canonical ensembles coincide. All the examples that we will discuss in the course will have a very large number of particles, N, and we can consider ourselves safely in the thermodynamic limit. For that reason, even in the canonical ensemble, we will often write E for the average energy rather than E. An Example: The Two State System Revisited We can rederive our previous results for the two state system using the canonical ensemble. It is marginally simpler. For a single particle with two energy levels, 0 and, the partition function is given by Z1 = n e\u2212\u03b2En = 1 + e\u2212\u03b2 = 2e\u2212\u03b2/2 cosh(\u03b2/2) We", " want the partition function for N such particles. But we saw in (1.23) that if we have independent systems, then we simply need to multiply their partition functions together. We then have from which we can easily compute the average energy Z = 2N e\u2212N \u03b2/2 coshN (\u03b2/2) E = \u2212 \u2202 \u2202\u03b2 log Z = N 2 (1 \u2212 tanh(\u03b2/2)) A bit of algebra will reveal that this is the same expression that we derived in the microcanonical ensemble (1.13). We could now go on to compute the heat capacity and reproduce the result (1.14). Notice that, unlike in the microcanonical ensemble, we didn\u2019t have to solve any combinatoric problem to count states. The partition function has done all that work for us. Of course, for this simple two state system, the counting of states wasn\u2019t di\ufb03cult but in later examples, where the counting gets somewhat tricker, the partition function will be an invaluable tool to save us the work. \u2013 21 \u2013 1.3.3 Entropy Recall that in the microcanonical ensemble, the entropy counts the (log of the) number of states with \ufb01xed energy. We would like to de\ufb01ne an analogous quantity in the canonical ensemble where we have a probability distribution over states with di\ufb00erent energies. How to proceed? Our strategy will be to again return to the microcanonical ensemble, now applied to the combined system + reservoir. In fact, we\u2019re going to use a little trick. Suppose that we don\u2019t have just one copy of our system S, but instead a large number, W, of identical copies. Each system lives in a particular state |n. If W is large enough, the number of systems that sit in state |n must be simply p(n)W. We see that the trick of taking W copies has translated the probabilities into eventualities. To determine the entropy we can treat the whole collection of W systems as sitting in the microcanonical ensemble to which we can apply the familiar Boltzmann de\ufb01nition of entropy (1.2). We must only \ufb01gure out how many ways there are of putting p(n)W systems into state |n for each |n. That\u2019s a simple combinatoric problem: the answer is \u2126 = W! n(p(n)W )! And the entropy is therefore S = kB log \u2126 = \u2212kBW n p(n) log p(n) (1.29) where we have used Stirling\u2019s formula to simplify the logarithms of factorials. This is the entropy for all W copies of the system. But we also know that entropy is additive. So the entropy for a single copy of the system, with probability distribution p(n) over the states is S = \u2212kB n p(n) log p(n) (1.30) This beautiful formula is due to Gibbs. It was rediscovered some decades later in the context of information theory where it goes by the name of Shannon entropy for classical systems or von Neumann entropy for quantum systems. In the quantum context, it is sometimes written in terms of the density matrix (1.24) as S = \u2212kBTr \u02c6\u03c1 log \u02c6\u03c1 When we \ufb01rst introduced entropy in the microcanonical ensemble, we viewed it as a function of the energy E. But (1.30) gives a very di\ufb00erent viewpoint on the entropy: \u2013 22 \u2013 it says that we should view S as a function of a probability distribution. There is no contradiction with the microcanonical ensemble because in that simple case, the probability distribution is itself determined by the choice of energy E. Indeed, it that the Gibbs entropy (1.30) reverts to the is simple to check (and you should!) Boltzmann entropy in the special case of p(n) = 1/\u2126(E) for all states |n of energy E. Meanwhile, back in the canonical ensemble, the probability distribution is entirely determined by the choice of temperature T. This means that the entropy is naturally a function of T. Indeed, substituting the Boltzmann distribution p(n) = e\u2212\u03b2En/Z into the expression (1.30), we \ufb01nd that the entropy in the canonical ensemble is given by S = \u2212 kB Z = kB\u03b2 Z n n e\u2212\u03b2En log e\u2212\u03b2En Z Ene\u2212\u03b2En + kB log Z As with all other important quantities, this can be elegantly expressed in terms of the partition function by S = kB \u2202 \u2202T (T log Z) (1.31) A Comment on the Microcanonical vs Canonical Ensembles The", " microcanonical and canonical ensembles are di\ufb00erent probability distributions. This means, using the de\ufb01nition (1.30), that they generally have di\ufb00erent entropies. Nonetheless, in the limit of a large number of particles, N \u2192 \u221e, all physical observables \u2014 including entropy \u2014 coincide in these two distributions. We\u2019ve already seen this when we computed the variance of energy (1.28) in the canonical ensemble. Let\u2019s take a closer look at how this works. The partition function in (1.21) is a sum over all states. We can rewrite it as a sum over energy levels by including a degeneracy factor Z = {Ei} \u2126(Ei)e\u2212\u03b2Ei The degeneracy factor \u2126(E) factor is typically a rapidly rising function of E, while the Boltzmann suppression e\u2212\u03b2E is rapidly falling. But, for both the exponent is proportional to N which is itself exponentially large. This ensures that the sum over energy levels is entirely dominated by the maximum value, E, de\ufb01ned by the requirement \u2126(E)e\u2212\u03b2E = 0 \u2202 \u2202E E=E \u2013 23 \u2013 and the partition function can be well approximated by Z \u2248 \u2126(E)e\u2212\u03b2E (This is the same kind of argument we used in (1.2.1) in our discussion of the Second Law). With this approximation, we can use (1.25) to show that the most likely energy E and the average energy E coincide: E = E (We need to use the result (1.7) in the form \u2202 log \u2126(E)/\u2202E = \u03b2 to derive this). Similarly, using (1.31), we can show that the entropy in the canonical ensemble is given by S = kB log \u2126(E) Maximizing Entropy There is actually a uni\ufb01ed way to think about the microcanonical and canonical ensembles in terms of a variational principle: the di\ufb00erent ensembles have the property that they maximise the entropy subject to various constraints. The only di\ufb00erence between them is the constraints that are imposed. Let\u2019s start with the microcanonical ensemble, in which we \ufb01x the energy of the system so that we only allow non-zero probabilities for those states which have energy E. We could then compute the entropy using the Gibbs formula (1.30) for any probability distribution, including systems away from equilibrium. We need only insist that all the probabilities add up to one: n p(n) = 1. We can maximise S subject to this condition by introducing a Lagrange multiplier \u03b1 and maximising S + \u03b1kB( n p(n) \u2212 1), \u2202 \u2202p(n) \u2212 n p(n) log p(n) + \u03b1 n p(n) \u2212 \u03b1 = 0 \u21d2 p(n) = e\u03b1\u22121 We learn that all states with energy E are equally likely. This is the microcanonical ensemble. In the examples sheet, you will be asked to show that the canonical ensemble can be viewed in the same way: it is the probability distribution that maximises the entropy subject to the constraint that the average energy is \ufb01xed. \u2013 24 \u2013 1.3.4 Free Energy We\u2019ve left the most important quantity in the canonical ensemble to last. It is called the free energy, F = E \u2212 T S (1.32) There are actually a number of quantities all vying for the name \u201cfree energy\u201d, but the quantity F is the one that physicists usually work with. When necessary to clarify, it is sometimes referred to as the Helmholtz free energy. The word \u201cfree\u201d here doesn\u2019t mean \u201cwithout cost\u201d. Energy is never free in that sense. Rather, it should be interpreted as the \u201cavailable\u201d energy. Heuristically, the free energy captures the competition between energy and entropy Immersed in a heat bath, energy that occurs in a system at constant temperature. is not necessarily at a premium. Indeed, we saw in the two-state example that the ground state plays little role in the physics at non-zero temperature. Instead, the role of entropy becomes more important: the existence of many high energy states can beat a few low-energy ones. The fact that the free energy is the appropriate quantity to look at for systems at \ufb01xed temperature is also captured by its mathematical properties. Recall, that we started in the microcanonical ensemble by de\ufb01ning entropy S = S(E, V ). If we invert this expression, then we can equally well think of energy as a function of entropy and volume: E = E(S, V", " ). This is re\ufb02ected in the \ufb01rst law of thermodynamics (1.16) which reads dE = T dS \u2212 pdV. However, if we look at small variations in F, we get dF = dE \u2212 d(T S) = \u2212SdT \u2212 pdV (1.33) This form of the variation is telling us that we should think of the free energy as a function of temperature and volume: F = F (T, V ). Mathematically, F is a Legendre transform of E. Given the free energy, the variation (1.33) tells us how to get back the entropy, Similarly, the pressure is given by S = \u2212 p = \u2212 \u2202F \u2202T V \u2202F \u2202V T \u2013 25 \u2013 (1.34) (1.35) The free energy is the most important quantity at \ufb01xed temperature. It is also the quantity that is most directly related to the partition function Z: F = \u2212kBT log Z (1.36) This relationship follows from (1.25) and (1.31). Using the identity \u2202/\u2202\u03b2 = \u2212kBT 2\u2202/\u2202T, these expressions allow us to write the free energy as F = E \u2212 T S = kBT 2 \u2202 \u2202T log Z \u2212 kBT \u2202 \u2202T (T log Z) = \u2212kBT log Z as promised. 1.4 The Chemical Potential Before we move onto applications, there is one last bit of formalism that we will need to introduce. This arises in situations where there is some other conserved quantity which restricts the states that are accessible to the system. The most common example is simply the number of particles N in the system. Another example is the electric charge Q. For the sake of de\ufb01niteness, we will talk about particle number below but all the comments apply to any conserved quantity. In both the microcanonical and canonical ensembles, we should only consider states that have a \ufb01xed value of N. We already did this when we discussed the two state system \u2014 for example, the expression for entropy (1.12) depends explicitly on the number of particles N. We will now make this dependence explicit and write S(E, V, N ) = kB log \u2126(E, V, N ) The entropy leads us to the temperature as 1/T = \u2202S/\u2202E and the pressure as p = T \u2202S/\u2202V. But now we have another option: we can di\ufb00erentiate with respect to particle number N. The resulting quantity is called the chemical potential, \u00b5 = \u2212T \u2202S \u2202N (1.37) Using this de\ufb01nition, we can re-run the arguments given in Section 1.2.2 for systems which are allowed to exchange particles. Such systems are in equilibrium only if they have equal chemical potential \u00b5. This condition is usually referred to as chemical equilibrium. \u2013 26 \u2013 To get a feel for the meaning of the chemical potential, we can look again at the \ufb01rst law of thermodynamics (1.16), now allowing for a change in particle number as well. Writing dS =... and rearranging, we have, dE = T dS \u2212 pdV + \u00b5dN (1.38) it is the energy cost to add one This tells us the meaning of the chemical potential: more particle to the system while keeping both S and V \ufb01xed. (Strictly speaking, an in\ufb01nitesimal amount of particle, but if we\u2019re adding one more to 1023 that e\ufb00ectively counts as in\ufb01nitesimal). If we\u2019re interested in electric charge Q, rather than particle number, the chemical potential is the same thing as the familiar electrostatic potential of the system that you met in your \ufb01rst course in Electromagnetism. There\u2019s actually a subtle point in the above derivation that is worth making explicit. It\u2019s the kind of thing that will crop up a lot in thermodynamics where you typically have many variables and need to be careful about which ones are kept \ufb01xed. We de\ufb01ned the chemical potential as \u00b5 = \u2212T \u2202S/\u2202N |E,V. But the \ufb01rst law is telling us that we can also think of the chemical potential as \u00b5 = \u2202E/\u2202N |S,V. Why is this the same thing? This follows from a general formula for partial derivatives. If you have three variables, x, y and z, with a single constraint between them, then \u2202x \u2202y z \u2202y \u2202z", " x \u2202z \u2202x y = \u22121 Applying this general formula to E, S and N gives us the required result \u2202E \u2202N S,V = \u2212 \u2202S \u2202N E,v \u2202E \u2202S N,V = \u00b5 If we work at constant temperature rather than constant energy, the relevant function is the free energy F (T, V, N ) = E \u2212 T S. Small changes are given by dF = \u2212SdT \u2212 pdV + \u00b5dN from which we see that the chemical potential can also be de\ufb01ned as \u00b5 = \u2202F \u2202N T,V 1.4.1 Grand Canonical Ensemble When we made the transition from microcanonical to canonical ensemble, we were no longer so rigid in our insistence that the system has a \ufb01xed energy. Rather it could freely \u2013 27 \u2013 exchange energy with the surrounding reservoir, which was kept at a \ufb01xed temperature. We could now imagine the same scenario with any other conserved quantity. For example, if particles are free to move between the system and the reservoir, then N is no longer \ufb01xed. In such a situation, we will require that the reservoir sits at \ufb01xed chemical potential \u00b5 as well as \ufb01xed temperature T. The probability distribution that we need to use in this case is called the grand canonical ensemble. The probability of \ufb01nding the system in a state |n depends on both the energy En and the particle number Nn. (Notice that because N is conserved, the quantum mechanical operator necessarily commutes with the Hamiltonian so there is no di\ufb03culty in assigning both energy and particle number to each state). We introduce the grand canonical partition function Z(T, \u00b5, V ) = e\u2212\u03b2(En\u2212\u00b5Nn) n (1.39) Re-running the argument that we used for the canonical ensemble, we \ufb01nd the probability that the system is in state |n to be p(n) = e\u2212\u03b2(En\u2212\u00b5Nn) Z In the canonical ensemble, all the information that we need is contained within the partition function Z. In the grand canonical ensemble it is contained within Z. The entropy (1.30) is once again given by while di\ufb00erentiating with respect to \u03b2 gives us S = kB \u2202 \u2202T (T log Z) E \u2212 \u00b5N = \u2212 \u2202 \u2202\u03b2 log Z (1.40) (1.41) The average particle number N in the system can then be separately extracted by N = 1 \u03b2 \u2202 \u2202\u00b5 log Z (1.42) and its \ufb02uctuations, 1 \u03b22 \u2206N 2 = \u22022 \u2202\u00b52 log Z = Just as the average energy is determined by the temperature in the canonical ensemble, here the average particle number is determined by the chemical potential. The grand canonical ensemble will simplify several calculations later, especially when we come to discuss Bose and Fermi gases in Section 3. \u2202N \u2202\u00b5 (1.43) 1 \u03b2 \u2013 28 \u2013 The relative size of these \ufb02uctuations scales in the same way as the energy \ufb02uctuations, \u2206N/N \u223c 1/N, and in the thermodynamic limit N \u2192 \u221e results from all three ensembles coincide. For this reason, we will drop the averaging brackets \u00b7 from our notation and simply refer to the average particle number as N. 1.4.2 Grand Canonical Potential The grand canonical potential \u03a6 is de\ufb01ned by \u03a6 = F \u2212 \u00b5N \u03a6 is a Legendre transform of F, from variable N to \u00b5. This is underlined if we look at small variations, d\u03a6 = \u2212SdT \u2212 pdV \u2212 N d\u00b5 (1.44) which tells us that \u03a6 should be thought of as a function of temperature, volume and chemical potential, \u03a6 = \u03a6(T, V, \u00b5). We can perform the same algebraic manipulations that gave us F in terms of the canonical partition function Z, this time using the de\ufb01nitions (1.40) and (1.41)) to write \u03a6 as \u03a6 = \u2212kBT log Z (1.45) 1.4.3 Extensive and Intensive Quantities There is one property of \u03a6 that is rather special and, at \ufb01rst glance, somewhat surprising. This property actually follows from very simple considerations of how di\ufb00erent variables change as we look at bigger and bigger systems. Suppose we have a system and we double it. That means that we double the volume V, double the number of particles N and double the energy E. What happens to all our other variables? We", " have already seen back in Section 1.2.1 that entropy is additive, so S also doubles. More generally, if we scale V, N and E by some amount \u03bb, the entropy must scale as S(\u03bbE, \u03bbV, \u03bbN ) = \u03bbS(E, V, N ) Quantities such as E, V, N and S which scale in this manner are called extensive. In contrast, the variables which arise from di\ufb00erentiating the entropy, such as temperature 1/T = \u2202S/\u2202E and pressure p = T \u2202S/\u2202V and chemical potential \u00b5 = T \u2202S/\u2202N involve the ratio of two extensive quantities and so do not change as we scale the system: they are called intensive quantities. \u2013 29 \u2013 What now happens as we make successive Legendre transforms? The free energy F = E \u2212 T S is also extensive (since E and S are extensive while T is intensive). So it must scale as F (T, \u03bbV, \u03bbN ) = \u03bbF (T, V, N ) Similarly, the grand potential \u03a6 = F \u2212 \u00b5N is extensive and scales as \u03a6(T, \u03bbV, \u00b5) = \u03bb\u03a6(T, V, \u00b5) (1.46) (1.47) But there\u2019s something special about this last equation, because \u03a6 only depends on a single extensive variable, namely V. While there are many ways to construct a free energy F which obeys (1.46) (for example, any function of the form F \u223c V n+1/N n will do the job), there is only one way to satisfy (1.47): \u03a6 must be proportional to V. But we\u2019ve already got a name for this proportionality constant: it is pressure. (Actually, it is minus the pressure as you can see from (1.44)). So we have the equation \u03a6(T, V, \u00b5) = \u2212p(T, \u00b5)V (1.48) It looks as if we got something for free! If F is a complicated function of V, where do these complications go after the Legendre transform to \u03a6? The answer is that the complications go into the pressure p(T, \u00b5) when expressed as a function of T and \u00b5. Nonetheless, equation (1.48) will prove to be an extremely economic way to calculate the pressure of various systems. 1.4.4 Josiah Willard Gibbs (1839-1903) \u201cUsually, Gibbs\u2019 prose style conveys his meaning in a su\ufb03ciently clear way, using no more than twice as many words as Poincar\u00b4e or Einstein would have used to say the same thing.\u201d E.T.Jaynes on the di\ufb03culty of reading Gibbs Gibbs was perhaps the \ufb01rst great American theoretical physicist. Many of the developments that we met in this chapter are due to him, including the free energy, the chemical potential and, most importantly, the idea of ensembles. Even the name \u201cstatistical mechanics\u201d was invented by Gibbs. Gibbs provided the \ufb01rst modern rendering of the subject in a treatise published shortly before his death. Very few understood it. Lord Rayleigh wrote to Gibbs suggesting that the book was \u201ctoo condensed and too di\ufb03cult for most, I might say all, readers\u201d. Gibbs disagreed. He wrote back saying the book was only \u201ctoo long\u201d. \u2013 30 \u2013 There do not seem to be many exciting stories about Gibbs. He was an undergraduate at Yale. He did a PhD at Yale. He became a professor at Yale. Apparently he rarely left New Haven. Strangely, he did not receive a salary for the \ufb01rst ten years of his professorship. Only when he received an o\ufb00er from John Hopkins of $3000 dollars a year did Yale think to pay America\u2019s greatest physicist. They made a counter-o\ufb00er of $2000 dollars and Gibbs stayed. \u2013 31 \u2013 2. Classical Gases Our goal in this section is to use the techniques of statistical mechanics to describe the dynamics of the simplest system: a gas. This means a bunch of particles, \ufb02ying around in a box. Although much of the last section was formulated in the language of quantum mechanics, here we will revert back to classical mechanics. Nonetheless, a recurrent theme will be that the quantum world is never far behind: we\u2019ll see several puzzles, both theoretical and experimental, which can only truly be resolved by turning on. 2.1 The Classical Partition Function For most of this section we will work in the canonical ensemble. We start by reformulating the idea of a partition function in classical mechanics. We\u2019ll consider a simple system \u2013 a single particle", " of mass m moving in three dimensions in a potential V (q). The classical Hamiltonian of the system3 is the sum of kinetic and potential energy, H = p2 2m + V (q) We earlier de\ufb01ned the partition function (1.21) to be the sum over all quantum states of the system. Here we want to do something similar. In classical mechanics, the state of a system is determined by a point in phase space. We must specify both the position and momentum of each of the particles \u2014 only then do we have enough information to \ufb01gure out what the system will do for all times in the future. This motivates the de\ufb01nition of the partition function for a single classical particle as the integration over phase space, Z1 = 1 h3 d3qd3p e\u2212\u03b2H(p,q) (2.1) The only slightly odd thing is the factor of 1/h3 that sits out front. It is a quantity that needs to be there simply on dimensional grounds: Z should be dimensionless so h must have dimension (length \u00d7 momentum) or, equivalently, Joules-seconds (Js). The actual value of h won\u2019t matter for any physical observable, like heat capacity, because we always take log Z and then di\ufb00erentiate. Despite this, there is actually a correct value for h: it is Planck\u2019s constant, h = 2\u03c0 \u2248 6.6 \u00d7 10\u221234 Js. It is very strange to see Planck\u2019s constant in a formula that is supposed to be classical. What\u2019s it doing there? In fact, it is a vestigial object, like the male nipple. It is redundant, serving only as a reminder of where we came from. And the classical world came from the quantum. 3If you haven\u2019t taken the Classical Dynamics course, you should think of the Hamiltonian as the energy of the system expressed in terms of the position and momentum of the particle. \u2013 32 \u2013 2.1.1 From Quantum to Classical It is possible to derive the classical partition function (2.1) directly from the quantum partition function (1.21) without resorting to hand-waving. It will also show us why the factor of 1/h sits outside the partition function. The derivation is a little tedious, but worth seeing. (Similar techniques are useful in later courses when you \ufb01rst meet the path integral). To make life easier, let\u2019s consider a single particle moving in one spatial dimension. It has position operator \u02c6q, momentum operator \u02c6p and Hamiltonian, \u02c6H = \u02c6p2 2m + V (\u02c6q) If |n is the energy eigenstate with energy En, the quantum partition function is Z1 = n e\u2212\u03b2En = n n|e\u2212\u03b2 \u02c6H|n (2.2) In what follows, we\u2019ll make liberal use of the fact that we can insert the identity operator anywhere in this expression. Identity operators can be constructed by summing over any complete basis of states. We\u2019ll need two such constructions, using the position eigenvectors |q and the momentum eigenvectors |p, 1 = dq |qq|, 1 = dp |pp| We start by inserting two copies of the identity built from position eigenstates, Z1 = n| dq |qq|e\u2212\u03b2 \u02c6H dq|qq|n n = dqdq q|e\u2212\u03b2 \u02c6H|q q|nn|q n But now we can replace q|q = \u03b4(q \u2212 q), to get n |nn| with the identity matrix and use the fact that Z1 = dq q|e\u2212\u03b2 \u02c6H|q (2.3) We see that the result is to replace the sum over energy eigenstates in (2.2) with a sum (or integral) over position eigenstates in (2.3). If you wanted, you could play the same game and get the sum over any complete basis of eigenstates of your choosing. As an aside, this means that we can write the partition function in a basis independent fashion as Z1 = Tr e\u2212\u03b2 \u02c6H \u2013 33 \u2013 So far, our manipulations could have been done for any quantum system. Now we want to use the fact that we are taking the classical limit. This comes about when we try to factorize e\u2212\u03b2 \u02c6H into a momentum term and a position term. The trouble is that this isn\u2019t always possible when there are matrices (or operators) in the exponent. Recall that, e \u02c6Ae \u02c6B = e \u02c6A+ \u02c6B+ 1", " 2 [ \u02c6A, \u02c6B]+... For us [\u02c6q, \u02c6p] = i. This means that if we\u2019re willing to neglect terms of order \u2014 which is the meaning of taking the classical limit \u2014 then we can write e\u2212\u03b2 \u02c6H = e\u2212\u03b2 \u02c6p2/2m e\u2212\u03b2V (\u02c6q) + O() We can now start to replace some of the operators in the exponent, like V (\u02c6q), with functions V (q). (The notational di\ufb00erence is subtle, but important, in the expressions below!), Z1 = = = = dq q|e\u2212\u03b2 \u02c6p2/2me\u2212\u03b2V (\u02c6q)|q dq e\u2212\u03b2V (q)q|e\u2212\u03b2 \u02c6p2/2m|q dqdpdpe\u2212\u03b2V (q)q|pp|e\u2212\u03b2 \u02c6p2/2m|pp|q 1 2\u03c0 dqdp e\u2212\u03b2H(p,q) where, in the \ufb01nal line, we\u2019ve used the identity q|p = \u221a 1 2\u03c0 eipq/ This completes the derivation. 2.2 Ideal Gas The \ufb01rst classical gas that we\u2019ll consider consists of N particles trapped inside a box of volume V. The gas is \u201cideal\u201d. This simply means that the particles do not interact with each other. For now, we\u2019ll also assume that the particles have no internal structure, so no rotational or vibrational degrees of freedom. This situation is usually referred to as the monatomic ideal gas. The Hamiltonian for each particle is simply the kinetic energy, H = p 2 2m \u2013 34 \u2013 And the partition function for a single particle is Z1(V, T ) = 1 (2\u03c0)3 d3qd3p e\u2212\u03b2p 2/2m (2.4) The integral over position is now trivial and gives d3q = V, the volume of the box. The integral over momentum is also straightforward since it factorizes into separate integrals over px, py and pz, each of which is a Gaussian of the form, So we have dx e\u2212ax2 = \u03c0 a Z1 = V 3/2 mkBT 2\u03c02 We\u2019ll meet the combination of factors in the brackets a lot in what follows, so it is useful to give it a name. We\u2019ll write The quantity \u03bb goes by the name of the thermal de Broglie wavelength, Z1 = V \u03bb3 2\u03c02 mkBT \u03bb = (2.5) (2.6) \u03bb has the dimensions of length. We will see later that you can think of \u03bb as something like the average de Broglie wavelength of a particle at temperature T. Notice that it is a quantum object \u2013 it has an sitting in it \u2013 so we expect that it will drop out of any genuinely classical quantity that we compute. The partition function itself (2.5) is counting the number of these thermal wavelengths that we can \ufb01t into volume V. Z1 is the partition function for a single particle. We have N, non-interacting, particles in the box so the partition function of the whole system is Z(N, V \u03bb3N (2.7) (Full disclosure: there\u2019s a slightly subtle point that we\u2019re brushing under the carpet here and this equation isn\u2019t quite right. This won\u2019t a\ufb00ect our immediate discussion and we\u2019ll explain the issue in more detail in Section 2.2.3.) \u2013 35 \u2013 Figure 8: Deviations from ideal gas law at sensible densities Figure 9: Deviations from ideal gas law at extreme densities Armed with the partition function Z, we can happily calculate anything that we like. Let\u2019s start with the pressure, which can be extracted from the partition function by \ufb01rst computing the free energy (1.36) and then using (1.35). We have p = \u2212 \u2202F \u2202V \u2202 \u2202V N kBT V = = (kBT log Z) (2.8) This equation is an old friend \u2013 it is the ideal gas law, pV = N kBT, that we all met in kindergarten. Notice that the thermal wavelength \u03bb has indeed disappeared from the discussion as expected. Equations of this form, which link pressure, volume and temperature, are called equations of state. We will meet many throughout this course. As the plots above show4, the ideal gas law is an extremely good description of gases at low densities. Gases deviate from this ideal behaviour as the densities increase and the interactions between atoms becomes", " important. We will see how this comes about from the viewpoint of microscopic forces in Section 2.5. It is worth pointing out that this derivation should calm any lingering fears that you had about the de\ufb01nition of temperature given in (1.7). The object that we call T really does coincide with the familiar notion of temperature applied to gases. But the key property of the temperature is that if two systems are in equilibrium then they have the same T. That\u2019s enough to ensure that equation (1.7) is the right de\ufb01nition of temperature for all systems because we can always put any system in equilibrium with an ideal gas. 4Both \ufb01gures are taken from the web textbook \u201cGeneral Chemistry\u201d and credited to John Hutchin- son. \u2013 36 \u2013 2.2.1 Equipartition of Energy The partition function (2.7) has more in store for us. We can compute the average energy of the ideal gas, E = \u2212 \u2202 \u2202\u03b2 log Z = 3 2 N kBT (2.9) There\u2019s an important, general lesson lurking in this formula. To highlight this, it is worth repeating our analysis for an ideal gas in arbitrary number of spatial dimensions, D. A simple generalization of the calculations above shows that Z = V N \u03bbDN \u21d2 E = D 2 N kBT Each particle has D degrees of freedom (because it can move in one of D spatial directions). And each particle contributes 1 2DkBT towards the average energy. This is a general rule of thumb, which holds for all classical systems: the average energy of each free degree of freedom in a system at temperature T is 1 2kBT. This is called the equipartition of energy. As stated, it holds only for degrees of freedom in the absence of a potential. (There is a modi\ufb01ed version if you include a potential). Moreover, it holds only for classical systems or quantum systems at suitably high temperatures. We can use the result above to see why the thermal de Broglie wavelength (2.6) can be thought of as roughly equal to the average de Broglie wavelength of a particle. Equating the average energy (2.9) to the kinetic energy E = p2/2m tells us that the average (root mean square) momentum carried by each particle is p \u223c mkBT. In quantum mechanics, the de Broglie wavelength of a particle is \u03bbdB = h/p, which (up to numerical factors of 2 and \u03c0) agrees with our formula (2.6). \u221a Finally, returning to the reality of d = 3 dimensions, we can compute the heat capacity for a monatomic ideal gas. It is CV = \u2202E \u2202T V = 3 2 N kB (2.10) 2.2.2 The Sociological Meaning of Boltzmann\u2019s Constant We introduced Boltzmann\u2019s constant kB in our original the de\ufb01nition of entropy (1.2). It has the value, kB = 1.381 \u00d7 10\u221223 JK \u22121 In some sense, there is no deep physical meaning to Boltzmann\u2019s constant. It is merely a conversion factor that allows us to go between temperature and energy, as re\ufb02ected \u2013 37 \u2013 in (1.7). It is necessary to include it in the equations only for historical reasons: our ancestors didn\u2019t realise that temperature and energy were closely related and measured them in di\ufb00erent units. Nonetheless, we could ask why does kB have the value above? It doesn\u2019t seem a particularly natural number. The reason is that both the units of temperature (Kelvin) and energy (Joule) are picked to re\ufb02ect the conditions of human life. In the everyday world around us, measurements of temperature and energy involve fairly ordinary numbers: room temperature is roughly 300 K; the energy required to lift an apple back up to the top of the tree is a few Joules. Similarly, in an everyday setting, all the measurable quantities \u2014 p, V and T \u2014 in the ideal gas equation are fairly normal numbers when measured in SI units. The only way this can be true is if the combination N kB is a fairly ordinary number, of order one. In other words the number of atoms must be huge, N \u223c 1023 (2.11) This then is the real meaning of the value of Boltzmann\u2019s constant: atoms are small. It\u2019s worth stressing this point. Atoms aren\u2019t just small: they\u2019re really really small. 1023 is an astonishingly large number. The number of grains of sand in all the beaches in the world is around 1018. The number of stars in our galaxy is about 1011. The number of stars in the entire visible", " Universe is probably around 1022. And yet the number of water molecules in a cup of tea is more than 1023. Chemist Notation While we\u2019re talking about the size of atoms, it is probably worth reminding you of the notation used by chemists. They too want to work with numbers of order one. For this reason, they de\ufb01ne a mole to be the number of atoms in one gram of Hydrogen. (Actually, it is the number of atoms in 12 grams of Carbon-12, but this is roughly the same thing). The mass of Hydrogen is 1.6 \u00d7 10\u221227 Kg, so the number of atoms in a mole is Avogadro\u2019s number, NA \u2248 6 \u00d7 1023 The number of moles in our gas is then n = N/NA and the ideal gas law can be written as pV = nRT where R = NAkB is the called the Universal gas constant. Its value is a nice sensible number with no silly power in the exponent: R \u2248 8 JK \u22121mol\u22121. \u2013 38 \u2013 2.2.3 Entropy and Gibbs\u2019s Paradox \u201cIt has always been believed that Gibbs\u2019s paradox embodied profound thought. That it was intimately linked up with something so important and entirely new could hardly have been foreseen.\u201d Erwin Schr\u00a8odinger We said earlier that the formula for the partition function (2.7) isn\u2019t quite right. What did we miss? We actually missed a subtle point from quantum mechanics: quantum particles are indistinguishable. If we take two identical atoms and swap their positions, this doesn\u2019t give us a new state of the system \u2013 it is the same state that we had before. (Up to a sign that depends on whether the atoms are bosons or fermions \u2013 we\u2019ll discuss this aspect in more detail in Sections 3.5 and 3.6). However, we haven\u2019t taken this into account \u2013 we wrote the expression Z = Z N 1 which would be true if all the N particles in the were distinguishable \u2014 for example, if each of the particles were of a di\ufb00erent type. But this naive partition function overcounts the number of states in the system when we\u2019re dealing with indistinguishable particles. It is a simple matter to write down the partition function for N indistinguishable particles. We simply need to divide by the number of ways to permute the particles. In other words, for the ideal gas the partition function is Zideal(N, V, T ) = 1 N! The extra factor of N! doesn\u2019t change the calculations of pressure or energy since, for each, we had to di\ufb00erentiate log Z and any overall factor drops out. However, it does change the entropy since this is given by, V N N!\u03bb3N (2.12) 1 = Z N S = \u2202 \u2202T (kBT log Zideal) which includes a factor of log Z without any derivative. Of course, since the entropy is counting the number of underlying microstates, we would expect it to know about whether particles are distinguishable or indistinguishable. Using the correct partition function (2.12) and Stirling\u2019s formula, the entropy of an ideal gas is given by, S = N kB log V N \u03bb3 + 5 2 (2.13) This result is known as the Sackur-Tetrode equation. Notice that not only is the entropy sensitive to the indistinguishability of the particles, but it also depends on \u03bb. However, the entropy is not directly measurable classically. We can only measure entropy di\ufb00erences by the integrating the heat capacity as in (1.10). \u2013 39 \u2013 The bene\ufb01t of adding an extra factor of N! was noticed before the advent of quantum mechanics by Gibbs. He was motivated by the change in entropy of mixing between two gases. Suppose that we have two di\ufb00erent gases, say red and blue. Each has the same number of particles N and sits in a volume V, separated by a partition. When the partition is removed the gases mix and we expect the entropy to increase. But if the gases are of the same type, removing the partition shouldn\u2019t change the macroscopic state of the gas. So why should the entropy increase? This is referred to as the Gibb\u2019s paradox. Including the factor of N! in the partition function ensures that the entropy does not increase when identical atoms are mixed5 2.2.4 The Ideal Gas in the Grand Canonical Ensemble It is worth brie\ufb02y looking at the ideal gas in the grand canonical ensemble. Recall that in such an ensemble, the gas is free to exchange both energy and particles with the outside reservoir. You could think of the system as some \ufb01xed", " subvolume inside a much larger gas. If there are no walls to de\ufb01ne this subvolume then particles, and hence energy, can happily move in and out. We can ask how many particles will, on average, be inside this volume and what \ufb02uctuations in particle number will occur. More importantly, we can also start to gain some intuition for this strange quantity called the chemical potential, \u00b5. The grand partition function (1.39) for the ideal gas is Zideal(\u00b5, V, T ) = \u221e N =0 e\u03b2\u00b5N Zideal(N, V, T ) = exp e\u03b2\u00b5V \u03bb3 From this we can determine the average particle number, Which, rearranging, gives N = 1 \u03b2 \u2202 \u2202\u00b5 log Z = e\u03b2\u00b5V \u03bb3 \u00b5 = kBT log \u03bb3N V (2.14) If \u03bb3 < V /N then the chemical potential is negative. Recall that \u03bb is roughly the average de Broglie wavelength of each particle, while V /N is the average volume taken 5Be warned however: a closer look shows that the Gibbs paradox is rather toothless and, in the classical world, there is no real necessity to add the N!. A clear discussion of these issues can be found in E.T. Jaynes\u2019 article \u201cThe Gibbs Paradox\u201d which you can download from the course website. \u2013 40 \u2013 up by each particle. But whenever the de Broglie wavelength of particles becomes comparable to the inter-particle separation, then quantum e\ufb00ects become important. In other words, to trust our classical calculation of the ideal gas, we must have \u03bb3 V /N and, correspondingly, \u00b5 < 0. At \ufb01rst sight, it is slightly strange that \u00b5 is negative. When we introduced \u00b5 in Section 1.4.1, we said that it should be thought of as the energy cost of adding an extra particle to the system. Surely that energy should be positive! To see why this isn\u2019t the case, we should look more closely at the de\ufb01nition. From the energy variation (1.38), we have \u00b5 = \u2202E \u2202N S,V So the chemical potential should be thought of as the energy cost of adding an extra particle at \ufb01xed entropy and volume. But adding a particle will give more ways to share the energy around and so increase the entropy. If we insist on keeping the entropy \ufb01xed, then we will need to reduce the energy when we add an extra particle. This is why we have \u00b5 < 0 for the classical ideal gas. There are situations where \u00b5 > 0. This can occur if we have a suitably strong repulsive interaction between particles so that there\u2019s a large energy cost associated to throwing in one extra. We also have \u00b5 > 0 for fermion systems at low temperatures as we will see in Section 3.6. We can also compute the \ufb02uctuation in the particle number, \u2206N 2 = 1 \u03b22 \u22022 \u2202\u00b52 log Zideal = N As promised in Section 1.4.1, the relative \ufb02uctuations \u2206N/N = 1/ small in the thermodynamic N \u2192 \u221e limit. \u221a N are vanishingly Finally, it is very easy to compute the equation of state in the grand canonical ensemble because (1.45) and (1.48) tell us that pV = kBT log Z = kBT e\u03b2\u00b5V \u03bb3 = kBT N (2.15) which gives us back the ideal gas law. \u2013 41 \u2013 Figure 10: Maxwell distribution for Noble gases: He, N e, Ar and Xe. 2.3 Maxwell Distribution Our discussion above focusses on understanding macroscopic properties of the gas such as pressure or heat capacity. But we can also use the methods of statistical mechanics to get a better handle on the microscopic properties of the gas. Like everything else, the information is hidden in the partition function. Let\u2019s return to the form of the single particle partition function (2.4) before we do the integrals. We\u2019ll still do the trivial spatial integral d3q = V, but we\u2019ll hold o\ufb00 on the momentum integral and instead change variables from momentum to velocity, p = mv. Then the single particle partition function is Z1 = m3V (2\u03c0)3 d3v e\u2212\u03b2mv2/2 = 4\u03c0m3V (2\u03c0)3 dv v2e\u2212\u03b2mv2/2 We can compare this to the original de\ufb01nition of the partition function: the sum over states of the probability of that state. But here too, the partition function is", " written as a sum, now over speeds. The integrand must therefore have the interpretation as the probability distribution over speeds. The probability that the atom has speed between v and v + dv is f (v)dv = N v2e\u2212mv2/2kBT dv (2.16) where the normalization factor N can be determined by insisting that probabilities sum to one, \u221e 0 f (v) dv = 1, which gives N = 4\u03c0 m 3/2 2\u03c0kBT \u2013 42 \u2013 This is the Maxwell distribution. It is sometimes called the Maxwell-Boltzmann distribution. Figure 10 shows this distribution for a variety of gases with di\ufb00erent masses at the same temperature, from the slow heavy Xenon (purple) to light, fast Helium (blue). We can use it to determine various average properties of the speeds of atoms in a gas. For example, the mean square speed is v2 = \u221e 0 dv v2f (v) = 3kBT m This is in agreement with the equipartition of energy: the average kinetic energy of the gas is E = 1 2mv2 = 3 2kBT. Maxwell\u2019s Argument The above derivation tells us the distribution of velocities in a non-interacting gas of particles. Remarkably, the Maxwell distribution also holds in the presence of any interactions. In fact, Maxwell\u2019s original derivation of the distribution makes no reference to any properties of the gas. It is very slick! Let\u2019s \ufb01rst think about the distribution of velocities in the x direction; we\u2019ll call this distribution \u03c6(vx). Rotational symmetry means that we must have the same distribution of velocities in both the y and z directions. However, rotational invariance also requires that the full distribution can\u2019t depend on the direction of the velocity; it can only depend on the speed v = v2 x + v2 z. This means that we need to \ufb01nd functions F (v) and \u03c6(vx) such that y + v2 F (v) dvxdvydvz = \u03c6(vx)\u03c6(vy)\u03c6(vz) dvxdvydvz It doesn\u2019t look as if we possibly have enough information to solve this equation for both F and \u03c6. But, remarkably, there is only one solution. The only function which satis\ufb01es this equation is \u03c6(vx) = Ae\u2212Bv2 x for some constants A and B. Thus the distribution over speeds must be F (v) dvxdvydvz = 4\u03c0v2F (v) dv = 4\u03c0A3v2e\u2212Bv2dv We see that the functional form of the distribution arises from rotational invariance alone. To determine the coe\ufb03cient B = m/2kBT we need the more elaborate techniques of statistical mechanics that we saw above. (In fact, one can derive it just from equipartition of energy). \u2013 43 \u2013 2.3.1 A History of Kinetic Theory The name kinetic theory refers to the understanding the properties of gases through their underlying atomic constituents. The discussion given above barely scratches the surface of this important subject. Kinetic theory traces its origin to the work of Daniel Bernoulli in 1738. He was the \ufb01rst to argue that the phenomenon that we call pressure is due to the constant bombardment of tiny atoms. His calculation is straightforward. Consider a cubic box with sides of length L. Suppose that an atom travelling with momentum vx in the x direction bounces elastically o\ufb00 a wall so that it returns with velocity \u2212vx. The particle experiences a change in momentum is \u2206px = 2mvx. Since the particle is trapped in a box, it will next hit the wall at a time \u2206t = 2L/vx later. This means that the force on the wall due to this atom is Summing over all the atoms which hit the wall, the force is F = \u2206px \u2206t = mv2 x L F = N mv2 x L where v2 gave in Maxwell\u2019s derivation above, we must have v2 and the pressure, which is force per area, is given be x is the average velocity in the x-direction. Using the same argument as we x = v2/3. Thus F = N mv2/3L p = N mv2 3L3 = N mv2 3V If this equation is compared to the ideal gas law (which, at the time, had only experimental basis) one concludes that the phenomenon of temperature must arise from the kinetic energy of the gas. Or, more precisely, one \ufb01nds the equipart", "ition result that we derived previously: 1 2mv2 = 3 2kBT. After Bernoulli\u2019s pioneering work, kinetic theory languished. No one really knew what to do with his observation nor how to test the underlying atomic hypothesis. Over the next century, Bernouilli\u2019s result was independently rediscovered by a number of people, all of whom were ignored by the scienti\ufb01c community. One of the more interesting attempts was by John Waterson, a Scottish engineer and naval instructor working for the East India Company in Bombay. Waterson was considered a crackpot. His 1843 paper was rejected by the Royal Society as \u201cnothing but nonsense\u201d and he wrote up his results in a self-published book with the wonderfully crackpot title \u201cThoughts on Mental Functions\u201d. \u2013 44 \u2013 The results of Bernouilli and Waterson \ufb01nally became accepted only after they were re-rediscovered by more established scientists, most notably Rudolph Clausius who, in 1857, extended these ideas to rotating and vibrating molecules. Soon afterwards, in 1859, Maxwell gave the derivation of the distribution of velocities that we saw above. This is often cited as the \ufb01rst statistical law of physics. But Maxwell was able to take things further. He used kinetic theory to derive the \ufb01rst genuinely new prediction of the atomic hypothesis: that the viscosity of a gas is independent of its density. Maxwell himself wrote, \u201dSuch a consequence of the mathematical theory is very startling and the only experiment I have met with on the subject does not seem to con\ufb01rm it.\u201d Maxwell decided to rectify the situation. With help from his wife, he spent several years constructing an experimental apparatus in his attic which was capable of providing the \ufb01rst accurate measurements of viscosity of gases6. His surprising theoretical prediction was con\ufb01rmed by his own experiment. There are many further developments in kinetic theory which we will not cover in this course. Perhaps the most important is the Boltzmann equation. This describes the evolution of a particle\u2019s probability distribution in position and momentum space as it collides with other particles. Stationary, unchanging, solutions bring you back to the Maxwell-Boltzmann distribution, but the equation also provides a framework to go beyond the equilibrium description of a gas. You can read about this in the lecture notes on Kinetic Theory. 2.4 Diatomic Gas \u201cI must now say something about these internal motions, because the greatest di\ufb03culty which the kinetic theory of gases has yet encountered belongs to this part of the subject\u201d. James Clerk Maxwell, 1875 Consider a molecule that consists of two atoms in a bound state. We\u2019ll construct a very simple physicist\u2019s model of this molecule: two masses attached to a spring. As well as the translational degrees of freedom, there are two further ways in which the molecule can move 6You can see the original apparatus down the road in the corridor of the Cavendish lab. Or, if you don\u2019t fancy the walk, you can simply click here: http://www-outreach.phy.cam.ac.uk/camphy/museum/area1/exhibit1.htm \u2013 45 \u2013 \u2022 Rotation: the molecule can rotate rigidly about the two axes perpendicular to the axis of symmetry, with moment of inertia I. (For now, we will neglect the rotation about the axis of symmetry. It has very low moment of inertia which will ultimately mean that it is unimportant). \u2022 Vibration: the molecule can oscillate along the axis of symmetry We\u2019ll work under the assumption that the rotation and vibration modes are independent. In this case, the partition function for a single molecule factorises into the product of the translation partition function Ztrans that we have already calculated (2.5) and the rotational and vibrational contributions, Z1 = ZtransZrotZvib We will now deal with Zrot and Zvib in turn. Rotation The Lagrangian for the rotational degrees of freedom is7 Lrot = 1 2 I( \u02d9\u03b82 + sin2 \u03b8 \u02d9\u03c62) The conjugate momenta are therefore p\u03b8 = \u2202Lrot \u2202 \u02d9\u03b8 = I \u02d9\u03b8, p\u03c6 = \u2202Lrot \u2202 \u02d9\u03c6 = I sin2 \u03b8 \u02d9\u03c6 from which we get the Hamiltonian for the rotating diatomic molecule, Hrot = \u02d9\u03b8p\u03b8 + \u02d9\u03c6p\u03c6 \u2212 L = p2 \u03b8 2I + p2 \u03c6 2I sin2 \u03b8 The rotational contribution to the partition function is then Zrot = = = 1 (2\u03c0)2 1 (", "2\u03c0)2 2IkBT 2 d\u03b8d\u03c6dp\u03b8dp\u03c6 e\u2212\u03b2Hrot 2\u03c0I \u03b2 \u03c0 0 d\u03b8 2\u03c0I sin2 \u03b8 \u03b2 2\u03c0 0 d\u03c6 7See, for example, Section 3.6 of the lecture notes on Classical Dynamics \u2013 46 \u2013 (2.17) (2.18) (2.19) From this we can compute the average rotational energy of each molecule, Erot = kBT If we now include the translational contribution (2.5), the partition function for a diatomic molecule that can spin and move, but can\u2019t vibrate, is given by Z1 = ZtransZrot \u223c (kBT )5/2, and the partition function for a gas of these object Z = Z N 1 /N!, from which we compute the energy E = 5 2N kBT and the heat capacity, CV = 5 2 kBN In fact we can derive this result simply from equipartition of energy: there are 3 translational modes and 2 rotational modes, giving a contribution of 5N \u00d7 1 2kBT to the energy. Vibrations The Hamiltonian for the vibrating mode is simply a harmonic oscillator. We\u2019ll denote the displacement away from the equilibrium position by \u03b6. The molecule vibrates with some frequency \u03c9 which is determined by the strength of the atomic bond. The Hamiltonian is then Hvib = p2 \u03b6 2m + 1 2 m\u03c92\u03b6 2 from which we can compute the partition function Zvib = 1 2\u03c0 d\u03b6dp\u03b6e\u2212\u03b2Hvib = kBT \u03c9 (2.20) The average vibrational energy of each molecule is now Evib = kBT (You may have anticipated 1 2kBT since the harmonic oscillator has just a single degree of freedom, but equipartition works slightly di\ufb00erently when there is a potential energy. You will see another example on the problem sheet from which it is simple to deduce the general form). Putting together all the ingredients, the contributions from translational motion, rotation and vibration give the heat capacity CV = 7 2 N kB \u2013 47 \u2013 Figure 11: The heat capacity of Hydrogen gas H2. The graph was created by P. Eyland. This result depends on neither the moment of inertia, I, nor the sti\ufb00ness of the molecular bond, \u03c9. A molecule with large I will simply spin more slowly so that the average rotational kinetic energy is kBT ; a molecule attached by a sti\ufb00 spring with high \u03c9 will vibrate with smaller amplitude so that the average vibrational energy is kBT. This ensures that the heat capacity is constant. Great! So the heat capacity of a diatomic gas is 7 2N kB. Except it\u2019s not! An idealised graph of the heat capacity for H2, the simplest diatomic gas, is shown in Figure 11. At suitably high temperatures, around 5000K, we do see the full heat capacity that we expect. But at low temperatures, the heat capacity is that of monatomic gas. And, in the middle, it seems to rotate, but not vibrate. What\u2019s going on? Towards the end of the nineteenth century, scientists were increasingly bewildered about this behaviour. What\u2019s missing in the discussion above is something very important:. The successive freezing out of vibrational and rotational modes as the temperature is lowered is a quantum e\ufb00ect. In fact, this behaviour of the heat capacities of gases was the \ufb01rst time that quantum mechanics revealed itself in experiment. We\u2019re used to thinking of quantum mechanics as being relevant on small scales, yet here we see that a\ufb00ects the physics of gases at temperatures of 2000 K. But then, that is the theme of this course: how the microscopic determines the macroscopic. We will return to the diatomic gas in Section 3.4 and understand its heat capacity including the relevant quantum e\ufb00ects. 2.5 Interacting Gas Until now, we\u2019ve only discussed free systems; particles moving around unaware of each other. Now we\u2019re going to turn on interactions. Here things get much more interesting. \u2013 48 \u2013 And much more di\ufb03cult. Many of the most important unsolved problems in physics are to do with the interactions between large number of particles. Here we\u2019ll be gentle. We\u2019ll describe a simple approximation scheme that will allow us to begin to understand the e\ufb00ects of interactions between particles. We\u2019ll focus once more on the monatomic gas. The ideal gas law is exact in the limit of no interactions between atoms. This is a good approximation when the density of atoms N/V is small. Corrections to the ideal gas law are", " often expressed in terms of a density expansion, known as the virial expansion. The most general equation of state is, p kBT = N V + B2(T ) N 2 V 2 + B3(2.21) where the functions Bj(T ) are known as virial coe\ufb03cients. Our goal is to compute the virial coe\ufb03cients from \ufb01rst principles, starting from a knowledge of the underlying potential energy U (r) between two neutral atoms separated by a distance r. This potential has two important features: \u2022 An attractive 1/r6 force. This arises from \ufb02uctuating dipoles of the neutral atoms. Recall that two permanent dipole moments, p1 and p2, have a potential energy which scales as p1p2/r3. Neutral atoms don\u2019t have permanent dipoles, but they can acquire a temporary dipole due to quantum \ufb02uctuations. Suppose that the \ufb01rst atom has an instantaneous dipole p1. This will induce an electric \ufb01eld which is proportional to E \u223c p1/r3 which, in turn, will induce a dipole of the second atom p2 \u223c E \u223c p1/r3. The resulting potential energy between the atoms scales as p1p2/r3 \u223c 1/r6. This is sometimes called the van der Waals interaction. \u2022 A rapidly rising repulsive interaction at short distances, arising from the Pauli exclusion principle that prevents two atoms from occupying the same space. For our purposes, the exact form of this repulsion is not so relevant: just as long as it\u2019s big. (The Pauli exclusion principle is a quantum e\ufb00ect. If the exact form of the potential is important then we really need to be dealing with quantum mechanics all along. We will do this in the next section). One very common potential that is often used to model the force between atoms is the Lennard-Jones potential, U (r) \u223c 12 r0 r 6 \u2212 r0 r (2.22) The exponent 12 is chosen only for convenience: it simpli\ufb01es certain calculations because 12 = 2 \u00d7 6. \u2013 49 \u2013 An even simpler form of the potential incorporates a hard core repulsion, in which the particles are simply forbidden from closer than a \ufb01xed distance by imposing an in\ufb01nite potential, U (r) = \u221e \u2212U0 6 r0 r r < r0 r \u2265 r0 (2.23) The hard-core potential with van der Waals attraction is sketched to the right. We will see shortly that the virial coe\ufb03cients are determined by increasingly dif\ufb01cult integrals involving the potential U (r). For this reason, it\u2019s best to work with a potential that\u2019s as simple as possible. When we come to do some actual calculations we will use the form (2.23). Figure 12: 2.5.1 The Mayer f Function and the Second Virial Coe\ufb03cient We\u2019re going to change notation and call the positions of the particles r instead of q. (The latter notation was useful to stress the connection to quantum mechanics at the beginning of this Section, but we\u2019ve now left that behind!). The Hamiltonian of the gas is H = N i=1 p2 i 2m + U (rij) i>j where rij = |ri \u2212 rj| is the separation between particles. The restriction i > j on the \ufb01nal sum ensures that we sum over each pair of particles exactly once. The partition function is then Z(N, V, T ) = 1 N! 1 (2\u03c0)3N N d3pid3ri e\u2212\u03b2H i=1 d3pi e\u2212\u03b2 j p2 j /2m i d3ri e\u2212\u03b2 j<k U (rjk) = = 1 N! 1 (2\u03c0)3N 1 N!\u03bb3N i \u00d7 i d3ri e\u2212\u03b2 j<k U (rjk) where \u03bb is the thermal wavelength that we met in (2.6). We still need to do the integral over positions. And that looks hard! The interactions mean that the integrals don\u2019t \u2013 50 \u2013 U(r)rr0 factor in any obvious way. What to do? One obvious way thing to try is to Taylor expand (which is closely related to the so-called cumulant expansion in this context) e\u2212\u03b2 j<k U (rjk) = 1 \u2212 \u03b2 j<k U (rjk) + \u03b22 2 j<k,l<m U (rjk)U (rlm) +... Unfortunately, this isn\u2019t so useful. We want each", " term to be smaller than the preceding one. But as rij \u2192 0, the potential U (rij) \u2192 \u221e, which doesn\u2019t look promising for an expansion parameter. Instead of proceeding with the naive Taylor expansion, we will instead choose to work with the following quantity, usually called the Mayer f function, f (r) = e\u2212\u03b2U (r) \u2212 1 (2.24) This is a nicer expansion parameter. When the particles are far separated at r \u2192 \u221e, f (r) \u2192 0. However, as the particles come close and r \u2192 0, the Mayer function approaches f (r) \u2192 \u22121. We\u2019ll proceed by trying to construct a suitable expansion in terms of f. We de\ufb01ne Then we can write the partition function as fij = f (rij) Z(N, V, T ) = 1 N!\u03bb3N = 1 N!\u03bb3N i i d3ri (1 + fjk) j>k d3ri 1 + j>k fjk + j>k,l>m fjkflm +... (2.25) The \ufb01rst term simply gives a factor of the volume V for each integral, so we get V N. The second term has a sum, each element of which is the same. They all look like d3ri f12 = V N \u22122 N i=1 d3r1d3r2 f (r12) = V N \u22121 d3r f (r) where, in the last equality, we\u2019ve simply changed integration variables from r1 and r2 to the centre of mass R = 1 2(r1 + r2) and the separation r = r1 \u2212 r2. (You might worry that the limits of integration change in the integral over r, but the integral over f (r) only picks up contributions from atomic size distances and this is only actually a problem close to the boundaries of the system where it is negligible). There is a term \u2013 51 \u2013 like this for each pair of particles \u2013 that is 1 can just call this a round 1 partition function is approximately 2N (N \u2212 1) such terms. For N \u223c 1023, we 2N 2. Then, ignoring terms quadratic in f and higher, the Z(N, V, T ) = V N N!\u03bb3N 1 + = Zideal 1 + N 2 2V d3r f (r) +... d3r f (r) +... N N 2V where we\u2019ve used our previous result that Zideal = V N /N!\u03bb3N. We\u2019ve also engaged in something of a sleight of hand in this last line, promoting one power of N from in front of the integral to an overall exponent. Massaging the expression in this way ensures that the free energy is proportional to the number of particles as one would expect: F = \u2212kBT log Z = Fideal \u2212 N kBT log 1 + N 2V d3r f (r) (2.26) However, if you\u2019re uncomfortable with this little trick, it\u2019s not hard to convince yourself that the result (2.27) below for the equation of state doesn\u2019t depend on it. We will also look at the expansion more closely in the following section and see how all the higher order terms work out. From the expression (2.26) for the free energy, it is clear that we are indeed performing an expansion in density of the gas since the correction term is proportional to N/V. This form of the free energy will give us the second virial coe\ufb03cient B2(T ). We can be somewhat more precise about what it means to be at low density. The exact form of the integral d3rf (r) depends on the potential, but for both the LennardJones potential (2.22) and the hard-core repulsion (2.23), the integral is approximately d3rf (r) \u223c r3 0, where r0 is roughly the minimum of the potential. (We\u2019ll compute the integral exactly below for the hard-core potential). For the expansion to be valid, we want each term with an extra power of f to be smaller than the preceding one. (This statement is actually only approximately true. We\u2019ll be more precise below when we develop the cluster expansion). That means that the second term in the argument of the log should be smaller than 1. In other words, N V 1 r3 0 The left-hand side is the density of the gas. The right-hand side is atomic density. Or, equivalently, the density of a substance in which the atoms are packed closely together. But we have a name for such substances \u2013 we call them liquids! Our expansion", " is valid for densities of the gas that are much lower than that of the liquid state. \u2013 52 \u2013 2.5.2 van der Waals Equation of State We can use the free energy (2.26) to compute the pressure of the gas. Expanding the logarithm as log(1 + x) \u2248 x we get p = \u2212 \u2202F \u2202V = N kBT V 1 \u2212 N 2V d3rf (r) +... As expected, the pressure deviates from that of an ideal gas. We can characterize this by writing pV N kBT = 1 \u2212 N 2V d3r f (r) (2.27) To understand what this is telling us, we need to compute d3rf (r). Firstly let\u2019s look at two trivial examples: Repulsion: Suppose that U (r) > 0 for all separations r with U (r = \u221e) = 0. Then f = e\u2212\u03b2U \u2212 1 < 0 and the pressure increases, as we\u2019d expect for a repulsive interaction. Attraction: If U (r) < 0, we have f > 0 and the pressure decreases, as we\u2019d expect for an attractive interaction. What about a more realistic interaction that is attractive at long distances and repulsive at short? We will compute the equation of state of a gas using the hard-core potential with van der Waals attraction (2.23). The integral of the Mayer f function is d3r f (r) = r0 0 d3r(\u22121) + \u221e r0 d3r (e+\u03b2U0(r0/r)6 \u2212 1) (2.28) We\u2019ll approximate the second integral in the high temperature limit, \u03b2U0 1, where e+\u03b2U0(r0/r)6 \u2248 1 + \u03b2U0(r0/r)6. Then d3r f (r) = \u22124\u03c0 r0 dr r2 + 4\u03c0U0 kBT \u221e r0 dr r6 0 r4 (2.29) = 4\u03c0r3 0 3 0 U0 kBT \u2212 1 Inserting this into (2.27) gives us an expression for the equation of state, pV N kBT = 1 \u2212 N V a kBT \u2212 b \u2013 53 \u2013 We recognise this expansion as capturing the second virial coe\ufb03cient in (2.21) as promised. The constants a and b are de\ufb01ned by a = 2\u03c0r3 0U0 3, b = 2\u03c0r3 0 3 It is actually slightly more useful to write this in the form kBT =.... We can multiply through by kBT then, rearranging we have kBT = 1 Since we\u2019re working in an expansion in density, N/V, we\u2019re at liberty to Taylor expand the last bracket, keeping only the \ufb01rst two terms. We get kBT = 2.30) This is the famous van der Waals equation of state for a gas. We stress again the limitations of our analysis: it is valid only at low densities and (because of our approximation when performing the integral (2.28)) at high temperatures. We will return to the van der Waals equation in Section 5 where we\u2019ll explore many of its interesting features. For now, we can get a feeling for the physics behind this equation of state by rewriting it in yet another way, p = N kBT V \u2212 bN \u2212 a N 2 V 2 (2.31) The constant a contains a factor of U0 and so capures the e\ufb00ect of the attractive interaction at large distances. We see that its role is to reduce the pressure of the gas. The reduction in pressure is proportional to the density squared because this is, in turn, proportional to the number of pairs of particles which feel the attractive force. In contrast, b only contains r0 and arises due to the hard-core repulsion in the potential. Its e\ufb00ect is the reduce the e\ufb00ective volume of the gas because of the space taken up by the particles. Figure 13: It is worth pointing out where some quizzical factors of two come from in b = 2\u03c0r3 0/3. Recall that r0 is the minimum distance that two atoms can approach. If we think of the each atom as a hard sphere, then they have radius r0/2 and volume 4\u03c0(r0/2)3/3. Which isn\u2019t equal to b. However, as illustrated in the \ufb01gure, the excluded volume around each \u2013 54 \u2013 r0r0/2 atom is actually \u2126 = 4\u03c0r3 0/3 = 2b. So why don\u2019t we have \u2126 sitting", " in the denominator of the van der Waals equation rather than b = \u2126/2? Think about adding the atoms one at a time. The \ufb01rst guy can move in volume V ; the second in volume V \u2212 \u2126; the third in volume V \u2212 2\u2126 and so on. For \u2126 V, the total con\ufb01guration space available to the atoms is 1 N! N m=1 (V \u2212 m\u2126 And there\u2019s that tricky factor of 1/2. Above we computed the equation of state for the dipole van der Waals interaction with hard core potential. But our expression (2.27) can seemingly be used to compute the equation of state for any potential between atoms. However, there are limitations. Looking back to the integral (2.29), we see that a long-range force of the form 1/rn will only give rise to a convergent integral for n \u2265 4. This means that the techniques described above do not work for long-range potentials with fall-o\ufb00 1/r3 or slower. This includes the important case of 1/r Coulomb interactions. 2.5.3 The Cluster Expansion Above we computed the leading order correction to the ideal gas law. In terms of the virial expansion (2.21) this corresponds to the second virial coe\ufb03cient B2. We will now develop the full expansion and explain how to compute the higher virial coe\ufb03cients. Let\u2019s go back to equation (2.25) where we \ufb01rst expressed the partition function in terms of f, Z(N, V, T ) = 1 N!\u03bb3N = 1 N!\u03bb3N i i d3ri (1 + fjk) j>k d3ri 1 + j>k fjk + j>k,l>m fjkflm +... (2.32) Above we e\ufb00ectively related the second virial coe\ufb03cient to the term linear in f : this is the essence of the equation of state (2.27). One might think that terms quadratic in f give rise to the third virial coe\ufb03cient and so on. But, as we\u2019ll now see, the expansion is somewhat more subtle than that. \u2013 55 \u2013 The expansion in (2.32) includes terms of the form fijfklfmn... where the indices denote pairs of atoms, (i, j) and (k, l) and so on. These pairs may have atoms in common or they may all be di\ufb00erent. However, the same pair never appears twice in a given term as you may check by going back to the \ufb01rst line in (2.32). We\u2019ll introduce a diagrammatic method to keep track of all the terms in the sum. To each term of the form fijfklfmn... we associate a picture using the following rules \u2022 Draw N atoms. (This gets tedious for N \u223c 1023 but, as we\u2019ll soon see, we will actually only need pictures with small subset of atoms). \u2022 Draw a line between each pair of atoms that appear as indices. So for fijfklfmn..., we draw a line between atom i and atom j; a line between atom k and atom l; and so on. For example, if we have just N = 4, we have the following pictures for di\ufb00erent terms in the expansion, f12 = f12f34 = f12f23 = f21f23f31 = We call these diagrams graphs. Each possible graph appears exactly once in the partition function (2.32). In other words, the partition function is a sum over all graphs. We still have to do the integrals over all positions ri. We will denote the integral over graph G to be W [G]. Then the partition function is Z(N, V, T ) = 1 N!\u03bb3N G W [G] Nearly all the graphs that we can draw will have disconnected components. For example, those graphs that correspond to just a single fij will have two atoms connected and the remaining N \u2212 2 sitting alone. Those graphs that correspond to fijfkl fall into two categories: either they consist of two pairs of atoms (like the second example above) or, if (i, j) shares an atom with (k, l), there are three linked atoms (like the third example above). Importantly, the integral over positions ri then factorises into a product of integrals over the positions of atoms in disconnected components. This is illustrated by an example with N = 5 atoms, W = d3r1d3r2d3r3f12f23f", "31 d3r4d3r5f45 \u2013 56 \u2013 123412341342134213452 We call the disconnected components of the graph clusters. If a cluster has l atoms, we will call it an l-cluster. The N = 5 example above has a single 3-cluster and a single 2-cluster. In general, a graph G will split into ml l-clusters. Clearly, we must have N l=1 mll = N (2.33) Of course, for a graph with only a few lines and lots of atoms, nearly all the atoms will be in lonely 1-clusters. We can now make good on the promise above that we won\u2019t have to draw all N \u223c 1023 atoms. The key idea is that we can focus on clusters of l-atoms. We will organise the expansion in such a way that the (l + 1)-clusters are less important than the l-clusters. To see how this works, let\u2019s focus on 3-clusters for now. There are four di\ufb00erent ways that we can have a 3-cluster, Each of these 3-clusters will appear in a graph with any other combination of clusters among the remaining N \u22123 atoms. But since clusters factorise in the partition function, we know that Z must include a factor U3 \u2261 d3r1d3r2d3r3 U3 contains terms of order f 2 and f 3. It turns out that this is the correct way to arrange the expansion: not in terms of the number of lines in the diagram, which is equal to the power of f, but instead in terms of the number of atoms that they connect. The partition function will similarly contain factors associated to all other l-clusters. We de\ufb01ne the corresponding integrals as l d3ri Ul \u2261 G i=1 G\u2208{l-cluster} (2.34) Notice that U1 is simply the integral over space, namely U1 = V. The full partition function must be a product of Ul\u2019s. The tricky part is to get all the combinatoric factors right to make sure that you count each graph exactly once. The sum over graphs G that appears in the partition function turns out to be W [G] = N! G {ml} l U ml l (l!)mlml! (2.35) \u2013 57 \u2013 132132132132132132132132+++ The product N!/ l ml!(l!)ml counts the number of ways to split the particles into ml l-clusters, while ignoring the di\ufb00erent ways to internally connect each cluster. This is the right thing to do since the di\ufb00erent internal connections are taken into account in the integral Ul. Combinatoric arguments are not always transparent. Let\u2019s do a couple of checks to make sure that this is indeed the right answer. Firstly, consider N = 4 atoms split into two 2-clusters (i.e m2 = 2). There are three such diagrams, f12f34 =, 2 so and f14f23 = the \ufb01nal result should be 3U 2 2. We can check this against the relevant terms in (2.35) which are 4!U 2. Each of these gives the same answer when integrated, namely U 2, f13f24 = 2 /2!22! = 3U 2 2 as expected. Another check: N = 5 atoms with m2 = m3 = 1. All diagrams come in the combinations U3U2 = 5 d3ri i=1 together with graphs that are related by permutations. The permutations are fully determined by the choice of the two atoms that sit in the pair: there are 10 such choices. The answer should therefore be 10U3U2. Comparing to (2.35), we have 5!U3U2/3!2! = 10U3U2 as required. Hopefully you are now convinced that (2.35) counts the graphs correctly. The end result for the partition function is therefore Z(N, V, T ) = 1 \u03bb3N {ml} l U ml l (l!)mlml! The problem with computing this sum is that we still have to work out the di\ufb00erent ways that we can split N atoms into di\ufb00erent clusters. In other words, we still have to obey the constraint (2.33). Life would be very much easier if we didn\u2019t have to worry about this. Then we could just sum over any ml, regardless. Thankfully, this is exactly what we can do if we work in the grand canonical ensemble where N is not \ufb01xed! The grand canonical ensemble is Z(\u00b5, V, T ) = N e\u03b2\u00b5N Z(N, V", ", T ) We de\ufb01ne the fugacity as z = e\u03b2\u00b5. Then we can write Z(\u00b5, V, T ) = N znZ(N, V, T ) = \u221e \u221e ml=0 l=1 z \u03bb3 mll 1 ml! Ul l! ml = \u221e l=1 exp Ulzl \u03bb3ll! \u2013 58 \u2013 +++ One usually de\ufb01nes bl = \u03bb3 V Ul l!\u03bb3l (2.36) Notice in particular that U1 = V so this de\ufb01nition gives b1 = 1. Then we can write the grand partition function as Z(\u00b5, V, T ) = \u221e l=1 exp V \u03bb3 blzl = exp V \u03bb3 \u221e l=1 blzl (2.37) Something rather cute happened here. The sum over all diagrams got rewritten as the exponential over the sum of all connected diagrams, meaning all clusters. This is a general lesson which also carries over to quantum \ufb01eld theory where the diagrams in question are Feynman diagrams. Back to the main plot of our story, we can now compute the pressure pV kBT = log Z = V \u03bb3 \u221e l=1 blzl and the number of particles N V = z V \u2202 \u2202z log Z = 1 \u03bb3 \u221e l=1 lblzl Dividing the two gives us the equation of state, pV N kBT = l blzl l lblzl (2.38) (2.39) The only downside is that the equation of state is expressed in terms of z. To massage it into the form of the virial expansion (2.21), we need to invert (2.38) to get z in terms of the particle density N/V. Equating (2.39) with (2.21) (and de\ufb01ning B1 = 1), we have \u221e l=1 blzl = = \u221e l=1 \u221e l=1 N V Bl l\u22121 \u221e m=1 mbmzm Bl \u03bb3(l\u22121) \u221e n=1 nbnzn l\u22121 \u221e m=1 mbmzm = 1 + B2 \u03bb3 (z + 2b2z2 + 3b3z3 +...) + B3 \u03bb6 (z + 2b2z2 + 3b3z3 +...)2 +... \u00d7 z + 2b2z2 + 3b3z3 +... \u2013 59 \u2013 where we\u2019ve used both B1 = 1 and b1 = 1. Expanding out the left- and right-hand sides to order z3 gives z + b2z2 + b3z3 +... = z + B2 \u03bb3 + 2b2 z2 + 3b3 + 4b2B2 \u03bb3 + B3 \u03bb3 z3 +... Comparing terms, and recollecting the de\ufb01nitions of bl (2.36) in terms of Ul (2.34) in terms of graphs, we \ufb01nd the second virial coe\ufb03cient is given by B2 = \u2212\u03bb3b2 = \u2212 U2 2V = \u2212 1 2V d3r1d3r2f (r1 \u2212 r2) = \u2212 1 2 d3rf (r) which reproduces the result (2.27) that we found earlier using slightly simpler methods. We now also have an expression for the third coe\ufb03cient, B3 = \u03bb6(4b2 2 \u2212 2b3) although admittedly we still have a nasty integral to do before we have a concrete result. More importantly, the cluster expansion gives us the technology to perform a systematic perturbation expansion to any order we wish. 2.6 Screening and the Debye-H\u00a8uckel Model of a Plasma There are many other applications of the classical statistical methods that we saw in this chapter. Here we use them to derive the important phenomenon of screening. The problem we will consider, which sometimes goes by the name of a \u201cone-component plasma\u201d, is the following: a gas of electrons, each with charge \u2212q, moves in a \ufb01xed background of uniform positive charge density +q\u03c1. The charge density is such that the overall system is neutral which means that \u03c1 is also the average charge density of the electrons. This is the Debye-H\u00a8uckel model. In the absence of the background charge density, the interaction between electons is given by the Coulomb potential U (r) = q2 r where we\u2019re using units in which 4\u03c00 = 1. How does", " the \ufb01xed background charge a\ufb00ect the potential between electrons? The clever trick of the Debye-H\u00a8uckel model is to use statistical methods to \ufb01gure out the answer to this question. Consider placing one electron at the origin. Let\u2019s try to work out the electrostatic potential \u03c6(r) due to this electron. It is not obvious how to do this because \u03c6 will also depend on the positions of all the other electrons. In general we can write, \u22072\u03c6(r) = \u22124\u03c0 (\u2212q\u03b4(r) + q\u03c1 \u2212 q\u03c1g(r)) (2.40) \u2013 60 \u2013 where the \ufb01rst term on the right-hand side is due to the electron at the origin; the second term is due to the background positive charge density; and the third term is due to the other electrons whose average charge density close to the \ufb01rst electron is \u03c1g(r). The trouble is that we don\u2019t know the function g. If we were sitting at zero temperature, the electrons would try to move apart as much as possible. But at nonzero temperatures, their thermal energy will allow them to approach each other. This is the clue that we need. The energy cost for an electron to approach the origin is, of course, E(r) = \u2212q\u03c6(r). We will therefore assume that the charge density near the origin is given by the Boltzmann factor, g(r) \u2248 e\u03b2q\u03c6(r) For high temperatures, \u03b2q\u03c6 1, we can write e\u03b2q\u03c6 \u2248 1 + \u03b2q\u03c6 and the Poisson equation (2.40) becomes \u22072 + 1 \u03bb2 D \u03c6(r) = 4\u03c0q\u03b4(r) where \u03bb2 D = 1/4\u03c0\u03b2\u03c1q2. This equation has the solution, \u03c6(r) = \u2212 qe\u2212r/\u03bbD r (2.41) which immediately translates into an e\ufb00ective potential energy between electrons, Ue\ufb00(r) = q2e\u2212r/\u03bbD r We now see that the e\ufb00ect of the plasma is to introduce the exponential factor in the numerator, causing the potential to decay very quickly at distances r > \u03bbD. This e\ufb00ect is called screening and \u03bbD is known as the Debye screening length. The derivation of (2.41) is self-consistent if we have a large number of electrons within a distance \u03bbD of the origin so that we can happily talk about average charge density. This means that we need \u03c1\u03bb3 D 1. \u2013 61 \u2013 3. Quantum Gases In this section we will discuss situations where quantum e\ufb00ects are important. We\u2019ll still restrict attention to gases \u2014 meaning a bunch of particles moving around and barely interacting \u2014 but one of the \ufb01rst things we\u2019ll see is how versatile the idea of a gas can be in the quantum world. We\u2019ll use it to understand not just the traditional gases that we met in the previous section but also light and, ironically, certain properties of solids. In the latter part of this section, we will look at what happens to gases at low temperatures where their behaviour is dominated by quantum statistics. 3.1 Density of States We start by introducing the important concept of the density of states. To illustrate this, we\u2019ll return once again to the ideal gas trapped in a box with sides of length L and volume V = L3. Viewed quantum mechanically, each particle is described by a wavefunction. We\u2019ll impose periodic boundary conditions on this wavefunction (although none of the physics that we\u2019ll discuss in this course will be sensitive to the choice of boundary condition). If there are no interactions between particles, the energy eigenstates are simply plane waves, \u03c8 = 1 \u221a V eik\u00b7x Boundary conditions require that the wavevector k = (k1, k2, k3) is quantized as ki = 2\u03c0ni L with ni \u2208 Z and the energy of the particle is 4\u03c022 2mL2 (n2 with k = |k|. The quantum mechanical single particle partition function (1.21) is given by the sum over all energy eigenstates, 2k2 2m 2 + n2 3) 1 + n2 En = = Z1 = e\u2212\u03b2En n The question is: how do we do the sum? The simplest way is to approximate it by an integral. Recall from the previous section that the thermal wavelength of the particle is de\ufb01ned to be 2\u03c02 mkBT \u03bb = \u2013 62 \u2013 The exponents that appear in the sum are all of the form \u223c", " \u03bb2n2/L2, up to some constant factors. For any macroscopic size box, \u03bb L (a serious understatement! Actually \u03bb L) which ensures that there are many states with En \u2264 kBT all of which contribute to the sum. (There will be an exception to this at very low temperatures which will be the focus of Section 3.5.3). We therefore lose very little by approximating the sum by an integral. We can write the measure of this integral as \u2248 d3n = V (2\u03c0)3 d3k = 4\u03c0V (2\u03c0)3 \u221e 0 dk k2 n where, in the last equality, we have integrated over the angular directions to get 4\u03c0, the area of the 2-sphere, leaving an integration over the magnitude k = |k| and the Jacobian factor k2. For future applications, it will prove more useful to change integration variables at this stage. We work instead with the energy, E = 2k2 2m \u21d2 dE = 2k m dk We can now write out integral as 4\u03c0V (2\u03c0)3 dk k2 = V 2\u03c02 dE 2mE 2 m 2 \u2261 dE g(E) (3.1) where g(E) = 3/2 V 4\u03c02 2m 2 E1/2 (3.2) is the density of states: g(E)dE counts the number of states with energy between E and E + dE. Notice that we haven\u2019t actually done the integral over E in (3.1); instead this is to be viewed as a measure which we can integrate over any function f (E) of our choosing. There is nothing particularly quantum mechanical about the density of states. Indeed, in the derivation above we have replaced the quantum sum with an integral over momenta which actually looks rather classical. Nonetheless, as we encounter more and more di\ufb00erent types of gases, we\u2019ll see that the density of states appears in all the calculations and it is a useful quantity to have at our disposal. 3.1.1 Relativistic Systems Relativistic particles moving in d = 3 + 1 spacetime dimensions have kinetic energy \u221a E = 2k2c2 + m2c4 (3.3) \u2013 63 \u2013 Repeating the steps above, we \ufb01nd the density of states is given by g(E) = \u221a V E 2\u03c023c3 E2 \u2212 m2c4 In particular, for massless particles, the density of states is 3.2 Photons: Blackbody Radiation g(E) = V E2 2\u03c023c3 (3.4) (3.5) \u201cIt was an act of desperation. For six years I had struggled with the blackbody theory. I knew the problem was fundamental and I knew the answer. I had to \ufb01nd a theoretical explanation at any cost, except for the inviolability of the two laws of thermodynamics\u201d Max Planck We now turn to our \ufb01rst truly quantum gas: light. We will consider a gas of photons \u2014 the quanta of the electromagnetic \ufb01eld \u2014 and determine a number of its properties, including the distribution of wavelengths. Or, in other words, its colour. Below we will describe the colour of light at a \ufb01xed temperature. But this also applies (with a caveat) to the colour of any object at the same temperature. The argument for this is as follows: consider bathing the object inside the gas of photons. In equilibrium, the object sits at the same temperature as the photons, emitting as many photons as it absorbs. The colour of the object will therefore mimic that of the surrounding light. For a topic that\u2019s all about colour, a gas of photons is usually given a rather bland name \u2014 blackbody radiation. The reason for this is that any real object will exhibit absorption and emission lines due to its particular atomic make-up (this is the caveat mentioned above). We\u2019re not interested in these details; we only wish to compute the spectrum of photons that a body emits because it\u2019s hot. For this reason, one sometimes talks about an idealised body that absorbs photons of any wavelength and re\ufb02ects none. At zero temperature, such an object would appear black: this is the blackbody of the title. We would like to understand its colour as we turn up the heat. To begin, we need some facts about photons. The energy of a photon is determined by its wavelength \u03bb or, equivalently, by its frequency \u03c9 = 2\u03c0c/\u03bb to be E = \u03c9 This is a special case of the relativistic energy formula (3.3) for massless particles, m = 0. The frequency is related to the (magnitude of", " the) wavevector by \u03c9 = kc. \u2013 64 \u2013 Photons have two polarization states (one for each dimension transverse to the direction of propagation). To account for this, the density of states (3.5) should be multiplied by a factor of two. The number of states available to a single photon with energy between E and E + dE is therefore g(E)dE = V E2 \u03c023c3 dE Equivalently, the number of states available to a single photon with frequency between \u03c9 and \u03c9 + d\u03c9 is g(E)dE = g(\u03c9)d\u03c9 = V \u03c92 \u03c02c3 d\u03c9 (3.6) where we\u2019ve indulged in a slight abuse of notation since g(\u03c9) is not the same function as g(E) but is instead de\ufb01ned by the equation above. It is also worth pointing out an easy mistake to make when performing these kinds of manipulations with densities of states: you need to remember to rescale the interval dE to d\u03c9. This is most simply achieved by writing g(E)dE = g(\u03c9)d\u03c9 as we have above. If you miss this then you\u2019ll get g(\u03c9) wrong by a factor of. The \ufb01nal fact that we need is important: photons are not conserved. If you put six atoms in a box then they will still be there when you come back a month later. This isn\u2019t true for photons. There\u2019s no reason that the walls of the box can\u2019t absorb one photon and then emit two. The number of photons in the world is not \ufb01xed. To demonstrate this, you simply need to turn o\ufb00 the light. Because photon number is not conserved, we\u2019re unable to de\ufb01ne a chemical potential for photons. Indeed, even in the canonical ensemble we must already sum over states with di\ufb00erent numbers of photons because these are all \u201caccessible states\u201d. (It is sometimes stated that we should work in the grand canonical ensemble at \u00b5 = 0 which is basically the same thing). This means that we should consider states with any number N of photons. We\u2019ll start by looking at photons with a de\ufb01nite frequency \u03c9. A state with N such photons has energy E = N \u03c9. Summing over all N gives us the partition function for photons at \ufb01xed frequency, Z\u03c9 = 1 + e\u2212\u03b2\u03c9 + e\u22122\u03b2\u03c9 +... = 1 1 \u2212 e\u2212\u03b2\u03c9 (3.7) We now need to sum over all possible frequencies. As we\u2019ve seen a number of times, independent partition functions multiply, which means that the logs add. We only need \u2013 65 \u2013 Figure 14: The Planck Distribution function (Source: E. Schubert, Light Emitting Diodes). to know how many photon states there are with some frequency \u03c9. But this is what the density of states (3.6) tells us. We have \u221e \u221e log Z = d\u03c9 g(w) log Z\u03c9 = \u2212 d\u03c9 \u03c92 log 1 \u2212 e\u2212\u03b2\u03c9 (3.8) V \u03c02c3 0 3.2.1 Planck Distribution 0 From the partition function (3.8) we can calculate all interesting quantities for a gas of light. For example, the energy density stored in the photon gas is E = \u2212 V \u03c02c3 However, before we do the integral over frequency, there\u2019s some important information contained in the integrand itself: it tells us the amount of energy carried by photons with frequency between \u03c9 and \u03c9 + d\u03c9 \u03c93 e\u03b2\u03c9 \u2212 1 log Z = \u2202 \u2202\u03b2 (3.9) d\u03c9 0 \u221e E(\u03c9)d\u03c9 = V \u03c02c3 This is the Planck distribution. It is plotted above for various temperatures. As you can see from the graph, for hot gases the maximum in the distribution occurs at a lower wavelength or, equivalently, at a higher frequency. We can easily determine where this maximum occurs by \ufb01nding the solution to dE(\u03c9)/d\u03c9 = 0. It is \u03c93 e\u03b2\u03c9 \u2212 1 (3.10) d\u03c9 \u03c9max = \u03b6 kBT where \u03b6 \u2248 2.822 solves 3 \u2212 \u03b6 = 3e\u2212\u03b6. The equation above is often called Wien\u2019s displacement law. Roughly speaking, it tells you the colour of a hot object. \u2013 66 \u2013 To compute the total energy in the gas of photons, we need to do the integration in (3.9). To highlight how the total energy depends on temperature, it is useful to perform", " the rescaling x = \u03b2\u03c9, to get E = V \u03c02c3 (kBT )4 3 \u221e 0 x3dx ex \u2212 1 The integral I = dx x3/(ex \u2212 1) is tricky but doable. It turns out to be I = \u03c04/15. (We will e\ufb00ectively prove this fact later in the course when we consider a more general class of integrals (3.27) which can be manipulated into the sum (3.28). The net result of this is to express the integral I above in terms of the Gamma function and the Riemann zeta function: I = \u0393(4)\u03b6(4) = \u03c04/15). We learn that the energy density E = E/V in a gas of photons scales is proportional to T 4, E = \u03c02k4 B 153c3 T 4 Stefan-Boltzmann Law The expression for the energy density above is closely related to the Stefan-Boltzmann law which describes the energy emitted by an object at temperature T. That energy \ufb02ux is de\ufb01ned as the rate of transfer of energy from the surface per unit area. It is given by Energy Flux = Ec 4 \u2261 \u03c3T 4 (3.11) where \u03c3 = \u03c02k4 B 603c2 = 5.67 \u00d7 10\u22128 Js\u22121m\u22122K \u22124 is the Stefan constant. The factor of the speed of light in the middle equation of (3.11) appears because the \ufb02ux is the rate of transfer of energy. The factor of 1/4 comes because we\u2019re not considering the \ufb02ux emitted by a point source, but rather by an actual object whose size is bigger than the wavelength of individual photons. This means that the photon are only emitted in one direction: away from the object, not into it. Moreover, we only care about the velocity perpendicular to the object, which is (c cos \u03b8) where \u03b8 is the angle the photon makes with the normal. This means that rather than \ufb01lling out a sphere of area 4\u03c0 surrounding the object, the actual \ufb02ux of photons from any point on the object\u2019s surface is given by 1 4\u03c0 2\u03c0 d\u03c6 \u03c0/2 0 0 d\u03b8 sin \u03b8 (c cos \u03b8) = c 4 \u2013 67 \u2013 Radiation Pressure and Other Stu\ufb00 All other quantities of interest can be computed from the free energy, F = \u2212kBT log Z \u221e V kBT \u03c02c3 = o d\u03c9 \u03c92 log 1 \u2212 e\u2212\u03b2\u03c9 We can remove the logarithm through an integration by parts to get, F = \u2212 = \u2212 = \u2212 \u03c93e\u2212\u03b2\u03c9 1 \u2212 e\u2212\u03b2\u03c9 x3 ex \u2212 1 dx \u221e d\u03c9 \u221e V 3\u03c02c3 0 V 1 \u03b244 3\u03c02c3 V \u03c02 453c3 (kBT )4 0 From this we can compute the pressure due to electromagnetic radiation, p = \u2212 \u2202F \u2202V T = E 3V = 4\u03c3 3c T 4 This is the equation of state for a gas of photons. The middle equation tells us that the pressure of photons is one third of the energy density \u2014 a fact which will be important in the Cosmology course. We can also calculate the entropy S and heat capacity CV. They are both most conveniently expressed in terms of the Stefan constant which hides most of the annoying factors, S = \u2212 \u2202F \u2202T V = 16V \u03c3 3c T 3, CV = \u2202E \u2202T V = 16V \u03c3 c T 3 3.2.2 The Cosmic Microwave Background Radiation The cosmic microwave background, or CMB, is the afterglow of the big bang, a uniform light that \ufb01lls the Universe. The intensity of this light was measured accurately by the FIRAS (far infrared absolute spectrophotometer) instrument on the COBE satellite in the early 1990s. The result is shown on the right, together with the theoretical curve for a blackbody spectrum at T = 2.725 K. It may look as if the error bars are large, but this is only because they have been multiplied by a factor of 400. If the error bars were drawn at the correct size, you wouldn\u2019t be able to to see them. \u2013 68 \u2013 This result is totally astonishing. The light has been traveling for 13.7 billion years, almost since the beginning of time itself. And yet we can understand it with ridiculous accuracy using such a simple calculation. If you\u2019re not awed by this graph then you have no soul. You can learn more about this in the lectures on cosmology. 3.2.3 The Birth of Quantum Mechanics The", " key conceptual input in the derivation of Planck\u2019s formula (3.10) is that light of frequency \u03c9 comes in packets of energy E = \u03c9. Historically, this was the \ufb01rst time that the idea of quanta arose in theoretical physics. Let\u2019s see what would happen in a classical world where light of frequency \u03c9 can have arbitrarily low intensity and, correspondingly, arbitrarily low energy. This is e\ufb00ectively what happens in the regime h\u03c9 kBT of the Planck distribution where the minimum energy \u03c9 is completely swamped by the temperature. There we can approximate 1 e\u03b2\u03c9 \u2212 1 \u2248 1 \u03b2\u03c9 and Planck\u2019s distribution formula (3.10) reduces to E(\u03c9) = V \u03c02c3 \u03c92kBT Notice that all hints of the quantum have vanished. This is the Rayleigh-Jeans law for the distribution of classical radiation. It has a serious problem if we try to extrapolate it to high frequencies since the total energy, E = \u221e 0 E(\u03c9)d\u03c9, diverges. This was referred to as the ultra-violet catastrophe. In contrast, in Planck\u2019s formula (3.10) there is an exponential suppression at high frequencies. This arises because when \u03c9 kBT, the \u2013 69 \u2013 temperature is not high enough to create even a single photon. By imposing a minimum energy on these high frequency modes, quantum mechanics ensures that they become frozen out. 3.2.4 Max Planck (1858-1947) \u201cA new scienti\ufb01c truth does not triumph by convincing its opponents and making them see the light, but rather because its opponents eventually die\u201d Max Planck Planck was educated in Munich and, after a brief spell in Kiel, moved to take a professorship in Berlin. (The same position that Boltzmann had failed to turn up for). For much of his early career, Planck was adamantly against the idea of atoms. In his view the laws of thermodynamics were inviolable. He could not accept that they sat on a foundation of probability. In 1882, he wrote \u201catomic theory, despite its great successes, will ultimately have to be abandoned\u201d. Twenty years later, Planck had changed his mind. In 1900, he applied Boltzmann\u2019s statistical methods to photons to arrive at the result we derived above, providing the \ufb01rst hints of quantum mechanics. However, the key element of the derivation \u2014 that light comes in quanta \u2014 was not emphasised by Planck. Later, when quantum theory was developed, Planck refused once more to accept the idea of probability underlying physics. This time he did not change his mind. 3.3 Phonons It is hard to imagine substances much more di\ufb00erent than a gas and a solid. It is therefore quite surprising that we can employ our study of gases to accurately understand certain properties of solids. Consider a crystal of atoms of the type shown in the \ufb01gure. The individual atoms are stuck fast in position: they certainly don\u2019t act like a gas. But the vibrations of the atoms \u2014 in other words, sound waves \u2014 can be treated using the same formalism that we introduced for photons. 3.3.1 The Debye Model Figure 15: Quantum mechanics turns electromagnetic waves into discrete packets of energy called photons. In exactly the same way, sounds waves in solids also come in discrete packets. They are called phonons. We\u2019ll take the energy of a phonon to again be of the form E = \u03c9 = kcs (3.12) \u2013 70 \u2013 where cs is now the speed of sound rather than the speed of light. The density of states for phonons is the same as that of photons (3.6) with two exceptions: we must replace the speed of light c with the speed of sound cs; and phonons have three polarization states rather than two. There are two transverse polarizations (like the photon) but also a longitudinal mode. The density of states is therefore g(\u03c9)d\u03c9 = 3V 2\u03c02c3 s \u03c92d\u03c9 There is one further important di\ufb00erence between phonons and photons. While light waves can have arbitrarily high frequency, sound waves cannot. This is because high frequency waves have small wavelengths, \u03bb = 2\u03c0cs/\u03c9. But there is a minimum waveit is not possible for sound length in the problem set by the spacing between atoms. waves to propagate through a solid with wavelength smaller than the atomic spacing because there\u2019s nothing in the middle there to shake. We will denote the maximum allowed phonon frequency as \u03c9D. The minimum wavelength, \u03bbD, should be somewhere around the lattice spacing between atoms, which is (V /N )1/3, so we expect", " that \u03c9D \u223c (N/V )1/3cs. But how can we work out the coe\ufb03cient? There is a rather clever argument to determine \u03c9D due to Debye. (So clever that he gets his initial on the frequency and his name on the section heading). We start by counting the number of single phonon states, \u03c9D 0 d\u03c9 g(w) = V \u03c93 D 2\u03c02c3 s The clever bit of the argument is to identify this with the number of degrees of freedom in the solid. This isn\u2019t immediately obvious. The number of degrees of freedom in a lattice of N atoms is 3N since each atom can move in three directions. But the number of single phonon states is counting collective vibrations of all the atoms together. Why should they be equal? To really see what\u2019s going on, one should compute the correct energy eigenstates of the lattice and just count the number of single phonon modes with wavevectors inside the \ufb01rst Brillouin zone. (You can learn about Brillouin zones in the Applications of Quantum Mechanics lecture notes.) But to get the correct intuition, we can think in in general the solid will have many phonons inside it and each of the following way: these phonons could sit in any one of the single-phonon states that we counted above. Suppose that there are three phonons sitting in the same state. We say that this state is occupied three times. In this language, each of the states above can be occupied \u2013 71 \u2013 an arbitrary number of times, restricted only by the energy available. If you want to describe the total state of the system, you need to say how many phonons are in the \ufb01rst state, and how many are in the second state and so on. The number of one-phonon states is then playing the role of the number of degrees of freedom: it is the number of things you can excite. The net result of this argument is to equate 3N = V \u03c93 D 2\u03c02c3 s \u21d2 \u03c9D = 6\u03c02N V 1/3 cs We see that \u03c9D is related to the atomic spacing (V /N )1/3 as we anticipated above, but now we have the coe\ufb03cient too. However, in some sense, the argument of Debye is \u201canswer analysis\u201d. It was constructed to ensure that we have the right high-temperature behaviour that we will see below. From the maximum frequency \u03c9D we can construct an associated energy scale, \u03c9D, and temperature scale, TD = \u03c9D kB This is known as the Debye temperature. It provides a way of characterising all solids: it is the temperature at which the highest frequency phonon starts to become excited. TD ranges from around 100K for very soft materials such as lead through to 2000K for hard materials such as diamond. Most materials have Debye temperatures around room temperature (\u00b1100K or so). Heat Capacity of Solids All that remains is to put the pieces together. Like photons, the number of phonons is not conserved. The partition function for phonons of a \ufb01xed frequency, \u03c9, is the same as for photons (3.7), Z\u03c9 = 1 + e\u2212\u03b2\u03c9 + e\u22122\u03b2\u03c9 +... = 1 1 \u2212 e\u2212\u03b2\u03c9 Summing over all frequencies, the partition function is then log Zphonon = \u03c9D 0 d\u03c9 g(\u03c9) log Z\u03c9 where the partition function Z\u03c9 for a single phonon of frequency \u03c9 is the same as that of a photon (3.7). The total energy in sound waves is therefore E = \u03c9D 0 d\u03c9 \u03c9g(\u03c9) e\u03b2\u03c9 \u2212 1 = 3V 2\u03c02c3 s \u03c9D 0 d\u03c9 \u03c93 e\u03b2\u03c9 \u2212 1 \u2013 72 \u2013 Figure 16: Experimental data for heat capacities. The solid line is the Debye prediction. (Source: D. Schroeder An Introduction to Thermal Physics) We again rescale the integration variable to x = \u03b2\u03c9 so the upper limit of the integral becomes xD = TD/T. Then we have E = 3V 2\u03c02(cs)3 (kBT )4 TD/T 0 dx x3 ex \u2212 1 The integral is a function of TD/T. It has no analytic expression. However, we can look in the two extremes. Firstly, for T TD we can replace the upper limit of the integral by in\ufb01nity. We\u2019re then left with the same de\ufb01nite integral that we appeared for photons, I = \u221e dx x3/(ex \u2212 1) = \u03c04/15. In this low-temperature regime, the heat capacity", " is proportional to T 3, CV = \u2202E \u2202T = 2\u03c02V k4 B 53c3 s T 3 (T TD) (3.13) It is often expressed in terms of the Debye temperature TD, so it reads CV = N kB 3 12\u03c04 5 T TD (3.14) In contrast, at temperatures T TD we only integrate over small values of x, allowing us to Taylor expand the integrand, TD/T 0 dx x3 ex \u2212 1 = TD/T 0 dx x2 +... = 3 TD T 1 3 +... \u2013 73 \u2013 This ensures that the energy grows linearly with T and the heat capacity tends towards a constant value CV = V k4 BT 3 D 2\u03c023c3 s = 3N kB (T TD) (3.15) This high-temperature behaviour has been known experimentally since the early 1800\u2019s. It is called the Dulong-Petit law. Debye\u2019s argument for the value of \u03c9D was basically constructed to reproduce the coe\ufb03cient 3N in the formula above. This was known experimentally, but also from an earlier model of vibrations in a solid due to Einstein. (You met the Einstein model in the \ufb01rst problem sheet). Historically, the real success of the Debye model was the correct prediction of the T 3 behaviour of CV at low temperatures. In most materials the heat capacity is dominated by the phonon contribution. (In metals there is an additional contribution from conduction electrons that we will calculate in Section 3.6). The heat capacity of three materials is shown in Figure 16, together with the predictions from the Debye model. As you can see, it works very well! The deviation from theory at high temperatures is due to di\ufb00erences between CV and Cp, the heat capacity at constant pressure. What\u2019s Wrong with the Debye Model? As we\u2019ve seen, the Debye model is remarkably accurate in capturing the heat capacities of solids. Nonetheless, it is a caricature of the physics. The most glaring problem is our starting point (3.12). The relationship E = \u03c9 between energy and frequency is \ufb01ne; the mistake is the relationship between frequency \u03c9 and wavevector (or momentum) k, namely \u03c9 = kcs. Equations of this type, relating energy and momentum, are called dispersion relations. It turns out that that the dispersion relation for phonons is a little more complicated. It is not hard to compute the dispersion relation for phonons. (You will, in fact, do this calculation in the Applications of Quantum Mechanics course). For simplicity, we\u2019ll work with a one dimensional periodic lattice of N atoms as shown in the \ufb01gure. The equilibrium position of each atom is xl = la and we impose periodic boundary conditions by insisting that xN +1 \u2261 x1. Let ul be the deviation from equilibrium, ul = xl \u2212la. If we approximate the bonds joining the atoms as har- Figure 17: \u2013 74 \u2013 monic oscillators, the Hamiltonian governing the vibrations is H = 1 2m \u02d9u2 i + i \u03b1 2 i (ui \u2212 ui+1)2 where \u03b1 is a parameter governing the strength of the bonds between atoms. The equation of motion is \u00a8ui = \u03b1 m (2ui \u2212 ui+1 \u2212 ui\u22121) This is easily solved by the discrete Fourier transform. We make the ansatz ul = 1 \u221a N k \u02dcukei(kla\u2212\u03c9kt) Plugging this into the equation of motion gives the dispersion relation \u03c9k = 2 \u03b1 m sin ka 2 To compute the partition function correctly in this model, we would have to revisit the density of states using the new dispersion relation E(k) = \u03c9k. The resulting integrals are messy. However, at low temperatures only the smallest frequency modes are excited and, for small ka, the sin function is approximately linear. This means that we get back to the dispersion relation that we used in the Debye model, \u03c9 = kcs, with the speed of sound given by cs = a\u03b1/m. Moreover, at very high temperatures it is simple to check that this model gives the Dulong-Petit law as expected. It deviates from the Debye model only at intermediate temperatures and, even here, this deviation is mostly negligible. 3.4 The Diatomic Gas Revisited With a bit of quantum experience under our belt, we can look again at the diatomic gas that we discussed in Section 2.4. Recall that the classical prediction for the heat capacity \u2014 CV = 7 2N kB \u2014 only agrees with experiment at very high temperatures. Instead, the data suggests that as the temperature is lowered, the vibrational modes and the rotational modes", " become frozen out. But this is exactly the kind of behaviour that we expect for a quantum system where there is a minimum energy necessary to excite each degree of freedom. Indeed, this \u201cfreezing out\u201d of modes saved us from ultra-violet catastrophe in the case of blackbody radiation and gave rise to a reduced heat capacity at low temperatures for phonons. \u2013 75 \u2013 Let\u2019s start with the rotational modes, described by the Hamiltonian (2.18). Treating this as a quantum Hamiltonian, it has energy levels E = 2 2I j(j + 1) j = 0, 1, 2,... The degeneracy of each energy level is 2j + 1. Thus the rotational partition function for a single molecule is \u221e (2j + 1)e\u2212\u03b22j(j+1)/2I Zrot = j=0 When T 2/2IkB, we can approximate the sum by the integral to get Zrot \u2248 \u221e 0 dx (2x + 1)e\u2212\u03b22x(x+1)/2I = 2I \u03b22 which agrees with our result for the classical partition function (2.19). In contrast, for T 2/2IkB all states apart from j = 0 e\ufb00ectively decouple and we have simply Zrot \u2248 1. At these temperatures, the rotational modes are frozen at temperatures accessible in experiment so only the translational modes contribute to the heat capacity. This analysis also explains why there is no rotational contribution to the heat capacity of a monatomic gas. One could try to argue this away by saying that atoms are point particles and so can\u2019t rotate. But this simply isn\u2019t true. The correct argument is that the moment of inertia I of an atom is very small and the rotational modes are frozen. Similar remarks apply to rotation about the symmetry axis of a diatomic molecule. The vibrational modes are described by the harmonic oscillator. You already computed the partition function for this on the \ufb01rst examples sheet (and, in fact, implicitly in the photon and phonon calculations above). The energies are E = \u03c9(n + 1 2 ) and the partition function is Zvib = e\u2212\u03b2\u03c9(n+ 1 2 ) = e\u2212\u03b2\u03c9/2 e\u2212\u03b2\u03c9n = n n e\u2212\u03b2\u03c9/2 1 \u2212 e\u2212\u03b2\u03c9 = 1 2 sinh(\u03b2\u03c9/2) At high temperatures \u03b2\u03c9 1, we can approximate the partition function as Zvib \u2248 1/\u03b2\u03c9 which again agrees with the classical result (2.20). At low temperatures \u03b2\u03c9 \u2013 76 \u2013 1, the partition function becomes Zvib \u2248 e\u2212\u03b2\u03c9/2. This is a contribution from the zeropoint energy of the harmonic oscillator. It merely gives the expected additive constant to the energy per particle, Evib = \u2212 \u2202 \u2202\u03b2 log Zvib \u2248 \u03c9 2 and doesn\u2019t contribute the heat capacity. Once again, we see how quantum e\ufb00ects explain the observed behaviour of the heat capacity of the diatomic gas. The end result is a graph that looks like that shown in Figure 11. 3.5 Bosons For the \ufb01nal two topics of this section, we will return again to the simple monatomic ideal gas. The classical treatment that we described in Section 2.2 has limitations. As the temperature decreases, the thermal de Broglie wavelength, 2\u03c02 mkBT \u03bb = gets larger. Eventually it becomes comparable to the inter-particle separation, (V /N )1/3. At this point, quantum e\ufb00ects become important. If the particles are non-interacting, there is really only one important e\ufb00ect that we need to consider: quantum statistics. Recall that in quantum mechanics, particles come in two classes: bosons and fermions. Which class a given particle falls into is determined by its spin, courtesy of the spinstatistics theorem. Integer spin particles are bosons. This means that any wavefunction must be symmetric under the exchange of two particles, \u03c8(r1, r2) = \u03c8(r2, r1) Particles with 1 tion, 2-integer spin are fermions. They have an anti-symmetrized wavefunc- \u03c8(r1, r2) = \u2212\u03c8(r2, r1) At low temperatures, the behaviour of bosons and fermions is very di\ufb00erent. All familiar fundamental particles such as the electron, proton and neutron are fermions. But an atom that contains an even number of fermions acts as a boson as long as we do not reach energies large enough to dislodge the constituent particles from their bound state. Similarly, an atom", " consisting of an odd number of electrons, protons and neutrons will be a fermion. (In fact, the proton and neutron themselves are not fundamental: they \u2013 77 \u2013 are fermions because they contain three constituent quarks, each of which is a fermion. If the laws of physics were di\ufb00erent so that four quarks formed a bound state rather than three, then both the proton and neutron would be bosons and, as we will see in the next two sections, nuclei would not exist!). We will begin by describing the properties of bosons and then turn to a discussion of fermions to Section 3.6. 3.5.1 Bose-Einstein Distribution We\u2019ll change notation slightly from earlier sections and label the single particle quantum states of the system by |r. (We used |n previously, but n will be otherwise occupied for most of this section). The single particle energies are then Er and we\u2019ll assume that our particles are non-interacting. In that case, you might think that to specify the state of the whole system, you would need to say which state particle 1 is in, and which state particle 2 is in, and so on. But this is actually too much information because particle 1 and particle 2 are indistinguishable. To specify the state of the whole system, we don\u2019t need to attach labels to each particle. Instead, it will su\ufb03ce to say how many particles are in state 1 and how many particles are in state 2 and so on. We\u2019ll denote the number of particles in state |r as nr. If we choose to work in the canonical ensemble, we must compute the partition function, Z = {nr} e\u2212\u03b2nrEr where the sum is over all possible ways of partitioning N particles into sets {nr} subject to the constraint that r nr = N. Unfortunately, the need to impose this constraint makes the sums tricky. This means that the canonical ensemble is rather awkward when discussing indistinguishable particles. It turns out to be much easier to work in the grand canonical ensemble where we introduce a chemical potential \u00b5 and allow the total number of particles N to \ufb02uctuate. Life is simplest it we think of each state |r in turn. In the grand canonical ensemble, a given state can be populated by an arbitrary number of particles. The grand partition function for this state is Zr = nr e\u2212\u03b2nr(Er\u2212\u00b5) = 1 1 \u2212 e\u2212\u03b2(Er\u2212\u00b5) Notice that we\u2019ve implicitly assumed that the sum above converges, which is true only if (Er \u2212 \u00b5) > 0. But this should be true for all states Er. We will set the ground state \u2013 78 \u2013 to have energy E0 = 0, so the grand partition function for a Bose gas only makes sense if \u00b5 < 0 (3.16) Now we use the fact that all the occupation of one state is independent of any other. The full grand partition function is then Z = 1 1 \u2212 e\u2212\u03b2(Er\u2212\u00b5) r From this we can compute the average number of particles, N = 1 \u03b2 \u2202 \u2202\u00b5 log Z = r 1 e\u03b2(Er\u2212\u00b5) \u2212 1 \u2261 nr r Here nr denotes the average number of particles in the state |r, nr = 1 e\u03b2(Er\u2212\u00b5) \u2212 1 (3.17) This is the Bose-Einstein distribution. In what follows we will always be interested in the thermodynamic limit where \ufb02uctuations around the average are negligible. For this reason, we will allow ourselves to be a little sloppy in notation and we write the average number of particles in |r as nr instead of nr. Notice that we\u2019ve seen expressions of the form (3.17) already in the calculations for photons and phonons \u2014 see, for example, equation (3.7). This isn\u2019t coincidence: both photons and phonons are bosons and the origin of this term in both calculations is the it arises because we sum over the number of particles in a given state rather same: than summing over the states for a single particle. As we mentioned in the Section on blackbody radiation, it is not really correct to think of photons or phonons in the grand canonical ensemble because their particle number is not conserved. Nonetheless, the equations are formally equivalent if one just sets \u00b5 = 0. In what follows, it will save ink if we introduce the fugacity z = e\u03b2\u00b5 (3.18) Since \u00b5 < 0, we have 0 < z < 1. \u2013 79 \u2013 Ideal Bose Gas Let\u2019s look again at a gas of non-relativistic particles, now through the eyes of quantum mechanics. The energy is E = 2k2 2m As explained in Section 3.1, we can replace", " the sum over discrete momenta with an integral over energies as long as we correctly account for the density of states. This was computed in (3.2) and we reproduce the result below: g(E) = 3/2 V 4\u03c02 2m 2 E1/2 (3.19) From this, together with the Bose-Einstein distribution, we can easily compute the total number of particles in the gas, N = dE g(E) z\u22121e\u03b2E \u2212 1, (3.20) There is an obvious, but important, point to be made about this equation. If we can do the integral (and we will shortly) we will have an expression for the number of particles in terms of the chemical potential and temperature: N = N (\u00b5, T ). That means that if we keep the chemical potential \ufb01xed and vary the temperature, then N will change. But in most experimental situations, N is \ufb01xed and we\u2019re working in the grand canonical ensemble only because it is mathematically simpler. But nothing is for free. The price we pay is that we will have to invert equation (3.20) to get it in the form \u00b5 = \u00b5(N, T ). Then when we change T, keeping N \ufb01xed, \u00b5 will change too. We have already seen an example of this in the ideal gas where the chemical potential is given by (2.14). The average energy of the Bose gas is, E = dE Eg(E) z\u22121e\u03b2E \u2212 1, (3.21) And, \ufb01nally, we can compute the pressure. In the grand canonical ensemble this is, pV = 1 \u03b2 log Z = \u2212 1 \u03b2 dE g(E) log 1 \u2212 ze\u2212\u03b2E We can manipulate this last expression using an integration by parts. Because g(E) \u223c E1/2, this becomes pV = 2 3 dE Eg(E) z\u22121e\u03b2E \u2212 1 = 2 3 E (3.22) \u2013 80 \u2013 This is implicitly the equation of state. But we still have a bit of work to do. Equation (3.21) gives us the energy as a function of \u00b5 and T. And, by the time we have inverted (3.20), it gives us \u00b5 as a function of N and T. Substituting both of these into (3.22) will give the equation of state. We just need to do the integrals... 3.5.2 A High Temperature Quantum Gas is (Almost) Classical Unfortunately, the integrals (3.20) and (3.21) look pretty \ufb01erce. Shortly, we will start to understand some of their properties, but \ufb01rst we look at a particularly simple limit. We will expand the integrals (3.20), (3.21) and (3.22) in the limit z = e\u03b2\u00b5 1. We\u2019ll \ufb01gure out the meaning of this expansion shortly (although there\u2019s a clue in the title of this section if you\u2019re impatient). Let\u2019s look at the particle density (3.20), N V = = = 1 4\u03c02 1 4\u03c02 1 4\u03c02 2m 2 2m 2 2m 2 3/2 \u221e 0 3/2 \u221e dE dE 3/2 0 z \u03b23/2 E1/2 z\u22121e\u03b2E \u2212 1 ze\u2212\u03b2EE1/2 1 \u2212 ze\u2212\u03b2E \u221e \u221a dx 0 xe\u2212x(1 + ze\u2212x +...) where we made the simple substitution x = \u03b2E. The integrals are all of the Gaussian type and can be easily evaluated by making one further substitution x = u2. The \ufb01nal answer can be conveniently expressed in terms of the thermal wavelength \u03bb, N V = z \u03bb3 3.23) Now we\u2019ve got the answer, we can ask what we\u2019ve done! What kind of expansion is z 1? From the above expression, we see that the expansion is consistent only if \u03bb3N/V 1, which means that the thermal wavelength is much less than the interparticle spacing. But this is true at high temperatures: the expansion z 1 is a high temperature expansion. At \ufb01rst glance, it is surprising that z = e\u03b2\u00b5 1 corresponds to high temperatures. When T \u2192 \u221e, \u03b2 \u2192 0 so naively it looks as if z \u2192 1. But this is too naive. If we keep the particle number N \ufb01xed in (3.20) then \u00b5 must vary as we change the temperature, and it turns out that \u00b5 \u2192 \u2212\u221e faster than \u03b2 \u2192 0. To see this,", " notice that to leading order we need z/\u03bb3 to be constant, so z \u223c T \u22123/2. \u2013 81 \u2013 High Temperature Equation of State of a Bose Gas We now wish to compute the equation of state. We know from (3.22) that pV = 2 3E. So we need to compute E using the same z 1 expansion as above. From (3.21), the energy density is E V = = = 2m 2 2m 2 1 4\u03c02 1 4\u03c02 3z 2\u03bb3\u03b2 3/2 \u221e dE 3/2 0 z \u03b25/2 z \u221a 2 4 E3/2 z\u22121e\u03b2E \u2212 1 dx x3/2e\u2212x(1 + ze\u2212x +...) \u221e 0 1 + +... (3.24) The next part\u2019s a little \ufb01ddly. We want to eliminate z from the expression above in favour of N/V. To do this we invert (3.23), remembering that we\u2019re working in the limit z 1 and \u03bb3N/V 1. This gives z = 1 \u2212 \u03bb3N V 1 \u221a 2 2 \u03bb3N V +... which we then substitute into (3.24) to get \u03bb3N \u03bb3N V +... and \ufb01nally we substitute this into pV = 2 Bose gas at high temperatures, 3E to get the equation of state of an ideal pV = N kBT 1 \u2212 +... \u03bb3N \u221a 2V 4 (3.25) which reproduces the classical ideal gas, together with a term that we can identify as the second virial coe\ufb03cient in the expansion (2.21). However, this correction to the pressure hasn\u2019t arisen from any interactions among the atoms: it is solely due to quantum statistics. We see that the e\ufb00ect of bosonic statistics on the high temperature gas is to reduce the pressure. 3.5.3 Bose-Einstein Condensation We now turn to the more interesting question: what happens to a Bose gas at low temperatures? Recall that for convergence of the grand partition function we require \u00b5 < 0 so that z = e\u03b2\u00b5 \u2208 (0, 1). Since the high temperature limit is z \u2192 0, we\u2019ll anticipate that quantum e\ufb00ects and low temperatures come about as z \u2192 1. \u2013 82 \u2013 Recall that the number density is given by (3.20) which we write as N V = 1 4\u03c02 2mkBT 2 3/2 \u221e 0 dx x1/2 z\u22121ex \u2212 1 \u2261 1 \u03bb3 g3/2(z) (3.26) where the function g3/2(z) is one of a class of functions which appear in the story of Bose gases, gn(z) = 1 \u0393(n) \u221e 0 dx xn\u22121 z\u22121ex \u2212 1 (3.27) These functions are also known as polylogarithms and sometimes denoted as Lin(z) = gn(z). The function g3/2 is relevant for the particle number and the function g5/2 appears in the calculation of the energy in (3.21). For reference, the gamma function has value \u0393(3/2) = \u03c0/2. (Do not confuse these functions gn(z) with the density of states g(E). They are not related; they just share a similar name). \u221a In Section 3.5.2 we looked at (3.26) in the T \u2192 \u221e limit. There we saw that \u03bb \u2192 0 but the function g3/2(z) \u2192 0 in just the right way to compensate and keep N/V \ufb01xed. What now happens in the other limit, T \u2192 0 and \u03bb \u2192 \u221e. One might harbour the hope that g3/2(z) \u2192 \u221e and again everything balances nicely with N/V remaining constant. We will now show that this can\u2019t happen. A couple of manipulations reveals that the integral (3.27) can be expressed in terms of a sum, gn(z) = = = = 1 \u0393(n) 1 \u0393(n) z dx zxn\u22121e\u2212x 1 \u2212 ze\u2212x dx xn\u22121e\u2212x \u221e m=0 zme\u2212mx 1 \u0393(n) 1 \u0393(n) \u221e m=1 \u221e m=1 zm zm mn dx xn\u22121e\u2212mx du un\u22121e\u2212u But the integral that appears in the last line above is nothing but the de\ufb01nition of the gamma", " function \u0393(n). This means that we can write gn(z) = \u221e m=1 zm mn \u2013 83 \u2013 (3.28) We see that gn(z) is a monotonically increasing function of z. Moreover, at z = 1, it is equal to the Riemann zeta function gn(1) = \u03b6(n) For our particular problem it will be useful to know that \u03b6(3/2) \u2248 2.612. Let\u2019s now return to our story. As we decrease T in (3.26), keeping N/V \ufb01xed, z and hence g3/2(z) must both increase. But z can\u2019t take values greater than 1. When we reach z = 1, let\u2019s denote the temperature as T = Tc. We can determine Tc by setting z = 1 in equation (3.26), Tc = 2\u03c02 kBm 1 \u03b6(3/2) 2/3 N V (3.29) What happens if we try to decrease the temperature below Tc? Taken at face value, equation (3.26) tells us that the number of particles should decrease. But that\u2019s a stupid conclusion! It says that we can make particles disappear by making them cold. Where did they go? We may be working in the grand canonical ensemble, but in the thermodynamic limit we expect \u2206N \u223c 1/ N which is too small to explain our missing particles. Where did we misplace them? It looks as if we made a mistake in the calculation. \u221a In fact, we did make a mistake in the calculation. It happened right at the beginning. Recall that back in Section 3.1 we replaced the sum over states with an integral over energies, V (2m)3/2 4\u03c023 \u2248 k dE E1/2 \u221a E in the integrand, the ground state with E = 0 doesn\u2019t Because of the weight contribute to the integral. We can use the Bose-Einstein distribution (3.17) to compute the number of states that we expect to be missing because they sit in this E = 0 ground state, n0 = 1 z\u22121 \u2212 1 (3.30) For most values of z \u2208 (0, 1), there are just a handful of particles sitting in this lowest state and it doesn\u2019t matter if we miss them. But as z gets very very close to 1 (meaning z \u2248 1\u22121/N ) then we get a macroscopic number of particles occupying the ground state. \u2013 84 \u2013 It is a simple matter to redo our calculation taking into account the particles in the ground state. Equation (3.26) is replaced by N = V \u03bb3 g3/2(z) + z 1 \u2212 z Now there\u2019s no problem keeping N \ufb01xed as we take z close to 1 because the additional term diverges. This means that if we have \ufb01nite N, then as we decrease T we can never get to z = 1. Instead, z must level out around z \u2248 1 \u2212 1/N as T \u2192 0. For T < Tc, the number of particles sitting in the ground state is O(N ). Some simple algebra allows us to determine that fraction of particles in the ground state is n0 N = 1 \u2212 V N \u03bb3 \u03b6(3/2) = 1 \u2212 3/2 T Tc (3.31) At temperatures T < Tc, a macroscopic number of atoms discard their individual identities and merge into a communist collective, a single quantum state so large that it can be seen with the naked eye. This is known as the Bose-Einstein condensate. It provides an exquisitely precise playground in which many quantum phenomena can be tested. Bose-Einstein Condensates in the Lab Figure 18: UFO=BEC Bose-Einstein condensates (often shortened to BECs) of weakly interacting atoms were \ufb01nally created in 1995, some 70 years after they were \ufb01rst predicted. These \ufb01rst BECs were formed of Rubidium, Sodium or Lithium and contained between N \u223c 104 \u2192 107 atoms. The transition temperatures needed to create these condensates are extraordinarily small, around Tc \u223c 10\u22127K. Figure 19 shows the iconic colour enhanced plots that reveal the existence of the condensate. To create these plots, the atoms are stored in a magnetic trap which is then turned o\ufb00. A picture is taken of the atom cloud a short time t later when the atoms have travelled a distance kt/m. The grey UFO-like smudges above are the original pictures. From the spread of atoms, the momentum distribution of the cloud is inferred", " and this is what is shown in Figure 19. The peak that appears in the last two plots reveals that a substantial number of atoms were indeed sitting in the momentum ground state. (This is not a k = 0 state because of the \ufb01nite trap and the Heisenberg uncertainty relation). The initial discoverers of BECs, Eric Cornell and Carl Wieman from Boulder and Wolfgang Ketterle from MIT, were awarded the 2001 Nobel prize in physics. \u2013 85 \u2013 Figure 19: The velocity distribution of Rubidium atoms, taken from the Ketterle lab at MIT. The left-hand picture shows T > Tc, just before the condensate forms. The middle and right-hand pictures both reveal the existence of the condensate. Low Temperature Equation of State of a Bose Gas The pressure of the ideal Bose gas was computed in (3.22). We can express this in terms of our new favourite functions (3.27) as p = 2 3 E V = kBT \u03bb3 g5/2(z) (3.32) Formally there is also a contribution from the ground state, but it is log(1 \u2212 z)/V which is a factor of N smaller than the term above and can be safely ignored. At low temperatures, T < Tc, we have z \u2248 1 and p = kBT \u03bb3 \u03b6(5/2) So at low temperatures, the equation of state of the ideal Bose gas is very di\ufb00erent from the classical, high temperature, behaviour. The pressure scales as p \u223c T 5/2 (recall that there is a factor of T 3/2 lurking in the \u03bb). More surprisingly, the pressure is independent of the density of particles N/V. 3.5.4 Heat Capacity: Our First Look at a Phase Transition Let\u2019s try to understand in more detail what happens as we pass through the critical temperature Tc. We will focus on how the heat capacity behaves on either side of the critical temperature. \u2013 86 \u2013 We\u2019ve already seen in (3.32) that we can express the energy in terms of the function g5/2(z), so the heat capacity becomes E V = 3 2 kBT \u03bb3 g5/2(z) = = 3 2 1 V dz dT CV V dE dT dg5/2 dz kBT \u03bb3 15kB 4\u03bb3 g5/2(z) + The \ufb01rst term gives a contribution both for T < Tc and for T > Tc. However, the second term includes a factor of dz/dT and z is a very peculiar function of temperature: for T > Tc, it is fairly smooth, dropping o\ufb00 at T = Tc. However, as T \u2192 Tc, the fugacity rapidly levels o\ufb00 to the value z \u2248 1 \u2212 1/N. For T < Tc, z doesn\u2019t change very much at all. The net result of this is that the second term only contributes when T > Tc. Our goal here is to understand how this contribution behaves as we approach the critical temperature. (3.33) Let\u2019s begin with the easy bit. Below the critical temperature, T < Tc, only the \ufb01rst term in (3.33) contributes and we may happily set z = 1. This gives the heat capacity, CV = 15V kB 4\u03bb3 \u03b6(5/2) \u223c T 3/2 (3.34) Now we turn to T > Tc. Here we have z < 1, so g5/2(z) < g5/2(1). We also have dz/dT < 0. This means that the heat capacity decreases for T > Tc. But we know that for T < Tc, CV \u223c T 3/2 so the heat capacity must have a maximum at T = Tc. Our goal in this section is to understand a little better what the function CV looks like in this region. To compute the second term in (3.33) we need to understand both g 5/2 and how z changes with T as we approach Tc from above. The \ufb01rst calculation is easy is we use our expression (3.28), gn(z) = \u221e m=1 zm mn \u21d2 d dz gn(z) = 1 z gn\u22121(z) (3.35) As T \u2192 Tc from above, dg5/2/dT \u2192 \u03b6(3/2), a constant. All the subtleties lie in the remaining term, dz/dT. After all, this is the quantity which is e\ufb00ectively vanishing for T < Tc. What\u2019s it doing at T > T", "c? To \ufb01gure this out is a little more involved. We start with our expression (3.26), g3/2(z) = N \u03bb3 V T > Tc (3.36) \u2013 87 \u2013 and we\u2019ll ask what happens to the function g3/2(z) as z \u2192 1, keeping N \ufb01xed. We know that exactly at z = 1, g3/2(1) = \u03b6(3/2). But how does it approach this value? To answer this, it is actually simplest to look at the derivative dg3/2/dz = g1/2/z, where g1/2(z) = 1 \u0393(1/2) \u221e 0 dx x\u22121/2 z\u22121ex \u2212 1 The reason for doing this is that g1/2 diverges as z \u2192 1 and it is generally easier to isolate divergent parts of a function than some \ufb01nite piece. Indeed, we can do this straightforwardly for g1/2 by looking at the integral very close to x = 0, we can write g1/2(z) = = = 1 \u0393(1/2) z \u0393(1/2) 2z 1 \u2212 z \u221a 0 dx dx 0 1 \u0393(1/2) x\u22121/2 z\u22121(1 + x) \u2212 1 + \ufb01nite x\u22121/2 (1 \u2212 z) + x 1 +... du 1 + u2 +... 0 where, in the last line, we made the substution u = x/(1 \u2212 z). So we learn that as z \u2192 1, g1/2(z) \u2192 z(1 \u2212 z)\u22121/2. But this is enough information to tell us how g3/2 approaches its value at z = 1: it must be g3/2(z) \u2248 \u03b6(3/2) + A(1 \u2212 z)1/2 +... for some constant A. Inserting this into our equation (3.36) and rearranging, we \ufb01nd that as T \u2192 Tc from above, N \u03bb3 V 3/(3/2) \u2212 1 A2 = 1 \u2212 \u2248 1 \u2212 B T Tc 2 \u03b6(3/2)2 A2 T \u2212 Tc Tc where, in the second line, we used the expression of the critical temperature (3.29). B is some constant that we could \ufb01gure out with a little more e\ufb00ort, but it won\u2019t be important for our story. From the expression above, we can now determine dz/dT as T \u2192 Tc. We see that it vanishes linearly at T = Tc. \u2013 88 \u2013 Putting all this together, we can determine the expression for the heat capacity (3.33) when T > Tc. We\u2019re not interested in the coe\ufb03cients, so we\u2019ll package a bunch of numbers of order 1 into a constant b and the end result is CV = 15V kB 4\u03bb3 g5/2(z) \u2212 b T \u2212 Tc Tc The \ufb01rst term above goes smoothly over to the expression (3.34) for CV when T < Tc. But the second term is only present for T > Tc. Notice that it goes to zero as T \u2192 Tc, which ensures that the heat capacity is continuous at this point. But the derivative is not continuous. A sketch of the heat capacity is shown in the \ufb01gure. Functions in physics are usually nice and Figure 20: Heat Capacity for a BEC smooth. How did we end up with a discontinuity in the derivative? In fact, if we work at \ufb01nite N, strictly speaking everything is nice and smooth. There is a similar contribution to dz/dT even at T < Tc. We can see that by looking again at the expressions (3.30) and (3.31), which tell us z = 1 + 1 n0 \u2212T /Tc)3/2 \u22121 (T < Tc) The di\ufb00erence is that while dz/dT is of order one above Tc, it is of order 1/N below Tc. In the thermodynamic limit, N \u2192 \u221e, this results in the discontinuity that we saw above. This is a general lesson: phase transitions with their associated discontinuities can only arise in strictly in\ufb01nite systems. There are no phase transitions in \ufb01nite systems. Super\ufb02uid Helium-4 A similar, but much more pronounced, discontinuity is seen in Helium-4 as it becomes", " a super\ufb02uid, a transition which occurs at 2.17 K. The atom contains two protons, two neutrons and two electrons and is therefore a boson. (In contrast, Helium3 contains just a single neutron and is a fermion). The experimental data for the heat capacity of Helium-4 is shown on the right. The successive graphs are zooming in on the phase transistion: the scales are (from left to right) Kelvin, milliKelvin and microKelvin. The discontinuity is often called the lambda transition on account of the shape of this graph. \u2013 89 \u2013 CVTTC3Nk /2B There is a close connection between BoseEinstein condensation described above and super\ufb02uids: strictly speaking a non-interacting BoseEinstein condensate is not a super\ufb02uid but super\ufb02uidity is a consequence of arbitrarily weak repulsive interactions between the atoms. However, in He-4, the interactions between atoms are strong and the system cannot be described using the simple techniques developed above. Figure 21: 4He. Something very similar to Bose condensation also occurs in superconductivity and super\ufb02uidity of Helium-3. Now the primary characters are fermions rather than bosons (electrons in the case of superconductivity). As we will see in the next section, fermions cannot condense. But they may form bound states due to interactions and these e\ufb00ective bosons can then undergo condensation. 3.6 Fermions For our \ufb01nal topic, we will discuss the fermion gases. Our analysis will focus solely on non-interacting fermions. Yet this simple model provides a (surprisingly) good \ufb01rst approximation to a wide range of systems, including electrons in metals at low temperatures, liquid Helium-3 and white dwarfs and neutron stars. Fermions are particles with 1 2-integer spin. By the spin-statistics theorem, the wavefunction of the system is required to pick up a minus sign under exchange of any particle, \u03c8(r1, r2) = \u2212\u03c8(r2, r1) As a corollory, the wavefunction vanishes if you attempt to put two identical fermions in the same place. This is a re\ufb02ection of the Pauli exclusion principle which states that fermions cannot sit in the same state. We will see that the low-energy physics of a gas of fermions is entirely dominated by the exclusion principle. We work again in the grand canonical ensemble. The grand partition function for a single state |r is very easy: the state is either occupied or it is not. There is no other option. Zr = n=0,1 e\u2212\u03b2n(Er\u2212\u00b5) = 1 + e\u2212\u03b2(Er\u2212\u00b5) \u2013 90 \u2013 So, the grand partition function for all states is Z = the average number of particles in the system r Zr, from which we can compute N = r 1 e\u03b2(Er\u2212\u00b5) + 1 \u2261 nr r where the average number of particles in the state |r is nr = 1 e\u03b2(Er\u2212\u00b5) + 1 (3.37) This is the Fermi-Dirac distribution. It di\ufb00ers from the Bose-Einstein distribution only by the sign in the denominator. Note however that we had no convergence issues in de\ufb01ning the partition function. Correspondingly, the chemical potential \u00b5 can be either positive or negative for fermions. 3.6.1 Ideal Fermi Gas We\u2019ll look again at non-interacting, non-relativistic particles with E = 2k2/2m. Since fermions necessarily have 1 2-integer spin, s, there is always a degeneracy factor when counting the number of states given by gs = 2s + 1 For example, electrons have spin 1 2 and, correspondingly have a degeneracy of gs = 2 which just accounts for \u201cspin up\u201d and \u201cspin down\u201d states. We saw similar degeneracy factors when computing the density of states for photons (which have two polarizations) and phonons (which had three). For non-relativistic fermions, the density of states is g(E) = 3/2 gsV 4\u03c02 2m 2 E1/2 We\u2019ll again use the notation of fugacity, z = e\u03b2\u00b5. The particle number is N = E = dE g(E) z\u22121e\u03b2E + 1 dE Eg(E) z\u22121e\u03b2E + 1 The average energy is And the pressure is pV = 1 \u03b2 dE g(E) log 1 +", " ze\u2212\u03b2E = 2 3 E \u2013 91 \u2013 (3.38) (3.39) At high temperatures, it is simple to repeat the steps of Section 3.5.2. (This is one of the questions on the problem sheet). Ony a few minus signs di\ufb00er along the way and one again \ufb01nds that for z 1, the equation of state reduces to that of a classical gas, pV = N kBT 1 + \u03bb3N \u221a 2gsV 4 +... (3.40) Notice that the minus signs \ufb01lter down to the \ufb01nal answer: the \ufb01rst quantum correction to a Fermi gas increases the pressure. 3.6.2 Degenerate Fermi Gas and the Fermi Surface In the extreme limit T \u2192 0, the Fermi-Dirac distribution becomes very simple: a state is either \ufb01lled or empty. 1 e\u03b2(E\u2212\u00b5) + 1 \u2212\u2192 1 0 for E < \u00b5 for E > \u00b5 It\u2019s simple to see what\u2019s going on here. Each fermion that we throw into the system settles into the lowest available energy state. These are successively \ufb01lled until we run out of particles. The energy of the last \ufb01lled state is called the Fermi energy and is denoted as EF. Mathematically, it is the value of the chemical potential at T = 0, \u00b5(T = 0) = EF (3.41) Filling up energy states with fermions is just like throwing balls into a box. With one exception: the energy states of free particles are not localised in position space; they are localised in momentum space. This means that successive fermions sit in states with ever-increasing momentum. In this way, the fermions \ufb01ll out a ball in momentum space. The momentum of the \ufb01nal fermion is called the Fermi momentum and is related to the Fermi energy in the usual way: kF = (2mEF )1/2. All states with wavevector |k| \u2264 kF are \ufb01lled and are said to form the Fermi sea or Fermi sphere. Those states with |k| = kF lie on the edge of the Fermi sea. They are said to form the Fermi surface. The concept of a Fermi surface is extremely important in later applications to condensed matter physics. We can derive an expression for the Fermi energy in terms of the number of particles N in the system. To do this, we should appreciate that we\u2019ve actually indulged in a slight abuse of notation when writing (3.41). In the grand canonical ensemble, T and \u00b5 are independent variables: they\u2019re not functions of each other! What this equation really means is that if we want to keep the average particle number N in the system \u2013 92 \u2013 \ufb01xed (which we do) then as we vary T we will have to vary \u00b5 to compensate. So a slightly clearer way of de\ufb01ning the Fermi energy is to write it directly in terms of the particle number Or, inverting, N = EF 0 dE g(E) = 3/2 gsV 6\u03c02 2m 2 E3/2 F EF = 2/3 2 2m 6\u03c02 gs N V (3.42) (3.43) The Fermi energy sets the energy scale for the system. There is an equivalent temperature scale, TF = EF /kB. The high temperature expansion that resulted in the equation of state (3.40) is valid at temperatures T > TF. In contrast, temperatures T < TF are considered \u201clow\u201d temperatures for systems of fermions. Typically, these low temperatures do not have to be too low: for electrons in a metal, TF \u223c 104K; for electrons in a white dwarf, TF > 107K. While EF is the energy of the last occupied state, the average energy of the system can be easily calculated. It is E = EF 0 dE Eg(E) = 3 5 N EF (3.44) Similarly, the pressure of the degenerate Fermi gas can be computed using (3.39), pV = 2 5 N EF (3.45) Even at zero temperature, the gas has non-zero pressure, known as degeneracy pressure. It is a consequence of the Pauli exclusion principle and is important in the astrophysics of white dwarf stars and neutron stars. (We will describe this application in Section 3.6.5). The existence of this residual pressure at T = 0 is in stark contrast to both the classical ideal gas (which, admittedly, isn\u2019t valid at zero temperature) and the bosonic quantum", " gas. 3.6.3 The Fermi Gas at Low Temperature We now turn to the low-temperature behaviour of the Fermi gas. As mentioned above, \u201clow\u201d means T TF, which needn\u2019t be particularly low in everyday terms. The number of particles N and the average energy E are given by, N = \u221e 0 dE g(E) z\u22121e\u03b2E + 1 (3.46) \u2013 93 \u2013 Figure 22: The Fermi-Dirac Distribution function at T = 0 and very small T. The distribution di\ufb00ers from the T = 0 ground state only for a range of energies kBT around EF. and E = \u221e 0 dE Eg(E) z\u22121e\u03b2E + 1 (3.47) where, for non-relativistic fermions, the density of states is g(E) = 3/2 gsV 4\u03c02 2m 2 E1/2 Our goal is to \ufb01rstly understand how the chemical potential \u00b5, or equivalently the fugacity z = e\u03b2\u00b5, changes with temperature when N is held \ufb01xed. From this we can determine how E changes with temperature when N is held \ufb01xed. There are two ways to proceed. The \ufb01rst is a direct approach on the problem by Taylor expanding (3.46) and (3.47) for small T. But it turns out that this is a little delicate because the starting point at T = 0 involves the singular distribution shown on the left-hand side of Figure 22 and it\u2019s easy for the physics to get lost in a morass of integrals. For this reason, we start by developing a heuristic understanding of how the integrals (3.46) and (3.47) behave at low temperatures which will be enough to derive the required results. We will then give the more rigorous expansion \u2013 sometimes called the Sommerfeld expansion \u2013 and con\ufb01rm that our simpler derivation is indeed correct. The Fermi-Dirac distribution (3.37) at small temperatures is sketched on the righthand side of Figure 22. The key point is that only states with energy within kBT of the Fermi surface are a\ufb00ected by the temperature. We want to insist that as we vary the temperature, the number of particles stays \ufb01xed which means that dN/dT = 0. I claim that this holds if, to leading order, the chemical potential is independent of temperature, so d\u00b5 dT T =0 = 0 \u2013 94 \u2013 (3.48) FEFET=0n(E)1ET << Let\u2019s see why this is the case. The change in particle number can be written as dN dT = = \u221e d dT \u221e 0 dE g(E) e\u03b2(E\u2212\u00b5) + 1 dE g(E) d dT 0 \u2248 g(EF ) \u221e 0 dE \u2202 \u2202T 1 e\u03b2(E\u2212\u00b5) + 1 1 e\u03b2(E\u2212EF ) + 1 There are two things going on in the step from the second line to the third. Firstly, we are making use of the fact that, for kBT EF, the Fermi-Dirac distribution only changes signi\ufb01cantly in the vicinity of EF as shown in the right-hand side of Figure 22. This means that the integral in the middle equation above is only receiving contributions in the vicinity of EF and we have used this fact to approximate the density of states g(E) with its value at g(EF ). Secondly, we have used our claimed result (3.48) to replace the total derivative d/dT (which acts on the chemical potential) with the partial derivative \u2202/\u2202T (which doesn\u2019t) and \u00b5 is replaced with its zero temperature value EF. Explicitly di\ufb00erentiating the Fermi-Dirac distribution in the \ufb01nal line, we have dN dT \u2248 g(EF ) \u221e 0 dE E \u2212 EF kBT 2 1 4 cosh2(\u03b2(E \u2212 EF )/2) \u2248 0 This integral vanishes because (E \u2212 EF ) is odd around EF while the cosh function is even. (And, as already mentioned, the integral only receives contributions in the vicinity of EF ). Let\u2019s now move on to compute the change in energy with temperature. This, of course, is the heat capacity. Employing the same approximations that we made above, we can write N,V \u2202E \u2202T \u221e \u221e 0 = dE Eg(E) \u2202 \u2202T dE EF g(EF ) + g", "(EF )(E \u2212 Ef ) 1 e\u03b2(E\u2212\u00b5) + 1 \u2202 \u2202T 1 e\u03b2(E\u2212EF ) + 1 3 2 CV = \u2248 0 However, this time we have not just replaced Eg(E) by EF g(EF ), but instead Taylor expanded to include a term linear in (E \u2212 EF ). (The factor of 3/2 comes about because Eg(E) \u223c E3/2. The \ufb01rst EF g(EF ) term in the square brackets vanishes by the same even/odd argument that we made before. But the (E \u2212 EF ) term survives. \u2013 95 \u2013 Writing x = \u03b2(E \u2212 EF ), the integral becomes CV \u2248 3 2 g(EF )T \u221e \u2212\u221e dx x2 4 cosh2(x/2) where we\u2019ve extended the range of the integral from \u2212\u221e to +\u221e, safe in the knowledge that only the region near EF (or x = 0) contributes anyway. More importantly, however, this integral only gives an overall coe\ufb03cient which we won\u2019t keep track of. The \ufb01nal result for the heat capacity is CV \u223c T g(EF ) There is a simple way to intuitively understand this linear behaviour. At low temperatures, only fermions within kBT of EF are participating in the physics. The number of these particles is \u223c g(EF )kBT. If each picks up energy \u223c kBT then the total energy of the system scales as E \u223c g(EF )(kBT )2, resulting in the linear heat capacity found above. Finally, the heat capacity is often re-expressed in a slightly di\ufb00erent form. Using (3.42) we learn that N \u223c E3/2 F which allows us to write, CV \u223c N kB T TF (3.49) Heat Capacity of Metals One very important place that we can apply the theory above it to metals. We can try to view the conduction electrons \u2014 those which are free to move through the lattice \u2014 as an ideal gas. At \ufb01rst glance, this seems very unlikely to work. This is because we are neglecting the Coulomb interaction between electrons. Nonetheless, the ideal gas approximation for electrons turns out to work remarkably well. From what we\u2019ve learned in this section, we would expect two contributions to the heat capacity of a metal. The vibrations of the lattice gives a phonon contribution which goes as T 3 (see equation (3.13)). If the conduction electrons act as an ideal gas, they should give a linear term. The low-temperature heat capacity can therefore be written as CV = \u03b3T + \u03b1T 3 Experimental data is usually plotted as CV /T vs T 2 since this gives a straight line which looks nice. The intercept with the CV axis tells us the electron contribution. The heat capacity for copper is shown in the \ufb01gure. \u2013 96 \u2013 We can get a rough idea of when the phonon and electron contributions to heat capacity are comparable. Equating (3.14) with (3.56), and writing 24\u03c02/5 \u2248 50, we have the the two contributions are equal when T 2 \u223c T 3 D/50TF. Ballpark \ufb01gures are TD \u223c 102K and TF \u223c 104K which tells us that we have to get down to temperatures of around 1K or so to see the electron contribution. For many metals, the coe\ufb03cient \u03b3 of the linear heat capacity is fairly close to the ideal gas value (within, say, 20% or so). Yet the electrons in a metal are far from free: the Coulomb energy from their interations is at least as important as the kinetic energy. So why does the ideal gas approximation work so well? This was explained in the 1950s by Landau and the resulting theory \u2014 usually called Landau\u2019s Fermi liquid theory \u2014 is the basis for our current understanding of electron systems. Figure 23: The heat capacity of copper (taken from A. Tari \u201cThe Speci\ufb01c Heat of Matter at Low Temperatures\u201d) 3.6.4 A More Rigorous Approach: The Sommerfeld Expansion The discussion in Section 3.6.3 uses some approximations but highlights the physics of the low-temperature fermi gas. For completeness, here we present a more rigorous derivation of the low-temperature expansion. To start, we introduce some new notation for the particle number and energy inte- grals, (3.46) and (3.47). We write, and N V E V = = = = 3/2 \u221e 0 2m 2 gs 4\u03c02 gs \u03bb3 f3/2(z) 3/2 \u221e", " 2m 2 gs 4\u03c02 gs 3 \u03bb3 kBT f5/2(z) 2 0 dE E1/2 z\u22121e\u03b2E + 1 dE E3/2 z\u22121e\u03b2E + 1 (3.50) (3.51) where \u03bb = 2\u03c02/mkBT is the familiar thermal wavelength and the functions fn(z) are the fermionic analogs of gn de\ufb01ned in (3.27), fn(z) = 1 \u0393(n) \u221e 0 dx xn\u22121 z\u22121ex + 1 (3.52) \u2013 97 \u2013 where it is useful to recall \u0393(3/2) = \u03c0/4 (which follows, of course, from the property \u0393(n + 1) = n\u0393(n)). This function is an example of a polylogarithm and is sometimes written as Lin(\u2212z) = \u2212fn(z). \u03c0/2 and \u0393(5/2) = 3 \u221a \u221a Expanding fn(z) We will now derive the large z expansion of fn(z), sometimes called the Sommerfeld expansion. The derivation is a little long. We begin by splitting the dx integral into two parts, \u0393(n)fn(z) = = = dx \u03b2\u00b5 0 \u03b2\u00b5 xn\u22121 z\u22121ex + 1 dx xn\u22121 1 \u2212 0 (log z)n n \u03b2\u00b5 \u2212 dx 0 \u221e + dx \u03b2\u00b5 1 1 + ze\u2212x xn\u22121 1 + ze\u2212x + xn\u22121 z\u22121ex + 1 \u221e + dx \u03b2\u00b5 \u221e dx \u03b2\u00b5 xn\u22121 z\u22121ex + 1 xn\u22121 z\u22121ex + 1 We now make a simple change of variable, \u03b71 = \u03b2\u00b5 \u2212 x for the \ufb01rst integral and \u03b72 = x \u2212 \u03b2\u00b5 for the second, \u0393(n)fn(z) = (log z)n n \u2212 \u03b2\u00b5 0 d\u03b71 (\u03b2\u00b5 \u2212 \u03b71)n\u22121 1 + e\u03b71 + \u221e 0 d\u03b72 (\u03b2\u00b5 + \u03b72)n\u22121 1 + e\u03b72 So far we have only massaged the integral into a useful form. We now make use of the approximation \u03b2\u00b5 1. Firstly, we note that the \ufb01rst integral has a e\u03b71 suppression in the denominator. If we replace the upper limit of the integral by \u221e, the expression changes by a term of order e\u2212\u03b2\u00b5 = z\u22121 which is a term we\u2019re willing to neglect. With this approximation, we can combine the two integrals into one, \u0393(n)fn(z) = (log z)n n + \u221e 0 d\u03b7 (\u03b2\u00b5 + \u03b7)n\u22121 \u2212 (\u03b2\u00b5 \u2212 \u03b7)n\u22121 1 + e\u03b7 We now Taylor expand the numerator, (\u03b2\u00b5 + \u03b7)n\u22121 \u2212 (\u03b2\u00b5 \u2212 \u03b7)n\u22121 = (\u03b2\u00b5)n\u22121 (1 + \u03b7/\u03b2\u00b5)n\u22121 \u2212 (1 \u2212 \u03b7/\u03b2\u00b5)n\u22121 = (\u03b2\u00b5)n\u22121 \u03b7 \u03b2\u00b5 = 2(n \u2212 1)(\u03b2\u00b5)n\u22122\u03b7 +... 1 + (n \u2212 1) +... \u2212 1 + (n \u2212 1) +... \u03b7 \u03b2\u00b5 From which we get \u0393(n)fn(z) = (log z)n n + 2(n \u2212 1)(log z)n\u22122 \u221e 0 d\u03b7 \u03b7 e\u03b7 + 1 \u2013 98 \u2013 We\u2019re left with a fairly simple integral. \u221e 0 d\u03b7 \u03b7 e\u03b7 + =1 d\u03b7 \u03b7e\u2212\u03b7 1 + e\u2212\u03b7 \u221e d\u03b7 \u03b7 e\u2212m\u03b7(\u22121)m+1 m=1 (\u22121)m+1 m2 \u221e 0 du ue\u2212u The integral in the last line is very simple: \u221e sum. There are a number of ways to manipulate this. We chose to write 0 du ue\u2212u = 1. We have only to do the 1 \u2212 1 22 + 1 32 \u2212 1 42... = 1 + = 1 \u2212 1 22 + 2 22 \u00b7 1 32 + 1 + \u2212 2 1 42 +... 1 1 32 +... 22 + 1 22 + 1 42 + 1 62 +... (3.53) This \ufb01nal", " sum is well known. It is 1/m2 = \u03b6(2) = \u03c02/6. (The original proof of this, by Euler, proceeds by Taylor expanding sin x/x, writing resulting sum as a product of roots, and then equating the x2 term). After all this work, we \ufb01nally have the result we want. The low temperature expan- sion of fn(z) is an expansion in 1/ log z = 1/\u03b2\u00b5, fn(z) = (log z)n \u0393(n + 1) 1 + \u03c02 6 n(n \u2212 1) (log z)2 +... (3.54) Back to Physics: Heat Capacity of a Fermi Gas Again The expansion (3.54) is all we need to determine the leading order e\ufb00ects of temperature on the Fermi gas. The number of particles (3.50) can be expressed in terms of the chemical potential as N V = gs 6\u03c023 (2m\u00b5)3/2 1 + 2 \u03c02 8 kBT \u00b5 +... (3.55) But the number of particles determines the Fermi energy through (3.43). For \ufb01xed particle number N, we can therefore express the chemical potential at \ufb01nite temperature in terms of the Fermi energy. After a small amount of algebra, we \ufb01nd \u00b5 = EF 1 \u2212 2 \u03c02 12 kBT EF +... \u2013 99 \u2013 We see that the chemical potential is a maximum at T = 0 and decreases at higher T. This is to be expected: recall that by the time we get to the classical gas, the chemical potential must be negative. Moreover, note that the leading order correction is quadratic in temperature rather than linear, in agreement with our earlier claim (3.48). The energy density can be computed from (3.51) and (3.54). It is E V gs = 10\u03c023 (2m)3/2\u00b55/2 1 + 2 5\u03c02 8 kBT \u00b5 +... Our real interest is in the heat capacity. However, with \ufb01xed particle number N, the chemical potential also varies with temperature at quadratic order. We can solve this problem by dividing by (3.55) to get E N = 3EF 5 1 + 2 5\u03c02 12 kBT EF +... Now the only thing that depends on temperature on the right-hand side is T itself. From this we learn that the heat capacity of a low temperature Fermi gas is linear in T, N,V But we now have the correct coe\ufb03cient that completes our earlier result (3.49). = N kB \u2202E \u2202T CV = T TF \u03c02 2 (3.56) 3.6.5 White Dwarfs and the Chandrasekhar limit When stars exhaust their fuel, the temperature T \u2192 0 and they have to rely on the Pauli exclusion principle to support themselves through the degeneracy pressure (3.45). Such stars supported by electron degeneracy pressure are called white dwarfs. In addition to the kinetic energy Ekinetic of fermions given in (3.43) the system has gravitational energy. If we make the further assumption that the density in a star of radius R is uniform, then 3 5 where GN is Newton\u2019s constant. In the problem set, you will minimise Egrav + Ekinetic to \ufb01nd the relationship between the radius and mass of the star, Egrav = \u2212 GN M 2 R R \u223c M \u22121/3 This is unusual. Normally if you throw some stu\ufb00 on a pile, the pile gets bigger. Not so for stars supported by degeneracy pressure. The increased gravitational attraction wins and causes them to shrink. \u2013 100 \u2013 As the star shrinks, the Fermi energy (3.43) grows. Eventually it becomes comparable to the electron mass me and our non-relativistic approximation breaks down. We can redo our calculations using the relativistic formula for the density of states (3.4) (which In the needs to be multiplied by a factor of 2 to account for the spin degeneracy). opposite regime of ultra-relativistic electrons, where E me, then we can expand the density of states as g(E) = V \u03c023c3 E2 \u2212 m2c4 2 +... from which we can determine the kinetic energy due to fermions, replacing the nonrelativistic result (3.44) by Ekinetic = V \u03c023c3 1 4 E4 F \u2212 m2c4 4 E2 F +... The Fermi energy can be expressed in", " terms of the particle number by N = V \u03c023c3 1 3 E3 F \u2212 m2c4 2 EF +... The total energy of the system is then given by, Egrav + EF = 3c 4 9\u03c0M 4 4m4 p 1/3 \u2212 3 5 GN M 2 1 R + O(R) where mp is the proton mass (and we\u2019re taking M = N mp as a good approximation to the full mass of the star). If the term above is positive then the star once again settles into a minimum energy state, where the 1/R term balances the term that grows linearly in R. But if the 1/R term is negative, we have a problem: the star is unstable to gravitational collapse and will shrink to smaller and smaller values of R. This occurs when the mass exceeds the Chandrasekhar limit, M > MC. Neglecting factors of 2 and \u03c0, this limit is MC \u223c c GN 3/2 1 m2 p (3.57) This is roughly 1.5 times the mass of the Sun. Stars that exceed this bound will not end their life as white dwarfs. Their collapse may ultimately be stopped by degeneracy pressure of neutrons so that they form a neutron star. Or it may not be stopped at all in which case they form a black hole. \u2013 101 \u2013 The factor in brackets in (3.57) is an interesting mix of fundamental constants associated to quantum theory, relativity and gravity. In this combination, they de\ufb01ne a mass scale called the Planck mass, M 2 pl = c GN When energy densities reach the Planck scale, quantum gravity e\ufb00ects are important meaning that we have to think of quantising spacetime itself. We haven\u2019t quite reached this limit in white dwarf stars. (We\u2019re shy by about 25 orders of magnitude or so!). Nonetheless the presence of the Planck scale in the Chandra limit is telling us that both quantum e\ufb00ects and gravitational e\ufb00ects are needed to understand white dwarf stars: the matter is quantum; the gravity is classical. 3.6.6 Pauli Paramagnetism \u201cWith heavy heart, I have decided that Fermi-Dirac, not Einstein, is the correct statistics, and I have decided to write a short note on paramagnetism\u201d Wolfgang Pauli, in a letter to Schr\u00a8odinger, reluctantly admitting that electrons are fermions We now describe a gas of electrons in a background magnetic \ufb01eld B. There are two e\ufb00ects that are important: the Lorentz force v \u00d7 B on the motion of electrons and the coupling of the electron spin to the magnetic \ufb01eld. We \ufb01rst discuss the spin coupling and postpone the Lorentz force for the next section. An electron can have two spin states, \u201cup\u201d and \u201cdown\u201d. We\u2019ll denote the spin state by a discrete label: s = 1 for spin up; s = \u22121 for spin down. In a background magnetic \ufb01eld B, the kinetic energy picks up an extra term Espin = \u00b5BBs (3.58) where \u00b5B = |e|/2m is the Bohr magneton. (It has nothing to do with the chemical potential. Do not confuse them!) Since spin up and spin down electrons have di\ufb00erent energies, their occupation num- bers di\ufb00er. Equation (3.50) is now replaced by 2m 2 2m 2 E1/2 e\u03b2(E+\u00b5BB\u2212\u00b5) + 1 E1/2 e\u03b2(E\u2212\u00b5BB\u2212\u00b5) + 1 3/2 \u221e 3/2 \u221e 1 4\u03c02 1 4\u03c02 N\u2191 V N\u2193 V dE dE = = 0 0 = = 1 \u03bb3 f3/2(ze\u2212\u03b2\u00b5BB) 1 \u03bb3 f3/2(ze+\u03b2\u00b5BB) \u2013 102 \u2013 Our interest is in computing the magnetization, which is a measure of how the energy of the system responds to a magnetic \ufb01eld, M = \u2212 \u2202E \u2202B (3.59) Since each spin has energy (3.58), the magnetization is simply the di\ufb00erence in the in the number of up spins and down spins, M = \u2212\u00b5B(N\u2191 \u2212 N\u2193) = \u2212 \u00b5BV \u03bb3 f3/2(ze\u2212\u03b2\u00b5BB) \u2212 f3/2(ze\u03b2\u00b5BB) At high temperatures, and suitably low magnetic \ufb01elds, we can approximate f3/2(z) \u2248 z as z \u2192 0, so", " that M \u2248 2\u00b5BV z \u03bb3 sinh(\u03b2\u00b5BB) We can eliminate the factor of z in favour of the particle number, N = N\u2191 + N\u2193 \u2248 2V z \u03bb3 cosh(\u03b2\u00b5BB) From which we can determine the high temperature magnetization, M \u2248 \u00b5BN tanh(\u03b2\u00b5BB) This is the same result that you computed in the \ufb01rst problem sheet using the simple two-state model. We see that, once again, at high temperatures the quantum statistics are irrelevant. The magnetic susceptibility is a measure of how easy it is to magnetise a substance. It is de\ufb01ned as \u03c7 \u2261 \u2202M \u2202B (3.60) Evaluating \u03c7 in the limit of vanishing magnetic \ufb01eld, we \ufb01nd that the magnetization is inversely proportional to the temperature, \u03c7(B = 0) = N \u00b52 B kBT This 1/T behaviour is known as Curie\u2019s law. (Pierre, not Marie). \u2013 103 \u2013 The above results hold in the high temperature limit. In contrast, at low temperatures the e\ufb00ects of the Fermi surface become important. We need only keep the leading order term in (3.54) to \ufb01nd M \u2248 3/2 \u00b5BV 6\u03c02 2m 2 (EF + \u00b5BB)3/2 \u2212 (EF \u2212 \u00b5BB)3/2 \u2248 3/2 \u00b52 BV 2\u03c02 2m 2 E1/2 F B where we\u2019re written the \ufb01nal result in the limit \u00b5BB EF and expanded to linear order in B. We could again replace some of the factors for the electron number N. However, we can emphasise the relevant physics slightly better if we write the result above in terms of the density of states (3.19). Including the factor of 2 for spin degeneracy, we have M \u2248 \u00b52 Bg(EF )B (3.61) We see that at low temperatures, the magnetic susceptibility no longer obeys Curie\u2019s law, but saturates to a constant \u03c7 \u2261 \u2202M \u2202B = \u00b52 Bg(EF ) What\u2019s happening here is that most electrons are embedded deep within the Fermi surface and the Pauli exclusion principle forbids them from \ufb02ipping their spins in response to a magnetic \ufb01eld. Only those electrons that sit on the Fermi surface itself \u2014 all g(EF ) of them \u2014 have the luxury to save energy by aligning their spins with the magnetic \ufb01eld. Notice that \u03c7 > 0. Such materials are called paramagnetic: in the presence of a magnetic \ufb01eld, the magnetism increases. Materials that are paramagnetic are attracted to applied magnetic \ufb01elds. The e\ufb00ect described above is known as Pauli paramagnetism. 3.6.7 Landau Diamagnetism For charged fermions, such as electrons, there is a second consequence of placing them in a magnetic \ufb01eld: they move in circles. The Hamiltonian describing a particle of charge \u2212e in a magnetic \ufb01eld B = \u2207 \u00d7 A is, H = 1 2m p + e A(r) 2 This has an immediate consequence: a magnetic \ufb01eld cannot a\ufb00ect the statistical physics of a classical gas! This follows simply from substituting the above Hamiltonian into the classical partition function (2.1). A simple shift of variables, p = p + e A shows that the classical partition function does not depend on B. This result (which was also mentioned in a slightly di\ufb00erent way in the Classical Dynamics course) is known as the Bohr-van Leeuwen theorem: it states that there can be no classical magnetism. \u2013 104 \u2013 Let\u2019s see how quantum e\ufb00ects change this conclusion. Our \ufb01rst goal is to understand the one-particle states for an electron in a constant background magnetic \ufb01eld. This is a problem that you will also solve in the Applications of Quantum Mechanics course. It is the problem of Landau levels. Landau Levels We will consider an electron moving in a constant magnetic \ufb01eld pointing in the \u02c6z direction: B = (0, 0, B). There are many di\ufb00erent ways to pick A such that B = \u2207 \u00d7 A gives the right magnetic \ufb01eld. These are known as gauge choices. Each choice will give rise to the same physics but, usually, very di\ufb00erent equations in the intervening steps. Here we make the following gauge choice, A = (\u2212By, 0,", " 0) (3.62) Let\u2019s place our particle in a cubic box with sides of length L. Solutions to the Schr\u00a8odinger equation \u02c6H\u03c8 = E\u03c8 are simply plane waves in the x and z directions, \u03c8(r) = ei(kxx+kzz)f (y) But the wavefunction f (y) must satisfy, \u2212 2 2m \u22022 \u2202y2 + 1 2m (kx \u2212 eBy)2 f (y) = Ef (y) (3.63) But this is a familiar equation! If we de\ufb01ne the cyclotron frequency, \u03c9c = eB m Then we can rewrite (3.63) as \u2212 2 2m \u22022 \u2202y2 + 1 2 m\u03c92 c (y \u2212 y0)2 f (y) = Ef (y) where y0 = kx/eB. This is simply the Schr\u00a8odinger equation for a harmonic oscillator, jiggling with frequency \u03c9c and situated at position y0. One might wonder why the harmonic oscillator appeared. The cyclotron frequency \u03c9c is the frequency at which electrons undergo classical Larmor orbits, so that kind of makes sense. But why have these become harmonic oscillators sitting at positions y0 which depend on kx? The answer is that there is no deep meaning to this at all! it is merely an artefact of our gauge choice (3.62). Had we picked a di\ufb00erent A giving rise to \u2013 105 \u2013 the same B, we would \ufb01nd ourselves trying to answer a di\ufb00erent question! The lesson here is that when dealing with gauge potentials, the intermediate steps of a calculation do not always have physical meaning. Only the end result is important. Let\u2019s see what this is. The energy of the particle is E = E + 2k2 z 2m where E is the energy of the harmonic oscillator, n + E = 1 2 \u03c9c n \u2208 Z These discrete energy levels are known as Landau levels. They are highly degenerate. To see this, note that kx is quantized in unit of \u2206kx = 2\u03c0/L. This means that we can have a harmonic oscillator located every \u2206y0 = 2\u03c0/eBL. The number of such oscillators that we can \ufb01t into the box of size L is L/\u2206y0. This is the degeneracy of each level, eBL2 2\u03c0 \u2261 \u03a6 \u03a60 where \u03a6 = L2B is the total \ufb02ux through the system and \u03a60 = 2\u03c0/e is known as the \ufb02ux quantum. (Note that the result above does not include a factor of 2 for the spin degeneracy of the electron). Back to the Diamagnetic Story We can now compute the grand partition function for non-interacting electrons in a magnetic \ufb01eld. Including the factor of 2 from electron spin, we have log Z = L 2\u03c0 dkz \u221e n=0 2L2B \u03a60 log 1 + z exp \u2212 \u03b22k2 z 2m \u2212 \u03b2\u03c9c(n + 1/2) To perform the sum, we use the Euler summation formula which states that for any function h(x), \u221e n=0 h(n + 1/2) = \u221e 0 h(x)dx + 1 24 h(0) +... We\u2019ll apply the Euler summation formula to the function, h(x) = dkz log 1 + exp \u2212 \u03b22k2 z 2m + \u03b2x \u2013 106 \u2013 So the grand partition function becomes log Z = V B \u03c0\u03a60 \u221e h(\u00b5 \u2212 \u03c9c(n + 1/2)) n=0 \u221e = = V B \u03c0\u03a60 V m 2\u03c022 0 \u00b5 \u2212\u221e h(\u00b5 \u2212 \u03c9cx) dx \u2212 dh(\u00b5) d\u00b5 +... V B \u03c0\u03a60 \u03c9c 24 +\u221e \u2212\u221e h(y) dy \u2212 (\u03c9c)2 24 dk \u03b2 e\u03b2(2k2/2m\u2212\u00b5) + 1 +... (3.64) The \ufb01rst term above does not depend on B. In fact it is simply a rather perverse way of writing the partition function of a Fermi gas when B = 0. However, our interest here is in the magnetization which is again de\ufb01ned as (3.59). In the grand canonical ensemble, this is simply M = 1 \u03b2 \u2202log Z \u2202B The second term in (3.64) is proportional to B2", " (which is hiding in the \u03c92 c term). Higher order terms in the Euler summation formula will carry higher powers of B and, for small B, the expression above will su\ufb03ce. At T = 0, the integrand is simply 1 for |k| < kF and zero otherwise. To compare to Pauli paramagnetism, we express the \ufb01nal result in terms of the Bohr magneton \u00b5B = |e|/2mc. We \ufb01nd M = \u2212 \u00b52 B 3 g(EF )B This is comparable to the Pauli contribution (3.61). But it has opposite sign. Substances whose magnetization is opposite to the magnetic \ufb01eld are called diamagnetic: they are repelled by applied magnetic \ufb01elds. The e\ufb00ect that we have derived above is known as Landau diamagnetism. \u2013 107 \u2013 4. Classical Thermodynamics \u201cThermodynamics is a funny subject. The \ufb01rst time you go through it, you don\u2019t understand it at all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don\u2019t understand it, but by that time you are used to it, so it doesn\u2019t bother you any more.\u201d Arnold Sommerfeld, making excuses So far we\u2019ve focussed on a statistical mechanics, studying systems in terms of their microscopic constituents. In this section, we\u2019re going to take a step back and look at classical thermodynamics. This is a theory that cares nothing for atoms and microscopics. Instead it describes relationships between the observable macroscopic phenomena that we see directly. In some sense, returning to thermodynamics is a retrograde step. It is certainly not as fundamental as the statistical description. Indeed, the \u201claws\u201d of thermodynamics that we describe below can all be derived from statistical physics. Nonetheless, there are a number of reasons for developing classical thermodynamics further. First, pursuing classical thermodynamics will give us a much deeper understanding of some of the ideas that brie\ufb02y arose in Section 1. In particular, we will focus on how energy \ufb02ows due to di\ufb00erences in temperature. Energy transferred in this way is called heat. Through a remarkable series of arguments involving heat, one can deduce the existence of a quantity called entropy and its implications for irreversibility in the Universe. This de\ufb01nition of entropy is entirely equivalent to Boltzmann\u2019s later de\ufb01nition S = kB log \u2126 but makes no reference to the underlying states. Secondly, the weakness of thermodynamics is also its strength. Because the theory is ignorant of the underlying nature of matter, it is limited in what it can tell us. But this means that the results we deduce from thermodynamics are not restricted to any speci\ufb01c system. They will apply equally well in any circumstance, from biological systems to quantum gravity. And you can\u2019t say that about a lot of theories! In Section 1, we brie\ufb02y described the \ufb01rst and second laws of thermodynamics as consequences of the underlying principles of statistical physics. Here we instead place ourselves in the shoes of Victorian scientists with big beards, silly hats and total ignorance of atoms. We will present the four laws of thermodynamics as axioms on which the theory rests. \u2013 108 \u2013 4.1 Temperature and the Zeroth Law We need to start with a handful of de\ufb01nitions: \u2022 A system that is completely isolated from all outside in\ufb02uences is said to be contained in adiabatic walls. We will also refer to such systems as insulated. \u2022 Walls that are not adiabatic are said to be diathermal and two systems separated by a diathermal wall are said to be in thermal contact. A diathermal wall is still a wall which means that it neither moves, nor allows particles to transfer from one system to the other. However, it is not in any other way special and it will allow heat (to be de\ufb01ned shortly) to be transmitted between systems. If in doubt, think of a thin sheet of metal. \u2022 An isolated system, when left alone for a suitably long period of time, will relax to a state where no further change is noticeable. This state is called equilibrium Since we care nothing for atoms and microstates, we must use macroscopic variables to describe any system. For a gas, the only two variables that we need to specify are pressure p and volume V : if you know the pressure and volume, then all other quantities \u2014 colour, smell, viscosity, thermal conductivity \u2014 are \ufb01xed. For other", " systems, further (or di\ufb00erent) variables may be needed to describe their macrostate. Common examples are surface tension and area for a \ufb01lm; magnetic \ufb01eld and magnetization for a magnet; electric \ufb01eld and polarization for a dielectric. In what follows we\u2019ll assume that we\u2019re dealing with a gas and use p and V to specify the state. Everything that we say can be readily extended to more general settings. So far, we don\u2019t have a de\ufb01nition of temperature. This is provided by the zeroth law of thermodynamics which states that equilibrium is a transitive property, Zeroth Law: If two systems, A and B, are each in equilibrium with a third body C, then they are also in equilibrium with each other Let\u2019s see why this allows us to de\ufb01ne the concept of temperature. Suppose that system A is in state (p1, V1) and C is in state (p3, V3). To test if the two systems are in equilibrium, we need only place them in thermal contact and see if their states change. For generic values of pressure and volume, we will \ufb01nd that the systems are not in equilibrium. Equilibrium requires some relationship between the(p1, V1) and (p3, V3). For example, suppose that we choose p1, V1 and p3, then there will be a special value of V3 for which nothing happens when the two systems are brought together. \u2013 109 \u2013 We\u2019ll write the constraint that determines when A and C are in equilibrium as which can be solved to give FAC(p1, V1; p3, V3) = 0 V3 = fAC(p1, V1; p3) Since systems B and C are also in equilibrium, we also have a constraint, FBC(p2, V2; p3, V3) = 0 \u21d2 V3 = fBC(p2, V2; p3) These two equilibrium conditions give us two di\ufb00erent expressions for the volume V3, fAC(p1, V1; p3) = fBC(p2, V2; p3) (4.1) At this stage we invoke the zeroth law, which tells us that systems A and B must also be in equilibrium, meaning that (4.1) must be equivalent to a constraint FAB(p1, V1; p2, V2) = 0 (4.2) Equation (4.1) implies (4.2), but the latter does not depend on p3. That means that p3 must appear in (4.1) in such a way that it can just be cancelled out on both sides. When this cancellation is performed, (4.1) tells us that there is a relationship between the states of system A and system B. \u03b8A(p1, V1) = \u03b8B(p2, V2) The value of \u03b8(p, V ) is called the temperature of the system. The function T = \u03b8(p, V ) is called the equation of state. The above argument really only tells us that there exists a property called temperature. There\u2019s nothing yet to tell us why we should pick \u03b8(p, V ) as temperature rather than, say \u03b8(p, V ). We will shortly see that there is, in fact, a canonical choice of temperature that is de\ufb01ned through the second law of thermodynamics and a construct called the Carnot cycle. However, in the meantime it will su\ufb03ce to simply pick a reference system to de\ufb01ne temperature. The standard choice is the ideal gas equation of state (which, as we have seen, is a good approximation to real gases at low densities), T = pV N kB \u2013 110 \u2013 4.2 The First Law The \ufb01rst law is simply the statement of the conservation of energy, together with the tacit acknowledgement that there\u2019s more than one way to change the energy of the system. It is usually expressed as something along the lines of First Law: The amount of work required to change an isolated system from state 1 to state 2 is independent of how the work is performed. This rather cumbersome sentence is simply telling us that there is another function of state of the system, E(P, V ). This is the energy. We could do an amount of work W on an isolated system in any imaginative way we choose: squeeze it, stir it, place a wire and resistor inside with a current passing through it. The method that we choose does not matter: in all cases, the change of the energy is \u2206E = W. However, for systems that are not isolated, the change of energy", " is not equal to the amount of work done. For example, we could take two systems at di\ufb00erent temperatures and place them in thermal contact. We needn\u2019t do any work, but the energy of each system will change. We\u2019re forced to accept that there are ways to change the energy of the system other than by doing work. We write \u2206E = Q + W (4.3) where Q is the amount of energy that was transferred to the system that can\u2019t be accounted for by the work done. This transfer of energy arises due to temperature di\ufb00erences. It is called heat. Heat is not a type of energy. It is a process \u2014 a mode of transfer of energy. There is no sense in which we can divide up the energy E(p, V ) of the system into heat and work. We can\u2019t write \u201cE = Q + W \u201d because neither Q nor W are functions of state. Quasi-Static Processes In the discussion above, the transfer of energy can be as violent as you like. There is no need for the system to be in equilibrium while the energy is being added: the \ufb01rst law as expressed in (4.3) refers only to energy at the beginning and end. From now on, we will be more gentle. We will add or subtract energy to the system very slowly, so that at every stage of the process the system is e\ufb00ectively in equilibrium and can be described by the thermodynamic variables p and V. Such a process is called quasi-static. \u2013 111 \u2013 For quasi-static processes, it is useful to write (4.3) in in\ufb01nitesimal form. Unfortunately, this leads to a minor notational headache. The problem is that we want to retain the distinction between E(p, V ), which is a function of state, and Q and W, which are not. This means that an in\ufb01nitesimal change in the energy is a total derivative, dE = \u2202E \u2202p dp + \u2202E \u2202V dV while an in\ufb01nitesimal amount of work or heat has no such interpretation: it is merely something small. To emphasise this, it is common to make up some new notation8. A small amount of heat is written \u2212dQ and a small amount of work is written \u2212dW. The \ufb01rst law of thermodynamics in in\ufb01nitesimal form is then dE = \u2212dQ + \u2212dW (4.4) Although we introduced the \ufb01rst law as applying to all types of work, from now on the discussion is simplest if we just restrict to a single method to applying work to a system: squeezing. We already saw in Section 1 that the in\ufb01nitesimal work done on a system is \u2212dW = \u2212pdV which is the same thing as \u201cforce \u00d7 distance\u201d. Notice the sign convention. When \u2212dW > 0, we are doing work on the system by squeezing it so that dV < 0. However, when the system expands, dV > 0 so \u2212dW < 0 and the system is performing work. Expressing the work as \u2212dW = \u2212pdV also allows us to underline the meaning of the new symbol \u2212d. There is no function W (p, V ) which has \u201cdW = \u2212pdV \u201d. (For example, you could try W = \u2212pV but that gives dW = \u2212pdV \u2212 V dp which isn\u2019t what we want). The notation \u2212dW is there to remind us that work is not an exact di\ufb00erential. Suppose now that we vary the state of a system through two di\ufb00erent quasi-static paths as shown in the \ufb01gure. The it is change in energy is independent of the path taken: dE = E(p2, V2) \u2212 E(p1, V1). In contrast, the work done \u2212dW = \u2212 pdV depends on the path taken. This simple observation will prove important for our next discussion. Figure 24: 8In a more sophisticated language, dE, \u2212dW and \u2212dQ are all one-forms on the state space of the system. dE is exact; \u2212dW and \u2212dQ are not. \u2013 112 \u2013 pVABPath IPath II 4.3 The Second Law \u201cOnce or twice I have been provoked and have asked company how many of them could describe the Second Law of Thermodynamics, the law of it was also negative. Yet I was asking entropy. The response was cold: something which is about the scienti\ufb01c equivalent of: \u2018Have you read a work of Shakespeare?\u2019 \u201d C", ".P.Snow (1959) C.P. Snow no doubt had in mind the statement that entropy increases. Yet this is a consequence of the second law; it is not the axiom itself. Indeed, we don\u2019t yet even have a thermodynamic de\ufb01nition of entropy. The essence of the second law is that there is a preferred direction of time. There are many macroscopic processes in Nature which cannot be reversed. Things fall apart. The lines on your face only get deeper. Words cannot be unsaid. The second law summarises all such observations in a single statements about the motion of heat. Reversible Processes Before we state the second law, it will be useful to \ufb01rst focus on processes which can happily work in both directions of time. These are a special class of quasi-static processes that can be run backwards. They are called reversible A reversible process must lie in equilibrium at each point along the path. This is the quasi-static condition. But now there is the further requirement that there is no friction involved. For reversible processes, we can take a round trip as shown to the right. Start in state (p1, V1), take the lower path to (p2, V2) and then the upper path back to (p1, V1). The energy is unchanged because dE = 0. But the total work done is non-zero: pdV = 0. By the \ufb01rst law of thermodynamics (4.4), the work performed by the system during the cycle must be equal to the heat absorbed by the system, \u2212dQ = pdV. If we go one way around the cycle, the system does work and absorbs heat from the surroundings; the other way round work in done on the system which then emits energy as heat. Figure 25: \u2013 113 \u2013 ppp12VV21V Processes which move in a cycle like this, returning to their original starting point, are interesting. Run the right way, they convert heat into work. But that\u2019s very very useful. The work can be thought of as a piston which can be used to power a steam train. Or a playstation. Or an LHC. The Statement of the Second Law The second law is usually expressed in one of two forms. The \ufb01rst tells us when energy can be fruitfully put to use. The second emphasises the observation that there is an arrow of time in the macroscopic world: heat \ufb02ows from hot to cold. They are usually stated as Second Law `a la Kelvin: No process is possible whose sole e\ufb00ect is to extract heat from a hot reservoir and convert this entirely into work Second Law `a la Clausius: No process is possible whose sole e\ufb00ect is the transfer of heat from a colder to hotter body It\u2019s worth elaborating on the meaning of these statements. Firstly, we all have objects in our kitchens which transfer heat from a cold environment to a hot environment: this is the purpose of a fridge. Heat is extracted from inside the fridge where it\u2019s cold and deposited outside where it\u2019s warm. Why doesn\u2019t this violate Clausius\u2019 statement? The reason lies in the words \u201csole e\ufb00ect\u201d. The fridge has an extra e\ufb00ect which is to make your electricity meter run up. In thermodynamic language, the fridge operates because we\u2019re supplying it with \u201cwork\u201d. To get the meaning of Clausius\u2019 statement, think instead of a hot object placed in contact with a cold object. Energy always \ufb02ows from hot to cold; never the other way round. The statements by Kelvin and Clausius are equivalent. Suppose, for example, that we build a machine that violates Kelvin\u2019s statement by extracting heat from a hot reservoir and converting it entirely into work. We can then use this work to power a fridge, extracting heat from a cold source and depositing it back into a hot source. The combination of the two machines then violates Clausius\u2019s statement. It is not di\ufb03cult to construct a similar argument to show that \u201cnot Clausius\u201d \u21d2 \u201cnot Kelvin\u201d. Figure 26: Our goal in this Section is to show how these statements of the second law allow us to de\ufb01ne a quantity called \u201centropy\u201d. \u2013 114 \u2013 FridgeKelvinNotColdHotQWQQHC Figure 27: The Carnot cycle in cartoon. 4.3.1 The Carnot Cycle Kelvin\u2019s statement of the second law is that we can\u2019t extract heat from a hot reservoir and turn it entirely into work. Yet, at \ufb01rst glance, this appears to be in contrast with what we know about reversible cycles. We have just seen that these necessarily have \u2212d", "Q = \u2212dW and so convert heat to work. Why is this not in contradiction with Kelvin\u2019s statement? The key to understanding this is to appreciate that a reversible cycle does more than just extract heat from a hot reservoir. It also, by necessity, deposits some heat elsewhere. The energy available for work is the di\ufb00erence between the heat extracted and the heat lost. To illustrate this, it\u2019s very useful to consider a particular kind of reversible cycle called a Carnot engine. This is series of reversible processes, running in a cycle, operating between two temperatures TH and TC. It takes place in four stages shown in cartoon in Figures 27 and 28. \u2022 Isothermal expansion AB at a constant hot temperature TH. The gas pushes against the side of its container and is allowed to slowly expand. In doing so, it can be used to power your favourite electrical appliance. To keep the temperature constant, the system will need to absorb an amount of heat QH from the surroundings \u2022 Adiabatic expansion BC. The system is now isolated, so no heat is absorbed. But the gas is allowed to continue to expand. As it does so, both the pressure and temperature will decrease. \u2022 Isothermal contraction CD at constant temperature TC. We now start to restore the system to its original state. We do work on the system by compressing the \u2013 115 \u2013 TTTHHCInsulatedInsulatedABCDAIsothermalAdiabaticcompressionAdiabaticcompressionIsothermalexpansionexpansion Figure 28: The Carnot cycle, shown in the p \u2212 V plane and the T \u2212 S plane. gas. If we were to squeeze an isolated system, the temperature would rise. But we keep the system at \ufb01xed temperature, so it dumps heat QC into its surroundings. \u2022 Adiabatic contraction DA. We isolate the gas from its surroundings and continue to squeeze. Now the pressure and temperature both increase. We \ufb01nally reach our starting point when the gas is again at temperature TH. At the end of the four steps, the system has returned to its original state. The net heat absorbed is QH \u2212 QC which must be equal to the work performed by the system W. We de\ufb01ne the e\ufb03ciency \u03b7 of an engine to be the ratio of the work done to the heat absorbed from the hot reservoir, \u03b7 = W QH = QH \u2212 QC QH = 1 \u2212 QC QH Ideally, we would like to take all the heat QH and convert it to work. Such an engine would have e\ufb03ciency \u03b7 = 1 but would be in violation of Kelvin\u2019s statement of the second law. We can see the problem in the Carnot cycle: we have to deposit some amount of heat QC back to the cold reservoir as we return to the original state. And the following result says that the Carnot cycle is the best we can do: Carnot\u2019s Theorem: Carnot is the best. Or, more precisely: Of all engines operating between two heat reservoirs, a reversible engine is the most e\ufb03cient. As a simple corollary, all reversible engines have the same e\ufb03ciency which depends only on the temperatures of the reservoirs \u03b7(TH, TC). \u2013 116 \u2013 THCTCTTHABCDVSTABDCp Proof: Let\u2019s consider a second engine \u2014 call it Ivor \u2014 operating between the same two temperatures TH and TC. Ivor also performs work W but, in contrast to Carnot, is not reversible. Suppose that Ivor absorbs Q H from the hot reservoir and deposits Q C into the cold. Then we can couple Ivor to our original Carnot engine set to reverse. The work W performed by Ivor now goes into driving Carnot. The net e\ufb00ect of the two engines is to extract Q H \u2212 QH from the hot reservoir and, by conservation of energy, to deposit the same amount Q statement tells us that we must have Q moved from the colder to hotter body. Performing a little bit of algebra then gives H \u2212 QH into the cold. But Clausius\u2019s H \u2265 QH; if this were not true, energy would be C \u2212 QC = Q Figure 29: Q C \u2212 Q H = QC \u2212 QH \u21d2 \u03b7Ivor = 1 \u2212 Q C Q H = QH \u2212 QC Q H \u2264 QH \u2212 QC QH = \u03b7Carnot The upshot of this argument is the result that we wanted, namely \u03b7Carnot \u2265 \u03b7Ivor The corollary is now simple to prove. Suppose that Ivor was reversible after all. Then we could use the same argument above to prove that \u03b7Ivor \u2265 \u03b7Carnot, so it must be true that \u03b7Ivor = \u03b7Carnot if Ivor is reversible. This means that", " for all reversible engines operating between TH and TC have the same e\ufb03ciency. Or, said another way, the ratio QH/QC is the same for all reversible engines. Moreover, this e\ufb03ciency must be a function only of the temperatures, \u03b7Carnot = \u03b7(TH, TC), simply because they are the only variables in the game.. 4.3.2 Thermodynamic Temperature Scale and the Ideal Gas Recall that the zeroth law of thermodynamics showed that there was a function of state, which we call temperature, de\ufb01ned so that it takes the same value for any two systems in equilibrium. But at the time there was no canonical way to decide between di\ufb00erent de\ufb01nitions of temperature: \u03b8(p, V ) or \u03b8(p, V ) or any other function are all equally good choices. In the end we were forced to pick a reference system \u2014 the ideal gas \u2014 as a benchmark to de\ufb01ne temperature. This was a fairly arbitrary choice. We can now do better. \u2013 117 \u2013 Q\u2019WQQHCColdHotHQ\u2019CIvorReverseCarnot Since the e\ufb03ciency of a Carnot engine depends only on the temperatures TH and TC, we can use this to de\ufb01ne a temperature scale that is independent of any speci\ufb01c material. (Although, as we shall see, the resulting temperature scale turns out to be equivalent to the ideal gas temperature). Let\u2019s now brie\ufb02y explain how we can de\ufb01ne a temperature through the Carnot cycle. The key idea is to consider two Carnot engines. The \ufb01rst operates between two temperature reservoirs T1 > T2; the second engine operates between two reservoirs T2 > T3. If the \ufb01rst engine extracts heat Q1 then it must dump heat Q2 given by Q2 = Q1 (1 \u2212 \u03b7(T1, T2)) where the arguments above tell us that \u03b7 = \u03b7Carnot is a function only of T1 and T2. If the second engine now takes this same heat Q2, it must dump heat Q3 into the reservoir at temperature T3, given by Q3 = Q2 (1 \u2212 \u03b7(T2, T3)) = Q1 (1 \u2212 \u03b7(T1, T2)) (1 \u2212 \u03b7(T2, T3)) But we can also consider both engines working together as a Carnot engine in its own right, operating between reservoirs T1 and T3. Such an engine extracts heat Q1, dumps heat Q3 and has e\ufb03ciency \u03b7(T1, T3), so that Q3 = Q1 (1 \u2212 \u03b7(T1, T3)) Combining these two results tells us that the e\ufb03ciency must be a function which obeys the equation 1 \u2212 \u03b7(T1, T3) = (1 \u2212 \u03b7(T1, T2)) (1 \u2212 \u03b7(T2, T3)) The fact that T2 has to cancel out on the right-hand side is enough to tell us that 1 \u2212 \u03b7(T1, T2) = f (T2) f (T1) for some function f (T ). At this point, we can use the ambiguity in the de\ufb01nition of temperature to simply pick a nice function, namely f (T ) = T. Hence, we de\ufb01ne the thermodynamic temperature to be such that the e\ufb03ciency is given by \u03b7 = 1 \u2212 T2 T1 \u2013 118 \u2013 (4.5) The Carnot Cycle for an Ideal Gas We now have two ways to specify temperature. The \ufb01rst arises from looking at the equation of state of a speci\ufb01c, albeit simple, system: the ideal gas. Here temperature is de\ufb01ned to be T = pV /N kB. The second de\ufb01nition of temperature uses the concept of Carnot cycles. We will now show that, happily, these two de\ufb01nitions are equivalent by explicitly computing the e\ufb03ciency of a Carnot engine for the ideal gas. We deal \ufb01rst with the isothermal changes of the ideal gas. We know that the energy in the gas depends only on the temperature9, E = 3 2 N kBT (4.6) So dT = 0 means that dE = 0. The \ufb01rst law then tells us that \u2212dQ = \u2212 \u2212dW. For the motion along the line AB in the Carnot cycle, we have QH = B A \u2212dQ = \u2212 B A \u2212d", "W = B A pdV = B A N kBTH V dV = N kBTH log (4.7) VB VA Similarly, the heat given up along the line CD in the Carnot cycle is QC = \u2212N kBTC log VD VC (4.8) Next we turn to the adiabatic change in the Carnot cycle. Since the system is isolated, \u2212dQ = 0 and all work goes into the energy, dE = \u2212pdV. Meanwhile, from (4.6), we can write the change of energy as dE = CV dT where CV = 3 CV dT = \u2212 N kBT V dV \u21d2 dT T = \u2212 After integrating, we have T V 2/3 = constant 2N kB, so N kB CV dV V 9A confession: strictly speaking, I\u2019m using some illegal information in the above argument. The result E = 3 2 N kBT came from statistical mechanics and if we\u2019re really pretending to be Victorian scientists we should discuss the e\ufb03ciency of the Carnot cycle without this knowledge. Of course, we could just measure the heat capacity CV = \u2202E/\u2202T |V to determine E(T ) experimentally and proceed. Alternatively, and more mathematically, we could note that it\u2019s not necessary to use this exact form of the energy to carry through the argument: we need only use the fact that the energy is a function of temperature only: E = E(T ). The isothermal parts of the Carnot cycle are trivially the same and we reproduce (4.7) and (4.8). The adiabatic parts cannot be solved exactly without knowledge of E(T ) but you can still show that VA/VB = VD/VC which is all we need to derive the e\ufb03ciency (4.9). \u2013 119 \u2013 Applied to the line BC and DA in the Carnot cycle, this gives THV 2/3 B = TCV 2/3 C, TCV 2/3 D = THV 2/3 A which tells us that VA/VB = VD/VC. But this means that the factors of log(V /V ) cancel when we take the ratio of heats. The e\ufb03ciency of a Carnot engine for an ideal gas \u2014 and hence for any system \u2014 is given by \u03b7carnot = 1 \u2212 QC QH = 1 \u2212 TC TH (4.9) We see that the e\ufb03ciency using the ideal gas temperature coincides with our thermodynamic temperature (4.5) as advertised. 4.3.3 Entropy The discussion above was restricted to Carnot cycles: reversible cycles operating between two temperatures. The second law tells us that we can\u2019t turn all the extracted heat into work. We have to give some back. To generalize, let\u2019s change notation slightly so that Q always denotes the energy absorbed by the system. If the system releases heat, then Q is negative. In terms of our previous notation, Q1 = QH and Q2 = \u2212QC. Similarly, T1 = TH and T2 = TC. Then, for all Carnot cycles 2 i=1 Qi Ti = 0 Figure 30: Now consider the reversible cycle shown in the \ufb01gure in which we cut the corner of the original cycle. From the original Carnot cycle ABCD, we know that QAB TH + QCD TC = 0 Meanwhile, we can view the square EBGF as a mini-Carnot cycle so we also have QGF TF G + QEB TH = 0 What if we now run along the cycle AEF GCD? Clearly QAB = QAE + QEB. But we also know that the heat absorbed along the segment F G is equal to that dumped along \u2013 120 \u2013 ABVpEGFDC the segment GF when we ran the mini-Carnot cycle. This follows simply because we\u2019re taking the same path but in the opposite direction and tells us that QF G = \u2212QGF. Combining these results with the two equations above gives us QAE TH + QF G TF G + QCD TC = 0 By cutting more and more corners, we can consider any reversible cycle as constructed of (in\ufb01nitesimally) small isothermal and adiabatic segments. Summing up all contributions Q/T along the path, we learn that the total heat absorbed in any reversible cycle must obey \u2212dQ T = 0 But this is a very powerful statement. It means that if we reversibly change our system from state A to state B, then the quantity B \u2212dQ/T is independent of the path taken. Either of the two paths shown in the \ufb01gure will give the same result: A Path I \u2212dQ T = Path II \u2212", "dQ T Given some reference state O, this allows us to de\ufb01ne a new function of state. It is called entropy, S Figure 31: S(A) = A 0 \u2212dQ T (4.10) Entropy depends only on the state of the system: S = S(p, V ). It does not depend on the path we took to get to the state. We don\u2019t even have to take a reversible path: as long as the system is in equilibrium, it has a well de\ufb01ned entropy (at least relative to some reference state). We have made no mention of microstates in de\ufb01ning the entropy. Yet it is clearly the same quantity that we met in Section 1. From (4.10), we can write dS = \u2212dQ/T, so that the \ufb01rst law of thermodynamics (4.4) is written in the form of (1.16) dE = T dS \u2212 pdV (4.11) \u2013 121 \u2013 pVABPath IPath II Irreversibility What can we say about paths that are not reversible? By Carnot\u2019s theorem, we know that an irreversible engine that operates between two temperatures TH and TC is less e\ufb03cient than the Carnot cycle. We use the same notation as in the proof of Carnot\u2019s theorem; the Carnot engine extracts heat QH and dumps heat QC; the irreversible engine extracts heat Q C. Both do the same amount of work W = QH \u2212 QC = Q H and dumps Q C. We can then write H \u2212 Q Q H TH \u2212 Q C TC = QH TH \u2212 QC TC = (Q H \u2212 QH) + (Q H \u2212 QH) \u2212 1 TC 1 TH 1 TH \u2212 1 TC \u2264 0 In the second line, we used QH/TH = QC/TC for a Carnot cycle, and to derive the inequality we used the result of Carnot\u2019s theorem, namely Q H \u2265 QH (together with the fact that TH > TC). The above result holds for any engine operating between two temperatures. But by the same method of cutting corners o\ufb00 a Carnot cycle that we used above, we can easily generalise the statement to any path, reversible or irreversible. Putting the minus signs back in so that heat dumped has the opposite sign to heat absorbed, we arrive at a result is known as the Clausius inequality, \u2212dQ T \u2264 0 We can express this in slightly more familiar form. Suppose that we have two possible paths between states A and B as shown in the \ufb01gure. Path I is irreversible while path II is reversible. Then Clausius\u2019s inequality tells us that \u2212dQ T = I \u2212dQ T \u21d2 I \u2212dQ T \u2212 II \u2212dQ T \u2264 0 \u2264 S(B) \u2212 S(A) (4.12) Figure 32: Suppose further that path I is adiabatic, meaning that it is isolated from the environment. Then \u2212dQ = 0 and we learn that the entropy of any isolated system never decreases, S(B) \u2265 S(A) (4.13) \u2013 122 \u2013 pVABPath IIReversibleIrreversiblePath I Moreover, if an adiabatic process is reversible, then the resulting two states have equal entropy. The second law, as expressed in (4.13), is responsible for the observed arrow of time in the macroscopic world. Isolated systems can only evolve to states of equal or higher entropy. This coincides with the statement of the second law that we saw back in Section 1.2.1 using Boltzmann\u2019s de\ufb01nition of the entropy. 4.3.4 Adiabatic Surfaces The primary consequence of the second law is that there exists a new function of state, entropy. Surfaces of constant entropy are called adiabatic surfaces. The states that sit on a given adiabatic surface can all be reached by performing work on the system while forbidding any heat to enter or leave. In other words, they are the states that can be reached by adiabatic processes with \u2212dQ = 0 which is equivalent to dS = 0. In fact, for the simplest systems such as the ideal gas which require only two variables p and V to specify the state, we do not need the second law to infer to the existence of an adiabatic surface. In that case, the adiabatic surface is really an adiabatic line in the two-dimensional space of states. The existence of this line follows immediately from the \ufb01rst law. To see this, we write the change of energy for an adiabatic process using (4.4) with \u2212dQ = 0, dE + pdV = 0 (4.", "14) Let\u2019s view the energy itself as a function of p and V so that we can write dE = \u2202E \u2202p dP + \u2202E \u2202V dV Then the condition for an adiabatic process (4.14) becomes \u2202E \u2202p dp + \u2202E \u2202V + p dV = 0 Which tells us the slope of the adiabatic line is given by dp dV = \u2212 \u2202E \u2202V + p \u2202E \u2202p \u22121 (4.15) The upshot of this calculation is that if we sit in a state speci\ufb01ed by (p, V ) and transfer work but no heat to the system then we necessarily move along a line in the space of states determined by (4.15). If we want to move o\ufb00 this line, then we have to add heat to the system. \u2013 123 \u2013 However, the analysis above does not carry over to more complicated systems where more than two variables are needed to specify the state. Suppose that the state of the system is speci\ufb01ed by three variables. The \ufb01rst law of thermodynamics is now gains an extra term, re\ufb02ecting the fact that there are more ways to add energy to the system, dE = \u2212dQ \u2212 pdV \u2212 ydX We\u2019ve already seen examples of this with y = \u2212\u00b5, the chemical potential, and X = N, the particle number. Another very common example is y = \u2212M, the magnetization, and X = H, the applied magnetic \ufb01eld. For our purposes, it won\u2019t matter what just that they exist. We need to choose three variables to variables y and X are: specify the state. Any three will do, but we will choose p, V and X and view the energy as a function of these: E = E(p, V, X). An adiabatic process now requires dE + pdV + ydX = 0 \u21d2 \u2202E \u2202p dp + \u2202E \u2202V + p dV + \u2202E \u2202X + y dX = 0 (4.16) But this equation does not necessarily specify a surface in R3. To see that this is not su\ufb03cient, we can look at some simple examples. Consider R3, parameterised by z1, z2 and z3. If we move in an in\ufb01nitesimal direction satisfying z1dz1 + z2dz2 + z3dz3 = 0 then it is simple to see that we can integrate this equation to learn that we are moving on the surface of a sphere, 1 + z2 z2 2 + z2 3 = constant In contrast, if we move in an in\ufb01nitesimal direction satisfying the condition z2dz1 + dz2 + dz3 = 0 (4.17) Then there is no associated surface on which we\u2019re moving. Indeed, you can convince yourself that if you move in a such a way as to always obey (4.17) then you can reach any point in R3 from any other point. In general, an in\ufb01nitesimal motion in the direction Z1dz1 + Z2dz2 + Z3dz3 = 0 has the interpretation of motion on a surface only if the functions Zi obey the condition Z1 \u2202Z2 \u2202z3 \u2212 \u2202Z3 \u2202z2 + Z2 \u2202Z3 \u2202z1 \u2212 \u2202Z1 \u2202z3 + Z3 \u2202Z1 \u2202z2 \u2212 \u2202Z2 \u2202z1 = 0 (4.18) \u2013 124 \u2013 So for systems with three or more variables, the existence of an adiabatic surface is not guaranteed by the \ufb01rst law alone. We need the second law. This ensures the existence of a function of state S such that adiabatic processes move along surfaces of constant S. In other words, the second law tells us that (4.16) satis\ufb01es (4.18). In fact, there is a more direct way to infer the existence of adiabatic surfaces which uses the second law but doesn\u2019t need the whole rigmarole of Carnot cycles. We will again work with a system that is speci\ufb01ed by three variables, although the argument will hold for any number. But we choose our three variables to be V, X and the internal energy E. We start in state A shown in the \ufb01gure. We will show that Kelvin\u2019s statement of the second law implies that it is not possible to reach both states B and C through reversible adi", "abatic processes. The key feature of these states is that they have the same values of V and X and di\ufb00er only in their energy E. To prove this statement, suppose the converse: i.e. we can indeed reach both B and C from A through means of reversible adiabatic processes. Then we can start at A and move to B. Since the energy is lowered, the system performs work along this path but, because the path is adiabatic, no heat is exchanged. Now let\u2019s move from B to C. Because dV = dX = 0 on this trajectory, we do no work but the internal energy E changes so the system must absorb heat Q from the surroundings. Now \ufb01nally we do work on the system to move from C back to A. However, unlike in the Carnot cycle, we don\u2019t return any heat to the environment on this return journey because, by assumption, this second path is also adiabatic. The net result is that we have extracted heat Q and employed this to undertake work W = Q. This is in contradiction with Kelvin\u2019s statement of the second law. Figure 33: The upshot of this argument is that the space of states can be foliated by adiabatic surfaces such that each vertical line at constant V and X intersects the surface only once. We can describe these surfaces by some function S(E, V, X) = constant. This function is the entropy. The above argument shows that Kelvin\u2019s statement of the second law implies the existence of adiabatic surfaces. One may wonder if we can run the argument the other way around and use the existence of adiabatic surfaces as the basis of the second law, dispensing with the Kelvin and Clausius postulates all together. In fact, we can almost do this. From the discussion above it should already be clear that the existence of \u2013 125 \u2013 EXVACB adiabatic surfaces implies that the addition of heat is proportional to the change in entropy \u2212dQ \u223c dS. However, it remains to show that the integrating factor, relating the two, is temperature so \u2212dQ = T dS. This can be done by returning to the zeroth law of thermodynamics. A fairly simple description of the argument can be found at the end of Chapter 4 of Pippard\u2019s book. This motivates a mathematically concise statement of the second law due to Carath\u00b4eodory. Second Law `a la Carath\u00b4eodory: Adiabatic surfaces exist. Or, more poetically: if you want to be able to return, there are places you cannot go through work alone. Sometimes you need a little heat. What this statement is lacking is perhaps the most important aspect of the second law: an arrow of time. But this is easily remedied by providing one additional piece of information telling us which side of a surface can be reached by irreversible processes. To one side of the surface lies the future, to the other the past. 4.3.5 A History of Thermodynamics The history of heat and thermodynamics is long and complicated, involving wrong turns, insights from disparate areas of study such as engineering and medicine, and many interesting characters, more than one of which \ufb01nd reason to change their name at some point in the story10. Although ideas of \u201cheat\u201d date back to pre-history, a good modern starting point is the 1787 caloric theory of Lavoisier. This postulates that heat is a conserved \ufb02uid which has a tendency to repel itself, thereby \ufb02owing from hot bodies to cold bodies. It was, for its time, an excellent theory, explaining many of the observed features of heat. Of course, it was also wrong. Lavoisier\u2019s theory was still prominent 30 years later when the French engineer Sadi Carnot undertook the analysis of steam engines that we saw above. Carnot understood all of his processes in terms of caloric. He was inspired by mechanics of waterwheels and saw the \ufb02ow of caloric from hot to cold bodies as analogous to the fall of water from high to low. This work was subsequently extended and formalised in a more mathematical framework by another French physicist, \u00b4Emile Clapeyron. By the 1840s, the properties of heat were viewed by nearly everyone through the eyes of Carnot-Clapeyron caloric theory. 10A longer description of the history of heat can be found in Michael Fowler\u2019s lectures from the University of Virginia: http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/HeatIndex.htm \u2013 126 \u2013 Yet the \ufb01rst cracks in caloric theory has already appeared before the turn of 19th century due to the work of Benjamin Thompson. Born in the English colony", " of Massachusetts, Thompson\u2019s CV includes turns as mercenary, scientist and humanitarian. He is the inventor of thermal underwear and pioneer of the soup kitchen for the poor. By the late 1700s, Thompson was living in Munich under the glorious name \u201cCount Rumford of the Holy Roman Empire\u201d where he was charged with overseeing artillery for the Prussian Army. But his mind was on loftier matters. When boring cannons, Rumford was taken aback by the amount of heat produced by friction. According to Lavoisier\u2019s theory, this heat should be thought of as caloric \ufb02uid squeezed from the brass cannon. Yet is seemed inexhaustible: when a cannon was bored a second time, there was no loss in its ability to produce heat. Thompson/Rumford suggested that the cause of heat could not be a conserved caloric. Instead he attributed heat correctly, but rather cryptically, to \u201cmotion\u201d. Having put a big dent in Lavoisier\u2019s theory, Rumford rubbed salt in the wound by marrying his widow. Although, in fairness, Lavoisier was beyond caring by this point. Rumford was later knighted by Britain, reverting to Sir Benjamin Thompson, where he founded the Royal Institution. The journey from Thompson\u2019s observation to an understanding of the \ufb01rst law of thermodynamics was a long one. Two people in particular take the credit. In Manchester, England, James Joule undertook a series of extraordinarily precise experiments. He showed how di\ufb00erent kinds of work \u2014 whether mechanical or electrical \u2013 could be used to heat water. Importantly, the amount by which the temperature was raised depended only on the amount of work, not the manner in which it was applied. His 1843 paper \u201cThe Mechanical Equivalent of Heat\u201d provided compelling quantitative evidence that work could be readily converted into heat. But Joule was apparently not the \ufb01rst. A year earlier, in 1842, the German physician Julius von Mayer came to the same conclusion through a very di\ufb00erent avenue of investigation: blood letting. Working on a ship in the Dutch East Indies, Mayer noticed that the blood in sailors veins was redder in Germany. He surmised that this was because the body needed to burn less fuel to keep warm. Not only did he essentially \ufb01gure out how the process of oxidation is responsible for supplying the body\u2019s energy but, remarkably, he was able to push this to an understanding of how work and heat are related. Despite limited physics training, he used his intuition, together with known experimental values of the heat capacities Cp and CV of gases, to determine essentially the same result as Joule had found through more direct means. \u2013 127 \u2013 The results of Thompson, Mayer and Joule were synthesised in an 1847 paper by Hermann von Helmholtz, who is generally credited as the \ufb01rst to give a precise formulation of the \ufb01rst law of thermodynamics. (Although a guy from Swansea called William Grove has a fairly good, albeit somewhat muddled, claim from a few years earlier). It\u2019s worth stressing the historical importance of the \ufb01rst law: this was the \ufb01rst time that the conservation of energy was elevated to a key idea in physics. Although it had been known for centuries that quantities such as \u201c 1 2mv2 + V \u201d were conserved in certain mechanical problems, this was often viewed as a mathematical curiosity rather than a deep principle of Nature. The reason, of course, is that friction is important in most processes and energy does not appear to be conserved. The realisation that there is a close connection between energy, work and heat changed this. However, it would still take more than half a century before Emmy Noether explained the true reason behind the conservation of energy. With Helmholtz, the \ufb01rst law was essentially nailed. The second remained. This took another two decades, with the pieces put together by a number of people, notably William Thomson and Rudolph Clausius. William Thomson was born in Belfast but moved to Glasgow at the age of 10. He came to Cambridge to study, but soon returned to Glasgow and stayed there for the rest of his life. After his work as a scientist, he gained fame as an engineer, heavily involved in laying the \ufb01rst trans-atlantic cables. For this he was made Lord Kelvin, the name chosen for the River Kelvin which \ufb02ows nearby Glasgow University. He was the \ufb01rst to understand the importance of absolute zero and to de\ufb01ne the thermodynamic temperature scale which now bears his favourite river\u2019s name. We presented Kelvin\u2019s statement of the second law of thermodynamics earlier in this Section. In Germany, Rudolph Clausius was developing the same ideas as Kelvin. But he managed to go further and, in 1865, presented the subtle therm", "odynamic argument for the existence of entropy that we saw in Section 4.3.3. Modestly, Clausius introduced the unit \u201cClausius\u201d (symbol Cl) for entropy. It didn\u2019t catch on. 4.4 Thermodynamic Potentials: Free Energies and Enthalpy We now have quite a collection of thermodynamic variables. The state of the system is dictated by pressure p and volume V. From these, we can de\ufb01ne temperature T, energy E and entropy S. We can also mix and match. The state of the system can just as well be labelled by T and V ; or E and V ; or T and p; or V and S... \u2013 128 \u2013 While we\u2019re at liberty to pick any variables we like, certain quantities are more naturally expressed in terms of some variables instead of others. We\u2019ve already seen examples both in Section 1 and in this section. If we\u2019re talking about the energy E, it is best to label the state in terms of S and V, so E = E(S, V ). In these variables the \ufb01rst law has the nice form (4.11). Equivalently, inverting this statement, the entropy should be thought of as a function of E and V, so S = S(E, V ). It is not just mathematical niceties underlying this: it has physical meaning too for, as we\u2019ve seen above, at \ufb01xed energy the second law tells us that entropy can never decrease. What is the natural object to consider at constant temperature T, rather than constant energy? In fact we already answered this way back in Section 1.3 where we argued that one should minimise the Helmholtz free energy, F = E \u2212 T S The arguments that we made back in Section 1.3 were based on a microscopic viewpoint of entropy. But, with our thermodynamic understanding of the second law, we can easily now repeat the argument without mention of probability distributions. We consider our system in contact with a heat reservoir such that the total energy, Etotal of the combined system and reservoir is \ufb01xed. The combined entropy is then, Stotal(Etotal) = SR(Etotal \u2212 E) + S(E) \u2248 SR(Etotal) \u2212 = SR(Etotal) \u2212 \u2202SR \u2202Etotal F T E + S(E) The total entropy can never decrease; the free energy of the system can never increase. One interesting situation that we will meet in the next section is a system which, at \ufb01xed temperature and volume, has two di\ufb00erent equilibrium states. Which does it choose? The answer is the one that has lower free energy, for random thermal \ufb02uctuations will tend to take us to this state, but very rarely bring us back. We already mentioned in Section 1.3 that the free energy is a Legendre transformation of the energy; it is most naturally thought of as a function of T and V, which is re\ufb02ected in the in\ufb01nitesimal variation, dF = \u2212SdT \u2212 pdV \u21d2 \u2202F \u2202T V = \u2212S, \u2202F \u2202V T = \u2212p (4.19) \u2013 129 \u2013 We can now also explain what\u2019s free about this energy. Consider taking a system along a reversible isotherm, from state A to state B. Because the temperature is constant, the change in free energy is dF = \u2212pdV, so F (B) \u2212 F (A) = \u2212 B A pdV = \u2212W where W is the work done by the system. The free energy is a measure of the amount of energy free to do work at \ufb01nite temperature. Gibbs Free Energy We can also consider systems that don\u2019t live at \ufb01xed volume, but instead at \ufb01xed pressure. To do this, we will once again imagine the system in contact with a reservoir at temperature T. The volume of each can \ufb02uctuate, but the total volume Vtotal of the combined system and reservoir is \ufb01xed. The total entropy is Stotal(Etotal, Vtotal) = SR(Etotal \u2212 E, Vtotal \u2212 V ) + S(E, V ) \u2202SR \u2202Vtotal \u2248 SR(Etotal, Vtotal) \u2212 E \u2212 V + S(E, V ) \u2202SR \u2202Etotal E + pV \u2212 T S T = SR(Vtotal) \u2212 At \ufb01xed temperature and pressure we should minimise the Gibbs Free Energy, G = F + pV = E + pV \u2212 T S (4.20) This is a Legendre transform of F, this time swapping volume for pressure: G = G(T", ", p). The in\ufb01nitesimal variation is dG = \u2212SdT + V dp In our discussion we have ignored the particle number N. Yet both F and G implicitly depend on N (as you may check by re-examining the many examples of F that we computed earlier in the course). If we also consider changes dN then each variations gets the additional term \u00b5dN, so dF = \u2212SdT \u2212 pdV + \u00b5dN and dG = \u2212SdT + V dp + \u00b5dN (4.21) While F can have an arbitrarily complicated dependence on N, the Gibbs free energy G has a very simple dependence. To see this, we simply need to look at the extensive properties of the di\ufb00erent variables and make the same kind of argument that we\u2019ve already seen in Section 1.4.1. From its de\ufb01nition (4.20), we see that the Gibbs free \u2013 130 \u2013 energy G is extensive. Therefore, It a function of p, T and N, of which only N is extensive. G(p, T, N ) = \u00b5(p, T )N (4.22) where the fact that the proportionality coe\ufb03cient is \u00b5 follows from variation (4.21) which tells us that \u2202G/\u2202N = \u00b5. The Gibbs free energy is frequently used by chemists, for whom reactions usually take place at constant pressure rather than constant volume. (When a chemist talks about \u201cfree energy\u201d, they usually mean G. For a physicist, \u201cfree energy\u201d usually means F ). We\u2019ll make use of the result (4.22) in the next section when we discuss \ufb01rst order phase transitions. 4.4.1 Enthalpy There is one \ufb01nal combination that we can consider: systems at \ufb01xed energy and pressure. Such systems are governed by the enthalpy, H = E + pV \u21d2 dH = T dS + V dp The four objects E, F, G and H are sometimes referred to as thermodynamic potentials. 4.4.2 Maxwell\u2019s Relations Each of the thermodynamic potentials has an interesting present for us. Let\u2019s start by considering the energy. Like any function of state, it can be viewed as a function of any of the other two variables which specify the system. However, the \ufb01rst law of thermodynamics (4.11) suggests that it is most natural to view energy as a function of entropy and volume: E = E(S, V ). This has the advantage that the partial derivatives are familiar quantities, \u2202E \u2202S V = T, \u2202E \u2202V S = \u2212p We saw both of these results in Section 1. It is also interesting to look at the double mixed partial derivative, \u22022E/\u2202S\u2202V = \u22022E/\u2202V \u2202S. This gives the relation \u2202T \u2202V S = \u2212 \u2202p \u2202S V (4.23) This result is mathematically trivial. Yet physically it is far from obvious. It is the \ufb01rst of four such identities, known as the Maxwell Relations. \u2013 131 \u2013 The other Maxwell relations are derived by playing the same game with F, G and H. From the properties (4.19), we see that taking mixed partial derivatives of the free energy gives us, The Gibbs free energy gives us While the enthalpy gives \u2202S \u2202V T = \u2202p \u2202T V \u2202S \u2202p T = \u2212 \u2202V \u2202T p \u2202T \u2202p S = \u2202V \u2202S p (4.24) (4.25) (4.26) The four Maxwell relations (4.23), (4.24), (4.25) and (4.26) are remarkable in that they are mathematical identities that hold for any system. They are particularly useful because they relate quantities which are directly measurable with those which are less easy to determine experimentally, such as entropy. It is not too di\ufb03cult to remember the Maxwell relations. Cross-multiplication always yields terms in pairs: T S and pV, which follows essentially on dimensional grounds. The four relations are simply the four ways to construct such equations. The only tricky part is to \ufb01gure out the minus signs. Heat Capacities Revisted By taking further derivatives of the Maxwell relations, we can derive yet more equations which involve more immediate quantities. You will be asked to prove a number of these on the examples sheet, including results for the heat capacity at constant volume, CV = T \u2202S/\u2202T |V, as well as the heat capacity at capacity at", " constant pressure Cp = T \u2202S/\u2202T |p. Useful results include, \u2202CV \u2202V T = T \u22022p \u2202T 2 V, \u2202Cp \u2202p T = \u2212T \u22022V \u2202T 2 p You will also prove a relationship between these two heat capacities, Cp \u2212 CV = T \u2202V \u2202T p \u2202p \u2202T V \u2013 132 \u2013 This last expression has a simple consequence. Consider, for example, an ideal gas obeying pV = N kBT. Evaluating the right-hand side gives us Cp \u2212 Cv = N kB There is an intuitive reason why Cp is greater than CV. At constant volume, if you dump heat into a system then it all goes into increasing the temperature. However, at constant pressure some of this energy will cause the system to expand, thereby doing work. This leaves less energy to raise the temperature, ensuring that Cp > CV. 4.5 The Third Law The second law only talks about entropy di\ufb00erences. We can see this in (4.10) where the entropy is de\ufb01ned with respect to some reference state. The third law, sometimes called Nernst\u2019s postulate, provides an absolute scale for the entropy. It is usually taken to be lim T \u21920 S(T ) = 0 In fact we can relax this slightly to allow a \ufb01nite entropy, but vanishing entropy density S/N. We know from the Boltzmann de\ufb01nition that, at T = 0, the entropy is simply the logarithm of the degeneracy of the ground state of the system. The third law really requires S/N \u2192 0 as T \u2192 0 and N \u2192 \u221e. This then says that the ground state entropy shouldn\u2019t grow extensively with N. The third law doesn\u2019t quite have the same teeth as its predecessors. Each of the \ufb01rst three laws provided us with a new function of state of the system: the zeroth law gave us temperature; the \ufb01rst law energy; and the second law entropy. There is no such reward from the third law. One immediate consequence of the third law is that heat capacities must also tend to zero as T \u2192 0. This follows from the equation (1.10) S(B) \u2212 S(A) = B A dT CV T If the entropy at zero temperature is \ufb01nite then the integral must converge which tells us that CV \u2192 T n for some n \u2265 1 or faster. Looking back at the various examples of heat capacities, we can check that this is always true. (The case of a degenerate Fermi gas is right on the borderline with n = 1). However, in each case the fact that the heat capacity vanishes is due to quantum e\ufb00ects freezing out degrees of freedom. \u2013 133 \u2013 In contrast, when we restrict to the classical physics, we typically \ufb01nd constant heat capacities such as the classical ideal gas (2.10) or the Dulong-Petit law (3.15). These would both violate the third law. In that sense, the third law is an admission that the low temperature world is not classical. It is quantum. Thinking about things quantum mechanically, it is very easy to see why the third law holds. A system that violates the third law would have a very large \u2013 indeed, an extensive \u2013 number of ground states. But large degeneracies do not occur naturally in quantum mechanics. Moreover, even if you tune the parameters of a Hamiltonian so that there are a large number of ground states, then any small perturbation will lift the degeneracy, introducing an energy splitting between the states. From this perspective, the third law is a simple a consequence of the properties of the eigenvalue problem for large matrices. \u2013 134 \u2013 5. Phase Transitions A phase transition is an abrupt, discontinuous change in the properties of a system. We\u2019ve already seen one example of a phase transition in our discussion of Bose-Einstein condensation. In that case, we had to look fairly closely to see the discontinuity: it was lurking in the derivative of the heat capacity. In other phase transitions \u2014 many of them already familiar \u2014 the discontinuity is more manifest. Examples include steam condensing to water and water freezing to ice. In this section we\u2019ll explore a couple of phase transitions in some detail and extract some lessons that are common to all transitions. 5.1 Liquid-Gas Transition Recall that we derived the van der Waals equation of state for a gas (2.31) in Section 2.5. We can write the van der Waals equation as p = kBT v \u2212 b \u2212 a v2 (5.1) where v = V /N is the volume per particle. equation written in terms", " of the particle density \u03c1 = 1/v. In the literature, you will also see this On the right we \ufb01x T at di\ufb00erent values and sketch the graph of p vs. V determined by the van der Waals equation. These curves are isotherms \u2014 line of constant temperature. As we can see from the diagram, the isotherms take three di\ufb00erent shapes depending on the value of T. The top curve shows the isotherm for large values of T. Here we can e\ufb00ectively ignore the \u2212a/v2 term. (Recall that v cannot take values smaller than b, re\ufb02ecting the fact that atoms cannot approach to arbitrarily closely). The result is a monotonically decreasing function, essentially the same as we would get for an ideal gas. In contrast, when T is low enough, the second term in (5.1) can compete with the \ufb01rst term. Roughly speaking, this happens when kBT \u223c a/v is in the allowed region v > b. For these low value of the temperature, the isotherm has a wiggle. Figure 34: \u2013 135 \u2013 bpvT>TT=TT<Tcccc At some intermediate temperature, the wiggle must \ufb02atten out so that the bottom curve looks like the top one. This happens when the maximum and minimum meet to form an in\ufb02ection point. Mathematically, we are looking for a solution to dp/dv = d2p/dv2 = 0. it is simple to check that these two equations only have a solution at the critical temperature T = Tc given by kBTc = 8a 27b (5.2) Let\u2019s look in more detail at the T < Tc curve. For a range of pressures, the system can have three di\ufb00erent choices of volume. A typical, albeit somewhat exagerated, example of this curve is shown in the \ufb01gure below. What\u2019s going on? How should we interpret the fact that the system can seemingly live at three di\ufb00erent densities \u03c1 = 1/v? First look at the middle solution. This has some fairly weird properties. We can see from the graph that the gradient is positive: dp/dv|T > 0. This means that if we apply a force to the container to squeeze the gas, the pressure decreases. The gas doesn\u2019t push back; it just relents. But if we expand the gas, the pressure increases and the gas pushes harder. Both of these properties are telling us that the gas in that state is unstable. If we were able to create such a state, it wouldn\u2019t hand around for long because any tiny perturbation would lead to a rapid, explosive change in its density. If we want to \ufb01nd states which we are likely to observe in Nature then we should look at the other two solutions. Figure 35: The solution to the left on the graph has v slightly bigger than b. But, recall from our discussion of Section 2.5 that b is the closest that the atoms can get. If we have v \u223c b, then the atoms are very densely packed. Moreover, we can also see from the graph that |dp/dv| is very large for this solution which means that the state is very di\ufb03cult to compress: we need to add a great deal of pressure to change the volume only slightly. We have a name for this state: it is a liquid. You may recall that our original derivation of the van der Waals equation was valid only for densities much lower than the liquid state. This means that we don\u2019t really trust (5.1) on this solution. Nonetheless, it is interesting that the equation predicts the existence of liquids and our plan is to gratefully accept this gift and push ahead to explore what the van der Waals tells us about the liquid-gas transition. We will see that it captures many of the qualitative features of the phase transition. \u2013 136 \u2013 bpT<Tcv The last of the three solutions is the one on the right in the \ufb01gure. This solution has v b and small |dp/dv|. It is the gas state. Our goal is to understand what happens in between the liquid and gas state. We know that the naive, middle, solution given to us by the van der Waals equation is unstable. What replaces it? 5.1.1 Phase Equilibrium Throughout our derivation of the van der Waals equation in Section 2.5, we assumed that the system was at a \ufb01xed density. But the presence of two solutions \u2014 the liquid and gas state \u2014 allows us to consider more general con\ufb01gurations: part of the system could be a liquid and part could be", " a gas. How do we \ufb01gure out if this indeed happens? Just because both liquid and gas states can exist, doesn\u2019t mean that they can cohabit. It might be that one is preferred over the other. We already saw some conditions that must be satis\ufb01ed in order for two systems to sit in equilibrium back in Section 1. Mechanical and thermal equilibrium are guaranteed if two systems have the same pressure and temperature respectively. But both of these are already guaranteed by construction for our two liquid and gas solutions: the two solutions sit on the same isotherm and at the same value of p. We\u2019re left with only one further requirement that we must satisfy which arises because the two systems can exchange particles. This is the requirement of chemical equilibrium, \u00b5liquid = \u00b5gas (5.3) Because of the relationship (4.22) between the chemical potential and the Gibbs free energy, this is often expressed as gliquid = ggas (5.4) where g = G/N is the Gibbs free energy per particle. Notice that all the equilibrium conditions involve only intensive quantities: p, T and \u00b5. This means that if we have a situation where liquid and gas are in equilibrium, then we can have any number Nliquid of atoms in the liquid state and any number Ngas in the gas state. But how can we make sure that chemical equilibrium (5.3) is satis\ufb01ed? Maxwell Construction We want to solve \u00b5liquid = \u00b5gas. We will think of the chemical potential as a function of p and T : \u00b5 = \u00b5(p, T ). Importantly, we won\u2019t assume that \u00b5(p, T ) is single valued since that would be assuming the result we\u2019re trying to prove! Instead we will show that if we \ufb01x T, the condition (5.3) can only be solved for a very particular value of pressure \u2013 137 \u2013 p. To see this, start in the liquid state at some \ufb01xed value of p and T and travel along the isotherm. The in\ufb01nitesimal change in the chemical potential is d\u00b5 = \u2202\u00b5 \u2202p T dp However, we can get an expression for \u2202\u00b5/\u2202p by recalling that arguments involving extensive and intensive variables tell us that the chemical potential is proportional to the Gibbs free energy: G(p, T, N ) = \u00b5(p, T )N (4.22). Looking back at the variation of the Gibbs free energy (4.21) then tells us that T N,T \u2202G \u2202p N = V \u2202\u00b5 \u2202p (5.5) = Integrating along the isotherm then tells us the chemical potential of any point on the curve, \u00b5(p, T ) = \u00b5liquid + p pliquid dp V (p, T ) N When we get to gas state at the same pressure p = pliquid that we started from, the condition for equilibrium is \u00b5 = \u00b5liquid. Which means that the integral has to vanish. Graphically this is very simple to describe: the two shaded areas in the graph must have equal area. This condition, known as the Maxwell construction, tells us the pressure at which gas and liquid can co-exist. Figure 36: I should confess that there\u2019s something slightly dodgy about the Maxwell construction. We already argued that the part of the isotherm with dp/dv > 0 su\ufb00ers an instability and is unphysical. But we needed to trek along that part of the curve to derive our result. There are more rigorous arguments that give the same answer. For each isotherm, we can determine the pressure at which the liquid and gas states are in equilibrium. The gives us the co-existence curve, shown by the dotted line in Figure 37. Inside this region, liquid and gas can both exist at the same temperature and pressure. But there is nothing that tells us how much gas there should be and how much liquid: atoms can happily move from the liquid state to the gas state. This means that while the density of gas and liquid is \ufb01xed, the average density of the system is not. It can vary between the gas density and the liquid density simply by changing the amount of liquid. The upshot of this argument is that inside the co-existence curves, the isotherms simply become \ufb02at lines, re\ufb02ecting the fact that the density can take any value. This is shown in graph on the right of Figure 37. \u2013 138 \u2013 bpT<Tcv Figure 37: The co-existence curve in red, resulting in constant pressure regions consisting of a harmonious mixture of vapour and liquid. To illustrate the physics of this situation, suppose that we sit at some \ufb01xed density \u03c1 = 1/v and cool the", " system down from a high temperature to T < Tc at a point inside the co-existence curve so that we\u2019re now sitting on one of the \ufb02at lines. Here, the system is neither entirely liquid, nor entirely gas. Instead it will split into gas, with density 1/vgas, and liquid, with density 1/vliquid so that the average density remains 1/v. The system undergoes phase separation. The minimum energy con\ufb01guration will typically be a single phase of liquid and one of gas because the interface between the two costs energy. (We will derive an expression for this energy in Section 5.5). The end result is shown on the right. In the presence of gravity, the higher density liquid will indeed sink to the bottom. Figure 38: Meta-Stable States We\u2019ve understood what replaces the unstable region of the van der Waals phase diagram. But we seem to have removed more states than anticipated: parts of the Van der Waals isotherm that had dp/dv < 0 are contained in the co-existence region and replaced by the \ufb02at pressure lines. This is the region of the p-V phase diagram that is contained between the two dotted lines in the \ufb01gure to the right. The outer dotted line is the co-existence curve. The inner dotted curve is constructed to pass through the stationary points of the van der Waals isotherms. It is called the spinodal curve. \u2013 139 \u2013 Figure 39: pvcT=TpvcT=Tvp The van der Waals states which lie between the spinodal curve and the co-existence curve are good states. But they are meta-stable. One can show that their Gibbs free energy is higher than that of the liquid-gas equilibrium at the same p and T. However, if we compress the gas very slowly we can coax the system into this state. It is known as a supercooled vapour. It is delicate. Any small disturbance will cause some amount of the gas to condense into the liquid. Similarly, expanding a liquid beyond the coexistence curve results in an meta-stable, superheated liquid. 5.1.2 The Clausius-Clapeyron Equation We can also choose to plot the liquid-gas phase diagram on the p \u2212 T plane. Here the co-existence region is squeezed into a line: if we\u2019re sitting in the gas phase and increase the pressure just a little bit at at \ufb01xed T < Tc then we jump immediately to the liquid phase. This appears as a discontinuity in the volume. Such discontinuities are the sign of a phase transition. The end result is sketched in the \ufb01gure to the right; the thick solid line denotes the presence of a phase transition. Figure 40: Either side of the line, all particles are either in the gas or liquid phase. We know from (5.4) that the Gibbs free energies (per particle) of these two states are equal, gliquid = ggas So G is continuous as we move across the line of phase transitions. Suppose that we sit on the line itself and move along it. How does g change? We can easily compute this from (4.21), dGliquid = \u2212SliquiddT + Vliquiddp \u21d2 dgliquid = \u2212sliquiddT + vliquiddp where, just as g = G/N an v = V /N, the entropy density is s = S/N. Equating this with the free energy in the gaseous phase gives dgliquid = \u2212sliquiddT + vliquiddp = dggas = \u2212sgasdT + vgasdp This can be rearranged to gives us a nice expression for the slope of the line of phase transitions in the p \u2212 T plane. It is dp dT = sgas \u2212 sliquid vgas \u2212 vliquid \u2013 140 \u2013 Tcliquidgascritical pointpT We usually de\ufb01ne the speci\ufb01c latent heat L = T (sgas \u2212 sliquid) This is the energy released per particle as we pass through the phase transition. We see that the slope of the line in the p \u2212 T plane is determined by the ratio of latent heat released in the phase transition and the discontinuity in volume. The result is known as the Clausius-Clapeyron equation, dp dT = L T (vgas \u2212 vliquid) (5.6) There is a classi\ufb01cation of phase transitions, due originally to Ehrenfest. When the nth derivative of a thermodynamic potential (either F or G usually) is discontinuous, we say we have an nth order phase transition. In practice, we nearly always deal with \ufb01rst, second and (very rarely) third order transitions", ". The liquid-gas transition releases latent heat, which means that S = \u2212\u2202F/\u2202T is discontinuous. Alternatively, we can say that V = \u2202G/\u2202p is discontinuous. Either way, it is a \ufb01rst order phase transition. The Clausius-Clapeyron equation (5.6) applies to any \ufb01rst order transition. As we approach T \u2192 Tc, the discontinuity diminishes and Sliquid \u2192 Sgas. At the critical point T = Tc we have a second order phase transition. Above the critical point, there is no sharp distinction between the gas phase and liquid phase. For most simple materials, the phase diagram above is part of a larger phase diagram which includes solids at smaller temperatures or higher pressures. A generic version of such a phase diagram is shown to the right. The van der Waals equation is missing the physics of solidi\ufb01cation and includes only the liquid-gas line. Figure 41: An Approximate Solution to the Clausius-Clapeyron Equation We can solve the Clausius-Clapeyron solution if we make the following assumptions: \u2022 The latent heat L is constant. \u2022 vgas vliquid, so vgas \u2212 vliquid \u2248 vgas. For water, this is an error of less than 0.1% \u2022 Although we derived the phase transition using the van der Waals equation, now we\u2019ve got equation (5.6) we\u2019ll pretend the gas obeys the ideal gas law pv = kBT. \u2013 141 \u2013 pTTcsolidgasliquid With these assumptions, it is simple to solve (5.6). It reduces to dp dT = Lp kBT 2 \u21d2 p = p0e\u2212L/kBT 5.1.3 The Critical Point Let\u2019s now return to discuss some aspects of life at the critical point. We previously worked out the critical temperature (5.2) by looking for solutions to simultaneous equations \u2202p/\u2202v = \u22022p/\u2202v2 = 0. There\u2019s a slightly more elegant way to \ufb01nd the critical point which also quickly gives us pc and vc as well. We rearrange the van der Waals equation (5.1) to get a cubic, pv3 \u2212 (pb + kBT )v2 + av \u2212 ab = 0 For T < Tc, this equation has three real roots. For T > Tc there is just one. Precisely at T = Tc, the three roots must therefore coincide (before two move o\ufb00 onto the complex plane). At the critical point, this curve can be written as pc(v \u2212 vc)3 = 0 Comparing the coe\ufb03cients tells us the values at the critical point, kBTc = 8a 27b, vc = 3b, pc = a 27b2 (5.7) The Law of Corresponding States We can invert the relations (5.7) to express the parameters a and b in terms of the critical values, which we then substitute back into the van der Waals equation. To this end, we de\ufb01ne the reduced variables, \u00afT = T Tc, \u00afv = v vc \u00afp = p pc The advantage of working with \u00afT, \u00afv and \u00afp is that it allows us to write the van der Waals equation (5.1) in a form that is universal to all gases, usually referred to as the law of corresponding states \u00afp = \u00afT \u00afv \u2212 1/3 8 3 \u2212 3 \u00afv2 Moreover, because the three variables Tc, pc and vc at the critical point are expressed in terms of just two variables, a and b (5.7), we can construct a combination of them \u2013 142 \u2013 Figure 42: The co-existence curve for gases. Data is plotted for N e, Ar, Kr,Xe, N2, O2, CO and CH4. which is independent of a and b and therefore supposedly the same for all gases. This is the universal compressibility ratio, pcvc kBTc = 0.375 (5.8) 3 8 = Comparing to real gases, this number is a little high. Values range from around 0.28 to 0.3. We shouldn\u2019t be too discouraged by this; after all, we knew from the beginning that the van der Waals equation is unlikely to be accurate in the liquid regime. Moreover, the fact that gases have a critical point (de\ufb01ned by three variables Tc, pc and vc) guarantees that a similar relationship would hold for any equation of state which includes just two parameters (such as a and b) but would most likely fail to hold for equations of state that included more than two parameters. Dubious as its", " theoretical foundation is, the law of corresponding states is the \ufb01rst suggestion that something remarkable happens if we describe a gas in terms of its reduced variables. More importantly, there is striking experimental evidence to back this up! Figure 42 shows the Guggenheim plot, constructed in 1945. The co-existence curve for 8 di\ufb00erent gases in plotted in reduced variables: \u00afT along the vertical axis; \u00af\u03c1 = 1/\u00afv along the horizontal. The gases vary in complexity from the simple monatomic gas N e to the molecule CH4. As you can see, the co-existence curve for all gases is essentially the same, with the chemical make-up largely forgotten. There is clearly something interesting going on. How to understand it? Critical Exponents We will focus attention on physics close to the critical point. It is not immediately \u2013 143 \u2013 obvious what are the right questions to ask. It turns out that the questions which have the most interesting answer are concerned with how various quantities change as we approach the critical point. There are lots of ways to ask questions of this type since there are many quantities of interest and, for each of them, we could approach the critical point from di\ufb00erent directions. Here we\u2019ll look at the behaviour of three quantities to get a feel for what happens. First, we can ask what happens to the di\ufb00erence in (inverse) densities vgas \u2212 vliquid as we approach the critical point along the co-existence curve. For T < Tc, or equivalently \u00afT < 1, the reduced van der Waals equation (5.8) has two stable solutions, \u00afp = 8 \u00afT 3\u00afvliquid \u2212 1 \u2212 3 \u00afv2 liquid = 8 \u00afT 3\u00afvgas \u2212 1 \u2212 3 \u00afv2 gas If we solve this for \u00afT, we have \u00afT = (3\u00afvliquid \u2212 1)(3\u00afvgas \u2212 1)(\u00afvliquid + \u00afvgas) gas\u00afv2 8\u00afv2 liquid Notice that as we approach the critical point, \u00afvgas, \u00afvliquid \u2192 1 and the equation above tells us that \u00afT \u2192 1 as expected. We can see exactly how we approach \u00afT = 1 by expanding the right right-hand side for small \u2261 \u00afvgas \u2212 \u00afvliquid. To do this quickly, it\u2019s best to notice that the equation is symmetric in \u00afvgas and \u00afvliquid, so close to the critical point we can write \u00afvgas = 1 + /2 and \u00afvliquid = 1 \u2212 /2. Substituting this into the equation above and keeping just the leading order term, we \ufb01nd \u00afT \u2248 1 \u2212 1 16 (\u00afvgas \u2212 \u00afvliquid)2 Or, re-arranging, as we approach Tc along the co-existence curve, vgas \u2212 vliquid \u223c (Tc \u2212 T )1/2 (5.9) This is the answer to our \ufb01rst question. Our second variant of the question is: how does the volume change with pressure as we move along the critical isotherm. It turns out that we can answer this question without doing any work. Notice that at T = Tc, there is a unique pressure for a given volume p(v, Tc). But we know that \u2202p/\u2202v = \u22022p/\u2202v2 = 0 at the critical point. So a Taylor expansion around the critical point must start with the cubic term, p \u2212 pc \u223c (v \u2212 vc)3 (5.10) This is the answer to our second question. \u2013 144 \u2013 Our third and \ufb01nal variant of the question concerns the compressibility, de\ufb01ned as \u03ba = \u2212 1 v \u2202v \u2202p T (5.11) We want to understand how \u03ba changes as we approach T \u2192 Tc from above. In fact, we met the compressibility before: it was the feature that \ufb01rst made us nervous about the van der Waals equation since \u03ba is negative in the unstable region. We already know = 0. So expanding for temperatures close to Tc, we that at the critical point \u2202p/\u2202v|Tc expect \u2202p \u2202v T ;v=vc = \u2212a(T \u2212 Tc) +... This tells us that the compressibility should diverge at the critical point, scaling as \u03ba \u223c (T \u2212 Tc)\u22121 (5.12) We now have three answers to three questions: (5.9), (5.10) and (5.12). Are they right?! By which I mean: do they agree with experiment? Remember that we\u2019re not sure that we can trust the van der Waals equation at the critical point so we should be nervous. However, there is also", " reason for some con\ufb01dence. Notice, in particular, that in order to compute (5.10) and (5.12), we didn\u2019t actually need any details of the van der Waals equation. We simply needed to assume the existence of the critical point and an analytic Taylor expansion of various quantities in the neighbourhood. Given that the answers follow from such general grounds, one may hope that they provide the correct answers for a gas in the neighbourhood of the critical point even though we know that the approximations that went into the van der Waals equation aren\u2019t valid there. Fortunately, that isn\u2019t the case: physics is much more interesting than that! The experimental results for a gas in the neighbourhood of the critical point do share one feature in common with the discussion above: they are completely independent of the atomic make-up of the gas. However, the scaling that we computed using the van der Waals equation is not fully accurate. The correct results are as follows. As we approach the critical point along the co-existence curve, the densities scale as vgas \u2212 vliquid \u223c (Tc \u2212 T )\u03b2 with \u03b2 \u2248 0.32 (Note that the exponent \u03b2 has nothing to do with inverse temperature. We\u2019re just near the end of the course and running out of letters and \u03b2 is the canonical name for this exponent). As we approach along an isotherm, p \u2212 pc \u223c (v \u2212 vc)\u03b4 with \u03b4 \u2248 4.8 \u2013 145 \u2013 Finally, as we approach Tc from above, the compressibility scales as \u03ba \u223c (T \u2212 Tc)\u2212\u03b3 with \u03b3 \u2248 1.2 The quantities \u03b2, \u03b3 and \u03b4 are examples of critical exponents. We will see more of them shortly. The van der Waals equation provides only a crude \ufb01rst approximation to the critical exponents. Fluctuations We see that the van der Waals equation didn\u2019t do too badly in capturing the dynamics It gets the qualitative behaviour right, but fails on precise of an interacting gas. quantitative tests. So what went wrong? We mentioned during the derivation of the van der Waals equation that we made certain approximations that are valid only at low density. So perhaps it is not surprising that it fails to get the numbers right near the critical point v = 3b. But there\u2019s actually a deeper reason that the van der Waals equation fails: \ufb02uctuations. \u221a This is simplest to see in the grand canonical ensemble. Recall that back in Section N, which allowed us to happily work in the 1 that we argued that \u2206N/N \u223c 1/ grand canonical ensemble even when we actually had \ufb01xed particle number. In the context of the liquid-gas transition, \ufb02uctuating particle number is the same thing as \ufb02uctuating density \u03c1 = N/V. Let\u2019s revisit the calculation of \u2206N near the critical point. Using (1.45) and (1.48), the grand canonical partition function can be written as log Z = \u03b2V p(T, \u00b5), so the average particle number (1.42) is N = V \u2202p \u2202\u00b5 T,V We already have an expression for the variance in the particle number in (1.43), \u2206N 2 = 1 \u03b2 \u2202N \u2202\u00b5 T,V Dividing these two expressions, we have \u2206N 2 N = 1 V \u03b2 \u2202N \u2202\u00b5 T,V \u2202\u00b5 \u2202p T,V = 1 V \u03b2 \u2202N \u2202p T,V But we can re-write this expression using the general relationship between partial derivatives \u2202x/\u2202y|z \u2202y/\u2202z|x \u2202z/\u2202x|y = \u22121. We then have \u2206N 2 N = \u2212 1 \u03b2 \u2202N \u2202V p,T 1 V \u2202V \u2202p N,T \u2013 146 \u2013 This \ufb01nal expression relates the \ufb02uctuations in the particle number to the compressibility (5.11). But the compressibility is diverging at the critical point and this means that there are large \ufb02uctuations in the density of the \ufb02uid at this point. The result is that any simple equation of state, like the van der Waals equation, which works only with the average volume, pressure and density will miss this key aspect of the physics. Understanding how to correctly account for these \ufb02uctuations is the subject of critical phenomena. It has close links with the renormalization group and conformal \ufb01eld theory which also arise in particle physics and string theory. You will meet some of these ideas in next year\u2019s Statistical Field Theory course. Here we will turn to a", " di\ufb00erent phase transition which will allow us to highlight some of the key ideas. 5.2 The Ising Model The Ising model is one of the touchstones of modern physics; a simple system that exhibits non-trivial and interesting behaviour. The Ising model consists of N sites in a d-dimensional lattice. On each lattice site lives a quantum spin that can sit in one of two states: spin up or spin down. We\u2019ll call the eigenvalue of the spin on the ith lattice site si. If the spin is up, si = +1; if the spin is down, si = \u22121. The spins sit in a magnetic \ufb01eld that endows an energy advantage to those which point up, EB = \u2212B N i=1 si (A comment on notation: B should be properly denoted H. We\u2019re sticking with B to avoid confusion with the Hamiltonian. There is also a factor of the magnetic moment which has been absorbed into the de\ufb01nition of B). The lattice system with energy EB is equivalent to the two-state system that we \ufb01rst met when learning the techniques of statistical mechanics back in Section 1.2.3. However, the Ising model contains an additional complication that makes the sysem much more interesting: this is an interaction between neighbouring spins. The full energy of the system is therefore, E = \u2212J ij sisj \u2212 B si i (5.13) The notation ij means that we sum over all \u201cnearest neighbour\u201d pairs in the lattice. The number of such pairs depends both on the dimension d and the type of lattice. \u2013 147 \u2013 We\u2019ll denote the number of nearest neighbours as q. For example, in d = 1 a lattice has q = 2; in d = 2, a square lattice has q = 4. A square lattice in d dimensions has q = 2d. If J > 0, neighbouring spins prefer to be aligned (\u2191\u2191 or \u2193\u2193). In the context of magnetism, such a system is called a ferromagnet. If J < 0, the spins want to antialign (\u2191\u2193). This is an anti-ferromagnet. In the following, we\u2019ll choose J > 0 although for the level of discussion needed for this course, the di\ufb00erences are minor. We work in the canonical ensemble and introduce the partition function Z = {si} e\u2212\u03b2E[si] (5.14) While the e\ufb00ect of both J > 0 and B = 0 is to make it energetically preferable for the spins to align, the e\ufb00ect of temperature will be to randomize the spins, with entropy winnning out over energy. Our interest is in the average spin, or average magnetization i The Ising Model as a Lattice Gas m = 1 N si = 1 N \u03b2 \u2202log Z \u2202B (5.15) Before we develop techniques to compute the partition function (5.14), it\u2019s worth pointing out that we can drape slightly di\ufb00erent words around the mathematics of the Ising model. It need not be interpreted as a system of spins; it can also be thought of as a lattice description of a gas. To see this, consider the same d-dimensional lattice as before, but now with particles hopping between lattice sites. These particles have hard cores, so no more than one can sit on a single lattice site. We introduce the variable ni \u2208 {0, 1} to specify whether a given lattice site, labelled by i, is empty (ni = 0) or \ufb01lled (ni = 1). We can also introduce an attractive force between atoms by o\ufb00ering them an energetic reward if they sit on neighbouring sites. The Hamiltonian of such a lattice gas is given by E = \u22124J ninj \u2212 \u00b5 ni where \u00b5 is the chemical potential which determines the overall particle number. But this Hamiltonian is trivially the same as the Ising model (5.13) if we make the identi\ufb01cation ij i si = 2ni \u2212 1 \u2208 {\u22121, 1} The chemical potenial \u00b5 in the lattice gas plays the role of magnetic \ufb01eld in the spin system while the magnetization of the system (5.15) measures the average density of particles away from half-\ufb01lling. \u2013 148 \u2013 5.2.1 Mean Field Theory For general lattices, in arbitrary dimension d, the sum (5.14) cannot be performed. An exact solution exists in d = 1 and, when B = 0, in d = 2. (The d = 2 solution is originally due to Onsager and is famously complicated! Sim", "pler solutions using more modern techniques have since been discovered). Here we\u2019ll develop an approximate method to evaluate Z known as mean \ufb01eld theory. We write the interactions between neighbouring spins in term of their deviation from the average spin m, sisj = [(si \u2212 m) + m][(sj \u2212 m) + m] = (si \u2212 m)(sj \u2212 m) + m(sj \u2212 m) + m(si \u2212 m) + m2 The mean \ufb01eld approximation means that we assume that the \ufb02uctuations of spins away from the average are small which allows us to neglect the \ufb01rst term above. Notice that this isn\u2019t the statement that the variance of an individual spin is small; that can never i = 1 and the variance (si \u2212 m)2 is be true because si takes values +1 or \u22121 so s2 always large. Instead, the mean \ufb01eld approximation is a statement about \ufb02uctuations between spins on neighbouring sites, so the \ufb01rst term above can be neglected when summing over ij. We can then write the energy (5.13) as Emf = \u2212J ij [m(si + sj) \u2212 m2] \u2212 B si i = 1 2 JN qm2 \u2212 (Jqm + B) si i (5.16) ij. The factor or 1/2 is there because where the factor of N q/2 in the \ufb01rst term is simply the number of nearest neighbour pairs ij is a sum over pairs rather than a sum of individual sites. (If you\u2019re worried about this formula, you should check it for a simple square lattice in d = 1 and d = 2 dimensions). A similar factor in the second term cancelled the factor of 2 due to (si + sj). We see that the mean \ufb01eld approximation has removed the interactions. The Ising model reduces to the two state system that we saw way back in Section 1. The result of the interactions is that a given spin feels the average e\ufb00ect of its neighbour\u2019s spins through a contribution to the e\ufb00ective magnetic \ufb01eld, Once we\u2019ve taken into account this extra contribution to Be\ufb00, each spin acts indepen- Be\ufb00 = B + Jqm \u2013 149 \u2013 Figure 43: tanh(Jqm\u03b2) for Jq\u03b2 < 1 Figure 44: tanh(Jqm\u03b2) for Jq\u03b2 > 1 dently and it is easy to write the partition function. It is Z = e\u2212 = e\u2212 1 2 1 2 \u03b2JN qm2 e\u2212\u03b2Be\ufb00 + e\u03b2Be\ufb00 N \u03b2JN qm2 2N coshN \u03b2Be\ufb00 (5.17) However, we\u2019re not quite done. Our result for the partition function Z depends on Be\ufb00 which depends on m which we don\u2019t yet know. However, we can use our expression for Z to self-consistently determine the magnetization (5.15). We \ufb01nd, m = tanh(\u03b2B + \u03b2Jqm) (5.18) We can now solve this equation to \ufb01nd the magnetization for various values of T and B: m = m(T, B). It is simple to see the nature of the solutions using graphical methods. B=0 Let\u2019s \ufb01rst consider the situation with vanishing magnetic \ufb01eld, B = 0. The \ufb01gures above show the graph linear in m compared with the tanh function. Since tanh x \u2248 x\u2212 1 3x3+..., the slope of the graph near the origin is given by \u03b2Jq. This then determines the nature of the solution. \u2022 The \ufb01rst graph depicts the situation for \u03b2Jq < 1. The only solution is m = 0. This means that at high temperatures kBT > Jq, there is no average magnetization of the system. The entropy associated to the random temperature \ufb02ucutations wins over the energetically preferred ordered state in which the spins align. \u2022 The second graph depicts the situation for \u03b2Jq > 1. Now there are three solutions: m = \u00b1m0 and m = 0. It will turn out that the middle solution, m = 0, is unstable. (This solution is entirely analogous to the unstable solution of the van der Waals equation. We will see this below when we compute the free energy). \u2013 150 \u2013 tanhmm0\u2212mtanhm0 For the other two possible solutions, m = \u00b1m0, the magnetization is non-zero. Here we see the e\ufb00ects", " of the interactions begin to win over temperature. Notice that in the limit of vanishing temperature, \u03b2 \u2192 \u221e, m0 \u2192 1. This means that all the spins are pointing in the same direction (either up or down) as expected. \u2022 The critical temperature separating these two cases is kBTc = Jq (5.19) The results described above are perhaps rather surprising. Based on the intuition that things in physics always happen smoothly, one might have thought that the magnetization would drop slowly to zero as T \u2192 \u221e. But that doesn\u2019t happen. Instead the magnetization turns o\ufb00 abruptly at some \ufb01nite value of the temperature T = Tc, with no magnetization at all for higher temperatures. This is the characteristic behaviour of a phase transition. B = 0 For B = 0, we can solve the consistency equation (5.18) in a similar fashion. There are a couple of key di\ufb00erences to the B = 0 case. Firstly, there is now no phase transition at \ufb01xed B as we vary temperature T. Instead, for very large temperatures kBT Jq, the magnetization goes smoothly to zero as m \u2192 B kBT as T \u2192 \u221e At low temperatures, the magnetization again asymptotes to the state m \u2192 \u00b11 which minimizes the energy. Except this time, there is no ambiguity as to whether the system chooses m = +1 or m = \u22121. This is entirely determined by the sign of the magnetic \ufb01eld B. In fact the low temperature behaviour requires slightly more explanation. For small values of B and T, there are again three solutions to (5.18). This follows simply from continuity: there are three solutions for T < Tc and B = 0 shown in Figure 44 and these must survive in some neighbourhood of B = 0. One of these solutions is again unstable. However, of the remaining two only one is now stable: that with sign(m) = sign(B). The other is meta-stable. We will see why this is the case shortly when we come to discuss the free energy. \u2013 151 \u2013 Figure 45: Magnetization with B = 0 and the phase transtion Figure 46: Magnetization at B = 0. The net result of our discussion is depicted in the \ufb01gures above. When B = 0 there is a phase transition at T = Tc. For T < Tc, the system can sit in one of two magnetized states with m = \u00b1m0. In contrast, for B = 0, there is no phase transition as we vary temperature and the system has at all times a preferred magnetization whose sign is determined by that of B. Notice however, we do have a phase transition if we \ufb01x temperature at T < Tc and vary B from negative to positive. Then the magnetization jumps discontinuously from a negative value to a positive value. Since the magnetization is a \ufb01rst derivative of the free energy (5.15), this is a \ufb01rst order phase transition. In contrast, moving along the temperature axis at B = 0 results in a second order phase transition at T = Tc. 5.2.2 Critical Exponents It is interesting to compare the phase transition of the Ising model with that of the liquid-gas phase transition. The two are sketched in the Figure 47 above. In both cases, we have a \ufb01rst order phase transition and a quantity jumps discontinuously at T < Tc. In the case of the liquid-gas, it is the density \u03c1 = 1/v that jumps as we vary pressure; in the case of the Ising model it is the magnetization m that jumps as we vary the magnetic \ufb01eld. Moreover, in both cases the discontinuity disappears as we approach T = Tc. We can calculate critical exponents for the Ising model. To compare with our discussion for the liquid-gas critical point, we will compute three quantities. First, consider the magnetization at B = 0. We can ask how this magnetization decreases as we tend towards the critical point. Just below T = Tc, m is small and we can Taylor expand (5.18) to get m \u2248 \u03b2Jqm \u2212 1 3 (\u03b2Jqm)3 +... \u2013 152 \u2013 Tm+1\u22121mTB<0B>0+1\u22121 Figure 47: A comparison of the phase diagram for the liquid-gas system and the Ising model. The magnetization therefore scales as m0 \u223c \u00b1(Tc \u2212 T )1/2 (5.20) This is to be compared with the analogous result (5.9) from the van der Waals equation. We see that the values of the exponents are the same in both cases. Notice that the derivative dm/dT becomes in\ufffd", "\ufffdnite as we approach the critical point. In fact, we had already anticipated this when we drew the plot of the magnetization in Figure 45. Secondly, we can sit at T = Tc and ask how the magnetization changes as we approach B = 0. We can read this o\ufb00 from (5.18). At T = Tc we have \u03b2Jq = 1 and the consistency condition becomes m = tanh(B/Jq + m). Expanding for small B gives m \u2248 B Jq + m \u2212 B Jq Jq + m \u2212 1 3 m3 + O(B2) So we \ufb01nd that the magnetization scales as m \u223c B1/3 (5.21) Notice that this power of 1/3 is again familiar from the liquid-gas transition (5.10) where the van der Waals equation gave vgas \u2212 vliquid \u223c (p \u2212 pc)1/3. Finally, we can look at the magnetic susceptibility \u03c7, de\ufb01ned as \u03c7 = N \u2202m \u2202B T This is analogous to the compressibility \u03ba of the gas. We will ask how \u03c7 changes as we approach T \u2192 Tc from above at B = 0. We di\ufb00erentiate (5.18) with respect to B to get \u03c7 = N \u03b2 cosh2 \u03b2Jqm 1 + Jq N \u03c7 \u2013 153 \u2013 TpTcTcliquidgascritical pointTBcritical point We now evaluate this at B = 0. Since we want to approach T \u2192 Tc from above, we can also set m = 0 in the above expression. Evaluating this at B = 0 gives us the scaling \u03c7 = N \u03b2 1 \u2212 Jq\u03b2 \u223c (T \u2212 Tc)\u22121 (5.22) Once again, we see that same critical exponent that the van der Waals equation gave us for the gas (5.12). 5.2.3 Validity of Mean Field Theory The phase diagram and critical exponents above were all derived using the mean \ufb01eld approximation. But this was an unjusti\ufb01ed approximation. Just as for the van der Waals equation, we can ask the all-important question: are our results right? There is actually a version of the Ising model for which the mean \ufb01eld theory is it is the d = \u221e dimensional lattice. This is unphysical (even for a string exact: theorist). Roughly speaking, mean \ufb01eld theory works for large d because each spin has a large number of neighbours and so indeed sees something close to the average spin. But what about dimensions of interest? Mean \ufb01eld theory gets things most dramatically wrong in d = 1. In that case, no phase transition occurs. We will derive this result below where we brie\ufb02y describe the exact solution to the d = 1 Ising model. There is a general lesson here: in low dimensions, both thermal and quantum \ufb02uctuations are more important and invariably stop systems forming ordered phases. In higher dimensions, d \u2265 2, the crude features of the phase diagram, including the existence of a phase transition, given by mean \ufb01eld theory are essentially correct. In fact, the very existence of a phase transition is already worthy of comment. The de\ufb01ning feature of a phase transition is behaviour that jumps discontinuously as we vary \u03b2 or B. Mathematically, the functions must be non-analytic. Yet all properties of the theory can be extracted from the partition function Z which is a sum of smooth, analytic functions (5.14). How can we get a phase transition? The loophole is that Z is only necessarily analytic if the sum is \ufb01nite. But there is no such guarantee when the number of lattice sites N \u2192 \u221e. We reach a similar conclusion to that of BoseEinstein condensation: phase transitions only strictly happen in the thermodynamic limit. There are no phase transitions in \ufb01nite systems. What about the critical exponents that we computed in (5.20), (5.21) and (5.22)? It turns out that these are correct for the Ising model de\ufb01ned in d \u2265 4. (We will brie\ufb02y sketch why this is true at the end of this Chapter). But for d = 2 and d = 3, the critical exponents predicted by mean \ufb01eld theory are only \ufb01rst approximations to the true answers. \u2013 154 \u2013 For d = 2, the exact solution (which goes quite substantially past this course) gives the critical exponents to be, m0 \u223c (Tc \u2212 T )\u03b2 with m \u223c B1/\u03b4 with \u03c7 \u223c (T \u2212 Tc)\u2212\u03b3 with \u03b2 = 1 8 \u03b4 =", " 15 7 4 \u03b3 = The biggest surprise is in d = 3 dimensions. Here the critical exponents are not known exactly. However, there has been a great deal of numerical work to determine them. They are given by \u03b2 \u2248 0.32, \u03b4 \u2248 4.8, \u03b3 \u2248 1.2 But these are exactly the same critical exponents that are seen in the liquid-gas phase transition. That\u2019s remarkable! We saw above that the mean \ufb01eld approach to the Ising model gave the same critical exponents as the van der Waals equation. But they are both wrong. And they are both wrong in the same, complicated, way! Why on earth would a system of spins on a lattice have anything to do with the phase transition between a liquid and gas? It is as if all memory of the microscopic physics \u2014 the type of particles, the nature of the interactions \u2014 has been lost at the critical point. And that\u2019s exactly what happens. What we\u2019re seeing here is evidence for universality. There is a single theory which describes the physics at the critical point of the liquid gas transition, the 3d Ising model and many other systems. This is a theoretical physicist\u2019s dream! We spend a great deal of time trying to throw away the messy details of a system to focus on the elegant essentials. But, at a critical point, Nature does this for us! Although critical points in two dimensions are well understood, there is still much that we don\u2019t know about critical points in three dimensions. This, however, is a story that will have to wait for another day. 5.3 Some Exact Results for the Ising Model This subsection is something of a diversion from our main interest. In later subsections, we will develop the idea of mean \ufb01eld theory. But \ufb01rst we pause to describe some exact results for the Ising model using techniques that do not rely on the mean \ufb01eld approximation. Many of the results that we derive have broader implications for systems beyond the Ising model. \u2013 155 \u2013 As we mentioned above, there is an exact solution for the Ising model in d = 1 dimension and, when B = 0, in d = 2 dimensions. Here we will describe the d = 1 solution but not the full d = 2 solution. We will, however, derive a number of results for the d = 2 Ising model which, while falling short of the full solution, nonetheless provide important insights into the physics. 5.3.1 The Ising Model in d = 1 Dimensions We start with the Ising chain, the Ising model on a one dimensional line. Here we will see that the mean \ufb01eld approximation fails miserably, giving qualitatively incorrect results: the exact results shows that there are no phase transitions in the Ising chain. The energy of the system (5.13) can be trivially rewritten as E = \u2212J N i=1 sisi+1 \u2212 B 2 N i=1 (si + si+1) We will impose periodic boundary conditions, so the spins live on a circular lattice with sN +1 \u2261 s1. The partition function is then Z =... N s1=\u00b11 sN =\u00b11 i=1 exp \u03b2Jsisi+1 + (si + si+1) \u03b2B 2 (5.23) The crucial observation that allows us to solve the problem is that this partition function can be written as a product of matrices. We adopt notation from quantum mechanics and de\ufb01ne the 2 \u00d7 2 matrix, si|T |si+1 \u2261 exp \u03b2Jsisi+1 + (si + si+1) \u03b2B 2 (5.24) The row of the matrix is speci\ufb01ed by the value of si = \u00b11 and the column by si+1 = \u00b11. T is known as the transfer matrix and, in more conventional notation, is given by T = e\u03b2J+\u03b2B e\u2212\u03b2J e\u03b2J\u2212\u03b2B e\u2212\u03b2J The sums over the spins si to multiply the matrices de\ufb01ned in (5.24) and the partition function becomes and product over lattice sites i in (5.23) simply tell us Z = Tr (s1|T |s2s2|T |s3... sN |T |s1) = Tr T N (5.25) \u2013 156 \u2013 where the trace arises because we have imposed periodic boundary conditions. To complete the story, we need only compute the eigenvalues of T to determine the partition function. A quick calculation shows that the two eigenvalues of T are \u03bb\u00b1 = e\u03b2J cosh \u03b2B \u00b1 e2\u03b2J cosh2 \u03b2B \u2212 2 sinh 2\u03b2J where, clearly, \u03bb\u2212 < \u03bb", "+. The partition function is then Z = \u03bbN + + \u03bbN \u2212 = \u03bbN + 1 + \u03bbN \u2212 \u03bbN + \u2248 \u03bbN + (5.26) (5.27) where, in the last step, we\u2019ve used the simple fact that if \u03bb+ is the largest eigenvalue then \u03bbN + \u2248 0 for very large N. \u2212 /\u03bbN The partition function Z contains many quantities of interest. In particular, we can use it to compute the magnetisation as a function of temperature when B = 0. This, recall, is the quantity which is predicted to undergo a phase transition in the mean \ufb01eld approximation, going abruptly to zero at some critical temperature. In the d = 1 Ising model, the magnetisation is given by m = 1 N \u03b2 \u2202 log Z \u2202B B=0 = 1 \u03bb+\u03b2 \u2202\u03bb+ \u2202B B=0 = 0 We see that the true physics for d = 1 is very di\ufb00erent than that suggested by the mean \ufb01eld approximation. When B = 0, there is no magnetisation! While the J term in the energy encourages the spins to align, this is completely overwhelmed by thermal \ufb02uctuations for any value of the temperature. There is a general lesson in this calculation: thermal \ufb02uctuations always win in one dimensional systems. They never exhibit ordered phases and, for this reason, never exhibit phase transitions. The mean \ufb01eld approximation is bad in one dimension. 5.3.2 2d Ising Model: Low Temperatures and Peierls Droplets Let\u2019s now turn to the Ising model in d = 2 dimensions. We\u2019ll work on a square lattice and set B = 0. Rather than trying to solve the model exactly, we\u2019ll have more modest goals. We will compute the partition function in two di\ufb00erent limits: high temperature and low temperature. We start here with the low temperature expansion. The partition function is given by the sum over all states, weighted by e\u2212\u03b2E. At low temperatures, this is always dominated by the lowest lying states. For the Ising model, \u2013 157 \u2013 we have Z = {si} \uf8eb exp \uf8ed\u03b2J \uf8f6 sisj \uf8f8 ij The low temperature limit is \u03b2J \u2192 \u221e, where the partition function can be approximated by the sum over the \ufb01rst few lowest energy states. All we need to do is list these states. The ground states are easy. There are two of them: spins all up or spins all down. For example, the ground state with spins all up looks like Each of these ground states has energy E = E0 = \u22122N J. The \ufb01rst excited states arise by \ufb02ipping a single spin. Each spin has q = 4 nearest neighbours \u2013 denoted by red lines in the example below \u2013 each of which leads to an energy cost of 2J. The energy of each \ufb01rst excited state is therefore E1 = E0 + 8J. There are, of course, N di\ufb00erent spins that we we can \ufb02ip and, correspondingly, the \ufb01rst energy level has a degeneracy of N. To proceed, we introduce a diagrammatic method to list the di\ufb00erent states. We draw only the \u201cbroken\u201d bonds which connect two spins with opposite orientation and, as in the diagram above, denote these by red lines. We further draw the \ufb02ipped spins as red dots, the un\ufb02ipped spins as blue dots. The energy of the state is determined simply by the number of red lines in the diagram. Pictorially, we write the \ufb01rst excited state as E1 = E0 + 8J Degeneracy = N \u2013 158 \u2013 The next lowest state has six broken bonds. It takes the form E2 = E0 + 12J Degeneracy = 2N where the extra factor of 2 in the degeneracy comes from the two possible orientations (vertical and horizontal) of the graph. Things are more interesting for the states which sit at the third excited level. These have 8 broken bonds. The simplest con\ufb01guration consists of two, disconnected, \ufb02ipped spins E3 = E0 + 16J Degeneracy = 1 2N (N \u2212 5) (5.28) The factor of N in the degeneracy comes from placing the \ufb01rst graph; the factor of N \u2212 5 arises because the \ufb02ipped spin in the second graph can sit anywhere apart from on the \ufb01ve vertices used in the \ufb01rst graph. Finally, the factor of 1/2 arises from the interchange of the two graphs", ". There are also three further graphs with the same energy E3. These are and E3 = E0 + 16J Degeneracy = N E3 = E0 + 16J Degeneracy = 2N \u2013 159 \u2013 where the degeneracy comes from the two orientations (vertical and horizontal). And, \ufb01nally, E3 = E0 + 16J Degeneracy = 4N where the degeneracy comes from the four orientations (rotating the graph by 90\u25e6). Adding all the graphs above together gives us an expansion of the partition function in power of e\u2212\u03b2J 1. This is Z = 2e2N \u03b2J 1 + N e\u22128\u03b2J + 2N e\u221212\u03b2J + (N 2 + 9N )e\u221216\u03b2J +... (5.29) 1 2 where the overall factor of 2 originates from the two ground states of the system. We\u2019ll make use of the speci\ufb01c coe\ufb03cients in this expansion in Section 5.3.4. Before we focus on the physics hiding in the low temperature expansion, it\u2019s worth making a quick comment that something quite nice happens if we take the log of the partition function, log Z = log 2 + 2N \u03b2J + N e\u22128\u03b2J + 2N e\u221212\u03b2J + 9 2 N e\u221216\u03b2J +... The thing to notice is that the N 2 term in the partition function (5.29) has cancelled out and log Z is proportional to N, which is to be expected since the free energy of the system is extensive. Looking back, we see that the N 2 term was associated to the disconnected diagrams in (5.28). There is actually a general lesson hiding here: the partition function can be written as the exponential of the sum of connected diagrams. We saw exactly the same issue arise in the cluster expansion in (2.37). Peierls Droplets Continuing the low temperature expansion provides a heuristic, but physically intuitive, explanation for why phase transitions happen in d \u2265 2 dimensions but not in d = 1. As we \ufb02ip more and more spins, the low energy states become droplets, consisting of a region of space in which all the spins are \ufb02ipped, surrounded by a larger sea in which the spins have their original alignment. The energy cost of such a droplet is roughly E \u223c 2JL \u2013 160 \u2013 where L is the perimeter of the droplet. Notice that the energy does not scale as the area of the droplet since all spins inside are aligned with their neighbours. It is only those on the edge which are misaligned and this is the reason for the perimeter scaling. To understand how these droplets contribute to the partition function, we also need to know their degeneracy. We will now argue that the degeneracy of droplets scales as Degeneracy \u223c e\u03b1L for some value of \u03b1. To see this, consider \ufb01rstly the problem of a random walk on a 2d square lattice. At each step, we can move in one of four directions. So the number of paths of length L is #paths \u223c 4L = eL log 4 Of course, the perimeter of a droplet is more constrained that a random walk. Firstly, the perimeter can\u2019t go back on itself, so it really only has three directions that it can move in at each step. Secondly, the perimeter must return to its starting point after L steps. And, \ufb01nally, the perimeter cannot self-intersect. One can show that the number of paths that obey these conditions is #paths \u223c e\u03b1L where log 2 < \u03b1 < log 3. Since the degeneracy scales as e\u03b1L, the entropy of the droplets is proportional to L. The fact that both energy and entropy scale with L means that there is an interesting competition between them. At temperatures where the droplets are important, the partition function is schematically of the form Z \u223c L e\u03b1Le\u22122\u03b2JL For large \u03b2 (i.e. temperature increases, one reaches the critical temperature low temperature) the partition function converges. However, as the kBTc \u2248 2J \u03b1 (5.30) where the partition function no longer converges. At this point, the entropy wins over the energy cost and it is favourable to populate the system with droplets of arbitrary sizes. This is the how one sees the phase transition in the partition function. For temperature above Tc, the low-temperature expansion breaks down and the ordered magnetic phase is destroyed. \u2013 161 \u2013 We can also use the droplet argument to see why phase transitions don\u2019t occur in d = 1 dimension. On a line, the boundary of any droplet always consists of just two points. This means that the energy cost to forming a droplet is always E = 2", "J, regardless of the size of the droplet. But, since the droplet can exist anywhere along the line, its degeneracy is N. The net result is that the free energy associated to creating a droplet scales as F \u223c 2J \u2212 kBT log N and, as N \u2192 \u221e, the free energy is negative for any T > 0. This means that the system will prefer to create droplets of arbitrary length, randomizing the spins. This is the intuitive reason why there is no magnetic ordered phase in the d = 1 Ising model. 5.3.3 2d Ising Model: High Temperatures We now turn to the 2d Ising model in the opposite limit of high temperature. Here we expect the partition function to be dominated by the completely random, disordered con\ufb01gurations of maximum entropy. Our goal is to \ufb01nd a way to expand the partition function in \u03b2J 1. We again work with zero magnetic \ufb01eld, B = 0 and write the partition function as \uf8eb \uf8f6 Z = {si} exp \uf8ed\u03b2J ij sisj \uf8f8 = {si} ij e\u03b2Jsisj There is a useful way to rewrite e\u03b2Jsisj which relies on the fact that the product sisj only takes \u00b11. It doesn\u2019t take long to check the following identity: e\u03b2Jsisj = cosh \u03b2J + sisj sinh \u03b2J = cosh \u03b2J (1 + sisj tanh \u03b2J) Using this, the partition function becomes Z = {si} ij cosh \u03b2J (1 + sisj tanh \u03b2J) = (cosh \u03b2J)qN/2 (1 + sisj tanh \u03b2J) (5.31) {si} ij where the number of nearest neighbours is q = 4 for the 2d square lattice. \u2013 162 \u2013 With the partition function in this form, there is a natural expansion which suggests itself. At high temperatures \u03b2J 1 which, of course, means that tanh \u03b2J 1. But the partition function is now naturally a product of powers of tanh \u03b2J. This is somewhat analogous to the cluster expansion for the interacting gas that we met in Section 2.5.3. As in the cluster expansion, we will represent the expansion graphically. We need no graphics for the leading order term. It has no factors of tanh \u03b2J and is simply That\u2019s simple. Z \u2248 (cosh \u03b2J)2N 1 = 2N (cosh \u03b2J)2N {si} Let\u2019s now turn to the leading correction. Expanding the partition function (5.31), each power of tanh \u03b2J is associated to a nearest neighbour pair ij. We\u2019ll represent this by drawing a line on the lattice: But there\u2019s a problem: each factor of tanh \u03b2J in (5.31) also comes with a sum over all spins si and sj. And these are +1 and \u22121 which means that they simply sum to zero, = sisj tanh \u03b2J si,sj sisj = +1 \u2212 1 \u2212 1 + 1 = 0 How can we avoid this? The only way is to make sure that we\u2019re summing over an even number of spins on each site, since then we get factors of s2 i = 1 and no cancellations. Graphically, this means that every site must have an even number of lines attached to it. The \ufb01rst correction is then of the form = (tanh \u03b2J)4 s1s2 s2s3 s3s4 s4s1 = 24(tanh \u03b2J)4 {si} There are N such terms since the upper left corner of the square can be on any one of the N lattice sites. (Assuming periodic boundary conditions for the lattice). So including the leading term and \ufb01rst correction, we have Z = 2N (cosh \u03b2J)2N 1 + N (tanh \u03b2J)4 +... We can go further. The next terms arise from graphs of length 6 and the only possibilities are rectangles, oriented as either landscape or portrait. Each of them can sit on \u2013 163 \u2013 ij1234 one of N sites, giving a contribution + = 2N (tanh \u03b2J)4 Things get more interesting when we look at graphs of length 8. We have four di\ufb00erent types of graphs. Firstly, there are the trivial, disconnected pair of squares = 1 2 N (N \u2212 5)(tanh \u03b2J)8 Here the \ufb01rst factor of N is the possible positions of the \ufb01rst square; the factor of N \u2212 5 arises because the possible location of the upper corner of the", " second square can\u2019t be on any of the vertices of the \ufb01rst, but nor can it be on the square one to the left of the upper corner of the \ufb01rst since that would give a graph that looks like which has three lines coming o\ufb00 the middle site and therefore vanishes when we sum over spins. Finally, the factor of 1/2 comes because the two squares are identical. The other graphs of length 8 are a large square, a rectangle and a corner. The large square gives a contribution = N (tanh \u03b2J)8 There are two orientations for the rectangle. Including these gives a factor of 2, Finally, the corner graph has four orientations, giving = 2N (tanh \u03b2J)8 = 4N (tanh \u03b2J)8 Adding all contributions together gives us the \ufb01rst few terms in high temperature expansion of the partition function Z = 2N (cosh \u03b2J)2N 1 + N (tanh \u03b2J)4 + 2N (tanh \u03b2J)6 (N 2 + 9N )(tanh \u03b2J)8 +... (5.32) + 1 2 \u2013 164 \u2013 There\u2019s some magic hiding in this expansion which we\u2019ll turn to in Section 5.3.4. First, let\u2019s just see how the high energy expansion plays out in the d = 1 dimensional Ising model. The Ising Chain Revisited Let\u2019s do the high temperature expansion for the d = 1 Ising chain with periodic boundary conditions and B = 0. We have the same partition function (5.31) and the same issue that only graphs with an even number of lines attached to each vertex contribute. But, for the Ising chain, there is only one such term: it is the closed loop. This means that the partition function is Z = 2N (cosh \u03b2J)N 1 + (tanh \u03b2J)N Figure 48: In the limit N \u2192 \u221e, (tanh \u03b2J)N \u2192 0 at high temperatures and even the contribution from the closed loop vanishes. We\u2019re left with Z = (2 cosh \u03b2J)N This agrees with our exact result for the Ising chain given in (5.27), which can be seen by setting B = 0 in (5.26) so that \u03bb+ = 2 cosh \u03b2J. 5.3.4 Kramers-Wannier Duality In the previous sections we computed the partition function perturbatively in two extreme regimes of low temperature and high temperature. The physics in the two cases is, of course, very di\ufb00erent. At low temperatures, the partition function is dominated by the lowest energy states; at high temperatures it is dominated by maximally disordered states. Yet comparing the partition functions at low temperature (5.29) and high temperature (5.32) reveals an extraordinary fact: the expansions are the same! More concretely, the two series agree if we exchange e\u22122\u03b2J \u2190\u2192 tanh \u03b2J (5.33) Of course, we\u2019ve only checked the agreement to the \ufb01rst few orders in perturbation theory. Below we shall prove that this miracle continues to all orders in perturbation theory. The symmetry of the partition function under the interchange (5.33) is known as Kramers-Wannier duality. Before we prove this duality, we will \ufb01rst just assume that it is true and extract some consequences. \u2013 165 \u2013 We can express the statement of the duality more clearly. The Ising model at tem- perature \u03b2 is related to the same model at temperature \u02dc\u03b2, de\ufb01ned as e\u22122 \u02dc\u03b2J = tanh \u03b2J (5.34) This way of writing things hides the symmetry of the transformation. A little algebra shows that this is equivalent to sinh 2 \u02dc\u03b2J = 1 sinh 2\u03b2J Notice that this is a hot/cold duality. When \u03b2J is large, \u02dc\u03b2J is small. Kramers-Wannier duality is the statement that, when B = 0, the partition functions of the Ising model at two temperatures are related by Z[\u03b2] = 2N (cosh \u03b2J)2N 2e2N \u02dc\u03b2J Z[ \u02dc\u03b2] = 2N \u22121(cosh \u03b2J sinh \u03b2J)N Z[ \u02dc\u03b2] (5.35) This means that if you know the thermodynamics of the Ising model at one temperature, then you also know the thermodynamics at the other temperature. Notice however, that it does not say that all the physics of the two models is equivalent. In particular, when one system is in the ordered phase, the other typically lies in the dis", "ordered phase. One immediate consequence of the duality is that we can use it to compute the exact critical temperature Tc. This is the temperature at which the partition function in singular in the N \u2192 \u221e limit. (We\u2019ll discuss a more re\ufb01ned criterion in Section 5.4.3). If we further assume that there is just a single phase transition as we vary the temperature, then it must happen at the special self-dual point \u03b2 = \u02dc\u03b2. This is kBT = 2J \u221a log( 2 + 1) \u2248 2.269 J The exact solution of Onsager con\ufb01rms that this is indeed the transition temperature. It\u2019s also worth noting that it\u2019s fully consistent with the more heuristic Peierls droplet 2 + 1) < log 3. argument (5.30) since log 2 < log( \u221a Proving the Duality So far our evidence for the duality (5.35) lies in the agreement of the \ufb01rst few terms in the low and high temperature expansions (5.29) and (5.32). Of course, we could keep computing further and further terms and checking that they agree, but it would be nicer to simply prove the equality between the partition functions. We shall do so here. \u2013 166 \u2013 The key idea that we need can actually be found by staring hard at the various graphs that arise in the two expansions. Eventually, you will realise that they are the same, albeit drawn di\ufb00erently. For example, consider the two \u201ccorner\u201d diagrams vs The two graphs are dual. The red lines in the \ufb01rst graph intersect the black lines in the second as can be seen by placing them on top of each other: The same pattern occurs more generally: the graphs appearing in the low temperature expansion are in one-to-one correspondence with the dual graphs of the high temperature expansion. Here we will show how this occurs and how one can map the partition functions onto each other. Let\u2019s start by writing the partition function in the form (5.31) that we met in the high temperature expansion and presenting it in a slightly di\ufb00erent way, Z[\u03b2] = (cosh \u03b2J + sisj sinh \u03b2J) {si} ij = {si} ij kij =0,1 Ckij [\u03b2J] (sisj)kij where we have introduced the rather strange variable kij associated to each nearest neighbour pair that takes values 0 and 1, together with the functions. C0[\u03b2J] = cosh \u03b2J and C1[\u03b2J] = sinh \u03b2J The variables in the original Ising model were spins on the lattice sites. The observation that the graphs which appear in the two expansions are dual suggests that it might be \u2013 167 \u2013 pro\ufb01table to focus attention on the links between lattice sites. Clearly, we have one link for every nearest neighbour pair. If we label these links by l, we can trivially rewrite the partition function as Z = Ckl[\u03b2J] (sisj)kl kl=0,1 l {si} Notice that the strange label kij has now become a variable that lives on the links l rather than the original lattice sites i. At this stage, we do the sum over the spins si. We\u2019ve already seen that if a given spin, say si, appears in a term an odd number of times, then that term will vanish when we sum over the spin. Alternatively, if the spin si appears an even number of times, then the sum will give 2. We\u2019ll say that a given link l is turned on in con\ufb01gurations with kl = 1 and turned o\ufb00 when kl = 0. In this language, a term in the sum over spin si contributes only if an even number of links attached to site i are turned on. The partition function then becomes Z = 2N Ckl[\u03b2J] kl l Constrained (5.36) Now we have something interesting. Rather than summing over spins on lattice sites, we\u2019re now summing over the new variables kl living on links. This looks like the partition function of a totally di\ufb00erent physical system, where the degrees of freedom live on the links of the original lattice. But there\u2019s a catch \u2013 that big \u201cConstrained\u201d label on the sum. This is there to remind us that we don\u2019t sum over all kl con\ufb01gurations; only those for which an even number of links are turned on for every lattice site. And that\u2019s annoying. It\u2019s telling us that the kl aren\u2019t really independent variables. There are some constraints that must be imposed.", " Fortunately, for the 2d square lattice, there is a simple way to solve the constraint. We introduce yet more variables, \u02dcsi which, like the original spin variables, take values \u00b11. However, the \u02dcsi do not live on the original lattice sites. Instead, they live on the vertices of the dual lattice. For the 2d square lattice, the dual vertices are drawn in the \ufb01gure. The original lattice sites are in white; the dual lattice sites in black. The link variables kl are related to the two nearest spin vari- Figure 49: ables \u02dcsi as follows: k12 = 1 2 (1 \u2212 \u02dcs1\u02dcs2) \u2013 168 \u2013 s2s54s1s~1s~4s~2s~33sk12s k13 = k14 = k15 = 1 2 1 2 1 2 (1 \u2212 \u02dcs2\u02dcs3) (1 \u2212 \u02dcs3\u02dcs4) (1 \u2212 \u02dcs1\u02dcs4) Notice that we\u2019ve replaced four variables kl taking values 0, 1 with four variables \u02dcsi taking values \u00b11. Each set of variables gives 24 possibilities. However, the map is not one-to-one. It is not possible to construct for all values of kl using the parameterization in terms of \u02dcsi. To see this, we need only look at k12 + k13 + k14 + k15 = 2 \u2212 = 2 \u2212 (\u02dcs1\u02dcs2 + \u02dcs2\u02dcs3 + \u02dcs3\u02dcs4 + \u02dcs1\u02dcs4) (\u02dcs1 + \u02dcs3)(\u02dcs2 + \u02dcs4) 1 2 1 2 = 0, 2, or 4 In other words, the number of links that are turned on must be even. But that\u2019s exactly what we want! Writing the kl in terms of the auxiliary spins \u02dcsi automatically solves the constraint that is imposed on the sum in (5.36). Moreover, it is simple to check that for every con\ufb01guration {kl} obeying the constraint, there are two con\ufb01gurations of {\u02dcsi}. This means that we can replace the constrained sum over {kl} with an unconstrained sum over {\u02dcsi}. The only price we pay is an additional factor of 1/2. Z[\u03b2] = 1 2 2N {\u02dcsi} ij C 1 2 [\u03b2j] (1\u2212\u02dcsi \u02dcsj ) Finally, we\u2019d like to \ufb01nd a simple expression for C0 and C1 in terms of \u02dcsi. That\u2019s easy enough. We can write Ck[\u03b2J] = cosh \u03b2J exp (k log tanh \u03b2J) = (sinh \u03b2J cosh \u03b2J)1/2 exp \u2212 \u02dcsi\u02dcsj log tanh \u03b2J 1 2 Substituting this into our newly re-written partition function gives Z[\u03b2] = 2N \u22121 (sinh \u03b2J cosh \u03b2J)1/2 exp {\u02dcsi} ij \u2212 1 2 \u02dcsi\u02dcsj log tanh \u03b2J \uf8f6 log tanh \u03b2J ij \u02dcsi\u02dcsj \uf8f8 = 2N \u22121(sinh \u03b2J cosh \u03b2J)N {\u02dcsi} \uf8eb exp \uf8ed\u2212 1 2 \u2013 169 \u2013 But this \ufb01nal form of the partition function in terms of the dual spins \u02dcsi has exactly the same functional form as the original partition function in terms of the spins si. More precisely, we can write where Z[\u03b2] = 2N \u22121(sinh 2\u03b2J)N Z[ \u02dc\u03b2] e\u22122 \u02dc\u03b2J = tanh \u03b2J as advertised previously in (5.34). This completes the proof of Kramers-Wannier duality in the 2d Ising model on a square lattice. The concept of duality of this kind is a major feature in much of modern theoretical physics. The key idea is that when the temperature gets large there may be a di\ufb00erent set of variables in which a theory can be written where it appears to live at low temperature. The same idea often holds in quantum theories, where duality maps strong coupling problems to weak coupling problems. The duality in the Ising model is special for two reasons: \ufb01rstly, the new variables \u02dcsi are governed by the same Hamiltonian as the original variables si. We say that the Ising model is self-dual. In general, this need not be the case \u2014 the high temperature limit of one system could look like the low-temperature limit of a very di\ufb00erent", " system. Secondly, the duality in the Ising model can be proven explicitly. For most systems, we have no such luck. Nonetheless, the idea that there may be dual variables in other, more di\ufb03cult theories, is compelling. Commonly studied examples include the exchange particles and vortices in two dimensions, and electrons and magnetic monopoles in three dimensions. 5.4 Landau Theory We saw in Sections 5.1 and 5.2 that the van der Waals equation and mean \ufb01eld Ising model gave the same (sometimes wrong!) answers for the critical exponents. This suggests that there should be a uni\ufb01ed way to look at phase transitions. Such a method was developed by Landau. It is worth stressing that, as we saw above, the Landau approach to phase transitions often only gives qualitatively correct results. However, its advantage is that it is extremely straightforward and easy. (Certainly much easier than the more elaborate methods needed to compute critical exponents more accurately). \u2013 170 \u2013 The Landau theory of phase transitions is based around the free energy. We will illustrate the theory using the Ising model and then explain how to extend it to di\ufb00erent systems. The free energy of the Ising model in the mean \ufb01eld approximation is readily attainable from the partition function (5.17), F = \u2212 1 \u03b2 log Z = 1 2 JN qm2 \u2212 N \u03b2 log (2 cosh \u03b2Be\ufb00) (5.37) So far in this course, we\u2019ve considered only systems in equilibrium. The free energy, like all other thermodynamic potentials, has only been de\ufb01ned on equilibrium states. Yet the equation above can be thought of as an expression for F as a function of m. Of course, we could substitute in the equilibrium value of m given by solving (5.18), but it seems a shame to throw out F (m) when it is such a nice function. Surely we can put it to some use! The key step in Landau theory is to treat the function F = F (T, V ; m) seriously. This means that we are extending our viewpoint away from equilibrium states to a whole class of states which have a constant average value of m. If you want some words to drape around this, you could imagine some external magical power that holds m \ufb01xed. The free energy F (T, V ; m) is then telling us the equilibrium properties in the presence of this magical power. Perhaps more convincing is what we do with the free energy in the absence of any magical constraint. We saw in Section 4 that equilibrium is guaranteed if we sit at the minimum of F. Looking at extrema of F, we have the condition \u2202F \u2202m = 0 \u21d2 m = tanh \u03b2Be\ufb00 But that\u2019s precisely the condition (5.18) that we saw previously. Isn\u2019t that nice! In the context of Landau theory, m is called an order parameter. When it takes nonzero values, the system has some degree of order (the spins have a preferred direction in which they point) while when m = 0 the spins are randomised and happily point in any direction. For any system of interest, Landau theory starts by identifying a suitable order parameter. This should be taken to be a quantity which vanishes above the critical temperature at which the phase transition occurs, but is non-zero below the critical temperature. Sometimes it is obvious what to take as the order parameter; other times less so. For the liquid-gas transition, the relevant order parameter is the di\ufb00erence in densities between the two phases, vgas \u2212 vliquid. For magnetic or electric systems, the order parameter is typically some form of magnetization (as for the Ising model) or \u2013 171 \u2013 the polarization. For the Bose-Einstein condensate, super\ufb02uids and superconductors, the order parameter is more subtle and is related to o\ufb00-diagonal long-range order in the one-particle density matrix11, although this is usually rather lazily simpli\ufb01ed to say that the order parameter can be thought of as the macroscopic wavefunction |\u03c8|2. Starting from the existence of a suitable order parameter, the next step in the Landau programme is to write down the free energy. But that looks tricky. The free energy for the Ising model (5.37) is a rather complicated function and clearly contains some detailed information about the physics of the spins. How do we just write down the free energy in the general case? The trick is to assume that we can expand the free energy in an analytic power series in the order parameter. For this to be true, the order parameter must be small which is guaranteed if we are close to a critical point (since", " m = 0 for T > Tc). The nature of the phase transition is determined by the kind of terms that appear in the expansion of the free energy. Let\u2019s look at a couple of simple examples. 5.4.1 Second Order Phase Transitions We\u2019ll consider a general system (Ising model; liquid-gas; BEC; whatever) and denote the order parameter as m. Suppose that the expansion of the free energy takes the general form F (T ; m) = F0(T ) + a(T )m2 + b(T )m4 +... (5.38) One common reason why the free energy has this form is because the theory has a symmetry under m \u2192 \u2212m, forbidding terms with odd powers of m in the expansion. For example, this is the situation in the Ising model when B = 0. Indeed, if we expand out the free energy (5.37) for the Ising model for small m using cosh x \u2248 1 + 1 2y2 +... we get the general form above with explicit expressions for F0(T ), a(T ) and b(T ), 4!x4 +... and log(1 + y) \u2248 y \u2212 1 2x2 + 1 FIsing(T ; m) = \u2212N kBT log 2 + N Jq 2 (1 \u2212 Jq\u03b2) m2 + N \u03b23J 4q4 24 m4 +... The leading term F0(T ) is unimportant for our story. We are interested in how the free energy changes with m. The condition for equilibrium is given by \u2202F \u2202m = 0 (5.39) But the solutions to this equation depend on the sign of the coe\ufb03cients a(T ) and 11See, for example, the book \u201cQuantum Liquids\u201d by Anthony Leggett \u2013 172 \u2013 Figure 50: Free energy when a(T ) > 0 Figure 51: Free energy when a(T ) < 0 b(T ). Moreover, this sign can change with temperature. This is the essence of the phase transitions. In the following discussion, we will assume that b(T ) > 0 for all T. (If we relax this condition, we have to also consider the m6 term in the free energy which leads to interesting results concerning so-called tri-critical points). The two \ufb01gures above show sketches of the free energy in the case where a(T ) > 0 and a(T ) < 0. Comparing to the explicit free energy of the Ising model, a(T ) < 0 when T > Tc = Jq/kB and a(T ) < 0 when T < Tc. When a(T ) > 0, we have just a single equilibrium solution to (5.39) at m = 0. This is typically the situation at high temperatures. In contrast, at a(T ) < 0, there are three solutions to (5.39). The solution m = 0 clearly has higher free energy: this is now the unstable solution. The two stable solutions sit at m = \u00b1m0. For example, if we choose to truncate the free energy (5.38) at quartic order, we have m0 = \u2212a 2b T < Tc If a(T ) is a smooth function then the equilibrium value of m changes continuously from m = 0 when a(T ) > 0 to m = 0 at a(T ) < 0. This describes a second order phase transition occurring at Tc, de\ufb01ned by a(Tc) = 0. Once we know the equilibrium value of m, we can then substitute this back into the free energy F (T ; m) in (5.38). This gives the thermodynamic free energy F (T ) of the system in equilibrium that we have been studying throughout this course. For the quartic free energy, we have F (T ) = F0(T ) F0(T ) \u2212 a2/4b T > Tc T < Tc (5.40) Because a(Tc) = 0, the equilibrium free energy F (T ) is continuous at T = Tc. Moreover, the entropy S = \u2212\u2202F/\u2202T is also continuous at T = Tc. However, if you di\ufb00erentiate the \u2013 173 \u2013 FmFm equilibrium free energy twice, you will get a term a 2/b which is generically not vanishing at T = Tc. This means that the heat capacity C = T \u2202S/\u2202T changes discontinuously at T = Tc, as be\ufb01ts a second order phase transition. A word of warning: if you want to compute equilibrium quantities such as the heat capacity, it\u2019s important that you", " \ufb01rst substitution in the equilibrium value of m and work with (5.40) rather than i (5.38). If you don\u2019t, you miss the fact that the magnetization also changes with T. We can easily compute critical exponents within the context of Landau theory. We need only make further assumptions about the behaviour of a(T ) and b(T ) in the vicinity of Tc. If we assume that near T = Tc, we can write b(T ) \u2248 b0, a(T ) \u2248 a0(T \u2212 Tc) (5.41) then we have m0 \u2248 \u00b1 a0 2b0 (Tc \u2212 T )1/2 T < Tc which reproduces the critical exponent (5.9) and (5.20) that we derived for the van der Waals equation and Ising model respectively. Notice that we didn\u2019t put any discontinuities into the free energy. Everything in F (T ; m) was nice and smooth. When Taylor expanded, it has only integer powers of m and T as shown in (5.38) and (5.41). But the minima of F behave in a non-analytic fashion as seen in the expression for m0 above. Landau\u2019s theory of phase transitions predicts this same critical exponent for all values of the dimension d of the system. But we\u2019ve already mentioned in previous contexts that the critical exponent is in fact only correct for d \u2265 4. We will understand how to derive this criterion from Landau theory in the next section. Spontaneous Symmetry Breaking As we approach the end of the course, we\u2019re touching upon a number of ideas that become increasingly important in subsequent developments in physics. We already brie\ufb02y met the idea of universality and critical phenomena. Here I would like to point out another very important idea: spontaneous symmetry breaking. The free energy (5.38) is invariant under the Z2 symmetry m \u2192 \u2212m. Indeed, we said that one common reason that we can expand the free energy only in even powers of m is that the underlying theory also enjoys this symmetry. But below Tc, the system must pick one of the two ground states m = +m0 or m = \u2212m0. Whichever choice it makes breaks the Z2 symmetry. We say that the symmetry is spontaneously broken by the choice of ground state of the theory. \u2013 174 \u2013 Spontaneous symmetry breaking has particularly dramatic consequences when the symmetry in question is continuous rather than discrete. For example, consider a situation where the order parameter is a complex number \u03c8 and the free energy is given by (5.38) with m = |\u03c8|2. (This is e\ufb00ectively what happens for BECs, super\ufb02uids and superconductors). Then we should only look at the m > 0 solutions so that the ground state has |\u03c8|2 = +m0. But this leaves the phase of \u03c8 completely undetermined. So there is now a continuous choice of ground states: we get to sit anywhere on the circle parameterised by the phase of \u03c8. Any choice that the system makes spontaneously breaks the U (1) rotational symmetry which acts on the phase of \u03c8. Some beautiful results due to Nambu and Goldstone show that the much of the physics of these systems can be understood simply as a consequence of this symmetry breaking. The ideas of spontaneous symmetry breaking are crucial in both condensed matter physics and particle physics. In the latter context, it is intimately tied with the Higgs mechanism. 5.4.2 First Order Phase Transitions Let us now consider a situation where the expansion of the free energy also includes odd powers of the order parameter F (T ; m) = F0(T ) + \u03b1(T )m + a(T )m2 + \u03b3(T )m3 + b(T )m4 +... For example, this is the kind of expansion that we get for the Ising model free energy (5.37) when B = 0, which reads FIsing(T ; m) = \u2212N kBT log 2 + JN q 2 m2 \u2212 N 2kBT (B + Jqm)2 + N 24(kBT )3 (B + Jqm)4 +... Notice that there is no longer a symmetry relating m \u2192 \u2212m: the B \ufb01eld has a preference for one sign over the other. If we again assume that b(T ) > 0 for all temperatures, the crude shape of the free energy graph again has two choices: there is a single minimum, or two minima and a local maximum. Let\u2019s start at suitably low temperatures for which the situation is depicted in Figure 52. The free energy once again has a double well,", " except now slightly skewed. The local maximum is still an unstable point. But this time around, the minima with the lower free energy is preferred over the other one. This is the true ground state of the system. In contrast, the point which is locally, but not globally, a minimum corresponds to a meta-stable state of the system. In order for the system to leave this state, it must \ufb01rst \ufb02uctuate up and over the energy barrier separating the two. \u2013 175 \u2013 Figure 52: The free energy of the Ising model for B < 0, B = 0 and B > 0. In this set-up, we can initiate a \ufb01rst order phase transition. This occurs when the coe\ufb03cient of the odd terms, \u03b1(T ) and \u03b3(T ) change sign and the true ground state changes discontinuously from m < 0 to m > 0. In some systems this behaviour occurs when changing temperature; in others it could occur by changing some external parameter. For example, in the Ising model the \ufb01rst order phase transition is induced by changing B. At very high temperature, the double well potential is lost in favour of a single minimum as depicted in the \ufb01gure to the right. There is a unique ground state, albeit shifted from m = 0 by the presence of the \u03b1(T ) term above (which translates into the magnetic \ufb01eld B in the Ising model). The temperature at which the meta-stable ground state of the system is lost corresponds to the spinodal point in our discussion of the liquid-gas transition. Figure 53: One can play further games in Landau theory, looking at how the shape of the free energy can change as we vary temperature or other parameters. One can also use this framework to give a simple explanation of the concept of hysteresis. You can learn more about these from the links on the course webpage. 5.4.3 Lee-Yang Zeros You may have noticed that the \ufb02avour of our discussion of phase transitions is a little di\ufb00erent from the rest of the course. Until now, our philosophy was to derive everything from the partition function. But in this section, we dumped the partition function as soon as we could, preferring instead to work with the macroscopic variables such as the free energy. Why didn\u2019t we just stick with the partition function and examine phase transitions directly? \u2013 176 \u2013 FFmFmmB<0B=0B>0Fm The reason, of course, is that the approach using the partition function is hard! In this short section, which is somewhat tangential to our main discussion, we will describe how phase transitions manifest themselves in the partition function. For concreteness, let\u2019s go back to the classical interacting gas of Section 2.5, although the results we derive will be more general. We\u2019ll work in the grand canonical ensemble, with the partition function Z(z, V, T ) = zN Z(N, V, T ) = zN N!\u03bb3N d3ri e\u2212\u03b2 j<k U (rjk) (5.42) N N To regulate any potential di\ufb03culties with short distances, it is useful to assume that the particles have hard-cores so that they cannot approach to a distance less than r0. We model this by requiring that the potential satis\ufb01es i U (rjk) = 0 for rjk < r0 But this has an obvious consequence: if the particles have \ufb01nite size, then there is a maximum number of particles, NV, that we can \ufb01t into a \ufb01nite volume V. (Roughly this number is NV \u223c V /r3 0). But that, in turn, means that the canonical partition function Z(N, V, T ) = 0 for N > NV, and the grand partition function Z is therefore a \ufb01nite polynomial in the fugacity z, of order NV. But if the partition function is a \ufb01nite polynomial, there can\u2019t be any discontinuous behaviour associated with a phase transition. In particular, we can calculate pV = kBT log Z (5.43) which gives us pV as a smooth function of z. We can also calculate log Z N = z \u2202 \u2202z which gives us N as a function of z. Eliminating z between these two functions (as we did for both bosons and fermions in Section 3) tells us that pressure p is a smooth function of density N/V. We\u2019re never going to get the behaviour that we derived from the Maxwell construction in which the plot of pressure vs density shown in Figure 37 exhibits a discontinous derivative. (5.44) The discussion above is", " just re-iterating a statement that we\u2019ve alluded to several times already: there are no phase transitions in a \ufb01nite system. To see the discontinuous behaviour, we need to take the limit V \u2192 \u221e. A theorem due to Lee and Yang12 gives us a handle on the analytic properties of the partition function in this limit. 12This theorem was \ufb01rst proven for the Ising model in 1952. Soon afterwards, the same Lee and Yang proposed a model of parity violation in the weak interaction for which they won the 1957 Nobel prize. \u2013 177 \u2013 The surprising insight of Lee and Yang is that if you\u2019re interested in phase transitions, you should look at the zeros of Z in the complex z-plane. Let\u2019s \ufb01rstly look at these when V is \ufb01nite. Importantly, at \ufb01nite V there can be no zeros on the positive real axis, z > 0. This follows follows from the de\ufb01ntion of Z given in (5.42) where it is a sum of positive quantities. Moreover, from (5.44), we can see that Z is a monotonically increasing function of z because we necessarily have N > 0. Nonetheless, Z is a polynomial in z of order NV so it certainly has NV zeros somewhere in the complex z-plane. Since Z (z) = Z(z), these zeros must either sit on the real negative axis or come in complex pairs. However, the statements above rely on the fact that Z is a \ufb01nite polynomial. As we take the limit V \u2192 \u221e, the maximum number of particles that we can \ufb01t in the system diverges, NV \u2192 \u221e, and Z is now de\ufb01ned as an in\ufb01nite series. But in\ufb01nite series can do things that \ufb01nite ones can\u2019t. The Lee-Yang theorem says that as long as the zeros of Z continue to stay away from the positive real axis as V \u2192 \u221e, then no phase transitions can happen. But if one or more zeros happen to touch the positive real axis, life gets more interesting. More concretely, the Lee-Yang theorem states: \u2022 Lee-Yang Theorem: The quantity 1 V \u0398 = lim V \u2192\u221e log Z(z, V, T ) exists for all z > 0. The result is a continuous, non-decreasing function of z which is independent of the shape of the box (up to some sensible assumptions such as Surface Area/V \u223c V \u22121/3 which ensures that the box isn\u2019t some stupid fractal shape). Moreover, let R be a \ufb01xed, volume independent, region in the complex z plane which contains part of the real, positive axis. If R contains no zero of Z(z, V, T ) for all z \u2208 R then \u0398 is a an analytic function of z for all z \u2208 R. In particular, all derivatives of \u0398 are continuous. In other words, there can be no phase transitions in the region R even in the V \u2192 \u221e limit. The last result means that, as long as we are safely in a region R, taking derivatives of with respect to z commutes with the limit V \u2192 \u221e. In other words, we are allowed to use (5.44) to write the particle density n = N/V as lim V \u2192\u221e n = lim V \u2192\u221e z \u2202 \u2202z p kBT = z \u2202\u0398 \u2202z \u2013 178 \u2013 However, if we look at points z where zeros appear on the positive real axis, then \u0398 will generally not be analytic. If d\u0398/dz is discontinuous, then the system is said to undergo a \ufb01rst order phase transition. More generally, if dm\u0398/dzm is discontinuous for m = n, but continuous for all m < n, then the system undergoes an nth order phase transition. We won\u2019t o\ufb00er a proof of the Lee-Yang theorem. Instead illustrate the general idea with an example. A Made-Up Example Ideally, we would like to start with a Hamiltonian which exhibits a \ufb01rst order phase transition, compute the associated grand partition function Z and then follow its zeros as V \u2192 \u221e. However, as we mentioned above, that\u2019s hard! Instead we will simply make up a partition function Z which has the appropriate properties. Our choice is somewhat arti\ufb01cial, Z(z, V ) = (1 + z)[\u03b1V ](1 + z[\u03b1V ]) Here \u03b1 is a constant which will typically depend on temperature, although we\u2019ll suppress this dependence", " in what follows. Also, [x] = Integer part of x Although we just made up the form of Z, it does have the behaviour that one would expect of a partition function. In particular, for \ufb01nite V, the zeros sit at z = \u22121 and z = e\u03c0i(2n+1)/[\u03b1V ] n = 0, 1,..., [\u03b1V ] \u2212 1 As promised, none of the zeros sit on the positive real axis.However, as we increase V, the zeros become denser and denser on the unit circle. From the Lee-Yang theorem, we expect that no phase transition will occur for z = 1 but that something interesting could happen at z = 1. Let\u2019s look at what happens as we send V \u2192 \u221e. We have \u0398 = lim V \u2192\u221e = lim V \u2192\u221e = 1 V 1 V log Z(z, V ) [\u03b1V ] log(1 + z) + log(1 + z[\u03b1V ]) \u03b1 log(1 + z) \u03b1 log(1 + z) + \u03b1 log z |z| < 1 |z| > 1 We see that \u0398 is continuous for all z as promised. But it is only analytic for |z| = 1. \u2013 179 \u2013 We can extract the physics by using (5.43) and (5.44) to eliminate the dependence on z. This gives us the equation of state, with pressure p as a function of n = V /N. For |z| < 1, we have p = \u03b1kBT log \u03b1 \u03b1 \u2212 n n \u2208 [0, \u03b1/2), p < kBT log 2 While for |z| > 1, we have p = \u03b1kBT log 2\u03b1n (2\u03b1 \u2212 n)2 n \u2208 (3\u03b1/2, 2\u03b1), p > kBT log 2 They key point is that there is a jump in particle density of \u2206n = \u03b1 at p = \u03b1kBT log 2. Plotting this as a function of p vs v = 1/n, we \ufb01nd that we have a curve that is qualitatively identical to the pressure-volume plot of the liquid-gas phase diagram under the co-existence curve. (See, for example, \ufb01gure 37). This is a \ufb01rst order phase transition. 5.5 Landau-Ginzburg Theory Landau\u2019s theory of phase transition focusses only on the average quantity, the order parameter. It ignores the \ufb02uctuations of the system, assuming that they are negligible. Here we sketch a generalisation which attempts to account for these \ufb02uctuations. It is known as Landau-Ginzburg theory. The idea is to stick with the concept of the order parameter, m. But now we allow the order parameter to vary in space so it becomes a function m(r). Let\u2019s restrict ourselves to the situation where there is a symmetry of the theory m \u2192 \u2212m so we need only consider even powers in the expansion of the free energy. We add to these a gradient term whose role is to captures the fact that there is some sti\ufb00ness in the system, so it costs energy to vary the order parameter from one point to another. (For the example of the Ising model, this is simply the statement that nearby spins want to be aligned). The free energy is then given by F [m(r)] = ddr a(T )m2 + b(T )m4 + c(T )(\u2207m)2 (5.45) where we have dropped the constant F0(T ) piece which doesn\u2019t depend on the order parameter and hence plays no role in the story. Notice that we start with terms quadratic in the gradient: a term linear in the gradient would violate the rotational symmetry of the system. \u2013 180 \u2013 We again require that the free energy is minimised. But now F is a functional \u2013 it is a function of the function m(r). To \ufb01nd the stationary points of such objects we need to use the same kind of variational methods that we use in Lagrangian mechanics. We write the variation of the free energy as \u03b4F = = ddr 2am \u03b4m + 4bm3 \u03b4m + 2c\u2207m \u00b7 \u2207\u03b4m ddr 2am + 4bm3 \u2212 2c\u22072m \u03b4m where to go from the \ufb01rst line to the second we have integrated by parts. (We need to remember that c(T ) is a function of temperature but does not vary in space so that \u2207 doesn\u2019t act on it). The minimum of the free energy is then determined by setting \u03b4F = 0 which means that we", " have to solve the Euler-Lagrange equations for the function m(r), c\u22072m = am + 2bm3 (5.46) The simplest solutions to this equation have m constant, reducing us back to Landau theory. We\u2019ll assume once again that a(T ) > 0 for T > Tc and a(T ) < 0 for T < Tc. Then the constant solutions are m = 0 for T > Tc and m = \u00b1m0 = \u00b1\u2212a/2b for T < Tc. However, allowing for the possibility of spatial variation in the order parameter also opens up the possibility for us to search for more interesting solutions. Domain Walls Suppose that we have T < Tc so there exist two degenerate ground states, m = \u00b1m0. We could cook up a situation in which one half of space, say x < 0, lives in the ground state m = \u2212m0 while the other half of space, x > 0 lives in m = +m0. This is exactly the situation that we already met in the liquid-gas transition and is depicted in Figure 38. It is also easy to cook up the analogous con\ufb01guration in the Ising model. The two regions in which the spins point up or down are called domains. The place where these regions meet is called the domain wall. We would like to understand the structure of the domain wall. How does the system interpolate between these two states? The transition can\u2019t happen instantaneously because that would result in the gradient term (\u2207m)2 giving an in\ufb01nite contribution to the free energy. But neither can the transition linger too much because any point at which m(r) di\ufb00ers signi\ufb01cantly from the value m0 costs free energy from the m2 and m4 terms. There must be a happy medium between these two. \u2013 181 \u2013 To describe the system with two domains, m(r) must vary but it need only change in one direction: m = m(x). Equation (5.46) then becomes an ordinary di\ufb00erential equation, d2m dx2 = am c + 2bm3 c This equation is easily solved. We should remember that in order to have two vacua, T < Tc which means that a < 0. We then have m = m0 tanh \u2212a 2c x where m0 = \u2212a/2b is the constant ground state solution for the spin. As x \u2192 \u00b1\u221e, the tanh function tends towards \u00b11 which means that m \u2192 \u00b1m0. So this solution indeed interpolates between the two domains as required. We learn that the width of the domain wall is given by \u22122c/a. Outside of this region, the magnetisation relaxes exponentially quickly back to the ground state values. We can also compute the cost in free energy due to the presence of the domain wall. To do this, we substitute the solution back into the expression for the free energy (5.45). The cost is not proportional to the volume of the system, but instead proportional to the area of the domain wall. This means that if the system has linear size L then the free energy of the ground state scales as Ld while the free energy required by the wall scales only as Ld\u22121. It is simple to \ufb01nd the parametric dependence of this domain wall energy without doing any integrals; the energy per unit area scales as \u2212ca3/b. Notice that as we approach the critical point, and a \u2192 0, the two vacua are closer, the width of the domain wall increases and its energy decreases. 5.5.1 Correlations One of the most important applications of Landau-Ginzburg theory is to understand the correlations between \ufb02uctuations of the system at di\ufb00erent points in space. Suppose that we know that the system has an unusually high \ufb02uctuation away from the average at some point in space, let\u2019s say the origin r = 0. What is the e\ufb00ect of this on nearby points? There is a simple way to answer this question that requires us only to solve the di\ufb00erential equation (5.46). However, there is also a more complicated way to derive the same result which has the advantage of stressing the underlying physics and the role played by \ufb02uctuations. Below we\u2019ll start by deriving the correlations in the simple manner. We\u2019ll then see how it can also be derived using more technical machinery. \u2013 182 \u2013 We assume that the system sits in a given ground state, say m = +m0, and imagine small perturbations around this. We write the magnetisation as If we substitute this into equation (5.46) and keep only terms linear in \u03b4m, we \ufb01nd m", "(r) = m0 + \u03b4m(r) (5.47) c\u22072\u03b4m + 2a c \u03b4m = 0 where we have substituted m2 (Recall that a < 0 in the ordered phase). We now perturb the system. This can be modelled by putting a delta-function source at the origin, so that the above equation becomes 0 = \u2212a/2b to get this result. c\u22072\u03b4m + 2a c \u03b4m = 1 2c \u03b4d(0) where the strength of the delta function has been chosen merely to make the equation somewhat nicer. It is straightforward to solve the asymptotic behaviour of this equation. Indeed, it is the same kind of equation that we already solved when discussing the Debye-H\u00a8uckel model of screening. Neglecting constant factors, it is \u03b4m(r) \u223c e\u2212r/\u03be r(d\u22121)/2 (5.48) This tells us how the perturbation decays as we move away from the origin. This In equation has several names, re\ufb02ecting the fact that it arises in many contexts. liquids, it is usually called the Ornstein-Zernicke correlation. It also arises in particle physics as the Yukawa potential. The length scale \u03be is called the correlation length \u03be = \u2212c 2a (5.49) The correlation length provides a measure of the distance it takes correlations to decay. Notice that as we approach a critical point, a \u2192 0 and the correlation length diverges. This provides yet another hint that we need more powerful tools to understand the physics at the critical point. We will now take the \ufb01rst baby step towards developing these tools. 5.5.2 Fluctuations The main motivation to allow the order parameter to depend on space is to take into the account the e\ufb00ect of \ufb02uctuations. To see how we can do this, we \ufb01rst need to think a little more about the meaning of the quantity F [m(r)] and what we can use it for. \u2013 183 \u2013 To understand this point, it\u2019s best if we go back to basics. We know that the true free energy of the system can be equated with the log of the partition function (1.36). We\u2019d like to call the true free energy of the system F because that\u2019s the notation that we\u2019ve been using throughout the course. But we\u2019ve now called the Landau-Ginzburg functional F [m(r)] and, while it\u2019s closely related to the true free energy, it\u2019s not quite the same thing as we shall shortly see. So to save some confusion, we\u2019re going to change notation at this late stage and call the true free energy A. Equation (1.36) then reads A = \u2212kBT log Z, which we write this as e\u2212\u03b2A = Z = e\u2212\u03b2En n We would like to understand the right way to view the functional F [m(r)] in this framework. Here we give a heuristic and fairly handwaving argument. A fuller treatment involves the ideas of the renormalisation group. The idea is that each microstate |n of the system can be associated to some speci\ufb01c function of the spatially varying order parameter m(r). To illustrate this, we\u2019ll talk in the language of the Ising model although the discussion generalises to any system. There we could consider associate a magnetisation m(r) to each lattice site by simply averaging over all the spins within some distance of that point. Clearly, this will only lead to functions that take values on lattice sites rather than in the continuum. But if the functions are suitably well behaved it should be possible to smooth them out into continuous functions m(r) which are essentially constant on distance scales smaller than the lattice spacing. In this way, we get a map from the space of microstates to the magnetisation, |n \u2192 m(r). But this map is not one-to-one. For example, if the averaging procedure is performed over enough sites, \ufb02ipping the spin on just a single site is unlikely to have much e\ufb00ect on the average. In this way, many microstates map onto the same average magnetisation. Summing over just these microstates provides a \ufb01rst principles construction of the F [m(r)], e\u2212\u03b2F [m(r)] = e\u2212\u03b2En n|m(r) (5.50) Of course, we didn\u2019t actually perform this procedure to get to (5.45): we simply wrote it down the most general form in the vicinity of a critical point with a bunch of unknown coe", "\ufb03cients a(T ), b(T ) and c(T ). But if we were up for a challenge, the above procedure tells us how we could go about \ufb01guring out those functions from \ufb01rst principles. More importantly, it also tells us what we should do with the Landau-Ginzburg free energy. Because in (5.50) we have only summed over those states that correspond to a particular \u2013 184 \u2013 value of m(r). To compute the full partition function, we need to sum over all states. But we can do that by summing over all possible values of m(r). In other words, Z = Dm(r) e\u2212\u03b2F [m(r)] (5.51) This is a tricky beast: it is a functional integral. We are integrating over all possible function m(r), which is the same thing as performing an in\ufb01nite number of integrations. (Actually, because the order parameters m(r) arose from an underlying lattice and are suitably smooth on short distance scales, the problem is somewhat mitigated). The result (5.51) is physically very nice, albeit mathematically somewhat daunting. It means that we should view the Landau-Ginzburg free energy as a new e\ufb00ective Hamiltonian for a continuous variable m(r). It arises from performing the partition function sum over much of the microscopic information, but still leaves us with a \ufb01nal sum, or integral, over \ufb02uctuations in an averaged quantity, namely the order parameter. To complete the problem, we need to perform the function integral (5.51). This is hard. Here \u201chard\u201d means that the majority of unsolved problems in theoretical physics can be boiled down to performing integrals of this type. Yet the fact it\u2019s hard shouldn\u2019t dissuade us, since there is a wealth of rich and beautiful physics hiding in the path integral, including the deep reason behind the magic of universality. We will start to explore some of these ideas in next year\u2019s course on Statistical Field Theory. \u2013 185 \u2013f a rigid rod. The masses are equal if and only if a force applied at the rod's midpoint causes them to accelerate at the same rate, so that the rod does not rotate. these delicate questions here. Based on your introductory course in general physics, you have a reasonably good idea what mass and force mean, and it is easy to describe how these parameters are defined and measured in many realistic situations. Mass The mass of an object characterizes the object's inertia\u2014its resistance to being accelerated: A big boulder is hard to accelerate, and its mass is large. A little stone is easy to accelerate, and its mass is small. To make these natural ideas quantitative we have to define a unit of mass and then give a prescription for measuring the mass of any object in terms of the chosen unit. The internationally agreed unit of mass is the kilogram and is defined arbitrarily to be the mass of a chunk of platinum\u2014iridium stored at the International Bureau of Weights and Measures outside Paris. To measure the mass of any other object, we need a means of comparing masses. In principle, this can be done with an inertial balance as shown in Figure 1.2. The two objects to be compared are fastened to the opposite ends of a light, rigid rod, which is then given a sharp pull at its midpoint. If the masses are equal, they will accelerate equally and the rod will move off without rotating; if the masses are unequal, the more massive one will accelerate less, and the rod will rotate as it moves off. The beauty of the inertial balance is that it gives us a method of mass comparison that is based directly on the notion of mass as resistance to being accelerated. In practice, an inertial balance would be very awkward to use, and it is fortunate that there are much easier ways to compare masses, of which the easiest is to weigh the objects. As you certainly recall from your introductory physics course, an object's mass is found to be exactly proportional to the object's weight 3 (the gravitational force on the object) provided all measurements are made in the same location. Thus two 3 This observation goes back to Galileo's famous experiments showing that all objects are accelerated at the same rate by gravity. The first modern experiments were conducted by the Hungarian physicist Eiityos (1848-1919), who showed that weight is proportional to mass to within Section 1.3 Mass and Force 11 objects have the same mass if and only if they have the same weight (when weighed at the same place), and a simple, practical way to check whether two masses are equal is simply to weigh them and see if their weights are equal. Armed with methods for comparing masses, we can easily set up a scheme to mea- sure arbitrary masses. First, we can build a large number of standard", " kilograms, each one checked against the original 1-kg mass using either the inertial or gravitational balance. Next, we can build multiples and fractions of the kilogram, again checking them with our balance. (We check a 2-kg mass on one end of the balance against two 1-kg masses placed together on the other end; we check two half-kg masses by verifying that their masses are equal and that together they balance a 1-kg mass; and so on.) Finally, we can measure an unknown mass by putting it on one end of the balance and loading known masses on the other end until they balance to any desired precision. Force The informal notion of force as a push or pull is a surprisingly good starting point for our discussion of forces. We are certainly conscious of the forces that we exert ourselves. When I hold up a sack of cement, I am very aware that I am exerting an upward force on the sack; when I push a heavy crate across a rough floor, I am aware of the horizontal force that I have to exert in the direction of motion. Forces exerted by inanimate objects are a little harder to pin down, and we must, in fact, understand something of Newton's laws to identify such forces. If I let go of the sack of cement, it accelerates toward the ground; therefore, I conclude that there must be another force \u2014 the sack's weight, the gravitational force of the earth \u2014 pulling it downward. As I push the crate across the floor, I observe that it does not accelerate, and I conclude that there must be another force \u2014 friction \u2014 pushing the crate in the opposite direction. One of the most important skills for the student of elementary mechanics is to learn to examine an object's environment and identify all the forces on the object: What are the things touching the object and possibly exerting contact forces, such as friction or air pressure? And what are the nearby objects possibly exerting action-at-a-distance forces, such as the gravitational pull of the earth or the electrostatic force of some charged body? If we accept that we know how to identify forces, it remains to decide how to measure them. As the unit of force we naturally adopt the newton (abbreviated N) defined as the magnitude of any single force that accelerates a standard kilogram mass with an acceleration of 1 m/s 2. Having agreed what we mean by one newton, we can proceed in several ways, all of which come to the same final conclusion, of course. The route that is probably preferred by most philosophers of science is to use Newton's second law to define the general force: A given force is 2 N if, by itself, it accelerates a standard kilogram with an acceleration of 2 m/s 2, and so a few parts in 109. Experiments in the last few decades have narrowed this to around one part in 1012. 12 Chapter 1 Newton's Laws of Motion = 2 N Figure 1.3 One of many possible ways to define forces of any magnitude. The lower spring balance has been calibrated to read 1 N. If the balance arm on the left is adjusted so that the lever arms above and below the pivot are in the ratio 1 : 2 and if the force F 1 is 1 N, then the force F2 required to balance the arm is 2 N. This lets us calibrate the upper spring balance for 2 N. By readjusting the two lever arms, we can, in principle, calibrate the second spring balance to read any force. on. This approach is not much like the way we usually measure forces in practice, 4 and for our present discussion a simpler procedure is to use some spring balances. Using our definition of the newton, we can calibrate a first spring balance to read 1 N. Then by matching a second spring balance against the first, using a balance arm as shown in Figure 1.3, we can define multiples and fractions of a newton. Once we have a fully calibrated spring balance we can, in principle, measure any unknown force, by matching it against the calibrated balance and reading off its value. So far we have defined only the magnitude of a force. As you are certainly aware, forces are vectors, and we must also define their directions. This is easily done. If we apply a given force F (and no other forces) to any object at rest, the direction of F is defined as the direction of the resulting acceleration, that is, the direction in which the Now that we know, at least in principle, what we mean by positions, times, masses, and forces, we can proceed to discuss the cornerstone of our subject \u2014 Newton's three body moves off. laws of motion. 4 The approach also creates the confusing appearance that Newton's second law is just a conse- quence of the definition of force. This is not really true: Whatever definition we choose for force, a large part of the second law is experimental. One advantage of defining forces with spring bal- ances", " is that it separates out the definition of force from the experimental basis of the second law. Of course, all commonly accepted definitions give the same final result for the value of any given force. Section 1.4 Newton's First and Second Laws; Inertial Frames 13 1.4 Newton's First and Second Laws; Inertial Frames In this chapter, I am going to discuss Newton's laws as they apply to a point mass. A point mass, or particle, is a convenient fiction, an object with mass, but no size, that can move through space but has no internal degrees of freedom. It can have \"translational\" kinetic energy (energy of its motion through space) but no energy of rotation or of internal vibrations or deformations. Naturally, the laws of motion are simpler for point particles than for extended bodies, and this is the main reason that we start with the former. Later on, I shall build up the mechanics of extended bodies from our mechanics of point particles by considering the extended body as a collection of many separate particles. Nevertheless, it is worth recognizing that there are many important problems where the objects of interest can be realistically approximated as point masses. Atomic and subatomic particles can often be considered to be point masses, and even macroscopic objects can frequently be approximated in this way. A stone thrown off the top of a cliff is, for almost all purposes, a point particle. Even a planet orbiting around the sun can usually be approximated in the same way. Thus the mechanics of point masses is more than just the starting point for the mechanics of extended bodies; it is a subject with wide application itself. Newton's first two laws are well known and easily stated: Newton's First Law (the Law of Inertia of forces, a particle moves with constant velocity v. and Newton's Second La For;in y particle of mass m, the net force F on the partici mass m times the particle's acceleration: In this equation F denotes the vector sum of all the forces on the particle and a is the particle's acceleration, dv a = \u2014 dt- v = d2r dt 2 Here v denotes the particle's velocity, and I have introduced the convenient notation of dots to denote differentiation with respect to t, as in v =r and a = v = f. 14 Chapter 1 Newton's Laws of Motion Both laws can be stated in various equivalent ways. For instance (the first law): In the absence of forces, a stationary particle remains stationary and a moving particle continues to move with unchanging speed in the same direction. This is, of course, exactly the same as saying that the velocity is always constant. Again, v is constant if and only if the acceleration a is zero, so an even more compact statement is this: In the absence of forces a particle has zero acceleration. The second law can be rephrased in terms of the particle's momentum, defined as p = mv. (1.18) In classical mechanics, we take for granted that the mass m of a particle never changes, so that Thus the second law (1.17) can be rephrased to say that p =my-,-- ma. In classical mechanics, the two forms (1.17) and (1.19) of the second law are com- pletely equivalent.5 Differential Equations When written in the form ml = F, Newton's second law is a differential equation for the particle's position r(t). That is, it is an equation for the unknown function r(t) that involves derivatives of the unknown function. Almost all the laws of physics are, or can be cast as, differential equations, and a huge proportion of a physicist's time is spent solving these equations. In particular, most of the problems in this book involve differential equations \u2014 either Newton's second law or its counterparts in the Lagrangian and Hamiltonian forms of mechanics. These vary widely in their difficulty. Some are so easy to solve that one scarcely notices them. For example, consider Newton's second law for a particle confined to move along the x axis and subject to a constant force Fo,. i(t) F m. This is a second-order differential equation for x (t) as a function of t. (Second-order because it involves derivatives of second order, but none of higher order.) To solve it 5 In relativity, the two forms are not equivalent, as we'll see in Chapter 15. Which form is correct depends on the definitions we use for force, mass, and momentum in relativity. If we adopt the most popular definitions of these three quantities, then it is the form (1.19) that holds in relativity. Section 1.4 Newton's First and Second Laws; Inertial Frames 15 one has only to integrate it twice. The first integration gives the velocity i(t)= I 1(0 dt = vo + fn where the constant of integration is the particle's initial velocity, and a second int", "e- gration gives the position x(t) = f (t) dt = x, v ot + \u2014Fo t2 2m where the second constant of integration is the particle's initial position. Solving this differential equation was so easy that we certainly needed no knowledge of the theory of differential equations. On the other hand, we shall meet lots of differential equations that do require knowledge of this theory, and I shall present the necessary theory as we need it. Obviously, it will be an advantage if you have already studied some of the theory of differential equations, but you should have no difficulty picking it up as we go along. Indeed, many of us find that the best way to learn this kind of mathematical theory is in the context of its physical applications. Inertial Frames On the face of it, Newton's second law includes his first: If there are no forces on an object, then F = 0 and the second law (1.17) implies that a = 0, which is the first law. There is, however, an important subtlety, and the first law has an important role to play. Newton's laws cannot be true in all conceivable reference frames. To see this, consider just the first law and imagine a reference frame \u2014 we'll call it 8 \u2014 in which the first law is true. For example, if the frame 8 has its origin and axes fixed relative to the earth's surface, then, to an excellent approximation, the first law (the law of inertia) holds with respect to the frame 8: A frictionless puck placed on a smooth horizontal surface is subject to zero force and, in accordance with the first law, it moves with constant velocity. Because the law of inertia holds, we call 8 an inertial frame. If we consider a second frame 8' which is moving relative to S with constant velocity and is not rotating, then the same puck will also be observed to move with constant velocity relative to 8'. That is, the frame 8' is also inertial. If, however, we consider a third frame 8\" that is accelerating relative to 8, then, as viewed from 8\", the puck will be seen to be accelerating (in the opposite direction). Relative to the accelerating frame 8\" the law of inertia does not hold, and we say that 8\" is noninertial. I should emphasize that there is nothing mysterious about this result. Indeed it is a matter of experience. The frame 8' could be a frame attached to a high-speed train traveling smoothly at constant speed along a straight track, and the frictionless puck, an ice cube placed on the floor of the train, as in Figure 1.4. As seen from the train (frame 8'), the ice cube is at rest and remains at rest, in accord with the first law. As seen from the ground (frame 8), the ice cube is moving with the same velocity as the train and continues to do so, again in obedience to the first law. But now consider conducting the same experiment on a second train (frame 8\") that is accelerating forward. As this train accelerates forward, the ice cube is left behind, and, relative to 8\", the ice cube accelerates backward, even though subject to no net 16 Chapter 1 Newton's Laws of Motion Figure 1.4 The frame 8 is fixed to the ground, while 8' is fixed to a train traveling at constant velocity v' relative to 8. An ice cube placed on the floor of the train obeys Newton's first law as seen from both and 8'. If the train to which 8\" is attached is accelerating forward, then, as seen in 8\", an ice cube placed on the floor will accelerate backward, and the first law does not hold in 8\". force. Clearly the frame S\" is noninertial, and neither of the first two laws can hold in S\". A similar conclusion would hold if the frame 8\" had been attached to a rotating merry-go-round. A frictionless puck, subject to zero net force, would not move in a straight line as seen in 8\", and Newton's laws would not hold. Evidently Newton's two laws hold only in the special, inertial (nonaccelerating and nonrotating) reference frames. Most philosophers of science take the view that the first law should be used to identify these inertial frames \u2014 a reference frame 8 is inertial if objects that are clearly subject to no forces are seen to move with constant velocity relative to 8. 6 Having identified the inertial frames by means of Newton's first law, we can then claim as an experimental fact that the second law holds in these same inertial frames? Since the laws of motion hold only in inertial frames, you might imagine that we would confine our attention exclusively to inertial frames, and, for a while, we shall do just that. Nevertheless, you should be aware that there are situations where it is necessary, or at least very convenient, to work in noninertial frames. The", " most important example of a noninertial frame is in fact the earth itself. To an excellent approximation, a reference frame fixed to the earth is inertial \u2014 a fortunate circumstance for students of physics! Nevertheless, the earth rotates on its axis once a day and circles around the sun once a year, and the sun orbits slowly around the center of the Milky Way galaxy. For all of these reasons, a reference frame fixed to the earth is not exactly inertial. Although these effects are very small, there are several phenomena \u2014 the tides and the trajectories of long-range projectiles are examples \u2014 6 There is some danger of going in a circle here: How do we know that the object is subject to no forces? We'd better not answer, \"Because it's traveling at constant velocity\"! Fortunately, we can argue that it is possible to identify all sources of force, such as people pushing and pulling or nearby massive bodies exerting gravitational forces. If there are no such things around, we can reasonably say that the object is free of forces. 7 As I mentioned earlier, the extent to which the second law is an experimental statement depends on how we choose to define force. If we define force by means of the second law, then to some extent (though certainly not entirely) the law becomes a matter of definition. If we define forces by means of spring balances, then the second law is clearly an experimentally testable proposition. Section 1.5 The Third Law and Conservation of Momentum 17 that are most simply explained by taking into account the noninertial character of a frame fixed to the earth. In Chapter 9 we shall examine how the laws of motion must be modified for use in noninertial frames. For the moment, however, we shall confine our discussion to inertial frames. Validity of the First Two Laws Since the advent of relativity and quantum mechanics, we have known that Newton's laws are not universally valid. Nevertheless, there is an immense range of phenom- ena \u2014 the phenomena of classical physics \u2014 where the first two laws are for all practical purposes exact. Even as the speeds of interest approach c, the speed of light, and relativity becomes important, the first law remains exactly true. (In relativity, just as in classical mechanics, an inertial frame is defined as one where the first law holds.)8 As we shall see in Chapter 15, the two forms of the second law, F = ma and F = it, are no longer equivalent in relativity, although with F and p suitably defined the second law in the form F = p is still valid. In any case, the important point is this: In the classical domain, we can and shall assume that the first two laws (the second in either form) are universally and precisely valid. You can, if you wish, regard this assumption as defining a model \u2014 the classical model \u2014 of the natural world. The model is logically consistent and is such a good representation of many phenomena that it is amply worthy of our study. 1.5 The Third Law and Conservation of Momentum Newton's first two laws concern the response of a single object to applied forces. The third law addresses a quite different issue: Every force on an object inevitably involves a second object \u2014 the object that exerts the force. The nail is hit by the hammer, the cart is pulled by the horse, and so on. While this much is no doubt a matter of common sense, the third law goes considerably beyond our everyday experience. Newton realized that if an object 1 exerts a force on another object 2, then object 2 always exerts a force (the \"reaction\" force) back on object 1. This seems quite natural: If you push hard against a wall, it is fairly easy to convince yourself that the wall is exerting a force back on you, without which you would undoubtedly fall over. The aspect of the third law which certainly goes beyond our normal perceptions is this: According to the third law, the reaction force of object 2 on object 1 is always equal and opposite to the original force of 1 on 2. If we introduce the notation F21 to denote the force exerted on object 2 by object 1, Newton's third law can be stated very compactly: 8 However, in relativity the relationship between different inertial frames \u2014 the so-called Lorentz transformation \u2014 is different from that of classical mechanics. See Section 15.6. LITO DIDLIOJO.1NWERMETO MB! IOTEK4 n M!J %. 18 Chapter 1 Newton's Laws of Motion F21 Figure 1.5 Newton's third law asserts that the reaction force exerted on object 1 by object 2 is equal and opposite to the force exerted on 2 by 1, that is, F12 = \u2014F21. object 1 exert orce F2 force F l ct 1 given s reaction This statement is illustrated in Figure 1.5, which you could think of as showing the force of the earth on the moon and the", " reaction force of the moon on the earth (or a proton on an electron and the electron on the proton). Notice that this figure actually goes a little beyond the usual statement (1.20) of the third law: Not only have I shown the two forces as equal and opposite; I have also shown them acting along the line joining 1 and 2. Forces with this extra property are called central forces. (They act along the line of centers.) The third law does not actually require that the forces be central, but, as I shall discuss later, most of the forces we encounter (gravity, the electrostatic force between two charges, etc.) do have this property. As Newton himself was well aware, the third law is intimately related to the law of conservation of momentum. Let us focus, at first, on just two objects as shown in Figure 1.6, which might show the earth and the moon or two skaters on the ice. In addition to the force of each object on the other, there may be \"external\" forces exerted by other bodies. The earth and moon both experience forces exerted by the sun, and both skaters could experience the external force of the wind. I have shown the net external forces on the two objects as Pi' and Fe2xt. The total force on object 1 is then and similarly law: (net force on 1) 12 Feixt F 1 F = - (net force on 2) F2 = F21 + Fe2xt. We can compute the rates of change of the particles' momenta using Newton's second = F1 = Fi2 + Feixt (1.21) Section 1.5 The Third Law and Conservation of Momentum 19 F 12 21 Figure 1.6 Two objects exert forces on each other and may also be subject to additional \"external\" forces from other objects not shown. and P2 =F2=F21-f-F2 xt. (1.22) If we now define the total momentum of our two objects as then the rate of change of the total momentum is just P = Pi \u00b1 P2, 1.3 = 1.3 1 + To evaluate this, we have only to add Equations (1.21) and (1.22). When we do this, the two internal forces, F 12 and F21, cancel out because of Newton's third law, and we are left with P = Feixt Fext Fext, (1.23) where I have introduced the notation Fe' denote the total external force on our two-particle system. The result (1.23) is the first in a series of important results that let us construct a theory of many-particle systems from the basic laws for a single particle. It asserts that as far as the total momentum of a system is concerned, the internal forces have no effect. A special case of this result is that if there are no external forces (Fe' = 0) then P = 0. Thus we have the important result: If Fext = 0, then P = const. (1.24) In the absence of external forces, the total momentum of our two-particle system is constant \u2014 a result called the principle of conservation of momentum. Multiparticle Systems We have proved the conservation of momentum, Equation (1.24), for a system of two particles. The extension of the result to any number of particles is straightforward in principle, but I would like to go through it in detail, because it lets me introduce some 20 Chapter 1 Newton's Laws of Motion oP 0 0 Figure 1.7 A five-particle system with particles labelled by a or fi = 1, 2,, 5. The particle a is subject to four internal forces, shown by solid arrows and denoted F 0 (the force on a by p). In addition particle a may be subject to a net external force, shown by the dashed arrow and denoted F aext. important notation and will give you some practice using the summation notation. Let us consider then a system of N particles. I shall label the typical particle with a Greek index a or /3, either of which can take any of the values 1, 2, -, N. The mass of particle a is m a and its momentum is pa. The force on particle a is quite complicated: Each of the other (N \u2014 1) particles can exert a force which I shall call Fan, the force on a by p, as illustrated in Figure 1.7. In addition there may be a net external force on particle a, which I shall call Fr. Thus the net force on particle a is (net force on particle a) = Fa = E Fan + Faext. (1.25) SO\u00ab Here the sum runs over all values of /3 not equal to a. (Remember there is no force Faa because particle a cannot exert a force on itself.) According to Newton's second law, this is the same as the rate of change", " of pa : oc, E Fas 17t. fiOcf (1.26) This result holds for each a = 1, -, N. Let us now consider the total momentum of our N-particle system, P = I Pa a where, of course, this sum runs over all N particles, a = 1, 2,, N. If we differen- tiate this equation with respect to time, we find or, substituting for Oa from (1.26), = E E Fap E Faext. a 00ce a (1.27) Section 1.5 The Third Law and Conservation of Momentum 21 The double sum here contains N (N \u2014 1) terms in all. Each term F ap in this sum can be paired with a second term F p, (that is, F12 paired with F21, and so on), so that E E F \u201ep = E (Fa, + F sa ). a f 3 0a a /3 >a The double sum on the right includes only values of a and /3 with a < p and has half as many terms as that on the left. But each term is the sum of two forces, (F aa Foa ), and, by the third law, each such sum is zero. Therefore the whole double sum in (1.28) is zero, and returning to (1.27) we conclude that P = E a Fext _ Fext (1.28) (1.29) The result (1.29) corresponds exactly to the two-particle result (1.23). Like the latter, it says that the internal forces have no effect on the evolution of the total momentum P \u2014 the rate of change of P is determined by the net external force on the system. In particular, if the net external force is zero, we have the Principle of Conservation of Momentum he net x orce F\"' on an N-particle'system is hero, t e total momentum P is As you are certainly aware, this is one of the most important results in classical physics and is, in fact, also true in relativity and quantum mechanics. If you are not very familiar with the sorts of manipulations of sums that we used, it would be a good idea to go over the argument leading from (1.25) to (1.29) for the case of three or four particles, writing out all the sums explicitly (Problems 1.28 or 1.29). You should also convince yourself that, conversely, if the principle of conservation of momentum is true for all multiparticle systems, then Newton's third law must be true (Problem 1.31). In other words, conservation of momentum and Newton's third law are equivalent to one another. Validity of Newton's Third Law Within the domain of classical physics, the third law, like the second, is valid with such accuracy that it can be taken to be exact. As speeds approach the speed of light, it is easy to see that the third law cannot hold: The point is that the law asserts that the action and reaction forces, F 12 (t) and F21 (t), measured at the same time t, are equal and opposite. As you certainly know, once relativity becomes important the concept of a single universal time has to be abandoned \u2014 two events that are seen as simultaneous by one observer are, in general, not simultaneous as seen by a second observer. Thus, even if the equality F 12 (t) = \u2014F21(t) (with both times the same) were true for one observer, it would generally be false for another. Therefore, the third law cannot be valid once relativity becomes important. 22 Chapter 1 Newton's Laws of Motion Figure 1.8 Each of the positive charges q 1 and q2 produces a magnetic field that exerts a force on the other charge. The resulting magnetic forces F 12 and F21 do not obey Newton's third law. Rather surprisingly, there is a simple example of a well-known force \u2014 the mag- netic force between two moving charges \u2014 for which the third law is not exactly true, even at slow speeds. To see this, consider the two positive charges of Figure 1.8, with q 1 moving in the x direction and q2 moving in the y direction, as shown. The exact calculation of the magnetic field produced by each charge is complicated, but a simple argument gives the correct directions of the two fields, and this is all we need. The moving charge q 1 is equivalent to a current in the x direction. By the right-hand rule for fields, this produces a magnetic field which points in the z direction in the vicinity of q2. By the right-hand rule for forces, this field produces a force F21 on q2 that is in the x direction. An exactly analogous argument (check it yourself) shows that the force F12 on q 1 is in the y direction, as shown. Clearly these two", " forces do not obey Newton's third law! This conclusion is especially startling since we have just seen that Newton's third law is equivalent to the conservation of momentum. Apparently the total momentum m iv i + m 2v2 of the two charges in Figure 1.8 is not conserved. This conclusion, which is correct, serves to remind us that the \"mechanical\" momentum my of particles is not the only kind of momentum. Electromagnetic fields can also carry momentum, and in the situation of Figure 1.8 the mechanical momentum being lost by the two particles is going to the electromagnetic momentum of the fields. Fortunately, if both speeds in Figure 1.8 are much less than the speed of light (v << c), the loss of mechanical momentum and the concomitant failure of the third law are completely negligible. To see this, note that in addition to the magnetic force between q i and q2 there is the electrostatic Coulomb force 9 km 21 r2, which does obey Newton's third law. It is a straightforward exercise (Problem 1.32) to show that the magnetic force is of order v 2 /c2 times the Coulomb force. Thus only as v approaches c \u2014 and classical mechanics must give way to relativity anyway \u2014 is the violation of 9 Here k is the Coulomb force constant, often written as k = 1/(47r E0). Section 1.6 Newton's Second Law in Cartesian Coordinates 23 the third law by the magnetic force important. 1\u00b0 We see that the unexpected situation of Figure 1.8 does not contradict our claim that in the classical domain Newton's third law is valid, and this is what we shall assume in our discussions of nonrelativistic mechanics. 1.6 Newton's Second Law in Cartesian Coordinates Of Newton's three laws, the one that we actually use the most is the second, which is often described as the equation of motion. As we have seen, the first is theoretically important to define what we mean by inertial frames but is usually of no practical use beyond this. The third law is crucially important in sorting out the internal forces in a multiparticle system, but, once we know the forces involved, the second law is what we actually use to calculate the motion of the object or objects of interest. In particular, in many simple problems the forces are known or easily found, and, in this case, the second law is all we need for solving the problem. As we have already noted, the second law, F (1.30) is a second-order, differential equation\" for the position vector r as a function of the time t. In the prototypical problem, the forces that comprise F are given, and our job is to solve the differential equation (1.30) for r (t). Sometimes we are told about r (t), and we have to use (1.30) to find some of the forces. In any case, the equation (1.30) is a vector differential equation. And the simplest way to solve such equations is almost always to resolve the vectors into their components relative to a chosen coordinate system. Conceptually the simplest coordinate system is the Cartesian (or rectangular), with unit vectors X, ST, and z, in terms of which the net force F can then be written as F= Fx X 1 F3,S7\u2018 and the position vector r as (1.31) (1.32) As we noted in Section 1.2, this expansion of r in terms of its Cartesian components is especially easy to differentiate because the unit vectors X, S7, are constant. Thus we can differentiate (1.32) twice to get the simple result F=Ii\u00b1j3ST+Ei. (1.33) 10 The magnetic force between two steady currents is not necessarily small, even in the classical domain, but it can be shown that this force does obey the third law. See Problem 1.33. \" The force F can sometimes involve derivatives of r. (For instance the magnetic force on a moving charge involves the velocity v = 1-.) Very occasionally the force F involves a higher derivative of r, of order n > 2, in which case the second law is an nth-order differential equation. 24 Chapter 1 Newton's Laws of Motion That is, the three Cartesian components ofr are just the appropriate derivatives of the three coordinates x, y, z of r, and the second law (1.30) becomes FX x f Fy Fz i = + m53 ST mE (1.34) Resolving this equation into its three separate components, we see that Fx has to equal mz and similarly for the y and z components. That is, in Cartesian coordinates, the single vector equation (1.30) is equivalent to the three separate equations: F = mr < > (1.35) Fx = Fy, = m5; Fz =", " This beautiful result, that, in Cartesian coordinates, Newton's second law in three dimensions is equivalent to three one-dimensional versions of the same law, is the basis of the solution of almost all simple mechanics problems in Cartesian coordinates. Here is an example to remind you of how such problems go. EXAMPLE 1.1 A Block Sliding down an Incline A block of mass in is observed accelerating from rest down an incline that has coefficient of friction,u and is at angle 8 from the horizontal. How far will it travel in time t? Our first task is to choose our frame of reference. Naturally, we choose our spatial origin at the block's starting position and the origin of time (t = 0) at the moment of release. As you no doubt remember from your introductory physics course, the best choice of axes is to have one axis (x say) point down the slope, one (y) normal to the slope, and the third (z) across it, as shown in Figure 1.9. This choice has two advantages: First, because the block slides straight down the slope, the motion is entirely in the x direction, and only x varies. (If we had chosen the x axis horizontal and the y axis vertical, then both x and y would vary.) Second, two of the three forces on the block are unknown (the normal force N and friction f; the weight, w = mg, we treat as known), and with our choice of axes, each of the unknowns has only one nonzero component, since N is in the y direction and f is in the (negative) x direction. We are now ready to apply Newton's second law. The result (1.35) means that we can analyse the three components separately, as follows: There are no forces in the z direction, so Fz = 0. Since Fz = mE, it follows that E = 0, which implies that z (or v z ) is constant. Since the block starts from rest, this means that z is actually zero for all t. With z = 0, it follows that z is constant, and, since it too starts from zero, we conclude that z = 0 for all t. As we would certainly have guessed, the motion remains in the xy plane. Since the block does not jump off the incline, we know that there is no motion in the y direction. In particular, ji = 0. Therefore, Newton's second law implies that the y component of the net force is zero; that is, Fy = 0. From Figure 1.9 we see that this implies that F = N \u2014 mg cos 0 = 0. Section 1.6 Newton's Second Law in Cartesian Coordinates 25 Figure 1.9 A block slides down a slope of incline O. The three forces on the block are its weight, w = mg, the normal force of the incline, N, and the frictional force f, whose magnitude is f =,u,N. The z axis is not shown but points out of the page, that is, across the slope. Thus the y component of the second law has told us that the unknown normal force is N = mg cos 9. Since f =,u,N, this tells us the frictional force, f = pung cos 9, and all the forces are now known. All that remains is to use the remaining component (the y component) of the second law to solve for the actual motion. The x component of the second law, Fx = mx, implies (see Figure 1.9) that Or wx - f =Ira mg sin 0 \u2014,umg cos\u00b0 = The m's cancel, and we find for the acceleration down the slope = g (sin 0 \u2014 \u00b5 cos 0). (1.36) Having found x, and found it to be constant, we have only to integrate it twice to find x as a function of t. First (Remember that x = 0 initially, so the constant of integration is zero.) Finally, = g (sin 0 \u2014 \u00b5 cos 6)t x(t) = 0 \u2014 cos t9)t 2 (again the constant of integration is zero) and our solution is complete. 26 Chapter 1 Newton's Laws of Motion Figure 1.10 The definition of the polar coordinates r and 0. 1.7 Two-Dimensional Polar Coordinates While Cartesian coordinates have the merit of simplicity, we are going to find that it is almost impossible to solve certain problems without the use of various non-Cartesian coordinate systems. To illustrate the complexities of non-Cartesian coordinates, let us consider the form of Newton's second law in a two-dimensional problem using polar coordinates. These coordinates are defined in Figure 1.10. Instead of using the two rectangular coordinates x, y, we label the position of a particle with its distance r from 0 and the angle 0 measured up from the x axis. Given the rectangular coordinates x", " and y, you can calculate the polar coordinates r and 0, or vice versa, using the following relations. (Make sure you understand all four equations.' 2 ) x = r cos y = r sin 0 r = x2 + y2 arctan(y/x) = (1.37) Just as with rectangular coordinates, it is convenient to introduce two unit vectors, which I shall denote by ii- and 0. To understand their definitions, notice that we can define the unit vector x as the unit vector that points in the direction of increasing x when y is fixed, as shown in Figure 1.11(a). In the same way we shall define i s- as the unit vector that points in the direction we move when r increases with 0 fixed; likewise, S is the unit vector that points in the direction we move when 0 increases with r fixed. Figure 1.11 makes clear a most important difference between the unit vectors x and Sr of rectangular coordinates and our new unit vectors r and 4. The vectors X and Sr are the same at all points in the plane, whereas the new vectors r and ofi change their directions as the position vector r moves around. We shall see that this complicates the use of Newton's second law in polar coordinates. Figure 1.11 suggests another way to write the unit vector r. Since r is in the same direction as r, but has magnitude 1, you can see that r r (1.38) This result suggests a second role for the \"hat\" notation. For any vector a, we can define fi as the unit vector in the direction of a, namely a = a/lal. 12 There is a small subtlety concerning the equation for 4): You need to make sure lands in the proper quadrant, since the first and third quadrants give the same values for y/x (and likewise the second and fourth). See Problem 1.42. Section 1.7 Two-Dimensional Polar Coordinates 27 y (a) (b) Figure 1.11 (a) The unit vector x points in the direction of increas- ing x with y fixed. (b) The unit vector r points in the direction of increasing r with 0 fixed; 3 points in the direction of increasing 0 with r fixed. Unlike x, the vectors r and 4) change as the position vector r moves. Since the two unit vectors 1- and 4) are perpendicular vectors in our two-dimensional space, any vector can be expanded in terms of them. For instance, the net force F on an object can be written F = Foi\u2018b. (1.39) If, for example, the object in question is a stone that I am twirling in a circle on the end of a string (with my hand at the origin), then Fr would be the tension in the string and Fo the force of air resistance retarding the stone in the tangential direction. The expansion of the position vector itself is especially simple in polar coordinates. From Figure 1.11(b) it is clear that r =. (1.40) We are now ready to ask about the form of Newton's second law, F = mr, in polar coordinates. In rectangular coordinates, we saw that the x component of F is just z, and this is what led to the very simple result (1.35). We must now find the components of r in polar coordinates; that is, we must differentiate (1.40) with respect to t. Although (1.40) is very simple, the vector r changes as r moves. Thus when we differentiate (1.40), we shall pick up a term involving the derivative of 1'. Our first task is to find this derivative of ii-. Figure 1.12(a) shows the position of the particle of interest at two successive times, t1 and t2 = t1 + At. If the corresponding angles 0 (t 1) and 0 (t2) are different, then the two unit vectors i(t 1) and 1- (t2) point in different directions. The change in i s- is shown in Figure 1.12(b), and (provided At is small) is approximately (Notice that the direction of Ai is perpendicular to r, namely the direction of 4).) If we divide both sides by At and take the limit as At -\u00f7 0, then Ail At \u2014> I dt and we find that (1.41) (1.42) Or AO 4) At 4). dt 28 Chapter 1 Newton's Laws of Motion (a) (b) Figure 1.12 (a) The positions of a particle at two successive times, t j and t2. Unless the particle is moving exactly radially, the corresponding unit vectors 1-(t 1 ) and 1.(t2 ) point in different directions. (b) The change Ai- in r is given by the triangle shown. (For an alternative proof of this", " important result, see Problem 1.43.) Notice that di/dt is in the direction of 0 and is proportional to the rate of change of the angle 0 \u2014 both of which properties we would expect based on Figure 1.12. Now that we know the derivative of r, we are ready to differentiate Equation (1.40). Using the product rule, we get two terms: and, substituting (1.42), we find for the velocity t, or v, dr = rr \u2014, dt v = r From this we can read off the polar components of the velocity: vr = r and v0 = rc = r w where in the second equation I have introduced the traditional notation w for the an- gular velocity (p. While the results in (1.44) should be familiar from your introductory physics course, they are undeniably more complicated than the corresponding results in Cartesian coordinates (v x = z and vy = 57). Before we can write down Newton's second law, we have to differentiate a second time to find the acceleration: a d. = \u2014r = \u2014 (rr d., dt dt rrP 0), where the final expression comes from substituting (1.43) for r. To complete the differentiation in (1.45), we must calculate the derivative of 0. This calculation is completely analogous to the argument leading to (1.42) and is illustrated in Figure 1.13. By inspecting this figure, you should be able to convince yourself that (1.43) (1.44) (1.45) (1.46) d4) dt Section 1.7 Two - Dimensional Polar Coordinates 29 (a) (b) Figure 1.13 (a) The unit vector at two successive times t1 and t2. (b) The change AO. Returning to Equation (1.45), we can now carry out the differentiation to give the following five terms: a = + i dt + (i4 + r.) \u2014d4) dt or, if we use (1.42) and (1.46) to replace the derivatives of the two unit vectors, a = \u2014 r.152)1- +(r \u00b1 24) isfi (1.47) This horrible result is a little easier to understand if we consider the special case that r is constant, as is the case for a stone that I twirl on the end of a string of fixed length. With r constant, both derivatives of r are zero, and (1.47) has just two terms: or a,--- ri4 a = \u2014rw2i + where w = denotes the angular velocity and a = is the angular acceleration. This is the familiar result from elementary physics that when a particle moves around a fixed circle, it has an inward \"centripetal\" acceleration r co2 (or v 2 I r) and a tangential acceleration, ra. Nevertheless, when r is not constant, the acceleration includes all four of the terms in (1.47). The first term, F in the radial direction is what you would probably expect when r varies, but the final term, 2i in the direction, is harder to understand. It is called the Coriolis acceleration, and I shall discuss it in detail in Chapter 9. Having calculated the acceleration as in (1.47), we can finally write down Newton's second law in terms of polar coordinates: F = ma Fr = m(F \u2014 42) { 1 = m (rs;6 21- (1.48) These equations in polar coordinates are a far cry from the beautifully simple equa- tions (1.35) for rectangular coordinates. In fact, one of the main reasons for taking the 30 Chapter 1 Newton's Laws of Motion trouble to recast Newtonian mechanics in the Lagrangian formulation (Chapter 7) is that the latter is able to handle nonrectangular coordinates just as easily as rectangular. You may justifiably be feeling that the second law in polar coordinates is so complicated that there could be no occasion to use it. In fact, however, there are many problems which are most easily solved using polar coordinates, and I conclude this section with an elementary example. EXAMPLE 1.2 An Oscillating Skateboard A \"half-pipe\" at a skateboard park consists of a concrete trough with a semicircu- lar cross section of radius R = 5 m, as shown in Figure 1.14. I hold a frictionless skateboard on the side of the trough pointing down toward the bottom and release it. Discuss the subsequent motion using Newton's second law. In particular, if I release the board just a short way from the bottom, how long will it take to come back to the point of release? Because the skateboard is constrained to move on a circular path, this prob- lem is most easily solved using polar coordinates with origin 0 at the center", " of the pipe as shown. (At some point in the following calculation, try writing the second law in rectangular coordinates and observe what a tangle you get.) With this choice of polar coordinates, the coordinate r of the skateboard is constant, r = R, and the position of the skateboard is completely specified by the angle 0. With r constant, the second law (1.48) takes the relatively simple form Fr = \u2014mR(P 2 Fo = m6. (1.49) (1.50) The two forces on the skateboard are its weight w = mg and the normal force N of the wall, as shown in Figure 1.14. The components of the net force F = w + N are easily seen to be Fr = mg cos 4) \u2014 N and Fo = \u2014mg sin 4). w = mg Figure 1.14 A skateboard in a semicircular trough of radius R. The board's position is specified by the angle measured up from the bottom. The two forces on the skateboard are its weight w = mg and the normal force N. Section 1.7 Two - Dimensional Polar Coordinates 31 Substituting for Fr into (1.49) we get an equation involving N, 0, and Fortunately, we are not really interested in N, and \u2014 even more fortunately \u2014 when we substitute for Fo into (1.50), we get an equation that does not involve N at all: \u2014mg sin 0 = m /? or, canceling the m's and rearranging, = \u2014 8: sin 0. (1.51) Equation (1.51) is the differential equation for 0(t) that determines the motion of the skateboard. Qualitatively, we can easily see the kind of motion that it implies. First, if 0 = 0, (1.51) says that = 0. Therefore, if we place the board at rest ( = 0) at the point 0 = 0, the board will never move (unless someone pushes it); that is, 0 = 0 is an equilibrium position, as you would certainly have guessed. Next, suppose that at some time, 0 is not zero and, to be definite, suppose that 0 > 0; that is, the skateboard is on the right-hand side of the half-pipe. In this case, (1.51) implies that < 0, so the acceleration is directed to the left. If the board is moving to the right it must slow down and eventually start moving to the left. 13 Once it is moving toward the left, it speeds up and returns to the bottom, where it moves over to the left. As soon as the board is on the left, the argument reverses (0 < 0, so 4 > 0) and the board must eventually return to the bottom and move over to the right again. In other words, the differential equation (1.51) implies that the skateboard oscillates back and forth, from right to left and back to the right. The equation of motion (1.51) cannot be solved in terms of elementary func- tions, such as polynomials, trigonometric functions, or logs and exponentials. 14 course is to use a computer to solve it numerically (see Problem 1.50). However, if the initial angle 0, is small, we can use the small angle approximation and, within this approximation, (1.51) becomes sin 0 = Thus, if we want more quantitative information about the motion, the simplest (1.52) (1.53) which can be solved using elementary functions. [By this stage, you have al- most certainly recognized that our discussion of the skateboard problem closely parallels the analysis of the simple pendulum. In particular, the small-angle 13 I am taking for granted that it doesn't reach the top and jump out of the trough. Since it was released from rest inside the trough, this is correct. Much the easiest way to prove this claim is to invoke conservation of energy, which we shan't be discussing for a while. Perhaps, for now, you could agree to accept it as a matter of common sense. 14 Actually the solution of (1.51) is a Jacobi elliptic function. However, I shall take the point of view that for most of us the Jacobi function is not \"elementary.\" 32 Chapter 1 Newton's Laws of Motion approximation (1.52) is what let you solve the simple pendulum in your intro- ductory physics course. This parallel is, of course, no accident. Mathematically the two problems are exactly equivalent.] If we define the parameter = R, then (1.53) becomes (1.54) (1.55) This is the equation of motion for our skateboard in the small-angle approxima- tion. I would like to discuss its solution in some detail to introduce several ideas that we'll be using again and again in what follows. (", "If you've studied differential equations before, just see the next three paragraphs as a quick review.) We first observe that it is easy to find two solutions of the equation (1.55) by inspection (that is, by inspired guessing). The function 0 (t) = A sin(cot) is clearly a solution for any value of the constant A. [Differentiating sin (cot) brings out a factor of w and changes the sin to a cos; differentiating it again brings out another w and changes the cos back to \u2014sin. Thus the proposed solution does satisfy 4 = \u2014co20.] Similarly, the function OW = B cos(wt) is another solution for any constant B. Furthermore, as you can easily check, the sum of these two solutions is itself a solution. Thus we have now found a whole family of solutions: (t) = A sin (cot) B cos(wt) (1.56) is a solution for any values of the two constants A and B. I now want to argue that every solution of the equation of motion (1.55) has the form (1.56). In other words, (1.56) is the general solution\u2014we have found all solutions, and we need seek no further. To get some idea of why this is, note that the differential equation (1.55) is a statement about the second derivative 4) of the unknown 0. Now, if we had actually been told what is, then we know from elementary calculus that we could find 0 by two integrations, and the result would contain two unknown constants \u2014 the two constants of integration \u2014 that would have to be determined by looking (for example) at the initial values of 0 and In other words, knowledge of would tell us that 0 itself is one of a family of functions containing precisely two undetermined constants. Of course, the differential equation (1.55) does not actually tell us \u2014 it is an equation for in terms of 0. Nevetheless, it is plausible that such an equation would imply that 0 is one of a family of functions that contain precisely two undetermined constants. If you have studied differential equations, you know that this is the case; if you have not, then I must ask you to accept it as a plausible fact: For any given second-order differential equation [in a large class of \"reasonable\" equations, including (1.55) and all of the equations we shall encounter in this book], the solutions all belong to a family of functions Principal Definitions and Equations of Chapter 1 33 containing precisely two independent constants \u2014 like the constants A and B in (1.56). (More generally, the solutions of an nth-order equation contain precisely n independent constants.) This theorem sheds a new light on our solution (1.56). We already knew that any function of the form (1.56) is a solution of the equation of motion. Our theorem now guarantees that every solution of the equation of motion is of this form. This same argument applies to all the second-order differential equations we shall encounter. If, by hook or by crook, we can find a solution like (1.56) involving two arbitrary constants, then we are guaranteed that we have found the general solution of our equation. All that remains is to pin down the two constants A and B for our skateboard. To do so, we must look at the initial conditions. At t = 0, Equation (1.56) implies that 0 = B. Therefore B is just the initial value of 0, which we are calling 0 o, so B = O.. At t = 0, Equation (1.56) implies that = coA. Since I released the board from rest, this means that A = 0, and our solution is (t) = cos(cot). (1.57) The first thing to note about this solution is that, as we anticipated on general grounds, 0 (t) oscillates, moving from positive to negative and back to positive periodically and indefinitely. In particular, the board first returns to its initial position 00 when wt = 27r. The time that this takes is called the period of the motion and is denoted z. Thus our conclusion is that the period of the skateboard's oscillations is = \u2014 = 27T 27r w R g (1.58) We were given that R = 5 m, and g = 9.8 m/s2. Substituting these numbers, we conclude that the skateboard returns to its starting point in a time r = 4.5 seconds. Principal Definitions and Equations of Chapter 1 Dot and Cross Products rs= rs cos 9 = rx sx ry sy r,s, [Eqs. (1.6) & (1.7)] rxs= (ry sz \u2014 rz sy, rz sx \u2014 rx sz, rx sy \u2014ry sx )=det rx [Eq.", " (1.9)] 1 r, sz sx y sY 34 Chapter 1 Newton's Laws of Motion Inertial Frames An inertial frame is any reference frame in which Newton's first law holds, that is, a nonaccelerating, nonrotating frame. Unit Vectors of a Coordinate System If 17, 0 are an orthogonal system of coordinates, then = unit vector in direction of increasing c with ri and fixed and so on, and any vector s can be expanded as s = Newton's Second Law in Various Coordinate Systems Vector Form Cartesian 2D Polar (r, F=mr IF, = m(F \u2014 r 2) F = \u2014 m + 24) ( (x, y, z) {Fx = mi F = mj; Fz = mE Cylindrical Polar (p, 0, z) Fr = M(16 142) F,, = m(pq5 + ZOO) = mE { Eq. (1.35) Eq. (1.48) Problem 1.47 or 1.48 Problems for Chapter 1 The problems for each chapter are arranged according to section number. A problem listed for a given section requires an understanding of that section and earlier sections, but not of later sections. Within each section problems are listed in approximate order of difficulty. A single star (*) indicates straightforward problems involving just one main concept. Two stars (**) identify problems that are slightly more challenging and usually involve more than one concept. Three stars (***) indicate problems that are distinctly more challenging, either because they are intrinsically difficult or involve lengthy calculations. Needless to say, these distinctions are hard to draw and are only approximate. Problems that need the use of a computer are flagged thus: [Computer]. These are mostly classified as *** on the grounds that it usually takes a long time to set up the necessary code \u2014 especially if you're just learning the language. SECTION 1.2 Space and Time 1.1* Given the two vectors b = + Sr and c = + find b c, 5b + 2c, b c, and b x c. 1.2 * Two vectors are given as b = (1, 2, 3) and c = (3, 2, 1). (Remember that these statements are just a compact way of giving you the components of the vectors.) Find b c, 5b \u2014 2c, b c, and b x c. 1.3 * By applying Pythagoras's theorem (the usual two-dimensional version) twice over, prove that the length r of a three-dimensional vector r = (x, y, z) satisfies r2 = x 2 + y2 + z2. Problems for Chapter 1 35 1.4 * One of the many uses of the scalar product is to find the angle between two given vectors. Find the angle between the vectors b = (1, 2, 4) and c = (4, 2, 1) by evaluating their scalar product. 1.5 * Find the angle between a body diagonal of a cube and any one of its face diagonals. [Hint: Choose a cube with side 1 and with one corner at 0 and the opposite corner at the point (1, 1, 1). Write down the vector that represents a body diagonal and another that represents a face diagonal, and then find the angle between them as in Problem 1.4.] 1.6 * By evaluating their dot product, find the values of the scalar s for which the two vectors b = x + s'Sr and c = x \u2014 sS7 are orthogonal. (Remember that two vectors are orthogonal if and only if their dot product is zero.) Explain your answers with a sketch. 1.7 * Prove that the two definitions of the scalar product r s as rs cos 9 (1.6) and E ri s, (1.7) are equal. One way to do this is to choose your x axis along the direction of r. [Strictly speaking you should first make sure that the definition (1.7) is independent of the choice of axes. If you like to worry about such niceties, see Problem 1.16.] 1.8 * (a) Use the definition (1.7) to prove that the scalar product is distributive, that is, r (u + v) = r u + r v. (b) If r and s are vectors that depend on time, prove that the product rule for differentiating products applies to r s, that is, that \u2014 (r s) r dt ds dr + s. dt dt 1.9 * In elementary trigonometry, you probably learned the law of cosines for a triangle of sides a, b, and c, that c2 = a 2 b2 \u2014 2ab cos 9, where 9 is the angle between the sides a and b.", " Show that the law of cosines is an immediate consequence of the identity (a + b) 2 = a2 b2 + 2a b. 1.10 * A particle moves in a circle (center 0 and radius R) with constant angular velocity w counter- clockwise. The circle lies in the xy plane and the particle is on the x axis at time t = 0. Show that the particle's position is given by r(t) = xR cos(wt) + y R sin(wt). Find the particle's velocity and acceleration. What are the magnitude and direction of the acceleration? Relate your results to well-known properties of uniform circular motion. 1.11 * The position of a moving particle is given as a function of time t to be r(t) = xb cos(wt) + ST\u2018c sin(wt), where b, c, and w are constants. Describe the particle's orbit. 1.12 * The position of a moving particle is given as a function of time t to be r(t) = xb cos(wt) + S7c sin(wt) + iv ot where b, c, v o and co are constants. Describe the particle's orbit. 1.13 * Let u be an arbitrary fixed unit vector and show that any vector b satisfies Explain this result in words, with the help of a picture. b2 (u b) 2 + (u x b) 2. 36 Chapter 1 Newton's Laws of Motion 1.14 * Prove that for any two vectors a and b, la + bi < (a + b). [Hint: Work out la 131 2 and compare it with (a + b)2.] Explain why this is called the triangle inequality. 1.15 * Show that the definition (1.9) of the cross product is equivalent to the elementary definition that r x s is perpendicular to both r and s, with magnitude rs sin 0 and direction given by the right-hand rule. [Hint: It is a fact (though quite hard to prove) that the definition (1.9) is independent of your choice of axes. Therefore you can choose axes so that r points along the x axis and s lies in the xy plane.] 1.16 ** (a) Defining the scalar product r s by Equation (1.7), r s = E ri si, show that Pythagoras's theorem implies that the magnitude of any vector r is r = r. (b) It is clear that the length of a vector does not depend on our choice of coordinate axes. Thus the result of part (a) guarantees that the scalar product r r, as defined by (1.7), is the same for any choice of orthogonal axes. Use this to prove that r s, as defined by (1.7), is the same for any choice of orthogonal axes. [Hint: Consider the length of the vector r 1.17 ** (a) Prove that the vector product r x s as defined by (1.9) is distributive; that is, that r x (u v) =--- (r x u) (r x v). (b) Prove the product rule \u2014(rxs)=rx d dt ds dr dt at xs. Be careful with the order of the factors. 1.18 ** The three vectors a, b, c are the three sides of the triangle ABC with angles a,,8, y as shown in Figure 1.15. (a) Prove that the area of the triangle is given by any one of these three expressions: area = -2 1a x bl = -2 lb X CI = -2 x al. (b) Use the equality of these three expressions to prove the so-called law of sines, that a sin a sin /3 sin y Figure 1.15 Triangle for Problem 1.18. 1.19 ** If r, v, a denote the position, velocity, and acceleration of a particle, prove that Problems for Chapter 1 37 d \u2014 dt [a (v x r)] = a (v x r). 1.20 ** The three vectors A, B, C point from the origin 0 to the three corners of a triangle. Use the result of Problem 1.18 to show that the area of the triangle is given by (area of triangle) = 2I(B x C) + (C x A) + (A x B)1. 1.21 ** A parallelepiped (a six-faced solid with opposite faces parallel) has one corner at the origin 0 and the three edges that emanate from 0 defined by vectors a, b, c. Show that the volume of the parallelepiped is la (b x 1.22 ** The two vectors a and b lie in the xy plane and make angles a and with the x axis", ". (a) By evaluating a b in two ways [namely using (1.6) and (1.7)] prove the well-known trig identity (b) By similarly evaluating a x b prove that cos(a \u2014 0) = cos a cos 0 + sin a sin,8. sin(a \u2014 0) = sin a cos /3 \u2014 cos a sin 8. 1.23 ** The unknown vector v satisfies b v = A and b x v = c, where A, b, and c are fixed and known. Find v in terms of A, b, and c. SECTION 1.4 Newton's First and Second Laws; Inertial Frames 1.24 * In case you haven't studied any differential equations before, I shall be introducing the necessary ideas as needed. Here is a simple excercise to get you started: Find the general solution of the first- order equation df/dt = f for an unknown function f(t). [There are several ways to do this. One is to rewrite the equation as df/f = dt and then integrate both sides.] How many arbitrary constants does the general solution contain? [Your answer should illustrate the important general theorem that the solution to any nth-order differential equation (in a very large class of \"reasonable\" equations) contains n arbitrary constants.] 1.25 * Answer the same questions as in Problem 1.24, but for the differential equation dfldt = \u20143f. 1.26 ** The hallmark of an inertial reference frame is that any object which is subject to zero net force will travel in a straight line at constant speed. To illustrate this, consider the following: I am standing on a level floor at the origin of an inertial frame S and kick a frictionless puck due north across the floor. (a) Write down the x and y coordinates of the puck as functions of time as seen from my inertial frame. (Use x and y axes pointing east and north respectively.) Now consider two more observers, the first at rest in a frame 8' that travels with constant velocity v due east relative to 8, the second at rest in a frame 8\" that travels with constant acceleration due east relative to S. (All three frames coincide at the moment when I kick the puck, and 8\" is at rest relative to 8 at that same moment.) (b) Find the coordinates x', y' of the puck and describe the puck's path as seen from 8'. (c) Do the same for 8\". Which of the frames is inertial? 1.27 ** The hallmark of an inertial reference frame is that any object which is subject to zero net force will travel in a straight line at constant speed. To illustrate this, consider the following experiment: I am 38 Chapter 1 Newton's Laws of Motion standing on the ground (which we shall take to be an inertial frame) beside a perfectly flat horizontal turntable, rotating with constant angular velocity co. I lean over and shove a frictionless puck so that it slides across the turntable, straight through the center. The puck is subject to zero net force and, as seen from my inertial frame, travels in a straight line. Describe the puck's path as observed by someone sitting at rest on the turntable. This requires careful thought, but you should be able to get a qualitative picture. For a quantitative picture, it helps to use polar coordinates; see Problem 1.46. SECTION 1.5 The Third Law and Conservation of Momentum 1.28 * Go over the steps from Equation (1.25) to (1.29) in the proof of conservation of momentum, but treat the case that N = 3 and write out all the summations explicitly to be sure you understand the various manipulations. 1.29 * Do the same tasks as in Problem 1.28 but for the case of four particles (N = 4). 1.30 * Conservation laws, such as conservation of momentum, often give a surprising amount of information about the possible outcome of an experiment. Here is perhaps the simplest example: Two objects of masses m 1 and m 2 are subject to no external forces. Object 1 is traveling with velocity v when it collides with the stationary object 2. The two objects stick together and move off with common velocity v'. Use conservation of momentum to find v' in terms of v, m 1, and m 2. 1.31 * In Section 1.5 we proved that Newton's third law implies the conservation of momentum. Prove the converse, that if the law of conservation of momentum applies to every possible group of particles, then the interparticle forces must obey the third law. [Hint: However many particles your system contains, you can focus your attention on just two of them. (Call them 1 and 2.) The law of conservation of momentum says that if there are no external forces on this pair of particles, then their total momentum must be constant. Use this to prove that F", "12 = \u2014 F21.] 1.32 ** If you have some experience in electromagnetism, you could do the following problem concerning the curious situation illustrated in Figure 1.8. The electric and magnetic fields at a point r 1 2 at r2 moving with constant velocity v 2 (with v2 << c) are l5 due to a charge q E(r i) = 1 q2 s and B(r 1, = \u2014 2 [to q2 v \\ 47r s 2 47r s 2 where s = r 1 \u2014 r2 is the vector pointing from r 2 to r1. (The first of these you should recognize as Coulomb's law.) If F712 and Fmi2ag denote the electric and magnetic forces on a charge q1 at r 1 with velocity v 1, show that Fr2ag < (v v2 /c2)Fie21. This shows that in the non-relativistic domain it is legitimate to ignore the magnetic force between two moving charges. 1.33 *** If you have some experience in electromagnetism and with vector calculus, prove that the magnetic forces, F12 and F21, between two steady current loops obey Newton's third law. [Hints: Let the two currents be I I and /2 and let typical points on the two loops be r 1 and r2. If dr 1 and dr2 are short segments of the loops, then according to the Biot\u2014Savart law, the force on dr 1 due to dr 2 is \u2014 \u2014 di% x (dr2 x g) it. /1 /2 47r s 2 where s = r 1 \u2014 r2. The force F12 is found by integrating this around both loops. You will need to use the \"B AC \u2014 CAB\" rule to simplify the triple product.] 15 See, for example, David J. Griffiths, Introduction to Electrodynamics, 3rd ed., Prentice Hall, (1999), p. 440. Problems for Chapter 1 39 1.34 *** Prove that in the absence of external forces, the total angular momentum (defined as L = ED, ra x pa ) of an N-particle system is conserved. [Hints: You need to mimic the argument from (1.25) to (1.29). In this case you need more than Newton's third law: In addition you need to assume that the interparticle forces are central; that is, Fap acts along the line joining particles ce and $. A full discussion of angular momentum is given in Chapter 3.] SECTION 1.6 Newton's Second Law in Cartesian Coordinates 1.35 * A golf ball is hit from ground level with speed v o in a direction that is due east and at an angle 6 above the horizontal. Neglecting air resistance, use Newton's second law (1.35) to find the position as a function of time, using coordinates with x measured east, y north, and z vertically up. Find the time for the golf ball to return to the ground and how far it travels in that time. 1.36 * A plane, which is flying horizontally at a constant speed v o and at a height h above the sea, must drop a bundle of supplies to a castaway on a small raft. (a) Write down Newton's second law for the bundle as it falls from the plane, assuming you can neglect air resistance. Solve your equations to give the bundle's position in flight as a function of time t. (b) How far before the raft (measured horizontally) must the pilot drop the bundle if it is to hit the raft? What is this distance if v o = 50 m/s, h = 100 m, and g ti 10 m/s 2? (c) Within what interval of time (+At) must the pilot drop the bundle if it is to land within +10 m of the raft? 1.37 * A student kicks a frictionless puck with initial speed v o, so that it slides straight up a plane that is inclined at an angle 0 above the horizontal. (a) Write down Newton's second law for the puck and solve to give its position as a function of time. (b) How long will the puck take to return to its starting point? 1.38 * You lay a rectangular board on the horizontal floor and then tilt the board about one edge until it slopes at angle 0 with the horizontal. Choose your origin at one of the two corners that touch the floor, the x axis pointing along the bottom edge of the board, the y axis pointing up the slope, and the z axis normal to the board. You now kick a frictionless puck that is resting at 0 so that it slides across the board with initial velocity (v ox, voy, 0). Write down Newton's second law using the given coordinates and then find how long the puck takes to return to the floor level and how far it is from 0 when it", " does so. 1.39 ** A ball is thrown with initial speed v o up an inclined plane. The plane is inclined at an angle above the horizontal, and the ball's initial velocity is at an angle 0 above the plane. Choose axes with x measured up the slope, y normal to the slope, and z across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance R = 2v 2 sin 0 cos(0 0)1(g cos t 0) from its launch point. Show that for given v o and 0, the maximum 0 possible range up the inclined plane is R. = vo /[g(1 + sin OA 1.40 *** A cannon shoots a ball at an angle 6 above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. (Use axes with x measured horizontally and y vertically.) (b) Let r(t) denote the ball's distance from the cannon. What is the largest possible value of 0 if r (t) is to increase throughout the ball's flight? [Hint: Using your solution to part (a) you can write down r 2 as x 2 + y 2, and then find the condition that r 2 is always increasing.] 40 Chapter 1 Newton's Laws of Motion SECTION 1.7 Two-Dimensional Polar Coordinates 1.41 * An astronaut in gravity-free space is twirling a mass m on the end of a string of length R in a circle, with constant angular velocity co. Write down Newton's second law (1.48) in polar coordinates and find the tension in the string. 1.42 * Prove that the transformations from rectangular to polar coordinates and vice versa are given by the four equations (1.37). Explain why the equation for 0 is not quite complete and give a complete version. and cb. solution.) 1.43 * (a) Prove that the unit vector r of two-dimensional polar coordinates is equal to = X cos 0 + 5r sin 0 (1.59) and find a corresponding expression for 4. (b) Assuming that 0 depends on the time t, differentiate your answers in part (a) to give an alternative proof of the results (1.42) and (1.46) for the time derivatives 1.44 * Verify by direct substitution that the function 0 (t) = A sin(wt) + B cos(wt) of (1.56) is a solution of the second-order differential equation (1.55), if, = \u2014co20. (Since this solution involves two arbitrary constants \u2014 the coefficients of the sine and cosine functions it is in fact the general 1.45 ** Prove that if v(t) is any vector that depends on time (for example the velocity of a moving particle) but which has constant magnitude, then v(t) is orthogonal to v(t). Prove the converse that if '(t) is orthogonal to v (t), then Iv (t) I is constant. [Hint: Consider the derivative of v2.] This is a very handy result. It explains why, in two-dimensional polars, di I dt has to be in the direction of 0 and vice versa. It also shows that the speed of a charged particle in a magnetic field is constant, since the acceleration is perpendicular to the velocity. 1.46 ** Consider the experiment of Problem 1.27, in which a frictionless puck is slid straight across a rotating turntable through the center 0. (a) Write down the polar coordinates r, 0 of the puck as functions of time, as measured in the inertial frame S of an observer on the ground. (Assume that the puck was launched along the axis 0 = 0 at t = 0.) (b) Now write down the polar coordinates r', 0' of the puck as measured by an observer (frame S') at rest on the turntable. (Choose these coordinates so that 0 and 0' coincide at t = 0.) Describe and sketch the path seen by this second observer. Is the frame S' inertial? 1.47 ** Let the position of a point P in three dimensions be given by the vector r = (x, y, z) in rectangular (or Cartesian) coordinates. The same position can be specified by cylindrical polar coordinates, p, 0, z, which are defined as follows: Let P' denote the projection of P onto the xy plane; that is, P' has Cartesian coordinates (x, y, 0). Then p and 0 are defined as the two-dimensional polar coordinates of P' in the xy plane, while z is the third Cartesian coordinate, unchanged. (a) Make a sketch to illustrate the three cylindrical coordinates. Give expressions", " for p, z in terms of the Cartesian coordinates x, y, z. Explain in words what p is (\"p is the distance of P from \"). There are many variants in notation. For instance, some people use r instead of p. Explain why this use of r is unfortunate. (b) Describe the three unit vectors p, 0, i and write the expansion of the position vector r in terms of these unit vectors. (c) Differentiate your last answer twice to find the cylindrical components of the acceleration a =r of the particle. To do this, you will need to know the time derivatives of j) and 0. You could get these from the corresponding two-dimensional results (1.42) and (1.46), or you could derive them directly as in Problem 1.48. Problems for Chapter 1 41 1.48 ** Find expressions for the unit vectors j), 4, and i of cylindrical polar coordinates (Problem 1.47) in terms of the Cartesian X, y, i. Differentiate these expressions with respect to time to find 01dt, clikldt, and dildt. 1.49 ** Imagine two concentric cylinders, centered on the vertical z axis, with radii R \u00b1 E, where E is very small. A small frictionless puck of thickness 2E is inserted between the two cylinders, so that it can be considered a point mass that can move freely at a fixed distance from the vertical axis. If we use cylindrical polar coordinates (p, 0, z) for its position (Problem 1.47), then p is fixed at p = R. while 0 and z can vary at will. Write down and solve Newton's second law for the general motion of the puck, including the effects of gravity. Describe the puck's motion. 1.50 *** [Computer] The differential equation (1.51) for the skateboard of Example 1.2 cannot be solved in terms of elementary functions, but is easily solved numerically. (a) If you have access to software, such as Mathematica, Maple, or Matlab, that can solve differential equations numerically, solve the differential equation for the case that the board is released from 0 \u00b0 = 20 degrees, using the values R = 5 m and g = 9.8 m/s2. Make a plot of 0 against time for two or three periods. (b) On the same picture, plot the approximate solution (1.57) with the same 00 = 20\u00b0. Comment on your two graphs. Note: If you haven't used the numerical solver before, you will need to learn the necessary syntax. For example, in Mathematica you will need to learn the syntax for \"NDSolve\" and how to plot the solution that it provides. This takes a bit of time, but is something that is very well worth learning. 1.51 *** [Computer] Repeat all of Problem 1.50 but using the initial value 0 0 = r/2. CHAPTER Projectiles and Charged Particles In this chapter, I present two topics: the motion of projectiles subject to the forces of gravity and air resistance, and the motion of charged particles in uniform magnetic fields. Both problems lend themselves to solution using Newton's laws in Cartesian coordinates, and both allow us to review and introduce some important mathematics. Above all, both are problems of great practical interest. 2.1 Air Resistance Most introductory physics courses spend some time studying the motion of projectiles, but they almost always ignore air resistance. In many problems this is an excellent approximation; in others, air resistance is obviously important, and we need to know how to account for it. More generally, whether or not air resistance is significant, we need some way to estimate how important it really is. Let us begin by surveying some of the basic properties of the resistive force, or drag, f of the air, or other medium, through which an object is moving. (I shall generally speak of \"air resistance\" since air is the medium through which most projectiles move, but the same considerations apply to other gases and often to liquids as well.) The most obvious fact about air resistance, well known to anyone who rides a bicycle, is that it depends on the speed, v, of the object concerned. In addition, for many objects, the direction of the force due to motion through the air is opposite to the velocity v. For certain objects, such as a nonrotating sphere, this is exactly true, and for many it is a good approximation. You should, however, be aware that there are situations where it is certainly not true: The force of the air on an airplane wing has a large sideways component, called the lift, without which no airplanes could fly. Nevertheless, I shall assume that f and v point in opposite directions; that is, I shall consider only objects for which the sideways force is zero, or at least small enough 43 44 Chapter 2 Projectiles and Charg", "ed Particles w = mg Figure 2.1 A projectile is subject to two forces, the force of gravity, w = mg, and the drag force of air resistance, f = \u2014 f (v)ir. to be neglected. The situation is illustrated in Figure 2.1 and is summed up in the equation of f. f = - f (v)v, (2.1) (2.2) (2.3) where V = v/ Iv I denotes the unit vector in the direction of v, and f (v) is the magnitude The function f (v) that gives the magnitude of the air resistance varies with v in a complicated way, especially as the object's speed approaches the speed of sound. However, at lower speeds it is often a good approximation to write' where flip and fquad stand for the linear and quadratic terms respectively, J f (v) = by + cv 2 = f; + f J J quad.flip = by and fquad = CV 2 The physical origins of these two terms are quite different: The linear term, f lip, arises from the viscous drag of the medium and is generally proportional to the viscosity of the medium and the linear size of the projectile (Problem 2.2). The quadratic term, fquad' arises from the projectile's having to accelerate the mass of air with which it is continually colliding; fquad is proportional to the density of the medium and the cross- sectional area of the projectile (Problem 2.4). In particular, for a spherical projectile (a cannonball, a baseball, or a drop of rain), the coefficients b and c in (2.2) have the form b =,8D and c = y D2 (2.4) where D denotes the diameter of the sphere and the coefficients 6 and y depend on the nature of the medium. For a spherical projectile in air at STP, they have the approximate values = 1.6 x 10-4 Ns/m 2 (2.5) I Mathematically, Equation (2.2) is, in a sense, obvious. Any reasonable function is expected to have a Taylor series expansion, f = a + by + cv2 For low enough v, the first three terms should give a good approximation, and, since f = 0 when v = 0 the constant term, a, has to be zero. Section 2.1 Air Resistance 45 and y = 0.25 Ns 2/m4. (2.6) (For calculation of these two constants, see Problems 2.2 and 2.4.) You need to remember that these values are valid only for a sphere moving through air at STP. Nevertheless, they give at least a rough idea of the importance of the drag force even for nonspherical bodies moving through different gases at any normal temperatures and pressures. It often happens that we can neglect one of the terms in (2.2) compared to the other, and this simplifies the task of solving Newton's second law. To decide whether this does happen in a given problem, and which term to neglect, we need to compare the sizes of the two terms: fquad.flip 2 C V by yD,8 v = (1.6 x 10 3 S ) Dv M2 (2.7) if we use the values (2.5) and (2.6) for a sphere in air. In a given problem, we have only to substitute the values of D and v into this equation to find out if one of the terms can be neglected, as the following example illustrates. EXAMPLE 2.1 A Baseball and Some Drops of Liquid Assess the relative importance of the linear and quadratic drags on a baseball of diameter D =7 cm, traveling at a modest v = 5 m/s. Do the same for a drop of rain (D =1 mm and v = 0.6 m/s) and for a tiny droplet of oil used in the Millikan oildrop experiment (D = 1.5,um and v = 5 x 10 -5 m/s). When we substitute the numbers for the baseball into (2.7) (remembering to convert the diameter to meters), we get fquad flin 600 [baseball]. (2.8) For this baseball, the linear term is clearly negligible and we need consider only the quadratic drag. If the ball is traveling faster, the ratio fquad/fain is even greater. At slower speeds the ratio is less dramatic, but even at 1 m/s the ratio is 100. In fact if v is small enough that the linear term is comparable to the quadratic, both terms are so small as to be negligible. Thus, for baseballs and similar objects, it is almost always safe to neglect A ir, and take the drag force to be f = (2.9) For the rain", "drop, the numbers give fquad fain 1 [raindrop]. (2.10) Thus for this raindrop the two terms are comparable and neither can be ne- glected \u2014 which makes solving for the motion more difficult. If the drop were 46 Chapter 2 Projectiles and Charged Particles a lot larger or were traveling much faster, then the linear term would be negligi- ble; and if the drop were much smaller or were traveling much slower, then the quadratic term would be negligible. But in general, with raindrops and similar objects, we are going to have to take both A in and fquad into account. For the oildrop in the Millikan experiment the numbers give fquad fun ti 10 [Millikan oildrop]. (2.11) In this case, the quadratic term is totally negligible, and we can take f = = (2.12) where the second, very compact form follows because, of course, vi = v. The moral of this example is clear: First, there are objects for which the drag force is dominantly linear, and the quadratic force can be neglected \u2014 notably, very small liquid drops in air, but also slightly larger objects in a very viscous fluid, such as a ball bearing moving through molasses. On the other hand, for most projectiles, such as golf balls, cannonballs, and even a human in free fall, the dominant drag force is quadratic, and we can neglect the linear term. This situation is a little unlucky because the linear problem is much easier to solve than the quadratic. In the following two sections, I shall discuss the linear case, precisely because it is the easier one. Nevertheless, it does have practical applications, and the mathematics used to solve it is widely used in many fields. In Section 2.4, I shall take up the harder but more usual case of quadratic drag. To conclude this introductory section, I should mention the Reynolds number, an important parameter that features prominently in more advanced treatments of motion in fluids. As already mentioned, the linear drag f lip can be related to the viscosity of the fluid through which our projectile is moving, and the quadratic term fquad is similarly related to the inertia (and hence density) of the fluid. Thus one can relate the ratio fquad/flin to the fundamental parameters 77, the viscosity, and 0, the density, of the fluid (see Problem 2.3). The result is that the ratio f quad/flin is of roughly the same order of magnitude as the dimensionless number R = DvQ/r), called the Reynolds number. Thus a compact and general way to summarize the foregoing discussion is to say that the quadratic drag fquad is dominant when the Reynolds number R is large, whereas the linear drag dominates when R is small. 2.2 Linear Air Resistance Let us consider first a projectile for which the quadratic drag force is negligible, so that the force of air resistance is given by (2.12). We shall see directly that, because the drag force is linear in v, the equations of motion are very simple to solve. The two forces on the projectile are the weight w = mg and the drag force f = \u2014by, as shown in Figure 2.2. Thus the second law, ml = F, reads ml = mg \u2014 by. (2.13) Section 2.2 Linear Air Resistance 47 Figure 2.2 The two forces on a projectile for which the force of air resistance is linear in the velocity, f = \u2014bv. An interesting feature of this form is that, because neither of the forces depends on r, the equation of motion does not involve r itself (only the first and second derivatives of r). In fact, we can rewriter as Y, and (2.13) becomes my = mg \u2014 bv, (2.14) a first-order differential equation for v. This simplification comes about because the forces depend only on v and not r. It means we have to solve only a first-order differential equation for v and then integrate v to find r. Perhaps the most important simplifying feature of linear drag is that the equation of motion separates into components especially easily. For instance, with x measured to the right and y vertically downward, (2.14) resolves into and mi)x = \u2014by, mi)y = mg \u2014 bv y. (2.15) (2.16) That is, we have two separate equations, one for v x and one for vy ; the equation for vx v y and vice versa. It is important to recognize that this happened only does not involve because the drag force was linear in v. For instance, if the drag force were quadratic, f = \u2014 cv 2i = \u2014cvv = -C, / V 2 + vy 2 V (2.17) then in (2.14)", " we would have to replace the term \u2014by with (2.17). In place of the two equations (2.15) and (2.16), we would have MI) X = \u2014 C,1V 2 X vy V X i)y = mg \u2014 c\\/vx vx vy. (2.18) Here, each equation involves both of the variables v x and vy. These two coupled differential equations are much harder to solve than the uncoupled equations of the linear case. Because they are uncoupled, we can solve each equation for linear drag separately and then put the two solutions together. Further, each equation defines a problem that is interesting in its own right. Equation (2.15) is the equation of motion for an object 48 Chapter 2 Projectiles and Charged Particles Figure 2.3 A cart moves on a horizontal frictionless track in a medium that produces a linear drag force. (a cart with frictionless wheels, for instance) coasting horizontally in a medium that causes linear drag. Equation (2.16) describes an object (a tiny oil droplet for instance) that is falling vertically with linear air resistance. I shall solve these two separate problems in turn. Horizontal Motion with Linear Drag Consider an object such as the cart in Figure 2.3 coasting horizontally in a linearly resistive medium. I shall assume that at t = 0 the cart is at x = 0 with velocity vx = vxo. The only force on the cart is the drag f = \u2014by, thus the cart inevitably slows down. The rate of slowing is determined by (2.15), which has the general form vx = \u2014kv x, (2.19) where k is my temporary abbreviation for k = blm. This is a first-order differential equation for vx, whose general solution must contain exactly one arbitrary constant. The equation states that the derivative of v x is equal to \u2014k times vx itself, and the only function with this property is the exponential function vx (t) = Ae -kt (2.20) which satisfies (2.19) for any value of the constant A (Problems 1.24 and 1.25). Since this solution contains one arbitrary constant, it is the general solution of our first- order equation; that is, any solution must have this form. In our case, we know that vx (0) = vxo, so that A = v xo, and we conclude that 1),(0 = Vxoe \u2014kt = Vxoe \u2014Ur where I have introduced the convenient parameter = 1/ k = mlb [for linear drag]. (2.21) (2.22) We see that our cart slows down exponentially, as shown in Figure 2.4(a). The parameter r has the dimensions of time (as you should check), and you can see from (2.21) that when t = r, the velocity is 1/e of its initial value; that is, r is the \"1/e\" time for the exponentially decreasing velocity. As t oo, the velocity approaches zero. To find the position as a function of time, we have only to integrate the velocity (2.21). Integrations of this kind can be done using the definite or indefinite integral. The definite integral has the advantage that it automatically takes care of the constant Section 2.2 Linear Air Resistance 49 0 xc (a) (b) t Figure 2.4 (a) The velocity vx as a function of time, t, for a cart moving horizontally with a linear resistive force. As t -+ oo, v, approaches zero exponentially. (b) The position x as a function of t for the same cart. As t -+ oc, x x o\u201e, vxo t. of integration: Since I), = dx/dt, r fo vx (t)dt = x(t) \u2014 x(0). 1 (Notice that I have named the \"dummy\" variable of integration t' to avoid confusion with the upper limit t.) Therefore x(t)= x(0) + f = 0 +[\u2014 vx0Te -t' t In the second line, I have used our assumption that x = 0 when t = 0. And in the last, I have introduced the parameter xoo = vxor, which is the limit of x(t) as t \u2014> co. We conclude that, as the cart slows down, its position approaches x,\u201e0 asymptotically, as shown in Figure 2.4(b). (2.23) (2.24) Vertical Motion with Linear Drag Let us next consider a projectile that is subject to linear air resistance and is thrown vertically downward. The two forces on the projectile are gravity and air resistance, as shown in Figure 2.5. If we measure y", " vertically down, the only interesting component of the equation of motion is the y component, which reads mi) y = mg \u2014 bvy. (2.25) With the velocity downward (v y > 0), the retarding force is upward, while the force of gravity is downward. If v y is small, the force of gravity is more important than the drag force, and the falling object accelerates in its downward motion. This will 50 Chapter 2 Projectiles and Charged Particles f.-bv w = mg y Figure 2.5 The forces on a projectile that is thrown ver- tically down, subject to linear air resistance. continue until the drag force balances the weight. The speed at which this balance occurs is easily found by setting (2.25) equal to zero, to give v y = mg/b or Vy = Vter where I have defined the terminal speed titer mg b [for linear drag]. (2.26) The terminal speed is the speed at which our projectile will eventually fall, if given the time to do so. Since it depends on m and b, it is different for different bodies. For example, if two objects have the same shape and size (b the same for both), the heavier object (m larger) will have the higher terminal speed, just as you would expect. Since Vter is inversely proportional to the coefficient b of air resistance, we can view titer as an inverse measure of the importance of air resistance \u2014 the larger the air resistance, the smaller ti ter, again just as you would expect. EXAMPLE 2.2 Terminal Speed of Small Liquid Drops Find the terminal speed of a tiny oildrop in the Millikan oildrop experiment (diameter D = 1.5 gm and density Q = 840 kg/m3). Do the same for a small drop of mist with diameter D = 0.2 mm. From Example 2.1 we know that the linear drag is dominant for these objects, so the terminal speed is given by (2.26). According to (2.4), b = j9D where = 1.6 x 10 -4 (in SI units). The mass of the drop is m = Q rc D3/6. Thus (2.26) becomes Vter = Q - D2 g 6 16 [for linear drag]. (2.27) Section 2.2 Linear Air Resistance 51 This interesting result shows that, for a given density, the terminal speed is proportional to D 2. This implies that, once air resistance has become important, a large sphere will fall faster than a small sphere of the same density. 2 Putting in the numbers, we find for the oildrop (840) x x (1.5 x 10_6)2 x (9.8) = Vter = 6 x (1.6 x 10 -4) 6.1 x 10-5 m/s [oildrop]. In the Millikan oildrop experiment, the oildrops fall exceedingly slowly, so their speed can be measured by simply watching them through a microscope. Putting in the numbers for the drop of mist, we find similarly that Vter = 1.3 m/s [drop of mist]. (2.28) This speed is representative for a fine drizzle. For a larger raindrop, the terminal speed would be appreciably larger, but with a larger (and hence also faster) drop, the quadratic drag would need to be included in the calculation to get a reliable value for titer. So far, we have discussed the terminal speed of a projectile (moving vertically), but we must now discuss how the projectile approaches that speed. This is determined by the equation of motion (2.25) which we can rewrite as mi) = \u2014b(v y titer ). (2.29) (Remember that titer = mg/b.) This differential equation can be solved in several ways. (For one alternative see Problem 2.9.) Perhaps the simplest is to note that it is almost the same as Equation (2.15) for the horizontal motion, except that on the right we now have (v y \u2014 titer) instead of v x. The solution for the horizontal case was the exponential function (2.20). The trick to solving our new vertical equation (2.29) is to introduce the new variable u = (vy \u2014 titer), which satisfies mu = \u2014bu (because titeris constant). Since this is exactly the same as Equation (2.15) for the horizontal motion, the solution for u is the same exponential, u = Ae -t ly. [Remember that the constant k in (2.20) became k = 1/t.] Therefore, V y \u2014 V ter = Ae -tai. When t = 0, v y = vyo, so A = vyo \u2014 titer and our final solution for v as a function of t is V (t)", " = V ter + ( vyo \u2014 V ter )e-t = yoe -t ir Vter ( 1 \u2014 IT ) - (2.30) (2.31) 2 We are here assuming that the drag force is linear, but the same qualitative conclusion follows for a quadratic drag force. (Problem 2.24.) 52 Chapter 2 Projectiles and Charged Particles 11 Y V ter t T 2T 3T Figure 2.6 When an object is dropped in a medium with linear resistance, vy approaches its terminal value Vter as shown. This second expression gives v y (t) as the sum of two terms: The first is equal to v yo t = 0, but fades away to zero as t increases; the second is equal to zero when when t = 0, but approaches vter as t oo. In particular, as t oo, V (t) V ter (2.32) just as we anticipated. Let us examine the result (2.31) in a little more detail for the case that v yo = 0; that is, the projectile is dropped from rest. In this case (2.31) reads V Y (t) = V ter (1 \u2014 e \u2014t it ). (2.33) This result is plotted in Figure 2.6, where we see that v y starts out from 0 and approaches the terminal speed, v y \u00b1 vter, asymptotically as t oo. The significance of the time r for a falling body is easily read off from (2.33). When t = r, we see that V y = V ter ( 1 \u2014 e-1) = 0. 63vter That is, in a time r, the object reaches 63% of the terminal speed. Similar calculations give the following results: time percent of vt, t 0 r 2r 3r 0 63% 86% 95% Of course, the object's speed never actually reaches v t\u201e, but r is a good measure of how fast the speed approaches v t\u201e. In particular, when t = 3r the speed is 95% of Net., and for many purposes we can say that after a time 3r the speed is essentially equal to vter Section 2.2 Linear Air Resistance 53,M=OMMM,22231faraAVV,Z EXAMPLE 2.3 Characteristic Time for Two Liquid Drops Find the characteristic times, r, for the oildrop and drop of mist in Example 2.2. The characteristic time r was defined in (2.22) as r = mlb, and vter was defined in (2.26) as vter = mg/b. Thus we have the useful relation Vter = gt (2.34) Notice that this relation lets us interpret vter as the speed a falling object would acquire in a time r, if it had a constant acceleration equal to g. Also note that, like vter, the time t is an inverse indicator of the importance of air resistance: When the coefficient b of air resistance is small, both v ter and r are large; when b is large, both v ter and r are small. For our present purposes, the importance of (2.34) is that, since we have already found the terminal velocities of the two drops, we can immediately find the values of r. For the Millikan oildrop, we found that v ter = 6.1 x 10 -5 m/s, therefore Vter - =- g 6.1 x 10 -5 9.8 = 6.2 x 10 -6 s [oildrop]. After falling for just 20 microseconds, this oildrop will have acquired 95% of its terminal speed. For almost every purpose, the oildrop always travels at its terminal speed. of its terminal speed. For the drop of mist of Example 2.2, the terminal speed was v ter = 1.3 m/s and so r = vt\u201eIg ti 0.13 s. After about 0.4 s, the drop will have acquired 95% Whether or not our falling object starts from rest, we can find its position y as a function of time by integrating the known form (2.30) of v y, V y (t) = Vter + (1/y0 \u2014 V ter)e \u2014th Assuming that the projectile's initial position is y = 0, it immediately follows that y(t) f v y (t1)dt' vtert (v y 0 vter)T ( 1 \u2014 e - tai) - (2.35) This equation for y(t) can now be combined with Equation (2.23) for x (t) to give us the orbit of any projectile, moving both horizontally and vertically, in a linear medium. 54 Chapter 2 Projectiles and Charged Particles 2.3 Tra", "jectory and Range in a Linear Medium We saw at the begining of the last section that the equation of motion for a projectile moving in any direction resolves into two separate equations, one for the horizontal and one for the vertical motion [Equations (2.15) and (2.16)]. We have solved each of these separate equations in (2.23) and (2.35), and we can now put these solutions together to give the trajectory of an arbitrary projectile moving in any direction. In this discussion it is marginally more convenient to measure y vertically upward, in which case we must reverse the sign of v t\u201e. (Make sure you understand this point.) Thus the two equations of the orbit become x(t) = V xo r (1 \u2014 e \u2014 t/t) y(t) (v yo + v ter, T (1\u2014 e \u2014 tit) \u2014 Vtert You can eliminate t from these two equations by solving the first for t and then substituting into the second. (See Problem 2.17.) The result is the equation for the trajectory: (2.36) (2.37) y Vyo Vter X + Nei.\"( in (1 vxo x vxo t This equation is probably too complicated to be especially illuminating, but I have plotted it as the solid curve in Figure 2.7, with the help of which you can understand some of the features of (2.37). For example, if you look at the second term on the right of (2.37), you will see that as x --> vxo r the argument of the log function approaches zero; therefore, the log term and hence y both approach \u2014oo. That is, the trajectory has a vertical asymptote at x = vxor, as you can see in the picture. I leave it as an exercise (Problem 2.19) for you to check that if air resistance is switched off (v t, and r both approach infinity), the trajectory defined by (2.37) does indeed approach the dashed trajectory corresponding to zero air resistance. Horizontal Range A standard (and quite interesting) problem in elementary physics courses is to show that the horizontal range R of a projectile (subject to no air resistance of course) is Rvac = 2vxo vyo g [no air resistance] (2.38) where Rvac stands for the range in a vacuum. Let us see how this result is modified by air resistance. solution of the equation The range R is the value of x when y as given by (2.37) is zero. Thus R is the Vyo Vter R v tert in (1 vxo R vxot = O. (2.39) Section 2.3 Trajectory and Range in a Linear Medium 55 Figure 2.7 The trajectory of a projectile subject to a linear drag force (solid curve) and the corresponding trajectory in a vacuum (dashed curve). At first the two curves are very similar, but as t increases, air resistance slows the projectile and pulls its trajec- tory down, with a vertical asymptote at x = V xo t. The horizontal range of the projectile is labeled R, and the corresponding range in vacuum Rvac. This is a transcendental equation and cannot be solved analytically, that is, in terms of well known, elementary functions such as logs, or sines and cosines. For a given choice of parameters, it can be solved numerically with a computer (Problem 2.22), but this approach usually gives one little sense of how the solution depends on the parameters. Often a good alternative is to find some approximation that allows an approximate analytic solution. (Before the advent of computers, this was often the only way to find out what happens.) In the present case, it is often clear that the effects of air resistance should be small. This means that both vt\u201e and r are large and the second term in the argument of the log function is small (since it has r in its denominator). This suggests that we expand the log in a Taylor series (see Problem 2.18): 'no - = - + 1E 2 + 1E 3 +.) (2.40) We can use this expansion for the log term in (2.39), and, provided t is large enough, we can surely neglect the terms beyond E 3. This gives the equation vterl R vterr [ R 1 R ) 2 + 1 R ) 31 = 0. (2.41) vxo r 2 1).\u201e,,T ) 3 vxo r vxo This equation can be quickly tidied up. First, the second term in the first bracket cancels the first term in the second. Next, every term contains a factor of R. This implies that one solution is R = 0, which is correct \u2014 the height y is zero when x", " = 0. Nevertheless, this is not the solution we are interested in, and we can divide out the 56 Chapter 2 Projectiles and Charged Particles common factor of R. A little rearrangement (and replacement of v t\u201e/r by g) lets us rewite the equation as This may seem a perverse way to write a quadratic equation for R, but it leads us quickly to the desired approximate solution. The point is that the second term on the right is very small. (In the numerator R is certainly no more than Rvac and we are assuming that r in the denominator is very large.) Therefore, as a first approximation we get This is just what we expected: For low air resistance, the range is close to Rvac. But with the help of (2.42) we can now get a second, better approximation. The last term of (2.42) is the required correction to Rvac ; because it is already small, we would certainly be satisfied with an approximate value for this correction. Thus, in evaluating the last term of (2.42), we can replace R with the approximate value R ti R vac, and we find as our second approximation [remember that the first term in (2.42) is just Rvac ] 2vxo vyo R = g 2 R 2. 3vxo r 2v xo v y0 R \u2014 Rvac g R Rvac (Rvac)2 2 3v t xo Rvac ( 1 4,, UY\u00b0 ) Uter (2.42) (2.43) (2.44) (To get the second line, I replaced the second Rvac in the previous line by 2viovyo / g and -cg by vt\u201e.) Notice that the correction for air resistance always makes R smaller than Rvac, as one would expect. Notice also that the correction depends only on the ratio vy,,/vt\u201e. More generally, it is easy to see (Problem 2.32) that the importance of air resistance is indicated by the ratio 04, of the projectile's speed to the terminal speed. If v /v t, << 1 throughout the flight, the effect of air resistance is very small; if vi vtor is around 1 or more, air resistance is almost certainly important [and the approximation (2.44) is certainly no good]. EXAMPLE 2.4 Range of Small Metal Pellets I flick a tiny metal pellet with diameter d = 0.2 mm and v = 1 m/s at 45\u00b0. Find its horizontal range assuming the pellet is gold (density Q. = 16 g/cm 3). What if it is aluminum (density Q ti 2.7 g/cm3)? In the absence of air resistance, both pellets would have the same range, 2vxo vyo Rvac = = 10.2 cm. Section 2.4 Quadratic Air Resistance 57 For gold, Equation (2.27) gives (as you can check) v t, ^ 21 m/s. Thus the correction term in (2.44) is 4 v y. 4 0.71 3 Vter 3 x 21 0.05. That is, air resistance reduces the range by 5% to about 9.7 cm. The density of aluminum is about 1/6 times that of gold. Therefore the terminal speed is one sixth as big, and the correction for aluminum is 6 times greater or about 30%, giving a range of about 7 cm. For the gold pellet the correction for air resistance is quite small and could perhaps be neglected; for the aluminum pellet, the correction is still small, but is certainly not negligible. 2.4 Quadratic Air Resistance In the last two sections we have developed a rather complete theory of projectiles subject to a linear drag force, f = \u2014by. While we can find examples of projectiles for which the drag is linear (notably very small objects, such as the Millikan oildrop), for most of the more obvious examples of projectiles (baseballs, footballs, cannonballs, and the like) it is a far better approximation to say that the drag is pure quadratic, f = \u2014cv4. We must, therefore, develop a corresponding theory for a quadratic drag force. On the face of it, the two theories are not so very different. In either case we have to solve the differential equation = mg f, (2.45) and in both cases this is a first-order differential equation for the velocity v, with f depending in a relatively simple way on v. There is, however, an important difference. In the linear case (f = \u2014by), Equation (2.45) is a linear differential equation, inas- much as the terms that involve v are all linear in v or its derivatives. In the quadratic case", ", Equation (2.45) is, of course, nonlinear. And it turns out that the mathemati- cal theory of nonlinear differential equations is significantly more complicated than the linear theory. As a practical matter, we shall find that for the case of a general projectile, moving in both the x and y directions, Equation (2.45) cannot be solved in terms of elementary functions when the drag is quadratic. More generally, we shall see in Chapter 12 that for more complicated systems, nonlinearity can lead to the astonishing phenomenon of chaos, although this does not happen in the present case. In this section, I shall start with the same two special cases discussed in Section 2.2, a body that is constrained to move horizontally, such as a railroad car on a horizontal track, and a body that moves vertically, such as a stone dropped from a window (both now with quadratic drag forces). We shall find that in these two especially simple cases the differential equation (2.45) can be solved by elementary means, and the solutions 58 Chapter 2 Projectiles and Charged Particles introduce some important techniques and interesting results. I shall then discuss briefly the general case (motion in both the horizontal and vertical directions), which can be solved only numerically. Horizontal Motion with Quadratic Drag Let us consider a body moving horizontally (in the positive x direction), subject to a quadratic drag and no other forces. For example, you could imagine a cycle racer, who has crossed the finishing line and is coasting to a stop under the influence of air resistance. To the extent that the cycle is well lubricated and tires well inflated, we can ignore ordinary friction, 3 and, except at very low speeds, air resistance is purely quadratic. The x component of the equation of motion is therefore (I'll abbreviate v x If we divide by v 2 and multiply by dt, we get an equation in which only the variable v appears on the left and only t on the right:4 m - = - CV. 2 dv dt m\u2014 = \u2014c dt. dv v 2 to v) (2.46) (2.47) This trick \u2014 of rearranging a differential equation so that only one variable appears on the left and only the other on the right \u2014 is called separation of variables. When it is possible, separation of variables is often the simplest way to solve a first-order differential equation, since the solution can be found by simple integration of both sides. Integrating Equation (2.47) we find v dv' m f vo v /2 = \u2014c f dt' 0 where vo is the initial velocity at t = 0. Notice that I have written both sides as definite integrals, with the appropriate limits, so that I shan't have to worry about any constants of integration. I have also renamed the variables of integration as v' and t' to avoid 3 As I shall discuss shortly, when the cyclist slows down to a stop, air resistance becomes smaller, and eventually friction becomes the dominant force. Nevertheless, at speeds around 10 mph or more, it is a fair approximation to ignore everything but the quadratic air resistance. 4 In passing from (2.46) to (2.47), I have treated the derivative dv/dt as if it were the quotient of two separate numbers, dv and dt. As you are certainly aware this cavalier proceeding is not strictly correct. Nevertheless, it can be justified in two ways. First, in the theory of differentials, it is in fact true that dv and dt are defined as separate numbers (differentials), such that their quotient is the derivative dv/dt. Fortunately, it is quite unnecessary to know about this theory. As physicists we know that dv/dt is the limit of A v / At, as both Av and At become small, and I shall take the view that dv is just shorthand for Av (and likewise dt for At), with the understanding that it has been taken small enough that the quotient dv/dt is within my desired accuracy of the true derivative. With this understanding, (2.47), with dv on one side and dt on the other, makes perfectly good sense. confusion with the upper limits v and t. Both of these integrals are easily evaluated, Section 2.4 Quadratic Air Resistance 59 and we find or, solving for v, ml \u2014 \u2014 = \u2014ct 1 vo 1 v v(t) = vo 1 \u00b1 cv.tIm 1 +tit 0 (2.48) (2.49) where I have introduced the abbreviation r for the combination of constants T = m C Vo [for quadratic drag]. (2.50) As you can easily check, t is a time, with the significance that when t = t the velocity is v = v0/2. Notice that", " this parameter t is different from the t introduced in (2.22) for motion subject to linear air resistance; nevertheless, both parameters have the same general significance as indicators of the time for air resistance to slow the motion To find the bicycle's position x, we have only to integrate v to give (as you should appreciably. check) x(t) = x. + f v(t') dt' = vo r ln (1 \u00b1 t/r), (2.51) if we take the initial position xo to be zero. Figure 2.8 shows our results for v and x as functions of t. It is interesting to compare these graphs with the corresponding graphs of Figure 2.4 for a body coasting horizontally but subject to a linear resistance. Superficially, the two graphs for the velocity look similar. In particular, both go to zero as t oo. But in the linear case v goes to zero exponentially, whereas in the quadratic case it does so only very slowly, like 1/ t. This difference in the behavior of v manifests itself quite dramatically in the behavior of x. In the linear case, we t t (a) (b) Figure 2.8 The motion of a body, such as a bicycle, coasting horizontally and subject to a quadratic air resistance. (a) The velocity is given by (2.49) and goes to zero like 1/ t as t -->- co. (b) The position is given by (2.51) and goes to infinity as t -->- oo. 60 Chapter 2 Projectiles and Charged Particles saw that x approaches a finite limit as t oo, but it is clear from (2.51) that in the quadratic case x increases without limit as t oo. The striking difference in the behavior of x for quadratic and linear drags is easy to understand qualitatively. In the quadratic case, the drag is proportional to v 2. Thus as v gets small, the drag gets very small \u2014 so small that it fails to bring the bicyle to rest at any finite value of x. This unexpected behavior serves to highlight that a drag force that is proportional to v 2 at all speeds is unrealistic. Although the linear drag and ordinary friction are very small, nevertheless as v 0 they must eventually become more important than the v 2 term and cannot be ignored. In particular, one or another of these two teams (friction in the case of a bicycle) ensures that no real body can coast on to infinity! Vertical Motion with Quadratic Drag The case that an object moves vertically with a quadratic drag force can be solved in much the same way as the horizontal case. Consider a baseball that is dropped from a window in a high tower. If we measure the coordinate y vertically down, the equation of motion is (I'll abbreviate v y to v now) my = mg \u2014 cv 2. (2.52) Before we solve this equation, let us consider the ball's terminal speed, the speed at which the two terms on the right of (2.52) just balance. Evidently this must satisfy c v 2 = mg, whose solution is vter = (2.53) For any given object (given in, g, and c), this lets us calculate the terminal speed. For example, for a baseball it gives (as we shall see in a moment) v t\u201e c=, 35 m/s, or nearly 80 miles per hour. We can tidy the equation of motion (2.52) a little by using (2.53) to replace c by mg I v t2e, and canceling the factors of m: This can be solved by separation of variables, just as in the case of horizontal motion: First we can rewrite it as g (1 V 2 vter dv 1\u2014 V2/V 2 r te = g dt. (2.54) (2.55) This is the desired separated form (only v on the left and only t on the right) and we can simply integrate both sides. 5 Assuming the ball starts from rest, the limits of 5 Notice that in fact any one-dimensional problem where the net force depends only on the velocity can be solved by separation of variables, since the equation my = F(v) can always be Section 2.4 Quadratic Air Resistance 61 integration are 0 and v on the left and 0 and t on the right, and we find (as you should verify \u2014 Problem 2.35) vter arctanh (\u2014 = t g v V ter (2.56) where \"arctanh\" denotes the inverse hyperbolic tangent. This particular integral can be evaluated alternatively in terms of the natural log function (Problem 2.37). However, the hyperbolic functions, sinh, cosh, and tanh, and their inverses arcsinh, ar", "ccosh, and arctanh, come up so often in all branches of physics that you really should learn to use them. If you have not had much exposure to them, you might want to look at Problems 2.33 and 2.34, and study graphs of these functions. Equation (2.56) can be solved for v to give To find the position y, we just integrate v to give v = vtertanh (\u2014. gt V ter )2 y (vte r g In cosh. L \\ vter (2.57) (2.58) While both of these two formulas can be cleaned up a little (see Problem 2.35), they are already sufficient to work the following example. EXAMPLE 2.5 A Baseball Dropped from a High Tower Find the terminal speed of a baseball (mass m = 0.15 kg and diameter D = 7 cm). Make plots of its velocity and position for the first six seconds after it is dropped from a tall tower. The terminal speed is given by (2.53), with the coefficient of air resistance c given by (2.4) as c = y D2 where y = 0.25 Ns 2/m4. Therefore vter = (0.15 kg) x (9.8 m/s 2 ) = 35 (2.59) m/s (0.251\\1-s 2/m4) x (0.07 m) 2 or nearly 80 miles per hour. It is interesting to note that fast baseball pitchers can pitch a ball considerably faster than vter. Under these conditions, the drag force is actually greater than the ball's weight! The plots of v and y can be made by hand, but are, of course, much easier with the help of computer software such as Mathcad or Mathematica that can make the plots for you. Whatever method we choose, the results are as shown in Figure 2.9, where the solid curves show the actual velocity and position while the dashed curves are the corresponding values in a vacuum. The actual velocity levels out, written as m dv I F(v) = dt. Of course there is no assurance that this can be integrated analytically if F(v) is too complicated, but it does guarantee a straightforward numerical solution at worst. See Problem 2.7. 62 Chapter 2 Projectiles and Charged Particles Figure 2.9 The motion of a baseball dropped from the top of a high tower (solid curves). The corresponding motion in a vacuum is shown with long dashes. (a) The actual velocity approaches the ball's terminal velocity v ter = 35 m/s as t oo. (b) The graph of position against time falls further and further behind the corresponding vacuum graph. When t = 6 s, the baseball has dropped about 130 meters; in a vacuum, it would have dropped about 180 meters. approaching the terminal value v t, = 35 m/s as t oo, whereas the velocity in a vacuum would increase without limit Initially, the position increases just as it would in a vacuum (that is, y = z gt 2), but falls behind as v increases and the air resistance becomes more important. Eventually, y approaches a straight line of the form y = vtert + const. (See Problem 2.35.) Quadratic Drag with Horizontal and Vertical Motion The equation of motion for a projectile subject to quadratic drag, resolves into its horizontal and vertical components (with y measured vertically upward) to give mr = mg \u2014 cv 2i = mg \u2014 cvv, mix = \u2014C V 2 X 2+ VV y X m 1)3, = \u2014mg \u2014 vx2 v; v y. (2.60) (2.61) These are two differential equations for the two unknown functions v x (t) and v y (t), but each equation involves both vx and vy. In particular, neither equation is the same as for an object that moves only in the x direction or only in the y direction. This means that we cannot solve these two equations by simply pasting together our two separate solutions for horizontal and vertical motion. Worse still, it turns out that the two equations (2.61) cannot be solved analytically at all. The only way to solve them is numerically, which we can only do for specified numerical initial conditions (that is, specified values of the initial position and velocity). This means that we cannot find the general solution; all we can do numerically is to find the particular solution corresponding to any chosen initial conditions. Before I discuss some general properties of the solutions of (2.61), let us work out one such numerical solution. Section 2.4 Quadratic Air Resistance 63 EXAMPLE 2.6 Trajectory of a Baseball The baseball of Example 2.5 is now thrown with velocity 30 m/s (about 70 mi/h) at", " 50\u00b0 above the horizontal from a high cliff. Find its trajectory for the first eight seconds of flight and compare with the corresponding trajectory in a vacuum. If the same baseball was thrown with the same initial velocity on horizontal ground how far would it travel before landing? That is, what is its horizontal range? We have to solve the two coupled differential equations (2.61) with the initial conditions vxo = vo cos 0 = 19.3 m/s and v y0 = vo sin 0 = 23.0 m/s and xo = yo = 0 (if we put the origin at the point from which the ball is thrown). This can be done with systems such as Mathematica, Matlab, or Maple, or with programming languages such as \"C\" or Fortran. Figure 2.10 shows the resulting trajectory, found using the function \"NDSolve\" in Mathematica. Several features of Figure 2.10 deserve comment. Obviously the effect of air resistance is to lower the trajectory, as compared to the vacuum trajectory (shown dashed). For example, we see that in a vacuum the high point of the trajectory occurs at t ti 2.3 s and is about 27 m above the starting point; with air resistance, the high point comes just before t = 2.0 s and is at about 21 m. In a vacuum, the ball would continue to move indefinitely in the x direction. The y (m) Figure 2.10 Trajectory of a baseball thrown off a cliff and subject to quadratic air resistance (solid curve). The initial velocity is 30 m/s at 50\u00b0 above the horizontal; the terminal speed is 35 m/s. The dashed curve shows the corresponding trajectory in a vacuum. The dots show the ball's position at one-second intervals. Air resistance slows the horizontal motion, so that the ball approaches a vertical asymptote just beyond x = 100 meters. 64 Chapter 2 Projectiles and Charged Particles effect of air resistance is to slow the horizontal motion so that x never moves to the right of a vertical asymptote near x = 100 m. The horizontal range of the baseball is easily read off the figure as the value of x when y returns to zero. We see that R ti 59 m, as opposed to the range in vacuum, Rvac ti 90 m. The effect of air resistance is quite large in this example, as we might have anticipated: The ball was thrown with a speed only a little less than the terminal speed (30 vs 35 m/s), and this means that the force of air resistance is only a little less than that of gravity. This being the case, we should expect air resistance to change the trajectory appreciably. This example illustrates several of the general features of projectile motion with a quadratic drag force. Although we cannot solve analytically the equations of motion (2.61) for this problem, we can use the equations to prove various general properties of the trajectory. For example, we noticed that the baseball reached a lower maximum height, and did so sooner, than it would have in a vacuum. It is easy to prove that this will always be the case: As long as the projectile is moving upward (v y > 0), the force of air resistance has a downward y component. Thus the downward acceleration is greater than g (its value in vacuum). Therefore a graph of v y against t slopes down from vy,, more quickly than it would in vacuum, as shown in Figure 2.11. This guarantees that vy reaches zero sooner than it would in vacuum, and that the ball travels less distance (in the y direction) before reaching the high point. That is, the ball's high point occurs sooner, and is lower, than it would be in a vacuum. Figure 2.11 Graph of v y against t for a projectile that is thrown upward (vyo > 0) and is subject to a quadratic resistance (solid curve). The dashed line (slope = \u2014g) is the corresponding graph when there is no air resistance. The projectile moves upward until it reaches its maximum height when vy = 0. During this time, the drag force is downward and the downward acceleration is always greater than g. Therefore, the curve slopes more steeply than the dashed line, and the projectile reaches its high point sooner than it would in a vacuum. Since the area under the curve is less than that under the dashed line, the projectile's maximum height is less than it would be in a vacuum. Section 2.5 Motion of a Charge in a Uniform Magnetic Field 65 I claimed that the baseball of Example 2.6 approaches a vertical asymptote as t \u2014> oo, and we can now prove that this is always the case. First, it is easy to convince yourself that once the ball starts moving downward, it continues to accelerate downward, with v y approaching \u2014 vt\u201e as 1 \u2014> oo. At the same time vx continues to decrease", " and approaches zero. Thus the square root in both of the equations (2.61) approaches v t\u201e. In particular, when t is large, the equation for v, can be approximated by say. The solution of this equation is, of course, an exponential function, v x = Ae \u2014kt, and we see that vx approaches zero very rapidly (exponentially) as t oo. This guarantees that x, which is the integral of v x, vx CVter vx = \u2014kv x m t x(t) = f v x (t f)dt', approaches a finite limit as t \u2014> oo, and the trajectory has a finite vertical asymptote as claimed. 2.5 Motion of a Charge in a Uniform Magnetic Field Another interesting application of Newton's laws, and (like projectile motion) an application that lets me introduce some important mathematical methods, is the motion of a charged particle in a magnetic field. I shall consider here a particle of charge q (which I shall usually take to be positive), moving in a uniform magnetic field B that points in the z direction as shown in Figure 2.12. The net force on the particle is just the magnetic force F = qv x B, (2.62) z B Figure 2.12 A charged particle moving in a uniform mag- netic field that points in the z direction. 66 Chapter 2 Projectiles and Charged Particles so the equation of motion can be written as mir = qv x B. (2.63) [As with projectiles, the force depends only on the velocity (not the position), so the second law reduces to a first-order differential equation for v.] As is so often the case, the simplest way to solve the equation of motion is to resolve it into components. The components of v and B are and from which we can read off the components of v x B: v x B = (v y B, \u2014v x B, 0). Thus the three components of (2.63) are v = (vx, vy, v z ) B= (0, 0, B), mi)x = q B v y mi)y = \u2014qBv x = 0. vZ = const, (.0 = qB m vx cov y 1 v y = \u2014coy x \u2014 The last of these says simply that v z, the component of the particle's velocity in the direction of B, is constant: a result we could have anticipated since the magnetic force is always perpendicular to B. Because vz is constant, we shall focus most of our attention on v, and v y. In fact, we can even think of them as comprising a two-dimensional vector (v x, vy ), which is just the projection of v onto the xy plane and can be called the transverse velocity, (vx, v y ) = transverse velocity. To simplify the equations (2.64) and (2.65) for v x and vy, I shall define the parameter which has the dimensions of inverse time and is called the cyclotron frequency. With this notation, Equations (2.64) and (2.65) become These two coupled differential equations can be solved in a host of different ways. I would like to describe one that makes use of complex numbers. Though perhaps (2.64) (2.65) (2.66) (2.67) (2.68) Section 2.5 Motion of a Charge in a Uniform Magnetic Field 67 Figure 2.13 The complex number ii = v x + i vy is repre- sented as a point in the complex plane. The arrow pointing from 0 to /7 is literally a picture of the transverse velocity vector (v x, v y ). not the easiest solution, this method has surprisingly wide application in many ar- eas of physics. (For an alternative solution that avoids complex numbers, see Prob- lem 2.54.) The two variables v x and vy are, of course, real numbers. However, there is nothing to prevent us from defining a complex number g = v, + ivy, (2.69) where i denotes the square root of \u20141 (called j by most engineers), i =,N,/ \u20141 (and ti is the Greek letter eta). If we draw the complex number ri in the complex plane, or Argand diagram, then its two components are v x and vy as shown in Figure 2.13; in other words, the representation of ri in the complex plane is a picture of the two- dimensional transverse velocity (v i, v y ). derivative. Using (2.68), we find that The advantages of introducing the complex number ri appear when we evaluate its r) = V, + ay = wvy \u2014 icor), = \u2014iw(v x + iv y ) or ii = \u2014icon", ". (2.70) We see that the two coupled equations for v x and vy have become a single equation for the complex number r7. Furthermore, it is an equation of the now familiar form U = ku, whose solution we know to be the exponential u = Ae kt. Thus we can immediately write down the solution for 77: II = Ae -i\" (2.71) Before we discuss the significance of this solution, I would like to review a few properties of complex exponentials in the next section. If you are very familiar with these ideas, by all means skip this material. 68 Chapter 2 Projectiles and Charged Particles 2.6 Complex Exponentials While you are certainly familiar with the exponential function ex for a real variable x, you may not be so at home with ez when z is complex.6 For the real case there are several possible definitions of ex (for instance, as the function that is equal to its own derivative). The definition that extends most easily to the complex case is the Taylor series (see Problem 2.18) ez = 1 Z \u2014 \u2014. (2.72) 2 Z 2! 3 Z 3! For any value of z, real or complex, large or small, this series converges to give a well- defined value for ez. By differentiating it, you can easily convince yourself that it has the expected property that it equals its own derivative. And one can show (not always so easily) that it has all the other familiar properties of the exponential function \u2014 for instance, that ezew = e (z+w ). (See Problems 2.50 and 2.51.) In particular, the function Ae k z (with A and k any constants, real or complex) has the property that d (Ae kz) = k (Ae k z). dz (2.73) Since it satisfies this same equation whatever the value of A, it is, in fact, the general solution of the first-order equation dfldz=kf. At the end of the last section, I introduced the complex number 1)(0 and showed that it satisfied the equation ri = \u2014icor). We are now justified in saying that this guarantees that ri must be the exponential function anticipated in (2.71). We shall be particularly concerned with the exponential of a pure imaginary number, that is, ei 8 where 0 is a real number. The Taylor series (2.72) for this function reads eit9 = 1+ i0 + (0)2 (0) 3 (0) 4 2! 3! 4! (2.74) Noting that i 2 = \u20141, i 3 = \u2014i, and so on, you can see that all of the even powers in this series are real, while all of the odd powers are pure imaginary. Regrouping accordingly, we can rewrite (2.74) to read \u2014 +... +i 9 -- +... 3! (2.75) The series in the first brackets is the Taylor series for cos 0, and that in the second brackets is sin 0 (Problem 2.18). Thus we have proved the important relation: ei0 = cos 0 + i sin 0. (2.76) 6 For a review of some elementary properties of complex numbers, see Problems 2.45 to 2.49. Section 2.6 Complex Exponentials 69 (a) (b) Figure 2.14 (a) Euler's formula, (2.76), implies that the complex num- ber eta lies on the unit circle (the circle of radius 1, centered on the origin 0) with polar angle 0. (b) The complex constant A = aei' lies on a cir- cle of radius a with polar angle S. The function 77 (t) = Ae -i cut lies on the same circle but with polar angle (3 \u2014 cot) and moves clockwise around the circle as t advances. This result, known as Euler's formula, is illustrated in Figure 2.14(a). Note especially that the complex number ei\u00b0 has polar angle 0 and, since cos t 0 + sin2 0 = 1, the magnitude of ei0 is 1; that is, ei9 lies on the unit circle, the circle with radius 1 centered at O. Our main concern is with a complex number of the form 17 = Ae -i \". The co- efficient A is a fixed complex number, which can be expressed as A = aei8, where a = IAI is the magnitude, and 3 is the polar angle of A, as shown in Figure 2.14(b). (See Problem 2.45.) The number 77 can therefore be written as 77 = Ae -i \" = aei8 e-iwt = aei(6-wt) (2.77) Thus i has the same magnitude as A (name", "ly a), but has polar angle equal to (6 \u2014 wt), as shown in Figure 2.14(b). As a function of t, the number 11 moves clockwise around the circle of radius a with angular velocity co. It is important that you get a good feel for the role of the complex constant A = aei6 in (2.77): If A happened to equal 1, then 17 would be just >7 = e -i \", which lies on the unit circle, moving clockwise with angular velocity w and starting from the real axis (r) = 1) when t = 0. If A = a is real but not equal to 1, then it simply magnifies the unit circle to a circle of radius a, around which ri moves with the same angular speed and starting from the real axis, at 17 = a when t = 0. Finally if A = aei5, then the effect of the angle 8 is to rotate ri through the fixed angle 8, so that ri starts out at t = 0 Armed with these mathematical results, we can now return to the charged particle with polar angle S. in a magnetic field. 70 Chapter 2 Projectiles and Charged Particles 2.7 Solution for the Charge in a B Field Mathematically, the solution for the velocity v of our charged particle in a B field is complete, and all that remains is to interpret it physically. We already know that v z, the component along B, is constant. The components (v x, vy ) transverse to B we have represented by the complex number 77 = vx + ivy, and we have seen that Newton's second law implies that 77 has the time dependence ri = Ae -iwt, moving uniformly around the circle of Figure 2.14(b). Now, the arrow shown in that figure, pointing from 0 to 77, is in fact a pictorial representation of the transverse velocity (v x, v y ). Therefore this transverse velocity changes direction, turning clockwise, with constant angular velocity w = q Blm and with constant magnitude. Because v z is constant, this suggests that the particle undergoes a spiralling, or helical, motion. To verify this, we have only to integrate v to find r as a function of t. That vz is constant implies that z(t) = zo vzo t. (2.78) The motion of x and y is most easily found by introducing another complex number where is the Greek letter xi. In the complex plane, is a picture of the transverse position (x, y). Clearly, the derivative of is 77, that is, = 77. Therefore, = x iy =J 77 dt f Ae -iwt dt i A \u2014e + constant. (2.79) If we rename the coefficient i A /w as C and the constant of integration as X + i Y, this implies that x + iy = Ce-c wt \u00b1 (X \u00b1 iY). By redefining our origin so that the z axis goes through the point (X, Y), we can eliminate the constant term on the right to give x iy = Ce -iwt (2.80) and, by setting t = 0, we can identify the remaining constant C as C = xo iyo. This result is illustrated in Figure 2.15. We see there that the transverse position (x, y) moves clockwise round a circle with angular velocity w = q B 1m. Meanwhile z as given by (2.78) increases steadily, so the particle actually describes a uniform helix whose axis is parallel to the magnetic field. 7 1 am assuming the charge q is positive; if q is negative, then w = qB/m is negative, meaning that the transverse velocity rotates counterclockwise. Principal Definitions and Equations of Chapter 2 71 Figure 2.15 Motion of a charge in a uniform magnetic field in the z direction. The transverse position (x, y) moves around a circle as shown, while the coordinate z moves with constant velocity into or out of the page. There are many examples of the helical motion of a charged particle along a magnetic field; for example, cosmic-ray particles (charged particles hitting the earth from space) can get caught by the earth's magnetic field and spiral north or south along the field lines. If the z component of the velocity happens to be zero, then the spiral reduces to a circle. In the cyclotron, a device for accelerating charged particles to high energies, the particles are trapped in circular orbits in this way. They are slowly accelerated by the judiciously timed application of an electric field. The angular frequency of the orbit is, of course, co = qB/m (which is why this is called the cyclotron frequency). The radius of the orbit is v my r = \u2014 = \u2014 = B. co q B q B (2.81)", " This radius increases as the particles accelerate, so that they eventually emerge at the outer edge of the circular magnets that produce the magnetic field. The same method that we have used here for a charge in a magnetic field can also be used for a particle in magnetic and electric fields, but since this complication adds nothing to the method of solution, I shall leave you to try it for yourself in Problems 2.53 and 2.55. Principal Definitions and Equations of Chapter 2 Linear and Quadratic Drags Provided the speed v is well below that of sound, the magnitude of the drag force f = \u2014f (v)i, on an object moving through a fluid is usually well approximated as f (v) = fii. + /quad where flin = by = 13Th) and fquad = cv2 = y D2V2. [Eqs. (2.2) to (2.6)] 72 Chapter 2 Projectiles and Charged Particles Here D denotes the linear size of the object. For a sphere, D is the diameter and, for a sphere in air at STP, j3 = 1.6 x 10 -4 1\\1-s/m2 and y = 0.25 I\\Ts 2/m4. The Lorentz Force on a Charged Particle F = q (E v x B). [Eq. (2.62) & Problem 2.53] Problems for Chapter 2 SECTION 2.1 Air Resistance Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). 2.1 * When a baseball flies through the air, the ratio fquad/fiin of the quadratic to the linear drag force is given by (2.7). Given that a baseball has diameter 7 cm, find the approximate speed v at which the two drag forces are equally important. For what approximate range of speeds is it safe to treat the drag force as purely quadratic? Under normal conditions is it a good approximation to ignore the linear term? Answer the same questions for a beach ball of diameter 70 cm. 2.2 * The origin of the linear drag force on a sphere in a fluid is the viscosity of the fluid. According to Stokes's law, the viscous drag on a sphere is flir, = 37r iiDv (2.82) where ri is the viscosity 8 of the fluid, D the sphere's diameter, and v its speed. Show that this expression reproduces the form (2.3) for Ain, with b given by (2.4) as b = 8D. Given that the viscosity of air at STP is 17 = 1.7 x 10-5 Ns/m 2, verify the value of,8 given in (2.5). 2.3 * (a) The quadratic and linear drag forces on a moving sphere in a fluid are given by (2.84) and (2.82) (Problems 2.4 and 2.2). Show that the ratio of these two kinds of drag force can be written as fquad/fiin = R/48,9 where the dimensionless Reynolds number R is R= Dvp (2.83) where D is the sphere's diameter, v its speed, and Q and 77 are the fluid's density and viscosity. Clearly the Reynolds number is a measure of the relative importance of the two kinds of drag. 1\u00b0 When R is 8 For the record, the viscosity of a fluid is defined as follows: Imagine a wide channel along which fluid is flowing (x direction) such that the velocity v is zero at the bottom (y = 0) and increases toward the top (y = h), so that successive layers of fluid slide across one another with a velocity gradient dvldy. The force F with which an area A of any one layer drags the fluid above it is proportional to A and to dv/dy, and )7 is defined as the constant of proportionality; that is, F =ri A dvldy. different for different shapes. 9 The numerical factor 48 is for a sphere. A similar result holds for other bodies, but the numerical factor is 1\u00b0 The Reynolds number is usually defined by (2.83) for flow involving any object, with D defined as a typical linear dimension. One sometimes hears the claim that R is the ratio f quad,/ \u201edu. Since f quad, lin = R /48 for a sphere, f this claim would be better phrased as \"R is roughly of the order of fquadf,,f;in.\" Problems for Chapter 2 73 very large, the quadratic drag is dominant and the linear can be neglected; vice versa when R is very small. (b) Find the Reynolds number for a steel ball bearing (diameter 2 mm) moving at 5 cm/s through glycerin (density 1.3", " g/cm 3 and viscosity 12 Ns/m 2 at STP). 2.4 ** The origin of the quadratic drag force on any projectile in a fluid is the inertia of the fluid that the projectile sweeps up. (a) Assuming the projectile has a cross-sectional area A (normal to its velocity) and speed v, and that the density of the fluid is Q, show that the rate at which the projectile encounters fluid (mass/time) is QA v. (b) Making the simplifying assumption that all of this fluid is accelerated to the speed v of the projectile, show that the net drag force on the projectile is pit v 2. It is certainly not true that all the fluid that the projectile encounters is accelerated to the full speed v, but one might guess that the actual force would have the form fquad = KOAV 2 (2.84) where K is a number less than 1, which would depend on the shape of the projectile, with K small for a streamlined body, and larger for a body with a flat front end. This proves to be true, and for a sphere the factor K is found to be K = 1/4. (c) Show that (2.84) reproduces the form (2.3) for fquad, with c given by (2.4) as c = y D2. Given that the density of air at STP is Q = 1.29 kg/m3 and that K = 1/4 for a sphere, verify the value of y given in (2.6). SECTION 22 Linear Air Resistance 2.5 * Suppose that a projectile which is subject to a linear resistive force is thrown vertically down with a speed v yo which is greater than the terminal speed v t\u201e. Describe and explain how the velocity varies with time, and make a plot of v y against t for the case that v yo = 2vt\u201e. 2.6 * (a) Equation (2.33) gives the velocity of an object dropped from rest. At first, when v y is small, air resistance should be unimportant and (2.33) should agree with the elementary result v y = gt for free fall in a vacuum. Prove that this is the case. [Hint Remember the Taylor series for ex = 1+ x + x 2 /2! + x 3 /3! +, for which the first two or three terms are certainly a good approximation when x is small.] (b) The position of the dropped object is given by (2.35) with v y0 = 0. Show similarly that this reduces to the familiar y = igt2 when t is small. 2.7 * There are certain simple one-dimensional problems where the equation of motion (Newton's second law) can always be solved, or at least reduced to the problem of doing an integral. One of these (which we have met a couple of times in this chapter) is the motion of a one-dimensional particle subject to a force that depends only on the velocity v, that is, F = F(v). Write down Newton's second law and separate the variables by rewriting it as m dv/F(v) = dt. Now integrate both sides of this equation and show that t = m fvo F(v). dv' Provided you can do the integral, this gives t as a function of v. You can then solve to give v as a function of t. Use this method to solve the special case that F(v) = Fo, a constant, and comment on your result. This method of separation of variables is used again in Problems 2.8 and 2.9. 2.8 * A mass m has velocity vo at time t = 0 and coasts along the x axis in a medium where the drag force is F(v) = \u2014cv 3/2. Use the method of Problem 2.7 to find v in terms of the time t and the other given parameters. At what time (if any) will it come to rest? 74 Chapter 2 Projectiles and Charged Particles 2.9 * We solved the differential equation (2.29), mi,y = \u2014b(vy \u2014 vmr), for the velocity of an object falling through air, by inspection \u2014 a most respectable way of solving differential equations. Never- theless, one would sometimes like a more systematic method, and here is one. Rewrite the equation in the \"separated\" form m dvY V y Vter \u2014b dt and integrate both sides from time 0 to t to find vy as a function of t. Compare with (2.30). 2.10 ** For a steel ball bearing (diameter 2 mm and density 7.8 g/cm 3) dropped in glycerin (density 1.3 g/cm3 and viscosity 12 Ns/m 2 at STP), the dominant drag", " force is the linear drag given by (2.82) of Problem 2.2. (a) Find the characteristic time t and the terminal speed v t\u201e. [In finding the latter, you should include the buoyant force of Archimedes. This just adds a third force on the right side of Equation (2.25).] How long after it is dropped from rest will the ball bearing have reached 95% of its terminal speed? (b) Use (2.82) and (2.84) (with K = 1/4 since the ball bearing is a sphere) to compute the ratio Jequadi, at the terminal speed. Was it a good approximation to neglect f J quad? 2.11 ** Consider an object that is thrown vertically up with initial speed v o in a linear medium. (a) Measuring y upward from the point of release, write expressions for the object's velocity v y (t) and position y(t). (b) Find the time for the object to reach its highest point and its position y max at that point. (c) Show that as the drag coefficient approaches zero, your last answer reduces to the well-known result ymax = zvo2 /g for an object in the vacuum. [Hint: If the drag force is very small, the terminal speed is very big, so v o/vm, is very small. Use the Taylor series for the log function to approximate ln(1 3) by 3 \u2014 Z 3 2. (For a little more on Taylor series see Problem 2.18.)] 2.12 ** Problem 2.7 is about a class of one-dimensional problems that can always be reduced to doing an integral. Here is another. Show that if the net force on a one-dimensional particle depends only on position, F = F(x), then Newton's second law can be solved to find v as a function of x given by [Hint: Use the chain rule to prove the following handy relation, which we could call the \"v dv I dx rule\": If you regard v as a function of x, then 2 V = V + \u2014 2 x F(x') dx'. 2 0 f x, = v \u2014 = dv dx 1 dv 2 2 dx (2.85) (2.86) Use this to rewrite Newton's second law in the separated form m d(v 2) = 2F (x) dx and then integrate from x o to x.] Comment on your result for the case that F(x) is actually a constant. (You may recognise your solution as a statement about kinetic energy and work, both of which we shall be discussing in Chapter 4.) 2.13 ** Consider a mass m constrained to move on the x axis and subject to a net force F = \u2014kx where k is a positive constant. The mass is released from rest at x = x0 at time t = 0. Use the result (2.85) in Problem 2.12 to find the mass's speed as a function of x; that is, dx I dt = g(x) for some function g(x). Separate this as dx I g(x) = dt and integrate from time 0 to t to find x as a function of t. (You may recognize this as one way \u2014 not the easiest \u2014 to solve the simple harmonic oscillator.) 2.14 *** Use the method of Problem 2.7 to solve the following: A mass m is constrained to move along the x axis subject to a force F (v) = \u2014Foe\" I V, where Fo and V are constants. (a) Find v(t) if the initial velocity is v o > 0 at time t = 0. (b) At what time does it come instantaneously to rest? (c) By integrating v(t), you can find x(t). Do this and find how far the mass travels before coming instantaneously to rest. Problems for Chapter 2 75 SECTION 2.3 Trajectory and Range in a Linear Medium 2.15 * Consider a projectile launched with velocity (v, o, vyo) from horizontal ground (with x measured horizontally and y vertically up). Assuming no air resistance, find how long the projectile is in the air and show that the distance it travels before landing (the horizontal range) is 2vxo v y \u201e/ g. 2.16 * A golfer hits his ball with speed v o at an angle 6 above the horizontal ground. Assuming that the angle 0 is fixed and that air resistance can be neglected, what is the minimum speed v o(min) for which the ball will clear a wall of height h, a distanced away? Your solution should get into trouble if the angle 0 is such that tan 0 < h 1 d. Explain. What is vo (min) if 6 = 25\u00b0, d = 50 m, and h = 2 m? 2.", "17 * The two equations (2.36) give a projectile's position (x, y) as a function of t. Eliminate t to give y as a function of x. Verify Equation (2.37). 2.18 * Taylor's theorem states that, for any reasonable function f (x), the value of f at a point (x 6) can be expressed as an infinite series involving f and its derivatives at the point x: f + 8) = f (x) + f'(x)8 + \u20141 f\u201d (x)8 2 + 1 r(x)6 3 \u00b1 (2.87) 2! 3 \u2014 where the primes denote successive derivatives of f (x). (Depending on the function this series may converge for any increment 3 or only for values of 8 less than some nonzero \"radius of convergence.\") This theorem is enormously useful, especially for small values of 3, when the first one or two terms of the series are often an excellent approximation.\" (a) Find the Taylor series for ln(1 + 8). (b) Do the same for cos 8. (c) Likewise sin 3. (d) And e8. 2.19 * Consider the projectile of Section 2.3. (a) Assuming there is no air resistance, write down the position (x, y) as a function of t, and eliminate t to give the trajectory y as a function of x. (b) The correct trajectory, including a linear drag force, is given by (2.37). Show that this reduces to your answer for part (a) when air resistance is switched off (r and v t, = gr both approach infinity). [Hint: Remember the Taylor series (2.40) for ln(1 \u2014 2.20 ** [Computer] Use suitable graph-plotting software to plot graphs of the trajectory (2.36) of a projectile thrown at 45\u00b0above the horizontal and subject to linear air resistance for four different values of the drag coefficient, ranging from a significant amount of drag down to no drag at all. Put all four trajectories on the same plot. [Hint: In the absence of any given numbers, you may as well choose convenient values. For example, why not take v xo = vyo = 1 and g = 1. (This amounts to choosing your units of length and time so that these parameters have the value 1.) With these choices, the strength of the drag is given by the one parameter v t\u201e = t, and you might choose to plot the trajectories for Vter = 0-351, 3, and oo (that is, no drag at all), and for times from t = 0 to 3. For the case that v ter = oo, you'll probably want to write out the trajectory separately.] 2.21 *** A gun can fire shells in any direction with the same speed v o. Ignoring air resistance and using cylindrical polar coordinates with the gun at the origin and z measured vertically up, show that 11 For more details on Taylor's series see, for example, Mary Boas, Mathematical Methods in the Physical Sci- ences (Wiley, 1983), p. 22 or Donald McQuarrie, Mathematical Methods for Scientists and Engineers (University Science Books, 2003), p. 94. 76 Chapter 2 Projectiles and Charged Particles the gun can hit any object inside the surface V 2 Z = g 2 p 2 2v 0 Describe this surface and comment on its dimensions. 2.22 *** [Computer] The equation (2.39) for the range of a projectile in a linear medium cannot be solved analytically in terms of elementary functions. If you put in numbers for the several parameters, then it can be solved numerically using any of several software packages such as Mathematica, Maple, and MatLab. To practice this, do the following: Consider a projectile launched at angle 0 above the horizontal ground with initial speed v 0 in a linear medium. Choose units such that v o = 1 and g = 1. Suppose also that the terminal speed vter = 1. (With v o = vter, air resistance should be fairly important.) We know that in a vacuum, the maximum range occurs at 0 = r/4 0.75. (a) What is the maximum range in a vacuum? (b) Now solve (2.39) for the range in the given medium at the same angle 0 = 0.75. (c) Once you have your calculation working, repeat it for some selection of values of 0 within which the maximum range probably lies. (You could try 0 = 0.4, 0.5,, 0.8.) (d) Based on these results, choose a smaller interval for 0 where you're sure the maximum lies and repeat the process. Repeat it again if necessary until you know the maximum range and the corresponding angle to two significant figures. Compare with", " the vacuum values. SECTION 2.4 Quadratic Air Resistance 2.23 * Find the terminal speeds in air of (a) a steel ball bearing of diameter 3 mm, (b) a 16-pound steel shot, and (c) a 200-pound parachutist in free fall in the fetal position. In all three cases, you can safely assume the drag force is purely quadratic. The density of steel is about 8 g/cm 3 and you can treat the parachutist as a sphere of density 1 g/cm 3. 2.24 * Consider a sphere (diameter D, density p sph) falling through air (density Q air ) and assume that the drag force is purely quadratic. (a) Use Equation (2.84) from Problem 2.4 (with K = 1/4 for a sphere) to show that the terminal speed is vter = 8 \u2014 3 D g Osph Qair (2.88) (b) Use this result to show that of two spheres of the same size, the denser one will eventually fall faster. (c) For two spheres of the same material, show that the larger will eventually fall faster. 2.25 * Consider the cyclist of Section 2.4, coasting to a halt under the influence of a quadratic drag force. Derive in detail the results (2.49) and (2.51) for her velocity and position, and verify that the constant r = m/cv o is indeed a time. 2.26 * A typical value for the coefficient of quadratic air resistance on a cyclist is around c = 0.20 N/(m/s) 2. Assuming that the total mass (cyclist plus cycle) is m = 80 kg and that at t = 0 the cyclist has an initial speed v o = 20 m/s (about 45 mi/h) and starts to coast to a stop under the influence of air resistance, find the characteristic time r = m/cv o. How long will it take him to slow to 15 m/s? What about 10 m/s? And 5 m/s? (Below about 5 m/s, it is certainly not reasonable to ignore friction, so there is no point pursuing this calculation to lower speeds.) Problems for Chapter 2 77 2.27 * I kick a puck of mass m up an incline (angle of slope = 0) with initial speed v o. There is no friction between the puck and the incline, but there is air resistance with magnitude f (v) = cv 2. Write down and solve Newton's second law for the puck's velocity as a function of t on the upward journey. How long does the upward journey last? 2.28 * A mass m has speed vo at the origin and coasts along the x axis in a medium where the drag force is F(v) = \u2014cv 312. Use the \"v dv/dx rule\" (2.86) in Problem 2.12 to write the equation of motion in the separated form m v dvl F (v) = dx, and then integrate both sides to give x in terms of v (or vice versa). Show that it will eventually travel a distance 2m lv o/c. 2.29 * The terminal speed of a 70-kg skydiver in spread-eagle position is around 50 m/s (about 115 mi/h). Find his speed at times t = 1, 5, 10, 20, 30 seconds after he jumps from a stationary balloon. Compare with the corresponding speeds if there were no air resistance. 2.30 * Suppose we wish to approximate the skydiver of Problem 2.29 as a sphere (not a very promising approximation, but nevertheless the kind of approximation physicists sometimes like to make). Given the mass and terminal speed, what should we use for the diameter of the sphere? Does your answer seem reasonable? 2.31 ** A basketball has mass m = 600 g and diameter D = 24 cm. (a) What is its terminal speed? (b) If it is dropped from a 30-m tower, how long does it take to hit the ground and how fast is it going when it does so? Compare with the corresponding numbers in a vacuum. 2.32 ** Consider the following statement: If at all times during a projectile's flight its speed is much less than the terminal speed, the effects of air resistance are usually very small. (a) Without reference to the explicit equations for the magnitude of v t\u201e, explain clearly why this is so. (b) By examining the explicit formulas (2.26) and (2.53) explain why the statement above is even more useful for the case of quadratic drag than for the linear case. [Hint: Express the ratio f 1 mg of the drag to the weight in terms of the ratio v /v ter.] 2.33 ** The hyperbolic functions cosh", " z and sinh z are defined as follows: cosh z = and sinh z = ez e' 2 ez \u2014 e \u2014 z 2 for any z, real or complex. (a) Sketch the behavior of both functions over a suitable range of real values of z. (b) Show that cosh z = cos(iz). What is the corresponding relation for sinh z? (c) What are the derivatives of cosh z and sinh z? What about their integrals? (d) Show that cosh2 z \u2014 sinh2 z = 1. (e) Show that f dx 1,\\/1\u00b1 x 2 = arcsinh x. [Hint: One way to do this is to make the substitution x = sinh z.] 2.34 ** The hyperbolic function tanh z is defined as tanh z = sinh z/ cosh z, with cosh z and sinh z defined as in Problem 2.33. (a) Prove that tanh z = tan(iz). (b) What is the derivative of tanh z? (c) Show that f dz tanh z = In cosh z. (d) Prove that 1 \u2014 tanh 2 z = sech2z, where sech z = 1/ cosh z. (e) Show that f dx/(1 \u2014 x 2) = arctanh x. 2.35 ** (a) Fill in the details of the arguments leading from the equation of motion (2.52) to Equations (2.57) and (2.58) for the velocity and position of a dropped object subject to quadratic air resistance. Be sure to do the two integrals involved. (The results of Problem 2.34 will help.) (b) Tidy the two equations by introducing the parameter r = vt\u201eIg Show that when t = r, v has reached 76% of its terminal value. What are the corresponding percentages when t = 2r and 3z? (c) Show that when t >> r, the position is approximately y v t ter const. [Hint: The definition of cosh x (Problem 2.33) - 78 Chapter 2 Projectiles and Charged Particles gives you a simple approximation when x is large.] (d) Show that for t small, Equation (2.58) for the position gives y ti Igt 2. [Use the Taylor series for cosh x and for ln(1 + 2.36 ** Consider the following quote from Galileo's Dialogues Concerning Two New Sciences: Aristotle says that \"an iron ball of 100 pounds falling from a height of one hundred cubits reaches the ground before a one-pound ball has fallen a single cubit.\" I say that they arrive at the same time. You find, on making the experiment, that the larger outstrips the smaller by two finger-breadths, that is, when the larger has reached the ground, the other is short of it by two finger-breadths. We know that the statement attributed to Aristotle is totally wrong, but just how close is Galileo's claim that the difference is just \"two finger breadths\"? (a) Given that the density of iron is about 8 g/cm 3, find the terminal speeds of the two iron balls. (b) Given that a cubit is about 2 feet, use Equation (2.58) to find the time for the heavier ball to land and then the position of the lighter ball at that time. How far apart are they? 2.37 ** The result (2.57) for the velocity of a falling object was found by integrating Equation (2.55) and the quickest way to do this is to use the integral f du I (1 \u2014 u 2) = arctanh u. Here is another way to do it: Integrate (2.55) using the method of \"partial fractions,\" writing 1 _ 1C1 \u2014 u which lets you do the integral in terms of natural logs. Solve the resulting equation to give v as a function of t and show that your answer agrees with (2.57). 2.38 ** A projectile that is subject to quadratic air resistance is thrown vertically up with initial speed vo. (a) Write down the equation of motion for the upward motion and solve it to give v as a function of t. (b) Show that the time to reach the top of the trajectory is trop = (v terl g) arctan (vo / vt\u201e). (c) For the baseball of Example 2.5 (with titer = 35 m/s), find t rop for the cases that v o = 1, 10, 20, 30, and 40 m/s, and compare with the corresponding values in a vacuum. 2.39 ** When a cyclist coasts to a stop, he is actually subject to", " two forces, the quadratic force of air resistance, f = \u2014cv 2 (with c as given in Problem 2.26), and a constant frictional force ft., of about 3 N. The former is dominant at high and medium speeds, the latter at low speed. (The frictional force is a combination of ordinary friction in the bearings and rolling friction of the tires on the road.) (a) Write down the equation of motion while the cyclist is coasting to a stop. Solve it by separating variables to give t as a function of v. (b) Using the numbers of Problem 2.26 (and the value f f, = 3 N given above) find how long it takes the cyclist to slow from his initial 20 m/s to 15 m/s. How long to slow to 10 and 5 m/s? How long to come to a full stop? If you did Problem 2.26, compare with the answers you got there ignoring friction entirely. 2.40 ** Consider an object that is coasting horizontally (positive x direction) subject to a drag force f = \u2014by \u2014 cv 2. Write down Newton's second law for this object and solve for v by separating variables. Sketch the behavior of v as a function of t. Explain the time dependence for t large. (Which force term is dominant when t is large?) 2.41 ** A baseball is thrown vertically up with speed v o and is subject to a quadratic drag with magnitude f (v) = cv 2. Write down the equation of motion for the upward journey (measuring y vertically up) and show that it can be rewritten as i) = \u2014g[l + (v/vt\u201e) 2 ]. Use the \"v dv dx rule\" (2.86) to write I) as v dvldy, and then solve the equation of motion by separating variables (put all terms involving v on one side and all terms involving y on the other). Integrate both sides to give y in terms of v, and hence v as a function of y. Show that the baseball's maximum height is Problems for Chapter 2 79 2 V ter in v 2 2 ter + vo Ymax = 2g 2 V ter (2.89) If vo = 20 m/s (about 45 mph) and the baseball has the parameters given in Example 2.5 (page 61), what is ym\u201e? Compare with the value in a vacuum. 2.42 ** Consider again the baseball of Problem 2.41 and write down the equation of motion for the downward journey. (Notice that with a quadratic drag the downward equation is different from the upward one, and has to be treated separately.) Find v as a function of y and, given that the downward journey starts at y m\u201e as given in (2.89), show that the speed when the ball returns to the ground is vt\u201evoilvL vo2. Discuss this result for the cases of very much and very little air resistance. What is the numerical value of this speed for the baseball of Problem 2.41? Compare with the value in a vacuum. 2.43 *** [Computer] The basketball of Problem 2.31 is thrown from a height of 2 m with initial velocity vo = 15 m/s at 45\u00b0 above the horizontal. (a) Use appropriate software to solve the equations of motion (2.61) for the ball's position (x, y) and plot the trajectory. Show the corresponding trajectory in the absence of air resistance. (b) Use your plot to find how far the ball travels in the horizontal direction before it hits the floor. Compare with the corresponding range in a vacuum. 2.44 *** [Computer] To get an accurate trajectory for a projectile one must often take account of several complications. For example, if a projectile goes very high then we have to allow for the reduction in air resistance as atmospheric density decreases. To illustrate this, consider an iron cannonball (diameter 15 cm, density 7.8 g/cm 3) that is fired with initial velocity 300 m/s at 50 degrees above the horizontal. The drag force is approximately quadratic, but since the drag is proportional to the atmospheric density and the density falls off exponentially with height, the drag force is f = c(y)v 2 where c(y) = y D2 exp(\u2014y/A) with y given by (2.6) and A ti 10, 000 m. (a) Write down the equations of motion for the cannonball and use appropriate software to solve numerically for x(t) and y(t) for 0 < t < 3.5 s. Plot the ball's trajectory and find its horizontal range. (b) Do the same calculation ignoring the variation of atmospheric density [that is, setting c(y) = c(0)], and yet again ignoring air resistance entirely. Plot all three trajectories for 0", " < t < 3.5 s on the same graph. You will find that in this case air resistance makes a huge difference and that the variation of air resistance makes a small, but not negligible, difference. SECTION 2.6 Complex Exponentials 2.45 * (a) Using Euler's relation (2.76), prove that any complex number z = x iy can be written in the form z = rei\u00b0, where r and 0 are real. Describe the significance of r and 0 with reference to the complex plane. (b) Write z = 3 + 4i in the form z = reie (c) Write z = 2e -i7113 in the form x iy. 2.46 * For any complex number z = x iy, the real and imaginary parts are defined as the real numbers Re(z) = x and Im(z) = y. The modulus or absolute value is lz I = \\/x 2 + y2 and the phase or angle is the value of 6 when z is expressed as z = re\u00b0. The complex conjugate is z* = x \u2014 iy. (This last is the notation used by most physicists; most mathematicians use -z-.) For each of the following complex numbers, find the real and imaginary parts, the modulus and phase, and the complex conjugate, 80 Chapter 2 Projectiles and Charged Particles and sketch z and z* in the complex plane: (a) z = 1 + i (c) z =,4e -in (b) z = 1 \u2014 (d) z 5eiwt. In part (d), w is a constant and t is the time. 2.47 * For each of the following two pairs of numbers compute z + w, z \u2014 w, zw, and z/w. (a) z = 6 + 8i and w = 3 \u2014 4i (b) z = 8e iTh/3 and w = 4e i'/ 6. Notice that for adding and subtracting complex numbers, the form x + iy is more convenient, but for multiplying and especially dividing, the form rei9 is more convenient. In part (a), a clever trick for finding z/w without converting to the form rei\u00b0 is to multiply top and bottom by w*; try this one both ways. 2.48 * Prove that lz I =,N/z*z for any complex number z. 2.49 * Consider the complex number z = e'9 = cos 8 + i sin 9. (a) By evaluating z 2 two different ways, prove the trig identities cos 26 = cos 2 9 \u2014 sin2 9 and sin 29 = 2 sin 8 cos 9. (b) Use the same technique to find corresponding identities for cos 39 and sin 39. 2.50 * Use the series definition (2.72) of ez to prove that 12 dez ldz = ez. 2.51 ** Use the series definition (2.72) of ez to prove that ezew = ez+w.. [Hint: If you write down the left side as a product of two series, you will have a huge sum of terms like zn Wm. If you group together all the terms for which n + m is the same (call it p) and use the binomial theorem, you will find you have the series for the right side.] SECTION 2.7 Solution for the Charge in a B Field 2.52 * The transverse velocity of the particle in Sections 2.5 and 2.7 is contained in (2.77), since + ivy. By taking the real and imaginary parts, find expressions for v x and vy separately. Based on these expressions describe the time dependence of the transverse velocity. 2.53 * A charged particle of mass m and positive charge q moves in uniform electric and magnetic fields, E and B, both pointing in the z direction. The net force on the particle is F = q(E + v x B). Write down the equation of motion for the particle and resolve it into its three components. Solve the equations and describe the particle's motion. 2.54 ** In Section 2.5 we solved the equations of motion (2.68) for the transverse velocity of a charge in a magnetic field by the trick of using the complex number ri = v, + i v y. As you might imagine, the equations can certainly be solved without this trick. Here is one way: (a) Differentiate the first of equations (2.68) with respect to t and use the second to give you a second-order differential equation for vx. This is an equation you should recognize [if not, look at Equation (1.55)] and you can write down its general solution. Once you know v x, (2.68) tells you v", " y. (b) Show that the general solution you get here is the same as the general solution contained in (2.77), as disentangled in Problem 2.52. 12 1f you are the type who worries about mathematical niceties, you may be wondering if it is permissible to differentiate an infinite series. Fortunately, in the case of a power series (such as this), there is a theorem that guarantees the series can be differentiated for any z inside the \"radius of convergence.\" Since the radius of convergence of the series for ez is infinite, we can differentiate it for any value of z. Problems for Chapter 2 81 2.55 *** A charged particle of mass m and positive charge q moves in uniform electric and magnetic fields, E pointing in the y direction and B in the z direction (an arrangement called \"crossed E and B fields\"). Suppose the particle is initially at the origin and is given a kick at time t = 0 along the x axis with vx = vxo (positive or negative). (a) Write down the equation of motion for the particle and resolve it into its three components. Show that the motion remains in the plane z = 0. (b) Prove that there is a unique value of vxo, called the drift speed v dr, for which the particle moves undeflected through the fields. (This is the basis of velocity selectors, which select particles traveling at one chosen speed from a beam with many different speeds.) (c) Solve the equations of motion to give the particle's velocity as a function of t, for arbitrary values of vx0. [Hint: The equations for (v x, vy ) should look very like Equations (2.68) except for an offset of v x by a constant. If you make a change of variables of the form U., = \u2014 vd, and u y = vy, the equations for (ux, u y ) will have exactly the form (2.68), whose general solution you know.] (d) Integrate the velocity to find the position as a function of t and sketch the trajectory for various values of vx.o. CHAPTER Momentum and Angular Momentum In this and the next chapter I shall describe the great conservation laws of momentum, angular momentum, and energy. These three laws are closely related to one another and are perhaps the most important of the small number of conservation laws that are regarded as cornerstones of all modern physics. Curiously, in classical mechanics the first two laws (momentum and angular momentum) are very different from the last (energy). It is a relatively easy matter to prove the first two from Newton's laws (indeed we already have proved conservation of momentum), whereas the proof of energy conservation is surprisingly subtle. I discuss momentum and angular momentum in this rather short chapter and energy in Chapter 4, which is appreciably longer. 3.1 Conservation of Momentum In Chapter 1 we examined a system of N particles labeled a = 1,, N. We found that as long as all the internal forces obey Newton's third law, the rate of change of the system's total linear momentum P = p i + p N = E pa is determined entirely by the external forces on the system: P = Fext (3.1) where Fext denotes the total external force on the system. Because of the third law, the internal forces all cancel out of the rate of change of the total momentum. In particular, if the system is isolated, so that the total external force is zero, we have the Principle of Conservation of Mo entum e net e t r e on de system is zero, the syst total t chani ornen urn P is cougar)t. 83 84 Chapter 3 Momentum and Angular Momentum If our system contains just one particle (N = 1), then all forces on the particle are external, and the conservation of momentum is reduced to the not very interesting statement that, in the absence of any forces, the momentum of a single particle is constant, which is just Newton's first law. However, if our system has two or more particles (N > 2), then momentum conservation is a nontrivial and often useful property, as the following simple and well-known example will remind you. EXAMPLE 3.1 An Inelastic Collision of Two Bodies Two bodies (two lumps of putty, for example, or two cars at an intersection) have masses m 1 and m 2 and velocities v 1 and v2. The two bodies collide and lock together, so they move off as a single unit, as shown in Figure 3.1. (A collision in which the bodies lock together like this is said to be perfectly inelastic.) Assuming that any external forces are negligible during the brief moment of collision, find the velocity v just after the collision. The initial total momentum, just before the collision, is = mivi m2v2 and the", " final momentum, just after the collision, is Pfin = m + m2v = (m 1 + m 2)v. (Notice that this last equation illustrates the useful result that, once two bodies have locked together, we can find their momentum by considering them as a single body of mass m 1 + m2.) By conservation of momentum these two momenta must be equal, Pfin = Pin, and we can easily solve to give the final velocity, M1V1 M2V2 V = m I -I- m2 (3.2) We see that the final velocity is just the weighted average of the original veloc- ities v 1 and v2, weighted by the corresponding masses m 1 and m 2. Figure 3.1 A perfectly inelastic collision between two lumps of putty. Section 3.2 Rockets 85 An important special case is when one of the bodies is initially at rest, as when a speeding car rams a stationary car at a stop light. With v 2 = 0, Equation (3.2) reduces to 1111 V = V1. m + m2 (3.3) In this case the final velocity is always in the same direction as v 1 but is reduced by the factor m i/ (m i + m 2). The result (3.3) is used by police investigating car crashes, since it lets them find the unknown velocity v 1 of a speeder who has rear-ended a stationary car, in terms of quantities that can be measured after the event. (The final velocity v can be found from the skid marks of the combined wreck.) This sort of analysis of collisions, using conservation of momentum, is an important tool in solving many problems ranging from nuclear reactions, through car crashes, to collisions of galaxies. 3.2 Rockets A beautiful example of the use of momentum conservation is the analysis of rocket propulsion. The basic problem that is solved by the rocket is this: With no external agent to push on or be pushed by, how does an object get itself moving? You can put yourself in the same difficulty by imagining yourself stranded on a perfectly frictionless frozen lake. The simplest way to get yourself to shore is to take off anything that is dispensible, such as a boot, and throw it as hard as possible away from the shore. By Newton's third law, when you push one way on the boot, the boot pushes in the opposite direction on you. Thus as you throw the boot, the reaction force of the boot on you will cause you to recoil in the opposite direction and then glide across the ice to shore. A rocket does essentially the same thing. Its motor is designed to hurl the spent fuel out of the back of the rocket, and by the third law, the fuel pushes the rocket forward. To analyse a rocket's motion quantitatively we must examine the total momentum. Consider the rocket shown in Figure 3.2 with mass m, traveling in the positive x direction (so I can abbreviate vx as just v) and ejecting spent fuel at the exhaust speed vex relative to the rocket. Since the rocket is ejecting mass, the rocket's mass m is steadily decreasing. At time t, the momentum is P(t) = my. A short time later at t dt, the rocket's mass is (m + dm), where dm is negative, and its momentum is (m + dm)(v dv). The fuel ejected in the time dt has mass (\u2014dm) and velocity Concerning the use of the small quantities like dt and dm, I recommend again the view that they are small but nonzero increments, with dt chosen sufficiently small that dm divided by dt is (within whatever we have chosen as our desired accuracy) equal to the derivative dm/dt. For more details, see the footnote immediately before Equation (2.47). 86 Chapter 3 Momentum and Angular Momentum ___,.... v Figure 3.2 A rocket of mass m travels to the right with speed v and ejects spent fuel with exhaust speed v ex rela- tive to the rocket. v \u2014 vex relative to the ground. Thus the total momentum (rocket plus the fuel just ejected) at t + dt is P(t + dt) = (m + dm)(v + dv) \u2014 dm(v \u2014 v ex ) = my + m dv + dm vex where I have neglected the doubly small product dm d v. Therefore, the change in total momentum is dP = P(t + dt) \u2014 P(t) = m dv + dm vex. (3.4) If there is a net external force F' (gravity, for instance), this change of momentum is F' di. (See Problem 3.11.) Here I shall assume that there are no external forces, so that P is constant and dP = 0. Therefore Dividing both sides", " by dt, we can rewrite this as m dv = \u2014dm v ex. Tr ) = \u2014 thvex (3.5) (3.6) where \u2014rh is the rate at which the rocket's engine is ejecting mass. This equation looks just like Newton's second law (mi) = F) for an ordinary particle, except that the product \u2014th vex on the right plays the role of the force. For this reason this product is often called the thrust: thrust = \u2014th Vex. (3.7) (Since rii is negative, this defines the thrust to be positive.) Equation (3.5) can be solved by separation of variables. Dividing both sides by m gives dm dv = \u2014 vex \u2014 If the exhaust speed v ex is constant, this equation can be integrated to give v \u2014 v o = vex ln(m o /m) (3.8) where vo is the initial velocity and m o is the initial mass of the rocket (including fuel and payload). This result puts a significant restriction on the maximum speed of the rocket. The ratio m o/m is largest when all the fuel is burned and m is just the mass Section 3.3 The Center of Mass 87 of rocket plus payload. Even if, for example, the original mass is 90% fuel, this ratio is only 10, and, since In 10 = 2.3, this says that the speed gained, v \u2014 v 0, cannot be more than 2.3 times v ex. This means that rocket engineers try to make v ex as big as possible and also design multistage rockets, which can jettison the heavy fuel tanks of the early stages to reduce the total mass for later stages. 2 3.3 The Center of Mass Several of the ideas of Section 3.1 can be rephrased in terms of the important notion of a system's center of mass. Let us consider a group of N particles, a = 1,, N, with masses m e, and positions ra measured relative to an origin O. The center of mass (or CM) of this system is defined to be the position (relative to the same origin 0) R= \u2014 mara = 1 N E a=1 t71 ir + m N r NM (3.9) where M denotes the total mass of all of the particles, M = E ma. The first thing to note about this definition is that it is a vector equation. The CM position is a vector R with three components (X, Y, Z), and Equation (3.9) is equivalent to three equations giving these three components, N X= E 77\u201e.\u201e,,, 1 M a=1 N 11 Y = \u2014 E mayor, m, a= zes. M a=1 Either way, the CM position R is a weighted average of the positions r 1,, rN, in which each position ra is weighted by the corresponding mass m a. (Equivalently, it is the sum of the ra, each multiplied by the fraction of the total mass at r a.) To get a feeling for the CM, it may help to consider the case of just two particles (N = 2). In this case, the definition (3.9) reads m R = m2r2 mi + m2 (3.10) It is easy to verify that the CM position has several familiar properties. For example, you can show (Problem 3.18) that the CM defined by (3.10) lies on the line joining the two particles, as shown in Figure 3.3. It is also easy to show that the distances of the CM from m 1 and m 2 are in the ratio m 2 / m i, so that the CM lies closer to the more massive particle. (In Figure 3.3 this ratio is 1/3.) In particular, if m 1 is much greater than m 2, the CM will be very close to r 1. More generally, going back to Equation (3.9) for the CM of N particles, we see that if m 1 is much greater than any of the other masses (as is the case for the sun as compared to all the planets), then m 1 ti M while ma \u00ab M for all other particles; this means that R is very close to r 1. Thus, for example, the CM of the solar system is very close to the sun. 2 Jettisoning the fuel tanks of stage 1 reduces the inital and final masses of stage 2 by the same amount. This increases the ratio m o/m when we apply (3.8) to stage 2. See Problem 3.12. 88 Chapter 3 Momentum and Angular Momentum Figure 3.3 The CM of two particles lies at the position (m i r i + m2r2)/M. You can prove that this lies on the line joining m 1 to m 2, as shown,", " and that the distances of the CM from m 1 and m 2 are in the ratio m 2/m 1. We can now write the total momentum P of any N-particle system in terms of the system's CM as follows: P = pa a a MR (3.11) where the last equality is just the derivative of the definition (3.9) of R (multiplied by M). This remarkable result says that the total momentum of the N particles is exactly the same as that of a single particle of mass M and velocity equal to that of the CM. We get an even more striking result when we differentiate (3.11). According to (3.1), the derivative of P is just F\"t. Therefore, (3.11) implies that Fext = MR. (3.12) That is, the center of mass R moves exactly as if it were a single particle of mass M, subject to the net external force on the system. This result is the main reason why we can often treat extended bodies, such as baseballs and planets, as if they were point particles. Provided a body is small compared to the scale of its trajectory, its CM position R is a good representative of its overall location, and (3.12) implies that R moves just like a point particle. Given the importance of the CM, you need to feel comfortable calculating the CM position for various systems. You may have had plenty of practice in introductory physics or in a calculus course, but, in case you didn't, there are several exercises at the end of this chapter. One important point to bear in mind is that when the mass in a body is distributed continuously, the sum in the definition (3.9) goes over to an integral 1 m f rdm= \u2014 1 f prdV R=\u2014 (3.13) where p is the mass density of the body, dV denotes an element of volume, and the integral runs over the whole body (that is, everywhere p 0). We shall be using similar integrals to evaluate the moment of inertia tensor in Chapter 10. Meanwhile, here is one example: Section 3.3 The Center of Mass 89 EXAMPLE 3.2 The CM of a Solid Cone Find the CM position for the uniform solid cone shown in Figure 3.4. It is perhaps obvious by symmetry that the CM lies on the axis of symmetry (the z axis), but this also follows immediately from the integral (3.13). For example, if you consider the x component of that integral, it is easy to see that the contribution from any point (x, y, z) is exactly cancelled by that from the point (\u2014x, y, z). That is, the integral for X is zero. Because the same argument applies to Y, the CM lies on the z axis. To find the height Z of the CM, we must evaluate the integral Z.- 1 f pzdV = 0 \u2014 f zdxdydz where I could take the factor p outside the integral since p is constant throughout the cone (as long as we understand the integral is limited to the inside of the cone) and I have changed the volume element dV to dx dy dz. For any given z, the integral over x and y runs over a circle of radius r = Rz I h, giving a factor of rrr2 R2z 2\" /12, so that = Z pyr R 2 fh Mh 2 0 z 3dz = pm R2 h4 3 h Mh 2 4 4 where in the last step I replaced the mass M by O times the volume or M = i pn- R 2h. We conclude that the CM is on the axis of the cone at a distance 4 h from the vertex (or ih from the base). Figure 3.4 A solid cone, centered on the z axis, with vertex at the origin and uniform mass density Q. Its height is h and its base has radius R. 90 Chapter 3 Momentum and Angular Momentum 3.4 Angular Momentum for a Single Particle In many ways the conservation of angular momentum parallels the conservation of ordinary (or \"linear\") momentum. Nevertheless, I would like to review the formalism in detail, first for a single particle and then for a multiparticle system. This will introduce several important ideas and some useful mathematics. The angular momentum \u00a3 of a single particle is defined as the vector r x p. (3.14) Here r x p is the vector product of the particle's position vector r, relative to the chosen origin 0, and its momentum p, as shown in Figure 3.5. Notice that because r depends on the choice of origin, the same is true of \u00a3: The angular momentum \u00a3 (unlike the linear momentum p) depends on the choice of origin, and we should, strictly speaking, refer to \u00a3 as the angular momentum relative to O. The time rate of change of is easily found", ": = \u2014 (r x p) =(r x p) (r x to). (3.15) d dt (You can easily check that the product rule can be used for differentiating vector products, as long as you are careful to keep the vectors in the right order. See Problem 1.17.) In the first term on the right, we can replace p by mi, and, because the cross product of any two parallel vectors is zero, the first term is zero. In the second term, we can replace p by the net force F on the particle, and we get (3.16) Here F (Greek capital gamma) denotes the net torque about 0 on the particle, defined as r x F. (Other popular symbols for torque are r and N.) In words, (3.16) says that the rate of change of a particle's angular momentum about the origin 0 is equal to the net applied torque about 0. Equation (3.16) is the rotational analog of the equation F for the linear momentum, and (3.16) is often described as the rotational form of Newton's second law. Figure 3.5 For any particle with position r relative to the origin O and momentum p, the angular momentum about 0 is defined as the vector t = r x p. For the case shown, t points into the page. Section 3.4 Angular Momentum for a Single Particle 91 Sun Figure 3.6 A planet (mass m) is subject to the central force of the sun (mass M). If we choose the origin at the sun, then r x F = 0, and the planet's angular momentum about 0 is constant. In many one-particle problems one can choose the origin 0 so that the net torque F (about the chosen 0) is zero. In this case, the particle's angular momentum about 0 is constant. Consider, for example, a single planet (or comet) orbiting the sun. The only force on the planet is the gravitational pull Gm M I r 2 of the sun, as shown in Figure 3.6. A crucial property of the gravitational force is that it is central, that is, directed along the line joining the two centers. This means that F is parallel (actually, antiparallel) to the position vector r measured from the sun, and hence that r x F = 0. Thus if we choose our origin at the sun, the planet's angular momentum about 0 is constant, a fact that greatly simplifies the analysis of planetary motion. For example, because r x p is constant, r and p must remain in a fixed plane; in other words, the planet's orbit is confined to a single plane containing the sun, and the problem is reduced to two dimensions, a result we shall exploit in Chapter 8. Kepler's Second Law One of the earliest triumphs for Newton's mechanics was that he was able to explain Kepler's second law as a simple consequence of conservation of angular momentum. Newton's laws of motion were published in 1687 in his famous book Principia. Nearly eighty years earlier, the German astronomer Johannes Kepler (1571-1630) had published his three laws of planetary motion. 3 These laws are quite different from Newton's laws in that they are simply mathematical descriptions of the observed motion of the planets. For example, the first law states that the planets move around the sun in ellipses with the sun at one focus. Kepler's laws make no attempt to explain planetary motion in terms of more fundamental ideas; they are just summaries \u2014 brilliant summaries, requiring great insight, but nonetheless just summaries \u2014 of the observed motions of the planets. All three of Kepler's laws turn out to be consequences of Newton's laws of motion. I shall derive the first and third of the Kepler laws in Chapter 8. The second we are ready to discuss now. 3 Kepler's first two laws appeared in his book Astronomia Nova in 1609 and the third in another book, Harmonices Mundi, published in 1619. 92 Chapter 3 Momentum and Angular Momentum Figure 3.7 The orbit of a planet with the sun fixed at O. Kepler's second law asserts that if the two pairs of points P. Q and 1)',Q' are separated by equal time intervals, dt = dt', then the two areas dA and dA' are equal. Kepler's second law is generally stated something like this: ach planet incases around the sun, cps c zt equ a n equal times. Keplers Second Law ne drawn rein i the planet to die s This rather curious statement is illustrated in Figure 3.7, which shows the path of a planet or comet \u2014 the law applies to comets as well \u2014 orbiting about the sun at the origin 0. (Throughout this discussion, I shall make the approximation that the sun is fixed; we shall see how to allow for the very small motion of the sun in Chapter 8.) The area \"swept", " out\" by the planet moving between any two points P and Q is just the area of the triangle 0 P Q. (Strictly speaking the \"triangle\" is the area between the two lines OP and 0 Q and the arc P Q. However, it is sufficient to consider pairs of points P and Q that are close together, in which case the difference between the arc P Q and the straight line P Q is negligible.) I shall denote the time elapsed between the planet's visiting P and Q by dt and the corresponding area of OP Q by dA. Kepler's second law asserts that if we choose any other pair of points P' and Q' separated by the same time interval (dt' = dt), then the area 0 P' Q' will be the same as 0 P Q, or dA' = dA. Equivalently we can divide both sides of this equality by dt and assert that the rate at which the planet sweeps out area, dAldt, is the same at all points on the orbit; that is, dA/dt is constant. To prove this result, we note first that the line OP is just the position vector r, and P Q is the displacement dr = v dt. Now, it is a well-known property of the vector Section 3.5 Angular Momentum for Several Particles 93 product that if two sides of a triangle are given by vectors a and b, then the area of the triangle is 4. 1a x bl. (See Problem 3.24.) Thus the area of the triangle OPQ is Replacing v by p/m and dividing both sides by dt, we find that dA = zIr x v dtl. dA dt 1 = 2m Ir x = 1 2m (3.17) where denotes the magnitude of the angular momentum e = r x p. Since the planet's angular momentum about the sun is conserved, this establishes that dA/dt is constant, which, as we have seen, is the content of Kepler's second law. An alternative proof of the same result adds some additional insight: It is a straightforward exercise to show that (Problem 3.27) = mr 2co (3.18) where w = is is the planet's angular velocity around the sun. And it is an equally simple geometrical exercise to show that the rate of sweeping out area is dA/dt = Zr 2w. (3.19) Comparison of (3.18) and (3.19) shows that is constant if and only if dA/dt is constant. That is, conservation of angular momentum is exactly equivalent to Kepler's second law. In addition, we see that as the planet (or comet) approaches closer to the sun (r decreasing) its angular velocity co necessarily increases. Specifically, co is inversely proportional to r2 ; for example, if the value of r at point P' is half that at P, then the angular velocity co at P' is four times that at P. It is interesting to note that our proof of Kepler's second law depended only on the fact that the gravitational force is central and hence that the planet's angular momentum about the sun is constant. Thus Kepler's second law is true for an object that moves under the influence of any central force. By contrast, we shall see in Chapter 8 that the first and third laws (in particular the first, which says that the orbits are ellipses with the sun at one focus) depend on the inverse-square nature of the gravitational force and are not true for other force laws. 3.5 Angular Momentum for Several Particles Let us next discuss a system of N particles, a = 1, 2,, N, each with its angular momentum lc, = ro, x pa (with all of the re, measured from the same origin 0, of course). We define the total angular momentum L as N N ta =>ra x pa L= ot=i ci=i (3.20) 94 Chapter 3 Momentum and Angular Momentum Differentiating with respect to t and using the result (3.16), we find that L=EQa = rc, x F, a a (3.21) where, as usual, Fa denotes the net force on particle a. This result shows that the rate of change of L is just the net torque on the whole system, an important result in its own right. However, my interest now is to separate the effects of the internal and external forces. As in Equation (1.25) we write Fa as (net force on particle a) = Fa = E Fas + raxt (3.22) 00ce where, as before, Fa18 denotes the force exerted on particle a by particle p, and FeaKt a by all agents outside our N-particle system. is the net force exerted on particle Substituting into (3.21), we find", " that = E ra x Fai3 + E ra x Faext (3.23) a 0Aa a Equation (3.23) corresponds to Equation (1.27) in our discussion of linear momen- tum back in Chapter 1, and we can rework it in much the same way as there, with one interesting additional twist. We can regroup the terms of the double sum, pairing each term aP with the corresponding term Pa, to give4 E E ra x Fa18 = E E(r a x Fa18 + r18 x Fsa). a /3 0a a 13>a If we assume that all the internal forces obey the third law (F as = \u2014Fsa ), then we can rewrite the sum on the right as (3.24) (3.25) E E(ra \u2014 rp ) x Fa18. a 13>a To understand this sum, we must examine the vector (ra \u2014 r18 ) = ra18, say. This is illustrated in Figure 3.8, where we see that ra18 is the vector pointing toward particle a from particle /3. If, in addition to satisfying the third law, the forces Fa18 are all central, then the two vectors ra18 and Fat point along the same line, and their cross product is zero. Returning to Equation (3.23), we conclude that, provided our various assumptions are valid, the double sum in (3.23) is zero. The remaining single sum is just the net external torque, and we conclude that L = (3.26) In particular, if the net external torque is zero, we have the 4 Be sure you understand what has happened here. For example, I have paired the term r 1 x F12 with the term r2 x F21. Section 3.5 Angular Momentum for Several Particles 95 Figure 3.8 The vector rah =(ra \u2014 rd points to particle a from particle /3. If the force F,,,6 is central (points along the line joining a and p), then rap and Fad are collinear and their cross product is zero. Principle of Conservation of Angular Momentum the net external torque on an N-particle system is /cm the system annular momentum i> =` Er x pa is constant, The validity of this principle depends on our two assumptions that all internal forces Foo are central and satisfy the third law. Since these assumptions are almost always valid, the principle (as stated) is likewise. It is of the greatest utility in solving many problems, as I shall illustrate shortly with a couple of simple examples. The Moment of Inertia Before discussing an example, it is worth noting that the calculation of angular momenta does not always require one to go back to the basic definition (3.20). As you probably recall from your introductory physics course, for a rigid body rotating about a fixed axis (for example, a wheel rotating on its fixed axle), the rather complicated sum (3.20) can be expressed in terms of the moment of inertia and the angular velocity of rotation. Specifically, if we take the axis of rotation to be the z axis, then L z, the z component of angular momentum, is just L z = Iw, where I is the moment of inertia of the body for the given axis, and w is the angular velocity of rotation. We shall prove and generalize this result in Chapter 10, or you can prove it yourself with the guidance of Problem 3.30. For now, I shall ask you to carry it over from introductory physics. In particular, as you may recall, the moments of inertia of various standard bodies are known. For example, for a uniform disk (mass M, radius R) rotating about its axis, I = 1M R 2. For a uniform solid sphere rotating about a diameter, I = s M R 2. In general, for any multiparticle system, I = E m apa2, where pa is the distance of the mass m a from the axis of rotation. 96 Chapter 3 Momentum and Angular Momentum EXAMPLE 3.3 Collision of a Lump of Putty with a Turntable A uniform circular turntable (mass M, radius R, center 0) is at rest in the xy plane and is mounted on a frictionless axle, which lies along the vertical z axis. I throw a lump of putty (mass m) with speed v toward the edge of the turntable, so it approaches along a line that passes within a distance s b of 0, as shown in Figure 3.9. When the putty hits the turntable, it sticks to the edge, and the two rotate together with angular velocity w. Find co. This problem is easily solved using conservation of angular momentum. Because the turntable is mounted on a frictionless axle, there is no torque on the table in the z direction. Therefore the z component of the external torque on", " the system is zero, and 1,, is conserved. (This is true even if we include gravity, which acts in the z direction and contributes nothing to the torque in the z direction.) Before the collision, the turntable has zero angular momentum, while the putty has t = r x p, which points in the z direction. Thus the initial total angular momentum has z component L in = tz r(mv) sin 0 = mvb. After the collision, the putty and turntable rotate together about the z axis with total moment of inertia 6 I = (m M I2)R 2, and the z component of the final angular momentum is L zfin = ICO. Therefore, conservation of angular momentum in the form L 1 = Lfin tells us that z z mvb = (m M/2)R 2w, Figure 3.9 A lump of putty of mass m is thrown with velocity v at a stationary turntable. The putty's line of approach passes within the distance b of the table's center 0. 5 In collision theory \u2014 the theory of collisions, usually between atomic or subatomic particles\u2014 the distance b is called the impact parameter. 6 This is m R 2 for the putty stuck at radius R plus 4 mR 2 for the uniform turntable. Section 3.5 Angular Momentum for Several Particles 97 or, solving for co, = m vb (rn + M12) R2. (3.27) This answer is not especially interesting. What is interesting is that we were able to find it with so comparatively little effort. This is typical of the conservation laws, that they can answer many questions so simply. The kind of analysis used here can be used in many situations (such as nuclear reactions) where an incident projectile is absorbed by a stationary target and its angular momentum is shared between the two bodies. Angular Momentum about the CM The conservation of angular momentum and the more general result (3.26), L = rext, were derived on the assumption that all quantities were measured in an inertial frame, so that Newton's second law could be invoked. This required that both L and Fe' be measured about an origin 0 fixed in some inertial frame. Remarkably, the same two results also hold if L and feat are measured about the center of mass \u2014 even if the CM is being accelerated and so is not fixed in an inertial frame. That is, \u2014L(about CM) = T\"t (about CM) (3.28) dt and hence, if fex t (about CM) = 0, then L(about CM) is conserved. We shall prove this result in Chapter 10, or you can prove it yourself with the guidance of Problem 3.37. I mention it now, because it allows a very simple solution to various problems, as the following example illustrates. EXAMPLE 3.4 A Sliding and Spinning Dumbbell A dumbbell consisting of two equal masses m mounted on the ends of a rigid massless rod of length 2b is at rest on a frictionless horizontal table, lying on the x axis and centered on the origin, as shown in Figure 3.10. At time t = 0, the left mass is given a sharp tap, in the shape of a horizontal force F in the y direction, lasting for a short time At. Describe the subsequent motion. There are actually two parts to this problem: We must find the initial motion immediately after the impulse, and then the subsequent, force-free motion. The initial motion is not hard to guess, but let us derive it using the tools of this chapter. The only external force is the force F acting in the y direction for the brief time At. Since P F= ext, the total momentum just after the impulse is P = F At. Since P = Mit. (with M = 2m), we conclude that the CM starts moving directly up the y axis with velocity = = F At/2m. While the force F is acting, there is a torque Fex t = Fb about the CM, and so, according to (3.28), the initial angular momentum (just after the impulse has 98 Chapter 3 Momentum and Angular Momentum X Figure 3.10 The left mass of the dumbbell is given a sharp tap in the y direction. ceased) is L = Fb At. Since L = ho, with I = 2mb2, we conclude that the dumbbell is spinning clockwise, with initial angular velocity w= F At I2mb. The clockwise rotation of the dumbbell means that the left mass is moving up relative to the CM with speed cob, and its total initial velocity is vieft= v crn + cob = F At /m. By the same token the right mass is moving down relative to the CM, and its total initial velocity is Vright = Van COb = 0. That is,", " the right mass is initially stationary, while the left one carries all the momentum F At of the system. The subsequent motion is very straightforward. Once the impulse has ceased, there are no external forces or torques. Thus the CM continues to move straight up the y axis with constant speed, and the dumbbell continues to rotate with constant angular momentum about the CM and hence constant angular velocity. Principal Definitions and Equations of Chapter 3 Equation of Motion for a Rocket = _thvex Fext. [Eqs. (3.6) & (3.29)] The Center of Mass of Several Particles R = m ares = m + m N rN M [Eq. (3.9)] where M is the total mass of all particles, M = E m a. Problems for Chapter 3 99 Angular Momentum angular momentum about 0 is For a single particle with position r (relative to an origin 0) and momentum p, the = r x p. [Eq. (3.14)] For several particles, the total angular momentum is Provided all the internal forces are central, L.E.ea = ro, x p, a=1 a=1 [Eq. (3.20)] = rext L [Eq. (3.26)] where Text is the net external torque. Problems for Chapter 3 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). SECTION 3.1 Conservation of Momentum 3.1 * Consider a gun of mass M (when unloaded) that fires a shell of mass m with muzzle speed v. (That is, the shell's speed relative to the gun is v.) Assuming that the gun is completely free to recoil (no external forces on gun or shell), use conservation of momentum to show that the shell's speed relative to the ground is v /(1 m/M). 3.2 * A shell traveling with speed v o exactly horizontally and due north explodes into two equal-mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed vo. What is the velocity of the other fragment? 3.3 * A shell traveling with velocity vo explodes into three pieces of equal masses. Just after the explosion, one piece has velocity v 1 = vo and the other two have velocities v2 and v3 that are equal in magnitude (v2 = v3) but mutually perpendicular. Find v2 and v3 and sketch the three velocities. 3.4 ** Two hobos, each of mass m h, are standing at one end of a stationary railroad flatcar with frictionless wheels and mass m t-c. Either hobo can run to the other end of the flatcar and jump off with the same speed u (relative to the car). (a) Use conservation of momentum to find the speed of the recoiling car if the two men run and jump simultaneously. (b) What is it if the second man starts running only after the first has already jumped? Which procedure gives the greater speed to the car? [Hint: The speed u is the speed of either hobo, relative to the car just after he has jumped; it has the same value for either man and is the same in parts (a) and (b).] 3.5 ** Many applications of conservation of momentum involve conservation of energy as well, and we haven't yet begun our discussion of energy. Nevertheless, you know enough about energy from your introductory physics course to handle some problems of this type. Here is one elegant example: An elastic collision between two bodies is defined as a collision in which the total kinetic energy of the two bodies after the collision is the same as that before. (A familiar example is the collision 100 Chapter 3 Momentum and Angular Momentum between two billiard balls, which generally lose extremely little of their total kinetic energy.) Consider an elastic collision between two equal mass bodies, one of which is initially at rest. Let their velocities be v i and v2 = 0 before the collision, and v'1 and v; after. Write down the vector equation representing conservation of momentum and the scalar equation which expresses that the collision is elastic. Use these to prove that the angle between vi and v; is 90\u00b0. This result was important in the history of atomic and nuclear physics: That two bodies emerged from a collision traveling on perpendicular paths was strongly suggestive that they had equal mass and had undergone an elastic collision. SECTION 3.2 Rockets 3.6 * In the early stages of the Saturn V rocket's launch, mass was ejected at about 15,000 kg/s, with a speed vex ti 2500 m/s relative to the rocket. What was the thrust on the rocket? Convert this to tons (1 ton ti 9000 newtons) and compare with the rocket's initial weight (about 3000 tons). 3.7 * The first couple of minutes of the launch of a space shuttle can be described very roughly as follows: The initial mass is", " 2 x 10 6 kg, the final mass (after 2 minutes) is about 1 x 10 6 kg, the average exhaust speed v ex is about 3000 m/s, and the initial velocity is, of course, zero. If all this were taking place in outer space, with negligible gravity, what would be the shuttle's speed at the end of this stage? What is the thrust during the same period and how does it compare with the initial total weight of the shuttle (on earth)? 3.8 * A rocket (initial mass m 0) needs to use its engines to hover stationary, just above the ground. (a) If it can afford to burn no more than a mass Am o of its fuel, for how long can it hover? [Hint: Write down the condition that the thrust just balance the force of gravity. You can integrate the resulting equation by separating the variables t and m. Take v\u201e to be constant.] (b) If vex ti 3000 m/s and A 10%, for how long could the rocket hover just above the earth's surface? 3.9 * From the data in Problem 3.7 you can find the space shuttle's initial mass and the rate of ejecting mass \u2014tic (which you may assume is constant). What is the minimum exhaust speed v ex for which the shuttle would just begin to lift as soon as burn is fully underway? [Hint: The thrust must at least balance the shuttle's weight.] of m is p maximum? 3.10 * Consider a rocket (initial mass m o) accelerating from rest in free space. At first, as it speeds up, its momentum p increases, but as its mass m decreases p eventually begins to decrease. For what value 3.11 ** (a) Consider a rocket traveling in a straight line subject to an external force F\" t acting along the same line. Show that the equation of motion is [Review the derivation of Equation (3.6) but keep the external force term.] (b) Specialize to the case of a rocket taking off vertically (from rest) in a gravitational field g, so the equation of motion becomes m U = vex + Fext. my = \u2014 Mv ex \u2014 mg. (3.29) (3.30) Assume that the rocket ejects mass at a constant rate, tit = \u2014 k (where k is a positive constant), so that m = mo \u2014 kt. Solve equation (3.30) for v as a function of t, using separation of variables (that is, rewriting the equation so that all terms involving v are on the left and all terms involving t on the right). (c) Using the rough data from Problem 3.7, find the space shuttle's speed two minutes into flight, assuming (what is nearly true) that it travels vertically up during this period and that g doesn't change appreciably. Compare with the corresponding result if there were no gravity. (d) Describe what would Problems for Chapter 3 101 happen to a rocket that was designed so that the first term on the right of Equation (3.30) was smaller than the initial value of the second. 3.12 ** To illustrate the use of a multistage rocket consider the following: (a) A certain rocket carries 60% of its initial mass as fuel. (That is, the mass of fuel is 0.6m 0.) What is the rocket's final speed, accelerating from rest in free space, if it burns all its fuel in a single stage? Express your answer as a multiple of v ex. (b) Suppose instead it burns the fuel in two stages as follows: In the first stage it burns a mass 0.3m o of fuel. It then jettisons the first-stage fuel tank, which has a mass of 0.1m o, and then burns the remaining 0.3m 0 of fuel. Find the final speed in this case, assuming the same value of v ex throughout, and compare. 3.13 ** If you have not already done it, do Problem 3.11(b) and find the speed v(t) of a rocket accelerating vertically from rest in a gravitational field g. Now integrate v(t) and show that the rocket's height as a function of t is y (t) = V ex t 1 2 gt 2 MP ex in ( M / m k Using the numbers given in Problem 3.7, estimate the space shuttle's height after two minutes. 3.14 ** Consider a rocket subject to a linear resistive force, f = \u2014by, but no other external forces. Use Equation (3.29) in Problem 3.11 to show that if the rocket starts from rest and ejects mass at a constant rate k = \u2014th, then its speed is given by k b v vex[l ( m 1. mo b I k SECTION 3.3 The Center of Mass 3.15 * Find the position of the center of mass", " of three particles lying in the xy plane at r 1 = (1, 1, 0), r2 = (1, \u20141, 0), and r 3 = (0, 0, 0), if m 1 = m 2 and m 3 = 10m 1. Illustrate your answer with a sketch and comment. 3.16 * The masses of the earth and sun are Me = 6.0 x 1024 and Ms ti 2.0 x 1030 (both in kg) and their center-to-center distance is 1.5 x 10 8 km. Find the position of their CM and comment. (The radius of the sun is R s 7.0 x 105 km.) 3.17 * The masses of the earth and moon are M,==, 6.0 x 1024 and Mm 7.4 x 1022 (both in kg) and their center to center distance is 3.8 x 10 5 km. Find the position of their CM and comment. (The radius of the earth is Re ti 6.4 x 103 km.) 3.18 ** (a) Prove that the CM of any two particles always lies on the line joining them, as illustrated in Figure 3.3. [Write down the vector that points from m 1 to the CM and show that it has the same direction as the vector from m l to m 2.] (b) Prove that the distances from the CM to m 1 and m 2 are in the ratio m 2/m l. Explain why if m 1 is much greater than m 2, the CM lies very close to the position of m 1. 3.19 ** (a) We know that the path of a projectile thrown from the ground is a parabola (if we ignore air resistance). In the light of the result (3.12), what would be the subsequent path of the CM of the pieces if the projectile exploded in midair? (b) A shell is fired from level ground so as to hit a target 100 m away. Unluckily the shell explodes prematurely and breaks into two equal pieces. The two pieces land 102 Chapter 3 Momentum and Angular Momentum at the same time, and one lands 100 m beyond the target. Where does the other piece land? (c) Is the same result true if they land at different times (with one piece still landing 100 m beyond the target)? 3.20 ** Consider a system comprising two extended bodies, which have masses M 1 and M2 and centers of mass at R 1 and R2. Prove that the CM of the whole system is at R= MiRI M2R2 + M2 This beautiful result means that in finding the CM of a complicated system, you can treat its component parts just like point masses positioned at their separate centers of mass \u2014 even when the component parts are themselves extended bodies. 3.21 ** A uniform thin sheet of metal is cut in the shape of a semicircle of radius R and lies in the xy plane with its center at the origin and diameter lying along the x axis. Find the position of the CM using polar coordinates. [In this case the sum (3.9) that defines the CM position becomes a two-dimensional integral of the form f r a dA where a denotes the surface mass density (mass/area) of the sheet and dA is the element of area dA = r dr d0.] 3.22 ** Use spherical polar coordinates r, 9, 4) to find the CM of a uniform solid hemisphere of radius R, whose flat face lies in the xy plane with its center at the origin. Before you do this, you will need to convince yourself that the element of volume in spherical polars is dV = r2dr sin 9 d0 dO. (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.) 3.23 *** [Computer] A grenade is thrown with initial velocity v o from the origin at the top of a high cliff, subject to negligible air resistance. (a) Using a suitable plotting program, plot the orbit, with the following parameters: v. = (4, 4), g = 1, and 0 < t < 4 (and with x measured horizontally and y vertically up). Add to your plot suitable marks (dots or crosses, for example) to show the positions of the grenade at t = 1, 2, 3, 4. (b) At t = 4, when the grenade's velocity is v, it explodes into two equal pieces, one of which moves off with velocity v -I- Av. What is the velocity of the other piece? (c) Assuming that Av = (1, 3), add to your original plot the paths of the two pieces for 4 < t < 9. Insert marks to show their positions at t = 5, 6, 7, 8, 9. Find some way to show clearly that the CM", " of the two pieces continues to follow the original parabolic path. SECTION 3.4 Angular Momentum for a Single Particle 3.24 * If the vectors a and b form two of the sides of a triangle, prove that z a x b I is equal to the area of the triangle. 3.25 * A particle of mass m is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially the particle is moving in a circle of radius ro with angular velocity wo, but I now pull the string down through the hole until a length r remains between the hole and the particle. What is the particle's angular velocity now? 3.26 * A particle moves under the influence of a central force directed toward a fixed origin 0. (a) Explain why the particle's angular momentum about 0 is constant. (b) Give in detail the argument that the particle's orbit must lie in a single plane containing 0. 3.27 ** Consider a planet orbiting the fixed sun. Take the plane of the planet's orbit to be the xy plane, with the sun at the origin, and label the planet's position by polar coordinates (r, 4)). (a) Show that the planet's angular momentum has magnitude = mr2w, where co = 4 is the planet's angular velocity about the sun. (b) Show that the rate at which the planet \"sweeps out area\" (as in Kepler's second law) is d A Idt = 4r2w, and hence that d A Idt = \u00a3/2m. Deduce Kepler's second law. Problems for Chapter 3 103 SECTION 3.5 Angular Momentum for Several Particles 3.28 * For a system of just three particles, go through in detail the argument leading from (3.20) to (3.26), L = rext, writing out all the summations explicitly. 3.29 * A uniform spherical asteroid of radius R o is spinning with angular velocity co o. As the aeons go by, it picks up more matter until its radius is R. Assuming that its density remains the same and that the additional matter was originally at rest relative to the asteroid (anyway on average), find the asteroid's new angular velocity. (You know from elementary physics that the moment of inertia is s MR 2.) What is the final angular velocity if the radius doubles? 3.30 ** Consider a rigid body rotating with angular velocity co about a fixed axis. (You could think of a door rotating about the axis defined by its hinges.) Take the axis of rotation to be the z axis and use cylindrical polar coordinates poi, Oa, z cy to specify the positions of the particles a = 1,, N that make up the body. (a) Show that the velocity of the particle a is paw in the 0 direction. (b) Hence show that the z component of the angular momentum f a of particle a is m apa2w. (c) Show that the z component L z of the total angular momentum can be written as L z = I co where I is the moment of inertia (for the axis in question), N,L ma pa2. a=1 (3.31) 3.31 ** Find the moment of inertia of a uniform disc of mass M and radius R rotating about its axis, by replacing the sum (3.31) by the appropriate integral and doing the integral in polar coordinates. 3.32 ** Show that the moment of inertia of a uniform solid sphere rotating about a diameter is s MR 2. The sum (3.31) must be replaced by an integral, which is easiest in spherical polar coordinates, with the axis of rotation taken to be the z axis. The element of volume is dV = r2dr sin 6) dB d/. (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.) 3.33 ** Starting from the sum (3.31) and replacing it by the appropriate integral, find the moment of inertia of a uniform thin square of side 2b, rotating about an axis perpendicular to the square and passing through its center. 3.34 ** A juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and throws it up so that, at the moment of release, it is horizontal, its CM is traveling vertically up at speed v o and it is rotating with angular velocity co o. To catch it, he wants to arrange that when it returns to his hand it will have made an integer number of complete rotations. What should vo be, if the rod is to have made exactly n rotations when it returns to his hand? 3.35 ** Consider a uniform solid disk of mass M and radius R, rolling without slipping down an incline which", " is at angle y to the horizontal. The instantaneous point of contact between the disk and the incline is called P. (a) Draw a free-body diagram, showing all forces on the disk. (b) Find the linear acceleration i) of the disk by applying the result L = tex t for rotation about P. (Remember that L = Ico and the moment of inertia for rotation about a point on the circumference is 2 MR 2. The condition that the disk not slip is that v = Rco and hence i) = Rt.) (c) Derive the same result by applying rext 104 Chapter 3 Momentum and Angular Momentum to the rotation about the CM. (In this case you will find there is an extra unknown, the force of friction. You can eliminate this by applying Newton's second law to the motion of the CM. The moment of inertia for rotation about the CM is 4 MR2.) 3.36 ** Repeat the calculations of Example 3.4 (page 97) for the case that the force F acts in a \"northeasterly\" direction at angle y from the x axis. What are the velocities of the two masses just after the impulse has been applied? Check your answers for the cases that y = 0 and y = 90\u00b0. 3.37 *** A system consists of N masses m a at positions ra relative to a fixed origin 0. Let ra denote the position of m a relative to the CM; that is, r at = ra \u2014 R. (a) Make a sketch to illustrate this last equation. (b) Prove the useful relation that > m a res = 0. Can you explain why this relation is nearly obvious? (c) Use this relation to prove the result (3.28) that the rate of change of the angular momentum about the CM is equal to the total external torque about the CM. (This result is surprising since the CM may be accelerating, so that it is not necessarily a fixed point in any inertial frame.) CHAPTER Energy This chapter takes up the conservation of energy. You will see that the analysis of energy conservation is surprisingly more complicated than the corresponding discus- sions of linear and angular momenta in Chapter 3. The main reason for the difference is this: In almost all problems of classical mechanics there is only one kind of linear momentum (p = my for each particle), and one kind of angular momentum (f = r x p for each particle). By contrast, energy comes in many different and important forms: kinetic, several kinds of potential, thermal, and more. It is the processes that trans- form energy from one kind to another that complicate the use of energy conservation. We shall see that conservation of energy is a quite subtle business, even for a system consisting of just a single particle. One manifestation of the relative difficulty of the discussion of energy is that we shall need some new tools from vector calculus, namely, the concepts of the gradient and the curl. I shall introduce these important ideas as we need them. 4.1 Kinetic Energy and Work As I have said, there are many different kinds of energy. Perhaps the most basic is kinetic energy (or KE), which for a single particle of mass in traveling with speed v is defined to be T = 1 mv 2 2 (4.1) Let us imagine the particle moving through space and examine the change in its kinetic energy as it moves between two neighboring points r 1 and r 1 + dr on its path as shown in Figure 4.1. The time derivative of T is easily evaluated if we note that v 2 = v v, so that \u2014 = -,m \u2014 (v v) = m(v v + v - V) = tn.@v. d T dt d dt (4.2) 105 106 Chapter 4 Energy Figure 4.1 Three points on the path of a particle: r 1, r 1 + dr (with dr infinitesimal) and r2. By the second law, the factor my is equal to the net force F on the particle, so that If we multiply both sides by dt, then since v dt is the displacement dr we find dT dt = F v. dT = F dr. (4.3) (4.4) The expression on the right, F dr, is defined to be the work done by the force F in the displacement dr. Thus we have proved the Work\u2014KE theorem, that the change in the particle's kinetic energy between two neighboring points on its path is equal to the work done by the net force as it moves between the two points.' So far we have proved the Work\u2014KE theorem only for an infinitesimal displace- ment dr, but it generalizes easily to larger displacements. Consider the two points shown as r 1 and r2 in Figure 4.1. We can divide the path between these points 1 and 2 into a large number of very small segments, to each", " of which we can apply the in- finitesimal result (4.4). Adding all of these results, we find that the total change in T going from 1 to 2 is the sum > F dr of all the infinitesimal works done in all the infinitesimal displacements between points 1 and 2: AT T2 \u2014 = F dr. (4.5) In the limit that all the displacements dr go to zero, this sum becomes an integral: EF dr ---> f F dr. 2 (4.6) 1 Two points that can be puzzling at first: The work F dr can be negative, if for example F and dr point in opposite directions. While the notion of a force doing negative work conflicts with our everyday notion of work, it is perfectly consisent with the physicist's definition: A force in the opposite direction to the displacement reduces the KE, so, by the work\u2014KE theorem, the corresponding work has to be negative. Second, if F and dr are perpendicular, then the work F dr is zero. Again this conflicts with our everyday sense of work, but is consistent with the physicist's usage: A force that is perpendicular to the displacement does not change the KE. Section 4.1 Kinetic Energy and Work 107 This integral, called a line integral, 2 is a generalization of the integral f f (x) dx over a single variable x, and its definition as the limit of the sum of many small pieces is closely analogous. If you feel any doubt about the symbol fl F dr on the right of (4.6), think of it as being just the sum on the left (with all the displacements infinitesimally small). In evaluating a line integral, it is usually possible to convert it into an ordinary integral over a single variable, as the following examples show. Notice that, as the name implies, the line integral depends (in general) on the path that the particle followed from point 1 to point 2. The particular line integral on the right of (4.6) is called the work done by the force F moving between points 1 and 2 along the path concerned. EXAMPLE 4.1 Three Line Integrals Evaluate the line integral for the work done by the two-dimensional force F = (y, 2x) going from the origin 0 to the point P = (1, 1) along each of the three paths shown in Figure 4.2. Path a goes from 0 to Q = (1, 0) along the x axis and then from Q straight up to P, path b goes straight from 0 to P along the line y = x, and path c goes round a quarter circle centered on Q. The integral along path a is easily evaluated in two parts, if we note that on OQ the displacements have the form dr = (dx, 0), while on QP they are dr = (0, dy). Thus Wa = f F dr = f F dr \u00b1 f F dr a 1 0 = f Fx (x, 0) dx + f Fy (1, y) dy 1 0 =0 + 2 J dy = 2. Figure 4.2 Three different paths, a, b, and c, from the origin to the point P = (1, 1). 2 Not an especially happy name for those of us who think of a line as something straight. However, there are curved lines as well as straight lines, and in general a line integral can involve a curved line, such as the path shown in Figure 4.1. 108 Chapter 4 Energy On the path b, x = y, so that dx = dy, and Wb = f F dr = f (F x dx Fy dy) = f (x 2x)dx = 1.5. 1 0 Path c is conveniently expressed parametrically as r (x, y) = (1 \u2014 cos 0, sin 0) where 8 is the angle between 0 Q and the line from Q to the point (x, y), with 0 < 8 < 7/2. Thus on path c and dr = (dx, dy) = (sin 0, cos 0) d0 = f F dr = f (Fx dx Fy dy) 7 /2 = f [sin 2 0 + 2(1 \u2014 cos 0) cos 6] d8 = 2 \u2014 7/4 = 1.21. Some more examples can be found in Problems 4.2 and 4.3 and, if you have never studied line integrals, you may want to try some of these. With the notation of the line integral, we can rewrite the result (4.5) as where I have introduced the notation W (1 -\u00b1 2) for the work done by F moving from point 1 to point 2. The result is the Work\u2014KE theorem for arbitrary displacements, large or small: The change in a particle's KE as it moves between points 1 and 2 is the work done by the", " net force. It is important to remember that the work that appears on the right of (4.7) is the work done by the net force F on the particle. In general, F is the vector sum of various separate forces F = \u00b1 \u00b1 F, Fi. i=i (For example, the net force on a projectile is the sum of two forces, the weight and air resistance.) It is a most convenient fact that to evaluate the work done by the net force F, we can simply add up the works done by the separate forces F 1,, Fn. This claim is easily proved as follows: W(1 \u2014\u203a- 2) = f F dr = f E Fi dr 2 2 2 Fi dr = w, (1_, 2). (4.8) _>. Section 4.2 Potential Energy and Conservative Forces 109 The crucial step, from the first line to the second, is justified because the integral of a sum of n terms is the same as the sum of the n individual integrals. The Work\u2013KE theorem can therefore be rewritten as T2 - = (1 -> 2). (4.9) i = In practice, one almost always uses the theorem in this way: Calculate the work W i n separate forces on the particle and then set A T equal to the sum done by each of the If the net force on a particle is zero, then the Work\u2013KE theorem tells us that the particle's kinetic energy is constant. This simply says that the speed v is constant, which, though true, is not very interesting, since it already follows from Newton's of all the Wi. first law. 4.2 Potential Energy and Conservative Forces The next step in the development of the energy formalism is to introduce the notion of potential energy (or PE) corresponding to the forces on an object. As you probably recall, not every force lends itself to the definition of a corresponding potential energy. Those special forces that do have a corresponding potential energy (with the required properties) are called conservative forces, and we must discuss the properties that distinguish conservative from nonconservative forces. Specifically, we shall find that there are two conditions that a force must satisfy to be considered conservative. To simplify our discussion, let us assume at first that there is only one force acting on the object of interest \u2014 the gravitational force on a planet by its sun, or the electric force qE on a charge in an electric field (with no other forces present). The force F may depend on many different variables: It may depend on the object's position r. (The farther the planet is from the sun, the weaker the gravitational pull.) It may depend on the object's velocity, as is the case with air resistance; and it may depend on the time t, as would be the case for a charge in a time-varying electric field. Finally, if the force is exerted by humans, it will depend on a host of imponderables hoW tired they are feeling, how conveniently they are situated to push, and so on. The first condition for a force F to be conservative is that F depends only on the position r of the object on which it acts; it must not depend on the velocity, the time, or any variables other than r. This sounds, and is, quite restrictive, but there are plenty of forces that have this property: The gravitational force of the sun on a planet (position r relative to the sun) can be written as F(r) = GmM r 2 which evidently depends only on the variable r. (The parameters G, m, M are constant for a given planet and given sun.) Similarly, the electrostatic force F(r) = qE(r) on a charge q by a static electric field E(r) has this property. Forces that do not satisfy this condition include the force of air resistance (which depends on the velocity), friction 110 Chapter 4 Energy Figure 4.3 Three different paths, a, b, and c, joining the same two points 1 and 2. (which depends on the direction of motion), the magnetic force (which depends on the velocity), and the force of a time-varying electric field E(r, t) (which obviously depends on time). The second condition that a force must satisfy to be called conservative concerns the work done by the force as the object on which it acts moves between two points r 1 and r2 (or just 1 and 2 for short), W (1\u2014* 2) = 2 F -dr. (4.10) Figure 4.3 shows two points, 1 and 2, and three different paths connecting them. It is entirely possible that the work done between points 1 and 2, as defined by the integral (4.10), has different values depending on which of the three paths, a, b, or c, the particle happens to follow. For example, consider the force of sliding friction as I push a heavy crate across the floor. This force", " has a constant magnitude, Ffric say, and is always opposite to the direction of motion. Thus the work done by friction as the crate moves from 1 to 2 is given by (4.10) to be Wfric (1 2) = \u2014 Ffii,L, where L denotes the length of the path followed. The three paths of Figure 4.3 have different lengths, and Whic (1 \u2014> 2) will have a different value for each of the three paths. On the other hand, there are forces with the property that the work W (1 --> 2) is the same for any path connecting the same two points 1 and 2. An example of a force with this property is the gravitational force, Fgrav = mg, of the earth on an object close to the earth's surface. It is easy to show (Problem 4.5) that, because g is a constant vector pointing vertically down, the work done in this case is Wgrav (1 2) = \u2014mgh, (4.11) where h is just the vertical height gained between points 1 and 2. This work is the same for any two paths between the given points 1 and 2. This property, the path Section 4.2 Potential Energy and Conservative Forces 111 independence of the work it does, is the second condition that a force must satisfy to be considered conservative, and we are now ready to state the two conditions: Conditions for a Force to be Conservative A fOCe F act: on a par isle is co se conservative ative if and only if it sa eS: MO' damns: i) F depends only on t p on r and not on the ve loci ty Or the tiff t car any other variable); thatF F(r). For any two mints I and 2, the work 2) don by F is the same for all paths between l and 2. The reason for the name \"conservative\" and for the importance of the concept is this: If all forces on an object are conservative, we can define a quantity called the potential energy (or just PE), denoted U(r), a function only of position, with the property that the total mechanical energy E = KE + PE = T U(r) (4.12) is constant; that is, E is conserved. To define the potential energy U (r) corresponding to a given conservative force, we first choose a reference point ro at which U is defined to be zero. (For example, in the case of gravity near the earth's surface, we often define U to be zero at ground level.) We then define U(r), the potential energy at an arbitrary point r, to be3 In words, U (r) is minus the work done by F if the particle moves from the reference point ro to the point of interest r, as in Figure 4.4. (We shall see the reason for the minus sign shortly.) Notice that the definition (4.13) only makes sense because of the property (ii) of conservative forces. If the work integral in (4.13) were different for different paths, then (4.13) would not define a unique function 4 U(r). 3 Notice that I have called the variable of integration r' to avoid confusion with the upper limit r 4 The definition (4.13) also depends on property (i) of conservative forces, but in a slightly subtler way. If F depended on another variable besides r (for instance, t or v), then the right side of (4.13) would depend on when or how the particle moved from r o to r, and again there would be no uniquely defined U(r). 112 Chapter 4 Energy Figure 4.4 The potential energy U(r) at any point r is de- fined as minus the work done by F if the particle moves from the reference point ro to r. This gives a well-defined function U (r) only if this work is independent of the path followed \u2014 that is, the force is conservative. EXAMPLE 4.2 Potential Energy of a Charge in a Uniform Electric Field A charge q is placed in a uniform electric field pointing in the x direction with strength Eo, so that the force on q is F = qE = qEoX. Show that this force is conservative and find the corresponding potential energy. The work done by F going between any two points 1 and 2 along any path is W(1 \u2014> 2) = f F -dr = qEo f dr = qE 0 f dx = qE o (x 2 \u2014 x i). (4.14) 2 2 2 1 1 This depends only on the two end points 1 and 2. (In fact it depends only on their x coordinates x 1 and x2.) Certainly, it is independent of the path, and the force is conservative. To define the corresponding potential energy U (r), we must first pick a", " reference point r o at which U will be zero. A natural choice is the origin, ro = 0, in which case the potential energy is U (r) = \u2014 W (0 \u2014k r) or, according to (4.14), U(r) = \u2014qE ox. We can now derive a crucial expression for the work done by F in terms of the potential energy U(r). Let r 1 and r2 be any two points as in Figure 4.5. If r o is the reference point at which U is zero, then it is clear from Figure 4.5 that W(ro r2) = W(ro \u2014> r1) W(r \u2014> r2 ) and hence W (r i r2) = W(ro \u2014> r2) \u2014 W (ro \u2014> r 1 ). (4.15) Each of the two terms on the right is (minus) the potential energy at the corresponding point. Thus we have proved that the work on the left is just the difference of these two potential energies: W(ri r2) = \u2014 [U(r2) \u2014 U(ri)] = \u2014 AU. (4.16) Section 4.2 Potential Energy and Conservative Forces 113 Figure 4.5 The work W (r i \u2014> r2) going from r 1 to r2 is the same as W (r o \u2014> r2) minus W(rc, --> r i). This result is inde- pendent of what path we use for either limb of the journey, provided the force concerned is conservative. The usefulness of this result emerges when we combine it with the Work\u2014KE theorem (4.7): Comparing this with (4.16), we see that or, moving the right side across to the left, 5 That is, the mechanical energy AT = W(r i \u2014> r2 ). AT = \u2014AU A(T + U) -= 0. E = T U (4.17) (4.18) (4.19) (4.20) does not change as the particle moves from r 1 to r2. Since the points r 1 and r2 were any two points on the particle's trajectory, we have the important conclusion: If the force on a particle is conservative, then the particle's mechanical energy never changes; that is, the particle's energy is conserved, which explains the use of the adjective \"conservative.\" Several Forces So far we have established the conservation of energy for a particle subject to a single conservative force. If the particle is subject to several forces, all of them conservative, our result generalizes easily. For instance, imagine a mass suspended from the ceiling by a spring. This mass is subject to two forces, the forces of gravity (Fgray 1 and the, spring (Fspr). The force of gravity is certainly conservative (as I've already argued), and, provided the spring obeys Hooke's law, Fspr is likewise (see Problem 4.42). We 5 We now see the reason for the minus sign in the definition of U. It gives the minus sign on the right of (4.18), which in turn gives the desired plus sign on the left of (4.19). 114 Chapter 4 Energy can define separate potential energies for each force, Ugrav for Fgrav and U U. for F Fspr, U gives (minus) the work done each with the crucial property (4.16) that the change in by the corresponding force. According to the Work\u2014KE theorem, the change in the mass's kinetic energy is A T = Wgrav W spr = Ugrav A Uspr), (4.21) where the second line follows from the properties of the two separate potential energies. Rearranging this equation, we see that A (T + Ugrav Uspr) = 0. That is, the total mechanical energy, defined as E =T Ugrav Uspr, is conserved. The argument just given extends immediately to the case of n forces on a particle, so long as they are all conservative. If for each force F i we define a corresponding potential energy Ui, then we have the Principle of Con ervation of Energy One Particle of>the n forces F n) acting on a Particle are conservative, each with its corresnding pot a energy U,(r), cmec hanical ene po E U T Ut ) U, (4.22) defin as n time. Nonconservative Forces If some of the forces on our particle are nonconservative, then we cannot define corre- sponding potential energies; nor can we define a conserved mechanical energy. Nev- ertheless, we can define potential energies for all of the forces that are conservative, and then recast the Work\u2014KE theorem in a form that shows how the nonconservative forces change the particle's mechanical energy. First, we divide the net force on the particle into two parts", ", the conservative part F., and the nonconservative part Fnc. For Fcons we can define a potential energy, which we'll call just U. By the Work\u2014KE theorem, the change in kinetic energy between any two times is A T = W = Wcons Wnc (4.23) The first term on the right is just \u2014 A U and can be moved to the left side to give A (T + U) = Wne. If we define the mechanical energy as E =T U, then we see that AE A(T + U) = Wnc- (4.24) Mechanical energy is no longer conserved, but we have the next best thing. The mechanical energy changes to precisely the extent that the nonconservative forces do work on our particle. In many problems the only nonconservative force is the force Section 4.2 Potential Energy and Conservative Forces 115 of sliding friction, which usually does negative work. (The frictional force f is in the direction opposite to the motion, so the work f dr is negative.) In this case ke is negative and (4.24) tells us that the object loses mechanical energy in the amount \"stolen\" by friction. All of these ideas are illustrated by the following simple example. EXAMPLE 4.3 Block Sliding Down an Incline Consider again the block of Example 1.1 and find its speed v when it reaches the bottom of the slope, a distance d from its starting point. The setup and the forces on the block are shown in Figure 4.6. The three forces on the block are its weight, w = mg, the normal force of the incline, N, and the frictional force f, whose magnitude we found in Example 1.1 to be f = pcmg cos 0. The weight mg is conservative, and the corresponding potential energy is (as you certainly recall from introductory physics, but see Problem 4.5) U = mgy where y is the block's vertical height above the bottom of the slope (if we choose the zero of PE at the bottom). The normal force does no work, since it is perpendicular to the direction of motion, so will not contribute to the energy balance. The frictional force does work Wfric fd = \u2014 Amgd cos 0. The change in kinetic energy is A T = Tf Ti = lmv2 and the change in potential energy is A U = Uf Ui = \u2014mgh = \u2014mgd sin 0. Thus (4.24) reads AT + AU = Wfric Or Solving for v we find im v2 \u2014 mgd sin 0 = mumgd cos 9. v = \\/2gd (sin 0 \u2014,u, cos 9). Figure 4.6 A block on an incline of angle 0. The length of the slope is d, and the height is h = d sin O. 116 Chapter 4 Energy As usual, you should check that this answer agrees with common sense. For example, does it give the expected answer when 9 = 900? What about 0 = 0? (The case 9 = 0 is a bit subtler.) 4.3 Force as the Gradient of Potential Energy We have seen that the potential energy U(r) corresponding to a force F(r) can be expressed as an integral of F(r) as in (4.13). This suggests that we should be able to write F(r) as some kind of derivative of U(r). This suggestion proves correct, though to implement it we shall need some mathematics that you may not have met before. Specifically, since F(r) is a vector [while U (r) is a scalar] we shall be involved in some vector calculus. Let us consider a particle acted on by a conservative force F (r), with corresponding potential energy U (r), and examine the work done by F(r) in a small displacement from r to r + dr. We can evaluate this work in two ways. On the one hand, it is, by definition, W(r r+dr) = F(r) dr = Fx dx + Fy dy + dz, (4.25) for any small displacement dr with components (dx, dy, dz). On the other hand, we have seen that the work W (r r+dr) is the same as (minus) the change in PE in the displacement: W(r r+dr) = \u2014dU = \u2014[U(r + dr) \u2014 U(r)] = \u2014[U(x + dx,y + dy,z + dz) \u2014 U(x,y,z)]. (4.26) In the second line, I have replaced the position vector r by its components to emphasize that U is really a function of the three variables (x, y, z). Now, for functions of one variable, a difference like that in (4.26) can be expressed in", " terms of the derivative: df = f (x + dx) \u2014 f (x) = df dx dx. (4.27) This is really no more than the definition of the derivative. 6 For a function of three variables, such as U(x, y, z), the corresponding result is dU = U(x + dx, y + dy, z + dz) \u2014 U(x, y, z) = \u2014 dx + \u2014 dy + \u2014 dz au ay au az au ax (4.28) where the three derivatives are the partial derivatives with respect to the three in- dependent variables (x, y, z). [For example, au/ax is the rate of change of U as x 6 Strictly speaking, this equation is exact only in the limit that dx 0. As usual, I take the view that dx is small enough (though nonzero) that the two sides are equal within our chosen accuracy target. Section 4.3 Force as the Gradient of Potential Energy 117 changes, with y and z fixed, and is found by differentiating U(x, y, z) with respect to x treating y and z as constants. See Problems 4.10 and 4.11 for some examples.] Substituting (4.28) into (4.26), we find that the work done in the small displacement from r to r + dr is W(r \u2014> r+dr) = \u2014 [\u2014ax + \u2014 dy + \u2014dz au ax au, B y au az (4.29) The two expressions (4.25) and (4.29) are both valid for any small displacement dr. In particular, we can choose dr to point in the x direction, in which case dy = dz = 0 and the last two terms in both (4.25) and (4.29) are zero. Equating the remaining terms, we see that Fx = \u2014au/ax. By choosing dr to point in the y or z directions, we get corresponding results for Fy and F2, and we conclude that au F =\u2014 x F Y au ay, F = \u2014 z aU az That is, F is the vector whose three components are minus the three partial derivatives of U with respect to x, y, and z. A slightly more compact way to write this result is this: F = \u2014x\u2014 \u2014 \u2014 \u2014 z\u2014.,s au ax,au y ay aU az Relationships like (4.31) between a vector (F) and a scalar (U) come up over and over again in physics. For example, the electric field E is related to the electrostatic potential V in exactly the same way. More generally, given any scalar f(r), the vector whose three components are the partial derivatives of f(r) is called the gradient of f, denoted Vf: (4.30) (4.31) (4.32),af,af,af Vf x\u2014 + y\u2014 + z\u2014. ay a z ax The symbol Vf is pronounced \"grad f\" The symbol V by itself is called \"grad,\" or \"del,\" or \"nabla.\" With this notation, (4.31) is abbreviated to This important relation gives us the force F in terms of derivatives of U, just as the definition (4.13) gave U as an integral of F. When a force F can be expressed in the form (4.33), we say that F is derivable from a potential energy. Thus, we have shown that any conservative force is derivable from a potential energy. 7 7 1 am following standard terminology here. Notice that we have defined \"conservative\" so that a conservative force conserves energy and is derivable from a potential energy. This is occasionally confusing, since there are forces (such as the magnetic force on a charge or the normal force on a sliding object) that do no work and hence conserve energy, but are not \"conservative\" in the sense defined here, since they are not derivable from a potential energy. This unfortunate confusion seldom causes trouble, but you may want to register it somewhere in the back of your mind. 118 Chapter 4 Energy EXAMPLE 4.4 Finding F from U The potential energy of a certain particle is U = Axy 2 B sin Cz, where A, B and C are constants. What is the corresponding force? To find F we have only to evaluate the three partial derivatives in (4.31). In doing this, you must remember that au/ax is found by differentiating with respect to x, treating y and z as constant, and so on. Thus au/ax = Ay2, and so on, and the final result is F = \u2014(x Ay 2 Sr 2Axy i BC cos Cz). It is sometimes convenient to remove the f from (4.32) and to write V=x\u2014 +y", "\u2014 +z\u2014. a ax a a y a az (4.34) In this view, V is a vector differential operator that can be applied to any scalar f and produces the vector given in (4.32). A very useful application of the gradient is given by (4.28), whose right-hand side you will recognize as V U dr. Thus, if we replace U by an arbitrary scalar f, we see that the change in f resulting from a small displacement dr is just This useful relation is the three-dimensional analog of Equation (4.27) for a function of one variable. It shows the sense in which the gradient is the three-dimensional equivalent of the ordinary derivative in one dimension. If you have never met the V notation before, it will take a little getting used to. Meanwhile, you can just think of (4.33) as a convenient shorthand for the three equations (4.30). For practice using the gradient, you could look at Problems 4.12 through 4.19. 4.4 The Second Condition that F be Conservative We have seen that one of the two conditions that a force F be conservative is that the work fi2 F dr which it does moving between any two points 1 and 2 must be independent of the path followed. You are certainly to be excused if you don't see how we could test whether a given force has this property. Checking the value of the integral for every pair of points and every path joining those points is indeed a formidable prospect! Fortunately, we never need to do this. There is a simple test, which can be quickly applied to any force that is given in analytic form. This test involves another of the basic concepts of vector calculus, this time the so-called curl of a vector. Section 4.4 The Second Condition that F be Conservative 119 It can be shown (though I shall not do so here 8) that a force F has the desired property, that the work it does is independent of path, if and only if V xF= 0 (4.36) everywhere. The quantity V x F is called the curl of F, or just \"curl F,\" or \"del cross F.\" It is defined by taking the cross product of V and F just as if the components of V, namely (a/ax, a/ay, a/az), were ordinary numbers. To see what this means, consider first the cross product of two ordinary vectors A and B. In the table below, I have listed the components of A, B, and A x B: vector x component y component z component A B A x Bx A Y B Y A z Bz A x B A y Bz \u2014 A z By A z Bx \u2014 A x B, A x B y \u2014 A y Bx The components of V x F are found in exactly the same way, except that the entries in the first row are differential operators. Thus, vector x component y component z component V F a/ax Fx a/ay Fy \u2014 a/az Fz \u2014 V X F ay z \u2014 y az x ax z ax Y ay x (4.37) (4.38) No one would claim that (4.36) is obviously equivalent to the condition that fi F dr is path-independent, but it is, and it provides an easily applied test for the 2 path-independence property, as the following example shows. EXAMPLE 4.5 Is the Coulomb Force Conservative? Consider the force F on a charge q due to a fixed charge Q at the origin. Show that it is conservative and find the corresponding potential energy U. Check that \u2014 V U = F. The force in question is the Coulomb force, as shown in Figure 4.7(a), F = kq r =\u2014r r 2 r 3 (4.39) where k denotes the Coulomb force constant, often written as 1/(470, and y is just an abbreviation for the constant kq Q. From the last expression we can read off the components of F, and using (4.38) we can calculate the components of V x F. For example, the x component is (V x F) x = Fz \u2014 a ay a _ a ( yz y \u2014 \u2014, ay r 3 az a ( yy) az r 3 (4.40) 8 The condition (4.36) follows from a result called Stokes's theorem. If you would like to explore this a little, see Problem 4.25. For more details, see any text on vector calculus or mathematical methods. I particularly like Mathematical Methods in the Physical Sciences by Mary Boas (Wiley, 1983), p. 260. 120 Chapter 4 Energy F Q (a) (b) Figure 4.7 (a) The Coulomb force F = yi./r 2 of the fixed charge Q on the charge q. (b) The work done by F as q", " moves from ro to r can be evaluated following a path that goes radially outward to P and then around a circle to r. The two derivatives here are easily evaluated: First, since az/ay = ay/az = 0, we can rewrite (4.40) as (V X F)\u201e = yz ( 1-r -3) \u2014 yy. (4.41) ay az Next recall that so that, for example, r (x 2 + y2 \u00b1 z 2 1/2 ) ar \u2014 ay y r (4.42) (Check this one using the chain rule.) We can now evaluate the two remaining derivatives in (4.41) to give (remember the chain rule again) (V x F)x = yz (-4. \u2014) \u2014 yy (-4 = 0. r r y r The other two components work in exactly the same way (check it, if you don't believe me), and we conclude that V x F = 0. According to the result (4.36), this guarantees that F satisfies the second condition to be conservative. Since it certainly satisfies the first condition (it depends only on the variable r), we have proved that F is conservative. (The proof that V x F = 0 is considerably quicker in spherical polar coordinates. See Problem 4.22.) The potential energy is defined by the work integral (4.13), U(r) = \u2014 f F(e) dr' ro (4.43) where ro is the (as yet unspecified) reference point where U (ro) = 0. Fortu- nately, we know that this integral is independent of path, so we can choose whatever path is most convenient. One possibility is shown in Figure 4.7(b), where I have chosen a path that goes radially outward to the point labeled P and then around a circle (centered on Q) to r. On the first segment, F(e) and Section 4.5 Time-Dependent Potential Energy 121 dr' are collinear, and F(r') dr' = (y I r'2)dr'. On the second, F(r') and dr' are perpendicular, so no work is done along this segment, and the total work is just that of the first segment, U(r) = \u2014 f dr' = 3\u00b1 -- Y. (4.44) r v r 0 r'2 r ro Finally, it is usual in this problem to choose the reference point r o at infinity, so that the second term here is zero. With this choice (and replacing y by kqQ) we arrive at the well-known formula for the potential energy of the charge q due to Q, U(r) = U(r) = (4.45) kqQ r Notice that the answer depends only on the magnitude r of the position vector r and not on the direction. To check V U let us evaluate the x component: (VU), = a ( kqQ) = ax r ) kqQ ar r2 ax (4.46) where the last expression follows from the chain rule. The derivative ar/ax is x/r [compare Equation (4.42)], so (VU)\u201e = \u2014kqQ\u2014 = \u2014Ft, x r3 as given by (4.39). The other two components work in exactly the same way, and we have shown that V U = \u2014F (4.47) as required. 4.5 Time-Dependent Potential Energy We sometimes have occasion to study a force F(r,t) that satisfies the second condition to be conservative (V x F = 0), but, because it is time-dependent, does not satisfy the first condition. In this case, we can still define a potential energy U (r, t) with the property that F = \u2014 V U, but it is no longer the case that total mechanical energy, E = T U, is conserved. Before I justify these claims, let me give an example of this situation. Figure 4.8 shows a small charge q in the vicinity of a charged conducting sphere (for example, a Van de Graaff generator) with a charge Q (t) that is slowly leaking away through the moist air to ground. Because Q(t) changes with time, the force that it exerts on the small charge q is explicitly time-dependent. Nevertheless, the spatial dependence of the force is the same as for the time-independent Coulomb force of Example 4.5 (page 119). Exactly the same analysis as in that example shows that V x F = 0. 122 Chapter 4 Energy Q (t) q F = kqQ(t) r2 Figure 4.8 The charge Q (t) on the conducting sphere is slowly leaking away, so the force on the small charge q varies with time, even if its position r is constant.", " Let me now justify the claims made above. First, since V x F(r, t) = 0, the same mathematical theorem quoted in connection with Equation (4.36) guarantees that the work integral fi2 F(r, t) dr (evaluated at any one time t) is path independent. This means we can define a function U(r, t) by an integral exactly analogous to (4.13), U(r, t) = \u2014 f F (r/, t) dr', (4.48) ro and, for the same reasons as before, F(r,t) = \u2014VU(r, t). (See Problem 4.27.) In this case, we can say the force F is derivable from the time-dependent potential energy U(r, t). So far everything has gone through just as before, but now the story changes. We can define the mechanical energy as E = T + U, but it is no longer true that E is conserved. If you review carefully the argument leading to Equation (4.19), you may be able to see what goes wrong, but we can in any case show directly that E =T + U changes as the particle moves on its path. As before, consider any two neighboring points on the particle's path at times t and t dt. Exactly as in (4.4), the change in kinetic energy is dT = \u2014 dt = (mir v)dt = F dr. (4.49) dT dt Meanwhile, U(r, t) = U (x, y, z, t) is a function of four variables (x, y, z, t) and dU = \u2014dx + \u2014dy + \u2014dz + \u2014dt. (4.50) aU ax au ay au at You will recognize the first three terms on the right as VU dr = \u2014F dr. Thus dU = \u2014F dr \u2014dt. (4.51) When we add this to Equation (4.49) the first two terms cancel, and we are left with d(T + U) = a \u2014dt. u at (4.52) Clearly it is only when U is independent of t (that is, au/at = 0) that the mechanical energy E = T + U is conserved. au az au at Section 4.6 Energy for Linear One-Dimensional Systems 123 Returning to the example of Figure 4.8, we can understand this conclusion and see what has happened to conservation of energy. Imagine that I hold the charge q stationary at the position of Figure 4.8, while the charge on the sphere leaks away. Under these conditions, the KE of q doesn't change, but the potential energy kq Q (t) / r slowly diminishes to zero. Clearly T U is not constant. However, while mechanical energy is not conserved, total energy is conserved: The loss of mechanical energy is exactly balanced by the gain of thermal energy as the discharge current heats up the surrounding air. This example suggests, what is true, that the potential energy depends explicitly on time in precisely those situations where mechanical energy gets transformed to some other form of energy or to mechanical energy of other bodies external to the system of interest. 4.6 Energy for Linear One-Dimensional Systems So far we have discussed the energy of a particle that is free to move in all three dimensions. Many interesting problems involve an object that is constrained to move in just one dimension, and the analysis of such problems is remarkably simpler than the general case. Oddly enough, there is some ambiguity in what a physicist means by a \"one-dimensional system.\" Many introductory physics texts start out discussing the motion of a one-dimensional system, by which they mean an object (a railroad car, for instance) that is confined to move on a perfectly straight, or linear, track. In discussing such linear systems, we naturally take the x axis to coincide with the track, and the position of the object is then specified by the single coordinate x. In this section I shall focus on linear one-dimensional systems. However, there are much more complicated systems, such as a roller coaster on its curving track, that are also one-dimensional, inasmuch as their position can be specified by a single parameter (such as the distance of the roller coaster along its track). As I shall discuss in the next section, energy conservation for such curvilinear one-dimensional systems is just as straightforward as for a perfectly straight track. To begin, let us consider an object constrained to move along a perfectly straight track, which we take to be the x axis. The only component of any force F that can do work is the x component, and we can simply ignore the other two components. Therefore the work done by F is the one-dimensional integral W (x i x 2) = f Fx (x) dx. x2 (4.", "53) If the force is to be conservative. Fx must satisfy the two usual conditions: (i) It must depend only on the position x [as I have already implied in writing the integral (4.53)]. (ii) The work (4.53) must be independent of path. The remarkable feature of one- dimensional systems is that the first condition already guarantees the second, so the latter is superfluous. To understand this property, you have only to recognize that in one dimension there is only a small choice of paths connecting any two points. Consider, for example, the two points A and B shown in Figure 4.9. The obvious path between points A and B is the path that goes from A directly to B (let's call this path 124 Chapter 4 Energy A B< x Figure 4.9 The path called ABC B goes from A past B and on to C, then back to B. \"AB\"). Another possibility, shown in the figure, is to go from A past B to C and then back to B (let's call this one \"ABCB\"). The work done along this path can be broken up as follows: W(ABCB) = W(AB) W(BC) + W(CB). Now, provided the force depends only on the position x [condition (i)] each increment of work going from B to C is exactly equal (but of opposite sign) to the corresponding contribution going from C to B. That is, the last two terms on the right cancel, and we conclude that W(ABC B) = W(AB), as required. One can of course concoct a path from A to B that doubles back and forth many times, but a little thought should convince you that any such path can be broken into a number of segments some of which together traverse the direct path AB exactly once, and all the rest of which cancel in pairs. Thus the work done on any path between A and B is the same as that on the direct path AB, and we have proved that in one dimension the first condition for a force to be conservative guarantees the second. Graphs of the Potential Energy A second useful feature of one-dimensional systems is that with only one independent variable (x) we can plot the potential energy U(x), and, as we shall see, this makes it easy to visualize the behavior of the system. Assuming all forces on the object are conservative, we define the potential energy as U (x) = \u2014 f F x (x) dx' xo (4.54) where Fx is the x component of the net force on the particle. For example, for a mass on the end of a spring obeying Hooke's law, the force is Fx = \u2014kx, and, if we choose the reference point x. = 0, Equation (4.54) gives the celebrated result for any spring obeying Hooke's law. Corresponding to the three-dimensional result F = \u2014 V U, we have the simpler result in one dimension (4.55) U = kx 2 2 F dU dx Section 4.6 Energy for Linear One-Dimensional Systems 125 Figure 4.10 The graph of potential energy U(x) against x for any one-dimensional system can be thought of as a picture of a roller coaster track. The force Fx = \u2014 dU ldx tends to push the object \"downhill\" as at x 1 and x2. At the points x3 and x4, where U(x) is minimum or maximum, dUldx= 0 and the force is zero; such points are therefore points of equilibrium. If we plot the potential energy against x as in Figure 4.10, we can easily see qualita- tively how the object has to behave. The direction of the net force is given by (4.55) as \"downhill\" on the graph of U(x) \u2014 to the left at x i and to the right at x 2. It follows that the object always accelerates in the \"downhill\" direction \u2014 a property that reminds one of the motion of a roller coaster, which also always accelerates downhill. This analogy is not an accident: For a roller coaster, U (x) is mgh (where h is the height above ground) and the graph of U(x) against x has the same shape as a graph of h against x, which is just a picture of the track. For any one -dimensional system, we can always think about the graph of U(x) as a picture of a roller coaster, and common sense will generally tell us the kind of motion that is possible at different places, as I now describe. At points, such as x 3 and x4, where dU I dx = 0 and U (x) is minimum or maximum, the net force is zero, and the object can remain in equilibrium. That is, the condition dU ldx = 0 characterizes points", " of equilibrium. At x 3, where d2 Uldx 2 > 0 and U(x) is minimum, a small displacement from equilibrium causes a force which pushes the object back to equilibrium (back to the left on the right of x 3, back to the right on the left of x3). In other words, equilibrium points where d2 U/dx 2 > 0 and U(x) is minimum are points of stable equilibrium. At equilibrium points like x4 where d2 U/dx 2 < 0 and U(x) is maximum, a small displacement leads to a force away from equilibrium, and the equilibrium is unstable. If the object is moving then its kinetic energy is positive and its total energy is necessarily greater than U(x). For example, suppose the object is moving some- where near the equilibrium point x = b in Figure 4.11. Its total energy has to be greater than U (b) and could, for example, equal the value shown as E in that fig- ure. If the object happens to be on the right of b and moving toward the right, its PE will increase and its KE must therefore decrease until the object reaches the turning point labeled c, where U (c) = E and the KE is zero. At x = c the object stops and, with the force back to the left, it accelerates back toward x = b. It can- not now stop until once again the KE is zero, and this occurs at the turning point a, where U (a) = E and the object accelerates back to the right. Since the whole 126 Chapter 4 Energy E Figure 4.11 If an object starts out near x = b with the energy E shown, it is trapped in the valley or \"well\" between the two hills and oscillates between the turning points at x = a and c where U(x) = E and the kinetic energy is zero. cycle now repeats itself, we see that if the object starts out between two hills and its energy is lower than the crest of both hills, then the object is trapped in the valley or \"well\" and oscillates indefinitely between the two turning points where U(x) = E. Suppose the cart again starts out between the two hills but with energy higher than the crest of the right hill though still lower than the left. In this case, it will escape to the right since E > U(x) everywhere on the right, and it can never stop once it is moving in that direction. Finally, if the energy is higher than both hills, the cart can escape in either direction. These considerations play an important role in many fields. An example from molecular physics is illustrated in Figure 4.12, which shows the potential energy of a typical diatomic molecule, such as HC1, as a function of the distance between the two atoms. This potential energy function governs the radial motion of the hydrogen atom (in the case of HC1) as it vibrates in and out from the much heavier chlorine atom. The zero of energy has been chosen where the two atoms are far apart (at infinity) and at rest. Notice that the independent variable is the interatomic distance r which, by its definition, is always positive, 0 < r < oo. As r -f 0, the potential energy gets very large, indicating that the two atoms repel one another when very close together (because of the Coulomb repulsion of the nuclei). If the energy is positive (E > 0) the H atom can escape to infinity, since there is no \"hill\" to trap it; the H atom can come in from infinity, but it will stop at the turning point r = a and (in the absence of any mechanism to take up some of its energy) it will move away to infinity again. On the other hand, if E < 0, the H atom is trapped and will oscillate in and out between the two turning points shown at r = b and r = d. The equilibrium separation of the molecule is at the point shown as r = c. It is the states with E < 0 that correspond to what we normally regard as the HC1 molecule. To form such a molecule, two separate atoms (with E > 0) must come together to a separation somewhere near r = c, and some process, such as emission of light, must remove enough energy to leave the two atoms trapped with E < 0. Section 4.6 Energy for Linear One-Dimensional Systems 127 Figure 4.12 The potential energy for a typical diatomic molecule such as HC1, plotted as a function of the distance r between the two atoms. If E > 0, the two atoms cannot approach closer than the turning point r = a, but they can move apart to infinity. If E < 0, they are trapped between the turning points at b and d and form a bound molecule. The equilibrium separation is r = c. Complete Solution of the Motion A third remarkable feature of one-dimensional conservative systems is that we", " can \u2014 at least in principle \u2014 use the conservation of energy to obtain a complete solution of the motion, that is, to find the position x as a function of time t. Since E = T U(x) is conserved, with U(x) a known function (in the context of a given problem) and E determined by the initial conditions, we can solve for T = 2ini2 = E \u2014 U(x) and hence for the velocity x as a function of x: = \u2014 U(x). (4.56) (Notice that there is an ambiguity in the sign since energy considerations cannot determine the direction of the velocity. For this reason, the method described here usually does not work in a truly three-dimensional problem. In one dimension, you can almost always decide the sign of i by inspection, though you must remember to do so.) Knowing the velocity as a function of x, we can now find x as a function of t, using separation of variables, as follows: We first rewrite the definition i = dx/dt as [Since z =.i(x), this separates the variables t and x.] Next, we can integrate between any initial and final points to give (4.57) dt = dx x tf = xf dx x1 1 X 128 Chapter 4 Energy This gives the time for travel between any initial and final positions of interest. If we substitute for i from (4.56) (and assume, to be definite, that z is positive) then the time to go from the initial x o at time 0 to an arbitrary x at time t is t = r dx' Jxo i(x') Im r dx' V 2 ix. E \u2014 U(x') (4.58) (As usual, I've renamed the variable of integration as x' to avoid confusion with the upper limit x.) The integral (4.58) depends on the particular form of U(x) in the problem at hand. Assuming we can do the integral [and we can at least do it numerically for any given U(x)], it gives us t as a function of x. Finally we can solve to give x as a function of t, and our solution is complete, as the following simple example illustrates. EXAMPLE 4.6 Free Fall I drop a stone from the top of a tower at time t = 0. Use conservation of energy to find the stone's position x (measured down from the top of the tower, where x = 0) as a function of t. Neglect air resistance. The only force on the stone is gravity, which is, of course, conservative. The corresponding potential energy is (Remember x is measured downward.) Since the stone is at rest when x = 0, the total energy is E = 0, and according to (4.56) the velocity is U(x) = \u2014mgx. 2 \u2014 N/E \u2014 U(x) = V2gx (a result that is well known from elementary kinematics). Thus t = fx dx' fx dx' Jo 1(x') = Jo V2gx' 2x g As anticipated, this gives t as a function of x, and we can solve to give the familiar result x = gt2. This simple example, involving the gravitational potential energy U(x) = \u2014mgx, can be solved many different (and some simpler) ways, but the energy method used here can be used for any potential energy function U(x). In some cases, the integral (4.58) can be evaluated in terms of elementary functions, and we obtain an analytic solution of the problem; for example, if U(x) = 4kx 2 (as for a mass on the end of a spring), the integral turns out to be an inverse sine function, which implies that x oscillates sinusoidally with time, as we should expect (see Problem 4.28). For some potential energies, the integral cannot be Section 4.7 Curvilinear One-Dimensional Systems 129 done in terms of elementary functions, but can nonetheless be related to func- tions that are tabulated (see Problem 4.38). For some problems, the only way to do the integral (4.58) is to do it numerically. 4.7 Curvilinear One-Dimensional Systems So far the only one-dimensional system I have discussed is an object constrained to move along a linear path, with position specified by the coordinate x. There are other, more general, systems that can equally be said to be one-dimensional, inasmuch as their position is specified by a single number. An example of such a one-dimensional system is a bead threaded on a curved rigid wire as illustrated in Figure 4.13. (Another is a roller coaster confined to a curved track.) The position of the bead can be specified by a single parameter, which we can choose as the distance", " s, measured along the wire, from a chosen origin 0. With this choice of coordinate, the discussion of the curved one-dimensional track parallels closely that of the straight track, as I now show. The coordinate s of our bead corresponds, of course, to x for a cart on a straight track. The speed of the bead is easily seen to be s, and the kinetic energy is therefore just as compared to the familiar 4mi 2 for the straight track. The force is a little more complicated. As our bead moves on the curved wire the net normal force is not zero; on the contrary, the normal force is what constrains the bead to follow its assigned curving path. (For this reason, the normal force is called the force of constraint.) On the other hand, the normal force does no work, and it is the tangential component Ftang of the net force that is our chief concern. In particular, it is fairly easy to show (Problem 4.32) that T = 2 Ftang = ms Figure 4.13 An object constrained to move on a curved track can be considered to be a one-dimensional system, with the position specified by the distance s (measured along the track) of the object from an origin 0. The system shown is a bead threaded on a stiff wire, bent into a double loop-the-loop. 130 Chapter 4 Energy (just as Fx = na on a straight track). Further, if all the forces on the bead that have a tangential component are conservative, we can define a corresponding potential energy U(s) such that Ftang = -LclU Ids, and the total mechanical energy E = T U(s) is constant. The whole discussion of Section 4.6 can now be applied to the bead on a curved wire (or any other object constrained to move on a one-dimensional path). In particular, those points where U(s) is a minimum are points of stable equilibrium, and those where U(s) is maximum are points of unstable equilibrium. There are many systems that appear to be much more complicated than the bead on a wire, but are nonetheless one-dimensional and can be treated in much the same way. Here is an example. EXAMPLE 4.7 Stability of a Cube Balanced on a Cylinder A hard rubber cylinder of radius r is held fixed with its axis horizontal, and a wooden cube of side 2b is balanced on top of the cylinder, with its center vertically above the cylinder's axis and four of its sides parallel to the axis. The cube cannot slip on the rubber of the cylinder, but it can of course rock from side to side, as shown in Figure 4.14. By examining the cube's potential energy, find out if the equilibrium with the cube centered above the cylinder is stable or unstable. Let us first note that the system is one-dimensional, since its position as it rocks from side to side can be specified by a single coordinate, for instance the angle 9 through which it has turned. (We could also specify it by the distance s of the cube's center from equilibrium, but the angle is a little more convenient. Either way the system's position is specified by a single coordinate, and our problem is definitely one-dimensional.) The constraining forces are the normal Figure 4.14 A cube, of side 2b and center C, is placed on a fixed horizontal cylinder of radius r and center 0. It is originally put so that C is centered above 0, but it can roll from side to side without slipping. Section 4.7 Curvilinear One-Dimensional Systems 131 and frictional forces of the cylinder on the cube; that is, these two forces constrain the cube to move only as shown in Figure 4.14. Since neither of these does any work we need not consider them explicitly. The only other force on the cube is gravity, and we know from elementary physics that this is conservative and that the gravitational potential energy is the same as for a point mass at the center of the cube; that is, U = mgh, where h is the height of C above the origin, as shown in Figure 4.14. (See Problem 4.6.) The length of the line shown as OB is just r b, while the length BC is the distance the cube has rolled around the cylinder, namely re. Therefore h = (r b) cos 0 + r0 sin 0 and the potential energy is U(0) = mgh = mg[(r b) cos 0 + r0 sin 0]. (4.59) To find the equilibrium position (or positions) we must find the points where dU I d0 vanishes. (Strictly speaking I haven't proved this very plausible claim yet for this kind of constrained system; I'll discuss it shortly.) The derivative is easily seen to be (check this for yourself ) dU de = mg [re cos 0 \u2014 b sin 0]. This vanishes at 0 =", " 0, confirming the obvious \u2014 that 0 = 0 is a point of equilibrium. To decide whether this equilibrium is stable, we have only to differentiate again and find the value of d2 U/d0 2 at the equilibrium position. This gives (as you should check) d2U d0 2 = mg (r \u2014 b) (4.60) (at 0 = 0). If the cube is smaller than the cylinder (that is, b < r), this second derivative is positive, which means that U(0) has a minimum at 0 = 0 and the equilibrium is stable; if the cube is balanced on the cylinder, it will remain there indefinitely. On the other hand, if the cube is larger than the cylinder (b > r), the second derivative (4.60) is negative, the equilibrium is unstable, and the smallest disturbance will cause the cube to roll and fall off the cylinder. Further Generalizations There are many other, more complicated systems that are still legitimately described as one dimensional. Such systems may comprise several bodies, but the bodies are joined by struts or strings in such a way that just one parameter is needed to describe the system's position. An example of such a system is the Atwood machine shown in Figure 4.15, which consists of two masses, m 1 and m 2, suspended from opposite ends of a massless, inextensible string that passes over a frictionless pulley. (To simplify the discussion, I shall assume the pulley is massless, although it is easy to allow for a mass of the pulley.) The two masses can move up and down, but the forces of the pulley on the string and the string on the masses constrain matters so that the mass m2 can move up only to the extent that m 1 moves down by exactly the same distance. 132 Chapter 4 Energy x m2g Figure 4.15 An Atwood machine consisting of two masses, m 1 and m 2, suspended by a massless inextensible string that passes over a massless, frictionless pulley. Because the string's length is fixed, the position of the whole system is specified by the distance x of m i below any convenient fixed level. The forces on the two masses are their weights m 1 g and m 2 g, and the tension forces FT (which are equal since the pulley and string are massless). Thus the position of the whole system can be specified by a single parameter, for example the height x of m 1 below the pulley's center as shown, and the system is again one-dimensional.9 Let us consider the energies of the masses m 1 and m 2. The forces acting on them are gravity and the tension in the string. Since gravity is conservative, we can introduce potential energies U 1 and U2 for the gravitational forces, and our previous considerations imply that in any displacement of the system, and A Ti + A Ul = witen A 72 + A u2 Wien (4.61) (4.62) where the terms W' denote the work done by the tension on m 1 and m 2. Now, in the absence of friction, the tension is the same all along the string. Thus, although the tension certainly does work on the two individual masses, the work done on m 1 is equal and opposite to that done on m 2, when m 1 moves down and m 2 moves an equal distance up (or vice versa). That is, wr w2ten. (4.63) 9 You may object, correctly, that the masses can also move sideways. If this worries you, we can thread each mass over a vertical frictionless rod, but these rods are actually unnecessary: As long as we refrain from pushing the masses sideways, each will remain in a vertical line of its own accord. Section 4.8 Central Forces 133 Thus, if we add the two energy equations (4.61) and (4.62), the terms involving the tension in the string cancel and we are left with A (Ti Ui T2 + U2) = O. That is, the total mechanical energy E = + Ui + T2 + U2 (4.64) is conserved. The beauty of this result is that all reference to the constraining forces of the string and pulley has disappeared. It turns out that many systems which contain several particles that are constrained in some way (by strings, struts, or a track on which they must move, etc.) can be treated in this same way: The constraining forces are crucially important in determining how the system moves, but they do no work on the system as a whole. Thus in considering the total energy of the system, we can simply ignore the constraining forces. In particular, if all other forces are conservative (as with our example of the Atwood machine), we can define a potential energy U oi for each particle a, and the total energy 4.6 apply. E_ (Ta + U\u201e)", " N 01=1 is constant. If the system is also one-dimensional (position specified by just one parameter, as with the Atwood machine), then all of the considerations of Section A careful discussion of constrained systems is far easier in the Lagrangian formula- tion of mechanics than in the Newtonian. Thus I shall postpone any further discussion to Chapter 7. In particular, the proof that a stable equilibrium normally corresponds to a minimum of the potential energy (for a large class of constrained systems) is sketched in Problem 7.47. 4.8 Central Forces A three-dimensional situation that has some of the simplicity of one-dimensional problems is a particle that is subject to a central force, that is, a force that is everywhere directed toward or away from a fixed \"force center.\" If we take the force center to be the origin, a central force has the form F(r) = (4.65) where the function f(r) gives the magnitude of the force (and is positive if the force is outward and negative if it is inward). An example of a central force is the Coulomb force on a charge q due to a second charge Q at the origin; this has the familiar form F(r) = kqQ r, r 2 (4.66) 134 Chapter 4 Energy which is obviously an example of (4.65), with the magnitude function given by f(r) = kqQ1r 2. The Coulomb force has two additional properties not shared by all central forces: First, as we have proved, it is conservative. Second, it is spherically symmetric or rotationally invariant; that is, the magnitude function f(r) in (4.65) is independent of the direction of r and, hence, has the same value at all points at the same distance from the origin. A compact way to express this second property of spherical symmetry is to observe that the magnitude function f(r) depends only on the magnitude of the vector r and not its direCtion, so can be written as f(r) = f(r) (4.67) A remarkable feature of central forces is that the two properties just mentioned always go together: A central force that is conservative is automatically spherically symmetric, and, conversely, a central force that is spherically symmetric is automati- cally conservative. These two results can be proved in several ways, but the most direct proofs involve the use of spherical polar coordinates. Therefore, before offering any proofs, I shall briefly review the definition of these coordinates. Spherical Polar Coordinates The position of any point P is, of course, identified by the vector r pointing from the origin 0 to P. The vector r can be specified by its Cartesian coordinates (x, y, z), but in problems involving spherical symmetry it is almost always more convenient to specify r by its spherical polar coordinates (r, 9, 0), as defined in Figure 4.16. The first coordinate r is just the distance of P from the origin; that is, r = jr I, as usual. The angle 9 is the angle between r and the z axis. The angle 0, often called the azimuth, is the angle from the x axis to the projection of r on the xy plane, as shown. 10 It is a simple exercise (Problem 4.40) to relate the Cartesian coordinates (x, y, z) to the polar coordinates (r, 9, q5) and vice versa. For example, by inspecting Figure 4.16 you should be able to convince yourself that x = r sin 0 cos 0, y = r sin 9 sin 0, and z = r cos 9. (4.68) A beautiful use of spherical coordinates, which may help you to visualize them, is to specify positions on the surface of the earth. If we choose the origin at the center of the earth, then all points on the surface have the same value of r, namely the radius of the earth.\" Thus positions on the surface can be specified by giving just the two angles (9, 0). If we choose our z axis to coincide with the north polar axis, then it is easy to see from Figure 4.16 that 9 gives the latitude of the point P, measured down from the north pole. (Since latitude is traditionally measured up from the equator, our angle 0 is often called the colatitude.) Similarly, 0 is the longitude measured east from the meridian of the x axis. 10 You should be aware that, while the definitions given here are those always used by physicists, most mathematics texts reverse the roles of 0 and 0. \" Actually the earth isn't perfectly spherical, so r isn't quite constant, but this doesn't change the conclusion that any position on the surface can be specified by giving 0 and 0. Section 4.8 Central Forces 135 Figure 4.16 The spherical polar coordinates (r, 0, 0) of a point", " P are defined so that r is the distance of P from the origin, 0 is the angle between the line OP and the z axis, and 0 is the angle of the line OQ from the x axis, where Q is the projection of P onto the xy plane. The statement that a function f (r) is spherically symmetric is simply the statement that, with r expressed in spherical polars, f is independent of 0 and 0. This is what we mean when we write f (r) = f (r), and the test for spherical symmetry is simply that the two partial derivatives of /a9 and of/a\u00a2 are both zero everywhere. The unit vectors r, 9, and 0 are defined in the usual way: First, r is the unit vector pointing in the direction of movement if r increases with 0 and 0 fixed. Thus, as shown in Figure 4.17, the vector r points radially outward, and is just the unit vector in the direction of r as usual. (On the surface of the earth, r points upward, in the direction of the local vertical.) Similarly, 9 points in the direction of increasing 0 with r and 0 fixed, that is, southward along a line of longitude. Finally, 0 points in the direction of increasing 0 with r and 0 fixed, that is, eastward along a circle of latitude. Since the three unit vectors 1', 9, and 4; are mutually perpendicular, we can evaluate dot products in spherical polars in just the same way as in Cartesians. Thus, if a = cioe + (10 b bo b + bo;/) and then (make sure you see this) a b = ark, ao be aobo. (4.69) Like the unit vectors of two-dimensional polar coordinates, the unit vectors and 0 vary with position, and, as was the case in two dimensions, this variability complicates many calculations involving differentiation, as we shall now see. 136 Chapter 4 Energy The Gradient in Spherical Polar Coordinates In Cartesian coordinates, we have seen that the components of V f are precisely the partial derivatives of f with respect to x, y, and z,,af,af,af Vf=x\u2014+y\u2014+z\u2014. ax ay az (4.70) The corresponding expression for V f in polar coordinates is not so straightforward. To find it, recall from (4.35) that, in a small displacement dr, the change in any function f (r) is df =V f dr. (4.71) To evaluate the small vector dr in polar coordinates, we must examine carefully what happens to the point r when we change r, 0, and 0: A small change dr in r moves the point a distance dr radially out, in the direction of I. As you can see from Figure 4.17, a small change dO in 6 moves the point around a circle of longitude (radius r) through a distance r dO in the direction of 0. (Note well the factor of r \u2014 the distance is not just d0.) Similarly, a small change d0 in 0 moves the point around a circle of latitude (radius r sin 6) through a distance r sin 0 d4. Putting all this together, we see that dr = drid-rd00\u00b1rsin0d\u00a24. Knowing the components of dr, we can now evaluate the dot product in (4.71) in terms of the unknown components of Vf, df = (V f), dr + (V n o r dO \u00b1 (V Do r sin 0 dO. (4.72) Figure 4.17 The three unit vectors of spherical polar co- ordinates at the point P. The vector r points radially out, 6 points \"south\" along a line of longitude, and 0 points \"east\" around a circle of latitude. Section 4.8 Central Forces 137 Meanwhile, since f is a function of the three variables r, 6, 4), the change in f is, of course, df = \u2014 dr + \u2014 dO + \u2014 dO. af ar af ae a f ao (4.73) Comparing (4.72) and (4.73), we conclude that the components of Vf in spherical polars are a f (Vf), = or, a little more compactly, (VA = r ae' and (Vf) 0 = (4.74) 1 of r sin e ao V f = r \u201eaf ar,slaf 0\u2014 r ae \u00b1 of r sin e ao (4.75) Similar considerations apply to the curl and other operators of vector calculus, all of which are markedly more complicated in spherical polar coordinates (and all other non-Cartesian coordinates) than in Cartesian coordinates. Since the formulas for these operators are very hard to remember, I have listed the more important ones inside the", " back cover. Proofs can be found in any textbook of vector calculus.' 2 Armed with these ideas, let us return to central forces. Conservative and Spherically Symmetric, Central Forces I claimed earlier that a central force is conservative if and only if it is spherically symmetric. This claim can be proved several different ways. The quickest proofs (though not necessarily the most insightful) use spherical polar coordinates. Let us assume first that the central force F(r) is conservative and try to prove that it must be spherically symmetric. Since it is conservative, it can be expressed in the form \u2014VU, which according to (4.75), has the form F(r) = \u2014VU = \u2014r\u2014 \u2014 0\u2014,au -lay ar r ae 1 au r sin0 Bo (4.76) Since F(r) is central, only its radial component can be nonzero, and the last two terms in (4.76) must be zero. This requires that au/a6 = au lao = 0; that is, U(r) is spherically symmetric, and (4.76) reduces to F(r) = \u2014r au. Br Since U is spherically symmetric (depends only on r), the same is true of a( for, and we see that the central force F(r) is indeed spherically symmetric. I shall leave the proof of the converse result, that a central force which is spherically symmetric is necessarily conservative, to the problems at the end of this chapter. (See Problems 4.43 12 See, for example, Mary L. Boas, Mathematical Methods in the Physical Sciences, John Wiley, 1983, p. 431. 138 Chapter 4 Energy and 4.44, but the simplest proof mimics almost exactly the analysis of the Coulomb force in Example 4.5.) The importance of these results is this: First, because a force F(r) that is central and spherically symmetric has a magnitude that depends only on r, it is nearly as simple as a one-dimensional force. Second, although F(r) is certainly not actually a one- dimensional force (its direction still depends on 0 and 0), we shall see in Chapter 8 that any problem involving this kind of force is mathematically equivalent to a certain related one-dimensional problem. 4.9 Energy of Interaction of Two Particles Almost all of our discussion of energy has focused on the energy of a single particle (or any larger object that can be approximated as a particle). It is now time to extend the discussion to systems of several particles, and I shall naturally start with just two particles. In this section, I shall suppose that the two particles interact via forces F12 (on particle 1 by particle 2) and F21 (on particle 2 by particle 1), but that there are no other, external, forces. In general, the force F12 could depend on the positions of both particles, so can be written as and by Newton's third law F12 = F12(r11 r2), F12 = \u2014 F21. As an example of such a two-particle system we could consider an isolated binary star, in which case the only two forces are the gravitational attraction of each star for the other. If we denote the vector pointing to star 1 from star 2 by r, as in Figure 4.18, the force F12 is just the familiar F 12 = Grnim 2, Gin im2 r = r. r3 r2 r r 1 \u2014 r2 Figure 4.18 The vector r pointing to particle 1 from particle 2 is just r = (r 1 \u2014 r2). Section 4.9 Energy of Interaction of Two Particles 139 The vector r can be written in terms of the two positions r 1 and r2. In fact, as can be seen in Figure 4.18, Thus the force F12, expressed as a function of r 1 and r2, is r = r 1 \u2014 r2. F12 = G m 1111 2 1ri \u2014 r2 1 3 (r 1 \u2014 r2). (4.77) A striking property of the force (4.77) is that it depends on the two positions r 1 and r2 only through the particular combination r 1 \u2014 r2. This property is not an accident, and is in fact true of any isolated two-particle system. The reason is that any isolated system must be translationally invariant: If we bodily translate the system to a new position, without changing the relative positions of the particles, the interparticle forces should remain the same. This is illustrated in Figure 4.19, which shows a pair of points r 1 and r2 and a second pair of points s 1 and s2, with s 1 \u2014 s2 = r 1 \u2014 r2. Since the two points r 1 and r2", " could be simultaneously translated to s 1 and s2, the force F12(r1, r2) must be the same as F 12 (s 1, s2) for any points satisfying r 1 \u2014 r2 = s 1 \u2014 s2. In other words, F 12 (r 1, r2) depends only on r 1 \u2014 r2, as claimed, and we can write F12 = F12(rl r2) (4.78) The result (4.78) greatly simplifies our discussion. We can learn almost everything about the force F 12 by fixing r2 at any convenient point. In particular, let us temporarily fix r2 at the origin, in which case (4.78) reduces to just F 12 (r 1 ). (This maneuver amounts to translating both particles until particle 2 is at the origin, and we know that the force is unaffected by any such translation.) With r 2 fixed, our discussion of the force on a single particle now applies. For example, if the force F12 on particle 1 is to be conservative, then it must satisfy V1 x F12 = 0 (4.79) r 1 1 2 r2 S1 s2 Figure 4.19 If r 1 \u2014 r2 = s i \u2014 s2, then two particles at r1 and r2 could be bodily translated to s 1 and s2 without affecting their relative positions. This means that the force between the particles at r 1 and r2 must be the same as that at s 1 and s2. 140 Chapter 4 Energy where V1 is the differential operator \u201e a a ayi + z a az1 F12 = with respect to the coordinates (x 1, y i, z 1) of particle 1. If (4.79) is satisfied, we can define a potential energy U(r 1) such that the force on particle 1 is This gives the force F 12 for the case that particle 2 is at the origin. To find it for particle 2 anywhere else we have only to translate back to an arbitrary position by replacing r 1 with r 1 \u2014 r2 to give F12 = - V1U(r1 (4.80) Notice that I don't have to change the operator V 1, since an operator like a/axi is unchanged by addition of a constant to x 1. To find the reaction force F21 on particle 2, we have only to invoke Newton's third law, which says that F21 = -F12. That is, we have only to change the sign of (4.80). We can re-express this by noticing that V1 U(r1 \u2014 r2) = \u2014V2 U(r 1 \u2014 r2 ), (4.81) where V2 denotes the gradient with respect to the coordinates of particle 2. (To prove this, invoke the chain rule. See Problem 4.50.) So, instead of changing the sign of (4.80) to find F21, we can simply replace V 1 by V2 to give Equations (4.80) and (4.82) are a beautiful result that generalizes to multiparticle systems. To emphasize what they say, let me rewrite them as F21 = -V2U(T1 - r2). (Force on particle 1) = (Force on particle 2) = \u2014 V 2 U. (4.82) (4.83) There is a single potential energy function U, from which we can derive both forces. To find the force on particle 1, we just take the gradient of U with respect to the coordinates of particle 1; to find the force on particle 2, we take the gradient with respect to the coordinates of particle 2. Before generalizing this result to multiparticle systems, let us consider the conser- vation of energy for our two-particle system. Figure 4.20 shows the orbits of the two particles. During a short time interval dt, particle 1 moves through dr 1 and particle 2 through dr2, and work is done on both particles by the corresponding forces. By the work\u2014KE theorem and similarly dT1 = (work on 1) = dr'F12 d T2 = (work on 2) = dr2 F21. Section 4.9 Energy of Interaction of Two Particles 141 2 r2 Figure 4.20 Motion of two interacting particles. During a short time interval dt, particle 1 moves from r 1 to r 1 + dr1 2 to r2 + dr2. and particle 2 from r Adding these, we find for the change in the total kinetic energy T = T1 + T2, dT = dT1 + dT2 = (work on 1) + (work on 2) = Wtot (4.84) where Wot = dr i F12 dr2 F21 denotes the total work done on both particles. Replacing F21 by \u2014F 12 and then replacing F12 with (", "4.80), we can rewrite W tot as Wtot = (dr1 \u2014 dr2) F12 = d(r 1 r2) [\u2014V 1 U(r1 \u2014 r2)]. (4.85) If we rename (r 1 \u2014 r2) as r, then the right side of this equation can be seen to be just (minus) the change in the potential energy, and we find that 13 Wtot = dr VU(r) = \u2014dU (4.86) where the last step follows from the property (4.35) of the gradient operator. It is worth pausing to appreciate this important result. The total work W tot is the sum of two terms, the work done by F12 as particle 1 moves through dr 1 plus the work done by F21 as particle 2 moves through dr 2. According to (4.86), the potential energy U takes both of these terms into account and W tot is simply \u2014dU. Returning to the total kinetic energy, we now see that according to (4.84) the change dT is just \u2014dU. Moving the term dU to the other side, we conclude that d(T + U) = O. That is, the total energy, E = T + U = + T2 U, (4.87) \" If you invoke the chain rule for differentiation, you can see that it makes no difference whether we write V1 U(r) or V U (r). 142 Chapter 4 Energy of our two-particle system is conserved. Note well that the total energy of our two particles contains two kinetic terms (of course), but only one potential term, since U accounts for the work done by both of the forces F12 and F21. Elastic Collisions Elastic collisions give a simple application of these ideas. An elastic collision is a collision between two particles (or bodies that can be treated as particles) that interact via a conservative force that goes to zero as their separation r 1 \u2014 r2 increases. Since the force goes to zero as I r 1 \u2014 r2 I oo, the potential energy U (r \u2014 r2) approaches a constant, which we may as well take to be zero. For example, the two particles could be an electron and a proton, or they could be two billiard balls. That the force between two billiard balls is conservative is not obvious, but it is a fact that billiard balls are manufactured so that they behave like almost perfect (that is, conservative) springs when they are forced together. It is certainly easy to think of other objects (such as lumps of putty), for which the interobject force is nonconservative, and the collisions of such objects are not elastic. In a collision, the two particles start out far apart, approach one another, and then move apart again. Because the forces are conservative, the total energy is conserved; that is, T U = constant (where, of course, T = T1 + T2). But when the particles are far apart, U is zero. Thus if we use the subscripts \"in\" and \"fin\" to label the situations well before and well after the particles come together, then conservation of energy implies that Tin = Tfin (4.88) In other words, an elastic collision can be characterized as a collision in which two particles come together and re-emerge with their total kinetic energy unchanged. However, it is important to remember that there is no principle of conservation of kinetic energy. On the contrary, while the particles are close together their PE is nonzero and their KE certainly is changing. It is only when they are well separated that the PE is negligible and conservation of energy leads to the result (4.88). The foregoing discussion may suggest that elastic collisions should be a very com- mon occurence. All that is needed is two particles whose interaction is conservative. In practice, elastic collisions are not as widespread as this seems to imply. The trouble comes from the requirement that it be two particles that enter and leave the collision. For example, if we fire one billiard ball at a second with sufficient energy, the two balls may shatter. Similarly, if we fire an electron with sufficient energy at an atom, the atom may fall apart or, at least, change the internal motion of its constituents. Even in the collision of two genuine particles, such as an electron and a proton, relativity tells us that, with sufficient energy, new particles can be created. Clearly, at high enough energy, the assumption that the two objects entering a collision can be approximated as indivisible particles eventually breaks down, and we cannot assume that collisions will be elastic, even if all the underlying forces are conservative. Nevertheless, at rea- sonably low energies there are many situations where collisions are perfectly elastic: Section 4.9 Energy of Interaction of Two Particles 143 At sufficiently low energy, collisions of an electron with an atom always are,", " and to a good approximation, the same is true of billiard balls. Elastic collisionS- provide several simple illustrations of the uses of conservation of energy and momentum, of which the following is one. EXAMPLE 4.8 An Equal -Mass, Elastic Collision Consider an elastic collision between two particles of equal mass, m 1 = m 2 = m (for example, two electrons, or two billiard balls), as shown in Figure 4.21. Prove that if particle 2 is initially at rest then the angle between the two outgoing velocities is 0 = 90\u00b0. Conservation of momentum implies that mv i vI2 or That the collision is elastic implies that zmvi = zmvi 2 + 4mAT22 or vi = vl + V2. 2 r2 V 1 -= V 1 + V2 r 2 Squaring (4.89), we find that 2 /2 V 1 = V 1 + 2v' v' v 2 r 2 2'and comparing the last two equations we see that V I 1 V 2i =- 0''that is, vl and v12 are perpendicular (unless one of them is zero, in which case the angle between them is undefined). This result was useful in atomic and nuclear physics; when an unknown projectile hit a stationary target particle, the fact that the two emerged traveling at 90\u00b0 was taken as evidence that the collision was elastic and the two particles had equal masses. 1 2 vi Figure 4.21 Elastic collision between two equal- mass particles. Particle 1 enters with velocity v 1 and collides with the stationary particle 2. The angle between the two final velocities vi and v 12 is 8. 144 Chapter 4 Energy 4.10 The Energy of a Multiparticle System We can extend our discussion of two particles to N particles fairly easily. The main complication is notational: The large number of E signs can make it hard to see clearly what is going on. For this reason, I shall start by considering the case of four particles (N = 4) and write out all of the various sums explicitly. Four Particles Let us consider, then, four particles, as shown in Figure 4.22. The particles can interact with each other (for example, they could be charged, so that each particle experiences the Coulomb force from the three others), and they may be subject to external forces, such as gravity or the Coulomb force of nearby charged bodies. In defining the energy of this system, the easy part is the kinetic energy T, which is, of course, the sum of four terms, T = Ti + T2 + T3 + T4, (4.90) one term TCY 2 = m a 0/2 V for each particle. i To define the potential energy, we must examine the forces on the particles. First, there are the internal forces of the four particles interacting with each other. For each pair of particles there is an action\u2014reaction pair of forces; for example, particles 3 and 4 produce the forces F34 and F43 shown in Figure 4.22. I shall take for granted that each of these interparticle forces Fad is unaffected by the presence of the other particles and any external bodies. For example, F34 is just the same as if particles 1 4 F43 1 F341.16----- 3 2 Figure 4.22 A system of four particles a = 1, 2, 3, 4. For each pair of particles, ap, there is an action\u2014reaction pair of forces, Fap and Fi5a, such as the pair F34 and F43 shown. In addition, each particle a may be subject to an external net force Fec,xt. The four particles could be charged dust motes floating in the air, with the forces Fad being electrostatic and Feaxt being gravity plus buoyancy of the air. Section 4.10 The Energy of a Multiparticle System 145 and 2 and all external bodies were removed. 14 Thus, we can treat the two forces F34 and F43 exactly as in Section 4.9. Provided the forces are conservative, we can define a potential energy U34 = U34 (r3 \u2014 r4) (4.91) and the corresponding forces are the appropriate gradients as in (4.83) F34 = \u2014 V3 U34 and F43 = \u2014 V4 U34. (4.92) There are in all six distinct pairs of particles, 12, 13, 14, 23, 24, 34, and for each pair we can define a corresponding potential energy U12,, U34 from which the corresponding forces are obtained in the same way. Each of the external forces Feuxt depends only on the corresponding position r a. (The force F1', for instance, depends on the position r 1, but not on r2, r3, r4.) Therefore, we can handle F aext exactly as we did the force on a", " single particle. In particular, if Fat is conservative, we can introduce a potential energy Ur t (ra ) and the corresponding force is given by Faext = Va Urt (ra ) (4.93) where, of course, V a denotes differentiation with respect to the coordinates of parti- We can now put all the potential energies together and define the total potential cle a. energy as the sum = Uint uext irf kt-)12, -r T T 13 m I TT I TT Li2,3 I TT 1-1 24 i u\\ L-1 34/ (urct uxt uxt u4ext). (4.94) In this definition, U 1nt is the sum over all six pairs of particles of the pairwise potential energies, U12,, U34, and Uext is the sum of the four potential energies, ulext, rirt u arising from the external forces. It is a fairly straightforward matter to show (see Problem 4.51 for more details) that the force on particle a is just (minus) the gradient of U with respect to the coordinates (x\u201e, y,, z\u201e). Consider, for instance, the gradient \u2014V, U. When \u2014 Vl acts on the first line of (4.94), its action on the first three terms, U12 + U,3 + U14 gives precisely the three internal forces, F,2 + F,3 + F14. Acting on the last three terms, U23 + U24 + U34, it produces zero, since none of these depend on r 1. When \u2014V1 acts on the second line of (4.94), its action on the first term, Ur, produces the external 14 This is quite a subtle point. I am, of course, not denying that the extra particles exert extra forces on particle 3. I claim only that the force of particle 4 on particle 3 is independent of the presence or absence of particles 1 and 2 and any external bodies. One could imagine a world where this claim was false (the presence of particle 1 could somehow change the force of 4 on 3), but experiment seems to confirm that in our world my claim is true. 146 Chapter 4 Energy force Frt. Acting on the last three terms it produces zero, since none of them depend on r 1. Accordingly, = F12 \u00b1 F13 + F14 + Feixt = (net force on particle 1). (4.95) In exactly the same way, we can prove that in general as expected. The second crucial property of our definition of potential energy U is that (provided all the forces concerned are conservative, so we can define U), the total energy, defined as E =T + U, is conserved. We prove this in the now familiar way (for more details, see Problem 4.52): Apply the work\u2014KE theorem to each of the four particles and add the results to show that, in any short time interval, dT = Wtot where Wtot denotes the total work done by all forces on all particles. Next show that W tot = \u2014dU, and conclude that dT = \u2014dU, and hence dE = dT + dU =O. That is, energy is conserved. N Particles The extension of these ideas to an arbitrary number of particles is now quite straight- forward, and I shall just write down the principal formulas. For N particles, labeled a = 1,, N, the total kinetic energy is just the sum of the N separate kinetic energies T = E Tc, =E -2m. V:* a a Assuming that all forces are conservative, for each pair of particles, a,8, we introduce the potential energy Uap that describes their interaction, and for each particle a we introduce the potential energy Ua\"t that describes the net external force on that particle. The total potential energy is then U = uint u ext uap uaext. (4.97) a /3>a a (Here the condition 13 > a in the double sum makes sure we don't double count the internal interactions Uo. For instance, we include U12 but not U21.) With the potential energy U defined in this way, the net force on any particle a is given by \u2014 Voi U, as in Equation (4.96), and total energy E = T + U is conserved. Finally, if any forces are nonconservative, we can define U as the potential energy pertaining to the conservative forces and then show that, in this case, dE = Wne where Wric is the work done by the nonconservative forces. Section 4.10 The Energy of a Multiparticle System 147 Rigid Bodies While the formalism of the last two sections is fairly general and complicated, you can perhaps take some comfort that most applications of the formalism are much simpler than the formalism itself. As one simple example,", " consider a rigid body, such as a golf ball or a meteorite, made up of N atoms. The number N is typically very large, but the energy formalism just developed usually turns out to be very simple. As you probably recall from elementary physics, the total kinetic energy of the N particles rigidly bound together is just the kinetic energy of the center-of-mass motion plus the kinetic energy of rotation. (I'll be proving this in Chapter 10, but I hope you'll accept it for now.) The potential energy of the internal, interatomic forces as given by (4.97) is If the interatomic forces are central (as is usually the case), then, as we saw in Section 4.8, the potential energy Uai3 actually depends on just the magnitude of ra \u2014 r,6 (not its direction). Thus we can rewrite (4.98) as u int = E E (las (ra \u2014 rs ). a p>a E E (las (Ira \u2014 rB I ). a )3>a (4.98) (4.99) Now, as a rigid body moves, the positions ra of its constituent atoms can, of course, move, but the distance Ira \u2014 ri3 I between any two atoms cannot change. (This is, in fact, the definition of a rigid body.) Therefore, if the body concerned is truly rigid, none of the terms in (4.99) can change. That is, the potential energy U int of the internal forces is a constant and can, therefore, be ignored. Thus, in applying energy considerations to a rigid body we can entirely ignore U int and have to worry only about the energy 1./' corresponding to the external forces. Since this latter energy is often a very simple function (see the following example), energy considerations as applied to a rigid body are usually very straightforward. EXAMPLE 4.9 A Cylinder Rolling down an Incline A uniform rigid cylinder of radius R rolls without slipping down a sloping track as shown in Figure 4.23. Use energy conservation to find its speed v when it reaches a vertical height h below its point of release. In accordance with the preceding discussion we can ignore the internal forces that hold the cylinder together. The external forces on the cylinder are the normal and frictional forces of the track and gravity. The first two do no work, and gravity is conservative. As you certainly recall from introductory physics, the gravitational potential energy of an extended body is the same as if all the mass were concentrated at the center of mass. (See Problem 4.6.) Therefore, uext m gy where Y is the height of the cylinder's CM measured up from any convenient reference level. The kinetic energy of the cylinder is T = Z M v 2 + 11 cot, where 148 Chapter 4 Energy Figure 4.23 A uniform cylinder starts from rest and rolls without slipping down a slope through a total vertical drop h= Yin \u2014 Yfin (with the CM coordinate Y measured vertically up). I is its moment of inertia, I = 4mR2, and w is its angular velocity of rolling, w = v/R. Thus the final kinetic energy is T = -3-MV2 4 and the initial KE is zero. Therefore, conservation of energy in the form A T = \u2014AU' implies that 2m v2 = _mg (yfin \u2014 Y in) = Mgh 4 and hence that the final speed is v = Principal Definitions and Equations of Chapter 4 Work\u2014KE Theorem The change in KE of a particle as it moves from point 1 to point 2 is AT -T2 \u2014= f F dr W(1 2) [Eq. (4.7)] 2 where T =lrn v2 and W(1 2) is the work which is done by the total force F on the particle and is defined by the preceding integral. Principal Definitions and Equations of Chapter 4 149 Conservative Forces and Potential Energy A force F on a particle is conservative if (i) it depends only on the particle's position, F = F(r), and (ii) for any two points 1 and 2, the work W(1 \u2014> 2) done by F is the same for all paths joining 1 and 2 (or equivalently, V x F = 0). [Sections 4.2 & 4.4] If F is conservative, we can define a corresponding potential energy so that U(r) \u2014 W (ro --> r) \u2014 f F(r') dr' [Eq. (4.13)] ro and F = \u2014VU. [Eq. (4.33)] If all the forces on a particle are conservative with corresponding potential energies U1,, Uri, then the total mechanical energy E=T Ul + [Eq. (4.22)] is constant. More generally if there are also nonconservative forces, 0E = Wnc, the work done by the nonconservative forces", ". Central Forces A force F(r) is central if it is everywhere directed toward or away from a \"force center.\" If we take the latter to be the origin, F(r) = f (r)i. [Eq. (4.65)] A central force is spherically symmetric [f (r) = f (r)] if and only if it is conservative. [Sec. (4.8)] Energy of a Multiparticle System If all forces (internal and external) on a multiparticle system are conservative, the total potential energy, U = utnt + uext E E ua, + E uaext [Eq. (4.97)] a p>a satisfies and (net force on particle a) = \u2014V a U [Eq. (4.96)] T U = constant. [Problem 4.52] 150 Chapter 4 Energy Problems for Chapter 4 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). SECTION 4.1 Kinetic Energy and Work 4.1 * By writing a b in terms of components prove that the product rule for differentiation applies to the dot product of two vectors; that is, 4.2 ** Evaluate the work done d \u2014 dt (a b) \u2014 b + a \u2014. da dt db dt W = f F dr = f (Fx dx Fydy) (4.100) P 0 0 by the two-dimensional force F = (x 2, 2xy) along the three paths joining the origin to the point P = (1, 1) as shown in Figure 4.24(a) and defined as follows: (a) This path goes along the x axis to Q = (1, 0) and then straight up to P. (Divide the integral into two pieces, fic; = f g + (b) On this path y = x 2, and you can replace the term dy in (4.100) by dy = 2x dx and convert the whole integral into an integral over x. (c) This path is given parametrically as x = t 3, y = t 2. In this case rewrite x, y, dx, and dy in (4.100) in terms of t and dt, and convert the integral into an integral over t. 4.3 ** Do the same as in Problem 4.2, but for the force F = (\u2014y, x) and for the three paths joining P and Q shown in Figure 4.24(b) and defined as follows: (a) This path goes straight from P = (1, 0) to the origin and then straight to Q = (0, 1). (b) This is a straight line from P to Q. (Write y as a function of x and rewrite the integral as an integral over x.) (c) This is a quarter-circle centered on the origin. (Write x and y in polar coordinates and rewrite the integral as an integral over 0.) 4.4 ** A particle of mass m is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially the particle is moving in a circle of radius ro with angular velocity w o, but I now pull the string down through the hole until a length r remains between the hole and the particle. (a) What is the particle's angular velocity now? (b) Assuming that I pull the string so slowly that we can approximate the particle's path by a (a) (b) Figure 4.24 (a) Problem 4.2. (b) Problem 4.3 circle of slowly shrinking radius, calculate the work I did pulling the string. (c) Compare your answer to part (b) with the particle's gain in kinetic energy. Problems for Chapter 4 151 SECTION 4.2 Potential Energy and Conservative Forces 4.5 * (a) Consider a mass m in a uniform gravitational field g, so that the force on m is mg, where g is a constant vector pointing vertically down. If the mass moves by an arbitrary path from point 1 to point 2, show that the work done by gravity is Wgray (1 \u2014> 2) = \u2014mgh where h is the vertical height gained between points 1 and 2. Use this result to prove that the force of gravity is conservative (at least in a region small enough so that g can be considered constant). (b) Show that, if we choose axes with y measured vertically up, the gravitational potential energy is U = mgy (if we choose U = 0 at the origin). 4.6 * For a system of N particles subject to a uniform gravitational field g acting vertically down, prove that the total gravitational potential energy is the same as if all the mass were concentrated at the center of mass of the system; that is, MgY a where M = E", " ma is the total mass and R = (X, Y, Z) is the position of the CM, with the y coordinate measured vertically up. [Hint: We know from Problem 4.5 that U, = 4.7 * Near to the point where I am standing on the surface of Planet X, the gravitational force on a mass m is vertically down but has magnitude my y 2 where y is a constant and y is the mass's height above the horizontal ground. (a) Find the work done by gravity on a mass m moving from r 1 to r2, and use your answer to show that gravity on Planet X, although most unusual, is still conservative. Find the corresponding potential energy. (b) Still on the same planet, I thread a bead on a curved, frictionless, rigid wire, which extends from ground level to a height h above the ground. Show clearly in a picture the forces on the bead when it is somewhere on the wire. (Just name the forces so it's clear what they are; don't worry about their magnitude.) Which of the forces are conservative and which are not? (c) If I release the bead from rest at a height h, how fast will it be going when it reaches the ground? 4.8 ** Consider a small frictionless puck perched at the top of a fixed sphere of radius R. If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy to find the puck's speed as a function of its height, then use Newton's second law to find the normal force of the sphere on the puck. At what value of this normal force does the puck leave the sphere?] 4.9 ** (a) The force exerted by a one-dimensional spring, fixed at one end, is F = \u2014kx, where x is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is U = 4kx2, if we choose U to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass m suspended from the other end and constrained to move in the vertical direction only. Find the extension x o of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form i ky 2 if we use the coordinate y equal to the displacement measured from the new equilibrium position at x = xo (and redefine our reference point so that U = 0 at y = 0). 152 Chapter 4 Energy SECTION 4.3 Force as the Gradient of Potential Energy 4.10 * Find the partial derivatives with respect to x, y, and z of the following functions: (a) f (x, y, z) = ax e bxy cy 2, (b)g(x, y, z) = sin (axyz 2), (c) h(x, y, z) = aexYlz, where a, b, and c are constants. Remember that to evaluate a flax you differentiate with respect to x treating y and z as constants. 2 4.11 * Find the partial derivatives with respect to x, y, and z of the following functions: (a) f (x, y, z) = aye 2byz cz 2, (b) g(x, y, z) = cos(axy 2z3), (c) h(x, y, z) = ar, where a, b, and c are constants and r = \\/x 2 + y2 + z2. Remember that to evaluate of/ax you differentiate with respect to x treating y and z as constants. 4.12 * Calculate the gradient Vf of the following functions, f (x, y, z): (a) f = x 2 + z3. (b) f = ky, where k is a constant. (c) f = r = -/x2 + y2 + z2. [Hint: Use the chain rule.] (d) f = 1/ r. 4.13 * Calculate the gradient Vf of the following functions, f (x, y, z): (a) f = ln(r ), (b) f = rn, (c) f = g(r), where r =,./x 2 + y2 + z2 and g(r) is some unspecified function of r. [Hint: Use the chain rule.] 4.14 * Prove that if f (r) and g(r) are any two scalar functions of r, then V (fg), fVg+ gVf 4.15 * For f (r) = x 2 + 2y 2 3z2, use the approximation (4.", "35) to estimate the change in f if we move from the point r = (1, 1, 1) to (1.01, 1.03, 1.05). Compare with the exact result. 4.16 * If a particle's potential energy is U(r) = k(x2 + y2 + z 2), ) where k is a constant, what is the force on the particle? 4.17 * A charge q in a uniform electric field E o experiences a constant force F = q Eo. (a) Show that this force is conservative and verify that the potential energy of the charge at position r is U (r) = \u2014qE0 r. (b) By doing the necessary derivatives, check that F = \u2014 V U. 4.18 ** Use the property (4.35) of the gradient to prove the following important results: (a) The vector Vf at any point r is perpendicular to the surface of constant f through r. (Choose a small displacement dr that lies in a surface of constant f. What is df for such a displacement?) (b) The direction of Vf at any point r is the direction in which f increases fastest as we move away from r. (Choose a small displacement dr = eu, where u is a unit vector and E is fixed and small. Find the direction of u for which the corresponding df is maximum, bearing in mind that a b = ab cos 9.) 4.19 ** (a) Describe the surfaces defined by the equation f = const, where f = x 2 + 4y2. (b) Using the results of Problem 4.18, find a unit normal to the surface f = 5 at the point (1, 1, 1). In what direction should one move from this point to maximize the rate of change of f? SECTION 4.4 The Second Condition that F be Conservative 4.20 * Find the curl, V x F, for the following forces: (a) F = kr; (b) F = (Ax, By e, Cz 3); (c) F = (Ay 2, Bx, Cz), where A, B, C and k are constants. 4.21 * Verify that the gravitational force \u2014GMini 1r 2 on a point mass m at r, due to a fixed point mass M at the origin, is conservative and calculate the corresponding potential energy. 4.22 * The proof in Example 4.5 (page 119) that the Coulomb force is conservative is considerably simplified if we evaluate V x F using spherical polar coordinates. Unfortunately, the expression for V x F in spherical polar coordinates is quite messy and hard to derive. However, the answer is given Problems for Chapter 4 153 inside the back cover, and the proof can be found in any book on vector calculus or mathematical methods. 15 Taking the expression inside the back cover on faith, prove that the Coulomb force F = y r 2 is conservative. 4.23 ** Which of the following forces is conservative? (a) F = k(x, 2y, 3z) where k is a constant. (b) F = k(y, x, 0). (c) F = k(\u2014y, x, 0). For those which are conservative, find the corresponding potential energy U, and verify by direct differentiation that F = \u2014 V U. 4.24*** An infinitely long, uniform rod of mass p, per unit length is situated on the z axis. (a) Calculate the gravitational force F on a point mass m at a distance p from the z axis. (The gravitational force between two point masses is given in Problem 4.21.) (b) Rewrite F in terms of the rectangular coordinates (x, y, z) of the point and verify that V x F = 0. (c) Show that V x F = 0 using the expression for V x F in cylindrical polar coordinates given inside the back cover. (d) Find the corresponding potential energy U. 4.25 *** The proof that the condition V x F = 0 guarantees the path independence of the work fi2 F dr done by F is unfortunately too lengthy to be included here. However, the following three exercises capture the main points: 16 (a) Show that the path independence of fi2 F dr is equivalent to the statement that the integral fr, F dr around any closed path F is zero. (By tradition, the symbol f is used for integrals around a closed path \u2014 a path that starts and stops at the same point.) [Hint: For any two points 1 and 2 and any two paths from 1 to 2, consider the work done by F going from 1 to 2 along the first path and then back to 1 along the second in the reverse direction.] (b) Stokes's theorem asserts that A, F dr = f (V x F) dA, where the integral on the right is a surface integral over", " a surface for which the path F is the boundary, and ii. and dA are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if V x F = 0 everywhere, then A, F dr = 0. (c) While the general proof of Stokes's theorem is beyond our scope here, the following special case is quite easy to prove (and is an important step toward the general proof): Let F denote a rectangular closed path lying in a plane perpendicular to the z direction and bounded by the lines x = B, x = B + b, y = C and y = C c. For this simple path (traced counterclockwise as seen from above), prove Stokes's theorem that home.] F dr = f (V x F) ill d A where it = z and the integral on the right runs over the flat, rectangular area inside F. [Hint: The integral on the left contains four terms, two of which are integrals over x and two over y. If you pair them in this way, you can combine each pair into a single integral with an integrand of the form Fx (x, C c, z) \u2014 F x (x, C, z) (or a similar term with the roles of x and y exchanged). You can rewrite this integrand as an integral over y of a Fx (x, y, z)/ay (and similarly with the other term), and you're SECTION 4.5 Time-Dependent Potential Energy 4.26 * A mass m is in a uniform gravitational field, which exerts the usual force F = mg vertically down, but with g varying with time, g = g (t). Choosing axes with y measured vertically up and defining U = mgy as usual, show that F = \u2014 V U as usual, but, by differentiating E = 4mv2 + U with respect to t, show that E is not conserved. 15 See, for example, Mathematical Methods in the Physical Sciences by Mary Boas (Wiley, 1983), p. 435. 16 For a complete discussion see, for example, Mathematical Methods, Boas, Ch. 6, Sections 8-11. 154 Chapter 4 Energy Figure 4.25 Problem 4.30 4.27 ** Suppose that the force F(r, t) depends on the time t but still satisfies V x F = 0. It is a mathematical fact (related to Stokes's theorem as discussed in Problem 4.25) that the work integral fi2 F(r, t) dr (evaluated at any one time t) is independent of the path taken between the points 1 and 2. Use this to show that the time-dependent PE defined by (4.48), for any fixed time t, has the claimed property that F(r, t) = \u2014 V U (r, t). Can you see what goes wrong with the argument leading to Equation (4.19), that is, conservation of energy? SECTION 4.6 Energy for Linear One-Dimensional Systems 4.28 ** Consider a mass m on the end of a spring of force constant k and constrained to move along the horizontal x axis. If we place the origin at the spring's equilibrium position, the potential energy is 2 kx 2. At time t = 0 the mass is sitting at the origin and is given a sudden kick to the right so that it moves out to a maximum displacement at x max = A and then continues to oscillate about the origin. (a) Write down the equation for conservation of energy and solve it to give the mass's velocity x in terms of the position x and the total energy E. (b) Show that E = 2kA 2, and use this to eliminate E from your expression for Use the result (4.58), t = f dxf li(x'), to find the time for the mass to move from the origin out to a position x. (c) Solve the result of part (b) to give x as a function of t and show that the mass executes simple harmonic motion with period 27,/m1k. 4.29 ** [Computer] A mass m confined to the x axis has potential energy U = kx4 with k > 0. (a) Sketch this potential energy and qualitatively describe the motion if the mass is initially stationary at x = 0 and is given a sharp kick to the right at t = 0. (b) Use (4.58) to find the time for the mass to reach its maximum displacement x max = A. Give your answer as an integral over x in terms of m, A, and k. Hence find the period r of oscillations of amplitude A as an integral. (c) By making a suitable change of variables in the integral, show that the period r is inversely proportional to the amplitude A. (d) The integral of part (b) cannot be evaluated in", " terms of elementary functions, but it can be done numerically. Find the period for the case that m = k = A = 1. SECTION 4.7 Curvilinear One-Dimensional Systems 4.30 * Figure 4.25 shows a child's toy, which has the shape of a cylinder mounted on top of a hemisphere. The radius of the hemisphere is R and the CM of the whole toy is at a height h above the floor. (a) Write down the gravitational potential energy when the toy is tipped to an angle 0 from the vertical. [You need to find the height of the CM as a function of 0. It helps to think first about the height of the hemisphere's center 0 as the toy tilts.] (b) For what values of R and h is the equilibrium at 0 = 0 stable? Problems for Chapter 4 155 Figure 4.26 Problem 4.34 4.31 * (a) Write down the total energy E of the two masses in the Atwood machine of Figure 4.15 in terms of the coordinate x and x. (b) Show (what is true for any conservative one-dimensional system) that you can obtain the equation of motion for the coordinate x by differentiating the equation E = const. Check that the equation of motion is the same as you would obtain by applying Newton's second law to each mass and eliminating the unknown tension from the two resulting equations. 4.32 ** Consider the bead of Figure 4.13 threaded on a curved rigid wire. The bead's position is specified by its distance s, measured along the wire from the origin. (a) Prove that the bead's speed v is just v (Write v in terms of its components, dx/dt, etc., and find its magnitude using Pythagoras's theorem.) (b) Prove that m's' = Ftang, the tangential component of the net force on the bead. (One way to do this is to take the time derivative of the equation v 2 = v v. The left side should lead you to :5' and the right to Ftang.) (c) One force on the bead is the normal force N of the wire (which constrains the bead to stay on the wire). If we assume that all other forces (gravity, etc.) are conservative, then their resultant can be derived from a potential energy U. Prove that Ftang = \u2014dU Ids. This shows that one- dimensional systems of this type can be treated just like linear systems, with x replaced by s and Fx by Ftang 4.33 ** [Computer] (a) Verify the expression (4.59) for the potential energy of the cube balanced on a cylinder in Example 4.7 (page 130). (b) Make plots of U(8) for b = 0.9r and b = 1.1r. (You may as well choose units such that r, m, and g are all equal to 1.) (c) Use your plots to confirm the findings of Example 4.7 concerning the stability of the equilibrium at 0 = 0. Are there any other equilibrium points and are they stable? 4.34 ** An interesting one-dimensional system is the simple pendulum, consisting of a point mass m, fixed to the end of a massless rod (length 1), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure 4.26. The pendulum's position can be specified by its angle 0 from the equilibrium position. (It could equally be specified by its distance s from equilibrium \u2014 indeed s = 10\u2014but the angle is a little more convenient.) (a) Prove that the pendulum's potential energy (measured from the equilibrium level) is U(0) = mgl(1 \u2014 cos 0). (4.101) Write down the total energy E as a function of 0 and 0. (b) Show that by differentiating your expression for E with respect to t you can get the equation of motion for 0 and that the equation of motion is just the familiar F = /a (where F is the torque, I is the moment of inertia, and a is the angular acceleration 0). (c) Assuming that the angle 0 remains small throughout the motion, solve for 0 (t) and show that the motion is periodic with period to = 270 / g. (4.102) 156 Chapter 4 Energy Figure 4.27 Problem 4.36 (The subscript \"o\" is to emphasize that this is the period for small oscillations.) 4.35 ** Consider the Atwood machine of Figure 4.15, but suppose that the pulley has radius R and moment of inertia I. (a) Write down the total energy of the two masses and the pulley in terms of the coordinate x and I. (Remember that the kinetic energy of a spinning wheel is i ho2.) (b) Show (what is true for any conservative one-dimensional", " system) that you can obtain the equation of motion for the coordinate x by differentiating the equation E = const. Check that the equation of motion is the same as you would obtain by applying Newton's second law separately to the two masses and the pulley, and then eliminating the two unknown tensions from the three resulting equations. 4.36 ** A metal ball (mass m) with a hole through it is threaded on a frictionless vertical rod. A massless string (length 1) attached to the ball runs over a massless, frictionless pulley and supports a block of mass M, as shown in Figure 4.27. The positions of the two masses can be specified by the one angle 9. (a) Write down the potential energy U(9). (The PE is given easily in terms of the heights shown as h and H. Eliminate these two variables in favor of 9 and the constants b and 1. Assume that the pulley and ball have negligible size.) (b) By differentiating U(0) find whether the system has an equilibrium position, and for what values of m and M equilibrium can occur. Discuss the stability of any equilibrium positions. 4.37 *** [Computer] Figure 4.28 shows a massless wheel of radius R, mounted on a frictionless, horizontal axle. A point mass M is glued to the edge of the wheel, and a mass m hangs from a string wrapped around the perimeter of the wheel. (a) Write down the total PE of the two masses as a function of the angle 0. (b) Use this to find the values of m and M for which there are any positions of equilibrium Describe the equilibrium positions, discuss their stability, and explain your answers in terms of torques. (c) Plot U (0) for the cases that m = 0.7M and m = 0.8M, and use your graphs to describe the behavior of the system if I release it from rest at 0 = 0. (d) Find the critical value of m/M on one side of which the system oscillates and on the other side of which it does not (if released from rest at 0 = 0). 4.38 *** [Computer] Consider the simple pendulum of Problem 4.34. You can get an expression for the pendulum's period (good for large oscillations as well as small) using the method discussed in connection with (4.57), as follows: (a) Using (4.101) for the PE, find 0 as a function of 0. Next use (4.57), in the form t = f d04, to write the time for the pendulum to travel from 0 = 0 to its maximum value (the amplitude) O. Because this time is a quarter of the period t, you can now write down the period. Show that Problems for Chapter 4 157 Figure 4.28 Problem 4.37 1 = to \u20147r f dO \u00b0 -isin2 (e13/2) \u2014 sin2(4)/2) 2 f 1 = r 0 du 7.,,/1-.2,/1- A2.2' (4.103) where ro is the period (4.102) (Problem 4.34) for small oscillations and A = sin(eD/2). [To get the first expression you will need to use the trig identity for 1 \u2014 cos 4) in terms of sin 2 (4)/2). To get the second you need to make the substitution sin(4)/2) = Au.] These integrals cannot be evaluated in terms of elementary functions. However, the second integral is a standard integral called the complete elliptic integral of the first kind, sometimes denoted K (A 2), whose values are tabulated 17 and are known to computer software such as Mathematica [which calls it EllipticK(A 2)]. (b) If you have access to computer software that knows this function, make a plot of r/r o for amplitudes 0 < < 3 rad. Comment. What becomes of r as the amplitude of oscillation approaches 7r? Explain. 4.39 *** (a) If you have not already done so, do Problem 4.38(a). (b) If the amplitude eD is small then so is A = sin el3/2. If the amplitude is very small, we can simply ignore the last square root in (4.103). Show that this gives the familiar result for the small-amplitude period, r = To = 2701g. (c) If the amplitude is small but not very small, we can improve on the approximation of part (b). Use the binomial expansion to give the approximation 1/- \u2018/1 \u2014 A 2u 2 ti 1 + 1A 2 u 2 and show that, in this approximation, (4.103) gives r = ro [l sin2 (43/2)]. What percentage correction does the second", " term represent for an amplitude of 45\u00b0? (The exact answer for elp = 45\u00b0 is 1.040 ro to four significant figures.) SECTION 4.8 Central Forces 4.40 * (a) Verify the three equations (4.68) that give x, y, z in terms of the spherical polar coordinates r, 9, (b) Find expressions for r, in terms of x, y, z. 17 See, for example, M.Abramowitz and I.Stegun, Handbook of Mathematical Functions, Dover, New York, 1965. Be warned that different authors use different notations. In particular, some authors call the exact same integral K (A). 158 Chapter 4 Energy Figure 4.29 Problem 4.44 4.41 * A mass m moves in a circular orbit (centered on the origin) in the field of an attractive central force with potential energy U = krn. Prove the virial theorem that T = nU 12. 4.42 * In one dimension, it is obvious that a force obeying Hooke's law is conservative (since F = \u2014kx depends only on the position x, and this is sufficient to guarantee that F is conservative in one dimension). Consider instead a spring that obeys Hooke's law and has one end fixed at the origin, but whose other end is free to move in all three dimensions. (The spring could be fastened to a point in the ceiling and be supporting a bouncing mass m at its other end, for instance.) Write down the force F(r) exerted by the spring in terms of its length r and its equilibrium length ro. Prove that this force is conservative. [Hints: Is the force central? Assume that the spring does not bend.] 4.43 ** In Section 4.8, I claimed that a force F(r) that is central and spherically symmetric is automati- cally conservative. Here are two ways to prove it: (a) Since F(r) is central and spherically symmetric, it must have the form F(r) = f (r)r. Using Cartesian coordinates, show that this implies that V x F = 0. (b) Even quicker, using the expression given inside the back cover for V x F in spherical polars, show that V x F = 0. 4.44 ** Problem 4.43 suggests two proofs that a central, spherically symmetric force is automatically conservative, but neither proof makes really clear why this is so. Here is a proof that is less complete but more insightful: Consider any two points A and B and two different paths ACB and ADB connecting them as shown in Figure 4.29. Path ACB goes radially out from A until it reaches the radius rB of B, and then around a sphere (center 0) to B. Path ADB goes around a sphere of radius rA until it reaches the line OB, and then radially out to B. Explain clearly why the work done by a central, spherically symmetric force F is the same along both paths. (This doesn't prove that the work is the same along any two paths from A to B. If you want you can complete the proof by showing that any path can be approximated by a series of paths moving radially in or out and paths of constant r.) 4.45 ** In Section 4.8, I proved that a force F(r) = f (r)i that is central and conservative is automat- ically spherically symmetric. Here is an alternative proof: Consider the two paths ACB and ADB of Figure 4.29, but with rB = rA + dr where dr is infinitesimal. Write down the work done by F(r) going around both paths, and use the fact that they must be equal to prove that the magnitude function f (r) must be the same at points A and D; that is, f (r) = f (r) and the force is spherically symmetric. Problems for Chapter 4 159 SECTION 4.9 Energy of Interaction of Two Particles 4.46 * Consider an elastic collision of two particles as in Example 4.8 (page 143), but with unequal masses, m 1 0 m 2. Show that the angle 0 between the two outgoing velocities satisfies 9 < 7r/2 if m i > m2, but 0 > r/2 if m i < m 2. 4.47 * Consider a head-on elastic collision between two particles. (Since the collision is head-on, the motion is confined to a single straight line and is therefore one-dimensional.) Prove that the relative velocity after the collision is equal and opposite to that before. That is, v 1 \u2014 v2 = \u2014(1/1 \u2014 v;), where v 1 and v2 are the initial velocities and vi and v'2 the corresponding", " final velocities. 4.48* A particle of mass m 1 and speed v 1 collides with a second particle of mass m 2 at rest. If the collision is perfectly inelastic (the two particles lock together and move off as one) what fraction of the kinetic energy is lost in the collision? Comment on your answer for the cases that m 1 <-< m 2 and that M2 << M 1' that on particle 2. 4.49 ** Both the Coulomb and gravitational forces lead to potential energies of the form U = y \u2014 r2 1, where y denotes ka 1q2 in the case of the Coulomb force and \u2014Gm 1m 2 for gravity, and r 1 and r2 1 U is the force on particle 1 and \u2014V 2 U are the positions of the two particles. Show in detail that \u2014V 4.50 ** The formalism of the potential energy of two particles depends on the claim in (4.81) that V1 U(r 1 \u2014 r2) = \u2014V 2 U(r 1 \u2014 r2). Prove this. (Use the chain rule for differentiation. The proof in three dimensions is notationally awkward, so prove the one-dimensional result that a ax, f (x i \u2014 x 2) = \u2014 \u2014 f (x i \u2014 x 2) a ax2 and then convince yourself that it extends to three dimensions.) SECTION 4.10 The Energy of a Multiparticle System 4.51 ** Write out the arguments of all the potential energies of the four-particle system in (4.94). For instance U = U(r 1, r2,, r 4), whereas U34 = U34 (r3 - r4). Show in detail that the net force on particle 3 (for instance) is given by \u2014V3 U. [You know that the separate forces, internal and external, are given by (4.92) and (4.93).] 4.52 ** Consider the four-particle system of Section 4.10. (a) Write down the work\u2014KE theorem for each of the four particles separately and, by adding these four equations, show that the change in the total KE in a short time interval dt is dT = Wtot where Wtot is the total work done on all particles by all forces. [This shouldn't take more than two or three lines.] (b) Next show that Wtot = \u2014dU where dU is the change in total PE during the same time interval. Deduce that the total mechanical energy E = T U is conserved. 4.53 ** (a) Consider an electron (charge \u2014e and mass m) in a circular orbit of radius r around a fixed proton (charge +e). Remembering that the inward Coulomb force ke g /r2 is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to \u20144 times its PE; that is, T = \u20141U and hence E = z U. (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius r around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches 160 Chapter 4 Energy from afar with kinetic energy T2. When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius r'. (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy E long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables T2, r, and r'. CHAPTER Oscillations Almost any system that is displaced from a position of stable equilibrium exhibits oscillations. If the displacement is small, the oscillations are almost always of the type called simple harmonic. Oscillations, and particularly simple harmonic oscillations, are therefore extremely widespread. They are also extremely useful. For example, all good clocks depend on an oscillator to regulate their time keeping: The first reliable clocks used a pendulum; the first accurate watches (historically crucial in navigation) used an oscillating balance wheel; modern watches use the oscillations of a quartz crystal; and today's most accurate clocks, such as the atomic clock at the National Institute for Standards and Technology in Boulder, Colorado, use the oscillations of an atom. In this chapter, we shall explore the physics and mathematics of oscillations. I shall begin with simple harmonic oscillations and then go on to", " damped oscillations (oscillations that die out because of resistive forces) and driven oscillations (oscillations that are maintained by an outside driving force, as in all clocks). The last three sections of this chapter describe the use of Fourier series in finding the motion of an oscillator driven by an arbitrary periodic driving force. 5.1 Hooke's Law As you are certainly aware, a mass on the end of a spring that obeys Hooke's law executes oscillations of the type that we call simple harmonic. Before we review the proof of this claim, let us first ask why Hooke's law is so important and appears so frequently. Hooke's law asserts that the force exerted by a spring has the form (for now we'll restrict ourselves to a spring confined to the x axis) Fx (x) = \u2014kx (5.1) where x is the displacement of the spring from its equilibrium length and k is a positive number called the force constant. That k is positive means that the equilibrium at x = 0 is stable: When x = 0 there is no force, when x > 0 (displacement to the right) the 161 162 Chapter 5 Oscillations force is negative (back to the left), and when x < 0 (displacement to the left) the force is positive (back to the right); either way, the force is a restoring force, and the equilibrium is stable. (If k were negative, the force would be away from the origin, and the equilibrium would be unstable, in which case we do not expect to see oscillations.) An exactly equivalent way to state Hooke's law is that the potential energy is U(x) = 2 kx 2. Consider now an arbitrary conservative one-dimensional system which is specified by a coordinate x and has potential energy U (x). Suppose that the system has a stable equilibrium position x = x0, which we may as well take to be the origin (x 0 = 0). Now consider the behavior of U(x) in the vicinity of the equilibrium position. Since any reasonable function can be expanded in a Taylor series, we can safely write U(x) = U(0) + c(o)x + 2u\"(0)x2+. (5.2) As long as x remains small, the first three terms in this series should be a good approximation. The first term is a constant, and, since we can always subtract a constant from U(x) without affecting any physics, we may as well redefine U (0) to be zero. Because x = 0 is an equilibrium point, U' (0) = 0 and the second term in the series (5.2) is automatically zero. Because the equilibrium is stable, U\"(0) is positive. Renaming U\"(0) as k, we conclude that for small displacements it is always a good approximation to take' U(x) = Ikx 2. (5.3) That is, for sufficiently small displacements from stable equilibrium, Hooke's law is always valid. Notice that if U\"(0) were negative, then k would also be negative, and the equilibrium would be unstable \u2014 a case we're not interested in just now. Hooke's law in the form (5.3) crops up in many situations, although it is certainly not necessary that the coordinate be the rectangular coordinate x, as the following example illustrates. EXAMPLE 5.1 The Cube Balanced on a Cylinder Consider again the cube of Example 4.7 (page 130) and show that for small angles B the potential energy takes the Hooke's-law form U(9) = 40 2. We saw in that example that U(0) = mg[(r b) cos 0 \u00b1 r0 sing]. If 0 is small we can make the approximations cos 9 ti 1 \u2014 0 2 /2 and sin 0 ti 0, so that case here. U(0) mg[(r b)(1 \u2014 0 2) + r02] = mg (r b) m g (r \u2014 b)0 2, 1 The only exception is if Un(0) happens to be zero, but I shall not worry about this exceptional Section 5.2 Simple Harmonic Motion 163 which, apart from the uninteresting constant, has the form 4/(0 2 with \"spring constant\" k = mg(r \u2014 b). Notice that the equilibrium is stable (k positive) only when r > b, a condition we had already found in Example 4.7. Figure 5.1 A mass m with potential energy U(x) = zkx 2 and total energy E oscillates between the two turning points at x = +A, where U(x) = E and the kinetic energy is zero. As discussed in Section 4.6, the general features of the motion of any one- dimensional system can be understood from a graph of U(x) against", " x. For the Hooke's-law potential energy (5.3), this graph is a parabola, as shown in Figure 5.1. If a mass in has potential energy of this form and has any total energy E > 0, it is trapped and oscillates between the two turning points where U(x) = E, so that the kinetic energy is zero and the mass is instantaneously at rest. Because U(x) is symmetric about x = 0, the two turning points are equidistant on opposite sides of the origin and are traditionally denoted x = \u00b1A, where A is called the amplitude of the oscillations. 5.2 Simple Harmonic Motion We are now ready to examine the equation of motion (that is, Newton's second law) for a mass m that is displaced from a position of stable equilibrium. To be definite, let us consider a cart on a frictionless track attached to a fixed spring as sketched in Figure 5.2. We have seen that we can approximate the potential energy by (5.3) or, equivalently, the force by Fx (x) = \u2014kx. Thus the equation of motion is mx= Fx = \u2014kx or = --x = \u2014co2 X m (5.4) where I have introduced the constant w =- 1,1 \u2014, rn which we shall see is the angular frequency with which the cart will oscillate. Although we have arrived at Equation (5.4) in the context of a cart on a spring moving along 164 Chapter 5 Oscillations x = 0 (equilibrium) Figure 5.2 A cart of mass m oscillating on the end of a spring. the x axis, we shall see eventually that it applies to many different oscillating systems in many different coordinate systems. For example, we have already seen in Equation (1.55) that the angle 4) that gives the position of a pendulum (or a skateboard in a half-pipe) is governed by the same equation, 4 = \u20140) 20, at least for small values of 0. In this section I am going to review the properties of the solutions of (5.4). Unfortunately, there are many different ways to write the same solution, all of which have their advantages, and you should be comfortable with them all. The Exponential Solutions Equation (5.4) is a second-order, linear, homogeneous differential equation 2 and so has two independent solutions. These two independent solutions can be chosen in several different ways, but perhaps the most convenient is this: x(t) = e iwt and x(t) = As you can easily check, both of these functions do satisfy (5.4). Further, any constant multiple of either solution is also a solution, and likewise any sum of such multiples. Thus the function x(t) = C ie iwt (5.5) is also a solution for any two constants C 1 and C2. (That any linear combination of solutions like this is itself a solution is called the superposition principle and plays a crucial role in many branches of physics.) Since this solution (5.5) contains two arbitrary constants, it is the general solution of our second-order equation (5.4). 3 any solution can be expressed in the form (5.5) by suitable choice of the Therefore, coefficients C 1 and C2. 2 Linear because it contains no higher powers of x or its derivatives than the first power, and homogeneous because every term is a first power (that is, there is no term independent of x and its derivatives). 3 Recall the result, discussed below Equation (1.56), that the general solution of a second-order differential equation contains precisely two arbitrary constants. Section 5.2 Simple Harmonic Motion 165 The Sine and Cosine Solutions The exponential functions in (5.5) are so convenient to handle that (5.5) is often the best form of the solution. Nevertheless, this form does have one disadvantage. We know, of course, that x (t) is real, whereas the two exponentials in (5.5) are complex. This means the coefficients C 1 and C2 must be chosen carefully to ensure that x (t) itself is real. I shall return to this point shortly, but first I shall rewrite (5.5) in another useful way. From Euler's formula (2.76) we know that the two exponentials in (5.5) can be written as e \u00b1iwt = cos(cot) i sin(cot). Substituting into (5.5) and regrouping we find that x(t) = (C1 + C2 ) cos(cot) i (CI \u2014 C2) sin(wt) = B i cos(wt) + B2 sin(wt) (5.6) (5.7) where B 1", " and B2 are simply new names for the coefficients in the previous line, B 1 - \u00b1 C2 and B2 -= i (C1 C2). The form (5.6) can be taken as the definition of simple harmonic motion (or SHM): Any motion that is a combination of a sine and cosine of this form is called simple harmonic. Because the functions cos(wt) and sin(cot) are real, the requirement that x (t) be real means simply that the coefficients B 1 and B2 must be real. We can easily identify the coefficients B 1 and B2 in terms of the initial conditions of the problem. Clearly at t = 0, (5.6) implies that x (0) = B 1. That is, B 1 is just the initial position x (0) = x 0. Similarly, by differentiating (5.6), we can identify coB 2 as the initial velocity v o. If I start the oscillations by pulling the cart aside to x = x0 and releasing it from rest (vo = 0), then B2 = 0 in (5.6) and only the cosine term survives, so that x(t) = cos(cot). (5.8) If I launch the cart from the origin (x o = 0) by giving it a kick at t = 0, only the sine term survives, and v. x(t) = \u2014 sin(wt). These two simple cases are illustrated in Figure 5.3. Notice that both solutions, like the general solution (5.6), are periodic because both the sine and cosine are. Since the argument of both sine and cosine is cot, the function x(t) repeats itself after the time r for which cot = 27. That is, the period is 27r = - = 2 (5.9) 166 Chapter 5 Oscillations (a) (b) Figure 5.3 (a) Oscillations in which the cart is released from x o at t = 0 follow a cosine curve. (b) If the cart is kicked from the origin at t = 0, the oscillations follow a sine curve with initial slope v o. In either case the period of the oscilla- tions is r = 2jr/w = k and is the same whatever the values of x o or vo. The Phase-Shifted Cosine Solution The general solution (5.6) is harder to visualize than the two special cases of Fig- ure 5.3, and it can usefully be rewritten as follows: First, we define yet another constant A =,\\/B12 + B22. (5.10) Notice that A is the hypotenuse to a right triangle whose other two sides are B I and B2. I have indicated this in Figure 5.4, where I have also defined 3 as the lower angle of that triangle. We can now rewrite (5.6) as x(t) = A \u2014 [ cos(wt) + sin(wt) ] B1 A = A[ cos 8 cos(cot) + sin 3 sin (wt)] = A cos(wt \u2014 O. (5.11) From this form it is clear that the cart is oscillating with amplitude A, but instead of being a simple cosine as in (5.8), it is a cosine which is shifted in phase: When t = 0 the argument of the cosine is \u20148, and the oscillations lag behind the simple cosine by the phase shift 6. We have derived the result (5.11) from Newton's second law, but, as so often happens, one can derive the same result in more than one way. In particular, Figure 5.4 The constants A and 6 are defined in terms of B l and B2 as shown. (5.11) can also be derived using the energy approach discussed in Section 4.6. (See Problem 4.28.) Section 5.2 Simple Harmonic Motion 167 Solution as the Real Part of a Complex Exponential There is still another useful way to write our solution, in terms of the complex exponentials of (5.5). The coefficients C 1 and C2 there are related to the coefficients B 1 and B2 of the sine-cosine form by Equation (5.7), which we can solve to give C 1 = z(B i - i B2 ) and i B2). (5.12) C2 = i (B1 Since B 1 and B2 are real, this shows that both C 1 and C2 are generally complex and that C2 is the complex conjugate of C 1, C2 = C (Recall that for any complex number z = x iy, the complex conjugate z* is defined as4 z* = x - iy.) Thus the solution (5.5) can be", " written as x(t) = C Cie -iwt (5.13) where the whole second term on the right is just the complex conjugate of the first. (See Problem 5.35 if this isn't clear to you.) Now, for any complex number z = x + iy, z z* = (x iy) (x - iy) = 2x = 2 Re z where Re z denotes the real part of z (namely x). Thus (5.13) can be written as x (t) = 2 Re C iei\". If we define a final constant C = 2C 1, we see from Equation (5.12) and Figure 5.4 C = B 1 - iB 2 = Ae -i8 (5.14) that and x(t) = Re Cei wt = Re Ae i(w\"). This beautiful result is illustrated in Figure 5.5. The complex number Ae i (w\") moves counterclockwise with angular velocity co around a circle of radius A. Its real part [namely x(t)] is the projection of the complex number onto the real axis. While the complex number goes around the circle, this projection oscillates back and forth on the x axis, with angular frequency w and amplitude A. Specifically, x(t) = A cos(wt - 6), in agreement with (5.11). 4 While most physicists use the notation z*, mathematicians almost always use E for the complex conjugate of z. 168 Chapter 5 Oscillations Figure 5.5 The position x (t) of the oscillating cart is the real part of the complex number Aei (t -S) As the latter moves around the circle of radius A, the former oscillates back and forth on the x axis with amplitude A. EXAMPLE 5.2 A Bottle in a Bucket A bottle is floating upright in a large bucket of water as shown in Figure 5.6. In equilibrium it is submerged to a depth do below the surface of the water. Show that if it is pushed down to a depth d and released, it will execute harmonic motion, and find the frequency of its oscillations. If do = 20 cm, what is the period of the oscillations? The two forces on the bottle are its weight mg downward and the upward buoyant force pg Ad, where Q is the density of water and A is the cross-sectional area of the bottle. (Remember that Archidemes' principle says that the buoyant force is pg times the volume submerged, which is just Ad.) The equilibrium depth do is determined by the condition mg = pgAdo. (5.15) Figure 5.6 The bottle shown has been loaded with sand so that it floats upright in a bucket of water. Its equilibrium depth is d = do. Suppose now the bottle is at a depth d x. (This defines x as the distance from equilibrium, always the best coordinate to use.) Newton's second law now Section 5.2 Simple Harmonic Motion 169 reads becomes mx = mg \u2014 pgA(d o x). By (5.15), the first and second terms on the right cancel, and we're left with \u2014pgAxlm. But again by (5.15), pgAlm= gldo, so the equation of motion = --x, 0 which is exactly the equation for simple harmonic motion. We conclude that the bottle moves up and down in SHM with angular frequency co =.\\/gIdo. A remarkable feature of this result is that the frequency of oscillations is indepen- dent of m, Q, and A; also, the frequency is the same as that of a simple pendulum of length 1 = do. If do = 20 cm, then the period is = \u2014 = 27T = / 0.20 m g V 9.8 m/s 2 = 0.9 sec. Try this experiment yourself! But be aware that the details of the flow of water around the bottle complicate the situation considerably. The calculation here is a very simplified version of the truth. Energy Considerations To conclude this section on simple harmonic motion, let us consider briefly the energy of the oscillator (the cart on a spring or whatever it is) as it oscillates back and forth. Since x (t) = A cos(wt \u2014 8), the potential energy is just Differentiating x (t) to give the velocity, we find for the kinetic energy U = 2kx 2 = IkA 2 cos2 (wt \u2014 8). T = 2 imi2 = 2 l inco2A2 sin2(wt \u2014 8) 1 2 kA 2 sine (cot \u2014 8) = - where the second line results from replacing w2 by k I m. We see that both U and T oscillate between 0 and 4kA 2, with their oscillations perfectly out of step \u2014 when U is maximum T is zero and vice versa", ". In particular, since sin e 0 + cos2 B = 1, the total energy is constant, E = T U = kA2, (5.16) as it has to be for any conservative force. 170 Chapter 5 Oscillations 5.3 Two-Dimensional Oscillators In two or three dimensions, the possibilities for oscillation are considerably richer than in one dimension. The simplest possibility is the so-called isotropic harmonic oscillator, for which the restoring force is proportional to the displacement from equilibrium, with the same constant of proportionality in all directions: F = \u2014kr. (5.17) That is, F.\u201e = \u2014kx, F y = \u2014ky (and F, = \u2014kz in three dimensions), all with the same constant k. This force is a central force directed toward the equilibrium position, which we may as well take to be the origin, as sketched in Figure 5.7(a). Figure 5.7(b) shows an arrangement of four identical springs that would produce a force of this form; it is easy to see that if the mass at the center is moved away from its equilibrium position it will experience a net inward force, and it is not too hard to show (Problem 5.19) that this inward force has the form (5.17) for small displacements r. 5 Another example of a two-dimensional isotropic oscillator is (at least approximately) a ball bearing rolling near the bottom of a large spherical bowl. Two important three-dimensional examples are an atom vibrating near its equilibrium in a symmetric crystal, and a proton (or neutron) as it moves inside a nucleus. Let us consider a particle that is subject to this type of force and suppose, for simplicity, that it is confined to two dimensions. The equation of motion, = F/m, separates into two independent equations: _ co2 \u2014602 Y (5.18) where I have introduced the familiar angular frequency w = fig m (which is the same in both x and y equations because the same is true of the force constants). Each of Y 0 71 r4,_ kr X > (a) (b) Figure 5.7 (a) A restoring force that is proportional to r defines the isotropic harmonic oscillator. (b) The mass at the center of this arrangement of springs would experience a net force of the form F = \u2014kr as it moves in the plane of the four springs. 5 1t is perhaps worth pointing out that one does not get a force of the form (5.17) by simply attaching a mass to a spring whose other end is anchored to the origin. these two equations has exactly the form of the one-dimensional equation discussed in the last section, and the solutions are [as in (5.11)1 Section 5.3 Two - Dimensional Oscillators 171 where the four constants A x, A y, 3x, and 8 are determined by the initial conditions of the problem. By redefining the origin of time, we can dispose of the phase shift 3 s, but, in general, we cannot also dispose of the corresponding phase in the y solution. Thus the simplest form for the general solution is x(t) = Ax cos(cot \u2014 3 s ) 1 y(t) = A y cos(wt y) x(t) = A s cos(cot) y(t) = A y cos(cot \u2014 6) (5.19) (5.20) where 8 = 6y \u2014 6 x is the relative phase of the y and x oscillations. (See Problem 5.15.) The behavior of the solution (5.20) depends on the values of the three constants A x, Ay, and 8. If either A s or Ay is zero, the particle executes simple harmonic motion along one of the axes. (The ball bearing in the bowl rolls back and forth through the origin, moving in the x direction only or the y direction only.) If neither A x nor A y 6. If 6 = 0, then x(t) and is zero, the motion depends critically on the relative phase y(t) rise and fall in step, and the point (x, y) moves back and forth on the slanting line that joins (A s, Ay ) to (\u2014A s, \u2014A y ), as shown in Figure 5.8(a). If 8 = 7r/2, then x and y oscillate out of step, with x at an extreme when y is zero, and vice versa; the point (x, y) describes an ellipse with semimajor and semiminor axes A s and Ay, as in Figure 5.8(b). For other values of 6, the point (x, y) moves around a slanting ellipse, as shown for the case 3 = 7r/4 in Figure 5.", "8(c). (For a proof that the path really is an ellipse, see Problem 8.11.) In the anisotropic oscillator, the components of the restoring force are propor- tional to the components of the displacement, but with different constants of propor- tionality: Fx = \u2014kx x, Fy = \u2014ky y, and = \u2014k zz. (5.21) (a) S =0 (b) 8 = 71-12 (c) 5 = 7/4 Figure 5.8 Motion of a two-dimensional isotropic oscillator as given by (5.20). (a) If 8 = 0, then x and y execute simple harmonic motion in step, and the point (x, y) moves back and forth along a slanting line as shown. (b) If 8 = 7r/2, then (x, y) moves around an ellipse with axes along the x and y axes. (c) In general (for example, 8 = 7/4), the point (x, y) moves around a slanted ellipse as shown. 172 Chapter 5 Oscillations An example of such a force is the force felt by an atom displaced from its equilibrium position in a crystal of low symmetry, where it experiences different force constants along the different axes. For simplicity, I shall again consider a particle in two dimensions, for which Newton's second law separates into two separate equations just as in (5.18): The only difference between this and (5.18) is that there are now different frequencies for the different axes, co, =.1k,Im and so on. The solution of these two equations is just like (5.20): _ co 2x 2 \u2014 60 Y Y = x(t) = Ax cos(cox t) y (t) = Ay COS(Wyt --- 6). (5.22) (5.23) Because of the two different frequencies, there is a much richer variety of possible motions. If cox /coy is a rational number, it is fairly easy to see (Problem 5.17) that the motion is periodic, and the resulting path is called a Lissajous figure (after the French physicist Jules Lissajous, 1822- 1880). For example, Figure 5.9(a) shows an orbit of a particle for which cox /coy = 2 and the x motion repeats itself twice as often as the y motion. In the case shown, the result is a figure eight. If cox /coy is irrational, the motion is more complicated and never repeats itself. This case is illustrated, with cox /coy = -s/2, in Figure 5.9(b). This kind of motion is called quasiperiodic: The motion of the separate coordinates x and y is periodic, but because the two periods are incompatible, the motion of r = (x, y) is not. (a) 60, = 2 coy (b) co, = V G coy Figure 5.9 (a) One possible path for an anisotropic oscillator with co, = 2 and coy = 1. You can see that x goes back and forth twice in the time that y does so once, and the motion then repeats itself exactly. (b) A path for the case cox = Vi and co = 1 from t = 0 to t = 24. In this case the path never repeats itself, although, if we wait long enough, it will come arbitrarily close to any point in the rectangle bounded by x = \u00b1Ax and y = \u00b1Ay. Section 5.4 Damped Oscillations 173 5.4 Damped Oscillations I shall now return to the one-dimensional oscillator, and take up the possibility that there are resistive forces that will damp the oscillations. There are several possibilities for the resistive force. Ordinary sliding friction is approximately constant in magni- tude, but always directed opposite to the velocity. The resistance offered by a fluid, such as air or water, depends on the velocity in a complicated way. However, as we saw in Chapter 2, it is sometimes a reasonable approximation to assume that the re- sistive force is proportional to v or (under different circumstances) to v 2. Here I shall assume that the resistive force is proportional to v; specifically, f -= \u2014by. One of my main reasons is that this case leads to an especially simple equation to solve, and the equation is itself a very important equation that appears in several other contexts and is therefore well worth studying. 6 Consider, then, an object in one dimension, such as a cart attached to a spring, that is subject to a Hooke's law force, \u2014kx, and a resistive force, \u2014b.. The net force on the object is \u2014b.i kx, and Newton's second law", " reads (if I move the two force terms over to the left side) \u00b1 kx = 0. (5.24) One of the beautiful things about physics is the way the same mathematical equation can arise in totally different physical contexts, so that our understanding of the equation in one situation carries over immediately to the other. Before we set about solving Equation (5.24), I would like to show how the same equation appears in the study of LRC circuits. An LRC circuit is a circuit containing an inductor (inductance L), a capacitor (capacitance C), and a resistor (resistance R), as sketched in Figure 5.10. I have chosen the positive direction for the current to be counterclockwise, and the charge q(t) to be the charge on the left-hand plate of the capacitor [with \u2014q (t) on the right], so that I (t) = (t). If we follow around the circuit in the positive direction, the electric potential drops by Li = L4 across the inductor, by RI = R4 across the resistor, and by q /C across the capacitor. Applying Kirchoff's second rule for circuits, we conclude that. 1 Rq \u2014q = O. (5.25) This has exactly the form of Equation (5.24) for the damped oscillator, and anything that we learn about the equation for the oscillator will be immediately applicable to the LRC circuit. Notice that the inductance L of the electric circuit plays the role of the mass of the oscillator, the resistance term R4 corresponds to the resistive force, and 1/ C to the spring constant k. 6 You should be aware, however, that although the case I am considering \u2014 that the resistive force f is linear in v \u2014 is very important, it is nevertheless a very special case. I shall describe some of the startling complications that can occur when f is not linear in v in Chapter 12 on nonlinear mechanics and chaos. 174 Chapter 5 Oscillations Figure 5.10 An LRC circuit. Let us now return to mechanics and the differential equation (5.24). To solve this equation it is convenient to divide by in and then introduce two other constants. I shall rename the constant b I m as 2p, b \u2014 m. 2P. (5.26) This parameter 8, which can be called the damping constant, is simply a convenient way to characterize the strength of the damping force \u2014 as with b, large 8 corresponds to a large damping force and conversely. I shall rename the constant k I m as 6002, that is, coo = (5.27) Notice that too is precisely what I was calling to in the previous two sections. I have added the subscript because, once we admit resistive forces, various other frequencies become important. From now on, I shall use the notation w o to denote the system's natural frequency, the frequency at which it would oscillate if there were no resistive force present, as given by (5.27). With these notations, the equation of motion (5.24) for the damped oscillator becomes Notice that both of the parameters f and too have the dimensions of inverse time, that is, frequency. Equation (5.28) is another second-order, linear, homogeneous equation [the last was (5.4)]. Therefore, if by any means we can spot two independent 7 solutions, x i (t) and x2 (t) say, then any solution must have the form Cix i (t) + C2x2 (t). What this 7 It is about time I gave you a definition of \"independent.\" In general this is a little complicated, but for two functions it is easy: Two functions are independent if neither is a constant multiple of the other. Thus the two functions sin(x) and cos(x) are independent; likewise the two functions x and x 2 ; but the two functions x and 3x are not. Section 5.4 Damped Oscillations 175 means is that we are free to play a game of inspired guessing to find ourselves two independent solutions; if by hook or by crook we can spot two solutions, then we have the general solution. In particular, there is nothing to stop us trying to find a solution of the form (5.29) (5.31) (5.32) (5.33) for which and x(t) = ert I = re' = r2 ert Substituting into (5.28) we see that our guess (5.29) satisfies (5.28) if and only if 2 r z i6r + co2 =0 0 (5.30) [an equation sometimes called the auxiliary equation for the differential equation (5.28)]. The solutions of this equation are, of course, r = - p \u00b1.3,", "52 - (002. Thus if we define the two constants r =\u2014,8 2 \u2014 o_) 2 0 r2 _p \\A82 2 then the two functions eri t and er2t are two independent solutions of (5.28) and the general solution is x(t) = Cle at C2er2t = e-Pt ( C ie -1132 '4 t + C2e - N/ P2- This solution is rather too messy to be especially illuminating, but, by examining various ranges of the damping constant /3, we can begin to see what (5.33) entails. Undamped Oscillation If there is no damping then the damping constant 13 is zero, the square root in the exponents of (5.33) is just i coo, and our solution reduces to x(t) = C i e 01 + C2e -iwc't, (5.34) the familiar solution for the undamped harmonic oscillator. Weak Damping Suppose next that the damping constant f3 is small. Specifically, suppose that fi < wo (5.35) 176 Chapter 5 Oscillations a condition sometimes called underdamping. In this case, the square root in the exponents of (5.33) is again imaginary, and we can write 2 (0 2 = i l/(0 2 /82 = o o where CO1 = -1(002 \u2014 /32 (5.36) The parameter co l is a frequency, which is less than the natural frequency co o. In the important case of very weak damping (/3 << coo), co l is very close to coo. With this notation, the solution (5.33) becomes x(t) = e (C iei w' t c2e -iw i t) (5.37) This solution is the product of two factors: The first, e - f t, is a decaying exponen- tial, which steadily decreases toward zero. The second factor has exactly the form (5.34) of undamped oscillations, except that the natural frequency co o is replaced by the somewhat lower frequency co l. We can rewrite the second factor, as in Equation (5.11), in the form A cos(co it \u2014 8) and our solution becomes x(t) = Ae - fi t cos(w i t \u2014 8). (5.38) This solution clearly describes simple harmonic motion of frequency co l with an exponentially decreasing amplitude Ae - fi t, as shown in Figure 5.11. The result (5.38) suggests another interpretation of the damping constant /3. Since,8 has the dimensions of inverse time, 1//3 is a time, and we now see that it is the time in which the amplitude function Ae - fi t falls to 1/e of its initial value. Thus, at least for underdamped oscillations, /3 can be seen as the decay parameter, a measure of the rate at which the motion dies out, (decay parameter) = /3 [underdamped motion]. x(t) Figure 5.11 Underdamped oscillations can be thought of as sim- ple harmonic oscillations with an exponentially decreasing amplitude Ae - fit. The dashed curves are the envelopes, \u00b1Ae -i3t. The larger the more rapidly the oscillations die out, at least for the case /3 < co o that Section 5.4 Damped Oscillations 177 we are discussing here. Strong Damping Suppose instead that the damping constant 13 is large. Specifically suppose that /3 > (00, (5.39) a condition sometimes called overdamping. In this case, the square root in the exponents of (5.33) is real and our solution is x(t) = C i e - (16 2-a)\u00b0)t C2e - (13+N/ P2-4) t (5.40) Here we have two real exponential functions, both of which decrease as time goes by (since the coefficients of t in both exponents are negative). In this case, the motion is so damped that it completes no bona fide oscillations. Figure 5.12 shows a typical case in which the oscillator was given a kick from 0 at t = 0; it slid out to a maximum displacement and then slid ever more slowly back again, returning to the origin only in the limit that t co. The first term on the right of (5.40) decreases more slowly than the second, since the coefficient in its exponent is the smaller of the two. Thus the long-term motion is dominated by this first term. In particular, the rate at which the motion dies out can be characterized by the coefficient in the first exponent, (decay parameter) = \u2014 \\/132 602 [overdamped motion]. (5.41) Careful inspection of (", "5.41) shows that \u2014 contrary to what one might expect \u2014 the rate of decay of overdamped motion gets smaller if the damping constant is made bigger. (See Problem 5.20.) 0 Figure 5.12 Overdamped motion in which the oscillator is kicked from the origin at t = 0. It moves out to a maximum displacement and then moves back toward 0 asymptotically as t \u2014> oo. Critical Damping The boundary between underdamping and overdamping is called critical damping and occurs when the damping constant is equal to the natural frequency, /3 = coo. This 178 Chapter 5 Oscillations case has some interesting features, especially from a mathematical point of view. When /3 = wo the two solutions that we found in (5.33) are the same solution, namely x(t) = e-)51' (5.42) [This happened because the two solutions of the auxiliary equation (5.30) happen to coincide when /3 = coo l This is the one case where our inspired guess, to seek a solution of the form x (t) = ert, fails to find us two solutions of the equation of motion, and we have to find a second solution by some other method. Fortunately, in this case, it is not hard to spot a second solution: As you can easily check, the function x(t) = to- i3t (5.43) is also a solution of the equation of motion (5.28) in the special case that /3 = co,. (See Problems 5.21 and 5.24.) Thus the general solution for the case of critical damping is x(t) = C l e - fi t + C2 t e - t. fi (5.44) Notice that both terms contain the same exponential factor e - Pt. Since this factor is what dominates the decay of the oscillations as t Do, we can say that both terms decay at about the same rate, with decay parameter (decay parameter) = /3 = co o [critical damping]. It is interesting to compare the rates at which the various types of damped oscil- lation die out. We have seen that in each case, this rate is determined by a \"decay parameter,\" which is just the coefficient of t in the exponent of the dominant expo- nential factor in x(t). Our findings can be summarized as follows: damping /3 decay parameter none under critical over /3 = 0 /3 < wo /3 = 0)0 /3 > (D. 0 18,8 13 \u2014 //32 \u2014 (02 o Figure 5.13 is a plot of the decay parameter as a function of /3 and shows clearly that the motion dies out most quickly when /3 = w o ; that is, when the damping is critical. There are situations where one wants any oscillations to die out as quickly as possible. For example, one wants the needle of an analog meter (a voltmeter or pressure gauge, for instance) to settle down rapidly on the correct reading. Similarly, in a car, one wants the oscillations caused by a bumpy road to decay quickly. In such cases one must arrange for the oscillations to be damped (by the shock absorbers in a car), and for the quickest results the damping should be reasonably close to critical. Section 5.5 Driven Damped Oscillations 179 co, Figure 5.13 The decay parameter for damped oscillations as a function of the damping constant /3. The decay parameter is biggest, and the motion dies out most quickly, for critical damping, with p = wo. 5.5 Driven Damped Oscillations Any natural oscillator, left to itself, eventually comes to rest, as the inevitable damping forces drain its energy. Thus if one wants the oscillations to continue, one must arrange for some external \"driving\" force to maintain them. For example, the motion of the pendulum in a grandfather clock is driven by periodic pushes caused by the clock's weights; the motion of a young child on a swing is maintained by periodic pushes from a parent. If we denote the external driving force by F(t) and if we assume as before that the damping force has the form \u2014by, then the net force on the oscillator is \u2014by \u2014 kx + F(t) and the equation of motion can be written as mz + bx + kx = F(t). (5.45) Like its counterpart for undriven oscillations, this differential equation crops up in several other areas of physics. A prominent example is the LRC circuit of Figure 5.10. If we want the oscillating current in that circuit to persist, we must apply a driving EMF, E(t), in which case the equation of motion for the circuit becomes 1,4 + + \u2014 q = E(t) 1", " (5.46) in perfect correspondence with (5.45). As before, we can tidy Equation (5.45) if we divide the equation by rn and replace blm by 2/3 and klm by w02. In addition, I shall denote F(t)/m by f (t) F(t), m (5.47) the force per unit mass. With this notation, (5.45) becomes 180 Chapter 5 Oscillations Linear Differential Operators Before we discuss how to solve this equation, I would like to streamline our notation. It turns out that it is very helpful to think of the left side of (5.48) as the result of a certain operator acting on the function x(t). Specifically, we define the differential D = \u2014 213\u2014 w 2. dt \u00b0 d 2 dt2 (5.49) The meaning of this definition is simply that when D acts on x it gives the left side Dx \u00b1 2/3X + coo2x. This definition is obviously a notational convenience \u2014 the equation (5.48) becomes operator of (5.48): just Dx = f (5.50) \u2014 but it is much more: The notion of an operator like (5.49) proves to be a powerful mathematical tool, with applications throughout physics. For the moment the impor- tant thing is that the operator is linear: We know from elementary calculus that the derivative of ax (where a is a constant) is just ax and that the derivative of x 1 + x2 is just x1 + 12. Since this also applies to second derivatives, it applies to the operator D: D(ax) = aDx and D(x 1 + x2) = Dx 1 + Dx 2. (Make sure you understand what these two equations mean.) We can combine these into a single equation: D(ax i + bx2) = aDx] bDx 2 (5.51) for any two constants a and b and any two functions x l (t) and x2 (t). Any operator that satisfies this equation is called a linear operator. We have actually used the property (5.51) of linear operators before. The equation (5.28) for a damped oscillator (not driven) can be written as Dx = 0. (5.52) The superposition principle asserts that if x i and x2 are solutions of this equation, then so is ax 1 + bx2 for any constants a and b. In our new operator notation, the proof is very simple: We are given that Dx 1 = 0 and Dx 2 = 0, and using (5.51) it immediately follows that D(ax i + bx2) = aDxi bDx 2 = 0 + = 0; that is, ax i + bx2 is also a solution. The equation (5.52), Dx = 0, for the undriven oscillator is called a homogeneous equation, since every term involves either x or one of its derivatives exactly once. The equation (5.50), Dx = f, is called an inhomogeneous equation, since it contains the inhomogeneous term f, which does not involve x at all. Our job now is to solve this inhomogeneous equation. Section 5.5 Driven Damped Oscillations 181 Particular and Homogeneous Solutions Using our new operator notation, we can find the general solution of Equation (5.48) surprisingly easily; in fact, we've already done most of the work. Suppose first that we have somehow spotted a solution; that is, we've found a function xp (t) that satisfies Dx P = f (5.53) We call this function x p (t) a particular solution of the equation, and the subscript \"p\" stands for \"particular.\" Next let us consider for a moment the homogeneous equation Dx = 0 and suppose we have a solution xh (t), satisfying Dx h = 0. (5.54) We'll call this function a homogeneous solution. 8 We already know all about the solutions of the homogeneous equation, and we know from (5.32) that x h must have the form xh (t) = Cier't C2er2t, (5.55) where both exponentials die out as t co. We're now ready to prove the crucial result. First, if x p is a particular solution satisfying (5.53), then x p + xh is another solution, for D(xp x h) = Dxp Dx h = f +0= f. Given the one particular solution x p, this gives us a large number of other solutions xp + xh. And we have in fact found all the solutions, since the function x h contains two arbitrary constants, and we", " know that the general solution of any second-order equation contains exactly two arbitrary constants. Therefore x p + xh, with xh given by (5.55) is the general solution. This result means that all we have to do is somehow to find a single particular solution xp (t) of the equation of motion (5.48), and we shall have every solution in the form x (t) = xp (t) x h (t). Complex Solutions for a Sinusoidal Driving Force I shall now specialize to the case that the driving force f (t) is a sinusoidal function of time, f (t) = fo cos(wt), (5.56) 8 Another common name is the complementary function. This has the disadvantage that it's hard to remember which is \"particular\" and which \"complementary.\" 182 Chapter 5 Oscillations where fo denotes the amplitude of the driving force [actually the amplitude divided by the oscillator's mass, since f (t) = F(t)/ m] and co is the angular frequency of the driving force. (Be careful to distinguish the driving frequency w from the natural frequency co o of the oscillator. These are entirely independent frequencies, although we shall see that the oscillator responds most when w w o.) The driving force for many driven oscillators is at least approximately sinusoidal. For example, even the parent pushing the child on a swing can be crudely approximated by (5.56); the driving EMF induced in your radio's circuits by a broadcast signal is almost perfectly of this form. Probably the chief importance of sinusoidal driving forces is that, according to Fourier's theorem, 9 essentially any driving force can be built up as a series of sinusoidal forces. of motion (5.48) takes the form Let us therefore assume the driving force is given by (5.56), so that the equation + 213.i + coo2x = fo cos(cot). (5.57) Solving this equation is greatly simplified by the following trick: For any solution of (5.57), there must be a solution of the same equation, but with the cosine on the right replaced by a sine function. (After all, these two differ only by a shift in the origin of time.) Accordingly, there must also be a function y(t) that satisfies \u00b1 2,85, + wo2 y = fo sin(cot). Suppose now we define the complex function z(t) = x(t) iy(t), with x(t) as its real part and y(t) as its imaginary part. If we multiply (5.58) by i and add it to (5.57), we find that +2& coo2z foeiwt. Although it may not yet look it, Equation (5.60) is a tremendous advance. Because of the simple properties of the exponential function, (5.60) is much easier to solve than either (5.57) or (5.58). And as soon as we find a solution z(t) of (5.60), we have only to take its real part to have a solution of the equation (5.57) whose solutions we actually want. In seeking a solution of (5.60), we are obviously free to try any function we like. In particular, let's see if there is a solution of the form z(t) = Ce iwt, (5.61) where C is an as yet undetermined constant. If we substitute this guess into the left side of (5.60), we get ( - 602 + 20a) coo2)Ce iwt = Lei t 9 Named for the French mathematician, the Baron Jean Baptiste Joseph Fourier, 1768-1830. See Sections 5.7-5.9. (5.58) (5.59) (5.60) Section 5.5 Driven Damped Oscillations 183 In other words, our guess (5.61) is a solution of (5.60) if and only if C = fo CO - CO2 \u00b1 2i,8w 0 (5.62) and we have succeeded in finding a particular solution of the equation of motion. Before we take the real part of z(t) = Ce\" t, it is convenient to write the complex coefficient C in the form C = Ae \u2014i6 (5.63) where A and 8 are real. [Any complex number can be written in this form; the particular notation is chosen to match (5.14).] To identify A and 6, we must compare (5.62) and (5.63). First, taking the absolute value squared of both equations we find that A 2 = cc* = 1'0 fo (002 w2 + 20 (0 (002 \u2014 (02 \u2014 2i p a) or A 2 = (", "0)2 (02 )2 \u00b1 4162(02. f. 2 (5.64) (Make sure you understand this derivation. See Problem 5.35 for some guidance.) We are going to see in a moment that A is just the amplitude of the oscillations caused by the driving force f (t). Thus the result (5.64) is the most important result of this discussion. It shows how the amplitude of oscillations depends on the various parameters. In particular, we see that the amplitude is biggest when co \u00b0 ti co, so that the denominator is small; in other words, the oscillator responds best when driven at a frequency w that is close to its natural frequency co., as you would probably have guessed. Before we continue to discuss the properties of our solution, we need to identify the phase angle 8 in (5.63). Comparing (5.63) and (5.62) and rearranging, we see that foeis = A(0)02 w2 + 200. Since fo and A are real, this says that the phase angle 8 is the same as the phase angle of the complex number (0)02 \u2014 w2) 2ifico. This relationship is illustrated in Figure 5.14, from which we conclude that 8 = arctan 2136\u00b0 (0 2 (02 (5.65) Our quest for a particular solution is now complete. The \"fictitious\" complex solution introduced in (5.59) is z(t) = = Ae i('-6) and the real part of this is the solution we are seeking, x(t) = A cos(cot \u2014 8) (5.66) where the real constants A and 8 are given by (5.64) and (5.65). 184 Chapter 5 Oscillations 2 Figure 5.14 The phase angle 8 is the angle of this triangle. The solution (5.66) is just one particular solution of the equation of motion. The general solution is found by adding any solution of the corresponding homogeneous equation, as given by (5.55); that is, the general solution is x(t) = A cos(wt \u2014 S) + C2e' 2t. (5.67) Because the two extra terms in this general solution both die out exponentially as time passes, they are called transients. They depend on the initial conditions of the problem but are eventually irrelevant: The long-term behavior of our solution is dominated by the cosine term. Thus the particular solution (5.66) is the solution with which we are usually concerned, and we shall explore its properties in the next section. Before we discuss an example of the motion (5.67), it is important that you be very clear as to the type of system to which (5.67) applies, namely, any oscillator for which both the restoring force (\u2014kx) and the resistive force (\u2014b1) are linear \u2014 a driven, damped linear oscillator, whose equation of motion (5.45) is a linear differential equation. Because nonlinear differential equations are often hard to solve, most mechanics texts have until recently focussed on linear equations. This created the false impression that linear equations were in some sense the norm, and that the solution (5.67) was the only (or, at least, the only important) way for an oscillator to behave. As we shall see in Chapter 12 on nonlinear mechanics and chaos, an oscillator whose equation of motion is nonlinear can behave in ways that are astonishingly different from (5.67). One important reason for studying the linear oscillator here is to give you a backdrop against which to study the nonlinear oscillator later on. The details of the motion (5.67) depend on the strength of the damping param- eter /3. To be specific, let us assume that our oscillator is weakly damped, with /3 less than the natural frequency co o (underdamping). In this case, we know that the two transient terms of (5.67) can be rewritten as in (5.38), so that You need to think very carefully about this potentially confusing formula. The second term on the right is the homogeneous or transient term, and I have added the subscript \"tr\" to distinguish the constants Atr and S t, from the A and S of the first term. The two constants A tr and St, are arbitrary constants; (5.68) is a possible motion of our system for any values of Atr and kr, which are determined by the initial conditions. Section 5.5 Driven Damped Oscillations 185 The factor e \u2014 fi t makes clear that this transient term decays exponentially and is indeed irrelevant to the long-term behavior. As it decays, the transient term oscillates with the angular frequency co l of the undriven (but still damped) oscillator,", " as in (5.36). The first term is our particular solution and its two constants A and 8 are certainly not arbitrary; they are determined by (5.64) and (5.65) in terms of the parameters of the system. This term oscillates with the frequency w of the driving force and with unchanging amplitude, for as long as the driving force is maintained. EXAMPLE 5.3 Graphing a Driven Damped Linear Oscillator Make a plot of x (t) as given by (5.68) for a driven damped linear oscillator which is released from rest at the origin at time t = 0, with the following parameters: Drive freqency w = 27., natural frequency coo = 5co, decay constant /3 = co o /20, and driving amplitude fo = 1000. Show the first five drive cycles. The choice of drive frequency equal to 27 means that the drive period is T = 27r/C0 = 1; this means simply that we have chosen to measure time in units of the drive period \u2014 often a convenient choice. That w o = 5w means that our oscillator has a natural frequency five times the drive frequency; this will let us distinguish easily between the two on a graph. That,f3 = wo /20 means that the oscillator is rather weakly damped. Finally the choice of fo = 1000 is just a choice of our unit of force; the reason for this odd-seeming choice is that it leads to a conveniently sized amplitude of oscillation (namely A close to 1). Our first task is to determine the various constants in (5.68) in terms of the given parameters. In fact this is easier if we rewrite the transient term of (5.68) in the \"cosine plus sine\" form, so that x(t) = A cos(cot \u2014 6) + e\u2014 $t [B i cos(w it) + B2 sin(w it)]. (5.69) The constants A and 8 are determined by (5.64) and (5.65), which, for the given parmeters, yield The frequency co l is A = 1.06 and 8 = 0.0208. (0,= - p2 = 9.987m, which is very close to wo, as we would expect for a weakly damped oscillator. To find B 1 and B2, we must equate x(0) as given by (5.69) to its given initial value xo, and likewise the corresponding expression for l o to the initial value vo. This gives two simultaneous equations for B 1 and B2, which are easily solved (Problem 5.33) to give B 1 = x o \u2014 A cos 8 and B2 = 1 (210 \u2014 (OA sin 8 + pB 1) (5.70) (Di or, with the numbers, including the initial conditions x o = vo = 0, B1 = \u20141.05 and B2 = \u20140.0572. 186 Chapter 5 Oscillations (a) (b) Figure 5.15 The response of a damped, linear oscillator to a si- nusoidal driving force, with the time t shown in units of the drive period. (a) The driving force is a pure cosine as a function of time. (b) The resulting motion for the initial conditons x o = vo = 0. For the first two or three drive cycles, the transient motion is clearly vis- ible, but after that only the long-term motion remains, oscillating sinusoidally at exactly the drive frequency. As explained in the text, the sinusoidal motion after t ti 3 is called an attractor. Putting all of these numbers into (5.69), we can now plot the motion, as shown in Figure 5.15, where part (a) shows the driving force f(t) = fo cos(wt) and part (b) the resulting motion x (t) of the oscillator. The driving force is, of course, perfectly sinusoidal with period 1. The resulting motion is much more interesting. After about three drive cycles (t 3), the motion is indistinguishable from a pure cosine, oscillating at exactly the drive frequency; that is, the transients have died out and only the long-term motion remains. Before t 3, however, the effects of the transients are clearly visible. Since they oscillate at the faster natural frequency wo, they show up as a rapid succession of bumps and dips. In fact, you can easily see that there are five such bumps within the first drive cycle, indicating that wo = 5w. Because the transient motion depends on the initial values xo and vo, different values of xo and vo would lead to quite different initial motion. (See Problem 5.36.) After a short time however (a couple of cycles in", " this example), the initial differences disappear and the motion settles down to the same sinusoidal motion of the particular solution (5.66), irrespective of the initial conditions. For this reason, the motion of (5.66) is sometimes called an attractor \u2014 the motions corresponding to several different initial conditions are \"attracted\" to the particular motion (5.66). For the linear oscillator discussed here, there is a unique attractor (for a given driving force): Every possible motion of the system, whatever its initial conditions, is attracted to the same motion (5.66). We shall see in Chapter 12 that for nonlinear oscillators there can be several different attractors and that for some values of the parameters the motion of Section 5.6 Resonance 187 an attractor can be far more complicated than simple harmonic oscillation at the drive The amplitude and phase of the attractor seen in Figure 5.15(b) depend on the parameters of the driving force (but not, of course, on the initial conditions). The dependence of the amplitude and phase on these parameters is the subject of the next frequency. section. 5.6 Resonance In the previous section we considered a damped oscillator that is being driven by a sinusoidal driving force (actually force divided by mass) f (t) = f0 cos(wt) with angular frequency co. We saw that, apart from transient motions that die out quickly, the system's response is to oscillate sinusoidally at the same frequency, w: x(t) = A cos(cot \u2014 6), with amplitude A given by (5.64), A2 = fo2 ( 0) 2 0)2)2 4fi2 w2 (5.71) and phase shift 8 given by (5.65). The most obvious feature of (5.71) is that the amplitude A of the response is proportional to the amplitude of the driving force, A a f0, a result you might have guessed. More interesting is the dependence of A on the frequencies co. (the natural frequency of the oscillator) and w (the frequency of the driver), and on the damping constant /3. The most interesting case is when the damping constant /3 is very small, and this is the case I shall discuss. With p small, the second term in the denominator of (5.71) is small. If w0 and w are very different, then the first term in the denominator of (5.71) is large, and the amplitude of the driven oscillations is small. On the other hand, if wo is very close to w, both terms in the denominator are small, and the amplitude A is large. This means that if we vary either w 0 or w, there can be quite dramatic changes in the amplitude of the oscillator's motion. This is illustrated in Figure 5.16, which shows A 2 as a function of w 0 with w fixed, for a rather weakly damped system (/3 = 0.1w). (Note that, because the energy of the system is proportional to A 2, it is usual to make plots of A 2 rather than A.) Although the behavior illustrated in Figure 5.16 is startlingly dramatic, the qualita- tive features are what you might have expected. Left to its own devices, the oscillator vibrates at its natural frequency w 0 (or at the slightly lower frequency co l if we allow for the damping). If we try to force it to vibrate at a frequency w, then for values of w close to wo the oscillator responds very well, but if w is far from w0, it hardly re- sponds at all. We refer to this phenomenon \u2014 the dramatically greater response of an oscillator when driven at the right frequency \u2014 as resonance. An everyday application of resonance is the reception of radio signals by the LRC circuit in your radio. As we saw, the equation of motion of an LRC circuit is exactly 188 Chapter 5 Oscillations 0 Figure 5.16 The amplitude squared, A 2, of a driven oscillator, shown as a function of the natural frequency coo, with the driving frequency co fixed. The response is dramatically largest when coo and co are close. the same as that of a driven oscillator, and LRC circuits show the same phenomenon of resonance. When you tune your radio to receive station KVOD at 90.1 MHz, you are adjusting an LRC circuit in the radio so that its natural frequency is 90.1 MHz. The many radio stations in your neighborhood are all sending out signals, each at its own frequency and each inducing a tiny EMF in the circuit of your radio. But only the signal with the right frequency actually succeeds in driving an appreciable current, which mimics the signal sent out by your favorite KVOD and reproduces its broadcast An example of a mechanical resonance of the kind discussed here is the behavior of a car driving on", " a \"washboard\" road that has worn into a series of regularly spaced bumps. Each time a wheel crosses a bump, it is given an upward impulse, and the frequency of these impulses depends on the car's speed. There is one speed at which the frequency of these impulses equals the natural frequency of the wheels' vibration on the springs l\u00b0 and the wheels resonate, causing an uncomfortable ride. If the car drives slower or faster than this speed, it goes \"off resonance\" and the ride is much sounds. smoother. Another example occurs when a platoon of soldiers marches across a bridge. A bridge, like almost any mechanical system, has certain natural frequencies, and if the soldiers happened to march with a frequency equal to one of these natural frequencies, the bridge could conceivably resonate sufficiently violently to break. For this reason, soldiers break step when marching across a bridge. The details of the resonance phenomenon are a bit complicated. For example, the exact location of the maximum response depends on whether we vary co o with co fixed or vice versa. The amplitude A is a maximum when the denominator, denominator = (0)02 \u2014 0)2) 2 4p20)2 (5.72) 10 It is the wheels (plus axles) that exhibit the resonant oscillations; the much heavier body of the car is relatively unaffected. Section 5.6 Resonance 189 of (5.71) is a minimum. If we vary wo with w fixed (as when you tune a radio to pick up your favorite station) this minimum obviously occurs when wo = w, making the first term zero. On the other hand, if we vary w with we, fixed (which is what happens in many applications), then the second term in (5.72) also varies and a straightforward differentiation shows that the maximum occurs when co = (02 = '0)02 2,82. (5.73) However, when,8 << wo (as is usually the most interesting case), the difference between (5.73) and w = wo is negligible. We have met so many different frequencies in this chapter that it may be worth pausing to review them. First there is w o the natural frequency of the oscillator (in the absence of any damping). Next when we added in a little damping, we found that the same system oscillated sinusoidally with frequency w 1 = 1/w02 \u2014,8 2 under an exponentially decaying envelope. Then we added a driving force with frequency w, which can, in principle, take on any value independently of the previous two. However, the response of the driven oscillator is biggest when w w o ; specifically, if we vary w with wo fixed, the maximum response comes when w = w 2 as defined by (5.73). To summarize: wo =./k/m = natural frequency of undamped oscillator 0) 1 = 2,82 = frequency of damped oscillator w = frequency of driving force (02 = \\/(0.2 \u2014 2,82 p = value of w at which response is maximum. In any case, the maximum amplitude of the driven oscillations is found by putting wo ti w in (5.71), to give A max fo 2Pwo (5.74) This shows that smaller values of the damping constant /3 lead to larger values of the maximum amplitude of oscillation, as illustrated in Figure 5.17.\" Width of the Resonance; the Q Factor You can see clearly from Figure 5.17 that if we make the damping constant p smaller, the resonance peak not only gets higher, but also gets narrower. We can make this idea more precise by defining the width (or full width at half maximum or FWHM) as the interval between the two points where A 2 is equal to half its maximum height. It 11 In this figure I chose to plot A 2 against w with wo fixed, rather than the other way around as in Figure 5.16. Note that the curves have very similar shapes either way. 190 Chapter 5 Oscillations Figure 5.17 The amplitude for driven oscillations as a function of the driving frequency (t) for three different values of the damp- ing constant p. Note how as,8 decreases the resonance peak gets higher and sharper. is a simple exercise (Problem 5.41) to show that the two half-maximum points are at co coo +,B, as in Figure 5.18. Thus the full width at half maximum is or, equivalently, the half width at half maximum is FWHM 2p HWHM The sharpness of the resonance peak is indicated by the ratio of its width 2fi to its position, coo. For many purposes, we want a very sharp resonance, so it is common practice to define a quality factor Q as the reciprocal of this ratio: (5.75) (5.76) (5.77) 3/ \\ coo \u2014 i wo+ Figure 5.18 The", " full width at half maximum (FWHM) is the distance between the points where A 2 is half its maximum value. Section 5.6 Resonance 191 A large Q indicates a narrow resonance, and vice versa. For example, clocks depend on the resonance in an oscillator (a pendulum, or a quartz crystal, for instance) to regulate the mechanism to move with very well-defined frequency. This requires that the width 2/3 be very small compared to the natural frequency co o. In other words, a good clock needs a high Q. The Q for a typical pendulum may be around 100; that for a quartz crystal around 10,000. Therefore quartz watches keep much better time than a typical grandfather clock. I2 There is another way to look at the quality factor Q. We saw that, in the absence of any driving force, the oscillations die out in a time of order 11p, (This was actually the time for the amplitude to drop to 1/e of its initial value.) The period of a single oscillation is, of course, (decay time) = 1//3. period = 27r/coo. (Remember we're assuming that,8 << wo, so we don't need to distinguish between coo and coi.) Thus we can rewrite the definition of Q as wo \u2014 = 2/3 1//3 27r/co. = decay time period (5.78) The ratio on the right is just the number of periods in the decay time. Thus, the quality factor Q is 7/- times the number of cycles our oscillator makes in one decay time. 13 The Phase at Resonance The phase difference 6 by which the oscillator's motion lags behind the driving force is given by (5.65) as 8 = arctan 2/3w CO 2 - 0)2 0 (5.79) Let us follow this phase as we vary w, starting well below a narrow resonance small). With w << coo, (5.79) implies that 8 is very small; that is, while w << coo the oscillations are almost perfectly in step with the driving force. (This was the case in Figure 5.15.) As w is increased toward co o, so 6 slowly increases. At resonance, where w = coo, the argument of the arctangent in (5.79) is infinite, so 8 = 7r/2 and the oscillations are 90\u00b0 behind the driving force. Once co > co o, the argument of the 12 Actually, both quartz watches and grandfather clocks keep much better time than this simple discussion would suggest. A good chronometer keeps the frequency very close to the center of the resonance. Thus the variability of the frequency is actually much smaller than the width of the resonance. Nevertheless, the stated conclusion is correct. 13 Yet another definition (and perhaps the most fundamental) is that Q = 27r times the ratio of the energy stored in the oscillator to the energy dissipated in one cycle. See Problem 5.44. 192 Chapter 5 Oscillations = 0.03 coo 0.3w, 6 7F - 7/2 - 0 Figure 5.19 The phase shift S increases from 0 through 7/2 to Jr as the driving frequency co passes through resonance. The narrower the resonance, the more suddenly this increase occurs. The solid curve is for a relatively narrow resonance (/3 = 0.03(00 Q = 16.7), and the dashed curve is for a wider resonance or (0= 0.3(00 or Q = 1.67). arctangent is negative and approaches 0 as co increases; thus 6 increases beyond 7/2 and eventually approaches 7. In particular, once co >> co., the oscillations are almost perfectly out of step with the driving force. All of this behavior is illustrated for two different values of )6 in Figure 5.19. Notice, in particular, that the narrower the resonance, the more quickly 6 jumps from 0 to 7. In the resonances of classical mechanics, the behavior of the phase (as in Fig- ure 5.19) is usually less important than that of the amplitude (as in Figure 5.18). 14 In atomic and nuclear collisions, the phase shift is often the quantity of primary inter- est. Such collisions are governed by quantum mechanics, but there is a corresponding phenomenon of resonance. A beam of neutrons, for example, can \"drive\" a target nucleus. When the energy of the beam equals a resonant energy of the system (in quantum mechanics energy plays the role of frequency) a resonance occurs and the phase shift increases rapidly from 0 to 7. 5.7 Fourier Series* * Fourier series have broad application in almost every area of modern physics. Nevertheless, we shall not be using them again until Chapter 16. Thus, if you are pressed for time you could omit the last three sections of this chapter on a first reading. In the last two sections,", " we have discussed an oscillator that is driven by a sinusoidal driving force f (t) = fo cos(cot). There are two main reasons for the importance of 14 The behavior of 8 can, nevertheless, be observed. Make a simple pendulum from a piece of string and a metal nut, and drive it by holding it at the top and moving your hand from side to side. The most obvious thing is that you will be most successful at driving it when your frequency equals the natural frequency, but you can also see that when you drive more slowly the pendulum moves in step with your hand, whereas when you move more quickly the pendulum moves oppositely to your hand. f(t) T -p Section 5.7 Fourier Series* 193 t (a) (b) Figure 5.20 Two examples of periodic functions with period T. (a) A rectangular pulse, which could represent a hammer hitting a nail with a constant force at intervals of r, or a digital signal in a telephone line. (b) A smooth periodic signal, which could be the pressure variation of a musical instrument. sinusoidal driving forces: The first is simply that there are many important systems in which the driving force is sinusoidal \u2014 the electrical circuit in a radio is a good example. The second is somewhat subtler. It turns out that any periodic driving force can be built up from sinusoidal forces using the powerful technique of Fourier series. Thus, in a sense that I shall try to describe, by solving the motion with a sinusoidal driver we have already solved the motion with any periodic driver. Before we can appreciate this wonderful result, we need to review some aspects of Fourier series. In this section I sketch the needed properties of Fourier series; 15 in the next we can apply them to the driven oscillator. Let us consider a function f (t) that is periodic with period r; that is, the function repeats itself every time t advances by the period 1\": f (t + -r) = f (t) whatever the value of t. We can describe a function with this property as being r - periodic. A simple example of a r -periodic function would be the force exerted on a nail by a hammer that is being swung at intervals of r, as sketched in Figure 5.20(a). Another could be the pressure exerted on your ear drum by a note played by a musical instrument, as sketched in 5.20(b). It is easy to think up many more periodic functions. In particular, there are lots of sinusoidal functions that are periodic with any given period: The functions cos(27rt/r), cos(47rt/r), cos(67rt/r), (5.80) are all r -periodic, as are the corresponding sine functions. (If t is increased by the amount T, each of these functions returns to its original value \u2014 see Figure 5.21.) We can write these sinusoidal functions a little more compactly if we introduce the angular frequency w = 27r/r,, in which case all of the functions of (5.80) and the corresponding sines can be written as cos(ncot) and sin(ncot) [n = 0, 1, 2, ]. (5.81) 15 As usual, I shall try to describe all of the theory that we shall be needing. For more details, see, for example, Mathematical Methods in the Physical Sciences by Mary Boas (Wiley, 1983), Ch. 7. 194 Chapter 5 Oscillations f(t) f(t) cos(277117) cos (47a/r) Figure 5.21 Any function of the form cos(2n7rt/r) (or the corresponding sine) is periodic with period r if n is an integer. Notice that cos(4rrt/r) also has the smaller period r/2, but this doesn't change the fact that it has period r as well. (If n = 0 the cosine function is just the constant 1 \u2014 which is certainly periodic \u2014 while the sine is 0 and not at all interesting.) That the sine and cosine functions (5.81) are all t -periodic is reasonably obvious. (Be sure you can see this.) What is truly amazing is that, in a sense, these sine and cosine functions define all possible r -periodic functions: In 1807 the French math- ematician Jean Baptiste Fourier (1768-1830) realized that every r -periodic function can be written as a linear combination of the sines and cosines of (5.81). That is, if f (t) is any 16 periodic function with period T then it can be expressed as the sum where the constants an and bn depend on the function f (t). This extraordinarily useful result is called Fourier's theorem, and the", " sum (5.82) is called the Fourier series for f (t) It is not hard to see why Fourier's theorem met with considerable surprise, and even skepticism, when he first published it. It claims that a discontinous function, such as the rectangular pulse of Figure 5.20(a), can be built up with sine and cosine functions that are continuous and perfectly smooth. Surprising or not, this turns out to be true, as we shall see by example shortly. Perhaps even more surprising, it is often the case that one gets an excellent approximation by retaining just the first few terms of a Fourier series. Thus, instead of having to handle a fairly disagreeable and possibly discontinuous function, we have only to handle a reasonably small number of sines and cosines. Before we discuss the application of Fourier's theorem to the driven oscillator, we need to look at a few properties of Fourier series. The proof of Fourier's theorem is difficult \u2014 indeed it was many years after Fourier's discovery before a satisfactory proof was found \u2014 and I shall simply ask you to accept it. However, once the result is accepted it is easy to learn to use it. In 16 As always with theorems of this kind, there are certain restrictions on the \"reasonableness\" of the function f (t), but certainly Fourier's theorem is valid for all of the functions we shall have occasion to use. particular, for any given periodic function f (t) it is easy to find the coefficients an and bn. Problem 5.48 gives you the opportunity to show that these coefficients are Section 5.7 Fourier Series* 195 given by and (asjti(i; 44) di.`.. (5.j34) Unfortunately the coefficients for n = 0 require separate attention. Since the term sin ncot in (5.82) is identically zero for n = 0, the coefficient b0 is irrelevant and we can simply define it to be zero. It is very easy to show (Problem 5.46) that Armed with these formulas for the Fourier coefficients, it is easy to find the Fourier series for any given periodic function. In the following example, we do this for the rectangular pulses of Figure 5.20(a). EXAMPLE 5.4 Fourier Series for the Rectangular Pulse Find the Fourier series for the periodic rectangular pulse f (t) shown in Figure 5.22 in terms of the period z, the pulse height fmax' and the duration of the pulse A. Using the values r = 1, fmax = 1, and AZ = 0.25, plot f (t), as well as the sum of the first three terms of its Fourier series, and the sum of the firs _ t eleven terms. Our first task is to calculate the Fourier coefficients an and bn for the given function. First according to (5.85) the constant term ac, is 1 ao = \u2014 r/2 f (t) dt fr/2 1Ar/2 -Ar/2 = \u2014 f fmax dt - f max Az (5.86) Chapter 5 Oscillations fmax \u20144- T -> \u2014112 7;2 AT AT/2 Figure 5.22 A periodic rectangular pulse. The period is t, the duration of the pulse is AT, and the pulse height is fmax. where the change in the limits of integration was allowed because the integrand f (t) is zero outside \u00b1 A T/2. Next, according to (5.83), all the other a coefficients (n > 1) are given by an = \u2014 f f (t) cos(nwt) dt r/2 2 \u2014r/2 2fmax At/2 t J\u2014At/2 cos(nwt) dt 4 fmax T 0 Ar/2 ( cos 2t 7/-n dt = sin 2f n n Ot -c ) (5.87) Notice that in passing from the second to the third line, I used a trick that is often useful in evaluating Fourier coefficients. The integrand on the second line, cos(ncot), is an even function; that is, it has the same value at any point t as at \u2014t. Therefore we could replace any integral from \u2014T to T by twice the integral from 0 to T. Finally the b coefficients are all exactly zero, for if you examine the integral (5.84) you will see that (in this case) the integrand is an odd function; that is, its value at any point t is the negative of its value at \u2014t. [Moving from t to \u2014t leaves f (t) unchanged, but reverses the sign of sin(nwt).] Therefore any integral from \u2014T to T is zero, since the left half exactly cancels the right. The required Fourier series is therefore f (t) = ao", " an cos(nwt) (5.88) 00 n=1 with the constant term a 0 given by (5.86) and all the remaining a coefficients (n > 1) by (5.87). If we put in the given numbers, these coefficients can all be evaluated and the resulting Fourier series is f (t) = f max [o.25 + 0.24 cos(27t) + 0.23 cos(47tt) + 0.20 cos(67t) + 0.16 cos(87t) + 0.12 cos(107rt) + 0.08 cos(127rt) + ] (5.89) Section 5.8 Fourier Series Solution for the Driven Oscillator* 197 f f(t) f f(t) (a) 3 terms (b) 11 terms Figure 5.23 (a) The sum of the first three terms of the Fourier series for the rectangular pulse of Figure 5.22. (b) The sum of the first 11 terms. The practical value of a Fourier series is usually greatest if the series con- verges rapidly, so that we get a reliable approximation by retaining just the first few terms of the series. Figure 5.23(a) shows the sum of the first three terms of the series (5.89) and the rectangular pulse itself. As you might expect, with just three smooth terms we do not get a sensationally accurate approximation to the original discontinuous function. Nevertheless, the three terms do a remarkable job of imitating the general shape. By the time we have included 11 terms, as in Figure 5.23(b), the fit is amazing. 17 In the next section, we shall use the method of Fourier series to solve for the motion of an oscillator driven by the periodic pulses of this example. We shall find the solution as a Fourier series, which con- verges so quickly that just the first 3 or 4 terms tell us most of what is interesting to know. 5.8 Fourier Series Solution for the Driven Oscillator* * This section contains a beautiful application of the method of Fourier series. Important as it is to understand this method, you can nevertheless omit the section without loss of continuity. In this section we shall combine our knowledge of Fourier series (Section 5.7) with our solution of the sinusoidally driven oscillator (Section 5.5) to solve for the motion of an oscillator that is driven by an arbitrary periodic driving force. To see how this works, let us return to the equation of motion (5.48) where x = x (t) is the position of the oscillator, /3 is the damping constant, w o is the natural frequency, and f = f (t) is any periodic driving force (actually force/mass) with period r. As before it is convenient to rewrite this in the compact form + 2,6.Z + coo2x = f Dx = f 17 Notice, however, that the Fourier series still has a little difficulty in the immediate neigh- borhood of the discontinuities in f (t). This tendency for the Fourier series to overshoot near a discontinuity is called the Gibbs phenomenon. 198 Chapter 5 Oscillations where D stands for the linear differential operator D = 2 \u201e d z p \u2014 dt dt2 2 co. \u00b0 The use of Fourier series to solve this problem hinges on the following observation: Suppose that the force f (t) is the sum of two forces, f (t) = f1 (t) f2 (t), for each of which we have already solved the equation of motion. That is, we already know functions x i (t) and x2 (t) that satisfy Dx t = fi and Dx 2 = f2. Then the solution 18 to the problem of interest is just the sum x(t) = x l (t) x 2 (t), as we can easily show: Dx = D(x l + x2) = Dxi + Dx2 = + f2 = f where the crucial second step is valid because D is linear. This argument would work equally well however many terms were in the sum for f (t), so we have the conclusion: If the driving force f (t) can be expressed as the sum of any number of terms f (t) = E fn(t) x(t) = xn (t). n and if we know the solutions x,, (t) for each of the individual forces f,i (t), then the solution for the total driving force f (t) is just the sum This result is ideally suited for use in combination with Fourier's theorem. Any periodic driving force f (t) can be expanded in a Fourier series of sines and cosines, and we already know the solutions for sinusoidal driving forces. Thus", ", by adding these sinusoidal solutions together, we can find the solution for any periodic driving force. To simplify our writing, let us suppose that the driving force f (t) contains only cosine terms in its Fourier series. [This was the case for the rectangular pulse of Example 5.4, and is true for any even function \u2014 satisfying f (\u2014t) = f (t) \u2014 because this condition guarantees that the coefficients of the sine terms are all zero.] In this case, the driving force can be written as 18 Strictly speaking we should not speak of the solution, since our second-order differential equation has many solutions. However, we know that the difference between any two solutions is transient \u2014 decays to zero \u2014 and our main interest is in the long-term behavior, which is therefore essentially unique. Section 5.8 Fourier Series Solution for the Driven Oscillator* 199 where f, denotes the nth Fourier coefficient of f (t), and w = 2n- / -c as usual. Now, each individual term f\u201ecos(ncot) has the same form (5.56) that we assumed for the sinusoidal driving force in Section 5.5 (except that the amplitude fc, has become fn and the frequency co has become nco). The corresponding solution was given in (5.66), 19 where from (5.64), and xn (t) = A n cos(nwt \u2014 8n ) A n = fn l 2 \u2014 n2w2)2 4182n2c02 6, = arctan 2,8nco co 2 0 n 2602 (5.91) (5.92) (5.93) from (5.65). Since (5.91) is the solution for the driving force fn cos(ncot), the solution for the complete force (5.90) is the sum This completes the solution for the long-term motion of an oscillator driven by a periodic driving force f (t). To summarize, the steps are: 1. Find the coefficients fn in the Fourier series (5.90) for the given driving force f (t) 2. Calculate the quantities A n and 8,\u201e as given by (5.92) and (5.93). 3. Write down the solution x (t) as the Fourier series (5.94). In practice, one usually needs to include surprisingly few terms of the solution (5.94) to get a satisfactory approximation, as the following example illustrates 2 0 EXAMPLE 5.5 An Oscillator Driven by a Rectangular Pulse Consider a weakly damped oscillator that is being driven by the periodic rectan- gular pulses of Example 5.4 (Figure 5.22). Let the natural period of the oscillator be r, = 1, so that the natural frequency is co o = 27r, and let the damping constant 19 The n = 0, constant term needs separate consideration. It is easy to see that for a constant force fo the solution is xo = fo/w02. This is actually exactly what you get if you just set n = 0 in (5.92) and (5.93). 20 The solution contained in Equations (5.92) to (5.94) can be written more compactly if you don't mind using complex notation. See Problem 5.51. 200 Chapter 5 Oscillations be /3 = 0.2. Let the pulse last for a time Ar = 0.25 and have a height f max = 1. Calculate the first six Fourier coefficients A n for the long-term motion x (t) of the oscillator, assuming first that the drive period is the same as the natural period, t = to = 1. Plot the resulting motion for several complete oscillations. Repeat these exercises for r = 1.5t0, 2.0r0, and 2.5ro. Before we look at any of these exercises, it is worth thinking of a real system this problem might represent. One simple possibility is a mass hanging from the end of a spring, to which a professor is applying regularly spaced upward taps at intervals r. An even more familiar example would be a child in a swing, to whom a parent is giving regularly spaced pushes \u2014 though in this case we need to be careful to keep the amplitude small to justify the use of Hooke's law. We are told to start by taking r = r o = 1; that is, the parent is pushing the child at exactly the natural frequency. The Fourier coefficients fn of the driving force were already calculated in Equations (5.86) and (5.87) of Example 5.4 (where they were called an ). If we substitute these into (5.92) for the coefficients A n and put in the given numbers (including r = r0 = 1), we", " find for the first six Fourier coefficients A o,, A5: Ao Al A2 A3 A4 A5 63 1791 27 5 0 1 (Since the numbers are rather small, I have quoted the values multiplied by 10 4 ; that is, A o = 63 x 10-4, A l = 1791 x 10 -4, and so on.) Two things stand out about these numbers: First, after A 1, they get rapidly smaller, and for almost all purposes it would be an excellent approximation to ignore all but the first three terms in the Fourier series for x(t). Second, the coefficient A l is vastly bigger than all the rest. This is easy to understand if you look at (5.92) for the coefficient A 1 : Since co = coo (remember the parent is pushing the child at the swing's natural frequency) and n = 1, the first term in the denominator is exactly zero, the denominator is anomalously small, and A l is anomalously big compared to all the other coefficients. In other words, when the driving frequency is the same as the natural frequency, the n = 1 term in the Fourier series for x (t) is at resonance, and the oscillator responds especially strongly with frequency coo. Before we can plot x (t) as given by (5.94), we need to calculate the phase shifts Sn using (5.93). This is easily done, though I shan't waste space displaying the results. We cannot actually plot the infinite series (5.94); instead, we must pick some finite number of terms with which to approximate x (t). In the present case, it seems clear that three terms would be plenty, but to be on the safe side I'll use six. Figure 5.24 shows x(t) as approximated by the sum of the first six terms in (5.94). At the scale shown, this approximate graph is completely indistinguishable from the exact result, which in turn is indistinguishable from a pure cosine21 with frequency equal to the natural frequency of the oscillator. The strong response at the natural frequency is just what we would expect. For 21 Actually it's a pure sine, but this is really cos(wt \u2014 8 1) with S i = 7r/2 as we should have expected because we're exactly on resonance. Section 5.8 Fourier Series Solution for the Driven Oscillator* 201 x(t) T 0.2 0.1 \u20140.1 1 2 3 4 5 6 Figure 5.24 The motion of a linear oscillator, driven by periodic rectangular pulses, with the drive period r equal to the natural period ro of the oscillator (and hence to = coo). The horizontal axis shows time in units of the natural period to. As expected the motion is almost perfectly sinusoidal, with period equal to the natural period. instance, anyone who has pushed a child on a swing knows that the most efficient way to get the child swinging high is to administer regularly spaced pushes at intervals of the natural period \u2014 that is, r = r o \u2014 and that the swing will then oscillate vigorously at its natural frequency. A driving force with any other period r can be treated in exactly the same way. The Fourier coefficients A o,, A5 for all of the values of r requested above are shown in Table 5.1. Table 5.1 The first six Fourier coefficients A, for the motion x (t) of a linear oscillator driven by periodic rectangular pulses, for four different drive periods r = to, 1.5ro, 2.0r0, and 2.5r0. All values have been multiplied by 104. r = 1.0 ro r = 1.5 ro r = 2.0 ro r = 2.5 ro A0 63 42 32 25 At A2 A3 A4 A5 1791 145 82 59 27 89 896 130 5 18 40 97 0 6 13 25 1 2 6 11 The entries in the four rows of this table deserve careful examination. The first row ( -r = ro) shows the coefficients already discussed, the most prominent feature of which is that the n = 1 coefficient is far the largest, because it is exactly on resonance. In the next row (r = 1.5r.), the n = 1 Fourier component has moved well away from resonance, and A l has dropped by a factor of 12 or so. Some of the other coefficients have increased a bit, but the net effect is that the oscillator moves much less than when r = r o. This is clearly visible in Figures 5.25(a) and (b), which show x(t) (as approximated by the first six terms of its Fourier series) for these two values of the drive period. 202 Chapter 5 Oscillations x(t) 0. 1- 0.1 t (a) T = To (b) T = 1.", "5r0 0.1 0.1 Figure 5.25 The motion of a linear oscillator, driven by periodic rectangular pulses, showing four different values of the drive period r. (a) When the oscillator is driven at its natural frequency (r = r o), the n = 1 term of the Fourier series is at resonance and the oscillator responds energetically. (b) When = 1.5ro, the response is feeble. (c) When r = 2.0r0, the n = 2 term is at resonance and the response is strong again. (d) When r = 2.5ro, the response is again weak. The third row of Table 5.1 shows the Fourier coefficients for a drive period equal to twice the natural period \u2014 the parent is pushing just once every second swing of the child. Now, with r = 2.0r0, the drive frequency is half the natural frequency (w = Ico0). This means that the n = 2 Fourier component, with fre- quency 2w = w0, is exactly on resonance, and the coefficient A2 is anomalously large. Once again we get a large response, as seen in Figure 5.25(c). Let us look a little closer at the case of Figure 5.25(c). It is, of course, a matter of experience that a perfectly satisfactory way to get a child swinging is to push once every two swings, although this naturally doesn't get quite the result of pushing once for every swing. If you look carefully at Figure 5.25(c), you will notice that the swings alternate in size \u2014 the even swings are slightly bigger than the odd ones. This too is to be expected: Because the oscillator is damped, the second swing after each push is bound to be a little smaller than the first. Finally, when r = 2.5r0, the n = 2 Fourier component is well past resonance and A2 is much smaller again. On the other hand, the n = 3 component is approaching resonance so that A3 is getting bigger. In fact, A2 and A3 are roughly Section 5.9 The RMS Displacement; Parseval's Theorem * 203 the same size, so that x (t) contains two dominant Fourier components and shows the somewhat kinky behavior seen in Figure 5.25(d). (Similar considerations apply to the case r = 1.5r 0, where the coefficients A l and A2 are both fairly large.) 5.9 The RMS Displacement; Parseval's Theorem * * The Parseval relation, which we introduce and apply in this section, is one of the most useful properties of Fourier series. Nevertheless, you could omit this section if pressed for time. In the last section we studied how the response of an oscillator varied with the frequency of the applied periodic driving force. We did this by solving for the motion x (t), using the method of Fourier series, for each of several interesting applied frequencies. It would be convenient if we could find a single number to measure the oscillator's response and then just plot this number against the driving frequency (or driving period). In fact there are several ways to do this. Perhaps the most obvious thing to try would be the oscillator's average displacement from equilibrium, (x). (I am using angle brackets ( ) to indicate a time average.) Unfortunately, since the oscillator spends as much time in any region of positive x as in the corresponding region of negative x, the average (x) is zero. 22 To get around this difficulty the most convenient quantity to use is the mean square displacement (x 2 ), and to give a quantity with the dimensions of length we usually discuss the root mean square or RMS displacement The definition of the time average needs a little care. The usual practice is to define ( ) as the average over one period r. Thus, x rms (x 2). 1 f 712 42 ) = \u2014 2 dt. X \u2014r/2 Because the motion is periodic, this is the same as the average over any integer number of periods, and hence also the average over any long time. (If this isn't clear to you, see Problem 5.54.) To evaluate the average (x 2 ), we use the Fourier expansion (5.94) of x(t) x (t ) = E An cos(ncot \u2014 8n) - 00 n=0 (In general, this series will contain sines as well as cosines, but in Example 5.5 the driving force contained only cosines, and for simplicity let us continue to assume this 22 Not quite true. The small constant term A o in (5.94) contributes a nonzero average (x) = Ao, but this does not reflect any oscillation, which is what we are trying to characterize. (5.95) (5.96) (5.", "97) 204 Chapter 5 Oscillations is the case. For the general case, see Problem 5.56.) Substituting for each of the factors x in (5.96), we get the appalling double sum (x 2 ) \u00b1 EE An cos(nwt \u2014 6,)A\u201e, cos(mwt \u2014 8, n ) dt. (5.98) f t/2 \u2014r/2 n m Fortunately, this simplifies dramatically. It is fairly easy to show (Problem 5.55) that the integral is just r/2 t r/2 cos(nwt \u2014 8n ) COS(111C0t \u2014 6 m ) dt = r/2 if m = n A 0 (5.99) r 0 ifm=n= 0 if m n. Thus, in the double sum (5.98), only those terms with m = n need to be retained, and we get the surprisingly simple result that (x 2) = AD \u00b1.1 A 2 n'2 n=1 (5.100) This relation is called Parseval's theorem 2 3 It has many important theoretical uses, but for our purposes its main application is this: Since we know how to calculate the coefficients A n, Parseval's theorem lets us find the response (x 2) of our oscillator. Moreover, by dropping all but some modest finite number of terms in the sum (5.100), we get an excellent and easily calculated approximation for (x 2 ), as the following example illustrates. EXAMPLE 5.6 The RMS Displacement for a Driven Oscillator Consider again the oscillator of Example 5.5, driven by the periodic rectangular pulses of Example 5.4 (Figure 5.22). Find the RMS displacement x rms = V(x 2) as given by (5.100) for this oscillator. Using the same numerical values as before (ro = 1, fi = 0.2, fmax = 1, A r = 0.25) and approximating (5.100) by its first six terms, make a plot of x m\u201e as a function of the drive period r for 0.25 < r < 5.5. We have already done all the calculations needed to write down a formula for xims = \\/(x 2 ). First, (x 2 ) is given by (5.100), where the Fourier coefficients An are given by (5.92) as A n = fn 1/((.02 \u2014 n2602)2 + 4/32n2(02 (5.101) 23 Remember that we made the simplifying assumption that our Fourier series contained only cosine terms. In general, the sum in (5.100) must include contributions 13,i2 from the sine terms as well. See Problem 5.56. Principal Definitions and Equations of Chapter 5 205 arms 0.1 - 0 1 2 4 T Figure 5.26 The RMS displacement of a linear oscillator, driven by periodic rectangular pulses, as a function of the drive period r \u2014 calculated using the first six terms of the Parseval expression (5.100). The horizontal axis shows r in units of the natural period r0. When r is an integral multiple of To the response is especially strong. and the Fourier coefficients fn of the driving force are given by (5.87) and (5.86) as = 2 fmax nn sin (7 rn Or t [for n > 1] (5.102) while fo = fmax A / r. Putting all these together gives the desired formula for x. (which I'll leave you to write down if you want to see it). If we now put in the given numbers, we are left with just one independent variable, the period of the driving force r. (Remember that w = 27/r.) Trun- cating the infinite series (5.100) after six terms, we arrive at an expression that is easily evaluated with the appropriate software (or even a programmable calcula- tor) and plotted as shown in Figure 5.26. This graph shows clearly and succinctly what we found in the previous example. As we increase the drive period t, the response of the oscillator varies dramatically. Each time r passes through an integer multiple of the natural period r o (that is, r = nto), the response exhibits a sharp maximum, because the nth Fourier component is at resonance. On the other hand, each successive peak is lower than its predecessor, since we elected to fix the width AT and height fmax of the pulses; thus as the drive period gets longer, the net effect of the force would be expected to get less. Principal Definitions and Equations of Chapter 5 Hooke's Law F = \u2014 k x <- > U = 4kx2 [Section 5.1] 206 Chapter 5 Oscillations Simple Harmonic", " Motion = \u2014(02x <=;. x(t) = A cos(cot \u2014 6), etc. [Section 5.2] Damped Oscillations If the oscillator is subject to a damping force \u2014by, then + 2/3k coo2x = 0 <===>. x (t) = Ae \u2014 fi t cos(co it \u2014 8) [Eqs. (5.28) & (5.38)] where p = 2b/m, coo =.Vklm, co l = 11w02 \u2014 p2, and the solution given here is for \"weak damping\" (p < (00 ). Driven Damped Oscillations and Resonance If the oscillator is also subject to a sinusoidal driving force F(t) = Info cos(cot), the long-term motion has the form x(t) = A cos(cot \u2014 8) [Eq. (5.66)] where fo2 A 2 = (0) 2 ___ (02)2 + 02(02 0 [Eq. (5.64)] and the phase shift 8 is given by (5.65). To this solution can be added a \"transient\" solution of the corresponding homogeneous equation, but this dies out as time passes. The long-term solution \"resonates\" (has a sharp maximum) when co is close to coo. Fourier Series If the driving force is not sinusoidal, but is still periodic, it can be built up as a Fourier series of sinusoidal terms, as in (5.90), and the resulting motion is the corresponding series of sinusoidal solutions, as in (5.94): x ( t). E An cos(ncot \u2014 6n ). 00 n=0 [Eq. (5.94)] The RMS Displacement The root mean square displacement Xmis = [Eqs. (5.95) & (5.96)] Problems for Chapter 5 207 is a good measure of the average response of the oscillator and is given by Parseval's theorem as xnns -./V \u00b1 An2. [Eq. (5.100)] 00 n=i Problems for Chapter 5 SECTION 5.1 Hooke's Law Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). 5.1 * A massless spring has unstretched length / 0 and force constant k. One end is now attached to the ceiling and a mass m is hung from the other. The equilibrium length of the spring is now / 1. (a) Write down the condition that determines 1 1. Suppose now the spring is stretched a further distance x beyond its new equilibrium length. Show that the net force (spring plus gravity) on the mass is F = \u2014 kx. That is, the net force obeys Hooke's law, when x is the distance from the equilibrium position \u2014 a very useful result, which lets us treat a mass on a vertical spring just as if it were horizontal. (b) Prove the same result by showing that the net potential energy (spring plus gravity) has the form U (x) = const ikx 2. 5.2 * The potential energy of two atoms in a molecule can sometimes be approximated by the Morse function, U (r) = ARe(R-01.5 \u2014 1)2 where r is the distance between the two atoms and A, R, and S are positive constants with S << R. Sketch this function for 0 < r < cc. Find the equilibrium separation rip, at which U (r) is minimum. Now write r = ro x so that x is the displacement from equilibrium, and show that, for small displacements, U has the approximate form U = const kx 2. That is, Hooke's law applies. What is the force constant k? 5.3 * Write down the potential energy U(0) of a simple pendulum (mass m, length 1) in terms of the angle 0 between the pendulum and the vertical. (Choose the zero of U at the bottom.) Show that, for small angles, U has the Hooke's law form U(0) = 4k02, in terms of the coordinate 0. What is k? 5.4 ** An unusual pendulum is made by fixing a string to a horizontal cylinder of radius R, wrapping the string several times around the cylinder, and then tying a mass m to the loose end. In equilibrium the mass hangs a distance lo vertically below the edge of the cylinder. Find the potential energy if the pendulum has swung to an angle 0 from the vertical. Show that for small angles, it can be written in the Hooke's law form U = 1142. Comment on the value of k. SECTION 5.2 Simple Harmonic Motion 5.5 * In Section 5.2 we discussed four equivalent ways to represent", " simple harmonic motion in one dimension: 208 Chapter 5 Oscillations x(t) = C C2e \u2014i \" = B1 cos(cot) + B2 sin(cot) (II) = A cos(cot \u2014 8) = Re Cei\" (I) (HI) (IV) To make sure you understand all of these, show that they are equivalent by proving the following implications: I II = III IV = I. For each form, give an expression for the constants (C1, C2, etc.) in terms of the constants of the previous form. 5.6 * A mass on the end of a spring is oscillating with angular frequency co. At t = 0, its position is xo > 0 and I give it a kick so that it moves back toward the origin and executes simple harmonic motion with amplitude 2x o. Find its position as a function of time in the form (III) of Problem 5.5. 5.7 * (a) Solve for the coefficients B 1 and B2 of the form (II) of Problem 5.5 in terms of the initial position xo and velocity vo at t = 0. (b) If the oscillator's mass is m = 0.5 kg and the force constant is k = 50 N/m, what is the angular frequency co? If x o = 3.0 m and vo = 50 m/s, what are B 1 and B2? Sketch x(t) for a couple of cycles. (c) What are the earliest times at which x = 0 and at which = 0? 5.8 * (a) If a mass m = 0.2 kg is tied to one end of a spring whose force constant k = 80 N/m and whose other end is held fixed, what are the angular frequency co, the frequency f, and the period r of its oscillations? (b) If the initial position and velocity are x o = 0 and vo = 40 m/s, what are the constants A and 6 in the expression x(t) = A cos(cot 6)? 5.9 * The maximum displacement of a mass oscillating about its equilibrium position is 0.2 m, and its maximum speed is 1.2 m/s. What is the period r of its oscillations? 5.10* The force on a mass m at position x on the x axis is F = \u2014Fo sink ax, where Fo and a are constants. Find the potential energy U(x), and give an approximation for U(x) suitable for small oscillations. What is the angular frequency of such oscillations? 5.11 * You are told that, at the known positions x 1 and x2, an oscillating mass m has speeds v 1 and v2. What are the amplitude and the angular frequency of the oscillations? 5.12 ** Consider a simple harmonic oscillator with period r. Let ( f ) denote the average value of any variable f (t), averaged over one complete cycle: (f) = f (t) dt. 1 f 0 (5.103) Prove that (T) = (U) = 4E where E is the total energy of the oscillator. [Hint: Start by proving the more general, and extremely useful, results that (sin 2 (cot \u2014 6)) = (cos2 (cot \u2014 8)) = 2. Explain why these two results are almost obvious, then prove them by using trig identities to rewrite sin e 6 and cost 0 in terms of cos(261 ).] 5.13 ** The potential energy of a one-dimensional mass m at a distance r from the origin is U(r)=Uo (L \u00b1A 211 ) for 0 < r < o, with U0, R, and X all positive constants. Find the equilibrium position r0. Let x be the distance from equilibrium and show that, for small x, the PE has the form U = const kx 2. What is the angular frequency of small oscillations? Problems for Chapter 5 209 Figure 5.27 Problem 5.18 SECTION 5.3 Two-Dimensional Oscillators 5.14 * Consider a particle in two dimensions, subject to a restoring force of the form (5.21). (The two constants kx and k may or may not be equal; if they are, the oscillator is isotropic.) Prove that its potential energy is l(kx,2 kyy 2 ). (5.104) 5.15 * The general solution for a two-dimensional isotropic oscillator is given by (5.19). Show that by changing the origin of time you can cast this in the simpler form (5.20) with 6 = (Sy 6x. [Hint: A change of origin of time is a change of variables from t to t' = t to. Make this change and choose the constant to appropriately, then rename t", "' to be t.] 5.16 * Consider a two-dimensional isotropic oscillator moving according to Equation (5.20). Show that if the relative phase is 6 = 7r/2, the particle moves in an ellipse with semimajor and semiminor axes A\u201e and A y. never repeats itself. 5.17 ** Consider the two-dimensional anisotropic oscillator with motion given by Equation (5.23). (a) Prove that if the ratio of frequencies is rational (that is, co x /coy = p I q where p and q are integers) then the motion is periodic. What is the period? (b) Prove that if the same ratio is irrational, the motion 5.18 *** The mass shown from above in Figure 5.27 is resting on a frictionless horizontal table. Each of the two identical springs has force constant k and unstretched length 10. At equilibrium the mass rests at the origin, and the distances a are not necessarily equal to lo. (That is, the springs may already be stretched or compressed.) Show that when the mass moves to a position (x, y), with x and y small, the potential energy has the form (5.104) (Problem 5.14) for an anisotropic oscillator. Show that if a <10 the equilibrium at the origin is unstable and explain why. 5.19 *** Consider the mass attached to four identical springs, as shown in Figure 5.7(b). Each spring has force constant k and unstretched length lo, and the length of each spring when the mass is at its equilibrium at the origin is a (not necessarily the same as 10). When the mass is displaced a small distance to the point (x, y), show that its potential energy has the form Ik'r 2 appropriate to an isotropic harmonic oscillator. What is the constant k' in terms of k? Give an expression for the corresponding force. SECTION 5.4 Damped Oscillations 5.20 * Verify that the decay parameter,8 \u2014 1,8 2 \u2014 coo for an overdamped oscillator (p > coo) de- creases with increasing p. Sketch its behavior for wo < p < 00. 5.21 * Verify that the function (5.43), x (t) = to \u2014 fi r, is indeed a second solution of the equation of motion (5.28) for a critically damped oscillator (,8 = (00). 5.22 * (a) Consider a cart on a spring which is critically damped. At time t = 0, it is sitting at its equilibrium position and is kicked in the positive direction with velocity v o. Find its position x (t) for 210 Chapter 5 Oscillations all subsequent times and sketch your answer. (b) Do the same for the case that it is released from rest at position x = xo. In this latter case, how far is the cart from equilibrium after a time equal to r o = 2n/wo, the period in the absence of any damping? 5.23 * A damped oscillator satisfies the equation (5.24), where F dmp = \u2014b1 is the damping force. Find the rate of change of the energy E = 4mi 2 + 4kx 2 (by straightforward differentiation), and, with the help of (5.24), show that dE/dt is (minus) the rate at which energy is dissipated by Fdrnp. 5.24 * In our discussion of critical damping (p = 600), the second solution (5.43) was rather pulled out of a hat. One can arrive at it in a reasonably systematic way by looking at the solutions for /3 < too and carefully letting /3 coo, as follows: For /3 < w o, we can write the two solutions as x 1 (t) = e-Pt cos(w i t) and x 2 (t) = e -13` sin(wit).' Show that as p wo, the first of these approaches the first solution for critical damping, x 1 (t) = e -Pt. Unfortunately, as /3 wo, the second of them goes to zero. (Check this.) However, as long as /3 w o, you can divide x 2 (t) by co l and you will still have a perfectly good second solution. Show that as p \u2014\u00b1 wo, this new second solution approaches the advertised to -13`. 5.25 ** Consider a damped oscillator with p < No. There is a little difficulty defining the \"period\" r 1 r1 is the time since the motion (5.38) is not periodic. However, a definition that makes sense is that between successive maxima of x (t). (a) Make a sketch of x (t) against t and indicate this definition of r on your graph", ". Show that r i = 27r/co 1. (b) Show that an equivalent definition is that r 1 is twice the time between successive zeros of x(t). Show this one on your sketch. (c) If /3 = w0 12, by what factor does the amplitude shrink in one period? 5.26 ** An undamped oscillator has period ro = 1.000 s, but I now add a little damping so that its period changes to r i = 1.001 s. What is the damping factor 8? By what factor will the amplitude of oscillation decrease after 10 cycles? Which effect of damping would be more noticeable, the change of period or the decrease of the amplitude? 5.27 ** As the damping on an oscillator is increased there comes a point when the name \"oscillator\" seems barely appropriate. (a) To illustrate this, prove that a critically damped oscillator can never pass through the origin x = 0 more than once. (b) Prove the same for an overdamped oscillator. 5.28 ** A massless spring is hanging vertically and unloaded, from the ceiling. A mass is attached to the bottom end and released. How close to its final resting position is the mass after 1 second, given that it finally comes to rest 0.5 meters below the point of release and that the motion is critically damped? 5.29 ** An undamped oscillator has period ro = 1 second. When weak damping is added, it is found that the amplitude of oscillation drops by 50% in one period r i. (The period of the damped oscillations is defined as the time between successive maxima, r i = 27r/w 1. See Problem 5.25.) How big is /3 compared to coo? What is r 1? 5.30 ** The position x(t) of an overdamped oscillator is given by (5.40). (a) Find the constants C 1 C2 in terms of the initial position x o and velocity vo. (b) Sketch the behavior of x (t) for the two and cases that v o = 0 and that x 0 = 0. (c) To illustrate again how mathematics is sometimes cleverer than we (and check your answer), show that if you let /3 0, your solution for x (t) in part (a) approaches the correct solution for undamped motion. 5.31 ** [Computer] Consider a cart on a spring with natural frequency w o = 2n, which is released from rest at xo = 1 and t = 0. Using appropriate graphing software, plot the position x (t) for 0 < t < 2 Problems for Chapter 5 211 and for damping constants,8 = 0, 1, 2, 4, 6, 27r, 10, and 20. [Remember that x(t) is given by different formulas for,8 < we, = coo, and /3 > coo l 5.32 ** [Computer] Consider an underdamped oscillator (such as a mass on the end of a spring) that is released from rest at position x o at time t = 0. (a) Find the position x(t) at later times in the form x(t) = e -fi t [B cos(w it) + B2 sin(w 1 t)]. That is, find B 1 and B2 in terms of xo. (b) Now show that if you let 13 approach the critical value wo, your solution automatically yields the critical solution. (c) Using appropriate graphing software, plot the solution for 0 < t < 20, with xo = 1, wo = 1, and /5 = 0, 0.02, 0.1, 0.3, and 1. SECTION 5.5 Driven Damped Oscillations 5.33 * The solution for x (t) for a driven, underdamped oscillator is most conveniently found in the form (5.69). Solve that equation and the corresponding expression for i, to give the coefficients B 1 B2 in terms of A, 8, and the initial position and velocity x o and vo. Verify the expressions given in and (5.70). 5.34 * Suppose that you have found a particular solution xp (t) of the inhomogeneous equation (5.48) for a driven damped oscillator, so that Dx p = f in the operator notation of (5.49). Suppose also that x (t) is any other solution, so that Dx = f. Prove that the difference x \u2014 x p must satisfy the corresponding homogeneous equation, D(x \u2014 x p) = 0. This is an alternative proof that any solution x of the inhomogeneous equation can be written as the sum of your particular solution plus a homogeneous solution; that is, x = xp xh. 5.", "35 ** This problem is to refresh your memory about some properties of complex numbers needed at several points in this chapter, but especially in deriving the resonance formula (5.64). (a) Prove that any complex number z = x iy (with x and y real) can be written as z = re\u00b0 where r and 0 are the polar coordinates of z in the complex plane. (Remember Euler's formula.) (b) Prove that the absolute value of z, defined as Izi = r, is also given by 1z1 2 = zz*, where z* denotes the complex conjugate of z, defined as z* = x \u2014 iy. (c) Prove that z* = (d) Prove that (zw)* = z*w* and that (1/z)* = 1/z*. (e) Deduce that if z = a/ (b ic), with a, b, and c real, then 1z1 2 = a 2/(b2 + C2 ). 5.36 ** [Computer] Repeat the calculations of Example 5.3 (page 185) with all the same parameters, but with the initial conditions xo = 2 and vo = 0. Plot x(t) for 0 < t < 4 and compare with the plot of Example 5.3. Explain the similarities and differences. 5.37 ** [Computer] Repeat the calculations of Example 5.3 (page 185) but with the following param- eters w = 27, coo = 0.25w, /3 = 0.2wo, fo = 1000 and with the initial conditions xo = 0 and vo = 0. Plot x(t) for 0 < t < 12 and compare with the plot of Example 5.3. Explain the similarities and differences. (It will help your explanation if you plot the homogeneous solution as well as the complete solution \u2014 homogeneous plus particular.) 5.38 ** [Computer] Repeat the calculations of Example 5.3 (page 185) but take the parameters of the system to be w = coo = 1, /3 = 0.1, and fo = 0.4, with the initial conditions x o = 0 and vo = 6 (all in some apppropriate units). Find A and 3, and then B 1 and B2, and make a plot of x (t) for the first ten or so periods. 212 Chapter 5 Oscillations 5.39 ** [Computer] To get some practice at solving differential equations numerically, repeat the calculations of Example 5.3 (page 185), but instead of finding all the various coefficients just use appropriate software (for example, the NDSolve command of Mathematica) to solve the differential equation (5.48) with the boundary conditions xo = vo = 0. Make sure your graph agrees with Figure 5.15. SECTION 5.6 Resonance 5.40 * Consider a damped oscillator, with fixed natural frequency coo and fixed damping constant 8 co. Show that the amplitude (not too large), that is driven by a sinusoidal force with variable frequency of the response, as given by (5.71) is maximum when co = \u2014 2/3 2. (Note that so long as the resonance is narrow this implies w ti wo.) 5.41 * We know that if the driving frequency w is varied, the maximum response (A 2) of a driven damped oscillator occurs at w ti wo (if the natural frequency is wo and the damping constant p c 0 o). Show that A 2 is equal to half its maximum value when w = coo \u00b1 8, so that the full width at half maximum is just 2,8. [Hint: Be careful with your approximations. For instance, it's fine to say w + coo ti 2wo, but you certainly mustn't say w \u2014 coo = 0.] 5.42 * A large Foucault pendulum such as hangs in many science museums can swing for many hours before it damps out. Taking the decay time to be about 8 hours and the length to be 30 meters, find the quality factor Q. 5.43 ** When a car drives along a \"washboard\" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscillates is each axle assembly, comprising the axle and its two wheels.) Find the speed of my car at which this oscillation resonates, given the following information: (a) When four 80-kg men climb into my car, the body sinks by a couple of centimeters. Use this to estimate the spring constant k of each of the four springs. (b) If an axle assembly (axle plus two wheels) has total mass 50 kg, what is the natural frequency of the assembly oscillating on its two springs? (c) If the bumps on a road are 80 cm apart, at about what speed would", " these oscillations go into resonance? 5.44 ** Another interpretation of the Q of a resonance comes from the following: Consider the motion of a driven damped oscillator after any transients have died out, and suppose that it is being driven close to resonance, so you can set w = co.. (a) Show that the oscillator's total energy (kinetic plus potential) is E = Zmw 2 A 2. (b) Show that the energy A Edis dissipated during one cycle by the damping force Fdmp is 2n- ini3wA 2. (Remember that the rate at which a force does work is Fv.) (c) Hence show that Q is 27r times the ratio E/ A Edis. 5.45 *** Consider a damped oscillator, with natural frequency co o and damping constant /3 both fixed, that is driven by a force F(t) = Fo cos(cot). (a) Find the rate P (t) at which F(t) does work and show that the average rate (P) over any number of complete cycles is m$w 2 A 2. (b) Verify that this is the same as the average rate at which energy is lost to the resistive force. (c) Show that as w is varied (P) is maximum when w = wo ; that is, the resonance of the power occurs at w = wo (exactly). SECTION 5.7 Fourier Series* 5.46 * The constant term ac, in a Fourier series is a bit of a nuisance, always requiring slightly special treatment. At least it has a rather simple interpretation: Show that if f (t) has the standard Fourier series (5.82), then cio is equal to the average ( f ) of f (t) taken over one complete cycle. Problems for Chapter 5 213 5.47 ** In order to prove the crucial formulas (5.83)0[5.85) for the Fourier coefficients an and bn, you must first prove the following: c/2 J r/2 cos(ncot) cos(mcot) d t = r/2 if 0 if m n. m = n 0 (5.105) (This integral is obviously r if m = n = 0.) There is an identical result with all cosines replaced by sines, and finally r/2 J\u2014r/2 cos(0 + 0) and so on.] cos(ncot) sin (mot) dt = 0 for all integers n and m, (5.106) where as usual w = 27r/r. Prove these. [Hint: Use trig identities to replace cos(0) cos(0) by terms like 5.48 ** Use the results (5.105) and (5.106) to prove the formulas (5.83)\u2014(5.85) for the Fourier coeffi- cients an and bn. [Hint: Multiply both sides of the Fourier expansion (5.82) by cos(mwt) or sin(mwt) and then integrate from \u2014z/2 to r/2.] 5.49 *** [Computer] Find the Fourier coefficients an and bn for the function shown in Figure 5.28(a). Make a plot similar to Figure 5.23, comparing the function itself with the first couple of terms in the Fourier series, and another for the first six or so terms. Take f max = 1. Figure 5.28 (a) Problem 5.49. (b) Problem 5.50 5.50 *** [Computer] Find the Fourier coefficients an and bn for the function shown in Figure 5.28(b). Make a plot similar to Figure 5.23, comparing the function itself with the sum of the first couple of terms in the Fourier series, and another for the first 10 or so terms. Take fmax = 1. SECTION 5.8 Fourier Series Solution for the Driven Oscillator* 5.51 ** You can make the Fourier series solution for a periodically driven oscillator a bit tidier if you don't mind using complex numbers. Obviously the periodic force of Equation (5.90) can be written as f = Re(g), where the complex function g is Show that the real solution for the oscillator's motion can likewise be written as x = Re(z), where g(t) = fneinwt. cx) n=0 z (t),Ecneinwt n=0 Chapter 5 Oscillations 214 and cn = fn (0 2 \u2014 n 2 co2 + 2i finco 0 This solution avoids our having to worry about the real amplitude A n and phase shift 5, separately. (Of course A n and 8n are hidden inside the complex number Ca.) 5.52 *** [Computer] Repeat all the calculations and plots of", " Example 5.5 (page 199) with all the same parameters except that /3 = 0.1. Compare your results with those of the example. 5.53 *** [Computer] An oscillator is driven by the periodic force of Problem 5.49 [Figure 5.28(a)], which has period r = 2. (a) Find the long-term motion x (t), assuming the following parameters: natural period v, = 2 (that is, co. = 7), damping parameter p = 0.1, and maximum drive strength fn.. = 1. Find the coefficients in the Fourier series for x(t) and plot the sum of the first four terms in the series for 0 < t < 6. (b) Repeat, except with natural period equal to 3. SECTION 5.9 The RMS Displacement; Parseval's Theorem * 5.54 * Let f (t) be a periodic function with period r. Explain clearly why the average of f over one period is not necessarily the same as the average over some other time interval. Explain why, on the other hand, the average over a long time T approaches the average over one period, as T --->- co. 5.55 ** To prove the Parseval relation (5.100), one must first prove the result (5.99) for the integral of a product of cosines. Prove this result, and then use it to prove the Parseval relation. 5.56 ** The Parseval relation as stated in (5.100) applies to a function whose Fourier series happens to contain only cosines. Write down the relation and prove it for a function x(t) [A\u201e cos(ncot \u2014 5,,,) Bn sin(nwt \u2014 3n ) ]. 00 n=0 5.57 ** [Computer] Repeat the calculations that led to Figure 5.26, using all the same parameters except taking /3 = 0.1. Plot your results and compare your plot with Figure 5.26. CHAPTER Calculus of Variations In many problems one needs to use non-Cartesian coordinates. Roughly speaking there are two classes of such problems. First, certain symmetries make it most advantageous to use special coordinates: Problems with spherical symmetry call out for the use of spherical polar coordinates; similarly, problems with axial symmetry are best treated in cylindrical polar coordinates. Second, when particles are constrained in some way, it is usually best to choose an appropriate, and usually non-Cartesian, coordinate system. For example, an object that is constrained to move on the surface of a sphere is probably best treated using spherical polar coordinates; if a bead slides on a curved wire, the best choice of coordinate may be just the distance along the curving wire from some convenient origin. Unfortunately, as we have seen, the expressions for the components of the accel- eration in non-Cartesian coordinates are quite messy, and the situation gets rapidly worse as we move on to more complicated systems. This makes Newton's second law difficult to use in non-Cartesian coordinates. We need an alternative (though ul- timately equivalent) equation of motion that works equally well in any coordinates, and the required alternative is provided by Lagrange's equations. The best way to prove \u2014 and to understand the great flexibility of \u2014 Lagrange's equations is to use a \"variational principle.\" Variational principles are important in many areas of mathematics and physics. It has proved possible to formulate almost every branch of physics \u2014 classical mechanics, quantum mechanics, optics, electro- magnetism, and so on \u2014 in variational terms. To the beginning student, accustomed to Newton's laws, a reformulation of classical mechanics in terms of a variational principle does not necessarily seem like an improvement. But because they allow a similar formulation of so many different subjects, variational methods have given a unity to physics and have played a crucial role in the recent history of physical theory. For this reason, I would like to introduce variational methods in a reasonably general setting. Therefore this short chapter is a brief introduction to variational problems in general. In the next chapter I shall apply what we learn here to establish the Lagrangian formulation of mechanics. If you are already familiar with the \"calculus of variations\" you could skip straight to Chapter 7. 215 216 Chapter 6 Calculus of Variations 6.1 Two Examples The calculus of variations involves finding the minimum or maximum of a quantity that is expressible as an integral. To see how this can arise, I would like to start with two simple, concrete examples. The Shortest Path between Two Points My first example is this problem: Given two points in a plane, what is the shortest path between them? While you certainly know the answer \u2014 a straight line \u2014 you probably have not seen a proof, unless you have studied the calculus of variations. The problem is illustrated in Figure 6.1, which shows the two", " given points, (x 1, y i) and (x2, y2), and a path, y = y(x), joining them. Our task is to find the path y(x) that has the shortest length and to show that it is in fact a straight line. The length of a short segment of the path is ds = \\/dx 2 + dy2, which, since dy = dy dx (x) dx, dx we can rewrite as ds = /dx 2 dy 2 + y' (x) 2 dx. (6.1) Thus the total length of the path between points 1 and 2 is L = f 2 ds = f -11 + y'(x)2 dx. x2 (6.2) xi This equation puts our problem in mathematical form: The unknown is the function y = y(x) that defines the path between points 1 and 2. The problem is to find the function y(x) for which the integral (6.2) is a minimum. It is interesting to contrast this with the standard minimization problem of elementary calculus, where the unknown y Y2 y1 - x1 x x2 Figure 6.1 A path joining the two points 1 and 2. The length of the short segment is ds = \\Idx 2 dy 2, and the total length of the path is L = fi2 ds. is the value of a variable x at which a known function f (x) is a minimum. Obviously our new problem is one stage more complicated than this old one. Before we set up the machinery to solve this new problem, let's consider another Section 6.1 Two Examples 217 example. Fermat's Principle A similar problem is to find the path that light will follow between two points. If the refractive index of the medium is constant, then the path is, of course, a straight line, but if the refractive index varies, or if we interpose a mirror or lens, the path is not so obvious. The French mathematician Fermat (1601-1665) discovered that the required path is the path for which the time of travel of the light is minimum. We can illustrate Fermat's principle using Figure 6.1. The time for light to travel a short distance ds is ds/v where v denotes the speed of light in the medium, v = c/n where n is the refractive index. Thus Fermat's principle says that the correct path between points 1 and 2 is the path for which the time 2 f 1 dt = (time of travel) = \u2014 = \u2014 n ds 2 ds I 1 C 2 l1 v is a minimum. If n is constant, then it can be taken outside the integral and the problem reduces to finding the shortest path between points 1 and 2 (and the answer is, of course, a straight line). In general, the refractive index can vary, n = n(x, y), and our problem is to find the path y(x) for which the integral 2 x2 x1 n(x, y) ds = f n(x, y'(x) 2 dx (6.3) is minimum. [In writing the last expression, I substituted (6.1) for ds.1 The integral that has to be minimized in connection with Fermat's principle is very similar to the integral (6.2) giving the length of a path; it is just a little more complicated, since the factor n(x, y) introduces an extra dependence on x and y. Similar integrals arise in many other problems. Sometimes we want the path for which an integral is a maximum, and sometimes we are interested in both maxima and minima. To get some idea of the possibilities, it is helpful to think again about the problem of finding maxima and minima of functions in elementary calculus. There we know that the necesary condition for a maximum or minimum of a function f (x) is that its derivative vanish, dfldx = 0. Unfortunately, this condition is not quite enough to guarantee a maximum or minimum. As you certainly recall from introductory calculus, there are essentially three possibilities, as illustrated in Figure 6.2. A point xo where dfldx is zero may be a maximum or a minimum or, if d2 f/dx 2 is also zero, it may be neither, as indicated in Figure 6.2(c). When df ldx = 0 at a point x o, but we don't know which of the three possibilities obtains, we say that xo is a stationary point of the function f (x), since an infinitesimal displacement of x from x o leaves f (x) unchanged (because the slope is zero). The situation for the problems of this chapter is very similar. The method I shall describe in the next section actually finds the path that makes an integral like (6.2) or", " 218 Chapter 6 Calculus of Variations xo (a) x xo (b) X, (c) Figure 6.2 If df/dx = 0 at xo, there are three possibilities: (a) If the second derivative is positive, then f (x) has a minimum at xo. (b) If the second derivative is negative, then f (x) has a maximum. (c) If the second derivative is zero, then there may be a minimum, a maximum, or neither (as shown). (6.3) stationary, in the sense that an infinitesimal variation of the path from its correct course doesn't change the value of the integral concerned. If you need to know that the integral is definitely minimum (or definitely maximum, or perhaps neither), you have to check this separately. Incidentally, we are now ready to explain the name of this chapter: Since our concern is how infinitesimal variations of a path change an integral, the subject is called the calculus of variations. For the same reason, the methods we shall develop are called variational methods, and a principle like Fermat's principle is a variational principle. 6.2 The Euler\u2014Lagrange Equation The two examples of the last section illustrate the general form of the so-called variational problem. We have an integral of the form where y(x) is an as-yet unknown curve joining two points (x 1, y i) and (x 2, y 2) as in Figure 6.1; that is, Y (xi) = YI and y (x 2) = y2. (6.5) Among all the possible curves satisfying (6.5) (that is, joining the points 1 and 2), we have to find the one that makes the integral S a minimum (or maximum or at least stationary). To be definite, I shall suppose that we wish to find a minimum. Notice that the function f in (6.4) is a function of three variables f = f (y, y', x), but because the integral follows the path y = y(x) the integrand f [y(x), y' (x), x] is actually a function of just the one variable x. Section 6.2 The Euler\u2014Lagrange Equation 219 Y(x) = y(x) + 17(x) (wrong) 2 y(x) (right) ty Y2 - y1 X1 Figure 6.3 The path y = y(x) between points 1 and 2 is the \"right\" path, the one for which the integral S of (6.4) is a minimum. Any other path Y (x) is \"wrong,\" in that it gives a larger value for S. Let us denote the correct solution to our problem by y = y(x). Then the integral S in (6.4) evaluated for y = y(x) is less than for any neighboring curve y = Y(x), as sketched in Figure 6.3. It is convenient to write the \"wrong\" curve Y(x) as where 17(x) (Greek \"eta\") is just the difference between the wrong Y(x) and the right y(x). Since Y (x) must pass through the endpoints 1 and 2, 17(x) must satisfy Y(x) = y(x) + 17(x) 77(x1) = /7(x2) = 0. (6.6) (6.7) There are infinitely many choices for the difference /7(x); for example, we could choose 77 = (x \u2014 x i) (x2 \u2014 x) or r7(x) = sin[7(x \u2014 x 1 )/(x2 \u2014 x i)]. The integral S taken along the wrong curve Y(x) must be larger than that along the right curve y(x), no matter how close the former is to the latter. To express this requirement, I shall introduce a parameter a and redefine Y(x) to be Y(x) = y(x) + aq(x). (6.8) The integral S taken along the curve Y(x) now depends on the parameter a, so I shall call it S(a). The right curve y(x) is obtained from (6.8) by setting a = 0. Thus the requirement that S is minimum for the right curve y (x) implies that S (a) is a minimum at a = 0. With this result, we have converted our problem to the traditional problem from elementary calculus of making sure that an ordinary function [namely S(a)] has a minimum at a specified point (a = 0). To ensure this, we must just check that the derivative dS/da is zero when a = 0. If we write out the integral S(a) in detail, it", " looks like this: S (a) = f f (Y, Y', x) dx x2 X2 = x.1 f + + x) dx. (6.9) 220 Chapter 6 Calculus of Variations To differentiate (6.9) with respect to a, we note that a appears in the integrand f, so we need to evaluate a flaa. Since a appears in two of the arguments of f, this gives two terms, namely (using the chain rule) af (y + alb y' + x) as af, =71\u2014 --r\" By, af, ay' and for dS/da (which has to be zero) dS da f' as dx = fx :2 \\ of + I ay' dx = 0. (6.10) This condition must be true for any )7(x) satisfying (6.7); that is, for any choice of the \"wrong\" path Y (x) = y(x) + ari(x). To take advantage of the condition (6.10), we need to rewrite the second term on the right using integration by parts' (remember that )7' means c/77/dx): f x2 )7'(x)-air dx =[17(x)-L aY I ay' L \u2014 jx 1 17(x)-d\u2014 L) dx. dx \\ay' I Because of the condition (6.7), the first term on the right (the \"endpoint term\") is zero. Thus2 f x2 171 (x) \u2014La dx = \u2014 ix2 )7(x) \u2014d ay' dx Jxi (L dx. j Substituting this identity into (6.10), we find that f x2 q(x) ( \u2014 af \u2014 \u2014 d Lf ) dx = O. ay dx ay' (6.11) (6.12) This condition must be satisfied for any choice of the function )7(x). Therefore, as I shall argue in a moment, the factor in large parentheses must be zero: for all x (in the relevant interval x 1 < x < x2). This is the so-called Euler\u2014Lagrange equation (named for the Swiss mathematician Leonhard Euler, 1707-1783, and the Italian-French physicist and mathematician Joseph Lagrange, 1736-1813), which lets If you are used to thinking of integration by parts in the form f v du = [uv] \u2014 f u dv, then you will find it helpful to recognize that another way to say the same thing is: f u'v dx = \u2014 f u v' dx. In words: In the integral f u'v dx, you can move the prime from the u to the v if you change the sign and add the endpoint contribution [u v]. 2 This is the simple form in which integration by parts often appears in physics: Provided the endpoint term [uv] is zero (as often happens), integration by parts lets you move the differentiation from the u to the v as long as you change the sign; that is, f u'v dx = \u2014 f uv'dx. Section 6.3 Applications of the Euler\u2014Lagrange Equation 221 us find the path for which the integral S is stationary. Before I illustrate its use, I need to discuss the step from (6.12) to (6.13), which is by no means obvious. Equation (6.12) has the form f q(x)g(x) dx = 0. I would certainly not claim that this condition alone implies that g(x) = 0 for all x. However, (6.12) holds for any choice of the function q(x), and if f (x)g(x) dx = 0 for any ri(x), then we can conclude that g(x) = 0 for all x. To prove this, we must assume that all functions concerned are continuous, but, as physicists, we would take for granted that this is the case. 3 Now, to prove the assertion, let us assume the contrary, that g(x) is nonzero in some interval between x 1 and x2. Then choose a function 7/ (x) that has the same sign as g(x) (that is, is positive where g is positive and ri is negative where g is negative). Then the integrand is continuous, satisfies q(x)g(x) > 0, and is nonzero at least in some interval. Under these conditions f Ti(x)g(x)dx cannot be zero. This contradiction implies that g(x) is zero for all x. This completes the proof of the Euler\u2014Lagrange equation. The procedure for using it is this: (1) Set up the problem so that the quantity whose stationary path you seek is expressed", " as an integral in the standard form x2 S = f x i f [y(x), (x), x]dx, (6.14) where f [y(x), y'(x), x] is the function appropriate to your problem. (2) Write down the Euler\u2014Lagrange equation (6.13) in terms of the function f[y(x), y'(x), x]. (3) Finally, solve (if possible) the differential equation (6.13) for the function y(x) that defines the required stationary path. I shall illustrate this procedure with a couple of examples in the next section. 6.3 Applications of the Euler\u2014Lagrange Equation Let us start with the problem that began this chapter, finding the shortest path between two points in a plane. EXAMPLE 6.1 Shortest Path between Two Points We saw that the length of a path between points 1 and 2 is given by the integral (6.2) as 2 L = f ds = f x2 11 \u00b1 yi2 dx. This has the standard form (6.14), with the function f given by f y', x) = (1 + y/2)1/2. (6.15) 3 The claimed result is clearly false if discontinuous functions are admitted. For instance, if we made g(x) nonzero at just one point, then f17 (x)g(x) dx would still be zero. 222 Chapter 6 Calculus of Variations To use the Euler\u2013Lagrange equation (6.13), we must evaluate the two partial derivatives concerned: of ay = 0 and f ay, = + y r2)1/2 (6.16) Since of/ay = 0, (6.13) implies simply that d af.0 dx ay' y r2 = c2(1 y t2), In other words, a f lay' is a constant, C. According to (6.16), this implies that or, with a little rearrangement, y'2 = constant. This implies that y'(x) is a constant, which we could call m. Integrating the equation y'(x) = m, we find that y(x) = mx b, and we have proved that the shortest path between two points is a straight line! A Note on Variables So far we have considered problems with two variables, which we have called x and y. Of these, x has been the independent variable, and y the dependent, through the relation y = y (x). Unfortunately, we are frequently forced \u2014 by convenience or tra- dition \u2014 to name the variables differently. For example, in a simple one-dimensional mechanics problem, the independent variable is the time t and the dependent variable is the position x = x (t). This means you will have to get used to seeing the Euler- Lagrange equation with the variables x and y replaced by an assortment of other variables, such as t and x. In the next example, the two variables are x and y, but the independent variable is y, and the roles of x and y in (6.13) and (6.14) will be exactly reversed. EXAMPLE 6.2 The Brachistochrone A famous problem in the calculus of variations is this: Given two points 1 and 2, with 1 higher above the ground, in what shape should we build a frictionless roller coaster track so that a car released from point 1 will reach point 2 in the shortest possible time? This problem is called the brachistochrone problem, from the Greek words brachistos meaning \"shortest\" and chronos meaning \"time.\" The geometry of the problem is sketched in Figure 6.4, where I have taken point 1 as the origin and I have chosen to measure y vertically down. The time to travel from 1 to 2 is time(1 2 ds 2) = f \u2014 v (6.17) where the speed at any height y is determined by conservation of energy to be v =,V2gy. (Problem 6.8.) Because this gives v as a function of y, it is convenient Section 6.3 Applications of the Euler\u2014Lagrange Equation 223 Figure 6.4 The brachistochrone problem is to find the shape of track on which a roller coaster released from point 1 will reach point 2 in the minimum possible time. to take y as our independent variable. That is, we shall write the unknown path as x = x (y). This means that the distance ds between neighboring points on the path has to be written as ds = /dx2 dy 2 = / x' (y) 2 + ldy (6.18) where a prime now denotes differentiation with respect to y; that is, x 1 (y) = dx/dy. Thus according to (", "6.17) the time of interest is time(1 2) = (6.19) 1 f Y2 3x1(y)2 + 1 dy. Jo Equation (6.19) gives the integral whose minimum we have to find. It is of the standard form (6.14), except that the roles of x and y have been interchanged, with the integrand To find the path that makes the time as small as possible, we have only to apply the Euler\u2014Lagrange equation (again with x and y interchanged) to this function, The function of (6.20) is independent of x, so the derivative of/ax is zero, and (6.21) tells us simply that a f lax' is a constant. Evaluating this derivative (and squaring it for convenience) we conclude that (6.20) (6.21) (6.22) \u201e f (x, x y) tx12 -I- 1 of ax d of dy ax' X /2 y(1 \u00b1 X12) = const = \u2014 1 2a X = y 2a \u2014 y where I have named the constant 1/2a for future convenience. This equation is easily solved for x' to give 224 Chapter 6 Calculus of Variations whence This integral can be evaluated by the unlikely looking substitution which gives (as you should check) x=f,1 2a \u2014 y y dy. y a(1 \u2014 cos 0) x = a f (1 \u2014 cos 0) ch9 (6.23) (6.24) = a(0 \u2014 sin 0) const. (6.25) The two equations (6.25) and (6.24) are parametric equations for the required path, giving x and y as functions of the parameter 0. We have chosen the initial point 1 to have x = y = 0, so we see from (6.24) that the initial value of 0 is zero. This in turn implies that the constant of integration in (6.25) is zero. Thus the final parametric equation for the path is x = a(0 \u2014 sin 0) and y = a(1 \u2014 cos 9) (6.26) with the constant a chosen so the curve passes through the given point (x 2, y 2). The curve (6.26) is plotted in Figure 6.5. In that figure I have continued the curve (with dashes) beyond the point 2 to show that the curve that solves the brachistochrone problem happens to be a cycloid \u2014 the curve traced out by a point on the rim of a wheel of radius a, rolling along the underside of the x axis (Problem 6.14). Another remarkable feature of this curve is this: If we release the cart from rest at point 2 and let it roll to the bottom of the curve (point 3 in the figure), the time to roll from 2 to 3 is the same whatever the position of 2, anywhere between 1 and 3. This means that the oscillations of a cart rolling back and forth on a cycloid-shaped track are exactly isochronous (period perfectly independent of amplitude), in contrast with the oscillations of a simple pendulum, which are only approximately isochronous, to the extent x 2a y Figure 6.5 The path for a roller coaster that gives the shortest time between the given points 1 and 2 is part of the cycloid with a vertex at 1 and passing through 2. The cycloid is the curve traced by a point on the rim of a wheel of radius a that rolls along the underside of the x axis. Point 3 is the lowest point on the curve. Section 6.3 Applications of the Euler\u2014Lagrange Equation 225 that the amplitude is small. (See Problem 6.25.) The isochronous property of the cycloid was actually used in the design of some clocks, one of which can be seen in the Victoria and Albert Museum in London. Maximum and Minimum vs. Stationary You have probably noticed that in neither example of this section did I check that the curves that we found actually gave a minimum value to the integral of interest \u2014 that the straight line between two points actually makes the path length minimum, not a maximum or just stationary. The Euler\u2013Lagrange equation guarantees only to give a path for which the original integral is stationary. The problem of deciding whether we have a minimum or maximum (or a stationary curve that is neither) is generally very difficult. In a few cases, it is easy to see which is the case. For instance, it really is obvious that a straight line gives the minimum distance between two points in a plane. In the case of the brachistochrone, it is not at all obvious that the path we found does yield a minimum time, though it is in fact true. To illustrate the variety of possibilities, consider the problem of finding the shortest path, or", " geodesic, between two points 1 and 2 on the surface of a globe. As you probably know, the answer is the great circle joining the two points. 4 Using the calculus of variations you can prove relatively easily that a great circle does indeed make the distance stationary: Using spherical polar coordinates, every point on the globe can be identified by the two angles 0 and 0. If you characterize a path as 0 = 4) (0) and set up an integral that gives the distance between 1 and 2 along this path, you can show that the Euler\u2013Lagrange equation for 4) (0) requires that the path follow a great circle. (See Problem 6.16 for details.) But you have to think a little carefully before deciding that this necessarily gives a minimum distance, since there are two different great- circle paths connecting any two points 1 and 2 on the globe: For simplicity consider two towns on the equator, Quito (near the Pacific coast of Ecuador) and Macapa (at the mouth of the Amazon on the Atlantic coast of Brazil). The \"right\" shortest path between these two is, of course, the great-circle path following the equator for about 2000 miles across South America. But a second possibility, which satisfies the Euler- Lagrange equation just as well, is to head west around the equator from Quito, across the Pacific, the African continent, and the Atlantic, arriving in Macapa some 22,000 miles later. You might guess that this path would be a maximum, but it is in fact neither maximum nor minimum: It is easy to construct nearby paths that are shorter, but it is also easy to find others that are longer. In other words, this second great-circle path gives neither a maximum nor a minimum. This second path is, of course, analogous to the horizontal point of inflection in elementary calculus. In this problem, luckily, it is obvious that the first path gives the true minimum. However, it should be clear that, in general, deciding what sort of stationary path the Euler\u2013Lagrange equation has given us can be tricky. 4 A great circle is the circle in which the globe intersects a plane through the globe's center. 226 Chapter 6 Calculus of Variations Fortunately for us, these questions are irrelevant for our purposes. We shall find that for the applications in mechanics all that matters is that we have a path which makes a certain integral stationary. It simply doesn't matter whether it gives a maximum, minimum, or neither. 6.4 More than Two Variables So far we have considered only problems with just two variables, the independent variable (usually x) and the dependent (usually y). For most applications in mechanics, we shall find that there are several dependent variables, though fortunately still only one independent variable, which is usually the time t. For a simple example where there are two dependent variables, we can go back to the problem of the shortest path between two points. When we found the shortest path between two points 1 and 2, we assumed that the required path could be written in the form y = y(x). Reasonable as this seems, it is easy to think of paths that cannot be written in this way, such as the path shown in Figure 6.6. If we want to be perfectly sure we have found the shortest path among all possible paths, we must find a method that includes these. The way to do this is to write the path in parametric form as x = x (u) and y = y(u), (6.27) where u is any convenient variable in terms of which the curve can be parameterized (for instance, the distance along the path). The parametric form (6.27) includes all of the curves considered before. [If y = y(x), just use x for the parameter u.] It also includes curves like that of Figure 6.6 and, in fact, all curves of interest. 5 The length of a small segment of the path (6.27) is ds = -1dx2 dy 2 x' (u) 2 yi(u) 2 du (6.28) where, as usual, a prime denotes differentiation with respect to the function's argu- ment; that is, x'(u) = dxldu and y'(u) = dyldu. Thus the total path length is u2 L = f x' (u) 2 y' (u) 2 du, (6.29) and our job is to find the two functions x (u) and y(u) for which this integral is This problem is more complicated than any we have considered before, because 5 1n case you are interested in mathematical niceties, I should say that in what follows I shall assume that all functions concerned are continuous and have continuous second derivatives. This assumption could be weakened a little, for example, by allowing some discontinuities in the minimum. derivatives. Section 6.4 More", " than Two Variables 227 Figure 6.6 This path between the two points 1 and 2 cannot be written as y = y(x) nor as x = x(y). It can be written in the parametric form (6.27). there are now two unknown functions x (u) and y(u). The general problem of this type is this: Given an integral of the form S f[x(u), y(u), x' (u), y' (u), u]du (6.30) U2 = L! 1 between two fixed points [x (u t ), y(u 0] and [x(u 2), y(u 2)], find the path [x (u), y(u)] for which the integral S is stationary. The solution to this problem is very similar to the one-variable case, and I shall just sketch it, leaving you to fill in the details. The upshot is that with two dependent variables, we get two Euler\u2014Lagrange equations. To prove this, we proceed very much as before. Let the correct path be given by x = x(u) and y = y(u), (6.31) and then consider a neighboring \"wrong\" path of the form x = x(u) + a(u) and y = y(u) +&(u) (6.32) (where is the Greek letter \"xi\"). The requirement that the integral S be stationary for the right path (6.31) is equivalent to the requirement that the integral S (a, 5), taken along the wrong path (6.32), satisfy as as \u20140 and 0 as as (6.33) when a = /3 = 0. These two conditions are the natural generalization of the condition (6.10) for the one-variable case. By an argument which exactly parallels that leading from (6.10) to (6.13), you can show that these two conditions are equivalent to the two Euler\u2014Lagrange equations (see Problem 6.26): of \u2014 \"f and ax du ax' of ay of d du ay' (6.34) 228 Chapter 6 Calculus of Variations These two equations determine a path for which the integral (6.30) is stationary, and, conversely, if the integral is stationary for some path, that path must satisfy these two equations. EXAMPLE 6.3 The Shortest Path between Two Points Again We can now solve completely the problem of the shortest path between two points. (That is, solve it including all possible paths, such as that in Figure 6.6.) From (6.29), we see that for this problem the integrand f is f (x, x', y, y', u) y'2. (6.35) Since this is independent of x and y, the two derivatives of/ax and of/ay on the left sides in (6.34) are zero. Therefore, the two Euler\u2014Lagrange equations imply simply that the two derivatives a f lax' and a f lay' are constants, of ax' t X \\/x 2 + y 2 = C1 and of ay 31'_Ix /2 + )) /2 = C2. (6.36) If we divide the second equation by the first and recognize that y'/x' is just the derivative dy I dx, we conclude that dy y 1 C2 dx x' C1 = m, (6.37) say. It follows that the required path is a straight line, y = mx + b. It is interest- ing that this proof using a parametric equation is not only better than our previous proof (in that the new proof includes all possible paths), it is also marginally easier. The generalization of the Euler\u2014Lagrange equation to an arbitrary number of dependent variables is straightforward, and doesn't need to be spelled out in detail. Here I would just like to sketch the way the Euler\u2014Lagrange equations will appear in the Lagrangian formulation of mechanics. The independent variable in Lagrangian mechanics is the time t. The dependent variables are the coordinates that specify the position, or \"configuration,\" of a system, and are usually denoted by q 1, q2,, qn. The number n of coordinates depends on the nature of the system. For a single particle moving unconstrained in three dimensions, n is 3, and the three coordinates q 1, q2, q3 could be just the three Cartesian coordinates x, y, z, or they might be the spherical polar coordinates r, B, 0. For N particles moving freely in three dimensions, n is 3N and the coordinates q 1,, qa could be the 3N Cartesian coordinates x1. For a double pendulum (two simple pendulums, with the second suspended from the bob of the first", ", as in Figure 6.7), there would be two coordinates q 1, q2, which could be chosen to be the two angles shown in Figure 6.7. Because the coordinates q 1,, qn can take on so many guises, they are often referred to as generalized coordinates. It is often useful to think of the n generalized coordinates as defining a point in an n-dimensional configuration space, each of whose points labels a unique position, or configuration, of the system. Section 6.4 More than Two Variables 229 Figure 6.7 A good choice of generalized coordinates to identify the position of a double pendulum is the pair of angles 8 and 602 between the pendulums and the vertical. The ultimate goal in most problems in Lagrangian mechanics is to find how the coordinates vary with time; that is, to find the n functions q 1 (t),, q n (t). One can regard these n functions as defining a path in the n-dimensional configuration space. This path is, of course, determined by Newton's second law, but we shall find that it can, equivalently, be characterized as the path for which a certain integral is stationary. This means that it must satisfy the corresponding Euler\u2013Lagrange equations (called just Lagrange equations in this context), and it turns out that these Lagrange equations are usually much easier to write down and use than Newton's second law. In particular, unlike Newton's second law, Lagrange's equations take exactly the same simple form in all coordinate systems. The integral S whose stationary value determines the evolution of the mechanical system is called the action integral. Its integrand is called the Lagrangian and depends on the n coordinates q1, q2,, qn, their n time derivatives 7,15 -.29, 4n and the time t, = 4, (41, 41,, 4n, 4n,t). (6.38) Notice that since the independent variable is t, the derivatives of the coordinates q, are time derivatives and are denoted, as usual, with dots as 4i. The requirement that the action integral S = f al,, qn, an, t)dt (6.39) be stationary implies n Euler\u2013Lagrange equations 3E., 841 d az az = d aG dt 8 41' aq2 dt 8 42' d and \u2014= \u2014. (6.40) aqn dt aqn These n equations correspond precisely to the two Euler\u2013Lagrange equations in (6.34) and are proved in exactly the same way. If these n equations are satisfied, then the action integral (6.39) is stationary; and if the action integral is stationary, then these n equations are satisfied. In the next chapter, you will see where these equations come from and how to use them. 230 Chapter 6 Calculus of Variations Principal Definitions and Equations of Chapter 6 The Euler\u2014Lagrange Equation An integral of the form x2 S = f xi f[y(x), y'(x), x] dx [Eq. (6.4)] taken along a path y = y (x) is stationary with respect to variations of that path if and only if y(x) satisfies the Euler\u2014Lagrange equation af _ d af =0 ay dx ay' [Eq. (6.13)] Several Variables If there are n dependent variables in the original integral, there are n Euler\u2014Lagrange equations. For instance, an integral of the form U 2 S = f f [x(u), y(u), Au), Au), u]du, with two dependent variables [x (u) and y(u)], is stationary with respect to variations of x(u) and y(u) if and only if these two functions satisfy the two equations of d of ax \u2014 du x' and of = d of ay du ay' [Eq. (6.34)] Problems for Chapter 6 SECTION 6.1 Two Examples Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). 6.1 * The shortest path between two points on a curved surface, such as the surface of a sphere, is called a geodesic. To find a geodesic, one has first to set up an integral that gives the length of a path on the surface in question. This will always be similar to the integral (6.2) but may be more complicated (depending on the nature of the surface) and may involve different coordinates than x and y. To illustrate this, use spherical polar coordinates (r, 0, 0) to show that the length of a path joining two points on a sphere of radius R is L = R f \\11+ sin2 0'(9)2 d0 (6.41) 02 e, if (01, 01", ") and (82, 02) specify the two points and we assume that the path is expressed as 0 = (0). (You will find how to minimize this length in Problem 6.16.) Problems for Chapter 6 231 Figure 6.8 Problem 6.3 6.2 * Do the same as in Problem 6.1 but find the length L of a path on a cylinder of radius R, using cylindrical polar coordinates (p, (1), z). Assume that the path is specified in the form 0 = (z). 6.3 ** Consider a ray of light traveling in a vacuum from point P1 to P2 by way of the point Q on a plane mirror, as in Figure 6.8. Show that Fermat's principle implies that, on the actual path followed, Q lies in the same vertical plane as P1 and P2 and obeys the law of reflection, that 0 1 = 02. [Hints: Let the mirror lie in the xz plane, and let P 1 lie on the y axis at (0, y i, 0) and P2 in the xy plane at (x2, y2, 0). Finally let Q = (x, 0, z). Calculate the time for the light to traverse the path P1 Q P2 and show that it is minimum when Q has z = 0 and satisfies the law of reflection.] 6.4 ** A ray of light travels from point P 1 in a medium of refractive index n 1 to P2 in a medium of index n 2, by way of the point Q on the plane interface between the two media, as in Figure 6.9. Show that Fermat's principle implies that, on the actual path followed, Q lies in the same vertical plane as P1 and P2 and obeys Snell's law, that n 1 sin 01 = n 2 sin 02. [Hints: Let the interface be the xz plane, and let P1 lie on the y axis at (0, h 1, 0) and P2 in the x, y plane at (x2, \u2014h 2, 0). Finally let Q = (x, 0, z). Calculate the time for the light to traverse the path P1 Q P2 and show that it is minimum when Q has z = 0 and satisfies Snell's law.] 6.5 ** Fermat's principle is often stated as \"the travel time of a ray of light, moving from point A to B, is minimum along the actual path.\" Strictly speaking it should say that the time is stationary, not minimum. In fact one can construct situations for which the time is maximum along the actual path. Here is one: Consider the concave, hemispherical mirror shown in Figure 6.10, with A and B at opposite ends of a diameter. Consider a ray of light traveling in a vacuum from A to B with one reflection at P, in the same vertical plane as A and B. According to the law of reflection, the actual Figure 6.9 Problem 6.4 232 Chapter 6 Calculus of Variations Figure 6.10 Problem 6.5 path goes via point Po at the bottom of the hemisphere (6 = 0). Find the time of travel along the path A PB as a function of 6 and show that it is maximum at P = Po. 6.6 ** In many problems in the calculus of variations, you need to know the length ds of a short segment of a curve on a surface, as in the expression (6.1). Make a table giving the appropriate expressions for ds in the following eight situations: (a) A curve given by y = y(x) in a plane, (b) same but x = x(y), (c) same but r = r (0), (d) same but 0 = 0(r); (e) curve given by 0 = (z) on a cylinder of radius R, (f) same but z = z(0); (g) curve given by 9 = 0(0) on a sphere of radius R, (h) same but 0 = 0(0). SECTION 6.3 Applications of the Euler\u2014Lagrange Equation 6.7 * Consider a right circular cylinder of radius R centered on the z axis. Find the equation giving 0 as a function of z for the geodesic (shortest path) on the cylinder between two points with cylindrical polar coordinates (R, cb i, z i) and (R, 02, z 2). Describe the geodesic. Is it unique? By imagining the surface of the cylinder unwrapped and laid out flat, explain why the geodesic has the form it does. 6.8 * Verify that the speed of the roller coaster car in Example 6.2 (page 222) is,I2gy. (Assume the wheels have negligible mass and neglect friction.) 6.9 * Find the", " equation of the path joining the origin 0 to the point P (1, 1) in the xy plane that makes the integral g (y'2 yy' + y 2) dx stationary. 6.10 * In general the integrand f (y, y', x) whose integral we wish to minimize depends on y, y', and x. There is a considerable simplification if f happens to be independent of y, that is, f = f (y', x). (This happened in both Examples 6.1 and 6.2, though in the latter the roles of x and y were interchanged.) Prove that when this happens, the Euler\u2014Lagrange equation (6.13) reduces to the statement that f = const. (6.42) Since this is a first-order differential equation for y(x), while the Euler\u2014Lagrange equation is generally second order, this is an important simplification and the result (6.42) is sometimes called a first integral of the Euler\u2014Lagrange equation. In Lagrangian mechanics we'll see that this simplification arises when a component of momentum is conserved. 6.11 ** Find and describe the path y = y(x) for which the integral.1 + y' 2 dx is stationary. \u2018/.7c,j 6.12** Show that the path y = y(x) for which the integral x0 \u2014 y' 2 dx is stationary is a sinh function. I..x, f.x x2 i Problems for Chapter 6 233 6.13 ** In relativity theory, velocities can be represented by points in a certain \"rapidity space\" in which the distance between two neighboring points is ds = [2/(1 \u2014 r2)]./dr2 r 2 d02, where r and 0 are polar coordinates, and we consider just a two-dimensional space. (An expression like this for the distance in a non-Euclidean space is often called the metric of the space.) Use the Euler\u2014Lagrange equation to show that the shortest distance from the origin to any other point is a straight line. 6.14 ** (a) Prove that the brachistochrone curve (6.26) is indeed a cycloid, that is, the curve traced by a point on the circumference of a wheel of radius a rolling along the underside of the x axis. (b) Although the cycloid repeats itself indefinitely in a succession of loops, only one loop is relevant to the brachistochrone problem. Sketch a single loop for three different values of a (all with the same starting point 1) and convince yourself that for any point 2 (with positive coordinates x 2, y2) there is exactly one value of a for which the loop goes through the point 2. (c) To find the value of a for a given point x2, y2 usually requires solution of a transcendental equation. Here are two cases where you can do it more simply: For x 2 = n- b, y2 = 2b and again for x2 = 27rb, y2 = 0 find the value of a for which the cycloid goes through the point 2 and find the corresponding minimum times. 6.15 ** Consider again the brachistochrone problem of Example 6.2 (page 222) but suppose that the car is launched from point 1 with a fixed speed v o. Show that the path of minimum time to the fixed point 2 is still a cycloid, but with its cusp (the top point of the curve) a height v, I2g above point 1. 6.16 ** Use the result (6.41) of Problem 6.1 to prove that the geodesic (shortest path) between two given points on a sphere is a great circle. [Hint: The integrand f (0, 0', 6) in (6.41) is independent of 0, so the Euler\u2014Lagrange equation reduces to afoot = c, a constant. This gives you 0 1 as a function of 6. You can avoid doing the final integral by the following trick: There is no loss of generality in choosing your z axis to pass through the point 1. Show that with this choice the constant c is necessarily zero, and describe the corresponding geodesics.] 6.17 ** Find the geodesics on the cone whose equation in cylindrical polar coordinates is z = Ap. [Let the required curve have the form 0 = 0(p).] Check your result for the case that 0. 6.18 ** Show that the shortest path between two given points in a plane is a straight line, using plane polar coordinates. 6.19 ** A surface of revolution is generated as follows: Two fixed points (x 1, y i ) and (x 2, y 2 ) in the x, y plane are joined by a curve y =", " y(x). [Actually you'll make life easier if you start out writing this as x = x(y).] The whole curve is now rotated about the x axis to generate a surface. Show that the curve for which the area of the surface is minimum has the form y = yo cosh[(x \u2014 xo)I y o], where xo and yo are constants. (This is often called the soap-bubble problem, since the resulting surface is usually the shape of a soap bubble held by two coaxial rings of radii y l and y?.) 6.20 ** If you haven't done it, take a look at Problem 6.10. Here is a second situation in which you can find a \"first integral\" of the Euler\u2014Lagrange equation: Argue that if it happens that the integrand f (y, y', x) does not depend explicitly on x, that is, f = f (y, y'), then Use the Euler\u2014Lagrange equation to replace of/ay on the right, and hence show that df of, of \u201e dx ay Y ay' Y df dx d dx,af ay' ) 234 Chapter 6 Calculus of Variations This gives you the first integral f \u2014 \u2014 af = const. ay' (6.43) This can simplify several calculations. (See Problems 6.21 and 6.22 for examples.) In Lagrangian me- chanics, where the independent variable is the time t, the corresponding result is that if the Lagrangian function is independent of t, then energy is conserved. (See Section 7.8.) 6.21 ** In Example 6.2 (page 222) we found the brachistochrone by exchanging the variables x and y. Here is a method that avoids that exchange: Write the time as in Equation (6.19) but using x as the variable of integration. Your integrand should have the form f (y, y', x) = -1(y 12 + 1)/y. Since this is independent of x, you can invoke the \"first integral\" (6.43) of Problem 6.20. Show that this differential equation leads you to the same integral for x as in Equation (6.23) and hence to the same curve as 6.22 *** You are given a string of fixed length l with one end fastened at the origin 0, and you are to place the string in the xy plane with its other end on the x axis in such a way as to enclose the maximum area between the string and the x axis. Show that the required shape is a semicircle. The area enclosed is of course f y dx, but show that you can rewrite this in the form fo f ds, where s denotes the distance measured along the string from 0, where f = \\ \u2014 y' 2, and y' denotes dy Ids. Since f does not involve the independent variable s explicitly, you can exploit the \"first integral\" (6.43) of Problem before. 6.20. 6.23 *** An aircraft whose airspeed is v o has to fly from town 0 (at the origin) to town P, which is a distance D due east. There is a steady gentle wind shear, such that v wind = V y X, where x and y are measured east and north respectively. Find the path, y = y(x), which the plane should follow to minimize its flight time, as follows: (a) Find the plane's ground speed in terms of v o, V, 0 (the angle by which the plane heads to the north of east), and the plane's position. (b) Write down the time of flight as an integral of the form fop f dx. Show that if we assume that y' and 0 both remain small (as is certainly reasonable if the wind speed is not too large), then the integrand f takes the approximate form f = (1 + I y'2)/(1 ky) (times an uninteresting constant) where k = V (c) Write down the Euler\u2014Lagrange equation that determines the best path. To solve it, make the intelligent guess that y(x) = Ax(D \u2014 x), which clearly passes through the two towns. Show that it satisfies the Euler- Lagrange equation, provided X = (-V4 2k 2 D2 \u2014 2)/ (k D2). How far north does this path take the plane, if D = 2000 miles, vo = 500 mph, and the wind shear is V = 0.5 mph/mi? How much time does the plane save by following this path? [You'll probably want to use a computer to do this integral.] 6.24 *** Consider a medium in which the refractive index n is inversely proportional to r2 ; that is, n = a I r2, where r is the distance", " from the origin. Use Fermat's principle, that the integral (6.3) is stationary, to find the path of a ray of light travelling in a plane containing the origin. [Hint: Use two- dimensional polar coordinates and write the path as 0 = 4> (r). The Fermat integral should have the form f f (0, 0', r) dr,, where f (0, 4>', r) is actually independent of 0. The Euler\u2014Lagrange equation therefore reduces to afoot= const. You can solve this for 0' and then integrate to give 0 as a function of r. Rewrite this to give r as a function of 0 and show that the resulting path is a circle through the origin. Discuss the progress of the light around the circle.] 6.25 *** Consider a single loop of the cycloid (6.26) with a fixed value of a, as shown in Figure 6.11. A car is released from rest at a point Po anywhere on the track between 0 and the lowest point P (that Problems for Chapter 6 235 Figure 6.11 Problem 6.25 is, Po has parameter 0 < 00 < 7). Show that the time for the cart to roll from Po to P is given by the integral time (Pc, P) = 1 \u2014 cos 0 cos 00 \u2014 cos 6 d6 and prove that this time is equal to 7 N/a I g. Since this is independent of the position of Po, the cart takes the same time to roll from P, to P, whether Po is at 0, or anywhere between 0 and P, even infinitesimally close to P. Explain qualitatively how this surprising result can possibly be true. [Hint: To do the mathematics, you have to make some cunning changes of variables. One route is this: Write = Tr \u2014 2a and then use the relevant trig identities to replace the cosines of 0 by sines of a. Now substitute sin a = u and do the remaining integral.] SECTION 6.4 More than Two Variables 6.26 ** Give in detail the argument that leads from the stationary property of the integral (6.30) to the two Euler\u2014Lagrange equations (6.34). 6.27 ** Prove that the shortest path between two points in three dimensions is a straight line. Write the path in the parametric form x = x(u), Y = Au), and z = z(u) and then use the three Euler\u2014Lagrange equations corresponding to (6.34). CHAPTER Lagrange's Equations The theoretical development of the laws of motion of bodies is a problem of such interest and importance that it has engaged the attention of all the most eminent mathematicians since the invention of dynamics as a mathematical science by Galileo, and especially since the wonderful extension which was given to that science by Newton. Among the successors of those illustrious men, Lagrange has perhaps done more than any other analyst to give extent and harmony to such deductive researches, by showing that the most varied consequences respecting the motions of systems of bodies may be derived from one radical formula; the beauty of the methods so suiting the dignity of the results as to make of his great work a kind of scientific poem. \u2014William Rowan Hamilton, 1834 Armed with the ideas of the calculus of variations, we are ready to set up the version of mechanics published in 1788 by the Italian\u2014French astronomer and mathematician Lagrange (1736-1813). The Lagrangian formulation has two important advantages over the earlier Newtonian formulation. First, Lagrange's equations, unlike Newton's, take the same form in any coordinate system. Second, in treating constrained systems, such as a bead sliding on a wire, the Lagrangian approach eliminates the forces of constraint (such as the normal force of the wire, which constrains the bead to remain on the wire). This greatly simplifies most problems, since the constraint forces are usually unknown, and this simplification comes at almost no cost, since we usually do not want to know these forces anyway. In Section 7.1, I prove that Lagrange's equations are equivalent to Newton's second law for a particle moving unconstrained in three dimensions. The extension of this result to N unconstrained particles is surprisingly straightforward, and I leave the details for you to supply (Problem 7.7). In the next few sections, I take up the harder, and more interesting, case of constrained systems. I begin with some simple examples and important definitions (such as degrees of freedom). Then, in Section 7.4, I prove Lagrange's equations for a particle constrained to move on a curved surface (leaving the general case to Problem 7.13). Section 7.5 offers several examples, some of which are distinctly easier to set up in the Lagrangian formulation than in the Newtonian. In 237 238 Chapter 7 Lagrange's Equations Section 7.6, I introduce the", " curious terminology of \"ignorable coordinates.\" Finally, after some summarizing remarks in Section 7.7, the chapter concludes with three sections on topics which, although very important, could be omitted on a first reading. In Section 7.8, I discuss how the laws of energy and momentum conservation appear in Lagrangian mechanics. Section 7.9 describes how Lagrange's equations can be extended to include magnetic forces, and Section 7.10 is an introduction to the idea of Lagrange multipliers. Throughout this chapter, except in Section 7.9, I treat only the case that all nonconstraint forces are conservative or can, at least, be derived from a potential energy function. This restriction can be significantly relaxed, but already includes most of the applications that you are likely to meet in practice. 7.1 Lagrange's Equations for Unconstrained Motion Consider a particle moving unconstrained in three dimensions, subject to a conserva- tive net force F(r). The particle's kinetic energy is, of course, T = mv 2 _ m li,2 2 - 2 - 2- m 2, 2, -r- y z and its potential energy is The Lagrangian function, or just Lagrangian, is defined as U = U(r) = U(x, y, z). = T \u2014 U. Notice first that the Lagrangian is the KE minus the PE. It is not the same as the total energy. You are certainly entitled to ask why the quantity T \u2014 U should be of any interest. There seems to be no simple answer to this question except that it is, as we shall see directly. Notice also that I am using a script for the Lagrangian l (to distinguish it from the angular momentum L and a length L) and that depends on the particle's position (x, y, z) and its velocity (5C ; that is, = C (x, y, z, x, 5), Z). Let us consider the two derivatives, and aC _ au- ax ax \" ac aT. Px. This notation gets into difficulty in field theories where the Lagrangian is often denoted by L, and \u00a3 is used for the Lagrangian density, but this won't be a problem for us. (7.1) (7.2) (7.3) (7.4) (7.5) Section 7.1 Lagrange's Equations for Unconstrained Motion 239 Differentiating the second equation with respect to time and remembering Newton's second law, F\u201e = fix (I take for granted that our coordinate frame is inertial), we see that 0L 0x d az dt ax (7.6) In exactly the same way we can prove corresponding equations in y and z. Thus we have shown that Newton's second law implies the three Lagrange equations (in Cartesian coordinates so far): az _ d az 0x dt ax'_ d az ay dt ay'and az d 8z \u2014 dt az (7.7) You can easily check that the argument just given works equally well in reverse, so that (for a single particle in Cartesian coordinates, at least) Newton's second law is exactly equivalent to the three Lagrange equations (7.7). The particle's path as determined by Newton's second law is the same as the path determined by the three Lagrange Our next step is to recognize that the three equations of (7.7) have exactly the form of the Euler\u2014Lagrange equations (6.40). Therefore, they imply that the integral S = f \u00a3 dt is stationary for the path followed by the particle. That this integral, called the action integral, is stationary for the particle's path is called Hamilton's principle 2 (after its inventor, the Irish mathematician, Hamilton, 1805-1865) and can be restated equations. as follows: The i actualpath which a particle follows between two points 1 and 2 in a given Hamilton's Principle b that the action Integra dt (7. o when t k along the ac ixal path. Although we have so far proved this principle only for a single particle and in Cartesian coordinates, we are going to find that it is valid for a huge class of mechanical systems and for almost any choice of coordinates. So far we have proved for a single particle that the following three statements are exactly equivalent: 2 Try not to be confused by the unlucky circumstance that Hamilton's principle is one possible statement of the Lagrangian formulation of classical mechanics (as opposed to the Hamiltonian formulation). 240 Chapter 7 Lagrange's Equations 1. A particle's path is determined by Newton's second law F = ma. 2. The path is determined by the three Lagrange equations (7.7), at least in Cartesian coordinates. 3. The path is determined by Hamilton's principle. Hamilton's principle has found generalizations in many fields outside", " classical mechanics (field theories, for example) and has given a unity to various diverse areas of physics. In the twentieth century it has played an important role in the formulation of quantum theories. However, for our present purposes its great importance is that it lets us prove that Lagrange's equations hold in more-or-less any coordinate system: Instead of the Cartesian coordinates r = (x, y, z), suppose that we wish to use some other coordinates. These could be spherical polar coordinates (r, 0, 0), or cylindrical polars (p, 0, z), or any set of \"generalized coordinates\" q1, q2, q3, with the property that each position r specifies a unique value of (q 1, q2, q3) and vice versa; that is, and qi = qi (r) for i = 1, 2, and 3, r = r(q1, q2, q3). (7.9) (7.10) These two equations guarantee that for any value of r = (x, y, z) there is a unique (q 1, q2, q3) and vice versa. Using (7.10) we can rewrite (x, y, z) and 57, in terms of (q 1, q2, q3) and (4 1, 42, q3). Next, we can rewrite the Lagrangian = 1mi -2 - U(r) in terms of these new variables as and the action integral as = Z(ql, q2, q3, 41, q2, q3) S = f(ql, q2, q3, 41, q2, q3) dt- t2 l Now, the value of the integral S is unaltered by this change of variables. Therefore, the statement that S is stationary for variations of the path around the correct path must still be true in our new coordinate system, and, by the results of Chapter 6, this means that the correct path must satisfy the three Euler-Lagrange equations, az d 8i dt 0L aq2 d dt.942'and d, (7.11) aq3 dt 84,3 with respect to the new coordinates q 1, q2, and q3. Since these new coordinates are any set of generalized coordinates, the qualification \"in Cartesian coordinates\" can be omitted from the statement (2) above. This result - that Lagrange's equations have the same form for any choice of generalized coordinates - is one of the two main reasons that the Lagrangian formalism is so useful. There is one point about our derivation of Lagrange's equations that is worth keeping at the back of your mind. A crucial step in our proof was the observation that (7.6) was equivalent to Newton's second law Fx = /9, which in turn is true only if the original frame in which we wrote down = T - U is inertial. Thus, although Section 7.1 Lagrange's Equations for Unconstrained Motion 241 Lagrange's equations are true for any choice of generalized coordinates q1, q2, q3 \u2014 and these generalized coordinates may in fact be the coordinates of a noninertial reference frame \u2014 we must nevertheless be careful that, when we first write down the Lagrangian = T \u2014 U, we do so in an inertial frame. We can easily generalize Lagrange's equations to systems of many particles, but let us first look at a couple of simple examples. EXAMPLE 7.1 One Particle in Two Dimensions; Cartesian Coordinates Write down Lagrange's equations in Cartesian coordinates for a particle mov- ing in a conservative force field in two dimensions and show that they imply Newton's second law. (Of course, we have already proved this, but it is worth seeing it worked out explicitly.) The Lagrangian for a single particle in two dimensions is =,C(x, y,, 57) = T \u2014 U = lm(i 2 + 572) \u2014 U (x, y). (7.12) To write down the Lagrange equations we need the derivatives az ax au F ax \" and aT = =mx, ax ax (7.13) with corresponding expressions for the y derivatives. Thus the two Lagrange equations can be rewritten as follows: d = ax az ay dt ax d az dt 8.9 <=> Fx = F = m3.7 F = ma. (7.14) Notice how in (7.13) the derivative aZ/ax is the x component of the force, and Oz.,/a.1 is the x component of the momentum (and similarly with the y components). When we use generalized coordinates q l, q2,, qn, we shall find that 8E.1,9q 1, although not", " necessarily a force component, plays a role very similar to a force. Similarly, azIa4i, although not necessarily a momentum component, acts very like a momentum. For this reason we shall call these derivatives the generalized force and generalized momentum respectively; that is, az = (ith component of generalized force) a q, (7.15) and = (ith component of generalized momentum). (7.16) a4i With these notations, each of the Lagrange equations (7.11) d aqi dt a4i 242 Chapter 7 Lagrange's Equations takes the form (generalized force) = (rate of change of generalized momentum) (7.17) I shall illustrate these ideas in the next example. EXAMPLE 7.2 One Particle in Two Dimensions; Polar Coordinates Find Lagrange's equations for the same system, a particle moving in two dimen- sions, using polar coordinates. As in all problems in Lagrangian mechanics, our first task is to write down the Lagrangian T \u2014 U in terms of the chosen coordinates. In this case we have been told to use polar coordinates, as sketched in Figure 7.1. This means the components of the velocity are v r = r and vo = 4, and the kinetic energy is T = lmv 2 r2 2). Therefore, the Lagrangian is (r, (p) = T \u2014 U = zm (I- 2 r 24;2) \u2014 U (r, 0). (7.18) Given the Lagrangian, we now have only to write down the two Lagrange equations, one involving derivatives with respect to r and the other derivatives with respect to 44. The r Equation The equation involving derivatives with respect to r (the r equation) is az ar d az dt ar tnrop \u2014 \u2014 \u2014ttnr ) =- mr. ;2 au Or d z dt (7.19) Since \u2014a U/ar is just Fr, the radial component of F, we can rewrite the r equation as Fr = M( ; - 42), (7.20) Figure 7.1 The velocity of a particle expressed in two-dimensional polar coordinates. Section 7.1 Lagrange's Equations for Unconstrained Motion 243 which you should recognize as Fr = mar, the r component of F = ma, first derived in Equation (1.48). (The term \u2014r 2 is the infamous centripetal accel- eration.) That is, when we use polar coordinates (r, 0), the Lagrange equa- tion corresponding to r is just the radial component of Newton's second law. (Note, however, that the Lagrangian derivation avoided the tedious calculation of the components of the acceleration.) As we shall see directly, the 0 equation works a bit differently and illustrates a remarkable feature of the Lagrangian approach. The (/) Equation The Lagrange equation for the coordinate 0 is or, substituting (7.18) for G, To interpret this equation, we need to relate the left side to the appropriate component of the force F = \u2014V U. This requires that we know the components of VU in polar coordinates: (If you don't remember this, see Problem 7.5.) The 0 component of the force is just the coefficient of in F = \u2014VU, that is, d dt ao au ao = \u2014(mr 0). 2. d dt VU au ar r r Bo o = au r ao (7.21) (7.22) (7.23) Thus the left side of (7.22) is r Fo, which is simply the torque F on the particle about the origin. Meanwhile, the quantity mr 2q on the right can be recognized as the angular momentum L about the origin. Therefore, the 0 equation (7.22) states that F dL dt (7.24) the familiar condition from elementary mechanics, that torque equals the rate of change of angular momentum. The result (7.24) illustrates a wonderful feature of Lagrange's equations, that when we choose an appropriate set of generalized coordinates the corresponding Lagrange equations automatically appear in a corresponding, natural form. When we choose r and 0 for our coordinates, the 0 equation turns out to be the equation for angular momentum. In fact, the situation is even better than this. Recall that I introduced the 244 Chapter 7 Lagrange's Equations a.c., aq5 ar., a o notion of generalized force and generalized momentum in (7.15) and (7.16). In the present case, the 0 component. of the generalized force is just the torque, (0 component of generalized force) = \u2014 = F (torque) (7.25) and the corresponding component of the generalized momentum is (0 component of generalized momentum) = = L (angular momentum). (7.26", ") With the \"natural\" choice for the coordinates (r and 0) the 0 components of the generalized force and momentum turn out to be the corresponding \"natural\" quantities, the torque and the angular momentum. Notice that the generalized \"force\" does not necessarily have the dimensions of force, nor the generalized \"momentum\" those of momentum. In the present case, the generalized force (0 component) is a torque (that is, force x distance) and the generalized momentum is an angular momentum (momentum x distance). This example illustrates another feature of Lagrange's equations: The 0 component azIao of the generalized force turned out to be the torque on the particle. If the torque happens to be zero, then the corresponding generalized momentum azIa.i. (the angular momentum, in this case) is conserved. Clearly this is a general result: The ith component of the generalized force is azIa4i. If this happens to be zero, then the Lagrange equation d 8q1 dt 84, says simply that the ith component azIa4i of the generalized momentum is constant. That is, if is independent of qi, the ith component of the generalized force is zero, and the corresponding component of the generalized momentum is conserved. In practice, it is often easy to spot that a Lagrangian is independent of a coordinate qi, and, if you can, then you immediately know a corresponding conservation law. We shall return to this point in Section 7.8. Several Unconstrained Particles The extension of the above ideas to a system of N unconstrained particles (a gas of N molecules, for instance) is very straightforward, and I shall leave you to fill in the details (Problems 7.6 and 7.7). Here I shall just sketch the argument for the case of two particles, mainly to show the form of Lagrange's equations for N > 1. For two particles, the Lagrangian is defined (exactly as before) as C = T \u2014 U, but this now means that,C(r i, r2, rl, t2) = \u2014 U(r i, r2). (7.27) As usual, the forces on the two particles are F 1 = \u2014V1 U and F2 = -V2U. Newton's second law can be applied to each particle and yields six equations, Flx = 15 1x, Fly = F2z = I52z Section 7.2 Constrained Systems; an Example 245 Exactly as in Equation (7.7), each of these six equations is equivalent to a correspond- ing Lagrange equation = d az ax 1 dt _ d ay, \u2014 dt ay,' az az2 d az (728) dt ail These six equations imply that the integral S = itt,2 dt is stationary. Finally, we can change to any other suitable set of six coordinates q 1, q2,, q 6. The statement that S is stationary must also be true in this new coordinate system, and this implies in turn that Lagrange's equations must be true with respect to the new coordinates: d az dt a4, aq1 az aq2 d az dt aq2 0J aq6 d az dt 0,46 (7.29) An example of a set of six such generalized coordinates that we shall use repeatedly in Chapter 8 is this: In place of the six coordinates of r 1 and r2, we could use the three coordinates of the CM position R = (m ir m 2r2)/(m + m 2 ) and the three coordinates of the relative position r = r 1 \u2014 r2. We shall find that this choice of coordinates leads to a dramatic simplification. For now, the main point is simply that Lagrange's equations are automatically true in their standard form (7.29) with respect to these new, generalized coordinates. The extension of these ideas to the case of N unconstrained particles is entirely straightforward, and I leave it for you to check. (See Problem 7.7.) The upshot is that there are 3N Lagrange equations az _ d az aqi dt [i =1,2,,3N], valid for any choice of the 3N coordinates q 1,, q3N needed to describe the N particles. 7.2 Constrained Systems; an Example Perhaps the greatest advantage of the Lagrangian approach is that it can handle systems that are constrained so that they cannot move arbitrarily in the space that they occupy. A familiar example of a constrained system is the bead which is threaded on a wire \u2014 the bead can move along the wire, but not anywhere else. Another example of a very constrained system is a rigid body, whose individual atoms can only move in such a way that the distance between any two atoms is fixed. Before I discuss the nature of constraints in general, I shall", " discuss another simple example, the plane pendulum. Consider the simple pendulum shown in Figure 7.2. A bob of mass m is fixed to a massless rod, which is pivoted at 0 and free to swing without friction in the xy plane. The bob moves in both the x and y directions, but it is constrained by the rod so that,/x 2 + y 2 =1 remains constant. In an obvious sense, only one of the coordinates is independent (as x changes, the variation of y is predetermined by the constraint equation), and we say that the system has only one degree of freedom. One way to express this would be to eliminate one of the coordinates, for instance by writing 246 Chapter 7 Lagrange's Equations Figure 7.2 A simple pendulum. The bob of mass m is con- strained by the rod to remain at distance / from 0. y = 1// 2 \u2014 x 2 and expressing everything in terms of the one coordinate x. Although this is a perfectly legitimate way to proceed, a simpler way is to express both x and y in terms of the single parameter 0, the angle between the pendulum and its equilibrium position, as shown in Figure 7.2. We can express all the quantities of interest in terms of 0. The kinetic energy is T im,2 = irn1 2 2. The potential energy is U = mgh where h denotes the height of the bob above its equilibrium position and is (as you should check) h =1(1 \u2014 cos 0). Thus the potential energy is U = mgl (1 \u2014 cos 0), and the Lagrangian is T \u2014 U = 1m1 2 (.1)2 \u2014 mg1(1 \u2014 cos 0). (7.30) Whichever way we choose to proceed \u2014 to write everything in terms of x (or y) or --- the Lagrangian is expressed in terms of a single generalized coordinate q and its time derivative 4, in the form L = (q, 4). Now, it is a fact (which I shall not prove just yet) that once the Lagrangian is written in terms of this one variable (for a system with one degree of freedom), the evolution of the system again satisfies Lagrange's equation (just as we proved for an unconstrained particle in the previous section.) That is, reads to give If we choose the angle 0 as our generalized coordinate, then Lagrange's equation d 84., dt a:q aq az ao d dt 0(1). \u2014mgl sin 0 = dt (m1 20) = m1 2if dt (7.31) (7.32) (7.33) The Lagrangian L is given by (7.30), and the needed derivatives are easily evaluated Section 7.3 Constrained Systems in General 247 Referring to Figure 7.2 you can see that the left side of this equation is just the torque F exerted by gravity on the pendulum, while the term m1 2 is the pendulum's moment of inertia I. Since 5 is the angular acceleration a, we see that Lagrange's equation for the simple pendulum simply reproduces the familiar result F = I a. 7.3 Constrained Systems in General Generalized Coordinates Consider now an arbitrary system of N particles, a = 1,, N with positions ra. We say that the parameters q 1,, (In are a set of generalized coordinates for the system if each position ra can be expressed as a function of q 1,, qn, and possibly the time t, = ro,(q i,, qn, t) [a = 1,, N], (7.34) and conversely each qi can be expressed in terms of the ra and possibly t, gi = qi (ri,, r N, t) [i = 1,, n]. (7.35) In addition, we require that the number of the generalized coordinates (n) is the smallest number that allows the system to be parametrized in this way. In our three- dimensional world, the number n of generalized coordinates for N particles is cer- tainly no more than 3N and, for a constrained system, is usually less \u2014 sometimes dramatically so. For example, for a rigid body, the number of particles N may be of order 10 23, whereas the number of generalized cordinates n is 6 (three coordinates to give the position of the center of mass and three to give the orientation of the body). To illustrate the relation (7.34), consider again the simple pendulum of Figure 7.2. There is one particle (the bob) and two Cartesian coordinates (since the pendulum is restricted to two dimensions). As we saw, there is just one generalized coordinate, which we took to be the angle 0. The analog of (7.34) is r (x, y) = (1 sin 0, /", " cos 0) (7.36) and expresses the two Cartesian coordinates x and y in terms of the one generalized coordinate 0. The double pendulum shown in Figure 7.3 has two bobs, both confined to a plane, so it has four Cartesian coordinates, all of which can be expressed in terms of the two generalized coordinates 0 1 and 02. Specifically, if we put our origin at the suspension point of the top pendulum, r 1 = (l i sin 01, /1 cos 01) = r1(01) (7.37) and r2 = (I i sin 01 + /2 sin 02, / 1 cos 0 1 + /2 cos 02) = r2(01, 02) (7.38) Notice that the components of r 2 depend on 0 1 and 02. 248 Chapter 7 Lagrange's Equations Figure 7.3 The positions of both masses in a double pendulum are uniquely specified by the two generalized coordinates 0 1 and 02, which can themselves be varied independently. In these two examples, the transformation between the Cartesian and the general- ized coordinates did not depend on the time t, but it is easy to think of examples in which it does. Consider the railroad car shown in Figure 7.4, which has a pendulum suspended from its ceiling and is being forced 3 to accelerate with a fixed accelera- tion a. It is natural to specify the position of the pendulum by the angle 0 as usual, but we must recognize that, in the first instance, this gives the pendulum's position relative to the accelerating, and hence non-inertial, reference frame of the car. If we wish to specify the bob's position relative to an inertial frame, we can choose a frame fixed relative to the ground, and we can easily express the position relative to this inertial frame in terms of the angle 0. The position of the point of suspension relative to the ground is (if we choose our axes and origin properly) just x s = Z at2, and the position of the bob is then easily seen to be r (x, y) = (1 sin 0 + lat 2, 1 cos 0) = r(0, t). (7.39) xs = at2 X IN a (given) Figure.7.4 A pendulum is suspended from the roof of a rail- road car that is being forced to accelerate with a fixed, known acceleration a. 3 The word \"forced\" is often used to describe a motion that is imposed by some outside agent and is unaffected by the internal motions of the system. In the present example, the \"forced\" acceleration of the car is assumed to be the same whatever the oscillations of the pendulum. Section 7.3 Constrained Systems in General 249 The relation between r and the generalized coordinate 0 depends on the time t, a possibility that I allowed for when writing (7.34). We shall sometimes describe a set of coordinates q l,, qn as natural if the relation (7.34) between the Cartesian coordinates ra and the generalized coordinates does not involve the time t. We shall find certain convenient properties of natural coordinates that do not generally apply to coordinates for which (7.34) does involve the time. Fortunately, as the name implies, there are many problems for which the most convenient choice of coordinates is also natural.4 Degrees of Freedom The number of degrees of freedom of a system is the number of coordinates that can be independently varied in a small displacement \u2014 the number of independent \"directions\" in which the system can move from any given initial configuration. For example, the simple pendulum of Figure 7.2 has just one degree of freedom, while the double pendulum of Figure 7.3 has two. A particle that is free to move anywhere in three dimensions has three degrees of freedom, while a gas comprised of N particles has 3N. When the number of degrees of freedom of an N-particle system in three dimen- sions is less than 3N, we say that the system is constrained. (In two dimensions, the corresponding number is 2N of course.) The bob of a simple pendulum, with one degree of freedom, is constrained. The two masses of a double pendulum, with two degrees of freedom, are constrained. The N atoms of a rigid body have just six degrees of freedom and are certainly constrained. Other examples are a bead constrained to slide on a fixed wire and a particle constrained to move on a fixed surface in three dimensions. In all of the examples I have given so far, the number of degrees of freedom was equal to the number of generalized coordinates needed to describe the system's configuration. (The double pendulum has two degrees of freedom and needs two generalized coordinates, and so on.) A system with this natural-seeming property is said to be holonomic. 5 That is, a holonomic system has n degrees of freedom and can be described by n generalized coordinates,", " q l,, qn. Holonomic systems are easier to treat than nonholonomic, and in this book I shall restrict myself to holonomic systems. You might imagine that all systems would be holonomic, or at least that nonholo- nomic systems would be rare and bizarrely complicated. In fact, there are some quite simple examples of nonholonomic systems. Consider, for instance, a hard rubber ball that is free to roll (but not to slide nor to spin about a vertical axis) on a horizontal table. Starting at any position (x, y) it can move in only two independent directions. Therefore, the ball has two degrees of freedom, and you might well imagine that its 4 Natural coordinates are sometimes called scleronomous, and those that are not natural, rheonomous. I shall not use these outstandingly forgettable names. Nonnatural coordinates are also sometimes called forced, since a time dependence in the relation (7.34) is usually associated with a forced motion, such as the forced acceleration of the car in Figure 7.4. 5 Many different definitions of \"holonomic\" can be found, not all of which are exactly equivalent. 250 Chapter 7 Lagrange's Equations Figure 7.5 The right triangle OPQ lies in the xy plane with sides OP and PQ of length c. If you roll a ball of circumference c around OPQ, it will return to its starting point with a changed orientation. configuration could be uniquely specified by two coordinates, x and y, of its center. But consider the following: Let us place the ball at the origin 0 and make a mark on its top point. Now, carry out the following three moves. (See Figure 7.5.) Roll the ball along the x axis for a distance equal to the circumference c, to a point P, where the mark will once again be on the top. Now roll it the same distance c in the y direction to Q, where the mark is again on top. Finally roll it straight back to the origin along the hypotenuse of the triangle OPQ. Since this last move has length,N,hc, it brings the ball back to its starting point, but with the mark no longer on the top. The position (x, y) has returned to its initial value, but the ball now has a different orientation. Evidently the two coordinates (x, y) are not enough to specify a unique configuration. In fact, three more numbers are needed to specify the orientation of the ball, and we need five coordinates in all to specify the configuration completely. The ball has two degrees of freedom but needs five generalized coordinates. Evidently it is a nonholonomic system. Although nonholonomic systems certainly exist, they are more complicated to an- alyze than holonomic systems, and I shall not discuss them further. For any holonomic system with generalized coordinates q l,, qn and potential energy U (q i,, qn, t) (which may depend on the time t as described in Section 4.5), the evolution in time is determined by the n Lagrange equations az aqi d az dt aqi [i = 1, -, n], (7.40) where the Lagrangian is defined as usual to be T \u2014 U. I shall prove this result in Section 7.4. 7.4 Proof of Lagrange's Equations with Constraints We are now ready to prove Lagrange's equations for any holonomic system. To keep things reasonably simple, I shall treat explicitly the case that there is just one particle. (The generalization to arbitrary numbers of particles is fairly straightforward \u2014 see Section 7.4 Proof of Lagrange's Equations with Constraints 251 Problem 7.13.) To be definite, I shall suppose the particle is constrained to move on a surface.6 This means that it has two degrees of freedom and can be described by two generalized coordinates q 1 and q2 that can vary independently. We must recognize that there are two kinds of forces on the particle (or particles, in the general case). First, there are the forces of constraint: For a bead on a wire the con- straining force is the normal force of the wire on the bead; for our particle, constrained to move on a surface, it is the normal force of the surface. For the atoms in a rigid body, the constraining forces are the interatomic forces that hold the atoms in place within the body. In general, the forces of constraint are not necessarily conservative, but this doesn't matter. One of the objectives of the Lagrangian approach is to find equations that do not involve the constraining forces, which we usually don't want to know anyway. (Notice, however, that if the constraining forces are nonconservative, Lagrange's equations in the simple unconstrained form of Section 7.1 certainly do not apply.) I shall denote the net constraining force on the particle by Fcstr, which", " in our case is just the normal force of the surface to which the particle is confined. Second, there are all the other \"nonconstraint\" forces on the particle, such as gravity. These are the forces with which we are usually concerned in practice, and I shall denote their resultant by F. I shall assume that the nonconstraint forces all satisfy at least the second condition for conservatism, so that they are derivable from a potential energy, U(r, t), and F = \u2014VU(r, t). (7.41) (If all the nonconstraint forces are actually conservative, then U is independent of t, but we don't need to assume this.) The total force on our particle is F tot = Fcstr + F. Finally, I shall define the Lagrangian, as usual, to be =T \u2014 U. (7.42) Since U is the potential energy for the nonconstraint forces only, this definition of excludes the constraint forces. This correctly reflects that Lagrange's equations for a constrained system cleverly eliminate the constraint forces, as we shall see. The Action Integral is Stationary at the Right Path Consider any two points r 1 and r2 through which the particle passes at times t 1 and t2. I shall denote by r(t) the \"right\" path, the actual path that our particle follows between the two points, and by R(t) any neighboring \"wrong\" path between the same two points. It is convenient to write R(t) = r(t) E(t), (7.43) 6 Actually, it is a bit hard to imagine how to constrain a particle to a single surface so that it can't jump off. If this worries you, you can imagine the particle sandwiched between two parallel surfaces with just enough gap between them to let it slide freely. 252 Chapter 7 Lagrange's Equations which defines E(t) as the infinitesimal vector pointing from r(t) on the right path to the corresponding point R(t) on the wrong path. Since I shall assume that both of the points r(t) and R(t) lie in the surface to which the particle is confined, the vector E(t) is contained in the same surface. Since both r(t) and R(t) go through the same endpoints, E(t) = 0 at t 1 and t2. Let us denote by S the action integral t2 S = f ti (R, t) dt, (7.44) taken along any path R(t) lying in the constraining surface, and by S o the correspond- ing integral taken along the right path r(t). As I shall now prove, the integral S is stationary for variations of the path R(t), when R(t) = r(t) or, equivalently, when the difference E is zero. Another way to say this is that the difference in the action 8S = S \u2014 S o (7.45) is zero to first order in the distance E between the paths, and this is what I shall prove. The difference (7.45) is the integral of the difference between the Lagrangians on = (R, t) - z(r, t). (7.46) If we substitute R(t) = r(t) + E(t) and Z(r, r, t) = T \u2014 U = 1mi- 2 \u2014 U(r, t), integrals the two paths, this becomes 7 = + C) 2 \u2014 1.2] \u2014 [U (r + C, t) - (r, t)] = m 1-4 \u2014 EVU 0(e 2), where 0(6 2) denotes terms involving squares and higher powers of E and E. Returning to the difference (7.45) in the two action integrals, we find that, to first order in E, 3S =dt f -C \u2014 c VU]dt. (7.47) t2 t, The first term in the final integral can be integrated by parts. (Recall that this just means moving the time derivative from one factor to the other and changing the sign.) The difference E is zero at the two endpoints, so the endpoint contribution is zero, and we get t2 SS = \u2014 f Jrni -FVU]dt. (7.48) 7 To understand the second term in the second line, recall that f(r E) \u2014 f(r) ti c Vf, for any scalar function f (r). Sec Section 4.3. Section 7.4 Proof of Lagrange's Equations with Constraints 253 Now, the path r (t) is the \"right\" path and satisfies Newton's second law. Therefore the term mi.' is just the total force on the particle, F rot F = - cstr", " + F. Meanwhile V U = \u2014F. Therefore, the second term in (7.48) cancels the second piece of the first, and we are left with f t2 t, SS E F cstr dt. (7.49) But the constraint force p ost, is normal to the surface in which our particle moves, while c lies in the surface. Therefore c F\u201e tr = 0, and we have proved that 6S = 0. That is, the action integral is stationary at the right path, as claimed. 8 The Final Proof We have proved Hamilton's principle, that the action integral is stationary at the path which the particle actually follows. However, we have proved it, not for arbitrary variations of the path, but rather for those variations of path that are consistent with the constraints \u2014 that is, paths which lie in the surface to which our particle is constrained. This means that we cannot prove Lagrange's equations with respect to the three Cartesian coordinates. On the other hand, we can prove them with respect to the appropriate generalized coordinates. We are assuming that our particle is confined by holonomic constraints to move on a surface, that is, a two-dimensional subset of the full three-dimensional world. This means that the particle has two degrees of freedom and can be described by two generalized coordinates, q 1 and q2, that can be varied independently. Any variation of q 1 and q2 is consistent with the constraints. 9 1 and q2 as Accordingly, we can rewrite the action integral in terms of q S = f ti Z(q 1, q2, 41, q2, t) dt, (7.50) (7.51) and this integral is stationary for any variations of q 1 and q2 about the correct path W i (t), q2 (t)]. Therefore, by the argument of Chapter 6 the correct path must satisfy the two Lagrange equations az = d az 8q 1 dt 841 and az _ d a42 \u2014 dt 8q2 The proof that I have given here applies directly only to a single particle in three dimensions, constrained to move on a two-dimensional surface, but the main ideas of the general case are all present. The generalization is, for the most part, relatively 8 The observation that the integrand in (7.49) is zero is really the crucial step in our proof. When you consider the generalization of the proof to an arbitrary constrained system (for instance, if you do Problem 7.13), you will find that there is a corresponding step and that the corresponding term is zero, for the same reason: The forces of constraint would do no work in a displacement that is consistent with the constraints. Indeed this is one possible definition of a force of constraint. 9 For example, if our surface is a sphere, centered at the origin, then the generalized coordinates q1, q2 could be the two angles 0, 0 of spherical polar coordinates. Any variation of B and 0 is consistent with the constraint that the particle remain on the sphere. 254 Chapter 7 Lagrange's Equations straightforward (see Problem 7.13), and meanwhile I hope that I have said enough to convince you of the truth of the general result: For any holonomic system, with n degrees of freedom and n generalized coordinates, and with the nonconstraint forces derivable from a potential energy U (q i,, q n, t), the path followed by the system is determined by the n Lagrange equations where is the Lagrangian = T \u2014 U and U = U (q i,, qn, t) is the total potential energy corresponding to all the forces excluding the forces of constraint. It was essential to our proof of Lagrange's equations that the nonconstraint forces be conservative (or, at a minimum, that they satisfy the second condition for conser- vatism) so that they are derivable from a potential energy, F = \u2014VU. If this is not true, then Lagrange's equations may not hold, at least in the form (7.52). An obvi- ous example of a force that does not satisfy this condition is sliding friction. Sliding friction cannot be regarded as a force of constraint (it is not normal to the surface) and cannot be derived from a potential energy. Thus, when sliding friction is present, Lagrange's equations do not hold in the form (7.52). Lagrange's equations can be modified to include forces like friction (see Problem 7.12), but the result is clumsy and I shall confine myself to situations where the equations (7.52) do hold. 7.5 Examples of Lagrange's Equations In this section I present five examples of the use of Lagrange's equations. The first two are sufficiently simple that they can be easily solved within the Newtonian formalism. My main purpose for including them is just to give you experience with using the Lagrangian approach. Nevertheless,", " even these simple cases show some advantages of the Lagrangian over the Newtonian formalism; in particular, we shall see how the Lagrangian approach obviates any need to consider the forces of constraint. The last three examples are sufficiently complex that solution using the Newtonian approach requires considerable ingenuity; by contrast, the Lagrangian approach lets us write down the equations of motion almost without thinking. The examples given here illustrate an important point to recognize about La- grange's equations: The Lagrangian formalism always (or nearly always) gives a straightforward means of writing down the equations of motion. On the other hand, it cannot guarantee that the resulting equations are easy to solve. If we are very lucky, the equations of motion may have an analytic solution, but, even when they do not, they are the essential first step to understanding the solutions and they often suggest a starting point for an approximate solution. The equations of motion can give simple answers to certain subsidiary questions. (For instance, once we have the equations of Section 7.5 Examples of Lagrange's Equations 255 motion, we can usually find the positions of equilibrium of a system very easily.) And we can always solve the equations of motion numerically for given initial conditions. The Lagrangian method is so important that it certainly deserves more than just five examples. However, the crucial thing is that you work through several examples yourself; therefore I have given plenty of problems at the end of the chapter, and it is essential that you work several of these as soon as possible after reading this section. EXAMPLE 7.3 Atwood's Machine Consider the Atwood machine first met in Figure 4.15 and shown again in Figure 7.6, in which the two masses m 1 and m 2 are suspended by an inextensible string (length 1) which passes over a massless pulley with frictionless bearings and radius R. Write down the Lagrangian \u00a3, using the distance x as generalized co- ordinate, find the Lagrange equation of motion, and solve it for the acceleration x. Compare your results with the Newtonian solution. Because the string has fixed length, the heights x and y of the two masses cannot vary independently. Rather, x y + r R = 1, the length of the string, so that y can be expressed in terms of x as y = \u2014x const. (7.53) Therefore, we can use x as our one generalized coordinate. From (7.53) we see that 57 =, so that the kinetic energy of the system is T = I m 1x 2 1 1 2 17 = -f k,m 1 + m2).\u00b12 x Figure 7.6 An Atwood machine consisting of two masses, m 1 and m 2, suspended by a massless inextensible string that passes over a massless, frictionless pulley of radius R. Because the string's length is fixed, the position of the whole system can be specified by a single variable, which we can take to be the distance x. 256 Chapter 7 Lagrange's Equations while the potential energy is U = \u2014rn i gx m 2gy = \u2014(m i \u2014 m 2)gx const. Combining these, we find the Lagrangian = T \u2014 U = 2(m l + m2)1 2 \u00b1 (m i \u2014 m2)gx, (7.54) where I have dropped an uninteresting constant. The Lagrange equation of motion is just or, substituting (7.54) for, (m1 \u2014 m2)g = (m1 + 711 2)x, which we can solve at once to give the desired acceleration az d ai ax dt ax = /12 1 \u2014 n2 2 m1 \u00b1 m2 g. By choosing m 1 and m2 fairly close together, one can make this acceleration much less than g, and hence much easier to measure. Therefore, the Atwood machine gave an early and reasonably accurate method for measuring g. The corresponding Newtonian solution requires us to write down Newton's second law for each of the masses separately. The net force on m 1 is m ig \u2014 Ft t is the tension in the string. (This is the force of constraint and needed where F no consideration in the Lagrangian solution.) Thus Newton's second law form i In the same way, Newton's second law for m 2 reads \u2014 Ft = mli. Ft \u2014 m 2g = (Remember that the upward acceleration of m 2 is the same as the downward acceleration of m 1.) We see that the Newtonian approach has given us two equa- tions for two unknowns, the required acceleration 3 and the force of constraint Ft. By adding these two equations, we can eliminate F t and arrive at precisely the equation (7.55) of the Lagrangian method and thence the same value (7.56) for 1. The Newtonian solution of the Atwood machine is", " too simple for us to get very excited by an alternative solution. Nevertheless, this simple example does illustrate how the Lagrangian approach allows us to ignore the unknown (and usually uninteresting) force of constraint and to eliminate at least one step of the Newtonian solution. (7.55) (7.56) is Section 7.5 Examples of Lagrange's Equations 257 EXAMPLE 7.4 A Particle Confined to Move on a Cylinder Consider a particle of mass m constrained to move on a frictionless cylinder of radius R, given by the equation p = R in cylindrical polar coordinates (p, 4), z), as shown in Figure 7.7. Besides the force of constraint (the normal force of the cylinder), the only force on the mass is a force F = \u2014kr directed toward the origin. (This is the three dimensional version of the Hooke's-law force.) Using z and 0 as generalized coordinates, find the Lagrangian. Write down and solve Lagrange's equations and describe the motion. Since the particle's coordinate p is fixed at p = R, we can specify its position by giving just z and 0, and since these two coordinates can vary independently the system has two degrees of freedom and we can use (z, 0) as generalized coordinates. The velocity has v p = 0, v4, = R(b, and I), = 1. Therefore the kinetic energy is T n v 2 = m (R 2 fp + The potential energy for the force F = \u2014kr is (Problem 7.25) U = 4 kr 2, where r is the distance from the origin to the particle, given by r 2 = R 2 + z2 (see Figure 7.7). Therefore U = Ik(R 2 z 2), and the Lagrangian is = Im(R 2c.62 + 12) \u2014 Ik(R 2 z 2). (7.57) Since the system has two degrees of freedom, there are two equations of motion. The z equation is az az d az dt az or \u2014 kz = (7.58) Figure 7.7 A mass m is confined to the surface of the cylinder p = R and subject to a Hooke's law force F = \u2014kr. 258 Chapter 7 Lagrange's Equations The 0 equation is even simpler. Since does not depend on 0, it follows that az.,130 = 0 and the 0 equation is just d dt aq a or 0 = dt, (7.59) The z equation (7.58) tells us that the mass executes simple harmonic motion in the z direction, with z = A cos(cot \u2014 8). The 0 equation (7.59) tells us that the quantity mR 2.. is constant, that is, that the angular momentum about the z axis is conserved \u2014 a result we could have anticipated since there is no torque in this direction. Because -p is fixed, this implies simply that cp is constant, and the mass moves around the cylinder with constant angular velocity cb, at the same time that it moves up and down in the z direction in simple harmonic motion. These two examples illustrate the steps to be followed in solving any problem by the Lagrangian method (provided all constraints are holonomic and the nonconstraint forces are derivable from a potential energy, as we are assuming): 1. Write down the kinetic and potential energies and hence the Lagrangian T \u2014 U, using any convenient inertial reference frame. 2. Choose a convenient set of n generalized coordinates q 1,, q n and find expressions for the original coordinates of step 1 in terms of your chosen generalized coordinates. (Steps 1 and 2 can be done in either order.) 3. Rewrite L in terms of q 1,, qn and q i,, 4n. 4. Write down the n Lagrange equations (7.52). As we shall see, these four steps provide an almost infallible route to the equations of motion of any system, no matter how complex. Whether the resulting equations can be easily solved is another matter, but even when they cannot, just having them is a huge step toward understanding a system and an essential step to finding approximate or numerical solutions. The next two examples illustrate how the Lagrangian approach can give the equa- tions of motion, almost effortlessly, for problems that would require considerable care and ingenuity using Newtonian methods. EXAMPLE 7.5 A Block Sliding on a Wedge Consider the block and wedge shown in Figure 7.8. The block (mass m) is free to slide on the wedge, and the wedge (mass M) can slide on the horizontal table, both with negligible friction. The block is released from the top of the wedge, with both initially at rest. If the wedge has angle a and the length of its sloping face is 1, how long does the block take to reach the bottom? The system has two degrees", " of freedom, and a good choice of the two generalized coordinates is, as shown, the distance q 1 of the block from the top of the wedge and the distance q2 of the wedge from any convenient fixed point on the table. The quantity we need to find is the acceleration :4 1 of the block relative to the wedge, since with this we can quickly find the time required to slide the length of the wedge. Our first task is to write down the Lagrangian, and Section 7.5 Examples of Lagrange's Equations 259 Figure 7.8 A block of mass m slides down a wedge of mass M, which is free to slide over the horizontal table. it is often safest to do this in Cartesian coordinates, and then rewrite it in terms of the chosen generalized coordinates. The kinetic energy of the wedge is just TM = 04-422 but that of the block is more complicated. The block's velocity relative to the wedge is 41 down the slope, but the wedge itself has a horizontal velocity 42 relative to the table. The velocity of the block relative to the inertial frame of the table is the vector sum of these two. Resolving into rectangular components (x to the right, y downward), we find for the velocity of the block relative to the table v = (vs, v y ) = (41 cosa + q2, 41 sin a). Thus the kinetic energy of the block is T m v 2 x v 2) y i m i, 2 _ 1_ 2 2 \"1 1 2 a + q iq2 cos a). L (I used the identity cos t a + sin2 a =1 to simplify this.) The total kinetic energy of the system is T = TM Tm = -21 (M ln)4 22 4tn (4 12 + 24,42 cos a). (7.60) The potential energy of the wedge is a constant, which we may as well take to be zero. That of the block is \u2014mgy, where y = ql sin a is the height of the block measured down from the top of the wedge. Therefore U = \u2014mgq i sin a and the Lagrangian is = T \u2014 U = 2(M m)4 22 lm(4 12 24,42 cos a) + mgqi sin a. (7.61) Once we have found the Lagrangian in terms of the generalized coordinates q 1 and q2, all we have to do is to write down the two Lagrange equations, one for q 1 and one for q 2, and then solve them. The q 2 equation (which is a little simpler) is ac aq2 d az dt 042 (7.62) 260 Chapter 7 Lagrange's Equations but, since L in (7.61) is clearly independent of q 2, this just tells us that the generalized momentum ala,72 is constant, A142+ m(42+ 41 cos a) = const (7.63) \u2014 a result you will recognize as conservation of the total momentum in the x direction (and something you could have written down without any help from Lagrange). The q 1 equation 8i d az 0.71 \u2014 dt a41 is more complicated, since neither derivative is zero. Sustituting (7.61) for we can write this as mg sin a = dt --m(q i 42 cos a). d = m(41 +.42 cos a). Differentiating (7.63) we see that which lets us eliminate 42 from (7.65) and solve for 4 1 : q2 = m M m q i cos ot, == 1 g sin a m cost M M (7.64) (7.65) (7.66) (7.67) Armed with this value for 4 1 we can quickly answer the original question: Since the acceleration down the slope is constant, the distance traveled down the slope in time t is 441t2, and the time to travel the length 1 is just V21/4 1, with 4 1 given by (7.67). More interesting than this answer is to check that the formula (7.67) for 41 agrees with common sense in various special cases. For example, if a = 90\u00b0, (7.67) implies that 4 1 = g, which is clearly right; and, if M oc, (7.67) implies that 41 \u2014> g sin ot, which is the well-known acceleration for a block on a fixed incline and clearly makes sense. I leave it as an exercise (Problem 7.19) to check that in the limit that M 0, our answers agree with what you could have predicted. EXAMPLE 7.6 A Bead on a Spinning Wire Hoop A bead of mass m is threaded on a frictionless circular wire hoop of radius R. The hoop lies in a vertical plane, which is forced to rotate about the hoop's vertical", " diameter with constant angular velocity (*/) = co, as shown in Figure 7.9. The bead's position on the hoop is specified by the angle 0 measured up from the vertical. Write down the Lagrangian for the system in terms of the Section 7.5 Examples of Lagrange's Equations 261 Figure 7.9 A bead is free to move around the frictionless wire hoop, which is spinning at a fixed rate w about its vertical axis. The bead's position is specified by the angle 0; its distance from the axis of rotation is p = R sin O. generalized coordinate 6 and find the equation of motion for the bead. Find any equilibrium positions at which the bead can remain with 9 constant, and explain their locations in terms of statics and the \"centrifugal force\" mco 2 p (where p is the bead's distance from the axis). Use the equation of motion to discuss the stability of the equilibrium positions. Our first task is to write down the Lagrangian. Relative to a nonrotating frame, the bead has velocity R6 tangential to the hoop and pw = (R sin 0)0) normal to the hoop (the latter due to the spinning of the hoop with angular velocity o)). Thus the kinetic energy is T = 4mv2 = ImR2 (62 co2 sine 9). The gravitational potential energy is easily seen to be U = mgR(1 \u2014 cos 0), measured from the bottom of the hoop. Therefore, the Lagrangian is z, = ImR2(62 + w2 sin2 0 \u2014 ) mgR(1 \u2014 cos0), (7.68) and the Lagrange equation is az _ d az ae dt 86 Or mR 2w2 sin g cos 0 \u2014 mg R sin g = m R 2o. Dividing through by m R2, we arrive at the desired equation of motion: = (co2 cos 0 \u2014 g I R) sin O. (7.69) Although this equation cannot be solved analytically in terms of elementary functions, it can, nevertheless, tell us lots about the system's behavior. To illustrate this, let us use (7.69) to find the equilibrium positions of the bead. An equilibrium point is any value of 0 \u2014 call it 00 \u2014 satisfying the following condition: If the bead is placed at rest (6 = 0) at 0 = 00, then it will remain at rest at 00. This condition is guaranteed if 9 = 0. (To see this, note that if 9 = 0, then 6 doesn't change and remains zero, which means that 9 doesn't change 262 Chapter 7 Lagrange's Equations and remains equal to 00.) Thus to find the equilibrium positions we have only to equate the right side of (7.69) to zero: (cot cos 0 \u2014 g I R) sin 0 = 0. (7.70) This equation is satisfied if either of the two factors is zero. The factor sin 9 is zero if 9 = 0 or Tr. Thus the bead can remain at rest at the bottom or top of the hoop. The first factor in (7.70) vanishes when cos y co2 R Since I cos 01 must be less than or equal to 1, the first factor can vanish only when cot > g/ R. When this condition is satisfied, there are two more equilibrium positions at 00 = \u00b1 arccos c0g2R J. (7.71) We conclude that when the hoop is rotating slowly (co t < g/R), there are just two equilibrium positions, at the bottom and top of the hoop, but when it rotates fast enough (co t > g I R), there are two more, symmetrically placed on either side of the bottom, as given by (7.71). 10 Perhaps the simplest way to understand the various equilibrium positions is in terms of the \"centrifugal force.\" In most introductory physics courses, the centrifugal force is dismissed as an abomination to be avoided by all right- thinking physicists. As long as we confine our attention to inertial frames, this is a correct (and certainly a safe) point of view. Nevertheless, as we shall see in Chapter 9, from the point of view of a noninertial rotating frame there is a perfectly real centrifugal force mco 2 p (perhaps more familiar as mv 2 I p), where p is the object's distance from the axis of rotation. Thus, taking the point of view of a fly perched on the rotating hoop, we can understand the equilibrium positions as follows: At the bottom or top of the hoop, the bead is on the axis of rotation and p = 0; therefore, the centrifugal force rnco2 p is zero. Furthermore, the force of gravity is normal to the hoop, so there is no force tending to move the bead along the wire and the bead remains at rest. The other two equilibrium points", " are a little subtler: At any position off the axis (such as that shown in Figure 7.9) the centrifugal force is nonzero and has a component pushing the bead outward along the wire; meanwhile the force of gravity has a component pulling the bead inward along the wire (provided the bead is below the halfway marks, 0 = \u00b17r/2). At either of the points given by (7.71), these two components are balanced (check this for yourself \u2014 Problem 7.28) and the bead can remain at rest. An equilibrium point 00 is not especially interesting unless it is stable \u2014 that is, the bead, if nudged a little away from 00, moves back toward 00. Using our m Notice that when w 2 = g/R the two extra positions given by (7.71) have just come into existence and coincide with the first point at the bottom with 6) = \u00b10. Section 7.5 Examples of Lagrange's Equations 263 equation of motion (7.69), we can easily address this issue, and I'll start with the equilibrium at the bottom, 9 = 0. As long as 0 remains close to 0, we can set cos 0 ti 1 and sin 9 ti 0 and approximate the equation by = (co2 \u2014 g 1 R)0 [9 near 0]. (7.72) If the hoop is rotating slowly (w 2 < g 1 R), this has the form 9 = (negative number)0. If we nudge the bead to the right (0 > 0), then since 0 is positive 0 is negative, and the bead accelerates to the left, that is, back toward the bottom. If we nudge it to the left, (9 < 0), then 9 becomes positive, and the bead accelerates to the right, which is again back toward the bottom. Either way, the bead returns toward the equilibrium, which is, therefore, stable. If we speed up the rotation of the hoop, so that w 2 > g 1 R, then the approxi- mate equation of motion (7.72) takes the form 0 = (positive number)0. Now a small displacement to the right makes B positive, and the bead accelerates away from the bottom. Similarly a displacement to the left makes 9 negative, and again the bead accelerates away from the bottom. Thus, as we increase w past the critical value where w 2 = g 1 R, the equilibrium at the bottom changes from stable to unstable. The equilibrium at the top (0 = 7r) is alway unstable (see Problem 7.28). This is easy to understand from our discussion of the centrifugal force. Near the top of the hoop, both the centrifugal and gravitational forces tend to push the bead away from the top, so there is no chance of a restoring force to pull it back to the equilibrium position. The other two equilibrium positions only exist when w 2 > g 1 R, and are easily seen to be stable: The equation of motion (7.69) is = (co2 cos\u00b0 \u2014 g 1 R) sin B. (7.73) To be definite, let us consider the equilibrium on the right with 0 < 9 < 7 r /2. At the equilibrium point, the term in parenthesis on the right of (7.73) is zero, while sin 0 is positive. If we increase 0 a little (bead moves up and to the right), sin 9 remains positive, but the term in parenthesis becomes negative. (Remember, cos 9 is a decreasing function in this quadrant.) Thus 9 becomes negative, and the bead accelerates back toward its equilibrium point. If we decrease 0 a little from the equilibrium, then 9 becomes positive, and again the bead accelerates back toward equilibrium. Therefore the equilibrium on the right is stable. As you would expect, a similar analysis shows that the same is true of the equilibrium on the left. We arrive at the following interesting story: When the hoop is rotating slowly (co2 < g 1 R), there is just one stable equilibrium, at 9 = 0. If we speed up the rotation, then as w passes the critical value where w 2 = g/R, this original 264 Chapter 7 Lagrange's Equations equilibrium becomes unstable, but two new stable equilibrium points appear, emerging from 0 = 0 and moving out to the right and left as we increase w still more. This phenomenon \u2014 the disappearance of one stable equilibrium and the simultaneous appearance of two others diverging from the same point \u2014 is called a bifurcation and will be one of our principal topics in Chapter 12 on chaos theory. It is interesting to note that the device of this example was used by James Watt (1736-1829) as a governor for his steam engines. The device rotated with the engine, and as the engine sped up the bead rose on the hoop. When the angular velocity co reached some predetermined maximum allowable value, the bead, arriving at a corresponding height, caused the supply", " of steam to be shut off. This example illustrates another strength of the Lagrangian method that was mentioned back in Section 7.1: The generalized coordinates can even be coordinates relative to a noninertial reference frame, just as long as the frame in which the Lagrangian = T \u2014 U was orginally written was inertial. In this example, the angle 0 was the polar angle of the bead, measured in the noninertial rotating frame of the hoop, but the Lagrangian (7.68) was defined as G = T \u2014 U with T and U evaluated in the inertial frame relative to which the hoop rotates. 11 In the next and final example of this section, we pursue the previous example of the bead on the rotating hoop, and obtain approximate solutions of the equation of motion in the neighborhood of the stable equilibrium points. EXAMPLE 7.7 Oscillations of the Bead near Equilibrium Consider again the bead of the previous example and use the equation of motion to find the bead's approximate behavior in the neighborhood of the stable equilibrium positions. When w2 < g/R, the only stable equilibrium is at the bottom of the hoop with 0 = 0. As long as 0 remains small, we can approximate the equation of motion (7.73) by setting sin 0 ti 0 and cos 0 \"=,' 1 to give = \u2014(g I R \u2014 0) 2)0 [0 near 0] = _c220 (7.74) where the second line introduces the frequency f2 = N/gIR\u2014co 2. As long as w2 < g/R, this defines f2 as a real positive number, and we recog- nize (7.74) as the equation for simple harmonic motion with frequency f2. We 11 Example 7.5 was another instance: The coordinate q 1 gave the position of the block relative to the accelerating frame of the wedge, but the kinetic energy T was evaluated in the inertial frame of the table. For another example, see Problem 7.30. Section 7.5 Examples of Lagrange's Equations 265 conclude that a bead which is displaced a little from the stable equilibrium at 0 = 0, executes harmonic motion with frequency Q, 49(t) = A cos(Qt \u2014 6). (7.75) If we speed up the rate of the hoop's rotation until cw t > g/R, then Q becomes pure imaginary, and, since cos is = cosh a, our solution (7.75) becomes a hy- perbolic cosine, which grows with time, correctly reflecting that the equilibrium at 0 = 0 has become unstable. Once co t > g/R, there are two stable equilibrium positions given by (7.71) and located symmetrically to the right and left of the bottom of the hoop. As you might expect, these behave in the same way, and to be definite I shall focus on the one on the right. Let us denote its position by 0 = 00, where, according to (7.70), 00 satisfies co2 cos 00 \u2014 gIR= 0. (7.76) Let us now imagine the bead placed close to 0 0 at 0 = 6 + e and investigate the time dependence of the small parameter E. Once again we can approximate the equation of motion (7.73), though this requires more care. If we approximate the factors of cos(0 0 + e) and sin(00 + E) by the first two terms of their Taylor series, cos(00 + E) = cos 00 \u2014 E sin 00 and sin(00 + E) = sin 00 + E cos 00 (7.77) then the equation of motion (7.73) becomes o = [co2 cos(00 + E) \u2014 g/R] sin(00 + E) [0 near 00[ = [co2 cos 00 \u2014 E cot sin 00 \u2014 gIR] [sin 00 + c cos 00 ]. (7.78) By (7.76) the first and third terms in the first square bracket cancel, leaving just the middle term \u2014E co2 sin 00. To lowest order in E we can drop the second term of the second bracket, and, since 0 is the same as E, we are left with = -E W2 sine 00 = \u2014Q12 E. (7.79) Here the second equality defines the frequency Q' = co sin 00, or, using (7.76), Q' = (02 2 g wR (7.80) (see Problem 7.26). Equation (7.79) is the equation for simple harmonic motion. Therefore, the parameter c oscillates about zero, and the bead itself oscillates about the equilibrium position 0 0 with frequency Qt. 266 Chapter 7 Lagrange's Equations 7.6 Generalized Momenta and Ignorable Coordinates As I have already mentioned, for any system with n generalized coordinates", " q, (i = 1,, n), we refer to the n quantities aZ/aqi = Fi as generalized forces and azia4i = pi as generalized momenta. With this terminology, the Lagrange equation, can be rewritten as az aqi d aL dt a4i'Fi = \u2014 pi. d dt (7.81) (7.82) That is, \"generalized force = rate of change of generalized momentum.\" In particular, if the Lagrangian is independent of a particular coordinate qi, then Fi = = 0 and the corresponding generalized momentum pi is constant. Consider, for example, a single projectile subject only to the force of gravity. The potential energy is U = mgz (if we use Cartesian coordinates with z measured vertically up), and the Lagrangian is = Z(x, y, z, X, 57, = + 572 + 1,2) mgz. (7.83) With respect to Cartesian coordinates, the generalized force is just the usual force (34., /ax = \u2014at /ax = F.\u201e, etc.) and the generalized momentum is just the usual mo- mentum (azia.i = mz = px, etc.) Because L is independent of x and y, it immedi- ately follows that the components px and py are constant, as we already knew. In general, the generalized forces and momenta are not the same as the usual forces and momenta. For instance, we saw in Equations (7.25) and (7.26) that in two- dimensional polar coordinates the 0 component of the generalized force is the torque, and that of the generalized momentum is actually the angular momentum. In any case, when the Lagrangian is independent of a coordinate qi the corresponding generalized momentum is conserved. Thus, if the Lagrangian of a two-dimensional particle is independent of 0, then the particle's angular momentum is conserved \u2014 another im- portant result (and one that is clear from the Newtonian perspective as well). When the Lagrangian is independent of a coordinate qt, that coordinate is sometimes said to be ignorable or cyclic. Obviously it is a good idea, when possible, to choose co- ordinates so that as many as possible are ignorable and their corresponding momenta are constant. In fact, this is perhaps the main criterion in choosing generalized coor- dinates for any given problem: Try to find coordinates as many as possible of which are ignorable. We can rephrase the result of the last three paragraphs by noting that the statement \",C is independent of a coordinate qi \" is equivalent to saying \"4, is unchanged, or invariant, when qi varies (with all the other qi held fixed).\" Thus we can say that if is invariant under variations of a coordinate qi then the corresponding generalized momentum pi is conserved. This connection between invariance of and certain conservation laws is the first of several similar results relating invariance under transformations (translations, rotations, and so on) to conservation laws. These results are known collectively as Noether's theorem, after the German mathematician Emmy Noether (1882-1935). I shall return to this important theorem in Section 7.8. Section 7.7 Conclusion 267 7.7 Conclusion The Lagrangian version of classical mechanics has the two great advantages that, unlike the Newtonian version, it works equally well in all coordinate systems and it can handle constrained systems easily, avoiding any need to discuss the forces of constraint. If the system is constrained, one must choose a suitable set of independent generalized coordinates. Whether or not there are constraints, the next task is to write down the Lagrangian in terms of the chosen coordinates. The equations of motion then follow automatically in the standard form 0L aqi d dt aqi [i =1,,n]. There is, of course, no guarantee that the resulting equations will be easy to solve, and in most real problems they are not, requiring numerical solution or at least some approximations before they can be solved analytically. Even in problems that are only moderately complicated, like the examples of Section 7.5, finding the equations of motion by Lagrange's method is remarkably easier than by using Newton's second law. Indeed, some purists object that the Lagrangian approach makes life too easy, removing the need to think about the physics. The Lagrangian formalism can be extended to include more general systems than those considered so far. One important case is that of magnetic forces, which I take up in Section 7.9. Dissipative forces, such as friction or air resistance, can sometimes be included, but it should be admitted that the Lagrangian formalism is primarily suited to problems where dissipative forces are absent or, at least, negligible. The final three sections of this chapter treat three advanced topics, all of which are centrally important", " in Lagrangian mechanics, but all of which could be omitted on a first reading. In Section 7.8, I give two more examples of the remarkable connection between invariance under certain transformations and conservation laws. This connection, known as Noether's theorem, is important in all of modern physics, but especially in quantum physics. Section 7.9 discusses how to include magnetic forces in Lagrangian mechanics, another topic of great importance in quantum theory. Finally, Section 7.10 introduces the method of Lagrange multipliers. This technique appears in many different guises in many areas of physics, but I shall restrict myself to some simple examples in Lagrangian mechanics. These last three sections are arranged to be self-contained and independent. You could study all of them, none of them, or any selection in between. 268 Chapter 7 Lagrange's Equations 7.8 More about Conservation Laws* * The material of this section is more advanced than the preceding sections, and you should feel free to omit it on a first reading. Be aware, however, that the material discussed here is needed before you read Section 11.5 and Chapter 13. In this section I shall discuss how the laws of conservation of momentum and energy fit into the Lagrangian formulation of mechanics. Since we derived the Lagrangian formulation from the Newtonian, anything that we already knew about conservation laws, based on Newtonian mechanics, will naturally still be true in Lagrangian me- chanics. Nevertheless, we can gain some new insights by examining the conservation laws from a Lagrangian perspective. Furthermore, much modern work takes the La- grangian formulation (based on Hamilton's principle, for example) as its starting point. In this context, it is important to know what can be said about conservation laws strictly within the Lagrangian framework. Conservation of Total Momentum We already know from Newtonian mechanics that the total momentum of an isolated system of N particles is conserved, but let us examine this important property from the Lagrangian point of view. One of the most prominent features of an isolated system is that it is translationally invariant; that is, if we transport all N particles bodily through the same displacement E, nothing physically significant about the system should change. This is illustrated in Figure 7.10, where we see that the effect of moving the whole system through the fixed displacement E is to replace every position r a by ra E, r 1 r i + c, r2 ---> r2 + c, rN -\u00b1 rN + E. (7.84) In particular, the potential energy must be unaffected by this displacement, so that U(r i + c,, r N + E, t) = U(r i,, r N, t) (7.85) r 1 + E r3 + E T2 + E T3 Figure 7.10 An isolated system of N particles is translationally invariant, which means that when every particle is transported through the same displacement E, nothing physically significant changes. Section 7.8 More about Conservation Laws * 269 or, more briefly, SU = 0 where 6U denotes the change in U under the translation (7.84). Clearly the velocities are unchanged by the translation (7.84). (Adding a constant E to all the I', doesn't change the fa.) Therefore ST = 0, and hence 8,C = 0 (7.86) under the translation (7.84). This result is true for any displacement E. If we choose E to be an infinitesimal displacement in the x direction, then all of the x coordinates x 1,, x N increase by E, while the y and z coordinates are unchanged. For this translation, the change in \u00a3 is 8,C)=E + az, 0-x1 =O. axN This implies that (7.87) Now using Lagrange's equations we can rewrite each derivative as where pax is the x component of the momentum of particle a. Thus (7.87) becomes az E 0. a=1 ax\u201e az ax\u201e d _ d dt a5c, pax dt \u2014Pax \u2014 dt a =1 d dt x \u2014n where Px is the x component of the total momentum P = Ea pa. By choosing the small displacement E successively in the y and z directions, we can prove the same result for the y and z components, and we reach the conclusion that, provided the Lagrangian is unchanged by the translation (7.84), the total momentum of the N- particle system is conserved. This connection between translational invariance of E., and conservation of total momentum is another example of Noether's theorem. Conservation of Energy Finally, I would like to discuss conservation of energy from the Lagrangian point of view. The analysis proves somewhat complicated, but introduces a number of ideas that are important in more advanced work, particularly in the", " Hamiltonian formulation of mechanics (Chapter 13). As time advances the function E. (q 1,, qn, 4,,, 4\u201e, t) changes, both because 270 Chapter 7 Lagrange's Equations t is changing, and because the q's and :9,'s change with the evolving system. Thus, by the chain rule, Z(qi,, qn, t) = \u2014 d t. qi aqi qi \u2014. (7.88) a i 4 at Now, by Lagrange's equation, I can replace the derivative in the first sum on the right by az = d az _ d dt 84i dt pi = =(pigi + pi4i) + - at dt ^ d dt. (PA) + at Meanwhile, the derivative in the second sum on the right of (7.88) is just the general- ized momentum pi. Thus, I can rewrite (7.88) as (7.89) (7.90) Now, for many interesting systems, the Lagrangian does not depend explicitly on time; that is, az/at. 0. When this is the case, the second term on the right of (7.89) vanishes. If we move the left side of (7.89) over to the right, we see that the time derivative of the quantity E pi 4i \u2014 \u00a3 is zero. This quantity is so important that it has its own symbol, = pig, \u2014L, n i =1 and is called the Hamiltonian of the system. With this terminology, we can state the following important conclusion: If the Lagrangian d7e of depend explicitly on tide lt is, Liar then the HamiltoniaiI is conserved. The discovery of any conservation law is a momentous event and is enough to justify saying that the Hamiltonian is an important quantity. In fact, it goes much further than this. As we shall see in Chapter 13, the Hamiltonian a is the basis of the Hamiltonian formulation of mechanics, in just the same way that C is the basis of Lagrangian mechanics. For the moment, the chief importance of our newly discovered Hamiltonian is that in many situations it is in fact just the total energy of the system. Specifically, we shall prove that, provided the relation between the generalized coordinates and Cartesians is time-independent, ra = ra(qt, qn), (7.91) Section 7.8 More about Conservation Laws * 271 the Hamiltonian g-C is just the total energy, g-C = T + U. (7.92) You may recall that we agreed in Section 7.3 to describe generalized coordinates that satisfy (7.91) as natural; thus, we can paraphrase the result (7.92) to say that, provided the generalized coordinates are natural, is just the total energy T U. To prove this, let us express the total kinetic energy T = i Ea mares in terms of the generalized coordinates q l,, qn. First, differentiating (7.91) with respect to t and using the chain rule, we find that 12 If we now form the scalar product of this equation with itself, we find a qi i=1 aq (7.93) 2 ra = v,arc, 4 ). ( ara k ) a4'04,, k where I have renamed the summation indices as j and k to avoid future confusion. The kinetic energy is now given as a triple sum, which I can reorganize and write as 13 (7.94) (7.95) (7.96) T = - a a \u2014 2 rijk`ljqk a j,k where A jk is shorthand for the sum A jk = A ik (q i,, q n ) =-- m arc( ) ( ara a aqi aqk a We can now evaluate the generalized momentum pi by differentiating (7.94) with respect to 4, (Problem 7.45), 191 = az aT = a4, 04, = EA,,4, Returning to Equation (7.90) for the Hamiltonian, we can rewrite the sum on the right as LPi 4i = E(E Ai141)4,=EA jAiiii _ 2T (7.97) 12 1f the relation (7.91) were explicitly time-dependent there would be one extra term in this expression for tci, namely arodat. This extra term would invalidate the conclusion (7.98) below that T U. 13 We can restate the result (7.94) to say that, provided the generalized coordinates are natural, the kinetic energy T is a homogeneous quadratic function of the generalized velocities 4i. This result plays an important role in several later developements. See, for instance, Section", " 11.5. 272 Chapter 7 Lagrange's Equations where the last step follows from (7.94). Therefore = p,4i \u2014L= 2T \u2014 (T \u2014 U)=T +U. (7.98) That is, provided the transformation between the Cartesian and generalized coordi- nates is time-independent, as in (7.91), the Hamiltonian X is just the total energy of the system. I have already proved that, provided the Lagrangian is independent of time, the Hamiltonian is conserved. Thus we now see that time independence of the Lagrangian [together with the condition (7.91)] implies conservation of energy. We can rephrase the time independence of G by saying that is unchanged by translations of time, t + E. Thus the result we have just proved is that invariance of \u00a3 under time translations is related to energy conservation, in much the same way that invariance of L under translations of space (r r E) is related to conservation of momentum. Both results are manifestations of Noether's famous theorem. 7.9 Lagrange's Equations for Magnetic Forces* * This section requires a knowledge of the scalar and vector potentials of electromagnetism. Although the ideas described here play an important role in the quantum-mechanical treatment of magnetic fields, they will not be used again in this book. Although I have so far consistently defined the Lagrangian as 1, = T \u2014 U, there are systems, such as a charged particle in a magnetic field, which can be treated by the Lagrangian method, but for which 1, is not just T \u2014 U. The natural question to ask is then: What is the definition of the Lagrangian for such systems? This is the first question I address. Definition and Nonuniqueness of the Lagrangian Probably the most satisfactory general definition of a Lagrangian for a mechanical system is this: General Definition o a L agrangian For a given ech uA:Et sysietx. lath nera1ii d coordinates q (q 1. q q i, q,\u201e qi q,4, the coord nates a Lagrangian and velocitir such that the Lagrange equabons di a4, In other words, a Lagrangian is any function for which Lagrange's equations are true for the system under consideration. Section 7.9 Lagrange's Equations for Magnetic Forces* 273 Obviously, for the systems that we have discussed so far, the old definition \u00a3 = T \u2014 U fits this new definition. But the new definition is much more general. In particular, it is easy to see that our new definition does not define a unique Lagrangian function. For example, consider a single particle in one dimension and suppose that we have found a Lagrangian L for this particle. That is, the equation of motion of the particle is Now let f (x,.i) be any function for which a,c _ d 0x dt ax 0f ax d of dt a = f (7.99) (7.100) (It is easy to think up such a function, for instance, f = xi.) If we replace in (7.99) by as Z then, by virtue of (7.100), the Lagrangian Z' gives exactly the same equation of motion The lack of uniqueness of the Lagrangian is similar to, though more radical than, the familiar lack of uniqueness in the potential energy (to which one can add any constant without changing any of the physical predictions). The crucial point is that any function \u00a3 which gives the right equation of motion has all of the features that we require of a Lagrangian (for instance, that the integral f dt is stationary at the right path) and so is just as acceptable as any other such function Z. If, for a given system, we can spot a function Z that leads to the right equation of motion, then we don't need to debate whether it is the \"right\" Lagrangian \u2014 if it gives the right equation of motion, then it is just as right as any other conceivable Lagrangian. Lagrangian for a Charge in a Magnetic Field Consider now a particle (mass m and charge q) moving in electric and magnetic fields E and B. The force on the particle is the well-known Lorentz force F = q(E + v x B), so Newton's second law reads mr = q (E x B). (7.101) To reformulate (7.101) in Lagrangian form, we have only to spot a function Z for which the three Lagrange equations are the same as (7.101). This can be done using the scalar and vector potentials, V(r, t) and A(r, t), in terms of which the two fields can be written i 4 E \u2014V V \u2014 and", " B=V x A. (7.102) A at 14 See, for example, David J. Griffiths, Introduction to Electrodynamics, (Prentice-Hall, 1999), p. 416-417. 274 Chapter 7 Lagrange's Equations I now claim that the Lagrangian function' s (r, t) = - 21Mi2 q(V \u2014 A) =.5? 2 i2) q(V \u2014 52A y \u2014.?1,) (7.104) has the desired property, that it reproduces Newton's second law (7.101). To check this, let us examine the first of the three Lagrange equations, (7.103) (7.105) (7.106) az d az, ax \u2014 dt. =mx + q A x. az 8.1 To see what this implies we have to evaluate the two derivatives of the proposed Lagrangian (7.104): az ax (ay ax.a x A ax.aA y.3 A ax ax ) and find When we differentiate this with respect to t, we must remember that A x A x (x, y, z, t). As t varies, x, y, z move with the particle and, by the chain rule, we d az dt ax aAx.aA x + aAx mi +q x\u2014ax x + 57 B y +z az at (7.107) Substituting (7.106) and (7.107) into (7.105), cancelling the two terms in 1, and rearranging, we find that Lagrange's equation (the x component, with the proposed Lagrangian) is the same as (ay aA q; (BA Y alAx 111 -5 = at x, ax ay ) q BAx aA z az ax (7.108) or, according to (7.102), mx = q (Ex + 543z \u2014 B y) (7.109) which you will recognize as the x component of Newton's second law (7.101). Since the y and z components work in the same way, we conclude that Lagrange's equations, with the proposed Lagrangian (7.104), are exactly equivalent to Newton's second law for the charged particle. That is, we have successfully recast Newton's second law for a charged particle into Lagrangian form with the Lagrangian (7.104). Using the Lagrangian (7.104), one can solve various problems involving charged particles in electric and magnetic fields. (See Problem 7.49 for an example.) Theoret- 15 Notice that you can, if you wish, write this as = T \u2014 U, but U is certainly not the usual PE, since it depends on the velocity 1; U is sometimes called a \"velocity-dependent PE,\" but notice that it is not true that the force on the charge is \u2014 V U. Section 7.10 Lagrange Multipliers and Constraint Forces* 275 ically, the most important conclusion of this analysis emerges when we evaluate the generalized momentum. For example, px = \u20147 = qA x. Since the y and z components work the same way, we conclude that (generalized momentum, p) = my + qA. (7.110) That is, the generalized momentum is the mechanical momentum my plus a magnetic term q A. This result is at the heart of the quantum theory of a charged particle in a magnetic field, where it turns out that the generalized momentum corresponds to the differential operator \u2014i hV (where h is Planck's constant over 27), so that the quantum analog of the mechanical momentum my is the operator \u2014ihV \u2014 qA. 7.10 Lagrange Multipliers and Constraint Forces* * The method of Lagrange multipliers is used in many areas of physics. Nevertheless, we shan't be using it again in this book, and you could omit this section without any loss of continuity. In this section, we discuss the method of Lagrange multipliers. This powerful method finds application in several areas of physics and takes on quite different appearances in different contexts. Here I shall treat only its application to Lagrangian mechanics, 16 and to keep the analysis simple I shall restrict the discussion to two-dimensional systems with just one degree of freedom. We have seen that one of the strengths of Lagrangian mechanics is that it can bypass all of the forces of constraint. However, there are situations where one actually needs to know these forces. For example, the designer of a roller coaster needs to know the normal force of the track on the car to know how strong to build the track. In this case, we can still use a modified form of Lagrange's equations, but the procedure is somewhat different: We do not choose generalized coordinates q 1,, qn all of which can be independently varied. (Remember that it was the", " independence of q 1,, that let us use the standard Lagrange equations without worrying about constraints.) Instead we use a larger number of coordinates and use Lagrange multipliers to handle the constraints. To illustrate this procedure, we'll consider a system with just two rectangular coordinates x and y, which are restricted by a constraint equation of the form 17 f (x, y) = const. (7.111) For example, we could consider a simple pendulum with just one degree of freedom (as in Figure 7.2). In treating this by the standard Lagrange approach one would use 16 For applications to other kinds of problems, see, for example, Mathematical Methods in the Physical Sciences by Mary Boas (Wiley, 1983), Ch. 4, Section 9 and Ch. 9, Section 6. 17 We'll see directly by example that some typical constraints can be put in this form. In fact, it is fairly easy to show that any holonomic constraints can be. 276 Chapter 7 Lagrange's Equations the one generalized coordinate 0, the angle between the pendulum and the vertical, and avoid any discussion of the constraints. If, instead, we choose to use the original rectangular coordinates x and y, then we must recognize that these coordinates are not independent; they satisfy the constraint equation f (x, y) =, / x 2 + y2 = where l is the length of the pendulum. We shall find that the method of Lagrange multipliers lets us accomodate this constraint, determine the time dependence of x and y, and find the tension in the rod. As a second example, consider the Atwood machine of Figure 7.6. In our previous treatment we used the one generalized coordinate x, the position of the mass m l, but we could instead use both coordinates x and y (the positions of both masses), provided we remember that the constancy of the string's length imposes the constraint that f(x, y) = x y = const. Here too, Lagrange multipliers will let us accomodate this constraint, solve for the time dependence of x and y, and find the constraint force, which is here the tension To set up our new method we start from Hamilton's principle. Our Lagrangian has the form (x, x, y, 52). (We could allow it to have an explicit dependence on t as well, but to simplify notation I shall assume it doesn't.) The proof of Hamilton's principle given in Section 7.4 applies even when the coordinates are constrained. (Indeed it was designed to allow for constraints.) Thus we can conclude as before that the action in the string. integral is stationary when taken along the actual path followed. If we denote this \"right\" path by x (t), y(t) and imagine a small displacement to a neighboring \"wrong\" path, S = f t, \u00a3(x, 1, y, 5,) dt x (t) y(t) x (t) + 8x (01 y(t) + 8y (t) 1 (7.112) (7.113) then, provided the displacement is consistent with the constraint equation, the action integral (7.112) is unchanged, SS = 0. To exploit this, we must write 8S in terms of the small displacements 6x and 8y: SS = f J (ax ax sx +\u2014az ai Sy + ay 85; ss,) dt. (7.114) The second and fourth terms can be integrated by parts, and we conclude that SS = f az d \u2014 ax \u2014dt ai az d ay dt a5, 8x dt f ( \u2014 \u2014 \u2014 \u2014) Sy dt = 0 (7.115) for any displacements 8x and 8y consistent with the constraints. Section 7.10 Lagrange Multipliers and Constraint Forces* 277 If (7.115) were true for any displacements, we could immediately prove two separate Lagrange equations, one for x and one for y. (Choosing Sy = 0, we would be left with just the first integral; since this has to vanish for any choice of Sx, the factor in parentheses would have to be zero, which implies the usual Lagrange equation with respect to x. And similarly for y.) This is exactly the correct conclusion for the case that there are no constraints. However, there are constraints, and (7.115) is only true for displacements Sx and Sy consistent with the constraints. Therefore, we proceed as follows: Since all points with which we are concerned satisfy f (x, y) = const, the displacement (7.113) leaves f (x, y) unchanged, so f = \u2014 Sx \u2014 Sy = 0 af ax af ay (7.116) for any displacement consistent with the constraints. Since this is zero, we can multiply it by an arbitrary function ).(t) \u2014 this is the Lagrange multiplier \u2014", " and add it to the integrand in (7.115), without changing the value of the integral (namely zero). Therefore SS = f (\u2014 + go\u2014 ax a ax d az \u2014 \u2014 \u2014 dt ax ) Sx dt az a f -8\u2014y gt)--ay \u2014 d ( dt ay Sy dt = 0 (7.117) for any displacement consistent with the constraints. Now comes the supreme cunning: So far A (t) is an arbitrary function of t, but we can choose it so that the coefficient of Sx in the first integral is zero. That is, by choice of the multiplier A (t), we can arrange that along the actual path of the system. This is the first of two modified Lagrange equations and differs from the usual equation only by the extra term involving A on the left. With the multiplier chosen in this way, the whole first integral in (7.117) is zero. Therefore the second integral is also zero (since their sum is), and this is true for any choice of Sy. (The constraint places no restriction on Sx or Sy separately \u2014 it only fixes Sx once Sy is chosen, or vice versa.) Therefore the coefficient of Sy in this second integral must also be zero and we have a second modified Lagrange equation, 278 Chapter 7 Lagrange's Equations We now have two modified Lagrange equations for the two unknown functions x (t) and y(t). This elegant result has been bought at the price of introducing a third unknown function, the Lagrange multiplier A (t). To find three unknown functions, we need three equations, but fortunately a third equation is already at hand, the constraint equation The three equations (7.118), (7.119), and (7.120) are sufficient, in principle at least, to determine the coordinates x(t) and y(t) and the multiplier A (t). Before we illustrate this with an example, there is one more bit of theory to develop. So far the Lagrange multiplier A (t) is just a mathematical artifact, introduced to help us solve our problem. However, it turns out to be closely related to the forces of constraint. To see this, we have only to look more closely at the modified Lagrange equations (7.118) and (7.119). The Lagrangian of our present discussion has the form E.) = 1.i 2 + 1m 2522 y) (In a problem like the simple pendulum, x and y are the two coordinates of a single mass and m 1 = m 2. In a problem like the Atwood machine, there are two separate masses and m 1 and m 2 are not necessarily equal.) Inserting this Lagrangian into (7.118), we find ay,af + X = ax ax (7.121) Now, on the left side \u20148U/ax is the x component of the nonconstraint force. (Re- member that U was defined as the potential energy of the nonconstraint forces.) On the right, m 1 z is the x component of the total force, equal to the sum of the nonconstraint and constraint forces. Thus m 1 x = \u2014a U/ax Fx\"'. Canceling the term \u2014a U/ax from both sides of (7.121), we reach the important conclusion that af ax Fcstr x (7.122) with a corresponding result for the y components. This then is the significance of the Lagrange multiplier: Multiplied by the appropriate partial derivatives of the constraint function f(x, y), the Lagrange multiplier ),.(t) gives the corresponding components of the constraint force. the example of the Atwood machine. Let us now see how these ideas work in practice by using the formalism to analyse Section 7.10 Lagrange Multipliers and Constraint Forces* 279 EXAMPLE 7.8 Atwood's Machine Using a Lagrange Multiplier Analyze the Atwood machine of Figure 7.6 (shown again here as Figure 7.11) by the method of Lagrange multipliers and using the coordinates x and y of the two masses. In terms of the given coordinates, the Lagrangian is T \u2014 U = z m 1 x2 + 1/122)1 2 in gX M2gY (7.123) and the constraint equation is f (x, y) = x y = const. (7.124) The modified Lagrange equation (7.118) for x reads and that for y is ac ax az, dy of 8x d = \u2014 dt ax of d a,c, dt as) or m i g (7.125) Or M2g =m2y. (7.126) These two equations, together with the constraint equation (7.124), are easily solved for the unknowns x(t), y(t), and )\\.(t). From (", "7.124) we see that 33 =, and then subtracting (7.126) from (7.125) we can eliminate and arrive at the same result as before, = m2)0mi m2). To better understand the two modified Lagrange equations (7.125) and (7.126), it is helpful to compare them with the two equations of the Newtonian solution. Newton's second law for m 1 is g \u2014 Ft = m 1 1 x Figure 7.11 The Atwood machine again. 280 Chapter 7 Lagrange's Equations where Ft is the tension in the string, and that for m 2 is These are precisely the two Lagrange equations (7.125) and (7.126), with the Lagrange multiplier identified as the constraint force m 2g \u2014 Ft = m23 = \u2014 Ft. [Two small comments: The minus sign occurs because both coordinates x and y were measured downward, whereas both tension forces are upward. In general, according to (7.122) the constraint force is afiax, but in this simple case, of/ax =11 7.50 through 7.52. You can find some more examples of the use of Lagrange multipliers in Problems Principal Definitions and Equations of Chapter 7 The Lagrangian The Lagrangian \u00a3 of a conservative system is defined as where T and U are respectively the kinetic and potential energies. T \u2014 U, [Eq. (7.3)] Generalized Coordinates The n parameters q l,, q, are generalized coordinates for an N-particle system if every particle's position rc, can be expressed as a function of q l,, qn (and possibly the time t) and vice versa, and if n is the smallest number that allows the system to be described in this way. [Eqs. (7.34) & (7.35)] If n < 3N (in three dimensions) the system is said to be constrained. The coor- dinates q 1,, qn are said to be natural if the functional relationships of the roe to q l,, qn are independent of time. The number of degrees of freedom of a system is the number of coordinates that can be independently varied. If the number of de- grees of freedom is equal to the number of generalized coordinates (in some sense the \"normal\" state of affairs), the system is said to be holonomic. [Section 7.3] Lagrange's Equations equations For any holonomic system, Newton's second law is equivalent to the n Lagrange _ d dt 04i [i =1,,n] [Sections 7.3 & 7.4] and the Lagrange equations are in turn equivalent to Hamilton's principle \u2014 a fact we used only to prove the Lagrange equations. [Eq. (7.8)] Problems for Chapter 7 281 Generalized Momenta and Ignorable Coordinates The ith generalized momentum pi is defined to be the derivative Pi az. ai If = 0, then we say the coordinate qi is ignorable and the corresponding generalized momentum is constant. [Section 7.6] The Hamiltonian The Hamiltonian is defined as 1C = 1 picIt - [Eq. (7.90)] If aziat= 0, then 31 is conserved; if the coordinates q 1,, qn are natural, 91 is just the energy of the system. Lagrangian for a Charge in an Electromagnetic Field The Lagrangian for a charge q in an electromagnetic field is (r, t) = \u2014 q(V \u2014 A). [Eq. (7.103)] Problems for Chapter 7 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). SECTION 7.1 Lagrange's Equations for Unconstrained Motion 7.1 * Write down the Lagrangian for a projectile (subject to no air resistance) in terms of its Cartesian coordinates (x, y, z), with z measured vertically upward. Find the three Lagrange equations and show that they are exactly what you would expect for the equations of motion. 7.2 * Write down the Lagrangian for a one-dimensional particle moving along the x axis and subject to a force F = \u2014 kx (with k positive). Find the Lagrange equation of motion and solve it. 7.3 * Consider a mass in moving in two dimensions with potential energy U (x, y) = Zkr 2, where r2 = x 2 + y 2. Write down the Lagrangian, using coordinates x and y, and find the two Lagrange equations of motion. Describe their solutions. [This is the potential energy of an ion in an \"ion trap,\" which can be used to study the properties of individual atomic ions.] 282 Chapter 7 Lagrange's Equations 7.4 * Consider a mass m moving in a frictionless plane that slopes at an angle a with", " the horizontal. Write down the Lagrangian in terms of coordinates x, measured horizontally across the slope, and y, measured down the slope. (Treat the system as two-dimensional, but include the gravitational potential energy.) Find the two Lagrange equations and show that they are what you should have expected. 7.5 * Find the components of V f (r, 0) in two-dimensional polar coordinates. [Hint: Remember that the change in the scalar f as a result of an infinitesimal displacement dr is df =V f dr.] 7.6 * Consider two particles moving unconstrained in three dimensions, with potential energy U(r 1, r2). (a) Write down the six equations of motion obtained by applying Newton's second law to each particle. (b) Write down the Lagrangian L (ri, r2, ri,t2) = T \u2014 U and show that the six La- grange equations are the same as the six Newtonian equations of part (a). This establishes the validity of Lagrange's equations in rectangular coordinates, which in turn establishes Hamilton's principle. Since the latter is independent of coordinates, this proves Lagrange's equations in any coordinate system. 7.7 * Do Problem 7.6, but for N particles moving unconstrained in three dimensions (in which case there are 3N equations of motion). 7.8 ** (a) Write down the Lagrangian L (x 1, x2, xl, i2) for two particles of equal masses, m 1 = m 2 = m, confined to the x axis and connected by a spring with potential energy U = -1kx 2. [Here x is the extension of the spring, x = (x 1 \u2014 x2 \u2014 1), where 1 is the spring's unstretched length, and I assume that mass 1 remains to the right of mass 2 at all times.] (b) Rewrite L in terms of the new variables X = i (x 1 + x2) (the CM position) and x (the extension), and write down the two Lagrange equations for X and x. (c) Solve for X (t) and x (t) and describe the motion. SECTION 7.3 Constrained Systems in General 7.9 * Consider a bead that is threaded on a rigid circular hoop of radius R lying in the xy plane with its center at 0, and use the angle 4) of two-dimensional polar coordinates as the one generalized coordinate to describe the bead's position. Write down the equations that give the Cartesian coordinates (x, y) in terms of and the equation that gives the generalized coordinate in terms of (x, y). 7.10 * A particle is confined to move on the surface of a circular cone with its axis on the z axis, vertex at the origin (pointing down), and half-angle a. The particle's position can be specified by two generalized coordinates, which you can choose to be the coordinates (p, 0) of cylindrical polar coordinates. Write down the equations that give the three Cartesian coordinates of the particle in terms of the generalized coordinates (p, 0) and vice versa. 7.11 * Consider the pendulum of Figure 7.4, suspended inside a railroad car, but suppose that the car is oscillating back and forth, so that the point of suspension has position x s = A cos wt, y s = 0. Use the angle as the generalized coordinate and write down the equations that give the Cartesian coordinates of the bob in terms of 0 and vice versa. SECTION 7.4 Proof of Lagrange's Equations with Constraints 7.12 * Lagrange's equations in the form discussed in this chapter hold only if the forces (at least the nonconstraint forces) are derivable from a potential energy. To get an idea how they can be modified to include forces like friction, consider the following: A single particle in one dimension is subject to various conservative forces (net conservative force = F = \u2014aU/ax) and a nonconservative force (let's Problems for Chapter 7 283 Figure 7.12 Problem 7.14 r az \u2014 + ax d az fric \u2014 8.i dt call it Ffric) Define the Lagrangian as E., = T \u2014 U and show that the appropriate modification is 7.13 ** In Section 7.4 [Equations (7.41) through (7.51)], I proved Lagrange's equations for a single particle constrained to move on a two-dimensional surface. Go through the same steps to prove Lagrange's equations for a system consisting of two particles subject to various unspecified constraints. [Hint: The net force on particle 1 is the sum of the total constraint force VI' and the total nonconstraint force F 1, and likewise for particle 2. The constraint forces come in many guises (the normal force of a surface", ", the tension force of a string tied between the particles, etc.), but it is always true that the net work done by all constraint forces in any displacement consistent with the constraints is zero \u2014 this is the defining property of constraint forces. Meanwhile, we take for granted that the nonconstraint forces are derivable from a potential energy U(r i, r2, t); that is, F 1 = \u2014V1 U and likewise for particle 2. Write down the difference SS between the action integral for the right path given by r 1 (t) and r2 (t) and any nearby wrong path given by r 1 (t) E 1 (t) and r2 (t) 62 (0. Paralleling the steps of Section 7.4, you can show that SS is given by an integral analogous to (7.49), and this is zero by the defining property of constraint forces.] SECTION 7.5 Examples of Lagrange's Equations 7.14 * Figure 7.12 shows a crude model of a yoyo. A massless string is suspended vertically from a fixed point and the other end is wrapped several times around a uniform cylinder of mass m and radius R. When the cylinder is released it moves vertically down, rotating as the string unwinds. Write down the Lagrangian, using the distance x as your generalized coordinate. Find the Lagrange equation of motion and show that the cylinder accelerates downward withi = 2g/3. [Hints: You need to remember from your introductory physics course that the total kinetic energy of a body like the yoyo is T = lmv 2 11(02, where v is the velocity of the center of mass, I is the moment of inertia (for a unform cylinder, I = im R 2 ) and co is the angular velocity about the CM. You can express w in terms of I.] 7.15 * A mass m i rests on a frictionless horizontal table and is attached to a massless string. The string runs horizontally to the edge of the table, where it passes over a massless, frictionless pulley and then hangs vertically down. A second mass m 2 is now attached to the bottom end of the string. Write down the Lagrangian for the system. Find the Lagrange equation of motion, and solve it for the acceleration of the blocks. For your generalized coordinate, use the distance x of the second mass below the tabletop. 284 Chapter 7 Lagrange's Equations 7.16 * Write down the Lagrangian for a cylinder (mass m, radius R, and moment of inertia I) that rolls without slipping straight down an inclined plane which is at an angle a from the horizontal. Use as your generalized coordinate the cylinder's distance x measured down the plane from its starting point. Write down the Lagrange equation and solve it for the cylinder's acceleration x. Remember that T = 4mv2 + z L02, where v is the velocity of the center of mass and co is the angular velocity. 7.17 * Use the Lagrangian method to find the acceleration of the Atwood machine of Example 7.3 (page 255) including the effect of the pulley's having moment of inertia I. (The kinetic energy of the pulley is 41(.02, where co is its angular velocity.) 7.18 * A mass m is suspended from a massless string, the other end of which is wrapped several times around a horizontal cylinder of radius R and moment of inertia I, which is free to rotate about a fixed horizontal axle. Using a suitable coordinate, set up the Lagrangian and the Lagrange equation of motion, and find the acceleration of the mass m. [The kinetic energy of the rotating cylinder is i /6)21 7.19 * In Example 7.5 (page 258) the two accelerations are given by Equations (7.66) and (7.67). Check that the acceleration of the block is given correctly in the limit M 0. [You need to find the components of this acceleration relative to the table.] 7.20 * A smooth wire is bent into the shape of a helix, with cylindrical polar coordinates p = R and z = X0, where R and X are constants and the z axis is vertically up (and gravity vertically down). Using z as your generalized coordinate, write down the Lagrangian for a bead of mass m threaded on the wire. Find the Lagrange equation and hence the bead's vertical acceleration E. In the limit that R 0, what is i? Does this make sense? 7.21 * The center of a long frictionless rod is pivoted at the origin, and the rod is forced to rotate in a horizontal plane with constant angular velocity w. Write down the Lagrangian for a bead threaded on the rod, using r as your generalized coordinate, where r, are the polar coordinates of the bead. (Notice that 0 is not an independent variable since it is", " fixed by the rotation of the rod to be 0 = wt.) Solve Lagrange's equation for r(t). What happens if the bead is initially at rest at the origin? If it is released from any point ro > 0, show that r(t) eventually grows exponentially. Explain your results in terms of the centrifugal force mw 2r. 7.22 * Using the usual angle 0 as generalized coordinate, write down the Lagrangian for a simple pendulum of length I suspended from the ceiling of an elevator that is accelerating upward with constant acceleration a. (Be careful when writing T; it is probably safest to write the bob's velocity in component form.) Find the Lagrange equation of motion and show that it is the same as that for a normal, nonaccelerating pendulum, except that g has been replaced by g + a. In particular, the angular frequency of small oscillations is./(g + a)/l. 7.23 * A small cart (mass m) is mounted on rails inside a large cart. The two are attached by a spring (force constant k) in such a way that the small cart is in equilibrium at the midpoint of the large. The distance of the small cart from its equilibrium is denoted x and that of the large one from a fixed point on the ground is X, as shown in Figure 7.13. The large cart is now forced to oscillate such that X = A cos wt, with both A and w fixed. Set up the Lagrangian for the motion of the small cart and show that the Lagrange equation has the form + wax = B cos wt where coo is the natural frequency coo =,/k/m and B is a constant. This is the form assumed in Section 5.5, Equation (5.57), for driven oscillations (except that we are here ignoring damping). Thus the system Problems for Chapter 7 285 x Figure 7.13 Problem 7.23 described here would be one way to realize the motion discussed there. (We could fill the large cart with molasses to provide some damping.) 7.24 * We saw in Example 7.3 (page 255) that the acceleration of the Atwood machine is x = (m \u2014 m 2)g/(m i + m2). It is sometimes claimed that this result is \"obvious\" because, it is said, the effective force on the system is (m 1 \u2014 m 2)g and the effective mass is (m 1 + m 2). This is not, perhaps, all that obvious, but it does emerge very naturally in the Lagrangian approach. Recall that Lagrange's equation can be thought of as [Equation (7.17)] (generalized force) = (rate of change of generalized momentum). Show that for the Atwood machine the generalized force is (m 1 \u2014 m 2)g and the generalized momentum (m i m 2).i. Comment. 7.25 * Prove that the potential energy of a central force F = \u2014krni (with n \u20141) is U = krn+1 1(n 1). In particular, if n = 1, then F = \u2014kr and U = tkr2 7.26 * In Example 7.7 (page 264), we saw that the bead on a spinning hoop can make small oscillations about any of its stable equilibrium points. Verify that the oscillation frequency S2' defined in (7.79) is equal to /w2 \u2014 (g 1 coR) 2 as claimed in (7.80). 7.27 ** Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley on frictionless bearings. The other end of this string supports a second similar pulley, over which passes a second string supporting a mass of 3m at one end and m at the other. Using two suitable generalized coordinates, set up the Lagrangian and use the Lagrange equations to find the acceleration of the mass 4m when the system is released. Explain why the top pulley rotates even though it carries equal weights on each side. 7.28 ** A couple of points need checking from Example 7.6 (page 260). (a) From the point of view of a noninertial frame rotating with the hoop, the bead is subject to the force of gravity and a centrifugal force mw 2p (in addition to the constraint force, which is the normal force of the wire). Verify that at the equilibrium points given by (7.71), the tangential components of these two forces balance one another. (A free-body diagram will help.) (b) Verify that the equilibrium point at the top (0 = tr) is unstable. (c) Verify that the equilibrium at the second point given by (7.71) (the one on the left, with 0 negative) is stable. 7.29", " ** Figure 7.14 shows a simple pendulum (mass m, length 1) whose point of support P is attached to the edge of a wheel (center 0, radius R) that is forced to rotate at a fixed angular velocity w. At t = 0, the point P is level with 0 on the right. Write down the Lagrangian and find the equation of motion for the angle 0. [Hint: Be careful writing down the kinetic energy T. A safe way to get the velocity right is to write down the position of the bob at time t, and then differentiate.] Check that your answer makes sense in the special case that co = 0. 286 Chapter 7 Lagrange's Equations Figure 7.14 Problem 7.29 7.30 ** Consider the pendulum of Figure 7.4, suspended inside a railroad car that is being forced to accelerate with a constant acceleration a. (a) Write down the Lagrangian for the system and the equation of motion for the angle 0. Use a trick similar to the one used in Equation (5.11) to write the combination of sin 0 and cos 0 as a multiple of sin( 0 + /3). (b) Find the equilibrium angle 0 at which the pendulum can remain fixed (relative to the car) as the car accelerates. Use the equation of motion to show that this equilbrium is stable. What is the frequency of small oscillations about this equilibrium position? (We shall find a much slicker way to solve this problem in Chapter 9, but the Lagrangian method does give a straightforward route to the answer.) 7.31 ** A simple pendulum (mass M and length L) is suspended from a cart (mass m) that can oscillate on the end of a spring of force constant k, as shown in Figure 7.15. (a) Write the Lagrangian in terms of the two generalized coordinates x and 0, where x is the extension of the spring from its equilibrium length. (Read the hint in Problem 7.29.) Find the two Lagrange equations. (Warning: They're pretty ugly!) (b) Simplify the equations to the case that both x and 0 are small. (They're still pretty ugly, and note, in particular, that they are still coupled; that is, each equation involves both variables. Nonetheless, we shall see how to solve these equations in Chapter 11 \u2014 see particularly Problem 11.19.) Figure 7.15 Problem 7.31 7.32 ** Consider the cube balanced on a cylinder as described in Example 4.7 (page 130). Assuming that b < r, use the Lagrangian approach to find the angular frequency of small oscillations about the top. The simplest procedure is to make the small-angle approximations to before you differentiate to get Lagrange's equation. As usual, be careful in writing down the kinetic energy; this is Z (1111.1 2 + 16 2), where v is the speed of the CM and / is the moment of inertia about the CM (2mb 2/3). The safe way to find v is to write down the coordinates of the CM and then differentiate. Problems for Chapter 7 287 7.33 ** A bar of soap (mass m) is at rest on a frictionless rectangular plate that rests on a horizontal table. At time t = 0, I start raising one edge of the plate so that the plate pivots about the opposite edge with constant angular velocity co, and the soap starts to slide toward the downhill edge. Show that the equation of motion for the soap has the form x \u2014 co 2x = \u2014g sin cot, where x is the soap's distance from the downhill edge. Solve this for x (t), given that x (0) = xo. [You'll need to use the method used to solve Equation (5.48). You can easily solve the homogeneous equation; for a particular solution try x = A sin cot and solve for A.] 7.34 ** Consider the well-known problem of a cart of mass m moving along the x axis attached to a spring (force constant k), whose other end is held fixed (Figure 5.2). If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency co =.\\/k/m. Using the Lagrangian approach, you can find the effect of the spring's mass M, as follows: (a) Assuming that the spring is uniform and stretches uniformly, show that its kinetic energy is 6 Mx 2. (As usual x is the extension of the spring from its equilibrium length.) Write down the Lagrangian for the system of cart plus spring. (Note: The potential energy is still ikx 2.) (b) Write down the Lagrange equation and show that the cart still executes SHM but with angular frequency co = Vkl(m M/3); that is, the", " effect of the spring's mass M is just to add M/3 to the mass of the cart. Figure 7.16 Problem 7.35 7.35 ** Figure 7.16 is a bird's-eye view of a smooth horizontal wire hoop that is forced to rotate at a fixed angular velocity co about a vertical axis through the point A. A bead of mass m is threaded on the hoop and is free to move around it, with its position specified by the angle (/) that it makes at the center with the diameter AB. Find the Lagrangian for this system using as your generalized coordinate. (Read the hint in Problem 7.29.) Use the Lagrange equation of motion to show that the bead oscillates about the point B exactly like a simple pendulum. What is the frequency of these oscillations if their amplitude is small? 7.36 *** A pendulum is made from a massless spring (force constant k and unstretched length / 0) that is suspended at one end from a fixed pivot 0 and has a mass m attached to its other end. The spring can stretch and compress but cannot bend, and the whole system is confined to a single vertical plane. (a) Write down the Lagrangian for the pendulum, using as generalized coordinates the usual angle (/) and the length r of the spring. (b) Find the two Lagrange equations of the system and interpret them in terms of Newton's second law, as given in Equation (1.48). (c) The equations of part (b) cannot be solved analytically in general. However, they can be solved for small oscillations. Do this and describe the motion. [Hint: Let 1 denote the equilibrium length of the spring with the mass hanging from it and 288 Chapter 7 Lagrange's Equations write r = 1 + E. \"Small oscillations\" involve only small values of c and 0, so you can use the small- angle approximations and drop from your equations all terms that involve powers of E or 0 (or their derivatives) higher than the first power (also products of E and 0 or their derivatives). This dramatically simplifies and uncouples the equations.] 7.37 *** Two equal masses, m l = m 2 = m, are joined by a massless string of length L that passes through a hole in a frictionless horizontal table. The first mass slides on the table while the second hangs below the table and moves up and down in a vertical line. (a) Assuming the string remains taut, write down the Lagrangian for the system in terms of the polar coordinates (r, 0) of the mass on the table. (b) Find the two Lagrange equations and interpret the 0 equation in terms of the angular momentum \u00a3 of the first mass. (c) Express in terms of t and eliminate 4 from the r equation. Now use the r equation to find the value r = ro at which the first mass can move in a circular path. Interpret your answer in Newtonian terms. (d) Suppose the first mass is moving in this circular path and is given a small radial nudge. Write r (t) = ro E (t) and rewrite the r equation in terms of E (t) dropping all powers of E (t) higher than linear. Show that the circular path is stable and that r(t) oscillates sinusoidally about r0. What is the frequency of its oscillations? 7.38 *** A particle is confined to move on the surface of a circular cone with its axis on the vertical z axis, vertex at the origin (pointing down), and half-angle a. (a) Write down the Lagrangian L in terms of the spherical polar coordinates r and 0. (b) Find the two equations of motion. Interpret the 0 equation in terms of the angular momentum tz, and use it to eliminate 4 from the r equation in favor of the constant f z. Does your r equation make sense in the case that = 0? Find the value ro of r at which the particle can remain in a horizontal circular path. (c) Suppose that the particle is given a small radial kick, so that r(t) = ro E(t), where E(t) is small. Use the r equation to decide whether the circular path is stable. If so, with what frequency does r oscillate about r 0? 7.39 *** (a) Write down the Lagrangian for a particle moving in three dimensions under the influence of a conservative central force with potential energy U (r), using spherical polar coordinates (r, 0, 4'). (b) Write down the three Lagrange equations and explain their significance in terms of radial accelera- tion, angular momentum, and so forth. (The 0 equation is the tricky one, since you will find it implies that the 0 component of varies with time, which seems to contradict conservation of angular mo- mentum. Remember, however, that to is", " the component of in a variable direction.) (c) Suppose that initially the motion is in the equatorial plane (that is, 00 = 7r/2 and 60 = 0). Describe the subsequent motion. (d) Suppose instead that the initial motion is along a line of longitude (that is, 4o = 0). Describe the subsequent motion. 7.40 *** The \"spherical pendulum\" is just a simple pendulum that is free to move in any sideways direction. (By contrast a \"simple pendulum\" \u2014 unqualified \u2014 is confined to a single vertical plane.) The bob of a spherical pendulum moves on a sphere, centered on the point of support with radius r = R, the length of the pendulum. A convenient choice of coordinates is spherical polars, r, 0, 0, with the origin at the point of support and the polar axis pointing straight down. The two variables 0 and 0 make a good choice of generalized coordinates. (a) Find the Lagrangian and the two Lagrange equations. (b) Explain what the 0 equation tells us about the z component of angular momentum tz. (c) For the special case that 0 = const, describe what the 0 equation tells us. (d) Use the 0 equation to replace cb by tz in the 0 equation and discuss the existence of an angle 00 at which 0 can remain constant. Why is this motion called a conical pendulum? (e) Show that if 0 = 00 E, with c small, then 0 oscillates about 00 in harmonic motion. Describe the motion of the pendulum's bob. 7.41 *** Consider a bead of mass m sliding without friction on a wire that is bent in the shape of a parabola and is being spun with constant angular velocity w about its vertical axis, as shown in Figure Problems for Chapter 7 289 7.17. Use cylindrical polar coordinates and let the equation of the parabola be z = kp 2. Write down the Lagrangian in terms of p as the generalized coordinate. Find the equation of motion of the bead and determine whether there are positions of equilibrium, that is, values of p at which the bead can remain fixed, without sliding up or down the spinning wire. Discuss the stability of any equilibrium positions you find. Figure 7.17 Problem 7.41 7.42 *** [Computer] In Example 7.7 (page 264), we saw that the bead on a spinning hoop can make small oscillations about its nonzero stable equilibrium points that are approximately sinusoidal, with frequency Q' =,Ico 2 \u2014 (g I co R) 2 as in (7.80). Investigate how good this approximation is by solving the equation of motion (7.73) numerically and then plotting both your numerical solution and the approximate solution (t) = 00 + A cos(Q't \u2014 8) on the same graph. Use the following numbers: g = R = 1 and (1)2 = 2, and initial conditions 6(0) = 0 and 0 (0) = 00 + E0, where E0 = 1\u00b0. Repeat with co = 10\u00b0. Comment on your results. 7.43 *** [Computer] Consider a massless wheel of radius R mounted on a frictionless horizontal axis. A point mass M is glued to the edge, and a massless string is wrapped several times around the perimeter and hangs vertically down with a mass m suspended from its bottom end. (See Figure 4.28.) Initially I am holding the wheel with M vertically below the axle. At t = 0, I release the wheel, and m starts to fall vertically down. (a) Write down the Lagrangian = T \u2014 U as a function of the angle 0 through which the wheel has turned. Find the equation of motion and show that, provided m < M, there is one position of stable equilibrium. (b) Assuming m < M, sketch the potential energy U(0) for -7 < < 47 and use your graph to explain the equilibrium position you found. (c) Because the equation of motion cannot be solved in terms of elementary functions, you are going to solve it numerically. This requires that you choose numerical values for the various parameters. Take M = g = R = 1 (this amounts to a convenient choice of units) and m = 0.7. Before solving the equation make a careful plot of U (0) against 0 and predict the kind of motion expected when M is released from rest at 0 = 0. Now solve the equation of motion for 0 < t < 20 and verify your prediction. (d) Repeat part (c), but with m = 0.8. 7.44 *** [Computer] If you haven't already done so, do Problem 7.29. One might expect that the rotation of the wheel would have little effect on the pendulum, provided the wheel is small and rotates slowly. (a) Verify this expectation by", " solving the equation of motion numerically, with the following numbers: Take g and 1 to be 1. (This means that the natural frequency N/g1 1 of the pendulum is also 1.) Take co = 0.1, so that the wheel's rotational frequency is small compared to the natural frequency of the pendulum; and take the radius R = 0.2, significantly less than the length of the pendulum. As initial conditions take 0 = 0.2 and 0 = 0 at t = 0, and make a plot of your solution 0 (t) for 0 < t < 20. Your graph should look very like the sinusoidal oscillations of an ordinary simple pendulum. Does the 290 Chapter 7 Lagrange's Equations period look correct? (b) Now plot (t) for 0 < t < 100 and notice that the rotating support does make a small difference, causing the amplitude of the oscillations to grow and shrink periodically. Comment on the period of these small fluctuations. SECTION 7.8 More About Conservation Laws* 7.45 ** (a) Verify that the coefficients A i j in the important expression (7.94) for the kinetic energy of any \"natural\" system are symmetric; that is, A i j = A 11. (b) Prove that for any n variables v 1,, v\u201e jkViVk dv.. j,k [Hint: Start with the case that n = 2, for which you can write out the sums in full. Notice that you need the result of part (a).] This identity is useful in many areas of physics; we needed it to prove the expression (7.96) for the generalized momentum pi. 7.46 ** Noether's theorem asserts a connection between invariance principles and conservation laws. In Section 7.8 we saw that translational invariance of the Lagrangian implies conservation of total linear momentum. Here you will prove that rotational invariance of,C implies conservation of total angular momentum. Suppose that the Lagrangian of an N-particle system is unchanged by rotations about a certain symmetry axis. (a) Without loss of generality, take this axis to be the z axis, and show that the Lagrangian is unchanged when all of the particles are simultaneously moved from (ra, Oa, 0e) to (re, Oa, Oa E) (same c for all particles). Hence show that E (x=i aoc,.0. (b) Use Lagrange's equations to show that this implies that the total angular momentum L, about the symmetry axis is constant. In particular, if the Lagrangian is invariant under rotations about all axes, then all components of L are conserved. 7.47 *** In Chapter 4 (at the end of Section 4.7) I claimed that, for a system with one degree of freedom, positions of stable equilibrium \"normally\" correspond to minima of the potential energy U (q). Using Lagrangian mechanics, you can now prove this claim. (a) Consider a one-degree system of N particles with positions ra = ra (q), where q is the one generalized coordinate and the transformation between r and q does not depend on time; that is, q is what we have now agreed to call \"natural.\" (This is the meaning of the qualification \"normally\" in the statement of the claim. If the transformation depends on time, then the claim is not necessarily true.) Prove that the KE has the form T = 4./44 2, where A = A(q) > 0 may depend on q but not on 4. [This corresponds exactly to the result (7.94) for n degrees of freedom. If you have trouble with the proof here, review the proof there.] Show that the Lagrange equation of motion has the form A(q).4 = \u2014 d \u2014 d \u2014q. dU 1 d A.2 q 2 q (b) A point go is an equilibrium point if, when the system is placed at go with 4 = 0, it remains there. Show that qo is an equilibrium point if and only if dU I dq = 0. (c) Show that the equilibrium is stable if and only if U is minimum at go. (d) If you did Problem 7.30, show that the pendulum of that problem does not satisfy the conditions of this problem and that the result proved here is false for that system. Problems for Chapter 7 291 SECTION 7.9 Lagrange's Equations for Magnetic Forces * 7.48 ** Let F = F (q 1,, q n ) be any function of the generalized coordinates (q 1,, qn ) of a system with Lagrangian (q i,, qn, 4 1,, 4\u201e, t). Prove that the two Lagrangians 4, and L I = dF I dt give", " exactly the same equations of motion. 7.49 ** Consider a particle of mass in and charge q moving in a uniform constant magnetic field B in the z direction. (a) Prove that B can be written as B= V x A with A = 413 x r. Prove equivalently that in cylindrical polar coordinates, A = 1Bp b. (b) Write the Lagrangian (7.103) in cylindrical polar coordinates and find the three corresponding Lagrange equations. (c) Describe in detail those solutions of the Lagrange equations in which p is a constant. SECTION 7.10 Lagrange Multipliers and Constraint Forces* 7.50 * A mass m 1 rests on a frictionless horizontal table. Attached to it is a string which runs hori- zontally to the edge of the table, where it passes over a frictionless, small pulley and down to where it supports a mass m 2. Use as coordinates x and y the distances of m i and m 2 from the pulley. These satisfy the constraint equation f (x, y)=--x-Fy=- const. Write down the two modified Lagrange equa- tions and solve them (together with the constraint equation) for x, 53, and the Lagrange multiplier X. Use (7.122) (and the corresponding equation in y) to find the tension forces on the two masses. Verify your answers by solving the problem by the elementary Newtonian approach. 7.51 * Write down the Lagrangian for the simple pendulum of Figure 7.2 in terms of the rectangular coordinates x and y. These coordinates are constrained to satisfy the constraint equation f y) \u00b1 y2 = 1. (a) Write down the two modified Lagrange equations (7.118) and (7.119). Comparing Vx2 these with the two components of Newton's second law, show that the Lagrange multiplier is (minus) the tension in the rod. Verify Equation (7.122) and the corresponding equation in y. (b) The constraint equation can be written in many different ways. For example we could have written 1(x, y) x2 + y 2 = *2. t Check that using this function would have given the same physical results. 7.52 * The method of Lagrange multipliers works perfectly well with non-Cartesian coordinates. Consider a mass m that hangs from a string, the other end of which is wound several times around a wheel (radius R, moment of inertia I) mounted on a frictionless horizontal axle. Use as coordinates for the mass and the wheel x, the distance fallen by the mass, and 4, the angle through which the wheel has turned (both measured from some convenient reference position). Write down the modified Lagrange equations for these two variables and solve them (together with the constraint equation) for.3j and and the Lagrange multiplier. Write down Newton's second law for the mass and wheel, and use them to check your answers for x and 4i. Show that X of/ax is indeed the tension force on the mass. Comment on the quantity X afoo. CHAPTER Two-Body Central-Force Problems In this chapter, I shall discuss the motion of two bodies each of which exerts a conservative, central force on the other but which are subject to no other, \"external,\" forces. There are many examples of this problem: the two stars of a binary star system, a planet orbiting the sun, the moon orbiting the earth, the electron and proton in a hydrogen atom, the two atoms of a diatomic molecule. In most cases the true situation is more complicated. For example, even if we are interested in just one planet orbiting the sun, we cannot completely neglect the effects of all the other planets; likewise, the moon\u2014earth system is subject to the external force of the sun. Nevertheless, in all cases, it is an excellent starting approximation to treat the two bodies of interest as being isolated from all outside influences. You may also object that the examples of the hydrogen atom and the diatomic molecule do not belong in classical mechanics, since all such atomic-scale systems must really be treated by quantum mechanics. However, many of the ideas I shall develop in this chapter (the important idea of reduced mass, for instance) play a crucial role in the quantum mechanical two-body problem, and it is probably fair to say that the material covered here is an essential prerequisite for the corresponding quantum material. 8.1 The Problem Let us consider two objects, with masses m i and m2. For the purposes of this chapter, I shall assume the objects are small enough to be considered as point particles, whose positions (relative to the origin 0 of some inertial reference frame) I shall denote by r1 and r2. The only forces are the forces F12 and F21 of their mutual interaction, which I shall assume is conservative and central. Thus the forces can be derived", " from a potential energy U(r i, r2). In the case of two astronomical bodies (the earth and 293 294 Chapter 8 Two-Body Central-Force Problems the sun, for instance) the force is the gravitational force Gm 01 211r 1 \u2014 r2 1 2, with the corresponding potential energy (as we saw in Chapter 4) For the electron and proton in a hydrogen atom, the potential energy is the Coulomb PE of the two charges (e for the proton and \u2014e for the electron), U(r1, r2 ) = Gm im2 1ri \u2014 r2 1. U(r1, r2) = ke 2,, Ir i \u2014 r2 1 (8.1) (8.2) where k denotes the Coulomb force constant, k = 1/47rEo. In both of these examples, U depends only on the difference (r 1 \u2014 r2), not on r 1 and r2 separately. As we saw in Section 4.9, this is no accident: Any isolated system is translationally invariant, and if U(r i, r2) is translationally invariant it can only depend on (r 1 \u2014 r2). In the present case there is a further simplification: As we saw in Section 4.8, if a conservative force is central, then U is independent of the direction of (r 1 \u2014 r2). That is, it only depends on the magnitude 11.1 \u2014 r2 1, and we can write variable U(r1, r2) = U(1r 1 \u2014 r2 1) (8.3) as is the case in the examples (8.1) and (8.2). To take advantage of the property (8.3), it is convenient to introduce the new r = r 1 \u2014 r2. (8.4) As shown in Figure 8.1, this is just the position of body 1 relative to body 2, and I shall refer to r as the relative position. The result of the previous paragraph can be rephrased to say that the potential energy U depends only on the magnitude r of the relative position r, U = U (r). (8.5) Figure 8.1 The relative position r = r 1 \u2014 r2 is the position of body 1 relative to body 2. Section 8.2 CM and Relative Coordinates; Reduced Mass 295 We can now state the mathematical problem that we have to solve: We want to find the possible motions of two bodies (the moon and the earth, or an electron and a proton), whose Lagrangian is = 1m 2q: \u2014 U(r). (8.6) Of course, I could equally have stated the problem in Newtonian terms, and I shall in fact feel free to move back and forth between the Lagrangian and Newtonian formalisms according to which seems the more convenient. For the present, the Lagrangian formalism is the more transparent. 8.2 CM and Relative Coordinates; Reduced Mass Our first task is to decide what generalized coordinates to use to solve our problem. There is already a strong suggestion that we should use the relative position r as one of them (or as three of them, depending on how you count coordinates), because the potential energy U(r) takes such a simple form in terms of r. The question is then, what to choose for the other (vector) variable. The best choice turns out to be the familiar center of mass (or CM) position, R, of the two bodies, defined as in Chapter 3 to be R = mirt m2r2 = /flirt m2r2 m1 + m2 (8.7) where as before M denotes the total mass of the two bodies: M mi + m2. As we saw in Chapter 3, the CM of two particles lies on the line joining them, as shown in Figure 8.2. The distances of the center of mass from the two masses m 2 and m 1 are in the ratio m it m2. In particular, if m 2 is much greater than m l, then the CM is very close to body 2. (In Figure 8.2, the ratio m1/m2 is about 1/3, so the CM is a quarter of the way from m 2 to m 1.) Figure 8.2 The center of mass of the two bodies lies at the position R = (m il% m 2r2)/M on the line joining the two bodies. 296 Chapter 8 Two-Body Central-Force Problems We saw in Section 3.3 that the total momentum of the two bodies is the same as if the total mass M = m 1 + m2 were concentrated at the CM and were following the CM as it moves: P = Mk. (8.8) This result has important simplifying consequences: We know, of course, that the total momentum is constant. Therefore", ", according to (8.8), R is constant; and this means we can choose an inertial reference frame in which the CM is at rest. This CM frame is an especially convenient frame in which to analyze the motion, as we shall see. I am going to use the CM position R and the relative position r as generalized coordinates for our discussion of the motion of our two bodies. In terms of these coordinates, we already know that the potential energy takes the simple form U U (r). To express the kinetic energy in these terms, we need to write the old variables r 1 and r2 in terms of the new R and r. It is a straightforward exercise to show that (see Figure 8.2) r i = R + M2 \u2014r and r2 = R \u2014 \u2014r. (8.9) m i M Thus the kinetic energy is T (in i. I 1 \u2014 2,\u201e, 2) m21 2 1 \u2014 2 (m i 1:1--71-1T + m 2 R ml r l 2 = 1 (mk2 M1M2 i.2) 2 M (8.10) The result (8.10) simplifies further if we introduce the parameter which has the dimensions of mass and is called the reduced mass. You can easily check that.1, is always less than both m i and m 2 (hence the name). If m 1 << m 2, then,t t is very close to m i. Thus the reduced mass for the earth\u2014sun system is almost exactly the mass of the earth; the reduced mass of the electron and proton in hydrogen is almost exactly the mass of the electron. On the other hand, if m i = m 2, then obviously = 4m Returning to (8.10), we can rewrite the kinetic energy in terms of\u00b5 as T = m \u00b1 1,42. 2 P\"' 2 (8.12) This remarkable result shows that the kinetic energy is the same as that of two \"fictitious\" particles, one of mass M moving with the speed of the CM, and the other Section 8.3 The Equations of Motion 297 of mass (the reduced mass) moving with the speed of the relative position r. Even more significant is the corresponding result for the Lagrangian: L =T\u2014 U = 1M142 (1,c42 \u2014 U (r)) = m + 'Crel. c (8.13) We see that by using the CM and relative positions as our generalized coordinates, we have split the Lagrangian into two separate pieces, one of which involves only the CM coordinate R and the other only the relative coordinate r. This will mean that we can solve for the motions of R and r as two separate problems, which will greatly simplify matters. 8.3 The Equations of Motion With the Lagrangian (8.13), we can write down the equations of motion of our two- body system. Because E., is independent of R, the R equation (really three equations, one each for X, Y, and Z) is especially simple, MR = 0 or R = const. (8.14) We can explain this result in several ways: First (as we already knew), it is a direct consequence of conservation of total momentum. Alternatively, we can view it as reflecting that is independent of R, or, in the terminology introduced in Section 7.6, the CM coordinate R is \"ignorable.\" More specifically, = 4MR 2 (which is the only part of E, that involves R) has the form of the Lagrangian of a free particle of mass M and position R. Naturally, therefore (Newton's first law), R moves with constant velocity. The Lagrange equation for the relative coordinate r is a little less simple but equally beautiful: E.,rel, the only part of E, that involves r, is mathematically indistinguishable from the Lagrangian for a single particle of mass itt and position r, with potential energy U(r). Thus the Lagrange equation corresponding to r is just (check it and see!) energy U(r). = \u2014VU(r). (8.15) To solve for the relative motion, we have only to solve Newton's second law for a single particle of mass equal to the reduced mass\u00b5 and position r, with potential The CM Reference Frame Our problem becomes even easier to. think about if we make a clever choice of reference frame. Specifically, because R = const, we can choose an inertial reference frame, the so-called CM frame, in which the CM is at rest and the total momentum 298 Chapter 8 Two-Body Central-Force Problems Figure 8.3 In the CM frame the center of mass is stationary at the origin. The relative position r is the position of particle 1 relative to particle 2; therefore, the position of particle 1 relative to the origin is r1 = (m 2/M)r. is", " zero. In this frame, R = 0 and the CM part of the Lagrangian is zero = 0). Thus in the CM frame = Zrei = \u2014 U(r) (8.16) and the problem really is reduced to a one-body problem. This dramatic simplifica- tion illustrates the curious terminology of the \"ignorable coordinate.\" Recall that a coordinate qi is said to be ignorable if a,C/aqi = 0. We see that, in the present case at least, the motion associated with the ignorable coordinate R really is something that we can ignore. It is worth taking a moment to consider what the motion looks like in the CM frame, as shown in Figure 8.3. The CM is stationary, and we naturally take it to be the origin. Both particles are moving, but with equal and opposite momenta. If m 2 is much greater than m 1 (as is often the case), the CM is close to m 2 and particle 2 has a small velocity. (In the figure, m2 = 3m 1 and hence v 2 = 3 v 1.) It is important to note that the relative position r is the position of particle 1 relative to particle 2, and is not the actual position of either particle. As shown in the picture, the position of particle 1 is actually r 1 = (m 2 /M)r. However, if m 2 >> m l, then the CM is very close to particle 2, which is almost stationary, and r 1 ti r; that is, r is very nearly the same thing as r 1. The equation of motion in the CM frame is derived from the Lagrangian Zrel of (8.16) and is just Equation (8.15). This is precisely the same as the equation for a single particle of mass equal to the reduced mass A, in the fixed central force field of the potential energy U(r). In the equations of this chapter, the repeated appearance of the mass au, will serve to remind you that the equations apply to the relative motion of two bodies. However, you may find it easier to visualize a single body (of mass g) orbiting about a fixed force center. In particular, if m2 >> m 1, these two problems are for practical purposes exactly the same. Moreover, if your interest actually is in a single body, of mass m say, orbiting a fixed force center, then you can use all of the same equations, simply replacing,u, with m. In any event, any solution for the relative coordinate r(t) always gives us the motion of particle 1 relative to particle 2. Equivalently, using the relations of Figure 8.3, knowledge of r(t) tells us the motion of particle 1 (or particle 2) relative to the CM. Section 8.3 The Equations of Motion 299 Conservation of Angular Momentum We already know that the total angular momentum of our two particles is conserved. Like so many other things, this condition takes an especially simple form in the CM frame. In any frame, the total angular momentum is L r 1 x p i r2 X p2 m ir i x m 2r2 x f2. (8.17) In the CM frame, we see from (8.9) (with R = 0) that r 1 = 1112 \u2014r and r2 \u2014 ---r. m 1 (8.18) Substituting into (8.17), we see that the angular momentum in the CM frame is L (m 2r x m ir x im2 m2 r x (8.19) where I have replaced m im2/M by the reduced mass pt. The most remarkable thing about this result is that the total angular momentum in the CM frame is exactly the same as the angular momentum of a single particle with mass,a and position r. For our present purposes the important point is that, because angular momentum is conserved, we see that the vector r x r is constant. In particular, the direction of r x r is constant, which implies that the two vectors r andr remain in a fixed plane. That is, in the CM frame, the whole motion remains in a fixed plane, which we can take to be the xy plane. In other words, in the CM frame, the two-body problem with central conservative forces is reduced to a two-dimensional problem. The Two Equations of Motion To set up the equations of motion for the remaining two-dimensional problem, we need to choose coordinates in the plane of the motion. The obvious choice is to use the polar coordinates r and 0, in terms of which the Lagrangian (8.16) is 14 07.2 r2(p) U(r). (8.20) Since this Lagrangian is independent of 0, the coordinate 0 is ignorable, and the Lagrange equation corresponding to 0 is just aE., const = \u00a3 [0 equation]. (8", ".21) Since,u,r2(i) is the angular momentum Q (strictly, the z component t z ), the 0 equation is just a statement of conservation of angular momentum. The Lagrange equation corresponding to r (often called the radial equation) is a,c d ar., ar = dt 300 Chapter 8 Two-Body Central-Force Problems or 442 dU dr [r equation]. (8.22) As we already saw in Example 7.2 [Equations (7.19) and (7.20)], if we move the centripetal term over over to the right, this is just the radial component of F = ma (or rather, F =!La, since p, has replaced m). 8.4 The Equivalent One-Dimensional Problem The two equations of motion that we have to solve are the 0 equation (8.21) and radial equation (8.22). The constant 2 (the angular momentum) in the 0 equation is determined by the initial conditions, and our main use for the 0 equation is to solve it for 40, rb \u2014 2' which will let us eliminate p from the radial equation in favor of the constant \u00a3. The radial equation can be rewritten as. = dU dr, Arq5 - = dU dr Fcf which has the form of Newton's second law for a particle in one dimension with mass,u and position r, subject to the actual force \u2014dUldr plus a \"fictitious\" outward centrifugal force' Fcf = \u00b5r 2. (8.25) In other words, the particle's radial motion is exactly the same as if the particle were moving in one dimension, subject to the actual force \u2014dU/dr plus the centrifugal force Fcf. We have now reduced the problem of the relative motion of two bodies to a single one-dimensional problem, as expressed by (8.24). Before we discuss what the solutions are going to look like, it is helpful to rewrite the centrifugal force, using the 0 equation (8.23) to eliminate q5 in favor of the constant \u00a3, Even better, we can now express the centrifugal force in terms of a centrifugal potential energy, Fcf = \u00a32. aux' Fcf = d \u00a32 dr 2,u,r2 dUcf dr (8.23) (8.24) (8.26) (8.27 1 This centrifugal force may be a little more familiar if I write it in terms of the azimuthal velocit vo = rci) as Fcf = p,v021r. Section 8.4 The Equivalent One-Dimensional Problem 301 where the centrifugal potential energy fief is defined as \u00a32, Ucf (r) = 2\u00b5r2. (8.28) Returning to (8.24), we can now rewrite the radial equation in terms of Ucf as where the effective potential energy Ueff (r) is the sum of the actual potential energy U (r) and the centrifugal (fa (r): According to (8.29), the radial motion of the particle is exactly the same as if the particle were moving in one dimension with an effective potential energy Ueff = U \u00b1 Uef. EXAMPLE 8.1 Effective Potential Energy for a Comet Write down the actual and effective potential energies for a comet (or planet) moving in the gravitational field of the sun. Sketch the three potential energies involved and use the graph of Ueff (r) to describe the motion of r. Since planetary motion was first described mathematically by the German astronomer Johannes Kepler, 1571-1630, this problem of the motion of a planet or comet around the sun (or any two bodies interacting via an inverse-square force) is often called The actual gravitational potential energy of the comet is given by the well- the Kepler problem. known formula where G is the universal gravitational constant, and m 1 and m 2 are the masses of the comet and the sun. The centrifugal potential energy is given by (8.28), so the total effective potential energy is U (r) = G m im 2 r Ueff (r) = \u00a32 Gm im2 +. 2 r 2,u, r (8.31) (8.32) The general behavior of this effective potential energy is easily seen (Figure 8.4). When r is large, the centrifugal term f 2/2,ur 2 is negligible compared to the gravitational term \u2014Gm im 2 1 r, and the effective PE, Ueff(r), is negative and 302 Chapter 8 Two-Body Central-Force Problems Figure 8.4 The effective potential energy Ueff (r) that governs the radial motion of a comet is the sum of the actual gravitational potential energy U(r) = \u2014Gm i m 2 Ir and the centrifugal term (Id = 22 /2\u00b5r2", ". For large r, the dominant effect is the attractive gravitational force; for small r, it is the repulsive centrifugal force. sloping up as r increases. According to (8.29), the acceleration of r is down this slope. [The roller coaster car accelerates down the track defined by Ueff (r).] Thus when a comet is far from the sun, F is always inward. When r is small, the centrifugal term t 2 /2p,r 2 dominates the gravitational term \u2014Gm im 2I r (unless \u00a3 = 0), and near r = 0, Ueff (r) is positive and slopes downward. Thus, as a comet gets closer to the suns eventually becomes outward, and the comet starts to move away from the sun again. The one exception to this statement is when the angular momentum is exactly zero, = 0, in which case (8.23) implies that q.5 = 0; that is, the comet is moving exactly radially, along a line of constant 4), and must at some time hit the sun. Conservation of Energy To find the details of the orbit we must look more closely at the radial equation (8.29). If we multiply both sides of that equation by i-, we find that In other words, d (1.2) \u2014dt 2. 11f ) = \u2014 (Jeff V dt ). d, 1 -2 Ueff (r) = const. (8.33) (8.34) This result is, in fact, just conservation of energy: If we write out Ueff as U \u00a32/2\u00b5r2 and replace \u00a3 by,u,r20, we see that 46..2 + Ueff(r) = 1107.2 + _41x2(i.,12 U(r) = E. (8.35) Section 8.4 The Equivalent One-Dimensional Problem 303 This completes the rewriting of the two-dimensional problem of the relative motion as an equivalent one-dimensional problem involving just the radial motion. We see that the total energy (which we knew all along is constant) can be thought of as the one- dimensional kinetic energy of the radial motion, plus the effective one-dimensional potential energy Ueff, since the latter includes the actual potential energy U and the kinetic energy 1p,r 2(j)2 of the angular motion. This means that all of our experience with one-dimensional problems, both in terms of forces and in terms of energy, can be immediately transferred to the two-body central-force problem. EXAMPLE 8.2 Energy Considerations for a Comet or Planet Examine again the comet (or planet) of Example 8.1 and, by considering its total energy E, find the equation that determines the maximum and minimum distances of the comet from the sun, if E > 0 and, again, if E < 0. In the energy equation (8.35) the term z F.cr2 on the left is always greater than or equal to zero. Therefore, the comet's motion is confined to those regions where E > Ueff. To see what this implies, I have redrawn in Figure 8.5 the graph of Ueff from Figure 8.4. Let us consider first the case that the comet's energy is greater than zero. In the figure I have drawn a dashed horizontal line at height E, labeled E > 0. A comet with this energy can move anywhere that this line is above the curve of Ueff (r), but nowhere that the line is below the curve. This means simply that the comet cannot move anywhere inside the turning point labeled rnnn, determined by the condition Ueff (rmin ) = E. (8.36) If the comet is initially moving in, toward the sun, then it will continue to do so until it reaches rmin, where i = 0 instantaneously. It then moves outward, and, since there are no other points at which 1- can vanish, it eventually moves off to infinity, and the orbit is unbounded. If instead E < 0, then the line drawn at height E (labeled E < 0) meets the curve of Ueff (r ) at the two turning points labeled r nnn and rni,\u201e and a comet with cA 0 Figure 8.5 Plot of the effective potential energy Ueff (r) against r for a comet. For a given energy E, the comet can only go where E > Ueff (r). For E > 0 this means it cannot go inside the turning point at r inin where Ueff = E. For E < 0 it is confined between the two turning points labeled rm., and rmax. 304 Chapter 8 Two-Body Central-Force Problems E < 0 is trapped between these two values of r. If it is moving away from the sun (i- > 0) it continues to do so until it reaches rmax, where 1- van", "ishes and reverses sign. The comet then moves inward until it reaches r mia, where 1- reverses again. Therefore, the comet oscillates in and out between r min and rmax. For obvious reasons, this type of orbit is called a bounded orbit. 2 Finally, if E is equal to the minimum value of Ueff (r) (for a given value of the angular momentum ), the two turning points r mm and rmax coalesce, and the comet is trapped at a fixed radius and moves in a circular orbit. In this example, I considered just the case of an inverse-square force, but many two- body problems have the same qualitative features. For example, the motion of the two atoms in a diatomic molecule is governed by an effective potential that was sketched in Figure 4.12 and looks very like the gravitational curve of Figure 8.5. Thus all of our qualitative conclusions apply to the diatomic molecule and many other two-body problems. In thinking about the radial motion of the two-body problem, you must not entirely forget the angular motion. According to (8.23), 4 = /,u,r2, and 4) is always changing, always with the same sign (continually increasing or continually decreasing). For example, as a comet with positive energy approaches the sun, the angle 4) changes, at a rate that increases as r gets smaller; as the comet moves away, q5 continues to change in the same direction, but at a rate that decreases as r gets larger. Thus the actual orbit of a positive-energy comet looks something like Figure 8.6. For the case of an inverse- square force (like gravity), the orbit of Figure 8.6 is actually a hyperbola, as we shall prove shortly, but the unbounded orbits (that is, orbits with E > 0) are qualitatively similar for many different force laws. For the bounded orbits (E < 0), we have seen that r oscillates between the two extreme values ram, and rmax, while 0 continually increases (or decreases, but let's suppose the comet is orbiting counter-clockwise, so that 0 is increasing). In the case of the inverse-square force, we shall see that the period of the radial oscillations happens to equal the time for 0 to make exactly one complete revolution. Therefore, the motion repeats itself exactly once per revolution, as in Figure 8.7(a). (We shall also see that, for any inverse-square force, the bounded orbits are actually ellipses.) For most other force laws, the period of the radial motion is different from the time to make one revolution, and in most cases the orbit is not even closed (that is, it never returns to its initial conditions). 3 Figure 8.7(b) shows an orbit for which r goes from rmin to I-max and back to rima in the time that the angle 0 advances by about 330\u00b0, and the orbit certainly does not close on itself after one revolution. 2 If we consider just one comet in orbit around the sun, then energy conservation implies that a bounded orbit (E < 0) can never change into an unbounded orbit (E > 0), nor vice versa. In reality a comet can occasionally come close enough to another comet or planet to change E, and the orbit can then change from bounded to unbounded or the other way. 3 Besides the inverse square force, the only important exception is the isotropic harmonic oscillator, for which the orbits are also ellipses, as discussed in Section 5.3. Section 8.5 The Equation of the Orbit 305 rmin Figure 8.6 Typical unbounded orbit for a positive-energy comet. Initially r decreases from infinity to r min and then goes back out to infinity. Meanwhile the angle 0 is continually increasing. rmin (a) (b) Figure 8.7 (a) The bounded orbits for any inverse-square force have the unusual property that r goes from rmin to rmax and back to \"-min in exactly the time that 0 goes from 0 to 360\u00b0. Therefore the orbit repeats itself every revolution. (b) For most other force laws, the period of oscillation of r is different from the time in which 0 advances by 360\u00b0, and the orbit does not close on itself after one revolution. In this example, r completes one cycle from rmin to rmax and back to rm. while 0 advances by about 330\u00b0. 8.5 The Equation of the Orbit The radial equation (8.29) determines r as a function of t, but for many purposes we would like to know r as a function of 0. For example, the function r = r (0) will tell us the shape of the orbit more directly. Thus we would like to rewrite the radial equation 306 Chapter 8 Two-Body Central-Force Problems as a differential equation for r in terms of 4. There are two tricks for doing this, but let me first", " write the radial equation in terms of forces: (8.37) (8.38) (8.39) where F(r) is the actual central force, F = \u2014dU/dr, and the second term is the centrifugal force. The first trick to rewriting this equation in terms of is to make the substitution = F(r) t2 pr 3 u = \u2014 or 1 r 1 = \u2014 and the second is to rewrite the differential operator d I dt in terms of d/d0 using the chain rule: d \u2014 d dt dt dtt d d Lug d,u,r2 \u2014 do (The third equality follows because \u00a3 = px 2(.P, and the last results from the change of variables u = 1/r.) Using the identity (8.39) we can rewriter on the left of the radial equation. First. r = \u2014(r ) = d dt tu 2 d (1) p, dcb u \u2014 = --- t du p dq5 and hence Or d. r = \u2014(r) dt p, d\u00a2 Lu 2 d \u00a3 du) \u00a32 u 2 d2 u p2 de (8.40) Substituting back into the radial equation (8.37) we find f2 u 2,42\u201e p2\u201e3 Lti4 =F+.1-(4 d02 For any given central force F, this transformed radial equation is a differential equation for the new variable u(0). If we can solve it, then we can immediately write down r as r = 1/u. In the next section, we shall solve it for the case of an inverse- square force and show that the resulting orbits are conic sections, that is, ellipses, parabolas, or hyperbolas. First, here is a simpler example. Section 8.5 The Equation of the Orbit 307 line. 1/ ro) EXAMPLE 8.3 The Radial Equation for a Free Particle Solve the transformed radial equation (8.41) for a free particle (that is, a particle subject to no forces) and confirm that the resulting orbit is the expected straight This example is probably one of the hardest ways of showing that a free particle moves along a straight line. Nevertheless, it is a nice check that the transformed radial equation makes sense. In the absence of forces, (8.41) is just u\"(0) = \u2014u(0) whose general solution we know to be u (0) = A cos(4 \u2014 3), (8.42) where A and 6 are arbitrary constants. Therefore, (renaming the constant A = 1 r (0) = = 0 u(0) cos(4 \u2014 6) (8.43) This unpromising-looking equation is in fact the equation of a straight line in polar coordinates, as you can see from Figure 8.8. In that picture Q is a fixed point with polar coordinates (r0, 6), and the line in question is the line through Q perpendicular to OQ. It is easy to see that the point P with polar coordinates (r, 0) lies on this line if and only if r cos(0 \u2014 6) = ro. In other words, Equation (8.43) is the equation of this straight line. Figure 8.8 The fixed point Q has polar coordinates (ro, 6) rel- ative to the origin 0. The point P with polar coordinates (r, 0) lies on the line through Q perpendicular to OQ if and only if r cos(0 \u2014 8) = ro. That is, the equation of this line is (8.43). 308 Chapter 8 Two-Body Central-Force Problems In the next section, I shall use the same transformed radial equation (8.41) to solve a much less trivial problem, finding the path of a comet or any other body held in orbit by an inverse-square force. 8.6 The Kepler Orbits Let us now return to the Kepler problem, the problem of finding the possible orbits of a comet or any other object subject to an inverse-square force. The two important examples of this problem are the motion of comets or planets around the sun (or earth satellites around the earth), in which case the force is the gravitational force \u2014Gm 1m2 /r 2, and the orbital motion of two opposite charges g l and q2, in which case the force is the Coulomb force km 21 r2. To include both cases and to simplify the equations, I shall write the force as (remember that u = 1/r) F (r) = \u2014 \u2014 = \u2014yu 2, Y r 2 where y is the \"force constant,\" equal to Gm 1m 2 in the gravitational case. 4 Thanks to our elaborate preparations, we can now solve the main problem very easily. Inserting the force (8.44) into", " the transformed radial equation (8.41), we find that u (0) must satisfy u\"(0) = -um+ yktif 2 Notice that it is a unique feature of the inverse-square force that the last term in this equation is a constant, since only in this case does the u 2 of the force cancel the 1/u 2 (8.44) (8.45) in (8.41). Because this last term is constant, we can solve (8.45) very easily: If we substitute the equation becomes w(0) = u(0) Yttg 2, w\"(0) = \u2014 w(0), which has the general solution w(0) = A cos(q5 \u2014 8), (8.46) where A is a positive constant and 8 is a constant that we can take to be zero by a suitable choice of the direction 4) = 0. Thus the general solution for u (0) can be written as u(0) = A cos \u2014 = (1 + E COSO (8.47) Ytt \u00a32 4 The constant y is positive for the gravitational force and for the force between two opposite charges. As discussed in Problem 8.31, for two charges of the same sign, y is negative. For now, we'll assume it is positive. Section 8.6 The Kepler Orbits 309 where E is just a new name for the dimensionless positive constant Af t /y\u00b5. Since u = 1/r, the constant y\u00b5/t 2 on the right has the dimensions [1/length], and I shall introduce the length C = \u2014 \u00a32 Y (8.48) in terms of which our solution becomes 1 c r(1 (0) = - (1 \u00b1 E cos 0) or r(0) = 1 + e cos 0 (8.49) This is our solution for r as a function of 0, in terms of the undetermined positive constant E and the length c = e 2 /y/i, (which is \u00a3 2 / Gm im 2p, in the gravitational problem). I shall now explore its properties, first for the bounded orbits and then for the unbounded. The Bounded Orbits The behavior of the orbit r(0) in (8.49) is controlled by the as-yet undetermined positive constant E. A glance at (8.49) shows this behavior is very different according as E < 1 or E > 1. If E < 1, the denominator of (8.49) never vanishes, and r (0) remains bounded for all 0. If E > 1 the denominator vanishes at some angle, and r (0) approaches infinity as 0 approaches that angle. Evidently the value E = 1 is the boundary between the bounded and unbounded orbits. I shall show shortly that this boundary corresponds exactly to the boundary between E < 0 and E > 0 discussed before. Meanwhile, let us start with the case that the constant c is less than 1. With E < 1, the denominator of r (0) in (8.49) oscillates as shown in Figure 8.9 between the values 1 \u00b1 E. Therefore, r (0) oscillates between rmin = 1+ E and rmax = 1 - E (8.50) with r = r min at the so-called perihelion when 0 = 0, and r = rmax at the aphelion when 0 = 7. Since r(0) is obviously periodic in 0 with period 27, it follows that r(27r) = r(0) and the orbit closes on itself after just one revolution. Thus the general appearance of the orbit is as in Figure 8.10. While the orbit shown in Figure 8.10 certainly looks like an ellipse, I have not yet proved that it really is. However, it is a reasonably easy exercise (see Problem 8.16) to rewrite (8.49) in Cartesian coordinates and cast it in the form (x + d) + y 2 2 a 2 = 1 b2 (8.51) 310 Chapter 8 Two-Body Central-Force Problems Figure 8.9 The denominator 1 + 6 cos 0 in Equation (8.49) for r(0) oscillates between 1 + E and 1 \u2014 E, and is periodic with period 27r. where (as you can easily check) a = 1 \u2014 6 2 b =, \u2014 62 and d = ac. (8.52) Equation (8.51) is the standard equation of an ellipse with semimajor and semiminor axes a and b, except that where we expect to see x we have x d. This difference reflects that our origin, the sun, is not at the center of the ellipse, but at a distance d from it, as shown in Figure 8.10. We can now identify the", " constant E, which started life as an undetermined constant of integration in (8.47). According to (8.52) the ratio of the major to minor axes is /1\u2014 6 2. \u2014 a (8.53) aphelion Figure 8.10 The bounded orbits of a comet or planet as given by Equation (8.49) are ellipses. The sun is at the origin 0, which is one focus of the ellipse (not the center C). The distances a and b are called the semimajor and semiminor axes. The parameter c = \u00a32/y\u00b5 introduced in (8.48) is the value of r when 0 = 90\u00b0. The points where the comet is closest and farthest from the sun are called the perihelion and aphelion. Section 8.6 The Kepler Orbits 311 Although you almost certainly don't remember it, this equation is the definition (or one possible definition) of the eccentricity of the ellipse. That is, this equation tells us that the constant E is the eccentricity. Notice that if E = 0, then b = a and the ellipse is a circle; if c 1, then b/a \u2014> 0 and the ellipse becomes very thin and elongated. Having identified the constant E as the eccentricity, we can now identify the position of the sun in relation to the ellipse. According to (8.52) the distance from the center C to the sun at 0 is d = ac, and (though again you may not remember it) ac. is the distance from the center to either focus of the ellipse. Thus the position of the sun is actually one of the ellipse's two focuses, and we have now proved Kepler's first law, that the planets (and comets whose orbits are bounded) follow orbits that are ellipses with the sun at one focus. EXAMPLE 8.4 Halley's Comet Halley's comet, named for the English astronomer Edmund Halley (1656- 1742), follows a very eccentric orbit with 6 = 0.967. At closest approach (the perihelion) the comet is 0.59 AU from the sun, fairly close to the orbit of Mercury. (The AU or astronomical unit is the mean distance of the earth from the sun, about 1.5 x 10 8 km.) What is the comet's greatest distance from the sun, that is, the distance of the aphelion? The given distance is rmin = 0.59 AU, and, according to (8.50), rmax/rmin (1 + E)/(1 \u2014 E). Therefore rmax \u2014 rmin = 1+ 6 1 \u2014 1.967 r min 0.033 min 60 rmin = 35 AU. This means that at its greatest distance Halley's comet is outside the orbit of Neptune. The Orbital Period; Kepler's Third Law We can now find the period of the elliptical orbits of the comets and planets. According to Kepler's second law (Section 3.4), the rate at which a line from the sun to a comet or planet sweeps out area is Since the total area of an ellipse is A = nab, the period is dA dt r A 27rab,a dA/dt \u00a3 If we square both sides and use (8.53) to replace b 2 by a 2 (1 \u2014 E 2), this becomes 2 = 47 2 a 4 ( 1 \u20ac 2)P,2 = 47 2 a 3c712 \u00a32 \u00a32 312 Chapter 8 Two-Body Central-Force Problems where in the last equality I used (8.52) to replace a(1 \u2014 E 2 ) by c. Since the length c was defined in (8.48) as e2 I yp,, this implies that r 2 = 47 2\" F\".,3\u201e (8.54) Finally, y is the constant in the inverse-square force law F = \u2014y/r2, and, for the gravitational force, y = Gm im 2 = GAM where M is the total mass, M = m 1 + m2. (Notice the handy identity that m im2 = AM.) In our case m 2 = Ms, the mass of the sun, which is very much greater than m l, the mass of the comet or planet. Thus, to an excellent approximation, M ti M s, and y = Gm lm 2 = GAM,. Therefore, the factor of p, in (8.54) cancels, and we find that This is Kepler's third law: Because the mass of the comet (or planet) has canceled out, the law says that for all bodies orbiting the sun, the square of the period is proportional to the cube of the semimajor axis. (For circular orbits, we can replace a 3 by r3.) The law applies equally", " to the satellites of any large body. For example, all satellites of the earth, including the moon, obey the same law [with Ms replaced by the earth's mass Me in (8.55)], and the same applies to all the moons of Jupiter. EXAMPLE 8.5 Period of a Low -Orbit Earth Satellite Use Kepler's third law to estimate the period of a satellite in a circular orbit a few tens of miles above the earth's surface. The period is given by (8.55) with Ms replaced by Me. Since the orbit is circular, we can replace a by r, and since the orbit is close to the earth's surface, r ti Re, the radius of the earth. Therefore 2 T = 472 Re3. GM, This simplifies if we recall that GM e /Re2 = g, the acceleration of gravity on the earth's surface, and we find that r = 27r \u2014 = 27r = 5070 s ti 85 min, (8.56) Re g 6.38 x 106 m 9.8 m/s2 in agreement with the well-known observation that low-orbit satellites circle the earth in about one and a half hours. Section 8.7 The Unbounded Kepler Orbits 313 Relation between Energy and Eccentricity Finally, we can relate the eccentricity E of the orbit to the energy E of the comet or other orbiting body. The simplest way to do this is to remember that, at its distance of closest approach rmin, the comet's energy is equal to the effective potential energy Ueff [Equation (8.36)], E = Ueff (rmin) = 22 rmin 2iurrain2 1,e2 2r min Armin 2y) (8.57) Now we know from (8.50) that r min = c/ (1 + E), and from its definition (8.48) that c = 2 2 /y,u. Therefore and, substituting into (8.57), 22 rmin \u2014 yu,(1 E) [(1 + c) \u2014 2y] yiu (1 + E) E = 22 2 = Y (E 2 \u2014 1). 2 2E2 (8.58) The calculations leading to (8.58) are equally valid for bounded and unbounded orbits, and they imply the following expected correlations: Negative energies (E < 0) correspond to eccentricities E < 1, which in turn correspond to bounded orbits. Posi- tive energies (E > 0) correspond to eccentricities E > 1, which in turn correspond to unbounded orbits. Equation (8.58) is a useful relation between the mechanical proper- ties E and f and the geometrical property E. It implies some interesting connections. For example, for a given value of the angular momentum E, the orbit of lowest pos- sible energy is the circular orbit with E = 0 (a connection which has an important counterpart in quantum mechanics). 8.7 The Unbounded Kepler Orbits In the previous section, we found the general Kepler orbit, as given by (8.49), r (0) = 1 + E cos 0 (8.59) and examined in detail the bounded orbits \u2014 those for which c < 1 or, equivalently, as we have seen, E < 0. In this section, I shall sketch the corresponding analysis of the unbounded orbits, with E > 1 and E > 0. The boundary between the bounded and unbounded orbits comes when E = 1 or E = 0. With E = 1, the denominator of (8.59) vanishes when 0 = +7r. Therefore, 314 Chapter 8 Two-Body Central-Force Problems hyperbola (\u20ac > 1) parabola (c = 1) \\ ellipse (0 < < 1) Figure 8.11 Four different Kepler orbits for a comet: a circle, an ellipse, a parabola, and a hyperbola. For clarity, the four orbits were chosen with the same values for and with the closest approaches all in the same direction. r(q0) oo as 0 \u00b17r. That is, if c = 1, the orbit is unbounded and goes off to infinity as the comet approaches 0 = +7r. Some elementary algebra, parallel to what led to (8.51), shows that with 6 = 1 the Cartesian version of (8.59) is y 2 = C 2 \u2014 2cx (8.60) which is the equation of a parabola. This orbit is shown (with the long dashes) in If E > 1 (or E > 0), the denominator of (8.59) vanishes at a value max determined Figure 8.11. by the condition E COS(Omax) = \u20141. Thus r(4)) ---> oc when cb ---> \u00b1Omax and the", " orbit is confined to the range of angles < 4 < 0max This gives the orbit the general appearance sketched in Figure 8.6. With c > 1 the Cartesian form of (8.59) is (Problem 8.30) \u2014 3) 2 a 2 y2 /32 = 1, (8.61) where you can easily identify the constants a,,6, and 3 (Problem 8.30). This is the equation of a hyperbola, and we have proved that, as anticipated, the positive energy Kepler orbits are hyperbolas. One such orbit is shown (with the smaller dashes) in Section 8.8 Changes of Orbit 315 Figure 8.11. Summary of Kepler Orbits Our results for the Kepler orbits can be summarized as follows: All of the possible orbits are given by Equation (8.59), and are characterized by the two constants of integration s E and c. The dimensionless constant 6 is related to the comet's energy by (8.58), r (0) = 1 \u00b1 c cos 0 E 2 Y 2f2 (E 2 \u2014 1). (8.62) (8.63) It is, as we have seen, the eccentricity of the orbit that determines the orbit's shape as follows: eccentricity energy orbit 6 = 0 E < 0 circle 0< E< 1 E< 0 ellipse E = 1 > 1 E = 0 parabola E > 0 hyperbola You can see from (8.62) that the constant c is a scale factor that determines the size of the orbit. It has the dimensions of length and is the distance from sun to comet when = 3r/2. It is equal to \u00a3 2 /yp, or, since y is the force constant Gm im2, \u00a32 c= Gm im 2p, (8.64) where m 1 is the mass of the comet, m 2 that of the sun, and p, is the reduced mass p, = m im 2 /(m i + m 2), which is exceedingly close to m 1 since m 2 is so large. 8.8 Changes of Orbit In this final section, I shall discuss how a satellite can change from one orbit to another. For example, a spacecraft wishing to visit Venus may want to transfer from a circular 5 Since Newton's second law is a second-order differential equation and the motion is in two dimensions, there are actually four constants of integration in all. The third is the constant S in (8.46) which we chose to be zero, forcing the axis of the orbit to be the x axis. The fourth is the comet's position on the orbit at time t = 0. 316 Chapter 8 Two-Body Central-Force Problems orbit close to the earth and centered on the sun to an elliptical orbit that will carry it to the orbit of Venus. Another example, and the one we shall discuss here, is an earth satellite wishing to change from one orbit about the earth to another, perhaps from a circular orbit to an elliptical orbit that will carry it to a higher altitude. The analysis of earth orbits is the same as that of orbits around the sun, except that the mass M, of the sun must be replaced by the mass Me of the earth, and the closest and furthest points from earth are called the perigee and apogee (instead of perihelion and aphelion for the sun). We shall confine attention to bounded, elliptical orbits, for which the most general form is r (0) = 1 \u00b1 E cos(0 \u2014 6) (8.65) (As long as we were interested in just one orbit, we could choose our x axis so that the angle 6 was zero. If we're interested in two arbitrary orbits, we cannot get rid of 6 in this way \u2014 anyway not for both.) Let us suppose that our spacecraft is initially in a orbit of the form (8.65) with energy E 1, angular momentum and orbital paramenters c 1, E 1, and 6 1. A common way to change orbits is for the spacecraft to fire its rockets vigorously for a brief time. To a good approximation we can treat this procedure as an impulse that occurs at a unique angle 00 and causes an instantaneous change of velocity by a known amount. From the known change in velocity, we can calculate the new energy E2 and angular momentum \u00a3 2. From (8.48) we can calculate the new value of c 2, and from (8.58) the new eccentricity E 2. Finally, because the new orbit must join onto the old one at the angle 0o, that is, r 1 (00) = r2 (00), we can find 6 2 from the equation Cl C2 1 + E i cos(00 \u2014 6 1) 1 + E2 COS(00 - 82) (8.66) This calculation, though straightforward in principle, is tedious, and not especially illuminating, in", " practice. To simplify the calculations and to better reveal the important features, I shall treat just one important special case. A Tangential Thrust at Perigee Let us consider a satellite that transfers from one orbit to another by firing its rockets in the tangential direction, forward or backward, when it is at the perigee of its initial orbit. By choice of our x axis, we can arrange that this occurs in the direction = 0, so that 00 = 0 and 6 1 = 0. Moreover, because the rockets are fired in the tangential direction, the velocity just after firing is still in the same direction, which is perpendicular to the radius from earth to the satellite. Therefore, the position at which the rockets are fired is also the perigee for the final orbit, 6 and 62 = 0 as well. Thus the equation (8.66) that assures the continuity of the orbit reduces to Cl C2 1 + El 1 + E2 (8.67) 6 Actually, it can be the perigee of the final orbit or the apogee, but we can treat both cases at once, as we shall see directly. Section 8.8 Changes of Orbit 317 Let us denote by X the ratio of the satellite's speeds just before and just after the firing of the rockets, v 2 = v 1. I shall call X the thrust factor; if A > 1, then the thrust was forward and the satellite sped up; if 0 < X < 1, then the thrust was backward and the satellite slowed down. (In principle, X could be negative, but this would represent a reversal of direction, an unlikely maneuver that I shan't consider.) At perigee the angular momentum is just \u00a3 = p,rv. The value of r does not change during the impulse, and I shall assume that the firing of the rockets changes the satellite's mass by a negligible amount. Under these assumptions, the angular momentum changes by the same factor as the speed: According to (8.48), the parameter c is proportional to \u00a3 2. Therefore, the new value of c is \u00a32 = )1 1. C2 = Substituting into (8.67) and solving for 6 2, we find for the new eccentricity E2 = A 2E1 + (),.2 - 1). Equation (8.70) contains almost all the interesting information about the new orbit. For example, if X > 1 (a forward thrust), it is easy to see that the new orbit has 6 2 > E1. Thus the new orbit has the same perigee as the old one, but has greater eccentricity and so lies outside the old orbit, as shown by the outer dashed curve in Figure 8.12(a). If we make X large enough, then the new eccentricity becomes greater than 1; in this case the new orbit is actually a hyperbola, and our spacecraft escapes from the earth. If we choose the thrust factor X < 1 (a backward thrust), then the new eccentricity is less than the old, 6 2 < E l, and the new orbit lies inside the old, as shown by the inner dashed curve in Figure 8.12(b). As we make X steadily smaller, eventually 6 2 move the satellite into a circular orbit. If we choose X still smaller, then 6 2 becomes negative. What does this signify? The parameter E started out as a positive constant, but the orbital equation r = c 1(1+ E cos 0) makes perfectly good sense with E < 0. (8.68) (8.69) (8.70) vanishes; that is, if we fire the rockets backward with just the right impulse, we can (a) Forward thrust (b) Backward thrust Figure 8.12 Changing orbits. The satellite's original orbit is shown as the solid curve, and the rockets are fired when the satellite is at the perigee P. (a) A forward impulse moves the satellite to the larger dashed elliptical orbit. (b) A backward impulse moves the satellite to the smaller dashed elliptical orbit. 318 Chapter 8 Two-Body Central-Force Problems The only difference is that the direction 0 = 0 is now the direction of maximum r and = 7 is that of minimum r; that is, the apogee and perigee have exchanged places. By administering a large enough backward thrust at P (the old orbit's perigee), we have transferred the satellite to a smaller orbit for which P is now the apogee. EXAMPLE 8.6 Changing between Circular Orbits A satellite's crew in a circular orbit of radius R 1 wishes to transfer to a circular orbit of radius 2R 1. It does this using two successive boosts, as shown in Figure 8.13. First it boosts itself at point P into an elliptical transfer orbit 2, just large enough to take it out to the required radius. Second, on reaching the required radius (at P', the apogee", " of the transfer orbit) it boosts itself into the desired circular orbit 3. By what factor must it increase its speed in each of these two boosts? That is, what are the required thrust factors A and A'? By what factor does the satellite's speed increase as a result of the whole maneuver? The initial circular orbit has c1 = R 1 and eccentricity E 1 = 0. The final orbit is to have radius R3 = 2R 1. According to (8.69), the transfer orbit has c 2 = X2 Ri and, according to (8.70), E 2 = (X2 \u2014 1), where A is the thrust factor of the first boost, at P. By the time the satellite reaches the point P', we want it to be at radius R3. Since PI is the apogee of the transfer orbit, this requires that R3 = c2 = A2 R1 A2 R1 1 \u2014 E2 1\u2014 2 \u2014 1) 2 \u2014 A 2 (8.71) which is easily solved for A to give A 2R3 Ri + R3 \u2014 1.15. (8.72) The satellite must boost its speed by about 15% to move into the required transfer orbit. Figure 8.13 Two successive boosts, at P and P', transfer a satellite from the smaller circular orbit 1 to a transfer orbit 2 and thence to the final circular orbit 3. Principal Definitions and Equations of Chapter 8 319 The second transfer occurs at P', the apogee of the transfer orbit. In Problem 8.33 you can show that the second thrust factor is A' = R 1 \u00b1 R3 2R1 ti 1.22; (8.73) that is, we need to boost the speed by 22% to move from the transfer orbit to the final circular orbit. It would be tempting to think that the overall change in speed, moving from the inital to the final orbit, was just the product X?.' of the two thrust factors, but this overlooks that the satellite's speed also changes as it moves around the transfer orbit. By conservation of angular momentum, it is easy to see that the speeds at the two ends of the transfer orbit satisfy v 2 (apo)R 3 = v2 (per)R i. Therefore, the overall gain in speed is given by v3 = A. ', v2(apo) v2 (per) V1 Rl \u00b1 R3 R1 2R3 2R 1 R3 R 1 \u00b1 R3 v1= - V1. R1 R3 (8.74) In the present case, R3 = 2R1 and hence v 3 = v 1/ That is, the final speed is actually less than the initial by a factor of This result [and more generally the result (8.74)] could have been anticipated. It is easy to show (Problem 8.32) that for circular orbits v a Thus doubling the radius necessarily required that the speed be reduced by a factor of, N/2. Principal Definitions and Equations of Chapter 8 The Relative Coordinate and Reduced Mass When rewritten in terms of the relative coordinate r = r 1 \u2014 r2 [Eq. (8.4)] and the CM coordinate R, the two-body problem is reduced to the problem of two independent particles, a free particle with mass M = m 1 + m2 and position R, and a particle with mass equal to the reduced mass = M1M2 ml + m2 [Eq. (8.11)] position r, and potential energy U (r). 320 Chapter 8 Two-Body Central-Force Problems The Equivalent One-Dimensional Problem The motion of the relative coordinate, with given angular momentum \u00a3, is equivalent to the motion of a particle in one (radial) dimension, with mass p, position r (with 0 < r < cc), and effective potential energy Ueff (r) = U(r) U cf (r) = U (r) [Eq. (8.30)] \u00a32 2pr2 where Ucf is called the centrifugal potential energy. The Transformed Radial Equation With the change of variables from r to u = 1/r and elimination of t in favor of 0, the equation of the one-dimensional radial motion becomes u\"(0) = \u2014 u(0) i2u(0)2 F. [Eq. (8.41)] The Kepler Orbits is For a planet or comet, the force is F = Gm im2/r 2 = y /r 2, and the solution of (8.41) r(0) = c 1-Fecos0 [Eq. (8.49)] where c =,e2 /y,u, and E is related to the energy by E \u2014 \u2014 1). \\ 2 2 11 Ze2 [Eq. (8.58)] This Kepler orbit is an ellipse, parabola,", " or hyperbola, according as the eccentricity E is less than, equal to, or greater than 1. Problems for Chapter 8 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). SECTION 8.2 CM and Relative Coordinates; Reduced Mass 8.1 * Verify that the positions of two particles can be written in terms of the CM and relative positions as r1 = R m 2r/M and r2 = R \u2014 m ir/M. Hence confirm that the total KE of the two particles can be expressed as T = 4 MR2 + 4,u,i2, where\u00b5 denotes the reduced mass \u00b5 = m 1m 2/ M. 8.2 ** Although the main topic of this chapter is the motion of two particles subject to no external forces, many of the ideas [for example, the splitting of the Lagrangian into two independent pieces = Zan as in Equation (8.13)] extend easily to more general situations. To illustrate this, consider the following: Two masses m 1 and m 2 move in a uniform gravitational field g and interact Problems for Chapter 8 321 via a potential energy U(r). (a) Show that the Lagrangian can be decomposed as in (8.13). (b) Write down Lagrange's equations for the three CM coordinates X, Y, Z and describe the motion of the CM. Write down the three Lagrange equations for the relative coordinates and show clearly that the motion of r is the same as that of a single particle of mass equal to the reduced mass it, with position r and potential energy U(r). 8.3 ** Two particles of masses m 1 and m 2 are joined by a massless spring of natural length L and force constant k. Initially, m2 is resting on a table and I am holding m 1 vertically above m 2 at a height L. At time t = 0, I project m 1 vertically upward with initial velocity v o. Find the positions of the two masses at any subsequent time t (before either mass returns to the table) and describe the motion. [Hints: See Problem 8.2. Assume that v o is small enough that the two masses never collide.] SECTION 8.3 The Equations of Motion 8.4 * Using the Lagrangian (8.13) write down the three Lagrange equations for the relative coordinates x, y, z and show clearly that the motion of the relative position r is the same as that of a single particle with position r, potential energy U(r), and mass equal to the reduced mass,u,. 8.5 * The momentum p conjugate to the relative position r is defined with components px = and so on. Prove that p = pti. Prove also that in the CM frame, p is the same as Pi the momentum of particle 1 (and also \u2014p 2). 8.6 * Show that in the CM frame, the angular momentum L i of particle 1 is related to the total angular momentum L by P 1 = (m 2/ M)L and likewise \u00a3 2 = (m 1 /M)L. Since L is conserved, this shows that the same is true of L 1 and \u00a3 2 separately in the CM frame. 8.7 ** (a) Using elementary Newtonian mechanics find the period of a mass m 1 in a circular orbit of radius r around a fixed mass m2. (b) Using the separation into CM and relative motions, find the corresponding period for the case that m 2 is not fixed and the masses circle each other a constant distance r apart. Discuss the limit of this result if m 2 oo. (c) What would be the orbital period if the earth were replaced by a star of mass equal to the solar mass, in a circular orbit, with the distance between the sun and star equal to the present earth\u2014sun distance? (The mass of the sun is more than 300,000 times that of the earth.) 8.8 ** Two masses m 1 and m 2 move in a plane and interact by a potential energy U(r) = ikr2. Write down their Lagrangian in terms of the CM and relative positions R and r, and find the equations of motion for the coordinates X, Y and x, y. Describe the motion and find the frequency of the relative motion. 8.9 ** Consider two particles of equal masses, m 1 = m 2, attached to each other by a light straight spring (force constant k, natural length L) and free to slide over a frictionless horizontal table. (a) Write down the Lagrangian in terms of the coordinates r 1 and r2, and rewrite it in terms of the CM and relative positions, R and r, using polar coordinates (r, 0) for r. (b) Write down and solve the Lagrange equations for the CM coordinates X, Y. (c) Write down the Lagrange equations for r and", " 0. Solve these for the two special cases that r remains constant and that remains constant. Describe the corresponding motions. In particular, show that the frequency of oscillations in the second case is co = '2k/m l. 8.10 ** Two particles of equal masses m 1 = m 2 move on a frictionless horizontal surface in the vicinity of a fixed force center, with potential energies U 1 = 1-kr? and U2 = z k r 22. In addition, they interact with each other via a potential energy U12 = i akr 2, where r is the distance between them and a and k are 322 Chapter 8 Two-Body Central-Force Problems positive constants. (a) Find the Lagrangian in terms of the CM position R and the relative position r = r i \u2014 r2. (b) Write down and solve the Lagrange equations for the CM and relative coordinates X, Y and x, y. Describe the motion. 8.11 ** Consider two particles interacting by a Hooke's law potential energy, U = Ikr2, where r is their relative position r = r 1 \u2014 r2, and subject to no external forces. Show that r(t) describes an ellipse. Hence show that both particles move on similar ellipses around their common CM. [This is surprisingly awkward. Perhaps the simplest procedure is to choose the xy plane as the plane of the orbit and then solve the equation of motion (8.15) for x and y. Your solution will have the form x = A cos cot B sin cot, with a similar expression for y. If you solve these for sin cot and cos cot and remember that sin e cos2 = 1, you can put the orbital equation in the form ax e 2bxy cy 2 = k where k is a positive constant. Now invoke the standard result that if a and c are positive and ac > b2, this equation defines an ellipse.] SECTION 8.4 The Equivalent One-Dimensional Problem 8.12 ** (a) By examining the effective potential energy (8.32) find the radius at which a planet (or comet) with angular momentum \u00a3 can orbit the sun in a circular orbit with fixed radius. [Look at dUeff I dr 1 (b) Show that this circular orbit is stable, in the sense that a small radial nudge will cause only small radial oscillations. [Look at d 2 Ueff I dr2 1 Show that the period of these oscillations is equal to the planet's orbital period. 8.13 *** Two particles whose reduced mass is\u00b5 interact via a potential energy U = Zkr2, where r is the distance between them. (a) Make a sketch showing U (r), the centrifugal potential energy Ucf (r), and the effective potential energy Ueff (r). (Treat the angular momentum \u00a3 as a known, fixed constant.) (b) Find the \"equilibrium\" separation r o, the distance at which the two particles can circle each other with constant r. [Hint: This requires that dUeif I dr be zero.] (c) By making a Taylor expansion of Ueff (r) about the equilibrium point ro and neglecting all terms in (r ro) 3 and higher, find the frequency of small oscillations about the circular orbit if the particles are disturbed a little from the separation ro. 8.14 *** Consider a particle of reduced mass\u00b5 orbiting in a central force with U = krn where kn > 0. (a) Explain what the condition kn > 0 tells us about the force. Sketch the effective potential energy Ueff for the cases that n = 2, \u20141, and \u20143. (b) Find the radius at which the particle (with given angular momentum.e) can orbit at a fixed radius. For what values of n is this circular orbit stable? Do your sketches confirm this conclusion? (c) For the stable case, show that the period of small oscillations about the circular orbit is rose =Toth/,N/n 2. Argue that if,N/n + 2 is a rational number, these orbits are closed. Sketch them for the cases that n = 2, \u20141, and 7. SECTION 8.6 The Kepler Orbits 8.15 * In deriving Kepler's third law (8.55) we made an approximation based on the fact that the sun's mass Ms is much greater than that of the planet m. Show that the law should actually read 1.2 = [47 2 /G 4s (m m)] a 3, and hence that the \"constant\" of proportionality is actually a little different for different planets. Given that the mass of the heaviest planet (Jupiter) is about 2 x 10 27 kg, while Ms is about 2 x 10 30 kg (and some planets have masses several orders of magnitude less than Jupiter), by what percent would you", " expect the \"constant\" in Kepler's third law to vary among the planets? 8.16** We have proved in (8.49) that any Kepler orbit can be written in the form 'JO) = c/(1 + E cos 0), where c > 0 and E > 0. For the case that 0 < E < 1, rewrite this equation in rectangular Problems for Chapter 8 323 coordinates (x, y) and prove that the equation can be cast in the form (8.51), which is the equation of an ellipse. Verify the values of the constants given in (8.52). 8.17 ** If you did Problem 4.41 you met the virial theorem for a circular orbit of a particle in a central force with U = krn. Here is a more general form of the theorem that applies to any periodic orbit of a particle. (a) Find the time derivative of the quantity G = r p and, by integrating from time 0 to t, show that G (t) \u2014 G(0) t =2(T) + (F r) where F is the net force on the particle and ( f) denotes the average over time of any quantity f. (b) Explain why, if the particle's orbit is periodic and if we make t sufficiently large, we can make the left-hand side of this equation as small as we please. That is, the left side approaches zero as t oo. (c) Use this result to prove that if F comes from the potential energy U = krn, then (T) = n(U)12, if now (f) denotes the time average over a very long time. 8.18 ** An earth satellite is observed at perigee to be 250 km above the earth's surface and traveling at about 8500 m/s. Find the eccentricity of its orbit and its height above the earth at apogee. [Hint: The earth's radius is Re ',-z= 6.4 x 106 m. You will also need to know GMe, but you can find this if you remember that GMe /RZ = g.] 8.19 ** The height of a satellite at perigee is 300 km above the earth's surface and it is 3000 km at apogee. Find the orbit's eccentricity. If we take the orbit to define the xy plane and the major axis in the x direction with the earth at the origin, what is the satellite's height when it crosses the y axis? [See the hint for Problem 8.18.] 8.20 ** Consider a comet which passes through its aphelion at a distance I-max from the sun. Imagine that, keeping riiiax fixed, we somehow make the angular momentum smaller and smaller, though not actually zero; that is, we let \u00a3 0. Use equations (8.48) and (8.50) to show that in this limit the eccentricity E of the elliptical orbit approaches 1 and that the distance of closest approach r mir, approaches zero. Describe the orbit with rmax fixed but f very small. What is the semimajor axis a? 8.21 *** (a) If you haven't already done so, do Problem 8.20. (b) Use Kepler's third law (8.55) to find the period of this orbit in terms of r max (and G and Ms). (c) Now consider the extreme case that the comet is released from rest at a distance rmax from the sun. (In this case is actually zero.) Use the technique described in connection with (4.58) to find how long the comet takes to reach the sun. (Take the sun's radius to be zero.) (d) Assuming the comet can somehow pass freely through the sun, describe its overall motion and find its period. (e) Compare your answers in parts (b) and (d). 8.22 *** A particle of mass m moves with angular momentum about a fixed force center with F(r) = k 1 r3 where k can be positive or negative. (a) Sketch the effective potential energy Ueff for various values of k and describe the various possible kinds of orbit. (b) Write down and solve the transformed radial equation (8.41), and use your solutions to confirm your predictions in part (a). 8.23 *** A particle of mass m moves with angular momentum \u00a3 in the field of a fixed force center with where k and A are positive. (a) Write down the transformed radial equation (8.41) and prove that the orbit has the form F(r) = -- + k 1' 2 A r3 r (0) = 1 + E COS(80) 324 Chapter 8 Two-Body Central-Force Problems where c, /3, and E are positive constants. (b) Find c and $ in terms of the given parameters, and describe the orbit for the case that 0", " < E < 1. (c) For what values of fi is the orbit closed? What happens to your results as A 0? \u00a32 < \u2014Am). 8.24 *** Consider the particle of Problem 8.23, but suppose that the constant A is negative. Write down the transformed radial equation (8.41) and describe the orbits of low angular momentum (specifically, 8.25 *** [Computer] Consider a particle with mass m and angular momentum \u00a3 in the field of a central force F = \u2014k/r5/2. To simplify your equations, choose units for which m = \u00a3 = k = 1. (a) Find the value ro of r at which Ueff is minimum and make a plot of Ueff (r) for 0 < r < 5ro. (Choose your scale so that your plot shows the interesting part of the curve.) (b) Assuming now that the particle has energy E = \u20140.1, find an accurate value of r\u201e,, n, the particle's distance of closest approach to the force center. (This will require the use of a computer program to solve the relevant equation numerically.) (c) Assuming that the particle is at r = rmir, when 0 = 0, use a computer program (such as \"NDSolve\" in Mathematica) to solve the transformed radial equation (8.41) and find the orbit in the form r = r(0) for 0 < < 77r. Plot the orbit. Does it appear to be closed? 8.26 *** Show that the validity of Kepler's first two laws for any body orbiting the sun implies that the force (assumed conservative) of the sun on any body is central and proportional to 1/r 2. 8.27 *** At time to a comet is observed at radius ro traveling with speed v o at an acute angle a to the line from the comet to the sun. Put the sun at the origin 0, with the comet on the x axis (at to) and its orbit in the xy plane, and then show how you could calculate the parameters of the orbital equation in the form r = c/[1 + E cos(0 \u2014 8)]. Do so for the case that ro = 1.0 x 10 11 m, vo = 45 km/s, and a = 50 degrees. [The sun's mass is about 2.0 x 10 30 kg, and G = 6.7 x 10-11 Nm2/s2.] SECTION 8.7 The Unbounded Kepler Orbits 8.28 * For a given earth satellite with given angular momentum f, show that the distance of closest approach rmin on a parabolic orbit is half the radius of the circular orbit. 8.29 ** What would become of the earth's orbit (which you may consider to be a circle) if half of the sun's mass were suddenly to disappear? Would the earth remain bound to the sun? [Hints: Consider what happens to the earth's KE and PE at the moment of the great disappearance. The virial theorem for the circular orbit (Problem 4.41) helps with this one.] Treat the sun (or what remains of it) as fixed. 8.30 ** The general Kepler orbit is given in polar coordinates by (8.49). Rewrite this in Cartesian coordinates for the cases that E = 1 and E > 1. Show that if E = 1, you get the parabola (8.60), and if E > 1, the hyperbola (8.61). For the latter, identify the constants a,,8, and 6 in terms of c and E. 8.31 *** Consider the motion of two particles subject to a repulsive inverse-square force (for example, two positive charges). Show that this system has no states with E < 0 (as measured in the CM frame), and that in all states with E > 0, the relative motion follows a hyperbola. Sketch a typical orbit. [Hint: You can follow closely the analysis of Sections 8.6 and 8.7 except that you must reverse the force; probably the simplest way to do this is to change the sign of y in (8.44) and all subsequent equations (so that F(r) = Ir2) and then keep y itself positive. Assume \u00a3 0.] Problems for Chapter 8 325 SECTION 8.8 Changes of Orbit 8.32 * Prove that for circular orbits around a given gravitational force center (such as the sun) the speed of the orbiting body is inversely proportional to the square root of the orbital radius. 8.33 ** Figure 8.13 shows a space vehicle boosting from a circular orbit 1 at P to a transfer orbit 2 and then from the transfer orbit at P' to the final circular orbit 3. Example 8.6 derived in detail the thrust factor required for the boost at P. Show similarly that the thrust factor required at P' is ),,,' =./(R", " i + R3)/2R i. [Your argument should parallel closely that of Example 8.6, but you must account for the fact that P' is the apogee (not perigee) of the transfer orbit. For example, the plus signs in (8.67) should be minus signs here.] 8.34 ** Suppose that we decide to send a spacecraft to Neptune, using the simple transfer described in Example 8.6 (page 318). The craft starts in a circular orbit close to the earth (radius 1 AU or astronomical unit) and is to end up in a circular orbit near Neptune (radius about 30 AU). Use Kepler's third law to show that the transfer will take about 31 years. (In practice we can do a lot better than this by arranging that the craft gets a gravitational boost as it passes Jupiter.) 8.35 *** A spacecraft in a circular orbit wishes to transfer to another circular orbit of quarter the radius by means of a tangential thrust to move into an elliptical orbit and a second tangential thrust at the opposite end of the ellipse to move into the desired circular orbit. (The picture looks like Figure 8.13 but run backwards.) Find the thrust factors required and show that the speed in the final orbit is two times greater than the initial speed. CHAPTER Mechanics in Noninertial Frames We saw in Chapter 1 that Newton's laws are valid only in the special class of inertial reference frames \u2014 frames that are neither accelerating nor rotating. The natural reaction to this realization is to resolve to treat all mechanics problems using only inertial frames, and this is in fact what we have done so far. Nevertheless, there are situations where it is very desirable to consider a noninertial frame. For example, if you are sitting in a car that is accelerating and you wish to describe the behavior of a coin that you are tossing in the air, it is very natural to want to describe it as seen by you, that is, in the accelerating frame of the car in which you are traveling. Another important example is a reference frame fixed to the earth. To an excellent approximation this is an inertial frame. Nevertheless, the earth is rotating on its axis and accelerating in its orbit, and, for both these reasons, a reference frame fixed to the earth is not quite inertial. In the majority of problems, the noninertial character of a frame fixed to the earth is totally negligible, but there are situations, for instance the firing of a long- range rocket, in which the earth's rotation has important consequences. Naturally, we would like to describe the motion relative to the earth-bound frame that we live in, but to do so, we must learn to do mechanics in noninertial frames. In the first two sections of this chapter, I shall describe the simple case of a reference frame that is accelerating but not rotating. In the remainder of the chapter, I shall discuss the case of a rotating reference frame. 9.1 Acceleration without Rotation Let us consider an inertial frame 8 and a second frame 8 that is accelerating relative to 8\u201e with acceleration A, which need not necessarily be constant. Notice that because the noninertial frame is the one we're really interested in, it's the one that I've called 8. Also I'm using capital letters for the acceleration (and velocity) of frame $ relative to frame 8.. The inertial frame S. could be a frame anchored to the ground. (We'll ignore the tiny acceleration of any earth-bound frame for now.) The frame 8 could 327 328 Chapter 9 Mechanics in Noninertial Frames etc.). that is, be a frame fixed in a railroad car that is moving relative to S o, with velocity V and acceleration A = V. Suppose now that a passenger in the car is playing catch with a tennis ball of mass m, and let us consider the motion of the ball, first as measured relative to S o. Because S o is inertial, we know that Newton's second law holds, so mfo = F, (9.1) where ro is the ball's position relative to S o and F is the net force on the ball, the vector sum of all forces on the ball (gravity, air resistance, the passenger's hand pushing it, Now consider the same ball's motion as measured relative to the accelerating frame S. The ball's position relative to S is r and, by the vector addition of velocities, its velocity r relative to S is related to r o by the velocity-addition formula:' (ball's velocity relative to ground) = (ball's velocity relative to car) + (car's velocity relative to ground). Differentiating and rearranging, we find that If we multiply this equation by m and use (9.1) to replace mi. by F, we find that ro = + V = A. mr = F\u2014mA. (9.2) (9.3) (9.4) This", " equation has exactly the form of Newton's second law, except that in addition to F, the sum of all forces identified in the inertial frame, there is an extra term on the right equal to \u2014mA. This means we can continue to use Newton's second law in the noninertial frame S provided we agree that in the noninertial frame we must add an extra force-like term, often called the inertial force: This inertial force experienced in noninertial frames is familiar in several everyday situations: If you sit in an aircraft accelerating rapidly toward takeoff, then, from your point of view, there is a force that pushes you back into your seat. If you are standing in a bus that brakes suddenly (A backward) the inertial force \u2014mA is forward and can make you fall on your face if you aren't properly braced. As a car goes rapidly 1 An important discovery of relativity is, of course, that velocities do not combine exactly ac- cording to the simple vector addition of (9.2); nevertheless, in the framework of classical mechanics (9.2) is correct. In the same way, I shall take for granted that the times measured in S and S o are the same, which is correct in classical, though not relativistic, mechanics. Section 9.1 Acceleration without Rotation 329 around a sharp curve, the inertial force experienced by the passengers is the so-called centrifugal force that pushes them outward. One can take the view that the inertial force is a \"fictitious\" force, introduced merely to preserve the form of Newton's second law. Nevertheless, for an observer in an accelerating frame, it is entirely real. In many problems involving objects in accelerated frames, much the simplest procedure is to go ahead and use Newton's second law in the noninertial frame, always remembering to include the extra inertial force as in (9.4). Here is a simple example: EXAMPLE 9.1 A Pendulum in an Accelerating Car Consider a simple pendulum (mass m and length L) mounted inside a railroad car that is accelerating to the right with constant acceleration A as shown in Figure 9.1. Find the angle 4, at which the pendulum will remain at rest relative to the accelerating car and find the frequency of small oscillations about this equilibrium angle. As observed in any inertial frame, there are just two forces on the bob, the tension in the string T and the weight mg; thus the net force (in any inertial frame) is F = T + mg. If we choose to work in the noninertial frame of the accelerating car, there is also the inertial force \u2014mA, and the equation of motion (9.4) is mf = T mg \u2014 mA. (9.6) A remarkable simplification in this problem is that, because the weight mg and the inertial force \u2014mA are both proportional to the mass m, we can combine these two terms and write me = T m(g \u2014 A) = T + m gaff (9.7) where gaff = g \u2014 A. We see that the pendulum's equation of motion in the accelerating frame of the car is exactly the same as in an inertial frame, except geff Figure 9.1 A pendulum is suspended from the roof of a railroad car that is accelerating with constant acceleration A. In the non- inertial frame of the car, the acceleration manifests itself through the inertial force \u2014mA, which, in turn, is equivalent to the re- placement of g by the effective geff = g \u2014 A. 330 Chapter 9 Mechanics in Noninertial Frames that g has been replaced by an effective geff = g \u2014 A, as shown in Figure 9.12 This makes the solution of our problem almost trivially simple. If the pendulum is to remain at rest (as seen in the car), thenr must be zero, and according to (9.7) T must be exactly opposite to mg eff. In particular, we see from Figure 9.1 that the direction of T (and hence of the pendulum) has to be 'eq = arctan(A /g). (9.8) The small-amplitude frequency of a pendulum in an inertial frame is well known to be co =,/g/L. Thus the period of our pendulum is obtained by replacing g by geff = \\/g 2 + A 2. That is, (0 = \\/g2 \u00b1 A2 L (9.9) It is worth comparing this solution with other, more direct methods. First we could certainly have found the equilibrium angle (9.8) working in an inertial frame anchored to the ground. In this frame, there are only two forces on the bob (T and mg). If the pendulum is to remain at rest in the railroad car, then (as seen from the", " ground) it must accelerate at exactly A. Therefore the net force T + mg must equal mA, and by drawing a triangle of forces you can easily convince yourself that this requires T to have the direction (9.8). On the other hand, to find the frequency of oscillations (9.9) directly in the ground- based frame requires considerable ingenuity, and does not give the insight of our noninertial derivation. We could also have derived both results using the Lagrangian method. This has the distinct advantage that you don't have to think about inertial and non- inertial frames \u2014 just write down the Lagrangian in terms of the generalized coordinate 0 and then crank the handle. However, finding the frequency (9.9) this way is quite clumsy, as you discovered if you did Problem 7.30. 9.2 The Tides A beautiful application of the result (9.4) is the explanation of the tides. As you probably know, the tides are the result of the bulges in the earth's oceans caused by the gravitational attraction of the moon and sun. As the earth rotates, people on the earth's surface move past these bulges and experience a rising and falling of the sea's level. It turns out that the most important contributor to this effect is the moon, and to simplify our discussion I shall, at first, ignore the sun entirely. I shall also assume, at first, that the oceans cover the whole surface of the globe. 2 This result, that the effect of being in an accelerated frame is the same as having an addi- tional gravitational force, is a cornerstone of general relativity, where it is called the principle of equivalence. Section 9.2 The Tides 331 (a) Wrong (a) Right Figure 9.2 Views of the earth and moon from high above the North Pole, with the earth rotating counterclockwise about the polar axis. (a) A plausible-sounding, but incorrect, explanation of the tides argues that the attraction of the moon causes the oceans to bulge toward the moon. As the earth rotates once a day, a person fixed on the earth would experience one high tide per day, not the observed two. (b) The correct explanation: The main effect of the moon's attraction is to give the whole earth (including the oceans) a small acceleration A toward the moon. As explained in the text, the residual effect is that the oceans bulge both toward and away from the moon as shown. As the earth rotates once a day, these two bulges cause the observed two high tides per day. A plausible-sounding, though entirely incorrect, explanation of the tides is illus- trated in Figure 9.2(a). According to this incorrect argument, the moon's attraction pulls the oceans toward the moon, producing a single bulge toward the moon. The trouble with this argument (beside that it is wrong) is that a single bulge would cause just one high tide per day, instead of the observed two. The correct explanation, illustrated in Figure 9.2(b), is more complicated. The dominant effect of the moon is to give the whole earth, including the oceans, an acceleration A toward the moon. This acceleration is the centripetal acceleration of the earth as the moon and earth circle around their common center of mass and is (almost exactly) the same as if all the masses that make up the earth were concentrated at its center. This centripetal acceleration of any object on earth, as it orbits with the earth, corresponds to the pull of the moon that the object would feel at the earth's center. Now any object on the moon side of the earth is pulled by the moon with a force that is slightly greater than it would be at the center. Therefore, as seen from the earth, objects on the side nearest the moon behave as if they felt a slight additional attraction toward the moon. In particular, the ocean surface bulges toward the moon. On the other hand, objects on the far side from the moon are pulled by the moon with a force that is slightly weaker than it would be at the center, which means that they move (relative to the earth) as if they were slightly repelled by the moon. This slight repulsion causes the ocean to bulge on the side away from the moon and is responsible for the second high tide of each day. We can make this argument quantitative (and probably more convincing) if we return to Equation (9.4). The forces on any mass m near the earth's surface are (1) the gravitational pull, mg, of the earth, (2) the gravitational pull, \u2014GM inmd/d2, of the moon (where M111 is the mass of the moon and d is the position of the object relative to the moon, as in Figure 9.3.), and (3) the net non-gravitational force F ng (for instance, 332 Chapter 9 Mechanics in Nonin", "ertial Frames X Figure 9.3 A mass m near the earth's surface has position r relative to the earth's center, and d relative to the moon. The vector do is the position of the earth's center relative to the moon's. the buoyant force on a drop of sea water in the ocean). Meanwhile the acceleration of our origin 0, the earth's center, is A = \u2014GM m do2 where do is the position of the earth's center relative to the moon. Putting all this together we find for (9.4) mi: = F \u2014 mA = (mg \u2014 GMm m Fng) GM.m- a\u00b0 d o2 or, if we combine the two terms that involve Mm, where the tidal force mi: = mg Ftid \u00b1 Fug id = \u2014GMmm d2 do2 a do) \u2014 Ft (9.10) (9.11) (9.12) is the difference between the actual force of the moon on m and the corresponding force if m were at the center of the earth. The entire effect of the moon on the motion (relative to the earth) of any object near the earth is contained in the tidal force Ftid of (9.12). At the point directly facing the moon, point P in Figure 9.4, the vectors d = MP and do = MO point in the same direction, but d < do. Thus the first term in (9.12) dominates, and the tidal force is toward the moon. At point R, facing directly away from the moon, again d (equal to MR) and do point in the same direction, but here d > do and the tidal force points away from the moon. At the point Q, the vectors d = MQ and do = MO point in different directions; the x components of the two terms in (9.12) cancel almost exactly, but only the first term has a y component. Thus at Q (and likewise at S) the tidal force Section 9.2 The Tides 333 moon Figure 9.4 The tidal force Ftid as given by (9.12) is outward (away from the earth's center) at points P and R, but inward (toward the earth's center) at Q and S. is inward, toward the earth's center, as shown in Figure 9.4. In particular, the effect of the tidal force is to distort the ocean into the shape shown in Figure 9.2(b), with bulges centered on the points P and R, giving the observed two high tides per day. Magnitude of the Tides The simplest way to find the height difference between high and low tides is to observe that the surface of the ocean is an equipotential surface a surface of constant potential energy. To prove this, consider a drop of sea water on the surface of the ocean. The drop is in equilibrium (relative to the earth's reference frame) under the influence of three forces: the earth's gravitational pull mg, the tidal force Ftid, and the pressure force Fp of the surrounding sea water. Since a static fluid cannot exert any shearing force, the pressure force F p must be normal to the surface of the ocean. (It is in fact just the buoyant force of Archimedes' principle.) Since the drop of water is in equilibrium, it follows that mg + Ftid must likewise be normal to the surface. Now, both mg and Ftid are conservative, so each can be written as the gradient of a potential energy: mg = \u2014 V Ueg and Ftid = V Utid where Ueg is the potential energy due to the earth's gravity and Utid is that of the tidal force, which is, by inspection of (9.l2), 3 Utid 1 in ( \u2014cl X d 2 (9.13) 3 According to (9.12), Ftid is the sum of two terms. The first is just the usual inverse square force with corresponding potential energy \u2014GM m m/d and the second is a constant vector pointing in the x direction, which gives the term \u2014GM. mx/d.2. 334 Chapter 9 Mechanics in Noninertial Frames Figure 9.5 The difference h between high and low tides is the difference between the lengths OP and OQ. This difference is much exaggerated here (since h turns out to be of order 1 meter), and both the lengths OP and OQ are very close to the radius of the earth, Re 6400 km. Also, r R e is actually about 60 times smaller than the earth\u2014moon distance OM = do. Thus, the statement that mg + Ftid is normal to the surface of the ocean can be rephrased to say that V ((leg Uttd) is normal to the surface, which in turn implies that U = (Ueg Uttd ) is constant on the surface. In other words, the surface of the ocean is an equipotential surface. Since U", " is constant on the surface, it follows that U (P) = U (Q) (see Figure 9.5) or The left side here is just Ueg (P) \u2014 Ue g (Q) = Utid(Q) \u2014 Utid(P) (9.14) Ueg (P) \u2014 Ueg (Q) = mgh (9.15) where h is the required difference between high and low tides (the difference between the lengths OP and OQ in Figure 9.5). To find the right side of (9.14) we must evaluate the two tidal potential energies, Uttd(Q) and Utid(P) from the definition (9.13). At point Q, we see that d = do2 r 2 (with r ti R e) and x = 0. Therefore, from (9.13), Und(Q) =- \u2014G Mmm 1.1d0 Re2 We can rewrite the square root as 1/d02 + Re = do -V1 (Re /do)2. Then, since R e /do << 1, we can use the binomial approximation (1 + E) -1/2 ti 1 \u2014 4\u20ac to give Utid(Q) m G M m 1 d o 2 Re 26/2 0 (9.16) Section 9.2 The Tides 335 At the point P, we see that d = do \u2014 R e and x = \u2014R e, and a similar calculation gives (Problem 9.5) Utid(P) GMm m ( 1 R e2 ) do do 2 * (9.17) (As you can check, one gets the same answer at the point R facing away from the moon, so that, in this approximation, the heights of the two daily tides should be the same.) Substituting (9.15), (9.16), and (9.17) into (9.14) we get mgh = GMmm 3R 2 e do 2d02 If we recall that g = GM,' Re2, this implies that h = 3 Mm R e4 2 Me d 3 o (9.18) Putting in the numbers (Mm = 7.35 x 1022 kg, Me = 5.98 x 1024 kg, R e = 6.37 x 106 m, and do = 3.84 x 108 m), we find for the height of the tides, due to the moon alone, h = 54 cm [moon alone]. (9.19) The height of the tides caused by the sun alone is also given by (9.18), but with Mm replaced by the mass of the sun, Ms = 1.99 x 103\u00b0 kg, and do replaced by the distance from the sun to the earth, 1.495 x 10 11 m. This gives h = 25 cm [sun alone]. (9.20) Although the sun's contribution to the tides is less than the moon's, it is certainly not neglible, and the two effects combine in an interesting way. First, consider a time when the earth, sun, and moon are approximately in line \u2014 either with the earth in the middle, as at full moon, or the moon in the middle as at new moon. (See Figure 9.6.) In this case the tidal forces due to the moon and sun reinforce each other (since the two bulges caused by the moon coincide with the two caused by the sun); thus, we would predict large tides (known as spring tides) with h given by the sum of (9.19) and (9.20), that is, h = 54 + 25 = 79 cm. On the other hand, if the sun, earth, and moon form a right angle, then the two tidal effects will cancel and we would predict smaller tides (known as neap tides) with height h = 54 \u2014 25 = 29 cm. Although the theory just presented is basically correct, especially in the middle of the larger oceans, the real situation involves many intriguing complications. Perhaps the most important complication is the effect of the continental land masses. So far I have pretended that the oceans cover the whole world, allowing the tidal forces of the moon and sun to collect the two bulges of Figure 9.5. But the presence of continents can affect our conclusions, leading sometimes to smaller and sometimes to larger tides. A small sea, such as the Black Sea or even the Mediteranean, that is shut off from the main oceans by land will obviously exhibit much smaller tides than we have calculated 336 Chapter 9 Mechanics in Noninertial Frames half moon small tides new moon large tides full moon large tides Figure 9.6 Four successive positions of the moon in its monthly orbit around the earth. At new moon and full moon, the tidal effects of the moon and sun reinforce each other and the tides are large (\"spring\" tides). At half moon, the two effects tend to cancel and the tides are", " small (\"neap\" tides). here. On the other hand, the tides moving across a large ocean can be blocked by the bordering continents and can build up to much greater heights. And tides entering narrow tapering estuaries can cause quite dramatic \"tidal bores.\" 9.3 The Angular Velocity Vector In the remainder of this chapter, we shall be discussing the motion of objects as seen in reference frames that are rotating (relative to inertial frames). Before we begin this discussion, I must introduce some concepts and notation for handling rotations. A detailed study of rotations is actually surprisingly complicated. Fortunately, we do not need many of the details, and some of the properties that are quite hard to prove are reasonably plausible and can be stated without proof. The rotating axes with which we shall be concerned are almost always axes fixed in a rigid body. The most important example is a set of axes fixed in the rotating earth, but we shall see several other examples in Chapter 10. In discussing the rotation of a rigid body, there are really just two situations that concern us: Sometimes the body is rotating about a point of the body that is fixed (in some inertial frame); for example, a wheel that is spinning about a fixed axle, or a pendulum swinging about a fixed pivot. If the rotating body has no fixed point (for example, a baseball that is spinning as it flies through the air), then we usually proceed in two steps: First, we find the motion of the center of mass, and then we analyze the rotational motion of the body relative to its CM. As soon as we restrict attention to the motion relative to the CM, we are in effect examining the motion in a reference frame in which the CM is fixed. Thus either way, our discussion of a rotating body concerns a body with one point effectively fixed. The crucial result concerning a body rotating about a fixed point is called Euler's theorem and states that the most general motion of any body relative to a fixed point 0 is a rotation about some axis through 0. Although this theorem is quite complicated Section 9.3 The Angular Velocity Vector 337 to prove, the result seems very natural and I hope you can accept it without proof. 4 It implies that to specify a rotation about a given point 0, we need give only the direction of the axis about which the rotation occurred and the angle through which the body rotated. Here, our concern is more with the rate of rotation, or angular velocity, and Euler's theorem implies that this can be specified by giving the direction of the axis of rotation and the rate of rotation about this axis. The direction of the axis of rotation can be specified by a unit vector u and the rate by the number co = dOldt. For example, a merry-go-round could be rotating about a vertical axis (u points vertically) at a rate of 10 rad/min (co = 10 rad/min). velocity vector It is often convenient to combine the unit vector u with co to form an angular = cou. (9.21) The single vector co specifies both the direction of the axis of rotation (namely, u, the direction of co) and the rate of rotation (namely, co, the magnitude of co). Actually we have not yet quite defined a unique vector co. For example, for the merry-go-round that is rotating about a vertical axis, the vector co points vertically, but does it point up or down? We remove this ambiguity using the right-hand rule: We choose the direction of co so that when our right thumb points along co, our right fingers curl in the direction of rotation. Another way to say this is that if you look along the vector co, you will see the body rotating clockwise. It is important to recognize that the angular velocity can change with time. If the speed of rotation is changing, then CO will be changing in magnitude, and if the axis of rotation is changing, then co will be changing in direction. For example, the angular velocity of a spacecraft that is tumbling out of control will usually change in both magnitude and direction. In this case co = co(t) is the instantaneous angular velocity at the time t. On the other hand, there are many interesting situations where co is constant (in magnitude and direction); for example, this is true (to an outstanding approximation) of the angular velocity of the earth spinning on its axis. A Useful Relation There is a useful relation between the angular velocity of a rigid body and the linear velocity of any point in the body. Consider, for example, the earth, rotating with angular velocity co about its center 0 (which I shall take to be stationary for the present discussion). Next, consider any point P fixed on (or in) the earth, for example the top of Mount Everest, with position r relative to 0. We can specify r by its spherical polar coordinates (r, 8, (p) with z axis pointing through the North Pole, so that 8 is the colatitude", " \u2014 the latitude measured down from the North Pole (instead of up from the equator, as is more usual with geographers). As the earth turns on its axis, the point P is dragged in an easterly direction around a circle of latitude, with radius p = r sin 8, as shown in Figure 9.7. This means that P moves with speed v = (or sin 0, and if you 4 You can find a proof in, for example, Classical Mechanics by Herbert Goldstein, Charles Poole, and John Safco (3rd ed., Addison-Wesley, 2002), Section 4-6. 338 Chapter 9 Mechanics in Noninertial Frames Figure 9.7 The earth's rotation drags the point P on the surface around a circle of latitude (radius p = r sin 6) with speed v = cop = wr sin 6 and hence velocity v = w x r. will check the direction in Figure 9.7, you will see that the vector velocity is co x r. You can easily see that this result is independent of the nature of the rotating body; that is, for any rigid body rotating with angular velocity co about an axis through 0, the velocity of any point P (position r) fixed on the body is This useful relation is, of course, a generalization of the familiar relation v = cor that you learned in introductory physics for the speed of a point on the perimeter of a wheel of radius r. It is perhaps worth emphasizing that there is a corresponding relation for any vector fixed in the rotating body. For example, if e is a unit vector fixed in the body, then its rate of change, as seen from the non-rotating frame, is de \u2014 dt = co x e, (9.23) a result we shall be using shortly. Addition of Angular Velocites A final basic property of angular velocities that is worth mentioning is that relative angular velocities add in the same way as relative translational velocities. We know (in the framework of classical mechanics) that if two frames 2 and 1 have relative velocity v21 and if a body 3 has velocity v 32 relative to frame 2, then the velocity of 3 relative to frame 1 is just the sum V31 = V32 + V21. (9.24) Section 9.4 Time Derivatives in a Rotating Frame 339 Suppose instead that frame 2 is rotating with angular velocity co 21 relative to frame 1 (both frames with the same origin 0) and that body 3 is rotating (about 0) with angular velocities &31 and 0)32 relative to frames 1 and 2. Now consider any point r fixed in body 3. Its translational velocities relative to frames 1 and 2 must satisfy (9.24). According to (9.22) this means that 0)31 x r = (0)32 x r) + (co21 x r) = (0)32 + 0)21 ) x r and, since this must be true for any r, it follows that (-1)31 = (1)32 te)21. (9.25) That is, angular velocities add in the same way as translational velocities. Notation for Angular Velocities In labeling angular velocities, I shall usually observe the following convention: I shall use the lower case letter co for the angular velocity of a body (such as a spinning top) whose motion is our primary object of interest. I shall use the capital letter 52 for the angular velocity of a noninertial, rotating reference frame relative to which we are calculating the motion of one or more objects. This distinction is consistent with the previous two sections, where I used capital A and V for the acceleration and velocity of a noninertial frame (relative to an inertial frame). In practice, to usually denotes an unknown, while St is usually a given, known angular velocity, such as the angular velocity of the earth as it rotates once a day. In the remainder of this chapter, we shall be concerned with the motion of objects as seen in a rotating reference frame, and, in accordance with this convention, I shall denote the angular velocity of that frame by 52. 9.4 Time Derivatives in a Rotating Frame We are now ready to consider the equations of motion for an object that is viewed from a frame S that is rotating with angular velocity S2 relative to an inertial frame So. While our conclusions will apply to any rotating frame, by far the most important example is a frame attached to the rotating earth, and this is the example that you can keep in mind. This being the case, let us pause to calculate the angular velocity of the earth, which rotates on its axis once every 24 hours. 5 Therefore, for a frame attached to the earth = 2m rad 24 x 3600 s 7.3 x 10-5 rad/s. (9.26) 5 Strictly speaking the period of rotation", " about the earth's axis is one sidereal day, the time to rotate once relative to the distant stars. This is shorter than the solar day by a factor of 365/366, but the difference is too small to worry about here. 340 Chapter 9 Mechanics in Noninertial Frames Y0 Figure 9.8 The frame S o defined by the three dashed axes is inertial. The frame S defined by the three solid axes shares the same origin 0, but is rotating with angular velocity St relative to S o. It is because this angular velocity is so small that we can often ignore it entirely. Nevertheless, we shall see that the earth's rotation does have measurable effects on the motion of projectiles, pendulums, and other systems. There are other noninertial effects (notably the tides), associated with the orbital motion of the earth and moon, but these are all much less important for the problems that we shall consider here, and I shall ignore them for now. I shall assume that the two frames S o and 3 share a common origin 0, as shown in Figure 9.8, so that the only motion of S relative to 8 0 is a rotation with angular velocity St. For example, the common origin 0 could be the center of the earth, S could be a set of axes fixed in the earth, and S o a set of axes with the same origin but with directions fixed relative to the distant stars. The frame S is convenient for us to use but is noninertial; the frame S o is relatively inconvenient, but is inertial. Let us now consider an arbitrary vector Q. This could be the velocity or position of a ball, the net force on an object, or any other vector of interest. Our first task is to relate the time rate of change of Q as measured in frame S o to the corresponding rate as measured in S. To distinguish these two rates of change, I shall temporarily use the = (rate of change of vector Q relative to inertial frame 3 0) following notation: (d Q dt) go and dQ \u2014 dt ( s = (rate of change of same vector Q relative to rotating frame S). To compare these two rates of change, I shall expand the vector Q in terms of three orthogonal unit vectors e l, e2, and e3 that are fixed in the rotating frame 8. (For Section 9.4 Time Derivatives in a Rotating Frame 341 instance, these three units vectors could point along the three solid axes shown in Figure 9.8.) Thus, 6 Q= Qiei + Q2e2 Q3e3 = E Qi ei 3 i= 1 (9.27) This expansion is chosen for the convenience of observers in the frame 8, since the unit vectors are fixed in that frame. Nonetheless, you should recognize that the expansion is equally valid in both frames. [Whichever frame we use, (9.27) is just an expansion of the one vector Q in terms of three othogonal vectors e 1, e2, and e3.1 The only difference is that for observers in S the vectors e 1, e2, and e3 are fixed, but as seen by observers in 80 the vectors e 1, e2, and e3 are rotating. Let us now differentiate the expansion (9.27) with respect to time. First, as seen in frame 8, the vectors e i are constant, and we get simply d()) dt ) v, dQ, e,. dt (9.28) [Since the expansion coefficients Q i in (9.27) are the same in either frame, we don't need to qualify the derivative on the right with a subscript S or S o l As seen in frame 8 0, the vectors e, vary with time. Thus, differentiating (9.27) in frame S o gives (dQ) dQ, dt ) 8 L7i t dt dei ) dt so (9.29) The derivative in the second term on the right is easily evaluated with the help of the \"useful relation\" (9.23). The vector e, is fixed in the frame 8, which is rotating with angular velocity S2 relative to S o. Therefore the rate of change of e i as seen in S o is given by (9.23) as (dei dt) so = S2 x ei. Thus, we can rewrite the second sum in (9.29) as / de EQA dt) so Q i (Stxei )=52x Qi ei =S2xQ. 6 Now is one of those moments when, as anticipated in Chapter 1, the notation e, with i = 1, 2, 3 is more convenient for our three unit vectors than X, if, and Z. For now, this is just because it lets us use the summation sign, E, in sums like (9.", "27). In Chapter 10, we'll find that the most convenient choice of rotating axes is to use the principal axes of the rotating body, and the notation e i works very naturally for them. Thus I shall mostly use this notation for the unit vectors fixed in a rotating body, and continue to use x, ST, and i for the nonrotating frame. 342 Chapter 9 Mechanics in Noninertial Frames Inserting this result into (9.29), and using (9.28) to replace the first sum, we find that This important identity relates the derivative of any one vector Q as measured in the inertial frame S. to the corresponding derivative in the rotating frame S. In the next section, I shall use it to find the form of Newton's second law in the rotating frame S. 9.5 Newton's Second Law in a Rotating Frame We are now ready to find the form of Newton's second law in the rotating frame S. To simplify matters, I shall assume that the angular velocity St of S relative to S. is constant, as is the case (to an outstanding approximation) for axes fixed to the earth. A rather surprising aspect of the statement that St is constant is that if it is true in one frame then it is automatically true in the other. This follows immediately from (9.30): Since St x52 = 0, the two derivatives of St are always the same; in particular, if one is zero, so is the other. Consider now a particle of mass m and position r. In the inertial frame S., the particle obeys Newton's second law in its normal form, / d2r\\ m t2) so =F (9.31) where as usual F denotes the net force on the particle, the vector sum of all forces identified in the inertial frame. The derivative on the left is, of course, the derivative evaluated by observers in the inertial frame S. However, we can now use Equation (9.30) to express this derivative in terms of the derivatives evaluated in the rotating frame S. First, according to (9.30) (dr) = (dr) + so dt 8 x ) r. Differentiating a second time, we find (d2r) d ) dr) cit 2 ) so ) so dt ) so _d ) Il dr) clt) so lAdt ) s = ( Applying (9.30) to the outside derivative on the right, we find ( 2 ) dt 2 )so 1 ( dr ) s L dt) s + x \u00b1 x [( dr ) + St x. (9.32) dt 8 Section 9.5 Newton's Second Law in a Rotating Frame 343 This rather messy result can be cleaned up. First, since our main concern is going to be with derivatives evaluated in the rotating frame 8, we'll revive the \"dot\" notation for these derivatives. That is, I shall use Q to denote = cl() dt 8 the derivative of any vector Q in the rotating frame 8. If we next note that, since Si is constant, its derivative is zero, and we group together two like terms, we can rewrite (9.32) as frame 8. (d 2t dt 2 ) 80 = + 212 x + S2 x (St x r) (9.33) where the dots on the right all indicate derivatives evaluated with respect to the rotating If we now substitute the result (9.33) into Newton's second law (9.31) for the inertial frame 8 0 and move two terms to the right, we find the form of Newton's second law for the rotating frame 8 to be where, as usual, F denotes the sum of all the forces as identified in any inertial frame. As with the accelerated frame of Section 9.1, we see that the equation of motion in a rotating reference frame looks just like Newton's second law, except that in this case there are two extra terms on the force side of the equation. The first of these extra terms is called the Coriolis force (after the French physicist G.G.de Coriolis, 1792-1843, who was the first to explain it), The second is the so-called centrifugal force I shall discuss these two terms in the next few sections. For now, the important point is that we can go ahead and use Newton's second law in rotating (and hence nonin- ertial) reference frames, provided we remember always to add these two \"fictitious\" 344 Chapter 9 Mechanics in Noninertiai Frames inertial forces to the net force F calculated for an inertial frame. That is, in a rotating frame,7 9.6 The Centrifugal Force We have just seen that in order to use Newton's second law in a rotating frame (such as a frame attached to the earth) we must introduce two inertial forces, the centrifugal and the", " Coriolis forces. To some extent, we can examine the two forces separately. In particular, the Coriolis force on an object is proportional to the object's velocity v = r relative to the rotating frame. Therefore, the Coriolis force is zero for any object that is at rest in the rotating frame, and it is negligible for objects that are moving sufficiently slowly. For the rest of this chapter, our main concern will be with the rotating frame of the earth, for which we can easily estimate the relative importance of the two inertial forces. Because both forces involve vector products, they depend on the directions of the various vectors, but for an order-of-magnitude estimate we can take Fcor mvC2 and Fcf mr Q2, where v is the object's speed relative to the rotating frame of the earth, that is, the speed as observed by us on the earth's surface. Therefore Fcor Fcf V RS2 V V (9.38) Here, in the middle expression, I have canceled a common factor of m Q and I have replaced r by the earth's radius R. (Remember that the origin is at the earth's center, so for objects near the earth's surface, r ti R.) In the last expression I have replaced RS-2 by V, the speed of a point on the equator as the earth rotates with angular velocity Q. Since V is approximately 1000 mi/h, (9.38) shows that for projectiles with v << 1000 mi/h it will be a good starting approximation to ignore the Coriolis force, and this is what I shall do in this section. 8 The centrifugal force is given by (9.36) as Fcf = m(St x r) x St. (9.39) 7 Our derivation of this important result hinged crucially on the relation (9.30) between time derivatives of a given vector in the rotating and nonrotating frames. If you find that relation confusing, you might prefer the alternative derivation based on the Lagrangian formalism and outlined in Problem 9.11. 8 As we shall see later, even when v << 1000 mi/h, the Coriolis force can have appreciable effects (for instance, with the Foucault pendulum). Nevertheless, it is certainly true that Fcor is small compared to Fa', and it makes sense to ignore the former at first. Section 9.6 The Centrifugal Force 345 Figure 9.9 The vector 52 x r is the velocity of an object as it is dragged eastward with speed Qp by the earth's rotation. Therefore, the centrifugal force, m (2 x r) x 2, points radially outward from the axis and has magnitude m Q2p. We can see what this looks like with the help of Figure 9.9, which shows an object on or near the earth's surface at colatitude O. The earth's rotation carries the object around a circle of latitude, and the vector Si x r (which is just the velocity of this circular motion as seen from a nonrotating frame) is tangent to this circle. Thus (St x r) x52 points radially outward from the axis of rotation in the direction of P, the unit vector in the p direction of cylindrical polar coordinates. The magnitude of (2 x r) x 52 is easily seen to be.2 2r sin B = Q 2p. Thus, Fcf = m Q2p P. (9.40) To summarize, from the point of view of observers rotating with the earth, there is a centrifugal force that is radially outward from the earth's axis and has magnitude m Q2p. If we momentarily let v = 52 x r denote the velocity associated with the earth's rotation (observed from a nonrotating frame), then its magnitude is v = Qp, and the centrifugal force takes the familiar form mv 2/p. Free-Fall Acceleration The free-fall acceleration that we call g is the initial acceleration, relative to the earth, of an object that is released from rest in a vacuum near the earth's surface. We can now see that this is actually a surprisingly complicated notion. The equation of motion (relative to the earth) is 9 Fgray Fcf (9.41) 9 I have defined g as the initial acceleration of a body released from rest to ensure that the Coriolis force is zero. When the object speeds up, we shall find that the Coriolis force eventually becomes important and the acceleration changes (although the effect is usually very small). 346 Chapter 9 Mechanics in Noninertial Frames where Fcf is given by (9.40) and Fgrav is the gravitational force Fgrav = r = mgo. GMm R2 (9.42) Here, M and R are the mass and radius of the earth, and 1- denotes the unit vector that", " points radially out from 0, the center of the earth. 10 The acceleration g, is defined by the second equality and could be called the \"true\" acceleration of gravity, inasmuch as it is the acceleration we would observe if there were no centrifugal effect. We see from (9.41) that the initial acceleration of a freely falling object is deter- mined by an effective force which is equal to the sum of two terms, Feff = Fgrav = mgo + m S22 R sin 9 p (9.43) where the last expression for Fcf comes from (9.40) with p replaced by R sin O. The two forces that make up the effective force are shown in Figure 9.10, from which it is clear that the free-fall acceleration is in general not equal to the true gravitational acceleration, either in magnitude or direction. Specifically, dividing (9.43) by m, we find for the free-fall acceleration g, g = go + Q2 R sin e p. (9.44) The component of g in the inward radial direction (the direction of \u20140 11 is grad = go \u2014 Q2 R sin2 0. (9.45) The second, centrifugal term is zero at the poles (9 = 0 or 7) and is largest at the equator, where its magnitude is easily found [using the value of Q from (9.26)] to be Q2 R = (7.3 x 10 -5 s-1 ) 2 x (6.4 x 106 m) ti 0.034 m/s 2. (9.46) Since go is about 9.8 m/s 2, we see that, because of the centrifugal force, the value of g at the equator is about 0.3% less than at the poles. 12 Although this difference is certainly small, it is easily measured with modern gravimeters that can measure g to about 1 part in 109. 1\u00b0 In claiming that the gravitational force is \u2014 (GMmIr 2)1, I am assuming that the earth is perfectly spherically symmetric, which, although a very good approximation, is not exactly true. Fortunately, all that matters is that Fgrav is certainly proportional to m, so it can always be written as mg,. For almost all purposes we can say that g o is in the direction of and it is certainly extremely close to that. 11 Strictly speaking this is the component of g in the direction of g o not \u2014 r. By the same token, the factor sin 2 9 should actually be sin 9 sin 9' where 9' is the angle between the line of g o and north (as opposed to 0, the angle between r and north). However, the difference (which is only because the earth is not perfectly spherically symmetric) is, for most,practical purposes, completely negligible. 12 The actual difference is more like 0.5%, the additional 0.2% being the result of the earth's bulge at the equator. Section 9.6 The Centrifugal Force 347 Figure 9.10 The free-fall acceleration (relative to the earth) of an object released from rest close to the earth's surface is the result of two force terms, the true gravitational force m go and the centrifugal, inertial force F cf, which points out from the axis of rotation. (The size of the centrifugal term is much exaggerated in this picture.) You can see from (9.44) and Figure 9.10 that the tangential component of g (the component normal to the true gravitational force) comes entirely from the centrifugal force and is gtang = SYR Sin 9 cos 6. (9.47) This tangential component of g is zero at the poles and at the equator, and is maximum at latitude 45\u00b0. The most striking feature of a nonzero value for gtang is that it means the free-fall acceleration is not exactly in the direction of the true gravitational force. As you can see from Figure 9.11, the angle between g and the radial direction is about a gtang/grad, and its maximum value (at 0 = 45\u00b0) is amax S2 2 R 2go 0.034 2 x 9.8 0.0017 rad 0.1\u00b0. (9.48) This angle a is the angle between the observed free-fall acceleration g and the true acceleration of gravity g o \u2014 what we might be tempted to call \"vertical.\" The value of a is actually rather difficult to measure. The direction of the observed g is easy (in principle, at least). To find the direction of g o, you might hope to use a plumb line, but a moment's thought should convince you that a plumb line is also subject to the centrifugal force and will hang in the direction of g, not that of g o. In fact, any attempt to", " find the direction of g o simply and directly winds up finding the direction of g. For this reason, in what follows I shall define \"vertical\" as the direction of a plumb line. Therefore, on those rare occasions when these tiny distinctions matter, \"vertical\" will mean \"in the direction of +g.\" By the same token, \"horizontal\" will mean \"perpendicular to g.\" 348 Chapter 9 Mechanics in Noninertial Frames Figure 9.11 Because of the centrifugal force, the free-fall accel- eration g has a nonzero tangential component (greatly exagger- ated here) and g deviates from the radial direction by the small angle a. 9.7 The Coriolis Force When an object is moving, there is a second inertial force that you must include when you want to use Newton's second law in a rotating frame. This is the Coriolis force (9.35) Fcor 2mi- x St = 2mv x 52 (9.49) where v = I. is the object's velocity relative to the rotating frame. There is a remarkable parallel between the Coriolis force and the well-known force qv x B on a charge q in a magnetic field B. Indeed, if we replace 2m by q and St by B, the former becomes exactly the latter. Although this parallel has no deep significance, it can often be a help in visualizing how the Coriolis force is going to affect a particle's motion. The magnitude of the Coriolis force depends on the magnitudes of v and 52 as well as their relative orientations. For the case that the rotating frame is the earth, we have seen in (9.26) that S2 ti 7.3 x 10 -5 s -1. For an object with v ti 50 m/s (a fast baseball, for example), the maximum acceleration the Coriolis force could produce (acting by itself and with v perpendicular to 52) would be ama\u201e = 2v C2 = 2 x (50 m/s) x (7.3 x 10 -5 S -1 ) ti 0.007 m/s2. Compared to the free-fall acceleration g = 9.8 m/s2, this is very small, though cer- tainly detectable if we were to take the trouble. Some projectiles, such as rockets and long-range shells, travel much faster than 50 m/s, and for them the Coriolis force is correspondingly more important. In addition, we shall see that there are systems, such as the Foucault pendulum, where the Coriolis force, though very small, can act for a long time and hence produce a large effect. Section 9.7 The Coriolis Force 349 Direction of the Coriolis Force Like the magnetic force qv x B, the Coriolis force 2mv x52 is always perpendic- ular to the velocity of the moving object, with its direction given by the right-hand rule. Figure 9.12 is an overhead view of a horizontal turntable that is rotating counter- clockwise relative to the ground. The angular velocity 52 points vertically up (out of the page in the figure). If we consider an object sliding or rolling across the turntable, it is easy to see that, whatever the object's position and velocity, the Coriolis force tends to deflect the velocity to the right. Similarly, if the turntable were rotating clock- wise, the Coriolis deflection would always be to the left. (Whether the object actually is deflected in the specified directions depends on what other forces are acting and how big they are.) We could imagine Figure 9.12 to be the Northern Hemisphere viewed from above the North Pole. (Since the earth rotates to the east, the angular velocity is directed as shown.) Thus we reach the conclusion that the Coriolis effect due to the earth's rotation tends to deflect moving bodies to the right in the Northern Hemisphere (and, of course, to the left in the Southern Hemisphere). 13 This effect is important to long- range gunners, who must aim to the left of their target in the Northern, and to the right in the Southern, Hemisphere. (See Problem 9.28.) An important example from Figure 9.12 Overhead view of a horizontal turntable that is rotating counterclockwise relative to an inertial frame. The turntable's an- gular velocity S2 points up out of the picture. As seen by observers on the turntable, the two objects sliding on the table are subject to Coriolis forces Feor = 2mv x St. Irrespective of the bodies' posi- tions and velocities, the Coriolis force always tends to deflect the velocity to the right. (If the direction of rotation were clockwise, then S2 would be into the page and", " the Coriolis force would tend to deflect moving objects to the left.) 13 Because the earth is three-dimensional (as opposed to a turntable, which is two-dimensional) the Coriolis effect is actually a little more complicated than this simple statement suggests. However, the statement above is certainly correct for objects moving parallel to the earth's surface and for low trajectory projectiles. 350 Chapter 9 Mechanics in Noninertial Frames low pressure Figure 9.13 A cyclone is the result of air moving into a low- pressure region and being deflected to the right (in the Northern Hemisphere) by the Coriolis effect. This causes a counterclock- wise flow with the inward pressure force balancing the outward Coriolis force. meteorology is the phenomenon of cyclones. These occur when the air surrounding a region of low pressure moves rapidly inward. Because of the Coriolis effect, the air is deflected to the right, as shown in Figure 9.13, and therefore begins to circulate counterclockwise (in the Northern Hemisphere \u2014 clockwise in the Southern). When this happens sufficiently violently, the result is a storm, known variously as a cyclone, hurricane, or typhoon. It is important to bear in mind that both the Coriolis and centrifugal forces are at root kinematic effects, resulting from our insistence on using a rotating frame of reference. In a few simple cases, it is actually easier (as well as instructive) to analyze the motion in an inertial frame and then transform the results to the rotating frame, as the following example illustrates. Nevertheless, the transformation between the two frames is usually so complicated that it is easier to work all the time in the rotating frame and to live with the \"fictitious\" Coriolis and centrifugal forces. EXAMPLE 9.2 Simple Motion on a Turntable Three observers A, B, and C are standing on a horizontal turntable with A at the center, C at the edge and B halfway between, as shown in Figure 9.14(a). The turntable is rotating counterclockwise (as seen from above) with angular velocity SI At time t = 0, A kicks a frictionless puck exactly toward B and C, but to his surprise the puck misses both B and C, the latter by an even bigger margin than the former. Explain these events from the points of view both of the observers on the rotating table and of an observer on the ground. The net force on the puck (as identified in any inertial frame) is zero. Thus in the rotating frame, the only two forces are the centrifugal and Coriolis forces. The former is always radially outward and has no bearing on the puck's deflection. The latter consistently deflects the puck's velocity to the right, just like an upward magnetic field acting on a positive charge. This causes the puck to follow the curved path shown in Figure 9.14(a). At time t 1 when it reaches Section 9.8 Free Fall and the Coriolis Force 351 (a) As seen in rotating frame (b) As seen in non-rotating frame Figure 9.14 (a) Three observers A, B, and C are in a line on a rotating turntable, with A at the center and C at the perimeter. Observer A kicks a puck toward B and C, but because of the Coriolis force the puck veers to the right and misses B and C. (b) The same experiment as seen by an observer on the ground. In this frame, the puck travels in a straight line, but by the time t 1 when it reaches the radius of B, observer B has moved to the left. By the time t 2 when the puck reaches the perimeter, C has moved even further to the left. the radius of B, it is a small distance to the right of B, and at time t2 when it arrives at the turntable's edge, it is about twice as far to the right of C. This explanation is correct and clear, but depends on our understanding the Coriolis force. By analyzing the same experiment in the ground-based frame, we can gain an additional understanding of why the deflection occurs. In the inertial frame of an observer on the ground, the net force on the puck is zero, and the puck follows a straight path, as shown in Figure 9.14(b). However, by the time t 1, when it should have been hitting B, observer B has moved to the left as the turntable rotates. By time t 2, when it should have been hitting C, observer C has moved even further to the left. As seen by the puck, B and C have moved to the left. Therefore, as seen by B and C, the puck has curved to the right as in Figure 9.14(a). This simple alternative explanation of the Coriolis effect is actually decep- tively simple. In general", ", the effects of the Coriolis and centrifugal forces are surprisingly complicated and are not nearly so easy to explain by reference to the nonrotating frame. (See Problems 9.20 and 9.24.) 9.8 Free Fall and the Coriolis Force Let us next consider the effect of the Coriolis force on a freely falling object, that is, an object falling in a vacuum, close to a point R on the earth's surface. For this analysis, we must also include the centrifugal force, so the equation of motion is mr =mgo +FCf + Fear (9.50) 352 Chapter 9 Mechanics in Noninertial Frames where, as before, mg o denotes the true force of the earth's gravity on the object. The centrifugal force is m (S2 x r) x S2, (where r is the object's position relative to the center of the earth), but to an outstanding approximation we can replace r by R (the position on the earth's surface where the experiment is being conducted). Thus, Fcf= m(52xR)x52. Returning to the equation of motion (9.50), you will recognize that the sum of the first two terms on the right is just mg, where g is the observed free-fall acceleration for an object released from rest at position R, as introduced in (9.44). In other words, we can omit the term Fef from (9.50), if we replace g o by the observed g at the location of our experiment. If we substitute 2mv x S2 for Fcor, the equation of motion becomes (after cancellation of a factor of m) = g x S2. (9.51) A simplifying feature of the equation (9.51) is that it does not involve the position r at all (only its derivatives ands). This means the equation will not change if we make a change of origin (since a change of origin amounts to adding a constant to r). Accordingly, I shall now choose my origin on the surface of the earth at the position R, as shown in Figure 9.15. With this choice of axes, we can resolve the equation of motion into its three components. The components ofr and S2 are = (.i, 5), ) z (up) Figure 9.15 Choice of axes for a free-fall experiment. The origin 0 is on the earth's surface at the experiment's location (position R relative to the center of the earth). The z axis points vertically up (more precisely, in the direction of \u2014g, where g is the observed free-fall acceleration), the x and y axes are horizontal (that is, perpendicular to g), with y pointing north, and x due east. The position of the falling object relative to 0 is r. Section 9.8 Free Fall and the Coriolis Force 353 and Thus, those ofr x 2 are = (0, Q sin 9, S2 cos 9). r x 52 = (S7S2 cos 0 \u2014 iS2 sin 0, cos 9, IS2 sin 0) (9.52) and the equation of motion (9.51) resolves into the following three equations: = 2S2(57cos 9 \u2014 sin 0) \u20142S2.i cos 0 E = \u2014g 2S2.i sin O. (9.53) We shall solve these three equations by making a succession of approximations that depend on the smallness of Q. First, because S2 is very small, we get a reasonable starting approximation if we ignore Q entirely. In this approximation, the equations reduce to = 0, y = 0, and E = \u2014g, (9.54) which are the equations of free fall solved in every introductory physics course. If the object is dropped from rest at x = y = 0 and z = h, then the first two equations imply that x, 5), x, and y all remain zero, while the last equation implies that z = \u2014gt and z = h \u2014 Igt 2. Thus our approximate solution is x = 0, y = 0, and z = h \u2014 2gt 2, (9.55) that is, the object falls vertically down with constant acceleration g. This approxima- tion is sometimes called the zeroth-order approximation because it involves only the zeroth power of Q (that is, it is independent of Q). It is well known to be a very good approximation, but it shows none of the effects of the Coriolis force. To get the next approximation, we argue as follows: The terms in (9.53) that involve Q are all small. Thus, it will be safe to evaluate these terms using our zeroth-order approximation for x, y, and z. Substituting (9.55) into the right side of (9.53), we get = 2S2gt sin 0, y", " = 0, and E = \u2014g. (9.56) The last two of these are exactly the same as in zeroth order, but the equation for x is new and is easily integrated twice to give x = -3 S2gt 3 sin 0, 1 (9.57) with y and z the same as in the zeroth approximation (9.55). This result is naturally called the first-order approximation (being good through the first power of S2). We can repeat this process again to get a second-order approximation and so on, but the first-order is good enough for our purposes. The striking thing about the solution (9.57) is that a freely falling object does not fall straight down. Instead the Coriolis force causes it to curve slightly to the east (positive x direction). To get an idea of the magnitude of the effect, consider an object 354 Chapter 9 Mechanics in Noninertial Frames dropped down a 100-meter-deep mine shaft at the equator, and let us find the total deflection by the time it hits the bottom. The time to reach the bottom is determined by the last of equations (9.55) as t =,/2h/g, and (9.57) gives for the total easterly deflection (putting 0 = 90\u00b0 and g ti 10 m/s2) X = (2h ) \"2 S2g \u2014 1 3 1 - 3 ;s--' x (7.3 x 10 -5 s -1 ) x (10 m/s 2) x (20 s2) 312 2.2 cm a small deflection, but certainly detectable. A small easterly deflection of this type was actually predicted by Newton and verified by his rival Robert Hooke (of Hooke's law fame, 1635-1703), although it was not properly explained until the Coriolis effect was understood. 9.9 The Foucault Pendulum As a final and striking application of the Coriolis effect, let us consider the Foucault pendulum, which can be seen in many science museums around the world and is named for its inventor, the French physicist Jean Foucault (1819-1868). This is a pendulum made of a very heavy mass m suspended by a light wire from a tall ceiling. This arrangement allows the pendulum to swing freely for a very long time and to move in both the east-west and north-south directions. As seen in an inertial frame, there are just two forces on the bob, the tension T in the wire and the weight mgo. In the rotating frame of the earth, there are also the centrifugal and Coriolis forces, so the equation of motion in the earth's frame is = T + mgo m (S2 x r)xSt+2m1 x S2. Exactly as in the previous section, the second and third terms on the right combine to give mg, where g is the observed free-fall acceleration, and the equation of motion becomes mr = T + mg + 2mr x St. (9.58) We can now choose our axes as in the previous section, so that x is east, y is north, and z vertically up (direction of - g), and the pendulum is as shown in Figure 9.16. I shall restrict our discussion to the case of small oscillations, so that the angle f3 between the pendulum and the vertical is always small. This allows two simplifying approximations: First, the z component of the tension T is well approximated by the magnitude; that is, 1\", = T cos 13 ti T. Second, it is not hard to see that, for small oscillations, Tz ti mg. 14 Putting these two approximations together, we can write T ti mg. (9.59) 14 Look at the z component of (9.58). In the limit of small oscillations, the term on the left and the last term on the right both approach zero, and you're left with 7; \u2014 mg = 0. Section 9.9 The Foucault Pendulum 355 Figure 9.16 A Foucault pendulum comprises a bob of mass m suspended by a light wire of length L from the point P on a high ceiling. The tension force on the bob is shown as T and its x and y components are Tx and Ty. For small oscillations the angle /3 is very small. We now need to examine the x and y components of the equation of motion (9.58). This requires that we identify the x and y components of T. If you look at Figure 9.16, you will see that, by similar triangles, Tx / T = \u2014x/L and similarly for Ty. Combining this with (9.59), we find that Tx = \u2014mgx/L and Ty = \u2014mgy/L.", " (9.60) The x and y components of g are, of course, zero, and the components of I- x St are given in (9.52). Putting all of these into (9.58), we find (after canceling a factor of m and dropping a term involving z, which is negligible compared to z or 5) for small oscillations).3i' = \u2014gxIL \u00b1 2.9S-2 cos 0 33 = \u2014gy I L \u2014 2.iS -2 cos 0. where as usual 0 denotes the colatitude of the location of the experiment. The factor g/L is just (pc:2,', where co o is the natural frequency of the pendulum, and S2 cos 0 is just Qz, the z component of the earth's angular velocity. Thus these two equations of motion can be rewritten as (9.61) (9.62) We can solve the coupled equations (9.62) using the trick, introduced in Chapter 2, of defining a complex number 1 \u2014 2Qz.5; + cox = 0 1 + 2S2,5c + o_)02y = 0. I = x iy. 356 Chapter 9 Mechanics in Noninertial Frames Recall that not only does this complex number contain the same information as the position in the xy plane, but a plot of in the complex plane is an actual bird's eye view of the pendulum's projected position (x, y). If we multiply the second equation of (9.62) by i and add it to the first, we get the single differential equation + 2i + coo2 q = 0. (9.63) This is a second-order, linear, homogeneous differential equation and so has exactly two independent solutions. Thus if we can find two independent solutions, we shall know that the most general solution is a linear combination of these two. As often happens, we can find two independent solutions by inspired guesswork: We guess that there is a solution of the form ri(t) (9.64) for some constant a. Substituting this guess into (9.63), we see immediately that it is a solution if and only if a satisfies a 2 2s-2zoi _ co 2 0 0 = Qz \\/s-2z2 wo2 or is ti^Z \u00b1 (00 (9.65) where the last line is an extremely good approximation since the earth's angular velocity Q is so very much smaller than the pendulum's coo. This gives us the required two independent solutions, and the general solution to the equation of motion (9.63) = e \u2014ic2 z (C l e i wo l C2 e \u2014i w.t ). (9.66) To see what this solution looks like, we need to fix the two constants C 1 and C2 by specifying the initial conditions. Let us suppose that at t = 0 the pendulum has been pulled aside in the x direction (east) to a position x = A and y = 0, and is released from rest (vx\u201e = vyo = 0). With these initial conditions, you can easily check that' s l = C2 = A/2, and our solution becomes r7(t) E_-- x(t) iy(t) = Ae \u2014ic2 z t cos coo t. At t = 0 the complex exponential is equal to one, and x = A, while y = 0. Because Q, << wo, the cosine factor in (9.67) makes many oscillations before the exponential changes appreciably from one. This implies that, initially, x(t) oscillates with angular frequency cot, between \u00b1A, while y remains close to zero. That is, initially, the pendulum swings in simple harmonic motion along the x axis, as indicated in Figure 9.17(a). C (9.67) 15 Actually, there is a small subtlety, in that these simple values depend on the (true) assumption that S-2, << wo, as you will see when you check them. Section 9.9 The Foucault Pendulum 357 y x = A cos co o t A / A \u2014A (a) t 0 (b)t>0 Figure 9.17 Overhead views of the motion of a Foucault pendulum. (a) For a while after being released, the pendulum swings back and forth along the x axis, with amplitude A and frequency wo. (b) As time advances, the plane of its oscillations slowly rotates with angular velocity equal to S2,, the z component of the earth's angular velocity. However, eventually the complex exponential e -iQz t begins to change, causing the complex number ri = x iy to rotate through an angle Q z t. In the Northern Hemisphere, where Q is", " positive, this means that the number x + iy continues to oscillate sinusoidally (due to the factor cos coot), but in a direction that rotates clockwise. That is, the plane in which the pendulum is swinging rotates slowly clockwise, with angular velocity Q 2, as indicated in Figure 9.17(b). In the Southern Hemisphere, where Q z is negative, the corresponding rotation is counterclockwise. If the Foucault pendulum is located at colatitude 9 (latitude 90\u00b0 \u2014 9), then the rate at which its plane of oscillation rotates is Q, = Q cos O. (9.68) At the North Pole (0 = 0), Q, = Q and the rate of rotation of the pendulum is the same as the earth's angular velocity. This result is easy to understand: As seen in an inertial (nonrotating) frame, a Foucault pendulum at the North Pole would obviously swing in a fixed plane; meanwhile, as seen in the same inertial frame, the earth is rotating counterclockwise (as seen from above) with angular velocity Q. Clearly then, as seen from the earth, the pendulum's plane of oscillation has to be rotating clockwise with angular velocity Q. At any other latitude, the result is much more complicated from an inertial point of view, but the rate of rotation of the Foucault pendulum is easily calculated from (9.68). At the equator (9 = 90\u00b0), Q, = 0 and the pendulum does not rotate. At a latitude around 42\u00b0 (the approximate latitude of Boston, Chicago, or Rome), Q, = Q cos 48\u00b0 ti 3S. Since Q equals 360\u00b0/day, 3 Q = 240\u00b0/day, and we see that in the course of 6 hours (a time for which a long, well-built pendulum will certainly continue to swing without significant damping), the pendulum's plane of motion will rotate through 60\u00b0 \u2014 an easily observable effect. 358 Chapter 9 Mechanics in Noninertial Frames 9.10 Coriolis Force and Coriolis Acceleration Recall that in Equation (1.48) of Chapter 1 we found the form of Newton's second law in two-dimensional polar coordinates, F = mie I Fr = m(F \u2014 4,2) F = m(rq + q5). (9.69) We can now understand the rather ugly last term in each of the two equations on the right in terms of the centrifugal and Coriolis forces. Consider a particle that is subject to a net force F and moves in two dimensions. (The exact same analysis works in three dimensions using cylindrical polar coordi- nates, but for simplicity I shall work in two dimensions.) Relative to any inertial frame with origin 0, the particle must satisfy (9.69). Now consider a noninertial frame 3' which shares the same origin 0 and is rotating at constant angular velocity Q, chosen so that Q = 4 at one chosen time t = to. That is, at the chosen instant t o, the frame 3' and the particle are rotating at the same rate. (For this reason, the frame 3' is sometimes called the co-rotating frame.) If the particle has polar coordinates (r', 0 1) relative to 3', then at all times (since 3 and 3' share the same origin), and at the time to r' = r = 0 since the frame 3' and the particle are rotating at the same rate at t = to. Newton's second law can be applied in the frame 3', provided we include the centrifugal and Coriolis forces. Thus F Fcor = (9.70) Let us write this equation in polar coordinates: The centrifugal force Fef is purely radial, with radial component r S22. (Remember that r' = r, so it makes no difference whether we write r or r'.) The Coriolis force Fcor is 2mv' x St, and, since v' is purely radial in the co-rotating frame, F ecr is in the 0' direction with 0' component \u20142mi- Q. Finally the term mi' on the right of (9.70) can be replaced by the analog of (9.69), except that in the co-rotating frame = 0 (at the chosen time to), so the terms containing ci will be absent. Putting all this together, we find for the equation of motion of the particle in the co-rotating frame, F + Fa \u00b1 Fcor = mi'' t i Fr + mrS22 = mi; I Fo \u2014 2mi-S2 = m4. (9.71) (Because the frame 3' is rotating at a constant\u2018rate, I could replace iii.' by 4 since they are equal.) Principal Definitions and Equations of", " Chapter 9 359 Let us now compare the equation of motion (9.69) for the inertial frame, with (9.71) for the co-rotating frame. The most important thing to recognize is that, because c.p, they are exactly the same equations for r and 0, although certain terms are distributed differently between the two sides. In (9.69) for the nonrotating frame, the only force terms on the left are the real net force, with components Fr and F,. On the right of (9.69), the acceleration contains the centripetal acceleration \u20141- c */).2 in its radial component and the Coriolis acceleration 24 in its 0 component. In (9.71) for the rotating frame, neither of these additional acceleration terms is present (because we arranged that 0' is zero), but instead they are reincarnated on the force side of the equations (with opposite signs, of course) as the centrifugal force m SZ2r in the radial equation and Coriolis force \u20142mi- Q in the 0 equation. Since the two versions of the equations are the same, it is clear that they are equally correct. In the inertial frame, the forces are simpler (no \"fictitious\" forces) but the accelerations are more complicated; in the rotating frame, it is the other way round. Which frame one chooses to use is dictated by convenience. In particular, when the observer is anchored to a rotating frame (as we earthlings are), it is generally more convenient to work in the rotating frame and to learn to live with the \"fictitious\" centrifugal and Coriolis forces. Principal Definitions and Equations of Chapter 9 Inertial Force in an Accelerating but Nonrotating Frame The motion of a body, as seen in a frame that has acceleration A relative to an inertial frame, can be found using Newton's second law in the form nil' = F + Finertial, where F is the net force on the body (as measured in any inertial frame) and Finertiai is an additional inertial force Finertial = \u2014 mA [Eq. (9.5)] The Angular Velocity Vector If a body is rotating about an axis specified by the unit vector u (direction given by the right-hand rule) at a rate co (usually measured in radians per second), its angular velocity vector is defined as (1) = WU. [Eq. (9.21)] The \"Useful Relation\" co is The velocity of a point r fixed in a rigid body that is rotating with angular velocity v = ot) x r. [Eq. (9.22)] 360 Chapter 9 Mechanics in Noninertial Frames Time Derivatives in a Rotating Frame If frame S has angular velocity 52 relative to frame S o, then the time derivatives of a single vector Q as seen in the two frames are related by (c/Q (dQ dt ) 80 dt /8 52 x Q. [Eq. (9.30)] Newton's Second Law in a Rotating Frame If frame S has angular velocity 52 relative to an inertial frame S o, then Newton's second law in the rotating frame takes the form mr = F + Fcor Fcf, [Eq. (9.37)] where F is the net force on the body (as measured in any inertial frame) and the inertial forces Fc0,. and Fa are the Coriolis and centrifugal forces, Fcor = 2mt x S2 and Fcf = m(52 x r) x 52. [Eqs. (9.35) & (9.36)] Free-Fall Acceleration The observed free-fall acceleration g (defined as the initial acceleration, relative to the earth, from rest) includes the \"true\" gravitational acceleration g, and the effect of the centrifugal force g = g, + (S2 x R) x 52. [Eq. (9.44)] \"Vertical\" is defined as the direction of g, and \"horizontal\" as perpendicular to g. Problems for Chapter 9 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). SECTION 9.1 Acceleration without Rotation 9.1 * Be sure you understand why a pendulum in equilibrium hanging in a car that is accelerating forward tilts backward, and then consider the following: A helium balloon is anchored by a massless string to the floor of a car that is accelerating forward with acceleration A. Explain clearly why the balloon tends to tilt forward and find its angle of tilt in equilibrium. [Hint: Helium balloons float because of the buoyant Archimedean force, which results from a pressure gradient in the air. What is the relation between the directions of the gravitational field and the buoyant force?] 9.2 * A donut-shaped space station (outer radius R)", " arranges for artificial gravity by spinning on the axis of the donut with angular velocity co. Sketch the forces on, and accelerations of, an astronaut standing in the station (a) as seen from an inertial frame outside the station and (b) as seen in the astronaut's personal rest frame (which has a centripetal acceleration A = to 2 R as seen in the inertial frame). What angular velocity is needed if R = 40 meters and the apparent gravity is to equal the usual value of about 10 m/s 2? (c) What is the percentage difference between the perceived g at a six-foot astronaut's feet (R = 40 m) and at his head (R = 38 m)? Problems for Chapter 9 361 SECTION 9.2 The Tides 9.3 ** (a) Consider the tidal force (9.12) on a mass m at the position P of Figure 9.4. Write d as (do \u2014 R e) = do (1 \u2014 R e/ do) and use the binomial approximation (1 \u2014 c) -2 1 + 2E to show that Fria \u2014(2G M inm R e l do3) X. Confirm the direction of the force shown in Figure 9.4 and make a numerical comparison of the tidal force with the gravitational force mg of the earth. (b) Do the corresponding calculations for the force at the point R. Compare this force with that of part (a) (magnitude and direction). 9.4 ** Do the same calculations as in Problem 9.3(a) but for the tidal force at the point Q in Figure 9.4. [In this case write d/d 2 = d/d3 and use the binomial approximation in the form (1 + c) -3 = 1 \u2014 3E.] 9.5 Irk Review the derivation of the tidal potential energy (9.16) of a drop of water at the point Q in Figure 9.5 and then give in detail the derivation of (9.17) for the tidal PE at the point P. 9.6*** Let h (8) denote the height of the ocean at any point T on the surface, where h (8) is measured up from the level at the point Q of Figure 9.5 and 6 is the polar angle TOR of T. Given that the surface of the ocean is an equipotential, show that h(8) = ho cost 8, where ho = 3 Mn, R:/ (2 Me clo3). Sketch and describe the shape of the ocean's surface, bearing in mind that h o << Re. [Hint: You will need to evaluate Utid ( T ) as given by (9.13), with d equal to the distance MT. To do this you need to find d by the law of cosines and then approximate d -1 using the binomial approximation, being very careful to keep all terms through order (Re /do) 2. Neglect any effects of the sun.] SECTION 9.4 Time Derivatives in a Rotating Frame 9.7 * (a) Explain the relation (9.30) between the derivatives of a vector Q in two frames S o and S for the special case that Q is fixed in the frame S. (b) Do the same for a vector Q that is fixed in the frame So and compare with your answer to part (a). SECTION 9.5 Newton's Second Law in a Rotating Frame 9.8 * What are the directions of the centrifugal and Coriolis forces on a person moving (a) south near the North Pole, (b) east on the equator, and (c) south across the equator? 9.9 * A bullet of mass m is fired with muzzle speed v o horizontally and due north from a position at colatitude 8. Find the direction and magnitude of the Coriolis force in terms of m, v o, 8, and the earth's angular velocity Q. How does the Coriolis force compare with the bullet's weight if v o = 1000 m/s and 0 = 40 deg? 9.10 ** The derivation of the equation of motion (9.34) for a rotating frame made the assumption that the angular velocity St was constant. Show that if St 0 then there is a third \"fictitious force,\" sometimes called the azimuthal force, on the right side of (9.34) equal to mr x 9.11 *** In this problem you will prove the equation of motion (9.34) for a rotating frame using the Lagrangian approach. As usual, the Lagrangian method is in many ways easier than the Newtonian (except that it calls for some slightly tricky vector gymnastics), but is perhaps less insightful. Let S be a noninertial frame rotating with constant angular velocity St relative to the inertial frame S o. Let both 362 Chapter 9 Mechanics in", " Noninertial Frames frames have the same origin, 0 = 0'. (a) Find the Lagrangian L = T \u2014 U in terms of the coordinates r and r of S. [Remember that you must first evaluate T in the inertial frame. In this connection, recall that vo = v + St x r.] (b) Show that the three Lagrange equations reproduce (9.34) precisely. SECTION 9.6 The Centrifugal Force 9.12 * (a) Show that to design a static structure in a rotating frame (such as a space station) one can use the ordinary rules of statics except that one must include the extra \"fictitious\" centrifugal force. (b) I wish to place a puck on a rotating horizontal turntable (angular velocity Q) and to have it remain at rest on the table, held by the force of static friction (coefficient p,). What is the maximum distance from the axis of rotation at which I can do this? (Argue from the point of view of an observer in the rotating frame.) the magnitude of a. 9.13 * Show that the angle a between a plumb line and the direction of the earth's center is well approximated by tan a = (R,Q2 sin 26)/(2g), where g is the observed free-fall acceleration and we assume the earth is perfectly spherically symmetric. Estimate the maximum and minimum values of 9.14 ** I am spinning a bucket of water about its vertical axis with angular velocity C2. Show that, once the water has settled in equilibrium (relative to the bucket), its surface will be a parabola. (Use cylindrical polar coordinates and remember that the surface is an equipotential under the combined effects of the gravitational and centrifugal forces.) 9.15 ** On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the North Pole and g = Ago at the equator (with 0 < A < 1). Find g(9), the free- fall acceleration at colatitude 9 as a function of 9. SECTION 9.7 The Coriolis Force 9.16 * The center of a long frictionless rod is pivoted at the origin and the rod is forced to rotate at a constant angular velocity Q in a horizontal plane. Write down the equation of motion for a bead that is threaded on the rod, using the coordinates x and y of a frame that rotates with the rod (with x along the rod and y perpendicular to it). Solve for x (t). What is the role of the centrifugal force? What of the Coriolis force? 9.17 * Consider the bead threaded on a circular hoop of Example 7.6 (page 260), working in a frame that rotates with the hoop. Find the equation of motion of the bead, and check that your result agrees with Equation (7.69). Using a free-body diagram, explain the result (7.71) for the equilibrium positions. 9.18 ** A particle of mass m is confined to move, without friction, in a vertical plane, with axes x horizontal and y vertically up. The plane is forced to rotate with constant angular velocity Q about the y axis. Find the equations of motion for x and y, solve them, and describe the possible motions. 9.19 ** I am standing (wearing crampons) on a perfectly frictionless flat merry-go-round, which is rotating counterclockwise with angular velocity Q about its vertical axis. (a) I am holding a puck at rest just above the floor (of the merry-go-round) and release it. Describe the puck's path as seen from above by an observer who is looking down from a nearby tower (fixed to the ground) and also as seen by me on the merry-go-round. In the latter case explain what I see in terms of the centrifugal and Coriolis forces. (b) Answer the same questions for a puck which is released from rest by a long-armed spectator who is standing on the ground leaning over the merry-go-round. Problems for Chapter 9 363 9.20 ** Consider a frictionless puck on a horizontal turntable that is rotating counterclockwise with angular velocity Q. (a) Write down Newton's second law for the coordinates x and y of the puck as seen by me standing on the turntable. (Be sure to include the centrifugal and Coriolis forces, but ignore the earth's rotation.) (b) Solve the two equations by the trick of writing ri = x iy and guessing a solution of the form /I = e-`at. [In this case \u2014 as in the case of critically damped SHM discussed in Section 5.4 you get only one solution this way. The other has the same form (5.43) we found for the second", " solution in damped SHM.] Write down the general solution. (c) At time t = 0, I push the puck from position ro = (x0, 0) with velocity v, = (vxo, vyo) (all as measured by me on the turntable). Show that x(t) = (xo v xo t) cos Qt (v yo Cbco)t sin Qt y(t) = \u2014(x 0 + v xo t) sin Qt + (v,,0 + Qxo)t cos Qt I (9.72) (d) Describe and sketch the behavior of the puck for large values of t. [Hint: When t is large the terms proportional to t dominate (except in the case that both their coefficients are zero). With t large, write (9.72) in the form x(t) = t (B i cos Qt B2 sin Qt), with a similar expression for y(t), and use the trick of (5.11) to combine the sine and cosine into a single cosine \u2014 or sine, in the case of y(t). By now you can recognize that the path is the same kind of spiral, whatever the initial conditions (with the one exception mentioned).] 9.21 ** When a puck slides on a rotating turntable, as in Problems 9.20 and 9.24, it can come instantaneously to rest. Sketch the shape of the path when this happens and explain. If you did Problem 9.24, comment on the relevance of this result to part (d) of that problem. 9.22 ** If a negative charge \u2014q (an electron, for example) in an elliptical orbit around a fixed positive charge Q is subjected to a weak uniform magnetic field B, the effect of B is to make the ellipse precess slowly \u2014 an effect known as Larmor precession. To prove this, write down the equation of motion of the negative charge in the field of Q and B. Now rewrite it for a frame rotating with angular velocity Si [Remember that this changes both d 2 r/dt 2 and drldt.] Show that by suitable choice of St you can arrange that the terms involvingr cancel out, but that you are left with one term involving B x (B x r). If B is weak enough this term can certainly be neglected. Show that in this case the orbit in the rotating frame is an ellipse (or hyperbola). Describe the appearance of the ellipse as seen in the original nonrotating frame. 9.23 ** Here is an unusual way to solve the two-dimensional isotropic oscillator \u2014 the motion of a particle subject to a force \u2014kr. Show that by choosing a suitable rotating reference frame, you can arrange that the centrifugal force exactly cancels the force \u2014kr. Recalling the analogy between the Coriolis and magnetic forces, you should be able to write down the general solution for the motion as seen in the rotating frame. If you write your solution in the complex form of Section 2.7, then you can transform back to the nonrotating frame by multiplying by a suitable rotating complex number. Show that the general solution is an ellipse. [See Problem 8.11 for some guidance on this last part.] 9.24 *** [Computer] Use a suitable plotting program (such as ParametricPlot in Mathematica) to plot the orbits (9.72) of the puck of Problem 9.20 on a rotating turntable with x o = Q = 1 and the following initial velocities vo : (a) (0, 1), (b) (0, 0), (c) (0, \u20141), (d) (-0.5, \u20140.5), (e) ( - 0.7, \u20140.7), (f) (0, \u20140.1). Comment on any interesting features. 364 Chapter 9 Mechanics in Noninertial Frames SECTION 9.8 Free Fall and the Coriolis Force 9.25 * A high-speed train is traveling at a constant 150 m/s (about 300 mph) on a straight, horizontal track across the South Pole. Find the angle between a plumb line suspended from the ceiling inside the train and another inside a but on the ground. In what direction is the plumb line on the train deflected? 9.26 ** In Section 9.8, we used a method of successive aproximations to find the orbit of an object that is dropped from rest, correct to first order in the earth's angular velocity Q. Show in the same way that if an object is thrown with initial velocity vo from a point 0 on the earth's surface at colatitude 0, then to first order in Q its orbit is x = vxot Q (v vo cos 0 \u2014 v zo sin 6)t2 4C", "2gt3 sin 0 y = vyo t \u2014 Q (vxo cos 0)t 2 z = vzot \u2014 Zgt 2 + (vxo sin 0)t 2. (9.73) [First solve the equations of motion (9.53) in zeroth order, that is, ignoring Q entirely. Substitute your zeroth-order solution for and z into the right side of equations (9.53) and integrate to give the next approximation. Assume that vo is small enough that air resistance is negligible and that g is a constant throughout the flight.] 9.27 ** In Section 9.8, we discussed the path of an object that is dropped from a very tall stepladder above the equator. (a) Sketch this path as seen from a tower to the north of the drop and fixed to the earth. Explain why the object lands to the east of its point of release. (b) Sketch the same experiment as seen by an inertial observer floating in space to the north of the drop. Explain clearly (from this point of view) why the object lands to the east of its point of release. [Hint: The object's angular momentum about the earth's center is conserved. This means that the object's angular velocity cb changes as it falls.] 9.28 ** Use the result (9.73) of Problem 9.26 to do the following: A naval gun shoots a shell at colatitude 0 in a direction that is a above the horizontal and due east, with muzzle speed v o. (a) Ignoring the earth's rotation (and air resistance), find how long (t) the shell would be in the air and how far away (R) it would land. If v o = 500 m/s and a = 20\u00b0, what are t and R? (b) A naval gunner spots an enemy ship due east at the range R of part (a) and, forgetting about the Coriolis effect, aims his gun exactly as in part (a). Find by how far north or south, and in which direction, the shell will miss the target, in terms of Q, vo, a, 0, and g. (It will also miss in the east\u2014west direction but this is perhaps less critical.) If the incident occurs at latitude 50\u00b0 north (0 = 40\u00b0), what is this distance? What if the latitude is 50\u00b0 south? This problem is a serious issue in long-range gunnery: In a battle near the Falkland Islands in World War I, the British navy consistently missed German ships by many tens of yards because they apparently forgot that the Coriolis effect in the southern hemisphere is opposite to that in the north. 9.29 ** (a) A baseball is thrown vertically up with initial speed v o from a point on the ground at colatitude 0. Use the solution (9.73) of Problem 9.26 to show that the ball will return to the ground a distance (4Q v o3 sin 0) / (3g 2 ) to the west of its launch point. (b) Estimate the size of this effect on the equator if vo = 40 m/s. (c) Sketch the ball's orbit as seen from the north (by an observer fixed to the earth). Compare with the orbit of a ball dropped from a point above the equator, and explain why the Coriolis effect moves the dropped ball to the east, but the thrown ball to the west. 9.30 *** The Coriolis force can produce a torque on a spinning object. To illustrate this, consider a horizontal hoop of mass m and radius r spinning with angular velocity w about its vertical axis at colatitude 0. Show that the Coriolis force due to the earth's rotation produces a torque of magnitude mw 2r 2 sin 0 directed to the west, where Q is the earth's angular velocity. This torque is the basis of the gyrocompass. Problems for Chapter 9 365 9.31 *** The Compton generator is a beautiful demonstration of the Coriolis force due to the earth's rotation, invented by the American physicist A. H. Compton (1892-1962, best known as author of the Compton effect) while he was still an undergraduate. A narrow glass tube in the shape of a torus or ring (radius R of the ring >> radius of the tube) is filled with water, plus some dust particles to let one see any motion of the water. The ring and water are initially stationary and horizontal, but the ring is then spun through 180\u00b0about its east\u2013west diameter. Explain why this should cause the water to move around the tube. Show that the speed of the water just after the 180\u00b0 turn should be 2QR cos 0, where Q is the earth's angular velocity, and 0 is the colatitude of the experiment. What would this speed be if R = 1 m and 19", " = 40\u00b0? Compton measured this speed with a microscope and got agreement within 3%. 9.32*** Do all parts of Problem 9.28, but find the distance by which the shell misses its target in both the north\u2013south and east\u2013west directions. [Hint: In this case you must recognize that the time of flight is affected by the Coriolis effect.] SECTION 9.9 The Foucault Pendulum 9.33 ** The general solution for the small-amplitude motion of a Foucault pendulum is given by (9.66). If at t = 0 the pendulum is at rest with x = A and y = 0, find the two coefficients C 1 and C2, and show that because Q << co o they are well approximated as C 1 = C2 = A/2, giving the solution (9.67). 9.34 *** At a point P on the earth's surface, an enormous perfectly flat and frictionless platform is built. The platform is exactly horizontal \u2014 that is, perpendicular to the local free-fall acceleration g p. Find the equation of motion for a puck sliding on the platform and show that it has the same form as (9.61) for the Foucault pendulum except that the pendulum's length L is replaced by the earth's radius R. What is the frequency of the puck's oscillations and what is that of its Foucault precession? [Hints: Write the puck's position vector, relative to the earth's center 0 as R r, where R is the position of the point P and r = (x, y, 0) is the puck's position relative to P. The contribution to the centrifugal force involving R can be absorbed into g p and the contribution involving r is negligible. The restoring force comes from the variation of g as the puck moves.] To check the validity of your approximations, compare the approximate size of the gravitational restoring force, the Coriolis force, and the neglected term m (St x r) x St in the centrifugal force. CHAPTER Rotational Motion of Rigid Bodies A rigid body is a collection of N particles with the property that its shape cannot change \u2014 the distance between any two of its constituent particles is fixed. A perfectly rigid body is, of course, an idealization, but an extremely useful one, and one to which many real systems are good approximations. In many ways a rigid body made up of N particles is much simpler than an arbitrary system of N particles: The arbitrary system requires 3N coordinates to specify its configuration, three coordinates for each of N particles. The rigid body requires only six coordinates, three to specify the position of the center of mass and three to specify the body's orientation. Further, we shall see that the motion of a rigid body can be divided into two separate simpler problems, the translational motion of the center of mass and the rotation of the body around the CM. I shall start the chapter with some general results, mostly related to the CM of the body. These results generalize the results we found at the beginning of Chapter 8 for two particles, and most of them apply to any system of N particles. However, I shall quickly specialize to the motion of a rigid body. Much the most interesting aspect of the latter is the rotational motion, and this is what will occupy us for most of the chapter. 10.1 Properties of the Center of Mass Consider a system of N particles a = 1,, N with masses m a and positions ra measured relative to a chosen origin 0. The center of mass of the system was defined in Chapter 3, Equation (3.9), to be the position (relative to the same origin 0) N E ma ra or R \u2014 m a=--1 --m f r dm 1 (10.1) where M denotes the total mass of all of the particles and the integral form is used when the system can be considered to be a continuous distribution of mass. 367 368 Chapter 10 Rotational Motion of Rigid Bodies The Total Momentum and the CM Several important parameters of the system's motion can be neatly expressed in terms of the CM. As we saw in Chapter 3, Equation (3.11), the total momentum is P =>f, pa = > mares a a (10.2) That is, the total momentum of the system is exactly the same as that of a single particle of mass equal to the total mass M and velocity equal to that of the CM. If we differentiate this result we see that P = MR or, since P equals the net external force F\"t on the system as we saw in Equation (1.29)], That is, the CM moves just as if it were a single particle of mass M subject to the net external force on the system. This result is the single most important justification for our treating extended objects like baseballs and comets as point particles. To the extent that these non", "point objects can be represented by their CM, they do move just like point particles. The Total Angular Momentum The role of the CM motion in a system's total angular momentum is more complicated, but equally crucial. The following argument does not depend on the system being a rigid body, but to be definite let us consider a rigid body made up of N pieces with masses m a, as sketched in Figure 10.1, where the body is shown as an ellipsoid. The position of m a relative to an arbitrary origin 0 is shown as ra and that of the CM relative to 0 as R. Also shown is the position r at of m a relative to the CM, which satisfies rai = R. The angular momentum f a of ma about the origin 0 is f a ra x pa = ra xmar a. Thus the total angular momentum relative to 0 is L a x ma t,. (10.4) (10.5) If we use (10.4) to rewrite both r a and ta, we find that L is the sum of four terms: L= R x m aR + >fd R x ra x m aR + x mares a Section 10.1 Properties of the Center of Mass 369 Figure 10.1 A rigid body (shown here as an ellipsoid) is made up of many small pieces, a = 1,, N. The mass of a typical piece is m a and its position relative to the origin 0 is ra. The position of the CM relative to 0 is R, and re,' denotes the position of m a relative to the CM, so that ra = R ra If we factor out of each of these four sums the terms that do not depend on a, we find (remember E ma = M) L = R x \u00b1 R x mcyroe' -1- ( ra ^ x + a x mares. (10.6) This expression can now be simplified dramatically. Notice first that the sum in parentheses in the third term on the right is the position of the CM relative to the CM (times M). This is, of course, zero (Problem 10.1): m ares =. 0 (10.7) Therefore the third term in (10.6) is zero. Differentiating this relation, we see that the sum in the second term in (10.6) is likewise zero. Thus all that remains of (10.6) is L=Rx1) \u00b11_:e xm a 0, a (10.8) The first term is the angular momentum (relative to 0) of the motion of the CM. The second is the angular momentum of the motion relative to the CM. Thus we can re-express (10.8) to say on of CM) + relative o CM). To illustrate this useful result consider the motion of a planet around the sun (which we can safely treat as fixed because it is so massive). In this case, (10.9) asserts that the total angular momentum of the planet is the angular momentum of the orbital motion 370 Chapter 10 Rotational Motion of Rigid Bodies of the CM around the sun, plus the angular momentum of its spinning motion around its CM, L = Lorb Lspin. (10.10) This division of the total angular momentum into its orbital and spin parts is especially useful because it is often true (at least to a good approximation) that the two parts are separately conserved. To see this, note first that since L orb = R x P, Lorb=---liXP\u00b1RA'.R xF\"t (since the first cross product is zero and P = Fext). That is, Lorb evolves just as if the planet were a point particle, with all its mass concentrated at its CM. In particular, if the force of the sun on the planet were perfectly central (Feat exactly collinear with R), then Lorb would be constant. In practice the force is not exactly central (since planets are not pefectly spherical and the sun's gravitational field is not perfectly uniform), but it is true to an excellent approximation. To find Lsp,\u201e, we can write Lsp,\u201e = L \u2014 Lorb. We already know that L = rext, so L = E ra x Feaxt = (r1a + R) x Fext a ra'x Fext R x F\"t. a (10.12) Subtracting (10.11) from (10.12) gives L spiri, Lspin L \u2014 Lorb r' x Fat a = rext (about CM); that is, the rate of change of L spip, the angular momentum about the CM, is just the net external torque, measured relative to the CM. [This natural-seeming result was mentioned without proof in Equation (3.28). What makes it a little surprising is that a reference frame attached to", " the CM is generally not an inertial frame. Surprising or not, the result is true and very useful.] Since the torque of the sun about the CM of any planet is very small, Lsp,,, is very nearly constant. Nevertheless, this useful conclusion, although an excellent approximation, is not exact. For instance, because of our own earth's equatorial bulge, there is a small torque on the earth due to the sun (and moon), and Lsp,,, is not quite constant. The slow changing of Lap,\u201e is responsible for the effect known as the precession of the equinoxes, the rotation of the earth's axis relative to the stars by some 50 arcseconds per year. There is a corresponding (though not exactly analogous) division of angular mo- mentum into its orbital and spin parts in quantum mechanics. For example, the angular momentum of the electron orbiting around the proton in a hydrogen atom is made up of two terms as in (10.10), and for much the same reasons each separate kind of an- gular momentum is almost perfectly conserved. Here too, this useful result is only approximately true: In this case there is a weak magnetic torque on the electron and neither the spin nor the orbital angular momentum is exactly conserved (although the total angular momentum is). Section 10.1 Properties of the Center of Mass 371 Kinetic Energy The total kinetic energy of N particles is T a=1 2. (10.13) As before, we can use (10.4) to replace ra by R+ ra, which gives f2 = a ece )2 = / 2 j_ i,/ I a and hence T= I 2 EmA 2 2Em a a2 + k-Em m ares (10.14) The sum in the last term on the right is zero by (10.7) and we find that T = MR2 m 2 a a (10.15) or axle. For a rigid body, the only possible motion relative to the CM is rotation. Thus we can rephrase this result to say that T = T (motion of CM) + T (rotation about CM). (10.17) This useful result says, for example, that the kinetic energy of a wheel rolling down the road is the translational energy of the CM plus the energy of the rotation about the From (10.14) we can derive an alternative and sometimes useful expression for the total kinetic energy. The derivation of (10.14) did not depend on R being the CM position, and (10.14) is actually valid for any point R fixed in the body. In particular, suppose we choose R to be a point of the body that happens to be at rest (even just instantaneously at rest). In this case the first and third terms on the right of (10.14) are both zero, and we find that T = Em it 2 \u2014 2 a a (10.18) This says that the total kinetic energy of a rigid body is just the rotational energy of the body relative to any point of the body that is instantaneously at rest. For example the kinetic energy of a rolling wheel can be evaluated as the energy of rotation about the point of contact with the road, since this point is instantaneously at rest. 372 Chapter 10 Rotational Motion of Rigid Bodies Potential Energy of a Rigid Body If all the forces on and within an N-particle rigid body are conservative, then, as we saw in Section 4.10, the total potential energy can be written as U = u ext U int (10.19) where Uext is the sum of all potential energies due to any external forces. (For example, if the body of interest is a baseball, U' could be the gravitational energy of all the particles that comprise the ball in the field of the \"external\" earth.) The internal potential energy U int is the sum of the potential energies for all pairs of particles, U int = Euo\u201e (1, co ) a \u2039p (10.20) where rah is the distance s between particles a and p. However, in a rigid body all of the interparticle distances r ah are fixed. Therefore, the internal potential energy is a constant and may as well be ignored. In other words, in discussing the motion of a rigid body we have to consider only the external forces and their corresponding potential energies. 10.2 Rotation about a Fixed Axis The results of the previous section show the importance of rotational motion. For example, the kinetic energy of any extended body flying through the air (you might think of a drum major's twirling baton) is the sum of two terms: the translational energy of the CM and the rotational energy of its spinning about the CM. The former we understand rather completely, but the latter we need to study. In most of the remainder of this chapter, we shall be focussing on rotational", " motion. In this section, we'll start with the special case of a body that is rotating about a fixed axis, such as the piece of wood shown in Figure 10.2 spinning on a fixed rod, and first calculate its angular momentum. Because the axis of rotation is fixed, we can agree to call it the z axis, with the origin 0 somewhere on the axis of rotation. As usual we imagine the body divided into many small pieces with masses m a (a = 1,, N) and the angular momentum is given by the usual formula L.Efa.Er xm a va a (10.21) 'Throughout this chapter I shall assume that all the internal forces are central. This guarantees that Uo depends only on the magnitude of rae,, not on its direction. It also ensures that the internal forces never contribute to changes in the total angular momentum. Section 10.2 Rotation about a Fixed Axis 373 Figure 10.2 An egg-shaped block of wood with a hole drilled through it is threaded on a rod fixed on the z axis. The block is spinning with angular velocity co. where the velocities v a are the velocities with which the pieces of the body are being carried in circles by the body's rotational velocity w. We saw in (9.22) that these are just va = w x ra. With our z axis along co, the components of co are co = (0, 0, w) ra = (x,\u201e ya, za ). Thus va = co x ra has components va = co x ra = (\u2014wy a, cox,\u201e 0) f a = ma ra x va = ma w(\u2014zaxa, \u2014zaya, x\u00ab 2 yc, ). (10.22) At last we are ready to calculate the total angular momentum of our spinning solid, and I shall start with the z component. If we put the z component of (10.22) into (10.21), we find that Now, the quantity y,:) is the same as pa2, where, as usual, p = /x2 + y2 denotes the distance of any point (x, y, z) from the z axis. Therefore, L z = E rria (x 2 + Ya2)(0. L z = Emap a2 w = Iz w, (10.23) (10.24) (10.25) and and finally where 374 Chapter 10 Rotational Motion of Rigid Bodies is the familiar moment of inertia about the z axis, as defined in every introductory physics course \u2014 the sum of all the constituent masses, each multiplied by the square of its distance from the z axis.2 Thus we have proved the familiar result that (angular momentum) = (moment of inertia) x (angular velocity). Note, however, that the angular momentum on the left of (10.24) is actually L z, and the moment of inertia is, of course, that for rotation about the z axis. To reinforce this gratifying result, let us calculate the kinetic energy of our rotating body. This is T = lEm\u201e\u201e2, or, since the speed of m a as it is carried in a circle around the z axis with angular velocity co is v a = paw, T a r\" a 0)2 2 z (10.26) another familiar result from introductory physics. So far we have met no surprises, but when we calculate the x and y components of L we find something unexpected: Substituting the x and y components of (10.22) into (10.21), we find for the x and y components of L: L x = \u2014 Enic,xaz\u201e,a) and L = \u2014 E (10.27) As we shall see in a moment the sums here are in general not zero, and we have the following surprising conclusion: The angular velocity co points in the z direction (the body rotates about the z axis), but, since L x and Ly can be nonzero, the angular momentum L may be in a different direction. That is, the angular momentum may not be in the same direction as the angular velocity, and the relation L = ho that you may have learned in introductory physics is generally not true! To better understand this rather unexpected conclusion, consider a rigid body that consists of a single mass m on the end of a massless rod, pivoted about the z axis at a fixed angle a, as in Figure 10.3. As this body rotates about the z axis, it is easy to see that the mass m has velocity v into the page (negative x direction) and hence that L = r x my is in the direction shown, at an angle (90\u00b0 \u2014 a) with the z axis. Clearly L y is not equal to zero, and, even though the body is rotating about the z axis, the angular momentum is not in that", " direction. In other words, L is not parallel to co. This example is worth pursuing a little further. It is clear from the picture that as the body rotates steadily about the z axis, the direction of L changes. (Specifically, L itself sweeps around the z axis.) Therefore, L 0, and a torque is required simply to keep the body rotating steadily. This conclusion, at first rather surprising, is actually easy to understand: The required torque is in the direction of L, which is out of the page (positive x direction) in Figure 10.3; that is, the torque must be counterclockwise. The easiest way to understand this is to put yourself in a frame that is rotating with 2 You may recall that, when actually calculating moments of inertia, we often replace the sum in (10.25) by an integral. For now, however, I shall continue to write moments of inertia as sums. Section 10.2 Rotation about a Fixed Axis 375 Figure 10.3 A rigid rotating body comprising a single mass m anchored to the z axis by a massless rod at a fixed angle a, shown at a moment when m happens to lie in the yz plane. As the body rotates about the z axis, m has velocity, and hence momentum, into the page (in the negative x direction) at the moment shown. Therefore the angular momentum L = r x p is directed as shown and is certainly not parallel to the angular velocity co. the body. In this frame, the mass m experiences a centrifugal force out from the z axis (to the right in the picture). Therefore a counterclockwise torque is required to prevent the rod that holds m from bending or breaking away from its anchor on the axis. When a body, such as the wheel of your car coasting along the highway, is rotating steadily about a fixed direction, we usually do not want to have to exert any torque on it. This means the body must be designed so that its angular momentum is parallel to its angular velocity. In the case of your car, this is guaranteed by the process of dynamical balancing of the wheels. If the wheels are not properly balanced, you are made quickly aware of it by a disagreeable vibration of the car. More generally, the question whether or not L and CO are parallel is an important issue throughout the study of rotating bodies and leads us to the important concept of principal axes, as we shall discuss in Section 10.4. The Products of Inertia We need to collect together our results for the angular momentum of a body rotating about the z axis and to streamline our notation. It is clear from (10.27) that L x and Ly are proportional to co, and the constants of proportionality are generally denoted by Ixz and /yz. Thus I shall rewrite (10.27) as L x Ixz w and L = I w yz y (10.28) 376 Chapter 10 Rotational Motion of Rigid Bodies where ixz= Ern. x 01 a z a and I = \u2014 yz m ayaza. (10.29) The two coefficients Ixz and /37, are called the products of inertia of the body. The rationale for this new notation is that /xz tells us the x component of L when c is in the z direction (and likewise for /37,). To conform with this notation, we have to rename 4, the old-fashioned moment of inertia about the z axis and call it /zz, Izz E mapa2 Erna(x,2 \u00b1ya2 ). (10.30) With this notation, we can say that, for a body rotating about the z axis, the angular momentum is L = (/xzw, /yz co, /zz o)). (10.31) Obviously, it is important to be able to calculate the coefficients /xz, /3,,, and /zz, for bodies of different shapes. I shall work out some examples in the remainder of this chapter and there are plenty more in the problems. Here are three simple examples for a start. EXAMPLE 10.1 Calculating Some Simple Moments and Products of Inertia Calculate the moment and products of inertia for rotation about the z axis of the following rigid bodies: (a) A single mass m located at the position (0, y0, z0) as shown in Figure 10.3. (b) The same as in part (a) but with a second equal mass placed symmetrically below the xy plane, as in Figure 10.4(a). (c) A uniform ring of mass M and radius po centered on the z axis and parallel to the xy plane, as in Figure 10.4(b). (a) (b) Figure 10.4 (a) A rigid body comprising two equal masses m held at equal distances above and below the xy plane and rotating about the z axis (shown at an instant when the", " two masses lie in the yz plane). (b) A uniform continuous ring of total mass M and radius po, centered on the z axis and parallel to the xy plane. (Example 10.1.) Section 10.2 Rotation about a Fixed Axis 377 (a) For the single mass of Figure 10.3, the sums in (10.29) and (10.30) each reduce to a single term and we find Ix, = 0, Iyz = \u2014112 Yo Z ol and izz = my02. That Iyz is nonzero confirms that L has a nonzero y component and hence that the angular momentum L is not in the same direction as the axis of rotation. The remaining parts (b) and (c) illustrate how when a rigid body has certain symmetries, the products of inertia can turn out to be zero. (b) For the two masses of Figure 10.4(a) each of the sums in (10.29) and (10.30) contains two terms and we find /\u201e = \u2014 x 01 CY = 0 z ci (10.32) because both masses have x a = 0, IyZ = \u2014 E m ayaz a = \u2014m[y oz o y o (\u2014z 0)] = 0, (10.33) and I\u201e = 2 2 ya ) = n2(l) n yo2 \u00b1 0 \u00b1 yo2) = 2myo2. The interesting case here is the product of inertia which is zero because the contribution of the first mass is exactly cancelled by that of the second mass, at the \"mirror image\" point with the opposite sign of z a. Clearly this will happen for any body that has reflection symmetry in the plane z =0;3 both of the products of inertia Ix, and Iyz will be zero because every term in each of the sums of (10.32) and (10.33) will be cancelled by another term with the opposite sign of za. With Ixz = /yz = 0, we see from (10.31) that when the body rotates about the z axis, the angular momentum L is also along the z axis. (c) Because the body in Figure 10.4(b) is a continuously distributed mass, we should in general evaluate the products and moment of inertia as integrals, but in this case we can see the answers without actually doing any integrals. Consider, first, the product of inertia Ix, as given by the sum in (10.32). Referring to Figure 10.4(b), it is easy to see that this sum is zero: Each contribution to the sum from a small mass m a at (xa, ya, z a) can be paired with the contribution from an equal mass diametrically across the circle at (\u2014x a, \u2014y a, z a). This second contribution, with the same value of z but the opposite value for x, exactly cancels the first, and we conclude that the whole sum Ixz is zero. By exactly the same argument, /yz = 0. The moment /zz is most easily evaluated in the form of the first sum in (10.30). Since all terms in the sum have the same value of pa (namely po), we can factor pa out of the sum and are left with izz = (E ma ) po2 = Mpgz. 3 We say that a body has reflection symmetry in the plane z = 0 if the mass density is the same at any point (x, y, z) as at the point (x, y, \u2014z), the reflection of (x, y, z) in a mirror located in the plane z =- 0. 378 Chapter 10 Rotational Motion of Rigid Bodies In this example, the two products of inertia were zero because the body was axially symmetric about its axis of rotation, 4 and it is easy to see that the same result holds quite generally: If a rigid body is axially symmetric about a certain axis (like a well-balanced wheel about its axle or a circular cone about its center line), then the two products of inertia for rotation about that axis are automatically zero, since the terms of the sums in (10.32) and (10.33) cancel in pairs. In particular, if a body is axially symmetric about a certain axis and is rotating about this symmetry axis, its angular momentum will be in the same direction. 10.3 Rotation about Any Axis; the Inertia Tensor So far we have considered only a body that is rotating about the z axis. In a certain sense this is quite general: Whatever axis a body is rotating about, we can choose to call it the z axis. Unfortunately, although this statement is true, it does not tell the whole story. First, we are often interested in bodies that are free to rotate about any axis a gy", "roscope's bearings allow it to rotate about any axis, and a projectile (such as a baseball or a drum major's baton thrown in the air) has the same freedom. When this is the case, the axis about which the body rotates may change with time. If this happens, then we can certainly choose as our z axis the axis of rotation at one instant, but a moment later our chosen z axis is almost certainly not the axis of rotation. For this reason alone, we must clearly examine the form of the angular momentum when a body is spinning about an arbitrary axis. The second reason that we need to consider rotation about an arbitrary axis is subtler, and I shall return to it later, but let me mention it briefly here. We have seen that in general the direction of the angular momentum of a spinning body is not the same as the axis of rotation. On the other hand, it sometimes happens that these two directions are the same. (For instance, we have seen that this is the case for an axially symmetric body rotating about its axis of symmetry.) When this is true, we say that the axis in question is a principal axis. We shall find for any given body, rotating about any given point, that there are three mutually perpendicular principal axes. Because much of our discussion of rotations is much easier when referred to these principal axes, we often wish to choose the principal axes as our coordinate axes. If we do this, then we are no longer at liberty to choose our z axis to coincide with an arbitrary axis of rotation. Again, we must learn to allow any axis to be the axis of rotation, and our first order of business is to calculate the angular momentum corresponding to such rotation. 4 We say that a body is axially, or rotationally, symmetric about an axis if the mass density is the same at all points on any circle centered on, and perpendicular to, the axis. In terms of cylindrical polar coordinates (p, 0, z) centered on the axis in question, the mass density is independent of 0. Alternatively, the mass distribution is unchanged by any rotation about the axis of symmetry. Section 10.3 Rotation about Any Axis; the Inertia Tensor 379 Angular Momentum for an Arbitrary Angular Velocity Let us then consider a rigid body rotating about an arbitrary axis with angular velocity (0= (wx, wy, Wz). Before we launch into the calculation of the angular momentum, let us pause to consider the kind of situation to which our calculations will apply. There are in fact two important cases that you could keep in mind: First, it sometimes happens that a rigid body has one fixed point, so that its only possible motion is a rotation about that fixed point. For example, the Foucault pendulum is fixed at its point of support in the ceiling, and its only possible motion is rotation about that fixed point. Again, a top spinning on a table can get its tip caught in a small dent in the table, and, from then on, it can only rotate about the fixed position of its tip. In either case, the magnitude and direction of the angular velocity co can change, but the rotation will always be about the fixed point, which we shall naturally take to be our origin. The second case that you could bear in mind is that of an object that has been thrown in the air. In this case there is certainly not any fixed point, but we have seen that we can analyze the motion in terms of the motion of the CM and the rotational motion relative to the CM. In this case, the motion that we are now analyzing is the motion relative to the CM, which we shall naturally take to be the origin. With these examples in mind, let us calculate the body's angular momentum, L = m a ra X Va = E ma res x (co x ra ). (10.34) We can evaluate this almost exactly as we did in the previous section for the case that co was along the z axis: For any position r we can write down the components of co x r and then r x (co x r), with the rather ugly result (there are several ways to do this, one of which is the so-called BAC \u2014 CAB rule \u2014 see Problem 10.19) r x (w x r) = ((y 2 2 + Z )CO, \u2014 XyWy \u2014 Xa0, \u2014 yxw x (z2 x2)wy \u2014 yzo) z, \u2014 zXCO, \u2014 zyco y (x 2 2\\ \\ y )(00 (10.35) Substituting (10.35) into (10.34), we can write down the three components of L as follows: L x Ly L z = = = Ixx wx Iyx( 0 x Izx cox Ixy coy lxz w z yy(Oy + Iyzwz Izy wy Izz wz (10.", "36) where the three moments of inertia, /xx, /yy, /zz, and the six products of inertia, /xy,, are defined in exact parallel with the definitions (10.29) and (10.30) of the previous section. For example, 380 Chapter 10 Rotational Motion of Rigid Bodies and similarly for Iyy and izz, and The rather clumsy result (10.36) can be streamlined in a couple of ways: If you don't mind replacing the subscripts x, y, and z with i = 1, 2, 3, then (10.36) takes the and so on. compact form This suggests another way to think about (10.36) since, as you may recognize, (10.39) is the rule for matrix multiplication. Thus, (10.36) can be rewritten in matrix form. First, we introduce the 3 x 3 matrix 3 Li = E /if I = [Ix, lyx I\u201e IXy Iyy 1\u201e Ix, lyz I\u201e (10.39) (10.40) which is called the moment of inertia tensor s or just inertia tensor. In addition, let us agree temporarily to think of three-dimensional vectors as 3 x 1 columns made up of their three components; that is, we write L= and w=. (10.41) L x L y L, \u2013 wx coy _wz (Notice that I am now using boldface for two kinds of matrices \u2014 square 3 x 3 matrices like I, and 3 x 1 column matrices like L and co that are really just vectors. You will quickly learn to distinguish the two kinds of matrix from the context.) With these notations, Equation (10.36) takes the very compact matrix form 5 The full definition of a tensor involves the transformation properties of its elements when we rotate our coordinate axes (see Section 15.17), but for our present purposes it is sufficient to say that a three-dimensional tensor is a set of nine numbers arranged as a 3 x 3 matrix as in (10.40). Section 10.3 Rotation about Any Axis; the Inertia Tensor 381 where the product on the right is the standard product of two matrices, the first a 3 x 3 square matrix and the second a 3 x 1 column. This beautiful result is our first example of the great usefulness of matrix algebra in mechanics. In many areas of physics \u2014 perhaps most especially in quantum mechan- ics, but also certainly in classical mechanics \u2014 the formulation of many problems in matrix notation is so much simpler than in any other way, that it is absolutely essential that you become familiar with the basics of matrix algebra. 6 An important property of the moment of inertia tensor (10.40) is that it is a symmetric matrix; that is, its elements satisfy =. (10.43) Another way to say this is that the matrix (10.40) is unchanged if we reflect it in the main diagonal \u2014 the diagonal running from top left to bottom right. Each element above the diagonal (for instance /,),) is equal to its mirror image (I yx) below the diagonal. To prove this property, you have only to look at the definition (10.38) to see that Ixy is the same as Iy.\u201e and similarly with all the off-diagonal elements /, 1 (the elements with i j). Yet another way to state this property is to define the transpose of any matrix A as the matrix A obtained by reflecting A in its main diagonal \u2014 the ij element of A is the ji element of A. Thus the result (10.43) means that the matrix I is equal to its own transpose, I = I. (10.44) This property \u2014 that the matrix I is symmetric \u2014 plays a key role in the mathematical theory of the moment of inertia tensor. EXAMPLE 10.2 Inertia Tensor for a Solid Cube Find the moment of inertia tensor for (a) a uniform solid cube, of mass M and side a, rotating about a corner (Figure 10.5) and (b) the same cube rotating about its center. Use axes parallel to the cube's edges. For both cases, find the angular momentum when the axis of rotation is parallel to x [that is, co = (co, 0, 0)] and also when co is along the body diagonal in the direction (1, 1, 1). (a) Because the mass is continuously distributed we need to replace the sums in the definitions (10.37) and (10.38) by integrals. Thus (10.37) becomes I = f dx dy dz 0 a fa fa 0 0 0(y2,z2) (10.45) where 0 = M la 3 denotes the mass density of the cube. This is the sum of two triple integ", "rals, each of which can be factored into three single integrals. For example, 6 The matrix operations that I shall assume you already know \u2014 matrix addition, multiplication, transposition, determinants, and a few more \u2014 can be found in Chapter 3 of Mary Boas, Mathemat- ical Methods in the Physical Sciences (Wiley, 1983). Some of the ideas I shall be developing in this and the next chapter are discussed in more detail in Chapter 10 of the same book. 382 Chapter 10 Rotational Motion of Rigid Bodies z 0 a Figure 10.5 A uniform solid cube of side a that is free to rotate about the corner 0. aa a a dz Q y 2 = (f d X) f y 2 dy) (f dz) dx dy fa fa. 0 a 0 0 = IQ a5 = 3 M a 2 ix, = 2 Ma (10.46) (10.47) The second term in (10.45) has the same value, and we conclude that with (by symmetry) the same values for Iyy and 4z The integral form of (10.38) for the off-diagonal elements of I is a f a f a xy = \u2014 f dx dy dz Qxy 0 0 0 = -- Q (f a X d X) y dy) (f (10.48) 0 0 a dz/ a 0 = \u2014 4 Qa5 = \u2014 1 M a2 with (again by symmetry) the same answer for all the other off-diagonal ele- ments. Putting all of these results together, we find for the moment of inertia tensor of a cube rotating about its corner I = \u2014 1 Ma 2 [ i Mat _ ima2 _ ima2 ima 2 _i ma 2 74 _i ma 2 _Ima2 3 M22 4 71 Mae 12 8 \u20143 \u2014 3 \u20143 8 \u20143 \u20143 \u20143 8 [about corner] (10.49) where the second, more compact, form follows from the rules for multiplying a matrix by a number. (Notice that, as expected, I is a symmetric matrix.) According to (10.42) the angular momentum L corresponding to an angular velocity w is given by the matrix product L = Iw, where the vectors L and w are understood as 3 x 1 columns made up of the three components of the vector Section 10.3 Rotation about Any Axis; the Inertia Tensor 383 concerned. Thus if the cube is rotating about the x axis, L = Ito = Mae 12 8 \u20143 \u20143 [ 3 8 \u20143 \u20143 3 8 w 0 0 Mae a) 8 = 12 \u20143w [ \u20143w (10.50) or, reverting to more standard vector notation, L = Ma2w (4, \u20141, 4 ). (10.51) As we have come to recognize is possible, we see that in this case L is not in the same direction as the angular velocity co = (a), 0, 0). If the cube is rotating about its main diagonal, then the unit vector in the direction of rotation is u = (1/ N/j) (1, 1, 1) and the angular velocity vector is = wu = (co/ N/j) (1, 1, 1). Thus according to (10.42), the angular momentum for this case is L Ico = = Ma2 12 [ 8 \u20143 \u20143 \u20143 \u20143 8 \u20143 \u20143 8 1 1 1 Ma2 12 Vj [ 2 2 2 = Ma2 6 co. (10.52) In this case, rotation about the main diagonal of the cube, we see that the angular momentum is in the same direction as the angular velocity. (b) If the cube is rotating about its center, then in Figure 10.5 we must move the origin 0 to the center of the cube. This means that all of the integrals in (10.45) and (10.48) run from \u2014a/2 to a/2 instead of 0 to a. Evaluating (10.45) for ixx [as in (10.46)] we find ixx = 4Ma2 (10.53) and likewise Iyy and /\u201e. When the limits in (10.48) are replaced by \u2014a/2 and a/2, both of the first two integrals are zero, and we conclude that Ixy = 0. As you can easily check, all of the off-diagonal elements of I work in the same way and are zero. We could actually have anticipated this vanishing of the off- diagonal elements of I based on Example 10.1. We saw there that if the plane z = 0 is a plane of reflection symmetry, then both 4, and /yz are automatically zero. (Every contribution from above the plane z = 0 was canceled by the contribution from", " the corresponding point below the plane.) Therefore, if two of the three coordinate planes x = 0, y = 0, and z = 0 are planes of reflection symmetry, this guarantees that all of the products of inertia are zero. For the cube (with 0 at its center) all three of the coordinate planes are planes of reflection symmetry, so it was inevitable that the off-diagonal elements of I turned out to be zero. 384 Chapter 10 Rotational Motion of Rigid Bodies Collecting results, we conclude that for a cube rotating about its center, the moment of inertia tensor is 4ma2 [o 1 0= 4Ma21 [about CM], (10.54) 1 0 0 0 0 1 where 1 denotes the 3 x 3 unit matrix10.55) We shall see that, because the moment of inertia tensor for rotation about the center of a cube is a multiple of the unit matrix, rotations about the center of a cube are especially easy to analyze. In particular, the angular momentum of the cube (rotating about its center) is L = Iw = 6 Ma2 1w = 6M a2, (10.56) which implies that the angular momentum L is in the same direction as the angu- lar velocity co, whatever the direction of w. This simple result is a consequence of the high degree of symmetry of the cube relative to its center. You will notice that (10.56) for rotation about any axis through the center of the cube agrees with (10.52) for rotation about the main diagonal through the corner of the cube. This is not an accident: The main diagonal through the center of the cube is exactly the same as the main diagonal through the corner. Therefore the angular momenta for rotations about these two axes have to agree. This example illustrated many of the features of a typical calculation of the moment of inertia tensor I, and you should certainly try some of the problems at the end of this chapter to get used to doing such calculations yourself. Meanwhile, here is one more example to illustrate the special features of a body that is axially symmetric. EXAMPLE 10.3 Inertia Tensor for a Solid Cone Find the moment of inertia tensor I for a spinning top that is a uniform solid cone (mass M, height h, and base radius R) spinning about its tip. Choose the z axis along the axis of symmetry of the cone, as shown in Figure 10.6. For an arbitrary angular velocity co, what is the top's angular momentum L? The moment of inertia about the z axis, /\u201e, is given by the integral /zz = f dVQ(x 2 + y 2) (10.57) where the subscript V on the integral indicates that the integral runs over the volume of the body, dV is an element of volume, and Q is the constant mass density O = M/ V = 3M/ (7r R 2h). This integral is easily evaluated in cylindrical Section 10.3 Rotation about Any Axis; the Inertia Tensor 385 Figure 10.6 A top comprising a uniform solid cone, of mass M, height h, and base radius R, spins about its tip. The radius of the cone at height z is r = Rz I h. (The top is not necessarily vertical, but, whatever its orientation, we choose the z axis along the axis of symmetry.) polar coordinates (p, 0, z) since x 2 + y 2 = p2. Thus the integral can be written p f dV p 2 Q f dz p dp p 2 (10.58) h o f 27r o fr o where the upper limit r of the p integral is the radius of the cone at height z, as shown in Figure 10.6. These integrals are easily carried out and give (Problem as7 10.26) /zz = AMR2. (10.59) Because of the top's rotational symmetry about the z axis, the other two moments of inertia, /xx and Iyy, are equal. (A rotation through 90\u00b0 about the z axis leaves the body unchanged but interchanges / xx and I yy. Therefore, /xx = yy.) To evaluate /xx we write Ixx = f dVQ(y 2 z 2) = f d17 py 2 f dV Qz 2. (10.60) 7 Here is one of those horrible moments when the traditional use of the Greek letter \"rho\" for density collides with its use for the cylindrical coordinate equal to the distance from the z axis. Notice that I am using two different versions of the letter, Q for density and p for the coordinate. Fortunately, this unlucky collision happens only very occasionally. If you need to refresh your memory about volume integrals in cylindrical coordinates, see Problem 10.26. 386 Chapter 10 Rotational Motion of Rigid Bodies The first integral here", " is the same as the second term in (10.57), and, by the rotatonal symmetry, the two terms in (10.57) are equal. Therefore, the first term of (10.60) is just /\u201e/2 or A MR2. The second integral in (10.60) can be evaluated using cylindrical polar coordinates, as in (10.58), and gives s Mh 2. Therefore Ixx = Iyy = M (R2 + 4h 2). (10.61) This leaves the off-diagonal products of inertia, / xy,, to calculate, but we can easily see that all of these are zero. The point is that because of the axial symmetry about the z axis, both of the planes x = 0 and y = 0 are planes of reflection symmetry. (See Figure 10.6.) By the argument given in Example 10.1, symmetry about the plane x = 0 implies that Ixy =.1\u201e = 0. Similarly, symmetry about y = 0 implies that Iy, = lyx = 0. Thus, symmetry about any two coordinate planes guarantees that all of the products of inertia are zero. Collecting results, we find that the moment of inertia tensor for a uniform cone (relative to its tip) is I = 20 R2 + 4h 2 0 0 0 0 0 0 2R2 0 0 0 0 A3 R2 \u00b1 4h 2 = [ 0 A2 0, (10.62) where the last form is just for convenience of discussion (and in the present case, it happens that Al = X 2). The most striking thing about this matrix is that its off- diagonal elements are all zero. A matrix with this property is called a diagonal matrix. [The inertia tensor (10.54) for a cube about its center was also diagonal, but since all its diagonal elements were equal, it was actually a multiple of the unit matrix 1, making it an even more special case.] The important consequence of I being diagonal emerges when we evaluate the angular momentum L for an arbitrary angular velocity w = (cox, wy, we): L = Ice) = (Xicox, A.30-)z) (10.63) While this may not look remarkable, notice that if the angular velocity co points along one of the coordinate axes, then the same is true of the angular momentum L. For example, if co points along the x axis, then co y = wz = 0 and (10.63) implies that L = Icy (A iwx, 0, 0) (10.64) and L also points along the x axis. Obviously, the same thing happens if co points along the y or z axes, and we see quite generally that if the inertia tensor I is diagonal, then L will be parallel to co whenever co points along one of the three coordinate axes. Section 10.4 Principal Axes of Inertia 387 10.4 Principal Axes of Inertia We have seen that in general the angular momentum of a body spinning about a point o is not in the same direction as the axis of rotation; that is, L is not parallel to co. We have also seen that, for certain bodies at least, there can be certain axes for which L and co are parallel. When this happens, we say that the axis in question is a principal axis. To express this definition mathematically, note that two nonzero vectors a and b are parallel if and only if a = Ab for some real number A. Thus we can define a principal axis of a body (about an origin 0) as any axis through 0 with the property that if co points along the axis, then L Act) (10.65) for some real number X. To see the significance of the number X in this equation, we can temporarily choose the direction of co to be the z direction, in which case L is given by (10.31) as L = (/\u201ezco, /yz co, /zz co). Since L is parallel to co, the first two components are zero, and we conclude that L = (0, 0, /\u201eco) = /zz co. Comparing with (10.65), we see that the number A in (10.65) is just the body's moment of inertia about the axis in question. To summarize, if the angular velocity co points along a principal axis, then L = Xco, where A is the moment of inertia about the axis in question. Let us review what we already know about the existence of principal axes. We saw at the end of the last section that if the inertia tensor I, with respect to a chosen set of axes, turns out to be diagonal, Xi 0 0 0 X2 0 0 0 X3, I = [ (10.66) then the chosen x, y, and z axes are principal axes. Conversely, if the", " x, y, and z axes are principal axes, then it is easy to see (Problem 10.29) that I must be diagonal, as in (10.66). The three numbers that I have denoted by A 1, X2, and X3 are the moments of inertia about the three principal axes and are called the principal moments. If a body has a symmetry axis through 0 (like the spinning top of Example 10.3), then that axis is a principal axis. Furthermore, any two axes perpendicular to the symmetry axis (like the x and y axes in Example 10.3) are also principal axes, since the inertia tensor with respect to these axes is diagonal. Again, if a body has two perpendicular planes of reflection symmetry through 0 (like the cube spinning about its center 8 in Example 10.2), then the three perpendicular axes defined by these two planes and 0 are principal axes. So far all our examples of principal axes have involved bodies with special sym- metries, and you may very reasonably be thinking that the existence of principal axes is somehow tied to a body's having some symmetry. In fact however, this is most 8 0f course the cube has three planes of reflection symmetry, but two are enough to guarantee the claim made here. For example, if the cone of Figure 10.6 were an elliptical cone, the z axis would no longer be an axis of rotational symmetry, but the planes x = 0 and y = 0 would still provide two planes of reflection symmetry, and the three coordinate axes would still be principal axes. 388 Chapter 10 Rotational Motion of Rigid Bodies emphatically not so. Although symmetries of a body make it much easier to spot the principal axes, it turns out that any rigid body rotating about any point has three principal axes: Existence of Principe Axe For any rigid body and any point 0 there are three perpendicular principa axes through 0, That is, there are three perpendicular axes through 0 with respect to which the inertia tensor I is diagonal and, hence, with the property that when the angular velocity co points along any one of these axes the same is true of the angular momentum This surprising result is the consequence of an important mathematical theorem which states that if I is any real symmetric matrix (such as the inertia tensor of some body with respect to any chosen set of orthogonal axes) then there exists another set of orthogonal axes (with the same origin) such that the corresponding matrix (call it I'), evaluated with respect to the new axes, has the diagonal form (10.66). This result, which is proved in the appendix, is extremely useful, since the discussion of rotational motion is much simpler if it can be referred to a set of principal axes, and the result guarantees that this can always be done. (It may be worth mentioning right away, however, that the principal axes of a rigid body are naturally fixed in the body. Thus when we choose the principal axes as our coordinate axes, we are committing ourselves to using a rotating set of axes.) While it is not essential to see a proof of the existence of principal axes, we certainly do need to know how to find the principal axes, and this is what I take up in the next section. Kinetic Energy of a Rotating Body It is naturally important to be able to write down the kinetic energy of a rotating body. I shall leave it as a challenging exercise (Problem 10.33) to show that T = to L 2 (10.67) In particular, if we use a set of principal axes for our coordinate system, then L (.X 1w1, A2w2, X3w3) and T = - 2 co2 1 1 A2W2 2 x 2\\ 3W3 j. (10.68) This important result is a natural generalization of Equation (10.26), T = 1 1 w 2, for rotation about a fixed z axis. We shall use the result in writing the Lagrangian for a spinning body in Section 10.9. Section 10.5 Finding the Principal Axes; Eigenvalue Equations 389 10.5 Finding the Principal Axes; Eigenvalue Equations Suppose that we wish to find the principal axes of a rigid body rotating about a point 0. Suppose further that, using some given set of axes, we have calculated the inertia tensor I for the body. If I is diagonal, then our axes already are the principal axes of the body, and there is nothing more to do. Suppose, however, that I is not diagonal. How are we to find the principal axes? The essential clue is Equation (10.65): If co points along a principal axis, then L must equal Xco (for some number X). Since L = Ice, it follows that w must satisfy the equation lio Xco (10.69) for some (as yet unknown) number X. The equation (10.69) has the form (mat", "rix) x (vector) = (number) x (same vector) and is called an eigenvalue equation. Eigenvalue equations are among the most important equations of modern physics and arise in many different areas. They always express the same idea, that some mathematical operation performed on a vector (co in our case) produces a second vector (ho here) that has the same direction as the first. A vector w that satisfies (10.69) is called an eigenvector and the corresponding number X, the corresponding eigenvalue. There are actually two parts to solving the eigenvalue problem of (10.69). Usually we want to know the directions of co for which (10.69) is satisfied (namely the direc- tions of the principal axes), and in most cases we also want to know the corresponding eigenvalues X (namely, the moments of inertia about the principal axes). In practice, we usually solve these two parts of the problem in the opposite order \u2014 first find the possible eigenvalues X and then the corresponding directions of co. The first step in solving the matrix equation (10.69) is to rewrite it. Since w = 1a) (where 1 is the 3 x 3 unit matrix), Equation (10.69) is the same as!co = aco, or, moving the right side over to the left, (I \u2014 )1)co = 0. (10.70) This is a matrix equation of the form Aco = 0, where A is a 3 x 3 matrix and coo is a vector, that is, a 3 x 1 column of numbers w x, wy, and co. The matrix equation Aco = 0 is really three simultaneous equations for the three numbers w x, coy, and w\u201e and it is a well-known property of such equations that they have a nonzero solution if and only if the determinant, det(A), is zero. 9 Therefore, the eigenvalue equation (10.70) has a nonzero solution if and only if det(I \u2014 X1) = 0. (10.71) This equation is called the characteristic equation (or secular equation) for the matrix I. The determinant involved is a cubic in the number X. Therefore, the equation 9 See, for example, Mary Boas, Mathematical Methods in the Physical Sciences (Wiley, 1983), page 133. 390 Chapter 10 Rotational Motion of Rigid Bodies is a cubic equation for the eigenvalues A and will, in general, have three solutions, A i, A2, and A3, the three principal moments. For each of these values of A, the equation (10.70) can be solved to give the corresponding vector w, whose direction tells us the direction of one of the three principal axes of the rigid body under consideration. If you have never seen this procedure for finding the principal axes (or, more generally, for solving eigenvalue problems), you will certainly want to see an example and to work through some yourself. Here is one as a start. EXAMPLE 10.4 Principal Axes of a Cube about a Corner Find the principal axes and corresponding moments for the cube of Example 10.2, rotating about its corner. What is the form of the inertia tensor evaluated with respect to the principal axes? Using axes parallel to the edges of the cube, we found the inertia tensor to be [Equation (10.49)] 8 \u20143 \u20143 I = p, \u20143 8 \u20143 \u20143 \u20143 8 (10.72) where I have introduced the abbreviation\u00b5 for the constant \u00b5 = Ma 2 / 12, which has the dimensions of moment of inertia. Since I is not diagonal, it is clear that our original chosen axes (parallel to the edges of the cube) are not the principal axes. To find the principal axes, we must find the directions of co that satisfy the eigenvalue equation Ito = Act). Our first step is to find the values of A (the eigenvalues) that satisfy the characteristic equation det(I \u2014 Al) = 0. Substituting (10.72) for I, we find that I \u2014 Al = [ \u2014 3p, 8p. \u2014 8p, \u20143,u, \u20143p, \u20143,u, \u20143p, \u20143p, 8p8p, \u2014 X \u20143p. \u20143,a = \u201434 8p, \u2014 X \u20143p. \u20143p, \u20143p, 8,u, \u2014 X. The determinant of this matrix is straightforward to evaluate and is det(I \u2014 Al) = (2p, \u2014 A)(11p, \u2014 A) 2. (10.73) Thus the three roots of the equation det(I \u2014 X1) = 0 (the eigenvalues) are X i = 2p and A2 = A3 = 11", "p. (10.74) Notice that in this case two of the three roots of the cubic (10.73) happen to be equal. Armed with the eigenvalues, we can now find the eigenvectors, that is, the directions of the three principal axes of our cube rotating about its corner. These are determined by Equation (10.70), which we must examine for each of the Section 10.5 Finding the Principal Axes; Eigenvalue Equations 391 eigenvalues X 1, X2, and X3 in turn (though in this case the last two are equal). Let us start with X i. With X = 1= 2\u00b5, Equation (10.70) becomes (I \u2014 X1)co =,u, \u20143 6 \u20143 = 0. (10.75) [ 6 \u20143 \u20143 \u20143 \u20143 6 [ co Wz This gives three equations for the components of co, 2cox \u2014 coy \u2014 coz = 0 \u2014co, + aoy \u2014 coz = 0 \u2014cox \u2014 coy + 2co, = 0. (10.76) Subtracting the second equation from the first we see that co x = coy, and the first then tells us that co x = co,. Therefore, cox = coy = coz, and we conclude that the first principal axis is in the direction (1, 1, 1) along the principal diagonal of the cube. If we define a unit vector e l in this direction, e 1 = -1-(1 1, 1), (10.77) then e l specifies the direction of our first principal axis. If co points along e l, then L = Icy = X ico. This says simply that the moment of inertia about this principal axis is X i = 2\u00b5 = 4Ma2. Thus our analysis of the first eigenvalue has produced this conclusion: One of the principal axes of a cube, rotating about its corner 0, is the principal diagonal through 0 (direction e l), and the moment of inertia for that axis is the corresponding eigenvalue 6 Ma e. The other two eigenvalues are equal (X2 = = 11\u00b5), so there is just one more case to consider. With X = 11A, the eigenvalue equation (10.70) reads (1 \u2014 ).1)co =,u, \u20143 [ \u20143 \u20143 = O. -3 \u20143 \u20143 \u20143 \u20143 \u20143 [ Wx W Wz This gives three equations for the components of co, but all three equations are actually the same equation, namely cox + coy + = 0. (10.78) This equation does not uniquely determine the direction of co. To see what it does imply, notice that cox + coy + coz can be viewed as the scalar product of co with the vector (1, 1, 1). Thus Equation (10.78) states simply that co e 1 = 0; that is, co needs only to be orthogonal to our first principal axis e l. In other words, any two orthogonal directions e 2 and e3 that are perpendicular to e l will serve as the other two principal axes, both with moment of inertia X2 = X3 = 1l\u00b5 = ii Mae. This freedom in choosing the last two principal axes is directly related to the circumstance that the last two eigenvalues X2 and X3 are equal; when all three eigenvalues are different, each one leads to a unique direction for the corresponding principal axis. 392 Chapter 10 Rotational Motion of Rigid Bodies Finally, if we were to re-evaluate the inertia tensor with respect to new axes in the directions e 1, e2, and e3, then the new matrix I' would be diagonal, with the principal moments down the diagonal, 1 0 A2 0 0 0 X3 I' = X 0 [0 = \u2014112 Ma e 2 0 0 0 11 0 0 0 11 For this reason the process of finding the principal axes of a body is described as diagonalization of the inertia tensor. The last paragraph of this example illustrates a useful point: By the time we have found the principal axes of a body with the corresponding principal moments, there is no need to re-evaluate the inertia tensor with respect to the new axes. We know that with respect to the principal axes it is bound to be diagonal, X 1 0 0 I' = 0 )\u2018.2 0, 0 0 X 3 (10.79) with the principal moments X 1, A2, and A3 down the diagonal. In general the three principal moments will all be different, in which case the directions of the three principal axes are uniquely determined and are automatically orthogonal (see Problem 10.38). As we saw in Example 10.4, it can happen that two of the principal moments are equal, in which", " case the corresponding two principal axes can have any direction that is orthogonal to the third axis. (This is what happened in the Example 10.4, and also what happens with any body that has rotational symmetry about an axis through 0.) If all three principal moments are the same (as with a cube or sphere about its center) then, in fact, any axis is a principal axis. For proofs of these statements about the uniqueness or otherwise of the principal axes, see Problem 10.38. 10.6 Precession of a Top due to a Weak Torque We now know enough about the angular momentum of a rigid body to start solv- ing some interesting problems. We'll start with the phenomenon of precession of a spinning top subject to a weak torque. Consider a symmetric top as shown in Figure 10.7. The axes labeled x, y, and z are fixed to the ground with the z axis vertically up. The top is pivoted freely at its tip 0, and it makes an angle 0 with the vertical. Because of the top's axial symmetry, its inertia tensor is diagonal, with the form Xi 0 0 0 A l 0 0 0 A3, I =[ (10.80) relative to the top's principal axes (namely, an axis along e 3, the symmetry axis, and any two orthogonal axes perpendicular to e3). Section 10.6 Precession of a Top due to a Weak Torque 393 Figure 10.7 A spinning top is made of a rod OP fastened nor- mally through the center of a uniform circular disc and freely pivoted at 0. The total mass is M, and R denotes the CM position relative to 0. The principal axes of the top are in the directions of the unit vector e 3 along the symmetry axis of the top, and any two orthogonal unit vectors e 1 and e2 perpendicular to e 3. The top's weight Mg produces a torque, which causes the angular momentum L to change. Let us suppose first that gravity has been switched off and that the top is spinning about its symmetry axis, with angular velocity w = we 3 and angular momentum L = A3Ce) =.A.,3coe3. (10.81) Since there is no net torque on the top, L is constant. Therefore, the top will continue indefinitely to spin about the same axis with the same angular velocity. 1\u00b0 Now let us switch gravity back on, causing a torque T = R x Mg, with magnitude RMg sin 0 and a direction which is perpendicular to both the vertical z axis and the axis of the top. Let us suppose further that this torque is small. (We can ensure this by arranging that any or all of the parameters R, M, or g are small compared with the other relevant parameters of the system.) The existence of a torque implies that the angular momentum starts to change, since T = L. The changing of L implies that w starts to change and that the components w 1 cot cease to be zero. However, to the extent that the torque is small, we can and expect co l and w2 to remain sma11. 11 This means that Equation (10.81) remains a good approximation. (That is, the main contribution to L continues to be the spinning about 1\u00b0 That the angular velocity remains constant is actually not completely obvious. However, we shall prove it later on, so please accept it as a reasonable claim for now. 11 Again, this plausible statement should be (and will be) proved, but let us accept it for now. 394 Chapter 10 Rotational Motion of Rigid Bodies e3.) In this approximation, the torque F is perpendicular to L (since F is perpendicular to e3), which means that L changes in direction but not in magnitude. From (10.81) we see that e3 starts changing in direction, while w remains constant. Specifically, the equation L = F becomes A3coe3 = R x Mg or, substituting R = Re3 and g (where i is the unit vector that points vertically upward), where. M gR e3 = X3o) z x e3 = 52 x e3 (10.82) 52 = MgR X36) z. (10.83) You will recognize (10.82) as saying that the axis of the top, e 3, rotates with angular velocity 52 about the vertical direction i. Our conclusion is that the torque exerted by gravity causes the top's axis to precess, that is, to move slowly around a vertical cone, with fixed angle 9 and with angular frequency Q = RMgP. 3co. 12 This precession, although surprising at first glance, can be understood in elementary terms: In the view of Figure 10.7, the gravitational torque is clockwise, and the torque vector IT is into the page. Since L = F, this requires that the change in L be into", " the page, which is exactly the direction of the predicted precession. This precession of the axis of a spinning body is an effect that you have almost certainly observed when playing with a child's top. The same effect shows up in several other situations. For example, the earth spins on its axis, much like a spinning top, and the axis of spin is inclined at an angle 9 = 23\u00b0 from the normal to the earth's orbit around the sun. Because of the earth's equatorial bulge, the sun and moon exert small torques on the earth and these torques cause the earth's axis to precess slowly (one complete turn in 26,000 years), tracing out a cone of half-angle 23\u00b0 around the normal to the orbital plane \u2014 a phenomenon known as the precession of the equinoxes. This means that in another 13,000 years the pole star will be some 46\u00b0 away from true north. 10.7 Euler's Equations We are now ready to set up the equations of motion (or at least one form of the equations of motion) for a rotating rigid body. The two situations to which our discussion will principally apply are these: (1) A body that is pivoted about one fixed point, like the spinning top of Section 10.6, and (2) a body without any fixed point, like 12 1 should perhaps emphasize again that the discussion here is an approximation, the criterion for its validity being that C2 <-< co. An exact analysis shows that, if launched as described here, the top will also make very small oscillations called nutations in the B direction, although in practice these are quickly damped out by friction. Section 10.7 Euler's Equations 395 Figure 1 0.8 Axes used to derive Euler's equations for a rotating rigid body (shown here as an egg). The axes labeled x, y, and z define an inertial frame, often called the space frame. The unit vectors, e l, e2, and e3 point along the principal axes of the body and define the rotating, noninertial body frame. If the body has no fixed point (like an egg thrown in the air) then 0 is normally chosen to be the CM of the body. If the body has a fixed pivot, then 0 is that fixed point, and we would generally choose 0 to be the origin of both reference frames. a drum major's flying baton, whose rotational motion about the CM we have chosen to examine. The equations that I shall derive are called Euler's equations (named for the same mathematician as the Euler\u2014Lagrange equations of Chapter 6) and can be regarded as the rotational version of Newton's second law in the form ma = F. As we shall see, there are some problems that can be easily solved using Euler's equations. However, many problems are more easily solved using the Lagrangian approach, which we'll take up in Section 10.10. Before we launch into the derivation of Euler's equations, there is a complication we must now face up to. To take advantage of our understanding of the inertia tensor, and particularly the principal axes, we naturally want to use the principal axes of the body as our coordinate axes. However, because the principal axes are fixed in the rotating body, this inevitably involves us in using axes that rotate. We shall, therefore, need to use the machinery of Chapter 9 for handling rotating reference frames. The notation that I shall use is illustrated in Figure 10.8. For an inertial frame, relative to which Newton's laws hold in their simple form, we'll use the axes labeled x, y, and z. This frame is traditionally called the space frame, presumably because it is fixed in space \u2014 that is, inertial. The rotating frame is defined by the three unit vectors e l, e2, and e3, fixed in the body and pointing along the principal axes of the body. This frame, fixed in the body, is called the body frame. If the angular velocity of the body is co, and the principal moments of the body are 1, A2, and A3, then the angular momentum, as measured in the body frame, is L X2602, X3(03), [in the body frame]. (10.84) 396 Chapter 10 Rotational Motion of Rigid Bodies Now, if I- is the torque acting on the body, we know that as seen in the space frame dL dt )space r. (10.85) We saw in Chapter 9 that the rates of change of any vector as seen in the two frames are related by (9.30) L d dt space =(\u2014 ) -1-coxL dL dt body =L+(ox L (10.86) where in the second line I have reintroduced the convention that a dot represents a time derivative evaluated in the rotating body frame (whose angular velocity is co,", " the angular velocity of the body itself). Substituting (10.86) into (10.85), we arrive at the equation of motion for the rotating body frame: L-FwxL=F. (10.87) This equation is called Euler's equation. Using (10.84), we can resolve Euler's equation into its three components: which are often referred to as Euler's equations. The three Euler equations determine the motion of a spinning body as seen in a frame fixed in the body. In general, they are difficult to use because the components r 1, r2, and I-, of the applied torque as seen in the rotating body frame are complicated (and unknown) functions of time. In fact, the main use of Euler's equations is in the case that the applied torque is zero, as I shall discuss in the next section. However, there are a few other cases where the torque is simple enough that we can get useful information from Euler's equations. For example, consider again the spinning top of Section 10.6. As we saw there, the gravitational torque on the top is always perpendicular to the axis e3, so r3 is always zero. Furthermore, because of the top's axial symmetry, the two moments of inertia A 1 and A2 are equal. Thus the third of the Euler equations (10.88) reduces to X3co3 = 0. That is, the component of co along the symmetry axis is constant, a result that I stated as reasonable, but did not prove, in the discussion of Section 10.6. 13 13 The other main unproved assertion of that discussion \u2014 that the components w 1 and w2 remain small for all times \u2014 can also be understood using Euler's equations. The components F 1 and F2 Section 10.8 Euler's Equations with Zero Torque 397 10.8 Euler's Equations with Zero Torque Let us now consider a rotating body subject to zero torque. In this case, Euler's equations (10.88) take the simple form = (X2 - X3) (2 63 X2 6.\u00b02 = (X3 - X1)W3 61) 1 X3 6.\u00b03 = (X1 - X2) 6 1 62 (10.89) I shall discuss these equations, first for the case that the three principal moments A l, A2, and A3 are all different, and then for the case that A i = X2 0 X3 (as with a spinning top). A Body with Three Different Principal Moments Let us suppose first that the principal moments of the body under consideration are all different. If at time t = 0 the body happens to be rotating about one of its principal axes (e3 say), then col = co2 = 0 at t = 0. Now, with co l = co2 = 0, the right sides of all three Euler equations (10.89) are zero. This shows that as long as co l and co2 are zero, all three components of co remain constant. That is, oh and co 2 remain zero and co 3 is constant. In other words, if the body starts out rotating about one of its principal axes, it will continue to do so, with constant angular velocity co. This statement applies, in the first instance, to the angular velocity as measured in the rotating body frame. However, with co along e 3, the angular momentum is L = )L 3co, and we know that L is constant as seen in any inertial frame. Thus our result applies equally in any inertial frame: If a body that is subject to no torque is spinning initially about any principal axis, it will continue to do so indefinitely with constant angular velocity. The converse of this result is also true. If at t = 0, the angular velocity is not along a principal axis, then co is not constant. To see this, notice first that if co is not along a principal axis, then at least two of its components are nonzero. If you look at the Euler equations (10.89) you will see that, with two components of co nonzero, at least one component of 61) must be nonzero. (For example, if co t and w2 are nonzero, then c'o3 0 0.) Therefore, with two of its components nonzero, co cannot be constant. We conclude that the only way a body with three different principal moments can rotate freely with constant angular velocity is by rotating about one of its principal axes. It is interesting to know if this kind of rotation is stable. That is, if a body is rotating about one of its principal axes and is given a very small kick, will it continue to rotate close to its original axis, or will its motion change completely? Let us suppose that the body is rotating about the axis e 3, with co l = co2 = 0. If now we give", " it a tiny kick, then co l and co2 will pick up nonzero values that are, at least initially, small. The question is whether the values of col and co2 will remain small or start to grow bigger. of the torque are nonzero, which is what drives w 1 and w2. Nonetheless, they are both small and oscillate rapidly as the top spins. It is reasonably easy to see that, under these conditions, the first two Euler equations (10.88) imply that w 1 and w2 also oscillate rapidly and with small amplitude. This is the required result. 398 Chapter 10 Rotational Motion of Rigid Bodies To answer this, we note from the third of the Euler equations (10.89) that as long as both coi and co2 are small, (o3 remains very small (\"small x small\"). Thus initially, at least, it is a good approximation to take (0 3 to be constant. In this case, the first two Euler equations read ithi = [(A2 \u2014 ), 3)603] (02 X26)2 = [(A3 \u2014 1*3l 6) 1 (10.90) where the coefficients in square brackets are (approximately) constant. These coupled first-order equations for co l and co2 are easily solved. Perhaps the simplest method is to differentiate the first equation once and then substitute the second, to give Col = [ (A3 X2) (A3 \u2014 ;L 1) 2 (.03 6)1. A1A2 (10.91) If the coefficient in square brackets is positive, the solution for co l is a sine or cosine in time, and ca l undergoes small oscillations, returning repeatedly to zero. According to the first equation of (10.90), co 2 is proportional to co l, so, under the same conditions, ct)2 undergoes similar small oscillations. To see what these conditions imply, notice that the coefficient in (10.91) is positive if A3 is larger than both Al and A2 or smaller than both Al and A2. Therefore, we have shown that if the body is spinning about either the principal axis with largest moment or that with smallest moment, the motion is stable against small disturbances. On the other hand if A3 lies between A l and A2, the coefficient in brackets in (10.91) is negative, and the solution for co l is a real exponential, which moves rapidly away from zero. 14 Since (02 a col, the same is true of co2, and we reach the following conclusion: For a freely spinning body, rotation about the principal axis with the intermediate moment (A3 less than X 1 but more than A2 or vice versa) is unstable. You can test this interesting claim with a book, held shut with a rubber band. If you throw it up giving it a spin about either the maximum or minimum axis, it will continue to spin stably about that axis. If you give it a spin about the intermediate axis, it will tumble wildly (and be much harder to catch). Motion of a Body with Two Equal Moments: Free Precession The complete solution of the Euler equations (10.89) for a freely rotating body with three different principal moments is possible, but complicated and unilluminating. If two of the three principal moments are equal (as with a spinning top), the correspond- ing problem is much easier and quite interesting. Let us, therefore, consider this case, 14 There is a small subtlety that may be worth mentioning: The solution for co l is a combination of two exponentials, el\"`, and you might think that the special case of a pure decaying exponential, e', was an exception to my claim. However, this solution is excluded by the initial conditions: If the body is given a small kick from co l = 0, then co l and d l must have the same sign, whereas for the pure decaying exponential they have opposite signs. Section 10.8 Euler's Equations with Zero Torque 399 and suppose that the first two moments are equal, Al = A2. The crucial simplifying feature of this case is that the third of the Euler equations (10.89) becomes That is, co3, the third component of the angular velocity (as measured in the body frame), is constant, Knowing that co3 is constant, we can now rewrite the first two Euler equations as cbi = (A1 \u2014 A3)(03 \u201e \u20142 = Qba)2 (Al \u2014 A3)(03 Cb2 = W1=\u2014Qbo)l A 1 A where I have defined the constant frequency (10.92) (10.93) (The subscript \"b\" stands for body, for reasons we'll see in a moment.) The coupled equations (10.92) for co l and cot are easily solved by the now familiar trick of setting col +", " iw2 = 17, which reduces (10.92) to (03 = 0. co3 = const. Q b = Al \u2014 X3 1 Qbq, \u20141 C2br 170e whence If we choose our axes so that co l = coo and cot = 0 at t = 0, then 1)0 = coo and, taking real and imaginary parts of )7, we find for the complete solution = (coo cos Q bt, \u2014co o sin Qbt, co3) (10.94) with coo and co3 both constant. The two components co l and (02 rotate with angular velocity Q b, while co3 remains constant. Since co o and co3 are constant, so is the angle a between co and e 3. Therefore, as seen in the body frame, co moves steadily around a cone, called the body cone with angular frequency S -2b given by (10.93), as indicated in Figure 10.9(a). The angular momentum L is given by L A1co2, A3w3) = (A I wo cos Qbt, \u2014. 1coc, sin Qbt, X3co3). (10.95) Comparison of (10.94) and (10.95) should convince you that the three vectors, co, L, and e3 lie in a single plane, with the angles between any two being constant in time. (We've shown this in the body frame, but this means that it's true in any frame.) Thus, as seen in the body frame, both co and L precess around e 3 at the same rate Qb. 400 Chapter 10 Rotational Motion of Rigid Bodies L (fixed) (a) Body frame (b) Space frame Figure 10.9 An axially symmetric body (shown here as a prolate spheroid or \"egg-shaped\" solid) is rotating with angular velocity co, not in the direction of any of the principal axes. (a) As seen in the body frame, both co and L precess about the symmetry axis, e 3, with angular frequency S-2b given by (10.93). (b) As seen in the space frame, L is fixed, and both co and e 3 precess about L with frequency S2, given by (10.96). To find what happens as seen in the space frame, note that in any inertial frame the vector L is constant. Therefore, as seen in the space frame the plane containing co, L, and e3 must rotate about L, and the two vectors co and e 3 precess about L, as shown in Figure 10.9(b). In particular, in the space frame, co traces out a cone, called the space cone, around which the body cone rolls. I leave it as a fairly difficult exercise (Problem 10.46) to show that the rate of precession of co around the space cone can be expressed in several ways, of which the simplest is this: Qs = A, (10.96) Note well that the free precession derived here has nothing to do with any external torques on the spinning body. On the contrary, we have derived it for a body subject to no external torque. An interesting example of this precession is provided by the earth. I have already mentioned that the sun and moon produce a small torque on the earth's equatorial bulge, and that this torque causes the precession of the equinoxes, the precession of the polar axis with a period of 26,000 years. But the equatorial bulge also means that the earth's moment of inertia about the polar axis is larger than the other two principal moments by about 1 part in 300. According to (10.93) this should imply a precession (unless the earth's rotation is perfectly aligned with the principal axis) with frequency S-2b = w3 /300. Since w3 represents one revolution per day, Qb should correspond to one revolution every 300 days. A tiny wobbling of the polar axis (by less than a second of arc) was discovered by the American amateur astronomer Seth Chandler (1846-1913). This Chandler wobble is apparently due to the free-body precession discussed here, although its period is more like 400 days, supposedly because the earth is not a perfectly rigid body. Section 10.9 Euler Angles* 401 10.9 Euler Angles* * Sections marked with an asterisk contain material that is usually somewhat more advanced and could be omitted if you are pressed for time. The trouble with the Euler equations (10.88) is that they refer to axes fixed in the body, and, except in a few very simple problems, such axes are hopelessly awkward to work with. We need to come up with equations of motion relative to a nonrotating space frame, and before we can do that we need a", " set of coordinates that specifiy the orientation of our body relative to such a frame. There are several ways to do this, all of them surprisingly cumbersome, but by far the most popular and useful is due to Euler (yet again!) and specifies the orientation of a rigid body by three Euler angles. In many applications, the body of interest rotates about a fixed point, and in this case (the only case we'll consider in detail) we naturally choose the fixed point to be the origin 0 of both the space and the body axes. As before, the basis vectors of the space frame will be called X, ST, and i. For the body frame, we'll use the principal axes of the body, with directions e l, e2, and e3. If two of the principal moments are equal, we'll take them to be numbers 1 and 2, so that A 1 = X2, and we'll refer to the third direction e3 as the axis of symmetry. Let us imagine the body oriented initially with its three axes lying along the corresponding space axes (e l along X and so on). We're going to see that by a sequence of three rotations, through angles 9, 0, and about three different axes, we can bring the body into any assigned orientation and that the angles (0,0, ilf) specify a unique orientation of the body. In particular, the angles 9 and q5 will be just the polar angles of the axis e 3 relative to the space frame. Step (a). Starting with the body axes aligned with the space axes, we first rotate the body through an angle about the axis i, as illustrated in the first frame of Figure 10.10. This rotates the first and second body axes in the xy plane. In particular, the second body axis now points in the direction labeled e2. Step (b). Next we rotate the body through an angle 9 about the new axis e2. This moves the body axis e 3 to the direction whose polar angles are 9 and O. Evidently our (a) Figure 10.10 Definition of the Euler angles 0, 0, and i. Starting with the body axes e l, e2, e3 X, y, i aligned, the three successive rotations bring the body axes to any and spaces axes prescribed orientation. 402 Chapter 10 Rotational Motion of Rigid Bodies first two steps can bring the body axis e 3 to any assigned orientation, and, with e 3 in position, the only remaining freedom is a rotation about e 3. Step (c). Finally, we'll rotate the body about e 3 through whatever angle * is needed to bring the body axes e 2 and e l into their assigned directions, as shown in the third frame of Figure 10.10. The three angles (0, 0, ik) are the Euler angles 15 that specify the body's orienta- tion. Before we can use them, we must calculate a few parameters in terms of them, starting with the angular velocity w. To find co, notice that we can regard the steps of Figure 10.10 as defining a sequence of four reference frames, starting with the space axes defined by 1, Sr, and z, and moving via two intermediate frames to wind up with the body axes defined by e 1, e2, and e3. To find the angular velocity of the body axes relative to the space axes, we have only to find the velocity of each of these frames relative to its predecessor and form their vector sum. [Remember that relative angular velocities add, as we observed in (9.25)1 As 0 varies, the frame defined by step (a) rotates with angular velocity co a = 0 relative to the space axes. Similarly the angular velocity of the frame defined by step (b) relative to its predecessor is co b = 9e2, and that of the body frame in step (c) is co c = fre3. Therefore, the required angular velocity of the body frame relative to the space frame is = 6)a + Wb Wc = (i)1 ee/2, \"Ce3. (10.97) This expresses co in terms of a rather messy mixture of unit vectors, but it is a simple matter to rewrite these vectors in terms of the unit vectors of any one frame (Problem 10.48). To find the angular momentum or kinetic energy, we need in general to find the components of ov relative to the principal axes e 1, e2, and e3. However, if we are content to consider just the symmetric case that A 1 = A 2, we don't even have to do this. The point is that with A1= X2 any two perpendicular axes in the plane of e 1 and e2 are also principal axes. Thus, instead of e 1, e2, and e3 we can use the axes e ll, e2, and e3 where ei and e2 are", " both shown in the second frame of Figure 10.10. This choice has the advantage that the last two terms in (10.97) need no conversion and, since (as you = (cos 0)e 3 \u2014 (sint9)&i, (10.98) should check) we find that 15 Beware of the many different conventions used in defining Euler's angles. The convention fol- lowed here is most popular in quantum mechanics but less so among authors of classical mechanics. It has the great advantage that B and are precisely the polar angles of the body axis e3. Section 10.10 Motion of a Spinning Top* 403 Knowing the angular velocity co with respect to a set of principal axes, we can immediately write down the angular momentum L and the kinetic energy T. The angular momentum is L = A2co2, A3co3) (relative to any set of principal axes). Thus in our case For future reference, note that the component of L along the body axis e3 is just L3 = ;L3CO3 = )n.3(c.fr COS 0). (10.101) Also, as you can check (Problem 10.49), the component along the space axis z is L z = sin2 6 + c/, cos 0) cos = ;4) sin2 0 + L3 cos 0 where in the second line I have used (10.101). Also for future reference, note that we can solve (10.103) for 4 in terms of 0, L, and L3: (10.102) (10.103) (10.104) L z \u2014 L 3 cos 0 sin2 9 We saw in (10.68) that the kinetic energy is T = z (7,,(0 12 + A.20) 2 + a.3q). Thus, for a body whose first two principal moments are equal, (10.99) gives We shall use this result in a moment to write down the Lagrangian for a spinning top. 10.10 Motion of a Spinning Top* Lagrange's Equations To illustrate the use of Euler angles, let us return to the symmetric spinning top discussed in Section 10.6 and shown in Figure 10.7. The motion of this system is most easily solved using the Lagrangian approach, so we'll start by writing down the Lagrangian,C, = T \u2014 U. The kinetic energy is given by (10.105), while the potential energy is U = MgR cos 0. Therefore the Lagrangian of the top is = i Ai((i52 sin2 0 + 62) \u00b1,A cos 69 2 \u2014 MgRcose. (10.106) 404 Chapter 10 Rotational Motion of Rigid Bodies With three generalized coordinates, there are three Lagrange equations. The 0 equation is p0 this gives az ao = X1.2 sin 0 cos 9 \u2014 A 3 (.fr + (I) cos 0) sin 0 MgR sin 0 [0 equation]. (10.107) The 0 and equations are simpler because neither cb nor '0 appears in so that both 0 and 0' are ignorable, and the corresponding generalized momenta are constants. For p* = = X 1(1) sin 2 0 \u00b1 X3(1fr cos 0) cos 0 = const [0 equation]. (10.108) Comparing with (10.102), we see that the generalized momentum p* is just the z component L z of the angular momentum, and the constancy of p0 is just a statement that L, is conserved \u2014 a result we could have anticipated since there is no torque about the z axis. Similarly, for p * we get \u2014 = A. 3 ( +/r cos 9) = const [lit equation]. (10.109) Comparing this with (10.101), we see that p* is the component of L along the body's symmetry axis e 3, and the constancy of p* tells us that L3 is conserved. An important consequence is that, since L 3 = A3w3, the component co 3 of the angular velocity is also constant (a result we proved back at the end of Section 10.7 using Euler's equations). Steady Precession As a first application of the Lagrange equations, let us investigate whether the top can exhibit a precession in which the top's axis moves around the z axis tracing a cone with constant angle 0. From (10.104) we see that if 0 is constant, then so is (1). That is, if the top's axis is to move around a cone of fixed angle 0, it has to do so at a constant angular velocity 0 = Q, say. Looking next at the equation (10.109), we see that, with 0 and 0 fixed, 1fr must also be constant. The rate Q at which", " the top precesses is determined by the 0 equation (10.107). If 9 is constant, the left side is zero, and, replacing 0 by Q, and (1k. cb cos 0) by co3, we find that S2 must satisfy X I S-22 cos 0 \u2014 A. 3w3C2 MgR 0. (10.110) This is a quadratic equation for Q. Thus, as long as the roots are real, there are two different rates Q at which the top can precess, for a given tilt 0 and given rate of spin co3. We can write down these two values of Q for any values of the other parameters, but the most interesting case is when the top is spinning rapidly and co 3 is large. In this case, it is easy to see that the two roots are real and that one root is much smaller than the other. The small root is (Problem 10.53) Q Mg R X30..)3 (10.111) Section 10.10 Motion of a Spinning Top* 405 This slow precession is precisely the motion predicted in Section 10.6 with Q given by (10.83). 16 The second, larger root of (10.110) (again asuming that w 3 is very large) is A3w3 X i cos 0 (10.112) (See Problem 10.53.) Notice that this faster precession does not depend on g, so we would expect to observe this one even in the absence of gravity. In fact, this precession is precisely the free precession predicted in Section 10.8 for a symmetric body moving in the absence of any torques. As you can check, the value of Q predicted here is the same as what was called Q s in (10.96). (See Problem 10.52.) Nutation In general, as a top precesses around the vertical axis (0 changing), the angle 0 varies as well. Thus as the axis swings around the vertical it also nods up or down in a motion called nutation, from a Latin word meaning to nod repeatedly. We can investigate the variation of 0 using the 0 equation (10.107). The first step is to use the 0 and 0' equations to eliminate the variables 0 and in favor of the constants p0 = 1,, and 13, L3. This gives a second-order, ordinary differential equation for 0, which can, at least in principle, be solved to give the time dependence of 0. To get a qualitative picture of the motion, a simpler procedure is to look at the total energy, E = T U, with T given by (10.105) and U = MgRcos9. In T, we can replace the variables 0 and in favor of the constants L z and L3, and we find (Problem 10.51) E = 410 2 Ueff (0) (10.113) where the effective potential energy U eff (0) is tido) = (Lz - COS 0) 2 L3 2X sin' 1 LA3 MgRcos0. (10.114) Equation (10.113) serves to emphasize that our problem has been reduced to a one- dimensional problem, involving just the coordinate O. We can predict the qualitative behavior of 0 by looking at a graph of Ueff (0). The coordinate 0 ranges from 0 to 7, and, because of the factor of sin e 0 in the denominator of the first term, Ueff( 0 ) approaches -Foo at the two ends, 0 = 0 and r. It is not hard to convince oneself that the graph of Ueff (0 ) has the \"U\" shape shown in Figure 10.11. From (10.113), it is clear that E > Ueff (0), so 0 is confined between the two turning points, 9 1 and 02, where E = Ueff(0) as shown in the figure. Evidently, 0 oscillates periodically, or \"nutates,\" between O i and 02 at the same time that the top's axis precesses around the vertical. 16 Here the denominator has a factor w3 where (10.83) has co, but this difference is irrelevant since both discussions assume that w3 is very large, so that (03 ti (0. 106 Chapter 10 Rotational Motion of Rigid Bodies Ueff 0 1 0 E 02 7/2 0 IT Figure 10.11 The effective potential energy (10.114) that deter- mines the time dependence of 0 for the symmetric top. Since E can be no less than (Jeff (0 ), the motion is confined to the interval 0 1 < 0 < 02. The details of the motion depend on just how 0 varies. According to (10.104), (i) = L, \u2014 L3 COS 0 )L i sin2 0 (10.115) which shows that there are", " two main possibilities: If L, is larger than L3 (in magni- tude), then (/) cannot vanish. Thus, although may vary, it can never change sign, so 0 changes in the same direction all the time (always increasing or always decreasing). Thus, the top precesses in a single direction, while its angle of tilt 9 oscillates between 0 1 and 02, producing the motion sketched in Figure 10.12(a). If L, is smaller than L3, then (1) would vanish at an angle 9o, such that L, \u2014 L3 cos 00 = 0. If this angle lies out- (a) (b) Figure 10.12 Nutation of a top. The top end of the spinning top moves on a sphere centered on the pivot at the bottom end. (a) Here never vanishes, and 0 always moves steadily in one direction while B oscillates between 0 1 and 02. (b) If cb changes sign, then 0 moves first forward and then backward while 0 oscillates. Principal Definitions and Equations of Chapter 10 407 side the range between 0 1 and 02 to which the motion is confined, then still cannot vanish and the motion is still as sketched in Figure 10.12(a). On the other hand, if the angle 00 lies between 0 1 and 02, then will change sign twice in each oscillation of O. In this case, the precession moves first in one direction, then in the other, and the overall motion is as sketched in Figure 10.12(b). Principal Definitions and Equations of Chapter 10 CM and Relative Motions L = L(motion of CM) + L (motion relative to CM). [Eq. (10.9)] and T = T (motion of CM) + T (motion relative to CM). [Eq. (10.16)] The Moment of Inertia Tensor The angular momentum L and angular velocity co of a rigid body are related by L = Ico [Eq.(10.42)] where L and co must be seen as 3 x 1 columns and I is the 3 x 3 moment of inertia tensor, whose diagonal and off-diagonal elements are defined as mo (ya z), etc. and /xy = \u2014 E m axa ya, etc. a a [Eqs. (10.37) & (10.38)] Principal Axes A principal axis of a body (about a point 0) is any axis through 0 with the property that if co points along the axis, then L is parallel to co; that is, L = Xco [Eq. (10.65)] for some real number X. For any body and any point 0, there are three perpendicular principal axes through 0. [Section 10.4 and Appendix] Evaluated with respect to its principal axes, the inertia tensor has the diagonal form Xi 0 0 0 x2 0. = 0 0 X3 [ [Eq. (10.79)] 408 Chapter 10 Rotational Motion of Rigid Bodies Euler's Equations If L denotes the rate of change of a body's angular momentum as seen in a frame fixed in the body (body frame), then it satisfies Euler's equations L + (t) x L = T. [Eqs. (10.87) & (10.88)] Euler's Angles The orientation of a rigid body can be specified by the three Euler angles 0, 0, '0 defined in Figure 10.10. [Sections 10.9 & 10.10] The Lagrangian for a rigid body spinning about a fixed pivot is = 4X 1 (ci)2 sine 0 6 2 ) + (i) cos 9) 2 \u2014 MgRcos0. [Eq. (10.106)] Problems for Chapter 10 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). SECTION 10.1 Properties of the Center of Mass 10.1 * The result (10.7), that E m a r = 0, can be paraphrased to say that the position vector of the CM relative to the CM is zero, and, in this form, is nearly obvious. Nevertheless, to be sure you understand the result, prove it by solving (10.4) for ea and substituting into the sum concerned. 10.2 * To illustrate the result (10.18), that the total KE of a body is just the rotational KE relative to any point that is instantaneously at rest, do the following: Write down the KE of a uniform wheel (mass M) rolling with speed v along a flat road, as the sum of the energies of the CM motion and the rotation about the CM. Now write it as the energy of the rotation about the instantaneous point of contact with the road and show that you get the same answer. (The energy of rotation is 410)2. The moment of inertia", " of a uniform wheel about its center is I = z MR 2. That about a point on the rim is /' = z MR2.) 10.3 * Five equal point masses are placed at the five corners of a square pyramid whose square base is centered on the origin in the xy plane, with side L, and whose apex is on the z axis at a height H above the origin. Find the CM of the five-mass system. 10.4 ** The calculation of centers of mass or moments of inertia usually involves doing an integral, most often a volume integral, and such integrals are often best done in spherical polar coordinates (defined back in Figure 4.16). Prove that f dV f(r) = f r 2dr f sin 0 d0 f d0 f (r, 0, 0). [Think about the small volume dV enclosed between r and r dr, 9 and 0 + d0, and 0 and 0 d0.] If the volume integral on the left runs over all space, what are the limits of the three integrals on the right? 10.5 ** A uniform solid hemisphere of radius R has its flat base in the xy plane, with its center at the origin. Use the result of Problem 10.4 to find the center of mass. [Comment: This and the next two Problems for Chapter 10 409 problems are intended to reactivate your skills at finding centers of mass by integration. In all cases, you will need to use the integral form of the definition (10.1) of the CM. If the mass is distributed through a volume (as here), the integral will be a volume integral with dm = p dV.1 10.6 ** (a) Find the CM of a uniform hemispherical shell of inner and outer radii a and b and mass M positioned as in Problem 10.5. [See the comment to Problem 10.5 and use the result of Problem 10.4.] (b) What becomes of your answer when a = 0? (c) What if b a? 10.7 ** A \"rounded cone\" is made by cutting out of a uniform sphere of radius R the volume with 0 < 00, where 8 is the usual angle measured from the polar axis and 00 is a constant between 0 and m. (a) Describe this cone and use the result of Problem 10.4 to find its volume. (b) Find its CM and comment on your results for the cases that 0o = 7 and 00 --\u00b1 0. 10.8 ** A uniform thin wire lies along the y axis between y = +L/2. It is now bent toward the left into an arc of a circle with radius R, leaving the midpoint at the origin and tangent to the y axis. Find the CM. [See the comment to Problem 10.5. In this case the integral is a one-dimensional integral.] Comment on your answer for the cases that R oo and that 27R = L. SECTION 10.2 Rotation about a Fixed Axis 10.9 * The moment of inertia of a continuous mass distribution with density p is obtained by converting the sum of (10.25) into the volume integral f p 2 dm f p2 0 d v (Note the two forms of the Greek. \"rho\": p = distance from z axis, p = mass density.) Find the moment of inertia of a uniform circular cylinder of radius R and mass M for rotation about its axis. Explain why the products of inertia are zero. 10.10 * (a) A thin uniform rod of mass M and length L lies on the x axis with one end at the origin. Find its moment of inertia for rotation about the z axis. [Here the sum of (10.25) must be replaced by an integral of the form f x 2,u dx where p, is the linear mass density, mass/length.] (b) What if the rod's center is at the origin? 10.11 ** (a) Use the result of Problem 10.4 to find the moment of inertia of a uniform solid sphere (mass M, radius R) for rotation about a diameter. (b) Do the same for a uniform hollow sphere whose inner and outer radii are a and b. [One slick way to do this is to think of the hollow sphere as a solid sphere of radius b from which you have removed a sphere of the same density but radius a.] 10.12 ** A triangular prism (like a box of Toblerone) of mass M, whose two ends are equilateral triangles parallel to the xy plane with side 2a, is centered on the origin with its axis along the z axis. Find its moment of inertia for rotation about the z axis. Without doing any integrals write down and explain its two products of inertia for rotation about the z axis. 10.13 ** A thin rod (of width zero, but not necessarily uniform) is pivoted freely at", " one end about the horizontal z axis, being free to swing in the xy plane (x horizontal, y vertically down). Its mass is m, its CM is a distance a from the pivot, and its moment of inertia (about the z axis) is I. (a) Write down the equation of motion L z = rz and, assuming the motion is confined to small angles (measured from the downward vertical), find the period of this compound pendulum. (\"Compound pendulum\" is traditionally used to mean any pendulum whose mass is distributed \u2014 as contrasted with a \"simple pendulum,\" whose mass is concentrated at a single point on a massless arm.) (b) What is the length of the \"equivalent\" simple pendulum, that is, the simple pendulum with the same period? 10.14 ** A stationary space station can be approximated as a hollow spherical shell of mass 6 tonnes (6000 kg) and inner and outer radii of 5 m and 6 m. To change its orientation, a uniform flywheel (radius 410 Chapter 10 Rotational Motion of Rigid Bodies 10 cm, mass 10 kg) at the center is spun up quickly from rest to 1000 rpm. (a) How long will it take the station to rotate by 10 degrees? (b) What energy was needed for the whole operation? [To find the necessary moment of inertia, you could do Problem 10.11.] 10.15 ** (a) Write down the integral (as in Problem 10.9) for the moment of inertia of a uniform cube of side a and mass M, rotating about an edge, and show that it is equal to 3 Ma e. (b) If I balance the cube on an edge in unstable equilibrium on a rough table, it will eventually topple and rotate until it hits the table. By considering the energy of the cube, find its angular velocity just before it hits the table. (Assume the edge does not slide on the table.) 10.16 ** Find the moment of inertia for a uniform cube of mass M and edge a as in Problem 10.15 and then do the following: The cube is sliding with velocity v across a flat horizontal frictionless table when it hits a straight very low step perpendicular to v, and the leading lower edge comes abruptly to rest. (a) By considering what quantities are conserved before, during, and after the brief collision, find the cube's angular velocity just after the collision. (b) Find the minimum speed v for which the cube rolls over after hitting the step. 10.17 ** Write down the integral for the moment of inertia of a uniform ellipsoid with surface (x /a ) 2 + (Y1b) 2 (z/c) 2 = 1 for rotation about the z axis. One simple way to do the integral is to make a change of variables to = xla,17 = y/b, and = z/c. Each of the two resulting integrals can be related to the corresponding integrals for a sphere (as in Problem 10.11). Do this. Check your answer for the case a = b =c. 10.18 *** Consider the rod of Problem 10.13. The rod is struck sharply with a horizontal force F which delivers an impulse F At = a distance b below the pivot. (a) Find the rod's angular momentum, and hence momentum, just after the impulse. (b) Find the impulse 17 delivered to the pivot. (c) For what value of b (call it bo) is ri = 0? (The distance bo defines the so-called \"sweet spot.\" If the rod were a tennis racket and the pivot your hand, then if the ball hits the sweet spot, your hand would experience no impulse.) SECTION 10.3 Rotation about Any Axis; the Inertia Tensor 10.19 * Verify that the components of the vector r x (co x r) are given correctly by Equation (10.35). Do this both by working with components and by using the so-called BAC \u2014 CAB rule, that is A x (B x C) = B(A C) \u2014 C(A B). 10.20 * Show that the inertia tensor is additive, in this sense: Suppose a body A is made up of two parts B and C. (For instance, a hammer is made up of a wooden handle wedged into a metal head.) Then IA = IB + I. Similarly, if A can be thought of as the result of removing C from B (as a hollow spherical shell is the result of removing a small sphere from inside a larger sphere), then I A = IB \u2014 lc. 10.21 ** The definition of the inertia tensor in Equations (10.37) and (10.38) has the rather ugly feature that the diagonal and off-diagonal elements are defined by completely different equations. Show that the two definitions can be combined into the single equation (which is", " slightly less messy in integral form) Iij = f Q(r26ii \u2014 rirj)dV Problems for Chapter 10 411 where 6 ii is the Kronecker delta symbol I 1 i = j 6ii 10 i j. (10.116) 10.22 ** A rigid body comprises 8 equal masses m at the corners of a cube of side a, held together by massless struts. (a) Use the definitions (10.37) and (10.38) to find the moment of inertia tensor I for rotation about a corner 0 of the cube. (Use axes along the three edges through 0.) (b) Find the inertia tensor of the same body but for rotation about the center of the cube. (Again use axes parallel to the edges.) Explain why in this case certain elements of I could be expected to be zero. 10.23 ** Consider a rigid plane body or \"lamina,\" such as a flat piece of sheet metal, rotating about a point 0 in the body. If we choose axes so that the lamina lies in the xy plane, which elements of the inertia tensor I are automatically zero? Prove that 4, = Ixx + iyy 10.24 ** (a) If I' denotes the moment of inertia tensor of a rigid body (mass M) about its CM, and I the corresponding tensor about a point P displaced from the CM by A = prove that /xx = /x7 mo 2 ^ 2 ) and / = yz yz \u2014 (10.117) and so forth. (These results, which generalize the parallel-axis theorem that you probably learned in introductory physics, mean that once you know the inertia tensor for rotation about the CM, calculating it for any other origin is trivially easy.) (b) Confirm that the results of Example 10.2 (page 381) fulfill the identities (10.117) [so that the calculations of part (a) of the example were actually unnecessary]. 10.25 ** (a) Find all nine elements of the moment of inertia tensor with respect to the CM of a uniform cuboid (a rectangular brick shape) whose sides are 2a, 2b, and 2c in the x, y, and z directions and whose mass is M. Explain clearly why you could write down the off-diagonal elements without doing any integration. (b) Combine the results of part (a) and Problem 10.24 to find the moment of inertia tensor of the same cuboid with respect to the corner A at (a, b, c). (c) What is the angular momentum about A if the cuboid is spinning with angular velocity w around the edge through A and parallel to the x axis? 10.26 ** (a) Prove that in cylindrical polar coordinates a volume integral takes the form f dV Pr), f p dp f c14) f dz f(p,O,z). (b) Show that the moment of inertia of a uniform solid cone pivoted at its tip and rotating about its axis is given by the integral (10.58), explaining clearly the limits of integration. Show that the integral evaluates to MR 2. (c) Prove also that 4, = M(R 2 4h 2 ) as in Equation (10.61). 10.27 *** Find the inertia tensor for a uniform, thin hollow cone, such as an ice-cream cone, of mass M, height h, and base radius R, spinning about its pointed end. 10.28 *** Find the moment of inertia tensor I for the triangular prism of Problem 10.12 with height h. (If you did Problem 10.12, you've already done about half the work.) Your result should show that I has the form we've found for an axisymmetric body. This suggests what is true, that three-fold symmetry about an axis (symmetry under rotations of 120 degrees) is enough to ensure this form. 412 Chapter 10 Rotational Motion of Rigid Bodies SECTION 10.4 Principal Axes of Inertia 10.29 * Prove that if the axes Ox, Oy, and Oz are principal axes of a certain rigid body, then the inertia tensor (with respect to these axes) is diagonal with the principal moments down the diagonal as 10.30 * Consider a lamina, such as a flat piece of sheet metal, rotating about a point 0 in the body. Prove that the axis through 0 and perpendicular to the plane is a principal axis. [Hint: See Problem 10.31 ** Consider an arbitrary rigid body with an axis of rotational symmetry, which we'll call i. (a) Prove that the axis of symmetry is a principal axis. (b) Prove that any two directions x and Sr perpendicular to i and each other are also principal axes. (c) Prove that the principal moments corresponding to these two", " axes are equal: A l = X2. 10.32 ** Show that the principal moments of any rigid body satisfy X3 < + X2. [Hint: Look at the integrals that define these moments.] In particular, if X2, then X3 < 2.11. (b) For what shape of body is X3 = X1 + X2? 10.33 k** Here is a good exercise in vector identities and matrices, leading to some important general results: (a) For a rigid body made up of particles of mass m a, spinning about an axis through the origin with angular velocity co, prove that its total kinetic energy can be written as Remember that v a = w x ra. You may find the following vector identity useful: For any two vectors in (10.66). 10.23.] a and b, written as (If you use the identity, please prove it.) (b) Prove that the angular momentum L of the body can be T = - ma[(wra) 2 \u2014 (co ra) 2]. (a x b) 2 a2, 2 D (a b) 2. L = >2, m a [cor,y 2 \u2014 ra (co ra )] T = lco L = 2 2 For this you will need the so-called BAC \u2014 CAB rule, that A x (B x C) = B(A C) \u2014 C(A B). (c) Combine the results of parts (a) and (b) to prove that Prove both equalities. The last expression is a matrix product; co denotes the 3 x 1 column of numbers cox, coy, coz, the tilde on iO denotes the matrix transpose (in this case a row), and I is the moment of inertia tensor. This result is actually quite important; it corresponds to the much more obvious result that for a particle, T = Or p. (d) Show that with respect to the principal axes, T = 4(x1,012 + A2,022 + x3w32), as in Equation (10.68). SECTION 10.5 Finding the Principal Axes; Eigenvalue Equations 10.34 * The inertia tensor I for a solid cube is given by (10.72). Verify that det(I \u2014 Al) is as given in (10.73). 10.35 ** A rigid body consists of three masses fastened as follows: m at (a, 0, 0), 2m at (0, a, a), and 3m at (0, a, \u2014a). (a) Find the inertia tensor I. (b) Find the principal moments and a set of orthogonal Problems for Chapter 10 413 principal axes. principal axes. 10.36 ** A rigid body consists of three equal masses (m) fastened at the positions (a, 0, 0), (0, a, 2a), and (0, 2a, a). (a) Find the inertia tensor I. (b) Find the principal moments and a set of orthogonal 10.37 *** A thin, flat, uniform metal triangle lies in the xy plane with its corners at (1, 0, 0), (0, 1, 0), and the origin. Its surface density (mass/area) is a = 24. (Distances and masses are measured in unspecified units, and the number 24 was chosen to make the answer come out nicely.) (a) Find the triangle's inertia tensor I. (b) What are its principal moments and the corresponding axes? 10.38 *** Suppose that you have found three independent principal axes (directions e l, e2, e3) and corresponding principal moments A 1, A2, A3 of a rigid body whose moment of inertia tensor I (not diagonal) you had calculated. (You may assume, what is actually fairly easy to prove, that all of the quantities concerned are real.) (a) Prove that if A i Ai then it is automatically the case that e i e j = 0. (It may help to introduce a notation that distinguishes between vectors and matrices. For example, you could use an underline to indicate a matrix, so that a is the 3 x 1 matrix that represents the vector a, and the vector scalar product a b is the same as the matrix product a b or b a. Then consider the number ii te j, which can be evaluated in two ways using the fact that both e, and e j are eigenvectors of I.) (b) Use the result of part (a) to show that if the three principal moments are all different, then the directions of three principal axes are uniquely determined. (c) Prove that if two of the principal moments are equal, A l = A2 say, then any direction in the plane of e l and e2 is also a principal axis with the", " same principal moment. In other words, when A l = A2 the corresponding principal axes are not uniquely determined. (d) Prove that if all three principal moments are equal, then any axis is a principal axis with the same principal moment. SECTION 10.6 Precession of a Top Due to a Weak Torque 10.39 * Consider a top consisting of a uniform cone spinning freely about its tip at 1800 rpm. If its height is 10 cm and its base radius 2.5 cm, at what angular velocity will it precess? SECTION 10.7 Euler's Equations 10.40 ** (a) A rigid body is rotating freely, subject to zero torque. Use Euler's equations (10.88) to prove that the magnitude of the angular momentum L is constant. (Multiply the i th equation by L i = X i coi and add the three equations.) (b) In much the same way, show that the kinetic energy of rotation Trot = z (x,(0,2 +x2(02 + ),,cop, as in (10.68), is constant. 10.41 ** Consider a lamina rotating freely (no torques) about a point 0 of the lamina. Use Euler's equations to show that the component of co in the plane of the lamina has constant magnitude. [Hint: Use the results of Problems 10.23 and 10.30. According to Problem 10.30, if you choose the direction e3 normal to the plane of the lamina, e 3 points along a principal axis. Then what you have to prove is that the time derivative of col co 22 is zero.] SECTION 10.8 Euler's Equations with Zero Torque 10.42 * I take a book that is 30 cm x 20 cm x 3 cm and is held shut by a rubber band, and I throw it into the air spinning about an axis that is close to the book's shortest symmetry axis at 180 rpm. What 414 Chapter 10 Rotational Motion of Rigid Bodies is the angular frequency of the small oscillations of its axis of rotation? What if I spin it about an axis close to the longest symmetry axis? 10.43 ** I throw a thin uniform circular disc (think of a frisbee) into the air so that it spins with angular velocity co about an axis which makes an angle a with the axis of the disc. (a) Show that the magnitude of co is constant. [Look at Equation (10.94).] (b) Show that as seen by me, the disc's axis precesses around the fixed direction of the angular momentum with angular velocity Q s = co-I4 \u2014 3 sine a. (The results of Problems 10.23 and 10.46 will be useful.) 10.44 ** An axially symmetric space station (principal axis e 3, and X i = X2) is floating in free space. It has rockets mounted symmetrically on either side that are firing and exert a constant torque r about the symmetry axis. Solve Euler's equations exactly for co (relative to the body axis) and describe the motion. At t = 0 take co = (cow, 0, 1\u00b030)- 10.45 ** Because of the earth's equatorial bulge, its moment about the polar axis is slightly greater that the other two moments, X3 = 1.00327A 1 (but X i = X 2). (a) Show that the free precession described in Section 10.8 should have period 305 days. (As described in the text, the period of this \"Chandler wobble\" is actually more like 400 days.) (b) The angle between the polar axis and co is about 0.2 arc seconds. Use Equation (10.118) from Problem 10.46 to show that as seen from the space frame the period of this wobble should be about a day. 10.46 *** We saw in Section 10.8 that in the free precession of an axially symmetric body the three vectors e 3 (the body axis), co, and L lie in a plane. As seen in the body frame, e 3 is fixed, and co and L precess around e 3 with angular velocity S2b = (03(X1 \u2014 X3)/X1 As seen in the space frame L is fixed and co and e3 precess around L with angular velocity In In this problem you will find three equivalent expressions for Q s. (a) Argue that St s = Sib + co. [Remember that relative angular velocities add like vectors.] (b) Bearing in mind that St b is parallel to e3 prove that Q s = co sin a/ sin 0 where a is the angle between e 3 and co and 6 is that between e 3 and L (see Figure 10.9). (c) Thence prove that = s sin a sin 0 = = L", ".1 1 2 + (x 2 ___ x2) sine 3 / 1 3 X (10.118) 10.47 *** Imagine that this world is perfectly rigid, uniform, and spherical and is spinning about its usual axis at its usual rate. A huge mountain of mass 10 -8 earth masses is now added at colatitude 60\u00b0, causing the earth to begin the free precession described in Section 10.8. How long will it take the North Pole (defined as the northern end of the diameter along co) to move 100 miles from its current position? [Take the earth's radius to be 4000 miles.] SECTION 10.9 Euler Angles 10.48 ** Equation (10.97) gives the angular velocity of a body in terms of an unholy mixture of unit vectors. (a) Find co in terms of X, ST', and i. (b) Do the same in terms of e l, e2, and e3. 10.49 ** Starting from Equation (10.100) for L, verify that L, is correctly given by Equations (10.102) and (10.103). 10.50 ** Equation (10.105) gives the kinetic energy in terms of Euler angles for a body with X 1 = X2 Find the corresponding expression for a body whose three principal moments are all different. Problems for Chapter 10 415 SECTION 10.10 Motion of a Spinning Top 10.51 * Verify that the energy of a symmetric top can be written as E 16 2 + U eff (0), where the effective potential energy is as given in (10.114). 10.52 ** Consider the rapid steady precession of a symmetric top predicted in connection with (10.112). (a) Show that in this motion the angular momentum L must be very close to the verti- cal. [Hint: Use (10.100) to write down the horizontal component Lhor of L. Show that if is given by the right side of (10.112), L hor is exactly zero.] (b) Use this result to show that the rate of precession Q given in (10.112) agrees with the free precession rate Q s found in (10.06). 10.53 ** In the discussion of steady precession of a top in Section 10.10, the rates Q at which steady precession can occur were determined by the quadratic equation (10.110). In particular, we examined this equation for the case that w 3 is very large. In this case you can write the equation as a0 2 bQ c = 0 where b is very large. (a) Verify that when b is very large, the two solutions of this equation are approximately \u2014c/b (which is small) and \u2014b/a (which is large). What precisely does the condition \"b is very large\" entail? (b) Verify that these give the two solutions claimed in (10.111) and (10.112). 10.54 *** [Computer] The nutation of a top is controlled by the effective potential energy (10.114). Make a plot of Ueff (0) as follows: (a) First, since the second term of Ueff (9) is a constant, you can ignore it. Next, by choice of your units, you can take MgR = 1 = A 1. The remaining parameters L, and L3 are genuinely independent parameters. To be definite set L z = 10 and L3 = 8 and plot Ueff (0) as a function of O. (b) Explain clearly how you could use your graph to determine the angle 0,\u201e at which the top could precess steadily with 0 = constant. Find 0,, to three significant figures. (c) Find the rate of this steady precession, Q = cb, as given by (10.115). Compare with the approximate value of Q given by (10.112). 10.55 *** The analysis of the free precession of a symmtric body in Section 10.8 was based on Euler's equations. Obtain the same results using Euler's angles as follows: Since L is constant you may as well choose the space axis i so that L = Li (a) Use Equation (10.98) for z to write L in terms of the unit vectors e' 1, e'2, and e3. (b) By comparing this expression with (10.100), obtain three equations for 0, 4, and. (c) Hence show that 0 and 4 are constant, and that the rate of precession of the body axis about the space axis i is Q s = L 0. 1 as in (10.96). (d) Using (10.99) show that the angle between (I) and e 3 is constant and that the three vectors L, co, and e 3 are always coplanar. 10", ".56 *** An important special case of the motion of a symmetric top occurs when it spins about a vertical axis. Analyze this motion as follows: (a) By inspecting the effective PE (10.114), show that if at any time 0 = 0, then L3 and L, must be equal. (b) Set L z = L3 = A30)3 and then make a Taylor expansion of (Jeff (0) about 0 = 0 to terms of order 0 2. (c) Show that if co 3 > corm\u201e = 21/Mg RX i/A 32, then the position 0 = 0 is stable, but if w3 < comir, it is unstable. (In practice, friction slows the top's spinning. Thus with w3 sufficiently fast, the vertical top is stable, but as it slows down the top will eventually lurch away from the vertical when co 3 reaches co mity ) 10.57 *** (a) Find the Lagrangian for a symmetric top whose tip is free to slide on a frictionless, horizontal table. For generalized coordinates use the Euler angles (0,0,0 plus X and Y, where (X, Y, Z) is the CM position relative to a fixed point on the table. (Note that the vertical position Z is not an independent coordinate, since Z = R cos 0.) (b) Show that the CM motion of (X, Y) separates completely from the rotational motion. (c) Consider the two possible rates of steady precession (10.111) and (10.112) (for given 0 and w3). How do these differ in the present case from their corresponding values when the tip is held at a fixed pivot? CHAPTER Coupled Oscillators and Normal Modes In Chapter 5 we discussed the oscillations of a single body, such as a mass on the end of a fixed spring. I now want to take up the oscillations of several bodies, such as the atoms that make up a molecule like CO 2, which we can imagine as a system of masses connected to one another by springs. If each mass were attached to a separate fixed spring, with no connections between the masses, then each would oscillate independently, as described in Chapter 5, and there would be nothing more to say. Thus our interest here is a system of masses that can oscillate and are connected to one another in some way \u2014 a system of coupled oscillators. A single oscillator has a single natural frequency, at which (in the absence of damping or driving forces) it will oscillate for ever. We shall find that two or more coupled oscillators have several natural (or \"normal\") frequencies and that the general motion is a combination of vibrations at all the different natural frequencies. Like the theory of rotating bodies in Chapter 10, the theory of coupled oscillators makes essential use of matrices, and many of the ideas you learned in Chapter 10 play an important role here. The most obvious applications of the ideas of this chapter are to the study of molecules, but there are many others, including acoustics, the vibrations of structures like bridges and buildings, and coupled electrical circuits. Throughout this chapter I shall assume that all of the forces with which we are concerned obey Hooke's law and hence that the equations of motion are all linear. While this is a special case, it is a very important special case, with many applications in mechanics and throughout physics. Nevertheless, you should bear in mind that the systems discussed here are a special case; we shall see in Chapter 12 how startlingly more complicated the motion of nonlinear oscillators can be. 11.1 Two Masses and Three Springs As a simple first example of coupled oscillators, consider the two carts shown in Figure 11.1. The carts move without friction on a horizontal track, between two fixed 417 418 Chapter 11 Coupled Oscillators and Normal Modes k1 --15-6-15Th 11 X 1 x2 equilibrium equilibrium Figure 11.1 Two carts attached to fixed walls by the springs labeled k 1 and k3, and to each other by k 2. The carts' positions x 1 and x2 are measured from their respective equilibrium positions. walls. Each is attached to its adjacent wall by a spring (force constants k 1 and k3), and the carts are attached to each other by a spring with force constant k 2. In the absence of the spring 2, the two carts would oscillate independently of each other. Thus it is spring 2 that \"couples\" the two oscillators. In fact, spring 2 makes it impossible that either cart move without the other moving as well: For example, if cart 1 is stationary and cart 2 moves, the length of spring 2 will change, which will produce a changing force on cart 1, causing it to move as well. 1 It is easy to find the equations of motion for the two carts, using either Newton's second law or Lagrange's equations. In general,", " Lagrange's equations are easier to write down, but in the present simple case, Newton's law may be a little more instructive. Suppose the two carts have moved distances x 1 and x2 (measured to the right) from their equilibrium positions. Spring 1 is now stretched by an amount x 1 and so exerts a force k 1x 1 to the left on cart 1. Spring 2 is more complicated since it is affected by the positions of both carts, but you can easily convince yourself that it is stretched by the amount x2 \u2014 x 1 and exerts a force k 2 (x2 \u2014 x 1 ) to the right on cart 1. Thus the net force on cart 1 is (net force on cart 1) = + k2 (x2 \u2014 x 1 ) = \u2014(k 1 + k2)x 1 + k2x2 (11.1) where the second line is just to show more clearly the dependence on the two variables x 1 and x2. You can find the net force on cart 2 in the same way, and the two equations of motion are m 1 1 1 = \u2014(k 1 + k2)x 1 + k2x2 1/22.2 = k2x1 - (k2 k3)x2. (11.2) Before we try to solve these two coupled equations, notice that they can be written in the beautifully compact matrix form 1 In the following discussion, it is simplest to assume that when the two carts are at their equilibrium positions the three springs are neither stretched nor compressed. (Their lengths are equal to their natural, unstretched lengths.) However, depending on the distance between the two walls, it could be that all three springs are compressed or all three are stretched. Fortunately, as you can easily check (Problem 11.1), none of the results of the next three sections is affected by these possibilities. Here I have introduced the (2 x 1) column matrix (or \"column vector\") Section 11.1 Two Masses and Three Springs 419 (11.4) which labels the configuration of our system. (It has 2 elements because the system has 2 degrees of freedom; for a system with n degrees of freedom it would have n elements.) I have also defined two square matrices, n0121 and K + kr, \u2014k2 \u2014k2 k2 k3 (11.5) The \"mass matrix\" M is (in this simple case, at least) a diagonal matrix, with the masses m 1 and m 2 down the diagonal. The \"spring-constant matrix\" K has nonzero off-diagonal elements, reflecting that the right sides of the two equations (11.2) couple x 1 and x2. Notice that the matrix equation (11.3) is a very natural generalization of the equation of motion of a single cart on a single spring: With just one degree of freedom, all three matrices x, M, and K are just (1 x 1) matrices, that is, ordinary numbers. The configuration x is the cart's position x, the mass matrix M is the cart's mass m, and K is the spring constant k. And the equation of motion (11.3) is just the familiar mx = \u2014kx. Notice also that both matrices M and K are symmetric, as will be true of all the corresponding matrices in this chapter. Although the symmetry of M and K does not play a very obvious role in the discussions here, it is in fact a key property of the underlying mathematics, as we shall see in the appendix. In trying to solve the equation of motion (11.3) we might reasonably guess that there could be solutions in which both carts oscillate sinusoidally with the same angular frequency w ; that is, x 1 (t) = oe i cos(wt \u2014 6 1 ) x 2 (t) = a2 cos(wt \u2014 8 2). (11.6) (11.7) In any event, there is nothing to stop us trying to find solutions of this form. (And we shall, in fact, succeed!) If there is a solution of this form, then there will certainly also be a solution of the same form but with the cosines replaced by sines: y1(t) = a i sin(cot \u2014 8 1 ) Y2(t) = a2sin(wt \u2014 82) and there is nothing to stop me combining these two solutions into a single complex z i (t) = x i (t) + iyi(t) = alei(cut-so = = ai ejw t (11.8) where al = ai e -i8 1, and, likewise, z2(t) = x2(t) + iy2(t) = a2ei(cot-82) = ate-is2eiwt = a2eiwt. (11", ".9) and and solution 420 Chapter 11 Coupled Oscillators and Normal Modes This trick of introducing a \"fictitious\" complex solution to the equation of motion is the same trick introduced in Section 5.5. I am not, of course, claiming that these complex numbers represent the actual motion of the two carts. The actual motion is given by the two real numbers (11.6) and (11.7). Nevertheless, for the right choices of a l, a2, and co, the two complex numbers (11.8) and (11.9) are (as we shall see) solutions of the equation of motion, and their real parts describe the actual motion of our system. The great advantage of the complex numbers is that, as you can see from the right sides of (11.8) and (11.9), they both have the same time dependence, given by the common factor ei '. This lets us combine the two complex solutions into a single (2 x 1) matrix solution of the form z(t) = [z i (t)] [al z 2 (t) a2 e = ae (11.10) where the column a is a constant, made up of two complex numbers, In seeking solutions of the equation of motion (11.3), we shall accordingly try for solutions z(t) of the complex form (11.10), bearing in mind that when we find such solutions the actual motion x(t) is equal to the real part of z(t), [ a2 [ cy le -i3 1 ate -i82 J a=. x(t) = Re z(t). When we substitute the form (11.10) into Equation (11.3), Mi = \u2014Kx, we obtain the equation --co2Ma e i wt = \u2014Ka eiwt or, cancelling the common exponential factor and rearranging, This equation is a generalization of the eigenvalue equation studied in Section 10.5. (In the usual eigenvalue equation, what we are calling co t is the eigenvalue, and where we have the matrix M the ordinary eigenvalue equation has the unit matrix 1.) It can be solved in almost exactly the same way. If the matrix (K \u2014 co 2M) has nonzero determinant, then the only solution of (11.11) is the trivial solution a = 0, corresponding to no motion at all. On the other hand, if Section 11.2 Identical Springs and Equal Masses 421 then there certainly is a nontrivial solution of (11.11) and hence a solution of the equations of motion with our assumed sinusoidal form (11.10). In the present case, the matrices K and M are (2 x 2) matrices, so the equation (11.12) is a quadratic equation for cw t and has (in general) two solutions for w2. This implies that there are two frequencies co at which the carts can oscillate in pure sinusoidal motion as in (11.10) [or, rather, (11.6) and (11.7) for the actual real motion]. 2 The two frequencies at which our system can oscillate sinusoidally (the so-called normal frequencies) are determined by the quadratic equation (11.12) for co t. The details of this equation depend on the values of the three spring constants and the two masses. While the general case is perfectly straightforward, it is not especially illuminating, and I shall discuss instead two special cases where one can understand more easily what is going on. I shall start with the case that the three springs are identical, and likewise the two masses. 11.2 Identical Springs and Equal Masses Let us continue to examine the two carts of Figure 11.1, but suppose now that the two masses are equal, m 1 = m 2 = m, and similarly the three spring constants, k i = k2 = k3 = k. In this case, the matrices M and K defined in (11.5) reduce to mo and K = 2k \u2014k 1 \u2014k 2k j. [ The matrix (K \u2014 co2M) of the generalized 3 eigenvalue equation (11.11) becomes (K 602m) = I 2k \u2014 mco 2 \u2014k L \u2014k 2k -- mw 2 (11.14) and its determinant is (K (02m) (2k in 602 ) 2 k2 = (k inc02)(3k m(02 ). det The two normal freqencies are determined by the condition that this determinant be zero and are therefore (0 = = 601 and co = These two normal frequencies are the frequencies at which our two carts can oscillate in purely sinusoidal motion. Notice that the first one, co l, is precisely the frequency of a single mass", " m on a single spring k. We shall see the reason for this apparent coincidence in a moment. 2 Since there are two solutions for w 2, you might think this would give four solutions for w = +,./o.)2. However, a glance at Equations (11.6) and (11.7) will convince you that +(t) and --co constitute the same frequency for the real motion. 3 From now on, I shall refer to (11.11) as the eigenvalue equation, omitting the \"generalized.\" 422 Chapter 11 Coupled Oscillators and Normal Modes Equation (11.15) tells us the two possible frequencies of our system, but we have not yet described the corresponding motions. Recall that the actual motion is given by the column of real numbers x(t) = Re z(t) where the complex column z(t) = ae i wr and a is made up of two fixed numbers, a =[ a i, a2 which must satisfy the eigenvalue equation (K \u2014 co2M)a = O. (11.16) Now that we know the possible normal frequencies, we must solve this equation for the vector a for each normal frequency in turn. The sinusoidal motion with any one of the normal frequencies is called a normal mode, and I shall start with the first normal mode. The First Normal Mode (K \u2014 co2M) of (11.14) becomes If we choose co equal to the first normal frequency, co l =,/k/m, then the matrix (Notice that this matrix has determinant 0, as it should.) Therefore, for this case, the eigenvalue equation (11.16) reads (K 4M) = [ \u2014kk k k [ 1 L -1 1 1 r a l L a2 0 al \u2014 a2 = 0 \u2014a 1 + a2 = 0. which is equivalent to the two equations Notice that these two equations are actually the same equation, and either one implies that a l = a2 = Ae -i8, say. The complex column z(t) is therefore and the corresponding actual motion is given by the real column x(t) = Re z(t) or z(t) = [ al l e iwit a2 L A A e i(wit-8) x(t) = [ x 1 (t) x2(t) r A] L A cos(wit \u2014 6). That is, x i (t) = A cos(co it \u2014 8) x2 (t) = A cos(co it \u2014 8) [first normal mode]. (11.18) Section 11.2 Identical Springs and Equal Masses 423 Figure 11.2 The first normal mode for two equal-mass carts with three identical springs. The two carts oscillate back and forth with equal amplitudes and exactly in phase, so that x i (t) = x2 (t), and the middle spring remains at its equilibrium length all the time. x1 Figure 11.3 In the first mode, the two positions oscillate sinusoidally, with equal ampitudes and in phase. We see that in the first normal mode the two carts oscillate in phase and with the same amplitude A, as shown in Figure 11.2. A striking feature of Figure 11.2 is that, because x i (t) = x 2 (t), the middle spring is neither stretched nor compressed during the oscillations. This means that, for the first normal mode, the middle spring is actually irrelevant, and each cart oscillates just as if it were attached to a single spring. This explains why the first normal frequency co l =,/kIrn is the same as for a single cart on a single spring. Another way to illustrate the motion in the first normal mode is just to plot the two positions x 1 and x2 as functions of t. This is shown in Figure 11.3. The Second Normal Mode The second normal frequency at which our system can oscillate sinusoidally is given by (11.15) as w2 =,/3k/m, which, when substituted into (11.14), gives Thus, for this normal mode, the eigenvalue equation (K \u2014 (022:1VI)a = 0 implies that F \u2014 k k 41V1) \u2014k \u20141( [11 1 1 [ ai ] =0 1 a 2 424 Chapter 11 Coupled Oscillators and Normal Modes Second mode x2= -X Figure 11.4 The second normal mode for two equal-mass carts with three identical springs. The two carts oscillate back and forth with equal amplitudes but exactly out of phase, so that x 2 (t ) = \u2014x 1 (t) at all times. X1 Figure 11.5 In the second mode, the two positions oscillate sinu- soidally, with equal ampitudes but", " exactly out of phase. which implies that a l + a2 = 0, or a l = \u2014a2 = Ae' s, say. The complex column z(t) is therefore and the corresponding actual motion is given by the real column x(t) = Re z(t) or z(t) = [ a2 eiwg = r i ]e (w2t--3) A \u2014 A x(t) [xi(t) Lx2(t) l r A L \u2014 A cos(co2t \u2014 8). That is, Figure 11.5. x 1 (t) = A cos(co2 t \u2014 3) x 2 (t) = \u2014A cos(co2 t \u2014 8) [second normal mode]. (11.20) We see that in the second normal mode the two carts oscillate with the same amplitude A but exactly out of phase, as shown in the picture of Figure 11.4 and the graphs of Notice that in the second normal mode, when cart 1 is displaced to the right, cart 2 is displaced an equal distance to the left, and vice versa. This means that when the outer two springs are stretched (as in Figure 11.4), the middle spring is compressed by twice as much. Thus, for example, when the left spring is pulling cart 1 to the left, the middle spring is pushing cart 1, also to the left, with a force that is twice as large. This means that each cart moves as if it were attached to a single spring with force constant 3k. In particular, the second normal frequency is co t =,/3k/m. Section 11.2 Identical Springs and Equal Masses 425 The General Solution We have now found two normal-mode solutions, which we can rewrite as x(t) = A l [ 1 cos(co it \u2014 6 1 ) and x(t) = A2 cos(W2t \u2014 62) 1 1 [ 1 \u2014 1 1 where co t and cot are the normal frequencies (11.15). Both of these solutions satisfy the equation of motion Mk = \u2014Kx for any values of the four real constants A 1, 61, A2, and 62. Because the equation of motion is linear and homogeneous, the sum of these two solutions is also a solution: X(t) = A1 [ cos(w1t \u2014 81) + A2 [ cos(w2t \u2014 82). (11.21) 11 1 1 1 Because the equation of motion is really two second-order differential equations for the two variables x 1(t) and x2 (t), its general solution has four constants of integration. Therefore the solution (11.21), with its four arbitrary constants, is in fact the general solution. Any solution can be written in the form (11.21), with the constants A 1, A2, 61, and 62 determined by the initial conditions. The general solution (11.21) is hard to visualize and describe. The motion of each cart is a mixture of the two frequencies, w 1 and w2. Since w2 = Ni -w1 the motion never repeats itself, except in the special case that one of the constants A 1 or A2 is zero (which gives us back one of the normal modes). Figure 11.6 shows graphs of the two positions in a typical nonnormal mode (with A l = 1, A2 = 0.7, 6 1 = 0, and 32 = 7r/2). About the only simple thing one can say about these graphs is that they certainly are not very simple! \u00a4 X1 Figure 11.6 In the general solution, both x 1 (t) and x2 (t) oscillate with both of the normal frequencies, producing a quite complicated non- periodic motion. Normal Coordinates We have seen that in any possible motion of our two-cart system, both of the co- ordinates x 1 (t) and x2 (t) vary with time. In the normal modes, their time depen- dence is simple (sinusoidal), but it is still true that both vary, reflecting that the two carts are coupled and that one cart cannot move without the other. It is possible to introduce alternative, so-called normal coordinates which, although less physi- cally transparent, have the convenient property that each can vary independently of 426 Chapter 11 Coupled Oscillators and Normal Modes the other. This statement is true for any system of coupled oscillators, but is espe- cially easy to see in the present case of two equal masses joined by three identical In place of the coordinates x 1 and x2, we can characterize the positions of the two carts by the two normal coordinates springs. and = (x1 + x2) = (x1 - x2). (11.22) (11.23) The physical significance of the original variables x 1 and x", "2 (as the positions of the two carts) is obviously more transparent, but t and b2 serve just as well to label the configuration of the system. Moreover, if you refer back to (11.18) for the first normal mode, you will see that in the first mode the new variables are given by i (t) = A cos(w it \u2014 S) 2(t) = [first normal mode], (11.24) whereas in the second mode, we see from (11.20) that i(t) = 0 2 (t) = A cos(co 2t \u2014 6)1 [second normal mode]. (11.25) In the first normal mode the new variable 1 oscillates, but 2 remains zero. In the second mode it is the other way round. In this sense, the new coordinates are independent \u2014 either can oscillate without the other. The general motion of our system is a superposition of both modes, and in this case both 1 and 2 oscillate, but t oscillates at the frequency co l only, and b2 at the frequency co t only. In some more complicated problems, these new normal coordinates represent a considerable simplification. (See Problems 11.9, 11.10, and 11.11 for some examples and Section 11.7 for further discussion.) 11.3 Two Weakly Coupled Oscillators In the last section we discussed the oscillations of two equal masses joined by three equal springs. For this system, the two normal modes were easy to understand and to visualize, but the nonnormal oscillations were much less so. A system where some of the nonnormal oscillations are readily visualized is a pair of oscillators which have the same natural frequency and which are weakly coupled. As an example of such a system, consider the two identical carts shown in Figure 11.7, which are attached to their adjacent walls by identical springs (force constants k) and to each other by a much weaker spring (force constant k 2 << k). We can quickly solve for the normal modes of this system. The mass matrix M is the same as before. The spring matrix K and the crucial combination (K \u2014 w2M) Section 11.3 Two Weakly Coupled Oscillators 427 Figure 117 Two weakly coupled carts. The middle spring which couples the two carts is much weaker than the outer two springs. that determines the eigenvalue problem are easily written down [starting with (11.5) for K]: k k 2 \u2014k 2 2 k k 2 \u2014k K = [ and (K \u2014 co2M) = [k k2 \u2014 mw2 \u2014k2 \u2014k 2 k + k2 \u2014 mco 2. (11.26) The determinant of (K \u2014 w 2M) is (k \u2014 mco 2)(k + 2k2 \u2014 mw2), and we conclude that the two normal frequencies are col = and 102 = k m k + 2k2 m (11.27) The first frequency is exactly the same as in the previous example, and we can easily see why. The motion in this first mode is, as you can check, the same motion as shown in Figure 11.2 for the first mode of the equal-spring case. The important point is that in this mode the two carts move together in such a way that the middle spring is undisturbed and hence irrelevant. Naturally we get the same frequency for this mode whatever the strength of the middle spring. In the second mode also, the motion is the same as for the corresponding mode of the equal-spring example \u2014 namely, the two carts oscillating exactly out of phase, both moving inward or both moving outward at any one time, as in Figure 11.4. But in this mode, the strength of the middle spring is, of course, relevant, and the second normal frequency w 2 as given by (11.27) depends on k 2. In the present case, w 2 is very close to co l, since k2 << k. To take advantage of this closeness, it is convenient to define wo to be the average of the two normal frequencies _ col + W2 2 wo Since co l and w2 are very close to each other, w o is very close to either, and for most purposes we can think of co o as essentially the same as co l =,/k/m. To show the small difference between co l and w2, I shall write WI = Wo 6 W2 = Wo E. and That is, the small number E is half the difference between the two normal frequencies. 428 Chapter 11 Coupled Oscillators and Normal Modes The two normal modes of the weakly coupled carts can now be written as z(t) C 1 [ 1 e 1 ] i(w \u2014ot and z(t) = C2 \u20141 el(w.+E)t. 1 1 1 1 Both of these satisfy the equation of motion", " for any values of the two complex numbers C 1 and C2. (It is convenient to continue to work with the \"fictitious\" complex solutions for a bit longer.) The sum of these two solutions is also a solution, z(t) = C 1 [ ]e1(wo \u2014 ')t + C2 [ _ 1 ei (w\u00b0+E)t (11.28) 1 and, since it contains four arbitrary real constants (the two complex constants C 1 and C2 are equivalent to four real constants), it is the general solution. The constants C 1 and C2 in (11.28) are determined by the initial conditions \u2014 the positions and velocities of the two carts at t = 0. To see some general features of the solution (11.28), it is helpful to factor it as z(t) = {C 1[1 ]e\u2014 lEt + C2 [ ei\"}ei'd (11.29) This expresses our solution as a product of two terms. The term in braces, { -}, is a (2 x 1) column matrix which depends on t. But because E is very small, this column varies very slowly compared to the second factor eiwot. Over any reasonably short time interval, the first factor is essentially constant and our solution behaves like z(t) = aelwot, with a constant. That is, over any short time interval, the two carts will oscillate sinusoidally with angular frequency wo. But if we wait long enough, the \"constant\" a will vary slowly, and the details of the two carts' motion will change. I shall illustrate this behavior in detail in a moment. Let us now examine the behavior of (11.29) for some simple values of the constants C1 and C2. First, if either C i or C2 is zero, the solution (11.29) reverts to one of the normal modes. (For instance, if C 1 = 0, the solution is the second normal mode.) A more interesting case is that C 1 and C2 are equal in magnitude, and, to simplify the discussion, I shall suppose C 1 and C2 are equal and real, C1 -= C2 =-. A/2, say. (The 2 is just for future convenience.) In this case (11.29) becomes Z(t ) = \u2014 A [e\u2014 i\" + 2 e'\" \u2014 e i\" e = A [ cos Et \u2014i sin Et e lw.t (11.30) To find the actual motion of the two carts, we must take the real part of this matrix, x(t) = Re z(t), whose two elements are the two positions, x i (t) = A cos Et cos wot x2 (t) = A sin Et sin coot. The solution (11.31) has a simple and elegant interpretation. Notice first that at time zero, x 1 = A, whereas 1 1 = x2 = x2 = 0. That is, our solution describes the motion Section 11.3 Two Weakly Coupled Oscillators 429 when cart 1 is pulled a distance A to the right and released at t = 0, with cart 2 stationary at its equilibrium position. Because E is very small, there is an appreciable interval (namely, 0 < t \u00ab 1/E) during which the functions in (11.31) that involve Et remain essentially unchanged, that is, cos Et ti 1 and sin Et ti 0. During this initial interval, the two positions, as given by (11.31), are just x i (t) A cos coot I x2(t) ti 0 [t ti 0]. (11.32) Initially, cart 1 oscillates with amplitude A and frequency co o, while cart 2 remains stationary. This simple state of affairs cannot last for ever. As soon as cart 1 begins to move, it starts to flex the weak middle spring, which starts to push and pull on cart 2. Although the force exerted by the middle spring is weak, it eventually starts to make cart 2 oscillate. This can be seen in (11.31), where the factor sin ct eventually becomes appreciable, and cart 2 starts to oscillate, also at the frequency co o. Notice that as the factor sin Et in x 2 (t) grows toward 1, the factor cos Et in x (t) shrinks toward zero, as it has to do to keep the total energy of the two oscillating carts constant. Eventually, when t = n- /2E, the factor sin Et reaches 1 (and cos Et reaches zero), and there is an interval when4 xi(t) 0 x 2 (t) ti A sin coot [t ti Jr/2E]. (11.33) Now that cart 2 is oscillating at maximum amplitude and cart 1 not at all, cart 2 starts", " to drive cart 1. Cart 1 begins to oscillate with increasing amplitude, and the amplitude of cart 2's oscillations begins to diminish again. This process, in which the two carts pass the energy back and forth from one to the other, continues indefinitely (or until dissipative forces \u2014 which we are ignoring \u2014 have removed all the energy). It is illustrated in Figure 11.8, which shows x i (t) and x2 (t) as given by (11.31) as functions of t for a couple of cycles of passing the energy from cart 1 to 2 and back to 1 again. If you have studied the phenomenon of beats, you have probably noticed the simi- larity of either graph in Figure 11.8 to a graph of beats. Beats are the result of superpos- ing two waves \u2014 sound waves, for example \u2014 with nearly equal frequencies. Because of the small difference in frequencies, the two waves move regularly in and out of phase (at any one location). This means that the resulting interference of the waves is alternately constructive and destructive, and a graph of the resultant signal looks just like either one of the graphs in Figure 11.8. To understand what is beating in the case of our two carts, we need to consider again the two normal coordinates of Equations (11.22) and (11.23), = + x2) and 2 = z (x 1 \u2014 x2). For the present solution 4 Notice that we can interpret what has happened in terms of the discussion of Section 5.6 on resonance. The weak spring has been driving cart 2 at the resonant frequency co o, so cart 2 should have responded by oscillating 7r/2 behind the driver. This is exactly what we see from (11.32) and (11.33), since sin coot is indeed 7r/2 behind cos coot. 430 Chapter 11 Coupled Oscillators and Normal Modes xl x2 IN\\A-41),t Figure 11.8 The positions x 1 (t) and x2 (t) of two weakly coupled oscillating carts if cart 1 is released from rest at x 1 = A > 0 and cart 2 at x 2 = 0. (11.31) these are [remember the trig identity cos B cos 0 + sin e sin 0 = cos(9 \u2014 0)] = 4A cos(coo \u2014 E)t = -A cos w it 4A cos(coo + E)t = 4A cos w2 t. 2(t) = (11.34) That is, the two normal coordinates oscillate with equal amplitudes, the first at the frequency co l and the second at the nearby frequency co t. Since x 1 (t) = 1 (t) + we see that x 1 (t) is the superposition of 1 (t) and 2,(t), and the waxing and waning of x 1 (t) is the result of beats between these two signals of nearly equal frequencies. The same applies to x 2 (t), except that, because x2 (t) = 1 (t) \u2014 2 (t), the moments of constructive interference for x 1 (t) are moments of destructive interference for x 2 (t) and vice versa, as is clearly seen in Figure 11.8. 11.4 Lagrangian Approach: The Double Pendulum The analysis of the two oscillating carts in the last three sections was based on Newton's second law. We could equally have derived the equations of motion using the Lagrangian formalism, although there is no particular advantage to doing so. However, we shall find that as we study systems of increasing complexity, the advantages of the Lagrangian approach rapidly become overwhelming. I shall start this section by rederiving the equations for the familiar two carts from their Lagrangian. I shall then do the same for another simple system with two degrees of freedom, the double pendulum. These two examples will pave the way for the general discussion in the next section. Lagrangian Approach for Two Carts on Three Springs Let us consider once more the two carts of Figure 11.1. We could write down the equations of motion (11.2) from Newton's second law as soon as we had identified the forces on each of the carts. To do the same thing with Lagrange's equations, we Section 11.4 Lagrangian Approach: The Double Pendulum have first to write down the kinetic and potential energies, T and U, and then the Lagrangian = T \u2014 U. The kinetic energy is just T _ \u2014 2 in.p. \u201e- 2 + 1 2 M2X 2. (11.35) To write down the potential energy, we must identify the extensions of the three springs as x i, x2 \u2014 x i, and \u2014x2, from which it immmediately follows that the potential energy is U = 11(4 + 2k2 (x \u2014 x", "2) 2 + 2k3x22 = + \u2014 k2x1x2 + (k2 k3)X 22 \u201e 2 b. (11.36) These results immediately give us the Lagrangian = T \u2014 U and thence the two Lagrange equations of motion: and d d dt ax l ax, or m lxl = \u2014 (ki + k2)x 1 + k2x2 = a,c a x2 2 or dt a x mnx2 = k2x 1 \u2014 (k2 k3)x2. These are precisely the two equations of motion (11.2), which we rewrote in the compact matrix form as MX = \u2014Kx. This alternative derivation of the same equations has no particular advantage for this simple system. Here is a second system, which is still very simple, but for which the Lagrangian approach is already distinctly more straightforward than the Newtonian. The Double Pendulum Consider a double pendulum, comprising a mass m 1 suspended by a massless rod of length L 1 from a fixed pivot, and a second mass m 2 suspended by a massless rod of length L2 from m i, as shown in Figure 11.9. It is a straightforward matter to write down the Lagrangian C as a function of the two generalized coordinates 0 1 and 02 shown. When the angle 0 1 increases from 0, the mass m 1 rises by an amount L 1 (1 \u2014 cos 0 1 ) and gains a potential energy = m ig L 1 (1 \u2014 cos 0i). U1 Similarly, as 02 increases from 0, the second mass rises by L 2 (1 \u2014 cos 0.2) but, in addition, its point of support (m 1) has risen by L 1 (1 \u2014 cos 01). Thus U2 = 2g [L 1 (1 \u2014 cos 01 ) + L 2 (1 \u2014 cos 02)]. The total potential energy is therefore U(01, 02) = (m m 2)g L 1 (1 \u2014 cos 0 1) m 2g L 2 (1 \u2014 cos 02). (11.37) 432 Chapter 11 Coupled Oscillators and Normal Modes Figure 11.9 A double pendulum. The velocity of m 2 is the vector sum of the two velocities shown, separated by an angle (p 2 \u2014 0 1 The velocity of m t is just L i(*pl in the tangential direction, as shown in Figure 11.9, so its kinetic energy is T1= _ IniL,12(i)12, The velocity of m 2 is the vector sum of two velocities, as I have indicated in Figure 11.9 \u2014 the velocity L24)2 of m 2 relative to its support m 1 plus the velocity L is(1)1 of its support. The angle between these two velocities is (0 2 \u2014 0 1), so the kinetic energy of m2 is T2 = I m2 [L i2 12 2L1L24)14)2 COS(4)1 \u2014 4)2) + L2) 22 ] and the total kinetic energy is T 21 (m1+ m2)L- 12. 12 in2 L 1 L 2q4 2cos(01 \u2014 02) + 4m2L22 (p22. (11.38) From (11.38) and (11.37) we can write down the Lagrangian = T \u2014 U and then the two Lagrange equations for 01 and 02. However, the resulting equations are too complicated to be especially illuminating and can certainly not be solved analytically. This situation is reminiscent of the simple pendulum, whose equation of motion = \u2014 g sin 4) is also unsolvable analytically, forcing us to solve it numerically or to make a suitable approximation. In that case, and in the present case, the simplest useful approximation is the small-angle approximation, which reduces the simple pendulum's equation to the solvable Lb = \u2014 g4. We are going to see that for almost all coupled oscillating systems, the exact equations are not analytically solvable, but that if we confine attention to small oscillations (certainly an important special case), the equations reduce to a standard form that is solvable. Returning to the equations of the double pendulum, let us assume that both angles 0 1 and 02 and the corresponding velocities,;. 1 and.7.52 remain small at all times. This lets us simplify the expressions for T and U by Taylor expanding them and dropping all terms that are of third power or higher in the four small quantities. In (11.38) for the kinetic energy, the only term that needs attention is the middle one. Since the Section 11.4 Lagrangian Approach: The Double Pendulum 433 factor cos(O i \u2014 40 is aleady multiplied by the doubly small product q.", "54,2, we can approximate the cosine by 1, to give T = +,\u201e, 2,,I., 2 2 \" 1 -I- in2)1-'1 M2L1L21:42 12m2 L 22,3,22. (11.39) In (11.37) for the potential energy, we must handle the cosines more carefully (since they are not already multiplied by any small quantities). The Taylor series for cos (/) gives the approximation cos cb ti 1 \u2014 0 2 /2, which reduces (11.37) to U = 2 (n1\u00b1m2)gL0 12 + 4fft 2g L 2 022. (11.40) Before we use these simplified expressions for T and U to give us the equations of motion, let us pause to examine what our assumption of small oscillations has achieved. The exact expression (11.38) for T was a transcendental function of the coordinates 01, 02 and velocities (b i, 42 ; the small-angle approximation reduced this to a homogeneous quadratic function s of the two velocities only. The exact expression (11.37) for U was a transcendental function of 0 1 and 02; the small- angle approximation reduced this to a homogeneous quadratic function of (p i and 02. We shall see that the same simplifications occur for a wide class of oscillating systems: The assumption that all oscillations are small reduces T to a homogeneous quadratic function of the velocities and U to a homogeneous quadratic function of the coordinates.6 The simplifying feature of these homogeneous quadratic forms for T and U is that when we differentiate them to get Lagrange's equations they reduce to homogeneous linear functions, yielding equations of motion that can always be easily We can now substitute the approximate expressions (11.39) and (11.40) for T and U into the Lagrangian = T \u2014 U and write down the two Lagrange equations of or (m 1 + in2)/41 + m2L1L24'2 = \u2014 (m1 + m2)gL1C (11.41) solved. motion for O f and 02: d az az dt a(b, and d ac dt az 802 or m m 2 L 22, = \u2014 m2gL2q52. m2gL202. (11.42) These two equations for cp i and 02 can be rewritten as a single matrix equation = \u2014KO (11.43) 5 A homogeneous quadratic function contains only second powers of its arguments \u2014 no first powers or constant terms, and no powers higher than two. 6 It is an almost unique feature of systems of masses connected by Hooke's-law springs that the exact expressions for T and U [as in (11.35) and (11.36)] already are in these simple forms, without our having to make any approximations. 434 Chapter 11 Coupled Oscillators and Normal Modes if we introduce the (2 x 1) column of coordinates and the two (2 x 2) matrices M = [ \\ 7- M2P-1 In2L 1L2 m2L1L2 111 2 L 22 and [ (m i m 2)gL i 0 0 m 2gL 2 ] (11.44) The matrix equation (11.43) is exactly analogous to (11.3) for the two carts on springs. In the present case, the \"mass\" matrix M is not actually made up of masses, but it still plays the role of inertia in the equation of motion (11.43). (That is, it multiplies the second derivatives of the coordinates.) Similarly the \"spring-constant\" matrix K is not actually made up of spring constants, but it plays the analogous role in the equation of motion. The procedure for solving the equations of motion (11.43) is exactly the same as for the two carts of Section 11.1. We first try to find solutions \u2014 normal modes \u2014 in which the two coordinates O l and 02 vary sinusoidally with the same angular frequency w. Exactly as before, any such solution 0 (t) can be written as the real part of a complex solution z(t) whose time dependence is just e`\"; that is, 0(t) = Re z(t) where z(t) = = e _twt a2 and the two components al, a2 of a are constants. Exactly as before, a function of this form satisfies the equation of motion (11.43) if and only if the frequency co and the column a satisfy the eigenvalue equation (K \u2014 w 2M)a = 0. This equation for a has a solution if and only if det(K \u2014 w 2M) = 0, a quadratic equation for", " w 2, which determines the two normal frequencies of the double pendulum. Knowing these two frequencies, we can go back and find the corresponding columns a, and we then know the two normal modes. Finally, the general motion of the system is just an arbitrary superposition of these two normal modes. Equal Lengths and Masses To simplify the discussion, let us now restrict our attention to the case that our double pendulum has equal masses, m 1 = m 2 = m, and equal lengths, L 1 = L2 = L, say. The equations tidy up appreciably if we recognize that,/gIL is the frequency of a single pendulum of the same length L. If we call this frequency then we can replace g everywhere by Lco,, and the two matrices M and K of (11.44) become (as you can check) m = in L2 1-21 I 1 and K= mL2 [ (11.45) 2,02 \u00b0 0 0 02]. \u201e Section 11.4 Lagrangian Approach: The Double Pendulum 435 The matrix (K \u2014 w 2M) of the eigenvalue equation is therefore (K _ co2m) [ 2(0)02 _ 0)2) \u2014co 2 2 \u2014 CO 2 kwo (0 ) 2 (11.46) The normal frequencies are determined by the condition det(K \u2014 w 2M) = 0, which 2(0) 2 ___ (02)2 w4 = w4 4 0) 2 (02 + 2w04 = 0 \\ 0 0 with the two solutions w 2 = (2 + -fi)co02. That is, the two normal frequencies are 2 col ( 2 ___,Ii) (0 2 0 and 2 W2 = (2 + )co 2 (11.47) (or co l ti 0.77coo and w2 ti 1.85w0) where coo =,/gIL is the frequency of a single pendulum of length L. Knowing the two normal frequencies we can now find the motion of the double pen- dulum in the corresponding normal modes, by solving the equation (K \u2014 w 2M)a = 0 with w = co l and w2 in turn. If we substitute w = w 1, as given by (11.47), into (11.46), gives given by we get 0)12m) = m L2(0.2(1-2 2 \u2014 1) [ _ \u2014 A/2. (K \u2014 1 Therefore, the equation (K \u2014 co i2M)a = 0 implies that a 2 =.411 1, and if we write a l = A le - ` s1, the two coordinates are (Mt) = [ 02(0 = Re ael wit \u2014 A1[1 ]cos(co t \u2014 3 \u2018N /2 1 1 ) [first mode]. (11.48) We see that in the first normal mode the two pendulums oscillate exactly in phase, with the amplitude of the lower pendulum.4 times that of the upper pendulum, as shown in Figure 11.10. = \\f 4)1 Figure 11.10 The first normal mode for a double pendulum with equal masses and equal lengths. The two angles cp i and 02 oscillate in phase, with the amplitude for 4)2 larger by a factor of 436 Chapter 11 Coupled Oscillators and Normal Modes Figure 11.11 The second normal mode for a double pendulum with equal masses and equal lengths. The two angles 0 1 and 02 oscillate exactly out of phase, with the amplitude for 0 2 larger by a factor of N/2. Turning to the second mode, we find from (11.46) and (11.47) that (K _ (022m) = _ mL 2c0 2 ( \u201e.. \u00b1 [ 2,N, / 1 2 The equation (K \u2014 w22M)a = 0 implies that a 2 = \u2014N/2a i, and if we write a l = A t e - ` s2, the two coordinates are = 0(0 1 [ 0 2 (0 -= Re ae lw2t -= A2 [ cos(w2t \u2014 82) [second mode]. (11.49) In this second mode, 02 (t) oscillates exactly out of phase with 01 (t), again with an amplitude that is times bigger than that of 0 1 (0, as shown in Figure 11.11. The general solution is an arbitrary linear combination of the two normal modes (11.48) and (11.49). 11.5 The General Case We have now studied in great detail the normal modes of two systems \u2014 a pair of carts attached to three springs and a double pendulum \u2014 and are ready to discuss the general case of a system with n degrees of freedom that is oscillating about a point of stable equilibrium.", " Since the system has n degrees of freedom, its configuration can be specified by n generalized coordinates, 7 q 1, -, qn. To avoid too much notational clutter, I shall now abbreviate the set of all n coordinates by a single boldface q, q = (q i,, q n ) 7 I take for granted that the system is holonomic, so that the number of degrees of freedom equals the number of generalized coordinates, as discussed in Section 7.3. Section 11.5 The General Case 437 [Thus for the two carts of Section 11.1, q would denote the two displacements q _-= (x 1, x2), and for the double pendulum, q = (4) 1, 02).] (Note well that q is not, in general, a three-dimensional vector; it is a vector in the n-dimensional space of the generalized coordinates q 1,, qn.) I shall assume that the system is conservative, so that it has a potential energy U(q i,,q n )=U(q) and Lagrangian T \u2014 U. The kinetic energy is, of course, T =^a 4ma ioe 2, where the sum runs over all the particles, a =1,-,N, that comprise the system. This has to be rewritten in terms of the generalized coordinates q = (q 1,-, q n ) using the relation between the Cartesian coordinates rot and the generalized coordinates ra = ra(qi, qn) (11.50) where I shall take for granted that this relation does not involve the time t explicitly. [Recall that, in the terminology of Section 7.3, generalized coordinates for which (11.50) does not involve the time are called \"natural.\"] We saw in detail in Section 7.8 that if we differentiate (11.50) with respect to t and substitute into the kinetic energy, we find that [compare Equation (7.94)] T = T (q, = EA,,,(q)4., 4k j,k where the coefficients A jk (q) may depend on the coordinates q. [Compare (11.38) for the case of the double pendulum.] Under our present assumptions, the Lagrangian has the general form (q, (I) = T(q, q) \u2014 U(q), where T(q, q) is given by (11.51), and U(q) is an as-yet unspecified function of the coordinates q. Our. final assumption on the system is that it is making small oscillations about a configuration of stable equilibrium. By redefining the coordinates if necessary, we can arrange that the equilibrium position is q = 0 (that is, q 1 = = qn = 0). Then, since we are interested only in small oscillations, we have only to concern ourselves with small values of the coordinates q, and we can use Taylor expansions of U and T about the equilibrium point q = 0. For U this gives U(q) = U(0) + E au qj + 2u aqi agiaqk q i qk + 2 j,k (11.52) where all derivatives are evaluated at q = 0. This can be much simplified. First, since U (0) is a constant we can simply drop it, by redefining the zero of potential energy. Second, since q = 0 is an equilibrium point, all of the first derivatives aula qj are zero. I shall rename the second derivatives as a2 ula qj aqk = K ik (which satisfy K ik = Kki since it makes no difference in which order we evaluate second derivatives). Finally, since the oscillations are small, I shall neglect all terms higher than second order in the small quantities q or q. This reduces U to U = U(q) - jkq j qk j,k (11.53) 438 Chapter 11 Coupled Oscillators and Normal Modes The kinetic energy is even simpler. Every term in (11.51) contains a factor 4 i 4k which is already second order in small quantities. Therefore, we can ignore everything but the constant term in the expansion of A ik (q). If we call this constant term Aik(0) = Mik, this reduces the kinetic energy to T = T (q) = jkgj gk j,k (11.54) and the Lagrangian to (q, q) = T (q) \u2014 U (q) (11.55) with T (q) given by (11.54) and U(q) by (11.53). Notice that the approximate forms (11.54) and (11.53) correspond to the approximations (11.39) and (11.40) for the dou- ble pendulum. Just like the latter approximations", ", they reduce the kinetic energy to a homogeneous quadratic function of the velocities 4 and the potential energy to a homogeneous quadratic function of the coordinates q. Just as with the double pendu- lum, this will guarantee that the equations of motion are solvable linear equations, but before we take up the equations, let's see one more simple example of this dramatic simplification of T and U that results from the assumption of small oscillations. EXAMPLE 11.1 A Bead on a Wire A bead of mass m is threaded on a frictionless wire that lies in the xy plane (y vertically up), bent in the shape y = f (x) with a minimum at the origin, as shown in Figure 11.12. Write down the potential and kinetic energies and their simplified forms appropriate for small oscillations about 0. This system has just one degree of freedom, and the natural choice of generalized coordinate is just x. With this choice, the potential energy is simply U = mgy = mg f (x). When we confine ourselves to small oscillations, we can Taylor expand f (x). Since f (0) = PO) = 0, this gives U(x) = mg f (x) ti 2mgf\"(0)x2. The kinetic energy is T = 4m (z 2 + )2), where, by the chain rule, 5) (x). Therefore, T = ;m[1 f (x) 2 ].i 2. Notice that the exact expression for T de- pends on x as well as x. However, since T already contains the factor i 2, when Figure 11.12 A bead threaded on a frictionless wire in the shape y = f (x). Section 11.5 The General Case 439 we make the small-oscillation approximation we can simply replace the term f t (x) by its value at x = 0 (namely zero) and we get T (x, = zm{1 f, (0 2, 2 ^, -5:MX. As expected, the small-oscillation approximation has reduced both U and T to homogeneous quadratic functions of x (for U) or x (for T). The Equation of Motion Returning to the approximate Lagrangian (11.55) for the general system, we can easily write down the equations of motion. Since there are n generalized coordinates qi, (i = 1,, n), there are n corresponding equations d az _ ac dt ag, \u2014 0g, [i = 1, n]. (11.56) To write these equations explicitly, we must differentiate the expressions (11.54) and (11.53) for T and U. If you have never tried differentiating sums like these, it may help to write them out explicitly at first. For example, for a system with just two degrees of freedom (n = 2), Equation (11.53) for U reads 2 j,k=1 K jk qj uk a = -1 (K 2 2 11q1 + v 12E11'12 m 1\u201821(12(11 + v \u201e 2\\ + v \u201e m /1- 22 E/2 ) 1 v \u201e 2 v 12viv2 + / 1.. 22q 2 v 2\\ (11.57) where the second line follows because K12 = K21. In this form we can easily differ- entiate with respect to either q 1 or q2. For example, au aql = K 11g1Kl2q2 with a corresponding expression for au 10,72, and quite generally (however many degrees of freedom there are) au, K i q [i = 1,,n]. (11.58) Since differentiation of the kinetic energy (11.54) works in exactly the same way, we can write down the n Lagrange equations (11.56): E mi;4; _ Ko i [i =- 1,, n}. (11.59) These n equations can be immediately grouped into a single matrix equation 440 Chapter 11 Coupled Oscillators and Normal Modes where q is the (n x 1) column q = qi qn \u2014 and M and K are the (n x n) \"mass\" and \"spring-constant\" matrices comprised of the numbers Mi./ and K 11 respectively. 8 The matrix equation (11.60) is, of course, the n-dimensional equivalent of the two- dimensional equations (11.3) for the pair of carts and (11.43) for the double pendulum, and it is solved in exactly the same way. We first seek normal modes with the now- familiar form q(t) = Re z(t), where z(t) = aeiwt and a is a constant (n x 1) column. These lead us to the eigenvalue", " equation which has solutions if and only if co satisfies the characteristic or secular equation This determinant is an nth degree polynomial in co t, so equation (11.63) has n solutions, which tell us the n normal frequencies of the system. 9 With co set equal to each of the normal frequencies in turn, Equation (11.62) determines the motion of the system in the corresponding normal mode. Finally, the general motion of the system is given by an arbitrary sum of the normal mode solutions (11.61). The general procedure outlined in the last three paragraphs is just what we have already discussed in detail for the examples of the two carts and the double pendulum. I shall go through one more example, this one with three degrees of freedom, in the next section, and you should certainly work some of the examples in the problems at the end of the chapter. 8 From this point in the calculation all we need is the two matrices M and K. Thus, in practice, there is actually no need to write down the Lagrangian, nor the Lagrange equations, since the matrices M and K can be read off directly from the approximate expressions (11.54) and (11.53) for T and U. 9 Two subtleties: First, it may happen that some of the roots of (11.63) are equal; this simply means that some of the normal modes have equal frequencies and presents no serious problem. Second \u2014 and much deeper \u2014 we need the n solutions of (11.63) for w 2 to be real and positive, in order that the normal frequencies be real. That this is actually so follows from properties of the matrices K and M as I shall discuss in the appendix. Section 11.6 Three Coupled Pendulums 441 11.6 Three Coupled Pendulums Consider three identical pendulums coupled by two identical springs as shown in Figure 11.13. As generalized coordinates it is natural to use the three angles shown as 01, 02, and 03, with the equilibrium position at 0 1 = 02 = 03 = 0. Our first task is to write down the Lagrangian for the system, at least for small displacements. The systematic, and perhaps the safest, procedure would be to write down the exact expressions for T and U and then make the small-angle approximations. In practice, finding the exact expressions can be very tedious. (In the present case, the potential energy of the springs depends on their extensions, and the exact expressions for these, good for any angles, are very cumbersome. See Problem 11.22.) It often happens that with care one can write down the small-angle approximations for T and U directly and save a lot of trouble, and this is what I shall do here. The kinetic energy of the three pendulums is easily seen to be T = lmL 2 (c.b12 (i)32) (11.64) which does not require any approximation. The gravitational potential energy of each pendulum has the form mgL(1 \u2014 cos 0) ;m gLO 2, where the last expression is the well-known small-angle approximation. Thus the total gravitational potential energy is Ugrav = 21 mg (,/,''K 2 th 2 2 ) (11.65) To find the potential energy of the two springs, we have to find how much each is stretched. For arbitrary values of the angles 0 this is a fairly messy affair, but for small angles the only appreciable stretching comes from the horizontal displacements of the pendulum bobs, each of which moves a distance of approximately L0 to the Figure 11.13 Three identical pendulums of lengths L and masses m are coupled by two identical springs with spring constants k. The gen- eralized coordinates are the three angles 0 1, 02, and 03. The springs' natural lengths are equal to the separation of the supports of the pen- dulums, so the equilibrium position is h = = 0, with all three pendulums hanging vertically. 442 Chapter 11 Coupled Oscillators and Normal Modes right. Thus, for example, the left spring is stretched by about L(0 2 \u2014 0 1), and the total spring potential energy is Uspr = 4kL2 [(02 - 002 + (03 - 02)2] 4a 2 (0,2 + 2022 + 032 - 20102 - 20203). (11.66) Before we combine these expressions for T and U to get the Lagrangian and the equations of motion, there is a useful device I would like to introduce. The equations we are going to derive involve several fixed parameters, (m, L, g, k), some of which are not especially interesting. The repeated writing of these parameters is, at the very least, an annoying chore and can easily lead to careless mistakes. Thus it is helpful to find some way to get rid of uninteresting parameters before we do any", " more calculation. A radical way to do this that is very popular with theoretical physicists is to choose a system of units such that the uninteresting parameters have the value 1 \u2014 a process sometimes described as choosing natural units. In the present problem, for example, we can choose m to be the unit of mass and L to be the unit of length. With this choice m and L naturally have the value 1, and they disappear from all subsequent work. This trick materially simplifies the trivial details of our calculations, reduces the danger of errors, and helps us to see the truly interesting features. The only serious disadvantage of using natural units is this: Once our calculations are complete, we sometimes want to know how our answers depend on the values of the parameters that have been suppressed. (What is the frequency of a certain normal mode, if L = 1.5 m?) To answer this kind of question, we have to put the banished parameters back into our answers. Although this process (of restoring the banished parameters) seems daunting to the beginner, it is usually fairly easy. For example, with m = L = 1, we are going to find that one of the normal frequencies is given by co2 = g. This implies that (in our system of units) the quantity glco2 has the value 1. But you will recognize that g / co2 has the dimensions of a length, and to say that a length has the value 1 in our units, is the same as saying that it has the value L in any system of units. Therefore, glco2 =L in general, and our answer is that co 2 = g/L whatever units we use. This puts the \"L\" back into our answer and lets us find co for any given value of L. Let us then choose units with m = L = 1, so that the kinetic and potential energies obtained from (11.64), (11.65), and (11.66) become T = + (1)22 + (4) (11.67) and g + + 03) + 4140i + 2022: (1)32 \u2014 2002 \u2014 20203). (11.68) We could now write down the Lagrangian and then the equations of motion, but there is actually no need to do this. We already know that the result will be the now-familiar matrix equation = \u2014KO (11.69) Section 11.6 Three Coupled Pendulums 443 where, in this case, 0 is the (3 x 1) column comprising the three angles 0 1, 02, and 03. The elements of the (3 x 3) matrices M and K can be read directly from (11.67) and (11.68) to give l =[ and K = g + 2k. (11.70) g + k \u2014k \u2014k 0 \u2014k g k 0 \u2014k The normal modes of our system have the familiar form q(t) = Re z(t) = Re aelw t, where a and w are determined by the eigenvalue equation (K \u2014 w2M)a = 0. (11.71) Our first step is to find the possible normal frequencies from the characteristic equation det(K \u2014 w2M) = 0, to which end we need to write down the matrix (K \u2014 w 2M), (K \u2014 w2M) = g + 2k \u2014 co 2 (11.72) g k \u2014 co 2 \u2014k 0 \u2014k \u2014k 0 \u2014 k g + k \u2014 co 2 The determinant is easily evaluated as det(K \u2014 co 2M) = (g \u2014 (02) (g + k (02)(g + 3k (02)9 so that the three normal frequencies are given by 2 CO = g, co 2 g k, and w 2 = g + 3k. 3 (11.73) Knowing the three normal frequencies, we can now find the three corresponding normal modes in turn. The first normal frequency has co l =. (This is in our units, where L = 1. As I have already mentioned, in arbitrary units it is co l =./g/L, which is the frequency for a single pendulum of length L. We'll see the reason for this coincidence in a moment.) If we substitute w i into Equation (11.72) for (K \u2014 co 2M), then the eigenvalue equation (11.71) implies (as you should check) that al = a2 = a3 =, [first mode] say. That is, in the first mode, 0 1 (t) = 02(t) = 03(t) = A cos(co \u2014 8) and the three pendulums oscillate in unison (equal amplitudes and phases), as shown in Figure 11.14(a). In this mode, the springs are neither compressed nor stretched, and their presence is irrelevant. Therefore, each pendulum oscillates just", " like a single pendulum, with frequency co l =.\u2018/gIL (or /k in our units). 1\u00b0 One has to think a little when reading off the elements of these matrices. The rule is this: If you ignore the factor z in front of (11.68), for example, then the diagonal element Kii is just the coefficient of 0, 2, while the off-diagonal element Kid is half the coefficient of 0,0j. To understand this, look at (11.57). 444 Chapter 11 Coupled Oscillators and Normal Modes (a) (b) (c) 4) 1 2 - (1) 1 4) 3 = (I) 1 41 CP 2 = 3 = 1 1 2 = 1 03-01 Figure 11.14 The three normal modes for three coupled pendulums. (a) In the first mode, the three pendulums oscillate in unison. Since neither spring is stretched or compressed, the frequency is just that for a single pendulum of the same length. (b) In the second mode, the outer two pendulums oscillate exactly out of phase, while the middle one doesn't move at all. (c) In the third mode, the outer two pendulums oscillate in unison, but the middle one is exactly out of phase and has twice the amplitude. If we substitute co = w2, the eigenvalue equation (11.71) implies that a l = \u2014a3 = Ae -is, but a2 = 0 [second mode]. Therefore, the outer two pendulums oscillate exactly out of phase, while the middle one sits at rest, as shown in Figure 11.14(b). Finally, substituting co = w3 into the eigenvalue equation (11.71) leads to the result al = = a3 = tie ` A S, [third mode] say. Thus, in the third mode, the outer two pendulums oscillate in unison, but the middle one oscillates with twice the amplitude and exactly out of phase, as shown in Figure 11.14(c). The general solution is an arbitrary linear combination of all three normal modes. 11.7 Normal Coordinates* * This section could be skipped if you are pressed for time. In Section 11.2, we found the normal modes for two equal masses joined by three identical springs. At the end of that section, I mentioned that one can replace the two coordinates x 1 and x2 by two \"normal coordinates\" = i (x + x2) and 2 = i (x 1 \u2014 x2). (11.74) These new coordinates have the property that each always oscillates at just one of the two normal frequencies \u2014 l at the frequency co l and S2 at the frequency co 2. In this section I will show that we can do the same thing for any system oscillating about a stable equilibrium point. If the system has n degrees of freedom, then it is described by n generalized coordinates q l,, qn and has n normal modes with frequencies co l,, co n. What I shall show is that we can introduce n new, normal, coordinates,, Sn, such that each normal coordinate s oscillates at just one frequency, namely the normal frequency col. Before I show this, I need to review and extend the discussion of Section 11.2 for the two carts. Section 11.7 Normal Coordinates* 445 Normal Coordinates for Two Carts on Springs The equations of motion for the positions x i and x2 of the two carts were given in (11.2), which I rewrite here (for the case of equal masses and equal spring con- stants) as A moment's inspection should convince you that if we add these two equations, we get an equation for t = z (x 1 + x2) alone, and if we subtract them, we get an equation for 2 = 1(x 1 \u2014 x2) alone: = \u20142kx1 k x2 = kx i \u2014 2kx2. m-5j2 mi = 171.2 = 1 1 (11.75) (11.76) These two equations are uncoupled, and show that each normal coordinate oscil- lates, as claimed, at a single frequency \u2014 1 at frequency co l =,/k/m and 2 at (02 = Ok/m. In other words, the normal coordinates behave just like the coordi- nates of two uncoupled oscillators \u2014 by going over to the normal coordinates, we have \"uncoupled\" the oscillations. Just as the equations (11.75) for x 1 and x2 can be rewritten as a single matrix equation Mk = \u2014Kx, so the two equations (11.76) for b l and 2 can be written as = with the important difference that the two matrices M' and K' are both diagonal: 1,,,, r m", " 01 L 0 and K ='[ k 0]. 0 3k (11.77) The transition from the original coordinates (x 1, x2) to the normal coordinates 2 ) is said to diagonalize the matrices M and K. That the new matrices are diagonal is precisely equivalent to the statement that the equations (11.76) for t and 2 are uncoupled and that 1 and 2 oscillate independently. We can define the two normal coordinates 1 and b2 differently, and more generally, in terms of the eigenvectors a that describe the motion of the normal modes and are determined by the eigenvalue equation (K \u2014 co 2M)a = 0. We saw in Section 11.2 that these two (2 x 1) columns are (for our two carts) 11 a(1) = [ and a = [ (11.78) 1 11 [Two important points: Each of these vectors contains an arbitrary multiplier A, but I now want to fix this, and the simplest choice here is to make A = 1; another, and sometimes better, choice is to normalize the vectors by putting in a factor of 1[4,. Second, each column a is made up of two numbers, which I have been calling a l and a2. But I am now discussing two different columns, a(l) and a (2), one for each normal 446 Chapter 11 Coupled Oscillators and Normal Modes mode, and I am using the parentheses in the subscripts to emphasize this distinction.] Now, it is easy to see that any (2 x 1) column vector can be written as a combination of the two vectors a (1) and a (2). (See Problem 11.33.) In particular, I can expand the column x in this way as x = i a (i) a(2) _ + i Si \u2014 (11.79) The first equality defines i and 2 as the coefficients in this expansion of x in terms of the eigenvectors a il) and a(2), but inspection of the last expression in (11.79) should convince you that it defines i and b2 to be precisely the normal coordinates of (11.74). That is, the normal coordinates, with the property that each oscillates independently at one of the normal frequencies, can be defined as the coefficients in the expansion of x in terms of the eigenvectors a il) and a (2). We shall see that this definition carries over naturally to the general case of oscillations of any system with n degrees of freedom. The General Case We can now easily introduce normal coordinates for an arbitrary oscillating system with n generalized coordinates q l,, qn. We know that such a system has n normal modes. In mode i, the column vector q oscillates sinusoidally, q(t) = a (i) cos(wi t \u2014 6i ) where the fixed column a (i) satisfies the eigenvalue equation Ka(i) = coi2Ma(i). (11.80) The columns a o),, a (n) are n independent, real\" (n x 1) columns, and any (n x 1) column can be expanded in terms of them; that is, the vectors a (1),, a (n) are a basis or complete set for the space of all (n x 1) vectors. (For a proof of these properties, see the appendix.) Thus any solution of the equations of motion q(t) can be expanded as q(t) = E i(t) a(i). i=1 (11.81) This definition of the normal coordinates i exactly parallels the new definition (11.79) for the system of two carts on springs, and we can now prove that it has the desired property that the different (t) oscillate independently, as fol- lows: The column q(t) satisfies the equation of motion Mq = \u2014Kg. \" That the vectors a u) are real is not obvious. In the case of the two carts, you can see from (11.78) that they are. In the general case, they are determined by the real eigenvalue equation (11.80) and it is fairly easy to prove that either they are real or they can be redefined so that they are. See the appendix. Principal Definitions and Equations of Chapter 11 447 If we replace q by its expansion (11.81), this equation becomes n E i (t) Ma (i) = \u2014 n i(t)Ka(i) = \u2014 E i(t ) coi2Ma (i) (11.82) i\u2014i where the last equality follows from the eigenvalue equation (11.80). Now, the n column vectors a (1),, a (n) are independent, and this property is unchanged when they are multiplied", " by the matrix M. Therefore, the n vectors Ma (1),, Ma (n) are also independent, and the equality (11.82) can only hold if all corresponding coefficients on each side are equal. 12 That is, * i (t) = This establishes that the normal coordinates i defined by (11.81) do indeed oscillate independently at the advertised frequencies. Principal Definitions and Equations of Chapter 11 The Equations of Motion in Matrix Form The configuration of a system with n degrees of freedom can be specified by an n x 1 column matrix q, comprising the n generalized coordinates q l,, qn. The equation of motion for small oscillations about a stable equilibrium (with the coordinates chosen so that q = 0 at equilibrium) has the matrix form Mq = \u2014Kq [Eq. (11.60)] where M and K are the n x n \"mass\" and \"spring -constant\" matrices. One way to find these matrices is to write the KE and PE of the system in the forms T = E M jk 41 4k and U= j,k j,k i k qi qk [Eqs. (11.54) & (11.53)] Normal Modes A normal mode is any motion in which all n coordinates oscillate sinusoidally with the same frequency to (a normal frequency) and can be written as q(t) = Re(a ei\") [Eq. (11.61)] where the constant n x 1 column a must satisfy the generalized eigenvalue equation (K \u2014 co2M)a = 0. [Eq. (11.62)] 12 The result I am using here is the analog of the familiar result in three-dimensional space that the equality E 3ti ei = E 3 \u00b5i et is only possible if A = \u00b5i for all i. For a proof of the independence of the vectors a (1),, a (n) see the appendix. 448 Chapter 11 Coupled Oscillators and Normal Modes For any system with n degrees of freedom and a stable equilibrium at q = 0, there are n normal frequencies co l,, con (some of which may be equal) and n independent corresponding eigenvectors a ( 1),, a (n). Any n x 1 column can be expanded in terms of these eigenvectors, and any solution of the equations of motion can be expanded in terms of the normal modes. Normal Coordinates When any solution is expanded in terms of the normal modes, the expansion coeffi- cients (t) are called normal coordinates, and each oscillates at the corresponding frequency coi. [Section 11.7] Problems for Chapter 11 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). SECTION 11.1 Two Masses and Three Springs 11.1 * In discussing the two carts of Figure 11.1, I mentioned that it is simplest to assume that when the two carts are in equilibrium the lengths L 1, L2, L3 of the three springs are equal to their natural, unstretched lengths / 1, /2, /3. However, this assumption is not needed, and the three springs could all be in tension (or compression) at the equilibrium position. (a) Find the relations among these six lengths (and the three spring constants k 1, k2, k3) required for the two carts to be in equilibrium. (b) Show that the net force on either cart is exactly as given in Equation (11.2), irrespective of how L 1, L2, L3 compare with / 1, /2, /3, just as long as x 1 and x2 are measured from the carts' equilibrium positions. 11.2 ** A massless spring (force constant k 1 ) is suspended from the ceiling, with a mass m 1 hanging from its lower end. A second spring (force constant k 2) is suspended from m l, and a second mass m 2 is suspended from the second spring's lower end. Assuming that the masses move only in a vertical direction and using coordinates y i and y2 measured from the masses' equilibrium positions, show that the equations of motion can be written in the matrix form My = \u2014Ky, where y is the 2 x 1 column made up of y i and y2. Find the 2 x 2 matrices M and K. SECTION 11.2 Identical Springs and Equal Masses 11.3 * Find the normal frequencies for the system of two carts and three springs shown in Figure 11.1, for arbitrary values of m 1 and m 2 and of k 1, k2, and k3. Check that your answer is correct for the case that m 1 = m 2 and k 1 = k2 = k3. 11.4 ** (a) Find the normal frequencies for the system of two carts and three springs shown in Figure 11.1, for", " the case that m i = m 2 and k 1 = k3, (but k2 may be different). Check that your answer is correct for the case that k 1 = k2 as well. (b) Find and describe the motion in each of the two normal modes in turn. Compare with the motion found for the case that k 1 = k2 in Section 11.2. Explain any similarities. 11.5 ** (a) Find the normal frequencies, co l and cot, for the two carts shown in Figure 11.15, assuming that m 1 = m 2 and k 1 = k2. (b) Find and describe the motion for each of the normal modes in turn. Problems for Chapter 11 449 k l k2 x 1 X2 Figure 11.15 Problems 11.5 and 11.6. 11.6 ** Answer the same questions as in Problem 11.5 but for the case that m 1 = m2 and k 1 = 3k2 /2. (Write k 1 = 3k and k2 = 2k.) Explain the motion in the two normal modes. 11.7 ** [Computer] The most general motion of the two carts of Section 11.2 is given by (11.21), with the constants A 1, A2, 31, and 62 determined by the initial conditions. (a) Show that (11.21) can be rewritten as x(t) = (B i cos w i t + Ci sin w i t) [ 1 ] + (B2 cos col t + C2 sin colt) [ 1 1. 1 1 This form is usually a little more convenient for matching to given initial conditions. (b) If the carts are released from rest at positions x 1 (0) = x 2 (0) = A, find the coefficients B 1, B2, C 1, and C2 and plot x 1 (t) and x2 (t). Take A = col = 1 and 0 < t < 30 for your plots. (c) Same as part (b), except that x 1 (0) = A but x2 (0) = 0. 11.8 ** [Computer] Same as Problem 11.7 but in part (b) the carts are at their equilibrium positions at t = 0 and are kicked away from each other, each with speed v o. In part (c), the carts start out at their equilibrium positions and cart 2 has speed v o to the right but cart 1 has initial speed 0. Take v o = co l = 1 and 0 < t < 30 for your plots. 11.9 ** (a) Write down the equations of motion (11.2) for the equal-mass carts of Section 11.2 with three identical springs. Show that the change of variables to the normal coordinates 1 = z (xi + x2) and 2 = z (x i \u2014 x2) leads to uncoupled equations for 1 and 2 (b) Solve for t and 2 and hence write down the general solution for x 1 and x2. (Notice how very simple this procedure is, once you have guessed what the normal coordinates are. For a simple symmetric system like this, you can sometimes guess the form of 1 and 2 by considering the symmetry \u2014 especially once you have some experience working with normal modes.) 11.10 *** [Computer] In general, the analysis of coupled oscillators with dissipative forces is much more complicated than the conservative case considered in this chapter. However, there are a few cases where the same methods still work, as the following problem illustrates: (a) Write down the equations of motion corresponding to (11.2) for the equal-mass carts of Section 11.2 with three identical springs, but with each cart subject to a linear resistive force \u2014by (same coefficient b for both carts). (b) Show that if you change variables to the normal coordinates 1 = 4 (xi + x2) and 2 = 4.(x 1 \u2014 x2), the equations of motion for t and 2 are uncoupled. (c) Write down the general solutions for the normal coordinates and hence for x 1 and x2. (Assume that b is small, so that the oscillations are underdamped.) (d) Find x 1 (t) and x 2 (t) for the initial conditions x 1 (0) = A and x 2 (0) = v 1 (0) = v2 (0) = 0, and plot them for 0 < t < 107r using the values A = k = m = 1, and b = 0.1. 11.11 *** (a) Write down the equations of motion corresponding to (11.2) for the equal-mass carts of Section 11.2 with three identical springs, but with each cart subject to a linear resistive force \u2014by (same coefficient b for both carts) and with a driving force F (t) = Fo cos cot", " applied to cart 1. (b) Show that if you change variables to the normal coordinates 1 = 1(x 1 + x2) and 2 = z (x 1 \u2014 x2), the equations of 450 Chapter 11 Coupled Oscillators and Normal Modes motion for c i and b2 are uncoupled. (c) Using the methods of Section 5.5, write down the general solutions. (d) Assuming that /3 = b/2m << coo, show that i resonates when co ti coo = Oclm and likewise b2 when co ti Nijcoo. (e) Prove, on the other hand, that if both carts are driven in phase with the same force Fo cos cot, only i shows a resonance. Explain. Figure 11.16 Problem 11.12 11.12 *** Here is a different way to couple two oscillators. The two carts in Figure 11.16 have equal masses m (though different shapes). They are joined by identical but separate springs (force constant k) to separate walls. Cart 2 rides in cart 1, as shown, and cart 1 is filled with molasses, whose viscous drag supplies the coupling between the carts. (a) Assuming that the drag force has magnitude /3m v where v is the relative velocity of the two carts, write down the equations of motion of the two carts using as coordinates x 1 and x2, the displacements of the carts from their equilibrium positions. Show that they can be written in matrix form as x + + (0\u201e2:x = 0, where x is the 2 x 1 column made up of x 1 and x2, coo =,/k/m, and D is a certain 2 x 2 square matrix. (b) There is nothing to stop you from seeking a solution of the form x(t) = Re z(t), with z(t) = aert. Show that you do indeed get two solutions of this form with r = icoo or r = \u201413 + iwi where co l = Jcoo \u2014 )52. (Assume that the viscous force is weak, so that,8 < coo.) (c) Describe the corresponding motions. Explain why one of these modes is damped but the other is not. SECTION 11.3 Two Weakly Coupled Oscillators 11.13 *** [Computer] Consider the two carts of Section 11.3, coupled by a weak spring and subject to a resistive force \u2014by (same force for each cart). (a) Write down the equations of motion for x 1 2 in the form (11.2) and show that if you change to the normal coordinates c i = z (x i + x2) and and x = 2 (x, - x2) the equations of motion for bi and 2 are uncoupled. (b) Solve for 1 (t) and 2 (t) assuming that the dissipative coefficient b is small (\"underdamped motion,\" as in Section 5.4), and hence write down the general solution for x i (t) and x 2 (t). (c) Suppose that cart 1 is held at x 1 = A and cart 2 at x 2 = 0, and they are released from rest at t = 0. Find and plot the two positions as functions of t for 0 < t < 80, using the values A = k = m = 1, k2 = 0.2, and b = 0.04. (In matching the initial conditions, take advantage of the fact that b << 1, and use a suitable computer program to make the plot.) Comment on your plots. SECTION 11.4 Lagrangian Approach: The Double Pendulum 11.14 ** Consider two identical plane pendulums (each of length L and mass m) that are joined by a massless spring (force constant k) as shown in Figure 11.17. The pendulums' positions are specified by the angles 0 1 and 02 shown. The natural length of the spring is equal to the distance between the two supports, so the equilibrium position is at 0 1 = 02 = 0 with the two pendulums vertical. (a) Write down the total kinetic energy and the gravitational and spring potential energies. [Assume that both angles remain small at all times. This means that the extension of the spring is well approximated by Problems for Chapter 11 451 Figure 11.17 Problem 11.14 L(02 \u2014 OA Write down the Lagrange equations of motion. (b) Find and describe the normal modes for these two coupled pendulums. 11.15 ** Write down the exact Lagrangian (good for all angles) for the double pendulum of Figure 11.9 and find the corresponding equations of motion. Show that they reduce to Equations (11.41) and (11.42) if both angles are small. 11.16 ** (a) Find the normal frequencies for", " small oscillations of the double pendulum of Figure 11.9 for arbitrary values of the masses and lengths. (b) Check that your answers are correct for the special case that m 1 = m 2 and L 1 = L2. (c) Discuss the limit that m 2 0. 11.17 ** (a) Find the normal freqencies and modes of the double pendulum of Figure 11.9, given that m 1 = 8m, m2 = m, and L 1 = L2 = L. (b) Find the actual motion 101 (t), 02 (t)i if the pendulum is released from rest with 0 1 = 0 and 02 = a. Is this motion periodic? 11.18 ** Two equal masses m are constrained to move without friction, one on the positive x axis and one on the positive y axis. They are attached to two identical springs (force constant k) whose other ends are attached to the origin. In addition, the two masses are connected to each other by a third spring of force constant k'. The springs are chosen so that the system is in equilibrium with all three springs relaxed (length equal to unstretched length). What are the normal frequencies? Find and describe the normal modes. Figure 11.18 Problem 11.19 11.19 *** A simple pendulum (mass M and length L) is suspended from a cart (mass m) that can oscillate on the end of a spring of force constant k, as shown in Figure 11.18. (a) Assuming that the angle cb remains small, write down the system's Lagrangian and the equations of motion for x and 0. (b) Assuming that m=M=L=g=1 and k = 2 (all in appropriate units) find the normal frequencies, and for each normal frequency find and describe the motion of the corresponding normal mode. 452 Chapter 11 Coupled Oscillators and Normal Modes 11.20 *** (a) A thin uniform rod of length 2b is suspended by two vertical light strings, both of fixed length 1, fastened to the ceiling. Assuming only small displacements from equilibrium, find the Lagrangian of the system and the normal frequencies. Find and describe the normal modes. [Hint: A possible choice of generalized coordinates would be x, the longitudinal displacement of the rod, and y i and y2, the sideways displacements of the rod's two ends. You'll need to find how high the two ends are above their equilibrium height and what angle the rod has turned through.] SECTIONS 11.5 and 11.6 The General Case & Three Coupled Pendulums 11.21 * Verify that if U = 4 Ei Ek Kikm k, where the coefficients Kik are all constant and satisfy Kii = K 1, then au/aqi E, Kiiq j, as claimed in Equation (11.58). 11.22 * Write down the exact potential energy of the three pendulums of Figure 11.13, good for all angles, small or large, and show that your answer reduces to (11.68) if all three angles are small. 11.23 ** Equation (11.73) gives the three normal frequencies of three coupled pendulums in natural units with L=m= 1. We have already seen that the value of w i2 in arbitrary units is g/L. Find the values of col' and w3 in arbitrary units. [Hint: Start by considering the quantity (022 (012. ] Figure 11.19 Problem 11.24 11.24 ** Two equal masses m move on a frictionless horizontal table. They are held by three identical taut strings (each of length L, tension T), as shown in Figure 11.19, so that their equilibrium position is a straight line between the anchors at A and B. The two masses move in the transverse (y) direction, but not in the longitudinal (x) direction. Write down the Lagrangian for small displacements, and find and describe the motion in the corresponding normal modes. [Hint: \"Small\" displacements have y i and y2 much less than L, which means that you can treat the tensions as constant. Therefore the PE of each string is just Td, where d is the amount by which its length has increased from equilibrium.] 11.25 ** Consider a system of carts and springs like that in Figure 11.1 except that there are three equal- mass carts and four identical springs. Solve for the three normal frequencies, and find and describe the motion in the corresponding normal modes. 11.26 ** A bead of mass m is threaded on a frictionless circular wire hoop of radius R and mass m (same mass). The hoop is suspended at the point A and is free to swing in its own vertical plane as shown in Figure 11.20. Using the angles q51 and 02 as generalized coordinates, solve for the normal frequencies of small oscillations, and find", " and describe the motion in the corresponding normal modes. [Hint: The KE of the hoop is.1. Iii, where I is its moment of inertia about A and can be found using the parallel axis theorem.] 11.27 ** Consider two equal-mass carts on a horizontal, frictionless track. The carts are connected to each other by a single spring of force constant k, but are otherwise free to move freely along the track. (a) Write down the Lagrangian and find the normal frequencies of the system. Show that one of the Problems for Chapter 11 453 Figure 11.20 Problem 11.26 normal frequencies is zero. (b) Find and describe the motion in the normal mode whose frequency is nonzero. (c) Do the same for the mode with zero frequency. [Hint: This one requires some thought. It isn't immediately clear what oscillations of zero frequency are. Notice that the eigenvalue equation (K \u2014 w2M)a = 0 reduces to Ka = 0 in this case. Consider a solution x(t) = a f (t), where f (t) is an undetermined function of t and use the equation of motion, MX = \u2014Kx, to show that this solution represents motion of the whole system with constant velocity. Explain why this kind of motion is possible here but not in the previous examples.] 11.28 ** A simple pendulum (mass M and length L) is suspended from a cart of mass in that moves freely along a horizontal track. (See Figure 11.18, but imagine the spring removed.) (a) What are the normal frequencies? (b) Find and describe the corresponding normal modes. [See the hint for Problem 11.27]. 11.29 *** A thin rod of length 2b and mass m is suspended by its two ends with two identical vertical springs (force constant k) that are attached to the horizontal ceiling. Assuming that the whole system is constrained to move in just the one vertical plane, find the normal frequencies and normal modes of small oscillations. Describe and explain the normal modes. [Hint: It is crucial to make a wise choice of generalized coordinates. One possibility would be r, 0, and a, where r and 0 specify the position of the rod's CM relative to an origin half way between the springs on the ceiling, and a is the angle of tilt of the rod. Be careful when writing down the potential energy.] 11.30 *** [Computer] Consider a system of carts and springs like that in Figure 11.1 except that there are four equal-mass carts and five identical springs. Solve for the four normal frequencies, and find and describe the motion in the corresponding normal modes. [This can be solved without the aid of a computer, but it would probably be worth your while to learn how to evaluate determinants and to solve the characteristic equation using suitable computer software.] 11.31 *** Consider a frictionless rigid horizontal hoop of radius R. Onto this hoop I thread three beads with masses 2m, in, and m, and, between the beads, three identical springs, each with force constant k. Solve for the three normal frequencies and find and describe the three normal modes. 11.32 *** As a model of a linear triatomic molecule (such as CO 2), consider the system shown in Figure 11.21, with two identical atoms each of mass m connected by two identical springs to a single atom of mass M. To simplify matters, assume that the system is confined to move in one dimension. (a) Write down the Lagrangian and find the normal frequencies of the system. Show that one of the 454 Chapter 11 Coupled Oscillators and Normal Modes Figure 11.21 Problem 11.32 normal frequencies is zero. (b) Find and describe the motion in the normal modes whose frequencies are nonzero. (c) Do the same for the mode with zero frequency. [Hint: See the comments at the end of Problem 11.271 SECTION 11.7 Normal Coordinates 11.33 * The eigenvectors a (1) and a(2) that describe the motion in the two normal modes of the two carts of Section 11.2 are given in (11.78). Prove that any (2 x 1) column x can be written as a linear combination of these two eigenvectors; that is, a (1) and a (2) are a basis of the space of (2 x 1) columns. 11.34 ** It is a crucial property of the eigenvectors, a (1),, a fro, describing the motion in the normal modes of an oscillating system that any (n x 1) column x can be written as a linear combination of the n eigenvectors; that is, the eigenvectors are a basis of the space of (n x 1) columns. This is proved in the appendix, but to illustrate it, do the following: (a", ") Write down the three eigenvectors a (1), a(2), and a(3) for the coupled pendulums of Section 11.6. (Each of these contains an arbitrary overall factor, which you can choose at your convenience.) Prove that they have the property that any (3 x 1) column x can be expanded in terms of them. (b) The expansion coefficients in this expansion are the normal coordinates 6 Find the normal coordinates for the three coupled pendulums, and explain the sense in which they describe the three normal modes of Figure 11.14. 11.35 ** Consider the two coupled pendulums of Problem 11.14. (a) What would be a natural choice for the normal coordinates S1 and 2? (b) Show that even if both pendulums are subject to a resisitive force of magnitude by (with b small), the equations of motion for t and 2 are still uncoupled. (c) Find and describe the motion of the two pendulums for the two modes. PART II CHAPTER 12 Nonlinear Mechanics and Chaos CHAPTER 13 Hamiltonian Mechanics CHAPTER 14 Collision Theory CHAPTER 15 Special Relativity CHAPTER 16 Continuum Mechanics Part I of this book contains the material that is in some sense essential. Part II contains mostly more advanced topics that could be considered optional \u2014 although all are of the greatest importance in modern physics. While the chapters of Part I were designed to be read pretty much in sequence, I have tried to make the five chapters of Part II mutually independent, so that you could pick and choose according to your tastes and available time. If you understand most of the material of Part I, you are ready to launch into any of the chapters of Part II. This is not to say that you have to have studied all of Part 1 before reading any of Part II. For example, you could read Chapter 12 on chaos as soon as you have studied Chapter 5 on oscillations. Similarly, you could read Chapter 13 on Hamiltonian mechanics immediately after Chapter 7 on Lagrangian mechanics, and likewise, Chapter 14 on collision theory immediately after Chapter 8 on two-body motion. CHAPTER Nonlinear Mechanics and Chaos One of the most fascinating and exciting discoveries of the last few decades has been the recognition that most systems whose equations of motion are nonlinear can exhibit chaos. This startling phenomenon, which shows up in many different areas \u2014 oscillating mechanical systems, chemical reactions, fluid flow, lasers, population growth, the spread of diseases, and many more \u2014 means that although a system obeys deterministic equations of motion (such as Newton's laws) its detailed future behavior may, as a practical matter, be unpredictable. The behavior of a chaotic mechanical system can be very complicated, and the need to describe this behavior has spawned a whole array of new ways to view the motion of such systems \u2014 state-space orbits, Poincare sections, bifurcation diagrams. Fortunately, there are systems that are complicated enough to exhibit chaos, but still simple enough not to need all these new tools for their description. In particular, a driven damped pendulum, whose equation of motion is nonlinear, can exhibit chaos but can be described in reasonably elementary terms. For this reason, I shall mostly focus on this system here. After a brief review of the difference between linear and nonlinear equations and of the properties of a damped linear oscillator, I shall describe in some detail the motion of a driven damped pendulum, using just straightforward graphs of its position against time. Once you are familiar with the basic phenomena, I shall give an introduction to some of the more sophisticated tools that you will need if you want to explore the rapidly expanding literature on chaos. To conclude this chapter I shall describe another system that can exhibit chaos \u2014 the so-called logistic map. Although this is not strictly part of mechanics, it shows many strong parallels with mechanical systems, and has the great advantage of a simplicity that allows an easy understanding of several aspects of its behavior. It also serves to illustrate some of the amazing universality of chaos. In planning this chapter, I felt strongly that it was important to convey a good understanding of a few topics, rather than a superficial glimpse of the whole field. In particular, chaos theory is divided into two broad areas \u2014 dissipative systems, such as the damped pendulum, and nondissipative, or \"Hamiltonian\" systems \u2014 and I decided to restrict myself entirely to the former. I know that some readers will question this 457 458 Chapter 12 Nonlinear Mechanics and Chaos decision. Many important applications of chaos theory (in astronomy and statistical mechanics, for instance) concern nondissipative systems, as did the pioneering work of Poincare. Nevertheless, it is my experience that the most accessible topics for the beginner concern dissipative systems, and, to keep the length of this chapter reasonably finite, I decided to treat only these. Please see this chapter as a sampler of the good things in chaos theory", "; it certainly makes no claim to tell the whole story. This chapter is very different from all the other chapters of this book. The theory of chaos is new and not at all elementary. (And parts of the theory have yet to be discovered!) It requires a much deeper understanding of differential equations than I am assuming in this book, and a proper exposition of chaos theory requires a whole book rather than one chapter. 1 Therefore, I shall restrict myself here to simply describing the fascinating main properties of chaotic motion, without much attempt to prove that the motion is as I claim. This is actually a reasonably satisfactory situation. Before you try to read any of the more advanced books, it is almost certainly a good thing to have some idea of what chaos involves and some familiarity with the tools used to describe it, and these are what I hope to communicate. 12.1 Linearity and Nonlinearity For a system to exhibit chaos its equations of motion must be nonlinear. We have noted examples of linear and nonlinear equations from time to time in this book, but let us review the two concepts now. A differential equation is linear if it involves the dependent variable or variables and their derivatives only linearly. The equation of motion of a cart (mass m) on a spring (force constant k), \u2014kx, (12.1) is a linear differential equation for the cart's position x. Similarly, the equations of motion for the two carts discussed in Chapter 11 [Equations (11.2) for example] are linear equations for the two carts' positions x 1 and x2. If we apply a driving force F (t) to the cart of Equation (12.1), the resulting equation, mz = \u2014kx F(t), (12.2) is still linear [though no longer homogeneous, since the \"inhomogeneous\" term F(t) does not involve the dependent variable x at all]. By contrast, the equation of motion for a simple pendulum (mass m, length L) is I4 = F or mL 2iio = \u2014mgL sin 0, (12.3) which is a nonlinear equation for 0, since sin 0 is not linear in 0. (If the oscillations are small, then sin 0 0, and the equation is well approximated by a linear equation; Several such books exist, of which my favorite is Nonlinear Dynamics and Chaos by Steven H. Strogatz, Addison-Wesley, Reading, MA (1994), but be warned, it takes eight chapters of mathematical preliminaries to get to the chaos. Section 12.1 Linearity and Nonlinearity 459 in general, however, the equation for the simple pendulum is definitely nonlinear.) Another example is the equation of motion of a single planet in the field of the sun, (12.4) which is a nonlinear equation for the variables r = (x, y, z) because the force term is nonlinear in x, y, and z. These two examples show that nonlinear equations are not especially unusual. On the contrary, many perfectly everyday systems have equations of motion that are nonlinear. So far in this book the main difference between linear and nonlinear differential equations has been that the former have been easily solved analytically, whereas most of the latter have been impossible to solve analytically. In fact, our experience in this regard reflects the true state of affairs: Almost all of the linear equations of mechanics are analytically solvable, and almost none of the nonlinear ones are. 2 This circum- stance is largely to blame for the failure until recently of scientists to recognize that chaos is an important and widespread phenomenon. Because nonlinear equations are so intractable, textbooks always focussed on linear problems. When nonlinear prob- lems had to be addressed, they were often solved using approximations that reduced then to linear problems. In this way, the astonishingly rich variety of complications that occur for nonlinear systems went almost completely unrecognized. The first per- son to notice some of the symptoms of chaos was the French mathematician Poincare (1854-1912) in his studies of the gravitational three-body problem \u2014 the motion of three bodies (such as the sun, earth, and moon) interacting via the gravitational force. The equation of motion for this system is nonlinear, like its two-body counterpart (12.4), and Poincare observed that it exhibits the phenomenon now called sensitivity to initial conditions that is one of the characteristics of chaotic motion, as we shall see. That Poincare's observation of chaos went nearly unnoticed by physicists until the 1970s is probably due to several factors. The discoveries of relativity (1905) and then of quantum mechanics (around 1925) diverted most physicists' attention away from classical mechanics. And the difficulty of solving nonlinear equations without the aid of computers certainly discouraged the pursuit of nonlinear problems. In any case it was only in the 1970s that computer solutions of", " various nonlinear problems 3 drew the attention of significant numbers of scientists (physicists and many others) to the phenomenon that we now call chaos. Nonlinearity is essential for chaos \u2014 if a system's equations of motion are linear, it cannot exhibit chaos. But nonlinearity does not guarantee chaos. For example, the equation (12.3) for a simple pendulum is nonlinear, but even when the amplitude is large (and the linear approximation is definitely not good) the simple pendulum never nonlinear equation is (12.4) for a planet, whose orbit 2 One of the rare examples of a solvable we found in Chapter 8. But notice that we did this by a cunning change of variables that reduced the nonlinear equation (8.3 7 3 The first such calculation, of atmospheri ) for r to the linear equation (8.45) for u. c convection, was made by the meteorologist Edward Lorenz at MIT in 1963, but this work did not attract widespread attention for another decade. For an exhaustive, but very readable, history of chaos theory see Chaos, Making a New Science by James Gleick, Viking-Penguin, New York (1987). 458 Chapter 12 Nonlinear Mechanics and Chaos decision. Many important applications of chaos theory (in astronomy and statistical mechanics, for instance) concern nondissipative systems, as did the pioneering work of Poincare. Nevertheless, it is my experience that the most accessible topics for the beginner concern dissipative systems, and, to keep the length of this chapter reasonably finite, I decided to treat only these. Please see this chapter as a sampler of the good things in chaos theory; it certainly makes no claim to tell the whole story. This chapter is very different from all the other chapters of this book. The theory of chaos is new and not at all elementary. (And parts of the theory have yet to be discovered!) It requires a much deeper understanding of differential equations than I am assuming in this book, and a proper exposition of chaos theory requires a whole book rather than one chapter. 1 Therefore, I shall restrict myself here to simply describing the fascinating main properties of chaotic motion, without much attempt to prove that the motion is as I claim. This is actually a reasonably satisfactory situation. Before you try to read any of the more advanced books, it is almost certainly a good thing to have some idea of what chaos involves and some familiarity with the tools used to describe it, and these are what I hope to communicate. 12.1 Linearity and Nonlinearity For a system to exhibit chaos its equations of motion must be nonlinear. We have noted examples of linear and nonlinear equations from time to time in this book, but let us review the two concepts now. A differential equation is linear if it involves the dependent variable or variables and their derivatives only linearly. The equation of motion of a cart (mass m) on a spring (force constant k), mx = \u2014kx, (12.1) is a linear differential equation for the cart's position x. Similarly, the equations of motion for the two carts discussed in Chapter 11 [Equations (11.2) for example] are linear equations for the two carts' positions x i and x2. If we apply a driving force F(t) to the cart of Equation (12.1), the resulting equation, mi = \u2014kx F(t), (12.2) is still linear [though no longer homogeneous, since the \"inhomogeneous\" term F(t) does not involve the dependent variable x at all]. By contrast, the equation of motion for a simple pendulum (mass m, length L) is I\u00a2 = F or = \u2014mgL sin 0, (12.3) which is a nonlinear equation for 0, since sin 0 is not linear in 0. (If the oscillations are small, then sin 0 ti 0, and the equation is well approximated by a linear equation; 1 Several such books exist, of which my favorite is Nonlinear Dynamics and Chaos by Steven H. Strogatz, Addison-Wesley, Reading, MA (1994), but be warned, it takes eight chapters of mathematical preliminaries to get to the chaos. Section 12.1 Linearity and Nonlinearity 459 in general, however, the equation for the simple pendulum is definitely nonlinear.) Another example is the equation of motion of a single planet in the field of the sun, mr = \u2014GmMilr 2, (12.4) which is a nonlinear equation for the variables r = (x, y, z) because the force term is nonlinear in x, y, and z. These two examples show that nonlinear equations are not especially unusual. On the contrary, many perfectly everyday systems have equations of motion that are nonlinear. So far in this book the main difference between linear and nonlinear differential equations has been that the former have been", " easily solved analytically, whereas most of the latter have been impossible to solve analytically. In fact, our experience in this regard reflects the true state of affairs: Almost all of the linear equations of mechanics are analytically solvable, and almost none of the nonlinear ones are a This circum- stance is largely to blame for the failure until recently of scientists to recognize that chaos is an important and widespread phenomenon. Because nonlinear equations are so intractable, textbooks always focussed on linear problems. When nonlinear prob- lems had to be addressed, they were often solved using approximations that reduced then to linear problems. In this way, the astonishingly rich variety of complications that occur for nonlinear systems went almost completely unrecognized. The first per- son to notice some of the symptoms of chaos was the French mathematician Poincare (1854-1912) in his studies of the gravitational three-body problem \u2014 the motion of three bodies (such as the sun, earth, and moon) interacting via the gravitational force. The equation of motion for this system is nonlinear, like its two-body counterpart (12.4), and Poincare observed that it exhibits the phenomenon now called sensitivity to initial conditions that is one of the characteristics of chaotic motion, as we shall see. That Poincare's observation of chaos went nearly unnoticed by physicists until the 1970s is probably due to several factors. The discoveries of relativity (1905) and then of quantum mechanics (around 1925) diverted most physicists' attention away from classical mechanics. And the difficulty of solving nonlinear equations without the aid of computers certainly discouraged the pursuit of nonlinear problems. In any case it was only in the 1970s that computer solutions of various nonlinear problems 3 drew the attention of significant numbers of scientists (physicists and many others) to the phenomenon that we now call chaos. Nonlinearity is essential for chaos \u2014 if a system's equations of motion are linear, it cannot exhibit chaos. But nonlinearity does not guarantee chaos. For example, the equation (12.3) for a simple pendulum is nonlinear, but even when the amplitude is large (and the linear approximation is definitely not good) the simple pendulum never 2 One of the rare examples of a solvable nonlinear equation is (12.4) for a planet, whose orbit we found in Chapter R. But notice that we did this by a cunning change of variables that reduced the nonlinear equation (8.37) for r to the linear equation (8.45) for u. 3 The first such calculation, of atmospheric convection, was made by the meteorologist Edward Lorenz at MIT in 1963, but this work did not attract widespread attention for another decade. For an exhaustive, but very readable, history of chaos theory see Chaos, Making a New Science by James Gleick, Viking-Penguin, New York (1987). 460 Chapter 12 Nonlinear Mechanics and Chaos exhibits chaos. On the other hand, if we add in a damping force \u2014by = \u2014b14 and a driving force F(t), (12.3) becomes the equation of the driven, damped pendulum: = \u2014mgL sin 4) \u2014 bL24) LF(t) (12.5) and this equation does lead to chaos for some values of the parameters. Loosely speaking, the requirement for chaos is that the equations of motion be nonlinear and somewhat complicated. Equation (12.3) for the simple pendulum is not sufficiently complicated, but Equation (12.5) for the driven damped pendulum is. Unfortunately, a discussion of precisely what is \"sufficiently complicated\" to produce chaos would be well beyond the scope of this book. 4 Another relatively simple example of a nonlinear system that exhibits chaos is the double pendulum of Section 11.4. In the small angle approximation, the equations of motion for the double pendulum are linear equations for the two angles 0 1 and 02 [see (11.41) and (11.42)], but in general they are nonlinear (see Problem 11.15), and they are sufficiently complicated to produce chaos. The driven damped pendulum and the double pendulum are two of the simplest mechanical systems that exhibit chaos. The driven damped pendulum has just one degree of freedom (one coordinate, 0, needed to specify its configuration), whereas the double pendulum has two (two coordinates, 0 1 and 02, needed). For this reason the driven damped pendulum is the simpler one to analyze and will be the main focus of our discussion here. What's Special about Nonlinearity? In the enormous set of all possible differential equations, the linear equations form a miniscule subset, with many simple properties that are not shared by the general non- linear equation. Thus it is really the linear equations which are \"special.\" Nevertheless, for the reasons already mentioned, many physicists are much more conversant with the linear case, and", " they are sometimes tempted to assume that familiar properties of linear equations will carry over to nonlinear equations. This dangerous assumption is frequently wrong. In particular, the main message of this chapter is that chaos, which never appears in linear systems, is a common occurrence in nonlinear systems. Unfor- tunately, the underlying theory of this particular difference is beyond the scope of this book, and we shall have to be content with seeing some simple examples of chaotic motion, without a detailed understanding of why chaos occurs. Here, I would like to mention just one huge difference between linear and nonlinear equations that high- 4 For the record, the criterion is this: As we shall see in Chapter 13, a set of second-order differential equations (like Newton's second law) for n variables can usually be rewritten as a set of first-order equations for N variables,, N where N > n, with the general form 6 =, for i = 1,, N. For instance, if we write (1) = w, the one equation (12.3) for the angle of the simple pendulum becomes two first-order equations, one for and one for w, namely (%) = w and ci) = (g I L) sin (/). When the right-hand sides of these equations are independent of t (as they are here) the equations are said to be autonomous. For a dissipative system to exhibit chaos, its equations of motion, when put in this standard autonomous form, must be nonlinear and have N variables with N > 3. Nondissipative systems need nonlinearity and N > 4. lights the importance of not letting the linear prejudice that many of us share mislead us in our study of nonlinear equations. Section 12.1 Linearity and Nonlinearity 461 Nonlinear Equations Don't Obey the Superposition Principle We saw in Chapter 5 that linear homogeneous equations satisfy the superposition principle \u2014 that any linear combination of solutions gives us another solution. We have used this result several times, particularly in Chapters 5 and 11, but let me refresh your memory with the example of a second-order equation of the form p(t)1. (t) + q (t)X (t) + r (t)x (t) = 0, (12.6) where x (t) is the unknown and p(t), q (t), and r (t) are known fixed functions. An example of such an equation is (12.1) for a cart on a spring.] Notice first that, because every term in this equation is linear in x(t) (or its derivatives), we can multiply through by any constant a and see at once that if x (t) is a solution, then so is ax (t). Second, if xi (t) and x 2 (t) are both solutions, then we can add the two corresponding equations, one for x i (t) and one for x2 (t), and conclude that x i (t) x2 (t) is also a solution. Thus any linear combination x(t) = aix i (t) a2x 2 (t) is also a solution of (12.6) \u2014 the result called the superposition principle. On the other hand, it is easy to see that neither of the arguments just given works if the equation is nonlinear. [Make sure you can see this. Suppose, for example, the last term in (12.6) was r (t),/ x(t); see Problem 12.31 Therefore, the superposition principle does not apply to nonlinear equations. An important consequence of the superposition principle that we have used repeat- edly is this : To find all the solutions of (12.6) we have only to find two independent solutions x i (t) and x 2 (t); then every solution can be expressed as a linear combination of x (t) and x2 (t). More generally, to find all the solutions of an nth order homoge- neous linear differential equation, we have only to find n independent solutions and then every solution can be expressed as a linear combination of these n solutions. Since the superposition principle does not apply to nonlinear equations, this dramatic simplification does not apply to nonlinear equations. There is a corresponding situation for inhomogeneous equations, such as (12.2) and (12.5), again as we saw in Chapter 5. If x p (t) is any one particular solution of a linear nth order inhomogeneous equation, then every solution can be written as xp (t) plus a linear combination of n independent solutions of the corresponding homogeneous equation. For nonlinear equations, there is no corresponding result (Problem 12.4). Thus every solution of any nth order linear equation (homogeneous or inhomogeneous) can be expressed simply in terms of n independent functions, but for nonlinear equations there", " is no such simple expression. With these general observations on nonlinear equations, let us take up the one nonlinear equation that we shall discuss in detail, Equation (12.5) for a driven damped 462 Chapter 12 Nonlinear Mechanics and Chaos pendulum. I shall first describe some features of its motion that we would expect (or at least that are not wholly unexpected), and I shall then take up the surprising features associated with the pendulum's chaotic motion. 12.2 The Driven Damped Pendulum DDP The equation of motion for the driven damped pendulum (or DDP) was given in (12.5). Since this equation is going to occupy us for several sections to come, I would like to make quite sure that you are clear where it came from and to tidy it up. The pendulum is sketched in Figure 12.1. The equation of motion is just I = F, where I is the moment of inertia and F is the net torque about the pivot. In this case I = m L 2, and the torque arises from the three forces shown in Figure 12.1. The resistive force has magnitude by and hence exerts a torque \u2014Lbv = \u2014bL 2. The torque of the weight is \u2014mgL sin 0, and that of the driving force is LF(t). Thus the equation of motion \u2014bLVP \u2014 mgL sin LF(t) (12.7) Throughout this chapter I shall assume that the driving force F(t) is sinusoidal, F(t) = Fo cos(cot) (12.8) where Fo is the drive amplitude (the amplitude of the driving force) and co the drive frequency. As I argued in Chapter 5, several real and interesting driving forces approximate this sinusoidal form quite closely, and it has proved possible to reproduce such sinusoidal forces with remarkable precision for experiments on chaos. Substituting into (12.7) and reorganizing a little, we find that b m g \u00b1 \u20144)\u00b1 L \u2014 sin 4) = cos wt. F0 m L (12.9) F is exactly as in (12.5). specifically that F(t) Figure 12.1 The three important forces on the driven damped pen- dulum are the resistive force with magnitude by, the weight mg, and the driving force F(t). (There is also a reaction force from the pivot at the top, but this contributes nothing to the torque.) Section 12.3 Some Expected Features of the DDP 463 In this equation, you will recognize the coefficient blm as the constant that we renamed as 2,8 in Chapter 5, b \u2014 w 2 0 - L where 13 was called the damping constant. Similarly the coefficient g I L is just wo2, where wo is the natural frequency of the pendulum. Finally, the coefficient Fo lmL must have the dimensions of (time) -2 ; that is, F.ImL has the same units as w 02. It is convenient to rewrite this coefficient as F.ImL = ycq. That is, we introduce a dimensionless parameter F. F. mLw 2 mg 0 (12.10) which I shall call the drive strength and is just the ratio of the drive amplitude F. to the weight mg. This parameter y is a dimensionless measure of the strength of the driving force. When y < 1, the drive force is less than the weight and we would expect it to produce a relatively small motion. (For instance, the drive force is insufficient to hold the pendulum out at 0 = 90\u00b0.) Conversely, if y > 1, the drive force exceeds the pendulum's weight, and we should anticipate that it will produce larger scale motions (for instance, motion in which the pendulum is pushed all the way over the top at = 7). Making all these substitutions, we get our final form of the equation of motion (12.9) for a driven damped pendulum This is the equation whose solutions we shall be studying for the next several sections. 12.3 Some Expected Features of the DDP Properties of the Linear Oscillator To appreciate the extraordinary richness of the chaotic motion of our driven damped pendulum, we must first review what sort of behavior we might expect, based on our experiences with linear oscillators. Specifically, if we release the pendulum near the equilibrium position 0 = 0 with a small initial velocity and if the drive strength is small, y \u00ab 1, we would expect cb to remain small at all times. Thus we should be 464 Chapter 12 Nonlinear Mechanics and Chaos y = 0.2 A AAAAAA V3 V V V V 1 \u20140.3 Figure 12.2 The motion of a DDP with a relatively weak drive strength of y = 0.2. The drive period was chosen to be r = 1, so that the hori- zontal axis gives the", " time in units of the drive period. You can see clearly that after about two cycles the motion has settled down to a perfectly sinusoidal motion with period equal to the drive period. able to approximate the term sin 0 in (12.11) as 0, and the equation of motion would become the linear equation 2/3c/5 + 0)020 = ycoo2 cos cot (12.12) which has exactly the form of (5.57) for the linear oscillator of Chapter 5. Thus the \"expected\" behavior of the driven damped pendulum, at least for a weak enough driving force, is just the behavior described in Section 5.5. This behavior can be quickly summarized: The initial behavior of the pendulum depends on the initial conditions, but any differences (or \"transients\") due to the initial conditions die out rapidly, and the motion approaches a unique \"attractor,\" in which the pendulum oscillates sinusoidally with exactly the frequency of the driving force: (t) = A cos(wt \u2014 6). (12.13) These predictions are nicely illustrated in Figure 12.2, which shows the actual motion of the driven damped pendulum for a fairly weak drive strength of y = 0.2. [Since the exact equation of motion (12.11) cannot be solved analytically, this and all subsequent plots of the motion of the DDP were made from numerical solutions of (12.11).5 ] The drive frequency was chosen to be w = 27, so that the drive period is r = /co = 1. This means that the horizontal axis shows time in units of the drive period.The natural frequency was chosen to be coo = 1.5w, so that the system is fairly close to resonance, since this is where chaotic motion is usually easiest to find. 6 The most striking feature of this plot is that after about two cycles the motion has settled down to a purely sinusoidal motion with exactly the period of the driving force, r = 1. The initial conditions chosen for this plot were that 0 = = 0 at t = 0. It is a fact (though not one that our one plot can show) that, whatever initial conditions we were to choose, the motion of a linear oscillator would always approach the same unique attractor as the initial transients die out. 5 All of these plots were made using Mathematica's numerical solver NDSolve. For many plots the default precision of 15 digits was more than sufficient, but where there was any reason for doubt, the precision was increased in integer steps until two successive calculations were indistinguishable. 6 The other parameters used were as follows: Damping constant 13 = coo/4, and initial conditions = =0 at t =. Section 12.3 Some Expected Features of the DDP 465 To summarize, for a linear damped oscillator, with a sinusoidal driving force: (1) There is a unique attractor which the motion approaches, irrespective of the chosen initial conditions. (2) The motion of this attractor is itself sinusoidal with frequency exactly equal to the drive frequency. Nearly Linear Oscillations of the DDP Let us now imagine increasing the drive strength so that the amplitude of oscillation increases to a value where the approximation is no longer satisfactory. As long as the amplitude is not too large, we would expect to get a satisfactory approximation by including just one more term in the Taylor series for sin 0 and writing sin 0 0 sin 0 \u2014 If we use this approximation in the exact equation of motion (12.11), we get the approximate equation + 2/3(6 + wo (0 \u2014 603) = y co 02 cos wt. (12.14) To the extent that the new nonlinear term involving 0 3 is small, we can anticipate that the solution of this equation will still be reasonably approximated (once the transients have died out) by an expression of the same form as before, (t) A cos(wt \u2014 8). When this is put into (12.14), the small term involving 0 3 contributes a term propor- tional to cos 3 (wt \u2014 8). Since COS3 1 X = 4- (cos 3x + 3 cos x) (12.15) (see Problem 12.5) there is now a small term on the left side of (12.14) proportional to cos 3(wt \u2014 8). Since the right side contains no terms with this time dependence, it follows that at least one of the terms on the left (0, (b, or and in fact all three) must. That is, a more exact expression for 0(t) must have the form 0(t) = A cos(wt \u2014 6) B cos3(cot \u2014 6), (12.16) with B much smaller than A. Therefore, we must anticipate that, as we increase the driving force and the amplitude increases, the solution", " will pick up a small term that oscillates with frequency 3w. We can repeat this argument: If we substitute the improved solution (12.16) back into (12.14), then the term in 0 3 will give even smaller terms of the form cos n (wt \u2014 8), with n an integer greater than 3. Therefore we must expect smaller corrections to (12.16) with frequencies nco, with n equal to various integers. Any term oscillating 466 Chapter 12 Nonlinear Mechanics and Chaos 7i12 \u2014712 (a) (b) Figure 12.3 (a) The motion of a DDP with drive strength y = 0.9 (and all other parameters the same as in Figure 12.2). After two or three drive cycles, the motion settles down to a regular oscillation, which has period equal to the drive period and looks at least approximately sinusoidal. (b) The solid curve is an enlargement of a single cycle of part (a), from t = 5 to 6. The dashed curve, which is a pure cosine with the same frequency, phase, and slope where they cross the axis, shows clearly that the actual motion is no longer perfectly sinusoidal; it is appreciably flatter at the extremes. with frequency equal to an integer multiple of co is called a harmonic of the drive frequency. Thus our conclusion is that, as the drive strength is increased and the nonlinearity becomes more important, the pendulum's motion will pick up various harmonics of the drive frequency co, the most important being the n = 3 harmonic already included in (12.16). The nth harmonic, with frequency nco, is periodic with period r\u201e = 27r/nco = r / n, where r = 27r/co is the drive period. Thus in one drive period, the nth harmonic repeats itself n times. In particular, in one drive period, every harmonic will have cycled back to its original value, and a motion that is made up of various harmonics will still be periodic with the same period as the driving force. The main difference between the motion implied by (12.16) (possibly with other harmonics included) and the motion (12.13) of the linear oscillator is that, with its extra term (or terms), (12.16) is no longer given by a single cosine function. We should be able to see this in a graph of 4) (t) against t, which must deviate slightly from a pure sinusoidal shape. However, in the regime we are considering, the coefficient B in (12.16) and the coefficients of any higher harmonics are all much smaller than A, and the difference between the actual motion and a pure cosine is quite hard to see. Figure 12.3(a) shows the motion of a damped driven pendulum with drive strength y = 0.9 (just below our rough boundary at y = 1 between weak and strong drive strengths). Just as in Figure 12.2, the motion has quickly settled down to steady oscillation with exactly the period of the driving force. At first glance the curve (after about t = 2) appears to be a pure cosine, but on closer examination you may be able to convince yourself that it is a little too flat at the crests and troughs. Figure 12.3(b) is an enlargement of one cycle of the motion (solid curve), with a superposed pure cosine with the same period and phase (dashed curve). This comparison shows clearly that the actual motion is no longer a single pure cosine? 7 The flattened shape at the extremes is nicely consistent with (12.16): Provided B and A have opposite signs, the second term in (12.16) reduces 40 (t) at the crests and troughs and increases it near where it crosses the axis. The behavior is also easy to understand physically: At the extremes, the restoring torque of gravity (mgL sin 0) is weaker than the linear approximation (mgL0), and the actual motion is less sharply curved. Section 12.4 The DDP: Approach to Chaos 467 The behavior of the DDP in the linear and nearly linear regimes can be quickly summarized: As the initial transients die out, the motion approaches a unique attractor, in which the pendulum oscillates with the same period as the driver. In the linear regime (driving strength y << 1) this limiting motion is given by a simple cosine function with frequency equal to the drive frequency (.0. In the not-quite-linear regime (y somewhat larger, but definitely not much greater than 1), the limiting motion is still periodic, with the same period, but it picks up some harmonics and is a sum of cosines with frequencies no) as in (12.16). As we shall see in the next section, we have only to increase the drive strength a little above y = 1, to encounter", " some dramatically different behavior. 12.4 The DDP: Approach to Chaos Let us now continue increasing the strength of our DDP's driver. Figure 12.4 shows the motion (0 against t) for all the same parameters and initial conditions as in the last two figures, except that I have increased the drive strength to y = 1.06, just a little above the rough boundary at y = 1 between weak and strong driving. The most striking thing about this plot is the dramatic oscillation of the initial transient motion. In the first three drive cycles, the pendulum swings from 0 = 0 to nearly 57 ; that is, it makes nearly two and a half counterclockwise rotations. In the next two cycles it swings back nearly to 0 = n and eventually settles down to more-or-less sinusoidal oscillations around 0 ti 27. (The position 0 = 27 is, of course, the same as 0 = 0, but the statement that 4) eventually centers on 27 is nonetheless meaningful, indicating that the pendulum has made one net counterclockwise rotation since t = 0.) It is impossible to be completely sure, based on a graph such as Figure 12.4, that the eventual motion really is exactly periodic. One way to examine this question more closely is to print out the positions 0 (t) at successive one-cycle intervals, t = to, to + 1, to + 2, to + 3, The larger we choose t o, the more closely these positions should agree with one another (if, the eventual motion is indeed periodic). For example, starting at t = 34 the positions (t) (as given by the same numerical solution on which Figure 12.4 was based) are Figure 12.4 The motion of a DDP with drive strength y = 1.06. The initial, rather wild, transients die out after about 9 drive cycles, and the motion settles down to an attractor with the same period as the driver. 468 Chapter 12 Nonlinear Mechanics and Chaos with all subsequent values equal to 6.0366. Evidently, to five significant figures, the motion has settled down to be perfectly periodic after 35 drive cycles. 8 Of course, it is possible that the motion does something nonperiodic in between the integer times shown, and certainly no one would accept our data as mathematical proof. Nevertheless, the evidence is overwhelming that for y = 1.06 (and with the initial conditions used) 0 (t) does approach an attractor that is periodic with the same period as the driver. In this respect, the motion shown for y = 1.06 is not much different from that for y = 0.9 as shown in Figure 12.3. However, the dramatic initial swings in Figure 12.4 are harbingers of interesting developments to come. Period Two Figure 12.5(a) corresponds exactly to the previous figure except that I have now increased the drive strength to y = 1.073. Again the most obvious feature is the wild initial oscillation, which now lasts for nearly 20 drive cycles before the motion settles down to steady oscillations that are at least approximately sinusoidal. However, if you look closely at these oscillations, you will notice that the crests and troughs (especially the troughs) are not all of the same height. Figure 12.5(b) is a many-fold enlargement of the same troughs between t = 20 and 30, and you can see clearly that the troughs alternate between two distinct heights. You might wonder if this alternation is itself a transient that will disappear after enough cycles, but this is not in fact so. A plot of the oscillations for 990 < t < 1000 looks exactly the same as that for 20 < t < 30. Another way to show this is to print out the numerical values of 0 (t) at one-cycle intervals. Starting at t = 30, this yields (t) 34 6.0366 35 6.0367 36 6.0366 37 38 6.0366 6.0366 39 6.0366 t (t) 30 \u20146.6438 31 \u20146.4090 32 \u20146.6438 33 \u20146.4090 34 \u20146.6438 35 \u20146.4090 8 Naturally, the motion takes longer to settle down to a constant if we insist on more significant figures. For example, it is not until 46 cycles have passed that 0 (t) starts repeating to six significant figures, after which 0 (t) = 6.03662 for t = 46, 47, 48, Section 12.4 The DDP: Approach to Chaos 469 20 25 30 vUUUJJ (a) \u20148.0 - \u20148.5 - (b) 20 22 24 26 28 30 Figure 12.5 (a) The first 30 cycles of a DDP with drive strength y = 1.073. The wild initial oscillations persist for nearly 20 drive cycles, after which the motion", " settles down to an attractor that is approximately sinusoidal. However, closer inspection shows that the crests and troughs of this attractor are not all of the same height. (b) An enlargement of the attractor for 20 < t < 30 showing just the troughs of part (a). The troughs alternate in height, repeating themselves once every two drive cycles. a pattern that repeats precisely forever. Evidently, by t = 30, the motion has settled down so that (t) has the value \u20146.6438 (to 5 significant figures) for all even values of t and has the distinct value \u20146.4090 for all odd values of t. This behavior means that the motion no longer repeats itself with the frequency of the driver. Rather, the motion is periodic with period equal to twice the drive period, and we say that the motion has period two. (In our units this last statement is literally true; in general it means that the period of the motion is two times the drive period.) It is important to recognise that this development is quite different from the appearance of the harmonics that we noticed in the case of nearly linear motion. A harmonic has frequency nco, an integer multiple of the drive frequency, and hence period equal to an integer submultiple of the drive period. What we have now found has period equal to an integer multiple of the drive period, and hence frequency co/n, which can be described as a subharmonic of the drive frequency. Looking at Figure 12.5(a) you can see that the motion is still very nearly sinusoidal with the period of the driver (period 1). Thus the dominant term in 4 (t) is still of the form A cos(a)t \u2014 8); nevertheless, (t) definitely contains a small subharmonic term with period 2. Period Three Although the attractor shown in Figure 12.5 has period two, the dominant behavior is still clearly of period one; that is, the new n = 2 subharmonic contributes only a small amount to the solution. If we increase the drive strength a little further, we find an 470 Chapter 12 Nonlinear Mechanics and Chaos (i) Figure 12.6 The motion of a DDP with drive strength y = 1.077. After little more than two drive cycles, the motion has settled down to a periodic attractor which repeats itself every three drive cycles (for example, the troughs come just before t = 5, 8, 11, 14, and so on); therefore, the attractor has period three. attractor for which the subharmonic is the dominant term. Figure 12.6 shows the first 15 cycles of the motion of our DDP with the drive strength increased to y = 1.077 (and all other parameters the same as before). In this case it is obvious at just a glance that the motion settles down to an attractor that repeats itself every three drive cycles and hence has period three. While it would be hard to question that this graph has period three, we can reinforce the conclusion by looking at the values of (t) at one-cycle intervals. Starting from t = 30 these are as follows: (t) 13.81225 7.75854 6.87265 13.81225 7.75854 6.87265 13.81225 7.75854 6.87265 30 31 32 33 34 35 36 37 38 with exactly the same pattern, repeating once every three drive cycles, continuing indefinitely. Evidently the solution has picked up a period-three term, and this term dominates the solution. More than One Attractor For a linear oscillator, with a given set of parameters, we proved in Section 5.5 that there is a unique attractor; that is, whatever the initial values of q5 and the eventual motion will always be the same, once the transients have died out. For a nonlinear oscillator, this is not the case, and the DDP with the drive strength y = 1.077 of Figure 12.6 furnishes a clear example. In Figure 12.7, I have shown the motion for a DDP with the same parameters as in Figure 12.6 (including the same drive strength), but with two different sets of initial conditions. The dashed curve is the same solution as Section 12.4 The DDP: Approach to Chaos 471 Figure 12.7 Two solutions for the same DDP, with the same drive strengths, but different initial conditions [0 (0) = (0) = 0 for the dashed curve, but 0 (0) = \u20147/2 and (1)(0) = 0 for the solid curve]. Even after the transients have died out, the two motions are totally different. in Figure 12.6, with the same initial conditions as we have used for every graph up to now, q5 (0) = (1)(0) = 0. The solid curve shows the motion of the", " exact same DDP, also with 5(0) = 0, but with 0 (0) = \u20147r/2; that is, for the solid curve, the pendulum was released from 90\u00b0 on the left. As you can clearly see, the two attractors (the curves to which the actual motions converge as the initial transients die out) are totally different. For the dashed curve, the attractor has period three, for the solid curve the eventual period is (as you can see if you look closely) actually two, with alternate troughs (and alternate crests) having slightly different heights. Evidently, for a nonlinear oscillator, different initial conditions can lead to totally different attractors. A Period-Doubling Cascade Having recognized that different initial conditions can lead to different attractors, we must anticipate that the evolution of the oscillations as we vary y may depend on the initial conditions that we choose. In the sequence of Figures 12.2 through 12.6, I used the initial conditions 0 (0) = 0 and (0) = 0 for all five pictures. It turns out that the new initial conditions 0 (0) = \u20147r/2 and QS (0) = 0, introduced in Figure 12.7, lead to a quite different and very interesting evolution. In Figure 12.8, I have shown the motion of the DDP for four successively larger values of y, all with these new initial conditions. The left-hand pictures show 0 (t) as a function of t for the first ten cycles of the driver. The first graph is for y = 1.06, the same value as was used in Figure 12.4, and, as in Figure 12.4, the motion settles down to a steady oscillation with period equal to the drive period; that is, the attractor has period one. To confirm this conclusion, the right-hand picture shows the same motion, but for 28 < t < 40 (by which time any initial transients have completely disappeared at the scale shown), with the vertical scale magnified to show clearly that successive oscillations are all of equal amplitude. For the second pair of graphs, the drive strength was increased to y = 1.078. At first glance, the motion looks very similar to that for y = 1.06, but on closer inspection you can see that the maxima and minima are not all of the same height. This is very visible 472 Chapter 12 Nonlinear Mechanics and Chaos y= 1.06 2_ Period 0 1 -2- y=1.078 2_ Period 0 2 -2- y= 1.081 2 _ Period 0 4 -2- y= 1.0826 Period 8 A(1 10 t -2- -2.5 IV t t 28 30 32 34 36 38 40 VII V -2- t -2.5 2 28 30 32 34 36 38 40 28 30 32 34 36 38 40 -2- -2.5 -2 - \u20142.5 1 8 28 30 32 34 36 38 40 Figure 12.8 A period-doubling cascade. The left-hand pictures show the first ten drive cycles of a DDP with successively larger drive strengths, as indicated on the left. All other parameters, including the initial conditions 4 (0) = \u2014n/2 and 0(0) = 0, are the same in all pictures. In each picture on the right I have enlarged the bottom of the corresponding motion on the left, to show more clearly the differences in extent of successive oscillations; these enlargements show 12 drive cycles, starting from t = 28, by which time the motion has settled down to a perfectly periodic attractor (at least at the scale shown). Each double-headed arrow shows one complete cycle of the corresponding motion; the periods of the four attractors are clearly seen to be 1, 2, 4, and 8, as indicated. in the enlargement on the right where you can see easily that the minima alternate between two distinct, fixed heights, so that the attractor now has period two. With y = 1.081, as in the third pair, the graph on the left again looks pretty much like its two predecessors, and it is hard to be sure just what is going on. One of the reasons is that we can't be sure that ten drive cycles (the number shown on the left) are long enough for all transients to have disappeared, but in the right-hand enlargement it is quite clear that the minima are alternating among four different values. That is, the period has doubled again to period four. In the last pair of pictures, with y = 1.0826, it is even harder to be sure what is happening in the left-hand picture, but the enlargement on the right makes clear that the motion eventually repeats once every eight drive cycles. That is, the attractor has period eight. The period-doubling cascade", " seen in these four pairs of pictures Section 12.4 The DDP: Approach to Chaos 473 III 1 1IIIIIIIIIIIIIIIIIIIIII 111111 A AOA AAAAA AAA MAP IIIIIIIIIII IIIIIIII I IIIIIIIIIIIII R Re 3.47 3.52 A 3.62 3.65 t AA r dl T(s) 0 50 100 150 200 Figure 12.9 A period-doubling cascade in convection of mercury in a small convection cell. The plots show the temperature at one fixed point in the cell as a function of time, for four successively larger temperature gradients as given by the parameter RIRc. continues. If we increase the drive strength further, we find motion with period 16, then 32, and so on to infinity. The period-doubling cascade of Figure 12.8 is a very striking phenomenon, but the quantitative differences between the four successive unenlarged graphs are quite small. You might guess that to build a driven damped pendulum sufficiently precise to observe these subtle differences would be very hard, and this is indeed the case. Nev- ertheless, such pendulums have been constructed and have been used to observe all of the effects described in this chapter, with amazing agreement between theory and experiment.9 Perhaps even more remarkable is that the phenomenon of period dou- bling is found in many completely different nonlinear systems \u2014 electrical circuits, chemical reactions, balls bouncing on oscillating surfaces, and many more. In each of these systems, there is a \"control parameter\" that can be varied (the driving strength of a DDP, a voltage in an electrical circuit, a flow rate in a chemical reaction). The 9 For a description of three of the commercially available \"chaotic pendulums\" see J. A. Black- burn and G. L. Baker, \"A Comparison of Commercial Chaotic Pendulums,\" American Journal of Physics, Vol. 66, p. 821, (1998). The Daedalon pendulum described there was used to get the data shown in Figure 12.32. 474 Chapter 12 Nonlinear Mechanics and Chaos behavior of the system is monitored as this parameter is varied, and it is found that the behavior exhibits period-doubling cascades. Figure 12.9 shows a cascade observed by Libchaber et al. 10 in the convection of mercury in a small box whose bottom is maintained at a slightly higher temperature than its top. This temperature difference is the control parameter and is measured by a number R, called the Rayleigh number. When R is very small, heat is conducted up with no convection. Then at a critical tem- perature difference, Rc, steady convection sets in, and, as R is increased still further, the convection becomes oscillatory. These oscillations can be observed by measuring the temperature at any fixed point in the cell, and Figure 12.9 shows four plots of the observed temperature (at one fixed point) against time, for four successively larger values of the control parameter R. The period doublings from 1 to 2, from 2 to 4, and from 4 to 8 are beautifully clear. Not only are period-doubling cascades observed in numerous different systems. In a sense that I shall describe directly, the cascades occur in the same way, a circumstance referred to as \"universality.\" The Feigenbaum Number and Universality Returning to the period-doublings of the DDP, you can see from the values of the drive strength y shown in Figure 12.8 that the doublings occur faster and faster as we increase y. To make this idea quantitative, we need to examine the threshold values of y at which the period actually doubles. For example, looking at the numbers in Figure 12.8, it seems clear that somewhere between y = 1.06 and 1.078, there must be a value yl where the period changes from 1 to 2. Finding where this threshold (or \"bifurcation point\") actually occurs is surprisingly hard, but it turns out that (to 5 significant figures) y l = 1.0663. Similarly, at y2 = 1.0793, the period changes from 2 to 4. If we let yn denote the threshold at which the period changes from 2n -1 to 2\", then the first few thresholds y r, are as shown in Table 12.1. In the last column of the table, I have shown the distances v n yn--1 between successive thresholds, which, as you can see, shrink geometrically, 11 each interval being about one fifth of its predecessor. In the late 1970's, the physicist Mitchell Feigenbaum (born 1944) showed not only that many different nonlinear systems undergo similar period-doubling cascades but that the cascades all show the same geometric acceleration; specifically, the intervals between the thresholds for", " the control parameter (the drive strength, in our case) satisfy 10 Reproduced with permission from A. Libchaber, C. Laroche, and S. Fauve, Journal de Physique-Lettres, vol. 43, p. 211 (1982). 11 A sequence of numbers, a l, a2, is geometric if a n\u00b1i = kan for some fixed number k. If k < 1, the geometric sequence goes to zero as n oo. Section 12.4 The DDP: Approach to Chaos 475 Table 12.1 The first four thresholds yn at which the period of the DDP [with the initial conditions 0 (0) = --z/2 and q (0) = 0] doubles from 1 to 2, 2 to 4, 4 to 8, and 8 to 16. The last column shows the widths of the intervals between successive thresholds. n period Yn interval 1 2 3 1 --> 2 1.0663 2 ---> 4 1.0793 4 ---> 8 1.0821 4 8 --> 16 1.0827 0.0130 0.0028 0.0006 where the constant 3 has the same value = 4.6692016 (12.18) for all such systems and is called the Feigenbaum number. 12 It is the widespread occurrence of period doubling and the fact that 6 has the same value for so many dif- ferent systems that has led to the phenomenon of period doubling being characterized as universal. I have written the Feigenbaum relation (12.17) with an \"approximately equal\" sign, because, strictly speaking, the relation holds only in the limit that n ---> oo. For many systems, however, the relation is a very good approximation for all values of n. (See Problems 12.11 and 12.29.) The Feigenbaum relation (12.17) implies that the intervals between successive thresholds approach zero rapidly, and hence that the thresholds themselves approach a finite limit yc, Yn Ye (as n oo). (12.19) Therefore, the sequence of thresholds yn satisfies Y1 < Y2 < < Yn < < Ye with infinitely many thresholds squeezed in the rapidly narrowing gap between y r, and yc. For our DDP, the limit y, is found to be yc = 1.0829. (12.20) We shall see that beyond the critical value y c, chaos sets in, so the period-doubling cascade is called a route to chaos. However, I should emphasize that there are systems that exhibit chaos without first going through a period-doubling cascade; that is, the period-doubling cascade is just one of several possible routes to chaos. 12 Actually there are two Feigenbaum numbers, and this one is often called Feigenbaum's delta. 476 Chapter 12 Nonlinear Mechanics and Chaos 12.5 Chaos and Sensitivity to Initial Conditions If we increase the drive strength y beyond the critical value y c = 1.0829, then our DDP begins to exhibit the behavior that has come to be called \"chaos.\" Figure 12.10 shows the first thirty drive cycles of the DDP with y = 1.105. The pendulum is obviously \"trying\" to oscillate with the period of the driver. Nevertheless, the actual oscillations wander around erratically and never repeat themselves exactly. Of course you might wonder if I have not given the oscillations time to settle down; perhaps at some later time they would converge to a periodic motion. In fact, however, a graph of any time interval is just as erratic, but never an exact repetition of any other interval. Even though the driving force is perfectly periodic and even after the transients have all died out, the long-term motion is definitely nonperiodic. This erratic, nonperiodic long-term behavior is one of the defining characteristics of chaos. The other defining characteristic is the phenomenon called sensitivity to initial conditions. Sensitivity to Initial Conditions The issue of sensitivity to initial conditions arises in connection with the following questions: Imagine two identical DDP's, with all parameters the same, but launched at t = 0 with slightly different initial conditions. [Perhaps the initial angles 4) (0) differ by a fraction of a degree.] As time goes by, do the motions of the two pendulums remain nearly the same? Do they perhaps get closer to one another? Or do they diverge and become more and more different? To make these questions more precise, let us denote the positions of the two pendulums by 0 1 (0 and 02(0. These two functions satisfy exactly the same equation of motion, but have slightly different initial conditions. If now AO (t) denotes the difference between our two solutions, AO(t) 02(0 - 01(t), (12.21) the issue is the time dependence of AO (t). Does AO (t) stay more-or-", "less constant? Does it decrease as time goes by? Or does it increase? For the linear oscillator discussed in Chapter 5, the answer is that AO (t) goes to zero, since we proved that all solutions of the equation of motion approach the 4) gr- y= 1.105 EI 0 p t Figure 12.10 Chaos. The first 30 drive cycles of the DDP with y = 1.105 are erratic and show no signs of periodicity. In fact the oscillations never do settle down to a regular periodic motion, and this erratic, nonperiodic long-term motion is one of the defining characteristics of chaos. Section 12.5 Chaos and Sensitivity to Initial Conditions 477 same attractor as t oo. Therefore the difference of any two solutions must approach zero. Further, the difference must approach zero exponentially. To see this, recall from Equation (5.67) that any solution has the form (t) = A cos(cot \u2014 6) + Cierlt c2er2t (12.22) where the cosine term is the same for all solutions, whereas the two decaying expo- nential terms have coefficients C 1 and C2 that depend on the initial conditions. This implies that when we take the difference of two solutions the cosine term drops out and we are left with AO (t) = B ier't + B2er2t, (12.23) where the constants B 1 and B2 depend on the two sets of initial conditions. The precise behavior of this difference depends on the relative sizes of the damping constant /3 and the natural frequency w o. In all of the examples so far in this chapter I have chosen /3 = 0.25wo, so that,8 < wo (the situation called underdamping). In this case, we saw in Section 5.4 that the coefficients r 1 and r2 have the form \u2014fi icoi. Some simple algebra then puts (12.23) in the form [compare Equation (5.38)] AO (t) = De - i3t cos(w it \u2014 8). (12.24) That is, AO (t) is the exponential e - Pt times an oscillatory cosine. There is a problem in trying to display the time dependence of a function like (12.24). The exponential factor decays so fast that one cannot easily show its range of values on a conventional graph. For example, with the values I have been using, = 0.25coo = 0.757 = 2.356, after just one drive cycle (t = 1) the exponential factor is e \u2014Pt = e -2.356 0.09, and AO (t) has diminished by an order of magnitude. If we wanted to plot AO (t) against t over 10 cycles, say, then AO (t) would shrink by about ten orders of magnitude \u2014 a range that cannot possibly be shown on a simple linear plot of AO (t) against t. As you probably know, the solution to this problem is to make a logarithmic plot; that is, we plot the log of A0 (t) against t. Actually since AO (t) can be negative, we must plot 1n1A0(t)1 against t. According to (12.24) this should obey In 1A0 = In D \u2014 fit + lnIcos(w it \u2014 8)1. (12.25) The first term on the right is a constant, and the second is linear in t with slope \u2014 /3. The third is a little complicated: Since 1 cos(w it \u2014 8)1 oscillates between 1 and 0, its natural log oscillates between 0 and \u2014oo. Thus a graph of In1A0(t)1 against t should bounce up and down (going to \u2014oo each time the cosine term vanishes), underneath an envelope that decreases linearly with slope \u2014fi. This is clearly visible in Figure 12.11, which shows a plot of log 1A.0 (t)I against t for the relatively weak driving strength y = 0.1, for which the linear approximation is certainly good. [I plotted the log to base 10 rather than the natural log, because the former is easier to interpret on a graph. Since log (x) is just the constant ln(10) times ln(x), this changes none of our qualitative predictions.] To plot this graph, I gave the first pendulum the same initial conditions as in Figures 12.8 and 12.10; the second pendulum was released with its 478 Chapter 12 Nonlinear Mechanics and Chaos y = 0.1 t 10-2 10-4 10_S 10-1\u00b0 10-12 \u2014 2 \u20144 \u20148 \u201410 \u201412 loglA0(01 \u20146 104 AcA(t)", " Figure 12.11 Logarithmic plot of AO (t), the separation of two identical DDP's, with a weak drive strength y = 0.1, that were released with initial positions that differ by 0.1 radians (or about 6\u00b0). The vertical axis on the left shows log I A0 (t) I, while that on the right shows I A0 (t) itself. The picture shows clearly that the maxima of log I AO (t) I decrease linearly, and hence that AO (t) decays exponentially. initial position 0.1 radians lower, so that the initial difference was AO (0) = 0.1 rad, or about 6 degrees. The most important feature of the plot is that the successive maxima of log I AO (t) I decrease perfectly linearly, confirming that AO (t) decays exponentially, dropping by about 10 orders of magnitude in the first ten drive cycles (as you can easily check from the graph). 13 So far we have proved that, in the linear regime, the separation AO (t) of two identical DDP's, launched with different initial conditions, decreases exponentially. This has an important practical consequence: In practice, we cannot possibly know the initial conditions of any system exactly. Therefore, when we try to predict the future behavior of our DDP we must recognize that the initial conditions we use may differ a little from the true initial conditions. This means that our predicted motion for t > 0 may differ from the true motion. But because AO (t) goes to zero exponentially, we can be sure that our error will never be worse than the initial error and will, in fact, rapidly approach zero. We can say that the linear oscillator is insensitive to its initial conditions. To achieve any prescribed accuracy in our predictions, we have only to ascertain the initial condition to this same accuracy. What happens as we increase the drive strength y out of the linear regime? Naturally, we can no longer depend on our proofs for the linear oscillator. However, we can reasonably expect that the difference AO (t) will continue to decay exponentially for at least some range of drive strengths. The question is \"How large is this range?\" and the answer is quite surprising: Provided the difference in initial conditions is sufficiently small, the difference AO (t) continues to decay exponentially for all values of y up to the critical value yc at which chaos sets in. For example, if y = 1.07 we 13 A second noteworthy feature is that the points where I z\\ (t) I vanishes (and, hence, log I AO (t) goes to \u2014oo) show up on the logarithmic plot as sharp downward spikes. This is because the plot- ting program can only sample a finite number of points and naturally misses the points where log 140 (t) I = \u2014oo. Instead it can only detect that there are points where log I AO WI has a pre- cipitous minimum, and this is what it shows. Section 12.5 Chaos and Sensitivity to Initial Conditions 479 y = 1.07 t 10 15 20 25 30 35 logiLlc/,(01 0 \u20144 - \u20146 - \u20148 - 10-8 Figure 12.12 Logarithmic plot of AO (t), the separation of two identical DDP's, with drive strength y = 1.07, that were released with initial posi- tions that differ by 0.1 radians (or about 6\u00b0). For the first 15 or so drive cycles, (t) holds fairly constant in amplitude, but then the maxima of log I (t) decrease linearly, implying that AO (t) decays exponentially. know that the motion has period 2 (for the initial conditions of Figures 12.8 through 12.11), and the motion is distinctly nonlinear; nonetheless, the difference AO (t) still decays exponentially, as is clearly visible in Figure 12.12, which shows the difference (t) for two solutions with the same initial conditions as in Figure 12.11. In this case, AO (t) remains pretty well constant in amplitude for the first 15 or 20 drive cycles, but then the crests of log I A (t) I drop perfectly linearly, indicating that AO (t) decays exponentially as t co. Notice, however, that the exponential decay is considerably slower than in the linear case: Here the amplitude drops by about 4 orders of magnitude in the last 25 cycles; in the linear case of Figure 12.11, it dropped by 10 orders of magnitude in just 10 cycles. Nevertheless, the main point is that AO (t) goes to zero exponentially, and, as in the linear regime, we can predict the future behavior of our DDP, confident that any uncertainties in our predictions will be not much larger (and usually much smaller) than our", " uncertainty in the initial conditions. If we now increase the drive strength past y, = 1.0829 into the chaotic regime, the picture changes completely. Figure 12.13 shows A (t) for the same DDP as in Figures 12.11 and 12.12, except that the drive strength is now y = 1.105, the same value used in our first plot of chaotic motion in Figure 12.10. The most obvious feature of this graph is that A (t) clearly grows with time. In fact, you will notice that to highlight the growth of A (t) I chose the initial difference to be just A (0) = 0.0001 radians. Starting from this tiny value, I AO I has increased in 16 drive cycles by more than 4 orders of magnitude to about I A Iti 3.5. From t = 1 through t = 16 (where the graph is leveling off), the maxima in Figure 12.13 grow almost perfectly linearly, implying that A (t) grows exponentially. 14 This exponential growth spells disaster for any attempt at accurate prediction of the DDP's 14 The eventual leveling of the curve is easily understood. You can see from Figure 12.10 that the angle OW [actually, 0 1 (0, but the same applies to 4, 2 (t)] oscillates between about 17. That is, neither 0 1 (0 nor 02 (0 ever exceeds magnitude n. Therefore their difference A (t) can never exceed 27. Thus the curve has to level off before AO (t) reaches 27. 480 Chapter 12 Nonlinear Mechanics and Chaos 0 - 1 \u20142 \u20143 \u20144 \u20145 log I AO(t) - 10 1 - lo-i - 10-2 - 10-3 - i o-4 - 10-5 - 10-6 Figure 12.13 The separation AO (t) of two identical pendulums, both with drive strength y = 1.105 and with an initial separation 64(0) = 10 -4 rad. After a small initial drop, the crests of log I Acb (t) I increase linearly, showing that AO (t) itself grows exponentially. long -term motion. For the present case, an error as small as 10 -4 radians in our initial conditions will have grown in 16 cycles to an error of about 3.5, or more than 7 radians. Thus an uncertainty of \u00b110 -4 radians in the initial conditions grows to an uncertainty Of fir, and an uncertainty of \u00b17t in the angle of a pendulum means that we have no idea at all where the pendulum is! I chose this example because it is especially dramatic. Nevertheless, in any chaotic motion, AO (t) grows exponentially for a while at least. Even if this growth levels out before AO (t) reaches 7, the exponential growth means that a tiny uncertainty in the initial conditions quickly grows into a large uncertainty in the predicted motion. It is in this sense that we say chaos exhibits extreme sensitivity to initial conditions, and this sensitivity is what can make the reliable prediction of chaotic motion a practical impossibility. The Liapunov Exponent What we have seen in the preceding three examples can be rephrased to say that the difference AO (t) between two identical DDP's released with slightly different initial conditions behaves exponentially: AO(t) Ke At (12.26) (where the symbol \"\u2014\" signifies that AO (t) may oscillate underneath an envelope with the advertized behavior, and K is a positive constant). The coefficient A in the exponent is called the Liapunov exponent. 15 If the long-term motion is nonchaotic (settles down to periodic oscillation) the Liapunov exponent is negative; if the long- term motion is chaotic (erratic and nonperiodic) the Liapunov exponent is positive. 15 Strictly speaking there are several Liapunov exponents, of which the one discussed here is the largest. Section 12.5 Chaos and Sensitivity to Initial Conditions 481 Higher Values of y So far we have seen that as we increased the drive strength y, the motion of our DDP became more and more complicated \u2014 from the linear regime, with its pure sinusoidal response, to the nearly linear regime, with the addition of harmonics, to the appearance of subharmonics and (for certain initial conditions at least) a period-doubling cascade, and finally to chaos. You might naturally anticipate that, if we were to increase y still further, the chaos would continue and intensify, but as usual our nonlinear system defies predictions. As y increases, the DDP actually alternates between intervals of chaos separated by intervals of periodic, nonchaotic motion. I shall illustrate this with just two examples. We saw that with y = 1.105, the DDP exhibits chaotic behavior (Figure 12.10) and exponential divergence of neighboring", " solutions (Figure 12.13). We have only to increase the drive strength to y = 1.13 to enter a narrow \"window\" of nonchaotic period-3 oscillation with exponential convergence of neighboring solutions, as shown in Figure 12.14. In part (a), you can see that within three drive cycles the motion settles into regular period-3 oscillations. Part (b) shows the separation AO (t) of two pendulums released with an initial difference AO (0) of 0.001 radians; in the first eight drive cycles, AO (t) actually increases, but from then on it decreases exponentially to zero, dropping by six orders of magnitude in the next twelve cycles. y= 1.13 5 10 15 20 0 11111111111111111111 /VS/VI, rrri\u201eInr \u20143 \u20146 \u20149 - (a) OW vs t (b) log (t)lvs t Figure 12.14 Motion of a DDP with y = 1.13. (a) The graph of (t) quickly settles down to oscillations of period 3 (same initial conditions as in Figures 12.8 and 12.10). (b) Logarithmic plot of the distance between two identical pendulums, the first with the same initial conditions as in part (a), the second with its initial angle 0.001 radians lower. After an initial modest increase, AO (t) goes to 0 exponentially. Figure 12.15 shows the corresponding two graphs for a drive strength of y = 1.503, where the motion has returned to being chaotic. 16 In part (a) we see a new kind of chaotic motion. The driving force is now strong enough to keep the pendulum rolling right over the top, and in the first 18 drive cycles the pendulum makes 13 complete clockwise rotations [0 (t) decreases by 267]. The motion here can be seen as a steady rotation at about one revolution per drive cycle, with an erratic 16 As we shall see, between the values y = 1.13 and 1.503, shown in Figures 12.14 and 12.15, the DDP has passed through several intervals of chaotic and nonchaotic motion, but I omit the details for now. 482 Chapter 12 Nonlinear Mechanics and Chaos (a) 4 (t) vs t (b) logl,A0(t) I vs t Figure 12.15 Motion of a DDP with y = 1.503. (Same initial conditions as in Figure 12.14.) (a) The graph of (t) against t oscillates erratically. In the first 18 cycles, it plunges to about \u2014267r ; that is, it makes about 13 complete clockwise rotations. It then starts to climb back but never actually repeats itself. (b) Logarthmic plot of the distance between two identical pendulums with an initial separation of 0.001 radians. For the first nine or ten cycles, AO (t) grows exponentially and then levels out. oscillation superposed. 17 At t = 18, the motion reverses to a more-or-less steady counterclockwise rotation with erratic oscillations superposed, and, as the picture suggests, the motion never settles down to be periodic. The logarithmic plot of Figure 12.15(b) shows the divergence of two pendulums with the same drive strength y = 1.503, but an initial separation of 0.001 radians. The separation of the two pendulums increases exponentially for the first 9 or 10 cycles and levels out by about t = 15. A dramatic feature of this divergence is that it is big enough to be seen in the conventional linear plot of Figure 12.16, which shows the actual positions 4 1 (t) and 02 (0 of the two pendulums. At first sight it is perhaps surprising that for the first 8.5 cycles the two curves are completely indistinguishable, but that the difference is then so abundantly visible. However, you can understand this striking behavior by reference to Figure 12.15(b), where you can see that until t 8.5 the separation AO (t), although growing rapidly, is nevertheless always less than about 1/3 radian \u2014 too small to be seen on the scale of Figure 12.16. By the time t ti 9.5, AO (t) has reached about 3 \u2014 which is easily visible on the linear plot \u2014 and is still climbing rapidly. Thus from t ti 9.5 the two curves are completely distinct. The main morals to be drawn from these last two examples are these: (1) Once the drive strength y of our DDP is past the critical value y c = 1.0829, there are intervals where the motion is chaotic and others where it is not. These intervals are often quite narrow, so that the chaotic motion comes and goes with startling rapidity. (2)", " The chaos can take on several different forms, such as the erratic \"rolling\" motion of Figure 17 1f you look closely you can see that for the first 7 cycles, the motion is very close to being a steady rotation of \u201427r per cycle, with a regular period-1 oscillation superposed. This type of motion is actually periodic, since a change of \u201427r brings the pendulum back to the same place. For some values of the drive strength, the long-term motion settles down exactly this way, a phenomenon called phase locking. (See Problem 12.17.) Section 12.6 Bifurcation Diagrams 483 Figure 12.16 Linear plot of the positions of the same two identi- cal DDP's whose separation AO (t) was shown in Figure 12.15(b) [AO (0) = 0.001 rad]. For the first eight and a half drive cycles, the two curves are indistinguishable; after this the difference is dramati- cally apparent. 12.15(a). (3) The erratic motion of chaos always goes along with the sensitivity to initial conditions associated with the exponential divergence of neighboring solutions of the equation of motion. 12.6 Bifurcation Diagrams So far, each of our pictures of the motion of the driven damped pendulum has shown the motion for one particular value of the drive strength y. To observe the evolution of the motion as y changes, we have had to draw several different plots, one for each value of y. One would like to construct a single plot that somehow displayed the whole story, with its changing periods and its alternating periodicity and chaos as y varies. This is the purpose of the bifurcation diagram. A bifurcation diagram is a cunningly constructed plot of q (t) against y as in Figure 12.17. Perhaps the best way to explain what this plot shows is just to describe in detail how it was made. Having decided on a range of values of y to display (from y = 1.06 to 1.087 in Figure 12.17) one must first choose a large number of values of y, evenly spaced across the chosen range. For Figure 12.17, I chose 271 values of y, spaced at intervals of 0.0001, y = 1.0600, 1.0601, 1.0602,, 1.0869, 1.0870. For each chosen value of y, the next step is to solve numerically the equation of motion (12.11) from t = 0 to a time tmax picked so that all transients have long since died out. To make Figure 12.17, I chose the same initial conditions as in the last few pictures, namely cb (0) = \u20147r/2 and (i) (0) = 0 18 18 Some authors like to superpose the plots for several different initial conditions. This gives a more complete picture of the many possible motions, but makes the plot harder to interpret. For simplicity, I chose to use just one set of initial conditions. 484 Chapter 12 Nonlinear Mechanics and Chaos 1.065 1.07 1.075 1.08 t = 501, 502,..., 600 4 (t) Figure 12.17 Bifurcation diagram for the driven damped pendulum for drive strengths 1.060 < y < 1.087. The period-doubling cascade is clearly visible: At yi = 1.0663 the period changes from 1 to 2, and at y2 = 1.0793 from 2 to 4. The next bifurcation, from period 4 to 8, is easily seen at y 3 = 1.0821, and that from 8 to 16 is just discernable at y4 = 1.0827. To the right of the critical value, ye = 1.0829, the motion is mostly chaotic, although at y = 1.0845 you can just make out a brief interval of period-6 motion. To understand our next move, recall that a good way to check for periodicity (or non-periodicity) is to examine the values 9) (to), (tc, + 1), 0 (to + 2), of 0 (t) for a large number of times at one-cycle intervals. If the motion is periodic with period n, these will repeat themselves after n cycles, otherwise not. Therefore, our next step is to use our solutions for 0 (t) to find the values of 0 (t) over a range of times at integer intervals from some chosen tmin to tmax (with train large enough that all transients have died out). For Figure 12.17, I found 0 (t) for 100 times, t = 501, 502,, 600. (Since this had to be done for 271 different values of y, there were in all 271 x 100 or nearly 30,000 calculations to do, and", " the whole process took several hours.) Finally, for each value of y, these hundred values of 0 (t) were drawn as dots on the plot of 0 against y. To see what this accomplishes, consider first a value of y such as y = 1.065 where we know that the motion has period 1. With the period equal to 1, the hundred successive values of 0 (t) are all the same, and the 100 dots all land at the same place in the plot of 0 against y. Thus what we see at any y for which the period is 1 is a single dot. From y = 1.06 till the threshold value y i = 1.0663 where the period doubles, our plot is therefore a single curve. At the threshold y i = 1.0663, the period changes to 2, and the positions 0 (501), 0 (502), 0 (503),, 0 (600) Section 12.6 Bifurcation Diagrams 485 now alternate between two different values. Therefore, these 100 points actually create exactly two distinct dots on the plot, and the single curve bifurcates at yi into two curves. At y2 = 1.0793 the period doubles again, to period 4, and each of the two curves bifurcates, giving four curves in all. The next doubling, to period 8, is easily seen (though I have not actually indicated it on the picture), and if you look closely you can just pick out some of the bifurcations to period 16. After this, the graph becomes a nearly solid confusion of points, and it is impossible to tell (from the graph, at least) the exact value y, where chaos begins, though it is clearly somewhere just below y = 1.083. Beyond this point, for the remainder of Figure 12.17, the motion is mostly chaotic, though you can see a small window at y = 1.0845, indicated by a vertical dashed line. (The window is especially noticeable in the upper section of the plot, where the dots are otherwise denser.) If you hold a ruler to this vertical line, you will see that at this particular value of y there are just six distinct points. That is, at y = 1.0845, the motion has returned briefly to being periodic, this time with period 6. A Larger View Figure 12.17 shows a rather small range of drive strengths (1.06 < y < 1.087) in great detail. Before we examine a larger ranges of drive strengths, we must cope with one small complication. As y increases, we have seen that the pendulum can start a \"rolling\" motion in which it makes many complete revolutions. In some cases, it can continue to \"roll\" indefinitely, so that 0 (t) eventually approaches \u00b1oc. Even if this rolling motion is perfectly periodic, the successive values 0(t0), o(to + 1), 0(t, + 2), never repeat themselves, since they increase by a multiple of 27r in each cycle. This renders a bifurcation plot, drawn exactly as in Figure 12.17, useless. The most obvious way to get around this difficulty is to redefine 0 so that it always lies in the range -7 <5 < 7r. Each time 0 increases past 7, we subtract 27, and each time it decreases past \u20147r, we add 27. With this modification, we can now draw a bifurcation diagram as before. However, keeping 0 between \u00b17 in this way has the disadvantage that it introduces a meaningless discontinuous jump into 0 (t), each time it passes \u00b17. A second, and sometimes simpler, way around the problem of the 27-ambiguity in 0 is to plot the values of the angular velocity (j)(t,), (j)(to + 1), + 2), (12.27) instead of the angular position 0 (to), The angular velocity (P is immune to the 27-ambiguity of 0 (since q.5 is unaffacted by the addition of any multiple of 27r to 0). Thus, if the motion is periodic with period n, then the values (12.27) will repeat themselves after n cycles, and otherwise not. Therefore, a bifurcation diagram drawn using the values of (\u00a2 instead of 0 will work just like Figure 12.17, even if the pendulum undergoes a rolling motion. 486 Chapter 12 Nonlinear Mechanics and Chaos a b c d e f 20 - 15 - 10- 5 - 0 - (t) 1. 3 1.4 1.5 -Y Figure 12.18 Bifurcation diagram showing values of (i) for the DDP with drive strengths 1.03 < y < 1.53. The intervals labelled a, b,, f across the top are as follows: (a) This interval is the same as was shown in much greater detail", " in Fig- ure 12.17. It starts with period 1, followed by a period-doubling cascade leading to chaos. (b) Mostly chaos. (c) Period 3. (d) Mostly chaos. (e) Period 1, followed by another period-doubling cascade. (f) Mostly chaos. Figure 12.18 is a bifurcation diagram drawn using values of (p over a range from just above y = 1.0 to just above y = 1.5. The first part of this picture, labelled (a) at the top, is the interval that was shown in much, greater detail in Figure 12.17, with a period-doubling cascade that starts from period 1 and ends in chaos. Section (b) is mostly chaos, although we already know that it contains some narrow windows of periodicity (most of which are completely hidden at the scale used here). Section (c) is very clearly period 3 and includes the value y = 1.13 that was shown in Figure 12.14. Section (d) is mostly chaos, while (e) starts with a long stretch of period 1, followed by another period-doubling cascade. Finally, section (f) is mostly chaos, although you can just pick out some windows of periodicity. \u201420 - f A (a) (b) 110 Figure 12.19 Motion of the DDP with drive strength y = 1.4. (a) The plot of,0(t) against t shows a periodic rolling motion in which q5 decreases by 27r in each drive cycle. (b) The plot of angular velocity q5(t) against t shows even more clearly that after about two drive cycles the motion becomes periodic, with 4(t) returning to the same value once each cycle. Section 12.7 State-Space Orbits 487 A striking feature of Figure 12.18 is the long interval of period-1 motion, from just below y = 1.3 to just above y = 1.4. This period-1 motion is actually a rolling motion, as you can see in Figure 12.19 which shows the motion for y = 1.4. In part (a), which shows q5 (t) as a function of t, you can see that the pendulum is rolling clockwise at a rate of one complete revolution per drive cycle (0 decreases by 207r in 10 cycles). That the motion really is periodic is even more evident in part (b), which shows (./)(t) as a function of t. After about two drive cycles, (t) is clearly periodic with period 1. 12.7 State-Space Orbits In the next two sections I give a brief introduction to the Poincare section, which is an important alternative way to view the motion of chaotic (and nonchaotic) systems. The Poincare section is a simplification of the so-called state-space orbit. This sim- plification is especially helpful for complicated multidimensional systems but can be introduced in the context of our one-dimensional driven damped pendulum. Thus I shall start in this section by describing state-space orbits for the DDP. In our discussion of the DDP we have focussed almost exclusively on the position (t) as a function of t. It turns out, however, that it is sometimes an advantage to follow both the position 0 (t) and the angular velocity..(t) as time evolves. In principle, if one knows 0 (t) for all t, then one can calculate (4. (t) by straightforward differentiation. Thus to follow (t) as well as 0 (t) is, in this sense, redundant. Nevertheless, following both variables can provide new insights into the motion, and this is what we shall now discuss. There is an immediate problem in plotting the two variables 0 (t) and.7.5(t) as func- tions of the third variable t, since this requires a three-dimensional plot \u2014 something which is hard to make, and not especially illuminating when made. The usual proce- dure is to draw the pair of values [0 (t), 4)(t)] as a point in a two-dimensional plane where the horizontal axis labels 0 and the vertical axis (b. (For reasons I'll discuss in a moment, this plane with coordinates 0 and ( is called state space.) As time passes, the point [OW, 4)(t)] moves in this two-dimensional space and traces out a curve, which is called a state-space orbit (or phase-space trajectory). Once you get used to interpreting these state-space orbits, you will find that they give a rather clear picture of the system's motion. As a first example, let us consider a DDP with y = 0.6 (a drive strength for which the linear approximation is still fairly good) and with our favorite initial conditions 0 (0) = \u20147", "/2 and (.1)(0) = 0. Figure 12.20 shows a conventional plot of OW against t for this case. To interpret this picture one has to know (as you certainly do) that the changing position 0 (t) is shown by the vertical displacement of the graph while the time t advances from left to right. With this understood, you can clearly see the motion starting at t = 0 with 0 (0) = \u20147r/2 and quickly approaching the expected sinusoidal attractor, with 0 (t) of the form 0 (t) = A cos(cot \u2014 3). Figure 12.21 shows the state-space orbit for the same DDP with the same initial conditions. Part (a) shows the first twenty cycles, 0 < t < 20. To interpret this picture, 488 Chapter 12 Nonlinear Mechanics and Chaos 7r/4 \u20147/2 Figure 12.20 Conventional plot of 0(t) against t for a DDP with drive strength y = 0.6. The motion quickly settles down to almost perfectly sinusoidal oscillation. (a) 0 t.15. 20 (b)5 Figure 12.21 State-space orbit for a DDP with drive strength y = 0.6. State space is the two-dimensional plane with coordinates 0 and (p ; the state-space orbit is just the path traced by the point [0(t), OW] as time passes. (a) The first 20 cycles, starting from the initial values 0 (0) = \u20147r/2 and 4 (0) = 0. The three dots labelled 0, 1, and 2 show the positions of [0 (t), (i)(t)] at t = 0, 1, and 2. The orbit spirals inward and rapidly approaches the period-one attractor, which appears as an ellipse in state space. (b) The same as (a) but with the first 5 cycles omitted so that only the elliptical attractor is seen. Between the times 5 < t < 20, the point [0 (t), (t)] moves 15 times around the same elliptical path. one has to know that as t advances the curve is traced, in the direction of the arrows, by the pair [0 (t), 0(t)]. With this understood, you can see clearly that the orbit starts out from 0(0) = \u20147r/2 and 0(0) = 0. Since the initial acceleration 0 is positive, 19 0(t) increases from the outset, and 0 (t) begins to increase as soon as 0 is nonzero. Thus the point [0 (t),.kt)] moves up initially, curving to the right. The oscillation of 0 (t) is evidenced by the back and forth, left-right, motion of the orbit; the oscillation of 0(t) by the up and down vertical motion. Eventually, as the transients die out, the motion approaches its long-term attractor, in which (in the linear approximation) we know that 0 (t) has the form 0 (t) = A cos(wt \u2014. (12.28) 19 As you can easily check, with the given initial conditions, both gravity and the drive force give the pendulum a positive acceleration at first. Section 12.7 State-Space Orbits 489 (a)Cit 20 (b) 5 t 5 20 Figure 12.22 State-space orbit for a DDP with drive strength y = 0.6 and initial conditions 0 (0) = q5 (0) = 0. (a) The first 20 cycles, starting from the origin and spiralling out toward the elliptical attractor. (b) In the 15 cycles 5 < t < 20, the orbit moves 15 times around the elliptical attractor to give exactly the same picture as in Figure 12.21(b). This implies that the angular velocity q(t) approaches the form c*/)(t) = \u2014wA sin(cot \u2014 8). (12.29) The two equations (12.28) and (12.29) are the parametric equations for an ellipse drawn clockwise in the plane of (0, (i)), with semimajor and semiminor axes A and wA. Thus, once the transients have died out, the point [0 (t), (t)] moves around this ellipse with angular frequency equal to the drive frequency co; that is, the state-space orbit completes one revolution per drive cycle. In Figure 12.21(a), the state-space orbit spirals in toward this ellipse, merging with it after about three cycles. [This already illustrates one small advantage of the state-space orbit over the conventional plot of 0 (t) against t: In the conventional Figure 12.20 the actual motion has become indistinguishable from the limiting sinusoidal motion after", " little more than 1 cycle; in the state-space plot of Figure 12.21(a) the actual and limiting orbits can be told apart for some three cycles. Thus the state-space orbit gives a more sensitive picture of the approach to the attractor.] Figure 12.21(b) is the same as part (a), except that I have omitted the first 5 cycles; that is, in part (b), 5 < t < 20 and only the elliptical attractor shows up. Since our main interest is usually in the limiting motion, state-space plots are usually drawn as in part (b), with enough initial cycles omitted so that only the limiting motion is visible. Figure 12.22 shows the state-space orbit for exactly the same DDP as in Figure 12.21, but with initial conditions 0 (0) = (P(0) = 0. In part (a) you can easily see that the orbit starts out with the stated initial conditions, and spirals outward, completing some 2.5 cycles before merging with the elliptical attractor. Part (b) shows the 15 cycles starting from t = 5, by which time the orbit is indistinguishable from its long- term attractor. In particular, Figure 12.22(b) is exactly the same as Figure 12.21(b), because for y = 0.6 all initial conditions lead to the same attractor. 490 Chapter 12 Nonlinear Mechanics and Chaos State Space I shall give a detailed discussion of state space in Chapter 13, but here is a brief expla- nation of the terminology. For our pendulum, state space (also called phase space) is the two-dimensional plane defined by the two variables 0, the angular position, and the angular velocity. This is to be contrasted with the one-dimensional configura- tion space defined by the one variable 0 that gives the position, or configuration of the system. More generally, the configuration space of an n-dimensional mechanical system is the n-dimensional space of its n position coordinates q 1,, qn, whereas state space is the 2n-dimensional space of the coordinates q 1,, qn and velocities q,, I shall discuss several properties and uses of state space in Chapter 13. Here I mention just one important feature: The \"state\" (or \"state of motion\" in full) of a mechanical system is often used to mean a specification of the motion (at any chosen time to) that is complete enough to determine uniquely the motion at all later times. That is, the state of a system defines the initial conditions needed to specify a unique solution of the equation of motion. For our pendulum, specification of the position 0 at time to is not sufficient to determine a unique solution, but specification of 0 and 0 is. That is, the two variables 0 and (P. define the state of the pendulum, and the space of all pairs (0, (b) is naturally called state space. A state-space orbit is simply the path traced in state space by the pair I0 (t), (.(t)] as time evolves. Natural as this name is, you must recognise that a state-space orbit is very different from the orbit of, say, a planet in ordinary space with coordinates r = (x, y, z). For example, a planet can have many different orbits passing through a single point r at a given time to. On the other hand, from what was just said about initial conditions, it follows that for any \"point\" (0, ) in state space, our pendulum has exactly one state-space orbit passing through (0, (b) at any given to. Another curious feature of state-space orbits concerns their direction of flow: Since the vertical axis represents the velocity 0, the motion at any point above the horizontal axis (, > 0) is always to the right (increasing 0), as seen in Figure 12.22. Similarly, the motion at any point below the horizontal axis has to be to the left. If an orbit crosses the horizontal axis then, since = 0, the orbit must be moving exactly vertically (0 not changing). All of these properties are illustrated in Figure 12.22. They imply that any closed state-space orbit, such as the elliptical attractor of Figure 12.22(b), is always traced in a clockwise direction. More State-Space Orbits As we increase the drive strength y, we know that the motion of our DDP undergoes various dramatic changes, some of which show up very nicely in plots of the state- space orbits. For instance, Figure 12.23 shows the state-space orbits for y = 1.078 and y = 1.081, both in the middle of the period-doubling cascade first shown in Figure 12.8. Both plots show forty cycles starting from t = 20, by which time all initial transients have completely died out. That is, both plots show the limiting", ", long-term motion and are to be compared with Figure 12.22(b). As in that picture, these new orbits move around the origin in more-or-less elliptical loops, but in both cases the Section 12.7 State-Space Orbits 491 (a) Period 2 (b) Period 4 Figure 12.23 State-space orbits showing the periodic attractors for ( a ) y = 1.078 with period 2 and (b) y = 1.081 with period 4. Both plots show the forty cycles from t = 20 to 60. In part (a), the orbit traces just two distinct loops twenty times each; in part (b) it traces four loops ten times each. Compare Figure 12.8 (middle two lines). orbit makes more than one loop before closing on itself. In part (a) there are two distinct loops, each of which lasts for one drive cycle, so that the motion repeats itself once every two cycles \u2014 that is, it has period two. In part (b) there are four distinct loops, indicating very clearly that the period has doubled again to period four. It is important to understand that it makes no difference how many cycles we plot in these two figures [as long as we start after the transients have died out, and plot at least two cycles in part (a) and four in part (b)]. I could have plotted from t = 20 to 100 or from 20 to 1000, and part (a) would still have shown the same two loops and part (b) the same four loops. Chaos If we increase the drive strength y a little further, we enter a region of chaos. Figure 12.24 shows the state-space orbit for drive strength y = 1.105, whose chaotic character was shown in Figures 12.10 and 12.13. Part (a) shows seven cycles from t = 14 to t = 21, and you can see clearly that in seven cycles the orbit fails to repeat or to close on itself. Thus if the motion is periodic, its period must be greater than 7. To decide whether it is periodic, we need to plot more cycles. In part (b), I have plotted from t = 14 to 200, and the plot has become an almost solid swath of black but has still not repeated itself. [The evidence for this last claim is that in a plot out to t = 400 (not shown) the curve moves into several of the remaining gaps of part (b); thus it has certainly not begun to repeat by t = 200.] Therefore, Figure 12.24 adds strong support to our conclusion that the motion never repeats itself and is in fact chaotic. The black swath of Figure 12.24(b) is very striking, but is too full of information to be of much use. We need a way to extract from this picture a smaller amount of information that might actually tell us more. The technique for doing this is the so- called Poincare section, but before we take this up, I want to give two more examples of state-space orbits. 492 Chapter 12 Nonlinear Mechanics and Chaos (a) 14 (b) 28 < t < 200 Figure 12.24 State-space orbits for a DDP with y = 1.105 showing the chaotic attractor. (a) In the seven cycles from t = 14 to 21 the orbit does not close on itself. (b) The same is true in the 186 cycles from t = 14 to 200, and by now it is pretty clear that the motion is never going to repeat itself and is in fact chaotic. State-Space Orbits for Rolling Motion We have already seen that for y = 1.4 our DDP executes a \"rolling motion,\" making a complete clockwise rotation once each drive cycle (Figure 12.19). The state-space orbit for this motion is shown in Figure 12.25. In this plot you can see clearly how, after a couple of cycles, the pendulum settles down to a periodic motion in which 0 decreases by 27 and the pendulum makes a complete clockwise rotation once per cycle. The plot of Figure 12.25 is a very satisfactory way of showing the state-space orbit over a small number of cycles. Sometimes, however, (if the motion is chaotic, for instance) one would like to show the orbit over a very long time interval \u2014 several hundred cycles, perhaps \u2014 and in this case 0 may range over many hundreds of complete revolutions. To show this, in the format of Figure 12.25, we would be forced to compress the scale on the 0 axis to the point where the motion would be completely indecipherable. The usual way around this difficulty is to redefine 0 so that it always 0 t :s 6 Figure 12.25 First six cycles of the state-space orbit for a DDP with y = 1.4, showing the periodic rolling motion, in which decreases by 27 in each cycle. The numbers 0, 1,, 6 indicate the state-space \"", "position\" (0, cb) at the times t = 0, 1,, 6. Section 12.7 State-Space Orbits (a) 0 (b) 1()-t-30 Figure 12.26 (a) The exact same orbit as in Figure 12.25 but with redefined so that it remains between -7 and 7. Each time 0 decreases to -7 (at A for example), the orbit disappears and reappears at +7r (at B for example). (b) By the time t = 10 the orbit has settled down to perfectly periodic motion, with each successive cycle lying exactly (on this scale) on top of its predecessor. lies between \u20147 and in Each time 0 decreases past \u20147 we add on 2n and each time it increases past 7 we subtract 27. (This is acceptable since any two values of 0 that differ by a multiple of 27 represent the same position of the pendulum.) With 0 redefined in this way, the state-space orbit of Figure 12.25 looks as shown in Figure 12.26(a). This new plot is not an obvious improvement on Figure 12.25 (though we shall see that it does have some advantages), but you should study it carefully to understand the relationship of the two kinds of picture. You can think of the new picture as being obtained from Figure 12.25 by cutting apart the intervals \u2014 37r < < - 7T, and \u201457 < (/) < \u2014 37r, and so on, and pasting them all back on top of the interval \u20147 < \u00a2 < 7c. In the resulting picture, 0 makes a discontinuous jump each time it arrives at 0 = \u00b17. For example, at about t = 0.7, 0 decreases to \u20147 at the point A and jumps to the point B. An advantage of Figure 12.26(a) over Figure 12.25 is that the new picture gives a more incisive test of the periodicity of the orbit. In the new picture, you can see that the orbit is approaching a periodic attractor, but it is also clear that in the interval 0 < t < 6 the orbit has definitely not reached the periodic attractor. (In fact you can just about see that there are 6 distinct loops.) On the other hand, by the time t = 10 the successive cycles are indistinguishable on the scale of these pictures. The twenty cycles shown in Figure 12.26(b) all disappear on the left at the same point C, reappear at D, and follow the exact same path back to C twenty times over. A disadvantage of either plot in Figure 12.26 is the spurious discontinuity each time 4) jumps from -7T to 7, as at points A and B, for example. We can get rid of these discontinuities (at least in our minds) if we imagine the plot cut out and rolled 494 Chapter 12 Nonlinear Mechanics and Chaos into a cylinder with the vertical lines 0 = \u00b17 glued together. In this way point A becomes the same as point B, and the state-space orbit moves continuously around the vertical cylinder. More Chaos As a final example of a state-space orbit, I have shown the orbit for a DDP with y = 1.5 in Figure 12.27. We already know that the motion is chaotic for this value of y, though for this picture I chose a smaller damping constant, )3 = coo/8 instead of the value /3 = w0 /4 that I used for all previous pictures in this chapter. As it turns out, this smaller damping makes the chaotic motion more wild and produces a Poincare section that is even more interesting and elegant (as I describe in the next section). With these parameters, the pendulum undergoes an erratic rolling motion, making many complete revolutions, first in one direction and then in the other. Thus we are forced to make a plot with 0 confined between \u20147 and 7 as in Figure 12.26 \u2014 but with dramatically different results. The motion does not repeat itself in the 190 cycles shown, with 10 < t < 200. (The evidence for this claim is that in a plot for 10 < t < 250 \u2014 not shown \u2014 the orbit moves into some of the unvisited regions of Figure 12.27. If it had begun to repeat itself before t = 200, it could not visit new ground after t = 200.) The dense tangle of threads running through Figure 12.27 lend strong support to the claim that for these parameters the motion is chaotic. Unfortunately, one could not claim that the picture sheds much light on the nature of chaotic motion. It is just too densely packed with information to convey any useful message. In the next section I describe the Poincare section, which is a technique for culling out of pictures like Figure 12.27 enough information to allow an interesting pattern to emerge. 10 -.5 200 Figure 12.27 The chaotic state-space orbit for a D", "DP with y = 1.5 and p = coo /8. In the 190 cycles shown, the motion does not repeat itself, and, in fact, it never does. Section 12.8 Poincare Sections 495 12.8 Poincare Sections For the periodic motion of a DDP, a state-space orbit is a descriptive way of viewing the pendulum's history. For chaotic motion, a state-space orbit conveys a sense of the dramatic nature of chaos, but is too full of information to be of much serious use. One way around this difficulty is a trick which we have used earlier and was suggested by Poincare: Instead of following the motion as a function of the continuous variable t, we look at the position just once per cycle at times t = to, to + 1, to + 2,. The Poincare section for a DDP is just a plot showing the pendulum's \"position\" 10 (t), (t)] in state space at one-cycle intervals t = to, to + 1, to \u00b1 2,, (12.30) with to usually chosen so that the initial transients have died out. 20 To illustrate this, consider the state-space orbit shown in Figure 12.28(a) for a DDP with y = 1.078 (and the damping constant restored to our usual value /3 = co o /4). The two loops of this orbit indicate that (as we already knew) the long-term motion has period two. To emphasize this I have drawn dots to show the position [0 (t), (A.(t)] at one-cycle intervals, t = 20, 21, 22,. Since the motion has period two, these alternate be- tween just two distinct positions and show up as just two dots. In a Poincare section one dispenses with the orbit and draws just the dots at one-cycle intervals as in Figure 12.28(b). When the motion is periodic, there is no particular advantage to the Poincare section over the complete state-space orbit of Figure 12.28(a), although the Poincare section does show the period very clearly. (A Poincare section with four dots would show period-four motion, and so on.) On the other hand, when the motion is chaotic, 10- cb \u2014 7r/2 \u201410 - 7r/2 (a) State-Space Orbit (a) Poincare Section Figure 12.28 (a) State-space orbit of a DDP with y = 1.078 for 20 < t < 60. The dots show the positions at t = 20, 21, 22, -, but, since the motion has period two, these alternate between just two fixed points. The right point shows the positions for t = 20, 22, ; the left one, those for t = 21, 23, - (b) In the corresponding Poincare section, one omits the orbit and draws only the dots showing the positions at t = 20, 21, 22,. The presence of just two dots is a clear indication of period-two motion. 20 For a multidimensional system, the Poincare section involves taking a two-dimesional slice, or section, through the multidimensional state space. Hence the word \"section.\" 496 Chapter 12 Nonlinear Mechanics and Chaos Figure 12.29 Poincare section for a pendulum with y = 1.5 and damping constant )6 = wo /8 for times 10 < t < 60000. This figure is made of nearly 60000 points showing the \"position\" [0 (t), OW] at one-cycle intervals, t = 10, 11,, 60000. The rectangular box indicates the region that is enlarged in Figure 12.30. no two cycles of the motion are the same, and the state-space orbit can be a real mess, as we saw in Figure 12.27. In this case, the Poincare section reveals some totally unexpected structures. To illustrate a Poincare section for chaotic motion, I chose the pendulum whose chaotic state-space orbit was shown in Figure 12.27. It is clear that, since this motion never repeats itself, the Poincare section will contain infinitely many points, and these infinitely many points will comprise a subset of the points of the full orbit. It is probably fair to say that no one could ever guess what this subset would look like, but with the aid of a high-speed computer we can find out. The result is shown in Figure 12.29. Although it is certainly not obvious exactly what this elegant figure signifies, it certainly is obvious that it signifies something. By selecting from Figure 12.27 just those points at one-cycle intervals, we have reduced the dense, and nearly solid, tangle of Figure 12.27 to the elegant curve of Figure 12.29. Actually, while Figure 12.29 looks like", " a relatively simple curve, it is not a curve at all, but rather a fractal. A fractal can be defined in various ways, but a characteristic feature of fractals is that when one enlarges the scale and zooms in on a portion of the picture, one uncovers further structures that are in some ways similar to the original picture (somewhat like a photograph of a person holding a photograph of a person holding a photograph... and so on). To illustrate this property of our Poincare section, I have zoomed in on the region indicated by the rectangular box at the bottom of Figure 12.29. Notice that (at the scale of Figure 12.29) this region comprises a prominent \"tongue\" pointing to the left near the left of the box, with a second tongue inside the first near the right of the box. Figure 12.30 is a fourfold enlargement of this box. This enlargement makes clear that the apparently single tongue on the right of the box of 12.29 is actually four tongues, while that on the left is actually at least five. Section 12.8 Poincare Sections 497 Figure 12.30 Enlargement of the small box at the bottom of the Poincare section of Figure 12.29. Each of the two tongues in the box of 12.29 is seen to be made up of several tongues. The box on the left is the region that is further enlarged in Figure 12.31. Figure 12.31 A further enlargement of the box at the left of the enlargement of Figure 12.30. Each of the five tongues in the box of 12.30 (except perhaps the innermost one) is seen to be made up of several tongues. This process of zooming in on successively smaller regions of the Poincare section can, at least in principle, be continued indefinitely. Figure 12.31 is a further fourfold enlargement of the region shown by the gray rectangle on the left of Figure 12.30. In this enlargement, we see that each of the five tongues on the left of Figure 12.30 (except perhaps the fifth one) actually consists of several separate tongues. This so-called self similarity of the figure is one of the characteristic features of a fractal. When the Poincare section of the motion of a chaotic system is a fractal, the long- term motion is said to be a strange attractor. It would unfortunately be well beyond the scope of this book to explain what it signifies that the Poincare section of a chaotic attractor is fractal, and indeed there is still much about this phenomenon that is not 498 Chapter 12 Nonlinear Mechanics and Chaos (b) Theory Figure 12.32 Poincare section for a DDP. (a) Experimental results using the Daedalon Chaotic Pendulum. (b) Theoretical prediction using the same parameters as in part (a). Courtesy of Professors H.J.T. Smith and James Blackburn and the Daedalon Corporation. understood. Nevertheless, it is undeniably fascinating that the strange geometrical structure of the fractal appears in our study of the long-term behavior of chaotic systems. This discovery has stimulated much research on both the physics of chaotic systems and the mathematics of fractals. To observe a strange attractor with a real pendulum would obviously be challeng- ing, but once again the experimentalists have risen to the challenge. Figure 12.32 shows a Poincare section made with the Daedalon chaotic pendulum. Part (a) shows the experimental results and part (b) the theoretical prediction (that is, a numerical solution of the equation of motion using the experimental values of the parameters). Considering the great subtlety of these graphs, the agreement is outstanding. 21 12.9 The Logistic Map As I have repeatedly emphasized, the phenomenon of chaos appears in many different situations. In particular, there are certain systems that can exhibit chaos, but whose equations of motion \u2014 called maps \u2014 are simpler than the equations of any mechan- ical system. Although these systems are not strictly part of classical mechanics, they are worth mentioning here, for several reasons: Because their equations of motion are simple, several aspects of their motion can be understood using quite elementary meth- ods. Any understanding of chaos that we get from studying these simpler systems can shed light on the corresponding behavior of mechanical systems. In particular, there is an intimate connection between these \"maps\" and the Poincare sections of mechan- ical systems. Finally, a discussion of chaos in this new context highlights the diversity of systems that exhibit the phenomenon. 21 There are, nevertheless, differences. One possible cause is the difficulty of making a drive motor that is perfectly sinusoidal. Section 12.9 The Logistic Map 499 Discrete Time and Maps In almost all problems of mechanics, one is concerned with the evolution of a system as time advances continuously. However, there are systems for which time is a discrete variable.", " The history of any event that occurs just once a year, such as the Super Bowl, is an example. The score at Super Bowl games is defined only at a sequence of discrete times starting in 1967 and spaced at one-year intervals: t = 1967, 1968, 1969,. (12.31) The attendance at your weekly lunch group is defined only for discrete times spaced a week apart. The total rainfall in the annual Indian monsoon is defined only for discrete times spaced a year apart. Even when a variable is defined as a function of continuous time, we may find that we need its values only at certain discrete times. For example, entymologists studying the population of a particular bug may have no interest in the population's day-by-day evolution; rather they may need to record the bug population just once a year, immediately after the year's new arrivals have hatched. Another example of this situation is the Poincare section for a mechanical system, as described in Section 12.8. Our ultimate interest is to know the state of the system for all (continuous) times, but we saw that it is sometimes useful to record just the state at discrete, one- cycle intervals. To the extent that we are prepared to settle for this smaller amount of information, the Poincare section reduces the problem of the pendulum (or whatever) to a discrete-time problem, and anything we can learn about discrete-time systems should shed light on the possible behavior of Poincare sections. In the case of the driven damped pendulum, we know that the state of the system, as given by the pair (t), 4)(01 at any time t determines uniquely the state at any later time. In particular, it determines the state [0 (t + 1), ci)(t + 1)] one cycle later. This means that there exists a function f (which we don't know, but which certainly exists) that, acting on any chosen pair WO, (b(t)], gives the corresponding pair (t + 1), (b(t + 1)]. That is, [0(t + 1), (i)(t + 1)] = f (t), (i)(t)1). (12.32) In the same way, we could imagine a bug species with the property that the population n t+i in year t + 1 is uniquely determined22 by the population n t in the preceding year t. Again this would imply the existence of a function f that carries any n t onto its corresponding n t\u00b1i: nt+t = f (lt). (12.33) 22 This is, of course, a very simplified model. In the real world, the population n 2+1 certainly depends on ti p but also on many other factors, such as the rainfall in year t, the supply of bug food, and the population of birds that like to eat the bug. Nevertheless, we can imagine a temperate island, with a constant supply of bug food and no bug predators, on which ti t+ i is uniquely determined by n, alone. 500 Chapter 12 Nonlinear Mechanics and Chaos We can call an equation of this form the growth equation for the population con- cerned. EXAMPLE 12.1 Exponential Population Growth The simplest example of a growth equation of the type (12.33) is the case where nt+1 is proportional to nt : nt+i = f (nt) = rnt. (12.34) That is, the function f (n) that gives next year's population in terms of this year's is just f (n) = rn (12.35) where the positive constant r could be called the growth rate or growth param- eter of the population. [For example if every bug alive this spring dies before next spring but leaves two surviving offspring, then the population would satisfy (12.34) with r = 2.] Solve the equation (12.34) for nt in terms of n o and discuss the long-term behavior of nt. The solution to (12.34) is easily seen by inspection. Observe that and n i = f (n o) = rno n2 = f (n = f (f (n o)) = r2no from which it is clear that t terms n t = f(n t _i) = f(f( f (no) )) = r t no- (12.36) We see that if r > 1, the population nt grows exponentially, approaching infinity as t oo. If r = 1, the population stays constant, and if r < 1 it decreases exponentially to zero. Before we discuss a more interesting growth equation, I need to introduce some terminology. In mathematics, the words \"function\" and \"map\" are used as almost exact synonyms. Thus we can say that Equation (12.33) defines nt+1 as a function", " of nt. Or we can say that (12.33) is a map in which f carries n t onto the corresponding 23 t+i, a relationship that we can represent thus: t+1 = f(nt) n (12.37) 23 The origin of this rather strange usage seems to be in cartography. A cartographer's map of the US, for example, establishes a correspondence between each actual point of the US and the corresponding point on a piece of paper, in somewhat the same way the function y = f (x) establishes a correspondence between each value x and the corresponding value y = f (x). Section 12.9 The Logistic Map 501 The whole sequence of numbers, no, n 1, n 2, can be written similarly as -->- (12.38) and is naturally described as an iterated map, or just map for short. For some reason the word \"map\" (as opposed to \"function\") is used almost universally to describe relationships like (12.37) between successive values of any discrete-time variable. The map (12.37) is a one-dimensional map since it carries the single number n t onto the single number n t +1. The corresponding relationship (12.32) for the Poincare section of a DDP defines a two-dimensional map, since it carries the pair of numbers [0 (t),.:\u00a2(t)] onto the pair [0 (t + 1),.:p(t + 1)]. The Logistic Map The exponential map of Example 12.1 with r > 1 is a reasonably realistic model for the initial growth of many populations, but no real population can grow exponentially for ever. Something \u2014 overcrowding or shortage of food, for example \u2014 eventually slows down the growth. There are many ways to modify the map (12.34) to give a more realistic model of population growth. One of the simplest is to replace the function f (n) = rn of (12.34) by f (n) = rn(1 \u2014 n1 N) (12.39) where N is a large positive constant, whose significance we shall see directly. That is, we replace the exponential map (12.34) with the so-called logistic map n t+i = f (n t ) = rnt (1 \u2014 n t /N). (12.40) As long as the population remains small compared to N, the term n t IN in (12.40) is unimportant, and our new map produces the same exponential evolution as the exponential map. But if n t grows toward N, the term n t IN becomes important, and the parenthesis (1 \u2014 nt I N) begins to diminish and \"kill off\" some of the excess growth. Thus this \"mortality factor\" (1 \u2014 n I N) in (12.39) produces exactly the expected slowing of the population growth as n becomes large, and overcrowding or starvation become important. In particular, if n t were to reach the value N, the parenthesis (1 \u2014 n t I N) in (12.40) would vanish, and next year's population n t+1 would be zero. If n t were greater than N, then n t+ I would be negative \u2014 which is impossible. In other words, a population governed by the logistic map (12.40) can never exceed the value N, the maximum or carrying capacity of the model. Notice that, because of the term involving n in the mortality factor, the logistic map (12.39) (unlike the exponential map) is nonlinear. It is this nonliriearity that makes possible the chaotic behavior of the logistic map. Figure 12.33 compares exponential and logistic growth, for a growth parameter r = 2 and an initial population no = 4. The upper curve (gray dots) shows the unending doubling of the exponential case; the lower curve (black dots) shows the growth predicted by the logistic map (12.40), with the same growth parameter r = 2 and with a carrying capacity N = 1000. As long as n remains small (much less than 1000), the 502 Chapter 12 Nonlinear Mechanics and Chaos 1000 500 Figure 12.33 Exponential and logistic growth, both with growth parameter r = 2. The gray dots show the exponential growth, increasing without limit; the black dots show the logistic growth, which eventually slows down and approaches an equilibrium at n = 500. The lines joining the dots are just to guide the eye. logistic growth is indistinguishable from pure exponential growth, but once n reaches 100 or so, the mortality factor visibly slows the logistic growth, which eventually levels off at around n = 500. Before we discuss the logistic map in any detail it is convenient to simplify it by changing variables from the population n to the relative population, x = n/N, (12.41) the ratio of the actual", " population n to its maximum possible value N. Dividing both sides of (12.40) by N, we see that x t obeys the growth equation where I have redefined the map f as a function of x to be Since the population n is confined to the range 0 < n < N, the relative population x = n/N is restricted to 0 < x < 1. (12.44) Within this range, the function x(1 \u2014 x) has a maximum of 1/4 (at x = 1/2). Thus, to guarantee that xt+1, as given by (12.42), never exceeds 1, we must limit the growth factor to 0 < r < 4. Therefore, we shall be studying the map (12.42) in the ranges 0 < x < 1 and 0 < r < 4. Section 12.9 The Logistic Map 503 r= 1.5 Figure 12.34 The relative population xt =n,IN for the logisitic map (12.42), with two different initial conditions for each of two different growth rates. (a) With the growth parameter r = 0.8, the population rapidly approaches zero whether xo = 0.1 or xo = 0.5. (b) With r = 1.5 and the same two initial condi- tions, the population approaches the fixed value 0.33. Before we look at some of the exotic aspects of the logistic map, let us first look at a couple of cases where it behaves just as one might expect. Figure 12.34(a) shows the logistic population for a growth parameter r = 0.8 and for two different initial values, xo = 0.1 and x0 = 0.5. You can see that in either case x t \u2014f 0 as t ----> co. In fact it is easy to see that as long as r < 1, the population eventually goes to zero whatever its initial value: From (12.42) we see that xt < rxt _ i and hence that xt < xo ; therefore, if r < 1, we conclude that x t \u2014+ 0 as t ---> oo. Figure 12.34(b) shows the logistic population for a growth parameter r = 1.5 and for the same two initial conditions. For x 0 = 0.1 we see that the population increases at first. On the other hand, for the larger initial value x 0 = 0.5 the mortality factor causes the population to decrease at first. In either case, it eventually levels out at x = 0.33. Fixed Points In both the cases shown in Figure 12.34 we can say that the logistic map has a constant attractor towards which the population eventually moves, namely x = 0 for any r < 1, and x = 0.33 for r = 1.5. If the population happens to start out equal to such a constant attractor, x 0 = x* say, then it simply remains fixed there for all time; that is, xt = x* for all t. This obviously happens if and only if f (x*) = (12.45) Any value x* which satisfies this equation is called a fixed point of the map f. These fixed points are analogous to the equilibrium points of a mechanical system, in that a system which starts at a fixed point remains there for ever. For a given map, we can solve the equation (12.45) to find the map's fixed points. For example, the fixed points of the logistic map must satisfy rx* (1 \u2014 x*) = x*, (12.46) 504 Chapter 12 Nonlinear Mechanics and Chaos X t + 1 xi Figure 12.35 Graphs of x against x (the 45\u00b0 line) and of the logistic function f (x) = rx(1 \u2014 x) (the two curves) for two choices of r, one less than and one greater than 1. The fixed points of the logistic map lie at the intersections of the line with the curve. When r < 1 there is just one intersection, at x* = 0; when r > 1 there are two intersections, one still at x* = 0 and the other at an x* > 0. which is easily solved to give x* = 0 or X -= (12.47) r \u2014 1 r The first solution is the fixed point x* = 0 that we have already noted. The second solution depends on the value of r. For r < 1 it is negative and hence irrelevant. For r =1 it coincides with the first solution x* = 0, but for r > 1 it is a distinct, second fixed point. For example, for r = 1.5 it gives the fixed point we have already noted at x* = 1/3. It is a fortunate circumstance that we can actually solve the equation (12.45) analytically to", " find the fixed points of the logistic map, but it is also instructive to examine the equation graphically, since graphical considerations give additional insight and can be applied to many different maps, some of which cannot necessarilly be solved analytically. To solve Equation (12.45) graphically, we just plot the two functions x and f (x) against x as in Figure 12.35 and read off the fixed points as those values x* where the two graphs intersect. When r is small, the curve of f (x) lies below the 45\u00b0 line of x against x, and the only intersection is at x = 0; that is, the only fixed point is x* = 0. When r is large the curve of f (x) bulges above the 45\u00b0 line and there are two fixed points. The boundary between the two cases is easily found by noting that the slope of f (x) at x = 0 is just r. Thus as we increase r, the curve moves across the 45\u00b0line (whose slope is 1) when r = 1. Thus for r < 1 there is just one fixed point at x* = 0; but when r > 1 there are two fixed points, one at x* = 0 and the other at an x* > 0. The advantage of this graphical argument is that it works equally well for any similar function f (x) as long as it is a single concave-down arch. [For example, f (x) could be the function f (x) = r sin(rrx) of Problem 12.23.] Section 12.9 The Logistic Map 505 A Test for Stability That x* is a fixed point (that is, an equilibrium value) for the logistic map guarantees that when the population starts out at x* it will stay there. By itself this is not enough to ensure that the value x* is an attractor for the map. We need to check in addition that x* is a stable fixed point, that is, that if the population starts out close to x* it will evolve towards x* not away from it. (This issue has an exact parallel in the study of equilibrium points of a mechanical system: If a system starts out exactly at an equilibrium point then it will \u2014 in principle \u2014 stay there indefinitely. But only if the equilibrium is stable will the system move back to equilibrium if disturbed a little away.) There is a simple test for stability, which we can derive as follows: If x t is close to a fixed point x*, we naturally write x t = X * + Et. (12.48) That is, we define Et as the distance of xt from the fixed point x*. If Et is small, this lets us evaluate xt\u00b1i as xt+i = f (x t ) = f (x* + Et ) f (x*) f'(x*)E t = x* A.E t (12.49) where in the last expression I have used the fact that x* is a fixed point [so that f (x*) = x*] and I have introduced the notation ) for the derivative of f (x) at x*, = f' (x*). (12.50) Now, according to (12.48), x t\u00b1i = x* E t+1. Comparing this with the last expression of (12.49) we see that Et+1 ^ /.Et (12.51) Because of this simple relation, the number X = f (x*) is called the multiplier or eigenvalue of the fixed point. It shows that if IXj < 1, then once x t is close to x*, successive values get closer and closer to x*. On the other hand, if IX I > 1, then when xt is close to x*, the succeeding values move away from x*. This is our required test for stability: Stability of Fixed Point x* E of xed point 0 map xi ); that f('* = If '`(x 1, then x is stable and acts attracto ) 1 > I hen x is unstable and acts as a e lle We can immediately apply this test to the two fixed points of the logistic map: Since f (x) = r x (1 \u2014 x), its derivative is (x) = r(1 \u2014 2x). 506 Chapter 12 Nonlinear Mechanics and Chaos 1.0 - 0.8 - 0.6 0.4 0.2 0 0........ 1 2 3 4 r Figure 12.36 The fixed points x* of the logistic map as functions of the growth parameter r. The solid curves show the stable fixed points and the dashed curves the unstable. Note how the fixed point x* = 0 becomes unstable at precisely the place (r = 1) where the nonzero fixed point first appears. At the fixed point x* = 0, this means the crucial derivative is therefore, the fixed", " point x* = 0 is stable for r < 1 but unstable for r > 1. At the fixed point x* = (r \u2014 1)/r, the derivative is ( ) r (x*) = 2 \u2014 r, so that this fixed point is stable for 1 < r < 3, but unstable for r > 3. These results are summarised in Figure 12.36, which shows the fixed-point values x* as functions of the growth parameter r with the stable fixed points shown as solid curves and the unstable as dashed curves. The arguments just given show exactly when each of the two fixed points becomes unstable and that the second one appears exactly when the first becomes unstable. On the other hand, it would be nice to have an argument that made clearer why the fixed points behave the way they do and showed more clearly (what is true) that the same qualitative conclusions apply to any other one-dimensional map with the same general features. Such an argument can be found by examining the graph of Figure 12.35. In that figure, we saw that the fixed points of the map correspond to intersections of the curve of f (x) against x with the 45\u00b0 line (x against x). When r is small, it was clear that there is only one such intersection at x* = 0. As r increased the curve bulged up more and eventually crossed the 45\u00b0 line producing a second intersection and hence a second, nonzero fixed point. We saw that this second intersection appears when the slope 1(0) of the curve at x = 0 is exactly 1 (that is, when it is tangent to the 45\u00b0 line). In light of our test for stability, this means the second fixed point has to appear at exactly the moment when the first one at x* = 0 changes from stable to unstable. Section 12.9 The Logistic Map 507 The First Period Doubling Figure 12.36 tells the whole story of the fixed points of the logistic map. The solid curves show the stable fixed points, which are the constant attractors. We see that for r < 1, the logistic population approaches 0 as t --+ oo; for 1 < r < 3, it approaches the other fixed point x* = (r \u2014 1)/r as t ---> oo. But what happens \u2014 and this proves the most interesting question \u2014 when r > 3? In the computer age this is easily answered. Figure 12.37 shows the first 30 cycles of the logistic population with a growth parameter of r = 3.2. The striking feature of this graph is that it no longer settles down to a single constant value. Instead, it bounces back and forth between the two fixed values shown as xa and x b, repeating itself once every two cycles. In the language developed for the driven damped pendulum, we can say that the period has doubled to period two, and this period-two limiting motion is called a two-cycle. Figure 12.37 A logistic population with growth parameter r = 3.2. The population never settles down to a constant value; rather, it os- cillates between two values, repeating itself once every two cycles. In other words, it has doubled its period to period two. We can understand the doubling of the period of the logistic map with the graphical methods already developed, although the argument is a bit more complicated. The essential observations are these: First, neither of the two limiting values x a and xb is a fixed point of the map f (x). Instead, f (x a) = xb and f (x b) = xa. (12.52) Let us, however, consider the double map (or second iterate map) g(x) = f (f (x)), (12.53) which carries the population xt onto the population two years hence, Xt+2 = g (XI). (12.54) It is clear from Figure 12.37 or Equations (12.52) that both x a and x b are fixed points of the double map g(x), 24 g(x\u201e) = x, and g (x b) = xb. (12.55) 24 These points are also called second-order fixed points. 508 Chapter 12 Nonlinear Mechanics and Chaos (a) (b) Figure 12.38 The logistic map f (x) and its second iterate g(x) = f (f (x)). (a) For a growth parameter r = 0.8, each function is a single arch which intersects the 45\u00b0 line just once, at the origin. (b) When r = 2.6, f (x) is higher than before, but still just a single arch; g(x) has developed two maxima with a valley in between. Both functions have acquired a second intersection with the 45\u00b0 line, at the same point marked x* Thus to study the two-cycles of", " the map f (x) we have only to examine the fixed points of the double map g(x) = f (f (x)), and for this we can use our understanding of fixed points. Before we do this, it is worth noticing that any fixed point of f (x) is automatically also a fixed point of g (x). (If x t+i = xt for all t, then certainly Xt+2 = X t.) Therefore, the two-cycles of f (x) correspond to those fixed points of g(x) that are not also fixed points of f (x). Since f (x) is a quadratic function of x, it follows that g(x) = f (f (x)) is a quartic function, whose explicit form can be written down and studied. However, we can gain a better understanding by considering its graph. When r is small, we know that f (x) is a single low arch (as in Figure 12.35) and you can easily convince yourself that g(x) is an even lower arch, as sketched in Figure 12.38(a), which shows both functions for a growth parameter r = 0.8. As r increases, both arches rise, and by the time r = 2.6 the function g(x) has developed two maxima as seen in Figure 12.38(b). (You can explore the reason for this development in Problem 12.26.) Also, both curves now intersect the 45\u00b0 line twice, once at the origin and once at the fixed point x* = (r \u2014 1)/r. That both curves intersect the 45\u00b0 line at the same points shows two things: First, as we already knew, every fixed point of f (x) is also a fixed point of g(x), and, second, (for the growth parameters shown in this figure) every fixed point of g(x) is also a fixed point of f (x); that is, there are no two-cycles yet. As we increase the growth parameter r still further, the two crests of the double map g(x) continue to rise, while the valley between them gets lower. (Again see Problem 12.26 for the reason.) Figure 12.39 shows the curves of f (x) and g(x) for growth parameters r = 2.8, 3.0, and 3.4. With r = 2.8 [part (a)], the double map g(x) still has just the same two fixed points as f (x), at x = 0 and at the point indicated as x*. By the time r = 3.4 [part (c)], the double map has developed two additional fixed points, shown as xa and xb ; that is, the logistic map now has a two-cycle. The threshold value at which the two-cycle appears is clearly the value for which the curve g(x) is tangent to the 45\u00b0 line \u2014 namely r = 3 for the logistic map, as in part (b) of the figure. If Section 12.9 The Logistic Map 509 Figure 12.39 The logistic map f (x) (solid curves) and its second iterate g(x) = f (f (x)) (dashed) for r = 2.8, 3.0, and 3.4. (a) With r = 2.8, the map g(x) has the same two fixed points as f (x), namely x = 0 and x = x* as shown. (b) When r reaches 3.0, the curve of g (x) is tangent to the 45\u00b0 line, and for any larger value of r, as in (c), the map g(x) has two extra fixed points labelled x a and x b. r < 3, the curve g(x) crosses the 45\u00b0 line just once, at x*; if r > 3 it crosses three times, once at x* and two more times, at a:, and x b (one above and one below x*). We already know that r = 3 is the threshold at which the fixed point x* becomes unstable. Thus, Figure 12.39 shows that the two-cycle appears at the moment when the \"one-cycle\" (that is, the fixed point) becomes unstable. Happily, we are now in a position to see why this has to be. We have already noted that the two-cycle appears precisely when the curve g(x) is tangent to the 45\u00b0 line at the point x*, that is, when (x*) = 1. (12.56) To see what this implies, let us evaluate the derivative g' (x) of the double map g(x) at either of the two-cycle fixed points, xa say, g (xa) = dx g(x) f (f (", "x)) \u2014 dx Xa Xa f f (X)) f I (X) xa = ft (xb) f'(xa). (12.57) Here in the last expression of the first line I have used the chain rule, and in the second line I have used the fact that f (x a) = xb. Let us apply this result to the birth of the two- cycle in Figure 12.39(b). At the moment of birth, x* = xa = xb, and we can combine (12.56) and (12.57) to give [f l (x*)]2 = 1. This means that I f' (x *) I = 1, and, by our test for stability, we see that the moment when the two-cycle is born is precisely the moment when the fixed point x* becomes unstable. We can use these same techniques to explore what happens as we increase r still further. For example, one can show (Problem 12.28) that the two-cycle that we have just seen appearing at r = 3 becomes unstable at r = 1 4- = 3.449 and is succeeded by a stable four-cycle. However, to keep this long chapter from growing totally out of 510 Chapter 12 Nonlinear Mechanics and Chaos Figure 12.40 The fixed-points and two-cycles of the logistic map as functions of the growth parameter r. The solid curves are stable and the dashed curves unstable. bounds, I shall just sketch briefly some highlights that can be found by exploring the logistic map numerically. Bifurcation Diagrams In Figure 12.36 we saw the complete history of the fixed points of the logistic map itself. If we add onto that picture the fixed points x a and x b of the double map g (x) = f (f (x)) (that is, the two-cycles of the logistic map) we get the graphs shown in Figure 12.40. These curves are reminiscent of the beginnings of the bifurcation diagram, Figure 12.17, for the driven damped pendulum. In fact, we can redraw Figure 12.40 using the same procedure as was used for Figure 12.17: First one picks a large number of equally spaced values of the growth parameter in the range of interest. (I chose the range 2.8 < r < 4, since this is where the excitement lies, and chose 1200 equally spaced values in this range.) Then for each value of r one calculates the populations x0, x 1, x2,, x t max, where tmax is some very large time. (I chose tmax = 1000.) Next one chooses a time tam, large enough to let all transients die out. (I chose tmifl = 900.) Finally, in a plot of x against r, one shows the values of x t for tima < t < tmax as dots above the corresponding value of r. The resulting bifurcation diagram for the logistic map is shown in Figure 12.41. The similarity of the bifurcation diagram of Figure 12.41 for the logistic map and Figure 12.17 for the driven pendulum is striking indeed. The interpretation of the two pictures is also similar. On the left of Figure 12.41 we see the period-one attractor for r < 3, followed by the first period doubling at r = 3. This is followed by the second doubling at r = 3.449, and a whole cascade of doublings that end in chaos near r = 3.570. With the help of this diagram we can predict the long-term behavior of the logistic population for any particular choice of r (though there is clearly much fine detail that cannot be distinguished at the scale of Figure 12.41). For example at r = 3.5, it is clear that the population should have period four, a claim that is borne out in Figure 12.42(a), which shows the twenty cycles from t = 100 to 120 for this value of r. Similarly, around r = 3.84, sandwiched between wide intervals of chaos, you can see a narrow window that appears to have period three, an observation borne Section 12.9 The Logistic Map 511 3.0 3.2 3.4 3.6 3.8 4.0 r Figure 12.41 Bifurcation diagram for the logistic map. A period-doubling cas- cade is clearly visible starting at r 1 = 3, with a second doubling at r2 = 3.449, and ending in chaos at r c = 3.570. Several windows of periodicity stand out clearly amongst the chaos, especially the period-three window near r = 3.84. The tiny rectangle near r = 3.84 is the region that is enlarged in Figure 12.44. 0.6 - 0.4 - 0.2 - 0 2 8 xl 0.", "5 (a) 100 120 110 (b) Figure 12.42 Long-term evolution of logistic populations with growth param- eters r = 3.5 and 3.84. (a) Period four. With r = 3.5, the twenty cycles 100 < t < 120 take on just four distinct values at intervals of four cycles. (The dashed horizontal line is just to highlight the constancy of every fourth dot.) (b) Period three. With r = 3.84, the population repeats every three cycles. out in Figure 12.42(b). As an example of chaos, Figure 12.43 shows the evolution of a population with r = 3.7; in the eighty cycles shown, there is no evidence of any repetition. From Figure 12.41 (and careful enlargements), one can read off the threshold values of r at which the period doublings occur. If we denote by rn the threshold at which the cycle of period 2n appears, these are found to be 512 Chapter 12 Nonlinear Mechanics and Chaos r = 3.7 0.5..'1\\11 1/AlT 0 100 120 140 160 180 Figure 12.43 Chaos. With r = 3.7, the eighty cycles, 100 < t < 180, of the logistic map show no tendency to repeat themselves. As you can check, the separation between successive thresholds shrinks geomet- rically, just like the corresponding intervals for the period doubling of the driven pendulum. In fact, (see Problem 12.29) the numbers (12.58) give a remarkable fit to the Feigenbaum relation (12.17) that we first met in connection with the DDP, with the same Feigenbaum constant (12.18). Another striking parallel with the driven pen- dulum is this (Problem 12.30): For those r for which the evolution is non-chaotic, if two populations start out sufficiently close to one another, their difference will con- verge exponentially to zero as t oo. For those r for which the evolution is chaotic, the same difference diverges exponentially as t oo. That is, the chaotic evolution of the logistic map shows the same sensitive dependence on initial conditions that we found for the driven pendulum. Perhaps the most striking feature of the logistic bifurcation diagram is that, when one zooms in on certain parts of the diagram, a perfect self similarity emerges. The small rectangle near r = 3.84 in Figure 12.41 has been enlarged many fold in Figure 12.44. Apart from the facts that this new picture is upside down and its scale is vastly different, it is a perfect copy of the whole original diagram of which it is a part. This is a striking example of the self similarity which appears in many places in the study of chaos and which we met in connection with the Poincare section of the DDP shown in Figures 12.29, 12.30, and 12.31. There are many other features of the logistic map and more parallels with the DDP, all worth exploring and some treated in the problems at the end of this chapter. Here, however, I shall leave the logistic map and close this chapter with the hope that you feel at home with some of the main features of chaos and the tools used to explore them. I hope too that your appetite has been whetted to explore this fascinating subject further.25 25 For a comprehensive history, with very little mathematics, see Chaos, Making a New Science by James Gleick, Viking-Penguin, New York (1987). For a quite mathematical, but highly readable, account of chaos in many different fields see Nonlinear Dynamics and Chaos by Steven H. Strogatz, Addison-Wesley, Reading, MA (1994). Two books which focus mostly on chaos in physical systems are Chaotic Dynamics: An Introduction by G.L. Baker and J.P.Gollub, Cambridge Universtiy Press, Cambridge (1996) and Chaos and Nonlinear Dynamics by Robert C. Hilborn, Oxford University Press, New York (2000). Principal Definitions and Equations of Chapter 12 513 x 3.84 3.855 Figure 12.44 A many-fold enlargement of the small rectangle in the logistic bifurcation diagram of Figure 12.41. This tiny section of the original diagram is a perfect, upside-down copy of the whole original. Note that this section is just one of three strands in the original; thus, although this diagram starts out looking just like period 1 doubling to period 2, it is actually period 3 doubling to period 6 and so on. Principal Definitions and Equations of Chapter 12 The Driven Damped Pendulum A damped pendulum that is driven by a sinusoidal force F (t) = Fo cos(cot) satisfies the nonlinear equation 2$ coo sin 4) = y co02 cos cot [Eq. (12.11)] where", " y = Fol mg is called the drive strength and is the ratio of the drive amplitude to the weight. Period Doubling For small drive strengths, (y 1) the long-term response, or attractor, of the pen- dulum has the same period as the drive force. But if y is increased past y i = 1.0663, for certain initial conditions and drive frequencies, the attractor undergoes a period- doubling cascade, in which the period repeatedly doubles, approaching infinity as y y, = 1.0829. [Section 12.4] Chaos If the drive strength is increased beyond yc, at least for certain choices of drive frequency and initial conditions, the long-tenn motion becomes nonperiodic, and we say that chaos has set in. As y is increased still further, the long-term motion varies, sometimes chaotic, sometimes periodic. [Sections 12.5 & 12.6] 514 Chapter 12 Nonlinear Mechanics and Chaos Sensitivity to Initial Conditions Chaotic motion is extremely sensitive to initial conditions. If two identical chaotic pendulums with identical drive forces are launched with slightly different initial conditions, their separation increases exponentially with time, however small the initial difference. [Section 12.5] Bifurcation Diagrams A bifurcation diagram is a plot of the system's position at discrete times, to, to + 1, to + 2, (more generally to, to + r, to + 2r, -) as a function of the drive strength (more generally the appropriate control parameter). [Figures 12.17 & 12.18] State-Space Orbits and Poincare Sections The state space for a system with n degrees of freedom is the 2n-dimensional space comprising the n generalized coordinates and the n generalized velocities. For the DDP, the points in state space have the form (0, 0). A state-space orbit is just the path traced in state space by a system as t evolves. A Poincare section is a state-space orbit restricted to discrete times t o, to + 1, to + 2, (and, when n > 2, to a subspace of fewer dimensions). [Sections 12.7 & 12.8] The Logistic Map The logistic map is a function (or \"map\") that gives a number x t at regular discrete intervals (for example, the relative population of a certain bug once each year) as x t _fi = rxt (1 \u2014 x t ). [Eq. (12.42)] Although this is not a mechanical system, it exhibits many of the features (period doubling, chaos, sensitivity to initial conditions) of nonlinear mechanical systems. [Section 12.9] Problems for Chapter 12 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). Warning: Even when the motion is nonchaotic, it can be very sensitive to tiny errors. In several of the computer problems you may need to increase your working precision to get satisfactory results. SECTION 12.1 Linearity and Nonlinearity 12.1 * Consider the nonlinear first-order equation 1 = 2V \u2014 1. (a) By separating variables, find a solution x i (t). (b) Your solution should contain one constant of integration k, so you might reasonably expect it to be the general solution. Show, however, that there is another solution, x 2 (t) = 1, that is Problems for Chapter 12 515 not of the form of x i (t) whatever the value of k. (c) Show that although x i (t) and x 2 (t) are solutions, neither Ax i (t), nor Bx 2 (t), nor x 1 (t) x 2 (t) are solutions. (That is, the superposition principle does not apply to this equation.) 12.2 * Here is a different example of the disagreeable things that can happen with nonlinear equations. Consider the nonlinear equation z = 21i. Since this is first-order, one would expect that specification of x (0) would determine a unique solution. Show that for this equation there are two different solutions, both satisfying the initial condition x (0) = 0. [Hint: Find one solution x i (t) by separating variables, but note that x 2 (t) = 0 is another. Fortunately none of the equations normally encountered in classical mechanics suffer from this disagreeable ambiguity.] 12.3 * Consider a second-order linear homogeneous equation of the form (12.6). (a) Write out a detailed proof of the superposition principle, that if x JO and x 2 (t) are solutions of this equation, then so is any linear combination a ix a 2x 2 (t), where a l and a2 are any two constants. (b) Consider now the nonlinear equation in which the third term of (12.6) is replaced by", " r (t),Vx (t). Explain clearly why the superposition principle does not hold for this equation. 12.4 * Consider an inhomogeneous second-order linear equation of the form p(t)x(t) q(t).i(t) r(t)x(t) = f (t). (12.59) Let x (t) denote a solution (a \"particular\" solution) of this equation and prove that any solution x (t) can be written as x(t) = x p (t) aix i (t) a2x 2 (t) (12.60) where x 1 (t) and x 2 (t) are two independent solutions of the corresponding homogeneous equation\u2014 that is, (12.59) with f (t) deleted. [Hint: Write down the equations for x(t) and xp (t) and subtract.] This result shows that to find all solutions of (12.59), we have only to find one particular solution and two independent solutions of the corresponding homogeneous equation. (b) Explain clearly why the result you proved in part (a) is not, in general, true for a nonlinear equation such as p(t).(t) q(t)x(t) r(t)Ax(t) = f (t). SECTION 12.3 Some Expected Features of the DDP 12.5 * Use Euler' s relation and the corresponding expression for cos 4) (inside the front cover) to prove the identity (12.15). 12.6 ** [Computer] (a) Use appropriate sofware to solve the equation (12.11) numerically, for a DDP with the following parameters: drive strength y = 0.9, drive frequency w = 2n-, natural frequency coo = 1.5o), damping constant p = 0)014, and initial conditions (0) = 4(0) = 0. Solve the equation and plot your solution for six cycles, 0 < t < 6, and verify that you get the result shown in Figure 12.3. (b) and (c) Solve the same equation twice more with the two different initial conditions 4)(0) = +7t/2 \u2014 both with..(0) = 0 \u2014 and plot all three solutions on the same picture. Do your results bear out the claim that, for this drive strength, all solutions (whatever their initial conditions) approach the same periodic attractor? 516 Chapter 12 Nonlinear Mechanics and Chaos SECTION 12.4 The DDP: Approach to Chaos 12.7 ** [Computer] Do all the same calculations as in Problem 12.6 but with a drive strength y = 1.06 and for 0 < t < 10. In part (a) verify that your results agree with Figure 12.4. Do your results suggest that there is still a unique attractor to which all solutions (whatever their initial conditions) converge? (At first sight the answer may appear to be \"No,\" but remember that values of 0 that differ by 27r should be considered to be the same.) 12.8 ** [Computer] Use a computer to find a numerical solution of the equation of motion (12.11) for a DDP with the following parameters: drive strength y = 1.073, drive frequency w = 27r, natural frequency wo = 1.5w, damping constant fi = wo /4, and initial conditions 0 (0) = 7r/2 and Q)(0) = 0. (a) Solve for 0 < t < 50, and then plot the first ten cycles, 0 < t < 10. (b) To be sure that the initial transients have died out, plot the ten cycles 40 < t < 50. What is the period of the long-term motion (the attractor)? 12.9 ** [Computer] Do the same calculations as in Problem 12.8 with all the same parameters except that 0 (0) = 0. Plot the first 30 cycles 0 < t < 30 and check that you agree with Figure 12.5. Plot the ten cycles 40 < t < 50 and find the period of the long-term motion. 12.10 ** [Computer] Explore the behavior of the DDP with the same parameters as in Problem 12.8, but with several different initial conditions. For example, you might keep 4(0) = 0 but try various different values for 0 (0) between \u2014n- and 7r. You will find that the initial behavior varies quite a lot according to the initial conditions, but the long-term motion is the same in all cases (as long as you remember that values of 0 that differ by multiples of 27r represent the same position). 12.11 ** Test how well the values of the thresholds yn given in Table 12.1 fit the Feigenbaum relation (12.17)", " as follows: (a) Assuming that the Feigenbaum relation is exactly true use it to prove that (Yn+1 Yn) = ( 1 / 6 ) n-1 (Y2 \u2014 Y1) and, hence, that a plot of lnly s\u2018 n+1 yn) against n should be a straight line with slope \u2014 In 8. (b) Make this plot for the three differences of Table 12.1. How well do your points seem to bear out our prediction? Fit a line to your plot (either graphically or using a least squares fit) and find the slope and hence the Feigenbaum number 6. [You would not expect to get very good agreement with the known value (12.18) for two reasons: You have only three points to plot and the Feigenbaum relation (12.17) is only approximate, except in the limit of large n. Under the circumstances, you will find the agreement is remarkable.] 12.12 ** Here is another way to look at the Feigenbaum relation (12.17): (a) Assuming that (12.17) is exactly true, prove that the thresholds )6, approach a finite limit y, and then prove that yp, = y, \u2014 where K is a constant. This means that a plot of y n against 6' should be a straight line. (b) Using the known value (12.18) of 8 and the four values of yn in Table 12.1 make this plot. Does it seem to fit our prediction? The vertical intercept of your graph should be y c. What is your value and how well does it agree with (12.20)? SECTION 12.5 Chaos and Sensitivity to Initial Conditions 12.13 * You can see in Figure 12.13 that for y = 1.105, the separation of two identical pendulums with slightly different initial conditions increases exponentially. Specifically AO starts out at 10 -4 and by t = 14.5 it has reached about 1. Use this to estimate the Liapunov exponent X as defined in (12.26), AO (t) K e At. Your answer should confirm that A > 0 for chaotic motion. 12.14 ** [Computer] Numerically solve the equation of motion (12.11) for a DDP with drive strength y = 1.084, and the following other parameters: drive frequency w = 27r, natural frequency wo = 1.5w, Problems for Chapter 12 517 damping constant 13 = w o /4, and initial conditions 0 (0) = 0(0) = 0. Solve for the first seven drive cycles (0 < t < 7) and call your solution 0 1 (t). Solve again for all the same parameters except that 0 (0) = 0.00001 and call this solution 02 (t). Let A0 (t) 02 (t) 0 1 (0 and make a plot of log I AO (t)i against t. With this drive strength the motion is chaotic. Does your plot confirm this? In what sense? 12.15 ** [Computer] Numerically solve the equation of motion (12.11) for a DDP with the following parameters: drive strength y = 0.3, drive frequency w = 27, natural frequency wo = 1.5w, damping constant 18 = woI4, and initial conditions 0 (0) = (i)(0) = 0. Solve for the first five drive cycles (0 < t < 5) and call your solution 01 (t). Solve again for all the same parameters except that 0 (0) = 1 (that is, the initial angle is one radian) and call this solution 02 (t). Let AO (t) = 02 (t) - 0 1 (t) and make a plot of log I AO (t)I against t. Does your plot confirm that A0 (t) goes to zero exponentially? Note: The exponential decay continues indefinitely, but A0 (t) eventually gets so small that it is smaller than the rounding errors, and the exponential decay cannot be seen. If you want to go further, you will probably need to crank up your precision. 12.16 ** Consider the chaotic motion of a DDP for which the Liapunov exponent is A = 1, with time measured in units of the drive period as usual. (This is very roughly the value found in Problem 12.13.) (a) Suppose that you need to predict 0 (t) with an accuracy of 1/100 rad and that you know the initial value 0 (0) within 10 -6 rad. What is the maximum time tmax for which you can predict 0 (t) within the required accuracy? This t max is sometimes called the time horizon for prediction within a specified accuracy. (b) Suppose that, with a vast expenditure of money and", " labor, you manage to improve the accuracy of your initial value to 10 -9 radians (a thousand-fold improvement). What is the time horizon now (for the same required accuracy of prediction)? By what factor has t max improved? Your results illustrate the difficulty of making accurate long-term predictions for chaotic motion. 12.17 ** [Computer] In Figure 12.15, you can see that for y = 1.503 the DDP \"tries\" to execute a steady rolling motion changing by 27 once each cycle, but that there is superposed an erratic wobbling and that the direction of the rolling reverses itself from time to time. For other values of y, the pendulum actually does approach a steady, periodic rolling. (a) Solve the equation of motion (12.11) for a drive strength y = 1.3 and all other parameters as in the first part of Problem 12.14, for 0 < t < 8. Call your solution 0 1 (t) and plot it as a function of t. Describe the motion. (b) It is hard to be sure that the motion is periodic based on this graph, because of the steady rolling through -27 each cycle. As a better check, plot 0 1 (t) \u00b1 2nt against time. Describe what this shows. This kind of periodic rolling motion is sometimes described as phase-locked. 12.18** [Computer] Since the rolling motion of Problem 12.17 is periodic (and hence not chaotic) we would expect the difference AO (t) between neighboring solutions (solutions of the same equation, but with slightly different initial conditions) to decrease exponentially. To illustrate this, do part (a) of Problem 12.17 and then find the solution of the same problem except that 0 (0) = 1. Call this second solution 02(0 and let AO (t) = 02 (t) - 0 1 (t). Make a plot of log I AO (t)I against t and comment. SECTION 12.7 State-Space Orbits 12.19 * Consider an undamped, undriven simple harmonic oscillator \u2014 a mass m on the end of a spring whose force constant is k. (a) Write down the general solution x (t) for the position as a function of time t. Use this to sketch the state-space orbit, showing the motion of the point [x(t),.i(t)] in the two- dimensional state space with coordinates (x, Explain the direction in which the orbit is traced as time advances. (b) Write down the total energy of the system and use conservation of energy to prove that the state-space orbit is an ellipse. 518 Chapter 12 Nonlinear Mechanics and Chaos 12.20 * Consider a weakly-damped, but undriven oscillator as described by Equation (5.28). The general motion is given by (5.38). (a) Use this to sketch the state-space orbit, showing the movement of (x, 5c) for 0 < t < 10 with the parameters b = 0, co l = 27r, and,8 = 0.5. (b) This system has a unique stationary attractor. What is it? Explain it, both with reference to your sketch and in terms of energy conservation. SECTION 12.9 The Logistic Map 12.21 ** Here is an iterated map that is easily studied with the help of your calculator: Let x t+1 = f (x t ) where f (x) = cos(x). If you choose any value for x o, you can find x 1, x2, x3, by simply pressing the cosine button on your calculator over and over again. (Be sure the calculator is in radians mode.) (a) Try this for several different choices of x o, finding the first 30 or so values of x t. Describe what happens. (b) You should have found that there seems to be a single fixed attractor. What is it? Explain it, by examining (graphically, for instance) the equation for a fixed point f (x*) = x* and applying our test for stability [namely, that a fixed point x* is stable if I f (x*)1 < 1]. 12.22 ** Consider the iterated map x t+i = f (x t ) where f (x) = x 2. (a) Show that it has exactly two fixed points of which just one is stable. What are they? (b) Show that x t approaches the stable fixed point if and only if \u20141 < x o < 1. The interval \u20141 < x o < 1 is called a basin of attraction since all sequences x o, x 1, that start in the \"basin\" are attracted to the same attractor. (c) Show that x t co if and only if Ixo l > 1.", " (Thus we could say the map has a second stable fixed point at x = oo and the basin of attraction for this fixed point is the set Ix o > 1.) For chaotic systems, the basins of attraction can be much more complicated than these examples and are often fractals. 12.23 ** [Computer] Consider the sine map xt\u00b1i = f (x t ) where f (x) = r sin(rrx). The interesting behavior of this map is for 0 < x < 1 and 0 < r < 1, so restrict your attention to these ranges. (a) Using a plot analogous to Figure 12.35, discuss the fixed points of this map. Show that the map has either one or two fixed points, depending on the value of r. Show that when r is small there is just one fixed point, which is stable. (b) At what value of r (call it ro) does the second fixed point appear? Show that ro is also the value of r at which the first fixed point becomes unstable. (c) As r increases, the second fixed point eventually becomes unstable. Find numerically the value r i at which this occurs. 12.24 ** [Computer] Consider the sine map of Problem 12.23. Using a progammable calculator (or a computer) you can easily find the first ten or twenty values of x t for any chosen inital value x o. Taking xo = 0.3, calculate the first 10 values of x t for each of the following values of the parameter r: (a) r = 0.1, (b) r = 0.5, (c) r = 0.78. In each case plot your results (x t against t) and describe the long-term attractor. If you did Problem 12.23, are your results here consistent with what you proved there? 12.25 ** [Computer] The sine map of Problem 12.23 exhibits period-doubling cascades just like the logistic map. To illustrate this, take x o = 0.8 and find the first twenty values x t for each of the following values of the parameter r: (a) r = 0.60, (b) r = 0.79, (c) r = 0.85, and (d) r = 0.865. Plot your results (as four separate plots) and comment. 12.26 ** The appearance of a two-cycle of the logistic map at the exact moment when the one-cycle becomes unstable follows directly from the behavior of the graphs of f (x) and g(x) = f (f (x)) as shown in Figures 12.38 and 12.39. The crucial point is that the function f (x) is a simple arch that gets steadily higher as we increase the control parameter r from 0; at the same time, f (f (x)) starts out as a simple arch which is lower than f (x), but developes two maxima [higher than f (x)] with a minimum [lower than f (x)] in between. Explain clearly in words why this behavior of f (f (x)) follows inevitably from that of f (x). [Hint: Since f (x) is symmetric about x = 0.5, you need only consider its behavior as x runs from 0 to 0.5. The advantage of this argument is that it applies to any map of the form f (x) = r0 (x) where (/)(x) has a single symmetric arch (such as the sine map of Problem 12.23).] Problems for Chapter 12 519 12.27** The two-cycles of the map f (x) correspond to fixed points of the second-iterate g(x) = f (f (x)). Thus the two values shown as x a and xb in Figure 12.37 are roots of the equation f (f (x)) = x. In the case of the logistic map this is a quartic equation, which is not too hard to solve: (a) Verify that for the logistic function f (x) x\u2014 f(f(x))=rx lx r \u2014 1) [r2 X 2 \u2014 r (r 1)x r l]. (12.61) Thus the fixed-point equation x \u2014 f (f (x)) = 0 has four roots. The first two are x = 0 and x = (r \u2014 1)/r. Explain these and show that the other two roots are r + 1,Ar 1)(r \u2014 3) Xa, xb 2r (12.62) Explain how you know that these are the two points of a two-cycle. (b) Show that for r < 3 these roots are complex and hence that there is no real two-cycle. (c) For r > 3 these two roots are real and there is a real", " two-cycle. Find the values of x a and xb for the case r = 3.2 and verify the values shown in Figure 12.37. 12.28 ** Equation (12.62) in Problem 12.27 gives the two fixed points of the two-cycle of the logistic map. One can observe this two-cycle only if it is stable, which will happen if I g' (x)I < 1, where g(x) is the double map g(x) = f (f (x)) and x is either of the values xa or xb. (a) Combine (12.62) with Equation (12.57) to find g' (xa). [Notice that because (12.57) is symmetric in x a and xb, you will get the same result whether you use x a or xb ; that is, the two points necessarily become stable or unstable at the same time.] (b) Show that the two-cycle is stable for 3 < r < 1 + Ni6-. This establishes that the threshold at which period 2 is replaced by period 4 is r 2 = 1 + = 3.449. 12.29 ** [Computer] The thresholds rn for period doubling of the logistic map are given by Equation (12.58). These should satisfy the Feigenbaum relation (12.17), at least in the limit that n cc (with y replaced by r, of course). Test this claim as follows: (a) If you have not done Problem 12.11 prove that the Feigenbaum relation (if exactly true) implies that (r,\u00b11 \u2014 rn ) = K / 6n. (b) Make a plot of ln(r\u201e \u00b1i \u2014 rn ) against n. Find the best-fit straight line to the data and from its slope predict the Feigenbaum constant. How does your answer compare with the accepted value 6 = 4.67? 12.30 ** [Computer] The chaotic evolution of the logistic map shows the same sensitivity to initial conditions that we met in the DDP. To illustrate this do the following: (a) Using a growth rate r = 2.6, calculate xt for 1 < t < 40 starting from x o = 0.4. Repeat but with the inital condition 4 = 0.5 (the prime is just to distinguish this second solution from the first \u2014 it does not denote differentiation) and then plot log Ix; \u2014 x t I against t. Describe the behavior of the difference x; \u2014 x t. (b) Repeat part (a) but with r = 3.3. In this case, the long term evolution has period 2. Again describe the behavior of the difference x; \u2014 x t. (c) Repeat parts (a) and (b), but with r = 3.6. In this case, the evolution is chaotic, and we expect the difference to grow exponentially; therefore, it is more interesting to take the two inital values much closer together. To be definite take x o = 0.4 and xo = 0.400001. How does the difference behave? 520 Chapter 12 Nonlinear Mechanics and Chaos 12.31 *** [Computer] When the evolution of the logistic map is non-chaotic, two solutions with the same r that start out sufficiently close, will converge exponentially. (This was illustrated in Problem 12.30.) This does not mean that any two solutions with the same r will converge. (a) Repeat Problem 12.30(a), with all the same parameters except r = 3.5 (a value for which we know the long-term motion has period 4). Does x; \u2014 x t approach zero? (b) Now do the same exercise but with the initial conditions x0 = 0.45 and xo = 0.5. Comment. Can you explain why, if the period is greater than 1, it is impossible that the difference x; \u2014 x t go to zero for all choices of initial conditions? 12.32 *** [Computer] Make a bifurcation diagram for the logistic map, in the style of Figure 12.41 but for the range 0 < r < 3.55. Take x0 = 0.1. Comment on its main features. [Hint: Start by using a very small number of points, perhaps just r going from 0 to 3.5 in steps of 0.5 and t going from 51 to 54. This will let you calculate for each of the values of r individually, and get the feel of how things work. To make a good diagram, you will then need to increase the number of points (r going from 0 to 3.55 in steps of 0.025, and t from 51 to 60, perhaps), and you will certainly need to automate the calculation of the large number of points.] 12.33 *** [Computer] Reproduce the logistic b", "ifurcation diagram of Figure 12.41 for the range 2.8 < r < 4. Take x0 = 0.1. [Hint: To make Figure 12.41 I used about 50,000 points, but you cer- tainly don't need to use that many. In any case, start by using a very small number of points, perhaps just r going from 2.8 to 3.4 in steps of 0.2 and t going from 51 to 54. This will let you calculate for each of the values of r individually, and get the feel of how things work. To make a good diagram, you will then need to increase the number of points (r going from 2.8 to 4 in steps of 0.025, and t from 500 to 600, perhaps), and you will certainly need to automate the calculation of the large number of points.] 12.34 *** [Computer] Make a bifurcation diagram for the sine map of Problems 12.23 and 12.25. This should resemble Figure 12.41 but for the range 0.6 < r < 1. Take x0 = 0.1. Comment on its main features. [Hint: Start by using a very small number of points, perhaps just r going from 0.6 to 0.8 in steps of 0.05 and t going from 51 to 54. This will let you calculate for each of the values of r individually, and get the feel of how things work. To make a good diagram, you will then need to increase the number of points (r going from 0.6 to 1 in steps of 0.005, and t from 400 to 500, perhaps), and you will certainly need to automate the calculation of the large number of points.] CHAPTER Hamiltonian Mechanics In the first six chapters of this book, we worked entirely with the Newtonian form of mechanics, which describes the world in terms of forces and accelerations (as related by the second law) and is primarily suited for use in Cartesian coordinate systems. In Chapter 7, we met the Lagrangian formulation. This second formulation is entirely equivalent to Newton's, in the sense that either one can be derived from the other, but the Lagrangian form is considerably more flexible with regard to choice of coordinates. The n Cartesian coordinates that describe a system in Newtonian terms are replaced by a set of n generalized coordinates q 1, q2,, (7,, and Lagrange's equations are equally valid for essentially any choice of q 1, q2,, qn. As we have seen on many occasions, this versatility allows one to solve many problems much more easily using Lagrange's formulation. The Lagrangian approach also has the advantage of eliminating the forces of constraint. On the other hand, the Lagrangian method is at a disadvantage when applied to dissipative systems (for example, systems with friction). By now, I hope you feel comfortable with both formulations and are familiar with the advantages and disadvantages of each. Newtonian mechanics was first expounded by Newton in his Principia Mathemat- ica, published in 1687. Lagrange published his formulation in his book Mechanique Analytique in 1788. In the early nineteenth century, various physicists, including La- grange, developed yet a third formulation of mechanics, which was put into a complete form in 1834 by the Irish mathematician William Hamilton (1805-1865) and has come to be called Hamiltonian mechanics. It is this third formulation of mechanics that is the subject of this chapter. Like the Lagrangian version, Hamiltonian mechanics is equivalent to Newtonian but is considerably more flexible in its choice of coordinates. In fact, in this respect it is even more flexible than the Lagrangian approach. Where the Lagrangian formalism centers on the Lagrangian function, the Hamiltonian approach is based on the Hamiltonian function 3-C (which we met briefly in Chapter 7). For most of the systems we shall meet, Ji is just the total energy. Thus one advantage of Hamilton's formalism is that it is based on a function, J-C, which (unlike the Lagrangian ) has a clear physical significance and is frequently conserved. The Hamiltonian approach is also 521 522 Chapter 13 Hamiltonian Mechanics especially well suited for handling other conserved quantities and implementing various approximation schemes. It has been generalized to various different branches of physics; in particular, Hamiltonian mechanics leads very naturally from classical mechanics into quantum mechanics. For all these reasons, Hamilton's formulation plays an important role in many branches of modern physics, including astrophysics, plasma physics, and the design of particle accelerators. Unfortunately, at the level of this book, it is hard to demonstrate many of the advantages of Hamilton's version over Lagrange's, and in this chapter I shall have to ask you to be content with learning the former as just an alternative to the latter \u2014 an alternative several of whose advantages I can mention but not explore", " in depth. If you go on to take a more advanced course in classical mechanics or to study quantum mechanics, you will certainly meet many of this chapter's ideas again. 13.1 The Basic Variables Because the Hamiltonian version of mechanics is closer to the Lagrangian than to the Newtonian and arises naturally from the Lagrangian, let us start by reviewing the main features of the latter, which centers on the Lagrangian function. For most systems of interest, \u00a3 is just the difference of the kinetic and potential energies, T \u2014 U, and in this chapter we shall confine attention to systems for which this is the case. The Lagrangian L is a function of the n generalized coordinates q 1,, qn, their n time derivatives (or generalized velocities) q i,, 4,, and, perhaps, the time: =, qn, 4 1,, 4,, T \u2014 U. (13.1) The n coordinates (q 1,, qn ) specify a position or \"configuration\" of the system, and can be thought of as defining a point in an n-dimensional configuration space. The 2n coordinates la,a1,, 4n, 41, 4n) define a point in state space, and specify a set of initial conditions (at any chosen time to) that determine a unique solution of the n second-order differential equations of motion, Lagrange's equations, az d aqi \u2014 dt a4i [i = 1,, n]. (13.2) For each set of initial conditions, these equations of motion determine a unique path or \"orbit\" through state space. You may also recall that we defined a generalized momentum given by Pi =. ai (13.3) If the coordinates (q 1,, qn ) are in fact Cartesian coordinates, the generalized momenta pi are the corresponding components of the usual momenta; in general, pi is not actually a momentum, but does, as we have seen, play an analogous role. The generalized momentum pi is also called the canonical momentum or the momentum conjugate to qi. Section 13.1 The Basic Variables 523 In the Hamiltonian approach, the central role of the Lagrangian \u00a3 is taken over by the Hamiltonian function, or just Hamiltonian, defined as J-C = - =t (13.4) The equations of motion, which we shall derive in the next two sections, involve derivatives of J-f rather than \u00a3 as in Lagrange's equations. We met the Hamiltonian function briefly in Section 7.8, where we proved that, provided the generalized coordinates (q 1,, qn ) are \"natural\" (that is, the relation between the q's and the underlying Cartesian coordinates is time independent), H is just the total energy of the system and is, therefore, familiar and easy to visualize. There is a second important difference between the Lagrangian and Hamiltonian formalisms. In the former we label the state of the system by the 2n coordinates (q1,, qn, 41,, 4\u201e), [Lagrange] (13.5) whereas in the latter we shall use the coordinates (q1, qn, Pi, Pn), [Hamilton] (13.6) consisting of the n generalized positions and the n generalized momenta (instead of the generalized velocities). This choice of coordinates has several advantages, a few of which I shall sketch and some of which you will have to take on faith. Just as we can regard the 2n coordinates (13.5) of the Lagrange approach as defining a point in a 2n-dimensional state space, so we can regard the 2n coordinates (13.6) of the Hamiltonian approach as defining a point in a 2n-dimensional space, which is usually called phase space! Just as the Lagrange equations of motion (13.2) determine a unique path in state space starting from any initial point (13.5), so (we shall see) Hamilton's equations determine a unique path in phase space starting from any initial point (13.6). A succinct way to state some of the advantages of the Hamiltonian formalism is that phase space has certain geometrical properties that make it more convenient than state space. Like Lagrange's approach, Hamilton's is best suited to systems that are subject to no frictional forces. Accordingly, I shall assume throughout this chapter that all the forces of interest are conservative or can at least be derived from a potential energy function. Although this restriction excludes many interesting mechanical systems, it still includes a huge number of important problems, especially in astrophysics and at the microscopic \u2014 atomic and molecular \u2014 level. I Many authors use the names \"state space\" and \"phase space\" interchangeably, but it is conve- nient to have different names for the different spaces, and I shall reserve \"state space\" for the space of positions and", " generalized velocities, and \"phase space\" for that of positions and generalized momenta. 524 Chapter 13 Hamiltonian Mechanics 13.2 Hamilton's Equations for One-Dimensional Systems To minimize notational complications, I shall first derive Hamilton's equations of motion for a conservative, one-dimensional system with a single \"natural\" generalized coordinate q. For example, you could think of a simple, plane pendulum, in which case q could be the usual angle 0, or a bead on a stationary wire, in which case q could be the horizontal distance x along the wire. For any such system, the Lagrangian is a function of q and g, that is, = \u00a3(q, q) = T (q, q) \u2014 U (q). (13.7) Recall that, in general, the kinetic energy can depend on q as well as 4, whereas, for conservative systems, the potential energy depends only on q. For example, for the simple pendulum (mass m and length L), = (o,.(p) = 1, 2 2 j:ML - mgL(1 \u2014 cos0), (13.8) where, in this case, the kinetic energy involves only not 0. For a bead sliding on a frictionless wire of variable height y = f (x), we saw in Example 11.1 (page 438) that = (x, 1) = T \u2014 U = lm[1 (x) 2]12 \u2014 mg f (x). (13.9) Here the x dependence of the kinetic energy came about when we rewrote the v in 2ni v2 in terms of the horizontal distance x. The two examples (13.8) and (13.9) illustrate a general result that we proved in Section 7.8 that the Lagrangian for a conservative system with \"natural\" coordinates (and in one dimension here) has the general form,C = (q, 4) = T \u2014 U = A (q) q 2 \u2014 U (q) (13.10) Notice that, while the kinetic energy can depend on q in a complicated way, through the function A(q), its dependence on q is just through the simple quadratic factor 4 2. As you can easily check by writing it down, Lagrange's equation for this Lagrangian is automatically a second-order differential equation for q. The Hamiltonian is defined by (13.4), which in one dimension reduces to /34 (13.11) In the discussion of Section 7.8, I offered some reason why one might perhaps expect a function defined in this way to be an interesting function to study. For now, let us just accept the definition as an inspired suggestion by Hamilton \u2014 a suggestion whose merit will appear as we proceed. 2 Given the form (13.10) of,C, we can calculate the 2 Actually, the change from to gf as the object of primary interest is an example of a mathematical maneuver, called a Legendre transformation, which plays an important role in several fields, most notably thermodynamics. For example, the change from the thermodynamic internal Section 13.2 Hamilton's Equations for One-Dimensional Systems 525 generalized momentum p as p = = A(q)q az 84 (13.12) so that pq = A(q)q 2 = 2T. Substituting into (13.11), we find that 9-C = p4 \u2014 G=2T \u2014(T \u2014 U)= T U. That is, the Hamiltonian G-C for the \"natural\" system considered here is precisely the total energy \u2014 the same result we proved for any \"natural\" system in any number of dimensions in Section 7.8. The next step in setting up the Hamiltonian formalism is perhaps the most subtle. In the Lagrangian approach, we think of as a function of q and 4, as is indicated explicitly in (13.10). Similarly, (13.12) gives the generalized momentum p in terms of q and 4. However, we can solve (13.12) for 4 in terms of q and p: 4 = p I A(q) = 4(q, p), (13.13) say. With 4 expressed as a function of q and p, let us now look at the Hamiltonian. Wherever 4 appears in H, we can replace it by 4(q, p), and 9-C becomes a function of q and p. In all its horrible detail, (13.11) becomes J-C(q, p) = p q(q, p) E.,(q,4(q, p)). (13.14) Our final step is to get Hamilton's equations of motion. To find these, we just evaluate the derivatives of 3-C(q, p) with respect to q and p. First, using the chain rule, we differentiate (13.14)", " with respect to q: = 84 _1- az + az 841 aq P aq aq 84 aq Now, in the third term on the right, you will recognize that az /84 = p. Thus, the first and third terms on the right cancel one another, leaving just ag-c aq aA aq d az dt 8q d dt P 131 (13.15) where the second equality follows from the Lagrange equation (13.2). This equation gives the time derivative of p (that is, /9) in terms of the Hamiltonian 3 -( and is the first of the two Hamiltonian equations of motion. Before we discuss it, let's derive the second one. Differentiating (13.14) with respect to p and using the chain rule, we find ag-f [. al az = q+ p\u2014 \u2014 ap aq ap. (13.16) energy U to enthalpy H is a Legendre transformation, closely analogous to Hamilton's change from \u00a3 to J-C. 526 Chapter 13 Hamiltonian Mechanics since the second and third terms in the middle expression cancel exactly. This is the second of the Hamiltonian equations of motion and gives q in terms of the Hamiltonian 3-C. Collecting them together (and reordering a bit), we have Hamilton's equations for a one-dimensional system: In the Lagrangian formalism, the equation of motion of a one-dimensional system is a single second-order differential equation for q. In the Hamiltonian approach, there are two first-order equations, one for q and one for p. Before we extend this result to more general systems or discuss any advantages the new formalism may have, let us look at a couple of simple examples. EXAMPLE 13.1 A Bead on a Straight Wire Consider a bead sliding on a frictionless rigid straight wire lying along the x axis, as shown in Figure 13.1. The bead has mass m and is subject to a conservative force, with corresponding potential energy U(x). Write down the Lagrangian and Lagrange's equation of motion. Find the Hamiltonian and Hamilton's equations, and compare the two approaches. Naturally, we take as our generalized coordinate q the Cartesian x. The Lagrangian is then Z(x, = T \u2014 U = 2mx 2 \u2014 U(x). The corresponding Lagrange equation is 8A ax d dt ax dU dx Or \u2014 \u2014 = mx (13.18) which is just Newton's F = ma, as we would expect. To set up the Hamiltonian formalism, we must first find the generalized momentum, p = = mx. az ax 0 in x Figure 13.1 A bead of mass m sliding on a frictionless straight wire. Section 13.2 Hamilton's Equations for One-Dimensional Systems 527 As expected, this is just the conventional mu momentum. This equation can be solved to give x = plm, which can then be substituted into the Hamiltonian to give (13.17) are 3-C = \u2014 = U (x)]= +U (x) n 2 m \u2014 n 2 2m 1,2 2m which you will recognize as the total energy, with the kinetic term -1.m.i 2 rewrit- ten in terms of momentum as p 2 / (2m). Finally, the two Hamilton equations = agi ap p m and = 8J-C_ \u2014 ax. dx The first of these is, from a Newtonian point of view, just the traditional definition of momentum, and, when we substitute this definition into the second equation, it gives us back ml = \u2014dU ldx again. As had to be the case, Newton, Lagrange, and Hamilton all lead us to the same familiar equation. In this very simple example, neither Lagrange nor Hamilton has any visible advantage over Newton. EXAMPLE 13.2 Atwood's Machine Set up the Hamiltonian formalism for the Atwood machine, first shown as Figure 4.15 and shown again here as Figure 13.2. Use the height x of m 1 measured downward as the one generalized coordinate. The Lagrangian is = T \u2014 U, where, as we saw in Example 7.3 (page 255), T \u2014 i(m 1 \u2014 2 m 2 )x.2 and U = \u2014(m 1 \u2014 m 2)gx. (13.19) x y Figure 13.2 An Atwood machine consisting of two masses, m 1 and m 2, suspended by a massless inextensible string that passes over a massless, frictionless pulley. Because the string's length is fixed, the position of the whole system is specified by the distance x of m 1 below any convenient fixed level. 528 Chapter 13 Hamiltonian Mechanics We can calculate the Hamiltonian 9-C either as 9-C = pz \u2014 G or, what is", " usually a little quicker (provided it is true 3), as 3-C = T U. Either way, we must first find the generalized momentum p=aLla.ic or, since U does not involve I, P = = (ni+ m2)1. aT ax We solve this to give I in terms of p as I = p 1(m i + m2), which we substitute into J-C to give 9-C as a function of x and p: aC = T U = 2 P 2 (m1 + m2) (m1 \u2014 m 2)gx. (13.20) We can now write down the two Hamilton equations of motion (13.17) as = 891 = oP mi+ m2 aaC and P = ax = (m1 \u2014 m2)g. Again, the first of these is just a restatement of the definition of the generalized momentum and, when we combine this with the second, we get the well-known result for the acceleration of the Atwood machine, x= MI in2 MI + M2 g. These two examples illustrate several of the general features of the Hamiltonian approach: Our first task is alway to write down the Hamiltonian 3-C (just as in the Lagrangian approach the first task is to write down IC). In the Hamiltonian approach there are usually a couple of extra steps, which are to write down the generalized momentum, to solve the resulting equation for the generalized velocity, and to express 3-C as a function of position and momentum. Once this is done, one can just turn the handle and crank out Hamilton's equations. In general, there is no guarantee that the resulting equations will be easy to solve, but it is a wonderful property of Hamilton's approach (like Lagrange's) that it provides an almost infallible way to find the equations of motion. 13.3 Hamilton's Equations in Several Dimensions Our derivation of Hamilton's equations for a one-dimensional system is easily ex- tended to multidimensional systems. The only real problem is that the equations can become badly cluttered with indices, so, to minimize the clutter, I shall use the ab- breviation introduced in Section 11.5: The configuration of an n-dimensional system 3 Remember that this second expression is true provided the generalized coordinate is \"natural,\" that is, the relation between the generalized coordinate and the underlying Cartesian coordinates is independent of time \u2014 a condition which is certainly met here (Problem 13.4). Section 13.3 Hamilton's Equations in Several Dimensions 529 is given by n generalized coordinates q l,, qi\u201e which I shall represent by a single bold-face q: momenta are q =(g1,, qn ). P = (Pi, Similarly, the generalized velocities become q = (41,, 4,), and the generalized It is important to remember that for now a bold-face q or p is not necessarily a three-dimensional vector. Rather q and p are n-dimensional vectors in the space of generalized positions or generalized momenta. Hamilton's equations follow directly from Lagrange's equations in their standard form. Thus to prove the former, we have only to assume the truth of the latter. To be specific, however, I shall make the same assumptions that we used in Chapter 7: I shall assume that any constraints are holonomic; that is, the number of degrees of freedom is equal to the number of generalized coordinates. I shall also assume that the nonconstraint forces can be derived from a potential energy function, though it is not essential that they be conservative (that is, the potential energy is allowed to depend on t). The equations that relate the N underlying Cartesian coordinates r 1,, r N to the n generalized coordinates q 1,, qn can depend on time; that is, it is not essential that the generalized coordinates be \"natural.\" These assumptions are enough to guarantee that the standard Lagrangian formalism applies, and will let us derive from it the Hamiltonian one. Thus our starting point is that there is a Lagrangian (q, 4, 0 T \u2014 U and that the evolution of our system is governed by the n Lagrange equations, aG aqi d aG dt 84, Li =1,, n]. (13.21) We shall define the Hamiltonian function as in (13.4), n i= 1 piqi \u2014 (13.22) where the generalized momenta are defined by Pi = az(4'4't) [i = 1,, n] (13.23) aqi as in (13.3). Just as in the one-dimensional case, our next step is to express the Hamiltonian as a function of the 2n variables q and p. To this end, note that we can view the equations (13.23) as n simultaneous equations for the n generalized velocities", " 4. We can in principle solve these equations to give the generalized velocities in terms of the variables p, q, and t: qi = qi(q1, qn, pi,, Pn, t) Ei = 1, nl 530 Chapter 13 Hamiltonian Mechanics or, more succinctly = 4 ( q, p,t). We can now eliminate the generalized velocities from our definition of the Hamilto- nian to give (again in agonizing detail) X = X (q, p, t) = E pi 4i (q, p, t) \u2014 \u00a3(q, 4 (q, p, t), t). (13.24) The derivation of Hamilton's equations now proceeds very much as in one dimen- sion, and I shall leave you to fill in the details (Problem 13.15). Following the same steps as led from (13.14) to (13.17), we differentiate J -C with respect to q, and then pi, and this leads to Hamilton's equations: Notice that for a system with n degrees of freedom, the Hamiltonian approach gives us 2n first-order differential equations, instead of the n second-order equations of Lagrange. Before we discuss an example of Hamilton's equations, there is one more derivative of J-C to consider, its derivative with respect to time. This is actually quite subtle. The function g-C (q, p, t) could vary with time for two reasons: First, as the motion proceeds, the 2n coordinates (q, p) vary, and this could cause g-C(q, p, t) to change; in addition, g-C(q, p, t) may have an explicit time dependence, as indicated by the final argument t, and this also can make gC vary with time. Mathematically, this means that dg-C I dt contains 2n + 1 terms, as follows: dg-C dt \\--Nn r a5-c i=1 L qi ag-c 15, 1 api J + ag-c. at (13.26) It is important to understand the difference between the two derivatives of g -C in this equation. The derivative on the left, Of I dt (sometimes called the total derivative), is the actual rate of change of g -C as the motion proceeds, with all the coordinates q1,, qn, Pi', p n changing as t advances. That on the right, as-c/at, is the partial derivative, which is the rate of change of g -C if we vary its last argument t holding all the other arguments fixed. In particular, if g -C does not depend explicitly on t, this partial derivative will be zero. Now it is easy to see that, because of Hamilton's equations (13.25), each pair of terms in the sum of (13.26) is exactly zero, so that we have the simple result dfl-C dt ag-c at Section 13.3 Hamilton's Equations in Several Dimensions 531 That is, g-C varies with time only to the extent that it is explicitly time dependent. In particular, if 3-C does not depend explicitly on t (as is often the case), then 3 -( is a constant in time; that is, the quantity g -C is conserved. This is the same result we derived in Section 7.8. 4 In Section 7.8, we proved a second result regarding time dependence: If the relation of the generalized coordinates q 1,, qn to the underlying rectangular coordinates is independent of t (that is, our generalized coordinates are \"natural\"), then the Hamiltonian g-C is just the total energy, J -C = T U. In the remainder of this chapter, I shall consider only the case that the generalized coordinates are \"natural\" and that g-C is not explicitly time-dependent. Thus it will be true from now on that g -C is the total energy and that total energy is conserved. Let us now work out an example of the Hamiltonian formalism for a system in two spatial dimensions. Unfortunately, like all reasonably simple examples, this does not exhibit any significant advantages of the Hamiltonian over the Lagrangian approach; rather, in this example, the Hamiltonian approach is just an alternative route to the same final equation of motion. EXAMPLE 13.3 Hamilton's Equations for a Particle in a Central Force Field Set up Hamilton's equations for a particle of mass m subject to a conservative central force field with potential energy U (r), using as generalized coordinates the usual polar coordinates r and 0. By conservation of angular momentum, we know that the motion is confined to a fixed plane, in which we can define the polar coordinates r and 0. The kinetic energy is given in terms of these generalized coordinates by the familiar expression T", " =2m(i 2 r2(1)2). (13.27) Since the equations relating (r, 0) to (x, y) are time-independent, we know that T U, which we must express in terms of r and 0 and the corresponding generalized momenta pr and po. These latter are defined by the relation s pi = m1.4 = aTia4i, which gives p aT and p = aT =mr2cb. (13.28) The momentum pr conjugate to r is just the radial component of the ordinary momentum my, but, as we first saw in Section 7.1 [Equation (7.26)], the 4 We proved there, in Equations (7.89) and (7.90), that J-C. is conserved if and only if does not depend explicitly on the time. These two conditions Of not explicitly time dependent or \u00a3 likewise) are equivalent, for, as you can easily check, 89flat =\u2014az/at. See Problem 13.16. 5 1n any problem where the potential energy U = U (q) is independent of the velocities q (as we are certainly assuming here), there is this small simplification that we can replace by T in the definition of p 532 Chapter 13 Hamiltonian Mechanics momentum po conjugate to q5 is the angular momentum. We must next solve the two equations (13.28) to give the velocities,' and ci) in terms of the momenta pr and po : Pr r = \u2014 m and = Po mr2 We can now substitute these into (13.27) and we arrive at the Hamiltonian, expressed as a function of the proper variables, = T U = 1 (pr2 (P ) +U (r). (13.29) D 2 r2 2 m We can now write down the four Hamilton equations (13.25). The two radial equations are. r = Pr = aPr m and fir = 2 = ax ar dU. \u2014 mr 3 dr (13.30) The first of these reproduces the definition of the radial momentum. If we substitute the first into the second, we obtain the familiar result that mr is the sum of the actual radial force (\u2014dUldr) plus the centrifugal force p021mr 3. [See Equations (8.24) and (8.26).] The two 0 equations are (93-c = = p app mr- o and 150 = = O. (13.31) The first of these reproduces the definition of p0. The second tells us what we already knew, that the angular momentum is conserved. As in the previous two examples, we see that the Hamiltonian formalism provides an alternative route to the same final equations of motion as we could find using either the Newtonian or Lagrangian approaches. This example illustrates the general procedure to be followed in setting up Hamil- ton's equations for any given system: 1. Choose suitable generalized coordinates, q 1,, qn. 2. Write down the kinetic and potential energies, T and U, in terms of the q's and's. 3. Find the generalized momenta p i,, pn. (We are now assuming our system is conservative, so U is independent of qi and we can use pi = aT 0.4i. In general, one must use pi = 4. Solve for the 4's in terms of the p's and q's. 5. Write down the Hamiltonian 9-C as a function of the p's and q's. [Provided our coordinates are \"natural\" (relation between generalized coordinates and underlying Cartesians is independent of time), 9 -C is just the total energy T U, but when in doubt, use 9-C = E pi 4i \u2014 Z. See Problems 13.11 and 13.12.] 6. Write down Hamilton's equations (13.25). Section 13.3 Hamilton's Equations in Several Dimensions 533 If you look back over the last example, you will see that the solution followed all of these six steps, and the same will be true of all later examples and problems. Before we do another example, let us compare these six steps with the corresponding steps in the Lagrangian approach. To set up Lagrange's equations, we follow the same first two steps (choose generalized coordinates and write down T and U). Steps (3) and (4) are unnecessary, since we don't have to know the generalized momenta, nor to eliminate the 4's in favor of the p's. Finally, one must carry out the analogs of (5) and (6); namely, write down the Lagrangian and Lagrange's equations. Evidently, setting up the Hamiltonian approach involves two small extra steps [Steps (3) and (4) above]", " as compared to the Lagrangian. Although both steps are usually quite straightforward, this is undeniably a small disadvantage of Hamilton's formalism. Now, here is another example. EXAMPLE 13.4 Hamilton's Equations for a Mass on a Cone Consider a mass m which is constrained to move on the frictionless surface of a vertical cone p = cz (in cylindrical polar coordinates p, 0, z with z > 0) in a uniform gravitational field g vertically down (Figure 13.3). Set up Hamilton's equations using z and 4) as generalized coordinates. Show that for any given solution there are maximum and minimum heights z ma, and zimn between which the motion is confined. Use this result to describe the motion of the mass on the cone. Show that for any given value of z > 0 there is a solution in which the mass moves in a circular path at fixed height z. Our generalized coordinates are z and cb, with p determined by the constraint that the mass remain on the cone, p = cz. The kinetic energy is therefore T = [p2 (p )2,2] [ (c2,2 (cz(b) 2] 1) Figure 13.3 A mass m is constrained to move on the surface of the cone shown. For clarity, the cone is shown truncated at the height of the mass, although it actually continues upward indefinitely. 534 Chapter 13 Hamiltonian Mechanics The potential energy is of course U = mgz. The generalized momenta are Pz = = m(c2 aT az and pas mc2z20- (13.32) These are trivially solved for z and and we can write down the Hamiltonian, g-C = T U = + mgz. (13.33) 1 [ PZ 2 P 2m (c2 + 1) c2z2 Hamilton's equations are now easily found: The two z equations are. z = \u2014 = Pz apz m (c2 + 1) The two 0 equations are and Pz = mg. (13.34) aJ-c az 2 po mc2z3 a9f 8130 = po mc2z2 and /5'95 = (13.35) ag-c 0. ao = The last of these tells us, what we could well have guessed, that po, which is just the z component of angular momentum, is constant. The easiest way to see that, for any given solution, z is confined between two bounds, z min and zmax, is to remember that the Hamiltonian function (13.33) is equal to the total energy, and that energy is conserved. Thus, for any given solution, (13.33) is equal to a fixed constant E. Now, the function J-C in (13.33) is the sum of three positive terms, and as z co the last term tends to infinity. Since a must equal the fixed constant E, there must be a z max which z cannot exceed. In the same way, the second term in (13.33) approaches infinity as z \u2014+ 0; so there has to be a ; inn > 0 below which z cannot go. In particular, this means that the mass can never fall all the way into the bottom of the cone at z = 0.6 The motion of the mass on the cone is now easy to describe. It moves around the z axis with constant angular momentum p = inc2z2 0 Since p0 1. is constant, the angular velocity varies \u2014 increasing as z gets smaller and decreasing as z gets bigger. At the same time, the mass's height z oscillates up and down between ; inn and zmax. (See Problems 13.14 and 13.17 for more details.) To investigate the possibility of a solution in which the mass stays at a fixed height z, notice that this requires that z = 0 for all time. This in turn requires that pz = 0 for all time, and hence j3 = 0. From the second of the z equations (13.34), we see that f3 z = 0 if and only if p0 \u00b1Vm2c2gz3. (13.36) 6 Two comments: It is easy to see that the second term in (13.33) is related to the centrifugal force; thus, we can say that the mass is held away from the bottom of the cone by the centrifugal force. Second, the one exception to this statement is if the angular momentum po = 0; in this case, the mass moves up and down the cone in the radial direction (4) constant) and will eventually fall to the bottom. Section 13.4 Ignorable Coordinates 535 If, for any chosen initial height z, we launch the mass with p, = 0 and po equal to one of these two values (going either clockwise or counterclockwise), then since /3, = 0, p,", " and hence z both remain zero, and the mass continues to move at its initial height around a horizontal circle. 13.4 Ignorable Coordinates So far we have set up Hamilton's formalism and have seen that it is valid whenever Lagrange's is. The former enjoys almost all the advantages and disadvantages of the latter, when either is compared with the Newtonian formalism. But it is not yet clear that there are any significant advantages to using Hamilton instead of Lagrange, or vice versa. As I have already mentioned, in more advanced theoretical work, Hamilton's approach has some distinct advantages, and, I shall try to give some feeling for a few of these advantages in the next four sections. In Chapter 7, we saw that if the Lagrangian happens to be independent of a coordinate qi, then the corresponding generalized momentum p, is constant. [This followed at once from the Lagrange equation,K qi = (d/dt)01 /84i, which can be rewritten as a /aq, = Pi. Therefore, if aZ/aq, = 0, it immediately follows that Pi = 01 When this happens, we say that the coordinate qi is ignorable. In the same way, if the Hamiltonian X is independent of qi, it follows from the Hamilton equation pi = \u20140-f/aqi that its conjugate momentum pi is a constant. We saw this in Equation (13.31) of the last example, where 3 -C was independent of 0, and the conjugate momentum pc (actually the angular momentum) was constant. The results of this and the last paragraph are in fact the same result, since it is easy to prove that 8,C/aq, = ---aH/aqi (Problem 13.22). Thus \u00a3 is independent of qi if and only if J-C is independent of qi. If a coordinate qi is ignorable in the Lagrangian then it is also ignorable in the Hamiltonian and vice versa. It is nevertheless true that the Hamiltonian formalism is more convenient for handling ignorable coordinates. To see this, let us consider a system with just two degrees of freedom and suppose that the Hamiltonian is independent of q2. This means that the Hamiltonian depends on only three variables, aC = uf (qi, Pi, P2). (13.37) For example, the Hamiltonian (13.29) for the central force problem has this property, being independent of the coordinate 0. This means that P2 = k, a constant that is determined by the initial conditions. Substitution of this constant into the Hamiltonian leaves = 9- (q 1, p i, k) which is a function of just the two variables q i and p i, and the solution of the mo- tion is reduced to a one-dimensional problem with this effectively one-dimensional Hamiltonian. More generally, if a system with n degrees of freedom has an ignorable coordinate qi, then solution of the motion in the Hamiltonian framework is exactly 536 Chapter 13 Hamiltonian Mechanics equivalent to a problem with (n \u2014 1) degrees of freedom in which qi and pi can be entirely ignored. If there are several ignorable coordinates, the problem is correspond- ingly further simplified. In the Lagrangian formalism, it is of course true that if qi is ignorable, then pi is constant, but this does not lead to the same elegant simplification: Supposing again that the system has two degrees of freedom and that q 2 is ignorable, then corresponding to (13.37) we would have = z(q1, 41, q2). Now, even though q2 is ignorable and p 2 is a constant, it is not necessarily true that q2 is constant. Thus the Lagrangian does not reduce cleanly to a one-dimensional function that depends only on q 1 and 4,. [For example, in the central-force problem, the Lagrangian has the faun (r, Y, q5), but, even though 0 is ignorable, q5 still varies as the motion proceeds, and the problem does not automatically reduce to a problem with one degree of freedom. 7 13.5 Lagrange's Equations vs. Hamilton's Equations For a system with n degrees of freedom, the Lagrangian formalism presents us with n second-order differential equations for the n variables q 1,, q,1. For the same system, the Hamiltonian formalism gives us 2n first-order differential equations for the 2n variables q 1,'' Pi''' That Hamilton could recast n second-order equations into 2n first-order equations is no particular surprise. In fact it is easy to see that any set of n second-order equations can be recast in this way: For simplicity, let us consider the case that there is just one degree of freedom, so that Lagrange's approach", " gives just one second-order equation for the one coordinate q. This equation can be written as where f is some function of its three arguments. Let us now define a second variable f (4, q, q) = 0 s = 4. (13.38) (13.39) In terms of this second variable, q = s and our original differential equation (13.38) becomes f(s,s, q) = 0. (13.40) We have now replaced the one second-order equation (13.38) for q with the two first- order equations (13.39) and (13.40) for q and s. 7 In this particular case, the difficulty is fairly easy to circumvent. In the discussion of Section 8.4, we wrote the Lagrange radial equation (8.24) in terms of the variable and then rewrote it eliminating in favor of po = f, which is constant. Section 13.5 Lagrange's Equations vs. Hamilton's Equations 537 Evidently, the fact that Hamilton's equations are first-order where Lagrange's are second-order is no particular improvement. However, the specific form of Hamilton's equations is in fact a big improvement. To see this, let us rewrite Hamilton's equations in a more streamlined form. First we can rewrite the first n of Equations (13.25) as 4i = \u2014 = fi (q,p) [i = 1,, n] (13.41) 091 apt where each ft is some function of q and p, and we can combine these n equations into a single n-dimensional equation f(q, p) (13.42) where the boldface f stands for the vector comprised of the n functions A In the same way we can rewrite the n equations for the p t in a similar form: = g (q, P) (13.43) where the boldface g stands for the vector comprised of the n functions g, = \u2014aHlaqi. Finally, we can introduce a 2n-dimensional vector z = (4, p) = (qi, qii, p i,, AO. (13.44) This phase-space vector or phase point z comprises all of the generalized coordinates and all of their conjugate momenta. Each value of z labels a unique point in phase space and identifies a unique set of initial conditions for our system. With this new notation, we can combine the two equations (13.42) and (13.43) into a single grand, 2n-component equation of motion = h(z) (13.45) where the function h is a vector comprising the 2n functions f 1,, ft, and g 1,, gn of Equations (13.42) and (13.43). Equation (13.45) expresses Hamilton's equations as a first-order differential equa- tion for the phase-space vector z. Furthermore, it is a first-order equation with the especially simple form:8 (first derivative of z) = (function of z). A large part of the mathematical literature on differential equations is devoted to equations with this standard form, and it is a distinct advantage of the Hamiltonian formalism that Hamilton's equations \u2014 unlike Lagrange's \u2014 are automatically of this form. Our combining of the n position coordinates q with the n momenta p to form a single phase space vector z = (q, p) suggests a certain equality between the position and momentum coordinates in phase-space, and this suggestion proves correct. We 8 If the Hamiltonian was explicitly time dependent, then (13.45) would take the (still very simple) form h(z,t). The form (13.45) without any explicit time dependence is said to be autonomous. 538 Chapter 13 Hamiltonian Mechanics have known since Chapter 7 that one of the strengths of the Lagrangian formalism is its great flexibility with respect to coordinates: Any set of generalized coordinates q = (q 1,, qn) can be replaced by a second set Q = (Q 1,, Qn ), where each of the new Qi is a function of the original (q 1,, qn ), Q = Q(q), (13.46) and Lagrange's equations will be just as valid with respect to the new coordinates Q as they were with respect to the old q. 9 We can paraphrase this to say that La- grange's equations are unchanged (or invariant) under any coordinate change in the n-dimensional configuration space defined by q = (q 1,, qn ). The Hamiltonian for- malism shares this same flexibility \u2014 Hamilton's equations are invariant under any coordinate change (13.46) in configuration space. However, the Hamiltonian formal- ism actually has a much greater flexibility and allows for certain coordinate changes in the 2n-dimensional phase space. We can consider changes of coordinates of", " the form Q = Q(q, p) and P = P(q, p), (13.47) that is, coordinate changes in which both the q's and the p's are intermingled. If the equations (13.47) satisfy certain conditions, this change of coordinates is called a canonical transformation, and it turns out that Hamilton's equations are invariant under these canonical transformations. Any further discussion of canonical transformations would carry us beyond the scope of this book, but you should be aware of their existence and that they are one of the properties that make the Hamiltonian approach such a powerful theoretical too1. 10 Problems 13.24 and 13.25 offer two examples of canonical transformations. 13.6 Phase-Space Orbits One can view the phase-space vector z = (q, p) of (13.44) as defining the system's \"position\" in phase space. Any point zo defines a possible initial condition (at any cho- sen time to), and Hamilton's equations (13.45) define a unique phase-space orbit or trajectory which starts from z o at to and which the system follows as time progresses. Since phase space has 2n dimensions, the visualization of these orbits presents some challenges unless n = 1. For example, for a single unconstrained particle in three di- mensions, n = 3, and the phase space is six-dimensional not something that most of us can visualize easily. There are various techniques, such as the Poincare section 9 Of course, this statement has to be qualified a little: The coordinates Q must be \"reasonable\" in the sense that each set Q determines a unique set q and vice versa, and the function Q(q) has to be suitably differentiable. 10 I should emphasize that there is no corresponding transformation in Lagrangian mechanics, which operates in the state space defined by the 2n-dimensional vector (q, 4). Since q is defined as the time derivative of q, there is no analog to (13.47) in which the q's and 4's get intermingled. Section 13.6 Phase-Space Orbits 539 Figure 13.4 One can imagine two orbits passing through the same point zo in phase space. However, Hamilton's equations guarantee that for any given point zo, there is a unique orbit passing through z o, so the two orbits must in fact be the same. described in Section 12.8, for viewing phase-space orbits in a subspace of fewer di- mensions than the full phase space, but here I shall give just two examples of systems for which n = 1, and the phase space is therefore only two-dimensional. Before we look at these examples, there is an important property of phase-space orbits that deserves mention right away: It is easy to see that no two different phase- space orbits can pass through the same point in phase space; that is, no two phase-space orbits can cross one another. For suppose two orbits pass through one and the same point zo, as in Figure 13.4. Now, from Hamilton's equations (13.45), it follows that for any point z o there can be only one distinct orbit passing through z o. Therefore the two orbits passing through zo have to be the same. Notice that this result excludes different orbits from passing through the same point even at different times: If one orbit passes through z o o today, yesterday, or tomorrow. 11 This today, then no different orbit can pass through z result \u2014 that no two phase-space orbits can cross \u2014 places severe restrictions on the way in which these orbits are traced out in phase space. It has important consequences in, for example, the analysis of chaotic motion of Hamiltonian systems. EXAMPLE 13.5 A One-Dimensional Harmonic Oscillator Set up Hamilton's equations for a one-dimensional simple-harmonic oscillator with mass m and force constant k, and describe the possible orbits in the phase space defined by the coordinates (x, p). The kinetic energy is T = lm.i 2 and the potential energy is U = 4kx2.nu,02x 2, if we introduce the natural frequency w =.Vklm. The generalized momentum is p = aT I a5c = mic, and the Hamiltonian (written as a function of x and p) is 9-C = T + U = P2 + 1 711(0.X 2 2. 2m 2 (13.48) 11 This is because our Hamiltonian is time-independent. If J{ is explicitly time-dependent, then we can assert only that no two orbits can pass through one point at the same time. 540 Chapter 13 Hamiltonian Mechanics Thus, Hamilton's equations give x arc ap p = \u2014 m and p = \u2014 = \u2014mw a3-c a The simplest way to solve these two equations is to eliminate p and get the familiar second-order equation.5c", "' = \u2014co2 x, with the equally familiar solution x = A cos(cot \u2014 6) and hence p = mz = \u2014mcoA sin(wt \u2014 8). (13.49) Phase space for the one-dimensional oscillator is the two-dimensional space with coordinates (x, p). In this space, the solution (13.49) is the parametric form for an ellipse, traced in the clockwise direction, as in Figure 13.5, which shows two phase-space orbits for the cases that the oscillator started out from rest at x = A (solid curve) and x = A/2 (dashed curve). That the orbits have to be ellipses follows from conservation of energy: The total energy is given by the Hamiltonian (13.48), whose initial value (for the solid curve with x = A and p = 0) is Imco2 A 2. Thus conservation of energy implies that 2 P + mco2x2 l in(02 A2 2m 2 2 Or x 2 2 2 + (mcoA)2 = 1. A (13.50) This is the equation for an ellipse with semimajor and semiminor axes A and mcoA, in agreement with (13.49). It is perhaps worth following one of the phase-space orbits of Figure 13.5 in detail. For the solid curve, the motion starts from rest with x at its maximum, at the point x = A, p = 0, shown as a dot in the figure. The restoring force causes m to accelerate back toward x = 0, so that x gets smaller while p becomes Figure 13.5 The phase space for a one-dimensional harmonic oscillator is a plane, with axes labelled by x (the position) and p (the momentum), in which the point representing the state of motion traces a clockwise ellipse. There is a unique orbit through each phase point (x, p). The outer orbit (solid curve) started from rest with x = A and p = 0; the inner one started from x = A / 2 and p =0. Section 13.6 Phase-Space Orbits 541 increasingly negative. By the time x has reached 0, p has reached its largest negative value p = \u2014mcoA. The oscillator now overshoots the equilibrium point, so x becomes increasingly negative, while p is still negative but getting less so. By the time x reaches \u2014A, p is again zero, and so on, until the oscillator gets back to its starting point x = A, p = 0 and the cycle starts over again. Notice that, in agreement with our general argument, the two orbits shown in Figure 13.5 do not cross one another. In fact, it is easy to see that no two ellipses with the form (13.50) can have any point in common unless they have the same value of A (in which case they are the same ellipse). In this simple one-dimensional case, the phase-space plot doesn't tell us anything we couldn't learn from the simple solution x = A cos(cot \u2014 6), but I hope you will agree that it does show some details of the motion rather clearly. If you have read Chapter 12, you will recognize that a phase-space orbit is closely related to the state-space orbit described in Section 12.7. The only difference is that the former traces the system's evolution in the (x, p) plane whereas the latter uses the (x, I) plane. In the present case there is almost no difference, since for one- dimensional motion along the x axis, p is proportional to I (specifically, p = Thus the two kinds of plot are identical except that the former is stretched by a factor of m in the vertical direction. Nevertheless, in the general 2n-dimensional case, the spaces defined by (q, p) and by (q, q) can be very different. As we shall see in the next section, the phase-space plot has some elegant properties not shared by the state-space one. It often happens that one needs to follow not just one orbit but several different orbits through phase space. For example, in the study of chaos we saw that it is of great interest to follow the evolution of two identical systems that are launched with slightly different initial conditions. If the motion is nonchaotic, then the two systems remain close together in phase space, but if the motion is chaotic they move apart so rapidly that their detailed motion is effectively unpredictable. In the next example, we look at four neighboring phase-space orbits of a particle falling under the influence of gravity. EXAMPLE 13.6 A Falling Body Set up the Hamiltonian formalism for a mass m constrained to move in a vertical line, subject only to the force of gravity. Use the coordinate x, measured downward from a convenient origin, and its conjugate momentum.", " Describe the phase-space orbits and in particular sketch the orbits from time 0 to a later time t for the following four different initial conditions at t = 0: (a) xo = p0 = 0 (that is, the mass is released from rest at x = 0); (b) x. = X, but po = 0 (the mass is released from rest at x = X); (c) xo = X and po = P (the mass is thrown from x = X with initial momentum p = P); 542 Chapter 13 Hamiltonian Mechanics (d) xo = 0 and po = P (the mass is thrown from x = 0 with initial momentum P = P). The kinetic and potential energies are T = lm.i 2 and U = \u2014mgx. (Remem- ber that x is measured downward.) The conjugate momentum is, of course, p = m5c and the Hamiltonian is = T U = p \u2014 mgx. (13.51) 2 2m To find the shape of the phase-space orbits, we don't have to solve the equations of motion, since conservation of energy requires that they satisfy 9 -C = const. For the Hamiltonian (13.51), this defines a parabola with the form x = kp 2+ const, with its symmetry axis on the x axis. To draw the four orbits asked for, it is helpful to solve the equations of motion p. x = \u2014 = \u2014 ap m and. = a3-c ax = mg. The second of these gives and the first then gives p = po mgt po 2 X = X, \u2014t \u2014gt m 2 \u2014 both results that are very familiar from elementary mechanics. Putting in the given initial conditions, one gets the four curves shown in Figure 13.6. As expected, the four orbits A o A,, Do D are parabolas, no one of which crosses any other. You can see that the initial rectangle A 0 B0 C0 D0 has evolved Figure 13.6 Four different phase-space orbits for a body moving vertically under the influence of gravity, with position x (measured vertically down) and momentum p. The four different initial states at time 0 are shown by the dots labeled A o, Bo, Co, and Do, with corresponding final states at a later time t labeled A, B, C, and D. Section 13.7 Liouville's Theorem * 543 into a parallelogram ABCD. However, it is easy to show that the area of the parallelogram is the same as that of the original rectangle. (Same base, A.Bo = AB, and same height. See Problem 13.27.) We shall see in the next section that these two properties (changing shape but unchanging area) are general properties of all phase-space orbits. In particular, that the area does not change is an example of the important result known as Liouville's theorem. 13.7 Liouville's Theorem* * This section contains material that is more advanced than most other material in this book. In particular, it uses the divergence operator and divergence theorem of vector calculus. If you haven't met these ideas before, you could, if you wish, omit this section. In Example 13.6 we saw that one can use a plot of phase space to track the motion of several identical systems evolving from various different initial conditions. In many problems, especially those of statistical mechanics, one has occasion to track the motion of an enormous number of identical systems. The state of each system can be labelled by a dot in phase space, and, if the number of these dots is large enough, we can view the resulting swarm of dots as a kind of fluid, with a density p measured in dots per volume of phase space. For example, in the statistical mechanics of an ideal gas, one wants to follow the motion of some 1023 identical molecules as they move inside a container. Each molecule is governed by the same Hamiltonian and moves in the same six-dimensional phase space with coordinates (x, y, z, p x, py, pr). Thus the state of the system at any one time can be specified by giving the positions of 10 23 23 dots form a swarm whose motion can (for many dots in this phase space; these 10 purposes) be treated like that of a fluid. For most of this section, you could bear this example in mind, though in specific cases I shall often specialize to a system with just one spatial dimension and hence two dimensions of phase space. The tracking of the motion of many identical systems by means of a cloud of dots in their phase space is illustrated in Figure 13.7. This picture could be viewed as a schematic representation of a multidimensional phase space, but let us, for simplicity, think of it as the two-dimensional phase space, with the coordinates z = (q, p) of a system with one degree of freedom. Each", " dot in the lower cloud represents the initial state of one system (for example, one molecule in a gas) by giving its position z = (q, p) in phase space at time to. Hamilton's equations determine the \"velocity\" with which each dot moves through phase space: (phase-space velocity) = = (4, /5) (13.52), ap. aq For each dot z in the initial cloud, there is a unique velocity i, and each dot moves off with its assigned velocity. In general, different dots will have different velocities, and the cloud can change its shape and orientation, as shown. On the other hand, as we shall prove shortly, the volume occupied by the cloud cannot change \u2014 the result called Liouville's theorem. 544 Chapter 13 Hamiltonian Mechanics Figure 13.7 The dots in the lower cloud label the states of some 300 identical systems at time t o. As time progresses each dot moves through phase space in accordance with Hamilton's equations, and the whole cloud moves to the upper position. To make this last idea more precise, we need to consider a closed surface in phase space such as that shown at the lower left of Figure 13.8. Each point on this closed surface defines a unique set of initial conditions and moves along the corresponding phase-space orbit as time advances. Thus the whole surface moves through phase space. It is easy to see that any point that is initially inside the surface must remain inside for all time: For suppose that such a point could move outside. Then at a certain moment it would have to cross the moving surface. At this moment we would have two distinct orbits crossing one another, which we know is impossible. By the same argument, any point that is initially outside the surface remains outside for all time. Thus the number of dots (representing molecules, for example) inside the surface is a constant in time. The main result of this section \u2014 Liouville's theorem \u2014 is that the volume of the moving closed surface in Figure 13.8 is constant in time. To prove this, we need to know the relationship between the rate of change of this kind of volume and the so-called divergence of the velocity vector. Figure 13.8 As time progresses, the closed surface at the lower left moves through phase space. Any point that is initially inside the surface remains inside for all time. Changing Volumes The two mathematical results that we need hold in spaces with any number of dimen- sions. We shall need to apply them in the 2n-dimensional phase space. Nevertheless, Section 13.7 Liouville's Theorem * 545 = iy& dA (a) (b) Figure 13.9 (a) The surface S moves with the fluid from its initial position (solid curve) at time t to a new position (dashed curve) at time t St. The change in V is just the volume between the two surfaces. (b) Enlarged view of the shaded volume of part (a). The vector n is a unit vector normal to the surface S, pointing outward. let us consider them first in the familiar context of our everyday three-dimensional space. We imagine a three-dimensional space filled with a moving fluid. The fluid at each point r is moving with velocity v. For each position r there is a unique velocity v, but v can, of course, have different values at different points r; thus, we can write v = v(r). This is exactly analogous to the situation in phase space, where each phase point z is moving with a velocity that is uniquely determined by its position in phase space, z = i(z). Let us now consider a closed surface S in the fluid at a certain time t. We can imagine marking this surface with dye so that we can follow its motion as it moves with the fluid. The question we have to ask is this: If V denotes the volume contained inside S, how fast does V change as the fluid moves? Figure 13.9(a) shows two successive positions of the surface S at two successive moments a short time St apart. The change in V during the interval St is the volume between these two surfaces. To evaluate this, consider first the contribution of the small shaded volume in Figure 13.9(a), which is enlarged in Figure 13.9(b). This small volume is a cylinder whose base has an area dA. The side of the cylinder is given by the displacement v St, so the cylinder's height is the component of v St normal to the surface. If we introduce a unit vector n in the direction of the outward normal to S, then the height of our cylinder is n v St and its volume is n v St dA. The total change in the volume inside our surface S is found by adding up all these small contributions to give SV = f nv St dA, (13.53) where the integral is a surface integral running over the whole", " of our closed surface S. Dividing both sides by St and letting St \u2014> 0, we get the first of our two key results: dV dt f nv dA. s (13.54) Figure 13.9(a) showed a fluid flow with the velocity v everywhere outward, as would be the case for the expanding air in a balloon whose temperature is rising. With v outward, the scalar product n v is positive (since n was defined as the outward 546 Chapter 13 Hamiltonian Mechanics normal) and the integral (13.54) is positive, implying that V is increasing, as it should. If v were everywhere inward (air in a balloon whose temperature is falling), then n v would be negative and V would be decreasing. In general, n v can be positive on parts of the surface S and negative on others. In particular, if the fluid is incompressible, so that V can't change, then the contributions from positive and negative values of n v must exactly cancel so that dVIdt can be zero. We have derived the result (13.54) for a surface S and volume V in three di- mensions, but it is equally valid in any number of dimensions. In an m-dimensional space both of the vectors n and v have m components and their scalar product is nv = n i v i + + n rn vm. With this definition, the result (13.54) is valid whatever the value of m. The Divergence Theorem The second mathematical result that we need is called the divergence theorem or Gauss's theorem. This is one of the standard results of vector calculus (similar to Stokes's theorem that we used in Chapter 4) and you can find its proof in any text on vector calculus, 12 although its proof is quite straightforward and instructive in various simple cases. (See Problem 13.37.) The theorem involves the vector operator called the divergence. For any vector v, the divergence of v is defined as v v av ax x + + ay av az ; (13.55) here v can be any vector (a force or an electric field, for instance), but for us it will always be a velocity. If you have not met the divergence operator before, you might like to practice with some of Problems 13.31, 13.32, or 13.34. The divergence theorem asserts that the surface integral in (13.54) can be expressed in terms of V v: nvdA= f V vdV. v (13.56) Here the integral on the right is a volume integral over the volume V interior to the surface S. This theorem is an amazingly powerful tool. It plays a crucial role in applications of Gauss's law of electrostatics. It often allows us to perform integrals that would otherwise be hard to evaluate. In particular, there are many interesting fluid flows with the property that V v = 0; for such flows, the integral on the left may be very awkward to evaluate directly, but thanks to the divergence theorem we can see immediately that the integral is in fact zero. Since this is how we shall be using the divergence theorem, let us look at an example of such a flow right away. 12 See, for example, Mary Boas, Mathematical Methods in the Physical Sciences (Wiley, 1983), p. 271. Section 13.7 Liouville's Theorem * 547 EXAMPLE 13.7 A Shearing Flow The velocity of flow of a certain fluid is v = kyx (13.57) where k is a constant; that is, v, = ky and vy = vz = 0. Describe this flow and sketch the motion of a closed surface that starts out as a sphere. Evaluate the divergence V v and show that the volume enclosed by any closed surface S moving with the fluid cannot change; that is, the fluid flows incompressibly. The velocity v is everywhere in the x direction and depends only on y. Thus all the points in any plane y = constant move like a sliding rigid plate \u2014 a pattern called laminar flow. 13 The speed increases with y, so each plane moves a little faster than the planes below it, in a shearing motion, as indicated by the three thin arrows in Figure 13.10. If we consider a closed surface moving with the fluid and initially spherical, then its top will be dragged along a little faster than its bottom and it will be stretched into an ellipsoid of ever-increasing eccentricity as shown. We can easily evaluate V v using the definition (13.55) V v = av ax ay ay Y + av Z _ aky az ax +0+0=0. (13.58) Now, combining our two results (13.54) and (13.56), we see that the rate of change of the volume inside any closed surface is y > x Figure 13.10 The fluid flow described by (", "13.57) is a laminar, shearing flow. The planes parallel to the plane y = 0 all move rigidly in the x direction with speed proportional to y. This shearing motion stretches the sphere into an ellipsoid. 13 From the latin lamina meaning a thin layer or plate, from which we also get the verb \"to laminate.\" 548 Chapter 13 Hamiltonian Mechanics Since we have shown that V v = 0 everywhere, it follows that dVIdt = 0 and the volume inside any closed surface moving with the fluid is constant for the flow of (13.57). If you have never met the divergence theorem before, you might want to try one or both of the Problems 13.33 and 13.37. Here let me say just one more thing about the significance of V v. If we apply (13.59) to a sufficiently small volume V, then V v will be approximately constant throughout the region of integration, and the right side of (13.59) becomes just (V v) V and (13.59) itself implies that V v = \u2014 1 dV V dt for any small volume V. Since dVIdt can be called the outward flow of v, we can say that V v is the outward flow per volume. If V v is positive at a point r, then there is an outward flow around r and any small volume around r is expanding (like the gas in a balloon that is heating up); if V v is negative, then there is an inward flow and any small volume around r is contracting (like the gas in a balloon that is cooling down). In our example, V v was zero, and any volume moving with the fluid was constant. The divergence generalizes easily to any number of dimensions. In an m- dimensional space with coordinates (x 1,, x,n ) the divergence of a vector v = (u 1,, v, n ) is defined as v v = \u2014 + + av, ax, avm axm and, with these definitions, the crucial result (13.59) takes exactly the same form, except that the integral is an integral over an m-dimensional region with volume element dV = dx i dx 2 - dx m. Liouville's Theorem We are finally ready to prove the main goal of this section \u2014 Liouville's theorem. This is a theorem about motion in phase space, the 2n-dimensional space with coordinates z = (q, p) (q1 ). To simplify the notation, I shall consider just a system with one degree of freedom, so that n = 1 and the phase space is just two- dimensional with phase points z = (q, p). The general case goes through in almost exactly the same way, as you can check for yourself by doing Problem 13.36. Each phase point z = (q, p) moves through the 2-dimensional phase space in accordance with Hamilton's equations, with velocity v = = (4, i,) = rt, ap 0g (13.60) We consider an arbitrary closed surface S moving through phase space with the phase points, as was illustrated in Figure 13.8. The rate at which the volume inside S changes is given by (13.59), where now the volume is a two-dimensional volume (really an area) and Section 13.7 Liouville's Theorem * 549 O v = a4 + = a aq Bp aq Bp + a ( Bp aq = 0, (13.61) which is zero because the order of the two differentiations in the double derivatives is immaterial. Since V v = 0, it follows that dV/dt = 0, and we have proved that the volume V enclosed by any closed surface is a constant as the surface moves around in phase space. This is Liouville's theorem. There is another way to state Liouville's theorem: We saw before that the number N of dots, representing identical systems, inside any given volume V cannot change. (No dot can cross the boundary S from the inside to the outside or vice versa.) We have now seen that the volume V cannot change. Therefore, the density of dots, p = N/ V cannot change either. This statement is sometimes paraphrased to say that the cloud of dots moves through phase space like an incompressible fluid. However, it is important to be aware what this statement means. The density p can, of course, be different at different phase points z = (q, p); all we are claiming is that as we follow a phase point along its orbit, the density at this point does not change. Unfortunately we cannot pursue here the consequences of Liouville's theorem, but I can just mention one example: We saw in Chapter 12 that when motion is chaotic, two identical systems that start out with nearly identical initial conditions move rapidly apart in phase space. Thus if we consider a small initial volume", " in phase space, such as is shown in Figure 13.8, and if the motion is chaotic, then at least some pairs of points inside the volume must move rapidly apart. But we have now seen that the total volume V cannot change. Therefore, as the volume grows in one direction, it must contract in another direction, becoming something like a cigar. Now it frequently happens that the region in which the phase points can move is bounded. (For example, conservation of energy has this effect for the harmonic oscillator of Figure 13.5.) In this case, as the volume V gets longer and thinner, it has to become intricately folded in on itself, adding another twist to the already fascinating story of chaotic motion. Finally, a couple of points on the validity of Liouville's theorem: First, in all of the examples of this chapter, we have assumed that the Hamiltonian is not explicitly de- pendent on time, a9-cot = 0, and that the forces are conservative and the coordinates \"natural,\" so that 3-C = T + U. However, none of these assumptions is necessary for the truth of Liouville's theorem. The proof given in this section depends only on the validity of Hamilton's equations, and any system which obeys Hamilton's equations also obeys Liouville's theorem. For example, a charged particle in an electromagnetic field obeys Hamilton's equations (Problem 13.18) and hence also Liouville's theorem, even though ofiat may be nonzero and 91 is certainly not equal to T + U. Second, Liouville's theorem applies to the Hamiltonian phase space with coordinates (q, p), and there is no corresponding theorem for the Lagrangian state space with coordi- nates (q, 4). This is one of the most important advantages of the Hamiltonian over the Lagrangian approach. 550 Chapter 13 Hamiltonian Mechanics Principal Definitions and Equations of Chapter 13 The Hamiltonian If a system has generalized coordinates q = (q 1,, q\u201e), Lagrangian and gener- alized momenta pi = 8,Cdlaqi, its Hamiltonian is defined as = i= 1 p i 4, \u2014 [Eq. (13.22)] always considered as a function of the variables q and p (and possibly t). Hamilton's Equations The time evolution of a system is given by Hamilton's equations qi = ag-c api and. pi = \u2014 ax a qi [i = 1,, n]. [Eq. (13.25)] Phase Space and Phase-Space Orbits The phase space of a system is the 2n-dimensional space with points (q, p) defined by the n generalized coordinates q, and the n corresponding momenta p,. A phase- space orbit is the path traced in phase space by a system as time evolves. [Sections 13.5 & 13.6] Liouville's Theorem If we imagine a large number of identical systems launched at the same time with slightly different initial conditions, the phase-space points that represent the systems can be seen as forming a fluid. Liouville's theorem states that the density of this fluid is constant in time (or, equivalently, that the volume occupied by any group of points is constant). [Section 13.7] Problems for Chapter 13 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). SECTION 13.2 Hamilton's Equations for One-Dimensional Systems 13.1 * Find the Lagrangian, the generalized momentum, and the Hamiltonian for a free particle (no forces at all) confined to move along the x axis. (Use x as your generalized coordinate.) Find and solve Hamilton's equations. 13.2 * Consider a mass m constrained to move in a vertical line under the influence of gravity. Using the coordinate x measured vertically down from a convenient origin 0, write down the Lagrangian Problems for Chapter 13 551 and find the generalized momentum p = 04.,181. Find the Hamiltonian J-C as a function of x and p, and write down Hamilton's equations of motion. (It is too much to hope with a system this simple that you would learn anything new by using the Hamiltonian approach, but do check that the equations of motion make sense.) 13.3 * Consider the Atwood machine of Figure 13.2, but suppose that the pulley is a uniform disc of mass M and radius R. Using x as your generalized coordinate, write down the Lagrangian, the generalized momentum p, and the Hamiltonian = p.x \u2014 L. Find Hamilton's equations and use them to find the acceleration 1. 13.4 * The Hamiltonian X is always given by g-C = pq \u2014 L (in one dimension), and this is the form you should use if in doubt. However, if your generalized coordinate q is \"natural\" (relation between q and the underlying Cartesian coordinates is independent", " of time) then J -C = T U, and this form is almost always easier to write down. Therefore, in solving any problem you should quickly check to see if the generalized coordinate is \"natural,\" and if it is you can use the simpler form J -C = T + U. For the Atwood machine of Example 13.2 (page 527), check that the generalized coordinate was \"natural.\" [Hint: There are one generalized coordinate x and two underlying Cartesian coordinates x and y. You have only to write equations for the two Cartesians in terms of the one generalized coordinate and check that they don't involve the time, so it's safe to use g -C = T U. This is ridiculously easy!] 13.5 ** A bead of mass m is threaded on a frictionless wire that is bent into a helix with cylindrical polar coordinates (p, 0, z) satisfying z = c0 and p = R, with c and R constants. The z axis points vertically up and gravity vertically down. Using 0 as your generalized coordinate, write down the kinetic and potential energies, and hence the Hamiltonian g -C as a function of 0 and its conjugate momentum p. Write down Hamilton's equations and solve for and hence E. Explain your result in terms of Newtonian mechanics and discuss the special case that R 13.6 ** In discussing the oscillation of a cart on the end of a spring, we almost always ignore the mass of the spring. Set up the Hamiltonian g-C for a cart of mass m on a spring (force constant k) whose mass M is not negligible, using the extension x of the spring as the generalized coordinate. Solve Hamilton's equations and show that the mass oscillates with angular frequency co = \\/k/(m + M/3). That is, the effect of the spring's mass is to add M/3 to m. (Assume that the spring's mass is distributed uniformly and that it stretches uniformly.) 13.7 *** A roller coaster of mass m moves along a frictionless track that lies in the xy plane (x horizontal and y vertically up). The height of the track above the ground is given by y = h(x). (a) Using x as your generalized coordinate, write down the Lagrangian, the generalized momentum p, and the Hamiltonian 9-C = p.z \u2014 \u00a3 (as a function of x and p). (b) Find Hamilton's equations and show that they agree with what you would get from the Newtonian approach. [Hint: You know from Section 4.7 that Newton's second law takes the form Ftang = ms, where s is the distance measured along the track. Rewrite this as an equation forl and show that you get the same result from Hamilton's equations.] SECTION 13.3 Hamilton's Equations in Several Dimensions 13.8 * Find the Lagrangian, the generalized momenta, and the Hamiltonian for a free particle (no forces at all) moving in three dimensions. (Use x, y, z as your generalized coordinates.) Find and solve Hamilton's equations. 13.9 * Set up the Hamiltonian and Hamilton's equations for a projectile of mass m, moving in a vertical plane and subject to gravity but no air resistance. Use as your coordinates x measured horizontally and y measured vertically up. Comment on each of the four equations of motion. 552 Chapter 13 Hamiltonian Mechanics 13.10 * Consider a particle of mass m moving in two dimensions, subject to a force F = + KS, where k and K are positive constants. Write down the Hamiltonian and Hamilton's equations, using x and y as generalized coordinates. Solve the latter and describe the motion. 13.11 * The simple form H = T U is true only if your generalized coordinates are \"natural\" (relation betweeen generalized and underlying Cartesian coordinates is independent of time). If the generalized coordinates are not \"natural,\" you must use the definition H = E pi qi \u2014 Z. To illustrate this point, consider the following: Two children are playing catch inside a railroad car that is moving with varying speed V along a straight horizontal track. For generalized coordinates you can use the position (x, y, z) of the ball relative to a point fixed in the car, but in setting up the Hamiltonian you must use coordinates in an inertial frame \u2014 a frame fixed to the ground. Find the Hamiltonian for the ball and show that it is not equal to T + U (neither as measured in the car, nor as measured in the ground-based frame). 13.12* Same as Problem 13.11, but use the following system: A bead of mass m is threaded on a frictionless, straight rod, which lies in a horizontal plane and is forced to spin with constant angular velocity co about a vertical axis through the midpoint of the rod. Find the Hamiltonian for the bead and show that it is not equal to", " T + U. 13.13 ** Consider a particle of mass m constrained to move on a frictionless cylinder of radius R, given by the equation p = R in cylindrical polar coordinates (p, 0, z). The mass is subject to just one external force, F = \u2014 kri, where k is a positive constant, r is its distance from the origin, and r is the unit vector pointing away from the origin, as usual. Using z and 0 as generalized coordinates, find the Hamiltonian H. Write down and solve Hamilton's equations and describe the motion. 13.14 ** Consider the mass confined to the surface of a cone described in Example 13.4 (page 533). We saw there that there have to be maximum and minimum heights z max and zmm, beyond which the mass cannot stray. When z is a maximum or minimum, it must be that i = 0. Show that this can happen if and only if the conjugate momentum p, = 0, and use the equation }C = E, where H is the Hamiltonian function (13.33), to show that, for a given energy E, this occurs at exactly two values of z. [Hint: Write down the function 3-C for the case that p, = 0 and sketch its behavior as a function of z for 0 < z < Do. How many times can this function equal any given E?] Use your sketch to describe the motion of the mass. 13.15 ** Fill in the details of the derivation of Hamilton's 2n equations (13.25) for a system with n degrees of freedom, starting from Equation (13.24). You can parallel the argument that led from (13.14) to (13.15) and (13.16), but you have 2n different derivatives to consider and lots of summations from i = 1 to n to contend with. 13.16 ** Starting from the expression (13.24) for the Hamiltonian, prove that a3flat [Hint: Consider first a system with one degree of freedom, for which (13.24) simplifies to 3-C (q, p, t) = p (q, p, t) \u2014 G (q, (q, p, t), t ).] 13.17 *** Consider the mass confined to the surface of a cone described in Example 13.4 (page 533). We saw that there are solutions for which the mass remains at the fixed height z = z o, with fixed angular velocity cb o say. (a) For any chosen value of po, use (13.34) to get an equation that gives the corresponding value of the height z o. (b) Use the equations of motion to show that this motion is stable. That is, show that if the orbit has z = z o E with c small, then E will oscillate about zero. (c) Show that the angular frequency of these oscillations is w = N/3 c/5 0 sin a, where a is the half angle of the cone (tan a = c where c is the constant in p = cz). (d) Find the angle a for which the frequency of oscillation w is equal to the orbital angular velocity ci5 0, and describe the motion for this case. Problems for Chapter 13 553 13.18 *** All of the examples in this chapter and all of the problems (except this one) treat forces that come from a potential energy U(r) [or occasionally U(r, t)]. However, the proof of Hamilton's equations given in Section 13.3 applies to any system for which Lagrange's equations hold, and this can include forces not derivable from a potential energy. An important example of such a force is the magnetic force on a charged particle. (a) Use the Lagrangian (7.103) to show that the Hamiltonian for a charge q in an electromagnetic field is g-c = (p \u2014 qA) 2/(2m) qV. (This Hamiltonian plays an important role in the quantum mechanics of charged particles.) (b) Show that Hamilton's equations are equivalent to the familiar Lorentz force equation mr = q (E v x B). SECTION 13.4 Ignorable Coordinates 13.19 * In Example 13.3 (page 531) we saw that if we write the Hamiltonian for a two-dimensional central force problem in terms of polar coordinates r and 0, then the coordinate 0 is ignorable. Write down the Hamiltonian for the same problem, but using rectangular coordinates x, y. Show that, with this choice, neither coordinate is ignorable. [The moral of this is that the choice of generalized coordinates calls for some care. In particular, you must look for any symmetries in a system and try to choose generalized coordinates to take advantage of them.] 13.20 * Consider a mass m moving in two dimensions, subject to a single force", " F that is independent of r and t. (a) Find the potential energy U(r) and the Hamiltonian a. (b) Show that if you use rectangular coordinates x, y with the x axis in the direction of F, then y is ignorable. (c) Show that if you use rectangular coordinates x, y with neither axis in the direction of F, then neither coordinate is ignorable. (Moral: Choose generalized coordinates carefully!) 13.21 ** Two masses m 1 and m 2 are joined by a massless spring (force constant k and natural length / 0) and are confined to move in a frictionless horizontal plane, with CM and relative positions R and r as defined in Section 8.2. (a) Write down the Hamiltonian g -C using as generalized coordinates X, Y, r, 0, where (X, Y) are the rectangular components of R, and (r, 0) are the polar coordinates of r. Which coordinates are ignorable and which are not? Explain. (b) Write down the 8 Hamilton equations of motion. (c) Solve the r equations for the special case that p0 = 0 and describe the motion. (d) Describe the motion for the case that p0 A 0 and explain physically why the r equation is harder to solve in this case. 13.22 ** In the Lagrangian formalism, a coordinate q, is ignorable if a Z, /49 qi = 0; that is, if is independent of qi. This guarantees that the momentum p, is constant. In the Hamiltonian approach, we say that q, is ignorable if g-f is independent of qi, and this too guarantees pi is constant. These two conditions must be the same, since the result \"p, = const\" is the same either way. Prove directly that this is so, as follows: (a) For a system with one degree of freedom, prove that 8g-co g = lag starting from the expression (13.14) for the Hamiltonian. This establishes that aJf/aq = 0 if and only if a,C/aq = 0. (b) For a system with n degrees of freedom, prove that a g -f/ aqi = \u2014a,C/aqi starting from the expression (13.24). 13.23 *** Consider the modified Atwood machine shown in Figure 13.11. The two weights on the left have equal masses m and are connected by a massless spring of force constant k. The weight on the right has mass M = 2m, and the pulley is massless and frictionless. The coordinate x is the extension of the spring from its equilibrium length; that is, the length of the spring is l e x where le is the equilibrium length (with all the weights in position and M held stationary). (a) Show that the total potential energy 554 Chapter 13 Hamiltonian Mechanics y le \u2014M= 2m Figure 13.11 Problem 13.23 (spring plus gravitational) is just U = Ikx 2 (plus a constant that we can take to be zero). (b) Find the two momenta conjugate to x and y. Solve for x and 52, and write down the Hamiltonian. Show that the coordinate y is ignorable. (c) Write down the four Hamilton equations and solve them for the following initial conditions: You hold the mass M fixed with the whole system in equilibrium and y = y o. Still holding M fixed, you pull the lower mass m down a distance x o, and at t = 0 you let go of both masses. [Hint: Write down the initial values of x, y and their momenta. You can solve the x equations by combining them into a second-order equation for x. Once you know x (t), you can quickly write down the other three variables.] Describe the motion. In particular, find the frequency with which x oscillates. SECTION 13.5 Lagrange's Equations vs. Hamilton's Equations 13.24 * Here is a simple example of a canonical transformation that illustrates how the Hamiltonian formalism lets one mix up the q's and the p's. Consider a system with one degree of freedom and Hamiltonian = H(q, p). The equations of motion are, of course, the usual Hamiltonian equations 4 = 83-781, and /3 = \u2014ag-f/aq. Now consider new coordinates in phase space defined as Q = p and P = \u2014q. Show that the equations of motion for the new coordinates Q and P are Q = 85f/a P and P = \u201483-co Q; that is, the Hamiltonian formalism applies equally to the new choice of coordinates where we have exchanged the roles of position and momentum. 13.25 *** Here is another example of a canonical transformation, which is still too simple to be of any real use, but does nevertheless illustrate the", " power of these changes of coordinates. (a) Consider a system with one degree of freedom and Hamiltonian J -C = H(q, p) and a new pair of coordinates Q and P defined so that q = 2P sin Q and p =,N/27P cos Q. (13.62) Prove that if agi/aq = \u2014/5 and Vf/ap = 4, it automatically follows that ax/aQ = \u2014P and ag-C/a P = Q. In other words, the Hamiltonian formalism applies just as well to the new coordinates as to the old. (b) Show that the Hamiltonian of a one-dimensional harmonic oscillator with mass m = 1 and force constant k = 1 is J-C = z (q2 p2). (c) Show that if you rewrite this Hamiltonian in terms of the coordinates Q and P defined in (13.62), then Q is ignorable. [The change of coordinates (13.62) was cunningly chosen to produce this elegant result.] What is P? (d) Solve the Hamiltonian equation for Q(t) and verify that, when rewritten for q, your solution gives the expected behavior. Problems for Chapter 13 555 SECTION 13.6 Phase-Space Orbits 13.26 * Find the Hamiltonian x for a mass m confined to the x axis and subject to a force Fx = \u2014kx 3 where k > 0. Sketch and describe the phase-space orbits. 13.27 ** Figure 13.6 shows some phase-space orbits for a mass in free fall. The points A o, Bo, Co, Do represent four different possible initial states at time 0, and A, B, C, D are the corresponding states at a later time. Write down the position x (t) and momentum p(t) as functions of t and use these to prove that ABC D is a parallelogram with area equal to the rectangle A 0 B0 C0 Do. [This is an example of Liouville's theorem.] 13.28 ** Consider a mass m confined to the x axis and subject to a force Fx = kx where k > 0. (a) Write down and sketch the potential energy U(x) and describe the possible motions of the mass. (Distinguish between the cases that E > 0 and E < 0.) (b) Write down the Hamiltonian 3-4x, p), and describe the possible phase-space orbits for the two cases E > 0 and E < 0. (Remember that the function J -C(x, p) must equal the constant energy E.) Explain your answers to part (b) in terms of those to part (a). SECTION 13.7 Liouville's Theorem* 13.29 * Figure 13.10 shows an initially spherical volume getting stretched into an ellipsoid by the shearing flow (13.57). Make a similar sketch for a volume that is initially spherical and centered on the origin. 13.30 * Figure 13.9 shows a fluid flow where the flow is everywhere outward (at least for all points on the surface S shown). This means that all contributions to the change 8V in volume are positive, and V is definitely increasing. Sketch the corresponding picture for the case that the flow is outward on the upper part of S and inward on the lower part. Explain clearly why the contributions n v 8t dA to the change in V from the lower part of S are negative, and hence that 8V can be of either sign, depending on whether the positive contributions outweigh the negative or vice versa. 13.31 * Evaluate the three-dimensional divergence V v for each of the following vectors: (a) v = kr, (b) v = k(z, x, y), (c) v = k(z, y, x), (d) v = k(x, y, \u20142z), where r = (x, y, z) is the usual position vector and k is a constant. 13.32 ** Evaluate the three-dimensional divergence V v for each of the following vectors: (a) v = (b) v = kxx, (c) v = kyi. We know that V v represents the net outward flow associated with v. In those cases where you found V v = 0, make a simple sketch to illustrate that the outward flow is zero; in those cases where you found V v 0, make a sketch to show why and whether the outflow is positive or negative. 13.33 ** The divergence theorem is a remarkable result, relating the surface integral that gives the flow of v out of a closed surface S to the volume integral of V v. Occasionally it is easy to evaluate both of these integrals and one can check the validity of the theorem. More often, one of the integrals is much easier to evaluate than the other, and the", " divergence theorem then gives one a slick way to evaluate a hard integral. The following exercises illustrate both of these situations. (a) Let v = kr, where k is a constant and let S be a sphere of radius R centered on the origin. Evaluate the left side of the divergence 556 Chapter 13 Hamiltonian Mechanics theorem (13.56) (the surface integral). Next calculate V v and use this to evaluate the right side of (13.56) (the volume integral). Show that the two agree. (b) Now use the same velocity v, but let S be a sphere not centered on the origin. Explain why the surface integral is now hard to evaluate directly, but don't actually do it. Instead, find its value by doing the volume integral. (This second route should be no harder than before.) 13.34 ** (a) Evaluate V v for v = kilr 2 using rectangular coordinates. (Note that 17 r 2 = r/r 3.) (b) Inside the back cover, you will find expressions for the various vector operators (divergence, gradient, etc.) in polar coordinates. Use the expression for the divergence in spherical polar coordinates to confirm your answer to part (a). (Take r 0.) 13.35 ** A beam of particles is moving along an accelerator pipe in the z direction. The particles are uniformly distributed in a cylindrical volume of length L o (in the z direction) and radius Ro. The particles have momenta uniformly distributed with pz in an interval p o Opt and the transverse momentum pi inside a circle of radius Ap 1. To increase the particles' spatial density, the beam is focused by electric and magnetic fields, so that the radius shrinks to a smaller value R. What does Liouville's theorem tell you about the spread in the transverse momentum p 1 and the subsequent behavior of the radius R? (Assume that the focusing does not affect either L o or Apz.) 13.36 ** Prove Liouville's theorem in the 2n-dimensional phase space of a system with n degrees of freedom. You can follow closely the argument around Equations (13.60) and (13.61). The only difference is that now the phase velocity v = z is a 2n-dimensional vector and V v is a 2n-dimensional divergence. 13.37 *** The general proof of the divergence theorem nvdA = f V vdV (13.63) is fairly complicated and not especially illuminating. However, there are a few special cases where it is reasonably simple and quite instructive. Here is one: Consider a rectangular region bounded by the six planes x = X and X + A, y = Y and Y B, and z = Z and Z C, with total volume V = ABC. The surface S of this region is made up of six rectangles that we can call S 1 (in the plane x = X), S 2 (in the plane x = X + A), and so on. The surface integral on the left of (13.63) is then the sum of six integrals, one over each of the rectangles S 1, S2, and so forth. (a) Consider the first two of these integrals and show that n vdA + f nv dA = f dy r li+B c dz [vx (X \u00b1 A, y, z) \u2014 v x (X, y, z)]. fs i s2 (b) Show that the integrand on the right can be rewritten as an integral of avx lax over x running from x = X to x = X + A. (c) Substitute the result of part (b) into part (a), and write down the corresponding results for the two remaining pairs of faces. Add these results to prove the divergence theorem (13.63). CHAPTER Collision Theory The collision experiment, or scattering experiment, is the single most powerful tool for investigating the structure of atomic and subatomic objects. In this type of experiment one fires a stream of projectiles, such as electrons or protons, at a target object\u2014 an atom or atomic nucleus, for example \u2014 and, by observing the distribution of \"scattered\" projectiles as they emerge from the collision, one can gain information about the target and its interactions with the projectile. Perhaps the most famous collision experiment was the discovery by Ernest Rutherford (1871-1937) of the structure of the atom: Rutherford and his assistants fired streams of a particles (the positively charged nuclei of helium atoms) at a thin layer of gold atoms in a sheet of gold foil; by measuring the distribution of the scattered a particles, they were able to deduce that most of the mass of an atom is concentrated in a tiny, positively charged \"nucleus\" at the center of the atom. Since that time, most discoveries in atomic and subatomic physics (the discoveries of the neutron, of nuclear fission and fusion, of qu", "arks, and many more) were made with the help of collision experiments, in which a stream of projectiles were directed at a suitable target and the outgoing particles carefully monitored. You could imagine doing a scattering experiment with larger objects \u2014 scattering one billiard ball off another, or even a comet off the sun \u2014 but in these cases there are usually easier ways to find out about the target. Thus the main application of collision theory is at the atomic level and below. Since the correct mechanics for atomic and subatomic systems is quantum mechanics, this means that the most widely used form of collision theory is quantum collision theory. Nevertheless, many of the central ideas of quantum collision theory \u2014 total and differential scattering cross sections, lab and CM reference frames \u2014 already appear in the classical theory, which gives an excellent introduction to these ideas without the complications of quantum theory. This, then, is the main purpose of this chapter, to give an introduction to the main ideas of collision theory in the context of classical mechanics. The main reason why collision theory is a rather complicated structure is that on the atomic and subatomic scale one cannot possibly follow the detailed orbit of the projectile as it interacts with the target. As we shall see, this means that we can 557 558 Chapter 14 Collision Theory learn very little by observing a single projectile. On the other hand, if we send in many projectiles, we can observe the number of them that get scattered in different directions, and we can learn lots from this. It is to handle the statistical distribution of the many scattered projectiles, in the many possible directions, that we have to introduce the idea of the collision cross section, which is the central concept of collision theory, both classical and quantum, 1 and is the main subject of this chapter. 14.1 The Scattering Angle and Impact Parameter Before we introduce the central concept of the scattering cross section, it is helpful to introduce two other important parameters, the scattering angle and the impact parameter; and before we do this, it is good to have a couple of simple collision experiments in mind. All collision experiments start with a projectile approaching a target from so far away that it moves essentially freely and its energy is purely kinetic. In Figure 14.1, a fixed target exerts a force on the projectile, and, as the latter gets close to the target, the orbit curves, so that the projectile is \"scattered\" and moves away in a different direction. An example of this sort of collision is the famous Rutherford experiment, in which the projectile and target were both positively charged particles and the force that caused the scattering was the Coulomb repulsion between them. In Figure 14.2, the target is a hard sphere, which exerts a force on the projectile only when they come into contact (a contact force); thus, the projectile travels in a straight line until it hits the target (if it does) and then bounces and moves off in a different direction. A familiar example of this kind of event is the collision of two billiard balls (though in this case the projectile and target would be the same size). With these two examples in mind, we can now define the first two key parameters of collision theory. The scattering angle 0 is defined as the angle between the incoming and outgoing velocities of the projectile, as indicated in both Figures 14.1 and 14.2. In the absence of any target, the scattering angle would, of course, be zero. Thus 0 = 0 corresponds to no scattering; for example, in Figure 14.2 the projectile could miss the target entirely and then 0 would be zero. The maximum possible value is 0 = 7r; in Figure 14.2, a head-on collision, in which the projectile comes in along the target's axis and bounces straight back, would give 0 = The impact parameter b is defined as the perpendicular distance from the pro- jectile's incoming straight-line path to a parallel axis through the target's center, as shown on the left in both figures. A second way to think about the impact parameter is illustrated toward the right in Figure 14.1: You can think of b as the distance that would be the distance of closest approach if there were no forces on the projectile, so that the orbit was just a straight line. In other words, the impact parameter tells Quantum mechanics is an intrinsically statistical theory, dealing with probabilities rather than definite predictable outcomes. Thus in quantum mechanics the need to discuss the distribution of different scattered directions is there from the outset \u2014 in contrast with the situation in classical mechanics, where the same need arises from the practical impossibility of observing the detailed orbit of a projectile. Nevertheless, the machinery for handling the problem \u2014 most notably the idea of the collision cross section \u2014 is very similar in the two theories. Section 14.1 The Scattering Angle and Impact Parameter 559 Figure 14.1 In this collision experiment the projectile approaches from the left moving like a free particle. When it begins to feel the force field of the target, its orbit curves and it moves away in a different direction. The", " scattering angle 9 is the angle between the initial and final velocities. The impact parameter b is the perpendicular distance from the incoming straight-line orbit to a parallel axis through the center of the target. Figure 14.2 A scattering experiment in which the projectile and target interact only through a contact force, so that the projectile is deflected only when it actually hits the target and bounces off it. A collison occurs only if the impact parameter b is less than the radius of the target. how closely the projectile was aimed at the target. If the impact parameter is very large, then the projectile will hardly feel the target, and 9 will be small. In fact, in Figure 14.2, 9 would be precisely zero for any value of b greater than the radius of the target. At the other extreme, the value b = 0 implies a head-on collision and often 2 9 = n. It is reasonably clear from Figures 14.1 and 14.2 that for a corresponds to given value of b there will be a unique corresponding value of 0, and we shall find, in fact, that the main theoretical task in classical collision theory is to find the functional relation 9 = 0 (b) between these two variables. In atomic and subatomic physics (where collision theory has its greatest applica- tion), the experimental status of the scattering angle is totally different from that of 2 In both our examples (the Rutherford experiment and hard-sphere scattering) b = 0 certainly implies 9 = Jr. On the other hand, if the force between the projectile and target is attractive, then a projectile with b = 0 will crash into the target, and \u2014 depending on the nature of the target \u2014 may never re-emerge, or may plow straight through and emerge with 9 = 0. 560 Chapter 14 Collision Theory scattered proton recoiling nucleus (a) (b) Figure 14.3 (a) In a cloud chamber, any moving charged particle leaves a vis- ible track which can be photographed for later examination. Here four protons have entered the chamber from the left, three of them passing straight through undeflected. The fourth was deflected when it hit a nitrogen nucleus and left the chamber near the top right. The recoiling nucleus can be seen moving down and to the right. (b) Tracing of the tracks involved in the collision. the impact parameter: The scattering angle 9 is rather easily measured, whereas the impact parameter can never be measured directly. Let us address these two points in turn. There are many ways to measure the scattering angle, one of the most transparent of which is illustrated in Figure 14.3. This could represent a photograph made in a cloud chamber of a collision of a proton with a nitrogen nucleus. These particles are of course far too small to be photographed directly, but the cloud chamber is one of several devices that let one record the track of a moving charged particle \u2014 even when the particle is far too small to be seen itself. In the cloud chamber, a charged particle moving through a cloud of supersaturated water vapor ionizes some of the atoms that it passes, and these ionized atoms cause some of the water vapor to condense, creating a visible trail, somewhat like the vapor trail left in the sky by some aircraft. In Fig- ure 14.3 you can clearly see the trail of the incoming proton. The nucleus which the proton eventually strikes is initially invisible, since it is not moving. However, when the proton strikes the nucleus, the proton's track makes an abrupt change of direction, and the nucleus recoils leaving its own track. From pictures like this, the scattering angle is easily measured. On the other hand, the impact parameter can never be directly measured. The problem is that impact parameters of interest are of atomic or subatomic size \u2014 around 0.1 nanometers or less. A cloud-chamber track like that of Figure 14.3 has a width of order 1 millimeter, some 10 million times greater than the largest impact parameters of interest. Obviously direct and meaningful measurements of the impact parameter are out of the question. As we shall see in the next section, it is the impossibility of measuring the impact parameter that leads us to the notion of the collision cross section. 14.2 The Collision Cross Section Let us imagine first that we were to observe a single collision like that shown in Figure 14.3. If we knew the impact parameter, then we could obviously deduce Section 14.2 The Collision Cross Section 561 something about the size of the target or the strength and range of the force that it exerts on the projectile. But given that we do not know the impact parameter, there is remarkably little we can learn from a single event like this; in fact, about the only thing we can conclude from this one event is that there is some kind of obstacle in the projectile's path. On the other hand, if we can observe many different collisions of similar projectiles and targets, then we can begin to", "s to apply. On the other hand, the targets must not be too numerous, or some targets may get hidden in the \"shadow\" of others, and the projectile may make several collisions. It was to avoid multiple collisions that Rutherford used a thin foil in his famous experiment. 562 Chapter 14 Collision Theory the probability that any one projectile makes a hit as it passes through the assembly on a random path is the ratio (probability of a hit) = area occupied by targets n tai.A.a total area = A = ntara. (14.1) If we now send in a beam containing a large number (call it Ninc) of incident projectiles, then the actual number of projectiles that get scattered (N\u201e) should be the product of the probability (14.1) and Ninc : This is the basic relation of collision theory. Since we can measure the numbers N sc and Nin, and the target density n tar, Equation (14.2) lets us find the size (or cross section) a of the target. In what follows, we shall see that the notion of the cross section gets generalized considerably and can become quite complicated, but the essential idea is always the same: By counting the number of scatterings (or reactions, or absoptions, or other processes) that result from a large number of similar collisions, one can use the analog of (14.2) to find the parameter a, which is always the effective area of the target for interacting with the projectile. EXAMPLE 14.1 Shooting Crows in an Oak Tree A hunter observes 50 crows settling randomly in an oak tree, where he can no longer see them. Each crow has a cross-sectional area a ti 2 ft2, and the oak has a total area (as seen from the hunter's position) of 150 square feet. If the hunter fires 60 bullets at random into the tree, about how many crows would he expect This situation closely parallels our simple scattering experiment. The target density is n tar = (number of crows)/(area of tree) = 50/150 = 1/3 ft -2. The number of incident projectiles is Nin, = 60, so, by the analog of (14.2), the expected to hit? number of hits is Nhit = Ninc ntar a = 60 x ft -2) x ft2) = 10. In practice, one often uses a steady stream of projectiles, and it may be more convenient to divide the incident number Nin, by the time At, to give the incident rate Rinc = Nind At. Similarly, the scattered rate is 1?\u201e = N\u201e/ At. Dividing both sides of (14.2) by At, we get the completely equivalent relation Rsc = Rine n for these rates. Since the cross section a is an area, the SI unit of cross section is, of course, the square meter. Atomic and nuclear cross sections are inconveniently small when Section 14.3 Generalizations of the Cross Section 563 measured in square meters. In particular, typical nuclear dimensions are around 10-14 m, so nuclear cross sections are conveniently measured in units of 10 -28 m2. This area has come to be called a barn (as in \"it's as big as a barn\"), 1 barn = 10 -28 m2. EXAMPLE 14.2 Scattering of Neutrons in an Aluminum Foil If 10,000 neutrons are fired through an aluminum foil 0.1 mm thick and the cross section of the aluminum nucleus is about 1.5 barns, 4 how many neutrons will be scattered? (Specific gravity of aluminum = 2.7.) The number of scatterings is given by (14.2), and we already know that NI= = 104 and a = 1.5 x 10 -28 m2. Thus all we need to find is the target density ntar, the number of aluminum nuclei per area of the foil. (Of course, the foil contains lots of atomic electrons as well, but these do not contribute appreciably to the scattering of neutrons.) The density of aluminum (mass/volume) is p = 2.7 x 103 kg/m3. If we multiply this by the thickness of the foil (t = 10-4 m), this will give the mass per area of the foil, and dividing this by the mass of an aluminum nucleus (m = 27 atomic mass units), we will have n tar : (2.7 x 103 kg/m3) x (10-4 m) = 6.0 x 1024 m-2. (14.3) n = gt tar m = 27 x 1.66 x 10 -27 kg Substituting into (14.2) we find for the number of scatterings N\u201e = N,\u201e, ntar a = (104) x (6.0 x 1024 m-2 ) x (1.5", " x 10-28 11112)= 9. Here, we used the given cross section a to predict the number Ns, of scatterings we should observe. Alternatively, we could have used the observed value of N\u201e to find the cross section a. 14.3 Generalizations of the Cross Section The relation (14.2), with its many generalizations, is the fundamental relation of collision theory. Theorists calculate the cross section a using assumed models of the target, and experimenters then use (14.2) to measure a and compare with the predicted value. However, the projectile and target, and their interactions, are generally much more complicated than the simple point projectile and hard-sphere target that we used in the last section. In this section, we'll look at a few slightly more interesting cases. 4 As we shall see shortly, the cross section of a target can be different for different projectiles. Thus, strictly speaking, I should say that the cross section of aluminum for scattering neutrons is about 1.5 barns. Moreover, the cross section can depend on the energy of the projectiles; the number given here is valid for energies from about 0.1 eV to about 1000 eV. 564 Chapter 14 Collision Theory Scattering of Two Hard Spheres Let us imagine a target that is a hard sphere of radius R2 and a projectile that is another hard sphere of radius R 1, as shown in Figure 14.5. The two spheres make contact if and only if the impact parameter b is less than or equal to the sum of the two radii, b < R1 + R2; that is, the center of the projectile must lie inside a circle centered on the target with radius R 1 + R2 and area a = r (R i + R2) 2. We can now go through exactly the same argument as in the last section, finding the probablility of any one projectile getting scattered, and thence the total number of projectiles scattered. The only difference is that the area of the target a = Jr R 2 must now be replaced by cr = (RI + R2) 2. Thus we arrive at the same conclusion except that now Nsc= Ninc ntar a a = IT (R + R2) 2. (14.4) (14.5) The main moral of this example is that we can continue to use the usual relation (14.4), but you should no longer see a as the cross section of the target. Rather, a is a property of the target and projectile and should be thought of as the effective area of the former for scattering the latter. In particular, the cross section of a particular target for scattering one kind of projectile may be very different from that for the same target with a different projectile. Figure 14.5 A hard-sphere projectile of radius R 1 approaches a hard- sphere target of radius R2, with impact parameter b. A collision occurs only if b < R1 + R2. EXAMPLE 14.3 Mean Free Path of an Air Molecule The N2 and 02 molecules in the air around us behave very much like hard spheres 0.15 nm. Use this to estimate the mean free path of an air molecule of radius R at S TP. The mean free path A of a molecule in a gas is an important parameter that determines several properties of the gas \u2014 the conductivity, viscosity, and diffusion rates, for example. It is defined as the average distance that a molecule travels between collisions with other molecules. To estimate this, let us follow one chosen molecule as it moves through the gas, starting immediately after a Section 14.3 Generalizations of the Cross Section 565 collision. To simplify our discussion, I shall assume that all the other molecules are stationary. (This approximation changes the answer a little, but we shall still get a reasonable estimate.) We can think of our chosen molecule as a projectile moving through a target assembly of all the other stationary molecules, and our problem is to find the average distance x it will travel before making a collision. The cross section for collisions is given by (14.5) as a = n (2R) 2 = 4 r R 2. If we imagine first a thin slice of the gas with thickness dx (in the direction of our projectile's velocity), then the \"target\" density of this slice is n = \u2014ax tar N V where N is the total number of molecules and V their total volume, so that N/ V is the number density (number/volume). Thus the probablity that our molecule will make a collision in any thin slice of thickness dx is given by (14.1) as prob(collision in dx) = Na dx. (14.6) V Let us denote by prob(x) the probability that the projectile travels a distance x without making any collisions. The probability that it travels a distance x without colliding and then does collide in the next dx is the product of prob(x) and", " (14.6): prob(first collision between x and x dx) prob(x) dx. (14.7) No- V On the other hand, this same probabilty is just prob(first collision between x and x dx) = prob(x) \u2014 prob(x dx) = \u2014 d \u2014 prob(x) dx. (14.8) d x Comparing (14.7) with (14.8), we get a differential equation for prob(x): el-pr dx ob(x) = prob(x), Na V from which we see that prob(x) decreases exponentially with x, prob(x) = e -(Na/ v)x. (14.9) [Here I have used the initial condition that prob(0) is obviously 1.] The mean free path is the average value of x (that is, A, = (x)), and, to find this average, we must multiply x by the probability (14.8) and integrate over all possible values of x: \u00b0\u00b0 (x ) = f x [Na e _o fal v)x]d x = Na. V V (14.10) 566 Chapter 14 Collision Theory At STP we know that 22.4 liters of air contain Avogadro's number of molecules (one mole). Therefore, V X = NA (47r R 2) (6.02 x 1023) x 47r(0.15 x 10 -9 m) 2 22.4 x 10-3 m3 = 1.3 x 10 -7 m = 130 nm. We see that the mean free path of an air molecule is considerably more than the inter-molecular spacing (around 3 nm) and vastly bigger than the molecular size (around 0.3 nm). Different Processes and Targets So far we have considered collisions in which the most that can happen is that the projectile is deflected by the target and emerges in a different direction. There are several other possibilities: Consider a collision between a point projectile and a target consisting of a ball of putty. If the putty is sufficiently absorbent, then any projectile that hits it will burrow in and never re-emerge. That is, the target will capture, or absorb, the projectile. Exactly our previous argument would then give the number of projectiles captured as Ncap = Ninc n tar a. We can easily make matters more complicated. For example, part of the target's surface could be absorbent and part hard. Projectiles that hit the absorbent surface would be captured and those that hit the hard part would be scattered (that is, re-emerge traveling in a different direction). In this case, we would have two separate relations analogous to (14.4), one to give the number of captures and the second for the number of scatterings: Ncap = Ninc ntar acap [capture] (14.11) and Nse = Ninc n tar a\u201e [scattering]. (14.12) Here acap is the area of that part of the target that absorbs the projectile and a s, the area of the part that scatters it. The total cross section of the target is, of course, atot = acap asc An example of a real collision in which both capture and scattering are possible is the collision of an electron and an atom, such as chlorine, that can capture an extra electron. In this case, we can no longer identify a particular area of the target that will capture the projectile, but it will still be true that the number of projectiles captured is proportional to Niac, the number of incident projectiles, and to n tar, the target density. Thus we can use the exact same equation (14.11) to define the capture cross section (reap as the relevant constant of proportionality. With this definition you can (and should) view acap as the effective area of the target for capturing the projectile. In the same way, we can use (14.12) to define the scattering cross section as, and then view a\u201e as the effective area of the target for scattering the projectile. Section 14.3 Generalizations of the Cross Section 567 In collisions of electrons and atoms, there are other possibilities besides scattering and capture. For example, if the incident electron has enough energy, it may be able to ionize the atom, knocking one or more of the atomic electrons free. The number of ionizations can be written as Nion = Nine ntar aion [ionization] (14.13) and this defines the ionization cross section as the effective area of the target atom for ionization by the incoming electron. Again, when a neutron collides with a nucleus of 235U, it can cause the nucleus to fission, splitting into two much smaller nuclei and releasing some 200 MeV of kinetic energy, and we can define the fission cross section ofis as the effective", " area of a 235 U nucleus for fission by neutron bombardment, satisfying the equation Nfis = Nific n ta,. o-fis There is one other classification that deserves mention. The word \"scattering\" is generally reserved for a process in which the projectile is deflected and moves off, leaving behind the same target \u2014 the same atom, the same nucleus, or whatever. This usage excludes processes like capture, in which the projectile does not emerge from the collision at all. It also excludes processes like ionization, in which an electron is knocked off the target atom, or like fission, in which the target nucleus is broken into pieces. If the internal motions of the target are left unchanged, the scattering is said to be elastic; if the internal motions of the target are changed by the collision, then the scattering is called inelastic. Consider, for example, the scattering of an electron by a stationary atom. To simplify our discussion, let us suppose that the atom is fixed (an excellent approximation, since an atom is so much heavier than an electron) and that the atomic electrons are initially in their ground state \u2014 the lowest possible energy levels and the atom's most stable state. When the incoming electron scatters off the atom, there are two possibilities: The electron may scatter elastically, emerging with its kinetic energy unchanged and leaving the target in its original state of internal motion. Or it can scatter inelastically, giving some of its kinetic energy to the atom and raising the target's internal motion to a higher energy leve1. 6 This latter process of atomic excitation was first observed by the German physicists James Franck (1882- 1964) and Gustav Hertz (1887-1975)and gave compelling evidence for the existence of atomic energy levels (and won them the 1925 Nobel Prize). When a projectile scatters off a target, we can, if we wish, distinguish between the two types of process, elastic and inelastic. The total number of scatterings N\u201e in a given experiment is the sum of the elastic and inelastic scatterings, Ns, -= Nel Ninei, and we can define corresponding cross sections satisfying Esc = ael For a given target and given projectile, we can enumerate all the possible outcomes of a collision \u2014 scattering, capture, ionization, fission, and so on \u2014 and for each 5 Recall that atoms can exist only in certain discrete \"energy levels!' Since an isolated atom eventually finds its way to the lowest energy level (ground state), the target atom in a collision is most often in its ground state. 6 In general, both projectile and target can move, and they may both have internal structure and hence energy of internal motion \u2014 as in the collision of two molecules in a gas. In this case, an elastic scattering is defined as one in which the internal motions of both projectile and target are unchanged. This means the total kinetic energies Tproj Ttar are the same before and after the encounter. +, 568 Chapter 14 Collision Theory outcome, we can define a corresponding cross section. The sum of all these partial cross sections is called the total cross section o-tot. For example, for electrons colliding with an atom, it may be that there are just three possible outcomes, scattering, capture, or ionization; in this case, the total cross section would be atot = asc + a + aion - cap By adding up the three equations (14.12), (14.11), and (14.13), we can see that the total cross section gives the total number of projectiles removed from the incident beam by all three possible processes: Ntot = Nine n tar Crtot [total]. That is, o-tot is the effective area of the target for interacting with the projectile in any of the possible ways. In most cases, the various cross sections we have defined are found to vary with the energy of the incident projectile. For an example that is easily understood, consider the ionization cross section for an electron colliding with an atom. To ionize the atom, the incoming electron must have the minimum energy needed to knock one of the atomic electrons free. If the incident electron's energy is less than this ionization energy, then ionization is impossible; therefore, N ion = 0, and the ionization cross section aion defined by (14.13) is exactly zero. Above the ionization energy, ionization is possible and both Nion and o-ton are normally nonzero. Obviously aion varies with energy. Although it is not always as obvious, we find in practice that almost all cross sections are likewise energy dependent. 14.4 The Differential Scattering Cross Section In defining the cross section a-sc in (14.12) we counted the total number of projectiles that were scattered, regardless of the particular direction in which they emerged. Obviously we could obtain more information if we chose to monitor these directions as well, and this leads us to the notion of the differential cross section,", " as we now discuss. To simplify matters, let us consider a collision where the only possible interaction is elastic scattering. For example, you could consider the scattering of a point projectile off a hard sphere (Figure 14.2) or the Rutherford scattering of a positively charged alpha particle off a heavy positive nucleus (Figure 14.1). In either case, there are just two possibilities: Either the projectile misses the target entirely and emerges unscattered, or it scatters elastically. 7 7 If the alpha particle of the Rutherford experiment had enough energy, it could also raise the nucleus to a higher energy level or knock it apart. However, at the low energies available to Rutherford, this was not a possibility, and we can confine our attention to these low energies for now. There is another subtlety in Rutherford scattering that is possibly worth mentioning: The Coulomb force F = kqQ1r2 of a truly isolated nucleus would extend all the way to infinity, and all alpha Section 14.4 The Differential Scattering Cross Section 569 If we are to monitor how many particles emerge in a given direction, we must agree on how to measure directions. It is standard to take the direction of the incident beam as our z axis, and then to specify the direction of any one scattered projectile by giving its polar angles 9 and 0. Since these angles form an infinite continuum, we cannot speak of the number of particles scattered in the exact direction (8, 0). Rather we must count the number of particles emerging in some narrow cone around (0, 0). To characterize the size of this narrow cone, we use the notion of solid angle, which is defined as follows. Solid Angle To understand the definition of the solid angle of a cone it helps to recall the definition of the ordinary angle between two lines in a plane. This is illustrated in Figure 14.6(a): If the two lines meet at 0, we draw a circle of any convenient radius r centered at 0. The two lines define an arc of length s on the circle, and we define the angle AO (in radians) as AO = s/r. (Since s is proportional to r, this definition is independent of our choice of r.) In a similar way, if a three-dimensional cone has its apex at 0, we draw a sphere of radius r centered at 0 as in Figure 14.6(b). The cone intersects the sphere in a spherical surface of area A (proportional to r2 ), and we define the solid angle of the cone as length = s (a) (b) Figure 14.6 (a) The ordinary two-dimensional angle AO subtended by an arc length s of a circle is defined as A9 = s 1 r where r is the radius of the circle and AO is in radians. (b) The solid angle AS2 of a cone subtended by an area A on a sphere is defined as AQ = A/r 2. Here r is the radius of the sphere and AS-2 is in steradians. particles, however large their impact parameter, would be deflected a tiny bit; in other words, all of the incident alphas would be scattered. In practice, however, the Coulomb force of the nucleus is always screened at large distances by the atomic electrons, and those alphas that pass outside the whole atom are not scattered. 570 Chapter 14 Collision Theory The unit of solid angle defined in this way is called the steradian, abbreviated sr. If the cone includes all possible directions, then since the area of the whole sphere is 47r 2, the solid angle is 47r. That is, the solid angle corresponding to all possible directions in three dimensions is 47 steradians, just as the ordinary angle corresponding to all directions in two dimensions is 27r radians. The cone shown in Figure 14.6(b) was a circular cone, but the definition 0S2 = A / r 2 works equally for any shape of cone. For instance, we shall need to consider the narrow cone with polar angles in the ranges 0 to 9 + de and 0 to 0 + d5; this cone intersects the sphere in a \"rectangular\" surface of area r 2 sin 0 de d(/), and so has solid angle The Differential Cross Section Armed with the notion of solid angle, we are ready to define the differential scattering cross section. We imagine our usual experiment, in which a large number Nine of projectiles are directed at a target assembly with density n tar. For any chosen cone of solid angle dS2 in any chosen direction (0,4)), we now monitor the number of projectiles scattered into this dQ, as sketched in Figure 14.7. We denote this number by Nsc (into dS2), and, by the familiar argument it must be proportional to Nine and to n tar, so can be written as Nsc (into dS2) = Nine ntar do- (into dS2) (", "14.16) where do- (into dS2) is the effective cross-sectional area of the target for scattering into the solid angle dS2. Since this is proportional to dS2, it is traditional to rewrite it as do- (into dS2) = ddS2 S-2 where the factor da I dQ is called the differential scattering cross section. In terms of it, we can rewrite (14.16) as (0, 0) Figure 14.7 Projectiles are incident from the left on a rectangular target assembly, and we monitor the number N\u201e(into dS2) emerging into a cone of solid angle dQ in the direction (0, (p). Section 14.4 The Differential Scattering Cross Section 571 where I have added the argument (0, 0) to emphasize that the differential cross section will (in general) depend on the direction of observation. Equation (14.17) can be taken as the definition of the differential cross section do-Ida It is the experimentalist's job to measure da I dS2 using (14.17), and it is the theorist's job to predict do- I dS2 based on some assumed model of the interactions between the projectile and the target. If we add up the numbers Nsc (into c/Q) for all possible solid angles c/ Q we will recover the total number of scatterings, N\u201e. That is, integrating (14.17) over all solid angles will give Nsc. Since Nsc = Nincntara where a is the total scattering cross section, we conclude that a = -2- d (0, 0) c/C2 = f sin 0 dB f 27r d0 (0, 0) (14.18) dS2 7r 0 da dQ 0 where the second form follows from (14.15). That is, the total cross section 8 is the integral over all solid angles of the differential cross section. EXAMPLE 14.4 Angular Distribution of Scattered Neutrons At an incident energy of several MeV (million electron volts), the differential cross section for scattering of neutrons off a heavy nucleus might have the form do- dS-2 (0, 0) o-\u201e,(1 + 3 cos 0 + 3 cos2 0) (14.19) where a0 is a constant that could be about 30 millibarns per steradian (mb/sr). Describe the angular distribution of the scattered neutrons and find the total scattering cross section. The most prominent feature of (14.19) is that it is independent of 0. (As we shall see in the next section, this is a fairly common occurrence.) This means that the distribution of scattered neutrons is axially symmetric, which makes visualization of the angular distribution much simpler, since we have only to worry about its dependence on 0. Figure 14.8 shows da I dS2 as a function of from 0 = 0 to 7. We see that da I d Q is largest at 0 = 0. That is, in this example at least, a particle that is scattered is most likely to be scattered near to the \"forward direction\" 0 = 0. (In quantum scattering, especially at high energies, this is often the case.) In this example, there is also a much smaller maximum in the backward direction at B = Tr, and the probability for scattering directly back at B = n is greater than for any other direction in the backward hemisphere 7/2 < 0 < 8 In general, one should say the total scattering cross section, asc ; here we are assuming that scattering is the only possible interaction, so they are the same. 572 Chapter 14 Collision Theory ) Figure 14.8 The differential cross section (14.19) for scattering of neutrons off a nucleus, plotted as a function of the scattering angle O. The vertical axis shows do- /d2 in units of millibarns per steradian. The total scattering cross section is found by integrating the differential cross section, as in (14.18). Since the integrand is independent of 0, the integration over 0 is trivial and gives a factor of 27r, so da 7r a =I \u2014 (0, cP)dS2 = 27rao f sin 0 dO (1 + 3 cos 0 + 3 cos 2 0) = 87rao = 754 mb. (14.20) 14.5 Calculating the Differential Cross Section To simplify the calculation of the differential cross section, I shall assume that the scattering is axially symmetric. This is certainly the case if the target is spherically symmetric (like the hard sphere of Figure 14.2 or any target which exerts a spherically symmetric force field), since spherical symmetry implies axial symmetry. It means that the differential cross section is independent of 0 and allows us to include all different values of 0", " in our discussion at the same time. We imagine a projectile incident on the target with impact parameter b. By calculating the projectile's trajectory, we can, in principle at least, find the corresponding scattering angle 8 = 0 (b) as a function of b. Alternatively, by solving for b, we can express b as a function of 0, that is, b = b(0). Let us next consider all those projectiles that approach the target with impact parameters between b and b + db. These are incident on the annulus (the shaded ring shape) shown on the left of Figure 14.9. This annulus has cross sectional area These same particles emerge between angles 8 and 8 dO in a solid angle do- = 27rb db. c/S2 = 271- sin& de, (14.21) (14.22) Section 14.5 Calculating the Differential Cross Section 573 Figure 14.9 All projectiles incident between b and b db are scattered be- tween angles B and B + dB. The area on which these particles impinge is do- = 27rb db, and the solid angle into which they scatter is dSZ = 27 sin 0 dB. as indicated in Figure 14.9. The differential cross section daldS2 is now found by simply dividing (14.21) by (14.22) to give where I have inserted the absolute value signs to ensure that do- IdS2 is positive. (Because 9 often decreases as b increases, db/d0 may be negative.) In summary: To calculate the differential cross section for a projectile scattering off a given target, we must first calculate the projectile's trajectory, to find the scattering angle B as a function of the impact parameter b (or vice versa). Then da/c1C2 is found by simply differentiating b with respect to 0, as in (14.23). EXAMPLE 14.5 Hard Sphere Scattering As a first example of the use of (14.23), find the differential cross section for scattering of a point projectile off a fixed rigid sphere of radius R. Integrate your result over all solid angles to find the total cross section. Our first task is to find the trajectory of a scattered projectile, as shown in Figure 14.10. The crucial observation is that when the projectile bounces off the hard sphere, its angles of incidence and reflection (both shown as a in the picture) are equal. (This \"law of reflection\" follows from conservation of energy and angular momentum \u2014 see Problem 14.13.) Inspection of the picture shows that the impact parameter is b = R sin a, and the scattering angle is 9 = n - 2a. Combining these two equations we find that b= R sin 7r 2 = R cos(0 2), (14.24) 574 Chapter 14 Collision Theory Figure 14.10 A point projectile bouncing off a fixed rigid sphere obeys the law of reflection, that the two adjacent angles labelled a are equal. The impact parameter is b = R sin a, and the scattering angle is 0 = - 2a. and from (14.23), we find the differential cross section do- b db R cos (0 / 2) R sin(0/2) dS2 \u2014 sin 0 dO sin 0 2 R2 = 4- (14.25) The most striking thing about this result is that the differential cross section is isotropic; that is, the number of particles scattered into a solid angle c/S2 is the same in all directions. To find the total cross section, we have only to integrate this result over all solid angles: du 2 dQ = R dQ 7rR2, c/S2 4 which is, of course, the cross-sectional area of the target sphere. 14.6 Rutherford Scattering Perhaps the most famous collision experiment of all time was Rutherford's experi- ment, in which he and his assistants observed the scattering of alpha particles off the gold nuclei in a thin gold foil and used the observed distribution to argue for the nu- clear model of the atom. According to this model, the force of a nucleus (charge Q) on an alpha (charge q) is F = kqQ y = r2 r2 (14.26) The alphas are scattered appreciably only if they approach close to the nucleus, well inside the orbiting atomic electrons. Therefore we can ignore the force of the latter, and (14.26) is the only force on the alphas. Therefore, as we saw in Chapter 8, the orbit of an alpha is a hyperbola, with the nucleus (which we'll treat as fixed for the moment) at its focus, as shown in Figure 14.11. If u denotes the unit vector pointing from the target to the alpha's point of closest approach, the orbit is symmetric about Section 14.6 Rutherford Scattering 575 Figure 14.11 The Rutherford scattering of an alpha particle off", " a fixed atomic nucleus. The orbit is a hyperbola, which is symmetric about the line labelled by the fixed unit vector u. The position of the particle can be labelled by its angle measured from u. As the particle moves away (t --> oo), --> Vio, and as t --> \u2014 oc, --> \u2014 iko. Therefore the scattering angle is 9 = it - 2*(). the direction of u, and it is convenient to label the alpha's position by the polar angle measured from u (see Figure 14.11). Let us denote by ifr o the limit of ik as the scattered alpha moves far away, so that the total angle subtended by the alpha's orbit is 2*o and the scattering angle is 9 = it \u2014 2*o. (14.27) Our job now is to relate the scattering angle 0 to the impact parameter b. We can do this by evaluating in two ways the change in the momentum of the projectile, AP = \u2014 (14.28) where p and p' are the momentum long before and long after the encounter. First, by conservation of energy, p and p' have equal magnitudes, so that the triangle shown in Figure 14.12 is isosceles, and I API = 2p sin(0/2). (14.29) On the other hand, from Newton's second law, Ap = f Fdt. Comparing Figures 14.12 and 14.11, you can see that Ap is in the same direction as the unit vector u. Thus = 2 p sin(0/2) Figure 14.12 The change in momentum of the projectile is Ap = \u2014 p. Since, Ipi =- Ip'l, it is easily seen that = 2p sin(0/2). 576 Chapter 14 Collision Theory the magnitude of Ap is given by the same integral, with F replaced by its component Fu in the direction of u, IAPI = f Fu dt. From Figure 14.11 you can see that Fu = (y/r 2) cos *. Using the now-familiar trick, we can write.dt = din*, where, since mr 2 * = = by (see Figure 14.11 again), we can replace * by bp I mr2. Putting all of this together, we find ApI *.3 y cos * d* v = r2 bp I mr 2 bp gym by'in 2 sin *,, = cos(0/2). (14.30) [To understand the limits in the integral, recall that as t +cc, so * \u00b1*0. In the last step I used (14.27) to replace * by (n - \u2014 9)/2 and hence sin *0 by cos(9/2).] Equating the two expressions (14.29) and (14.30) for I Ap I, we can solve for b to give b = ym cos(0/2) y p2 sin(9/2) m v2 = \u2014 cot(0/2) (14.31) where in the last equality I replaced p by my, and v is the projectile's incident speed. Having found the impact parameter b as a function of the scattering angle 0, we can now use the result (14.23) to give the differential cross section do 1 b \u2014 sin 0 db d0 11 cot (0/2) 2 sin(0/2) cos(0/2) my 2 mv 2 2 sin2 (0/2) or, replacing y by kq Q, where E is the energy of the incident projectiles, E = Zmv 2. This is the celebrated Rutherford scattering formula. It gives the differential cross section for scattering of a charge q, with energy E, off a fixed target of charge Q. While it is still today, nearly a century after its derivation by Rutherford, a much-used result, its great historical importance is that it was used to prove the existence of the atomic nucleus, as we now discuss briefly.9 9 Since the atom is a microscopic system, for which quantum, not classical, mechanics should be used, you may be surprised that the classical Rutherford formula worked so well for Rutherford and his assistants. It is one of the most amazing accidents in the history of physics that the quantum formula for scattering of two charged particles agrees exactly with Rutherford's classical formula. (This is certainly not true for other force laws.) Section 14.6 Rutherford Scattering 577 The Experiment of Geiger and Marsden The best known and most important Rutherford-scattering experiment was performed by Rutherford's assistants Hans Geiger (inventor of the Geiger counter, 1882-1945) and Ernest Marsden (1889-1970) and published in 1913. Their goal was to test Rutherford's \"planetary\" model of the atom, according to which most of the atomic mass is concentrated in a tiny,", " positively charged nucleus. 10 As we have seen, this model leads to the cross section (14.32) for scattering of alpha particles, with several very specific predictions: The scattering probablility should be inversely proportional to sin4 0 /2, inversely proportional to the energy squared E 2, and proportional to the nuclear charge squared ( Q 2). Geiger and Marsden were able to verify all of these predictions with amazing precision, and hence to contribute to the rapid acceptance of Rutherford's nuclear atom. They used alpha particles coming from radon gas (\"radium emanation\" as it was called then), with energy around 6.5 MeV. (1 MeV = 10 6 electron volts, and 1 eV = 1.6 x 10 -19 joules.) They directed a narrow \"pencil\" of these at a thin metal foil and counted the scattered particles using a small zinc sulphide screen. Any alpha particle striking this screen caused a tiny flash of light or \"scintillation,\" which could be observed through a microscope. In this way, it was possible to count up to about 90 alpha particles per minute (a job needing great patience and concentration!). To observe the angular dependence of the scattering, they could swing the screen and microcope around to angles in the range 5\u00b0 < 8 < 150\u00b0. To test the dependence on incident energy, they passed the incident particles through thin sheets of mica, to slow them down and hence vary their energy. And to test the dependence on nuclear charge, they used various different target foils (gold, platinum, tin, silver, copper, and aluminum). EXAMPLE 14.6 Angular Dependence To isolate its angular dependence write the Rutherford cross section (14.32) as do- dQ (0) = a0 (E) sin4 0/2 (14.33) and find o-0 (E) for scattering of 6.5 MeV alphas off gold. Find the differential cross section at 150\u00b0 and 5\u00b0 (Geiger and Marsden's largest and smallest angles). Find the number of alphas they would have had to count in a minute assuming the following values: The number of incident alphas in one minute, N mc = 6 x 108 ; the thickness of the gold foil, t = 1 Am; area of zinc sulphide screen = 1 mm 2 ; and distance of screen from target =1 cm. Make a useable plot of the differential cross section as a function of scattering angle 9. 1\u00b0 Initially, the sign of the nuclear charge (positive or negative) was not clear, but it was soon found to be positive, with an equal negative charge carried by the orbiting electrons. 578 Chapter 14 Collision Theory The charge of the alpha particle is q = 2e and that of the gold nucleus is Q = 79e, so ao(E) (2 x 79 x ke 2 )2 4E This is easily evaluated in SI units, though a slicker way is to use the useful combination ke g = 1.44 MeVfm (where fm stands for femtometer or 10 -15 m). Either way, we find that ao = 76.6 x 10-30 m2 /sr = 0.766 barns/sr. Substituting into (14.33) we get da dQ (150\u00b0) = 0.88 barns/sr and (5\u00b0) = 2.1 x 10 5 barns/sr. (14.34) dQ The huge difference between these \u2014 more than 5 orders of magnitude \u2014 presents considerable practical difficulties, as we shall see. Before we can sub- stitute into (14.17) to give the actual numbers counted we need to calculate n tar and c/S2. As usual, we can find n tar in terms of the density of gold (specific gravity 19.3) and its atomic mass (197): or (19.3 x 103 kg/m3) x (10-6m) = 5.90 x 10 n tar \u2014 197 x 1.66 x 10 -27 kg 22 m-2. Geiger and Marsden's screen had area A = 1 mm 2 and was at a distance r = 10 mm from the target. Therefore, it subtended a solid angle dS2 = \u2014 = 0.01 sr. A r 2 Putting all of this together, we find for the number of alphas hitting their screen at 150\u00b0 in a minute Nsc (at 150\u00b0) = N;,,\u00b0 ntar \u2014 (150\u00b0) A2 da c/S2 = (6 x 108) x (5.90 x 1022 m-2) x (0.88 x 10 -28 m2/sr) x (0.01 sr) = 31, same calculation gives a number that they could count easily and accurately. On the other hand, the Nsc (at 5\u00b0) = 7.5 x 10 6, a number that", " they could not possibly count or even estimate. Obviously measur- ing the cross section at small angles required them to use a much, much weaker source than at large angles. Because of the huge variation of the cross section as the scattering angle varies, a straightforward linear plot of da/dQ is not especially useful. If we choose a scale to show the small angles, the cross section for large angles will Section 14.7 Cross Sections in Various Frames* 579 Figure 14.13 Semilog plot of the Rutherford differential cross section as a function of angle O. The dots are the measurements of Geiger and Marsden. appear to be zero; if we arrange to show the large angles, the cross section for small angles will disappear off scale. The solution to this is to make a semilog plot, that is, to graph the log of the cross section against O. This gives the curve shown in Figure 14.13, where you can see clearly the variation by more than 5 orders of magnitude between 15\u00b0 and 180\u00b0. The dots in this figure are the original data of Geiger and Marsden and show clearly why Rutherford's model of the atom gained such quick acceptance. 14.7 Cross Sections in Various Frames* * As usual, sections marked with an asterisk can be omitted on a first reading. For the most part, we have so far discussed collisions in which the target particle is fixed. While this is an excellent approximation if the target is very heavy compared to the projectile (as in scattering of electrons off an atom, for example), we must nevertheless recognize that there is no such thing as a truly fixed particle, and we must learn to treat collisions of two particles both of which can move. Fortunately, we already know how to do this: If we observe the motion in the CM frame (the reference frame where the center of mass is at rest), then the motion of the relative coordinate r = r l \u2014 r2 is precisely the same as that of a single particle with mass equal to the reduced mass iu = mi m2 /M. Thus, if we view the collision in the CM frame, then our problem is reduced back to the motion of a single \"equivalent particle\" in a fixed force field. The only remaining difficulty is this: The CM frame is usually not the frame in which we do experiments. Thus we must learn how to relate cross sections calculated in the CM frame to their corresponding values in the lab frame, the frame of the laboratory in which the experiment is to be performed. In particular, we are going to want to find the relation between the differential cross sections (do-/ dQ) of the CM frame and the corresponding (daldC2__) lab measured in the lab. 580 Chapter 14 Collision Theory The CM Variables Before we take up the problem of translating between frames, I need to mention two more elegant features of the CM frame. First, recall that the Lagrangian for two particles, when written in terms of the CM and relative coordinates, has the form (8.13) = mk2 ith2 u ( r ). Differentiating with respect to the three components of t, we find that the general- ized momentum corresponding to r is (14.35) (14.36) P = Pi% That is (as you would probably have guessed) the momentum for the relative motion is just that of a single particle of mass p, and velocity 1.. The second property concerns the momenta of the two particles, as measured in the CM frame. Recall from (8.9) that ri = R + 2 r and r2 = R \u2014 \u2014r. (14.37) We can differentiate these to get the two velocities. In particular, in the CM frame, R = 0, so we find that r l = (m 2 /M)i. Multiplying by m 1, we find for the momentum of the projectile p i = \u00b5r = p. That is, in the CM frame, the projectile's momentum p i is the same as the momentum p of the relative motion. In the same way we can prove that p 2 = \u2014p, so that The first equality here confirms that the total momentum in the CM frame is zero. The second is useful in the evaluation of cross sections: In measuring the differential cross section, we count the number of times the projectile emerges with its momentum p i i = p (in the CM frame), this is the same thing as inside some solid angle clS2. Since p the number of times the relative momentum p emerges in d Q. Therefore we can find the differential cross section in the CM frame just as if a single particle of mass it were scattering off a fixed target. Thus, for example, the Rutherford formula (14.32), for scattering of one particle off a fixed target, also gives the differential cross section for scattering of two charged particles in their CM frame, provided we replace m by it. General Relation between Cross Sections in Different Frames", " In the CM frame the projectile and target approach one another with equal and opposite momenta. In the lab frame of the traditional collision experiment (such as the Rutherford experiment) the target is initially at rest. In many modern colliding- beam experiments the projectile and target are both moving, in opposite directions. Section 14.7 Cross Sections in Various Frames* 581 In all of these cases, the initial momenta are collinear, and to simplify matters, I shall restrict our discussion to this case. 11 To see how the cross section transforms between two different frames, we have only to look at its definition. Let us start with the total cross section, which was defined in (14.2) so that (We'll continue to assume that the only possible outcome of a collision is elastic scat- tering, so we don't have to worry about other processes like absorption or ionization.) We can use this same definition in either frame. Thus we define the CM total cross section ac. by define alab by where all four quantities are measured in the CM frame. In exactly the same way we (14.39) (14.40) Nsc = Nine ntar Nem = Nem,cm Sc me \"tar 'cm Nlab = N.lab \u201e,lab \u201e sc inc ''tar 'lab N sc m N sl a b c c where all four quantities are measured in the lab frame. While any particular scattering event may look very different as viewed from the two different frames, the total number of events must be the same in either frame. Therefore, In the same way, the number of incident particles is the same, as seen in either frame, so that Nicnin c = Niinacb. The target density n tar is the density of target particles (number/area) seen from the incident direction, as illustrated in Figure 14.4. Since this is unaffected by any forward (or backward) motion of the target, ntc = nt. arm Comparing now (14.39) and (14.40), we see that each of the first three terms of (14.39) is equal to the corresponding term in (14.40). Therefore the final terms must also be equal, and we have the elegantly simple result that the total scattering cross sections ar in the CM and lab frames are equal, acm = alab [total scattering cross section]. (14.41) If other outcomes were possible, such as absorption or ionization, then exactly the same argument would lead to exactly corresponding results; for example, the total absorption cross section is the same in the CM and lab frames. The differential cross section is a little more complicated. The scattering angle, measured as gem in the CM frame, will generally have a different value Slab in the lab frame, and a given solid angle measured as dQ em in the one will be measured as dQlab 11 Some colliding beams are not perfectly collinear. Some experiments in atomic physics use beams that intersect at a large angle, and in the collisions of molecules in a gas, the two particles can approach one another at any angle. Although such oblique collisions are not especially difficult to handle, I shall consider only collinear collisions here. 582 Chapter 14 Collision Theory in the other. Apart from these complications, we can use the same argument as before. The definition of the differential cross section (14.17), Nsc (into dQ) = Ntac nrar \u2014 (14.42) da dQ (K2, can be used in either frame. Just as before, the numbers N tnc and n tar have the same values in either frame. Further, the number scattered in any chosen solid angle, called dS2 cm, in the CM frame is the same as the number going into the corresponding solid angle, called dC2 \u2014lab 9 in the lab frame. Thus, as before, the first three terms in (14.42) have the same values in both frames, and the same must therefore be true of the final product; that is or ( do- ) da \" \" 14 / cm \"\"lab \" lab ( do- ) = do- ) dQ c\u201e, S.-2 1 lab WS-2 /cm dS-21ab (14.43) (14.44) We see that the differential cross sections are not the same in the two frames, but only because a given solid angle has different values (dS2, n, and dC2lab,1 according to the \u2014 frame used. Since dQ = sin B de dO \u2014d (cos 0) d4, and since the azimuthal angle 4 of the outgoing momentum is the same in both frames, we can rewrite (14.44) in the more useful, if perhaps less transparent, form (The absolute value signs are needed since both cross sections are by definition positive, whereas the derivative on the right can sometimes be negative.) The problem of", " transforming the differential cross section from the CM to the lab frame is now reduced to the kinematic problem of finding Ocirt in terms of Blab, or vice versa, and then taking the indicated derivative. 14.8 Relation of the CM and Lab Scattering Angles* * As usual, sections marked with an asterisk can be omitted on a first reading. To relate the CM and lab scattering angles, we need to look at the momenta of the particles in both frames. Let us add subscripts \"cm\" and \"lab\" for these, and use a Section 14.8 Relation of the CM and Lab Scattering Angles * 583 P lab I P lab2 1 \u2014P 1 2 Blab Plabl (a) CM frame (b) lab frame Figure 14.14 (a) An elastic collision as seen in the CM frame. The two particles approach the CM (shown as a cross) with equal and opposite momenta. (b) The same collision as seen in the lab frame, where particle 2 is initially at rest. prime to indicate the outgoing values (and \"unprime\" for the incoming). The CM momenta are given by (14.38) as Parii = \u2014 Pcm2 = P (initial) (14.46) and or = \u2014 Picm2 = P (final) (14.47) where, as usual, p and p' denote the relative momentum (p = p,t). These values are illustrated in Figure 14.14(a). Notice that, by conservation of energy, all four momenta have equal magnitudes in the CM frame. To be definite, I shall confine our discussion to the case that the \"lab frame\" is the traditional lab frame, in which particle 2 (the target) is initially at rest. The various momenta as seen in this frame are illustrated in Figure 14.14(b). To find these momenta, we can return to the two equations (14.37): r1 =- R \u2014r M2 and r2 = R \u2014 mi \u2014r. (14.48) Since particle 2 is initially at rest, the second of these implies that = = P = M M2 m2 (14.49) This is the velocity of the center of mass, as seen in the lab frame, and allows us to relate any of the lab momenta to their corresponding CM values. In particular, differentiating the first of Equations (14.48), we find that Plabl = M 111.1 \u201e = M1 M2 Plab 1 = )47) P (14.50) 584 Chapter 14 Collision Theory Plab 1 Figure 14.15 The relation of initial and final momenta in the CM and lab frames. The number is the mass ratio m l/m 2 and was chosen to be 0.5 for this picture. The momenta shown as p and p' are the initial and final relative momenta, which are the same as p erm and picrni. where I have introduced the important mass ratio (14.51) (14.52) (14.53) In exactly the same way, the final momentum D /lab 1 i S - P'labl = XP + The two results (14.50) and (14.52) are illustrated in Figure 14.15, where the lines BC and BD represent the initial and final momenta of particle 1 in the CM frame, while AC and AD are the corresponding lab values. By dropping a perpendicular from the point D to the line AC, you can check (Problem 14.25) that tan Olab = sin Oc. + cos Ocm which tells us kb in terms of Om. Before we use this result to give us the lab cross section in terms of the corresponding CM value, let us use Figure 14.15 to establish a few other results. Because 1pi Ip1 1, the point D lies on a circle with center B and radius p, as indicated. It is easy to see that, unless O m = 0 or fr, Slab is always less than ecni, as you might expect. The details of Figure 14.15 depend on the relative sizes of the two masses: Suppose first that X < 1 (that is, the projectile is lighter than the target, m 1 < m2). In this case, the point A lies inside the circle as shown in Figure 14.15. (That figure was drawn using a mass ratio A = 0.5.) If O cm = 0, then the point D coincides with C and Slab is zero as well. If we imagine 0,,, increasing continuously from 0 to 7, the point D moves continuously around the semicircle from C to E, with \u00b0lab always less than Om, until Ocm = rr, at which point \u00b0lab is also equal to m. (That is, if the projectile bounces straight back in the CM frame, the same is true", " in the lab \u2014 at least if A < 1.) This behavior is illustrated in Figure 14.16, which is a graph of O iab c,,, for a mass ratio A = 0.5. against O Section 14.8 Relation of the CM and Lab Scattering Angles* 585 7r 7r/2 - Blab 0 0 Bcm 0/2 7T Figure 14.16 The lab scattering angle Bl ab as a function of the CM angle 0,,\u201e (14.53) for a mass ratio of 0.5 (that is, m i = 0.5m 2). The two angles are equal at 0 and 7r, but Blab < Ocm everywhere else. If A = 1 (that is, the projectile and target have equal masses), the behavior of e lab Bcm is surprisingly different. In this case, (14.53) reduces to as a function of Blab = le cm (14.54) (see Problem 14.27). Thus, as Bcm varies from 0 to 7, Blab runs from 0 to 7/2; in particular, in the lab frame of an equal-mass collision, the scattering angle can never exceed 90\u00b0. If A > 1, the situation is different again, as you can explore in Problem 14.31. To find the lab differential cross section, we need to find the derivative d(cos 9,\u201e,) I d(cos Olab) in (14.45). Using (14.53) it is a reasonably straightforward exercise (Prob- lem 14.26) to show that d(cos Blab) 1 + A cos Ocin d(cos 0,,,) (1 2A cos Ocm )n. 2) 3/2 Substituting into (14.45), we find that (do. ) (do- ) (1 + 2A cos Ocm )\u2018..2)3/2 dQ lab d S-2 ern I 1 + X cos Oem I (14.55) (14.56) EXAMPLE 14.7 Hard Sphere Scattering Again Find the CM and lab differential cross sections for scattering of a point projectile of mass m 1 off a hard sphere of radius R and mass m 2 = 2m 1, and plot each as a function of the appropriate scattering angle. The CM cross section is the same as that for a particle with mass equal to the reduced mass,u,, scattering off a fixed target. In Example 14.5 (page 573) we found the latter to be just R 2 /4. Therefore, (do- ) dQ cm 4 586 Chapter 14 Collision Theory Figure 14.17 (a) The differential cross section for scattering off a hard sphere is isotropic in the CM frame. (b) In the lab frame, it is peaked in the forward direction. and the lab cross section can be written down immediately from (14.56) (with = 0.5). The two cross sections are plotted as functions of their respective angles in Figure 14.17. 12 As we already knew, the CM cross section is isotropic. The lab cross section is markedly skewed in the forward direction. Principal Definitions and Equations of Chapter 14 The Scattering Angle and Impact Parameter The scattering angle is the angle B by which a projectile is deflected in its encounter with a target. The impact parameter is the distance b by which the projectile would have missed the center of the target if it had been undeflected. [Section 14.1] The Collision Cross Section The cross section aoc for a particular outcome \"oc\" (elastic scattering, absorption, reaction, fission) is defined by Noc = Nincntaraoc [Sections 14.2 & 14.3] where In10, is the number of outcomes of the type considered, N inc is the number of incident projectiles, and n tar is the density (number/area) of targets. The Differential Cross Section The differential cross section \u2014 (0, 0) for scattering in a direction (0,0) is de- do- fined fined by 12 Equation (14.56) gives the lab cross section as a function of the CM angle B um. To express it as an explicit function of B lab, one would have to solve (14.53) for Ocni in terms of Blab. To make the plot of Figure 14.17(b), a much simpler procedure is to treat both B lab and (do- I dQ) iab as functions of the parameter Ocm and make a parametric plot with O cm running from 0 to 7. Problems for Chapter 14 587 N\u201e (into dQ) = Ninc nt [Eq. (14.17)] da \u2014 ar (IQ (0, 0) dS2. Calculating the Differential Cross Section If you can find the scattering angle 0 as a function of", " the impact parameter b (or vice versa), then do dS-2 b db sin 0 d0 [Eq. (14.23)] The Rutherford Formula by the Rutherford formula The differential cross section for scattering a charge q off a fixed charge Q is given da (IQ kqQ -E sin20/2) ) 2 [Eq. (14.32)] The CM and Lab Cross Sections The lab frame is generally understood to be the frame in which the target is at rest; the CM frame is that in which the CM is at rest. The differential cross sections in the two frames satisfy ( do- d (cos 0cm ) 2 Ilab cm d (cos Olab) [Eq. (14.45)] Problems for Chapter 14 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). SECTION 14.2 The Collision Cross Section 14.1 * A blueberry pancake has diameter 15 cm and contains 6 large blueberries, each of diameter 1 cm. Find the cross section a of a blueberry and the \"target\" density n tar (number/area) of berries in the pancake, as seen from above. What is the probability that a skewer, jabbed at random into the pancake, will hit a berry (in terms of a and n tar, and then numerically)? 14.2 * (a) A certain nucleus has radius 5 fm. (1 fm = 10 -15 m.) Find its cross section a in barns. (1 barn = 10-28 m2.) (b) Do the same for an atom of radius 0.1 nm. (1 nm = 10 -9 m.) 14.3 * A beam of particles is directed through a tank of liquid hydrogen. If the tank's length is 50 cm and the liquid density is 0.07 gram/cm3, what is the target density (number/area) of hydrogen atoms seen by the incident particles? 14.4 ** The cross section for scattering a certain nuclear particle by a copper nucleus is 2.0 barns. If 109 of these particles are fired through a copper foil of thickness 10 gm, how many particles are 588 Chapter 14 Collision Theory scattered? (Copper's density is 8.9 gram/cm 3 and its atomic mass is 63.5. The scattering by any atomic electrons is completely negligible.) 14.5 ** The cross section for scattering a certain nuclear particle by a nitrogen nucleus is 0.5 barns. If 10 11 of these particles are fired through a cloud chamber of length 10 cm, containing nitrogen at STP, how many particles are scattered? (Use the ideal gas law and remember that each nitrogen molecule has two atoms. The scattering by any atomic electrons is completely negligible.) 14.6 ** Our definition of the scattering cross section, A r sc = Nincn tar a 9 applies to an experiment using a narrow beam of projectiles all of which pass through a wide target assembly. Experimenters sometimes use a wide incident beam, which completely engulfs a small target assembly (the beam of photons from a car's headlamp, directed at a small piece of plastic, for example). Show that in this case N sc = ninc NtarG where n in, is the density (number/area) of the incident beam, viewed head-on, and Ntar is the total number of targets in the target assembly. SECTION 14.4 The Differential Scattering Cross Section 14.7 * Calculate the solid angles subtended by the moon and by the sun, both as seen from the earth. Comment on your answers. (The radii of the moon and sun are R m = 1.74 x 10 6 m and R s = 6.96 x 108 dm = 3.84 x 108 m and ds = 1.50 x 10 11 m.) m. Their distances from earth are 14.8 * In their famous experiment, Rutherford's assistants, Geiger and Marsden, detected the scattered alpha particles using a zinc sulphide screen, which produced a tiny flash of light when struck by an alpha particle. If their screen had area 1 mm 2 and was 1 cm from the target, what solid angle did it subtend? 14.9 * By integrating the element of solid angle (14.15), dS2 = sin 0 dO dO, over all directions, verify that the solid angle corresponding to all directions is 47r steradians. 14.10 * By evaluating the necessary integral, verify the result (14.20) for the total cross section of Example 14.4 (page 571). (This is very easy if you change variables to u = cos 0.) 14.11 ** The differential cross section for scattering 6.5-MeV alpha particles at 120\u00b0 off a silver nucleus is about 0.5 barns/sr. If a total of 10 10 alphas impinge on a silver foil of thickness 1 itm and if we", " detect the scattered particles using a counter of area 0.1 mm 2 at 120\u00b0 and 1 cm from the target, about how many scattered alphas should we expect to count? (Silver has a specific gravity of 10.5, and atomic mass of 108.) 14.12 *** [Computer] In quantum scattering theory, the differential cross section is equal to the absolute value squared of a complex number f (6), called the scattering amplitude: da (K2 = If(0)1 2. (14.57) The scattering amplitude can in turn be written as an infinite series f (0) = \u2014 E(ze + h P e=o ei4 sin 8.e Pe (cos 0). (14.58) Here h = 1.05 x 10 -34J s is called \"h bar\" and is Planck's constant divided by 27r, and p is the momentum of the incident projectile. The real numbers 8 e are called the phase shifts and depend on Problems for Chapter 14 589 the nature of the projectile and target and on the incident energy. And Pt (cos 0) is the so-called Legendre polynomial. (P0 = 1, P1 = cos0, etc.) The partial-wave series (14.58) is especially useful at low energies where only a few of the phase shifts are different from zero. (a) Write down this series for the case of 10-MeV neutrons (mass m = 1.675 x 10 -27 kg) scattering off a certain heavy nucleus for which 8 = \u201430\u00b0, 6 1 = 150\u00b0, and all other phase shifts are negligible. (b) Find an expression for the differential cross section (in terms of h, m, the incident energy E, and the two nonzero phase shifts) and plot it for 0 < 0 < 180\u00b0. (c) Find the total scattering cross section. SECTION 14.5 Calculating the Differential Cross Section 14.13 ** In deriving the cross section for scattering by a hard sphere, we used the \"law of reflection,\" that the angles of incidence and reflection of a particle bouncing off a hard sphere are equal, as in Figure 14.10. Use conservation of energy and angular momentum to prove this law. (The definition of \"hard-sphere scattering\" is that a projectile bounces with its kinetic energy unchanged. That the force is spherically symmetric implies, as usual, that angular momentum about the sphere's center is conserved.) 14.14 ** One can set up a two-dimensional scattering theory, which could be applied to puck projectiles sliding on an ice rink and colliding with various target obstacles. The cross section a would be the effective width of a target, and the differential cross section da la would give the number of projectiles scattered in the angle d0. (a) Show that the two-dimensional analog of (14.23) is da Id() = Idbld0I. (Note that in two-dimensional scattering it is convenient to let 0 range from -11\" to 7r.) (b) Now consider the scattering of a small projectile off a hard \"sphere\" (actually a hard disk) of radius R pinned down to the ice. Find the differential cross section. (Note that in two dimensions, hard \"sphere\" scattering is not isotropic.) (c) By integrating your answer to part (b), show that the total cross section is 2R as 14.15 *** [Computer] Consider a point projectile moving in a fixed, spherical force whose potential expected. energy is U(r) = \u2014U\u00b0 { 0 (R < r) (0 < r < R) (14.59) where U0 is a positive constant. This so-called spherical well represents a projectile which moves freely in either of the regions r < R and R < r, but, when it crosses the boundary r = R, receives a radially inward impulse that changes its kinetic energy by +Uo (+Uo going inward, \u2014Uo going outward). (a) Sketch the orbit of a projectile that approaches the well with momentum po and impact parameter b < R. (b) Use conservation of energy to find the momentum p of the projectile inside the well (r < R). Let denote the momentum ratio = po / p and let d denote the projectile's distance of closest approach to the origin. Use conservation of angular momentum to show that d = fib. (c) Use your sketch to prove that the scattering angle 0 is 0 = 2 (arcsin \u2014 \u2014 arcsin. (14.60) R This gives 0 as a function of b, which is what you need to get the cross section. The relation depends on the momentum ratio, which in turn depends on the incoming momentum po and the well depth U0. Plot 0 as a function of b for the case that = 0.5. (d) By differentiating 0 with respect to b, find an 590 Chapter", " 14 Collision Theory expression for the differential cross section as a function of b, and make a plot of d6IdS2 against 0 for the case that = 0.5. Comment. [Hint: To plot as a function of 6 you don't need to solve for b in terms of 6; instead, you can make a parametric plot of the point (6, da / c1S2) as a function of the parameter b running from 0 to R.] (e) By integrating dalc1S-2 over all directions, find the total cross section. SECTION 14.6 Rutherford Scattering 14.16 * One of the specific predictions of Rutherford's model of the atom was that the cross section should be inversely proportional to E 2 or, equivalently, to v 4. To test this, Geiger and Marsden varied the speed v by passing the incident alphas through thin sheets of mica to slow them down. According to Rutherford's prediction, the product of N s, and v4 should be the same whatever the incident speed (provided all other variables were held constant). Add a row showing N s,v4 to this table of their data and see how well Rutherford's prediction was confirmed. Number of mica sheets 0 1 2 3 5 6 Counts, Ns, (per min) 24.7 29 33.4 44 101 255 Speed, v (arbitrary units) 1 0.95 0.90 0.85 0.77 0.69 0.57 4 81 14.17 * Another specific prediction of Rutherford's model of the atom was that the cross section should be proportional to the nuclear charge squared, that is, to Z 2, where Z is the atomic number, the number of protons in the nucleus. To test this, Geiger and Marsden counted the number of scatterings off various different targets (holding all other variables fixed), with the following results: Target Gold Platinum Tin Silver Copper Aluminum Nse Z 1319 79 1217 78 467 50 420 47 152 29 26 13 Add a row to this table to show the ratio N\u201e/Z2 and see how well Rutherford's prediction was confirmed. (At the time the atomic number was not known with certainty, nor was it well understood. Rutherford had guessed, correctly, that the nuclear charge was roughly equal to half the atomic mass, and this is what they used in place of Z.) The relatively poor agreement for the case of aluminum is probably due to our neglect of the target recoil, which is more important for the lighter targets.) 14.18 ** One the first observations that suggested his nuclear model of the atom to Rutherford was that several alpha particles got scattered by metal foils into the backward hemisphere, 7/2 < 0 < r - an observation that was impossible to explain on the basis of rival atomic models, but emerged naturally from the nuclear model. In an early experiment, Geiger and Marsden measured the fraction of incident alphas scattered into the backward hemisphere off a platinum foil. By integrating the Rutherford cross section (14.33) over the backward hemisphere, show that the cross section for scattering with 6 > 90\u00b0 should be 47 t co (E). Using the following numbers, predict the ratio Ns,(0 > 90\u00b0)/Nin,: thickness of platinum foil ti 3 pm, density = 21.4 gram/cm3, atomic weight = 195, atomic number = 78, energy of incident alphas = 7.8 MeV. Compare your answer with their estimate that \"of the incident a particles about 1 in 8000 was reflected\" (that is, scattered into the backward hemisphere). Small as this fraction is, it was still far larger than any rival model of the atom could explain. 14.19 ** An important simplification in our derivation of the Rutherford cross section was that the projectile's orbit is symmetric about the direction u of closest approach. (See Figure 14.11.) Prove that this is true of almost any conservative central force, as follows: (a) Assume that the effective potential (actual plus centrifugal) behaves as in Figure 8.4; that is, it approaches zero as r coo and approaches Problems for Chapter 14 591 +oo as r --> 0. 13 Use this to prove that any projectile that comes in from infinity must reach a minimum value rmin and then move out to infinity again. (b) This implies that a projectile must visit any value of r between rrnin and infinity exactly twice, once on the way in and again on the way out. Prove that the values of i- at these two points are equal and opposite, and that the values of 1.fr are equal (where tfr is the polar angle defined in Figure 14.11). (c) Use these results to prove that the orbit is symmetric under reflection about the direction u. 14.20 ** The derivation of the Rutherford cross section was made simpler by the fortuitous cancellation of the factors of r in the integral (", "14.30). Here is a method of finding the cross section which works, in principle, for any central force field: The general appearance of the scattering orbit is as shown in Figure 14.11. It is symmetric about the direction u of closest approach. (See Problem 14.19.) If ir is the projectile's polar angle, measured from the direction u, then --> +0. as t --> +oo and the scattering angle is 6 = rc \u2014 21/1f. (See Figure 14.11 again.) The angle iir o is equal to f Ili dt taken from the time of closest approach to oo. Using the now-familiar trick you can rewrite this as f dr. Next rewrite 1p in terms of the angular momentum C and r, and rewrite i- in terms of the energy E and the effective potential Ueff defined in Equation (8.35). Having done all this you should be able to prove that 0 =7r. \u2014 2 co (b / r2 ) dr fr. 11 \u2014 (b/r) 2 \u2014 U(r)/E (14.61) Provided this integral can be evaluated, it gives 0 in terms of b, and hence the cross section (14.23). For examples of its use, see Problems 14.21, 14.22, and 14.23. 14.21 ** Use the general relation (14.61) from Problem 14.20 to rederive the relation (14.24) for scattering by a hard sphere. Rutherford scattering. 14.22 *** Use the general relation (14.61) from Problem 14.20 to rederive the relation (14.31) for 14.23 *** Consider the scattering of a particle with energy E by a fixed, repulsive 1/r 3 force field, with potential energy U = y/r2. Use the relation (14.61) from Problem 14.20 to find 0 in terms of b and hence show that the differential cross section is y 7r 2 (7r \u2014 0) dS-2 E 0 2 (27r \u2014 0) 2 sin6 (14.62) To refresh your memory as to how to find r i.nin you might look at Figure 8.5 (the case E > 0). You should be able to solve your equation for 0 in terms of b to get b in terms of 0 and thence the cross section. SECTION 14.8 Relation of the CM and Lab Scattering Angles* 14.24 ** Consider the scattering of two particles of equal mass (for example, scattering of protons off protons). In this case, elab = 40,m. (See Problem 14.27.) (a) Use this result in (14.45) to prove that 13 Although this behavior is definitely the norm, there are a few force fields for which it is not true. If the actual potential energy is strongly attractive near r = 0 (for example, U (r) = \u20141/r 3), then it dominates the centrifugal potential near r = 0, and the effective potential does not approach +oc as r -+ 0. Our argument also breaks down for the special case that b = 0, in which case the projectile may smash directly into the target. 592 Chapter 14 Collision Theory when the projectile and target have equal masses (da dS2 lab ) do. 6/2 4 cos viab ( cm (14.63) (14.64) (b) Write down the lab cross section for the scattering of two equal-mass hard spheres. (We know that in the CM frame, the differential cross section is R 2 /4 where R = R 1 + R2.) By integrating over all directions, verify that the total cross section in the lab frame is 7 R 2, as it has to be. 14.25 ** Using Figure 14.15, prove Equation (14.53), that tan Blab = sin 0,. X \u00b1 cos Bcm 14.26 ** Using suitable trig identities, rewrite (14.64) to give cos Blab as a function of cos 9,., and verify the derivative (14.55) that was essential for finding the relation between the CM and lab cross sections. 14.27 ** (a) By setting the mass ratio X = 1 in Equation (14.64) of Problem 14.25, prove that in equal- mass scattering, Olab = 10cm. (b) Redraw Figure 14.15 for the case that A = 1 (m 1 = m2) and explain why the maximum value of Ow, is 7/2. 14.28 ** It is often interesting to know about the momentum of the recoiling target particle in the lab frame. Let us denote by *lab the recoil angle, defined as the angle between the recoil momentum Plab2 and the incident direction (the angle below the dashed line in Figure 14.14). _(a) Show that", " in Figure,1ab, 14.15 the recoil momentum is represented by the vector DC. Deduce that (7r \u2014 Ocm)/2. (b) Show that, in the special case of equal masses (m = m2), Blab = 7r/2; that is, the angle between the two outgoing particles in an equal-mass, elastic collision is 90\u00b0. (c) Prove this last result directly using just conservation of momentum and energy (in an elastic collision). 14.29 ** An elastic collision is defined as one in which the total kinetic energy of the two particles is the same before and after the collision. (a) Show that in the CM frame, the individual kinetic energies of the two particles are separately conserved in an elastic collision. (b) Explain clearly why the same result is obviously not true in the lab frame. (Think about the energy of the target particle.) (c) Let A E denote the energy gained by the target particle in the collision (and hence the energy lost by the projectile). Using Figure 14.15, show that the fractional energy lost by the projectile (in the lab frame) is D E E = (1 + A) 2 4X 2 sin (Oc./2) where, as usual, A is the mass ratio m 1 / ni 2. (Note that in Figure 14.15 the line DC represents the recoil momentum of the target.) (d) For a given mass ratio X, what sort of collision gives the largest fractional energy loss? What value of X maximizes this energy loss? (Your answer is important in situations where one wants a particle to lose energy as quickly as possible \u2014 as in a nuclear reactor, for example.) 14.30 *** If you have not already done so, do Problem 14.24. (a) Now consider the scattering of two equal-mass hard spheres, A and B, with B initially stationary. Write down the standard expression (14.42) for the number of projectiles A scattered into a solid angle dO at a chosen angle O. Call this number N (A into d Q at 0). Now suppose that we monitor for the number of target particles B recoiling into the same solid angle dS2 at the same angle 0. Find N(B into dSl at 0), the number of B's that will be observed. How does this compare with the number of A's? Problems for Chapter 14 593 14.31*** [Computer] Consider the elastic scattering of a projectile that is heavier than the target, that is, m 1 > m2 or X > 1. (a) Draw the analog of Figure 14.15 for this case. Show clearly that there are two different values of the CM angle O cm corresponding to each value of Blab. (b) What are the two CM angles that correspond to Old, = 0? In terms of this example, explain why there is this two-fold ambiguity in Ocn, when m 1 > m 2. (c) Plot Blab as a function of O cm for the case that X = 2. (d) Use your picture from part (a) to find an expression for the maximum possible value of () lab for a given value of Check that your answer is correct for the case that X = 1. CHAPTER Special Relativity From its publication in 1687 until 1905, Newtonian mechanics reigned supreme. It was applied to more and more systems, almost always with complete success. In those rare instances where Newtonian ideas appeared to fail, it was found that some complication had been overlooked, and, when this complication was included, Newton could again account for all the observations. 1 Newton's formulation was supplemented with new ideas (such as the notion of energy) and recast in different guises (by Lagrange and Hamilton), but the foundations seemed unshakeable. Then, toward the end of the nineteenth century, a few observations were made that seemed inconsistent with the classical, Newtonian, ideas. Heroic efforts were made to bring these observations into line with classical physics, but in 1905, Albert Einstein (1879-1955) published his first paper on the theory that we now call relativity, in which he showed that particles with speeds approaching the speed of light require a completely new form of mechanics, as I describe in this chapter. Even at slower speeds, Newtonian mechanics is only an approximation to the new \"relativistic mechanics,\" but the difference is usually so small as to be undetectable. In particular, at the speeds usually encountered on earth, Newtonian mechanics is completely satisfactory, which explains why it is still a crucial and interesting part of physics (and justifies the other 15 chapters of this book).2 I Perhaps the greatest such triumph for Newton was the prediction and discovery of the planet Pluto: Calculations of the orbit of Uranus (taking account of the other known planets, and based, of course, on Newtonian mechanics) disagreed with the observed position by some 1.5", " minutes of arc. In 1846, it was shown independently by the English astronomer John Couch Adams (1819-1892) and the Frenchman Urbain Leverrier (1811-1877) that this discrepancy could be explained by the presence of a hitherto unnoticed planet outside the orbit of Uranus. Within a few months, the new planet, now called Pluto, was discovered by the German Johann Galle (1812-1910) at exactly its predicted position. 2 In writing this chapter on relativity (particularly in the opening sections and the problems), it was sometimes difficult to resist borrowing ideas from the relativity chapters of Modern Physics, by Chris Zafiratos, Michael Dubson, and myself (second edition, Prentice Hall, 2003). I am grateful to Prentice Hall for giving me permission to do so. 595 596 Chapter 15 Special Relativity 15.1 Relativity Let us first consider the significance of the name \"relativity.\" A moment's thought should convince you that most physical measurements are made relative to a chosen reference system. That the position of a particle is r = (x, y, z) means that its position vector has components (x, y, z) relative to some chosen origin and a chosen set of axes. That an event occurs at time t = 5 s means that t is 5 seconds relative to a chosen origin of time, t = 0. If we measure the kinetic energy T of a car, it makes a big difference whether T is measured relative to a reference frame fixed on the road or to one fixed in the car. Almost all measurements require the specification of a reference frame, relative to which the measurement is to be made, and we can refer to this fact as the relativity of measurements. The theory of relativity is the study of the consequences of the relativity of measurements. At first thought, this would seem unlikely to be a very interesting topic, but Einstein showed that a careful study of how measurements depend on the choice of coordinate system revolutionizes our whole conception of space and time, and requires a complete rethinking of Newtonian mechanics. Einstein's relativity is really two theories. The first, called special relativity, is \"special\" in that it focuses primarily on unaccelerated frames of reference. The second, called general relativity, is \"general\" in that it includes accelerated reference frames. Einstein found that the study of accelerated frames leads naturally to a theory of gravitation, and general relativity turns out to be the relativistic theory of gravity. In practice, general relativity is required only in situations where its predictions differ appreciably from those of Newtonian gravity. These include the study of the intense gravity of black holes, of the large-scale universe, and of the effect of the earth's gravity on the extremely accurate time measurements needed for the global positioning system. In nuclear and particle physics, where we consider particles that move near the speed of light, but where gravity is usually completely negligible, special relativity is normally all that is needed. In this chapter, I shall treat only the special theory of relativity. 3 15.2 Galilean Relativity Many of the ideas of relativity are present in classical physics, and we have in fact met several in earlier chapters. Let us review these ideas and recast some of them in a form more suitable for our discussion of Einstein's relativity. As we discussed in Chapter 1, Newton's laws hold in many different reference frames, namely, the so-called inertial frames, any one of which moves at constant velocity relative to any other. We can rephrase this to say that, in classical physics, 3 To cover general relativity would require another book. Some good references are: R. Geroch, General Relativity from A to B, University of Chicago Press, 1978; I. R. Kenyon, General Relativity, Oxford University Press, 1990; B.F.A.Schutz, A First Course in General Relativity, Cambridge University Press, 1985; and James B. Hartle, Gravity: An Introduction to Einstein's General Relativity, Addison Wesley, 2003. Section 15.2 Galilean Relativity 597 Newton's laws are invariant (that is, unchanged) as we transfer our attention from one inertial frame to another. The classical transformation from one frame to a second, moving at constant velocity relative to the first, is called the Galilean transformation, so a compact way to say the same result is that Newton's laws are invariant under the Galilean transformation. Let us first review this claim. The Galilean Transformation For simplicity, consider first two frames 8 and 8' that are oriented the same way; that is, the x' axis is parallel to the x axis, y' parallel to y, and z' parallel to z. Suppose further that the velocity V of 8' relative to S is along the x axis. It was a fundamental assumption of Newtonian mechanics that there is a single universal time t. Thus if the observers in S and 3' agree to synchronize their clocks", " (and to use the same unit of time), then t' = t. Finally, we can choose our origins 0 and 0' so that they coincide at the time t = t' = 0. This configuration is illustrated in Figure 15.1, where S is a frame fixed to the ground. (We'll assume that a frame fixed to the earth is inertial \u2014 that is, we'll ignore the slow rotation of the earth.) The frame 8' is fixed in a train that is traveling with velocity V along the x axis. Consider now some event, such as the explosion of a small firecracker. As measured by observers in S this occurs at position r = (x, y, z) and time t; as measured in 8' it occurs at r' = (x', y', z') and time t'. Our first (and very simple) task is to establish the mathematical relation between the coordinates (x, y, z, t) and (x', y', z', t'). A moment's inspection of Figure 15.1 should convince you that x' = x \u2014 Vt, and that = y and z' = z. By the classical assumption concerning time, t' = t, so the required relations are = x \u2014 Vt = = z t. (15.1) These four equations are called the Galilean transformation. They give the coor- dinates (x', y', z', t') of any event as measured in S' in terms of the corresponding coordinates (x, y, z, t) of the same event as measured in 8. They are the mathematical expression of the classical ideas about space and time. 8 0 x Vt Figure 15.1 The frame S is fixed to the ground, while 5' is fixed in a railroad car traveling with constant velocity V in the x direction. The two origins coincide, 0 = 0', at time t = t' = 0. The star indicates an event, such as a small explosion. 598 Chapter 15 Special Relativity The Galilean transformation (15.1) relates the coordinates measured in two frames arranged with corresponding axes parallel and with relative velocity along the x axis, as shown in Figure 15.1 \u2014 an arrangement we can call the standard configuration. This is not, of course, the most general configuration. For example, if the relative velocity V is in an arbitrary direction, it is easy to see that (15.1) can be rewritten compactly as r' = r \u2014 Vt and t' = t. (15.2) This is still not the most general form of the Galiliean transformation, since we could rotate the axes, so that corresponding axes were no longer parallel, and we could displace the origins 0 or 0' and the origins of time. However, (15.2) is general enough for our present purposes. Using the Galilean transformation (15.2) we can immediately relate the velocities of an object, as measured in the two frames. If v(t) = i(t) is the velocity of the object as measured in S and v'(t) is likewise for 8' then by differentiating (15.2) we find immediately that (remember that V is constant) V = V - V. (15.3) This is the classical velocity-addition formula, which asserts that, according to the ideas of classical physics, relative velocities add (or subtract) according to the normal rules of vector arithmetic. Galilean Invariance of Newton's Laws To prove the invariance of Newton's laws under the Galilean transformation, suppose that the second law holds in frame 8; that is, that F = ma, with all three variables measured in 8. Now it is an experimental fact (at least in the domain of classical mechanics) that measurements of the mass of any object give the same results in all inertial frames. Thus the mass m' measured in 8' is the same as that measured in 8, and in' = m. The proof that the same is true for the net force depends, to some extent, on one's definition of force. If we take the view that forces are defined by the readings on spring balances, then it is clear that the force F' measured in 5' is the same as that measured in 5, and F' = F. Finally, differentiating (15.3) with respect to time (and remembering that V is constant, by assumption) we see that a' = a. We have now proved that each of the variables F', m', and a' of frame 5' is equal to the corresponding variable F, m, and a of frame 8'. Therefore, if it is true that F = ma, it is also true that F' = m'a'. That is, Newton's second law is invariant under the Galilean transformation. I leave it as an exercise (Problem 15.1) to prove that the same is true of the", " first and third laws. The invariance of the laws of mechanics under the Galilean transformation was known to Galileo, who used it to argue that no experiment could tell whether the earth was \"really\" moving or \"really\" at rest, and hence that Kepler's sun-centered view of the solar system was just as reasonable as the traditional earth- centered view. Section 15.2 Galilean Relativity 597 Newton's laws are invariant (that is, unchanged) as we transfer our attention from one inertial frame to another. The classical transformation from one frame to a second, moving at constant velocity relative to the first, is called the Galilean transformation, so a compact way to say the same result is that Newton's laws are invariant under the Galilean transformation. Let us first review this claim. The Galilean Transformation For simplicity, consider first two frames 8 and 8` that are oriented the same way; that is, the x' axis is parallel to the x axis, y' parallel to y, and z' parallel to z. Suppose further that the velocity V of 8' relative to S is along the x axis. It was a fundamental assumption of Newtonian mechanics that there is a single universal time t. Thus if the observers in S and 8' agree to synchronize their clocks (and to use the same unit of time), then t' = t. Finally, we can choose our origins 0 and 0' so that they coincide at the time t = t' = 0. This configuration is illustrated in Figure 15.1, where 8 is a frame fixed to the ground. (We'll assume that a frame fixed to the earth is inertial \u2014 that is, we'll ignore the slow rotation of the earth.) The frame 8' is fixed in a train that is traveling with velocity V along the x axis. Consider now some event, such as the explosion of a small firecracker. As measured by observers in S this occurs at position r = (x, y, z) and time t; as measured in 8' it occurs at r' = (x', y', z') and time t'. Our first (and very simple) task is to establish the mathematical relation between the coordinates (x, y, z, t) and (x', y', z', t'). A moment's inspection of Figure 15.1 should convince you that x' = x \u2014 Vt, and that y' = y and z' = z. By the classical assumption concerning time, t' = t, so the required relations are x \u2014 Vt yr = y z' = z = t. ( 15.1 These four equations are called the Galilean transformation. They give the coor- dinates (x', y', z', t') of any event as measured in 8' in terms of the corresponding coordinates (x, y, z, t) of the same event as measured in 8. They are the mathematical expression of the classical ideas about space and time. Figure 15.1 The frame S is fixed to the ground, while S' is fixed in a railroad car traveling with constant velocity V in the x direction. The two origins coincide, 0 = 0', at time t = t' = 0. The star indicates an event, such as a small explosion. 598 Chapter 15 Special Relativity The Galilean transformation (15.1) relates the coordinates measured in two frames arranged with corresponding axes parallel and with relative velocity along the x axis, as shown in Figure 15.1 \u2014 an arrangement we can call the standard configuration. This is not, of course, the most general configuration. For example, if the relative velocity V is in an arbitrary direction, it is easy to see that (15.1) can be rewritten compactly as rl = r \u2014 Vt and ti = t. (15.2) This is still not the most general form of the Galiliean transformation, since we could rotate the axes, so that corresponding axes were no longer parallel, and we could displace the origins 0 or 0' and the origins of time. However, (15.2) is general enough for our present purposes. Using the Galilean transformation (15.2) we can immediately relate the velocities of an object, as measured in the two frames. If v(t) i(t) is the velocity of the object as measured in 3 and vi(t) is likewise for 8' then by differentiating (15.2) we find immediately that (remember that V is constant) v \u2014 V. (15.3) This is the classical velocity-addition formula, which asserts that, according to the ideas of classical physics, relative velocities add (or subtract) according to the normal rules of vector arithmetic. Galilean Invariance of Newton's Laws To prove the invariance of Newton's laws under the Galilean transformation, suppose that the second law holds in frame 8; that is, that F = ma, with all three", " variables measured in 3. Now it is an experimental fact (at least in the domain of classical mechanics) that measurements of the mass of any object give the same results in all inertial frames. Thus the mass tit' measured in 3' is the same as that measured in 8, and m' = m. The proof that the same is true for the net force depends, to some extent, on one's definition of force. If we take the view that forces are defined by the readings on spring balances, then it is clear that the force F' measured in 8' is the same as that measured in 8, and F' = F. Finally, differentiating (15.3) with respect to time (and remembering that V is constant, by assumption) we see that a' = a. We have now proved that each of the variables F', rn', and a' of frame 8' is equal to the corresponding variable F, m, and a of frame 8'. Therefore, if it is true that F = ma, it is also true that F' = m'a'. That is, Newton's second law is invariant under the Galilean transformation. I leave it as an exercise (Problem 15.1) to prove that the same is true of the first and third laws. The invariance of the laws of mechanics under the Galilean transformation was known to Galileo, who used it to argue that no experiment could tell whether the earth was \"really\" moving or \"really\" at rest, and hence that Kepler's sun-centered view of the solar system was just as reasonable as the traditional earth- centered view. Section 15.2 Galilean Relativity 599 Galilean Relativity and the Speed of Light While Newton's laws are invariant under the Galilean transformation, the same is not true of the laws of electromagnetism. Whether we write them in their compact form as Maxwell's four equations, or in their original form (as Coulomb's law, Faraday's law, and so on), they can be true in one inertial frame, but if they are, and if the Galilean transformation were the correct relation between different inertial frames, then they could not be true in any other inertial frame. By far the quickest way to verify this claim is to recall that Maxwell's equations imply that light (and, more generally, any electromagnetic wave) propagates through the vacuum in any direction with speed C = 3.00 x 10 8 m/s, (15.4) 1 'N/E13 110 where co and p. 0 are the permittivity and permeability of the vacuum. Thus if Maxwell's equations hold in frame 8, then light must travel at the same speed c in any direction, as measured in S. But now consider a second frame 8', traveling at speed V along the x axis of 8, and imagine a beam of light traveling in the same direction. In S the light's speed is v = c. Therefore, in S' its speed is given by the classical velocity-addition formula (15.3) as VI C \u2014 V, as shown on the left of Figure 15.2. Similarly, a beam of light traveling to the left will have speed v = c in 8, but v' = c + V in 8'. Depending on its direction, any beam of light will have speed v' (as measured in 8') that varies anywhere between c \u2014 V and c + V. Therefore, Maxwell's equations cannot hold in the inertial frame 8'. S 8'= c+ V V Figure 15.2 Two frames S and S' in the standard configuration with relative velocity V. Two beams of light approach the car from opposite directions. If, as measured in 8, the light has speed c in either direction, then the classical velocity-addition formula implies that, as measured in 8', it has speed c \u2014 V traveling to the right, and c + V traveling to the left. If the Galilean transformation were the correct transformation between inertial frames, then although Newton's laws would hold in all inertial frames, there could only be one frame in which Maxwell's equations hold. This supposed unique frame, in which light would travel at the same speed in all directions, is sometimes called the ether frame.4 4 The origin of the name is this: It was assumed that light must propagate through a medium, in much the same way that sound travels through the air. Since no one had ever detected this medium 600 Chapter 15 Special Relativity The Michelson\u2014Morley Experiment The state of affairs just described, with the laws of mechanics valid in all inertial frames, but the laws of electromagnetism valid in a unique frame, was well under- stood toward the end of the nineteenth century. It was regarded by some (most notably Einstein) as unpleasing, and it was eventually shown by Einstein to be wrong. Nev- ertheless, it was logically consistent, and most physicists took for granted", " that there could be only one frame in which the speed of light had the same value c in all di- rections. Since the earth travels at a considerable speed in a continually changing direction around the sun, it seemed obvious that the earth must spend most of its time moving relative to the ether frame and hence that the speed of light as measured on earth should be different in different directions. The effect was expected to be very small. (The earth's orbital speed is V ti 3 x 104 m/s, large by terrestrial standards, but very small compared to c = 3 x 108 m/s. Thus the fractional variation, between c \u2014 V and c -I- V, was expected to be very small.) Nevertheless, in 1880, the Amer- ican physicist Albert Michelson (1852-1931), later assisted by the chemist Edward Morley (1838-1923), devised an interferometer that should have easily detected the expected differences in the speed of light. To their surprise and dismay they found absolutely no variation. Their experiments, and many different experiments with the same objective, have been repeated and have never found any reproducible evidence of variations in the speed of light relative to the earth. With hindsight, it is easy to draw the right conclusion: Contrary to all expectations, the speed of light is the same in all directions relative to an earth-based frame, even though the earth has different velocities at different times of year. In other words, it is not true that there is only one frame in which light has the same speed in all directions. This conclusion is so surprising that is was not taken seriously for twenty years. Instead, several ingenious theories were advanced to explain the Michelson\u2014Morley result while preserving the idea of a unique ether frame. For example, the so-called ether-drag theory held that the ether \u2014 the medium through which light was sup- posed to propagate \u2014 was dragged along with the earth, in much the same way the atmosphere is dragged along. This would imply that earth-bound observers are at rest relative to the ether and should measure the same speed of light in all directions. However, the ether-drag theory was incompatible with the phenomenon of stellar aber- ration.5 None of these alternative theories was able to explain all of the observed facts (at least, not in a reasonable and economical way), and today almost all physicists accept that there is no unique ether frame and that the speed of light is a universal constant, with the same value in all directions in all inertial frames. The first person to and since light could travel through seemingly empty space, the medium clearly had most unusual properties, and was named \"ether\" after the Greek for the stuff of the heavens. The \"ether frame\" was the frame in which the supposed ether was at rest. 5 The ether-drag theory would require that light entering the earth's envelope of ether would be bent. This would contradict stellar abberation, in which the apparent position of any one star moves around a small circle as the earth moves around its circular orbit \u2014 in a way that makes clear that the light from the star travels in a straight line as it approaches the earth. Section 15.3 The Postulates of Special Relativity accept this surprising idea whole-heartedly was Einstein, as we now discuss. In par- ticular, we shall see that the universality of the speed of light requires us to reject the Galilean transformation and the classical picture of space and time on which it was based. This, in turn, will require us to modify much of our Newtonian mechanics. 15.3 The Postulates of Special Relativity The special theory of relativity is based on the acceptance of the universality of the speed of light, as suggested by the Michelson\u2014Morley experiment. 6 Einstein proposed two postulates, or axioms, expressing his conviction that all the laws of physics should hold in all inertial frames, and from these postulates, he developed his special theory of relativity. we mean by an inertial frame: Before we discuss the postulates of relativity, it would be good to agree on what Definition of an inertial Frame An inertial frame is arty reference frame (that is a system coordinates x and time t) in which all the laws of physics hold in their w ual form. Notice that I have not yet specified what \"all the laws of physics\" are. Following Einstein, we shall use the postulates of relativity to help us decide what the laws of physics could be. (As always, the ultimate test will be whether they agree with experiment.) It will turn out that one of the classical laws that carries over into relativity is the law of inertia, Newton's first law. Thus our newly defined inertial frames are in fact the familiar \"unaccelerated\" frames, where an object subject to no forces travels with constant velocity. As before, a frame fixed to the earth is (to a good approximation) inertial; a frame fixed to an accelerating rocket", " or a spinning turntable is not. The big difference between the inertial frames of relativity and those of classical mechanics is the mathematical relation between different frames. In relativity, we shall find that the classical Galilean transformation must be replaced by the so-called Lorentz transformation. Notice also that I have specified that an inertial frame is one where the physical laws hold \"in their usual form.\" As we saw in Chapter 9, one can sometimes modify physical laws so that they hold in noninertial frames as well. (For example, by introducing the centrifugal and Coriolis forces, we could use Newton's second law in a rotating frame.) It is to exclude such modifications that I added the qualifier \"in their usual form.\" 6 Whether Einstein actually knew about the Michelson\u2014Morley result when he was formulating his theory is not clear. There is some evidence that he did, but it seems clear that his main motivation was the conviction that Maxwell's equations should hold in all inertial frames. Whether he knew or not affects neither Einstein's amazing accomplishment nor the importance of the Michelson\u2014Morley result as beautifully clear evidence in favor of Einstein's assumptions. 602 Chapter 15 Special Relativity The first postulate of relativity asserts the existence of many different inertial frames, traveling at constant velocity relative to one another: First Postulate of Relativi an inertial fra h and if a second frame oyes with t^o stant velocity etativ then also an inertial fram e Another way to say this is that the laws of physics are invariant as we transfer our attention from one frame to a second one moving at constant velocity relative to the first. This is what we proved for the laws of mechanics, but we are now claiming it for all the laws of physics. Another popular statement of the first postulate is that \"there is no such thing as absolute motion.\" To understand this, consider two frames, S attached to the earth and 8' attached to a rocket coasting at constant velocity relative to the earth. A natural question is whether there is any meaningful sense in which we could say that S is really at rest and 8' is really moving (or vice versa). If the answer were \"yes,\" then we could say that S is absolutely at rest and that anything moving relative to S is in absolute motion. However, this would contradict the first postulate of relativity: All of the laws observable by scientists in S are equally observable by scientists in 8'; any experiment that can be performed in S can equally be performed in 8'. Therefore, no experiment can show which frame is really moving. Relative to the earth, the rocket is moving; relative to the rocket, the earth is moving; and this is all we can say. Yet another statement of the first postulate is that among all the inertial frames, there is no preferred frame. The laws of physics single out no one frame as being in any way more special than any other. The second postulate specifies one of the laws that holds in all inertial frames: The speed cif li ht (in vacuum) s the same value in i ry fl ee t on in a Second Postulate of Relativity This is, of course, the Michelson\u2014Morley result. Although the second postulate flies in the face of our everyday experience, it is by now a firmly established experimental fact. As we explore the consequences of Ein- stein's postulates we are going to encounter several surprising predictions, all of which seem to contradict our experience (for example, the phenomenon called time dilation, described in the next section). If you have difficulty accepting these predictions, there are two points to bear in mind: First, they are all logical consequences of the sec- ond postulate. Thus, once you have accepted the latter (surprising, but indisputably true), you have to accept all of its logical consequences, however counterintuitive they may seem. Second, all of these surprising phenomena (including the second postu- late itself) have the subtle property that they become important only when objects travel with speeds comparable to the speed of light. In everyday life, with all speeds Section 15.4 The Relativity of Time; Time Dilation 603 much less than c, these phenomena simply do not show up. In this sense, none of the surprising consequences of Einstein's postulates really conflict with our everyday experience. 15.4 The Relativity of Time; Time Dilation Measurement of Time in a Single Frame We are going to find that the second postulate forces us to abandon the classical notion of a single universal time. Instead, we shall find that the time of any one event, as measured in two different inertial frames, is in general different. This being the case, we need first to be quite clear what we mean by time, as measured in a single frame. I shall take for granted that we have at our disposal lots of reliable tape measures and clocks. The clocks need not be identical", ", but they must have the property that, when brought together at the same point, at rest in the same inertial frame, they agree with one another. Let us now consider a single inertial frame 8, with origin 0. We can station a chief observer at 0 with one of our clocks, and she can easily time any nearby event, such as a small explosion, since she will see it essentially instantaneously. To time an event farther away from the origin is harder, since light from the event has to travel to 0 before she can sense it. If she knew how far away the event occurred, then she could calculate how long the signal took to reach her (she knows that light travels at speed c) and subtract this from the time of arrival to give the time of the event. A simpler way to proceed (in principle anyway) is to employ a large number of helpers stationed at regular intervals throughout the region of interest and each with his own clock. The helpers can measure their distances from 0, and we can check that their clocks are synchronized with the clock at 0 by having the chief observer send out a light signal at an agreed time (on her clock). Each helper can calculate the time taken by the signal to reach him and (allowing for this transit time) check that his clock agrees with the clock at 0. With enough helpers, stationed closely enough together, there will be a helper close enough to any event to time it essentially instantaneously. Once he has timed it, he can, at his leisure, inform everyone else of the result by any convenient means (such as a telephone). In this way, any event can be assigned a unique and well-defined time t as measured in the frame S. In what follows, I shall assume that any inertial frame S comes with a set of rectangular axes Oxyz and a team of helpers stationed at rest throughout 8 and equipped with synchronized clocks. This allows us to assign a position (x, y, z) and a time t to any event, as observed in the frame 8. Time Dilation Let us now compare measurements of times made by observers in two different inertial frames. Consider our familiar two frames, 8 anchored to the ground and 8' traveling with a train in the x direction at speed V relative to 8. We now examine a thought experiment (or gedanken experiment, from the German) in which an observer on the train sets off a flashbulb on the floor of the train. The light travels to the roof, where 604 Chapter 15 Special Relativity (a) (b) Figure 15.3 (a) The thought experiment as seen in frame 8'. The light travels straight up and down again, and the flash and beep occur at the same place. (b) As seen in 8, the flash and beep are separated by a distance V At. Notice that in 8 two observers are needed to time the two events, since they occur in different places. it is reflected back and returns to its starting point, where it strikes a photocell and causes an audible \"beep.\" We wish to compare the times, At and At', as measured in the two frames, between the flash as the light leaves the floor and the beep as it returns. As seen in the frame 5', our experiment is shown in Figure 15.3(a). If the height of the train is h, then, as seen in 5', the light travels a total distance 2h at speed c (second postulate) and so takes a time At' = 2h (15.5) This is the time between the flash and the beep, as measured by an observer in 8' (provided, of course, his clock is reliable). As seen in 8, our experiment is shown in Figure 15.3(b). In particular, the same beam of light is seen to travel along the two sides AB and BC of a triangle. If At is the time between the flash and the beep (as measured in 8), the side AC has length V At. Thus the triangle AB D has sides7 h, V At/2, and c At 12. (Notice that this is where we use the second postulate, that the speed of light is c in either frame.) Therefore, (c. At/2)2 h 2 + + (V At 12) 2, which we can solve to give At = 2h,/c2 _ V2 2h C 1 \u2014 132 (15.6) where I have introduced the useful abbreviation which is just the speed V measured in units of c. 7 I take for granted that the height of the train is the same in either frame. We'll prove this shortly. Section 15.4 The Relativity of Time; Time Dilation 605 The striking thing about the two results (15.5) and (15.6) is that they are not equal. The time between the same two events (the flash and the beep) has different values as measured", " in the two different inertial frames. Specifically, At = At' \u2014 /3 2 (15.8) We derived this result for a thought experiment with a flash of light reflected back to its source by a mirror on the ceiling of the railroad car, but the conclusion applies to any two events that occur at the same place in the train. Suppose, for instance, an observer at rest in 8' were to shout \"Good\" and a moment later \"Grief.\" In principle, we could ignite a flashbulb at the \"good,\" and arrange a mirror which would reflect the light back to arrive at the moment of \"grief.\" Therefore, the relation (15.8) must apply to these two events, the \"good\" and the \"grief.\" Since the timing of the two events cannot depend on whether we actually did the experiment with the light and the beeper, we conclude that the relation (15.8) must apply to any two events that occur at the same place in the frame 8'. You should avoid thinking that the clocks in one of our frames are somehow running incorrectly \u2014 on the contrary, it was essential to our argument that all clocks, in both frames, were running correctly. Further, it makes no difference what particular kinds of clock we used, so the conclusion (15.8) applies to all (accurate) clocks. That is, time itself, as measured in the two frames, is different in accordance with (15.8). As we shall discuss shortly, this surprising conclusion has been verified repeatedly. If the frame 8' is actually at rest (relative to 8), then V = 0, so /3 = 0, and (15.8) reduces to At' = At. That is, there is no difference in the times unless 8' is actually moving relative to 8. Moreover, at normal terrestrial speeds, V << c, so,8 << 1 and the denominator in (15.8) is very close to one. That is, at the speeds of our everyday experience, the two times are very nearly equal \u2014 so close that it would be almost impossible to detect any difference, as the following example shows. EXAMPLE 15.1 Time Differences for a Jet Plane Suppose that the pilot of a jet traveling at a steady V = 300 m/s arranges to set off a flashbulb at intervals of exactly one hour (as measured in his reference frame). If we arrange two observers on the ground to check this, what would they measure for the time At between two successive flashes? (Take the ground to be an inertial frame; that is, ignore effects of the earth's rotation.) The required interval is given by (15.8) with At' = 1 hour and 13 = V I c = 10-6. So At= At' 1h \\/1 \u2014 /3 2 -\u2018/1 \u2014 10 -12 1 h x (1 + x 10 -12) = 1 h 1.8 x 10 -9 s 606 Chapter 15 Special Relativity where in going to the second line, I have used the binomial approximation. 8 In this experiment, the time difference is less than 2 nanoseconds (1 ns = 10 -9 s). It is not hard to see why classical physicists had failed to detect such differences! As we increase V, the difference between the times in (15.8) gets bigger, and if we let V approach c, we can make the difference as large as we please. For example, if V = 0.99c, then /3 = 0.99 and (15.8) gives At' 7At. Speeds this high are routinely achieved by the accelerators at particle-physics labs, and the predicted time difference is precisely confirmed. If we put V = c (that is,,8 = 1) in (15.8), we would get the absurd result At'.--- At/0, and if we put V > c (that is, 8 > 1), we would get an imaginary value for At'. These results suggest that V must always be less than c, V < c, a suggestion that proves correct and is one of the most profound results of relativity: The relative speed of two inertial frames can never equal or exceed c. That is, the speed of light, in addition to being the same in all inertial frames, is also the universal speed limit for the relative motion of any two inertial frames. The factor 1/11 \u2014 /3 2 occurs so often in relativity, it usually given its own name, y, It is useful to remember that this new factor always satisfies y > 1, and as p 1 (that In terms of the parameter y the result (15.8) can be written a little more com- is, V -\u00b1 c) y co. pactly as At = y At' > At'. (15.10) The asymmetry of this result (that At' is never more than", " At) seems at first glance to violate the postulates of relativity, since it suggests a special role for the frame S' \u2014 namely, that S' is the special frame in which the time interval is minimum. However, this is just as it should be, since in our thought experiment S' is special, because it is the frame where the two events in question (the flash and the beep) occur at the same place. (This asymmetry was implicit in Figure 15.3, which showed one observer measuring At', but two measuring At.)-To emphasize this asymmetry, the time At' is often renamed Ato and (15.10) rewritten as 8 This is a nice example of a calculation where one almost has to use the binomial approximation, since most calculators cannot tell the difference between 1 and 1 \u2014 10-12. Section 15.4 The Relativity of Time; Time Dilation 607 The subscript on A to is to emphasize that Ato is the time elapsed on a clock at rest in the special frame where the two events in question occurred at the same place. This time is often called the proper time between the two events. In (15.11), At is the corresponding time measured in any frame and is always greater than or equal to the proper time Ato. For this reason, the effect implied by (15.11) is called time dilation and can be loosely stated by saying that a moving clock is observed to run slow. As measured by observers on the ground, a clock in the moving train is found to run slow. Finally, I should emphasize the fundamental symmetry between any two inertial frames. We chose to do our thought experiment in a way that gave the frame 8' a special role. (It was the frame in which the flash and beep occurred at the same place.) But we could have done the experiment the other way round, with the flashbulb, mirror and beeper at rest on the ground, and in this case, we would have found the opposite effect, that At' = y At. The advantage of writing the time-dilation formula in the form (15.11) is that it avoids the problem of remembering which is frame 8, and which 8'; the subscript on At, always flags the proper time \u2014 the time measured in the frame in which the two events were at the same place. Evidence for Time Dilation Time dilation was predicted in 1905 but was not experimentally verified until 1941, by B. Rossi and D. B. Hal1.9 The problem was, of course, to get a clock traveling sufficiently fast to show a measurable dilation. Rossi and Hall exploited the natural clocks that come with unstable subatomic particles, which decay (on average) after a definite time, characteristic of the particle. The lifetime of an unstable particle can be specified by its half-life, t112, the time in which half of a large number of the particles will decay. The muon is an unstable particle that is created in the earth's upper atmosphere when cosmic ray particles (mostly protons and alpha particles) from outer space collide with atmospheric atoms. Many of these muons have speeds quite close to the speed of light, and they live long enough to find their way down to the earth's surface. The muon had been discovered in 1935 by Carl Anderson in his studies of cosmic rays. By 1941 its half-life was known to be about t 112 = 1.5,u,s, meaning that half of a sample of muons at rest would decay in this time. If time dilation is correct, the half-life for a moving muon (as measured by earth-bound observers) should be larger by the factor y as in (15.11). For example, if the muon had speed 0.8c, then y = 1.67, and the muon's half-life should be tv2 (at speed 0.8c) = 1.67 x t u2 (at rest) = 2.5,u,s. Rossi and Hall were able to separate out cosmic-ray muons according to their speed and they could find their half-lives by measuring how many of them survived the journey through the atmosphere. Although their measurements had quite large ex- perimental errors, they were nonetheless good enough to verify Einstein's prediction (15.11) and to exclude the classical assumption of a single universal time. 9 B. Rossi and D. B. Hall, Physical Review, vol. 59, p. 223 (1941). 608 Chapter 15 Special Relativity A test of time dilation using man-made clocks had to await the development of superaccurate atomic clocks. In 1971 four portable atomic clocks were synchronized with a reference clock at the U. S. Naval Observatory in Washington DC and then flown around the world in a jet plane and returned to the Naval Observatory. The observed discrepancy between the reference clock and the portable clocks was (273 + 7) ns (averaged over", " the four clocks) in excellent agreement with the predicted value (275 \u00b1 21) ns. 10 Tests of time dilation \u2014 using both the natural clocks of unstable particles and man-made atomic clocks \u2014 have been repeated with ever-increasing precision, and there is now no doubt that the relativity of time, as embodied in (15.11), is true. Another important test that is carried out thousands of times every day is the Global Positioning System (GPS). This system, which is used by airplanes, ships, cars, and hikers to find their positions within a few meters, times the arrival of signals from 24 GPS overhead satellites at the observer's receiver and calculates the receiver's position from the known positions of the satellites. To find the position within a few meters requires an accuracy of a few nanoseconds, which requires that allowances be made for the relativistic differences between the times of the satellite and earth- bound reference frames. The success of the GPS is a daily tribute to the correctness of relativity.'' 15.5 Length Contraction The postulates of relativity have forced us to the conclusion that time is relative \u2014 the time between two given events is different when measured in different inertial frames \u2014 and, even more important, this conclusion is born out by experiment. This, in turn, implies that the length of an object is likewise dependent on the frame in which it is measured. To see this, we'll conduct a second thought experiment with the train of Figure 15.3, this time measuring its length. For an observer (let's call him Q) on the ground (frame 8) the simplest procedure is probably to measure the time At for the train to pass him and calculate the length as 12 / = VAL (15.12) 1\u00b0 See J. C. Hafele and R. E. Keating, Science, vol. 177, p. 166 (1972). Two trips were made, one going west and the other going east, both with satisfactory results. The numbers quoted here are for the more decisive westward trip. This experiment was actually a test of general, as well as special, relativity, since the predicted discrepancy has an appreciable contribution from gravitational effects. 11 For a readable account of the large role of relativity in the GPS, see N. Ashby, Physics Today, May 2002, p. 41. As described there, there are important contributions from general, as well as special, relativity. Thus, the success of the GPS is a test of both theories. 12 With so many of the familiar classical ideas being questioned, you are entitled to ask if it is legitimate to use the classical formula (15.12). However, this is just the definition of velocity (velocity = distance/time), and is certainly valid in any one reference frame (as long as we measure all quantities in this same frame). Section 15.5 Length Contraction To find the length of the train as measured in the train's rest frame, an observer on the train could simply use a long tape measure. However, for comparison with (15.12), it is convenient to use a different method. We can station two observers on the train, one at the front and another at the back, and have them record the times at which they pass the observer Q on the ground. The difference At' between these two times is the time (as measured in frame 8') for the train to pass observer Q, so the length of the train (again as measured in 8') is just 1' = VAt'. (15.13) Notice that we are making an important assumption here, that the speed of frame 3 relative to 3' is the same as the speed V of S' relative to 8. (The relative velocities are in opposite directions, but their magnitudes are the same.) This is true in classical mechanics, and it is also true in relativity, where it follows from the two postulates. The details of the argument require some care, but the gist is this: Consider the transformation from frame 8 to 3'. We'll denote it by (8 \u2014+ 8') temporarily. Suppose that, before making this transformation, we were to rotate our axes through 180'about the y (or z) axis, then make the transformation, and then rotate back again. The effect of the rotations is to reverse the direction of the x axis (and finally rotate it back again). The net effect of all three operations is precisely the transformation (8' 8). Since the rotations certainly don't change any speeds, we've proved that the speed of 3' relative to 8 is the same as that of 8 relative to 8'. Comparing (15.12) with (15.13), we see that, since the times At and At' are unequal, the same has to be true of the lengths / and 1'. To quantify the difference, we must be careful to get the relation between At and At' the right way around. These two times are the times (as measured in S and 8')", " between two events: \"front of train opposite observer Q\" and \"back of train opposite observer Q.\" These two events occur at the same place in frame 8, so At is the proper time, and At' = y At. Inserting this into (15.13) and comparing with (15.12), we see that = yl or (15.14) The length of the train as measured in S is less than that measured in 8' (unless V = 0). Like time dilation, the effect (15.14) is asymmetric, reflecting the asymmetry of the experiment. The frame 3' is special, since it is the unique frame where the object being measured (the train) is at rest. [We could, of course, have done the experiment the other way round. If we had measured the length of a building that is at rest on the ground, then the roles of 1 and would have been reversed.] To avoid confusion as to which frame is which, it is common to rewrite (15.14) as 610 Chapter 15 Special Relativity where 1\u00b0 denotes the length of an object measured in the object's rest frame (the frame in which the object is at rest), while 1 is the length in any frame. The length lo is called the object's proper length. Since 1 <10 (if V 0 0), this difference in lengths is called the length contraction (or the Lorentz contraction, or Lorentz\u2013Fitzgerald contraction, after the two physicists \u2014 the Dutch Hendrik Lorentz, 1853-1928, and the Irish George Fitzgerald, 1851-1901\u2014 who first suggested there must be some such effect.) The result can be loosely paraphrased by saying that a moving body is observed to be contracted. Like time dilation, length contraction is a real effect, well established by exper- iment. Since the two effects are so intimately connected, any evidence for one can be taken as evidence for the other. In particular, the decay of a high-speed unstable particle, when viewed in the particle's rest frame, can be interpreted as clear evidence for length contraction. (See Problem 15.12.) Lengths Perpendicular to the Relative Velocity The length contraction just derived applies to lengths in the direction of the relative velocity, such as the length of a train in the direction of motion. It is easy to see that there can be no analogous contraction or expansion of lengths perpendicular to the motion, such as the height of the train. Suppose for example there were a contraction and imagine two observers Q standing at rest in 8 and Q' in 8'. Suppose further that Q and Q' are equally tall (when at rest) and that Q' is holding a knife exactly level with the top of his head. If there is a contraction, then as measured by Q, observer Q' will be shortened as he rushes past, and Q will be scalped, or worse, as the knife goes by. But, unlike our previous thought experiments, this experiment is completely symmetric between the two frames: There is just one observer in each frame, and the only difference is the direction of the relative velocities. Therefore, it must also be that, as seen by Q', it is Q who is contracted; so the knife misses Q, and Q is not scalped. The assumption of a contraction has led us to a contradiction and there can be no contraction. A similar argument excludes the possibility of expansion, and, in fact, the knife just scrapes past Q as seen in either frame. We conclude that lengths perpendicular to the relative motion are unchanged. The length-contraction formula (15.15) applies only to lengths parallel to the relative velocity. 15.6 The Lorentz Transformation According to the classical notions of space and time, we saw that the mathematical relation between coordinates in two inertial frames 8 and 8' is the Galilean transfor- mation (15.1). In relativity, this cannot be the correct relation. (For example, time dilation contradicts the equation t = t'.) However, we can deduce the correct relation using an argument similar to the one that we used in connection with Figure 15.1 to derive the Galilean result. We imagine two frames, S attached to the ground and 8' attached to a train moving with speed V relative to S. We imagine, further, the explo- sion of a firecracker, which leaves a burn mark on the wall of the railroad car at a point Section 15.6 The Lorentz Transformation 611 V measured in 8 Figure 15.4 The coordinate x' is the horizontal distance, measured in 8', between the origin 0' and the burn mark at P'. The distances x and Vt are both measured in 8 at the time t (measured in 8) of the explosion. P'. The coordinates of this explosion are (x, y, z, t) as measured by observers in S and (x', y', z", "', t') in 8'. Our object is to find formulas for x', y', z', and t' in terms of x, y, z, and t. The thought experiment is illustrated in Figure 15.4, which is just like Figure 15.1 except that we now know we must be very careful to identify the frames (S or 8') relative to which the various distances are measured. Since lengths perpendicular to the relative velocity are the same in both frames, we can immediately write Y = Y and z = z (15.16) exactly as with the Galilean transformation. The coordinate x' is the horizontal dis- tance between the origin 0' and the burn mark at P', as measured in 8'. The same distance as measured in 8 is x \u2014 V t, since x and Vt are the distances from 0 to P' and from 0 to 0' at the instant t of the explosion (measured in 8). Therefore, by the length-contraction formula (15.15) (x' is the proper length here) or x \u2014Vt =x l ly y (x \u2014 V t). (15.17) This is the third of the four equations that we need. Notice that if V << c then y ti 1 and (15.17) reduces to the Galilean relation x' = x \u2014 V t. Finally, to get an equation for t' we can use a simple trick. We could repeat the previous argument with the roles of 8 and 8` exchanged. That is, we could let the explosion burn a mark at a point P on a wall fixed in S. Arguing as before, we would get the result x = y(x' + Vt'). (15.18) (Notice that we could get this result directly from (15.17) by exchanging the primed and unprimed variables and replacing V by \u2014 V.) Substituting (15.17) into (15.18), we can eliminate x' and solve for t', to give (as you should check) = y(t \u2014 Vx/c 2). (15.19) This is the required equation for t'. When V << c, we can neglect the second term and y 1, so (15.19) reduces to the Galilean relation t' = t. 612 Chapter 15 Special Relativity Collecting together the results (15.16), (15.17), and (15.19), we get the required four equations: These four equations are called the Lorentz transformation or the Lorentz\u2014Einstein transformation, in honor of Lorentz, who first proposed them, and Einstein, who first interpreted them correctly. The Lorentz transformation gives the coordinates (x', y', z', t') of an event, as measured in 8', in terms of its coordinates (x, y, z, t) as measured in S. It is the correct relativistic version of the classical Galilean trans- formation (15.1). If we wanted to know the coordinates (x, y, z, t) in terms of (x', y', z', t'), we could solve the four equations (15.20), but a simpler way is just to exchange primed and unprimed variables and replace V by \u2014 V. Either way, the result is the inverse Lorentz transformation x = y (xl vt') y = y' z z t = y (t' Vx'/c 2). (15.21) The Lorentz transformation expresses all of the properties of space and time that follow from the postulates of relativity. Using it, one can calculate all of the kinematic relations between measurements made in different inertial frames. There are several examples of its use in the problems at the end of this chapter and here are a couple more. EXAMPLE 15.2 Rederiving Length Contraction Use the Lorentz transformation to rederive the length contraction formula (15.15). (Note that this will not give an alternative derivation of length contrac- tion, since length contraction was used in deriving the Lorentz transformation. Rather we shall just get a consistency check.) Consider our usual two frames, 3 fixed to the ground and 3' fixed to a train traveling along the x axis with speed V relative to 3. We wish to compare the lengths of the train as measured in and 8'. The measurement in 8' is easy, since the train is at rest in this frame. An observer can, at his leisure, measure the x' Section 15.6 The Lorentz Transformation 613 1 =x2 - x1. x2 = y (x2 \u2014 Vt2) x l = y (x i \u2014 Vt 1 ). coordinates xi and x; of the back and front of the train, and its length is just the difference /' = xz \u2014 x;. This length is the proper length of the train, so 10", "'1/ -x11. (15.22) The measurement in S is harder since the train is moving. We could, with enough care, station two observers Q 1 and Q2 beside the track so that the back of the train passes Q 1 at the exact same instant (t 1 = t2) that the front passes Q2. The length as measured in S is then just Now, applying the Lorentz transformation (15.20) to the event \"front of train passes Q2\" we get and, for the event \"back of train passes Q 1,\" Subtracting and remembering that t 2 = t1, we find /0 = x;, \u2014 = y (x 2 \u2014 x 1) = yl or / = to /y, which is the length contraction (15.15). Our next example is one of the many seeming paradoxes of relativity. EXAMPLE 15.3 A Relativistic Snake A relativistic snake, of proper length 100 cm, is traveling across a table at V = 0.6c. To tease the snake, a physics student holds two cleavers 100 cm apart and plans to bounce them simultaneously on the table so that the left one lands just behind the snake's tail. The student reasons as follows: \"The snake is moving with /3 = 0.6, so its length is contracted by the factor y = 5/4 (check this) and its length measured in my frame is 80 cm. Therefore, the cleaver in my right hand bounces well ahead of the snake, which is unhurt.\" This scenario is shown in Figure 15.5. Meanwhile the snake reasons thus: \"The cleavers are approaching me at /3 = 0.6, so the distance between them is contracted to 80 cm, and I shall certainly be cut to pieces when they fall.\" Use the Lorentz transformation to resolve this paradox. Let us choose frames S and 8' in the usual way. The student is at rest in 8, with the cleavers at x i, = 0 and x R = 100 cm. The snake is at rest in 8', with its tail at x' = 0 and its head at x' = 100. To resolve the dispute, we must find where and when the two cleavers fall, as observed in S and in 8'. In S the cleavers fall simultaneously at t = 0. At this time the snake's tail is at x = 0. Since his length is 80 cm, his head has to be at x = 80 cm. [You can 614 Chapter 15 Special Relativity xi = Figure 15.5 The snake paradox, as seen in the student's frame 8. The cleavers fall simultaneously at time t = 0. check this, if you want, using the transformation equation x' = y (x \u2014 V t); with x = 80 cm and t = 0, this gives the correct value x' = 100 cm.] As observed in S, the experiment is as shown in Figure 15.5. The right cleaver falls comfortably ahead of the snake, the student is right, and the snake is unharmed. What is wrong with the snake's reasoning? To answer this, we must examine the coordinates and times at which the two cleavers bounce, as observed in 8'. The left cleaver falls at t L = 0 and x i, = 0. According to the Lorentz transfor- mation (15.20), the coordinates of this event, as seen in S' are and tL = y(t i, \u2014 Vx L /c2) = 0 xL = y(x L \u2014 Vt L ) = 0. As expected, the left cleaver falls just behind the snake's tail, at time ti = 0, as shown in Figure 15.6(a). So far there are no surprises. However, the right cleaver falls at tR = 0 and x R = 100 cm. Therefore, as seen in 8', it falls at a time given by the Lorentz transformation as tR =y(tR \u2014Vx R 1c2), \u20142.5 ns. (Check the numbers yourself.) The crucial point is that, as seen in 8', the two cleavers do not fall at the same time. Since the right cleaver falls before the left one, it does not necessarily hit the snake, even though they are only 80 cm apart = 0 e / xL = 0 x' = 100 x 'R = 125 Figure 15.6 The snake paradox, as measured in the snake's frame 8'. The cleavers move to the left with speed V, and the right one falls 2.5 ns before the left one. Even though the cleavers are only 80 cm apart, this lets them land 125 cm apart. Section 15.7 The Relativistic Velocity-Addition Formula (in this frame). In fact, the position at which the right cleaver falls", " is given by the Lorentz transformation as x' = y(x R \u2014 Vt R ), 125 cm. The right cleaver does indeed miss the snake! The resolution of this paradox, and many similar paradoxes, is that two events that are simultaneous in one frame are not necessarily simultaneous in a different frame \u2014 an effect sometimes called the relativity of simultaneity. As soon as we recognize that the two cleavers fall at different times in the snake's frame, there is no longer any problem understanding how they can both contrive to miss the snake. 15.7 The Relativistic Velocity-Addition Formula As our next, and very important, application of the Lorentz transformation, let us use it to derive the relativistic velocity-addition formula. This formula is the answer to the following question: If an object \u2014 an electron, a baseball, a planet \u2014 is moving with velocity v relative to an inertial frame 8, how can we calculate its velocity v' relative to some other frame 8'? In classical physics, the answer to this question is the classical velocity-addition formula: If V denotes the velocity of 8' relative to 8, then v' = v \u2014 V. (Presumably, whoever named this formula wrote it as v = v' + V.) For the special case that the axes of 8 and 8' are parallel and V is in the x direction (our \"standard\" configuration), this becomes vx = vX \u2014 V, vy = vY' and vz = V,. (15.23) Our task now is to find the corresponding relativistic result. Consider a particle moving with position r(t) or r'(t'), as seen in S or 3'. The definition of the velocity v is the derivative v = \u2014 dr dt (15.24) where dr = r2 \u2014 r 1 is the infinitesimal displacement between the positions at times t 1 2 = t1 dt. Now, we can write down the Lorentz transformation for (x 2, y2, z2, 12) and t and (x 1, y i, z 1, ti), and taking differences, we find dx' = y(dx \u2014 Vdt), dy' = dy, dz' = dz, dt' = y(dt \u2014 V dx c 2). (15.25) (Notice that dr and dt satisfy exactly the same transformation equations as r and t. This is because the Lorentz transformation turned out to be linear.) Using the definition (15.24), we can write down the components of v', and substituting (15.25) we find for v' x dx' = \u2014 = y(dx \u2014 Vdt) dt) x dt' y(dt \u2014 V dx1c2) 616 Chapter 15 Special Relativity or, canceling the factors of y and dividing top and bottom by dt, Similarly, v, = dy' = dy Y dt' y(dt \u2014 V dx1c 2) Dividing top and bottom by dt, we find for v; (and similarly viz) Notice that 1/ even though dy' = dy. This is because dt' dt. Notice also that Y vY y is the factor pertaining to the speed V of 3' relative to ; that is y = 1'1 \u2014 V 2 /c2. The three equations in (15.26) and (15.27) are the relativistic velocity-addition formulas or the relativistic velocity transformation. If all velocities are much less than c, then y ti 1 and we can ignore the second term in the denominators, and we recover the classical results (15.23). However, when the velocities concerned approach c, the relativistic velocity transformation can have some surprising results, as the following examples illustrate. EXAMPLE 15.4 Adding Two Velocities Close to c A rocket travelling at speed 0.8c relative to the earth shoots forward bullets with speed 0.6c (relative to the rocket). What is the bullets' speed relative to the earth? If we choose frames in the usual way, with S fixed to the earth and 3' fixed to the rocket, then V = 0.8c and v' = 0.6c. Our task is to find v. The classical answer is of course v = v' + V = 1.4c. The relativistic answer is given by the inverse of (15.26), which we can find by the usual trick of exchanging primed and unprimed variables and reversing the sign of V. The result (from which I omit the subscripts x since all velocities are in the x direction) is v = I)/ + V 1 + v'V/c2 0.6c + 0.8c 1 + 0.8 x 0.6 1.4 1.48 =", " c 0.95c. (15.28) The striking thing about this is that when we \"add\" 0.8c to 0.6c we get an answer that is less than c. In fact, it is fairly easy to prove that for any velocity with v' < c, Section 15.8 Four-Dimensional Space\u2014Time; Four-Vectors 617 the corresponding v is also, automatically, less than c. (See Problem 15.43.) That is, anything that travels with speed less than c in one frame has speed less than c in all frames. EXAMPLE 15.5 Adding Two Velocities One of Which Equals c The rocket of Example 15.4 shoots forward a signal (a pulse of light, for instance) with speed c relative to the rocket. What is the signal's speed relative to the earth? Here v' = c, so (15.28) becomes 'V/c2 1 + V/c = c. (15.29) That is, a signal traveling in the x direction with speed c relative to SI also has speed c relative to 3. This result is actually true whatever the direction of travel, as you can prove in Problem 15.43. It asserts that the speed of light is invariant under the Lorentz transformation, in obedience to the second postulate of relativity (which led us to the Lorentz transformation in the first place). 15.8 Four-Dimensional Space\u2014Time; Four-Vectors The Lorentz transformation (15.20) mixes up space and time, in the sense that each of the equations for x' and t' involves both x and t. The Russian\u2014German mathematician Hermann Minkowski (1864-1909) suggested that this mixing of space and time implies that time should be combined with the three spatial coordinates to form a four-dimensional space\u2014time, on which the Lorentz transformations act as a kind of rotation. Before we examine this suggestion, let us review a couple of facts about the ordinary rotations of our everyday three-dimensional space. Rotations of Ordinary Three-Dimensional Space In discussing the vectors of ordinary three-dimensional space, it will be convenient to change our notation a bit: For any choice of orthogonal axes, I shall use the notation mentioned in Chapter 1, with the three unit vectors labeled e l, e2, e3. The components of a general vector q we'll call q l, q2, q3, so that q = q ie i q2e2 q3e3 = E qi ei 3 (15.30) where, as usual, qi = ei q. (15.31) 618 Chapter 15 Special Relativity To conform with this notation, I shall, from now on, rename the position vector r = (x, y, z) as x = (x 1, x2, x3). Now consider a rotation which carries the axes defined by e l, e2, e3 into a second set with unit vectors e' 1, e'2, e'3. The components q; of the same vector with respect to the new axes are easily found: r t qi = e i q = e' i E q i ej = E(e'i e i )qi (15.32) 3 j=1 3 i=1 This equation expresses each of the coordinates q; in the new coordinate system as a sum over the coordinates q j in the old system. (That is, it does for rotations what the Lorentz transformation does when we pass between frames in relative motion.) The coefficients in this sum are the scalar products e' i e i of the unit vectors of the new and old systems. 13 We can express (15.32) more compactly if we adopt the matrix notation of Chapter 10: Let R denote the 3 x 3 square matrix with elements Rij = e'i e 3 (15.33) and let q and q' denote the 3 x 1 columns made up of the coordinates qi q2 q = [ q3 ql 17,/; qs [ and qt (15.34) [As in Chapter 10, it is good to stay a little relaxed about this notation. When there is any danger of confusion, we'll agree that q is a column matrix as in (15.34), but, when there is no such danger, we'll continue to call q a \"vector\" and even write it as the row (q 1, q2, q3).] With these notations, the rotation (15.32) takes the compact form = Rq (15.35) The effect of rotating our axes is to multiply the column q of coordinates qi by a certain 3 x 3 rotation matrix R. EXAMPLE 15.6 A Simple Rotation about One Axis Consider a set of rectangular axes with the x ix", "2 plane chosen horizontal and the x3 axis vertically upward, and suppose we rotate these axes about the x 2 axis through an angle 0, to give a new set of axes as shown in Figure 15.7. Find the rotation matrix R for this rotation. The required matrix R is easily written down using (15.33). Inspection of Figure 15.7 let's one evaluate the necessary scalar products to give 13 The numbers e', e j are often called the direction cosines of the new axes with respect to the old, since e', e j = cos Oil where Oi j is the angle between e' i and e. Section 15.8 Four-Dimensional Space\u2014Time; Four-Vectors 619 Figure 15.7 The primed axes are obtained from the unprimed by a counter-clockwise rotation about the x 2 axis (into the page) through an angle O. The x2 direction is unaffected by this rotation and the unit vectors e 2 and e'2 both point into the page. (15.36) (15.37) But surely such people would be dissuaded from this view when they noticed The effect of the rotation R on the coordinates of any point is x' = Rx or R= [ 0 cos 0 \u2014 sine 0 1 0 sin 0 0 cos 0 xi = (cos 0)x 1 + (sin 0)x 3 x2 = x2 X /3 = (- sin 0) x 1 + (cos 0)x 3. One of the best arguments for regarding the three numbers x 1, x2, x3 as coordinates in a single three-dimensional space is that rotations can mix them up as in (15.37). One could imagine people taking the view that vertical distances (x3) were somehow fundamentally different from horizontal ones (x 1 or x2). 14 that (15.37) mixes x i and x3 together (and, for 0 = Tr/2, simply exchanges their roles). We shall now argue similarly that Lorentz transformations are a kind of rotation that mixes the space and time coordinates in a four-dimensional space- time. Lorentz Transformations as \"Rotations\" of Space\u2014Time A glance at the Lorentz transformation (15.20) should convince you that it mixes x and t in somewhat the same way that the rotation (15.37) mixes x 1 and x3. We can make this parallel surprisingly close by polishing our notation. First, we'll rename our space coordinates x 1, x2, x3 as above, and we'll introduce a fourth coordinate X4 = Ct (15.38) 14 Bizarre as such a view may appear, it does seem to be endorsed by some standard practices. For example, people in the business of storing water behind dams measure the volume of stored water in acrefeet, with horizontal areas measured in acres, but vertical depths in feet. 620 Chapter 15 Special Relativity where the factor c guarantees that x 4 has the same dimensions as x 1, x2, and x 3. Recalling the definition fi = V/c, we can rewrite the Lorentz transformation (15.20) as We can further improve the parallel with the rotation (15.37) if we note that since y > 1 we can define an \"angle\" 4> such that y = cosh (/). Some simple algebra (Problem 15.30) should convince you that this makes y$ = sinh 4> and (15.39) be- comes x l = yx 1 \u2014 y f3x 4 X2 = x2 X 3 = X3 X4 = \u2014y,8x 1 yx 4 xi = (cosh 0)x 1 \u2014 (sinh 4>)x4 X2 = x2 X3 = x3 X41 = (- sinh 0)x 1 + (cosh 4>)x4 (15.39) (15.40) and our parallel is as close as it can be. It is important to understand that no one would claim that the Lorentz transformation (15.40) mixes x 1 and x4 in exactly the way that the rotation (15.37) mixes x 1 and x3 \u2014 the trig functions of (15.37) have become hy- perbolic functions in (15.40) (and a sign has changed). Nevertheless, the parallel is close and is a powerful argument for regarding x 4 = ct as the fourth coordinate in a four-dimensional space\u2014time or just four-space. Four-Vectors The four numbers x 1, x2, x3 and x4 = ct constitute a vector in four-dimensional space\u2014time. Such vectors are called four-vectors to distinguish them from the three- dimensional, vectors, such as the position three-vector x = (x 1, x2, x3). Unfortunately, several different notations are used for four-vectors. I shall use", " ordinary italic letters for four-vectors; for example, X = (X1,'X2, x3, x4) = (x, Ct) for the position\u2014time vector just discussed. We shall be meeting several other four- vectors (for example, the four-momentum p, to be defined shortly). My notation for an arbitrary four-vector will be q = (q 1, q2, q3, q4) (q, q4), where the bold face q, comprising the first three components of q, is called the spatial component of q and the fourth component q4 is called the time component. 15 15 This notation has two drawbacks you should be alert to: (1) Since we'll still be using italic symbols for various scalars (for example, m for mass), you will need to tell from the context whether an italic symbol is a four-vector or a scalar. (2) We can no longer use the convention that q is the Section 15.8 Four-Dimensional Space\u2014Time; Four-Vectors 621 As with three-vectors, it is often convenient to understand four-vectors to be 4 x 1 columns, as x = and q (15.41) xl X2 X3 _ X4 _ q 1 q2 q3 _q4 _ With this notation, the Lorentz transformation (15.39) can be written in matrix form x' =A x (15.42) where A is the 4 x 4 matrix A= y 0 0 _ \u2014 y,8 0 1 0 0 0 0 1 0 \u2014 yfi 0 0 y _ [standard boost]. (15.43) This is not the most general Lorentz transformation. It is the transformation between two frames in what we have called standard configuration, with corresponding axes parallel and with the velocity of 3' relative to CS along the x axis. For many purposes, this standard transformation is the only one we need to consider, but we should take a moment to discuss more general transformations. Any Lorentz transformation which leaves corresponding axes parallel is called a pure boost or just boost, since all it does is \"boost\" us from one frame to another traveling at constant velocity relative to the first, without any rotation. The general transformation involves some rotation as well. If the transformation is a pure rotation (no relative motion, just a change of orientation) then of course t' = t and only the three spatial coordinates get changed. Thus we can write a pure rotation in the form (15.42), where the 4 x 4 matrix A has the block form A= AR = R [pure rotation] (15.44) 0 0 0 0 0 0 1 where R is the 3 x 3 matrix of the given rotation. (If you've never seen this kind of block matrix before, write out the equation x' = ARx in all its detail, and notice how it rotates the three spatial coordinates but leaves the fourth component unchanged, so t' = t.) If we want to write down a pure boost AB in an arbitrary direction u, we can con- struct it from a couple of rotations plus a standard boost: First we rotate so that our new x 1 axis points along the required direction u; next we make the standard boost (15.43); magnitude of the three-vector q. Instead, we'll just use 1(11 (though I'll continue to use r for the magnitude of the position vector, r = 622 Chapter 15 Special Relativity and then we rotate back to our original orientation. Finally, any Lorentz transforma- tion A can be expressed as the product of a boost followed by a suitable rotation, A = A R A B. 16 For some practice at handling different Lorentz transformations, see A four-vector is defined as anything that transforms in the same way as the space\u2013time vector x = (x, ct) under all Lorentz transformations (15.42). The formal Problems 15.32 to 15.34. definition is this: Definition of a Four-Vec o hi each inertial frame a four-vector is s y a set of four numbers q = q q4 ) such that the values in trn rarnes and 8' an.:. related by the equation qR Act Is the Lorentz t sforrn \"Tian connecting S and Obviously the space\u2013time vector x = (x, ct) fits this definition, and we shall find several more examples in the next few sections (including the four-momentum p mentioned above). The great merit of the notion of four-vectors is that it often allows one to check with almost no effort whether a proposed physical law is relativistically invariant. Suppose for example we believe that there should be a law of the form q = p (15.45) where we know thatg and p are four-vectors. (The law of conservation", " of momentum has this form, p fin = pin, as we shall see.) Suppose further that the law is true in one frame 8. Since the corresponding values in any other frame 8' are q' = Aq and p' = Ap, we have only to multiply both sides of (15.45) by A and we see that q' = p'. That is, the truth of our proposed four-vector law (15.45) in one frame S assures its truth in any other frame 8'. (Of course, this doesn't guarantee that the law is true \u2014 only experiment can test that but it does guarantee that the law would be consistent with the postulates of relativity.) Any single quantity that is invariant under rotations is called a rotational scalar or a three-scalar; for example, the mass m of an object is a three-scalar, and so is the time t. In the same way, any single quantity that is invariant under Lorentz transformations is called a Lorentz scalar or four-scalar. For example, we shall see that the mass m (if suitably defined) is a four-scalar; that is, the mass of any object has the same value in all inertial frames. On the other hand, the time t is not a four-scalar; rather, as we have already seen, it is the fourth component of a four-vector. 16 Actually we still don't have the most general transformation, since the spatial origins of our two frames still coincide at t = t' = 0. This can be taken care of by shifting one of the origins, but this additional possibility need not concern us here. Section 15.9 The Invariant Scalar Product 623 15.9 The Invariant Scalar Product A set of transformations \u2014 such as the set of all rotations in three-space, or all Lorentz transformations in four-space \u2014 can often be characterized by the quantities that they leave invariant. Our main interest here is, of course, in the Lorentz transfor- mations, but, for a little guidance as to what we should expect, let us look first at rotations. The Invariant Scalar Product in Three-Space One of the most obvious properties of rotations in three-space is that they do not change the length of any vector. If we define S = X X, 2 -t- X2 1 2 X +x3 (15.46) as the length squared of any vector x, then the value of s = x x in one coordinate system is the same as its value s' = x' x' in any other system obtained from the first by rotation. (In the terminology just introduced, s is a rotational scalar.) Since this is true for any x, we can replace x by x = a + b (where a and b are any two other vectors) and the invariance of (a + b) 2 implies that aa + 2ab + bb = a'a' + 2a'b' + b' b'. (15.47) Canceling the terms that we already know to be equal, we find that ab a'b'. (15.48) In other words, invariance of the length of any three-vector under rotation implies that the scalar product of any two three-vectors is invariant. We shall employ a similar argument for the scalar product of four-space in a moment. The Invariant Scalar Product in Four-Space We can construct a scalar product in four-space with several of the properties of its three-dimensional analog. For any four-vector x = (x 1, x2, x3, x4) (x, ct), let us define S 2 X 2 2 + X2 + X3 - X4 15.49) This s is obviously a generalization of the three-dimensional length squared, but note well the minus sign on the fourth term. (It is because of this minus sign that Lorentz transformations of space\u2013time are not exactly analogous to rotations of ordinary space.) The quantity s is invariant under Lorentz transformations, as we can 624 Chapter 15 Special Relativity easily prove: Consider first the standard boost (15.39). Under this transformation, the quantity s becomes /2 S = X 1 / 2 + X2 /2 + X3 /2 - X4 = y2(xi i3x4)2 + x22 x32 + y 2 j3x + x4 2 = y 2(1 182)4 + x22 + x32 y2(1 /32)x42 = s where the last equality follows because y 2 (1 \u2014,8 2) = 1. Therefore, the quantity s is invariant under the standard boost. But we have seen that any Lorentz transformation can be built up", " from a standard boost and rotations, and s is certainly unchanged by a rotation (since r 2 and t are separately invariant under rotation). Therefore s is invariant under any Lorentz transformation. Before we discuss the significance of the new invariant quantity s, we can use it to define an invariant scalar product in four-space. For any two four-vectors x = (x 1, x2, x 3, x4) and y = (y i, y2, y 3, y4), we define r (Again, note well the minus sign on the fourth component \u2014 this \"scalar product\" is a little different from the usual scalar product in ordinary space.) Obviously the invariant s of (15.49) is just s = x x. And, as with rotations, the argument leading from (15.47) to (15.48) implies that because x x is invariant for any one four-vector x, the scalar product x y is invariant for any two four-vectors x and y. We shall see that the scalar product x y plays as big a role in relativity as the ordinary product a b does in classical physics. The scalar product of any four-vector x with itself is often written as x x = x 2, and can be called the \"invariant length squared\" of x, but you must not be misled by this terminology into thinking that x 2 is positive. On the contrary, x 2 can obviously be positive, negative, or zero. For future reference, we can rewrite the invariance of x y as a property of the Lorentz matrices A: We know that x y = x' y' whatever the values of x and y, provided x' = Ax and y' = Ay where A is any Lorentz transformation. Therefore, we can say that, for any Lorentz transformation A, x y = (Ax) (Ay) (15.51) for any two columns of four numbers x and y. To understand where the scalar product came from, consider an experiment in which a flash bulb at the origin x = 0 is fired at time t = 0 in a frame 3. The light 17 Be warned! Physicists are fairly evenly divided between those who use the definition (15.50) and those who put a minus sign in front of the whole expression. Both conventions have their advantages and disadvantages. Section 15.10 The Light Cone from the flash will spread out at speed c, so that at any later time t it occupies the sphere r 2 = c2t2. Using our new notation, we can say that the spreading wavefront is located by the condition x x = r 2 c2t2 = 0. Now, the invariance of x x implies that x x = 0 if and only if x' x' = 0 in any other frame 8'. Therefore a spherical wave spreading with speed c as seen in 8 will be a spherical wave spreading with speed c in 8', and vice versa. We see that the invariance of the scalar product x x is a reflection of the second postulate, that the speed of light is the same in all inertial frames. 15.10 The Light Cone The scalar product x x lets us divide space\u2013time into five physically distinct regions. To help visualize this, it is convenient to ignore one of the spatial dimensions (x 3 say), so we can plot the remaining two spatial dimensions horizontally and x 4 = ct vertically up, as in Figure 15.8. (Mathematically, this amounts to confining our attention to the \"plane\" x3 = 0.) Consider again the light from a flashbulb fired at the space-time origin 0 (that is, fired at x = 0 when t = 0). As time passes, the light moves outward in the x ix, plane on an expanding circle with r 2 = c2 t2, and this sweeps out the upper half- cone shown in Figure 15.8. This cone, called the forward light cone, is therefore the set of all points in space\u2013time that would be visited by light released from the origin 0. Mathematically, it is the set of all space\u2013time points x = (x, ct) satisfying and t > 0. X4 = ct forward light cone backward light cone \u2014 elsewhere absolute past Figure 15.8 The light cone is defined by the condition x x r 2 c2t 2 0 and divides space\u2013time into five distinct parts: the forward and backward light cones, with t > 0 and t < 0 respectively; the interiors of the forward and backward light cones, called the absolute future and the absolute past; and the outside of the cone, labeled \"elsewhere.\" 626 Chapter 15 Special Relativity The lower half-cone shown in Figure 15.8 is called the backward light cone and is the set of all space\u2014time points x = (", "x, ct) with the property that light released from x could subsequently pass through the origin, 0. The whole light cone (forward and backward) is made up of the straight lines representing the path of any light ray that passes through the origin. Since light travels at speed c, and x4 was cunningly chosen to be x4 = ct, these lines have slope 1, so the surface of the light cone (if drawn to scale) makes an angle of 45\u00b0with the time axis. Since the light cone is defined by the condition that x x = 0 and since x x is invariant (has the same value in all frames), it follows that the light cone is itself an invariant concept. That is, observers in any two frames will always agree as to which points lie on the light cone. Interior of the Light Cone; Future and Past Consider next a space\u2014time point P, with coordinates x = (x, ct), that lies inside the forward light cone. This obviously has t > 0 and r2 < c2t2 or x4 > 0, and or x x < 0). (15.52) These two conditions have a remarkable consequence: Notice first that since t > 0, we can assert that any event that occurs at P is later than any event at 0, at least as observed in the frame S in which our coordinates x are measured. But what about some other frame 8'? To answer this, note that the second condition in (15.52) is just that x x < 0, and we know that x x is invariant under Lorentz transformations. Therefore x' x' is also negative, and the second condition is satisfied in 8' as well. To see that the first condition is also satisfied in 8', suppose first that 8' is related to 8 by the standard Lorentz boost (15.39), under which x4 = y (x4 \u2014,6x1). (15.53) Now, we know that I,B I < 1 and the second condition in (15.52) guarantees that Ix ] I < x4. Therefore x 14 > 0, and the first condition is also satisfied in 8'. Since any Lorentz transformation can be made up of a standard boost and rotations (and rotations don't change x 4 at all), we conclude that both conditions (15.52) hold in all frames if they hold in one frame. In other words, the statement that P lies inside the forward light cone is a Lorentz-invariant statement. In particular, if P lies inside the forward light cone, then all observers will agree that an event that occurs at P is later than one at 0. For this reason, the inside of the forward light cone is often called the absolute future \u2014 \"absolute\" because all observers agree that P is in the future of 0. In a similar way, if P lies inside the backward light cone, then P is earlier than 0 as measured by all inertial observers, and this region is called the absolute past. (See Problem 15.39.) So far we have considered the light cone with its vertex at the space\u2014time origin 0. This is defined by the light rays which happen to pass through 0. If we considered instead light which passes through some other space\u2014time point, Q, this would define a light cone with its vertex at Q \u2014 the light cone of Q. This cone would look just like Figure 15.8, except that the vertex would be at the arbitrary point Q rather Section 15.10 The Light Cone 627 x4 = ct X2 Figure 15.9 The light cone of an arbitrary space\u2014time point Q with coordi- nates x Q = (xQ, ct Q ) is made up of all light rays that pass through Q. The point shown as P lies outside the light cone of Q. than the origin 0, as shown in Figure 15.9. Any point P on this cone must satisfy (Xp \u2014 x0 2 = c2 (tp \u2014 tQ ) 2, so that (x p \u2014 x0 2 = 0 (P on light cone of Q). (15.54) The inside of the forward light cone of Q is the absolute future of Q, all of whose points are later than Q as seen by all inertial observers. That is, for any point P inside the forward light cone of Q, all observers agree that tp > t Q. Exterior of the Light Cone; Space-Like Vectors The situation is entirely different for a point P that lies outside the light cone, as in Figure 15.9. First, the condition for P to be outside is that (Xp \u2014 XQ) 2 > C (tp \u2014 tO 2 2, (15.55) or, equivalently (Xp \u2014 XQ) 2 > 0 (P outside light cone of Q). (15.56) This condition is", " symmetric between P and Q. Thus if P is outside the light cone of Q, then Q is outside the light cone of P, and vice versa. It is clear from Figure 15.9 that there are points P outside the light cone of Q, with P later than Q (that is, tp > tQ ), and others for which P and Q are simultaneous (tp = tQ ), and still others with P earlier that Q (tp < tQ ). There is nothing remarkable about this claim; it is a straightforward consequence of the geometry of Figure 15.9. What is remarkable is the following proposition (which I shall prove directly): Let P be any given space\u2014time point outside the light cone of a second given Proposition point Q. Then but and (1) there exist frames 8 in which tp > tQ (2) there also exist frames 8' in which tp = t'Q (3) there also exist frames 8\" in which tp\" < tQ\". 628 Chapter 15 Special Relativity This startling proposition implies that the time ordering of any two given events, each outside the other's light cone, can be different in different frames: Where one observer says that event A occurred before event B, a second observer can find them the other way around (and a third can find them to be simultaneous). This has profound implications related to the notion of causality: If one event A (an explosion, for instance) is the cause of another event B (the collapse of a distant building), then A must obviously occur first in time, since causes always precede their effects. According to our proposition, if the space\u2014time point P is outside the light cone of Q, then neither Q nor P is unambiguously first in time. (In some frames it's one, and in some frames it's the other.) Therefore, nothing that happens at Q can be the cause of anything that happens at P, nor the other way around. Now, any kind of signal traveling from Q to P would have to travel with speed greater than c. [This follows from (15.55).] Conversely, if a signal emanating from Q had speed greater than c, then it could travel to some point P outside Q's light cone. It follows that no causal influence can travel faster than the speed of light. 18 Because the region outside the light cone of Q is completely immune to anything that happens at Q, this region is sometimes called the \"elsewhere\" of Q. To simplify the proof of our proposition, let us put the point Q at the origin 0, and abbreviate the coordinates of P to x = (x, ct). (The general case is no harder; it is just a bit messier notationally.) By making a rotation if necessary, we can put x on the positive x 1 axis, so that x = (x 1, 0, 0, x4). (15.57) Now let us assume that statement (1) above is true (so that x 4 > 0), and prove statements (2) and (3). [Obviously one of the statements must be true, and you can easily check that our arguments work equally well starting from either (2) or (3).] Let us now make the standard boost (15.39) to a new frame S' in which y (x4 \u2014 f3x 1 ). x4 (15.58) Since P is outside the light cone of 0, x2 > c2t 2 which for the vector (15.57) means that x 1 > x4. Therefore we can choose,8 = x4 /x i < 1, and, according to (15.58), x41 = 0. That is, t' = 0, and we have proved statement (2) above for the case that Q = 0. If P was outside the light cone of an arbitrary point Q, the corresponding boost would give a vector x'p \u2014 x lc, with its fourth component equal to zero, and hence t. Q = t''again as required. A four-vector whose fourth component is zero can be described as a pure-space vector, and one which can be brought into this form by a Lorentz transformation is called space-like. That is, a four-vector is space-like if there is a frame in which it is a pure-space vector, with zero fourth component. With this terminology, we can say 18 1f the causal signal could have speed greater than c, then it could travel from Q to some P outside Q's light cone, but we have just seen that this is impossible. Section 15.10 The Light Cone 629 that the outside of the light cone is made up of all space-like vectors. Similarly, we can rephrase the result about causal relations to say that, if the separation x p \u2014 x Q of two points P and Q is space-like, then nothing that happens at P can influence", " what happens at Q, nor vice-versa. To prove statement (3) [from the assumed truth of statement (1)] we have only to look at the transformation (15.58) again. Since x 1 > x4 we can choose 6 a little bigger than x4/x 1 but still less than 1, and, with this choice, we get a frame (S\", say) in which t\" < 0 (or, in general, tpll < ) as required. Time-Like Vectors An argument similar to that just given for space-like vectors (Problem 15.44) shows that if a four-vector q lies inside the light cone (that is q q < 0), then there exists a frame 8' in which it has the pure-time form q' = (0, 0, 0, q4). Naturally, therefore, we describe vectors inside the light-cone as being time-like. These can then be subdivided into forward time-like vectors (with q4 > 0) and backward time-like (with q4 < 0). An important example of a forward time-like vector is the displacement four-vector dx of any material particle in a time dt. As we shall discuss shortly, a material particle can be defined as any particle with positive mass (m > 0). Equivalently (as we shall see), it is any particle for which, at any given time, there exists a rest frame; that is, a frame in which the particle is at rest, with v = 0. It is a matter of experience that all of the normal constituents of matter \u2014 electrons, protons, neutrons \u2014 have this property, and likewise all composites, such as atoms, molecules, baseballs, and stars. 19 Suppose now that, between the times t and t dt, a material particle moves from x to x dx, and consider the four-vector displacement In the particle's rest frame, dx = 0, and dx has the pure-time form dx = (0, 0, 0, cdt). Therefore, dx is time-like in all frames, and dx 2 < 0. Since dx = (dx, c dt) = (v, c)dt. dx 2 = (v2 \u2014 c2) dt2 < 0, we conclude that v2 < c2 in all frames; that is, material particles cannot travel with speeds greater than or equal to the speed of light. Notice that this is the third sense in which we have proved the speed of light acts as a universal speed limit: (1) The relative speed of any two inertial frames is always less than c. (2) In any one inertial frame the speed of any causal signal is always less than or equal to c. And now (3) in any inertial frame, the speed of any material particle is less than c. 19 As we shall discuss in Section 15.16, the only common particle which does not have this property is the photon, the particle of light. Since this travels at speed c in all frames, it certainly does not have a rest frame. Naturally, we do not regard a photon as a material particle. 630 Chapter 15 Special Relativity 15.11 The Quotient Rule and Doppler Effect As a beautiful application of the properties of four-vectors, we next discuss the Doppler effect \u2014 the change in frequency of a wave due to the motion of the wave's source or the observer. Before we do this, we need to derive one more important property of four-vectors, the quotient rule. The Quotient Rule Suppose that we find a quantity k that is specified by four numbers k = (k 1, k2, k3, k4) in every inertial frame S. It is naturally tempting to think that k is a four-vector, but this is not necessarily so. For example, in discussing the motion of an object of mass m, charge q, volume V, and temperature T, we could define k (m, q, V, T), and this set of four numbers would be defined in every frame but is fairly obviously not a four-vector; that is, its value in one frame S is not related to the value in another frame 8' by the Lorentz transformation k' = Ak. Although this example may seem a bit artificial, it does show clearly that not every quantity k with four components is necessarily a four-vector. On the other hand, k is a four vector if it satisfies the conditions of the following theorem: The Quotient Ru e S uppo that is known to be a four-vecto and that in every inertial.t k k2, k3, k4) is a set of four nu suppose further that for ever every ue o x the quantity 0 = k x= k l \u00b1 12x2 k 3x3 k 4x 4 is found to have e sar", "ne value in ai frames (that is, k x is a four-scalar), then k is a four- The proof of this rule is surprisingly simple: First, that 0 is a four-scalar implies that k x = k' x'. (15.59) But from (15.51) we know that for any Lorentz transformation A k x (Ak) (Ax). Now, by assumption, x is a four-vector, so we can replace Ax by x', to give k x = Akx'. (15.60) Comparing (15.59) and (15.60), we see that k' x' = (Ak) x'. Section 15.11 The Quotient Rule and Doppler Effect 631 This equation is true for any choice of x'. If we choose x' = (1, 0, 0, 0), then it tells us that the first component of k' is equal to the first component of Ak, and continuing in this way we can show that all four components are equal, so that = Ak, which shows that k (the \"quotient\" of the scalar 0 and the vector x) is indeed a four- vector. Armed with this quotient rule, let us return to the Doppler effect. Doppler Effect When you think of the Doppler effect, you probably think of the Doppler shift of sound. The sound of a police siren rushing toward us has a higher pitch, and then \u2014 as it passes us and speeds away \u2014 a lower pitch, than when the siren is stationary; as a train speeds past a crossing bell, the passengers hear first a higher pitch as the train approaches the bell, then a lower pitch as the train moves away. There is a corresponding effect with light and all other forms of electromagnetic waves. The famous \"red shift\" of light from stars is used routinely by astronomers to find how fast a star is moving away from us (and, indirectly, how far away the star is); in the \"Doppler cooling\" of atoms, the Doppler shift of laser light \"seen\" by a moving atom is used to selectively slow down fast moving atoms and hence bring groups of atoms to very low temperatures; and on the highway, the Doppler shift of radar bounced off the front of your car is used by the police to measure your speed. To derive the formula for the Doppler shift of light we must work relativistically. (Light travels at the speed of light!) Armed with our knowledge of four-vectors, we shall find that the derivation is surprisingly easy. Any sinusoidal plane wave has the form = A cos(k x \u2014 cot \u2014 8). (15.61) Here the nature of the function 0 depends on the wave under consideration; for a sound wave, it could be taken to be the pressure change produced by the sound; for light, it could be any component of the electromagnetic field. The vector k is called the wave vector; its direction is the direction of propagation of the wave and its magnitude is 11(1 = 27/X, where is the wavelength; w is the angular frequency, co = 27 -c v, where v is the ordinary frequency; and 8 is a (usually not very interesting) constant phase shift. The speed of the wave is v = co/NI; for light in a vacuum, this is of course c, so w = ciki. Our main concern with the plane wave (15.61) is the phase k x \u2014 wt. It is impossible to resist writing this as a four-dimensional scalar product where as always x = (x, ct) and k denotes the wave four-vector, k x \u2014 wt = k x, k = (k, co/c). (15.62) (15.63) 632 Chapter 15 Special Relativity source (frame 8') Figure 15.10 The Doppler experiment. A source of light is moving along the x 1 axis with speed V relative to the frame S. The observer in frame S sees the light from the source traveling at an angle 9 with the x 1 axis. The light's frequency is w as measured in S and w' = w o as measured in the source's rest frame 5'. To prove that k defined this way really is a four-vector, we note that the phase k x at any point x determines the position on the wave relative to the troughs or crests of the wave. Since this has to be the same in any frame, it follows that k x is a four- scalar, and since x is certainly a four-vector, the quotient rule guarantees that k is a four-vector as its name implies. Since the fourth component of k is w /c and since we now know how k transforms, we are ready to find the frequency of a light signal as measured in a frame relative to which", " the source is moving. The experiment we have in mind is shown in Figure 15.10. An observer at rest in the frame 8 observes a railroad car moving at speed V along the x 1 axis. The car is emiting light of angular frequency w' = wo, as measured in the car's rest frame S'. If the light reaching the observer travels at an angle 0 with the x 1 axis, we want to know its frequency co, as measured by the observer. The wave four-vector k of the light reaching the observer has the form k = (k, k4) where k4 = (o/c = Ike. According to the standard Lorentz boost, Setting k41 = a/c, k4 = w/c, and k 1 = 11( I cos 9 = (w/c) cos 0, we find that = y (k4 \u2014 = yw(1 \u2014 cos 0). Solving for w and replacing a! by wo, we get the relativistic Doppler formula for light where wo is the frequency of the light in the rest frame of the source, w is the frequency observed in a frame where the source has velocity V, and 9 is the angle between V and the observed light. Section 15.12 Mass, Four-Velocity, and Four-Momentum 633 15.12 Mass, Four-Velocity, and Four-Momentum You may be becoming impatient that in eleven sections we have so far discussed only the kinematics of relativity. In fact, this reflects a truth about relativity \u2014 that many of its most interesting features, such as time dilation and the impossibility of causal signals traveling faster than the speed of light, are purely kinematic. Nevertheless, it is high time we took up relativistic dynamics, and this is what we now do. In this section, I shall introduce the relativistic definitions of the mass and mo- mentum of an object. It is important to recognize that there is really no such thing as the \"correct\" definition of a concept like mass or momentum when we move into the terra incognita of a new subject like relativity. Like Humpty Dumpty, we are, in principle, entitled to define words however we want. 2\u00b0 Nevertheless, there are certain requirements of reasonableness that we can hope to impose: Any definition of mo- mentum should coincide as closely as possible with the nonrelativistic definition in the domain where the latter has proved useful \u2014 namely when the speed of the object is much less than c. And we would like our new definitions to share with their non- relativistic counterparts any properties that seem essential to the concept concerned. For example, we shall seek a definition of relativistic momentum with the property that the total momentum of an isolated system is conserved. Mass in Relativity There are, in fact, two different definitions of mass in relativity, both of which meet our requirements of reasonableness, and both with their supporters. The definition I shall use can be described as the invariant mass and is favored by the majority of practising physicists. The other, called the variable mass, is favored mostly by popularizers of relativity, since it makes some ideas seem easier at first. I shall describe the variable mass briefly later, but throughout this chapter I shall use the invariant mass, which is defined as follows: Given any object at rest (or moving with speed much, much less than c), we know that the nonrelativistic definition of mass produces a well-defined and useful quantity. To emphasize its definition, this mass is often called the rest mass. To define the mass of the same object traveling at high speed, we shall adopt the following, embarrassingly simple definition: Definition of Invariant Mass in ofate l t hatever its sp ed, is defined tri be its rest ma If observers in an inertial frame S see an object sailing by at half the speed of light and want to know its mass, they must somehow bring the object to rest (or move themselves into a frame moving with the object) and then measure its mass using 20 \"When / use a word,\" Humpty Dumpty said in a rather scornful tone, \"it means just what I choose it to mean \u2014 neither more nor less:' Lewis Carroll, Alice's Adventures in Wonderland. 634 Chapter 15 Special Relativity any convenient technique of nonrelativistic mechanics. The equivalence of all inertial frames guarantees that this procedure will produce the same answer in all frames, so the resulting mass can be called the invariant mass. However, since it is the only definition we shall be using, we shall generally call it just the mass. Since the mass defined this way has the same value in all frames, it is a Lorentz scalar. The Proper Time of a Body Before we take up the definition of relativistic momentum, it is convenient to introduce two more important kinematic quantities. The three-dimensional", " position x(t) of a body at time t defines a point x = (x(t), ct) in space\u2014time, and, as time advances, this point traces a path, called the body's world line. We have seen that the separation dx between neighboring points x and x + dx on the world line of a material body is a time-like vector. This means that there is a frame (namely the body's rest frame) where the separation is pure time-like, with the form dx o = (0, 0, 0, c dto). (The subscript \"o\" indicates the rest frame.) Since the two positions in three-space are equal (x o = xo + dxo), the time dto is the proper time between the two points on the body's world line. To find this proper time, we don't actually have to go to the rest frame. In any other frame, the separation has the form dx = (v dt, c dt) and since dx 2 = dx 2, it follows that \u2014c 2dt = (1) 2 \u2014 C2)dt 2, which we can solve for 02 0 dto to give dto = dt,/1 \u2014 v 2 /c2 = (15.65) dt y(v) where y (v) is the familiar y factor 1/V1 \u2014 v 2/c2, calculated for the body's speed v. [You will recognize (15.65)as the time-dilation formula (15.11), so we didn't really need to go through this calculation.] We can apply (15.65) in any frame S and will, of course, get the same value for dto. That is, the proper time dto is a Lorentz scalar, which makes it a convenient quantity to work with, as we shall see. The Four-Velocity We have seen that the three-dimensional velocity v of a body transforms according to the rather complicated velocity-addition formulas (15.26) and (15.27). The reason for the complication is easy to see: The three-velocity v = dx/dt is the quotient of a three-vector dx and the fourth component dt of a four-vector. So, little wonder that it transforms awkwardly. Having recognized the problem, we can easily construct a related vector which transforms more simply. If we considered u = dx/dto instead of dx/dt, then at least the denominator would be a scalar. In fact, while we're about it, we may as well consider the four-vector u = \u2014 dx (dx dt dt = o dto dto c\u2014) (15.66) Section 15.12 Mass, Four-Velocity, and Four-Momentum 635 Since this four-velocity is the quotient of a four-vector and a four-scalar, it clearly is a four-vector. If we use (15.65) to replace dto by dt I y, we find that dx dt_ ) dt c dt = y(v, c). u (15.67) The most prominent feature of this result is that the three-velocity v is not the spatial part of the four-velocity u (which is why I called the latter u rather than v). However, if our body is moving much slower than c, then y ti 1 and the spatial part of the four-velocity is indistinguishable from the ordinary three-velocity v. As we shall see directly, the fact that u is a four-vector makes it useful in our efforts to construct a relativistic mechanics. Relativistic Momentum We are now ready to address the next definition in our relativistic mechanics \u2014 the definition of the momentum p of a body with mass m and velocity v. We obviously want our definition to agree with the classical definition (p = mv), at least at nonrela- tivistic speeds, Iv! << c. What else we ask of our definition depends on what classical property of momentum we regard as so important that it should carry over to relativity. It would be hard to name the single most important property of momentum in classical mechanics, but the conservation of momentum is surely a strong candidate, and we shall look for a definition of p with the property that the total momentum P = E p of an isolated system of bodies is conserved. To be consistent with the postulates of relativity, this law, if true at all, must be true in all inertial frames. The simplest possibility would be that we could continue to use the classical defi- nition p = my, but we can dismiss this possibility fairly easily. With a little ingenuity one can construct a thought experiment in which the total classical momentum E my is conserved in one frame 8, but not in a second frame 8'. One example, shown", " in Figure 15.11, is an elastic collision of two equal-mass particles a and b. As seen in frame S, the two particles approach the origin with equal and opposite velocities in the x 1x2 plane, and emerge with the x 2 components of their velocities reversed. Ob- viously, the total classical momentum, as measured in 8, is zero (E my = 0) before and after the collision, and classical momentum is conserved. Figure 15.11(b) shows the same experiment, as seen in a frame 8' which travels along the x 1 axis of S with speed V equal to the component v 1 of particle a, so that particle a travels straight up and then down the x2 axis as seen in 8'. Using the relativistic velocity transformations (15.26) and (15.27), one can find the four velocities as measured in 8'. Since these calculations, although reasonably straightforward, are quite messy, I shall leave them as an exercise for the reader (Problem 15.54), but the important conclusion is easily stated: When we substitute the velocities of 8' we find that the total classical momen- tum is not conserved in frame21 8'; that is, E in E mv'n Evidently, if we were fi 21 The reason is actually fairly easy to see. The transformation (15.27) of the y component of velocity depends on the x component. Since particles a and b have different x components of velocity 636 Chapter 15 Special Relativity x1 Frame 8' (a) (b) Figure 15.11 An elastic collision between two equal-mass particles, a and b. (a) In frame 8, the incoming particles approach with equal but opposite velocities and emerge with their x2 components reversed. The total classical momentum E my is zero before and after the collision. (b) The frame 3' has velocity equal to the x 1 component of a's initial velocity in 8. Using the relativistic velocity-addition formula, it is easy to show that the total classical momentum E my' is not conserved in this frame. to adopt the classical definition of momentum, a law of conservation of momentum would be inconsistent with the postulates of relativity. The problem with the classical definition of momentum, p = my, derives from the awkward transformation of the three-velocity v, and this suggests a more promising approach to our problem. Instead of using the three-velocity v, suppose we used the four-velocity u to define the four -momentum of any object of mass m as p.mu. (; y t.t.tv dehniti u (15.0). [The last expression follows from (15.67).] Since m is a four-scalar and u is a four- vector, this defines p as a four-vector. If, in the usual way, we write then this defines the three -momentum, p, as the spatial part of the four-vector p or, comparing with (15.68), (15.69) If our object is traveling slowly (1v1 << c), then y ti 1 and our new definition of p agrees with the classical one, p = my. In general, however, the two definitions differ by a factor of y, and it is the new definition (15.70) that proves useful. in 8, their y components wind up with different magnitudes in 8'. Thus E invy is nonzero and actually changes sign when the velocities reverse in the collision. Section 15.12 Mass, Four-Velocity, and Four-Momentum 637 What about conservation of momentum? Since we are now saddled with a four- dimensional momentum vector, it seems clear that conservation of momentum, if it is to be true at all, should be a four-dimensional law, which we could write as E pfin = spin. (15.71) This is really four equations. The first three would be the law of conservation of the newly defined three-momentum, and the fourth would be the conservation of something else, namely the fourth component E p4. Obviously, we need to find out quickly what this fourth component is, but I shall give this important question a section of its own. Briefly, though, we shall find that the fourth component of the new four-momentum is the energy (actually E c), so that the law (15.71) is a wonderfully compact combination of the old laws of momentum and energy conser- vation. Here, what I want to emphasize is this: Since the four-momentum p is a four-vector, the same is true of both sides of (15.71). Therefore, if (15.71) is true in one frame 3, it is automatically true in all frames; that is, our proposed law of conservation of four- momentum is compatible with the", " postulates of relativity. Whether the law is actually true must, of course, be decided by experiment. As you have no doubt guessed, the verdict is clear: Countless experiments have shown that the total four-momentum of an isolated system is constant. Variable Mass Some physicists like to rewrite the definition (15.70) of the relativistic three- momentum by introducing a variable mass mvar = Y(v)m. (15.72) With this definition the three-momentum becomes P = mvarv. (15.73) This has the advantage that it makes the relativistic momentum look like its nonrela- tivistic counterpart, p = my. Nevertheless, it has important disadvantages, which have led the majority of practising physicists to avoid the use of the variable mass. First, it is not necessarily a good idea to make a new definition look like its older counterpart when there are, in fact, important differences. Second, the introduction of the variable mass fails to achieve a complete parallel with classical mechanics. For example, it is not true that the relativistic kinetic energy (which we shall define in the next section) is equal to zmvarv2, nor is it true (in general) that F = mvara. (See Problems 15.59 and 15.79.) Third, unlike the invariant mass, the variable is not a Lorentz scalar. For all of these reasons, I shall not use the variable mass here. 638 Chapter 15 Special Relativity 15.13 Energy, the Fourth Component of Momentum The conservation of four-momentum means that, for any isolated system, there are four conserved quantities. The first three comprise the newly defined three-momentum p. But what about the fourth component? We are going to find that, for any freely moving object, the fourth component of the four-momentum defined by (15.68) is the energy divided by c. This is such an important result that I'll state it as a formal definition and then discuss its justification and consequences. Accordingly, we make the definition Definition of Relativistic Energy The energy!. of a freely moving object obj ct with four-mome u xl p = (p, p4) 4c = yr; (15.74) where the second expression follows from (15.68), which implies that p4 = ymc. With this definition, we can rewrite the four-momentum p as p = (p, E/c), (15.75) which explains why the four-momentum p is also called the momentum \u2014energy four -vector. In partial justification of the definition (15.74), notice first that (15.74) does at least have the dimensions of energy, namely [mass x speed 2]. Next, let us look at E for a nonrelativistic object to see if it looks like the nonrelativistic energy. With v << c, we can expand y using the binomial series to give Y = [1 \u2014 (v /0 2] -1/2 = 1 + 2 (v/c) 2, (15.76) so that (15.74) becomes E ti mc 2 lmv 2 (15.77) provided v << c. In nonrelativistic mechanics, mass was believed to be absolutely conserved, so the term m c 2 would have been considered to be constant. Since the zero of energy was arbitrary, a classical physicist would have interpreted (15.77) to say that the newly defined E is just the classical kinetic energy plus an irrelevant constant. To illustrate the result (15.77), consider for a moment an elastic collision. Suppose, for example, that two atoms a and b, with masses m ian and m ibn and nonrelativistic speeds vain and vbin collide and re-emerge with final masses man and m fibn and speeds van and vfin. (Of course, in classical mechanics, the initial and final masses would be the same, but we'll use different labels to avoid prejudging this question.) In any two- body collision, conservation of the newly defined relativistic energy (15.74) implies that Ein Ein = Efin Efin a b a b Section 15.13 Energy, the Fourth Component of Momentum 639 or, if the collision is nonrelativistic, [m a c2 4main (vain ) [rnbin c2 m in, \\ 2] ( 2] + 2 b 1b ) fi n C 2 1 fin/ fin\\ + \u20142MaVa ) 111 a = [ [ fin 2 111 b C 1 fin, fin.2] \u20142 n1 bb ) Regrouping terms, we can write this as miry, + T in = mfin c2 Tfin (15.78) where M in denotes the initial total", " mass, T in the initial total kinetic energy, and so on.22 Now, according to classical ideas, mass is conserved, so M in =Mfin. Therefore, the two mass terms in (15.78) would cancel, leaving just T in = Tfin. That is, the total kinetic energy would be conserved \u2014 which is precisely what we know to be true in an elastic collision. Thus, in the context of nonrelativistic elastic collisions, conservation of the newly defined relativistic energy (15.74) coincides with the familiar conservation of classical energy. This is perhaps the simplest and strongest single argument for regarding the definition (15.74) as an appropriate generalization of the classical notion of energy (together, of course, with the experimental fact that energy defined this way is conserved). The argument of the last paragraph gives us a reassuringly familiar result in the case of elastic collisions. However, it is going to lead us to the first big surprise of our relativistic mechanics. We know that, even in the context of nonrelativistic mechanics, there are inelastic processes, in which total kinetic energy is not conserved. For example, in the case of our two atoms, the collision could disturb the internal motion of the atomic electrons, changing the internal energy of one (or both) of the atoms. In this case we know that the atoms would emerge with changed total kinetic energy, so that T fin 0 T in. (That such processes can occur is a well-established experimental fact; the Franck\u2013Hertz experiment, in which electrons collided inelastically with mercury atoms, was a famous example.) Now, the argument leading to (15.78) depends only on the (true) assumption that relativistic energy is conserved, so (15.78) must apply to any possible nonrelativistic collision. In particular, in an inelastic collision with T fin 0 Tin, (15.78) implies that the total mass of the two atoms has to change, M fin A Alm. If we imagine an inelastic collision in which one of the atoms has its internal energy changed while the other is completely unchanged, then for the first atom we can say this: If the atom gains internal energy (is \"excited\"), then T fin < Tin and hence, from (15.78), Mfin > M in; that is, when an atom gains internal energy it has to gain mass. Conversely, if the atom loses internal energy, it has to lose mass. If relativistic energy is conserved (as it is), then it follows logically that mass cannot be conserved. The first question that must be addressed is why this nonconservation 22 By \"kinetic energy\" I mean here the kinetic energy im v 2 of the translational motion of either atom as a whole. Of course, an atom may also have kinetic energy of its electrons as they orbit the nucleus, but we shall include that as part of the atom's internal energy. 640 Chapter 15 Special Relativity of mass had not been discovered much sooner. To answer this, notice that (15.78) can be written as AMc2 = \u2014AT. (15.79) The brief answer to our questions is this: By everyday standards, c 2 is an exceedingly large quantity, so that, even if A T is fairly large, AM = A T/c 2 is still very small \u2014 in most cases, unobservably so \u2014 as the following example illustrates. EXAMPLE 15.7 Mass Change in the Franck\u2014Hertz Experiment In their famous experiment in 1914, James Franck and Gustav Hertz fired electrons through a container of mercury vapor. The mercury atom has a state whose internal energy is 4.9 eV (1 eV = 1.6 x 10 -19 J) higher than the atom's normal \"ground state.\" In some of the collisions between the electrons and the mercury atoms, a mercury atom was excited into this state, with the result that the final kinetic energy of the emerging particles was 4.9 eV less than the initial; that is, A T = \u20144.9 eV. By how much was the mass of the mercury atom increased by the collision? According to (15.79), the increase of mass is AM = \u2014 = AT 4.9 eV = 8.7 x 10 -36 kg. c2 c2 (Check the conversion and the arithmetic yourself.) The electron emerges with its mass unchanged (it is still just an electron), so all of this mass increase goes to the mercury atom (which is now an excited mercury atom). The increase is fantastically small by everyday standards, but the real question is this: How big is the mass increase compared to the original mass of the mercury atom, which is 200.6 atomic mass units or 3.3 x 10 -25 kg? This fractional change in the mass of the atom is Am _ x 10 -36 m", " 3.3 x 10 -25 = 2.6 x 10-11. This fractional change is far too small to be detected by any direct measurement of masses. collision The energy released in typical chemical reactions is also of order a few eV per atom. For example, the burning of hydrogen in oxygen can be thought of as an inelastic H2 \u00b1 H2 -F 02 -> H2O + H2O (15.80) in which the total kinetic energy of the final two molecules is about 5 eV more than that of the original three. 23 Conservation of relativistic energy requires that the outgoing 23 It is not a coincidence that the energy release (or loss) in chemical reactions is about the same as that in the Franck\u2014Hertz experiment of Example 15.7. In both cases the change originates in Section 15.13 Energy, the Fourth Component of Momentum 641 water molecules have less total mass than the initial hydrogens and oxygen, but, again, the difference is far too small to be detected by direct mass measurements. In nuclear reactions, the kinetic energy released can be much greater. For example, in the neutron-induced fission n+ 235U_\u00f7 9O r 143B a n n n, the kinetic energy increases by about 200 MeV, and the fractional loss of mass is about 1 part in 1000 \u2014 still not very large, but large enough to measure directly for many nuclear reactions. As we shall see later there are processes in which the mass change is even larger, but the evidence from nuclear physics is already sufficient to confirm the relativistic prediction (15.79) beyond reasonable doubt. Mass Energy We have seen that the term mc 2 in the result (15.77), E mc 2 + lmv2, is certainly not the \"irrelevant constant\" that a classical physicist would have taken it to be. In fact, in relativity, unlike classical mechanics, the energy does not contain an arbitrary constant. This is because we certainly want the four-momentum p = (p, E lc) to be a four-vector, and the addition of a constant to E would destroy this desirable property. (See Problem 15.66.) Bearing this in mind, let us look again at the relativistic definition of the energy of an object, E = ymc 2. Even if the object is at rest, with y = 1, the object still has some energy, given by E = mc2 (perhaps the most famous equation in all of physics). This energy is naturally called the rest energy of the object or, since it is associated with the mass m, the mass energy. The concept of mass energy lets us interpret the inelastic processes discussed above as processes in which some mass energy is converted into kinetic energy or vice versa. In the processes of atomic and nuclear physics, this conversion typically involves only a tiny fraction of the total mass energy, but there are processes in which there is 100% conversion. For example, in a collision between an electron (e - ) and its \"antiparticle,\" the positron e+, both particles can be annihilated, e - + e+ \u2014> radiation with 100% of their mass energy becoming the energy of electromagnetic radiation. When an object is moving, y > 1 and its energy E = ymc 2 is greater than its rest energy mc2. This suggests that we define a quantity T by the equation E = mc2 + T. (15.81) This T is the additional energy that an object has by virtue of its motion and is naturally called the kinetic energy, T = E \u2014 mc 2 = (y \u2014 1)mc2. (15.82) differences in the energy levels of the electrons in the atoms or molecules, and these differences are almost always of order a few eV. 642 Chapter 15 Special Relativity When the object is moving slowly, we have seen that T ti lmv 2 [this follows from (15.77)], but in general the nonrelativistic result is incorrect and we must use the relativistic definition (15.82). Three Useful Relations There are three useful relations among the parameters m, v, p, and E that characterize an object's motion. First, since p = ym(v, c) and also p = (p, E/c), we see at once that This relation lets you find the velocity of an object if you know its three-momentum p and energy E. Consider next the invariant \"length squared\" p2 = p p. In the object's rest frame p has the form p = (0, 0, 0, mc), so that p p = \u2014(mc) 2. Since both sides of this equation are invariant, it immediately follows that the same relation holds in any frame: For any object with four-momentum p and mass m, in any inertial frame. This relation is well worth memorizing and can greatly", " simplify several calculations, as we shall see. Finally, it is sometimes useful to rewrite the result (15.84) in terms of the three- momentum and the energy. Since p = (p, E Ic), (15.84) becomes p 2 \u2014 (E/c) 2 = \u2014 (mc) 2 or This shows that the three quantities E, mc 2, and 10 c are related like the sides of a right triangle, with E as the hypotenuse, as indicated in Figure 15.12. At this stage there is no deep geometrical significance to this statement, but it does give an easy way to remember and visualize the relation (15.85). If the speed is much less than c, then y ti 1, so E ti mc 2 ; in this case, the hypotenuse and base of the triangle are nearly equal, so T <<mc2 and the triangle is very low (height << base). On the other hand, if v is very close to c, then y >> 1, so E >> mc2 ; in this case, T >> mc2, so the energy is mostly kinetic, and the triangle is very tall (height >> base) with E IP I c. Section 15.13 Energy, the Fourth Component of Momentum 643 Iplc Figure 15.12 The three parameters E, mc2, and 'plc are related like the sides of a right triangle with E as the hypotenuse. EXAMPLE 15.8 Energy and Momentum of an Electron The rest energy of an electron is about 0.5 MeV (actually 0.511, but for many purposes 0.5 is good enough). What is the electron's mass in the SI unit (the kilogram) and in MeV/c 2? If its kinetic energy is T = 0.8 MeV, what is its total energy E and what is the magnitude I pl of its three-momentum in MeV/c? What is its speed? The given rest energy tells us that mc 2 = 0.5 MeV. (15.86) Solving for m, converting eV to joules, and putting in the value of c, we find (as you can check) that m 9 x 10 -31 kg (more precisely, 9.11 x 10 -31 ). Straightforward as this calculation is in kilograms, it is even easier, and often much more convenient, in MeV/c 2. We simply divide both sides of (15.86) by c2 and we have the answer directly: m = 0.5 MeV/c2. Evidently, the mass in MeV/c 2 is numerically the same as mc 2 in MeV. This is so simple that it takes a little getting used to. If you've never used the unit MeV/c2 before, just keep reminding yourself that the statement m = 0.5 MeV/c2 mc 2 = 0.5 MeV. The reason this is is precisely equivalent to the statement that such a convenient way to specify masses is that our real concern is often not with the mass itself, but rather with the corresponding energy mc 2, and the most convenient unit for the latter is often the MeV. If T = 0.8 MeV, then clearly E = T mc 2 = 1.3 MeV, and by the \"useful relation\" (15.85) Iplc = 1E2 \u2014 (mc2)2 =.11.32 \u2014 0.5 2 MeV = 1.2 MeV. Once again, we could get an answer in SI units by making the necessary conversions, but a simpler, and often more convenient, course is just to divide both sides by c to give IPI = 1.2 MeV/c. 644 Chapter 15 Special Relativity Finally, according to the useful relation (15.83), the electron's dimensionless speed,8 = v/c is that is, v = 0.92c. = Ipic 1.2 MeV E 1.3 MeV = 0.92; Notice how nicely the factors of c cancel if we measure masses in MeV/c 2 and momenta in MeV/c when using the relations (15.83) and (15.85). This is because m and p enter these relations only through the combinations mc 2 and pc. For some practice at using these relations and the new units see Problems 15.61 to 15.63. 15.14 Collisions The laws of conservation of energy and momentum play a key role in the analysis of collisions. In this section, I shall illustrate this claim with three examples. EXAMPLE 15.9 Collision of Two Lumps of Putty A relativistic ball of putty, with mass m a, energy Ea, and velocity va, collides with a stationary ball of mass m b, as shown in Figure 15.13. If the two", " balls fuse to form a single lump, what is the lump's mass m and with what velocity v does it move off? To find the final mass, we have only to recall that the invariant \"length squared\" of the four-momentum of any object is \u2014m 2 c2. If we denote the final four-momentum by p fin, then (pfin) 2 = (15.87) By conservation of momentum-energy, p fin = pin, where pin is the total initial momentum; that is, pin = pa + Pb, from which (Pin) 2 = (Pa + Pb)2 = Pa2 Pb + 2Pa Pb = 2c2 ___ inby ___ 2Eamb (15.88) before after Va mb V Figure 15.13 Two balls of putty collide and form a single lump. Section 15.14 Collisions 645 where the last term comes about because pb = (0, 0, 0, m bc). Comparing (15.87) and (15.88), we find that the mass of the final lump is m = m 2 + in 2 2Eam b /c2. a Notice that if the original motion was nonrelativistic, then Ea R= mace, and we recover the nonrelativistic result m = ma + mb ; but, in general, m > ma + mb. According to the useful relation (15.83), the final velocity is v = pfinc2 /Efin. By conservation of four-momentum, we can replace the components of p fin by those of p in to give V -= pa C2 ya ina Va Ea + m bc2 y am a m b (15.89) where ya denotes the y factor for the incoming ball a. Notice that if va \u00ab c, this reduces to the familiar nonrelativistic result v = mava /(ma m b ). The CM Frame In nonrelativistic mechanics, we have seen that a very useful concept is that of the CM or center-of-mass frame \u2014 the frame in which the center of mass of a system is at rest. Alternatively, this frame can be characterized as the frame in which the total momentum is zero, P = E p = 0. (So you can think of \"CM\" as standing for \"center of momentum\" if you like.) This alternative definition carries over directly to relativistic mechanics. 24 We have seen that the four-momentum p of any material particle is forward time-like (lies inside the forward light cone). 25 Now, it is a simple exercise to prove (Problem 15.69) that the sum of any number of forward time-like vectors is itself forward time-like. Therefore, the total momentum P = E p of any collection of particles is also time-like, and this guarantees that there exists a frame in which P has the form P = (0, 0, 0, P4). Naturally, we define this frame, in which the total three-momentum P = 0, to be the CM frame of the system. It often happens that a collision problem is very easy to solve in the CM frame. Thus, if we need to solve the same problem in some other frame 3, the simplest procedure is often to transform from 8 to the CM frame, solve the problem there, and then transform back to 3, as the following two examples illustrate. 24 Oddly enough, the notion of center of mass does not carry over satisfactorily into relativity. Thus, it is better to think of the CM frame as the center-of-momentum frame. 25 So far, we are considering only material particles, that is, particles with mass m > 0. We shall soon discuss the case of massless particles, for which p lies on the light cone. Fortunately (with one small exception) the same results apply even when some of the particles are massless. See Problems 15.88 and 15.89. 646 Chapter 15 Special Relativity EXAMPLE 15.10 An Elastic Head-On Collision Consider an elastic head-on collision between a projectile, with mass m a and velocity va, and a stationary target of mass m b, as shown in Figure 15.14(a). (That the collision is head-on means that the two particles emerge from the collision both moving along the line of the incident velocity v a, as shown). What is the final velocity v b of the target particle b? Let us denote by S the lab frame, in which the experiment actually occurs (with b initially at rest), and take the direction of the incident velocity to be the x axis. To solve the problem directly in frame 8, we would write down the equations of conservation of energy and momentum and solve for the requested final velocity. Unfortunately, the equations are extremely messy, and a much simpler course is to transform to the CM frame 8", "'. In the CM frame the two incoming three-momenta are equal and opposite, and it is easy to see (Problem 15.68) that the collision simply reverses them both. Thus our procedure will be this: (1) Transform p b to the CM frame 8'. (2) Reverse its spatial part p b. And (3) transform back to S and calculate the velocity. Before we do this, we need to find the velocity of the CM frame 8' relative to S. Since the total four-momentum is Ea \u00b1mbc 2) P = (Pa, c the required (dimensionless) velocity $ to transform to the CM frame is pac # = Ea + rn bc2 (15.90) (In the nonrelativistic limit, with pa ti ma y, and Ea ti m ace, this corresponds to a velocity v = m a va /(m a m b), which is the velocity of the center of mass, as expected.) We can now follow our three steps to solve the problem. In the lab frame, the initial four-momentum of the target b is pbin = (0, 0, 0, m bc) [lab frame, initial]. (15.91) after 0 b 0 b a a (a) Lab frame 8 before after (b) CM frame S' a b Figure 15.14 (a) An elastic head-on collision as seen in the lab frame 8, where the target b is initially at rest. (b) In the CM frame 8', all of the three-momenta have the same magnitude. The only effect of the collision is to reverse the three- momentum of each particle. (The arrows represent momenta.) Section 15.14 Collisions 647 Applying the standard Lorentz boost, with velocity (15.90), we find the corre- sponding CM momentum Pb\"n =YinbC(\u2014)6, 0, 0, 1) [CM frame, initial]. (15.92) In the CM frame the collision simply reverses the spatial components of this momentum. Thus the corresponding final momentum is Ifin pb = ymbc(P, 0, 0, 1) [CM frame, final]. (15.93) Finally, transforming back to the lab frame, we find pen = y 2m bc (2/3, 0, 0, (1 $ 2)) [lab frame, final]. (15.94) The corresponding dimensionless velocity is just the ratio 2$/(1 + 13), so the actual final velocity of the target is 2/3 vb = 1+ )62 C (15.95) with /3 given by (15.90). The answer (15.95), although easily found, is not especially illuminating in general. In the special case that the two masses are equal, it is easy to show (Problem 15.73) that (15.95) reduces to Vb=V a ; that is, the target b emerges with precisely the velocity of the incoming projectile a, and the projectile therefore comes to a dead stop. This behavior is well known to students of nonrelativistic mechanics (and to billiards players) and shows that, in the case that m a = mb, the relativistic result (15.95) agrees exactly with the familiar nonrelativistic one. Whether or not the two masses are equal, it is easily shown that in the limit va \u00ab c, the relativistic result (15.95) approaches the corresponding nonrelativis- tic one. (See Problem 15.73.) Threshold Energies Most of the elementary particles that have been discovered in the last seventy years or so were found when they were produced in collisions of other particles. For example, the negative pion, 7 \u2014, can be created in a collision of a proton and a neutron, Similarly, the first antiproton to be observed was produced by a proton\u2014proton colli- sion in the reaction P+P --->- 13 +P+P+P. (15.96) (The negatively charged antiproton p is the \"antiparticle\" of the proton, with the same mass but opposite charge.) A quantity of great concern to any experimenter hoping to observe this kind of reaction is the threshold energy, defined as the minimum energy of the initial particles for which the reaction can occur. 648 Chapter 15 Special Relativity Let us consider a reaction of the form ad-b--->c1+ + g. In the traditional collision experiment, one of the original particles (let's say b) was usually at rest \u2014 defining what we have been calling the lab frame. Thus, the concern was to know the threshold energy for a reaction of this kind in the lab frame. At first glance this would seem a simple matter to calculate. The minimum possible energy of the final", " particles is just their total rest energy E m finc2, the energy they have when all at rest. Surely, then, the threshold energy for the reaction is just E mfinc2. Unfortunately this pausible argument is wrong. The trouble is that in the lab frame the total initial three-momentum is nonzero. (Particle b is at rest, so a has to be moving to bring in the necessary energy.) Conservation of three-momentum requires that the final three-momentum be nonzero, and so the final particles cannot all be at rest. Therefore, the threshold energy is more than just E m finc2. But what is it? The easiest way to answer this question is to observe that in the CM frame, where the total three-momentum is zero, all of the final particles can be at rest. Thus Ecm >E m finc2 (15.97) and the equality here is possible, with all final particles at rest. We can now find the threshold energy in the lab, by comparing the total four-momenta in the two frames. In the CM frame the total four-momentum has the form P cm = (0, 0, 0, Ecm/c). In the lab frame, it is P - lab = Pa + Pb, where pa = (pa, Ea /c) and ph = (0, 0, 0, m bc) are the momenta of the two orginal particles. Now by the invariance of the scalar product, 2,2 /,,, 2 += Pb) 2 + b PbPbcml 2Pa P \u2014M a2C2 Inb2C 2Eam b p 2 so m = lab' 2 P or, solving for Ea, Ea E 2 ___ m 2 c4 m 2 c4 cm a 2m bc2 Inserting the minimum value of Ecm from (15.97), we find for the minimum energy of the projectile a in the lab frame E min (E mfio m 2 2 m 2 a = a b C2. 2m b (15.98) A famous example of the use of this equation was in the design of the experiment to verify the existence of the antiproton using the reaction (15.96). In this reaction E m fin = 4m p, while m a = mb = mp, so the minimuum energy (15.98) is 7m pc2 ; that is, the minimum kinetic energy for protons to produce antiprotons by the reaction (15.96) was 6m pc2 ti 5600 MeV. The reaction in question was first observed at Berkeley, using protons accelerated to this energy by a machine called the Bevatron, which had been specifically designed to accelerate protons to about 6000 MeV, just enough more than the threshold to be sure to do the job. Section 15.15 Force in Relativity 649 An important feature of (15.98) is that the leading term is proportional to (E m fin ) 2 Thus if the particle one is hoping to produce is very heavy, Er may be prohibitively large. For example, the particle called the * (or J/*) has a mass of about 3100 MeV/c 2 and was discovered in the reaction e+ e - where the positron and electron have mass about 0.5 MeV/c 2 each. Putting these numbers into (15.98), we see that, if this reaction was to be produced by firing positrons at stationary electrons (or vice versa), the minimum incident energy would have had to be a fantastic Emin = 107 MeV \u2014 well out of reach of any electron or positron accelerator in existence today. The way around this seemingly hopeless obstacle was to use colliding beams, with the electrons and positrons approaching one another with approximately equal and opposite momenta. That is, the experiment was done in the CM frame. From (15.97), you can see that the threshold in this case is only Ecn, = 3100 MeV. In this experiment, the advantages of this much smaller threshold energy far outweighed the disadvantages of having to work with two beams of high-energy particles. 15.15 Force in Relativity We have not yet introduced the concept of force into our relativistic mechanics. One of the reasons for this is that force plays a much smaller role in relativity than in nonrelativistic mechanics. Another is that the concept of force is much more complicated in relativity. The most obvious complication is that (like several other parameters, such as mass and velocity) force can be defined in several different ways. A second complication arises from the possibility that the rest mass of an object can change. As we have seen, an inelastic collision of an electron with an atom can give the atom additional internal energy and increase its rest mass. For a macroscopic example of the same effect, imagine", " holding a flame under a metal object; the heat absorbed increases the object's internal energy and its rest mass. Like most introductory texts, I shall avoid the complication of such \"heat-like forces\" by confining attention to forces that do not change the rest masses of the objects on which they act. 26 Fortunately, these include many of the important forces in special relativity, including the Lorentz force F = q(E v x B) (15.99) on a charge q in electric and magnetic fields E and B. 26 For a careful and clear discussion of \"heat-like\" forces, see the excellent book of Wolfgang Rindler, Introduction to Special Relativity, Oxford University Press, second edition, 1991, but be warned that Rindler tends to use m to denote the variable mass (what we would call ym). 650 Chapter 15 Special Relativity Of the several conceivable definitions of force in relativity, the single most useful is probably the three-force defined as where p denotes the relativistic three-momentum p = y m v. This is not, of course, the same as the nonrelativistic force, since p is not the nonrelativistic momentum, but it does have the essential merit that it agrees with the nonrelativistic definition when v << c (and y ti 1). A second property that recommends the definition (15.100) is that experiment shows that, with this definition, the force on a charge q in an electromagnetic field is given by the Lorentz equation (15.99). Thirdly, with the definition (15.100), we can prove the analog of the work-KE theorem, as follows: Recall the useful relation (15.85), that E 2 = (pc)2 + on c2. \\ 2. ) Differentiating both sides with respect to time, we see that (remember we're assuming that the rest mass m doesn't change, so there is no term in dm/dt, but see Problem 15.85) dE dt E\u2014 =pct \u2014 = pc 2 P 2 F dt or, dividing both sides by E and recalling that pct /E = v, dE dt =v F. (15.101) Multiplying both sides by dt we find that dE = F dx, (15.102) where dx denotes the displacement dx = v dt. Finally, since E = mc2 T and we are assuming that the mass m does not change, we find that dT =F dx (15.103) which is precisely the work-KE theorem, generalized to include relativistic energies and forces. EXAMPLE 15.11 Motion with a Constant Force An object of fixed rest mass m is acted on by a uniform, constant force F (for example, the force on a charge in a uniform electrostatic field) and is released from rest at the origin at t = 0. Find the object's three-momentum p, its three- velocity v, and its position x, all as functions of time. Section 15.15 Force in Relativity 651 Integrating (15.100) (with F constant), we find immediately p = Ft. (15.104) From (15.85) it is easy to see that y2 = 1 + p2 /(mc) 2, so y = N/1 (Ft/mc) 2 and v= p my Ft tn-j1 (Ft/mc) 2 (15.105) When t is small, we can neglect the second term inside the square root, and we recover the nonrelativistic answer v = Ft/ m, but when t gets large, the second term in the square root dominates and we find that v approaches c, without ever quite reaching it. This is consistent with our knowledge that no material particle can have speed greater than or equal to the speed of light. To find the object's position x, we have only to integrate (15.105) to give x _F (flic) 2 (11 ( Ft 2 M F ) (15.106) As you can easily check, when t is small, this reduces to the familiar nonrela- tivistic result x = 1Ft 2 /m (that is, lat 2); when t --> co, it is asymptotic to (ct + const), in the direction of F, as the speed approaches the speed of light (Problem 15.82). Potential Energy It can happen that, at least in one frame 8, the force F on an object is the gradient of a function U(x); that is, F = \u2014 V U (x) and the force is conservative. This is the case, for example, for a charge q moving in an electrostatic field. When this happens, the work done on the object as it moves through a displacement dx is F dx = \u2014 V U dx = \u2014", "dU. Combining this with the work\u2014KE theorem (15.103), we find that dT = \u2014dU or d(T + U) = 0; that is, just as in nonrelativistic mechanics, if the force on an object is conservative, T + U is conserved. The Four-Force The three-force F = dp/dt is not the spatial part of a four-vector. (The trouble is that, although dp is the spatial part of a four-vector, dt is not a scalar.) In this respect, the three-force is like the three-velocity v = dxldt, and the transformation of F from one 652 Chapter 15 Special Relativity velocity, it is easy to see how to define a four-force that is closely related to the three- force. We can define the four-force on an object as the derivative of p with respect to the proper time to measured along the object's world line: (There is no widely accepted notation for the four-force, but K is one of the several notations used.) Since dp is a four-vector and dto is a four-scalar, K is automatically a four-vector. Since dto = dt/y, we can rewrite K as K (K, K4) = y ( = y (F, v F/c) (15.108) dp 1 dE dt c dt where the last equality follows from (15.101). We see that the spatial part of the four- force is y times the three-force F, just as the spatial part of the four-velocity u is the y times the usual three-velocity v, as in (15.67). The advantages of the four-force stem from its being a four-vector. This means that its transformation from one frame to another is just the familiar Lorentz transfor- mation. It also means that the Lorentz invariance of any physical law formulated in terms of the four-force is easy to check. The main disadvantage of the four-force is that it gives the time derivative of momentum with respect to the proper time, where the three-force gives the derivative with respect to the time of any one inertial frame. Since our main interest is usually in the motion of an object in terms of the time in one particular frame (as opposed to the proper time of the moving object), the three-force is, in this respect, more useful. 15.16 Massless Particles; the Photon A surprising consequence of relativity is the possibility of particles with zero mass, m = 0. In nonrelativistic mechanics, the notion of a massless particle makes no sense at all. The definitions p = my and T = 4m v2 show clearly that a particle with m = 0 would have no momentum and no kinetic energy and, hence, would presumably be nothing at all. At first glance, the same argument might seem to apply in relativity: If we let m 0 in the relativistic definitions p = ymv and E = ymc2 (15.109) we would seem to get the same conclusion \u2014 that a massless particle would have no momentum or energy and hence would not exist. Let us shelve this difficulty for a moment and look at the two relations (15.85) and (15.83) E2 = (mc 2) 2 (pc) 2 and (15.110) v pc c E Section 15.16 Massless Particles; the Photon 653 If there were to be a particle with m = 0 (and if these relations still applied) then the first relation would become E = Iplc [if m = 0] (15.111) and this, combined with the second relation, would imply that the particle's speed had to be c, m = 0. v = c [if m = 0]. (15.112) In other words, if there were to be a massless particle, then the usual relations of relativistic mechanics would require that it always travel with speed c. If we return now to our original definitions (15.109) of p and E, we see that as m 0, so v c and y co. Thus for a massless particle, the two definitions would take the form oo x 0, which is undefined and hence does not actually rule out the existence of particles with Evidently, relativistic mechanics has room for massless particles, always traveling at speed c. Whether such particles exist is, of course, a question for experiment, and experiment tells us unambiguously that they do. The photon is the particle that carries the energy and momentum of electromagnetic waves; and experiment shows that, for a photon, E and p do satisfy (15.111) and that photons do always travel (no surprise!) at the speed of light. 27 Notice that with m", " = 0, the four-momentum of a photon satisfies p2 = 0 ; (15.113) its invariant length squared is zero. We have seen that the four-momentum of a material particle (that is, a particle with in > 0) is always forward time-like. By contrast, that of any massless particle lies on the forward light cone and is forward light-like. Since the definitions (15.109) are no longer meaningful, the question naturally arises how the energy and momentum of a massless particle are defined. In principle, at least, they can be defined using the conservation laws. Consider, for example, the emission of a photon by an atom X: X* X + y (15.114) where X* denotes an excited state of the atom, and X its ground state, and y is the standard symbol for a photon. Since we already know how to define and measure the energy and momentum of either state of the atom, we can find the corresponding values for the photon from conservation of energy and momentum. This definition must, of course, be checked for consistency. Would a different process yield the same answers 27 It used to be thought that the neutrino was another example of a massless particle, but current evidence shows that its mass, though small, is definitely nonzero, perhaps about 10 -6 times the electron mass. (For comparison, the experimental limit on a possible photon mass is of order 10 -20 times the electron mass.) On theoretical grounds it is generally assumed there must be a massless particle called the graviton, which does for gravity what the photon does for electromagnetism, but there is no direct evidence for the graviton. 654 Chapter 15 Special Relativity for the same photon? For example, suppose we allowed the photon of (15.114) to collide with a second atom Y and eject an electron (the photoelectric effect): y + Y Y \u00b1 + e (15.115) where 17 + denotes the positive ion of Y with one electron removed. Using this process, we could measure the energy and momentum of the photon, and these second measurements should yield the same answers as the first. Experiment has shown repeatedly that the energy and momentum of the photon are consistently defined in this way. In fact there is a second way to find the energy and momentum of the photon. One of the first discoveries (due to Max Planck and Einstein) in the unfolding of quantum mechanics was that the energy of a photon is related to the frequency of its associated electromagnetic wave by the famous relation E = loco (15.116) where h is Planck's constant (actually the original Planck constant h divided by 27r, h = /27r = 1.05 x 10 -34 Js) and omega is the angular frequency of the wave. Similarly, the momentum of a photon is given by p = Pik (15.117) where k is the wave vector of the wave. Thus, E and p can be found by measuring the frequency and wave vector of the corresponding wave. It is a pleasing fact that the two relations (15.116) and (15.117) can be combined into a single four-vector relation. Since p = (p, E I c) and the wave four-vector is k = (k, co/c), the two relations imply that 270., p = hk. (15.118) Since both sides of this equation are four-vectors, this relation is relativistically invariant; that is, the two quantum relations (15.116) and (15.117) are consistent with the principles of relativity. The relation (15.117) is often rewritten in terms of the wavelength A. Since lk I = I P I = hlki = 27rh, A = \u2014. h A (15.119) In this form the relation is often called the de Broglie relation in honor of the French physicist Louis de Broglie (1892-1987), who first proposed that this relation should apply to the quantum wave associated with any particle \u2014 not just photons. For future reference, I'll rewrite a photon's four-momentum as follows: p = hk = h, (k, 642 ) = \u2014hc\u00b0 (k, 1) (15.120) where the last equality holds since Ike = co/c, so k = (co/c)k. Section 15.16 Massless Particles; the Photon 655 The Compton Effect Historically, the most influential and persuasive evidence for the existence of massless photons obeying the relations of the last few paragraphs was the experiment of the American physicist Arthur Compton (1892-1962) in 1923. Compton fired X-ray photons at stationary electrons 28 and measured the increase in wavelength of the scattered photons. Classical theory would require that the wavelength of the scattered radiation should be exactly the same as that of the incident waves, but an increase in wavelength is easily", " explained if the radiation is carried by particle-like photons. When a photon collides with an electron, the electron recoils, taking some of the photon's original energy. Thus the emerging photons must have less energy and hence less momentum than those in the incident beam. According to (15.119) less momentum means longer wavelength, and the increase in wavelength is explained. Using the relations of the last few paragraphs, Compton was able to calculate the expected shift in wavelength and his experiment triumphantly confirmed his calculations. The Compton experiment is shown schematically in Figure 15.15. The photon comes in from the left with four-momentum p yo and emerges at angle 0 with four- momentum py, where according to (15.120) Pyo h co c (ko, 1) and p, (k, 1). (15.121) hw c The electron's initial four-momentum is po = (0, 0, 0, mc) (15.122) Pyo Figure 15.15 A photon, labeled y, with four-momentum p yo collides with a stationary electron. The photon emerges at an angle 0 with four- momentum py, and the electron recoils with four-momentum p. 28 The target electrons were actually the valence electrons in the carbon atoms of a piece of graphite. Thus the electrons were certainly not perfectly stationary, but their kinetic energy (a few eV) was negligible compared to the energy of the X-ray photons (many thousands of eV). By the same token, the binding of the electrons (binding energy ti a few eV) was unimportant with such high photon energies. 656 Chapter 15 Special Relativity and it recoils with four-momentum p. Now, by conservation of four-momentum, Po + Py o = P Pr or Po \u00b1 (Py o Py ) = P. Squaring both sides we find that Po2 2P0 (Py o Py ) (Py20 2 Py0 Py Py) = P2 Since po2 = p2 = \u2014m2c2, these two terms cancel, and since p y20 = py2 = 0, these two terms drop out, and we are left with Po (P y o Py ) = Py o P y. (15.123) Substitution of (15.122) and (15.121), followed by a little algebra, yields (as you should check) or coo \u2014 w = (1 \u2014 cos 0)(00.) mc 2 1 Co 1 h wo mc 2 (1 cos 0). Finally, replacing co by 2n- c1A., we find the desired shift in wavelength, A X = X \u2014 X0 = \u2014 (1 \u2014 cOs 0), (15.124) h MC where I have replaced 27r h by the original Planck constant h. This is the celebrated Compton formula for the shift in the wavelength of scattered radiation. That Comp- ton's data agreed with this prediction at several angles gave strong support to the idea that the energy and momentum of radiation are carried by photons obeying the laws of relativistic mechanics, but with m = 0. 15.17 Tensors* *As usual, sections marked with an asterisk can be omitted on a first reading. In the next and final section of this chapter, I shall give a very brief account of the relativistic form of electromagnetic theory. Unfortunately, even this brief introduction requires some knowledge of the properties of tensors in four-dimensional space\u2014 time, and these four-tensors are the main subject of the present section. A complete account of four-tensors requires a rather elaborate machinery of \"covariant\" and \"contravariant\" vectors, but, for our present purposes we can manage without this formalism. Nevertheless, you should be aware that if you wish to pursue the study of relativity much further, you will need to master this more elaborate machinery. 29 I shall start my account by examining the transformation properties of three- Here dimensional vectors and tensors. Section 15.17 Tensors* 657 Vectors and Tensors in Three Dimensions A three-vector a is characterized by its three components (in each frame) and by its transformation properties under rotations, as in (15.35), a' = Ra, (15.125) where a and a' denote the columns made up of the three components in each of the two frames, and R is the (3 x 3) rotation matrix connecting them. In detail, this matrix equation reads a. = E R. a. IJ (15.126) where the sum runs from j = 1 to 3. For future reference we need to establish an important property of the rotation matrices R. We know, of course, that any rotation leaves the scalar product a b of any two vectors invariant; that is, a b = a' b', with a' given by (", "15.125), and likewise b'. Now, in matrix notation, the scalar product is a b =- (15.127) where we must now insist that b denotes the column of numbers b 1, b2, b3 and a the row of numbers a l, a2, a3. Thus, in matrix notation, the equation a b = a' b' reads ab = (Ra)(Rb) =a(RR)b. (15.128) [Here I used the result that (Ra) = aR \u2014 see Problem 15.94.] Since this must hold for any choices of a and b, it is easy to show (Problem 15.95) that RR = 1, (15.129) where, as before, 1 denotes the 3 x 3 unit matrix. Any matrix satisfying this condition is said to be an orthogonal matrix, so we have proved that rotations are given by 3 x 3 orthogonal matrices. A three-dimensional tensor T comprises nine elements Tit (in each three- dimensional Cartesian reference frame), where i and j both take values from 1 to 3. (Strictly speaking, a tensor with nine elements Tii is a second-rank tensor. More generally, a tensor of rank n has 3n elements, but we shall only be concerned with 29 For a clear and reasonably simple account see David J. Griffiths, Introduction to Electrody- namics, third edition (Prentice Hall, 1999), pp. 501 and 535. 658 Chapter 15 Special Relativity the case n = 2.) To find the transformation properties of a tensor, let us consider the simple example of a tensor with elements Tij = ai b j (15.130) where ai and b j are the components of any two vectors. (For example, a could be the position of a particle and b its velocity.) This obviously has the requisite nine elements and is, in a sense, the prototypical tensor. The tranformation of the tensor (15.130) follows immediately from that of the two vectors from which it is constructed: T.'. = i j 1 j = (ERik ak )( k k,1 RikR This transformation is the defining characteristic of a tensor. It closely parallels the transformation (15.126) for a vector, except that the tensor, with its two indices, gets two rotation matrices, one for each index. We can write the transformation (15.131) in matrix form if we note that (R) 31 = (R) 11, where R denotes the tranpose of R, so that (15.131)can be written (15.131) (15A32) Any three-dimensional (second-rank) tensor transforms according to this equation, and any set of nine elements which transforms in this way is a tensor. One of the most important operations one can perform with tensors is to multiply them by vectors. For example, the angular momentum L of a rigid body is given by the product L = Ice) of the moment of inertia tensor I and the angular velocity vector (t). It is important that any product of this form is, as we would expect, a vector, and this is easy to prove: Let T be any tensor and a any vector, and in every reference frame let the column of three numbers b be defined as b = Ta. To show that, with this definition, b is a vector, we use the properties (15.132) and (15.125) of T and a as follows: = T'a' = (RTR)(Ra) = RT(RR)a = RTa = Rb, (15.133) where the fourth equality follows because, according to (15.129), RR = 1. We con- clude that b, which is clearly a column of three numbers, transforms like a vector; that is, b = Ta is a vector. We can turn this argument around and prove (Problem 15.99) that if a and b are known to be vectors and in every frame it is found that b = Ta, where T is a 3 x 3 array of numbers (in every frame), then T satisfies (15.132) and is therefore a tensor. Section 15.17 Tensors* 659 Vectors and Tensors in Four-Dimensional Space\u2014Time The discussion of vectors and tensors in four-dimensional space\u2014time closely parallels that just given for three dimensions, with the only complications coming from the minus sign in the invariant scalar product. A four-vector is given by a column a of four numbers (in each frame), which transform under the Lorentz transformation a' = Aa as we move from one inertial frame 8 to another, 8'. This transformation leaves invariant the scalar product, which we can write in matrix notation as a b = a", "2b2 a3b3 \u2014 a 4b4 = aGb (15.134) where b denotes the column for the four-vector b, a is the row for a, and G is the 4 x 4 matrix G +1 0 0 0 -- 0 +1 0 0 0 0 +1 0 0 0 0 \u20141 (15.135) This metric matrix, inserted between a and b in (15.134), simply changes the sign of the fourth component of b and so inserts the needed minus sign in the scalar product. The scalar product (15.134) is invariant when we replace a and b by Aa and Ab, and exactly the argument that led to (15.129) shows (Problem 15.98) that AGA = G (15.136) the relativistic analog of the condition RR = 1 for rotations. The set of all 4 x 4 matrices that satisfy the condition (15.136) is called the Lorentz group, since all Lorentz transformations must satisfy this condition. A four-tensor (strictly speaking a four-tensor of rank 2) is defined as a set of sixteen numbers TAI) (defined for every inertial frame 8), where the indices it and v run from 1 to 4, which, when formed into a 4 x 4 matrix T, satisfy T' = AT (15.137) \u2014 a property that exactly parallels Equation (15.132) for three-tensors. Just as we form the scalar (or dot) product (15.134) of two four-vectors by inserting the matrix G between the two appropriate matrices, so we can form a dot product of a tensor and a vector in the same way: T. a = T Ga. (15.138) It is a straightforward exercise to show that, if T is any tensor and a any vector, then b = T. a is a four-vector. [The proof parallels (15.133) for three dimensions \u2014 see Problem 15.96.] Conversely, you can prove (Problem 15.99) a \"quotient rule\" that if a and b are known to be four-vectors and if b = T- a in every frame 8, then T (the \"quotient\" of b and a) is a four-tensor. Armed with these definitions and properties of four-tensors, we are ready for our brief venture into relativistic electrodynamics. 660 Chapter 15 Special Relativity 15.18 Electrodynamics and Relativity That light travels at speed c in all directions is a consequence of the laws of classical electromagnetism, and special relativity grew out of the realization that the speed of light c is the same in all inertial frames. These two observations suggest that classical electromagnetism might already be consistent with the principles of relativity. The simplest way to prove this suggestion is to show that the familiar laws of electrody- namics can be written in terms of four-scalars, four-vectors, and four-tensors, so that their invariance under Lorentz transformations is self-evident. Here I shall do this for just one law, the centrally important Lorentz-force law F = q (E v x B). (15.139) In the process, we shall find the transformation rules for the electric and magnetic fields E and B. Before we address the properties of the fields E and B, you need to know that it is an experimental fact that the charge q of any particle has the same value in all inertial frames; that is, q is a Lorentz scalar. I shall take the view that the Lorentz equation (15.139) is an observed fact, valid in all inertial frames (which it definitely is). In the form (15.139), it certainly does not look relativistically invariant, and our task is to rewrite it so that it does. Our first clue for how to do this is to notice that (15.139) defines F as a linear function of v. The next, and very natural, step is to rewrite this linear relation in terms of the four-force K and the four-velocity u. Recall that K= (IF, and u = (yv, yc). (15.140) yv F) c ) Multiplying both sides of (15.139) by y, you can see that K is a linear function of u. The simplest such relation would have the form K u, where would be some as-yet unknown four-tensor (and I have inserted a separate factor of q since K is obviously proportional to q). In matrix form, this relation would read K = qJ'Gu [where K and u must now be seen as 4 x 1 columns, and G is the metric matrix (15.135)]. The 16 elements", " of the matrix J\"G can be found by writing out the components of K one at a time. For example, from (15.140) and (15.139), the first component of K is K 1 -= yq(E1 v 2 B3 \u2014 v 3 B2) = q[B3 u 2 \u2014 B 2 u 3 (E1/c)u4]. (15.141) The coefficients of u l, -, u4 are just the first row of the matrix J'G in the proposed relation K = O'Gu. Proceeding in this way, we find the whole matrix TG to be J'G - 0 -B3 B3 0 B2 \u2014B 1 -B2 Ei/C B1 0 E2/C E3/c _ E 1 /c E2/c E3/c 0 _ (15.142) [Compare the first row of this matrix with the coefficients in (15.141); for more details, see Problem 15.104.] Finally, since G 2 = 1 (the 4 x 4 unit matrix), we can multiply by G on the right to give the electromagnetic field tensor Section 15.18 Electrodynamics and Relativity 661 0 B3 \u2014B2 \u2014E1/c \u2014B3 0 B2 \u2014B1 B1 \u2014E2/c 0 \u2014E3/c E1/c E2/c E3/c 0 (15.143) in terms of which the Lorentz force takes the beautifully simple form We are taking the view that the Lorentz-force law is an experimental fact, valid in all inertial frames. In the equation (15.144), K and u are known to be four-vectors, and the charge q is a scalar. It follows from the quotient rule quoted below Equation (15.138) that T is a four-tensor, which will let us find the behavior of the electric and magnetic fields under any Lorentz transformation. 3\u00b0 For future reference, notice that 3 is an antisymmetric tensor; that is, the matrix Y is antisymmetric, satisfying 3 = Lorentz Transformation of Electric and Magnetic Fields The field tensor Y specifies the fields E and B in any given inertial frame 8. Since 3 is a four-tensor, its value in any other frame 8' is given by (15.137) as = AT A. (15.145) For any given Lorentz transformation A, it is a straightforward, though tedious, exercise to work out the right side of (15.145), and comparing the result to the definition (15.143) of we can write down the transformed fields. For example, for the standard boost with velocity v along the x 1 axis, one finds (Problem 15.105) The most striking feature of the transformations (15.146) is that they mix up the electric and magnetic fields. A configuration whose fields are purely electric in one frame 8 (B = 0 everywhere, as for any static charge distribution), will inevitably have nonzero magnetic components in some other frames 8' (B' 0). Thus we can say that 30 There are many different ways to arrive at the field tensor Y, some of which define 3 to be a four-tensor. In such an approach, the Lorentz-force equation (15.144) automatically has the form \"four-vector = four-vector,\" which guarantees the Lorentz invariance of the Lorentz-force equation. 662 Chapter 15 Special Relativity in relativity the existence of electric fields requires the existence of magnetic fields and vice versa. An important advantage of knowing the transformation properties of the elec- tromagnetic fields is this: In seeking the fields due to a certain charge and current distribution in a frame 8, it may be possible to find a frame 8' in which the fields are more easily evaluated. If this happens, then our simplest course may be to write down the fields in S' and then transform them back to the original frame 8, as in the following example. EXAMPLE 15.12 Fields of a Long Straight Current Find the E and B fields of an infinitely long, uniform line charge with density X (measured in coulombs/meter), placed on the z axis of frame S and traveling with speed v in the +z direction. The moving line charge constitutes a current I = v along the z axis, so our problem is to find the combined fields of a line charge and a line current. Let us recognize first that this can be done by elementary methods, without leaving the frame 8: Using Gauss's law we can show that the E field of the line charge is E = /p radially outward from the z axis. (Here k = 1/47r E0 is the Coulomb force constant and p is the perpendicular distance from the z axis, that is, the first of", " the coordinates p, 0, z of cylindrical polar coordinates.) Similarly, using Ampere's law, we can show the B field of the current is B = (u0/27r)Ilp in the direction given by the right-hand rule, where,u 0 is the so-called permeability of space. We can express these two well-known results compactly using the unit vectors of cylindrical polar coordinates: E 2kX '13 p and / B \u2014 \u20140. 271- p (15.147) Both of these fields are sketched in Figure 15.16(a). While the derivation using Gauss's and Ampere's laws is perfectly straight- forward, it is instructive to rederive the same results by transforming to a frame 8' traveling with the charges. In 8', there is no current, so the only field is the radial electric field, E' = I p', as shown in Figure 15.16(b). This field is in the direction of the unit vector (x' I p', y' I p', 0), so can be written as E' = P' = 2kX' p /2 (x l, y', 0). (15.148) Before we transform this back to the original frame 8, we must recognize that the charge densities X and X' are not equal: The total charge contained in any given segment of the z axis must be the same in either frame (invariance of charge), so that X Az = A'Az', but, because of length contraction, Az = Az' I y. Therefore X = y X'. (15.149) Section 15.18 Electrodynamics and Relativity 663 (a) Frame 8 (b) Frame 5' Figure 15.16 The fields produced by a line charge on the z axis. (a) In frame 8, the line charge is moving up, out of the page. This constitutes a current, which produces a B field looping around the z axis \u2014 in addition to the E field, which is radially out from the z axis. (b) The frame S' is the rest frame of the charges, so there is no current and hence no B field \u2014 just the radial E field. We must now transform the fields E', given by (15.148), and B' = 0 back to frame S. To do this we note first that 8' is traveling along the z axis of S (not the x axis, as in the standard boost). Thus we must first rewrite (15.146) for a boost along the z axis. We must then find the inverse of this transformation (since we want the unprimed fields in terms of the primed). The result is, as you can easily check, E1 = y(E-E 13cB2), E2 = y(E'2 \u2014 PcB), B 1 = y(.B \u2014 8 Elc), B2 = y(13;,8 c), E3 = E'3 B3 = B. (15.150) Substituting B' = 0 and the components of E' from (15.148), we find that E = y (x,y,0)= 21(A' p 2 2kX \u201e p. (15.151) In writing the first equality I used the fact that x and y, and hence p = -/x 2 + y 2, are invariant under a boost in the z direction; in the second, I replaced y X' by X. This agrees exactly with the E field found in (15.147). Similarly, substituting B' = 0 and (15.148) into the expressions for B in (15.150), we find the magnetic field B = y,8 ( y, x,. 210: c2 If we make the replacements y).' = X, )9 = v/c, k/c 2 = 1/ (47 E oc2) =,u,o 147r,, and (\u2014y/p, x/p, 0) = 0, this becomes B = XV di p 27r p (15.152) 664 Chapter 15 Special Relativity and, since )\u2018.v = I, the current, this is exactly the same as the B field in (15.147). The remarkable feature of this derivation of B is that it made no reference to Ampere's law. Gauss's law in the frame 8', combined with the Lorentz transformation of the fields, has given us the result that we normally see as an expression of Ampere's law. With this striking example of the behavior of the electromagnetic field under the Lorentz transformation, I must end our brief foray into relativistic electrodynamics. You can explore a few more aspects in the problems at the end of this chapter,", " and after that you could read the excellent books of Griffiths and Jackson. 31 Principal Definitions and Equations of Chapter 15 Time Dilation If two events, as observed in frame 8 0, occur at the same place and are separated by a time At\u00b0, then the time between them as measured in any other frame S is At = y At0 [Eq. (15.11)] where y = 1/ N/1 \u2014 p2, p = v lc, and V is the speed of 8 relative to 8 0. Length Contraction If, as observed in frame 8 0, a body is at rest and has length / 0, then its length measured in a frame S traveling with velocity V in the direction of the length is / = /0 /y. [Eq. (15.15)] Lengths perpendicular to V are unchanged. The Lorentz Transformation The coordinates of any one event as measured in two frames (in standard configura- tion) are related by the Lorentz transformation: x' = y (x \u2014 Vt) Y' z = z t' = y(t \u2014 Vx/c 2). [Eq. (15.20)] 31 Chapter 12 of David J. Griffiths, Introduction to Electrodynamics, (third edition, Prentice Hall, 1999) is at approximately the level of this book but naturally emphasizes electrodynamics much more heavily. J. D. Jackson's Classical Electrodynamics (third edition, John Wiley, 1998) is a graduate text, which you could tackle after reading Griffiths' book. Principal Definitions and Equations of Chapter 15 665 The inverse Lorentz transformation is obtained by exchanging primed and unprimed variables and changing the sign of V. The Velocity-Addition Formula The velocities of a single object as measured in two frames (in standard configuration) are related by the velocity -addition formula / = V x v x \u2014 V 1 \u2014 v.\u201e, V/c2'V/ = 3' v Y y( 1 \u2014 vx V/c2) y(1 \u2014 vx V/c2) and v' = Z vz [Eqs. (15.26) & (15.27)] Four-Vectors If we rewrite the coordinates (x, y, z) as (x 1, x2, x3) and introduce x4 = ct, then the four-vectors x = (x 1, x 2, x3, x4) label points in a four-dimensional space\u2014time. If we agree to arrange the components of x in a 4 x 1 column, then Lorentz transformations become \"rotations\" of the form x' = Ax, where A is a 4 x 4 matrix. A four-vector is any set of four numbers, q = (q 1, q2, q3, q4) (one set for each inertial frame) which transform this way, q' = Aq [Section 15.8] The Invariant Scalar Product The scalar product of two four-vectors x and y is defined as x = xiYi x2Y2 x3Y3 x4Y4 [Eq. (15.50)] and is invariant under all Lorentz transformations. The scalar product of a vector with itself is often written as x x = x 2. The Light Cone The light cone of a point Q in space\u2014time consists of all light rays through Q; equivalently, it contains all points P with (x p \u2014 x0 2 = 0. [Section 15.10] The Relativistic Doppler Effect Light from a source traveling with velocity V relative to frame S is observed at an angle 9 (0 = angle between V and the ray of light). If the frequency of the light, as measured in the source's rest frame is coo the frequency observed in S is coo = y(1 \u2014 cos 0). [Eq. (15.64)] 666 Chapter 15 Special Relativity Mass, Four-Velocity, Momentum, and Energy The (invariant) mass of an object is defined to be its rest mass. The four-velocity is u = \u2014 = y(v, c). [Eqs. (15.66) & (15.67)] dx dto The four-momentum is p = mu = (ymv, ymc) = (p, E I c). [Eqs. (15.68), (15.70), & (15.75)] Three Useful Relations = pc/E, p p = \u2014(mc) 2, and E2 = (mc2) 2 (pc) 2. [Eqs. (15.83)\u2014(15.85)] Three-Force and Four-Force The three-force F and four-force K on a particle are F \u2014 and K", " = [Eqs. (15.100) & (15.107)] dp dt dp dto. Massless Particles With m = 0, a massless particle has E = Iplc, v = c, and p 2 = 0. [Eqs. (15.111)\u2014(15.113)] Transformation of the Electromagnetic Fields Under the standard boost, the electric and magnetic fields transform as follows: = E1, E2 = y (E2 \u2014 8cB 3), E 13 = y (E3 13cB 2) 13= B1, B2 = y (B 2 \u00b1,E3/c), 13= y (B3 \u2014,8E 21c) [Eq. (15.146)] Problems for Chapter 15 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). SECTION 15.2 Galilean Relativity 15.1 * Using arguments similar to those of Section 15.2, prove that Newton's first and third laws are invariant under the Galilean transformation. Problems for Chapter 15 667 15.2 ** Consider a classical inelastic collision of the form A -I- B D. (For example, this could be a collision such as Na + Cl \u2014> Na+ + Cl \u2013 in which two neutral atoms exchange an electron and become oppositely charged ions.) Show that the law of conservation of classical momentum is invariant under the Galiliean transformation if and only if total mass is conserved \u2014 as is certainly true in classical mechanics. (We shall find in relativity that the classical definition of momentum has to be modified and that total mass is not conserved.) SECTION 15.4 The Relativity of Time; Time Dilation 15.3 * A low-flying earth satellite travels at about 8000 m/s. What is the factor y for this speed? As observed from the ground, by how much would a clock traveling at this speed differ from a ground-based clock after one hour (as measured by the latter)? What is the percent difference? 15.4 * What is the factor y for a speed of 0.99c? As observed from the ground, by how much would a clock traveling at this speed differ from a ground-based clock after one hour (one hour as measured by the latter, that is)? 15.5 * A space explorer A sets off at a steady 0.95c to a distant star. After exploring the star for a short time, he returns at the same speed and gets home after a total absence of 80 years (as measured by earth-bound observers). How long do A's clocks say that he was gone, and by how much has he aged as compared to his twin B who stayed behind on earth? [Note: This is the famous \"twin paradox.\" It is fairly easy to get the right answer by judicious insertion of a factor of y in the right place, but to understand it, you need to recognize that it involves three inertial frames: the earth-bound frame 5, the frame 8' of the outbound rocket, and the frame 8\" of the returning rocket. Write down the time dilation formula for the two halves of the journey and then add. Notice that the experiment is not symmetrical between the two twins: A stays at rest in the single inertial frame 5, but B occupies at least two different frames. This is what allows the result to be unsymmetrical.] 15.6 * When he returns his Hertz rent-a-rocket after one week's cruising in the galaxy, Spock is shocked to be billed for three weeks' rental. Assuming that he traveled straight out and then straight back, always at the same speed, how fast was he traveling? (See note to Problem 15.5.) 15.7 ** The muons created by cosmic rays in the upper atmosphere rain down more-or-less uniformly on the earth's surface, although some of them decay on the way down, with a half-life of about 1.5,u,s (measured in their rest frame). A muon detector is carried in a balloon to an altitude of 2000 m, and in the course of an hour detects 650 muons traveling at 0.99c toward the earth. If an identical detector remains at sea level, how many muons should it register in one hour? Calculate the answer taking account of the relativistic time dilation and also classically. (Remember that after n half-lives, 2' of the original particles survive.) Needless to say, the relativistic answer agrees with experiment. 15.8 ** The pion (n- \u00b1 or 7r \u2013 ) is an unstable particle that decays with a proper half-life of 1.8 x 10 -8 s. (This is the half-life measured in the pion's rest frame.) (a) What is the pion's half-life measured in a frame S where it", " is traveling at 0.8c? (b) If 32,000 pions are created at the same place, all traveling at this same speed, how many will remain after they have traveled down an evacuated pipe of length d = 36 m? Remember that after n half-lives, 2' of the original particles survive. (c) What would the answer have been if you had ignored time dilation? (Naturally it is the answer (b) that agrees with experiment.) 668 Chapter 15 Special Relativity 15.9 ** One way to set up the system of synchronized clocks in a frame 8, as described at the beginning of Section 15.4, would be for the chief observer to summon all her helpers to the origin 0 and synchronize their clocks there, and then have them travel to their assigned positions very slowly. Prove this claim as follows: Suppose a certain observer is assigned to a position P at a distance d from the origin. If he travels at constant speed V, when he reaches P how much will his clock differ from the chief's clock at 0? Show that this difference approaches 0 as V 0. 15.10 *** Time dilation implies that when a clock moves relative to a frame 8, careful measurements made by observers in S will find that the clock is running slow. This is not at all the same thing as saying that a single observer in 8 will see the clock running slow, and this latter statement is not always true. To understand this, remember that what we see is determined by the light as it arrives at our eyes. Consider an observer standing close beside the x axis as a clock approaches her with speed V along the axis. As the clock moves from position A to B, it will register a time A t\u201e, but as measured by the observer's helpers, the time between the two events (\"clock at A\" and \"clock at B\") is At = y Ato. However, since B is closer to the observer than A is, the light from the clock at B will reach the observer in a shorter time than will the light from A. Therefore, the time At see see between the observer's seeing the clock at A and seeing it at B is less that At. (a) Prove that Ats\u201e = At (1 \u2014 /3) = At o (which is less than A t o). Prove both equalities. (b) What time will the observer see once the clock has passed her and is moving away? The moral of this problem is that you must be careful how you state or think about time dilation. It's fine to say \"Moving clocks are observed, or measured, to run slow,\" but it is definitely wrong to say \"Moving clocks are seen to run slow.\" SECTION 15.5 Length Contraction be 80 cm. What is v? 15.11 * As a meter stick rushes past me (with velocity v parallel to the stick), I measure its length to 15.12 ** Consider the experiment of Problem 15.8 from the point of view of the pions' rest frame. What is the half-life of the pions in this frame? In part (b), how long is the pipe as \"seen\" by the pions and how long does it take to pass the pions? How many pions remain at the end of this time? Compare with the answer to Problem 15.8 and describe how the two different arguments led to the same result. 15.13 ** (a) A meter stick is at rest in frame S o, which is traveling with speed V = 0.8c in the standard configuration relative to frame S. (a) The stick lies in the xoyo plane and makes an angle Bo = 60\u00b0 with the xo axis (as measured in S o). What is its length 1 as measured in 8, and what is its angle B with the x axis? [Hint: It may help to think of the stick as the hypotenuse of a 30-60-90 triangle of plywood.] (b) What is 1 if 19 = 60\u00b0? What is 80 in this case? 15.14 *** Like time dilation, length contraction cannot be seen directly by a single observer. To explain this claim, imagine a rod of proper length lo moving along the x axis of frame S and an observer standing away from the x axis and to the right of the whole rod. Careful measurements of the rod's length at any one instant in frame S would, of course, give the result / = / o/ y. (a) Explain clearly why the light which reaches the observer's eye at any one time must have left the two ends A and B of the rod at different times. (b) Show that the observer would see (and a camera would record) a length more than 1. [It helps to imagine that the x axis is marked with a graduated scale.] (c) Show that if the observer is standing close beside the track, he will", " see a length that is actually more than lo ; that is, the length contraction is distorted into an expansion. Problems for Chapter 15 669 SECTION 15.6 The Lorentz Transformation 15.15 * Solve the Lorentz transformation equations (15.20) to give x, y, z, t in terms of x', y', z', t'. Verify that you get the inverse Lorentz tranformation (15.21). Observe that you could have found the same result by interchanging primed and unprimed variables and changing V to \u2014V. 15.16 * Consider two events that occur at positions r 1 and r2 and times tl and t2. Let Ar = r2 \u2014 r 1 2 \u2014 t1. Write down the Lorentz transformation for r 1 and t 1, and likewise for r 2 and t2, and and At = t deduce the transformation for Ar and At. Notice that differences Ar and At transform in exactly the same way as r and t. This important property follows from the linearity of the Lorentz transformation. 15.17 * Consider two events that occur simultaneously at t = 0 in frame 3, both on the x axis at x = 0 and x = a. (a) Find the times of the two events as measured in a frame S' traveling in the positive direction along the x axis with speed V. (b) Do the same for a second frame 3\" traveling at speed V but in the negative direction along the x axis. Comment on the time ordering of the two events as seen in the three different frames. This startling result is discussed further in Section 15.10. 15.18 ** Use the inverse Lorentz transformation (15.21) to rederive the time-dilation formula (15.8). [Hint: Consider again the thought experiment of Figure 15.3, with the flash and the beep that occur at the same positions as seen in frame 3'.] 15.19 ** A traveler in a rocket of proper length 2d sets up a coordinate system S' with its origin 0' anchored at the exact middle of the rocket and the x' axis along the rocket's length. At t' = 0 she ignites a flashbulb at 0'. (a) Write down the coordinates xF, tF and x/3', tB for the arrival of the light at the front and back of the rocket. (b) Now consider the same experiment as observed from a frame S relative to which the rocket is traveling with speed V (with S and 3' in the standard configuration). Use the inverse Lorentz transformation to find the coordinates x F, tF and xB, tB for the arrival of the two signals. Explain clearly in words why the two arrivals are simultaneous in 3' but not in S. This phenomenon is called the relativity of simultaneity. SECTION 15.7 The Relativistic Velocity-Addition Formula 15.20 * Newton's first law can be stated: If an object is isolated (subject to no forces), then it moves with constant velocity. We know that this is invariant under the Galilean transformation. Prove that it is also invariant under the Lorentz transformation. [Assume that it is true in an inertial frame 3, and use the relativistic velocity-addition formula to show that it is also true in any other 3'.] 15.21 * A rocket traveling at speed 4c relative to frame S shoots forward bullets traveling at speed ic relative to the rocket. What is the speed of the bullets relative to 3? 15.22 * A rocket is traveling at speed 0.9c along the x axis of frame 3. It shoots a bullet whose velocity v' (measured in the rocket's rest frame 3') is 0.9c along the y' axis of 3'. What is the bullet's velocity (magnitude and direction) as measured in 5? 15.23 * As seen in frame 3, two rockets are approaching one another along the x axis traveling with equal and opposite velocities of 0.9c. What is the velocity of the rocket on the right as measured by observers in the one on the left? [This and the previous two problems illustrate the general result that in relativity the \"sum\" of two velocities that are less than c is always less than c. See Problem 15.43.] 670 Chapter 15 Special Relativity 15.24 * A robber's getaway vehicle, which can travel at an impressive 0.8c, is pursued by a cop, whose vehicle can travel at a mere 0.4c. Realizing that he cannot catch up with the robber, the cop tries to shoot him with bullets that travel at 0.5c (relative to the cop). Can the cop's bullets hit the robber? 15.25 * A rocket is traveling at speed V along the x axis of frame 8. It emits a signal (for example", ", a pulse of light) that travels with speed c along the y' axis of the rocket's rest frame 8'. What is the speed of the signal as measured in 8? 15.26 * Two objects A and B are approaching one another, traveling in opposite directions along the x axis of frame S with speeds v A and v B. At time t = 0, they are at positions x = 0 and x = d. Write down their positions for an arbitrary time t and show that they meet at time t = d/(v A v B ). Notice that this implies that the relative velocity of the two objects is V A vB as measured in the frame 8 in which they are both moving. This may seem surprising at first thought, since we can clearly choose values of V A and v B for which this relative velocity is larger than c.32 15.27 *** Frame 8' travels at speed V 1 along the x axis of frame 8 (in the standard configuration). Frame 8\" travels at speed V2 along the x' axis of frame 8' (also in the standard configuration). By applying the standard Lorentz transformation twice find the coordinates x\", y\", z\", t\" of any event in terms of x, y, z, t. Show that this transformation is in fact the standard Lorentz transformation with velocity V given by the relativistic \"sum\" of V 1 and V2. 15.28 *** The relativistic velocity-addition formula is the answer to the following question: If u is the velocity of an inertial observer B relative to an observer A, and v is the velocity of C relative to B, what is the velocity w of C relative to A? Let us denote the answer by w = \"u v.\" In classical physics, this is just the ordinary vector sum of u and v; in relativity, it is given by the inverse of the velocity addition formulas (15.26) and (15.27) (at least for the case that u points along the x axis). Taking u = (u, 0, 0) and v = (0, v, 0), write down the components of \"u + v\" and also of \"v + u.\" [Be careful to distinguish between the y factors y u and y, pertaining to u and v.] Show that \"u + v\" A \"v + u,\" but that the two vectors have equal magnitudes and differ only by a rotation about the z axis. This rotation is sometimes called the Wigner rotation and is the cause of the so-called Thomas precession, which has an important effect on the fine structure of atomic energy levels. SECTION 15.8 Four-Dimensional Space\u2014Time; Four-Vectors 15.29 * (a) Find the 3 x 3 matrix R(0) that rotates three-dimensional space about the x 3 axis, so that e l rotates through angle 0 toward e2. (b) Show that [R(0)] 2 = R(20), and interpret this result. 15.30 * The \"angle\" 0 introduced in connection with Equation (15.40) has several useful properties. For any speed v < c (with corresponding factors,8 and y) we can define 0 so that y = cosh 4). Defined in this way, 0 is called the rapidity corresponding to v. Prove that sinh 4 = py and that tanh = 13. 15.31 * Here is a handy property of the rapidity introduced in Problem 15.30: Suppose that observer B has rapidity 0 1 as measured by A and that C has rapidity 4)2 as measured by B (with both velocities along the x axis). That is, the speed of B relative to A has S u = tanh O l and so on. Prove that the rapidity of C as measured by A is just 4) = 41 + 02. 32 Nevertheless, this does not violate any principle of relativity. We shall see that no single object can have speed greater than c relative to any inertial frame, but there is nothing to prohibit two objects from having relative speed greater than c as measured in a frame where both objects are moving. Problems for Chapter 15 671 15.32 * In Section 15.8, I claimed that the 4 x 4 matrix AR corresponding to a pure rotation has the block form (15.44). Verify this claim by writing out the separate components of the equation x' = A Rx and showing that the spatial part (x 1, x2, x3) is rotated, while x4 is unchanged. 15.33 ** (a) By exchanging x 1 and x2, write down the Lorentz transformation for a boost of velocity V along the x2 axis and the corresponding 4 x 4 matrix AB2. (b) Write down the 4 x 4 matrices AR + 1x2 plane through +7/2, with the angle of rotation measured and AR_ that represent", " rotations of the x counterclockwise. (c) Verify that AB2 = A R_ A B l AR\u00b1, where AB1 is the standard boost along the x 1 axis, and interpret this result. 15.34 ** Let AB(0) denote the 4 x 4 matrix that gives a pure boost in the direction that makes an angle 0 with the x 1 axis in the x 1x2 plane. Explain why this can be found as A B (0) = A R (-0)A B (0)AR (0), where A R (0) denotes the matrix that rotates the x 1 x2 plane through angle 0 and A B (0) is the standard boost along the x 1 axis. Use this result to find A B (0) and check your result by finding the motion of the spatial origin of the frame S as observed in 8'. 15.35 ** Prove the following useful result, called the zero-component theorem: Let q be a four-vector, and suppose that one component of q is found to be zero in all inertial frames. (For example, q4 = 0 in all frames.) Then all four components of q are zero in all frames. SECTION 15.9 The Invariant Scalar Product 15.36 * We have seen that the scalar product x x of any four-vector x with itself is invariant under Lorentz transformations. Use the invariance of x x to prove that the scalar product x y of any two four-vectors x and y is likewise invariant. 15.37 * Verify directly that x' y' = x y for any two four-vectors x and y, where x' and y' are related to x and y by the standard Lorentz boost along the x 1 axis. 15.38 ** As an observer moves through space with position x(t), the four-vector (x(t), ct) traces a path through space\u2014time called the observer's world line. Consider two events that occur at points P and Q in space\u2014time. Show that if, as measured by the observer, the two events occur at the same time t, then the line joining P and Q is orthogonal to the observer's world line at the time t; that is, (x p \u2014 x Q ) dx = 0, where dx joins two neighboring points on the world line at times t and t + dt. SECTION 15.10 The Light Cone 15.39 * Suppose that a point P in space\u2014time with coordinates x = (x, x4 ) lies inside the backward light cone as seen in frame S. This means that x - x < 0 and x 4 < 0 at least in frame S. Prove that these two conditions are satisfied in all frames. Since this means that all observers agree that t < 0, this justifies calling the inside of the backward light cone the absolute past. 15.40 * Show that the statement that a point x in space\u2014time lies on the forward light cone is Lorentz invariant. 15.41 * In the proposition on page 627, it is obvious that at least one of the three statements has to be true. In the proof given there, I showed that if statement (1) is true, then so are statements (2) and (3). To complete the proof, show that (2) implies (1) and (3). [Strictly speaking you should also check that (3) implies (1) or (2), but this is so similar to the argument already given that you needn't bother.] 15.42 * Prove that if x is time-like and x y = 0, then y is space-like. 672 Chapter 15 Special Relativity 15.43 * (a) Show that if a body has speed v < c in one inertial frame, then v < c in all frames. [Hint: Consider the displacement four-vector dx = (dx, c dt), where dx is the three-dimensional displacement in a short time dt.] (b) Show similarly that if a signal (such as a pulse of light) has speed c in one frame, its speed is c in all frames. 15.44 ** (a) Show that if q is time-like, there is a frame 8' in which it has the form q' (0, 0, 0, (4). (b) Show that if q is forward time-like in one frame 8, then it is forward time-like in all inertial frames. SECTION 15.11 The Quotient Rule and Doppler Effect 15.45 * The quotient rule derived at the start of Section 15.11 is only one of several similar quotient rules. Here is another. Suppose that k and x are both known to be four-vectors and that in every inertial frame k is a multiple of x. That is, k = Ax", " in frame 8, and k' = A'x' in frame 8', and so on. Then the factor A (the \"quotient\" of k and x) is in fact a four-scalar with the same value in all frames, A = A'. Prove this quotient rule. 15.46 * (a) Show that in the case that the source is approaching the observer head on, the Doppler formula (15.64) can be rewritten as w = w0 V (1 + 13)1(1- p). (b) What is the corresponding result for the case that the source is moving directly away from the observer? 15.47 * Consider the tale of the physicist who is ticketed for running a red light and argues that because he was approaching the intersection, the red light was Doppler shifted and appeared green. How fast would he have to have been going? (A rea ;--'=, 650 nm and Agree\u201e 530 nm.) 15.48 ** The factor y in the Doppler formula (15.64), which can be ascribed to time dilation, means that even when 0 = 90\u00b0 there is a Doppler shift. (In classical physics there is no Doppler shift when 6 = 90\u00b0 and the source has zero velocity in the direction of the observer.) This transverse Doppler shift is therefore a test of time dilation, and has yielded some very accurate tests of the theory. However, except when the source is moving very close to the speed of light, the transverse shift is quite small. (a) If V = 0.2c, what is the percentage shift when 6 = 90\u00b0? (b) Compare this with the shift when the source approaches the observer head-on. SECTION 15.12 Mass, Four-Velocity, and Four-Momentum 15.49 * Show that the four-velocity of any object has invariant length squared u u = \u2014c 2 15.50 * For any two objects a and b, show that the scalar product of their four-velocities is u a ub = \u2014 C 2 Y (vrei), where y (v) denotes the usual y factor, y (v) = 1/V1 \u2014 v 2 /c2, and vrei denotes the speed of a in the rest frame of b or vice versa. 15.51 ** (a) For the collision shown in Figure 15.11, verify that all four components of the total four- momentum pa + Pb [with the individual momenta defined relativistically as in (15.68)] are conserved in the frame S of part (a). (b) In two lines or less, prove that total four-momentum is conserved in the frame 8' of part (b). [This problem does not, of course, prove that the law of conservation of four- momentum is generally true, but it does at least show that the law is consistent with the collision of Figure 15.11.] 15.52 ** (a) Suppose that the total three-momentum P = E p of an isolated system is conserved in all inertial frames. Show that if this is true (which it is), then the fourth component P4 of the total four-momentum P = (P, P4) has to be conserved as well. (b) Using the zero-component theorem of Problem 15.35, you can prove the following stronger result very quickly: If any one component of the total four-momentum P is conserved in all frames, then all four components are conserved. Problems for Chapter 15 673 15.53 ** For any two objects a and b, show that Pa Pb = maEb = mbEa = mambc2 Y (vrel) where m a is the mass of a, and Eb is the energy of b in a's rest frame, and vice versa, and v ro is the speed of a in the rest frame of b (or vice versa). 15.54 *** (a) Using the correct relativistic velocity-addition formulas make a table showing the four velocities as seen in frame S' of the collision of Figure 15.11(b) in terms of the initial velocity of a in S. [Give the latter some simple name, such as v a 0).] (b) Add a column showing the total classical momentum m a y', m b vb before and after the collision, and show that the y component of the classical momentum is not conserved in 8'. 15.55 *** Since the four-velocity u = y (v, c) is a four-vector its transformation properties are simple. Write down the standard Lorentz boost for all four components of u. Use these to deduce the relativistic velocity-addition formula for v. SECTION", " 15.13 Energy, the Fourth Component of Momentum 15.56 * When oxygen combines with hydrogen in the reaction (15.80) about 5 eV of energy is released (that is, the kinetic energy of the two final molecules is 5 eV more than that of the initial three molecules). (a) By how much does the total rest mass of the molecules change? (b) What is the fractional change in total mass? (c) If one were to form 10 grams of water by this reaction, what would be the total change 15.57 * When a radioactive nucleus of astatine 215 decays at rest, the whole atom is torn into two in in mass? the reaction 215At 211,, 151 + He. 4 The masses of the three atoms are (in order) 214.9986, 210.9873, and 4.0026, all in atomic mass units. (1 atomic mass unit = 1.66 x 10 -27 kg = 931.5 MeV/c2.) What is the total kinetic energy of the two outcoming atoms, in joules and in MeV? 15.58 * (a) What is a particle's speed if its kinetic energy T is equal to its rest energy? (b) What if its energy E is equal to n times its rest energy? 15.59 * If one defines a variable mass m var = ym, then the relativistic momentum p = ymv becomes m vary which looks more like the classical definition. Show, however, that the relativistic kinetic energy is not equal to im var v2. 15.60 * A particle of mass m a decays at rest into two identical particles each of mass m b. Use conservation of momentum and energy to find the speed of the outgoing particles. 15.61 * A particle of mass 3 MeV/c 2 has momentum 4 MeV/c. What are its energy (in MeV) and speed (in units of c)? 15.62 * A particle of mass 12 MeV/c 2 has a kinetic energy of 1 MeV. What are its momentum (in MeV/c) and its speed (in units of c)? 15.63 * (a) What is a mass of 1 MeV/c 2 in kilograms? (b) What is a momentum of 1 MeV/c in kgm/s? 674 Chapter 15 Special Relativity 15.64 * As measured in the inertial frame 8, a proton has four-momentum p. Also as measured in 8, an observer at rest in a frame 8' has four-velocity u. Show that the proton's energy, as measured by this observer, is \u2014u p. 15.65 ** The relativistic kinetic energy of a particle is T = (y \u2014 1)m c2. Use the binomial series to express T as a series in powers of /3 = v/c. (a) Verify that the first term is just the nonrelativistic kinetic energy, and show that to lowest order in /3 the difference between the relativistic and nonrelativistic kinetic energies is 3,8 4mc2 /8. (b) Use this result to find the maximum speed at which the nonrelativistic value is within 1% of the correct relativistic value. 15.66 ** In nonrelativistic mechanics, the energy contains an arbitrary additive constant \u2014 no physics is changed by the replacement E E+ constant. Show that this is not the case in relativistic mechanics. [Hint: Remember that the four-momentum p is supposed to tranform like a four-vector.] SECTION 15.14 Collisions 15.67 * Two balls of equal masses (in each) approach one another head-on with equal but opposite velocities of magnitude 0.8c. Their collision is perfectly inelastic, so they stick together and form a single body of mass M. What is the velocity of the final body and what is its mass M? 15.68 * Consider the elastic, head-on collision of Example 15.10, in which two particles (masses m a and m b) approach one another traveling along the x axis, collide, and emerge traveling along the same axis. In the CM frame (by its definition) p ian = \u2014pibn. Use conservation of momentum and energy to prove that p afin = \u2014pia; that is, the momentum of particle a (and likewise b) just reverses itself in the CM frame. 15.69 * (a) Show that the four-momentum of any material particle (m > 0) is forward time-like. (b) Show that the sum of any two forward time-like vectors is itself forward time-like, and hence that the sum of any number of forward time-like vectors is itself forward time-like. 15.", "70 * (a) Use the results of Problem 15.69 to prove that for any number of material particles there exists a CM frame, that is, a frame in which the total three-momentum is zero. (b) Relative to an arbitrary frame 8, show that the velocity of the CM frame is given by /3 = E pc/ E E. 15.71 * One way to create exotic heavy particles is to arrange a collision between two lighter particles ad-b-->d\u00b1ed- +g where d is the heavy particle of interest and e,-, g are other possible particles produced in the reaction. (A good example of such a process is the production of the particle in the process e+ e+ in which there are no other particles e,-, g.) (a) Assuming that and is much heavier that any of the other particles, show that the minimum (or threshold) energy to produce this reaction in the CM frame is Ecm ti m dc2. (b) Show that the threshold energy to produce the same reaction in the lab frame, where the particle b is initially at rest, is Elab md2c2 /2m b. (c) Calculate these two energies for the process e+ e- \u2014>, with m e ', 0.5 MeV/c 2 and m * P-', 3100 MeV/c2. Your answers should explain why particle physicists go to the trouble and expense of building colliding-beam experiments. 15.72 * A mad physicist claims to have observed the decay of a particle of mass M into two identical particles of mass m, with M < 2m. In response to the objections that this violates conservation of energy, he replies that if M was traveling fast enough it could easily have energy greater than 2mc 2 and hence could decay into the two particles of mass m. Show that he is wrong. [He has forgotten that both Problems for Chapter 15 675 energy and momentum are conserved. You can analyse this problem in terms of these two conservation laws, but it is much simpler to go to the rest frame of M.] 15.73 ** Consider the head-on elastic collision of Example 15.10, in which the final velocity vb of particle b is given by (15.95). (a) Show that, in the special case that the masses are equal (m a = mb), v b = va, the initial velocity of particle a. Show that in this case the final velocity of a is zero. [This result for equal-mass collisions is well known in classical mechanics; you have now shown that it extends to relativity.] (b) Show that in the nonrelativistic limit (15.95) reduces to v b = 2va m a / (m a m b). By doing the necessary nonrelativistic calculations, show that this agrees with the nonrelativistic answer for elastic head-on collisions. 15.74 ** A particle a traveling along the positive x axis of frame S with speed 0.5c decays into two identical particles, a b b, both of which continue to travel on the x axis. (a) Given that m a = 2.5m b, find the speed of either b particle in the rest frame of particle a. (b) By making the necessary transformation on the result of part (a), find the velocities of the two b particles in the original frame S. 15.75 ** A particle of unknown mass M decays into two particles of known masses m a = 0.5 GeV/c 2 m b = 1.0 GeV/c 2, whose momenta are measured to be pa = 2.0 GeV/c along the x 2 axis and and pb = 1.5 GeV/c along the x 1 axis. (1 GeV = 10 9 eV.) Find the unknown mass M and its speed. 15.76 ** Particle a is pursuing particle b along the x 1 axis of a frame S. The two masses are m a and m b and the speeds are I), and v b (with I), > v b). When a catches up with b they collide and coalesce to form a single particle of mass m and speed v. Show that M2 = mQ M b 2 Ll anl bY (V a ))/ (V b)(1 \u2014 V a l) b I C 2) and find v. 15.77 *** Consider the elastic head-on collision of Example 15.10, in which particle a collides with a stationary particle b. Assuming that ma A m b, show that the final kinetic energy of particle a satisfies Tafin < (ma mo2c2/2mb. [Hint: Look at the CM frame where you can show that the four-vector pin \u2014 pbin is time-like, so that (p afin \u2014 pbin) 2 < O.] (b) The result", " of part (a) implies that if Tain is large, almost all this incoming energy is lost to b. This is quite different from the nonrelativistic situation. Prove that in nonrelativistic mechanics the proportion of kinetic energy retained by a is fixed, independent of Tr. Specifically, Tafin = Tanana mb)2/(ma mo2. 15.78 *** Consider the elastic collision shown in Figure 15.17. In the lab frame 8, particle b is initially at rest; particle a enters with four-momentum pa and scatters through an angle 8; particle b recoils at an angle 'ilr. In the CM frame 8', the two particles approach and emerge with equal and opposite momenta, and particle a scatters through an angle 0'. (a) Show that the velocity of the CM frame relative to the lab frame is V = pa c2 / (Ea m bc2). (b) By transforming the final momentum of a back from the CM to the lab frame, show that tan 6 = sin 0' yv (cos 0' V/vt) (15.153) where va` is the speed of a in the CM frame. (c) Show that in the limit that all speeds are much smaller than c, this result agrees with the nonrelativistic result (14.53) (where X = m a /m b). (d) Specialize now to the case that m a = m b. Show that, in this case, V/ v cc = 1, and find a formula like (15.153) for tan Vr. 676 Chapter 15 Special Relativity a lab frame 8 CM frame 8' Figure 15.17 Problem 15.78. (e) Show that the angle between the two outgoing momenta is given by tan(6 = 2/(6y v sin 0'). Show that in the limit that V <.< c, you recover the well-known nonrelativistic result that 9 + = 90\u00b0. SECTION 15.15 Force in Relativity 15.79 * Consider an object of mass m (which you may assume is constant), acted on by a force F. From the definition (15.100) prove that F = yma + (F v)v/c 2, where a = dv/dt is the object's acceleration. Notice that it is certainly not true in relativity that F = ma. Nor is it true that F = m vara, where m var is the variable mass mvar = Ym, except in the special case that F happens to be perpendicular to v. In general, F and a are not even in the same direction. 15.80 * A particle of mass m and charge q moves in a uniform, constant magnetic field B. Show that if v is perpendicular to B, the particle moves in a circle of radius r =11)103 1. (15.154) [This result agrees with the nonrelativistic result (2.81), except that p is now the relativistic momentum p = ymv.] 15.81 * An electron (mass 0.5 MeV/c 2) moves with speed 0.7c in a circular path in a magnetic field of 0.02 teslas. Using the relativistic result (15.154) of Problem 15.80, find the radius of the electron's orbit. What would your answer have been if you used the classical definition of momentum? [Needless to say, the relativistic result is confirmed by experiment, and this gave some of the first evidence of the correctness of relativistic mechanics. 33] 15.82 * (a) Verify the result (15.106) for the position of a particle moving in a uniform electric field, by integrating the expression (15.105). (b) When t is small, the particle should be moving slowly and 33 In a paper in Gottingen Nachrichten, p. 143 (1901), Walter Kaufmann showed that the electron's \"apparent mass\" (what we would call its variable mass) in a magnetic field seemed to increase with speed in rough accord with the relativistic formula m vai. = y (v)m. Notice that this predated Einstein's first paper on relativity by some four years. Problems for Chapter 15 677 (15.106) should agree with the nonrelativistic result x = -lat 2. Verify that it does. (c) Show that when t is large, x = F(ct + const) and explain this result. 15.83 * Starting from the definition (15.100) of the force F on an object, prove that the transformation of the components of F as we pass from a frame S to a second frame 8', traveling at speed V in the standard configuration relative to 8, is F 1 f3F 71c 1 - pd vc'F'", " = 2 - F2, = F3 y(1 - Pvilc) Y(1 \u2014 Pvilc) (15.155) where 3 fi (V) and y = y ( V) relate to the relative speed of the two frames and v is the velocity of the object as measured in S. 15.84 ** A mass m is thrown from the origin at t = 0 with initial three-momentum po in the y direction. If it is subject to a constant force F, in the x direction, find its velocity v as a function of t, and by integrating v find its trajectory. Check that in the nonrelativistic limit the trajectory is the expected parabola. 15.85 ** We have seen that there are processes in which the mass of an object varies with time. (a) Starting from (15.85), prove that dmIdto = \u2014 u K1c 2, where to is the object's proper time, u is its four-velocity, and K is the four-force on the object. (b) This means that the necessary and sufficient condition that a force doesn't change an object's mass is that u K = 0. It is an experimental fact that if a charged particle is at rest in an electromagnetic field (even instantaneously) then dE/dt = 0. Use this to argue that electromagnetic forces do not cause a particle's mass to change. SECTION 15.16 Massless Particles; the Photon 15.86 * The neutral pion rt-0 is an unstable particle (mass m = 135 MeV/c 2 ) that can decay into two photons, 7r \u00b0 y + y. (a) If the pion is at rest, what is the energy of each photon? (b) Suppose instead that the pion is traveling along the x axis and that the photons are observed also traveling along the x axis, one forward and one backward. If the first photon has three times the energy of the second, what was the pion's original speed v? 15.87 * A neutral pion (Problem 15.86) is traveling with speed v when it decays into two photons, which are seen to emerge at equal angles B on either side of the original velocity. Show that v = c cos 8. 15.88 * Two particles a and b with masses m a = 0 and m b > 0 approach one another. Prove that they have a CM frame (that is, a frame in which their total three-momentum is zero). [Hint: As you should explain, this is equivalent to showing that the sum of two four-vectors, one of which is forward light-like and one forward time-like, is itself forward time-like.] 15.89 * Show that any two zero-mass particles have a CM frame, provided their three-momenta are not parallel. [Hint: As you should explain, this is equivalent to showing that the sum of two forward light-like vectors is forward time-like, unless the spatial parts are parallel.] 15.90 ** The first positrons to be observed were created in electron\u2014positron pairs by high-energy cosmic-ray photons in the upper atmosphere. (a) Show that an isolated photon cannot convert to an electron\u2014positron pair in the process y e+ + e \u2014. [Show that this process inevitably violates conservation of four-momentum.] (b) What actually occurs is that a photon collides with a stationary nucleus with the result y + nucleus --> e + + e\u2014 + nucleus. 678 Chapter 15 Special Relativity Convince yourself that the formula (15.98) can be used to find the minimum energy for a photon to induce this reaction. [The derivation of (15.98) assumed that the incident particle had m > 0.] Show that, provided the mass of the nucleus is much greater than that of the electron, the minimum photon energy to induce this reaction is approximately 2m ec2. [This is exactly the energy one would have calculated for the process y \u2014>- e+ e \u2014 and shows that the role of the nucleus is just as a \"catalyst\" that can absorb some three-momentum.] 15.91 ** An excited state X* of an atom at rest drops to its ground state X by emitting a photon. In atomic physics it is usual to assume that the energy Ey of the photon is equal to the difference in energies of the two atomic states, A E = (M* \u2014 M)c 2, where M and M* are the rest masses of the ground and excited states of the atom. This cannot be exactly true, since the recoiling atom X must carry away some of the energy D E. Show that in fact Ey = A E[1 \u2014 A E/(2M*c 2)]. Given that A E is of order a few eV, while the lightest atom has M", " of order 1 GeV/c 2, discuss the validity of the assumption that E = 15.92 ** A positive pion decays at rest into a muon and neutrino, n-,u+ + v. The masses involved are m, = 140 MeV/c2, m = 106 MeV/c2, and m, = 0. (There is now convincing evidence that m y is not exactly zero, but it is small enough that you can take it to be zero for this problem.) Show that the speed of the outgoing muon has /3 = (m,2 \u2014 m L2)/(m n2 m A2). Evaluate this numerically. Do the same for the much rarer decay mode 7+ e+ v, (m e = 0.5 MeV/c2). 15.93 *** Consider a head-on elastic collision between a high-energy electron (energy Eo and speed i5oc) and a photon of energy Ey0. Show that the final energy Ey of the photon is E E \u00b0 2 + (1 \u2014 13(0E0 1 E yo 1 + [Hint: Use (15.123).] Show that Ey < E0, but that if /3,, 1, then E y I E0 1; that is, a very high- energy electron loses almost all its energy to the photon in a head-on collision. What fraction of its original energy would the electron retain if E e, ti 10 TeV and the photon was in the visible range, Eyo 3 eV? (Remember that the mass of the electron is about 0.5 MeV/c 2 ; 1 TeV = 10 12 eV.) SECTION 15.17 Tensors 15.94 * Prove that for any two matrices A and B, where A has as many columns as B has rows, the transpose of A B satisfies (A B) = 13' A. 15.95 * By making suitable choices for the n-dimensional vectors a and b, show that if aCb = abb for any choices of a and b (where C and D are n x n matrices), then C = D. 15.96 * Prove that if T and a are respectively a four-tensor and a four-vector, then b = Ta = T Ga is a four-vector; that is, it transforms according to the rule b' = Ab. 15.97 * (a) A tensor T is said to be symmetric if TA, = Tvii. Prove that if T is symmetric in one inertial frame, then it is symmetric in all inertial frames. (b) T is antisymmetric if Tii, Prove that if T is antisymmetric in one inertial frame, then it is antisymmetric in all inertial frames. (An example of the latter property is the electromagnetic field tensor, which is antisymmetric in all frames.) 15.98 ** (a) Use the invariance of the scalar product a b = aGb to prove that the 4 x 4 Lorentz transformation matrices A must satisfy the condition (15.136), AGA = G. (b) Verify that the standard Lorentz boost (15.43) does satisfy this condition. Problems for Chapter 15 679 15.99 ** A useful form of the quotient rule for three-dimensional vectors is this: Suppose that a and b are known to be three-vectors and suppose that for every orthogonal set of axes there is a 3 x 3 matrix T with the property that b = Ta for every choice of a, then T is a tensor. (a) Prove this. (b) State and prove the corresponding rule for four-vectors and four-tensors. 15.100 *** (a) The statement that V is a vector operator means that if 0 (x) is any scalar, then the three components of V0 = (80/8x 1, 80/ax2, 80/ax3) transform according to the three-vector transformation law (15.126). Prove this last statement. [Hint: Remember the chain rule, that 80/8x, = Ej (axyax i )ao/axi.i.i (b) Prove that in four-dimensional space\u2014time, if 0 is any four-scalar, the quantity0 4) defined with the components I=14) ao ao, _ao) ax2'ax3 ax4l (15.156) (note well the minus sign on the fourth component) is a four-vector. This result is crucial in writing down Maxwell's equations for the electromagnetic field. SECTION 15.18 Electrodynamics and Relativity 15.101 * (a) Prove that E B and E2 \u2014 c2 B 2 are", " both invariant under any Lorentz tranformation. [Use the transformation equations (15.146) to prove the required results for the standard boost and then explain why if either quantity is invariant under the standard boost then it is invariant under any Lorentz transformation.] Use these results to prove the following two propositions: (b) If E and B are perpendicular in frame 3, then they are perpendicular in any other frame 8', and (c) if E > cB in a frame 3, then there cannot exist a frame in which E = 0. 15.102 * (a) Starting from the transformation equations (15.146) for the standard boost along the x 1 3 axis. (b) Write down the inverse of this transformation axis, find the corresponding boost along the x and then verify the results (15.151) and (15.152) for the fields of a moving line charge. 15.103 * (a) Using the transformation equations (15.146), show that if E = 0 in frame 3, then E' = v x B' in 3'. (b) Similarly, show that if B = 0 in frame 3, then B' = \u2014v x E'/c 2. 15.104 ** We defined the electromagnetic field tensor by the equation K = q3 u qTGu, where K is the four-force on a charge q and u is its four-velocity. (a) Starting from the Lorentz force (15.139) write down the four components of K [as in (15.141)]. (b) Use these to find the matrix TG and show that the tensor a' has the form claimed in (15.143). 15.105 ** Since T is a four-tensor it has to transform according to the rule (15.145), = AajC. Using the form (15.143) for a' and the standard Lorentz boost for A, find the matrix 9' and verify the transformation equations (15.146) for the electromagnetic fields. 15.106 ** Derive the Lorentz-force law from Coulomb's law as follows: (a) If a charge q is at rest in frame 3', then Coulomb's law tells us that the force on q is F' = qE'. Use the inverse of the force transformation (15.155) in Problem 15.83 to write down the force F as seen in S. (Answer in terms of E' for now.) (b) Now use the field transformation (15.146) to rewrite your answer in terms of E and B and show that F = q(E v x B). 15.107 ** It is a result well known in classical electromagnetism that one can introduce a three-scalar potential 0 and a three-vector potential A such that the fields E and B can be written as E = \u2014 Wp \u2014 \u2014 aA and B=V x A. (15.157) At 680 Chapter 15 Special Relativity In relativity these potentials are combined to form a single four-potential A = (A, 0/c). Prove that = q 4 A \u2014 q u A i, where q is the four-dimensional gradient operator defined in (15.156) of Problem 15.100. (If we accept that A really is a four-vector, this gives an alternative proof that is a four tensor.) 15.108 ** Consider an electric charge distribution, with charge density Q, moving with velocity v relative to a frame S. (a) Show that p= ypo, where Q 0 is the charge density in the rest frame. (Notice that v can vary with position, so different parts of the distribution will have different rest frames, but that's all right.) (b) The three-current density is defined as J = Qv. Show that the four-current density, defined as J = (J, cp), is a four-vector. (c) It is a well-known result in electromagnetism that conservation of charge implies the so-called equation of continuity, V J.90/3t = 0 (where V J = 8J, /axi is the so-called divergence of J). Show that this condition is equivalent to the manifestly invariant condition q J = 0, where q is the four-dimensional gradient defined in (15.156) of Problem 15.100. 15.109 *** Two equal charges q are moving side-by-side in the positive x direction in frame S. The distance between them is r and their speed is v. Find the force on either charge due to the other in two ways: (a) Find the force in their rest frame 3' and transform back to 3, using the force transformation (15.155) of Problem 15.83. Note that the force in S is less than in the rest frame. (b) Find the electric and magnetic", " fields in S' and thence in 3, using the field transformation (15.146). Use these fields (in 3) to write down the Lorentz force on either charge in 3. Note that in S there is an attractive electric force and a repulsive magnetic force. As )3 1 they become nearly equal and their resultant approaches 15.110 *** A charge q is moving with constant speed v along the x axis of frame 3 with position uti. (a) Write down the electric and magnetic fields in the charge's rest frame 3'. (b) Use the inverse of the field transformation (15.146) to write down the electric field in the original frame S. [In the first instance, you will find E in terms of the primed variables x', y', z', t', but you can use the standard Lorentz transformation to eliminate them in favor of x, y, z, t.] Show that the field at position r and zero. time t is E kq(1 \u2014 $ 2) R (1 \u2014 S2 sine 6) 3/2 R 2 (15.158) where R = r \u2014 vti is the vector pointing from the charge's position to the point of observation r, and 8 is the angle between R and the x axis. (c) Sketch the behavior of the field strength as a function of for fixed R, and make a sketch of the electric field lines at one fixed time t. 15.111 *** Two of Maxwell's four equations read V x B\u2014 C^ at = accoJ 1 co and V E= \u2014 Q (15.159) where J and Q are the current and charge densities that gave rise to the fields. Show that these two equations can be written as the single four-vector equation q = \u2014g o \", where q is the four- dimensional gradient operator introduced in Problem 15.100, J is the four-current (J, cQ), and the scalar product q J' = CHAPTER Continuum Mechanics We can divide classical mechanics into three main areas, in order of increasing complexity. (1) The mechanics of point masses. Occasionally, these point masses are elementary particles, such as the electron, whose mass is (as far as we know) concentrated at a point; but usually they are extended objects whose mass is certainly not localized at a point but which, for the purposes at hand, can be approximated as if it were. Thus in treating the flight of baseballs, or finding the orbits of the planets, it is an often an excellent approximation to treat them as point masses. In this case, the configuration of any system is given by a finite set of coordinates, three for each point mass. (2) The mechanics of rigid bodies. Here we acknowledge that the mass of interest is spread out over some nonzero volume, but we assume that the relative positions of the various parts of any one body are fixed; that is, all bodies are rigid. As we saw, we have to allow for the possible rotational motion of such bodies, but the configuration of any system is still specified by a discrete and finite set of coordinates; for example, for a single rigid body we need just six coordinates, three for the CM position and three more for the body's orientation. The notion of a rigid body is an idealization \u2014 all real bodies can be deformed \u2014 but for many systems it is a reasonable and extremely useful approximation. (3) Finally, there is continuum mechanics, in which we acknowledge that the mass in a system can be distributed over some region and that the relative positions of the various parts can change in a continuous, but otherwise arbitrary, manner. Clearly the motion of any fluid \u2014 the flow of air past an airplane wing or of water down a pipe -- is a problem in continuum mechanics. But so also is the motion of a solid when the independent motions of its parts are important, as for example in the flexing of a heavily loaded steel beam or the vibration of the earth's crust in response to an earthquake. In continuum mechanics a system comprises a continuously infinite number of parts, and the specification of its configuration requires an infinite number of coordinates. So far in this book we have treated only the first two topics, the mechanics of point masses and rigid bodies \u2014 what we could call discrete mechanics \u2014 with a discrete, finite number of coordinates. This final chapter is intended as the briefest of introductions to continuum mechanics. A thorough introduction would require 681 682 Chapter 16 Continuum Mechanics another book, but I hope that I can give at least a sense of some of the central ideas. Specifically, we'll see how the passage from discrete to continuum mechanics changes the ordinary differential equations of the former to partial differential equations. We'll see how these partial differential equations often lead to the wave equation, which governs the behavior of sound waves in liquids and gases, of seismic waves in the earth's crust, and of many other waves, most notably electromagnetic waves such as light and microwaves. The first three sections", " deal with one-dimensional continuous systems, and then, in Section 16.4 we move out into three dimensions. Perhaps the biggest complication in three dimensions is that the forces and displacements involve two tensors, the stress and strain tensors. One of the main objectives of this chapter is to introduce these two important concepts. Section 16.7 introduces the stress tensor for fluids and solids, and Section 16.8 the strain,tensor for solids. Section 16.9 gives the relation between the two tensors, the generalized Hooke's law. Sections 16.10 and 16.11 derive the equation of motion for an elastic solid and use it to analyze the longitudinal and transverse waves in a solid. The last two sections are a very brief introduction to the mechanics of inviscid fluids. In Section 16.12, we'll derive the equation of motion and the so-called continuity equation, and in Section 16.13 we'll use them to analyze the possible waves in a fluid. Before we get started, I should emphasize that the notion of a continuous distri- bution of matter is itself an idealization. The properties of the air flowing in a wind tunnel certainly appear to vary continuously and smoothly from place to place. For example, we are used to assuming that air has a density p(r) which gives the mass p(r)dV in a small volume d V, and, while p(r) may certainly vary with r, we assume that it does so smoothly. However, we know perfectly well that, when viewed under a super-microscope that could resolve fractions of nanometers, the air would be seen to consist of individual molecules, and the density p(r) would vary wildly between large values near to each molecule and zero in the huge spaces in between. Fortunately, the scale of these wild variations is tiny compared to the scales of normal interest. For ex- ample, even if we were interested in regions as small as 1 mm 3, this volume contains some 10 16 molecules. Thus the density p(r) that we actually work with is the density averaged over this huge number of molecules and does indeed vary smoothly with r. The idea that, at a scale of millimeters or more, matter can be treated as continuous, with parameters such as the density being averages over many molecules, is called the continuum hypothesis. The success of continuum mechanics is ample justification for this hypothesis, which we shall adopt throughout this chapter. 16.1 Transverse Motion of a Taut String As our first example of a continuous system, let us consider a taut string lying along the x axis. In equilibrium, we assume that the string lies exactly on the x axis, but now suppose that it is undergoing a small motion (perhaps an oscillatory motion) in the y direction. One simple way to specify the string's configuration at any one time is to give its displacement u (x) from the x axis, as shown in Figure 16.1(a). Specifically, at any one time, a small element of the string whose equilibrium position was at x on Section 16.1 Transverse Motion of a Taut String 683 (a) (b) Figure 16.1 (a) The position of a continuous string, at any one time, is specified by the function u(x) which gives the string's displacement from its equilibrium position on the x axis. (b) A set of n point masses joined by massless strings has its configuration specified by the discrete set of displacements u 1 with i = 1,, n. the axis is now located a distance y = u (x) above the axis. This scheme needs little explanation, but is worth contrasting with a related discrete system, namely a set of n point masses m l,, ru n joined by a taut massless string, which lies in equilibrium on the x axis. If these masses are allowed to move in the y direction as in Figure 16.1(b), their configuration can be specified by their displacements u l,, u\u201e from the axis. Where the discrete system is specified by these n variables u,, with i = 1,, n, our continuous system is specified by the continuous function u (x). The role of the discrete index i attached to u, is now played by the continuous variable x in u(x). Where the index i specifies which of the n masses is at position y = u,, the variable x specifies which of the infinitely many pieces of the string is at y = u (x ). If the systems of Figure 16.1 are moving, the displacements u depend on the time t. In the discrete case they become u 1 (t) and in the continuous case we must write u (x, t), a function of two variables. In the discrete case, Newton's second law becomes a set of ordinary differential equations for the u i (t) (for example the coupled differential equations of Chapter 11). In the continuous case, Newton's law becomes", " a partial differential equation for u(x, t), involving partial derivatives with respect to both x and t, as we now show. To explore the motion of the string, we shall apply Newton's second law to a small segment AB of the string, between x and x dx as shown in Figure 16.2. To simplify our discussion we shall ignore gravity, and we shall assume that the displacement u (x, t) remains so small for all x and all t, that the string remains nearly parallel to the x axis. This guarantees that the string's length is essentially unchanged and hence that the tension T remains the same for all x and all t. The net force on the segment AB is then just F\"t = F 1 + F2, where F 1 and F2 are the tension forces due to the adjacent sections of string, as shown in Figure 16.2. If 0 denotes the angle between the string and the x axis, the x component of this net force is Fxnet = T cos(4) + dO) \u2014 T cos but, since 0 and 0 + dO are both small, both cosines are very close to 1 and Fr is negligible, consistent with our assumption that the motion is in the y direction only. On the other hand, the y component is certainly not negligible: re' = T sin(0 dO) \u2014 T sin 0 = T cos 4) d( (16.1) 684 Chapter 16 Continuum Mechanics Figure 16.2 The two forces on the small element AB of string are the tension forces F 1 and F2 exerted by the adjacent sections of string. Since 0 is small we can replace cos 0 by 1, and we can write d0 = (80/ax)dx. [The derivative is a partial derivative since 0 = (x, t) depends on x and t.] Finally, again since 0 is small, 0 = au/ax, the slope of the string. Therefore, By Newton's second law, Fynet = ma y, where ay is the acceleration ay = a2 u/at2 and m is the mass of the segment AB, equal to\u00b5 dx if we use iu to denote the linear mass density of the string. Therefore, F net Tdx = T dx. ax 8u 2 ax2 Fnet A2\u201e ate dx. (16.2) (16.3) Equating the /two expressions (16.2) and (16.3), we arrive at the equation of motion of our taut string: Here I have introduced the important constant where T is the tension in our string and A is its linear mass density (mass/length). The equation of motion (16.4) is called the one -dimensional wave equation since its solutions are waves traveling along the string, as we shall see. As anticipated, it is a partial differential equation, involving derivatives with respect to x and t. The constant Section 16.2 The Wave Equation 685 C has the dimensions of speed (as you should check) and is the speed with which the waves travel. The wave equation (16.4) governs the motion of many different waves \u2014 waves on a string, waves of sound or light, seismic waves, and many more. Therefore, I shall give it a new section of its own. 16.2 The Wave Equation We are going to show that there are just three kinds of solution of the wave equation (16.4): (1) a disturbance u (x, t) that travels rigidly along the string from left to right; (2) a disturbance u(x, t) that travels rigidly along the string from right to left; and (3) any combination of these two. The proof of this claim is startlingly simple, although it depends on a trick that you probably wouldn't think of right away. We change variables from x and t to = x \u2014 ct and = x + ct. (16.6) It is a straightforward exercise (Problem 16.4) to show that 82u ate \u2013 c\u2014 = \u2014 4C- - - 2 a au a2u aX 2 ari so, written in terms of the new variables, the wave equation (16.4) becomes simply (16.7) (16.8) To solve this equation, let us temporarily write aulaq = h, so that (16.8) becomes ahln= 0. This states simply that h does not depend on c, although it can, of course, depend on 77. Therefore, we can write h = h(17), and we now have a au _ 0 ) nal au = h(q) ari For any given value of c we can integrate this equation to give u = f h + \"constant,\" where the \"constant\" may be different for different values of If we call this \"constant\" f (0 and set the integral f h (r)) dr", ") = g (q), then we have proved that every solution of (16.8) must have the form u= f + g(q). (16.9) By substituting into the left side of (16.8), you can see that a function of this form is a solution of (16.8) for any choice of the two functions f () and g(77). Thus, (16.9) is the general solution of (16.8). Reverting to the original variables x and t, we have shown that the general solution of the wave equation (16.4) has the form u(x, t) = f (x \u2014 ct) g(x ct) (16.10) 686 Chapter 16 Continuum Mechanics ct f(x) s-, f(x \u2014 ct) Figure 16.3 Motion of the wave (16.11). At time 0, the disturbance is given by u = f (x). At a later time t, it is given by u = f (x \u2014 ct), which has the same shape but has moved rigidly to the right by a distance ct. where f and g are any two functions. To see what these solutions represent, let us consider first the case that the function g = 0, so that our solution is just u(x, t) = f (x \u2014 ct). (16.11) What does this solution look like? Notice first that at time t = 0, the solution is u (x, 0) = f (x); that is, the function f (x) is just the disturbance at time t = 0. Figure 16.3 shows a possible such function. The solid curve, with a large maximum at x = 0 and a small dip on the left, shows the function f (x), the shape of the disturbance at t = 0. At a later time t, the disturbance is given by f (x \u2014 ct). Since f (x) has its maximum at x = 0, it follows that f (x \u2014 ct) has its maximum when x \u2014 ct = 0. Therefore, the maximum that was at x = 0 is now at x = ct. Since a similar argument applies to any point of the curve (for example, the minimum on the left), we conclude that the whole disturbance has moved bodily to the right by a distance ct. That is, the disturbance is a wave traveling rigidly to the right with speed c. A similar argument shows that a solution of the form u(x, t) = g(x ct) repre- sents a wave traveling rigidly to the left with speed c, and the general solution (16.10) is a superposition of two waves, one traveling to the right and the other to the left. The functions f and g that appear in the general solution (16.10) are determined by the initial conditions of any particular problem. As you might guess, to determine a particular solution we need to specify the position u and the initial velocity it = au/at at one initial time, as in the following example. EXAMPLE 16.1 Evolution of a Triangular Wave A short segment of a long taut string is pulled aside and released from rest at t = 0, so that its initial displacement is u(x, 0) = u o (x) (16.12) where u o (x) is the triangular function shown in Figure 16.4(a). Find the distur- bance u (x, t) for any later time t. The solution must have the form (16.10), where the two functions f and g are to be determined by the initial conditions. The given initial displacement (16.12) implies that f (x) + g(x) = uo(x). (16.13) Section 16.2 The Wave Equation 687 t = 0 uo(x) (a) (b) E t > 0 Figure 16.4 (a) The initial displacement of a string at t = 0 is given by the triangular function u o (x). (b) At any later time, the wave consists of two triangles, each half as high as the original, one traveling to the right and one to the left. This does not, by itself, determine f and g separately, and we must also look at the initial velocity. Differentiating (16.10) with respect to t, we see that the initial velocity of the string (in the y direction, of course) is r au [at = \u2014cf/(x) cgi(x) where the prime denotes differentiation of a function with respect to its ar- gument. In our case, the string is released from rest, so f'(x) \u2014 g' (x) = 0. Integrating with respect to x, we conclude that' f (x) \u2014 g(x", ") = 0. (16.14) Solving (16.13) and (16.14), we conclude that f (x) = g(x) = 111 0 (x) and the actual disturbance (16.10), at any time t, is u(x, t) = f (x \u2014 ct) g(x ct) = luo (x \u2014 ct) ltt o (x ct). (16.15) Our original triangle has separated into two triangles, half as high, traveling outward in opposite directions with speed c, as shown in Figure 16.4(b). It is interesting to let the solution (16.15) evolve backward to times t < 0. At these times it represents two triangles approaching the origin from opposite sides. When t is close to 0, the two triangles meet and start to interfere. At t = 0 they overlap exactly and interfere to produce a triangular wave twice their individual heights, and then, as t increases past 0, they separate again and move apart as in Figure 16.4(b). 1 Strictly speaking there should be a constant of integration in (16.14), but as you can easily check, it cancels out of u = f g, so we can just as well choose it to be zero. 688 Chapter 16 Continuum Mechanics An important special case of the solution (16.10) is the case that the functions f and g are sinusoidal. If g = 0, then the disturbance takes the form u(x, t) = A sin[k(x \u2014 ct)] = A sin(kx \u2014 wt) (16.16) where A and k are arbitrary constants, and co = kc. This is a sinusoidal wave traveling to the right with amplitude A, wave number k (or wavelength A = 27/k) and angular frequency co (or period r = 27/co). If we replace x \u2014 ct with x + ct, we obtain a similar sinusoidal wave u(x, t) = A sin[k(x + ct)] = A sin(kx + cot) (16.17) traveling to the left. The sum of these two solutions is itself a solution, u(x, t) = A sin(kx \u2014 wt) + A sin(kx + cot) = 2A sin(kx) cos(cot). (16.18) (Use the relevant trig identities to check this.) This solution has the remarkable property that it is not traveling at all (neither to right nor left). Instead it simply oscillates up and down like cos(cot), with amplitude (at any one point x) equal to 2A sin(kx). In particular, at those points (the nodes) where kx is an integer multiple of 7 (kx = nn- ), the string does not move at all, as shown in Figure 16.5. We see that by superposing two carefully chosen traveling waves we have formed a standing wave. As we shall see in the next section, these standing waves play an important role in the oscillations of a finite length of string and are, in fact, the continuum analog of the normal modes of a system of coupled oscillators. Figure 16.5 The standing wave (16.18) at three successive times, t = 0 (solid curve), t = r/4 (long dashes), and t = r/2 (short dashes), where r is the period. The small dots on the x axis are the nodes, where kx = nir and the string does not move at all. Half way between any two successive nodes is an antinode, where the string oscillates up and down with maximum amplitude 2A. 16.3 Boundary Conditions; Waves on a Finite String* * As usual, sections marked with an asterisk can be omitted on a first reading. So far we have been assuming, implicitly, that our string is infinitely long, or at least that it is so long that we can ignore any effects of its ends. Real strings are, of course, finite in length and have ends. The motion of the string itself is governed Section 16.3 Boundary Conditions; Waves on a Finite String * 689 by the same wave equation (16.4) as before, but the existence of the ends imposes additional boundary conditions on its solutions. These boundary conditions vary with the nature of the string's ends; for instance, the string may be tied down at an end, or it may simply flap freely, and the boundary conditions appropriate to these two situations are quite different. Here we shall consider just one type of boundary condition and one method of solving for it. Specifically, we'll consider a string that is tied down at both ends (at x = 0 and x = L), and we shall solve this problem by a method analogous to our discussion of normal modes in Chapter", " 11. Normal Modes The problem that we have to solve is this: For 0 < x < L, the displacement u(x, t) of our string must satisfy the wave equation (16.4) 32 u ate C 2 02u axe' (16.19) with initial conditions that fix the position u and velocity it at t = 0; in addition, it must satisfy the boundary conditions at x = 0 and x = L that u(0, t) = u(L, t) = 0 (16.20) for all times t. Of the several ways to solve this problem, we shall follow the approach of Chapter 11; that is, we shall start by looking for solutions that vary sinusoidally in time, with the form u(x, t) = X (x) cos(cot \u2014 8), (16.21) where the function X (x) and the constants co and 8 are to be determined. As usual, there is nothing to stop us seeking solutions with this form, and, as before, we shall find that such solutions do exist and that any solution of our problem can be written Substitution of the assumed form (16.21) into the wave equation (16.19) reduces in terms of them. the latter to the form \u2014 (0 2 X (x) cos(cot \u2014 8) = c2 cos(wt \u2014 8) dX (2x) 2 d x Or where d2 X (x) d x 2 = k- X (x) co. k = \u2014 (16.22) (16.23) 690 Chapter 16 Continuum Mechanics We see that the assumption of the sinusoidal time dependence (16.21) has reduced the partial differential equation (16.19) to the ordinary differential equation 2 (16.22) \u2014 an equation, moreover, whose solutions we can easily write down. The general solution of (16.22) is just X (x) = a cos(kx) + b sin(kx) (16.24) and this yields a solution of the wave equation (16.19) for any choice of the constants a and b. However, we have still to satisfy the boundary conditions (16.20), which require that X (0) = X (L) = 0. (16.25) The condition that X (0) = 0 requires simply that the coefficient a in (16.24) be zero, so that X (x) = b sin(kx). Thus, the condition that X (L) = 0 requires that either b = 0 or that sin(kL) = 0. In the former case, our solution is identically zero, and the string doesn't move \u2014 a solution, but a trivial one. If sin (k L) = 0, then kL must be an integer multiple of Tr, and we get a nontrivial solution, u(x, t) = sin(kx)A cos(cot \u2014 6), (16.26) where the boundary conditions have forced k to have one of the values k = k\u201e = n 1 [n = 1, 2, 3, ]. (16.27) By (16.23), co = ck, so the corresponding frequency w must have the form w = con = C [n = 1, 2, 3, ]. (16.28) We conclude that there are indeed solutions in which the string oscillates sinu- soidally with a single frequency w, provided w has one of the values (16.28). This result is reminiscent of Chapter 11, where we found that a system of n coupled os- cillators could oscillate in any of various sinusoidal normal modes with frequencies co l,, con. The main difference is that the systems of Chapter 11 had a finite number of degrees of freedom and an equal number of normal frequencies. Here, our string has an infinite number of degrees of freedom and an infinite number of normal frequencies as in (16.28). Figure 16.6 shows the three normal modes of lowest frequency for our string \u2014 the fundamental and the first two overtones. If you compare these pictures with Figure 16.5, you will see that each of the normal modes of our finite string is just a section of a standing wave on an infinite string. That our finite string is fixed at its two ends means that the points x = 0 and x = L must be nodes, which requires that the length L must equal an integer number of half wavelengths, L = nA 12. Since = 27/ k, this explains the condition (16.27) that k =n7r1L. The allowed frequencies (16.28) of our string are all integer multiples of the lowest frequency, con = nw i. They are, among many other things, the frequencies at which any 2 Our method of solution here is closely related to the method of separation of variables. See", " Problem 16.9. Section 16.3 Boundary Conditions; Waves on a Finite String* 691 n n= Figure 16.6 The three lowest-frequency normal modes (16.26) of a string of length L fixed at both ends. In each picture the solid curve, the long dashes, and the short dashes show the string at three successive times, separated by a quarter cycle. The n = 1 mode is called the fundamental. stringed musical instrument, such as a piano or guitar, can vibrate. The corresponding modes, including the fundamental, are called the harmonics of the string, since they \"harmonize\" well (that is, make a pleasing sound to the ear) when played together. The General Solution The normal modes (16.26) determine all possible motions of our finite string, in the sense that any possible motion can be expanded in terms of the normal-mode solutions. To see this we shall need to use some of the properties of Fourier series described in Section 5.7. First, let us note that any motion of the string is given by a function u(x, t) that satisfies the wave equation (16.19) and the boundary conditions (16.20), and is determined by its initial position u(x, 0) = u o (x) and velocity a (x, 0) = o (x). To see that any such solution can be expanded in terms of the normal modes, we first rewrite the normal-mode solution (16.26) in the \"sine plus cosine\" form as u(x, t) = sin kn x(Bn cos con t + CT, sin coa t). (16.29) Our claim is that any possible motion can be written as a linear combination of these normal-mode solutions: u(x, t) E sin knx (B,, cos con t C sin con t. ). (16.30) 00 n=1 To prove this, we note first that this linear combination certainly satisfies the wave equation and the boundary conditions that u vanish at x = 0 and x = L. At time t = 0, the claimed solution is just u (x, = E 13\u201e sin kn x. (16.31) 00 n=1 692 Chapter 16 Continuum Mechanics This is a Fourier sine series, and the coefficients B, can be chosen so the u(x, 0) matches any given initial value u o (x). 3 Similarly, the velocity of the proposed solution (16.30) is (x, 0) -= E 0\u201e C\u201e sin knx (16.32) 00 n=1 and we can choose the coefficients C, so that this is equal to any given initial velocity /Jo (x). We conclude that the proposed solution satisfies the equation of motion and the boundary conditions, and that by choice of the coefficients B, and Cr, can match any given initial conditions. Therefore, any possible motion of the string can be expanded in terms of the normal modes as in (16.30). Since all of the frequencies in (16.30) are integer multiples of the lowest frequency (con = nco l), every term in (16.30) is periodic with period r =-27/0)1. Therefore, every possible motion of our finite string is periodic with this period. [Of course, if certain coefficients in (16.30) are zero, the motion may be periodic with a shorter period as well, but every solution has the period of the fundamental mode.] EXAMPLE 16.2 A Triangular Wave on a Finite String A string of length L = 8 is fixed at both ends. It is given a small triangular displacement, as in Figure 16.7 and released from rest at t = 0. Find the Fourier coefficients Bn and Cn in the expansion (16.30) and using some reasonable finite number of terms to approximate the infinite series, plot the position of the string at five equally spaced times from t = 0 to t = r/2, where r is the period of the motion. Since the string is initially at rest, all of the coefficients C, are zero. The coefficients B\u201e are given by the integral L 2 B, = \u2014 u o (x) sin dx. nL L 0 (16.33) [This is nearly, but not quite, the standard formula (5.84); for details, see Problem 16.13.] It is easy to see that this is zero when n is even. When n is odd, we can write n = 2m + 1 and you can check (Problem 16.10) that B2m\u00b1 = 32 (2m + 1)2n2 (1\u2014co s (2m + 1)n- 8 (16.34) Putting these coefficients into the expansion (16.30) and choosing some reasonable finite number of term, we can", " get a good approximation for the displacement u (x, t) for all times t. Using just the first five or so terms, we get a moderate approximation, but since it is very little trouble (for a computer) to use more terms, I chose to use the sum of the first twenty. The results are shown in 3 There is a small complication that I am ignoring here. The series (16.31) is not the usual Fourier series since it is missing any cosine terms. However, it contains twice as many sine terms as the usual Fourier series since kr, = tur I L (as opposed to the usual 2nn- I L), and one can prove that this series can be used to expand any (reasonable) function on the interval 0 < x < L. See Problem 16.13. Section 16.3 Boundary Conditions; Waves on a Finite String* 693 x = 0 x = 8 Figure 16.7 A string is released from rest at t = 0 in the triangular position shown. Figure 16.8. Each of these five plots deserves careful attention. The first shows the initial displacement of Figure 16.7, as approximated by the first 20 terms of its Fourier series. The approximation is very good, although it fails to reproduce the sharp turn at the apex. (Obviously, no finite sums of sines or cosines can actually have an instantaneous change of slope.) In the second picture, the initial triangle has split into two separate triangles, traveling in opposite directions. This is exactly the behavior we saw in Example 16.1 (Figure 16.4); because neither of the waves has reached the boundaries at x = 0 and L, the motion is so far unaffected by the presence of the boundaries. Skipping for a moment over the third picture, you can see in the fourth that each triangle has been reflected by the walls, and is now traveling back toward the center, although they are now inverted. In the third picture, the original and the reflected waves are both present and are interfering destructively to produce zero net displacement. Finally, in the last picture, the two reflected waves have coalesced momentarily into a single inverted triangle. If we were to follow the motion further, we would see the two reflected waves continue on until they hit the opposite walls and reflect again. (See Problem 16.11.) t = 0 t = T/8 t = T/4 t = 37/8 t= T/2 Figure 16.8 Five successive snapshots of a string that was released from the initial position of Figure 16.7, calculated using the first twenty nonzero terms of the Fourier series (16.30). The first picture shows the initial position (approxi- mated by 20 terms of its Fourier series). The four succeeding pictures show the position at intervals of -r/8, where -r is the period of the fundamental. 694 Chapter 16 Continuum Mechanics 16.4 The Three-Dimensional Wave Equation We live in a three-dimensional world, and the wave equation (16.4) a2 u at2 =c 2 a2u axe (16.35) needs to be generalized to three dimensions. It is not hard to guess what the proper gen- eralization should be. If p = p(x, y, z, t) = p(r, t) denotes some sort of disturbance in a three-dimensional system (for example, the pressure in a sound wave traveling through the air), then we would surely guess that the appropriate generalization of (16.35) should be a2 p ate 2 (a2 p a2 p a2p ) = - C a x 2 a y2 az2 (16.36) We shall meet a couple of examples of disturbances that do indeed satisfy this three- dimensional wave equation later in this chapter. In particular, I shall prove in Section 16.13 that the pressure 4 in any inviscid fluid (for example, air) is an example, for which the wave speed c is given by c= BM 00 (16.37) where BM denotes the bulk modulus of the fluid, and po is the equilibrium density. (I'll define the bulk modulus shortly. For the moment, just see it as a parameter that characterizes the resistance of the fluid to compression.) It is usual to streamline the notation in the wave equation (16.36): If, as usual, we take the view that V is the \"vector\" with components v Ca a y'az ) ax' a then the scalar product of V with itself is obviously You may well have met this differential operator before, perhaps in your study of electromagnetism. It plays a huge role in many subjects \u2014 electromagnetism, quantum mechanics, fluid mechanics, elasticity, thermodynamics, and more \u2014 and 4 Strictly speaking the pressure p discussed throughout this section is the incremental pressure, the difference between the total pressure and the equilibrium atmospheric pressure.", " Section 16.4 The Three-Dimensional Wave Equation 695 is called the Laplacian for its role in Laplace's equation of electrostatics. With this notation, we can rewrite the three-dimensional wave equation (16.36) as Plane Waves The equation (16.39) has many, many solutions, of which the simplest are the so- called plane waves. A simple example of a plane wave is a solution of (16.39) [or (16.36)] which is independent of y and z, Obviously a disturbance with this form has the same value at all points in any plane x,--- constant. If we substitute this form into (16.36), the derivatives with respect to y and z drop out, and we are left with the one-dimensional wave equation p(r, t) = p(x, t). 82, at e 2.92P ax e whose most general solution we already know to be p = f (x \u2014 ct) g(x ct). In particular, a solution p = f (x \u2014 ct) is a plane disturbance (p constant in any plane perpendicular to the x axis) that is traveling in the x direction with speed c (hence the name \"plane wave\"). Similarly, a solution of the form p = f (y \u2014 ct) or f (z \u2014 ct) is a plane wave traveling in the y or z direction. More generally, if n denotes an arbitrary unit vector, then a disturbance of the form p= f (n r \u2014 ct) (16.40) satisfies the wave equation (16.39), is constant in any plane perpendicular to n, and travels in the direction n with speed c. (See Problem 16.15.) If the function f is a sinusoidal function, f () = cos kc, say, the wave (16.40) is the sinusoidal plane wave p = cos[k(n r \u2014 ct)], (16.41) whose crests lie in planes perpendicular to n and travel with speed c in the direction of n, as shown in Figure 16.9. This kind of wave is easier to visualize in two dimensions. You could think, for example, of \"plane\" waves on the surface of a pond. A plane wave is a mathematical idealization that never occurs in practice, since no real disturbance can be constant over an infinite plane. Nevertheless, it is frequently a very useful approximation. The light shining on us from the sun is well approximated as a plane wave, as is the sound from a distant explosion. 696 Chapter 16 Continuum Mechanics Figure 16.9 The sinusoidal plane wave (16.41). The wave crests (or wavefronts) are planes perpendicular to the unit vector n and they travel with speed c in the direction of n. Spherical Waves Another important solution of the three-dimensional wave equation is a spherical wave, for example, a sound wave traveling radially outward from a small, omnidi- rectional loudspeaker. If we assume that such a wave is spherically symmetric, then it must have the form p = p(r, t). (That is, in spherical polar coordinates, p is inde- pendent of B and 0.) It is not hard to show that for a function of this form v2 p 1 8 2 ( r are \u2018 rPi. (16.42) The obvious way to prove this is to evaluate the left side using the definition (16.38) of V 2 ; the simplest is to look up the expression for V2 in polar coordinates inside the back cover. See Problem 16.16.] Therefore, the wave equation becomes a2p 8t2 2 21 3 = c --(rp) r art a2 ate (rp) = c2\u2014 (rp) a2 3r2 or, multiplying both sides by r, (16.43) We see that for a spherical wave the function rp(r, t) satisfies the one-dimensional wave equation with respect to r and t. Therefore the general solution has the form rp(r, t) = f (r \u2014 ct) g(r ct). In particular, if the function g is zero, the disturbance has the form p(r, t) = \u2014 f (r \u2014 ct). (16.44) 1 r The factor f (r \u2014 ct) represents a disturbance traveling rigidly outward. Since this is what one might have guessed for a radially spreading wave, the question is, \"Why the Section 16.5 Volume and Surface Forces 697 factor of 1/ r?\" To answer this we need a result that you may recall from an introductory physics class. The intensity of any wave in three dimensions is defined as the power delivered by the wave to unit area perpendicular to the direction of propagation, and the intensity of a sound wave is proportional to p", " 2. (For guidance in proving this, see Problem 16.37.) Thus the factor 1/ r in (16.44) implies that the intensity is proportional to 1/r 2, which is exactly what is required for conservation of energy: At a distance r from the source, the energy of the wave is spread out over an area 47 r 2. Therefore, as the wave moves radially outward, the intensity has to fall off like 1/r 2 to keep the total power constant. If the function f in (16.44) is sinusoidal, f = cos say, then (16.44) repre- sents a sinusoidal wave whose crests are traveling radially outward from the origin with speed c, as illustrated in Figure 16.10. Figure 16.10 A spherical wave. The wave crests are spheres moving outward from the origin. To help visualize what this shows, it may help to think of it as a two-dimensional wave, such as the ripples created on a pond by a stick which moves in and out of the water at the origin. 16.5 Volume and Surface Forces Our next objective is to see how to find the equation of motion of a three-dimensional continuous system. In general, this is a very complicated problem, and the details depend strongly on the precise nature of the system. For example, the equations of motion of a fluid are very different from those of an elastic solid. Nevertheless, there are some reasonably simple general principles that apply to many different continuous systems, and these are what I shall describe next. As you would probably guess, we find the equation of motion of a continuous body by applying Newton's second law to an arbitrary small element d V of the body. (I use the word \"body\" here to mean any chunk of matter, solid, liquid, or gas.) This is exactly analogous to what we did in Section 16.1, where we applied Newton's second law to a length dx of the one-dimensional string, but the three-dimensional case is naturally more complicated. We need first to discuss the geometry of the volume element d V and then the specification of the forces on, and resulting displacement of, dV. 698 Chapter 16 Continuum Mechanics Elements of Volume The shape of the volume element on which we focus is arbitrary. It could be spherical, or rectangular (in the shape of a brick), or anything else our whim dictates. For simplicity, you could have in mind a simple rectangular volume, as in Figure 16.11. The volume on which we focus will usually be an infinitesimal volume, as the label dV is intended to suggest. The surface that bounds d V could be the actual boundary of the continuous body (for example, the walls of the cylinder containing a gas), but it will usually be an \"imaginary\" surface, that is, an arbitrary surface interior to the whole body. The whole bounding surface is naturally a closed surface that divides all of space into exactly two parts, the \"inside\" (namely, d V) and the \"outside\" (everything else). This means that we can specify the orientation of any part of the surface, such as the face S in Figure 16.11, by the unit vector n, normal to the surface and pointing outward from dV. 5 Figure 16.11 A small element of volume d V of a continuous body can have any shape, but a convenient choice is the rectangular shape shown here. The orientation of any part S of the surface (the right-hand end here) is specified by a unit vector n that points outward from dV. Forces on the Volume Element The two most important kinds of force on a volume element dV of a continuous system are called volume forces and surface forces. An example of a volume force is the force of gravity, F = QgdV, where o is the mass density of the material and g is the acceleration of gravity. A second example is the electrostatic force F = QE dV of an electric field E on a material with charge density p. The definition of a volume force is simply that it is a force proportional to the volume dV. Volume forces are generally the result of an external field (such as gravity), and to be definite I shall usually assume that the only volume force is that of gravity. In any event, the body forces are almost always known and well understood. Therefore, our main concern is with the surface forces. A familiar example of a surface force is the force p dA of the pressure p in a fluid on a small surface element of area dA. The definition of a surface force is a force proportional to the area dA of the surface on which it acts. Surface forces are generally 5 When one is concerned with a nonclosed surface, the orientation of a small part S can still be specified by a unit vector n that is normal to S, but this leaves an ambiguous sign (since n and \u2014 n both fit the definition). Fortunately, in this chapter S", " will always be a part of a closed surface, so we can insist, simply and unambiguously, that n point outward. Section 16.5 Volume and Surface Forces 699 (a) Pressure (b) Tension (c) Shear Figure 16.12 Three different surface forces on the face S of a rectangular volume. (a) Hydrostatic pressure acts normally to the surface and inward, so that F = \u2014pndA, with the minus sign since F is inward whereas n is outward. (b) A simple tension in the direction normal to S. (c) By definition, a shearing force acts tangentially to S. F S ----------- 1\\ T - - - dx Figure 16.13 A shearing force F applied to the face S of the rectangular solid of Figure 16.12 (seen here from directly in front). If the opposite face is held fixed, the shear produces the motion shown, in which the planes parallel to S all move in the direction of F, changing the originally rectangular cross section into a parallelogram. The distances dx and dy are used to define the shearing strain in Equation (16.54) below. the result of intermolecular forces of the molecules just outside the surface acting on those just inside. Figure 16.12 shows three important special cases of surface forces, a pressure force, a simple tension, and a shearing force. Notice that both the pressure force and the tension act normally to the surface S, whereas the shearing force, by definition, acts tangentially to S. The tendency of a shearing force is to produce the shearing motion shown in Figure 16.13. When Is Pressure Isotropic? To conclude this section, I shall prove a result that you probably learned in an introductory physics class, that the pressure in any static fluid acts equally in all directions, or, briefly, that the pressure is isotropic. Actually, the result is a bit more general than this, and I shall prove it in its greater generality. A characteristic property of any fluid is that it can support no shearing forces in equilibrium, and the absence of shearing forces is in fact the essential feature that leads to the isotropy of pressure. I shall prove in a moment that in any substance where there are no shearing forces the pressure is isotropic. Clearly the result applies to any static fluid, but it also applies 700 Chapter 16 Continuum Mechanics Figure 16.14 The surfaces S 1 and S2 (seen edge on) are normal to the two arbitrary unit vectors n 1 and n2. They are identical rectangles and, together with S3, form an isosceles prism (seen end on). The three forces F 1, F2, and F3 are the surface forces on the three faces and are normal to the surfaces since, by assumption, there are no shearing forces. to a moving fluid, provided there are no shearing forces. Since the cause of shearing forces in fluids is viscosity, we can say that pressure is isotropic even in moving fluids, provided the viscosity is zero. Of course, there are very few fluids whose viscosity is exactly zero, but there are plenty of situations where the viscosity is small enough to be negligible, the most obvious example being any flow in which all speeds are very slow. Our result is obviously very useful, but, more important, the method of proof has many other applications, as we shall see. Let us consider, then, any medium in which there are no shearing forces, and let n 1 2 be any two directions. At any particular point in the medium, let us construct and n two small, equal rectangular surfaces, S i normal to n1 and S2 normal to n2, so as to form a small triangular prism, as in Figure 16.14. The surface forces on the three faces shown are normal to the faces and can be written as F 1 = \u2014 p in i d A i and so on, where we write the pressures on the three faces as p i, /32, and p3 to allow for the possibility that they might be different. (Our aim is to prove that, in fact, p i = P2 = p3.) These pressures can, of course, vary from point to point in the fluid, but by considering a small enough volume we can ensure that they vary by a negligible amount within our volume. We are now ready to apply Newton's second law to our small prism. The mass of the prism is m = QdV. The net force on the prism is F 1 + F2 + F3 + Fvoi where Fvoi denotes the total volume force (for example, the weight F voi = Qg dV). Thus the equation F = ma becomes 6 F + F2 \u00b1 F3 + Fv01 = ma which we can rewrite as F i \u00b1 F2 \u00b1 F3 = ma \u2014 F vo1.", " (16.45) 6 Strictly speaking we should include the two pressure forces on the two ends of our prism, but we shall be concerned only with the components of this equation in the plane of the picture in Figure 16.14, so we can ignore these. Section 16.6 Stress and Strain: The Elastic Moduli 701 Now comes the supreme act of cunning. Equation (16.45) applies to the small prism of Figure 16.14, but it would certainly also apply to a smaller prism. Let us therefore shrink our prism by a factor of A in all three directions. The three surface terms on the left of (16.45) are proportional to area, so they will decrease by a factor of A 2. Both the mass and volume force on the right are proportional to d V and must decrease by a factor of A 3. Thus the counterpart of (16.45) for our smaller prism is A2(F1 + F2 + F3) = A3 (ma \u2014 Fvol) or, dividing both sides by A2, (F1 + F2 + F3) = JL(ma \u2014 F wa ) (16.46) This equation holds for any value of A (smaller than 1). In particular, we can let A approach zero, and we reach the surprising conclusion that the three surface terms must sum to zero by themselves, F 1 \u00b1 F2 \u00b1 F3 = O. (16.47) It is easy to check that, because the triangle in Figure 16.14 is isosceles, this requires that F 1 and F2 must have equal magnitudes, F1 = F2. (Just take components perpendicular to F3 to check this.) Since F1 = p idA l and F2 = p2dA2, and d A l = dA2, we conclude that Pi = Pz (16.48) Since the directions ni and n2 were arbitrary, we have proved that pressure is inde- pendent of direction in any medium where there are no shearing forces. In particular, pressure is isotropic in any static fluid and also in any moving fluid that has negligible viscosity. 16.6 Stress and Strain: The Elastic Moduli As we shall see in the next section, the surface forces inside a continuous body (solid, liquid, or gas) can be expressed in terms of a three-dimensional tensor called the stress tensor. In Section 16.8, we'll see that the resulting displacements of the body can be expressed in terms of a second tensor called the strain tensor. Finally, before we can write down the equation of motion we need to establish the relationship between the two tensors. [This last statement is the continuum analog of the familiar requirement that to write down the equation of motion of a mass on a spring, we need to know Hooke's law (F = kx), which relates the tension in the spring (F) to the extension (x).} The general theory of the stress and strain tensors is quite complicated, so in this section I shall mention a few simple special cases before we plunge into the general case. 702 Chapter 16 Continuum Mechanics Stress Since any surface force F is proportional to the area A of the surface on which it acts, it is natural to consider the ratio F/A, and this ratio is called the stress. As we shall see in the next section, in general we need to discuss a stress tensor, but a simple example that lacks this complication is the pressure force in a static fluid, for which the stress is just the pressure: F A tension area stress = = pressure, p [in a static fluid]. (16.49) Similarly, you may recall that the stress in a wire or rod subjected to a simple tension is defined as stress = [for a wire in tension] (16.50) where the area is the cross-sectional area of the wire. For a simple shearing force, like that in Figure 16.13, the shearing stress is defined as shearing force stress = area [for a shear]. (16.51) However complicated the situation, we shall find that the stress (or any component of the stress tensor) can be defined as the ratio of a surface force (or one of its components) to the area of the surface on which it acts. In particular, stress always has the dimensions of [force/area]. Strain The result of a stress is almost always a deformation, or change in the dimensions, of the body on which the stress acted \u2014 a change in the volume of a liquid, or the length of a wire, for instance. When this change is expressed as a fractional change it is called the strain. For example, in a static fluid, the strain would be the fractional change in volume, d V V dl I strain = \u2014 [in a static fluid]. (16.52) For a wire under", " tension, the strain would be the fractional change in length, strain = [for a wire in tension]. (16.53) For the simple shearing force of Figure 16.13, the strain is defined as strain = dy [for a shear] dx (16.54) where dy is the displacement in the direction of shear and dx is the perpendicular distance across which the shear occurs (see Figure 16.13). Section 16.6 Stress and Strain: The Elastic Moduli 703 Relation of Stress to Strain: The Elastic Moduli When the stresses in a medium are not too large, we would expect that the resulting strain would be linear in the stress. In the case of a stretched wire, this relation is written as stress = (Young's modulus) x strain or A = YM dl \u2014 (16.55) F A where YM is Young's modulus for the material of the wire. 7 [If you rewrite this equation as F = (A YM/ 1)dl, you will recognize it as Hooke's law, with the force constant k = (A YM/ 1). The advantage of writing it in terms of Young's modulus is that YM, unlike k, is characteristic of the material and independent of the dimensions A and /1 In Equation (16.55), dl is the extension caused by an increment dF in the tension. For any material subject to hydrostatic pressure only, a small increase dp in pressure will cause a change in volume given by stress = (bulk modulus) x strain or dp = \u2014BM (16.56) where BM is the bulk modulus for the material, and the minus sign is because an increase in pressure causes a decrease in volume. For a shearing stress, stress = (shear modulus) x strain or = SM (16.57) A where SM is the shear modulus for the material. To summarize this section, stress characterizes the surface forces in a continuous medium, d V V dy dx while strain characterizes the resulting deformation, 7 There are almost as many notations for the various elastic moduli as there are books on the subject. The notations used here are unconventional, but you will, I hope, be able to remember which is which. 704 Chapter 16 Continuum Mechanics While stress always has the units of pressure (force/area), strain is always dimen- sionless.8 The various elastic moduli (Young's, bulk, and shear) are the ratios of stress to the corresponding strain, 16.7 The Stress Tensor In this section I shall derive the general expression for the surface force on a small area dA of a closed surface S in a continuous medium. As usual, we shall use n to denote the unit outward normal to S at the location of dA. To streamline the notation, I shall define a vector dA in the direction of n, with magnitude dA. That is, dA = ndA. (16.61) This vector tells us the orientation and size of the small piece of surface under consideration. Our first task is to show that the surface force F(dA) on the surface element specified by dA is in fact a linear function of dA, that is, that F(Ai dAi + X2 dA2) = XiF(dAi) X2F(dA2) (16.62) where X i and A2 are any two real numbers and dA i and dA2 are any two vectors. The force F(dA) is independent of the precise shape of surface element. On the other hand, it is proportional to the area dA, so F (X dA) = AF(dA) (16.63) for any positive number X (not too large). If we were to replace dA by \u2014dA, this would interchange the inside and outside of our surface, and by Newton's third law this would change the sign of the surface force; that is, F(\u2014dA) = \u2014F(dA). Therefore, Equation (16.63) actually holds for negative, as well as positive, values of X. Let us next consider any two small vectors dA t and dA2. At any point in our continuous medium, consider two small rectangular surfaces touching along one common edge, with orientations and areas given by dA t and dA2, as shown in Figure 16.15. Consider now the triangular prism defined by these two rectangles and the third rectangle labeled as dA 3 in the figure. This prism has two remarkable properties. First, 8 There is no very obvious way to remember which is stress and which strain. One possibility is this: In everyday language we say \"stress causes strain\" and likewise \"force causes deformation.\" Thus \"stress\" goes with \"force\" and \"str", "ain\" with \"deformation.\" Alternatively, note that alphabet- ically, \"stress\" comes after \"strain\" just as \"force\" comes after \"deformation.\" Section 16.7 The Stress Tensor 705 F(dA2) Figure 16.15 The arbitrary small vectors dA 1 and dA2 define two rectangular surfaces (seen edge on) which meet at one common edge (bottom right). These two rectangles define a triangular prism (seen end on) whose third rectangular face is labeled by the vector dA 3. The surface forces on the three rectangular faces are shown as F(dA1 ) and so on. because the three edges shown form a closed triangle, the same is true of the three vectors dA 1, dA2, and dA3. Therefore dA i + dA2 + dA3 = O. Second, by the same argument as we used to prove the isotropy of pressure in an inviscid fluid [Equation (16.47)], we can prove that 9 F(dA 1) + F(dA 2) + F(dA3) = 0. Exploiting these last two equations in turn, we find that F(dA 1 + dA2 ) = F(\u2014dA 3) \u2014F(dA3) F(dA 1) + F(dA2 ). (16.64) Finally, combining (16.63) and (16.64), we can immediately verify (16.62), and we have proved that the force F(dA) is linear in dA. It is a fundamental result of linear algebra, that if one vector (F in this case) is a linear function of a second vector (dA), then the components of the first are related to those of the second by a linear relation of the form l\u00b0 Fi (dA) = Eau dA; (16.65) 3 f=i 9 I am again ignoring the forces on the two triangular ends. If we denote the two ends by dA 4 5, then dA4 = \u2014dA5, so, by (16.63), F(dA4) = \u2014F(dA5) and these two forces cancel each and dA other. to The proof is quite easy: Suppose that u is a linear function of v. Since u i = e1 u and v = ei vi, it follows that u,(v) = e1 u( e j vi ) = j [ei u(e Av.!, which has the advertized form (16.65) with aif = e, u(ei). 706 Chapter 16 Continuum Mechanics or, in matrix form, In this second relation (16.66), / denotes a (3 x 3) matrix, 11 made up of the nine numbers ail of (16.65). The matrix E defines a second-rank, three-dimensional tensor called the stress tensor. The stress tensor can, of course, vary from point to point in the medium, but its significance at each point r is this: For every point r in the medium (and at each given time t), there is a unique (3 x 3) matrix E which gives the force on any surface element dA at r via the relation (16.66). The Elements of the Stress Tensor The mathematical significance of the stress tensor I and its elements aq cannot be more succinctly expressed than in the relations (16.66) and (16.65), but to get a feel for their physical significance it helps to look at some special cases. For example, let us consider a small area dA normal to the x axis, for which dA = e 1 dA. Since only one of the components of dA is nonzero (namely, the first component), the sum in (16.65) reduces to a single term. For example, the x component of (16.65) reads F1 (on area dA normal to e l) Turning this around, we can say that a ll is the first component of the force per area on a surface perpendicular to the first (x) axis. In the same way, we conclude that aii is the ith component of the force per area on a surface perpendicular to the ith axis. To put this another way, a diagonal element aii of the stress tensor, E, gives the normal component of the force per area on a surface perpendicular to the ith axis. The off-diagonal elements a, (i j) can be similarly interpreted. Consider again the case of a small area dA normal to the x axis, for which the second component of (16.65) reads F2 (on area dA normal to e l ) = 621 dA. Evidently a21 is the second (y) component of the force per area on", " a surface perpen- dicular to the first (x) axis. A similar argument applies to 6 31, and we can say that a21 and a31 are the two components of the tangential or shearing force per area on a surface perpendicular to the first axis. More generally the six off-diagonal elements (i j) tell us the six shearing forces on the three coordinate planes through the point under consideration. 11 Don't confuse the boldface capital Greek \"sigma,\" I, in (16.66) with the summation sign in (16.65). Section 16.7 The Stress Tensor 707 The Stress Tensor Is Symmetric The six off-diagonal elements a ii are not actually independent, since they are equal in pairs. Specifically, the stress tensor / is symmetric, so that a u = 6i1, as we can now prove using, yet again, the argument introduced at the end of Section 16.5 to prove the isotropy of pressure in an inviscid fluid. This time, we'll consider a small prism, whose axis is in the z direction and whose cross section is a square, parallel to the xy plane, as shown in Figure 16.16. The prism's angular momentum about its axis satisfies dL 3 = r3, dt (16.67) where F3 is the z component of the net torque on the prism. The four forces (actually components of forces) that contribute to F3 are the shearing forces shown in the figure as Fa, Fb, Fc, and Fd. From (16.65), we see that Fa = a12 dA, while Fb = 621 d A. The forces Fc. and Fd are equal in magnitude to Fa and Fb respectively, but in the opposite directions. Thus the total torque F3 is = Fbi \u2014 Fal = (621 1712)1 dA. (16.68) Using the now-familiar trick, we next reduce our prism by a factor of X in all three directions. In (16.67) this reduction multiplies F3 by a factor of A 3, but L3 by a factor of X 4. If we divide by X 3 and let X \u2014* 0, we see that F 3 must in fact be zero. According to (16.68), this implies that 621 = a12. Similar arguments take care of the other off- diagonal elements, and we conclude that Cr = ti a1 (16.69) for all i and j. That is, the stress tensor I is symmetric. Figure 16.16 End view of a square prism with its axis parallel to the z axis. The four forces that contribute to the prism's rotation about its axis are shown as Fa, Fb, Fc, and Fd. The square ends have side / and the four faces that are seen end-on here have area dA. 708 Chapter 16 Continuum Mechanics EXAMPLE 16.3 The Stress Tensor in a Static Fluid Write down the stress tensor E in a static fluid at a point where the pressure is p. We know that in a static fluid there are no shearing forces and that, at any given point, the pressure is the same in all directions. Thus the surface force on a small surface element labeled by dA is just F(dA) = \u2014pdA where p is a constant (at any one point of the fluid), independent of dA. (The minus sign is because dA points outward, whereas the pressure force is inward.) Comparing this with the definition (16.66) of the stress tensor, we see that I= \u2014pl (16.70) where 1 is the (3 x 3) unit matrix. This beautifully simple result expresses succinctly that in a static fluid (and also a moving fluid provided the viscosity is negligible) the only surface force is the pressure force, which is normal to the surface and independent of the surface's orientation. 12 EXAMPLE 16.4 A Numerical Example of Stress At a certain point P in a continuous medium the stress tensor has the value \u20141 2 0 [ 2 \u20142 0. 0 0 1 (16.71) A small surface element at P has area dA and is parallel to the plane x y z= 0. What is the force on this surface element and what is the angle between this force and the normal to the surface element? To be definite, take P to be in the positive octant (x, y, z all positive), and assume that the outside of the surface is the side away from the origin. Notice first that I have not specified the units of the components of 1, but they would of course be the units of pressure (force/area). If our medium was the water at the bottom of a river they might be kilopascals; for a rock in the earth's crust", ", they might be megapascals. The force on our surface element is given by (16.66) with dA = n d A, where n is the unit normal to the surface. We are told that the surface is parallel to the plane x y z = 0, so n 12 This example suggests a neat alternative proof of the isotropy of the pressure force. The absence of any shearing forces means that I must be diagonal (all off-diagonal elements equal to zero) with respect to any choice of axes. It is easy to show that the only tensor with this property is a multiple of the unit matrix. 13 There are several ways to see this. One simple one is to note that the plane is given in the form f (x, y, z) = constant, and it is a standard result of vector calculus that the vector V f is normal to a Section 16.8 The Strain Tensor for a Solid 709 n = 1 1 1 ] 0 [ (16.72) Therefore the force on the surface element is F(dA) = dA I n = \u2014 dA (16.73) 1 0 1 The angle 0 between the force and the normal is given by cos 0 = F n IFl lni 2/3 /2,/2/3 x 1 So 0 = arccos(12/3) = 35.3\u00b0. This last illustrates the obvious fact that, in the presence of shearing forces, the force on a surface element is not necessarily normal to the surface. 16.8 The Strain Tensor for a Solid In the last section we saw how the stress tensor expresses the surface forces inside a continuous medium, solid, liquid, or gas. This section presents a corresponding discussion of the strain tensor as a description of the displacements within the medium. Unfortunately, in this case, the analysis of solids is quite different from that of fluids, and for simplicity I shall confine myself to the discussion of solids. 14 To specify the configuration of a continuous solid we must give the position of each of its continuously many constituent pieces. A convenient way to do this is to specify that the particular small volume dV that was \"originally\" at position r is now at position r u(r). The \"original\" position could be its equilibrium position or just the position at some convenient initial time to. Either way, the vector u(r) is the displacement needed to move the piece from its reference position r to its current position. At first glance, you might think that u(r) would be a good measure of the strain of the body, but it isn't hard to see that this is not so. Consider for example, the possibility that u(r) = uo is just a constant (independent of r). Such a displacement simply moves our whole body rigidly through the vector uo and requires no internal stresses at all. Stresses arise not so much from displacement of our solid as from distortion, and distortions require that different parts of the body are displaced by different amounts, surface of this form (see Problem 4.18). Therefore, n = \u00b1Vf11V f I. Since n must point away from the origin, the plus sign applies, and it is easy to check that this gives the result (16.72). 14 It is easy to see that there has to be a big difference between solids and fluids. For example, a change in the shape of a solid certainly constitutes a strain and usually entails appreciable stresses; but a change in shape of a fluid (transferring milk from a square carton to a round bowl, for example) does not normally constitute a strain since it requires no stresses. 710 Chapter 16 Continuum Mechanics r + dr r + dr (a) (b) Figure 16.17 (a) In a rigid translation, all points in the body are displaced by the same amount. That is, u(r) is the same for all r, and the separation dr of any two neighboring points is unchanged. (b) Any distortion of the body requires that u(r) vary from point to point. Here the points r and r + dr move by different amounts, and their separation changes from dr to dr + du. as illustrated in Figure 16.17(b). Figure 16.17(a) shows a rigid translation, with u(r) the same for all r, so that the separation, dr, of any two neighboring points remains the same. In part (b), u(r) and u(r + dr) are different, and the separation of the two neighboring points changes from dr to dr + du. The change du in their separation can be expressed in terms of the derivatives of u with respect to r: du i au ar (16.74) or, in matrix notation, du = D dr (16.75) where D is the derivatives matrix (or derivatives tensor) made up of the", " partial derivatives auj iari : The elements of the derivatives matrix D tell us how rapidly the displacement u(r) varies as we move around inside the solid, and you might reasonably guess that D would be a good measure of strain. Unfortunately there is one more complication to discuss. We have already noted that a rigid translation of our solid should not count as a strain, and the same is true of any rigid rotation. Thus, we must examine what form D would take for a rigid rotation and then, for an arbitrary displacement given by D, somehow subtract out of D that part which corresponds to a rigid rotation, to leave something that truly represents what we want to mean by strain. In what follows we shall be concerned only with small strains (meaning that all of the derivatives in D are much less than 1). So let us consider a small rigid rotation Section 16.8 The Strain Tensor for a Solid 711 which we can label by a vector 0 = Bu, where the unit vector u identifies the axis of rotation and 0 is the (small) angle of rotation. It is not hard to calculate the resulting displacement of any point r from scratch, but we can save a little trouble by recalling Equation (9.22), v = w x r, for the velocity of a point r in a rigid body rotating with angular velocity co. Multiplying both sides by a small time dt, we find that the displacement u(r) is 15 u(r) = v dt = co dt xr=Oxr (16.77) since 0 = w dt. If you write out the components of this equation and differentiate, you can easily check that du = D dr as in (16.75) where [0 D = \u2014 03 93 0 02 \u2014 01 \u201402 0 1 0 _ [any small rotation]. (16.78) That is, for any small rotation given by the vector 0 = (01, 02, 03) the derivatives matrix is given by the antisymmetric matrix (16.78). (A matrix M is antisymmetric if M = \u2014M.) Conversely, any antisymmetric matrix has the form (16.78), so any antisymmetric matrix (with all its elements small) is the derivatives matrix for a small rotation. Therefore, if the derivatives matrix (16.76) is found to be antisymmetric, it corresponds to a rotation and should not be considered a strain. To exploit the result of the last paragraph, we need to use an elementary theorem from matrix theory, that any square matrix M can be written as the sum of two matrices, one of which is antisymmetric and one symmetric, as we can easily verify from the following obvious identity: m = 2 (M \u2014 1\u20184) +2 (M +M). (16.79) Since the first of these is clearly antisymmetric and the second symmetric, this proves the claimed theorem. The derivatives matrix, D, of any displacement can be decomposed in this way as D=A+E. (16.80) Here A is the antisymmetric part of D; it represents a rigid rotation and does not contribute to the strain. The second term E is called the strain tensor; 16 it is the symmetric part of D, 15 R is important that dt, and hence 0, be small. Otherwise, v will change appreciably during the rotation. 16 The terminology here is hopelessly nonuniform. The \"strain tensor\" has many slightly differ- ent definitions and is denoted by many different symbols. About the best one can say of the usage here is that at least some other authors use it. 712 Chapter 16 Continuum Mechanics where D is the derivatives matrix, as defined in (16.76). The strain tensor defined in this way is a good measure of strain as the following two examples illustrate. EXAMPLE 16.5 Dilatation The strain tensor E at a certain point P in a solid is a multiple of the unit matrix, E = el (16.82) where e is a small number, positive or negative. Describe the displacement of points in the neighborhood of P (which, for convenience, we can take to be the origin). Since we are not interested in overall rigid displacements or rotations, we may as well assume that the point P does not move and that the immediate neighborhood of P is unrotated. In this case, the antisymmetric part A of D in (16.80) is zero and D = E. Thus a point at position dr (relative to P) is displaced to dr + du, as illustrated in Figure 16.18, where du = E dr = e dr. That is, the point dr is moved to (1 + e)dr. Since this statement is independent of the direction of dr, we conclude that the whole sphere of any small radius dr is enlarged, or", " dilated, in all directions by the factor 1 + e, and we refer to the strain (16.82) as a spherical strain or a dilatation. If e is positive, the sphere is actually enlarged; if e is negative, it is reduced. Figure 16.18 The strain (16.82) moves the point at dr radially out to (1 + e)dr. Thus any small sphere centered on P is just dilated by a factor of 1 e in all directions. For future reference, notice that since any volume is stretched by a factor of (1 + e) in all three directions, volumes are increased by a factor of (1 + e) 3 3e. (Remember that e << 1.) In other words, the dilatation E = el results in 1+ a fractional increase of 3e in any small volume; that is, Section 16.8 The Strain Tensor for a Solid 713 EXAMPLE 16.6 A Shearing Strain The strain tensor E at a certain point P in a solid has the form 16.84) (with y << 1) or, if we denote the elements of E by Eu, then E12 = 621 = y while all other cu are zero. Describe the displacement of points in the neighborhood of P (which we can take to be the origin again). As before we'll assume that there is no overall translation or rotation, so that E is the same as the derivatives matrix D, whose only nonzero elements are 8u 1 are au2 ar l = y. This implies that if we move out along the r2 axis, the only component of u that changes is u 1. Therefore, any point on the r2 axis is displaced sideways, in the r i 1 axis is displaced direction, as shown in Figure 16.19. Similarly, a point on the r upward in the r2 direction. The net effect is a shear in which the two axes are tilted toward one another as shown in the figure. (This is if y is positive; if y is negative, they tilt the other way.) The angle through which both axes tilt is equal to the parameter y, as long as y is small. du i = y dr2 2 y du2 = y dri r, Figure 1 6.1 9 Under the strain (16.84), points on the r 2 axis move in the r 1 direction and vice versa. The result is a shear in which the two axes tilt as shown. We see from this last example that the off-diagonal elements eu of the strain tensor are associated with shearing strains. In the same way, the diagonal elements are associated with stretching along the axes. For instance if E 11 is nonzero, then points on the r 1 axis are displaced along the axis (in addition to any sideways displacement), and the whole axis is stretched by a factor 1 + E 11. For this reason, the three diagonal elements E1 1, E22, and \u20ac 33 can be called the stretching elements of E. 714 Chapter 16 Continuum Mechanics Decomposition of the General Strain Tensor The last two examples lead us to our final maneuver with the strain tensor. We've seen that if E is diagonal and its diagonal elements are equal, Ell = E 22 = E33 = e, so that E = el, then the corresponding deformation is the simple dilatation of Example 16.5. Even if a given strain tensor E does not meet these conditions, you might guess that we could define e as its average stretch, that is, the average of the three diagonal elements, SUM e = 4611+ E22 + E33) 3 (16.85) and that the simple dilatation el would bear some useful relation to the original tensor E. Before I show that this is the case, I should mention that, in the theory of matrices, the sum of the diagonal elements is an important concept called the trace of the matrix. That is, for any (n x n) matrix M with elements m id, we define its trace, tr M, as the tr M = m i, + +Trinn (16.86) Thus another way to state the definition (16.85) of the average stretch e of any strain tensor E is that it is 1/3 of the trace: e= 3 tr E (16.87) The matrix el is a pure dilatation, which naturally changes the volume of any small region around the point of interest, and it can be shown (Problem 16.24) that it changes the volume by the same amount as the original strain E. Therefore, if we write E=e1 (16.88) we have expressed E as the sum of two separate strains, the first of which is a pure dilatation that changes volumes by the same amount as E and the second of which gives the same shearing strains as E but causes no change", " of volumes. We can call the first term, el, the spherical part of E. The second term, E' = E \u2014 el, is sometimes called the strain deviator or deviatoric part of E, presumably because it is the amount by which E deviates from the corresponding pure dilatation. Mathematically we can characterize the decomposition (16.88) by saying that the first term is a multiple of the unit matrix with the same trace as E and the second has zero trace. We shall see in the next section that this decomposition of E plays an essential role in the relation between strain and stress. EXAMPLE 16.7 A Numerical Example of Strain The strain tensor at a certain point in a solid is found to be \u20140.01 0.02 0.05 0.02 0.03 0.04. E = [ 0.05 0.04 0.04 (16.89) Section 16.9 Relation between Stress and Strain: Hooke's Law 715 Decompose this strain as in (16.88) into its spherical and deviatoric parts. The average stretch is easily seen to be e = 3 tr E = 0.02 and E' is then found by subtraction as E' = E \u2014 el. So e I = 0.02 0 0 0.02 0 0 0 0 0.02 and E' = 0.01 0.04.. (16.90) \u20140.03 0.02 0.05 0.02 0.05 0.04 0.02 It is easy to check that the original strain tensor E is indeed equal to el + E'. Notice that the trace of el is the same as that of E, as it should be, while the trace of E' is zero. 16.9 Relation between Stress and Strain: Hooke's Law The final step in writing down the equation of motion for a continuous solid is to find the relation between the stress and strain tensors, E and E. This relation, sometimes called the constitutive equation, corresponds to Hooke's law for a mass on a spring, expressing the force (or stress) in terms of the extension (or strain). It is reasonable to assume that, at least for small disturbances, the relation between stress and strain should be linear, and this is what I shall assume here. There certainly are plenty of examples of materials that fit this assumption reasonably well \u2014 a chunk of rubber or metal, or even the rock of the earth's crust. When the required relation is linear, it is called the generalized Hooke's law. To simplify the discussion still further (and this is a huge simplification), I shall assume that the solid is isotropic, which implies that the relation between I and E is independent of our choice of axes, or rotationally invariant. We wish to express the stress tensor / as a function E = f (E) of the strain tensor E. The function f must be linear, and it must be rotationally invariant. Linearity is a familiar property. To be rotationally invariant means this: If R denotes any rotation of our coordinate axes and MR the result of rotating a matrix M by the rotation R, then it must be true that f (ER) = [f (E)] R (16.91) for any strain E and any rotation R. That is, the stress corresponding to the rotated strain ER (left side of the equation) must be the same as the result of rotating the stress corresponding to E (right side of the equation). Now, we have seen that any strain tensor can be decomposed as the sum of its spherical and deviatoric parts: E=e1+E'. (16.92) This decomposition has two important properties. First, it is rotationally invariant; that is, when we rotate axes, each part rotates separately into the corresponding part of the rotated tensor. (The spherical part of E rotates into the spherical part of ER, and 716 Chapter 16 Continuum Mechanics likewise the deviatoric part.) Second, it is impossible to decompose E any further and retain this property.'? The crucial result, the proof of which is unfortunately beyond the scope of the mathematics I am assuming here, is this: If E = f (E) where the function f is linear and rotationally invariant, and if E is decomposed as in (16.92), then the most general possible form of the function f is this: where a and f3 are two constants (which depend on the material of which our solid is made) and e = tr E as usual. 18 The relation (16.93) is called the generalized Hooke's law, or just Hooke's law, and any solid that obeys it is called an elastic solid. It is often convenient to rewrite Hooke's law (16.93) in terms of E (rather than E'", " = E \u2014 el), to give l9 We can solve this for E in terms of E in two steps. Taking the trace of (16.93) we find that tr I = 3ae, so e = tr 1/3a. Substituting this into (16.94) we find that As you might guess, the constants a and /3 are related to the elastic moduli introduced at the end of Section 16.6 [Equations (16.55) to (16.57)]. Let's start with the bulk modulus. 17 1n the language of group theory, the two parts in (16.92) are irreducible. For a discussion of the necessary group theory see Chapter 10 of Mathematics for Scientists and Engineers by Harold Cohen, Prentice Hall (1992) or Chapter 16 of Mathematical Methods of Physics by Jon Mathews and R. L. Walker, W. A. Benjamin (1970). 18 Within the framework of group representations, the proof is amazingly simple: The linear function f must commute with all rotations. The decomposition (16.92) splits the space of all symmetric matrices into two irreducible subspaces (of dimensions 1 and 5 respectively). By Schur's lemma, the restriction of f to either of these irreducible subspaces can be at most multiplication by a scalar (which we can call a or 13) and we have proved (16.93). 19 This equation is often written as I = 3Xel + 2,uE, in which case A and,u, are called the Lame constants. Section 16.9 Relation between Stress and Strain: Hooke's Law 717 Bulk Modulus Imagine a solid subject only to an isotropic pressure, p (no shear stresses). In this case we know that the stress tensor has the simple form E = \u2014pl. Substituting this into Hooke's law (16.95) we find that E= 107 [\u2014a + (a \u2014 13)]131 \u2014 121 -,8 a (16.96) That is, the strain tensor E is also a multiple of the unit matrix, which we could write as E = el, with e= \u2014pla. But we know from (16.83) that e= idV I V. Comparing these two expressions for e, we conclude that p = dV I V, and comparing this with the definition (16.56) of the bulk modulus, we see that a = 3BM. (16.97) Shear Modulus Let us consider next the simple shearing strain E given in (16.84) of Example 16.6 and illustrated in Figure 16.19. Since this has zero trace, e = 0, and Hooke's law (16.94) reduces to the simple form E_ /3E. In particular, the stress responsible for the shear is F \u2014 A = al2 = /36 12 PY (16.98) We need to compare this with the definition (16.57) of the shear modulus. Unfortu- nately, (16.57) refers to the strain of Figure 16.13, which is not quite the same as that of Figure 16.19. Specifically, in Figure 16.19 both axes tilt inward by an angle y, whereas in Figure 16.13 the x axis turns through angle 8 while the y axis is un- changed. A moment's thought should convince you that the displacement of Figure 16.13 is a combination of a simple shear, as in Figure 16.19, followed by a rotation to bring the y axis back to its original direction. This means that the angle 9 of Figure 16.13 equals twice the angle y of Figure 16.19, 0 = 2y. Putting this into the definition (16.57) of the shear modulus, we find that dy dx A = SM\u2014 = SW+ = 2SM y, and comparing with (16.98), we conclude that the constant /3 is just twice the shear modulus, f3 = 2 SM. (16.99) 718 Chapter 16 Continuum Mechanics Young's Modulus We can similarly identify Young's modulus in terms of the constants a and /3, but I shall leave this as an exercise (Problem 16.27). The result is that YM = 3afi BM SM 9 2a + /3 3BM + SM' (16.100) where the last expression results from substituting (16.97) for a and (16.99) for /3. An interesting feature of this result is that it shows that only two of the three elastic moduli are independent. For instance, if we know BM and SM we can calculate YM. (See Problem 16.26.) 16.10 The Equation of Motion for an Elastic Solid Let us consider an infinitesimal volume dV of", " an elastic solid. Its mass is g dV, where Q is its density, and its position is r + u(r, t), where r is its equilibrium position and u(r, t) is its displacement from equilibrium. [The displacement u(r, t) depends on r and t, but the equilibrium position r of any given piece of the solid is fixed and independent of t.] Newton's second law applied to this volume reads Q dV \u2014 =Fv oi+ f ear (16.101) a 22u a t where F.I denotes the volume force and F., the net surface force on dV. The volume force is proportional to dV, and to be definite, I shall assume it is just the force of gravity, so that Fv\u201ei= g gdV. (16.102) We know that the surface force on any small element of the surface of our volume is EdA, where as usual E is the 3 x 3 stress tensor and dA is the vector labeling the surface element (considered to be a 3 x 1 column vector in the matrix product). Thus the net surface force is Fsur = f I dA (16.103) where the integral runs over the closed surface of the volume under consideration. To cast this integral in a more convenient form we need to use the divergence theorem. This theorem was introduced in Chapter 13 on Hamiltonian mechanics, but in case you didn't read that chapter I'll review the main ideas here. (For more details, see Section 13.7 and Problems 13.31 to 13.34.) The divergence theorem states that any surface integral of the form f v dA over a closed surface S is equal to a certain volume integral, namely v dA = f vdV. s (16.104) Section 16.10 The Equation of Motion for an Elastic Solid 719 Here v is any vector, V is the volume enclosed by the surface S, and V v is the divergence of v, v v aVz \u2014 ax \u00b1Y ay aVz aZ 3. 1=1 av 1 1 (16.105) where I have introduced the convenient shorthand a..== a arj Rewritten in terms of components, the divergence theorem (16.104) reads f EvidA ; = E a.v dV'(16.106) and the ith component of the surface force (16.103) is (Fsur)i =v;i (16.107) For each fixed value of i (1, 2, or 3), this has the form of the left side of (16.106), so can be replaced by a term with the form of the right side, We can write this in more compact vector form if we introduce a vector V E defined so that its ith component is2\u00b0 (Fsur)i = f E ai cri; dV. (V 3 EaJaJi Fsur = f VEd V. (16.108) (16.109) (16.110) Note well that since V is a vector operator while E is a tensor, V E is a vector. Armed with this notation we can write (16.108) as This result is valid for the surface force on any volume, small or large. However, our interest is in an infinitesimal volume dV. For a small enough volume, the integrand is constant and the result becomes just Fsur = V EdV (16.111) for any small volume dV. 20 For an arbitrary tensor M, the elements m u and m Ji aren't necessarily the same, and we have to be careful of the order of the indices i and j. Luckily E is symmetric, so it doesn't matter which order we write them in. 720 Chapter 16 Continuum Mechanics Let us now return to the equation of motion (16.101) and substitute (16.102) for Fvoi and (16.111) for Fsur. When we do this, every term acquires a factor of d V, which we can cancel to give a2u \u2014at2 \" \u00b1 V (16.112) This important equation is easy to understand. The left side represents the ma of ma = F. The first term on the right represents the force of gravity (or more generally the body force) and the second the surface force. Before we can use this equation of motion, we must use Hooke's law to replace the stress tensor E by the strain tensor E. In the form (16.94), Hooke's law reads = (a \u2014 p)el \u00b1 PE (16.113) or, in terms of its components, crii = (a \u2014 fi)ecS ii Pep (16.114) where Sp denotes the Kronecker delta symbol 6. 1 if j =i 0 if / t (16.115) Since cii", " = i (aj u i + ai tt j ), the average stretch e is e = Ecii. \u201e Putting these results into (16.114), we find a..=(a- - p 3 P)6pv u +.. 2 13(a. + au) u IJ JI' This lets us evaluate V / for use in (16.112): (V.)= a.cr i t (a \u2014 /3) 3 s..a J i.u)+ fiEa.a.u.+ 1 13Ea.a.u. JIJ 2 JI' Each of the terms in this ugly result simplifies. In the first term, notice that Ei SJia; = ai. In the second term, E 81 8,141 =ai Ej aj u i =ai V u, and in the third, E i aj a = J V 2. Therefore v.E=1(CY \u2014 p)v(v.u)+ 13v(v.u)+1/3v 2u + 4,8>v(v 113V 2U = (BM + ISM)V(V u) SMV2u (16.116) where in the last line I have used (16.97) and (16.99) to rewrite a and /3 in terms of the bulk and shear moduli, BM and SM. Section 16.11 Longitudinal and Transverse Waves in a Solid 721 Finally we are ready to write the equation of motion of an elastic solid in a usable form. Sustituting (16.116) into (16.112), we find In the next section, we'll use this equation, often called the Navier equation (after the French engineer Claude Navier, 1785-1836), to derive the two main kinds of wave in an elastic solid. 16.11 Longitudinal and Transverse Waves in a Solid It is well known that there are two main kinds of wave in an elastic solid\u2013longitudinal and transverse. To show this, we'll examine the Navier equation (16.117). We'll assume that gravity is unimportant and set g = 0. (This is usually an excellent ap- proximation, one exception being very slow \u2014 r 200 s \u2014 free oscillations of the earth, for which gravity is important.) Without loss of generality, we'll look for a plane wave propagating in the x (that is, r 1) direction, so that u depends only on x and t. Let us first examine the possibility of a longitudinal disturbance, for which the displacement u would be in the direction of propagation, u = [ux (x, t), 0, 0]. In this case, V u = aux jax and the only nonzero component of V (V u) is its x component, which is 8 2 u x /ax 2. The only nonzero component of V 2u is likewise the x component, which is also equal to 8 2u x /ax 2. Putting all of this together in the equation of motion (16.117), we obtain This is the wave equation, with wave speed 8 2 ux Q ate = (BM + '1SM) 8 ux 2 axe BM + iSM Clong = Q (16.118) (16.119) We conclude that longitudinal waves are indeed possible, with speed c iong given by (16.119). If instead we look for a transverse (or shear) wave, traveling in the x direction but with the displacement in the y direction, then u would have the form u = [0, u y (x, t), 0]. 722 Chapter 16 Continuum Mechanics In this case V u = 0, while V2u has only a y component equal to 8 2uy /a.t2, so (16.117) becomes a2u ate Y \u2014SM a2u axe C an = SM Q This is the wave equation, with wave speed (16.120) and we conclude that transverse waves are possible, with speed Can given by (16.120). Notice that Cl ong > Ctran. Therefore, if longitudinal and transverse signals set out simultaneously from some source, the longitudinal one will arrive first at a distant detector. For instance, it is well known in the earth sciences that the longitudinal waves from a distant earthquake arrive before the transverse ones, and this gives seismologists a way to measure how far away an earthquake or explosion occurred. For this reason, longitudinal waves are also called primary or P waves and transverse secondary or S waves. EXAMPLE 16.8 Waves in Rock The elastic moduli of the material of the earth's crust vary, but representative values would be BM 40 GPa and SM ti 25 GPa. (These are the approximate values for granite, whose density is about 2.7 x 10 3 kg/m3, and 1", " GPa = 10 9 Pa = 109 N/m2.) What would be the speed of longitudinal and transverse seismic waves in rock with these values? According to (16.119), the longitudinal speed is (40 \u00b1 4 x 25) x 109 N/m2 Clong = 2.7 x 103 kg/m3 ti 5.25 km/s. Similarly, from (16.120) we find the transverse speed to be Ctran 25 x 109 N/m2 2.7 x 103 kg/m3 ti 3.0 km/s. A striking feature of the formula (16.120) for the transverse wave speed is that if the shearing modulus SM is zero \u2014 as it is in fluids \u2014 then c u.., is zero. This suggests (correctly) that fluids cannot support transverse waves. 21 A beautiful application of this result is that tranverse seismic waves (unlike longitudinal) are found not to propagate through the center of the earth, showing that some region near the earth's center (namely, the outer core) is liquid. 21 0f course this argument is not quite watertight since our derivation of (16.120) assumed a solid medium. Nevertheless the suggested conclusion is correct, and correct for essentially the right reason: Transverse waves require a transverse (shearing) restoring force, which a fluid cannot supply. Section 16.12 Fluids: Description of the Motion* 723 16.12 Fluids: Description of the Motion* * As usual, sections marked with an asterisk can be omitted on a first reading. In the last four sections we have focussed primarily on the motion of solid contin- uous media. In the final two sections of this chapter, I would like to give a brief introduction to the motion of fluids. Unfortunately, the analysis of a general fluid \u2014 particularly a viscous fluid \u2014 is complicated, and, to keep this chapter from growing unconscionably long, I shall restrict myself mostly to inviscid or ideal fluids, that is, fluids whose viscosity is negligible. One might argue that to neglect viscosity is to throw the baby out with the bathwater, but the fact is that there are many problems in fluid motion where it is reasonable to neglect viscosity, and, more important, all of the tools needed to discuss inviscid fluids are needed for the analysis of viscous fluids, so the discussion here is really an essential preliminary to any subsequent study of viscous fluids. Furthermore, several ideas discussed here \u2014 the convective derivative and the equation of continuity, in particular \u2014 are equally applicable to either case. Material versus Spatial Descriptions So far we have analyzed what is happening in a continuous medium by specifying that the piece of material that was originally at position r is now at position r u(r, t). This approach is sometimes called the material description, since it focuses on a particular piece of the material. It turns out that in discussing fluids it is often more convenient not to follow the individual pieces of fluid, but rather to specify what is happening at each fixed point in space. Thus we might give the velocity v(r, t), the density Q (r, t), and so forth, of the fluid at each fixed point r (and time t). This approach is often called the spatial description. 22 Some advantages of the spatial approach when discussing a fluid are reasonably obvious: With a solid, each material piece of the solid usually has a well-defined equilibrium position r, and this is what we use to label each piece [giving the piece's displacement u(r, t) from r]. In a fluid, the pieces of the fluid do not generally have a unique equilibrium position. We could, of course, use the piece's initial position as a reference, but in a problem of fluid flow, the pieces usually stray inconveniently far from their initial positions. Thus in discussing a fluid, it is usually more convenient simply to focus on what is happening at each fixed point in space \u2014 the spatial description. The Material Derivative The main disadvantage of the spatial description emerges most clearly when we try to use Newton's second law. The acceleration a in F = ma is, of course, the acceleration of any small piece of the fluid. Unfortunately, if v(r, t) is the fluid's velocity at a 22 The material and spatial descriptions are often called the Lagrangian and Eulerian descrip- tions, respectively \u2014 names that are neither historically accurate (both methods being due to Euler) nor especially easy to remember. 724 Chapter 16 Continuum Mechanics point r, then a is not just a v/a t. The partial derivative av/at is the rate of change of the fluid velocity at the fixed point r, whereas the acceleration we want is the rate of change of v of a given piece of fluid as it moves. This same distinction applies to any other parameter and is, perhaps, easier to visualize in the case of a", " scalar, such as the density or temperature. Consider, for example, the density of a small element of the fluid. If at time t the element is at position r, then the required density is Q (r, t). But a short time dt later, the element will have moved to a new position r + dr, where dr = v dt, so the density is now Q(r + dr, t + dt). (Note well how, if we wish to follow the material element, we must evaluate the new density at the new time t + dt and the new position r \u00b1 dr.) Thus the change in the density of our volume element is clQ Q(r \u00b1 dr, t + dt) \u2014 Q(r, t) = \u2014dt + dr VQ aQ at a = \u2014dt + (vdt) VQ. Dividing both sides by dt we find for the time derivative of Q This derivative is called the material derivative since it gives the rate of change of as we follow the motion of a material piece of the fluid. 23 We can apply a similar argument to other parameters of the fluid. For example, dp/dt (defined in exactly the same way) would be the rate of change of the pressure as we follow the motion of a material element of fluid. In particular, we can examine each component of the fluid's velocity and putting the three components together find This derivative is, of course, the acceleration of the volume element of fluid. Armed with this result, we are ready to write down the equation of motion for an inviscid fluid. Equation of Motion for an Inviscid Fluid Consider now a small volume element dV of fluid with mass p dV and acceleration given by (16.122). Newton's second law implies that 23 Other names for the material derivative are total or convective derivative. Some authors use the symbol D/Dt (instead of d/dt) to emphasize the special character of this derivative. Section 16.12 Fluids: Description of the Motion * 725 dV \u2014 = F = Fvoi Fsur. (16.123) dv dt I shall assume that the only volume force is gravity, so that F ool = p dVg, where g is the acceleration of gravity. We know from (16.111) that the surface force is Fsur = V EdV and from (16.70) that E = \u2014 pl. (This last was derived for a static fluid, but the argument needed only that the viscosity be negligible.) To evaluate V I, we'll take components: (V I), = E a, cyji = \u2014 E a, (ps,i) \u2014a, p. Therefore, V E = \u2014 V p. Putting all these results together in (16.123), we see that every term contains a factor of dV, and, canceling this factor, we arrive at the equation of motion for an inviscid fluid, Bernoulli's Theorem As a first simple application of the equation of motion (16.124), let us derive a familiar result from most introductory physics courses, Bernoulli's theorem. This theorem is usually stated for the case of steady flow of an incompressible, inviscid fluid, and this is the case I'll consider here. That the flow is steady means that the parameters p, v, and o at any fixed point r are constant; that is, the partial derivatives ap/at and so on are all zero. That the fluid is incompressible means that the density doesn't change, do/dt = 0. (Notice that in the case of the density, both B oot and 4/dt are zero.) I shall consider the case that g is uniform and in the negative z direction, so that we can write g = \u2014 V (gz). 24 Let us make this replacement in (16.124), and then dot the whole equation with v. This gives pv \u2014 + QV V (gz) v -V p = 0. (16.125) dv dt We can simplify all three terms in this equation. For the first term, note that v dv Idt = 4d (v 2)Idt. For the second and third terms, note that v V f = df/dt \u2014 of/at, for any function f. For the functions of interest here, of/at = 0, so v V f = df I dt. With these replacements, (16.125) becomes 1 dv 2 0 \u2014 dt \u2014 2 + d(gz) dp dt \u2014 = 0. dt (16.126) 24 Even if g is not uniform, it is certainly conservative, so we can always introduce a function cl) (called the gravitational potential) such that g = \u2014V O. To avoid introducing unfamiliar symbols, I decided to treat the common case that (", "1) = gz. 726 Chapter 16 Continuum Mechanics Finally, since dQ/dt = 0, we can bring the factors of Q inside the derivatives and we find that d ( 1 dt 2 pv - + pgz p) = O. (16.127) This asserts that the quantity ill = z Q21 Qgz p is constant as we move along with 2 any material element of the fluid. In other words, W is constant along any streamline, which is precisely the content of Bernoulli's theorem for steady, incompressible, inviscid flow. Because the term 40, v 2 appears with the pressure p in the Bernoulli and is associated with the motion, IQ v2 is sometimes called the dynamic pressure. The Equation of Continuity The conservation of mass implies an important relation, called the equation of conti- nuity, between the density and velocity of any fluid, viscous or inviscid. If you have taken a course in electromagnetism, you may have met a corresponding relation that reflects the conservation of charge. To prove the relation, consider a small volume d V of fluid. As this volume moves, its mass pd V cannot change. Therefore, d \u2014 dt p dt dt (p dV) = dV \u2014 + p\u2014(dV) = 0. (16.128) The rate of change of a moving volume was evaluated in Equation (13.59) of Section 13.7. (If you did not read Chapter 13, you could read just the proof of this result, starting at the subsection \"Changing Volumes\" on page 544.) For any volume V (small or large) the result is For an infinitesimal volume d V, this reduces to dV dt d \u2014 dt f V vdV. V (dV) = V v dV Inserting this in (16.128) and canceling the common factor of dV, we arrive at the equation of continuity, or equivalently (see Problem 16.34) This relation plays a crucial role in fluid dynamics, as does the corresponding relation (with 0 replaced by the charge density) in electrodynamics. We shall use it in the next section in deriving the speed of sound in a fluid. Section 16.13 Waves in a Fluid* 727 16.13 Waves in a Fluid* *As usual, sections marked with an asterisk can be omitted on a first reading. Armed with the equations of motion and continuity, we can now discuss the possibility of waves in an inviscid fluid. We imagine a fluid which undergoes a small disturbance from equilibrium, so that it acquires a small (presumably oscillatory) velocity v and its pressure and density become P = Po --F. 11 = Q.+ Q'. (16.131) (16.132) and to give Here p 0 and po are the equilibrium values, and p' and p' are small increments. Notice first that in equilibrium, the equation of motion (16.124) implies simply that Dog \u2014 V p\u201e = 0. (16.133) For the actual situation, we insert (16.131) and (16.132) into the equation of motion (Q.+ o') (\u2014 +y vy) = (po + \u2014 v(po + p'). ay at This ugly equation simplifies. First, by the equilibrium condition (16.133), the first and third terms on the right cancel exactly. Second, by the assumption that the disturbance is small, we can drop any terms that are second order or higher in the small quantities v, p', or p', or their derivatives. Thus we can ignore the term v Vv, and likewise the terms involving p' on the left. This leaves us with ay at = g VP' (16.134) Finally, it is not hard to show (Problem 16.38) by putting in realistic numbers that the term 0'g on the right is negligible compared to V pi. Thus the equation of motion for our small disturbance is just 728 Chapter 16 Continuum Mechanics We can treat the equation of continuity (16.130) in a similar way. Inserting (16.131) and (16.132), we get (as you should check) = \u2014Q,,V v \u2014 v V Q o. at (16.136) For essentially the same reasons that the first term on the right of (16.134) was negligible, the last term here can be neglected (see Problem 16.38 again), so the equation of continuity becomes The equations of motion (16.135) and continuity (16.137) give two equations for the three variables v, p', and Q'. We get a third equation by looking at the definition (16.56) of the bulk modulus, p = BM (\u2014dV I V). As stated, this relates the pressure to the", " corresponding change of volume, but it applies equally to a change in pressure, in which case it would read dp = BM (\u2014d V IV). (For a moment we'll call the volume of interest V, so dV can be its increment.) Now, the mass of a given element of fluid cannot change, so the quantity QV must be a constant. Therefore Q dV + V clQ = 0, or dV I V = \u201441p. Combining these two results we see that Now, in our case, the change of pressure dp is what we've been calling p'. Likewise, 4 is what we've been calling Q', and the original density is Q\u201e. Therefore, in our current notation (16.138) (16.139) We can use this last result to eliminate Q i from the equation of continuity (16.137) to give we obtain If we differentiate this with respect to t and invoke the equation of motion (16.135), 821/ ate = BM V = V2p t av BM at Q0 where in the second expression I interchanged the order of the spatial and time derivatives, and in the third expression I used the equation of motion (16.135). This result is the three-dimensional wave equation with wave speed dp = BM \u2014 do. Q = BM (Z. Q0 ap r at = \u2014BM V v. Section 16.13 Waves in a Fluid* 729 C BM Q. (16.140) I'll show in just a moment that the waves in a fluid are necessarily longitudinal. Since longitudinal waves in a continuous medium are what we generally call sound waves, we have proved that the speed of sound in a fluid is c =.03M/po. EXAMPLE 16.9 Speed of Sound in Water Given that the bulk modulus of water is 2.2 GPa, what is the speed of sound in The density of water is, of course, 1000 kg/m 3, so the speed of sound as given water? by (16.140) is c = 2.2 x 109 N/m2 103 kg/m3 = 1.5 km/s. To show that waves in a fluid are longitudinal, let us imagine a wave traveling in the direction of a unit vector n, so that f (n r \u2014 ct). (16.141) According to the equation of motion (16.135), this implies that a v \u2014 \u2014 1, V p at 00 1 \u2014 \u2014 00 V f (n r \u2014 ct) = -- f t (n r \u2014 ct) n 00 where f' denotes the derivative of f with respect to its argument. This relation can be immediately integrated to give the velocity of the fluid: 25 v = -- n f ft (n r \u2014 ct) dt = \u2014 11 Qo f (n r \u2014 ct) = (16.142) coo CQ0 (If you don't see the integration here, try the change of variables = n r \u2014 ct.) This result has two important features. First, we see that the fluid velocity is proportional to the pressure p'. In particular, in a sinusoidal wave, the velocity will oscillate in phase with p'. Second, the fluid velocity is in the direction of propagation; that is, the wave is longitudinal. As we anticipated in Section 16.11, a fluid cannot support transverse waves. 25 In the third expression, I have omitted a constant of integration. This \"constant\" could depend on r (though certainly not on t) and could, in fact, be nonzero. For example, we could imagine a disturbance that included a constant uniform velocity superposed on the oscillatory wave motion. However, such a time-independent velocity is not part of what we would describe as the wave, so we can take it to be zero in our discussion of the wave motion. 730 Chapter 16 Continuum Mechanics Principal Definitions and Equations of Chapter 16 The One-Dimensional Wave Equation The one-dimensional wave equation is a2u 2 =c n2\u201e axe x 2 [Eq. (16.4)] The general solution is u(x, t) = f (x \u2014 ct) + g(x ct) [Eq. (16.10)] the first term of which represents a disturbance traveling to the right and the second one traveling to the left. The Three-Dimensional Wave Equation 2 D2 p. = c a2P ate [Eq. (16.39)] Stress, Strain, and the Elastic Moduli stress =, strain = fractional deformation. [Eqs. (16.58) & (16.59)] force area elastic modulus (YM, BM, SM) = [Eq. (16.60)] stress corresponding strain The Stress Tensor a small element of area dA is The stress tensor is a 3", " x 3 symmetric matrix E defined so that the surface force on F(dA) = EdA. [Eq. (16.66)] The Strain Tensor for a Solid If u(r) denotes the displacement of an element of solid from its original position r, the strain tensor is the 3 x 3 symmetric matrix E with elements Eij = \u2014 1 (atli 2 ari at j ) arl [Eq. (16.81)] which can be decomposed as the sum of two terms: E = el + E' [Eq. (16.88)] where e = 3trE. Principal Definitions and Equations of Chapter 16 731 Generalized Hooke's Law For an isotropic solid, the most general relation between the stress and strain tensors is E = ael + OE' [Eq. (16.93)] where the constants a and /3 can be related to the three elastic moduli. [Eqs. (16.97), (16.99) & (16.100)] Equation of Motion for an Elastic Solid The equation of motion for an isotropic elastic solid is the Navier equation: a 2u Q ate = pg + (BM + 3sm) v(v u) + SMV2u. [Eq. (16.117)] Waves in a Solid or secondary or S) waves are The speeds of longitudinal (or primary, or P) waves and of transverse (or shear, BM + 1SM Clong = p and Ctran = [Eqs. (16.119) & (16.120)] SM. 0 The Material Derivative in a Fluid The time rate of change of a parameter (the density, temperature, or velocity) as we follow a material element of a moving fluid is the material derivative cl n dt _ at + v \\'' [Eq. (16.121)] Equation of Motion for an Inviscid Fluid dv p dt = pg \u2014 V p. [Eq. (16.124)] Equation of Continuity Conservation of mass implies the equation of continuity: a o at +v ( Qv). o. [Eq. (16.130)] 732 Chapter 16 Continuum Mechanics Waves in a Fluid The speed of longitudinal waves in a fluid is C BM p. [Eq. (16.140)] but a fluid cannot support transverse waves. Problems for Chapter 16 Stars indicate the approximate level of difficulty, from easiest (*) to most difficult (***). SECTION 16.1 Motion of a Taut String 16.1 * Verify that the quantity c = bt that appears in the wave equation for a string does indeed have the units of a speed. 16.2 ** The wave equation (16.4) is the equation of motion for a continuous string, as illustrated in Figure 16.1(a). You can obtain this equation as the limit as n oo of the equations for the n discrete masses of Figure 16.1(b). You need to be careful with the limiting process. As n oo, the spacing b between the masses (see Figure 16.20) and the individual masses m must both go to zero in such a way that the linear mass density m lb approaches p, the density of the continuous string. You can guarantee this by taking m = ktb. Write down Newton's second law for the position u i of the ith mass and show that it goes over to the wave equation as b 0. SECTION 16.2 The Wave Equation 16.3 * Let f () be an arbitrary (twice differentiable) function. Show by direct substitution that f \u2014 ct) is a solution of the wave equation (16.4). 16.4* Show that if we make the change of variables = x \u2014 ct and rl = x ct, then, as in (16.7), r2 \" \u2014 a2 ax2 82 u ate \u2014 4C2 a a \" a a q. 16.5 * (a) Show that u = g(x ct) is a solution of the wave equation (16.4) for any twice differentiable function g(0. (b) Argue clearly that this solution represents a disturbance that travels undistorted to the left. b Figure 16.20 Problem 16.2 Problems for Chapter 16 733 16.6 * There is a small flaw in Example 16.1 (page 686). In Equation (16.14) I omitted a constant of integration, so the equation should really have read f (x) \u2014 g(x) = k. Show that this still gives the same final answer (16.15). 16.7 ** [Computer] Make plots of the two triangular waves of Example 16.1 (page 686) at several closely spaced times and then animate them. Describe the motion. For the", " purposes of the plot you may as well take the speed c, the height of the triangle at time 0, and the half width of the base all equal to 1. Make your plots for lots of times ranging from t = \u20144 to 4. 16.8 ** [Computer] Make plots similar to Figure 16.5 of the standing wave (16.18) for several equally spaced times from t = 0 to r, the period. Take 2A = 1 and k = w = 27r. Animate your pictures and describe the motion. SECTION 16.3 Boundary Conditions; Waves on a Finite String 16.9 ** The motion of a finite string, fixed at both ends, was determined by the wave equation (16.19) and the boundary conditions (16.20). We solved these by looking for a solution that was sinusoidal in time. A different, and rather more general, approach to problems of this kind is called separation of variables. In this approach, we seek solutions of (16.19) with the separated form u(x, t) = X (x)T (t), that is, solutions that are a simple product of one function of x and a second of t. [As usual, there's nothing to stop us trying to find a solution of this form. In fact, there is a large class of problems (including this one) where this approach is known to produce solutions, and enough solutions to allow expansion of any solution.] (a) Substitute this form into (16.19) and show that you can rewrite the equation in the form T\"(t)I T (t) = c 2 X\"(x)/ X (x). (b) Argue that this last equation requires that both sides of this equation are separately equal to the same constant (call it K). It can be shown that K has to be negative. 26 Use this to show that the function T (t) has to be sinusoidal \u2014 which establishes (16.21) and we're back to the solution of Section 16.3. The method of separation of variables plays an important role in several areas, notably quantum mechanics and electromagnetism. 16.10 ** Using the integral (16.33), show that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for n even and given by (16.34) for n odd. 16.11 ** [Computer] Make plots similar to Figure 16.8 of the wave of Example 16.2 but from t = 0 to r, the period, and for more closely spaced times. Animate your pictures and describe the motion. 16.12 ** Consider a semi-infinite string, fixed at the origin x = 0 and extending far out to the right. Let f () be a function that is localized around the origin, such as the function of Figure 16.4(a). (a) Describe the wave given by the function f (x ct) for a large negative time to. (b) One way to solve for the subsequent motion of this wave on the semi-infinite string is called the method of images and is as follows: Consider the function u = f (x ct) \u2014 f (\u2014x ct). (The second term here is called the \"image.\" Can you explain why?) Obviously this satisfies the wave equation for all x and t. Show that it coincides with the given wave of part (a) at the initial time t o and everywhere on the semi-infinite string. Show also that it obeys the boundary condition that u = 0 at x = 0. (c) It is a fact that there is a unique wave that obeys the wave equation and any given initial and boundary conditions. Therefore the wave of part (b) is the solution for all times (on our semi-infinite string). Describe the motion on the semi-infinite string for all times. 26 Actually this isn't hard to show. Look at the equation X\"(x)/X(x) = K/c 2. You can show that if K > 0 there are no solutions satisfying the boundary conditions that X (0) = X (L) = 0. 734 Chapter 16 Continuum Mechanics 16.13 ** In connection with Equation (16.31), I claimed that any function on the interval 0 < x < L can be expanded in a Fourier series containing just sine functions. This is at first sight very surprising since one is used to the claim that the general Fourier series requires sines and cosines. In this problem, you'll prove this surprising claim. Let f (x) be any function defined for 0 < x < L. We can define a function f (x) for all x by setting it equal to the given function in the original interval and requiring that f (\u2014x) \u2014 f (x) and f (x 2L) = f (x). (16.143) for all x", ". Prove that this defines a function which is (1) periodic with period 2L, (2) odd, and (3) the same as the original f (x) on the original interval. Write down the ordinary Fourier expansion for this new f (x) and show that the coefficients of the cosine terms are all zero. This establishes the possibility of expanding the original function in terms of sines alone. 27 Bearing in mind that the period of the new function is 2L, write down the standard formula (5.84) for the expansion coefficients and show that your answer agrees with (16.33). The Fourier sine series is especially convenient for discussing functions that are zero at the end points x = 0 and L. 16.14 *** [Computer] A taut string of length L = 1 is released from rest at t = 0, with initial position u(x, 0) 2x [0 x 4] 2(1 \u2014 x) < x < 1]. (16.144) Take the wave speed on the string to be c = 1. (a) Sketch this initial shape and find the coefficients B, in its Fourier sine series (16.31). (b) Make plots of the sum of the first several terms for several closely spaced times between t = 0 and r, the period. Animate your plots and describe the motion. SECTION 16.4 The Three-Dimensional Wave Equation 16.15 ** Let f () be any function with first two derivatives f V) and f\"(0, and let n be an arbitrary fixed unit vector. (a) Show that V f (n r \u2014 ct) = nf'(n r \u2014 ct). (b) Hence show that f (n r \u2014 ct) satisfies the three-dimensional wave equation (16.38). (c) Argue that f (n r \u2014 ct) represents a signal that is constant in any plane perpendicular to n (at any fixed time t) and propagates rigidly with speed c in the direction of n. 16.16 ** Let f (r) be any spherically symmetric function; that is, when expressed in spherical polar coordinates, (r, 8, 0), it has the form f (r) = f (r), independent of 8 and 0. (a) Starting from the definition (16.38) of V 2, prove that V2 f = - (rf). 1 8, \u2014 0r2 r (b) Prove the same result using the formula inside the back cover for V 2 in spherical polar coordinates. (Obviously, this second proof is much simpler, but the hard work is hidden in the derivation of the formula for V 2.) only sines. 27 But note that it has the form E Bn sin(n7rx I L). The usual Fourier series has sines and cosines, but their argument is 2n7 rx /L. Thus the new Fourier sine series has, in a sense, twice as many terms to make up for having Problems for Chapter 16 735 SECTION 16.6 Stress and Strain: the Elastic Moduli 16.17 ** In Section 16.1 we derived the wave equation for transverse waves in a taut string. Here you will examine the possibility of longitudinal waves in the same string. Suppose that an element of string whose equilibrium position is x is displaced a short distance in the x direction to x u(x, t). (a) Consider a short piece of string of length 1 and use the definition (16.55) of Young's modulus YM to show that the tension is F = A YM au/ax, where A is the cross sectional area of the string. [If the string is already in tension in its equilibrium position, this F is the additional tension, that is, F = (actual \u2014 equilibrium).] (b) Now consider the forces on a short section of string dx and show that u obeys the wave equation with wave speed c =,/YM/Q where p is the density (mass/volume) of the string. SECTION 16.7 The Stress Tensor 16.18 * Figure 16.15 is an end view of a triangular prism, whose three faces are labeled by the vectors dA 1, etc. (The magnitude of dA 1 is the area of the corresponding surface and the direction is normal to it. There are two more faces parallel to the plane of the paper, but these do not concern us.) The ends of the three faces form a closed triangle. Expain clearly why this implies that dA dA 2 dA3 = 0. 16.19 * Let n 1 and n2 be any two unit vectors and P a point in a continuous medium. F(n 1 dA) is the surface force on a small area d A at P with unit outward normal n 1, so n2 F(n i d", "A) is the component of that force in the direction of n 2. Prove Cauchy's reciprocal theorem that n2 F(n 1 dA) = n 1 F(n 2 dA). 16.20 ** It is found that the stress tensor at any point (x, y, z) in a certain continuous medium has the form (with an unspecified, convenient choice of units) XZ z2 0 = [ z2 \u2014y 0 _ (16.145) Find the surface force on a small area dA of the surface x 2 + y 2 + 2z2 = 4 at the point (1, 1, 1). 16.21 ** At any given point P of a continuous medium, the surface forces are given by the stress tensor, which is a real symmetric matrix L It is a well-known theorem of linear algebra (see the appendix) that any such matrix can be brought into diagonal form by a suitable rotation of the Cartesian coordinate axes. Use this to prove that at any point P there are three orthogonal directions (the principal stress axes at P) with the property that the surface force on any surface normal to one of these directions is exactly normal to the surface. 16.22 *** Show that if the stress tensor E is diagonal (all off-diagonal elements zero) with respect to any choice of orthogonal axes, then it is in fact a multiple of the unit matrix. This gives an alternative and elegant proof that if there are no shearing stresses (in any coordinate system) then the pressure forces are independent of direction. To do this problem, you need to know how the elements of a tensor transform as we rotate our coordinate axes, as described in Section 15.17. Assume that with respect to one set of axes E is diagonal but that not all three diagonal elements are equal. (For example, 6 11 a33.) It is not hard to come up with a rotation \u2014 that of Equation (15.36) will do \u2014 such that in the rotated system a1 3 0. 736 Chapter 16 Continuum Mechanics SECTION 16.8 The Strain Tensor for a Solid 16.23 * An important tool in the development of the strain tensor was the decomposition (16.79) of a matrix M into its antisymmetric and symmetric parts. Prove that this decomposition is unique. [Hint: Show that if M = MA + Ms where MA and Ms are respectively antisymmetric and symmetric, then MA = 4 (M - M) and Ms = 4 (M + M).] 16.24 * Write out the components of the displacement (16.77), u(r) = 0 x r, for a small rotation 0 and verify that the derivatives matrix is given by Equation (16.78). 16.25 *** At a certain point P (which you can choose to be your origin) in a continuous solid, the strain tensor is E. Assume for simplicity that whatever displacements have occurred left P fixed and the neighborhood of P unrotated. (a) Show that the x axis near P is stretched by a factor of (1 + E11). (b) Hence show that any small volume around P has changed by dV I V = tr E. This shows that any two strains that have the same trace dilate volumes by the same amount. In the decomposition E = el + (16.88), the spherical part el changes volumes by the same amount as E itself, while the deviatoric part E' doesn't change volumes at all. SECTION 16.9 Relation between Stress and Strain: Hooke's Law 16.26 * The table below gives the three elastic moduli for several materials. According to (16.100) Young's modulus for any given material can be calculated if we know the bulk and shear moduli. Using the data for BM and SM, calculate YM for each of the materials and compare with the given values in the third column. (The densities will be needed for Problem 16.32.) Elastic Moduli (in GPa) and Densities (in g/cm 3) Material BM SM YM Iron Steel Sandstone 90 140 17 Perovskite 270 150 390 Water 2.2 40 100 7.8 80 200 7.8 6 0 16 0 1.9 4.1 1.0 16.27 *** Consider a taut wire or rod lying along the x axis. To define Young's modulus YM we apply a pure tension along the axis; that is, a stress with a ll > 0 and all other at] = 0. (a) Use Equation (16.95) to write down the corresponding strain tensor E. (b) Argue from the definition (16.55) of Young's modulus to show that YM = 1/E11. (c) Combine these two results to verify the expression (16", ".100) for YM, showing in particular that YM = 9 BM SM/ (3 BM + SM). 16.28 *** Consider again the wire or rod of Problem 16.27. In general, when one stretches the wire longitudinally it will contract in the transverse directions. The ratio of the transverse fractional contrac- tion to the longitudinal fractional stretch is called Poisson's ratio (after the French mathematician and student of Laplace and Lagrange, 1781-1840) and is denoted by the Greek letter \"nu,\" v. (a) Show that v = \u2014 E22/E11. (b) Use the method of Problem 16.27 to show that v = (3 BM \u2014 2 SM)/ (6 BM + 2 SM). (c) Calculate Poisson's ratio for the five materials listed in Problem 16.26. Comment on its value for materials with BM >> SM. 16.29 *** When we change our coordinate axes, the strain tensor changes in accordance with Equation (15.132), which we can rewrite as ER = RER, where R is the (3 x 3) orthogonal rotation matrix. Use the property (15.129) of orthogonal matrices to show that tr ER = tr E; that is, the trace of any tensor is rotationally invariant. Use this result to show that the decomposition E = el + E' is rotationally invariant, in the sense described below Equation (16.92). Problems for Chapter 16 737 SECTION 16.10 The Equation of Motion for an Elastic Solid 16.30 * If 8i, denotes the Kronecker delta symbol (16.115) and a is a vector with components di 1, 2, 3), prove that E j = a,. In the same way, show that E j sji aj = 8\u201e a result we used in (j -= proving the important identity (16.116). SECTION 16.11 Longitudinal and Transverse Waves in a Solid 16.31 * A seismograph records the signals arriving from a distant earthquake. If the S waves arrive 12 minutes after the P waves, how far away was the earthquake? Use the speeds found in Example 16.8 16.32 * [Computer] Using appropriate software, calculate the speeds of longitudinal and transverse waves in the five materials listed in Problem 16.26. Arrange for the software to give you a nice readable (page 722). table of values. SECTION 16.12 Fluids: Description of the Motion 16.33 * Write down the equation of motion (16.124) as applied to a static fluid. Assuming that g is uniform and p is constant (independent of r), prove the well-known result from introductory physics that the pressure difference between two points r 1 and r2 is just Ap = ggh, where h is the vertical 16.34 ** Equations (16.129) and (16.130) are two different forms of the equation of continuity. Prove difference in elevations of r 1 and T.,. that they are equivalent. SECTION 16.13 Waves in a Fluid 16.35 * A crucial step in showing that the waves in a fluid are necessarily longitudinal was the integral in (16.142). For an arbitrary function f (0, with derivative f V), prove that f f (n r \u2014 ct)dt = \u2014 f (n r \u2014 ct)/c. 16.36 ** To find the speed of sound in air using the result (16.140) requires a little care. (Even the great Newton got this one wrong!) The trouble is to decide on the correct value of the bulk modulus of air. Because the vibrations are so rapid, there is no time for heat transfer and the air expands and contracts adiabatically, so that pVY = constant, where y is the so-called \"ratio of specific heats,\" y = 1.4 for air. (a) Show that the bulk modulus is BM = yp. (b) Use the ideal gas law, pV = nRT to show that the density is go = pM/RT, where M is the average molecular mass of air (M';,\u2018-', 29 grams/mole). (c) Put these results together to show that the speed of sound is c=,/y R TIM. Find the speed of sound at 0\u00b0C, and compare with the accepted value of 331 m/s. 16.37 ** Show that the intensity I of a sound wave is proportional to the square of the pressure increment p'. To do this consider a small sliver of fluid normal to the direction of propagation with area dA. Write down the rate at which this sliver does work on the fluid just in front of it, then divide by dA to show that the time-average intensity is (I) = (p12)1cgo. 738", " Chapter 16 Continuum Mechanics 16.38 *** A crucial step in deriving the wave equation for waves in a fluid was the neglect of the first term on the right of Equation (16.134). (a) Justify this by using (16.139) to rewrite the right side of (16.125) as Q'g \u2014 BMVQ'/Q0. Argue that the ratio of the first to the second term is of order ge oA/BM, where A is a typical distance over which Q' varies. (A good choice for A would be the wavelength of the proposed wave \u2014 of order a centimeter or at most a few meters.) Using the values for water (BM = 2 GPa, etc.) show that the first term is negligible. (You would also reach the same conclusion for air.) (b) Show with a similar argument that the second term on the right of (16.136) is negligible. APPENDIX Diagonalizing Real Symmetric Matrices A.1 Diagonalizing a Single Matrix In Chapter 10 we met the moment of inertia tensor I for a rigid body. With respect to an arbitrary set of orthogonal axes, I is a real symmetric 3 x 3 matrix which gives the body's angular momentum L in terms of its angular velocity co as L = Ico. We defined a principal axis as an axis with the property that if co points along the axis, then L is parallel to co, that is L = Ito = Act), (A.1) for some number X. We saw that if a body has three orthogonal principal axes, then with respect to these axes I has the diagonal form Xi I = [ 0 0 0 X2 0 0 0 X3 1 (A.2) where X 1, X2, and X3 are the moments of inertia about the three principal axes. Conversely, if I has this diagonal form, then the axes with respect to which I was calculated were principal axes. For this reason, the process of finding the principal axes is often referred to as diagonalization of the inertia tensor. I claimed in Section 10.4 that any rigid body, spinning about any origin 0, does have three orthogonal principal axes. It is the main purpose of this appendix to prove that claim. The process of diagonalizing a matrix comes up over and over again in many different branches of physics. For example, if you read Chapter 16, you know that both the stress and strain tensors are given by real, symmetric matrices, and it is frequently convenient to find axes with respect to which one of these is diagonal. [For example, the axes with respect to which the stress tensor (at a given point P) is diagonal are called the principal axes of stress and have the tidy property that the stress along each of these axes is a pure stretch.] In quantum mechanics, probably 739 740 Appendix: Diagonalizing Real Symmetric Matrices the most important thing to do with the operator that represents any given dynamical variable is to diagonalize it. 1 To emphasize the generality of the process, I shall denote the matrix that we wish to diagonalize by A, and since we frequently have occasion to diagonalize an n x n matrix, where n is any integer, not necessarily 3, we'll suppose for now that A is an arbitrary n x n symmetric, real matrix. Nevertheless, the example that you may want to keep in mind is that A = I, the 3 x 3 moment of inertia tensor of a rigid body. In the context of classical mechanics, the matrices that we wish to diagonalize are almost always tensors (moment of inertia tensor, stress tensor, strain tensor, etc.), and in this section I shall assume that A is the matrix representing an n-dimensional tensor. Before we prove our main result, let us pause to consider the effect of changing the axes with respect to which the tensor of interest is evaluated. In general, if the n x n matrix A represents an arbitrary n-dimensional tensor (with respect to a given set of axes), then the matrix A' that represents the same tensor with respect to a different set of axes is given by the orthogonal transformation A' = RAR, where R is the orthogonal rotation matrix that relates the two sets of axes, as discussed in connection with Equation (15.132). Fortunately, if you haven't yet studied the orthogonal transformations of tensors, you can still follow our main proof, if you are content to consider just the case that the matrix A is the moment of inertia tensor A = I. This matrix was defined by the sums (10.37) and (10.38) (or the corresponding integrals, for a continuous body). If we change our axes, the set of coordinates x, y, z is replaced by a different set x', y', z', and using these new coordinates naturally leads to", " a different 3 x 3 matrix I', and this is all you need to know about the relation of the two matrices I and I'. We are now ready to prove the following important theorem: Diagonalization of a Rear Symmetric Tenso If A is area symmetric n x n matrix representing an n-dimensions tensor, then ere exist rz orthogonal unit vectors e with the properties that (I) each e is an igerivector of A, that is for some real eigenvalu and (2) with rest t the axes defined by s n unit. vectors. the tensor is repreN d by the atrix Aei 0 0 A.3) (A.4 1 In quantum mechanics, the dynamical variables are represented by complex, Hermitian ma- trices (rather than real, symmetric matrices). However, the problem of diagonalizing them is very similar in the two cases. Here I shall discuss just the case of real symmetric matrices. Section A.1 Diagonalizing a Single Matrix 741 Before we prove this, notice that since the n unit vectors e l,, e n are mutually orthogonal, they are certainly linearly independent. Therefore, any vector in the n- dimensional space can be expanded in terms of them; that is, e l,, e n form an orthonormal basis of the space in which they lie. The proof of this result proceeds in several steps: Step 1. A has at least one eigenvalue and corresponding eigenvector. We have seen repeatedly that the eigenvalue equation (A.3) requires that det(A \u2014 Al) = 0. This determinant is an nth degree polynomial in X so it certainly has at least one zero, X = X i, say.2 Now, it is a well-known result of linear algebra 3 that, if det(B) = 0, then there exists at least one nonzero vector a such that Ba = 0. Therefore, since det(A \u2014 A 1 1) = 0, there exists at least one eigenvector a such that Aa = A la. (A.5) Step 2. The eigenvalue is real. Nothing we have said so far guarantees that the eigenvalue X i and eigenvector a are real. To show that X 1 is, consider the following: If we multiply (A.5) on the left by the row A* (that is, the complex conjugate of the transpose of the column a), we find that = a*Aa. a*a (A.6) Now, it is easy to see that both terms in this fraction are real: First, a*a = \u2014 E lai l 2 > 0. Meanwhile the numerator is a*Aa = (5*Aa) - = aAa* = (a*Aa)* which shows that a*Aa is real. [For the first equality, I used the fact that the left side is a 1 x 1 matrix and hence equal to its transpose; for the second I used the well-known result that (mnp) - = fulfil and that our given matrix A is symmetric; in the last, I used the fact that A is real.] Since both numerator and denominator in (A.6) are real (and the denominator nonzero), it follows that the eigenvalue X 1 is real. (Notice that this argument applies to any eigenvalue of A; thus, any eigenvalue of a real symmetric matrix is real.) Step 3. The eigenvector can be taken to be real. One might be tempted to expect that the eigenvectors of a real matrix are necessarily real, but this is actually 2 111 general, an nth degree polynomial has n zeroes, but some (or even all) of these can be equal. Nevertheless, we're certainly safe in claiming that there is at least one zero. 3 See, for example, Mathematical Methods for Scientists and Engineers by Donald A. McQuarrie (University Science Books, 2003), page 434, or Mathematical Methods in Physical Sciences by Mary Boas (Wiley, 1983), page 133. 742 Appendix: Diagonalizing Real Symmetric Matrices false, for, if a real vector a satisfies (A.5), then so too does i a, which is certainly not real. Thus, an eigenvector of A can in general be complex. However, taking the complex conjugate of (A.5) and remembering that A and Al are real, we see that, if a is an eigenvector, so is a*. This in turn means that both of the vectors a + a* and i (a \u2014 a*) are likewise. Since both of these are real and at least one is nonzero, we have shown that for every e", "igenvalue there is at least one real eigenvector. Therefore, we can, without loss of generality, assume that the eigenvector a is real. Step 4. Choose a new basis including the eigenvector. Our next step is to normalize the real eigenvector a and to choose a new orthonormal basis with this normalized eigenvector as its first unit vector. That is, we define the unit vector which also satisfies the eigenvalue equation (A.5), e l = \u2014 a I al Ae 1 = X 1 e 1, (A.7) (A.8) and then choose n \u2014 1 more unit vectors orthogonal to e l and to each other to define a new set of orthogonal axes. 4 With respect to this new basis, the vector e l is represented by a column whose first entry is a 1 and all of whose other entries are zero. The eigenvalue equation (A.8) implies that (with respect to our new basis) the first column of the matrix representing A has Alfor its first entry and zeroes for all the rest. Since the matrix is symmetric, it follows that it has the form (new matrix A with respect to new basis) = 0 0 A l (A.9) X 0 0 where A l is an (n \u2014 1) x (n \u2014 1) real symmetric matrix. Step 5. Repeat steps 1 through 4 on the matrix A l. The matrix A 1 can be viewed as acting on the (n \u2014 1)-dimensional subspace orthogonal to our first new basis vector e l. It has at least one real eigenvalue X2 and a corresponding eigenvector, which we can take to be real and normalize to give our second unit vector e2. We now choose an orthonormal basis comprising e l, e2, and n \u2014 2 other unit 4 In three dimensions, it is easy to see that this is always possible. Given e 1 we just choose any two unit vectors in the plane perpendicular to e 1. In n dimensions the argument is essentially the same; to make it watertight, one can use the Gram-Schmidt orthogonalization procedure. See Mathematical Methods for Scientists and Engineers by Donald A. McQuarrie (University Science Books, 2003), page 448. Section A.2 Simultaneous Diagonalization of Two Matrices 743 vectors, and with respect to this second new basis, the matrix representing our tensor has the form (new matrix A with respect to second new basis A2 (A.10) where A2 is an (n \u2014 2) x (n \u2014 2) real symmetric matrix. Step 6. Repeat steps 1 through 4 on A2, then A3, etc. After n \u2014 3 further repetitions, the matrix representing our tensor will have the diagonal form claimed in (A.4), and this completes our proof. A.2 Simultaneous Diagonalization of Two Matrices In Chapter 11 we saw that a system with n degrees of freedom, oscillating about a position of stable equilibrium, obeys an equation of motion of the form Mij = \u2014Kq, (A.11) where q is a column of n generalized coordinates, and M and K are n x n real symmetric matrices, called the mass and spring-constant matrices respectively. In what follows it is important that both M and K are positive definite matrices. To see what this means, consider first the matrix K: According to (11.53), the potential energy is U = z qKq. This is zero at the equilibrium position q = 0 and, since the equilibrium is stable, U must be greater than 0 for any q 0 0. Therefore, the matrix K must have the property that qKq > 0 for any q not equal to zero \u2014 the defining property of a positive definite matrix. Similarly, according to (11.54), the kinetic energy is T = 2044, and this must be positive for any q 0; that is, M also must be positive definite. We defined a normal mode as any motion in which all n coordinates oscillate sinusoidally at the same frequency w, so that q(t) = Re(ae iwt), and we saw that a normal mode is possible if and only if w and a satisfy Ka = w2Ma. (A.12) I claimed (and in some specific examples we saw explicitly) that there are n indepen- dent solutions a of this generalized eigenvalue equation and hence that any possible motion can be expressed as a linear combination of the normal modes. 744 Appendix: Diagonalizing Real Symmetric Matrices To prove this claim, notice that if we expand a solution of (A.11) in terms of the n independent solutions a of (A.12), then the new generalized coordinates q' satisfy5 -r qi 2 r qi", ". (A.13) Comparing (A.13) with (A.11), we see that we have to prove that there exists a basis of the n-dimensional configuration space with respect to which the matrices M and K are both diagonal, with the forms M' = 1 = [ ; and K' = (A.14) 1 0 0 1 : 1 2 co i 0 0 1 w 2 n In particular, in the new basis, the mass matrix is just the unit matrix. We shall prove this, although we shall see that in general the new basis is not orthonormal. The proof leans heavily on our previous proof and, like it, proceeds in several steps. Step 1. Diagonalize M. Since M is real and symmetric, we can find an n x n orthogonal matrix R which diagonalizes M. That is, M' = RMR is diagonal, with the form my = = (A.15) 0. Pi o. If we define q' = Rq and K' = RKR, then, with respect to the new coordinates, the eigenvalue equation (A.12) becomes Ica' = w2M'a'. Step 2. Rescale coordinates so that M\" = 1. In terms of the new coordinates q', the kinetic energy is 1-/ T=-f. qMq'= 1. 2 (A.16) Since this must be positive for any ef 0, all of the numbers A i in (A.15) must be positive. Therefore, we can scale each of the coordinates q; up by a factor of,Au i. Specifically, we'll define new coordinates q\"' = qi Aui. If, in addition, we define a diagonal (though not orthogonal) matrix 1/'nfiTt. S = [ 0 1/,/,tt n (A.17) and set M\" = SM'S = 1 and K\" = (A.18) then, with respect to these new coordinates, the mass matrix is just the unit matrix, the kinetic energy has the simple form T= 4 E q\" 2, and, most important, since M\" = 1, 5 Notice that the coordinates that I am here calling q; are precisely the normal coordinates that I introduced in Section 11.7 (where I called them 0. Section A.2 Simultaneous Diagonalization of Two Matrices 745 the generalized eigenvalue equation Ka = co2Ma has become the ordinary eigenvalue equation K\"a\" = co2 a\". (A.19) Step 3. Diagonalize K\". According to Section A.1, there is an orthogonal matrix T which diagonalizes K\". That is, if we define K\"' = TK\"T and Mill = TM\"T = 1, (A.20) then both Kll' and M''' are diagonal matrices, with M'\" still equal to 1. This proves the existence of the required n eigenvectors with the advertised properties. 6 6 One small point: Since the eigenvalues are supposed to be the squares of the normal frequencies, it is essential that they be positive. This is assured since K and hence K\"' are positive definite. Further Reading Texts on classical mechanics at about the level of this book Ralph Baierlein, Newtonian Mechanics (McGraw-Hill, 1983) Vernon Barger and Martin Olsson, Classical Mechanics: A Modern Perspective (2nd edition, McGraw-Hill, 1995) Grant Fowles and George Cassiday, Analytical Mechanics (6th edition, Saunders, 1999) Longman, 1996) T. W. B. Kibble and F. H. Berkshire, Classical Mechanics (4th edition, Keith Symon, Mechanics (3rd edition, Addison-Wesley, 1971) Stephen Thornton and Jerry Marion, Classical Dynamics of Particles and Systems (5th edition, Thomson, 2004) More advanced texts on classical mechanics Louis Hand and Janet Finch, Analytical Mechanics (Cambridge University Press, 1998) \u2014 an undergraduate text, but distinctly more advanced than any of the above. Herbert Goldstein, Charles Poole, and John Safko, Classical Mechanics (3rd, edition, Addison-Wesley, 2002) \u2014 an astonishingly successful and enduring graduate text, first published in 1950. Books on mathematical methods Mary Boas, Mathematical Methods in the Physical Sciences (2nd edition, Wiley, 1983) \u2014 an undergraduate text, beautifully written and comprehensive; still one of the best around. Donald McQuarrie, Mathematical Methods for Scientists and Engineers (University Science Books, 2003) \u2014 a new entry at the undergraduate level, 747 748 Further Reading which has received glowing reviews \u2014 readable and very comprehensive, with 1161 pages. Jon Mathews and R. L. Walker, Mathematical Methods of Physics (2nd edition, W. A. Benjamin, 1970) \u2014 a", " graduate text, but very accessible. Tables of integrals and other mathematical formulas M. Abramowitz and I. Stegun, Handbook of Mathematical Functions (Dover, H. B. Dwight, Tables of Integrals and Other Mathematical Data (4th edition, Alan Jeffrey, Handbook of Mathematical Formulas and Integrals (2nd edition, 1965). MacMillan, 1961) Academic Press, 2000) Books on Chaos James Gleick, Chaos, Making a New Science (Viking-Penguin, 1987) \u2014 a highly readable, nontechnical history of chaos theory. Gregory Baker and Jerry Gollub, Chaotic Dynamics: An Introduction (2nd edition, Cambridge University Press, 1996) \u2014 a pioneering undergraduate text on chaos, which naturally covers much more ground than here. Steven H. Strogatz, Nonlinear Dynamics and Chaos (Addison-Wesley, 1994) \u2014 a beautiful, mathematical account of many aspects of chaos theory. Books on relativity Albert Einstein, Relativity (15th edition, Crown, 1961) \u2014 a readable account by the great man himself. Wolfgang Rindler, Introduction to Special Relativity (Oxford University Press, 1991) \u2014 a very nice account that goes a bit further than here, by one of the James Hartle, Gravity: An Introduction to Einstein's General Relativity (Addison-Wesley, 2003) \u2014 an excellent account of the general theory written experts. for undergraduates. C. W. Misner, K. S. Thorne, and J. A. Wheeler, Gravitation (Freeman, 1970) \u2014 a classic, and still the most comprehensive text on general relativity. Books on continuum mechanics Gerard Middleton and Peter Wilcock, Mechanics in the Earth and Environ- mental Sciences (Cambridge University Press, 1994) D. S. Chandrasekharaiah and Lokenath Debnath, Continuum Mechanics Lawrence E. Malvern, Introduction to the Mechanics of a Continuous Medium (Academic Press, 1994) (Prentice Hall, 1969) Answers for Odd-Numbered Problems Chapter 1 1.1 b + c = + + 5b + 2c = + 5\"/ + b c 1, b xc=x\u2014y\u2014 z. 1.11 The particle moves counterclockwise around the ellipse (x/b) 2 (y/c) 2 = 1 in the xy plane, 1.5 6 = arccos,j2/3 = 0.615 rad or 35.3\u00b0. making one complete orbit in a period 271(o. 1.23 v = \u2014 b x c)/b 2. 1.25 Any solution has the form f (t) = Ae -3t, which contains one arbitrary constant. 1.27 As seen from the ground the puck travels straight across the turntable passing through the center 0. As seen by an observer sitting on the turntable, the puck follows a curving path as shown. As seen from ground As seen from turntable 1.35 The position is r = (vot cos 0, 0, vot sin 9 \u2014 2 gt2). The time to return to ground is t = (2v o sin 9)/g, and the distance traveled is (21) 02 sin 9 cos 9)/g. 1.37 (a) If we measure x straight up the slope, x = vot \u2014 Zgt 2 sin 0. (b) Time to return, t = 2v ol (g sin 9). 1.39 x = v o t cos 0 \u2014 \u201eIgt 2 sin 0, y = v ot sin 0 \u2014 ; g t 2 cos O, z = 0. 1.41 Tension = no2 R (or mv 2 /R). 749 750 Answers for Odd-Numbered Problems 1.47 (a) p \\tx 2 + y 2, 0 = arctan (y Ix) (chosen to lie in the correct quadrant), and z is the same as in Cartesians. The coordinate p is the perpendicular distance from P to the z axis. If we use r for the coordinate p, then r is not the same thing as irk and r is not the unit vector in the direction of r [see part (b)]. (b) The unit vector 1'6 points in the direction of increasing p (with 0 and z fixed), that is, directly away from the z axis; / is tangent to a horizontal circle through P centered on the z axis (counterclockwise, seen from above); i is parallel to the z axis. r = pi) + (c) ap = p \u2014 13(1)2, ao + 2,4, 1.49 1= \u00b1 (Pt and z z o v ort \u2014 Zgt 2. 1.51 In the picture, the solid curve is a numerical solution of the differential equation (found with Mathematica's NDSolve).", " The dashed curve is the small-oscillation approximation (1.57) with the same initial condition (0\u201e = n/2). Considering how large the initial angle is, the small-angle approximation does remarkably well. The only significant discrepancy is that the approximation oscillates somewhat too fast, as one would expect. (For large amplitudes, the true period is a little longer.) Chapter 2 2.1 The two forces are about equal when v ti 1 cm/s, and if v >> 1 cm/s the linear force is negligible. For a beachball, the corresponding speed is about 1 mm/s. 2.3 (b) R ti 0.01, and it is very safe to neglect the quadratic drag. 2.5 Answers for Chapter 2 751 vy Vter T 2.7 If F = Fo, a constant, v = v o + at, where a = Im. 2.11 (a) vy (t)= \u2014v ter ter.Vo + ter (, v le - t/r and y(t) = d \u2014Vtert (V0 + Vter )T (1 \u2014 (b) ttop = r ln(1 vo /vt\u201e) and Ymax = [ v0 Vter in( 1 Vo/Vt\u201e)]T- 2.13 v = \u2014 x 2, where co =,/k/m, and x(t) = xo cos cot. 2.15 Time of flight, t =2vyo lg. V yo x \u2014 o 2 (\u2014 ). x X xo 2.19 (a) y in free fall). 2.23 (a) Terminal speed = 22 m/s (ballbearing), (b) 140 m/s (steel shot), (c) 107 m/s (parachutist 2.27 Velocity, v(t) = V ter tan - t); (time to top) = arctan arctan, where vo Vter c vt\u201e m m cvt\u201e vo Vter Vter =.1mg sin(9)/c. 2.29 time (sec) 1 5 10 20 30 actual speed (m/s) speed in vacuum (m/s) 9.7 9.8 37.7 49.0 48.1 98.0 50.0 50.0 196.0 294.0 0 0 0 2.31 (a) Terminal speed, V ter = 20.2 m/s. (b) Time to ground, t = 2.78 s (2.47 in vacuum), and speed at ground, v = 17.7 m/s (24.2 in vacuum). 2.33 (a) (b) sinh(z) = sin(iz). (c) d cosh(z) dz = sinh(z), = cosh(z), d sinh(z) f cosh(z) dz = sinh(z), sinh(z) dz = cosh(z). dz f 752 Answers for Odd-Numbered Problems 2.35 (b) At t = 2r and 3r, the speed v is 96% and 99.5% of its terminal value. 2.39 (a) t = m Vffrc (arctan \u2014 arctan v. ffr ffr (b) v (m/s) 15 10 5 0 t (s) 6.3 18.4 48.3 142 The corresponding times if we neglect friction are (from Problem 2.26) 6.7, 20.0, 60.0, and oo. To neglect friction, compared to the quadratic air resistance, is quite good at higher speeds, but terrible at very low speeds. 2.41 The velocity is v(y) = (v2 + vt er 2 )e-2gy/vte2r Ymax = 17.7 m, compared with 20.4 in the vacuum. V 2 where v r\u201e = s/mg/c; ter' 2.43 (a) The solid curve is the true trajectory, the dashed is that in a vacuum. (m) (b) The true range is 17.7 m and the range in vacuum 24.8 m. 2.45 (b) z = 3 + 4i = 5e\u00b0.927i ; (c) z = 2e -L7113 = 1 \u2014 i 2.47 (a)z + w = 9 + 4i, z \u2014 w = 3 + 12i, zw = 50, z/w = \u20140.56 + 1.92i ; (b)z + w = (4 + 20) + (4-0 + 2)i, z \u2014 w = (4 \u2014 + (4,0 \u2014 2)i, zw =", " 32i, z/w = + i. 2.49 (b) cos 39 = cos 0 (cos2 9 \u2014 3 sin2 0) and sin 30 = sin 0(3 cos2 9 \u2014 sin2 0). 2.53 mV, = qBv y, mi) \u2014q Bv\u201e, mi),= qE. The motion of x and y is the same as in Figure 2.15, clockwise motion around a circle at constant angular velocity w = q B 1m. Meanwhile, z = zo + vzot + 4az t2, where az = qE/m. The particle moves in a helix or spiral of constant radius around the z axis, with an increasing pitch as the motion in the z direction accelerates. 2.55 (a) i)x = cov y, v y = \u2014ce)(vx \u2014 EIB), and vZ = 0. (b) vdr = E/B. (c) vx = v dr + x \u2014 The transverse velocity (v x, vy ) goes steadily around a circle of radius (vxo \u2014 vdr), with a constant v dr, cos cot, vy = \u2014(v xo \u2014 v d,.) sin wt, and vz = 0. drift vd, in the x direction superposed. (d) x = vdrt + R sin cot and y = R (cos cot \u2014 1) where R = ( v dr)/w. This trajectory is a cycloid, whose precise appearance depends on the initial velocity vxo, as illustrated below for seven different values of v xo. Notice, in particular, that if vxo =vdr R = 0 and the charge drifts straight through the fields, as we already knew. (The values of vxo are shown as multiples of vdr.) Answers for Chapter 3 753 v\u201e, 1.5 12 2.55 (cont) y vxo = 3 vxo= 1 4- x 6 12 -4- y Chapter 3 of the initial direction. 3.3 The vectors v2 and v3 have equal magnitudes v 2 = v3 = vo, and are at 45\u00b0 on either side 2 v0 3.7 Final speed, v 2100 m/s. Thrust 2.5 x 107 N, a little bigger than the initial weight 3.11 (b) and (c) v = v ex ln(m o /m) \u2014 gt ti 900 m/s, compared to 2100 m/s in zero gravity. 2.0 x 107 N. 3.9 Minimum exhaust speed 2400 m/s. 3.13 Height ti 4.0 x 10 4 m. 3.15 CM position, R-- (1/6, 0, 0). 754 Answers for Odd-Numbered Problems 3.15 (cont) y m3 CM X m l M2 3.17 The CM is about 4.6 x 10 3 km from the earth's center. 3.19 (a) The CM would follow the same parabola as the unexploded shell. (b) The second piece hits the gun that fired it. (c) No. 3.21 X = Z = 0 and Y 4R/37r. 3.23 Velocity of second piece is v \u2014 Av. The CM (empty circles) is at the midpoint of the line joining the two fragments and clearly continues on the same parabola as the grenade followed before the explosion. -20 3.25 Final angular velocity, w = coo (rol r)2 3.29 Final angular velocity, co = wo (R o/R) 5 = coo/32. 3.31 Moment of inertia, I = 1MR2. 3.33 Moment of inertia, I = 3Mb 2. 3.35 (a) (b) and (c) Either way, i) = 3g sin y. 3.37 (a) Answers for Chapter 4 755 CM Chapter 4 4.7 (a) W(r i (b) 4.3 (a) W = 0, (b) W = 1, (c) W = 7/2. r2) = \u2014 (my/ 3)(Y2.Y 13); U(r) = (my/ 3)y3. (c) vfin = V2yh 3/3. 4.9 (b) xo = mg I k. 4.11 4.13 is \u2014n). function f ay 2 2byz + cz 2 of/ax 0 of/ay lay 2bz f 1 az 2by + 2cz cos(axy 2z3) \u2014ay 2z 3 sin(axy2z 3) \u20142axyz 3 sin (axy 2z 3) \u20143axy 2z 2 sin(axy 2z3) ar a", "-1 x2 + y2 + z2 ax I r ay 1 r azlr function ln(r) rn g(r) aox x I r2 nxrn -2 g'(r)x I r a lay y 1_2 nyrn -2 alaz z 1 r2 nzrn-2 (r)y I r g' (r)z I r 4.15 Using (4.35) we get Af ti 0.44, compared with the exact Af = 0.45 (to two figures). 4.19 (a) The surface x 2 + 4y2 = K is an elliptical cylinder, centered on the z axis, with \"radius\" V17 in the x direction and half that in the y direction. (b) The unit normal to the surface is n = (1, 4, 0)/ 17 (or \u2014n). The direction of maximum increase is n (and maximum decrease 4.21 The gravitational potential energy is U(r) = \u2014GMm/r. 4.23 (a) F is conservative and U = \u2014 414x2 2y2 + 3z2). (b) F is conservative and U = \u2014kxy. (c) F is not conservative. 756 Answers for Odd-Numbered Problems 4.29 (a) E A (b) The time to reach A is t (0 A) = on I2k/ A4 - x4. The period is 4t(0 A). 4.31 (a) Dropping all uninteresting constants, we have E = z (m 1 + m2)1.2 \u2014 (m 1 \u2014 m 2) g x (d) r = 3.71. 4.33 (b) (c) For b < r there can be two further equilibrium points (symmetrically placed on either side of 0 = 0), both of which are unstable. 4.35 (a) E = 1(m 1 + m2 + I I R2).Z 2 \u2014 (m i \u2014 m 2)gx (plus a constant that we may as well drop). (b) Equation of motion is (m 1 + m2 + I I R2)3C = (m 1 \u2014 m 2)g, either way. 4.37 (a) U(0) = Mg R(1 \u2014 cos 0) \u2014 mg RO. (b) There are equilibrium positions only if m < M. If m = M, there is one equilibrium (unstable) at 0 = 90\u00b0. If in < M, there are two positions determined by the condition m = M sin 0, which has two solutions symmetrically placed above and below 0 = 90\u00b0. The lower position is stable, the upper unstable. [See the pictures of part (c)]. (c) 0 (1) IT m = 0.8M If m = 0.7M, the wheel swings up to a maximum 0 < 7r, then swings back to 0 = 0, and oscillates indefinitely. If m = 0.8M, the wheel swings past 0 = 7 and continues to rotate counterclockwise until the string runs out. (d) The critical value of m/M is 0.725. 4.39 (c) If (I) = 45\u00b0, this approximation gives t = 1.037r 0, which represents a 3.7% correction to the small-amplitude approximation (r 0), and is itself within 0.3% of the exact answer (1.040r0). Answers for Chapter 5 757 4.51 U(r i, r2, r3, r4) = [U12(r1 r2) + Un (r i \u2014 r3) + L/14 (r t \u2014 r4) + U23(r2 \u2014 r3) r4 ) U34(r3 \u2014 r4) \u25aa U24(r2 \u25aa [uie. xt (ro urt (r2) u3Xt (r3) u4xt (r4 ) 1. 4.53 (b) E = + T2 + + U2 + U12 = ;m2v22 \u2014ke 2 \u2014 + \u2014 \u2014 l r1 C 1 r2 1 r12 (c) Long before: E = T1 + T2 + Ui + 0 + 0 = T2 - -. Long after: E' = Ti' + 7-,+ 0 + U21 + 0 = T 11 \u2014 By conservation of energy, T i' = T2 + 4 ke2 \u2014 ( \u2014 ke 2 2r ke2 2r' 1 r 1 r' Chapter 5 5.3 U(0) = mgl(1 \u2014 cos 0) and k = mgl. 5.5 (a) B i = C I + C2 and B2", " = i ( CI \u2014 C2). (b) A = B? + B22 and 6 = arctan(B2/B 1), chosen in the right quadrant. (c) C = Ae -t 8. (d) C 1 = C/2 and C2 = C */2. 5.7 (a) 13 1 = xo and B2 = Vo /W. (b) = 10s -1, B 1 = 3m, B2 = 5 m. x(t) (c) The first time x = 0 is t = 0.26 s; the first time z = 0 is t = 0.10 s. 5.9 Period, r = 1.05 s. 5.11 5.13 A = 2 2 X2 2 2 i 1.)2 \u2014 X 2 V 1 \u2014 V2 2 and co = 2 2 \u2014 1.) 2 2 X2 2\" X i r = and w 11 2U. mXR2 5.17 (a) If the fraction p lq is in its lowest terms, r = 27p/co x. 5.19 k' = 2k(2a \u2014 l o)/a. 5.23 dE I dt = i(mi + kx). 758 Answers for Odd-Numbered Problems 5.25 (a) and (b) (c) With p = w0 /2, the amplitude shrinks by a factor of 0.026 in one period (much more than in the picture, for which p was chosen to be w 0/10 and the shrinkage factor is about 0.53). 5.29 Ti = 1.006 sec, and /3 = 0.110w 0. 5.31 Each picture shows x(t) as a function of t for the value of /3 indicated. )3=0 0= 1 p = 2 0 = 4 /3=6 0 = 27 /3 = 10 0 = 20 5.37 A = 26.9, 8 = 3.04 rad, B 1 = 26.7, and B2 = \u20146.18. -1 x(t) 25 0 -25 The solid curve is the actual motion; the dashed curve is the transient, homogeneous solution. 5.43 (a) k,==, 4 x 104 N/m. (b) f,'-'-' 6 Hz. (c) v,'', 5 m/s or roughly 10 mph. 5.49 a0 = fm,\u201e/2 and an = 4f,../(m7) 2, for n odd but 0 for n even (n > 0); bn = 0 for all n. Answers for Chapter 6 759 2 terms tf 6 terms The left picture shows the sum of the first two terms (the constant term plus the first cosine) and the \"sawtooth function\" itself. The right picture shows the sum of the first six terms. This follows the sawtooth so closely that it is hard to tell them apart except at the corners. 5.53 A0 = 1/2(002, 4/71272\\/(6002 \u2014 n 2(02)2 + (2j3nw) 2 for n odd but zero for n even (> 0). (a) With to = 2, coo = 7, and the first four coefficients A n (n = 0, 1, 2, 3) are 0.0507, 0.6450, 0, and 0.0006. x( t) 0.5 - 0 \u20140.5 1 11 111 1 2 3 4 5 6 t (a) (b) (b) With To = 3, coo = 27/3, and the first four coefficients A n (n = 0, 1, 2, 3) are 0.1140, 0.0734, 0, and 0.0005. 5.57 The left picture shows the data for this problem; the right shows the data of Figure 5.26 (though drawn here to a slightly different scale). Notice that the resonances with /3 = 0.1 are twice as high and half as wide as those for 8 = 0.2. Chapter 6 6.3 Time to travel from P1 via Q to P2 is 1x 2 + y i2 + z2 + \\/(x \u2014 x 2) 2 + y22 + z2 /c. 6.5 Time to travel from A via P to B is 14,(R I c) cos(6/2). ( 760 Answers for Odd-Numbered Problems 6.7 0 = az + b, where the constants a and b are chosen so the path passes through the given endpoints. In general there are many different paths of this form. 6.9 y = sinh(x)/ sinh(1). 6.11 The path is a parabola, x = C", " (y \u2014 D) 2 /4C, with C and D constant. 6.17 p = p0/ cos[(0 \u2014 qu o)/ \\/1 + ),.2]. With A = 0, the cone becomes a plane, and this equation becomes the equation of a straight line. 6.23 (a) v = \\Ay() cos 0 + V y) 2 (v 0 sin 0)2. (c) ymax = 366 miles; (time saved) = 27 minutes. Chapter 7 7.1 = 4,n(i 2 + 5,2 + \u2014mg = mz. ) mgz. The three Lagrange equations are 0 = mx, 0 = mj;, and 7.3 = In2(.0 +.92\\ \u2014 ) 4k(x 2 + y 2). The two Lagrange equations are m.W = \u2014kx and my = \u2014ky.. This is the isotropic oscillator of Section 5.3. 7.5 f (VA = and (V = 1 f \u2014 r a 7.7 (a) m aic, = [a = 1,, N], (b) L = Ea \u2014 U(r1,, r N ). 7.9 x = R cos 0, y = R sin 0, and 0 = arctan(y/x), chosen to lie in the correct quadrant. 7.11 x = A cos wt + 1 sin 4), y = 1 cos 0, and 0 = arctan[(x \u2014 A cos cot) I y]. 7.15 = z (m1 + m2)1 2 + m2gx, a = gm2/(m1 + m2). 7.17 = g(m i \u2014 m 2)I(m i + m 2 + //R 2). 7.21 = 4m(i- 2 + r2w2), and r = Aewt Be \u2014 \" 7.23 C = \u2014 Act) sin cot) 2 \u2014 4kx 2. 7.27 (Acceleration of mass 4m) = g/7 downward. 7.29,C = Zm [R 2co2 / 202 + 2R1c4 sin(4) \u2014 cot)] \u2014 mg(R sin cot \u2014 1 cos 4)) and /it) = \u2014 g sin + co2 R cos(0 \u2014 cot). 7.31 (a) C = (m + 111).i 2 Al (.1, 22 + Lc.p cos 0) \u2014 Ikx 2 + MgL cos 0. The x and 0 equations are (m + Mg + ML( cos 0 \u2014 oio 2 sin 0) = \u2014kx and M (a \u00b1 x cos 0) = \u2014Mg sin 0. (b) With x and 0 both small, these become On + + Ma = \u2014kx and M + = \u2014 M gc/). 7.33 x(t) = xo cosh cot + (g I2co2) (sin cot \u2014 sinh cot). 7.35 = 4mR2 [w2 + 4 + (0\\2 ) lar frequency is w. 2(1)(\u00a2 + co) cos 44 For small oscillations about B, the angu- 7.37 (a) = imr24)2 \u2014 mgr. (b) The r and 4) equations are mrii) 2 \u2014 mg = 2mi' and mr 2V) = const. (c) ro = / (m20,0. (d) Angular frequency =,/3/2 tImr02. j Answers for Chapter 7 761 7.39 (a) (r2 r 262 r2 sine 0 (i02\\ \u2014 ) U (r). (b) The r, 0, and equations are mr = mr (6 2 + sin2 0 4)2 ) \u2014 U/ar d \u2014 t (mr 29) = mr2 sin 0 cos 0 % 22 d \u2014 d t (mr 2 sin2 0 (*p) = o. (c) The motion remains in the equatorial plane 0 = 7r/2, consistent with our knowledge that the motion is confined to a plane. (d) The motion remains in the longitudinal plane 4) = 0 0. 7.41 G = (1 + 4k2p2)/5 (152 + p2602 4k2 p2i) 2) _ mg, 2, Kp 4k2pi5 2 = (w2 2gk)p. and the equation of motion is The bottom of the wire, p = 0, is an equilibrium, which is stable if w 2 < 2gk, but unstable if w2 > 2gk. If w2 = 2gk, the bead", " is in equilibrium at any p, but the equilibrium is unstable (except 7.43 (a) = (M m)RW M g R(1 \u2014 cos 4)) + mg Rq5, and the equation of motion is (M at p = 0). m)Rcb = \u2014M g sin (/) + mg. (b) Notice that there are actually several equilibriums, separated by one or more complete revolutions. m = 0.7 7r/2 (d) U )) m = 0.8 7r/2 7r )0 157r 107r 57r 10 15 20 10 15 20 762 Answers for Odd-Numbered Problems 7.49 (b),C = 1742 + A = (o 2 p24)2,\u00b12) 4qB 2 / p qo and the three Lagrange equations are ma)2 03,0(b, d (inp2s, iqBp2) = 0, and mE = 0. dt 7.51 \u00a3 (x, y) = 4m(i 2 + 5?2) mgy. (a) The two modified Lagrange equations are 1 =mx and mg + A\u2014 =my. Chapter 8 8. gt 2 + Ma) 2 sin tot and yi = v 0 Ni t \u2014 lgt 2 m Miv\u00b0 sin tot. co 2 8.7 (a) Period r = 2n- r3121.1Gm 2. (b) r = 27r 3/2 /.../GM. These two answers are the same in the limit that m 2 oo. (c) r = 0.71 years. 8.9 (a) Z = 4m(i2 \u00b1 Y2) + 11,4,7.2 + )24,2. - ) 1. k(r \u2014 L) 2. (b) MX = 0 and MY = 0, with solutions R = Ro + Rot. (c) The r and 0 equations are Ai' = A4 2 \u2014 k(r \u2014 L) and p,r20 = const. If r = const, then 0 = const, and r = L + \u00b5r02/k. If 0 = const, then r = L + A cos(wt \u2014 8), where w =,/k/tt = V2k I m i. 8.13 (a) PE ro (b) ro = (E2/ k\u00b5) 114. (c) Angular frequency of oscillations, w =,/4k/kt. 8.15 Percent variation,==,' 0.1%. 8.19 Eccentricity, E = 0.17; (height when on y axis) = 1424 km. 8.21 (a) If 0, then a rmax/2. (b) (c) t = (7/2 4./2GM)(rmax) 3/2 (d) and (e) = (7[V2GM)(rmax) 3/2. = (27r/N/2GM)(rmax ) 3/2 = 2r(t, 0). 8.23 (b) = + mA/f 2 and c=f2f3 21mk. (c) The orbit is closed if 0 is a rational number, p lq (where p and q are integers). If A \u2014> 0, the orbit becomes a Kepler ellipse. Answers for Chapter 9 763 8.25 (a) ro = 1. 0.1 0 \u20140.1 \u20140.2 (b) You can see from the picture on the left that if E \u20140.1, the inner turning point is at about rrnin = 0.7. If we use this as a starting value for any equation-solving program (such as Mathematica's FindRoot), we find that this root of the equation Ueff(r) = \u20140.1 is actually rmin = 0.6671. (c) Obviously the orbit shown on the right has not closed after 3.5 revolutions, and it clearly won't close for a long time. (In fact, it never does, but this is harder to prove.) 8.27 c = 8.87 x 107 km, E,--- 0.753, B = 1.72 rad. 8.29 The new orbit would be a parabola, tangent to the old circular orbit at the point at which the great disappearance occurred. The earth would be just not bound. 8.31 The sketch shows the path of the relative position r = (x, y), that is, the orbit of particle 1 as seen from particle 2. y 2 8.35 First thrust factor =.4/5 ; second = 1/5/8. Chapter 9 9.1 (Angle of tilt) = arctan(A/g)", " with vertical. 9.3 (a) Fttd1 mg ti 1.1 x 10 -7. (b) Same magnitude, opposite direction. 9.9 Fox- = 2m v\u201eS'2 cos 0 due east; Fcorl mg 0.011. 9.13 The maximum value of a is about 0.1'; the minimum is zero. 764 Answers for Odd-Numbered Problems 9.15 g = go \\ / cos2 9 + A2 sine 9. 9.19 (a) As seen from the ground, the puck moves in a straight line. As seen from the merry-go- round, its initial acceleration is radially outward; as it speeds up, it curves to its right and spirals outward from the center. (b) As seen from the ground, it remains stationary. As seen from the merry-go-round, it moves in a clockwise circle centered on the axis of the merry-go-round. 9.21 9.27 9.29 This is what happens in Problem 9.24(d). 9.25 Angle = 0.13\u00b0 to left. East x (east) 9.31 v = 0.11 mm/s. 9.33 Chapter 10 10.3 R = (0, 0, H15). 10.5 R = (0, 0, 3R/8). 10.9 I = Z MR2. (a) View from earth (b) View from space up & down down only z ( up) Cl = 1+ and C2 = A 2 wo (1 - 2 coo. 10.7 (a) V = 2 R 3 (1 \u2014 cos 0 0); (b) R = (0, 0, Z) where Z = 3R 1 \u2014 cos 200 16 1 \u2014 cos 00 3 Answers for Chapter 10 765 (a) I(solid) = \u2014 M R` 10.11 ; (b) I(hollow) = M i 2 5 2 b5 \u2014 as 5 b3 \u2014 a 3. 10.13 (a) a).,/mgal I; (b) / = I 1(ma). 10.15 co =,N/3,g(N/2 \u2014 1)/2a. 10.17 /z, = 15-M (a2 + b2). 10.23 All products of inertia involving z are automatically zero, Ix, = /y, = /,x = lzy = 0. 10.25 (a) and (b) The inertia tensors Ic\u201e, about the CM and I A about A are 1 I, ICM = - IY1 b2 \u00b1 c,2 0 0 C2 + a2 0 0 0 0 a 2 \u00b1 b 2 and 3 1 3 IA = \u2014 M [ [ 4(b2 + c2) \u20143ba \u20143ca \u20143ab 4(c2 + a2) \u20143ac \u2014 3bc \u20143cb 4(a 2 + b2) [ (R 2 \u00b1 2h 2) I= \u2014 M 1 4 0 0 (R 2 + 2h 2) 0. 0 0 0 2R 2 (c) L = i Mco (4(b 2 + c2), \u20143ab, \u20143ac). I =--- ma 2 [10 b) The principal moments are A l = 10ma 2, A2 = 7ma2, and A3 = 5ma 2. The corresponding principal directions are e l = (1, 0, 0), e2 =,i (0, 1, 1), and e3 = *i. (0, 1, \u20141). (b) A l = 1, A2 = 3, and A3 = 4; e l = -,,15. (1, 1, 0), e 2 =,,+ (1, \u20141, 0), and e 3 = (0, 0, 1). 10.39 Q ',-z,-' 21 rad/s or about 200 rpm. 10.47 About 1010 years. 10.53 b 2 >> 4ac. 10.57 (a),L, = 4m(i2 + Y 2) + Ixcm(4)2 sine 0 + 62\\ + ) + 1AT (ik. + c/). cos 0) 2 \u2014 M g R cos 6) where Alm and ),.7 are the two principal moments about the CM. (c) The larger precession rate is bigger about the CM than about the tip. According to (10.111) the smaller rate is unchanged. [But note that (10.111) is an approximation; if we keep the next term in the approximation, we find that the smaller rate is slightly reduced.] 10.27 10.35 (a) 10.37 (a) 766 Answers for Odd-Numbered Problems Chapter 11 11.1 k 1", " (1 1 \u2014 L 1) = k2 (12 \u2014 L2) = k3(13 \u2014 L 3). 11.3 w2 = 2.11.2 Imi (k2 k3) m2 (k1 k2) \u00b1 I/m /2. (k 2 k3) 2 m2 (k 1 k2) 2 \u2014 2mim2 (k2k3 k3ki kik2 \u2014 I. 11.5 (a) Let m 1 = m2 = m and k 1 = k2 = k, and wo = Vk/m. Then the normal frequencies are 3\u2014 vs 2 = 0.62coo and (02 = = 1.62coo. 3 -F 2 (b) In mode 1, the two carts oscillate in phase, with the amplitude of m 2 equal to 1.62 times that of m l. In mode 2, they oscillate exactly out of phase, with the amplitude of m 2 equal to 0.62 times that of m l. 11.7 (b) B i = A and B2 = CI = C2 = O. x1 t 1 AAAAA VVVVV30 (b) 0 (c) (c) B 1 = B2 = A/2 and C 1 = C2 = 0. 11.9 (b) 1 = A l cos(w it \u2014 6 1) and 2 = A2 cos(w2t \u2014 32), where co l =,/k/m and w2 = N/3k I m, and A 1, 31, A2, 32 are all arbitrary constants. Hence 11.11 (a) (c) Xi = Al cos(w1t \u2014 31) + A2 cos(w2t \u2014 62) X2 = Ai COS(Cfht \u2014 31) \u2014 A2 cos(w2t \u2014 62). mi l = \u20142kx 1 kx 2 \u2014 mi 2 = kx 1 \u2014 2kx 2 \u2014 F o cos cot 1 (t) = A 1 cos(cot \u2014 6 1) B ie -13` cos(co it \u2014 ri ) 2 (t) = A2 cos(wt \u2014 32) + B2e -13t cos(w2t \u2014 62') where the constants A 1, A2, 6 i, 82 are given by Eqs. (5.64) and (5.65) (except that now f o = Fo/2m and, in the case of A2 and 62, co,' is replaced by 3(1)02); the constants B 1, B2, 81, 82tr in the transient terms are arbitrary and determined by the initial conditions, and w 1 = 0.)02 \u2014 /3 2, while w2 =.13wo2 \u2014 02. 11.13 (b) Let p = b/2m, co l = -1k I m \u2014 fi 2, and w2 = j(k 2k 2)/ m \u2014 16 2. Then = e -$t (Bi cos co lt + C i sin w i t) and 2 = fi t (B 2 cos wet + C2 sin wet) where B 1, C1, B2, C2 are arbitrary constants. (c) With the given initial conditions (and,8 << 1) x i = fit (cos co lt + cosw2t) and x2 = 2Ae 4t (cos co lt \u2014 cosw2t). Answers for Chapter 11 767 xi A i AA1i_ V 1FV M... V V A 11.15 E., = T \u2014 U where T and U are given in (11.38) and (11.37). The 0 1 equation is (m l + m2)Li 81 + m 2 L 1 L262 cos(0 1 \u2014 02)-}- m 2 L 1 L 2622 sin(0 1 \u2014 02) = \u2014(m 1 \u00b1 m 2)gL 1 sin 0 1 and the 02 equation is m2 L 1 L 201 cos(0 1 \u2014 02) + m 2 L 2262 \u2014 m 2 L 1L 2612 sin(01 \u2014 02) = \u2014m 2g L 2 sin 02. 11.17 (a) (0 2 \u2014 4) 2 an \u2014 4 coo d (D \u2014 4) 2 \u2014 2 o' For the first mode, a = A ]; for thesecond, a = A[ ]. 23'1 2 3 (b) [ ] = \u2014a [ 1 cos colt \u2014 [ 1 cos colt I, which is not periodic. 2 1 3 \u2014 3 11.19 (a) = -1(m + M)I 2 + ML1.6 + IML 26 2 \u2014 (ikx 2 + ;Mg /9 2). The x and 0 equations are", " (m + Mg M Lo = \u2014kx and ML 28 = \u2014MgLO. (b) With the given numerical values, the normal frequencies are (Di = \u2014./2 = 0.77 and co2 = + A/2 = 1.85. In the first mode, the cart and bob oscillate in phase (both moving to the right and then both moving to the left), with the bob's amplitude (of motion relative to the cart) A/2 times bigger than the cart's. In the second mode, the cart and bob oscillate exactly out of phase, again with the amplitude of the bob equal to A/2 times that of the cart. 11.23 W22 =--+ 1\u00b1 and L m 2 g W3 =\u2014+ 3\u2014 L m. k 11.25 If (00 =.,/klm, then the normal frequencies are co l = coo -12 \u2014./27, (02 = and 0)3 = (D0-12 Ah. 768 Answers for Odd-Numbered Problems In the first mode, all three carts oscillate in phase, with al = a3 = a21,1; in the second, the middle cart is stationary, while the first and third oscillate exactly out of phase, with al = \u2014a3 and a2 = 0; in the third, the left and right carts oscillate in phase, while the middle one is exactly out of phase, with al = a3 = \u2014a2I 11.27 (a) G = Zm(xl ) - k(x i \u2014 x 2) 2. The normal frequencies are co l = 0 and cot = co\u2018A (with coo =,/k/m). (b) In the second mode, the two carts oscillate with equal amplitudes, but exactly out of phase. (c) In the first mode, x 1 = x2 = x. + vot; that is, they move at constant velocity with the spring at its equilibrium length. 11.29 coi = N/2k/m, cwt = N/6k1m, and co3 =.1g/r., where ro is the equilibrium value of r. 11.31 The three normal frequencies are 0, N/2 coo, and 0 coo. 11.35 (a) 1 = 01 + 02 and b2 = 0 1 \u2014 02 (or any convenient multiple of these). (c) The first mode has 0 1 = 02 = Ae - d* cos(w it \u2014 8), and the second 0 1 = \u201402 = Ae - fi t cos(co2t \u2014 8), where (0 1 = /32, (02 = /32, and 13 = 2m. 2k m Chapter 12 12.7 12.1 x 1 (t) = (t + k) 2 + 1, for any constant k. y = 1.073 - r 20 30 t'27r -3r 12.11 (b) The least-squares line has slope -1.54, giving 8 = e 1 '54 = 4.66, compared with the correct value 4.67. ln(Yn +1 - Yn) Answers for Chapter 12 769 12.13 X ti 0.64. 12.15 12.17 -57r -107r -157r The picture confirms that AO (t) decreases exponentially. +2771 7T 0 2 4 6 8 t 12.19 (a) x = A cos(cot - 8), where w =,/k/m, and A and 8 are arbitrary constants. (b) E = -1kx 2 = const. 12.21 (a) xo XI x2 x3 x4 x5 0 3 1.00 0.54 0.86 -0.99 0.55 0.85 0.65 0.66 100 0.86 0.65 0.80 0.70 0.793 0.791 0.765 X28 0.7391 0.7391 0.7391 X29 0.7391 0.7391 0.7391 x30 0.7391 0.7391 0.7391 770 Answers for Odd-Numbered Problems (b) There is a single, stable attractor at x* = 0.739085. 12.23 (a) (b) r0 = 1/7 = 0.318. (c) r1 = 0.720. 12.25 0 10 20 r = 0.865 0 0 10 20 12.27 (c) xi, = 0.5130 and xb = 0.7995. 12.29 From the slope of the least-squares fit, we get 8 = 4.71. Answers for Chapter 13 771 12.31 0. n -4- -8- -12-", " (a) 12.33 See Figure 12.41. 10 20 30 40.%.V.V.%%... 10 I 20 I 30 I 40 I t N loglx'-x 0 -4- -8- -12.. (b) Chapter 13 m2 + m). 13.1 9-C = p 2 /2m. The Hamilton equations are x = plm and p = 0, with solutions p= po = const, and x = xo -1- vot, where vo = &/m. 13.3 The Hamiltonian is J-C = p2 12(m i + m2 + ;M) - (m 1 - m 2)gx. The Hamilton equations area = p (m i + m2 + 1M) and P = (m - m2)g, and the acceleration isi = g(m i - m2)/(m + 13.5 The Hamiltonian and the two Hamilton equations are = 2m (c 2 + R2) + mgc0, = m (c2 R2)' g 2 /(c2 R2 ) Combining the last two gives E = c = - 2 P 2 P 13.7 (a) 9- C = + mgh(x). (b) The Hamilton equations are 2m[1 + h'(x)2] i = p m[1 + h'(x)2] and p - p2h'(x)h\"(x) in,h, (x). m[1 + W(0 21 2 6 and 15 = -mgc. 13.9 The Hamiltonian is H = (px2 + py2)/2m + mgy. The Hamilton equations are x = px/m, Px = 0,.5Y = py /m, and Py = -mg. 13.11 The Hamiltonian is J-C = p x V mgz (with x measured along the tracks 2 2 2 Px Py Pz 2m and z vertically up). 13.13 9._c = 2m 1 ( 7z2, 4_ L- 49 n 2 R 2 + 2 \u2018 Y ik(R2 + z2,-,), z = A cos(cot - 6), and (/) = const. 13.17 (a) zo = [p42)/(m2c28)p/3 (d) a = arcsin(1/0) = 35.3\u00b0. 13.19 = 1 (p2 py2) u(s/x2 + y2). 2m 13.21 (a) (px2 P2) + j- (132 + 2M = Y 2/1 2 n 1- 2 2 _ is not. (c) r = lo + A cos(wt - 8), where w =,/k/ ju and A and 6 are constants. ik(r - l0)2. X, Y, and (/) are ignorable; r 772 Answers for Odd-Numbered Problems 13.23 (b) The two conjugate momenta are px = m(x + 5,) and py = + 4Y). 3-C = \u2014 [i (px \u2014 py)2 + /3,2] + z kx 2. (c) x = xo cos cot and y = yo ixo u \u2014 cos wt), where 1 2m 13.25 (c) P = H. (d) Q = t + const. w = N/4k/3m. 13.29 13.31 (a) 3k, (b) 0, (c) k, (d) 0. 13.33 (a) LHS = 4n- kR3 = RHS. (b) Same. 14.1 a = 0.79 cm2, n tar = 0.034 cm-2, probability = 0.027. 14.3 (Number density) = 2.1 x 10 28 atoms/m 2. 14.7 O \u2014 moon \u2014moon = 6.45 x 10 -5 sr, and -- sun = 6.76 x 10-5 sr. Chapter 14 14.5 Nsc = 2.7 x 107. 14.11 N\u201e ';=,' 29. 14.15 (a) (b) \\//3 2 2mUo. (c) See left picture. Answers for Chapter 15 773 (d) do- dQ (sin 0)d0 I db where de db = 2 1 n,/ for 0 < 0 < ()max = 7T \u2014 arcsin for 0 > Omax, da I dS-2 = 0. (e) o-tot = 7/- R 2. 14.17 N\u201e/Z 2 = 0.21, 0.20, etc. 14.", "31 (b) If \u00b0lab = 0, then O cm could be 0 or Tr. cm(deg) 90 180 (c) (d) elab(max) = arcsin(1/) ). Chapter 15 10 -8 %. aged. dilation. 15.11 v = IC 5 \" 15.3 y = 1 + 3.56 x 10 -1\u00b0 ; difference = (Ara \u2014 At) = \u20141.28 its; percent difference = \u20143.56 x 15.5 A's clocks read, and A has aged by, 25 years, compared with the 80 years by which B has 15.7 Number expected, N = 420, taking account of time dilation; but N = 29, ignoring time 15.13 (a) 1 = 91.7 cm; (b) l = 83.2 cm. 15.17 (a) If we call the two events 1 and 2, then, as observed in 3', ti = 0 but t2 = \u2014y,6a1c. (b) As observed in 8\", ti = 0 but t2' = 15.19 (a) = d, =d/c, xii3 = \u2014d, tii3 =d1c. (b) xF = y(1 + 13)d, tF = y (1 + P)d lc, xB = \u2014y(1 \u2014,6)d, t B = y(1 \u2014 153)d I c 15.23 The velocity of the right rocket relative to the left one is 0.994c to the left. 15.21 v = 0.91c. 15.25 (Speed as measured in 8) = c. 15.29 (a) [ cos 9 sin 0 0 R(0) = \u2014 sin 0 cos 0 0. 0 0 1 774 Answers for Odd-Numbered Problems 15.33 (a) xi = x;, = X3 = X1 X3 y (x2 - fix4) X /4 y (x4 - f3x2 ) whence AB2 = 0 [ 1 0 0 Y 0 -yfi 0 0 0 0 1 0 0 \u2014 YP y - (b) AR+ = - 0 - and AR = - 0 - 15.47 v = 0.20c. 15.55 Since u is a four-vector ui = y(V)[u l - )6(V)u 4], d2 = u2, u 3 = U3'1/4 = y(v)C114 \u2014 P( v)ud 15.57 T(Bi) T(He) = 1.3 x 10 -12 J = 8.1 MeV. 15.61 E 5 MeV; v = 0.8c. 15.63 1 MeV/c2 = 1.78 x 10-3\u00b0 kg; 1 MeV/c = 5.34 x 10 -22 kgm/s. 15.65 (b) v. = 0.12c. 15.67 vf = 0; M = 14m. 15.75 M = 2.95 GeV/c 2 ; v = 0.65c. 15.81 r(rel) = 8.3 cm; r(nonrel) = 5.9 cm. 15.71 (c) The minimum energies are E cn, = 3100 MeV, but Elab ;\u2018-'- 9.6 x 106 MeV. 15.93 If E denotes the final energy of the electron, E/Eo ti 0.002. 15.109 (a) If we choose axes so that the two particles lie in the xy plane, then in their rest frame 8', the force of one on the other is F' = (0, kq 2 I r'2, 0). According to (15.155) this transforms to F = (0, kq 2 lyr2, 0) in S. (b) In 8' the fields of one charge at the position of the other are E' = (0, kq/r (2, 0) and B' = (0, 0, 0). In 8, they are E = (0, ykq/ r 2, 0) and B = (0, 0, y )60 2 Icr2); these produce a force F = qv x B which is easily seen to be the same as in part (a). Chapter 16 16.7 Answers for Chapter 16 775 4 t = 0 -4 t = 0.5 t = 1.0 -4 4 -.5 7-\u203a:\\ t = -2.0 (//7.\\\\I -4 t = -1.5 /\\ t = -1.0 -4 t = -0.5 t = T/2 t = 5T//8 t= T 16.11", " 16.27 2a + /3 E= 611 [ 3\u00b01/3 0 0, 16.31 distance = 5040 km. Index All entries are identified by their page number. In addition, when a reference is to a whole section or chapter, I have indicated the section or chapter in parenthesis; for example, (Sec.1.1) or (Ch.1). Similarly, when a reference is primarily to a figure, example, problem, or footnote, I have added a parenthesis, such as (Fig.1.2), (Ex.1.3), (Pr.1.4), or just (Ft.). Coriolis, and Coriolis force, 358 374 Autonomous equation, 460 (Ft.), Absolute future, 626 Conservation of angular Atwood machine, 131-133 Absolute motion, non-existence momentum by Lagrange's equations, 255 Accelerating reference frame, 327 in terms of Euler angles, 403 in terms of CM and relative (Ex.7.3) coordinates, 369 L = Icy, 380 axis, 373 L, = I,co, for rotation about z not necessarily parallel to co, (Ex.7.8) double, 285 (Pr.7.27) energy of, 155 (Pr.4.31) including pulley, 156 (Pr.4.35) using Hamilton, 527 (Ex.13.2) using Lagrange multiplier, 279 of several particles, 93-95 of single particle, t, 90 537 (Ft.) Auxiliary equation, 175 of two bodies in CM frame, 299 Axial symmetry, 378 total, L, 93-95 Angular velocity, of earth's spin, Backward light cone, 626 339 Balance, inertial, 10 of, 602 Absolute past, 626 (Sec.9.1) Acceleration, centripetal, 29 Coriolis, 29 (Sec.9.10) free-fall, 345-347 in 2D polar coordinates, 29 in Cartesian coordinates, 23 Action integral, 239 Addition, of vectors, 6 Air resistance, 43-65 of angular velocities, 338 Angular velocity vector, 336 Barn, 563 (Sec.9.3) addition of, 338 comparison of linear and Anisotropic oscillator, 171-172 quadratic, 45 linear, 44, 46 (Secs.2.2-2.4) Aphelion, 309 Apogee, 316 quadratic, 44, 57 (Secs.2.4-2.5), Approach to chaos, for DDP, 467 on spinning hoop, oscillations Angular momentum, 90 (Secs.3.4\u2014 Attractor, 186 (Sec.12.4) 73 (Pr.2.4) 3.5) about CM, 97 as orbital plus spin, 370 conservation of, See DDP has several attractors, (Ex.13.1) 470-471 for logistic map, 503 strange, 497 Bernoulli's theorem, 725- 726 fi, damping constant, 174 Basin of attraction, 518 (Pr.12.22) Bead, on rotating rod, 284 (Pr.7.21) on spinning hoop, 260 (Exs.7.6 & 7.7) of, 264 (Ex.7.7) on wire, using Hamilton, 526 777 778 Index Bifurcation, 264 of DDP, 474 Bifurcation diagram, Block, (Ex.4.3) Chaos, 457 (Ch.12) and sensitivity to initial Complete elliptic intergal, 157 (Pr.4.38) conditions, for DDP, 479-483 Complete solution of 1D for DDP, 483-486 (Figs.12.17\u2014 and sensitivity to initial motion,using energy, 127 12.18) for logistic map, 510-513 conditions, for logistic map Complex exponentials, 68-69 Complex numbers, 79 (Prs.2.45\u2014 on incline, 24 (Ex.1.1), 115 for DDP, 476 used for charge in magnetic field, 519 (Pr.12.29) criterion for, 460 (Ft.) for logistic map, 511-512 2.51) 67-71 Compton, Characteristic equation, 389 effect, 655-656 sliding on wedge, 258 (Ex.7.5) (Fig.12.43) BM = bulk modulus, 703 for air, 737 (Pr.16.36) in terms of Hooke's law constant Body cone, for free precession, Charge in magnetic field, 65 inertia tensor of, 384 (Ex.10.3) a, 717 399 Body frame, 395 Boost, 621 Bottle in a bucket, 168 (Ex.5.2) Boundary conditions, 689 Bounded orbits, 304, 309", "-312 Brachistochrone, 222 (Ex.6.2), 234 (Pr.6.21) (Pr.6.25) Bulk modulus, See BM Calculus of variations, 215 (Ch.6) definition of, 218 with several variables, 226 (Sec.6.4) Canonical momentum, 522 Characteristic time, r, for linear drag, 48, 52 for quadratic drag, 59 (Secs.2.5-2.7) helical motion of, 70-71 Classical, definition of force, 11 definition of mass, 10 mechanics, 3 Cloud chamber, 560 generator, 365 (Pr.9.31) Cone, CM of, 89 (Ex.3.2) Configuration space, 522 Conjugate momentum, 522 Conservation, of angular momentum, 91, 96, 299 of energy for multiparticle system, 146 140-142 definition of momentum, 14 of energy, 114 acceleration related to external of energy in Lagrangian force, 88 defined as integral, 88 frame, See CM frame mechanics, 269-272 of momentum, 18, 21, 83 Conservation laws, in Lagrangian of earth & moon, 101 (Pr.3.17) mechanics, 268 (Sec.7.8) of earth & sun, 101 (Pr.3.16) Conservative force, 109-111 of solid cone, 89 (Ex.3.2) velocity related to total momentum, 88 central implies spherical symmetry, 137, 158 (Pr.4.45) Coulomb force as example, 119 is a cycloid, 224, 233 (Pr.6.14) CM, center of mass, 87, 295, 367 of energy for two particles, isochronous property of, 234 (Sec.10.1) Canonical transformation, 538 CM frame, 296 examples of, 554 (Prs.13.24\u2014 for two-body motion, 297-298 (Ex.4.5) defined, 111 13.25) Carrying capacity, 501 relativistic, 645 Colatitude, 9, 134 Cauchy's reciprocal theorem, 735 Collision, second condition for, 118-119 Constitutive equation, 715 Constrained systems, 245, 249 (Pr.16.19) Causality, 628 elastic, 99 (Pr.3.5), 142 Constraint equation, 275 elastic, equal mass, 143 (Ex.4.8) Constraint force, 251 Center of mass, See CM elastic, relativistic, 646 eliminated in Lagrangian Central force, 18, 91, 133 (Sec.4.8) (Ex.15.10) approach, 237 conservative implies spherical of putty lump with turntable, 96 related to Lagrange multiplier, symmetry, 137, 158 (Pr.4.45) (Ex.3.3) spherical symmetry implies of relativistic putty, 644 conservative, 158 (Prs.4.43 & (Ex.15.9) 278 Continuity equation, 726 Continuum hypothesis, 682 perfectly inelastic, 84 (Ex.3.1), Continuum mechanics, 681 Central-force, two-body motion, 159 (Pr.4.48) (Ch.16) See Two-body, central-force unequal masses, elastic, 159 Core (outer) of earth is liquid, 722 Centrifugal force, 262, 343, 344 Collision theory, 557 (Ch.14) contribution to g, 345-347 Centripetal acceleration, 29 Chandler wobble, 400 See also, Scattering, Cross Coriolis force, 343, 348 (Sec.9.7) and Coriolis acceleration, 358 Complementary function, 181 (Ft.) (Sec.9.10) (Pr.4.46) quantum, 557 section Coriolis acceleration, 29 and Coriolis force, 358 (Sec.9.10) 4.44) motion (Sec.9.6) Index 779 compared to magnetic force, 348 Damping constant 13, 174 Dot product, 6 effect on free fall, 351 (Sec.9.8) of DDP, 463 Coulomb force is conservative, DDP, 462 (Sec.12.2) equivalence of two definitions of, 35 (Pr.1.7) 119 (Ex.4.5) approach to chaos, 467 Double map, f (f (x)), 507 Coupled differential equations, 47 Coupled oscillators, 417 (Ch.11) damped, 449 (Pr.11.10) damped and driven, 449 (Pr.11.11) expected properties of, 463 (Sec.12.4) (Sec.12.3) rolling motion of, 481- 482 (Fig.12.15), 486-487 matrix equation of motion for, (", "Fig.12.19) 418-419, 439 weak coupling, 426 (Sec.11.3) with n degrees of freedom, 436 Decay parameter, 176, 177, 178 Decomposition of strain tensor, 714 Degrees of freedom, 249 Coupled pendulums, 441 V, del, 117 Double pendulum, 430 (Sec.11.4) Drag, See Air resistance Drive frequency, co, of DDP, 462 vs. natural frequency coo, 182 Drive strength, y, of DDP, 463 Driven damped oscillations (linear), 179 (Secs.5.5-5.6) complex form of, 182 Fourier series for, 197 (Sec.5.8) Driven damped pendulum, See Covariant & contravariant vectors, V2 = Laplacian, 694-695 Dumbbell, sliding & spinning, 97 as differential operator, 118 DDP (Sec.11.5) (Sec.11.6) 656 Derivatives matrix, D, 710 for a small rotation, 711 (Ex.3.4) Dynamical balancing, of car Deviatoric part E' of strain tensor, wheels, 375 Critical damping, 177-178 as limit of weak damping, 210-211 (Prs.5.24 & 5.32) 714 Cross product, 7 Diagonal matrix, 386 e l, e2, e3, unit vectors, 5 Cross section, 560 (Secs.14.2\u2014 Diagonalization of matrices, 739 Eccentricity, E, of Kepler orbits, 14.3) capture, 566 definition of, 562 of a single matrix, 739 related to energy, 313 Effective PE, for central force, 301 311 differential, See Differential of inertia tensor, 392 Eigenvalue, 389 cross section of two matrices, 743 (Sec.A.2) of fixed point, 505 elastic & inelastic, 567 Diatomic molecule, PE of, (Appendix) (Sec.A.1) 126-127 Eigenvalue equation, 389 generalized, for coupled fission, 567 in various frames, 579 ionization, 567 ionization, is zero below ionization energy, 568 scattering, 566 total, 568 Differential cross section, 568 (Secs.14.4-14.5) oscillators, 420 Eigenvector, 389 calculation of, 572 (Sec.14.5) Elastic collision, 99 (Pr.3.5), 142 definition of, 570-571 energy lost in lab frame, 592 for hard sphere scattering, 573 (Pr.14.29) (Ex.14.5) Crows in oak tree, 562 (Ex.14.1) Cube, balanced on cylinder, 130 for Rutherford scattering, 576 in lab, related to CM, 582, 585 PE near equilibrium, 162 (Ex.4.7) (Ex.5.1) Curl, of vector, 119, 152-153 (Prs.4.22 & 4.25) Curvilinear 1D systems, 129 (Sec.4.7) in various frames, 579 Differential equations, 14 coupled, 47 general solution of, 32 Differential operator, 180 Differentiation, of vectors, 7 equal mass, 143 (Ex.4.8) relativistic, 646 (Ex.15.10) unequal masses, 159 (Pr.4.46) Elastic modulus = stress/strain, Electric and magnetic forces, relative strengths of, 38 703-704 (Pr.1.32) Electric field, of charge with constant velocity, 680 Cyclic coordinates, 266 partial, 116-117, 152 (Prs.4.10 (Pr.15.110) See also Ignorable coordinates & 4.11) Electrodynamics, relativistic, 660 Cycloid, 224, 233 (Pr.6.14) See also Brachistochrone Cyclone, 350 Dilatation, 712 (Ex.16.5) Discrete time, 499 Divergence, V - v, 546 (Sec.15.18) Electromagnetic, four-vector current density, 680 Cyclotron frequency, 66, 71 as outward flow/volume, 548 (Pr.15.108) Cylinder on incline, 147 (Ex.4.9) in n dimensions, 548 Cylindrical polar coordinates, 40 Divergence theorem, 546 (Pr.1.47) proof of, 556 (Pr.13.37) Doppler effect, 631-632 four-vector potential, 680 (Pr.15.107) momentum, 22 Electromagnetic field, Damped oscillations, 173 (Sec.5.4) transverse, 672 (Pr.", "15.48) Lorentz transformation of, 661 Electromagnetic field (continued) Fixed point, 503 780 Index of moving line charge, 662 (Ex.15.12) tensor, 661 Elliptic integral, 157 (Pr.4.38) Elliptical orbits, of planets, 309 Elsewhere, 628 Energy, 105 (Ch.4) conservation of, See Conservation of energy kinetic, See Kinetic energy mass, 641 mechanical, 113 of 1D linear systems, 123 of comet, related to eccentricity, (Sec.4.6) 313 eigenvalue of, 505 multiplier of, 505 stable, 505 Force, using energy, 128 (Ex.4.6) Free precession, of axially symmetric body, 398-400 Frequency, drive, of DDP, 462 as gradient of PE, 116-117 central, See Central force driving co vs. natural coo, 182 natural, wo, of DDP, 463 centrifugal, See Centrifugal force Frisbee, precession of, 414 conservative, See Conservative (Pr.10.43) force Full width at half maximum, 189 constraint, See Constraint force Fundamental mode, 690 Coriolis, See Coriolis force FWHM, 189 definition of, 11 derivable from PE, 117 fictitious, 329 heat-like, 649 inertial, 328 g, contribution of centrifugal force Galilean invariance, of Newton's to, 345-347 laws, 598 of multiparticle system, 144 nonconservative, 114 Galilean relativity, 596 (Sec.15.2) (Sec.4.10) of SHM, 169 of two particles, 138-142 on particle a, Fa = \u2014V, 146 Galilean transformation, 597 relativistic, 649 (Sec.15.15) spherically symmetric, 134 y = drive strength, of DDP, 463 y = force constant, for Kepler potential, See Potential energy surface, 698-699 relativistic, 638 rest, 641 threshold, 647 tidal, 332 volume, 698 Force constant, y, for Kepler Gauss's theorem, See Divergence Equation of continuity, 726 problem, 308 theorem Equation of motion, 23 for elastic solid, 721 for inviscid fluid, 725 Forced coordinates, 249 (Ft.) Forward light cone, 625 Gedanken experiment = thought experiment, 603 Foucault pendulum, 354 (Sec.9.9) Geiger & Marsden, 577 problem, 308 y = V1 \u2014 13 2, 606 r = torque, 90 Fourier series, 192 (Secs.5.7-5.9) Generalized coordinates, 240, See also Newton's second law Four-force, 652 Equilibrium of 1D system, when Fourier coefficients, integrals for, dUldx = 0, 125 Ether drag, 600 195 Euler angles, 401 (Sec.10.9) definition of, 194 Eulerian description of fluid, 723 for driven oscillator, 197 Euler-Lagrange equation, 220 with two dependent variables, (Ft.) 227 10.8) for rectangular pulse, 195 (Sec.5.8) (Ex.5.4) Fourier sine series, 692, 734 Fourier's theorem, 194 Euler's equations, 394 (Secs.10.7\u2014 (Pr.16.13) with zero torque, 397 (Sec.10.8) Four-momentum, 636 Euler's formula, 68-69 Even function, 198 Four-scalar, 622 Four-space, 620 Four-vector, 620 data, 590 (Prs.14.16-14.17) General solution,of second-order differential equation, 32 247-249 forced, 249 (Ft.) natural, 249 Generalized force, 241 0 component = torque, 244 Generalized Hooke's law, 716 Generalized momentum, 241, 266 0 component = angular momentum, 244 Geodesic, on cone, 233 (Pr.6.17) on cylinder, 232 (Pr.6.7) Feigenbaum relation, for DDP, current density, 680 (Pr.15.108) 474-475 definition of, 622 for logistic map, 519 (Pr.12.29) electromagnetic potential, 680 on sphere, 225, 233 (Pr.6.16) Gibbs phenomenon, 197 (Ft.) go = \"true\" acceleration of gravity, Fermat's principle, 217 and law of reflection, 231 (Pr.6.3) (Pr.15.107) Four-velocity, 634 Fractal, 496 Fictitious force, 329 frame First integral, of Euler-Lagrange Free fall, equation, 232 (Prs.6.10", " & acceleration, 345-347 GPS, importance of time dilation 346 for, 608 in spherical polar coordinates, 4.15 & 4.18) 136-137 and Snell's law, 231 (Pr.6.4) Frame of reference, See Reference Gradient, V, 117, 152 (Prs.4.12\u2014 6.20) and Coriolis force, 351 (Sec.9.8) Graviton, 653 (Ft.) Index 781 Growth equation, 499-500 block on, 24 (Ex.1.1), 115 of rotation about any axis, 388, Half-life, 607 Half-pipe and skateboard, 30 Independent functions, 174 (Ft.) Inelastic collision, 84 (Ex.3.1), Half width at half maximum, 190 159 (Pr.4.48) (Ex.4.3) 412 (Pr.10.33) relativistic, 641 rotational, in terms of Euler angles, 403 Halley's comet, 311 Inertia tensor, 378 (Sec.10.3) Hamiltonian, X, 270, 521 (Ch.13) definition of, 380 total, as orbital plus spin, 371 Kronecker delta symbol, 8ii, 411 as energy for natural system, 270-272, 525 definition of, 523; 529 for charge in magnetic field, 553 (Pr.13.18) not equal to energy for non-natural systems, 552 (Prs.13.11 & 13.12) diagonalization of, 392 for lamina, 411 (Pr.10.23) for solid cone, 384 (Ex.10.3) for solid cube, 381 (Ex.10.2) symmetry of, 381 Inertial balance, 10 Inertial force, 328 (Pr.10.21) \u00a3 = angular momentum, 90 L = total angular momentum, Lagrange multipliers, 275 93-95 (Sec.7.10) Hamilton's equations, 521 Inhomogeneous equation, 180 Lagrange's equations, 237 (Ch.7) Inertial frame, 9, 15, 601 related to constraint forces, 278 Integral, line, 107 and conservation laws, 268 compared with Lagrange's, 536 Invariance, (Ch.13) (Sec.13.5) derivation of, in ID, 525-526 for 1D oscillator, 539 (Ex.13.5) for 1D systems, 524 (Sec.13.2) transformation, 598 rotational, 134 translational, 139 for central force, 531 (Ex.13.3) Invariant, for free fall, 541 (Ex.13.6) mass, 633 for mass on cone, 533 (Ex.13.4) length squared,.x 624 (Sec.7.8) (Sec.13.5) (Sec.7.9) (Sec.7.1) modified, 277 for magnetic forces, 272 for unconstrained motion, 238 of Newton's laws under Galilean compared with Hamilton's, 536 in several dimensions, 528 scalar product, 623 (Sec.15.9) with constraints, 250 (Sec.7.4) scalar product, definition in 4D, with Lagrange multipliers, 277 Inverse Lorentz transformation, Lagrangian description of fluid, See also Lagrangian 624 612 (Ex.14.5) Inviscid fluid, 723 (Secs.16.12\u2014 Lagrangian, lab & CM cross sections for, 585 16.13) for charge in magnetic field, Irreducible subspaces, 716 (Ft.) Isotropic oscillator, 170-171 Isotropy of shear-free pressure, 699 Iterated map, 501 723 (Ft.) 273-274 general definition of, 272 = T \u2014 U, 238 nonuniqueness of, 272-273 of top, in terms of Euler angles, 403 KE, See Kinetic energy See also Lagrange's equations Kepler orbits, 308 (Secs.8.6-8.7) = mass ratio, 584 are elliptical, 309 changes of, 315 (Sec.8.8) eccentricity of, 311, 313 hyperbolic, 314-315 parabolic, 314 A = thrust factor, 317 Lame constants, 716 (Ft.) Lamina, 547 (Ft.) inertia tensor of, 411 (Pr.10.23) principal axis of, 412 (Pr.10.30) (Sec.13.3) See also Hamiltonian, Jf Hamilton's principle, 239 Hard-sphere scattering, 573 (Ex.14.7) Harmonic, 466 of finite string, 691 Heat-like force, 649 Hobos on flatcar, 99 (Pr", ".3.4) Holonomic systems, 249 Homogeneous equation, 180 Homogeneous solution, 181 Hooke's law, 161 (Sec.5.1) generalized, for solid, 716 Horizontal, definition of, 347 Hurricane, 350 HWHM, 190 Hyperbolic functions, 61, 77 See also Bounded orbits, (Prs.2.33-2.34) Unbounded orbits Laminar flow, 547 Laplacian, V 2, 694-695 Hyperbolic Kepler orbits, 314-315 Kepler problem, 301 Larmor precession, 363 (Pr.9.22) See also Two-body, central-force Latitude, 134 Ideal fluid, 723 (Secs.16.12-16.13) motion Law of inertia, 13 Ignorable coordinates, 266 Kepler's first law, 311 See also Newton's first law in Hamiltonian mechanics, 535 Kepler's second law, 91-93 Law of reflection, in hard-sphere (Sec.13.4) Kepler's third law, 311-312 scattering, 589 (Pr.14.13) Impact parameter, b, 558-560 Kinetic energy, 105 Legendre transformation, 524 (Ft.) Incline, of rotation about a fixed axis, Length contraction, 608 (Sec.15.5) and rolling cylinder, 147 (Ex.4.9) 374 cannot be seen, 668 (Pr.15.14) Length contraction (continued) Map, 500 Moment of inertia tensor, See 782 Index formula, 609 Liapunov exponent, 480 Lift, 43 Light cone, 625 (Sec.15.10) forward & backward, 625-626 double, f (f (x)), 507 iterated, 501 logistic, See Logistic map second iterate, f (f (x)), 507 sine, 518 (Prs.12.23-12.25) Line integral, 107 Mass, Linear drag, 44, 46 (Secs.2.2-2.4) change of, in Franck-Hertz compared to quadratic, 45 horizontal motion with, 48 experiment, 640 (Ex.15.7) classical definition of, 10 trajectory of projectile with, 54 vertical motion with, 49 energy, 641 invariant, 633 Linear operator, 180 is proportional to weight, 10 of photon, related to wave vector, Linearity & nonlinearity, 458 matrix, 419, 440 (Sec.12.1) nonconservation of, in relativity, relativistic, 636 Liouville's theorem, 543 (Sec.13.7) proof of, 548-549 Lissajous figure, 172 639-640 ratio, A, 584 relativistic definition of, 633 Logistic map, 498 (Sec.12.9) variable, 633, 637 total, in terms of CM velocity, 88 Momentum-energy, 638 See also Four-momentum Morse potential energy, 207 bifurcation diagram for, 510- Massless particles, 652 (Sec.15.16) (Pr.5.2) chaos, 511-512 (Fig.12.43) Material description of fluid, Multiplication, of matrices, 380 Material derivative, 724 = reduced mass, 296 sensitivity to initial conditions, for standard boost, 621 Multistage rockets, 101 (Pr.3.12) 513 defined, 502 (Pr.12.29) Feigenbaum relation for, 519 519 (Pr.12.30) Longitude, 0, 134 Longitudinal wave, in solid, 721 in string, 735 (Pr.16.17) Lorentz-Fitzgerald contraction, 610 Lorentz force, 660 Lorentz scalar, 622 Lorentz transformation, 610 (Sec.15.6) equations, 612 inverse, 612 standard, 621 LRC circuit, 173 driven, 179 723 Matrix, diagonal, 386 multiplication, 380 orthogonal, 657 positive definite, 743 rotation (3D), 618 rotation (4D), 621 trace of, 714 unit, 1, 384 564 (Ex.14.3) Mechanical energy, 113 Mechanics, classical, 3 See also Length contraction Maxwell's equations, in four- Natural units, 442 vector form, 680 (Pr.15.111) Navier equation, 721 Mean free path, of air molecules, Neap tides, 335 of electromagnetic fields, 661 continuum, 681 (Ch.16) Hamiltonian, See Hamilton's Lagrangian, See Lagrange's (Sec, 1.6) equations equations Magnetic and electric forces, nonlinear, 457 (Ch.12) relative strengths of, 38 quantum, 3 (Pr.1.32) Method of images, 733 (Pr.16", ".12) rotational form of, 90 Magnetic field, charge in, 65 Metric matrix, G, 659 validity of, 17 (Secs.2.5-2.7) Magnetic force, MeV/c, 643 MeV/c2, 643 600 and Lagrange's equations, 272 Michelson-Morley experiment, and conservation of momentum, (Sec.7.9) (Pr.1.33) between two current loops, 38 Minkowski, 617 Moment of inertia, 95 18 invalid in relativity, 21 validity of, 21 violates Newton's third law, 22 I\u201e 373-374 violated by magnetic forces, 22 Newton's third law, 17 (Sec.1.5) Inertia tensor Momentum, angular, See Angular momentum canonical, 522 classical definition of, 14 conjugate, 522 conservation of, See Conservation of momentum electromagnetic, 22 generalized, See Generalized momentum 654 Multiplier, Lagrange, 275 (Sec.7.10) of fixed point, 505 n, unit vector normal to surface, 698 Natural coordinates, 249 Natural frequency, coo, 174 of DDP, 463 vs. driving frequency, w, 182 Neutrino, 653 (Ft.) Newton's first law, 13 validity of, 17 Newton's laws, 3 (Ch.1) Newton's second law, 13 in 2D polar coordinates, 29 in Cartesian coordinates, 23 in rotating frame, 342 (Sec.9.5) in rotating frame, using Lagrange, 361 (Pr.9.11) Index 783 Noether's theorem, 267, 272 and angular momentum, 290 Foucault, 354 (Sec.9.9) in an accelerating car, 329 (Pr.7.46) (Ex.9.1) Nonconservative force, 114 simple, See Simple pendulum internal, of rigid body, 147 kx 4, 154 (Pr.4.29) of 1D linear systems, 124-126 of charge in electric field, 112 Nonholonomic systems, 249-250 spherical, 288 (Pr.7.40) (Ex.4.2) Noninertial frame, 15, 327 (Ch.9) Perigee, 316 Nonlinear mechanics, 457 (Ch.12) Perihelion, 309 Nonlinearity, 458 (Sec.12.1) Normal coordinates, 425, 444 Period-doubling cascade, for logistic map, 510 (Sec.11.7), 454 (Prs.11.33\u2014 in convection of mercury, 11.35) Normal frequencies, 421 Normal modes, 417 (Ch.11) defined, 422 of finite string, 689-691 473-474 (Fig.12.9) of DDP, 471-475 Period three, of DDP, 469 Period two, of DDP, 468 of diatomic molecule, 126-127 of many particles, 145, 146 of simple pendulum, 155 (Pr.4.34) of spring, 151 (Pr.4.9) of two charges, 121 time-dependent, 121, 154 (Pr.4.27) Precession, free, 398-400 Numerical solution, for trajectory of logistic map, 507 free, using Euler angles, 415 of baseball, 63 (Ex.2.6) Periodic function, definition of, Nutation, of top, 405 193 (Pr.10.55) of equinoxes, 394 Phase-locked motion, 517 coo = natural frequency, 174, 463 (Pr.12.17) One-dimensional systems, Phase point, z, 537 graphs of PE for, 124-126 Phase shift, energy of, 123 (Secs.4.6 1.7) near resonance, 191 of frisbee, 414 (Pr.10.43) of top due to weak torque, 392 (Sec.10.6) of top, using Euler angles, 404 Preferred frame, non-existence of, Operator, linear, 180 Orthogonal matrix, 657 Oscillations, 161 (Ch.5) scattering, 588 (Pr.14.12) 602 Phase space, 523 Pressure is isotropic if no shearing Phase-space orbit, 538 (Sec.13.6) forces, 699 coupled, See Coupled oscillators for 1D oscillator, 539 (Ex.13.5) Primary, P, wave, 722 damped, 173 (Sec.5.4) for free fall, 541 (Ex.13.6) Principal axes of inertia, 387 driven by rectangular pulse, 199 Phase-space vector, z, 537 (Ex.5.5) driven damped (linear), 179 (Secs.5.5-5.6) unit vector, derivative of, 28 (Sec.10.4", ") existence of, 388 finding, 389 (Sec.10.5) driven damped, complex form & 1.48) (Ex.10.4) of, 182 in spherical polars, 135-136 of lamina, 412 (Pr.10.30) in cylindrical polars, 40 (Prs.1.47 for cube about corner, 390 in two dimensions, 170 (Sec.5.3) of bead on spinning hoop, 264 in 2D, 26 Principal moments (of inertia), 387 Principal stress axes, 735 (Ex.7.7) overdamped, 177 underdamped, 176-177 Photon, 652 (Sec.15.16) relation between p & k, 654 Pion, decay of, 667 (Pr 15.8) (Pr.16.21) Product, Planck's constant, 588 (Pr.14.12), cross, of two vectors, 7 Partial derivative, 116-117, 152 cylindrical, 40 (Pr.1.47) Q factor, 190 Overdamped oscillations, 177 Overtones, 690 654 Plane wave, 695 Pluto, discovery of, 595 (Ft.) Parabolic Kepler orbits, 314 Poincare, 459 Parallel-axis theorem, generalized, section, 495 (Sec.12.8) 411 (Pr.10.24) Parseval's theorem, 204 Point mass, 13 Polar coordinates, (Prs.4.10 & 4.11) Partial fractions, 78 (Pr.2.37) in 2D, 26 spherical, 134-135 Partial-wave series, 589 (Pr.14.12) Positive definite matrix, 743 Particle, 13 Particular solution, 181 PE, See Potential energy Pendulum, double, 430 (Sec.11.4) driven damped, See DDP Potential energy, defined, 111 effective, for two-body motion, 301 dot, of two vectors, 6, 35 (Pr.1.7) of inertia, 375-376 of vector and scalar, 6 Proper length, 610 Proper time, 607 of an object, 634 Quadratic drag, 44, 57 (Secs.2.4\u2014 2.5), 73 (Pr.2.4) compared to linear, 45 horizontal and vertical motion with, 62 horizontal motion with, 58 trajectory of baseball with, 63 in uniform gravitational field, (Ex.2.6) 151 (Prs.4.5 & 4.6) vertical motion with, 60 784 Index Quantum mechanics, 3 Quasiperiodic motion, 172 Quotient rule, 630 Quantum collision theory, 557 19, unit vector, 40 (Prs.1.47 and Scleronomous coordinates, 249 1.48) Rigid body, 147 RLC circuit, 173 driven, 179 rotation of, 367 (Ch.10) (Ft.) Secondary, S, wave, 722 Secular equation, 389 Self similarity, of fractal, 497 derivative of, 27 RMS, displacement, 203 of logistic bifurcation diagram, unit vector of 2D polars, 26 for driven oscillator, 204 (Ex.5.6) 512 unit vector, of spherical polars, Rockets, 85 (Sec.3.2) multistage, 101 (Pr.3.12) Saturn V, 100 (Pr.3.6) Sensitivity to initial conditions, for DDP, 480 (Fig.12.13) for logistic map, 519 (Pr.12.30) for two-body motion, 299-300 space shuttle, 100 (Prs.3.7 & 3.9) Separation of variables, thrust of, 86 Rolling, of DDP, 481-482 for first-order differential equation, 58, 60 (Ft.), 73 135-136 Radial equation, transformed, 306 Range, of baseball with quadratic drag, (Fig.12.15), 486-487 (Pr.2.7) 63 (Ex.2.6) (Fig.12.19) for wave equation, 733 (Pr.16.9) of projectile with linear drag, 54 Root mean square, See RMS Shear modulus, See SM Rapidity, 670 (Prs.15.30-15.31) Rotating reference frame, 339 Shear wave, in solid, 721-722 Rectangular pulse, (Secs.9.4-9.10) driving an oscillator, 199 Newton's second law in, 342 (Ex.5.5) (Sec.9.5) Shearing flow, 547 (Ex.13.7) SHM, See Simple harmonic motion Fourier series for, 195 (Ex.5.4) time derivatives in, 339 (Sec.9.4) Shortest path between two points, Reduced mass,", " it, 296 Reference frame, 9 Rotation, 367 (Ch.10) 221 (Ex.6.1), 228 (Ex.6.3) about a fixed axis, 372 (Sec.10.2) in three dimensions, 235 accelerating, 327 (Sec.9.1) about any axis, 378 (Sec.10.3) (Pr.6.27) Rotationally invariant force, 134 Rutherford formula, 576 Rutherford scattering, 557, 574 Simple pendulum, Relativistic electrodynamics, 660 angular dependence of, 577 PE of, 155 (Pr.4.34) body, 395 inertial, See Inertial frame noninertial, 15, 327 (Ch.9) Rotation matrix, four-dimensional, 621 three-dimensional, 618 rotating, See Rotating reference Rotational scalar, 622 frame space, 395 Reflection symmetry, 377 Relative position, r, 294 (Sec.15.18) Relativistic snake, 613 (Ex.15.3) (Sec.14.6) (Ex.14.6) Relativity, 596 Saturn V rocket, 100 (Pr.3.6) Galilean, 596 (Sec.15.2) Scalar, general, 596 of simultaneity, 615, 669 (Pr.15.19) of time, See Time dilation special, See Special relativity Lorentz, 622 rotational, 622 Scalar product, 6 equivalence of two definitions of, 35 (Pr.1.7) Simple harmonic motion, 163 (Sec.5.2) as real part of complex exponential, 167 definition of, 165 energy of, 169 exact period of, 156 (Pr.4.38) second approximation for period, 157 (Pr.4.39) Simultaneity, 615, 669 (Pr.15.19) Simultaneous diagonalization, of two matrices, 743 (Sec.A.2) Sine map, 518 (Prs.12.23-12.25) Skateboard in half-pipe, 30 SM = shear modulus, 703 in terms of Hooke's law constant Snake, relativistic, 613 (Ex.15.3) Snell's law and Fermat's principle, Resonance, 187 (Sec.5.6) Scattering, /3, 717 on washboard road, 188, 212 amplitude (quantum), 588 (Pr.5.43) phase shift near, 191 width of, 189 Rest energy, 641 Rest frame, 610 Reynolds number, 46, 72 (Pr.2.3) Rheonomous coordinates, 249 (Ft.) (Pr.1.47) (Pr.14.12) angle, in lab related to CM, 584 231 (Pr.6.4) angle, 9, 558 elastic & inelastic, 567 Soap-bubble problem, 233 (Pr.6.19) of neutrons off aluminum, 563 Solid angle, 569 (Ex.14.2) Space cone, for free precession, of two hard spheres, 564 400 Rutherford, See Rutherford Space frame, 395 p = distance from z axis, 40 scattering Space shuttle, 100 (Prs.3.7 & 3.9) Sthur's lemma, 716 (Ft.) Space-like vector, 628 Stretching elements, \u20ac ii, of strain (Pr.4.27) Space-time, 617 (Sec.15.8), 620 Spatial description of fluid, 723 Special relativity, 595 (Ch.15) postulates of, 601 Stationary path, 217-218 Stationary point, 217 Steradian, sr, 570 Stokes's law, 72 (Pr.2.2) with quadratic drag, 60 6, unit vector, of spherical polar coordinates, 135-136 Three-force, 650 Strain = fractional deformation, Three-momentum, relativistic, 636 Index 785 Speed, 721 of longitudinal wave in solid, of longitudinal wave in string, 735 (Pr 16.17) of sound in air, 737 (Pr.16.36) of sound in fluid, 729 of sound in water, 729 (Ex.16.9) of transverse wave in solid, 722 684 Speed of light, and Michelson-Morley experiment, 600 as speed limit for causal influences, 628 702-703 Strain tensor, E, definition of, 711 decomposition of, 714 for a solid, 709 (Sec.16.8) for dilatation, 712 (Ex.16.5) for shear, 713 (Ex.16.6) stretching elements, Eh, 713 Stress & strain tensors, relation between, 715 (", "Sec.16.9) Stress = force/area, 702-703 Stress tensor, I, 704 (Sec.16.7) definition of, 706 in a static fluid, 708 (Ex.16.3) of transverse waves on string, Strange attractor, 497 as speed limit for inertial frames, is symmetric, 707 606 714 as speed limit for material tensor, 713 particles, 629 String, invariance of, 672 (Pr.15.43) non-invariance under Galilean transformation, 599 longitudinal motion of, 735 (Pr.16.17) transverse motion of, 682 Spherical part el of strain tensor, (Sec.16.1) Spherical pendulum, 288 (Pr.7.40) Spherical polar coordinates, 134-135 gradient in, 136-137 Strong damping, 177 Subharmonic, 469 Sum, of vectors, 6 Superposition principle, 164 not true for nonlinear equations, Spherical strain, 712 (Ex.16.5) Spherical wave, 696-697 461 Surface force, 698-699 Spherically symmetric force, 134 Sweet spot, 410 (Pr.10.18) central implies conservative, 158 Symmetry, (Prs.4.43-4.44) Spinning hoop with bead, 260 (Exs.7.6 & 7.7) Spring tides, 335 Spring-constant matrix, 419, 440 sr, steradian, 570 Stability, of freely spinning body, 398 of fixed points, 505 Stable equilibrium, when d2 U/dx 2 > 0, 125 Standard boost, 621 Standard configuration, 598 Standing wave, 688 State (or state of motion), 490 State space, 490, 522, State-space orbit, 487 (Sec.12.7) defined, 490 axial, 378 of inertia tensor, 381 of stress tensor, 707 under reflection, 377 Target density, s tar' 561 r = characteristic time, for linear drag, 48, 52 for quadratic drag, 59 Taylor's series, 75 (Pr.2.18) Tensor, 656 (Sec.15.17) electromagnetic, 661 four-dimensional, 659 inertia, See Inertia tensor three-dimensional, 657-658 Terminal speed, of baseball, 61 (Ex.2.5) with linear drag, 50 Three-scalar, 622 Threshold energy, 647 Thrust, of rocket, 86 Thrust factor, X, 317 Tidal force, 332 Tides, 330 (Sec.9.2) neap, 335 spring, 335 Time, classical view of, 9 discrete, 499 horizon, 517 (Pr.12.16) in relativity, 603 (Sec.15.4) proper, 607 proper, of an object, 634 Time-dependent PE, 121, 154 Time derivatives, in rotating frames, 339 (Sec.9.4) Time dilation, 603 (Sec.15.4) cannot be seen, 668 (Pr.15.10) evidence for, 607-608 for jet plane, 605 (Ex.15.1) formula, 606 Time-like vector, 629 Top, motion of, using Euler angles, 403 (Sec.10.10) nutation of, 405 precession of, due to weak torque, 392 (Sec.10.6) precession of, using Euler angles, 404 Torque, F, 90 Trace of a matrix, 714 Transformed radial equation, for two-body motion, 306 Transients, 184 Translational invariance, 139 Transpose A of matrix A, 381, 658 Transverse wave, in solid, 721-722 on string, 682 (Secs.16.1-16.3) Triangle inequality, 36 (Pr.1.14) Turning point, 125 for radial motion of comet, 303 Twin paradox, 667 (Pr.15.5) Two-body, central-force motion, 293 (Ch.8) closed and unclosed orbits, 304 equation for relative motion, 297 differentiation of, 7 transformed radial equation, 306 35 (Pr.1.7) Two-cycle, for logistic map, 507 unit, See Unit vectors two definitions of dot product, 786 Index Two-body, central-force motion (continued) effective PE for, 301 equivalent 1D problem, 300 (Sec.8.4) Lagrangian for, 297 radial equation for, 299-300 relative motion lies in fixed plane, 299 Two-dimensional, oscillator, 170 (Sec.5.3) polar coordinates, 26 Typhoon, 350 Unbounded orbits, 303, 313 (Sec.8.7) 176-177 Unit matrix, 1, 384 Unit vectors, el, e2", ", e3, 5 j, k, 4 and 3, 26 r and 3, derivatives of, 27-28 135-136 ;5, 3,1, 40 (Prs.1.47 and 1.48) X, S7, 4 Units, natural, 442 Universality, of period doubling, 475 Unstable equilibrium, when d2 U/dx 2 < 0, 125 vdv/dx rule, 74 (Pr.2.12) Variable mass, 633, 637 Variational principle, 215, 218 Vector, 4 cross product, 7 curl of, 119 dot product, 6 product, 7 scalar product,6 sum, 6 times scalar, 6 triangular, on finite string, 692 (Ex.16.2) triangular, on infinite string, 686 (Ex.16.1) Wave equation, for string, 684 in terms of Laplacian, 695 one-dimensional, 685 (Sec.16.2) solution by separation of variables, 733 (Pr.16.9) three-dimensional, 694 (Sec.16.4) Velocity, four dimensional, 634 in 2D polar coordinates, 28 Weak damping, 175-177 Weakly coupled oscillators, 426 (Sec.11.3) Velocity-addition formula, Weight is proportional to mass, 10 classical, 328, 598 relativistic, 616 Width of resonance, 189 Work, Vertical, definition of, 347 as change in PE, 112 (Pr.8.17) Viscosity, 7), 72 (Ft.) Volume force, 698 Vter, terminal speed, 50, 60 Washboard road, 188, 212 (Pr.5.43) Wave, in fluid, 727 (Sec.16.13) in fluid is longitudinal, 729 in rock, 722 (Ex.16.8) in solid, 721 (Sec.16.11) longitudinal, in solid, 721 on string, 682 (Secs.16.1\u2014 16.3) plane, 695 spherical, 696-697 standing, 688 in infinitesimal displacement, 106 Work-KE theorem, 108 infinitesimal form, 106 World line, 634, 671 (Pr.15.38) 1, unit vector, 4 '3'7, unit vector, 4 YM = Young's modulus, 703 in terms of Hooke's law constants a &,8, 718 related to BM & SM, 718 Young's modulus, See YM Yoyo, 283 (Pr.7.14) 1, unit vector, 4 Zero-component theorem, 671 transverse or shear, 721-722 (Pr.15.35) Underdamped oscillations, Virial theorem, 158 (Pr.4.41), 323 done by force, 107 Miscellaneous Data (for use in some end-of-chapter problems) Solar System (mass of earth) = 5.97 x 1024 kg (radius of earth) = 6.38 x 106 m (mass of moon) = 7.35 x 1022 kg (radius of moon) = 1.74 x 106 m (mass of sun) = 1.99 x 103\u00b0 kg (radius of sun) = 6.96 x 108 m (earth-moon distance) = 3.84 x 10 8 m (earth-sun distance) = 1.50 x 10 11 m Ideal Gases Avogadro's number, NA = 6.02 x 1023 particles/mole Boltzmann's constant, k = 1.38 x 10-23 J/K = 8.62 x 10-5eV/K Gas constant, R = 8.31 J/(moleK) = 0.0821 literatm/(moleK) STP = 0\u00b0C and 1 atm (Volume of 1 mole of gas at STP) = 22.4 liters Conversion Factors Area: 1 barn = 10-28 m2 Energy: 1 eV = 1.60 x 10-19 J 1 cal =4.184J Length: 1 inch = 2.54 cm 1 mile = 1609 m Mass: 1 u (atomic mass unit) = 1.66 x 10-27 kg = 931.5 MeV/c2 1 lb (mass) = 0.454 kg 1 MeV/c2 = 1.074 x 10-3 u = 1.783 x 10-3\u00b0 kg Momentum: 1 MeV/c = 5.34 x 10-22 kgm/s A Few More Constants Speed of light, c = 3.00 x 108 m/s Planck's constants: h = 6.63 x 10 -34 Js and h = 1.05 x 10-34 Js Vacuum permeability, p.o = 4n x 10-7 N/A2 Coulomb", " force constant, k = 1/(47rE0) = 8.99 x 109 Nm zicz co = 8.85 x 10-12 c2gNm2) Vacuum permittivity, Vector Identities A (B x C) = B (C x A) = C (A x B) A x (B x C) = B(A C) - C(A B) [BAC - CAB rule] Vector Calculus V f =x \u201e af af \u201e af + y \u2014 z- ay az ax \u201e af ar af ap = of rsm9 a(/). af r 80 af p 92) \u2014 \u2014 Z af az [Cartesian] [spherical polars] [cylindrical polars] V x A = (\u2014a A, - ( 2-A, - \u2014A z) ay az Y aZ 8 aX --a A \u201e \u2014a A x ) ax ay ( [Cartesian] =i 1 [ a (sin 0A 0 ) r sin 0 ao a\u00a2 19- [ r sin A, - - \u2014 r (rA0)] a ra 449 Ar] +, p r Lp 0(/) [\u2014a (rA 9 )\u2014 r ar A, - Aol az j + is r-a Ai, - 2-A21 az - ap i [spherical polar] + i - [\u2014a (pito) - Tia A p] P aP [cylindrical polar] v A = \u2014A x + \u2014A ) + \u2014A z [Cartesian] = \u2014 \u2014 (r-A,)+ r sin 9 80 (sin 0,49) + r sin 0 a A, [spherical polars] a ay a az 3x a 1 a a 1 a r2 dr a pap =- \u2014 (PAp)\u00b1 \u2014A0+ - p a az \u2014Az 2 821 821 021 V f = aX 2 &W.?' aZ2 1 r2 sin e 00 a (sin 0 8f ae )\u00b1 r2 sin2 9 ao2 021 = (11)+ 1 a2 ar2 I a an 1 32f a2f P2 ao2 + 8z 2 P aP aP [cylindrical polars] [Cartesian] [spherical polars] [cylindrical polars] PHYSICS Classical Mechanics John R. Taylor UNIVERSITY OF COLORADO, BOULDER \"A superb text. The clarity and readability of the book is so much better than anything else on the market, that I predict this will soon be the most widely used book on the subject in all American universities. I judge it to be at least ten times better, maybe more, than the other popular classical mechanics books on the market right now.\" \u2014American Journal of Physics \"Taylor's book is unique among classical mechanics texts. It comprehensively covers the field at the Sophomore/Junior level. At the same time, it is immensely readable, a quality that comparable texts lack.\" \u2014Jonathan Friedman, Amherst College \"Many of my students thought that Taylor's Classical Mechanics was the clearest textbook that they had ever used.\" \u2014Joel Fajans, University of California, Berkeley John Taylor has brought to his new book, Classical Mechanics, all of the insight that made his Introduction to Error Analysis an international best-seller. With unusual' clarity, Classical Mechanics covers the topics normally found in books at this level, and features a large selection of interesting problems \u2014744 in all\u2014at the end of each chapter, ranging from simple exercises to challenging computer projects. John Taylor as \"Mr. Wizard\" on his bed of nails John R. Taylor is the author of three best-selling textbooks, including Introduction to Error Analysis, which has been translated into nine languages, and Modern Physics. He is Professor of Physics and Presidential Teaching Scholar at the University of Colorado in Boulder, where he has won numerous teaching awards, served as Associate Editor of the American Journal of Physics, and received an Emmy Award for his television series called \"Physics 4 Fun.\" University Science Books www.uscibooks.com About the cover: Taken in 1918 or 1919 on a highway in France near the WW I front, the photo shows a classic Model T automobile in mechanical difficulties. An old song called \"Get Out and Get Under\" immortalizes this driver's dilemma as he investigates beneath the running board. ISBN 1-891389-22-X 9 0 0 0 9 781891 389221f this formula is found in Stodola's classic treatise on steam dQJm = d (11 +.PrJ + v 212 + gz) turbines, first published in the German language in 1903. The first person to invent a theory simultaneously involving tbc ideas of conservation and conversion of energy was the young French military engineer Nicolas Leonard Sadi Camot (", " 1796-1832). The strikingly original ideas of Carnot.'s work make it among the most brilliant new departures in thcoretica.l physics. Salli Carnot was the son of Napoleon's general, Lazare Carnot. and was educated at the fomous Ecole Polytcchnique in Paris. Between 1794,:nd 1830, Ecole Polytechnique had such famous teachers as Lagrange, Fourier, Laplace, Ampere, Cauchy, Coriolis, Poisson, Gay-Lussac. and Poiseuillc. A.Her his formal education Carnot chose a career as an army officer. Britain was then a powerful military force, primarily as a result of the industrial revolution broughi about by the steam engine. French technology was not developing as fast as Britain's. Carnot was convinced that France's inadequate utilization of steam power had made it militarily inferior. He began 10 study the fundamentals of steam engine technology, and in 1824 he published the results of his study in the form of a Ill\" ii. Introduction -=17 brochure \"Reflection on the Motive Power of Heat and on Machines Fitted to Develop that Power\". Carnot was trained in the basic principles of hydraulics, pumps and water wheels at the Ecole Polytecbnique. During Carnot' s time. caloric theory of heat was siill persisting, and the water wheel as the major source of mechanical power was gradually getting replaced by the steam engine. Carnot conceived that the power of a steam engine was released, as the heat.fluid or caloric fell from the high temperature of the boiler to the lower temperature of the condenser,.in much the same way that water falls through a water wheel to produce mechanical shaft work output. Carnot stated, \"The motive power of a water wheel depends on its height and the quantity ofliquid. The motive power of heat also depends on the quantity of caloric used and on the height of its fall, i.e., the difference of temperatures of the bodies between which the exchange of caloric is made\". Till Carnot's time thermodynamics was developed primarily on an empricial basis provided by chemistry. Carnot approached an engineering problem, the efficiency of heat engines, In terms of entirely new concepts with the steam engine serving as the stimulus. Carnot observed that the existence of temperature d.i fferences is a necessary condition for producing mechanical work by means of a heat engine. He simplified the problem to its bare essentials and stipulates, that this system, consisti1.1g essentially of a working substance. should exchange heat with its surroundings only at two fixed temperatures. In order to conceptualize such a situation, he i.ntroduces the idea of heat reservoirs. Two important conclusions emerged from Carnot's work: 1. No one could build a water wheel that woo.Id. produce a continuous work output unless water actually entered and exited the wheel. If water with a certain kinetic and potential energy entered ihe wheel, then the same amount of water with a lower energy must also exit the wheel. It is thus impossible to make a water wheel that converts all the energy of the inlet water into shaft work output. There must be an outflow of water from the wheel. If this idea is extended to a steam engine by replacing the water by heat fluid caloric-, it can be concluded that when caloric at a certain energy le~\u00b7el (temperature) enters a work producing heat engine, it must also exit the heat engine at a low energy level (temperature). Thus a continuously operating heat engine that converts all of its caloric (heat) input direclly i.nto work ouiput i~ not possible. This is very close to the Kelvin-Planck statement of second law as it is known today. 2. The maximum efficiency of a water wheel was independent of the type of the liquid and depended only on the inJet and outlet flow energies. The maximum efficiency of the steam engine (or any heat engine) depends only on the temperatures of the high and low temperature lhennal reservoirs of the engine and is independent of the working fluid. To achieve the maximum ctl1ciency there must not be any mechanical friction or other losses of any kind. Only at the age of 36, Sadi Carnot died of cholera following an attack of scarlet fever. The significance ofCamot's work was not recognized until 1850, when Rudolf Clausius ( 1822-1888) and William Thomson ( 1824-1907) worked out a II I'II \u2022 I II 18=- Basic and Apf!litd Tl,mnod1namin clear fonnulation of the conservation of energy principle. Carnot's first conclusion was then called the second law of thermodynamics by Clausius, and Thomson used Carnot's second conclusion to develop the concept of absolute temperature scale. Thermodynamics is thus said to have originate<!", " from. the \"clumsy puffing of the early steam engines\" and is often called \"the daughter of steam engine\". Carnot's ideas were so revolutionary that they were largely ignored. Soon after Carnot's death, Emile Clapeyron (1799-1864), a French mining engineer, strengthened Carnot's ideas by using more precise mathematical derivation. Clapeyron constiucted its thennodynamic cycle by deducing that it must be composed of two reversible isothermal processes and two reversible adiabatic processes. It is now known as Carnot's cycle. It was the first heat engine cycle to be conceptualized. No other heat engine can equal its efficiency. Clapeyron was later able to derive a relation for the enthalpy change of the liquid to vapour phase (hrg) in terms of pressure, temperature and specific volume. This provided the first equation, now called the Clausius-Clapeyron equation, representing the first order phase transition, which could be used to estimate a property that is not directly measurable in terms of properties that arc directly measurable. Clapeyron's equation is now most easily derived from one of Maxwell's equations. WilJiam Thomson (1824-1907), who bc,came a professor of natural philosophy at the University of Glasgow i.n 1848 at the age of 24 only, rejected the caloric theory of heat and for the first time used the tenns \"thermodynamics\" and \" mec.hanical energy\". Apart from the deducti.on of the absolute temperature scale, Thomson worked with Joule from 1852 to 1862 in a series of experiments to measure the temperature of gas in a controlled expansion and propounded the Joule-Thomson cffe<:t for real gases. Rudolf Julius Emanuel Clausius (1822-1888) realised. that there were two distinct Jaws at work, the first law due to Joule and Mayer and the second law as expounded by Carnot. He defined the internal energy U. Although both Kelvin and Clausius use<I the function Qt(jT for some years, Clausius recognized the value of this function and to describe it he coined the word \"entropy\" from the Greek word \"tropee\" meaning \"transfonnation\" and assigned it the symbol S. Clausius in 1865, summarised the first and second laws ofthennodynamics in the following words: \"' Die Energie dcr Welt isl konstant. Die Entropie der Welt strebt einem Maximum zu'' whioh is translated as \"The energy of the world is constant. The entropy of the world tends toward a maximum\". The world here means the universe, the system and the surroundings together. These statements made a strong impression upon a young student, Max Karl Ernst Ludwig Planck (1858-1 947). He was educated at the universities of Munich and Berlin. In his autobiography be stated,, \"One day l happimed to come..,. lnlroduction -=19 across the treatises ofRudolfClausius, whose lucid style and enlightening clarity of reasoning made an enormous impression on me, and I became deeply absorbed in his articles, with an ever increasing enthusiasm. I appreciated especially his exact fonnulation of the two laws of thermodynamics, ahd the sharp distinction, which he was the first to esiablish between them\". In 1897, Planck1 demonstrated lhe close connection between the second law and lhe concept of reversibility. He stated the second law as the impossibility of a cyclic device which prodnccs positive work and exchanges heat with a single reservoir. Similar statement was also made by Kelvin, and is now recognized as Kelvin,Planck siatement of second law. Poincare2 in 1908, extended the work of Planck and prescribed a complete structure of classical thermodynamics. The property, entropy, plays a steller role in thennodynarnics. It was introduced via the concept of heat engines. In 1909. the Greek mathematician Caratbeodory proved the existence of entropy function mathematically without ihe aid of Carnot engines and refrigerators. Caratheodory' s statement of second law may be stated as: \"In the neighbourhood of any arbitrary state PO of a pbysical system, there exist neighbouring states which are not accessible from P0 along quasi-static adiabatic paths\". From the standpoint of the engineer and physicist it is entirely mathematical in form and devoid of physical insight. William John Macquom Rankine (1820-1872) defined the thennodynamic efficiency of a beat engine and showed the usefulness of p-v diagrams as related to work. He wrote the first text book on thermodynarnics3, and was the first to work out the thermodynamic cycle for the adiabastic cylinder steam engine, now known as Rankine cycle for a vapour", " kPa. Take the density of mercury as 13.6 x HY kfm3. density of water as l.03 kg/m\u00b7, the barometer reading as 76.1 cm Hg. and g 11$ 9.806 m/sl. Hg T.:.\u00b7 5.::-x T ~ ii \u00b7.\u00b7::. -.. ~ Fig. P.1.8 Iii It Jntroduction 1.9 A vacuum gauge mOWJled on a condenser n:adi. 0.66 m Hg. What is the abtolutc prcuwe in the c:otldenser in kPa when the atmospheric pressure is 101.3 kPa? A.IIS. 8.8 kP11 1.10 The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of a building are 730 and 760 mm Hg, respectively, determine the height of the building. Assume an average air density of 1.18 kg/m3 \u2022 Iii I, Temperature 2.1 Zeroth Law of Thermodynamics The property which distinguishes thermodynamics from other sciences is tem perature. One might say that temperature bears as important a relation to ther modynamics as force does to statics or velocity does to dynamics. Tempera ture is associated with the ability to distinguish hot from cold. When two bodies at different temperatures are brought into c,ontact, after some time they attain a common temperature and are then said to exist in thermal equilibrium. When a body A is i11 thermal eq11ilibri11m with a body B, and also separately with a body C. then B and C will be in thermal eq11ilibri11m with each other. This is known as the zeroth law of thermodynamics. It is the basis of tem perature measurement. ln order to obtain a quantitative measure of temperature, a reference body is wed, and a certain physical characteristic of this body which changes with temperature is selected. The changes in the selected characteristic may be taken as an indication of change in temperature. The selected characteristic is ca.lied the thermometric property, and the reference body which is used in the detenni nation of temperature is called the thermometer. A very common thennometer consists of a small amount of mercury in an evacuated capillary tube. In this case the extension of the mercury in the tube is u~ed as the thermometric prop erty. There are five different kinds ofthennomcter, each with its own thmnomctric property, as shown in Table 2.1. 2.2 Measurem.eot of Temperature-the Reference Points The temperature of a system is a property that detennines whether or not a system is in thennal equilibrium with other systems. If a body is at, say, 70\u00b0C, it will be 70\u00b0C, whether measured by a mercury-in-glass thermometer, resist- I I I\u2022 I '. 111 ance thennometer or constant volume gas tbennometer. If Xis the tbennomet ric property, let us arbitrarily choose for the temperature common to the ther mometer and to all systems in thermal equilibrium with it the following linear function of X: -=25 Table 2.1 11inmomtlm and 11itrmomtlric Prop~ties Thern,o,neter I. Constant volume gas thermometer 2 Constant pressure gas thennometer 3. Eleclrical resistance thcnnomcicr 4. Thermocouple S. Mercury-in-glass thennometer Tller,nometr/C' property. Pressure Volume Resistance Thennal e.m.f. Length Symbol p \" R \u00a3 L 8 (X) = aX, where a is an arbitrary comtant. If X1 corresponds to 6 (X1), lhen X2 will correspond to that is 9(Xi) \u00b7 X2 x, (2.1) Two temperatures on the linear X scale are to each other as lhe ratio of the corresponding X's. 2.2.1 Method in Use Before 1954 The thennometer is first placed in contact with the system whose temperature 6 (X) is to be measured, and then in contact with an arbitrarily chosen standard system in an easily reproducible state where the temperature is 8 (X1). Thus 6(X1) _ X1 6(X) -7 (2.2) Then lhe thennometer at the temperature 8 (X) is placed in contact with another arbitrarily chosen standard system in another easily reproduciblci state where lhe temperature is 8 (X2). It gives 8(X2) _ X2 8( X) - X (2.3) From Eqs (2.2) and (2.3) 6(X1) -8(X", ".si, alld A.pplitd 11imnodynamia 2.5 Gas Thermometers A schematic diagram of a constant volume gas thennometer is given in Fig. 2.1. A small amount of gas is enclosed in bulb B which is in communication via the capillary tube C with one li.mb of the mercury manometer M. The other limb of the mercury manometer is open to the atmosphere and can be moved vertically to adjusl the mercury levels so that the mercury just touches lip L of the capillary. The pressure in the bulb is used as a thennometric property and is given by where Po is lhe atmospheric pressure and PM is lhe density of mercury. P\"'Po+ PMZg Po lh- L~ M-n-\u00b7 F!eX.ible tubl11g Fig. 2.1 Corutont iYJlumt gas IAmnometn When the bulb is brought in contact with the system whose tempera!Ure is to be measured, the bulb, in course of time, comes in thennal CQuilibrium with the system. The gas in the bulb expands, on being heated, pushing the mercury downward. The flexible limb of the manometer is then adjusted so that the mercury again touches the lip L. The difference in mercury level Z is recorded and lhe pressure p of the gas in the bulb is estimated. Since the volwne of the trapped gas is constaD.t, from the ideal gas equation, llT= X...dp R (2.10) i.e. the temperature in.crease is proportional to the pressure increase. In a constant pressure gas thcnnometer, the mercury levels have to be adjusted to keep Z constant, and the volume of gas V, which would vary with the temperature of the system, becomes the thcnnometric property. llT=.E..llV R (2.11) I!!!I ii I + II Ttmptraturt ~29 i.e. the temperature increase is proportional to the observed volume increase. The constant volume gas thermometer is, however, mostly in use. since it is simpler in construction and easier to operate. \u00b7 2.6 Ideal Gas Temperature Let us suppose that the bulb of a constant volume gas lbennometcr contains an amount of gas snch that when the bulb is surrounded by water at its triple point, the pressure Pt is I 000 mm Hg. Keeping the volume V constant, let the following procedure be conducted: (a) Su1TOund the bulb with steam condensing at I atm, determine the gas pressure p and calculate 8=273.16 _I!_ 1000 (b) Remove some gas from the bulb so that when it is SUITOunded by water at its ttiple point, tbe presstue p1 is 500 mm Hg. Determine the new values of p and then e for steam condensing at I atm. 8 = 273.16 _f!_ 500 (c) Continue reducing the amount of gas in the bulb so that p, and p have smaller and smaller values, e.g., Pt having, say, 250 mm Hg, 100 mm Hg, and so on. At each value of Pt calculate the corresponding 8. (d) Plot 8 vs. p, and extrapolate the curve to the axis where Pt = 0. Read from the graph lim8 p,-+O The graph, as shown in rig. 2.2, i.ndicatcs that although the rcad.ings of a constant volume gas thennometer depend npon the nature of the gas, all gases indicate the same temperature asp, i.r lowered and nrade to t1pproach zero. A similar series of tests may be conducted with a constant pressure gas thermometer. The constant pressure may first be taken to be I 000 mg Hg, then 500 nun Hg, etc. and at each value of p, the volumes of gas V and V1 may be recorded when the bulb is surrounded by steam condensing at 1 atm aod the ttiple point of water, respectively. The correspondfog value of 8 may be calcu lated from 6'=' 273.16..!:'... V. and 8 vs. p may be plotted, similar to fig. 2.2. It is found from the experiments that all gases indicate the same value of() asp approach zero. Since a real gas, as used in the bulb, behaves as an ideal gas as pressure approaches zero (which would be explained later in Chapter 10), the ideal gas temperature T is defined by either of the two equations \"I' I II BaJic and Applied Thmnodyr,amiCJ -Pb mm Hg Fig. 2.2 Ideal gas tcnpenturt for steam point T= 273.16 lim _e_ P, Pt~ 0", " = 273.16 lim X. V. p-+0 where 9 has been replaced by T to denote this particular temperature scale, the ideal gas temperature scale. 2.7 Celsius Temperature Scale The Celsius temperature scale employs a degree of the same magnitude as that of the ideal gas scale, but its zero point is s.hit\\ed, so that the Celsius temperature of the triple point of water is 0.01 degree Celsius or 0.01\u00b0C. If t denotes the Celsius temperature, then t = T- 273.lS0 Thus the Celsius temperature t, at which steam condenses at I atm. pressure t, = T, - 273.IS0 = 373.15 - 273.IS \"' 100.00\u00b0c Similar measurements for ice points show this temperature on the Celsius scale to be 0.00\u00b0C. The only Celsius temperature which is fixed by definition is that of the triple point. 2.8 Electrical Resistance Thermometer 1n the resistance thermometer (Fig. 2.3) the change in resistance of a metal wire due to its change in temperature is the thermometric property. The wire, fre- 1 I \u20221, ;,I Tn,,.perature -=31. quently platinum, may be incorporated in a Wheatstone bridge circuit. The plati num resistance thermometer measures temperature to a high degree of accu racy and sensitivity, which makes it suitable as a standard for the calibration of other thermometer... In a restricted range, the following quadratic equation is often used R = Ro(l + At + Br) where R0 is the resistance of the platinum wire when it is swrounded by melting ice and A and B are constants. - - - -~ /./ Fig. 2.3 Resistance thermometer 2.9 Thermocouple A thermocouple circuit made up from joining two wires A and B made of dissimilar metals is shown in Fig. 2.4. Due to the Seeback effect, a net e.m.f. is generated in the circuit which depends on the difference in temperature between the hot and cold junctions and is, therefore, a thermometric property of the circuit. This e.m.f. can be measured by a microvoltmeter to a high degree of Wire A f( __ w_ 1re_8~ Test Junction I I I\u00b7 ~ / :-\u00b7\u00b7\u00b7... \u00b7\u2022,....... ;., To potentiometer 1----s----,..+--'--i Copperwin,s Ice-water mixture Fig. 2.4 'lllermocouplt I I,,, ill I 1,,'\" 32=- accuracy. The choice of metals depeods largely on the temperature range to be investigated, and. copper-constantan, chromel-alumel and platinum-platinum rhodium are typical combinations in use. A thermocouple is calibrated by measuring the thennal e.m.f. at various known tempera'tures. the reference junction being kept at 0\u00b0C. The results of such measureme.1tts on most thennocouples can usually be represented by a cubic.equation of the form.,\u2022 e=a+bt+cr+df where \u00a3 is the thermal e.m.f. and the constants a, b, c and dare different for each thermocouple. The advantage of a thermocouple is that it comes to thermal equilibrium wiih the system, whose temperature is to be measured, quite rapidly, because its mass is small. 2.10 International Practical Temperature Scale An international temperature scale was adopted at the Seventh General Conference on Weights and Measures held in 1927. It was not to replace the Celsius or ideal gas scales, but to provide a scale that could ~e easily and rapidly used to calibrate scientific and industrial instruments. Slight refinements were incorporated into the scale in revisions adopted in 1948, 1954, 1960 and 1968. The international practical scale agrees with the Celsius scale at the defining fixed points listed in Table 2.2. The temperarure inten'al from the oxygen point to the gold point is divided into three main parts, as given below. Table 2.2 Tt111pvatures of Fixttl Points Normal boiling point of oxygen Triple point of waler (standard) Normal boiling point of water Normal boiling point of sulphur (Normal melting point of zinc-suggested as an alternative to the sulphur point) Normal melting point of antimony Noffllal melting point of silver Normal melting point of gold -182.97 +0.01 100.00 444.60 419.SO 630.50 960.80 1063.00 (a) From O to 6606 C A platinum resistan.ce. thermometer with a plattnum wire whose diameter must lie between 0.05 and 0.20 mm is used, and the tempera ture is given by the eqnation R = RJ..", " junction at t\u00b0C on gas thennometer scale and reference junction at ice point is given by e = 0.20 t-5 x 1~ r2 mV The millivoltmeter is calibrated at ice and steam points. What will this thermome&er read in a place where the gas thermometer n:ads 50\u00b0C? Solutio11 At ice point., when t = 0\u00b0C. e = 0 m V At steam point, when t = 100\u00b0C, e = 0.20 x 100 - 5 x 10-1 x (100)2 At t= S0\u00b0C, When the gas thermometer reads S0\u00b0C, the thermocouple will read = ISmV e= 0.20 x SO- S x 10-1 (50)2 = 8.75 mV lOO X 8.75, or 58.33\u00b0C IS Ans. REVIEW QUESTIONS 2.1 What is the zeroth law of thermodynamics? 21 Define thermometric property. 2.3 What is a thermometer? 2.4 Wh.at is a fixed point? 2.5 How many fixed points were used prior to 1954? What are these? 2.6 What is the standard fixed point in thcnnomctery? Define it. 2.7 Why is a gas chosen as the standard thermometric subsiance? 2.8 What is an ideal gas? 2.9 What is the difference between the universal gas conslllllt and a characteristic gas constant? 2. IO What is a constant volume gas lhermomc:u:r? Why is it pn:fem:d to a (;()IIS!alrt pmlllwe gaa thermometer? 2.11 What do you understand by the ideal gas temperature scale? 2.12 How can the ideal gas temperature for 1hc steam point be measured\u00b7? 2.13 What is the Celsius temperature scale? 2.14 What is the advantage of a thcnnocouple in temperature measurement? 2.15 How does the resistance ihennometer measure tempemture? 2.16 What is the need of the international practical temperature scale? PROBLEMS 2.1 The limiting value of the ratio of the pressure of gas at the steam point and al the triple point of water when the gas is kepi at constant volume is found to be 1.36605. What is the ideal gas temperature of the steam point? 2.2 In a c-ODStant volume gas thermometer the following pairs of pressures read\u00b7 ings were taken at the boiling poinl of water and the boiling point of sulphur, respectively: Water b.p. Sulphur b.p. 50.0 96.4 200 387 300 S82 JOO 193 Iii I, 36=- Basic and.4ppli,d Tlinmodynamia The numbers are the gas pressures, mm Hg, each pair being taken with the same amount of gas in the thermometer, but the successive _pairs being taken with different amounts of gas in the tbennonM:tcr. Plot the ratio of Si,,p.: Hpb.p. against the reading at the water boiling point, and extrapolate the plot to zero pressure at the water boiling point.. This gives the ratio of Sb.p.: H20b.p. on a gas thennometer operating at zero gas pressure. i.e., an ideal gas thermometer. What is the boiling point of sulphur on the gas scale, from your plot? A.n.r. 445\u00b0C 23 The resistance of a platinum wire is found IO be 11,000 ohms at the ice point, 1 S.247 ohms at the steam point, and 28.887 ohms at the sulphur point. Find the c:oostanta A aoo JJ in the equation R = R<)(l +.4t + Bl) and plot R apinst I in the rang,: O to 660\u00b0C. 2.4 When the n:fm:nce junction of a thermocouple is kept at the ice point and the test junction is the Celsius tempen11we,. and e.m.f. \u00a3 of the 1hmnocouple is given by the equation e \u2022ar + br where a= O \u2022.:!O mV/deg, and b = - SO x 10--' mV/deg2 (a) Compute the e.m.f. when t = - 100\u00b0C, 200\u00b0C, 400\u00b0C, and 500\u00b0C, and draw graph of \u00a3 against, in this range. (b) Suppose the e.m.f. e is taken as a ihennomeiric property and thal a temperature scale 1\u2022 is dc:fined by tbc linear equation.,\u2022 =a e+ll and th", " A = cross-sectional area of lhe cylinder = : d, where D is the cylinder diameter and L = stroke of piston, or leng1h of cylinder. Let Nbe tbe revolutions per min.ute (r.p.m.) of the cr,mkshaft. In a two stroke cycle, the engine cycle is completed in two strokes of the piston or in. one revolution of the crankshaft. ln a four-stroke cycle, the engine cycle is completed in four strokes of the piston or two revolutions of the cranksha.ft. For a two-stroke engine, work done in one minute= Pm ALN, and for a four stroke engine. work done in one minute = p., ALN/2. The power developed inside the cylinder of the engine is called indicated po'K'er {IP), Pm.AL( Nor li)n 2 60 kW IP= (3.8) where Pm is in kPa and II is the number of cylinders in the engine. The power available at the crankshaft is always less than this value (IP) due to friction, etc. and is calleJ tbe broke power (BP} or shaft power (SP). If ct> is the angular velocity of the crankshaft in radian/sec, then BP= T t\u00bb (3.9) where Tis the torque transmitted to the crankshaft in mN. BP= 21CTN 60 where N is the number of revolutions per minute (11>m). The mechanical efficiency of the engine, l'lmecb\u2022 is defined as,,,_~ = w BP (3.10) (3.11) An engine is said to be double-aciing. if the working fluid is made to work on both s.ides of the piston. Such an engine theoretically develops twice the amount of work developed in a single-acting engine. Most reciprocating steam engines are double-acting, and so are many ma.tine diesel engines. Internal combustion engines for road transpon are always single-1\\ct.ing. 3.4 Other Types of Work Transfer There are forms of work other than pd V or displacement work. The following 11re the additional types of work transfer which may get involved in system surroundings interactions. (a) Electrical Work When a cwrent flows through a resistor (Fig. 3.13), taken as a system. there is work transfer into the system. This is because the cum:nt can drive a motor. the motor can drive a pulley and the pulley can raise a weight.!,,, I!'Work and Htal Transfer -=45 ~m y Fig. 3.13 Electrieal work The current now, J, in amperes, is given by / = dC d1' where C is the charge in coulombs and r is time in seconds. Thus dC is the charge crossing a boundary during time dr. If Eis the voltage potential. the work is dW =E\u00b7 dC =Eidt 2 W\"\" J \u00a3/dT I The electrical power will be W = lim dW =El dr.... o df This is the rate at which work is transferred. {3.12) (3.13) (b) Shaft Worlr. When a shaft, taken as the system (Fig. 3.14), is rotated by a motor, there is work transfer into the system. This is because the shaft can rotate a pulley which can raise a weight. lfT is the torque applied to the shaft and d9 is the angular displacement of the sha.ll. the shan work is and the shaft power is 2 W= J Td8 I (3.14) {3.15) where cu is the angular velocity and Tis considered a constant in this case. Sy,11,m boundary Motor T L_,. N Shaft Fig. 3.14 Sl,aft work (c) Paddle\u00b7Wheel Work or Stirring Work As the weight is lowered. and the paddle wheel turns (fig. 3.15), there is work transfer into the tluid system, I,,, 1,1 I,,, l\\1a1cri~ 46=- &uic 11,ul Appli,d 171trmodyn11mics which gets stirred. Since the volume of the system remains constant, J pd V = O. If m is the mass of the weight lowered through a distauce dz and Tis the torque transmitted by the shaft in rotating through an angle dB, the differential work transfer to the fluid is given by dW = mgdz = Td8 and the total work transfer is 2 W= J mgdz = j W'dz = j Td(J 2 2 (3.15) where W' is the weight lowered. I I", " wire 2 W= f pdV i W= f EdC= f Eld\u00b7r 2 I I 2 W= f Td9 I 2 W=- J C1d4 I'W=-1 fdL I (3.25) I II 1\\ I II Work and Htal Transfer -=49 Magnetised solid W=- J Hdl 2 I It may be noted in the above expressions thnt the work is equal to the integral of the product of an intensive propeny and the change in its related extensive property. These expressions are valid only for infinitesimally slow quasi-static proccssc.s. There are some other forms of work which can be identified in processes Chat are not quasi-static, for eKample, lhe work done by shearing forces in a process involving friction in a viscous fluid. 3.5 Free Expansion with Zero Work Transfer Work u:amfcr is identified only at the boundaries ofa system. It is a boundary phenomenon, and a form of energy in transit crossing the boundary. Let us consider a gas separated from the vacuum by a partition (Fig. 3.17). Let the partition be removed. The gas rushes to fill the entire volume. The expansion of a gas against vacuum is calledfree expansion. Ifwe neglect the work associated with the removal of partition, and consider the gas and vacuum together as our system (fig. 3. 17a), there is no work transfer involved here, since no work crosses the system boundary. and h.eoce 2 J <tW= 0, although J p dV.e 0 2 I I If only the gas is taken as the system (fig. 3.17), when the partition is removed there is a change in the volume of the gas, and one is tempted to 2. calculate lhe work from lhe expression f pd JI. However, this is not a qua.sisratic I process, although the initial and final end states arc in equilibrium. Therefore, the work cannot be calculated from this relation. The two end. states can be located on lhe p-1' diagram and tb.ese are joined by a dotted line (Fig. 3.17c) to indicate that the process had occurred. However, if the vacuum space is divided into a large number of small volumes by panirions and the partitions are removed one by one slowly (Fig. 3.17d), then every state passed through by the system is an equilibrium state and the work done can then be, estimated from the relation J pdf'(Fig. 3.17e), Yet, in free expansion ofa gas, 2 I there is no resistance to the Ouid at the system boundary as the volume of the gas increases to till up the vacuum space. Work is done by a system to overcome some resistance. Since vacuum does not offer any resistance, there is no work: transfer involved in l'ree expansion. \u2022 I,, Ill 1'\" 50=- &sit: and Applied Tfinmodynamics 1'''', 2.... _.. V (c) 1'''' ',,..... 2 (d) Fig. 3.17 Free o;pa,ssion - v (e) 3.6 Net Work Done by a System Olten different fonns of a work transfer occur simultaneously during a process executed by a system. When all these worlc interactions have been evaluated, the total or net work done by the system would be equal to the algebraic sum of these as given below w IDlal = w dl,ploccmtal + w.tar + w clc,;criul + w t1unJli +... 3. 7 Heat Transfer Heat is defined as the form of energy that is transferred across a boundary by virtue of a temperature difference. The temperature difference is the 'potential' or 'force' and heat transfer i.s the ''llux'. The iransfer of heat between two bodies in direct contact is called conduc tion. Heat may be transferred between two bodies separated by empty space or gases by the mechanism of radiation through electro.magnetic waves. A third method of heat transfer is convection which refers to the transfer of beat be tween a wait,and a fluid system in motion. The direction of heat transfer is taken from the high temperature system to the low temperature system. Heat flow into a system is taken to be positive, and heat flow out ofa system is taken as negative (Fig. 3.13). The symbol Q is used for heat transfer, i.e., the quantity of heat transfc1TCd within a certain time. Heat is a fonn of energy in transit (like work transfer). It is a boundary phenomenon, since it occurs only at the boundary of a system, Energy transfer by virtue of temperature difference only is called heat transfer. All other energy interactions may be termed as work transfer. I I! ii", " I + II 52=- &sic and Applied 171mnoa,namits Quasl-S!8tic t 2 2 - Az1la: Heet Ira~ Fig. 3.19 Rlprtsenl4tio,a of work lta,ufn- and luat transfn- in quaJi\u00b7Jtatic prottuu on p\u00b7u atld T\u00b71t \"'\"'ditllJUS Just like displacement work,!he heat transfer can also be written as tbe integral of the product of the intensive property T and the differential change of an extensive property, say X (Fig.3. l 9b). 2 Q1_2 = j It Q = j TdX 2 (3.26) I I lt must also be valid for a quasi-static process only, and the heat transfer involved is represented by the area under the path 1- 2 in T- X plot (Fig. 3. 19b). Heat transfer is, therefore, a path function, i.e., the amount of heat transferred when a system changes from a state I to a state 2 depends on the path the system follows (Fig. 3.19b). Therefore, IJQ is an inexact differential. Now, where Xis an extensive property and dX is an exact differential. dQ=TdX dX= l.1Q T (3.27) To make (f Q integrable, i.e., an exact differential, it must be multiplied by an integrating factor which is, in this case, I IT. The extensive property Xis yet to be defined. It has been introduced in Chapter 7 and it is called 'entropy'. 3.9 Specific Heat and Latent Heat The specific heat of a substance is defined as the amount of beat required to raise a unit mass of the substance through a unit rise in temperature. The symbol c will be used for specific heat. c = _g_ J/lcg K m\u00b7!lt I I +! It jp11 Work and I/tat Transfer -=53 where Q is the amowit of heat transfer (J), m, the mass of the substance (kg), and tt, the rise in temperature (K.). Since heat is not a property, as explained later, so the specific heat is qualified with the process through which exchange of heat is made. For gases, if the proces_-. is at constant pressure, it is cp, and if the process is at constant volume, it is c \u2022. For solids and liquids, however. the specific heat does not depend on the process. An elegant manner of defining specific heats, cv and cP' in temis of properties is given in Secs 4.5 and 4.6. The product of mass and specific heat (me) is called the heat capacity of the substance. The capital letter C, CP or C_., is used for heat capacity. The latent heat is the amount of.heat transfer required to cause a phase change in unit mass of a substance at\" c:,mstant pressure and temperature. There are three phases in which matter can exist: solid. liquid. and vapour or gas. The late11t heat of fusion (/fJ is the amount of heat transferred to melt unit mass of solid into liquid. or to freeze unit mass of liquid to solid. The latem heat of,) is the quantity of heat required to vaporize unit mass ofliquid vaporization (/,.\u2022 1 into vapour, or condense unit mass of vapour into liquid. The late11t heat of s11hlimation (/,uh) is the amount of heat transferred to convert unit mass of solid to vapour or vice versa. lru is not much affected by pressure, whereas lw,p is highly sensitive to pressure. 3.10 Points to Remember Regarding Heat Transfer and Work Transfer (a) Heat transfer and work transfer are the energy interactions. A closed system and its surroundings can interact in, two ways: by heat transfer and by work transfer. Thermodynamics st,1dies how these interactions bring about property changes in a system. (b) The same effect in a closed system can be brought about either by heat transfer or by work transfcL Whether heat transfer or work transfer has taken place depends on what constitutes the system. (c) Both heat transfer and work transfer are boundary phenomena. Both are observed at the boundaries of the system, and both represent energy crossing the boundarie.s of the system. (d) ft is wrong to say 'total heat' or 'heal content' of a closed system, because heat or work is not a property of the system. Heat, like work. cannot be stored by the system. Both heat and work are the energy in transit. ( e) Heat transfer is the energy interaction due to temperature difference only. All other energy interactions may he termed as work transfer.", " work transfer observed at the boundaries of a system, and (b) energy in storage, where energy is stored eithermacroscopical(y by virtue of motion, position or configuration of the system, or microscopically in the molecules or atoms constituting the system. 4.1 First Law for a Closed System Undergoing a Cycle The traru.ferofheat and the performance of work may both cause the same effect in a system. Heat and work are different forms of.the same entity, called energy, which is conserved. Energy which enters a system as heat may leave the system as work, or energy which enters the system as work may leave as beat. Let us consider a closed system which consists of a known mass of water contained in an adiabatic vessel having a thermometer and a paddle wheel, as shown in.Fig. 4.1. Let acerfai..o amount ofwork W1_2 be done upon the system by the paddle wheel. The quantity of work can be measured by the fall of weight which drives th.c paddle wheel through a.Pulley. The system was initially at temperature 11, the same as that of atmosphere, and after work transfer let the temperature rise to t2. The pressure is always I atrn. The process 1-2 undergone by lhe system is shown in Fig. 4.2 in generalized thermodynamic coordinatesX, Y. Let the insulation now be removed. Tbe system and the surroundings interact by heat transfer till the system returns to the original temperature t 1, attaining the condition of thennal equilibrium with the atmosphere. The amount ofheat transfer Q2_ 1 from the system during this process, 2-1, shown in Fig. 4.2, can be estimated. The system thus executes a cycle, which consists of a definite amount of work input W1_2 to the system followed by the transfer of an amount of heat Q2_1 from the system. It has been found that this W1_2 is always proportional to the beat Q2-1, and the constant of proportionality is called the Joule's equivalent I I +I I! I h I B41ic and Applied Tlttnnod:Jnamics Adiabatic _, Fig. <4.1.Adiabatic work or the mechanical equivalent of heat. In the simple example given here, there are only two energy transfer quantities as the system performs a thermodynamic cycle. If the. cycle involves many more beat and work quantities, the same result will be found. Expressed algebraically. (t W)<y,:lc ~ J (tQ).y<lc (4.1) where J is the Joule's equivalent. This is also expressed in the form where lhc symbol f denotes the cyclic integral for the closed path. This is the first law for a closed system undergoing a cycle. It is accepted as a general law of nature, since no violation of it has ever been demonstrated. In the S.I. system of units, bolh heat and work are measured in lhe derived unit of energy, the Joule. The constant of proportionality, J, is therefore unity (J= I Nm/J). The first law of thennodynamics owes much to J.P. Joule who, during the period 1840-1849, carried out a series of experiments to investigate the equivalence of work and heat. ln one of these experiments, Joule used an apparatus similar to the one shown in Fig. 4.1. Work was transferred to the r:~~-\u00b7 1 - x Fig. 4.2 Cycle compkted lry a systcm with two nie,gy illteractions: adiabatic work transfcr w,_z!ollowtd lry luat transfer (b.1 I I +j+,,, \"'First Law ofThm,codyMmics -=65 measured mass of water by means of a paddle wheel driven by the falling weight. The rise in the temperature of water was recorded. Joule also used mercury as the fluid system, and later a solid system of metal blocks which absorbed work by friction when rubbed against each other. Other experiments involved the supplying of work in an electric current. In every case, he fonnd the same ratio (.I) between the amount of work and the quantity of heat that would produce identical effects in the system. Prior to Joule, beat was considered to be an invisible fluid flowing from a body of higher calorie to a body of lower calorie, and this was known as the caloric theory of heat. It was Joule who first established that heat is a form of energy, and thus laid the foundation of the first law of thermodynamics.,.2 Fint Law for a Closed System Undergoing a Change of State The expression (l:W)cycle = (!Q)cyde applies only to systems undergoing cycles,", " and the algebraic summation of all energy transfer across system boundaries is zero. But if a system undergoes a change of state during which both heat transfer and work transfer are involved, the net energy transfer will be stored or accumulated within the system. If Q is the amount of heat transferred to the system and W is the amount of work transferred from the system during the process (Fig. 4.3), the net energy transfer (Q - W} will be stored in the system. Energy in storage is neither heat nor work, and is given the name internal energy or simply, the energy of the system. Therefore Q-W=M where Mis the increase in the energy of the system or (4.2) Here Q, W, and Mare all expressed in the same units ( in joules). Energy may be stored by a system in different modes, as explained in Article 4.4. Q=M+W If there are more energy transfer qnantitics involved in the process, as shown in Fig. 4.4, the first law gives (Q2 + Q) - Q 1) = l:J.E + (W2 + W3 - W1 - Jf'4) a ~e ---~w Surroundings O; ~ I 3 ~ ~-L\\ -;~l I System,. / \\.~ /.. Q 1 W1'W2 w. Surroundings Fig. 4.3 Heat atu:l work inlnact{ons of a S]Slnn witlt its J11rroundings in a process Fig. 4.4 S,itm\u00b7ntrroundi\"81 inttraenon in a proceu inoolDing many mngy jlru:u 11 h 4 I I Basie aNI.Applied Tllm,iot!yMmia Energy is thus conserved in the operation. The first law is a particular fonnulation of the principle of the conservation of energy. Equation (4.2) may also be considered as the definition of energy. This definition docs not give an absolute value of energy E, but only the change of energy AE for the process. It can, however, be shown that the energy has a definite value at every state of a syscem and is, therefore, a property of the system. 4.3 Energy-A Property of the System Consider a system which chang.es its state from state I to state 2 by following the path A, and returns from slate 2 to siate 1 by following the path B (Fig. 4.5). So the system undergoes a cycle. Writing the first law for pathA and for path B Q,. = 1:i.E,. + w,. QB=AEe+ Wa The processes A and B together constitute a cycle, for which or ~ (l:W)eycle = (l:Q)cy.lc w,. + W8 = Q11 + Q8 ~-~=~-~ (4.3) (4.4) (4.S) From equations (4.3}, (4.4), and (4.5), ii yields (4.6) Similarly, bad the system returned from state 2 to state I by following the path AEA = -AE9 C instead of path B AEA =-AEc (4.7) Prom equations (4.6) and (4.7) AEa =Mc (4.8) Therefore, it is seen that the change in energy between two states of a system is the same, whatever path the system may follow in undergoing that change of state. If some arbitrary value of energy is assigned to state 2, the value of energy at state I is fixed independent of the path the system follows. Therefore, eoergy has a definite value for every state of the system. Hence,.it is a point function and a property of the system. - v Fig. 4.5 Energy- a p,optrty of a system \"' I ol First Law of Tlurmotlywamia -=67 The energy Eis an extensive property. The specific energy, e = Elm (Jlkg). is an intensive property. The cyclic integral of any property is um, because the final state is identical with the initial state. f dE = 0, f dJI = 0, etc. So for a cyi:;le, the equation (4.2) reduces to equation (4.1). 4.4 Dill'erent Forms of Stored Energy The symbol E refers to lhe total energy stored in a system. Basically there are two modes in which energy may be stored in a system: (a) Macroscopicenergymode (b) Microscopic energy mode The macroscopic energy mode includes the macroscopic kinetic energy and potential energy of a system. let us consider a Ouid element of mass m having the ceotre", " o.fthe system. If the specific heat of a substance is defined in tenns of heat transfer, then Cv=(!~l Since Q is not a property, this definition does not imply that c v is a property of a subslance. Therefore, this is not the appropriate method of defining the specific heat, although (dQ)v = dll. Siru:e 11, T, and v are properties, c. is a property of the system. The product mcv = Cy is called the heat capacity at constant volume (J/K.). 4.6 Enthalpy The enthalpy of a substance, h, is defined as h=u+pr, It is an intensive property of a system (Id/kg). (4.24) Internal energy change is equal to the heat transferred in a constant volume process involving no work oihcr than pdY work. From equation (4.22), it is possible to derive an expression for the heat transfer in a constant pressure process involving no work other than pdV work. ln such a process in a closed stationary system of unit mass of a pure substance dQ=du+pdv At constant pressure pdv = d(pv) ( d Q)p.. du + d(ptl) (d\"Q)P = d(11 + pr,) (d\"Q)p = dh or where h = 11 + pv is the specific enthalpy, a property of the system. or (4.25) Heat transferred at constant pressure increases the enthalpy of a system. For an ideal gas, the enthalpy becomes h=u+RT ~2~ Since the internal energy of an ideal gas depends only on the temperarure (Eq. 4.1 I), the enthalpy of an ideal gas also depends on the temperature only, i.e. h = f(T) only (4.27) Tola) enthalpy H = mh Also and H= U+pJI h = Him (Jlkg) 4.7 Specific Heat at Constant Pressure The specific heat at constant pressure cP is defined as the rate of change of enthalpy with respect to temperature when the pressure is held constant \u2022 :,. :11\u00b7: II \u2022 C - ( ; } \" ) p ar p -=71 (4.28) Sin~ J,, T and p are propenies, so cP is a propeny of the system. Like '\\-, cp should not be defined in terms of heat transfer of consiant pressure, although (dQ)p=dh. For a constant prcsgwc process (~l,)p = f cP \u00b7dT 1j The first law for a closed stationaty system of unit mass Again d:Q=du+pdv ll=u+pt, dh =du+ pdV + vdp = dQ+ vdp dQ =dh-vdp.. or :. From equations (4.19) and (4.20) (dQ)p =dh (Q)p = (&lt)p r. (Q)p = ]cpdT 1j (4.29) (4.30) c,, is a property of lhe system, just like c.,. The heat capacity at constant pressure cp is equal to mcp (J/K). 4.8 Energy of an Isolated System An isolated system is one in which there is no interaction of the system with the SWTOWl.dings. For an isolated system, d Q = 0, (t W = 0. The first law gives or The energy of an isolated system is always constant. dE=O \u00a3=constant 4,9 Perpetual Motion Machine of the First Kind-PMM1 The first law states the general principle of the conservation of energy. Energy i., neither created nor destroyed. but only gets transformed from one form to another. There can be no machine which would continuously supply mechanical work without some other fonn of energy disappearing simultaneously (Fig. 4.8). Such a fictitious machine is called a perpetual motion machine of the first kind, or in brief, PMMI. A PMM I is thus impossible. \"I''II 72=- 811Jic and Applitd 77urmodynamia The wnverse oflhe above statement is also true, i.e., there can be no machine which would continuously consume work without some other form of energy appearmg ~mmhancously (Fig. 4.9). Q Q I Engine ] -,.. W E}- w Fig. U A PMMI Fig. 4.9 Tl1t conoerst of PMMI 4.10 Limitations of the First Law The first law deals with the amounts of energy of various forms transferred between the system and its surroundings and with changes in the energy stored in \u2022", " the system. It treats work and heat interactions as equivalent forms of energy in transit and does not indicate the possibility of a spontaneous process proceeding in a certain direction. It is the second law which assigns a quality to different fonns of energy, and also indicates lhe direction of any spontaneous process. Example 4,1 A stationary mass of gas is compressed without friction from an initial state of0.3 m3 and 0.105 MPa to a final state of0.15 m3 and 0.105 MPa, the pressure remaining constant during the process. Th.ere is a transfer of 3 7.6 kJ of beat from the gas during the process. How much does the internal energy of the gas change? Solution First law for a stationary system in a process gives Q=llU+ W or Here (l) r. W1-2 = jpdY= p(Yi - J-i) Yi =0.105 (0.15-0.30) MJ =- lS.75 kJ Q,_2 = -37.6 kJ :. Substituting in equation (I) -37.6 lcJ = U2 - U1 - IS.75 kJ U2 - U1 = - 21.85 kJ Ans. \u2022 I, I I I,. First law of Tlurmod:yruunics ~73 Toe internal energy of lhe gas decreases by 2 l.85 kJ in the process. Example,.2 When a system is taken from state a to state b, in Fig. Ex. 4.2, along path acb, 84 kJ of beat flow into the system, and the system does 32 kJ of w'ork. (a) How much will the heat that flows into the system.l!ong path a db be, if th:e work done is 10.5 k.J? (b) When the system is returned from b to a along the curved path, the work done on the system is 21 kJ. Does the system absorb or liberate heat, and how mucb of lhe heat is absorbed or liberated? (c) If u. = 0 and Ud = 42 kJ, find the heat absorbed in the processes ad and db. Solution Webave (a) (b) - - v Fig. Ex. U Q,cb = 84 kJ watb = 32 kJ Q..:0 = ub - u. + w acb Ub - U1 = 84 - 32 = S2 kJ Q,db = ub - u. + w lldb = S2 + 10.5 = 62.S kJ Q1,..a = U1 - Ub + W1,..a =-52-21 =-73 kJ Toe system liberates 73 kJ of heat. (c).Now W ldb = W..i+ wdl, = W..i= 10.S kJ Q.d = U4 - u. + W..i = 42 - 0 + JO.S = 52.S kJ Q.db = 62.5 kJ \"' Q..i + Qdb Qdl> = 62.S - 52.S = 10 kJ Ans. Ans. Ans. Ans. Example 4.3 A piston and cylinder machine contains a fluid system which passes through a complete cycle of four processes. During a cycle, the sum of all heat transfers is -170 kJ. The system completes I 00 cycles per min. Complete the I! I! I 74=- Basit and Applied 11imnodynamics following table showing I.he method for each item, and compute I.he net rate of work output in kW. Process a-b b--c c-d d-a Solution Process a-b: Q {kJ/mill) 0 21,000 -2,100 W (Ir.I/min) 2,170 0 i!E (k.llmin) -36,600 Process b-c: Process c-d: Process d-a: Q=il.E+ W 0=.0\u00a3+2170 llE = -2170.lcJ/min Q=il.E+ W 21,000 = AE+ 0 AE = 21,000 lcJ/min Q=AE+W -2100--36,600+ W W = 34,500 lcJ/min I,Q = - 170 kJ cycle The system completes l 00 cycles/min. Qab + Qt.: + Q.d + Qda = - 17,000 lcJ/min 0 + 21,000- 2,100 + Qd& = - 17,000 Qda = - 35,900 lcJ/min Now f dE = 0, since cyclic integral of any property is zero. tiEIH) + tiEk + AEo...i+ ti.E", ".25 kJ, W = - 56.25 kJ 4.7 The same liquid as in Problem 4.6 is stirred in a conducting chamber. During the process 1.7 kJ of beat arc transferred from the liquid to the surroundings, while the temperature of the liquid is rising by 15\u00b0C. Find /J..E and Wfor ihe process. A.ns. tJ.E = 56.25 kJ, W = S1.9S kJ 4.8 The properties of a certain fluid an: related as follows u = 196 +0.718 t pV z 0.287 (I+ 273) where u is the ~i>ecific internal energy (kJ/kg), t is in \u00b0C, p is pressure (kN/m2 and vis specific volume (m3/kg). For this fluid, find c. and cP ), Ans. 0.718, 1.005 kJ/kg K 4.9 A system composed of2 kg of the above.fluid expands in a frictionless piston and cylinder machine from an inil.ial state of I MPa, IOO\"C to a final temperature of 30\u00b0C. If then: is no heat transfer, ftnd lhe net work for the process. An.,. IOO.S2 kJ 4.10 If all lhe work in the expansion of Problem 4.9 is done on the moving piston, show that the equation representing the path of the expansion in the pv-plane is given by pvu = constant. 4.11 A stationary system consistingof2 kg ofthenuid of Problem 4.8 expands in ao 2 \"'constant. The initial conditions are I MPa \u00b7 adiabatic process according to pv 1 \u2022 and 200\u00b0C, and the final pressure is 0.1 MPa. Find W and tJ. U for the process. Why is the wort 11'8.Mfer not equal to J pd r? Ans. ff's 216.83, l!U O - 216.83 kJ, j pdV ; 434.4 kJ 4.12 A mixture of gases expands at constant pressure from I MPa, 0.03 m3 to 0.06 m1 with 84 kJ positive heat transfer. There is no work other than that done on a piston. Find.1. E for the gaseous mixture. Ans. S4 kJ The same mixture expands through the same siaie path while a stirring device does 21 kJ of work on the system. Find t.E, W, and Q for the process. Ans. 54 kJ. - 21 kJ, 33 kJ d I I II I I II Fust Law ofTAnmodynamics -=79 4.13 A mass of 8 kg gas expands within a flexible container so that the p-v relationship is of the fonn pvi.2 = const. The initial pressun: is I 000 kPa and the initial volume is I m3 \u2022 The final pressure is 5 kPa. If specific internal energy of the gas decreases by 40 kl/kg, find the heat transfer in magnitude and direction. Ans.+ 2615 lcJ 4.14 A gas of mass I.S kg undergoes a quasi-static expansion which follows n relationship, p = o + bV, where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 ml and 1.20 ml. The specific internal energy of the gas is given by the relation 11 ; 15 pv - 8HJ/lcg where p is in kPa and v is in m3/lcg. Calculate the net heat transfer and the maximum illtemal energy oflhe gas anaincd during expansion. Ans. 660 kJ, 503.3 kJ 4.15 The heat capacity at constan1 pressure of a unain system is a function of temperature only and may be expressed as C = 2.093 + 41\u00b787 J/K P I+ 100 where t is the temperature of the system in \u00b0C. The system is heated while it is mainillined at a pressure of l atmosphere until its volume increases from 2000 cm3 to 2400 cm1 and its tempcratun: increases from 0\u00b0C to 100\u00b0C. {a) Find the magnitude of the heat interaction. (b) How much docs the int.emal energy of the system increase? Ans. (a) 238.32J (b) 197.79 J 4.16 An imaginary engine receives heat and docs work on a slowly moving piston at such rotes that the cycle of operation of I kg of working fluid can be represented as a circle 10 cm in diameter on a p-v diagram on which I cm = 300 k.Pa and I cm = 0.1 m3", "/kg. (a) How much work is done by each kg of working lluid for each cycle of operation? (b) The thennal efficiency of an engine is def med as the ratio of work done and heat input in a cycle. If the heat rejected by the engine in a cycle is I 000 kJ per kg of working fluid, what would be its thermal efficiency'? Ans, (a) 2356.19 kJ/kg, (b) 0.702 4.17 A gas undergoes a thennodynamic cycle con~isting of three processes beginning at an i.niiial state where p, = l bar, V1 = I. 5 m3 and U1 = 512 kJ. The processes arc as follows: (i) Process 1-2: Compression withpV = constant top2 = 2 bar, U2 = 690 kJ (ii) Process 2--3: W2l \u00b7= 0. Q23 = - 150 kJ, and (iii) Process 3-1: W31 = + 50kJ. Neglecting KE and PE changes, dcten:ni11e the heat interactions Q12 and Q31 \u2022 Attr. 74 kJ. 22 kJ 4.18 A gas undergoes a thermoy1u:mic cycle cons.isting of the follO\\ving processes: (i) Process 1- 2: Constant pressure p = 1.4 bar. 111 = 0.028 m\\ W12 = 10.5 kJ. (ii) Process 2- 3: Compression with pV = constant.. U3 = U1, (iii) Process 3- 1: Constani volume, U1 - U3 = - 26.4 lcJ. There are no significani changes in KE and PE. (a) Sketch the cycle on ap-V diagram. (b) Calculate the net work for the 80~ Basic and Appl~d T1tnmodynamie1 4. 19 4.20 4.21 4.22 c,d\u2022 cycle in kJ. (c) Calculate the heat transfer for process 1-2 (d) Show thai IQ\u2022IW. Ans. (b) - 8.28 kJ, (c) 36.9 kJ -,.1< A certain gas of mass 4 kg is contained within a piston cylinder assembly. The gas undergoes a process for which pVI.> = constant. The initial state is given by 3 bar, 0.1 nl. The change in internal energy of the gas in the process is 112 - u 1 = - 4.6 kJ/kg. Find the net heat transfer for the process when. the final volume is 0.2 m 3\u2022 Neglect the changes in KE and PE. Ans. - 0.8 lLI An electric gener,llur coupled to a windmill produces an aver.ige electrical power output of5 kW. The power is used to charge a storage battery. Heat.transfer from the battery to the surroundings occurs at a constant rate of 0.6 kW. Determine the total amount of energy stored i.n the battery in 81, of operation. Ans. 1.27 x lOs kJ A gas in a piston-cylinder assembly undergoes two processes in series. From state I to state 2 then: is energy transfer by heat to the gas of 500 kJ. and the gas does work on the piston amounting 800 kJ. The second process. from state 2 to state 3. is a i:onstant pressure compression at 400 kPa, during which there is a heat iransfer from the gas amounting 450 kJ. The following data are also known: U1 - 2000 kJ and U3 ~ 3500 kJ. Neglecting changes in KE and PE, calculate the change in volume oft.he gas during process 2- 3. Ans, - 5.625 nl Air is contained in a rigid well-insulated rank with a volume of 0.2 m3 \u2022 The tank is fitted with a paddle wheel which transfers energy to the air at a constant rate of 4 W for20 min. The initial density of the air is 1.2 kg/m3. If no changes in KE or PE occur, determine (a) the specific volume at ihe final state. (b) the change in specific internal energy of the air. A,u. (a) 0.833 m3Jkg. (b) 20 ltJ/kg fll I First Law Applied to Flow Processes 5.1 Control Volume For any system and in any process, the first law can be written as Q=tt.E+ W where\u00a3 represents all fonns of energy stored in the system. For a pure. substance \u00a3=EK+ Ep+ U where EK is the K.E., Ep the P.E., and U the residual energy stored", " N/m2 v 1, v2-specific volwne, m3/kg u1, ur-specific internal energy, J/kg V 1, V2- velocity, mis 2 1, Z2-elevation above an arbitrary datum, m d'Q -net rate of heat transfer th.rough the control surface, J/s df dW. d,r\" -net rate ofworlt t:ran.sfer through the control surface, J/s exclusive of work done at sections I and 2 in transferring the Ouid through the control surface. t-time,s. Subscripts I and 2 refer to the inlet and exit sections. I I!I!! I 84=- Basic an4 Applied Thmt!odyaamics 5.3.1 Mass Balanu By the conservation of mass, if there is no accumulation of mass within the control volume., the mass flow rate entering must equal the mass flow rate leaving, or or WI \"'W2 A1V1 = A2V2 VI V2 This e<juation is known as the equation of,\u00b7ontinuity. 5.3.2 Eftetgy Balance (5.2) In a flow process, the wor.k transfer may be of two types: the external work and the flow work. The external work refers to all the work transfer across the control surface other than that due to normal fluid forces. In engineering thennodynanucs the only kinds of external work of importance are shear ( shaft or stirring),wor.k and electrical work. In Fig. 5.2 the only extema.l work occurs in the form of shai\\ work, Wx. Theilow work, as discussed in Sec. 3.4, is the displacement work done by the fluid of mass dm.1 at the inlet section 1 and that of mass dm2 at the exit section 2, which are (-p1 v1dm1) and (+p2v2dm2) respectively. Therefore, the total work transfer is given by W = w. - P1t11dm1 + P2'U2dm2 (5.3) ln the rate form, or \u00b7...., (5.4) Since there is no accumulation of energy, by the conservation of energy, the tolal rate of flow of all energy streams entering the control volume must equal the total rate of flow of all energy streams leaving the control volume. This may be \u00b7 expressed in the following equation. w,e, + ltQ = w2e2 + d.W df df Substituting for \";' from Eq. (S.4) w1e1 + - ltQ df = w2e2 +.::..:..:..L - WtP11:11 + w2P2V 2 ltW: df where e1 ande2 refer to I.he energy camed into or ou.t of the control volume with unit mass of fluid. (5.S) I I! ii I FirJt Law Applied to Flow ProceJJtJ -.:::=85 The specific energy e is given by e=ek+e-P+u y2 = 2 +Zg+u Substituting the expression fore in Eq. (5.5) or where And,since Dividing Eq. (5.7) by!7 ) dW, =w2 ~ +T+Zig + dr\" Vf ( It= u+ pv.. let W = W1 = W2 = - dm df (5.6) (5.7) (5.8) Equations (5.7) and (5.8) are known as steady flow energy equations (S.F.E.E. ), for a single stream of fluid entering and a single stream of fluid leaving the control volume. All lhe tenns in Eq. (5.8) represent energy flow per unit mass of fluid (11kg) whereas all the terms in.Eq. (5.7) represent energy flow per unit time {11kg). The basis of energy flow per wiit mass is usually more convenien.t when only a single stream of fluid enters and leaves a control volwne. When more than one fluid stream is involved the basis of energy flow per unit time is more convenient Equation (5.8) can be written in the following form, Q-W, =(h2 -h 1)+ w-\u00b7w 2 + g(.l:?-Z1) (5.9) where Q and W1 refer to energy transfer per unit mass. In the differential fonn, the SFEE becomes 1 I +\u2022 nl h I II 86=- rtQ- crw. = dh + vdv +", " yT1, i.e., the temperature inside the vessel becomes independent of time and is equal to yT1 throughout the charging process. (ii) The charging process will stop when pressure inside the vesel reaches that of the main. The charging time can be found by setting p = p 1 in the pressure relation moRr1i moR,.., P,-Po\"'------e gT1 aY aV By rearrangement, r = -.!. hi [\u2022 - (Pt - Po )-a_V_] moRr1i a REvIEw gUESTIONS S.l Explain the system approach and the control volume approach in the analysis of a flow process. 5.2 What is a steady flow process? What is steady state? S.3 Write the steady flow energy equation for- a single stream entering and a single stream leaving a control volume and explain the various terms in it. S.4 Give the differential form ofihe S.F.E.E. S.S Under what conditions docs the S.F.E.E. reduce to Euler's equation? S.6 How does Bernoulli's equation compare with S.F.E.E.? S. 7 What will be the velociiy of a Ouid leaving a nozzle, lfthe velocity of approach is very small? 5.8 Show that the enthalpy of a fluid before throttling is equal 10 that a\u00b7fter throttling. S.9 Write the general energy equation for a variable now process. 5.10 What is the system technique in a bottle-filling process? 5.11 Explain the control volume technique in a variable now process. I I II\u2022 111 1 \u00b7 I! I \"-1 First IAw Applied u, Flow Ptoeust1 -=107 PllOBLEMS 5.1 A blower handles I kgfs of air at 20\u00b0C and consumes a power of IS kW. The inlet and outlet velocities of air arc I 00 mis and I SO mis respectively. Find the exit air temperature, assuming adiabatic conditions. Take cp of air as 1.005 kJ/kg-K. Ans. 28.38\u00b0C 5.2 A turbine operates under steady flow conditions, receiving steam at the following state: pressure 1.2 MPa, temperature I 88\u00b0C, enthalpy 2785 kJ/kg, velocity 33.3 mis and elevation 3 m. The steam leaves the turbine at the following state: pressure 20 kPa, enthalpy 2512 kl/kg, velocity I 00 mis, and elevation Om. Heat is lost to the surroundings at the rate of0.29 kJ/s. If lhe rate of steam flow through the turbine is 0.42 kg/s, what is the power output of the tuJbine in kW? Ans. 112.51 kW 5.3 A nozzle is a device for increasing the velocity of a steadily flowing stream. At the inlet to a certain nozzle, the enthalpy of the fluid passing is 3000 kJ/kg and the velocity is 60 mis. At the discharge end, the enthalpy is 2762 kJ/kg. The noule is horizontal and there is negligible heat loss from it. (a) Find the velocity at exit from the nozzle. (b) If the inlet area is 0.1 m2 and the specific volume at inlet is 0.187 m3/kg, find the mass flow rate. (c} If the specific volume at the nozzle exit is 0.498 rihkg, find the exit area of the nozzle. Ans. (a) 692.S mis, (b) 32.08 kg/s (c) 0.023 m2 5.4 In an oil cooler, oil flows steadily through a bundle of metal tubes submerged in a steady steam of colling water. Under steady flow conditions, the oil enters at 90\u00b0C and leaves at 30\u00b0C, while the water enters at 25\u00b0C and leaves at 70\u00b0C. The enthalpy of oil at t\u00b0C is given by h = I.68, + 1 o.s x ao\u00b7\u2022 1 kl/kg What is the cooling water flow required for cooling 2. 78 kg/s of oil? Ans. 1.473 kg/s S.S A thermoelectric generator consists of a series of semiconductor elements (Fig. P. S.S), heated on one side and cooled on the other. Electric cum:nt flow is produced as a result of energy transfer as heat In a particular experiment the Fig. P.5.5 1 I, r I,1,,. l>.la,Cria 108=- /JaJic a,u/ Applitd Tlurmotlynamies current was measured to be 0.5 amp and the electrostatic potenti3l at", " {I) was 0.8 volt above tbat at (2). Energy transfer as heat to the hot side of the generator was taking place at a r.itc of S.S wans. Determine the rate ol'energy transfer as heal from the cold side and the energy conversion efficiency. At!$. Q1 \u2022 S. l watts, 11 - 0.073 S.6 A turbocompressor deliver., 2.33 m3/s of air at 0.276 MPa. 43\u00b0C which is heated at th.is pressure 10 430\u00b0C and finally eitpanded in a turbine whicb delivers 1860 kW. During the expansion, there is a beat transfer of 0.09 MJls to the surroundings. Calculate the turbine exhaust temperature if changes in kinetic and potential energy are negligible. Take cp a 1.005 kjlkgK S.8 A11s. IS7\u00b0C S.7 A reciprocating air compressor takes in 2 m1/min at 0.11 MPa, 20\u00b0C which it delivers at 1.5 M.Pa. I. I 1\u00b0C to an aflercooler whc.ro the air is cooled at constant pressure to 2S\u00b0C. The power absorbed by the compressor is 4. l5 kW. Determine the heat transfer in (a) the compressor, and (b) the cooler. State your assumptions. Ans. - 0.17 kJ/s, - 3.76 ld/s. l.n a water cooling tower air enters at a height of I m above the ground level and leaves at a height of 7 m. The inlet and outlet velocities are 20 mis and 30 mis respectively. Water entei:s al a height of8 m and leaves at a height ofO.S m. The velocity of water al entry and ex.it are 3 mis and I mis n:spcctively. Water temperatures arc 80\u00b0C and 50\u00b0C at the entry and exit respectively. Air temperatures are 30\u00b0C and 70\u00b0C at ihe entry and exit respectively. Tiie colling tower is well insulated and a fan of 2.25 kW drives the air through the cooler. Find the amount of air per second required for I kg/s of water flow. The values of er of air and water are 1.005 and 4.1871kg K respectively.,411.r. 3.16 kg 5. 9 Air at 101.325 kPa, 20\u00b0C is iaken into a gas iurblne power plant al a velocity of 140 mis through an opening of 0.1 S m2 cross-sectional area. The air is compressed heated. expanded through a rurbine, and ex.baustcd at 0.18 MPa, 150-C through an opening of 0.10 mi cross-sectional area. T11e power output is 375 kW. Calculate the net amount of heat added to the air in kJ/lcg. Assume that air obeys the law pt\u2022 ~ 0.287 (t + 273) whe1e p is the pressure in kPa v is the spccilk volume in m3/kg, and, is the temperature in \u2022c. Take er= l.005 kJ/kg K..4ns. 150.23 kj/kg S.10 A gas llows steadily through a rotary compressor. The gas enters the compressor at a temperature of I 69 C, a pressure of I 00 kPa. and an enthalpy of 39 l.2 kJ1kg. The gas leaves the compressor at a temperature of2459 C, a pressure of 0.6 MPa. and an entha.lpy of 534.5 kJ/Jcg. There is tJO heat transfer to or from the gas as it flows through the compressor. (a) Evaluate the eKtemal work done per unit mass of gas assuming the gas velocities at entry and exit to be negligible. (b) Evaluate the cxtemal work done per uo.it mass of gas when the gas velocity at entry is 80 m/ s and that at ex.it is 160 mis. A11s. 143.3 kJ/kg, 152.9 kJ/kg S.11 The steam supply to an engine comprises two streams which mix before entering the engine. One stream is supplied at the rate of 0.01 kg/s with an enthalpy of 2952 kJ/kg and a velocity of 20 mis. The other stream is supplied at ihe rate of 0.1 kg/s with an enthalpy of2569 kJ/kg and a velocity of 120 mis. At the c:x.it \"I' Finl l..llw Applid to Flow Proussts -=109 from the engine the nuid leaves as two streams, one of", " the temperature difference, there will be heat leakage Q 1 from the body to the surroundings. The body will be maintained at the co.nstant temperature t 1, if heat is discharged into the body at the same rate at which heat leaks out of the body. The heat is extracted from lhe low temperature reservoir, which is nothing but the atmosphere, and discharged into the high temperature body B, with the expenditure of work W ln a cyclic device called a heat pump. The working fluid operat.cs in a cycle tlowing through the evaporator \u00a3 1\u2022 compressor C1, condenser C2 and expander \u00a3 2, similar to a refrigerator, but the attention is here focussed on the high temperature body B. Here QI and W arc of primary interest, and the COP is defined as COP=ja_ w [COP] = Q. H.P. Q. -Q_z (6.7) From equations (6.6) and (6.7), it is found that t6.8) The COP of a heat pump is greater than the COP of a refrigerator by unity. [COPJ11.P. = (COP],..,+ I Equation (6.8) expresses a very interesting feature of a heat pump. Q1 = [COPJH.P. W Since =(COPre,+ l)W (6.9) Q1 is always greater than W. For an electtical resis1ance heater, if Wis the electrical energy consumption, then the heat transferred to the space at steady state is W only, i.e., Q1 = W. I I!! ii I Stcond Law of Tlurmodynamitt -=119 A I kW eleclric heater can give I kW ofheatat steadysta.teand nothing more. In other words, I kW of work (l1igh grade energy) dissipates to give l kW of heat (low grade energy), which is thennodynamically inefficient. However, if this electrical e.oergy Wis used to drive the compressor of a heat pump, the heat supplied Q1 will always be more than W, or Q1 > W. Thus, a heat pump provides a thermodynamic advantage over direct heating. For heat to flow from a cooler to a hotter body, W cannot be zero, and hence, the COP (both for refrigerator and heal pomp) canot be infinity. Therefore, W>O, and COP <oo. 6.7 Equivalence of Kelvin-Planck and Clausius Statements At first sight, Kelvin-Planck's and Clausius\u00b7 statements may appear to be unconnected, but it can easily be shown that they are virtually two parallel statements oJ tbe second law and are equivalent in all respects. The equivalence of the two statements will be proved if it can be shown that the violation. of one statement implies the violation of the second, and vice versa. (a) Let us first consider a cyclic heat pump P which transfers heat from a low temperature reservoir (f:i) to a high temperature reservoir (i1) with no other effect, i.e., with no expenditure of work, violating Clausius statement (Fig. 6.9). HotR-*att1 W=O\u00b7\u00b7 01 ~{\"II._/ p 01 I \u00b7~ I Cold Reservoir at t2 Flg. 6.9 Violation ef tlu Clausil/.S staltmtnl Let us assume a cyclic beat engine.E oper;iting between the same thcnnal energy reservoirs, producing W0ct in one cycle. The rate of working of the heat engine is such that it draws an amount of heat Q1 fror\u2022 the hot reservoir equal to that discharged by the heat pamp. Then the hot reservoir may be eliminated and the heat Q1 discharged by the heat pump is fed t.o the heat engine. So we see that the heat pump P and the heat engine E acting together constitute a heat engine :,, : 120=- Basit and Applitd 11ttmodJ'IIIJmia operating in cycles and producing net work while exchanging heat only with one body at a single fixed temperature. This violates the Kelvin-Planck statement (b) Let us now consider a perpetual motion machine of the second kind(\u00a3) which produces net work in a cycle by exchanging heat with only one thermal energy reservoir (att1) and thus violates the Kelvin-.Planck statement (Fig. 6. JO). Let us assume a cyclic heat pump (P) extracting heat Q2 from a low temperature reservoir at t2 and discharging heat to the high temperature reservoir at t 1 with the expenditure of work W equal to what the PMM2 delivers in a complete cycle. So E and P together constitute a heat pump working in cycles and producing the sole effect of transferring heat from a lower to a higher temperature body, thus violating ihe Clausius statement", ". Fig. 6.10 Violation of the Ktlvin-Plantk stalemrnl 6.8 Reversibility and Irreversibility The second law of thennodynamics enables us to divide all processes into two classes: (a) Reversible or ideal process. (b) Jm:versible or natural process. A reversible process is one which is perfonned in such a way that at the conclusion of the process. both the system and the surroundings may be restored to their initial states, without producing any changes in the rest ofthc universe. Let the state of a system be represented by A (Fig. 6.1 1), and let the system be taken to stateB by following the pathA-8. lfthe system and also the surroundings are restored to their initial states and no change in the universe is produced. then the process A.-B will be a reversible process. In the reverse process, the system has to be taken from state B to A by fol.lowing the same path.8- A. A reversible process sl1ould not leave any trace or relic to show that the process had ever occurred. \"I' Suond Law of 17urmodynamil;s -=121 f (a) - - J t (b) Fig. 6.11 Rromihlt fn'<>ass A reversible process is carried out infinitely slowly with an infinitesimal gradient, so that every state passed through by the system is an equilibrium state. So a reversible proce.ss coincides with a quasi.static process. Any natural process carried out with a finite gradient is an irreversibl.e process. A reversible process, which consists of a succession of equilibrium states, is an idealized hypothetical process, approached only as a limit. It is said lo be an asymptote to reality. All spontaneous processes are irreversible. 6.9 Causes of Irreversibility Broken eggs, spilt milk, bumt boats, the wasted ye.ars of indolence that the locusts have eaten are merely proverbial metaphors for irreresibility. The irreversibility of a process may be due to either one or both of the following: ( a) Lack of equilibrium during the process. (b) Involvement of dissipative effects. 6.9.1 J\"eversilJility due to Lack of Equilibrium. The lack of equilibrium (mechanical, thermal or chemical) between the system and its surroundirtgs, or between two systems, or iwo parts of the same system, causes a spontaneous change which is irreversible. The followin_g are specific examples in this regartl: (a) Heat Transfer th.rough a Finite Temperature Difference A heat transfer process approaches reversibility as the temperature difference between two bodies approaches zero. We define a reversible heat transfer _process as one in which beat is transferred through an infinitesimal temperature difference. So to transfer a finite amount of heat through an infinitesimal temperatnre difference would require an infinite amount of time, or infinite area. All actual heat transfer processes are through a finite temperature difference and are, therefore, irreversible, and the greater the temperature difference, the greater is the irreversibility. We can demonstrate by the second law that the heat transfer through a finite temperature difference is irreversible. Let us assume that a source alt A and a si.nk 1,1 Ii I II 122=- BaJic and Applfrd Tltennodynamics att8 (tA > to) arc available, and let QA-B be the amount of heat flowing from A to B (Fig. 6.12). Let us assume an engine operating between.A and B, taking heat Q1 from A and discharging heat Q2 to B. Let the heat transfer process be reversed, and Qa-A be the heat flowing fromB to A, and let the rate of working of the engine be such that Q2 = QB-A (Fig. 6.13). Then the sink B may be eliminated. The net result is that E. produces net work Win a cycle by exchanging heat only with A, thus violating the Kelvin-Planck statement. So the heat transfer process QA-B is irreversible, and Q11_,... is not possible. Source A. t\"' Source A, IA o, Sink 8, t8 Sink 8, fs Fig. 6.12 Htal trans.for l/lro11glt a finite Fig. 6.13 te111peraturt diffmnu lleaJ tran.rfer through a finite t#llptralllrt diffn-, u irrroersihk (b) La.ck of Pressure Equilibrium within the Interior of the System or betwee.n the System and the SurroW1dlu~ When there eitists a difference in pressure between the system and the surroundings, or within the system", " itself, then both th.e system and its surroundings or the system alone, will undergo a change of state which will cease only when mechanical equilibrium i.s eatablished. The revese of this process is not possible spontaneously without producing any other effect. That the reverse process will violate the se.cond law becomes obvious from the following illustration. (c) Free Expansion Let us consider an insulated container (Fig. 6.14) which is divided. into two compartments A and B by a thin. diaphragm. Compartment A contains a mass of gas, while compartment B is completely evacuated. If the diaphragm is punctured, the gas in A wm expand into B until the pressures in A and B become equal. This is known as free or unrestrained ex_pansion. We can demonstrate by the second law, that the process of free expansion is irreversible. To prove this, let us assume that free expansioo. is reversible, and that the gas in B returns into A wiih an increase in pressure, and B becomes evacuated a.,; I \u2022 I'i l l! ii'Second I.Aw ofTl,mnodynamia -=123 before (Fig. 6.15). There is no other effect. Let us install an engine (a machine, not a cyclic beat engine) between A and B, and permit the gas to expand through the engine from A to B. The engine develops a work output Wat the expense of the internal energy of the gas. The internal energy of the gas (system) in B can be restored to its initial value by heat transfer Q (= W) from a source. Now, by the use of the reversed free expansion, the system can be restored to the initial state of high pressure in A and vacuum inB. The net result is a cycle, in which we observe that net work output W is accomplished by exchanging heat with a single reservoir. This violates the Kelvin-Planck statement. Hence, free expansion is irreversible. The same argument will hold if the compartment B is not in vacuum but at a pressure lower than!hat in compartment A (case b). r w Fig. 6.14 Frtt txparuion Fig. 6.15 Suond fuw demonstrates that frtt txpa,uilm is irmmibu 6.9.2 J,m,ersi/Jility dill! to Dissipative Effects The irreversibility of a process may be due to the dissipative effects in which work is done without prodncing an equivalent increase in the kinetic or potential energy of any system. The transformation of work into molecular internal energy either of the system or of the reservoir takes place through the agency of such phenomena as friction, viscosity, ioelasticity, electrical resistance, and magnetic hysteresis. These effects are known as dissipative effects, and work is said to be dissipated. Dissipation of energy means the transition of ordered macroscopic motion into chaotic molecular motion, the reverse of which is not possible without violating second Jaw. (a) Friction Friction is always present in moving devices. Friction may be reduced by suitable lubrication, but it can never be completely eliminated. If this were possible, a movable device could be kept in continual motion without violating either of the two laws of thermodynamics. The continual motion of a movable device in the complete absenc,e of friction is known as perpetual motion of tire third kind. That friction ma.kes a process irreversible can be demonstrated by the second Jaw. Let us consider a system consisting of a flywheel and a brake block (Fig. 6.16). The flywheel was rotating with a certain rpm, and it was brought to 4,111 Matcria 124=- Basie and Af,plitd Thmnodynamics rest by applying the friction brake. The distance moved by the brake block is ve.ry small, so work transfer is very nearly e.qual to zero. Ifllle braking process occurs very rapidly, there is little heat transfer. Using suffix 2 after braking and suffix l before. braking. and applying the first law, we have ~fJ-~, : 'i ---\u00b7\u00b7---\u00b7-----J '- SY9tem boundary Fig. 6.16 /mvnsibility dut to dissipativt tjfut lik1 friction Q,-2 = E2 - E, + W1-2 O= \u00a32 -\u00a31 +O. E2 =E1 (6.10) The energy of the system (isolated) remains constant. Since the energy may exist in the forms of kinetic. poten1ial. and molecular internal energy, we ha ve mV2 U2 + - -2 2 mV:2 - + mZzg\"\" U1 + -. - 1 2 - + mZ1g Since lhc wheel is brought to rest, V 2 = 0, end.", "!here is no change in P.E. mV'.2 V2 =V1+ - -1 2 (6.11) Therefore, the molecular internal energy of the system (i.e., of the brake and the wheel) increases by the absorption of the K.E. of the wheel. The reverse process. i.e., the conversion of chis increase in molecular internal energy into.K..E. within the system to cause the wheel to rotate is not possible. To prove it by the second law, let us assume ihai it is possible, and imagine the following cycle with three processes: Process A: Initially, the wheel and the brake are at high temperature as a result.of the absorption of the K.c. of the wheel, and the flywheel is at rest. Let the flywheel now start rotating al a particular rpm at the expense of the internal energy of the wheel and brake, the temperature of which will then decrease. Process 8: Let the flywllecl be brought to rest by using its K.E. in rnising weights, with no change in ternpcrdtuTe. Process C: Now let beat be supplied from a source to the flywheel aml the brnke, to restore the system to its inilial state. Therefore, the processes A, B. and C together constitute a cycle producing work by 1:xchanging heat with a single reservoir. This violates tbe Kelvin-Planck statement, and it will become a PMM2. So the braking process. i.e., the transfonnation ofK.E. m10 molecular internal energy. 1s irre1,ersible. I l l, I II Second Law of 17itrmodynamia -=125 (b) Paddle-Wheel Work Transfer Work may be transferred into a system in an insulated container by means of a paddle wheel (Fig. 6.17) which is also known as stirring work. Here work transferred is dissipated adiabatically into an increase in the molecular internal energy of the system. To prove the irreversibility of the process, let us assume that the same amount of work is deli vercd by lhe system at the expense of its molecular internal energy, and the temperature of the system goes down (Fig. 6.18). The system is brought back to its initial state by heat transfer from a source. These two processes together constitute a cycle in which there is work output and the system exchanges heat with a single reservoir. It becomes a PMM2, and hence the dissipation of stirring work to internal energy is irrevers ible. Fig. 6.17 Adiabatic work traruftt Fig. 6.18 l11romibility dut to di.uipation of st11ing work into intttnal tnergy (c) Transfer of Electricity through a Resistor The flow of electric current through a wire represents work transfer, because the curre\u00b7nt can drive a motor which can raise a weight. Taking the wire or the resistor as the system (Fig. 6.19) \u00b7 and writing the first law Q,.2\"' U2 - u, + w,.2 Here both W1_2 and Q1\u2022 2 are negative. W1-2 = U2 - u, + Q, - 2 (6.12) A part of the work transfer is stored as an increase in the internal energy of the wire (to give an increase in its temperature), and the remainder leaves the system as heat. At steady state, the internal energy and hence the temperature of the resistor become constant with respect to time and (6.13) ~ I ~\"\"':\"\"'.\"\" - --,-'vvvvvvvv~ Resistor (sysLem) I w,\" o Fig. 6.19 lmnmibility due to di.uipation oftltctrical work into i11ttrnal tntrgy 1, t I r 11 H 126~ &uic and Applied T1rmnodyll4miCJ The reverse process, i.e., the conversion of heat Q,_2 into electrical work W1_2 of the same magnitude is not possible. Let us assume that this is possible. Then heat Q1_2 will be absorbed and equal work W1-2 will be delivered. But this will become a PMM2. So the dissipation of electrical work into internal energy or heat is irreversible. 6.10 Conditions for Reversibility A natural process is irreversible because the conditions for mechanical, thermal and chemical equilibrium are not satisfied, and the dissipative effects, in which work is transformed into an increase in internal energy, are present. For a process to be reversible, it must not possess these features. If a process is performed quasi-statically, the system passes through states ofthennodynamic equilibrium, which may be traversed as well in one direction as in the opposite direction. ff there are no dissipative effects, all", " the work done by the system during the performa11ce of a process in one directio11 can be returned to the system d11ring the reverse process. A process will be reversible when it is performed in such a way that the system is at all times infinitesimally 11ear a state of thermodynamic equilibrium and in the absence of dissipative effect of any form. Reversible processes are, therefore, purely ideal, limiting cases of actual processes. 6.11 Carnot Cycle A reversible cycle is an ideal hypothetical cycle in which all the processes constituting the cycle are reversible. Carnot cycle is a reversible cycle. For a stationary system, as in a piston and cylinder machine, the cycle consists of the following four succc-ssive processes (Fig. 6.20): ~ AdlabaUc cover (BJ ~ -~ / i---- Sink, 12 \u00b7- System'- Adlabellc Fig. 6.20 Comol luat mginNtaliona,y ry1ttm I I +j+.., 11,11 Sewnd law o/T1urmodyn.amia -=127 (a) A reversible isothermal process in which heat Q1 enters the system at t 1 reversibly from a constant temperature source at t 1 when the cylinder cover is in contact with the diathennic cover A. The int.emal energy oftbe system increases. From First law, (6.14) (for an ideal gas only, U1 \"' Ui) (b )A reversible adiabatic process in which the diathermic cover A is replaced by the adiabatic cover B, and work W8 is done by the system adiabatically and reversibly at the expense of its internal energy, and the temperature of the system decreases from t 1 to t2\u2022 Using the first law, (6.1 S) (c) A reversible isothermal process in which Bis replaced by A and heat Q2 leaves the system at r2 to a constant temperature sink at 12 reversibly, and the internal energy of the system further decreases. From the first law, -Qi\"' U4 - U3 - WJ--4 (6.16) only for an ideal gas, U) = U4 (d) A reversible adiabatic process in which B again replaces A, and work WP is done upon che system reversibly and adiabatically, and Che internal energy of the system increases and the temperature rises from t2 to t 1\u2022 Applying the first law, (6.17) Two reversible isotherms and two reversible adiabatics constitute a Carnot cycle, which is represented in p-v coordinates in Fig. 6.21. Summing up Eqs (6.14) to (6.J7), or Q1 - Q2 = (W,_2 + W2_3) - < wl-4 + w 4--1) r Qnet= r wl>t'I ()'de C)'(lc A cyclic heat engine operating on the Carnot cycle is called a Carnot heat engine. Q. t,r Rev. isolherm (t1) Fig. 6.. 21 Carnot CJClt Iii h I+ 128=- Basic and Applittf 1Jtmruxiynamics For a steady flow system, the Carnot cycle is represented as shown in Fig. 6.22. Here heat 0 1 is transferred to the system reversibly and isothennally at t 1 in the heat exchanger A, work WT is done by the system reversibly and adiabatically in the turbine (8), then heal 02 is transferred from the system reversibly and isothermally at t2 in the heat exchanger (C), and then work Wp is done upon the system reversibly and adiabatically by the pump (D). To satisfy the condiiions for the Carnot cycle, there must not be any friction or heat transfer in the pipelines through which the working fluid flows. (1)! 'j\"..., - - -1-0-,-----' Sourot!, t1 ;, -1- ; -Flow l t 11 11 - ; /S~ i-\"\" boundary Heat exchanger (A) Pump(O) i i! \u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7-\u00b7\u00b7\u00b7\u00b7,\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7-----to, \u00b7\u00b7\u00b7\u00b7--\u00b7--- \u00b7----\u00b7'1 (3 ) 1-Flow Heat &xd'langer (C) lz, G) j sink,t2 6.12 Reversed Heat Engin.e Since all the processes of the Carnot cycle are rt:ven;ible, it is possible to imagine that the processes are individually reversed and carried out in reverse order. When a reversible process is reversed., all tht: energy transfers associated wid1 the process are reversed in direction. but", "/s what is the least power necessary to pump this heat out continuously? So/11tion Freezer temperature, Ambient air temperature, T2 =-15 + 273.. 258 K T1 = 30 + 273 = 303 K The refrigeraior cycle removes heat from the freezer at the same rate at which heat leaks into it (Fig. Ex. 6.2). I Ambient air T1 = 303 K L __ ~ __ ___, w Freezer r2 = 258 K 0 2 = 1.75 kJ/s Fig. Ez. 6.2 For minimum power requirement ~=~ 7i 1j 1 I +\u2022 nl h! II UO=- Basie and Applitd T1uro1ody11amies QI = ~;; X 303 = 2.06 kJ/s W=Q,-Q2 = 2.06 - I. 75 = 0.31 kJ/s =0.31 kW Example 6.3 A reversible heat engine operates between two.reservoirs at temperatures of 600\u00b0C and 40\u00b0C. The engine drives a revers.ible refrigerator which operates between resetvoirs at temperarures of 40\u00b0C and - 20\u00b0C. The heat transfer to the heat engine is 2000 kJ and the net work output of the combined engine refrigerator plant is 360 kJ. (a) Evaluate the heat transfer to tbe refrigerimt and the net heat transfer to the reservoir at 40\u00b0C. {b) Reconsider (a) given that the efficien.cy of the heat engine and the COP of the refrigerator are each 40% of their maximum possible values. Solution by (a) Maximum efficiency of the heat engine cycle (Fig. Ex. 6.3) is given t W=360kJ Fig. Ex. 6.3,. 'IIIIIX = 1 - Ti = 1 - fil = 1 - 0.3S8 = 0.642 1j 87) Again Jt; Qi =0.642 H'1 = 0.642 X 2000 = 1284 kJ Maximum COP oftbe refrigerator cycle (COP)m:u = _!}___ = Tz - 7j 253 313- 253 Also Since COP=~ =4.22 ff; W1 \u2022\u2022 W2 = W = 360 kJ... 4.22 I II 1\\ I II --= 141 W2 = W1 - W = I 284 - 360 = 924 kJ Q4 = 4.22 X 924 = 3899 kJ Q3 = Q4 + W2 = 924 + 3899 = 4823 kJ Qi= Q, - w, = 2000-1284 = 716 kJ Heat rejection to the 40\u00b0C reservoir (b) Efficiency of the actual heat engine cycle = Q2 + Q) = 716 + 4823 = 5539 Ju Ans. (a) ''1 = 0.4 l'.lmJJJt = 0.4 X 0.642 W1 = 0.4 X 0.642 X 2000 = 513.6 kJ W2 = 513.6-360 = 153.6 kJ COP of the actual refrigerator cycle COP= ~ = 0.4 x 4.22 = 1.69 Hi Therefore Q4 = 153.6 x l.69 = 259.6 kJ Q:, = 259.6 + 153.6 = 413.2 kJ Q2 = Q, - w, = 2000-513.6 = 1486.4 kJ Ans. (b) Heat rejected to the 40\u00b0C reservoir = Q2 + Q3 = 413.2 + 1486.4 = 1899.6 kJ Ari.,. (b) Example 6.4 Which is the more efTec:tive way to increase the efficiency of a Carnot engine: to increase T1, keeping T2 constant; or to decrease T2, keeping T1 constant? Solution The efficiency of a Carnot engine is given by JJ=l-.fi 1j If 12 is constant ( :~ t? = ;~ As T1 increases, l'.I increases, and the slope( :i, ) decreases (Fig. Ex. 6.4.1 ). I Tz If T1 is constant, ((Jr/) =-..l As T2 decreases, ri increases, but the slope ( :ii t 07i T1 7i (Fig. Ex. 6.4.2). remains constant 1 \u2022\u00b7 h I 1 1111' \u2022 ;\u00b0li.1a1crit1 142=- Basic and Applied 17urmodyNJ.mics 1.0 \u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7,\u00b7- - -\u00b7- \u00b7- \u00b7 \u00b7 1.0 I:.\" t -r, Fig. Ex. 6.U 1~ - r2 Fig. Ex. 6.4.2 o1j Tz ( ~11", " a reversible cycle. I - d\"2 ~ [1 - d(h J dQ dQ n:v I I,, ill I I II 160=- &uie and Applied TTurmodynamia or or Since or c!Q S tQi, for any process AB, n:vcrsible or ineversible. T 7; For a reversible process Hence, for any process AB Then for any cycle f (!is fds (7.7) (7.8) Since entropy is a property lllld the cyclic integral of any property is zero Th.is equation ii known as Che intqUality of Clausius. It provides the criterion (7.9) of th'1 reversibility of o cycle. If f 4f = 0, the cycle is reversible, f 4f < 0, Che cycle is irreversible and possible \u2022 d I'\" Entropy f 'i > 0, the ~le is impossible, since it violales the second law. 7.7 Entropy Change In an Irreversible Procesa For any process undergone by a system, we have from Eq. (7.8) tQ < ds T - or (7.10) This is further clarified if we consider the cycles as shown in Fig. 7. t I, 2 A~/7 /.' C 1 \\~ Flg. 7.11 EntrOf!! drangt in an imvmihk process s where A and B are re\\lenibte processes and C is an irreversible process. For the rnemble cycle consisting of A and B or 2 dQ I dQ J-=-J2 T 1 T B A (7.11) For the irreversible cycle consisting of A and C, by the inequality of Clausius, f dQ = j~+ J~ <O I T 2 T T A From Eqs (7.11) and (7.12), C (7.12) I I,, ill I II I I II 162=- Basic and Applied Thmrwdynamics I _ J dQ + J dQ <tQ I dQ [r>[r Since the path B is reversible, B C (7.13} (7.14) I I / dQ = / dS 2 T B 2 B Since entropy is a property,entropy changes for the paths Band C would be the same. Therefore, I I /dS.,,/dS (7.15) From Eqs (7.13) to (7.15), 2 B 2 C I I f dS> f ltQ 2 T C Thus, for any irreversible process, 1 C whereas for a reversible process dS> ltQ T dS.. dQ.,,,. T Therefore, for the general case, we can write or (7.16) The equality sign holds good for a reversible process and lhe inequality sign for an im:versible proce&S. Entropy -=163 7.8 Entropy Principle For any infut.itesimal process undergone by a system, we have from Eq. ( 7. I 0) for the total mass For an isolated system which does not undergo any energy interaction with the surroundings, 4 Q = 0. Therefore, for an isolated system For a reversible process, or dS,... =O S= constant For an irreversible process dS;so > 0 (7.17) It is thus proved that the entropy of an isolated system can never decrease. It always increases and remains constant only when the process is reversible. This is known as the principle of increase of entropy, or simply the entropy principle. It is the quantitative general statement of second law from the macroscopic viewpoint. An isolated system can always be formed by including any system and its surroundings within a single boundary (Fig. 7.12). Sometimes ihe origiual system which is then only a part of the isolated system is called a'subsyste,n'. - System ~, - ~ ~ ~.... 0 - Svrroundings isolated (oomposite) system Flg. 7.12 isolated :rystnn The system and the surroundings together (the univcn;e or the isolated system) include everything which is affected by the process. For all possible processes that a system in the given surroundings can undergo or d.~univ <!: 0 dS,ys + dS,urr <!: 0 (7.18) Eutropy may decrease locally at some region within the isolated system, but it must be compensated by a greater increase of entropy somewhere within the system so that the net effect of an irreversible process is an entropy increase of I I +! It!111 the whole system. The entropy increase of an isolated system is a measure of the extent of irreversibility of the process undergone by the system. Rudolf Clausius summarized the first and second laws of thermodynamics in t", " the heat capacity of lhe liquid. (7.24) which is positive. 7.10 Entropy T.ransfer Mechanisms Entropy can. be transferred to or from a system in two fonns: heat transfer and mass flow. In contrast, energy is transferred by work also. Entropy transfer is recognised at the system boundary as entropy crosses the boundary, and it represents the entropy gained or lost by a system during a process. The only form of entropy interaction associaled with a fixed mass or closed system is heat transfer, and thns the entropy transfer for an adiabatic closed system is zero. It is being explained below in more details: (a) Heat Transfer Since dS = dQ,..,, when heat is added lo a system d Q is T positive, and the entropy of the system increases. When beat is removed from the system, dQ is negative, and the entropy of the system decreases. Heat ttansferred to the system of fix.ed mass increases the internal energy of the system, as a result of which the molecules (of a gas) move with higher kine1ic energy and collide more frequently, and so the disorder in the system increases. Heat is thus regarded as disorganised or disordered energy transfer which increases molecular chaos (see Sec. 7.16). If beat Q flows reversibly from the system to the surroundings at T0 (Fig. 7.21), lhc entropy increase of the sur:ronodings is 6S =JL To sun The entropy of the system is reduced by &S \u2022Y\u2022..,_.Q_ To The temperatnre of the boundary where heat transfer oc.curs is the constant temperature T0\u2022 lt may be said that the system has Jost t-ntropy to the surroundings. AJtcmatively. one may state lhat the surroundings have gained 1 I 1\u2022 :11: i1, 172=- Ba.Jic an.ii.Applitd 17termodynamiCJ entropy from the system. Therefore, there is entropy transfer from the system to the su1Toundings along with heat flow. In other words, since the beat inflow increases the molecular disorder, there is flow of disorder along with heat. The sign of entropy transfer is the same as the sign of heat transfer: positive, if into the system, and negati1\u2022e, if out of the system. Fig. 7.21 Entropy trarisfer along witlt. luatjlow On the other hand, there is no entropy transfer associated with work. In Fig. 7.22, the system delivers work to a flywheel, where energy is stored in a fully recoverable form. The flywheel molecules are simply put into rotation around the axis in a perfectly organised manner, and there is no dissipation and hence no entropy increase of the flywheel. The same can be said about work transfer in the compression of a spring or in the raising of a weight by a certain height. There is thus no entropy transfer along with work. Ir work is dissipated adiabatically into internal energy increase of the system (Subsection 7.9.5), there is an entropy increase in the system, but there is as such no entropy transfer to it. Work is thus entropy-free, and no entropy is transferred with work. Energy is transfen:ed with both heat and work, whereas entropy is transferred only with heat. The first law of thermodynamics makes no distinction between heat transfer 11J1d work. It considers them as equals. The distinction between heat transfer and work is brought about by the second law: an energy interaction which is accompanied by entropy transfer is heat transfer, and an energy interaction which is not accompanied by entropy transfer is work. Thus, only energy is exchanged during work interaction, whereas both ene:-gy and entropy are exchanged duri,rg heat transfer. (b) Mus Flow Mass contains entropy as well as energy, and the entropy and energy of a system are proportional to the mass. When the mass of a system is doubled, so are the entropy and energy of the system. Both entropy and energy are carried into or out of a system by streams nf matter, and the rates of entropy and energy transport into or out of a system are proportional to the mass flow rate. Closed systems do not involve any mass flow and thus any entropy transport. When an amount of mass m enters or leaves a system, an entropy of amount ms. s being the specific entropy, accompanies it. Therefore, the entropy of a system increases by ms when the mass of amount m enters it, and decreases by the same amount when it leaves it at the same state. I I 'I+ d I \u2022 1-.lalcria EntTofJY -=173 F l~ Fig. 7.22 No ttllropy transfer awng with work transfer 7.11 Entropy Generation in a Closed System The entropy of", ",~. account, respectively, for rotes of entropy transfer into and out of the CV accompanying mass flow. The rate of entropy increase of the control volume cxceds, or is equal 10, the net I Ill I I II 178=.- Basic and Applied Tlt.nmodynamics rate of c.ntropy transfer into it. The differenc.e is the entropy generated within the control volume due to ineversibility. Hence, the rate of entropy generation is given by s,. as - I, m\u00b7s\u00b7 + I, m s - Q 11'11 df T \u2022 C. I I i (7.35) By the second Jaw, sgtn ~ o lf the process is reyersible, S800 = 0. For an irreversible process, s,.\u2022 > 0. The magnitude of S gen q11:3nti fies th~ irreversibility of the process. If systems \u2022 0) 8 it can be ~aid that the system A A and B operate so that (Sgc.JA > (S8 operates more irreversibly than system B. The unit of Sg\u2022\u2022 is WIK. At steady state, the continuity equation gives \" the energy equation becomes o= Q- w.,, + ~m;(1r+!:+gz)_ - I,m.(1a + V 2 2 e I I 2 + gz) e and the entropy equation reduces to o = Q + I,m,s, - I,mc-1., + sscn T i \u2022 (7.36) (7.37) (7.38) These equations often must be solved simultaneously, together with appropriate propeny relations. Mass and energy are conserved quantities, but entropy is not generally conserved. The rate al which entropy is transferred out must exc.eed the rate at which entropy enters the CV, the difference being the rate of entropy generated within the CV owing to irreversibilities. For one-inlet and one-exit control volumes, the entropy equation becomes O= % + m(s1 -s2) + s,.., (7.39) 7.13 First and Second Laws Combined By the second law 4Qn:v= TdS and by the first law, for a closed non-flow system, dQ=dU+ pdV \"I''II Again, the enthalpy Entropy TdS=dU+pdV H=U+pV dH=dU+pdV+ Vdp = TdS+ Vdp TdS = dH - Vdp -=179 (740) (7.41) Equations (7.40) and {7.41) are the thermodynamic equations relating the properties of the system. Let us now examine the following equations as obtained from lhe first and second laws: (a) (t Q = dE + d: W- This equation holds good for any process, reversible or irreversible, and for any system. (b) (t Q = dU + pd\u00b7 W- This equation holds good for any process undergone by a closed stationary system. (c) dQ = dU + pdV- This equation holds good for a closed system when only pdV-work is present. This is true only for a reversible (quasi-static) process. (d) ct Q = TdS-This equation is true only for a reversible process. (e) TdS = dU + pdV- This equation holds good for any process reversible or irreversible, undergone by a closed system, since ii is a relation among properties which arc independent of the path. (f) TdS = dH - Vdp-This equation also relates only the propenies of a system. There is no path function tem1 in the equation. Hence the equation holds good for any process. The use of the term 'irreversible process' is doubtful, since no irreversible path or process can be plotted on thennodynamic coordinates. It is more logical to state tbat 'the change of state is irreversible, rather than say 'it is an irreversible process'. A natural process whic.h is inherently irreversible is indicated by a dotted line connecting the initial and!iDal states, both of which arc in equilibrium. The dotted line has no other meaning, since it can be drawn in any way. To determine lhe entropy change for a real process, a known reversible path is made to connect the two end states, and integration is performed on this path using either equation (e) or equation (f), as gi.ven above. Therefore, the entropy change of a system between two identifiable equilibrium states is the same whether the intervening process is reversible or the change of state is irreversible. 7.14 Reversible Adiabatic Work in a Steady Flow System ln. the differential form, the", " gas is due to the coordinated motion of all tbe molecules with the same average velocity in the same direction. The potential energy is due ro the vantage position taken by the molecules or displacements of molecules from their normal positions. Heat or thermal energy is due to the random thermal moti.on of molecules in a completely disorderly.fashion and tbe average velocity is ze.ro. Orderly energy cao be readily converted into dis.orderly energy. e.g., mechanical and electrical energies into internal energy (and then heat) by friction and Jou.le effect. Orderly energy can also be convened into one another. But there are nmural limitatio11s on the conversi<m of disorderly energy imo orderly energy. as delineated by the sec(lt1d law. When work is dissipated into inte.mal energy, the disorderly motion of molecules is increased. Two gases, when mixed, represent a higher degree of disorder than when they are separated. An irreversible process always tends to take the system (isolated) to a state of greater disorder. It is a tendency oo the part of nature to proceed to a state of greater disorder. An isolated system always tends to a state of greater entropy. So there is a close link between entropy and disorder. It may be stated roughly ;,I''II EntropJ -=183 that the entropy of a system is a measure of the degree of molecular dis{)rder exi.fting in the system. When heat is imparted to a system, the disorderly motion of molecules increases, and so the entropy of the system increases. The reverse occurs when heat is removed from the system. Ludwig Boltzmann (1877) introduced statistical concepts to define disorder by attaching to each state a thermodynamic probability, expressed by the quantity rv; which is greater the more disordered the state is. The increase of entropy.implies tha.1 the system proceeds by itself from one state to another with a higher them1odynamic probability (or disorder number). An irreversible process goes on until the most probable state (equHibrium state when Wis maximum) corresponding to the maximum value of entropy is reached. Boltzmann assumed a functional relation between S and W. While entropy is additive, probability is multiplicative. If the two parts A and B of a system in equilibrium are considered, the entropy is the sum and the thennodyna.mic probability is the product S=S,. + S8 Again, w~ w,..w8 S = S (W), S,. = S( Jf'.,.), and S8 = S(W8) S(W) ~ S(Jf',.W8) = S(W,.) + S(W8 ) which is a well-known functional equaton for the logarithm. Thus the famous relation is reached S=Kln W (7.48) where K is a constant, known as Boltzmann constant. 111is is engraved upon Boltzmann's tombstone in Vienna. When W = l, which represents the greatest order, S = 0. This occurs only at T = OK. This state cannot be reached in a finite number of operations. This is the Nemst-Stmon statement of third law of thermodynamics. In the case of a gas, IV increases due to an increase in volume V or temperature T. In the reversible adiabatic expansion of a gas the increase in disorder due to an increase in volume is just compensated by the decrease in disorder due to a decrease in temperature, so that fhe disorder number or entropy remains constant. 7.17 Absolute Entropy It is important to note that one is interested only in the amount by which the entropy of the system changes in going from an initial to a final slate, and not in the value of absolute entropy. In cases where it is necessary, a zero value of entropy of the system at an arbitrarily chosen standard state is assigned, and the entropy changes arc calculated with reference to this standard state. 7.18 Entropy and Informatio.n Theory The starting point of information theory is the concept of uncertainty. Lei us define an event as an occurrence which can result in one of the many possible \u2022 :,. :11\u00b7, :1. Basic and Applitd 11tnmodyr,amics outcomes. The outcome of the event is known only after it has occurred, and before its occurrence we do not know which one of the several possible outcomes will actually result. We are thus uncertain with regard to the outcome before the occurrence of the event. After the event has occurred, we are no longer uncertain about it. lf we know or can assign a probability to each one of the outcomes, then we will have some information as 10 which one of the outcomes is most likely to occur. For example, let us consider the throwing of a dice and try to guess the result. Each event of turning l, 2, 3, 4, 5", " and 6 bas a probability of 1/6. If it is told that lhe result is odd, lhe probability of a correct guess is 1/3. If it is told further that the number is not a 3, the probability of the correct guess becomes 1/2. rt is ihus seen that the smaller the probability, greater is the uncertainty. The amount of information conveyed by a message increases as the amount of uncertainty regarding the message becomes greater. The more it is known about the message a source will produce, the less the uncertainty, and less the in(ormation conveyed. The entropy of communication theory is a measJJre of this uncertainty conveyed by a message from u source. As stated earlier, in information theory a value of uncertainty is associated with each outcome of an event. Let us denote the uncertainty about an outcome whose probability is p. The knowledge of Ibis uncertainty depends on certain characteristics given below: 1. The uncertainty u aboo.t an event A with possible outcomes N1, N2, \u2022\u2022\u2022 depends upon the probabilities p 1, p2, \u2022\u2022\u2022 of these outcomes, or: 11 = f(p,, Pi,... ) If p, is the probability of the i\u2022th outcome: \"= /(pi) 2. The wicertainty \u00b7u is a mo)\\otonic.fwlction of the probability of the outcome p and it decreases with increasing probability, or: du/dp < 0 3. When the probability p = I, the uncenainty 11 = 0. 4. The uncertainty about two independent events A and B taken together as one should be the sum of lhe uncertainties about A and B iaken separately, \"(A, B) =\"A+ \"e In other words, uncertainty is an additive property. In the case of probabilities, however, p(A., B) =PA\u00b7 Pa The fun.ctional relationship between u and p can be derived. as in Eq. (7.48), (7.49) \"' Entropy -=185 where K is a constant. Since pis always less than 1, u is always positive. S. If the event has a very large number of outcomes, an average value of uncertainty for the outcomes of the event is significant. If p 1, p 2, p 3, \u2022.\u2022 be the probabilities of the outcomes of an event with uncertainties u1, u1, 113,..., then the average value of the uncertainty for the event as a whole is given by the expectation value of tlie uncertainty <u>, which is written as: <11> = L P; 11, i Since p/s are probabilities, I.A= 1 i Again,. since u; = - K In Pi, therefore, <si> = I, P; 11; = - K I, P; In Pi ; i (7.50) The measure of uncertainty can be used to de.fine the amount of infonnation contained in a message. The infonnation in a message has the effect of reducing or eliminating uncertainties. We, therefore, define the information I in a mtts.:sage as the decrease in uncertainty as a result of m::eiving the me&liage, or (7.51) where 111 and u2 arc the uncertainties before a.nd after receiving the message respectively about the outcome with which the message is concerned. Using Eq. (7.49), /=11,-112 (7.52) where p 1 and p2 are the probabi.lities before and after receiving the message respectively. l ( the message removes the uncertainty completely giving complete information about the outcome, u2 = 0 and I= u1\u2022 lo genera.I, information is equal to the uncertainty. When we are dealing with many outcomes with their associated uncertainties, we can define the average or expected value of the infonnation, <I>= <u>\"' -KI,P. lnp; i (7.53) The eKpected value ofinfonnation is also called entropy in information cheory and is desiguated by th.c symbol S, so that S =-KI,p; lnp; I Th.is is known as Shannon \u00b7.r forniula. (7.54) Iii It 186~ &uit a11d Applitd '11u:rmotf]ttamics 7.18.1 Statistical Formalism Let Y1, V2, \u2022\u2022\u2022 Jlc are the values of the various outcomes, and p 1, p 2, \u2022\u2022\u2022, Pc arc the probabilities associated. Let V is known as the average value of all the outcomes, which is the same as the expected value. The equations of constraint are:!pi= I 'I.p;V, = <V-> = V (7.55) If p;'s are given, then V can be estimated from the above equation.", " But if V is given and p'; s are required, then we have c unknowns, viz., p 1, p2, \u2022\u2022\u2022, p~, and only two equations. If c > 2, the problem is indetenninate. If one set of p'; s is chosen arbitrarily, some outcomes are certainly overemphasized. The problem is then how to avoid bias in selecting a certain set of p', s. Tbe answer according 10 E.T. Jaynes is to assign that set of values of p'; s which is consistent with the given infonnation and which maximizes the uncertainty. This is the principle of minimum prejudice, enunciated by Jaynes in the following words: \"The le\"st prejudiced or most unbiased estimate of probabilities is rhut assignment which maximizes the entropy S, subject to the given i11formation \", The entropy S represents the uncertainty of an evenl, which is to be maximized: S \"\" - K!. p, In p, (7.54) subject to the two constraints of Eq. {7.55). Lagrange's method of undeter\u00b7 mined multipliet'! (see Chap. 21) will be used for the solution. Differentiating Eqs {7.54) and (7.55) Idp, =:0 Ip; dpi =O (7.56) (7.57) dS = - K [I. lnp, dp, + L:lp,) = 0 I. In P; dp; = 0 or, (7.58) where V,s are held constant and dS = 0 for S to be maximum. Multiplying Eq. (7.56) by,land Eq. (7.57) by {J and adding to Eq. (7.58). I(ln P; + A + PY;] dp1 = 0 where,land fJ are the Lagrange's multipliers. Since dp;'s are non-zero, (7.59) or In p, + A + /W; = 0 P; = e-i. e\u00b7flV, The Eq. (7.60) represents a set of c equations: P1 = e\u00b7>. e-llV1 P2=e->.e-l1V2 Pc = e\u00b7). e\u00b7P v. The above equations are the desired unbiased set of p'; s. (7.60) (7.61) iii I EntrfJPJ -=187 Again. C.~, LP; = I,e-.le-/ill', = I _. ~ = :,Ee-llv, I or.t = In I,e-pv, From Eq. (7.60}, e-pv, Pi= I.e-/JV, Thus \u2022.t has been estimated. Using Eq. (7.SS) LY;e-Pv, _ I.Pi f'; = L -fJV = V e ;'(7.62) (7.63) (7.64) Since P is known, {j can be detennined. The entropy can be expressed in temts of l and P as: S = - K 1: P; [-.t - /jV,) = KA.l:pi + K fJE p; Y; S = K.t + KP <Y> The above procedure is referred to as the Jaynes\u00b7 formalism. (7.65} 7.18.2 Information Theory Applud to a System of Particles Let us consider a system having a large number of particles. According to the quantum theory (see Chap. 19). the energy that the system can have is discretely distributed. The system cannot have any energy, bui only certain values of energy levels. Let us denote by P; the _pcobability or the energy level E:;. A high probability signifies that the corresponding energy level is more frequently attained by the system, i.e,, the system can be found for longer durations or time in that energy level. The problem is to determine the most probable state or the system subje.ct to the constraints imposed by the nature of the system. Usually, the average energy of the system, -which is also the expectation energy <E> is known by physical measurements. Therefore, and <E> =!p;E; = E l:p, = I In stati.stical thermodynamics, one proceeds to determ.ine the numbe.r of microstates corresponding to the most probable rnacrostaie of the system (see Chap. 19). corresponding to the thermodynamic equilibrium state when W is maximized to yield S = K In W. The entropy signifies the uncertainty inherent in lhe system., Ill I, 188=-", " the reservoir would decrease so that the net entropy change of me water (system) and the resevoir together would be positive definite. Water is being heated irreversibly, and to find the entropy change of water, we have lo assume a reversible path between the end states which are at equilibrium. ~ servoir 373K,J / ) Q t-\u00b7) ~ / Fig. Ex. 7.2 (.6 ('\\ = f 4Q = J mcdT :: lffC In 1i. \"\"\"\u2022\"\" T. I T T T1 =Ix 4.187 In 121 = l.'.305 kJ/K 273'I,,,!I I! I Errlropy -=193 The reservoir temperature remains constant irrespective of the amount of heat withdrawn from it. Amount of heat absorbed by the system from the reservoir, Q = J X 4.187 X (373 - 273) = 418.7 kJ :. Entropy change of the reservoir (~S) = _ _Q. = _ 418.7 = -1.122 kJ/K 373 \"\" T :. Entropy change of the universe (M)univ = (ds),,,.,_ + (6S)M = 1.305 - 1.122 = 0. 183 kJ/K (b) Wamr is being heated in two stages, first from 273 K to 323 K by bringing it in contact wilh a reservoir at 323 K. a.nd lhen from 323 K to 373 K by bringing it in contact of a second reservoir at 373 K.. nu. dT 3'3lC dT (M),..11n = J me-+ J mc323K T Ans.{a) 21) lC T =4.187 (1n ;;; +In;;;) =(0.1673 +0.1441)4.187 = 1.305 kJ/K (M} = _ 1 X 4.187x (323 - 273) = -0.647 kJ/K 323 = _ IX 4.187 X (373 - 323) = _ O.S6 kJ/K 373 real (Mk. 11 (M)w,, V = (~S)\"\"ter + (4S)re, I + (6S)~ 11 = 1.305 - 0.647 - 0.56 = 0.098 kJ/K Aris.(b) (c) The entropy change of the universe would be Jess and less if the water is heated in more and more stages, by bringing the water in contact successively with more and more heat reservoirs, each succeeding reservoir being at a higher temperature than the preceding one. When water is heated in in.finite steps, by bringing it in contact with an infinite number of reservoirs in succession, so that at any instant the tempera ture difference between the water and the reservoir in contact is infinitesimally small, then the net entropy change of the universe would be zero, and the water would be reversibly heated. Example 7.3 One kg ofice. at - 5\u00b0C is ellposed to the atmosphere which is at 20\u00b0C. The ice melts and comes into thermal equilibrium with the atmosphere. (a) Determine the entropy increase of the universe. (b) What is lhe minimum amount of work necessary to convert the water back into ice at - 5\u00b0C? cP of ice is 2.093 kJ/kg Kand the latent heat of fusion of ice is 333.3 kJ/kg.,, lol I'\" 194=-.Basic and Af>plitd Tltmnodynam.ia Solutio11 Heat absorbed by ice Q from the atmosphere (Fig. Ex. 7.3.1) Fig. F.z. 7.3.1 = Heat absorbed in solid phase + Latent heat + Heat absomed in liquid phase = 1 X 2.093 X (0 -- (--5)) + 1 X 333.3 + 1 X 4.187 X (20 -- 0) = 421.S kJ Entropy change of the atmospher. (.:\\S) = _ Q =.... 427.5 = -1.46 kJ/1( ann T 293 Entropy cbS11ge of the system (ice) as it gets heated from -S\u00b0C to 0\u00b0C (ASJsyt<em = J mcp - dT T = l x 2.093 In - = 2.093 x 0.0186. 2B 268 2 \" 268 = 0.0389 kJ/K En11opy change of the system as ice mel15 at 0\u00b0C to become water at 0\u00b0C (6S11),y,tem = 333.3 273 = 1.22 kJ/ K Entropy change of water as it gets heated from 0\u00b0C to 20\u00b0C (M111>-", " summarize the first and second laws of thermody\u00b7 n.am.ics? 7.13 Show that the transfer of heat through a finiie. tentpera1ure diffe.rence. is irreversible. 7.l4 Show that the a diabatlc mixing of two fluids is irreversible. 7.lS What is the maxmium work obtainable from two finite bodies at temperatures T1 and T2? 7.16 Determine the maximum work obiaioable by using one finite body at temperature T and a thermal energy reservoir at temperature T0, T > T0 \u2022 7.17 What arc the causes of entropy incrcllSC? 7.18 Why is an iscntropic process not necessarily an adiabatic process? 7.19 What is the reversible adiabatic work for a steady flo111 system when K.E. and P.E. changes arc negligibly srnaJI? How is it difTeren.t from that for a closed. stationary system? 7.20 Under what conditions is the wor.k done equal to (a) J p dv, (b) - J v dp? 7.21 Why are ihe equations TdS:dU+pdY.. TdS.~ dH - Vdp valid for any process between two equilibrium cod states? 7.22 Why is the second law called a directional law of nature? 7.23 How is entropy related io molecular disorder in a system? 7.24 Show ihat entropy varies logarithmically with the disorder number. 7.25 What do you undeniand by perfect order? 7.26 Give the Nemst-Simon statement of the third law of thermodynamics. 7.27 Why does entropy n:main constant in a reversible adiabatic process? 7.28 What do you understand by the postulator:y approach of thermodynamics? 7.29 What do you understand by 'lost work'? Entropy ~205 7.30 The amount of entropy generation quantifies the intrinsic irrcversiblity of a process. Explain. 7.31 Show that s,,,. is not a lhennodynamic propeny. 7.32 Give the expression for the eniropy gcncmion rate for a cootrol volume of a steady now system. 7.33 What is the entropy generation in the isothermal dissipation of work? 7.34 What is the entropy generation in the adiabatic dissipation of work? 7.35 What do you understand by entropy transfer? Why is entropy transfer associated with heat transfer and not with work lransfer? 7.36 What is the relation between probability and uncertainty of an ev.ent'! How is entropy defined in communication theory? 7.37 State the.five characteristics on which the uncenainty of 11D event depends. What is the expecution value of uncenainty? 7.38 Define infonnation and explain its relation with entropy. What is Shannon's formula? 7.39 What is bias? State and explain the principle of minimum prejudice. 7.40 Explain the procedure of Jaynes' formalism to prove: S = KA+ X.fj<V> 7.41 Explain how information theory is applied to a system of particles. What is panition function? 7.42 Explain the relation of infonnation theory and classical thermodynamics. 7.43 How do the heat transfer and the lost work affect changes in p, and hence the cniropy of a system? 7.44 Since infonnarion theocy considers heat and W\"Ork ai: derived quantities, show that for a reversible process: (a) O\" W. = - I.p. dt; = p dJI (b) d: Q + [d: w, - It W).. I.E; dp, {c) O\" Q, = TdS 7.45 Explain how entropy is a fwidlLID.l!rttal concept in info.nnation lhc:ory and not a derived fwK:tion as in classical lhennodynamics. PROBLEMS 7.1 On the basis of the first law fill in the blank spaces in the following table of imaginary beat engine cycles. On the basis of the second law classify each cycle as reversible, irreversible, or impossible. Cycle Temperat1,1re Raie of Heat Flow Rate of work Efficiency So1,1rce 327\u00b0C 1000\u00b0C 750 K 700K Sink 27\u00b0C 100\u00b0C 300 K 300K Supply Rejection 420 k.J/s 230 k.J/s... kJ/min 4.2 MJ/min... kJ/s 3 kW... kJ/s... kW (a) (b) (c) (d) outpiit... kW... kW 26 kW 2 lc", "W 65% 600/o 7.2 TIii: latent heat of bion of water at o\u2022c is 335 kJ/kg. How much does the cniropy of 1 kg of ice change 1111 it melts into water in uch of lhe following ways: I I!!I ii I + II 206~ Barie and Applied Tfltrmodynamics {a) Heat is supplied reversibly to a mixture of ice and water at 0\u00b0C. (b) A mixtun: of ice nnd water at o\u2022c is stirred by a paddle wheel. 1:J Two kg of water at 80\u00b0C are mixed adiabatically with 3 kg of water at 30\u00b0C io a consla.Qt pressure procen of I atmosphere. Find the increase in the entropy of lhe total mass of water due to the mixing process Ccp of waier = 4.187 kJ/kgK). Ans. 0.0592 U/K 7.4 In a Carnot cycle, heat is supplied at 3S0\u00b0C and reject~\"<! at 27\u00b0C. The working fluid is water which, while receiving heat, evaporates from liquid at 350\u00b0C to sieam at 350\u00b0C. The associated entropy change is 1.44 kJ/kg K. (a) If the cycle operates on a siationaiy mass of l kg of water. how much is the work done per cycle, and how much is the beat supplied.! lb) If the cycle operates in steady flow with. a power output of 20kW, what is the steam flow rate? Ans. (a) 465.12, 897.12 kJ/kg, (b) 0.043 kg/s 1.5 A heat engine receives reversibly 420 kJ/c-ycle of heat from a source at 327\u00b0C, and rejec,is hear reversibly to sink at 27\u00b0C. There are n.o other heat transfers. For each of the three hypothetical amounts of heat rejected, in (a). {b). and (c) below, compute the cyclic integral of i!Qlr. From these results show which case is irreversible, which reversible, and which impossible: (a) 210 kJ/cycle rejected, (b) IOS kJ/cycle rejected, (c) 315 kJ/cycle rejected. Ans. (a) Reversible, (b) Impossible, (c) Irreversible 7.6 In Fig. P. 7.6, abctJ represents a Carnot cycle bounded by two reversible adiabatics and two reversible isothenns at temperatures T1 and T2 (T1 > T2). The oval figure is a reversible cycle, where heat is absorbed at temperatures less th:in, or equal to, r 1, and rejected at temperatures greater than, or equal to, T1\u2022 Provo that the efficiency of the oval cycle is less than that of the Carnot cycle. - -... v Fig. P. 7.6 7.7 Water is heated ai a constant pressure of 0.7 MPa. The boliag pomus 164.97\u00b0C. The init.ial temperature of water is 0\u00b0C. The latent heat of evaporation is 2066.3 kJ/kg. Find the increase of entropy of water, if the final state is steam. A11s. 6.6967 kJ/kg K 7.8 Ooe kg of air initially at 0. 7 MPa. 20\u00b0C changes to 0.35 MPa. 60\u00b0C by the three reversible oon\u2022llow processes, as shown in Fig. P. 7.8. Process t-,i-2 co1isists of a constant pressure eitpansion followed by a constant volume cooling, process 1 \u00b7b-2 an isothermal expansion followed by II constal\\t pn!ssure expansion, and process l-c-2 an adiabatic expansion followed by a constant volume heating. Determine the changes of internal energy, enthalpy, and,, Entr~py -=207 entropy for each process, and find the work transfer and heat transfer for each process. Takes cP \u2022 1.005 and c. c 0. 718 kJ/kg, K and assume the specific heats to be constant. Also assume for air pv = 0.287 T, where pis the pressure in kPa. V the specific volume in m3ikg; and Tthe temperature in K. r Rev. lso~rmal r 0,35Mpa. eo\u2022c v Fig. P. 7.8 7.9 Ten grammes of water at 20\u00b0C is convened into ice at -1o\u2022c at constant atmospheric pressure. Assuming tbe specific heat of liquid water 10 remain constant at 4.2 J/gX", "kg. Ans. 104.9 J!I( 7.33 A thermally insulated 50-obm resistor carries a cum:nt of I A for 1 s. The initial temperature of the resistor is I 0\u00b0C. Its mass is 5 g :and its specific heat is 0.85 J/g K. (a) What is the change in entropy of the resistor? (b) What i.s the change in entropy of the universe? Ans. (a) 0.173 J/K (b) 0.173 J/K 7.34 The value of cP for a certain substance can be represented by cP =a+ bT. (a) Determine the heat absorbed and the increase in entropy of a mass m of the substance when its temperature is increased at constant pressure from T1 to Tz. (b} Find the increase in the molal specific entropy of copper, when the temperature is increased at constant pressure from 500 to 1200 K.. Given. for copper: when T= 500 K, cP - 2S.2 x ta3 and when T \u2022 1200 K. cP \u2022 30.1 x l<Y J/kmolK. AIIS. (a)m[ a(T2 -1j )+ f<Tf - 7j2 >] 111[ aln i + b(~ - 7j)] (b) 23.9 kJ/k.mol K 7.35 An iron block of unknown mass at ss\u2022c is dropped into an insulated tank that contains O, I m3 of water at 20\u00b0C. At the same time a paddle-wheel driven by a 200 W motor is activated to stir the water. Thennal equilibrium is established after 20 min when the final temperature is 24\u00b0C. Detennine the ma.u of the iron block and the entropy generated during the process. Ans. 52.2 kg, 1.285 kJ/K 7.36 A piston-cylinder device contains 1.2 kg of nitrogen gas at 120 kPa and 27\u00b0C. The gas is now compressed slowly in a polytropic process during which pVIJ = constant.The process ends when the volume is reduced by one-half. Dctennine the entTopy ch.Inge of nitrogen during this process. Ans. - 0.0615 kJ/K. Iii I'II 212=- Basic and Applied 17amnodynamics 7.37 Air enters a compressor at ambient conditions of96 kPa and 17\u00b0C with a low velocity and exits at I MPa, 327\u00b0C. and 120 mis. The compressor is cooled by the ambient air at 17\u00b0C at- a rate of 1.500 kJ/min. The power input to the compres.sor is 300 kW. Oct.ermine (a) the mass now rate of air and (b) the raie of entropy generation. Ans. (a) 0.85 l kg.ts, (b) 0.144 kW/K 7.38 A gearbox operating at steady slate receives 0. I kW along the input shaft and delivers 0.09.5 kW along the output shaft. The outer surface of the gearbox is at so\u2022c. For the gearbox, determine (a) the rate. of heat transfer, (b) the rate at which entropy is produced. Atts. (a)-0.005 kW.(b) l.54x 10-skW/K 7.39 At steady state, an electric inotor develops power along its output shaft at the r.ite of 2 kW while dr.awing 20 amperes at 120 volts. The outer surface of the motor is at S0\u00b0C. For the motor. determine the rate of heat transfer and the rate of entropy generation. Ans. - 0.4 kW, 1.2.4 x io-\u00b73 kW/K 7.40 Show that the minimum theoretical work input required by a refrigeration cycle to bring two.finite bodies from the same initial temperature to the final temperatures of T1 1111d T2 (T2 < T1) is given by W mil = 111c (2(T1 Ti)'12 - T1 - T1) 7.41 A rigid tank coniains an ideal gas at 40\u00b0C that is being ~1irred by a paddle wheel. The paddle wheel docs 200 kJ of work on the ideal gas. It is observed that the temperature of the ideal gas remains constant during this process as l rewlt of heat transfer between the syslem and the surroundings at 25\u00b0C. Delermine (a) Ille eno:opy change of the ideal gas and (b) the total entropy generation. Ans. (a) 0, (b) 0.671 kJ/K 7.42 A cylindrical rod of length L insulated on its lateral surface is initially in contact", ", when beat loss occurs at a higher temperature T1 than when the same heat loss oocurs al a lower temperature T2\u2022 Therefore, a heat loss of I kJ at, say, I 000\u00b0C is more harmfol than the same heat loss of I kJ at, say, 100\u00b0C. Adequate insulation must be provided for high temperature fluids (T>> T0) to prevent the precious beat loss. This many oot be so important for low temperature fluids (T- T0), since the loss of available energy from such fluids would be low. (Similarly, insulation must be provided adequately for very low temperature fluids (T <<To) to prevent heat gain from surroundings and preserve available energy.) The available eneq,,y or exergy of a fluid at a higher temperature T1 is more than that at a lower temperature T2, and decreases as the temperature decreases. When the fluid reaches the ambient temperature, its eKergy is zero. The second law, therefore, affixes a quality to energy of a system at any state. The quality of energy of a gas at. say, I 000\u00b0C is superior to that at. say, I 00\u00b0C, since the gas at 1000\u00b0C has the capacity of doing more work than the gas at I 00\u00b0C, under the same environmental conditions. An awareness of this energy quality as of energy quantity is essential for lhe efficient use of our energy resoun:es and for energy conservation. The concept ofavailable energy or exergy provides a useful meas11re of this energy quality.!\u2022 \"''!t'~I H Availal>k EMrgf, E.tergy and /rrnmil,ilil:y -=221 8.3.1 Lau, of Degradation of Enert:'J The available ene.rgy of a system decreases as its temperature o.r p.ressure decreases and approaches that of the surroundings. When heat is transferred from a system. it.s temperature decreases and hence the quality of its energy deteriorates. The degradation is more for energy loss at a higher temperature than lhat at a lower temperature. Quantity-wise the energy loss may be the same. but quality-wise the losses are different. While tire first law swtes that energy i.v always conserved quantity-wise, the seco,uJ lmv emphasizes that energy always degrades quality-wise. When a gas is throttled adiabatically from a high to a low pressure, the enthalpy (or energy per unit mass) remains the same, but there is a degradation of energy or available work. The same ho?ds good for pressure drop due to friction of a tluid flowing through an insulated pipe. lf the first law is the law of conservation of energy, the second law is called the law of degmdlllio11 of en.ergy. Energy is always conserved, but its quality is always degrade<l. Article 8.2. L which shows how energy gets degraded by thermal irreversibility and produces less useful work can be explained in a little different w11y. Let two bodies I and 2 of constant heat capacities C1 and C2 be at temperatures T1 and T2 (T1 > T~). These arc connected by a rod and a small quantity of heat Q tlows from I to 2. The total change of entropy is: /!,S = /!,S 1 + 1!,S2 = Q[_!_ -....!..] > 0 r2 7i lsince T1 > T2) The entropy will continue to increase till Lhcnnal equ.ilibrium is reached. Let us now suppose that instead of allowing heat Q to lfow from I to 2, we used it to operate a Carnot engine and obtain mechanical work. with T0 as the sink temperature. The ma.1timum work obtainable is: W1=Q[1- ~ J If, however. we first\" allow Q to flow from l to 2 and then use it co operate the Carnot engine, we obtain: W2=Q[1-~]<W1 Thus, in the course of the irreversible heat conduction che energy has become degraded to the extent that the u~eful work has been decreased by AW= W1 - W2 = 70 /!,S The increase in entropy in an irreversible change is chus o mellSW'C of the extent to wb.ich energy becomes degraded in that change. Conversely, in order to extract lh.c maximum work from a system, changes must be pcrfonned in a reversible manner so that total entropy (/!,S\u2022)' + l!,S,..,'} is conserved. lt is worth pointing that if the two bodies were allowed to reach thennal equilibrium (a) by heal conduction and (b) by operating a Carnot engine betw", " only with the surroundings at T0, p0\u2022 8.5.1 Juuersibh Work in a Steady Flow Process For a steady flow process dm 1 =dm1 =dm d[ U- T0 S+ m;2 + mgzl \"'O and Equation (8.24) reduces to 4'Waw,; = c1m[( h1 - Tos1 + ;_ 2 + gz1 )-(1r2 -Tos2 + VJ + gz2 )] (8.25) For total mass flow, ihe integral fonn of Eq. (8.25) becomes If' max= ( H1 - foS1 + m ;,2 + mgz, )-( H2 - ToS:i + m ;/ + mg::2) (8.26) The expression (H -T0S) is called the Keenan function, B. wmu = ( 2 ). mV1 (!Ji + -2- + \"'8%1 - Bi+ -2- + mc:i2 mVf ) = lf1 - lf2 where ff IS called the availability function of a steady flow process given by mV2 lfl=B+ -2-+mgz (8.27) On a unit mass basis, Wmax=(h1-ToJ'1+ ;_ 2 +gz1)-(h2-Tos2+ V! +gz2) I I!!I ii I + II Aoai/41,/t ~. &err, and Irr=iliility -=227 lf K.E. and P.E. changes are neglected, Eqs (8.27) and (8.28) reduce to (8.28) (8.29) (8.30) and per unit mass w_=B1 -B2 -(H1 - T~1)-(H2 -ToS-i) = (H1 - H2) - To(S1 - S2) w INllt = b1 - b2 \u00b7 = (/,1 - n;i) - 70 (s1 - s;i) 8.5.2 Rlversiblt Work in a Closed System For a closed system, Equation (8.24) then becomes dm 1 = dmz = 0 dW\"'\"=-d[U-foS+ m;2 +mgzl where =- d(E- ToS)0 mV2 E=U+--+mgz 2 For a change of state of the system from the initial slate I to the final state 2, Wmu =\u00a31 -\u00a32 - T0 (S1-S2) = (E1 - ToS1)-(E2-T~;) Ifthc K.E. and P.E. changes arc neglected, Eq. (8.31) reduces to Wmu = <U1 - ToS1)-(U2 - ToS2) For unit mass of fluid, w_ = (u 1-u2)-T0 (s1 -.s:J.. (u, - f <>11) -(112 - Tof2) (8.31) (8.32) (8.33) 8.6 Useful Work All of the work Wofthe system with a flexible boundary would not be available for delivery, since a certain portion of it would be spent in pushing out the atmosphere (Fig. 8.12). The useful work is defined as the actual work delivered by a system less the work performed on. the atmosphere. If 111 and V2 are the initial and final volume of the system andp0 is the atmospheric pressure, then the work done on the atmosphere is p0 (V2 - V1). Therefore, the useful work Wu becomes I I!! I II I 228=- BtJJit and Applied 111mnod:ynamin _..\u00b7@-;., ( \\ Initial Boundaly i>o,T0 \\,, \u2022\u2022\u2022\u2022\u2022 \u2022 ____ _.\u2022\u2022\u2022\u2022\u2022\u2022..-4--- Final Boundary Surroondin911 (Atmosphere) Flf. 8.12 Work t{ont by a clostd systtm in pus/ting out tltt atmosplurt Similarly, the maximum useful work will be (WJ- -W--po(Vz- Y1) (8.35) In differential form (dWJ,_ =dW- -podY (8.36) Jn a steady now system, the volume of the system does not change. Hence, the maximum useful work would remain the same, i.e., no work is done on the atmosphere, or (8.37) But in the case of an unsteady-flow open system or a closed system, the vol", "wne of the system changes. Hence, when a system exchanges heat only with the atmosphere, the maximum useful work becomes (d W11)mu = ctW,_-p0 dY Substiruting dW rrw< from Eq. (8.24), 2 (dWJmax = dm1 ( h1 -Tosi+ ~ + g:1 )-dm2 ( \"2 - Tosi+ V! + gz2) -d[ U + p0Y - T0S+ m; 2 + mgz l This is the maximum useful worlc: for an unsteady open system. For tbe closed system, Eq. (8.38) reduces to (dWJ\"\"\"' =-d[ U + p0 Y-ToS+ '\"; =-d [E+PQ JI-ToSL, 2 + mgz l (WJ,,.. =\u00a3, -Ez+Po (Y, - Yz)- To(S1-Si> IfK.E, and P.E. changes are neglected, Eq. (8.40) becomes (8.38) (8.39) (8.40) (W.,),,.. = U1 - U2 + Po (Y1 - Y2) - T0 (S1 - S2) (8.41) This can also be written in the following form ( w.. )...... = (U, + Po Y, - ToS1)- ( U2 + PoY2 - T oS'2) (8.42) \"I''\" -=229.. ~, -'2 where \u2022 is called the C1vailabiliiy functio11 for a closed system given by The useful work per unit mass becomes,.. u+p0Y-T~ (Wu)max.:o (111 + PoV1 - To,1\"1)- (uz + Por>2 - To,1\"2) (8.43) 8.6.1 Muimum Useful Work Obtainal>k when the System &c.hanges Heat u,ith a T1Jmnal Rum,oir in, Addition to the Atmosphere If the open system discussed in Sec. 8.5 exchanges heat with a thennal energy reservoir at temperature TR in addition to the atmosphere, the maximum useful work will be increased by d QR ( l - ~ ), where i! QR is the beat m:eived by the system. For a steady!low process, (WJmu'\"' w_ = ( H1 - ToS1 + m:l + mgz1) 10 ) - H2-To.s':i+-2-+111gz2 +Q11. l-Ta. mV} ) ( ( = \\V1 - \u00a5'2 + QR( 1 - ~ ) (8.44) For a closed system {Wu)mu = W\"'..,. -po (Vi - Y1)+ QR ( 1- ~) or (W11)mu = \u00a3 1 - E2 + Po (V1 - Y2) - T0(S1 - S2) + QR ( 1- ~ ) (8.45) lfK.E. and P.E. changes are neglected, then for a steady/low process: (W11),.... ~ (H1 - H2) - T0 (S1 - S2 ) + Q11. ( 1- ~) (8.46) and for a closed system: (W11)ow\"' U, - U2 + Po<J'1 - V2)-TJ..S1 -Si)+ Q11. ( 1- ~) (8.47) 8. 7 Dead State If the state of a system departs from that of the sll.rfoundings, an opportunity exists for producing work (Fig. 8.13). However, as the system changes its state I I! ii I + 11 230=- Basi, a,uJ Applied 'Inlrmod]namics towards that of the surroundings, Lhis opponunity diminishes, and it ceases to exist when the two arc in equilibrium with each other. When the system is in e<1uilibrium with Lhe swtoundings, it must be in pressure and temperature equilibrium with the surroundings, i.e., at p0 anq T0. Jt must also be in chemical e({Uilibrium. with the surroundings, i.e., th.ere should not be any chemical reaction or mass transfer. The system must have zero velocity and minimum potential energy. This state of the system is known as the dead state, which is designated by affixing subscript 'O' to the properties. Any change in the state of the system from the dead state is a measure of the available work", " that can be e1ttracted from it. Farther the initial point of the sysiem from the dead state in te.lTllS of p, t either above or below it, higher will be the available energy or exergy of the system (Fig. 8.13). AU spontaneous processes terminate at the dead state. Q. t Dead State. t 1I Availabillly t 0 - - - --T - -Po A~iability lsothelm 81 T0 1' (a) (b) Fig. 8.13 ADailal>u work of a 11stttr1 dtCTeosts /lS its s/!Jtt apprflO&MS Po, To 8.8 Availability Whenever useful work is obtained during a process in which a finite system undergoes a change of state, the process must terminate when the pressure and temperature of the system have become equal to the pressure and temperature of the surroundings,p0 and T0, i.e., when the system has reached the dead state. An air engine operating with compressed air taken from a cylinder will continue to deliver work till the pressure of air in the cylinder becomes equal to that of the sur.roundings,p0. A certain quantity of exhaust gases from an internal combustion engine used as the high temperature source of a beat engine will deliver work until the temperature of the gas becomes equal to that oflhe surroundings, T0\u2022 The availability (A) of a given system is de.lined as the maxim11m use/ti/ work (total work minus pdV work) that is obtainable in a process in which the system comes to equilibrium with its surrormdings. Availability is thus a composite property depending on the state ofboth the systetn and surroundings. 8.8.1 A,ailulility i\u2022 a Steady Flow Process The rcvenible {maximum) work associated with a steady flow process for a single flow is given by Eq. (8.26) \"I' o II W-=(H,-ToS1 +m;i2 +mgz1)-(H2 -ToS2 +m:f +mgr2 ) ~231 With a given state for the mass entering the control volume, the maximum useful work obtainable (i.e., the availability) would be when this mass leaves the control volume in equilibrium with the surroundings (i.e., at the dead state). Since there is no change in volume, no work will be done on the atmosphere. Let us designate the initial state of the mass entering the C.V. with parameters having no subscript and the final dead state of the mass leaving the C.V. with parameters having subscript 0. The maximum work or availability, A, would be A~(H-ToS+ mt +mgz)-(H0 -T~0 +mg%o)=~-\u00a5o (8.48) where yi is called the availability function for a steady flow system and VO= 0. This is the availability of a system at any state as it enters a C. V. in a steady now process. The availability per unil mass would be a=(lr-To.s+ ~ 2 +gz)-(lr0 -TQS0 +gz)=~-% (8.49) If subscripts I and 2 denote the states ofa system entering and leaving a C.V., the decrease in availability or maximum work obtainable for the given system\u00b7 surronndings combination would be w_ =a,-a2= 11'1- ~2 = [(11, -Tosi+ i,2 + g.:1 )-<ho -Toso+ gzo)] -[(1rz-To.so+ Vf +gz2)-<1io-ro1o+aro>].. (h1 - h2)- To (s1 - s2) + v?-vf 2 + g(:1 - z2) (8.50) lfK.E. and P.E. changes an neglected. Wm,..= (h1 - TQS1)-(lri-TQSi) =b, -b2 where bis the specific Keenan function. If more than one flow into and out of the C. V. is involved. w...... = l:m; V,; - l:m.,Y,., j j 8.8.2 AvaUabilily in a Nonjlow P7otess Let us consider a closed system and denote its initial state by parameters without any subscript and the final dead state with subscript 'O'. The availabiliiy of the system A, i.e., the maximum useful work obtainable as the system reaches the dead state, is g,iven by Eq. (8.40). \u2022 I \u2022\u2022, I! I!!", " I 232=- &uic and Applied 11tmnodynamics A =(WJnw: = E-E0 + p0 (V- V0)-To(S-S0) = ( U + mt + mg.z: )-(U0 + mg.z:t>) + p0( V- V0) - T0(S- S0) (8.51) If K.. E. and P.E. changes are neglected and for unit mass, the availability becomes \"= u - u0 + p0(v- v0) - T0 (s - s0) = (u + Por, - T&1) - (uo-PoVo -ToJo) = ~- ~ (8.52) where~ is the availability function of the closed system. If the system undergoes a change of state from I to 2, the decrease in availability will be a=(~,-'\u00b0)-(~-~) -~,-~ = (u 1 - u2) + p0 (v1 - r,2) - T0(s1 - s2) (8.53) This is the max.imum useful work obtainable under the given surroundings. 8.9 Availability In Chemical Reactions In many chemical reactions the reactants are often in pressure and temperature equilibrium with the surroundings (before the reaction takes place) and so are the products after the reaction. An internal combustion engine can be cited as an example of such a process if we visualize the products being cooled to atmospheric temperature T0 before being discharged from the engine. (a) Let us first consider a system which is in temperature equilibrium with the surroundings before and after the process. The maximum work obtainable during a change of state is given by Eq. (8.31 ), w....,. = \u00a3 1 - E2 - T0(S1 - S2) = ( U1 + m;f + mg.z:1 ) - ( U2 + m;J + mgz2 )-To(S1 -S2) If.K.E. and P.E. changes are neglected, w _ = U1 - U2 - T0 (S1 - S2) Since the initial and final temperatures of the system are the same as that of che surroundings, T1 = T2 = T0 = T, say, then (WT)IIIIL'< = (U1 - Uz)r- 1lS1 - S:Jr Let a property called Helmholtzfunction F'be defined by the relation F= U-TS Then for two equilibrium states I and 2 at the same temperature T, (F1 - F,i>f = (U1 - U1)r- TtS1 - S2>,, (8.54) (8.55) (8.56) \"I' AllflilaMt Enn-gy, E.xtTg'/ and /1m'el'si6ility ~233 Fro.m Eqs (8.54) and (8.56), or (Wr)- = (F, - F2tr W1 S (F1 - F1h (8.57} (8.58) The wo.r.k done by a sysiem in any process between two equilibrium states at the same temperature during which the system exchanges heat only with the environment is equal to or Jess 1han the decrease in the.Helmholtz functio.n of the system during the process. The maximum work is done when the process is reversible and the equality sign holds. If the process is irreversible, the work is less than the maximum. (b) Let us now consider a system which is in both pressure and temperature equilibrium with the surroundings before and after the process. When the volume of the system increases some work is done by the system against lhe surroundings (pdV work), and this is not available for doing useful work. The availability of the system, as defined by Eq. (8.51), neglecting the K.E. and P.E. changes, can be expressed in the form A= (W.Jmu = (U + PoY- To,S)-(Uo + Po'Yo -ToSo) The maximum work obtainable during a change of slate is the decrease in availability of the system, as given by Eq.(8.SJ) for unit mass. (Wu)mu =A, -A2 = ~1- ~2 = (U, - U2) + PoW1 - Y2) - To(S1 - S2) lfthe initial and final equilibrium states of the system are at the same pressure and temperature of the surroundings, say p 1 = p 2 ; Po = p, and T1 = T2 = T0 = T. Then, cw.. >.... = (U1", " T0 = 298 K. Ans. 0.956. 0.879 8.12 Air tnters an adiabatic compressor at atmospheric conditions of I bar, J 5\u00b0C and leaves at 5.5 bar. The mass tlow rate is 0.0 I kg/s and the efficiency of the compressor is 75%. After leaving the compressor, the air is cooled to 40\u00b0C in an 111 -=275 al\\crcooler. Calculate (a) the power required lo drive the comprc!>-sor, and (b) the rate of irreversibility for the overall process ( compressor and cooler). A11s. (a) 2.42 kW, (b) I kW 8.13 In a rotary compressor, air enters at I.I bar. 21\u00b0C where ii is compressed adiabatically to 6.6 bar, 2S0\u00b0C. Calculate the irreversibility and the entropy production for unit mass now rate. The atmosphere is at 1.03 bar. 20\u00b0C. Neglect the K.E. changes. AltS. 19 kJ/kg, 0.064 kJ/kg K 8.14 In a steam boiler, the hot gases from a fire transfer heat to waler which vaporizes at a constant temperature of242.6\u00b0C (3.5 MPa). The gases are cooled from I JOO to 4 30\u00b0C and have an average specific heat, cP = I ;046 kJ/kg K ove.r this temperature range. The latent beat of vaporization of steam at 3.S MPa is I 753.7 kJ!kg. Iftbe steam generation rate is 12.6 kg/sand there is negligible heat loss ti:om the boiler, calculate: (a) the rate of heat transfer, (b) the rate of loss of eKergy of the gas. (c) the rate of gain of exer1,,y of the. steam. and (d) ihc rate of entropy generation. Take T0 = 21 \u00b0C. Ans. (a) 22096 kW, (b) 15605.4 kW (c) 9501.0 kW, (d) 2.0.76 kW/K S.J 5 An economizer. a gas-to-water finned tube heat exchanger. receives 67.:S kg/s of ga.\\ cP = 1.0046 kJ/kg K, and 51.1 kg/s of water, c r = 4.186 kJ/kg K. The water rises io temperature from 402 to 469 K, where the gas falls in temperature from 682 to470 K. There are no changes of kinetic energy, andp0 = 1.03 bar and T0 = 289 K. Determine: (a) rate of change of availability of the water, (b) the rate of change ofavailabitity of the gas. and (c) the rate of entropy generation. A11s. (a) 4802.2 kW, (b) 7079,8 kW. (c) 7.73 kW/K 8.16 The exhaust gases from a gas turbine arc used to heat water in an adiabatic counterflow heat exchanger. The gases are cooled from 260 to 120\u00b0C, while water enters at 65\u00b0C. The flow rates of the gas and water a~ 0.38 kg/s and 0.50 kg/s respectively. The constant pressure speciJic heals for the gas and waler are 1.09 and 4.186 kJ/kg K respectively. Calculate the l\"ate of exergy loss due to heat iransfer. Take T0 \"' 35\u00b0C. AIIS. I 1.92 kW 8.17 The exhaust from a gas turbine at 1.12 bar. 800 K nows steadily into a heat exchanger which coo.ls the gas to 700 K without significant pressure drop. The heat transfer from the gas heats an air now at constant pressure, which enters ihe heat exchanger at 470 K. The mas~ flow rate of air is twice that of the gas and the surroundings are al 1.03 bar, 20\u00b0C. Detem,ine: (a) the decrease in availability of the eithau~t ga.ses, and (b) 1hc to1al entropy production per ks of gas. (c) Wbat arrangement would be nec.essary lo make the heat trJnsfer reve~ibl.e and how much would this increase the power output of the plant per kg of turbine gas\"! Take cP for exhaust gas as 1.08 and for air as I.OS kJ/kg K. Neglect heat transfer to the surroundings and the changes in kinetic and potential energy. A11s. (a) 66 kJ/kg. (b", ") 0.0731 kJ/kg K. (c) 38.7 kJ/kg 8.18 An air prehcater is used to beat up the air used for combustion by cooling the outgoing products of combustion from a furnace. The rate of flow of the products is IO kg/s, and the products are cooled from '300\u00b0C 10 200\u00b0c. and for the products at this temperature cp = 1.09 kJ/kg K. The rate of air flow is 9 kgls, the initial air temperature is 40\u00b0C, and for the air cP = I.005 kJ/kg K. ( a) What is the initial and final availability of the products? \"' 276=- lJasic and Applied Tlttrmodynamics (b) Whal is the irreversibility for this process? (c) If ihe heat transfer from ibc products were to tnke place reversibly through heat engines, what would be the finally temperature of the air'! What power would be developed by the heal engine.s? Take T0 = 300 K. Ans {a) 85.97, 39.68 kJ/kg, (b) 256.S kW; {c) 394.41 K, 353.65 kW 8.19 A massof2 kg of air in a vessel expands from 3 bar, 70\"C to I bar, 40\"C. while receiving 1.2 kJ of heat from a reservoir at 120\u00b0C. The environment is al 0.98 bar, 27\u00b0C. Calculate the maximum work and the work done on lhc atmosphere.,fas. 177 kJ, 112.S k.J 8.20 Air entefS the compressor of a gas turbine al I bar, 30\u00b0C and leaves ihe compressor at 4 bar. Tlte compressor has an efficiency of82%. calculate per kg of air (a) the work of compression, (b) the reversible work of compression. and (c) the im:versibilicy. [For air, use Ti... (1!L)r-11r Ii Pi where Ti, is the temperature of air after iscnlropic compression and y= 1.4. The compressor efficiency is defined as (T2, - T1)!(T2 - T1), where T2 is ihe actual temperature of air after compressional.} Ans. (a) 180.5 kJlkg, (b) 159.5 lu/kg (c) 21 kJ/kg 8.21 A mass of 6.98 kg of air is in a vessel al 200 kPa, 27\u00b0G. Beat is traosfcrred to the air from a reservoir at 72 7\u00b0C until ihc temperature of air rises io 327\u00b0C. The enviror.u:nent is at IOOkPa. t 7\u2022c. Determine (a) the iui1ial and final availability of air, (b) the maximum useful work associated with the process. A11s. (a) 103.S. 621.9 kJ (b) 582 kJ 8.22 Air enters an adiabatic compressor in steady '/low at 140 kPa, 17\u00b0(' and 70 mis and leaves it at 350 kPa, 127\u00b0C and 110 mis. The environment is at 100 kPa. 7\u00b0C. Calculate per kg or air {a) the: actual amount of work required, (b) ibc minimum work required, and (c) the irreversibility of the process. AltS. (a) 114.4 kJ, (b) 97.3 kJ, (c) 17.1 kJ 8.23 Air ex-pands in a turbine adiabatically from 500 kPa, 400 K and 150 m/s to 100 kPa, 300 Kand 70 m/s. The environment is at 100 kPa. 17\u00b0C. Calculate per kg of air (a) the maximum work output, (b) the actual work output. and (c) the irreversibility. Ans. (a) 159 kJ, (b) 109 kJ. (c) 50 kJ 8.24 Calculate the specific e,ccrgy of air for a stale ai 2 bar, 393.15 K whe.n the surroundings are at I bar. 293.1 SK. Take cP = I and R = 0.287 kJ/kg K. An.r. 72.31 kJ/kg 8.25 Calculate lhe specific exergy of CO2 (er = 0.8659 andR = 0.1889 k.J/kg K) for a state at 0.7 bar,268.15 Kand forthc environment at LO bar and 293.15 K. An.s. - 18", " substance of constant chemical composition throughout its mass. It is a one-component system. It may exist in one or more phases. 9.1 Jrv Diagram for a Pure Substance Assume a unit mass of ice (solid water) at -I 0\u00b0C and I atm contained in a cylinder and piston machine (Fig. 9.1 ). Let the ice be heated slowly so that its temperature i:; always uniform. The changes which occur in the mass of water would be traced as the temperature is increased while the pressure is held constant. Let the state changes of water be plotted on p-v coordinates. The distinct regimes of heating, as shown in Fig. 9.2, are: 1 aim Fig. 9.1 Healing of H20 at a co,utanl pms,m of 1 aim 1-2 The temperature of ice increases from - I 0\u00b0C to 0\u00b0C. The volume of ice would increase, as would be the case for any solid upon heating. At state 2, i.e. 0\u00b0C, the ice would start melting. 2-J lee melts into water at a constant temperature of 0\u00b0C. At state 3, the melting process ends. There is a decrease in volume, which is a peculiarity of water. I I,.,!,I'II 280=- 81JJ1t and Ap;lied Tlttrmodynamiu 3-4 The temperature of water iacrcases, upon heating, from 0\u00b0C to 100\u00b0C. The volume of water increases bcc:ause of thermal expansion. 4-5 The water starts boiling at state 4 and boiling ends at state 5. This,phase change from liquid to vapour occurs at a constant temperature of 100\u00b0C (the pressure being constant at I atm). There is a large increase in volume. 5-6 The vapour is heated to, say, 250\u00b0C (state 6). The volume of vapour increases from v5 to v6\u2022 Water existed in the solid phase between 1 and 2, in the liquid phase between 3 and 4, and io the gas phase beyond 5. Between 2 and 3, the solid changed intO' the liquid \u00b7phase by absorbing the latent heat of fusion and between 4 and 5, the liquid changed into the vapour phase by absorbing the latent heat of vaporization, both at constant temperature and pressure. The states 2, 3, 4 and 5 are known as saturation states. A saturation state is a state from which a change of phase may occur without a change of pressure or temperature. State 2 is a saturated,folid state because a solid can change into liquid at constant.Pressure and temperature from state 2. States 3 and 4 are both saturated liquid states. In state 3, the liquid is saturated with respect to solidification, whereas in state 4, the liquid is saturated with respect to vaporizaiion. State 5 is a saturated vapour state, because from state 5, the vapour can condense into I iquid without a change of.pressure or temperature. If the heating of ice at -10\u00b0C to steam at 250\u00b0C were done at a constant pressure of 2 atm, similar regimes of heating would have been obtained with similar saturation states 2. 3, 4 and 5, as shown in Fig. 9.2. All the state changes of the system can similarly be plotted oo thep-vcoordinates, when it is heated at ditfe.rcnt constant pressures. All the saturated solid states 2 at various pressures are joined by a line., as shown in Fig. 9.3. t :.:... ; ~ ; ;-i i-~-~- ~ r7 6 -\u00b7\u00b7\u00b7- - -~,~-~ 3 4 1 2 Fig. 9.2 Cliangu in tlu aolumt a/ wattT during lualing at co,u~nt pressure - -... v Similarly, all the saturated liquid states 3 with respect to solidification, all the saturated liquid states 4 with respect to vaporization, and all the saturated vapour states 5, are joined together. Figure 9.4 shows state changes of a pure substance other ihan water whose volume increases on melting. The line passing through all the saturated solid states 2 (Figs 9.3 and 9.4) is called the saturated solid line. The lines passing through all the saturated liquid states 3 and 4 with respect to solidification and vaporization respectively are,,'II Propertfa of Pim Sul\u00bbtanas -=281 known as the saturated liquid lines. and the line passing through all the snturated vapour states 5, is the saturaJed vapo1ir line. The saturated liquid line with respect to vaporization and the saturated vapour line incline towards each other and form what is known as the saruratio,i or i\u2022apm1r dome. The two lines meet al", " Similarly It= hr+ x hr8 S = sr+.XSfg \"= u,+.x \"fg P ~Pt+.x Pr, However, Volume fraction of vapour or voidage, a= v/v m=mr+m8 pv = P(Or + p8 v8 = Prf..:o-vJ + p8 v11 p = (1 - a)pr+ ap1 (9.6) (9.7) (9.8) (9.9) (9.9a) 9.7 Steam Tables The properties of water are arranged in the steam tables as functions of pressure and temperature. Separate tables are provided to give the properties of water in the saturation states and in the liquid and vapour phases. The internal energy of saturated water at tl,e triple point (t = 0.0 I 0 C) is arbitmrily chosen to be zero. Since Ir = u + pv, the enthalpy of saturated wa.ter at 0.0 I \u00b0C ls slightly positive because of the small value of (pv) tcnn. The entropy of saturated water is also chosen to be zero at the triple point. 9.7.1 Saturation States When a liquid and its vapour are in equilibrium at a certain pressure and temperature, only the pressure or the temperature is sufficient to identify the saturation state. If the pressure is given, the temperature ofihe mixture gets fixed, which is known as the saturation temperaiure, or if the temperature is given, the saturation pressure gets fix.ed. Saturated liquid or the saturated vapour has only one indepe.ndent variable, i.e. only one propeny js required to be known to fix up the state. Tables A. I (a) and A. l (b) in the appendix give the properties of saturat\u00b7 ed liquid and saturated vapour. In Table A. l(a), the independent variable is temperature. At a particular temperature, the values of saturation pressurep, and Vr, v,, hr, hrg, /,g,.r, and s, arc gi.ven, where vi, hr, and sr refer to the saturated, hrs\u2022 and sr, h8 and sg refer to the saturated v<:1pour state; andt1r liquid states; v8 8 8 refer to the changes in the property values during :vaporation ( or condensation) at that temperature, whe~ vr, = v8 - v,and s,8 In Table A. l(b), the independent variable is pressure. At a particular pressure, the values of saturation temperature t, and llr, v8, sr, and \u00b7\\tare given. Depending upon whether the pressure or tl1e temperature is given, either Table A. l(a) or Table A. l(b) can be conveniently used for computing the properties of saturation states. : s1 -s,., Ii,. lie,, h1 If data arc required for intermediate temperatures or pressures, linear interpolation is normally accurate. The reason for the two iables is to reduce the amount of interpolation required. Iii I I II 294=- Basic arrd Applied TheTmodynamics 9.7.2 Liquid-oapour Miztutts Let us consider a mixture of saturated liquid water and water vapour in equilibrium at pressure p and temperature I. The composition of the mixture by mass will be given by its quality x, and. its state will be within the vapour dome (Fig. 9.18). The properties of the mixture are as given in Aniclc 9.6, i.e. V \"''t't+ ZVr1 \"... ur+..t \"r, h = he+ z hr, s =s,+x s,8 where Vr, \"ts\u2022 ur, u,,. hr, Itri, s, and s.rg are the saturation properties at the given pressure and temperature. s, s So - s Fig. 9.18 Property in two plum region I fp or t and the quality of the mixture are given, the propenies of the mixh1re (v, u, h, ands) can be evaluated from the above equations. Sometimes, instead of quality, one of the above propenies, say, speci.fic volwne v, and pressure or temperature arc given. In that case, the quality of the mixture x has to be calculated from the given v and p or t and then x being known, other properties are evaluated. 9.7.3 Superheated Vapour When the tem,perature of the vapour is greater than the saturation tempe!'llture corresponding to the given pressure, the vapour is said to be superheated (state I in Fig. 9. I 9). The difference", " or its qual.ity has to be evaluated first. Now V =vr+ XVtg 0.09 = 0.001177 + X (0.09963 - 0.001177) 11 I!! '!1 I I II Proptrtia of Purt S11bstancu ~303 or At x = 0.904 or 90.4% 2 MPa, 111 = 908.79 and h1, = 1890.7 kJ/lcg s, = 2.4474 and sr, = 3.8935 kJfkg IC. h = hr+ x hre \"'908.79 + 0.904 x 1890.7 = 2618.79 kJ/kg s=sr+><s,, = 2.4474 + 0.904 >< 3.8935 = 5.9534 kJfkg K Ans. Ans. Exiunple 9., Find the enthalpy, entropy. and volume of steam at 1.4 MPa, 3so0c. Solution Atp = 1.4 MPa, in Table A. l(b), 1$11 = 195.07\u00b0C. Therefore, the state of steam must be in the superheated region. In Table A.2, for properties of superheated steam, at 1.4 MPa, 350\u00b0C t1 = 0.2003 m3/lcg h:: 3149.5 kJ/kg s = 7.1360 kJfkg K and at 1.4 MPa, 400\u00b0C v = 0.2178 m3/kg Ii = 3257.5 kJ/'kg s = 7.3026 kJ/kg K :. By interpolation at u MPa, 380\u00b0c V = 0.2)08 m3/kg Ii:: 3214.3 kJ/kg s = 7.2360 kJ/kg IC. Ans. Ezample 9.5 A vessel of volume 0.04 m3 contains a mixture of saturated water and sahlrated steam at a temperarure of250\u00b0C. The mass of the liquid present is 9 kg. Find the pressure, the mass. the specific volume, the enthalpy, th.e entropy, and the internal energy. Soluti<Jn From Table A.l(a), at 250\u00b0C p... 1 = 3.973 MPa Volume ofliquid, Volume of vapour, :. Mass of vapour v, = 0.0012512 m3/kg, v8 = 0.05013 m3/lcg 1,, = 1085.36 kJ/kg, - 1716.2 kJ/kg h, 8 s(t = 3.2802 kJ/'kg K sr = 2.7927 kJ/kg K. Vr = mr v, = 9 X 0.0012512 =0.01126 m1 v, = 0.04 - O.ot 126 -0.02874m3 v, 0.02874 m, = -;;- = O.OS013 = O.S7S kg f I,, Ill I '11,, 11 304=- &uie and Applitd Tllen,,odynamia :. Toi.al mass of mixture, m \"'m,+ m, = 9 + 0.575 = 9.575 kg Quality of mixture, 0.575 =0.06 9.515 v =v,+xvr, = 0.0012512 + 0.06 (0.05013 - 0.0012512) =0.00418 m31kg la = hr+ zhr, = 1085.)6 + 0.06 X 1716.2 = 1188.32 lcJ/kg I ~s,+:ur, '\"'2.7927 + 0.06 X 3.2802 = 2.9895 k:Jlkg K u=h-pv X 0.00418 = 1188.32 - 3.973 X 103 \"\"1171.72 k:Jlkg 11,= 1080.39 and ur, = 1522.0 k:Jlkg U '\"'11,+ xw,. = 1080.39 + 0.06 x 1522 = 1071.71 kJ/kg Also, at 250\u00b0C, Ans. Ans. Ans. Ans. Ans. Ans. Eumple 9.6 Steam initially at 0.3 MPa, 2S0\u00b0C is cooled at constant volume. (a) At what temperature will the steam become saturated vapour? (b) What is the quality at 80\u00b0C? What is the heat transferred per kg of steam in cooling from 2S0\u00b0C to 80\u00b0C? &>lution At 0.3 MPa. tut= 133", ".SS\u00b0C Since t >,..,.the state would be in the superheated region (Fig. Ex. 9.6). From Table A.2, for properties of superheated steam, at 0.3 MPa, 250\u00b0C v = 0.7964 m31kg h = 2967.6 kJ/kg Q. l 1> - Fig. EL 9.6 I I 11! ii I + II Pr~,rtiu of Pure Sarutanw -=305 In Table A.I when when v1 = 0.8919,,.., = 120\u00b0C v, = 0.7706, t.,., = 125\u00b0C = 0.7964,,..,. by linear interpolation, would be 123.9\u00b0. Therefore, when v8 Steam would bet:ome saturated vapour at t = 123.9\u00b0C At 80\u00b0C, vr= 0.001029 m3/kg, v8 = 3.407 m3/kg, lar= 334.91 kJ/kg, I,,,= 2308.8 kJ/kg, Pu,= 47.39 kPa VJ = Vi = 0.7964 m3/lcg = Vf80\"C + X2Vfg80\"C = 0.001029 + X2 (3.407 - 0.001029) x = 0. 79539,,. 0.234 2 3.40597 li2 = 334.91 + 0.234 X 2308.8 = 875.9 k.J/kg li2 = 2967.6 kJ/kg From the first law ofthecmodynamics Ans. (a) Am. (b) or 4Q=du+pdv (d'Q)y = du Q,.2 = \"2 - u, \"' (l,2 - P2 V2)- (h1 - P, v,) = (h2 - h1) + v(p, - P2) = 875.9- 2967.6) + 0.7964 (300 - 47.39) = 2091.7 + 201.5 = - 1890.2 kJ/kg Ans. (c).E:sample 9.7 Steam initially at 1.5 MPa. 300\"C e1tpands reversibly and.. adiabatically in a sicam twbine to 40\u00b0C. Determine the ideal work output of the turbine per leg of steam. Solution The steady flow energy equation for the control volume, as shown in Fig. Ex. 9. 7. I, gives ( other energy terms being neglected) \"\u2022 =h2+ W W=h 1 -h2 \u2022CO-C Fig. EL 9.7.1 I!I It I 306=- Basic and Applied Thennodynamia Work is done by ste,am at the expense of a fa) in its enthalpy value. The process is reversible and adiabatic, so it is isentropic. The process is shown on the T-s and h- s diagrams in Fig. Ex. 9.7.2. From Table A.l{a), at 40\u00b0C \u00b7 P q t = 7.384 kPa, s, = 0.5725, ands,,= 7.6845 lcJ/kg K h, = 167.57, and h, 8 = 2406.7 kJ/kg Atp = 1.5 MPa, t = 300\u00b0C, from the tabulated prope;ties of superheated steam (Table A.2) \"1 = 6.9189 ltJ/kg K h, = 3037.6 ltJ/kg / - s (a) Fig. EL 9.7.2 (bl Since s, =s2 6.9189 =sc+ ~.ffl40'C = 0.5725 + X2 X 7.6845 x2 = ~:~:S = 0.826 or 82.6% l,l = l,f4'1'C + Xi Jiff4/1'C... 167.57 + 0.826 X 2406.7 = 2152.57 lcJ/kg W = \"\u2022 - h2 = 3037.6 -2)52.57 = 885.03 lcJ/kg Am. Example 9.8 Steam at 0.8 MPa, 250\"C and flowing at the rate of I kg/s passes into a pipe canying wet steam at 0.8 MPa, 0.95 dry. After adiabatic mixing the tlow rate is 2.3 kg/s. Determine the condition of steam after mixing. The mixture is now expanded in a frictionless nozzle isentropically to a pressure of 0.4 MPa.", " = 2327.03 kJJkg U2 = h2 - P2 V2 = h2 = 2327.03 kJ/kg It -It = S X 2706.7 + 10 X 2327.03 3 \"' IS Now for the mixture \"'2453.6 kJ/kg :::: U3 lt3 = 2453.6 kJ/lcg = U3 I I,, ill I II I I II 314=- Basi, and Applitd T1iermod1namics V3 = 0.496 m3/kg From the Mollier diagram, with the given values of h and v, point 3 after mixing is fixed (Fig. Ex. 9.13.2). x3 = 0.870 s) = 6.29 kJ/kg K Ans. V3 = 0.496 m3/kg./C l Flg. Ex. 9.13.2 p 3 = 3.5 bar s4 =sg0.l.Mh = 7.1271 kJ/kg K S2 =s111.SMP& + 0.8s,ao.SMl'lt = 1.8607 + 0.8 x 4.9606 = 5.8292 kJlkg K A.ns. Entropy change during the process =m3 s3 -{m 1 s1 + m2 s2) = 15 X 6.298 -(5 X 7.127l + 10 X 5.8292) = 0.43 kJ/kg Ans. E:itample 9.14 Steam generated at a pressure of 6 MPa and a temperature of 400\u00b0C is supplied io a turbine via a throttle valve which reduces the pressure to S MPa. Expansion in the turbine is adiabatic to a pressure of 0.2 MPa, the isentropic efficiency (actual enthalpy drop/isentropic enthalpy drop) being 82%. The surroundings are at 0.1 MPa, 20\u00b0C. Determine the availability of steam before and after the throttle valve and at the turbine exhaust, and calculate the specific work output from the turbine. The K.E. and P.E. changes are negligible. Sol11tion Steady flow availability IJI is given by ip= (h - lr0)-T0(s - s0) +.! Vf + g(Z - Z0) 2 where subscript O refers to the surroundings. Since the K.E. and P.E. changes are negligible 'Iii = Availability of steam before throttling = (111 -110) - TJs1 -s0 ) \u2022 I,, 111'\" Properties of Purt Substances -=315 At 6 MPa, 400\u00b0C (Fig. El(.. 9.14) h, = 3177.2 kJ/kg s 1 = 6.5408 kJ/kg K 6MPa,400\u00b0C 6MPa 0.2 MPa (al - s (bl Fig. Ex. 9.U At20\u00b0C h0 = 83.96 kJ/kg s0 = 0.2966 kJ/kg K Y't = (3 177.2 - 83.96) - 293 (6.5408 - 0.2966) = 3093.24 - 1829.54 = 1263.7 kJ/kg Ans. Now h1 = h2, for throttling At h = 3177.2 kJ/kg and p = 5 MPa, from the superheated steam table lz = 390\u00b0C s2 = 6.63 kJ/kg K by linear interpolation l/12 = Availability of steam after throttling } = (/12 - ho) - To(sz - so) = (3177.2 - 83.96) - 293 (6.63 - 0.2966) = 3093.24 - 1855.69 = 1237.55 kJ/kg Decrease in availability due to throttling = 1/11 - Y,2 = 1263.7 - 1237.55 = 26.15 kJ/kg Now Si= SJs = 6.63 = 1.5301 + X35 (7.127 1 - 1.5301) X3 = ~ =0,91 12 s 5.5970 h3, = 504.7 + 0.9112X2201.9 = 251 J.07 kJ/kg h2 - /13, = 3177.2- 2511.07 = 666.13 kJ/kg h2- h3 = hi5(h1 -h3,) = 0.82 X 666.13 = 546.2 kJ/kg 316~ Basic and Applttd 11tmno,Jy'1tlfflta IJ3 = 2631 kl/", "kg= S04.7 + Xi X 2201.7 2126.3.\u00a53 = 2201.7 = 0.966 S3 = 1.5301 + 0.966 X 5.S97 '\"6.9368 IYJ = Availability of steam at turbine exhaust lr0) - T0(s3 - S0) = (lr3 = {2631 - 83.96) - 293 (6.~368 - 0.2966) = 2S47.04 - 1945.58 = 60 I. 46 kJ/lcg Specific work output from the turbine Ans. The work done is less than the loss of availability of steam. between states 2 = lt2 -h3 = 3177.2-2631 = S46.2 kJfkg and 3, because of the irreversibility accounted for by the isentropic efficiency. Eumple 9.15 A steam turbine receives 600 kg/h of steam at 25 bar, 350\u00b0C. At a certain. stage of the turbine, steam at the rate of 150 kg/his extracted at 3 bar, 200\u00b0C. The remaining steam leaves the!:Urbine at 0.2 bar. 0.92 dry. During the expansion process, there is heat transfer from the turbine to the sur.roundings at the rate of lO kJ/s. Evaluate per kg of steam entering the turbine (a) the availability of steam entering ffld leaving the turbine, (b) the ma,cimwn work, and (c) the irreversibility. The atmosphere is at 30\u00b0C. Solution At 25 bar, 350\u00b0C At 30\u00b0C, At 3 bar, 200\u00b0C At 0.2 bar (0.92 dry) IJI = 3125.87 kJ/kg s 1 = 6.8481 kJfkg K h0 = 125.79 kJ/kg s0 = soo\u2022c = 0.4369 kJ/kg K lr2 = 2865.5 kJ/kg S2 =7.3115 kJfkg K Irr= 251.4 kl/kg hr,= - 2358.3 k:J/kg s, = 0.8320 lcJ/kg K's1 = 7.9085 kJ/kg K h3 = 251.4 + 0.92 X 2358.3 = 2421.04 kJ/kg S3 = 0.8320 + 0.92 >< 7.0765\"' 7.3424 kJ/kg K Th.estates of steam are shown in fig. Ex... 9.15. (a) Availability of steam entering the turbine '1'1 = (lt1 - lt0) - T0 (s1 -so) = (3125.87 - 125.79) - 303 (6.8481 - 0.4369) = 3000.08 - 194,2.60 = 1057.48 kJ/kg Ans. Availability of steam leaving the turbine at state 2, '1'2 = (hz - ho) - To(s2 - so) I I!!I ii I + II PropntitS of PK,1 SalJSlaru:11 -=317 m, =450kg/h 1>3=0.2 ber l'.3 = 0.92 0.25kg Fig.&. 9.15 = (2865.5 - 125.79)-303 (7.3115 -0.4369) = 2739.71 -2083.00 = 656.71 kJ/kg Availability of steam leaving the turbine at state 3, '1'3 = (h3 - lt0) - T0 (s3 -s0) = (2421.04- 125.79)-303 (7.3524 -0.4369) = 199.85 kJ/kg (b) Maximum work per kg of steam entering the turbine Wrcv = Y'1 - '1'2- -'1'3 = 1057,48 -0.25 X 656.71- 0.75 X 199.85 ffl2 - M3 \"'1 = 743.41 kJ/kg m1 Ans. ( c) (m:versibility I= TJ.w2 s2 + w3 s3 -w1 s,) -Q = 303 (ISO x 7.3115 +450>< 7.3424- 600 x 6.8481)-(- lOx 3600) = 303 (1096.73 + 3304.08 - 4108.86) = 36000 = 124,460.85 kJ/h = 124.461 MJ/h = 124 = 207.44 kJ/kg Ans. 103 \u00b7 461 X 600 Ezample", " 9.16 Determine the exergy of (a) 3 kg of water at I bar and 90\u00b0C, (b) 0.2 kg of steam at 4 MP a, 500\u00b0C and ( c) 0.4 kg of wet steam at 0.1 bar and 0.85 quality, (d) 3.kgofice at I bar- 10\u00b0C. Assume a dead state of I bar and 300 K. So/uJion At the dead state of 1 bar, 300 K. u0 = 113.1 k.1/kg, 1,0 = 113.2.kJ/kg K. v0 = O.OOIOOS m3/kg,.r0 = 0.395 k.1/kg, = m[(u + p0v- T0 s)- (u0 + p0v0 - T0 s0)1 Exergy of the system: Now, uo + Po'Oo - Toso= ho- Toso = 113.2 - 300 x 0.395 =-5.3 k.1/kg I I 1 11! 11 11+ 318=- Bast, and Applied TlurmodJMmio (a) For water at I bar, 90\u00b0C Since U = 376.9 k.Jfkg, Ji= 377 kJfkg, V = 0.001035 m3fkg s = l.193 kl/kg K. p=po, u +p0v-To5 =u+pv-T0 s =lr-T0 :r \"'377 - 300 X l.193 = 19.J kJ/kg Hence, ; \"'3(19.1 - (- 5.3)] = 3 X 24.4 (b) At p = 4 MPa, t = S00\u00b0C = 73.2 kJ II= 3099.8, h = 3446.3 kl/ltg, V\"' 0.08637 m3Jltg s = 7.090 kJ/kgK u + p0v- To5 = 3099.8 + 100 x 0.08637- 300 x 7.090 = 981.4 kJ/kg, = 0.2 [981.4 -(-5.3)J\"' 197.34 kJ (c) At 0.1 bar, 0.85 quality, Ans. Ans. U = 192 + 0.85 X 2245 = 2100.25 kJ/kg h = 192 + 0.85 X 2392\"' 2225.2 kl/kg s = 0.649 + 0.85 x 7.499 = 7.023 kJlltg K V = 0.001010 + 0.85 X )4.67 = 12.47 m3Jltg u + p0v-T0 f = 2100.25 + 100 x 12.47 - 300 x 7.023 = 1240.4 kJ/kg, = 0.4( 124-0.4 - (- 5.3 )) = 498.3 kJ Ans. (d) Since p = p0, At 100 kPa, - 10\u00b0C, I = U - U0 + pJ.,Y - Vo) - T0{S - S0). ~ H - H0 - V(p - p0) - T0(S - So) \u00b7 = m((h - 1,0) - T0(s - s0)] h = - 354.1 kJlltg ands= - 1.298 kJlltg K ~-= 3[- 354. I - 113.2 - 300 (- 1.298 - 0.0395)) = 81.2 kJ. Ans. Example 9.17 A flow of hot water at 90\u00b0C is used to heat relatively cold water at 25\u00b0C to a temperature of 50\u00b0C in~ heat exchanger. The cold water flows at the rate of l kg/s. When the heat exchanger is operated in the parallel mode, the exit temperature of the hot water streau: must not be less than 60\u00b0C. In the counterflow operatfon, the exit temperature of hot water can be as low as 35\u00b0C. Compare the second law efficiency and the rate of exergy destruction in the two modes of operation. Take 70 = 300 K. Solution Given: th 1 = 90\u00b0C, 'ci = 25\u00b0C, tc1 = 60\u00b0C, 111 0 = I kg/s, T0 = 300 K.'\" Properties of Pllre Substances -=319 The two modes of operation of ( a) parallel flow and (b) counterflow are shown in Fig. Ex. 9.17. - ~Ao (8) - Ao {b) Fig. Ex. 9.17 In parallel flow mode (a", "), th 2 = 60\u00b0C. Neglecting any heat loss, \"'h ch(th1 -th1) = m, c.<tci - 'c,>,nh (90- 60) = 1c50 - 25) ti,h = 0.833 kg/s t112 = 3S\u00b0C, In counterflow mode, ti1h (90 - 35) = I (50 - 25) mb = ~ = 0.454 kg/s 55 Thus, the counterllow arrangement uses significantly less hot water. Assuming that the hot water stream on exit from the heat exchanger is simply dumped into the drain, the exergy flow rate of the hot water stream at entry is \u00b7 considered as the exergy input rate to the process. an = mh((h 1 - h0) - T0(s1 - s0)] At 300 Kor 27\u00b0C, h0 = 113.2 kJ/kg and s0 = 0.395 kJ/kg K At 90\u00b0C, \", = 376.92 kJ/kg, s, = I.1925 kJ/kgK or,= 0.833 [(376.92 - 113.2) -300 (l.1925 -0.395)) = 0.833 (263.72 - 239.25) = 20.38 kW Parallel flow: At60\u00b0c, At 2s0c, At so\u2022c, Rate of exergy gain: 112 = 25 I.I 3 kJ/kg, h3 = 104.89 kJ/kg, 1,4 = 209.33 kJ/kg, S2 =- 0.8312 kJ/kg K S3 = 0.3674 kJ/kg K S4 = 0.7038 kJ/kg I( 320=- = 1[(209.33 - 104.89)- 300(0.7038 - 0.3674)) = 104.44 - 100.92 = 3.52 kW ( 11n) = 3.Sl = 0.172 or 17.2% P 20.38 Rate of exergy loss by hot water:.. m11[(/i 1 - hi)- T0(s1 - si)) = 0.833 [(376.92 - 2Sl.13) - 300(1.1925 -0.8312)) = 0.833 (125.79 - 108.39) = 14.494 kW Rate of irreversibility or exergy destruction: = 14.494- 3.52 = 10.974 kW If the hot water stnam is not dumped to the drain, '1 =...l:E... = 0.243 or 24.3% ll.l' 14.494 Ans. Counterflow: At 35\u00b0C, lr2 - 146.68 kJ/kg, s2 = 0.5053 lcJ/kg K Rate of exergy gain of cold water= mb [(/r4 -/i3)-T0(s4 -,3)) = 3.52 kW (same u in parallel flow) Rate of exergy input (if exiting hot water is dumped to the swroundi.ngs) \"'0.454 (263.72 - 239.25) = 11.11 kW 17 = 3 U,C 52 \u00b7 11.IJ = 0.3168 or 31.68% Rate of exergy loss of hot water: = mb((/i1 - /ii)- T0{s1 -sz)) - 0.454 [(376.92 - 146.68)- 300(1.1925 -O.SOS3)) = 0.454 (230.24- 206.16) = 10.94 kW 7firc= \u00b7 Rate of irreversibility or exergy destruction: 3 52 \u00b7 10.94 =0.3217or32.17% Ans. = 10.94- 3.52 = 7.42 kW Ans. The second law efficiency for the counterflow\u00b7 arrangement is significantly higher and the rate of irreversibility is substantially lower compared to the pa.rullel flow arrangement. Example 9.18 A small geothennal well in. a remote desert. area produces SO kg/h of saturated steam vapour at I S0\u00b0C. The envirorunent temperature is 45\u00b0C. This geothennal steam will be suitably used to produce cooling for homes at 23\u00b0C. The steam will emerge from this system as saturated liquid at I atm. Estimate the maximum cooling rate that could be provided by such a system. I I!I!! I Solution The energy balance of the control volume as shown in Fig. Ex. 9.18 gives: -=321 The entropy balance is: Sl\"D = [ ~: + W S1 ] - [ f + ~ S1 ] where Tis the", " 2) The steam initially at 10 bar, 40% quality is heated at constant volume until the pressure rises to 35 bar; (2-3). It is then expanded isothermally to 10 bar: (3-1) Jt is finally cooled. at constant pressure bac.k to its initial staie. Sketch the cycle on T-s coordinates, and calculate the work done, the heat transferred. and the change of entropy for each of the three processes. What is the thennal efficiency of the cycle?.4,u. O; 1364 kJ; 2.781 kJ/K, 367.5 kJ; 404.6 kJ; 0.639 kJ/K; - 209.1 kJ; - 1611 kJ; - 3.419 kJ/K 8.93% 9.26 Detennine the excrgy per unit mass for the steady flow of each of the following: (al steam at 1.5 MPa, soo\u2022c (b) air at 1.5 MPa, 500\u00b0C (c) water at 4 MPa. 300 K (d) air at 4 MPa, 300 K (e) air at 1.5 MPa, 300K A11s. (a) 1220 kJ/kg, (b) 424 kJ/kg, (c) 3.85 kJ/kg; (d) 316 kJ/kg. (el 232 kJ/lcg 9.27 A liquid (cp = 6 kJ/lcg K) is heated ai an appro:i1imately constant pressure from 298 K to 90\u00b0C by passing it through tubes immersea in a furnace. The mass now rate is 0.2 kgls. Detennine (a) the heating load in kW. (b) the exergy production rate in kW corresponding to the temperature rise of the fluid. Ans. (a) 78 kW, (b) 7.44 kW 9.28 A Oow of hot water at 80\u00b0C is used to heat cold water from 20\u00b0C to 4S\u00b0C in a heat exchanger. The cold water flows at the rate of 2 kg/s. When operated in parallel mode. the exit temperature of hot water stream cannot be less than 55\u00b0C, while. in the counterflow mode. it can be as low as 30\u00b0C. Assuming the,11' ii. Pro/mties of P,,,t Substancrs -=327 surroundings are at 300 K. compare the second law efficiencies for the two modes of operation. 9.29 Water at 90\u00b0C is nowing in a pipe. The pressure of the water is 3 bar, the mass flow rate is IO kg/s. the velocity is 0.5 mis and the elevation of the pipe is 200 m above the exit plane of the pipeline (ground level). Compute (a) the thermal exergy flux. (b) die pressure exergy flu,t, (c) the exergy flux from KE, (d) the exergy nux from PE, (e) total exergy flux of the stream. Ans, (a) 260 kW. (b)2.07kW, (c) 1.25 x 10- l kW, (d) 19.6 kW, (c) 282 kW 9.30 A cylinder fitted with a piston contains 2 kg steam at 500 kPa, 400\u00b0C. Find the entropy change and the work done when the steam expands to a final pressure of 200 kPa in each of the following ways: (a) adiabatically and reversibly, (b) adiabatically and irreversibly to an equilibrium temperature of300\u00b0C. Am. (a) O. 386.7 kJ, (b) 0.1976 kJ/K. 309.4 kJ 9.31 Steam expands isentropically in a nozzle from I MPa, 250\u00b0C to IO kPa. The steam flow rate is l kg/s. Neglecting the KE of steam at inlet to the nozzle. find the velocity of steam at exit from the nozzle and the exit area of the nozzle. Ans. 1223 mis, 100 cm2 9.32 Hot helium gas at 800\u00b0C is supplied 10 a steam generator and. is cooled to 450\u00b0C while serving as a heat source for the generation of steam. Water enters the steam generator at 200 bar, 250\u00b0C and leaves as superheated steam at 200 bar. 500\u00b0C. The temperature of the surroundings is 27\u00b0C. For I kg helium. detem1ine (a) the maximum work that could be produced by the heat removed from belium, (b) the mass of steam generated per kg of helium, (c) lbeaclual work done in the steam cycle per kg of helium. {d) the net change for", " by Tds = du + pdr, or ds= du +E.dv T T The internal energy u is assumed to be a functi.on of Tand v, i.e. (10.8) or From Eqs (10.8) and (10.9) u =f(T, v) du=(!; )v dT+(!: t dv ds = 1.(E!.) dT + 1.[( du) + p]dv r atlT rarv Again, let s-f(T, v) ds = (~) dT +(!!.) dv av T ar V Comparing Eqs ( 10.10) and (10.11) (10.9) (10.10) (10.11) (J 0.12) I I I! II I Pr11pertin of Gam and Gas MiJtturts -=333 Differentiating Eq. (10.12) with respect to v when Tis constant 1 a2u d2s dTdv =r aTdv Differentiating Eq. (10.13) with respect to Twhen v is constant d2 s at1aT T avdT + T dT \" T2 dv 1 I d2 u I (dp) I (d\") _..f!_ fl (10.13) (10.14) (10.IS) From Eqs {10.14) and (10.15) I d2 r ar.av r av\u00b7dT (au) +p=r(dp) ar V av T u _ I d2u + l ( dp ) r ar V.i or For an ideal gas [JfJ=RT v( dp) =R dT V (!~ )v.. = =; From Eqs (10.16) and (10.17) (du) =O dV T Therefore, u does not change when v changes at constant temperature. Similarly, if u.. f(T, p), it can be shown that ( : ; t = 0. Therefore, u does not change with p either, when T remains constant. u does not change unless T changes. Then only for an ideal gas. This is known asJ011le'slaw. If u = /(T, t1) u == I (T) (10.19) di.I = ( du ) dT + ( du ) dv dT v Since the last tenn is zero by Eq. (I 0.18), and by definition dv T!,, \"'l (du) r ;n, 1 p r 2 (10.16) (10.17) (10.18) 33\u2022=- Basic alld Applied Tltmnodynami,s c \"\"(dll) dT, V \u2022 (10.20) The equation du \"' Cv dT holds good for an ideal gas for any process, whereas du= c.,d.T for any other substance it is true for a constant volume process only. Since c. is constant for an ideal gas, llu = cv llT The enthalpy of any substance is given by For an ideal gas Therefore \u00b7 only for an ideal gas Now Since R is a constant h=u+pv II= u+RT \"=f(n d/1 =du+ RdT 6/1=6u+R6T \"'Cv llT + RllT =(cv + R)i1T Since h is a function of T only, and by definition C =(~) dT p dh =cpdT llJt = cp t.T From Eqs (10.22) and (10.23) P or (10.21) (10.22) (10.23) (10.24) Cp =cv+ R c, - Cv = R or (10.25) The Eq. d/1 00 cP dTh.olds good for an ideal gas, even when pressure changes, but for any other substance, this is true only for a constant pressure chan.ge. The ratio of c pie, is of importance in ideal gas computations, and is designated by the symbol y. i.e. or From Eq. (10.25) (y- l)c.=R \u2022 I \u2022\u2022, 11! I!I I and Propnties of Gases and Gas Miirtura Cv = r ~ I I kJ/kg K C = --1..!i_ p r-1 -=335 {10.26) If R = i issubstituted in Eq. (10.26). \u00b5 P P and Cv = l'Cv = (cy)mo1 = R 1 :", ".~ in m1 + m2 + \u00b7 \u00b7 \u00b7 + fflc (10.100) Toe gas constant of lhe mixture is thus the weighted mean, on a mass basis, of the gas constants of the components. Toe total DWl8 of gas mixture m is m =m1 +... +m. If\u00b5 denotes the equivalent molecular weight of the mixture having n tolal number of moles. 0[ n\u00b5 = n1 \u00b5 1 + n2\u00b5 2 +... + n.,~ /J \"'X1/J1 + X2JJ.l + \u2022\u2022\u2022 + XJJc /J \"'IX'l:JlK. (10.101) A quantity called the partial volume of a component of a mixture is the volume that the component alone would occupy at the pressure and temperature of the mixture. Designating the partial volumes by V1, V2, etc. pV1 = m1R1T,pl'2 = m2R2T,..., pJlc = mcR.T p(V1 + Jl2 +... +Ye)~ (m1R1 + mzR.2 +... + mtft.)T or From Eqs (10.98), (10.99) and (10.102) JI= 1'1 + Y2 +... + Ve Toe total volume is thus equal to the sum oflhe partial volumes. The specific volume of the mixture, v, is given by or JI v= -= - - - - - m m1 + m2 + \u00b7\u00b7 \u00b7 + me JI.l = \"'1 + \"'2 + \u00b7 \u00b7 \u00b7 + ll'lc t1 V (10.102) (10.103) 1, 11!,!1, Propn-ties of Ga= and Gas Mixt11rts -=363 or l.:..l.+....L+\u00b7\u00b7\u00b7+..l. ll2 V tie V1 (10.104) where v1, v2, \u2022\u2022\u2022 denote specific volumes of the components. each component occupying the total volume. Therefore, the density of the mixture p = P1 + P2 +... + Pc (10.105) 10.10 Internal Ene.rgy, Enthalpy and Specific Heats of Gas Mixtures When gases at equal pressures and temperatures are mixed adiabatically without work, as by inter-diffusion in a constant volume container, the first law requires that the internal energy of the gaseous system remains constant, and experiments show that the temperature remains constant. Hence the internal energy of a.mixture of gases is equal to the sum of the internal energies of the individual components, each taken at the temperature and volume of the mixture (i.e. sum of the 'partial' internal energies). This is also true for any of the thennodynamic propenies likeH, C.,. Cp, S, F and G and is known as Gibbs theorem. Therefore, on a mass basis \u00b7 mum= m 1u 1 + m2u2 +... + mcllc ''n. \"' m,111 + m2u2 +... + m.u. m1+m2 +\u00b7\u00b7\u00b7+ me ( I 0.106) which is the average specific internal energy of the mixture. Similnrly, the total enthalpy of a gas mixture is the sum of the 'partial' enthalpies an \u00b7 d mh,,. = m1h1 + miJ12 +... + mJrc,. = m1h1 +m1h2 +\u00b7\u00b7\u00b7+mch\u2022 n.,. m1+m2 +\u00b7\u00b7\u00b7+m0 From the definitions of specific heats, it follows that m1Cv1 + m2Cv2 + \u00b7 \u00b7 \u00b7 + fflcCv c m1 + m2 + \u00b7 \u00b7 \u00b7 + me C = \"\" c = m1CP1 + m2cr2 +... + mcepc pm m1 + m2 +... +me and (10.107) (10.108) (10.109) 10.11 Entropy of Gas Mixtures Gibbs theorem states that lhe total cn\\ropy of a mixture of gases is the sum of the panial entropies. The partial entropy of one of the gases of a mixture is the entropy 1,1 It 364=- Basic and Applied Tllmnodynamit:S that the gas would have if it occupied the whole volume alone at the same temperature. Let. us imaginge a number of inert ideal gases separated from one another by suitable partitions, all the gases being at the same temperature T and pressure p. The total entropy (initial) Si= n1s1 + n,s2 +... + \"e'c = Ina:.s.,: From property relation Tds = d", "umeof0.6 m1 \u2022 lf the air expands according to ihe law. pt\u2022\" z con slant, between the same end states, calculate the heat given 10, or extracted from, the air during the expansion, and show that it is approximately equal to tbe cbanie of entropy multiplied by the mcWl absolute temper.a~. I 0.15 0.5 kg of air, initially at 25\u00b0C, is h.eated reversibly at constant pressure Wllil the volume is doubled, and is ihen heated reversibly at consiant volume until the pressure is doubled. For the total path, find the work transfer, the heat transfer, and the change of entropy. 10. I 6 An ideal gas cycle of three processes uses Argon (Mot. wt. 40) as a working substance. Process 1-2 is a reversible adiabatic expansion from 0.014 m3, 700 kPa, 280\u00b0C to O.OS6 m3\u2022 Process 2-3 is a reversible isothennal process. Process 3-1 is a constant pressure process in which beat transfer is zero. Sketch lhe cycle in thep-v and T- s planes. and find (a) the work iransfcr in process 1- 2, (b) the work transfer in process 2- 3, and (c) the net work of the cycle. Take y# 1.67. Ans. (a} 8.85 kJ (b) 8.96 lcJ (c) 5.82 kJ 10.17 A ga.~ occupies 0.024 m3 at 700 kPa and 95\u00b0C. lt is expanded in the non-now process according to the law piP ~ constant to a press= of 70 kPa after which it is he.ited at. constant pressure back to its original temperature. Sketch the process on the p-v and T- s diagrams, and calculate for the whole process the work done, the heat transferrcd, and the change of entropy. Take CP = 1.04 7 and c,, = 0.775 kJ/kg K for the gas. 10.18 0.5 kg of air at 600 kPa receives an addition of heat at con&'ta.nt volume so that its temperature rises from I 1o\u2022c to 650\u00b0C. II then expands in a cyclinder polytropiecally to its original temperature and the index of expansion is 1.32. Finally, it is compressed iso1hermaly to its original volwne. Calculate (a) the change of entropy during each of the three stag~-s, (b) the pressures at lhe end of constant volume heat addjtion and at the end of e.xpao.sioo. Sketch tbe processes on the p- v and T- s diagrams. I 0.19, 0.S kg of helium and O.S kg of nitrogen are mixed at 20\u00b0C and ata total pressure of 100 kPa, Fi.nd (a) the volume of the mixture, (b) the panial volumes of the components, (c) thepanial pressures of the components. (d) the mole fractions of the components, (e) the specific beats cP and cv of the mixture, and (0 the gas constant of the mixture. Ans. (a) 3.481 m3 (b) 3.045, 0.436 m3 (c) 87.S, 12.S kPa (dJ 0.875, 0. 125 (el 3.11, 1.921 lcJ!k (f)J p!9 kJ/kgX. Propmits of GOJts atld G.u Mix/um -=387 I 0.. 20 A gaseous mixture consists of I kg of oxygen and 2 kg of ni1rogen at a pressure of ISO k.Pa and a temperature of20\u00b0C. Detennine the changes in intcmal energy. enthalpy and entropy of the mixture when the mixture is heated to a temperature of I 00\u00b0C (a) at constant volume. and (b) at constant pressure. 10.21 A closed rigid cyclinder is divided by a diaphragm into rwo equal compartments, each of volume OJ m3. Each compartment contains air at a temperature of20\u00b0C. The pressure i.n one compartment is 2.5 MPa and in the other compartment is I MPa. The diaphragm i.s ruptured so that the air in both the compartments mixes IO bring the pressure to a unifonn value throughout the cylinder which is insulated. Find the ne1 change of entropy for tbe mixing process. 10.22 A vessel is divided into three compartments (a), (b), and (c) by two partitions. Pan (a) contains", " compress the ai.r adiabatically from the same initial state 10 the same final pressure 31ld the isenuopic efficiency 9fthc compressor. Ans. - 22.S kJ/kg, - 190 kJ/q. 84.4% Ill\" ii. 390=- Basie and Applied T/tmr,odjna,,,ics 10.38 A slow-speed reciprocating air compressor with a water jacket for cooling approximates a quasi-static compression process following a path pvl.l = const. If air enters at a temperature of20\u00b0C and a pressure of I bar, and is compressed to 6 bar at a rate of 1000 kg/h, determine the discharge temperature of air, the power required and the heat transferred per kg. Ans. 443 K. 51.82 kW, 36 kJ/kg I 0.39 A single-acting two-stage reciprocating air compressor with complete intercooling delivers 6 kg/min ai lS bar pressure. Assume an iniake-condition of I bar and l 5\u00b0C and that the compression and expansion processes are (?.Olytropic with 11 ~ 1.3. Calculate: (a) the power required. (b) tho isothermal efficiency. Ans. (a) 26.1 S kW (b) 8S.6% I 0.40 A two-stage air compressor receives 0.238.r.r? Is of air at I bar and 27\u00b0C and discharges \u00b7it at IO bar. The polytropic index of compression is 1.35. Determine (a) ihe mini.mum power necessary for compression. (b) the power needed for single-siage compression to the same pressure, (c) the maximum temperature for (a) and (b), and (d) the heat removed in the intcrcooler. Ans. (a) 63.8 kW, (b) 74.9 kW, (c) 404.2 K, 544.9 K. (d) 28.9 kW I0.41 A mass of an ideal gas exists initially at a pressure of 200 kPa, temperature 300 K. and specific volume 0.5 m3/kg. The value of yis 1.4. (it) Dctennine the specific heats of the gas. (b) What is the change in entropy when the gas is expanded to pressure 100 k.Pa accordinf. to the law ptF 3 = const1 (c) What will be the entropy change if the path is pv1 \u2022 = const. (by the application of a cooling jacket during the process)? (d) What is the inference you can draw from this example? Ans. (a) 1.166, 0.833 kJ/kg K, (b) 0.044 kJ/kg K (c) - 0.039 kJ/kg K (d) Entropy increases when 11 < rand decreases when n > r 10.42 (a) A closed system of 2 kg of air lnitially at pressure 5 attn and temperatun:. 227\u00b0C, cKpands reversibly to pressure 2 atm following the law pv1.2s \"' const. Assuming air as an ideal gas, determine the work done and the heat transferred. Ans. 193 kJ, 72 kJ (b) If the sys&em does lhe.sa.nie apa.nsion in a \u00a7tea.dy flow process, what is the work done by the system'? Ans. 24 l kJ I 0.43 Air contained in a cylinder fitted with a piston is compressed rever.;ibly acc-0rding to the Jaw pv1.2s = const. The mass of air in the cylinder is 0, t kg. The initial pressure is I 00 kPa and the initial temperal\\lfC 20\u00b0C. The final volume is I /8 of the initial volume. Determine the work and tbe heat transfer. AIU. - 22.9.kJ, - 8. 7 kJ t 0.44 Air is contaiocd in a cylinder fitted wiih a frictionless piston. Initially the cylinder contains O.S ml of air at 1.S bar, 20\u00b0C. the air is then compressed reversibly according to the law pv\" = constant until the final pressure is 6 bar, at which point the temperature is l20\u00b0C. Determine: (a) the polytropic index n. (b) the final volume of air, (c) the work done on the air and the heat transfer. and (d) the net change in entropy J 0.45 The specific heat at constant pressure for air is given br Ans. (a) l.2685, {b) 0.1676 m) (c) - 95.3 kJ", "kg.K respectively. The above gas mixture is lhen reversibly compressed to a pressure of 6 bar 2 = constant, (f) Detennine the work and heat according to the law p11 interactions in magui'tude and sense, and (g) the change in entrol'Y. Ans. (a) 22.8S82 (b) 0.3637 kJ/kgK (c) PHi = 0.S7J4,J'N = 1.1428, Pco.: 0.2858 bar (d) 1.384 (e) 0.3476 kJ/kgK (t) - 70.455 kl, - 33.772 kJ (g) -0.1063 kJIK. 10.66 A four cylinder single-stage air comp.ressor has a bore of 200 mm and a stroke of 300 mm and runs at 400 rpm. Al a working pressure of 721.3 kPa it delivers 3.1 m3 of air per min at 270\u00b0C. Calculate (a) the mas.~ flow rate, (b) the m,e air delivery (FAD) (c) eJfective swept volume, ( d) volumetric efficiency. Tllke lhe inlet condition as that of the free air at IO I.3 kPa. 21 \u2022c. \u2022 1 \u00b7 1 Ans. (,l) 0.239 kg.ls (b) 0.199 m3/s (c) 0.299 ml, (d) 79.2% 10.67 A single stage reciprocating air compressor bas a swept volume of 2000 cm3 and l1lilS at 800 rpm. lt operates on a pressure ratio of 8, with a clearance of.So/o of the swept volume. Assume NTP room conditions, and at inlet, and polytropic compression and e:itp:insion with n - 1.25. Calculate (a) the indlcatcd power, (b) volumetric efficiency, (c) mass flow rate, (d) the free air delivery FAD, (e) isothennal efficiency, (I) actual power needed to drive the compressor, if the mechanical efficiency is 0.85. Ans. {a) S.47 kW (b) 78.6% (c) 1.54 kg/min (d) 1.26 m3/min, (e) 80.7%. ( f) 6.44 kW I 0.<>8 A two-stage single-act.ing reciprocating compressor 1akes in air at the rate of 0.2 m3/s. Intake pressure and temperature arc 0.1 MPa and J6\u00b0C respectively. The air is compressed to a ftnal pressure of0.7 MPa. The intcm1ediaie pressure is ideal. and intertooling is perfect. The compression indell is 1.25 and the compressor mn.~ al 10 rps. Neglecting clearance, determine (a) the intermediate pressure, (b) the touil volwneof each cylinder, (c) the power required to drive the compressor,. ( d) the rate of heat absorption in the intcrcoolcr. Ans. (a) 0.264 MPa (b) 0.0076 m3 h.p. and 0.02 m3 l.p cylinder (c) 42.8 kW (d) 14.95 kW 10.6.9 A 3-stage single-acting air compressor running in an abnosphcrc at 1.013 bar and IS\u00b0C has a free air delivery of 2.83 m3/min. The suction pressure and temperature arc 0.98 bar and 32\u00b0C n:spectively. The delivery pressure is to be 72 bar. Calculate the indicate power required, assuming complete lntereooling, n = 1.3 and that ihe compressor is designed for minimwn worL. What wiU be the heat loss to the intercoole~? Ans. 25.568 kW, 13.78 kW Iii I, Proptrlia of Casts arrd Gas Mixtures -=395 10.70 Predict the pressure of nitrogen gas at T = I 7S Kand v = 0.00375 m31kg on the basis of (a) the ideal gas equation of siate, (b) the van der Waals equation of state, (c) the Beattie-Bridgeman equation of state and (d) the Benedict-Webb-Rubin equation of stale. Compare the values obtained with the experimentally detennined value oft 0,000 kPa. Ans. (a) 13,860 kPa (b) 9468 kPa (c) 1.0,110 kPa (d) 10,000 kPa I 0. 71 The pressure", " t1 Q is the heat interaction involved, then the entropy change dSA=....!Q...,dS9=-__!g_ T-dT T+dT dS=dSA+dSe=d'4_T+dT-T-dT =--;,r\u00b7C'JQ,[ I I ] 2\u00b7dT So there is a decrease in entropy for the isolated system of A and B together. It is thus clear that the variation in temperature dT cannot take place. Th.e system, therefore, exists in a stable equilibrium condition. Perturbation of the state produces an absurd situation and the system must revert to the original stable state. It may be observed: If for all the possible variations in state of the isolated system, there is a negative change in entropy, then the system is in stable equilibrium. (dS)u,v > 0 (dS)u,v = 0 ( dS)u,v < 0 Spontaneous change Equilibrium Criterion of slability (11.47) Similarly (dG}p,T < 0, (dF)r,v < 0 (dG)p,T = 0, (dfh.v = 0 (dG)p,T > 0, (df.'h_v > 0 Spontaneous change Equilibrium Criterion of stability (l 1.48) A system is in a state of stable equilibrium if, for any finite variation of the system at constant T and p, G increases, i.e. the stable equilibrium state corresponds to the minimum value of G. A system is said to be in a state of neu1ral equilibrium when the thermodynamic criterion of equilibrium (G, F, S, U or H) remains at constant value for all possible variations of finite magnitnde. If perturbed, the s.ystem does not revert to the original state. For a system at constant T and p, the criterion of neutral equilibrium is 6GT,p = 0 Similarly SFy,v = 0, SH5, P = 0, SUs.v = 0, SSu,v = o A system is in a state of unstable equilibrium when the thennodynamic criterion is neither an extremum nor a cons1ant value for all possible variations in the system. If ihe system is in unstable equilibrium, there will be a spontaneous \u00b7 change accompanied by '\u00b7 \u2022 6Gr,p < 0, 6Fr.v < 0, 6U5,v < 0, 6Hs,p < 0, 6Su,v > 0 A system is in a state of metustuble equilibrium if it is stable to small but not to large disturbances. A mixture of oxygen and hydrogen is in a metastable II I equilibrium. A little sparlc may st.art a chemical reaction. Such a mixture is not in its most stable state, even though in the absence of a spark it appears to be stable. Figure 11.14 shows different types of equilibrium together with their mechanical analogies. Shas been used as the criterion for equilibrium. Si.bit Unstable Composition Metastable Fig. 11.14 Types of e9uilibrium 11.15 Local Equilibrium Conditions Let an arbitrary division of an isolated system be considered, such that Then for equilibrium, it must satisfy the condition S=S1 +S,, U= U1 + Ui (<SSJ..i.v = 0 to first order in small displacements (otherwise {jS could be made positive because of higher o.rder terms). Now to the first order in a very small change TdS=dU+pdY {j,S= ( :i, t 6U1 + ( :i2 t 6U2 + ( :~ l SY,+ ( :~ l 6Y2 (!it= ~.(;it=~ s1s-.l.su, +..l.su2 + lisr, + P2 6Y2 7j 6U1 =- 6U2 and 6V1 =- 6V2 6 1S = (.1. -..l.) 6 U1 + (l!J_ -11) 8 V1 + Second order tenns 7i 7i 1i 1i T2 7j Now Again When i; ;;,s = 0, at equilibn,im T1 = Tz, Pi =pz I I ti!! I TltmnodyMmic Rtlations, bfuilil,rium and Tltird uzw 11.16 Conditions of Stability At equilibrium, S = Smu., F = F min\u2022 G = Gmin\u2022 and SS = 0, SF= O; SG = 0; these are necessary but not sufficient conditions for equilibrium. To prove that Sis a maximum, and G or Fa minimum, it must satisfy S2S< 0, S2F", " the quality I! I! I Vapour PrJwtr Cyelts -='69 Q, l l (8) -(b) ---- s (c) Fig. 12.1' &heal cycle of steam at turbine exhaust within a pe.m1issible limit. The quality improves from X4\u20225 to x4, by the use of reheat. By increasing the number of reheats, still higher steam pressures could be used, but the mechanical stresses increase at a higher proportion than the increase in pressure, because of the prevailing high temperature. The cost and fabrication difficulties wilt also increase. In that way, the maximum steam pressure gets fixed, and more than two reheats have not yet been used so far. In Fig. 12.14, only ideal processes have been considered. The irreversibilities in the expansion and compression processes have been considered in the example given later. 12.7 Ideal Regenerative Cycle In order to increase the mean temperature of heat addition (Tm 1), aitention was so far confined to increasing the amouut of beat supplied at high temperatures, such as increasing superheat, using higher pressure and temperature of steam, and using reheat. The mean temperature of beat addition can also be increased by decreasiog the amount of heat added at low temperatures. rn a saturated steam Rankine cycle (Fig. 12.15}, a considerable part of the total heat supplied is in the liqu1d phase when heating up water from 4 to 4', at a temperature lower than T1, the maximum temperature of the cycle. For maximum efficiency, all heat should \u00b7 be supplied at T1, and fccdwater should enter the boiler at state 4'. This may be accomplished in what is known as an ideal regenerative cycle, the flow diagram l'.>fwhich is shown in Fig. 12.16 and the corresponding.T'-s diagram in Fig. 12.17. T t -. s Fig. 12.15 Simplt Rankine Cycle The unique feature of the ideal regenerative cycle is that the condensate, after leaving the pump circulates around the turbine casing, counterflow to the direction of vapour flow iu the turbine (Fig. 12. 16). Thus. it is possible to transfer beat from the vapour as it flows through the turbine to the liquid flowing around the turbine. Let us assume that this is a reversible heat transfer, i.e., at each point the temperature of the vapour is only i'nfinitesimally higher thau the temperature of the liquid. The process 1-2' (Fig. 12.17) thus represents reversible expansion of steam in the turbine with reversible heat rejection. For any small step in the process of heating the water, \u00b7.., Vapour Power Cycles -=471 Fig. 12.16 ltftal rtgtntratiDt cyclt\u00b7hasi, schtmt... l - -~s Fig. 12.17 ltual rtgruratiDt cycle an T\u00b7s pltit and 1:J.T(water) = -1:J.T (steam) /ls (water) = -1:J..s (steam) Then the slopes of lines 1-2' and 4'-3 (Fig. 12.17) will be identical at every temperature and the lines will be identical in contour. Areas 4-4' -b-a-4 and 2'-1-d-c-2' are not only equal but congruous. Therefore, all the heat added from an external source (Q1} is at the constant temperature T1, and all the heat rejected (Q2} is at the constant temperature T2, both being reversible. Then Q, = Ir, - lr4\u2022 = T1(S1 - S4\u2022) Q2 = lr2\u2022 - lr3 = T2(S2\u2022 - S3) Since,12=-. &sic a,ul ApplieJ 17imnodynami,:s or.,, -s., =s2\u2022 -s3 tJ=l-~=J- :i; Qi 1j The efficiency oflhe. ideal regenerative cycle is thus equal to the Carnot cycle efficiency. Writing lhe steady flow energy equation for lhe twbine or,., - W'r - llr + \"\u2022 - \"\u2022\u00b7 = O Wr=(h1 -Jir)-(h4,-IIJ The, pump work remains the same as in the Rankine cycle, i.e. (12.17) Wp=114 -113 The net work output of the ideal regenerative cycle is lb.us less, and hence its steam rate wiU be more, although it is more efficient, when compared with the Rankine cycle. However, the cycle is not practicable for the following reasons: (a)", "h7 \"'l = (I - m1) Ii-, - h6 h, - h6 (12.24) (12.25) From Eqs (12.24) and (12.25), m I and m2 can be evaluated. Equations (12.24) atid ( 12.25) (811 also be written altemati vely as 118 ) = m1 (h2 - (1 - m1 )(h9 - 119) (I - m1 - mv (h, -h6) = m2 (h3 - 111) Energy gain of fe<:dwater.. Energy given off by vapo~ in condensation Heaters have been assumed to be adequately insulated, end th.ere is no beat ga.in from, or heat loss to, the sunoundings., lol I '\"'\" Vapour~ Cycle$ -=475 Path 1-2-3-4 in Fig. 12.19 represents the states of a decreasing mass of. fluid. For 1 kg of steam, the states would be represented by the path l-2'-3'-4'. From equation ( l 2.18 ), Wr = (h 1 - h2) + (I - m 1)(h2 - h,) + (l -m1 - m2}(h3 - h4) where = (Ji, - h2) + (h2, - h3,) + (h3' - h4,) (12.26) C 12.27) (12.28) The cycle l-2-2'-3'-3\"-4'-S-6-7-8-9-10-1 represents 1 kg ofworlcing fluid. The heat released by steam condensing from 2 to 2' ntilized in heating up the water from 8 to 9. (l - m1) (.h2 - h3) = 1 Cli2, - h3,) ( l - m1 - m2) (h) - h 4) = I (hr - h4,) Similarly From equations (12.26). (12.29) and (12.30) l(h3\u2022 - Jr3.. ) = l(h7 - h6) (12.29) (12.30) Wr = (h 1 - h4,) - (hi - h2,} - h3, - hr) = (h 1 - h4,) - (h9 - h8) - (h7 - h6) (12.31) The similarity of equations (12.17) and (12.31) can be noticed. It is seen that the stepped cycle 1- 2'- 3' - 4'- 5- 6- 7- 8- 9- J O approximates the ideal regenera\u00b7 tive cycle in Fig. 1'2.17, and that a greater number of stages would give a closer approximation (Fig. 12.20). Thus the heating of feedwater by steam 'bled' from the turbine, known as regeneration, carnotizes the Rankine cycle.._ i Fig. 12.20 Rlgninativ, gel, will,. 11UJny sys af ftedwater lteating - s The heatrejectedQ2 in the cycle decreases from(h4 - h5) to(l,4, - l,5). There is also loss in work output by the amount (Area under 2- 2' + Arca under 3 '- 3\" Area under 4 - 4'), as shown by the hatched area in Fig. 12.19(b). So the steam \u00b7\u00b7 rate increases by regeneration, i.e.. more steam has to circulate per hour to produce unit shaft output. Tbe enthalpy-entropy diagram of a regenerative cycle is shown in Fig. 12.21. /JaJi\u00a3 and Applied Tlwmwdynamics <C t ---~, Fig. 12.21 Rtgtntrati'Df: cyclt 9n h\u00b7s diocram 12.9 Reheat-Regenerative Cycle The reheating of steam is adopted when the vaporization pressure is high. The effect of reheat alone on the thermal efficiency of the cycle is very small. Regeneration or the heating up of feedwater by steam extracted from the turbine has a marked effect on cycle efficiency. A modem steam power plant is equipped with both. Figures 12.22 and 12.23 give the flow and T-s diagrams of a steam plant with reheat and three stages of feedwatcr heating. Here W1 = (1, 1 - h2) + (l - 111 1) (h2 - h3) + (I - 1111) (h4 - h5) + (I -1111 -1112) (h,-h6) + (I -111 1 -m2", " because there has always been the possibility of improving steam cycles by increasing pressure and temperature, and by using reheat and regeneration. Over the above, mercury is expensive, l.i.inited in snpply, and highly toxic. The mercury-steam cycle represents the two-'fluid cycles. The mercury cycle is called the topping cycle and the steam cycle is called the bottoming cycle. If a sulphur dioxide cycle is added to it in the low temperature range, so that the heat released during the condensation of steam is utilized in fonniog sulphur dioxide vapour which expands in another turbine, then thi: mercury-steam-sulphur dioxide cycle is a three-flu.id or tertiary cycle. Similarly, other liquid metals, apart from mercwy, like sodium or potassium, may be considered for a working fluid in the topping cycle. Apart from S02 other refrigerants (ammonia, freons, etc.) may be considered as working fluids for the bottoming cycle. Since the possibilities of improving steam cycles are dimioishing., and the incentives to reduce fuel cost are very much increasing, coupled cycles, like the mercury-steam cycle, may receive more favourable consideration in the near future. Iii I, 486=- 12.14 Thermodynamics of Coupled Cycles If two cycles are coupled in series where heat lost by one is absorbed by the other (Fig. 12.31), as in the mercury-steam binary cycle, let 71 1 and 'h be the efficiencies of the toppi\u00b7ng and bottom cycles respectively, and Tl be the overall efficiency of the combined cycle. T/i = I- Qz and 112.. 1- Q, Q1 Q1 m, Bottom ~ - & cycle f Fig. 12.31 lwq Npou, cydu,oupkd in series or Now Q2 = Q,(l - '11) and Ql = Q2.0 - 112.) '1 = 1 - Q, = 1- Q2(l - 1Ji) Qi Qi = 1- Qi(l-1J1)(l-1}2) Qi = 1 -(I - q1)(1-T/J If there are n cycles coupled in series, the overall efficiency would be given by \" 11 = I - fl (1 - '1;) J=I i.e. or 11 = 1 - (1-11 1) (l -112HI -1h)... (I -1J.,) 1 -17 = (l - t1 1Hl - n2)(1 - 113 ) \u2022\u2022\u2022 (l - nn>\u00b7 Total Joss = Product of losses in all the cycles. For two cycles coupled in series Tl= 1 -0 -111HI -112> = J -(1- '11-1'12 + f11Tlz) = n, + 112- '11112 11 = 17, + '12 - 111112 or This!hows that the overall efficiency of two cycles coupled in series equals the swn of the individual efficiencies minus their product. 1, \"''!t I Vapogr Power Cyclts By combining two cycles in series. even ifindividual efficiencies are low, it is possible to have a fairly h.igh combined. efficiency, which cannot be attained by a single cycle. For exam.pie, if 11 1 = 0.50 and 172 = 0.40 11 = 0.5 + 0.4 - 0.S X 0.4 - 0.70 It is almost impossible to achieve such a high efficiency in a single cycle. 12.15 Process Heat and By-Product Power There are several industries, such as paper mills, textile mills, chemical factories, dying plants, rubber manufacturing plants. sugar factories, etc., where saturated steam at the desired temperature is required for heating, drying, etc. For constant temperature heating (or drying), steam is a very good medium, since isothermal conditiou can be maintained by allowing saturated steam to condeuse at that temperature and utilizing the latent heat released for heating purposes. Apart from the, process heat, the factory also needs power to drive various machines, for lighting., and for other purposes. Formerly it was the practice to generate steam for power purposes at a moderate pressure and to generate separately saturated steam for process work at a pres.sure which gave the desired heating temperature. Having two separate units for process heat and power is wasteful, for of the total heat supplied to the steam for power purposes, a greater part will normally be carried away by the cooling water in the condenser. By modifying the initial steam pressW'e and exhaust pressure, it is possible to generate- the requin:d power and make available for process work the required quantity of exhaust steam at the desired temperature. In Fig. 12.32, the exhaust steam from the turbine is utilized for process heating, the process heater", ":e gas. The rate of ex.ergy dec~e of the gas stream, a1i = faergy input rate= 73 - 1.83 = 71.l 7::: 71.2 MW The rate of exergy increase of steam = Exergy utilization rate Now, oru = w; [h1 - h4 - T0(s1 -s4)] h 1 = (h1) 40 \"\"'\"' 2801 kJ/kg, h3 = 169 kJ/kg S3 = S4 = 0.576 k.J/kgK, h4 = 172.8 k.J/kg S1 = S2 = 6.068 k.J/kgK, h2 = 1890.2 kJ/kg WT= h, - h2 = 2801 - 1890.2 = 910.8 kJ/kg Wr = h, - h3 = 172.8 - 169 = 3.8 kJ/kg Q, = h, - h4 = 2801 - 172.8 = 2628 k.J/kg Q2 = h2- h3: 1890.2- 169 = 1721 kJ/kg W0, 1 = W1 - Wp = Q1 - Q2 = 907 kJ/kg Q1 =w1 x2628= lOOx 103 kW w, = 38 kg/s or. \"' 38 [2801 - 172.8 - 300 (6.068 - 0.576)) = 37.3 MW Rate of exergy destruct.ion in the steam generator = Rate of exergy decrease of gases - Rate of exe-rgy increase of steam. i = a,, -ofu = 71.2- 37.3\"\" 33.9 MW Rate of useful mechanical power output W net = 38 x 907 = 34.5 MW Ex.ergy now rate of wet st~am to the condenser ai;, = w1 [h2 - h3 - T0(s2 -s3)] = 38 [1890- 169- 300 (6.068 - 0.576)] = 2.8 MW Th.is is the exergy loss to the surroUJ1dings. The energy and exergy balances arc shnwn in Fig. Ex. 12.7 (b) and (c). The second law efficiency is given by 11 = Useful exergy output = fil. \"'0.473 or 47.3% 0 Exergy input 73 AM. 1, ti It I Vapour Po111\" Cy,us ~503 Example 12.8 In a steam power plant, the condition of steam at turbine inlet is 80 bar, 500\u00b0C and the condenser pressure is 0.1 bar. The heat source comprises a stream of exhaust gases from a gas turbine discharging at 560\u00b0C and I atm pressnre. The minimum temperature allowed for the exhaust gas stream is 4 50 K. The mass flow rate of the hot gases is such that the heat inpnt rate to the steam cycle is I 00 MW. The ambient condition is given by 300K and l atm. Determine 1Ji, work ratio and 1Jn of the following cycles: (a) basic Rankine cycle, without superheat, (b) Rankine cycle with superheat, (c) Rankine cycle with reheat such that steam expands in the h.p. turbine until it exits as dry saturated vapour, (d) ideal regenerative cycle, with the exit temperature of the exhaust gas stream taken as 320\u00b0C, because the saturation temperaiure of steam at 80 bar is close to 300\u00b0C. Solution For the Jirst law analysis of each cycle-, knowledge of the Jr values at each of the states indicated in Fig. Ex. 12.8 is required. (a) Basic Rankine cycle (Fig. I 2.8a): By usual procedure with the help of steam tab.I es, Ans.,f,u, h 1 = 2758, h2 = 1817, h'J\"' 192 and h4 = 200 kJllcg Wr = 111 -h2 = 941 kJlkg, WP=\"\u2022 - h3 = 8 kJllcg Q1 = 111 - h4 = 2SS8 kJlkg. W oet = 933 kJ/kg 171 = w.., ~ 933 = 0.365 or 36.5% Q, 2558 Work ratio= WT - Wp = 933 \"'0.991. T,=833:T 9411 ; - - - - \\. To ~ T 2 - s (a),.. t (b} T T - s (c) Fig. EL 12.8 - >- $ (d) Ill I, -&uic anJ Applitd Thmnodynamits Power output= 711Q", " 162.7 Id/kg h8 = 762.81 kJ/kg 1,1 - h2 = 0.785 (h1 - h2,) \"\"0.785 X 275 = 215.9 kJ/kg. h1 =3285 - 215.9 = 3069.I kJ/kg Ene.rgy balance for the heater gives \u00b7 m x h4 + (l - m)'11 = l x h 8 m x 3072.5 + (1-m) x 162.7\"\" 1 x 762.81 m = 600.1 l = 0.206 leg/leg steam flow at twbine inlet. 2909.8 Ans. (c) 1lc:,,:lc= (lr1 -hz)+(h3 -114 )+(1-m)(h,. -h5 ) (h1 -hg)+(h3 -hz) 215.9 + 207.S + 0.794 X 703.4 2522.2 + 210.9.. 931.9 0.359.2 or 35.92% 2733.1 \u00b7 Ans. (d) t Example 12.12 A binary-vapour cycle operates on mercury and steam. Saturated mercury vapour at 4.5 bar is supplied to the mercury turbine, from wh.ich it exhausts at 0.04 bar. The mercury condenser generates saturated steam at 15 bar whlch is e xpanded in a steam turbine to 0.04 bar. (a) Find the overall efficiency of the cycle. (b) If 50,000 kglh of steam flows through the steam turbine, what is the flow through the mercury turbine? (c) Assuming that all processes are reversible, what is the useful work done in the binary vapour cycle for the specified steam t1ow? (d) lfthe steam leavi.ug the mercury condenser is superheated to a temperature of300\u00b0C in a superheater located in the mercury boiler, and if the iutemal efficiencies of the mercury and steam turbines are 0.85 and 0.87 respectively, calculate the overall efficiency of the cycle. The properties of saturated mercury are given below p (bar) t(\"C) lr1 lr 1 {kJlkg) Sf Sg {kJ/kg K) VJ V1 (m3 /kg) 4 \u2022.S 0.04 450 216.9 62.93 29.98 355.98 329.85 0.1352 0.S397 79.9 X 10-4 0.068 0.6925 76.5 x 10~ 5.178 0.0808 Iii I, Vapour Powtt Cycles -.= 511 S0lutio11 The cycle is shown in Fig. Ex. 12.12. For the mercury cycle, h6 = 355.98 kI/ltg s1 = 0.5397 kI/ltg JC = s0 =sf+.i:b sr, = 0.0808 + xb (0.6925 - 0.0808) X = 0.4589 = 0.75 0.6117 b hb = 29.98 + 0.75 X 299.87 = 254.88 kJ/kg (WT)m = ha - hb = 355.98 - 254.88 = 101.l kJ/ltg (Wp)m = hr Ire= 76.5 x 10-6 x 4.46 x 100.. 3.41 >< 10-2 kJ/kg WQOI ~ IOI.I kJ/kg Q, = h. -/rd= 355.98- 29.98 = 326 kI/Jcg '1 = w \u2022\u2022\u2022 = lOl.l \"'0.31 or31% m Q1 326 - - - s Fig. Ex. 12.12 For the steam cycle h, = 2792.2 kJ/kg s1 = 6.4448 kJ/kgK = s2 =.rr + x2 s,112 = 0.4226 + X2(8.4746 - 0.4226) X = 6.0222 = 0.748 2 8.0520 h2 = 12J.46 + 0.748 >< 2432.9 = 1941.27 kJ/kg. (Wr)s1 = h1 - lr2 = 2792.2 - 1941.27 (W,)51 = hr h 3 = 0.001 >< 14.96 x 100 = 1.496 k:Ilkg = 1.5 kJ/kg = 850.93 kI/kg h4 = 121.46 + 1.5 = 122.96 kI/kg 512~ Bam and Applud Tlurmod7namics Q, = Ii, - It", " rate is 150 kg/h. Determine (a) the turbine isentropic efficiency. (b) the net work output (c) the cycle efficiency, and (d} the area of solar collector needed if the collectors pick up 0.58 kW/ni2. Ans. (a)0.767, {b) 15.51 kW, (c) 14.7%, (d) 182.4m2 12.4 In a reheat cycle, the initial steam pressure and the maxjmum temperature are lSO bar and S50\u00b0C respectively. If ihe condenser pressure is 0.1 bar and the moisture at the condenser inlet is 5%, and ffi-Uming.ideal ptOCCsses, detelllline (a) the reheat pressure-, (b) the cycle efficiency, and (c} the steam rate. Ans. 13.5 bar, 43.6%, 2.05.kg/kWh 12.5 In a nuclear power-plant heat is transferred in the reactor io liquid sod'ium. The liquid sodium i.s then pumped to a heat exchanger where heat is transferred to steam. The steam leaves this heat exchanger as saturated vapour al SS bar. and is then superheated in an external gas-fired superheator to 650\u00b0C. The steam then e.nters the turbine, which has one extraction point at 4 bar, where steam flows to an open feedwater heater. The turbine efficiency is 75% and the c.ondenser temperature is 40\u00b0C. Determine the heat transfer in the reactor and in the supemcatcr to produce a power output of 80 MW. 12.6 In a reheat cycle, steam at 500\u00b0C expands in a h.p. turbine till it is saturated vapour. II is reheated at constant pressure to 400\u00b0C a11d then expands in a l.p. turbine to 40\u00b0C. If the maximum moisture coo.tent at the turbine f!Xhaust is limited to 15%, find (a) the reheat pressure. (b) the prcs.~ure of steam at the inlet to the h.p. 1urbine, (c) the net specific work output, (d) the cycle efficiei1cy, and (e) the steam rate. Assume all ideal processes. What would have been the quality, the work output, and the.cycle efficiency without the reheating of steam? Assume tbat the other conditioos remain the same. 12.7 A regenerative cycle operates with steam supplied at 30 bar and 300\u00b0C. and condenser pressure of0.08 bar. The extraction points for two heaters (one closed and one open) are at 3.5 bar and 0.7 bar respectively. Calculate the thermal efficiency oflhe plant, neglecting pump work 12.8 The net power output of the turbine in an ideal reheat-regeneratiw cycle is 100 MW. Steam enters the high-pressure (H.P.) turbine at 90 bar, 550\u00b0C. After ellpansion to 7 bar. some of the steam goe..~ to an open heatcJ and the balance is reheated to 400\u00b0C, alkr which it expands to 0.07 bar. (a) What is the steam now rate to the H.P. turbine'? (b) V.'hat is the total pwnp work? (c) Calculate the cycle efficiency. (d) lftbcro is a J0\u00b0C rise in the temperature of the cooling water, what is the rate ofnow of the cooling water in the conden~er? (e) If the velocity of the steam flowing from the turbine to the condenser is limited to a maximum of 130 mis. find the diameter of the connecting pipe. 12.9 A mercury cycle is superposed on the steam cycle operating between the boiler outlet condition of 40 bar, 400\u00b0C and the condenser temperature of 40\u00b0C. The heal releaso!d by mercury condensing at 0.2 bar is used to impart the latent hc:at of vaporizatton to the water in the steam cycle. Mercury enters the mercury turbine...,. 516=- Ba.Jit and Applied 11urmotlynamies as satwated vapour at 10 bar. Compllle (a) kg ofmercwy cin:ulated per kg of water, and (b) the efficiency of the combined cycle. The propeny values of saturated men:w:y a.n: given below v, p (bar) t (\"C) h1 (Id/kg} h 1 s1 s1 (k.J/kg K) 10 0.2 SlS.5 277.3 72.23 38.35 363.0 0.1478 0.5167 336.", " 0.957 d1y at 17 bar being throttled to 3.5 bar. The remaining steam flows through a H.P. turbine which exhausts at a pressure of3.5 bar. The exhaust steam mixes with the process steam before entering the L.P. turbine which develops 1337.5 kW. At the exhaust the pressure is 0.3 bar. and the steam is 0.912 dry. Draw a line diagram of the plant and detennine (a) steam quality at the exhaust from the H.P. turoine, (b) the power developed by the H.P. turbine, and (c) the isentropic efficiency of the 1-1.P. turbine. Ans. {a) 0.96, (b) 1125 kW, {c) 77% 12.20 In a cogeneralion plan!, the power load is 5.6 MW and the healing load is 1.163 MW. Steam is generated at 40 bar and 500\u00b0C and is expanded isentropically through a turbine to a condenser at 0.06 bar. The beating load is supplied by eximciing steam from ihe turbine al 2 bar which condensed in the process heater to saturated liquid al 2 bar and then puruped back to lhe boiler. Compute (a) the steam generation capacity of the boiler in tonoeslb, (b) the heat input to the boiler in MW, and (c) the heat rejected to the condenser in MW. Ans. (a) 19.07 t/h. (b) 71.57 MW. a.nd (c) 9.607 MW 12.21 Steam is supplied to a pass-out turb.ine at 35 bar, 350\u00b0C and dry saturated process sh~am is required at 3.5 bar. The low pressure stage exhausts at 0.07 bar and the condition line mny be assumed to be s1raight (the condition line is the locus passing through the slates of steam leaving the various stages of the inrbine). If the power required is I MW and the maximum process load is 1.4 kW, estimate!he maxfmum steam flow through the high and low pressure stages. Assume that the steam just condenses in the process plant. Ans. 1.543 and 1.182 kg/s \"'' Vapoar Power Cyclts ~519 12.22 Geolhennal energy from a natural geyser cilll be obtained as a continuous supply of steam 0.87 dry at 2 bar and at a flow rate of 2700 kg/h. This is utilized in a mixed-pressure cycle to augment!he superheated exhaust fcom a high pressure turbine of83% internal efficiency, which is supplied with 5500 kg/h of steam at 40 bar and 500\u00b0C. The mixing process is adiabatic and the mixture is expanded lo a condcn9Cr pressure of 0.10 bar in a low pressure turbine of 78% internal efficiency. Dete.nnmc the power output and!he thermal e'fficie.ocy of the plant. Ans. I 745 kW. 35% 12.23 Ina study for a space project it is thought that the condensation ofa working fluid might be possible at --40\u00b0C. A binary cycle is proposed, using Refrigerant-12 as the low temperarure fluid, and water as the high temperature fluid. Sieam is generated at 80 bar, 500\u00b0C and expands in a turbine or 81 % isentropic efficiency to 0.06 bar, at which pressure it is condensed by the generation of dry soturated refrigerant vapour at 30\u00b0C from saturated Hquid at - 40\u00b0C. The isentropic efficiency of the R-12 turbine is 83o/o. Determine the mass ratio ofR-12 to water and tl1e efficiency of the cycle. Neglect all losses. Ans. 10.86; 44.4% 12.24 Steam is generated at 70 bar, S00\u00b0C and expands in a turbine to 30 bar with an isentropic efficiency of77%. At this condition it is mixed w.i1h twice its mass of steam at 30 bar, 400\u00b0C. The mixture the.n expands wi.th an isentropic efficiency of 80% to 0.06 bar. At a point in the expansion where the pressure is S bar, steam is bled for fcedwater heating in a din:ct contact heater. which raise.~ the feedwater to the saturation temperature of the bled steam. Ca.lculate the mass of stearn bled \u00b7 per kg of high pressw-e steam and the cycle efficiency. AI; BS.Sume lb.at the L.P. expansioo (Dl)dition line in might. Ans. O.S3", ",_4 =-RT2 1n V4 Q4_1 = c.(T, - T2); W 4-1 = 0 Due to heat tmnsfers at constant volume processes, the efficiency of the Stirling cycle is less than that of the Carnot cycle. Howe\\'er, if a regenerative arrangement is used such that Q2_3 = Q4_1, i.e., the area under 2-3 is equal to the area under 4-1, then the cycle efficiency becomes Gas Powrr Cyelu --=523 t - I I (a) (b) Fig. 13.2 Slirlirig cyelt R1j 1n ~ \u00b7- RT2 1n.!l. f1 = V1 R7j In..!1 v, V\u2022 - 7j - J; 7j (13.2) So, the regenerative Stirling cycle has the same efficiency as the Carnot cycle. 13.3 Ericsson Cycle (1850) The Ericsson cycle (Fig. 13.3) is made up of: 4 Q, t t -1 ) (a) - s (b) Fig. 13.3 Eria1011 cyek Two reversible isolhenns and two reversible isobars. For 1 kg of ideal gas Q1_2 = W1_2 = RT1 In 1!J_ P2 Q2_3 = cp(T2 - T1); W2_3 = P2(V3 - v2) = R (T2 - T,) h I c,1 \u2022 524=- Basic and Applied 1lrmnodynamfcs Q3-<1 = W3~ = - RT2 In J!J.. Pi Q._1 = c,(T1 - T4); W4_1 = p 1(v1 - v\u2022) = R(T1 - T2) Since part of tbe heat is transfeaed at constant pressure and part at constant temperature, the efficiency of the Ericsson cycle is less than that of the Carnot cycle. But with ideal regeneration, Q,._3 = Q4-.1 so that all the heat is added from the external source at T1 and all the heat is rejected to an external sink at T2, the efficiency of the cycle becomes equal to the Carnot cycle efficiency, since fl= l - ~ = 1Q. RT,, In J!J.. R1j In }!J.. Pi Pl = 1- 1i 7j (13.3) The regenerative, Stirling and Ericsson cycles have the same efficiency as the Carnot cycle, but much less back work. Hot air engines working on these cycles have been successfully operated. But it is difficult to transfer beat to a gas at high rates since the gas film has a very low thermal conductivity. So there has not been much progress in the development of hot air engines. However, since the cost of internal combustion engine fuels is getting excessive, these may find a field of use in the near future. 13.4. Air Standard Cycles Internal combustion engines (Fig. 13.4) in which the combusri.on of fuel occurs in the engine cylinder itself are non-cyclic heat engines. The temperature due to the evolution of heat because of lhe combustion of fuel inside the cylinder is so high that the cylinder is cooled by water circulation around it -to avoid rapid deterioration. The work'ing fluid, the fuel-air mixture, undergoes pennanent chemical change due to combustion, and the products of combustion after doing work are thrown out of the engine, and a fresh charge is taken. So the working flnid does not undergo a complete thermodynamic eye.le. To simplify the analysis ofl.C. engines, air standard cycles are conceived. In an air standard cycle, a certain mass of air operates in a complete thermodynamic cycle, where heat is added and rejected with external heat reservoirs, and all the processes in the cycle are reversible. Air is assumed to behave as an ideal gas, and its specific heats are as&11.med to be constant. These air standard cycles are so conceived that they correspond to the operations ofintemal combustion engines. 13.5 Otto Cycle (1876) One very common type of intema.l combustion engines is the Spark Ignition (S. [.) \u00b7 engine used in automobiles. The 0110 cycle is the air standard cycle of such an engine. The sequence of processes in the elementary operation of the S.l. engine is given below, with rcfercnc.e to Fig. 13.S{a, b) where the sketches of the engine and the indicator diagram are given.'II Gas Powrr Cydts -=525 Sparil plug or fuel injedor [ i, db~77[1; c1eeran0& volume Top /j deadC111rtter T,\u00b7,", "1... F - - - - Bon, - - /, Stroke r-a~ ~~~~~~~~ ~I Q....-1.. l~I._ - Bottom dead center Cylinder wall Pis I.on - ----1'-, r--'-.-r--' \\ ~! rr0 \\ -~ --- --i~ / Crank mechanism / Fig. 13.4 Nomrncklturt far rtciprocatin& piJton-c,lin.dn ttrginu. Fuet\u2022\u2022ir mixture -=I -- ----r- Combustion products (e) Inlet valve Exhaust valve i~ --1> 2, 6 1 (b) Fig. 13.5 (a) S.J. ntgiu, (b) Jndiaitor diagram rlltf l 11 526=- &sic and Applitd Thermodyna.mia Process f- 2, Intake. The inlet valve is open, the piston moves to the right, admitting fuel\u2022air mixture into the cylinder at constant pressure. Process 2- 3. Compression. Both the valves are closed, the piston compresses the combustible mixture to I.be mini.mum vo.lume. Proce.ss :3- 4, Combustion. The mixture is then ignited by means of a spark, combustion takes place, and there is an increase in temperature and pressure. Process 4-5, Expansion. The products of combustion do work on the piston which moves to the right, and the pressure and temperature of the gases decrease. Process J - 6. Blow-down. The exhaust valve opens, and the pressure drops to the initial pressure. Process 6- J, Exhaust. With the exhaust valve open, the piston moves inwards to expel the combustion products from the cylinder at constant pressure. The series of processes as described above constitute a mechcmica/ cycle, and not a th.ennodynamic cycle. The cycle is c,ompleted in four strokes of the piston. Figure 13.5 (c) shows the air standard cycle (Otto cycle) corresponding 10 the above engine. It consists of: Two reversible adiabatics and two reversible isochores. Q, t -0 Fig. 13.5 (e) Otto 9ek - s Air is compressed in process 1-2 reversibly and adiabatically. Heat is then added to air reversibly at constant volume in process 2- 3. Work is done by air in. expanding reversibly and adiabatically in process 3-4. Heat is then rejected by air reversibly at constant volume in process4- I, and the system (air) comes back to its initial state. Heat transfer processes have been substituted for the combustion and blow-down processes of the engine. The intake and exhaust processes of the engine cancel each other. Let m be the fi.xed mass of air undergoing the cycle of operations as described above. Heat supplied Q1 = Q2_3 = me. (T1 - T2) Heat rejected Q2 \"' Q.._1 \"' me. (T4 - T1) Efficiency Tl\"' 1 - Qi = I - mc.(T. -1j} Q1 mc.(1; - 7;) \"' 528=- Basic a,id Applied Tlinmotlynamics burns spontaneously. The rate of burning can, to some extent, be controlled by the \u00b7 rate of injection of fuel. An engine operating in this way is called a compression ignition (C.l) engine. The sequence of processes in the elementary operation of a C.I. engine, shown in Fig. 13.6, is given below. f -ti (bl (a} Flg. 13.6 (a) C.L tngint (b) /nd(cator diagram Process 1-1. Intake. The air valve is open. The piston moves out admitting air intc;, the cylinder at constant pressure. Process 2-3. Compression. The air is then compressed by the piston to the mi\u00b7nimum volume with all the valves closed. Process 3-4. Fuel injection and combustion. The fuel valve is open, fuel is sprayed into the hot air, and combustion takes place at constant pressure. Process 4-5. Expansion. The combustion products expand, doing work on the piston which moves out to the maximum volume. Process 5- 6. Blow-down. The exhaust valve opens, and the pressure drops to the illitial pressure. Pressure 6-1, Exhaust. With the exhaust valve open, the piston moves towards,. the cylinder cover driving away the combustion products from the cylinder at constant pressure.. The above processes constitute an engine cycle, which is completed in four ~ strokes of the-piston or two revolutions of the crank shaft. / '. ing to the C.l. engine, as described above. The cycle is composed of: Figure 13. 7 shows", "1r~\"t-l lil+ T1] (13.IS) To find (rp)opt dw.,.. ~ - cp - [ r. ( r -, ),-1+111r1-1, 3 --r- rp r.( v -1) 11-,1111-\u2022,]- 0 I 7 rp - - T ( Y -1 )r.(l'Y)-2 = T ( Y - I )\u00b7r.-1/Y 3 y r I r p 7 -<11tH11'!)+2 = 1j 7j p 11 I!! '!1 I I II or (r \\ p,cp! = Gas Power Cyclts -=543 T, )y/2.(y-l)....!. 1j ( T. (r) = ~ rmln ( From Eqs ( 13.14) and (13.16) l' cpl )112(y-t) (r p)cp1 = J(rp ),,_ (13.16) (13.17) Substituting the values of(rp)op, in Eq. t13.15) nrt 1 (1i) r-1 W. =(W. \\ =c T, -T.1j) r r-1 - 1j - - - - \u00b7 - - + 1j r. 2<r -1> r 7j 2(7 -1)...Vinu r > ] or = cp[T3 - 2J1j 1j + T1] (Wnct>mu =cp(JTir,.,, -JTm'in )2 f1cy,:le - - I _ l y-liy rp - - I - J Tmin -T. mu (13.18) {13.19) Considering the cycles 1-2' -3' -4' and 1-2\" -3'' -4\" (Fig. 13.17), it is obvfous that lo obtain a reasonable work capacity, a certain reduction in efficiency must be accepted. 13.9.5 Effett of lntercooling and Reheating on Brayton C,ck The efficiency of the Brayton cycle may often be increased by the use of staged compression with intcrcooling, or by using staged heat supply, ca.lied reheat. Let the compression process be divided into two stages. Air, after being compressed in the first stage, is cooled to the initial temperature in a heat exchanger, called an intercooler, and theu compressed further in the second stage {Fig. 13.20). J-2'- 5-6 is the ideal cycle without intercooling, having a single stage compression, 1-2-3-4-6 is the cycle with intercooling, having a two-stage compression. The cycle 2-3-4-2' is thus added to the basic cycle 1-2' -5-6. There is more work capacity, since the included area is more. There is more heat supply also. For the cycle 4-2'-2-3, T\"' 1 is lower and T m2 higher (lower rp) than those of the basic cycle l- 2'- 5-6. So the efficiency of the cycle reduces by staging the compression and intercooling. But if a regenerator is used, t.he low temperature heat addition (4-2') may be obtained by recovering energy from the exhaust gases from the turbine. So there may be a net gain in efficiency when intercooling is adopted in conjunction with a regenerator. Similarly, let the total heat supply be given in two stages and the expansion process be divided in stages in two turbines {T1 and T2) with intennediate reheat, as shown in Fig. 13.21. 1-2-3-4' is the cycle with a single-stage heat supply \u2022 I,,!!I I!! I s\u00ab~ having no reheat, wiih total expansion. in one turbine only. l-2-3-4-5-6 is the cycle with a single-stage reheat, having the expansion divided into two stages. With the basic cycle, the cycle 4-S-6-4 is added because of reheat. The work capacity increases, but the heat supply also increases. (a) f l.,' ------ - -. s (b)'- - -~6 -\" (c) Fig. 13.20 Effect of inkrcooli\"l on Brayton cycfl In the cycle 4-S-6-4', rP is lower than in the basic cycle 1\u00b72-3-4', so its efficiency is lower. Therefore, the efficiency of the cycle decreases with the use of reheat. But T6 is greater than T'", " Fuel energy input - 322\u00b75 x 43,300 = 353525 kW 39.5 =353.5 MW ffo,<1. = ~~ 3 5 = 0.537 or 53.7% REvlEw Q,UESnONS A 11.1. Ans. 13.l What are cyclic and non-cyclic heal engines? Give examples. 13.2 What ae the four processes which constitute the Stirling cycle'! Show that the regenerative Stirling cycle has the same efficiency as the Carnot cycle. State the four processes thal constitute the Ericsson cycle. Show thal the regenerative Ericsson cycle has the same efficiency as the Carnot cycle. 13.3 13.4 Mention the merits and demerits of the Stirling and Ericsson cycles. 13.5 What is an air standard cycle'? Why are such cycles conceived'? 13.6 What is a spark ignition engine? What is the air standard cycle of such an engine? What are,its four processes? Show tl1at the efficiency of the Otto cycle depends only on the compression ratio. 13.7 l3.8 How is the compression ratio of an SI engine fixed'! 13.9 What is a compression ignition engine? Why is the compression ratio of such an engine more than lhat of an SI engine? l3.10 State lhe four processes oflhe Diesel cycle. 13.11 Explain the mi.xed or dual cycle. 13.12 For lhe\u00b7 same compression ratio and heat rejection. wh.ich cycle is most efficient: Otto, Diesel or Dual? Explain with p-v and r~.r diagrams. \u00b7 13.13 With the help of p-v and T-s diagrams. show 1ha1 for the same maximum pressure and temperature of the cycle and the same heat rejection, 'l0lff<1 > 11 Doal > l10i,o 13.14 What are the three basic components of a gas lu!bine plant? Whal is the air standard cycle of such a plant? What are the processes it consists of? I I,, lol I'\" 510=- Basic and Applied Thmnodynamics 13. l S Show that the efficiency of the Brayton cycle depends only on the pressure ratio. 13.16 What is t~e application of the closed cycle gas turbine plant? 13.17 Discuss the merits and demerits of Brayton and Otto cycles applied to reciprocating and rotating plants. \u00b7 13.18 W~t,is the' effect of regeneration on Brayton cycle efficiency? Define the effectiveness of a regenerator. 13.19 Whal is the effect of irrcvcrsibilitics in turbine and compressor on Brayton cycle efficiency? 13.20 Explain the effect of pressure ratio on the net output and efficiency of a Brayton cycle. 13.21 Derive the expression of optimum pressure ratio for maximwn net work output in an ideal Brayton cycle. What is the corresponding cycle efficiency. 13.22 Explain the effects of: (a) intercooling, and (b) reheating, on Brayton cycle. I 3.23 Whal is a free shaft tuzbine? 13.24 With the help of flow and T-s diagrams explain the ai.r standard cycle for a jet propulsion planL 13.25 Wiih the help of a neat sketch explain the operation of a turbojet engine. How is the thrust developed in this engine? Why does a commercial airplane fly at high altitudes? 13.26 Define. propulsive power aod propulsive efficiency. 13.27 Why m regenerators and intercoolers not used in aircraft engines? What is after\u00b7 burning? Why is it used? 13.28 Explain the working of a turbofan engine with the help of a n~t sketch. Define \"bypass.ratio\". How does it influence the engine th.Nst? 13.29 How does a turboprop engine differ from a turbofan engine? 13.30 What is a ramjet? How is the thrust produced here? 13.J I Whal is a rocket? How is ii propelled? 13.32 Explain the advantages and disadvantages of a gas twbine plant for a utility system. 13.33 What arc the advantages of a combined gas turbine-steam tuzbine power plant? 13.34 With the help of flow and T'-s diagrams explain the operation of a combined GT- ST plant. Why is supplementary firing often used? PROBLEMS 13.1 In a Stirling cycle the volume varies between 0.03 and 0.06 m3, the maximum p.n:ssurc is 0.2 MPa, and the temperature varies between S40\u00b0C and 270\u00b0C. The working fluid is air (an ideal gas). (a) Find the efficiency and the work done per cycle for the ~imp le cycle. (", "b) Find the efficiency and the wodt done per cycle for the cycle with an ideal regenerator, and compare with the Carnot cycle having the same isothermal beat supply process and the same temperature nmgc.,tns, (a) 27.7%, 53.7 kl/kg, {b) 33.2% 13.2 An Ericsson cycle operating with an ideal regenerator wow between I JOO K and 288 K. The pressure at the begiMing of isothennal compression is 1.013 bar. Detenninc (a) the compressor and tuzbine work per kg of air, and (b) the cycle efficiency. Aiu. (a) Wr \u2022 46S kJ/kg, We - 121.8 kJ/kg (b) 0.738 111 I i1, -=511 13.3 Plot the efficiency of the air standard Otto cycle as a function of the compression ratio for compression ratios from 4 to 16. 13.4 Find the air standard efficitncies for Otto cycles with a compression ratio of 6 using ideal gases having specific heat ratios 1.3. 1.4 aud 1.67. \\\\'hat are the advantages and disadvantages of using helium as the working nuid? 13.5 An engine equipped with a cylinder having a bore of IS cm and a stroke of 45 cm \u2022 compute lhc air operates on an Otto cycle. If the clearance volume i.~ 2000 cm3 standard efficiency. Ans. 47.4% 13.6 In an air stand3rd Otto cycle the compression ratio is 7. and compression begins at 35\u00b0C, 0.1 MPa. The maximum temperature of the cycle is I 100\u00b0C. Find (a) the temperature and pressure at the cardinal pcims of the cycle, (b) the heat supplied per kg ofair, (c) the work done per kg ofair, (d) the cycle efficiency, and (_e) the n1.e.p. of the cycle. 13.7 An engine working on the Otto cycle has an air siandard cycle clliciency of 56% and rejects.544 kJ/kg of air. The pressure and temperature of air at the beginning of compression are 0.1 MPa aod 60\u00b0C respectively. Compute (a) the compression ratio of the engine. (b) the work done per kg of air, (c) the pressure und temperarurc at the end of compression, and (d) the maximum pressure in the cycle. 13.8 For an air standard Diesel cycle with a compression ratio of 15 plot the efficiency as a function of the cut-off ratio for cut-off ratios frt'm l to 4. Compare with the results of Problem I :u, 13.9 Ju an air standard Diesel cycle, lhe compression ratio is IS. Compression begins at 0.1 MPa. 4-0\u00b0C. The heat added is t.67S MJ/kg. Find ta) the maximum temperature of the cycle, (b) the work done per kg of air, l c_) thc cycle etlicicncy. (d) the temperature at the end of the i.sentropic e11paosloo, (e) the cut-off ratio. (f) the maximwn pressure of the cycle. and (g) the m.e.p. of the cyde. 13.10 Two engines are to operate on Ott.o and Die~cl cycles with the following data: Maximum temperature 1400 X, exhaust temperature 700 K. State ofai.r at the beginning of compression 0.1 MPa. 300 K. Estimate the compression ratios. the maximum pressures, dliciencies. and mtc of work outputs (for I kg/min of air) of the respective cycles. Ans. Otto-,\u00b7~ ~ 5.656, Pmex ~ 2,64 MPa, IV z 2872 kJ/kg, 1J = 50% Oicscl--,\u00b7t =- 7.456, p.,,,. = 1.665 M.Pa. 1\u00a5= 446.45 kJ/kg, 1J O 60.8% 13.1 I An air standard limited pressul'f: cycle has a compression ratio of IS and compression begins at 0.1 MPa, 40~c. The maximum pressure is limited to 6 Ml'a and the heat added is 1.675 MJ/kg. Compute (a) the heat supplied \u2022 \u00b7 con~1an1 volume per kg of air, {b) the heat supplied al constant pressure per kg of air, (c) the work done per kg ohir, (d) the cycle efficiency, (c) the tcmpt\"alurc at the cod of the consiant volume heating process. (t) the cut-off ratio. and (g) the", " m.e.p. of the cycle. A.11s. (a) 23S kJ/kg, (b) 1440 kJ/kg, (c> 1014 kJlkg, (d} 60.5%, (e) 12S2 K, t{) 2.144 (g) 1.21 MPa 13.12 In an ideal cycle for an internal combustion engine the pressure and temperature at the beginning of adiabatic compression arc respectively 0.1 1 MPa and l I 5\u00b0C, the compression ratio being 16. At the end of compression heat is added to the working fluid. first. at constant volume, aod then at cons1ani pressure reversibly.!ii I 512=- Bosi, and Applud Thermodynamits The working fluid is then apanded adiabatically and reven;ibly to the origina.l volume. If the working lluid is air and the maximum pressure and temperature are respectively 6 MPa and 2000\u00b0C, determine, per kg of air (a) the p.ressure, temperature, volume. and entropy of the air at the five cardinal points of the cycle (take s1 as the entropy of air at the beginning of compression), a.od (b) the work output and efficiency of the cycle. 13.13 Show that tbe air standard efficiency for a cycle comprising two coJUtant pressure processes and two isothermal processes (all re\\'cnible) is given by - {7j - T:z) In {rpiY- lliT f'/ - 7j(l + In (rp)'T-l)IY - T,] where T1 and T2 arc the maximum and minimwn temperatures of the cycle, and rr is the pressure ratio. 13.14 Obiain an expression for the specific work done by an engine working on the Otto cycle in terms of the maximum and minimum temperatures of the cycle, the. compression ratio rk, and constants of the working fluid (assumed to be an ideal gas). Hence show that the compression ratio for maximum specific work output is given by - ( Tmill )1120-y> 't- - -r.... 13. IS A dual combustion cycle operates with a volumetric compression r,llio 'k = 12, and with a cut-off ratio 1.615. The maximum pressure is given by p,.., = 54pi, where p 1 is \u00b7the pressure before compression. Assuming indices of compression and ei1pansion of 1.35, show that the m.e.p. of the cycle Pm= IOp, Hence evaluate (e) tcmperarures at cenlinal point, wilh T1 = 335 K. and (b) cycle efficiency. Ans. (a) T2 = 805 K,p2 -= 29.2p1, T) = 1490 K, T4 = 2410 K, T5 = 1200 K, (b) '1 = 0.67 13.16 Recalculate (a) the temperatures at the cardinal points, (b) the m.e.p., and (c) the cycle efficiency when the cycle of Problem 13.15 is a Diesel cycle with the same compression ratio and a cut-off ratio such as to give an expansion curve coincident with the lower part of that of the dual cycle of Problem 13.15. Ans. (a)T2 = 80SK, T3 = 1970K,T4 = 1142 K(b}6.82p 1,(c) 17=0.513 13.17 In an air standard Brayton cycle the compression ratio i.s 7 and the maximum temperature of the cycle is 800\u00b0C. The compression begins at 0.1 MPa, 35\u00b0C. Compare the maximum specific volume and the maximum pressure with the Otto cycle of Problem 13.6. Fin<l (a) the heat supplied per kg of air. (b) the net work done per kg of air, (c) the cycle efficiency, :uid (d) the temperature at the end of the expansion process.. 13.18 A gas turbine plant operates on the Brayton cycle between the temperatures 27\u00b0C and 800\u00b0C. (a) Find the pressure ratio at which the cycle efficiency approaches the Carnot cycle efficiency. (b) find the pressure ratio at which the work done per kg of air is maximum, and ( c) compare the efficiency at thi.s pressure ratio with the Carnot efficiency for the given temperatures.,,'II B1JSic and Applitd 171n111od7namio all prwesses a.re ideal. Determine (a) pressure p 5\u2022 (b) the net work per kg and mass flow rate, (c) temper.iture T3 and cycle lbettl'.lal efficiency, and (d)", " pressure ratio across a single stage means low volumetric efficiency. Also, with dry comp.ression the high pressure ratio results in high compressor discharge temperature whic.b may damage tbe refrigerant. To reduce the wod, of compression and improve the COP, multistage compression with inten:ooling may be adopted. Since the intercooler tempera~ may be below the temperature of available cooling water used for the condenser, the refrigerant itself may be used as the inte~ooling medium. Figure 14.11 shows a two-stage compression system with a direct contact heat exchanger. As shown in Sec. l 0.4, for minimum work, the intercooler pressw-e p, is the geometric mean of the evaporator and condenser pressures,p1 andp2, or P;\"\"\".JP1\u00b7P2 By making an energy balance oft.be direct contact beat exchanger, m2h2 + m1itt, - ~h1 + m1h) '\"1 \"',,,__,,,,,,, -ht, ffl2 The desired refrigerating effect determines the now rate in the low pressure loop, m2, a& given below mz(h, - \"8} = i4000 x P 3600 where Pis the capacity, in toMes of refrigeration. m. 3.89 Pk I 2= - - - gs h, -Its Figure 14.12 shows a two-stage vapour compression system with a flash chamber intercooler, where the vapour from the flash chambc.r (state 9) mixes with the vapour from the LP compressor (state 2) to form vapour at state 3, which enters the HP compressor. 14.3.5 Refrigerants The most widely used refrigerants now-a-days are a group of halogenated hydrocarbons, or chlorofluorocarbons (CFCs) marketed under the various proprietary names of freon, genetron, arcton, isotron, frigen, and, so on. These are either methane-based or ethane-based, where the hydroge.n atoms are replaced by II l l ii \u2022 588=- Ba.sit: and Applied '171ermadynamics \\~ I H.P. l.P. Compre$$or Condenser lc=== = =l Flash _.,,-_.,,.,,-_-_-_.,,.,,-_.,,-_-_-_-_.,,-_} chamber \u2022:::\u2022:-::~::::\u00b7~::--:-:-J (a) f -h (b) Fig. U..12 Two-1ta,e oapour comprmion system with a flash intercaoln third digit indicates the number of fluorine atoms, all other atoms in the hydrocarbon being chlorine. For example, R-110 is <;Cl6, R-113 is <;Cl3F3, R-142 is <;H3CIF2, and so on. The use of these refrigerants is now discouraged, since these, being largely insoluble in water, move up, react with ozone in the ozone layer (which protects the earth from pernicious ultraviolet rays, and deplete it. Ii was realized in mid-seventies that the CFCs not only allow more ultraviolet radiation into the earth's atmosphere, but also prevent the infrared radiation from escaping the eanh to outer space, which contributes to the greenhouse effect and I I +j+ \"'I,I -=589 hence, global warming. As a result, the use of some CFCs is banned (by Montreal Protocol, t 987) and phased out in many countries. Fully halogenated CFCs (such as R- 11, R- 12 and R-115) do the most dama.ge to the ozone layer. The partially halogenated refrigerants such as R-22 have about 5% of the ozone depleting potential (ODP) ofR-12. CFCs, friendly to the ozone layer that protecl~ the earth fr~m ultraviolet rays and which do not contribute to the greenhouse effect are being developed. The chlorine free R-134,a, a recent finding, is presently replacing R-12, the most widely used refrigerant, particularly in domestic refrigerators and fre.ezers and automotive air conditioners. Two important parameters that need. to be considered in the selection of a refrigerant are the temperatures of the two media (the refrigerated space and the environment), with which the refrigerant exchanges heat. To have reasonable heat transfer rate, a temperature d.i ffercnce of 5 to I 0\u00b0C should be maintained between the refrigerant and the medium. If a space is to be maintained at -10\u00b0C, e.g. the refrigerant should evaporate at about - 20\u00b0C (Fig. 14.", "13), the saturation pressure at which should be above atmospheric pressure to prevent any air leakage into the system. Again, the temperature of the refrigerant in the condenser should be above the cooling medium by aboul I0\u00b0C, as shown int.he figure, the saturation pressul'e at which must be below the crilical pressure of the refrigerant. If a single refrigerant cannot.meet the temperature requirements (-20\u00b0C to 50\u00b0C range), two cycles with two different refrigerants can be used in series (Fig. 14.14). Such a coupled cycle makes a cascade refrigeration system. Othe.r desirable characteristics of a refrigerant are that it should be nontoxic, noncorrosive, nonflammable, and chemically stable, should have a large enthalpy of vaporization to minimize the mass flow, and should be available at low cost. T 10\"C... c / 12,-' 10\"C -\u00b7 r\u00b7\u00b7- \u00b7\\ \u00b7\u00b7\u00b7 40\"C \u00b7\u00b7--\u00b7\u00b7.. \u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7r\u00b7\u00b7\u00b7 _t_ 3'\\. \\,, -10\"C \u00b7\u00b7~,..t-:\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7.. \u00b7- \u00b7---~... -..,... -.. - s Fig. U.13 \u2022\u2022, ; I!'Rlfrigrraiion Cyek, -=591 Ammonia is widely used in food refrigeration facilities such as the cooling of fresh fruits, vegetables, meat and fish, refrigeration of beverages and dairy products such as beer, wine, milk and cheese, freezing\u00b7 of ice cream and ice production, low temperature refrigeration in the pharmaceutical and other process industries. The advantages of ammonia are its low cost, higher COPs and thus lower energy costs, greater detectability in the event of a leak, no effect on the ozone layer, and more favourable thennodynamic and transport properties and thus higher heat transfer coefficients requiring smaller and lower cost heat exchangers. The major drawback of ammonia is its toxicity which makes it unsuitable for domestic use. Other nuids used as refrigerants are sulphur dioxide, methyl chloride, ethyl chloride, hydrocarbons like propane, bulalle, ethane, ethylene, etc. carbon dioxide, air and water. U.4 Absorption Refrigeration Cycle'The absorption refrigeration system is a heat operated unit which uses a refrigerant that is alternately absorbed and liberated from the absorbent. In the basic absorption system, the compressor in the vapour compression cycle is replaced by an absorber-generator assembly involving less mechanical work. Figure 14.15 gives the basic absorption refrigeration cycle, in which ammonia is the refrigerant and water is the absorbent. This is known as the aqua-ammonia absorption system. Ammonia vapour is vigorously absorbed in water. So when low-pressure ammonia vapour from the evaporator comes in contact in the absori>er with the weak solution (the concentration of ammonia in water isfow) coming from the generator, it is readily absorbed, releasing the latent heat of condensation. The Steam or \u2022ectr1dty-----;r--~ -.. Generator Condena8' I\u00a7 1 \u00b7 I Cooling water NH3 Liquid NH3 vapour Absorber Q,. Cooling waler Evaporator Brine (NaCl or CaClz In waler) flg. U.15 Vapo\"r a6sorption refrigeration plDnt-Jlow diagrar R.efriltTatjqn Cycks -=593 The final reduction in the percentage of water vapour in the ammonia going to the condenser occurs in the rectifier which is a water-cooled heat exchanger which condenses water vapour and returns it to the generator through the drip line, as shown in Fig. 14.16. The use of a suction-line heat exchanger is to reduce QA and increase Qn, thus achieving a double benefit. In the absorber the weak solution is &1)rayed to expose a larger surface area so as to accelerate the rate of absorption of ammonia vapour. There is anotb.er absorption refrigeration system, namely, lithium bromide water vapour absorption (Fig. 14.17). Here the refrigerant is water and the absorbent is the solution oflithium bromide salt in water. Since water cannot be cooled below 0\u00b0C, it can be used as a refrigerant in air conditionin11, units. Lithium bromide solution has a strong affinity for water vapour because of its very low vapour pressure. It absorbs water vapour as fast as it is released in the evaporat.or. Hz()Vap0ur Oc 7.ScmHgabs - - - Refrigeration L...;:::===t--... C)vaporator Wal.er 13 c load", "cp 3 X 14000 = 48.6 kW 0.24x3600 Ans. Ans.... I 4 We Fig, Ex.U.7 o; (a) - s (b) I I!! ii I + II 612=- 14.28 How can a heat pump be used for (a) space heating (b) year-round air conditioning? 14.29 How is a reversed Brayton cycle used for refrigeration? 14.30 Why is the COP ora gas cycle refrigeration system low? 14.31 Why is gas cycle refrigeration preferred in aircraft? 14.32 What is the principle of the Linde-Hampson system for liquefaction of air? 14.33 Derive the expressions of liquid yield and the minimum wor.k requirement.in a Linde-Hampson system. 14.34 How does Claude cycle differ from a Linde-Hampson cycle in the context of the liquefaction of air. 14.35 With the help of now andp-h diagrams, explain how dry ice is produced. PR.OBI.EMS 14.1 A refrigerator using R-134a operates on an ideal vapour compression cycle between 0.12 and 0. 7 MPa. The mass now of refrigerant is 0.05 kgls. Determine (a) the rate of heat mnoval from the refrigerated space, (b) the power input to ihe compressor, (c) the n.te of heat rejection to the environment, and (d) the COP, Ans. (a) 7.35 kW. (b) 1.85 kW, (c) 9.20 kW, (d)\u20223:97 l 4.2 A Refrigerant-I 2 vapour compression cycle has a refrigeration load of 3 tonnes. The cvaporu.lor o.nd condenser tcmpcn.rures an: - 20\u00b0C and 40\u00b0C respectlvely. Find (a) the refrigerant now rate in kg/s, (b) the volume flow rate handled by the compn:ssor in mJ/s, (c) the work input to the compressor in kW, (d) the heal rejected in the condenser in kW, and (c) the iS1.\"lltropic discharge temperature. lfthere is 5 C deg. of superbeating of vapour before it enters the compressor, and 5 C deg. subcooling of liquid before it flows through the expansion valve, detcm1ine the above quantities. 14.3 AS tonne R-12 plant maintains a cold store al - 15\u00b0C. The refrigerant flow rate is 0.133 kgls. The vapour lea.ves ihc evaporator witli 5 C deg. superheat. Cooling water is available in pleniy a.I 2S\u00b0C. A suction line heat exchanger subcools the refrigerant before throttling. Find (a) the compressor discharge temperature, (b) 1he COP, (c) tbe amou111 of subcooling in C deg., and (d) the cylinder dimensions of the compressor, if the speed is 900 rpm, stroke-to-bore ratio is l.2, and volumetric efficiency is 9S%. Allow approximately S\u00b0C temperature difference in the evaporator and condenser. Ans. (a) 66\u00b0C, (b) 4.1 (c) 125\u00b0C, (d) 104.S mm, 125 mm 14.4 A vapour compression refrigeration system uses R-12 and operates between pressure limits ofO. 745 and 0.15 MPa. The vapour cn1ering the compressor has a temperature of - 10\u00b0C and the liquid leaving the condenser is at 28\u00b0C. A refrigen.ting load of 2 kW is required. Detem1ine lhe COP and the swept volume of the compressor jf it has a volumetric efficiency of76% and :runs at 600 rpm. J Ans. 4.15, 243 cm 14.5 A food-freezing system i:i:quires 20 tonnes of refrigeration at an evaporator temperature of - 3 s\u2022c and a condenser temperature of 2S0C. The refrigerant, R- 12, is subcooled 4\u00b0C before entering the expansion valve, and the vapour is 614=- 14.14 A heat pump is 1ousean R-12 cycle to operate between outdoor air at- 1\u00b0C and air in a domestic heating system at 40\u00b0C. The temperature difference in the evaporator and the condenser is 8\u00b0C. The compressor efficiency is 80\u00b0A., and the compression begins with saturated vapour. The expansion begins with.saturated liquid. The combined efficiency of the motor and belt drive is 75%. lfthe required heat supply to the w", "ann air is 43.6 kW. what will be the electrical load in kW? 14.1 S An ideal (Carnot) refrigeration system operates between the temperature limiis of- 30\u00b0C and 25\u00b0C. Find the ideal COP and the power required from an extemal source to absorb 3.89 kW at low temperature. If the system operates as a heat pwnp, determine the COP and the power required to discharge 3.89 kW at high temperatUre. 1.4.16 An ammonia-absorption system has an evaporator temperature of- 12\u00b0C and a condenser temperature of S0\u00b0C. The generator temperature is 150\u00b0C. In this cycle, 0.42 kJ is uansferrcd to the ammonia in the evaporator for each kJ transferred to the ammonia solution in ibe generator from the high temperature soim:e. It is desired to compare the performance of this cycle with the performance of a similar vapour compression cycle. For this, it is assumed that a reservoir is available at I 50\u00b0C, and that heat is uansfcrrcd from this reservoir-to a revetsible engine which rejects beat to the surroundings at 2.S\u00b0C. This work is the.n used to drive an ideal vapour compression system with ammon.ia as th~ refrigerant. Compare the amount of refrigeration that can be achieved per kJ from the high temperature source in this case with the 0.42 k.J that can be achieved in the absorption system. 14.17 An R-12 plant is to cool milk from 30\u00b0C to I \u00b0C involving a refrigeration capacity of 10 ioMes. Cooling water for the condenser is available al 2s\u2022c and a 5 C deg. rise in its 1emperature is allowable. Determine lhe suitable condensing and evaporating temperatures. providing a minimum of S C deg. differential, and calculate the theoretical power required in kW and the cooling water requirement in kg/s. Also, find the percentage of na~b gas at the end of the throttling. Assume a 2 C deg. subcooling in the liquid refrigerant leaving the condenser. 14.18 The following daia pertain 10 an air cycle refrigeration system for an aircraft: Capacity 5 lonnes Cabin air inlet temperatW?: 1 s\u2022c and outlet temperature 25\u00b0C Pressure ratio ac.ross the compressor S The aircraft is flying at 0.278 km/s where the ambienl condi1ions are o\u2022c and 80 kPa. Find the COP and the cooling effectiveness of the heat exchanger. The cabin is al 0.1 MPa, and the cooling turbine powers the circulating fans. 14. 19 A waler cooler supplies chilled wi.t..:r at 7\u00b0C when water is supplied to it at 27\u00b0C at a rate of 0. 7 litres/min., while the power consumed amounts to 200 watts. Compare the COP of this refrigeration plant with that of ihe ideal refrigeration cycle for a similar situation. 14.20 A refrigerating plant of 8 lonncs capacity has an evaporation temperature of - 8\u00b0C and condenser temperatt1rc of 30\u2022c. The refrigerant, R-12, is ~'Ubcooled S\u00b0C before entering the expansion valve and the vapour is superheated 6\u00b0C before leaving the evaporator coil. The compression of the refrigerant is isentropic. If there is a suction pressure drop of 20 kPa through the valves. and discharge pn.-ssure drop of 10 kPa through the valves, detecmine the COP of the plant, ;.i. Psychrometrics The properties of the mixtures of ideal gases were presented in Chapter IO. The name 'psychrometrics \u2022 is given to the study of the propenies of airwater vapour mixtures. Atmospheric air is considered to be a mixture of dry air and water vapour. The control of moisture (or water vapour) content in the atmosphere is essential for the satisfactory operation of many processes involving hygroscopic materials like paper and textiles, and is important in comfort air conditioning. 15.1 Properties of Atmospheric Air Dry air is a mechanical mjxture of the gases: oxygen, nitrogen, carbon dioxide, hydrogen, argon, neon, krypton, helium, ozone, and xenon. However, oxygen and nitrogen make up the major pan of the combination. Dry air is considered to consist of2l% oxygen and 79% nitrogen by volume, and 23% oxygen and 77% nitrogen by mass. Completely dry air does not exist in nature. Water vapour in varying amounts is diffused through it. If pa andp., arc the partial pressnrcs of dry air and water vapour respectively, then 'by Dalton's law of partial pressures p", ".+p..,=p where p is the atmospheric pressure. :. Mole-fraction of dry air,.x. = I!.!_ = Pa p and mole fraction of water vapour, xw = p..., =p., p (\u00b7: p = l atJn.) Since Pw is very small, the saturation temperature of water vapour atp., is less than atmospheric temperature, '\u2022rm (Fig. IS. l ). So the water vapour in air exists in the superheated state, and air is said to be unsaturated.! I I! 'I 1,, 618=-... i Dew point temperalura t4 _., Fig. 15.1 Sllllu of waler uapour in mixture Relative humidity (R.H., ~) is defined as the ratio of partial pressure of water vapour,pw, in a mixture to the saturation pressure,Ps, of pure water. at the same temperature of the mixture (Fig. IS. I) R.H.(f)=~ P, If water is injected into unsaruroted air in a container, water will evaporate, which will increase the moisture content of the air, andpw will increase. This will continue till air becomes saturated at that temperatw-e, and there will be no more evaporation of water. For saturated air, the relative humidity is 100%. Assuming water vapour as an ideal gas Pw V =mw RK20T= n.,, RT 2oT\"' 111 RT and Ps J' = m, RH where JI is the volwne and T the temperature of air, the subscripts w and s indicating the unsaturated and saturated states of air respectively. ~= Pw = \"'\u2022 P. m, mass of water vapour in a given \u00b7 volume of air at temperatw-e T mass of water vapour when the same volume of air is saturated at temperature T =~=Xw x. n, Specific humidity or humidity ratio, W, is defined as the mass of water vapour (or moisture) per unit mass of dry air in a mixture of air and water vapour. If G = mass of dry air m = mass of water vapour W-=!!!. G \"' Specific bumidi ty is the maximum when air is saturated at temperature T, or -=619 W =W=!!!J_ s G ma.t If dry air and water vapour behave as ideal gases Pw Jl=mRw T P. f'= GR.T W=!!..:..&.....l!:!_=~ 8.3143/28.96 G Pa Rv, p-pw 8.3143/18..., W =0.622~ p- Pw (IS.I) where p, is the atmospheric pressure. IC p.., is constant, W remains cons1ant. If air is saturated at temperature T W = w. = 0.622 ___&__ p-p, where p. is the saturation pressure of water vapour at temperature T. T11e degree of saturatio11, \u00b5, is the ratio of the actual specific humidity and the saturated specific humidity, both at the same temperature r. \u00b5=Jf..= W. 0.622~ p-p.,, 0.622 ___J!J_ p- P, = Pw. p- P, P, p- p., If ~ = p., = 0, p., = 0, :x.., = 0, W = 0, i.e. for dry air, \u00b5 = 0 P, If ~= 100%,p..,=p,, W= W,,\u00b5= I Therefore,\u00b5 varies between O and l. If a mixture of air and superheated ( or unsaturated) water vapour is cooled at constant pressure, the partial pressure of each constituent remains constant until the water vapour reaches the saturated state. Further cooling causes condensation. The temperature at which water vapour starts condensing is called the dew poinr temperatiire, tdp, of the mixture (Fig. 15.1 ). It is equal to the saturation temperarure at the panial pressure, Pw, of the water vapour in the mixture. Dry bulb temperature (dbt) is the temperature recorded by the thcnnometer with a dry bulb. \u00b7 Wet bulb temperarure (wbt) is the temperature recorded by a the.nnometer when the bulb is enveloped by a cotton wick saturated with water. As lhe air stream flows past it, some water evaporates, taking the latent heat from the water soaked wick, thus decreasing its temperature. Energy is then transferred to the I I,, II L \u2022'\" 620=- /JaJic and. Applied 17imnodynamia wick from ihe", " The equations G1 + G2 = G3 G 1W1 + G2W2 = G3W3 G1111 + G21r2 = GiJ,3 result in Wi-~ = Jii-h3 =.\u00a7_ lf'.i - If\\ Gz Iii - lr1 From the psychrometric chart W1 = 0.0058, W2 \"' 0.0187 kg vap.lkg dry air (a) Fig. Ex. 15.5 20\u00b0c -oar (b) I!I It I Ps]drromdria h1 = 35, h2 \"'90 k.J/kg dry air 0.0187- ~ =~=.!. 2 W3 - O.OOS8 G2 W3 = 1. >< 0.187 +.!. x 0.0058 = 0.0144 kg vap./kg dry air 3 3 Again Iii-hi=~=.!. h3 -Iii G2 2 /r3 = 1_ /r2 = l 1,1 = _! X 90 + l X 35 3 3 J... 71.67 k:J/kg dry air 3 Final condition of air is given by W3 =0.0144 k8 vapJkg dry air lt3 = 71.67 k.Jlkg dry air Example 15,6 Saturated air at 21 \u00b0C is passed through a drier so lhat ita final relative humidity is 20%. The drier uses silica gel adsorbent. The air is lhc:n passed lhrough a cooler until its fmal temperature is 21 \u00b0C without a change in specific hwnldity. Find out (a) the temperature of air at the end of the d!ying process, (b) the heat rejected in kJ/kg dry air during the cooling process; \u00b7(c) Iii: relative humidity at the end of the cooling process, (d) the dew point temperature at the end of the drying pr~ss. and ( c) the moisture removed during the drying process in kg vapJkg dry air. Solution From the psychrometric chart (Fig.Ex. 15.6) 11.2\"C 21\"C 38.5-C OBT Flg. Es. 15.6 T2 = 38.s<'C /r3 = 60.S - 42.0 = 18.5 k.J/kg dry air /r1- Ans. (a) A,u. (b) I 11 II I PJ,rhromtlrin -=639 Example 15,8 Water at 30\u00b0C flows into a cooling tower at the rate of 1.15 kg per kg of air. Air enters the tower at a dbt of20\u00b0C and a relative humidity of 60% and leaves it at a dbt of 28\u00b0C and 90% relative humidity. Make-up water is supplied at 20\u00b0C. Determine: (i) the temperature of water leaving the tower, (ii) the fraction of water evaporated, and (iii) the approach and range of the cooling tower. Solution Properties of air entering and leaving the tower (Fig. 15.13) are twbl = J5.2\u00b0C t,.1,2 = 26.7\u00b0C h 1 = 43 kJ/kg dry air h2 = 83.5 kJ/kg dry air W1 = 0.0088 kg water vapour/kg dry air W2 = 0.0213 kg water vapour/kg dry air Enthalpies of the water entering the tower and the make-up water are li,.3 = 125.8 kJ/kg lim = 84 kJ/kg From the energy balance Eq. ( 15.5), hw3 - hw4 = _Q_ [(h2 - /, 1) - (W~ - W,)liwJ mw 1= \u00b7 1.15 [(83.5 -43)-(0.0213 - 0.0088) 84} Temperature drop of water =34.2 kJ/kg t 3 = t. = w w4 34 2 \u00b7 4.19 = 30 - t, \"\"\u2022 lw,1 = 2 l.8\u00b0C Approach= tw4 = l...t.1 = 21.8-15.2 = 6.6\u00b0C Range = lw3 - t..,,4 = 30 - 2 I.8 = 8.2\u00b0C Ftllcrion of water evaporated, :i: = G (J1'2 - W1) = 1(0.0213 - 0.0088) = 0.0 I 25 kg/kg dry Bir Example 15.9 Water from a cooling system is itself to be cooled in a cooling tower at a rate of 2.78 kg/s. The water enters the tower at 65", "\u00b0C and leaves a collecting tank at the base at 3 8\u00b0C. Air flows through 1he tower, entering the base at 15\u00b0C, 0.1 MPa, 55% RH, and leaving the top at 35\u00b0C, 0.1 MPa, saturated. MaJce\u00b7up water enters the collecting tank at 14\u00b0C. Determine lhe air flow rate into the tower in m3/s and the make-,up water flow rate in kg/s. Solution Figure Ex. 15.9 shows the flow diagram of the cooling tower. From the steam tables. at at I 5\u00b0C, p,., = 0.01705 bar, h8 = 2528.9 kJ/kg 35\u00b0C.p,., = 0.05628 bar. h,, = 2565.3 kJ/kg 640=- Hot satul8ted a.- {35\u00b0C, 100% R... ) 0 1---- - -~ I Nr {15\u00b0C, 55% RH) l ee;;e;;e;;e;e;e;;e;s,;s Jee;e;,-1.:::;::..,, \"\u00b7--\u00b7-----~-\u00b7--------\u00b7--------------\u2022\u2022 \u2022\u2022 \u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022\u2022 \u2022 \u00a9 Coldwa1er \u00a9 (30\u00b0C, m,.) Flg. Ex. 15,9 0.55.. u\u2022c f1 = (p p) Pwt = 0.55 X 0.01705 - 0.938 X 10\"1 bar f2 = - 1.00 Pw (p.)i,\u2022c Pwi = 0.05628 bar W1 = 0.622 _&_.. 0.622 X 0.938 X 10\u00b72 1.00 - 0.00938 p - Pw = 0.00589 kg vap.Jkg dry air W2 = 0.622 x l.:~~~::628 = 0.0371 kg vapJdry air Make-up water ~W1 -W1 = 0.0371 - O.OOS89 = 0.o3 121 -\"8 vap.Jkg dry air Energy balance gives For l kg of dry air cia (t2 :. 1.005 (35 - JS)+ 0.0371 X 2565.3- 0.00589 X 2528.9 t 1) + Wzl,2 - W11i1 + m..., (l,4 -1,3) -(W2 - W1)1J, = 0 + M.,.. (- 35) 4.187 -0.03121 X 4.187 X 14 = 0 Basic and Applied Thmnodynamics PROBLEMS 15.1 An air-wall!r vapour mixture at 0.1 Mpa, 30\u00b0C, 80% RH has a volume of SO m3 \u2022 Calculate the specific humidity, dew point, wbt. mass of dry air, and mass of water vapour. If the mixture is cooled at conswit pressure to s\u2022c. calculate the amount of water vapour condensed. IS.2 A sling psychromeler reads 40\u00b0C dbt and 36\u00b0C wbt. Find the humidity ratio, relative humidity, dew point temperature, specific volume, and enthalpy of air. 15.3 Calculate the amount of heat removed per kg of dry air if the initial condition of air is 35\u00b0C. 70% RI-I, and the final condition is 25\u00b0C, 60\"/o RH. I S.4 Two streams of air 25\u00b0C, 50% RH and 25\u00b0C, 60\"/o RH are mixed ad iabaiically to obtain 0.3 kg/s of dry air at 30\u00b0C. Calculate the amounts of air drJwn from both the streams and the humidity ratio of the mix.ed air. I S.S Air at 40\u00b0C dbtand 27\u00b0C wbt is 10 be cooled and dehumidified by passing it over a rcfrigeran1-lilled coil to give a final condition of I 5\u00b0C and 90% RH. Find the amounts of heal and moisture removed per kg of dry air. 15.6 An air-water vapour mixture enters a beater-humidifier unit a1 5\u00b0C, I 00 k.Pa, SO% RH. The flow rate of dry air is 0.1 kg/s. Liquid water at I0\u00b0C is sprayed into the mixture a1 tbc ra1e of 0.002 kg/s. The mixture leaves the unit at 30\u00b0C, I 00 kPa. Calculate (a) the relative humidity at the outlet, and (b) the n1te of heal transfer to the unit. 15.7 A laboratory has a volume of 470 m1, and is to be", ". 15.13 A grain dryer consists of a vcnical cylindrical hopper through which hot airis blown. The air enters the ba.~ at 1.38 bar, 65\u00b0C, SO% RH. At the top, saturated air is discharged into the atmosphere at I.03S bar, 60\u00b0C. Estimate the moisture picked up by I kg of dry air, a.nd th.e total enthalpy change between the eotering and leaving streams ell,pressed per unit mass of dry air. Ans. 0.0864 kJ/kg air, 220 kJ/kg air 15.14 Air ente!'ll a counterflow cooling tower al a rate of JOO m3/s at 30\u00b0C dbt and 40% relative humidity. Air leaves at tbe top of the tower at 32\u00b0C and 90% relative humidity. Water enters the tower at 35\u00b0C and the water flow rate is 1.2 times the mass llow rate of air. Make-up water is supplied ai 20\u00b0C. Whal arc ihc range and approach or the tower? At what rate is heat. absorbed from the load by the stream of water on its way back to the top of the tower? What percentage of the water now rate must be supplied as make-up water to replace the water evaporated in.lo the air stream'! An4. Range a 8.7\u00b0C, Approach ~ 6.3\u00b0C, Q = SOOS kW, % make-up = t.39\"A. II I'Reactive Systems In this chapter we shall study the thermodynamics of mixtures that may be undergoing chemical reaction. With every chemical reaction is associated a chemical equation which is obtained by balancing the atoms of each of the atomic species involved in the reaction. The initial constituents which stan the reaction arc called the reactants, and the final constituents which arc formed by chemical reaction with the rearrangement of the atoms and electrons are called the products. The reaction between lhe reactants, hydrogen and oxygen, to fonn the product water can be expressed as I H2+ 2 02 ~ H20 (16.1) The equation indicates that one mole of hydrogen and half a mole of oxygen combine to form one mole of water. The reaction can also proceed in the reverse direction. The coefficients I, f \u2022 I in the chemfoal Eq. (16.1) arc called stoichiometric coefficients. 16.1 Degree of Reaction Let us suppose that we have a mixture of four substances, A 1, A2, A 3 and A 4, capable of undergoing a reaction of the type v1 A 1 + v2 A 2 ~ V3A3 + V4A4 where the v\u2022s arc the stoichiometric coefficients. Starting with arbitrary amounts of both initial and final constituents, let us imagine that the reaction proceeds completely to the right with the disappearance of at least one of the initial constituents, say, A 1\u2022 Then the original nwnber of moles of the initial constituents is given in the fonn I I I\u2022 I! &activt S,stffll.S -=645 \"\u2022 (original) = no v, n2 (originsJ) = \"o V2 + N2 where n0 is an arbitra.ry positive number, and N2 is the residue (or excess) of A2, i.e., the number of moles of A 2 which cannot combine. lf the reaction is assumed to proceed completely to the left with the disappearance of the fmal constituent A3, then!13 ( original) = no V3 n4 (original) = n0 V4 + N4 where n0 is an arbitrary positive number and N4 is the excess number of moles of A4 left after the reaction is complete from right to left. For a reaction that has occurred completely to the left, there is a maximum amount possible of each initial constituent, and a minimum amount possible of each final constituent, so that 111 (max) = n0 v1 no V1 (Numberofmc.lcaofA 1 fonocd byQtQliuJ l'HClioP) (Original numberofmolesoU,l (no V3 A3 + \"o V4 A4 -t no V1 A, + no V2 A2) =(no+ no) v, = (no Vi+ Nz) (Orisillll munm- of molciofA2) = (n0 + no) v2 + N2 n3 (min)= 0 (The constituent A3 completely disappears by reaction) n4 (min) = N4 (The excess number ofmoles of A4 that are left after the reaction (N11111'bcrof111olnofA2 fomied bychnnic.lR*lion", ") \"o Vz + is complete to the left) Similarly, if the reaction is imagined to proceed completely to the right, there is a minimum emowit of each initial constituent, and a maximwn amount of each final corL!ltit:u.ent, so that n1 (min) =O n 2 (min) =N2 n3 (max)= no V3 + no V3 (A.moclal formed by cllem.ieel mac1ion) (n0 v1 A1 +,\u202211 v2 A2 -t n0 v3 A3 + n0 v4 A4) (Origia.l omounl) == (no + no)V3 n4 (max)= (n0 + no) v4 + N4 Let us suppose that the reaction proceeds partially either to the right or lo the.left to the extent that th.en: are n1 moles of A 1, n2 moles of A2, n3 moles of.43, and n4 moles of.44. The degree ( or advancement) of reaction \u00a3 is defined in terms of any one of the initial constituents, say,A1, as the fraction \"1 (max)- \"1 E = _....._........,.......... _..,__ n1 (max) - \"t (min) 1,1 It.Basic and Applitd J'hmnodynamics It is seen that when 111 = 111 (max), E = 0, the reaction will start from left to right. When 111 = n1 (min), e = I, reaction is complete from left to right. The degree of feaction can thus be written in the form ~./ ~ \u00a3= (no+ no)V1 -n, (no+ no)V1 Therefore n, = {no + no)v, - (no+ no} v,e = n (at start) - n (consumed) = Number of moles of.41 at start - number of moles of A 1 conswned in the reaction = (110 + n0) v 1 (I - e) 112 = n (at stan) - n (consumed) = (no+ no) V2 + N2 - (no+ no) V2f. = (110 + no) V2 (I - e) + N2 113..,, (at stan) + n (formed), = 0 + (no + no) V3 \u00a3 =(no+ no) V3E n4 = n (at start) + 11 (formed) = N4 + (Ito + llo) V4\u00a3 = (n0 + n0) V4E + N4 (16.2) The number of moles of the constituents change during a chemical reaction, not iodepe.ndently but restricted by the above relations. These equations arc the equations of constraint. The n's are functions of e only. In a homogeneous system, in a given reaction, the mole fraction x's are also functions of e only, as iUustrated below. Let us take the reaction I Hz+ 2 02~ H20 io which 110 moles of hydrogen combine with n0/2 moles of oxygen to fonn n0 rnoles of water. The n's and x's as functions of e are shown in the table given below. n X. n, ~ \"o (1- \u00a3) 111 2(1- \u00a3) x1~tn=J-e A A1.. H2 A2\u202202 V v1 = I I V2\u20222 112 = 1(1- f) 2 A3=H20 V3= I n1 = nr,E r.it =!!!.(3 - e) 2 1-\u00a3 :s-2=-3-\u00a3 2E x3=-3-\u00a3,, I,, \"' I \" luadi~ Systnns -=6.,1 ':. if the reaction is imagined to advance to an infinitesimal extent, the degree of reaction changes from E to E + de, and the various n \u2022 s will change by the amounts dn1 = - {i,o + lf'o) v, de dn2 =-(no+ no) V2 de dn3 = (no + nv) V3 d.t dn4 = (n0 + n;,) v4 d.t dlsi di,2 - - = - - = - - = -- = (n0 + 110) d\u00a3 -v1 -V2 dn 3 v3 d 114 V 4, or which shows that the dn's are proportional to the v's. 16.2 Reaction Equtlibrlum Let us consider a homogeneous phase having arbitrary amounts of the constituents. A", " 1,..42, A3 and A4, capable of undergoing the reaction v, A1 + V2A2 ~V3.43 + V4A4 The phase is at uniform temperature T and pressure p. The Gibbs fimction of the mixture is G = \u00b5 1n1 + 1,\u00a3zn2 + \u00b5,713 + P4n4 where then' s are the number of moles of the constituents at any moment, and the \u00b5.'s arc the chemical potentials. Let us imagine that the reaction is allowed to take place at coustant T and p. The degree of reaction changes by an infinitesimal amount from e to e + dE. The change in the Gibbs fimction is dGr. P =!\u00b5t dnt = \u00b5, dn, + \u00b5z dn2 + \u00b53 dn3 + \u00b54 dn4 The equations of constraint in differential form arc dn1 = - (n0 + n'0) v1 de, dn3 = (n0 + n'0) v3 d\u00a3 dn2 = - (n0 + n'0) v2 de, dn4 = (n0 + n'0) v4 de On substitutio.n. dGr. P - (n0 + n'o) (- v1\u00b5 1 - V'1}t2 + v3\u00b5 3 + v~ d\u00a3 (16.3) When the reaction proceeds spontaneously to the right, d\u00a3 is posifrve, and since dGT, P < 0 (v1\u00b5. 1 + v~) > (v)ll3 + v.JJ.i) If l: \\Vl1;. = ( v3\u00b5> + v.\u00b54) - ( v1\u00b5 1 + Y21,\u00a3z), then it ia negative for the reaction to the right. When the reaction proceeds spontaneously to the left, de is negative (v1\u00b51 + v~) < (v)l,l3 + v~) i.e., l:ytllt is positive. At equilibrium, lhe Gibbs function will be minimwn, and 1, \"'\u2022!!'6'8=- Barie and.Applied Tliermodynamits V1P1 + V2#1::i = V)}l) + V,Jl4 (16.4) which is called the tuJUiltio11 of reactio11 equilibrium. Therefore, it is Che value of l:VkA, which c:au.ses or forces Che spont.aneous reaction and iB called the 'chemical ajfinizy'. 16.3 I..aw of Mass Action For a homogeneous phase chemical reaction at constant temperature and pressure, when the constituents are ideal gases, the chemical potentials are given by the expressioos of the type 14. =in:~+ I.op+ In x.J where the f s are functions oftemperature only (Article I 0.11). Substituting in the equation of reaction equilibrium (16.4) v1 (~1 +!up+ In x1) + v2 (~ + lnp + In x2) = V3 ( ~ + In p + In X3) + V4 ( ~4 + In p + In X4) On reaminging V3 l.o.C3 + v,.1.iu.- Vi llu, - V2 IIU1 + (V3 + V,e - v, - Vi) lnp =-(V3 f3+ v4 ~.- v1~ 1 - v2\u00a2-l> Denoting In K = - (v3;, + V4~4 - v,~. - V1\u00a2-l> where K, known as the equilibrium constant, is a function oftemperarure only V4 ] V).C3 V1 \u00b7.C4 Vi X1 \u2022X2 [ v>+v,-v1-v2=K p \u00a3~\u00a3\u2022 (16.S) This equation is called the law of mass action. K has the dimension of pressure raised to the (v3 + v4 - v1 - vi}th power. Here the x's are the values of mole fractions at equilibrium when the degree of reaction is ~The law of mass action can also be written in this form i>{'\u00b7p;\u2022 p~'.,;:_2 where the p's are the partial pressures. =K 16.4 Heat of Reactton The equilibrium constant K is defined by the expression In K = - (v3;, + V4;. - V1f1 - v2~ 652=- The equilibrium constant is given by \u00a32 =ln--e-\u00b7p 1-E! Since the three gases are", " changes in K.E. and P.E., then the energy equation becomes or H11.=H, I, n,ii; = I, lte jje R p or I, ni [ii,0 + Mil - I, ne [.iir\u00b0 + MiJ R P (16.23) For such a process, lhe temperature of the products is called the adiabatic flame temperature which is the muimum temperature achieved for the given reactants. The adiabatic flame temperature can be controlled by the amount of excess air supplied; it is the maximum with a stoichiometric mixture. Since the maximum permissible temperature in a gas turbine is fixed from metallurgical considerations, close control of the temperature of the products is achieved by controlling the excess air. For a given reaction the adiabatic flame temperature is computed by trial and error. The eneq,,y of the reactants HR being known, a suitable temperature is chosen for the products so that the energy of the products at that temperature becomes equal to the energy of the reacmnts. 16.16 Enthalpy and Internal Energy of Combustion: Heating Value The entlialpy of combustion is defined as the difference between the enthalpy of the products and the enthalpy of the react.ants when complete combustion occurs at a given temperature and pressure. Therefore or ;;RP =Hp\u00b7-HR h~p = I, ne[ii? + t1ii]., - I, II; (iif + ~ii]; P R (16.24) where ii'RP is the enthalpy of combustion (kJ/k.g or kJ/kgmol) of the fuel. The values of the enthalpy or combustion of different hydrocarbon fu.els at 25\u00b0C, 1 atm. are given in Table 16.4. The internal enerro1 of combustion. uRP\u2022 is defined in a similar way. = L ne [ii?+ 11ii - pvJe - Ln;[iif + Mi - pv], r R If all the gaseous constituents are considered ideal gases and the volume of liquid and solid considered is BSSwned to be negligible compared to gascoua volume,, i,I I I II B.uit a,ul Applittl 17,mnody,camia ii'a, = jjRP - RT (nga...,us productt - ngaseouaroactani.) (16.25) In the case of a constant pressure or steady 'flow process, ihc negative of the enthalpy of combustion is frequently called the heating va/11e at constant. pressure, which represents the heat transferred from the chamber during combustion at constant pressure. Similarly, the negative of ihe internal energy of combustion is sometimes designated as ihe heating value at constant volume in ihe case of combustion. because it represents the amount of heat transfer in the constant volume process. The higher heating value (HHY) or higher calorific value (HCY) is the beat transfened when H20 in the products is in ihe liquid state. The lower heating value (LHY) or lower calo.ri fie value (LCY) is the heat transferred in the reaction when H20 in the products is in the vapour state. Therefore LHV = HHV -n%o \u00b7 hr, where '''Mio is th.c mass of water formed in the reaction. 16.17 Absolute Entropy and the Third Law of Thermodynamic. So far only th.e first law aspects of chemical reactions have been discussed. The second law analysis of chemical reactions needs a base for the enlropy of various substances. The entropy of substances at the absolute zero of temperature, called absolute entropy, is dealt with by the third. law of thermodynamics formulated in the early twentieth century primarily by W.H. Nemst (1864-1941) and Max Planck (I 853-l 947). The third law states that the entropy of a perfect crystal is zero at the absolute zero of temperature and it represents the maximum degree of order. A substance not having a perfect crystalline structure and possessing a degree of randomness such as a solid solution or a glassy solid, has a finite value of entropy at absolute zero. The third law (see Chapter \u00b711) provides an absolute base from which the entropy of each substance can be mcasu.red. The entropy relative to this base is referred to as the absolute entropy. Table 16.3 gives the absolute entropy of various substances at the standard state 25\u00b0C, I atm. For any other state sT.p = s~ + (A.i\">r.1a.iro.-.T.p where :si refers to the absolute entropy at l atm. and temperature T, and (Ash, 1 atm..... T. P refers", "n]\"2... ~ [p!L i f... V3 Y4 n[i,.;: where By logarithmic.differentiation, In= 111 + n:a + n3 + 114 and &v= v3 + v4 - v1 - v2\u2022 dK _ Liv dp + Liv dI.n = v, dn3 + v. dn4 _ Vi d111 _ v, dn2 K n3 \"2 Now, I:n n, p \"1 n1 =(n0 + n'o) v 1 (1-e) n2 = (no+ n' o) V2 (1 - \u00a3) 113 =(no+ n'o) V3\u00a3 114 - (n0 + n'o) v 4e :tn = (n0 + n'o) (v 1 + v2 + \u00a3Av) dn1 V1 d In= dn 1 + dn2 + dn3 + dn4,.. (n0 + n'o) Av dt - - = - -= -= - =(n0 +n 0)de dn 4 V4 dn3 V3 dn2 V2, (l) (2) (3) Again, From Eqs (1), (2) and (3), dK -Liv dp +Av(n0+n'0)i1vde K p = V 3 (n0 + n'o) v,dE (n0 + n'o) V 4dt + V 4..:....::'---....::.;;......;;:'-- n3 \"\u2022 -vi (no+ n' o) V1dt -(no+ n' o) V2dE -Vz......;.--\"--\"'-'-..::...- n1 \u2022 \"2 dK - Liv dp - (n0 + n' o> [ Vi + v! + ~ + Vi - (LiV) n3 n1 n 2 n4 In K p 2 J dt ~ II 678=- &sic and Applied Thmnodynamia...!L1n[z;>x;\u2022J \u2022 \u2022 dE x~I.x;z &ample 16.5 t=t\u2022 (a) At any value of e, = (n0 +n'0 ).(V1 +V1)(V3+V4) Proved. l:111t E (l - \u00a3) Starting with v 1 moles ofA1 and v2 moles of A2, show that: G = \u00a3 (v3\u00b53 + VJt4 - v.\u00b5. - ViJJ2] + V1\u00b51 + Vzlli (b) At equilibrium, G(min) = V1Jl1e + ViJJ2, where the subscript e denotes an equilibrium value. (} ( G - ( ) c R T (d) Ate=O, [. ) Y) v, ] nun = E In~ - In X3e %4\u2022 xfJ z;~ xf\u2022 z;2 \\14 VJ G- (}(min) RT (e) Ate= J, Solution (a) where, Gr.i, = Jl11t1 + \"'1\"1 + Jl3n1 + JJ..ar14 n1 = V1(1 - e). \"2 = V2(l - e), 1'13 = V3\u00a3 and 114 = V4E. Gr.p = \u00b51V1 (1- \u00a3) + JLiV2(1-e) + Jl3V3E+ \u00b54V4\u00a3 = E{V3\u00b53 + V4\u00b54 - V1.U1 - V2\u00b5iJ Proved..Proved. (b) _Ateq~librium, [! ~ l.p = V'J)l) + VJl., - V1Jl1 - V2J.'2 = 0. G(min) = V1Jlie + V2Jl2e (c) We have from (a) and (b), G-G(min) = [v3\u00b5 3 + v4\u00b54 - v1\u00b5 1 - v2 Jl2] + v1\u00b5 1 + V2Jli For an ideal gas, - V1\u00b51c - V2Jl1c \u00b5t =RT(~+ lnp+lnzt] luactive Systems -=679.XV3.XV4] +ln-3_.C_ xr\u2022 x;2 + v1RT (;1 + lnp + ln.r1) + v2RT", " has been given regarding the size of the heat exchanger for heating or cooling of a Ouid. If we consider a steel block heated in a furnace. to be allowed to cool in room air, we can estimate the amount of heat lost by the block in cooling by energy balance. But how Jong the cooling process will take place cannot be answered by thermodynamics. It is the science of heat transfer wh.ich is concerned with the estimation of the rate at which heat is transferred, the duration of heating or cooling for a certain heat duty W1d the surface area required to accomplish that heatdoty. There are three modes in which heat may be transferred: (a) conduction, (b) convection and (c) radiation. Conduction refers to the transfer ofheat between two bodies or two pans of the same body through molecules which are more or less stationary. In liquids and gases conduction results from the transpori of energy by molecular motion near the wall and in solids it takes place by a combination of lattice vibration and electron transport. ln general, good electrical conductors are also good thcnnal conducto.rs. Convection heat transfer occurs because of the motion of a fluid past a heated surface-the faster the motion, the greater the heat transfer. The convection heat transfer is usually assumed to be proportional to the surface area in contact with the fluid and the difference in temperature of the surface and fluid. Thus. Q = hA (Tw-TrJ where h is called the convection heat transfer coefficient, which is a strong fonction of both floid properties and fluid velocity (W/m?K). El.tmmJ.J of Hrat Tra,ufrr ~727 Radiation heat transfer is the result of electromagnetic radiation emitted by a surface because of the temperature of the surface. This dilTcrs from other forms of electromagnetic radiation such as radio, television, X-rays and r-rays which arc not related to temperature. 18.2 Conduction Heat Transfer Fourier's law of heat conduction states that the rate of heat flux is linearly proportional to th.e temperature gradient. For one dimensional or unidi~tional heat conduction, dt q oo dx or dt q=-K dx where q is the rate of heat flux in W/m2, dlldx is the temperature gradient in x-direction, and K is the constant of proponionnlity which is a property of the material through which heat is being conducted and is known as thermal conductivity. q is a vector quantity. The negative sign is being used because beat flows from a high to a low temperature region and the slope dt/dr is negative. (18.1) Fig. 18.1 Heat roruluction tlrrougla a tDall For a fini1e temperature difference (t 1 - (Fig. 18.1) t2) across a wall of thickness x - K 12 - t, - K r, - 12 WI If A is the surface area normal to heat now, then the rate ofheat transfer f2 -,, Q =q\u00b7A=-KA-x or Q \"\"KA. - 1 2 Watts t -t - X (18.2) The dimension ofthennal conducrivity is W/mK. Since dr/dr. =q/K for the same q, ifK is low (i.e., for an insulator), d1/dx will be lllfge, i.e., there will be a large temperature difference across the wall, and if K is high (i.e., for a conductor), dr/dr will be small, or there will be a small temperature difference across the wall.,, It'728=- 18.2.1 Iusuta\u2022ce Coac,pt Heat flow has an analogy in the flow of electricity. Otun's Jaw states that the i;wrent I flowing through a wire (Fig. 18.2) is proportional to the potential difference\u00a3, or I=.!_ R where 1/R is the constant of proportionality, and R is known as the resistance of the wire, which is a property of the material. Since the temperature difference and heat flux in conduction are similar to the potential difference and electric current respectively, the heat conduction rate through the wall may be written as. i. ti- Fig. 18.2 EJ\u00abtrical wistanct =,Pt Q =_KA tz - t1 = '1 -'2 = '1 - '2 R x/KA z where R = x/KA is the thermal resistance to heat flow offered by the wall (Fig. 18.3). For a composite wall, as shown in Fig. 18.4, there arc two resistances in series. The slope of the temperature profile depends on the thermal conductivity of the material. I; is the interface temperature", " Oiot number and Fourier number? What is their physical significance? 18.13 What do you understand by natural conve.ction and forced convection? 18.14 How is heat transfer coefficient defined. What is its dimension? 18.1 S What ere the lhR:e Rsi.stanccs offered to heat transfer from one fluid to another Lhrough a clean wall? 18. i 6 What is Raynolds number? What is its critical value when the flow through a Nbe becomes turbuleni? 18.17 Whai are Prandtl number and Nussdt number? 18.18 For fully developed laminar flow in a iubc, what are the values of Nusseh number (a) for constant wall temperilture, (b) for constanl wall heat flux7 18.19 What are the expressions of Nusselt number for (a) laminar flow and (b) turbulent flow. over a!lat plate'! 18.20 What is Dittus-Boelter equation? Where is it used? 18.21 What is Grashof number? When does it become signi.ficant? 18.22 What is a heat exchanger\"! 18.23 Why are counterflow heat exchangers superior lo parallel flow heat exchangers'? 18.24 What is log-mean temperature difference? 18.25 Why do the directions of flows of ihe two fluids in a heal exchanger become immaterial when one of the two fluids undergoes phase change'! l 8.26 When is e-NTU method convenient to use in heat e11changcr analysis'! 18.27 Define (a) effectiveness, (b) heat capacity ratio and (c) NTU, in reganl to a heat exchanger. I 8.28 Derive the expression for effective11ess in a (a) parallel flow heat exchanger. (b) counter-now heat exchanger. 18.29 What is the elltpression for effectiveness when one of the fluids undergoes phase change? 18.30 Find the expression for effectiveness of a balanced heat exchanger with equal heat capacities. 18.3 I Define absorptivity. reflectivity and trartsmissivity. 18.32 What is emis.,ivity? Whal is Kirchhoff's law? 18.33 What is a black body? 18.34 What is the Stcfan\u2022Boh:zmann law? 18.35 What is the view factor? Why is it significant in radiant he-<11 exchange between two bodies? 18.36 What is the n.\"Ciprocity theo.rem'l..,. 768=- Dasi, and Applied Thmnodynamit.1 18.37 What is a gray body'? 18.38 What are meant by (a) monochromatic emissivc power, (b) total cmissive power? 18.39 What is Planck's law of thermal radiaiion? Explain its importance. 18.40 Define radiosity and irradiation for a gray body. 18.41 Explain (a) surface resistance and (b) space resistance in radiant energy chaqge between two gray bodies. 18.42 Give the radiation network for two gray surfaces anti derive the view factor. 18.43 What do you mean by ''floating node\"'! 18.44 How would you define the heat transfer coefficient for combined convection and radiation? PROBLEMS 18.1 A room has a brick wall 25 cm in thickness. The insldc air is at 2S\u00b0C and ihe outside air is at - I S\u00b0C. The heat transfer coeflicients on the inside and outside are 8.72 and 28 W/m2 K respectively. The thennal conductivity of brick is 0.7 W/mK.. Find the rate of heat transfer through the wall and the inside surface temperature. 18.2 For the wall in the above problem, ii is proposed. to reduc.e the heat transfer by fixing an insulaling board {K = 0.05 W/mK), 2.S cm in thickness, 10 the inside surface. Find the rate of heat transfer and the inside surface temperature. HU Sheets of brass and steel, each I cm thick. are placed in contact. The outer surface of brass is kept at I 00\u00b0C and the outer surface of steel i~ kept at o\u2022c. What is the temperature of the common interface'! The thermal conductivitie.s of brnss and steel are in the ratio 2: I. Ans. 66.7\u00b0C 18.4 In.a pipe carrying steam, the oulsidc surface (15 cm OD) is at 300\u00b0C. The pipe is to be covered with insulation (K = 0.07 W/mK) such that the outside surface temperature does not exceed 60\u00b0", " rod. Only three of the four L's are independent, for it can be proved that, if the departUre from equilibrium is not great, L12 =L21 This is known as Onsager reciprocal relation. 20.3 Phenomenological Laws (20.8) A large number of phenomenological laws exist, which describe Ute irreversible processes in the form ofproportionalities, e.g. Fo11rier's law between heat flow and temperature gradient,Fick'slaw between flow of matter of a component in a mixture and its concentration gradient, Ohm's law between electrical current and potential gradient, Newton's law between shearing force and velocity gradient, the chemical reaction law between reaction rate and chemical potential. The causes which are responsible to the occurrence of these irreversible phenomena, such as, the temperature gradient, potential gradient, concentration gradient, and chemical affinity arc called the generalized forces, denoted by Xh = I, 2...., 11). The irreversible phenomena, such as heat flow, electrical current flow, dilTusion, chemical reaction rate, et.c. caused by thefi>rces arc called.fluxe~\u00b7, symbolized by.Ji(i\"' I, 2,..., 11). A thermodynamic force may be defined as a quantity which measures the extent to which the system is displaced from equilibrium. When two or more of these phenomena occur simultaneously, they interfere and give rise to new effects. The examples of such cross-phenomena are: ( 1) The two reciprocal phenomena of thermoelectricity arising from the interference of heat conduction and electrical conduction. viz., Seebeck effect and Peltier effect. (2) The coupling of diffusion and heat conduction giving rise to thermal diffusion, called the Sorer effect (concentration gradient fonncd as a result of a temperature gradient) and its inverse phenomenon, theD1ifo11r effect(temperature difference arising when a concentration gradient exists). Two coupled transport processes can be e-xpressed in the generalized form (20.9) J 1 =L11 X1 + L12X2 J2 = L21X1 + L12X2 (20.10) For two primary processes (say, heat conduction and flow of electricity) the basic or primary laws will be of the form: J 1 = l 11 X1, for process I alone, say, heat conduction, where J 1 = JQ, X1 = dT!dx and L11 is the thermal conductivity, and, J2 = L2i)(2, for process 2 alone, say electrical flow, whereJ2 = J 1,.X2 = dE/dt and \u00a3.i2 is the electrical conductivity. Iii I 830~ Basic aNi Applied Thermodynamics If process 2 influences process 1 and vice versa, J 1 = L12 X2 (the quantity of heat now J 1 due to electrical potential X1); J2 = L21 X1 (the electrical current flow J2 due to temperature gradient X2). The latter two processes are called coupled processes, and the coefficientsL12 and.Li I are called coupling coefficients. The first digit in the subscripts of the L's refers to the flux and the second digit refers to the force. If L 12 = L21 = O,. lhe fluxes are dependent only on the primary forces and are uncoupled. 20.4 Rate of Entropy Generation: Principle of Superposition The rate of entropy generation is the product of forces and fluxes. From Eq. (20.5), (20.11) where or liT T liE a.ndX2 ~ ~ The rate of entropy generation in irreversible steady state coupled processes is thus the sum of the entropy generation for each of the processes, i.e. (20.11 a) This is known as the principle of superposition as stated by Verschaffelt er= J1X1 + J~2 + J~3 +... (Prigogine, 1961). Let us consider a control volume (fig. 20.3) in which the properties vary from point lo point and wilh time. Toe internal energy: u \"\"j{z, y, z:, t) Fig. 20.3 Htal a11ti work flow i11 a mlllrol oolu!M! I!! ii I + II lrrrvnsihk 17urmodynamics -=833 or, By Kirchhoff's law, divJ1= 0 div Jw = E div J1 + J1 grad E divJw -.fa grad E By substituting in Eq. (20.19), Therefore, p :; = - div J0 -J1 grad E as - div JQ p-----T 01 JI grad E T =-div Jo -Jo grad T - J, grad E T T2 T", "20.56) The Peltier coefficient is called the Peltier emf, since its unit is volts. The Peltier emf at a junction depends 011 temperature and the materials of the junction. If Ohmic (or Joulean) heating is neglected, then Eq. (20.55) reduces to: JQ =Ji T(S~ -S~ =J1 Ku If the current is reversed.. in direction (Fig. 22. 7), (20.57) JQ = (JQ)a- (JQ),. = J1 T (S~ - S~) (20.58) Peltier heat is, then!fore, absorbed in the junction to keep it isothermal at = -,rA. B JI iemperature T. Peltier effect is thus rellef'sible. 8 Lo Fig. 20. 7 Iuomal of nmml i11 tflt jundion of diJsillWDr maurialJ, nl I I II -=843 Thomaon Effect Let us consider an element of length Ill of a rod or a wire through which heat is flowing steadily by conduction due to a temperature difference AT (Fig. 20.8). If an electric current I is sent through the element, it is found that the temperature distribution in the element is altered by an amount that is not entirely due to Joule effect. The rate, at which the electrical energy is dissipated into internal energy increase of the element, is greater or less than the f R beating, the difference depending on the magnitude and direction of the current, on the temperature, and on the material. This phenomenon is known as the Thomson effect. Allowing for Joule effect, the heat that must be supplied or extracted laterally at all places along the element to restore the initial temperature distribu.tion (without electric current} is called the Thomson heat. x - - -... Flg. 20.8 Coupltd flow of Atat and tuctricil'j in an elnnnit Let JQ represent the beat that must be removed laterally per unit area in unit time from the element carrying a current to restore the element to its original temperature distribution. If M is the potential difference across the clement, the rate at which electrical work is done on the element is/t:.E. The rate at which heat is transponed per unit area by the electrical current into the element is (JQ>r+AT = J1(Q*)r + AT, and the rate at which. heat is transported bl the current out of the element per unit area is (JQ}r = J1(Q\u2022)r, both in kW/m. The conduction heat flowing into and out of the clement is the same and need not be considered here. By first law, JQ = J, t:.E + J,(Q\")r + Ar - J,(Q*)y = J, [A\u00a3+ (T+ An (S*)r.,1. r- TS*] (S*)r+ AT =S* + dT AT dS Now, Therefore,,, I 11 H 1,I 844=- Basie afld Applied Thmnodynamies or + ds\u2022 (~r)2 - rs\u2022] dT Jo =J{~E + T ~s; \u00b74T + s\u2022 llT] neglecting the small quantity ds\u2022 (ii7)2\u2022 dT From Eq. (20.50), or Also, Substituting tlE i.n Eq. (20.59), (20.59) (20.60) J'Q=J,[:!l...llx-S*llT+T dS* llT+s\u00b7~r] A. J 2 =-t!i.x+J,T dT A.T dS* dT The current density J 1 is given by l:!.E dE J,=-A.- =A.llx dx Therefore, the Joulcan heat per unit area is: J1A.E\u00b7J1= J,r~ = 1 llx From Eq. (20.61), (20.61) {20.62) This is the excess heat that must be removed per unit area from the element laterally, over and above the Joulean heat, and is the Thomso11 heat. The quantity T dS \u2022 A.Tis expressed in volts and is called the Thomson emf. Thomson heat is dT. proportional to J1 as well as A..T and the constant of proportionality is called the Thomson coefficient, a. If beat is added to keep the same temperature, a is positive. If heat is removed, a is negative. So, /rra,mible Tlttrmodynamics", " motion of its molecules. The gas is assumed to consist of a large number of identical, discrete particles called molecules, a molecule being the smallest un.it having the same chemical properties as the substance. 21.1 Molecular Model The kinetic theory is based on certain assumptions regarding the molecular nature oflhe gas. I. Any finite volume of a gas consists of a very large number of molecules. The number of molecules i.n I kgmol of a gas is 6.023 x I 026, which is the Avogadro's number N0. At standard conditions of 760 mm Hg and 0\u00b0C, I kgmol of gas has a volume of 22.4 m3, so that there are approximately 3 x 1016 molecules in 1 mm1 ofvolwne. 2. The molecules are like identical hard spheres moving about continuously in random directions. They are separated from one another by distances about JO times the molecular diameier (2 to 3 x 10-10 m). 3. The molecules move only in straight line paths, the directions of which change only by collisions. 4. Collisions of molecules are perfectly elastic, so that. there is no decrease in kinetic energy during a collision. S. The molecules arc disiributed uniformly throughoui the container. lftbere are N molecules in a container of volume V, Lhe average number of molecules per unit volume, n, is equal to N/V and dN = ndV, where the Ii, M \" Kinttie Tlieory of Gases and Distribution of Moltr:ular Yel\u00abities -= 861 volume clement dV is small compared to the dimensions ofihe container, but large enough to contain many molecules. 6. All directions of molecular velocities are equally probable. 21.2 Distribution of Molecular Velocities in Direction Let us imagine that to each molecule is attached a vector representing the magnitude and direction of its velocity: When all these velocity vectors are transferred to a common origin, we have a distribution in velocity space (Fig. 21. l ). Let us now construct a sphere of arbitrary radius r with its centre at the origin (Fig. 21.2). The velocity vectors, extended if necessary, intersect the surface of the sphere at as many points as there are molecules. The average number of these velocity points per unit area is NI( 4 tr?). An element of area on the surface of the sphere of radius r in an arbitrary direction specified with reference to a polar coordinate system by the angle 8 and ~ (Fig. 21.2) is given by: Fig. 21.1 Vtl\u00abilJ D\u00abtors of moltr:ules in oelocitJ spare dA =? sin 8\u00b7d9\u00b7d~ The number of molecules having velocities in a direction between 9 and 9 + d 9 ; and ; + d;, which is denoted by d2 N8\u2022, is d2N.... =_!!__?sin 9\u00b7d8\u00b7 d; \"\" 4Rr2 =Ji.sin 8\u00b7dB\u00b7d9 41l' (21.1) f h., 862=- /JaJic and Applitd 17termodynamia Fig. 21.2 Elrmtnlal area on tit, surface of a sphere in 11tlodty spact When we divide the two sides of the equation by volume V and substitute n -= NIV, we get: (21.2) The magnitude of the molecular velocity is the speed. Not all the molecules have the same speed. The speed of the molecules can vary from zero to the speed oflight. However, for mathematical convenience, we will assume the molecular speed varying from zero to infinity. 21.3 Molecular Collisions with a Stationary Wall Any surface in contact with a gas is constantly bombarded by the gas molecules from all directions and with all speeds. Let us consider th.e area dA of such a surface (Fig. 21.3) with the angles 6 and 4' so specified. The nurober of molecules travelling in the 64' direction and with a cenain velocity v are designated by dne;,\u00b7 The collision of any one of these molecnles with area dA is called a 64Jt) collision. The slanted cylinder (Fig. 21.3) has edges in the direction 8, 4' and a length vdt, equal to the distance travelled in the 84' direction with speed v. The number of IJ;v collisions with area dA in time di equals the nwnber of 64'V molecules in this volume. Let dnv represent the number of molecules per unit volume with speeds between v and v + dv.. From Eq. (21.2), the number of 94Jt) molecules per unit volume is: d)n...., = - 1-dn sin 9-d9\u00b7d\"' V \" 4Jl' The volume of the", "Fig, 21. 7) so that the surface area of a sphere represents, at an instant of time, all molecules of equal velocities. Each molecule has a representative point in velocity space. The number of molecules whose velocities lie between v and Fig. 21.7 ~tlocity S/J4U v + dv would be represented by the spherical strip of thickness dv, and be denoted by dNv. Since the total number of molecules N is very forge, the strip dv, although small. still contains a large number of molecules. Let dN.,, represent the number of molecules whose x-component velocities lie between v, and v, + dv... Then the fr.iction (dNv.)IN is a func.tion of the rnagnjtude ofvA and the distance dvx, or dN. N = tci,.)dv. (21.26) \"' Kinttic 'INo,y of Gaus and Distribution of Moltcular Ye/Qatits -= 871 wbere/(v.J is called the distribution function fou-corupooent of velocity. Similarly, and dNv __, =f(r, )chi N y y dN ---1 = f(v) dv N z z (21.27) {21.28) Some of the molecules of dNv, have y-component velocities lying between Vy and Vy+ dvy, and let this be represented by d2 N.,.,.,. A differential of second order <t\" N~ v has been used, since this is a small fracti~n of an already small fraction, but still large enough to contain many molecules. Since the number of molecules is large, the following relation holds good......., d 2 N v,,vy = _.:.i._ dN dNvx N (21.29) or d2N v \u2022\u2022 v1 \"\u2022 \"\"dN dN...l. Vy N = Nf (vJf(vy) dr,1 dvy (21.30) Similarly, d3Nv., v.,. v, n:present the number of molecules whose velocity components would lie between Vx 8Dd Vx + dvx, Vy and Vy + dvy and v 1 and 'fJ1 + dv1 \u2022 Therefore, (21.31) The number of representative points per unit volume, or the density of poi.tits in d)Nv V V = N/(Vx)/(Vy)/('fJJ dt\\ doy chi,.,,.,. t velocity space, repree;ented by(). is: p= \u2022\u00b7 '' \u2022 =N/(vJ/(vy)/(vJ d3N,, v,, dv.. dvyd'fJ1 (2 1.32) Since the velocity distribution is isotropic, the density is the same in any volume element so that: dp = Nf(v.)/(v1)/('fJJ dvx + NJ ~v.J/'(vy)/('fJJ dvy + Nf (oJ/('fJy)/'(vJ dt>t = O f'('o.) d'fJ + /'{vy) dv + /'(v1 tdv =O /(v.) /(vy) /('c,~) \u2022 \u2022 Y or (21.33) At a radial distance v from the origin, v1 = v: + v 2 + v 2 = constant \u2022 y \" v,dv, + vydl?y + vi<f11i = 0 (21.34) To solve the Eq. (21.33) subject to the constraint of the equation (21.34), Lagrange's method of undetermined multipliers is applied. Multiplying Eq. (21.34) by,t and adding to Eq. (21.33),, Iii It 872=- where.tis called the Lagrange's multiplier. Since u,.. Vy and v\" a.re now independent variables, the coefficients of dv,v dvy and dv1 are individually equated to zero. (21.35) lt f'{v.) +,tu = O /('u.) r<v.,) /(v.,) + lv., = 0 /'(Va) +,\\v = 0 /(TJ,) z From Eq. (21.36), or Similarly, -(A~)12 1 /(v1)= a\u00b7 e /(vJ = a\u00b7 e-().~)'2 and (21.41) where a is the constant of integration. The symmetry provides the same integration constant for", " 2375.3 2382.2 2389.2 2396.0 2402.9 2409.8 2416.6 2423.4 2430.1 2436.8 2443.5 2450.1 2456.6 2463.1 2469.5 2475.9 2482.2 2488.4 2494.5 2500.6 2506.S hr hr, 0.00 20.98 41.99 62.98 83.94 IO<I.87 1:zs.n 146.66 167.54 188.42 209.31 230.20 2S1.1 I 272.03 292.96 313.91 334.88 355.88 376.90 397.94 419.02 2SOIJ 2489.6 2477.7 2465.9 2454.1 2442.3 2430.5 2418.6 2406.7 2394.8 2382.7 2370.7 2358.5 2346.2 2333.8 2321.4 2308.8 2296.0 2283.2 2270.2 2257.0 It, 2SOl.3 2510.5 2519.7 2528.9 2538.1 2547.2 2556.2 2565.3 2574.3 2583.2 2592.1 2600.9 2609.6 2618.2 2626.8 2635.3 2643.7 2651.9 2660.1 2668.1 2676.0 s, 0.0000 0.0761 0.1510 0.2245 0.2966 03673 0.4369 0.5052 0.5724 0.6386 0.7037 0.7679 0.8311 0.8934 0.9548 1.0154 1.0752 1.1342 1.1924 1.2500 1..3068 \u00a311ap. Vapour sr,.fg 9.IS62 8.9496 8.7498 8.5569 8.3706 8.1905 8.0164 7.8478 7.6845 7.5261 7.3725 7.2234 7.0784 6.9375 6.8004 6.6670 6.5369 6.4102 6.2866 6.1659 6.0480 9.1562 9.0257 8.9007 8.7813 8.6671 8.5579 8.4533 8.3530 82569 8.1647 8.0762 7 \u2022. 9912 7.0015 7.830') 7.7552 7.6824 7.6121 7.5444 7.4790 7.4158 7.3548 \u2022 Adapted from Joseph H. Keenan. Frederick G. Keyes, Philip G. Hill, and Joan G. Moore, Steam Tables, John Wiley and Sons, New York, 1969. i... [ ;,,,1! T\u2022b.le A.l.2 Sa111rattd wattr: Pressurt Ta.bk Specific Vo/11mc. m1/kg lntema/ E:.'nerg_v. kJ/kg &tha/J~v. kJ!kg E11rropy, kl/kg K Evap. Vapo11r Pressure Temp, \u2022c T Ir Pa p 0.6113 1.0 l.S 2.0 2.5 3.0 4.0 s.o 7.S 10.0 15.0 20.0 25.0 30.0 40.0 50.0 75.0 MPa 0.100 0.125 0.150 0.175 0.200 0.22S 0.250 0.01 6.98 13.03 17.50 21.08 24.08 28.96 32.88 40.29 45.81 53.97 60.06 64.97 69.10 75.87 81.33 91.77 99.62 IOS.99 111.37 116.06 120.23 124.00 lI7.43 Sat. Liquid v, 0.001000 0.001000 0.001001 0.001001 0.001002 0.001003 0.001004 OJ.'J01005 0.001008 0.001010 0.001014 0.001017 0.()01020 0.001022 0.001026 0.001030 0.001037 0.001043 0.()()1048 0.001053 0.001057 0.001061 0.001064 0.001067 : - Sat. Vapour v, 206.132 129.208 87.980 67.004 54.2S4 45.665 34.800 28.193 19.238 14", ".674 I0.022 7.649 6.204 5.229 3.993 3.240 2.217 Sar. u, Sar. liquid Evap. Vapour \", u,, 2375.3 237S.J 2355.7 2385.0 2338,6 2393.3 2399.5 2326.0 2404.4 2315.9 2408.5 2307.5 2293.7 2415.2 2420.5 2282.7 2430.5 2261.7 2437.9 2246.1 2448.7 2222.8 2456.7 2205.4 2463.1 2191.2 2468.4 2179.2 2477.0 2159.5 2483.8 2143.4 24%.7 2112.4 0 29.29 S4,.70 73.47 88.47 101.03 121.44 137.79 168.76 191.79 225.90 251.35 271.88 289.18 317.51 340.42 384.29 Sat. Liquid hr 0.00 29.29 54.70 73.47 88.47 101.03 121.44 137.79 168.77 191.81 225.91 251.38 271.90 289.21 3l7.SS 340.47 334.36 Sat. E1\u2022ap. Yapour Liquid h, 2SOl.3 2514.2 2525.3 2533.5 2540.0 2545.5 25S4.4 2561.4 2574.8 2584.6 2599.1 2609.7 2618.2 2626.3 2636.7 2645.9 2663.0 Sat. s, 0 O.IOS9 0.1956 0.2607 0.3120 0.3545 0.4226 0.4763 0.5763 0,6492 0.7548 0.8319 0.8930 0.9439 1.0258 1.0910 1.2129 hri 2S01.3 2484.9 2470.6 2460.0 2451.6 2444.5 24329 2423.7 2406.0 2392.. 8 2373.l 2358.3 2346.3 2336.1 2319.2 2305.4 2278.6 l.6940 1.3749 1.1593 1.0036 0.8857 0.7933 0.7187 417.33 444.16 466.92 486.78 504.47 S20A5 535.08 2088.7 2069.3 2052,7 2038.l 2025.0 2013.1 2002.1 2506.1 2513.5 2519.6 2524.9 2529.5 2533.6 2S37.2 417.44 444.30 46W8 48697 504.68 520.69 SJS.34 2258.0 2241.1 2226.5 2213.6 2202.0 2191.3 2181.S 2675.5 2685.3 2693.5 2700..5 2706.6 2712.0 2716.9 1.3025 1.3739 1.4335 1.4848 1.5300 l.570S I.W72 s,, 9.IS62 8.8697 8.6322 8.4629 8.3311 8.2231 8.0520 7.9187 7.6751 7.50!0 7.2536 7.0766 6.9383 6.8247 6.6441 6.5029 6.2434 6.0568 5.9104 5.7897 5.6868 5.5970 5.5173 S.4455 Sat. s. 9.IS62 89756 8.8278 8.7236 8.6431 8,5775 8.4746 8.3950 8.25.1 4 8.1501 s.. 0084 7.9085 7,8313 7.7686 7.6700 7.5939 7.4563 ~ OI II i:,, r ~-.. g_ :l a. ~ ~ SI.!, t ;I i'l' 7.3593 7.2843 7.2232 7.1717 7.1271 7.0878 7.0526 ----- ~cljlc l\"olu111e.,,,'Ag Internal Ener,ry. lcJ/kg Ltllrtipy. k.Jlkg Entropy. k.Jlkg K MP a p Sat. Pressure Temp. oc liquid Vr T 2.50 223.9') 0.001 197 0.001207 2", ".15 229.12 0.001216 3.00 233.90 0.001226 3.25 238.38 0.001235 3.50 242.60 0.001252 250.40 4.0 5.0 0.001286 263.99 0.001319 275.64 6.0 0.001351 7.0 285.88 0.001384 295.06 8.0 9.0 0.001418 303.40 0.001452 IO.O 311.06 0.001489 318.15 11.0 0.001527 324.75 12.0 0.001567 330.93 13.0 0.00.1611 336.75 14.0 1;.o 0.001658 342.24 0.00l711 347.43 16.0 17.0 352.37 0.001770 357.06 O.iXJl840 18.0 0.001924 361.54 19.0 0.002035 365.81 20.0 0.002206 369.89 21.0 22,0 0.002808 373.80 0.003155 22.09 374.14 Sat. Vapour V I o.oms 0.07275 0.06668 0.06152 0.05707 0.049778 0.039441 0.032440 0.027370 0.02351& 0.0204!!4 0.018026 0.015987 0.014263 0.012780 0.011485 0.010338 0.009306 0.008365 0.007490 0.006657 0.005834 0.004953 0.003526 0.0031S5 =- ~ Ur, Sat. Sat. Sat. Sat. Sat. liquid Evap. Vapour Uquitl Evap. Vapour Liquid lit 962.09 2603.1 985.97 2603.8 1008.41 2604.1 1029.60 2604.0 1049.73 2603.7!087.29 2602.3 1154.21 2597.1 1213.32 2589.7 2580.5 1266.97 2569.8 1316.61 1363.23 2557.8 1407.53 2544.4 2529.7 1450.05 2513.7 1491.24 1531.46 2496.1 1571.08 2476.8 1610.45 2455.4 1650.00 2431.7 1690.25 2405.0 1731.97 2374.3 1776.43 2338.1 1826.18 22.93.1 2230.7 1888.30 2081.4 2034.92 2029.6 209').26 Ir# 2803.1 2803.9 2804.1 2804.0 2803.4 2801.4 2794.3 2784.3 2772.1 2757.9 2742.1 2724:7 2705.6 2684.8 2662.2 2637.5 2610.5 2580.6 2547.2 2509,1 2464..5 2409.7 2334.7 2159.0 2099.3 \"rs 1644.0 1621.2 159'>.3 1578.4 1558.3 1520.0 1449.3 1384.3 Dao 12<,4.3 1207.3 1151.4 1096.1 1040.8 985.0 928.2 869.8 809.1 744.8 675.4 598.2 507.6 388.7 108.2 0 Sr 2.5546 2.6018 2,6456 2.6866 2.7252 2.7963 2.9201 3.0266 3.1210 3.2067 3.2857 3.3595 3.4294.).4961 3.S604.3.6231 3.6847 3.47(,() 3.8078 3.8713 3.9387 4.0137 4.1073 4.3307 4.4297 \"r 9S9.09 982.65 1004.76 1025.62 1045.41 1082.28 1147.78 1205.41 1257.51 1305.54 1350.47 1393.00 14.33.<,8 1472.92 1511.09 1548.53 1585.58 1622.63 16(,0.16 1698.86 1739.87 1785.47 1841.97 1973.16", " 2029.58 ''rs 1841.0 1817.9 1795.7 1774.4 1753.7 1714.1 1640.1 1571.0 1505.1 1441.3 l'.378.9 1317.1 1255.5 1193.6 1130.8 1066.5 1000.0 930.6 856.9 777.1 688.1 583.6 446.4 124.0 0 Evap. Vapour Sat..i, 6.2574 6.2208 6.1869 6.1551 6.1252 6.0700 5.97.33 5.8891 S.8132 5.7431 5.6771 5.6140 5.5527 5.4923 5.4323 5.3716 5.3097 S.2454 S.1776 5.1044 5.0227 4.9269 4.8015 4.S224 4.4297 s,, 3.7028 3.6190 3.5412 3.4685 3.4000.3.2737 3.0532 2.8625 2.6922 25365 2..3915 2.2545 2.1233 1.9962 1.8718 1.7485 l.6250 J.49')5 1.3698 1.2330 1.0841 0.9132 0.6942 0.1917 0 <D t,O QO II ;;;\u2022 8'.. l i [.. ;;I i1,!. :,.. s\u00b7 ;I -~\u00b7 14.674 14.869 17.196 19.513 21.825 24.136 26.445 31.063 35.679 40.295 44.911 49.526 54.141 58.7S7 63.372 67.987 72.603 P \"' 10 kPa (4S.8J) ll 2437.9 2443.9 2515.5 2587.9 2(,61.) 2736.0 2812.1 2.968.9 3112.3 3302.S 3479.6 3<163.8 3855.0 4053.0 4257.5 4467.9 4683.7 h s 2584.6 8.1501 2592.6 8.1749 2687.5 8.4479 2783.0 8.6881 2879.5 8.9037 9,1002 2..977.3 3076.5 9.281 2 3279.5 9.ftJ76 3489.0 9.8977 370S.4 10.1608 392l!.7 Hl.4028 10.6281 4159.t 10.8395 4396.4 11.0392 4640.6 11.2287 4891.2 I. 1.4()()0 5147.8 11.58 10 S409.7........ - \u2022r.,.. -.......... -r\"'-\u00b7 P = 50 kPa (81.33) V 3.240 u 2483.8 3.418 3.889 4.356 4.821 5.284 6.209 7.134 8.058 8.981 9.904 10.828 11.751 12.674 13.597 14.521 2511.6 2585.6 2659.8 2735.0 2811.3 2968.4 3131.9 3302.2 3479.5 3663.7 3854.9 4052.9 4257.4 4467.8 4683.6 h 2645.9 2682.5 2780.1 2877.6 2976.0 3075.5 3278.9 3488.6 3705.1 3928.5 4158.9 4396.3 4640.5 4891.1 5147.7 5409.6 s 7.5939 7.6947 7.9400 8.157') 8.3555 8.5372 8.8641 9.1545 9.4177 9.6599 9.8852 I0.0967 101964 10.4851! 10.6662 10.8382 P \"' /00 kPa (99.62) \" 1.6940 ll 2506. I h 2675.S s 7.3593 1.6958 l,9364 2.1723 2.4060 2.6388 3.1026 3.5655 4.0278 4.4899 49517 5.4135 S.8753 6,3370 6.7986 7.2603 2506.6 2582.7 2658.0 2733.7 2810.4 2967.8", " 3131.5 3301.9 3479.2 3663.5 3854.8 4052.8 4257.3 4467.7 4683.5 7.3614 2676.2 7.6133 2776.4 7.8342 2875.3 2974.3 8.0332 3074.3 8.2157 8.S434 3278.1 8.8341 3488.1 3704.7 9.0975 9.3398 3928.2 9.5652 4158.7 9.7767 4396.1 4640.3 9.9764 4890.9 10,1658 5147.6 10.3462 5409.5 10.S182 ~ ~ ~.... \u00a7''P = 2(/(J kPa (120.23) P = 300 k.Po (/33,5.5) P ~ 400 /cPa (143.63) 2529.5 0.88573 2576.9 0.959(14 1.08034 2654.4 1.19880 2731.2 1.31616 2808.6 1.54930 29(,6.7 1.78139 3U0.7 2.01297 3301.4 2.24426 3478.8 2.47539 3663.2 2706.6 2768.8 2870.5 2971.0 3071.8 3276.5 3487.0 3704..0 3927.7 4158.3 7.1271 7.2795 7.5066 7.7085 7.89\"26 8.2217 8.5132 8.7769 9.0194 9.2450 ll.60582 2.543.6 0.63388 2570.8 0.71629 2650.7 0.79636 2728.7 0.87529 2806.7 1.03151 2965.5 1.18669 3130.0 1.34136 3300.8 1.49573 3478.4 1.64994 3662.9 2725.3 2761.0 2865.5 2967.6 3069.3 3275.0 3486.0 3703.2 3927.1 4157.8 6.9918 7.0778 7.3115 7.5165 7.7022 8.0329 8.3250 8.5892 8.8319 9.0575 0.46246 0.47084 0.53422 059512 0.65484 0.77262 0.88934 l.00555 J.12147 1.2372.2 2553.6 2564.5 2646.8 2726.1 2804-.8 2964.4 3129.2 3300.2 34n,9 3662.5 2738.5 2752,8 2860.5 29642 3066.7 3273.4 3484.9 3702.4 3926.5 4157.4 6.8958 6.9.299 7.1706 7.3788 7.5661 7.8984 8.1912 8.4557 8.6987 8.9244 11. (0 ~ (0 T Sat. so 100 ISO 200 250 300 400 500 600 700 800 900 1000 1100 1200 1300 Sat. ISO 200 250 300 400 500 600 700 800 - P = 1.20 MPa (187.99) P = 1.40 MPa (195.07) T 600 700 800 900 1000 1100 1200 1300 Sat. 225 250 300 350 400 500 (iOO 700 8(1() 900 1000 1100 1200 13QO a - P = 1.00 MPa (179.91) V u 0.40109 3296.8 o.44n9 3475.4 0.49432 3660.5 0.54075 3852.S 0.58712 40505 0.63345 4255.I 0.67977 4465.6 0.72608 4681.3 I, 3697.9 3923.1 41S4.8 4392.9 4637.6 4888.5 5145.4 5407.4 s 8.0289 8.2731 8.4996 8.7118 8.9119 9.1016 9.2821 9.4S4,2 V u 0.33393 3295.6 0.37294 3474.5 0.41177 3659.8 0.4S051 3851.6 0.48919 4050.0 0.52783 42S4.6 0.56646 4465.1 '4680.9 0.60507 h 3696.3 3922.0 4153.9 4392.", "2 4637.0 4888.0 5144.9 5406.9 P = 1.60 MPa (201.40) P = 1.80 MPa (207.15) 0.12380 2595.9 0.13287 2644.6 0.14184 2692.3 0.15862 2781.0 0.17456 2866.0 0.19005 2950.1 0.22029 3119.5 0.24998 3293.3 0.27937 3472.7 0.30859 3658.4 o.33m 38S0.S 0.36678 4049.0 0.39581 4253.7 0.42482 4464.2 0.45382 4679.9 2794.0 2857.2 2919.2 3034,8 3145.4 3254.2 3471.9 3693.2 3919.7 4152.1 4390.8 4635.8 4887.0 5143.9 S406,0 6.4217 6.5518 6.6732 6.8844 7.0693 7.2373 7.5389 7.8080 8.0535 8.2808 8.4934 8.6938 8.8837 9.0642 9.2364 0.11042 2598.4 2636.6 0.1 1673 0.12497 2686.0 1).140'11 2776.8 0.15457 2862.9 0.16847 2947.7 0.19550 3117.8 0.22199 3292.l 0.24818 3471.9 0.27420 3657.7 0.30012 384,9.9 0.32598 4048.4 0.35180 4253.2 0.37761 4463.7 0.40340 4679.4 2797.1 2846.7 2911.0 3029.2 3141.2 3250.9 3469.7 3691.7 3918.6 4151.3 4390.1 4635.2 4886.4 5143.4 S405.6 s 7.9434 8.1881 8.4149 8.6272 8.8274 9.0171 9.1977 9.3698 6.3793 6.4807 6.6066 6.8226 7.0099 7.1793 7.4824 7.7523 7.9983 8.2258 8.4386 8.6390 8.8290 9.00% 9.1817 V 0.28S96 031947 0.35281 0.38606 0.41924 0.45239 0.48552 0.518(>4 u 3294.4 3473.6 3659.1 3851.0 4049.5 4254.1 4464.6 4680.4 h 3694.8 3920.9 4153.0 4391.5 4636.4 4887.5 5144.4 S406.S s 7.87!0 8.1160 8.3431 8.5555 8.7558 8.9456 9.1262 9.2983 I' = 2.00 Ml'a (212.42) 0.09963 0.10377 0.11144 0.12$47 0.13857 0.15120 0.17568 0.19960 0.22323 0.24668 0.27004 0.29333 0.31659 0.33984 0.36306 2600.3 2628.3 2679.6 2772.6 2859.8 2945.2 3116.2 3290.9 3471.0 3657.0 3849.3 4047.9 4252.7 4463.2 4679.0 6.3408 2799.5 6.4146 2835.8 6.5452 2902.S 6.7663 3023.5 6.9562 3137.0 7.1270 3247.6 3467.6 7.4316 7.702j 3690.1 ;.9487 3917.S 8.1766 41S0.4 8.3895 4389.4 8.5900 4634.6 4885.9 8.7800 5142.9 8..9606 9.1328 5405.1.. ~..... ~ e 11 $ - T 800 900 1000 1100 1200 1300 Sat. 350 400 450 500 550 600 650 700 800 900 1000 1100 1200 1300 - 0.05409 0.05950 0.06485 0.07016 0.07S44 0.08072.ot0338.011470.015649", ".018446.020800.022927.024911.026797.028612.032096.035457.038748.042001.045233.()48455 P ~ 9.00 MPa (JOJ.40) P = J.0.00 MPo (JI 1.06) V II,, 4119.4 4364.7 4613.9 4867.7 5126.2 5389.2 3632.5 3829.2 4030.3 4236.3 4447.2 4662.7 $ V 7.4597 7.6782 7.8821 8.0739 8.2556 8.42!B 0.04859 0.05349 0.05832 0.06312 0.06789 0.07265 u 3629.0 3826.3 4027.8 4234,.0 4444.9 4660.4,, 4114.9 4361.2 4611.0 4865.1 5123.8 5387.0 s 7.4077 7.6272 7.8315 8.0236 8.2054 8.3783 P = 15 MPa (342.24) P : 17.5 MPa (354.75) P = 12.50 MPa (327.89) V u It.r 0.03869 0.04267 0.04658 0.050<!5 0.05430 0.05813 3620.0 3819.1 4021.6 4228.2 4439.3 4654.8 4103.7 4352.5 4603.8 4858.8 5118.0 5381.4 P = 20 MPa (365.81) 2455.4 2520.4 2740.7 2879.5 2996.5 3104.7 3208.6 3310.4 3410.9 3611.0 3811.9 4015.4 4222.6 44.33.8 4649_1 2610.S 2692.4 2975.4 3156.2 3308.5 3448.6 3582.3 3712J 3840.1 4092.4 4343.8 4596.6 4852.6 5112.3 5375.9.0079204 5.3097 5.4420.0017139.0124477 5.88!0.0151740 6.1403.(1173585 6.3442.0192877 6.5198.0210640 6.6775.0227372 6.8223 6.9572.0243365 7.2<MO... 0273&49.0303071 7.4279.0331580 7.6347 7.8282.0359695.0387605 8.0108.0415417 8.1839 2390.2 1632.0 2685.0 2844,2 2970.3 308).8 3191.5 3296.0 3398.8 3601.9 3804.7 4009.3 4216.9 4428.3 46435 2528.8 1662.0 2<J02.8 3109.7 3274.0 3421.4 3560.1 3693.9 3824.7 4081.1 4335.1 45895 4846.4 S106.6 5370.S 5.1418 3.7612 5.7212 6.0182 6.2382 6.4229 6.5866 6.7356 6.8736 7.1245 7.3507 75588 7.7530 7.9359 8.1093.0058342.001fl640.0099423.0126953.0147683.0165553.0181781.0196929.0211311.0238532.0264463.0289666.0314471.0339071.0363574 2293.1 1612.3 2619.2 2806.2 2942.8 3062.3 3174.0 3281.S 3386.5 3592.7 3797.4 4003.1 4211.3 4422.8 4638.0 2400.7 1645.6 2818.1 3060.1 3238.2 3393.5 3537.6 3675.3 3809.1 4069.8 4326.4 4582.5 4840.2 5101.0 5365.1 7.2965 7.5181 7.7237 7.9165 8.0087 8.2717 4.9'269 3.7280 5.5539 5.9016 6.14", "00 6.3347 6.5048 6.6582 6.7993 7.0544 7.2830 7.49'25 7.6874 7.8706 8.0441 $' l... ff II \"\u00b0 Cl> <I, T 375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300 375 400 425 450 500 600 6SO 700 800 900 1000 1100 1200 1300 V.001973.006004.007882.009162.011124.012724.014138.OIS433.016647.018913.021045.023102.025119.0271.15.029101.0016406.0019077.0025319.0036931.0056225.008()1)43.0090636.()0()9415.0115228.0129626.0143238.0156426.0169403.0182292 P= 15 MPa u h 1798.6 2430.l 2609.2 2720.7 2884.3 3017.5 3137.9 325L.6 3361.4 3574.3 3783.0 3990.9 4200.2 4412.0 4626.9 1847.9 2580.2 2806.3 2949.7 3162.4 3335.6 3491.4 3637.5 3777.6 4047.1 4309.1 4S68.5 4828.2 5089.9 53S4.4 P O 40 MPa 1677.1 18S4.5 2006.8 2365.1 2678.4 3022.6 3158.0 3283.6 35i7.9 3739.4 3954.6 4167.4 4380.1 4S94.3 1742,7 1930.8 2198.1 2512.8 2903:3 3346.4 3520.6 3681.3 3978.8 4257.9 4527.6 4793.l 5057.7 5323.5 s 4.0319 5.1418 5.4722 5.6743 5.9592 6.1764 6.3602 6.5229 6.6707 6.9345 7.1679 7.3801 7.5165 7.7604 7.9342 3.8289 4.1134 4.5028 4.9459 S.4699 6.0113 6.2054 6.3750 6.6662 6.9150 7.1356 7.3364 7.5224 7.6969 II.001789,002790.OOS304.006735.008679.010168.011446.012596.01366i.015623.017448.019196.02()1)()3.022589.0242()6.0015593.0017309.0020071.0024862.0038924.0061123.0069657.0077274.1)090761.0102831.0114113.0124966.0135606.0146159 P; 30MPa u h 1737.8 2067.3 24S5.l 2619.3 2820.7 2970.J 3100.5 3221.0 3335.8 3555.6 3768,5 3978.8 41892 4401.3 4616.0 1791.4 2151.0 2614.2 2821.4 3081.0 3275.4.3443.9 3598.9 3745.7 4024.3 4291.9 45S4.7 4816.3 5079.0 5344.0 P = 50MPt1 1638.6 1788.0 1959.6 2159.6 2525.5 2942.0 3()()3.6 3230.S 3479.8 3710.3 3930.5 4145.7 4359.1 4572.8 1716.S 1874.6 2060.0 2283.9 2720.1 3247.6 3441.8 3616.9 3933.6 4224.4 4501.1 4770.6 S037.2 5303.6 s 3.9303 4.4728 S.1503 5.4423 5.7904 6.0342 6.2330 6.4057 6.5(,()6 6.8332 7.0717 7.2867 7.4845 7.6691 7.8432 3.7638 4.0030 4.273'3 4.5883 5.1725 5.8177", " 6.0342 6.2189 6.5290 6.7882 7.0146 7.2183 7.4058 7.5807 V.001700.002100.003428.004962.006927.008345.009527.010575.011533.013278.014883.016410.017895.019360.020815.0015027.0016335.0018165.0020850.0029557.0048345.OOS5953.0062719.0074588.0085083.0094800.0104091.0113167.0122155 P; 35 MPa u h 1702.9 1914.0 2253.4 2498.7 2751.9 2920.9 3062.0 3189.8 3309.9 3536.8 37S4.0 3966.7 4178.3 4390.7 4605.1 P = 60MPa 1762.4 1987.5 2373.4 2672.4 2994.3 3213.0 3395.5 3559.9 3713.5 4001.5 4274.9 4541.1 4804.6 5068.4 5333.6 1609.3 1745.3 1892.7 2053.9 2390.5 2861.1 3028.8 3177,3 3441.6 3681.0 3906.4 4124.1 43382 4551.4 1699.5 1843.4 2001.7 2179.0 2567.9 3151.2 3364,.6 3553.6 3889.1 4191.5 4475.2 4748.6 5017.2 5284.3 s 3.8721 4.2124 4.7747 5.1962 5.6281 5.9025 6.1178 6.3010 6.4631 6.7450 6.9886 7.2063 7.4056 7.5910 7.7652 3.7140 3.9317 4.1625 4.4119 4.9320 5.6451 5.8829 6.0824 6.4110 6.6805 6.9126 7.1194 7.3082 7.4837 ~ - \"\u00b0 c.,, 0, 11 g, ;:;-.. ~ ~ ~ ~... ~ i., \"'.. ~ a\u00b7 :11 P -= 5.00 MPa (263.99) P m 10.00 MPa (311.06) P-= 15.00 MPo (342.U) Table A.U ComflrtSJtd li911id Watn V.0012859 JXXYR11.0009995.0010056.00101.49.0010268.0010410.0010576.0010768.0010988.0011240.0011530.0011866.0012264.0012748 u 1147.78 0.03 83.64 166.93 250.21 333.69 417.50 SOl.79 586.74 672.61 759.62 848.08 938.43!031.34 1127.92 Ir s 11S4.21 2.9201 5.02 0.0001 88.64 0..2955 171.95 0.5706 255.28 0.8284 338.83 1.0719 422.71 1J030 507.07 1.5232 592.13 1.7342 678.10 1.9374 765.24 2.1341 853.85 2.3254 944.36 2.5128 1037.47 2.6978 1134.30 2.8829 V.0014524.0009952.0009972.0010034.0010127.0010245.0010385.0010549.0010737.0010953.0011199.0011480.0011805.0012187.0012645.0013216.0013972 T Sat. 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 =- ~ u $ I,. 139).00 1407.53 3.359S 10.05 0.0003 O.IO 93.32 0.2945 83.35 176.36 0.5685 166.33 259.47 0.8258 249.34 1.0687 342.81 332.56 426.48 1.2m 416.09 1.5188 510.61 500.o? 1.7291 595.40 584.67 670.11 681.07 1.9316 767.", "83 2.1274 756.63 844,.49 855.97 2.3178 934.07 945.88 2.5038 1025.94 1038.13 2.6872 1121.03 1133.68 2.8698 3.0547 1220.90 1234.11 1328.34 1342.31 3.2468 V.0016S81.0009928,()()()9950.0010013.0010105.0010222.0010361.0010522.00!0707.0010918.0011159.0011433.0011748.0012114.0012550.0013084\u00b7.0013770.0014724.0016311 u ISSS.58 0.15 83.05 165.73 248.49 331.46 414.72 498.39 582.64 667.@ 753.74 841.04 929.89 1020.82 1114.59 1212.47 1316.58 1431.0S )567.42 h s 1610.4S 3.6847 15.04 0.0004 97.97 0.2934 180.75 0.5665 263.65 0.8231 346.79 1.0655 430.26 1.2954 514.17 I.S144 598.70 1.7241 684.07 1.9259 770.48 2.1209 858.18 2.3103 947.52 2.4952 1038.99 2,6nO 1133.41 2.8575 1232.. ()9 3.(1392 1337.23 3.2259 1453.13 3.4246 IS91.88 3.6S45 ::.. ~ s....;\u2022 ~ 11 co \"' ~ Table A.2 '11tmnodyna,nk Propntia of Rtfrig,rant\u00b7 12\u2022 (Di&ldo,odijluoromelliane) Table A.2.1 &tllrlUld R,/iil,mml\u00b7 1 Z S~cific Yoiu11te Sat. Liqufd v, C11tJlg 0.608 0.612 0.617 0.622 0.627 0.632 0.637 0.642 0.648 0.654 0.6S9 Sat. Vapour v, /ffJ/kg 4.4\\S545 3.037316 2.L38345 1.537651 1.127280 0.841166 0.637910 0.491000 0.383105 O.J02682 0.241910 Sat. Liquid hr -43.243 -38.968 -34.688 -30.401 -26. 103 - 21.793 -17.469 -13.129 -s.n2 -4.396 -0.000 Enthalpy Evap. hr, Ulkg 189.618 187.608 185.612 183.625 181.640 179.651 177.653 175.641 173.611 171.5S8 169.479 Sat. Vapour h, 146.375 148.640 IS0.924 153.224 155.536 157.857 160.184 162.512 164.840 167.163 169.479 Temperature oc I ~&sure MP a p -90 -85 -80 -75 -70 -65 -60 -ss -so -45 -40 0.0028 0.0042 0.0062 0.0088 0.0123 0.0168 0.0226 0.0300 0.0391 O.OS04 0.0642 Entropy Sat. Liquid.,, Sat. Vapour '\u2022 kJ/1:g K -0.2084 -0.!8S4 -0.1630 - 0.1411 -0. 1197 -0.0987 -0.0782 -O.OS81 -0.0384 -0.0190 -0.0000 0.8268 0.8116 0.7979 0.7855 0.7744 0.7643 0.7552 0.7470 0.7396 0.7329 0.72b9 \u2022 Adapled from FuMDJnentafs of Cla,1fcal T1tem1ody,wmlcs by -G. J. Van Wylen and R. SO!llllllg. Johll Wiley, New Yol'k 1976, P. 667-673 (with the k.il.ld permiaaion of the publisben, John Wiley & Sons, Inc, Nt\\lll Yort). =i I 11 Table A..2.", "2 S11fJtrh,a1,d &frigerant-12 V,,,1kg h J:.1/kg s /rJ/kg K V,,t1/kg h Ulkg s lrJ/kg K V mlkg It /cJ kg s /cJ kg K 0.341857 0.356227 0.370508 0.384716 0.398863 0.412959 0.427012 0.441030 0.455017 0.468978 0.482917 0.496838 0.088608 0.092550 0.096418 0.100228 0.10398.9 0.107710 0.JJl397 0.115055 0.118690 0.122304 0.0.S MPa 181.042 186.757 192.567 198.471 204.469.210.557 216.733 222.997 229..344 2J5.774 242.282 248.868 (J.20 Ml'a 189.669 195.878 202.135 208.446 214.814 221.243 227.735 134.291 24-0.910 247.593 0.7912 0.81.33 0.8350 0.8562 0.8770 0.8974 0.9175 0.9372 0.9565 0.9755 0.9942 1.0126 0:7320 0.7543 0.7700 0.7972 0.8178 0$381 0.8578 0.8772 0.8962 0.9149 0.167701 0.175222 0.182647 0.189994 0.197277 0.204506 0.211691 0.218839 0.225955 0.233044 0240111 0.247159 0.069752 0.073024 0.076218 0.079350 0.082431 0.085470 0.088474 0.091449 0.094398 0.097327 0.10 MPa 179.861 185.707 191.628 197.628 203.707 209.866 216.104 222.421 228.815 235.285 241.829 248.446 0.25 MPa 188.644 194.969 201.322 207.715 214.153 220.642 227.185 233.785 24-0.443 247.160 0.7401 0.7628 0.7849 0.8064 0.8275 0.8482 0.8684 0.8883 0.9078 0.9269 0.9457 0.%42 0.7139 0.7366 0.7587 0.7801 0.8010 0.8214 0.8413 0.8608 0.8800 0.8987 0.15 MPa 184.619 190.660 196.762 202.927 209.160 215.463 221.835 228.277 234.789 241.371 248.020 O.JO MPa 187..583 194.034 200.490 206,969 213.480 220.030 226.627 233.27J 2J9.271 246.723 0.114716 0.119866 0.124932 0.129930 0.134873 0.139768 0.144625 0.149450 0.154247 0.159020 0.163774 0.057150 0.059984 0.062734 0.065418 0.068049 0.070635 0.073185 0.075750 0.078200 0.080673 0.7318 0.7543 0.7763 0.19n 0.8186 0.8390 0.8591 0.8787 0.8980 0.9169 0.9354 0.6984 0.7216 0.7440 0.7658 0.7869 0.8075 0.8276 0.8473 0.866S 0.8853 oc -20.0 -10.0 0.0 IO.O 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 :i.. ~ s ~ ~ II ~... <C - t 100.0 110.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0 110.0 40.0 50.0 60.0 70.0 80", ",0 90.0 100.0 110.0 so.o <iO.O V 0.125901 0.1294&3 0.04S836 0.047971 0.050046 0.052072 0.054059 0,056014 0.057941 0.059846 0.061731 0.063600 0.026761 0.028100 0.029387 0.030632 0.03184\u00b73 0.033027 0.034189 0.035332 It 254.339 261.147 {UOMPa 198.762 20S.428 212.095 21s.m 225.488 232.230 239.012 245.837 2S2.707 259.624 0.70 MPa 207.S80 214,745 221.854 228.931 235.997 243.066 250.146 257.247 1.00 MPa /I 0.9332 0.9S12 0.7199 0.7423 0.7639 0.7849 0.8054 0.8253 0.8448 0.8638 0.8825 0.9008 0.7148 0.7373 0.7590 o.n99 0.8002 0.8199 0.8392 0.8579 \" 0.100238 0.103134 0.035646 0.037464 0.039214 0.040911 0.042565 0.044-184 0.045774 0.047340 0.048886 0.050415 It 253.936 260.770 O.JO MPa 196.935 203.814 210.6S6 217.484 224.315 232.161 238.031 244.932 251.869 258.S45 0.80 MPa 205.924 0.022830 213.290 0.024068 220.558 0.025247 227.766 0.026380 234.941 0.027477 0.028545 242.101 0.029588 \u2022.. 249.260 256.428 0.030612 1.20 MPa /I 0.9171 0.9352 0.6999 0.7230 0.7452 0.7fi67'0.7875 0.8077 0.8275 0.!1467 0.8656 0.8840 0.7016 0.7248 0.7469 0.7682 0.7888 0.8088 0,8283 \u2022 0.8472 V 0.83127 0.08SS66 0.030422 0.031966 0.033450 0.034887 0.036285 0.037653 0.038995 0.040316 0.041619 0.019744 0.020912 0.022012 0.023062 0.024072 0.025051 0.026005 0.026937 It 253.530 260.391 0.60 MPa 202.116 209.154 216.141 223.104 230.062 237.027 244.00') 251.016 258.053 0.90 MPa 204.170 211.765 218.21.2 226.564 233.8S6 141.113 248.355 255.593 UOMPa II 0.9038 0.9220 0.7063 0.7291 0.7511 0.7723 0.7929 0.8129 0.8324 0.8514 0.8700 0.6~ 0.7131 0.7358 0.7575 0.7785 0.7987 0.8184 0.8376 0.018366 0.019410 210.162 217.810 0.7021 0.7254 0.014483 0.015463 ~661 214.80:S 0.6812 0.7060 0.012579 21 l.457 0.6876 =- : t,:I 11 r. \"' \"'i ~'r.\u00a7. l: :!I 5\u00b7 -.,,.... Table A.3 T/tm,,od]lf41Nit Pr<ifHrlits of Rlfrigrra11J\u00b7 12 (M1111\"'M1mxiijbioromtthaM) Table A.3.1 Sat11r,ud Jufripront\u00b7ZZ Specific volumt(r,h/,.g) Entholpy(lcJ/kg) Entropy(kJ/kg K) t II p Abs. Press. Tcmf>. MP a oc - 70 -65 -60 -SS -50 - 45 -40 -3S -30 - 25 -20 -", "IS -10 -5 0 5 10 15 20 2.S 30 0.0205 0.0280 0.0375 0.0495 0.0644 0.0827 0.1049 0.1317 0.1635 O.ZOlO 0.2448 0.2957 0.3543 0.4213 0.4976 0.5838 0.6807 0.7891 0.9099 1.0439 1.1919 Sat. Liquid i\u2022, 0.0()()670 0.000676 0.000682 0.000689 0.000695 0.000702 0.(J007{J9 0.000717 0.000725 0.000733 0.000741 0.000750 0.000759 0.()()()76.~ 0.00()778 0.000789 0.()()()800 0.000812 0.<n)824 0.000838 0.000852 Evap. Vr, 0.940268 0.704796 0.536470 0.414138 0.323862 0.256288 0.205036 0.165683 0..135120 0.111126 0.092102 0.076876 0.064581. 0.054571 0.046357 0.039567 0.033914 0.029176 0.025179 0.021787 0.018890 Sat. J'apour v. 0.940')3 0.705478 0.537152 0.414827 0.324557 0.256990 0.205745 0.166400 0.. 135844 0.111859 O.O'J2843 0.077625 0.065340 0.055339 0.047135 0.040356 0.034714 0.029987 0.026003 0.0\"-2624 0.019742 Sat. Liqui<I hr - 30.607 -2.S.658 -20.6S2 -15.585 -10.4511 -5.262 0 5.328 I0.72S 16.191 21.728 27.334 33.012 38.762 44.586 50.485 56.463 62.523 68.670 74.910 81.250 El\u2022ap.,,, 249.425 246.925 244JS4 241.703 238.965 236.IJ2 233.198 230.156 227.001 223.727 220327 216.798 213.132 209.323 205.364 201.246 196.960 192.495 187.836 182.968 177.869 Vapour \"\u00b7 218.180 221.267 223.702 226.117 228.509 230.870 233.197 235.4ll4 237.726 239.918 242.055 244.132 246.144 248.085 249.949 2.Sl.731 253.423 255.018 2S6.S06 257.817 2S9.l 19 Sat. Liquid.~ - 0.1401 -0.1161 -0.0924 -0.0689 -0.0457 -0.0227 0 0.0225 0.0449 0.0670 0.0890 0.1107 0.1324 0.1538 0.1751 0.1963 0.2173 0.2382 0.2590 0.2797 0.3004 Evap..fr,. 1.2277 1.1$62 1.1463 1.1079 1.0708 1.0349 1.0002 0.9664 0.9335 0.9015 0.8703 0.8398 0.8099 0.7806 0.7518 0.7235 0.6956 0.6680 0.6407 0.6137 O.S867 ~ - tl, e ):,\u00b7.. \"'.... :,.. ~.... ~ 0.9889 r t ; a\u00b7 Sat Vaf>')11r s 1.0876 1.0701 1.0540 1.0390 1.0251 1.0122 1.(1()02 0.97ll4 0.9685 0.9593 0.9505 0.942.2 0.9344 0.9269 0.9197 0.9129 0.9062 0.8997 0..8934 0.8871 TeMp. oc V nl\"l kg -40 -30 - 20 - 10 0 10 20 30 4() so 60 70 80 90 -:?O -10 0 10 0.44()633 0.460641 0.480543 0.500357 0.520095 0.539", "77[ 0.559393 0.578970 0.598507 0.618011 0.637485 0.656935 0.676362 0.69Snt 0.115203 0.120647 0.120003 0.131286 It kJ! kg 0.05 MPa 234.724 240.602 246.586 252.676 258.874 265,180 271.S94 278.l 15 284.743 291.478 :?98.319 305.265 312.314 319.465 0.10 MPa 243.140 249.574 256.069 262.633 = - Table A.J,2 Suprt/1111/td Rtfrigtr<ml\u00b722 $ k.Jlkg K V mJ/kg h kJlkg 0.10 MP\" 233.337 239.3-59 245.466 251.665 257.956 264.345 270.831 277.416 284.101 290.887 297.772 304.757 311.842 319.026 0.25 MPa $ kJ!kg K V,nJ/kg 1.00523 1.03052 1.0.5513 1.07914 1.10261 1.1.2558 1.14809 1.17017 1.19187 1.21320 1.23418 1.25484 J.27519 l.29524 0.148723 O. l5.S851 0.162879 0.169823 0.176699 0.183516 0.190284 0.197011 0.203702 0.210362 0.21691)7 0.223608 0.230200 h kJlkg 0.15 MPt, 238.078 244.319 250.61 1 257.022 263.496 270.057 276.709 283.452 290.289 297.220 304.246 311.368 318.584 O.JO MPa s kJ!kg K 0.98773 1.01288 l.03733 1.06116 1.08444 1.10721 1.12952 1.15140 1.17289 1.19402 1.21479 113525 l.2SS40 0.216331 0.226754 0.237064 0.247279 0.257415 0.267485 0.277500 0187467 0197394 0.307287 0.317149 0.326986 0.336801 OJ46S96 0.095280 0.099689 0.104022 248.492 2.sS.097 261.755 0.98231 1.00695 1.03089 0.078344 0.082128 0.085832 247.382 254.104 260.861 0.96170 0.98677 1.01106 1.07616 1.10084 1.12495 1.14855 1.17166 1.19433 1.21659 1.23846 1.25998 1.28114 1.30199 1.32253 1.34278 1.3627S 0.981114 1.00676 1.03091! I.OS458 \"' t 11 ttr ~-.. l. :i i r,!. i::!I a\u00b7 Temp. \u2022c V 11l!kg 20 30 40 50 60 70 80 90 100 110 0 10 20 30 40 so (i() 70 80 90 100 110 120 130 0.136509 0.141681 0.146809 0.151902 0.156963 0.161997 0.167008 0.171999 0.176972 0.181931 0.060131 0.063060 0.065915 0.0687!0 0.071455 0.0741(,{J 0.076830 0.079470 0.082085 \u00b7 o.084679 0.087254 0.089813 0.092358 0.1\u00bb489() Table A.3.2 Superheated Rtfrigerant\u00b722 \" kl/ kg 0.20 MPu 269.273 275.992 282.796 289.686 296.664 303.731 3!0.890 318.139 325.480 332.912 0.40 MPa 252.051 259.023 266.010 273.029 280.092 287.20') 294.386 301.630 308.944 316.332 323.796 331.339 338.961 346.664 s k.flkg K \\I,r,3/kg.1.07763 1.10016 1.12224 1.14390 1", ".1651(, 1.18607 1.20663 1.22687 1.24681 1.26646 0.95359 0.97866 1.00291 1.02646 1.04938 1.07175 L00362 1.11504 1.13605 1.15668 1.17695 1.19690 1.21654 l.23S88 0.108292 0.112508 0.116681 0.120815 0.124918 0.128993 0.133044 0.137075 0.141089 0.145086 0.1)49355 0.0517SI 0.054081 0.056358 0.058590 0.060786 0.062951 0.065090 0.067206 0.069303 0.071384 0.073450 0.07SS03 \" k)l kg 1).25 MPt1 268.476 275.267 282.132 289.076 296.102 303.212 310.409 317.692 325.063 332.522 II.JO,\\If Pa 257.108 264.295 271.483 278.690 285.930 293.215 300.552 307.949 315.410 322.939 330.53() 338.213 345.963 s k.Jlkg K V m1/kg Ii kl/kg s k.flkg K 1.05421 1.07699 1.09927 1.12109 1.14250 1.16353 1.18420 ).20454 l.22456 J.24428 0.95223 0.97717 1.00128 1.02467 1.04743 1.06963 1.09133 l.11257 1.13340 1.15386 1.17395 1.19373 1.21319 0.089469 0.093051 0.0%588 0.100085 0.103550 0.106986 0.110399 0.113790 0.117164 0.120522 0.040180 0.042280 0.044307 0.046276 0.048198 0.050081 0.051931 0.053754 0.055553 0.057332 0.059094 0.060842 0.062576 0.30 MPa 267.6(,7 274.531 281.460 288.460 295.535 302.689 309.924 317.241 324.643 332.129 0.60 MPa 255.109 262.517 269.888 277.250 284.622 292.020 299.456 306.938 314A7S 322071 329.731 337.458 345.255 1.03468 1.05771 1.08019 1.10220 1.12376 1.14491 1.16569 1.18612 1.20623 1.22603 0.9294.S 0.95517 0.97989 1.00378 1.02695 1.04950 1.07149 1.09298 1.11403 1.13466 1.15492 1.17482 1.19441 =- ~ ~ ~..... \" \" ~.., 11 11 co.... --1 Temp. \u2022c V mJ/kg 20 30 40 50 (iO 70 80 90 100 110 120 130 140 150 30 40 50 (iO 70 80 90 100 0.035487 0.037305 0.039059 0.040763 0.042424 0.044052 0.045650 0.047224 0.048778 0.050313 0.051834 0.053341 O.OS4836 0.056321 0.024600 0.025995 0.027323 0.028601 0.029836 0.031038 0.032213 0.033364 Table A.3.2 Suprrlttattd &frignant-ZZ h JcJ/ Jcg 0.70 MPa 260.661 268,240 275.769 0.70 MPu 283.282 290.800 298339 305.912 313.527 321.192 328.914 336.696 344.541 352.454 360.435 1.00 MPa 262..912 271.042 279.046 286.973 294.859 302.727 31059<) 318.488 J k.J/Jcg K V mJ/kg 0.93565 0.96105 0.98549 1.00'}10 1.03201 1.05431 1.07606 1.09732 1.11815 1.", "13856 1.15861 1.17832 l.l9770 1.21679 0.913S8 0.93996 0.96512 0.98928 1.01260 1.03520.1.05718 1.07861 0.030366 0.032034 0.033632 0.035175 0.036674 0.038136 0.039568 0.040974 0.042359 0.043725 0.045076 0.046413 0.047738 0.049052 0.020851 0.022051 0.023191 0.024282 0.025336 0.026359 0.027357 \" kJ /lcg 0.80 M'Pa 258.737 266.533 274.243 0.80 M'Pa 281.907 289.553 297.202 304.868 312.565 320.303 328.087 335.925 343.821 351.778 359.799 1.10 MPa 267.602 276.011 284.263 292.415 300.508 308.570 316.623 s kJ/kg K V m\u00b71/ kg h kJ/kg s kJ/kg K 0.91787 0.94402 0.96905 0.99314 1.01644 1.03906 l.06108 l.08257 l.10359 1.12417 1.14437 1.16420 l.l8369 1.20288 0.91411 0.94055 0.96570 0.98981 1.01305 1.03556 I.OS744 0.026355 0.027915 0.029397 0.030819 0.032193 0.033528 0.034832 0.036108 0.037363 0.038598 0.039817 0.041022 0.042215 0.043398 0.017120 0.018247 0.019299 0.020295 0.021248 0.022167 0.023058 0.90 Pa 256.713 264.760 272.670 0.90.Mpa 280.497 288.278 296.042 303.807 311.590 319.401 327.251 335.147 343.094 351.097 359.159 1.40 MPa 263.861 272.766 281.401 289.858 298.202 306.473 314.703 0.90132 0.92831 0.95398 0.97859 1.00230 1.02526 1.04757 1.06930 1.09052 1.11128 J.13162 1.15158 1.17119 1.19047 0.89010 0.91809 0.94441 0.96942 0.99339 1.01649 1.03884 - ID.. co 11 ~ ;:;\u2022.. :0... ~ 'S ~ A. ;;l \" ; ~.... :,I C> 5\u00b7 Tt1\"'P\u00b7 \u2022c V m 1/ kg JOO 110 120 130 140 150 160 170 180 190 200 90 100 110 120 130 140 150 160 170 180 190 200 210 220 0.011598 0.012208 0.012788 0.013343 0.013880 0.014400 0.014907 0.015402 0.015887 0.016364 0.016834 0.005037 0.005804, 0.0.06405 0.006924 0.007391 0.007822 0.008226 0.()()8610 0.008978 0.009332 0.009675 0.010009 0.0!0335 0,0106S4,,. Ir.JI kg 2.50 1\\-/Pa 302.935 312.261 321.400 330.412 339.336 348.205 357.040 365.860 374.679 383.508 392.354 4.00 MPa 265.629 280.997 293.748 305.273 316.080 326.422 336.446 346.246 355.885 365.409 374.853 384.240 393.593 402.925 $ kJ/ kg K \\I 11/Jkg 0.95939 0.98405 1.00760 1.03023 1.05210 1.07331 1.09395 1.1 1408 l.13376 1.15303 1.17192 0.82544 0.86721 0.90094 0.93064 0.95778 0.98312 1.00710 1.02999 l.OS199 1.07324 1.", "09386 1..11391 1.13347 I.JS259 0.009098 0.009674 0.010211 0.010717 0.011200 0.011665 0.012114 0.012550 0.012976 0.013392 0.013801 0.003334 0.004255 0.004851 0.005335 0.005757 0.()06139 0.006493 0.006826 0.007142 0.007444 0.007735 0.008018 0.008292 Ir kJ/kg J.00 MPa 296.663 306.744 316.470 325.955 335.270 344.467 353.584 362.647 371.679 380.695 389.708 5. 00 MP<1 253.042 275.919 291.362 304.469 316.379 327.563 338.266 348.633 358.760 368.713 378.537 388.268 397.932 s kJ/kg K V 111Jl kg Ir kJ/kg :J lcJ/kg K 0.92881 0.95547 0.98053 1.00435 1.02718 1.04918 1.07047 1.09116 1.11131 J.13()9').l.15024 0.78005 0.84064 0.88045 0.91337 0.94256 0.96931 0.99431 1.01797 1.04057 1.06230 1.08328 1.10363 1.12343 0.007257 0.007829 0.008346 0.008825 0.009276 0.()(1')704 0.010114 0.010510 0.010894 0.011268 0.011634 0.002432 0.003333 0.0038()() 0.004345 0.004728 0.005071 0.005386 0.005680 0.005958 0.006222 o.0064n 0.006722 3.50 MPa 289.504 300.640 311.129 321.196 330.976 340.554 349.989 359.324 368.590 377.810 387.004 6.00 MPa 243.278 212.385 290.253 304.757 317.633 329.553 340.849 351.715 362.271 372.602 382.764 392.801 0.89872 0.92818 0.95520 0.98047 l.00445 1.02736 1.04940 1.07071 1.09138 1.11151 1.13115 0.74674 0.82185 0.86675 0.902.10 0.93310 0.96094 0.98673 1.01098 1.03402 1.05609 1.07734 l.O'n90 - ID ~ C 11 r...:\u00b7 l ~ \"6?:, \" I'>. ;l..... -!} i \"'.. \u00b7;a 6' Temp. oc - 25 -20 - 10 0 IO 20 30 40 50 60 70 ~ 90 100 0 10 20 JO 40 50 60 70 80 90 100 110 120 V m1/kg 0.19400 0.19860 0.20765 0.21652 0.22527 0.23393 0.24250 0.25102 0.25948 0.26791 0.27631 0.28468 0.29303 0.30136 0.082637 0.086584 0.090408 0.090139 0.007798 0.1 01401 0.104958 0.108480 0.1 11972 0. 115440 0.118888 0.122318 0.125734 h kJ/kg 0./1} MPa 383.212 387.215 395.270 403.413 411.668 420,048 428.564 437.223 446.029 454.986 464.096 473.359 482.777 492.349 u.t3 117Pa 399.579 408.357 417.15 1 425.997 434.925 443.953 453.094 462..359 471.754 481.285 490.955 500.766 510.720 TallleA.. U S11p,rlllcld R\u00b7TJ4a s k)/kg K V m\u00b71/kg h fcJ! kg s kJ/kg K V mJ/kg 1.75058 1.76655 l.79775 1.82813", " 1.85780 1.88689 l.91S45 1.94355 1.97123 1.99853 2.02S47 2.05208 2.07837 2.10437 1.7421W 1.77440 1.80492 1.83460 1.86357 1.89195 1.91980 1.94720 1.97419 2.00080 2.02707 2.05302 2.07866 0.13603 0.14222 0.14828 0.15424 0.16011 0.16592 0.17168 0.17740 0.18308 0.18874 0.19437 0.19999 0.071110 0.0744 15 0.077620 0.080748 0.083816 0.086838 0.089821 0.092774 0.095702 0.098609 0.101498 0.104371 0./5 MPa 393.839 402.187 410.602 419.111 427.730 436.473 445.350 454.366 463.525 472.831 482.285 491.888 0.30 MPa 407.171 416.124 425.0% 434.124 443.234 452.442 461.763 471.206 480.777 490.482 500324 510.304 1.76058 1.79171 1.82197 l.85150 1.88041 1.90879 1.93669 1.964 16 1.99125 2.01798 2.04438 2.07046 1.75637 1.78744 1.81754 1.84684 1.87547 1.90354 1.93110 1.95823 1.98495 2.01131 2.03734 2.06305 0.10013 O. IOSO I 0.10974 0.11436 0.11889 0.12335 0.12776 0.13213 0.13646 0.14076 0.14504 0. 14930 0.051681 0.054362 0.056926 0.059402 0.061812 0,064 169 0.066484 0.068767 0.071022 0.073254 0.075468 0.077665 h lc}lkg 0.20 MPo I Id/kg K 392.338 400.911 409.500 418.145 426.875 435.708 444.658 453.735 462.946 4n.296 481.788 491.424 0.40 M.Po 404.651 413.965 423.216 432.46.5 441.751 451.104 460.545 470.088 479.745 489.523 4.99.428 S09.464 l.73276 1.7(,474 1.79562 1.82563 1.85491 1.88357 1.91171 1.93937 1.96661 1.99346 2.01997 2J>4614 1.72611 1.75844 1.78947 1.81949 1.84868 l.8nJ8 1.90510 l.93252 t.95948 1.98604 2.012.23 2.03809 ~ - \"I:>.. ~ ;s... ~\u00b7.. 11 ~ w T~mp. \"C \",,,J;kg 20 30 40 50 60 70 80!JO 100 110 120 130 140 40 50 (,() 70 80!JO HX) 110 120 130 140 ISO 0.042256 0.044457 0.046557 0.048581 0.050547 0.052467 0.054351 0.056205 0.058035 0.059845 0.061639 0.063418 0.065184 0.027113 0.028611 0.030024 0.031375 0.032678 0.033944 0.035180 0.036392 0.037584 0.038760 0.039921 0.041071 h lt./llrg O.JOMPo 41J.64S 421.221 430.720 440.205 449.718 459.290 468.942 478.690 488.546 498.518 508.613 518.835 529.187 0.80 MPo 424.860 435.114 445.223 455.270 465.308 475.375 485.499 495.698 505.988 516.379 526.880 537.496 s kl", "/kg I( V m1/lcg 1.73420 1.76632 l.79715 1.82696 J.85596 l.~26 1.91199 1.93921 t.96598 J.99235 2.01836 2.04403 2.06940 1.74457 1.77680 \u00b7 1.80761 1.83732 1.86616 1.89427 1.92177 1.94874 1.97525 2.00135 2.02708 2.05247 0.036094 0.037958 0.039735 M41447 0.043108 0.044730 0.046319 0.047883 0.049426 0.05@51 0.052461 0.053958 0.023446 0.024868 0.026192 0.027447 0.028649 0.029810 0.030940 0.032043 0.033126 0.034190 0.035241 0.036778 h lrJl/cg 0.60MPo 419.@3 428.881 438.589 448.279 457.994 467.764 477.611 487.550 497.594 507.750 518.026 528.425 0.90 MPa 422.642 433.235 443.595 453.835 464.025 474.216 484.441 494.726 505.088 515.542 526.096 536.760 I 1,;J/Ag K. V \"'}/kg 1.746!0 1.77786 1.80838 1.83791 1.86664 1.89471 1.92220 1.94920 1.97576 2.00193 2.02774 2.05322 1.72943 1.76273 l.7C),t31 1.82459 l.85387 l.88232 1.910!0 1.93730 1.96399 t.99()25 2.01611 2.04161 0.030069 0.031781 0.033392 0.034929 0.036410 0.0371148 0.039251 0.040627 0.041980 0.0433 14 0.044633 0.045938 0.020473 0.021S49 0.023110 0.024293 0.0254 17 0.026497 0.027543 0.028561 0.029556 0.030533 0.031495 0.032444 \" kJl/cg 0.70 MPo 416.809 426.933 436.895 446.782 456.655 4(,6.554 476.507 486,535 496.654 506.875 517.207 527.656 1-00 MP\" 420.249 431243 441.890 452.345 462.703 473.027 4&3.361 493.736 504.175 514.694 52S.30S 536.017 s kl/kg I( 1.72770 1.76056 1.79187 1.82201 1.85121 1.87964 1.90743 1.93467 1.96143 1.98777 2.01372 2.03932 1.71479 1.74936 1.78181 1.81273 1.84248 l.87131 1.89938 1.92682 1.95371 1.98013 2.00613 2.03175 - <O (n.. II ;:;\u00b7 r.. t. ~ a: ~ \"\".. ~....... ~ ~ l! 5\u00b7 TeMp. \"C V 11.11kg 90 100 110 120 130 140 150 160 170 180 190 200 0.005755 0.006653 0.. 001339 0.007924 0.008446 0.008926 0.009375 0.009801 0.010208 0.010601 0.010982 0.0113S3 Ii Ulkg J.OMPa 436.193 453.731 468.500 482.043 494.915 507.388 519.618 531.704 543.713 555.690 567,670 579.678 s k.1/kg /( I' 1111/kg h kJ/kg J.50 MPa s k.llig K V m1!kg Ii kJ/kg 4.0 MPa s k.lllrg K l.69')50", " l.74717 1.78623 1.82113 1.85347 1.88403 1.91328 1.94151 1.96892 l.99S6S 2.02180 2.04745 0.004839 0.005667 0.006289 0.006813 0.007279 0.007706 0.008103 0.008480 0.008839 0.009185 0.009S19 440.433 459.211 474.697 488.771 502.079 514.928 S27.496 S39.890 SS2.185 564.430 576.665 1:70386 1.75355 1.79346 1.82881 1.86142 1.8.9216 1.92151 1.94980 1.97724 2.00397 2.03010 0.004277 0.005005 O.OOS559 0.006027 0.006444 0.006825 0.007181 0.007517 0.007837 0.008145 446.844 465.987 481.865 496295 S\u00ae.925 523.072 535.917 S48.S73 S61.117 573.601 1.71480 1.76415 1.80404 1.83940 l.87200 1.90271 1.93203 1.96028 1.98766 2.01432 <.A O> '\u00b0 II r.. ;:;\u00b7 ~ ~ ~ [ r.. ~ I i:I' Table A.& 1Tl#mod],i,a111\u00a5 Pn\u00a5rtiu wf A1111nonia Table A.5.1 Saturated A,n,nonia Tabk S~cifit: vo/umt (mJ l1rg) Enthalpy (k.llkg) T (OC) -50 -48 -46 -44 -42 -40 -38 -36 -34 -32 -30 -28 -26 -24 -22 -20 -18 p (kPa) 40.88 45.96 51.55 57.69 64.42 71.77 79.80 88.54 98.05 10837 119.55 131.64 144.70 158.78 173.93 190.22 207.71 Sat. liquid \"' 0.001424 0.001429 0.001434 0.001439 0.001444 0.001449 0.001454 0.001460 0.001465 0.001470 0.001476 0.001481 0.001487 0.001492 0.001498 0.001.s04 0.001510 - Sat. vapour, vg 2.62S4 2.3533 2.1140 1.9032 1.7170 1.5521 1.4058 12757 1.1597 1.0562 0.9635 0.8805 0.8059 0.7388 0.6783 0.6237 O.S743 Sat. liquid, Its -44.3 -JS.S -26.6 -17.8 -8.9 0.0 8.9 17.8 26.8 35.7 44.7 53.6 62.6 71.6 80.7 89.7 98.8 \u00a3vap.\u2022 htg 1416.7 1411.3 1405.8 1400.3 1394.7 1389.0 1383.3 13n.6 1371.8 1365.9 1360.0 1354.0 1.347.9 1)41.8 1335.6 1329.3 1322.9 Sat. vapour, hg 1372.4 1375.8 1379.2 1382.5 1385.8 1389.0 1392.2 1395.4 1398.5 1401.6 1404.6.1407.6 1410.5 1413.4 1416.2 1419.0 142).7 Entropy (kl/kg K) Sat. vapour sg 6.lS6l 6.1149 6.0746 6.0352 5.9967 5.9589 5.9220 5.8858 5.8504 5.8156 5.7815 5.7481 5.7153 5.6831 5.6S15 5.6205 5.S900 Sat. liquid, st -0.1942 -0.1547 - 0.1156 -0.0768 -0.0382 0.0000 0.0380 0.0757 0.1132 0.1504 0", ".1873 0.2240 0.2605 0.2967 0.3327 0.3684 0.4040.... ~ \"6... :;;\u2022 ~ 11 IO u, \" Abs. Prtss. (kPa) (Sat. temp.. \u00b0C) so (-46.54) 75 (-39.18) 100 (-33.01) 125 (-29.08) ISO (-25.23) 200 (-18.86) 2S0 (- 13.67) JOO (-9.23) \\t Ir s V Ir.~ V It s V It s V Ir II V Ir s V It s V Ir s i\u2022. - Table A.5.2 Superheated Ammonia Table Temperature (\"C) JO 40 10 0 /0 60 80 /00.1478.1 1499.2 1.7591 1476.1 1.3145 1474.1 1.6163 1604.1 1.2631 1452.2 l.6659 1626.0 1520.4 1541.7 l.8932 1518.9 1540.3 1.41<,() 1.4664 1517.3 1538.9 - 10 2.S48l 2.6482 2.7479 2.8479 2.9464 3.0453 1457.0 - 10 70 50 3.2427 3.3413 3.4397 3.1441 2.4474 1649.7 1627.8 1435.8 1563.0 1584.5 1606.1 6.3256 6.4077 6.4865 65625 6.6360 6.7073 6.776(, 6.8441 6.9099 6.9743 7.0372 1.9597 2.0261 J.6233 1,6915 1.8263 2.2903 2.\u00ae33 2.1584 2.2244 1433.0 1454.7 1497.S 1583.4 1648.9 1626.9 l<i05.I 1561.8 6.1190 6.2028 6.2828 6.3597 6.4339 6.5058 6.5756 6.6434 6.7096 6.7742 6.8373 1.3654 1.5664 1.5165 1.2110 1.8145 1.7155 1582.2 1430.1 1560.5 1495.7 1692.6 1648.0 6.5003 0.5668 6.6316 6.6950 5.9695 6.0552 6.1366 6.2144 61894 6.3618 6.4321 6.8177 1.4501 1.2107 1.0059 1.0476 0.9635 1.1703 1.0889 1.1297 1.3309 13707 1.2509 1.2909 1603.0 1625.0 1493.9 1427.2 IS59.3 1472.0 1449.8 1515.7 1537.5 1691.8 1581.1 1647.2 6.3887 6.1006 6.1763 6.2494 6.3201 S.8512 5.9389 6.0217 6.7072 6.4555 6.5206 6.5842 0.7984 0.8344 0.8&)7 0.9045 6.9388 0.9729 1.1408 l.207.2. 1.1074 1.0740 l.0405 1.0068 1424.I 1624.1 1536.1 1691.1 1602.0 1580.0 1558.0 1514'1 1469.8 1646.3 5.7526 5.8424 5.926' 6.0066 6.0831 6.4295 6.4933 6.6167 6.1568 6.2280 6.2970 6.3641 0.6738 0.7001 0.7261 0.9035 1510.9 1533.2 1448.1 1689.6 6.0(1)1 5.8559 S.9342 6.4732 05354 0.5568 0.5780 0.5989 0.6196 0.6401 0.6605 0.6809 0.7212 16881 1575.4 1620.3 1642.8 1597.8 1484.5 6.3613 6.0368 6.1052 6.1717 6.2365 0.5316 0.5488 0.5658 0.5997 1686.7 1641.1 16", "18.4 1595.7 6.2693 6.0114 6.0785 6.1437 0.4689 0.4837 0.5129 0.4541 0,6199 0.6471 1465.5 1442.0 5.6863 5.7737 0.4910 0.5135 1436.6 1461.0 1552.9 5.56()() 5.6517 5.7465 5.8165 5.8928 5.9661 \u00b70.4792 0.4968 0.5113 0.4243 0.4430 0.4613 1456.3 1573.0 1504.2 1527.4 1480.6 1550.3 5.5193 5.6366 S.7186 5.7963 S.8707 5.9423 0.3<,()5 0.3770 0.3929 0.4086 0.4239 OA391 0.7519 0.7774 0.8029 0.8282 0.8533 1555.5 1644.6 6.0813 6.1512 6.2189 6.2849 6.3491 1599.9 1622.2 IS07.6 1530.3 1577.7 1447.3 1492.1 <D ~ 11 ~...:\u00b7 it t t.... i:'., ~,.!} g l1 6' Alu. pnss. (kPa) (Sat. temp., \u00b0C) 350 (- 5.35) 400 (- t.89) 450 (1.26) 500 (4.14) 600 (9.29) 700 (13.81) 800 (17.86) 900 (21.S4I \" h s V =- Temperatunt (\"C) JO 50 50 40 20 80 1468.1 1637.6 1542.2 1614.5 1521.3 1612.6 1515.0 1591.5 IS44.9 1635.8 1518.2 1493.6 1497.2 70 1616.5 - 20 - JO /0 1478.5 30 1521.4 80 1639.3 20 1590.7 60 JOO 0 IS47.6 1570.7 1685.2 1593.6 145L5 5.4(,()() 5.S502 S.6312 5.7135 5.7800 5.8615 5.9314 5.9990 6.0647 6.1910 0.3125 0.3274 0.3417 0.3556 0.3692 0.3826 0.3959 0.4090 0.4220 0.4478 1472.4 1683.7 1568.3 1446.5 5.4735 5.5591 5.6405 5.7173 5.7907 5.8613 5.9296 5.9957 6.1228 5.3803 0.2752 0.2887 0.3017 0.3143 0.3266 0.3387 0.3506 0.3624 0.3740 0.3971 1682.2 1589.3 1.565.9 1441.3 5.7989 5.8678 5.9345 6Jl623 5.3078 5.4042 5492.6 5.5752 5.6532 5.7275./80 140 120 100 0.3975 0.3357 0.3565 0.3771 1680.7 1774.7 1727.5 '6.0079 6.1301 6.2472 0.2957 0.3130 0.3302 0.2508 0.2600 0.2691 1772.4 1724.9 1677.7 1606.6 1582.7 1558.5 6.1541 5.5697 5.6436 5.7144 5.7826 5.9129 6.0363 0.2821 0.2369 0.2522 0.2672 0.2212 0.2291 1722.4 1770.2 1674.6 1626.8.1602.6 1578.2 5.6303 5.6997 5.8316 5.9562 6.0749 5.5582 70 60 40 0.3251 02698 0.2813 0.2926 0.3036 0.3144 1587.1 1610.6 1634.0 1489.9 1563.4 1539.5 5.7425 5.8120 5.8793 5.4314 55157 5.5950 5.67040.2781 0.2217 0.2317 0.2414", " 1630.4 1482.4 1533.8 1508.6 5.3222 5,4!02 5.4923 0.1874 0.1963 0.2048 0.2131 1553.4 1474.5 5.2259 5.3179 5.4029 5.4826 0.1696 0.1773 0.1848 0.1920 0.1991 0.1615 IS98.6 1466.3 1548.3 1495.0 5.1387 5.2351 0.2060 0.2196 0.2329 0.2459 0.2589 1768.0 1816.4 1623.1 1719.8 IS22.2 5.3232 5.4053 5.4827 5.5562 5.6268 S.7603 S.8861 6.0057 6.1202 0.2178 0.2294 0.1488 0.1559 0.1627 0.1693 0.1757 0.1820 0.1942 0.2061 1814.4 176S.7 1717.1 1488.0 S.IS93 5.2S08 S.3354 S.4147 5.4897 5.5614 S.6968 S.8237 S.9442 6.0S94 0.1321 0.1388 0.14S0 O.ISII O.IS70 0.1627 0.1739 0.1847 0.19S4 0.20S8 0.2162 IS16.2 1543.0 1619.4 IS94.4 IS73.7 1501.9 1671.6 1668.S 1528.1 IS69.I 160 :.. ~... \" \"' s\u00b7 00 11 OI - 96'=- Basir a,ul Applied Tlumtodynamia Tt\"'P\u00b7 K 250 300 350 400 450.500 550 600 650 700 750 800 900 1000 CarlHm diazide, C01 0.791 0.602 1.314 0.&46 0.657 1.288 0.895 0.706 1.268 0.939 0.750 1.252 0.978 0.790 1.239 1.014 0.825 1.229 1.046 0.857 1.220 1.07S 0.886 1.213 1.102 0.913 1.207 1.126 0.937 1.202 1.148 0.959 1.197 1.169 0980 1.193 1.204 1.015 1.186 1.181 1.045 1.234 H)vlroge11, H2 Te,,,p. K Carbon monoxide, CO 9.</27 1.416 250 1.039 0.743 1.400 14.0SI 1.040 0.744 I0.183 1.405 300 1.39') 14.307 1.043 0.746 1.398 14A27 10.302 1.400 350 1.395 14.476 10.352 1398 400 J.047 0.751.1.392 14.501 10.3n 1.398 450 1.054 0.757 I0.389 1.397 500 1.063 0.767 1.387 14.513 1.075 0.778 1.382 14.530 I0.-405 1.396 550 1.087 0.790 1.376 14.S46 10.422 1.396 600 1.100 0.803 1.370 14.571 10.447 1.395 650 1.113 0.816 1.364 14.604 10.480 1.394 700 1.126 0.829 J.3S8 14.645 10.521 1.392 750 1.139 0.842 1.353 14.695 10.570 1.390 800 1.163 0.866 1.343 14.822 10.698 1.385 900 1.185 0.888 1.335 14.983 10.859 1.380 1()00 Source: Adapted from K. Wark, Thermodynamics, 4th ed., McGraw;Hill. New York, 1983, as based oo ''Tables of Tbennal Propenies of Gases;\u00b7 NBS Circular 564, 1955. I ill I! I Carbon Dioiide (COz> (h1Jiu ~ - J9J511 k.Jllr:mo/ M e 44.01 (h\" - hf98) kJ/lwwl - 9364 - 6456 -3414 0 67 4008 8314 12916 li765 22815 28041 33405 38894 44484 50158 55907 6", " dnswe,~ \\\u00b7 Sl8dlustrat1ons l 664 1eview quen,on.s Tata M,-<lra.. ~Hill i Jlnr.011\",.,, (?t.\u2022 tf('l,r1C1.-./IIN C,.nqor11tj4'\\ t,/hich refers to the mass per unit volume is the ratio of mass and volume occupied. Its units are kg/m3. Density = (Mass/Volume) Fig. 1.7 Bourdan tube for pressure measurement The specific volume is the volume per unit mass of the substance. It is defined by ratio of the volume occupied and the mass of substance. Its units are m3/kg. Specific volume = (Volume/Mass) Density or specific volume conform to the definitive specification of a thermodynamic property and are capable of getting associated with other properties such as temperature, pressure and internal Fundamental Concepts and Definitions ____________________________________________ 11 energy. Also, the volume occupied by a material is a measure of distance between molecules and thus indicates their molecular energy. Weight is actually the force due to gravity acting on any substance. Thus, it is the product of mass and gravitational acceleration. Its units are Newtons. Weight = (mass \u00d7 gravitational acceleration) Specific weight of a substance is the ratio of weight of substance and volume of substance. Specific weight = (Weight/Volume) = (density \u00d7 gravitational acceleration) Specific gravity is defined as the ratio of the density of any substance and standard density of some reference substance. For solids and liquids the water density at some specified temperature say 0\u00baC or 4\u00baC is taken as standard density. 1.11 ENERGY AND ITS FORMS \u201cEnergy is usually defined as the ability to do mechanical work\u201d. It is indeed quite difficult to precisely define the \u201cenergy\u201d. We feel energy at every moment and can sense it very oftenly. Another broader definition of energy says that \u201cenergy refers to the capacity for producing effects.\u201d Total energy at any moment may be the algebraic summation of the different forms of energy. Conversion of energy from one to other is also possible. In thermodynamics we are primarily interested in studying the change in total energy of a system. Thus, for analysis relative value of energy is considered instead of absolute value. Energy can be classified in the following general categories; (a) Energy in transition: It refers to the energy that is in process of transition between substances or regions because of some driving potential, such as difference or gradient of force, or of temperature, or of electrical potential etc. For example heat, work etc. (b) Energy stored in particular mass: It refers to the potential and kinetic energy associated with masses that are elevated or moving with respect to the earth. Apart from above broad classification the energy can also be categorised into various forms. (i) Macroscopic energy: It refers to the energy possessed by a system considered at macro- scopic level such as kinetic energy, potential energy etc. (ii) Microscopic energy: It refers to the energy defined at molecular level. Summation of energy at molecular level or microscopic energy results in internal energy. Some of the popular forms of energy are described below : Potential energy: This type of energy is based on relative position of bodies in a system, i.e. elevation in a gravitational field. Potential energy for a mass m at elevation z is given as : P.E. = m.g.z Here g is the gravitational acceleration and elevation is measured from some reference point. Kinetic energy: It is based on the relative movement of bodies. For a mass m moving with certain velocity c it could be mathematically expressed as; K.E. = (1/2) m.c2 Internal energy: Internal energy of a system is the energy associated with the molecular structure at molecular level. Let us study fall of a \u2018weight\u2019 from certain height on the floor. Upon hitting the floor, \u2018weight\u2019 comes to dead stop and its potential energy and kinetic energy both reduce to zero. Question arises, where does the vanishing potential energy and kinetic energy go upon \u2018weight\u2019 touching the floor. If we touch the point of contact between \u2018weight\u2019 and floor, it is observed that both these points are slightly hotter than before impact. Thus, it is obvious that the energy has changed its form from potential and kinetic to internal energy and causes rise in temperature of \u2018weight\u2019 and floor at the points of contact. 12 _________________________________________________________ Applied Thermodynamics Internal energy is sum of energy associated with molecules which may have translational, vibrational and rotational motions etc. and respective energies causing these motions. Internal energy may be thus comprising of sensible energy, latent energy, chemical energy, nuclear energy etc. \u2018Sensible energy\u2019 refers to the kinetic energy associated with molecules. \u2018Latent energy\u2019 refers to", " the energy associated with phase of a substance. \u2018Chemical energy\u2019 refers to the energy associated with molecular bonds. \u2018Nuclear energy\u2019 refers to the energy associated with the bonds within nucleus of atom itself. Total energy of a system may be given as summation of different forms of energy at a moment. Mathematically; T.E (Total energy) where = K.E + P.E + I.E K.E = Kinetic energy P.E = Potential energy I.E = Internal energy Some different forms of energy interaction and associated work interactions with block diagram are given in table 1.4. Table 1.4 Some forms of energy and the associated work interactions S. No. Macroscopic form of energy Governing equation Energy interaction Work interaction Block diagram 1. 2. 3. 4. Kinetic energy (translation) Kinetic energy (rotational) F = m \u00b7 dV dt T = J \u00b7 \u03c9 d dt Spring stored energy (translational) F = kx \u2206E = ( 2 V 2 V\u2212 1 2 2 1 = \u2013 F \u00b7 dx m\u00b7 ) 1 \u2206E = 2 ( \u03c9 \u03c9\u2212 2 1 2 2 J\u00b7 ) = \u2013 T \u00b7 d\u03b8 F x m \u03b8 T J k\u00b7 = \u2013 F \u00b7 dx F k F = 0 x F \u2206E = ( x\u2212 2 x 2 2 1 1 2 ) Spring stored energy (rotational) T = K \u00b7 \u03b8 K\u00b7 1 \u2206E = 2 ) ( \u03b8 \u03b8\u2212 2 2 1 2 = \u2013 T \u00b7 d\u03b8 5. Gravitational F = mg \u2206E = mg\u00b7 = \u2013 F \u00b7 dz energy (Z2 \u2013 Z1) 6. Electrical energy (capacitance) u = q c 7. Electrical energy \u03c6 = L \u00b7 i (inductance) 2q c = \u2013 u \u00b7 dq 1 2 cu2 \u2206E = 1 2 = Li2 = \u2013 i \u00b7 d\u03c6 1 \u2206 Fundamental Concepts and Definitions ____________________________________________ 13 1.12. HEAT AND WORK When two systems at different temperatures are brought into contact there are observable changes in some of their properties and changes continue till the two don\u2019t attain the same temperature if contact is prolonged. Thus, there is some kind of energy interaction at the boundary which causes change in temperatures. This form of energy interaction is called heat. Thus \u2018heat\u2019 may be termed as the energy interaction at the system boundary which occurs due to temperature difference only. Heat is observable in transit at the interface i.e. boundary, it can not be contained in a system. In general the heat transfer to the system is assigned with positive (+) sign while the heat transfer from the system is assigned with negative (\u2013) sign. Its units are Calories. In thermodynamics the work can be defined as follows: \u201cWork shall be done by the system if the total effect outside the system is equivalent to the raising of weight and this work shall be positive work \u201d. In above definition the work has been defined as positive work and says that there need not be actual raising of weight but the effect of the system behaviour must be reducible to the raising of a weight and nothing else. Its units are N. m or Joule. Heat and work are two transient forms of energy. Let us look at a piston cylinder mechanism (closed system), where high pressure air is filled inside the cylinder fitted with a piston exerting force against some resistance. As the piston moves a distance say \u2018l\u2019, the work would be done. It can be reduced to the raising of weight by replacing this resisting system by a frictionless pulley and lever such that a weight W is raised, Fig. 1.8. For example, if an electrical battery and resistance is considered as a system, then this system shall do work when electric current flows in an external resistance as this resistance could be replaced by an ideal frictionless motor driving a frictionless pulley and raising a weight. Here, also in reference to work it is obvious that the work is the entity which is available at the boundary of system, thus work can not be stored rather it is energy interaction in transit at the boundary. From the thermodynamic definition of work the sign convention established as positive work shall be the one which is done by the system while the negative work shall be the one that is done upon the system. Fig. 1.8 Thermodynamic work 14 _________________________________________________________ Applied Thermodynamics 1.13 GAS LAWS Thermodynamic analysis relies largely upon the gas laws, which are known as Boyle\u2019s law (1662) and Charle\u2019s law (1787). Boyle\u2019s law says that if temperature of a gas is held constant then its molar volume is inversely proportional to the pressure. Mathematically it can be related as p v = constant. Here p is the pressure and v is the", " molar volume of gas, i.e. volume per mole. Charle\u2019s law says that for the pressure of gas held constant the volume of gas is directly proportional to the temperature of gas. Mathematically it can be given as v /T = constant, where T is the temperature of the gas. It also says that if the molar volume of gas is held constant, the pressure of gas is directly proportional to temperature, i.e. p/T = constant. Figure 1.9 shows the graphical representation. Fig 1.9 Graphical representations of gas laws at constant temperature and at constant pressure Boyle\u2019s and Charle\u2019s law when combined together result in, p v /T = constant or p v = R T, where R is the universal gas constant. 1.14 IDEAL GAS Engineering thermodynamics deals with different systems having gaseous working fluids. Some gases behave as ideal gas and some as non-ideal gas. Based on the experimental methods various equations of state of gases have been developed. For perfect gas the ideal gas equation shows that p v = R T, where R is the universal gas constant and can be related as R = R /M, here R is the characteristic gas constant and M is the molar mass or molecular weight of the substance, v is volume per mole. Universal gas constant has value given as 8.31441 kJ/k mol.K. or pV= m RT, where m is mass of the substance, V is the volume of substance, i.e, where \u2018n\u2019 is no. of moles. Gas constant is also related to specific heats at constant pressure and volume as follows, R = cp \u2013 cv Upon plotting the variables P, V, T for an ideal gas on three mutually perpendicular axes, the three dimensional entity generated is called P-V-T surface and can be used for studying the thermodynamic properties of ideal gas. Figure 1.10 shows the typical P-V-T surface for an ideal gas. Fundamental Concepts and Definitions ____________________________________________ 15 For certain gases the molecular weight and gas constant are given in table 1.5. Fig. 1.10 P-V-T surface for ideal gas Table 1.5 Gas Molecular weight, kg/kmol Gas constant, kJ/kg.K Air Carbon dioxide Hydrogen Helium Nitrogen Oxygen Steam 28.97 44.01 2.016 4.004 28.01 32.00 18.02 0.287 0.189 4.124 2.077 0.297 0.260 0.461 1.15 DALTON\u2019S LAW, AMAGAT\u2019S LAW AND PROPERTY OF MIXTURE OF GASES Dalton\u2019s law of partial pressures states that the \u201ctotal pressure of a mixture of gases is equal to the sum of partial pressures of constituent gases.\u201d Partial pressure of each constituent can be defined as the pressure exerted by the gas if it alone occupied the volume at the same temperature. Thus, for any mixture of gases having \u2018j\u2019 gases in it, the mathematical statement of Dalton\u2019s law says, if and p = p1 + p2 + p3 +..... + pj V = V1 = V2 = V3 =..... = Vj T = T1 = T2 = T3 =..... Tj Dalton\u2019s law can be applied to both mixture of real gases and ideal gases. (a \u2192, constituent gases Mixture V T, (b) m V 1, Fig. 1.11 (a) Dalton\u2019s law of partial pressures, (b) Amagat\u2019s law Let us take mixture of any three, perfect gases, say, 1, 2, 3 in a container having volume \u2018V\u2019 and temperature T. 16 _________________________________________________________ Applied Thermodynamics Equation of state for these gases shall be, p1V = m1R1T ; p2 V = m2 \u00b7 R2 \u00b7 T, p3V = m3 \u00b7 R3 \u00b7 T The partial pressures of three gases shall be, \u00b7 \u00b7 3 m R T V 3 p1 = 1 1m R T V, p2 = 2 2\u00b7 \u00b7 m R T V, p3 = From Dalton\u2019s law; p = p1 + p2 + p3 = (m1R1 + m2R2 + m3R3) \u00b7 T V or, it can be given in general form as, pV = where i refers to constituent gases Amagat\u2019s law of additive volumes states that volume of a gas mixture is equal to the sum of volumes each gas would occupy at the mixture pressure and temperature. V = V1 + V2 + V3........... + Vj p = p1 = p2 = p3........ pj T = T1 = T2 = T3 =", "........ Tj Mass balance upon mixture yields m = m1 + m2 + m3 or m = j \u2211 = i 1 m i From above the gas constant for the mixture can be given as; R = or, in general form Mole fraction xi of a constituent gas can be defined as the ratio of number of moles of that constituent to the total number of moles of all constituents. Thus mole fractions of three gases, if number of moles of three gases are n1, n2 and n3; x1 = + n 1 x2 = + n 1 x3 = + Fundamental Concepts and Definitions ____________________________________________ 17 or, in general xi = Total no. of moles, n i n\u2211 i n = n1 + n2 + n3 or Sum of mole fractions of all constituent equals to 1, ix\u2211 = \u2211 in n = 1 Number of moles of any constituent gas, ni = n \u00b7 xi For Mi being the molecular weight of a constituent gas, the mass mi of that constituent shall be mi = ni \u00b7 Mi or, mi = n \u00b7 xi \u00b7 Mi and the total mass m, shall be m = im\u2211 = n. ix\u2211 \u00b7 Mi Molecular weight of mixture shall be: ix\u2211 \u00b7 Mi M = = m n 1.16 REAL GAS When a gas is found to disobey the perfect gas law, i.e. the equation of state for ideal gas, then it is called \u2018real gas\u2019. Real gas behaviour can also be shown by a perfect gas at the changed thermodynamic states such as high pressure etc. Deviation of real gas from ideal gas necessitates the suitable equation of state which can be used for interrelating the thermodynamic properties P, V, and T. From the kinetic theory of gases it is obvious that the ideal gas equation of state suits the gas behaviour when intermolecular attraction and volume occupied by the molecules themselves is negligibly small in reference to gas volume. At high pressures intermolecular forces and volume of molecules both increase and so the gas behaviour deviates from ideal gas to real gas. A number of mathematical equations of state suggested by Van der-Waals, Berthelot, Dieterici, Redlich-Kwong, Beattie-Bridgeman and Martin-Hou etc. are available for analysing the real gas behaviour. Dalton\u2019s law and Amagat\u2019s law can also be used for real gases with reasonable accuracy in conjunction with modified equations of state. As the ideal gas equation does not conform to the real gas behaviour in certain ranges of pressures and temperatures, so the perfect gas equation should be modified using compressibility factor for the gas at given pressure and temperature. Such modified form of equations shall be; P v = Z \u00b7 R \u00b7 T Here Z is the compressibility factor, a function of pressure and temperature. Thus, compressibility factor is like a correction factor introduced in ideal equation of state for suiting the real gas behaviour. Compressibility factor is an indication of deviation of a gas from ideal gas behaviour and can be mathematically given as; 18 _________________________________________________________ Applied Thermodynamics or Z = \u0192(P, T) v actual v ideal Z = Here, v ideal = RT P i.e. Z = 1 for ideal gases while Z can be greater than or less than unity. Individual graphical representations are available for getting the compressibility factor as shown in Fig 1.12. Compressibility factor charts are available for different substances. Compressibility factors for various substances can also be shown on a generalized compressibility chart using reduced properties. Reduced properties are non-dimensional properties given as ratio of existing property to critical property of substance. Such as reduced pressure is ratio of pressure of gas to critical pressure of gas. Similarly, reduced temperature can be given by ratio of temperature of gas to critical temperature of gas. Reduced pressure, pR = Reduced temperature, TR = p p c T T c (a) Oxygen (b) Carbon dioxide Fig. 1.12 Compressibility factors, Z Fundamental Concepts and Definitions ____________________________________________ 19 where Pc and Tc denote critical pressure and critical temperature respectively. These reduced pressure and reduced temperature are used for getting the generalized compressibility chart of the form, Z = \u0192 (pR, TR) where Z for all gases is approximately same. This similar behaviour of compressibility correction factor for different gases in respect to reduced pressures and temperatures is called \"principle of corresponding states.\" Fig. 1.13 shows a generalized compressibility chart. In generalized compressibility chart a set of curves is fitted through a set of experimentally determined Z values plotted against reduced Fig. 1.13 (a) Generalized compressibility chart, pR \u2264 1.0 Fig. 1.13 (b) Generalized compressibility chart, pR \u2264 10.0 20 _________________________________________________________ Applied Thermodynamics Fig", ". 1.13 (c) Generalized compressibility chart, 10 \u2264 pR \u2264 40 pressure and reduced temperatures for several gases. On the generalized compressibility chart it could be seen that at \u201cvery small pressures the gases behave as an ideal gas irrespective of its temperature\u201d and also at \u201cvery high temperatures the gases behave as ideal gas irrespective of its pressure\u201d. 1.17 VANDER\u2019 WAALS AND OTHER EQUATIONS OF STATE FOR REAL GAS Vander\u2019 Waals suggested the equation of state for real gas in 1873. The equation has been obtained applying the laws of mechanics at molecular level and introducing certain constants in the equation of state for ideal gas. This equation agrees with real gas behaviour in large range of gas pressures and temperatures. Vander\u2019 Waals equation of state for real gas is as follows = RT where \u2018a\u2019 is the constant to take care of the mutual attraction amongst the molecules and thus accounts for cohesion forces. Gas Helium Hydrogen Oxygen Air Carbon dioxide Table 1.6 Vander\u2019 Waals constant Constant a, N.m4/( kg. mol)2 Constant b, m3/kg.mol 34176.2 \u00d7 102 251.05 \u00d7 102 1392.5 \u00d7 102 1355.22 \u00d7 102 3628.50 \u00d7 102 2.28 \u00d7 10\u20132 2.62 \u00d7 10\u20132 3.14 \u00d7 10\u20132 3.62 \u00d7 10\u20132 3.14 \u00d7 10\u20132 \uf8eb \uf8ec \uf8ed a 2 v \uf8f6 \uf8f7 \uf8f8 Fundamental Concepts and Definitions ____________________________________________ 21 Constant \u2018b\u2019 accounts for the volumes of molecules which are neglected in perfect gas equation, thus it denotes \u201cco-volume\u201d, Mathematically, = a 27 2 2 R T c 64 Here, pc, Tc are critical point pressures and temperatures having values as given in appendix. Thus these constants \u2018a\u2019 & \u2018b\u2019 are determined from behaviour of substance at the critical point. In general it is not possible to have a single equation of state which conforms to the real gas behaviour at all pressures and temperatures. A few more equations of state for real gas as suggested by various researchers are as follows. Redlich-Kwong equation of state for real gas, = p RT \u2212 \u00b7 where a = 0.4278 Berthelot equation of state for real gas, \uf8eb \uf8ec \uf8ec \uf8ed 2.5 2 \u00b7 c R T p c \uf8f6 \uf8f7 \uf8f7 \uf8f8 and b = 0.08664 \uf8eb \uf8ec \uf8ed \u00b7 R T c p c \uf8f6 \uf8f7 \uf8f8 p = RT \u2212 where \uf8eb \uf8ec \uf8ec \uf8ed a = 27\u00b7 2 3 \u00b7 R T c p c 64 \u00b7 \uf8f6 \uf8f7 \uf8f7 \uf8f8 and \uf8f6 \uf8f7 \uf8f8 Here a and b refer to the constants as suggested in respective equations. Beattie-Bridgeman equation of state given in 1928, for real gas has five constants determined experimentally. It is \uf8eb \u2212\uf8ec 0 1 \uf8ed a v \uf8f6 \uf8f7 \uf8f8 and = B B \uf8eb \u2212\uf8ec 0 1 \uf8ed b v \uf8f6 \uf8f7 \uf8f8 where, Constants used in Beattie \u2013 Bridgeman equation are given in Table 1.7 when p is in k pa, v is in m3/k mol, T is in K, and R = 8.314 k pa m3/k mol.K. Table 1.7. Beattie -Bridgeman constants Gas Helium Hydrogen Oxygen Air Carbon dioxide A0 2.1886 20.0117 151.0857 131.8441 507.2836 a 0.05984 \u20130.00506 0.02562 0.01931 0.07132 B0 0.01400 0.02096 0.04624 0.04611 0.10476 b c 0.0 \u20130.04359 0.004208 \u20130.001101 0.07235 40 504 4.80 \u00d7104 4.34 \u00d7 104 6.60 \u00d7 105 22 _________________________________________________________ Applied Thermodynamics Virial equations of state propose a form of equation which can be suitably modified and used for real gases. These equations of state are in the form, pv RT pv RT = A0 + A1. p + A2. p2 + A3. p3 +..... = B0 + 1B +....... or Where A0, A1, A2, A3,.........and B0, B1, B2, B3....... are called the \"virial coefficients\" and depend upon temperature alone", ". Virial equations of state can be suitably modified based on experimental P, v, T data for the real gas analysis. Virial constants can be calculated if the suitable model for describing the forces of interaction between the molecules of gas under consideration is known. 1. Find out the pressure difference shown by the manometer deflection of 30 cm of Mercury. Take local acceleration of gravity as 9.78 m/s2 and density of mercury at room temperature as 13,550 kg/m3. EXAMPLES Solution: From the basic principles of fluid statics, Pressure difference = \u03c1\u00b7gh = 13550 \u00d7 30 \u00d7 10\u20132 \u00d7 9.78 = 39755.70 Pa Ans. 2. An evacuated cylindrical vessel of 30 cm diameter is closed with a circular lid. Estimate the effort required for lifting the lid, if the atmospheric pressure is 76 cm of mercury column (Take g = 9.78 m/s2) Solution: Effort required for lifting the lid shall be equal to the force acting upon the lid. Thus, effort required = Pressure \u00d7 Area = (76 \u00d7 10\u20132 \u00d7 13550 \u00d7 9.78) \u00d7 (3.14 \u00d7 (30 \u00d7 10\u20132)2/4) = 7115.48 N Ans. 3. Calculate the actual pressure of air in the tank if the pressure of compressed air measured by manometer is 30 cm of mercury and atmospheric pressure is 101 kPa. (Take g = 9.78 m/s2) Solution: Pressure measured by manometer on the tank is gauge pressure, which shall be = \u03c1.g.h = (13550 \u00d7 9.78 \u00d7 30 \u00d7 10\u20132) = 39755.70 Pa = 39.76 kPa Actual pressure of air = Gauge pressure + atmospheric pressure = 39.76 + 101 = 140.76 kPa Ans. 4. Determine gauge pressure at a depth of 1 m in a tank filled with oil of specific gravity 0.8. Take density of water as 1000 kg/m3 and g = 9.81 m/s2. Fundamental Concepts and Definitions ____________________________________________ 23 Solution: Density of oil = Specific gravity \u00d7 Density of water \u03c1 \u03c1 Gauge pressure = (\u03c1 oil = 0.8 \u00d7 1000 oil = 800 kg/m3 oil \u00d7 g \u00d7 h) = 800 \u00d7 9.81 \u00d7 1 = 7848 N/m2 or Gauge pressure = 7.848 kPa. Ans. 5. Calculate the gas pressure using a mercury manometer with one limb open to atmosphere as shown in Fig. 1.14. Barometer reading is 76 cm and density of mercury is 13.6 \u00d7 103 kg/m3. Take g = 9.81 m/s2. Solution: Figure shows that the difference of height in mercury columns is 40 cm. In reference to level AB the pressure exerted by gas, pgas can be written as sum of atmospheric pressure and pressure due to mercury column at AB pgas = (\u03c1 mercury \u00d7 9.81 \u00d7 40 \u00d7 10\u20132) + Atmospheric pressure = (13.6 \u00d7 103 \u00d7 9.81 \u00d7 40 \u00d7 10\u20132) + (13.6 \u00d7 103 \u00d7 9.81 \u00d7 76 \u00d7 10\u20132) = 154762.56 N/m2 Pgas = 154.76 kPa Ans. or Fig. 1.14 6. 1 kg of water falls from an altitude of 1000 m above ground level. What will be change in the temperature of water at the foot of fall, if there are no losses during the fall. Take specific heat of water as 1 kcal/kg\u00b7K Solution: Initially when water is at 1000 m, it shall have potential energy in it. This potential energy shall get transformed upon its fall and change the temperature of water. By law of energy conservation Potential energy = Heat required for heating water \u00d7 \u00d7 1 9.81 1000 4.18 = = 1 \u00d7 1 \u00d7 103 \u00d7 \u2206T or \u2206T = 2.35\u00baC Change in temperature of water = 2.35\u00baC Ans. 24 _________________________________________________________ Applied Thermodynamics 7. A spring balance is used for measurement of weight. At standard gravitational acceleration it gives weight of an object as 100 N. Determine the spring balance reading for the same object when measured at a location having gravitational acceleration as 8.5 m/s2. Solution: At standard gravitational acceleration, mass of object = Spring balance reading = Gravitational force in mass = 10.194 kg 100 9.81 = 10.194 \u00d7 8.5 = 86.649 N = 86.65 N Ans. 8. An incompressible gas in the cylinder of 15 cm diameter is used to support a piston, as shown. Manometer indicates a difference of 12 cm of Hg column for the gas in cylinder. Estimate the mass of piston that can be supported by the gas. Take density of mercury as 13.6 \u00d7 103 kg/", "m3. Solution: Piston shall be supported by the gas inside, therefore, let mass of piston be \u2018m\u2019 kg. Weight of piston = Upward thrust by gas m..81 = (12 \u00d7 10\u20132 \u00d7 13.6 \u00d7 103 \u00d7 9.81) \u00d7 m = 28.84 kg Mass of piston = 28.84 kg. Ans. Fig. 1.15 4 \u00d7 (15 \u00d7 10\u20132)2 9. Determine pressure of steam flowing through a steam pipe when the U-tube manometer connected to it indicates as shown in figure 1.16. During pressure measurement some steam gets condensed in manometer tube and occupies a column of height 2 cm (AB) while mercury gets raised by 10 cm (CD) in open limb. Consider barometer reading as 76 cm of Hg, density of mercury and water as 13.6 \u00d7 103 kg/m3 and 1000 kg/m3 respectively. Solution: Let us make pressure balance at plane BC. psteam + pwater, AB = patm + pHg, CD psteam = patm + pHg, CD \u2013 pwater, AB patm = (13.6 \u00d7 103 \u00d7 76 \u00d7 10\u20132 \u00d7 9.81) patm = 101396.16 N/m2 pwater, AB = (1000 \u00d7 2 \u00d7 10\u20132 \u00d7 9.81) pwater, AB = 196.2 N/m2 Fig. 1.16 Fundamental Concepts and Definitions ____________________________________________ 25 pHg, CD = (13.6 \u00d7 103 \u00d7 10 \u00d7 10\u20132 \u00d7 9.81) pHg, CD = 13341.6 N/m2 Substituting for getting steam pressure, or psteam = 101396.16 + 13341.6 \u2013 196.2 psteam = 114541.56 N/m2 psteam = 114.54 kPa Ans. 10. A vessel has two compartments \u2018A\u2019 and \u2018B\u2019 as shown with pressure gauges mounted on each compartment. Pressure gauges of A and B read 400 kPa and 150 kPa respectively. Determine the absolute pressures existing in each compartment if the local barometer reads 720 mm Hg. Solution: Atmospheric pressure from barometer = (9810) \u00d7 (13.6) \u00d7 (0.720) = 96060 Pa = 96.06 kPa Absolute pressure in compartment A, Pabs, A = Pgauge, A + Patm = 400 + 96.06 = 496.06 kPa Absolute pressure in compartment B, Pabs, B = Pgauge, B + Patm = 150 + 96.06 = 246.06 kPa Absolute pressure in compartments A & B = 496.06 kPa & 246.06 kPa Ans. Fig. 1.17 11. Determine the air pressure in a tank having multifluid manometer connected to it, with the tube open to atmosphere as shown in figure. Tank is at an altitude where atmospheric pressure is 90 kPa. Take densities of water, oil and mercury as 1000 kg/m3, 850 kg/m3 and 13600 kg/m3 respectively. Fig. 1.18 26 _________________________________________________________ Applied Thermodynamics Solution: It is obvious that the lengths of different fluids in U-tube are due to the air pressure and the pressure of air in tank can be obtained by equalizing pressures at some reference line. Strating from point (1) the pressure can be given as under for equilibrium, p1 + \u03c1 Given : oil \u00b7 g \u00b7 h2= patm + \u03c1 mercury \u00b7 g \u00b7 h3 oil = 850 kg/m3, \u03c1 mercury = 13600 kg/m3 water \u00b7 g \u00b7 h1 + \u03c1 water = 1000 kg/m3, \u03c1 \u03c1 h1 = 0.15 m, h2 = 0.25 m, h3 = 0.40 m, patm = 90 kPa Substituting we get p1 = 139.81 kPa Air pressure = 139.81 kPa Ans. 12. Estimate the kinetic energy associated with space object revolving around earth with a relative velocity of 750 m/s and subjected to gravitational force of 4000 N. Gravitational acceleration may be taken as 8 m/s2. Solution: Mass of object = Gravitational force Gravitational acceleration = 4000 8 = 500 kg Kinetic energy = 1 2 \u00d7 500 \u00d7 (750)2 = 140625000 J Kinetic energy = 1.4 \u00d7 108 J Ans. 13. Determine the molecular weight of a gas if its specific heats at constant pressure and volume are cp = 2.286 kJ/kg K and cv = 1.768 kJ/kg K. Solution: Gas constant shall be, R = cp \u2013 cv = 0.518 kJ/kg.K Molecular weight of gas = R R = Universal gas constant Characteristic gas constant =", " 8.3143 0.518 = 16.05 kg/k mol Ans. 14. A perfect gas at pressure of 750 kPa and 600 K is expanded to 2 bar pressure. Determine final temperature of gas if initial and final volume of gas are 0.2 m3 and 0.5 m3 respectively. Fundamental Concepts and Definitions ____________________________________________ 27 Solution: Initial states = 750 \u00d7 103 Pa, 600 K, 0.2 m3 Final states = 2 bar or 2 \u00d7 105 Pa, 0.5 m3. Using perfect gas equation, p V 1 1 T 1 0. 10 T 2 T2 = 400 K = \u00d7 0.5 \u00d7 3 \u00d7 750 10 600 Final temperature = 400 K or 127\u00ba C Ans. 15. A vessel of 5 m3 capacity contains air at 100 kPa and temperature of 300K. Some air is removed from vessel so as to reduce pressure and temperature to 50 kPa and 7\u00baC respectively. Find the amount of of air removed and volume of this mass of air at initial states of air. Take R = 287 J/kg.K for air. Solution: Initial states : 100 \u00d7 103 Pa, 300 K, 5 m3 Final states : 50 \u00d7 103 Pa, 280 K, 5 m3 Let initial and final mass of air be m1 and m2. From perfect gas equation of air, m1 = p V 1 1 RT ; m2 = 1 p V 2 2 RT 2 m1 = \u00d7 \u00d7 3 100 10 5 \u00d7 287 300 ; m2 = \u00d7 \u00d7 3 50 10 5 \u00d7 287 280 Mass of removed, (m1 \u2013 m2) \uf8eb \uf8ec \uf8ec \uf8ed \u00d7 3 \u00d7 100 10 \u00d7 287 300 \u00d7 \u00d7 50 10 5 \u00d7 287 280 \uf8f6 \uf8f7 \uf8f7 \uf8f8 m1 \u2013 m2 = 2.696 kg Volume of this mass of air at initial states i.e 100 kPa and 300 K RT 2 1 p 1 \u00d7 \u00d7 2.696 287 300 3 \u00d7 100 10 = Volume = 2.32 m3 Mass of air removed = 2.696 kg Volume of air at initial states = 2.32 m3 Ans. 16. A cylindrical vessel of 1 m diameter and 4 m length has hydrogen gas at pressure of 100 kPa and 27\u00baC. Determine the amount of heat to be supplied so as to increase gas pressure to 125 kPa. For hydrogen take Cp = 14.307 kJ/kg.K, Cv = 10.183 kJ/kg K. 28 _________________________________________________________ Applied Thermodynamics Solution: Assuming hydrogen to be perfect gas let initial and final states be denoted by subscript 1 and 2, Here V2 = V1 \u00b7 T2 = 375 K As it is constant volume heating so, heat supplied, p T 1 2 p 1 T2 = = = 3 \u00d7 \u00d7 125 10 \u00d7 100 10 300 3 Q = m \u00b7 Cv (T2 \u2013 T1) From perfect gas characteristics, R = Cp \u2013 Cv R = 4.124 kJ/kg \u00b7 K Mass of hydrogen, m = Heat added, Heat to be supplied 3 \u03c0 \u00d7 \u00d7 \u00d7 100 10 p V 1 1 RT = m = 0.254 kg Q = 0.254 \u00d7 10.183 \u00d7 (375 \u2013 300) (0.5) \u00d7 4.124 10 300 \u00d7 1 2 3 \u00d7 4 = 193.99 kJ Ans. 17. Two cylindrical vessels of 2 m3 each are inter connected through a pipe with valve in-between. Initially valve is closed and one vessel has 20 kg air while 4 kg of air is there in second vessel. Assuming the system to be at 27\u00baC temperature initially and perfectly insulated, determine final pressure in vessels after the valve is opened to attain equilibrium. Solution: When the valve is opened then the two vessels shall be connected through pipe and transfer of air shall take place in order to attain equilibrium state. After attainment of equilibrium total mass of air shall be 24 kg. Final total volume = 2 \u00d7 2 = 4 m3 Using perfect gas equation. pV = mRT For air, Substituting values, p = p = mRT V R = 287 J/kg K \u00d7 \u00d7 24 287 300 4 = 516600 N/m2 Final pressure = 516.6 kPa Ans. 18. Determine the pressure of 5 kg carbon dixoide contained in a vessel of 2 m3 capacity at 27\u00ba C, considering it as (i) perfect gas (ii) real gas. Fundamental Concepts and Definitions ____________________________________________ 29 Solution: Given : Volume, Temperature, V = 2 m3, Universal gas constt. = 8.314 kJ/kg. K T = 27\u00baC = (273 + 27) K T = 300 K m = 5 kg Mass, Let pressure exerted be \u2018p\u2019. (i) Considering it as perfect gas, pV = mR", " CO2 T CO2R = Universal gas constt. Molecular weight of CO 2 CO2R = 3 \u00d7 8.314 10 44.01 CO2R Substituting in perfect gas equation, \u00d7 = 188.9 J/kg \u00b7 K \u00d7 5 188.9 300 2 Pressure = 1.417 \u00d7 105 N/m2 Ans. (ii) Considering it as real gas let us use Vander-Waals equation; = 141675 N/m2 ) = RT where \u2018 v \u2019 is molar specific volume and constants \u2018a\u2019 and \u2018b\u2019 can be seen from Table 1.6. R = 8.314 \u00d7 103 Molar specific volume, v = \u00d7 2 44.01 5 v =17.604 m3/kg \u00b7 mol Vander-Waals Constant, a = 3628.5 \u00d7 102 N. m4/(kg \u00b7 mol)2 b = 3.14 \u00d7 10\u20132 m3/kg \u00b7 mol Substituting values in Vander Waals equation, \uf8eb +\uf8ec p \uf8ec \uf8ed \u00d7 3628.5 10 2 (17.604) 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 (17.604 \u2013 3.14 \u00d7 10\u20132) = (8.314 \u00d7 103 \u00d7 300) p + 1170.86 = 141936.879 p = 140766.019 N/m2 Pressure = 1.408 \u00d7 105 N/m2 For CO2 as perfect gas = 1.417 \u00d7 105 N/m2 For CO2 as real gas = 1.408 \u00d7 105 N/m2 Ans. (using Vander-Waals equation) 30 _________________________________________________________ Applied Thermodynamics 19. Determine the specific volume of steam at 17672 kPa and 712 K considering it as (a) perfect gas, (b) considering compressibility effects. Take critical pressure = 22.09 MPa, critical temperature =647.3 K, Rsteam = 0.4615 kJ/kg\u00b7K. Solution: (a) Considering steam as perfect gas, Sp. volume = steam \u00b7 T R p = \u00d7 0.4615 712 17672 Specific volume = 0.0186 m3/kg Ans. (b) Considering compressibility effects, the specific volume can be given by product of compressibility factor \u2018Z\u2019 and \u201cspecific volume when perfect gas\u201d. Reduced pressure = p Critical pressure = 17672 3 22.09\u00d710 Reduced pressure = 0.8 Reduced temperature = Critical temperature T = 712 647.3 Reduced temperature = 1.1 From generalized compressibility chart compressibility factor \u2018Z\u2019 can be seen for reduced pressure and reduced temperatures of 0.8 and 1.1. We get, Z = 0.785 Actual specific volume = 0.785 \u00d7 0.0186 = 0.0146 m3/kg. Ans. 20. A spherical balloon of 5 m diameter is filled with Hydrogen at 27\u00baC and atmospheric pressure of 1.013 bar. It is supposed to lift some load if the surrounding air is at 17\u00baC. Estimate the maximum load that can be lifted. Solution: Balloon filled with H2 shall be capable of lifting some load due to buoyant forces. Volume of balloon = \uf8f6 \uf8f7 \uf8f8 = 65.45 m3 Mass of H2 in balloon can be estimated considering it as perfect gas. Gas constant for H2 = 3 \u00d7 8.314 10 2 = 4.157 \u00d7 103 J/kg \u00b7 K Fundamental Concepts and Definitions ____________________________________________ 31 Mass of H2 in balloon = \u00b7 V P balloon balloon \u00b7H T R H 2 2 = 1.013 10 \u00d7 5 \u00d7 4.157 10 \u00d7 3 65.45 \u00d7 300 2Hm Volume of air displaced = Volume of balloon = 5.316 kg = 65.45 m3 Mass of air displaced = 1.013 10 \u00d7 \u00d7 5 65.45 \u00d7 + (17 273) R air Rair = 0.287 kJ/kg. K mair = 1.013 10 \u00d7 5 \u00d7 0.287 10 \u00d7 3 65.45 \u00d7 290 mair = 79.66 kg Load lifting capacity due to buoyant force = mair \u2013 2Hm = 79.66 \u2013 5.316 = 74.344 kg Ans. 21. A pump draws air from large air vessel of 20 m3 at the rate of 0.25 m3/min. If air is initially at atmospheric pressure and temperature inside receiver remains constant then determine time required to reduce the receiver pressure to 1 4 th of its original value. Solution: Let volume of receiver be V, m3 and volume sucking rate of pump be v m3/min, then theoretically problem can be modelled using perfect gas equation. Here p is pressure in receiver and T is temperature in vessel. pV = mRT Here pressure \u2018p\u2019 and temperature", " \u2018T \u2019 shall change with respect to time t. Differentiating perfect gas equation with respect to time. V \u00b7 dp dt = RT dm dt Here \uf8eb \uf8ec \uf8ed dm dt \uf8f6 \uf8f7 \uf8f8 is mass extraction rate from receiver in kg/min. This mass rate can be given using perfect gas equation when volume flow rate (m3/min) is given as v. So. dm dt \u2212 = pv RT (\u2013ve as mass gets reduced with time) Substituting, V\u00b7 dp dt = \u2013 RT\u00b7 pv RT 32 _________________________________________________________ Applied Thermodynamics V\u00b7 dp dt t dt\u222b 0 = \u2013 pv = \u2013 tV dp \u222b ln Here final pressure, p2 = p 1 4, V = 20 m2, v = 0.25 m3/min So time ln time = 20 0.25 ln (4) = 110.9 minutes = 110.9 minutes Ans. 22. In 5 kg mixture of gases at 1.013 bar and 300 K the various constituent gases are as follows, 80% N2, 18% O2, 2% CO2. Determine the specific heat at constant pressure, gas constant for the constituents and mixture and also molar mass of mixture taking \u03b3 = 1.4 for N2 and O2 and \u03b3 = 1.3 for CO2. Universal gas constant = 8314 J/kg \u00b7 K Solution: Gas constants for constituent gases shall be, 2NR = 8314 mol. wt. of N = 2 8314 28 2OR = 8314 mol. wt. of O = 2 8314 32 = 296.9 J/kg \u00b7 K = 259.8 J/kg \u00b7 K 2COR = 8314 mol. wt. of CO = 2 8314 44 = 188.9 J/kg. K Gas constant for mixture, Rmixture \uf8f6 \uf8f7 \uf8f8 CO 2 M \uf8f6 \uf8f7 \uf8f8. R CO 2 = (0.80 \u00d7 296.9) + (0.18 \u00d7 259.8) + (0.02 \u00d7 188.9) Rmixture = 288.06 J/kg. K Specific heat at constant pressure for constituent gases. p NC,. NR 2 = \uf8eb \uf8ec \uf8ed 1.4 0.4 \uf8f6 \uf8f7 \uf8f8 \u00d7 296.9 = 1.039 kJ/kg. K Fundamental Concepts and Definitions ____________________________________________ 33 p OC, 2 = p COC, 2 = pC, mixture = \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed \u03b3 \u2212 \u03b3 1 \u03b3 \u2212 \u03b3 1 \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8. OR 2 = \uf8eb \uf8ec \uf8ed 1.4 0.4 \uf8f6 \uf8f7 \uf8f8 \u00d7 259.8 = 0.909 kJ/kg. K. COR 2 = \uf8eb \uf8ec \uf8ed 1.3 0.3 \uf8f6 \uf8f7 \uf8f8 \u00d7 188.9 = 0.819 kJ/kg. CO 2 M \uf8f6 \uf8f7 \uf8f8. C P CO, 2 pC, mixture = (0.80 \u00d7 1.039) + (0.18 \u00d7 0.909) + (0.02 \u00d7 0.819) = 1.0276 kJ/kg. K Molar mass of mixture = \u2211 xi. Mi = Total mass of mixture Total no. of moles xi = n i n\u2211, here Mi = mol. wt. of ith constituent. i No. of moles of constituent gases, 2Nn = Nm 2 Mol. wt. \u00d7 0.8 5 28 = N 2 = 0.143 2On = Om 2 Mol. wt. = O 2 \u00d7 0.18 5 32 = 0.028 2COn = COm 2 Mol. wt. CO 2 = \u00d7 0.02 5 44 = 0.0023 Total mole numbers in mixture = 2COn = (0.143 + 0.028 + 0.0023) 2Nn 2On + + Mole fractions of constituent gases, \u2211ni = 0.1733 2Nx = 2Ox =.143 0.1733 = 0.028 0.1733 = 0.825 = 0.162 n CO 2 \u2211 n i Molecular wt. of mixture = Molar mass of mixture = \u2211xi. Mi 0.0023 0.1733 = 0.0133 2COx = = = (0.825 \u00d7 28) + (0", ".162 \u00d7 32) + (0.0133 \u00d7 44) = 28.87 kg/kmol Ans. 34 _________________________________________________________ Applied Thermodynamics 23. A gas mixture comprises of 18% O2, 75% N2 and 7% CO2 by volume at 0.5 MPa and 107\u00baC. For 5 kg mass of mixture carry out gravimetric analysis and determine the partial pressure of gases in mixture. Solution: Mole fraction of constituents \u21d2 xi = n i n = V i V where \u2018ni\u2019 and \u2018Vi\u2019 are no. of moles and volume fraction of constituent while \u2018n\u2019 and V are total no. of moles and total volume of mixture. 2Ox = 2Nx = 2COx = 0.18 1 0.75 1 0.07 1 = 0.18 = 0.75 = 0.07 Molecular weight of mixture = Molar mass = (0.18 \u00d7 32) + (0.75 \u00d728) + (0.07 \u00d7 44) = 29.84 Gravimetric analysis refers to the mass fraction analysis. Mass fraction of constituents = im m = iV V \u00d7 ( / ) Mol. wt. of constituent Mol. wt. of mixture Mole fraction of O2 = \u00d7 0.18 32 29.84 = 0.193 Mole fraction of N2 = \u00d7 0.75 28 29.84 = 0.704 \uf8fc \uf8f4 \uf8f4 \uf8fd Ans. \uf8f4 \uf8f4\uf8fe \u00d7 0.07 44 29.84 Mole fraction of CO2 = Partial pressures of constituents = Volume fraction \u00d7 Pressure of mixture Partial pressure of O2 = 0.18 \u00d7 0.5 = 0.09 MPa Partial pressure of N2 = 0.75 \u00d7 0.5 = 0.375 MPa = 0.104 Ans. \uf8fc \uf8fd \uf8fe Partial pressure of CO2 = 0.07 \u00d7 0.5 = 0.35 MPa 24. A steel insulated tank of 6 m3 volume is equally divided into two chambers using a partition. The two portions of tank contain N2 gas at 800 kPa and 480 K and CO2 gas at 400 kPa and 390 K. Determine the equilibrium temperature and pressure of mixture after removing the partition. Use \u03b3 = 1.4 for N2, \u03b3 = 1.3 for CO2. Solution: Since tank is insulated so adiabatic mixing can be considered. Let us consider N2 and CO2 to behave as perfect gas. No. of moles of N2 Fundamental Concepts and Definitions ____________________________________________ 35 2Nn = No of moles of CO2 \u00d7 \u00d7 800 10 3 \u00d7 8314 480 = 0.601 p CO 2. R T Total no. of moles of mixture, 2COn =. V CO 2 CO 2 = 3 \u00d7 \u00d7 400 10 3 \u00d7 8314 390 = 0.370 n = 2 CO 2 n N n+ = 0.601 + 0.370 = 0.971 Specific heat for N2 at constant volume, v NC, 2 = R N 2 \u2212\u03b3 ( N 2 (8314 / 28) (1.4 1)\u2212 = 1) v NC = 742.32 J/kg. K, 2 Specific heat for CO2 at constant volume, v COC, 2 = R \u03b3 ( CO 2 CO 2 \u2212 (8314 / 44) (1.3 1)\u2212 = 1) v COC, 2 = 629.85 J/kg. K Mass of N2 = 2Nn \u00d7 Mol. wt. of N2 = 0.601 \u00d7 28 = 16.828 kg Mass of CO2 Let us consider the equilibrium temperature of mixture after adiabatic mixing at T. Applying \u00d7 Mol. wt. of CO2 = 0.370 \u00d7 44 = 16.28 kg. = CO2n energy conservation principle : T T\u2212 ( T T\u2212 ( ).. N2, N2 vC vC + CO2m N2m. {16.828 \u00d7 742.32 (T \u2013 480)} + {16.28 \u00d7 629.85 (T \u2013 390)} = 0 22745.7. T = 9995088.881 Equilibrium temperature, T = 439.4 K Ans. Equilibrium pressure of mixture, Tmixture = 439.4 K, Vmixture = 6 m3, CO2 = 0 CO2. ) pmixture = mixture.n R T. V mixture = \u00d7 0.971 8314 439.4 6 \u00d7 Equilibrium pressure = 591.205 kPa Ans. 25. 2 kg of Hydrogen and 3 kg of Helium are mixed together in an insulated container at atmospheric pressure and 100 K temperature. Determine the specific heat of final mixture if specific heat at constant pressure is 11", ".23 kJ/kg. K and 5.193 kJ/kg. K for H2 and He respectively. Solution: Two gases are non reacting, therefore, specific heat of final mixture can be obtained by following 36 _________________________________________________________ Applied Thermodynamics for adiabatic mixing. Cp, mixture = He + c + p He, ) He m Substituting values, = \u00d7 (2 11.23) \u00d7 (3 5.193) + + (3 2) Cp, mixture = 7.608 kJ/kg. K Ans. 26. A mixture of 18 kg hydrogen, 10 kg nitrogen and 2 kg of carbon dioxide is put into a vessel at atmospheric conditions. Determine the capacity of vessel and the pressure in vessel if it is heated upto twice of initial temperature. Take ambient temperature as 27\u00baC. Solution: Gas constant for mixture can be obtained as;. 2 R CO 2 ) + m m CO 2 CO ) ( m H 2. Rmixture = + 2HR = 2NR = 2COR = 8.314 2 8.314 28 8.314 44 kJ/kg. K kJ/kg. K kJ/kg. K 2HR 2NR 2COR = 4.15 kJ/kg. K = 0.297 kJ/kg. K = 0.189 kJ/kg. K Rmixture = + \u00d7 (18 4.15 10 0.297 2 0.189) 30 + \u00d7 \u00d7 Rmixture = 2.606 kJ/kg. K Considering mixture to be perfect gas; Capacity of vessel Vmixture = m mixture. R mixture. T p Here, p = 101.325 kPa \u00d7 \u00d7 30 2.606 300.15 101.325 Vmixture = Capacity of vessel = 231.58 m3 Ans. For constant volume heating, final pressure shall be, Fundamental Concepts and Definitions ____________________________________________ 37 final pfinal = pinitial \u00d7 T T initial pfinal = 101.325 \u00d7 2 = 202.65 kPa Ans. 27. Determine the ratio of exit to inlet diameter of a duct having heating element in the mid of duct. Atmospheric air enters the duct at 27\u00baC and gets heated up to 500 K before leaving the duct. The kinetic and potential energy changes during passage of air through duct are negligible. Solution: Said air heating process is the case of constant pressure process. Let inlet state be \u20181\u2019 and exit state \u20182\u2019. Therefore, by Charle\u2019s law volume and temperature can be related as \u00d7 \u00d7 \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8 Velocity at 2 Velocity at 1 = T 2 T 1 \u2206K.E = 0, so or Since or Exit to inlet diameter ratio = 500 300.15 = 1.29 = 1.29 Ans. 28. A vessel of 2 m3 volume contains hydrogen at atmospheric pressure and 27\u00baC temperature. An evacuating pump is connected to vessel and the evacuation process is continued till its pressure becomes 70 cm of Hg vacuum. Estimate the mass of hydrogen pumped out. Also determine the final pressure in vessel if cooling is carried up to 10\u00baC. Take atmospheric pressure as 76 cm of Hg and universal gas constant as 8.314 kJ/kg. K Solution: For hydrogen, gas constant, R = 8.314 2 R = 4.157 kJ/kg. K Say initial and final states are given by \u20181\u2019 and \u20182\u2019. Mass of hydrogen pumped out shall be difference of initial and final mass inside vessel. Final pressure of hydrogen = Atm. pr. \u2013 Vacuum pr. 38 _________________________________________________________ Applied Thermodynamics Therefore, pressure difference = 76 \u2013 6 = 76 \u2013 70 = 6 cm of Hg. = 70 cm of Hg. = 70 76 \u00d7 101.325 kPa = 93.33 kPa Mass pumped out = \u2212 p V 1 1 RT 1 p V 2 2 RT 2 ; here V1 = V2 = V and T1 = T2 = T. = V RT \u2212 ( 93.33 10 \u00d7 4.157 300.15 10 \u00d7 3 During cooling upto 10\u00baC, the process may be considered as constant volume process. Say the state before and after cooling are denoted by suffix 2 and 3. = 0.15 kg. Ans. Therefore, p3 = T 3 T 2. p 2 283.15 300.15 \u00d7 \u00d7 6 101.325 76 = Final pressure after cooling = 7.546 kPa. Ans. 1.1 Define thermodynamics and discuss different approaches to study of thermodynamics. 1.2 Write short notes on the following: Thermodynamic properties, state, path, process, closed system, isolated system, open system, extensive and intensive properties. 1.3 What is meant by quasi-static process? Also discuss its physical significance. 1", ".4 Describe thermodynamic equilibrium of a system. 1.5 State thermodynamic definition of work. Also differentiate between heat and work. 1.6 What is energy? What are different forms of it? 1.7 Explain the concept of continuum. 1.8 Define perfect gas. 1.9 Differentiate between characteristic gas constant and universal gas constant. 1.10 State the Dalton's law of partial pressures and assumptions for it. 1.11 What is meant by real gas? Why ideal equation of state cannot be used for it? 1.12 Write equations of state for real gas. 1.13 Define conpressibility factor. Fundamental Concepts and Definitions ____________________________________________ 39 1.14 Write Boyle\u2019s law and Charle's law. 1.15 Determine the absolute pressure of gas in a tank if the pressure gauge mounted on tank reads 120 kPa [221.3 kPa] pressure. 1.16 What shall be the volume of a fluid having its specific gravity as 0.0006 and mass as 10 kg? [16.67 m3] 1.17 Determine the pressure of compressed air in an air vessel, if the manometer mounted on it shows a pressure of 3 m of mercury. Assume density of mercury to be 13.6 \u00d7 103 kg/m3 and atmospheric [501.25 kPa] pressure as 101 kPa. 1.18 Calculate the kinetic energy of a satellite revolving around the earth with a speed of 1 km/s. Assume [254.8 MJ] 1.19 If the gauge pressure of oil in a tube is 6.275 kPa and oil\u2019s specific gravity is 0.8, then determine depth [80 cm] 1.20 Determine the work required for displacing a block by 50 m and a force of 5 kN. [250 kJ] 1.21 Determine the barometer reading in millimetres of Hg if the vacuum measured on a condenser is 74.5 [760 mm] acceleration due to gravity as 9.91 m/s2 and gravitational force of 5 kN. of oil inside tube. cm of Hg and absolute pressure is 2.262 kPa. 1.22 Determine the absolute pressures for the following; (i) Gauge pressure of 1.4 MPa (ii) Vacuum pressure of 94.7 kPa Take barometric pressure as 77.2 cm of Hg and density of mercury as 13.6 \u00d7 103 kg/m3. m from its unstretched length of 0.05 m. as perfect gas. Also determine the mass using compressibility charts. barometric pressure as 101 kPa and specific gravity of sea water as 1.025. [1.5 MPa, 8.3 kPa] 1.23 Determine the pressure acting upon surface of a vessel at 200 m deep from surface of sea. Take [2.11 MPa] 1.24 A vacuum gauge gives pressure in a vessel as 0.1 bar, vacuum. Find absolute pressure within vessel in bars. Take atmospheric pressure as 76 cm of mercury column, g = 9.8 m/s2, density of mercury = 13.6 g/cm3. [0.91 bar] 1.25 Determine the work done upon a spring having spring constant of 50 kN/m. Spring is stretched to 0.1 [0.0625 kJ] 1.26 Determine the mass of oxygen contained in a tank of 0.042 m3 at 298 K and 1.5 \u00d7 107 Pa considering it [8.25, 8.84] 1.27 What will be specific volume of water vapour at 1 MPa and 523 K, if it behaves as ideal gas? Also [0.241 m3/kg, 0.234 m3/kg] 1.28 Calculate the pressure of CO2 gas at 27\u00baC and 0.004 m3/kg treating it as ideal gas. Also determine the [14.17 MPa, 6.9 MPa] 1.29 Determine molecular weight and gas constant for a mixture of gases having 65% N2, 35% CO2 by mole. [33.6 kg/k mol. 0.247 kJ/kg. K] 1.30 Considering air as a mixture of 78% N2, 22% O2 by volume determine gas constant, molecular weight, [0.2879 kJ/kg. K, 28.88 kg/K mol, 1.0106 kJ/kg. K, 0.722 kJ/kg. K] 1.31 What minimum volume of tank shall be required to store 8 kmol and 4 kmol of O2 and CO2 respectively [149.7 m3] 1.32 Two tanks A and B containing O2 and CO2 have volumes of 2 m3 and 4 m3 respectively. Tank A is at 0.6 MPa, 37\u00baC", " and tank B is at 0.1 MPa and 17\u00baC. Two tanks are connected through some pipe so as to allow for adiabatic mixing of two gases. Determine final pressure and temperature of mixture. determine the same considering generalized compressibility chart. pressure using Van der Waals equation of state. Cp and Cv for air at 25\u00baC. at 0.2 MPa, 27\u00baC? [0.266 MPa, 30.6\u00baC] 1.33 Determine the molecular weight and gas constant for some gas having CP = 1.968 kJ/kg. K, Cv = 1.507 [18.04 kg/kmol, 0.461 kJ/kg. K] kJ/kg. K. 40 _________________________________________________________ Applied Thermodynamics 2 2.1 INTRODUCTION Thermodynamics is the branch of science which deals with the energy interactions. In order to find whether energy interactions are taking place or not some measurable mathematical parameters are needed. These parameters are called thermodynamic properties. Out of number of thermodynamic properties discussed earlier the \u2018temperature\u2019 is one property. One is well familiar with the qualitative statement of the state of a system such as cold, hot, too cold, too hot etc. based on the day to day experience. The degree of hotness or coldness is relative to the state of observer. For example, let us take an iron bar. Obviously the bar shall have intial temperature equal to the room temperature. Now let us heat this metal bar. Observations at the molecular level show that upon heating the molecular activity inside the bar gets increased. This may be attributed to the more agitated state of molecules as energy is given to them in the form of heating of the bar. From the physiological sensations it can be felt that this has resulted in increase in the degree of hotness of the bar. This qualitative indication of the relative hotness can be exactly defined by using thermodynamic property known as temperature. If this hot bar is brought in contact with another bar at room temperature one can feel that after some time the two bars which were initially at high and low temperatures attain the same temperature which is lying between the two temperatures. It is indicative of the fact that there has been exchange of some entity between two bars resulting in the attainment of final equilibrium temperature. This state of attainment of common equilibrium temperature is also termed as the state of thermal equilibrium. Thus, the temperature becomes a potential indicator of the energy interactions in the systems. A look at the history shows that for quantitative estimation of temperature a German instrument maker Mr. Gabriel Daniel Fahrenheit (1686-1736) came up with idea of instrument like thermometer and developed mercury in glass thermometer. Here he observed that height of mercury column used to change as the bulb of thermometer was brought in the environments having different degrees of hotness. In the year 1742, a Swedish astronomer Mr. Anders Celsius described a scale for temperature measurement. This scale later on became very popular and is known as Centigrade Scale. For caliberation of these measuring instruments some reference states of different substances were used initially and the relative state of temperature of the substance could be quantified. Later on with the passage of time things were standardised and internationally acceptable temperature scales and instruments were developed. 2.2 PRINCIPLE OF TEMPERATURE MEASUREMENT AND ZEROTH LAW OF THERMODYNAMICS After the identification of \u2018Temperature\u2019 as a thermodynamic property for quantification of the energy interactions the big question was its estimation. Based on the relative degree of coldness/hotness concept it was concluded that the absolute value of temperature is difficult to be described. Hence it was mooted Zeroth Law of Thermodynamics ___________________________________________________ 41 to make temperature estimations in reference to certain widely acceptable known thermal states of the substances. Temperature is thus the intensive parameter and requires reference states. These acceptable known thermal states are such as the boiling point of water commonly called steam point, freezing point of water commonly called ice point etc. These easily reproducible and universally acceptable states of the substance are known as reference states and the temperature values assigned to them are called reference temperatures. Since these reference points and reference temperatures maintain their constant value, therefore these are also called fixed points and fixed temperatures respectively. A list of these fixed points is given in Table 2.1. Table 2.1 Some fixed points used for International Practical Temperature Scale Sl. No. Reference State Temperature \u00b0C 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. Ice point Steam point Triple point of water Triple point of hydrogen Triple point of oxygen Oxygen point (normal boiling point) Silver point (normal freezing point) Gold point (normal freezing point) Zinc point (normal freezing point) Neon point (normal boiling point) Sulphur point (normal boiling point) 0 100 0.010 \u2013259.34 \u2013218.79", " \u2013182.96 961.93 1064.43 419.58 \u2013246.05 444.60 The methodology adopted was to first develop a temperature measurement system which could show some change in its characteristics (property) due to heat interactions taking place with it. Such systems are called thermometers, the characteristics of property which shows change in its value is termed thermometric property and the substance which shows change in its thermometric property is called thermometric substance. Science that deals with the temperature and its measurement is called thermometry. For example in case of clinical thermometer the mercury in glass is the thermometric substance and since there is change in length of mercury column due to the heat interactions taking place between the thermometer and the body whose temperature is to be measured, therefore the length is the thermometric property. Thus, the underlying principle of temperature measurement is to bring the thermometer in thermal equilibrium with the body whose temperature is to be measured, i.e. when there is no heat interaction or the state when two (thermometer and body) attain same temperature. In this process it is to be noted that thermometer is already caliberated using some standard reference points by bringing thermometer in thermal equilibrium with reference states of the substance. Zeroth law of thermodynamics states that if the bodies A and B are in thermal equilibrium with a third body C separately then the two bodies A and B shall also be in thermal equilibrium with each other. This is the principle of temperature measurement. Block diagram shown in Fig. 2.1a and 2.1b show the zeroth law of thermodynamics and its application for temperature measurement respectively. Thermal equilibrium by Zeroth law Body A Body B Thermal equilibrium Thermal equilibrium Body C Fig. 2.1a Zeroth law of thermodynamics 42 _________________________________________________________ Applied Thermodynamics Body whose temperature is to be measured Thermal equilibrium (for temperature measurement) Reference body and states Thermometers Thermal equilibrium (for caliberation) Fig. 2.1b Application of Zeroth law for temperature measurement 2.3 TEMPERATURE SCALES Number of temperature measuring scales came up from time to time. The text ahead gives a brief idea of the different temperature scales used in thermometry. Different temperature scales have different names based on the names of persons who originated them and have different numerical values assigned to the reference states. (a) Celsius Scale or Centigrade Scale Anders Celsius gave this Celsius or Centigrade scale using ice point of 0\u00b0C as the lower fixed point and steam point of 100\u00baC as upper fixed point for developing the scale. It is denoted by letter C. Ice point refers to the temperature at which freezing of water takes place at standard atmospheric pressure. Steam point refers to the temperature of water at which its vaporization takes place at standard atmospheric pressure. The interval between the two fixed points was equally divided into 100 equal parts and each part represented 1\u00baC or 1 degree celsius. (b) Fahrenheit Scale Fahrenheit gave another temperature scale known as Fahrenheit scale and has the lower fixed point as 32 F and the upper fixed point as 212 F. The interval between these two is equally divided into 180 part. It is denoted by letter F. Each part represents 1 F. (c) Rankine Scale Rankine scale was developed by William John MacQuorn Rankine, a Scottish engineer. It is denoted by letter R. It is related to Fahrenheit scale as given below. TR = TF + 459.67 (d) Kelvin Scale Kelvin scale proposed by Lord Kelvin is very commonly used in thermodynamic analysis. It also defines the absolute zero temperature. Zero degree Kelvin or absolute zero temperature is taken as \u2013273.15\u00baC. It is denoted by letter K. Zeroth Law of Thermodynamics ___________________________________________________ 43. 0\u00baC 0 0 1 \u00baR. 7 6 1 7 6 \u00baF 2 1 2 Steam point Triple point of water Ice point Absolute zero i s u s e C l Detailed discussion on Kelvin scale has been done in chapter 4 along with absolute thermodynamic temperature scale. Mathematically, it is related to the different temperature scales as follows, Fig. 2.2 Different temperature scales CT 100 KT 100 = = 32 FT \u2212 180 = KT \u2212 273.15 100 = RT \u2212 491.67 180 RT 180 2.4 TEMPERATURE MEASUREMENT For measurement of temperature number of thermometers are available using different thermometric properties of the thermometric substances. Length, volume, pressure, resistance, e.m.f. etc. are the commonly used thermometric properties for thermometers. Different thermometers developed using these thermometric properties are given below. (a) Liquid Thermometer Liquid thermometers are those thermometers that employ liquids as the thermometric substance and the change in volume of liquid with heat interaction is the characteristics used for temperature measurement. Commonly used liquids in such thermometers are Mercury and Alcohol. Fig. 2.3 shows the mercury in glass thermometer. In this the change in volume of the", " mercury results in the rise or fall in the level of mercury column in the glass tube. Out of the two liquids mercury is preferred over alcohol as it has low specific heat and hence absorbs little heat from body. Mercury is comparatively a good conductor of heat. Mercury can be seen in a fine capillary tube conveniently. Mercury does not wet the wall of the tube. Mercury has a uniform coefficient of expansion over a wide range of temperature and remains liquid over a large range as its freezing and boiling points are \u201339\u00baC and 357\u00b0C respectively. Thick glass wall Thin glass wall Bulb of large volume having mercury Capillary of small volume Fig. 2.3 Mercury in glass thermometer 44 _________________________________________________________ Applied Thermodynamics (b) Gas Thermometers Thermometers using gaseous thermometric substance are called gas thermometers. Gas thermometers are advantageous over the liquid thermometers as the coefficient of expansion of gases is more compared to liquids therefore these are more sensitive. Also thermal capacity of a gas is low compared to liquid so even a small change can also be recorded accurately. Gas thermometers are not suitable for routine work as they are large, cumbersome and can be used only in certain fixed conditions. These are used mainly for calibration and standardization purpose. Main types of gas thermometers are discussed ahead. (i) Constant volume gas thermometer : Fig. 2.4 shows a typical constant volume gas thermometer having a glass bulb \u2018B\u2019 connected to glass tube. Other end of glass tube is connected to mercury reservoir through a rubber tube. There is a fixed marking \u2018M\u2019 over the glass tube. Difference in levels of mercury in reservoir with reference to mark \u2018M\u2019 is seen on the scale. Bulb \u2018B\u2019 is generally filled with 1/7th of its volume by mercury so as to compensate for expansion of bulb \u2018B\u2019. This is done so as to keep volume of air in bulb upto the fixed mark \u2018M\u2019. P h M Tubing B Bulb Fig. 2.4 Constant volume gas thermometer Initially the bulb \u2018B\u2019 is kept in melting ice and reservoir level is suitably adjusted so that mercury level is at mark \u2018M\u2019. Corresponding to this difference in level of reservoir and mark \u2018M\u2019 of hi height, the ice point pressure shall be, Pi = P + (hi \u00b7 \u03c1\u00b7 g) = P0 Bulb is kept at the boiling water (steam point) and again the reservoir is adjusted so as to keep mercury at the fixed mark. For difference in mercury levels between mark \u2018M\u2019 and reservoir level being hs the pressure corresponding to steam point shall be Ps = P + (hs \u00b7 \u03c1 \u00b7 g) = P100 Now for the bulb \u2018B\u2019 kept in the bath whose temperature is to be measured, again the reservoir is to be adjusted so as to keep mercury level at mark \u2018M\u2019. At this state if the difference in mercury levels is ht, then the pressure shall be, For a fixed volume, the pressure variation with respect to temperature can be given as, Pt = P + (ht \u00b7 \u03c1 \u00b7 g) P = P0 (1 + \u03b1 \u00b7 t) Similarly making appropriate substitutions one can give the temperature t as follows \u00d7 ) 100 h i \u2212 ) h i (ii) Constant pressure gas thermometer : These thermometers are based on the principle that, pressure remaining constant the volume of a given mass of gas is directly proportional to its absolute temperature. Fig. 2.5 shows a constant pressure gas thermometer having a silica bulb \u2018B\u2019 connected to Zeroth Law of Thermodynamics ___________________________________________________ 45 the reservoir \u2018R\u2019 containing mercury through a connecting tube \u2018A\u2019, compensating bulb \u2018C\u2019 having a compensating tube with volume equal to the connecting tube. Manometer tube contains sulphuric acid. A Silica bulb B C R Reservoir Manometer Compensating bulb Fig. 2.5 Constant pressure gas thermometer Initially the reservoir is filled with mercury upto zero marking and the stop cock is closed. The bulbs \u2018B\u2019 \u2018R\u2019 and \u2018C\u2019 are immersed in melting ice. Tubes are sealed when the pressure on the two sides as shown by manometer is the same i.e. the pressure in silica bulb \u2018B\u2019 and compensating bulb \u2018C\u2019 are same. When the pressure on two sides of the manometer containing sulphuric acid is same the acid level in two limbs shall be same. This way the pressure of gas and air can be maintained same. Now let us assume the silica bulb \u2018B\u2019 to have definite number of molecules of air. Also, the compensating bulb and compensating tube contain the same number of molecules of air. If the silica bulb is immersed in the", " environment whose temperature is to be measured and compensating bulb being kept in melting ice. Both connecting tube and compensating tubes are at the room temperature and the air in silica bulb attains temperature equal to the temperature to be measured. (c) Electrical resistance thermometer Electrical resistance thermometer first developed by Siemen in 1871, also known as \u2018Platinum Resistance Thermometer\u2019 works on the principle of change in resistance of the thermometric substance (platinum) with temperature. Thus resistance is the thermometric property used in these thermometers. It consists of a pure platinum wire wound in a double spiral on a mica plate. Two ends of the platinum wire are connected to the copper leads (for low temperatures) or platinum leads (for high temperatures). Principle of Wheatstone bridge is employed in these thermometers, as shown in Fig. 2.6. It has a set of compensating leads having exactly similar resistance as leads used. Platinum wire and the compensating leads are enclosed in a sealed glazed porcelain tube having binding terminals at the top. The resistance of wire can be mathematically related as Galvano meter Resistance (platinum) Fig. 2.6 Electrical resistance thermometer, (principle of wheatstone bridge) where a and b are the constants having their values depending upon the nature of material used. Rt = R0 \u00b7 (1 + a \u00b7 t + b \u00b7 t2) 46 _________________________________________________________ Applied Thermodynamics Using fixed points of ice point and steam point the temperature can be mathematically obtained by substituting the different parameters in the following \u2212 \u00d7 ) 100 ) R i, where Ri & Rs are resistance values for ice and steam points (d) Thermoelectric Thermometer Thermo electric thermometer works on the principle of Seebeck effect. Seebeck effect says that a current flows or e.m.f. is produced in a circuit of two dissimilar metals having one junction as hot while other as cold junction. Current produced in this way is called thermo electric current while the e.m.f. produced is called thermo e.m.f. Measurement of temperature is being done by knowing the e.m.f. produced which is the thermometric property here. is In such type of thermometer a sensitive connected with galvanometer thermocouple as shown in Fig. 2.7. One junction is kept at ice point and other in oil bath having any temperature. Upon heating the oil bath it is seen that the thermal e.m.f. is produced by Seebeck effect. Temperature of the oil bath is measured by some calibrated thermometer of any other type. Further the temperature of oil bath is changed to known temperatures and for different temperatures the e.m.f. is noted and a graph is plotted between the temperature of bath and e.m.f. Now for using this thermocouple the cold junction shall still be maintained at the ice point while the hot junction may be kept in contact with the bath whose temperature is to be measured. To get the temperature depending upon the e.m.f. available the caliberated graph is used and corresponding temperature noted from there. In these thermometers the potentiometers may also be used as shown in Fig. 2.8. Here also the one junction is maintained at ice point while the other junction is put at the temperature to be measured. The potentiometer wire is directly calibrated to measure temperature. Galvanometer Cu Fe Oil bath Cu Ice Fig. 2.7 Thermoelectric thermometer using a galvanometer Battery Rh Galvanometer Ice cold Standard cell Hot Fig. 2.8 Thermoelectric thermometer using a potentiometer Here the length of the potentiometer wire at which the balance point is obtained is used for getting temperature. 1. Determine the human body temperature in degree celsius (\u00b0C) if the temperature in Fahrenheit is 98.6\u00b0F. EXAMPLES Zeroth Law of Thermodynamics ___________________________________________________ 47 Solution: Degree Celsius and Fahrenheit are related as below, T (\u00b0C) = T F \u00ba ( \u2212 ) 32 1.8 Substituting values. T (\u00b0C) = \u2212 98.6 32 1.8 = 37\u00b0C Temperature in degree celsius shall be 37\u00b0C. Ans. 2. A temperature scale is being developed using the following relation. t = a \u00b7 ln(p) + b\uf8eb \uf8ec \uf8ed 2 \uf8f6 \uf8f7 \uf8f8 where \u2018p\u2019 is thermometric property and \u2018a\u2019 and \u2018b\u2019 are constants. Determine celsius temperature corresponding to thermometric property of 6.5, if ice point and steam point give thermometric property value of 3 and 8. Solution: For Ice point; t = 0\u00b0C and p = 3 For Steam point; t = 100\u00b0C and p = 8 Using thermometric relation, 0 = a l", "n(3) + b\uf8eb \uf8ec \uf8ed 2 \uf8f6 \uf8f7 \uf8f8 100 = a ln(8) + b 2 Solving the above two equations, we get a = 101.95 b = 224 Thus, t = 101.95. ln(p) + \uf8eb \uf8ec \uf8ed 224 2 \uf8f6 \uf8f7 \uf8f8 For t = 101.95 ln(p) + 112 p = 6.5, t = 302.83\u00b0C Ans. 3. In a thermoelectric thermometer for t\u00b0C temperature, the emf is given as; E = 0.003 \u00b7 t \u2013 5 \u00d7 10\u20137 \u00b7 t2 + 0.5 \u00d7 10\u20133, volts Thermometer is having reference junction at ice point and is calibrated at ice point and steam points. What temperature shall be shown by the thermometer for a substance at 30\u00b0C? Solution: At ice point; t = 0\u00b0C, E0 = 0.5 \u00d7 10\u20133, volts At steam point, t = 100\u00b0C, E100 = 0.0265, volts t = 30\u00b0C When Thus temperature shown by this thermometer; E30 = 9.14 \u00d7 10\u20133 volts \uf8eb \uf8ec \uf8ed t = E E 30 100 \u2212 \u2212 E 0 E 0 \uf8f6 \uf8f7 \uf8f8 \u00d7 (T100 \u2013 T0) 48 _________________________________________________________ Applied Thermodynamics \uf8eb \uf8ec \uf8ec \uf8ed = \u00d7 9.14 10 \u2212 3 \u2212 \u00d7 0.5 10 \u2212 3 \u2212 3 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u00d7 100 \u2212 0.0265 0.5 10 \u00d7 = 33.23\u00b0C Ans. 4. Estimate the % variation in temperature reading from a thermocouple having its test junction in gas and other reference junction at ice point. The temperature of gas using gas thermometer is found 50\u00b0C. Thermocouple is caliberated with emf varying linearly between ice point and steam point. When thermocouple\u2019s test junction is kept in gas at t\u00b0C and reference junction at ice point, the e.m.f. produced in millivolts is, e = 0.18 \u00b7 t \u2013 5.2 \u00d7 10\u20134 \u00d7 t2, millivolts. Solution: As ice point and steam points are two reference points, so at ice point having t = 0\u00b0C, e.m.f. = 0 at steam point having t = 100\u00b0C, e.m.f. = 12.8 mV at gas temperature of 50\u00b0C, e.m.f. = 7.7 mV Since e.m.f. variation is linear so, temperature at e.m.f. of 7.7 mV; \u00d7 \u2212 (100 0) 7.7 (12.8 0) \u2212 = = 60.16\u00b0C Temperature of gas using thermocouple = 60.16\u00b0C % variation in temperature reading with respect to gas thermometer reading of 50\u00b0C. \u2212 = 60.16 50 50 = 20.32% Ans. \u00d7 100 5. In an unknown temperature scale freezing point of water is 0\u00b0X and boiling point of water is 1000\u00b0X. Obtain a conversion relation between degrees X and degree celsius. Also determine the absolute zero in degree X. Solution: Let the conversion relation be X = aC + b where C is temperature in degree celsius, a & b are constants and X is temperature in \u00b0X. At freezing point, temperature = 0\u00b0C, 0\u00b0X or, \u21d2 At boiling point, temperature = 100\u00b0C, 1000\u00b0 1000 = a \u00b7 100 + b \u21d2 Conversion relation a = 10 X Absolute zero temperature in \u00b0C = \u2013 273.15\u00b0C Absolute zero temperature in \u00b0X = \u2013 2731.5\u00b0X = 10. C Ans. \u2013 2731.5\u00b0X Ans. Zeroth Law of Thermodynamics ___________________________________________________ 49 2.1 State Zeroth law of thermodynamics. 2.2 Explain, how the Zeroth law of thermodynamics can be used for temperature measurement. 2.3 Write short notes on the following: Thermometry, thermometric substance, thermometric property, Constant volume gas thermometer. 2.4 Sketch and explain the working of constant pressure thermometer. 2.5 Write equivalence amongst different temperature scales. Also write brief note on each of them. 2.6 Obtain triple point of water in Fahrenheit, Rankine and Kelvin scale. 2.7 Heating of a body causes its temperature to change by 30\u00b0F. Find out the increase in temperature in \u00b0R and \u00b0C. 2.8 Temperature of an object changes by 10\u00b0C. What is the change in temperature in \u00b0R, \u00b0F. (", "\u201318\u00b0R, \u201318F) 2.9 Prove that the difference between the two temperatures in Celsius scale is same as that in Kelvin scale. 2.10 On some temperature scale 0\u00b0C is equivalent to 100\u00b0B and 100\u00b0C is equivalent to 300\u00b0B. Determine the (50\u00b0C) 2.11 During temperature measurement of a body it is seen that the same numerical reading is obtained in (419.67\u00b0R) 2.12 Write a generic computer program for conversion of temperature in \u00b0C, K, \u00b0F and \u00b0R into one another. Celsius and Fahrenheit scales. What is the temperature in degree Rankine? temperature in \u00b0C corresponding to 200\u00b0B. 50 _________________________________________________________ Applied Thermodynamics 3 3.1 INTRODUCTION Let us take water in a container and heat it from the bottom. What will happen? Container and the water inside shall start getting heated up. This heating is being sensed by either touching it or by measuring its initial and final temperatures. What has caused it to happen so? Let us take bicycle wheel and paddle it very fast, Chained wheel starts rotating very fast. Now let us apply the brake. The wheel gets stopped. What has made it to stop? Also, if we touch the brake shoe and observe its temperature it shall show that the brake shoe has got heated up. Why has it happened so? Answer for the above question lies in the energy interactions. The heating up of the container and water has been caused by the heat being added through the burner flame. Heat available in the flame is being transferred to the container resulting into temperature rise of the container and water. The fast rotation of the bicycle wheel by paddling has been due to work done in the form of paddling and causing wheel to rotate. Subsequent application of the brake has brought wheel to rest as the driving potential with the wheel is gradually dissipated due to the friction between the brake shoe and wheel. Thus the energy of wheel gets transferred to brake shoe bringing it to rest and heating up of the brake shoe (observed by the rise in temperature of brake shoe). Thus, it is obvious that there is some entity which is responsible for the above phenomenon. This entity is called the energy which is available in different forms, some times causing the wheel rotation, heating up of water etc. Similar to the cases discussed above one may look at other real life systems and understand phenomenon occurring in them. It can be concluded that it is the energy interaction in some form which is responsible for occurrence of such phenomenon. These energy interactions only permit the transformation of energy from one form to other while making the creation and destruction of energy impossible. Hence, it is true that \u201cenergy can neither be produced nor destroyed, it can only change it\u2019s form\u201d. The prevailing law of \u201cenergy conservation\u201d also states the same thing. Here in this chapter we shall look into the first law of thermodynamics, heat, work and its definition, flow and non-flow work, their estimation, steady and unsteady flow processes, their analysis and limitations of the first law of thermodynamics. 3.2 THERMODYNAMIC PROCESSES AND CALCULATION OF WORK Thermodynamic processes can be precisely categorized as cyclic process and non-cyclic process. The cyclic process is the one in which the initial and final states are identical i.e. system returns to its initial states after occurrence of process. The non cyclic process is the one in which the initial and final states are different i.e. the occurrence of process is accompanied by the state change. Thermodynamic work and its explanation has already been given in Article 1.12. First Law of Thermodynamics _____________________________________________________ 51 Let us consider a system consisting of a tank filled with water and fitted with a stirrer at room temperature, Fig. 3.1. Work can be transferred to the system by the stirrer and the temperature of water shall rise. When stirring stops, the system shall cool down till it reaches to the room temperature. Thus, the process is cyclic as the initial and final states are identical. Let us now take a cylinder having piston and gas filled inside. If the gas is made to expand due to heating, the piston shall undergo displacement and say the piston displacement is dx. If the force exerted by gas on face of piston is F and the cross section area of piston is A, then the displacement work done may be given by : dW = F \u00b7 dx For the gas pressure being p, the force may be given by F = p \u00b7 A. Substituting for F, dW = p \u00b7 A \u00b7 dx Stirrer Tank Temperature rises upon stirring, temperature restores its original value when stirring is stopped. Fig. 3.1 Cyclic process or, dW = p \u00b7 dV, where dV is the elemental change in volume or the volumetric displacement. If the total displacement of piston is given by L then the total work", " can be had by integrating the above dW with respect to x for displacement L, or with respect to volume for volume change. W = \u222b p\u00b7dV = \u222b p \u00b7 A\u00b7dx Now, let us examine whether the work estimated above is in conformity to thermodynamic definition of work or not. If the piston displacement is transferred to a suitable link then the weight can be raised, thus it satisfies thermodynamic definition of work. What about the nature of process? cyclic or non cyclic. It is obvious that the initial and final states are not identical therefore, it is a non-cyclic process. Thus, the work W as defined above refers to thermodynamic work for a non-cyclic process. Thermodynamic processes can be further classified based on the thermodynamic constraints under which they occur. Different types of thermodynamic processes are as detailed below. (i) Constant pressure process or isobaric process: It refers to the thermodynamic process in which there is no change in pressure during the process. Such type of processes are also known as isobaric processes. To understand let us take a cylindrical vessel having gas in it. It has a piston above it. Piston is free to reciprocate in the cylinder. Under normal situation piston shall be subjected to atmospheric pressure. Now, let heat be added to cylinder from bottom of cylinder. Due to heat addition, presuming energy transfer taking place reversibly and system always remaining in equilibrium, the gas shall try to expand. Expansion of gas results in raising up of the piston and it attains a new state say 2. Process is shown on p-V diagram in Fig. 3.2. 52 _________________________________________________________ Applied Thermodynamics Piston Gas Cylinder Pressure Isobaric process 1 2 p = p r Heating V1 V2 Volume The work involved in the raising of piston shall be given by, Fig. 3.2 Isobaric process V 2 W1\u20132 = \u222b P\u00b7dV = P \u00b7 (V2 \u2013 V1) V 1 Mathematically from the first law of thermodynamics, it can be given that, dQ = dU + dW 2 \u222b dQ = 2 \u222b dU + 1 1 Q1\u20132 = m cv 2 \u222b dW 1 (T2 \u2013 T1) + P(V2 \u2013 V1) = m cv (T2 \u2013 T1) + mR(T2 \u2013 T1) Substituting for cv, i.e. R \u03b3 \u2212 cv = ( 1) Q1\u20132 = mR (T2 \u2013 T1) \uf8f1 \uf8f2 \uf8f3 1 \u2212 \u03b3 ( 1) + \uf8fc \uf8fd 1 \uf8fe (ii) Constant volume process or isochoric process: When a fluid undergoes a thermodynamic process in a fixed enclosed space such that the process occurs at constant volume, then the process is called constant volume process or isochoric process. Let us consider heating of a gas in fixed enclosure at constant volume. On p\u2013V diagram this process is represented by a vertical line as shown in Fig. 3.3. Area under the process line is zero which indicates that there is rise in pressure but there is no work done as there is no change in volume. Work involved shall be, e r u s s e r p1 p2 P 2 1 W1\u2013 dV = 0 V Volume Fig. 3.3 Isochoric process From first law of thermodynamics, First Law of Thermodynamics _____________________________________________________ 53 dQ = dU + dW 2 \u222b dQ = 2 \u222b dU + 2 \u222b dW = 2 \u222b dU + 0 1 1 1 1 or Thus, it indicates that the effect of heat addition in constant volume process is to increase the Q1\u20132 = U2 \u2013 U1 = mcv (T2 \u2013 T1) temperature and consequently the internal energy of system. (iii) Constant temperature process or isothermal process: Thermodynamic process in which the temperature remains constant is called constant temperature or isothermal process. In this case the gas or vapour may be heated at constant temperature and there shall be no change in internal energy. The work done will be equal to the amount of heat supplied, as shown ahead. For a perfect gas during isothermal process; p1V1 = p2V2 = Constant, or, P = 1 1PV V so work involved, W1\u20132 = V 2 \u222b P \u00b7 dV V 1 V 2 \u222b 1 1p V V W1\u20132 = V 1 W1\u20132 = P1V1 ln r dV = P1V1 ln where r = ratio of final and initial volumes. By first law of thermodynamics p1 p2 T = T1 2 2 V1 V2 Volume Fig. 3.4 Is", "othermal process 2 \u222b dQ = 2 \u222b dW + 2 \u222b dU 1 1 1 Q1\u20132 = W1\u20132 + (U2 \u2013 U1) = W1\u20132 + 0 U2 \u2013 U1 = mcv (T2 \u2013 T1), and T1 = T2 as (iv) Adiabatic process: An adiabatic process is the thermodynamic process in which there is no heat interaction during the process, i.e. during the process, Q = 0. In these processes the work interaction is there at the expense of internal energy. If we talk of adiabatic expansion then it shall mean that work is done at the cost of its own internal energy. The adiabatic process follows the law PV \u03b3 = constant where \u03b3 is called adiabatic index and is given by the ratio of two specific heats. Thus, it is obvious that adiabatic expansion shall be accompanied by the fall in temperature while temperature will rise during adiabatic compression. The adiabatic expansion process is shown on Fig. 3.5. Work done during expansion shall be, V 2 \u222b P \u00b7 dV, where PV \u03b3 = constant, therefore solving after substitution. Work shall be, W1\u20132 = V 1 W1\u20132 = \u2212 PV ) 54 _________________________________________________________ Applied Thermodynamics From first law of thermodynamics 2 \u222b dQ = 2 \u222b dU + 1 1 2 \u222b dW 1 Q1\u20132 = (U2 \u2013 U1) \u21d2 0 = mcv W1\u20132 = mcv (T2 \u2013 T1) + (T1 \u2013 T2) p1 1 e r u s s e r P p2 \u03b3 PV = constant 2 V1 dV V2 Volume Fig. 3.5 Adiabatic expansion (v) Polytropic process: Polytropic process is the most commonly used process in practice. In this, the thermodynamic process is said to be governed by the law PVn = constant where n is the index which can vary from \u2013 \u221e to + \u221e. Figure 3.6 shows some typical cases in which the value of n is varied and the type of process indicated for different values of n. Thus the various thermodynamics processes discussed above are special cases of polytropic process. Work interaction in case of polytropic process can be given as, W1\u20132 = V 2 \u222b p \u00b7 dV V 1 whereP1V1 Solving the above, we get n = P2V2 n = constant W1\u2013, isobaric process n = 1, isothermal process n = 2, polytropic process, isochoric process n = \u221e From first law of thermodynamics, Fig. 3.6 Polytropic process 2 \u222b dQ = 2 \u222b dU + 1 1 2 \u222b dW 1 Q1\u20132 = (U2 \u2013 U1 Q1\u20132 = mcv (T2 \u2013 T1) + mR ) ) cv = R \u03b3 \u2212 or R = cv 1 (\u03b3 \u2013 1) or or also Q1\u20132 = mcv {(T2 \u2013 T1) + \u03b3 \u2212 ( \u2212 ( n 1) 1) \u00b7 (T1 \u2013 T2)} Thus heat transfer during a polytropic process for a perfect gas; Q1\u20132 = mcv (T2 \u2013 T1) \u00d7 \u03b3 \u2212\uf8eb \uf8ec \u2212\uf8ed 1 n n \uf8f6 \uf8f7 \uf8f8 or also, substituting for cv Q1\u20132 = \uf8f6\u2212 \uf8eb n\u03b3 \uf8f7 \uf8ec \uf8f8 \u00d7 W \u2212\uf8ed \u03b3 1 First Law of Thermodynamics _____________________________________________________ 55 (vi) Hyperbolic process: Hyperbolic process is the one in which product of pressure and volume remains constant during the process. The curve for such an expansion process is a rectangular hyperbola and hence this is known as hyperbolic expansion. For a perfect gas PV T = Constant, if T is also constant then it means that for a perfect gas the hyperbolic process shall also be isothermal process. Figure 3.7 shows hyperbolic expansion process between 1 and 2. Work done during process shall be W1\u20132 = 2 1 \u222b p \u00b7 dV and p1V1 = p2V2 = constant W1\u2013 dV = p1V1 ln V 2 V 1 or, W1\u20132 = p1V1 ln r, where r = V 2 V, ratio of final and initial volumes from first law of thermodynamics, 1 2 \u222b dQ = 1 2 \u222b dU + 1 2 1 \u222b dW = (U2 \u2013 U1) + p1V1 ln \ufffd", "\ufffd\ufffd \uf8ec \uf8ed V 2 V 1 \uf8f6 \uf8f7 \uf8f8 p1 1 e r u s s e r P p2 PV = constant 2 V1 dV V2 Volume Fig. 3.7 Hyperbolic expansion (vii) Free Expansion: Free expansion, as the name implies refers to the unrestrained expansion of a gas. Let as take an insulated tank having two compartments separated by a partition, say A and B. Let us assume that compartment A is filled with gas while B is having vacuum. If now the partition is removed and gas allowed to occupy the whole volume of tank, then the gas expands to fill the complete volume space. New pressure of gas will be lesser as compared to initial pressure of gas occupying the compartment A. A close look at the expansion process shows that the expansion due to removal of partition is unresisted expansion due to gas expanding in vacuum. This is also known as free expansion. The reverse of free expansion is impossible and so it is an irreversible process. Partition of removable type Insulated tank Vacuum Gas A B Fig. 3.8 Free expansion 56 _________________________________________________________ Applied Thermodynamics During free expansion no work shall be done by the gas or on the gas due to no boundary displacement in the system. Also in the above there shall be no heat interaction as tank is insulated. From first law of Wfree expansion = 0 thermodynamics, \u2206Q = \u2206U + \u2206W O = \u2206U + O or, UA + B = UA, i.e. initial and final internal energies are same, which means for a perfect gas initial and final temperatures of gas are same. Table 3.1 Thermodynamic processes Process Sl. No. Governing equations Heat interaction 1. Isobaric p = constant process 2. Isochoric process T 2 T 1 = v 2 v 1 index n = 0 V = constant p T 1 1 T p 2 2 index, n = \u221e = q = cp \u00d7 (T2 \u2013 T1) q = cv \u00d7 (T2 \u2013 T1) 3. Isothermal T = constant process p1V1 = p2V2 index, n = 1 q = p1V1 \u00d7 ln \uf8eb \uf8ec \uf8ed V 2 V 1 \uf8f6 \uf8f7 \uf8f8 Displacement work or non flow work during state change from 1 to 2 W = 2 \u222b 1.p dV W = p(V2 \u2013 V1) W = 0 W = P1V1 ln. Adiabatic \u03b3 p V 1 1 = process \uf8eb = \uf8ec \uf8ed index, n = \u03b3 First Law of Thermodynamics _____________________________________________________ 57 5. Polytropic process n p V 1 1 n p V \u2212\uf8eb \uf8ec Cn = cv \u2212\uf8ed 1 where Cn is specific heat for polytropic process. 6. Hyperbolic pV = constant process but not necessarily T = constant Unresisted expansion 7. Free expansion in adiabatic conditions \u2212\uf8eb \u03b3 \uf8ec q = cv T2 \u2013 T1) or, q = \uf8f6\u2212 \uf8eb n\u03b3 \uf8f7 \uf8ec \u03b3 \u2212\uf8ed 1 \uf8f8 \u00d7 work q = cv (T2 \u2013 T1) + RT1 ln \uf8eb \uf8ec \uf8ed = RT1 ln \uf8eb \uf8ec \uf8ed.3 NON-FLOW WORK AND FLOW WORK Work interaction taking place in a system can be classified as flow work or non-flow work based on the nature of process. Two basic types of processes are (i) Flow process (ii) Non flow process 3.3.1 Flow Process Flow process is the one in which fluid enters the system and leaves it after work interaction, which means that such processes occur in the systems having open boundary permitting mass interaction across the system boundary. Figure 3.9 shows the flow process occurring in a system. Flow processes can be further classified into steady flow and non-steady flow processes. Examples of engineering systems having steady flow processes are flow through nozzles, turbines, compressors etc. and the examples of nonsteady flow processes are the filling or emptying of any vessel. Flow process shown indicates various energy and mass interactions taking place across the system boundary. 58 _________________________________________________________ Applied Thermodynamics Heat addition Flow in Control volume Flow out Wort Control system boundary Fig. 3.9 Flow process As the mass interaction takes place here so for every mass fraction crossing the boundary, work is done upon or by the system. Thus, a portion of work is always required to push the fluid mass into or out of the system. This amount of work is called flow work, or, \u201cwork required for causing flow of fluid to or from the system is called flow work\ufffd", "\ufffd. Here in the control volume shown say, some mass of fluid element is to be pushed into the control volume. Fluid mass can be injected into the control volume with certain force, say F. The force required for pushing (F) owing to the pressure P of fluid element may be quantified as; F = P.A, where A is cross-section area of the passage. For injecting entire mass of fluid, the force F must act through a distance L. Thus, work done in injecting the fluid element across the boundary will be, Control volume P L Virtual piston Fig. 3.10 Flow work W = F\u00b7L = P\u00b7A\u00b7L, (kJ) w = P\u00b7v, (kJ/kg) or This work is the flow work. Thus, flow work per unit mass can be given as the product of pressure and specific volume. It is also referred to as flow energy or convected energy or transport energy. 3.3.2 Non-Flow Processes Non-flow process is the one in which there is no mass interaction across the system boundaries during the occurrence of the process. Figure 3.11 shows block diagram of a piston-cylinder arrangement in which a constant mass of fluid inside the cylinder is shown to undergo compression. Thus, during compression the type of process shall be non-flow process and the work interaction shall be non-flow work. Say, the force exerted by piston is F, and cross-section area of piston being A, the elemental work done in compressing along the length dL shall be dW = F \u00b7 dL If pressure of fluid is P then F = P \u00b7 A. so The total work done in piston displacement, from 1 to 2 shall be, dW = P \u00b7 A \u00b7 dL First Law of Thermodynamics _____________________________________________________ 59 2 \u222b dW = 2 \u222b P \u00b7 A \u00b7 dL 1 1 2 \u222b dW = 2 \u222b P \u00b7 dV 1 1 W1\u20132 = 2 \u222b P \u00b7 dV 1 or dL F 1 L 2 Fig. 3.11 Non-flow process Thus, this is called the non-flow work or displacement work. 3.4 FIRST LAW OF THERMODYNAMICS Benjamin Thompson (Count Runsford) 1753-1814 discovered the equivalence of work and heat in the course of manufacturing canon (1797) by boring solid metal submerged in the water. He was intrigued by the water boiling because of mechanical work of boring, as no heat had been added to the water. In his words, \u201cis it possible that such a quantity of heat as would have caused five pounds of ice cold water to boil could have been furnished by so inconsiderable a quantity of metallic dust merely in consequence of a change in its capacity for heat?\u201d Other experiments later discovered more evidence until some fifty years after the above experiment. Let as take a bicycle, tyre pump and use it for inflating the bicycle tyre. It is observed that the pump becomes hotter during use. This phenomenon of heating of pump is obviously not from heat transfer but because of the work done. Although the heating of pump could also be realized by heat transfer. It indicates that some effects can be caused equivalently by heat or work and that there exist some relationship between heat and work. James Prescott Joule (1818-1889) an English scientist and one time student assistant to John Dalton (1766-1844) with assistance from Lord Kelvin showed conclusively that mechanical work and heat are equivalent. For example, let us take a closed system which permits work interaction and heat interaction both, as in case of stirring in a container, fig. 3.12. As a result of stirring it is seen that the temperature of water gets raised up. This rise in temperature can be accounted by quantifying the amount of heat supplied for raising this temperature. Thus, it is obvious that for any closed system undergoing a cycle \u222b W = J \u00b7 \u222b Q, where J is Joule\u2019s constant. i.e., the net heat interaction is proportional to the work interaction. Also the constant is known as \u201cJoule\u2019s mechanical equivalent of heat\u201d. Joule\u2019s constant is described as; Insulated tank W Q = J = 4.18 Thus, J is a numerical conversion factor which could be Joules Calories unity if the heat is also given in joules. For any cyclic process in the closed system the relationship between heat and work shall be, (if the consistent units are used) \u222b \u03b4q = \u222b \u03b4W. Stirrer Water Fig. 3.12 Closed system 60 _________________________________________________________ Applied Thermodynamics Thus first law of thermodynamics states that \u201cin a closed system undergoing a cyclic process, the net work done is proportional to the net heat taken from the surroundings\u201d or \u201cfor any cycle of a closed system the net heat transfer equals", " the net work\u201d. First law of thermodynamics can\u2019t be proved but it is supported by a large number of experiments and no exceptions have been observed. It is therefore termed as the law of nature. Mathematical expression for the first law of thermodynamics can be rearranged and it shall be, \u222b (\u03b4q \u2013 \u03b4W) = 0 which shows that the quantity (\u03b4q \u2013 \u03b4W) is a thermodynamic property. For non-cyclic process: Let us now take up a system undergoing a non-cyclic process where transfer of heat and work take place and there is some change in the state of system i.e. initial and final states are different. Figure 3.13 shows the non-cyclic process occurring between states 1 and 2. The change in state is accomplished by the energy interactions. If we assume the system to have the heat interaction \u2206Q and work interaction \u2206W, then from the basic principles it can be said that : Energy lost = Energy gained as the energy can neither be created nor destroyed. Therefore, between states 1\u20132 one can write energy balance as, Q1\u20132 \u2013 W1\u20132 = U1\u20132 where Q1\u20132, W1\u20132 and U1\u20132 are the heat, work and stored energy values. This stored energy is called as internal energy for a system having negligible electrical, magnetic, solid distortion and surface tension effects. General expression based on above can be given as follows : \u2206Q \u2013 \u2206W = \u2206U 1 p 2 V Fig. 3.13 Non cyclic process or 2 \u222b dQ \u2013 2 \u222b dW = 1 1 2 \u222b dU 1 or, for elemental interactions; dQ \u2013 dW = dU dQ = dU + dW Thus, the first law of thermodynamics for non-cyclic processes can be given by \u222b dQ = \u222b dU + \u222b dW Above equations make it obvious that the internal energy change in the closed system during any non-cyclic process is obtained by subtracting the net amount of work done by the system from the net amount of heat added to the system i.e. \u2206U = Q \u2013 W. Actually, there is no absolute value of internal energy of any system. Therefore its value may be taken to be zero for any particular state of the system and absolute value in reference to arbitarily assumed state may be easily defined. Mathematically, it can be shown that the internal energy is a thermodynamic property, as explained ahead. Let us consider the noncyclic process following paths A, B and C in the directions as shown in Fig. 3.14. As the processes A & B and A & C constitute a thermodynamic cycle starting and finishing at state 1, the first law of thermodynamics for cyclic process can be employed, 1 C p A B V 2 Fig. 3.14 Two different thermodynamic cycles. First Law of Thermodynamics _____________________________________________________ 61 \u222b (\u03b4Q \u2013 \u03b4W) = 0 For the cycle following path 1\u2013A\u20132\u2013B\u20131, the first law of thermodynamics says, \u222b (\u03b4Q \u2013 \u03b4W) = 0 1\u2013A\u20132\u2013B\u20131 or 2 \u222b (\u03b4Q \u2013 \u03b4W) + 1 \u222b (\u03b4Q \u2013 \u03b4W) = 0 1, A 2, B which can be rewritten as, 2 \u222b (\u03b4Q \u2013 \u03b4W) = \u2013 1 \u222b (\u03b4Q \u2013 \u03b4W) 1, A 2, B or 2 1, A \u222b (\u03b4Q \u2013 \u03b4W) = \u222b (\u03b4Q \u2013 \u03b4W) 2 1, B (i) Also, for the cycle following path 1\u2013A\u20132\u2013C\u20131, the first law of thermodynamics can be applied as, \u222b (\u03b4Q \u2013 \u03b4W) = 0 1\u2013A\u20132\u2013C\u20131 2 \u222b (\u03b4Q \u2013 \u03b4W) + 1 \u222b (\u03b4Q \u2013 \u03b4W) = 0 1, A 2, C 2 \u222b (\u03b4Q \u2013 \u03b4W) = \u2013 1 \u222b (\u03b4Q \u2013 \u03b4W) 1, A 2, C 2 \u222b (\u03b4Q \u2013 \u03b4W) = 2 \u222b (\u03b4Q \u2013 \u03b4W) 1, A 1, C or or or (ii) From equations (i) & (ii) it is obvious that 2 \u222b (\u03b4Q \u2013 \u03b4W) = 2 \u222b (\u03b4Q \u2013 \u03b4W) = 2 \u222b (\u03b4Q \u2013 \u03b4W) 1, A 1, B 1, C which shows that (\u03b4Q \u2013 \u03b4W) is some property as it is independent of the path being", " followed. Also, it can be rewritten as, 2 \u222b \u03b4U = 2 \u222b \u03b4U = 2 \u222b \u03b4U 1, A 1, B 1, C or \u2206U1\u20132, A = \u2206U1\u20132, B = \u2206U1\u20132, C which means the change in internal energy is independent of the path followed and therefore internal energy is a thermodynamic property. 62 _________________________________________________________ Applied Thermodynamics 3.5 INTERNAL ENERGY AND ENTHALPY Let us take a mass at certain elevation in earth\u2019s gravitational field and make it move with certain velocity. Energy considerations say that the mass shall have the potential energy (P.E = mgz) and kinetic energy (K.E = (1/2). mC2) stored in it. Similarly, several other forms of energy such as due to magnetic, electrical, solid distortion and surface tension effects can be estimated as the contributory components of stored energy. Difference of heat and work interactions yield the stored energy as given below; E = Q \u2013 W. If the energy at macroscopic level as discussed above could be separated from the total stored energy E, then the amount of energy left shall be called internal energy. Mathematically, Internal energy, U = (Stored energy) \u2013 (Kinetic energy) \u2013 (Potential energy) \u2013 (Magnetic energy) \u2013 (Electrical energy) \u2013 (Surface tension energy) \u2013 (Solid distortion energy). Therefore, stored energy is summation of internal energy, potential energy, kinetic energy, magnetic, electrical, surface tension, solid distortion etc. types of energy. For the situation when magnetic, electric, surface tension, solid distortion effects are negligible, the stored energy shall be; or, E = U + KE + PE E = U + 2 mC 2 + mgz or, on unit mass basis; e = u + 2 C 2 + gz and the change in stored energy relative to some reference state shall be given as, \u2206E = \u2206U + \u2206KE + \u2206PE. Enthalpy (H) of a substance at any point is quantification of energy content in it, which could be given by summation of internal energy and flow energy. Enthalpy is very useful thermodynamic property for the analysis of engineering systems. Mathematically, it is given as, H U PV + = On unit mass basis, the specific enthalpy could be given as, h = u + pv A look at expression of enthalpy shows that as we can\u2019t have absolute value of internal energy, the absolute value of enthalpy can not be obtained. Therefore only change in enthalpy of substance is considered. For certain frequently used substances such as steam, the enthalpy values of steam are available in tabulated form in Steam Tables at different thermodynamic states. From the definition of enthalpy; or For a constant pressure process, dp = 0. h = u + pv dh = du + p \u00b7 dv + v \u00b7 dp. or, dh = du + pdv dh = dqp = constt (From first law of thermodynamics) First Law of Thermodynamics _____________________________________________________ 63 3.6 SPECIFIC HEATS AND THEIR RELATION WITH INTERNAL ENERGY AND ENTHALPY Specific heats of the substance refer to the amount of heat interaction required for causing unit change in temperature of the unit mass of substance. This unit change in temperature may be realized under constant volume and constant pressure conditions separately. Therefore, the above heat value obtained with heat interaction occurring under constant volume conditions is called specific heat at constant volume, denoted as cv. Whereas the above heat value obtained with heat interaction occurring under constant pressure conditions is called specific heat at constant pressure, denoted as cp. Mathematically, the heat interaction causing \u2206T temperature change in m mass of substance can be given as, For isochoric conditions; and for isobaric conditions or Qv = m \u00b7 cv \u00b7 \u2206T Qp = m \u00b7 cp \u00b7 \u2206T vQ m T\u2206 \u00b7 cv = or, cp = For getting the specific heat values, substituting m = 1, \u2206T = 1, pQ m T\u2206 \u00b7 The specific heat at constant volume can also be given as the partial derivative of internal energy with respect to temperature at constant volume. c Q= v v and c p Q= p Thus or, \u2202\uf8eb \uf8f6 u \uf8f7\u2202\uf8ed \uf8ec \uf8f8 T v \uf8eb \uf8ec \uf8ed du dT \uf8f6 \uf8f7 \uf8f8v cv = cv = Also from first law of thermodynamics, on unit mass basis at constant volume, dv = 0 or dq = du + pdv dq = du dq =", " cv \u00b7 dT = du = c v dq dT, for v = constant Specific heat at constant pressure can be given as the partial derivative of enthalpy with respect to temperature at constant pressure. Mathematically: or \u2202\uf8eb h \uf8ec \u2202\uf8ed T \uf8eb \uf8ec \uf8ed dh dT \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8 p p cp = cp = 64 _________________________________________________________ Applied Thermodynamics From definition of enthalpy, at unit mass basis. or at constant pressure, dp = 0 h = u + pv dh = du + pdv + vdp dh = du + pdv substituting from first law of thermodynamics dq = du + pdv or dh = dq dq = cp \u00b7 dT = dh = c p dq dT, for p = constant Let us try to establish relationship between cp and cv. From enthalpy definition, at unit mass basis h = u + pv h = u + RT {for ideal gas} or Taking partial derivative, dh = du + RdT Also we know for an ideal gas, cp dh = cp \u00b7 dT; du = cv \u00b7 dT Substituting dh and du or or cp \u00b7 dT = cv cp = cv \u2212 pc \u00b7 dT + R \u00b7 dT + R = R c v Difference of specific heats at constant pressure and volume is equal to the gas constant for an ideal gas. Also the ratio of specific heats at constant pressure and volume could be given as \u03b3, = \u03b3 pc cv Combining above two relations of cp and cv we get, R = ( \u20131) \u03b3 R\u03b3. \u03b3 ( \u20131) and cv cp = 3.7 FIRST LAW OF THERMODYNAMICS APPLIED TO OPEN SYSTEMS Let us consider an open system as shown in Fig. 3.15 having inlet at section 1\u20131 and outlet at section 2\u20132. The cross-section area, pressure, specific volume, mass flow rate, energy at section 1\u20131 and 2\u20132 are Inlet Section 1\u20131 = A1, p1, v1, m1, e1 Section 2\u20132 = A2, p2, v2, m2, e2 Control boundary Q 1 1 Open system Fig. 3.15 Outlet 2 2 W First Law of Thermodynamics _____________________________________________________ 65 Open system is also having heat and work interactions Q, W as shown in figure above. Applying the energy balance at the two sections, it can be given as, Energy added to the system + Stored energy of the fluid at inlet = Stored energy of the fluid at outlet Quantifying the various energies; Energy of fluid at inlet shall comprise of stored energy and flow energy as given here. Similarly, energy of fluid at outlet shall comprise of stored energy and flow energy, = m1(e1 + p1v1) The energy added to the system shall be the net energy interaction due to heat and work interactions. = Stored energy + Flow energy = m2 (e2 + p2v2) Writing energy balance, mathematically; = Q \u2013 W Q \u2013 W + m1 (e1 + p1v1) = m2 (e2 + p2v2) or If the mass flow rates at inlet and exit are same, then Q + m1(e1 + p1v1) = W + m2(e2 + p2v2) On unit mass basis Q + m(e1 + p1v1) = W + m(e2 + p2v2) q + e1 + p1v1 = w + e2 + p2v2 Thus, Heat + (Stored energy + Flow energy)1 = Work + (Stored energy + Flow energy)2 Stored energy at inlet and outlet can be mathematically given as, e1 = u1 + 2 C 1 2 + gz1 and 2 C 2 2 where C1 and C2 are velocities at inlet and exit, u1 and u2 are internal energy at inlet and outlet, z1 and z2 are elevations of inlet and exit. e2 = u2 + + gz2 3.8 STEADY FLOW SYSTEMS AND THEIR ANALYSIS Steady flow refers to the flow in which its properties at any point remain constant with respect to time. Steady system is the system whose properties are independent of time, i.e. any property at a point in system shall not change with time. Let us take an open system having steady flow. Figure 3.16 shows steady flow system having inlet at section 1\u20131, outlet at section 2\u20132, heat addition Q and work done by the system W", ". Control boundary Q 1 System In 1Z1 2 2 Out W Fig. 3.16 Steady flow system,, p C A v 2 2, 2 2 Z2 Datum 66 _________________________________________________________ Applied Thermodynamics At section 1\u20131 At section 2\u20132 Pressure, (N/m2) Sp volume, (m3/kg) Velocity, (m/s) Elevation, (m) Cross-section area, (m2) Mass flow rate, (kg/s) Internal energy, (J/kg) p1 v1 C1 z1 A1 m1 u1 p2 v2 C2 z2 A2 m2 u2 As described in earlier article the energy balance when applied to open system results in Substituting for e1 and e2 Q + m1(e1 + p1v1) = W + m2(e2 + p2v2) Q + m1 \uf8eb \uf8ec \uf8ec \uf8ed + u 1 2 C 1 2 + + gz 1 p v 1 1 = W + m2 \uf8eb \uf8ec \uf8ec \uf8ed + u 2 2 C 2 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 + + gz 2 p v 2 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 and from definition of enthalpy, therefore, h1 = u1 + p1v1 h2 = u2 + p2v2 + + gz + gz 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 Above equation is known as steady flow energy equation (S.F.E.E.). If the mass flow rates at inlet and exit are same, i.e. m1 = m2 = m 2 C 1 2 Q + m then, h 1 + \uf8eb \uf8ec \uf8ec \uf8ed + gz + gz 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 or, on unit mass basis the S.F.E.E. shall be + gz + gz 2 where q = Q m, w = W m The steady flow energy equation can be used as a tool for carrying out thermodynamic analysis of engineering system with suitable modifications. Special Case : Such as for any system of perfectly insulated type, Q = 0 The steady flow energy equation gets modified to; h1 + 2 C 1 2 + gz1 = w + h2 + 2 C 2 2 + gz2 First Law of Thermodynamics _____________________________________________________ 67 Application of Continuity equation results in, or, m1 = m2 For any system having more than one inlets, outlets and energy interactions the example is shown below. W1 Q1 W2 Q3 Q2 In Out 1 1 2 2 Steady flow system 3 3 4 4 In Out Datum Fig. 3.17 Example of steady flow system Salient properties at different sections are tabulated as under Pressure, (N/m2) Sp. volume, (m3/kg) Mass flow rate, (kg/s) Internal energy, (J/kg) Velocity, (m/s) Elevation, (m) Cross-section area, (m2) Section 1\u20131 Section 2\u20132 Section 3\u20133 Section 4\u20134 p1 v1 m1 u1 C1 z1 A1 p2 v2 m2 u2 C2 z2 A2 p3 v3 m3 u3 C3 z3 A3 p4 v4 m4 u4 C4 z4 A4 Net heat added, Q = Q1 \u2013 Q2 + Q3 Net work done, W = W1 + W2 Applying steady flow energy equation on the system as shown in Fig 3.17; Q + m1 \uf8eb \uf8ec \uf8ec \uf8ed + u 1 2 C 1 2 + + gz 1 \uf8f6 \uf8f7 v + m3 \uf8f7 \uf8f8 + + gz + m2 \uf8eb \uf8ec \uf8ec \uf8ed + u 2 2 C 2 2 + + gz 2 \uf8f6 \uf8f7 p v + m4 \uf8f7 + + gz 4 p v 4 4 \uf8f6 \uf8f7 \uf8f7 \uf8f8 Substituting enthalpy values, h1, h2, h3, h4 and for Q and W + gz 1 + 2 C 2 2 + gz + gz 3 + gz 4 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f8 Case 1 If the inlet and exit velocities are negligible, then KE1 = KE2 = KE3 = KE4 = 0 68 _________________________________________________________ Applied Thermodynamics and S.F.E.E. is modified to Q1 \u2013 Q2 +", " Q3 + m1(h1 + gz1) + m3(h3 + gz3) = W1 + W2 + m2(h2 + gz2) + m4(h4 + gz4) Case 2 If there is no change in elevation and mass flow rates at all inlets and outlets are same, then, m1 = m2 = m3 = m4 = m Q1 \u2013 Q2 + Q3 + m. h1 + m h3 = W1 + W2 + m h2 + m h4 or, on unit mass basis q1 \u2013 q2 + q3 + h1 + h3 = w1 + w2 + h2 + h4 3.9 FIRST LAW APPLIED TO ENGINEERING SYSTEMS Here the first law of thermodynamics applied to different engineering systems is discussed. It is assumed in general that the processes are of steady flow type and so the steady flow energy equation can be directly used with modifications in it. (a) Turbine: It is the device in which the high temperature and high pressure fluid is expanded to low temperature and pressure resulting in generation of positive work at turbine shaft. Thus, turbine is a work producing device. Turbines using gas as working fluid are called gas turbine where as turbines using steam are called steam turbines. Expansion in turbine is assumed to be of adiabatic type so that the maximum amount of work is produced. Assuming change in kinetic energy, potential energy to be negligible, the steady flow energy equation can be modified and written between 1 and 2 as, i.e., Q = 0 and total energy interaction is available in the form of work O + mh1 = WT + mh2 WT = m(h1 \u2013 h2) Turbine work = m(h1 \u2013 h2) = m cp(T1 \u2013 T2) Here m is mass flow rate and T1, T2 are temperatures at inlet and outlet. Fluid out 2 Q = 0 WT 1 Fluid in Fig. 3.18 Turbine (b) Compressor: Compressor is a work absorbing device used for increasing the pressure of a fluid. Pressure of a fluid is increased by doing work upon it, which is accompanied by increase in temperature depending on the gas properties. First Law of Thermodynamics _____________________________________________________ 69 WC 2 High pressure Fluid out 1 Low pressure fluid in Fig. 3.19 Compressor For compression of a gas adiabatic process is used as in this there is no heat loss and so minimum work requirement. Let us assume change in kinetic energy and potential energy to be negligible between 1 & 2 and also flow to be of steady type. Applying steady flow energy equation in modified form: Q = 0 \u2206KE = 0 \u2206PE = 0 Wc = (\u2013ve) work for compression mh1 = \u2013 Wc + mh2 Wc = m(h2 \u2013 h1) or Adiabatic compression work = m(h2 \u2013 h1) = mcp (T2 \u2013 T1) Here T1, T2 are temperatures at inlet and outlet and m is mass flow rate. (c) Pump: A pump is used for pumping liquid or suction of liquid. In case of pump the following assumptions can be made for using S.F.E.E. (i) Heat transfer is zero, Q = 0 (ii) Change in internal energy is zero, \u2206U = 0 Therefore \uf8eb \uf8ec \uf8ec \uf8ed + gz + gz 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u2013 Wpump or, Wpump = m{(p2v2 \u2013 p1v1) + C 2 2 2 C\u2212 1 2 + g(z2 \u2013 z1)} Wpump 1 In Fig. 3.20 Pump 2 Out 70 _________________________________________________________ Applied Thermodynamics (d) Boiler: Boiler is the engineering device used for steam generation at constant pressure. Heat is supplied externally to the boiler for steam generation depending upon state of steam desired. Boiler may be assumed similar to a closed vessel having no work interaction, no change in kinetic energy, no change in potential energy. i.e. Applying steady flow energy equation W = 0, \u2206KE = 0, \u2206PE = 0. or Qboiler + m(h1) = m(h2) Qboiler = m(h2 \u2013 h1) = m cp (T2 \u2013 T1) Q boiler 2 Steam out 1 Water in Fig. 3.21 Boiler (e) Condenser: Condenser is the device used for condensing vapour into liquid at constant pressure. It is a type of heat exchanger in which another cool fluid is used for condensing the vapours into liquid. Heat exchange between the hot fluid and cold fluid takes place indirectly as", " cold fluid passes through the tubes and hot vapours are outside tubes in the shell. Steam in 1 Cold fluid in Fluid out 2 Condensate out Fig. 3.22 Condenser Steady flow energy equation can be applied with the following assumptions : (i) No work interaction, W = 0 (ii) No change in kinetic energy, \u2206KE = 0 (iii) No change in potential energy, \u2206PE = 0 Heat lost by steam, Q = m (h1 \u2013 h2) ( f ) Nozzle: Nozzle is the engineering device in which expansion of fluid takes place and pressure drops simultaneously. Thus in nozzle the velocity of fluid increases from inlet to exit. In case of subsonic flow the nozzle has converging cross-section area in the duct where as in supersonic flow the nozzle has diverging cross-section area in the duct. Let us take a converging cross-section area duct as shown in Fig. 3.23. Flow through the nozzle may be analysed with following assumptions: (i) No heat interaction, i.e. Q = 0, during passage through duct. (ii) No work interaction, i.e. W = 0, during passage through duct. First Law of Thermodynamics _____________________________________________________ 71 (iii) No change in elevation from 1 to 2, i.e. \u2206PE = 0. Applying S.F.E.E on nozzle, h1 + 2 C 1 2 C 2 2 2 C\u2212 1 2 = h2 + 2 C 2 2 = h1 \u2013 h2 C2 = 2 C 1 + 2( h 1 \u2212 h 2 ) or, or, exit shall be: ( c T p 1 In case, the velocity at inlet to nozzle is very small, then C1 may be neglected and velocity at nozzle C2 = T 2 2 ) 2 C 1 + \u2212 C2 = 2 ( pc T 1 T\u2212 2 ) 1 2 2, C h2 Fluid out Fluid in 1, C h1 1 Nozzle Fig. 3.23 Nozzle (g) Throttling: Throttling refers to passage of a fluid through some restricted opening under isenthalpic conditions. Thus in the figure shown below the fluid passes through a restriction from section 1 to 2 and undergoes drop in its pressure and increase in volume, but during this passage enthalpy remains constant, such that h1 = h2. Based on above throttling process the device called \u201cthrottle valve\u201d has been developed in which pressure drop is realized without involving any work and heat interaction, change in kinetic energy and potential energy. Temperature may drop or increase during the throttling process and shall depend upon the Joule-Thomson coefficient, a property based on characteristic of substance. \uf8eb \uf8f6\u2202 T \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p Joule-Thomson coefficient \u00b5 = =constt. and if \u00b5 = 0, Temperature remains constant \u00b5 > 0, Temperature decreases. \u00b5 < 0 Temperature increases. h Restricted opening 1 2 Fig. 3.24 Throttling process 72 _________________________________________________________ Applied Thermodynamics (h) Combustion chamber: Combustion chambers are commonly used in gas turbine installations, in which fuel is injected at high pressure into a chamber having high pressure, high temperature air in it and ignited for heat release at constant pressure. Mass balance yields: m2 = m1 + mf where m1, m2, mf are mass flow rates at 1, 2, and f shown in figure. S.F.E.E. may be applied with assumptions of \u2206KE = 0, \u2206PE = 0, W = 0 Here, Q = mf \u00d7 Calorific value of fuel or Q = mf \u00d7 CV substituting for Q Q + m1 h1 + mf hf = m2 h2 mf CV + m1 h1 + mf hf = m2h2 f Fuel mf 1 m1 Air in 2 m2 Combustion products, out Fig. 3.25 Combustion chamber (i) Adiabatic mixing: Adiabatic mixing refers to mixing of two or more streams of same or different fluids under adiabatic conditions. Let us consider two streams of same fluid with mass flow rates m1 and m2 to get mixed together adiabatically. Assumptions for applying S.F.E.E shall be; (i) No heat interaction, Q = 0 (ii) No work interaction, W = 0 (iii) No change in kinetic energy, \u2206KE = 0 (iv) No change in potential energy, \u2206PE = 0. Thus, or or m1 \u00b7 h1 + m2 \u00b7 h2 = m3 \u00b7 h3. m1 \u00b7 cp \u00b7 T1 + m2 \u00b7 cp \u00b7 T2 = m3 \u00b7 cp \u00b7 T", "3 = By mass balance, m1 + m2 = m3 m h1 1, m h2 2, 1 1 2 2 3 3 m h3 3, 3 Fig. 3.26 Adiabatic mixing First Law of Thermodynamics _____________________________________________________ 73 3.10 UNSTEADY FLOW SYSTEMS AND THEIR ANALYSIS In earlier discussions, for a steady flow system, it has been assumed that the properties do not change with time. However, there exist a number of systems such as filling up of a bottle or emptying of a vessel etc. in which properties change continuously as the process proceeds. Such systems can not be analysed with the steady state assumptions. Unsteady flow processes are also known as transient flow processes or variable flow processes. Let us take example of filling up of the bottle. The bottle is filled up gradually, therefore it is case of an unsteady system. By conservation of mass, the unsteady process over a period of time \u2018dt\u2019 can be expressed as following in generic form. (Mass entering the control volume in time dt) \u2013 (Mass leaving the control volume in time dt) = Net change in mass in control volume in time dt. If the mass flow rate at inlet and exit are given as mi, me then and also, i dm dm \u2212 e dt dt = \u2211 mi \u2013 \u2211 me = (mfinal \u2013 minitial)cv cdm dt v Fig. 3.27 Filling of the bottle By the conservation of energy principle applied on control volume for time \u2018t\u2019, energy balance yields; Net energy interaction across the boundary in time dt + Energy entering into control volume in time dt \u2013 Energy leaving out of control volume in time dt = Change in energy in control volume in time dt Mathematically, it can be given as: (Q \u2013 W) + \u2211 Ei \u2013 \u2211 Ee = \u2206Ecv where Ei = + gz i ). dt 0 t 2 C e 2 Thus, the above mass balance and energy balance can be used for analysing the unsteady flow systems with suitable assumptions. Ee = ( m h e e gz dt ). \u222b + + 0 e It may be assumed that the control volume state is uniform and fluid properties are uniform and steady at inlet and exit. Simplified form of energy balance written above can be given as + gz + gz e ) = (mfinal. ufinal \u2013 minitial. uinitial)cv If the changes in kinetic energy and potential energy are negligible, then energy balance gets modified as; Q \u2013 W + \u2211 mi \u00b7 hi \u2013 \u2211 me \u00b7 he = (mfinal \u00b7 ufinal \u2013 minitial \u00b7 uinitial)cv 74 _________________________________________________________ Applied Thermodynamics Case 1: Let us now use the energy and mass balance to the unsteady flow process of filling up a bottle as shown in Figure 3.27. Bottle is initially empty and connected to a pipe line through valve for being filled. Let us denote initial state of system by subscript 1 and final state by 2. Initially as bottle is empty, so m1 = 0 From mass balance \u2211 mi \u2013 \u2211 me = (m2 \u2013 0)cv Here there is no exit from the bottle so me = 0 hence, or, \u2211 mi = m2 mi = m2 Mass entered into bottle = Final mass inside the bottle Applying the energy balance assuming change in kinetic and potential energy to be negligible, treating bottle filling process to be occurring in insulated environment, and no work interaction, we get Q \u2248 0, W \u2248 0, \u2206KE \u2248 0, \u2206PE \u2248 0, Initial internal energy in bottle = 0 Mass leaving = 0 0 = \u2013 \u2211 mi \u00b7 hi + (m2\u00b7u2)cv mi \u00b7 hi = m2u2 hi = u2 as mi = m2 Enthalpy of fluid entering bottle = Final internal energy of fluid in bottle. If fluid is ideal gas, then cp\u00b7Ti = cv \u00b7T2 or also or where T2 = \u03b3 \u00b7Ti pc cv = \u03b3 Fig. 3.28 Emptying of bottle Case 2: Let us now take a case of emptying of bottle. Arrangement is shown in Fig. 3.28. Initially bottle has mass m1 and finally as a result of emptying, say mass left is m2 after some time. Applying mass balance, (as mass entering is zero), or, or or 0 \u2013 \u2211 me = (m2 \u2013 m1)cv \u2211 me = (m1 \u2013 m2)cv me = (m1 \u2013 m2)cv Total mass leaving the bottle = (Mass reduced in bottle) Applying energy balance, with the assumptions given below; (i) No heat interaction i.e. Q = 0 (ii)", " No work interaction i.e. W = 0 (iii) No change in kinetic energy i.e. \u2206KE = 0 (iv) No change in potential energy i.e. \u2206PE = 0 \u2013 \u2211 me \u00b7 he = (m2 u2 \u2013 m1u1)cv (\u2013 me \u00b7 he) = (m2u2 \u2013 m1u1)cv or, Substituting for \u2018me\u2019 we get (m2 \u2013 m1)cv \u00b7 he = (m2u2 \u2013 m1u1)cv First Law of Thermodynamics _____________________________________________________ 75 In case of complete emptying, m2 = 0 and so, u= 1 eh 3.11 LIMITATIONS OF FIRST LAW OF THERMODYNAMICS First law of thermodynamics based on law of energy conservation has proved to be a powerful tool for thermodynamic analysis. But over the period of time when it was applied to some real systems, it was observed that theoretically first law stands valid for the processes which are not realizable practically. It was then thought that there exist certain flaws in first law of thermodynamics and it should be used with certain limitations. Say for example let us take a bicycle wheel and paddle it to rotate. Now apply brake to it. As a result of braking wheel comes to rest upon coming in contact with brake shoe. Stopping of wheel is accompanied by heating of brake shoe. Examining the situation from Ist law of thermodynamics point of view it is quite satisfying that rotational energy in wheel has been transformed into heat energy with shoe, thus causing rise in its temperature: Now, if we wish to introduce the same quantity of heat into brake shoe and wish to restore wheel motion then it is not possible simply, whereas theoretically first law permits the conversion from heat to work (rotation of wheel in this case) as well. Therefore, it is obvious that Ist law of thermodynamics has certain limitations as given below: (i) First law of thermodynamics does not differentiate between heat and work and assures full convertibility of one into other whereas full conversion of work into heat is possible but the vice-versa is not possible. (ii) First law of thermodynamics does not explain the direction of a process. Such as theoretically it shall permit even heat transfer from low temperature body to high temperature body which is not practically feasible. Spontaneity of the process is not taken care of by the first law of thermodynamics. Perpetual motion machine of the first kind (PMM-I) is a hypothetical device conceived, based on violation of First law of thermodynamics. Let us think of a system which can create energy as shown below. Q = 0 PMM-I ( )a W 0\u2260 Q 0\u2260 PMM-I W = 0 ( )b Fig. 3.29 PMM-I, based on violation of Ist law of thermodynamics Here a device which is continuously producing work without any other form of energy supplied to it has been shown in (a), which is not feasible. Similarly a device which is continuously emitting heat without any other form of energy supplied to it has been shown in (b), which is again not feasible. Above two imaginary machines are called Perpetual Motion Machines of 1st kind. 76 _________________________________________________________ Applied Thermodynamics 1. Figure shows a system comprising of gas in cylinder at pressure of 689 kPa. EXAMPLES Paddle wheel Piston Gas Cylinder Fig. 3.30 Fluid expands from a volume of 0.04 m3 to 0.045 m3 while pressure remains constant. Paddle wheel in the system does a work of 4.88 kJ on the system. Determine (a) work done by system on the piston (b) the net amount of work done on or by the system. Solution: (a) It is a closed system. If the pressure on face of piston is uniform, then the work done on piston can be obtained as, W = p 2 \u222b dV 1 = 689 \u00d7 103 (0.045 \u2013 0.04) Work done on piston, W = 3445 J or 3.445 kJ Work done on piston = 3.445 kJ Ans. (b) Paddle work done on the system = \u2013 4.88 kJ Net work of system\u21d2 Wnet = Wpiston + Wpaddle = 3445 \u2013 4880 Wnet = \u20131435 J Work done on system = 1435 J or 1.435 kJ. Work done on system = 1.435 kJ Ans. 2. A gas at 65 kPa, 200\u00b0C is heated in a closed, rigid vessel till it reaches to 400\u00b0C. Determine the amount of heat required for 0.5 kg of this gas if internal energy at 200\u00b0C and 400\u00b0C are 26.6 kJ/kg and 37.8 kJ/kg respectively. Solution: Given", " m = 0.5 kg u1 = 26.6 kJ/kg u2 = 37.8 kJ/kg As the vessel is rigid therefore work done shall be zero. From first law of thermodynamics; W = 0 Q = U2 \u2013 U1 + W = m(u2 \u2013 u1) + 0 Q = 0.5 (37.8 \u2013 26.6) Q = 5.6 kJ Heat required = 5.6 kJ Ans. Q =? W = 0 Gas in closed rigid vessel Fig. 3.31 First Law of Thermodynamics _____________________________________________________ 77 3. Carbon dioxide passing through a heat exchanger at a rate of 50 kg/hr is to be cooled down from 800\u00b0C to 50\u00b0C. Determine the rate of heat removal assuming flow of gas to be of steady and constant pressure type. Take cp = 1.08 kJ/kg K. Solution: 1 2 T1 = 800 \u00baC T2 = 50 \u00baC Fig. 3.32 Given, Writing down the steady flow energy equation. m = 50 kg/hr q + h1 + 2 C 1 2 + gz1 = h2 + 2 C 2 2 + gz2 + w Here let us assume changes in kinetic and potential energy to be negligible. During flow the work interaction shall also be zero. Hence or q = h2 \u2013 h1 Q = m (h2 \u2013 h1) = m \u00b7 cp \u00b7 (T2 \u2013 T1) = 50 \u00d7 1.08 \u00d7 (750) = 40500 kJ/hr Heat should be removed at the rate of 40500 kJ/hr Ans. 4. A completely evacuated cylinder of 0.78 m3 volume is filled by opening its valve to atmosphere and air rushing into it. Determine the work done by the air and by surroundings on system. Solution : Total work done by the air at atmospheric pressure of 101.325 kPa, \u222b cylinder W =. p dv. p dv \u222b air + = 0 + p.\u2206v, it is \u2013ve work as air boundary shall contract Work done by surroundings on system = + 79.03 kJ \uf8fc Work done by air = \u2013 101.325 \u00d7 0.78 = \u2013 79.03 kJ \uf8fd \uf8fe Ans. 5. A system comprising of a gas of 5 kg mass undergoes expansion process from 1 MPa and 0.5 m3 to 0.5 MPa. Expansion process is governed by, p.v1.3 = constant. The internal energy of gas is given by, u = 1.8 pv + 85, kJ/kg. Here \u2018u\u2019 is specific internal energy, \u2018p\u2019 is pressure in kPa, \u2018v\u2019 is specific volume in m3/kg. Determine heat and work interaction and change in internal energy. Solution: Given mass of gas, m = 5 kg, pv1.3 = constant Assuming expansion to be quasi-static, the work may be given as, W = m\u222b p.dv \u2212 \u2212 = 1 From internal energy relation, change in specific internal energy, \u2206u = u2 \u2013 u1 = 1.8 (p2v2 \u2013 p1v1), kJ/kg 78 _________________________________________________________ Applied Thermodynamics Total change, Between states 1 and 2, \u2206U = 1.8 \u00d7 m \u00d7 (p2v2 \u2013 p1v1), kJ \u2206U = 1.8 \u00d7 (p2V2 \u2013 p1V1) or \u21d2 p1V1 p1V1 1.3 = p2V2 1.3 = p2V2 1.3 1.3 V2 = (0.5). \uf8eb \uf8ec \uf8ed V2 = 0.852 m3 1 0.5 1/1.3 \uf8f6 \uf8f7 \uf8f8 Total change in internal energy, \u2206U = \u2013133.2 kJ Work, W = \u00d7 \u2212 \u00d7 (0.5 0.852 1 0.5) 10 \u2212 (1 1.3) \u00d7 3 From first law, W = 246.67 kJ \u2206Q = \u2206U + W = \u2013133.2 + 246.7 \u2206Q = 113.5 kJ Heat interaction = 113.5 kJ Work interaction = 246.7 kJ Change in internal energy = \u2013133.2 kJ \uf8fc \uf8fd \uf8fe Ans. 6. A gas contained in a cylinder is compressed from 1 MPa and 0.05 m3 to 2 MPa. Compression is governed by pV1.4 = constant. Internal energy of gas is given by; where p is pressure in kPa and V is volume in m3. U = 7.5 pV \u2013 425, kJ", ". Determine heat, work and change in internal energy assuming compression process to be quasi- static. Also find out work interaction, if the 180 kJ of heat is transferred to system between same states. Also explain, why is it different from above. Solution: Final state, volume V2 = \uf8eb \uf8ec \uf8ed p 1 p 2 1/1.4 \uf8f6 \uf8f7 \uf8f8 \u00b7 V1 = \uf8eb \uf8ec \uf8ed 1/1.4 \uf8f6 \uf8f7 \uf8f8 1 2 V2 = 0.03 m3 \u00b7 0.05 Change in internal energy, For quasi-static process, \u2206U = U2 \u2013 U1 = (7.5 p2V2 \u2013 7.5 p1V1) = 7.5 \u00d7 103 (2 \u00d7 0.03 \u2013 1 \u00d7 0.05) \u2206U = 75 kJ First Law of Thermodynamics _____________________________________________________ 79 Work, W = 2 \u222b 1 \u00b7p dV 2 0.03 1 0.05) 10 \u2212 (1 1.4) \u00d7 3 = = From first law of thermodynamics, Heat interaction \u2206Q = \u2206U + W W = 25 kJ, (\u2013ve) = 75 + (\u201325) = 50 kJ Heat = 50 kJ Work = 25 kJ (\u2013ve) Ans. \uf8fc \uf8fd \uf8fe Internal energy change = 75 kJ If 180 kJ heat transfer takes place, then from Ist law, \u2206Q = \u2206U + W Since end states remain same, therefore \u2206U, i.e. change in internal energy remains unaltered. or 180 = 75 + W W = 105 kJ This work is different from previous work because the process is not quasi-static in this case. Ans. 7. Determine the heat transfer and its direction for a system in which a perfect gas having molecular weight of 16 is compressed from 101.3 kPa, 20\u00b0C to a pressure of 600 kPa following the law pV1.3 = constant. Take specific heat at constant pressure of gas as 1.7 kJ/kg.K. Solution: Characteristic gas constant, R = Universal gas constant Molecular weight or or For polytropic process,, J/kg.K = 3 8.3143\u00d710 16 = 519.64, J/kg.K = 0.51964, kJ/kg.K R = 0.520, kJ/kg.K Cv = Cp \u2013 R = 1.7 \u2013 0.520 Cv = 1.18, kJ/kg.K \u03b3 = C p C v = 1.7 1.18 = 1.44 V2 = \uf8eb \uf8ec \uf8ed p 1 p 2 1/1.3 \uf8f6 \uf8f7 \uf8f8 \u00b7V1 80 _________________________________________________________ Applied Thermodynamics or Work, For polytropic process Heat, T2 = T1 \u2212 1.3 1 1. T2 = 293 \u00b7 0.231 \uf8eb \uf8ec \uf8ed 600 101.3 \uf8f6 \uf8f7 \uf8f8 T2 = 441.9 K T\u2212 ( R T 1 2 \u2212 (1.3 1) W = ) W = 258.1 kJ/kg Q = \uf8f6\u2212 \uf8eb n\u03b3 \uf8f7 \uf8ec \uf8f8 \u00b7 W = \u2212\uf8ed \u03b3 1 = 82.12, kJ/kg (+ve) Ans. \u2212 1.44 1.3 \u2212 1.44 1 \uf8eb \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f8 \u00d7 258.1 8. In a nozzle air at 627\u00b0C and twice atmospheric pressure enters with negligible velocity and leaves at a temperature of 27\u00b0C. Determine velocity of air at exit, assuming no heat loss and nozzle being horizontal. Take CP = 1.005 kJ/kg.K for air. Solution: Applying steady flow energy equation with inlet and exit states as 1, 2 with no heat and work interaction and no change in potential energy. h1 + 2 C 1 2 Given that, C1 \u2248 0, negligible inlet velocity C2 = = h2 + 2 C 2 2 h\u2212 2 h 1 2( ) Exit velocity, C2 = 2\u00b7 \u00b7( pC T 1 T\u2212 2 ) Given, T1 = 900 K, T2 = 300 K \u00d7 or 3 2 1.005 10 (900 300) \u00d7 \u2212 C2 = C2 = 1098.2 m/s Exit velocity = 1098.2 m/s. Ans. 9. An air compressor requires shaft work of 200 kJ/kg of air and the compression of air causes increase in enthalpy of air by 100 kJ/kg of air.", " Cooling water required for cooling the compressor picks up heat of 90 kJ/kg of air. Determine the heat transferred from compressor to atmosphere. Solution: Work interaction, W = \u2013 200 kJ/kg of air Increase in enthalpy of air = 100 kJ/kg of air Total heat interaction, Q = Heat transferred to water + Heat transferred to atmosphere. Writing steady flow energy equation on compressor, for unit mass of air entering at 1 and leaving at 2. h1 + 2 C 1 2 + gZ1 + Q = h2 + 2 C 2 2 + gZ2 + W 82 _________________________________________________________ Applied Thermodynamics Solution: Let mass of steam to be supplied per kg of water lifted be \u2018m\u2019 kg. Applying law of energy conservation upon steam injector, for unit mass of water lifted. Energy with steam entering + Energy with water entering = Energy with mixture leaving + Heat loss to surroundings. m 2 \uf8ee \uf8ef \uf8f0 (50) 2 + \u00d7 (720 10 2 \u00d7 4.18) \uf8f9 \uf8fa \uf8fb + 1 \u00d7 [(24.6 \u00d7 103 \u00d7 4.18) + (9.81 \u00d7 2)] K.E. Enthalpy Enthalpy P.E = (1 + m) \uf8ee \uf8ef \uf8f0 \u00d7 (100 10 3 \u00d7 4.18) + 2 (25) 2 \uf8f9 \uf8fa \uf8fb + [m \u00d7 12 \u00d7 103 \u00d7 4.18] Enthalpy K.E Heat loss m [3010850] + [102847.62] = (1 + m). (418312.5) + m[50160] Upon solving, m = 0.124 kg steam/kg of water Steam supply rate = 0.124 kg/s per kg of water. Ans. 12. An inelastic flexible balloon is inflated from initial empty state to a volume of 0.4 m3 with H2 available from hydrogen cylinder. For atmospheric pressure of 1.0313 bar determine the amount of work done by balloon upon atmosphere and work done by atmosphere. Solution: Balloon initially empty Balloon after being inflated Fig. 3.34 Here let us assume that the pressure is always equal to atmospheric pressure as balloon is flexible, inelastic and unstressed and no work is done for stretching balloon during its filling. Figure 3.34 shows the boundary of system before and after filling balloon by firm line and dotted line respectively. Displacement work, W = \u222b cylinder.p dV + \u222b.p dV balloon \u222b.p dV = 0 as cylinder shall be rigid. = 0 + p \u00b7 \u2206V = 0 + 1.013 \u00d7 105 \u00d7 0.4 = 40.52 kJ Work done by system upon atmosphere = 40.52 kJ \uf8fc \uf8fd \uf8fe Work done by atmosphere = \u2013 40.52 kJ Ans. 13. In a steam power plant 5 kW of heat is supplied in boiler and turbine produces 25% of heat added while 75% of heat added is rejected in condenser. Feed water pump consumes 0.2% of this heat added First Law of Thermodynamics _____________________________________________________ 83 for pumping condensate to boiler. Determine the capacity of generator which could be used with this plant. Solution: Given, so, Qadd = 5000 J/s WT = 0.25 \u00d7 5000 = 1250 J/s Qrejected = 0.75 \u00d7 5000 = 3750 J/s Wp = (\u2013) 0.002 \u00d7 5000 = 10 J/s Capacity of generator = WT \u2013 WP = 1250 \u2013 10 = 1240 J/s or 1240 W = 1.24 kW Ans. Qadd WT (+ve) Boiler Turbine Generator Condenser Wp (\u2013ve) Qrejected Feed pump Fig. 3.35 14. In a gas turbine installation air is heated inside heat exchanger upto 750\u00b0C from ambient temperature of 27\u00b0C. Hot air then enters into gas turbine with the velocity of 50 m/s and leaves at 600\u00b0C. Air leaving turbine enters a nozzle at 60 m/s velocity and leaves nozzle at temperature of 500\u00b0C. For unit mass flow rate of air determine the following assuming adiabatic expansion in turbine and nozzle, (a) heat transfer to air in heat exchanger (b) power output from turbine (c) velocity at exit of nozzle. Take cp for air as 1.005 kJ/kg\u00b0K. Solution: In heat exchanger upon applying S.F.E.E. with assumptions of no change in kinetic energy, no work interaction, no change in potential energy, for unit mass flow rate of air, h1 + Q1\u20132 = h2 Heat transfer to air in heat exchanger Q1\u20132 = 726.62 kJ Ans. Q1\u20132 = h2", " \u2013 h1 Q1\u20132 = Cp \u00b7 (T2 \u2013 T1) 27 \u00b0C Air 1 750 \u00b0C 2 Heat exchanger Fig. 3.36 84 _________________________________________________________ Applied Thermodynamics In gas turbine let us use S.F.E.E., assuming no change in potential energy, for unit mass flow rate of air h2 + 2 C 2 2 = h3 + 2 C 3 2 + WT WT = (h2 \u2013 h3 \uf8f6 \uf8f7 \uf8f7 \uf8f8 = Cp(T2 \u2013 T3.005 (750 \u2013 600) + 2 50 \uf8eb \uf8ec \uf8ec \uf8ed 2 60 \u2212 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u00d7 10\u20133 Power output from turbine = 150.2 kJ/s Ans. 600 \u00b0C, 60 m/s 3 WT Gas turbine 2 50 m/s 750 \u00b0C Fig. 3.37 Applying S.F.E.E. upon nozzle assuming no change in potential energy, no work and heat interac- tions, for unit mass flow rate, h3 + 2 C 3 2 2 C 4 2 = h4 + 2 C 4 2 = (h3 \u2013 h4) + 2 C 3 2 = Cp(T3 \u2013 T4) + 2 C 3 2 = 1.005 (600 \u2013 500) + \uf8eb \uf8ec \uf8ec \uf8ed 260 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u00d7 10\u20133 = 102.3 2 C 4 2 C4 = 14.3 m/s Velocity at exit of nozzle = 14.3 m/s Ans. 600 \u00b0C 60 m/s 3 500 \u00b0C 4 Nozzle Fig. 3.38 First Law of Thermodynamics _____________________________________________________ 85 15. One mol of air at 0.5 MPa and 400 K, initially undergoes following processes, sequentially (a) heating at constant pressure till the volume gets doubled. (b) expansion at constant temperature till the volume is six times of initial volume. Determine the work done by air. Solution: For constant pressure heating, say state changes from 1 to 2 Wa = 2 \u222b 1 p dV 1 It is given that so Wa = p1 (V2 \u2013 V1) V2 = 2V1 Wa = p1 V1 Wa = RT1 For subsequent expansion at constant temperature say state changes from 2 to 3. Also given that Work, Temperature at 2 can be given by perfect gas considerations as, or Total work done by air, = RT1 + RT2 ln (3) = RT1 + 2RT1 ln (3) = RT1 (1 + 2 ln 3) = 8.314 \u00d7 400 (1 + 2 ln 3) Work done = 10632.69 kJ Ans. 16. Determine the work done by gas for the arrangement shown in Fig. 3.39. Here spring exerts a force upon piston which is proportional to its deformation from equilibrium position. Spring gets deflected due to heating of gas till its volume becomes thrice of original volume. Initial states are 0.5 MPa and 0.5 m3 while final gas pressure becomes 1 MPa. Atmospheric pressure may be taken as 1.013 \u00d7 105 Pa. Solution: Let stiffness of spring be k and it undergoes a deflection by \u2018x\u2019 along x-axis. Force balance at any equilibrium position of piston shall be, p. A = patm \u00b7 A + kx, here x shall be linear displacement of piston due to expansion of gas. Let volume of gas change from V0 to some value V. Then, x =, V0 is volume of gas when spring is at its natural length. \u2212 V V 0 A RT V dV = RT2 ln V 3 V 2 V 3 V = 6, so 1 V 3 V = 3 2 pdV 3 \u222b 2 3 \u222b Wb = = 2 Wb = RT2 ln (3 T2 = 2 T1 W = Wa + Wb 86 _________________________________________________________ Applied Thermodynamics p \u00b7 A = patm \u00b7 \uf8f6 \uf8f7 \uf8f8 or (p \u2013 patm Work done by gas between initial and final states, W = f \u222b i. p dV W = patm (Vf \u2013 Vi = patm (Vf \u2013 Vi) + = (Vf \u2013 Vi) patm + Vf \u2013 Vi) patm + \uf8f1 \uf8f2 \uf8f3 k 2 2 A (( ))( Vf \u2013 Vi) \uf8f1 \uf8f2 \uf8f3 P atm k 2 2 A (( )) \uf8fc \uf8fd \uf8fe from above force balance, (pf \u2013 patmpi \u2013 patm or W = (Vf \u2013 Vi) = (Vf \u2013", " Vi atm \uf8eb \uf8ec \uf8ed p f \u2212 p atm atm + 2 \uf8fc \uf8fd \uf8fe, substituting pressure and volume values, Vf = 3Vi W = 0.75 \u00d7 106 J Ans. y x Atm. pr. x Fig. 3.39 First Law of Thermodynamics _____________________________________________________ 87 17. A closed insulated container has frictionless and smooth moving insulated partition as shown in Fig. 3.40 such that it equally divides total 1 m3 of volume, when both the gases are at initial pressure of 0.5 MPa and ambient temperature of 27\u00b0C. Subsequently the nitrogen is heated using electrical heating element such that volume of N2 becomes 3/4 of total volume of container. Determine, (i) final pressure of hydrogen, (ii) Work done by partition, (iii) Workdone by N2 and H2 (iv) Heat added to N2 by electric heater. Insulated Electric heating element Take,p NC 2 = 1.039 kJ/kg. K,,p HC 2 = 14.307 kJ/kg. K, 2NR = 0.2968 kJ/kg. K, 2HR = 4.1240 kJ/kg. K N2 H2 Frictionless moving partition Solution: With the heating of N2 it will get expanded while H2 gets compressed simultaneously. Compression of H2 in insulated chamber may be considered of adiabatic type. Adiabatic Index of compression for H2 can be obtained as, Fig. 3.40,p HC \uf8f6 \uf8f7 \uf8f7\u2212 \uf8f7\u2212 1 \uf8f8 14.307 = 4.124 \u03b3 H2 = 1.405 Adiabatic Index of expansion for N2,,p NC 2 = 2NR \uf8eb \uf8ec \uf8ec \uf8ed \u03b3 N \u03b3 N 2 2 \uf8f6 \uf8f7 \uf8f7\u2212 \uf8f7\u2212 1 \uf8f8 1.039 = 0.2968 \u03b3 2N = 1.399 (i) For hydrogen.405, V1 = 0.5 m3 2H Here \u03b3 = p1 = 0.5 \u00d7 106 Pa, V2 = 0.25 m3 Final pressure of H2 = 0.5 \u00d7 106 \uf8eb \uf8ec \uf8ed 0.5 0.25 1.405 \uf8f6 \uf8f7 \uf8f8 = 1.324 MPa Ans. (ii) Since partition remains in equilibrium throughout hence no work is done by partition. It is a case similar to free expansion. Partition work = 0. Ans. (iii) Work done upon H2, 2HW = p V\u2212 2 2 1) PV 1 1 \u2212\u03b3 ( H 2 88 _________________________________________________________ Applied Thermodynamics Here p1 = 0.5 \u00d7 106 Pa, p2 = 1.324 \u00d7 106 Pa, V1 = 0.5 m3, V2 = 0.25 m3. \u2212 ( )0.081 10 0.405 = (\u2013) 2 \u00d7 105 J Ans. Work done by hydrogen, 2HW = \u00d7 6 Work done by N2 = Work done upon H2 Work done by nitrogen = + 2 \u00d7 105 J Ans. (iv) Heat added to N2 can be obtained using first law of thermodynamics as 2NQ 2NW \u21d2 2NU\u2206 = + 2NQ = mcv(T2 \u2013 T1) + 2NW Final temperature of N2 can be obtained considering it as perfect gas. Therefore, T2 = p2 = Final pressure of N2 which will be equal to that of H2 as the partition is free and frictionless. T2 = Final temperature of N2 = p2 = 1.324 \u00d7 106 Pa. 1.324 10 \u00d7 0.5 10 \u00d7 \u00d7 6 \u00d7 0.75 300 \u00d7 6 0.5 = 1191.6 K mass of N2, m = p V 1 1 RT = 1 6 \u00d7 \u00d7 0.5 10 \u00d7 0.2968 10 3 0.5 \u00d7 300 = 2.8 kg. Specific heat at constant volume, Cv = Cp \u2013 R \u21d2 Heat added to N2, N2Q = {2.8 \u00d7 (1191.6 \u2013 300) \u00d7 0.7422 \u00d7 103} + 2 \u00d7 105 = 0.7422 kJ/kg. K., NCv 2 = 2052.9 kJ Ans. 18. A cylinder of 2 m3 has air at 0.5 MPa and temperature of 375\u00b0K. Air is released in atmosphere through a valve on cylinder so as to run a frictionless turbine. Find the amount of work available from turbine assuming no heat loss and complete kinetic energy being used for running turbine. Take Cp, air = 1.003 kJ/kg", ". K, Cv, air = 0.716 kJ/kg. K, Rair = 0.287 kJ/kg. K. Solution: Let initial states and final states of air inside cylinder be given by m1, p1, V1, T1, and m2, p2, V2, T2 respectively. It is a case of emptying of cylinder. Initial mass of air, m1 = p V 1 1 P T = 9.29 kg. 1. air For adiabatic expansion during release of air through valve from 0.5 MPa to atmospheric pressure T2 = T 1 = 375 \uf8eb \uf8ec \uf8ec \uf8ed \u00d7 5 1.013 10 6 \u00d7 0.5 10 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u2212 1.4 1 1.4 First Law of Thermodynamics _____________________________________________________ 89 Final mass of air left in tank, m2 = T2 = 237.65 K p V 2 2 RT 2 5 1.013 10 \u00d7 \u00d7 (0.287 10 m2 = \u00d7 3 \u00d7 2 237.65) = 2.97 kg Writing down energy equation for unsteady flow system (m1 \u2013 m2m1 \u2013 m2) \uf8f6 \uf8f7 \uf8f7 \uf8f8 2 C 2 = m1u1 \u2013 m2u2 = (m1 u1 \u2013 m2u2) \u2013 (m1 \u2013 m2)h2 T1 \u2013 m2 Cv Kinetic energy available for running turbine T2) \u2013 (m1 \u2013 m2) \u00b7 Cp \u00b7 T2 = (m1 Cv = (9.29 \u00d7 0.716 \u00d7 103 \u00d7 375) \u2013 (2.97 \u00d7 0.716 \u00d7 103 \u00d7 237.65) \u2013 {(9.29 \u2013 2.97) \u00d7 1.003 \u00d7 103 \u00d7 237.65} = 482.54 \u00d7 103 J Amount of work available = 482.54 kJ Ans. 19. A rigid and insulated tank of 1 m3 volume is divided by partition into two equal volume chambers having air at 0.5 MPa, 27\u00b0C and 1 MPa, 500 K. Determine final pressure and temperature if the partition is removed. Solution: Using perfect gas equation for the two chambers having initial states as 1 and 2 and final state as 3. p1 = 0.5 \u00d7 106 Pa, V1 = 0.5 m3, T1 = 300 K n1 = p V 1 1 RT 1 = 6 \u00d7 \u00d7 0.5 10 \u00d7 8314 300 0.5 n1 = 0.1002 and or 2 n2 = p V 2 2 RT where p2 = 1 \u00d7 106 Pa, V2 = 0.5 m3, T2 = 500 K \u00d7 \u00d7 1 10 0.5 \u00d7 8314 500 n2 = 0.1203 = 6 For tank being insulated and rigid we can assume, \u2206U = 0, W = 0, Q = 0, so writing \u2206U, \u2206U = n1 Cv(T3 \u2013 T1) + n2 Cv(T3 \u2013 T2) = 0 T3 = 409.11 K Using perfect gas equation for final mixture, ( p3 = + ) n RT.75 MPa Final pressure and temperature = 0.75 MPa, 409.11 K Ans. 90 _________________________________________________________ Applied Thermodynamics 20. An evacuated bottle of 0.5 m3 volume is slowly filled from atmospheric air at 1.0135 bars until the pressure inside the bottle also becomes 1.0135 bar. Due to heat transfer, the temperature of air inside the bottle after filling is equal to the atmospheric air temperature. Determine the amount of heat transfer. [U.P.S.C., 1994] Solution: Initial system boundary P = 1.0135 bar atm Valve Initial system boundary Evacuated bottle Final system boundary after filling Displacement work; Heat transfer, Fig. 3.41 W = 1.0135 \u00d7 105 \u00d7 (0 \u2013 0.5) W = \u2013 0.50675 \u00d7 105 Nm Q = 0.50675 \u00d7 105 Nm Heat transfer = 0.50675 \u00d7 105 Nm Ans. 21. A compressed air bottle of 0.3 m3 volume contains air at 35 bar, 40\u00b0C. This air is used to drive a turbogenerator sypplying power to a device which consumes 5 W. Calculate the time for which the device can be operated if the actual output of the turbogenerator is 60% of the maximum theoretical output. The ambient pressure to which the tank pressure has fallen is 1 bar. For air, pC Cv = 1.4. [U.P.S.C. 1993] Solution: Here turbogenerator is fed with compressed air from a compressed air bottle. Pressure inside bottle gradually", " decreases from 35 bar to 1 bar. Expansion from 35 bar to 1 bar occurs isentropically. Thus, for the initial and final states of pressure, volume, temperature and mass inside bottle being given as P1, V1, T1 & m1 and P2, V2, T2 & m2 respectively. It is transient flow process similar to emptying of the bottle, Given: P1 = 35 bar, T1 = 40\u00b0C or 313 K 1 V1 = 0.3 m3; V2 = 0.3 m3 P2 = 1 bar. First Law of Thermodynamics _____________________________________________________ 91 T2 = T1 \u2212 \uf8f6 \uf8f7 \uf8f8 T2 = 113.22 K By perfect gas law, initial mass in bottle, m1 = PV 1 1 RT = 1 2 \u00d7 \u00d7 35 10 0.3 \u00d7 0.287 313 m1 = 11.68 kg P V 2 2 RT = m2 = 0.923 kg m2 = 2 2 \u00d7 \u00d7 1 10 \u00d7 0.287 113.22 0.3 Final mass in bottle, Energy available for running turbo generator or work; W + (m1 \u2013 m2) h2 = m1 u1 \u2013 m2 u2 W = (m1u1 \u2013 m2u2) \u2013 (m1 \u2013 m2) h2 T1 \u2013 m2 cv T2) \u2013 (m1 \u2013 m2) \u00b7 cp \u00b7 T2 Taking cv = 0.718 kJ/kg. K and cP = 1.005 kJ/kg \u00b7 K = (m1 cv W = {(11.68 \u00d7 0.718 \u00d7 313) \u2013 (0.923 \u00d7 0.718 \u00d7 113.22)} This is the maximum work that can be had from the emptying of compressed air bottle between \u2013 {(11.68 \u2013 0.923) \u00d7 1.005 \u00d7 113.22} W = 1325.86 kJ given pressure limits. Turbogenerator\u2019s actual output = 5 kJ/s Input to turbogenerator = 5 0.6 = 8.33 kJ/s. Time duration for which turbogenerator can be run; \u2206t = 1325.86 8.33 \u2206t = 159.17 sec. Duration \u2248 160 seconds Ans. 22. 3 kg of air at 1.5 bar pressure and 77\u00b0C temperature at state 1 is compressed polytropically to state 2 at pressure 7.5 bar, index of compression being 1.2. It is then cooled at constant temperature to its [U.P.S.C. 1992] original state 1. Find the net work done and heat transferred. Solution: Different states as described in the problem are denoted as 1, 2 and 3 and shown on p-V diagram. Process 1-2 is polytropic process with index 1.2 So, or T2 = T1 \u2212 \uf8f6 \uf8f7 \uf8f8 92 _________________________________________________________ Applied Thermodynamics P = Constant 3 2 1.2 PV = Constant 7.5 bar P 1.5 bar PV = Constant 1 V Fig. 3.42 At state 1, or, For process 1-2, or, \u2212 1.2 1 1.2 \uf8f6 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ed = 350. 7.5 1.5 T2 = 457.68 K P1V1 = mRT1 \u00d7 5 1.5 10 3 10 \u00b7 V1 = 3 \u00d7 0.287 \u00d7 350 V1 = 2.009 \u2248 2.01 m3 1.2 2V = 1.2 PV 1 1 P 2, V2 = 1.2 \u00d7 \uf8f1 \uf8f4 1.5 (2.01) \uf8f2 \u00d7 \uf8f4 7.5 10 \uf8f3 5 \u00d7 5 10 1 \uf8fc \uf8f4 1.2 \uf8fd \uf8f4 \uf8fe V2 = 0.526 m3 gets modified as, 3 Process 2-3 is constant pressure process, so \u21d2 V3 = \u00b7V T 2 3 T 2 or, Here process 3-1 is isothermal process, so T1 = T3 \u00d7 0.526 350 V3 = 457.68 V3 = 0.402 m3 or, During process 1-2 the compression work; W1\u20132 = =. m R T 1 \u2212.( 1 \u2212 T 2 ) n \u00d7 3 0.287(457.68 350) \u2212 (1 1.2) \u2212 W1\u20132 = \u2013 463.56 kJ First Law of Thermodynamics _____________________________________________________ 93 Work during process 2-3, Work during process 3-1, W2\u20133 = P2 (V3 \u2013 V2) = 7.5 \u00d7 105 (0.402 \u2013 0.526", ") = \u2013 93 kJ W3\u20131 = P3V3 ln \uf8eb \uf8ec \uf8ed V 1 V 3 \uf8f6 \uf8f7 \uf8f8 = 7.5 \u00d7 105 \u00d7 0.402 \u00d7 ln \uf8eb \uf8ec \uf8ed 2.01 0.402 \uf8f6 \uf8f7 \uf8f8 Net work, W3\u20131 = 485.25 kJ Wnet = W1\u20132 + W2\u20133 + W3\u20131 = \u2013 463.56 \u2013 93 + 485.25 Network = \u2013 71.31 kJ Ans. \u2013ve work shows work done upon the system. Since it is the cycle, so Wnet = Qnet \u03c6 dW = \u03c6 dQ = \u2013 71.31 kJ Heat transferred from system = 71.31 kJ Ans. 23. A compressed air bottle of volume 0.15 m3 contains air at 40 bar and 27\u00b0C. It is used to drive a turbine which exhausts to atmosphere at 1 bar. If the pressure in the bottle is allowed to fall to 2 bar, [U.P.S.C. 1998] determine the amount of work that could be delivered by the turbine. Solution: = 0.718 kJ/kg K, \u03b3 = 1.4 Initial mass of air in bottle \u21d2 m1 = Final mass of air in bottle \u21d2 m2 = cp = 1.005 kJ/kg. K, cv \u00d7 \u00d7 0.287 300 \u00d7 40 10 2 0.15 = p V 1 1 RT 1 m1 = 6.97 kg p V 2 2 RT, m2 = 2 \u00d7 \u00d7 2 10 0.15 \u00d7 0.287 127.36 \u2212 1.4 1 1.4 = \uf8f6 \uf8f7 \uf8f8 \uf8eb 2 \uf8ec \uf8ed 40 T2 = 127.36 K, m2 = 0.821 kg. Energy available for running of turbine due to emptying of bottle, = (m1 cv = {(6.97 \u00d7 0.718 \u00d7 300) \u2013 (0.821 \u00d7 0.718 \u00d7 127.36)} T2) \u2013 (m1 \u2013 m2) \u00b7 cp \u00b7 T2 T1 \u2013 m2 cv \u2013 {(6.97 \u2013 0.821) \u00d7 1.005 \u00d7 127.35} = 639.27 kJ. Work available from turbine = 639.27 kJ Ans. 94 _________________________________________________________ Applied Thermodynamics 3.1 Define the first law of thermodynamics. Also give supporting mathematical expression for it. 3.2 How the first law of thermodynamics is applied to a closed system undergoing a non-cyclic process? 3.3 Show that internal energy is a property. 3.4 Explain the following : (a) Free expansion (b) Polytropic process (c) Hyperbolic process Also obtain expressions for work in each case. 3.5 Show that for a polytropic process. \uf8f6\u2212 \uf8eb n\u03b3 \uf8f7 \uf8ec \u03b3 \u2212\uf8ed 1 \uf8f8 W Q = where Q and W are heat and work interactions and n is polytropic index. 3.6 Derive the steady flow energy equation. 3.7 Explain a unsteady flow process. 3.8 Show that for a given quantity of air supplied with a definite amount of heat at constant volume, the rise in pressure shall be directly proportional to initial absolute pressure and inversely proportional to initial absolute temperature. undergoes an increase in pressure upto 4.13 bar in a closed vessel? 3.9 How much work is done when 0.566 m3 of air initially at a pressure of 1.0335 bar and temperature of 7\u00b0C [0] 3.10 An ideal gas and a steel block are initially having same volumes at same temperature and pressure. Pressure on both is increased isothermally to five times of its initial value. Show with the help of P\u2013V diagram, whether the quantities of work shall be same in two processes or different. If different then which one is greater. Assume processes to be quasi-static. and delivering 25 kWh of mechanical work. Is this engine possible? 3.11 An inventor has developed an engine getting 1055 MJ from fuel and rejecting 26.375 MJ in exhaust [No] 3.12 For an ideal gas the pressure is increased isothermally to \u2018n\u2019 times its initial value. How high would the gas be raised if the same amount of work were done in lifting it? Assume process to be quasi-static. 3.13 A system\u2019s state changes from a to b as shown on P\u2013V diagram c a P b d V Fig. 3.43 Along path \u2018acb\u2019 84.4 kJ of heat flows into the system and system does 31.", "65 kJ of work. Determine heat flow into the system along path \u2018adb\u2019 if work done is 10.55 kJ. When system returns from \u2018b\u2019 to \u2018a\u2019 following the curved path then work done on system is 21.1 kJ. How much heat is absorbed or rejected? If internal energy at \u2018a\u2019 and \u2018d\u2019 are 0 and 42.2 kJ, find the heat absorbed in processes \u2018ad\u2019 and [63.3 kJ, \u2013 73.85 kJ, 52.75 kJ, 10.55 kJ] \u2018db\u2019. 3.14 A tank contains 2.26 m3 of air at a pressure of 24.12 bar. If air is cooled until its pressure and temperature becomes 13.78 bar and 21.1\u00b0C respectively. Determine the decrease of internal energy. [\u2013 5857.36 kJ] First Law of Thermodynamics _____________________________________________________ 95 3.15 Water in a rigid, insulating tank is set in rotation and left. Water comes to rest after some time due to viscous forces. Considering the tank and water to constitute the system answer the following. (i) Is any work done during the process of water coming to rest? (ii) Is there a flow of heat? (iii) Is there any change in internal energy (U)? (iv) Is there any change in total energy (E)? [No, No, Yes, No] 3.16 Fuel-air mixture in a rigid insulated tank is burnt by a spark inside causing increase in both temperature and pressure. Considering the heat energy added by spark to be negligible, answer the following : (i) Is there a flow of heat into the system? (ii) Is there any work done by the system? (iii) Is there any change in internal energy (U) of system? (iv) Is there any change in total energy (E) of system? done by hydrogen if the atmospheric pressure is 101.325 kPa. volume changes from 6 to 4 m3. Here pressure \u2018p\u2019 is in Pa and volume \u2018V\u2019 is in m3. [No, No, No, No] 3.17 Calculate the work if in a closed system the pressure changes as per relation p = 300. V + 1000 and [\u2013 5000J] 3.18 Hydrogen from cylinder is used for inflating a balloon to a volume of 35m3 slowly. Determine the work [3.55 MJ] 3.19 Show that the work done by an ideal gas is mRT1, if gas is heated from initial temperature T1 to twice of initial temperature at constant volume and subsequently cooled isobarically to initial state. 3.20 Derive expression for work done by the gas in following system. Piston-cylinder device shown has a gas initially at pressure and volume given by P1, V1. Initially the spring does not exert any force on piston. Upon heating the gas, its volume gets doubled and pressure becomes P2. Fig. 3.44 Piston-cylinder arrangement 3.21 An air compressor with pressure ratio of 5, compresses air to 1 4 th of the initial volume. For inlet temperature to be 27\u00b0C determine temperature at exit and increase in internal energy per kg of air. [101.83\u00b0C, 53.7 kJ/kg] 3.22 In a compressor the air enters at 27\u00b0C and 1 atm and leaves at 227\u00b0C and 1 MPa. Determine the work done per unit mass of air assuming velocities at entry and exit to be negligible. Also determine the additional work required, if velocities are 10 m/s and 50 m/s at inlet and exit respectively. [200.9 kJ/kg, 202.1 kJ/kg] 3.23 Turbojet engine flies with velocity of 270 m/s at the altitude where ambient temperature is \u201315\u00b0C. Gas temperature at nozzle exit is 873 K and fuel air ratio is 0.019. Corresponding enthalpy values for air and gas at inlet and exit are 260 kJ/kg and 912 kJ/kg respectively. Combustion efficiency is 95% and calorific value of fuel is 44.5 MJ/kg. For the heat losses from engine amounting to 21 kJ/kg of air [613.27 m/s] determine the velocity of gas jet at exit. 3.24 Oxygen at 3MPa and 300\u00b0C flowing through a pipe line is tapped out to fill an empty insulated rigid tank. Filling continues till the pressure equilibrium is not attained. What shall be the temperature of the oxygen inside the tank? If \u03b3 = 1.39. [662.5\u00b0C] 3.25 Determine work done by fluid in the thermodynamic cycle comprising of following processes :", " (a) Unit mass of fluid at 20 atm and 0.04 m3 is expanded by the law PV1.5 = constant, till volume gets doubled. (b) Fluid is cooled isobarically to its original volume. (c) Heat is added to fluid till its pressure reaches to its original pressure, isochorically. [18.8 kJ] 96 _________________________________________________________ Applied Thermodynamics 3.26 An air vessel has capacity of 10 m3 and has air at 10 atm and 27\u00b0C. Some leakage in the vessel causes air pressure to drop sharply to 5 atm till leak is repaired. Assuming process to be of reversible adiabatic [45.95 kg] type determine the mass of air leaked. 3.27 Atmospheric air leaks into a cylinder having vacuum. Determine the final temperature in cylinder when inside pressure equals to atmospheric pressure, assuming no heat transferred to or from air in cylinder. [144.3\u00b0C] 3.28 Determine the power available from a steam turbine with following details; Steam flow rate = 1 kg/s Velocity at inlet and exit = 100 m/s and 150 m/s Enthalpy at inlet and exit = 2900 kJ/kg, 1600 kJ/kg Change in potential energy may be assumed negligible. [1293.75 kW] 3.29 Determine the heat transfer in emptying of a rigid tank of 1m3 volume containing air at 3 bar and 27\u00b0C initially. Air is allowed to escape slowly by opening a valve until the pressure in tank drops to 1 bar [76.86 kJ] pressure. Consider escape of air in tank to follow polytropic process with index n = 1.2 3.30 A pump is used for pumping water from lake at height of 100 m consuming power of 60 kW. Inlet pipe and exit pipe diameters are 150 mm and 180 mm respectively. The atmospheric temperature is 293 K. Determine the temperature of water at exit of pipe. Take specific heat of water as 4.18 kJ/kg.K [293.05K] 3.31 Air at 8 bar, 100\u00b0C flows in a duct of 15 cm diameter at rate of 150 kg/min. It is then throttled by a valve upto 4 bar pressure. Determine the velocity of air after throttling and also show that enthalpy remains [37.8 m/s] constant before and after throttling. 3.32 Determine the power required by a compressor designed to compress atmospheric air (at 1 bar, 20\u00b0C) to 10 bar pressure. Air enters compressor through inlet area of 90cm2 with velocity of 50 m/s and leaves with velocity of 120 m/s from exit area of 5 cm2. Consider heat losses to environment to be 10% [50.4 kW] of power input to compressor. 4 4.1 INTRODUCTION Earlier discussions in article 3.11 throw some light on the limitations of first law of thermodynamics. A few situations have been explained where first law of thermodynamics fails to mathematically explain non-occurrence of certain processes, direction of process etc. Therefore, need was felt to have some more law of thermodynamics to handle such complex situations. Second law came up as embodiment of real happenings while retaining the basic nature of first law of thermodynamics. Feasibility of process, direction of process and grades of energy such as low and high are the potential answers provided by IInd law. Second law of thermodynamics is capable of indicating the maximum possible efficiencies of heat engines, coefficient of performance of heat pumps and refrigerators, defining a temperature scale independent of physical properties etc. 4.2 HEAT RESERVOIR Heat reservoir is the system having very large heat capacity i.e. it is a body capable of absorbing or rejecting finite amount of energy without any appreciable change in its\u2019 temperature. Thus in general it may be considered as a system in which any amount of energy may be dumped or extracted out and there shall be no change in its temperature. Such as atmosphere to which large amount of heat can be rejected without measurable change in its temperature. Large river, sea etc. can also be considered as reservoir, as dumping of heat to it shall not cause appreciable change in temperature. Heat reservoirs can be of two types depending upon nature of heat interaction i.e. heat rejection or heat absorption from it. Heat reservoir which rejects heat from it is called source. While the heat reservoir which absorbs heat is called sink. Some times these heat reservoirs may also be called Thermal Energy Reservoirs (TER). 4.3 HEAT ENGINE Heat engine is a device used for converting heat into work as it has been seen from nature that conversion from work to heat may take place easily but the vice-versa is not simple to be realized. Heat and work have been categorized as two forms of energy of low grade and high grade type. Conversion of high grade of energy to low grade", " of energy may be complete (100%), and can occur directly whereas complete conversion of low grade of energy into high grade of energy is not possible. For converting low grade of energy (heat) into high grade of energy (work) some device called heat engine is required. Thus, heat engine may be precisely defined as \u201ca device operating in cycle between high temperature source and low temperature sink and producing work\u201d. Heat engine receives heat from source, transforms 98 _________________________________________________________ Applied Thermodynamics some portion of heat into work and rejects balance heat to sink. All the processes occurring in heat engine constitute cycle. T1, Source Q1 HE W Q(= \u20131 Q2) Q2 T2, Sink Fig. 4.1 Heat engine Block diagram representation of a heat engine is shown above. A practical arrangement used in gas turbine plant is also shown for understanding the physical singnificance of heat engine. 1 WC C 4 Thigh Qadd Heat exchanger 1, Source 2 WT T G 3 C T G : Compressor : Turbine : Generator Qrejected Tlow Heat exchanger 2, Sink Fig. 4.2 Closed cycle gas turbine power plant Gas turbine installation shows that heat is added to working fluid from 1\u20132 in a \u2018heat exchanger 1\u2019 and may be treated as heat supply by source. Working fluid is expanded in turbine from 2\u20133 and produces positive work. After expansion fluid goes to the \u2018heat exchanger 2\u2019 where it rejects heat from it like heat rejection in sink. Fluid at state 4 is sent to compressor for being compressed to state 1. Work required for compression is quite small as compared to positive work available in turbine and is supplied by turbine itself. Therefore, heat engine model for it shall be as follows, Thigh Source Q = Q 1 add HE \u2013W WT ( C) = W Q rejected 2= Q Tlow Sink Fig. 4.3 Heat engine representation for gas turbine plant Efficiency of heat engine can be given by the ratio of net work and heat supplied. \u03b7 heat engine = Net work Heat supplied = W Q 1 Second Law of Thermodynamics ___________________________________________________ 99 For gas turbine plant shown and Also since it is operating in cycle, so; W = WT \u2013 WC Q1 = Qadd therefore, efficiency of heat engine can be given as; WT \u2013 WC = Qadd \u2013 Qrejected \u03b7 heat engine = \u2212 W W T C Q add Q add \u2212 Q rejected Q add = \u03b7 heat engine \u2212 1 = Q rejected Q add 4.4 HEAT PUMP AND REFRIGERATOR Heat pump refers to a device used for extracting heat from a low temperature surroundings and sending it to high temperature body, while operating in a cycle. In other words heat pump maintains a body or system at temperature higher than temperature of surroundings, while operating in cycle. Block diagram representation for a heat pump is given below: B o dy at pu m p Q 2 Low tem p. s urrou ndings T 2 Fig. 4.4 Heat pump As heat pump transfers heat from low temperature to high temperature, which is non spontaneous process, so external work is required for realizing such heat transfer. Heat pump shown picks up heat Q2 at temperature T2 and rejects heat Q1for maintaining high temperature body at temperature T1. For causing this heat transfer heat pump is supplied with work W as shown. As heat pump is not a work producing machine and also its objective is to maintain a body at higher temperature, so its performance can\u2019t be defined using efficiency as in case of heat engine. Performance of heat pump is quantified through a parameter called coefficient of performance (C.O.P). Coefficient of performance is defined by the ratio of desired effect and net work done for getting the desired effect. C.O.P. = Desired effect Net work done For heat pump : Net work = W Desired effect = heat transferred Q1 to high temperature body at temperature, T1. 100 _________________________________________________________ Applied Thermodynamics also so (COP)HP = 1Q W W = Q1 \u2013 Q2 (COP) = HP Q 1 Q Q\u2212 1 2 Refrigerator is a device similar to heat pump but with reverse objective. It maintains a body at temperature lower than that of surroundings while operating in a cycle. Block diagram representation of refrigerator is shown in Fig 4.5. Refrigerator also performs a non spontaneous process of extracting heat from low temperature body for maintaining it cool, therefore external work W is to be done for realizing it. Block diagram shows how refrigerator extracts heat Q2 for maintaining body at low temperature T2 at the expense of work W and rejects heat to high temperature surroundings. T1 High temp. surroundings Q1 T < T2 1 R W R : Refrigerator Q2 Body, T2 Fig. 4.5 Refrigerator Performance of refrigerator is also quantified by coefficient of performance, which could be defined as:", " Here or (COP)refrigerator = Desired effect Net work = 2Q W W = Q1 \u2013 Q2 (COP)refrigerator = Q 2 Q Q\u2212 1 2 COP values of heat pump and refrigerator can be interrelated as: (COP)HP = (COP)refrigerator + 1 4.5 STATEMENTS FOR IIND LAW OF THERMODYNAMICS Rudolph Julius Emmanuel Clausius, a German physicist presented a first general statement of second law of thermodynamics in 1850 after studying the work of Sadi Carnot. It was termed as Clausius statement of second law. Lord Kelvin and Max Planck also came up with another statement of second law which was termed as Kelvin-Planck statement for second law of thermodynamics. Thus, there are two statements of second law of thermodynamics, (although they are equivalent as explained ahead). Clausius statement of second law of thermodynamics: \u201cIt is impossible to have a device that while operating in a cycle produces no effect other than transfer of heat from a body at low temperature to a body at higher temperature.\u201d Above statement clearly indicates that if a non spontaneous process such as transferring heat from low temperature body to high temperature body is to be realized then some other effects such as Second Law of Thermodynamics ___________________________________________________ 101 external work requirement is bound to be there. As already seen in case of refrigerator the external work is required for extracting heat from low temperature body and rejecting it to high temperature body. Kelvin-Planck statement of second law of thermodynamics: \u201cIt is impossible for a device operating in a cycle to produce net work while exchanging heat with bodies at single fixed temperature\u201d. It says that in order to get net work from a device operating in cycle (i.e. heat engine) it must have heat interaction at two different temperatures or with body/reservoirs at different temperatures (i.e. source and sink). Thus, above two statements are referring to feasible operation of heat pump/refrigerator and heat engine respectively. Devices based on violation of IInd law of thermodynamics are called Perpetual motion machines of 2nd kind (PMM-II). Fig 4.6 shows such PMM-II. Source, T1 Q1 High temp. body T1 2= Q Q1 HE W Q = 1 HP W = 0 Q2 = 0 Sink, T2 ( )a Q2 Low temp. reservoir T2 ( )b Fig. 4.6 Perpetual Motion Machine of IInd kind PMM-II shown in Fig. 4.6a, refers to a heat engine which produces work while interacting with only one reservoir. PMM-II shown in Fig. 4.6b, refers to the heat pump which transfers heat from low temperature to high temperature body without spending work. 4.6 EQUIVALENCE OF KELVIN-PLANCK AND CLAUSIUS STATEMENTS OF IIND LAW OF THERMODYNAMICS Kelvin-Planck and Clausius statements of IInd law of thermodynamics are actually two different interpretations of the same basic fact. Here the equivalence of two statements has been shown. For establishing equivalence following statements may be proved. (a) System based on violation of Kelvin-Planck statement leads to violation of Clausius statement. (b) System based on violation of Clausius statement leads to violation of Kelvin-Planck statement. The exaplanation for equivalence based on above two is explained ahead. (a) Let us assume a heat engine producing net work while exchanging heat with only one reservoir at temperature T1, thus based on violation of Kelvin Planck statement. Let us also have a perfect heat pump operating between two reservoirs at temperatures T1 and T2. Work requirement of heat pump may be met from the work available from heat engine. Layout shown ahead explains the proposed arrangement. Source, T1 Q1 HE T > T1 2 W Q3 W HP Q2 Sink, T2 Fig. 4.7 System based on violation of Kelvin Planck statement 102 _________________________________________________________ Applied Thermodynamics If heat pump takes input work from output of heat engine then, Q3 = Q2 + W W = Q1 Q3 = Q1 + Q2 and or Combination of heat engine and heat pump shall thus result in an equivalent system working as heat pump transferring heat from low temperature T2 to high temperature T1 without expense of any external work. This heat pump is based on violation of Clausius statement and therefore not possible. Hence, it shows that violation of Kelvin Planck statement leads to violation of Clausius statement. Source, T1 Q3 HP due to HE and HP together Q2 Sink, T2 Fig. 4.8 Equivalent system (b) Let us assume a heat pump which operating in cycle transfers heat from low temperature reservoir to high temperature reservoir without expense of any work, thus based on violation", " of Clausius statement. Source, T1 = Q Q 2 1 Q = Q3 1 HP W = 0 T > T1 2 HE W = Q \u2013 Q3 4 Q2 Q2 Sink, T2 Fig. 4.9 System based on violation of Clausius statement Heat pump transfers heat Q1 to high temperature reservoir while extracting heat Q2 from low temperature reservoir. Mathematically, as no work is done on pump, so Q2 = Q1 Let us also have a heat engine between same temperature limits of T1 and T2 and produce net work W. Heat engine receives heat Q3 from source which may be taken equal to Q1. Let us now devise for heat rejected from heat pump be given directly to heat engine. In such a situation the combination of heat pump and heat engine results in equivalent heat engine which produces work \u2018W\u2019 while exchanging heat with only one reservoir at temperature T2. Arrangement is shown by dotted lines. This type of equivalent system is producing work as a result of only one heat interaction and thus violation of Kelvin Planck statement. Second Law of Thermodynamics ___________________________________________________ 103 Thus, it shows that violation of Clausius statement also causes violation of Kelvin Planck statement. Hence from (a) and (b) proved above it is obvious that the Clausius and Kelvin-Planck statements are equivalent. Conceptually the two statements explain the basic fact that, (i) net work can\u2019t be produced without having heat interactions taking place at two different temperatures. (ii) non spontaneous process such as heat flow from low temperature body to high temperature body is not possible without spending work. 4.7 REVERSIBLE AND IRREVERSIBLE PROCESSES Reversible processes as described in chapter 1 refer to \u201cthe thermodynamic processes occurring in the manner that states passed through are always in thermodynamic equilibrium and no dissipative effects are present.\u201d Any reversible process occurring between states 1\u20132 upon reversal, while occurring from 2\u20131 shall not leave any mark of process ever occurred as states traced back are exactly similar to those in forward direction. Reversible processes are thus very difficult to be realized and also called ideal processes. All thermodynamic processes are attempted to reach close to the reversible process in order to give best performance. Thermodynamic process which does not fulfil conditions of a reversible process are termed irreversible processes. Irreversibilities are the reasons causing process to be irreversible. Generally, the irreversibilities can be termed as internal irreversibility and external irreversibility. Internal irreversibility is there because of internal factors whereas external irreversibility is caused by external factors at the system-surrounding interface. Generic types of irreversibilities are due to; (i) Friction, (ii) Electrical resistance, (iii) Inelastic solid deformations, (iv) Free expansion (v) Heat transfer through a finite temperature difference, (vi) Non equilibrium during the process, etc. (i) Friction: Friction is invariably present in real systems. It causes irreversibility in the process as work done does not show equivalent rise in kinetic or potential energy of the system. Fraction of energy wasted due to frictional effects leads to deviation from reversible states. (ii) Electrical resistance: Electrical resistance in the system also leads to presence of dissipation effects and thus irreversibilities. Due to electric resistance dissipation of electrical work into internal energy or heat takes place. The reverse transformation from heat or internal energy to electrical work is not possible, therefore leads to irreversibility. (iii) Inelastic solid deformation: Deformation of solids, when of inelastic type is also irreversible and thus causes irreversibility in the process. If deformation occurs within elastic limits then it does not lead to irreversibility as it is of reversible type. (iv) Free expansion: Free expansion as discussed earlier in chapter 3, refers to the expansion of unresisted type such as expansion in vacuum. During this unresisted expansion the work interaction is zero and without expense of any work it is not possible to restore initial states. Thus, free expansion is irreversible. (v) Heat transfer through a finite temperature difference: Heat transfer occurs only when there exist temperature difference between bodies undergoing heat transfer. During heat transfer if heat addition is carried out in finite number of steps then after every step the new 104 _________________________________________________________ Applied Thermodynamics state shall be a non-equilibrium state. In order to have equilibrium states in between, the heat transfer process may be carried out in infinite number of steps. Thus, infinitesimal heat transfer every time causes infinitesimal temperature variation. These infinitesimal state changes shall require infinite time and process shall be of quasi-static type, therefore reversible. Heat transfer through a finite temperature difference which practically occurs is accompanied by irreversible state changes and thus makes processes irreversible. (vi) Non equilibrium during the process: Irreversibilities are introduced due to lack of thermodynamic equilibrium during the", " process. Non equilibrium may be due to mechanical inequilibrium, chemical inequilibrium, thermal inequilibrium, electrical inequilibrium etc. and irreversibility are called mechanical irreversibility, chemical irreversibility, thermal irreversibility, electrical irreversibility respectively. Factors discussed above are also causing non equilibrium during the process and therefore make process irreversible. Comparative study of reversible and irreversible processes shows the following major differences. Difference between reversible and irreversible processes Reversible process Irreversible process (i) Reversible process can not be realized (i) All practical processes occurring are in practice irreversible processes (ii) The process can be carried out in the reverse direction following the same path as followed in forward direction (iii) A reversible process leaves no trace of occurrence of process upon the system and surroundings after its' reversal. (iv) Such processes can occur in either directions without violating second law of thermodynamics. (v) A system undergoing reversible processes has maximum efficiency. So the system with reversible processes are considered as reference systems or bench marks. (vi) Reversible process occurs at infinitesimal rate i.e. quasi-static process. (vii) System remains throughout in thermodynamic equilibrium during occurrence of such process. (viii) Examples; Frictionless motion, controlled expansion and compression, Elastic deformations, Electric circuit with no resistance, Electrolysis, Polarization and magnetisation process etc. (ii) Process, when carried out in reverse direction follows the path different from that in forward direction. (iii) The evidences of process having occurred are evident even after reversal of irreversible process. (iv) Occurrence of irreversible processes in either direction is not possible, as in one direction it shall be accompanied with the violation of second law of thermodynamics. (v) System having irreversible processes do not have maximum efficiency as it is accompanied by the wastage of energy. (vi) Irreversible processes occur at finite rate. (vii) System does not remain in thermodynamic equilibrium during occurrence of irreversible processes. (viii) Examples; Viscous fluid flow, inelastic deformation and hysteresis effect, Free expansion, Electric circuit with resistance, Mixing of dissimilar gases, Throttling process etc. Second Law of Thermodynamics ___________________________________________________ 105 4.8 CARNOT CYCLE AND CARNOT ENGINE Nicholas Leonard Sadi Carnot, an engineer in French army originated use of cycle (Carnot) in thermodynamic analysis in 1824 and these concepts provided basics upon which second law of thermodynamics was stated by Clausius and others. Carnot cycle is a reversible thermodynamic cycle comprising of four reversible processes. Thermodynamic processes constituting Carnot cycle are; (i) Reversible isothermal heat addition process, (1\u20132, Qadd) (ii) Reversible adiabatic expansion process (2\u20133, Wexpn +ve) (iii) Reversible isothermal heat release process (3\u20134, Qrejected) (iv) Reversible adiabatic compression process (4\u20131, Wcompr \u2013ve) Carnot cycle is shown on P\u2013V diagram between states 1, 2, 3 4, and 1. A reciprocating piston- cylinder assembly is also shown below P\u2013V diagram. Process 1 \u20132 is isothermal heat addition process of reversible type in which heat is transferred to system isothermally. In the piston cylinder arrangement heat Qadd can be transferred to gas from a constant temperature source T1 through a cylinder head of conductor type. First law of thermodynamics applied on 1\u20132 yields; Qadd = U2 \u2013 U1 + W1\u20132 1 P Reversible adiabatics Qadd Wcompr. 4 2 Wexpn Qrejected V Reversible isothermals 3 Cylinder head of insulated type System Piston Insulated cylinder Cylinder head of conducting type Fig. 4.10 Carnot cycle 106 _________________________________________________________ Applied Thermodynamics Qadd Heat exchanger 1 T1 1 WCompr C 4 2 3 Wexpn T C T : Compressor : Turbine Qrejected T3 Heat exchanger 2 Fig. 4.11 Gas turbine plant: Carnot heat engine For the perfect gas as working fluid in isothermal process no change in internal energy occurs, therfore U2 = U1 and Qadd = W1\u20132 Process 2\u20133 is reversible adiabatic expansion process which may be had inside cylinder with cylinder head being replaced by insulating type cylinder head so that complete arrangement is insulated and adiabatic expansion carried out. During adiabatic expansion say work Wexpn is available, From first law of thermodynamics; Q2\u20133 = 0 or Process 3\u20134 is reversible isothermal heat rejection for which cylinder head of insulating type may be 0 = (U3 \u2013 U2) + Wexpn Wexpn = (U2 \u2013 U3) replaced by conducting type as", " in 1\u20132 and heat (Qrejected) be extracted out isothermally. From first law of thermodynamics applied on process 3\u20134, \u2013Qrejected = (U4 \u2013 U3) + (\u2013W3\u20134) for perfect gas internal energy shall remain constant during isothermal process. Thus, U3 = U4 or \u2013Qrejected = \u2013W3\u20134 Qrejected = W3\u20134 Process 4\u20131 is the reversible adiabatic compression process with work requirement for compression. In the piston cylinder arrangement cylinder head of conducting type as used in 3\u20134 is replaced by insulating type, so that the whole arrangement becomes insulated and adiabatic compression may be realized, From first law applied on process 4\u20131 Q4\u20131 = 0 For adiabatic process; \u21d2 0 = (U1 \u2013 U4) + (\u2013Wcompr) = W compr \u2212 ( U U 1 4 ) or Efficiency of reversible heat engine can be given as; \u03b7 Net work Heat supplied Here, Net work = Wexpn \u2013 Wcompr rev, HE = Second Law of Thermodynamics ___________________________________________________ 107 and heat is supplied only during process 1\u20132, therefore Substituting in the expression for efficiency. heat supplied = Qadd Also for a cycle so Hence \u03b7 rev, HE = W expn \u2212 W compr Q add W\u2211 = Q\u2211 cycle cycle Wnet = Qadd \u2013 Qrejected \u03b7 rev, HE \u2212 1 = Q rejected Q add As the heat addition takes place at high temperature, while heat rejection takes place at low temperature, so writing these heat interactions as Qhigh, Qlow we get, \u03b7 rev, HE = \u2212 1 \u03b7 Carnot = \u2212 1 Q Q low high Q Q low high Piston-cylinder arrangement shown and discussed for realizing Carnot cycle is not practically feasible as; (i) Frequent change of cylinder head i.e. of insulating type and diathermic type for adiabatic and isothermal processes is very difficult. (ii) Isothermal heat addition and isothermal-heat rejection are practically very difficult to be realized (iii) Reversible adiabatic expansion and compression are not possible. (iv) Even if near reversible isothermal heat addition and rejection is to be achieved then time duration for heat interaction should be very large i.e. infinitesimal heat interaction occurring at dead slow speed. Near reversible adiabatic processes can be achieved by making them to occur fast. In a piston-cylinder reciprocating engine arrangement such speed fluctuation in a single cycle is not possible. Carnot heat engine arrangement is also shown with turbine, compressor and heat exchangers for adiabatic and isothermal processes. Fluid is compressed in compressor adiabatically, heated in heat exchanger at temperature T1, expanded in turbine adiabatically, cooled in heat exchanger at temperature T3 and sent to compressor for compression. Here also following practical difficulties are confronted; (i) Reversible isothermal heat addition and rejection are not possible. (ii) Reversible adiabatic expansion and compression are not possible. Carnot cycle can also operate reversibly as all processes constituting it are of reversible type. Reversed Carnot cycle is shown below; 108 _________________________________________________________ Applied Thermodynamics 1 Qadd Wcompr. P 4 2 T1 = constant Wexpn 3 T3 = constant Qrejected V Heat engine cycle in reversed form as shown above is used as ideal cycle for refrigeration and called \u201cCarnot refrigeration cycle\u201d. Fig. 4.12 Reversed Carnot cycle 4.9 CARNOT THEOREM AND ITS COROLLARIES Carnot theorem states that \u201cany engine cannot have efficiency more than that of reversible engine operating between same temperature limits.\u201d Different corollaries of Carnot theorem are, (i) Efficiency of all reversible engines operating between same temperature limits is same. (ii) Efficiency of a reversible engine does not depend on the working fluid in the cycle. Using Clausius and Kelvin Planck statements, the Carnot theorem can be proved easily. Let us take two heat engines HEI and HEII operating between same temperature limits T1, T2 of source and sink as shown in Fig. 4.13a. Source, T1 Q1, I W I HEI Q2, I Source, T1 Source, T1 Q1, II Q1, I Q1, II Q 1, I Q = Q 1, II 1, I HEII W II W I W II HEI HEII W I HEI W II HEII Q2, II Q2, I ) \u2013 (W W I II Q2, II Q2, I \u2013 (W W I II ) Q2, II Sink, T2 ( )a Sink, T2 ( )b Fig", ". 4.13 Proof of Carnots theorem Sink, T2 ( )c Arrangement shown has heat engine, HE1 getting Q1,I from source, rejecting Q2, I and producing work WI. Heat engine, HEII receives Q1,II, rejects Q2,II and produces work WII. Efficiency of engines HEI, HEII WI = Q1, I \u2013 Q2, I WII = Q1, II \u2013 Q2, II \u03b7 IHE = W I Q 1,I Second Law of Thermodynamics ___________________________________________________ 109 \u03b7 IIHE = W II Q 1,II Now let us assume that engine HE1 is reversible engine while HEII is any engine. As per Carnot\u2019s theorem efficiency of HEI (reversible engine) is always more than that of HEII. Let us start with violation of above statement, i.e., efficiency of HEII is more than that of HEI \u03b7 \u03b7 < IHE IIHE or W I Q 1,I < W II Q 1,II Let us take the heat addition to each engine to be same i.e. Hence as shown in Figure 4.13b. Q1, I = Q1, II WI < WII Also we have assumed that engine HE1 is of reversible type, so let us operate it in reversed manner, Let us also assume that the work requirement of reversed heat engine, HE1 be fed by work output WII of the heat engine HEII. Since WII is more than W1, a net work (WII \u2013 WI) shall also be available as output work after driving HE1. Also since Q1, I and Q1, II are assumed to be same, the heat rejected by reversed HE1 may be supplied to heat engine, HEII as shown in figure 4.13c by dotted lines. Thus, it results into an equivalent heat engine which produces net work (WII \u2013 WI) while heat interaction takes place with only one reservoir at temperature, T2. This is a violation of Kelvin Planck statement, so the assumption made in beginning that efficiency of reversible engine is less than that of other engine, is not correct. Hence, it is established that out of all heat engines operating within same temperature limits, the reversible engine has highest efficiency. Similarly for showing the correctness of corollaries of Carnot theorem the heat engines and their combinations be considered like above and proved using Kelvin-Planck and Clausius statements. 4.10 THERMODYNAMIC TEMPERATURE SCALE After the Carnot's theorem and its corollary were stated and verified, it was thought to have a thermodynamic temperature scale, independent of thermometric substance and principles of thermometry. Such a temperature scale can be developed with the help of reversible heat engine concept and is called thermodynamic temperature scale. Defining thermodynamic temperature scale refers to the assigning of numerical values to different temperatures using reversible heat engines. From the previous discussions on heat engines it is obvious that the efficiency of a reversible heat engine depends on the temperatures of reservoir with which heat interaction takes place. Mathematically, it can be easily given by any unknown function \u2018\u0192\u2019; where Thigh and Tlow are the two temperatures of high temperature source and low temperature sink. \u03b7 rev, HE = \u0192 (Thigh, Tlow) or \u03b7 rev, HE = 1 \u2013 Q Q low high = \u0192 (Thigh, Tlow) Unknown function \u2018\u0192\u2019 may be substituted by another unknown function, say \u03c6 110 _________________________________________________________ Applied Thermodynamics Source, Thigh Qhigh HE W Qlow Sink, Tlow (a) Source, T1 Q1 HE1 W1 Q2 T2, Temperature Hypothetical reservoir HE2 W2 Q3 Sink, T3 (b) Fig. 4.14 Reversible heat engine and its combinations Q high Q low = \u03c6 (Thigh, Tlow) Thus, some functional relationship as defined by \u2018\u03c6\u2019 is established between heat interactions and temperatures. Let us now have more than one reversible heat engines operating in series as shown in Figure 4.14b, between source and sink having T1 and T3 temperatures. In between an imaginary reservoir at temperature T2 may be considered. From the above for two reversible heat engines; Q 1 Q = \u03c6 (T1, T2) 2 Q 2 Q = \u03c6 (T2, T3) 3 Combination of two heat engines may be given as shown here, Source, T1 Q1 HE W Q3 Sink, T3 Fig. 4.15 Equivalent heat engine for two reversible heat engines operating in seriesT1, T3) 3 and or Second Law of Thermodynamics ___________________________________________________ 111 T1, T3) = \u03c6 (T1, T2). \u03c6 (T2, T3)", " Above functional relation is possible only if it is given by another function \u03c8 as follows. Q 1 Q = 3 or Thus, \u03c6 (T1, T2) = \u03c6 (T2, T3T1, T3 ) ) Lord Kelvin based upon his observations proposed that the function \u03c8 (T) can be arbitrarily chosen based on Kelvin scale or absolute thermodynamic temperature scale as; \u03c8 (T) = Temperature T in Kelvin Scale Therefore where T1, T2, T3 are temperatures in absolute thermodynamic scale. Here heat absorbed and heat rejected is directly proportional to temperatures of reservoirs supplying and accepting heats to heat engine. For a Carnot heat engine or reversible heat engine operating between reservoirs at temperature T and triple point of water, Tt 273.16 or = T 273.16. Q Q t 112 _________________________________________________________ Applied Thermodynamics Here for a known Q and Qt values the temperature T can be defined. Thus, heat interaction acts as thermometric property in thermodynamic temperature scale, which is independent of thermometric substance. It may be noted that negative temperatures cannot exist on thermodynamic temperature scale. Let us now have a large number of reversible heat engines (Carnot engines) operating in series as shown in Figure 4.16. Source, T1 Q1 HE1 W1 Q2 HE2 W2 Q3 HE3 W3 Q4 HE4 W4 Q5 HE5 W5 Q6 Qn HEn Wn Qn + 1 Tn + 1 Sink All temperature are in Kelvin From thermodynamic temperature scale for different engines, Fig. 4.16 Series of reversible heat engines For nth engine Second Law of Thermodynamics ___________________________________________________ 113 Q n Q n+1 = T n T n+1 Here work output from each engine shall continuously diminish the heat supplied to subsequent heat engine. Let us assume work outputs from \u2018n\u2019 engines to be same; i.e. or or W1 = W2 = W3 = W4 =... = Wn (Q1 \u2013 Q2) = (Q2 \u2013 Q3) = (Q3 \u2013 Q4) =... = (Qn \u2013 Qn + 1) (T1 \u2013 T2) = (T2 \u2013 T3) = (T3 \u2013 T4) =... = (Tn \u2013 Tn + 1) It is obvious that for a large number of heat engines the heat rejected by nth engine shall be negligible i.e for very large value of n, Qn + 1 or \u2192 0 for Lim n \u2192 \u221e, Lim Qn + 1 \u2192 0 Thus, from thermodynamic temperature scale when heat rejection approaches zero, the temperature of heat rejection also tends to zero as a limiting case. But in such a situation when heat rejection is zero, the heat engine takes form of a perpetual motion machine of 2nd kind, where work is produced with only heat supplied to it. Thus, it leads to violation of Kelvin-Planck statement. Hence it is not possible. Also it can be said that \u201cit is impossible to attain absolute zero temperature in finite number of operations.\u201d There exists absolute zero temperature on thermodynamic temperature scale, but cannot be attained without violation of second law of thermodynamics. This fact is popularly explained by third law of thermodynamics. Carnot cycle efficiency can now be precisely defined as function of source and sink temperatures. \u03b7 carnot = 1 \u2013 Q Q low high \u03b7 carnot = 1 \u2013 T low T high Thus, it is seen that Carnot cycle efficiency depends only upon lower and higher temperatures. Carnot cycle efficiency is high for small values of sink temperature (Tlow) and larger values of source temperature (Thigh). Therefore for maximum efficiency, Carnot cycle must operate between maximum possible source and minimum possible sink temperatures. EXAMPLES 1. Using IInd law of thermodynamics show that the following are irreversible (i) Free expansion. (ii) Heat transfer through finite temperature difference. Solution: (i) Let us consider a perfectly insulated tank having two compartments divided by thin wall. Compartment I has gas while II has vacuum. When wall is punctured then gas in I expands till pressure in I and II gets equalised. Let us assume that free expansion is reversible i.e. the gas in II returns into I and original states are restored. When gas is allowed to expand, say it produces work W from a device D due to expansion. This work W is available due to change in internal energy of gas. Internal energy of gas can be restored by adding equivalent heat Q to it from a source as shown. This whole arrangement if consolidated can be 114 _________________________________________________________ Applied Thermodynamics treated as a device which is producing work while exchanging heat with single body. Thus, it is violation of IInd law of thermodynamics, therefore the assumption that free expansion is reversible is incorrect. Free expansion is irreversible. Gas I Vacuum II I II Q Source", " D W Fig. 4.17 Free expansion (ii) For showing that the heat transfer through finite temperature difference is irreversible, let us start with the fact that such heat transfer is reversible. Let us take a heat source (T1) and sink (T2) and assume that a heat Q1\u20132 flows from T1 to T2. Let us have a heat engine operating between T1 and T2 as shown and producing work W. Let us reverse heat transfer process from T2 to T1 i.e. Q2\u20131, as assumed. Let us assume Q2 = Q2\u20131. This assumption paves the way for eliminating sink. Let us now remove sink and directly supply Q2 as Q2\u20131 (= Q2). This results in formation of a heat engine which produces work while exchanging heat with single reservoir, the violation of IInd law of thermodynamics. (Kelvin Planck statement). Source, T1 T > T2 1 Q1 Source, T1 Q1 Q1\u20132 HE W Q2\u20131 HE W Q2 Sink, T2 Q2 Sink, T2 Fig. 4.18 Heat transfer through a finite temperature difference Hence, assumption that heat transfer through finite temperature is reversible, stands incorrect. Therefore, heat transfer through finite temperature difference is irreversible. 2. Determine the heat to be supplied to a Carnot engine operating between 400\u00baC and 15\u00baC and producing 200 kJ of work. Solution : To find out In Carnot engine from thermodynamic temperature scale; Q1 =? and work = Q1 \u2013 Q2 T1 400\u00b0C Q1 HE 200 kJ Q2 15 \u00b0C T2 Fig. 4.19 Second Law of Thermodynamics ___________________________________________________ 115 Thus and 673 288 Q 1 Q = Q1 \u2013 Q2 = 200 kJ 2 (1) (2) From equations 1 and 2, upon solving and Q1 = 349.6 kJ Q2 = 149.6 kJ Heat to be supplied = 349.6 kJ Ans. 3. A refrigerator operates on reversed Carnot cycle. Determine the power required to drive refrigerator between temperatures of 42\u00baC and 4\u00baC if heat at the rate of 2 kJ/s is extracted from the low temperature region. Solution: T 1 (273 + 42) K Q1 R W Q2 (273 + 4) T2 K Fig. 4.20 To find out, Given : and W =? T1 = 315 K, T2 = 277 K Q2 = 2 kJ/s From thermodynamic temperature scale or or = 315 Q 1 2 277 Q1 = 2.274 kJ/s Power/Work input required = Q1 \u2013 Q2 = 2.274 \u2013 2 Power required = 0.274 kJ/s Power required for driving refrigerator = 0.274 kW Ans. 4. A reversible heat engine operates between two reservoirs at 827\u00baC and 27\u00baC. Engine drives a Carnot refrigerator maintaining \u201313\u00baC and rejecting heat to reservoir at 27\u00baC. Heat input to the engine is 2000 kJ and the net work available is 300 kJ. How much heat is transferred to refrigerant and total heat rejected to reservoir at 27\u00baC? Solution: Block diagram based on the arrangement stated; 116 _________________________________________________________ Applied Thermodynamics T1, 827 \u00b0C 2000 kJ Q1 HE Q2 27 \u00b0C \u201313 \u00b0C T3 Q3 R Q4 W E W R 300 kJ T2 Low temperature reservoir Fig. 4.21 We can write, for heat engine1) (2) (3) 1100 300 Substituting Q1 = 2000 kJ, we get Q2 = 545.45 kJ Also For refrigerator, Q 1 Q = 2 WE = Q1 \u2013 Q2 = 1454.55 kJ Also, and or Equations (1) & (2) result in, From equations (1) & (3), Q 260 3 Q = 300 4 WR = Q4 \u2013 Q3 WE \u2013 WR = 300 WR = 1154.55 kJ Q4 \u2013 Q3 = 1154.55 Q3 = 7504.58 kJ Q4 = 8659.13 kJ Total heat transferred to low temperature reservoir = Q2 + Q4 = 9204.68 kJ Heat transferred to refrigerant = 7504.58 kJ \uf8fc \uf8fd \uf8fe Total heat transferred to low temperature reservoir = 9204.68 kJ Ans. 5. In a winter season when outside temperature is \u20131\u00baC, the inside of house is to be maintained at 25\u00baC. Estimate the minimum power required to run the heat pump of maintaining the temperature. Assume heating load as 125 MJ/h. Solution: Also we know COPHP = 1Q W = Q 1 Q Q\u2212 1 2 = 1 \uf8eb \u2212\uf8ec 1 \uf8ed Q", " 2 Q 1 \uf8f6 \uf8f7 \uf8f8 Second Law of Thermodynamics ___________________________________________________ 117 Thus Also therefore or, 298.15 272.15 Q 1 Q = COPHP = 11.47 2 COPHP = 1Q W, Substituting Q1 W = 10.89 MJ/h W = 3.02 kW Minimum power required = 3.02 kW Ans. 2 5\u00baC Q 1 = 12 5 M J/ h W H P Q 2 \u2013 1\u00baC Fig. 4.22 6. A cold storage plant of 40 tonnes of refrigeration capacity runs with its performance just 1 4 th of its Carnot COP. Inside temperature is \u201315\u00baC and atmospheric temperature is 35\u00baC. Determine the power [Take : One ton of refrigeration as 3.52 kW] required to run the plant. Solution: Cold storage plant can be considered as a refrigerator operating between given temperatures limits. Capacity of plant = Heat to be extracted = 140.8 kW Carnot COP of plant = ( 308 1 258.15 ) 1\u2212 = 5.18 COP = 5.18 4 = 1.295 COP = 2Q W, hence W = 108.73 kW. Power required = 108.73 kW Ans. Actual Also actual 35\u00b0C Q1 R W Q2 = 140.8 kW \u201315\u00b0C Fig. 4.23 7. What would be maximum efficiency of engine that can be had between the temperatures of 1150\u00baC and 27\u00baC? Solution: Highest efficiency is that of Carnot engine, so let us find the Carnot cycle efficiency for given temperature limits. \uf8eb \uf8ec \uf8ed \u03b7= 1 \u2013 + 273 27 + 273 1150 \u03b7= 0.7891 or 78.91% Ans. \uf8f6 \uf8f7 \uf8f8 8. A domestic refrigerator maintains temperature of \u2013 8\u00baC when the atmospheric air temperature is 27\u00baC. Assuming the leakage of 7.5 kJ/min from outside to refrigerator determine power required to run this refrigerator. Consider refrigerator as Carnot refrigerator. Solution: Here heat to be removed continuously from refrigerated space shall be 7.5 kJ/min or 0.125 kJ/s. For refrigerator, C.O.P. shall be, 118 _________________________________________________________ Applied Thermodynamics or 0.125 W 265 \u2212 (300 265) = W = 0.0165 kJ/s. Power required = 0.0165 kW Ans. 9. Three reversible engines of Carnot type are operating in series as shown between the limiting temperatures of 1100 K and 300 K. Determine the intermediate temperatures if the work output from engines is in proportion of 3 : 2 : 1. 27\u00b0C Q R W 7.5 kJ/min \u2013 8\u00b0C 7.5 kJ/min Fig. 4.23 Solution: Here, W1 : W2 : W3 = 3 : 2 : 1 Efficiency of engine, HE1, for HE2 engine, for HE3 engine, W 1 Q = 1 \uf8eb \u2212\uf8ec 1 \uf8ed T 2 1100 \uf8f6 \uf8f7 \uf8f8 \u21d2 Q1 = W 2 Q = 2 \uf8eb \u2212\uf8ec 1 \uf8ed T 3 T 2 \uf8f6 \uf8f7 \uf8f8 1100. (1100 W 1 T\u2212 2 ) W 3 Q = From energy balance on engine, HE1 3 \uf8eb \u2212\uf8ec 1 \uf8ed 300 T 3 \uf8f6 \uf8f7 \uf8f8 Above gives, Q1 = W1 + Q2 \u21d2 Q2 = Q1 \u2013 W1 1Q = \uf8f1 \uf8f2 \uf8f3 1100 W 1 \u2212 (1100 \uf8f2 \u2212 1100 \uf8f3 \uf8fc \uf8fd \uf8fe T 2 Substituting Q2 in efficiency of HE2 W 2 T 2 \u2212 1100 \uf8eb \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f8 T 2 W 1 \uf8eb \u2212\uf8ec 1 \uf8ed T 3 T 2 \uf8f6 \uf8f7 \uf8f8 = or or or \u2212 1100 T 2 \uf8f6\uf8eb \uf8f7\uf8ec \uf8f8\uf8ed 1100 \u2212 T 3 \u2212 T 2 \uf8f6 \uf8f7 \uf8f8 2200 \u2013 2T2 = 3T2 \u2013 3T3 \uf8f1 \uf8f4 2 =\uf8f2 \uf8f4 3 \uf8f3 \uf8eb \uf8ec \uf8ed T 2 1100 \u2212 \u2212 3 3 = 2200 Second Law of Thermodynamics ___________________________________________________ 119 Energy balance on engine HE2 gives, Substituting in efficiency of HE2, Q2 = W2 + Q3 or or W 2 W Q W2. T2 = (W2 + Q3) (T2 \u2013 T3) W T 2 3 T\u2212 ( T 2 3 Q3 = ) Substit", "uting Q3 in efficiency of HE3, \u2212 300 \uf8f6 \uf8eb \uf8f7 \uf8ec \uf8f8 \uf8ed \u2212 300 3T3 \u2013 T2 = 600 \u2212 300 \u2212 T 3 T2 T3 1100 K Q1 HE1 W1 Q2 HE2 W2 Q3 HE3 W3 Q4 300 K Fig. 4.25 Solving, equations of T2 and T3, T3 = 433.33 K Intermediate temperatures: 700 K and 433.33 K Ans. T2 = 700 K 10. A Carnot engine getting heat at 800 K is used to drive a Carnot refrigerator maintaining 280 K temperature. Both engine and refrigerator reject heat at some temperature, T, when heat given to engine is equal to heat absorbed by refrigerator. Determine efficiency of engine and C.O.P. of refrigerator. Solution: Efficiency of engine, For refrigerator, COP W Q 1 \uf8eb \uf8ec \uf8ed = 800 T\u2212 800 \uf8f6 \uf8f7 \uf8f8 It is given that so, from engine From refrigerator, 280 = 3Q T \u2212 ( W Q1 = Q3 = Q 280) W Q Q W \uf8eb \uf8ec \uf8ed = 800 T\u2212 800 \uf8f6 \uf8f7 \uf8f8 = 280 T \u2212 280 120 _________________________________________________________ Applied Thermodynamics 800 K 280 K Q1 HE Q2 Q3 R Q4 W T, K Fig. 4.26 From above two \uf8eb \uf8ec \uf8ed Q W \uf8f6 \uf8f7 \uf8f8 may be equated, T \u2212 280 280 800 T\u2212 800 = Temperature, T = 414.8 K Efficiency of engine = C.O.P. of refrigerator = \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed \u2212 800 414.8 800 \uf8f6 \uf8f7 \uf8f8 280 \u2212 414.8 280 \uf8f6 \uf8f7 \uf8f8 = 0.4815 Ans. = 2.077 Ans. 11. 0.5 kg of air executes a Carnot power cycle having a thermal efficiency of 50%. The heat transfer to the air during isothermal expansion is 40 kJ. At the beginning of the isothermal expansion the pressure is 7 bar and the volume is 0.12 m3. Determine the maximum and minimum temperatures for the cycle in Kelvin, the volume at the end of isothermal expansion in m3, and the work and heat transfer for each of the four processes in kJ. For air cP = 1.008 kJ/kg. K, cv= 0.721 kJ/kg. K. [U.P.S.C. 1993] Solution: Given : \u03b7 carnot = 0.5, m = 0.5 kg P2 = 7 bar, V2 = 0.12 m3 Let thermodynamic properties be denoted with respect to salient states; Carnot efficiency \u03b7 Carnot = 1 \u2013 T 1 T 2 2 40 kJ 3 T 1 4 S Fig. 4.27 Second Law of Thermodynamics ___________________________________________________ 121 or, T 1 T = 0.5 2 T2 = 2T1 or, Corresponding to state 2, P2 V2 = mRT2 7 \u00d7 105 \u00d7 0.12 = 0.5 \u00d7 287 \u00d7 T2 T2 = 585.36 K Heat transferred during process 2-3 (isothermal expansion), Q23 = 40 kJ Q23 = W23 = P2V2 ln \uf8eb \uf8ec \uf8ed.5 \u00d7 0.287 \u00d7 585.36 ln V\uf8eb 3 \uf8ec \uf8ed 0.12 \uf8f6 \uf8f7 \uf8f8 40 = mRT2 ln \u00d7 V3 = 0.1932 m3 T 2 2 T1 = 292.68 K T1 = \u2212 = pc c\u03c5 P1 = 0.613 bar 1.008 0.721 P1 V1 = mRT1, \u03b3 = 1.398 Temperature at state 1, During process 1\u20132, Thus, Heat transferred during process 4 \u2013 1 (isothermal compression) shall be equal to the heat transferred during process 2 \u2013 3 (isothermal expansion). 0.613 \u00d7 105 \u00d7 V1 = 0.5 \u00d7 287 \u00d7 292.68 V1 = 6.85 \u00d7 10\u20134 m3 For isentropic process, dQ = 0 dW = dU During process 1 \u2013 2, isentropic process, W12 = \u2013mcv (T2 \u2013 T1) Q12 = 0, W12 = \u20130.5 \u00d7 0.721 (585.36 \u2013 292.68) W12 = \u2013 105.51 kJ, (\u2013ve work) During process 3 \u2013 4, isentropic process, W34 = \u2013mcv (T4 \u2013 T", "3) Q34 = 0, W34 = + 0.5 \u00d7 0.721 \u00d7 (585.36 \u2013 292.68) W34 = + 105.51 kJ (+ve work) Ans. Process Heat transfer Work interaction 40 kJ 0 \u2013 40 kJ \u2013 105.51, kJ 40 kJ + 105.51, kJ \u2013 40 kJ 122 _________________________________________________________ Applied Thermodynamics Maximum temperature of cycle = 585.36 kJ Minimum temperature of cycle = 292.68 kJ Volume at the end of isothermal expansion = 0.1932 m3 12. A reversible engine as shown in figure during a cycle of operation draws 5 mJ from the 400 K reservoir and does 840 kJ of work. Find the amount and direction of heat interaction with other reservoirs. [U.P.S.C. 1999] 200 K Q3 300 K Q2 HE W = 840 kJ Fig. 4.28 400 K Q1 = 5 mJ Solution: Let us assume that heat engine rejects Q2 and Q3 heat to reservoir at 300 K and 200 K respectively. Let us assume that there are two heat engines operating between 400 K and 300 K temperature reservoirs 1Q\u2032\u2032 from and between 400 K and 200 K temperature reservoirs. Let each heat engine receive reservoir at 400 K as shown below: 1Q\u2032 and 400 K Q'1 HE' Q2 300 K Q\" Q' Q MJ HE' W = 840 kJ Q3 300 K Thus, also, and Fig. 4.29 Assumed arrangement 1 = Q1 = 5 \u00d7 103 kJ Q\u2032 1 + Q\u2032\u2032 \u2032 Q 1 Q 2 \u2032\u2032 Q 1 Q 3 = = 400 300 400 200, or, Q\u2032 1 = 4 3 Q2 or, Q\u2032\u2032 1= 2Q3 Substituting Q\u2032 1 and Q\u2032\u2032 1 4 3 Also from total work output, Q\u2032 Q2 + 2Q3 = 5000 1 + Q\u2032\u2032 1 \u2013 Q2 \u2013 Q3 = W 5000 \u2013 Q2 \u2013 Q3 = 840 Q2 + Q3 = 4160 Q3 = 4160 \u2013 Q2 Second Law of Thermodynamics ___________________________________________________ 123 Substituting Q3, 4 3 Q2 + 2(4160 \u2013 Q2) = 5000 4 3 Q2 \u2013 2 Q2 = 5000 \u2013 8320 = \u2013 3320 Q\u2212 22 3 Q2 = 4980 kJ Q3 = \u2013 820 kJ and Negative sign with Q3 shows that the assumed direction of heat Q3 is not correct and actually Q3 heat will flow from reservoir to engine. Actual sign of heat transfers and magnitudes are as under: 200 K 300 K 400 K Q3 = 820 kJ Q2 = 4980 kJ HE Q1 = 5 mJ W = 840 kJ Fig 4.30 Q2 = 4980 kJ, from heat engine Q3 = 820 kJ, to heat engine Ans. 13. A heat pump working on a reversed Carnot cycle takes in energy from a reservoir maintained at 3\u00baC and delivers it to another reservoir where temperature is 77\u00baC. The heat pump drives power for it's operation from a reversible engine operating within the higher and lower temperature limits of 1077\u00baC and 77\u00baC. For 100 kJ/s of energy supplied to the reservoir at 77\u00baC, estimate the energy taken from the [U.P.S.C. 1994] reservoir at 1077\u00baC. Solution: Arrangement for heat pump and heat engine operating together is shown here. Engine and pump both reject heat to the reservoir at 77\u00baC (350 K). For heat engine. \u03b7 E = 1 \u2013 350 1350 = W Q 1 0.7407 = 1 \u2212 Q Q 2 Q 1 0.7407 = 1 \u2013 Q 2 Q 1 Q2 = 0.2593 Q1 COPHP = Q 4 Q Q\u2212 4 3 77 \u00b0C or 350 K W Fig. 4.31 Q4 HP Q3 3\u00b0C or 276 K Q2 HE Q1 1077 \u00b0C or 1350 K For heat pump 124 _________________________________________________________ Applied Thermodynamics COPHP = 350 \u2212 350 276 = Q 4 Q Q\u2212 4 3 \u21d2 Work output from engine = Work input to pump Q4 = 1.27Q3 Q1 \u2013 Q2 = Q4 \u2013 Q3 \u21d2 Q1 \u2013 0.2593 Q1 = Q4 \u2013 Q 4 1.27 Also it is given that Q2 + Q4 = 100 Substituting Q2 and Q4 as function of Q1 in following expression, Q2 + Q4 = 100 0.2593 Q1 + Q 1 0.287 = 100 Q1 = 26.71 kJ Energy taken by engine from reservoir at 1077\u00baC = 26.71 kJ Ans. 14. A reversible engine is used for only driving a reversible refrigerator. Engine is supplied 2000 kJ", "/s heat from a source at 1500 K and rejects some energy to a low temperature sink. Refrigerator is desired to maintain the temperature of 15\u00baC while rejecting heat to the same low temperature sink. Determine the temperature of sink if total 3000 kJ/s heat is received by the sink. Solution: Let temperature of sink be Tsink K. Given: Since complete work output from engine is used to run refrigerator so, Qsink, HE + Qsink, R = 3000 kJ/s 2000 \u2013 Qsink, HE = Qsink, R \u2013 QR Also for engine, For refrigerator, QR = 3000 \u2013 2000 = 1000 kJ/s Source 1500 K 2000 1500 = Q sink, HE T sink \u21d2 Qsink, HE = 4 3 Tsink, 2000 kJ/s RQ 288 = Q sink,R T sink \u21d2 Qsink, R = 1000 T sink 288 Substituting Qsink, HE and Qsink, R values. 1000 4 3 Tsink + T sink 288 = 3000 15 \u00b0C or 288 K QR R Qsink, R HE Qsink, HE W Tsink Fig. 4.32 \u21d2 Temperature of sink = 351.28\u00baC Ans. Tsink = 624.28 K 15. A reversible heat engine runs between 500\u00baC and 200\u00baC temperature reservoirs. This heat engine is used to drive an auxiliary and a reversible heat pump which runs between reservoir at 200\u00baC and the body at 450\u00baC. The auxiliary consumes one third of the engine output and remaining is consumed for driving heat pump. Determine the heat rejected to the body at 450\u00baC as fraction of heat supplied by reservoir at 500\u00baC. Second Law of Thermodynamics ___________________________________________________ 125 Solution: Let the output of heat engine be W. So W 3 is consumed for driving auxiliary and remaining W 2 3 is consumed for driving heat pump for heat engine, \u03b7 = W Q = 1 \u2212 1 473 773 W Q = 0.3881 1 COP of heat pump = 723 \u2212 723 473 = Q 3 2 / 3 W \u21d2 Substituting W, 2.892 = 33 Q 2 W Q 3 Q = 0.7482 1 Ratio of heat rejected to body at 450\u00baC to the heat supplied by the reservoir = 0.7482 Ans. T1 = 773 K T3 = 723 K W 3 Auxiliary Q1 HE Q2 2W 3 Q3 HP Q'2 T2 = 200\u00ba C or 473 K Fig. 4.33 16. A reversible heat engine operates between a hot reservoir at T1 and a radiating surface at T2. Heat radiated from the surface is proportional to the surface area and temperature of surface raised to power 4. Determine the condition for minimum surface area for a given work output. Solution: Heat rejected = Heat radiated from surface at T2 = K. A. A is surface area and K is proportionality constant. 4 2T, where 2Q W = T 2 T\u2212 2 T 1 T1 Q1 HE W = ( Q Q1 \u2013 ) 2 Q2 T2 Fig. 4.34 126 _________________________________________________________ Applied Thermodynamics.. In order to have minimum surface area the denominator in above expression of A should be maximum i.e. 2T (T1 \u2013 T2) should be maximum. Differentiating with respect to T2. T 2 3 d dT {( 2 3 2T (T1 \u2013 T2))} = 0 3T1. 2 2T \u2013 4 3 2T = 0 \u21d2 Taking second differential T 2 T = 1 3 4 \u21d2 T2 = T1. 3 4 2 d 2 dT 2 { 3 2T. (T1 \u2013 T2)} = 6T1. T2 \u2013 12 2 2T Upon substitution it is \u2013ive so the minimum surface area \uf8eb \uf8ec \uf8ed is the condition for { 3 2T (T1 \u2013 T2)} to be maximum and so T 2 T 1 = 3 4 Ans. 17. A cold body is to be maintained at low temperature T2 when the temperature of surrounding is T3. A source is available at high temperature T1. Obtain the expression for minimum theoretical ratio of heat supplied from source to heat absorbed from cold body. Solution: Let us consider a refrigerator for maintaining cold body and also a reversible heat engine for driving refrigerator to operate together, Fig 4.35. To obtain; Q 1 Q 3 For heat engine, For refrigerator, Combining the above two: W Q 1 3Q ) ) Ratio of heat supplied from source to heat absorbed from cold body = T T T.( 1 2 3 T T T.( 1 3 \u2212 \u2212 2 ) ) Ans. Second Law of Thermodynamics ___________________________________________________ 127 T2 Q3 R Q4 T1 Source Q1 HE Q2 W T", "3, Sink Fig. 4.35 18. A heat pump is run by a reversible heat engine operating between reservoirs at 800\u00b0C and 50\u00b0C. The heat pump working on Carnot cycle picks up 15 kW heat from reservoir at 10\u00b0C and delivers it to a reservoir at 50\u00b0C. The reversible engine also runs a machine that needs 25 kW. Determine the heat received from highest temperature reservoir and heat rejected to reservoir at 50\u00b0C. 800\u00b0C or 1173K Q1 HE 10\u00b0C or 283K Q = 15 kW 3 WHE WHP HP Q4 Q2 25 kW 50\u00b0C or 323 K Fig. 4.36 Solution: Schematic arrangement for the problem is given in figure. For heat engine, \u03b7 HE = HEW Q 1 = \u2212 1 323 1173 HEW Q 1 = 0.7246 WHP = Q4 \u2013 Q3 = Q4 \u2013 15 COP = 323 \u2212 (323 283) = Q 4 4 15\u2212 Q Q4 = 17.12 kW WHP = 17.12 \u2013 15 = 2.12 kW WHE = WHP + 25 WHE = 27.12 kW \u03b7 HE = 0.7246 = HEW Q 1 Q1 = 37.427 kW Q2 = Q1 \u2013 WHE \u21d2 For heat pump, \u21d2 \u21d2 \u21d2 Since, \u21d2 \u21d2 \u21d2 128 _________________________________________________________ Applied Thermodynamics = 37.427 \u2013 27.12 Q2 = 10.307 kW Hence heat rejected to reservoir at 50\u00b0C \u21d2 = Q2 + Q4 = 10.307 + 17.12 = 27.427 kW Ans. Heat received from highest temperature reservoir = 37.427kW Ans. 19. Two insulated tanks are connected through a pipe with closed valve in between. Initially one tank having volume of 1.8m3 has argon gas at 12 bar, 40\u00b0C and other tank having volume of 3.6m3 is completely empty. Subsequently valve is opened and the argon pressure gets equalized in two tanks. Determine, (a) the final pressure & temperature (b) the change of enthalpy and (c) the work done considering argon as perfect gas and gas constant as 0.208 kJ/kg. K Solution: Total volume, V = V1 + V2 = 5.4 m3 By perfect gas law, p1V1 = mRT1 12 \u00d7 102 \u00d7 1.8 = m \u00d7 0.208 \u00d7 313 Argon 1.8 m3 \u21d2 m = 33.18 kg (1) By gas law for initial and final state, Empty initially 3.6 m3 (2) Valve Fig. 4.37 p1V1 = pfinal\u00d7Vfinal 12 \u00d7 102 \u00d7 1.8 = pfinal \u00d7 5.4 Final pressure \u21d2 pfinal = 400 kPa or 4 bar Ans. Here since it is insulated system and it has no heat transfer so, there will be no change in internal energy, hence there will be no change in temperature. Also by Ist law of thermodynamics, since there is no heat transfer due to system being insulated and no work due to frictionless expansion; Final temperature = 313K. dq = du + dw \u21d2 du = 0 i.e. Tinitial = Tfinal Change in enthalpy = 0 Ans. Work done = 0 Ans. 4.1 State the Kelvin Planck and Clausius statements of 2nd law of thermodynamics. 4.2 Show the equivalence of two statements of 2nd law of thermodynamics. 4.3 Write short notes on the following: Heat reservoir, Heat engine, Heat pump and refrigerator. 4.4 Explain the reversible and irreversible processes. 4.5 Describe Carnot cycle and obtain expression for its efficiency as applied to a heat engine. Second Law of Thermodynamics ___________________________________________________ 129 4.6 Why Carnot cycle is a theoretical cycle? Explain. 4.7 Show that coefficient of performance of heat pump and refrigerator can be related as; COPRef = COPHP \u2013 1 4.8 State Carnot theorem. Also prove it. 4.9 Show that the efficiencies of all reversible heat engines operating between same temperature limits are same. 4.10 Show that efficiency af an irreversible engine is always less than the efficiency of reversible engine operating between same temperature limits. 4.11 Assume an engine to operate on Carnot cycle with complete reversibility except that 10% of work is required to overcome friction. For the efficiency of reversible cycle being 30%, what shall be the efficiency of assumed engine. For same magnitude of energy required to overcome friction, if machine operated as heat pump, then [27%, 2.12] what shall be ratio between refrigerating effect and work required. 4.12 A Carnot engine operating between certain temperature limits has an efficiency of 30%. Determine the ratio of refrigerating effect and work required for operating the cycle as a heat pump between", " the same [2.33] temperature limits. 4.13 An inventor claims to have developed an engine that takes in 1055 mJ at a temperature of 400K and rejects 42.2 MJ at a temperature of 200 K while delivering 15kWh of mechanical work. Check whether [Engine satisfies Ist law but violates 2nd law] engine is feasible or not. 4.14 Determine which of the following is the most effective way to increase Carnot engine efficiency (i) To increase T2 while keeping T1 fixed. (ii) To decrease T1 while keeping T2 fixed. [If T1 is decreased] 4.15 A refrigerator has COP one half as great as that of a Carnot refrigerator operating between reservoirs at temperatures of 200 K and 400 K, and absorbs 633 KJ from low temperature reservoir. How much [1899 kJ] heat is rejected to the high temperature reservoir? 4.16 Derive a relationship between COP of a Carnot refrigerator and the efficiency of same refrigerator when operated as an engine. Is a Carnot engine having very high efficiency suited as refrigerator? 4.17 Calculate COP of Carnot refrigerator and Carnot heat pump, if the efficiency of the Carnot engine [5, 6] 4.18 For the reversible heat engines operating in series, as shown in figure 4.36. Show the following, if work between same temperature limits is 0.17. output is twice that of second. 3T2 = T1 + 2T3 T1 Q1 HE1 W1 Q2 T2 HE2 W2 Q3 T3 Fig. 4.36 4.19 A domestic refrigerator is intended to freeze water at 0\u00baC while water is available at 20\u00baC. COP of refrigerator is 2.5 and power input to run it is 0.4 kW. Determine capacity of refrigerator if it takes 14 [10 kg] minutes to freeze. Take specific heat of water as 4.2 kJ/kg. \u00baC. 130 _________________________________________________________ Applied Thermodynamics air at 30\u00baC. Determine the temperature maintained inside the cold storage. 4.20 A cold storage plant of 49.64 hp power rating removes 7.4 MJ/min and discharges heat to atmospheric [\u201340\u00baC] 4.21 A house is to be maintained at 21\u00baC from inside during winter season and at 26\u00baC during summer. Heat leakage through the walls, windows and roof is about 3 \u00d7 103 kJ/hr per degree temperature difference between the interior of house and environment temperature. A reversible heat pump is proposed for realizing the desired heating/cooling. What minimum power shall be required to run the heat pump in reversed cycle if outside temperature during summer is 36\u00baC? Also find the lowest environment temperature during winter for which the inside of house can be maintained at 21\u00baC. [0.279 kW, 11\u00baC] 4.22 Estimate the minimum power requirement of a heat pump for maintaining a commercial premises at 22\u00baC when environment temperature is \u20135\u00baC. The heat load on pump is 1 \u00d7 107 kJ/day. 4.23 A reversible engine having 50% thermal efficiency operates between a reservoir at 1527\u00baC and a reservoir at some temperature T. Determine temperature T in K. 4.24 A reversible heat engine cycle gives output of 10 kW when 10 kJ of heat per cycle is supplied from a source at 1227\u00baC. Heat is rejected to cooling water at 27\u00baC. Estimate the minimum theoretical number of [75] cycles required per minute. 4.25 Some heat engine A and a reversible heat engine B operate between same two heat reservoirs. Engine A has thermal efficiency equal to two-third of that of reversible engine B. Using second law of thermodynamics show that engine A shall be irreversible engine. 4.26 Show that the COP of a refrigeration cycle operating between two reservoirs shall be, COPref = \uf8eb \uf8ec \uf8ed 1 \u03b7 max \u2212 1 \uf8f6 \uf8f7 \uf8f8, if \u03b7 max refers to thermal efficiency of a reversible engine operating between same temperature limits. 4.27 A heat pump is used for maintaining a building at 20\u00baC. Heat loss through roofs and walls is at the rate of 6 \u00d7 104 kJ/h. An electric motor of 1 kW rating is used for driving heat pump. On some day when environment temperature is 0\u00baC, would it be possible for pump to maintain building at desired [No] temperature? 4.28 Three heat engines working on carnot cycle produce work output in proportion of 5 : 4 : 3 when operating in series between two reservoirs at 727\u00b0C and 27\u00b0C. Determine the temperature of intermediate [435.34\u00b0C, 202\u00b0C] reservoirs. 4.29 Determine the power required for running a heat pump which has to maintain temperature of 20\u00b0C when atmospheric temperature is \u201310\u00b0C. The heat losses through the walls of", " room are 650 W per unit [2 kW] temperature difference of inside room and atmosphere. 4.30 A heat pump is run between reservoirs with temperatures of 7\u00b0C and 77\u00b0C. Heat pump is run by a reversible heat engine which takes heat from reservoir at 1097\u00b0C and rejects heat to reservoir at 77\u00b0C. Determine the heat supplied by reservoir at 1097\u00b0C if the total heat supplied to reservoir at 77\u00b0C is [25.14 kW] 100 kW. 4.31 A refrigerator is used to maintain temperature of 243K when ambient temperature is 303K. A heat engine working between high temperature reservoir of 200\u00b0C and ambient temperature is used to run this refrigerator. Considering all processes to be reversible, determine the ratio of heat transferred from [0.69] high temperature reservoir to heat transferred from refrigerated space. 5 5.1 INTRODUCTION Till now the detailed explanation of Zeroth law, first law and second law of thermodynamics have been made. Also we have seen that the first law of thermodynamics defined a very useful property called internal energy. For overcoming the limitations of first law, the second law of thermodynamics had been stated. Now we need some mathematical parameter for being used as decision maker in respect of feasibility of process, irreversibility, nature of process etc. Here in this chapter a mathematical function called \u2018entropy\u2019 has been explained. \u2018Entropy\u2019 is the outcome of second law and is a thermodynamic property. Entropy is defined in the form of calculus operation, hence no exact physical description of it can be given. However, it has immense significance in thermodynamic process analysis. 5.2 CLAUSIUS INEQUALITY Let us take any reversible process 1\u20132 as shown on P\u2013V diagram. Let us also have a reversible adiabatic process 1\u20131\u2032 followed by reversible isothermal process 1\u2032\u20132\u2032 and a reversible adiabatic process 2' \u2013 2, as approximation to the original process 1 \u2013 2 such that area under 1 \u2013 2 equals to that under 1\u20131\u2032\u20132\u2032 \u20132. By first law of thermodynamics for process shown by 1\u20132. 1 1\u2032 p Reversible adiabatic Reversible isothermal 2\u2032 2 V Fig. 5.1 Reversible adiabatic, isothermal and reversible process First law on 1\u20131\u2032\u20132\u2032\u20132 processes; (Heat and work are path functions and internal energy is point Q1\u20132 = (U2 \u2013 U1) + W1\u20132 function) As already assumed that so Q1\u20131\u2032\u20132\u2032\u20132 = (U2 \u2013 U1) + W1\u20131\u2032\u20132\u2032\u20132 W1\u20132 = W1\u20131\u2032\u20132\u2032\u20132 Q1\u20131\u2032\u20132\u2032\u20132 = Q1\u20132 132 _________________________________________________________ Applied Thermodynamics In the path 1 \u2013 1\u2032 \u20132\u2032 \u20132 during adiabatic processes 1 \u2013 1\u2032 & 2 \u20132\u2032 there is no heat interaction so the total heat interaction in 1 \u20132 is getting occurred during isothermal process 1\u2032 \u2013 2\u2032 only. Hence, it is always possible to replace any reversible process by a series of reversible adiabatic, reversible isothermal and reversible adiabatic processes between the same end states provided the heat interaction and work involved remains same. If the number of reversible adiabatic and reversible isothermal processes is quite large then the series of such processes shall reach close to the original reversible process. Let us undertake this kind of substitution for the processes in a reversible cycle a 1 p 3 Reversible isothermal 5 2 4 7 V Reversible adiabatic 6 b 8 Reversible cycle Q1\u20132 1 2 3 4 Q3\u2013 4 Magnified view Fig. 5.2 A reversible cycle replaced by reversible adiabatics and reversible isotherms Figure 5.2 shows replacement of original processes in cycle a \u2013 b \u2013 a by adiabatic and isothermals. This shall result in a number of Carnot cycles appearing together in place of original cycle. Two Carnot cycles thus formed are shown by 1 \u20132 \u2013 4 \u20133 and 5 \u2013 6 \u2013 8 \u2013 7. Magnified view of first Carnot cycle is also shown separately where heat supplied at high temperature is Q1 \u20132 and heat rejected at low temperature is Q3 \u20134. From thermodynamic temperature scale; For Carnot cycle 1 \u2013 2 \u2013 4 \u2013 3, Q Q \u2212 1 2 \u2212 3 4 or Q T \u2212 1 2 For Carnot cycle 5 \u2013 6 \u2013 8 \u2013 7 Now taking sign conventions for heat added and rejected and Entropy _______________________________________________________________________ 133 Hence, if there are \u2018n\u2019 number of Carnot cycles replacing the original reversible cycle, then Q T 5\u20136 or, it can be given as summation of the ratio of heat interaction (Q) to", " the temperature (T) at which +... = + + \uf8f6 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ed it occurs, being equal to zero. Q \u2211 = 0 T If number of Carnot cycles is very large, then the zig-zag path formed due to replacing adiabatics and isotherms shall reach very close to original cycle. In such situation the cyclic integral of Q T may be given in place of above. or \uf8eb \u222b dQ \uf8ec \uf8ed T \uf8f6 \uf8f7 \uf8f8 rev = 0 Here it indicates that \uf8eb \uf8ec \uf8ed dQ T \uf8f6 \uf8f7 \uf8f8 is some thermodynamic property. Above expression developed for a reversible heat engine cycle also remains valid for internally reversible engines. In case of internally reversible engines T shall be temperature of working fluid at the time of heat interaction. \u222b \uf8eb \uf8ec \uf8ed dQ T \uf8f6 \uf8f7 \uf8f8 = 0 int, rev Let us now try to find out what happens to dQ T there be a reversible and irreversible heat engine operating between same temperature limits, such that heat added to them is same. when we have an irreversible engine cycle. Let \u222b From Carnot\u2019s theorem for both reversible and irreversible heat engine cycles, irrev rev > \u03b7 \u03b7 \uf8eb \u2212\uf8ec 1 \uf8ed > \uf8f6 \uf8f7 \uf8f8 rev Q rejected Q add \uf8f6 \uf8f7 \uf8f8 irrev \uf8eb \u2212\uf8ec 1 \uf8ed Q rejected Q add Q \uf8eb \uf8ec \uf8ed rejected Q \uf8f6 \uf8f7 \uf8f8 > Q \uf8eb \uf8ec \uf8ed rejected Q \uf8f6 \uf8f7 \uf8f8 add For same heat added, i.e. Qadd, rev = Qadd, irrev = Qadd irrev add rev Qrejected, irrev > Qrejected, rev Q rejected, irrev Q rejected, rev > 1 or or or For absolute thermodynamic temperature scale, \uf8eb \uf8ec \uf8ec \uf8ed Q add Q rejected \uf8f6 \uf8f7 \uf8f7 \uf8f8 rev = T add T rejected, upon substitution we get, 134 _________________________________________________________ Applied Thermodynamics Q rejected, irrev Q add > T rejected T add or or Q rejected, irrev T rejected > add Q T add Upon substituting sign convention, we get add Q T add + Q rejected, irrev T rejected Q add, irrev T add + Q rejected, irrev T rejected < 0 < 0 If it is given in the form of cyclic integral. or \u222b \uf8eb \uf8ec \uf8ed dQ T \uf8f6 \uf8f7 \uf8f8 irrev < 0 Now combining for reversible and irreversible paths it can be given as; \u222b \uf8eb \uf8ec \uf8ed dQ T \uf8f6 \uf8f7 \uf8f8 \u2264 0 This is called Clausius inequality. here, \u222b \u222b \u222b \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed dQ T dQ T dQ for reversible cycle < 0, for irreversible cycle > 0, for impossible cycle 5.3 ENTROPY\u2013A PROPERTY OF SYSTEM From Clausius inequality mathematically it is shown that for a reversible cycle. \u222b \uf8eb \uf8ec \uf8ed dQ T \uf8f6 \uf8f7 \uf8f8 rev = 0 Let us take a reversible cycle comprising of two processes A and B as shown and apply Clausius inequality. b \u222b a path A dQ T + a \u222b b path B dQ T = 0 Entropy _______________________________________________________________________ 135 or b \u222b a path A b \u222b a path A dQ T dQ T a \u222b = \u2013 b path B dQ T b \u222b a path A dQ T = a B Rev. cycle p A b V Fig. 5.3 Reversible cycle Hence, it shows that \uf8eb \uf8ec \uf8ed dQ T \uf8f6 \uf8f7 \uf8f8 is some property and does not depend upon path followed. This thermodynamic property is called \u201centropy\u201d. Entropy is generally denoted by \u2018S\u2019 or \u2018\u03c6\u2019. Thus, the energy interactions in the form of heat are accompanied by entropy changes. Writing it as function of entropy change. b \u222b a path A dQ T = b \u222b a path A dQ T = Sb \u2013 Sa \u222b \uf8eb \uf8ec \uf8ed dQ T \uf8f6 \uf8f7 \ufffd", "\ufffd rev = dS\u222b or Since entropy is point function and depends only upon end states therefore entropy change for any process following reversible or irreversible path shall be same. (Sb \u2013 Sa)rev, path = (Sb \u2013 Sa)irrev, path b \u222b a dQ rev T = \u2206 S rev, path = \u2206 S irrev,path Entropy is an extensive property and has units J/K. Specific entropy may be given on unit mass basis; s = S m (J/kg K) Entropy, as obvious from definition is defined for change in entropy value, therefore absolute value of entropy cannot be defined. Entropy is always given as change, i.e. relative values can be expressed. Let us now have two thermodynamic cycles a \u2013 b \u2013 a following paths (a \u2013 R \u2013 b, b \u2013 R \u2013 a) and (a \u2013 R \u2013 b, b \u2013 I \u2013 a). 136 _________________________________________________________ Applied Thermodynamics a p R R Here : denotes reversible path : denotes irreversible path R I b I V Fig. 5.4 Reversible and irreversible cycle We have from Clausius inequality. For reversible cycle a \u2013 R \u2013 b \u2013 R \u2013 a \u222b \uf8eb \uf8ec \uf8ed dQ T \uf8f6 \uf8f7 \uf8f8 rev = 0 b \u222b a R, dQ T + a \u222b b R, dQ T = 0 For irreversible cycle a \u2013 R \u2013 b \u2013 I \u2013 a \u222b \uf8eb \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f8 dQ T < 0 b \u222b a R, dQ T + a \u222b b I, dQ T < 0 b \u222b a R, dQ T < \u2013 a \u222b b I, dQ T Also from definition of entropy. \u222b \uf8eb \uf8ec \uf8ed dQ T \uf8f6 \uf8f7 \uf8f8 rev = dS\u222b so from above b \u222b a R, dQ T b dS\u222b a = \uf8eb \uf8ec \uf8ed dQ T \uf8f6 \uf8f7 \uf8f8 rev = dS For reversible cycle. or or or Entropy _______________________________________________________________________ 137 b \u222b a, R dQ T = \u2013 a \u222b b, R dQ T Substituting it in expression for irreversible cycle. \u2013 a \u222b b, R dQ T < \u2013 a \u222b b I, dQ T also from definition of entropy. or a dS\u222b \u2013 b a \u222b b, I dQ T < \u2013 a \u222b b, I dQ T < a dS\u222b b or, in general it can be given as, \uf8eb \uf8ec \uf8ed dQ T \uf8f6 \uf8f7 \uf8f8 irrev < dS Combining the above two we get inequality as following, where \u2265 dS dQ T dS = dS > dQ T dQ T for reversible process for irreversible process Mathematical formulation for entropy (dQrev = T \u00b7 dS) can be used for getting property diagrams between \u201ctemperature and entropy\u201d (T \u2013 S), \u201centhalpy and entropy\u201d (h \u2013 S). 1 T dA Tds dq = = ds 2 Area under process curve on T\u2013S diagram gives heat transferred, for internally reversible process S Fig. 5.5 T-S diagram dQint, rev = T \u00b7 dS Q int, rev 2 = \u222b 1 \u00b7 T dS 138 _________________________________________________________ Applied Thermodynamics 5.4 PRINCIPLE OF ENTROPY INCREASE By second law, entropy principle has been obtained as, \u2265 dS dQ T For an isolated system: dQ = 0, therefore dSisolated \u2265 0 for a reversible process dSisolated = 0 i.e. Sisolated = constant for an irreversible process dSisolated > 0 which means the entropy of an isolated system always increases, if it has irreversible processes, In general form dS isolated \u2265 0 It may be concluded here that the entropy of an isolated system always increases or remains constant in case of irreversible and reversible processes respectively. This is known as \u201cPrinciple of entropy increase\u201d of \u201centropy principle\u201d. Universe which comprises of system and surroundings may also be treated as isolated system and from entropy principle; dSUniverse \u2265 0 Which means that entropy of universe either increases continuously or remains constant depending upon whether processes occurring in it are of \u201cirreversible\u201d or \u201creversible\u201d type respectively. Since, therefore or Universe = System + Surrounding. dSsystem + dSsurrounding \u2206Ssystem + \u2206Ssurrounding \u2265 0 \u2265 0 Since most of the processes occurring generally are of irreversible type in universe, so it can be said that in", " general entropy of universe keeps on increasing and shall be maximum after attaining the state of equilibrium, which is very difficult to attain. In the above expression, system and surroundings are treated as two parts of universe (an isolated system). So the total entropy change during a process can be given by sum of \u201centropy change in system\u201d and \u201centropy change in surroundings\u201d. This total entropy change is also called \u201centropy generation\u201d or \u201centropy production\u201d. Entropy generation will be zero in a reversible process. Therefore \u2206Stotal = Sgen = \u2206Ssystem + \u2206Ssurrounding For closed systems In case of closed systems there is no mass interaction but heat and work interactions are there. Entropy change is related to heat interactions occurring in system and surroundings. Total entropy change. or Entropy generated \u2206Stotal = Sgen = \u2206Ssystem + \u2206Ssurrounding For system changing it's state from 1 to 2 i.e. initial and final state. \u2206Ssystem = S2 \u2013 S1 Entropy _______________________________________________________________________ 139 For surroundings, entropy change depends upon heat interactions. \u2206Ssurrounding = Q surrounding T surrounding Surrounding System State change of system from 1 to 2 Tsurrounding Qsurrounding Fig. 5.6 Closed system \u2013 \u2206Stotal = Sgen = (S2 \u2013 S1) + Q surrounding T surrounding or or where m is mass in system and s1 and s2 are specific entropy values at initial and final state, then S2 \u2013 S1 = m(s2 \u2013 s1) S gen = ( m s 2 \u2212 + s 1 ) Q surrounding T surrounding For open systems In case of open systems the mass interactions also take place along with energy interactions. Here mass flow into and out of system shall also cause some entropy change, so a control volume as shown in figure is to be considered. Entropy entering and leaving at section i \u2013 i and o \u2013 o are considered. Mass flow carries both energy and entropy into or out of control volume. Entropy transfer with mass flow is called \u201centropy transport\u201d. It is absent in closed systems. Surrounding Control volume i i Inlet Fig. 5.7 Open system o o Outlet Qsurrounding If control volume undergoes state change from 1 to 2, then entropy change in control volume shall be (S2 \u2013 S1) while entropy entering and leaving out may be given as Si and So respectively. By principle of entropy increase, total entropy change shall be, \u2206Stotal = Sgen = (S2 \u2013 S1) + (So \u2013 Si) + Q surrounding T surrounding Entropy entering and leaving out may be given as sum of entropy of all mass flows into and out of system in case of uniform flow process. 140 _________________________________________________________ Applied Thermodynamics Therefore Si = \u2211 mi. si So = \u2211 mo. so where mi and mo are mass flows into and out of system, and si and so are specific entropy associated with mass entering and leaving. Substituting, \u2206 S total = S gen = ( surrounding T surrounding In case of steady flow process since properties do not change with respect to time during any process, therefore within control volume there shall be no change in entropy. S1 = S2 i.e. Total entropy change or entropy generation for this case shall be; \u2206Stotal = Sgen = (So \u2013 Si) + Q surrounding T surrounding \u2206 S total = S gen ( = \u2211. surrounding T surrounding \u2265 0. In all the cases discussed above \u2206Stotal \u2265 0 or Sgen Entropy generated can be taken as criterion to indicate feasibility of process as follows; * If Sgen or \u2206Stotal = 0 then process is reversible. * If Sgen or \u2206Stotal > 0 then process is irreversible. * If Sgen or \u2206Stotal < 0 then process is impossible. One thing is very important about entropy generated that Sgen is not a thermodynamic property and it\u2019s value depends on the path followed whereas entropy change of system (S2 \u2013 S1) is a point function and so thermodynamic property. It is because of the fact that entropy change is cumulative effect of entropy transfer/change in system and surroundings. 5.5 ENTROPY CHANGE DURING DIFFERENT THERMODYNAMIC PROCESSES Isothermal process Let us find out entropy change for isothermal heat addition process. As isothermal process can be considered internally reversible, therefore entropy change shall be; or \u2206Sa \u2013 b = \uf8eb \uf8ec \uf8ed b \u222b a dQ T \uf8f6 \uf8f7 \uf8f8 \u2206Sa \u2013 b = 1 T b dQ\u222b a or Q \u2212 a b T where Qa \u2013 b is total heat interaction during state change a \u2013 b at temperature T", ". \u2212\u2206 S a b = Entropy _______________________________________________________________________ 141 Isentropic process It is the process during which change in entropy is zero and entropy remains constant during process. Isentropic process T a b s Fig. 5.8 Isentropic process It indicates that when \u2206Sa \u2013 b = 0. then which means there is no heat interaction during such process and this is adiabatic process. Qa \u2013 b = 0 Hence, it can be said that \"a reversible isentropic process shall be adiabatic, where as if isentropic process is adiabatic then it may or may not be reversible\u2019\u2019. Thus, adiabatic process may or may not be reversible. It means in reversible adiabatic process all states shall be in equilibrium and no dissipative effects are present along with no heat interaction whereas in adiabatic process there is no heat interaction but process may be irreversible. Finally, it can be concluded that an adiabatic process may or may not be isentropic whereas a reversible adiabatic process is always isentropic. An adiabatic process of non isentropic type is shown below where irreversibility prevails, say due to internal friction. T d' d a b c b' s Fig. 5.9 Isentropic and non-isentropic processes Here a \u2013 b is reversible adiabatic expansion of isentropic type. Non-isentropic or adiabatic expansion is shown by a \u2013 b'. Isentropic expansion efficiency may be defined as ratio of actual work to ideal work available during expansion. \u03b7 isen, expn = Actual work in expansion Ideal work in expansion \u03b7 isen,expn = 142 _________________________________________________________ Applied Thermodynamics Similarly, isentropic and non-isentropic compression process are shown as c \u2013 d and c \u2013 d' respectively. Isentropic compression efficiency can be defined on same lines as, \u03b7 isen, compr = Ideal work in expansion Actual work in expansion \u03b7 isen, compr = For ideal gases Combination of first and second law yields; Tds = du + pdv also we know du = cv \u00b7 dT and for perfect gas p = RT v substituting for du and R T \u00b7 ds = cv \u00b7 dT + or or \u00b7 dv RT v Rd v v ds = cv \u00b7 dT T + s2 \u2013 s1 = 2 \u222b 1 c v \u00b7 + dT T 2 \u222b 1 v Rd v If cv is function of temperature then ( ) \u00b7 If specific heat is constant then, 1 dT T + R \u00b7 ln ln T 2 T 1 + R \u00b7 ln v 2 v 1 Also combination of Ist and IInd law yields following using; h = u + pv, or dh = du + pdv + vdp substituting or entropy change T \u00b7 ds = dh \u2013 v dp dh = cp \u00b7 dT, and v = RT p T \u00b7 ds = cp \u00b7 dT \u2013 v \u00b7 dp Rdp p ds = cp \u00b7 dT T \u2212 s2 \u2013 s1 = 2 \u222b 1 pc dT T \u2212 2 \u222b 1 Rdp p Entropy _______________________________________________________________________ 143 If specific heat is function of temperature then If specific heat is constant, then s 2 \u2212 s 1 = 2 \u222b 1 ( ) pc T dT T \u2212 R ln ln \u2212 R ln T 2 T 1 p 2 p 1 Above expressions given in enclosed box may be suitably used for getting the change in entropy. Polytropic Process Entropy change in a polytropic process having governing equation as pvn = constant, can be obtained as below, For polytropic process between 1 and 2, n p=v p 1 1 n v 2 2 or Also, from gas laws \uf8f6 \uf8f7 \uf8f8 = Above two pressure ratios give \uf8f6 \uf8f7 \uf8f8 Substituting \uf8eb \uf8ec \uf8ed v 2 v 1 \uf8f6 \uf8f7 \uf8f8 in the entropy change relation derived earlier. s2 \u2013 s1 = c v ln T 2 T 1 + R ln v 2 v 1 = c v ln \uf8eb \uf8ec \uf8ed T 2 T 1 \uf8f6 \uf8f7 \uf8f8 + R ln 1 \u20131 n \uf8eb \uf8ec \uf8ed T 1 T 2 \uf8f6 \uf8f7 \uf8f8 For perfect gas + \uf8eb R \uf8ec \u2212\uf8ed n 1 \uf8f6 \uf8f7 \uf8f8. ln T 1 T 2 s2 \u2013 s1 = v c \uf8f6 \uf8f7 \uf8f8 ln T 2 T 1 \uf8eb \ufffd", "\ufffd \uf8ed R = cp \u2013 cv R = \u03b3. cv \u2013 cv 144 _________________________________________________________ Applied Thermodynamics Substituting R in entropy change R = cv (\u03b3 \u20131) s2 \u2013 s1 = cv ln \uf8eb \uf8ec \uf8ed ) 1) ln \uf8eb \uf8ec \uf8ed T 1 T 2 \uf8f6 \uf8f7 \uf8f8 = cv ln s2 \u2013 s1 = cv ln \uf8f1 \u2212\uf8f2 1 \uf8f3 \u03b3 \u2212 \uf8eb \uf8ec \u2212\uf8ed \uf8ec \u2212\uf8ed Entropy change in polytropic process \u00b7 ln \uf8f6 \u2212 \u03b3 \uf8eb n \uf8f7\uf8ec \u2212\uf8ed.6 ENTROPY AND ITS RELEVANCE Entropy has been introduced as a property based on the concept of IInd law of thermodynamics and derived from the thermodynamics involved in heat engines. A large number of definitions are available for entropy. To understand entropy let us take some gas in a closed vessel and heat it. Upon heating of gas the motion of gas molecules inside the vessel gets increased. State of molecular motion inside vessel depends upon the quantum of heat supplied. If we measure new kinetic energy of gas molecules, it is found to be larger than that initially. Also, the rate of intermolecular collision and randomness in molecular motion gets increased. In nutshell it could be said that heating has caused increase in energy level of gas molecules and thus resulting in increased disorderness inside the vessel. Higher is the energy with molecules, higher shall be the degree of disorderness. Entropy is closely defined using the degree of disorderness. It is said that greater is the molecular disorderness in the system greater shall be entropy. Mathematically, it can be supported by greater entropy value due to large heat supplied (dQ/T). Thus \u201centropy can be defined as a parameter for quantifying the degree of molecular disorderness in the system\u201d. \u201cEntropy is a measure of driving potential available for occurrence of a process\u201d. Entropy is also an indicator of the direction of occurrence of any thermodynamic process. Mathematically, it has been seen from second law of thermodynamics that entropy of an isolated system always increases. Therefore, a process shall always occur in such a direction in which either entropy does not change or increases. In general almost all real processes are of irreversible type so entropy tends to increase. As entropy cannot be defined absolutely so the change in entropy should always have a positive or zero value. 5.7 THERMODYNAMIC PROPERTY RELATIONSHIP Different thermodynamic properties such as P, V, T, U, H, S etc. can be related to one another using the combination of mathematical forms of first law, second law of thermodynamics and definitions of properties. Here specific values of properties are related. For a non-flow process in closed system. or dq = du + dw dq = du + p \u00b7 dv Entropy _______________________________________________________________________ 145 Also, for a reversible process from definition of entropy, by second law we can write Combining above two, dq = Tds = Tds du + p d \u00b7 v From definition of enthalpy, specific enthalpy h = u + pv dh = du + p\u00b7dv + v\u00b7dp substituting from above + v Above relations may be used for getting the variation of one property with the other, such as for constant pressure process, dh = T\u00b7ds = dh T ds dp \u00b7 \uf8eb \uf8ec \uf8ed dh ds \uf8f6 \uf8f7 \uf8f8 = T p =constt. or or which means slope of constant pressure line on enthalpy \u2013 entropy diagram (h \u2013 s) is given by temperature. Also from above two relations T ds \u00b7 = c dT \u00b7 v + p d \u00b7 v Substituting for dh and rearranging, dh = T\u00b7ds + v \u00b7 dp \u2212 v Tds dp = \u00b7p c dT or For a constant pressure process above yields \uf8eb \uf8ec \uf8ed dT ds \uf8f6 \uf8f7 \uf8f8 = T c p p =constt. It gives the slope of constant pressure line on T \u2013 s diagram. Similarly, for a constant volume process, {as du = cv {as dh = cp \u00b7 dT} \u00b7 dT} \uf8eb \uf8ec \uf8ed dT ds \uf8f6 \uf8f7 \uf8f8 v = T c v =const. It gives the slope of constant volume line on T \u2013 s diagram. It can be concluded from the above mathematical explanations for slope that slope of constant volume line is more than the slope of constant pressure line as cp > cv. Constant volume lines T Constant pressure lines Fig. 5.10 T-", "s diagram showing isobaric and isochoric process. s 146 _________________________________________________________ Applied Thermodynamics 5.8 THIRD LAW OF THERMODYNAMICS \u2018Third law of thermodynamics\u2019, an independent principle uncovered by \u2018Nernst\u2019 and formulated by \u2018Planck\u2019, states that the \u201cEntropy of a pure substance approaches zero at absolute zero temperature.\u201d This fact can also be corroborated by the definition of entropy which says it is a measure of molecular disorderness. At absolute zero temperature substance molecules get frozen and do not have any activity, therefore it may be assigned zero entropy value at crystalline state. Although the attainment of absolute zero temperature is impossible practically, however theoretically it can be used for defining absolute entropy value with respect to zero entropy at absolute zero temperature. Second law of thermodynamics also shows that absolute zero temperature can\u2019t be achieved, as proved earlier in article 4.10. Third law of thermodynamics is of high theoretical significance for the sake of absolute property definitions and has found great utility in thermodynamics. 1. Calculate the change in entropy of air, if it is throttled from 5 bar, 27\u00baC to 2 bar adiabatically. EXAMPLES Solution: Here p1 = 5 bar, T1 = 300 K. p2 = 2 bar, cp air = 1.004 kJ/kg.K R = 0.287 kJ/kg.K Entropy change may be given as; s2 \u2013 s1 = \uf8f1 \uf8f2 \uf8f3 c p ln T 2 T 1 \u2212 R ln \uf8fc \uf8fd \uf8fe p 2 p 1 for throttling process h1 = h2 i.e. or Hence, cpT1 = cpT2 T1 = T2 Change in entropy = 1.004 ln (1) \u2013 0.287 ln Change in entropy = 0.263 kJ/kg.K Ans. = 0.263 kJ/kg.. Find the change in entropy of steam generated at 400\u00baC from 5 kg of water at 27\u00baC and atmospheric pressure. Take specific heat of water to be 4.2 kJ/kg.K, heat of vaporization at 100\u00baC as 2260 kJ/kg and specific heat for steam given by; cp = R (3.5 + 1.2T + 0.14T2), J/kg.K Solution: Total entropy change = Entropy change during water temperature rise (\u2206S1). + Entropy change during water to steam change (\u2206S2) + Entropy change during steam temperature rise (\u2206S3) Q 1 T where Q1 = m cp \u00b7 \u2206T \u2206S1 = 1 Heat added for increasing water temperature from 27\u00baC to 100\u00baC. = 5 \u00d7 4.2 \u00d7 (100 \u2013 27) = 1533 kJ Entropy _______________________________________________________________________ 147 therefore, \u2206S1 = 1533 300 = 5.11 kJ/K Entropy change during phase transformation; \u2206S2 = Q 2 T 2 Here Q2 = Heat of vaporization = 5 \u00d7 2260 = 11300 kJ Entropy change, \u2206S2= 11300 373.15 = 30.28 kJ/K. Entropy change during steam temperature rise; \u2206S3 = 673.15 \u222b 373.15 dQ T Here dQ = mcp \u00b7 dT; for steam R = Therefore, cp for steam = 0.462 (3.5 + 1.2 \u00b7 T + 0.14T 2) \u00d7 10\u20133 = 0.462 kJ/kg.K 8.314 18 = (1.617 + 0.5544 T + 0.065 T 2) \u00d7 10\u20133 or \u2206S3 = 673.15 \u222b 373.15 5 \u00d7 10\u20133 \u00d7 \uf8eb \uf8ec \uf8ed 1.617 T + + 0.5544 0.065 T \uf8f6 \uf8f7 \uf8f8 dT = 51843.49 \u00d7 10\u20133 kJ/K \u2206S3 = 51.84 kJ/K Total entropy change = 5.11 + 30.28 + 51.84 = 87.23 kJ/K Ans. 3. Oxygen is compressed reversibly and isothermally from 125 kPa and 27\u00baC to a final pressure of 375 kPa. Determine change in entropy of gas? Solution: Gas constant for oxygen: R = 8.314 32 = 0.259 kJ/kg.K For reversible process the change in entropy may be given as; \u2206s = cp ln T 2 T \u2013 R ln 1 p 2 p 1 Substituting values of initial & final states \u2206s = \u2013 R ln \uf8eb \uf8ec \uf8ed", " 375 125 \uf8f6 \uf8f7 \uf8f8 = \u20130.285 kJ/kg.K Entropy change = \u2013 0.285 kJ/kg. K Ans. 4. Determine the change in entropy of universe if a copper block of 1 kg at 150\u00baC is placed in a sea water at 25\u00baC. Take heat capacity of copper as 0.393 kJ/kg K. 148 _________________________________________________________ Applied Thermodynamics Solution: Entropy change in universe \u2206Suniverse = \u2206Sblock + \u2206Swater where \u2206Sblock = mC. ln T 2 T 1 Here hot block is put into sea water, so block shall cool down upto sea water at 25\u00baC as sea may be treated as sink. Therefore, and T1 = 150\u00baC or 423.15 K T2 = 25\u00baC or 298.15 K \uf8eb \uf8ec \uf8ed \u2206Sblock = 1 \u00d7 0.393 \u00d7 ln = \u2013 0.1376 kJ/K 298.15 423.15 \uf8f6 \uf8f7 \uf8f8 Heat lost by block = Heat gained by water = \u2013 1 \u00d7 0.393 \u00d7 (423.15 \u2013 298.15) = \u2013 49.125 kJ Therefore, Thus, \u2206Swater = 49.125 298.15 \u2206Suniverse = \u2013 0.1376 + 0.165 = 0.165 kJ/k = 0.0274 kJ/k or 27.4 J/K Entropy change of universe = 27.4 J/K Ans. 5. Determine change in entropy of universe if a copper block of 1 kg at 27\u00baC is dropped from a height of 200 m in the sea water at 27\u00baC. (Heat capacity for copper= 0.393 kJ/kg.K) Solution: \u2206Suniverse = \u2206Sblock + \u2206Ssea water Since block and sea water both are at the same temperature so, \u2206Suniverse = \u2206Ssea water Conservation of energy equation yields; Q \u2013 W = \u2206U + \u2206PE + \u2206KE Since in this case, W = 0, \u2206KE = 0, \u2206U = 0 Q = \u2206PE Change in potential energy = \u2206PE = mgh = 1 \u00d7 9.81 \u00d7 200 = 1962 J Q = 1962 J 1962 300 Entropy change of universe = 6.54 J/kg.K Ans. \u2206Suniverse = \u2206Ssea water = = 6.54 J/kg K 6. Determine entropy change of universe, if two copper blocks of 1 kg & 0.5 kg at 150\u00baC and 0\u00baC are joined together. Specific heats for copper at 150\u00baC and 0\u00baC are 0.393 kJ/kg K and 0.381 kJ/kg K respectively. Solution: Here, \u2206Suniverse = \u2206Sblock 1 + \u2206Sblock 2 Two blocks at different temperatures shall first attain equilibrium temperature. Let equilibrium temperature be T\u0192. Entropy _______________________________________________________________________ 149 Then from energy conservation. 1 \u00d7 0.393 \u00d7 (423.15 \u2013 T\u0192) = 0.5 \u00d7 0.381 \u00d7 (T\u0192 \u2013 273.15) T\u0192 = 374.19 K Hence, entropy change in block 1, due to temperature changing from 423.15 K to 374.19 K. \u2206S1 = 1 \u00d7 0.393 \u00d7 ln \uf8eb \uf8ec \uf8ed 374.19 423.15 \uf8f6 \uf8f7 \uf8f8 = \u2013 0.0483 kJ/K Entropy change in block 2 \u2206S2 = 0.5 \u00d7 0.381 \u00d7 ln \uf8eb \uf8ec \uf8ed 374.19 273.15 \uf8f6 \uf8f7 \uf8f8 = 0.0599 kJ/K Entropy change of universe = 0.0599 \u2013 0.0483 = 0.0116 kJ/K Entropy change of universe = 0.0116 kJ/K Ans. 7. A cool body at temperature T1 is brought in contact with high temperature reservoir at temperature T2. Body comes in equilibrium with reservoir at constant pressure. Considering heat capacity of body as C, show that entropy change of universe can be given as \uf8f6 \uf8f7 \uf8f8 \u2212 ln \uf8f9 \uf8fa \uf8fb T 1 T 2 Solution: Since body is brought in contact with reservoir at temperature T2, the body shall come in equilibrium when it attains temperature equal to that of reservoir, but there shall be no change in temperature of the reservoir. Entropy change of universe \u2206Suniverse = \u2206Sbody+ \u2206Sreservoir \u2206Sbody = C ln T 2 T 1 \u2206Sreservoir = \u2212 as, heat gained by body = Heat lost", " by reservoir Thus, = C (T2 \u2013 T1Suniverse = C ln or, rearranging the terms, \u2206Suniverse = Hence proved ln \uf8eb \uf8ec \uf8ed T 1 T 2 \uf8f6 \uf8f7 \uf8f8 8. Determine the rate of power loss due to irreversibility in a heat engine operating between temperatures of 1800 K and 300 K. Engine delivers 2 MW of power when heat is added at the rate of 5 MW. 150 _________________________________________________________ Applied Thermodynamics Solution: For irreversible operation of engine Rate of entropy generation = 1800 Q 2 T 2 W = Q1 \u2013 Q2 = 5 \u00d7 106 \u2013 Q2 W = 2 MW = 2 \u00d7 106 W Q2 = 3 \u00d7 106 W entropy generated = \u00d7 6 10 + \u2212\uf8eb \uf8f6 5 3 \uf8ec \uf8f7 \uf8ed \uf8f8 1800 300 \u2206Sgen = 7222.22 W/K \u2206Sgen = T2 \u00d7 = 300 \u00d7 7222.22 = 2.16 \u00d7 106 W = 2.16 MW. Also, given so Therefore, Work lost or 1800 K T1 Q1 HE W Q2 300 K T2 Fig. 5.11 Work lost = 2.16 MW Ans. 9. A system at 500 K and a heat reservoir at 300 K are available for designing a work producing device. Estimate the maximum work that can be produced by the device if heat capacity of system is given as; C = 0.05 T 2 + 0.10T + 0.085, J/K Solution: System and reservoir can be treated as source and sink. Device thought of can be a Carnot engine operating between these two limits. Maximum heat available from system shall be the heat rejected till it\u2019s temperature drops from 500 K to 300 K. System T1 = 500 K Q1 HE W Q2 Reservoir T2 = 300 K Fig. 5.12 Therefore, Maximum heat Q1 = T 2 \u222b T 1 \u00b7 C dT Entropy _______________________________________________________________________ 151 300 \u222b (0.05 T 2 + 0.10T + 0.085) dT Q1 = 500 Entropy change of system, \u2206Ssystem = Q1 = 1641.35 \u00d7 103 J 300 \u222b C 500 dT T = \u2013 4020.043 J/K \u2206Sreservoir = \uf8eb \uf8ec \uf8ec \uf8ed = 1641.35 10 \u00d7 3 \u2212 W 300 \uf8f6 \uf8f7 \uf8f7 \uf8f8 Also, we know from entropy principle \u2265 0 \u2206Suniverse \u2206Suniverse = \u2206Ssystem + \u2206Sreservoir and Thus, upon substituting (\u2206Ssystem + \u2206Sreservoir) \u2265 0 \uf8f1 \uf8f4 \uf8f2 \uf8f4 \uf8f3 \u2212 4020.043 \uf8eb + \uf8ec \uf8ec \uf8ed 1641.35 10 \u00d7 3 \u2212 W 300 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe \u2265 0 or or 1451.123 \u2013 W 300 \u2265 0 1451.123 \u2265 W 300 W \u2264 435337.10 W \u2264 435.34 kJ Hence Maximum work = 435.34 kJ For the given arrangement, device can produce maximum 435.34 kJ of work. Ans. 10. Determine the change in enthalpy and entropy if air undergoes reversible adiabatic expansion from 3MPa, 0.05 m3 to 0.3 m3. Solution: For reversible adiabatic process, governing equation for expansion, PV1.4 = Constt. Also, for such process entropy change = 0. Initial state : 3MPa, 0.05 m3 Final state : 0.3 m3.4 \uf8f6 \uf8f7 \uf8f8 or V = 1 \uf8f6 1.4 1.4 \uf8f7 \uf8f7 \uf8f8 1 1p V p \uf8eb \uf8ec \uf8ec \uf8ed Using we get p2 = 0.244 MPa 152 _________________________________________________________ Applied Thermodynamics From first law, second law and definition of enthalpy; or, for adiabatic process of reversible type, dS = 0. dH = T\u00b7dS + Vdp dH = V\u00b7dp 2 dH\u222b 1 2 \u222b 1 = \u00b7V dp Substituting V, and actual states H2 \u2013 H1 = 244 \u222b \uf8eb \uf8ec \uf8ec \uf8ed \u00d7 3000 0.05 p 1 \uf8f6 1.4 1.4 \uf8f7 \uf8f7 \uf8f8 dp or 3000 \u2206H = 268.8 kJ Enthalpy change = 268.8 kJ. Entropy change = 0 Ans.", " 11. During a free expansion 2 kg air expands from 1 m3 to 10m3 volume in an insulated vessel. Determine entropy change of (a) the air (b) the surroundings (c) the universe. Solution: During free expansion temperature remains same and it is an irreversible process. For getting change in entropy let us approximate this expansion process as a reversible isothermal expansion. (a) Change in entropy of air \u2206Sair = m.R ln V 2 V 1 = 2 \u00d7 287 ln \uf8eb \uf8ec \uf8ed \u2206Sair = 1321.68 J/K 10 1 \uf8f6 \uf8f7 \uf8f8 = 1321.68 J/K Ans. (b) During free expansion on heat is gained or lost to surroundings so, Entropy change of surroundings, \u2206Ssurroundings = 0 = 0 Ans. (c) Entropy change of universe \u2206Suniverse = \u2206Sair + \u2206Ssurroundings = 1321.68 J/K = 1321.68 J/K Ans. 12. Determine the change in entropy of 0.5 kg of air compressed polytropically from 1.013 \u00d7 105 0.8 MPa and 800 K following index 1.2. Take Cv = 0.71 kJ/kg. K. Solution: Let initial and final states be denoted by 1 and 2. aP to For polytropic process pressure and temperature can be related as Entropy _______________________________________________________________________ 153 or \uf8eb \uf8ec \uf8ec \uf8ed Temperature after compression = 1128.94 K T2 = 800 \u00d7 \u00d7 0.8 10 6 5 1.013 10 \u00d7 \u2212 1.2 1 1.2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 Substituting in entropy change expression for polytropic process, (s2 \u2013 s1) = Cv \u2212\uf8eb n \uf8ec \u2212\uf8ed n \u03b3 1 \uf8f6 \uf8f7 \uf8f8. ln T 2 T 1 = 0.71 \u00d7 103 \uf8eb \uf8ec \uf8ed \u2212 1.2 1.4 \u2212 1.2 1 \uf8f6 \uf8f7 \uf8f8 ln \uf8eb \uf8ec \uf8ed 1128.94 800 \uf8f6 \uf8f7 \uf8f8 = \u2013244.54, kJ/kg. K Total entropy change = m (s2 \u2013 s1) = 0.5 \u00d7 244.54 \u2206S = 122.27 J/K Ans. 13. A heat engine is working between the starting temperature limits of T1 and T2 of two bodies. Working fluid flows at rate \u2018m\u2019 kg/s and has specific heat at constant pressure as Cp. Determine the maximum obtainable work from engine. Solution: In earlier discussions we have seen that in order to have highest output from engine, it should operate in reversible cycle and satisfy following relation, \u222b dQ T = 0 Let us assume that the two bodies shall attain final temperature of T\u0192 and engine shall then get stopped. so, T \u0192 \u222b T 1 mC p + dT T T \u0192 \u222b T 2 mC p dT T = 0 mC p \uf8f1 \uf8f4 \uf8f2 \uf8f4 \uf8f3 ln \uf8eb \uf8ec \uf8ed T \u0192 T 1 \uf8f6 \uf8f7 \uf8f8 + ln \uf8eb \uf8ec \uf8ed or or mCp \u00b7 ln \uf8eb \uf8ec \uf8ec \uf8ed \uf8eb \uf8ec \uf8ec \uf8ed 2 T \u0192 2\u00b7 T T 1 2 T \u0192 2 Here, mCp \u2260 0, so, ln or T\u0192 = 2\u00b7T T 1 Maximum work = Qsupplied \u2013 Qrejected = mCp(T1 \u2013 T\u0192) \u2013 mCp(T\u0192 \u2013 T2) = mCp{T1 \u2013 2T\u0192 + T2} 154 _________________________________________________________ Applied Thermodynamics = mCp{T1 \u2013 2 \u00b7 2\u00b7T T + T2} 1 Maximum work = mCp { \u2013T 1 T 2 }2 Ans. 14. A heat engine operates between source at 600 K and sink at 300 K. Heat supplied by source is 500 kcal/s. Evaluate feasibility of engine and nature of cycle for the following conditions. (i) Heat rejected being 200 kcal/s, (ii) Heat rejected being 400 kcal/s (iii) Heat rejected being 250 kcal/s. Solution: Clausius inequality can be used for cyclic process as given below; consider \u20181\u2019 for source and \u20182\u2019 for sink. (i) For Q2 = 200 kcal/s \u222b dQ \ufffd", "\ufffd dQ T = 500 600 \u2212 200 300 = 0.1667 As \u222b dQ T > 0, therefore under these conditions engine is not possible. Ans. (ii) For Q2 = 400 kcal/s \u222b dQ T = 500 600 \u2212 400 300 = \u2013 0.5 Here \u222b dQ T < 0, so engine is feasible and cycle is irreversible Ans. (iii) For Q2 = 250 kcal/s \u222b dQ T = 500 600 \u2212 250 300 = 0 Here, \u222b dQ T = 0, so engine is feasible and cycle is reversible. Ans. 15. Along a horizontal and insulated duct the pressure and temperatures measured at two points are 0.5 MPa, 400 K and 0.3 MPa, 350 K. For air flowing through duct determine the direction of flow. Solution: Let the two points be given as states 1 and 2, so, Let us assume flow to be from 1 to 2 p1 = 0.5 MPa, T1 = 400 K p2 = 0.3 MPa, T2 = 350 K So entropy change \u2206s1\u20132 = s1 \u2013 s2 = Cp ln For air, Hence R = 0.287 kJ/kg. K Cp = 1.004 kJ/kg. K \uf8f6 400 \uf8f7 \uf8f8 350 s1 \u2013 s2 = 1.004 ln \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed T 1 T 2 \uf8f6 \uf8f7 \uf8f8 \u2013 R ln \uf8eb \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f8 p 1 p 2 \u2013 0.287 ln \uf8eb \uf8ec \uf8ed 0.5 0.3 \uf8f6 \uf8f7 \uf8f8 Entropy _______________________________________________________________________ 155 = \u2013 0.01254 kJ/kg. K or s1 \u2013 s2 = 0.01254 kJ/kg.K It means s2 > s1 hence the assumption that flow is from 1 to 2 is correct as from second law of thermodynamics the entropy increases in a process i.e. s2 \u2265 s1. Hence flow occurs from 1 to 2 i.e. from 0.5 MPa, 400 K to 0.3 MPa & 350 K Ans. 16. An ideal gas is heated from temperature T1 to T2 by keeping its volume constant. The gas is expanded back to it's initial temperature according to the law pvn = constant. If the entropy change in the two processes are equal, find the value of \u2018n\u2019 in terms of adiabatic index \u03b3. [U.P.S.C. 1997] Solution: During constant volume process change in entropy \u2206S12 = mcv. ln T 2 T 1 Change in entropy during polytropic process, \u2206S23 = mcv Since the entropy change is same, so \u2206S12 = \u2206S23 \u03b3 \u2212\uf8eb \uf8ec \u2212\uf8ed n n 1 \uf8f6 \uf8f7 \uf8f8 ln T 2 T 1 mcv ln T 2 T = mcv 1 \u03b3 \u2212\uf8eb \uf8ec \u2212\uf8ed n n 1 \uf8f6 \uf8f7 \uf8f8 ln Ans. or 17. A closed system executed a reversible cycle 1\u20132\u20133\u20134\u20135\u20136\u20131 consisting of six processes. During processes 1\u20132 and 3\u20134 the system receives 1000 kJ and 800 kJ of heat, respectively at constant temperatures of 500 K and 400 K, respectively. Processes 2\u20133 and 4\u20135 are adiabatic expansions in which the steam temperature is reduced from 500 K to 400 K and from 400 K to 300 K respectively. During process 5\u20136 the system rejects heat at a temperature of 300 K. Process 6\u20131 is an adiabatic compression process. Determine the work done by the system during the cycle and thermal efficiency of the cycle. [U.P.S.C. 1995] Solution: 1 500 K Q12 = 1000 kJ 2 T 400 K 300 K 6 Q34 = 800 kJ 4 5 Q56 3 S Fig. 5.13 Heat added = Q12 + Q14 Total heat added = 1800 kJ 156 _________________________________________________________ Applied Thermodynamics For heat addition process 1\u20132, Q12 = T1 \u00b7 (S2 \u2013 S1) 1000 = 500 \u00b7 (S2 \u2013 S1) or, For heat addition process 3\u20134, Q34 = T3 \u00b7 (S4 \u2013 S3) 800 = 400 \u00b7 (S4 \u2013 S3) S2 \u2013 S1 = 2 or, Heat rejected in process 5\u20136 S4 \u2013 S3 = 2 Q56 = T5 \u00b7 (S5 \u2013 S6) = T5 \u00b7 {", "(S2 \u2013 S1) + (S4 \u2013 S3)} Q56 = 300 \u00b7 {2 + 2} = 1200 kJ Net work done = Net heat = (Q12 + Q34) \u2013 Q56 Wnet = 1800 \u2013 1200 Wnet = 600 kJ W net Heat added Thermal efficiency of cycle = = 600 1800 = 0.3333 or 33.33% Work done = 600 kJ Thermal efficiency = 33.33% Ans. 18. A reversible heat engine has heat interaction from three reservoirs at 600 K, 700 K and 800 K. The engine rejects 10 kJ/s to the sink at 320 K after doing 20 kW of work. The heat supplied by reservoir at 800 K is 70% of the heat supplied by reservoir at 700 K then determine the exact amount of heat interaction with each high temperature reservoir. 800 K 700 K 600 K Q'= 0.7Q \" 1 1 Q \"1 Q1 Q '''1 HE W = 20 kW Q2 = 10 kW 320 K Fig. 5.14 Solution: Let heat supplied by reservoir at 800 K, 700 K and 600 K be Q'1, Q\"1, Q'''1. Here, \u21d2 Also given that, Q1 \u2013 Q2 = W Q1 = 30 kJ/s Q'1 = 0.7 Q''1 Q'''1 = Q1 \u2013 (0.7 Q\"1 + Q\"1) Entropy _______________________________________________________________________ 157 For reversible engine Q'''1 = Q1 \u2013 1.7 Q''1 \u21d2'Q 1 800 + \" Q 1 700 + \"'Q 1 600 \u2212 Q 2 320 = 0 \u21d2'Q 0.7 1 800 + \" Q 1 700 ( Q 1 + \" 1 Q\u2212 1.7 600 ) \u2212 10 320 = 0 \u21d2 Q\"1 = 88.48 kJ/s Q'1 = 61.94 kJ/s Q'''1 = \u2013 120.42 kJ/s Heat supplied by reservoir at 800 K = 61.94 kJ/s Heat supplied by reservoir at 700 K = 88.48 kJ/s Heat supplied to reservoir at 600 K = 120.42 kJ/s Ans. 19. A rigid insulated tank is divided into two chambers of equal volume of 0.04 m3 by a frictionless, massless thin piston, initially held at position with a locking pin. One chamber is filled with air at 10 bar & 25\u00b0C and other chamber is completely evacuated, Subsequently pin is removed and air comes into equilibrium. Determine whether the process is reversible or irreversible. Consider, R = 0.287 kJ/ kg.K and cv = 0.71 kJ/kg.K. Solution: Let us assume process to be adiabatic and so the heat interaction would not be there. Also in view of this expansion being frictionless expansion there would be no work done, i.e., W = 0, Q = 0. Let initial and final states be indicated by subscripts 1 and 2. Using first law of thermodynamics; \u21d2 \u21d2 dQ = dW + dU 0 = 0 + mcv(T2 \u2013 T1) T1 = T2 = 298 K Volume changes are, V2 = 2V1 = 0.08 m3 Using gas laws, p1V1 = p2V2 \u21d2 p2 = 0.5p1 = 5 bar Initial Mass of air, m1 = p V 1 1 RT 1 \u21d2 = m 1 2 \u00d7 10 10 \u00d7 \u00d7 0.287 298 0.04 = 0.4677 kg Change of entropy, (S2 \u2013 S1) = mRln \uf8eb \uf8ec \uf8ed V 2 V 1 \uf8f6 \uf8f7 \uf8f8 + mc ln S2 \u2013 S1) = 0.4677 {0.287 \u00d7 ln2 + 0.71 ln1) = 0.09304 kJ/K From reversibility/irreversibility considerations the entropy change should be compared with \uf8f6 \uf8f7 \uf8f8. \uf8eb \uf8ec \uf8ed 2 \u222b 1 dQ T rev 158 _________________________________________________________ Applied Thermodynamics In this case; \uf8eb \uf8ec \uf8ed 2 \u222b 1 dQ T \uf8f6 \uf8f7 \uf8f8 rev = 0, while entropy change = 0.09304 kJ/K. Here (S2 \u2013 S1) > \uf8eb \uf8ec \uf8ed 2 \u222b 1 dQ T \uf8f6 \uf8f7 \uf8f8 rev, which means the process is irreversible. Ans. 20. Two tanks A and B are connected through a pipe with valve in between. Initially valve is closed and tanks A and B contain 0.6 kg of air at 90\u00b0C, 1 bar and 1 kg of air at 45\u00b0C, 2 bar respectively", ". Subsequently valve is opened and air is allowed to mix until equilibrium. Considering the complete system to be insulated determine the final temperature, final pressure and entropy change. Solution: In this case due to perfectly insulated system, Q = 0, Also W = 0 Let the final state be given by subscript f \u2032 and initial states of tank be given by subscripts \u2018A\u2019 and \u2018B\u2019. pA = 1 bar, TA = 363 K, mA = 0.6 kg; TB = 318K, mB = 1kg, pB = 2 bar \u2206Q = \u2206W + \u2206U 0 = 0 + {(mA + mB) + Cv.Tf \u2013 (mA.CvTA) \u2013 (mB.Cv.TB)} ( Tf = ). 0.6 363 1 318) + (0.6 1) Using gas law for combined system after attainment of equilibrium, Tf = 334.88 K, Final temperature = 334.88 K Ans. ( m pf = + ). m RT f A B ( V +V A ) B VA = A m RT A p A ; V B = m RT B B p B VA = 0.625 m3; VB = 0.456 m3 \u00d7 (1 0.6) 0.287 334.88 + (0.625 0.456) pf = + \u00d7 = 142.25 kPa \u21d2 Final pressure = 142.25 kPa Ans. Entropy change; \u2206S = {((mA + mB).sf) \u2013 (mA.sA + mBsB)} \u2206S = {mA(sf \u2013 sA) + mB (sf.\u2013 sB)} = \uf8f1 \uf8f4 \uf8f2 \uf8f4 \uf8f3 \uf8eb \uf8ec m C A \uf8ed ln p T f T A \u2212 R ln \uf8ed ln p T f T B \u2212 R ln p f p B \uf8f6 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe Considering Cp = 1.005 kJ/kg.K Entropy _______________________________________________________________________ 159 \u2206S = \uf8f1 \uf8f2 \uf8f3 0.6 1.005ln \uf8eb \uf8ec \uf8ed 334.88 363 \u2212 0.287 ln 142.25 100 \uf8f6 \uf8f7 \uf8f8 + \uf8eb 1 1.005ln \uf8ec \uf8ed 334.88 318 \u2212 0.287ln 142.25 200 \uf8fc \uf8f6 \uf8fd \uf8f7 \uf8f8\uf8fe \u2206S = { \u2013 0.1093 + 014977} = 0.04047 kJ/K Entropy produced = 0.04047 kJ/K Ans. 21. Three tanks of equal volume of 4m3 each are connected to each other through tubes of negligible volume and valves in between. Tank A contains air at 6 bar, 90\u00b0C, tank B has air at 3 bar, 200oC and tank C contains nitrogen at 12 bar, 50oC. Considering adiabatic mixing determine (i) the entropy change when valve between tank A and B is opened until equilibrium, (ii) the entropy change when valves between tank C; tank A and tank B are opened until equilibrium. Consider RAir = 0.287 kJ/kg.K, gAir = 1.4, RNitrogen = 0.297 kJ/kg. K and gNitrogen = 1.4. Solution: Let states in tanks A, B & C be denoted by subscripts A, B & C respectively. (i)When tank A and B are connected; Cv,Air = ( R r Air Air \u2212 = 0.718kJ/kg.K 1) After adiabatic mixing let the states be denoted by subscript \u2018D\u2019. Internal energy before mixing = Internal energy after mixing Air + = (, Air T. D, Air \u21d2 Using gas laws, TD = mA = mB = + + = = (. Air A. p V B B R T Air 10 4 \u00d7 0.287 363 = 23.04kg 2 \u00d7 \u00d7 3 10 4 \u00d7 0.287 573 Final temperature \u21d2 TD = \uf8f1 \uf8f2 \uf8f3 \u00d7 (23.04 363) + + (23.04 7.29) \u00d7 (7.29 573) = 7.29kg \uf8fc \uf8fd \uf8fe = 413.47 K Final pressure, pD=. Air D D R T m V D = pD = +.( R T m Air.287 413.47 (23.04 7.29) + (4 4) + = 449.89 kPa pD = 4.4989 bar 160 ________________________________", "_________________________ Applied Thermodynamics Entropy change, \u2206\u2206S = (SD \u2013 SA) + (SD \u2013 SB) \u2206 Air, ln T D T A \u2212. m R A Air ln + ) Air, ln T D T B \u2212. m R B Air ln ) \uf8fc \uf8fd \uf8fe p D p B \u2206S = \uf8f1\uf8eb \uf8f2\uf8ec \uf8ed \uf8f3 \u00d7 23.04 1.005ln \uf8eb \uf8ec \uf8ed 413.47 363 \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8 \u2212 \uf8eb \uf8ec \uf8ed \u00d7 23.04 0.287 ln \uf8eb \uf8ec \uf8ed 449.89 600 \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8 + \uf8eb \uf8ec \uf8ed 7.29 1.005ln \u00d7 \uf8eb \uf8ec \uf8ed 413.47 573 \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ed \u2013 7.29 0.287 ln \u00d7 \uf8eb \uf8ec \uf8ed 449.89 300 \uf8f6 \uf8f7 \uf8f8 \uf8fc \uf8f6 \uf8fd \uf8f7 \uf8f8\uf8fe = {3.014 + 1.904 + (\u2013 2.391) \u2013 0.848} Entropy chnage, \u2206S = 1.679 kJ/K Ans. (ii) After the three tanks A, B, and C are interconnected then the equilibrium will be attained amongst three. Equilibrium between A & B will result in state D as estimated in part (i) above. Thus it may be considered as the mixing of state D and nitrogen in tank C. Let the final state attained be \u2018F\u2019. After adiabatic mixing the final gas properties (as a result of mixing of air, state D and nitrogen, state C) may be estimated as under mc =. p V c c R Nitrogen. T c = 2 \u00d7 \u00d7 12 10 4 \u00d7 0.297 323 mc = 50.04 kg ; mD = mA + mB = 30.33 kg Cv, Nitrogen = \u03b3 R Nitrogen Nitrogen = 0.297 \u20131 (1.4 1) \u2212 = 0.7425 kJ/kgK Cp,Nitrogen = \u03b3 Nitrogen.Cv, Nitrogen = 1.4 \u00d7 0.7425 = 1.0395 kJ/kg.K mF = (mD + mc) = (23.04 + 7.29 + 50.04) = 80.37 kg Cv,F = {(mD.Cv, Air + mc.Cv, Nitrogen)/(mD + mC)} = 0.733 kJ/kg.K RF =. m R D Air ( Nitrogen ) Entropy _______________________________________________________________________ 161 = \uf8f1 \uf8f2 \uf8f3 \u00d7 (30.33 0.287 + \u00d7 50.04 0.297 80.37 \uf8fc \uf8fd \uf8fe RF = 0.293 kJ/kg.K By first law of thermodynamics, \u2206Q = \u2206W + \u2206U; here \u2206Q = 0, \u2206W = 0 Internal energy before mixing = Internal energy after mixing + m D m C T ). F mD.Cv,Air.TD + mc.Cv,Nitrogen.Tc = ( \u21d2 (30.33 \u00d7 0.718 \u00d7 413.47) + (50.04 \u00d7 0.7425 \u00d7 323) = 80.37 \u00d7 0.733 \u00d7 TF \u21d2 TF = 356.55 K; Final temperature after mixing = 356.55 K Final pressure after mixing; C v F pF = 80.37 0.293 356.55) + + (4 4 4) \u00d7 Entropy change after mixing, \u2206S = (SF \u2013 SD) + (SF \u2013 SC) pF = 699.68 kPa; Final pressure. \u2206S = \uf8f1 \uf8f2 m C D \uf8f3 ln p Air, T F T D \u2212 R Air ln Nitrogen, ln T F T C \u2212 R Nitrogen ln \uf8fc \uf8fd \uf8fe p F p C = \uf8f1 \uf8f2 30.33 1.005 ln \uf8f3 356.55 413.47 \u2212 0.287 ln 699.68 449.89 \uf8fc \uf8fd \uf8fe + 50.04 1.0395ln \uf8f1 \uf8f2 \uf8f3 356.55 323 \u2212 0.297 ln 699.", "68 1200 \uf8fc \uf8fd \uf8fe = \u2013 8.359 + 13.158 \u2206S = 4.799 kJ/kg.K Ans. 5.1 Discuss the significance of Clausius inequality. 5.2 Define the \u2018entropy\u2019. Also explain how it is a measure of irreversibility? 5.3 Explain the difference between isentropic process and adiabatic process. 5.4 How does the second law of thermodynamics overcome limitations of first law of thermodynamics? 5.5 Show that entropy of universe is increasing. 5.6 Is the adiabatic mixing of fluids irreversible? If yes, explain. 5.7 Why does entropy generally increase? Explain. 5.8 Explain the entropy principle and apply it to a closed system. 5.9 How the feasibility of any process can be ensured? 5.10 Give the third law of thermodynamics. 5.11 Explain why the slope of constant volume line is more than the slope of constant pressure line on T\u2013S diagram. 5.12 Explain, whether the arrangement shown below for a reversible engine is feasible. If no then why? Give the correct arrangement. 162 _________________________________________________________ Applied Thermodynamics 20 K 400 K 0.82 mJ 5 mJ HE 840 kJ 4.98 mJ 300 K Fig. 5.15 5.13 Using second law of thermodynamics check the following and also indicate nature of cycle. (i) Heat engine receiving 1000 kJ of heat from a reservoir at 500 K and rejecting 700 kJ heat to a sink at 27\u00baC. (ii) Heat engine receiving 1000 kJ of heat from a reservoir at 500 K and rejecting 600 kJ of heat to a sink at 27\u00baC. (i) Possible, irreversible cycle (ii) Possible, reversible cycle 5.14 Determine the change in entropy of air during it's heating in a perfectly insulated rigid tank having 5 kg of air at 2 atm. Air is heated from 40\u00baC to 80\u00baC temperature. atmosphere at 27\u00baC during its operation. 5.15 Calculate change in entropy of air during the process in which a heat engine rejects 1500 kJ of heat to [5 kJ/K] 5.16 Determine the final temperature and total entropy change during a process in which metal piece of 5 kg at 200\u00baC falls into an insulated tank containing 125 kg of water at 20\u00baC. Specific heat of metal = 0.9 [21.53\u00baC, 0.592 kJ/K] kJ/kg.K, Specific heat of water = 4.184 kJ/kg.K. 5.17 Show that for air undergoing isentropic expansion process; ds = c p d v v + c v dp p 5.18 Determine the change in entropy of air, if it is heated in a rigid tank from 27\u00baC to 150\u00baC at low pressure. [246.8 J/kg.K] 5.19 An electrical resistance of 100 ohm is maintained at constant temperature of 27\u00baC by a continuously flowing cooling water. What is the change in entropy of the resistor in a time interval of one minute? [0] 5.20 A water tank of steel is kept exposed to sun. Tank has capacity of 10 m3 and is full of water. Mass of steel tank is 50 kg and during bright sun temperature of water is 35\u00baC and by the evening water cools down to 30\u00baC. Estimate the entropy change during this process. Take specific heat for steel as 0.45 kJ/ [5.63 kJ/K] kg.K and water as 4.18 kJ/kg.K. 5.21 Heat engine operating on Carnot cycle has a isothermal heat addition process in which 1 MJ heat is supplied from a source at 427\u00baC. Determine change in entropy of (i) working fluid, (ii) source, (iii) total [1.43 kJ/K, \u2013 1.43 kJ/K, 0] entropy change in process. 5.22 A system operating in thermodynamic cycle receives Q1 heat at T1 temperature and rejects Q2 at temperature T2. Assuming no other heat transfer show that the net work developed per cycle is given as, W cycle = gen Entropy _______________________________________________________________________ 163 where Sgen is amount of entropy produced per cycle due to irreversibilities in the system. 5.23 A rigid tank contains 5 kg of ammonia at 0.2 MPa and 298 K. Ammonia is then cooled until its pressure drops to 80 kPa. Determine the difference in entropy of ammonia between initial and final state. [\u201314.8 kJ/K] 5.24 Determine the change in entropy in each of the processes of a thermodynamic cycle having following processes; (i) Constant pressure cooling from 1 to 2, P1 = 0.5 MPa, V", "1 = 0.01 m3 (ii) Isothermal heating from 2 to 3, P3 = 0.1 MPa, T3 = 25\u00baC, V3 = 0.01 m3 (iii) Constant volume heating from 3 to 1. Take Cp = 1 kJ/kg. K for perfect gas as fluid. [\u20130.0188 kJ/kg. K, 0.00654 kJ/kg. K, 0.0134 kJ/kg. K] 5.25 Conceptualize some toys that may approach close to perpetual motion machines. Discuss them in detail. 5.26 Heat is added to air at 600 kPa, 110\u00b0C to raise its temperature to 650\u00b0C isochorically. This 0.4 kg air is subsequently expanded polytropically up to initial temperature following index of 1.32 and finally compressed isothermally up to original volume. Determine the change in entropy in each process and pressure at the end of each process. Also show processes on p-V and T-s diagram, Assume [0.2526 kJ/K, 0.0628 kJ/K, 0.3155 kJ/K Cv = 0.718 kJ/kg.K, R = 0.287 kJ/kg.K 1445 kPa, 38.45 kPa] 5.27 Air expands reversibly in a piston-cylinder arrangement isothermally at temperature of 260\u00b0C while its volume becomes twice of original. Subsequently heat is rejected isobarically till volume is similar to original. Considering mass of air as 1 kg and process to be reversible determine net heat interaction and total change in entropy. Also show processes on T-s diagram. [\u2013 161.8 kJ/kg, \u2013 0.497 kJ/kg.K] 5.28 Ethane gas at 690 kPa, 260\u00b0C is expanded isentropically up to pressure of 105 kPa, 380K. Considering initial volume of ethane as 0.06 m3 determine the work done if it behaves like perfect gas. compressed up to 690 kPa following p.V. 1.4 = constant. Also determine the change in entropy and heat transfer if the same ethane at 105 kPa, 380K is [0.8608 kJ/K, 43.57 kJ] 5.29 Determine the net change in entropy and net flow of heat from or to the air which is initially at 105 kPa, 15\u00b0C. This 0.02 m3 air is heated isochorically till pressure becomes 420 kPa and then cooled isobarically [\u2013 0.011kJ/K, \u2013 6.3 kJ] back up to original temperature. 5.30 Air initially at 103 kPa, 15\u00b0C is heated through reversible isobaric process till it attains temperature of 300\u00b0C and is subsequently cooled following reversible isochoric process up to 15\u00b0C temperature. [101.9 kJ, 0.246 kJ/K] Determine the net heat interaction and net entropy change. 5.31 Calculate the entropy change when 0.05 kg of carbon dioxide is compressed from 1 bar, 15\u00b0C to 830 kPa pressure and 0.004m3 volume. Take Cp = 0.88 kJ/kg.K. This final state may be attained following [0.0113 kJ/K] isobaric and isothermal process. 5.32 Two insulated tanks containing 1 kg air at 200 kPa, 50\u00b0C and 0.5 kg air at 100 kPa, 80\u00b0C are connected through pipe with valve. Valve is opened to allow mixing till the equilibrium. Calculate the amount of [0.03175 kJ/K] entropy produced. 164 _________________________________________________________ Applied Thermodynamics 6 6.1 INTRODUCTION Engineering systems have an inherent requirement of some substance to act as working fluid i.e. transport agent for energy and mass interactions. Number of working fluids are available and are being used in different systems suiting to the system requirements. Steam is also one of such working fluids used exhaustively because of its favourable properties. In thermal power plants steam is being extensively used. Water has capability to retain its chemical composition in all of its\u2019 phases i.e. steam and ice, and also it is almost freely available as gift of nature. Pure substance refers to the \u201csubstance with chemical homogeneity and constant chemical composition.\u201d H2O is a pure substance as it meets both the above requirements. Any substance, which undergoes a chemical reaction, cannot be pure substance. 6.2 PROPERTIES AND IMPORTANT DEFINITIONS Pure substance as defined earlier is used for operating various systems, such as steam is used for power generation in steam power plants. Hence, for thermodynamic analysis thermodynamic properties are required. Pressure and temperature", " are the properties that can be varied independently over wide range in a particular phase. Therefore, the behaviour of properties of pure substance have to be studied and mathematical formulations be made for their estimation. Various dependent properties discussed ahead shall be enthalpy, internal energy, specific volume, entropy etc. Some of terms used in discussion ahead are given as under. (a) Sensible heating: It refers to the heating of substance in single phase. It causes rise in temperature of substance. In case of cooling in above conditions it shall be called sensible cooling. (b) Latent heating: It is the heating of substance for causing its phase change without any change in it\u2019s temperature. If heat is extracted for causing phase change without any change in its temperature it will be called latent cooling. (c) Normal boiling point: It is the temperature at which vapour pressure equals to atmospheric pressure and at this temperature phase change from liquid to gas begins. (d) Melting point: It is the temperature at which phase change from solid to liquid takes place upon supplying latent heat. (e) Saturation states: Saturation state of a substance refers to the state at which its phase transformation takes place without any change in pressure and temperature. These can be saturated solid state, saturated liquid state and saturated vapour state. For example saturated Thermodynamic Properties of Pure Substance ________________________________________ 165 vapour state refers to the state of water at which its phase changes to steam without varying pressure and temperature. (f) Saturation pressure: It is the pressure at which substance changes its phase for any given temperature. Such as at any given temperature water shall get converted into steam at a definite pressure only, this pressure is called saturation pressure corresponding to given temperature. For water at 100\u00b0C the saturation pressure is 1 atm pressure. (g) Saturation temperature: It refers to the temperature at which substance changes its phase for any given pressure. For water at 1 atm pressure the saturation temperature is 100\u00b0C. (h) Triple point: Triple point of a substance refers to the state at which substance can coexist in solid, liquid and gaseous phase in equilibrium. For water it is 0.01\u00b0C i.e. at this temperature ice, water and steam can coexist in equilibrium. Table 6.1 given below gives triple point data for number of substances. Table 6.1 Triple point and critical point Triple point Critical point Pressure, kPa 0.611 5.1 7.0 0.15 12.5 6.1 517 1.65 \u00d7 10\u20137 Temperature, \u00b0C Pressure, MPa Temperature, \u00b0C 0.01 \u2013 271 \u2013 259 \u2013 219 \u2013 210 \u2013 78 \u2013 57 \u2013 39 22.12 0.23 1.3 5.0 3.4 11.3 7.39 18.2 374.15 \u2013 268 \u2013 213 \u2013 119 \u2013 147 132 31 899 Substance Water Helium Hydrogen Oxygen Nitrogen Ammonia Carbondioxide Mercury (i) Critical states: \u201cCritical state refers to that state of substance at which liquid and vapour coexist in equilibrium.\u201d In case of water at 22.12 MPa, and 374.15\u00b0C the water and vapour coexist in equilibrium, thus it is the highest pressure and temperature at which distinguishable water and vapour exist together. Data for critical state of many substances is given in the table 6.1. Specific volume at critical point for water is 0.00317 m3/kg. (j) Dryness fraction: It is the mass fraction of vapour in a mixture of liquid and vapour at any point in liquid-vapour mixture region. It is generally denoted by \u2018x\u2019. It is also called quality of steam. (k) Compressed liquid or subcooled liquid: Liquid at temperature less than saturation temperature corresponding to a given pressure is called compressed liquid or subcooled liquid. Degree of subcooling is given by the temperature difference between liquid temperature and saturation temperature of liquid at given pressure. Degree of subcooling = Saturation temperature at given pressure \u2013 Temperature of liquid. (l) Superheated steam: Steam having temperature more than the saturation temperature corresponding to given pressure is called superheated steam. Amount of superheating is quantified by degree of superheating. Degree of superheating is given by difference between temperature of steam and saturation temperature at given pressure. Degree of superheating = Temperature of steam \u2013 Saturation temperature at given pressure. 166 _________________________________________________________ Applied Thermodynamics 6.3 PHASE TRANSFORMATION PROCESS Let us study phase transformation from ice to steam by taking ice at \u201320\u00b0C in an open vessel i.e. at atmospheric pressure, and heat it from bottom. Salient states passed through the phase change are as given under. Melting point of ice is 0\u00b0C and boiling point of water is 100\u00b0C for water at 1 atmospheric pressure. 1 atm ( )a", " Ice at \u201320 \u00b0C p c2 c1 c c3 d2 a2 b2 d1 a1 b1 b a d d3 a3 b3 e2 e1 e e3 f2 f1 f f3 p2 p1 1 atm p3 V b ( ) Volume change with pressure Fig. 6.1 Phase transformation process Say initial state is given by \u2018a\u2019 at \u201320\u00b0C and 1 atmospheric pressure. (i) Upon heating the ice its temperature increases from \u201320\u00b0C to 0\u00b0C while being in solid phase. Temperature increase is accompanied by increase in volume and new state \u2018b\u2019 is attained. This heating is sensible heating as heating causes increase in temperature in same phase. (ii) After ice reaches to 0\u00b0C, the melting point, it is ready for phase transformation into water. Further heat addition to it causes melting and now water at 0\u00b0C is available. This heating is called latent heating and heat added is called latent heat. New state attained is \u2018c\u2019 and volume gets reduced due to typical characteristic of water. As defined earlier state \u2018b\u2019 is called saturation solid state as phase can change here without any change in pressure and temperature. State \u2018c\u2019 is called saturated liquid state with respect to solidification. (iii) Further heating of water at 0\u00b0C shall cause increase in its temperature upto 100\u00b0C. This heat addition is accompanied by increase in volume and state changes from \u2018c\u2019 to \u2018d\u2019 as shown on pV diagram. Here typical behaviour of water from 0 to 4\u00b0C is neglected. This heating is sensible heating in liquid phase. State \u2018d\u2019 is called saturated liquid state with respect to vaporization. Thus, there are two saturated liquid states \u2018c\u2019 and \u2018d\u2019 depending upon direction of transformation. (iv) Water at 100\u00b0C and 1 atmosphere is ready for getting vaporized with supply of latent heat of vaporization. Upon adding heat to it the phase transformation begins and complete liquid gradually gets transformed into steam at state \u2018e\u2019. This phase change is accompanied by large increase in volume. Heating in this zone is called latent heating. State \u2018e\u2019 is called saturated vapour state or saturated steam state. (v) Steam at 100\u00b0C upon heating becomes hotter and its temperature rises. Say, the heating causes temperature rise upto 200\u00b0C. This increase in temperature is also accompanied by increase in volume up to state \u2018f\u2019 as shown on p-V diagram. This heating is sensible heating in gaseous phase. Similar phase transformations can be realized at different pressures and such salient states be identified and marked. Joining all saturated solid states at different pressures results in a locii, which is called \u201csaturated solid line.\u201d Similarly, joining all saturated liquid states with respect to solidification and saturated liquid states with respect to vaporization results in two \u2018saturated liquid lines\u2019. Locii of all saturated vapour states at different pressure generates \u2018saturated vapour line\u2019. The lines thus obtained are shown in Fig. 6.2 detailing p-V diagram for water. Point at which \u201csaturated liquid line\u201d with respect Thermodynamic Properties of Pure Substance ________________________________________ 167 P SL Sat solid line Sat liquid lines L CL S Liquid + Vapour LV Critical Point ( )CP Sat vapour line Isotherms V Solid + Vapour SV Vapour Triple point line V Fig. 6.2 P-V diagram for water to vaporization meets with \u201csaturated vapour line\u201d is called \u201cCritical point\u201d. Critical point is also sometimes referred to as \u201cCritical state\u201d. P SL Sat liquid lines Sat Solid line L S Liquid + Vapour, LV Critical Point ( )CP Sat vapour line Isotherms V Solid + Vapour, SV Triple point line V Fig. 6.3 P-V diagram for Carbon dioxide On P-V diagram region marked S shows the solid region, mark SL shows solid-liquid mixture region, mark LV shows liquid-vapour mixture region, mark CL shows compressed liquid region. Triple point line is also shown and it indicates coexistence of solid, liquid and gas in equilibrium. Region marked SV and lying below triple point line is sublimation region where solid gets transformed directly into vapour upon supply of latent heat of sublimation. P-V diagram for substance which has different characteristics of contraction upon freezing is also shown here, Fig. 6.3. Carbon dioxide is such substance. Difference in p-V diagrams for two different substances (water and CO2) may be understood from here. 6.4 GRAPHICAL REPRESENTATION OF PRESSURE, VOLUME AND TEMPERATURE Graphical representations based on variation of thermodynamic properties can be obtained from the study of actual", " phase transformation process. In earlier discussion the variation of pressure and volume 168 _________________________________________________________ Applied Thermodynamics during phase transformation from ice to steam has been shown and explained. On these lines the standard graphical representations in terms of p-V, T-V, p-T and p-V-T for water can be obtained. p-V diagram as obtained has already been discussed in article 6.2. (i) T-V diagram: It gives variation of temperature with volume. Let us look at different steps involved in phase transformation and how are the temperature variations. a \u2013 b: Temperature rises from \u201320\u00b0C to 0\u00b0C with volume increase, (ice) b \u2013 c: Temperature remains constant at 0\u00b0C due to phase change and volume decreases, (ice to water) f d e 200 \u00b0C 100 \u00b0C T 0 \u00b0C \u201320\u00b0C c a b V Fig. 6.4 Temperature-Volume variation at 1 atm pressure c \u2013 d: Temperature increases from 0\u00b0C to 100\u00b0C and volume increases, (water). d \u2013 e: Temperature remains constant at 100\u00b0C, phase changes from liquid to vapour and volume increases, (water to steam). e \u2013 f: Temperature increases from 100\u00b0C to 200\u00b0C and volume increases, (steam). Similar to above, the T-V variations at different pressures can be obtained and identical states i.e. saturation states be joined to get respective saturation lines as shown in Fig. 6.5 ahead Critical point P = constant P = constant 1 2 Liquid-vapour region LV V Vapour region Saturated liquid line Saturated vapour line V Fig. 6.5 T-V diagram for water (ii) P-T diagram: It is the property diagram having pressure on Y-axis and temperature on X-axis. This can also be obtained by identifying and marking salient states during phase transformation and subsequently generating locii of identical states at different pressures. For the phase change discussed in article 6.3, pressure and temperature variation shall be as described ahead. This phase change occurs at constant pressure of 1 atm. Thermodynamic Properties of Pure Substance ________________________________________ 169 a \u2013 b: Temperature rises from \u201320\u00b0C to 0\u00b0C, phase is solid. b \u2013 c: Temperature does not rise, phase changes from ice to water (solid to liquid) c \u2013 d: Temperature rises from 0\u00b0C to 100\u00b0C, phase is liquid d \u2013 e: Temperature does not rise, phase changes from liquid to gas (water to steam) e \u2013 f: Temperature rises from 100\u00b0C to 200\u00b0C, phase is gas (steam)., d e2 2, d e1 1 d, e f2 f1 f p2 p1 1 atm a2 a1 a b,c2 2 1, b c1 b, c \u2013 20 \u00b0C p T Fig. 6.6 P-T variation for phase transformation at constant pressures Fusion curve Vaporisation curve CP p S L V Triple point 0.01 \u00b0C T Fig. 6.7 P-T diagram for water (iii) P-V-T surface: This is the three dimensional variation describing three thermodynamic properties P, V and T. As we know that for defining a state at least two properties are needed, therefore on P-V-T surface also, by knowing any two the third can be seen and continuous variation of these properties is available. Here pressure, volume and temperature are taken on mutually perpendicular axis and surface obtained is depicted below with all salient points. P-V-T surface shall be different for different substances depending on their characteristics. Here P-V-T surface for two substances having opposite characteristics are given in Fig. 6.8. P Critical point Liquid Solid LV Vap SV T Triple point line V (A) P-V-T surface for water (which expands upon freezing.) 170 _________________________________________________________ Applied Thermodynamics P Liquid Critical point Solid LV Gas SV Vap T V Triple point line (B) P-V-T surface for Carbon dioxide (which contracts upon freezing.) Fig. 6.8 P-V-T surface for water and CO2 6.5 THERMODYNAMIC RELATIONS INVOLVING ENTROPY Entropy change during phase transformation process can be studied with temperature at any given pressure, based on discussions in article 6.3. Entropy changes for every state change are estimated and plotted on T-S diagram for 1 atm. pressure and \u2018m\u2019 mass of ice. a \u2013 b: Temperature changes from \u201320\u00b0C to 0\u00b0C. Phase is solid Entropy change \u2206Sa \u2013 b = Sb \u2013 Sa = Tb \u222b Ta dQ \u2212 a b T Here and or dQa \u2013 b = m \u00b7 cp, ice \u00b7 dT Ta = 253 K, Tb = 273 K. \u2206Sa \u2013 b = Tb \u222b Ta", " m c dT,ice. p T b \u2013 c: Temperature does not change. It is constant at 0\u00b0C. Entropy change, Here, \u2206Sb \u2013 c = Sc \u2013 Sb = \u2206Qb \u2013 c = Latent heat of fusion/melting of ice at 1 atm \u2212\u2206 Q b c T b Tb = 273 K c \u2013 d: Temperature rises from 0\u00b0C to 100\u00b0C. Phase is liquid. Entropy change, \u2206Sc \u2013 d = Sd \u2013 Sc = Td \u222b Tc dQ \u2212 c d T Thermodynamic Properties of Pure Substance ________________________________________ 171 Here or, dQc \u2013 d = m \u00b7 cp,water \u00b7 dT and Tc = 273 K, Td = 373 K. \u2206Sc \u2013 d = Td \u222b Tc m c p dT,water T d \u2013 e: Temperature does not change. It is constant at 100\u00b0C. Entropy change, \u2212\u2206 Q d e T d Here, \u2206Qd \u2013 e = latent heat of vaporization at 1 atm. and \u2206Sd \u2013 e = Se \u2013 Sd = Td = 373 K. e \u2013 f: Temperature rises from 100\u00b0C to 200\u00b0C. Phase is gas (steam) Entropy change, \u2206Se \u2013 f = Sf \u2013 Se = T f \u222b Te dQ \u2212 e f T Here, dQe \u2013 f = m \u00b7 cp,steam \u00b7 dT and Te = 373 K, Tf = 473 K or, \u2206Se \u2013 f = T f \u222b Te m c p dT,steam. T Above entropy change when plotted on T\u2013S axis result as below. 200 C\u00b0 100 \u00b0C T 0 C \u00b0 \u00b0 \u201320 C 1 atm pressure f d e b c a Sc\u2013d s Sb\u2013c Sa\u2013b Se\u2013f Sd\u2013e Fig. 6.9 Temperature-entropy variation for phase change at 1 atm. On the similar lines, (as discussed above) the T\u2013S variation may be obtained for water at other pressures. In the T\u2013S diagram different important zones and lines are earmarked. Out of whole T\u2013S diagram, generally the portion detailing liquid, liquid-vapour zone, vapour zone are of major interest due to steam undergoing processes and subsequently condensed and evaporated. Both T\u2013S diagrams are shown here. 172 _________________________________________________________ Applied Thermodynamics S T Critical state SL L P2 = C P1 = C LV SV s T L P = Const P P2 1> V Triple point line Critical point tc = 374.15 \u00b0C P = 221.2 bar Saturated vapour line P= Const LV Saturated liquid line Sfg V P increasing Sf Sg Fig. 6.10 T\u2013s diagram for water Fig. 6.11 T\u2013s diagram showing liquid and vapour regions 6.6 PROPERTIES OF STEAM For thermodynamic analysis the following thermodynamic properties of steam are frequently used. \u201cSaturation temperature (T), saturation pressure (P), specific volume (\u03bd), enthalpy (h), entropy (s), internal energy (u)\u201d Let us look at T\u2013S diagram below. Saturated liquid line Saturated vapour line CP p3 d3 Super heated region Subcooled region b3 T3 a3 b2 T2 L T T1 a2 a1 b1 xk j3 e3 c3 k3 k2 j2 e2 c2 e1 k1 j1 xj s d2 V p2 p1 d1 say, x k x j x e = 0.10 = 0.80 = 0.90 c1 xe Constant dryness fraction lines Fig. 6.12 T\u2013s diagram Thermodynamic properties and nomenclature used is indicated inside bracket along with property. Discussion is based on unit mass of steam/mixture. T\u2013S diagram for 2-phases i.e. liquid and vapour has saturated liquid line and saturated vapour line meeting at critical point. Three constant pressure lines corresponding to pressure P1, P2 and P3 are shown. Let us take a constant pressure line for pressure p1 which has states a1, b1, k1, j1, e1, c1, d1 shown upon it. Region on the left of saturated liquid line is liquid region. Region enclosed between saturated liquid line and saturated vapour line is liquid-vapour mixture Thermodynamic Properties of Pure Substance ________________________________________ 173 region or also called wet region. Region on the right of saturated vapour line is vapour region. All the states lying on saturated liquid line are liquid (water) states shown as b1, b2, b3 at different pressures. States a1, a2 and a3 are the states lying in subcooled", " region. Compressed liquid or subcooled liquid exists at a1, a2 and a3 at pressures p1, p2 and p3. Degree of sub cooling at a1 = Saturation temperature for pressure p1 \u2013 Temperature at a1 = (T1 \u2013 Ta1) where T1, T2, T3 are saturation temperatures at pressures p1, p2 and p3. At constant pressure p1 when we move towards right of state b1 then the phase transformation of water into steam (vapour) just begins. This conversion from liquid to vapour takes place gradually till whole liquid gets converted to vapour. Phase transformation into vapour gets completed at c1, c2, c3 at pressures p1, p2 and p3 respectively. States c1, c2, c3 are called saturated vapour states and substance is completely in vapour phase at these points. Beyond state c1, at pressure p1 it is vapour phase and sensible heating shall cause increase in temperature. State d1 is the state of steam called superheated steam. Superheated steam exists at d1, d2, and d3 at pressures p1, p2 and p3. Degree of superheat at d1 = Td1 \u2013 T1 Similarly, degree of superheat at d2 = Td2 \u2013 T2, degree of superheat at d3 = Td3 \u2013 T3. As pressure is increased upto critical pressure then constant pressure line is seen to become tangential to critical point (CP) at which water instantaneously flashes into vapour. In the wet region states k1, j1 and e1 are shown at pressure p1. At state b1 mixture is 100% liquid At state k1 mixture has larger liquid fraction, say 90% liquid fraction and 10% vapour fraction. At state j1 mixture has say 20% liquid fraction and 80% vapour fraction. At state e1 mixture has say 10% liquid fraction and 90% vapour fraction. At state c1 mixture is 100% vapour. Similarly, from explanation given above the states b2, k2, j2, e2, c2 and b3, k3, j3, e3 and c3 can be understood at pressures p2 and p3 respectively. At any pressure for identifying the state in wet region, fraction of liquid and vapour must be known, for identifying state in subcooled region, degree of subcooling is desired and for identifying state in superheated region, degree of superheating is to be known. For wet region a parameter called dryness fraction is used. Dryness fraction as defined earlier can be given as, x: Dryness fraction; x = Mass of vapour Mass of liquid + Mass of vapour Dryness fraction values can be defined as follows; Dryness fraction at state b1 = 0 Dryness fraction at state k1 = 0.10 \uf8f1 \uf8f2 \uf8f3 0.10 (m) (0.9 + 0.1)m as assumed = \uf8fc \uf8fd \uf8fe Dryness fraction at state j1 = 0.80 Dryness fraction at state e1 = 0.90 Dryness fraction at state c1 = 1.00 Thus, saturated liquid line and saturated vapour line are locii of all states having 0 and 1 dryness fraction values. 174 _________________________________________________________ Applied Thermodynamics At critical point dryness fraction is either 0 or 1. For different pressures the locii of constant dryness fraction points may be obtained and it yields constant dryness fraction lines corresponding to xk(0.10), xj(0.80), xe(0.90), as shown by dotted lines. Let us use subscript \u2018f \u2019 for liquid states and \u2018g\u2019 for vapour states. Therefore, enthalpy corresponding to saturated liquid state = hf, Enthalpy corresponding to saturated vapour state = hg. Similarly, entropy may be given as sf and sg. Specific volume may be given as vf and vg. Internal energy may be given as uf and ug. At any pressure for some dryness fraction x, the total volume of mixture shall comprise of volume occupied by liquid and vapour both. Total volume, V = Vf + Vg. Similarly, Total mass, m = mf + mg, i.e. mass of liquid and mass of vapour put together substituting for volume, where m is total mass and v is specific volume of mixture m\u00b7v = mf \u00b7 vf + mg.vg. or or From definition, \uf8eb \uf8ec v = 1 \u2013 \uf8ed gm m \uf8eb \uf8ec v = 1 \u2013 \uf8ed gm m \uf8eb \uf8ec v = 1 \u2013 \uf8ed gm m \uf8f6 \uf8f7 \uf8f8 \uf8f6 \ufffd", "\ufffd\ufffd \uf8f8 \uf8f6 \uf8f7 \uf8f8 \u00b7 vf + \uf8eb \uf8ec \uf8ed gm m \uf8f6 \uf8f7 \uf8f8 vg \uf8eb \uf8ec \uf8ed gm m \uf8f6 \uf8f7 \uf8f8 \u00b7 vg \u00b7 vf + \u00b7 vf + \uf8eb \uf8ec \uf8ed gm m \uf8f6 \uf8f7 \uf8f8 \u00b7 vg x = m g m+ g m f = gm m or, or, or, here vfg indicates change in specific volume from liquid to vapour v = (1 \u2013 x) \u00b7 vf + x \u00b7 vg v = vf + x \u00b7 (vg \u2013 vf) v = vf + x \u00b7 vfg. Similarly, enthalpy, entropy and internal energy may be defined for such states in wet region. i.e. or here hfg is difference in enthalpy between saturated liquid and vapour states. Actually hfg is energy or heat required for vaporization or heat to be extracted for condensation i.e. latent heat. h = hf + x (hg \u2013 hf) h = hf + x \u00b7 hfg Similarly, s = sf + x \u00b7 sfg where sfg = sg \u2013 sf u = uf + x \u00b7 ufg where ufg = ug \u2013 uf. Thermodynamic Properties of Pure Substance ________________________________________ 175 6.7 STEAM TABLES AND MOLLIER DIAGRAM Steam being pure substance has its unique and constant properties at different pressures and temperatures. Therefore, thermodynamic properties can be estimated once and tabulated for future use. Steam table is a tabular presentation of properties such as specific enthalpy, entropy, internal energy and specific volume at different saturation pressures and temperatures. Steam table may be on pressure basis or on temperature basis. The table on pressure basis has continuous variation of pressure and corresponding to it : saturation temperature (Tsat), enthalpy of saturated liquid (hf), enthalpy of saturated vapour (hg), entropy of saturated liquid (sf), entropy of saturated vapour (sg), specific volume of saturated liquid (vf), specific volume of saturated vapour (vg), internal energy of saturated liquid (uf), internal energy of saturated vapour (ug) are given on unit mass basis, i.e. as shown in table 6.2. Similar to above the temperature based table which gives continuous variation of temperature and corresponding to it saturation pressure and other properties as hf, hg, hfg, sf, sg, sfg, vf, vg, uf, ug and ufg are given. Similarly, steam properties for superheated steam are also estimated and tabulated at some discrete pressures for varying degree of superheat. Super heated steam table are available for getting enthalpy, entropy, specific volume and internal energy separately. Example of superheated steam table for enthalpy is given here: Table 6.2 Pressure based steam table Enthalpy Entropy Specific volume Internal energy hf hg hfg sf sg sfg vf vg kJ/kg kJ/kg kJ/kg kJ/kg\u00b0K kJ/kg\u00b0K kJ/kg\u00b0K m3/kg m3/kg uf kJ/kg ug kJ/kg ufg kJ/kg Table 6.3 Temperature based steam table Enthalpy Entropy Specific volume Internal energy Pressure Sat. Temp. Tsat \u00b0C Temperature Sat. Pressure \u00b0C kPa hf hg hfg sf sg sfg vf vg kJ/kg kJ/kg kJ/kg kJ/kg\u00b0K kJ/kg\u00b0K kJ/kg\u00b0K m3/kg m3/kg uf kJ/kg ug kJ/kg ufg kJ/kg Table 6.4 Superheated steam table for enthalpy Enthalpy values for varying degree of superheat (kJ/kg) Pressure Sat. temp. \u00b0C (Tsat) kPa T1 T3 Here T1, T2, T3, T4, T5, T6, T7... are more than Tsat and have increasing value of degree of superheat. Steam tables as discussed above are available in appendix in this text book. Mollier diagram is the enthalpy-entropy (h\u2013s) diagram for steam. This diagram is obtained on the T6 T2 T4 T5 T7 basis of following equation depending upon the phase transformation as discussed earlier. Tds = dh \u2013 vdp. (First and second law combined) For constant pressure \uf8eb \uf8ec \uf8ed dh ds \uf8f6 \uf8f7 \uf8f8 p = T 176 _________________________________________________________ Applied Thermodynamics Enthal", "py entropy diagram as obtained for all phases of water is as given in Figure 6.13 here. Generally, liquid and vapour region is only of interest in engineering systems, so mostly used portion of h\u2013s diagram is taken out and shown in Fig. 6.14. It is popularly known as mollier diagram or mollier chart. Critical point L SL S h LV SV V Saturated vapour line Triple point line Saturated liquid line s Fig. 6.13 Enthalpy-entropy diagram for all phases P =221.2 bar cr Critical point Sat liq. line p = c p = c C : Constant Constant temperature lines h hg ht = st sfg s LV sg V T = c T = c Saturated vapour line Fig. 6.14 h\u2013s diagram (Mollier diagram) Different significant lines such as saturated liquid line, saturated vapour line, isobaric lines, isothermal lines, constant specific volume lines, constant dryness fraction lines are shown upon Mollier diagram. Nature of variation of different lines can be explained from the real behaviour of substance and mathematical expression based on combination of first and second law. Such as, why isobaric lines diverge from one another? This is due to the increase in saturation temperature with increase in pressure. Slope of isobar is equal to saturation temperature as shown in the beginning, therefore it also increases with increasing pressure. Thermodynamic Properties of Pure Substance ________________________________________ 177 Why isothermal lines are not visible in wet region? It is because constant temperature lines and constant pressure lines coincide upon in wet region. For every pressure there shall be definite saturation temperature which remains constant in wet region. Mollier chart is also given in appendix, at the end of this book. 6.8 DRYNESS FRACTION MEASUREMENT Dryness fraction is the basic parameter required for knowing the state of substance in liquid-vapour mixture region (wet region). For any pressure the dryness fraction varies from 0 to 1 in the wet region i.e. from saturated liquid to saturated vapour. Dryness fraction being ratio of mass of vapour and total mass of substance can be conveniently estimated if these two mass values are known. It may also be termed as \u2018quality\u2019 of steam or \u2018dryness factor\u2019. Dryness fraction = Mass of vapour Total mass i.e. (mass of vapour + mass of liquid) Here we shall be looking into standard methods available for dryness fraction measurement. These are; (i) Throttling calorimeter (ii) Separating calorimeter (iii) Separating and throttling calorimeter (iv) Electrical calorimeter. Const. pressure p1 CP h 1 p2 2 Saturated vapour line Constant dryness fraction lines s Fig. 6.15 Throttling process on h\u2013s diagram (i) Throttling calorimeter: In this the throttling action is utilised for getting dryness fraction. If a mixture is throttled, then upon throttling its enthalpy remains constant before and after throttling. Let us look upon states of substance on h\u2013s diagram, before and after throttling. Wet mixture being at state \u20181\u2019 initially, attains a new state \u20182\u2019 upon being throttled upto pressure p2. This state at the end of throttling lies in the superheated region such that, Say, dryness fraction at state 1 is x, then enthalpy at this point can be given as h1 = h2 h1 = hf at p + x \u00d7 hfg at p 1 1 178 _________________________________________________________ Applied Thermodynamics In the above expression h f p at 1 and h fg p at 1 can be seen from steam table if pressure of wet steam is known. Also the enthalpy at state 2 (end of throttling) can be seen from superheated steam table if pressure and temperature at \u20182\u2019 are known. Substituting in, h1 = h2 = h2 + x \u00d7 hfg at p hf at p 1 1 are all known as explained above., hfg at p 1 1 Here h2, hf at p Therefore, x = h 2 \u2212 h f p at 1 h fg p at 1 Now the arrangements are to be made for, (a) measurement of pressure of wet steam in the begining, (b) throttling of wet mixture such that state at the end of throttling lies in superheated region, (c) measurement of pressure, temperature of throttled steam. Arrangement used in throttling calorimeter is as shown in Fig. 6.16. Steam main Throttle valve Sampling bulb P1 T2 Temperature measurement after throttling T2 Pressure measurement before throttling Exhaust to condenser Pressure of steam after throttling Fig. 6.16 Throttling cal", "orimeter (ii) Separating calorimeter: In this type of calorimeter the known mass of wet mixture is collected through a sampling bulb and sent to a separating chamber. Separating chamber has the series of obstacles, and zig-zag path inside it so that when mixture passes through them the liquid particles get separated due to sudden change in direction of flow and gravity action. Liquid thus separated out is collected in a collection tank and is measured. Thus by knowing the two mass values dryness fraction can be estimated as; Dryness fraction = {(Total mass) \u2013 (Mass of liquid)} Total mass Layout of separating calorimeter is given in Fig. 6.17. Thermodynamic Properties of Pure Substance ________________________________________ 179 Steam main Wet mixture Valve Sampling bulb Perfectly insulated Vapour out Separating chamber Valve Liquid Collection tank Fig. 6.17 Separating calorimeter (iii) Separating and throttling calorimeter: Some times when the wet mixture is extremely wet then upon throttling state of steam is unable to become superheated. In such situations mixture is first passed through separating calorimeter to reduce liquid fraction in it and subsequently this less wet mixture is passed through throttling calorimeter. In such calorimeters the arrangement is as shown below. Here excessively wet steam is first sent to separating calorimeter where its wetness is reduced by. Less wet steam is sent to throttling calorimeter and its separating out some liquid fraction, say mass mf1 dryness fraction estimated as x2. Then at the end this throttled steam (of superheated type) is made to pass through condenser. Mass of condensate is measured from that collected in condensate tank, say m2. Thus m2 is total mass of steam sent from separating calorimeter to throttling calorimeter. As mass of then total mass of wet steam under liquid collected in collection tank of separating calorimeter is mf1 examination is (mf1 + m2). Dryness fraction at section 1\u20131 shall be; x1 = (Mass of vapour at 1 \u2013 1) Total mass Steam main Valve Throttle valve Thermometer Sampling bulb 1 1 1' 1' 2 2 Pr. gauge Valve Condenser Cold water in out Collection tank Condensate tank Separating calorimeter Throttling calorimeter Fig. 6.18 Separating and throttling calorimeter 180 _________________________________________________________ Applied Thermodynamics Mass of vapour at 1\u20131 shall be similar to mass of vapour entering at 2\u20132. Mass of vapour at 2\u20132 = x2 \u00d7 m2 Hence, dryness fraction at 1\u20131, x1 = \u00b7 x m 2 2 m+ 2 1 f m Separating and throttling processes occurring are also shown on h\u2013s diagram in Fig. 6.19. p1 h 1' 1 p2 2 x2 x1 s Fig. 6.19 Separating and throttling together on h\u2013s diagram (iv) Electrical Calorimeter: In electrical calorimeter too the principle employed is similar to that of throttling calorimeter. Here also wet mixture is brought to the superheated state by heating and not by throttling. For known amount of heat added and the final enthalpy for superheated steam being known, one can find out the initial enthalpy. For mass m of mixture, heat Qadd added by heater, and the enthalpies before and after heating being h1, h2, steady flow energy equation may be written as; mh1 + Qadd = mh2. Steam main Electrical heater Valve Insulation Sampling bulb 1 1 2 2 Temperature measurement Pressure measurement Qadd Exhaust steam collection (m.kg) Fig. 6.20 Electrical calorimeter Here, for electrical heater Qadd = V.I, where V and I are voltage and current. h2 is known, as mixture is brought to superheated state and pressure and temperature measured, locate enthalpy from superheated steam table, also h1 = hf at p1 + x. hfg at p1, here hf at p1 & hfg at p1 can be seen from steam table. Now h1 being known, for known m, h2, Qadd, dryness fraction \u2018x\u2019 can be easily obtained. Thermodynamic Properties of Pure Substance ________________________________________ 181 EXAMPLES 1. Derive the expressions for the following : (a) Work of evaporation or external work of evaporation (b) True latent heat (c) Internal energy of steam (d) Entropy of water (e) Entropy of evaporation (f) Entropy of wet steam (g) Entropy of superheated steam Solution: (a) Work of evaporation: This is the work done due to evaporation of", " water to steam as phase transformation from water to steam is accompanied by increase in volume. Work of evaporation can be estimated as; = p (vg \u2013 vf), for unit mass For very low pressures of steam generation, where vf <<< vg. Work of evaporation = p\u00b7vg Work of evaporation for wet steam with dryness fraction \u2018x\u2019 = p \u00b7x \u00b7vg. (b) True latent heat: Latent heat causing the phase transformation from water to steam is accompanied with change in volume as well. Therefore, latent heat shall have two components i.e. (i) true latent heat causing phase change and (ii) work of evaporation due to volume increase. Mathematically, for unit mass, True latent heat = hfg \u2013 p(vg \u2013 vf). (c) Internal energy of steam: For a given mass of steam the total heat energy with steam can be said to comprise of (i) Sensible heat (ii) True latent heat (iii) Work of evaporation Out of above three the third component gets consumed in doing work. The internal energy of steam shall consist of first two components. For the enthalpy \u2018h\u2019 of steam, (for unit mass) Internal energy, u = h \u2013 p(vg \u2013 vf) neglecting vf for low pressures, u = h \u2013 p\u00b7vg. u = h \u2013 p\u00b7xvg for wet steam for super heated steam, Hence h = hg + cp superheat (Tsuperheat \u2013 Tsat) (d) Entropy of water: Entropy change of water as discussed in article 6.5 can be given as : u = {hg + cp superheat \u00b7 (Tsuperheat \u2013 Tsat)} \u2013 pvsuperheat 2 ds\u222b T 2 \u222b c = p water T 1 for constant specific heat of water 1 (s2 \u2013 s1) = cp water ln dT T \uf8eb \uf8ec \uf8ed T 2 T 1 \uf8f6 \uf8f7 \uf8f8 182 _________________________________________________________ Applied Thermodynamics Absolute entropy may be given in reference to absolute zero temperature s = cp water \u00b7 ln \uf8eb \uf8ec \uf8ed T 273.15 \uf8f6 \uf8f7 \uf8f8 (e) Entropy of evaporation: During evaporation heat absorbed is equal to latent heat of evaporation. Therefore, for unit mass sevaporation = fgh T sat During incomplete evaporation (for wet steam) sevaporation = \u00b7 x h fg T sat (f) Entropy of wet steam: Entropy of wet steam = Entropy of water + Entropy of evaporation For unit mass, swet = cp water ln T 2 T 1 + x \u00b7 fgh T 2 (g) Entropy of super heated steam: Entropy change during constant pressure heating for superheating unit mass of the steam. = cp steam ln \uf8eb \uf8ec \uf8ed T sup T sat \uf8f6 \uf8f7 \uf8f8 Total entropy of super heated steam, starting with water at temperature T1. ssuperheat = cp water ln \uf8eb \uf8ec \uf8ed T sat T 1 \uf8f6 \uf8f7 \uf8f8 + fgh T sat + cp steam \u00b7 ln \uf8eb \uf8ec \uf8ed T super heat T sat \uf8f6 \uf8f7 \uf8f8. 2. Throttling calorimeter has steam entering to it at 10 MPa and coming out of it at 0.05 MPa and 100\u00b0C. Determine dryness fraction of steam. Solution: During throttling, At state 2, enthalpy can be seen for superheated steam at 0.05 MPa and 100\u00b0C. Thus, At state 1, before throttling h2 = 2682.5 kJ/kg h1 = h2 hf at10MPa = 1407.56 kJ/kg hfg at10MPa = 1317.1 kJ/kg h1 = hf at10MPa + x1 hfg at10MPa = h2 2682.5 = 1407.56 + (x1 \u00b7 1317.1) x1 = 0.968 Thermodynamic Properties of Pure Substance ________________________________________ 183 10 MPa 0.05 MPa 1 2 h s Fig. 6.21 Dryness fraction is 0.968. Ans. 3. Determine internal energy of steam if its enthalpy, pressure and specific volumes are 2848 kJ/kg, 12 MPa and 0.017 m3/kg. Solution: Internal energy, u = h \u2013 pv = (2848 \u2013 12 \u00d7 103 \u00d7 0.017) kJ/kg = 2644 kJ/kg Internal energy = 2644 kJ/kg Ans. 4. Determine entropy", " of 5 kg of steam at 2 MPa and 300\u00b0C. Take specific heat of super heated steam as 2.1 kJ/kg.K. Solution: Steam state 2 MPa and 300\u00b0C lies in superheated region as saturation temperature at 2 MPa is 212.42\u00b0C and hfg = 1890.7 kJ/kg. Entropy of unit mass of superheated steam with reference to absolute zero. Substituting values = cp water In T\uf8eb \uf8ec \uf8ed sat 273.15 \uf8f6 \uf8f7 \uf8f8 + fgh,2MPa T sat + cp superheat ln T super heat T sat = 4.18 ln \uf8eb \uf8ec \uf8ed 485.57 273.15 \uf8f6 \uf8f7 \uf8f8 + \uf8eb \uf8ec \uf8ed 1890.7 485.57 \uf8f6 \uf8f7 \uf8f8 + \uf8ee \uf8ef \uf8f0 2.1 ln \uf8eb \uf8ec \uf8ed 573.15 485.57 \uf8f6 \uf8f7 \uf8f8 \uf8f9 \uf8fa \uf8fb Entropy of 5 kg of steam = 33.23 kJ/K = 6.646 kJ/kg.K. Entropy of steam = 33.23 kJ/K Ans. 5. Water in a pond boils at 110\u00b0C at certain depth in water. At what temperature the water shall boil if we intend to boil it at 50 cm depth from above mentioned level. Solution: Boiling point = 110\u00b0C, pressure at which it boils = 143.27 kPa (from steam table, sat. pressure for 110\u00b0C) At further depth of 50 cm the pressure = 143.27 \u2013 ((103 \u00d7 9.81 \u00d7 0.50) \u00d7 10\u20133) = 138.365 kPa. 184 _________________________________________________________ Applied Thermodynamics Boiling point at this depth = Tsat, 138.365 From steam table this temperature = 108.866 = 108.87\u00b0C Boiling point = 108.87\u00b0C Ans. 6. Water-vapour mixture at 100\u00b0C is contained in a rigid vessel of 0.5 m3 capacity. Water is now heated till it reaches critical state. What was the mass and volume of water initially? Solution: In a rigid vessel it can be treated as constant volume process. Since final state is given to be critical state, then specific volume at critical point, v1 = v2 At 100\u00b0C saturation temperature, from steam table v2 = 0.003155 m3/kg Thus for initial quality being x1 v f = 0.001044 m3/kg, vg = 1.6729 m3/kg or v1 = vf 100\u00b0C + x1 \u00b7 vfg100\u00b0C 0.003155 = 0.001044 + x1 \u00d7 1.671856 x1 = 0.0012627 Mass of water initially = Total mass \u00b7 (1 \u2013 x1) Total mass of fluid = V v = 2 0.5 0.003155 = 158.48 kg Mass of water = 158.48 kg Volume of water = 158.48 \u00d7 0.001044 = 0.1652 m3 Mass of water = 158.48 kg, Ans. Volume of water = 0.1652 m3 7. Determine slope of an isobar at 2 MPa and 500\u00b0C on mollier diagram. Solution: On mollier diagram (h \u2013 s diagram) the slope of isobaric line may be given as \uf8eb \uf8ec \uf8ed dh ds \uf8f6 \uf8f7 \uf8f8 p = const = Slope of isobar From Ist and IInd law combined; Tds = dh \u2013 vdp for constant pressure \uf8eb \uf8ec \uf8ed dh ds \uf8f6 \uf8f7 \uf8f8 = T p = const Here temperature T = 773.15 K hence slope = \uf8eb \uf8ec \uf8ed dh ds \uf8f6 \uf8f7 \uf8f8 = 773.15 p = const Slope = 773 15 Ans.. Thermodynamic Properties of Pure Substance ________________________________________ 185 8. Determine enthalpy, specific volume, entropy for mixture of 10% quality at 0.15 MPa. Solution: Given, x = 0.10 At 0.15 MPa, from steam table; Enthalpy at Specific volume, hf = 467.11 kJ/kg, hg = 2693.6 kJ/kg v f = 0.001053 m3/kg, vg = 1.1593 m3/kg sf = 1.4336 kJ/kg.K, sg = 7.2233 kJ/kg.K x = 0.10 h = hf", " + x.hfg = 467.11 + {0.10 \u00d7 (2693.6 \u2013 467.11)} h = 689.759 kJ/kg v = vf + x.vfg = 0.001053 + {0.10 \u00d7 (1.1593 \u2013 0.001053)} Entropy, v = 0.116877 m3/kg s = sf + x.sfg = 1.4336 + {0.10 \u00d7 (7.2233 \u2013 1.4336)} s = 2.01257 kJ/kg.K h = 689.759 kJ/kg v = 0.116877 m3/kg s = 2.01257 kJ/kg.K Ans. 9. In a piston-cylinder arrangement the steam at 1.0 MPa, 80% dryness fraction, and 0.05 m3 volume is heated to increase its volume to 0.2 m3. Determine the heat added. Solution: Given; Initial states, 1: P1 = 1.0 MPa, V1 = 0.05 m3, x1 = 0.80 Final state, 2: V2 = 0.2 m3, P2 = 1 MPa Work done during constant pressure process, W = P1(V2 \u2013 V1) W = 1000 (0.2 \u2013 0.05) W = 150 kJ V 1 v 1 Mass of steam = From steam table at P1; vf = 0.001127 m3/kg v g = 0.19444 m3/kg uf = 761.68 kJ/kg ufg = 1822 kJ/kg v 1 = vf + x1 vfg v 1 = 0.15578 m3/kg 0.05 0.15578 Hence, mass of steam = = 0.32097 kg so Specific volume at final state = V 2 mass of steam v 2 = 0.62311 m3/kg 186 _________________________________________________________ Applied Thermodynamics Corresponding to this specific volume the final state is to be located for getting the internal energy at final state at 1 MPa hence state lies in superheated region, from the steam table by interpolation we get temperature as; v2 > vg1 MPa State lies between temperature of 1000\u00b0C and 1100\u00b0C so exact temperature at final state = 1000 + \u00d7 100 (0.62311 \u2013 0.5871) (0.6335 \u2013 0.5871) = 1077.61\u00b0C Thus internal energy at final state, 1 MPa, 1077.61\u00b0C; u2 = 4209.6 kJ/kg Internal energy at initial state, From first law of thermodynamics; Q \u2013 W = \u2206U u1 = uf + x1.ufg = 761.68 + 0.8 \u00d7 1822 u1 = 2219.28 kJ/kg Q = (U2 \u2013 U1) + W = m(u2 \u2013 u1) + W = 0.32097 (4209.6 \u2013 2219.28) + 150 Q = 788.83 kJ Heat added = 788.83 kJ Ans. 10. Steam at 800 kPa and 200\u00b0C in a rigid vessel is to be condensed by cooling. Determine pressure and temperature corresponding to condensation. Solution: Here steam is kept in rigid vessel, therefore its\u2019 specific volume shall remain constant. Specific volume at initial state, 800 kPa, 200\u00b0C, v1, It is superheated steam as Tsat = 170.43\u00b0C at 800 kPa. From superheated steam table; v1 = 0.2404 m3/kg. At the begining of condensation, specific volume = 0.2404 m3/kg. v2 = 0.2404 m3/kg This v2 shall be specific volume corresponding to saturated vapour state for condensation. Thus v2 = vg = 0.2404 m3/kg Looking into steam table vg = 0.2404 m3/kg shall lie between temperatures 175\u00b0C (vg = 0.2168 m3/kg) and 170\u00b0C (vg = 0.2428 m3/kg) and pressures 892 kPa (175\u00b0C) and 791.7 kPa (170\u00b0C). By interpolation, temperature at begining of condensation T2 = 175 \u2013 (175 \u2013 170) \u00d7 (0.2404 \u2013 0.2168) (0.2428 \u2013 0.2168) T2 = 170.46\u00b0C Similarly, pressure, p2 = 892 \u2013 (892 \u2013 791.7) \u00d7 (0.2404 \u2013 0.2168) (0.2428 \u2013 0.2168) p2 = 800.96 k", "Pa Pressure and temperature at condensation = 800.96 kPa and 170.46\u00b0C Ans. Thermodynamic Properties of Pure Substance ________________________________________ 187 11. Feed water pump is used for pumping water from 30\u00b0C to a pressure of 200 kPa. Determine change in enthalpy assuming water to be incompressible and pumping to be isentropic process. Solution: From Ist and IInd law; Tds = dh \u2013 vdp for isentropic process, ds = 0 hence i.e. Corresponding to initial state of saturated liquid at 30\u00b0C; from steam table; dh = vdp (h2 \u2013 h1) = v1 (p2 \u2013 p1) Therefore p1 = 4.25 kPa, vf = v1 = 0.001004 m3/kg (h2 \u2013 h1) = 0.001004 (200 \u2013 4.25) (h2 \u2013 h1) = 0.197 kJ/kg Enthalpy change = 0.197 kJ/kg Ans. 200 kPa p1 T 30 \u00b0C 2 1 s Fig. 6.22 12. A rigid vessel contains liquid-vapour mixture in the ratio of 3:2 by volume. Determine quality of water vapour mixture and total mass of fluid in vessel if the volume of vessel is 2 m3 and initial temperature is 150\u00b0C. Solution: From steam table at 150\u00b0C, vf = 0.001091 m3/kg, vg = 0.3928 m3/kg Volume occupied by water = 1.2 m3 Volume of steam = 0.8 m3 Mass of water \u21d2 mf = 1.2 0.001091 = 1099.91 kg Mass of steam \u21d2 mg = 0.8 0.3928 = 2.04 kg Total mass in tank = mf + mg = 1103.99 kg Quality or Dryness fraction, x = 2.04 1103.99 = 0.001848 Mass = 1103.99 kg, Quality = 0.001848 Ans. 188 _________________________________________________________ Applied Thermodynamics 13. Steam turbine expands steam reversibly and adiabatically from 4 MPa, 300\u00b0C to 50\u00b0C at turbine exit. Determine the work output per kg of steam. Solution: From SFEE on steam turbine; Initially at 4 MPa, 300\u00b0C the steam is super heated so enthalpy from superheated steam table or W = (h1 \u2013 h2) Mollier diagram. Reversible adiabatic expansion process has entropy remaining constant. On Mollier diagram the state 2 can be simply located at intersection of constant temperature line for 50\u00b0C and isentropic expansion line. h1 = 2886.2 kJ/kg, s1 = 6.2285 kJ/kg.K Else from steam tables at 50\u00b0C saturation temperature; Here Hence enthalpy at state 2, hf = 209.33 kJ/kg, sf = 0.7038 kJ/kg.K hfg = 2382.7 kJ/kg, sfg = 7.3725 kJ/kg.K s1 = s2, let dryness fraction at 2 be x2, 6.2285 = 0.7038 + x2 \u00d7 7.3725 x2 = 0.7494 h2 = hf + x2.hfg = 209.33 + 0.7494 \u00d7 2382.7 h2 = 1994.93 kJ/kg Steam turbine work = (2886.2 \u2013 1994.93) = 891.27 kJ/kg Turbine output = 891.27 kJ/kg Ans. T 50 \u00b0C 4 MPa, 300 \u00b0C p1 1 2 s Fig. 6.23 14. In a closed vessel the 100 kg of steam at 100 kPa, 0.5 dry is to be brought to a pressure of 1000 kPa inside vessel. Determine the mass of dry saturated steam admitted at 2000 kPa for raising pressure. Also determine the final quality. Solution: It is a constant volume process. Volume of vessel V = (Mass of vapour) \u00d7 (Specific volume of vapour) Initial specific volume, v1 v1 = vf, 100kPa + x1.vfg, 100kPa at 100 kPa from steam table; hf, 100 kPa = 417.46 kJ/kg Thermodynamic Properties of Pure Substance ________________________________________ 189 uf, 100 kPa = 417.36 kJ/kg vf, 100 kPa = 0.001043 m3/kg hfg, 100 kPa = 2258 kJ/kg, ufg, 100 kPa = 2088.7 kJ/kg, vg, 100 kPa = 1.6940 m3/kg", " x1 = 0.5, v1 = 0.8475 m3/kg given T 2 1 S Fig. 6.24 Enthalpy at 1, h1 = 417.46 + 0.5 \u00d7 2258 = 1546.46 kJ/kg Thus, volume of vessel = (100 \u00d7 0.5) \u00d7 (0.8475) V = 42.375 m3 Internal energy in the beginning = U1 = m1 \u00d7 u1 = 100 (417.36 + 0.5 \u00d7 2088.7) Let the mass of dry steam added be \u2018m\u2019, Final specific volume inside vessel, v2 U1 = 146171 kJ v 2 = vf, 1000 kPa + x2.vfg, 1000 kPa At 2000 kPa, from steam table, vg, 2000 kPa = 0.09963 m3/kg ug, 2000 kPa = 2600.3 kJ/kg hg, 2000 kPa = 2799.5 kJ/kg Total mass inside vessel = Mass of steam at 2000 kPa + Mass of mixture at 100 kPa V v 2 = v V g 2000 kPa + V v 1 42.375 v 2 = 42.375 0.09963 + 42.375 0.8475 v 2 = 0.089149 v 2 = vf, 1000 kPa + x2.vfg, 1000 kPa substituting At 1000 kPa from steam table, hf, 1000 kPa = 762.81 kJ/kg, hfg, 1000 kPa = 2015.3 kJ/kg vf, 1000 kPa = 0.001127 m3/kg vg, 1000 kPa = 0.19444 m3/kg x2 = 0.455 it gives For adiabatic mixing 190 _________________________________________________________ Applied Thermodynamics (100 + m) \u00b7 h2 = 100 \u00d7 h1 + m \u00d7 hg, 2000 kPa (100 + m) \u00b7 (762.81 + 0.455 \u00d7 2015.3) = {100 \u00d7 (1546.46)} + {m \u00d7 2799.5} It gives upon solving; (100 + m) (1679.77) = 154646 + (2799.5)m m = 11.91 kg Mass of dry steam at 2000 kPa to be added = 11.91 kg Quality of final mixture = 0.455 Ans. 15. In a condenser the following observations were made, Recorded condenser vacuum = 71.5 cm of Mercury Barometer reading = 76.8 cm of Mercury Temperature of condensation = 35\u00b0C Temperature of hot well = 27.6\u00b0C Mass of condensate per hour = 1930 kg Mass of cooling water per hour = 62000 kg Inlet temperature = 8.51\u00b0C Outlet temperature 26.24\u00b0C Determine the state of steam entering condenser. Steam Cooling water Hot well Condensate Fig. 6.25 Solution: From Dalton\u2019s law of partial pressure the total pressure inside condenser will be sum of partial pressures of vapour and liquid inside. Condenser pressure = \uf8f1 \uf8f2 \uf8f3 \u2212 76.8 71.5 (73.55) \u00d7 101.325 \uf8fc \uf8fd \uf8fe kPa Partial pressure of steam corresponding to 35\u00b0C from steam table; = 7.30 kPa = 5.628 kPa Enthalpy corresponding to 35\u00b0C from steam table, hf = 146.68 kJ/kg hfg = 2418.6 kJ/kg Let quality of steam entering be \u2018x\u2019. From energy balance; mw(26.24 \u2013 8.51) \u00d7 4.18 = 1930 (146.68 + x \u00d7 2418.6 \u2013 4.18 \u00d7 27.6) Thermodynamic Properties of Pure Substance ________________________________________ 191 or which gives 62000 (17.73 \u00d7 4.18) = 1930 (31.31 + x \u00d7 2418.6) x = 0.97 Dryness fraction of steam entering = 0.97 Ans. 16. In a vertical vessel of circular cross section having diameter of 20 cm water is filled upto a depth of 2 cm at a temperature of 150\u00b0C. A tight fitting frictionless piston is kept over the water surface and a force of 10 kN is externally applied upon the piston. If 600 kJ of heat is supplied to water determine the dryness fraction of resulting steam and change in internal energy. Also find the work done. Solution: Heating of water in vessel as described above is a constant pressure heating. Pressure at which process occurs = Force area + atmospheric pressure = \uf8eb \uf8ec \uf8ec \uf8ec \uf8ec \uf8ed 10 2 (0.2) \u03c0 4 + 101.3 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7", " \uf8f8 kPa = 419.61 kPa Volume of water contained = \u03c0 4 \u00d7 (0.2)2 \u00d7 (0.02) Volume = 6.28 \u00d7 10\u20134 m3 Mass of water = 6.28 \u00d7 10\u20134 \u00d7 103 = 0.628 kg Constant pressure T 1 2 S Fig. 6.26 Heat supplied shall cause sensible heating and latent heating. Hence, Enthalpy change = Heat supplied 600 = {(hf at 419.6 kPa + x.hfg at 419.6 kPa) \u2013 (4.18 \u00d7 150)} \u00d7 0.628 600 = {(612.1 + x.2128.7) \u2013 627} \u00d7 0.628 Dryness fraction x = 0.456 Dryness fraction of steam produced = 0.456 Ans. Internal energy of water, initially U1 = mh1 \u2013 p1V1 = (0.628 \u00d7 4.18 \u00d7 150) \u2013 (419.61 \u00d7 6.28 \u00d7 10\u20134) U1 = 393.5 kJ 192 _________________________________________________________ Applied Thermodynamics Finally, internal energy of wet steam Here U2 = mh2 \u2013 p2V2 V2 = m\u00b7x\u00b7vg at 419.61 kPa = 0.456 \u00d7 0.4435 \u00d7 0.628 = 0.127 m3 Hence U2 = (0.628 \u00d7 1582.8) \u2013 (419.61 \u00d7 0.127) U2 = 940.71 kJ Change in internal energy = U2 \u2013 U1. Change in internal energy = 547.21 kJ Ans. Work done = P \u00b7 (V2 \u2013 V1) = 419.61 \u00d7 (0.127 \u2013 6.28 \u00d7 10\u20134) Work done = 53.03 kJ Work done = 53.03 kJ Ans. 17. In a separating and throttling calorimeter the total quantity of steam passed was 40 kg and 2.2 kg of water was collected from separator. Steam pressure before throttling was 1.47 MPa and temperature and pressure after throttling are 120\u00b0C and 107.88 kPa. Determine the dryness fraction of steam before entering to calorimeter. Specific heat of superheated steam may be considered as 2.09 kJ/kg.K. Separating calorimeter 40 kg 2 1.47 MPa 2.2 kg Fig. 6.27 Throttling calorimeter 3 107.88 kPa 120 \u00b0C Solution: Consider throttling calorimeter alone, Degree of superheat = 120 \u2013 101.8 = 18.2\u00b0C Enthalpy of superheated steam = (2673.95 + (18.2 \u00d7 2.09)) Enthalpy before throttling = Enthalpy after throttling = 2711.988 kJ/kg or 840.513 + x2.1951.02 = 2711.988 x2 = 0.9592 For separating calorimeter alone, dryness fraction, x1 = \u2212 40 2.2 40 x1 = 0.945 Overall dryness fraction = (x1.x2) = (0.945 \u00d7 0.9592) = 0.9064 Dryness fraction : 0.9064 Ans. 18. A rigid vessel is divided into two parts A and B by means of frictionless, perfectly conducting piston. Initially, part A contains 0.4 m3 of air (ideal gas) at 10 bar pressure and part B contains 0.4 m3 of wet steam at 10 bar. Heat is now added to both parts until all the water in part B is evaporated. At this condition the pressure in part B is 15 bar. Determine the initial quality of steam in part B and the total [U.P.S.C. 1995] amount of heat added to both parts. Thermodynamic Properties of Pure Substance ________________________________________ 193 Solution: Here heat addition to part B shall cause evaporation of water and subsequently the rise in pressure. Final, part B has dry steam at 15 bar. In order to have equilibrium the part A shall also have pressure of 15 bar. Thus, heat added Q = V(P2 \u2013 P1) = 0.4(15 \u2013 10) \u00d7 102 Q = 200 kJ Final enthalpy of dry steam at 15 bar, h2 = hg at 15 bar h2 = 2792.2 kJ/kg Let initial dryness fraction be x1. Initial enthalpy, h1 = hf at 10bar + x1\u00b7hfg at 10 bar h1 = 762.83 + x1\u00b72015.3 Heat balance yields, h1 + Q = h2 (762.83 + x1\u00b72015.3) + 200 = 2792.2 x1 = 0.907 Heat added = 200 kJ Initial quality = 0", ".907 Ans. 19. A piston-cylinder contains 3 kg of wet steam at 1.4 bar. The initial volume is 2.25 m3. The steam is heated until its\u2019 temperature reaches 400\u00b0C. The piston is free to move up or down unless it reaches the stops at the top. When the piston is up against the stops the cylinder volume is 4.65 m3. Determine the [U.P.S.C. 1994] amounts of work and heat transfer to or from steam. Solution: From steam table, specific volume of steam at 1.4 bar = 1.2455 m3/kg = vg at 1.4 bar Specific volume of wet steam in cylinder, v1 = Dryness fraction of initial steam, x1 = 2.25 3 = 0.75 m3/kg 0.75 1.2455 = 0.602 Initial enthalpy of wet steam, h1 = hf at 1.4 bar + x1 \u00b7 hfg at 1.4 bar = 457.99 + (0.602 \u00d7 2232.3) \u21d2 h1 = 1801.83 kJ/kg At 400\u00b0C specific volume of steam, v2 = 4.65 3 = 1.55 m3/kg For specific volume of 1.55 m3/kg at 400\u00b0C the pressure can be seen from the steam table. From superheated steam tables the specific volume of 1.55 m3/kg lies between the pressure of 0.10 MPa (specific volume 3.103 m3/kg at 400\u00b0C) and 0.20 MPa (specific volume 1.5493 m3/kg at 400\u00b0C). Actual pressure can be obtained by interpolation; P2 = 0.10 + \uf8eb \u2212 0.20 0.10 \uf8ec \u2212 (1.5493 3.103) \uf8ed P2 = 0.199 MPa \u2248 0.20 MPa Saturation pressure at 0.20 MPa = 120.23\u00b0C \uf8f6 \uf8f7 \uf8f8 \u00d7 (1.55 \u2013 3.103) 194 _________________________________________________________ Applied Thermodynamics Finally the degree of superheat = 400 \u2013 120.23 = 279.77\u00b0C Final enthalpy of steam at 0.20 MPa and 400\u00b0C, h2= 3276.6 kJ/kg Heat added during process = m (h2 \u2013 h1) Internal energy of initial wet steam, u1 = uf at 1.4 bar + x1.ufg at 1.4 bar = 3 \u00d7 (3276.6 \u2013 1801.83) \u2206Q = 4424.31 kJ u1 = 457.84 + (0.607 \u00d7 2059.34) u1 = 1707.86 kJ/kg Internal energy of final state, u2 = uat 0.2 MPa, 400\u00b0C u2 = 2966.7 kJ/kg Change in internal energy \u21d2 \u2206U = m(u2 \u2013 u1) = 3 \u00d7 (2966.7 \u2013 1707.86) \u2206U = 3776.52 kJ From first law of thermodynamics, Work done \u2206W = \u2206Q \u2013 \u2206U = 4424.31 \u2013 3776.52 Work done, \u2206W = 647.79 kJ Heat transfer = 4424.31 kJ Work transfer = 647.79 kJ Ans. 20. An insulated vessel is divided into two compartments connected by a valve. Initially, one compartment contains steam at 10 bar, 500\u00b0C, and the other is evacuated. The valve is opened and the steam is allowed to fill the entire volume, achieving a final pressure of 1 bar. Determine the final temperature, in \u00b0C, the percentage of the vessel volume initially occupied by steam and the amount of entropy produced, [U.P.S.C. 1993] in kJ/kg. K. Solution: Here throttling process is occurring therefore enthalpy before and after expansion remains same. Let initial and final states be given by 1 and 2. Initial enthalpy, from steam table. h1 at 10 bar and 500\u00b0C = 3478.5 kJ/kg s1 at 10 bar and 500\u00b0C = 7.7622 kJ/kg.K v1 at 10 bar and 500\u00b0C = 0.3541 m3/kg Finally pressure becomes 1 bar so the final enthalpy at this pressure (of 1 bar) is also 3478.5 kJ/kg which lies between superheat temperature of 400\u00b0C and 500\u00b0C at 1 bar. Let temperature be T2, hat 1 bar, 400\u00b0C = 3278.2 kJ/kg hat 1 bar, 500\u00b0C = 3488.1 kJ/kg h2 = 3478.5", " = hat 1 bar, 400\u00b0C + ( h at 1 bar,500\u00baC h\u2212 \u2212 (500 400) at 1 bar,400\u00baC ) (T2 \u2013 400) Thermodynamic Properties of Pure Substance ________________________________________ 195 3478.5 = 3278.2 + \uf8f1 \uf8f2 \uf8f3 \u2212 3488.1 3278.2 100 \uf8fc \uf8fd \uf8fe (T2 \u2013 400) T2 = 495.43\u00b0C, Final temperature = 495.43\u00b0C Ans. Entropy for final state, s2 = sat 1 bar, 400\u00b0C + ( s at 1 bar,500\u00baC at 1 bar,400\u00ba C s\u2212 \u2212 (500 400) ) (495.43 \u2013 400) s2 = 8.5435 + \uf8f1 \uf8f2 \uf8f3 \u2212 8.8342 8.5435 100 \uf8fc \uf8fd \uf8fe \u00d7 (95.43) s2 = 8.8209 kJ/kg. K Change in entropy, \u2206s = 8.8209 \u2013 7.7622 = 1.0587 kJ/kg. K Change in entropy = 1.0587 kJ/kg K Ans. Final specific volume, v2 = vat 1 bar, 400\u00b0C ( v + at 1 bar,500\u00baC at 1 bar,400\u00ba C v\u2212 \u2212 (500 400) ) \u00d7 (95.43) = 3.103 + \uf8f1 \uf8f2 \uf8f3 \u2212 3.565 3.103 100 \uf8fc \uf8fd \uf8fe \u00d7 95.43 v 2 = 3.544 m3/kg Percentage volume occupied by steam = 0.3541 3.544 \u00d7 100 = 9.99% Percentage of vessel volume initially occupied by steam = 9.99% Ans. 21. Determine the maximum work per kg of steam entering the turbine and the irreversibility in a steam turbine receiving steam at 2.5 MPa, 350\u00b0C and rejecting steam at 20 kPa, 0.92 dry. During the expansion the one-quarter of initial steam is bled at 30 kPa, 200\u00b0C. Consider the heat loss during expansion as 10kJ/s and atmospheric temperature as 30\u00b0C. Solution: For the states shown on turbine in the figure, the steam table can be used to get following values: At 2.5 MPa, 350\u00b0C, h1 = 3126.3 kJ/kg, s1 = 6.8403 kJ/kg.K 2.5 MPa, 350\u00b0C, 1kg/s 1 Steam Turbine 3 0.75kg/s 2 20 kPa. 0.92 dry 30 kPa, 200\u00b0C 0.25kg/s Fig. 6.28 196 _________________________________________________________ Applied Thermodynamics At 30 MPa, 200\u00b0C, At 20 kPa, 0.92 dry, \u21d2 h2 = 2878.6 kJ/kg. s2 = 8.5309 kJ/kg.K hf = 251.40kJ/kg, hg = 2609.7kJ/kg sf = 0.8320 kJ/kg.K, sfg = 7.0766 kJ/kg.K h3 = 251.40 + 0.92 \u00d7 (2609.7 \u2013 251.40) h3 = 2421.04 kJ/kg s3 = 0.8320 + (0.92 \u00d7 7.0766) s3 = 7.3425 kJ/kg.K At atmospheric temperature, Availability of steam entering turbine, hf at 30\u00b0C = h0 = 125.79 kJ/kgs; sfat 30\u00b0C = s0 = 0.4369kJ/kg.K A1 = (h1 \u2013 h0) \u2013 T0(s1 \u2013 s0) = (3126.3 \u2013 125.79) \u2013 303(6.8403 \u2013 0.4369) = 1060.28 kJ/kg Availability of steam leaving turbine at state 2 & 3, A2 = (h2 \u2013 h0) \u2013 T0(s2 \u2013 s0) = (2878.6 \u2013 125.79) \u2013 303(8.5309 \u2013 0.4369) A2 = 300.328 kJ/kg A3 = (h3 \u2013 h0) \u2013 T0(s3 \u2013 s0) = (2421.04 \u2013 125.79) \u2013 303(7.3425 \u2013 0.4369) Maximum work per kg of steam entering turbine for A3 = 202.85 kJ/kg Wmax = 1 \u00d7 A1 \u2013 m2A2 \u2013 m3A3 Wmax = A1 \u2013 (0.25)A2 \u2013 0.75A3 = 10", "60.28 \u2013 (0.25 \u00d7 300.328) \u2013 (0.75 \u00d7 202.85) Irreversibility, I = T0 (m2 \u00d7 s2 + m3 \u00d7 s3 \u2013 m1s1) \u2013 Q. Wmax = 833.06 kJ/kg Ans. = 303 (0.25 \u00d7 8.5309 + 0.75 \u00d7 7.3425 \u2013 6.8403) \u2013 (\u2013 10) Irreversibility, I = 252.19 kJ/s Ans. 22. Determine the change in availability due to throttling of steam from 6 MPa and 400\u00b0C to 5 MPa when surroundings are at 100 kPa and 20\u00b0C. The changes in KE and PE may be considered negligible. Solution: From steam tables Initially at 6 MPa, 400\u00b0C, h1 = 3177.2 kJ/kg s1 = 6.5408 kJ/kg.K Thermodynamic Properties of Pure Substance ________________________________________ 197 After throttling at 5 Mpa, Hence at 5 MPa and h2 = h1; in view of throttling process. Superheated Steam table gives, h2 = 3177.2 kJ/kg at 5 MPa, hat 350\u00b0C = 3068.4 kJ/kg at 5 MPa, hat 400\u00b0C = 3195.7 kJ/kg Hence by interpolation at 5 MPa, enthalpy of 3177.2 kJ/kg will be at T2 = 350 + \u2212 (400 350) \u00d7 \u2212 (3177.2 3068.4) \u2212 (3195.7 3068.4) T2 = 392.7\u00b0C After throttling steam will be at 5 MPa, 392.7\u00b0C. By interpolation Entropy, s2 = 6.6172 kJ/kg.K 6MPa 1 5MPa 2 h s Fig. 6.29 At dead state of 20\u00b0C, hf at 20\u00b0C = h0 = 83.96 kJ/kg sf at 20\u00b0C = s0 = 0.2966 kJ/kg Hence availability at state 1, \u21d2 A1 = (h1 \u2013 h0) \u2013 T0 (s1 \u2013 s0 ) A1 = (3177.2 \u2013 83.96) \u2013 293(6.5408 \u2013 0.2966) + 0 + 0 A1 = 1263.68 kJ/kg Availability of steam after throttling, A2 = (h2 \u2013 h0) \u2013 T0 (s2 \u2013 s0 ) 198 _________________________________________________________ Applied Thermodynamics A2 = (3177.2 \u2013 83.96) \u2013 293 (6.6172 \u2013 0.2966) A2 = 1241.30 kJ/kg Change in availability = A2 \u2013 A1 = (1241.30 \u2013 1263.68) = \u2013 22.5 kJ/kg = 22.5 kJ/kg, decrease. Ans. 23. A parallel flow heat exchanger has hot water flowing at 95\u00b0C for heating cold water at 15\u00b0C to 45\u00b0C. Hot water flows at the rate of 800 gm/sec and the temperature of this hot water stream should not be less than 50\u00b0C at exit. Estimate the second law efficiency and rate of exergy destruction considering dead state temperature of 25\u00b0C. Solution: Let hot stream and cold stream be shown to enter at section 1\u20131 and leave at 2 \u2013 2 Given; TH1 TC1 = 95\u00b0C, TH2 = 15\u00b0C, TC2 = 50\u00b0C, mH = 800 gm/s = 45\u00b0C mH = 0.8kg/s Hot, TH1 Cold, TC1 1 1 2 2,TH2 mH,T mC Heat exchanger - Parallel flow T T T T Length along heat exchanger Fig. 6.30 For parallel flow heat exchanger as shown in figure; mH \u00d7 Cp,H.(TH1 \u21d2 \u21d2 \u2013 TH2 ) \u2013 TC1 ) = mc \u00d7 Cp,c.(TC2 0.8 \u00d7 (95 \u2013 50) = mc (45 \u2013 15) mc = 1.2 kg/s Second law efficiency = (Rate exergy increase in cold stream) (Rate of exergy input to heat exchanger through hot stream) The exergy entering (input) through hot water stream, AH1 = mH{(hH1 \u2013 h0) \u2013 T0(sH1 \u2013 s0)} Thermodynamic Properties of Pure Substance ________________________________________ 199 h0 = 104.89 kJ/kg, s0 = 0.3674 kJ/kg.K saturated liquid yields. hH1 = hf at 95\u00b0C = 397.96 kJ/kg Using steam tables At 25\u00b0", "C At 95\u00b0C, (for hot stream) At 50\u00b0C, (for hot stream) At 45\u00b0C, (for cold stream) At 15\u00b0C, (for cold stream) = sf at 95\u00b0C = 1.2500 kJ/kg.K = hf at 50\u00b0C = 209.33 kJ/kg.K = sf at 50\u00b0C = 0.7038 kJ/kg.K = hf at 45\u00b0C = 188.45 kJ/kg.K = sf at 45\u00b0C = 0.6387 kJ/kg.K = hf at 15\u00b0C = 62.99 kJ/kg.K = sf at 15\u00b0C = 0.2245 kJ/kg.K sH1 hH2 sH2 hc2 sc2 hc1 sc1 AH1 AH1 Rate of exergy input through hot water stream, = 0.8 \u00d7 {(397.96 \u2013 104.89) \u2013 298(1.25 \u2013 0.3674)} = 24.04 kJ/s Rate of exergy increase in cold stream; \u2206Ac = mc{(hc2 \u2206Ac = 1.2{(188.45 \u2013 62.99) \u2013 298(0.6387 \u2013 0.2245)} \u2206Ac = 2.43 kJ/s \u2206 ) \u2013 T0(sc2 \u2013 hc1 \u2013 sc1 )} Second law efficiency = A c A H 1 = 2.43 24.04 = 0.1011 Rate of exergy loss in hot stream, or 10.11% Ans. \u2013 hH2 \u2206AH = mH{(hH1 ) \u2013 T0(sH1 \u2206AH = 0.8{(397.96 \u2013 209.33) \u2013 298(1.25 \u2013 0.7038)} \u2206AH = 20.69 kJ/s \u2013 sH2 )} Hence exergy destruction = \u2206AH \u2013 \u2206AC = 20.69 \u2013 2.43 = 18.26 kJ/s Ans. 6.1 Discuss generation of steam from ice at \u20135\u00b0C at 1 atm with the help of T\u2013S and P\u2013V diagrams. 6.2 What is meant by mollier diagram? Explain. 6.3 Write short notes on the following; Sensible heating, Latent heating, Critical point, Triple point 6.4 Discuss different zones on T\u2013V diagram for steam. 200 _________________________________________________________ Applied Thermodynamics 6.5 Derive the expression for enthalpy change during steam generation from feed water to superheated steam. 6.6 Discuss the throttling calorimeter for dryness fraction measurement. 6.7 Give a neat sketch of \u201cseparating and throttling calorimeter\u201d for dryness fraction measurement. 6.8 Sketch the throttling and superheating processes on h\u2013s and T\u2013S diagrams. 6.9 Determine the final condition of steam if it is passed through a reducing valve which lowers the [0.87] 6.10 Determine the final condition of steam, workdone, heat transferred and change in entropy if 0.5 kg of pressure from 2 MPa to 1 MPa. Assume initial state of steam to be 15% wet. steam at 1 MPa and 0.8 dry is heated at constant pressure until its volume gets doubled. [408.6\u00b0C, 77.5 kJ, 453.5 kJ, 0.895 kJ/K] 6.11 Determine the state of substance if 3346 kJ of heat is added to wet steam in a closed rigid vessel of 3m3 volume containing 5 kg of wet steam at a pressure of 200 kPa till its pressure become 304 kPa. [Dry] 6.12 Complete the following table from steam table. Pressure (MPa) Temperature (\u00b0C) Enthalpy (kJ/kg) Quality (x) Specific volume (m 3/kg) Entropy (kJ/kg.K) (a) (b) (c) (d) (e) 1 \u2013 10 20 15 \u2013 250.4 \u2013 700 800 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 0 0.8 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 6.5865 \u2013 \u2013 \u2013 \u2013 (a) 179.9\u00b0C, 762.8 kJ/kg, 1, 0.1944 m3/kg. (b) 4 MPa, 1087.31 kJ/kg, 1.252 m3/kg, 2.7964 kJ/kg.K (c) 311.06\u00b0C, 2461.33 kJ/kg, 0.01442 m3/kg, 5.1632 kJ/kg.K (d) 3809 kJ/kg, 1, 0.", "02113 m3/kg, 6.7993 kJ/kg. (e) 4092.4 kJ/kg, 1, 0.0321 m3/kg, 7.204 kJ/kg.K. 6.13 Determine the pressure in a rigid vessel and volume of rigid vessel if it contains 500 kg of water at 65\u00b0C. [25 kPa, 0.51 m3] 6.14 Estimate the change in volume of water and the total heat required for its\u2019 vaporization in a boiler producing saturated steam at 75 kPa. One kg feed water is supplied to boiler as saturated water. [2.22 m3, 2.28 MJ] 6.15 Determine enthalpy, entropy and specific volume for following cases (i) Steam at 4 MPa and 80% wet. (ii) Steam at 10 MPa and 550\u00b0C. (iii) Steam at 8 MPa and 295\u00b0C. Also estimate the above properties using Mollier diagram and quantify the percentage variation [1430.13 kJ/kg, 3.45 kJ/kg.K, 0.011 m3/kg] [3500.9 kJ/kg, 6.76 kJ/kg.K, 0.036 m3/kg] [2758 kJ/kg, 5.74 kJ/kg.K, 0.024 m3/kg] 6.16 Determine the temperature of steam at 20 MPa if its specific volume is 0.0155m3/kg. [520\u00b0C] 6.17 Steam undergoes reversible adiabatic expansion in steam turbine from 500 kPa, 300\u00b0C to 50 kPa. Determine the work output per kg of steam turbine and quality of steam leaving steam turbine. [357.64 kJ/kg, 0.98] 6.18 Steam flowing through two pipelines at 0.5 MPa are mixed together so as to result in a mixture flowing at 2.2 kg/s and mass flow ratio of two is 0.8. One stream has quality of 0.8. Determine the temperature of second stream so as to result in the final mixture having dryness fraction of 0.994. [300\u00b0C approx.] 6.19 A steam turbine operates with isentropic efficiency of 90%. Turbine handles 6 kg/s of steam at 0.980 MPa and 200\u00b0C and leaves at 0.294 MPa. Determine the power developed in hp and change of entropy [1660 hp, 0.050 kJ/kg.K] from inlet to exit. Thermodynamic Properties of Pure Substance ________________________________________ 201 6.20 A boiler is fed with water velocity of 2m/s, 1.96 MPa, 100\u00b0C. Steam is produced at 400\u00b0C temperature and comes out with velocity of 50 m/s. Determine the rate at which heat should be supplied per kg of [2824.8 kJ/kg] steam for above operation of boiler. 6.21 A steam nozzle is supplied steam at 1 MPa, 200\u00b0C and 100 m/s. Expansion upto 0.3 MPa occurs in the nozzle. Assuming isentropic efficiency of nozzle to be 0.9 determine final steam velocity. 6.22 Combined separating and throttling calorimeter is used to determine quality of steam. Following observations are made; Steam inlet pressure = 1.4 MPa Pressure after throttling = 0.1 MPa Temperature after throttling = 120\u00b0C Water collected in separator = 0.45 kg Steam condensed after throttling = 6.75 kg Take specific heat of superheated steam = 2.1 kJ/kg.K Also find the limiting quality of steam to be measured by above throttling calorimeter alone assuming [0.90, 0.94] that separating calorimeter is not there. 6.23 Steam at 400kPa, dryness fraction of 0.963 is isentropically compressed till it becomes dry saturated. This one kg steam is then heated isobarically till the initial volume is attained and subsequently steam is restored to initial state following isochoric cooling. Determine the net work and net heat interac[29.93 kJ/kg, 29.93 kJ/kg] tions. Also show processes on T-s diagram. 6.24 Wet steam at 1 MPa, 0.125m3 volume and enthalpy of 1814 kJ is throttled up to 0.7 bar pressure. Determine the final state of steam, initial mass and dryness fraction considering cp = 2.1 kJ/kg.K [101.57\u00b0C, 0.675kg, 0.953] 6.25 Steam initially at 5 bar, 0.6 dry is isochorically heated till its pressure becomes 10 bar. This 15 kg steam is expanded up to", " 3 bar following pv1.3 = constant. Subsequently steam is cooled at constant pressure till its dryness fraction becomes half of that existed after second process. Determine the heat, work and entropy change in three processes. [I process: 13.38 MJ, 0, 30,285 kJ/K. II process: \u2013 1.25MJ, 2.73 MJ, \u2013 2.99kJ/K III process: \u2013 15.22 MJ, \u2013 1.28 MJ, 37.4 kJ/K] 6.26 Determine the heat transfer and change in entropy in each process when steam at 20 bar, 250\u00b0C expands till it reaches 4 bar following pv1.35 = constant and subsequently heated at constant volume [\u2013 319.36 kJ/kg, & \u2013 0.725 kJ/kg.K till its pressure becomes 8 bar. 764.95kJ/kg & 1.65 kJ/kg.K] 6.27 A closed vessel of 0.6 m3 initially has steam at 15 bar, 250\u00b0C. Steam is blown off till pressure drops up to 4 bar. Subsequently vessel is cooled at constant pressure till it becomes 3 bar. Considering the expansion of gas to be isentropic during blow-off determine heat transferred during cooling process. [\u2013 620.38 kJ] 6.28 Determine the heat transferred when steam is taken out isobarically from a boiler tank till boiler is left with 80% water only. Volume of boiler tank is 10m3 and initially it has equal volumes of steam and water [1.75 \u00d7 106 kJ] at 10 bar. 6.29 Determine the temperature of steam at 1.5 MPa having mass of 50 gm and stored in vessel with volume of 0.0076 m3. Vessel is cooled until pressure in vessel becomes 1.1 MPa. Determine the temperature at which steam will be just dry saturated during cooling process. Also determine the final dryness [250\u00b0C, 191.6\u00b0C, 0.85, 18.63 kJ] fraction and total heat rejected. 6.30 Calculate the dryness fraction of steam after throttling when it is throttled from 1.4 MPa to 1 MPa & 423K. Also determine the final condition of steam if this pressure drop takes place in closed vessel of 0.56 m3 [0.98, 0.298] volume and heat is lost by conduction and radiation. 202 _________________________________________________________ Applied Thermodynamics 7 7.1 INTRODUCTION In the present civilization the use of energy resources has increased tremendously. Fast depleting fossil fuel reserves have inevitably gathered the attention of one and all to think and devise for optimum energy utilization. In order to optimally use energy, the efforts are required for identification and elimination of the sources of inefficiency during it\u2019s use, which obviously requires in depth study and analysis. A look into the laws of thermodynamics shows that the first law of thermodynamics bases upon the series of experiments done by James Joules, demonstrating the bidirectional numerical equivalence of converting work into heat while second law of thermodynamics exhibits a unidirectional equivalence between work and heat, i.e. for a given amount of heat the equivalent amount of work cannot be obtained whereas vice-a-versa may be there. Thus, the concept of quality of energy came into existence and work is considered as high grade of energy and heat as low grade of energy. Other forms of high grade energy are electrical energy, wind energy, tidal energy etc. and low grade energy may be heat from nuclear reactions, heat from combustion of fuel etc. Engineers have been using the first law of thermodynamics stating the energy conservation, therefore it could be concluded that energy can not be destroyed and exists with matters in all forms everywhere. It is now quite convincing to understand that the scarcity of energy resources and energy crisis is a paradox. Still in real life we find scarcity of energy, as in practice one is interested in the ability to feed, drive machines and occurrence of energy processes etc. Such discussions gave birth to the concept of \u2018available energy\u2019 and \u2018unavailable energy\u2019 or a concept of \u2018maximum work\u2019. This concept became very important in phenomenological thermodynamics, as it referred to the possibilities of performing work in real conditions. G. Gouy and A. Stodola pioneered in the studies pertaining to effect of ambient temperature upon the obtainable work and law of the loss of maximum work. The law of the loss of maximum work says that the work obtained is always less than the maximum obtainable work due to the irreversibility in thermal processes. Available energy concept came out of these propositions. Quality of energy, its convertibility into other forms and capability to perform work etc. are quantitatively defined using availability analysis. New term \u2018exergy\u2019 was introduced by Z. Rant in 1956 so as to differentiate it", " from energy. \u2018Exergy\u2019 analysis or \u2018availability\u2019 analysis has capability to identify and quantify the causes of thermodynamic imperfections in thermodynamic processes and thus indicate about the possibilities of improving the processes. It is preferred over energy analysis as energy analysis can not detect majority of thermodynamic imperfections. Such as the irreversible heat transfer, throttling and adiabatic combustion etc. do not have any energy loss but make the quality of energy inferior. Energy entering with fuel, electricity, flowing streams of matter and so on can be Availability and General Thermodynamic Relations __________________________________ 203 accounted for in products and by products. Energy cannot be destroyed. The idea that something can be destroyed is useful but should not be applied to \u2018energy\u2019, however it could be applied to another variable \u2018exergy\u2019. Moreover, it is exergy and not energy that properly gauges the quality (utility), say one kJ of electricity generated by a power plant versus one kJ in plant cooling water stream. Electricity obviously has greater quality and the greater economic value. These phenomenon can be evaluated by second law analysis easily. Exergy analysis could be integrated with principles of engineering economics to determine the potential for cost effective improvement of existing systems. Exergy and costing principles can also be used at initial design stage to develop systems that are \u2018optimized in annualized cost\u2019, \u2018sparing in use of fossil fuels\u2019 and \u2018environmentally friendly\u2019. Let us see a electricity generating power cycle as shown. 100 units 25 units, electricity 75 units, surroundings (a) Energy basis 100 units 68-70 units 25 units, electricity 7\u20135 units, surroundings (b) Exergy basis Fig. 7.1 Energy and exergy basis Here Fig. 7.1(a) represents energy basis indicating that out of 100 energy units entering with fuel, 25 energy units are obtained as electricity and the remaining 75 units are discharged to surroundings. On exergy basis it may also be considered that 100 units of exergy enter with fuel are 25 units of exergy exit along with the electricity. For remaining 75 units it is seen that 68-70 units of exergy are destroyed within the plant due to irreversibilities and only 5-7 units are discharged to surroundings. Here it is worth noting from exergy basis that out of 75 units of energy considered to be discharged to surroundings in energy basis actually only 5-7 units are discharged to surroundings and the majority 68-70 units are lost due to irreversibilities. The loss of energy due to irreversibilities can be minimized by minimizing or eliminating the causes of irreversibility. Exergy analysis thus shows that significant performance improvement can come only by identifying and correcting the sources of inefficiency within system as the discharge is a minor area of concern. 7.2 AVAILABILITY OR EXERGY From earlier discussions, it is obvious that energy can be conveniently categorised as low grade energy and high grade energy. Also, the second law of thermodynamics prohibits the complete conversion of low grade energy into high grade energy. The portion of low grade energy that can be converted is called \u2018available energy\u2019 or \u2018exergy\u2019 or \u2018availability\u2019 and the portion of energy not available for conversion is called \u2018unavailable energy\u2019 or \u2018anergy\u2019. Mathematically; Anergy = Energy \u2013 Exergy. 204 _________________________________________________________ Applied Thermodynamics \u201cExergy can be quantified as the amount of work obtainable by bringing some matter to the state of thermodynamic equilibrium with common components of natural surroundings through reversible processes, thus involving interaction only with above mentioned components of nature.\u201d As per Moran and Sciubba (1994), the \u201cexergy refers to the maximum theoretical work that can be extracted from a combined system comprising of \u2018system\u2019 and \u2018environment\u2019 as the system passes from a given state to equilibrium with the environment\u2014that is, system changes its\u2019 state to the dead state at which combined system possesses energy but no exergy.\u201d Rickert defined \u201cexergy as the shaft work or electrical energy necessary to produce a material in its specified state from materials common in the environment in a reversible way, heat being exchanged only with environment.\u201d Exergy is an extensive property whose value is fixed by the state of system once the environment has been specified. Exergy can also be represented on an intensive basis i.e. per unit mass or per mole basis. For all states of the system exergy shall be numerically greater than or equal to zero. Exergy \u2265 0 Exergy as defined above is a measure of departure of the state of a system from that of environment. For state at emperature T and environment at temperature T0 the difference (T ~ T0) shall decide the value of exergy i.e. greater the difference, the greater shall be exergy value. This exergy can be of basically two types i.", "e. chemical exergy and thermomechanical exergy. Thermomechanical exergy can be further classified as physical, kinetic and potential exergy. Physical exergy is the work obtainable by taking the substance by reversible physical processes from its initial states pressure \u2018p\u2019 and temperature \u2018T \u2019 to the state determined by the temperature and pressure of environment. Kinetic exergy is equal to the kinetic energy, when the velocity is considered relative to the surface of the earth. Potential exergy is equal to the potential energy when it is evaluated with respect to the average level of the surface of the earth in the locality of the process under consideration. Chemical exergy refers to the work that can be obtained by taking a substance state at environmental pressure and temperature to the state of thermodynamic equilibrium with environment and bring system to restricted dead state. Thermomechanical exergy refers to the maximum theoretical work obtainable as system passes from some given state to the restricted dead state. Thermal exergy is defined as the sum of \u2018physical exergy\u2019 and \u2018chemical exergy\u2019. Rant defined, \u201cexergy as that part of energy which could be fully converted into any other kind of energy\u201d. Exergy is function of state parameters of matter under consideration and of the state parameters of common components of environment as exergy results from the possibility of interaction between matter under consideration and common components of environment. \u2018Environment\u2019 here refers to the region or part of surroundings whose intensive properties do not change significantly with the occurrence of processes under consideration, while \u2018surroundings\u2019 comprise of everything that is not included in system. Environment is considered to be large and homogeneous in terms of pressure and temperature. Environment is regarded free of irreversibilities. All significant irreversibilities are present in the system and its\u2019 immediate surroundings. Irreversibilities present within system are called \u2018internal irreversibilities\u2019 while \u2018external irreversibilities\u2019 are those present in its\u2019 immediate surroundings. \u2018Dead state\u2019 refers to the state at which system and the environment are at mechanical, thermal and chemical equilibrium. Thus neither there can be any spontaneous change within the system or within the environment, nor any spontaneous interaction between the two. Dead state being a limiting state is also called \u2018restricted dead state\u2019. At dead state the system is at same temperature and pressure as that of its\u2019 surroundings and shall have no kinetic energy or potential energy relative to surroundings. A system Availability and General Thermodynamic Relations __________________________________ 205 shall thus have zero exergy (availability) at dead state and yield maximum possible work only when it follows a reversible process from its\u2019 state to the state of its\u2019 surroundings (dead state). Exergy or availability thus quantifies the maximum theoretical work available without violation of any laws of thermodynamics. An engine operating with heat reservoir at T1 and supplying Q1 amount of heat and the environment temperature being T0 shall give maximum amount of work when it operates between T1 and T0 Maximum efficiency, \u03b7 max = 1 \u2013 Maximum work = Q1. \u03b7 T 0 T = \u03b7 1 max = Availability rev Let us consider a \u2018combined system\u2019 and find work done. Combined system comprises of control system and environment. Contents of control system do not mix with environment or have any reaction with environment. Maximum work is available when control system changes its state from initial state to dead state. W Control system Q Control surface of combined system Wc Environment at & T p 0 0 Fig. 7.2 Combined system Let the control system have heat and work interaction Q and W with environment. Let us have only work interactions, Wc across the control surface of combined system. Let us use following nomenclature, A = availability or exergy E = energy, KE = kinetic energy, Subscript: PE = potential energy p = pressure Q = heat S = entropy T = temperature U = internal energy W = work c = combined system e = environment s = control system 0 = dead state i = initial state Here for combined system \u2206Vc = 0, where Vc = Vs + Ve, although Vs of system or Ve of environment may change but total volume of combined system shall remain constant. Total work interaction of combined system can be given by total energy change of combined system. Wc = \u2013 \u2206Ec Energy change of combined system = Energy change in control system + Energy change in environment \u2206Ec = \u2206Es + \u2206Ee 206 _________________________________________________________ Applied Thermodynamics Energy change in control system, \u2206Es = Energy of system at dead state i.e., final state \u2013 Energy of system at initial state Energy of system at dead state, as energy E = U + KE + PE Es, at dead state = U0, as KE = PE = 0 Energy of system at initial state, Es,initially = Es,i \u2206Es = U0 \u2013", " Es,i Energy change in environment, shall be due to heat interaction and the work associated with its\u2019 volume change (pdv work). For example expansion inside a piston cylinder arrangement shall have piston also displacing the volume of environment (pdV work is boundary work). Change in extensive properties, internal energy, entropy, volume of environment can be given by first law of thermodynamics. Hence, work interaction of combined system, \u2206Ee = \u2206Ue = T0 \u2206Se \u2013 p0.\u2206Ve Also, we have seen that for combined system Wc = \u2013 {(U0 \u2013 Es,i) + (T0 \u2206Se \u2013 p0 \u2206Ve)} \u2206Vc = 0 \u2206Vs + \u2206Ve = 0 or, Here, \u2206Vs = change in volume of system = (Final volume of system at dead state \u2013 Initial volume) \u2206Vs = \u2013 \u2206Ve Substituting in Wc, \u2206Vs = V0 \u2013 Vs,i Wc = {(Es,i \u2013 U0) + p0(Vs,i \u2013 V0) \u2013 T0 \u2206Se} Total entropy change of combined system shall be due to irreversibilities within the combined system; \u2206Sc = \u2206Ss + \u2206Se \u2206Sc = (S0 \u2013 Ss,i) + \u2206Se \u2206Se = (\u2013 S0 + Ss,i) + \u2206Sc or, Substituting \u2206Se in work, Wc, Wc = {(Es,i \u2013 U0) + p0(Vs,i \u2013 V0) \u2013 T0 ((\u2013 S0 + Ss,i) + For the combined system total entropy change shall be either zero for reversible process or more \u2206Sc)} than zero for irreversible process; Mathematically, \u2206 Sc Hence, \u2265 0 \u2264 {(Es,i \u2013 U0) + p0(Vs,i \u2013 V0) \u2013 T0(Ss,i \u2013 S0)} For reversible processes there will be no entropy generation, i.e. \u2206Sc = 0 and work shall be maximum Wc only when the combined system is internally reversible from all respects. Wc, max = {(Es,i \u2013 U0) + p0(Vs,i \u2013 V0) \u2013 T0(Ss,i \u2013 S0)} Thus, In general terms for any initial state of system, which is having all reversible processes. Wc, max = {(E \u2013 U0) + p0(V \u2013 V0) \u2013 T0(S \u2013 S0)} Availability or exergy, A = Wc, max = {(E \u2013 U0) + p0(V \u2013 V0) \u2013 T0(S \u2013 S0)} Availability or exergy cannot be less than zero as the maximum work interaction can not be less than zero. Availability and General Thermodynamic Relations __________________________________ 207 Also it can be given as, Wc = A \u2013 T0. \u2206Sc Above expression shows that some work done by combined system gets lost i.e. the irreversibilities causing entropy production keep work below its\u2019 maximum value. Availability or exergy is not conserved like energy. Exergy gets destroyed by irreversibilities when the control system changes to dead state and no work is done by combined system as in case of spontaneous change. Availability destruction is proportional to entropy generation due to irreversibilities in processes. Irreversibility can be given as the product of dead state temperature and entropy generation due to irreversible process. I = T0. \u2206Sc and Wc = A \u2013 I and I = A \u2013 Wc Above discussion indicates that the maximum work shall be obtained when a process takes place in reversible manner. But in fact almost all the processes in real life occur in irreversible manner, so some portion of energy is always unavailable. As irreversible processes are continuously increasing therefore unavailable energy is also gradually increasing. This phenomenon is also called principle of degradation of energy or law of degradation of energy. Availability, A = {(E \u2013 U0) \u2013 T0(S \u2013 S0) + p0(V \u2013 V0)}, kJ Availability per unit mass, \u03c9 = {(e \u2013 u0) \u2013 T0(s \u2013 s0) + p0(v \u2013 v0)}, kJ/kg Availability or exergy is thus a measure of departure of state of system from that of environment. Thus, it is an attribute of system and environment together. However, once the environment is specified, a numerical value can be assigned to availability in terms of property values for system only. Hence, exergy can be regarded as property of the system. 7.3 AVAILABILITY ASSOCIATED WITH HEAT AND WORK Let us consider a reversible heat engine having heat transfer from environment to control system and vice-a-versa. Environment", ", T0 (\u2013 d )Q0 Reversible H.E. d W = (\u2013 d Q 0 ) \u2013 d Q (d )Q Control system T ( )a Q (\u2013 Environment, T0 dQ0 Reversible H.E. d W = Q (\u2013 d ) \u2013 d Q0 (\u2013 d )Q T > T 0 Q (\u2013 d ) > 0 Q ) > 0 (d 0 Control system T ( )b Let us consider a reversible heat engine transferring heat \u03b4Q to the control system at temperature T from environment at temperature T0. From second law of thermodynamics, Fig. 7.3 208 _________________________________________________________ Applied Thermodynamics \u03b4 Q T \uf8eb \uf8ec \uf8ed \u2212\u03b4 Q 0 T 0 \uf8f6 \uf8f7 \uf8f8 = \u03b4W = \uf8eb \uf8ec \uf8ed T 0 T \u2212 1 \uf8f6 \uf8f8.\u03b4Q \uf8f7 work, Let us now consider a reversible heat engine transferring heat \u03b4Q from control system to environment at T0. From second law of thermodynamics, so work, \u03b4 Q 0 T 0 Q \u2212\u03b4\uf8eb \uf8ec \uf8ed T \uf8f6 \uf8f7 \uf8f8 = \u03b4W = \uf8eb \u2212\uf8ec 1 \uf8ed T 0 T \uf8f6 \uf8f8 (\u2013 \u03b4Q\u03b4Q \uf8f7 = Availability associated with heat transfer : Let us consider a control system at dead state interacting with other system and there is heat interaction Q in control system. Let the final state of control system be given by \u2018f \u2019. Due to heat interaction Q the control system may get heated up or cooled so that the final state is different from that of environment. Control system\u2019s temperature may increase from T0 to Tf or decrease from Tf to T0 but in every case availability shall increase. Availability of control system at final state gives maximum work that will be available from the combined system (control system + environment) as control system returns to dead state. Work available from a reversible heat engine when control system gets heated or cooled by environment, Wmax = \u222b \uf8f7 \uf8f8 \u03b4Q Availability associated with heat transfer = f \u222b 0 \uf8eb \u2212\uf8ec 1 \uf8ed T 0 T \uf8f6 \uf8f7 \uf8f8 \u03b4Q = \u222b If there is irreversibility present within control system due to internal irreversibilities, then availability change from initial to final state can be given as = \u222b Since, here control system\u2019s initial state was dead state having zero availability so the change in availability \u2206 Availability and General Thermodynamic Relations __________________________________ 209 Availability associated with work : Let us consider a control system initially at dead state. Control system has adiabatic compression occurring in it due to work interaction with some other system. \u2013 W work is done on control system and it attains some final state, \u2018f \u2019. Availability in this case shall be the maximum work available from the combined system of control system and environment as control system returns to the dead state. If the work W is done by the control system as it returns from final state \u2018f \u2019 to dead state and the change in volume Vf to V0 takes place by displacing the environment (pdV work), then availability associated with work, Aw = \u2206Aw = [W \u2013 p0(Vf \u2013 V0)] In case no boundary work is there, then Vf = V0 Aw = W = \u2206Aw Here it is also the availability change as system is returning to dead state. In case there is availability loss due to internal irreversibilities then change in availability, Aw = [W \u2013 p0(Vf \u2013 V0)] \u2013 I = \u2206Aw Similarly, availability associated with kinetic energy and potential energy can be given as, = KEA 2 mV 1 2 ; availability with K.E. and PEA = mgz ; availability with P.E. Generalized availability equation : A general availability equation for a control system having heat and work interactions with other systems can be obtained using earlier formulations. Let us consider a control system interacting with other systems and also having irreversibilities causing availability destruction in it. For elemental change during a process the energy balance can be given as, dE = \u03b4Q \u2013 \u03b4W. Total entropy change, dS = \u03b4 Q T + \u03b4Sirrev where T is temperature on control surface having \u03b4Q heat transfer and \u03b4Sirrev is entropy generated due to irreversibilities Energy equation can be rewritten as, Entropy equation can be rewritten as, dE + p0dV = \u03b4Q \u2013 \u03b4W + p0dV + T0.\u03b4Sirrev Combining modified forms of energy and entropy equations by subtracting one from other, T0 dS", " = \u03b4 0\u00b7Q T T dE + p0dV \u2013 T0\u00b7dS = \u03b4Q \u2013 \u03b4W + p0dV \u2013 \u03b4 0\u00b7Q T T \u2013 T0\u00b7\u03b4Sirrev dE + p0dV \u2013 T0\u00b7dS = \uf8eb \u2212\uf8ec 1 \uf8ed T 0 T \uf8f6 \uf8f8 \u00b7\u03b4Q \u2013 (\u03b4W \u2013 p0 \uf8f7. dV) \u2013 T0\u00b7\u03b4Sirrev 210 _________________________________________________________ Applied Thermodynamics We have already seen in earlier article that the change in availability can be given as, Hence, dA = dE + p0.dV \u2013 T0.dS \u2212 dV. 0 \u2212 ) \u03b4. S T 0 irrev dA = \uf8eb \uf8ec \uf8ed T0\u00b7\u03b4Sirrev = I Here, For any process in control system between states 1 and 2, availability change can be given as, \u2206A 1 \u20132 = 2 \uf8f6 \uf8f7 \uf8f8 \uf8eb \u2212\uf8ec 1 \uf8ed T \u222b 0 dQ T 1 Availability associated with heat transfer \u2013 \u2212 \u2206 Availability associated Irreversibility with work interaction ) Generally, for any control mass in control system the availability change can be given as \uf8f6 \u00b7 \uf8f7 \uf8f8 dQ Above availability change can also be given on per unit time basis. 7.4 EFFECTIVENESS OR SECOND LAW EFFICIENCY Performance of engineering systems are generally measured using efficiency as defined by first law of thermodynamics. Efficiency as defined by first law uses energy for its\u2019 quantification. Second law efficiency or effectiveness or exergetic efficiency is an analogous parameter defined using availability. Energy balance for a system with steady state yields, Energy in = Energy output + Energy loss Availability equation shall yield, Availability in = (Availability output + Availability loss + Availability destruction due to Mathematically, irreversibility) (by first law), Efficiency \u03b7 = Energy out in product (= Output) Energy in \u03b7 = Energy input \u2013 Energy loss Energy input \u03b7= \u2212 1 Energy loss Energy input (by second law), Effectiveness, \u03b5 = Availability output Availability in \u03b5 = 1 \u2013 \uf8ee \uf8ef \uf8f0 Availability loss + Availability destruction due to irreversibility Availability in \uf8f9 \uf8fa \uf8fb Effectiveness can also be given as the ratio of thermal efficiency to the maximum possible thermal efficiency (reversible processes) under same conditions. Availability and General Thermodynamic Relations __________________________________ 211 \u03b5 = \u03b7 th \u03b7 th,rev For work producing systems, effectiveness = For work absorbing systems, effectiveness = \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed Useful work Reversible work or maximum work Reversible work or maximum work Useful work \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8 For refrigerators and heat pumps, effectiveness = \uf8eb \uf8ec \uf8ed COP COP rev. In general terms, Second law efficiency or effectiveness = \uf8f6 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ed Availability used Availability supplied \uf8f6 \uf8f7 \uf8f8 7.5 SECOND LAW ANALYSIS OF STEADY FLOW SYSTEMS Let us consider a steady flow system as shown, Surroundings Q m c, 1 1h z, 1,1, 1 1 W 2 2,2 m c, 2, h z, 2 2 Fig. 7.4 Steady flow system Section 1\u20131 and 2\u20132 refer to inlet and exit respectively. Steady flow system interacts with surroundings at P0 and T0. Steady flow energy equation can be given as, Q + m1 \uf8eb \uf8ec \uf8ec \uf8ed + h 1 2 c 1 2 + gz 1 \uf8f6 \uf8f7 \uf8f7 \uf8f8 = W + m2 \uf8eb \uf8ec \uf8ec \uf8ed + h 2 2 c 2 2 + gz 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 Entropy generated, Sgen = m2s2 \u2013 m1s1 + Q surr T 0 Heat transfer from control system, (\u2013) = Heat gained by surrounding (+); \u2013 Q = Qsurr, from above two equations, substitution for Q yields, \u2013T0 Sgen + m1 \uf8eb \uf8ec \uf8ec \uf8ed + h 1 2 c 1 2 + gz + m2 \uf8eb \uf8ec \uf8ec \uf8ed + h 2 2 c 2 2 + gz 2 \u2212 s T 2 0 \uf8f6 \uf8f7 \uf8f7 \uf8f8 212 _________________________________________________________ Applied Thermodynamics or, W = m1 \uf8eb \uf8ec", " \uf8ec \uf8ed + h 1 2 c 1 2 + gz \u2013 m2 + gz 2 \u2212 T s 0 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u2013 T0 \u00b7 Sgen W can be quantified as above. This W shall be actual work available from system. Here entropy generation due to irreversibilities in processes reduce W and so for fully reversible processes Sgen = 0 and we get maximum available work; Wmax = m1 \uf8eb \uf8ec \uf8ec \uf8ed + h 1 2 c 1 2 + gz \u2013 m2 + gz 2 \u2212 T s 0 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 In general terms, actual work available in this kind of systems where boundary work (p.dV) is absent can be given as under, W = \u2211mi ( h i + 2 c i 2 + gz i \u2212 0\u00b7 T s i ) \u2013 \u2211me \uf8eb \uf8ec \uf8ec \uf8ed + h e 2 c e 2 + gz e \u2212 0\u00b7 T s e \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u2013 T0 \u00b7 Sgen where subscript \u2018i\u2019 and \u2018e\u2019 refer to inlet and exit in system. For no irreversibilities present or for reversible processes, W max = \u2211 \uf8eb \uf8ec + gz + gz e \u2212 \u00b7 T s e 0 \uf8f6 \uf8f7 \uf8f7 \uf8f8 Expression 1 Change in availability, \u2206A = \u2211mi \uf8f1 \uf8f4 \uf8f2 \uf8f4 \uf8f3 ( + gz me \u00b7 \uf8f1 \uf8f4 \uf8f2 \uf8f4 \uf8f3 ( + gz ) \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe on unit mass basis, \u2206\u03c9 = {T0 \u00b7 \u2206s \u2013 \u2206h \u2013 \u2206K.E. \u2013 \u2206P.E.} kJ/kg where \u2206K.E. and \u2206P.E. refer to kinetic energy and potential energy changes in system. It indicates that the change in availability can be given by the difference of fluid stream availability at inlet and exit. \u201cFluid stream availability\u201d can be defined in respect to dead state as, c0 = 0, z0 = 0 \u03c8 = (h \u2013 h0) + 2 c 2 + gz \u2013 T0(s \u2013 s0) kJ/kg + gz, kJ/kg. Here underlined terms are called physical exergy, c2/2 is kinetic exergy and gz is potential exergy \u2206A = Wmax = \u2211mi i \u2013 \u2211me. \u03c8 \u03c8 e Change of availability can be obtained using stream availability as described above. \u201cStream availability\u201d is quantification of availability at a point. Availability and General Thermodynamic Relations __________________________________ 213 Irreversibility rate in steady flow process can be given as, I = T0.Sgen, (kW) Exergy (Availability) and energy can be compared based upon their characteristics as given below. Exergy (availability) Energy 1. Exergy does not follow the law of 1. Energy follows the law of conservation. conservation. 2. It is function of states of the matter under consideration and the \u2018environment\u2019. 3. It is estimated with respect to the state of reference imposed by environment. 4. Exergy always depends upon pressure. 5. Exergy increases with temperature drop at low temperatures. For constant pressure processes exergy attains minimum value at the temperature of environment. 2. It is function of the state of matter under consideration. 3. It may be calculated based upon the assumed state of reference. 4. In case of ideal gas energy does not depend upon pressure. 5. Energy increases with rise of temperature. 6. Exergy has positive value for ideal vacuum. 6. Energy is zero for an ideal vacuum. 7.6 GENERAL THERMODYNAMIC RELATIONS Objective of this section is to develop mathematical relations for estimation of various thermodynamic properties such as u, h, s etc. for a compressible system. Thermodynamic properties such as pressure, volume and temperature (P, V, T) etc. can be directly measured experimentally while some other properties can not be measured directly and require thermodynamic relations for their determination. These thermodynamic relations are the basis for getting useful thermodynamic properties. Important mathematical relations : To define state of a simple compressible system of known mass and composition one requires minimum two independent intensive properties. Thus, all intensive properties can be determined through functions of the two independent intensive properties such as, p = p(T, v), u = u(T, v), h = h(T, v)... Above are functions of two independent variables and can be given in general as", ", z = z(x, y), where x, y are independent variables. Exact differential : In earlier discussions we have seen that the differential of any property should be exact. Therefore, let us review calculus briefly. Exact differential of any function z shall be as given below for z being continuous function of x and y. or, dz = \uf8eb \uf8ec \uf8ed \u2202 z \u2202 x \uf8f6 \uf8f7 \uf8f8 dx \uf8eb + \uf8ec \uf8ed \u2202 z \u2202 y \uf8f6 \uf8f7 \uf8f8 x dy y dz = M \u00b7 dx + N \u00b7 dy where, M = \u2202\uf8eb z \uf8ec \u2202\uf8ed x \uf8f6 \uf8f7 \uf8f8 y, N = \uf8eb \uf8f6\u2202 z \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 y x i.e. M is partial derivative of z with respect to x when variable y is held constant and N is partial derivative of z with respect to y when variable x is held constant. Here, since M and N have continuous first partial derivative therefore, order of differentiation is immaterial for properties and second partial derivative can be given as, 214 _________________________________________________________ Applied Thermodynamics \u2202 \u2202 y \uf8ee \uf8ef \uf8ef \uf8f0 \u2202\uf8eb z \uf8ec \u2202\uf8ed \uf8ef \uf8ec \u2202\uf8ef \uf8ed or, \uf8eb \u2202 M \uf8ec \u2202\uf8ed y \uf8f6 \uf8f7 \uf8f8 = x \u2202\uf8eb N \uf8ec \u2202\uf8ed x \uf8f6 \uf8f7 \uf8f8 y Thus, the test of exactness for any property shall be = \u2202\uf8eb N \uf8ec \u2202\uf8ed x \uf8f6 \uf8f7 \uf8f8 y Reciprocity relation and cyclic relation : Let us consider three variables x, y, z such that any two of these are independent variables. Thus, we can write x = x(y, z); y = y(x, z) In differential form, dx = \uf8eb \uf8f6\u2202 x \uf8f7 \uf8ec \uf8f8 \u00b7 dy + \u2202\uf8ed y z \u2202\uf8eb \uf8f6 x \uf8f7\u2202\uf8ed \uf8ec \uf8f8 \u00b7 dz; dy = z y \u2202\uf8eb y \uf8ec \u2202\uf8ed x \uf8f6 \uf8f7 \uf8f8 dx + z \u2202\uf8eb y \uf8ec \u2202\uf8ed z \uf8f6 \uf8f7 \uf8f8 dz x Combining above two relations we get \uf8ee \u2212\uf8ef 1 \uf8ef \uf8f0 \uf8eb \uf8ec \uf8ed \uf8f6\u2202 x \u2202 y \u2202\uf8eb y \u2202\uf8ed x \uf8f7 \uf8ec \uf8f8 z \uf8f6 \uf8f7 \uf8f8 z \uf8f9 \uf8fa \uf8fa \uf8fb dx = \uf8ee \uf8ef \uf8ef \uf8f0 \uf8eb \uf8ec \uf8ed \uf8f6\u2202 \uf8f9 \uf8fa \uf8fa \uf8fb dz As x and z are independent variables so let us keep z constant and vary x, i.e. dz = 0 and dx \u2260 0 which yields reciprocity relation as, \uf8ee \u2212\uf8ef 1 \uf8ef \uf8f0 \uf8eb \uf8ec \uf8ed \uf8f6\u2202 x \u2202 y \u2202\uf8eb y \u2202\uf8ed \uf8f6\u2202 x \u2202 y \u2202\uf8eb y \u2202\uf8ed x \uf8f7 \uf8ec \uf8f8 z \uf8eb \uf8ec \uf8ed or, or, \uf8eb \uf8ec \uf8ed \uf8f6\u2202 \uf8ec \u2202\uf8ed x \uf8f6 \uf8f7 \uf8f8 z Reciprocity relation Similarly, let us keep x constant and vary z i.e. dx = 0, dz \u2260 0 which shall be possible only when; \uf8ee \uf8ef \uf8ef \uf8f0 \uf8eb \uf8ec \uf8ed \uf8f6\u2202 or \uf8eb \uf8ec \uf8ed \uf8f6\u2202 Cyclic relation Availability and General Thermodynamic Relations __________________________________ 215 7.6.1 Gibbs and Helmholtz Functions For a simple compressible system of fixed chemical composition thermodynamic properties can be given from combination of first law and second law of thermodynamics as, du = T \u00b7 ds \u2013 pdv dh = T \u00b7 ds + vdp Gibbs function (g) and Helmholtz function (f) are properties defined as", " below. Gibbs function, g = \u2212 h T s \u00b7, on unit mass basis i.e. specific Gibb\u2019s function also, \u2212 G H T S = \u00b7 Helmholtz function, \u0192 = \u2013 u T \u00b7 s, on unit mass basis i.e. specific Helmholtz function = F U \u2013 T \u00b7 S also, In differential form Gibbs function can be given as below for an infinitesimal reversible process dg = dh \u2013 T \u00b7 ds \u2013 s \u00b7 dT dg = \u2212 vdp sdT for a reversible isothermal process, 2 \u222b 1 dg = 2 \u222b 1 vdp 2 2 \u2212 = or, also dG Vdp SdT for reversible isothermal process; \u222b 1 For a \u201creversible isobaric and isothermal process\u201d, dp = 0, dT = 0 dG = 0 i.e. \u2018Gibbs function\u2019 is also termed as \u2018Gibbs free energy\u2019. For a reversible isobaric and isothermal process Gibbs free energy remains constant or Gibbs function of the process remains constant. Such reversible thermodynamic processes may occur in the processes involving change of phase, such as sublimation, fusion, vaporization etc., in which Gibbs free energy remains constant. G = constant dG\u222b Vdp = 1 \u2018Helmholtz function\u2019 is also called \u2018Helmholtz free energy\u2019. For any infinitesimal reversible process Helmholtz function can be given in differential form as, or, or, d\u0192 = du \u2013 T \u00b7 ds \u2013 sdT df = \u2212 pdv \u2212 sdT dF = \u2212 \u2212 pdV SdT For a reversible isothermal process d\u0192 = \u2013 pdv or, or, 2 \u222b 1 df = \u2013 2 dFp dv pdV For a reversible isothermal and isochoric process, dT = 0, dV = 0 or, df = 0 dF = 0 216 _________________________________________________________ Applied Thermodynamics or, F = Constant Above concludes that the Helmholtz free energy remains constant during a reversible isothermal and isochoric process. Such processes may occur during chemical reactions occurring isothermally and isochorically. 7.6.2 Maxwell Relations Differential equations of thermodynamic properties, u, h, f and g can be given as function of p, T, v, s as below: du = T \u00b7 ds \u2013 pdv dh = T \u00b7 ds + vdp df = \u2013 pdv \u2013 s \u00b7 dT dg = vdp \u2013 s \u00b7 dT Above equations can be used for defining the functions u, h, f, g based upon analogy with, dz = M \u00b7 dx + N \u00b7 dy, for z = z(x, y) and \uf8eb \u2202 M \uf8ec \u2202\uf8ed y \uf8f6 \uf8f7 \uf8f8 \u2202\uf8eb N \uf8ec \u2202\uf8ed x \uf8f6 \uf8f7 \uf8f8. y = x From above four equations for properties to be exact differentials, we can write functions; u = u(s, v) h = h(s, p) f = f(v, T) g = g(p, T) For differential of function \u2018u\u2019 to be exact; \u2202\uf8eb T \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 s = \u2013 \u2202\uf8eb p \uf8ec \u2202\uf8ed s \uf8f6 \uf8f7 \uf8f8 v For differential of function \u2018h\u2019 to be exact; For differential of function \u2018f\u2019 to be exact; For differential of function \u2018g\u2019 to be exact; \uf8eb \uf8f6\u2202 T \uf8f7 \uf8ec \uf8f8 = \u2202\uf8ed p s \u2202\uf8eb v \uf8ec \u2202\uf8ed s \uf8f6 \uf8f7 \uf8f8 p \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 = v \u2202\uf8eb s \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 T \u2202\uf8eb v \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p = \u2013 \uf8eb \uf8f6\u2202 s \uf8f7 \uf8ec \u2202\uf8ed p \uf8f8 T Above four conditions for exact differentials of thermodynamic properties result into \u201cMaxwell relations\u201d. Thus, \u201cMaxwell relations\u201d are : \u2202\uf8eb T \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 = \u2013 s \u2202\uf8eb p \ufffd", "\ufffd\ufffd \u2202\uf8ed s \uf8f6 \uf8f7 \uf8f8 v \uf8f6\u2202 \uf8eb T \uf8f7 \uf8ec \uf8f8 = \u2202\uf8ed p s \u2202\uf8eb v \uf8ec \u2202\uf8ed s \uf8f6 \uf8f7 \uf8f8 p \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 = v \u2202\uf8eb s \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 T Availability and General Thermodynamic Relations __________________________________ 217 \u2202\uf8eb v \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 = \u2013 p \uf8eb \uf8f6\u2202 s \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p T Maxwell relations have large significance as these relations help in estimating the changes in entropy, internal energy and enthalpy by knowing p, v and T. Some applications of these equations are discussed in subsequent articles. 7.6.3 Clapeyron Equation Let us look upon phase change at fixed temperature and pressure and estimate changes in specific entropy, internal energy and enthalpy during phase change. Let us start with one of Maxwell relations From earlier discussions on pure substances we have seen that during phase transformation at some temperature the pressure is saturation pressure. Thus pressure is also independent of specific volume and can be determined by temperature alone. Hence, psat = f (Tsat) \uf8f6 \uf8f7 \uf8f8 = v \u2202\uf8eb p \uf8ec \u2202\uf8ed T dp dT \uf8f6 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ed or sat Here \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 sat is the slope of saturation curve on pressure-temperature (p \u2013 T) diagram at some point determined by fixed constant temperature during phase transformation and is independent of specific volume. Substituting in the Maxwell relation. \u2202\uf8eb s \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 T = \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 sat Pressure Solid Liquid P Slope = dp d T S a t Vapour Fig. 7.5 Pressure-temperature diagram for pure substance T Temperature Thus, during vaporization i.e. phase transformation from liquid to vapour state, above relation can be given as, dry vapour \u222b sat. liquid ds = dry vapour \u222b sat. liquid \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 \u00b7 dv sat 218 _________________________________________________________ Applied Thermodynamics Using notations for dry vapour and saturated liquid it can be given as, or, (sg \u2013 sf) = \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 sat = \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 sat \u00b7 (vg \u2013 vf \uf8f6 \uf8f7 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 sat \uf8eb = \uf8ec \uf8ec \uf8ed s v fg fg \uf8f6 \uf8f7 \uf8f7 \uf8f8 From differential form of specific enthalpy, dh = T\u00b7ds + v\u00b7dp for phase change occurring at constant pressure and temperature, for saturated liquid to dry vapour transformation, dh = T\u00b7ds (hg \u2013 hf) = T\u00b7(sg \u2013 sf) hfg = T\u00b7sfg Substituting fgh T in place of entropy sfg in \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 sat \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 sat \uf8eb = \uf8ec \uf8ec \uf8ed h fg \u00b7 T v fg \uf8f6 \uf8f7 \uf8f7 \uf8f8 Clapeyron equation Above equation is called Clapeyron equation. It can be used for determination of change in enthalpy during phase change i.e. hfg from the p, v and T values which can be easily measured. Thus, Clapeyron equation can also be used for \u201csublimation process\u201d or \u201c melting occurring at constant temperature and pressure\u201d just by knowing slope of saturation curve on p-T diagram, temperature and change in specific volume. Hence, for initial state \u20181\u2019 getting", " transformed into final state \u20182\u2019 due to phase transformation at constant pressure and temperature, general form of Clapeyron equation: \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 sat \uf8eb \uf8ec \uf8ed h 12 12\u00b7 T v \uf8f6 \uf8f7 \uf8f8 = At low pressure during liquid-vapour transformation it is seen that specific volume of saturated liquid state is very small as compared to dry vapour state, i.e. vf <<< vg. Also at low pressure the substance in vapour phase may be treated as perfect gas. Therefore, Clapeyron equation can be modified in the light of two approximations of \u201cvf being negligible compared to vg at low pressures\u201d and \u201cideal gas equation of state during vapour phase at low pressure, \uf8eb \uf8ec \uf8ed = v g RT p \uf8f6 \uf8f7 \uf8f8 \u201d. Clapeyron equation thus becomes, Clausius-Clapeyron equation as given here, \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 sat ( \u2212 ) Availability and General Thermodynamic Relations __________________________________ 219 or, or, or, \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed dp dT dp dT \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8 \u2212 h h g \u00b7( f T RT p / ) = sat \uf8eb \uf8ec \uf8ed \u00b7 fgh RT p 2 \uf8f6 \uf8f7 \uf8f8 = sat \uf8eb \uf8ec \uf8ed dp p \uf8f6 \uf8f7 \uf8f8 sat = h \uf8eb \u00b7fg \uf8ec \uf8ed dT 2 RT \uf8f6 \uf8f7 \uf8f8 sat Clausius-Clapeyron equation Integrating between two states 1 and 2 ln \uf8eb \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f8 p 2 p 1 sat = \uf8eb fgh \uf8f6 \uf8f7 \uf8f8 sat Clausius-Clapeyron equation is thus a modified form of Clapeyron equation based upon certain approximations and is valid for low pressure phase transformations of liquid-vapour or solid-vapour type. 7.6.4 General Relations for Change in Entropy, Enthalpy, Internal Energy and Specific Heats Let us now derive expressions for changes in entropy, enthalpy, internal energy and specific heats as a function of thermodynamic properties, p, v and T. For defining a state any two of the properties amongst the p, v, and T may be regarded as independent properties. Let us take (T, p) and (T, v) as two sets of independent properties for defining other dependent properties. Temperature and Pressure (T, p) as Independent Properties : By considering T and p as independent properties, dependent property say entropy can be given as, Writing differential form of entropy function, s = s(T, p) ds = \u2202\uf8eb \uf8f6 s \uf8f7\u2202\uf8ed \uf8ec \uf8f8 \u00b7 dT + T p \uf8eb \uf8f6\u2202 s \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p T \u00b7 dp From Maxwell relations the partial derivative \uf8f6\u2202 \uf8eb s \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p T can be substituted by \u2013 \u2202\uf8eb \uf8f6 v \uf8f7\u2202\uf8ed \uf8ec \uf8f8 T p as, = ds \uf8eb \uf8ec \uf8ed \u2202 s \u2202 T \uf8f6 \uf8f7 \uf8f8 p \u00b7 dT \u2212 \uf8eb \uf8ec \uf8ed \u2202 v \u2202 T \uf8f6 \uf8f7 \uf8f8 p \u00b7 dp Similarly, specific enthalpy can be given as function of T and p; h = h(T, p) Writing differential form; dh = \u2202\uf8eb h \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 dT + p \uf8eb \uf8f6\u2202 h \uf8f7 \uf8ec \uf8f8 \u00b7 dp \u2202\uf8ed p T We have already seen that specific heat at constant pressure can be given as function of specific enthalpy and temperature at constant pressure. Cp = \u2202\uf8eb h \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 P 220 _________________________________________________________ Applied Thermodynamics Substituting Cp in dh, = dh C p \u00b7 dT \uf8f6\u2202 \uf8eb h + \uf8ec", " \uf8f7 \u2202\uf8ed \uf8f8 p T \u00b7 dp From definition of enthalpy, first and second law combined, dh = T \u00b7 ds + vdp Substituting dh and ds from above, \uf8f6\u2202 \uf8eb h \uf8f7 \uf8ec \uf8f8 \u00b7 dp = T \u00b7 \u2202\uf8ed p T \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 dT \u2013 T p \u2202\uf8eb v \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 \u00b7 dp + v\u00b7 dp p Cp \u00b7 dT + or, \uf8f1 \uf8f4 \uf8f2 \uf8f4 \uf8f3 \uf8eb \uf8ec \uf8ed \uf8f6\u2202 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe = dp \uf8f1 \uf8f4 \uf8f2 dT p \uf8f4 \uf8fe Here T and p are considered to be independent variables so let us keep pressure constant and vary temperature i.e. dp = 0, dT \u2260 0. It yields in modified form of above underlined equation as \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p \u2212 C p \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe dT = 0 or, or, CP = T\u00b7 \u2202\uf8eb \uf8f6 s \uf8f7\u2202\uf8ed \uf8ec \uf8f8 T p \u2202\uf8eb \uf8f6 s \uf8f7\u2202\uf8ed \uf8ec \uf8f8 = T p pC T Substituting \u2202\uf8eb \uf8f6 s \uf8f7\u2202\uf8ed \uf8ec \uf8f8 T p in ds expression we get, = ds C dT p T \u2212 \uf8ec \u2202\uf8eb v \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 \u00b7 dp p Similarly, temperature can be kept constant and pressure varied independently as, dT = 0, dp \u2260 0 Above underlined equation gets modified as, \uf8f1 \uf8eb \uf8f4 \uf8f2 \uf8ec \uf8ed \uf8f4 \uf8f3 \uf8f6\u2202 \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p \u2212 v \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe dp = 0 or, Substituting \uf8f6\u2202 \uf8eb h \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p T or, \uf8eb \uf8ec \uf8ed \uf8f6\u2202 \u2202\uf8eb v \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p in expression for dh we get = dh C dT p + \uf8f1 \uf8f4 \u2212\uf8f2 v T \uf8f4 \uf8f3 \u2202 \uf8eb v \uf8ec \u2202\uf8ed dp Availability and General Thermodynamic Relations __________________________________ 221 h2 \u2013 h1 = T 2 \u222b T 1 C dT p + \u222b p 2 p 1 \uf8f1 \uf8f4 \u2212\uf8f2 v T \uf8f4 \uf8f3 \u2202 \uf8eb v \uf8ec \u2202\uf8ed dp Temperature and Specific Volume (T, v) as independent property: Considering T and v as independent properties the dependent properties can be given as, u = u(T, v) Writing differential form of specific internal energy. = du \uf8eb \uf8ec \uf8ed \u2202 u \u2202 T \uf8f6 \uf8f7 \uf8f8 v \u00b7 dT + \uf8eb \uf8ec \uf8ed \u2202 u \u2202 v \uf8f6 \uf8f7 \uf8f8 T dv From definition of specific heat at constant volume, Cv = \u2202\uf8eb u \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 v or, = du C dT \u00b7v + \uf8ec \u2202\uf8eb u \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 T dv Writing specific entropy as function of T and v, s = s(T, v) = ds \uf8eb \uf8ec \uf8ed \u2202 s \u2202 T \uf8f6 \uf8f7 \uf8f8 v + dT \uf8eb \uf8ec \uf8ed \u2202 \u2202 s v \uf8f6 \uf8f7 \uf8f8 T dv From Maxwell relations, \u2202\uf8eb s \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 = T \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7", " \uf8f8 ; substituting in above we get v From I and II law combined, Substituting du and ds in above equation, ds = \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 dT + du = T \u00b7 ds \u2013 pdv v \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 dv v Cv dT + \u2202\uf8eb u \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 dv = T T \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 dT + T \u00b7 v \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 dv \u2013 pdv v Rearranging terms dv \uf8fe = \uf8f1 \uf8f2 dT \uf8fe v As T and v are considered independent variables therefore let us keep T as constant and v as variable, i.e. dT = 0, dv \u2260 0. It yields \uf8fc \uf8fd \uf8fe dv = 0 or 222 _________________________________________________________ Applied Thermodynamics Similarly, let us keep v constant and T as variable i.e dv = 0 dT \u2260 0. It yields \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 v \u2212 C v \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe dT = \uf8ec \u2202\uf8ed \u2202\uf8eb s = \uf8ec \u00b7 T \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 v or, \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 v = C v T or, or, Let us now substitute \u2202\uf8eb u \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 T in the differential function du which yields, = du C dT \u00b7 \u00b7 \uf8ec \u2202\uf8ed dv \uf8f4 \uf8fe For any state change from 1 to 2 we can get change in internal energy, as dT \u00b7 \uf8f7\u2202\uf8ed \uf8ec \uf8f8 T v \u2212 \uf8fc \uf8f4 \uf8fd p dv \uf8f4 \uf8fe Also, let us substitute \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 v in the expression of entropy change \u2018ds\u2019. It results in, = ds C dT v T + \uf8ec \u2202\uf8eb p \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 dv v Thus, number of expressions are available for getting the change in h, u and s, which are summarized as under. = dh C dT p = du C dT v + \uf8f1 \uf8f4 \u2212\uf8f2 v T \uf8f4 \uf8f3 \u2202 \uf8eb v \uf8ec \u2202\uf8ed dp + \uf8f1 \uf8f4 \uf8f2 T \uf8f4 \uf8f3 \u2202 \uf8eb p \uf8ec \u2202\uf8ed dv \uf8f4 \uf8fe = ds C dT p \u00b7 T \u2212 \uf8ec \u2202\uf8eb v \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 \u00b7 dp p = ds C dT v T + \uf8ec \u2202\uf8eb p \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 \u00b7 dv v Availability and General Thermodynamic Relations __________________________________ 223 From above two expressions of entropy change, the difference between Cp and Cv values can be obtained as, (Cp \u2013 Cv)dT = T \u00b7 \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 v dv + T \u2202\uf8eb v \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p dp Let us write from equation of state, the function p = p(T, v) Differential, dp = \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 dT + v \u2202\uf8eb p \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 T dv Substituting dp in above equation of (Cp \u2013 Cv) dT we get, (Cp \u2013 Cv) dT = T \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \ufffd", "\ufffd\ufffd \uf8f8 dv + T v \u2202\uf8eb v \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 \u00b7 p \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 dT + T v \u2202\uf8eb v \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p \u2202\uf8eb p \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 dv T or \uf8f9 \uf8fa \uf8fa \uf8fb dv dT = Since T and v are independent so let us keep T as constant and v as variable i.e dT = 0 dv \u2260 0 \uf8f9 \uf8fa \uf8fa \uf8fb dv = 0 or, \u2202\uf8eb p \uf8ec \u2202\uf8ed Similarly keeping v as constant and T variable i.e., dv = 0, dT \u2260 dT = 0 or Substituting for \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 v from above in (Cp \u2013 Cv) we get In single phase region the specific volume can also be given as function of T & p and the differential of function v shall be, v = v (T, p) 224 _________________________________________________________ Applied Thermodynamics or, dv = \u2202\uf8eb v \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p dT + \uf8eb \uf8f6\u2202 v \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p T dp The above differential form of specific volume indicates that it depends upon partial derivatives of specific volume with respect to temperature and pressure. Partial derivatives of v with respect to temperature can be related to \u201cvolume expansivity\u201d or \u201ccoefficient of volume expansion\u201d as below, Volume expansivity, \u03b2 1 = \uf8ec v \u2202\uf8eb v \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p Partial derivative of specific volume with respect to pressure can be related to \u201cisothermal compressibility\u201d, \u03b1 as below. Isothermal compressibility, \u03b1 = \u2212 1 v \uf8eb \u2202 v \uf8ec \u2202\uf8ed p \uf8f6 \uf8f7 \uf8f8 T Inverse of isothermal compressibility is called \u201cisothermal bulk modulus\u201d, \uf8f6 \uf8f7 \uf8f8 T Thus, volume expansivity gives the change in volume that occurs when temperature changes while pressure remains constant. Isothermal compressibility gives change in volume when pressure changes while temperature remains constant. These volume expansivity and isothermal compressibility are thermodynamic properties. B T = \u2013 v \u2202\uf8eb p \uf8ec \u2202\uf8ed v Similarly, the change in specific volume with change in pressure isentropically is also called \u201cisentropic compressibility\u201d or \u201cadiabatic compressibility\u201d, Mathematically \u03b1 s = \u2212 1 v \uf8eb \u2202 v \uf8ec \u2202\uf8ed p \uf8f6 \uf8f7 \uf8f8 s. Reciprocal of isentropic compressibility is called \u201cisentropic bulk modulus,\u201d Bs = \u2013 v Substituting \u03b2 and \u03b1 in (Cp \u2013 Cv) expression; \u2202\uf8eb p \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 s Cp \u2013 Cv = \u2013 T(\u03b22 \u00b7 v2) \uf8eb \uf8ec \uf8ed \uf8f6\u2212 1 \uf8f7 \u00b7 v\u03b1 \uf8f8 or, C p C Mayer relation Above difference in specific heat expression is called \u201cMayer relation\u201d and helps in getting significant conclusion such as, The difference between specific heats is zero at absolute zero temperature i.e. specific heats at constant pressure and constant volume shall be same at absolute zero temperature (T = 0 K). Specific heat at constant pressure shall be generally more than specific heat at constant volume i.e., Cp \u2265 Cv. It may be attributed to the fact that \u2018\u03b1\u2019 the isothermal compressibility shall always be Availability and General Thermodynamic Relations __________________________________ 225 +ve and volume expansivity \u2018\u03b2\u2019 being squared in (Cp \u2013 Cv) expression shall also be +ve. Therefore (Cp \u2013 Cv), shall be either zero or positive value depending upon magnitudes of v, T, \u03b2 and \u03b1. For incompressible substances having dv = 0, the difference (Cp \u2013 Cv) shall be nearly zero. Hence, specific heats at constant pressure and at constant volume are identical. Let us obtain expression for ratio of", " specific heats. Earlier we have obtained Cp and Cv as below, \uf8f6 \uf8f7 \uf8f8 Cp = T \u00b7 or, = \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 \u2202\uf8eb s \uf8ec \u2202\uf8ed T p pC T p and vC T By cyclic relation we can write for p, T and s properties; Cv = T \u00b7 \uf8f6 \uf8f7 \uf8f8 or v \u2202\uf8eb s \uf8ec \u2202\uf8ed T = \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 v \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 \uf8f6\u2202 \uf8eb T \uf8f7 \uf8ec \uf8f8 \u00b7 \u2202\uf8ed p s \u2202\uf8eb \uf8f6 p \uf8f7\u2202\uf8ed \uf8ec \uf8f8 s T \u00b7 p = \u2013 1 or, \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p = \uf8eb \uf8ec \uf8ed \uf8f6\u2202 T \uf8f7\u2202 \uf8f8 p \u2212 1 \u2202\uf8eb p \uf8ec \u2202\uf8ed s \u00b7 s \uf8f6 \uf8f7 \uf8f8 T Similarly for s, T and v properties we can write using cyclic relation1 \u2202\uf8eb s \uf8ec \u2202\uf8ed \uf8f6 \uf8f7 \uf8f8 or Substituting in the relation for (Cp/T) and \uf8eb \uf8ec \uf8ed vC T \uf8f6 \uf8f7 \uf8f8. \uf8eb \uf8ec \uf8ed pC T \uf8eb \uf8ec \uf8ed vC T \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8 and = \uf8eb \uf8ec \uf8ed \uf8f6\u2202 T \uf8f7\u2202 \uf8f8 p \u2212 1 \u2202\uf8eb p \u00b7 \uf8ec \u2202\uf8ed Taking ratio of two specific heats \uf8f6\u2202 \u2202\uf8eb p \u00b7 \uf8ec \u2202\uf8ed 226 _________________________________________________________ Applied Thermodynamics or \uf8f6 \uf8f7 \uf8f8 = \u2202\uf8eb T \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 \u2202\uf8eb v \uf8ec \u2202\uf8ed s \uf8f6 \uf8f7 \uf8f8 T \u00b7 s \u00b7 \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 s \u00b7 \uf8eb \uf8f6\u2202 s \uf8f7 \uf8ec \u2202\uf8ed By chain rule of calculus we can write, \uf8f6\u2202 \uf8eb v \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p T = \u2202\uf8eb v \uf8ec \u2202\uf8ed s \uf8f6 \uf8f7 \uf8f8 T \u00b7 \uf8eb \uf8f6\u2202 s \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p T and s Upon substitution in specific heat ratio we get, \u2202\uf8eb p \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 = s \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 \u00b7 \u2202\uf8eb T \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 s or, \uf8eb \uf8ec \uf8ed C p C v \uf8f6 \uf8f7 \uf8f8 = \uf8eb \uf8ec \uf8ed \uf8f6\u2202 v \uf8f7 \u2202 p \uf8f8 T \u00b7 \u2202\uf8eb p \uf8ec \u2202\uf8ed \uf8ec \u2202\uf8ed = Isothermal compressibility Isentropic compressibility Thus, ratio of specific heats at constant pressure and constant volume can be given by the ratio of isothermal compressibility and isentropic compressibility. 7.6.5 Joule-Thomson Coefficient Joule-Thomson coefficient is defined as the rate of change of temperature with pressure during an isenthalpic process or throttling process. Mathematically, Joule-Thomson coefficient (\u00b5) can be given as, \u00b5 = \uf8f6\u2202 \uf8eb T \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p h It is defined in terms of thermodynamic properties and is itself a property. Joule-Thomson coefficient gives slope of constant enthalpy lines on temperature\u2014", "pressure diagram. Thus, it is a parameter for characterizing the throttling process. Slope of isenthalpic line may be positive, zero or negative, i.e. \u00b5 > 0, \u00b5 = 0 and \u00b5 < 0 respectively. Mathematically evaluating the consequence of \u00b5 we see, \u2013 for \u00b5 > 0, temperature decreases during the process. \u2013 for \u00b5 = 0, temperature remains constant during the process. \u2013 for \u00b5 < 0, temperature increases during the process. Joule-Thomson expansion can be shown as in Fig. 7.6. Here gas or liquid is passed through porous plug for causing isenthalpic process. Valve put near exit is used for regulating pressure after constant enthalpy process i.e. p2. Availability and General Thermodynamic Relations __________________________________ 227 Inversion line Inversion states or points Porous plug Inlet p T1, 1 p T2, 2 Valve Exit T \u00b5<0 \u00b5 < 0 1, T p1 Heating region p \u2202T \u2202p = slope, \u00b5 h = Constant lines Fig. 7.6 Joule-Thomson expansion If pressure p2 is varied then the temperature variation occurs in the isenthalpic manner as shown in T-p diagram. This graphical representation of isenthalpic curve gives the Joule-Thomson coefficient by its slope at any point. Slope may be positive, negative or zero at different points on the curve. The points at which slope has zero value or Joule-Thomson coefficient is zero are called \u201cinversion points\u201d or \u201cinversion states\u201d. Temperature at these inversion states is called \u201cinversion temperature\u201d. Locii of these inversion states is called \u201cinversion line\u201d. Thus, inversion line as shown divides T-p diagram into two distinct region i.e. one on the left of line and other on the right of line. For the states lying on left of the inversion line temperature shall decrease during throttling process while for the states on right of inversion line throttling shall cause heating of fluid being throttled. Temperature at the intersection of inversion line with zero pressure line is called \u201cmaximum inversion temperature\u201d. 7.6.6 Chemical Potential In case of multicomponent systems such as non-reacting gas mixtures the partial molal properties are used for describing the behaviour of mixtures and solutions. Partial molal properties are intensive properties of the mixture and can be defined as, where X is extensive property for multi component system in single phase. Xi = \u2202\uf8eb \uf8f6 X \uf8f7\u2202\uf8ed \uf8ec \uf8f8 (T, p, n) i.e. function of temperature, pressure and no. of moles of each component nk refers to the all n values with varing k values and are kept constant except for ni. In multicomponent systems the partial molal Gibbs function for different constituents are called \u201cchemical potential\u201d for particular constituent. Chemical potential, \u00b5 can be defined for ith component as, \u00b5 i = \u2202\uf8eb \uf8f6 G \uf8f7\u2202\uf8ed \uf8ec \uf8f8 n i T p n, k, where G, ni, nk, T and P have usual meanings. Chemical potential being a partial molal property is intensive property. Also, it can be given as ) Thus, for non reacting gas mixture the expression for internal energy, enthalpy, Helmholtz function can be given using G defined as above, 228 _________________________________________________________ Applied Thermodynamics Internal energy, U = TS \u2013 pV + j \u2211 = i 1 n \u00b5 i i Enthalpy, H = TS + Helmholtz function, F = \u2013 pV + Writing differential of G considering it as function of (T, p, n1, n2,... nj) G = G(T, p, n1, n2, n3...nj) dG = \uf8f6\u2202 \uf8eb G \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p,T n dp + \u2202\uf8eb G \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8,p n dT + j \u2211 = i 1 \u2202\uf8eb \uf8f6 G \uf8f7\u2202\uf8ed \uf8ec \uf8f8 n i T p n, k, dni From definition of Gibbs function dG = Vdp \u2013 SdT, for T = constant, V = for pressure as constant, \u2013 S = Therefore, \uf8f6\u2202 \uf8eb G \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p,T n \u2202\uf8eb G \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8,p n = dG Vdp \u2013 Sd", "T j + \u2211 = 1 i ( \u00b5 i \u00b7 dn i ) Also from ) we can write differential as, dG = j \u2211 = 1 i ( \u00b7 n d\u00b5 \u00b7 dn i ) From two differential of function G we get, Vdp \u2212 SdT j = \u2211 = 1 i ( \u00b7 n d\u00b5 i i ) Above equation is also called Gibbs-Duhem equation. 7.6.7 Fugacity From earlier discussions for a single component system one can write, G = n \u00b7 \u00b5 Availability and General Thermodynamic Relations __________________________________ 229 or, \u00b5 = G n or \u21d2 Chemical potential for pure substance = Gibbs function per mole. g = G n = \u00b5 For Gibbs function written on unit mole basis, For constant temperature v = \uf8eb \uf8f6\u2202 \u00b5 \uf8f7 \uf8ec \u2202\uf8ed p \uf8f8 T If single component system is perfect gas then, v = RT p or, or \uf8f6\u2202 \uf8eb \u00b5 \uf8f7 \uf8ec \uf8f8 = \u2202\uf8ed p T RT p T\u00b5 = RT ln p + constant Here chemical potential may have any value depending upon the value of pressure. Above mathematical formulation is valid only for perfect gas behaviour being exhibited by the system. For a real gas above mathematical equation may be valid if pressure is replaced by some other property called \u2018fugacity\u2019. Fugacity was first used by Lewis. Fugacity denoted by \u2018 \u2019 can be substituted for pressure in above equation, For constant pressure using v = \u00b5 = RT ln + Constant \uf8eb \uf8f6\u2202\u00b5 \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 Tp and above equation, we get RT \uf8eb \u2202 ln \uf8ec \u2202\uf8ed p \uf8f6 \uf8f7 \uf8f8 T = v Thus, for a limiting case when ideal gas behaviour is approached the fugacity of a pure component shall equal the pressure in limit of zero pressure. \uf8eb lim \uf8ec p\u2192 \uf8ed p 0 \uf8f6 \uf8f7 \uf8f8 = p For an ideal gas For real gas, equation of state can be given using compressibility factor as, p v = ZRT = 1 or, v = ZRT p Substituting the fugacity function, ZRT p = RT \u2202 \uf8eb \uf8ec \uf8ed ln \u2202 p \uf8f6 \uf8f7 \uf8f8 T 230 _________________________________________________________ Applied Thermodynamics \uf8eb \u2202 ln \uf8ec \u2202\uf8ed p \uf8f6 \uf8f7 \uf8f8 T Z p = \uf8eb \u2202 \uf8ec \u2202\uf8ed ln ln \uf8f6 \uf8f7 Tp \uf8f8 = Z. Here as p \u2192 0 the Z \u2192 1 \u00b5 T = R T ln + constant or, Also we have seen or, or, Integrating between very low pressure p* and high pressure p. (d\u00b5)T = R T d (ln )T dgT = R Td (ln )T * g \u222b g dg T = * \u222b RT d(ln )T or g = g* + RT ln Here for very low pressure, * = p* or, g = g* + RT ln \uf8eb \uf8ec \uf8ed * \uf8f6 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ec \uf8ed * p \uf8f6 \uf8f7 \uf8f7 \uf8f8 When low pressure is 1 atm then the ratio \uf8eb \uf8ec \uf8ed * \uf8f6 \uf8f7 \uf8f8 is called \u201cactivity\u201d. EXAMPLES 1. Steam at 1.6 MPa, 300\u00baC enters a flow device with negligible velocity and leaves at 0.1 MPa, 150\u00baC with a velocity of 150 m/s. During the flow heat interaction occurs only with the surroundings at 15\u00baC and steam mass flow rate is 2.5 kg/s. Estimate the maximum possible output from the device. Solution: Let us neglect the potential energy change during the flow. Surroundings at 15\u00b0C 2 p 2 T 2 = 0.1 MPa = 150\u00b0C Flow device 1 p 1 T 1 = 1.6 MPa = 300\u00b0C Fig. 7.7 Availability and General Thermodynamic Relations __________________________________ 231 Applying S.F.E.E., neglecting inlet velocity and change in potential energy, Wmax = (h1 \u2013 T0 s1 \uf8f6 \uf8f7 \uf8f7 \uf8f8 Wmax = (h1 \u2013 h2) \u2013 T0(s1 \u2013 s2) \u2013 2 C 2 2 From steam tables, Given; h1 = hat 1.6 MPa", ", 300\u00baC = 3034.8 kJ/kg s1 = sat 1.6 MPa, 300\u00baC = 6.8844 kJ/kg \u00b7 K h2 = hat 0.1 MPa, 150\u00baC = 2776.4 kJ/kg s2 = sat 0.1 MPa, 150\u00baC = 7.6134 kJ/kg \u00b7 K T0 = 288 K Wmax = (3034.8 \u2013 2776.4) \u2013 288(6.8844 \u2013 7.6134) \u2013 2 (150) 2 \u00d7 10\u20133 Maximum possible work = 2.5 \u00d7 457.1 kJ/s = 1142.75 kW = 457.1 kJ/kg Maximum possible work = 1142.75 kW Ans. 2. Two tanks A and B contain 1 kg of air at 1 bar, 50\u00baC and 3 bar, 50\u00baC when atmosphere is at 1 bar, 15\u00baC. Identify the tank in which stored energy is more. Also find the availability of air in each tank. Solution: In these tanks the air stored is at same temperature of 50\u00baC. Therefore, for air behaving as perfect gas the internal energy of air in tanks shall be same as it depends upon temperature alone. But the availability shall be different. Both the tanks have same internal energy Ans. Availability of air in tank A = {E \u2013 U0} + p0(V \u2013 V0) \u2013 T0(S \u2013 S0) = m {(e \u2013 u0) + p0(v \u2013 v0) \u2013 T0(s \u2013 s0)} \uf8eb \uf8ec RT p RT ln p T T 0 \u2212 R ln \uf8f6 \uf8f7 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe ln T T 0 + T R 0 ln \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe p p 0 For tank A, m = 1 kg, cv = 0.717 kJ/kg \u00b7 K, T = 323 K, R = 0.287 kJ/kg \u00b7 K, T0 = 288 K, p0 = 1 bar, cp = 1.004 kJ/kg \u00b7 K AvailabilityA = 1 {0.717 (323 \u2013 288) + 0.287 (1 \u00d7 323 \u2013 288) \u2013 (288 \u00d7 1.004) ln \uf8eb \uf8ec \uf8ed 323 288 \uf8f6 \uf8f7 \uf8f8 + 288 \u00d7 0.287 ln 1} = 1.98 kJ 232 _________________________________________________________ Applied Thermodynamics For tank B, T = 323 K, p = 3 bar AvailabilityB = 1{ \u2212 0.717 (323 288) 0.287 + \uf8eb \uf8ec \uf8ed 1 3 \u00d7 \u2212 323 288 \uf8f6 \uf8f7 \uf8f8 \u2013 \uf8eb \uf8ec \uf8ed 288 1.004 ln \u00d7 323 288 \uf8f6 \uf8f7 \uf8f8 + \uf8eb \uf8ec \uf8ed 288 0.287 ln \u00d7 \uf8f6 \uf8f7 \uf8f8 = 30.98 kJ 3 1 Availability of air in tank B is more than that of tank A. Availability of air in tank A = 1.98 kJ Ans. Availability of air in tank B = 30.98 kJ 3. 15 kg/s steam enters a perfectly insulated steam turbine at 10 bar, 300\u00baC and leaves at 0.05 bar, 0.95 dry with velocity of 160 m/s. Considering atmospheric pressure to be 1 bar, 15\u00baC. Determine (a) power output, (b) the maximum power for given end states, (c) the maximum power that could be obtained from exhaust steam. Turbine rejects heat to a pond having water at 15\u00baC. Solution: From steam tables, Enthalpy at inlet to turbine, h1 = 3051.2 kJ/kg s1 = 7.1229 kJ/kg \u00b7 K 15 kg/s 10 bar, 300\u00b0C 1 W 1 bar, 15\u00b0C 2 0.05 bar, 0.95 Fig. 7.8 Enthalpy at exit of turbine, h2 = hat 0.05 bar, 0.95 dry s2 = sf at 0.05 bar + (0.95 \u00d7 sfg at 0.05 bar) s2 = 0.4764 + (0.95 \u00d7 7.9187) s2 = 7.999 kJ/kg \u00b7 K Similarly, h2 = hf at 0.05 bar + (0.95 \u00d7 hfg at 0.05 bar) = 137.82 + (0.95 \u00d7 2423.7) h2 = 2440.34 kJ/kg Neglecting the change in potential energy and velocity", " at inlet to turbine, the steady flow energy equation may be written as to give work output. w = (h1 \u2013 h2) \u2013 2 V 2 2 Availability and General Thermodynamic Relations __________________________________ 233 w = (3051.2 \u2013 2440.34) \u2013 \uf8eb \uf8ec \uf8ec \uf8ed 2 (160) 2 \u2212 3 \u00d7 10 \uf8f6 \uf8f7 \uf8f7 \uf8f8 w = 598.06 kJ/kg Power output = m.w = 15 \u00d7 598.06 = 8970.9 kW Power output = 8970.9 kW Ans. Maximum work for given end states, wmax = (h1 \u2013 T0 \u00b7 s1\u00b7 T s \uf8f6 \uf8f7 \uf8f7 \uf8f8 wmax = (3051.2 \u2013 288 \u00d7 7.1229) \u2013 2440.34 + (160) 2 \u00d7 10 \u2212 3 2 \uf8eb \uf8ec \uf8ec \uf8ed \u2212 \u00d7 288 7.999 \uf8f6 \uf8f7 \uf8f7 \uf8f8 wmax = 850.38 kJ/kg Wmax = m\u00b7wmax = 15 \u00d7 850.38 = 12755.7 kW Maximum power output = 12755.7 kw Ans. Maximum power that could be obtained from exhaust steam shall depend upon availability with exhaust steam and the dead state. Stream availability of exhaust steam, Aexhaust = + h0 \u2013 T0 s0) = (h2 \u2013 h0) + 2 V 2 2 \u2013 T0 (s2 \u2013 s0) Approximately the enthalpy of water at dead state of 1 bar, 15\u00baC can be approximated to saturated liquid at 15\u00baC. h0 = hf at 15\u00baC = 62.99 kJ/kg s0 = sf at 15\u00baC = 0.2245 kJ/kg \u00b7 K Maximum work available from exhaust steam = (2440.34 \u2013 62.99) + \uf8eb \uf8ec \uf8ec \uf8ed 2 (160) 2 \u2212 3 \u00d7 10 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u2013 288 (7.999 \u2013 0.2245) Maximum power that could be obtained from exhaust steam Aexhaust = 151.1 kJ/kg = m \u00d7 Aexhaust = 15 \u00d7 151.1 = 2266.5 kW Maximum power from exhaust steam = 2266.5 kW Ans. 234 _________________________________________________________ Applied Thermodynamics 4. 5 kg of steam, initially at elevation of 10 m and velocity of 25 m/s undergoes some process such that finally it is at elevation of 2m and velocity of 10 m/s. Determine the availability corresponding to the initial and final states. Also estimate the change in availability assuming environment temperature and pressure at 25\u00baC and 100 kPa respectively. Thermodynamic properties u, v, s are as under. Dead state of water Initial state Final state u0 = 104.86 kJ/kg v0 = 1.0029 \u00d7 10\u20133 m3/kg s0 = 0.3673 kJ/kg \u00b7 K u1 = 2550 kJ/kg v1 = 0.5089 m3/kg s1 = 6.93 kJ/kg \u00b7 K u2 = 83.94 kJ/kg v2 = 1.0018 \u00d7 10\u20133 m3/kg s2 = 0.2966 kJ/kg \u00b7 K Solution: Availability at any state can be given by + gz \uf8f9 \uf8fa \uf8fa \uf8fb Availability at initial state, A1 = 5 \uf8ee \uf8f0 (2550 104.86) 10 \u00d7 \u2212 3 + 100 10 (0.5089 1.0029 10 \u00d7 3 \u2212 \u00d7 \u2212 3 ) \u2212 298(6.93 \u2013 0.3673) 10 \u00d7 3 + 2 (25) 2 + \u00d7 (9.81 10) \uf8f9 \uf8fa \uf8fa\uf8fb A1 = 2704.84 kJ. = 2704.84 kJ Availability at initial state Ans. Availability at final state A2 = 5 \uf8ee \uf8f0 3 (83.93 104.86) 10 \u2212 \u00d7 + 100 10 (1.0018 10 \u00d7 \u00d7 3 \u2212 3 \u2212 1.0029 \u00d7 \u2212 3 3 10 ) 298(0.2966 0.3673) 10 \u2212 \u2212 \u00d7 2 + + (10) 2 \uf8f9 \u00d7 \uf8fa (9.81 2) \uf8fa\uf8fb A2 = 1.09 kJ Availability at final state = 1.09 kJ Ans. Change in availability: A2 \u2013 A1 = 1.09 \u2013 2704.84 = \u2013 2703.75 kJ Hence availability decreases by 2703.75 kJ Ans. Availability and General Thermodynamic Relations __________________________________ 235 5. For a", " steady flow process as shown below, prove that irreversibility, I = T0 Sgen\u2019 where T0 and P0 are temperature and pressure at dead state. Q 1 at temperature T 1 1 m 2 m Fig. 7.9 Solution: Let us assume changes in kinetic and potential energy to be negligible. Let us use subscript 1 for inlet and 2 for outlet. From first law of thermodynamics; Q1 + m1h1 = m2h2, here m1 = m2 = m or Q1 = m(h2 \u2013h1) From second law of thermodynamics, Sgen + Q 1 T 1 + ms1 = ms2; m(s2 \u2013 s1) = Q 1 T 1 + Sgen From availability considerations in control volume, \uf8eb \u2212\uf8ec 1 \uf8ed \uf8f6 \uf8f7 \uf8f8 T 0 T 1 Q1 + ma1 = m\u00b7a2 + I or, \uf8eb \u2212\uf8ec 1 \uf8ed \uf8f6 \uf8f7 \uf8f8 T 0 T 1 Q1 + m{(h1 \u2013 h0) \u2013 T0(s1 \u2013s0)} = m{(h2 \u2013 h0) \u2013 T0(s2 \u2013 s0)} + I or, Upon substituting from above equations, m(s2 \u2013 s1 or it can be given that I T 0 = Sgen or, = I T S 0 \u00b7 gen Hence proved. 6. Exhaust gases leave an internal combustion engine at 800\u00baC and 1 atmosphere, after having done 1050 kJ of work per kg of gas in engine. (cp of gas = 1.1 kJ/kg \u00b7 K). The temperature of surrounding is 30\u00baC (i) How much available energy per kg of gas is lost by throwing away the exhaust gases? (ii) What is the ratio of the lost available exhaust gas energy to engine work? Solution: Loss of available energy = Irreversibility = T0 \u00b7 \u2206Sc Here, T0 = 303 K = Temperature of surroundings \u2206Sc = \u2206Ss + \u2206Se 236 _________________________________________________________ Applied Thermodynamics Change in entropy of system = 1050 + (273 800) = 0.9786 kJ/kg \u00b7 K Change in entropy of surroundings = pc\u2212 \u2212 \u00b7(800 30) + (273 30) \u2212 \u00d7 1.1 770 303 = = \u2013 2.7954 kJ/kg \u00b7 K Loss of available energy = 303 (\u2013 2.7954 + 0.9786) Loss of available energy = 550.49 kJ/kg = \u2013 550.49 kJ/kg Ratio of lost available exhaust gas energy to engine work = 550.49 1050 = 0.524 1 Ans. 7. 10 kg of water undergoes transformation from initial saturated vapour at 150\u00baC, velocity of 25 m/s and elevation of 10 m to saturated liquid at 20\u00baC, velocity of 10 m/s and elevation of 3 m. Determine the availability for initial state, final state and change of availability considering environment to be at 0.1 MPa and 25\u00baC and g = 9.8 m/s2. Solution: Let us consider velocities and elevations to be given in reference to environment. Availability is given by A = m \uf8ee \uf8ef\uf8f0 (u \u2013 u0) + P0(v \u2013 v0) \u2013 T0(s \u2013 s0) + 2 C 2 + gz \uf8f9 \uf8fa\uf8fb Dead state of water, u0 = 104.88 kJ/kg v0 = 1.003 \u00d7 10\u20133 m3/kg s0 = 0.3674 kJ/kg \u00b7 K For initial state of saturated vapour at 150\u00baC. u1 = 2559.5 kJ/kg, v1 = 0.3928 m3/kg, s1 = 6.8379 kJ/kg \u00b7 K For final state of saturated liquid at 20\u00baC, u2 = 83.95 kJ/kg, v2 = 0.001002 m3/kg, s2 = 0.2966 kJ/kg \u00b7 K Substituting in the expression for availability Initial state availability, A1 = 10 \u00d7 [(2559.5 \u2013 104.88) + (0.1 \u00d7 103 \u00d7 (0.3928 \u2013 0.001003) \u2013 (298.15 \u00d7 (6.8379 \u2013 0.3674)) + \uf8eb \uf8ec \uf8ec \uf8ed 2 (25) 2 \u2212 3 \u00d7 10 \uf8f6 \uf8f7 \uf8f7 \uf8f8 + (9.81 \u00d7 10 \u00d7 10\u20133)] Final state availability A1 = 5650.28 kJ A2 = 10[(83.95 \u2013 104.88 + (0.1 \u00d7 103 \u00d7 (0", ".001002 \u2013 0.001003)) \u2013 (298.15 \u00d7 (0.2966 \u2013 0.3674)) + \uf8eb \uf8ec \uf8ec \uf8ed 2 (10) 2 \u00d7 \u2212 3 10 \uf8f6 \uf8f7 \uf8f7 \uf8f8 + (9.81 \u00d7 3 \u00d7 10\u20133) \uf8f9 \uf8fa\uf8fb Availability and General Thermodynamic Relations __________________________________ 237 A2 = 2.5835 kJ 2.58 kJ Change in availability, \u2206A = A2 \u2013 A1 = 2.58 \u2013 5650.28 = \u2013 5647.70 kJ Initial availability = 5650.28 kJ Final availability = 2.58 kJ Change in availability = Decrease by 5647.70 kJ Ans. 8. A steam turbine has steam flowing at steady rate of 5 kg/s entering at 5 MPa and 500\u00baC and leaving at 0.2 MPa and 140\u00baC. During flow through turbine a heat loss of 600 kJ/s occurs to the environment at 1 atm and 25\u00baC. Determine (i) the availability of steam at inlet to turbine, (ii) the turbine output (iii) the maximum possible turbine output, and (iv) the irreversibility Neglect the changes in kinetic energy and potential energy during flow. Solution: Let inlet and exit states of turbine be denoted as 1 and 2. At inlet to turbine, p1 = 5 MPa, T1 = 500\u00baC, h1 = 3433.8 kJ/kg, s1 = 6.9759 kJ/kg \u00b7 K At exit from turbine. At dead state, p2 = 0.2 MPa, T2 = 140\u00baC, h2 = 2748 kJ/kg, s2 = 7.228 kJ/kg \u00b7 K Availability of steam at inlet, A1 = m[(h1 \u2013 h0) \u2013 T0 (s1 \u2013 s0)] p0 = 101.3 kPa, T0 = 25\u00baC, h0 = 104.96 kJ/kg, s0 = 0.3673 kJ/kg \u00b7 K A1 = 5 [(3433.8 \u2013 104.96) \u2013 298.15 (6.9759 \u2013 0.3673)] A1 = 6792.43 kJ Availability of steam at inlet = 6792.43 kJ Ans. Applying first law of thermodynamics Q + mh1 = mh2 + W. W = m(h1 \u2013 h2) \u2013 Q = 5(3433.8 \u2013 2748) \u2013 600 W = 2829 kJ/s Turbine output = 2829 kW Ans. Maximum possible turbine output will be available when irreversibility is zero. Wrev = Wmax = A1 \u2013 A2 = m [(h1 \u2013 h2) \u2013 T0(s1 \u2013 s2)] = 5[(3433.8 \u2013 2748) \u2013 298.15 (6.9759 \u2013 7.228)] Wmax = 3053.18 kJ/s Maximum output = 3053.18 kW Ans. 238 _________________________________________________________ Applied Thermodynamics Irreversibility can be estimated by the difference between the maximum output and turbine output. I = Wmax \u2013 W = 224.18 kJ/s Irreversibility = 224.18 kW Ans. 9. Show that the sublimation line and vaporization lines have different slopes at triple point on the phase diagram of water. Solution: It is desired to show that slope of sublimation line shown by 0\u20131 is different than vaporization line 1 \u2013 2. Fusion line e 2 n lin o riz a tio p a V 1 Triple point Sublimation line T Fig 7.10 P 0 To show the slope let us find steam states, By Clapeyron equation. dp dT values at triple point 1. Here i, f, g subscripts refer to ice, water and \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 \u2212 \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 \u2212 1 2 s ig v ig = s fg \u2212 \u2212 0 1 v For triple point state sig = sif + sfg vig = vif + vfg and Substituting in above slope difference expression fg \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 \u2212 \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 \u2212 if v if + + s v fg fg \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u2212 s v fg fg = s if \u00b7 v fg + = s", " ( fg v if \u2212 s v fg fg \u2212 s ) \u00b7 fg v fg if v fg \u00b7 v + fg v \u00b7 v if v \u00b7 s if ( v if fg + v \u2212 s fg ) \u00b7 v = fg It is seen that vif <<< vfg but the order of sif being less than sfg is not very small as compared to vif <<< vfg. Neglecting smaller terms by order of magnitude fg \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 \u2212 0 1 \u2212 \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 \u2212 1 2 = sif v fg Availability and General Thermodynamic Relations __________________________________ 239 Here, sif and vfg both are positive quantities so the ratio \uf8eb \uf8ec \uf8ec \uf8ed sif v fg \uf8f6 \uf8f7 \uf8f7 \uf8f8 is also positive and hence difference of slopes between sublimation line and vaporization line is positive. Thus, it shows that slope of sublimation line and vaporization line are different. 10. Obtain the expression for change in internal energy of gas obeying the Vander Waals equation of state. Solution: Van der Waals equation of state can be given as under, \uf8eb +\uf8ec p \uf8ed a 2 v \uf8f6 \uf8f8 (v \u2013 b) = RT \u21d2 p = \uf8f7 RT \u2212 v b \u2212 a 2 v \u21d2 RT v b\u2212 = p + a 2 v Differentiating this equation of state, partially w.r.t. T at constant volume, \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 = v R v b\u2212 General expression for change in internal energy can be given as under, du = Cv dT + \uf8f1 \uf8f2 T \uf8f3 \u2202\uf8eb \uf8f6 p \uf8f7\u2202\uf8ed \uf8ec \uf8f8 T v \u2212 p \uf8fc \uf8fd \uf8fe dv Substituting in the expression for change in internal energy du = Cv \u00b7 dT + \uf8f1 \uf8f2 \uf8fc \uf8fd \uf8fe dv Substituting for \uf8eb RT \uf8ec \u2212\uf8ed v b \uf8f6 \uf8f7 \uf8f8 is expression of du, du = Cv \u00b7 dT + \uf8f1 \uf8f2 \uf8f3 + p a 2 v \u2212 p \uf8fc \uf8fd \uf8fe dv du = Cv \u00b7 dT + a 2 v dv The change in internal energy between states 1 and 2, 2 \u222b 1 du = u2 \u2013 u1 = 2 vC\u222b 1 dT \u2013 dT \uf8f6 \uf8f7 \uf8f8 Ans. 11. 500 kJ of heat is removed from a constant temperature heat reservoir maintained at 835 K. Heat is received by a system at constant temperature of 720 K. Temperature of the surroundings, the lowest available temperature is 280 K. Determine the net loss of available energy as a result of this irreversible [U.P.S.C. 1992] heat transfer. 240 _________________________________________________________ Applied Thermodynamics Solution: Here, T0 = 280 K, i.e surrounding temperature. Availability for heat reservoir = T0 \u00b7 \u2206Sreservoir = 280 \u00d7 500 835 = 167.67 kJ/kg \u00b7 K Availability for system = T0 \u00b7 \u2206Ssystem 500 720 = 280 \u00d7 Net loss of available energy = (167.67 \u2013 194.44) = \u2013 26.77 kJ/kg \u00b7 K = 194.44 kJ/kg \u00b7 K Loss of available energy = 26.77 kJ/kg \u00b7 K Ans. 12. Steam flows through an adiabatic steady flow turbine. The enthalpy at entrance is 4142 kJ/kg and at exit 2585 kJ/kg. The values of flow availability of steam at entrance and exit are 1787 kJ/kg and 140 kJ/kg respectively, dead state temperature T0 is 300 K, determine per kg of steam, the actual work, the maximum possible work for the given change of state of steam and the change in entropy of steam. [U.P.S.C. 1993] Neglect changes in kinetic and potential energy. Solution: Here dead state is given as 300 K and the maximum possible work for given change of state of steam can be estimated by the difference of flow availability as given under: Wmax = 1787 \u2013 140 = 1647 kJ/kg Actual work from turbine, Wactual = 4142 \u2013 2585 Actual work = 1557 kJ/kg Actual work = 1557 kJ/kg Maximum possible work =", " 1647 kJ/kg Ans. 13. What shall be second law efficiency of a heat engine having efficiency of 0.25 and working between reservoirs of 500\u00baC and 20\u00baC? Solution: Reversible engine efficiency, \u03b7 rev = 1 \u2013 min T T max = 1 \u2013 293 773 = 0.6209 Second law efficiency = \u03b7 \u03b7 rev = 0.25 0.6209 = 0.4026 or 40.26% = 40.26% Ans. 14. An adiabatic cylinder of volume 10 m3 is divided into two compartments A and B each of volume 6 m3 and 4 m3 respectively, by a thin sliding partition. Initially the compartment A is filled with air at 6 bar and 600 K, while there is a vacuum in the compartment B. Air expands and fills both the compartments.Calculate the loss in available energy. Assume atmosphere is at 1 bar and 300 K. [U.P.S.C. 1997] Availability and General Thermodynamic Relations __________________________________ 241 Solution: A 6 bar, 600 K m 6 3 B m4 3 Fig 7.11 Here T0 = 300 K, P0 = 1 bar VA = 6 m3, VB = 4 m3 P1 = 6 bar, T1 = 600 K Initially, V1 = VA = 6m3, and finally, V2 = VA + VB = 10m3 Expansion occurs in adiabatic conditions. Temperature after expansion can be obtained by considering adiabatic expansion. \u03b3 \uf8f6 \uf8f7 \uf8f8 T2 = 600 \u2212 (1.4 1) \uf8eb \uf8ec \uf8ed 6 10 \uf8f6 \uf8f7 \uf8f8 = 489.12 K Mass of air, m = PV 1 1 RT 1 = 5 \u00d7 \u00d7 6 10 6 \u00d7 287 600 = 20.91 kg Change in entropy of control system, (S2 \u2013 S1) = mcv ln T 2 T 1 + mR ln V 2 V 1 \u2206SS = (S2 \u2013 S1) = 20.91 \uf8f1 \uf8f2 \uf8f3 0.718 ln \u00d7 \uf8eb \uf8ec \uf8ed 489.12 600 \uf8f6 \uf8f7 \uf8f8 + 0.287 ln \uf8eb \uf8ec \uf8ed 10 6 \uf8f6 \uf8f7 \uf8f8 \uf8fc \uf8fd \uf8fe = \u20132.01 \u00d7 10\u20133 kJ/K Here, there is no change in entropy of environment, \u2206Se = 0 Total entropy change of combined system = \u2206Sc = \u2206Ss + \u2206Se Loss of available energy = Irreversibility = T0 \u00b7 \u2206Sc = \u2013 2.01 \u00d7 10\u20133 kJ/K = 300 \u00d7 (\u20132.01 \u00d7 10\u20133) = \u2013 0.603 kJ Loss of available energy = 0.603 kJ Ans. 15. Prove that ideal gas equation satisfies the cyclic relation. Solution: Ideal gas equation, Pv = RT Let us consider two variable (v, T) to be independent and P as dependent variable. 242 _________________________________________________________ Applied Thermodynamics P = P(v, T) = RT v By cyclic relation Let us find the three partial derivatives separately and then substitute. \u2202\uf8eb P \uf8ec \u2202\uf8ed v \uf8f6 \uf8f7 \uf8f8 T = \u2212 RT, \u2202\uf8eb T \uf8ec \u2202\uf8ed \uf8f6 \uf8f7 \uf8f8 = \u2212 RT Pv = \u2013 1 Hence proved. Substituting \u2212\uf8eb \uf8ec \uf8ed RT 2 v 16. A heat engine is working between 700\u00ba C and 30\u00baC. The temperature of surroundings is 17\u00baC. Engine receives heat at the rate of 2\u00d7104 kJ/min and the measured output of engine is 0.13 MW. Determine the availability, rate of irreversibility and second law efficiency of engine. Solution: Availability or reversible work, Wrev = \u03b7 rev \u00b7 Q1 = \uf8eb \u2212\uf8ec 1 \uf8ed 303 573 \uf8f6 \uf8f7 \uf8f8 \u00d7 2 \u00d7 104 = 1.38 \u00d7 104 kJ/min Rate of irreversibility, I = Wrev \u2013 Wuseful 4 \uf8eb \uf8ec \uf8ec \uf8ed \u00d7 1.38 10 60 = \u2212 3 \u00d7 0.13 10 \uf8f6 \uf8f7 \uf8f7 \uf8f8 = 100 kJ/s Second law efficiency = W useful W rev = 3 \u00d7 0.13 10 4 \u00d7 1.38 10 60 \uf8eb \uf8ec \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f7 \uf8f8 = 0.5652 or 56.52% Availability = 1.38 \u00d7 104 k", "J/min, Rate of irreversibility = 100 kW, Second law efficiency = 56.52% Ans. 17. A rigid tank contains air at 1.5 bar and 60\u00baC. The pressure of air is raised to 2.5 bar by transfer of heat from a constant temperature reservoir at 400\u00baC. The temperature of surroundings is 27\u00baC. Determine [U.P.S.C. 1998] per kg of air, the loss of available energy due to heat transfer. Solution: Loss of available energy = Irreversibility = T0 \u00b7 \u2206Sc T0 = 300 K, Change in entropy of system = \u2206Ss \u2206Sc = \u2206Ss + \u2206Se Availability and General Thermodynamic Relations __________________________________ 243 Change in entropy of environment/surroundings = \u2206Se Here heat addition process causing rise in pressure from 1.5 bar to 2.5 bar occurs isochorically. Let initial and final states be given by subscript 1 and 2, T1 = 333 K, T2 =?, P1 = 1.5 bar, P2 = 2.5 bar T2 = \u00d7 2.5 333 1.5 = 555 K Heat addition to air in tank, Q = m \u00b7 cp \u00b7 \u2206T = 1\u00d7 1.005 \u00d7 (555 \u2013 333) Q = 223.11 kJ/kg \u2206Ss = Q T = 1 223.11 333 = 0.67 kJ/kg \u00b7 K \u2212 Q \u2212 223.11 673 \u2206Se = T reservoir \u2206Sc = 0.67 \u2013 0.332 \u2206Sc = 0.338 kJ/kg \u00b7 K Loss of available energy = 300 \u00d7 (0.338) = = \u2013 0.332 kJ/kg \u00b7 K = 101.4 kJ/kg Loss of available energy = 101.4 kJ/kg Ans. 18. Using the Maxwell relation derive the following T \u00b7 ds equation, T \u00b7 ds = Cp \u00b7 dT \u2013 T\u00b7 \u2202\uf8eb v \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 dp p [U.P.S.C. 1998] Solution: Let s = s(T, p) \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 ds = p \u00b7 dT + \uf8f6\u2202 \uf8eb s \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p T \u00b7 dp or, T\u00b7ds = T\u00b7 Using Maxwell\u2019s relation, \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 \u00b7dT + T\u00b7 p \uf8f6\u2202 \uf8eb s \uf8ec \uf8f7 \u2202\uf8ed \uf8f8 p T \u00b7 dp \uf8f6\u2202 \uf8eb s \uf8f7 \uf8ec \u2202\uf8ed \uf8f8 p T = \u2013 \u2202\uf8eb v \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p and Substitution yields, T\u00b7 \u2202\uf8eb s \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p = Cp \u00b7 T ds C dT T p \u00b7 = \u2212 \u00b7 \u2202\uf8eb v \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 p \u2202 p \u00b7 Hence proved 244 _________________________________________________________ Applied Thermodynamics 19. Determine the enthapy of vaporization of water at 200\u00baC using Clapeyron equation. Compare it with tabulated value. Solution: Clapeyron equation says, hfg = T \u00b7 vfg \u00b7 \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 sat From steam tables Let us approximate, vfg = (vg \u2013 vf)at 200\u00baC = (0.12736 \u2013 0.001157) = 0.126203 m3/kg \uf8eb \uf8ec \uf8ed dp dT \uf8f6 \uf8f7 \uf8f8 sat, 200\u00baC \u2206\uf8eb p \uf8ec \u2206\uf8ed T \uf8f6 \uf8f7 \uf8f8 = = sat, 200\u00baC P sat at 205\u00baC P\u2212 \u2212 (205 195) sat at 195\u00baC \u2212 (1.7230 1.3978) 10 = = 0.03252 MPa/\u00baC Substituting in Clapeyron equation, hfg = (273 + 200) \u00d7 0.126203 \u00d7 0.03252 \u00d7103 = 1941.25 kJ/kg Calculated enthalpy of vaporization = 1941.25 kJ/kg. Enthalpy of vaporization from steam table = 1940.7", " kJ/kg. Ans. 20. Determine hfg of R \u201312 refrigerant at \u2013 10\u00baC using both Clapeyron equation and the ClapeyronClausius equation. Give the deviation in %. Take Psat at \u2013 5\u00baC = 260.96 kPa Psat at \u2013 15\u00baC = 182.60 kPa. vg at \u2013 10\u00baC = 0.07665 m3/kg vf at \u2013 10\u00baC = 0.00070 m3/kg R = 0.06876 kJ/kg \u00b7 K hfg at \u2013 10\u00baC = 156.3 kJ/kg from tables. Solution: By Clapeyron equation hfg = T \u00b7 vfg \uf8f1 \uf8f2 \uf8f3 dP dT \uf8fc \uf8fd \uf8fe sat = T \u00b7 (vg \u2013 vf) \u2206\uf8f1 P \uf8f2 \u2206\uf8f3 T \uf8fc \uf8fd \uf8fe = (\u2013 5 + 273) \u00d7 (0.07665 \u2013 0.0007) \u00d7 \uf8eb \uf8ec \uf8ed = 268 \u00d7 0.07595 \u00d7 hfg = 159.49 kJ/kg \u2212 (260.96 182.60) 10 P sat at 5\u00ba \u2212 P \u2212 \u2212 C sat at 15\u00ba \u2212 \u2212 \u2212 ( 5 ( 15)) C \uf8f6 \uf8f7 \uf8f8 Availability and General Thermodynamic Relations __________________________________ 245 By Clapeyron-Clausius equation, ln \uf8eb \uf8ec \uf8ed \uf8f6 P 2 \uf8f7 \uf8f8 P 1 sat = \uf8eb \uf8ec \uf8ed ln \u2212 P sat at 5\u00ba P \u2212 sat at 15\u00ba C C \uf8f6 \uf8f7 \uf8f8 = fgh R fgh sat \uf8ee \uf8ef \uf8f0 1 1 \u2212 T \u2212 sat at 15\u00ba C T sat at 5\u00ba \u2212 C \uf8f9 \uf8fa \uf8fb \uf8eb \uf8ec \uf8ed ln 260.96 182.60 \uf8f6 \uf8f7 \uf8f8 = fgh 0.06876 hfg = 169.76 kJ/kg \u00d7 \u21d2 % deviation from Clapeyron equation \uf8ee \uf8ef \uf8f0 1 + \u2212 ( 15 273) \u2013 1 \u2212 + ( 5 273) \uf8f9 \uf8fa \uf8fb \uf8eb \uf8ec \uf8ed \u2212 169.76 159.49 159.49 \uf8f6 \uf8f7 \uf8f8 \u00d7 100 = = 6.44% hfg by Clapeyron equation = 159.49 kJ/kg hfg by Clapeyron-Clausius equation = 169.76 kJ/kg % deviation in hfg value by Clapeyron-Clausius equation compared to the value from Clapeyron equation = 6.44% Ans. 21. Determine the volume expansivity and isothermal compressibility of steam at 300 kPa and 300\u00baC. Solution: Volume expansivity = Isothermal compressibility = Let us write v T = v and T v = P v. The differences may be taken for small pressure and temperature P changes. Volume expansivity, 1 v v T = 300 kpa 1 v at 300 kpa, 300\u00baC \uf8ee \uf8ef \uf8f0 \u2212 v v C 250\u00ba 350\u00ba \u2212 (350 250) C \uf8f9 \uf8fa \uf8fb 300 kpa 1 0.8753 0.9534 0.7964 100 300 kpa = = Volume expansivity = 1.7937 \u00d7 10\u20133 K\u20131 Ans. 246 _________________________________________________________ Applied Thermodynamics \u2212 1 = v at 300kpa, 300\u00baC \uf8ee \uf8ef \uf8f0 \u2212 v 250kpa v 350kpa \u2212 (350 250) \uf8f9 \uf8fa \uf8fb \u00b0 300 C = \u20131 0.8753 0.76505 1.09575 100 = 3.778 \u00d7 10\u20133 KPa\u20131 Isothermal compressibility = 3.778 \u00d7 10\u20133 kPa\u20131 Ans. 22. An evacuated tank of 0.5 m3 is filled by atmospheric air at 1 bar till the pressure inside tank becomes equal to atmospheric temperature. Considering filling of tank to occur adiabatically determine final temperature inside the tank and also the irreversibility and change in entropy. Take atmospheric temperature as 25\u00baC. Solution: Filling of the tank is a transient flow (unsteady flow) process. For the transient filling process, considering subscripts \u2018i\u2019 and \u2018f\u2019 for initial and final states, hi = uf cp Ti = cv Tf Tf = Tf = c p c v Ti 1.005 0.718 \u00d7 298.15 Inside final temperature, Tf =", " 417.33 \u00b7 K Ans. Change in entropy \u2206Sgen = (Sf \u2013 Si) + \u2206Ssurr = cp ln T f T i + 0 = 1.005 \u00d7 ln 417.33 298.15 Change in entropy \u2206Sgen = 0.3379 kJ/kg \u00b7 K Ans. Irreversibility, I = T0 \u00b7 \u2206Sgen = 298.15 \u00d7 0.3379 Irreversibility, I = 100. 74 kJ/kg Ans. 23. A closed vessel stores 75 kg of hot water at 400\u00baC. A heat engine transfers heat from the hot water to environment which is maintained at 27\u00baC. This operation of heat engine changes temperature of hot water from 400\u00baC to 27\u00baC over a finite time. Determine the maximum possible work output from engine. Take specific heat of water as 4.18 kJ/kg\u00b7 K. Availability and General Thermodynamic Relations __________________________________ 247 Solution: Here the combined closed system consists of hot water and heat engine. Here there is no thermal reservoir in the system under consideration. For the maximum work output, irreversibility = 0 Therefore, d dt (E \u2013 T0 S) = Wmax or Here Therefore, Wmax = (E \u2013 T0 S)1 \u2013 (E \u2013 T0 S)2 E1 = U1 = m cp T1, E2 = U2 = m cp T2 T1 = 400 + 273 = 673 K, T2 = 27 + 273 = 300 K = T0 Wmax = mcp (T1 \u2013 T2) \u2013 T0(S1 \u2013 S2) \uf8eb \uf8ec \uf8ed = mcp (T1 \u2013 T2) \u2013 T0 \u00b7 m\u00b7cp ln \uf8f6 \uf8f7 \uf8f8 T 1 T 2 \uf8f1 \uf8f2 \uf8f3 = 75 \u00d7 4.18 \u00d7 = 40946.6 kJ (673 300) 300 \u2212 \u2212 \u00d7 ln \uf8eb \uf8ec \uf8ed 673 300 \uf8f6 \uf8f7 \uf8f8 \uf8fc \uf8fd \uf8fe Maximum work = 40946.6 kJ Ans. 24. In a steam turbine the steam enters at 50 bar, 600\u00baC and 150 m/s and leaves as saturated vapour at 0.1 bar, 50 m/s. During expansion, work of 1000 kJ/kg is delivered. Determine the inlet stream availability, exit stream availability and the irreversibility. Take dead state temperature as 25\u00baC. Solution: h1 = hat 50 bar, 600\u00baC = 3666.5 kJ/kg, s1 = sat 50 bar, 600\u00baC = 7.2589 kJ/kg \u00b7 K h2 = hg at 0.1 bar = 2584.7 kJ/kg, s2 = sg at 0.1 bar = 8.1502 kJ/kg \u00b7 K Inlet stream availability = = h 1 2 c 1 2 \u2013 T0s1 3666.5 2 (150) 10 3 2 \u2013 (298 \u00d7 7.2589) Input stream availability is equal to the input absolute availability. = 1514.59 kJ/kg Exit stream availability = = \uf8eb \uf8ec \uf8ec \uf8ed h 2 2 c 2 2 \u2013 T0 s2 2584.7 + (50) 2 \u00d7 10 \u2212 3 2 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u2013 (298 \u00d7 8.1502) = 157.19 kJ/kg 248 _________________________________________________________ Applied Thermodynamics Exit stream availability is equal to the exit absolute availability. Wrev = 1514.59 \u2013 157.19 = 1357.4 kJ/kg Irreversibility = Wrev \u2013 W = 1357.4 \u2013 1000 = 357.4 kJ/kg This irreversibility is in fact the availability loss. Inlet stream availability = 1514.59 kJ/kg Exit stream availability = 157.19 kJ/kg Irreversibility = 357.4 kJ/kg Ans. 7.1 Define \u2018available energy\u2019 and \u2018unavailable energy\u2019. 7.2 What do you understand by second law efficiency? How does it differ from first law efficiency? 7.3 What is meant by a dead state? Discuss its\u2019 importance. 7.4 Define availability. Obtain an expression for availability of closed system. 7.5 Differentiate between useful work and maximum useful work in reference to the availability. 7.6 What do you understand by Gibbs function? How does it differ from the availability function? 7.7 Describe the Helmholtz function. 7.8 What are Maxwell relations? Discuss their significance? 7.9 Describe Clapeyron equation. 7.10 What do you understand by Joule-Thom", "son coefficient? Explain. 7.11 Describe chemical potential. 7.12 Write short notes on the following: (i) Clapeyron-Clausius equation, (ii) Volume expansivity (iii) Fugacity, (iv) Second law analysis of engineering systems. 0.145 m3 initial volume to 0.58 m3 final volume. constant temperature while the temperature of surroundings is 22\u00baC. 7.13 Determine the loss of availability when 1 kg air at 260\u00baC is reversibly and isothermally expanded from [70.56 kJ/kg] 7.14 Determine the entropy generation and decrease in available energy when a heat source of 727\u00baC transfers heat to a body at 127\u00baC at the rate of 8.35 MJ/min. Consider the temperature of sink as 27\u00baC. [12.54 kJ/K \u00b7 min, 3762 kJ] 7.15 Determine the available energy of furnace having the gases getting cooled from 987\u00baC to 207\u00baC at [\u2013518.1 kJ/kg] 7.16 Determine the available amount of energy removed and the entropy increase of universe when 5 kg air at 1.38 bar, 500 K is cooled down to 300 K isobarically. The temperature of surroundings may be taken [\u2013268.7 kJ. 3.316 kJ/K] as 4\u00baC. 7.17 Determine the entropy change, unavailable energy and available energy for the process in which 2 kg air is heated isobarically so as to cause increase in its temperature from 21\u00baC to 315\u00baC. Take T0 = 10\u00baC. [1.393 kJ/K, 394.2 kJ, 196.6 kJ] 7.18 Steam enters in a steam turbine at 60 bar, 500\u00baC and leaves at 0.1 bar, 0.89 dry with a flow rate of 3.2652 [1286.2 kJ/s] 7.19 Determine the available portion of heat removed from 2.5 kg air being cooled from 2.1 bar, 205\u00baC to 5\u00baC [\u2013 97.2 kJ] at constant volume. The heat is rejected to surroundings at \u2013 4\u00baC. \u00d7 104 kg/hr. Determine the loss of available energy. Availability and General Thermodynamic Relations __________________________________ 249 7.20 Prove that heat is an inexact differential. {Q (T, s)}. 7.21 Derive an expression for change in entropy of a gas obeying Vander Waals equation of state. 7.22 Determine the coefficient of thermal expansion and coefficient of isothermal compressibility for a gas obeying Vander Waals equation of state. engine has efficiency of 0.40. 7.23 Determine the second law efficiency of a heat engine operating between 700\u00baC and 30\u00baC. The heat [55.74%] 7.24 Determine the amount of heat that can be converted to the useful work if total heat at 1000 kJ is [624.84 kJ] 7.25 Determine the change in availability of air contained in an insulated vessel of 20 \u00d7 103 cm3. The initial state of air is 1 bar. 40\u00baC. The air is heated so as to arrive at temperature of 150\u00baC. The temperature of [0.027 kJ] surrounding environment may be considered as 20\u00baC. available at 500\u00baC. The temperature of environment is 17\u00baC. 7.26 Determine the enthalpy of vaporization of water at 50\u00baC using the Clapeyron equation. equation as compared to value in steam table. [2396.44 kJ/kg] 7.27 Determine the % variation in the enthalpy of vaporization of steam at 500 kPa using Clapeyron [0.201%] 7.28 Air enters a compressor at 40\u00baC, 500 kPa for being compressed upto 2000 kPa. Consider the compression to be at constant temperature without internal irreversibilities. Air flows into compressor at 6 kg/min. Neglecting the changes in kinetic energy and potential energy determine the availability transfers accompanying heat and work and irreversibility. Take T0 = 25\u00baC, P0 = 1 bar. The control volume may be taken as under, (i) Control volume comprises of compressor alone (ii) Control volume comprises of compressor and its immediate surroundings so that heat transfer occurs at T0. [(i) \u2013 0.597 kJ/s, \u201312.46 kJ/s, 0.597 kJ/s (ii) 0 kJ/s, \u201312.46 kJ/s, 0.597 kJ/s] 7.29 Steam expands in a cylinder from 22 bar, 450\u00baC to 4.5 bar, 250\u00baC. The environment may be considered to be at 1 bar, 25\u00baC. Determine (", "i) the availability of steam at inlet, exit and change in availability. (ii) the irreversibilities in the expansion process. [196 kJ, 132 kJ, \u201364 kJ, 5.5 kJ] 7.30 In a steam power cycle steam enters at 60 bar, 500\u00baC into turbine and leaves at 0.04 bar. The isentropic efficiency of turbine is 85% and that of pump is 70%. Considering the environment to have T0 = 25\u00baC, P0 =1 bar, determine the second law efficiency of constituent components i.e. boiler, turbine, condenser [67.6%, 99%, 84.4%, 65%] and pump. 250 ________________________________________________________ Applied Thermodynamics 8 8.1 INTRODUCTION Thermodynamic cycles can be primarily classified based on their utility such as for power generation, refrigeration etc. Based on this thermodynamic cycles can be categorized as; (i) Power cycles, (ii) Refrigeration and heat pump cycles. (i) Power cycles: Thermodynamic cycles which are used in devices producing power are called power cycles. Power production can be had by using working fluid either in vapour form or in gaseous form. When vapour is the working fluid then they are called vapour power cycles, whereas in case of working fluid being gas these are called gas power cycles. Thus, power cycles shall be of two types, (a) Vapour power cycle, (b) Gas power cycle. Vapour power cycles can be further classified as, 1. Carnot vapour power cycle 2. Rankine cycle 3. Reheat cycle 4. Regenerative cycle. Gas power cycles can be classified as, 1. Carnot gas power cycle 2. Otto cycle 3. Diesel cycle 4. Dual cycle 5. Stirling cycle 6. Ericsson cycle 7. Brayton cycle Here in the present text Carnot, Rankine, reheat and regenerative cycles are discussed. (ii) Refrigeration and heat pump cycles: Thermodynamic cycles used for refrigeration and heat pump are under this category. Similar to power cycles, here also these cycles can be classified as \u201cair cycles\u201d and \u201cvapour cycles\u201d based on type of working fluid used. 8.2 PERFORMANCE PARAMETERS Some of commonly used performance parameters in cycle analysis are described here. Thermal efficiency: Thermal efficiency is the parameter which gauges the extent to which the energy input to the device is converted to net work output from it. Vapour Power Cycles ___________________________________________________________ 251 Thermal efficiency = Net work in cycle Heat added in cycle Heat rate: Heat rate refers to the amount of energy added by heat transfer to cycle to produce unit net work output. Usually energy added may be in kcal, unit of net work output in kW h and unit of heat rate may be in kcal/kW h. It is inverse of thermal efficiency. Back work ratio: Back work ratio is defined as the ratio of pump work input (\u2013ve work) to the work produced (+ve work) by turbine. Back work ratio = W pump W turbine Generally, back work ratio is less than one and as a designer one may be interested in developing a cycle which has smallest possible back-work ratio. Small back-work ratio indicates smaller pump work (\u2013ve work) and larger turbine work (+ve work). Work ratio: It refers to the ratio of net work to the positive work. Mathematically, work ratio = W net W turbine Specific steam consumption: It indicates the steam requirement per unit power output. It is generally given in kg/kW. h and has numerical value lying from 3 to 5 kg/kW. h Specific steam consumption = 3600 W net, kg/kW.h 8.3 CARNOT VAPOUR POWER CYCLE Carnot cycle has already been defined earlier as an ideal cycle having highest thermodynamic efficiency. Let us use Carnot cycle for getting positive work with steam as working fluid. Arrangement proposed for using Carnot vapour power cycle is as follows. 1 \u2013 2 = Reversible isothermal heat addition in the boiler 2 \u2013 3 = Reversible adiabatic expansion in steam turbine 3 \u2013 4 = Reversible isothermal heat rejection in the condenser 4 \u2013 1 = Reversible adiabatic compression or pumping in feed water pump T 2 3 1 4 s Boiler B 2 (Steam) 1 Steam turbine ST 3 Condenser Feed pump Water Water 4 Fig 8.1 Carnot vapour power cycle Fig 8.2 Arrangement for Carnot cycle Assuming steady flow processes in the cycle and neglecting changes in kinetic and potential energies, thermodynamic analysis may be carried out. 252 ________________________________________________________ Applied Thermodynamics Thermal efficiency = Net work Heat added Net work = Turbine work \u2013 Compression/Pumping work For unit mass flow. W = (h2 \u2013 h3) \u2013 (h1 \u2013 h4)", " Heat added, Qadd = (h2 \u2013 h1) ) h 3 ( h 2 \u03b7 Carnot = Here heat rejected, Qrejected = (h3 \u2013 h4 or \u03b7 Carnot = 1 \u2013 Q rejected Q add Also, heat added and rejected may be given as function of temperature and entropy as follows: Also, s1 = s4 and s2 = s3 Qadd = T1 \u00d7 (s2 \u2013 s1) Qrejected = T3 \u00d7 (s3 \u2013 s4) Therefore, substituting values: \u03b7 Carnot = 1 \u2013 T 3 T 1 or \u03b7 Carnot = 1 \u2013 minimum T T maximum Let us critically evaluate the processes in Carnot cycle and see why it is not practically possible. 1\u20132: Reversible Isothermal Heat Addition Isothermal heat addition can be easily realised in boiler within wet region as isothermal and isobaric lines coincide in wet region. But the superheating of steam can\u2019t be undertaken in case of Carnot cycle as beyond saturated steam point isothermal heat addition can\u2019t be had inside boiler. This fact may also be understood from T\u2013S diagram as beyond 2 the constant pressure line and constant temperature lines start diverging. It may be noted that boiler is a device which generates steam at constant pressure. 2\u20133: Reversible Adiabatic Expansion Saturated steam generated in boiler at state \u20182\u2019 is sent for adiabatic expansion in steam turbine upto state 3. During this expansion process positive work is produced by steam turbine and a portion of this work available is used for driving the pump. 3\u20134: Reversible Isothermal Heat Rejection Heat release process is carried out from state 3 to 4 in the condenser. Condenser is a device in which constant pressure heat rejection can be realized. Since expanded steam from steam turbine is available in wet region at state 3. Therefore, constant temperature heat rejection can be had as constant temperature and constant pressure lines coincide in wet region. Heat rejection process is to be limited at state 4 which should be vertically below state 1. Practically it is very difficult to have such kind of control. Vapour Power Cycles ___________________________________________________________ 253 4\u20131: Reversible Adiabatic Compression (Pumping) Carnot cycle has reversible adiabatic compression process occurring between 4 and 1, which could be considered for pumping of water into boiler. In fact it is very difficult for a pump to handle wet mixture which undergoes simultaneous change in its phase as its pressure increases. Above discussion indicates that Carnot vapour power cycle is merely theoretical cycle and cannot be used for a practical working arrangement. Also the maximum efficiency of Carnot cycle is limited by maximum and minimum temperatures in the cycle. Highest temperature attainable depends upon metallurgical limits of boiler material. 8.4 RANKINE CYCLE Rankine cycle is a thermodynamic cycle derived from Carnot vapour power cycle for overcoming its limitations. In earlier discussion it has been explained that Carnot cycle cannot be used in practice due to certain limitations. Rankine cycle has the following thermodynamic processes. 1 \u2013 2 = Isobaric heat addition (in boiler) 2 \u2013 3 = Adiabatic expansion (in turbine) 3 \u2013 4 = Isobaric heat release (in condenser) 4 \u2013 1 = Adiabatic pumping (in pump) T \u2013 S, h \u2013 S and P \u2013 V representations are as shown below. 1 4 T p1 p2 Fig. 8.3 T\u2013s, h\u2013S and P\u2013V representations of Rankine cycle Practical arrangement in a simple steam power plant working on Rankine cycle is shown ahead. Thus in Rankine cycle, isothermal heat addition and heat rejection processes have been replaced by isobaric processes. Realization of \u2018isobaric heat addition\u2019 and \u2018heat rejection\u2019 in \u2018boiler\u2019 and \u2018condenser\u2019 respectively is in conformity with nature of operation of these devices. Isobaric heat addition can be had in boiler from subcooled liquid to superheated steam without any limitations. Boiler B 2 (Steam) 1 Steam turbine ST 3 Condenser Feed pump Water Water 4 Fig. 8.4 Simple steam power plant layout 254 ________________________________________________________ Applied Thermodynamics Let us understand the arrangement. 1 \u2013 2: High pressure water supplied by feed pump is heated and transformed into steam with or without superheat as per requirement. This high pressure and temperature steam is sent for expansion in steam turbine. Heat added in boiler, for unit mass of steam. Qadd = (h2 \u2013 h1) 2 \u2013 3: Steam available from boiler is sent to steam turbine, where it's adiabatic expansion takes place and positive work is available. Expanded steam is generally found to lie in wet region. Expansion of steam is carried out to the extent of wet steam having dryness fraction above 85% so as to", " avoid condensation of steam on turbine blades and subsequently the droplet formation which may hit hard on blade with large force. Turbine work, for unit mass, Wturbine = (h2 \u2013 h3). 3 \u2013 4: Heat rejection process occurs in condenser at constant pressure causing expanded steam to get condensed into saturated liquid at state 4. Heat rejected in condenser for unit mass, Qrejected = (h3 \u2013 h4) 4 \u2013 1: Condensate available as saturated liquid at state 4 is sent to feed pump for being pumped back to boiler at state 1. For unit mass, Pump work Wpump = h1 \u2013 h4. Here pumping process is assumed to be adiabatic for the sake of analysis whereas it is not exactly adiabatic in the pump. From first and second law combined together; dh = T \u00b7 ds + v \u00b7 dp. Here in this adiabatic pumping process. ds = 0 dh = v \u00b7 dp. (h1 \u2013 h4) = v4 (p1 \u2013 p4) (h1 \u2013 h4) = v4 (p1 \u2013 p3). {as p3 = p4} Therefore or or Wpump = v4(p1 \u2013 p3) Rankine cycle efficiency can be mathematically given by the ratio of net work to heat added. \u03b7 Rankine = \u2212 W pump W turbine Q add h 3 ( h 2 In the above expression, the enthalpy values may be substituted from steam table, mollier charts and \u03b7 Rankine = ( h 1 ) h 1 h 2 h 4 ( ) ) \u2212 \u2212 \u2212 \u2212 by analysis for getting efficiency value. Rankine cycle efficiency may be improved in the following ways: (a) By reducing heat addition in boiler, which could be realized by preheating water entering into it. (b) By increasing steam turbine expansion work, i.e. by increasing expansion ratio within limiting dryness fraction considerations. (c) By reducing feed pump work. (d) By using heat rejected in condenser for feed water heating. etc. Irreversibilities and losses in Rankine cycle: In actual Rankine cycle there exist various irreversibilities and losses in its\u2019 constituent components and processes in them. In Rankine cycle the major irreversibility is encountered during the expansion through turbine. Irreversibilities in turbine significantly reduce the Vapour Power Cycles ___________________________________________________________ 255 expansion work. Heat loss from turbine to surroundings, friction inside turbine and leakage losses contribute to irreversibilities. Due to this irreversible expansion there occurs an increase in entropy as compared to no entropy change during reversible adiabatic expansion process. This deviation of expansion from ideal to actual process can be accounted for by isentropic turbine efficiency. Ideal expansion in steam turbine is shown by 2\u20133 on T\u2013S representation. Actual expansion process is shown by 2\u20133'. 11' 2 T 4 33' S Fig. 8.5 Rankine cycle showing non-ideal expansion and pumping process Isentropic turbine efficiency, \u03b7 isen, t = or, \u03b7 isen, t = W t,actual W t,ideal = \uf8eb \uf8ec \uf8ed h 2 h 2 \u2212 \u2212 h 3' \u2212 2 3 ; Actually, W2\u20133\u2032 < W2\u20133 Another important location for irreversibilities is the pump. During pumping some additional work is required to overcome frictional effects. Ideally pumping is assumed to take place with no heat transfer during pumping whereas actually it may not be so. Thus the pumping process as shown by ideal process 4\u20131 gets\u2019 modified to 4\u20131' which is accompanied by increase in entropy across the pump. Isentropic efficiency of pump is a parameter to account for non-idealities of pump operation. Isentropic efficiency of pump is defined by; or, \u03b7 isen, p = W p,ideal W p,actual = \u03b7 isen' ; Actually, W4\u20131 < W4\u20131\u2032 Thus, it indicates that actually pump work required shall be more than ideal pump work requirement. Apart from the turbine and pump irreversibilities explained above there may be other sources of inefficiency too. These turbine and pump irreversibilities accounted for by isentropic efficiency of turbine and pump are called external irreversibility. Sources of internal irreversibilities are heat transfer from system to surroundings, frictional pressure loss in rest of components etc. There also occurs the steam pressure drop due to friction between pipe surface and working fluid. 8.5 DESIRED THERMODYNAMIC PROPERTIES OF WORKING FLUID Working fluid being used in vapour power cycles must have following desirable properties. Generally water is used as working fluid in vapour power cycles as it is easily available in abundance and satisfies most of requirements. Other working fluids may be mercury, sulphur dioxide", " and hydrocarbons etc. (i) Working fluid should be cheap and easily available. 256 ________________________________________________________ Applied Thermodynamics (ii) Working fluid should be non-toxic, non-corrosive and chemically stable. (iii) Fluid must have higher saturation temperature at moderate pressures as it shall yield high efficiency because most of heat will be added at high temperature. Thus, mean temperature of heat addition shall be high even at moderate pressure. (iv) Working fluid should have smaller specific heat so that sensible heat supplied is negligible and Rankine cycle approaches to Carnot cycle. In case of fluid having small specific heat hatched portion shown in Fig. 8.6 will be absent or minimum. (v) Saturated vapour line should be steep enough so that state after expansion has high dryness fraction. 1'234' = Carnot cycle 1234 = Rankine cycle 1 1' 2 4 4' 3 T S Fig. 8.6 Carnot cycle and Rankine cycle emphasizing for small specific heat of fluid (vi) Working fluid density should be high so that the size of plant becomes smaller. (vii) Working fluid should have its' critical temperature within metallurgical limits. (viii) It should show significant decrease in volume upon condensation. (ix) Working fluid should have its' freezing point much below atmospheric pressure so that there is no chance of freezing in condenser. 8.6 PARAMETRIC ANALYSIS FOR PERFORMANCE IMPROVEMENT IN RANKINE CYCLE Let us carry out study of the influence of thermodynamic variables upon Rankine cycle performance. (i) Pressure of steam at inlet to turbine: Steam pressure at inlet to turbine may be varied for same temperature of steam at inlet. Two different pressures of steam at inlet to turbine, also called throttle pressure are shown in Fig. 8.7. Comparative study shows that for back pressure and steam inlet temperatures being same the increase in steam inlet pressure from p'1 to p1 is accompanied by the reduction in net heat added as shown by hatched area A2'3'37 and increase in net heat added by the amount shown by area A1'1271'. Generally two areas A2'3'37 and A1'1271' are nearly same which means that the increment in net heat added due to increasing throttle pressure from p1' to p1 is accompanied by decrease in net heat addition and the net heat added remains same as at lower throttle pressure p1'. But increase in throttle pressure to p1 also causes reduction in the heat rejected. At pressure p1' heat rejected is given by area A43'6'54 while at pressure p1 heat rejected is given by area A43654 3 3' 6 6' P1' P3 P1 2' T 1' 7 1 2 4 5 s Fig. 8.7 Rankine cycle showing two different throttle pressures A43654 < A43'6'54 (Heat rejected)cycle 1234 < (Heat rejected)cycle 1'2'3'4 Cycle efficiency is given by \u03b7 cycle = 1\u2013 Heat rejected Heat added Vapour Power Cycles ___________________________________________________________ 257 Hence, it is obvious that increasing steam pressure at inlet to steam turbine is accompanied by increase in cycle thermal efficiency. But this increase in pressure increases wetness of steam as shown by states 3 and 3\u2032 i.e. x3 < x3, where x is dryness fraction. This increase in wetness of expanding steam decreases the adiabatic efficiency of turbine and also increases the chances of erosion of steam turbine blades. Therefore, as there are two contrary consequences of increasing throttle pressure so a compromise is to be had. Normally, to avoid erosion of turbine blades the minimum dryness fraction at turbine exhaust should not go below 0.88. (ii) Temperature of steam at inlet to turbine: Increasing temperature of steam at inlet to turbine may also be called as superheating of steam at inlet to turbine. The two Rankine cycles having different degree of superheating are shown in Fig. 8.8. For two different temperatures of steam at inlet to turbine i.e. T2 and T2' while T2 < T2' the comparison of two cycles 12341 and 12'3'41 shows the effect of increasing temperature at inlet to turbine. Rankine cycle represented on T\u2013S diagram show that increasing temperature from T2 to T2' causes increase in net work by amount as shown by area A22'3'32. This increase in steam temperature is also accompanied by increased heat addition as shown by area A22'6'62. It is seen that this ratio of increase in net work to increase in heat addition is more than similar ratio for rest of cycle so the net effect is to improve the cycle thermal efficiency. It may also be stated that this increase in steam temperature from T 2' T 2 T 1 4 5 P1 2' 2 P", "3 3 3' 6 6' s Fig. 8.8 Effect of increasing temperature at inlet to turbine T2 to T2' i.e. increase in degree of superheat increases mean temperature of heat addition, which increases thermal efficiency. With increased steam temperature the state of steam after expansion becomes more dry i.e. increase in temperature from 2 to 2' makes steam more dry after expansion i.e. from 3 to 3' having dryness fraction x3' > x3. This hotter steam supply to turbine is also advantageous from the specific work point of view. The work done per unit mass gets increased by superheating steam at inlet to turbine. Therefore, one is always interested in realizing highest possible temperature of steam, provided it is within metallurgical temperature limits. Highest practical steam temperature at turbine inlet presently is 650\u00baC. (iii) Pressure at the end of expansion: Let us see the influence of pressure at the end of expansion from steam turbine. This pressure may also be called exhaust pressure or back pressure or condenser pressure. Rankine cycle with two different exhaust pressures is shown in Fig. 8.9, while the maximum pressure and temperature remains same. 258 ________________________________________________________ Applied Thermodynamics With the lowering of back pressure from p3 to p3' Rankine cycles get modified from 12341 to 1'23'4'1'. This reduction in back pressure causes increment in net work as shown by area A1'1433'4'51' and also the heat addition increases by the amount shown by area A1'166'1'. It is seen that the two areas are such that the thermal efficiency of cycle increases by lowering back pressure as increase in heat addition is more than increase in heat rejection. This lowering back pressure is accompanied by increase in wetness of steam from 3 to 3' i.e. dryness fraction x3 > x3'. Practically there exists limitation of dryness fraction after expansion to avoid erosion of turbine blades so lowering exhaust pressure is limited by it, inspite of showing improvement in thermal efficiency. 1 1' T 5 4 4' 66' P1 P3 P3' 2 3 3' S Fig. 8.9 Effect of varying exhaust pressure (iv) Temperature of feed water at inlet to boiler: Temperature of feed water at inlet to boiler may be increased by employing some means such as feed water heating. This increase in feed water temperature reduces the heat requirement in boiler for getting desired state at inlet to steam turbine. Thus, with the reduced heat addition the thermal efficiency gets increased. Different approaches practically used for improving the Rankine cycle performance have resulted into modified forms of Rankine cycle also called as Reheat cycle, Regenerative cycle etc. 8.7 REHEAT CYCLE Schematic of reheat cycle is as shown in Fig. 8.10. Reheat cycle is based on the simple fact of realizing high efficiency with higher boiler pressure and yet avoid low quality of steam at turbine exhaust. Here steam generated in boiler is supplied to high pressure steam turbine at state 2 and is expanded upto state 3. This steam is sent to boiler for being reheated so that its temperature gets increased, normally this temperature after reheating may be equal to temperature at inlet of high pressure steam turbine. Steam after reheating is supplied to subsequent turbine at state 4, say to low pressure steam turbine. Steam is now expanded upto the exhaust pressure say state \u20185\u2019. Expanded steam is subsequently sent to condenser and condensate at state \u20186\u2019 is pumped back to the boiler employing feed pump at state \u20181\u2019. Thus, it is possible to take advantage of high steam pressure at inlet to steam turbine as the problem of steam becoming excessively wet with increasing steam pressure could be regulated by reheating during the expansion. Expansion occurs in two stages one begining at high pressure and other occurring at low pressure with reheating in between. The principal advantage of reheat is to increase the quality of steam at turbine exhaust. Vapour Power Cycles ___________________________________________________________ 259 3 2 4 Boiler HPST LPST 1 HPST : High pressure steam turbine LPST : Low pressure steam turbine 5 Condenser Feed pump Fig 8.10 Reheat cycle 2 3 T 1 6 6 4 5 Fig 8.11 T\u2013S representation for reheat cycle S Secondary advantage of reheating is marginal improvement in thermal efficiency when steam pressure is above 100 bar. At low steam pressure reheating does not show gain in cycle thermal efficiency and even the efficiency may be less than that of Rankine cycle due to mean temperature of heat addition being lower. Generally, with modern high pressure boilers and supercritical boilers reheating is essentially employed. Reheating is disadvantageous from economy of plant perspective as the cost of plant increases due to arrangement for reheating and increased condensation requirements due to increased dryness fraction of steam after expansion. Thermodynamic analysis of reheat cycle as shown on T\u2013S diagram may", " be carried out for estimation of different parameters as below, Total turbine work output = WHPST + WLPST Net work, Wnet = (Total turbine work output) \u2013 (Pump work) Wnet = WHPST + WLPST \u2013 Wp where different works for ms mass of steam are, HP steam turbine, WHPST = ms \u00b7 (h2 \u2013h3) 260 ________________________________________________________ Applied Thermodynamics LP steam turbine, WLPST = ms \u00b7 (h4 \u2013h5) Feed Pump, Wp = (h1 \u2013h6) \u00b7 ms Wnet = {(h2 \u2013 h3) + (h4 \u2013 h5) \u2013 (h1 \u2013 h6)} \u00b7 ms Heat supplied for ms mass of steam; Qadd = (h2 \u2013 h1) \u00b7 ms + ms \u00b7 (h4 \u2013 h3) W Q Cycle thermal efficiency, \u03b7 Reheat = net add \u03b7 Reheat {( h 2 = \u2212 ) h 3 {( )} h 3 h 6 )} Specific work output, W reheat = {( )} Generally not more than two stages of reheat are practically employed. Theoretically, the improvement in efficiency due to second reheat is about half of that which results from single reheat. Also more number of reheat stages shall result into superheated steam at turbine exhaust. Thus, mean temperature of heat rejection gets raised and efficiency drops. 8.8 REGENERATIVE CYCLE Regenerative cycle is a modified form of Rankine cycle in which it is devised to increase mean temperature of heat addition so that cycle gets close to Carnot cycle in which all heat addition occurs at highest possible temperature. In regenerative cycle the feed water is heated up so as to reduce the heat addition in boiler and heat addition occur at hotter feed water temperature. Theoretically regenerative cycle arrangement is as shown in Fig. 8.12. B 1 B : Boiler 2 ST 3 ST : Steam Turbine Condenser 4 p1 5 2 Feed pump p3 4' 3 3 Fig 8.12 Schematic for theoretical regenerative cycle and T-s representation. Vapour Power Cycles ___________________________________________________________ 261 Theoretical arrangement shows that the steam enters the turbine at state 2 (temperature T2) and expands to (temperature T3) state 3. Condensate at state 5 enters the turbine casing which has annular space around turbine. Feed water enters turbine casing at state 5 and gets infinitesimally heated upto state 1 while flowing opposite to that of expanding steam. This hot feed water enters into boiler where steam generation occurs at desired state, say 2. Feed water heating in steam turbine casing is assumed to occur reversibly as the heating of feed water occurs by expanding steam with infinitesimal temperature difference and is called \u201cregenerative heating\u201d. This cycle is called regenerative cycle due to regenerative heating employed in it. Regenerative heating refers to the arrangement in which working fluid at one state is used for heating itself and no external heat source is used for this purpose. Here feed water picks up heat from steam expanding in steam turbine, thus the expansion process in steam turbine shall get modified from 2-3' ideally to 2-3. Heat picked up by feed water for getting heated up from state 5 to 1 is shown by hatched area A17651 on T-S diagram. Under ideal conditions for cent per cent heat exchange effectiveness the two areas i.e. A29832 indicating heat extraction from steam turbine and A17651 indicating heat recovered by feed water shall be same. Thus, T-S representation of regenerative cycle indicates that the cycle efficiency shall be more than that of Rankine cycle due to higher average temperature of heat addition. But there exists serious limitation regarding realization of the arrangement described above. Limitations are due to impossibility of having a steam turbine which shall work as both expander for getting work output and heat exchanger for feed water heating. Also with the heat extraction from steam turbine the state of expanded steam at exhaust pressure shall be extremely wet, hence not desired. Due to these limitations the regenerative cycle is realized employing the concept of bleeding out steam from turbine and using it for feed water heating in feed water heaters. Feed Water Heaters: The feed water heater refers to the device in which heat exchange occurs between two fluids i.e. steam and feed water either in direct contact or indirect contact. Direct contact feed water heater is the one in which bled steam and feed water come in direct contact. These are also called open feed water heater. Bled steam b Mixture of feed water & Bled steam c Open feed water heater or, direct contact feed water heater a Feed water d Bled steam b c a Feed water Closed feed water heater or, Indirect contact feed water heater or, surface type feed water heater Fig. 8.13 Feed water heaters In open feed water heater two fluids i.e., bled steam and feed", " water are at same pressure and adiabatic mixing is assumed to take place. Normally, it is considered that the mixture leaves open feed water heater as saturated liquid. Energy balance upon it shall be as follows, ma \u00b7 ha + mb \u00b7 hb = (ma + mb) \u00b7 hc where subscripts a,b and c are for feed water, bled steam and mixture of the two as shown in Figure 8.13. Indirect contact feed water heater as shown in Figure 8.13 is the one in which two fluid streams i.e. bled steam and feed water do not come in direct contact, but the heat exchange between two streams 262 ________________________________________________________ Applied Thermodynamics occurs indirectly through metal interface. These are also called closed feed water heaters. In these feed water heaters since two fluids do not contact each other so they may be at different pressures. In these the arrangement comprises of steel, copper or brass tubes of solid drawn type placed in a shell. The heat transfer takes place through tube surface. Feed water flows inside the tube and is heated by extracted steam from outside. Steam enters in the shell and comes in contact with the tubes and then condenses over tubes. Steam condenses and trickles down and is collected in shell. Figure 8.14 gives schematic of surface type feed water heater. Performance of feed water heater is quantified using a parameter called \u201cterminal temperature difference\u201d. Terminal temperature difference (T.T.D.) refers to the difference of temperature between temperature of feed water outlet and saturation temperature of steam entering the heater. Terminal temperature difference = (Feed water outlet temperature \u2013 Saturation temperature of steam entering heater) T.T.D. has its value lying around 5\u20138\u00baC. T.T.D. shall be zero in desuperheater type heaters where superheated steam is used for feed water heating upto saturation temperature of steam. In feed water heaters where steam pressure is quite high, the condensate from heater is expanded in an expander called \u2018flash tank\u2019 or \u2018drain expander\u2019 or \u2018drain cooler\u2019. In this flash tank some portion of condensate gets converted into steam which is further used for heating feed water. Feed water outlet Feed water inlet A Steam inlet A A A Air vent Inside of feed water heater Condensate out Fig. 8.14 Surface type feed heater Direct contact heaters or open type heaters are more efficient than indirect contact type due to direct contact between two fluids. Feed water can achieve saturation temperature corresponding to the pressure of heating steam. In this case the terminal temperature difference is zero. During heating the non-condensable gases dissolved in water get released and are thrown out through vent passage. Deaerator is a type of open feed water heater. Schematic is shown in subsequent article on deaerator. Deaerator: Deaerator is a type of open type feed water heater employed for the removal of dissolved oxygen and carbon dioxide from the feed water. The dissolved oxygen when not removed gets disintegrated into nascent oxygen at high temperature and pressure and forms iron oxide upon coming in contact with metal. This iron oxide formed causes pitting on the metal surface. At high temperature and pressure, dissolved CO2 combines with metal and forms carbolic acid which causes mild pitting on Vapour Power Cycles ___________________________________________________________ 263 metal surfaces. Therefore, it becomes necessary to remove dissolved gases from the feed water. Although feed water treatment plant is there but still the impurities may creep in along with the make up water added to compensate for loss of water from system due to leakages at valve, pipe flanges, steam valve spindles and boiler blow down etc. Normally added water is 3-5% of total boiler feed. Deaeration of water is based on the principle of decreasing partial pressure of gas for removal of dissolved gases. Henery's law and Dalton's law of partial pressure may be considered in this reference for understanding the phenomenon. Henery's law states that, \u201cthe mass of gas dissolved in a definite mass of liquid at a given temperature is directly proportional to partial pressure of gas in contact with liquid\u201d. It is good for the gases having no chemical reaction with water. Decrease in partial pressure of gas in water is achieved by increasing the vapour pressure by heating the water. Here feed water is heated by low pressure steam for heating it upto its saturation temperature. Feed water entering deaerator is broken into small particles so as to increase contact area for better heat exchange with high temperature steam. Constructional detail of deaerator is shown in Fig. 8.15 which has basically deaerator head and storage tank. Water enters deaerator head from top on to a distributor plate. The water trickles down from upper most tray to the bottom trays through tiny holes in these trays. Steam enters storage tank from one end and enters deaerator head after passing through", " water collected in tank. Steam gradually heats feed water flowing downward with its\u2019 portion getting condensed and remaining steam flowing along with liberated gases out of deaerator head. Steam and gas mixture vented out from the top of deaerator head may be used for preheating make up water or feed water entering deaerator, if economical. The deaerated water is collected in storage tank below the deaerator head. A bubbler line is also provided in deaerator tank for fast heating of deaerator system during start up of unit. Bubbler line is a perforated pipe laid at bottom of storage tank through which steam for heating the water is supplied. Air vent Steam Condensate sprayers Condensate inlet Condensate Storage tank Bled steam Deaerated condensate Fig. 8.15 Deaerator Feed heater arrangements: In regenerative cycle, feed water heaters of different types are employed. There are some generic arrangements frequently used in these cycles. The arrangements are discussed below with two bleed points from where m1 and m2 masses of steam are bled out at pressures p1 and p2 and expansion occurs upto pressure p3. Total steam flowing is taken as 1 kg. (i) Surface type heaters method: This employs surface type feed water heaters and the arrangement for them is as shown in Fig. 8.16. Here two surface heaters are used for showing the arrangement. Condensate of the bled steam is drained out using drain pump and sent in the main line at high pressure. This arrangement is also called as drain pump method. 264 ________________________________________________________ Applied Thermodynamics Bled steam m p1 1, (1\u2013 )m1 1 kg Bled steam m p2 2, Expanded steam m \u2013 m p ), 2 3 (1 \u2013 1 Boiler feed pump m1 m2 (1 \u2013 m \u2013 m1 2 ) Drain pump Drain pump Condensate extraction pump p 3 > > p 2 p 1 Fig. 8.16 Arrangement in drain pump method or surface type feed water heaters method (ii) Open type heater method: In this arrangement the open type feed water heaters are employed as shown. Here due to contact of two fluids, there occurs mixing of bled steam with water and is taken out using pump for being sent to next open feed water heater. (iii) Surface type heaters with hot well: This arrangement employs a hot well with surface type heaters. Bled steam condensate leaving surface type heaters is sent to hot well from where it is picked up by pump and flown through heaters for getting heated up. Arrangement is shown in Fig. 8.18. Bled steam m p1 1, Bled steam m p2 2, Expanded steam m \u2013 m p 3 1 (1 \u2013 ), 2 1 kg (1\u2013 )m1 m1 Boiler feed pump Pump Condensate extraction pump (C.E.P.) Fig. 8.17 Arrangement with open feed water heaters Expanded steam ), m \u2013 m1 2 3p (1 \u2013 Bled steam m p1 1, Bled steam m p2 2, 1 kg Feed pump m2 C.E.P. (1 \u2013 m \u2013 m1 2 ) m1 Fig 8.18 Arrangement with surface type feed water heaters and hotwell Hot well Vapour Power Cycles ___________________________________________________________ 265 (iv) Cascade method: This arrangement is shown in Fig. 8.19. Here bled steam condensate is throttled and cascaded into low pressure surface heaters. Bled steam condensate from last heater is sent to hotwell from where it is picked up and pumped through surface type heaters. For lowering the pressure of condensate before mixing the traps may also be used. A trap allows the liquid to be throttled to a lower pressure and also traps the vapour. Bled steam m p1 1, Bled steam m p2 2, Feed pump 1 kg C.E.P. m1 Throttle valve (Trap) Throttle valve (Trap) + (m m1 ) 2 (1 \u2013 m \u2013 m1 2 ) Fig. 8.19 Cascade method Thermodynamics of regenerative cycle: Schematic of regenerative cycle with single feed water heater of open type is shown in Fig. 8.20. Arrangement shows that steam is bled out from turbine at state 6 and fed into feed heater. 1 kg 2 Boiler 1 Turbine m, kg, 6 3, (1 \u2013 ), kgm Open feed water heater Condenser 4 5 Condensate extraction pump 7 1 kg Feed pump Fig. 8.20 Schematic for regenerative cycle with one open feed water heater 266 ________________________________________________________ Applied Thermodynamics Feed water leaving at state 7 as shown on T-S diagram is being pumped upto boiler pressure through feed pump. T-S diagram indicates that the", " amount of heat picked up by feed water is shown by hatched area A1751'. In case of absence of bleeding and feed heater the feed water will enter into boiler at state 1' as compared to state 1 when regenerative heating is employed. Thus, advantage of hotter feed water to boiler can be realized by bleeding expanding steam from turbine. Regeneration can be seen in the feed water heating as the bled steam gets mixed with feed water at state 5 thus resulting into hot feed water. 1 kg m, kg (1 \u2013 m ), kg 2 6 3 1 1' 5 7 4 T s Fig. 8.21 T-s representation for regenerative cycle with one open feed water heater As described earlier the bleeding offers advantage in terms of increased cycle efficiency due to increased mean temperature of heat addition. Hotter feed water also offers advantage in terms of reduced thermal stresses in boiler due to reduced temperature difference and less tendency of condensation of sulphur dioxide. Bleeding of steam causes reduced mass flow in condenser thereby reducing size of condenser. Bleeding is also disadvantageous because the work done per unit mass of steam gets reduced, thus increasing cost of the plant. Boiler capacity is to be increased for a given output. Here it can be concluded that if the number of feed heaters be increased then their could be substantial increase in feed water temperature, thus offering a cycle having high mean temperature of heat addition, close to Carnot cycle. But it shall be accompanied by reduced work output and increased cost of the plant. Generally, the number of feed water heaters employed lies between 3 to 8 with average temperature rise in each heater being 10\u201315\u00baC. For example, if there are six heaters then first two may be surface type or indirect contact type followed by open type or direct contact feed water heater which shall also act as deaerator followed by three surface type feed water heaters. For the regenerative cycle considered, with unit mass of steam leaving boiler and \u2018m\u2019 kg of steam bled out for feed water heating: Steam turbine work = (h2 \u2013 h6) + (1 \u2013 m) \u00b7 (h6 \u2013 h3) Pump work = (1 \u2013 m) \u00b7 (h5 \u2013 h4) + 1 \u00b7 (h1 \u2013 h7) Net work = {(h2 \u2013 h6) + (1 \u2013 m) \u00b7 (h6 \u2013 h3)} \u2013 {(1 \u2013 m) \u00b7 (h5 \u2013 h4) + (h1 \u2013 h7)} Heat added = 1 \u00b7 (h2 \u2013 h1) Hence, regenerative cycle efficiency = Net work Heat added Vapour Power Cycles ___________________________________________________________ 267 h 1 Example: Regenerative cycle with two surface type heaters, (Fig. 8.22) \u03b7 regenerative = \u2212 )} {(1 h 3 \u2212 ) ( h 2 {( h 2 \u2212 h 6 ) + \u2212 (1 )( m h 6 \u2212 \u2212 )( )} Let us carry out thermodynamic analysis for 1 kg of steam generated in boiler at pressure p1 and masses of bled steam be m6 and m7 at pressure p6 and p7. Steam turbine work = {1 \u00b7 (h2 \u2013 h6) + (1 \u2013 m6) \u00b7 (h6 \u2013 h7) + (1 \u2013 m6 \u2013 m7) \u00b7 (h7 \u2013 h3)} Total pumping work = {(1 \u2013 m6 \u2013 m7) \u00b7 (h5 \u2013 h4) + m7 \u00b7 (h10 \u2013 h9) + m6 \u00b7 (h14 \u2013 h8)} Bled steam leaving surface heaters 1 and 2 are at state 8 and 9 which are saturated liquid states at respective pressure, i.e., h8 = h at 7 p 6 Heat added in boiler = 1 \u00b7 (h2 \u2013 h1) at p \u2019 f f h9 = h Applying heat balance on two surface heaters we get, On surface heater 1, m6 \u00b7 h6 + (1\u2013 m6) \u00b7 h12 = m6 \u00b7 h8 + (1 \u2013 m6) \u00b7 h13 On surface heater 2, m7 \u00b7 h7 + (1\u2013 m6 \u2013 m7) \u00b7 h5 = m7 \u00b7 h9 + (1 \u2013 m6 \u2013 m7) \u00b7 h11 At the point of mixing of output from surface heater and bled condensate the heat balance yields. 1 \u00b7 h1 = (1 \u2013 m6) \u00b7 h13 + m6 \u00b7 h14 h1 = (1 \u2013 m6) \u00b7 h13 + m6 \u00b7 h14 and (1 \u2013 m6 \u2013 m7) \u00b7 h11 + m7 \u00b7 h10 = (1 \u2013 m6) \u00b7 h12 Upon the pumps 1, 2 and 3; On pump 1, and v4(p1 \u2013 p3) = h5 \u2013 h4 h5 = v4 (p1 \u2013 p3) + h4", " 2 Boiler 1 6 7 3 Condenser SH1 13 12 11 SH2 5 4 14 8 10 9 Pump 1 SH = Surface heater Pump 3 Pump 2 Fig. 8.22 (a) Schematic of regenerative cycle with two surface type heaters 268 _________________________________________________________ Applied Thermodynamics 13 1 8 p1 2 m6 m1 6 7 p6 p7 p3 (1 \u2013 m \u2013 m6 ) 7 3 5 9 4 T Fig. 8.22 Schematic and T-S representation for regenerative cycle with two surface type heaters s (b) T-s diagram On pump 2, On pump 3, v9(p1 \u2013 p7) = h10 \u2013 h9 h10 = v9 (p1 \u2013 p7) + h9 v8(p1 \u2013 p6) = h14 \u2013 h8 h14 = v8 (p1 \u2013 p6) + h8 Above different enthalpy expressions can be used for getting enthalpy values at salient points of interest in order to get the net work and cycle efficiency of this arrangement. 8.9 BINARY VAPOUR CYCLE Generally, water is used as working fluid in vapour power cycle as it is found to be better than any other fluid if looked from the point of view of desirable characteristics of working fluid. Water is poor in respect to the following desired characteristics of working fluid. Fluid should have critical temperature well above the highest temperature set by metallurgical limits of construction material. Fluid should have a saturation pressure at the maximum cycle temperature that poses no strength problems and a saturation pressure at the minimum cycle temperature that posses no difficulty of leakage from atmosphere. In respect to above properties water is found to exhibit poor characteristics as; Water has critical temperature of 374\u00baC which is about 300 \u00baC less than the temperature limits set by metallurgical properties. The saturation pressure of water is quite high even at moderate temperatures so it does not have desirable properties at higher temperatures. Therefore in high temperature region a substance which has low saturation pressure should be used and the fluid should have its\u2019 critical temperature well above metallurgical limits of about 600\u00baC. Therefore, it can be concluded that no single working fluid satisfies all the desirable requirements of working fluid, different working fluids may have different attractive features in them, but not all. So let us think of striking a combination of any two working fluids which are well suited together such as mercury and water. In such cases two vapour cycles operating on two different working fluids are put together and the arrangement is called binary vapour cycle. Mercury has comparatively small saturation pressures at high temperature values but due to excessively low pressure values and large specific volume values at low temperatures it cannot be used alone Vapour Power Cycles ___________________________________________________________ 269 as working fluid. Mercury also does not wet the surface in contact so there is inefficient heat transfer although 0.002% of solution of magnesium and potassium is added to give it wetting property of steel. Steam is used with mercury for overcoming some limitations of mercury. Thus in combination of mercury-steam, the mercury is used for high pressures while steam is used for low pressure region. Layout for mercury-steam binary vapour cycle is shown on Fig. 8.23 along with it\u2019s depiction on T-S diagram. Here, mercury vapour are generated in mercury boiler and sent for expansion in mercury turbine and expanded fluid leaves turbine and enters into condenser. From condenser the mercury condensate is pumped back into the mercury boiler. In mercury condenser the water is used for extracting heat from mercury so as to condense it. The amount of heat liberated during condensation of mercury is too large to evaporate the water entering mercury condenser. Thus, mercury condenser also acts as steam boiler. For superheating of steam an auxilliary boiler may be employed or superheating may be realized in the mercury boiler itself. MT : Mercury turbine ST : Steam turbine 1 M erc ury b oile r Stea m M T 2 M erc ury c yc le M erc ury c on d e ns e r 3 P um p d a S T b e Stea m cy c le Stea m C on d en s e r c 4 Mercury-steam binary vapour cycle P um p M erc ury cy c le : 1 23 Stea m cy c le : a bc d ea Thermodynamic cycle for mercury-steam binary vapour cycle s Fig. 8.23 Schematic of binary vapour cycle 270 _________________________________________________________ Applied Thermodynamics Net work from cycle shall be, Wnet = WMT + WST \u2013 \u2211Wpump Work from mercury turbine, WMT = mMT \u00b7 (h1 \u2013 h2) Work from steam turbine, WST = mST \u00b7 (ha \u2013 hb) Pump work = mMT \u00b7 (h4 \u2013 h3) + mST \u00b7 (hd \u2013 hc", ") Heat added to the cycle, Qadd = mMT \u00b7 (h1 \u2013 h4) + mST \u00b7 (ha \u2013 he) Binary Cycle efficiency, binary = W Q net add { m MT ( ST { ( h m 1 MT \u2212 \u2212 \u2212 + MT ( m h a ST \u2212 \u2212 \u2212 ) h 3 )} h e m h d ST ( \u2212 h c )} 8.10 COMBINED CYCLE Combined cycle refers to the combination of two cycles operating in synergy. The thermodynamic cycles operating together in the form of combined cycles have capability to operate in isolation too for producing work output. These different cycles have to operate on different fluids. Among different combined cycles the gas/steam combination is popular. The gas/steam combined cycles have combination of Brayton cycle and Rankine cycle. Exhaust gases from gas turbine in Brayton cycle are sent to heat recovery steam generator (HRSG) or waste heat recovery boiler (WHRB) for generation of steam to be expanded in steam turbine in Rankine cycle. High temperature cycle in combined cycle is called topping cycle and low temperature cycle is called bottoming cycle. Thus, in combined cycle the heat rejected by higher temperature cycle is recovered, in lower temperature cycle such as in heat recovery steam generator for generation of steam which subsequently runs steam turbine and augments the work output. In different combined cycles the topping cycles could be Otto cycle, Brayton cycle and Rankine cycle while Rankine cycle is generally used as bottoming cycle. Fig. 8.24 shows the layout of a typical gas/steam combined cycle. Combined cycle could have various arrangements depending upon the alterations in topping cycle and bottoming cycle arrangements. In the shown layout there is simple gas turbine cycle, compression of air occurs between states 1 and 2. Subsequently heat addition and expansion occurs in combustion chamber and gas turbine through processes 2-3 and 3-4 respectively. Exhaust gases from gas turbine enter into heat recovery steam generator (HRSG) at state 4 and leave at state 5. Steam generated at state 6 from HRSG is sent to steam turbine for expansion and thus steam turbine work output augments work output of gas turbine. Expanded steam enters condenser at state 7 and condensate is sent back to HRSG at state 12 after passing it through deaerator. For ma, mf and ms being flow rates of air, fuel and steam respectively thermodynamic analysis is carried out as under. Vapour Power Cycles ___________________________________________________________ 271 (m + m ) f a 5 Stack gases Drum GT Exhaust 4 (m + m ) f a HRSG mf fuel CC 2 C 3 GT 1 ma Atm. Air. 6 ms ST Condenser : Gas Turbine BFP : Boiler feed pump : Compressor C CC : Combustion chamber CEP : Condensate extraction pump GT HRSG: Heat recovery steam generator ST ma mf ms m\u2032 s : Steam turbine : Mass flow rate of air, kg/s : Mass flow rate of fuel, kg/s : Mass flow rate of steam, kg/s : Mass flow rate of bled steam, kg/s 10 m s' 7 8 CEP 9 ms 12 Deaerator 11 BFP Approach temperature (a) Layout of gas/steam combined cycle 2'2 5 12 9 11 Pinch point temp. difference T 3 4'4 6 10 8 7 7' S (b) T-s diagram representation for combined cycle. Fig. 8.24 (a) Layout of gas/steam combined cycle (b) T-S diagram representation for combined cycle 272 _________________________________________________________ Applied Thermodynamics Thus the work requirement in compressor, Wc = ma (h2 \u2013 h1) Heat addition in combustion chamber, for fuel having calorific value CV Qadd = mf \u00d7 CV Energy balance upon combustion chamber yields, ma \u00d7 h2 + mf \u00d7 CV = (ma + mf) h3 Work available from gas turbine WGT = (ma + mf) (h3 \u2013 h4) Net work from topping cycle, Wtopping = WGT \u2013 Wc Work available from steam turbine, for bled steam mass flow rate for deaeration being ms. WST = {ms(h6 \u2013 h7) + (ms \u2013 m\u2032 s ) (h7 \u2013 h10)} Pump works WCEP = (ms \u2013 ms WBFP = ms.(h12 \u2013 h11) \u2032) (h9 \u2013 h8) Net work available from bottoming cycle Wbottoming = WST \u2013 WCEP \u2013 WBFP Hence total work output from combined cycle Wcombined = Wtopping + Wbottoming Thermal efficiency of combined cycle, \u03b7 combined = W combined Q add Thermal efficiency of topping cycle (gas turbine cycle), \u03b7 topping = W topping Q add We can see that work output of gas turbine cycle is less than combined cycle work output, while the heat addition remains same. Thus, thermal", " efficiency of combined cycle is more than gas turbine cycle (topping cycle), As, Wcombined > Wtopping \u21d2 \u03b7 combined > \u03b7 topping 8.11 COMBINED HEAT AND POWER Combined heat and power refers to the arrangement in which cycle produces work (power) along with heat utilization for process heating. There exist number of engineering applications where both power and process heat are simultaneously required. Such arrangement is also called cogeneration. Cogeneration may be defined as the arrangement of producing more than one useful form of energy. Food processing Vapour Power Cycles ___________________________________________________________ 273 industry and chemical industry are the industries where steam is required for different processes and cogeneration is an attractive option for getting electricity alongwith process steam. Schematic of a cogeneration plant is shown in Fig. 8.25. Qadd Boiler 9 1 WT Pressure reducing valve Turbine 5 6 Process Heating Pump 8 7 2 3 Condenser 4 Pump Fig. 8.25 Schematic for cogeneration Cogeneration arrangement is popularly used in cold countries for district heating where, in this arrangement the power plant supplies electricity along with steam for process needs, such as space heating and domestic water heating. Thermodynamic analysis of the cogeneration arrangement shows; Heat added in boiler Qadd = m1 (h1 \u2013 h9) Heat used in process heating Qprocess = {m5 (h5 \u2013 h7) + m6 (h6 \u2013h7)} Turbine work: WT = {(m1 \u2013 m5) \u00b7 (h1 \u2013h6) + (m1 \u2013 m5 \u2013 m6) \u00b7 (h6 \u2013 h2)} Pump work: WP = {(m5 + m6) \u00b7 (h8 \u2013h7) + (m1 \u2013 m5 \u2013 m6) \u00b7 (h4 \u2013 h3)} When there is no process heating then; m5 = 0 and m6 = 0 8.12 DIFFERENT STEAM TURBINE ARRANGEMENT In certain applications simple steam turbines are unable to meet specific requirements. Back pressure turbine, pass out or extraction turbine and mixed pressure turbines are such special purpose turbines whose details are given ahead. (a) Back pressure turbine: Back pressure turbine is the one in which steam is not expanded upto lowest pressure in steam turbine, instead steam leaves the turbine at higher pressure which is appropriate for the process steam/heating requirement. Thus, in back pressure turbine expansion is limited to high back pressure and steam leaving turbine goes for process heating. Generally, steam leaving turbine at high back pressure will be superheated. Since steam is to be used for process heating so the rate of heat transfer should be high. Superheated steam is not suitable for heating because of small rate of heat transfer therefore superheated steam should be desuperheated and brought to saturated steam state as saturated steam has high rate of heat transfer and also the control of temperature is convenient. Thus, back pressure turbine has the provision of desuperheating as shown in Fig. 8.26. The steam leaving tubine enters into desuperheater where it is transformed into saturated steam. Saturated steam is subsequently sent for process heating where it gets condensed and condensate is sent back to the boiler through pump. A by pass valve is also provided so that if there is no power requirement then whole steam may be sent for process heating through desuperheater by closing turbine valve and opening by pass valve. 274 _________________________________________________________ Applied Thermodynamics (b) Pass out or extraction turbine: Pass out turbine refers to the steam turbine having provision for extraction of steam during expansion. Such provision is required because in combined heat and power requirement the steam available from back pressure turbine may be more than required one or the power produced may be less than the required value. Pass out turbine has arrangement for continuous extraction of a part of steam at the desired pressure for process heating and left out steam goes into low pressure section of turbine through a pressure control valve. In the low pressure section of turbine, a control mechanism is provided so that the speed of turbine and pressure of steam extracted remains constant irrespective of the variations in power produced and process heating. 1 Boiler By-pass valve Turbine valve 2 Turbine 3 Water Pump 8 Fig. 8.26 Back pressure turbine 7 Pressure control valve Steam exhaust Steam in Extracted steam Fig. 8.27 Pass out turbine The pass out turbines have to operate under widely varying load so its efficiency is quite poor. For facilitating the operation of pass out turbine from no extraction to full steam extraction conditions, nozzle control geverning or throttle control governing are used. (c) Mixed pressure turbine: These are the turbines which have capability of admitting steam at more than one pressures and subjecting multiple pressure steam streams to expand. Generally, mixed pressure turbines utilize high pressure steam from a boiler and also low pressure steam from exhaust of a noncondensing engine or some auxiliary of the plant.", " Vapour Power Cycles ___________________________________________________________ 275 Steam out Mixed Pressure Turbine High pressure steam in Low pressure steam in Fig 8.28 Mixed pressure turbine Mixed pressure turbines are preferred when steam at single pressure is not available in desired quantity for producing required power. These mixed pressure turbines actually have more than one turbines in one cylinder. EXAMPLES 1. A Carnot cycle works on steam between the pressure limits of 7 MPa and 7 kPa. Determine thermal efficiency, turbine work and compression work per kg of steam. Solution: T-s representation for the Carnot cycle operating between pressure of 7 MPa and 7 kPa is shown in Fig. 8.29 Enthalpy at state 2, h2 = hg at 7 MPa h = 2772.1 kJ/kg Entropy at state 2, s2 = sg at 7 MPa s2 = 5.8133 kJ/kg \u00b7 K Enthalpy and entropy at state 3, For process 2 \u20131, s1 = s2\u00b7 Let dryness fraction at state 1 be x1. h3 = h\u0192 at 7 MPa = 1267 kJ/kg s3 = s\u0192 at 7 MPa = 3.1211 kJ/kg \u00b7K 3 2 7 MPa 7 kPa T 4 1 s Fig. 8.29 s1 = s2 = s\u0192 at 7 kPa + x1 \u00b7 s\u0192g at 7 kPa 5.8133 = 0.5564 + x1 \u00b7 7.7237 x1 = 0.6806 Enthalpy of state 1, h1 = h\u0192 at 7 kPa + x1\u00b7 h\u0192g at 7 kPa = 162.60 + (0.6806 \u00d7 2409.54) h1 = 1802.53 kJ/kg Let dryness fraction at state 4 be x4, For process 4\u20133, s4 = s3 = s\u0192 at 7 kPa + x4\u00b7 s\u0192g at 7 kPa 3.1211 = 0.5564 + x4\u00b7 7.7237 276 _________________________________________________________ Applied Thermodynamics x4 = 0.3321 Enthalpy at state 4, h4 = h\u0192 at 7 kPa + x4\u00b7 h\u0192g at 7 kPa = 162.60 + (0.3321 \u00d7 2409.54) h4 = 962.81 kJ/kg Thermal efficiency = Net work Heat added Expansion work per kg = h2 \u2013 h1 = (2772.1 \u2013 1802.53) = 969.57 kJ/kg Compression work per kg = h3 \u2013 h4 = (1267 \u2013 962.81) = 304.19 kJ/kg (+ve) Heat added per kg = h2 \u2013 h3 = (2772.1 \u2013 1267) = 1505.1 kJ/kg (\u2013ve) Net work per kg = (h2 \u2013 h1) \u2013 (h3 \u2013 h4) = 969.57 \u2013 304.19 = 665.38 kJ/kg Thermal efficiency = 665.38 1505.1 = 0.4421 or 44.21% Thermal efficiency = 44.21% Turbine work = 969.57 kJ/kg (+ve) Compression work = 304.19 kJ/kg (\u2013ve) Ans. 2. A steam power plant uses steam as working fluid and operates at a boiler pressure of 5 MPa, dry saturated and a condenser pressure of 5 kPa. Determine the cycle efficiency for (a) Carnot cycle (b) Rankine cycle. Also show the T-s representation for both the cycles. Solution: From steam tables: At 5 MPa h\u0192, 5MPa = 1154.23 kJ/kg, s\u0192, 5 MPa = 2.92 kJ/kg \u00b7 K hg, 5MPa = 2794.3 kJ/kg, sg, 5 MPa = 5.97 kJ/kg \u00b7 K At 5 kPa h\u0192, 5kPa = 137.82 kJ/kg, s\u0192, 5kPa = 0.4764 kJ/kg \u00b7 K hg, 5kPa = 2561.5 kJ/kg, sg, 5kPa = 8.3951 kJ/kg \u00b7 K v\u0192, 5kPa = 0.001005 m3/kg As process 2-3 is isentropic, so s2 = s3 and 1 6 4 5 T 2 3 5 MPa 5 kPa S Fig. 8.30 Hence enthalpy at 3, Enthalpy at 2, s3 = s\u0192, 5kPa + x3 \u00b7 s\u0192g, 5kPa = s2 =", " sg, 5MPa x3 = 0.694 Carnot cycle : 1\u20132\u20133\u20134\u20131 Rankine cycle : 1\u20132\u20133\u20135\u20136\u20131 h3 = h\u0192, 5kPa + x3\u00b7 h\u0192g, 5kPa h3 = 1819.85 kJ/kg h2 = hg, 5MPa = 2794.3 kJ/kg Vapour Power Cycles ___________________________________________________________ 277 Process 1-4 is isentropic, so s1 = s4 s1 = 2.92 = 0.4764 + x4\u00b7 (8.3951 \u2013 0.4764) x4 = 0.308 h4 = 137.82 + (0.308 \u00d7 (2561.5 \u2013 137.82)) h4 = 884.3 kJ/kg Enthalpy at 4, Enthapy at 1, h1 = h\u0192 at 5 MPa h1 = 1154.23 kJ/kg Carnot cycle (1-2-3-4-1) efficiency: \u03b7 carnot = Net work Heat added \u2212 ( ) \u2212 \u2212 {(2794.3 1819.85) \u2212 (2794.3 1154.23) \u2212 (1154.23 884.3)} = = or \u03b7 carnot = 0.4295 \u03b7 carnot = 42.95% Ans. In Rankine cycle, 1-2-3-5-6-1 Pump work, h6 \u2013 h5 = v\u0192, 5(p6 \u2013 p5) = 0.001005 (5000 \u2013 5) h6 \u2013 h5 = 5.02 h5 = h\u0192 at 5kPa = 137.82 kJ/kg Hence h6 = 137.82 + 5.02 = 142.84 kJ/kg h6 = 142.84 kJ/kg Net work in Rankine cycle = (h2 \u2013 h3) \u2013 (h6 \u2013 h5) = 974.45 \u2013 5.02 = 969.43 kJ/kg Heat added = h2 \u2013 h6 = 2794.3 \u2013 142.84 = 2651.46 kJ/kg Rankine cycle efficiency = 969.43 2651.46 \u03b7 Rankine = 0.3656 or \u03b7 Rankine = 36.56% Ans. 278 _________________________________________________________ Applied Thermodynamics 3. A steam turbine plant operates on Rankine cycle with steam entering turbine at 40 bar, 350\u00baC and leaving at 0.05 bar. Steam leaving turbine condenses to saturated liquid inside condenser. Feed pump pumps saturated liquid into boiler. Determine the net work per kg of steam and the cycle efficiency assuming all processes to be ideal. Also show cycle on T-s diagram. Also determine pump work per kg of steam considering linear variation of specific volume. Solution: From steam table h2 = hat 40 bar, 350\u00baC = 3092.5 kJ/kg s2 = sat 40 bar, 350\u00baC = 6.5821 kJ/kg \u00b7K h4 = h\u0192 at 0.05 bar = 137.82 kJ/kg s4 = s\u0192 at 0.05 bar = 0.4764 kJ/kg v4 = v\u0192 at 0.05 bar = 0.001005 m3/kg 350\u00b0C 1 4 T 40 bar 0.05 bar 2 3 s Fig. 8.31 Let dryness fraction at state 3 be x3, For ideal process, 2-3, s2 = s3 s2 = s3 = 6.5821 = s\u0192 at 0.05 bar + x3 \u00b7 sfg at 0.05 bar 6.5821 = 0.4764 + x3 \u00b7 7.9187 x3 = 0.7711 h3 = h\u0192 at 0.05 bar + x3 \u00b7 h\u0192g at 0.05 bar = 137.82 + (0.7711 \u00d7 2423.7) h3 = 2006.74 kJ/kg For pumping process h1 \u2013 h4 = v4 \u00b7 \u2206p = v4 \u00d7 (p1 \u2013 p4) h1 = h4 + v4 \u00d7 (p1 \u2013 p4) = 137.82 + (0.001005 \u00d7 (40 \u2013 0.05) \u00d7 102) Pump work per kg of steam = (h1 \u2013 h4) = 4.02 kJ/kg h1 = 141.84 kJ/kg Net work per kg of steam = (Expansion work \u2013 Pump work) per kg of steam = (h2 \u2013 h3) \u2013 (h1 \u2013 h4) Vapour Power Cycles ___________________________________________________________ 279 = 1081.74 kJ/kg Cycle efficiency = Net work Heat added = 1081", ".74 h\u2212 ) ( h 1 2 1081.74 \u2212 (3092.5 141.84) = = 0.3667 or 36.67% Net work per kg of steam = 1081.74 kJ/kg Cycle efficiency = 36.67% Pump work per kg of steam = 4.02 kJ/kg Ans. 4. A steam power plant running on Rankine cycle has steam entering HP turbine at 20 MPa, 500\u00baC and leaving LP turbine at 90% dryness. Considering condenser pressure of 0.005 MPa and reheating occurring upto the temperature of 500\u00baC determine, (a) the pressure at wich steam leaves HP turbine (b) the thermal efficiency Solution: Let us assume that the condensate leaves condenser as saturated liquid and the expansion in turbine and pumping processes are isentropic. From steam tables, h2 = hat 20 MPa, 500\u00baC = 3238.2 kJ/kg s2 = 6.1401 kJ/kg \u00b7 K h5 = hat 0.005 MPa, 0.90 dry h5 = h\u0192 at 0.005 MPa, + 0.9 \u00d7 h\u0192g at 0.005 MPa = 137.82 + (0.9 \u00d7 2423.7) h5 = 2319.15 kJ/kg s5 = s\u0192 at 0.005 MPa, + 0.9 \u00d7 s\u0192g at 0.005 MPa = 0.4764 + (0.9 \u00d7 7.9187) s5 = 7.6032 kJ/kg \u00b7 K h6 = h\u0192 at 0.005 MPa = 137.82 kJ/kg 20 MPa 2 4 0.005 MPa It is given that temperature at state 4 is 500\u00baC and due to isentropic process s4 = s5 = 7.6032 kJ/kg \u00b7K. The state 4 can be conveniently located on Mollier chart by the intersection of 500\u00baC constant temperature line and entropy value of 7.6032 kJ/kg \u00b7 K and the pressure and enthalpy obtained. But these shall be approximate. T 1 6 3 5 s Fig. 8.32 280 _________________________________________________________ Applied Thermodynamics The state 4 can also be located by interpolation using steam table. The entropy value of 7.6032 kJ kg \u00b7 K lies between the superheated steam states given under, p = 1.20 MPa, sat 1.20 MPa and 500\u00baC = 7.6759 kJ/kg \u00b7 K p = 1.40 MPa, sat 1.40 MPa and 500\u00baC = 7.6027 kJ/kg \u00b7 K By interpolation state 4 lies at pressure = 1.20 + \u2212 (1.40 1.20) \u2212 (7.6027 7.6759) = 1.399 MPa \u2248 1.40 MPa (7.6032 \u2013 7.6759) Thus, steam leaves HP turbine at 1.4 MPa Enthalpy at state 4, h4 = 3474.1 kJ/kg For process 2-3, s2 = s3 = 6.1401 kJ/kg \u00b7 K. The state 3 thus lies in wet region as s3 < sg at 1.40 MPa. Let dryness fraction at state 3 be x3. s3 = s\u0192 at 1.4 MPa + x3 \u00b7 sfg at 1.4 MPa 6.1401 = 2.2842 + x3 \u00b7 4.1850 x3 = 0.9214 h3 = h\u0192 at 1.4 MPa + x3 \u00b7 h\u0192g at 1.4 MPa = 830.3 + (0.9214 \u00d7 1959.7) = 2635.97 kJ/kg Enthalpy at 1, h1 = h6 + v6(p1 \u2013 p6) = h\u0192 at 0.005 MPa + v\u0192 at 0.005 MPa (20 \u2013 0.005) \u00d7 103 = 137.82 + (0.001005 \u00d7 19.995 \u00d7 103) h1 = 157.91 kJ/kg Net work per kg of steam = (h2 \u2013h3) + (h4 \u2013 h5) \u2013 (h1 \u2013 h6) = 1737.09 kJ/kg Heat added per kg of steam = (h2 \u2013 h1) = 3080.29 kJ/kg Thermal efficiency = Net work Heat added = 1737.09 3080.29 = 0.5639 or 56.39% Pressure of steam leaving HP turbine = 1.40 MPa Ans. Thermal efficiency = 56.39% 5. In a steam turbine installation running on ideal Rankine cycle steam leaves the boiler at 10 MPa", " and 700\u00baC and leaves turbine at 0.005 MPa. For the 50 MW output of the plant and cooling water entering and leaving condenser at 15\u00baC and 30\u00baC respectively determine (a) the mass flow rate of steam in kg/s (b) the mass flow rate of condenser cooling water in kg/s Vapour Power Cycles ___________________________________________________________ 281 (c) the thermal efficiency of cycle (d) the ratio of heat supplied and rejected (in boiler and condenser respectively). Neglet K.E. and P.E. changes. Solution: From steam table At inlet to turbine, h2 = hat 10 MPa, 700\u00baC h2 = 3870.5 kJ/kg s2 = 7.1687 kJ/kg \u00b7 K For process 2-3, s2= s3 and s3 < s\u0192 at 0.005 MPa so state 3 lies in wet region. Let dryness fraction at state 3 be x3. s3 = 7.1687 = s\u0192 at 0.005 MPa + x3 \u00b7 s\u0192g at 0.005 MPa 7.1687 = 0.4764 + (x3 \u00d7 7.9187) x3 = 0.845 h3 = h\u0192 at 0.005 MPa + x3 \u00b7 hfg at 0.005 MPa = 137.82 + (0.845 \u00d7 2423.7) h3 = 2185.85 kJ/kg h4 = h\u0192 at 0.005 MPa = 137.82 kJ/kg For pumping process, (h1 \u2013 h4) = v4 \u00d7 (p1 \u2013 p4) 10 MPa 2 0.005 MPa 1 4 T 3 S Fig. 8.33 v4 = v\u0192 at 0.005 MPa = 0.001005 m3/kg h1 = 137.82 + (0.001005 \u00d7 (10 \u2013 0.005)) \u00d7 102 h1 = 138.82 kJ/kg Net output per kg of steam, wnet = (h2 \u2013 h3) \u2013 (h1 \u2013 h4) = (3870.5 \u2013 2185.85) \u2013 (138.82 \u2013 137.82) Mass flow rate of steam, ms = = 29.69 kg/s wnet = 1683.65 kJ/kg \u00d7 50 10 1683.65 3 282 _________________________________________________________ Applied Thermodynamics 3, h3 = 2185.85 kJ/kg 15 \u00b0C, water 30 \u00b0C, water 4, h4 = 137.82 kJ/kg Fig. 8.34 By heat balance on condenser, for mass flow rate of water being mw kg/s. (h3 \u2013 h4) \u00d7 ms = mw \u00b7 Cp, w (Tw, out \u2013 Tw, in) 29.69 \u00d7 (2185.85 \u2013 137.82) = mw \u00d7 4.18 (15) mw = 969.79 kg/s The heat added per kg of steam qadd = (h2 \u2013 h1) = 3731.68 kJ/kg Thermal efficiency = w q net add = 1683.65 3731.68 = 0.4512 or 45.12% Ratio of heat supplied and rejected = (.822 Mass of flow rate of steam = 29.69 kg/s Mass flow rate of condenser cooling water = 969.79 kg/s Thermal efficiency = 45.12% Ratio of heat supplied and rejected = 1.822 Ans. 6. A regenerative Rankine cycle has steam entering turbine at 200 bar, 650\u00baC and leaving at 0.05 bar. Considering feed water heaters to be of open type determine thermal efficiency for the following conditions; (a) there is no feed water heater (b) there is only one feed water heater working at 8 bar (c) there are two feed water heaters working at 40 bar and 4 bar respectively. Also give layout and T-s representation for each of the case described above. Solution: Case (a) When there is no feed water heater Thermal efficiency of cycle = From steam tables Vapour Power Cycles ___________________________________________________________ 283 h2 = hat 200 bar, 650\u00baC = 3675.3 kJ/kg s2 = sat 200 bar, 650\u00baC = 6.6582 kJ/kg \u00b7 K h4 = h\u0192 at 0.05 bar = 137.82 kJ/kg v4 = v\u0192 at 0.05 bar = 0.001005 m3/kg. h\u0192 at 0.05 bar = 137.82 kJ/kg, h\u0192g at 0.05 bar = 2423.7 kJ/kg s\u0192 at 0.05 bar = 0.4764 kJ/", "kg \u00b7 K, s\u0192g at 0.05 bar = 7.9187 kJ/kg \u00b7 K For process 2 \u2013 3, s2= s3. Let dryness fraction at 3 be x3. s3 = 6.6582 = s\u0192 at 0.05 bar + x3 \u00b7 s\u0192g at 0.05 bar 6.6582 = 0.4764 + x3 \u00b7 7.9187 x3 = 0.781 h3 = h\u0192 at 0.05 bar + x3 \u00b7 hfg at 0.05 bar = 2030.73 kJ/kg For pumping process 4-1, h1 \u2013 h4 = v4 \u00b7 \u2206p h1 \u2013 137.82 = 0.001005 \u00d7 (200 \u2013 0.05) \u00d7 102 Thermal efficiency of cycle = h1 = 157.92 kJ/kg \u2212 \u2212 (3675.3 2030.73) \u2212 (3675.3 157.92) \u2212 (157.92 137.82) 200 bar, 650 \u00b0C 2 Boiler ST 1 3 0.05 bar Condenser T 4 Feed pump 200 bar 0.05 bar 1 4 2 3 s Fig. 8.35 Layout and T-s diagram, (Q 6.a) = 0.4618 or 46.18% Case (b) When there is only one feed water heater working at 8 bar Here, let mass of steam bled for feed heating be m kg For process 2-6, s2 = s6 = 6.6582 kJ/kg \u00b7 K Let dryness fraction at state 6 be x6 s6 = sf at 8 bar + x6 \u00b7 sfg at 8 bar 284 _________________________________________________________ Applied Thermodynamics 200 bar, 650 \u00b0C 2, 1 kg Boiler 1 ST 8 bar 6 3 Condenser T 4 CEP OFWH m kg 5 7 (1 \u2013 ) kgm FP 1 7 5 4 200 bar 2 8 bar m kg 0.05 bar (1 \u2013 m ) kg 6 3 S CEP = Condensate Extraction Pump FP = Feed Pump ST = Steam Turbine OFWH = Open Feed Water Heater Fig. 8.36 Layout and T-s diagram, (Q 6,b) From steam tables, h\u0192 at 8 bar = 721.11 kJ/kg v\u0192 at 8 bar = 0.001115 m3/kg, h\u0192g at 8 bar = 2048 kJ/kg s\u0192 at 8 bar = 2.0462 kJ/kg \u00b7 K, s\u0192g at bar = 4.6166 kJ/kg \u00b7 K Substituting entropy values, h6 = h\u0192 at 8 bar + x6 \u00b7 h\u0192g at 8 bar = 721.11 + (0.999 \u00d7 2048) = 2767.06 kJ/kg Assuming the state of fluid leaving open feed water heater to be saturated liquid at 8 bar. x6 = 0.999 h7 = h\u0192 at 8 bar= 721.11 kJ/kg. For process 4-5, h5 = h4 + v4 \u00d7 (8 \u2013 0.05) \u00d7 102 = 137.82 + (0.001005 \u00d7 7.95 \u00d7 102) = 138.62 kJ/kg Applying energy balance at open feed water heater, m \u00d7 h6 + (1 \u2013 m) \u00d7 h5 = 1\u00d7 h7 (m \u00d7 2767.06) + ((1 \u2013 m) \u00d7 138.62) = 721.11 m = 0.2216 kg For process 7-1, h1 = h7 + v7 (200 \u2013 8) \u00d7 102; here h7 = h\u0192 at 8 bar, 7v = fv at 8 bar h1 = h7 + (200 \u2013 8) \u00d7 102 = 721.11 + (0.001115 \u00d7 192 \u00d7 102) fv at 8 bar h1 = 742.518 kJ/kg + \u2212 \u2212 (1 ( ) h 6 h 2 Thermal efficiency of cycle = )\u00b7( m h 6 \u2212 \u2212 \u2212 ) {( )( )} \u2212 (3675.3 2767.06) {(1 0.2216) \u2212 = \u00d7 + \u2212 (1 0.2216) \u2212 \u00d7 (138.62 137.82) \u2212 (3675.3 742.518) \u2212 (2767.06 2030.73) \u2212 + \u2212 (742.518 721.11)} Vapour Power Cycles ___________________________________________________________ 285 Thermal efficiency of cycle = 0.4976 or 49.76% Case (c) When there are two feed water heaters working at 40 bar and 4 bar Here, let us assume the mass of steam at 40 bar, 4 bar to be m1 kg, and m2 kg respectively. For process 2\u201310", "\u20139\u20133, s2 = s10 = s9 = s3 = 6.6582 kJ/kg \u00b7K At state 10. s10 > sg at 40 bar (6.0701 kJ/kg \u00b7 K) so state 10 lies in superheated region at 40 bar pressure. From steam table by interpolation, T10 = 370.36\u00baC so, h10 = 3141.81 kJ/kg Let dryness fraction at state 9 be x9 so, s9 = 6.6582 = sf at 4 bar + x9 \u00b7 sfg at 4 bar 6.6582 = 1.7766 + x9 \u00d7 5.1193 x9 = 0.9536 h9 = hf at 4 bar + x9 \u00d7 hfg at 4 bar = 604.74 + 0.9536 \u00d7 2133.8 h9 = 2639.53 kJ/kg Assuming the state of fluid leaving open feed water heater to be saturated liquid at respective pressures i.e. h11 = hf at 4 bar = 604.74 kJ/kg, v11 = 0.001084 m3/kg = vf at 4 bar h13 = hf at 40 bar = 1087.31 kJ/kg, v13 = 0.001252 m3/kg = vf at 40 bar For process 4\u20138, i.e. in CEP. For process 11-12, i.e. in FP2, h8 = h4 + v4 \u00d7 (4 \u2013 0.05) \u00d7 102 = 137.82 + (0.001005 \u00d7 3.95 \u00d7 102) h8 = 138.22 kJ/kg h12 = h11 + v11(40\u20134) \u00d7 102 = 604.74 + (0.001084 \u00d7 36 \u00d7 102) h12 = 608.64 kJ/kg 200 bar, 650 \u00b0C 2, 1 kg Boiler 1' ST 40 bar 10 9 (1 \u2013 3 m m1 \u2013 2 ) Condenser m1 4 bar m2 OFWH 1 OFWH 2 13 12 11 4 8 CEP FP1 (1 \u2013 )m1 FP2 286 _________________________________________________________ Applied Thermodynamics 1\u00a2 200 bar 2 40 bar T m1 10 12 13 8 11 4 m2 m m1 \u2013 2 ) 9 (1\u2013 3 4 bar 0.05 bar Fig. 8.37 Layout and T-s diagram. (Q6.c) S For process 13-1' i.e. in FP1, h'1= h13 + v13(200 \u2013 40) \u00d7 102 = 1087.31 + (0.001252 \u00d7 160 \u00d7 102) (m1 \u00d7 3141.81) + (1 \u2013 m1) \u00d7 608.64 = 1087.31 m1 = 0.189 kg Applying energy balance an open feed water heater 1 (OFWH1) h'1 = 1107.34 kJ/kg (m1 \u00d7 h10) + (1 \u2013 m1) \u00d7 h12 = 1 \u00d7 h13 (m1 \u00d7 3141.81) + (1 \u2013 m1) \u00d7 608.64 = 1087.31 m1 = 0.189 kg Applying energy balance an open feed water heater 2 (OFWH2) m2 \u00d7 h9 + (1 \u2013 m1 \u2013 m2) h8 = (1 \u2013 m1) \u00d7 h11 (m2 \u00d7 2639.53) + (1 \u2013 0.189 \u2013 m2) \u00d7 138.22 = (1\u2013 0.189) \u00d7 604.74, m2 = 0.151 kg Thermal efficiency of cycle, \u03b7 = {( h 2 \u2212 h 10 ) + \u2212 (1 )( m h 1 10 \u2212 h 9 ) + \u2212 (1 \u2212 )( m m h 9 2 )' \u2212 )} { W CEP + W FP 1 + W FP 2 } WCEP = (1\u2013 m1 \u2013 m2) (h8 \u2013 h4) = 0.264 kJ/kg steam from boiler W = (h1' \u2013 h13) = 20.03 kJ/kg of steam from boiler FP1 W FP2 = (1 \u2013 m1) (h12 \u2013 h11) = 3.16 kJ/kg of steam from boiler W W + FP1 WCEP + { \u2212 (3675.3 3141.81) FP2 = 23.454 kJ/kg of steam from boiler + \u2212 (1 0.189)(3141.81 2639.53) \u2212 + \u2212 } (1 0.189 0.151)(2639.53 2030.73) {23.454} \u2212 \u2212 \u2212 \u03b7 = \u2212 (3675.3 1107.34) = 0.5137 or 51.37% Cycle thermal efficiency", " \u03b7 a = 46.18% \u03b7 b = 49.76% \u03b7 c = 51.37% Ans. Hence it is obvious that efficiency increases with increase in number of feed heaters. Vapour Power Cycles ___________________________________________________________ 287 7. A reheat cycle has steam generated at 50 bar, 500\u00baC for being sent to high pressure turbine and expanded upto 5 bar before supplied to low pressure turbine. Steam enters at 5 bar, 400\u00baC into low pressure turbine after being reheated in boiler. Steam finally enters condenser at 0.05 bar and subsequently feed water is sent to boiler. Determine cycle efficiency, specific steam consumption and work ratio. Solution: From steam table, h2 = hat 50 bar, 500\u00baC = 3433.8 kJ/kg s2 = sat 50 bar, 500\u00baC = 6.9759 kJ/kg \u00b7K s3 = s2 = 6.9759 kJ/kg \u00b7 K Since s3 > sg at 5 bar SO state 3 lies in superheated region at 5 bar, By interpolation from steam tables, T3 = 183.14\u00baC at 5 bar, h3 = 2818.03 kJ/kg h4 = h at 5 bar, 400\u00baC = 3271.9 kJ/kg s4 = s at 5 bar, 400\u00baC = 7.7938 kJ/kg \u00b7 K T 1 6 50 bar 5 bar 4 2 3 0.05 bar 5 s Fig. 8.38 For expansion process 4-5, s4 = s5 = 7.7938 kJ/kg \u00b7 K Let dryness fraction at state 5 be x5. s5 = sf at 0.05 bar + x5 \u00d7 sfg at 0.05 bar 7.7938 = 0.4764 + x5 \u00d7 7.9187 x5 = 0.924 h5 = hf at 0.05 bar + x5 \u00d7 hfg at 0.05 bar h5 = 137.82 + 0.924 \u00d7 2423.7 = 2377.32 kJ/kg h6 = hf at 0.05 bar = 137.82 kJ/kg h6 = vf at 0.05 bar = 0.001005 m3/kg For process 6-1 in feed pump, h1 = h6 + v6 \u00d7 (50 \u2013 0.05) \u00d7 102 h1 = 137.82 + 0.001005 \u00d7 (49.95 \u00d7 102) h1 = 142.84 kJ/kg 288 _________________________________________________________ Applied Thermodynamics Cycle efficiency = W Q net add WT = (h2 \u2013 h3) + (h4 \u2013 h5) = (3433.8 \u2013 2818.03) + (3271.9 \u2013 2377.32) = 1510.35 kJ/kg Wpump = (h1 \u2013 h6) = 142.84 \u2013 137.82 = 5.02 kJ/kg Wnet = WT \u2013 Wpump = 1505.33 kJ/kg Qadd = (h2 \u2013 h1) = 3433.8 \u2013 142.84 = 3290.96 kJ/kg Cycle efficiency = 1505.33 3290.96 = 0.4574 or 45.74% We know, 1 hp = 0.7457 kW Specific steam consumption = \u00d7 0.7457 3600 1505.33 = 1.78 kg/hp \u00b7 hr Work ratio = Net work Positive work = net W W = T 1505.33 1510.35 = 0.9967 Cycle efficiency = 45.74%, Specific steam consumption = 1.78 kg/hp. hr. Work ratio = 0.9967. Ans. 8. In a steam power plant the high pressure turbine is fed with steam at 60 bar, 450\u00baC and enters low pressure turbine at 3 bar with a portion of steam bled out for feed heating at this intermediate pressure. Steam finally leaves low pressure turbine at 0.05 bar for inlet to condenser. Closed feed heater raises the condensate temperature to 115\u00baC. Bled steam leaving closed feed heater is passed through trap to mix with condensate leaving condenser. Consider actual alternator output to be 30 MW, boiler efficiency as 90% and alternator efficiency of 98%. Determine, (a) the mass of steam bled for feed heating, (b) the capacity of boiler in kg/hr. (c) the overall thermal efficiency of plant Also give layout and T-s diagram. Solution: From steam tables, At state 2, h2 = 3301.8 kJ/kg, s2 = 6.7193 kJ/kg \u00b7 K h5 = hf at 0.05 bar = 137.82 kJ/kg, v5 = vf at 0.05 bar = 0.", "001005 m3/kg Let mass of steam bled for feed heating be m kg/kg of steam generated in boiler. Let us also assume that condensate leaves closed feed water heater as saturated liquid i.e. h8 = hf at 3 bar h8 = 561.47 kJ/kg s2 = s3 = s4 = 6.7193 kJ/kg \u00b7 K For process 2\u20133\u20134, Let dryness fraction at state 3 and state 4 be x3 and x4 respectively. s3 = 6.7193 = sf at 3 bar + x3 \u00b7 sfg at 3 bar Vapour Power Cycles ___________________________________________________________ 289 = 1.6718 + x3 \u00d7 5.3201 x3 = 0.949 s4 = 6.7193 = sf at 0.05 bar + x4 \u00b7 sfg at 0.05 bar = 0.4764 + x4 \u00d7 7.9187 x4 = 0.788 60 bar, 450 \u00b0C 2 Boiler 1 HPT LPT 3 bar 3 4 0.05 bar 1 1 kg Condenser 7 T m kg m (1 \u2013 ) kg 56 8 9 s 7 FP 6 8 5 9 Fig. 8.39 Layout and T-s diagram 60 bar 3 bar 0.05 bar 2 3 4 Thus, h3 = hf at 3 bar + x3 \u00b7 hfg at 3 bar = 561.47 + (0.949 \u00d7 2163.8) = 2614.92 kJ/kg h4 = hf at 0.05 bar + x4 \u00b7 hfg at 0.05 bar = 137.82 + (0.788 \u00d7 2423.7) = 2047.6 kJ/kg Assuming process across trap to be of throttling type so, h8 = h9 = 561.47 kJ/kg. Assuming v5 = v6, Pumping work (h7 \u2013 h6) = v5 \u00b7 (60 \u2013 0.05) \u00d7 102 (h7 \u2013 h6) = 6.02 kJ/kg For mixing process between condenser and feed pump, (1 \u2013 m) \u00b7 h5 + m \u00b7 h9 = 1 \u00b7 h6 (1 \u2013 m) \u00d7 137.82 + m \u00d7 561.47 = h6 Therefore, Applying energy balance at closed feed water heater; h6 = 137.82 + m \u00d7 423.65 h7 = h6 + 6.02 = 143.84 + m \u00d7 423.65 m \u00d7 h3 + (1 \u2013 m)h7 = m \u00b7 h8 + (4.18 \u00d7 115) (m \u00d7 2614.92) + (1 \u2013 m) (143.84 + m \u00d7 423.65) = m \u00d7 561.47 + 480.7, m = 0.144 kg Steam bled for feed heating = 0.144 kg/kg steam generated. Ans. The net power output Wnet = (h2 \u2013 h3) + (1 \u2013 m) (h3 \u2013 h4) \u2013 (1\u2013 m) \u00d7 (h7 \u2013 h6) = 1167.28 kJ/kg steam generated Mass of steam required to be generated = 3 30 \u00d710 0.98 W\u00d7 net 290 _________________________________________________________ Applied Thermodynamics = 3 30 \u00d710 \u00d7 0.98 1167.28 = 94428 kg/hr = 26.23 kg/s Capacity of boiler required = 94428 kg/hr Ans. Overall thermal efficiency = W Q net add Qadd = ) ( h\u2212 h 2 1 0.90 = \u2212 (3301.8 4.18 115) 0.90 \u00d7 = 3134.56 kJ/kg Overall thermal efficiency = 1167.28 3134.56 = 0.3724 or 37.24% Overall thermal efficiency = 37.24% Ans. 9. A steam power plant has expansion occurring stages in three stages with steam entering first stage at 30 bar, 400\u00b0C and leaving first stage at 6 bar for being sent to second stage with some steam being bled out for feed heating in closed feed water heater. Steam leaves second stage at 1 bar and enters third stage with some more steam being bled out for feed heating in closed feed water heater. Steam finally leaves third stage at 0.075 bar after complete expansion and enters condenser. Condensate temperature is 38\u00baC at inlet to second heater, temperature of feed water after first heater and second heater is 150\u00baC and 95\u00baC respectively. Bled steam is condensed to saturated liquid with no undercooling in each of feed heater. Drain from first heater is passed through steam trap into second feed heater and combined drain from second heater is pumped by drain pump into feed line after second heater. Considering efficiency ratio of turbine as 0.8 and turbine output of 15 MW determine the capacity of drain pump. Neglect drain", " pump work. Solution: 30 bar, 400 \u00b0C 2 Boiler HPT IPT 1 6 bar 3 4 LPT 1bar 5 (1 \u2013 m m1 \u2013 ) 2 Condenser T m1 m2 13 CFWH-1 CFWH-2 10 9 6 6' CEP 12 11 8 7 FP 13 6 10 8 9 11 7 12 6' s Fig. 8.40 Layout and T-s diagram 30 bar 2 6 bar 3 3' 1 bar 0.075 bar 4 4' 5 5' Vapour Power Cycles ___________________________________________________________ 291 At inlet to first turbine stage, h2 = 3230.9 kJ/kg, s2 = 6.9212 kJ/kg \u00b7K For ideal expansion process s2 = s3 By interpolation, T3 = 190.97\u00baC from superheated steam tables at 6 bar h3 = 2829.63 kJ/kg actual state at exit of first stage h'3 = h2 \u2013 0.8 \u00d7 (h2 \u2013 h3) h'3 = 2909.88 kJ/kg Actual state 3' shall be at 232.78\u00baC, 6 bar, so s'3 = 7.1075 kJ/kg \u00b7 K For second stage s'3 = s4; By interpolation, s4 = 7.1075 = sf at 1 bar + x4 \u00b7 sfg at 1 bar 7.1075 = 1.3026 + x4 \u00b7 6.0568 x4 = 0.958 h4 = hf at 1 bar + x4 \u00b7 hfg at 1 bar = 417.46 + (0.958 \u00d7 2258) h4 = 2580.62 kJ/kg Actual enthalpy at exit from second stage, h4' = h3' \u2013 0.8 (h3' \u2013 h4) h4' = 2646.47 kJ/kg Actual dryness fraction, x4' \u21d2 h4' = hf at 1 bar + x4' \u00b7 hfg at 1 bar x4' = 0.987, Actual entropy, s4' = 7.2806 kJ/kg \u00b7 K For third stage, s4' = s5 = 7.2806 = sf at 0.075 bar + x5 \u00b7 sfg at 0.075 bar x5 = 0.8735 h5 = 2270.43 kJ/kg Actual enthalpy at exit from third stage, h5' = h4' \u2013 0.8(h4' \u2013 h5) h5' = 2345.64 kJ/kg Let mass of steam bled out be m1 and m2 kg at 6 bar, 1 bar respectively. By heat balance on first closed feed water heater, (see schematic arrangement) h11 = hf at 6 bar = 670.56 kJ/kg m1 \u00d7 h3' + h10 = m1 \u00b7 h11 + 4.18 \u00d7 150 (m1 \u00d7 2829.63) + h10 = (m1 \u00b7 670.56) + 627 h10 + (2159.07) m1 = 627 By heat balance on second closed feed water heater, (see schematic arrangement) h7 = hf at 1 bar = 417.46 kJ/kg m2 \u00b7 h4\u00b4 + (1\u2013m1 \u2013 m2) \u00d7 4.18 \u00d7 38 = (m1 + m2) \u00b7 h7 + 4.18 \u00d7 95 \u00d7 (1 \u2013 m1\u2013m2) (m2 \u00b7 2646.47) + (1\u2013 m1 \u2013 m2) \u00d7 158.84 = ((m1 + m2) \u00b7 417.46) + (397.1 \u00d7 (1 \u2013 m1 \u2013 m2)) m2 \u00d7 2467.27 \u2013 m1 \u00d7 179.2 \u2013 238.26 = 0 292 _________________________________________________________ Applied Thermodynamics, m h1 3 ', m h2 4'13 150 \u00b0C 10, 1 kg 9 95 \u00b0C 6 (1 \u2013 m m1 \u2013 2 ) 1 kg, 10 9,(1 \u2013 m m1 \u2013 2 ) ( m m1 + ) 2 8, m h1 11 7 ( m m h 7 ), + 2 1 Heat balance at point of mixing. h10 = (m1 + m2) \u00b7 h8 + (1 \u2013 m1 \u2013 m2) \u00d7 4.18 \u00d7 95 Neglecting pump work, h7 = h8 h10 = m2 \u00d7 417.46 + (1 \u2013 m1 \u2013 m2) \u00d7 397.1 1 kg, 10 9 (1 \u2013 m m1 \u2013 ) 2 ( m m1 + ) 2 8 Substituting h10 and solving we get, m1 = 0.1293 kg m2 = 0.1059 kg/kg of steam generated. Turbine output", " per kg of steam generated, wT = (h2 \u2013 h3') + (1 \u2013 m1) (h3' \u2013 h4') + (1 \u2013 m1 \u2013 m2) \u00b7 (h4' \u2013 h5') wT = 780.446 kJ/kg of steam generated. Rate of steam generation required = 3 \u00d7 15 10 780.446 = 19.22 kg/s or = 69192 kg/hr Capacity of drain pump i.e. FP shown in layout = (m1 + m2) \u00d7 69192 = 16273.96 kg/hr Capacity of drain pump = 16273.96 kg/hr Ans. 10. A steam power plant has steam entering at 70 bar, 450\u00baC into HP turbine. Steam is extracted at 30 bar and reheated upto 400\u00baC before being expanded in LP turbine upto 0.075 bar. Some portion of steam is bled out during expansion in LP turbine so as to yield saturated liquid at 140\u00baC at the exit of open feed water heater. Considering HP and LP turbine efficiencies of 80% and 85% determine the cycle efficiency. Also give layout and T-s diagram. Solution: At inlet to HP turbine, h2= 3287.1 kJ/kg, s2 = 6.6327 kJ/kg.K Vapour Power Cycles ___________________________________________________________ 293 2 3 4 Boiler HPT 1 IPT 5 6 7 Condenser OFWH FP 8 CEP 9 450 \u00b0C 400 \u00b0C 328.98 \u00b0C 1 T 98 7 S 70 bar 30 bar 2 4 3' 3 CEP = Condensate extraction pump FP = Feed pump OFWH = Open feed water heater 0.075 bar 5 6 5' 6' By interpolation state 3 i.e. for insentropic expansion between 2 \u2013 3 lies at 328.98\u00baC at 30 bar. Fig. 8.41 Layout and T-S diagram Actual enthalpy at 3', h3' = h2 \u2013 0.80 (h2 \u2013 h3) h3 = 3049.48 kJ/kg. h3' = 3097 kJ/kg Enthalpy at inlet to LP turbine, h4 = 3230.9 kJ/kg, s4 = 6.9212 kJ \u00b7 K For ideal expansion from 4-6, s4 = s6. Let dryness fraction at state 6 be x6. s6 = 6.9212 = s\u0192 at 0.075 bar + x6 \u00b7 s\u0192g at 0.075 bar x6 = 0.827 h6 = h\u0192 at 0.075 bar + x6 \u00b7 h\u0192g at 0.075 bar = 2158.55 kJ/kg For actual expansion process in LP turbine. h6' = h4 \u2013 0.85 (h4 \u2013 h6) h6' = 2319.4 kJ/kg 294 _________________________________________________________ Applied Thermodynamics Ideally, enthalpy at bleed point can be obtained by locating state 5 using s5 = s4. The pressure at bleed point shall be saturation pressure corresponding to the 140\u00baC i.e. from steam table p5 = 3.61 bar. Let dryness fraction at state 5 be x5. s5 = 6.9212 = s\u0192 at 140\u00baC + x5 \u00b7 sfg at 140\u00baC h5 = h\u0192 at 140\u00baC + x5 \u00b7 h\u0192g at 140\u00baC \u21d2 h5 = 2712.38 kJ/kg \u21d2 x5 = 0.99 Actual exthalpy h5' = h4 \u2013 0.85 (h4 \u2013 h5) = 2790.16 kJ/kg Enthalpy at exit of open feed water heater, h9 = h\u0192 at 30 bar = 1008.42 kJ/kg Specific volume at inlet of CEP, v7 = 0.001008 m3/kg, Enthalpy at inlet of CEP, h7 = 168.79 kJ/kg For pumping process 7-8 Applying energy balance at open feed water heater. Let mass of bled steam be m kg per kg of steam generated. h8 = h7 + v7 (3.61 \u2013 0.075) \u00d7 102 h8 = 169.15 kJ/kg m \u00d7 h5' + (1 \u2013 m) \u00b7 h8 = h9 (m \u00d7 2790.16) + ((1 \u2013 m) \u00b7 169.15) = 1008.42, 5 m h'5 m = 0.32 kg/kg of steam generated For process on feed pump, 9 \u2013 1, v9 = v\u0192 at 140\u00baC = 0.00108 h1 = h9 + v9 \u00d7 (70 \u2013 3.61) \u00d7 102 h1 = 1015.59", " kJ/kg Net work per kg of steam generated, OFWH (1 \u2013 ),m h8 8 Wnet = (h2 \u2013 h3') + (h4 \u2013 h5') + (1 \u2013 m) \u00b7 (h5' \u2013 h6') \u2013 {(1 \u2013 m) (h8 \u2013 h7) + (h1 \u2013 h9)} = 181.1 + 440.74 + 320.117 \u2013 {0.2448 + 7.17} 1 kg, h9 9 Heat added per kg of steam generated, Wnet = 934.54 kJ/kg steam generated qadd = (h2 \u2013h1) + (h4 \u2013 h3') qadd = 2262.51 + 133.9 = 2396.41 kJ/kg of steam generated Thermal efficiency, \u03b7 = net W q add 934.54 2396.41 \u03b7 = 0.3899 or 38.99% = Thermal efficiency = 38.99% Ans. 11. A steam power plant works on regenerative cycle with steam entering first turbine stage at 150 bar, 500\u00baC and getting expanded in three subsequent stages upto the condenser pressure of 0.05 bar. Some steam is bled out between first and second stage for feed heating in closed feed water heater at 10 bar with the saturated liquid condensate being pumped ahead into the boiler feed water line. Feed water leaves closed feed water heater at 150 bar, 150\u00baC. Steam is also taken out between second and third stages at 1.5 bar for being fed into an open feed water heater working at that pressure. Saturated liquid at 1.5 bar leaves open feed water heater for being sent to closed feed water heater. Considering mass flow rate of 300 kg/s into the first stage of turbine determine cycle thermal efficiency and net power developed in kW. Also give lay out and T-s representation. Vapour Power Cycles ___________________________________________________________ 295 Solution: Enthalpy of steam entering ST1, h2 = 3308.6 kJ/kg, s2 = 6.3443 kJ/kg \u00b7 K For isentropic expansion 2-3-4-5, s2 = s3 = s4 = s5 Let dryness fraction of states 3, 4 and 5 be x3, x4 and x5 \u21d2 \u21d2 \u21d2 s3 = 6.3443 = s\u0192 at 10 bar + x3 \u00b7 s\u0192g at 10 bar x3 = 0.945 h3 = 2667.26 kJ/kg s4 = 6.3443 = s\u0192 at 1.5 bar + x4 \u00b7 s\u0192g at 1.5 bar x4 = 0.848 \u21d2 h4 = 2355.18 kJ/kg s5 = 6.3443 = s\u0192 at 0.05 bar + x5 \u00b7 s\u0192g at 0.05 bar x5 = 0.739 \u00b7 h5 = 1928.93 kJ/kg h6 = hf at 0.05 bar = 137.82 kJ/kg v6 = 0.001005 m3/kg = vf at 0.05 bar h7 = h6 + v6 (1.5 \u2013 0.05) \u00d7 102 = 137.96 kJ/kg 2 Boiler ST1 ST2 ST3 1 3 4 CFWH 11 12 FP2 9 10 OFWH FP1 Condenser 5 6 7 8 CEP 150 bar 2 CEP = Condensate extraction pump FP = Feed pump CFWH = Closed feed water heater OFWH = Open feed water heater T 1 12 11 9 10 8 7 6 10 bar 1.5 bar 0.05 bar 3 4 5 S Fig. 8.42 Layout and T-s diagram 296 _________________________________________________________ Applied Thermodynamics h8 = h\u0192 at 1.5 bar = 467.11 kJ/kg, v8 = 0.001053 m3/kg = v\u0192 at 1.5 bar h9 = h8 + v8(150 \u2013 1.5) \u00d7 102 = 482.75 kJ/kg h10 = h\u0192 at 150 bar = 1610.5 kJ/kg v10 = 0.001658 m3/kg = v\u0192 at 150 bar h12 = h10 + v10 (150 \u2013 10) \u00d7 102 = 1633.71 kJ/kg Let mass of steam bled out at 10 bar, 1.5 bar be m1 and m2 per kg of steam generated. Heat balance on closed feed water heater yields, m h1, 3 3 150 \u00b0C, 11 (1 \u2013 )m1 9 (1 \u2013 ),m h1 9, m h1 10 10 m1 \u00b7 h3 + (1 \u2013 m1) h9 = m1 \u00b7 h10 + (1 \u2013 m1)", " \u00d7 4.18 \u00d7 150 (m1 \u00d7 2667.26) + (1 \u2013 m1) \u00d7 482.75 = (m1 \u00d7 1610.5) + (627 \u00b7 (1 \u2013 m1)) m1 = 0.12 kg/kg of steam generated. Heat balance on open feed water can be given as under m2 \u00b7 h4 + (1 \u2013 m1 \u2013 m2) \u00b7 h7 = (1 \u2013 m1) \u00b7 h8 (m2 \u00d7 2355.18) + (1 \u2013 m1 \u2013 m2) \u00d7 137.96 = (1 \u2013 m1) \u00d7 467.11 (m2 \u00d7 2355.18) + (1 \u2013 0.12 \u2013 m2) \u00d7 137.96 = (1 \u2013 0.12) \u00d7 467.11 m2 = 0.13 kg/kg of steam, m h2 4 4 7 (1 \u2013 m m h ), 2 \u2013 1 7 For mass flow rate of 300 kg/s \u21d2 m1 = 36 kg/s, m2 = 39 kg/s For mixing after closed feed water heater, (1 \u2013 ), m h1 8 8 Net work output per kg of steam generated = h1 = (4.18 \u00d7 150) \u00b7 (1 \u2013 m1) + m1 \u00d7 h12 = 747.81 kJ/kg ST1w + ST3w \u2013 {wCEP + wFP + ST2w + w } FP2 wnet = (h2 \u2013 h3) + (1 \u2013 m1) (h3 \u2013 h4) + (1 \u2013 m1 \u2013 m2) (h4 \u2013 h5) \u2013 {(1 \u2013 m1 \u2013 m2) \u00b7 (h7 \u2013 h6) + (1 \u2013 m1) \u00b7 (h9 \u2013 h8) + (m1 \u00b7 (h12 \u2013 h10))} wnet = 641.34 + 274.63 + 319.69 = {0.105 + 13.76 + 2.7852} wnet = 1219.00 kJ/kg of steam generated. Heat added per kg of steam generated. qadd = (h2 \u2013 h1) = 2560.79 kJ/kg Cycle thermal efficiency, \u03b7 = net w q add = 0.4760 or 47.6% Net power developed in kW = 1219 \u00d7 300 = 365700 kW Cycle thermal efficiency = 47.6% Net power developed = 365700 kW Ans. Vapour Power Cycles ___________________________________________________________ 297 12. A steam power plant has expansion of steam leaving boiler at 100 bar, 500\u00baC occurring in three stages i.e. HPT, IPT and LPT (high pressure, intermediate pressure and low pressure turbine) upto condenser pressure of 0.075 bar. At exit of HPT some steam is extracted for feed heating in closed feed water heater at 20 bar and remaining is sent to IPT for subsequent expansion upto 4 bar. Some more quantity of steam is extracted at 4 bar for feed heating in open feed water heater and remaining steam is allowed to expand in low pressure turbine upto condenser pressure. Feed water leaves closed feed water heater at 100 bar and 200\u00baC. The condensate leaving as saturated liquid at 20 bar is trapped into open feed water heater. The state of liquid leaving open feed water heater may be considered saturated liquid at 4 bar. For a net power output of 100 MW determine thermal efficiency and steam generation rate in boiler. (a) Modify the above arrangement by introducing reheating of steam entering IPT at 20 bar upto 400\u00baC. Obtain thermal efficiency of modified cycle and compare it with non-reheat type arrangement. Solution: At inlet to HPT, h2 = 3373.7 kJ/kg, s2 = 6.5966 kJ/kg \u00b7 K For isentropic expansion between 2-3-4-5, s2 = s3 = s4 = s5 State 3 lies in superheated region as s3 > sg at 20 bar. By interpolation from superheated steam table, T3 = 261.6\u00baC. Enthalpy at 3. h3 = 2930.57 kJ/kg. Since s4 < sg at 4 bar so states 4 and 5 lie in wet region. Let dryness fraction at state 4 and 5 be x4 and x5. for state 5, s4 = 6.5966 = s\u0192 at 4 bar + x4 \u00b7 s\u0192g at 4 bar x4 = 0.941 h4 = h\u0192 at 4 bar + x4 \u00b7 h\u0192g at 4 bar = 2612.65 kJ/kg s5 = 6.5966 = s\u0192 at 0.075 bar + x5 \u00b7 sfg at 0.075 bar x5 = 0.784 100 bar, 500", "\u00b0C 2 Boiler 1 HPT 3 r a b 0 2 IPT LPT 4 bar 4 0.075 bar 5 100 bar, 200\u00b0C CFWH OFWH Condenser 6 7 CEP 8 FP 10 Trap 9 11 298 _________________________________________________________ Applied Thermodynamics Non reheat cycle : 1-2-3-4-5-6-7-8-9-10-11-1 Reheat cycle : 1-2-3-3'-4'-5'-6'-7-8-9-10-11-1 20 bar 3' 100 bar 2 3 4 bar 4' 0.075 bar 1 9 10 T 8 7 6 11 4 5 5' S Fig. 8.43 Layout and T-s diagram h5 = h\u0192 at 0.075 bar + x5 \u00b7 hfg at 0.075 bar = 2055.09 kJ/kg Let mass of steam bled at 20 bar, 4 bar be m1 and m2 per kg of steam generated respectively. h10 = h\u0192 at 20 bar = 908.79 kJ/kg, h8 = h\u0192 at 4 bar = 604.74 kJ/kg At trap h10 = h11 = 908.79 kJ/kg At condensate extraction pump, (CEP), h7 \u2013 h6 = v6 (4 \u2013 0.075) \u00d7 102 v6 = v\u0192 at 0.075 bar = 0.001008 m3/kg h6 = h\u0192 at 0.075 bar = 168.79 kJ/kg \u21d2 h7 = 169.18 kJ/kg At feed pump, (FP), h9 \u2013 h8 = v8 (20 \u2013 4) \u00d7 102 h8 = h\u0192 at 4 bar = 604.74 kJ/kg v8 = v\u0192 at 4 bar = 0.001084 m3/kg h9 = 604.74 + (0.001084 \u00d7 16 \u00d7 102) = 606.47 kJ/kg Let us apply heat balance at closed feed water heater, m1 \u00b7 h3 + h9 = m1 \u00b7 h10 + 4.18 \u00d7 200 (m1 \u00d7 2930.57) + 606.47 = (m1 \u00d7 908.79) + 836 m1 = 0.114 kg m1 3 200 \u00b0C, 1 9 1 kg m1 10 Vapour Power Cycles ___________________________________________________________ 299 Applying heat balance at open feed water, m1 h11 + m2 \u00b7 h4 + (1 \u2013 m1 \u2013 m2) \u00b7 h7 = h8 (m1 \u00b7 908.79) + (m2 \u00d7 2612.65) + ((1 \u2013 m1 \u2013 m2) \u00b7 169.18) = 604.74, m2 = 0.144 kg, m h2 4 4 m h1, 11 11 7 (1 \u2013 1kg h8 8 Net work per kg steam generated, m m h ), 2 \u2013 1 7 wnet = (h2 \u2013 h3) + (1 \u2013 m1) \u00b7 (h3 \u2013 h4) + (1 \u2013 m1 \u2013 m2) (h4 \u2013 h5) \u2013 {(1 \u2013 m1 \u2013 m2) \u00b7 (h7 \u2013 h6) + (h9 \u2013 h8)} = 443.13 + 281.67 + 413.7 \u2013 {0.288 + 1.73} Heat added per kg steam generated, wnet = 1136.48 kJ/kg qadd = (h2 \u2013 h1) = (3373.7 \u2013 4.18 \u00d7 200) = 2537.7 kJ/kg Thermal efficiency = Steam generation rate = net w q add = 0.4478 or 44.78% 3 \u00d7 100 10 w net = 87.99 kg/s Thermal efficiency = 44.78% Steam generation rate = 87.99 kg/s Ans. (a) For the reheating introduced at 20 bar up to 400\u00baC: The modified cycle representation is shown on T-s diagram by 1-2-3-3'-4'-5'-6-7-8-9-10-11 Boiler 1 100 bar, 500\u00b0C 2 HPT IPT LPT 3 3' 4' 20 bar, 400\u00b0C CFWH OFWH 11 9 7 5' 6 10 Trap 8 FP CEP Fig. 8.44 Reheat cycle 300 _________________________________________________________ Applied Thermodynamics At state 2, At state 3, At state 3', h2 = 3373.7 kJ/kg, s2 = 6.5966 kJ/kg \u00b7 K h3 = 2930.57 kJ/kg h3' = 3247.6 kJ/kg s3' = 7.1271 kJ/kg \u00b7 K At state 4' and 5'", " s3' = s4' = s5' = 7.1271 kJ/kg \u00b7 K From steam tables by interpolation state 4' is seen to be at 190.96\u00baC at 4 bar Let dryness fraction at state 5' be x5'. h4' = 2841.02 kJ/kg \u21d2 s5' = 7.1271 = s\u0192 at 0.075 bar + x5' \u00b7 s\u0192g at 0.075 bar x5' = 0.853 h5' = h\u0192 at 0.075 bar + x5' \u00b7 h\u0192g at 0.075 bar h5' = 2221.11 kJ/kg Let mass of bled steam at 20 bar and 4 bar be m1', m2' per kg of steam generated. Applying heat balance at closed feed water heater. \u21d2 m1' \u00b7 h3 + h9 = m1' \u00b7 h10 + 4.18 \u00d7 200 m1' = 0.114 kg 1 9 3 10 4\u2032 8 7 11 Applying heat balance at open feed water heater (0.114 \u00d7 908.79) + (m2' \u00b7 2841.02) + (1 \u2013 0.114 \u2013 m2') \u00b7 169.18 = 604.74 m1' \u00b7 h11 + m2' \u00b7 h4' + (1 \u2013 m1' \u2013 m2') \u00b7 h7 = h8 m2' = 0.131 kg Net work per kg steam generated wnet = (h2 \u2013 h3) + (1 \u2013 m1') \u00b7 (h3' \u2013 h4') + (1 \u2013 m1' \u2013 m2') (h4' \u2013 h5') \u2013 {(1 \u2013 m1' \u2013 m2') \u00b7 (h7 \u2013 h6) + (h9 \u2013 h8)} wnet = 443.13 + 360.22 + 468.03 \u2013 {0.293 + 1.73} wnet = 1269.36 kJ/kg Heat added per kg steam generated, qadd = (h2 \u2013 h1) + (1 \u2013 m1') (h3' \u2013 h3) = 2537.7 + 280.88 qadd = 2818.58 kJ/kg Vapour Power Cycles ___________________________________________________________ 301 Thermal efficiency \u03b7 = netw q add = 0.4503 or 45.03% % Increase in thermal efficiency due to reheating = \uf8eb \uf8ec \uf8ed \u2212 0.4503 0.4478 0.4478 \uf8f6 \uf8f7 \uf8f8 \u00d7 100 = 0.56% Thermal efficiency of reheat cycle = 45.03% % Increase in efficiency due to reheating = 0.56% Ans. 13. In a binary vapour cycle working on mercury and steam, the mercury vapour is generated dry saturated at 8.45 bar and expanded upto 0.07 bar in mercury turbine. The condenser or mercury cycle is used for generating steam at 40 bar, 0.98 dry. The steam is superheated separately upto 450\u00baC and then supplied into steam turbine for being expanded upto 0.075 bar. Two closed feed water heaters are used by bleeding out steam at 8 bar and 1 bar so as to provide feed water leaving at 150\u00baC and 90\u00baC respectively. Condensate leaves feed water heater as saturated liquid at respective pressures and is mixed with the hot feed water leaving the respective feed heater. The turbine running on mercury has capability of converting 85% of available heat into work. The enthalpies of mercury may be taken as, enthalpy of dry saturated vapour at 8.45 bar = 349 kJ/kg enthalpy after isentropic expansion to 0.07 bar = 234.5 kJ/kg enthalpy of saturated liquid at 0.07 bar = 35 kJ/kg Assume feed water to enter at 150\u00baC into mercury condenser. Neglect pump work for getting efficiency. Determine the steam generation rate per kg of mercury and efficiency of cycle. Solution: For mercury cycle, Insentropic heat drop = 349 \u2013 234.5 = 114.5 kJ/kg Hg Actual heat drop = 0.85 \u00d7 114.5 = 97.325 kJ/kg Hg Heat rejected in condenser = [349 \u2013 97.325 \u2013 35] Heat added in boiler = 349 \u2013 35 = 314 kJ/kg = 216.675 kJ/kg For steam cycle, Enthalpy of steam generated = hat 40 bar, 0.98 dry = 2767.13 kJ/kg Enthalpy of steam at inlet to steam turbine h2 = h at 40 bar, 450\u00baC = 3330.3 kJ/kg Entropy of steam at inlet to steam turbine, s2 = 6.9363 kJ", "/kg \u00b7 K Therefore, heat added in condenser of mercury cycle = hat 40 bar, 0.98 dry \u2013 hfeed at 40 bar = 2767.13 \u2013 4.18 \u00d7 150 = 2140.13 kJ/kg Therefore, mercury required per kg of steam = 2140.13 Heat rejected in condenser = For isentropic expansion, s2 = s3 = s4 = s5 = 6.9363 kJ/kg \u00b7 K = 9.88 kg per kg of steam 2140.13 216.675 State 3 lies in superheated region, by interpolation the state can be given by, temperature 227.07\u00baC at 8 bar, h3 = 2899.23 kJ/kg 302 _________________________________________________________ Applied Thermodynamics 8.45 bar a m1 m2 CFWH1 15 14 CFWH2 12 11 10 13 9 FP1 FP2 Regenerative arrangement 7 6 T 0.07 bar 1 14 11 10 13 12 8 7 9 6 steam 40 bar b b\u2032 2 8 bar 1 bar 0.075 bar 3 4 5 m1 m2 S Fig. 8.45 T-s diagram showing expansion and bleeding State 4 lies in wet region, say with dryness fraction x4, s4 = 6.9363 = s\u0192 at 1 bar + x4 \u00d7 s\u0192g at 1 bar h4 = h\u0192 at 1 bar + x4 \u00b7 hfg at 1 bar = 2517.4 kJ/kg \u21d2 x4 = 0.93 Let state 5 lie in wet region with dryness fraction x5, s5 = 6.9363 = s\u0192 at 0.075 bar + x5 \u00b7 sfg at 0.075 bar x5 = 0.828 h5 = 2160.958 kJ/kg Let mass of steam bled at 8 bar and 1 bar be m1 and m2 per kg of steam generated. h7 = h6 + v6(1 \u2013 0.075) \u00d7 102 = h\u0192 at 0.075 bar + v\u0192 at 0.075 bar (1 \u2013 0.075) \u00d7 102 = 168.79 + 0.001008 \u00d7 (1 \u2013 0.075) \u00d7 102 h7 = 168.88 kJ/kg h9 = h\u0192 at 1 bar = 417.46 kJ/kg h13 = h\u0192 at 8 bar = 721.11 kJ/kg Applying heat balance on CFWH1, T1 = 150\u00baC and also T15 = 150\u00baC m1 \u00d7 h3 + (1 \u2013 m1) \u00d7 h12 = m1 \u00d7 h13 + (4.18 \u00d7 150) \u00d7 (1 \u2013 m1) (m1 \u00d7 2899.23) + (1\u2013 m1) \u00d7 h12 = (m1 \u00d7 721.11) + 627 \u00b7 (1 \u2013 m1) Applying heat balance on CFWH2, T11 = 90\u00baC m2 \u00d7 h4 + (1 \u2013 m1 \u2013 m2) \u00d7 h7 = m2 \u00d7 h9 + (1 \u2013 m1 \u2013 m2) \u00d7 4.18 \u00d7 90 (m2 \u00d7 2517.4) + (1 \u2013 m1 \u2013 m2) \u00d7 168.88 = (m2 \u00d7 417.46) + 376.2 (1 \u2013 m1 \u2013 m2) Heat balance at mixing between CFWH1 and CFWH2, (1 \u2013 m1 \u2013 m2) \u00d7 4.18 \u00d7 90 + m2 \u00d7 h10 = (1 \u2013 m1) \u00d7 h12 376.2 (1 \u2013 m1 \u2013 m2) + m2 \u00d7 h10 = (1 \u2013 m1) \u00d7 h12 For pumping process 9\u201310, h10 = h9 + v9(8 \u20131) \u00d7 102 Vapour Power Cycles ___________________________________________________________ 303 Solving above equations, we get h10 = h\u0192 at bar + v\u0192 at 1 bar (7 \u00d7 102) h10 = 417.46 + 0.001043 \u00d7 700 = 418.19 kJ/kg m1 = 0.102 kg per kg steam generated m2 = 0.073 kg per kg steam generated. Pump work in process 13\u201314, h14 \u2013 h13 = v13 \u00d7 (40 \u2013 8) \u00d7 102 = 0.001252 \u00d7 32 \u00d7 102 h14 \u2013 h13 = 4.006 kJ/kg Total heat supplied = (9.88 \u00d7 314) + (3330.3 \u2013 2767.13) = 3665.49 kJ/kg of steam Net work per kg of steam, wnet = wmercury + wsteam = {9.88 \u00d7 97.325} + {(h2 \u2013 h3) + (1 \u2013 m1) \u00b7 (h3 \u2013 h4) + (1 \u2013", " m1 \u2013 m2) \u00b7 (h4 \u2013 h5) \u2013 (1 \u2013 m1 \u2013 m2) (h4 \u2013 h6) \u2013 m2 (h10 \u2013 h9) \u2013 m1 (h14 \u2013 h13)} = {961.571} + {431.07 + 342.88 + 294.06 \u2013 0.074 \u2013 0.053 \u2013 0.408} wnet = 2029.046 kJ/kg Thermal efficiency of binary vapour cycle = 2029.046 3665.49 = 0.5536 or 55.36% Thermal efficiency = 55.36% Ans. 14. A steam power plant has mixed pressure turbine of output 2500 hp with high pressure steam entering at 20 bar, 300\u00baC and low pressure steam entering at 2 bar and dry saturated. The steam leaves turbine at 0.075 bar and efficiency ratio of both HP and LP stages are 0.8. The Willan\u2019s line for both are straight line and steam consumption at no load is 10% of full load. Determine the HP steam required for producing 1000 hp if low pressure steam is available at the rate of 1.5 kg/s. Solution: This is a mixed pressure turbine so the output of turbine shall be sum of the contributions by HP and LP steam streams. For HP: At Inlet of HP steam, \u21d2 h1 = 3023.5 kJ/kg, s1 = 6.7664 kJ/kg \u00b7 K Ideally, s2 = s1 = 6.7664 s2 = 6.7664 = s\u0192 at 0.075 bar + x3 \u00d7 s\u0192g at 0.075 bar x3 = 0.806 \u21d2 20 bar, 300\u00b0C 2 bar, dry saturated 2 h3HP = h\u0192 at 0.075 bar + x3 \u00b7 hfg at 0.075 bar = 2108.03 kJ/kg 1 Actual enthalpy drop in HP = (h1 \u2013 h3HP) \u00d7 0.8 = 732.38 kJ/kg For LP: At inlet of LP steam h2 = 2706.7 kJ/kg, s2 = 7.1271 kJ/kg \u00b7 K Enthalpy at exit, h3LP = 2222.34 kJ/kg Actual enthalpy drop in LP = (h1 \u2013 h3LP) \u00d7 0.8 = 387.49 kJ/kg Mixed ST 0.075 bar 3 304 _________________________________________________________ Applied Thermodynamics HP steam consumption at full load = \u00d7 2500 0.7457 732.38 = 2.54 kg/s HP steam consumption at no load = 0.1 \u00d7 2.54 = 0.254 kg/s LP steam consumption at full load = \u00d7 2500 0.7457 387.49 = 4.81 kg/s LP steam consumption at no load = 0.1 \u00d7 4.81 = 0.481 kg/s The problem can be solved geometrically by drawing Willan\u2019s line as per scale on graph paper and finding out the HP stream requirement for getting 1000 hp if LP steam is available at 1.5 kg/s. or, Analytically the equation for Willan\u2019s line can be obtained for above full load and no load conditions for HP and LP separately. Willan's line for HP: y = mx + C, here y = steam consumption, kg/s x = load, hp yHP = mHP \u00b7 x + CHP 2.54 = mHP \u00b7 2500 + CHP 0.254 = mHP \u00b7 0 + CHP \u21d2 CHP = 0.254 and mHP = 9.144 \u00d7 10\u20134 \u21d2 yHP = 9.144 \u00d7 10\u20134 \u00b7 xHP + 0.254 Willan's line for LP: yLP = mLP \u00b7 xLP + CLP 4.81 = mLP \u00b7 2500 + CLP 0.481 = mLP \u00b7 CLP \u21d2 CLP = 0.481, mLP = 1.732 \u00d7 10\u20133 yLP = 1.732 \u00d7 10\u20133 \u00b7 xLP + 0.481 Total output (load) from mixed turbine, x = xHP + xLP For load of 1000 hp to be met by mixed turbine, let us find out the load shared by LP for steam flow rate of 1.5 kg/s y kg/s 0 LP HP x load, hp Fig. 8.46 Tentative representation of Willan\u2019s line 1.5 = 1.732 \u00d7 10\u20133 \u00b7 xLP + 0.481 xLP = 588.34 hp Since by 1.5 kg/s of LP steam only 588.34 hp output contribution is made so remaining (1000 \u2013 588.34 = 411.66 hp), 411.66 hp should be contributed by HP steam. By Willan\u2019s line", " for HP turbine, So, HP steam requirement = 0.63 kg/s yHP = (9.144 \u00d7 10\u20134 \u00d7 411.66) + 0.254 = 0.63 kg/s HP steam required = 0.63 kg/s Ans. 15. A steam power plant installation has steam leaving boiler at 40 bar, 300\u00baC and expanding in HP turbine upto 2 bar. Half of steam leaving HP turbine is sent for process heating and remaining enters a separator where all the moisture is removed. Dry steam from separator is sent to low pressure LP turbine at 2 bar and gets expanded upto 0.075 bar. The drain (moisture) of separator gets mixed with condensate from process heater and combined flow enters the hot well at 90\u00baC. Trap is provided at exit of both process heater and separator. Condensate extraction pump extracts condensate from condenser and sends it to hot well at 40\u00baC. Neglecting pump work and radiation losses etc. determine temperature of water leaving hotwell and heat transferred per kg in process heater. Also find out thermal efficiency of installation and give layout. Vapour Power Cycles ___________________________________________________________ 305 Solution: Let us carry out analysis for 1 kg of steam generated in boiler. Enthalpy at inlet to HPT, h2 = 2960.7 kJ/kg, s2 = 6.3615 kJ/kg \u00b7 K State at 3 i.e. exit from HPT can be identified by s2 = s3 = 6.3615 kJ/kg \u00b7 K Let dryness fraction be x3, s3 = 6.3615 = s\u0192 at 2 bar + x3 \u00b7 s\u0192g at 2 bar \u21d2 x3 = 0.863 h3 = 2404.94 kJ/kg If one kg of steam is generated in boiler then at exit of HPT, 0.5 kg goes into process heater and 0.5 kg goes into separator Boiler 1 40 bar, 300\u00b0C 2 HPT IPT 2 bar Process heater 4 Separator 7 3 5 9 10 Trap 6 Condenser Hotwell 8 11 12 CEP 13 FP Fig. 8.47 Layout Mass of moisture retained in separator = (1 \u2013 0.863) \u00d7 0.5 = 0.0685 kg Therefore, mass of steam entering LPT = 0.5 \u2013 0.0685 = 0.4315 kg Total mass of water entering hot well at 8 (i.e. from process heater and drain from separator) = 0.5685 kg Let us assume the temperature of water leaving hotwell be T\u00baC. Applying heat balance for mixing; (0.5685 \u00d7 4.18 \u00d7 90) + (0.4315 \u00d7 4.18 \u00d7 40) = (1 \u00d7 4.18 \u00d7 T), T = 68.425\u00baC Temperature of water leaving hotwell = 68.425\u00baC Ans. Applying heat balanced on trap 0.5 \u00d7 h7 + 0.0685 \u00d7 h\u0192 at 2 bar = (0.5685 \u00d7 4.18 \u00d7 90) h7 = 358.59 kJ/kg Therefore, heat transferred in process heater = 0.5 \u00d7 (h3 \u2013 h7) 306 _________________________________________________________ Applied Thermodynamics Heat transferred per kg steam generated = 1023.175 kJ/kg steam generated Ans. = 1023.175 kJ/kg steam generated For state 10 at exit of LPT, s10 = s3 = s2 = 6.3615 kJ/kg \u00b7 K Let dryness fraction be x10 \u21d2 \u21d2 s10 = 6.3615 = s\u0192 at 0.075 bar + x10 \u00b7 s\u0192g at 0.075 bar x10 = 0.754 h10 = hf at 0.075 bar + x10 \u00b7 hfg at 0.075 bar h10 = 1982.91 Net work output, neglecting pump work per kg of steam generated, Heat added in boiler per kg steam generated, qadd = (h2 \u2013 h1) wnet = (h2 \u2013 h3) \u00d7 1 + 0.4315 \u00d7 (h3 \u2013 h10) = 555.76 + 182.11 wnet = 737.87 kJ/kg steam generated = (2960.7 \u2013 4.18 \u00d7 68.425) qadd = 2674.68 kJ/kg Thermal efficiency = net w q add = 0.2758 or 27.58% Thermal efficiency = 27.58% Ans. 16. In a steam power plant operating on Rankine cycle, the steam enters the turbine at 70 bar and 550\u00baC with a velocity of 30 m/s. It discharges to the condenser at 0.20 bar with a velocity 90 m/s. If the steam [U.P.S.C. 1992] flow rate", " is 35 kg/s, find the thermal efficiency and net power produced. Solution: From steam tables, h1 = 3530.9 kJ/kg, s1 = 6.9486 kJ/kg \u00b7 K Assuming isentropic expansion in nozzle, s1 = s2 = 6.9486 kJ/kg \u00b7 K 70 bar, 550\u00b0C, 35 kg/s Boiler 1 Steam Turbine 0.20 bar 2 Condenser T 3 550\u00b0 4 3 70 bar 1 2 0.02 bar 4 (a) Condensate extraction and feed pump (b) s Fig. 8.48 Schematic and representation on T-s diagram Vapour Power Cycles ___________________________________________________________ 307 Let dryness fraction at state 2 be x2 then; s2 = s\u0192 at 0.2 bar + x2 \u00d7 s\u0192g at 0.2 bar 6.9486 = 0.8320 + x2 \u00b7 (7.0766) Dryness fraction at state 2, x2 = 0.864 Hence, h2 = hf at 0.2 bar + x2\u00b7 hfg at 0.2 bar = 251.40 + (0.864 \u00d7 2358.3) h2 = 2288.97 kJ/kg Considering pump work to be of isentropic type, \u2206h34 = v3 \u00d7 \u2206p34 From steam stable, or v3 = v\u0192 at 0.2 bar = 0.001017 m3 /kg \u2206h34 = 0.001017 \u00d7 (70 0.20) 10 \u00d7 \u2212 5 3 10 Pump work, WP = \u2206h34 = 7.099 kJ/kg Turbine work, WT = \u2206h12 = (h1 \u2013 h2) = (3530.9 \u2013 2288.97) WT = 1241.93 kJ/kg Net work = WT \u2013 WP = 1241.93 \u2013 7.099 Wnet = 1234.831 kJ/kg Power produced = mass flow rate \u00d7 Wnet = 35 \u00d7 1234.831 = 43219.085 kJ/s Net Power = 43.22 MW Ans. Heat supplied in boiler = (h1 \u2013 h4), kJ/kg Enthalpy at state 4, h4 = h3 + \u2206h34 = h\u0192 at 0.2 bar + \u2206h34 = 251.40 + 7.099 Total Heat supplied to boiler = 35 \u00d7 (3530.9 \u2013 258.49) h4 = 258.49 kJ/kg = 114534.35 kJ/s Thermal efficiency = Net work Heat supplied = 0.3773 Thermal efficiency = 37.73% Ans. 17. The following data refers to a steam turbine power plant employing one stage of regenerative feed heating: State of steam entering HP stage : 10 MPa, 600\u00baC State of steam entering LP stage: 2 MPa, 400\u00baC Condenser pressure: 10 KPa 308 _________________________________________________________ Applied Thermodynamics The correct amount of steam is bled for feed heating at exit from the HP stage. Calculate the mass of steam bled per kg of steam passing through the HP stage and the amount of heat supplied in the boiler [U.P.S.C. 1993] per second for an output of 10 MW. Neglect pump work, Solution: 1 Boiler HPT LPT 2 7 6 5 Feed pump Condenser T 3 4 CEP 7 5 6 4 S Fig. 8.49 Possible arrangement and T-s representation 10 MPa 1 2 MPa 2 10 kPa 3 From steam tables: h1 = 3625.3 kJ/kg, s1 = 6.9029 kJ/kg \u00b7 K Due to isentropic expansion, s1 = s2 = s3 = 6.9029 kJ/kg \u00b7 K At state 2, i.e. at pressure of 2 MPa and entropy 6.9029 kJ/kg \u00b7 K. By interpolating state for s2 between 2 MPa, 300\u00baC and 2 MPa, 350\u00baC from steam tables, h2 = 3105.08 kJ/kg For state 3, i.e. at pressure of 0.01 MPa entropy, s3 lies in wet region as s3 < sg at 0.01 MPa. Let dryness fraction be x3 at this state s3 = s\u0192 at 0.01 MPa + x3 \u00b7 s\u0192g at 0.01 MPa 6.9029 = 0.6493 + x3 \u00d7 7.5009 Enthalpy at state 3, x3 = 0.834 h3 = hf at 0.01 MPa + x3 \u00b7 h\u0192g at 0.01 MPa = 191.83 + (0.834 \u00d7", " 2392.8) h3 = 2187.43 kJ/kg Let the mass of steam bled be mb per kg of steam from exit of HP for regenerative feed heating. Considering state at exit from feed heater being saturated liquid the enthalpy at exit of feed heater will be, h\u0192 at 2 MPa. For adiabatic mixing in feed heater, for one kg of steam leaving boiler, the heat balance yields, h6 = h\u0192 at 2 MPa = 908.79 kJ/kg While neglecting pump work, (1 \u2013 mb) \u00b7 h5 + mb \u00b7 h2 = h6 Substituting in heat balance on the feed heater, h5 = h4 = h\u0192 at 0.01 MPa = 191.83 kJ/kg (1 \u2013 mb) \u00b7 191.83 + mb \u00b7 3105.08 = 908.79 Vapour Power Cycles ___________________________________________________________ 309 mb = 0.246 kg per kg of steam entering HP turbine Steam bled per kg of steam passing through HP stage = 0.246 kg Ans. Let mass of steam leaving boiler be m kg/s. Output = 10 \u00d7 103 = m(h1 \u2013 h2) + m(1 \u2013 mb) (h2 \u2013 h3) 10 \u00d7 103 = m{(3625.3 \u2013 3105.08) + (1 \u2013 0.246) (3105.08 \u2013 2187.43)} Neglecting pump work, Heat supplied to boiler, m = 8.25 kg/s h7 = h6 = 908.79 kJ/kg Q7\u20131 = m(h1 \u2013 h7) Q7\u20131 = 8.25 (3625.3 \u2013 908.79) = 22411.208 kJ/s Heat added = 22411.21 kJ/s Ans. 18. Steam enters the first stage of a turbine at 100 bar, 500\u00baC and expands isentropically to 10 bar. It is then reheated to 500\u00baC and expanded in the second stage to the condenser pressure of 0.1 bar. Steam is bled from the first stage at 20 bar and fed to a closed feed water heater. Feed water leaves the closed heater at 100 bar, 200\u00baC (enthalpy = 856.8 kJ/kg), while the condensate is supplied to the open heater into which steam is bled at 4 bar pressure. Saturated liquid at 4 bar exits from the open heater and enters the closed heater. The net output of the turbine is 50 MW. Assuming the turbine and pump processes to be isentropic, determine the mass of steam bled at each feed water heater per kg of steam entering the first stage, the mass of steam entering the first stage per second, and the thermal efficiency. [U.P.S.C. 1995] Solution: From steam table, at inlet to first stage of turbine, h1 = hat 100 bar, 500\u00baC = 3373.7 kJ/kg s1 = sat 100 bar, 500\u00baC = 6.5966 kJ/kg \u00b7 K 1 kg 2 1 )m6 (1 \u2013 3 Boiler 1 kg m6 HPT LPT 6 8 m8 m m6 \u2013 (1 \u2013 4 ) 8 Condenser CFWH 7' 5 4' 11 m6 10 7 OFWH CEP 9 1 kg FP 500 \u00b0C 11 10 100 bar 7 20 bar 10 bar T 5 9 4' 4 bar 0.1 bar 7' S 1 6 2 3 8 4 Fig. 8.50 Arrangement and T-s representation 310 _________________________________________________________ Applied Thermodynamics Due to isentropic expansion, s1 = s6 = s2 and s3 = s8 = s4 State at 6 i.e. bleed state from HP turbine, Temperature by interpolation from steam table = 261.6\u00baC At inlet to second stage of turbine, h6 = 2930.572 kJ/kg h3 = hat 10 bar, 500\u00baC = 3478.5 kJ/kg s3 = sat 10 bar, 500\u00baC = 7.7622 kJ/kg \u00b7 K At exit from first stage of turbine i.e. at 10 bar and entropy of 6.5966 kJ/kg \u00b7 K Temperature by interpolation from steam table at 10 bar and entropy of 6.5966 kJ/kg \u00b7 K T2 = 181.8\u00baC h2 = 2782.8 kJ/kg h8 = 3188.7 kJ/kg State at 8, i.e. bleed state from second stage of expansion, i.e. at 4 bar and entropy of 7.7622 kJ/kg \u00b7 K Temperature by interpolation from steam table, T8 = 358.98\u00baC 359\u00baC State at", " 4 i.e. at condenser pressure of 0.1 bar and entropy of 7.7622 kJ/kg. K the state lies in wet region. So let the dryness fraction be x4. s4 = s\u0192 at 0.1 bar + x4 \u00b7 s\u0192g at 0.1 bar 7.7622 = 0.6493 + x4 \u00b7 7.5009 x4 = 0.95 h4 = h\u0192 at 0.1 bar + x4 \u00b7 h\u0192g at 0.1 bar = 191.83 + (0.95 \u00d7 2392.8) h4 = 2464.99 kJ/kg h11 = 856.8 kJ/kg, h9 = h\u0192 at 4 bar h9 = 604.74 kJ/kg Given, Considering pump work, the net output can be given as, where Wnet = WHPT + WLPT \u2013 (WCEP + WFP) WHPT = {(h1 \u2013 h6) + (1 \u2013 m6) (h6 \u2013 h2)} per kg of steam from boiler. WLPT = {(1 \u2013 m6) + (h3 \u2013 h8) (1 \u2013 m6 \u2013 m8) (h8 \u2013 h4)} per kg of steam from boiler. For closed feed water heater, energy balance yields; m6 \u00b7 h6 + h10 = m6 \u00b7 h7 + h11 Assuming condensate leaving closed feed water heater to be saturated liquid, h7 = h\u0192 at 20 bar = 908.79 kJ/kg h7 = h7' = 908.79 kJ/kg Due to throttline, For open feed water heater, energy balance yields, m6 \u00b7 h7' + m8 \u00b7 h8 + (1 \u2013 m6 \u2013 m8) \u00b7 h5 = h9 For condensate extraction pump, h5 \u2013 h4' = v4'\u00b7 \u2206p h5 \u2013 h\u0192 at 0.1 bar = v\u0192 at 0.1 bar \u00b7 (4 \u2013 0.1) \u00d7 102 h5 \u2013 191.83 = (0.001010) \u00d7 (3.9 \u00d7 102) h5 = 192.224 kJ/kg For feed pump, h10 \u2013 h9 = v9 \u00b7 \u2206p h10 \u2013 604.74 = v\u0192 at 4 bar \u00d7 (100 \u2013 4) \u00d7 102 h10 \u2013 604.74 = 0.001084 \u00d7 96 \u00d7 102 h10 = 615.15 kJ/kg Vapour Power Cycles ___________________________________________________________ 311 Substituting in energy balance upon closed feed water heater, m6 \u00d7 2701.2 + 615.15 = m6 \u00d7 908.79 + 856.8 m6 = 0.135 kg per kg of steam from boiler. Substituting in energy balance upon feed water heater, m6 \u00b7 h7' + m8 \u00b7 h8 + (1 \u00b7 m6 \u2013 m8) \u00b7 h5 = h9 (0.135 \u00d7 908.79) + (m8 \u00d7 3188.7) + (1 \u2013 0.135 \u2013 m8) \u00d7 192.224 = 604.74 m8 = 0.105 kg per kg of steam from boiler Let mass of steam entering first stage of turbine be m kg, then WHPT = m {(h1 \u2013 h6) + (1 \u2013 m6) (h6 \u2013 h2)} = m {(3373.7 \u2013 2930.572) + (1 \u2013 0.135)} (2930.572 \u2013 2782.8) WHPT = m{570.95}, kJ Also, WLPT = {(1\u2013 m6) (h3 \u2013 h8) + (1 \u2013 m6 \u2013 m8) \u00b7 (h8 \u2013 h4)}, per kg of steam from boiler WLPT = {(1 \u2013 0.135) (3478.5 \u2013 3188.7) + (1 \u2013 0.135 \u2013 0.105) \u00b7 (3188.7 \u2013 2464.99)} WLPT = m {800.69} kJ Pump works (negative work) WCEP = m \u00b7 (1 \u2013 m6 \u2013 m8) (h5 \u2013 h4') = m \u00b7 (1 \u2013 0.135 \u2013 0.105) \u00b7 (192.224 \u2013 191.83) WCEP = {0.299 m} WFP = m {h10 \u2013 h9} = m {615.45 \u2013 604.74} WFP = {10.71 m} Net output Wnet = WHPT + WLPT \u2013 WCEP \u2013 WFP 50 \u00d7 103 = {570.95 m + 800.69 m \u2013 0.299 m \u2013 10.71 m}", " m = 36.75 kg/s Heat supplied in boiler, Qadd = m(h1 \u2013 h11) = 36.75 (3373.7 \u2013 856.8) = 92496.075 kJ/s Thermal efficiency = W Q net add = 3 \u00d7 50 10 92496.075 = 0.54056 or 54.06% Mass of steam bled at 20 bar = 0.135 kg per kg of steam entering first stage Mass of steam bled at 4 bar = 0.105 kg per kg of steam entering first stage Mass of steam entering first stage = 36.75 kg/s Thermal efficiency = 54.06% Ans. 312 _________________________________________________________ Applied Thermodynamics 19. A steam power plant installation has reheating and regenerative feed water heating employing a surface type feed heater and other contact type feed heater on high pressure side and low pressure side respectively. Steam enters HP turbine at 100 bar, 803 K and leaves high pressure turbine at 25 bar from where some steam is bled for feed heating in high pressure surface type heater and remaining is reheated up to 823 K and then expanded in low pressure turbine up to 0.05 bar pressure. The contact type feed heater is supplied with steam bled at 6 bar from LP steam turbine. There occurs throttling pressure loss of 3 bar in reheater. Surface type feed heater sends the drain to contact type feed heater from where the total feed is sent to surface type feed heater employing a boiler feed water pump as saturated water at pressure of 100 bar. Determine the amounts of steam bled off, overall thermal efficiency and specific steam consumption in kg/kwh. Considering tubring efficiency pump efficiency, generator efficiency, and mechanical efficiency as 0.85, 0.90 & 0.95 respectively and plant output as 120 MW. Consider discharges of drains at saturated liquid state at respective pressures in feed heaters. Also show how the processes on T-s and h-s diagrams along with line sketch of arrangement. Solution: From steam table, At 100 bar, 803 K the state of inlet steam At inlet to LP steam turbine at 22 bar, 823K h1 = 3450.02 kJ/kg, s1 = 6.6923 kJ/kgK h3 = 3576.99 kJ/kg, s3 = 7.52411 kJ/kg.K For exit from HP turbine, s1 = s2 Using Mollier diagram Boiler 100 bar, 803 K 12 h2 = 3010, kJ/kg.K. 22 bar, 823 K 2 1 HP Turbine 3 LP Turbine 120 MW 4, 0.05bar 25 bar 7 is similar to 2 9 6 bar B.F.P. 11 8 Condenser 5 C.E.P. 6 Contact type feed heater 10 Surface type feed heater BFP : Boiler feed pump CEP : Condensate extraction pump Vapour Power Cycles ___________________________________________________________ 313 1kg 2 1'1 1 1 1 m kg1 8 T m kg2 (' 10 5 S 100 bar 1 2'7' 2 7 9 4 25 bar 22 bar 6 bar 0.05 bar h 3 9' 4' 100 bar 1 2, 7 2'7' 25 bar 3 9 4' 4 m2 (1 \u2013 m \u2013 m ) 2 1 6 bar 0.05 bar m1 S Fig. 8.51 Let us consider 1 kg of steam generated in boiler and bled fractions be m1 & m2. Considering turbine efficiency, \u03b7 h turb = 1 h 1 \u2032\u2212 h 2 \u2212 h 2 0.85 = h \u2032\u2212 3450.02 \u2212 (3450.02 3010) 2 \u21d2 \u2032 = 3076.003 kJ/kg h2 From Mollier diagram, considering isentropic expansion in LP turbine Considering turbine efficiency, 0.85 = h9 = 3175 kJ/kg h4 = 2300kJ/kg \u2032\u2212 h 9 \u2212 h 9 h 3 h 3 \u21d2 Also \u21d2 From steam table 0.85 = h \u2032\u2212 ) (3576.99 \u2212 (3576.99 3175) 9 \u2032 = 3235.29 kJ/kg h9 0.85 = \u2212 \u2212 \u2032 3576.99 ) \u2212 (3576.99 2300) \u2032 h 4 \u2032 = 2491.5 kJ/kg h4 h5 = hf at 0.05 bar = 137.82 kJ/kg, v s = vf at 0.05 bar = 0.001005 m3/kg 314 _________________________________________________________ Applied Thermodynamics h8 = hf at 25 bar = 962.11 kJ/kg, h12 = hf at 100 bar = 1407.56 kJ/kg, Energy balance at surface type feed heater yields; m1 \u00d7 h7 +", " h11 \u2032 = h12 + m1h8 \u21d2 m1 = ( \u2032 \u2212 \u2212 h 11 ( h 8 ) h 12 ) h 7 In this arrangement, h2 = h7 \u21d2 m1 = Energy balance at contact type heater yields m1 \u00d7 h8 + m2 \u00d7 h9 + (1 \u2013 m1 \u2013 m2)h6 for process 5\u20136 in condensate extraction pump, \u2032 = 1 \u00d7 h10 \u2032 \u2212 \u2212 ( h 11 h 8 ( ) h 12 \u2032 ) h 2 h6 \u2013 h5 = v5. \u2206p = v5 \u00d7 (p6 \u2013 p5) h6 \u2013 137.82 = 0.001005 \u00d7 (6 \u2013 0.05) \u00d7 102 \u21d2 Considering pump efficiency \u03b7 h6 = 138.42 kJ/kg \u2212 pump = ) ( ( ) \u2212 (138.42 137.82) 0.90 = ( \u2032 \u2212 h 6 \u2032 = 138.49 kJ/kg 137.82) From steam tables, h6 h10 = hf at 6 bar = 670.56 kJ/kg v10 = vf at 6 bar = 0.001101 m3/kg h8 = hf at 25 bar = 962.11 kJ/kg Substituting in heat balance of contact type feed heater, (m1 \u00d7 962.11) + (m2 \u00d7 3175) + ((1 \u2013 m1 \u2013 m2) \u00d7 138.49) = 670.56 (m1 \u00d7 823.62) + (m2 \u00d7 3036.51) = 532.07 For process 10-11 in boiler feed pump; (h11 \u2013 h10) = v10.\u2206p = 0.001101 \u00d7 (100 \u2013 6) \u00d7 102 h11 = 670.56 + (0.001101 \u00d7 94 \u00d7 102) h11 = 680.91 Considering pump efficiency, 0.90 = \u2212 h 11 \u2032 \u2212 h 11 h 10 h 10 h11 \u2032 = 682.06 kJ/kg Vapour Power Cycles ___________________________________________________________ 315 Substituting values, m1 = \u2212 (682.06 1407.56) \u2212 (962.11 3076.003) = 0.343 kg/kg of inlet steam to HP turbine Also substitution yields; (0.343 \u00d7 823.62) + (m2 \u00d7 3036.51) = 532.07 m2 = 0.082 kg/kg of inlet steam to HP turbine \u2032) Work from HP turbine, WHP = (h1 \u2013 h2 WHP = (3450.02 \u2013 3076.003) = 374.017 KJ/kg Work from LP turbine, \u2032) + (1 \u2013 m1 \u2013 m2) (h9 WLP = (1 \u2013 m1) (h3 \u2013 h9 WLP = (1 \u2013 0.343)(3576.99 \u2013 3235.29) + (1 \u2013 0.343 \u2013 0.082) \u00d7 (3235.29 \u2013 2491.5) WLP = 652.18 kJ/kg \u2032) \u2032\u2013h4 Pump work = WCEP + WBFP = (1 \u2013 m1 \u2013 m2) (h6 \u2032 \u2013 h5) + ( h11 \u2032\u2013 h10) Wpump = (1 \u2013 0.343 \u2013 0.082) (138.49 \u2013 137.82) + (682.06 \u2013 670.56) = 0.38525 + 11.5 = 11.885 kJ/kg Net work, Wnet = WHP + WLP \u2013 Wpump = 374.017 + 652.18 \u2013 11.885 = 1014.312 kJ/kg Specific steam consumption = 3600 W \u00d7 \u03b7 net gen \u00d7 \u03b7 mech = 3600 \u00d7 (1014.312 0.95 0.95) \u00d7 Specific steam consumption = 3.93 kg/kw.h Ans. Overall thermal efficiency = = W net \u00d7 \u03b7 mech \u00d7 \u03b7 \u2212 ( h 1 h 12 ) + \u2212 (1 )( m h 1 3 gen \u2212 \u2032 h 2 ) \u00d7 1014.312 0.95 0.95 \u00d7 \u2212 (345.02 1407.56) + \u2212 (1 0.343)(3576.99 3076.003) \u2212 Overall thermal efficiency, \u03b7 overall = 0.3656 or 36.56% Ans. Mass of steam required = specific steam consumption \u00d7 120 \u00d7 103 = 471600 kg/hr Mass of steam bled from HP turbine = 0.343 \u00d7 471600 = 161758.8 kg/hr Ans. Mass of steam held from LP turbine = 0.082 \u00d7 471600 = 38671.2 kg/hr Ans. 316 ________________________________________________________", "_ Applied Thermodynamics 20. In an installation where the heating of 14000 kW and electrical power both are required, a back pressure turbine is employed. Back pressure turbine has initial steam state of 20 bar, 350\u00b0C and exhaust steam leaves at 1.5 bar. The efficiency ratio of turbine is 0.75 and in this installation the condensate drain from heater is fed back to boiler at condensing temperature. Determine the power available to generator. Solution: The states of steam at inlet and exit of back pressure turbine are shown here. From steam tables: 20 bar 1 350\u00b0C r a 1. 5 b h 2' 2 s Fig. 8.52 At 20 bar, 350\u00b0C h1 = 3137.0 kJ/kg, s1 = 6.9563 kJ/kg.K s1 = s2 = 6.9563 kJ/kg.K, let dryness fraction be x2 at '2' s2 = 6.9563 kJ/kg.K = sf at 1.5 bar + x2 sfg at 1.5 bar \u21d2 \u21d2 6.9563 = 1.4336 + x2 \u00d7 7.2233 x2 = 0.765, Enthalpy at '2', h2 = hf at 1.5 bar + x2.hfg at 1.5 bar h2 = 467.11 + (0.765 \u00d7 2226.5) h2 = 2170.38 kJ/kg.K For given efficiency ratio of turbine; 0.75 = \u2212 ) ( h 1 h 1 h2\u2032 = 2412.04 kJ/kg.K. h \u2032 2 ) h 2 \u2212 ( Enthalpy of condensed steam = hf at 1.5 bar = 467.11 kJ/kg.K In this back pressure turbine the steam at state h2\u2032 will be available for heaitng till it becomes condensed steam with enthalpy hf at 1.5 bar Thus mass of steam required for heating action. = (Heating to be done) \u2212 h ( h \u2032 2 f at 1.5 bar ) Vapour Power Cycles ___________________________________________________________ 317 = 14000 \u2212 (2412.04 467.11) = 7.198 kg/s Hence power available to generator \u2032) = mass flow rate of steam \u00d7 (h1 \u2013 h2 = 7.198 \u00d7 (3137 \u2013 2412.04) = 5218.26 kW 21. In a back pressure turbine installation the steam is supplied at pressure of 20 bar, 350\u00b0C and the exhaust steam from turbine at pressure of 2 bar is used for process heating. Condensate from heater is returned to boiler at 120\u00b0C. Considering turbine isentropic efficiency of 0.8, boiler efficiency of 80% and heat required for heating being 9000 kW determine the power generated and boiler heat requirement. Solution: From steam table, 20 bar 1 350\u00b0C 2 bar h 2' 2 s Fig. 8.53 h1 = 3137 kJ/kg s1 = 6.9563 kJ/kg.K. At 2 bar pressure, let dryness fractions fraction be x2. s2 = s1 = 6.9563 = sf at 2 bar + x2.sfg at 2 bar \u21d2 hence, 6.9563 = 1.5301 + (x2 \u00d7 5.597) x2 = 0.969 h2 = hf at 2 bar + x2.hfg at 2 bar = 504.70 + (0.969 \u00d7 2201.9) h2 = 2638.34 kJ/kg Considering turbine insentripic efficiency h 0.8 = 1 h 1 \u2032\u2212 h 2 \u2212 h 2 = h \u2032\u2212 2 3137 \u2212 (3137 2638.34) \u21d2 h2\u2032 = 2738.07 kJ/kg 318 _________________________________________________________ Applied Thermodynamics Let us consider the condensate going to boiler to be saturated liquid at 120\u00b0C. Let the mass flow rate of steam be ms kg/s. Thus, heat supplied for process heating = 9000 kW \u21d2 9000 = ms (h2\u2032 \u2013 hf at 120\u00b0C) 9000 = ms(2738.07 \u2013 503.71) ms = 4.03 kg/s Thus, power developed = ms \u00d7 (h1 \u2013 h2\u2032) = 4.03(3137 \u2013 2738.07) Power develped = 1607.68 kW Total heat consumption in in boiler = (h1 \u2013 hf at 120\u00b0C) \u00d7 ms = (3137 \u2013 503.71) \u00d7 4.03 = 10612.16 kW Actual heat consumption in boiler = 10612.16 0.8 = 13265.2 kJ/s 22. A pass out turbine is fed with steam at 20 bar, 250\u00b0", "C. The pass out steam pressure is 5 bar and the exhaust pressure is 0.075 bar. In this installation the total power required is 4500 kW and heat load is 15000 kW. The efficiency of HP and LP turbine stages is 0.8 in each stage. Considering the reduction in total steam consumption to result in steam consumption rate of 10 kg/s, determine the new power output if the heat load remains same. Also assume the nozzle control governing and throttle governing at the inlets of HP turbine and LP turbine respectively. Solution: For first part of expanssion in HP stage. From steam table h1 = 3137 kJ/kg s1 = 6.9563 kJ/kg.K. h It indicates that state of steam after expansion in HP stage lies in superheated region. From superheated steam table, degree of superheat may be estimated. At 5 bar, 20 bar 1 2 2' 5 bar P3'' 3'' 0.075 bar 4' 2' 4 3' 3 s Fig. 8.54 sg at 5 bar, 151.86\u00b0C = 6.8213 kJ/kg.K sat 5 bar, 200\u00b0C = 7.0592 kJ/kg.K \u00d7 (200 151.86) \u2212 T2 = T2 = 179.18\u00b0C \u2212 (6.9563 6.8213) \u2212 (7.0592 6.8213) + 151.86 Vapour Power Cycles ___________________________________________________________ 319 hg at 5 bar, 151.86\u00b0C = 2748.7 kJ/kg, hat 5 bar, 200\u00b0C = 2855.4 kJ/kg, h2 = 2748.7 + \u2212 (2885.4 2748.7) (200 151.86) \u2212 \u00d7 \u2212 (179.18 151.86) h2 = 2813.41 kJ/kg. Considering turbine efficiency, 0.8 = ( h 1 ( h 1 \u2032\u2212 h 2 \u2212 h 2 ) ) 0.8 = h \u2032\u2212 2 ) (3137 \u2212 (3137 2813.41) h2\u2032 = 2878.13 kJ/kg Heat available for process heating = h2\u2032 \u2013 hf at 5 bar = 2878.13 \u2013 640.23 = 2237.9 kJ/kg Let mass flow rate of steam be ms, HP to HP turbine and ms,heat for process heating after HP. Thus, mass flow rate of steam required for process heating of 15000 kW ms, heat = 15000 2237.9 = 6.70 kg/s. The h-s diagram shows two different arrangements. One arrangement shown by (1 \u2013 2\u2032 \u2013 3\u2032) indicates the case of no extraction in turbine. While (1 \u2013 2\u2032 \u2013 3\u2032\u2032 \u2013 4\u2032) shows the case of passout turbine. For the first case Using turbine efficiency, 0.8 = ( T 1 ( T 1 \u2032\u2212 T 2 \u2212 T 2 ) ) \u21d2 \u21d2 \u2032 = T1 \u2013 0.8 (T1 \u2013 T2) = 350 \u2013 0.8(350 \u2013 179.18) \u2032 = 213.34\u00b0C T2 T2 s2\u2032 = 6.8213 + \u2212 (7.0592 6.8213) (200 151.86) \u2212 \u00d7 \u2212 (213.34 151.86) \u2032 = 7.125 kJ.kg.K s2 s3 = s2 \u2032 = 7.125 kJ/kg.K Let dryness fraction at 3 be x3 s3 = sf at 0.075 bar + x3 \u00d7 sfg at 0.075 bar 7.125 = 0.5754 + (x3 \u00d7 7.675) \u21d2 x3 = 0.853 320 _________________________________________________________ Applied Thermodynamics \u21d2 h3 = hf at 0.075 bar + x3.hfg at 0.075 bar = 168.79 + (0.853 \u00d7 2406) h3 = 2221.11 kJ/kg Using LP turbine stage efficiency, 0.8 = 0.2878.13 \u2212 (2878.13 2221.11) h \u2032\u2212 3 ) Power produced by HP stage, = ms,HP \u00d7 (h1 \u2013 h2\u2032) h3\u2032 = 2352.5 kJ/kg = ms,HP (3137 \u2013 2878.13) P,HP = 258.87 ms,HP Power produced by LP stage, = (ms,HP \u2013 ms,heat) (h2\u2032 \u2013 h3\u2032) = (ms,HP \u2013 6.7) (2878.13 \u2013 2352.51) P,LP = (ms,HP \u2013 6.7)525.62 Total power produced = PHP + PLP = 258.87 ms,HP + 525.62 (ms,", "HP \u2013 6.7) 4500 = (784.49 ms,HP \u2013 3521.65) ms,HP = 10.23 kg/s For the IInd case when steam consumption changes Since total steam consumption is reduced to 10 kg/s, i.e., ms,HP = 10kg/s, but heat load remains save, so steam required for process heating would not change and ms,heat remains as 6.7 kg/s. Thus ms,HP, new = 10 kg/s & ms, heat, new = 6.7 kg/s. Also it is given in problem that the nozzle control governing employed at entrance of HP turbine stage therefore pressure at inlet to turbine remains same as 20 bar. Hence power produced by HP turbine = 10 \u00d7 (h1 \u2013 h2\u2032) = 10 \u00d7 (3137 \u2013 2878.13) = Power, HP, new = 2588.7 kW Also the problem says that LP steam turbine stage is having throttle control governing, so by symmetry of triangles on pressure \u2013 steam consumption graph. Here, = p \u2032 2 M \u2032 2 p \u2032\u2032 3 M \u2032\u2032 3 M3 \u2032\u2032 = (ms,HP,new \u2013 ms,heat, new) = 10 \u2013 6.7 = 3.3 kg/s. Vapour Power Cycles ___________________________________________________________ 321 M2 \u2032 = (ms,HP \u2013 ms,heat) = 10.23 \u2013 6.7 = 3.53 kg/s. \u2032\u2032 = p3 \u00d7 5 3.3 3.53 = 4.67 bar \u2032 = h3 Also, in view of throttling, h2 From steam table this state of steam at 4.67 bar pressure and enthalpy of 2878.13 kJ/kg can be located as it lies in super heated state by interpolation. Also it can be directly seen from mollier diagram, which would however yield approximate value. After locating this state on mollier diagram considering iseentropic expansion the state 4 can be located and enthalpy value be known. \u2032\u2032 = 2878.13 kJ/kg. At 4.67 bar, 2878.13 kJ/kg let temperature be T3 By interpolation, \u2032\u2032 At 4.67 bar, 200\u00b0C,h = 2857.08 kJ/kg s = 7.096kJ/kg.K At 4.67 bar, 250\u00b0C, h = 2961.86 kJ/kg, s = 7.30654 \u2032\u2032 = T3 \u2212 (250 200) \u00d7 \u2212 (2878.13 2857.08) \u2212 (2961.86 2857.08) \u2032\u2032 = 210.04\u00b0C T3 \u21d2 \u2032\u2032 = s3 7.096 + \u2212 (7.30654 7.096) \u2212 (250 200) \u00d7 (210.04 200) \u2212 \u2032\u2032 = 7.138 kJ/kg.K. s3 For isentropic expansion in LP turbine stage, s3 At 0.075 bar and entropy value of 7.138 kJ/kg.K the state of steam lies in wet region, hence let \u2032\u2032 = s4 us consider dryness fraction at state 4 as x4. s4 = 7.138 = sf at 0.075 bar + x4.sfg at 0.075 bar 7.138 = 0.5764 + (x4. 7.6750) x4 = 0.855 \u21d2 Enthalpy at state 4 h4 = hf at 0.075 bar + x4.hfg at 0.075 bar h4 = 168.79 + (0.855 \u00d7 2406) = 2225.92 kJ/kg Considering isentropic efficiency of LP turbine, \u21d2 0.8 = 0.8 = \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed h \u2032\u2032 3 h \u2032\u2032 3 \u2212 \u2212 h \u2032 4 h 4 \uf8f6 \uf8f7 \uf8f8 2878.13 \u2212 2878.13 2225.92 h \u2032\u2212 4 \uf8f6 \uf8f7 \uf8f8 \u21d2 h4\u2032 = 2356.36 kJ/kg 322 _________________________________________________________ Applied Thermodynamics Power produced by LP turbine = M3\u2032\u2032 \u00d7 (h3\u2032\u2032 \u2013 h4\u2032) = 3.3 \u00d7 (2878.13 \u2013 2356.36) PowerLP, new = 1721.84 kW. Total power produced = Power, new = 2588.7 + 1721.84 = 4310.54 kw 23. Determine the steam consumption rate and the heat available for process heating from the pass out turbine installation having steam entering HP turbine stage at 20 bar, 350\u00b0C and the steam pressure before first stage no", "zzles is 17.5 bar. Steam is available for process heating at 5 bar pressure, the pressure before LP turbine's first stage nozzles is 4 bar and the exhaust pressure is 0.075 bar. Condensed steam temperature after process heating is 90\u00b0C. Electric power developed is 8 mW, alternator efficiency is 97.5%, turbine's mechanical losses are 50 kW, HP & LP turbine efficiency are 80% each and condenser discharge is 8 kg/s. Solution: All the salient states are tentatively shown on h-s diagram. Steam table can be used to get various enthalpy values at salient states. Also mollier diagram can be used for quick but approximate analysis. h1 = h20bar, 350\u00b0C = 3137 kJ/kg. s1 = 6.9563 kJ/kg.K \u2032 = h1 = 3137 kJ/kg. h1 h 20 bar 17.5 bar 1 1' 2' 4 2 3 4' 5 bar 4 bar 0.075 bar s Fig. 8.55 Let us assume total steam consumption rate be m1 kg/s and mh be the mass flowrate of steam for process heating and m3 kg/s be steam flow rate to LP stage for expansion. Let us find out remaining enthalpy and other properties through mollier diagram. entropy \u2032 = s2 = 7 kJ/kg.K. s1 h2 = 2830 kJ/kg. Considering HP stage efficiency, \u2212 h \u2032 0.8 = 1 \u2212\u2032 h 1 h \u2032 2 h 2 Vapour Power Cycles ___________________________________________________________ 323 \u2212 ) 0.8 = h \u2032 2 (3137 \u2212 (3137 2830) h3 = h2\u2032 = 2891.4 kJ/kg \u21d2 h2\u2032 = 2891.4 kJ/kg From mollier diagram, s3 = s4 = 7.1 kJ/kg.K h4 = 2210 kJ/kg Considering LP stage efficiency, 0..8 = \u2212 (2891.4 \u2212 (2891.4 2210) h 4 ) \u2032 For Process heating M1 Mh M3 4,4' 2,2' 3 HP LP 1 1' Fig. 8.56 Schematic for pass out turbine \u21d2 h4\u2032 = 2346.28 kJ/kg Consider discharge of 8 kg/s to be the steam flow rate through LP turbine, i.e., m3 = 8 kg/s. The power made available to alternator can be estimated by considering electric power developed. Power available to alternator = 8 0.975 = 8.2051 MW Total power produced = 8.2051 \u00d7 103 + 50 = 8255.1 kW Power produced by LP turbine = m3 \u00d7 (h3 \u2013 h4\u2032) = 5451.2 kW. Power produced by HP turbine = 8255.1 \u2013 5451.2 = 2803.9 kW Mass flow rate through HP turbine \u21d2 m1 = 2803.9 \u2212\u2032 h h \u2032 2 1 ( ) m1 = 2803.9 \u2212 (3137 2891.4) Total steam consumption rate. m1 = 11.42 kg/s Heat available for process heating = mh \u00d7 (h2 \u2013 hf at 90\u00b0C) = (m1 \u2013 m3) (2830 \u2013 376.92) = (11.42 \u2013 8) (2830 \u2013 376.92) = 8389.53 kW 24. A mixed pressure turbine has high pressure steam entering at 12 bar, 300\u00b0C, low pressure steam entering at 2 bar, dry and saturated and exhausts steam at 0.075 bar Turbine delivers 720 kW and mechanical efficiency is 90%. The efficiency of HP and LP turbine stages is 80% each. Determine the steam consumption rate in kg/kW.h at inlet of both HP and LP modes. 324 _________________________________________________________ Applied Thermodynamics 12 bar 300\u00b0C 2 bar 2 0.075 bar 3' 4' 3 Fig. 8.57 Solution: The expansion processes are shown on h-s diagram. from steam table, for HP steam, 1 4 s \u21d2 h1 = hat 12 bar,300\u00b0C = 3045.8 kJ/kg s1 = 7.0317 kJ/kg.K., s4 = s1 s4 = sf at 0.075 bar + x4.sfg at 0.075 bar 7.0317 = 0.5764 + x4.7.6750 x4 = 0.841 h4 = hf at 0.075 bar + x4. hfg at 0.075 bar = 168.79 + (0.841 \u00d7 2406) h h4 = 2192.24 kJ/kg Considering efficiency, 0.8 = h 1 h 1 \u2032\u2212", " h 4 \u2212 h 4 \u21d2 \u2032 = 2362.95 kJ/kg h4 From steam turbine, for LP steam, h2 = hg at 2 bar = 2706.7 kJ/kg s2 = sg at 2 bar = 7.1271 kJ/kg.K Let dryness fraction at 3 be x3. s3 = s2 = sf at 0.075 bar + x3. sfg at 0.075 bar = 7.1271 = 0.5764 + (x3 \u00d7 7.6750) x3 = 0.854 h3 = hf at 0.075 bar + x3.hg at 0.075 bar = 168.79 + (0.854 \u00d7 2406) h3 = 2223.51 kJ/kg Considering LP efficiency, 0.8 = ( h 2 ( h 2 \u2032\u2212 h 3 \u2212 h 3 ) ) 0.8 = h \u2032\u2212 3 (2706.7 \u2212 (2706.7 2223.51) ) \u2032 = 2320.15 kJ/kg h3 For given turbine output of 720 kW and mechanical efficiency of 90% the power developed in turbine = 720 0.9 = 800 kW Let the steam consumption rate (in kg/s) of HP and LP steam be mHP & mLP and power produced Vapour Power Cycles ___________________________________________________________ 325 (in kW) be PHP and PLP. PHP = mHP \u00d7 (h1 \u2013 h4\u2032) HP steam consumption, kg/kWh = LP steam consumption, kg/kWh = HP m P HP = 3600 \u2212 h \u2032 h 4 1 ( ) LP m P LP = 3600 \u2212 h \u2032 h 3 2 ( ) = 5.27 kg/kWh = 9.31 kg/kWh. 25. Determine the power output from a mixed pressure turbine fed with 2 kg/s of HP steam at 16 bar, 300\u00b0C and 1.5 kg/s of LP steam at 2 bar dry saturated. The expansion efficiency is 90% for both and the exhaust pressure is 0.075 bar. Consider that low pressure steam is admitted through a throttle value and there is no throttling of high pressure steam. This mixed pressure turbine develops 3 MW power full output when dry and saturated steam at 2 bar is supplied and leaves at 0.075 bar. Neglect the throttling losses. Solution: Let us assume that pressure after mixing of high pressure and low pressure streams is p2 bar. 16 bar 300\u00b0C 2 bar P2 2' 0.075 bar h 2 3 4 3' 5' 5 From steam tables; s Fig. 8.58 mHP = 2kg/s, mLP = 1.5 kg/s h1 = hat 15 bar, 300\u00b0C = 3034.8kJ/kg s1 = sat 10 bar, 300\u00b0C = 6.8844 kJ/kg.K = s3 Using expansion efficiency; h 0.9 = 1 h 1 Let dryness fraction at state 3 be x3 so. \u2212 \u2212 h \u2032 3 h 3 s3 = 6.8844 = sfat 2bar + x3. sfg at 2bar 6.8844 = 1.5301 + (x3 \u00d7 5.597) 326 _________________________________________________________ Applied Thermodynamics x3 = 0.9566 h3 = hf at 2 bar + x3. hfg at 2 bar = 504.7 + (0.9566 \u00d7 2201.9) h3 = 2611.04 kJ/kg In this case it is given that this turbine develops 3 MW when LP steam is supplied at 2 bar, dry saturated and expanded upto 0.075 bar. h3 \u2032 = h1 \u2013 (0.9 \u00d7 (h1 \u2013 h3)) = 2653.42 kJ/kg Let mass flow rate of steam in this case be ms kg/s. In this case the enthalpy, hin = hg at 2 bar = 2706.7 kJ/kg = h2 = h2 sin = sg at 2 bar = 7.1271 kJ/kg.K. Let dryness fraction at exit state be xout. sout = sin = sf at 0.075 bar + xout sfg at 0.075 bar 7.1271 = 0.5764 + xout. 7.6750. xout = 0.8535 hout = hf at 0.075 bar + xout.hfg at 0.075 bar = 168.79 + (0.8535 \u00d7 2406) hout = 2222.31 kJ/kg Considering expansion efficiency \u2212 \u2212 \u2032 = 0.9 = h in h in hout\u2032 = 2270.75 kJ/kg h out h out (2706.7 h \u2212 ) \u2032", " out \u2212 (2706.7 2222.31) ms \u00d7(hin \u2013 hout\u2032) = 3 \u00d7 103 ms = 3000 \u2212 (2706.7 2270.75) = 6.88 kg/s. 2 bar 0.075 bar h in out' out s Fig. 8.59 Vapour Power Cycles ___________________________________________________________ 327 Mixed pressure turbine is being fed with HP steam too along with LP steam. This is normally done when power available cannot be met by LP steam. During this mixed pressure operation it is quite possible that pressure of steam at exit of HP turbine may not be same as LP steam pressure of 2 bar and let us assume that the steam pressure at exit from HP steam be p2. In this case LP steam has to be throttled up to pressure p2 while maintaining speed constant. Thus, for throttle governing considerations. p 2 m+ ) LP ( m HP = 2 \u00d7 2 10 sm p2 = \u00d7 2 10 2 \u00d7 + (2 1.5) 6.88 = 1.017 \u00d7 102 kPa For state after mixing; ( m HP h4 = Power output from mixed pressure turbine ) \u00d7 h \u2032 3 m ( HP + \u00d7 ( m m LP ) LP \u00d7 h 2 \u2032 ) = 2676.25 kJ/kg For pressure of 1.017 bar and enthalpy of 2676.25 kJ/kg the iseutropic expansion upto 0.075 bar can be seen. Power = mHP(h1 \u2013 h3\u2032) + (mHP + mLP)(h4 \u2013 h5\u2032) h5 = 2290 kJ/kg Considering expansion efficiency, 0.9 = h 4 h 4 \u2032\u2212 h 5 \u2212 h 5 h5\u2032 = 2328.63 kJ/kg Power = 2(3034.8 \u2013 2653.42) + (2 + 1.5)(2676.25 \u2013 2328.63) = 1979.43 kW. 8.1 Explain Carnot vapour power cycle and its relevance. 8.2 Give limitations of Carnot vapour power cycle and explain how Rankine cycle helps in overcoming them. 8.3 Discuss the limitations of maximum and minimum temperatures in a steam power cycle. 8.4 What should be the desired properties of a working fluid in a Rankine cycle? 8.5 Compare the efficiency and specific steam consumption of ideal Carnot cycle with Rankine cycle for given pressure limits. 8.6 Discuss the effect of pressure of steam at inlet to turbine, temperature at inlet to turbine and pressure at exit from turbine upon Rankine cycle performance. 8.7 Describe reheat cycle and compare it with simple Rankine cycle. 8.8 What do you understand by ideal regenerative cycle? Why is it not possible in practice? Also give actual regenerative cycle. 328 _________________________________________________________ Applied Thermodynamics 8.9 Compare performance of regenerative cycle with simple Rankine cycle. 8.10 What do you understand by binary vapour power cycles? 8.11 Discuss advantages of binary vapour cycles over single vapour cycle. 8.12 What is mixed pressure turbine? 8.13 Describe pass out turbines and back pressure turbines. 8.14 Discuss briefly the factors limiting thermal efficiency of a steam power plant. determine cycle efficiency considering it to work on Carnot cycle. 8.15 Describe combined heat and power cycles. 8.16 A steam engine working on Rankine cycle operates between 1.96 MPa, 250\u00baC and 13.7 kPa. If engine consumes steam at the rate of 0.086 kg per second, determine Rankine cycle efficiency, neglecting [58.8%] pump work. Also, find Rankine cycle efficiency considering pump work. 8.17 In a steam power plant working on boiler pressure of 80 bar and condenser pressure of 0.075 bar [44.9%] 8.18 A reheat cycle operates between 80 bar and 0.075 bar. Temperature of steam entering turbine is 600\u00baC and first stage of expansion is carried out till the steam becomes dry saturated. Subsequently steam is reheated upto the initial temperature at inlet. Neglecting pump work determine efficiency and specific [42.5%, 141 kg/hp-hr] steam consumption in kg/hp\u00b7hr. 8.19 A steam turbine installation of 60 MW capacity has steam entering turbine at 7 MPa, 500\u00baC with steam bleeding at 2 MPa and 0.2 MPa for feed heating. Remaining steam at 2 MPa is reheated upto 480\u00baC. Steam finally leaves turbine at 36 mm Hg (absolute). Give layout, mass fraction of steam bled out per kg of steam generated, cycle efficiency and mass flow rate of steam entering turbine. rate considering turbine internal efficiency of 70% and load requirement of 800 kW. [0.159, 0.457, 46.58 kg/s] 8.20 In a steam turbine", " steam enters at 1.4 MPa, 320\u00baC and exhausts at 175 kPa. Determine the steam flow [2.6 kg/s] 8.21 During trial of turbine in steam power plant the steam is found to enter turbine at 2.1 MPa, 260\u00baC with flow rate of 226 kg/s. Mechanical efficiency is 95% and turbine output is 100 MW. The exhaust pressure is 140 kPa and condensate temperature is 70\u00baC. Unfortunately the old boiler fails and is to be replaced by another boiler generating steam at 4.9 MPa, 320\u00baC. This unit may drive an additional turbine with an internal efficiency of 90% which exhausts at 2.1 MPa. A part of exhaust is reheated to above trial conditions, including flow rate and remainder heats feed water to 205\u00baC. Determine (a) total plant output, (b) new total steam flow rate, kg/s, (c) old [14.9 MW, 293 kg/s, 94.3%, 22.9%] turbine internal efficiency, (d) the overall thermal efficiency. 8.22 A steam turbine has steam entering at 40 bar, 450\u00baC and leaving at 0.06 bar. Turbine develops 4000 hp at 3000 rpm with expansion occurring in two stages and reheating in between at 2 bar and both stages give equal output. The stages have efficiency ratio of 0.8. Determine thermal efficiency of cycle. [32.2%] 8.23 A regenerative cycle has steam supplied at 28 bar, 300\u00baC in first stage. A steam fraction is bled out at 3 bar for feed heating. The feed heater drains are pumped into feed line downstream of heater at the same temperature as the bled steam. The steam expansion may be considered isentropic throughout. Condenser works at 0.15 bar. Determine mass of steam bled per kg of steam generated and thermal [0.143 kg/kg steam generated, 31.2%] efficiency. Also give layout and T-s diagram. 8.24 A steam power plant has steam generated at 140 bar, 400\u00baC and throttled to 70 bar before supplied to turbine. Steam expands isentropically to the pressure of 15 bar and at this pressure some quantity of steam is bled out for feed heating which leaves as condensate at 40\u00baC. Remaining steam is reheated at constant pressure by mixing it with the throttled steam taken from boiler in such amount that the resultant mixture has temperature of 250\u00baC. This mixture then expands isentropically upto condenser Vapour Power Cycles ___________________________________________________________ 329 pressure of 0.075 bar. Determine horse power developed by turbine for steam generation rate of 50 [14400 hp] ton/hr. Also give layout and T-s diagram. 8.25 Steam is required for process heating at 3 bar, dry saturated along with power output of 1 MW. A back pressure turbine with 70% internal efficiency is used for this purpose. For the steam supplied at the rate of 2.78 kg/s determine the pressure and temperature of steam to be generated from boiler. [37.3 bar, 344\u00baC] 8.26 A steam turbine has steam being generated at 30 bar, 400\u00baC for being expanded upto 0.04 bar with bleeding of some steam at 3 bar for feed heating in closed feed water heater. Feed water leaves feed heater at 130\u00baC and the condensate leaving feed heater as saturated liquid is sent to drain cooler where it gets cooled to 27\u00baC and finally discharged in hotwell. Condensate from condenser is also discharged in hotwell from where feed water is picked by a pump and passed through drain cooler and subsequently closed feed water heater. Assuming no losses and negligible pump work determine the thermal efficiency of cycle and mass of steam bled out per kg of steam generated. Draw layout and T[38.82%, 0.167 kg/kg steam] s diagram. 8.27 In a steam turbine plant steam enters turbine at 20 bar, 250\u00baC and leaves at 0.05 bar. During expansion some amount of steam is bled out at 5 bar, 1.5 bar and 0.3 bar for feed heating in closed feed water heaters. Condensate leaves each feed heater as saturated liquid and is passed through traps into next feed heater at lower pressure. Combined drain at 0.3 bar is cooled in drain cooler upto condenser temperature and then put into hot well which also collects condensate coming from condenser, Determine, (a) the masses of steam bled per kg of steam generated (b) the thermal efficiency of cycle (c) the net work output per kg of steam generated (d) the layout and T-s diagram. Neglect pump work. [0.088 kg, 0.0828 kg, 0.046 kg, 35.63%, 807 kJ/kg ] 8.28 A", " binary vapour power cycle works on mercury and steam such that dry saturated mercury vapour at 4.5 bar is supplied to mercury turbine and leaves at 0.04 bar. Steam is generated as dry saturated at 15 bar and supplied to steam turbine for being expanded upto condenser pressure of 0.04 bar. Determine thermal efficiency of cycle. For mercury take, h\u0192 at 4.5 bar = 62.93 kJ/kg, hg at 4.5 bar = 355.98 kJ/kg, vf at 0.04 bar = 0.0000765 m3/kg h\u0192 at 0.04 bar = 29.98 kJ/kg, hg at 0.04 bar = 329.85 kJ/kg sg at 4.5 bar = 0.5397 kJ/kg \u00b7 K, sg at 0.04 bar = 0.6925 kJ/kg \u00b7 K, s\u0192 at 0.04 bar = 0.0808 kJ/kg \u00b7 K 330 _________________________________________________________ Applied Thermodynamics 9 9.1 INTRODUCTION In the preceding chapter brief description is made for thermodynamic cycles comprising of both power and refrigeration cycles. In the present chapter the thermodynamic cycles operating on gas as working fluid and producing power are studied. Gas power cycles are commonly used for power production as in gas turbine plants, internal combustion engines and other applications. Here performance prediction based on thermodynamic analysis is presented for ideal cycles of different types used for gas turbine installations and internal combustion engines. 9.2 AIR-STANDARD CYCLE Air standard cycle refers to thermodynamic cycle being studied with certain assumptions, so as to use the principles of thermodynamics conveniently. It is the most simplified form of thermodynamic cycle under consideration. General assumptions made for a cycle to be air-standard cycle are as follows: (i) Air is the working fluid and behaves as a perfect gas. (ii) Working fluid does not get changed in its mass and composition. (iii) Thermodynamic processes constituting cycle are reversible. (iv) There is no heat loss from system to surrounding and vice-versa. (v) During heat addition process, heat is assumed to be supplied from a high temperature source. (vi) During heat rejection process, heat is assumed to be rejected to a low temperature sink. (vii) Specific heats of working fluid do not change throughout the cycle. After incorporating above assumptions in thermodynamic cycle analysis, the analysis done is called air standard cycle analysis. Results obtained from such analysis disagree quite largely from the actual performances due to over simplification, but to begin with, such analysis is necessary. Gradually the assumptions made may be removed and analysis becomes realistic. Air standard or ideal efficiency can be given as follows: Ideal work done i.e. (heat supplied \u2013 heat rejected) Heat supplied = air\u03b7 = 1 \u2212 heat rejected heat supplied Mean effective pressure (m.e.p.) is the parameter which indicates the average pressure throughout the cycle. Mathematically, it is given as follows: m.e.p. = work done per cycle stroke volume Gas Power Cycles ______________________________________________________________ 331 Graphically it is given by mean height of P \u2013 V diagram. Air standard cycle analysis for Carnot gas power cycle has already been discussed in chapter 4. Otto Cycle : This is a modified form of Carnot cycle in order to make it a realistic cycle. Otto cycle has two constant volume and two adiabatic processes as shown below. 3 2 Qadd p Wcompr. V Wexpn. T 4 1 Qreject 2 1 S 3 4 Fig 9.1 p-V and T-S representations of Otto cycle Thermodynamic processes constituting Otto cycle are 1 \u2013 2 = Adiabatic compression process, (\u2013ve work, Wcompr) 2 \u2013 3 = Constant volume heat addition process (+ve heat, Qadd) 3 \u2013 4 = Adiabatic expansion process, (+ve work, Wexpan) 4 \u2013 1 = Constant volume heat rejection process (\u2013ve heat, Qreject) In order to have an engine based on Otto cycle let us find out the relevance of above processes. Spark ignition type internal combustion engines are based on this cycle. Process 1 \u2013 2, adiabatic compression process can be realized by piston moving from volume V1 to V2 and therefore compressing air. Process 2 \u2013 3, heat addition process can be undertaken in constant volume manner with piston at volume V2 and heat added to woking fluid. Heat addition is practically realized by combustion of fuel and air. As a result of heat addition the compressed air attains state 3 and it is allowed to expand from 3\u20134 adiabatically. After expansion air is brought back to original state 1 by extracting heat from it at volume V1. Internal combustion engine based on Otto cycle is explained ahead. Let us find air-standard thermal efficiency of Otto cycle. Compression ratio for the cycle", " shown can be given by the ratio of volumes of air before and after compression. It is generally denoted by r. For unit mass of air and properties at states given with subscript 1, 2, 3, 4, we can write = = Heat added during 2\u20133, constant volume process Heat rejected during 4\u20131, constant volume process qadd = cv \u00d7 (T3 \u2013 T2) Air standard efficiency of Otto cycle qrejected = cv \u00d7 (T4 \u2013 T1) \u03b7 otto = Net work Heat added For a cycle, 332 _________________________________________________________ Applied Thermodynamics Net work = Heat added \u2013 Heat rejected Substituting in the expression for efficiency; = cv {(T3 \u2013 T2) \u2013 (T4 \u2013 T1)} c {( T 3 v \u03b7 otto = ) \u2212 )} \u2212 \u2212 ( T 4 ) T 2 or and \u03b7 otto = 1 ) ) For perfect gas, by gas laws\u2212 1 \uf8f6 \uf8f7 \uf8f8 \u03b3 from above Substituting in the expression for \u2212 \u2212 otto or or \u03b7 otto =1 \u2013 1 \u03b3\u2212 1 r Air standard efficiency of Otto cycle thus depends upon compression ratio (r) alone. For As compression ratio increases, the Otto cycle efficiency increases as shown in Fig. 9.2. r = 1, \u03b7 otto = 0 \u03b3 = 1.67 (Theoretical monotomic gas) \u03b3 = 1.40 (Air standard) \u03b3 = 1.20 100 80 60 40 20 \u03b7 % 0 4 8 12 16 20 24 28 Compression ratio Fig. 9.2 Otto cycle efficiency vs. compression ratio Gas Power Cycles ______________________________________________________________ 333 Mean effective pressure for Otto cycle can be estimated by knowing the area of diagram (work) and length of diagram (volume). Mean effective pressure (m.e.p.) Net Work (W) Volume change ( V)\u2206 Net work can be obtained from the area enclosed by cycle on P\u2013V diagram. Thus, Net work = Area of diagram = \uf8f1 \uf8f2 \uf8f3 ) \uf8fc \uf8fd \uf8fe = or \uf8f1 \uf8f4 1 \uf8f2 \u2212 \uf8f4 1 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe If Then Therefore, Net work V2 = V3 1, i.e. unity clearance volume, V1 = r and V4 = r \uf8f1 \uf8f4 1 \uf8f2 \u2212 \uf8f4 1 \uf8f6 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe Also, for adiabatic expansion and compression processr)\u03b3 4 Substituting in expression for Net work, Net work = 1 1)\u03b3 \u2212 ( {p4 \u00b7 r \u00b7 (r\u03b3 \u2013 1 \u20131) \u2013 p1 \u00b7 r (r\u03b3 \u2013 1 \u2013 1)} or or r \u03b3 \u2212 = ( (r\u03b3 \u2013 1 \u20131) \u00b7 (p4 \u2013 p1) Writing expression for m.e.p. as ratio of area of diagram and length of diagram (V1 \u2013 V2). \u2212 \u2212 (m.e.p.)Otto = 1)\u00b7( p 4 \u03b3 )\u00b7( p 1 1, 1 Let us write ratio Therefore (m.e.p.)Otto = ( r r \u03b3\u2212 \u2212 1 \u2212 r ( 1)\u00b7( 1) \u03b3\u2212 \u2212 1 \u2212 r ( 1)\u00b7 \u00b7( p 1 \u03b3 \u2212 1)\u00b7( 1) \u03b1 \u2212 1) 334 _________________________________________________________ Applied Thermodynamics Otto cycle has limitations such as isochoric heat addition and rejection is difficult in piston cylinder arrangement. Also, adiabatic compression and expansions are difficult to be realized in practice. Diesel cycle : Diesel cycle is modified form of Otto cycle. Here heat addition process is replaced from constant volume type to constant pressure type. In a piston cylinder arrangement heat addition with piston at one position allows very little time for heat supply in Otto cycle. By having heat addition at constant pressure the sufficient time is available for heat supply in Diesel cycle. Compression ignition engines work based on Diesel cycles. Thermodynamic processes constituting Diesel cycle are as given below. 1 \u2013 2 = Adiabatic compression, (\u2013ve work, Wcompr) 2 \u2013 3 = Heat addition at constant pressure (+ve heat, Qadd) 3 \u2013 4 = Adiabatic expansion, (+ve work, Wexpn) 4 \u2013 1 = Heat rejection at constant volume (\u2013ve heat, Qrejected) Constant pressure line 3 4 Constant volume line Qadd 2 3 p Wexpn. T Wcompr. V 4 1 Qrejected 2 1 S Fig 9.3 p \u2013V and T\u2013S representation for Diesel cycle P \u2013 V and T \u2013 S representations for the cycle are shown in Fig.", " 9.3. Thermodynamic analysis of the cycle for unit mass of air shows; Heat added = cp (T3 \u2013 T2) Heat rejected = cv (T4 \u2013 T1) V 1 V 2 Let us assume; Compression ratio, r = Cut off ratio, \u03c1= Expansion ratio = V 3 V 2 V 4 V 3 Air standard efficiency for Diesel cycle may be given as, \u03b7 diesel = heat added \u2013 heat rejected heat added ( ) Gas Power Cycles ______________________________________________________________ 335 ( T 4 ( T 3 Using perfect gas equation and governing equation for thermodynamic process 1 \u20132; \u03b7 diesel = 1p V \u03b3 p V 2 2 and Combining above two, we get \u03b3 \uf8f6 \uf8f7 \uf8f8 \u21d2 also as or 1 T 2 T = (r)\u03b3 \u2013 1 T2 = T1 \u00b7 r\u03b3 \u2013 T3 = T2 \u00b7 \u03c1 T3 = T1 \u00b7 r\u03b3\u20131 \u00b7 \u03c1 or or Also for adiabatic process 3 \u2013 4 combining the following: We get, or \u00b7 and \u2212 1 \u03b3\u2212 1 \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8 T4 = T1 \u00b7 \u03c1 \u00b7 r\u03b3\u20131 \u00d7 T4 = T1 \u00b7 \u03c1\u03b3 \u03c1 \u2212 \u03b3 1 \u2212 \u03b3 1 r Gas Power Cycles ______________________________________________________________ 337 or (m.e.p.)Diesel = 1) r \u2212 1 \u03b3 \u03b3 \u03c1 \u00b7( \u2212 1) \uf8f9 \uf8fb \u03b3 ( \u2212 1)( r \u2212 1) Dual cycle : It is also called \u2018mixed cycle\u2019 or \u2018limited pressure cycle.\u2019 Dual cycle came up as a result of certain merits and demerits associated with Otto cycle and Diesel cycle due to heat addition occurring at constant volume and constant pressure respectively. Dual cycle is the combination of Otto cycle and Diesel cycle in which heat addition takes place partly at constant volume and partly at constant pressure. Thermodynamic processes involved in Dual cycle are given as under. 1 \u2013 2 = Adiabatic compression (\u2013ve work, Wcompr) 2 \u2013 3 = Heat addition at constant volume (+ve heat, Qadd, v) 3 \u2013 4 = Heat addition at constant pressure (+ve heat, Qadd, \u03c1) 4 \u2013 5 = Adiabatic expansion (+ve work, Wexpn) 5 \u2013 1 = Heat rejection at constant volume (\u2013ve heat, Qrejected) Qadd,p 4 Qadd,v Wexpn. 3 2 p 5 1 Qrejected Wcompr. V T 2 1 S 3 4 5 Fig 9.5 P\u2013V and T\u2013S representations of dual cycle Let us assume the following for thermodynamic analysis: Clearance volume = Unity Compression ratio, r = Cut-off ratio, \u03c1 = Pressure ratio during heat addition For unit mass of air as working fluid throughout the cycle. Total Heat added = Heat added at constant volume (2 \u20133) + Heat added at constant pressure (3 \u2013 4) Heat rejected, Qrejected = cv (T5 \u2013 T1) Air standard efficiency for Dual cycle can be given as; Qadd = cv (T3 \u2013 T2) + cp (T4 \u2013 T3) 338 _________________________________________________________ Applied Thermodynamics \u03b7 dual = Heat added \u2013 Heat rejected Heat added ( ) or or or or or From gas laws applied to process 2\u20133 T2 = T3 \u00d7 p 2 p 3 T2 = 3T T4 = T3 \u00d7 V 3 T4 = \u03c1 \u00b7 T3 \u2212 1 \uf8f6 \uf8f7 \uf8f8 T5 = T5 = T 4 \u03b3T 4 \u2212 \u03b3 1 r T5 = \u03b3 \u03c1 \u00b7T 3 \u2212 \u03b3 1 r For process 3\u20134, For adiabatic process 4\u20135 Substituting T4 Gas Power Cycles ______________________________________________________________ 339 For adiabatic process 1\u20132 or Substituting for T2 \u03b3 \uf8f6 \uf8f7 \uf8f8 T1 = T 2 \u03b3\u2212 1 r T1 = T 3 \u03b3\u03b1 \u2212 1 \u00b7 r Substituting for T1, T2, T4, T5 in expression for efficiency. \u03b3 \u03b1 \u03c1 \u00b7 \u2212 + 1) 1 \u22121 \u03b3 \u03b1 ( dual \u03b1\u03b3 \u03c1 \u00b7 ( \uf8f9 \uf8fa \uf8fb \u2212 1) For unity cut off ratio i.e. absence of 3 \u2013 4 process, cycle becomes equal to Otto cycle. i.e. for \u03c1 = 1, \u03b7 dual = 1 \u2013 For the pressure ratio \u03b1 being unity, cycle gets modified to Diesel cycle. 1 \u03b3\u2212 = \u03b7 1 otto r i.e. for \u03b1 = 1, \u03b7 dual = 1 \u2013 1 \u03b3\u03b3 \u00b7 r \u2212 1", " \uf8eb \uf8ec \uf8ec \uf8ed Mean effective pressure of dual cycle can be obtained as follows: Net work = Area enclosed in P \u2013 V diagram \u03b3\u03c1 \u03c1 ( \uf8f6\u2212 1 \uf8f7 \uf8f7\u2212 1) \uf8f8 = \u03b7 diesel = p3 (V4 \u2013 V3 or or \u2212 \u2212 Making suitable substitutions for p1, p2 and p5, it gets modified as \u2212 \u2212 1) = p3(\u03c1 \u2013 1) \u03b1 ( p 3 \u03c1 \u2212 p 3 \uf8ee \u03b1\u03b3 \u03c1 \u00b7 ( \uf8f0 p 1 \u00b7( \u00b7 \u2212 1) \uf8f9 \uf8fb Net work = Substituting for mean effective pressure \u03b1 \u03b3 ( \u2212 1) (m.e.p.)dual = Area of diagram Length of diagram 1 r \u2212 ( 1 \uf8ee \u03b1\u03b3 \u03c1 \u00b7 ( \uf8f0 \u2212 + 1) \u03b1 ( \u2212 \u2212 1) r \u2212 1 \u03b3 \u03b3 \u03b1 \u03c1 \u00b7( \u00b7 \u2212 1) \u03b1 \u03b3 ( \u2212 1) \uf8f9 \uf8fb \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe Substituting p3 as function of p1 (m.e.p.)dual = 1\u00b7 p r \u03b3 \uf8ee \u03b1\u03b3 \u03c1 \u00b7 ( \uf8f0 \u2212 + 1) \u03b1 ( \u2212 \u2212 1) r \u2212 1 \u03b3 \u03b3 \u03b1 \u03c1 \u00b7( \u00b7 \u2212 1) \uf8f9 \uf8fb \u03b3 ( \u2212 1)( r \u2212 1) 340 _________________________________________________________ Applied Thermodynamics 9.3 BRAYTON CYCLE Brayton cycle, popularly used for gas turbine power plants comprises of adiabatic compression process, constant pressure heat addition, adiabatic expansion process and constant pressure heat release process. A schematic diagram for air-standard Brayton cycle is shown in Fig. 9.6. Simple gas turbine power plant working on Brayton cycle is also shown here Fig. 9.6 Brayton cycle on P\u2013V and T\u2013S diagram F C.C 3 2 C WC 1 G WT 4 HE1 2 C WC 1 Qadd WT 3 G 4 HE2 Qrejected C : Compressor CC : Combustion chamber (a) Open type F : Fuel G : Generator HC : Heat exchanger (b) Closed type Fig. 9.7 Simple gas turbine plant Thermodynamic cycle shows following processes: 1-2 : Adiabatic compression, involving (\u2013ve) work, WC in compressor. 2-3 : Constant pressure heat addition, involving heat Qadd in combustion chamber or heat exchanger. 3-4 : Adiabatic expansion, involving (+ve) work, WT in turbine. 4-1 : Constant pressure heat rejection, involving heat, Qrejected in atmosphere or heat exchanger. In the gas turbine plant layout shown process 1\u20132 (adiabatic compression) is seen to occur in compressor, heat addition process 2\u20133 occurs in combustion chamber having open type arrangement and in heat exchanger in closed type arrangement. Process 3\u20134 of adiabatic expansion occurs in turbine. In open type arrangement exhaust from turbine is discharged to atmosphere while in closed type, heat rejection occurs in heat exchanger. In gas turbine plant of open type, air entering compressor gets compressed and subsequently brought up to elevated temperature in combustion chamber where fuel is added to high pressure air and combustion occurs. High pressure and high temperature combustion products are sent for expansion in turbine where its\u2019 expansion yields positive work. Expanded combustion Gas Power Cycles ______________________________________________________________ 341 products are subsequently discharged to atmosphere. Negative work required for compression is drawn from the positive work available from turbine and residual positive work is available as shaft work for driving generator. In gas turbine plant of closed type the working fluid is recycled and performs different processes without getting contaminated. Working fluid is compressed in compressor and subsequently heated up in heat exchanger through indirect heating. High pressure and high temperature working fluid is sent for getting positive work from turbine and the expanded working fluid leaving turbine is passed through heat exchanger where heat is picked up from working fluid. Thus, the arrangement shows that even costly working fluids can also be used in closed type as it remains uncontaminated and is being recycled. Air standard analysis of Brayton cycle gives work for compression and expansion as; WC = m1 \u00b7 (h2 \u2013 h1) WT = m3 \u00b7 (h3 \u2013 h4) for air standard analysis, m1 = m3, where as in actual cycle m3 = m1 + mf, in open type gas turbine m3 = m1, in closed type gas turbine For the fuel having calorific value CV the heat added in air standard cycle; Qadd = m1(h3 \u2013 h2), whereas Qadd = mf \u00d7 CV for actual cycle. Net work = WT \u2013 WC Wnet = {m3 (h3 \u2013 h4", ") \u2013 m1(h2 \u2013 h1)} Air standard cycle efficiency = W Q net add {( )} \u2212 \u2212 ( h 2 ) h 2 \u03b7 Brayton = \uf8f1 \u2212\uf8f2 ) ) \uf8fc \uf8fd \uf8fe \uf8f1 \uf8f4 \u2212\uf8f2 1 \uf8f4 \uf8f3 \u03b7 Brayton = T 1 T 2 and (h3 \u2013 h2) = cp (T3 \u2013 T2) ( T 4 ( T 3 \u2212 \u2212 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe ) ) as, (h4 \u2013 h1) = cp (T4 \u2013 T1) For processes 1\u20132 and 3\u20134, which are of isentropic type, \u2212 \u03b3 1 and ) ( \u03b3 Let the pressure ratio be, so, 2 p p = rr T 3 T 4 1) \u03b3 \u2212 \u03b3 and ( ( )r T 3 T = 4 1) \u03b3 \u2212 \u03b3 342 _________________________________________________________ Applied Thermodynamics \uf8eb \uf8ec \uf8ed \uf8f6\u2212 ) ) \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe in cycle efficiency: Making substitution for \u03b7 Brayton = \uf8eb \u2212\uf8ec \u2013 \uf8eb \uf8ec \uf8ed \uf8f6\u2212 Air standard Brayton cycle efficiency: \u03b7 Brayton = \u2212 1 1 \u2212 \u03b3 1 \u03b3 r Thus, it is obvious from the expression of efficiency that it depends only on pressure ratio (r) and nature of gas (\u03b3). For pressure ratio of unity, efficiency shall be zero. For a particular gas the cycle efficiency increases with increasing pressure ratio. Here the variation of efficiency with pressure ratio is shown for air (\u03b3 = 1.4) and monoatomic gas as argon (\u03b3 = 1.66). % 100 80 60 40 20 \u03b3 = 1.66 \u03b3 = 1.4 0 0 2 4 6 8 10 12 14 16 Pressure ratio, r Fig. 9.8 Efficiency vs. pressure ratio in simple cycle Specific work output for the plant can be given as below: or, w = {(h3 \u2013h4) \u2013 (h2 \u2013h1)}, as m1 = 1 w = cp {(T3 \u2013 T4) \u2013 (T2 \u2013T1)} w c T = 1p \uf8f1 \uf8eb \uf8f4 \uf8f2 \uf8ec \uf8f4 \uf8ed \uf8f3 \uf8f6 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe Gas Power Cycles ______________________________________________________________ 343 \uf8f1 \uf8eb \uf8f4 \uf8f2 \uf8ec \uf8f4 \uf8ed \uf8f3 \uf8f6 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe = In the cycle, states 3 and 1 refer to the states having highest temperature and pressure and lowest temperature and pressure states. T3 is also referred as turbine inlet temperature while T1 is ambient T temperature in case of open type arrangement. Let the ratio 3 T 1 Therefore, upon making substitutions, T be given by some constant \u2018t\u2019 i.e. 3 T 1 = t. as, w c T 1p \uf8f1 \uf8f4 \uf8f2 \u2212 and \uf8f6 \uf8f7 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe Here \uf8eb \uf8ec \uf8ec \uf8ed w c T 1p \uf8f6 \uf8f7 \uf8f7 \uf8f8 is a non-dimensional form of specific work output. In general for a given arrangement T1 being ambient temperature remains fixed (mostly atmospheric temperature). So specific work output may be said to depend upon \u2018t\u2019 (temperature ratio) and pressure ratio \u2018r\u2019. Specific work output shall be zero for unity pressure ratio value i.e. for r = 1, w c T 1p = 0. For certain value of temperature ratio specific work output shall increase with increasing pressure ratio value. Graphical pattern showing variation of specific work output with pressure ratio is shown below for certain values of \u2018t\u2019..6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 t = 5 4 3 2 00 2 4 6 8 10 12 14 Pressure ratio, r Fig. 9.9 Specific work output vs. pressure ratio for simple cycle 344 _________________________________________________________ Applied Thermodynamics The temperature ratio \u2018t\u2019 is generally used as design parameter. Higher is the \u2018t\u2019 value higher shall be specific work output. Maximum value of \u2018t\u2019 depends upon the maximum temperature which the highly stressed parts of turbine can withstand for the required working life (also called metallurgical limit). In the modern", " gas turbine value of \u2018t\u2019 ranges from 3.5 to 6. Further higher values of turbine inlet temperature and thus \u2018t\u2019 value can be realized with improved gas turbine blade cooling. Specific work output vs. pressure ratio variation shows that there exist some optimum pressure ratio value at which specific work output is maximum at given \u2018t\u2019 value. Let us differentiate expression for specific work output with respect to \uf8eb \uf8ec \uf8ed \uf8f6\u2212 \u03b3 1 \uf8f7 \uf8f8. \u03b3 r \uf8eb \uf8ec \uf8ed \uf8f6\u2212 \u03b3 1 \uf8f7 \u03b3 \uf8f8 = a, t = constant Let r w = cp T1 \uf8f1 \uf8f4 \uf8f2 = cp T1 \uf8f1 \uf8f2 ) \uf8fc \uf8fd \uf8fe dw da dw da = cp T1 \uf8f1 \uf8f2, so t 2 a t = a2 \u20131 = 0 Equating a = t, or = \u03b3 t r / 2( \u03b3 \u2212 1) \u03b3 \u2212 1 \u03b3 r = \u22121 \u03b3 \u03b3 \u22121 \u03b3 \u03b3 \u22c5 r = t t or, r Thus, Substituting for \u22121 \u03b3 \u03b3 r and t; or= 4 Thus, the specific work output is seen to be maximum for a given \u2018t\u2019 value when the pressure ratio is such that temperature at compressor outlet and turbine outlet are equal. For all values of pressure ratio \u2018r\u2019 lying between 1 and t \u03b3/2(\u03b3\u20131) temperature T4 is greater than T2. In order to reduce heat transfer from external source for improving efficiency a heat exchanger can be incorporated. This concept of using heat exchanger gives birth to modified form of simple gas turbine cycle with heat exchange. This modified form of cycle is also called regenerative gas turbine cycle. Simple gas turbine cycles with modifications of reheat and intercooling are also used for improved performance. Gas Power Cycles ______________________________________________________________ 345 9.4 REGENERATIVE GAS TURBINE CYCLE In earlier discussions it is seen that for the maximization of specific work output the gas turbine exhaust temperature should be equal to compressor exhaust temperature. The turbine exhaust temperature is normally much above the ambient temperature. Thus, their exist potential for tapping the heat energy getting lost to surroundings with exhaust gases. Here it is devised to use this potential by means of a heat exchanger called regenerator, which shall preheat the air leaving compressor before entering the combustion chamber, thereby reducing the amount of fuel to be burnt inside combustion chamber (combustor). Regenerative air standard gas turbine cycle shown ahead in Fig. 9.10 has a regenerator (counter flow heat exchanger) through which the hot turbine exhaust gas and comparatively cooler air coming from compressor flow in opposite directions. Under ideal conditions, no frictional pressure drop occurs in either fluid stream while turbine exhaust gas gets cooled from 4 to 4' while compressed air is heated from 2 to 2'. Assuming regenerator effectiveness as 100% the temperature rise from 2\u20132' and drop from 4 to 4' is shown on T-S diagram. 3 4' Regenerator 2' 2 C 1 Fuel 4 T CC 3 T 2' Heat exchange 4' 4 2 1 G S Fig. 9.10 Regenerative air standard gas turbine cycle. Regenerator effectiveness, \u03b5 = h 2'h 4 Thus, thermodynamically the amount of heat now added shall be h 2 h 2 \u2212 \u2212, where \u2018h\u2019 refers to specific enthalpy values. Qadd, regen = m(h3 \u2013 h2') where as without regenerator the heat added; Qadd = m(h3 \u2013 h2) Here it is obvious that, Qadd, regen < Qadd. This shows an obvious improvement in cycle thermal efficiency as every thing else remains same. Net work produced per unit mass flow is not altered by the use of regenerator. Air standard cycle thermal efficiency, \u03b7 regen = \u2212 ( )\u2032 h 2 \u00b7( c T p 3 \u03b7 regen = \u2212 ) ) Thus, regenerative gas turbine cycle efficiency approaches to Carnot cycle efficiency at r = 1 which is because in this limiting situation the Carnot cycle requirement of total heat addition and rejection at maximum and minimum cycle temperature is satisfied. Variation of efficiency with cycle pressure ratio for regenerative cycle is shown in Fig. 9.11 for fixed \u2018t\u2019 values. 346 _________________________________________________________ Applied Thermodynamics Cycle efficiency is seen to decrease with increasing pressure ratio till pressure ratio becomes equal \uf8eb \u03b3 \uf8ec 2t \uf8ed \uf8f6\u2212 1 \uf8f7 \u03b3 \uf8f8. In earlier discussions it has been shown that at this pressure ratio value the specific work to r = output becomes maximum. Same specific work output variation is true for this case too. For", " further higher values of pressure ratio the heat exchanger shall cool the air leaving compressor and lower the cycle efficiency than simple cycle efficiency. This is because of which the constant \u2018t\u2019 lines are not extended beyond the point where they meet efficiency curve for simple cycle 100 80 60 40 20 10 12 14 16 Pressure ratio, r Fig. 9.11 Efficiency vs. pressure ratio for regenerative gas turbine cycle. Thus, it can be concluded that for getting improvement in efficiency by incorporating regenerator the pressure ratio \u2018r\u2019 appreciably less than the optimum one for maximum specific work output should be used. Also maximum turbine inlet temperature should be used. For ideal heat exchange in regenerator with 100% effectiveness, Upon substitution it yields T2' = T4 \u03b7 regen = \u2212 ( regen = ) ) From thermodynamic processes or, or \uf8f6\u2212 Gas Power Cycles ______________________________________________________________ 347 Also we know T 3 T = t; 1 \uf8eb \uf8ec \uf8ed \uf8f6\u2212 = or, t Substituting in expression for efficiency, \u03b7 regen = 1 \u2013 \uf8eb \uf8ec \uf8ed \uf8f6\u2212 \u03b3 1 \uf8f7 \u03b3 \uf8f8 r t Thus, the expression for regenerative cycle efficiency shows that efficiency of regenerative gas turbine cycle depends upon maximum temperature (T3) in cycle and efficiency increases with increasing \u2018t\u2019 value or turbine inlet temperature (T3) at constant cycle pressure ratio. Also the efficiency decreases with increasing pressure ratio for fixed \u2018t\u2019 value, contrary to the behaviour of efficiency shown in case of simple gas turbine cycle where efficiency increases with increasing pressure ratio. At i.e. r = 1, \u03b7 regen = 1 \u2013 regen = \u03b7 \u03b7 Carnot at min T max 9.5 REHEAT GAS TURBINE CYCLE Reheat gas turbine cycle arrangement is shown in Fig. 9.12. In order to maximize the work available from the simple gas turbine cycle one of the option is to increase enthalpy of fluid entering gas turbine and extend its expansion upto the lowest possible enthalpy value. C : Compressor CC : Combustion chamber G : Generator f : Fuel HPT : High pressure turbine LPT : Low pressure turbine RCC : Reheat combustion chamber f f 2 CC 3 4 RCC C HPT 1 5 LPT 6 G T Reheating 5 6 3 4 4' 2 1 S Fig. 9.12 Reheat gas turbine cycle 348 ________________________________________________________ Applied Thermodynamics This can also be said in terms of pressure and temperature values i.e. inject fluid at high pressure and temperature into gas turbine and expand upto lowest possible pressure value. Upper limit at inlet to turbine is limited by metallurgical limits while lower pressure is limited to near atmospheric pressure in case of open cycle. For further increasing the net work output the positive work may be increased by using multistage expansion with reheating in between. In multistage expansion the expansion is divided into parts and after part expansion working fluid may be reheated for getting larger positive work in left out expansion. For reheating another combustion chamber may be used. Here in the arrangement shown ambient air enters compressor and compressed air at high pressure leaves at 2. Compressed air is injected into combustion chamber for increasing its temperature upto desired turbine inlet temperature at state 3. High pressure and high temperature fluid enters high pressure turbine (HPT) for first phase of expansion and expanded gases leaving at 4 are sent to reheat combustion chamber (reheater) for being further heated. Thus, reheating is a kind of energising the working fluid. Assuming perfect reheating (in which temperature after reheat is same as temperature attained in first combustion chamber), the fluid leaves at state 5 and enters low pressure turbine (LPT) for remaining expansion upto desired pressure value. Generally, temperature after reheating at state 5 is less than temperature at state 3. In the absence of reheating the expansion process within similar pressure limits goes upto state 4'. Thus, reheating offers an obvious advantage of work output increase since constant pressure lines on T-S diagram diverge slightly with increasing entropy, the total work of the two stage turbine is greater than that of single expansion from state 3 to state 4'. i.e., (T3 \u2013 T4) + (T5 \u2013T6) > (T3 \u2013 T4\u2032). Here it may be noted that the heat addition also increases because of additional heat supplied for reheating. Therefore, despite the increase in net work due to reheating the cycle thermal efficiency would not necessarily increase. Let us now carry out air standard cycle analysis. Net work output in reheat cycle, Wnet, reheat = WHPT + WLPT \u2013 WC WHPT = m(h3 \u2013 h4), WLPT = m(h5 \u2013 h6), WC = m(h2 \u2013", " h1) assuming Wnet, reheat = m{(h3 \u2013 h4) + (h5 \u2013 h6) \u2013 (h2 \u2013 h1)} Wnet, reheat = m cp {(T3 \u2013 T4) + (T5 \u2013 T6) \u2013 (T2 \u2013 T1)} T3 = T5 i.e. perfect reheating Wnet, reheat = m cp {(2T3 \u2013 T4) \u2013 T6 \u2013 (T2 \u2013 T1)} \uf8f1 \uf8eb \uf8f4 \uf8f2 \uf8ec \uf8f4 \uf8ed \uf8f3 Wnet, reheat = m cp T1 \uf8f6 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe Here \uf8eb \uf8ec \uf8ed Let pressure ratio for HP turbine be r1 and for LP turbine be r2 then; T 3 T = t; pressure ratio, r = 1 \uf8eb \u03b3 \uf8ec \u03b3 \u2212\uf8ed = r1 \u00d7 r2 \uf8eb \u03b3 \uf8ec \u03b3 \u2212\uf8ed 1 r1 = \uf8eb \uf8ec \uf8ed T 3 T 4 \uf8f6 \uf8f7 \uf8f8 \uf8eb \u03b3 \uf8ec \u2212\uf8ed \u03b3 1 \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8, r2 = \uf8eb \uf8ec \uf8ed or \uf8eb \uf8ec \uf8ed \uf8f6\u2212 \uf8f6 \uf8f7 \uf8f8 = Gas Power Cycles ______________________________________________________________ 349 Let then \u2212 net, reheat mc T 1 p W net, reheat mc \uf8f6\u2212 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \u2212 \uf8eb \uf8ec \uf8ec \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed \uf8f6\u2212 \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \u2212 \uf8eb \uf8ec \uf8ec \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed \uf8f6\u2212 \u03b3 1 \uf8f7 \u03b3 \uf8f8 r \u2212 1 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8f4 \uf8fd \uf8f4 \uf8f4 \uf8fe dw dr = Equating dw dr 1 to zero, 1 \u03b3\u2212 \u03b3 2 1/ r 1 = 1 \uf8f6\u2212 \uf8f6\u2212 1 \uf8f7 \uf8f8 = \u03b3 \uf8eb \uf8ec 2 1r \uf8ed \uf8f6\u2212 \u03b3 1 \uf8f7 \u03b3 \uf8f8 or, r1 = r = r2 Thus, expansion pressure ratio for high pressure turbine and low pressure turbine should be same for maximum work output from the cycle. With such optimum division the expression for work output and efficiency can be given as \uf8eb \uf8ec \uf8ed \uf8f6\u2212 \uf8eb \uf8ec \uf8ed \uf8f6\u2212 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8f4 \uf8fd \uf8f4 \uf8f4 \uf8fe Let, \uf8eb \uf8ec \uf8ed \uf8f6\u2212 \u03b3 1 \uf8f7 \u03b3 \uf8f8 r = c then, \uf8eb \uf8ec \uf8ec \uf8ed W net, reheat mc ) \uf8fc \uf8fd \uf8fe W net, reheat mc \uf8fe Efficiency for reheat cycle \u03b7 reheat = ( m h 3 \u2212 W net, reheat + ( ) m h 5 h 2 \u2212 h 4 ) 350 ________________________________________________________ Applied Thermodynamics W net, reheat \u2212 ( ) T T 5 2 + {( T 3 = mc p \u2212 T 4 )} \u03b7 reheat = Let us simplify denominator first, mc net, reheat \uf8eb \uf8f6 \uf8ec \uf8f7 \uf8ed \uf8f8 We know Also Substituting, = m cp T1 \uf8f1 \uf8eb \uf8f4 \uf8f2 \uf8ec \uf8f4 \uf8ed \uf8f3 T5 = T3 and T 3 T = t 1 = m cp T1 \uf8f1 \uf8eb \uf8f4 \uf8f2 \uf8ec \uf8f4 \uf8ed \uf8f3 \u2212 cp T1 \uf8f1 \uf8eb \uf8f4 \uf8ec \uf8f2 \uf8ec \uf8ec \uf8f4 \uf8ed \uf8f3 \uf8eb \uf8ec \uf8ed \uf8f6\u2212 cp T1 \uf8f6\u2212 Thus, \uf8f1 \uf8f2 t Denominator = m cp T1 Substituting in expression for efficiency for optimum division of pressure ratio", "; \u03b7 reheat = W net, reheat \uf8f1 \uf8f2 1 2 t \uf8f3 \u2212 \u2212 c mc T p \uf8fc \uf8fd \uf8fe t c \u03b7 reheat = \uf8f1 \uf8f2 \uf8f3 \u2212 352 ________________________________________________________ Applied Thermodynamics reducing compression work. It is based on the fact that for a fixed compression ratio higher is the inlet temperature higher shall be compression work requirement and vice-a-versa. Schematic for intercooled gas turbine cycle is given in Fig. 9.15. Thermodynamic processes involved in multistage intercooled compression are shown in Figs. 9.16, 9.17. First stage compression occurs in low pressure compressor (LPC) and compressed air leaving LPC at \u20182\u2019 is sent to intercooler where temperature of compressed air is lowered down to state 3 at constant pressure. In case of perfect intercooling the temperatures at 3 and 1 are same. Intercooler is a kind of heat exchanger where heat is picked up from high temperature compressed air. The amount of compression work saved due to intercooling is obvious from p-V diagram and shown by area 2342'. Area 2342' gives the amount of work saved due to intercooling between compression. Intercooler Q 2 LPC 1 fuel CC 5 4 3 HPC 6 T G CC : Combustion chamber LPC : Low pressure compressor HPC : High pressure compressor T : Turbine G : Generator Fig. 9.15 Gas turbine cycle with intercooling'' = Isothermal process 1 \u2013 2 = Adiabatic compression 3 \u2013 4 = Adiabatic compression 1 V Fig. 9.16 Intercooled compression 5 6 T 2' 2 1 4 3 S Fig. 9.17 T-S diagram for gas turbine cycle with intercooling Some large compressors have several stages of compression with intercooling between stages. Use of multistage compression with intercooling in a gas turbine power plant increases the net work produced because of reduction in compressor work. Intercooled compression results in reduced temperature at the end of final compression. T-S diagram for gas turbine cycle with intercooling shows that in the absence of intercooling within same pressure limits the state at the end of compression would be 2' while with perfect intercooling this state is at 4 i.e., T2' > T4. The reduced temperature at compressor 356 ________________________________________________________ Applied Thermodynamics a T 2 2s 1 4 4s b Fig. 9.22 Effect of irreversibilities and losses in gas turbine cycle. Isentropic efficiency of turbine and compressor can be mathematically given as below: Isentropic efficiency of turbine, \u03b7 isen 4s \uf8fc \uf8fd \uf8fe i.e. \u03b7 isen, t = Actual expansion work Ideal expansion work Isentropic efficiency of compressor, \u03b7 isen, c = \uf8f1 \uf8f2 \uf8f3 sh 2 h 2 \uf8fc\u2212 h 1 \uf8fd \u2212 h \uf8fe 1 i.e. \u03b7 isen, c = Ideal compressor work Actual compressor work Other factors causing the real cycle to be different from ideal cycle are as given below: (i) Fluid velocities in turbomachines are very high and there exists substantial change in kinetic energy between inlet and outlet of each component. In the analysis carried out earlier the changes in kinetic energy have been neglected whereas for exact analysis it cannot be. (ii) In case of regenerator the compressed air cannot be heated to the temperature of gas leaving turbine as the terminal temperature difference shall always exist. (iii) Compression process shall involve work more than theoretically estimated value in order to overcome bearing and windage friction losses. Different factors described above can be accounted for by stagnation properties, compressor and turbine isentropic efficiency and polytropic efficiency. Stagnation properties: Stagnation properties are properties corresponding to stagnation states. Stagnation state of a fluid refers to the state at which fluid is adiabatically brought to the state of rest without work transfer. By using stagnation properties for themodynamic analysis the effect of kinetic energy variation is taken care and analysis becomes realistic. Let us write down the steady flow energy equation considering no heat and work interactions between states 1 and state 0. Let state \u20182\u2019 be the stagnation state. Generally stagnation states are denoted by using subscript \u20180\u2019. At stagnation state velocity of fluid shall be zero. h1 + 2 c 1 2 = h0 + 0, or, h0 = h1 + 2 c 1 2 Gas Power Cycles ______________________________________________________________ 357 or, h= + h 0 2 c 2 When the fluid is perfect gas, then where T0 = stagnation temperature or total temperature T = static temperature, c = velocity 2 c 2 p c = dynamic temperature When a gas stream is slowed down", " and the temperature rises then there is simultaneous rise in pressure. The stagnation pressure p0 is defined similar to T0 with certain modification that the gas is brought to rest isentropically Stagnation pressure can be mathematically given as, 0p p = \uf8eb \uf8ec \uf8ed 0T T 1) \u03b3 \u03b3\u2212 /( \uf8f6 \uf8f7 \uf8f8 Let us use stagnation properties for an isentropic compression process between states 1 and 2. Stagnation pressure ratio, p 02 p 01 = = \u00d7 p 02 p 2 \u00d7 p 1 p 01 p 2 p 1 \uf8eb \uf8ec \uf8ed T 02 T 2 \u00d7 T 1 T 01 \u00d7 \uf8f6 \uf8f7 \uf8f8 T 2 T 1 \u03b3 \u03b3\u2212 /( 1) \uf8eb \u03b3 \uf8ec \u03b3 \u2212\uf8ed 1 \uf8f6 \uf8f7 \uf8f8 p 02 p 01 \uf8eb = \uf8ec \uf8ed T 02 T 01 \uf8f6 \uf8f7 \uf8f8 Thus, it shows p0 and T0 can be used similar to static values. Stagnation pressures and temperatures can be used with static values to determine combined thermodynamic and mechanical state of stream. Representation of compression process on T-S diagram using static and stagnation properties gives better conceptual information. T p02 02 02' 2' 2 01 1 S p2 C 2 2 (2 )Cp p01 p1 C 2 1 (2 )Cp Fig. 9.23 Compression process representation using static and stagnation states 358 ________________________________________________________ Applied Thermodynamics 9.11 COMPRESSOR AND TURBINE EFFICIENCY Isentropic efficiency of compressor and turbine as defined earlier can also be given in terms of stagnation properties. Isentropic compressor efficiency, \u03b7 isen, c = W W ideal actual \u2206 h 0'\u2206 h 0 = In terms of temperature values, assuming mean specific heat value over a range of temperature, the efficiency can be given as, \u03b7 isen, c = T 02'T 02 \u2212 \u2212 T 01 T 01 Isentropic turbine efficiency, \u03b7 isen, t = W W actual ideal \u03b7 isen, t = T 03 T 03 \u2212 \u2212 T 04 T 04'p03 p3 p04 p4 T 03 3 04' 4' 04 4 s Fig. 9.24 Expansion process representation using static and stagnation states Polytropic efficiency: In case of multistage compression and expansion the isentropic efficiency of complete machine as defined earlier may vary with pressure ratio. Actually it is found that compressor isentropic efficiency decreases with pressure ratio while turbine isentropic efficiency increases with pressure ratio, therefore use of fixed typical values of these efficiencies is not proper. Let us consider an axial flow compressor comprising of number of stages having equal stage efficiency for all constituent stages as \u03b7 s. Let isentropic temperature rise in a stage be \u2206Ts' while actual temperature rise is \u2206Ts. Let isentropic temperature rise in whole compressor be \u2206T' while actual temperature rise be \u2206T. For a stage, \u03b7 s ='s \u2206 T \u2206 T s Gas Power Cycles ______________________________________________________________ 359 2 2' T T 1 S Multi stage compression Fig. 9.25 Isentropic efficiency of compression, \u03b7 isen, c = From stage efficiency expression, \u2206 'T \u2206 T 3 4 4' S Multi stage expansion \u2211 \u2206Ts = \u2211 \uf8eb \uf8ec \uf8ed 'Ts = \u2211 \u2206 T's \u03b7 s It is obvious that, \u2211 \u2206Ts = \u2206T for multistage compressor \u2206T = \u2211 \u2206 T's \u03b7 s Substituting from compressor isentropic efficiency in stage efficiency expression \u2206 \u03b7 'T = \u2211 \u2206 T's \u03b7 s isen, c From the nature of constant pressure lines on T-S diagram it could be concluded that constant \u03b7 isen, c = \u03b7 s \uf8eb \u2206 T \uf8ec \uf8ec \u2211 \u2206\uf8ed'' T s \uf8f6 \uf8f7 \uf8f7 \uf8f8 or so, or pressure lines are of diverging type. Therefore, \u2206T' < \u2211 \u2206Ts \u03b7 Hence, With the increasing pressure ratio this difference between \u2211 \u2206Ts \u03b7< s isen, c'with increasing pressure ratio the \u03b7'and \u2206T' goes on increasing and so isen, c goes on reducing despite constant stage efficiency. This 360 ________________________________________________________ Applied Thermodynamics phenomenon may be attributed to the fact that the increase in temperature due to friction in a stage results in more work requirement on next stage. This effect is also termed as the \u2018preheat\u2019", " effect. For multistage expansion in turbine: Let expansion in turbine be of multistage tpye with each stage having equal stage efficiency of \u03b7 s and actual stage temperature drop as \u2206Ts. Let overall temperature drop in turbine be \u2206T' for isentropic expansion and \u2206T for actual expansion while turbine isentropic efficiency is \u03b7 s and stage isentropic temperature drop as \u2206T' isen, t. Stage efficiency of turbine, \u03b7 s = Cumulatively \u2206 \u2206 T s'T s \u2211 \u2206Ts \u2032 = T s \u2211 \u2206 \u03b7 s It is obvious from Fig. 9.25 showing multistage expansion that, \u2211 \u2206Ts = \u2206T \u2206 T \u03b7 s \u2211 \u2206Ts \u2032 = So, or, Turbine isentropic efficiency, \u03b7 isen, t. ='Combining two expressions for turbine isentropic efficiency and stage efficiency, it yields, \u2206 \u2206 T T isen, t = \u03b7 \u03b7 s \u00b7 \u2032 \u2211 \u2206 sT \u2206 \u2032 T It may also be seen that due to diverging nature of constant pressure lines on T-S diagram, \u2206T \u2032 < \u2211 \u2206Ts \u2032. \u03b7 \u03b7> s Therefore, With increasing pressure ratio this difference between \u2206T \u2032 and \u2211 \u2206Ts' goes on increasing and thus isen, t with increasing pressure ratio \u03b7 isen, t goes on increasing despite constant stage efficiency. This may be attributed to the fact that frictional reheating in one stage is partially recovered as work in next stage. Graphical pattern showing behaviour of \u03b7 isen, c and \u03b7 isen, t is shown in Fig. 9.26 95 90 85 80 75 Turbine Compressor Fig. 9.26 Isentropic efficiency of compressor and turbine with pressure ratio Pressure ratio r Gas Power Cycles ______________________________________________________________ 361 Thus, it is seen that isentropic efficiency of turbine and compressor are not suitable parameters to account for non-idealities in turbines and compressors. Hence the polytropic efficiency, also called small-stage efficiency is used for turbines and compressors. \u201cPolytropic efficiency is defined as the isentropic efficiency of an elemental stage in the compression or expansion process such that it remains constant throughout the whole process.\u201d Polytropic efficiency for compressor and turbine may be poly, c and \u03b7 denoted as \u03b7 poly, t. In an elemental stage let the isentropic temperature rise be dT' while actual temperature rise is dT. For compressor, \u03b7 poly, c = \u00b4dT dT Isentropic compression processes shall follow; T \u03b3\u2212 \u03b3 1/ p = constant Taking log of both sides, log T = Differentiating partially, \uf8eb \uf8ec \uf8ed \uf8f6\u2212 1\u03b3 \uf8f7 \u03b3 \uf8f8 log p + log (constant) \u00b4dT T \uf8eb \uf8ec \uf8ed \uf8f6\u2212 1\u03b3 \uf8f7 \u03b3 \uf8f8 dp p = Substituting dT' from polytropic efficiency expression, \u03b7 poly, c dT T \uf8eb \uf8ec \uf8ed \uf8f6\u2212 1\u03b3 \uf8f7 \u03b3 \uf8f8 dp p = Integrating between states 1 and 2; \u03b7 poly, c ln \uf8eb \uf8ec \uf8ed T 2 T 1 \uf8f6 \uf8f7 \uf8f8 = ln \uf8eb \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f8 ln \uf8eb \uf8ec \uf8ed \u03b7 poly \uf8eb \uf8ec \uf8ed ln or Let us write \uf8eb \uf8ec \uf8ec \uf8ed \uf8eb \uf8ec \uf8ec \uf8ed \u2212 \u03b3 1 \u03b3 \u03b7 \u00b7 poly, \u03b3\u03b7 \u00b7 poly, c \uf8f6 \uf8f7 \uf8f7 \uf8f8 = \u2212\uf8eb \uf8ec \uf8ed 1n n \uf8f6 \uf8f7 \uf8f8 then) n Hence \u2018n\u2019 is new index of non isentropic compression process and may be called polytropic index. Thus, polytropic efficiency of compressor yields modified index of compression to account for nonisentropic compression. 362 ________________________________________________________ Applied Thermodynamics \u2212 1n \u00b7 poly, c \uf8f6 \uf8f7 \uf8f7 \uf8f8 Similarly for polytropic efficiency of expansion in turbine. In elemental stage of turbine the isentropic temperature rise dT\u00b4 and actual temperature rise dT occurs. \u03b7 poly, t = \uf8eb \uf8ec \uf8ed dT dT \u00b4 \uf8f6", " \uf8f7 \uf8f8 Expansion in elemental stage is governed by Taking log and differentiating partially, T \u2212 = Constant \u03b3 1 \u03b3 p \uf8f6\u2212 1\u03b3 \uf8f7 \uf8f8 \u03b3 Substituting from polytropic efficiency of turbine, dT T \uf8eb \uf8ec \uf8ed = \u2032 dp p dT poly,t \u00b7 T \u03b7 dT T \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed = = \uf8f6\u2212 1\u03b3 \uf8f7 \u03b3 \uf8f8 dp p \u03b3 ( \u2212 \u03b7 poly, t 1) \u00b7 \u03b3 \uf8f6 \uf8f7 \uf8f8 \u00b7 dp p Integrating between states 3 and \u00b7( \u03b7 poly,t \u03b3 \u2212 1) \uf8f6 \uf8f7 \uf8f7 \uf8f8 Here also the polytropic index for non-isentropic expansion process may be given as, ( n \u2212 1 \u2212 1n n \uf8eb = \uf8ec \uf8ed \u03b7 \u03b3 poly,t \u00b7( \u03b3 \u2212 1) \uf8f6 \uf8f7 \uf8f8 Thus, the polytropic efficiency may be assumed constant over a range of pressure ratio as it takes care of variation of isentropic efficiency with pressure ratio. 9.12 ERICSSON CYCLE From earlier discussions, we have seen that the thermal efficiency of gas turbine power plant may be increased by regeneration, reheat and intercooling. But there exists limit to the number of stages practically. If a large number of multistages of compression with intercooling and large number of multistages of expansion with reheating and regeneration are employed then the cycle is called Ericsson cycle. Ericsson Gas Power Cycles ______________________________________________________________ 363 cycle is shown below in Fig. 9.27. Here 1-2 is isothermal compression while 2-3 is constant pressure heat addition process. 3-4 is isothermal expansion and 4-1 is constant pressure heat rejection process. 3-4 is isothermal expansion and 4-1 is constant pressure heat rejection process.Heat addition and heat rejection processes may be considered to occur in heat exchanger having 100% effectiveness. T Tmax 3 4 T Tmax 3 4 Tmin 2 p c = 1 S p c = Tmin 2 Fig. 9.27 Ericsson cycle p c = p c = 1 S The Ericsson engine involves; (i) a turbine through which gas expands isothermally doing work and absorbing heat from an energy reservoir at Tmax. (process 3-4). (ii) a compressor that compresses gas isothermally while heat is rejected from the gas at Tmin (process 1-2) to the energy reservoir at Tmin. (iii) a counter flow heat exchanger which is used as regenerator and the gas coming from compressor is heated in process 2-3 while the gas leaving turbine is cooled in process 4-1. In this type of regenerator the desired heat exchange is possible only when an infinitesimal temperature difference exists between two gas streams at any section and regenerator should operate reversibly. Although such regenerator is not practically possible. Thus, all processes in Ericsson cycle are reversible processes and cycle approaches to Carnot cycle. Energy reservoir at Tmin Regenerator (Heat exchanger) Wc C C 2 1 Energy reservoir at Tmax 3 T T WT 4 Fig. 9.28 Schematic arrangement for Ericsson cycle Heat added, Qadd = RTmax \u00b7 ln V 4 V = RTmax \u00b7 3 p 3 p 4 364 _________________________________________________________ Applied Thermodynamics Heat rejected, Qrejected = \u2013 RTmin \u00b7 ln V 2 V = RTmin \u00b7 ln 1 p 2 p 1 Here, Also, p2 = p3 & p1 = p4, so Wnet = Qadd \u2013 Qrejected p 3 p = 4 p 2 p 1 Ericsson cycle efficiency, \u03b7 Ericsson = Substitution yields, Q add \u2212 Q Q rejected add \u03b7 Ericsson = min \u2212 T T max T max T T max min \u03b7 Ericsson = 1 \u2013 = \u03b7 Carnot 9.13 STIRLING CYCLE Stirling cycle consists of four reversible processes as given below: (i) Isothermal compression from state 1 to 2 at temperature Tmin. (ii) Constant volume heat addition from state 2 to 3. (iii) Isothermal expansion from state 3 to 4 at temperature Tmax. (iv) Constant volume heat rejection from state 4 to 1. Here also similar to Ericsson cycle a regenerator with 100% effectiveness is employed. Regenerator facilitates for the heat rejected during process 4\u20131 to be used as heat input in process 2-3. Efficiency of Stirling cycle is similar to that of Carnot cycle. P 3 2 T = c 4 Tmax T = c Tmin 1 T Tmax 3 V = c Tmin 2 1", " 4 V = c V s Fig. 9.29 Stirling cycle For a regenerator having 100% effectiveness the heat added in process 2-3 and heat rejected in process 4-1 will be same. If we consider regenerator as part of system then heat interactions with surroundings shall be taking place at temperatures Tmin and Tmax only. Heat added in cycle will be during process 3\u20134 at temperature Tmax and heat rejected in cycle will be during process 1-2 at temperature Tmin. Gas Power Cycles ______________________________________________________________ 365 Heat added, Qadd = RTmax \u00b7 ln Heat rejected, Qrejected = RTmin \u00b7 ln \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed V 4 V 3 V 1 V 2 \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8 Here, V4 = V1 and V3 = V2 so Stirling cycle thermal efficiency, \u03b7 stirling = W Q net add = \u2013Q add Q rejected Q add = 1 \u2013 RT min \u00b7ln RT max \u00b7ln \uf8eb \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed V 1 V 2 V 4 V 3 \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f8 stirling = 1 \u2013 min \u03b7 T T max = \u03b7 Carnot In case of regenerator operating with effectiveness less than 100% the heat added and rejected in processes 2-3 and 4-1 will not be same and some heat would be lost. Let us consider regenerator effectiveness as \u2018\u2208\u2019. Heat added, Qadd = RTmax \u00b7 ln Heat rejected, Qrejected = RTmin \u00b7 ln \uf8eb \uf8ec \uf8ed 1 \u2013 \u2208) \u00b7 Cv ( T max T\u2212 min ) + (1 \u2013 \u2208) \u00b7 Cv ( T max T\u2212 min ) Modified value of Stirling cycle efficiency, \u03b7\u2032 stirling = R T max EXAMPLES \u2212 ( R T r )\u00b7ln T min max + \u2212 \u2208 (1 ) ( C T v \u00b7ln r \u2212 T min ) max 1. A four stroke SI engine has the compression ratio of 6 and swept volume of 0.15 m3. Pressure and temperature at the beginning of compression are 98 kPa and 60\u00baC. Determine the pressure, volume and temperatures at all salient points if heat supplied to it is 150 kJ/kg. Also find out entropy change, work done, efficiency and mean effective pressure of cycle assuming cp = 1 kJ/kg \u00b7 K, cv = 0.71 kJ/kg \u00b7 K. Also plot the cycle on T-S diagram. 366 _________________________________________________________ Applied Thermodynamics 3 4 3 2 p 4 1 Swept volume V T 2 1 S Fig. 9.30 Solution: SI engines operate on Otto cycle. Consider working fluid to be perfect gas. Here Given: Therefore, \u03b3= c p c v = 1.4 cp \u2013 cv = R = 0.29 kJ/kg \u00b7 K. P1 = 98 kPa T1 = 60 + 273.15 = 333.15 K Q2-3 = 150 kJ/kg V 1 V = 2 V2 = 0.03 m3 + 0.15 V 2 V 2 Compression ratio, r = = 6 Total cylinder volume = 0.18 m3 = V1 From perfect gas law, PV = mRT From state 1 to 2 by PV\u03b3 = constant m = PV 1 1 RT = 0.183 kg. 1 1 1PV \u03b3 = 2 2P V \u03b3 or P2 = P1 \u00d7 \uf8eb \uf8ec \uf8ed V 1 V 2 \u03b3 \uf8f6 \uf8f7 \uf8f8 = 1204.03 kPa Also From heat addition process 2 \u2013 3 Also from PV yields, T2 = 682.18 K Q2-3 = mcv (T3 \u2013 T2) 150 = 0.183 \u00d7 0.71 (T3 \u2013 682.18) T3 = 1836.65 K Gas Power Cycles ______________________________________________________________ 367, P3 = 3241.64 kPa For adiabatic expansion 3-4, and Hence and from 3 3P V \u03b3 4 4P V \u03b3 = V4 = V1 P4 = \u03b3 \u00b7P V 3 3 \u03b3 V 1 = 263.85 kPa ; T4 = 896.95 K Entropy change from 2-3 and 4-1 are same, and can be given as, S3 \u2013 S2 = S4 \u2013 S1 = mcv ln T 4 T = 0.183 \u00d7 0.71 \u00d7 ln 1 \uf8eb \uf8ec \uf8ed 896.95 333.15 \uf8f6 \uf8f7 \uf8f8 Entropy change \u2206S3-", "2 = \u2206S4-1 = 0.1287 kJ/K. Heat rejected, Q4-1 = mcv \u00d7 (T4 \u2013T1) = 0.183 \u00d7 0.71 (896.95 \u2013333.15) Q4\u20131 = 73.25 kJ Net Work = (Q2-3) \u2013 (Q4-1) = 150 \u2013 73.25 Net Work = 76.75 kJ Efficiency = Net work Heat added = 76.75 150 = 0.5117 or \u03b7= 51.17% Mean effective pressure = Work Volume change = 76.75 0.15 = 511.67 kPa m.e.p. = 511.67 kPa Ans. 2. In a Diesel engine during the compression process, pressure is seen to be 138 kPa at th of stroke and 7 8 1.38 MPa at th of stroke. The cut-off occurs at 1 15 compression ratio assuming indicated thermal efficiency to be half of ideal efficiency, mechanical efficiency as 0.8, calorific value of fuel = 41800 kJ/kg and \u03b3 = 1.4. Also find fuel consumption bhp/hr. Solution: As given th of stroke. Calculate air standard efficiency and 1 8 VA = V2 + VB = V2 + 7 8 1 8 (V1 \u2013 V2) (V1 \u2013 V2) 368 _________________________________________________________ Applied Thermodynamics 2 3 p 1.38 MPa B 138 kPa A 4 1 V Fig. 9.31 and also or pA \u00b7 AV \u03b3 = pB \u00b7 BV / \uf8f6 \uf8f7 \uf8f8 P B P A 1380 138 1/1..18 Also substituting for VA & VB +.18 It gives r = 19.37 V 1 V = 19.37, Compression ratio = 19.37 2 As given; Cut off occurs at \uf8f1 \uf8f2 \uf8f3 V V\u2212 1 2 15 \uf8fc \uf8fd \uf8fe volume or Cut off ratio, V3 = \uf8f1 \uf8f2 \uf8f3 V V\u2212 1 2 15 \uf8fc \uf8fd \uf8fe + V2 \u03c1= V 3 V = 2.22 2 Gas Power Cycles ______________________________________________________________ 369 Air standard efficiency for Diesel cycle = 1 \u2013 1 \u03b3 \u03b3\u2212 1 \u00b7 r \uf8ee \uf8ef \uf8f0 \u03b3\u03c1 \uf8f9\u2212 1 \uf8fa \u2212 \u03c1 1 \uf8fb = 0.6325 \u03b7 air standard = 63.25% Overall efficiency = air standard efficiency \u00d7 0.5 \u00d7 0.8 Fuel consumption, bhp/hr = = 0.6325 \u00d7 0.5 \u00d7 0.8 = 0.253 or 25.3% \u00d7 \u00d7 75 60 60 \u00d7 \u00d7 0.253 41800 10 2 = 0.255 kg Compression ratio = 19.37 Air standard efficiency = 63.25% Fuel consumption, bhp/hr = 0.255 kg Ans. 3. In an IC engine using air as working fluid, total 1700 kJ/kg of heat is added during combustion and maximum pressure in cylinder does not exceed 5 MPa. Compare the efficiency of following two cycles used by engine: (a) cycle in which combustion takes place isochorically. (b) cycle in which half of heat is added at constant volume and half at constant pressure. Temperature and pressure at the beginning of compression are 100\u00baC and 103 kPa. Compression and expansion processes are adiabatic. Specific heat at constant pressure and volume are 1.003 kJ/kg \u00b7 K and 0.71 kJ/kg \u00b7 K. Solution: 1-2-3-4 = cycle (a) 1-2\u00b4-3\u00b4-4\u00b4-5 = cycle (b) c Here p c v and R = 0.293 kJ/kg \u00b7 K Let us consider 1 kg of air for perfect gas, = \u03b3 = 1.4 or PV = mRT V1 = mRT 1 p 1 \u00d7 \u00d7 1 0.293 373.15 103 = V1 = 1.06 m3 at state 3, p3V3 = mRT3 or T3 = V\u00d7 5000 3 \u00d7 1 0.293 = 17064.8 V2 4' 3 3' 2' p 4 5 1 2 V Fig. 9.32 for cycle (a) and also for cycle (b) T\u2032 3 =17064.8 V\u2032 2 370 _________________________________________________________ Applied Thermodynamics (a) For Otto cycle or From gas law or Q23 = cv(T3 \u2013T2); given Q23 = 1700 kJ/kg 1700 = 0.71(T3 \u2013T2) T3 \u2013T2 = 2328.77, or T2 = T3 \u2013 2328.77 2328.77) T \u2212 ( 3 p", " 2 2328.77) T \u2212 3 \u00d7 5000 V 2 T 3 5000 T 3, {as V2 = V3} Substituting T3 as function of V2 p 2 V \u2212 2 (17064.8 2328.77) = 5000 (17064.8 )V 2 or Also or upon solving it yields 5000(17064.8 2328.77) \u2212 V 2 (17064.8 ) V 2 p2 = \u03b3 = 1 1p V \u03b3 2 2p V 103 \u00d7 (1.06)1.4 = 5000(17064.8 2328.77) \u2212 V 2 (17064.8 ) V 2 \u00b7 ( 1.4 2V ) 381.4V2 = 17064.8 2.4 2V \u2013 2328.77 1.4 2V V2 = 44.7 2.4 2V \u2013 6.1 1.4 2V 44.7 1.4 2V 2V \u2013 0.136 0.4 1.4 \u2013 6.1 0.4 2V \u2013 1 = 0 2V \u2013 0.022 = 0 or By hit and trial it yields Thus, compression ratio Otto cycle efficiency, or (b) For mixed or dual cycle Given: V2 = 0.18 m3 V 1 V = r = 2 1.06 0.18 = 5.89 1 \u03b3\u2212 = 0.5083 \u03b7 otto = 1\u2013 \u03b7 otto = 50.83% r 1 cp (T4' \u2013 T3') = cv (T3' \u2013 T2') = 1700 2 = 850 Gas Power Cycles ______________________________________________________________ 371 or or Also \u21d2 T3' \u2013 T2' = 1197.2 T2' = T3' \u2013 1197.' 1197.2 ) = 5000\u00b7V T 3' 2'or P 2'T \u2212 ( 1197.2) 3' Also we had seen earlier that = 5000 T 3'T3\u2032 = 17064.8 V2\u2032 P \u2032 2 \u2032 \u2212 V 2 (17064.8 1197.2) 5000 (17064.8 = )V \u2032 2 or P2\u2032 = For adiabatic process 1-2' 5000(17064.8 \u2212 V \u2032 2 (17064.8 V \u2032 2 1197.2) ) \u03b3 = 1 1PV \u03b3 2 2P V \u2032 \u2032 103 \u00d7 (1.06)1.4 = (V2\u2032)1.4 \u00b7 \uf8eb \uf8ec \uf8ed 5000 \u2212 350.78 V 2 \u2032 \uf8f6 \uf8f7 \uf8f8 111.75 V2\u2032 = 5000 2.4 2'V \u2013 350.78 1.4 2'V 1 = 44.74 1.4 2'V \u2013 3.14 0.4 2'V 1.4 2'V \u2013 0.07 0.4 2'V \u2013 0.022 = 0 or or By hit and trial Therefore upon substituting V2' V2' = 0.122 m3 P2' = 2124.75 kPa T3' = 2082 K T2' = 884.8 K From constant pressure heat addition cp (T4' \u2013 T3') = 850 1.003 (T4' \u2013 2082) = 850 T4' = 2929..122 2929.5 V4' = 2082 V4' = 0.172 m3 Also or {here p3' = p4' and V2' = V3'} 372 _________________________________________________________ Applied Thermodynamics Using adiabatic formulations \u03b3T 4 T 5 = \uf8f6 \uf8f7 \uf8f8 {here V5 = V1} T5 = 2929.5 \u00d7 0.4 \uf8eb \uf8ec \uf8ed 0.172 1.06 \uf8f6 \uf8f7 \uf8f8 or Heat rejected in process 5 \u2013 1, Q5\u20131 = cv (T5 \u2013T1) T5 = 1415.4 K Efficiency of mixed cycle = = 0.71 (1415.4 \u2013 373.15) Q5\u20131 = 739.99 kJ \u2212 1700 739.99 1700 = 0.5647 or 56.47% Otto cycle = 50.83%; \u03b7 \u03b7 mixed cycle = 56.47% Ans. 4. In an air standard Brayton cycle the minimum and maximum temperature are 300 K and 1200 K, respectively. The pressure ratio is that which maximizes the net work developed by the cycle per unit mass of air flow. Calculate the compressor and turbine work, each in kJ/kg air, and thermal efficiency of the [U.P.S.C. 1994] cycle. Solution: Maximum temperature, T3 = 1200 K Minimum temperature, T1 = 300 K Optimum", " pressure ratio for maximum work output, \u03b3 \u03b3\u2212 2( 1) rp = \uf8eb \uf8ec \uf8ed T max T min \uf8f6 \uf8f7 \uf8f8 \u03b3 \u03b3\u2212 1) 2( \uf8f6 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ed rp = T 3 T 1 rp = 11.3 2 p p = rp = 11.3, For process 1\u20132 \u2212\uf8ed 1 \uf8f7 \uf8f8 For process 3\u20134, T2 = 300 \u00b7 ( T2 = 600.21 K 11.3 \u2212 )1.4 1 1. \uf8f6 \uf8f7 \uf8f8, \u21d2 T4 = 599.79 K Gas Power Cycles ______________________________________________________________ 373 Heat supplied, Q23 = cp (T3 \u2013 T2) = 602.79 kJ/kg P = P2 3 P = P1 4 3 4 T 2 1 S Fig. 9.33 Compressor work, WC = cp \u00b7(T2 \u2013 T1) = 1.005 \u00d7 (600.21 \u2013 300) = 301.71 kJ/kg Turbine work WT = cp \u00b7 (T3 \u2013 T4) = 1.005 (1200 \u2013 599.79) = 603.21 kJ/kg Thermal efficiency = Net work Heat supplied = \u2212 W W T c Q 23 = 603.21 \u2013 301.71 602.79 = 0.5002 Compressor work = 301.71 kJ/kg, Turbine work = 603.21 kJ/kg, Thermal efficiency = 50.02% Ans. 5. A gas turbine unit receives air at 1 bar, 300 K and compresses it adiabatically to 6.2 bar. The compressor efficiency is 88%. The fuel has a heating value of 44186 kJ/kg and the fuel-air ratio is 0.017 kg fuel/kg of air. The turbine internal efficiency is 90%. Calculate the work of turbine and compressor per kg of air compressed and thermal efficiency. For products of combustion cp = 1.147 kJ/ kg K, \u03b3 = 1.33. [U.P.S.C. 1992] Solution: Gas turbine cycle is shown by 1\u20132\u20133\u20134 on T-S diagram, Given: P1 = 1 bar, P2 = P3 = 6.2 bar, F/A ratio = 0.017 T1 = 300 K, \u03b7 Heating value of fuel = 44186 kJ/kg For process 1\u20132 being isentropic, compr. = 88%, \u03b7 turb = 90% 6.2 bar 1 bar 4' 2' T 3 4 2 300 T2 = 505.26 K 1 S Fig. 9.34 Considering compressor efficiency, \u03b7 compr = \u2212 \u2212 = 533.25 K, 0.88 = \u2212 (505.26 300) \u2212 300) ( 'T 2 Actual temperature after compression, T'During process 2\u20133 due to combustion of unit mass of air compressed the energy balance shall be as under, Heat added = mf \u00d7 Heating value 374 _________________________________________________________ Applied Thermodynamics \uf8eb \uf8ec \uf8ed m f m a \uf8f6 \uf8f7 \uf8f8 or Here, = ((ma + mf) \u00b7 cp, comb \u00b7 T3) \u2013 (ma \u00b7 cp, air \u00b7 T2\u2032) \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u2013 (cp, air \u00d7 533.25) \uf8f6 \uf8f7 \u00b7 c \uf8f8, comb \u00b7 \u00d7 44186 = m f m a = 0.017, cp, comb = 1.147 kJ/kg \u00b7 K, cp, air = 1.005 kJ/kg \u00b7 K Upon substitution (0.017 \u00d7 44186) = ((1 + 0.017) \u00d7 1.147 \u00d7 T3) \u2013 (1.005 \u00d7 533.25) For expansion 3\u20134 being T3 = 1103.37 T4 = 1103.37 \u00d7 0.33 1.33 \uf8eb \uf8ec \uf8ed 1 6.2 \uf8f6 \uf8f7 \uf8f8 Actual temperature at turbine inlet considering internal efficiency of turbine, T4 = 701.64 K \u2212 \u2212 \u03b7 turb = T'4 T 4 T 3 T 3 T4\u2032 = 741.81 K ; 0.90 = (1103.37 \u2212 (1103.37 701.64) 'T\u2212 4 ) Compressor work, per kg of air compressed = cp, air \u00b7 (T2\u2032 \u2013 T1) WC = 1.005 \u00d7 (533.25 \u2013 300) WC = 234.42 kJ/kg or air Compressor work = 234.42 kJ/kg of air Ans. Turbine work, per kg", " of air compressed = cp, comb \u00b7 (T3 \u2013 T4\u2032) = 1.147 \u00d7 (1103.37 \u2013 741.81) Turbine work = 414.71 kJ/kg of air Ans. WT = 414.71, kJ/kg of air Net work = WT \u2013 WC = (414.71 \u2013 234.42) Wnet = 180.29, kJ/kg of air Heat supplied = 0.017 \u00d7 44186 = 751.162 kJ/kg of air. Thermal efficiency = W net Heat supplied = 180.29 751.162 = 24% Thermal efficiency = 24% Ans. 6. In a Brayton cycle gas turbine power plant the minimum and maximum temperature of the cycle are 300 K and 1200 K. The compression is carried out in two stages of equal pressure ratio with intercooling 376 _________________________________________________________ Applied Thermodynamics Compressor work, WC = WC1 + WC2 Using Isentropic efficiency of compressor, = cp (T2\u2032 \u2013 T1) + cp(T4\u2032 \u2013 T3) = cp (T2\u2032 \u2013 T1) + cp(T4\u2032 \u2013 T1) \u03b7 isen\u2032 T 3 4 = T 4 T 4' \u2212 \u2212 T 1 T 1 Thus, ( WC = cp \u00b7 T 2 \u03b7 \u2212 T 1 ) isen, c ( + cp T 4 \u03b7 \u2212 T 1 ) isen, c p c \u03b7 isen, c \u03b3 ( \u2212 1 isen, c = WC = \u03b3 ( \u2212 1 + \uf8f6 \uf8f7 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ec \uf8ed \uf8f6 \uf8f7\u2212 1 \uf8f7 \uf8f8 Turbine work, Using isentropic efficiency of turbine, \u03b7 WT = cp \u00b7 (T5 \u2013 T6\u2032) T 5 T 5 isen, T = \u2032\u2212 T 6 \u2212 T 6 WT = cp \u00b7 (T5 \u2013 T6) \u03b7 isen, T = cp \u00b7 T5 \u00b7 \u03b7 isen, T \uf8eb \uf8ec \u2212\uf8ec pr Net work Wnet = WT \u2013 TC = cp \u00b7 T5 \u00b7 \u03b7 isen, T \u00b7 \uf8eb \uf8ec \u2212\uf8ec 1 \uf8ec \uf8ed 1 \u2212 \u03b3 1 \u03b3 pr \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \u2013 isen, c Differentiating new work with respect to rp. Putting, net d W d r p net d W d r p = cp \u00b7 T5 \u00b7 \u03b7 isen, T \u00b7 \uf8eb \uf8ec \uf8ed \uf8f6\u2212 1\u03b3 \uf8f7 \u03b3 \uf8f8 \u03b3 \u2212 1 2 \u03b3 \u2013 \u00b7 pr \uf8f6\u2212 \uf8eb pc isen, c \u03b3 \u2212 \u2212 1 2pr \u03b3 = 0 cp \u00b7 T5 \u00b7 \u03b7 isen, T pr isen Gas Power Cycles ______________________________________________________________ 377 ) \u2212 \u03b3 3(1 \u03b3 2 pr = T 1 T \u03b7 \u00b7 5 isen, c \u03b7 isen, T \u21d2 rp = \uf8eb \uf8ec \uf8ec \uf8ed T 5 \u00b7 T 1. \u03b7 isen, c \u03b3 2 \u2212 \u03b3 3(1 ) \uf8f6 \uf8f7 \uf8f7 \uf8f8 \u03b7 isen, T Substituting known values, rp, opt = rp, opt = \uf8eb \uf8ec \uf8ec \uf8ed \uf8eb \uf8ec \uf8ed T 1 T 5 \u00b7 \u03b7 isen, c \u03b7 isen, T \u03b3 2 \u2212 \u03b3 3(1 ) \uf8f6 \uf8f7 \uf8f7 \uf8f8 300 1200 0.85 0.90 \u00d7 \u00d7 \u00d7 2 1.4 \u2212 3(1 1.4) \uf8f6 \uf8f7 \uf8f8 = 13.6 Overall optimum pressure ratio = 13.6 Ans. 7. An air compressor has eight stages of equal pressure 1.35. The flow rate through the compressor and its\u2019 overall efficiency are 50 kg/s and 82% respectively. If the air enters the compressor at a pressure of 1 bar and temperature of 313 K, determine; (i) State of air at the exit of compressor (ii) Polytropic or small stage efficiency (iii) Efficiency of each stage (iv) Power required to drive the compressor assuming overall efficiency as 90%. [U.P.S.C. 1992] Solution: (i) Theoretically state of air at exit can be determined by the given stage pressure ratio of 1.35. Let pressure at inlet to first stage be P1 and subsequent intermediate pressure be P2, P3, P4, P5, P6, P7, P8, and exit pressure being P9. Therefore.", "35 or P 9 P = (1.35)8 = 11.03 1 Theoretically, the temperature at exit of compressor can be predicted considering isentropic compression of air (\u03b3 = 1.4 = ( 11.03 )0.4 1.4 T9 = 621.9 K Considering overall efficiency of compression 82% the actual temperature at compressor exit can be obtained \u2212 T 9 T 9, actual T 1 \u2212 T 1 = 0.82 378 _________________________________________________________ Applied Thermodynamics T9, actual = 689.71 K Let the actual index of compression be \u2018n\u2019, then \uf8eb \uf8ec \uf8ed T 9, actual T 1 \uf8f6 \uf8f7 \uf8f8 \u2212 1 n \uf8f6 nP 9 \uf8f7 P \uf8f8 1 \uf8eb \uf8ec \uf8ed = \uf8eb \uf8ec \uf8ed 689.71 313 \u2212 1 n n ) 11.03 \uf8f6 \uf8f8 = ( \uf8f7 n = 1.49 State of air at exit of compressor, Pressure = 11.03 bar Temperature = 689.71 K Ans. (ii) Let polytropic efficiency be \u03b7 polytropic for compressor, then; \u2212 1n n \uf8eb \uf8ec \uf8ed \u03b3 \uf8f6\u2212 1 \uf8f7 \u03b3 \uf8f8 \u00d7 = 1 \u03b7 polytropic \u2212 \uf8eb \uf8ec \uf8ed \u2212 1.4 1 (1.49 1) 1.4 1.49 \u03b7 polytropic = 0.8688 or 86.88% \uf8f6 \uf8f7 \uf8f8 \u00d7 \u03b7 = 1 polytropic Polytropic efficiency = 86.88% Ans. (iii) Stage efficiency can be estimated for any stage. Say first stage. Ideal state at exit of compressor stage \u21d2 \u2212 \uf8f6 \uf8f7 \uf8f8 Actual temperature at exit of first stage can be estimated using polytropic index 1.49. T2 = 341.05 K \u2212 1 n n T 2, actual \uf8f6 \uf8f7 \uf8f8 T2, actual = 345.47 K Stage efficiency for first stage, \u03b7 s, actual ) T 1 \u2212 T 1 ) (341.05 313) (345.47 313) \u2212 \u2212 = = 0.8638 or 86.38% Actual temperature at exit of second stage, \u2212 1 n n T 3, actual T 2, actual = \uf8eb \uf8ec \uf8ed P 3 P 2 \uf8f6 \uf8f7 \uf8f8 Gas Power Cycles ______________________________________________________________ 379 T3, actual = 381.30 K Ideal temperature at exit of second stage \u2212 \u03b3 1 \u03b3 T 3 T 2, actual = \uf8eb \uf8ec \uf8ed P 3 P 2 \uf8f6 \uf8f7 \uf8f8 T3 = 376.43 K Stage efficiency for second stage, \u03b7 s,2 = \u2212 T 3 T 3, actual T 2, actual \u2212 T 2, actual = \u2212 376.43 345.47 \u2212 381.3 345.47 Actual temperature at exit of third stage, \u03b7 s,2 = 0.8641 or 86.41% \u2212 1 n n T 4, actual T 3, actual = \uf8eb \uf8ec \uf8ed P 4 P 3 \uf8f6 \uf8f7 \uf8f8 T4, actual = 420.85 K Ideal temperature at exit of third stage, \u2212 \u03b3 1 \u03b3 T 4 T 3, actual = \uf8eb \uf8ec \uf8ed P 4 P 3 \uf8f6 \uf8f7 \uf8f8, T4 = 415.47 K Stage efficiency for third stage, \u03b7 s, 3 = \u2212 T 4 T 4, actual T 3, actual \u2212 T 3, actual \u2212 (415.47 381.3) \u2212 (420.85 381.3) = = 0.86396 or 86.40% Stage efficiency = 86.4% Ans. (iv) From steady flow energy equation, WC = 9 9 dw\u222b,dh\u222b 1 1 dh = dq + vdp = and dh = du + pdv + vdp for dq = 0 in adiabatic process dh = vdp WC = 9 \u222b 1 vdp Here for polytropic compression Pv1.49 = constant i.e. n = 1.49 WC = \uf8eb n \uf8ec \u2212\uf8ed n \uf8f6 \uf8f7 \uf8f8 1 mRT \uf8fa \uf8fa \uf8fb 380 _________________________________________________________ Applied Thermodynamics \uf8eb \uf8ec \uf8ed 1.49 0.49 \uf8f6 \uf8f7 \uf8f8 \u00d7 50 \u00d7 0.287 \u00d7 313 = \uf8eb \uf8ec \uf8ed 11.03 1.0 \u2212 1.49 1 \uf8f6 1", ".49 \uf8f7 \uf8f8 \uf8ee \uf8ef \uf8ef \uf8ef \uf8f0 \uf8f9 \uf8fa\u2212 1 \uf8fa \uf8fa \uf8fb WC = 16419.87 kJ/s Due to overall efficiency being 90% the actual compressor work = WC, actual = 18244.30 kJ/s Power required to drive compressor = 18244.30 kJ/s Ans. cW 0.90 8. A thermodynamic cycle has following processes occurring sequentially; Adiabatic compression from 1 to 2 Constant volume heat addition from 2 to 3 (explosion) Adiabatic expansion from 3 to 4 Isothermal heat rejection from 4 to 1 Obtain expression for air standard cycle efficiency considering compression ratio of \u2018r\u2019 and expansion ratio of \u2018e\u2019. Solution: Let us analyze for 1 kg of air, Given; V2 = V3, T1 = T4, For process 4\u20131, P1V1 = P4V4 = RT1 Heat added = cv(T3 \u2013T2) Heat rejected in isothermal process, 4\u20131 = P4V4 ln \uf8eb \uf8ec \uf8ed Fig. 9.36 Net work = Heat added \u2013 Heat rejected \uf8f6 \uf8f7 \uf8f8 Wnet = cv(T3 \u2013 T2) \u2013 P4V4 ln \uf8eb \uf8ec \uf8ed e r Efficiency, \u03b7 = Net work Heat added = ( ln \uf8eb \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f8 ) RT 1 ln ( c T v 3 As, R \u03b3 \u2212 cv = ( 1), substituting in expression for efficiency, Gas Power Cycles ______________________________________________________________ 381 \u03b7 =1 \u2013 \u03b3 ( \u2212 1)\u00b7 T 1 ln ( ) ln \u03b7 =) ln \uf8eb \uf8ec \uf8ed \u2212 For process 3\u20134, \uf8eb \uf8ec \uf8ed V 4 V 3 \u03b3 = e\u03b3\u20131 For process 1\u20132, V 1 V 2 Substituting in expression of \u2212 1 \uf8f6 \uf8f7 \uf8f8 = r\u03b3\u20131 \u03b3 ( \u2212 1) ln \u2212 Ans. 9. In a gas turbine installation air is supplied at 1 bar, 27\u00baC into compressor having compression ratio of 8. The air leaving combustion chamber is heated upto 1100 K and expanded upto 1 bar. A heat exchanger having effectiveness of 0.8 is fitted at exit of turbine for heating the air before its inlet into combustion chamber. Assuming polytropic efficiency of the compressor and turbine as 0.85 and 0.90 determine cycle efficiency, work ratio and specific work output of plant. Take cp = 1.0032 kJ/kg \u00b7K for air. poly, T = 0.90 poly, c = 0.85, \u03b7 Solution: \u03b3 = 1.4, \u03b7 Using polytropic efficiency the index of compression and expansion can be obtained as under, Let compression index be nc \uf8eb \uf8ec \uf8ec \uf8ed \u03b3 \u2212 \u03b3 \u03b7 \u00b7 \uf8f6\u2212 1c n \uf8f7 n \uf8f8 c 8 bar poly Let expansion index be nT. \u03b7 c = 1.506 \uf8eb \uf8ec \uf8ed \u21d2 For process 1\u20132 \u03b3 poly,T\u00b7 ( \u03b3 n T n T \u03b7 = \uf8eb \uf8f6\u2212 1 \uf8ec \uf8f7 \uf8ed \uf8f8 nT = 1.346 \u2212 1) \uf8f6 \uf8f7 \uf8f8 1 bar 5 2 2' T 4 4' 6 1 S Fig. 9.37 T-S representation 382 _________________________________________________________ Applied Thermodynamics \u2212 1 nc nc 300 \u2212 1.506 = ( )1.506 1 8 T2 = 603.32 K T3 = 1100 K nT nT \u2212.346 1 1.346 T 4 1100 \uf8eb \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f8 1 8 = T4 = 644.53 K Using heat exchanger effectiveness, \u03b5 = 0..8 = 603.32 \u2212 (644.53 603.32) HE = Heat exchanger C = Compressor CC = Combustion chamber T = Turbine 2 5 3 CC 4 Fig. 9.38 Layout of regenerative cycle T5 = 636.28 K Heat added in combustion chambers, qadd = cp (T3 \u2013 T5) qadd = 1.0032 (1100 \u2013 636.28) qadd = 465.204 kJ/kg Compressor work WC = cp (T2 \u2013 T1) = 1.0032 (603.32 \u2013 300) = 304.29 kJ", "/kg Turbine work, WT = cp (T3 \u2013 T4) = 1.0032 (1100 \u2013 644.53) = 456.93 kJ/kg Cycle efficiency = \u2212 W W T C q add = \u2212 456.93 304.29 465.204 = 0.3281 or 32.81% Work ratio.334 Specific work output = WT \u2013 WC = 152.64 kJ/kg Cycle efficiency = 32.81%, Work ratio = 0.334, Ans. Specific work output = 152.64 kJ/kg 10. A gas turbine plant has air being supplied at 1 bar, 27\u00baC to compressor for getting compressed upto 5 bar with isentropic efficiency of 85%. Compressed air is heated upto 1000 K in combustion chamber where also occurs a pressure drop of 0.2 bar. Subsequently expansion occurs to 1 bar in turbine. Determine isentropic efficiency of turbine, if thermal efficiency of plant is 20%. Gas Power Cycles ______________________________________________________________ 383 Neglect the air property variation throughout cycle. Take \u03b3 = 1.4 Solution: T1 = 300 K, P1 = 1 bar, P2 = 5 bar, \u03b7 For process 1\u20132' c = 0.85, T3 = 1000 \uf8f6 \uf8f7 \uf8f8 \u21d2 T2\u2032 = 300 ( )(1.4 1) 1.4 5 \u2212 For process 3\u20134' T2\u2032 = 475.15 K \u2032 \u2212 T 2 \u2212 T 2 T2 = 506.06 K \u03b7 c = 0.85 = T 1 T 1 '.4 1 1.4 = \uf8eb \uf8ec \uf8ed 1 4.8 \uf8f6 \uf8f7 \uf8f8 T4\u2032 = 1000 \u00d7 \u2212 1.4 1 1.4 \uf8eb \uf8ec \uf8ed 1 4.8 \uf8f6 \uf8f7 \uf8f8 T4\u2032 = 638.79 K 4.8 bar 3 1 bar 4 4' T 2 2' 1 5 bar S Fig. 9.39 T-S representation Compressor work per kg WC = cp (T2 \u2013 T1) = 1.0032 \u00d7 (506.06 \u2013 300) = 206.72 kJ/kg Turbine work per kg, WT = cp(T3 \u2013 T4) = 1.0032 \u00d7 (1000 \u2013 T4), kJ/kg Net output, Wnet = WT \u2013 WC = {206.72 \u2013 (1.0032 (1000 \u2013 T4))}, kJ/kg Heat added, qadd = cp (T3 \u2013 T2) = 1.0032 (1000 \u2013 506.06) = 495.52 kJ/kg Thermal efficiency, \u03b7 = \u21d2 0.20 = net W q add T4 = 892.73 K Therefore, isentropic efficiency of turbine, \u03b7 T = \u03b7 T = 0.2969 or 29.69% 206.71 (1.0032 (1000 \u2212 \u00d7 \u2212 T 4 )) 495.52'\u2212 (1000 892.73) \u2212 (1000 638.79) = Turbine isentropic efficiency = 29.69% Ans. 11. A gas turbine plant has air supplied at 1 bar, 27\u00baC for being compressed through pressure ratio of 10. Compression of air is achieved in two stages with perfect intercooling in between at optimum pressure. The maximum temperature in cycle is 1000 K and compressed air at this temperature is sent for expansion in two stages of gas turbine. First stage expansion occurs upto 3 bar and is subsequently reheated upto 995 K before being sent to second stage. Fuel used for heating in combustion chamber has calorific value of 42,000 kJ/kg. Considering cp = 1.0032 kJ/kg. K throughout cycle determine, net output, thermal efficiency and air fuel ratio when air flows into compressor at 30 kg/s. Take isentropic efficiency of compression and expansion to be 85% and 90% respectively. Gas Power Cycles ______________________________________________________________ 385 \u21d2 T6 = 738.04 K T7 = 995 K For expansion in 7\u20138', \u2212 \u03b3 1 \u03b3'\uf8f6 \uf8f7 \uf8f8 \u21d2 T8' = 995 \u2212 1.4 1 1.4 \uf8eb \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f8 1 3 = 726.95 K Considering expansion efficiency, 0.90 = \u21d2 T8 = 753.75 K Expansion work output per kg air = cp (T5 \u2013T6) + cp (T7 \u2013 T8) WT = 1.0032 {(1000 \u2013 738.04) + (995 \u2013 753.75)} WT = 514.85 kJ/kg Heat added per kg", " air = cp(T5 \u2013 T4) + cp(T7 \u2013 T6) qadd = 1.0032 {(1000 \u2013 437.36) + (995 \u2013 738.04)} qadd = 822.22 kJ/kg 822.22 42000 = 0.01958 Fuel required per kg of air, mf = Air-fuel ratio = 1 0.01958 Net output = WT \u2013 WC = 239.26 kJ/kg = 51.07 Output for air flowing at 30 kg/s, = 239.26 \u00d7 30 = 7177.8 kW Thermal efficiency = \u2212 W W T C q add = 239.26 822.22 = 0.2909 or 29.09% Thermal efficiency = 29.09%, Net output = 7177.8 kW, A/F ratio = 51.07 Ans. 12. A regenerative-reheat cycle has air entering at 1 bar, 300 K into compressor having intercooling in between the two stages of compression. Air leaving first stage of compression is cooled upto 290 K at 4 bar pressure in intercooler and subsequently compressed upto 8 bar. Compressed air leaving second stage compressor is passed through a regenerator having effectiveness of 0.80. Subsequent combustion chamber yields 1300 K at inlet to turbine having expansion upto 4 bar and then reheated upto 1300 K before being expanded upto 1 bar. Exhaust from turbine is passed through regenerator before discharged out of cycle. For the fuel having heating value of 42000 kJ/kg determine fuel-air ratio in each combustion chamber, total turbine work and thermal efficiency. Consider compression and expansion to be isentropic and air as working fluid throughout the cycle. Solution: T1 = 300 K, P1 = 1 bar, P2 = P3 = 4 bar, T3 = 290 K, T6 = 1300 K, P6 = P4 = 8 bar, T8 = 1300 K, P8 = 4 bar Gas Power Cycles ______________________________________________________________ 387 Heat added per kg air, qadd = cp (T6 \u2013 T5) + cp (T8 \u2013 T7) WT = 660.85 kJ/kg qadd = 765.45 kJ/kg Total fuel required per kg of air = 765.45 42000 Net work Wnet = WT \u2013 WC = 450.88 kJ/kg = 0.0182 Cycle thermal efficiency, \u03b7 = net W q add = 450.88 765.45 = 0.5890 or 58.9% Fuel required per kg air in combustion chamber 1, = Fuel required per kg air in combustion chamber 2, = T 5 \u2212 ( pc T 6 42000 T 7 \u2212 ( pc T 8 42000 ) ) = 0.0126 = 0.0056 Fuel-air ratio in two combustion chambers = 0.0126, 0.0056 Total turbine work = 660.85 kJ/kg Cycle thermal efficiency = 58.9% Ans. 13. A Stirling engine operates between temperature limits of 700 K and 300 K with compression ratio of 3. During heat addition process total 30 kJ/s heat is added. The regenerator efficiency is 90% and the pressure at beginning of compression is 1 bar. The number of cycles per minute is 100. Considering specific heat at constant volume as 0.72 kJ/kg\u00b7K, determine the brake output. Take R = 29.27 kJ/kg \u00b7 K. Also determine stroke volume. Solution: Work done per kg of air = R(T2 \u2013 T1). ln r = 29.27 (700 \u2013 300) In 3 W = 12862.55 kJ/kg Heat added per kg of air = RT2 ln r + (1 \u2013 \u03b5) cv \u00b7 (T2 \u2013 T1) = (29.27 \u00d7 700 \u00d7 ln 3) + {(1 \u2013 0.9) \u00d7 0.72 \u00d7 (700 \u2013 300)} q = 22538.27 kJ/kg For 30 kJ/s heat supplied, the mass of air/s = 30 22538.27 = 1.33 \u00d7 10\u20133 kg/s Mass of air per cycle = \u2212\u00d7 3 1.33 10 100 = 1.33 \u00d7 10\u20135 kg/cycle Brake output = 12862.55 \u00d7 1.33 \u00d7 10\u20133 = 17.11 kW 388 _________________________________________________________ Applied Thermodynamics 700 K 2 3 T 300 K 1 4 S Stroke volume, V = mRT P = Fig. 9.42 Stirling cycle ( )5 \u2212\u00d7 1.33 10 \u00d7 1 10 \u00d7 29.27 300 \u00d7 2 = 0.00116 m3 Brake output = 17.11 kW Stroke volume = 0.00116 m3 Ans. 14. A gas turbine installation operates with fixed maximum and minimum temperatures T3 and T1 respectively. Show that the", " optimum specific work output shall be obtained at same overall pressure ratio for each of following arrangement. Also give expression for this pressure ratio for air as working fluid throughout. (i) there is single stage compression followed by two stages of expansion in turbine. The expansion ratio in two stages is equal and reheating is done upto the maximum temperature at inlet of second stage of expansion. (ii) there occurs compression in two stages of equal compression ratio with intercooling upto the minimum cycle temperature at inlet to second stage of compression followed by single stage expansion in turbine. Take isentropic efficiency of compressor and turbine stages as \u03b7 Solution: Let the overall pressure ratio be r. For arrangement (i) c and \u03b7 T respectively. CC RH 3 C : Compressor CC : Combustion chamber T : Turbine RH : Reheater 5 T3 3 5 2 2' 4 4' 6 6' T 1 6 T1 1 S Fig. 9.43 Layout and T-S diagram for (i) arrangement Gas Power Cycles ______________________________________________________________ 389 In compressor T2\u2032 = T1 \u00b7 r \u03b7 c ='\u21d2 T2 = T1 \uf8f1 \uf8f4 +\uf8f2 \uf8fc \uf8f4\u2212 1 \uf8fd \uf8f4 \uf8f4 \uf8fe Let expansion ratio in turbine stages be rT = For first stage expansion r \u2212 \u03b3 1 \u03b3'\uf8f6 \uf8f7 \uf8f8 T4\u2032 = T3 \u2212 \u03b3 1 \u03b3 \uf8eb \uf8ec \uf8ed 1 Tr \uf8f6 \uf8f7 \uf8f8 Isentropic efficiency of turbine' \u21d2 T4 = T3 \u2013 \u03b7 T (T3 \u2013 T4') \u21d2 T4 = T3 \uf8f1 \uf8f4 \uf8f2 1 \uf8f4 \uf8f3 \u2212 \uf8eb \uf8ec \u03b7 1T \uf8ec \uf8ec \uf8ed \u2212 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe 1 \u2212 \u03b3 1 \u03b3 Tr For second stage expansion, T5 = T3 \u2212 \u03b3 1 \u03b3 T 5 T 6' = \uf8eb \uf8ec \uf8ed P 5 P 6 \uf8f6 \uf8f7 \uf8f8, T6\u2032 = T3 \u00b7 Isentropic efficiency' 5 T6 = T5 \u2013 \u03b7 T (T5 \u2013 T6\u2032) \uf8eb \uf8ec \u03b7 1T \uf8ec \uf8ec \uf8ed T6 = T5 \uf8f1 \uf8f4 \uf8f2 1 \uf8f4 \uf8f3 \u2212 \u2212 Tr 1 \u2212 \u03b3 1 \u03b3 Net work per kg of air \u2212 \u03b3 1 \u03b3 \uf8eb \uf8ec \uf8ed 1 Tr \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe = T3 \uf8f1 \uf8f4 \uf8f2 1 \uf8f4 \uf8f3 \u2212 \uf8eb \uf8ec \u03b7 1T \uf8ec \uf8ec \uf8ed \u2212 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe 1 \u2212 \u03b3 1 \u03b3 Tr Gas Power Cycles ______________________________________________________________ 391 For first stage compressor \uf8f6 \uf8f7 \uf8f8, T2\u2032 = T1 \u00b7 ( cr \u2212 \u03b3 1 ) \u03b3 Isentropic compression efficiency, \u03b7 c = For second stage compressor, Ta = T1 ', \u21d2 T2 = T1 \uf8f1 \uf8f4 +\uf8f2 \uf8f6 \uf8f7 \uf8f8, Tb\u2032 = T1 ( cr \u2212 \u03b3 1 ) \u03b3 Isentropic efficiency of compression, \u03b7 c = For expansion in turbine, T b' T b \u2212 \u2212 T a T a \u21d2 Tb = T1 \uf8f1 \uf8f4 +\uf8f2'= \uf8eb \uf8ec \uf8ed P 3 P 4 \uf8f6 \uf8f7 \uf8f8 \u21d2 T4 Isentropic expansion efficiency' \u21d2 T4 = T3 Net work per kg of air 1T \uf8ec \uf8ec \uf8ed \u2212 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe 1 \u2212 \u03b3 1 \u03b3 r Wnet = {cp (T2 \u2013 T1) + cp(Tb \u2013 Ta)} \u2013{cp(T3 \u2013 T4)} = cp \uf8f1 \uf8f4 \uf8f4 \uf8f2 \uf8f6 \uf8f7\u2212 \uf8fc \uf8f4 \uf8f4 \uf8fd \uf8f4 \ufffd", "\ufffd\ufffd \uf8fe \u2013 cp \uf8f1 \uf8f4 \uf8f2 1T \uf8ec \uf8ec \uf8ed \u2212 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f7 \uf8f8 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe 1 \u2212 \u03b3 1 \u03b3 r Differentiating Wnet with respect to \u2018r\u2019 and equating it to zero. \uf8f6 \u03b7 \u03b7 \u00b7T c \uf8f7 T \uf8f8 1 = 0, yields, dW dr \u2212 1) \u03b3 ( )r = \uf8eb \uf8ec \uf8ed T 3 net 31 \uf8f6 \uf8f7 \uf8f8 Ans. Hence it is proved that the optimum pressure ratio remains same for both arrangements as shown here. 392 _________________________________________________________ Applied Thermodynamics 15. In a gas/steam combined cycle power plant exhaust from gas turbine leaves finally through HRSG at 420 K and generates steam in heat recovery steam generator (HRSG) at 6 MPa and 400\u00baC. Ambient air at 1 bar, 17\u00baC enters the compressor operating at pressure ratio of 10. Turbine inlet temperature of gas turbine is 1400 K. The steam turbine operates with condenser pressure of 15 kPa. Consider air as working fluid throughout gas turbine cycle and combined cycle output to be 37.3 MW. Determine the overall efficiency, mass flow rate of steam per kg of air. Take cp, air = 1.0032 kJ/kg \u00b7 K Solution: C : Compressor CC : Combustion chamber T : Turbine FP : Feed pump ST : Steam turbine a CC 3 1 1 bar, 17\u00b0C Gas turbine cycle condenser Steam turbine cycle 3 4 Steam turbine cycle 15 kPa a b Gas turbine cycle 2 1 d c S T Fig. 9.45 Layout and combined T-S diagram (approximate representation) Gas Power Cycles ______________________________________________________________ 393 In gas turbine cycle \u21d2 T2 = 290 \u00d7 ( 10 \u2212 )1.4 1 1.4 = 559.9 K T3 = 1400 K \u21d2 T4 = 1400 \u2212 1.4 1 1.4 \uf8eb \uf8ec \uf8ed 1 10 \uf8f6 \uf8f7 \uf8f8 = 725.13 K Compressor work per kg, WC = cp(T2 \u2013 T1) = 1.0032 \u00d7 (559.9 \u2013 290) = 270.76 kJ/kg Turbine work per kg, WT = cp(T3 \u2013 T4) = 677.03 kJ/kg Heat added in combustion chamber per kg, qadd = cp(T3 \u2013 T2) = 842.79 kJ/kg Net gas turbine output, Wnet, GT = WT \u2013 WC = 406.27 kJ/kg air Heat recovered in HRSG for steam generation per kg of air qHRSG = cp(T4 \u2013 T5) = 306.11 kJ/kg At inlet to steam turbine, ha = 3177.2 kJ/kg, sa = 6.5408 kJ/kg \u00b7 K For expansion in steam turbine, sa = sb Let dryness fraction at state b be x. sb = 6.5408 = sf at 15 kPa + x \u00b7 sfg at 15 kPa x = 0.7976 hb = hf at 15 kPa + x \u00b7 hfg at 15 kPa = 2118.72 kJ/kg At exit of condenser, hc = hf at 15 kPa = 225.94 kJ/kg, vc = 0.001014 m3/kg At exit of feed pump, hd \u21d2 hd \u2013 hc = vc \u00b7 (6 \u00d7 103 \u2013 15) \u00d7 102 hd = 606.88 kJ/kg Heat added per kg of steam = ha \u2013 hd = 2570.32 kJ/kg Mass of steam generated per kg of air = q HRSG h\u2212 h d a ( ) = 0.119 kg steam per kg air. Net steam turbine cycle output, Wnet, ST = (ha \u2013 hb) \u2013 (hd \u2013 hc) Wnet, ST = 677.54 kJ/kg Steam cycle output per kg of air = Wnet, ST \u00d7 0.119 = 80.63 kJ/kg air Total combined cycle output = (Wnet, GT + Wnet, ST) = 486.9 kJ/kg air Combined cycle efficiency, \u03b7 cc = + W net,ST ( W net,GT q add ) = 0.5777 or 57.77% In the absence of steam cycle, Gas turbine cycle efficiency, \u03b7 GT = W net,GT q add \u03b7 GT = 0.4821 or 48.21%", " 394 _________________________________________________________ Applied Thermodynamics Thus, efficiency is seen to increase in combined cycle upto 57.77% as compared to gas turbine offering 48.21% efficiency. Overall efficiency = 57.77% Steam per kg of air = 0.119 kg steam/kg air Ans. 16. In an I.C. engine operating on the dual cycle (limited pressure cycle), the temperature of the working fluid (air) at the beginning of compression is 27\u00baC. The ratio of the maximum and minimum pressures of the cycle is 70 and the compression ratio is 15. The amount of heat added at constant volume and at constant pressure are equal. Compute the air standard thermal efficiency of the cycle. State three main reasons why the actual thermal efficiency is different from the theoretical value. [U.P.S.C. 1993] Solution: T1 = 27\u00baC = 300 K P 4 P 1 = 70 P 3 P 1 = Compression ratio, V 1 V 2 = 15 = V 1 V 3 Heat added at constant volume = Heat added at constant pressure 4 3 2 P 5 1 V Fig. 9.46 Q23 = Q34 m \u00b7 cv(T3 \u2013 T2) = m \u00b7 cp\u00b7(T4 \u2013 T3) (T3 \u2013 T2) = \u03b3 \u00b7(T4 \u2013 T3) For process 1\u20132; and \uf8eb \uf8ec \uf8ed For process 2\u2013 300 = (15)0.4 T2 = 886.25 = (15)1. P2 = P1 \u00b7 (15)1.4 P2 = 44.3 P1 P 2 P = 3 T 2 T 3 Gas Power Cycles ______________________________________________________________ 395 = 886.25 T 3 44.3 P 1 70 P 1 T3 = 1400.39 or, Using equal heat additions for processes 2\u20133 and 3\u20134, (T3 \u2013 T2) = \u03b3 (T4 \u2013 T3) (1400.39 \u2013 886.25) = 1.4(T4 \u2013 1400.39) For process 3\u20134, T4 = 1767.63 15 \uf8eb \uf8ec \uf8ed V 1 V 4 \uf8f6 \uf8f7 \uf8f8 \u00d7 \uf8eb \uf8ec \uf8ed 1400.39 1767.63 \uf8f6 \uf8f7 \uf8f8 = V 1 V = 11.88 4 V 5 V = 11.88 4 For process 4\u20135 \uf8f6 \uf8f7 \uf8f8 or11.88)1.4 or, 5 T 4 T = (11.88)0.4 5 T5 = 1767.63 0.4 (11.88) = 656.85 Air standard thermal efficiency = 1 \u2013 Heat rejected Heat added = \u00b7( 656.85 300) \u2212 1.4 (1767.63 1400.39) \u2212 \u00d7 + \u2212 (1400.39 886.25) 396 _________________________________________________________ Applied Thermodynamics Air standard thermal efficiency = 0.6529 Air standard thermal efficiency = 65.29% Ans. Actual thermal efficiency may be different from theoretical efficiency due to following reasons; (a) Air standard cycle analysis considers air as the working fluid while in actual cycle it is not air throughout the cycle. Actual working fluid which are combustion products do not behave as perfect gas. (b) Heat addition does not occur isochorically in actual process. Also combustion is accompanied by inefficiency such as incomplete combustion, dissociation of combustion products, etc. (c) Specific heat variation occurs in actual processes where as in air standard cycle analysis specific heat variation is neglected. Also during adiabatic process theoretically no heat loss occurs while actually these processes are accompanied by heat losses. 9.1 What do you mean by air standard cycles? Discuss its\u2019 significance. 9.2 Derive the expression for Carnot cycle efficiency. 9.3 Discuss limitations of Carnot cycle and explain Otto cycle, Diesel cycle and Dual cycle in light of these limitations. 9.4 Derive expression for efficiency of Otto cycle, Dual cycle and Diesel cycle. 9.5 What is meant by mean effective pressure? Discuss its\u2019 relevance. 9.6 Obtain expressions for mean effective pressure of Otto cycle, Diesel cycle and Dual cycle. 9.7 Discuss gas turbine cycles. Obtain efficiency of Brayton cycle. 9.8 What are possible modifications in gas turbine cycle? Explain. 9.9 Explain different irreversibilities and losses in gas turbine cycle. 9.10 Define polytropic efficiency and its significance mathematically for compressors. 9.11 Obtain optimum pressure ratio condition for minimum compressor work requirement in two stage perfect intercooled compression. 9.12 Compare the influence of reheating, regeneration and intercooling on performance of gas turbine cycle. heat added if cycle efficiency is 79.4% and the minimum pressure of cycle is 100 kPa. 9", ".13 Discuss Ericsson cycle and obtain its efficiency. 9.14 Obtain efficiency of Stirling cycle. 9.15 Write short notes on compressor and turbine efficiencies. 9.16 An Otto cycle operates between maximum and minimum pressures of 600 kPa and 100 kPa. The minimum and maximum temperatures in the cycle are 27\u00baC and 1600 K. Determine thermal efficiency of [48%] cycle and also show it on T-s and P-V diagram. 9.17 In an air standard Carnot cycle operating between temperatures of 57\u00baC and 1327\u00baC, determine the [300 kJ/kg] 9.18 Determine air standard efficiency for a diesel engine having L/D ratio of 1.6 and bore as 25 cm. Take the clearance volume and pressure at end of suction as 1.2 litres and 1 atm. respectively. Also obtain mean [63.5, 6 bar] effective pressure. Volume after combustion is 1.8 times clearance volume. 9.19 An engine operates on Dual cycle with a compression ratio of 15. At the end of suction the air is available at 1 atm and 27\u00baC. Total heat added is 430 kJ/kg. Heat supply is in ratio of (0.536:1) for heat supply at constant volume and constant pressure. Determine cycle efficiency and mean effective [65%, 0.342 MPa] pressure. 9.20 A regenerative cycle has heat exchanger effectiveness of 75% and pressure ratio of 4 with compression occurring in two stages of equal pressure ratio with intercooling back to initial temperature of 15\u00baC. Maximum temperature in cycle is 650\u00ba C and expansion occurs with efficiency of 0.88. Compression process has isentropic efficiency of 0.85 in each stage. Considering air as working fluid throughout [34.6%] the cycle determine efficiency of cycle. Gas Power Cycles ______________________________________________________________ 397 9.21 A gas turbine installation has intercooling, regeneration and reheating with air entering compressor at 100 kPa, 290 K and compressed to 0.41 MPa. Subsequently it is cooled until temperature drops to 13\u00baC in intercooler and finally compressed to 0.75 MPa. Regenerator effectiveness is 0.70 and turbine inlet temperature (TIT) is 1350 K and expands to 0.41 MPa where it is reheated upto 1350 K. The exhaust pressure is 0.1 MPa. Determine overall efficiency considering expansion and compression processes [54.3%] to be of isentropic type. Take air as working fluid throughout. 9.22 In an actual gas turbine the compressor requires 300 kJ/kg of work to quadruple the inlet pressure. For inlet air temperature of 100\u00baC determine, the compressor exit air temperature and compressor efficiency. [671.6 K, 60.6%] 9.23 In a gas turbine installation the exhaust from gas turbine is sent for process heating at 200 kPa. Turbine produces power just sufficient to drive the compressor. Turbine inlet temperature is 815\u00baC and air is supplied to compressor at 1 bar, 17\u00baC. Determine compressor pressure ratio for air as working [3] fluid throughout. 9.24 A Brayton cycle producing 75 kW is designed for maximum work. The compressor inlet conditions are 100 kPa and 27\u00baC. Compression ratio is 5.5. For air standard Brayton cycle determine the turbine inlet [794 K, 38.6%, 0.631 kg/s] temperature, cycle efficiency and air flow rate. 9.25 A gas turbine cycle has reheating and heat exchanger employed. The air is supplied at 1 bar, 15\u00baC into compressor where it is compressed upto 4 bar with isentropic efficiency of 82%. Turbine inlet temperature is 700\u00baC and expansion occurs in two stages with equal pressure ratio upto 1 bar. The reheating between two stages occurs upto 700\u00baC. Isentropic efficiency of expansion is 85%. Exhaust from low pressure turbine is passed into a heat exchanger having 0.75 effectiveness. Heat exchanger heats the discharge from compressor before being supplied to combustion chamber. Considering air [32.1%] as woking fluid throughout determine cycle efficiency. 9.26 In a gas turbine plant air is supplied at 1 bar, 290 K to compressor having isentropic efficiency of 0.84 and pressure ratio of 5. The air flow rate to compressor is 111600 kg/hr. Compressed air is passed through a heat exchanger and it leaves exchanger at 360\u00baC, from where it is supplied to combustion chamber. Fuel having calorific value of 41800 kJ/kg is supplied at the rate of 890 kg/hr in combustion chamber. The combustion products expand from 4.7 bar to 1.95 bar, 410\u00baC in turbine. Take cp air = 1.0032 kJ/kg K,", " \u03b3 air = 1.4, cp comb, products = 1.1286 kJ/kg K, \u03b3 comb, products = 1.34. Determine hp developed. [3000 hp] 9.27 Show that the necessary condition for positive work output from an open cycle gas turbine can be given by, c \u00d7 \u03b7 \u03b7 The simple gas turbine cycle has compressor with isentropic efficiency of \u03b7 c and pressure ratio of \u2018r\u2019. Turbine has isentropic efficiency of \u03b7 T and expansion ratio similar to that of compressor. The maximum and minimum temperature are Tmax and Tmin and air is working fluid throughout with \u03b3 as ratio of specific heat. T \u00d7 Tmax > Tmin. r \u03b3 \u2212 1 \u03b3 9.28 In a gas turbine installation based on Brayton cycle the highest and lowest temperatures are TH and TL respectively. The pressure ratio and heat input from high temperature reservoir vary such that temperature TH and TL remain constant. Show that the expression for net maximum work output is, Wmax =..29 A gas turbine installation has double stage compressor and double stage turbine having pressure ratio across each stage as 8:1 in both compressor and turbine. Pressure at inlet to compressor is 1 bar and temperature entering each compressor is 293K and temperature 398 _________________________________________________________ Applied Thermodynamics entering each turbine is 1373K. Considering ideal regenerator in cycle determine cycle thermal efficiency and turbine work output. [0.613, 1.235 MJ/kg] 9.30 An ideal Stirling cycle running on hydrogen works between temperature of 723\u00b0C and 23\u00b0C while highest and lowest pressure are 30 bar and 5 bar respectively. Considering it to be closed cycle determine the heat transfer to regenerator per cycle, net work per cycle and cycle efficiency. [3.67 MJ, 0.85 MJ, 70%] 9.31 An ideal Eriscsson cycle is running on helium. Helium is at 1.5 bar, 500\u00b0C at the start of compression process and expansion process occurs at a pressure of 35 bar. Temperature of low temperature reservoir is 298 K. Determine the thermal efficiency and net work of cycle. [61.5%, 31.075MJ] 9.32 Determine thermal efficiency of cycle, backwork ratio and net power produced from a gas turbine installation having compressor admitting air at 1 bar, 300K & 5 m3/s volume flow rate. Air is compressed in two stages up to 12 bar with intercooling to 300K between stages at a pressure of 350 kPa. Turbine inlet temperature is 1127 \u00b0C and expanssion occurs in two stages with reheat to 1340K between stages at a pressure of 3.5 bar. Compressor and turbine stage efficiencies are 87% and 85% respectively. Also find the % gain in efficiency if a regenerator with effectiveness of 80% is employed in plant. Show the process on T-S diagram. [29.8%, 0.441, 2.17 MW, 59.66%] 9.33 A gas turbine runs at pressure ratio of 7 and maximum temperature of 1000K. Air enters compressor at temperature of 288 K. The isentropic efficiency of turbine and compressor are 90% & 85% respectively. Find the percentage reduction in efficiency compared to ideal Brayton cycle. Also find net work output and back work ratio of gas turbine. Show the [32.85%, 132.5 kJ/kg, 0.656] processes on T-S diagram. 10 10.1 INTRODUCTION Every real life system requires energy input for its\u2019 performance. Energy input may be in the form of heat. Now question arises from where shall we get heat? Traditionally heat for energy input can be had from the heat released by fuel during combustion process. Fuels have been provided by nature and the combustion process provides a fluid medium at elevated temperature. During combustion the energy is released by oxidation of fuel elements such as carbon C, hydrogen H2 and sulphur S, i.e. high temperature chemical reaction of these elements with oxygen O2 (generally from air) releases energy to produce high temperature gases. These high temperature gases act as heat source. In this chapter the detailed study of fuels and their combustion is being made. Air fuel ratio: It refers to the ratio of amount of air in combustion reaction with the amount of fuel. Mathematically, it can be given by the ratio of mass of air and mass of fuel. Molecular wt. of air Molecular wt. of fuel Mass of air Mass of fuel no. of moles of air no. of moles of fuel AF = Fuel-air ratio is inverse of Air-fuel ratio. Theoretical air-fuel ratio can be estimated from stoichio- metric combustion analysis for just complete combustion. Equivalence ratio: It is the ratio of actual fuel-air ratio to the theoretical fuel-air ratio for complete combustion. Fuel-air mixture", " will be called lean mixture when equivalence ratio is less than unity while for equivalence ratio value being greater than unity the mixture will be rich mixture. Theoretical air: Theoretical amount of air refers to the minimum amount of air that is required for providing sufficient oxygen for complete combustion of fuel. Complete combustion means complete reaction of oxygen present in air with C, H2, S etc. resulting into carbon dioxide, water, sulphur dioxide, nitrogen with air as combustion products. At the end of complete reaction there will be no free oxygen in the products. This theoretical air is also called \u201cstoichiometric air\u201d. Excess air: Any air supplied in excess of \u201ctheoretical air\u201d is called excess air. Generally excess air is 25 to 100% to ensure better and complete combustion. Flash point and Fire point: Flash point refers to that temperature at which vapour is given off from liquid fuel at a sufficient rate to form an inflammable mixture but not at a sufficient rate to support continuous combustion. Fire point refers to that temperature at which vaporization of liquid fuel is sufficient enough to provide for continuous combustion. These temperatures depend not only on the fuel characteristics but also on the rate of heating, air movement over fuel surface and means of ignition. These temperatures are specified in reference to certain standard conditions. Although flash point and fire point temperatures are defined in relation with ignition but these temperatures are not measure of ignitability of fuel but of the initial volatility of fuel. 400 _________________________________________________________ Applied Thermodynamics Adiabatic flame temperature: Adiabatic flame temperature refers to the temperature that could be attained by the products of combustion when the combustion reaction is carried out in limit of adiabatic operation of combustion chamber. Limit of adiabatic operation of combustion chamber means that in the absence of work, kinetic and potential energies the energy released during combustion shall be carried by the combustion products with minimum or no heat transfer to surroundings. This is the maximum temperature which can be attained in a combustion chamber and is very useful parameter for designers. Actual temperature shall be less than adiabatic flame temperature due to heat transfer to surroundings, incomplete combustion and dissociation etc. Wet and dry analysis of combustion: Combustion analysis when carried out considering water vapour into account is called \u201cwet analysis\u201d while the analysis made on the assumption that vapour is removed after condensing it, is called \u201cdry analysis\u201d. Volumetric and gravimetric analysis: Combustion analysis when carried out based upon percent- age by volume of constituent reactants and products is called volumetric analysis. Combustion analysis carried out based upon percentage by mass of reactants and products is called gravimetric analysis. Pour point: It refers to the lowest temperature at which liquid fuel flows under specified condi- tions. Cloud point: When some petroleum fuels are cooled, the oil assumes cloudy appearance. This is due to paraffin wax or other solid substances separating from solution. The temperature at which cloudy appearance is first evident is called cloud point. Composition of air: Atmospheric air is considered to be comprising of nitrogen, oxygen and other inert gases. For combustion calculations the air is considered to be comprising of nitrogen and oxygen in following proportions. Molecular weight of air is taken as 29. Composition of air by mass = Oxygen (23.3%) + Nitrogen (76.7%) Composition of air by volume = Oxygen (21%) + Nitrogen (79%) Enthalpy of combustion: Enthalpy of combustion of fuel is defined as the difference between the enthalpy of the products and enthalpy of reactants when complete combustion occurs at given temperature and pressure. It may be given as higher heating value or lower heating value. Higher heating value (HHV) of fuel is the enthalpy of combustion when all the water (H2O) formed during combustion is in liquid phase. Lower heating value (LHV) of fuel refers to the enthalpy of combustion when all the water (H2O) formed during combustion is in vapour form. The lower heating value will be less than higher heating value by the amount of heat required for evaporation of water. HHV = LHV + (Heat required for evaporation of water) It is also called calorific value of fuel and is defined as the number of heat units liberated when unit mass of fuel is burnt completely in a calorimeter under given conditions. Enthalpy of formation: Enthalpy of formation of a compound is the energy released or absorbed when compound is formed from its elements at standard reference state. Thus enthalpy of formation shall equal heat transfer in a reaction during which compound is formed from its\u2019 elements at standard reference state. Enthalpy of formation will have positive (+ ve) value if formation is by an endothermic reaction and negative (\u2013 ve) value if formation is by an exothermic reaction. Standard reference state: It refers", " to thermodynamic state at which the enthalpy datum can be set for study of reacting systems. At standard reference state, zero value is assigned arbitrarily to the enthalpy of stable elements. Generally, standard reference state is taken as 25\u00b0C and 1 atm, Tref = 25\u00b0C = 298.15 K, pref = 1 atm i.e. Dissociation: It refers to the combustion products getting dissociated and thus absorbing some of energy. Such as, the case of carbon dioxide getting formed during combustion and subsequently getting dissociated can be explained as below, Combustion: C + O2 CO2 + Heat Fuels and Combustion __________________________________________________________ 401 Dissociation: Heat + CO2 C + O2 Thus generally, dissociation has inherent requirement of high temperature and heat. 10.2 TYPES OF FUELS \u2018Fuel\u2019 refers to a combustible substance capable of releasing heat during its combustion. In general fuels have carbon, hydrogen and sulphur as the major combustible chemical elements. Sulphur is found to be relatively less contributor to the total heat released during combustion. Fuels may be classified as solid, liquid and gaseous fuel depending upon their state. Solid fuel: Coal is the most common solid fuel. Coal is a dark brown/black sedimentary rock derived primarily from the unoxidized remains of carbon-bearing plant tissues. It can be further classified into different types based upon the composition. Composition can be estimated using either \u201cproximate analysis\u201d or by \u201cultimate analysis\u201d. Proximate analysis is the one in which the individual constituent element such as C, H2, S, N2 etc. are not determined rather only fraction of moisture, volatile matter, ash, carbon etc. are determined. Thus proximate analysis is not exact and gives only some idea about the fuel composition. Proximate analysis of coal gives, various constituents in following range, Moisture 3\u201330%, Volatile matter 3\u201350%, Ash 2\u201330% and Fixed carbon 16\u201392%. In \u201cultimate analysis\u201d the individual elements such as C, H2, N2, S and ash etc. present in the fuel are determined on mass basis. Thus, it gives relative amounts of chemical elements constituting fuel. In general the percentage by mass of different elements in coal lies in the following range: Carbon Hydrogen Oxygen Nitrogen Sulphur Ash : : : : : : 50 to 95% 2 to 50% 2 to 40% 0.5 to 3% 0.5 to 7% 2 to 30% Different types of coal available are listed in the table hereunder. These are typical values for particular types of coal samples and may change for different types of coal. Table 10.1 Different types of coal Sl. No. 1. 2. 3. 4. Type % by mass Moisture Volatile matter in dry coal Ultimate analysis, % by mass C H2 O2 N2 + S2 Ash Lower calorific value, kcal/kg Peat Lignite Bituminous coal Anthracite coal 20 15 2 1 65 50 25 4 43.70 56.52 6.42 5.72 44.36 31.89 1.52 4.00 1.62 4.25 3200 2450 74.00 5.98 13.01 2.26 4.75 7300 90.27 3.30 2.32 1.44 2.97 7950 Liquid fuels: Fuels in liquid form are called liquid fuels. Liquid fuels are generally obtained from petroleum and its by-products. These liquid fuels are complex mixture of different hydrocarbons, and obtained by refining the crude petroleum oil. Commonly used liquid fuels are petrol, kerosene diesel, aviation fuel, light fuel oil, heavy fuel oil etc. Various liquid fuels of hydrocarbon family lie in Paraffin (CnH2n + 2 \u2013 chain structure), Olefins (CnH2n \u2013 chain structure), Napthalene (CnH2n \u2013 ring structure), Benzene (CnH2n \u2013 6 \u2013 ring structure), 402 _________________________________________________________ Applied Thermodynamics Nepthalene (CnH2n \u2013 12 \u2013 ring structure) category. Percentage by volume composition of some of liquid fuels is given below. Table 10.2 Composition of liquid fuels Sl.No. Fuel Carbon % by volume Hydrogen Sulphur 1. 2. 3. 4. 5. 6. Petrol Kerosene Diesel Benzole Light fuel oil Heavy fuel oil 85.5 86.3 86.3 91.7 86.2 86.1 14.4 13.6 12.8 8.0 12.4 11.8 0.1 0.1 0.9 0.3 1.4 2.1 Liquid fuels offer following advantages over solid fuel. (i) Better mixing of fuel and air is possible with liquid fuel. (ii", ") Liquid fuels have no problem of ash formation. (iii) Storage and handling of liquid fuels is easy compared to solid fuels. (iv) Processing such as refining of liquid fuels is more convenient. Gaseous fuels: These are the fuels in gaseous phase. Gaseous fuels are also generally hydrocarbon fuels derived from petroleum reserves available in nature. Most common gaseous fuel is natural gas. Gaseous fuels may also be produced artificially from burning solid fuel (coal) and water. Some of gaseous fuels produced artificially are coal gas, producer gas etc. Volumetric analysis of gaseous fuels is presented in Table 10.3. Gaseous fuels offer all advantages as there in liquid fuels except ease of storage. Table 10.3 Composition of gaseous fuels Sl.No. Fuel 1. 2. 3. Natural gas Coal gas Producer gas H2 \u2014 53.6 12 O2 \u2014 0.4 \u2014 10.3 CALORIFIC VALUE OF FUEL N 2 3 6 52 % by volume CO CH4 C2H4 93 \u2014 25 2.6 \u2014 0.4 C2H6 C4H8 3 \u2014 \u2014 \u2014 3 \u2014 CO 2 \u2014 3 4 1 9 29 During combustion, the chemical energy of fuel gets transformed into molecular kinetic or molecular potential energy of products. This energy associated with combustion also called calorific value of fuel is very important to be known for thermodynamic design and calculations of combustion systems. \u201cBomb calorimeter\u201d is one of the ways to get the heating value of solid and liquid fuels when burnt at constant volume. Different types of bomb calorimeters as given by At water, Davis, Emerson, Mahler, Parr, Peters and Williams are available. Bomb calorimeter as given by Emerson is discussed here. For getting the heating value of gaseous fuel the gas calorimeter is also discussed here. 10.4 BOMB CALORIMETER Emerson\u2019s bomb calorimeter is shown in Fig. 10.1 here. Its major components are bomb, bucket, stirrer, crucible or fuel pan, jacket, thermometer etc. A known quantity of fuel under investigation is kept in the crucible. Crucible has a electric coil with d.c. supply in it. Bomb is charged with oxygen under Fuels and Combustion __________________________________________________________ 403 pressure. Bomb is surrounded by a bucket containing water to absorb the heat released as fuel burns. Bomb also has an outer jacket with dead air space surrounding the bucket to minimise heat loss to surroundings. When electricity is flown into coil the fuel gets ignited. Bomb is actually a strong shell capable of withstanding about 100 atmosphere pressure. Inner bomb wall surface is lined with enamel and external wall surface is plated so as to prevent corrosion due to high temperature combustion products. Different operations while using it for calorific value measurement are as given ahead. First weigh the empty calorimeter bucket and fill it with a definite quantity of water at temperature about 2\u20133\u00b0C less than jacket water temperature. Charge bomb with oxygen at high pressure without disturbing the fuel kept in crucible. Ensure that there are no leaks in bomb and after ensuring it as leak proof place it in bomb jacket. Install thermometer and stirrer. Start stirrer for 3 to 4 minutes for temperature uniformity in bucket. Take temperature readings at definite interval, say after every five minutes. Fig. 10.1 Bomb calorimeter Active the coil so as to fire fuel. Record temperature every 30 seconds until maximum temperature is reached and after attaining maximum temperature read temperature after every 5 minutes. These temperatures are required to account for the heat exchange with jacket water. Later on remove bomb from calorimeter, release the gases and dismantle the bomb. Collect and weigh the iron fuse wire which remains. For getting accurate result the bomb should be washed with distilled water and washings titrated to obtain the amount of nitric acid and sulphuric acid formed. Corrections are made for small amount of heat transfer which occurs. For making the calculations a curve of temperature is plotted with time. Determine the rate of temperature rise before firing. Also, determine the rate of temperature drop after the maximum temperature is reached. Cooling correction may be added to the measured temperature rise. Heat balance may be applied as, 404 _________________________________________________________ Applied Thermodynamics Heat released by fuel during combustion + Heat released by combustion of fuse wire = Heat absorbed by water and calorimeter For getting heat absorbed by calorimeter the water equivalent of calorimeter is used which can be determined by burning a fuel of known calorific value. Generally benzoic acid and napthalene having calorific value of 6325 kcal/kg and 9688 kcal/kg are being used for finding water equivalent of calorimeter. Mathematically, mfuel CVfuel + mfw CVfw = (mw + mc) R ) \u22c5 \u2212 m R m CV ( m + \u22c5", " w c fw fw CVfuel = \uf8f1 \uf8f4 \uf8f2 \uf8f4 \uf8f3 m fuel \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe where mfuel and mfw are mass of fuel and mass of fuse wire, mw and mc are mass of water in calorimeter and water equivalent of calorimeter, R is correct temperature rise, CVfuel is higher calorific value of fuel, CVfw is calorific value of fuse wire. Calorific value of fuel estimated is higher calorific value as the water formed during combustion is condensed. 10.5 GAS CALORIMETER Schematic of gas calorimeter is shown in Fig.10.2 It is used for estimating the heating value of gaseous fuels. It has burner with arrangement to regulate and measure flow rate and pressure of gaseous fuel. Combustion products pass through tubes which are surrounded by flowing water. Here the volume flow rate of gas flowing through calorimeter is measured. Water conditions are adjusted so as to cool the products of combustion to ambient air temperature. Rate of water flow through the calorimeter is measured and its\u2019 temperature rise is determined. If the heat exchange between the calorimeter and its\u2019 surroundings is neglected then the heat received by water shall be equal to the heating value of fuel. For precise estimation of heating value of gaseous fuels the procedure as specified by ASTM is to be followed. 10.6 COMBUSTION OF FUEL Combustion of fuel refers to the chemical reaction that occurs between fuel and air to form combustion products with energy release i.e. oxidation of combustible fuel results into energy and products of combustion. In other words, during chemical reaction the bonds within fuel molecules get broken and atoms and electrons rearrange themselves to yield products. During complete combustion carbon present in fuel transforms into carbon dioxide, hydrogen into water, sulphur into sulphur dioxide and nitrogen into nitrogen oxides. Thus it is obvious that the total mass before combustion and after combustion remains constant although the elements exist in form of different chemical compounds in reactants and products. Generic form of combustion equations shall be as follows; (i) C + O2 CO2 On mass basis, (12) + (32) (44) 8 3 1 kg (C) + kg(CO2) 1 mol (C) + 1 mol (O2) 1 mol (CO2) kg(O2) 11 3 Fuels and Combustion __________________________________________________________ 405 (ii) (iii) (iv) 2C + O2 2CO On mass basis, (24) + (32) (56) Fig. 10.2 Gas calorimeter 1 kg (C) + 4 3 kg (O2) 7 3 kg (CO) On mole basis, 2 mol (C) + 1 mol (O2) 2 mol (CO) 2CO + O2 2CO2 On mass basis, 56 + 32 88 1 kg (CO) + 11 7 On mole basis, 2 mol (CO) + 1 mol (O2) 2 mol (CO2) kg (O2) kg (CO2) 4 7 2H2 + O2 2H2O On mass basis, 4 + 32 36 1 kg (H2) + 8 kg (O2) 9 kg (H2O) On mole basis, 2 mol (H2) + 1 mol (O2) 2 mol (H2O) (v) S + O2 SO2 On mass basis, 32 + 32 64 406 _________________________________________________________ Applied Thermodynamics 1 kg (S) + 1 kg (O2) 2 kg (SO2) On mole basis, 1 mol (S) + 1 mol (O2) 1 mol (SO2) (vi) CH4 + 2O2 CO2 + 2H2O On mass basis, 16 + 64 44 + 36 1 kg (CH4) + 4 kg (O2) On mole basis, 1 mol (CH4) + 2 mol (O2) 1 mol (CO2) + 2 mol (H2O) kg (CO2) + kg (H2O) 11 4 9 4 (vii) 2C2H6 + 7O2 4CO2 + 6H2O On mass basis, 60 + 224 176 + 108 1 kg (C2H6) + 56 15 kg (O2) 44 15 kg (CO2) + 27 15 kg (H2O) On mole basis, 2 mol (C2H6) + 7 mol (O2) 4 mol (CO2) + 6 mol (H2O) In general, for any hydrocarbon\u2019s complete combustion, (viii) a C8H18 + b O2 d CO2 + e H2O Equating C, H and O on both sides of above equation, a 8 = d", " a 18 = e 2 b 2 = (d 2) + e or, d = 8a e = 9a or, or, 2b = 2d + e, or 2b = 16a + 9a Above yield, b = 12.5a, d = 8a, e = 9a In order to avoid fraction let us round it off by multiplying by 2, 2a C8H18 + 25a O2 16a CO2 + 18a H2O Combustion equation shall now be, On mass basis, 200 57 1 kg (C8H18) + On mole basis, 2C8H18 + 25O2 16CO2 + 18H2O 228 + 800 704 + 324 176 57 kg (CO2) + kg (O2) 81 57 kg (H2O) 2 mol (C8H18) + 25 mol (O2) 16 mol (CO2) + 18 mol (H2O) Theoretically, the calorific value of fuel can be determined based upon the fuel constituents and the heat evolved upon their oxidation. Heat evolved during oxidation of some of the fuel constituents are as under, Fuel constituent Higher calorific value, kcal/kg Lower calorific value, kcal/kg Table 10.4 C C CO H2 S 8100 2420 2430 34400 2200 \u2014 \u2014 \u2014 29000 \u2014 Products of oxidation CO2 CO CO2 H2O SO2 Fuels and Combustion __________________________________________________________ 407 In any fuel containing carbon (C), hydrogen (H2), oxygen (O) and sulphur (S) the higher calorific value of fuel can be estimated using respective calorific values for constituents. In the fuel oxygen is not present in free form but is associated with hydrogen. On mass basis, H2 + O H2O (18) (16) (2) 1 kg (H2) + 8 kg (O) 9 kg (H2O) 1 8 kg (H2) + 1 kg (O) 9 8 kg (H2O) Above chemical equation indicates that with every unit mass of oxygen, 1 8 of oxygen mass shall be the mass of hydrogen associated with it. Thus, the free hydrogen available for oxidation (combustion) shall be only H where H refers to hydrogen mass and O refers to oxygen mass. O 8 The higher calorific value of fuel can thus be given as under using the mass fractions of constituent elements known. If percentage mass fractions of fuel constituents are given by C, H, O and S then. H.C.V. of fuel = 1 100 [8100\u00b7C + 34,400 H + 2220\u00b7S] kcal/kg O 8 Lower calorific value of fuel can be given by, L.C.V. of fuel = (H.C.V. of fuel) \u2013 (Heat carried by water vapour formed per kg of fuel burnt) The amount of latent heat carried depends upon the pressure at which evaporation takes place and quantity of water vapour formed. Generally, the evaporation is considered to take place at saturation temperature of 15\u00b0C and the latent heat of water vapour at this saturation temperature is 588.76 kcal/kg. During combustion of fuel the water shall be formed due to hydrogen present in fuel, therefore mass of water vapour can be given by the mass fraction of hydrogen. Thus, L.C.V. of fuel = H.C.V. of fuel 9 H. 588 76 100, kcal/kg 10.7 COMBUSTION ANALYSIS From earlier discussions it is seen how the combustion of a fuel can be written in the form of chemical reactions for the oxidation of different elements constituting fuel. Based on chemical reactions the mass of oxygen required per kg of element can be estimated. From these the oxygen requirement per kg of fuel constituents and subsequently mass of air required for the estimated oxygen requirement can be known. Let us see how air required for complete burning of a fuel having 85.5% carbon, 12.3% hydrogen and 2.2% ash is calculated; Mass of oxygen required for constituents; 1 kg of carbon shall require 8 3 kg of oxygen 1 kg of hydrogen shall require 8 kg of oxygen Ash shall not undergo oxidation. 408 _________________________________________________________ Applied Thermodynamics Table 10.5 Air requirement for fuel (on mass basis) Mass of constituents per kg of fuel (A) Oxygen required per kg of constituent (B) Oxygen required per kg of fuel (C) = (A) (B) C = 0.855 H2 = 0.123 Ash = 0.022 8/3 8 \u2014 2.28 0.984 \u2014 (C) = 3.264 Mass of air required, (theoretical, for air having 23% oxygen, 77% nitrogen) D = C 100 23 =. 3 264 100 23 = 14.19 Above calculations show", " that for the given fuel composition 14.19 kg of minimum air shall be theoretically required for complete combustion of one kg of fuel. Combustion analysis can be carried out either on mass basis or volume basis. Similar kind of analysis can be made on volume basis. Let us take heptane (C7H16) as the fuel and consider its\u2019 complete combustion. Table 10.6 Air required for fuel (on volume basis) Consti- Proportuent tional mass, A Proportional volume, Percentage volume, B = A. mol wt. C = B B O2 required per m3 of D 100 O2 required per 100 m3 of element, E = D C element Air required for 100 m3 of fuel (theoretical for air having 21% O2 and 79% N2), F E 21 100 = C H2 12 7 = 84 7 2 8 = 16 8 B = 15 46.67 53.33 1 0.5 46.67 26.665 E = 73.335 100 =. 73 335 21 = 349.21 Hence volume of air required is 349.21 m3 for 100 m3 of fuel. Volumetric analysis of combustion can be transformed into gravimetric analysis and vice versa. Such conversion is required because the volumetric analysis is available experimentally. Let us take an exhaust gas sample containing CO, CO2, N2 and O2. Let their volume fractions be X1, X2, X3 and X4 respectively. Fuels and Combustion __________________________________________________________ 409 Table 10.7 Conversion from volume based analysis to mass based analysis Constituent Volume of Constituent per m3, V Molecular weight, M Proportional Mass of constituent masses, X = M.V per kg of dry exhaust gases, CO CO2 N2 O2 X1 X2 X3 X4 28 44 28 32 28.X1 44.X2 28.X3 32.X4 M V. X 1 m = 28 X X X 44 X X 28 X X X 32 3 2 4 Total of proportional masses X = 28X1 + 44X2 + 28X3 + 32X4 10.8 DETERMINATION OF AIR REQUIREMENT (a) From combustion equations of fuel constituents (percentage basis) such as C, H, O and S as described earlier the total oxygen required per kg of fuel for complete combustion can be given as; = 1 100 8 3 H 8C O 8 S As air is taken to have 23% of oxygen by mass, therefore total air requirement per kg of fuel shall be; 8 3 = 1 100 C 8 H S O 8 100 23 Minimum air required per kg of fuel for complete combustion = 1 23 8 3 8C H O 8 S (b) Air requirement per kg of fuel can also be determined if the volumetric analysis of dry flue gases and mass fraction of carbon per kg of fuel is known. Let us take the dry flue gases having mass fractions as defined earlier as, kg of CO and kg of CO2. 1 kg of flue gas has 44 2X X 28 1X X Now the total mass of carbon present in CO and CO2 present in 1 kg of flue gas can be estimated and equated with mass of carbon present per kg of fuel. Mass of carbon per kg of flue gas = Mass of carbon in CO per kg flue gas Mass of carbon in CO per kg flue gas = 28 X 1 X + Mass of carbon in CO2 per kg flue gas 12 X 1 X 12 28 410 _________________________________________________________ Applied Thermodynamics Mass of carbon in CO2 per kg flue gas = Total carbon per kg of flue gas = 12 44 12 X 2 X X 2 X 44 X 2 X 12 12 X 1 X = 12 X (X1 + X2)kg per kg of flue gas. Now if we look at the fact that from where carbon is coming in flue gases then it is obvious that carbon is available only in fuel. Let us assume that carbon present in fuel completely goes into flue gases. Let us also assume that fuel does not contain nitrogen so what ever nitrogen is there in flue gas it will be because of nitrogen present in air. Let the mass of flue gases formed per kg of fuel after combustion be \u2018mg\u2019. The mass of carbon in per kg fuel say C can be equated to total mass of carbon calculated above. Mass fraction of carbon in fuel = Total mass of carbon in mg mass of flue gases. X 1 X 2 12 X C = mg 12 X X 2 C X 1 Mass of flue gases, mg = due to combustion of unit mass of fuel shall be Mass of air supplied per kg of fuel can be known from the nitrogen fraction present in flue gases as only source of nitrogen is air. 0.77 kg of N2 is available in 1 kg of air. From flue gas analysis is the", " mass of nitrogen available in per kg of flue gas so total mass of nitrogen present in flue gases (mg) \uf8f1 \uf8eb \uf8f4 28 \uf8f2 \uf8ec \u03a3\uf8f4 \uf8ed \uf8f3 X X From air composition 0.77 kg N2 is present per kg of air so the mass of air supplied per kg of flue Mass of air supplied per kg of fuel, kg air per kg of flue gas gas formed, 28 3X X 28 3X X 1 0 77. \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe \uf8f6 \uf8f7 \uf8f8 m = =. \u22c5 3 g 28 3X X mg 0 77. Substituting for mg, = \uf8eb \uf8ec \uf8ed 3 28 \u03a3 X X \uf8f6 \uf8f7 \uf8f8 \u00d7 1 0.77 \uf8f1 \uf8f4 \u00d7 \uf8f2 \uf8f4 \uf8f3 ( 12 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe Mass of air supplied per kg of fuel = 3 28 X X 1. 9 24 C X 2 where X1, X2, X3 are volume fractions of CO, CO2 and N2 present in unit volume of flue gas and C is mass fraction of carbon present in unit mass of fuel. For complete combustion of fuel there shall be no CO and only CO2 shall contain carbon i.e. X1 = 0. (c) Generally for ensuring complete combustion of fuel, excess air is supplied. In case of combustion if there is incomplete combustion of carbon resulting into formation of carbon monoxide, then Fuels and Combustion __________________________________________________________ 411 additional oxygen shall be required for converting carbon monoxide (CO) into carbon dioxide (CO2). kg of CO and From earlier discussions let us consider unit mass of flue gas containing kg of O2. The oxygen present in fuel remains in association with hydrogen. For combustion of 28 1X X 32 4X X CO into CO2 the mass of oxygen required is in 1 kg flue gas 4 7 kg per kg of CO. Thus oxygen required for CO present = 4 7 1X 28 X kg of oxygen Therefore, excess oxygen available per kg of flue gas formed, 16 1X X = = 32 16 X 4 X X 1 X = 16 X (2X4 \u2013 X1) Excess oxygen supplied per kg of fuel burnt = 16 mg 1 X 1 C 12 16 23. 69. Hence excess air supplied per kg of fuel Excess air supplied per kg of fuel burnt = 10.9 FLUE GAS ANALYSIS Flue gas analysis refers to the determination of composition of exhaust gases. Flue gas analysis can be done theoretically and experimentally. Here experimental method of flue gas analysis is described. Various devices available for measuring the composition of products of combustion (flue gas) are Orsat Analyzer, Gas chromatograph. Infrared analyzer and Flame ionisation detector etc. Data from these devices can be used to determine the mole fraction of flue gases. Generally this analysis is done on dry basis which may also be termed as \u201cdry product analysis\u201d and it refers to describing mole fractions for all gaseous products except water vapour. 412 _________________________________________________________ Applied Thermodynamics Orsat analyzer: It is also called as Orsat apparatus and is used for carrying out volumetric analysis of dry products of combustion. Schematic of apparatus is shown in Fig.10.3 It has three flasks containing different chemicals for absorption of CO2, O2 and CO respectively and a graduated eudiometer tube connected to an aspirator bottle filled with water. Fig. 10.3 Orsat analyzer Flask I is filled with NaOH or KOH solution (about one part of KOH and 2 parts of water by mass). This 33% KOH solution shall be capable of absorbing about fifteen to twenty times its own volume of CO2. Flask II is filled with alkaline solution of pyrogallic acid and above KOH solution. Here 5 gm of pyrogallic acid powder is dissolved in 100 cc of KOH solution as in Flask I. It is capable of absorbing twice its own volume of O2. Flask III is filled with a solution of cuprous chloride which can absorb CO equal to its\u2019 volume. Cuprous chloride solution is obtained by mixing 5 mg of copper oxide in 100 cc of commercial HCl till it becomes colourless. Each flask has a valve over it and C1, C2, C3 valves are put over flasks I, II and III. All the air or any other residual gas is removed from eudiometer by lifting the aspirator bottle and opening main value. The flue gas for analysis is taken by opening the main valve (three way valve) while valves C1, C2 and C3 are closed. 100 cc of flue gas may be taken into", " eudiometer tube by lowering aspirator bottle until the level is zero and subsequently forced into flasks for absorbing different constituents. Aspirator bottle is lifted so as to inject flue gas into flask I with only valve C1 in open state where CO2 present shall be absorbed. Aspirator bottle is again lowered and reading of eudiometer tube taken. Difference in readings of eudiometer tube initially and after CO2 absorption shall give percentage of CO2 by volume. Similar steps may be repeated for getting O2 and CO percentage by volume for which respective flask valve shall be opened and gas passed into flask. Thus Orsat analyzer directly gives percentage by volume of constituents. In case of other constituents to be estimated the additional flasks with suitable chemical may be used. The remaining volume in eudiometer after absorption of all constituents except N2 shall give percentage volume of N2 in flue gas. As in combustion of hydrocarbon fuel the H2O is present in flue gases but in orsat analysis dry flue gases are taken which means H2O will be condensed and separated out. Therefore the percentage Fuels and Combustion __________________________________________________________ 413 by volume of constituents estimated shall be on higher side as in actual product H2O is there but in dry flue gas it is absent. Orsat analyzer does not give exact analysis. 10.10 FUEL CELLS Fuel cell refers to a device having fuel and oxidizer in it which undergoes controlled chemical reaction to produce combustion products and provide electric current directly. In fuel cells the fuel and oxidizer react in stages on two separate electrodes i.e. anode (+ ive electrode) and cathode (\u2013 ive electrode). Two electrodes are separated by electrolyte in between. The chemical reaction is carried out to produce electric power without moving parts or the use of intermediate heat transfers as in power cycles. Thus fuel cell does not work on any cycle. Fuel cells based on hydrogen-oxygen fuel cells have been used to provide power to the spacecrafts. Fuel cells based on natural gas are also under process of development. Fuel cells are also being developed to power automobiles. Let us look at hydrogen-oxygen fuel cell in detail. Figure 10.4 shows the schematic of hydrogen-oxygen fuel cell. Here H2 supplied diffuses through the porous anode and reacts on anode surface with OH\u2013ions resulting into H2O and free electrons, the reaction for it is given in figure. Fig. 10.4 Hydrogen-Oxygen fuel cell Free electrons liberated enter the circuit while water goes into electrolyte. Oxygen supplied combines with water in electrolyte and electrons coming from electric circuit to produce OH\u2013 ions and H2O as per chemical reaction given in figure. OH\u2013 ions are transported through the electrolyte. Overall fuel cell has chemical reaction as Thus in hydrogen-oxygen fuel cell electricity and water are produced. H2 + 1 2 O2 H2O EXAMPLES 1. Coal having following composition by mass is burnt with theoretically correct amount of air. 86% C, 6% H, 5% O, 2% N, 1% S Determine the air-fuel ratio. 414 _________________________________________________________ Applied Thermodynamics Solution: Combustion equation for the coal (100 kg of coal) can be given as under; 86 12 C 6 1 H 5 16 O 2 14 N 1 32 S + n(O2 + 3.76 N2) From above equation C, H, O, S, N can be equated on both the sides as under, a CO2 + b SO2 + d N2 + e H2O C; 7.16 = a H; 6 = 2e O; (0.3125 + 2n) = 2a + 2b + e N; (0.1429 + 3.76 2n) = 2d S; 0.03125 = b Solving we get a = 7.16 b = 0.03125 d = 32.163 e = 3 n = 8.535 Amount of air required shall be [8.535 (4.76)] kg mol per 100 kg of coal. Air-fuel ratio =.535 8 28.97. 4 76 100 = 11.77 kg air per kg of fuel Air-fuel ratio = 11.77 Ans. 2. One kg C8H18 fuel is supplied to an engine with 13 kg of air. Determine the percentage by volume of CO2 in dry exhaust gas considering exhaust gas to consist of CO2, CO and N2. Solution: Combustion equation in mol. basis for one kg of fuel supplied shall be as under. (C8H18) + n(0.21 O2 + 0.79 N2) a CO2 + b CO + d N2 + e H2O 1 114 Equating the coefficients on both sides, C; 8 114 = 0.0702 = a + b O2; 0.", "21 n = a + b 2 e 2 H; 18 114 = 0.1579 = 2e N2; 0.79 n = d. Also it is given that 13 kg of air per kg of fuel is supplied, therefore n = 13 28.97 = 0.4487 Solving above following are available, a = 0.0393 b = 0.0309 Fuels and Combustion __________________________________________________________ 415 d = 0.3545 e = 0.07895 n = 0.4487 Constituents of dry exhaust gas shall be CO2, CO and N2 as indicated, Therefore dry exhaust gas = (a + b + d) = (0.0393 + 0.0309 + 0.3545) = 0.4247 Percentage by volume of CO2 in dry exhaust gas = = a b a d 100. 0 0393 100 0 4247. % by volume of CO2 = 9.25% Ans. = 9.25% 3. In a boiler the coal having 88% C, 3.8% H2, 2.2% O2 and remaining ash is burnt in the furnace. It is found that CO2 going with flue gases constitute to be 12% and temperature of flue gases is 260C. The flue gas sample is analyzed using Orsat apparatus at room temperature. Determine the percentage of CO2 that would be there for complete combustion of fuel. Solution: Actual percentage of CO2 in flue gases = 12% Constituent mass per kg of fuel C = 0.88 H2 = 0.038 Combustion product Molecular weight (x) (y) Volume fraction z = (x)/(y) CO2 0 88. 44 12 CO2 44 H2O. 0 038 18 2 H2O CO2 H2O 0.0733 18 0.019 Exhaust gases shall comprise of CO2, H2O, O2 and N2 Actual percentage of CO2 = CO 100 H O O CO N 2 2 2 2 2 12 = 0.0733 100 0.0733 0.019 O N2 2 (O2 + N2) = 0.5185 Orsat apparatus analyzes only dry flue gas so the percentage of CO2 can be obtained as, % CO2 = CO 100 O 2 2 N CO 2 = 2. 0 0733 100. 0 0733 0.5185 = 12.38% % CO2 = 12.38% Ans. 416 _________________________________________________________ Applied Thermodynamics 4. Determine the percentage analysis of combustion products by mass and by volume when gravimetric analysis of a hydrocarbon fuel indicates 86% C and 14% H2. Excess air supplied is 50% for combustion. Consider air to have 23.2% of oxygen by mass and remaining as nitrogen, molecular weights of carbon, oxygen, nitrogen and hydrogen are 12, 16, 14 and 1 respectively. Also estimate the change in internal energy of per kg of products when cooled from 2100C to 900C when the internal energies of combustion products are as under, TC 2100 900 CO2 2040 677 Internal energy, kJ/kg H2O 3929 1354 N2 1823 694 O2 1693 635 Solution: Theoretical mass of oxygen required per kg of fuel depending upon its\u2019 constituents C + O2 CO2 12 kg + 32 kg 44 kg or, 1 kg + 8 3 kg 11 3 kg or, For hydrogen, 0.86 kg + 2.29 kg 3.15 kg 2H2 + O2 2H2O 4 kg + 32 kg 36 kg 1 kg + 8 kg 9 kg 0.14 kg + 1.12 kg 1.26 kg Hence, minimum mass of oxygen required kg per kg of fuel = 0 86 As it is given that 50% excess air is supplied so the excess oxygen = 3.41 0.5 = 1.705 kg Total oxygen supplied = 3.41 + 1.705 = 5.115 kg. 0 14 8. 8 3 = 3.41 kg Constituents of exhaust gases per kg of fuel, CO2 = H2O = 9 0.14 = 1.26 kg 5 115..2 23 Excess O2 = 1.705 kg N2 = 76.8 = 16.932 kg 0.86 = 3.15 kg 11 3 For one kg of fuel the composition of combustion products is as under, Exaust gas constituents (x) CO2 = 3.15 kg H2O = 1.26 kg N2 = 16.932 kg O2 = 1.705 kg Total = 23.047 Gravimetric analysis % 100 x x g = 13.67 5.47 73.47 7.39 100% Molecular weight, y Proportional volume, g y z = Volumetric analysis %, 100 z z v = 44 18 28 32 0.311 0.304 2.", "624 0.231 3.47 8.963 8.761 75.619 6.657 100% Fuels and Combustion __________________________________________________________ 417 Change in internal energy for given temperature variation shall be estimated as under, on gravimetric basis Mass fraction of constituents of exhaust gases (g) CO2 = 0.1367 H2O = 0.0547 N2 = 0.7347 O2 = 0.0739 Change in internal energy per kg of constituents, (b) kJ/kg 2040 \u2013 677 = 1363 3929 \u2013 1354 = 2575 1823 \u2013 694 = 1129 1693 \u2013 635 = 1058 Change in internal energy (c) = (g b), kJ 186.32 140.85 829.48 78.19 Total change in internal energy = 1234.84 kJ/kg of exhaust gases Change in internal energy = 1234.84 kJ/kg of exhaust gases Ans. 5. Determine the percentage excess air supplied to boiler for burning the coal having following composition on mass basis, C: 0.82 H2: 0.05 O2: 0.08 N2: 0.03 S: 0.005 moisture: 0.015 Volumetric analysis of dry flue gases shows the following composition, CO2 = 10% CO = 1% N2 = 82% O2 = 7% Solution: For constituent components combustion reactions are as under. Out of given constituents only C, H2, and S shall require oxygen for burning reaction. C + O2 CO2 12 kg + 32 kg 44 kg 8 3 kg 1 kg + kg 11 3 2H2 + O2 2H2O 4 kg + 32 kg 36 kg 1 kg + 8 kg 9 kg S + O2 SO2 32 kg + 32 kg 64 kg 1 kg + 1 kg 2 kg Net mass of oxygen required per kg of coal = {(0.82 = 2.512 kg 8 3 ) + (0.05 8) + (0.005 1) \u2013 (0.08)} 418 _________________________________________________________ Applied Thermodynamics Mass of air required per kg of coal, considering oxygen to be 23% in the air by mass, = 100 23 2.512 = 10.92 kg air per kg of coal Volume of constituent per Molecular wt. Proportional Mass per kg of mass of flue gas Mass of carbon per kg of dry flue gas mol. of dry (b) constituents d = flue gas % (a) c = (a b) c c CO2 = 10 CO = 1 N2 = 82 O2 = 7 44 28 28 32 440 28 2296 224 (c) = 2988 0.147 0.0094 0.768 0.075. 0 147 12 44. 0 0094 12 28 = 0.04009 = 0.00403 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 Total carbon per kg of dry flue gas = 0.04412 Carbon is given to be 0.82 kg in per unit mass of coal and the mass of carbon per unit mass of dry flue gas is 0.04412 kg so the mass of dry flue gases per kg of coal shall be = Carbon mass per kg of dry flue gases Carbon mass per kg of coal = 0.82 0.04412 = 18.59 kg Therefore, the mass of CO per kg of coal = 18.59 0.0094 = 0.1747 kg The mass of excess O2 per kg of coal (i.e. unutilized O2) = 18.59 0.075 = 1.394 kg The CO produced in combustion products per kg of coal shall further require O2 for its\u2019 complete burning to CO2. 2CO + O2 2CO2 56 kg + 32 kg 88 kg 1 kg + 4 7 kg 11 7 kg The mass of O2 required for complete burning of 0.1747 kg CO per kg of coal shall be = 0.1747 4 7 = 0.998 kg. Out of excess O2 coming out with dry flue gases 0.0998 kg of O2 shall be utilized for complete burning of CO. Thus, the net excess O2 per kg of coal = 1.394 \u2013 0.0998 = 1.2942 kg O2 Hence, excess air required for 1.2942 kg O2 = = 5.627 kg air 1.2942 100 23 Fuels and Combustion __________________________________________________________ 419 % excess air = 5 627..92 10 100 = 51.53% % excess air = 51.53% Ans. 6. C2H6 burns completely with air when the air-fuel ratio is 18 on mass basis. Determine the percent excess or percent deficiency of air as appropriate and the dew point temperature of combustion products when cooled at 1 atm. Solution: Given air-fuel ratio on mass basis is first transformed into the air-fuel ratio on m", "olar basis. A/F ratio on molar basis = No.of moles of air No.of moles of fuel = Massof air Mol.wt.of air Massof fuel Mol.wt.of fuel = Mass of air Mol.wt.of fuel \u00d7 Mass of fuel Mol.wt.of air = (A/F ratio on mass basis) Mol.wt.of C H Mol.wt.of air 2 6 = 18 30 29 A/F ratio on molar basis = 18.62 Chemical reaction for complete combustion may be given as, Also, for complete combustion there will be C2H6 + n(O2 + 3.76 N2) aCO2 + bN2 + dO2 + eH2O 18.62 4.76 moles of air for each mole of fuel, So, n = 3.912 Equating coefficient C: 2 = a H: 6 = 2e Solving we get, a = 2 b = 14.71 d = 0.412 e = 3 O: 2n = 2a + 2d + e N: (3.76 2n) = 2b Complete combustion reaction may be given as, C2H6 + 3.912 (O2 + 3.76 N2) 2CO2 + 14.71N2 + 0.412O2 + 3H2O C2H6 + 3.5(O2 + 3.76 N2) 2CO2 + 13.16 N2 + 3H2O or, Hence air-fuel ratio (theoretical) on mol. basis 3.5 4 76 1 = 16.66 (A/F) =. Since theoretical air-fuel ratio is less than actual so it means excess air is supplied. Percentage of excess air = = A F actual A F 18 62 16.66 16.66. theoretical 100 A F theoretical 100 420 _________________________________________________________ Applied Thermodynamics = 11.76% Percentage of excess air = 11.76% Ans. Total amount of mixture = (a + b + d + e) Partial pressure of water vapour P = = (2 + 14.71 + 0.412 + 3) = 20.122 kg/mol. of fuel e d e b a = 0.151 bar 1.013 From steam table saturation temperature corresponding to 0.151 bar is seen to be 54C. Dew point temperature = 54C Ans. 7. Obtain the volumetric composition of combustion products obtained after combustion of C7H16 (heptane) being burnt with 50% excess air. Also obtain the average molecular weight and specific volume of combustion products at S.T.P. Consider the volume per kg mol. at S.T.P. to be 22.4 m3 and air to have 21% O2 and 79% N2 by volume. Solution: Here excess air supplied shall result in unutilized O2 with combustion products. *Excess O2 supplied = 0.5 73.33 = 36.665 m3 Total O2 supplied = 73.33 + 36.665 = 109.995m3 **Total N2 supplied = (109.995) 79 = 413.79m3 21 Total volume of combustion products = {46.67 + 53.33 + 36.665 + 413.79} = 550.455 m3 Average molecular weight of combustion products can be estimated by summation of mass of constituents per mol. of combustion products. Constituents of combustion products Fraction of constituents per unit products a a( ) b = Molecular weight (c) Mass of constituents per mol. of combustion (d = b \u00d7 c) Average molecular weight of combustion products c = (d) CO2(= 46.67) H2O(= 53.33) O2(= 36.665) N2(= 413.79) 46.67. 550 455 53 33. 550 455. 36.665 550 455. 413 79.. 550 455 = 0.0848 = 0.0969 = 0.0666 = 0.7517 44 18 32 28 3.7312 1.7442 2.1312 21.0476 28.65 Constituents Proportional Proportional % Volume of C7H16 volume mass (a) (b) = a mol wt. (c) = b 100 b O2 required per m3 of fuel O2 required for 100 m3 of fuel % Combustion product per100 m3 of fuel (d) (e) = c d CO2 H2O O2 N 2 12 7 = 84 2 8 = 16 7 8 C (mol. wt. = 12) H2 (mol. wt. = 2) 46.67 53.33 1 (C + O2 CO2) (1 mol 1 mol 1 mol.) 0.5 (2H2 + O2 2H2O) 2", " mol 1 mol 2 mol 46.67 46.67 26.66 53.33 (b) = 15 73.33 46.67 53.33 36.66*5 413.79 422 _________________________________________________________ Applied Thermodynamics Specific volume = 22.4 28.65 = 0.7818 m3/kg Average mol. wt. = 28.65 Specific volume = 0.7818 m3/kg Ans. Mass of fuel having 0.864 kg of carbon = 0.864 Mass of carbon per kg of fuel = 0 864.. 0 847 = 1.02kg It shows that one mol of dry flue gases shall require burning of 1.02 kg of fuel and 29.89 kg of air. So the air-fuel ratio = % Excess air = = 29.3 29 89. 1 02.. 29 3 15 15 100 = 95.33% Air fuel ratio = 29.3 Ans. % excess air = 95.33% Here the calculations above have been based on balancing the carbon present. These can also be done based on oxygen-hydrogen balance as under; Alternate approach (Not very accurate) Mass of oxygen present in one mol of dry flue gas = Mass of oxygen in CO2 + Mass of O2 appearing with dry flue gas 32 + (0.108 32) = 3 168. 44 = 5.76 kg Mass of oxygen present with nitrogen in air = 22.96 23.2 76.8 = 6.94 kg Here, the mass of oxygen supplied in air (6.94 kg) is more than oxygen present in dry flue gases (5.76 kg) which indicates that some oxygen gets consumed in formation of water. Orsat analysis analyzes dry flue gas. Mass of oxygen forming water (H2O) = 6.94 \u2013 5.76 = 1.18 kg 2H2 + O2 2H2O 4 kg + 32 kg 36 kg 1 kg + 8 kg 9 kg Mass of hydrogen in fuel = Mass of fuel =. 1 18 8. 0 1475. 0 153 = 0.1475 kg = 0.964 kg Thus one mol of dry flue gases are generated when 0.964 kg of fuel is burnt with 29.89 kg of air. 8. During Orsat analysis of the combustion products of an engine running on diesel (C12H26) the CO2 and O2 are found to be 7.2% and 10.8% respectively and rest is N2 by volume. Determine the airfuel ratio and percentage excess air considering air to have O2 and N2 in proportion of 23.2% and 76.8% respectively by mass. Fuels and Combustion __________________________________________________________ 423 Solution: There are two approaches for getting the air-fuel ratio and subsequently excess air. One is by carbon-balancing and other by balancing of oxygen-hydrogen. Minimum air required for complete combustion of 1 kg of fuel can be obtained considering the composition of fuel i.e. C12H26. Oxygen required per kg of constituents Oxygen required per kg of fuel (c) = a b 0.847 = 2.259 8 3 Mass of constituent per kg of fuel, C12H26 (a) C = 12 12 12 12 26 1 = 0.847 H2 = 26 1 12 12 26 1 (b) 8 3 8 Minimum air required per kg of fuel for complete combustion 100 23.2 3.483 8 0.153 = 1.224 = 15 kg = 0.153 Total = 3.483 Dry flue gases contain 7.2% of CO2, 10.8% O2 and 82% N2 by volume. This volumetric composition is to be converted to gravimetric (mass basis) i.e. mol fractions may be estimated as mol is unit of volume. Mass of constituent in one mol of dry flue gases = Volume fraction Mol. wt. of constituent Mass of CO2 in one mol of dry flue gases = 0.072 44 = 3.168 kg Mass of N2 in one mol of dry flue gases = 0.82 28 = 22.96 kg Corresponding to N2 the mass of air per mol of dry flue gases can be obtained as, 22 = 100.96 76.8 = 29.89 kg Carbon present in one mol of dry flue gases = 3 168. = 0.864 kg 12 44 Therefore, from the carbon present in one mol of dry flue gases the amount of fuel containing this much of carbon can be calculated. So, air fuel ratio = = 31.0 % excess air = 100 = 106.67%. 29 89.964 0 31 15 15 Air-fuel ratio and % excess air calculated differ from those estimated using balancing of carbon. Generally, carbon balancing is used for getting accurate results. 9. Fuel having 88% C and 12% H2 is burnt using 98% of", " air compared to theoretical air requirement for combustion. Determine the following considering that H2 is completely burnt and carbon burns to CO and CO2 leaving no free carbon. (a) % analysis of dry flue gases by volume (b) % heat loss due to incomplete combustion. Consider gross calorific values in kJ/kg as, C to CO2 = 34694 kJ/kg, C to CO = 10324 kJ/kg, H2 = 143792 kJ/kg 424 _________________________________________________________ Applied Thermodynamics Solution: Mass of constituent Oxygen required per kg of fuel per kg of constituent Oxygen required per kg of fuel (a) C = 0.88 H2 = 0.12 (b) 8 3 8 c = (a b) 2.35 0.95 Total 3.30 Minimum air required per kg of fuel for complete combustion 100 23.2 (c) \u00d7 3.30 100 23.2 = 14.27 kg Actual air supplied for burning = 0.98 14.27 = 13.98 kg Less amount of air supplied = 14.27 \u2013 13.98 = 0.29 kg. Due to this amount of air complete carbon could not get burnt into CO2 and CO is there. Let us write burning of carbon to yield CO and CO2 2C + O2 2CO 24 kg + 32 kg 56 kg 1kg + 1.33 kg 2.33 kg C + O2 CO2 12 kg + 32 kg 44 kg 1 kg + 2.67 kg 3.67 kg Thus the amount of air saved by burning C to CO instead of CO2, per kg of carbon = (2.67 \u2013 1.33) 100 23.2 = 5.78 kg Thus the amount of carbon burnt to CO per kg of fuel shall be = 0.29. 0 88. 5 78 = 0.044 kg Out of 0.88 kg C per kg of fuel only 0.044 kg C gives CO while remaining (0.88 \u2013 0.044 = 0.836) 0.836 kg C shall yield CO2. Volumetric analysis of combustion products is present as under. Mass of Constituents of Proportional dry flue gas constituent Volume Percentage volume b 100 b (c) = per kg of fuel (a) 0.836 0.044 44 12 28 12 = 3.06 = 0.103 0.768 13.98 = 10.74 CO2 CO N2 in mol a. Mol wt (b) =. 3 06 44 = 0.069 15.11% = 0.0037. 0 103 28. 10 74 28 (b) = 0.4567 = 0.384 0.81% 84.08% 100% Fuels and Combustion __________________________________________________________ 425 Volumetric analysis of dry flue gas = 15.11% CO2, 0.81% CO, 84.08 % N2 Calorific value of fuel in actual case due to incomplete burning of carbon. Ans. = (0.836 34694) + (0.044 10324) = 29458.44 kJ Theoretical calorific value of fuel with complete burning of carbon = (0.88 34694) = 30530.7 kJ Theoretical calorific value of fuel ={(0.88 34694) + (0.12 143792)} Percentage heat loss due to incomplete combustion = 47785.76 kJ = 30530 72. 47785 76 29458 44.. 100 = 2.24% % Heat loss = 2.24% Ans. 10. During production of gas the air and steam are passed through an incandescent coal bed. The coal is seen to have 95% of carbon and remaining as incombustible. The gas produced has hydrogen, nitrogen and carbon monoxide. Determine, the steam required per kg of coal and total air required per kg of coal when the heat of formation for steam is 147972 kJ/kg of hydrogen and for carbon monoxide it is 10324 kJ/kg of carbon. Also obtain volumetric analysis of gas. Take temperature of water as 20C at 1 atm pressure. Take air to have 23.2% O2 and 76.8% N2 by mass. Solution: Combustion equation can be written, separately for reaction with air and with steam as both are passed through. Reaction with air 2C + O2 2CO 24 kg + 32 kg 56 kg kg of O2 7 3 4 3 1 kg of C + kg of CO + 10324 kJ...(i) Reaction with steam C + H2O CO + H2 12 kg + 18 kg 28 kg + 2 kg 1 kg C + 3 2 kg H2O 7 3 kg CO + 1 kg C + 3 2 kg H2O 7 3 kg CO + 1 6 1 6 kg H2 + {10324 \u2013", " 1 6 147972 \u2013 3 2 100 20. 4 18 2253 Sensible heat Latent heat kg H2 \u2013 182191 kJ...(ii) 426 _________________________________________________________ Applied Thermodynamics released provided equation (ii) is multiplied by Two equations for burning with air and reaction with steam can be balanced in terms of heat 0.057 kg C + 0.085 kg H2O (steam) 0.1322 kg CO + 0.0094 kg H2 \u2013 10324 kJ Thus, total carbon used from reaction (i) and (iii) with no heat release can be given as, throughout, so equation (ii) gets modified into, 10324 182191...(iii) = (1 + 0.057) kg C = 1.057 kg carbon. Steam required per kg of coal = Total air required per kg of coal = 0 085. 0. 1 057.95 = 0.0764 kg 4 3 0.95 1 057. 100 23.2 = 5.16 kg Steam required per kg of coal = 0.0764 kg Air required per kg of coal = 5.16 kg Ans. Total carbon monoxide = (2.333 + 0.132) = 2.465 kg Total hydrogen = 0.0094 kg Total nitrogen = 4 3 76.8.2 23 = 4.414 kg Volumetric analysis Constituents Mol. wt. (a) CO = 2.465 kg H2 = 0.0094 kg N2 = 4.414 kg (b) 28 2 28 Proportional volume in mol. (c) = a b 0.088 0.0047 0.1576 (c) = 0.2503 % volume of gas (d) = 100 c c 35.16%, CO 1.88%, H2 62.96%, N2 100% % by volume = 35.16% CO, 1.88% H2, 62.96% N2 11. Determine the higher and lower calorific values of coal for which following observations are Ans. made in bomb calorimeter. Mass of coal sample = 1 gm Mass of water in bomb calorimeter = 2.5 kg Initial temperature of water = 20C Maximum recorded temperature of water = 22.6C Water equivalent of apparatus = 750 gm Cooling correction = + 0.018C Consider coal to have 5% H2 in it. Solution: Using the cooling correction the corrected rise in temperature of water can be obtained as, = (22.6 \u2013 20) + 0.018 = 2.618\u00b0C Fuels and Combustion __________________________________________________________ 427 Heat supplied to calorimeter and water = (2500 + 750) 2.618 = 8508.5 Cal/gm Higher calorific value of coal = 8508.5 Cal/gm In order to get lower calorific value the heat carried by the steam formation is to be excluded from higher calorific value. 2H2 + O2 2H2O 1 kg H2 + 8 kg O2 9 kg H2O (Mass of steam formed per kg of coal) = 0.05 9 = 0.45 gm steam per gm of coal. From steam table hfg at 20C = 586 kcal/kg \uf8eb \uf8ec \uf8ec \uf8ed Lower heating value of coal = 8508.5 \u2013 0.45 \u00d7 \u00d7 586 10 3 3 10 \uf8f6 \uf8f7 \uf8f7 \uf8f8 = 8244.8 cal/gm Higher heating value = 8508.5 kcal/kg Lower heating value = 8244.8 kcal/kg Ans. 12. Composition of a fuel by volume is as follows: H2 \u2013 52% CH4 \u2013 20% CO \u2013 16% CO2 \u2013 3% O2 \u2013 2% N2 \u2013 7% Higher calorific value of fuel constituents H2, CO and CH4 may be taken as 28424 kJ/m3, 27463 kJ/m3, 87780 kJ/m3 at 1 bar and 0C respectively. The latent heat of one kg of water for transformation to steam at 1 bar, 0C may be taken as 2445 kJ. Air may be considered to have 21% O2 by volume. Determine (a) (b) (c) the volume of air required for complete combustion of 1 m3 of fuel. the volumetric analysis of dry flue gases considering 20% excess air. the lower calorific value of fuel. Solution: Combustion reactions for constituents are, 2H2 + O2 2H2O, 1 mol 0.5 mol 1 mol CH4 + 2O2 2H2O + CO2, 1 mol 2 mol 2 mol 1 mol 2CO + O2 2CO2, 1 mol 0.5 mol 1 mol Minimum air required = 0.72 100 21 = 3", ".43 m3 Volume of air required for complete combustion = 3.43 m3 Ans. Volumetric analysis Constituents O2 required per per m3 of fuel m3 of constituents (a) of fuel (b) O2 required per m3 of fuel (c) = a b H2 = 0.52 CH4 = 0.20 CO = 0.16 CO2 = 0.03 O2 = 0.02 N2 = 0.07 0.5 2 0.5 \u2014 \u2014 \u2014 0.26 0.40 0.08 \u2014 (\u2013 0.02) \u2014 Combustion products per m3 of fuel CO2 \u2014 0.20 0.16 0.03 \u2014 H2O 0.52 0.40 \u2014 \u2014 \u2014 O2 \u2014 \u2014 \u2014 \u2014 0.144 N 2 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u03a3CO2 = 0.39 \u2014 \u03a3H2O = 0.92 \u2014 \u03a3O2 = 0.144 3.322 \u03a3N2 = 3.322 Total combustion product per m3 of fuel = 4.776 m3 Total dry flue gases per m3 of fuel = 3.856 m3 Excess air supplied is 20% so total volume of air supplied = Minimum air required + excess air supplied Fuels and Combustion __________________________________________________________ 429 Excess air supplied = 0.2 3.43 = 0.686 m3 Total air supplied = 3.43 + 0.686 = 4.116 m3 air per m3 of fuel **Oxygen present in excess air = 0.72 0.2 = 0.144 m3, This will be with combustion products Nitrogen present in total air = 79. 4 116 21 = 3.252 m3 **Since 0.07 m3 N2 per m3 of fuel is already present in fuel so total nitrogen available in combus- tion products = 3.252 + 0.07 = 3.322 m3 per m3 of fuel. 0.20 % by volume of CO2 =. 0 03. 0 16. 3 856 100 = 10.12% % by volume of O2 = % by volume of N2 =. 0 144 100. 3 856 3 322 100.. 3 856 = 3.73% = 86.15% Volumetric composition of dry flue gases = 10.12% CO2, 3.73% O2, 86.15% N2 Ans. Higher calorific value may be obtained from the values given for constituents H2, CH4 and CO. = {(0.52 28424) + (0.20 27463) + (0.16 87780)} = 34317.88 kJ/m3 Steam produced during combustion of 1 m3 of fuel. msteam =.92 0. 22 4 = 0.0411 kg Lower calorific value = Higher calorific value \u2013 (msteam = {34317.88 \u2013 (0.0411 2445)} = 34217.39 kJ/m3 latent heat) Lower calorific value = 34217.39 kJ/m3 Ans. 13. Determine the volume of air supplied for complete combustion of one m3 of fuel gas. If 40% excess air is supplied then find the percentage contraction in volume after the products of combustion have been cooled. The composition of fuel gas by volume is, H2 = 20%, CH4 = 3%, CO = 22%, CO2 = 8%, N2 = 47%.Consider air to have 21% O2 and 79% N2 by volume. Solution: Combustion reactions for constituents are 2H2 + O2 2H2O 1 mol 0.5 mol 1 mol CH4 + 2O2 2H2O + CO2 1 mol 1 mol 2 mol 2 mol 2CO + O2 2CO2 1 mol 1 mol 0.5 mol 430 _________________________________________________________ Applied Thermodynamics Volumetric analysis (without excess air) Constituents per m3 of fuel gas (a) H2 = 0.20 CH4 = 0.03 CO = 0.22 CO2 = 0.08 N2 = 0.47 O2 required per m3 of fuel constituents (b) 0.5 2 0.5 \u2014 \u2014 O2 required per m3 of fuel (c) = a b 0.1 0.06 0.11 \u2014 \u2014 O2 = 0.27 Combustion products per m3 of fuel CO2 \u2014 0.03 0.22 0.08 \u2014 CO2 = 0.33 H2O 0.2 0.06 \u2014 \u2014 \u2014 O2 \u2014 \u2014 \u2014 \u2014 \u2014 H2O = 0.26 \u2014 N2 = 0.47 N 2 \u2014 \u2014 \u2014 0.47 Minimum volume of air required = 100 0.27 21 = 1.286 m3 air per m3 of fuel. Minimum air required =1.286 m3/m", "3 of fuel Ans. With excess air there shall be additional amount of O2 and N2 present in combustion products. With 40% excess air the dry flue gas shall comprise of following: (i) CO2 = 0.33 m3 per m3 of fuel. (ii) N2 = 0.47 + nitrogen from total air (minimum air + excess air) = 0.47 + 1.286. 1 4 79 100 = 1.89 m3 per m3 of fuel (iii) O2 = oxygen from excess air = 21 1.286. 0 4 100 = 0.108 m3 per m3 of fuel Total volume of dry flue gas formed = (0.33 + 1.89 + 0.108) m3 per m3 of fuel = 2.328 m3 per m3 of fuel. Volume before combustion = Volume of fuel gas + Volume of air supplied % Contraction in volume = 100 = 20.52% = (1 + 1.286 \u00b4 1.5)m3 = 2.929 m3 2.929 2. 2 328.929 % contraction in volume = 20.52% Ans. 14. A hydrocarbon fuel when burned with air gave the following Orsat analysis, CO2: 11.94%, O2: 2.26%, CO: 0.41%, N2: 83.39% the air-fuel ratio on mass basis the percent of carbon and hydrogen in the fuel on mass basis, and Determine, (i) (ii) (iii) percentage of theoretical air supplied. Assume air to have 21% oxygen. Fuels and Combustion __________________________________________________________ 431 Solution: Orsat analysis gives, CO2 = 11.94%, O2 = 2.26%, CO = 0.41%, N2 = 83.39% H2O = 100 \u2013 (11.94 + 2.26 + 0.41 + 83.39) H2O = 2% Let hydrogen fuel be \u2018CaHb\u2019. CaHb + d O2 + On mass basis, 79 21 d N2 e CO + f CO2 + 28e + 44f + 9b + 32g + 28h b 2 H2O + g O2 + h N2 Mass of constituents of exhaust gas, CO2 = 44f, CO = 28e, H2O = 9b, O2 = 32g, N2 = 28h Exhaust gas mass = (28e + 44f + 9b + 32g + 28h) As per analysis, 44f = 0.1194 (28e + 44f + 9b + 32g + 28h) 28e = 0.0041 (28e + 44f + 9b + 32g + 28h) 9b = 0.02 (28e + 44f + 9b + 32g + 28h) 32g = 0.0226 (28e + 44f + 9b + 32g + 28h) 28b = 0.8339 (28e + 44f + 9b + 32g + 28h) Also, 12 a = 12e + 12f, or, a = e + f 79 21 d 28 = 28h h = 3.76d Solving above we get a = 19.53e b = 15.18e d = 54.09e f = 18.53e g = 4.82e h = 203.37e Air-fuel ratio = 32 d 12 28 = 29.77 d 79 21 a b Air-fuel ratio = 29.77 Ans. Carbon fraction = Hydrogen fraction = 12 12 a a b b 12 a b = 0.9392 = 0.0608 % Carbon = 93.92% Ans. 432 _________________________________________________________ Applied Thermodynamics Percentage theoretical air supplied = % Hydrogen = 6.08% 32 d 79 21 79 21 28 d 100 32 d d 28 12 a b = 96.75 % Theoretical air supplied = 96.75% Ans. 15. Obtain the stoichiometric reaction equation for CH4 being burnt with (i) (ii) (iii) 100% theoretical air 200% excess air 20% less than theoretical air requirement Consider air to have 21% oxygen and 79% nitrogen by mass. CH4, Solution: Let number of air molecules required for one molecule of CH4 be \u2018n\u2019. Combustion equation for CH4 + n O 2 79 21 N 2 a CO2 + b N2 + d H2O From above equation the C, H2 and O2 can be equated as under; C, 1 = a, or a = 1 O2, n = a + d 2, or 2n = 2a + d H2, 2 = d, or d = 2 N2, n 79 21 It gives = b, or b = 3.76 n n = 2, a =", " 1 b = 7.52, d = 2 Stoichiometric combustion equation shall be, For 100% theoretical air CH4 + 2(O2 + 3.76 N2) CO2 + 7.52N2 + 2H2O Ans. When 200% of excess air is added then total air supplied will be 300% i.e. (200% excess + 100% theoretical). Due to this excess O2 shall be there in combustion products. Thus modified form of combustion equation shall be, CH4 + 3 2 (O2 + 3.76 N2) a CO2 + b N2 + d H2O + e O2 Equating C, H2, O2 and N2 we get O2; a + C H2; d = 2 N2; b = 6 3.76 Fuels and Combustion __________________________________________________________ 433 It yields, a = 1 b = 22.56 d = 2 e = 4 Now the combustion equation with 200% excess air shall be; With 200% excess air; Ans. CH4 + 6(O2 + 3.76N2) CO2 + 22.56N2 + 2H2O + 4O2 When there is 20% less air then actual air supplied will be 20% less than 100% theoretical air. Due to less than theoretical air supplied there will be incomplete combustion and CO will be present with combustion products. CH4 + 0.8 2 (O2 + 3.76N2) aCO2 + bCO + dN2 + eH2O Equating C, H2, O2 and N2 we get C; a + b = 1 = 1.6 e b 2 2 H2; e = 2 N2; d = 1.2 3.76 = 4.512 O2; a + It yields, a = 0.2 b = 0.8 d = 4.512 e = 2 Hence for 20% less air the combustion equation shall be as under For 20% less air. CH4 + 1.6 (O2 + 3.76N2) 0.2CO2 + 0.8CO + 4.512N2 + 2H2O 16. Determine the adiabatic flame temperature for the combustion of carbon monoxide (CO) with 150 percent theoretical amount of oxygen to form CO2. The reactants enter the steady flow reactor at 25C, 1 atm and the products are CO2 and excess O2. Enthalpy of O2, CO and CO2 are 0 kJ/kg mole, \u2013 110418 kJ/kg and \u2013 393,137 kJ/kg mole respectively. The constant pressure specific heats of CO2 and O2 at 1 atm, may be assumed to be 56.43 and 36.5 kJ/kg mole K, respectively.[U.P.S.C. 1993] Ans. Solution: Combustion equation of carbon monoxide yields, 2CO + 3 2 O2 2CO2 + 1 2 O2 Since the process is adiabatic so the total enthalpy of reactants and products shall remain same, Enthalpy of reactants = 2 (\u2013 110,418) + 1 2 (0) = \u2013 220,836 kJ/kg mole 434 _________________________________________________________ Applied Thermodynamics Enthalpy of products for adiabatic flame temperature to be T\u00b0K. = [2 (\u2013 393,137) + (56.43 \u00d7 (T \u2013 298)) + ( 1 2 \u00d7 36.5 \u00d7 (T \u2013 298))] Enthalpy of reactants = Enthalpy of products \u2013 220,836 = [2 \u00d7 (\u2013 393,137) + (56.43 \u00d7 (T \u2013 298)) + ( 1 2 \u00d7 36.5 \u00d7 (T \u2013 298))} T = 7869.48 K. Adiabatic flame temperature = 7869.48 K Ans 10.1 What do you understand by fuel? 10.2 Describe different types of fuel. 10.3 What do you mean by endothermic and exothermic reactions? Explain. 10.4 Describe proximate analysis and ultimate analysis and their relevance. 10.5 What is meant by stoichiometric air-fuel ratio? 10.6 Describe calorific values of fuel and its\u2019 measurement. 10.7 What is excess air? 10.8 10.9 Describe fuel cells. 10.10 Define adiabatic flame temperature. 10.11 A coal sample has 90% carbon, 3.3% hydrogen, 3% oxygen, 0.8% nitrogen. 0.9% sulphur and 2% ash. For 50% excess air supply determine percentage composition of dry flue gases by volume and minimum oxygen required. Explain working of Orsat analyzer for flue gas analysis. 10.13 For a hydrocarbon fuel having 84.8", "% carbon, 15.2% hydrogen by mass determine [13% CO2, 7.1.% O2, 80% N2, minimum oxygen = 2.64 kg] 10.12 A coal sample has 85% carbon, 6% hydrogen, 6% oxygen and 3% ash by mass. What will be the minimum quantity of air required per kg of coal? Take air to have 23% O2 and 77% N2 by mass. [11.70 kg] the stoichiometric air-fuel ratio. Also obtain the volumetric composition of combustion products with 15% excess air being supplied. [15.11, 12.51% CO2, 2.89% O2, 84.60% N2] 10.14 For a coal fuel sample having 86% carbon, 4% hydrogen, 4.5% oxygen and 5.5% ash by mass determine minimum air required per kg of coal. Also find volumetric composition of combustion products when the actual air supplied for combustion is 1.5 times the minimum air required for complete combustion. [11.16 kg air/kg of fuel, 12.52% CO2, 80.47% N2, 7.01% O2] 10.15 During the combustion of coal having 22% carbon, 3.5% hydrogen, 6.5% oxygen and remaining ash by mass, the dry flue gas has volumetric composition of 8% CO2, 1.5% CO, 80.5% N2 and 10% O2. Calculate the minimum air required per kg of coal and % excess air. Consider air to have 23% and 77% of oxygen and nitrogen respectively by mass. [3.48 kg air, 43.3% excess air] 10.16 A liquid fuel has 86% C and 14% H2 by mass. During combustion actual air supplied is 10% less than the theoretically required for complete combustion. Obtain the volumetric composition of dry flue gases considering that all hydrogen gets burnt and carbon gets burnt to form carbon monoxide and carbon dioxide with no free carbon left. [11.39% CO2, 4.78% CO, 83.82% N2] Fuels and Combustion __________________________________________________________ 435 10.17 A producer gas sample has 2% CH4, 2% O2, 14% H2, 22% CO, 5% CO2 and 55% N2 by volume. Determine the minimum air requirement per m3 of gas and the volumetric analysis of combustion products for 40% excess air supply. Take air to have 21% oxygen by volume. [0.952 m3 air, 14.7% CO2, 81.25% N2, 4.05% O2] 10.18 Determine air fuel ratio and percentage excess air for a liquid fuel sample whose ultimate analysis shows 85% carbon and 15% hydrogen. It produces dry exhaust gases having 11.5% CO2, 1.2% CO, 0.9% O2, 86% N2 by volume after combustion. [15.07, 12.47% excess air] 10.19 C10 H22 is burnt with air fuel ratio of 13 by mass. Combustion is such that complete of hydrogen burns into water and there is no free oxygen and no free carbon. Also the carbon dioxide and carbon monoxide are present in combustion products. Determine volumetric analysis of combustion products. [7.68% CO2, 6.25% CO, 15.32% H2O, 70.75% N2] 10.20 A hydrocarbon fuel has 86% carbon and 14% hydrogen. The heat evolved by 1 kg carbon burning to CO2 is 33800 kJ and 1 kg hydrogen burning is 142000 kJ. For 25% of excess air supply determine the mass of air required and higher calorific value. [18.52 kg air, 48948 kJ/kg fuel] 10.21 Determine air fuel ratio for combustion of C3H8 (Propane) with 200 percent theoretical air supplied. [31.62] 10.22 Transform the volumetric analysis as given into gravimetric analysis. 15% CO, 2.2% CO, 1.6% O2, 81.2% N2. [21.6% CO2, 2.2% CO, 1.7% O2, 74.5% N2] 10.23 A coal sample has 66% C, 5.9% H2, 1% S, 19.9% O2, 1.5% N2, 5.6% ash by mass and calorific value of 29000 kJ/kg. When this coal is burnt with 30% excess air the temperature of flue gases leaving is 300C and ambient temperature is 17C. Considering combustion to be complete and partial pressure of moisture to be 0.07 bar in flue gases and", " energy accompanied being 3075 kJ/kg of steam, determine, the air-fuel ratio by mass. the volumetric analysis of combustion products. the heat carried away by flue gases as percentage of heat released by coal. (i) (ii) (iii) [11.55, 19.37% O2, 4.25% H2O, 0.16% SO2, 4.91% O2, 71.31% N2, 17.36%] 10.24 Determine the higher and lower calorific values of gas at atmospheric pressure and 15C using the following observations from a gas calorimeter for any fixed time. Atmospheric pressure = 76 cm Hg Gas burnt = 0.015 m3 Cooling water collected = 11 kg Condensate collected = 0.01 kg Cooling water temperature rise = 6C Gas pressure above atmosphere = 4.2 cm of water Gas temperature = 17C Latent heat of steam at atmospheric pressure = 2440 kJ/kg Density of mercury = 13600 kg/m3. [Higher calorific value = 18470 kJ/m3, Lower calorific value = 18445.6 kJ/m3] 10.25 Determine the net and gross calorific values per kg of mixture at constant pressure for stoichiometric mixture of air and C6H6 (benzene) vapour at 25C. Enthalpy of combustion for C6H6 at 25C is \u2013 3169500 kJ/kmol and the water is present in vapour phase in the combustion products. [2861 kJ/kg, 2981 kJ/kg] 436 _________________________________________________________ Applied Thermodynamics 11 11.1 INTRODUCTION Steam is extensively used for various applications such as power production, industrial processes, work interaction, heating etc. With the increasing use of steam in different engineering systems the steam generation technology has also undergone various developments starting from 100 B.C. when Hero of Alexandria invented a combined reaction turbine and boiler. Boiler, also called steam generator is the engineering device which generates steam at constant pressure. It is a closed vessel, generally made of steel in which vaporization of water takes place. Heat required for vaporization may be provided by the combustion of fuel in furnace, electricity, nuclear reactor, hot exhaust gases, solar radiations etc. Earlier boilers were closed vessels made from sheets of wrought iron which were lapped, riveted and formed into shapes of simple sphere type or complex sections such as the one shown in Fig. 11.1. It is the \u2018Wagon boiler\u2019 of Watt developed in 1788. Fig. 11.1 Wagon boiler of Watt, (1788) According to A.S.M.E. (American Society of Mechanical Engineers, U.S.A.) code a boiler is defined as a combination of apparatus for producing, furnishing or recovering heat together with the apparatus for transferring the heat so made available to water which could be heated and vaporised to steam form. Boilers and Boiler Calculations ___________________________________________________ 437 Boiler technology got revolutionized during second world war, when the need arose for the boilers to supply steam to field installations. Field requirements were critical as the boiler installation and commissioning should take place in minimum time. Therefore the \u2018Package boilers\u2019 which were complete with all auxiliaries as one unit came up and gradually transformed into modern boiler having lot of accessories and mountings. Thus in a boiler other than heat supplying unit, shell and tubes, a number of other devices are used for its control, safe and efficient operation. Devices which are mounted on boiler for its control and safe operation are called \u201cmountings\u201d while devices which are mounted on boiler for improving its performance are called \u201caccessories\u201d. Thus boiler mountings are necessary while boiler accessories are optional. 11.2 TYPES OF BOILERS Boilers are of many types. Depending upon their features they can be classified as given under: (a) Based upon the orientation/axis of the shell: According to the axis of shell boiler can be classified as vertical boiler and horizontal boiler. (i) Vertical boiler has its shell vertical. (ii) Horizontal boiler has its shell horizontal. Inclined boiler has its shell inclined. (iii) (b) Based upon utility of boiler: Boilers can be classified as (i) Stationery boiler, such boilers are stationery and are extensively used in power plants, industrial processes, heating etc. (ii) Portable boiler, such boilers are portable and are of small size. These can be of the following types, Locomotive boiler, which are exclusively used in locomotives. Marine boiler, which are used for marine applications. (c) Based on type of firing employed: According to the nature of heat addition process boilers can be classified as, (i) Externally fired boilers, in which heat", " addition is done externally i.e. furnace is outside (ii) the boiler unit. Such as Lanchashire boiler, Locomotive boiler etc. Internally fired boilers, in which heat addition is done internally i.e. furnace is within the boiler unit. Such as Cochran boiler, Bobcock Wilcox boiler etc. (d) Based upon the tube content: Based on the fluid inside the tubes, boilers can be, (i) Fire tube boilers, such boilers have the hot gases inside the tube and water is outside surrounding them. Examples for these boilers are, Cornish boiler, Cochran boiler, Lancashire boiler, Locomotive boiler etc. (ii) Water tube boilers, such boilers have water flowing inside the tubes and hot gases surround them. Examples for such boilers are Babcock-Wilcox boiler, Stirling boiler, La-Mont boiler, Benson boiler etc. (e) Based on type of fuel used: According to the type of fuel used the boilers can be, (i) Solid fuel fired boilers, such as coal fired boilers etc. (ii) Liquid fuel fired boilers, such as oil fired boilers etc. (iii) Gas fired boilers, such as natural gas fired boilers etc. (f) Based on circulation: According to the flow of water and steam within the boiler circuit the boilers may be of following types, (i) Natural circulation boilers, in which the circulation of water/steam is caused by the density difference which is due to the temperature variation. 438 _________________________________________________________ Applied Thermodynamics (ii) Forced circulation boilers, in which the circulation of water/steam is caused by a pump i.e. externally assisted circulation. (g) Based on extent of firing: According to the extent of firing the boilers may be, (i) Fired boilers, in which heat is provided by fuel firing. (ii) Unfired boilers, in which heat is provided by some other source except fuel firing such as hot flue gases etc. (iii) Supplementary fired boilers, in which a portion of heat is provided by fuel firing and remaining by some other source. 11.3 REQUIREMENTS OF A GOOD BOILER Different requirements of a good boiler are given below. In general boiler is supposed to generate large quantity of steam at desired pressure and temperature quickly and efficiently. (a) (b) It should be capable of generating steam at desired rate at desired pressure and temperature with minimum fuel consumption and cost. It should have sufficient steam and water storage capacity to meet fluctuation in demand and to prevent fluctuation in steam pressure or water level. (c) Boiler should have a constant and thorough circulation of water. (d) (e) Boiler should have capability to get started quickly from cold. (f) Its construction should be simple and have good workmanship for the ease of inspection and repairs i.e. easily accessible parts. It should be equipped with all necessary mountings. (g) Boiler should have its heating surface nearly at right angle to the current of hot gases for good heat transfer. (h) There should be minimum frictional power loss during flow of hot gases and water/steam i.e. pressure drop throughout the system should be minimum. (i) Tubes should be so designed so as to have minimum soot deposition and good strength against wear. Boiler should have a mud drum to receive all impurities. (j) Boiler should have strength to withstand excessive thermal stresses. (k) Boiler should occupy less floor area and space. Boilers may be selected for a particular applications considering above general requirements and constraints, if any. For deciding the boiler for any application, generally following criterion are made; (i) Steam pressure requirement (ii) Steam temperature requirement (iii) Steam generation rate (iv) Initial cost and constraints (v) Running and maintenance costs (vi) Availability of fuel and water (vii) Inspection and maintenance requirements. 11.4 FIRE TUBE AND WATER TUBE BOILERS Fire tube boilers are those boilers in which hot gases (combustion products) flow inside the tubes and water surrounds them. Water extracts heat for its phase transformation from the hot gases flowing inside the tubes, thus heat is indirectly transferred from hot gas to water through a metal interface. Such boilers came up in eighteenth century and were extensively used for steam generation in variety of applications. With the passage of time and coming up of another types of boilers the fire tube Boilers and Boiler Calculations ___________________________________________________ 439 Fig. 11.2 Fire tube boiler Fig. 11.3 Water tube boiler 440 _________________________________________________________ Applied Thermodynamics boilers have lost their charm to some extent due to limitations in terms of steam pressure. Fire tube boilers are used for applications having small steam requirement. Different types of fire tube boilers have been discussed ahead. Water tube boilers are those boil", "ers in which water flows inside the tubes and hot gases surround them. This type of boilers came up as a solution to the problem of explosion faced in fire tube boilers when the pressure and steam generation capacity were increased. In such boilers the shell behaved as heated pressure vessel subjected to internal pressure which set up tensile stresses (hoop stress) in walls. Mathematically, this stress can be given as, Hoop stress = P D 2 t thickness of shell wall. where P is internal working pressure, D is diameter of shell and t is Above expression shows that if \u2018P\u2019 (pressure) increases then either \u2018D\u2019 (diameter) should be decreased or \u2018t\u2019 (thickness) be increased to keep stress within acceptable limits. While increasing thickness the mass of boiler and cost of manufacturing both increase therefore the reduction of \u2018D\u2019 (diameter) is an attractive option. This became the basis for water tube boilers in which small diameter of tube facilitated quite high pressure steam generation. Such boilers came up in late eighteenth and nineteenth century. George Babcock and Stephen Wilcox gave straight-tube boiler of water tube boiler type in 1867 which was subsequently modified and developed as present \u2018Babcock and Wilcox boiler\u2019. Water tube boilers may be further classified based on type of tubes employed. These can be Straight water tube boilers and Bent water tube boilers. Straight water tube boilers are those in which tubes carrying water are straight from one end to the other end. At the two ends headers are provided. In general water comes down from drum into down header and after passing through tubes get heated and evaporated to steam which is carried back to drum through upcomer header or riser. Circulation of water is caused by the density difference as density of feed water is more than density of hot water/wet/ dry steam due to lower temperature of feed water. Fig. 11.4 Relative position of boiler sections Bent water tube boilers are those in which bent tubes are employed for carrying water. Bent water tubes are advantageous over straight water tubes in many respects. Bent tubes offer better access into boiler and ease of inspection and maintenance. Also tube arrangement can be modified so as to maximize heating surface and exposure of tubes to hot gases. Circulation is better in case of bent tube boilers as compared to straight tube, since the orientation of tubes in case of former is generally at inclination from vertical while for later it is horizontal. Stirling Boilers and Boiler Calculations ___________________________________________________ 441 boiler is one such boiler. In water tube boilers the heat distribution generally occurs amongst economiser tubes, evaporator tubes, superheater tubes. Hottest gases are designed to come in contact with superheater tubes. The evaporator tubes are in between superheater and economizer tubes. Relative position of three sections shall be as shown here. A comparative study between fire tube and water tube boiler is presented below to understand relative merits of one over the other. Advantages of fire tube boilers (a) Fire tube boilers are more flexible and can meet sudden steam demand without much pressure fluctuations. It is because of the large volume of water contained by these boilers and heat energy stored in hot water. It may be noted that energy stored in a definite volume of water at given pressure and temperature shall be more than that stored in same volume of steam at same thermodynamic states. (b) Fire tube boilers are less sensitive to the failure of feed water supply as they have large capacity of water stored. Such feed water supply failure is very damaging in water tube boilers due to small storage capacity. (c) Fire tube boilers are rigid and simple in construction, therefore have great reliability and less initial cost. Number of parts in fire tube boilers is less than those in water tube boilers so maintenance cost is also small. Since thickness of boiler shell is large enough so the problems of pitting and erosion are less. Also the large drum of boiler provides ample water space and desired conditions for dry steam generation. Advantages of water tube boilers (a) Steam generation rate is large in water tube boilers as compared to fire tube boilers due to small quantity of water contained, large heating surface, better circulation of water etc. Water tube boilers are made in bigger sizes with very high limit to maximum output due to smaller drum, circulation etc. (b) Maximum pressure of steam generation is quite high in water tube boilers (125 bar and above) compared to fire tube boilers (up to 20 bar) due to fluid flowing through tubes of small diameter and diameter of drum being relatively small. In case of explosion the steam generation may not stop in water tube boilers as the place of explosion in tubes can be plugged easily. While in fire tube boilers the explosion is very dangerous due to large quantity of water flashing into steam. (c) (d) Water tube boilers are easy to fabricate and", " transport due to the small size of drum. The shell of fire tube boiler shall be nearly twice or thrice of the shell of water tube boiler for same power. (e) Water tube boilers are generally externally fired and various parts of boiler are more readily accessible for cleaning, inspection and maintenance, compared to fire tube boilers. Characteristics of fire tube and water tube boilers are tabulated as under Table 11.1 Characteristics Fire tube boiler (a) Steam Pressure (b) Unit output (c) Fuel It is limited to 20\u201330 bar. In case of waste heat boilers, it can be more. Limited to about 20 MW. within design limits. All commercial fuels and treated waste can be used. Water tube boiler It is virtually unlimited within metallurgical and design limits. It is virtually unlimited Any fuel can be used. Also the furnace size is large. 442 _________________________________________________________ Applied Thermodynamics Characteristics Fire tube boiler Water tube boiler (d) Erection (e) Efficiency It is packaged ready for work site. Normally 80\u201385%, gross calorific value, but can be further increased using accessories. (f) Application Generally for heat supply. (g) Inspection requirement. Frequent inspection requirement. It is more than in water tube It is to be shop assembled or erected at site. Normally 85\u201390%, gross calorific but can be further increased using accessories. Generally for power and heat supply together. Inspection requirement is less than in fire tube boiler. boilers. Composite boilers: These are the boilers developed as combination of fire tube and water tube boilers so as to derive advantages of both designs. In these boilers there is no restriction in pressure and output capabilities. Such boilers are usually transported to site in sections/modules and assembled at site. Generic arrangement in composite boiler is shown in Fig. 11.5. Fig. 11.5 Composite boiler 11.5 SIMPLE VERTICAL BOILER Simple vertical boiler shown in Fig. 11.6 has a vertical boiler shell of cylindrical shape. It has fire box of cylindrical type inside the shell. Vertical passage of tubular type called uptake is provided over fire box for exhaust of flue gases. Cross tubes are provided for improving water circulation and increasing heating surface. At the bottom of fire box a fire grate is provided for burning fuel. Total heating surface area is about 7\u201310 times grate area. Man hole and hand holes are provided in the shell for access to inside of shell. Hot gases raising from fire grate go upwardly and heat the water contained in shell and tubes. Steam generated in shell can be tapped through a steam stop valve placed on the crown of shell. Such boilers have steam generation capacity up to 1000 kg per hour and maximum steam pressure up to 10 bar. Size of the boiler ranges from 0.6 m diameter to 2 m diameter and height from 1.2 m to 4 m high. Boiler efficiency is nearly 50%. Boilers and Boiler Calculations ___________________________________________________ 443 Fig. 11.6 Simple vertical boiler 11.6 COCHRAN BOILER This is a fire tube boiler of vertical type and came up as a modification over the simple vertical boiler in order to maximize heating surface. Total heating surface area is 10\u201325 times the grate area. It has cylindrical shell with hemispherical crown. Hemispherical geometry offers maximum volume space for given mass of material and is also very good for strength and maximization of radiant heat absorption. Figure. 11.7 shows the schematic of Cochran boiler with various mountings upon it. Fire box is also of hemispherical form. Flue gases flow from fire box to refractory material lined combustion chamber through a flue pipe. Incomplete combustion if any can get completed in combustion chamber and hot gases subsequently enter into tubes. After coming out of fire tubes hot gases enter into smoke box having chimney upon it. As the fire box is separately located so any type of fuel such as wood, paddy husk, oil fuel etc. can be easily burnt. These boilers are capable of generating steam up to pressure of 20 bar and steam generating capacity from 20 kg/hr to 3000 kg/hr. Boilers have dimensions ranging from 1 m diameter and 2 m height to 3 m diameter and 6 m height. Efficiency of such boilers ranges between 70 and 75%. 444 _________________________________________________________ Applied Thermodynamics Fig. 11.7 Cochran boiler 11.7 LANCASHIRE BOILER It is a horizontal fire tube boiler. General arrangement in the boiler is shown in Fig. 11.8. Boiler is mounted on a brickwork setting with front end of shell sloping about 1 : 250 for emptying the shell. It has a circular shell connected to end plates supported by gusset plates. Two fire tubes run throughout the length of the boiler. Fire tubes are of diameter less than half the diameter of shell and diameter of fire tubes is reduced as shown to have access to lower", " side of boiler. Fire bridge is provided to prevent fuel from falling over the end of furnace. Fire bridge also helps in producing a better mixture of air and gases for perfect combustion by partly enveloping the combustion space. Hot gases start from grate area, enter into fire tubes and come out at back of boiler from where these gases flow towards the front of boiler through bottom flue. Upon reaching the front these hot gases flow through the side flues and enter the main outlet. Outlet passage may also be used commonly by more than one boilers. About 85% of actual heat transferred is transferred through surface of fire tubes while 15% is transferred through bottom and side flues. Boilers and Boiler Calculations ___________________________________________________ 445 Plan, elevation and side views of Lancashire boiler shown in figure explain the furnace, different firetubes, bottom flues, side flues etc. Dampers are provided at the end of side flues for regulating the pressure difference (draught) for exit of burnt gases. Other mountings and accessories are shown in the elevation of Lancashire boiler. Fig. 11.8 Lancashire boiler Working pressure in these boilers are in the range of 0.7 MPa to 2 MPa and efficiency of the boiler is about 65%\u201370%. Size of these boiler depends upon size of shell which may be 2 m to 3 m in diameter and 6m to 10m in length. 446 _________________________________________________________ Applied Thermodynamics 11.8 CORNISH BOILER This is a horizontal fire tube boiler having single flue gas tube. General arrangement is very similar to Lancashire boiler. Water surrounds the flue gas tube in the shell. Hot flue gases after passing through the tube are divided into two portions at the end of boiler and pass through side flue passages to reach upto the front of boiler and then enter into bottom flue gas passage for escaping out through chimney after traversing the entire length of bottom passage. Hot gases thus traverse complete length of passage from end to end of boiler thrice i.e. through main flue gas tube, side flues and bottom flues. Heat transfer is more from side flues than bottom flue due to sedimentation in bottom. These boilers are generally capable of producing steam up to the rate of 1350 kg/hr and maximum steam pressure up to 12 bar. Shell is generally of length 4 to 7 m and diameter 1.2 to 1.8 m. Fig. 11.9 Cornish boiler 11.9 LOCOMOTIVE BOILERS These boilers were invented for getting steam to run a steam engine used in locomotives. These are fire tube type of boilers. It has basically three parts i.e. smoke box, shell and fire box. Figure 11.10 shows a general arrangement in locomotive boiler. Inside fire box the fuel (coal) is burnt over the grate. For feeding fuel the fire hole is used. Hot gases produced in fire box are diverted by fire brick arch and enter into the fire tubes surrounded with water. Steam produced gets collected in a steam drum fitted on top of the shell. Arrangement for super heating is there in these boilers. As shown the wet steam goes through inlet headers of superheater and after passing through tubes, it returns to the outlet header of superheater and is taken out for steam engine. A very large door is provided at the end of smoke box so as to facilitate cleaning and maintenance of complete boiler. As it is a moving boiler, therefore, its chimney is completely eliminated. For expelling the burnt gases (draught) the exhaust steam coming out from steam engine is being used. Thus it is an artificial draught used in these boilers for expelling burnt gases Fig. 11.10 Locomotive boiler 448 _________________________________________________________ Applied Thermodynamics 11.10 NESTLER BOILERS This is fire tube type of fired horizontal axis boiler. Figure 11.11 shows the schematic arrangement in Nestler boiler. The boiler shell comprising of two thick mild steel plates with large number of fire tubes fitted between two plates is shown here. A bigger diameter furnace tube extending from burner end to other end is used for carrying hot flue gases from one smoke box to the other smoke box. At the rear end smoke box chimney is provided for rejection of exhaust gases. Fig. 11.11 Nestler boiler Hot gases pass through furnace tube and enter into rear end smoke box and pass through fire tubes to front end smoke box for final discharge through chimney. Water surrounding tubes get transformed into steam and gets collected in steam space. Here furnace oil, a dark coloured highly viscous oil is used for firing in furnace. Oil is first heated up to 80C by electric heater before being supplied to burner for injection into furnace tube. Blower is employed for atomization of furnace oil into furnace. Such boilers are capable of generating steam upto pressure of 10\u201311 bar. 11.11 BABCOCK", " AND WILCOX BOILER It is a water tube boiler suitable for meeting demand of increased pressure and large evaporation capacity or large sized boiler units. Figure 11.12 shows the Babcock and Wilcox boiler. It has three main parts: (i) Steam and water drum (ii) Water tubes (iii) Furnace. Steam and water drum is a long drum fabricated using small shells riveted together. End cover plates can be opened as and when required. Mountings are mounted on drum as shown. Drum is followed by water tubes which are arranged below drum and connected to one another and drum through headers. Header in which water flows from drum to tubes is called down take header while headers in which flow is from tubes to drum is called uptake header. Soot deposition takes place in mud box which is connected to downtake header. \u201cBlow off cock\u201d for blowing out the sediments settled in mud box is shown in figure. Superheater tubes are also shown in the arrangement, which are U-shape tubes placed horizontally between drum and water tubes. Superheating of steam is realized in superheater tubes. Boilers and Boiler Calculations ___________________________________________________ 449 Fig. 11.12 Babcock and Wilcox boiler Below the superheater and water tubes is the furnace, at the front of which fuel feed hopper is attached. Mechanical stoker is arranged below the hopper for feeding fuel. Bridge wall and baffles made of fire resistant bricks are constructed so as to facilitate hot gases moving upward from the grate area, then downwards and again upwards before escaping to the chimney. A smoke box is put at the back of furnace through which smoke goes out via chimney, put at top of smoke box. A damper is used for regulating pressure difference (draught) causing expulsion of hot gases. The complete boiler unit with all mountings and accessories is suspended by steel slings from girders resting on steel columns. It is done so as to permit free expansion and contraction of boiler parts with temperature. 11.12 STIRLING BOILER This is a water tube boiler in which bent tubes are connected to three or four drums together. These bent tubes are inclined as shown in Fig. 11.13. Bent tubes are advantageous in respect to flexibility in maximizing heating surface and no requirement of headers. For deflecting the hot gases the baffles are provided. Here three steam drums and one mud drum is shown. Feed water enters the first steam drum located near the exit passage and pass through tubes to mud drum and subsequently gets raised through tubes into other steam drum. Thus the water is circulated so as to pick maximum heat from hot gases. Steam generated is collected in upper portion of steam drums from where it can be extracted out. Hot gases coming from furnace area travel across the boiler and go out of exit passage after transferring heat contained by them. Such boilers are capable of generating steam upto maximum pressure of 60 bar and steam gen- eration rate up to 50,000 kg/hr. 450 _________________________________________________________ Applied Thermodynamics Fig. 11.13 Stirling boiler 11.13 HIGH PRESSURE BOILER High pressure boilers generally operate in supercritical range. Need of such boilers is felt because high pressure and temperature of steam generated in boiler improves plant efficiency. These boilers have Fig. 11.14 High pressure boiler with natural circulation Boilers and Boiler Calculations ___________________________________________________ 451 forced circulation of water/steam in the boiler. This forced circulation is maintained by employing suitable pump. The steam drum is of very small size and in some cases it may be even absent too. This is because of using forced circulation. In case of natural circulation drum size has to be large. Schematic of high pressure boiler is shown in figure 11.14. In fact the high pressure boilers have been possible because of availability of high temperature resistant materials. Here direct heating of water tubes is done by the excessively hot gases present in fire box. The fire box has large volume as otherwise exposed water tubes shall melt. Heat is picked by number of parallel tubes containing water. These parallel tubes appear as if it is a wall due to close spacing of tubes. Water circulation circuit is shown in line diagram. High pressure boilers may have natural circulation in case the steam pressure desired lies between 100 and 170 bar and size is not constraint. High pressure boilers have capability of generating larger quantity of steam per unit of furnace volume. High pressure boilers are disadvantageous from safety point of view and therefore, stringent reliability requirements of mountings is there. 11.14 BENSON BOILER It is a water tube boiler capable of generating steam at supercritical pressure. Figure 11.15. shows the schematic of Benson boiler. Mark benson, 1992 conceived the idea of generating steam at supercritical pressure in which water flashes into vapour without any latent heat requirement. Above critical point the water transforms into steam in the absence of boiling and without any change in volume i.e. same density. Contrary", " to the bubble formation on tube surface impairing heat transfer in the normal pressure boilers, the supercritical steam generation does not have bubble formation and pulsations etc. due to it. Steam generation also occurs very quickly in these boilers. As the pressure and temperatures have to be more than critical point, so material of construction should be strong enough to withstand thermal stresses. Feed pump has to be of large capacity as pressure inside is quite high, which also lowers the plant efficiency due to large negative work requirement. Benson boilers generally have steam generation pressure more than critical pressure and steaming rate of about 130\u2013135 tons/hr. Thermal efficiency of these boilers is of the order of 90%. Fig. 11.15 Benson boiler 452 _________________________________________________________ Applied Thermodynamics 11.15 LOEFFLER BOILER This a forced circulation type boiler having both direct and indirect type of heat exchange between superheated steam/water and hot gases/steam respectively. Schematic arrangement of a Loeffler boiler is shown in Fig. 11.16. Here the hot combustion gases emerging out of furnace are firstly used for superheating of steam and secondly for reheating/economiser sections. Steam generation is realized through the superheated steam being injected into evaporator drum. Saturated steam thus generated in evaporator drum as a result of mixing of superheated steam and water is picked up by steam circulation pump. This pump forces saturated steam at high pressure through superheater tubes where the hottest flue gases from furnace superheat steam coming from evaporator. Flue gases subsequently pass through reheater/economiser sections as shown. Superheated steam coming out of superheater section is partly taken out through steam main and remaining is injected into evaporator drum. Generally superheated steam is divided in proportion of 1 : 2 for steam main and evaporator drum respectively. Feed water to the boiler is pumped by feed pump through the economiser section to evaporator drum. Generally steam generated is at pressure of about 120 bar and temperature of 500C. Fig. 11.16 Schematic arrangement in Loeffler Boiler Loeffler boiler is advantageous in many respects such as there is no possibility of soot deposition in evaporator section. Also by the use of higher pressure steam the heat transfer rate gets improved. This boiler is much compact as compared to other natural circulation boilers. 11.16 VELOX BOILER Velox boiler is a fire tube boiler having forced circulation. Boiler has gas turbine, compressor, generator, feed pump, circulation pump etc. as its integral components. Thus Velox boiler unit is a compact steam generating plant. Figure 11.17 shows the line diagram of Velox boiler unit. Boiler unit has a compressor supplying high pressure air at about 3 bar into the oil burner so as to produce combustion products at high pressure and thus have hot flue gases flowing through fire tubes at very high velocity of the order of supersonic velocity. Flue gases flowing at supersonic velocity Boilers and Boiler Calculations ___________________________________________________ 453 Fig. 11.17 Velox boiler unit facilitate very high rate of heat interaction between two fluids, generally of the order of 2 107 kcal/m3 of combustion volume. Combustion space is lined with concentric vertical tubes having hot flue gases passing through the inner tube and water surrounding it in outer tube. Hot flue gases pass through superheater section and subsequently enter into gas turbine for its expansion. Gas turbine drives the compressor used for producing compressed air. Expanded gases coming out of gas turbine at about 100\u2013125 m/sec enter into economiser where feed water picks up heat from gas turbine exhaust. Hot feed water coming out of economiser is sent into steam/water drum from where water is circulated through vertical concentric tubes using a circulating pump. During the water flow in combustion volume space it partially gets transformed into steam and the mixture is injected tangentially into drum. Tangential discharge of mixture forms a circulatory flow (vortex) causing steam release due to centrifugal action, thus separation of water/steam. Steam is subsequently passed through superheater section while water is again circulated using circulation pump. Steam passes through steam headers after superheating. Surplus energy, if any in gas turbine is used by alternator attached to it which supplements the electricity requirement in various auxiliary devices. Velox boilers are very flexible and capable of quick starting. Overall efficiency of the boiler unit is about 55\u201360%. Boiler is capable of handling maximum of 100 tons/hr water which is limited by the limitation of maximum power requirement in compressor. 11.17 LA MONT BOILER This is a water tube boiler having forced circulation. Schematic showing the arrangement inside boiler is given in Fig. 11.18. Boiler has vertical shell having three distinct zones having water tubes in them, namely evaporator section, superheater section and economiser section. 454 _________________________________________________________ Applied Thermodynamics Fig.", " 11.18 La Mont boiler Feed water is fed from feed pump to pass through economiser tubes. Hot water from economiser goes into drum from where hot feed water is picked up by a circulating pump. Centrifugal pump may be steam driven or of electric driven type. Pump increases pressure and water circulates through evaporation section so as to get converted into steam and enters back to drum. Steam available in drum enters into superheater tubes and after getting superheated steam leaves through steam main. 11.18 FLUIDIZED BED BOILER Fluidized bed combustion is the recent development and large number of boilers are coming up with this arrangement. Schematic of fluidized bed combustion is shown in Fig. 11.19 for explaining the principle of fluidized combustion. Here a bed of inert, refractory sand type material is forced to get fluidized by the air passing through it. Air used for fluidization is heated before being sent into the bed. Auxiliary fuel which is generally gas gets burnt above or within the bed so as to cause bed temperature to go up to about 650C. When suitable temperature level is attained then coal is fed on it or into it for being burnt. The burning of auxiliary fuel is stopped at the moment when burning of coal becomes self sustainable. Boilers and Boiler Calculations ___________________________________________________ 455 Fig. 11.19 Horizontal fire tube boiler with fluidized bed The supply of coal and air are governed by the demand on the boiler. Maximum bed temperature generally reaches up to 950C as this temperature control avoids clinker formation and emission of undesirable salts. For maintaining temperature of bed the arrangement is made for cooling of bed by water tubes and also by supplying excess air for cooling. Sometimes the low temperature flue gases leaving boiler are recirculated for bed cooling. Fluidized bed combustion offers advantage of using any kind of fuel i.e. solid, liquid or gaseous fuel. Also in this type of combustion the use of dolomite or lime stone as bed material helps in retaining sulphur in fuel. The clinker formation and emission of undesired substances is also avoided as the combustion can be controlled up to 950C. Due to large quantity of both combustible and incombustible material present on the bed there occurs the problems of erosion in bed tubes and surroundings and also large burden on bed etc. Fluidized bed combustion is used in both fire tube and water tube boilers but the water tube boiler offers advantage of greater flexibility in design of furnace shape and allowing for greater freeboard in which entrained particles can drop back into bed. Air velocity is generally limited to 2.5 m/s as beyond this the possibility of incomplete combustion increases. Fluidized bed boilers may have different types of fluidized beds such as, (i) Shallow beds, which have bed depth up to about 30 cm. Due to such small depth the tubes for cooling can\u2019t be used in bed and excess air or recirculated flue gases are used for cooling. Boilers and Boiler Calculations ___________________________________________________ 457 Fig. 11.21 Line diagram for single pressure HRSG 01 02 03 04 05 06 07 08 09 10 11 12 13 Fig. 11.22 Multi pressure HRSG Heat recovery steam generators may generate steam at single pressure or at multiple pressures. The schematic for single pressure and multi pressure steam generation are shown in Figs. 11.21, 11.22. Circulation system in such boilers may be natural circulation or forced circulation. Natural circulation HRSG typically consist of vertical tubes and horizontal flow arrangement. Circulation is maintained by the density difference between cold water in downcomer and hot steam-water mixture in evaporator tubes. Forced circulation HRSG are characterized by horizontal tubes with vertical gas flow and use of pumps to circulate steam-water mixture inside the tubes. Due to vertical arrangement of forced circulation HRSG\u2019s, the plan area required is less. Therefore, where available floor space is limited, vertical waste heat boilers are suitable. Different aspects like performance, start up time and field erection requirements are generally same for both forced and natural circulation boilers but the operation and maintenance costs are higher for forced circulation HRSG due to presence of circulation pump. Generally horizontal natural circulation HRSG\u2019s are preferred choice, not withstanding their disadvantage in respect of space requirements. However, in applications where space constraints exist, vertical HRSG\u2019s with natural circulation have been recently developed. HRSG\u2019s generating steam at multipressure are attractive as they extract heat effectively and efficiently. Each pressure level of steam generation requires an economiser, an evaporator and a super- 458 _________________________________________________________ Applied Thermodynamics heater as shown. The positioning of these various heat exchangers in the gas stream is critical as general design philosophy is to exchange heat from gas to fluid at the highest temperature difference available. This is best accomplished by making gas and steam/water temperature gradient nearly", " parallel to each other. In single pressure HRSG the superheater, evaporator and economiser are placed in descending order along the gas path while in multi pressure HRSG this general order is maintained but various sections may be interchanged so that a nearly parallel temperature gradient may be achieved. Figure. 11.23 details temperature variation along the length of HRSG. Fig. 11.23 Temperature variation in HRSG HRSG has three basic parameters of pinch point, approach temperature and allowable gas side pressure drop through heat recovery system which effect the effectiveness of heat exchange. Pinch Boilers and Boiler Calculations ___________________________________________________ 459 point is the difference between the gas temperature leaving the evaporator section of system and the saturation temperature corresponding to the steam pressure in that section. Approach temperature is the difference between the saturation temperature of fluid and inlet temperature of fluid. In general it is seen that lowering the pinch point shows an increase in total heat recovered in system. However, lowering pinch point shall require more heat exchange surface and result in increase in cost and gas side draught losses. Unit investment cost is higher at low pinch point. Thus optimum design is obtained based upon it. Generally pinch point of 8\u201310C is used in view of above aspects. Lowering the approach temperature can result in more steam production at that pressure level but high approach temperature offers high level of stability. Higher approach temperatures in economiser section will eliminate the probability of steam formation in economiser at lower loads or during start up. The allowable gas side pressure drop through HRSG influences the design and cost of unit. Very low pressure drop results in large heat exchange surface and low gas velocities. Very small gas velocity produces higher uncertainty of design with consequence of unstable performances. Generally allowable pressure drop is 250 mm to 300 mm of H2O. Stack gas temperature (temperature of exhaust gases leaving HRSG) selection depends upon the need to effectively recover the heat and also prevent corrosion etc. in stack. The HRSG may also be of supplementary fired type when heat input by exhaust gases is insufficient for steam generation of desired quantity and quality. In supplementary fired HRSG burners are also put in HRSG for increasing its steam generation capacity. 11.20 BOILER MOUNTINGS AND ACCESSORIES Boiler mountings and accessories have been defined earlier and shown on the different boilers. Different mountings are (i) Water level indicator (ii) Safety valves (iii) High steam and low water safety valves (iv) Fusible plug (v) Pressure gauge (vi) Stop valve (vii) Feed check valve (viii) Blow off cock (ix) Manhole and mud box Various boiler accessories are: (i) Superheater (ii) Economiser (iii) Air preheater (iv) Feed pump Water level indicator: It is used for knowing the level of water in boiler as water level inside boiler should not go below a certain limit. General arrangement is shown in Fig. 11.24 with the different parts in it. It has two tubes one is front glass tube while other is metal tube. Water level is seen through glass tube which is made strong enough to withstand high steam pressure and temperature. Two control cocks Fig. 11.24 Water level indicator 460 _________________________________________________________ Applied Thermodynamics are provided for regulating steam and water passage from boiler to glass tube. For blow off purpose a blowing cock is also provided as shown. In case of breakage of glass tube the possibility of accident is prevented by providing two balls. As glass tube breaks the rush of water and steam carries the two balls with it and closes the openings for glass tube, thus water and steam flowing out can be prevented. Number of other types of water level indicators are also available. Safety valve: Its function is to prevent the steam pressure from exceeding a limiting maximum pressure value. Safety valve should operate automatically by releasing excess steam and bring pressure down within safe limits. These are of different types such as \u2018dead weight safety valve\u2019, \u2018lever safety valve\u2019 \u2018spring loaded safety valve\u2019 etc. Figure 11.25 gives the general description of \u2018dead weight safety valve\u2019. Fig. 11.25 Dead weight safety valve It has a large vertical pipe on the top of which a valve seat is fixed. Valve rests upon this valve seat. A weight carrier is hung on the top of valve upon which cast iron rings enclosed in cast iron cover are placed in weight carrier as dead weight. When the pressure of steam exceeds the total weight of valve, it is lifted and falls back as steam pressure gets reduced. High steam and low water safety valve: This is a combined form of safety valve and low water level indicator. Figure. 11.26 shows the high steam and low water safety valve. It prevents from excessive pressure as it has a simple lever safety valve loaded by two weights as shown. Low water safety arrangement is activated through float put in boiler shell and prevents from overheating due to", " low water. It has two distinct valves as shown in Fig. 11.26. When the pressure inside goes beyond limiting value then high steam valve gets lifted up and excess pressure of steam is released. When the water level goes below critical level then low water valve gets raised up as it is fulcrumed and is linked to float on one end and balance weight on other end. With raising of low water valve the hemispherical valve shifts up from valve seat and steam escapes out with hissing sound. This hissing sound is loud enough to warn Boilers and Boiler Calculations ___________________________________________________ 461 attendant about the low water level in boiler. When the water level is correct then high steam valve simply acts as the dead weight safety valve. A drain pipe is also provided so that the steam getting condensed can be drained out. Condensation of steam is possible due to throttle of steam during its release from valves. Fig. 11.26 High steam and low water safety valve Fusible plug: It is a safety device used for preventing the level of water from going down below a critical point and thus avoid overheating. Fusible plug is mounted at crown plate of combustion chamber. Fig. 11.27 Fusible plug Fusible plug has gun metal body and a copper plug put with fusible metal at interface of copper plug and gun metal body. As water level goes down the heat available from furnace could not be completely utilized for steam formation and so the overheating may cause melting of fusible metal. Fusible metal is a low melting point metal. Thus upon melting of lining the copper plug falls down and 462 _________________________________________________________ Applied Thermodynamics water falls from this opening onto furnace and thus quenches fire. Pressure gauge: It is mounted at front top. Generally Bourdon type pressure gauge are being used for pressure measurement. Pressure is continuously monitored so as to avoid occurrence of over shooting of boiler pressure. Although safety devices to protect boiler against pressure rising beyond a limit are provided but pressure gauges are also used for monitoring pressure. Stop valve: It regulates the flow of steam from the boiler as shown in Fig 11.28. This is generally mounted on highest part of boiler shell and performs function of regulating the flow of steam from boiler. Stop valve generally has main body of cast steel, valve, valve seat and nut etc. are of brass. Stop valve can be easily operated by rotating the hand wheel which causes lifting or lowering of spindle, thus causing opening or closing of valve. Fig. 11.28 Stop valve Feed check valve: It is a non return valve at the end of delivery pipe from feed water pump and is placed on boiler shell slightly below normal water level. Figure 11.29 shows the arrangement in a feed check valve. It has a check valve whose opening and closing are regulated by the position of spindle. By hand wheel rotation the position of spindle can be altered suitably. Feed check valve permits unidirec- Fig. 11.29 Feed check valve Boilers and Boiler Calculations ___________________________________________________ 463 tional flow of water from feed pump to be boiler shell. Under normal running the pressure of feed water coming from pump is more than pressure inside the boiler and so the feed water continues to enter the shell. While during the non working of feed pump the pressure in boiler shell is more and so the check valve gets closed. Blow off cock: It is used for periodical cleaning by discharging the water and sediments from bottom of boiler. Figure 11.29 shows the blow off cock. Blow off cock is fitted to the bottom of boiler shell. Blow off cock has a plug of conical type put into the mating casing. Plug position is altered for opening and closing the flow. Plug has rectangular opening which when comes in line with inlet and outlet passage then blow off cock is open and when opening is not in line then cock is closed. Plug is rotated by spindle. Fig. 11.30 Blow off cock Blow off cock also helps in regulating the salt concentration as frequent draining helps in throwing out the salt deposited over period of time. Opening blow off cock removes deposited sediments in boiler. Manhole and mud box: Manhole provides opening for cleaning, inspection and maintenance purpose. Mud box is a collection chamber (as shown in Babcock and Wilcox boiler) for collecting the mud. Superheater: Its purpose is to super heat steam and is a type of heat exchanger in which steam flows inside tubes and hot gases surround it. Figure 11.31 shows the smooth tube hairpin type superheater (Sudgen\u2019s superheater) and convective and radiant superheater. 464 _________________________________________________________ Applied Thermodynamics Fig. 11.31 Superheater In hair pin superheater the steam generated is passed through isolating valve to U-shaped steel tubes. Superheated steam leaves superheater through tube connected to steam stop valve. Hot gases from fire tube are diverted over superheater tubes by dam", "per as shown. These hot gases upon passing over steel tubes leave boiler through bottom flue. The convective and radiant superheater as shown has two set of tubes picking up heat through convection and radiation. Economizer: It is also a heat recovery device in which feed water is heated from heat available with exhaust gases. Thus hot feed water available from economizer lowers the fuel requirement in Fig. 11.32 Economizer Boilers and Boiler Calculations ___________________________________________________ 465 combustion. It is also a type of heat exchanger having exhaust gas and feed water as two fluids. General arrangement in economizer is shown in Fig. 11.32. Economizer also helps in removal of dissolved gases by preheating of water and thus minimizes tendency of corrosion and pitting. Hotter feed water also reduces thermal strain in boiler parts. Economizer is located in the boiler structure so as to expose the economizer surface to hot gases. Its location varies with the boiler designs. Typical economizer called Green\u2019s economizer as shown in Fig. 11.32 has vertical pipes of cast iron fitted with two headers at bottom and top respectively. Feed water passes through bottom header, economizer tubes and top header to boiler. Thus economizer is simply a heat exchanger where heat is transferred from hot flue gases to water inside the tubes through metal interface. Top header is also provided with a safety valve so as to avoid explosion due to excessive pressure of water developing inside economizer tubes. Bottom header is also provided with a blow off valve so as to throw out the sediments deposited in feed water. Economizer is also provided with scrapers fitted to clean pipes from the deposition of soot carried by the flue gases. Continuous scrapping is always desired so as to maximize heat transfer rate. Economizer also has a by pass provided so that flue gases can be diverted when economizer is out of full or part operation due to failure or cleaning purpose or feed water temperature control. Air preheater: It is used for recovering the heat going along with exhaust gases by the air before being sent to furnace. Heat is recovered by passing exhaust gases through an air to air heat exchanger as shown in Fig. 11.33. Air preheaters are generally placed after economizer and before chimney. Air when preheated before supply to furnace/combustion chamber helps in achieving \u2018faster rate of combustion\u2019, \u2018possibility of burning inferior quality coal/fuel\u2019 and \u2018increased rate of evaporation from boiler\u2019 etc. Air preheaters are of tubular type, plate type and regenerative type. This classification of air preheaters bases upon the kind of arrangement used for heat exchange between two fluids. Generally, tubular type air preheater are generally used in small boilers. Tubular air preheater has hot flue gases passing inside tubes and air blown over these tubes. Fig. 11.33 Tubular air preheater In case of plate type air preheater there are number of plates having air and flue gases flowing through alternative spacings. In regenerative type air preheater there is a wire mesh rotor which is alternatively heated and cooled by the hot flue gases and air to be used for combustion. Feed pump: Feed pump is used for sending water into boiler at the pressure at which steam generation takes place. It is generally of three types i.e. centrifugal pump, reciprocating pump and injectors. A reciprocating type feed pump is shown in Fig. 11.34. In boilers the pumps raise feed water pressure to the value more than the highest operating pressure of boiler. Pumps also have capability to deliver feed water in excess to the maximum evaporation rate of boiler. This excess capacity of feed pump is generally 15\u201320% of maximum continuous rating and is required to meet one or more of following situations. (i) Sometimes excessive steam demand may occur. 466 _________________________________________________________ Applied Thermodynamics (ii) Since boilers are to be blown out frequently to remove depositions and salts, therefore excess capacity is required. (iii) Malfunctioning of boiler may cause carrying away of water with steam, thereby causing water shortage in boiler. (iv) Over a period of time pump capacity decreases and so excess pump capacity is desired. Fig. 11.34 Reciprocating type pump, Duplex feed pump Now a days multistage centrifugal type pumps are used, which have favourable pressure/volume characteristics. Steam trap: Steam traps perform function of catching steam getting condensed in the form of condensate (water) due to partial condensation of steam in pipes, steam jackets etc. Figure 11.35 shows a bucket type steam trap. Water available due to partial condensation enters steam trap at inlet A. Steam Fig. 11.35 Bucket type steam trap trap casing already has water in it and bucket keeps on floating. As the water level in steam trap casing rises to the extent that water overfl", "ows from the bucket, due to excess weight of water in bucket the bucket sinks down and discharge valve opens causing water to leave through outlet passage B. After sufficient water is drained out the weight of water in bucket reduces and the bucket starts floating again. This rise of bucket closes the discharge valve again. Boilers and Boiler Calculations ___________________________________________________ 467 11.21 BOILER DRAUGHT Draught refers to the pressure difference created for the flow of gases inside the boiler. Boiler unit has a requirement of the expulsion of combustion products and supply of fresh air inside furnace for continuous combustion. The obnoxious gases formed during combustion should be discharged at such an height as will render the gases unobjectionable. A chimney or stack is generally used for carrying these combustion products from inside of boiler to outside, i.e. draught is created by use of chimney. Draught may be created naturally or artificially by using some external device. Draught can be classified as below: In this the pressure difference is created naturally without using any positive displacement device. Artificial draught is created using some external assistance causing forced displacement of gases. It can be created either by using mechanical devices or steam. Artificial draught can be of induced type, forced type or combination of two types. Thus the draught in boiler may be said to be required for, \u2018providing and maintaining the supply of sufficient air for combustion\u2019, \u2018expulsion of combustion products from furnace region\u2019 and \u2018discharge of burnt gases to atmosphere\u2019. The amount of draught required shall depend upon, \u2018type of boiler\u2019, \u2018rate of fuel burning\u2019, \u2018rate at which combustion products are produced\u2019 and \u2018the air requirement rate\u2019. As the pressure difference is very small so draught is measured in \u2018mm\u2019 of water. Mathematically, pressure due to 1 mm of water column is equivalent to 1 kgf/m2. 11.22 NATURAL DRAUGHT It is produced employing chimney. The natural draught is produced by a chimney due to the fact that the hot gases inside the chimney are lighter than the outside cold air i.e. density difference of hot gases inside chimney and cold atmospheric air. Thus in a boiler unit the combustion products (hot) rise from fuel bed through chimney, and are replaced by fresh air (cold) entering the grate. It means that amount of draught produced by a chimney depends upon flue-gas temperature. Intensity of draught produced by chimney also depends upon height of chimney. Draught produced by a taller chimney is large as the difference in weight between the column of air inside and that of air outside increases with height. Generally draught is less than 12 kgf/m2 in chimneys. 468 _________________________________________________________ Applied Thermodynamics In stricter terms the word \u2018chimney\u2019 is used for brick or concrete structure and \u2018stack\u2019 is used for metallic one. Chimneys are generally made of steel, brick or reinforced concrete. Steel chimneys or stacks are most desirable for smaller boiler units due to small initial cost, ease of construction and erection. On account of small space requirement as compared to other stacks, self sustaining steel stacks are used in some large power plants. Steel stacks have problem of rust and corrosion, so painting requirements are quite stringent. Brick chimneys are required where permanent chimney with longer life is required. Such chimneys have inherent disadvantages of leakages etc. across the construction, therefore careful construction is required. Leakage of air across chimney wall effects intensity of draught. Brick chimneys are constructed of round, octagonal, or square section. Generally brick chimney has two walls with air space between them and inner wall having fire brick lining. Concrete chimneys are used due to the absence of joints, light weight and space economy as compared with brick chimneys. Also the reinforced concrete chimney is less expensive compared to brick chimney along with minimum chances of leakage across walls. Calculations: As it is obvious from earlier discussion that the vertical duct called chimney creates natural draught so estimation of height of chimney is very important. Figure 11.36 shows the schematic of chimney in a boiler unit. During no working of boiler the pressure inside boiler is atmospheric pressure. Pressure at outlet of chimney will be less than atmospheric pressure due to altitude difference. Fig. 11.36 Schematic of chimney During boiler operation the chimney shall be filled with hot gases and the pressure at bottom of chimney (pb) shall be summation of \u2018pressure at chimney outlet\u2019, (p0) and \u2018pressure due to hot gas column of height H\u2019. Pressure at bottom of chimney = Pressure at outlet + Pressure due to hot gas column pb = p0 + g g H g is density of hot gases. where However, the pressure at grate level remains unchanged", ". Pressure difference between the grate level pressure and bottom of chimney causes flow of gases. This pressure difference is also called static draught. Let us consider the combustion of fuel in furnace. Combustion products are released as a result of this combustion process. Fuel may be considered to be comprising of hydrocarbons. Fuel + Air Combustion products + Heating value In the hydrocarbon fuel major constituents are carbon, hydrogen, nitrogen etc. As a result of complete combustion carbon gets transformed into carbon dioxide and hydrogen yields steam. The combustion products have major fraction of carbon dioxide and steam, but the volume of steam is negligible compared to volume of combustion product. The volume of combustion products can be taken equal to volume of air supplied, measured at same temperature and pressure. Boilers and Boiler Calculations ___________________________________________________ 469 Let us assume various properties as, Ta = Atmospheric temperature, K Tg = Average temperature of hot gases inside chimney, K T0 = Absolute zero temperature, 273.15K a = Density of air at absolute temperature, 1.293 kg/m3 g = Density of hot gases inside chimney, kg/m3 p = Pressure difference, draught in Pa p = Pressure difference, draught in kgf/m2 hg = Equivalent height of hot gas column to produce draught, p in \u2018metres\u2019. hw = Equivalent height of water column in \u2018mm\u2019 to produce draught, p m = Mass of air supplied per kg of fuel C = Hot gas velocity in chimney, m/s Mg = Discharge rate through chimney, kg/s Total mass of combustion products (hot gases) for one kg of fuel = (m + 1), kg Using the assumption regarding combustion products, Volume of hot gases at temperature, Tg = Volume of air supplied at temperature, Tg = = Massof air Density of air at Tg Pressure of hot gases in chimney at grate level = Density of hot gases height of chimney = gravitational acceleration H g Mass of hot gases Volume of hot gases at Tg = Density of air at Ta Height g T 0 a T a H g = = = = Pressure due to cool air (outside) column of height H at grate level 470 _________________________________________________________ Applied Thermodynamics a = T H g 0 T a Natural Draught produced = Difference of pressures due to cool air column and hot gas column of height \u2018H\u2019. Or T0 H g 1 T a Draught in kgf/m2, p = a T0, Pa, kgf/m2 1 T g 1 T g Substituting values for T0, a and rounding off values we get, p = 353, kgf/m2 Height of hot gas column equivalent to draught produced, hg = \u2206 \u2032 p, kgf m 2 Density of hot gases, kg m at hg = H m m 1 1, metres As the 1 mm of water column exerts a pressure of 1 kgf/m2 so the draught in terms of water column can be given by, hw = a T0 H Thus natural draught can be mathematically given by p, p, hg and hw. Actually in boilers this draught requirement is different from that theoretically estimated due to the draught losses. The magnitude of these losses varies from boiler to boiler due to different arrangements within them. Some of generic losses shall be because of:, mm of water frictional losses due to resistance offered by passage surface roughness, different equipments as grate, superheater, air preheater, economiser etc. through which gas passes. Boilers and Boiler Calculations ___________________________________________________ 471 pressure losses in bends, baffles, supports etc. kinetic energy required with gases for moving at certain velocity throughout. Therefore while designing the chimney due considerations should be made for overcoming above losses. These losses constitute about 20% of the static draught produced. Hot gas velocity in chimney: Assuming chimney to be frictionless the hot gas velocity in chimney could be given using the equivalent hot gas column height; C = 2g hg, m/s For the chimney having friction losses, the hot gas velocity shall be lesser. If the equivalent pressure loss due to friction in hot gas column is given by hf then the velocity of hot gases, C = 2g /s f h h g Substituting values for \u2018g\u2019 C = 4.43 h g 1 h f h g or C = K hg where K is a constant and its value is available for different types of chimneys as given below. It depends upon the friction loss fraction. K = 0.825 for brick chimney, and K = 1.1 for steel chimney Diameter of chimney: Diameter of chimney could be estimated from the mass flow rate of hot gases through chimney and its velocity", ". Chimney cross-sectional area, A = Discharge rate Velocity of hot gases Density of hot gas Diameter of chimney = \u00d7 4 \u03c0 M g \u22c5 \u03c1 C g Discharge through chimney: Mass flow rate of hot gases through chimney could be obtained as, Mg = Cross-sectional area Velocity of hot gas Density Mg = A C g Mg = A g K hg For hot gas pressure pg, the density of gas g can be given using perfect gas approximation, g = p g R T g. Here R is gas constant 472 _________________________________________________________ Applied Thermodynamics so Mg = A p K h g g R T g Discharge through chimney can be mathematically maximized for certain conditions. A look at expression of discharge given earlier shows that for a particular chimney. h g 1 T g Mg or Mg = Constant Differentiating discharge with respect to hot gas temperature and equating it to zero for optimum condition, g d M d T g = 0 we get, or T g T a = Tg = 2 1m m 2 1m m Ta Upon substituting Tg value in 2 d M g 2 = 0 we see that it is condition for maxima of Mg. d T g Thus discharge through chimney is maximum for the hot gas temperature given by 1m m Ta Tg = 2 This hot gas temperature shall be slightly more than twice of ambient temperature for maximum discharge rate through chimney. For the condition of maximum discharge, draught can be obtained as, 1 2 Ta pfor max. discharge = a T0 H g pfor max. discharge = a T H g 0 2 T a In terms of water column, hw, for max. discharge = a 2 T H 0 T a mm of water column. Boilers and Boiler Calculations ___________________________________________________ 473 Maximum discharge could be obtained upon substituting optimum temperature, Tg Mg, max = For a smooth and frictionless chimney, 2 Mg, max =, kg/s Above expression for maximum discharge rate can also be used for getting the height of chimney for a given gas flow rate. Efficiency of chimney: It has been explained in natural draught that the pressure difference is created due to density difference caused by temperature difference. Hence it is obvious that the flue gases should leave at quite high temperature for creating required density difference. Thus the flue gases leave with sufficient heat energy, which could be used in boiler if some other mechanism is employed for exhaust of flue gases such as artificial draught. Therefore, efficient chimney should have such a design so that flue gases leave at lowest possible temperature. Efficiency of chimney is quantification of the cost of natural draught in terms of energy, i.e. the large amount of usable energy going along waste hot gases. Normally this efficiency of chimney is less than 1 percent. Chimney efficiency is defined as the ratio of \u201cenergy with unit mass of gas in natural draught\u201d and \u201cthe extra heat carried by same mass of gas due to high temperature in natural draught as compared to that in artificial draught\u201d. Chimney efficiency = Energy with unit mass of gas in natural draught Extra heat carried away in natural draught compared to artifical draught by unit mass of gas Let us assume that the temperature of flue gases in artificial draught is Tg,a which will be less than Tg in natural draught. Let the specific heat of hot flue gases be Cp,g, J/kg K. So, energy with unit mass of hot flue gas in nature draught = 1 hg g, joule = 9.81 H m 1 m 1 T g T a Extra heat carried by unit mass of hot gas in natural draught compared to artificial draught = Cp, g 1 (Tg \u2013 Tg, a) = Cp, g 1 (Tg \u2013 Tg, a), joule 81 H C p g, chimney = Expression for efficiency of chimney shows that it is directly proportional to the height of chimney, but the efficiency is found to be very small even for very tall chimney. It is seen that about more than 20% of total heat released gets lost along with hot flue gases in case of natural draught. Therefore the artificial draught becomes more economical in the situation where cost of harnessing the extra heat carried with flue gases is less than the cost of energy lost in natural draught. 474 _________________________________________________________ Applied Thermodynamics 11.23 ARTIFICIAL DRAUGHT Artificial draught refers to the externally created draught employing some equipments for it. Its requirement is felt, when the natural draught becomes insufficient for exhaust of flue gases. In general it is seen that for draught requirements being more than 40 mm of water, the natural draught does not work and becomes highly uneconomical", ". In the modern large power plants this draught produced by chimney is insufficient and requires some artificial method. Also the size of boiler units in use today forbid the use zof natural draught as the flue gas handling capacity is limited. In case of natural draught the fuel rate upto (20 kg/hr per m2 of grate area could be handled while with artificial draught it goes up to 300 kg/hr per m2 of grate area. Apart from these limitations the economy of using artificial draught over natural draught beyond a limit also make it attractive. For same steam generation the fuel consumption gets reduced by up to 15% with use of artificial draught in a boiler. Artificial draught may be produced either by mechanical means such as fans, blowers etc. or by using steam jet for producing draught. Thus artificial draught can be classified as, (i) Mechanical draught (ii) Steam jet draught. Artificial draught systems do not require tall chimney/stack, but small stack is always required for discharge of flue gases to certain height in atmosphere for minimizing pollution. 11.23.1 Mechanical Draught Mechanical draught produced using fans, blowers etc. could be of forced type, induced type or the combination of the two. Line diagram showing the arrangements is shown in Fig. 11.37. (i) Forced draught: It is the arrangement in which high pressure air is delivered to the furnace so as to force flue gases out through stack. Air under pressure may be fed to stokers or grate for which a fan/blower is put at the bottom of furnace. As due to pressurised air the pressure inside furnace becomes more than atmospheric pressure so it should be properly sealed, otherwise gas may leak through the cracks in setting into the boiler unit. Also the flames from furnace may flare out upon opening the fire door, so it should be equipped with dampers to shut off air supply when furnace doors are opened. It is obvious from here that the fan in case of forced draught shall handle fresh atmospheric air. Induced draught: Induced draught is the one in which the suction created on furnace side draws flue gases and throws them out through small chimney/stack. Fan is located at base of chimney in induced draught so as to reduce pressure at fuel bed below atmospheric pressure. The fan in induced draught shall handle hot flue gases. Power required to drive the fan/blower in case of induced draught is less than that in case of forced draught fan. Mathematically it can be given as below. For volume of fluid handled being V (m3/s) at pressure of p, the power required shall be, (ii) = p V mech In case of induced draught fan, as hot flue gases are to be handled, so, total mass handled by fan for mf kg of fuel burnt per unit time = mf + m mf From earlier article, density of hot flue gases = = mf (1 + m), kg/s 1 m m T g a T 0 Boilers and Boiler Calculations ___________________________________________________ 475 Therefore, volume handled by fan = = Vinduced = Mass Density Power required = p V induced mech Induced draught fan power requirement = p m m T g f T 0 a mech In case of forced draught the fan handles air at atmospheric temperature so, total mass of air handled = mf m, kg/s Volume of air at absolute zero temperature = Volume of air at atmospheric temperature = m3/s Power required in forced draught fan = Forced draught fan power requirement = m mf mech 0 p m m T a f T 0 mech a Comparing the two power requirements, Power required in induced draught Power required in forced draught = T g T a = More than 1 As Tg > Ta so the power requirement in induced draught is more than that of forced draught. Fig. 11.37 Mechanical draught (iii) Balanced draught: Sometimes it is seen that forced draught or induced draught alone is not suitable for boiler unit due to their own inherent limitations. In these situations a combination of forced draught and induced draught is being used. Such combined arrangement for 476 _________________________________________________________ Applied Thermodynamics draught is called balanced draught. Here both forced draught fan/blower and induced draught fan/blower are employed. Forced draught fan ensures complete supply of air for proper combustion after overcoming all resistances while induced draught fan takes care of post combustion resistances, thus ensuring complete removal of flue gases. Pressure variation in balanced draught shows the pressure values through out boiler unit. Fig. 11.38 Pressure variation in balanced draught (iv) Comparison of forced and induced draught: A comparison of forced and induced draught system shows that, Power requirement is more in case of induced draught fan blower as compared to forced draught because of hot fl", "ue gases being handled by induced draught fan and atmospheric air being handled by forced draught fan. Induced draught fan size may be upto twice of the fan size in forced draught. Material of induced draught fan should be capable of handling hot gases laden with solid particles (ash). As the induced draught fan/blower withstands high temperature, so the fan cooling requirements are very stringent as compared to forced draught. Leakage of atmospheric air into furnace due to sub atmospheric pressure inside the induced draught system causes dilution of flue gases. In case of forced draught, the pressure inside is more than atmospheric pressure so chances of leakage of furnace gases out to atmosphere are there. This may cause blow out upon opening of furnace doors in case of forced draught. Combustion rate is better in case of forced draught system due to better air dispersion across the grate, as compared to induced draught system. Maintenance is easy in forced draught fan as compared to induced draught due to their locations. (v) Advantages of mechanical draught: Mechanical draught has various advantages in comparison with natural draught. Such as, Better fuel economy due to complete utilization of heat in boiler, i.e. flue gases could be cooled to lowest possible temperature before exhaust. Better combustion rate as compared to natural draught system, because of better distribution and mixing of air and fuel. Combustion and evaporation rates can be easily regulated in case of mechanical draught by varying operating states of fan/blower. Boilers and Boiler Calculations ___________________________________________________ 477 Inferior quality fuel may also be used in mechanical draught. Height of chimney/stack used in mechanical draught is much smaller as compared to natural draught. Boiler plant efficiency gets improved with mechanical draught. Efficiency of mechanical draught is much more than that of natural draught. Fuel burning per unit grate area per unit time is up to 300 kg/m2\u00b7hr in mechanical draught as compared to upto 100 kg/m2\u00b7hr in natural draught. Mass of flue gases handled is more in mechanical draught as compared to natural draught. 11.23.2 Steam Jet Draught Artificial draught may also be produced by steam jets. Draught produced using steam jets may also be of induced type or forced type. In case of forced steam jet draught, the steam jet is located so as to force air and steam up through the bed, i.e. steam-jet is placed before the grate. In case of induced draught the steam jet is placed near stack with direction to smoke box so as to induce air across the grate and throw it out from stack. Steam jet draught is simple, cheap and requires minimum attention. With forced draught, the use of steam jet also prevents the formation of clinkers, thus permitting for low grade fuels in furnace. Schematic for steam jet draught is shown in Fig. 11.39. Steam jet draught is disadvantageous because of the fact that steam is not available initially when boiler unit just starts. Steam for draught is taken from boiler unit itself. Initially high pressure steam may be supplied externally. Steam jet draught is used in locomotive boilers. Various merits of steam jet draught are listed below, Fig. 11.39 Schematic for steam jet draught. For operation of steam jet draught steam is required, which is available without much investment. Low grade fuels can also be used in furnace with steam jet draught. Steam jet draught is simple, economical, requires least attention and minimum space. Use of steam and its mixing with ash/burnt residuals prevent formation of clinkers. 11.24 EQUIVALENT EVAPORATION From earlier discussions it is seen that there exists a large variety of the boilers in terms of their arrangement, efficiency, steam generation rate, steam condition, type of fuel used, firing method and draught etc. For comparing one boiler with other any of the above parameters can not be considered as they are interdependent. Therefore, for comparing the capacity of boilers working at different pressures, temperatures, different final steam conditions etc. a parameter called \u201cequivalent evaporation\u201d can be 478 _________________________________________________________ Applied Thermodynamics used. Equivalent evaporation actually indicates the amount of heat added in the boiler for steam generation. Equivalent evaporation refers to the quantity of dry saturated steam generated per unit time from feed water at 100C to steam at 100C at the saturation pressure corresponding to 100C. Sometimes it is also called equivalent evaporation from and at 100C. Thus, mathematically it could be given as, Equivalent evaporation = Mass of steam generated per hour \u00d7 ( Heat supplied to generate steam in boiler ) Heat supplied for steam generation at 100 C from water at 100 C i.e. Latent heat ( \u00b0 \u00b0 ) Heat supplied for generating steam at 100C from", " water at 100C at respective saturation pressure is 538.9 or 539 kcal/kg. For a boiler generating steam at \u2018m\u2019 kg/hr at some pressure \u2018p\u2019 and temperature \u2018T\u2019, the heat supplied for steam generation = m (h \u2013 hw), where h is enthalpy of final steam generated and hw is enthalpy of feed water. Enthalpy of final steam shall be; h =hf + hfg = hg, for final steam being dry saturated steam hf, hfg and hg are used for their usual meanings. h = hf + x \u00d7 hfg, for wet steam as final steam. h = hg + cp, sup. steam (Tsup \u2013 Tsat), for superheated final steam. m h hw 538.9 Equivalent evaporation, (kg/kg of fuel) = Equivalent evaporation is thus a parameter which could be used for comparing the capacities of different boilers. 11.25 BOILER EFFICIENCY Boiler efficiency quantifies, how effectively the heat is being used in boiler. Thus it could be given by the ratio of heat actually used for steam generation and total heat available due to combustion of fuel in boiler. Boiler efficiency = Heat used in steam generation Total heat available due to fuel burning ( m h \u00d7 m \u2212 ) h w C.V. f = here mf is the mass of fuel burnt per hour, C.V. is calorific value of fuel used (kcal/kg), m is mass of steam generated per hour and enthalpies h and hw are that of final steam and feed water, kcal/kg. Generally high heating value of fuel is used as calorific value of fuel. 11.26 HEAT BALANCE ON BOILER Heat balance on boiler refers to the accounting for total heat released inside boiler and its distribution. Total heat available inside boiler is due to burning of fuel and can be quantified by the product of mass of fuel and heating value of fuel. Heat distribution can be given comprising of the following, based on unit mass of fuel burnt. Boilers and Boiler Calculations ___________________________________________________ 479 (a) Heat used for generation of steam: Heat taken for generation of steam per kg of fuel burnt shall be; Qsteam = msteam (h \u2013 hw) Here msteam is mass of steam generated per kg of fuel burnt, h is enthalpy of final steam produced and hw is enthalpy of feed water. Qsteam = Equivalent evaporation 539, kcal/kg of fuel burnt (b) Heat lost due to incomplete combustion: Combustion is said to be complete when the carbon and hydrogen present in fuel get burnt into carbon dioxide and water vapour respectively. If the carbon burns into carbon monoxide instead of carbon dioxide then combustion is said to be incomplete. This incomplete combustion can be easily made complete by supplying excess air and devising for its fast mixing with fuel inside combustion chamber. Heat loss due to incomplete combustion can be quantified by knowing the additional heat that would be available with complete combustion. Heat loss due to incomplete combustion = Heat released when carbon burns into CO2 \u2013 Heat released when carbon burns into CO. During complete combustion of carbon into CO2, 3.38 104 kJ of heat is released while with incomplete combustion of carbon into CO, 1.012 104 kJ of heat is released by burning one kg, of carbon. Hence heat loss due to incomplete combustion of one kg of carbon shall be = (3.38 104 \u2013 1.012 104), kJ = 2.368 104 kJ per kg of carbon In order to get the estimate of total heat loss, it is first required to get the mass of carbon burnt to CO instead of CO2. For which the mass fraction of CO present in flue gas mixture comprising of CO and CO2 is required. Fraction of CO in flue gases = f CO f CO2 f CO where fCO and fCO2 are percentage by volume of CO and CO2 present in flue gases. f CO f fc f Mass of carbon burnt into CO instead of CO2 = CO where fC is fraction of carbon present in per kg of fuel. Qincomplete = Heat loss due to incomplete combustion fc 2.368 104, kJ/kg of fuel = f CO f CO f 2 CO CO 2 (c) Heat loss to dry flue gases: A large portion of heat getting lost goes along with flue gases. Flue gases leaving boiler comprises of dry flue gases and steam. Heat loss with dry flue gases can be given by, Qdry flue = mdfg Cpg (Tg \u2013 Ta) where mdfg is mass of dry flue gas per kg of fuel, Cpg is specific heat of dry flue gas and Tg and Ta are temperature of flue gas and air entering combustion chamber", ". (d) Heat loss to steam in flue gases: Steam is produced due to burning of hydrogen present in fuel into water vapour. Heat lost with steam in flue gases shall be Qsteam in flue = ms (hs1 \u2013 hf1) 480 _________________________________________________________ Applied Thermodynamics where ms is mass of steam produced per kg of fuel, hs1 and hf1 are \u201centhalpy values of steam at gas temperature Tg and partial pressure of vapour in flue gas\u201d and \u201centhalpy of water at mean boiler temperature\u201d respectively. (e) Heat lost in unburnt fuel: Some portion of heat may get lost in unburnt fuel, which could be given by the product of mass of unburnt fuel per kg of fuel and its calorific value, as Qunburnt = mubf CV where mubf is mass of unburnt fuel per kg of fuel and CV is calorific value of fuel. (f) Heat loss due to moisture in fuel: Moisture present in fuel shall also cause the loss of heat. This moisture shall get evaporated and superheated as fuel is burnt. For evaporation and superheating of moisture latent and sensible heat requirement shall be met from heat available in boiler due to burning of fuel. Mathematically, for unit mass of fuel burnt, it can be given as, Qmoisture = mmoist (hs2 \u2013 hf2) where mmoist is mass of moisture per kg of fuel burnt, hs2 is enthalpy of final steam produced and hf2 is enthalpy of water at boiler furnace temperature. (g) Heat loss due to convection, radiation and other unaccountable losses: In a boiler heat also gets lost due to convection, radiation from the boiler\u2019s surface exposed to atmosphere. The heat loss may also be there due to unconsumed hydrogen and hydrocarbon etc. Exact quantification of these losses is not possible, therefore these can be estimated by the difference of total heat available and cumulative heat loss described from (a) to (f). Heat loss due to convection, radiation and other unaccounted losses = Heat released by per kg fuel burnt \u2013 Heat loss components described from (a) to (f). Qunaccounted = (mf CV) \u2013 (Qsteam + Qincomplete + Qdry flue + Qsteam in flue + Qunburnt + Qmoisture) Heat balance sheet can be drawn after the above different components are quantified. It gives a quick account of heat released and its distribution for unit mass of fuel burnt or unit time. Heat balanced sheet shall be as given below. Table 11.2 Heat balance sheet on per minute basis (unit time basis) Total Heat (kcal) Supplied percentage Heat distribution (kcal) percentage, % Q = mf CV 100% (a) Heat used for generation 100 Q stream Q Qsteam = msteam (h \u2013 hw) (b) Heat lost due to incomplete 100 Q incomplete Q combustion, Qincomplete = f co f co 2 f co fc 2.368 104 4.18 (c) Heat lost due to dry flue gas, Qdry flue = mdfg Cpg (Tg \u2013 Ta) (d) heat loss of steam in flue gas, 100 Q dry flue Q 100 Q steam in flue Q (contd.) Boilers and Boiler Calculations ___________________________________________________ 481 Total Heat (kcal) Supplied percentage Heat distribution kcal Qsteam in flue = ms (hs1 \u2013 hf1) (e) Heat lost in unburnt fuels Qunburnt = mubf CV (f) Heat loss due to moisture in fuel Qmoisture = mmoist (hs2 \u2013 hf2) (g) Heat loss due to convection, radiation and other unaccounted losses,Qunaccounted = Q \u2013 Qsteam \u2013 Qincomplete \u2013 Qdry flue\u2013 Qsteam in flue \u2013 Qunburnt \u2013 Qmoisture percentage, % 100 100 Q unbrunt Q Q moisture Q Q unaccounted 100 Q 11.27 BOILER TRIAL Boiler trial refers to running the boiler under test conditions for its performance estimation. It gives the steam generation capacity of boiler, thermal efficiency of plant and heat balance sheet of the boiler. Under trial the boiler is run for quite long durations so as to attain steady state. Generally the boilers are run for 4 to 6 hours duration for the boilers of oil fired type and coal fired types. Duration of boiler run for attaining steady state changes from boiler to boiler. Observations are taken after the boiler attains steady state for a duration ranging from 10\u201315 minutes. Measurements are made for fuel supply, combustion analysis, steam generation rate and its quality/state, flue gas and their analysis, temperature and pressure at salient locations", " and all other measurements as required for heat balance sheet preparation. EXAMPLES 1. A boiler has chimney of 30 m height to produce natural draught of 12 mm of water column. Ambient air temperature is 27C and boiler furnace requires 20 kg of air per kg of fuel for complete combustion. Determine minimum temperature of burnt gases leaving chimney. Solution: Given: m = 20 kg air per kg of fuel, H = 30 m, Ta = 273 + 27 = 300 K Draught in terms of water column, hw = a T0 H Substituting a, T0 and rounding off m hw = 353 20 1 20 m, mm of water 1 T g, mm of water Tg 1 1 T a 1 300 12 = 353 30 Tg = 432.86 K Temperature of burnt gases = 432.86 K Ans. 482 _________________________________________________________ Applied Thermodynamics 2. Determine the height of chimney required in a boiler having natural draught equivalent to 20 mm of water. The flue gases are at temperature of 300C, atmospheric air temperature is 27C and 18 kg air per kg of fuel is required in boiler. Solution: Given: m = 18 kg air per kg of fuel, hw = 20 mm, Ta = 300K, Tg = 573 K Draught in terms of water column, hw = 353 H 20 = 353 H m 1 m 1 T a 1 300 18 1 18 1 T g 1 573 mm of water H = 37.99 m Height of chimney = 37.99 m Ans. 3. A boiler house has natural draught chimney of 20 m height. Flue gases are at temperature of 380C and ambient temperature is 27C. Determine the draught in mm of water column for maximum discharge through chimney and also the air supplied per kg of fuel. Solution: Given: Height of chimney, H = 20 m, Tg = 653 K, Ta = 300 K For maximum discharge condition, T g T a = 2 1m m m 1 2 1 = 653 300 m = 11.32 kg air per kg of fuel Air supplied = 11.32 kg/kg of fuel Ans. Draught in mm of water column hw = 353 20 1 300. 11 32 1. 11 32 1 653 hw = 11.77 mm of water Draught = 11.77 mm of water Ans. 4. A boiler may have waste gases leaving the installation when artificial draught is used at 150C. The natural draught chimney is of 60 m height. The hot gases within chimney are at temperature of 300C and air requirement is 19 kg per kg of fuel burnt. The atmospheric air is at 17C temperature and mean specific heat of hot gases is 1.0032 kJ/kg K. The calorific value of fuel burnt is 32604 kJ/kg. Determine (i) (ii) (iii) the draught produced in mm of water the efficiency of chimney the extra heat carried away by flue gases per kg of fuel. Boilers and Boiler Calculations ___________________________________________________ 483 Solution: Given: Tg = 300 + 273 = 573 K, Ta = 17 + 273 = 290 K, Tg, a = 150 + 273, H = 60 m, m = 19 kg air per kg of fuel, Tg, a = 423 K Draught in mm of water column hw = 353 H 1 T a = 353 60 m 1 m 19 1 19 1 290 1 573 1 T g = 34.13 mm Draught = 34.13 mm of water column Ans. Chimney efficiency = 9 81. H c pg 9 81 60. =. 1 0032 19 573 290 19 1 573 423 1 1 103 = 3.431 10\u20133 or 0.3431% Chimney efficiency = 0.3431% Ans. Extra heat carried by flue gases per kg of fuel = (19 + 1) Cp, g (Tg \u2013 Tg, a) = 20 1.0032 (573 \u2013 423) = 3009.6 kJ per kg of fuel Extra heat carried away by flue gases per kg of fuel burnt = 3009.6 kJ Ans. 5. Determine the temperature of hot flue gases, natural draught produced and efficiency of chimney for maximum discharge through chimney having height of 80 m. Boiler furnace is supplied with 20 kg air per kg of fuel. The minimum temperature of hot gases with artificial draught is 110C. Temperature of surroundings is 27C and specific heat of flue gases is 1.0032 kJ/kg K. Solution: Given: Ta = 27 + 273 = 300 K, Tg, a = 110 + 273 = 383 K, m = 20, H = 80m, cp, g = 1.0032 kJ/kg K For maximum discharge through chimney, = 2 T g", " T a 1m m Tg = 300 2 20 1 20 = 630 K Natural draught produced in mm of water column, 484 _________________________________________________________ Applied Thermodynamics hw = 353 H = 353 80 1 T a m 1 m 1 T g hw = 47.07 mm of water H. 9 81 1 300 1 630 20 1 20 T g m T a T g 630 300 630 = 3.167 10\u20133 or 0.3167% m 1 T g a, 9 81 80. 20 21. 1 0032 383 p g, = c 1 1 103 Efficiency of chimney = Hot gas temperature in chimney = 630 K Natural draught = 47.07 mm of water Chimney efficiency = 0.3167% Ans. 6. In a boiler installation coal is burnt at the rate of 2.5 103 kg/hr and hot gases are generated at the rate of 20 kg per kg of coal burnt. Hot gases are at the temperature of 327C and ambient air temperature is 27C. Different pressure losses in furnace grate, flues, bends and chimney are measured to be 7 mm, 6 mm, 3 mm, and 2 mm of water respectively. Hot gases leaving chimney have velocity equivalent to 1.2 mm of water column. Considering the actual natural draught to be 90% of theoretical draught determine the height and diameter of chimney. Solution: Given: Tg = 327 + 273 = 600 K, Ta = 27 + 273 = 300 K, m + 1 = 20, m = 19 kg air/kg fuel Pressure head required to overcome different losses and velocity head shall be sum of all losses, given h = 7 + 6 + 3 + 2 = 18 mm of water. Theoretically the draught of 18 mm of water is to be produced but as mentioned, the actual draught will be more than theoretical draught produced. Actual natural draught, hw = 18 0.9 = 20 mm of water. Let height of chimney be H, hw =353 H \u2234 20 = 353 H H = 35.88 20 19 1 300 1 600 Height of chimney = 35.88 m Ans. Density of hot gases Boilers and Boiler Calculations ___________________________________________________ 485 = 353 T g g = 353 600 m 1 m 19 1 19 = 0.619 kg/m3 Velocity of hot gases, C = Height of hot gases column, hg = H 2 g hg m = 35.88 1 m T g T a 19 20 1 600 300 1 Mass flow rate of hot gases, Velocity, hg = 32.29 m 10 2 3600.5 Mg = 3 20 = 13.89 kg/s C = Mg = D =. 9 81 32.29 = 25.17 m/s 2 g hg = 2 D2 C g 4 4 M c g Diameter of chimney, D = 4 13 89.. 25 17 0 619. Diameter of chimney = 1.06 m Ans. D = 1.06 m 7. Determine the capacity of motor required for running induced draught fan and forced draught fan required for maintaining draught of 50 mm of water column. Consider hot gases to leave boiler at 300C, coal being burnt per hour at the rate of 2000 kg/hr, air supplied at the rate of 19 kg/kg of coal, ambient air temperature of 27C and mechanical efficiency as 90%. Solution: Power required in FD fan = P m M T 1 T 0 mech P m M T Power required in 1D fan = T 0 Given, T1 = 300 K, T0 = 273 K, T = 573 K, m = 19 kg/kg coal, M = 2000 kg/hr = 0.556 kg/s, = 1.293 kg/m3, hwater = 50 mm or 50 kgf/m2, mech = 0.90 mech P = 50 kgf/m2 or P = 490.5 N/m2 486 _________________________________________________________ Applied Thermodynamics Substituting in expression for power outputs, For FD fan, power = For 1D fan, power = 490 5 19.293 1 = 4893.11 W = 4.89 kW 490 5 19 1.293 300 0 273.556 0.90 0 273.556 0.90 573 = 9345.84 W Power for FD fan = 4.89 kW Power for 1D fan = 9.35 kW Ans. = 9.35 kW 8. A boiler installation is seen to have operating parameters as given under when natural draught system, forced draught system and induced draught systems are used. Considering specific heat of hot gases as 1.0032 kJ/kg K determine, (i) (ii) the ratio of power required for induced and forced draught system. the ratio of heat carried away with flue gases in artificial draught and natural draught. Hot gas", " temperature, C Temperature of surroundings, C Mass of air required per kg of fuel Solution: Natural draught Artificial draught Forced Induced 327 177 2 7 2 0 Brake power for induced draught Brake power for forced draught T g T a Heat carried by hot flue gases in artificial draught for unit mass of fuel burnt 273 273 177 27 = 1.5 = = Qg, a d = (20 + 1) 1.0032 (177 \u2013 27) = 3160.08 kJ per kg of fuel burnt Heat carried by hot flue gases in natural draught for unit mass of fuel burnt Qg, ad = (25 + 1) 1.0032 (327 \u2013 27) = 7824.96 kJ per kg of fuel burnt Thus it shows that significantly large amount of heat is lost in natural draught as compared to artificial draught. Ratio of heat carried away in artificial and natural draught =. 3160 08.96 7824 = 0.404 Ratio of power required = 1.5 Ratio of heat carried away = 0.404 Ans. 9. Determine the actual evaporation per kg of coal and the equivalent evaporation if during boiler trial of one hour duration following observations are made: Feed water supply temperature: 27C Mean steam generation pressure: 10 bar, Boilers and Boiler Calculations ___________________________________________________ 487 Dryness fraction of steam generated: 0.95 Feed water supplied: 2500 kg/hr Coal burnt: 275 kg/hr Mass of water in boiler after trial = 300 kg less than that at commencenent of trial. Solution: From steam table at 10 bar, hf = 762.81 kJ/kg, hg = 2778.1 kJ/kg hfg = 2015.29 kJ/kg Enthalpy of steam steam generated, h = hf + x hfg = 762.81 + 0.95 2015.29 h = 2677.34 kJ/kg Mass of water evaporator per hour = 2500 + 300 = 2800 kg/hr Water evaporated per kg of coal = 2800 275 = 10.18 kg per kg of coal. Actual evaporation = 10.18 kg per kg of coal Ans. Equivalent evaporation = 10 18. 2677 34. 2257 = 12.08 kg per kg of coal Equivalent evaporation = 12.08 kg per kg of coal Ans. 10. A boiler is being tested for 24 hours and during this trial steam at average pressure of 10 bar, dry saturated is produced from 15 ton of water consuming 1.5 ton of coal. Composition of coal has 3% moisture and 4% ash. Feed water is added at 35\u00b0C. Determine, (i) (ii) (iii) the boiler efficiency, the equivalent evaporation per kg of dry coal, and the equivalent evaporation per kg of combustible present in coal. Solution: Heat supplied per kg of water for steam generation = 2778.1 \u2013 4.18 35 Enthalpy of steam generated = hg at 10 bar = 2778.1 kJ/kg Steam generated per kg of coal = = 10 kg Boiler efficiency = = 0.8744 or 87.44% 3 = 2631.8 kJ/kg 15 10.5 10 1. 2631 8 10 30 1 103. 3 Boiler efficiency = 87.44% Ans. Equivalent evaporation per kg of dry coal = 10 2257 = 12.02 kg 2631 8.. 1 0 03 Equivalent evaporation per kg of combustible present in coal =. 12 02 0 0.93.97 488 _________________________________________________________ Applied Thermodynamics Equivalent evaporation per kg of dry coal = 12.02 kg Equivalent evaporation per kg of combustible = 12.53 kg Ans. = 12.53 kg 11. During the boiler trial for 24 hours following observations were made. Steam generation: 16 bar, dry saturated. Coal consumed: 10,000 kg Rate of steam generation: 2500 kg/hr Feed water temperature: 27C Total heating surface area: 3000 m2 Total grate area: 4m2 Calorific value of coal: 28000 kJ/kg Determine: (i) (ii) (iii) (iv) the mass of coal burnt per m2 of grate per hour the equivalent evaporation from and at 100C per kg of coal the equivalent evaporation from and at 100C per m2 of total heating surface per hour the boiler efficiency Solution: Coal burnt per hour =, 10 000 24 = 416.67 kg/hr Coal burnt per m2 of grate per hour = 416.67 4 Equivalent evaporation from and at 100C per kg of coal can be obtained by the ratio of heat added to steam per kg of coal burnt and latent heat from and at 100C = 104.17 kg/m2 grate surface per hr =", " Heat added to steam per kg of coal burnt Latent heat from and at100 C Rate of steam generated per kg of coal = = 5.99 6 kg steam/kg coal Heat added to steam per kg of coal = 6 [hg at 16 bar \u2013 4.18 27} 2500 416.67 = 6{2794 \u2013 112.86} = 16086.84 kJ Latent, heat from and at 100C = 2257 kJ/kg Equivalent evaporation from and at 100C per kg of coal = 16086.84 2257 = 7.13 kg Equivalent evaporation from and at 100C per m2 of total surface per hour =. 7 13 416.67 3000 = 0.99 kg Boiler efficiency = \u00d7 7.13 2257 28000 = 0.5747 or 57.47% Boilers and Boiler Calculations ___________________________________________________ 489 Mass of coal burnt per m2 of grate per hour = 104.17 kg Equivalent evaporation from and at 100C per kg of coal = 7.13 kg Equivalent evaporation from and at 100C per m2 of total heating surface per hour = 0.99 kg Ans. Boiler efficiency = 57.47% 12. Steam is generated in a boiler at 30 bar 300C at the rate of 11 kg/s with feed water entering economiser at 100C. During one hour test 5000 kg fuel is used in boiler. Calorific value of fuel is 35000 kJ/kg. For the feed water being supplied to boiler to be at 27\u00b0C determine; (i) (ii) (iii) the equivalent evaporation per kg of fuel the boiler efficiency the percentage of fuel energy utilised in economiser Solution: Mass of steam generated per kg of fuel = 7.92 kg/kg fuel 11 3600 5000 Heat added to per kg steam per kg of fuel = (Enthalpy of steam at 30 bar, 300C \u2013 Enthalpy of feed water) = (2993.5 \u2013 (4.18 27)) = 2880.64 kJ Equivalent evaporation per kg of fuel = Boiler efficiency =.92 7 2880 64. 2257 10 11 2257 35000. = 10.11 kg = 0.6519 or 65.19% Heat utilised in economiser per kg of fuel = 7.92 4.18 (100 \u2013 27) Percentage of energy utilised in economiser = = 0.069, or 6.9% = 2416.71 kJ 2416.71 35000 Equivalent evaporation per kg of fuel = 10.11 kg Boiler efficiency = 65.19% Percentage of energy utilised in economiser = 6.9% Ans. 13. A boiler is capable of generating 8 kg steam per kg of fuel at 400C. Feed water is supplied at pressure of 30 bar, 40C and leaves economiser at 150C for entering to evaporator. Steam leaves evaporator with dryness fraction of 0.98 and enters the superheater. Fuel used has calorific value of 29000 kJ/kg. Considering no pressure loss inside the boiler determine boiler efficiency and fraction of heat given to steam in each section of boiler. Solution: From steam table; hf at 30 bar = 1008.42 kJ/kg, hg at 30 bar = 2804.2 kJ/kg Enthalpy of final steam generated = hat 30 bar, 400C = 3230.9 kJ/kg Heat supplied per kg of fuel = 1 29000 kJ/kg = 29000 kJ/kg Heat to be added for per kg steam generation = hat 30 bar, 400\u00b0C \u2013 hwater at 40\u00b0C = 3230.9 \u2013 4.18(40) 490 _________________________________________________________ Applied Thermodynamics = 3063.7 kJ/kg Boiler efficiency = = 0.8452 84.52% 29000 Heat added in economiser per kg of steam generated = 4.18 (150 \u2013 40) 8 3063 7. % Fraction of heat in economiser = 459 8. 3063 7. = 15.01% = 459.8 kJ/kg 100 Heat added in evaporator per kg of steam generated = (hf at 30 bar + 0.98 hfg at 30 bar) \u2013 (4.18 150) = 1008.42 + (0.98 1795.78) \u2013 627 = 2141.28 kJ/kg % Fraction of heat in evaporator =.28 2141. 3063 7 100 = 69.89% Heat added in super heater per kg of steam generated, by difference = (3063.7 \u2013 459.8 \u2013 2141.28) = 462.62 kJ/kg % Fraction of heat in superheater = Boiler efficiency = 84.52%. 462 62 30", "63 7. 100 = 15.1% % fraction of heat in economiser, evaporator and superheater = 15.01%, 69.89% and 15.1% Ans. 14. During a boiler trial the volumetric analysis of flue gases (combustion products) going across the economiser section are 8.4% CO2, 11.3% O2 and 80.3% N2 and 8% CO2, 11.4% O2 and 80.6% N2 respectively at inlet and exit of economiser section of boiler. Feed water enters economiser at 20C and leaves at 125C at the rate of 3 kg/s. Temperature of flue gases at inlet and exit of economiser are 425C and 300C respectively. Coal is supplied to boiler at the rate of 18 kg/min and the coal used has 80% of carbon by mass in it. Considering atmospheric air to be at 15C and specific heat of flue gases as 1.05 kJ/kg C determine, (i) (ii) (iii) the heat released by flue gases in economiser section, the amount of air leaked into economiser, the heat gained by feed water in economiser. Solution: Given: 0.8 kg of C in 1 kg of coal. Table 11.3 gives the mass of dry flue gas at inlet and exit of economiser as 23.65 kg and 24.78 kg respectively. Thus, there is increase in mass of dry flue gases from inlet to exit. This increase in flue gases may be attributed to the leakage of air. Therefore, air leakage in economiser per kg of coal = 24.78 \u2013 23.65 = 1.13 kg air per kg coal Heat rejected by flue gases during its\u2019 passage through economiser shall be given by the difference of heat entering with flue gas and air leakage and heat leaving with flue gases at exit of economiser. Table 9.3 Volumetric and gravimetric analysis of flue gas in economiser Constituents Mol. Wt. Volume per m3, (b) Proportional mass (c = a b) (a) 44 32 28 CO2 O2 N2 Total At inlet At exit At inlet At exit 0.084 0.113 0.08 0.114 3.696 3.616 3.52 0.803 1.00 0.806 1.00 22.484 29.796 22.568 29.736 C per kg of flue gases, 12 d = c c 44 At inlet 0.03383 At exit 0.03228 3.648 Dry flue gas per kg Dry flue gas per kg of of coal =. 0 8. 0 03383 = 23.65 kg coal =. 0 8. 0 03228 = 24.78 kg 492 _________________________________________________________ Applied Thermodynamics It may be assumed that the specific heat of leakage and flue gases are same. Heat entering economiser with flue gases and leakage = (23.65 1.05 425) + (1.13 1.05 15) = 10571.61 kJ Heat leaving economiser with flue gases = (24.78 1.05 300) = 7805.7 kJ Thus, heat lost in economiser per kg of coal = 10571.61 \u2013 7805.7 Heat picked by feed water in economiser per kg of coal = 2765.91 kJ T = mw Cp,w 3 60 18 = 4.18 (125 \u2013 20) = 3971 kJ per kg of coal Heat released by flue gases = 2765.91 kJ per kg of coal Air leakage = 1.13 kg air per kg of coal Heat gained by feed water = 3971 kJ per kg of coal Ans. 15. Draw up a heat balance sheet for the boiler in kJ per kg of dry coal and also determine the boiler efficiency and effectiveness of air heater for the following data obtained during boiler trial. The boiler has economiser and air preheater as two accessories in it. Atmospheric air temperature: 15C Steam generation: 40 bar, 400C Steam generated per kg of coal = 8 kg Feed water temperature at inlet to economiser = 27C Feed water temperature at exit of economiser = 137C Moisture in coal burnt = 1.5% Flue gas temperature entering air heater =300C Flue gas temperature leaving air heater and entering chimney = 150C Temperature of air entering boiler furnace = 120C Dry coal composition by mass = 84% C, 4% H2, 7% O2 and remainder ash Dry flue gas composition by volume = 12.5% CO2, 7.5% O2, 80% N2 Datum temperature = 15\u00b0C Calorific value of coal = 32600 kJ/kg For air and dry flue gas, cp =1.00", "32 kJ/kg K Partial pressure of vapour in flue gas = 0.075 bar Specific pressure of vapour = 2.0064 kJ/kg K Determine the boiler efficiency and the efficiency of heat exchange in air heater. Also prepare heat balance sheet in kJ per kg of dry coal. Boilers and Boiler Calculations ___________________________________________________ 493 Solution: Here the schematic of economiser, boiler and air heater are as shown below: Let us now determine the mass of dry flue gas per kg of dry coal for which the mass of carbon per kg of dry flue gas can be obtained by the following analysis. Fig. 11.40 Composition by volume of dry flue gas Molecular weight CO2 = 0.125 O2 = 0.075 N2 = 0.80 44 32 28 Mass of constituents (a) 0.125 44 = 5.5 0.075 32 = 2.4 0.80 28 = 22.4 a = 30.3 Mass per kg of dry flue gas, b = a a 0.1815 0.0792 0.7393 Mass of carbon per kg of dry flue gas. 0 1815 12 44 = 0.0495 The dry coal is given to have C as 0.84 kg per kg of coal. Therefore, the mass of dry flue gas per kg of coal = carbon present per kg of coal carbon present per kg of dry flue gas For the given coal composition the H2O produced during combustion = 0.04 9 = 0.36 Given coal has 0.05 kg ash per kg of coal. Amount of air supplied for combustion of one kg of dry coal = 0.84 0.0495 = 16.97 kg dry flue gas per kg of coal = 16.97 \u2013 (1 \u2013 ash content \u2013 H2O formed) = 16.97 \u2013 (1 \u2013 0.05 \u2013 0.36) = 16.38 kg Moisture per kg of dry coal = Moisture in coal burnt per kg of coal Mass of dry coal per kg of coal fired = 0.015 1. 0 015 = 0.0152 kg 494 _________________________________________________________ Applied Thermodynamics Hence, the total moisture per kg of dry coal = 0.36 + 0.0152 = 0.3752 kg Steam generated per kg of dry coal = 8. 1 0 015 = 8.12 kg steam From steam tables Enthalpy of steam generated = hat 40 bar, 400C = 3213.6 kJ/kg At partial pressure of vapour i.e. 0.075 bar, saturation temperature is 40.29C and hg at 0.075 bar = 1168.79 kJ/kg For the given datum temperature of 15C the heat available with steam, dry flue gas, moisture in flue gas, feed water etc. can be estimated as under. Heat available with steam = 8.12 (3213.6 \u2013 (4.18 15)) = 25585.31 kJ Heat carried by dry flue gases = (16.97 1.0032 (150 \u2013 15) Heat carried by moisture in flue gas = 2298.28 kJ = 0.3752 {(168.79 \u2013 (4.18 15)) + 2.0064 (150 \u2013 40.29)} = 122.39 kJ Heat available with feed water = 8.12 (27 \u2013 15) 4.18 = 407.29 kJ Heat available with one kg of coal = 32600 kJ/kg The heat balance sheet may be prepared for datum state of 15C as under per kg of dry coal. Heat Supplied/kg of coal Heat Utilized/kg of coal Quantity, kJ% Quantity, kJ% Heat available with coal Heat available with feed water 32600 98.77% (a) Heat utilized by steam (b) Heat carried by dry 25585.31 2298.28 77.51% 6.96% 407.29 1.23% (c) Heat carried by moisture flue gas in flue gas 122.39 0.37 Total 33007.29 100% (d) Heat loss due to radiation etc. 5001.31 33007.29 15.16 100% Boiler efficiency can be obtained as under, Heat utilized by steam per kg of coal = 8.12 (3213.6 \u2013 (4.18 27)) = 25178.01 kJ Heat supplied by fuel = 32600 kJ/kg coal Boiler efficiency =. 25178 01 32600 = 77.23% For getting the efficiency of heat exchange in air heater the accounting may be made for total heat available and heat utilized. Heat utilized by air = 16.38 1.0032 (120 \u2013 15) = 1725.4 kJ/kg coal Heat available in air heater = Heat available with dry flue gas + Heat supplied by the moisture. = = {", "16.97 1.0032 (300 \u2013 137)} + {0.3752 2.0064 (300 \u2013 137)} 2897.67 kJ/kg coal Boilers and Boiler Calculations ___________________________________________________ 495 Efficiency of heat exchange in air heater =. 1725 4. 2897 67 = 59.54% Boiler efficiency = 86.75% Efficiency of heat exchange in air heater = 59.54% Ans. 16. In a boiler the coal utilized has composition by mass as 85% C, 5% H2, 6% ash and remaining oxygen. The combustion results in flue gases at 200C temperature and composition by volume of dry flue gas as 11% CO2, 1% CO, 8% O2 and 80% N2. The temperature of air is 20C and the pressure of flue gas is 1.5 bar. Consider the specific heat of dry flue gas as 1.0032 kJ/kg, specific heat of superheated steam 2.05 kJ/kg.K, air to have 23% O2 by mass and calorific value of 1 kg CO getting burnt to CO2 as 10,000 kJ/kg. Determine, total air supplied per kg of coal, (a) (b) heat carried away by moist flue gas per kg of coal, and (c) the partial pressure of steam in hot flue gas. Solution: Here for getting the mass of air supplied per kg of coal the combustion analysis may be carried out as under. Composition by volume of dry flue CO2 = 0.11 CO = 0.01 O2 = 0.08 N2 = 0.80 Total = 1.00 Molecular weight Mass of constituents (a) Mass per kg of dry flue gas (b) = a a 44 28 32 28 0.11 44 = 4.84 0.01 28 = 0.28 0.08 32 = 2.56 0.80 28 = 22.4 0.1611 0.0093 0.0085 0.7457 a = 30.04 b = 0.9246 Mass of carbon per kg of dry flue gas = 0.0439 0.0093 0.1611 12 44 12 28 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 Total mass of carbon = 0.0478 = 0.0039 Mass of dry flue gas per kg of coal = = Massof carbon per kg of coal given as 0.85 Massof carbon per kg of dry flue gas 0.85 0.0478 = 17.78 kg/kg of coal H2O generated during combustion = 0.05 9 = 0.45 kg/kg of coal Mass of air supplied per kg of coal = 17.78 \u2013 (1 \u2013 ash \u2013 H2O) = 17.78 \u2013 (1 \u2013 0.06 \u2013 0.45) = 17.29 kg/kg of coal Total air supplied per kg of coal = 17.29 kg Ans. 496 _________________________________________________________ Applied Thermodynamics For the pressure of flue gas being 1.5 bar the partial pressure of steam can be given as No.of H O mol in flue gas Total no.of mol in flue gas 2 Total pressure 0.45 18 = 17.78 30.04 The enthalpy of this steam from steam table,. 0 45 18 1.05 = 0.04 bar hg at 0.04bar = 2554.4 kJ/kg, Tsat = 28.96C Therefore, heat in vapour = 0.45 {(2554.4 \u2013 4.18 20) + 2.05 (200 \u2013 28.96)} = 1269.64 kJ/kg of coal Heat in dry flue gas = 1.0032 (17.78 (200 \u2013 20)) = 3210.64 kJ/kg of coal Heat in CO =.28 0. 30 04 17.78 10000 = 1657.26 kJ/kg of coal Total heat carried by most flue gas = 1269.64 + 3210.64 + 1657.26 = 6137.54 kJ/kg of coal Partial pressure of steam = 0.04 bar Heat carried by moist flue gas per kg of coal = 6137.54 kJ Ans. 17. A boiler unit generates steam at 20 bar, 300\u00b0C from feed water supplied to boiler at 50\u00b0C. Coal used in boiler has calorific value of 30,000 kJ/kg and is used at rate of 600 kg/hr for steam generation rate of 5000 kg/hr. Determine the overall efficiency of boiler and the equivalent evaporation of boiler unit at 100\u00b0C in kg/hr. Determine the saving of coal in kg/hr if an economiser fitted to boiler to raise feed water temperature up to 75\u00b0C increases the overall efficiency of boiler unit by 5% for all other things remaining same. Solution: Steam generation", " per unit coal burnt per hour = 5000 600 At 20 bar, 300\u00b0C, the enthalpy of final steam, msteam = 8.33kg steam/kg of coal Enthalpy of feed water, hfinal = 3023.5 kJ/kg hwater = 209.33 kJ/kg Overall efficiency of boiler = \u2212 8.33(3023.5 209.33) 30000 = 0.7814 = 78.14% Equivalent evaporation of boiler unit Boilers and Boiler Calculations ___________________________________________________ 497 m steam ( h\u2212 ) water h final 2257 \u2212 8.33(3023.5 209.33) 2257 = = = 10.386 kg steam per kg of coal Equivalent evaporation of boiler unit at 100\u00b0C in kg/hr = (10.386 \u00d7 600) kg/hr = 6231.6 kg/hr After fitting economiser the enthalpy of feed water, hwater = 313.93 kJ/kg Modified overall efficiency of boiler unit = 78.14 + 5 = 83.14% Let the coal consumption be mcoal kg per hour. Modified overall efficiency of boiler unit = 0.8314 = (3023.5 313.93) 5000 \u2212 \u00d7 m coal \u00d7 30,000 mcoal = 543.17 kg/hr Saving of coal = 600 \u2013 543.17 = 56.83 kg/hr Ans. 18. A boiler generates 5000 kg/hr steam at 20 bar, 0.98 dry from feed water supplied to it at 60\u00b0C. Boiler sums on coal supplied at the rate of 600 kg/hr and air supplied at the rate of 16 kg per kg coal. The calorific value of coal is 30000 kJ/kg and boiler room temperature is 20\u00b0C. Considering 86% of heat being lost with flue gases. Determine temperature of flue gases leaving boiler. Take specific heat of flue gases as 1.005 kJ/kg\u00b7K. Solution: Mass of steam generated per kg of coal = 5000 600 = 8.33 kg steam per kg coal Enthalpy of final steam produced at 20 bar, 0.98 dry hfinal = hf at 20 bar + 0.98 \u00d7 hfg at 20 bar = 908.79 + (0.98 \u00d7 1890.7) = 2761.67 kJ/kg Enthalpy of feed water, hwater = hf at 60\u00b0 C = 251.13 kJ/kg Heat utilized for steam generation = 8.33(2761.67 \u2013 251.13) = 20912.8 kJ per kg of coal For the given coal, the heat lost per kg of coal = 30,000 \u2013 20912.8 = 9087.2 kJ per kg coal 498 _________________________________________________________ Applied Thermodynamics Heat lost with flue gases = 0.86 \u00d7 9087.2 = 7814.9 kJ per kg coal Let the temperature of flue gases leaving boiler be Tgas. Heat lost with flue gases = 7814.9 = (mflue + mair) \u00d7 cp gas (Tgas \u2013 293) 7814.19 = (16 + 1) \u00d7 1.005 \u00d7 (Tgas \u2013 293) Tgas = 750.37 K = 477.37\u00b0C Ans. 19. In a boiler unit forced draught fan delivers ambient air at 20\u00b0C with velocity of 20 m/s. The draught lost through grate is 30 mm of water column. Determine the power required to drive the fan if fan's mechanical efficiency is 80% and coal is burnt at the rate of 1000 kg per hour and air is supplied at the rate of 16 kg per kg of coal. Ambient pressure and density of air may be taken as 1.01325 bar and 1.29 kg/m3. Solution: Total draught loss = Pressure equivalent to velocity head + Draught loss through grate. Pressure equivalent to velocity head = 1 2 2 V\u03c1 1 = \u00d7 2 \u00d7 1.29 (20) 2 = 258 N/m2 Since 1 mm of water column is equal to 9.81 N/m2; so pressure equivalent to velocity head = 258 9.81 = 26.29 mm of water Hence total draught loss = 26.29 + 30 = 56.29 mm of water column Pressure required, p = 56.29 \u00d7 9.81, N/m2 = 552.21 N/m2 Forced draught fan power requirement = = \u00b7 \u00b7 \u00b7 p m m T f a T\u03c1 \u00b7 a \u03b7 \u00b7 mech o a \u00d7 552.21 1000 16 293 \u00d7 1.29 273 0.80 3600 \u00d7 \u00d7 \u00d7 \u00d7, W F.D. fan, power = 2552.39 W or 2.55 kW Ans. 20. A boiler unit has 45 m high chimney through which flue gases at 630", "K flow. Air requirement is 15 kg air per kg of fuel burnt and ambient temperature is 300 K. Determine the draught produced in mm of water column, equivalent draught in metre of hot gas column and temperature of chimney gases for maximum discharge in a given time and the draught produced. Also find the efficiency of chimney if the minimum temperature of artificial draught is 150\u00b0C and mean specific heat of flue gas is 1.005 kJ/kg.K. If net calorific value of fuel is 30,000 kJ/kg then determine the percentage of extra heat spent in natural draught. Solution: Given Tg = 630K, Ta = 300K, Tg, a = 150 + 273 = 423K, H = 45m, m = 15 kg air/kg fuel Draught in mm of water column = hw = 353 \uf8f9 \uf8fa \uf8fa \uf8fb hw = \u00d7 353 45 \uf8ee \uf8ef \uf8f0 1 300 \uf8eb \u2212 \uf8ec \uf8ed + 15 1 15 \uf8f6 \uf8f7 \uf8f8. 1 630 \uf8f9 \uf8fa \uf8fb Boilers and Boiler Calculations ___________________________________________________ 499 hw = 26.06 mm of water column Ans. Draught in metres of hot gas column = hg = \uf8fe hg = 45 \uf8f1 \uf8eb \uf8f2 \uf8ec \uf8ed \uf8f3 \uf8f6 630 \uf8f7 \uf8f8 15 1 300 15 +. \u2212 \uf8fc \uf8fd 1 \uf8fe Temperature of chimney gases for maximum discharge, hg = 43.59 metres of hot gas column Ans. Tg,max = T a \uf8eb \u00b72\u00b7 \uf8ec \uf8ed m + 1 m \uf8f6 \uf8f7 \uf8f8 \u21d2 Tg,max = 300 2 \u00d7 \u00d7 \uf8eb \uf8ec \uf8ed + 15 1 15 \uf8f6 \uf8f7 \uf8f8 = 640K. Ans. Draught produced for condition of maximum discharge. hw = 353 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe \u00d7 353 45 = \uf8f1 \uf8f2 \uf8f3 1 300 \u2212 \uf8eb \uf8ec \uf8ed + 15 1 15 \uf8f6 \uf8f7 \uf8f8 \u00d7 1 630 \uf8fc \uf8fd \uf8fe = 26.05 mm = 26.05 mm of water column Ans. Efficiency of chimney = 9.81 \u00d7 Tm \uf8f6 g \uf8f7+ \uf8f8.81 45 \u00d7 \uf8f1 \uf8eb \uf8f2 \uf8ec \uf8ed \uf8f3 \u00d7 \uf8f6 \uf8f7 \uf8f8 15 + 15 1 \u2212 630 300 \u00d7 \u2212 \uf8fc \uf8fd 1 \uf8fe 3 1.005 (630 423) 10 \u00d7 = = 2.0556 \u00d7 10\u20133 Efficiency of chimney = 0.2056% Ans. Extra heat carried away by the flue gases per kg of fuel = (15 + 1) \u00d7 cp.g \u00d7 (Tg \u2013 Tg,a) = (15 + 1) \u00d7 1.005 \u00d7 (630 \u2013 423) = 3328.56 kJ 500 _________________________________________________________ Applied Thermodynamics % Heat spent in natural draught = 3328.56 30000 \u00d7 100 percentage extra heat carried in natural draught = 11.09% Ans. 21. In a boiler installation dry flue gases are formed at the mean temperature of 630 K when outside air temperature is 300 K. Air is consumed at the rate of 15 kg air per kg of coal and coal is required at the rate of 1600 kg per hour. Actual draught may be taken as 60% of theoretical draught. Determine height of chimney, if the various draught losses are 14 mm of water column. Solution: Theoretical draught = 14 0.6 = 23.33 mm of water column We know hw = 353 \u00d7 \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe \u21d2 \u21d2 23.33 = 353 H \uf8f1 \uf8f2 \uf8f3 1 300 \u2212 \uf8eb \uf8ec \uf8ed + 15 1 15 \uf8f6 \uf8f7 \uf8f8 1 630 \uf8fc \uf8fd \uf8fe H = 40.29 m Height of chimney = 40.29 m Ans. 22. A boiler unit has forced draught fan maintaining draught of 100 mm of water column while discharging 30 m3/s through outlet section of 1.8m2 area. Ambient temperature is 300 K and assume mass of 1 m3 of air at NTP as 1.293 kg to find out power of motor of forced draught fan if fan efficiency is 85%. Determine the power consumption if FD fan is substituted by ID fan of similar efficiency considering flue gas temperature of 150\u00b0C", ". Solution: Considering the discharge rate, the velocity of air through outlet = 30 1.8 = 16.67m/s = f. m m \u03c1 a Pressure created due to the gases flowing at 16.67 m/s = 1 2 2 V\u03c1 1 = \u00d7 2 \u00d7 1.293 (16.67) 2 = 179.66 N/m2 or 18.31 mm of water Total draught = Static draught + Draught due to discharge = (100 + 18.31)mm of water = 118.31 mm of water column Power of motor of forced draught fan = \u22c5 \u22c5 mech Boilers and Boiler Calculations ___________________________________________________ 501 = 118.31 9.81 300 30 \u00d7 \u00d7 \u00d7 0.85 273 10 \u00d7 \u00d7 3 = 45.01 kW Power consumption of induced draught (ID) fan = PowerFD \u00d7 T,g a T a = 45.01 \u00d7 = 423 300 63.46 kW Ans. 23. In a boiler installation feed water enters at 30\u00b0C and leaves economiser section at 110\u00b0C for being fed into boiler. Steam generated in boiler at 20 bar, 0.98 dry and fed to super heater where its' temperature is raised up to 300\u00b0C. For the coal with calorific value of 30,500 kJ/kg and steam generation rate of 10kg/kg of coal burnt determine the energy received per kg of water and steam in economiser, boiler and superheater section as fraction of energy supplied by coal. Take Cp,feed water = 4.18 kJ/kg\u00b7K, Cp,superheated steam = 2.093 kJ/kg\u00b7K. Solution: From steam table, Fig. 11.41 h4 = 3023.5 kJ/kg hf at 20 bar = 908.79 kJ/kg hfg at 20 bar = 1890.7 kJ/kg For feed water, h3 = 908.79 + (0.98 \u00d7 1890.7) = 2761.7 kJ/kg h1 = hf at 30\u00b0C = 125.79 kJ/kg Total heat supplied = (h4 \u2013 h1) = (3023.5 \u2013 125.79) = 2897.71 kJ/kg Heat consumed in economiser = Cp, feed water \u00d7 (110 \u2013 30) = 4.18 \u00d7 (110 \u2013 30) = 334.4 kJ/kg steam = 334.4 \u00d7 8 = 2678.4 kJ/kg coal Heat consumed in boiler = h3 \u2013 h2 = 2761.7 \u2013 (Cp feed water \u00d7 110) = 2301.9 kJ/kg steam = 2301.9 \u00d7 8 = 18415.2 kJ/kg coal Heat consumed in super heater = h4 \u2013 h3 = 3023.5 \u2013 2761.7 = 261.8 kJ/kg steam = 261.8 \u00d7 8 = 2094.4 kJ/kg coal Heat provided by burning of coal = 30,500 kJ/kg coal Fraction of energy consumed in economiser = 2678.4 30,500 = 0.0878 or 8.78% Ans. 502 _________________________________________________________ Applied Thermodynamics Fraction of energy consumed in boiler = 18415.2 30500 = 0.6038 or 60.38% Ans. Fracton of energy consumed in superheater = 2094.4 30500 = 0.0687 or 6.87% Define boiler. 11.1 Classify the boilers and briefly describe each type of them. 11.2 Enlist the requirements of a good boiler. 11.3 Differentiate between fire tube and water tube boilers. 11.4 Describe briefly a vertical boiler. Also give its\u2019 neat sketch. 11.5 Sketch and completely label a Lancashire boiler. Also explain its\u2019 working. 11.6 11.7 Sketch and describe working of Locomotive boilers. 11.8 What do you understand by high pressure boilers? 11.9 Explain working of Bacock and Wilcox boiler. 11.10 Sketch and describe a Stirling boiler. 11.11 Sketch and describe working of Loeffler boiler. 11.12 Differentiate between mountings and accessories. 11.13 Classify mountings into safety fittings and control fittings. 11.14 Describe superheater, economiser and air preheater with neat sketches. Also indicate suitable location of these on a boiler with line diagram. 11.15 Write short notes on, water level indicator, safety valves, fusible plug, feed check valve, pressure gauge, stop valve and blow off cock. 11.16 Define boiler draught and also classify it. 11.17 Describe functions of chimney in a boiler. 11.18 Obtain the expression for the natural draught in terms of height of water column.", " Also state the assumption made. 11.19 Derive the condition for maximum discharge through a chimney in natural draught. 11.20 Compare natural draught with artificial draught. 11.21 Describe briefly different types of mechanical draught. 11.22 Compare the power requirements for forced draught and induced draught. 11.23 Define equivalent evaporation and also give its significance. 11.24 Describe balanced draught. 11.25 What is meant by boiler trial? Explain. 11.26 Determine the height of chimney required to produce draught equivalent to 16.7 mm of water column for the flue gases at 300C and ambient temperature of 20C. Take the air requirement to [30 m] be 20 kg/kg of fuel. 11.27 Calculate the draught produced in mm of water by chimney of 35 m height, flue gas temperature of 643 K, boiler house temperature of 307 K and air supplied at 18.8 kg per kg of coal. [20 mm] 11.28 Show that the height of chimney required for producing a draight equivalent to 15 mm of water can not be less than 30 m, if the flue gas temperature is 250C, ambient temperature is 20C and minimum 18 kg air per kg of fuel is required. Boilers and Boiler Calculations ___________________________________________________ 503 11.29 In a boiler installation the height of chimney is 38 m and the hot gas temperature is 570 K and ambient air temperature is 300 K. Considering the diameter of chimney to be 2m, air/fuel ratio [3012 kg/min] being 19 determine the mass flow rate of flue gases flowing out. 11.30 Determine the percentage of excess air supplied in a boiler where coal having 80% C, 6% moisture, 14% ash by mass is burnt inside. Hot gas temperature is 570 K and ambient air temperature is 300 K. The height of chimney is 30 m to produce draught equivalent to 1.5 cm of water column. [20.1%] 11.31 For the maximum discharge condition through a chimney having height of 15 m determine the [9.2 mm] 11.32 Determine the efficiency of chimney which operates with hot gas temperature of 370C in natural draught and if it is operated with artificial draught the hot gas temperature drops down to 150C. The height of chimney is 45 m, air/fuel ratio is 24 and ambient temperature is 35C, Take Cp for hot gases as 1.004 kJ/kg K. [0.2%] draught in mm of water when ambient air temperature is 15C. 11.33 In a boiler chimney the different pressure losses are measured to be as under, Pressure loss on grate and boiler flues = 6 mm of water in each. Pressure loss in bends and chimney duct = 2 mm of water in each. Pressure head causing flow of hot gases = 1.2 mm of water. Determine the height and diameter of chimney if air/coal ratio is 14, average coal consumption is 2000 kg/hr, hot gas temperature is 300C, ambient air temperature is 310 K and actual draught is 0.8 [44.8 m, 1.6 m] times theoretical draught. 11.34 Determine the power required for a forced draught and induced draught fan in a boiler having coal consumption of 25 kg/min and air/fuel ratio of 19 by mass. The ambient air temperature and hot gas temperature at exit are 30C and 140C respectively. Draught required is equivalent to 32.2 mm [4.16 hp, 5.97 hp] of water and mechanical efficiency of motor/fan is merely 70%. 11.35 Determine the capacity of motor required for running forced draught fan to maintain draught of 50 mm of water when the hot flue gas temperature and boiler house temperature are 175C and 30C respectively. The boiler has coal requirement of 25.4 kg/hr and the flue gases are produced at the rate of 19 kg per kg of coal burnt. Considering mechanical efficiency of 78% compare the power requirement of forced draught and induced draught arrangement. [50.5 hp, FD requires 33% less power than 1D] 11.36 A boiler has equivalent evaporation of 1300 kg/hr from and at 100C. The fuel consumption per hour is 159.5 kg/hr and the boiler efficiency is 72%. Determine actual equivalent evaporation if feed is supplied at 110C to generate steam at 100 kg/m2 per hour and 15 bar, 200C. Also find calorific value of coal burnt and the grate area. [1255 kg/hr, 25498 kJ/kg, 12.6 m2] 11.37 Determine equivalent evaporation per kg of coal and efficiency of boiler when it consumes coal at the rate of 52 kg/hr and steam is generated at 7 bar, 0", ".907 dry, 541.7 kg/hr from feed water supplied at 40C. [11.075 kg per kg of coal, 83.3%] 11.38 During a boiler trial for 24 hour the following is observed; Steam generated = 160000 kg Mean steam pressure = 12 bar State of steam generated = 0.85 Feed water temperature = 30C Coal burnt = 16000 kg 504 _________________________________________________________ Applied Thermodynamics C.V. of coal = 33400 kJ/kg Determine equivalent evaporation from and at 100C and efficiency of boiler. [10.45 kg/kg of coal, 70.65%] 11.39 A boiler has air supplied at 20C and flue gas temperature of 220C. The coal burnt has composition by mass of 85% C, 3.9% H, 1.4% O and remaining ash. Volumetric composition of dry flue gas is 12.7% CO2, 1.4% CO, 4.1.% O2, 81.8% N2. The dew point temperature of wet flue gases is 50C. Calorific value of coal may be taken as 10157.4 kJ/kg. Take Cp of dry flue gas as 1.0032 kJ/kg K, Cp of superheated steam as 1.5048 kJ/kg K. Determine the heat carried away by dry flue gas, heat carried by moisture from combustion and heat loss due to incomplete combustion of carbon for per kg of coal burnt. [3080.7 kJ/kg, 969.8 kJ/kg, 2027.3 kJ/kg] 11.40 Draw heat balance sheet for a boiler whose 24 hour trial yields following; Ambient temperature = 35C Coal burnt = 19200 kg, Ash collected = 1440 kg, Combustion in ash = 54 kg, Steam generated = 198000 kg, Moisture in coal = 2.16%, Steam temperature = 198.9C, Feed water temperature = 150C, Flue gas temperature = 315C, Coal composition by mass = 83.39% C, 4.56% H2, 5.05% O2, 0.64% S, 1.03% N2, 5.33% ash Flue gas analysis by volume = 12.8% CO2, 6.4% O2, 0.2% CO, 80.6% N2 The partial pressure of steam = 0.07 bar, Specific heat of steam = 2.0064 kJ/kg K [Heat supplied in steam generation = 66.62% Heat carried by dry flue gas = 13.38%, Heat carried by moisture = 3.76%, Heat carried by ash = 0.28%, Unaccounted heat = 15.96%] 11.41 Steam is generated at 100000 kg/hr from a boiler at 100 bar, 500C with feed water supplied at 160C. The fuel burnt in boiler has calorific value of 21000 kJ/kg and boiler efficiency is 88%. Determine rate of fuel burnt per hour and percentage of total heat absorbed in economiser, evaporator and superheater. [14592 kg/hr, 27.09%, 48.84%, 24.07%] 11.42 Compare the steam generation capacities of two boilers A and B for which operating parameters are, Steam generation pressure = 14 bar, feed water temperature = 27C, Specific heat of feed water = 41.8 kJ/kg K, Specific heat of steam = 2.1 kJ/kg K, State of steam = 0.9 dry For boiler A: For boiler B: Boiler efficiency = 73% Calorific value of coal = 34000 kJ/kg State of steam = 240C Boiler efficiency = 79.5% Calorific value of coal = 46000 kJ/kg [For boiler A 10 kg/kg of coal burnt, for boiler B 14 kg/kg of coal burnt] Boilers and Boiler Calculations ___________________________________________________ 505 11.43 A boiler generates steam at 15 bar, 0.95 dry from feed water entering at 65C, 2040 kg/hr. Dry coal burnt is 232 kg/hr and mean boiler house temperature is 25.3C and flue gas temperature is 440C. Consider partial pressure of steam as 0.07 bar. Composition of dry coal by mass: 83% C, 6% H2, 5% O2, 6% ash. Composition of dry flue gas by volume: 10.50% CO2, 1.3% CO, 7.67% O2, 80.53% N2 Higher calorific value of dry coal = 34276 kJ/kg Specific heat of dry flue gas = 0.9948 kJ/kg K Determine (a) total mass of flue gases per kg of coal, (b)", " percentage of excess air supplied, (c) heat balance sheet per kg of dry coal [17.58 kg/kg of coal, excess air = 49.51%, Heat taken by steam = 62.02%,Heat taken by dry flue gas = 21.1%, Heat by moisture = 5.32%, Unaccounted = 11.56%] 506 ________________________________________________________ Applied Thermodynamics 12 12.1 INTRODUCTION Steam engine is a device which is especially designed to transform energy, originally dormant or potential, into active and usefully available kinetic energy. Origin of reciprocating steam engine dates back to BC 200. Steam engine is an exceedingly ingenious, but unfortunately, still very imperfect device for transforming heat obtained by chemical combination of combustible substance with air into mechanical energy. In steam engine the mechanical effect is seen due to the expansion of steam which is generated in boiler and supplied to steam engine. Steam engines have been successfully used in the mill, driving locomotive or steam boat, pumps, fans, blowers, small electricity generators, road rollers etc. A steam engine plant shall have boiler, condenser, feed pump along with steam engine. Steam generated in boiler is sent to steam engine where it is expanded upto certain pressure. Steam leaving engine are fed to condenser where steam gets converted into condensate (water) which is sent back to boiler through feed pump. Figure 12.1 shows the schematic of simple steam engine plant. Fig. 12.1 Schematic for simple steam engine plant Here in this chapter the philosophy of steam engine, its construction and working, compound steam engine etc. are described in detail. 12.2 CLASSIFICATION OF STEAM ENGINES Steam engines can be classified in the following ways. Selection of particular steam engine depends upon the requirements and constraints of an application and features of the engine (i) Based on axis of engine: Steam engine can be classified as horizontal or vertical steam engine based upon the orientation of cylinder axis. Engine having vertical axis cylinder is called vertical steam engine and engine having horizontal axis cylinder is called horizontal steam engine. Steam Engine _________________________________________________________________ 507 (ii) Based on speed of engine: Speed of steam engine can be considered for classifying steam engine as below: (a) High speed engine\u2014having speed of 250 rpm and above. (b) Medium speed engine having speed lying between 100 to 250 rpm. (c) Slow speed engine having speed less than 100 rpm. (iii) Based on type of steam action: Depending upon whether the steam acts on both sides of piston or only on one side of piston the engine can be classified as double acting steam engine or single acting steam engine. In case of single acting engine out of two piston strokes i.e. forward stroke and backward stroke, only one stroke is used for work output. While in double acting engine both the forward and backward strokes of piston are used for work output alternatively i.e. expansion of steam occurs from both sides of piston alternatively. Theoretically, it may be said that power developed from double acting engine will be nearly double of power developed from single acting steam engine. Power output from double acting engine is not exactly twice of output from single acting engine because only one side complete piston face is available for action on other side piston face area available for action is less due to presence of connecting rod, pin etc. (iv) Based on expansive or non expansive type of working: Steam engines can be of non expansive type in which steam is injected throughout the piston stroke i.e. piston movement is caused due to high pressure steam forcing piston from one end to other end throughout and not due to expansion of steam. Such type of steam engines are called non-expansive steam engine. P-V diagram showing expansive and non expansive engine is given in Fig. 12.2. Expansive type steam engine has steam injection for only a part of piston stroke and the expansion occurs in remaining part of stroke. The cut off ratio in case of non expansive engine is unity while for expansive engine it is less than unity. Expansive engines offer greater efficiency as compared to non expansive engine but the work out put per stroke from expansive engine is less as compared to non expansive engine. Expansive engines are used where efficiency is important or engine operation is required on intermittent basis. For large power requirements such as in case of boiler feed pumps, hoist engines etc. the non-expansive engines are required. Fig. 12.2 Expansive and non-expansive engine on P-V diagram (v) Based on type of exhaust from engine: Exhaust from steam engine may either go to atmosphere or to condenser. In certain applications where expansion in steam engine is 508 ________________________________________________________ Applied Thermodynamics extended up to sub atmospheric pressure then the exhaust steam from engine is sent to condenser. Such type of engine whose exhaust goes to condenser are termed as \u2018Condensing engine\u2019. The steam engines whose exhaust steam is delivered to atmosphere is called \u2018non-cond", "ensing engine\u2019. Steam engine of condensing type generally deliver steam at 0.05 bar inside condenser. Steam transforms into condensate in condenser and condensate is sent again to boiler through feed pump. Due to lower back pressure in condensing engine the work output from such engine is more. Non condensing engine delivers steam at pressure more than atmospheric pressure or near atmospheric pressure and this steam can not be re-used for the engine cycle. (vi) Based on number of stages: Depending upon the number of stages of expansion in steam engine, it may be termed as single stage, two stage, three stage, four stage engine etc. (vii) Based on number of cylinders: Steam engine may have steam expanding in single cylinder and engine is called \u201csimple steam engine\u201d. Steam engine may have the expansion of steam occuring in more than one cylinder such as expansion occurring in high pressure cylinder and low pressure cylinder etc. Such steam engine having expansion in more than one cylinders sequentially are termed as \u201cCompound steam engine\u201d. Compound steam engine having three cylinders may have one cylinder called as high pressure cylinder, second cylinder as intermediate pressure cylinder and third cylinder as low pressure cylinder. (viii) Based on type of governing: Steam engine may have different types of governing as throttle governing and cut-off governing. Engine may be called as throttle governed engine or cutoff governed engine. (ix) Based on type of application: Steam engine may be of following types depending upon applications as; (a) Stationary engine (b) Marine engine (c) Locomotive engine 12.3 WORKING OF STEAM ENGINE Schematic of simple steam engine is shown in Fig. 12.3 along with important components labelled upon it. Fig.12.3 Schematic of simple steam engine Steam Engine _________________________________________________________________ 509 Simple steam engine shown is a horizontal double acting steam engine having cylinder fitted with cylinder cover on left side of cylinder. Cylinder cover has stuffing box and gland through which the piston rod reciprocates. One end of piston rod which is inside cylinder has piston attached to it. Piston has piston rings upon it for preventing leakage across the piston. Other end of piston rod which is outside cylinder has cross head attached to it. Cross head slides in guide ways so as to have linear motion in line with engine axis. Cross head is connected to the small end of connecting rod by the gudgeon pin. Big end of connecting rod is mounted over crank pin of the crank. Reciprocating motion of piston rod is transformed into rotary motion of crankshaft by cross head, connecting rod and crank. Cross head transmits the motion of piston rod to connecting rod. Cross head guide ways bear the reaction force. Crank is integral part of crank shaft mounted on bearings. Crankshaft has fly wheel mounted on one end and pulley mounted on other end. Crankshaft also has an eccentric with eccentric rod mounted on it. Eccentric performs function of converting rotary motion of crankshaft into reciprocating motion of valve rod. Other end of eccentric rod transmits motion to valve rod which passes through stuffing box fitted in steam chest. Valve rod controls the movement of D-slide valve inside the seam chest. Dslide valve opens and closes the exhaust and inlet passages from steamchest to engine cylinder. Steam chest has two openings one for inlet of live steam and other for exit of dead or expanded steam. Live steam refers to the steam having sufficient enthalpy with it for doing work in steam engine. Dead steam refers to the steam having insufficient enthalpy with it and does not have capability to produce work. High pressure and high temperature steam (live steam) enters from main inlet passage into steam chest. D-slide valve occupies such a position that passage (port 1) from the steam chest to engine cylinder gets opened. High pressure steam enters cylinder and forces piston towards other dead centre. Linear motion of piston is transformed into rotation of crankshaft through crosshead, connecting rod, gudgeon pin and crank. When piston reaches other dead centre then the corresponding displacement of valve rod causes shifting of D-slide valve such that other passage (port 2) from steam chest to cylinder gets opened and passage 1 comes in communication with the exhaust passage. Thus the live steam enters from steam chest to cylinder through passage 1 and dead steam leaves from cylinder to exhaust passage through passage 2. Steam forces piston from inner dead centre to outer dead centre and from other side of piston dead steam leaves from exhaust passage simultaneously. Since it is double acting steam engine so both strokes produce shaft work due to steam being injected on both sides of piston alternatively. If it is a single acting steam engine then steam injection takes place on one side of piston only for producing power and return stroke occurs due to inertia of flywheel. Flywheel mounted on crankshaft overcomes the fluctuations in speed, if any due to its high inertia. Pulley mounted on shaft is used for", " transmitting power. Apart from the components shown in figure the engine has throttle valve, governor mechanism, oil pump, relief valves etc. for its proper functioning. Different major components of steam engine are described as below. (i) Piston and Piston rod: Piston is a cylindrical part made of cast iron, cast steel or forged steel depending upon application. Piston has rectangular grooves upon it to prevent leakage across the piston-cylinder interface. Piston is fitted on the piston rod by nut. Generally piston rod is tapered for its\u2019 portion inside piston and has threaded portion at end with provision of tightening nut upon it as shown in Fig. 12.4. 510 ________________________________________________________ Applied Thermodynamics (ii) Piston rings: Piston rings are fitted over piston in circumferential grooves. Piston rings are generally two or three in number so as to prevent leakage. Rings are split and in case of single ring leakage may occur across the point of split of ring. Hence more than one rings are put upon piston with staggered split ends so that leakage may not occur across these openings. Rings have sufficient elasticity in them so as to exert continuous force on cylinder wall for preventing leakage. Rings may be of cast iron and are turned slightly more than bore of cylinder and split at some point. Fig. 12.4 Piston with piston rod Fig. 12.5 Piston rings (iii) Connecting rod: Connecting rod as shown in Fig. 12.6 has big end and small end fitted upon crank pin and cross head respectively. It generally has I-section or circular section and is made of steel by forging. Connecting rod has relative motion occurring at big end and small end so there is requirement of lubrication at these locations. There is suitable passage inside connecting rod for carrying lubricant under pressure to the respective locations. Lubricant generally passes from main bearing section to the crankshaft and subsequently to crank pin. Crank pin has holes over it from where lubricant oozes out and lubricates the mating parts of big end and pin. Crank pin also forces lubricant through connecting rod to the gudgeon pin on the cross head. Connecting rod has a through hole from big end to small end for this purpose. Steam Engine _________________________________________________________________ 511 Fig. 12.6 Connecting rod Fig. 12.7 Crank pin, crank and crank shaft 512 ________________________________________________________ Applied Thermodynamics (iv) Crank and Crank shaft: Figure 12.7 shows the crank pin, crank and crank shaft as integral part. Reciprocating motion of cross head is transmitted to crank pin through connecting rod and can be transformed into rotary motion of crank shaft due to typical geometry. Connecting rod has its big end mounted on crank pin and the size of stroke depends upon the distance between the shaft and crank pin. Distance between the crank pin axis and axis of crankshaft is called radius of crank. Stroke of piston is double of the radius of crank as one complete rotation of crank shaft shall occur due to displacement of piston between two dead centres. Different types of crank arrangements used in steam engines are of overhung crank and disc crank as shown in Fig. 12.7. The overhung crank has crank pin mounted on crank as a cantilever supported element. Crank pin is shrunk fit on the crank i.e. heating for expanding the hole and subsequent cooling for shrunk fit. Disc crank has a cast iron disc in place of crank, upon which the crank pin is mounted. In case of crank shaft of heavier engines having bearing support on both ends their are two discs having pin in between. (v) Stuffing box: Stuffing box is a kind of packing box through which the reciprocating shaft/ rod passes and leakage across the reciprocating member is prevented. Stuffing box and different parts of it are shown in Fig. 12.8. Body of stuffing box is integral casting with cylinder cover and has gland to be tightened upon it with packing in between. Subsequently the packing is compressed by tightening gland by nuts and studs. Fig. 12.8 Stuffing box Leakage is prevented due to tightening of gland causing packing to be under pressure. Generally greased hempen rope, or asbestos or rubber ropes are used as packing material. Pressure exerted by packing can be changed by degree of tightening of studs and nuts. Metal packings, are also sometimes used in case of high pressure steam engines. Stuffing box are placed at the location where piston rod passes through cylinder head and where the valve rod passes through the steam chest. (vi) Crosshead and guideways: Piston rod is connected to connecting rod through cross head in between. Piston rod always does linear motion while connecting rod has continuously varying positions depending upon the position of crankshaft. Thus inspite of oblique position of connecting rod at any instant the piston rod has to be always in line with cylinder axis. Cross head and guides are the intermediate members which permit this", " transfer of motion from piston rod to connecting rod. Steam Engine _________________________________________________________________ 513 Since directions of forces acting in linearly moving piston rod and oblique moving connecting rod are different so they result into a third resultant force which is countered by guides. During both forward stroke and backward stroke these resultant forces are of different types thereby requiring guides on both sides of cross head. Fig. 12.9 Crosshead and guides (vii) Eccentric: Eccentric performs the function of transforming only rotary motion of crankshaft into reciprocating motion of valve rod. Operating principle of eccentric is quite similar to that of crank as both transform one type of motion into other. Important difference between the eccentric and crank is that eccentric can not convert reciprocating motion into rotary motion as done by crank. Figure 12.10 shows the eccentric which has three basic parts called eccentric sheave, eccentric strap and eccentric rod. There is relative motion between the sheave and strap. Arrangement of lubrication is also provided at the interface of sheave and strap. Sheave is fixed on the shaft using a key. Sheave may be of solid (single piece) type or of split type. Split sheaves are preferred when eccentric is to be used between two cranks and single piece eccentric can not be slided on the shaft. Sheaves are made out of cast iron and steel. After placing sheave upon shaft the eccentric strap is tightened through bolts and nuts. Distance between centre of sheave to centre of crankshaft is called \u201cradius of eccentric\u201d and the total linear displacement of valve is called \u201cthrow of eccentric\u201d. Throw of eccentric is equal to twice the radius of eccentric. Fig. 12.10 Eccentric 514 ________________________________________________________ Applied Thermodynamics (viii) Slide valve: Slide valve is provided in steam chest for quantitative regulation of supply of live steam into cylinder and exhaust of dead steam from cylinder. Slide valve is operated by the valve rod connected to eccentric rod and eccentric. Slide valves are of two types namely D-slide valve and Piston valve. Figure 12.11 shows the D-slide valve having shape of alphabet-D. Valve has its\u2019 two ends resting upon the steam chest having three openings comprising of two between cylinder and chest and one between chest and exhaust port. Amongst three ports two outer ports are called steam ports at head end and crank end respectively. Third port in middle is called exhaust port. In one position of D-slide valve as shown in figure both passages connecting steam chest with cylinder are closed due to overlap of valve over passages. Distance by which valve overlaps the ports in its middle position is called \u2018lap\u2019. Every valve while resting on port shall have overlaps on both sides and are called as \u2018steam lap\u2019 or \u2018outside lap\u2019 on one side and on other side it is called \u2018exhaust lap\u2019 or \u2018inside lap\u2019. When valve moves towards cylinder head then the port 2 is open for live steam to enter into cylinder from crank end. Also port 1 and port 2 are enclosed in D-slide valve and dead steam coming out from cylinder port 1 leaves out through exhaust port 3. When valve moves from cylinder head end to crank end then the port 1 is open for live steam to enter into cylinder while dead steam shall leave cylinder from port 2 to port 3. The cut off of steam occurs till the valve keeps ports in open state, allowing entry of live steam (high pressure and high temperature steam). The size of lap decides the point of cut off. In case of no lap the admission of steam and cut off shall occur at the dead centres. Also the exhaust of steam shall occur at dead centres. Due to non linear motion of connecting rod and eccentric rod the steam lap and exhaust laps are not equal. Ports are made open somewhat earlier than the stipulated position of piston so as to reduce throttling losses. Thus valve opens the port even when piston is at dead centre. This distance by which valve covers the steam port when piston is at dead centre is called \u2018lead\u2019. Similarly the \u2018angle of advance\u2019 can be obtained for lead. Amount of valve opening/closing can be modified by the valve gear mechanism and thus allowing for variable expansion in cylinder. Fig. 12.11 D-slide valve Steam Engine _________________________________________________________________ 515 D-slide valve has simple construction and ease of repair and maintenance. But these valves have significant amount of frictional losses and excessive wear of valve seat. There occurs large amount of throttling due to gradual opening/closing of ports. Also there are large condensation losses as the ports 1 and 3 are alternatively used by live steam (high pressure and temperature steam) and dead steam (low pressure and temperature steam) thereby causing alternate heating and cooling of steam chest ports. These valves are not suited for super- heated steam as due to valve and seat being of flat type can not be properly lubricated and so high", " temperature steam may cause excessive thermal stresses and distortion of valve/valve seat. Piston valve: Piston valve has piston-cylinder kind of arrangement for regulating the opening/ closing of different ports. These piston valves have lesser frictional resistance compared to D-slide valve as D-slide valves are highly pressed down against valve seat in steam chest and have larger frictional losses. Schematic of piston valve for double acting steam engine is shown in Fig. 12.12. Fig. 12.12 Piston valve Here similar to D-slide valve having D-shape, a piston valve with neck in between performs the similar function. Two pistons (P1 and P2) on same valve rod cover and uncover the ports 1 and 3 for the purpose of injecting high pressure steam and rejecting low pressure steam alternately. Working principle is similar to that of D-slide valve. For example when piston is near the cylinder head end then it shall uncover port 2 for live steam to enter into cylinder. Expanded steam or low pressure steam on other side of piston shall find its way out to exhaust through port 1 and 3. Similarly during piston valve\u2019s position near crank end the port 1 acts as inlet port and port 2 acts as exhaust port. These piston valves also suffer from disadvantages of throttling of steam due to slow opening/closing of ports. Also the points of admission of steam, cut-off and exhaust etc. are not independently adjustable. 12.4 THERMODYNAMIC CYCLE Steam engine works on the modified Rankine cycle. P-V representation of cycle is shown in Fig. 12.13 for steam engine with and without clearance volume. Let us take steam engine without clearance in which live steam enters the cylinder at state 1 and steam injection continues up to state 2. Point 2 is showing the state at cut-off. Subsequently steam is expanded up to state 3 theoretically and stroke gets completed. The expansion process is of hyperbolic type. Hyperbolic expansion process is one having the P-V = constant. In actual practice this expansion is not continued upto 3 due to the fact that positive work available from engine in later part of stroke is 516 ________________________________________________________ Applied Thermodynamics much less than negative work required for maintaining piston movement. Hence expansion process is terminated even before this piston reaches the extreme position (dead centre). Fig. 12.13 Modified Rankine cycle on P-V diagram for steam engines Here expansion process is shown to be terminated at state 5 and exhaust port is opened at this point, thereby causing sudden drop of pressure as shown in 5-6. Exhaust of dead steam occurs from 6 to 4. It is obvious from P-V diagram that the terminating expansion process before piston reaching dead centre shortens the expansion stroke length although actual work output from engine does not get affected. This modified form of cycle is called as \u2018modified Rankine cycle\u2019. In case of steam engine with clearance volume the modified Rankine cycle is as shown by 125641 while for engine without clearance volume it is given by 125641. Practically steam engine always has clearance volume but for the ease of mathematical analysis it may be considered to be without clearance volume. Therefore such cycle without clearance is also termed as hypothetical cycle. T-S representation of the modified Rankine cycle is shown in Fig. 12.14. Fig. 12.14 T-S representation of modified Rankine cycle From p-V diagram the work output from steam engine can be mathematically given by knowing the area enclosed by cycle. Theoretical work in case of hypothetical cycle, i.e. without clearance. W = Area enclosed by 125641 on P-V diagram W = Area 1290 + Area 25689 \u2013 Area 4680 W = p1(V2 \u2013 V1) + p1 V2 log V 5 V 2 \u2013 p3 (V5 \u2013 V1) Steam Engine _________________________________________________________________ 517 Here expansion ratio, r = V 5 V 2 and V1 = 0 as zero clearance is there. W = (p1 \u00d7 V2) + (p1 \u00d7 V2 ln r) \u2013 p3 V5 Work output neglecting clearance volume; Wwithout CV = p1 V2 (1 + ln r) \u2013 p3 V5 For steam engine cycle considering clearance volume cycle work shall be, W = Area 125641 on p-V diagram W = Area 1290 + Area 2589 \u2013 Area 4680 \u2013 Area 1144 \u2013 (p3 V6) \u2013 (p1 \u2013 p3) V1 V 5 V 2 W = (p1 V2) + p1 V2 ln W = (p1 V2) + p1 V2 ln r \u2013 (p3 V6) \u2013 (p1 \u2013 p3) V1 V5 = V6 As Wwith CV = (p1 V2) (1 + ln r) \u2013 (p3 V5) \u2013 (p1 \u2013 p3", ") V1 Let us denote pressure p3 as back pressure pb and clearance volume as VC then expressions for cycle work without and with clearance volume are as follows; Wwithout CV = p1 V2 (1 + ln r) \u2013 (pb rV2) rV2) \u2013 (p1 \u2013 pb) VC Mean effective pressure for the hypothetical cycle can be obtained using the work obtained earlier. Wwith CV = p1 V2 (1 + ln r) \u2013 (pb Mean effective pressure = In case of without clearance volume or, mepwithout CV = mepwithout CV = Cycle work Stroke volume without CV V 5 W 1 p V 1 2 ln p b r V 2 r r V 2 mepwithout CV = ln r 1 1 p r \u2013 pb In case of cycle with clearance volume, mepwith CV = = W with CV Stroke volume 1 p V 1 2 ln r mepwith CV = p V 1 2 1 ln rV 518 ________________________________________________________ Applied Thermodynamics 12.5 INDICATOR DIAGRAM Indicator diagrams for steam engine are shown in Fig. 12.15. Hypothetical indicator diagram shown by cycle abcde and the actual indicator diagram for an engine are given here. Hypothetical indicator diagram is obtained considering all processes in cycle to be ideal and assuming no heat loss and pressure drop etc. Fig. 12.15 Hypothetical and actual indicator diagram Hypothetical indicator diagram shows the indicator diagram having steam admission beginning at a and continuing up to b. Expansion occurs between b and c in hyperbolic manner. At c there occurs sudden pressure drop due to opening of exhaust valve up to d. Piston travels from d to e and steam injection begins at a and thus cycle gets completed. In this hypothetical diagram all processes are considered to occur as per their theoretical assumptions. While actual diagram is based upon the actual occurrence of all processes. This is a reason why actual diagram shown by abcde differs from hypothetical indicator diagram. In hypothetical diagram it is assumed that there occurs no pressure drop, valves open and close instantaneously, expansion occurs following hyperbolic process and admission of steam and its exhaust occur at end of strokes etc. The actual indicator diagram differs from hypothetical indicator diagram because of the following factors. (ii) (i) Practically there shall be pressure drop due to friction, throttling and wire drawing etc. Friction, throttling and wire drawing occur in valves and ports connecting steam chest and cylinder thereby causing pressure drop. Inlet and exit valves (ports 1, 2 and 3) can never be opened or closed instantaneously, which means that there shall always be some time required for completely opening or closing the valves. Therefore, in order to ensure entry of maximum amount of high pressure and temperature (live steam) steam it is desired to advance the opening of inlet valve and suitably modify the operation of valves for remaining processes. Thus it may be said that actually the inlet, cut off and release occur gradually. (iii) Expansion in steam engine does not occur hyperbolically in actual process due to varying heat interactions. This causes shift in expansion line from bc in hypothetical engine to the actual expansion line as shown in actual indicator diagram (bc). It may be understood that due to condensation of steam the expansion follows some other law. (iv) Exhaust of steam begins at c in order to overcome limitation of gradual opening of exhaust valve. Also the sudden pressure drop is not possible. Actual pressure drop occurs during cd in non-instantaneous manner upto back pressure value. Steam Engine _________________________________________________________________ 519 (v) (vi) In actual cycle the exhaust pressure will always be above the back pressure in order to overcome the friction and throttling losses during exhaust. In actual indicator diagram exit of steam is stopped at e, thereby causing trapping of some steam which gets subsequently compressed in cylinder till piston moves towards inner dead centre and inlet port is opened. This residual steam getting compressed offers cushioning effect. This compression of residual steam is represented on actual indicator diagram by ea. (vii) Actual engine shall always have clearance volume contrary to hypothetical diagram based on no clearance volume assumption. Clearance volume is required, because, the piston can not be allowed to collide with cylinder head. Clearance volume offers cushioning effect due to steam occupying it and \u2018clearance volume also allows for accumulation of condensate (water) with excessive pressure rise which may otherwise cause damage to cylinder head\u2019. Thus it is seen that the actual indicator diagram is modified form of hypothetical indicator diagram due to practical limitations. The amount by which actual indicator diagram differs from hypothetical indicator diagram is quantified by parameter called \u2018diagram factor\u2019. Diagram factor is the ratio of area enclosed in actual indicator diagram to the area enclosed in hypothetical indicator diagram. Diagram factor has value less than unity. Designer shall always intend to have engine having diagram factor close to unity. Diagram factor = Area enclosed in", " actual indicator diagram Area enclosed in hypothetical indicator diagram. 12.6 SATURATION CURVE AND MISSING QUANTITY In a steam engine it could be seen that the same passage acts as the passage for inlet of live steam (high pressure and high temperature steam) and exit of dead steam (low pressure and low temperature steam). During the steam admission stroke passage walls, valve face/port and cylinder walls become hotter and in exhaust stroke these surfaces become cooler due to low temperature steam passing through passages. Subsequently when high temperature steam again enters the engine cylinder then the hot steam comes in contact with cool surfaces which lead to condensation of a portion of fresh steam entering. Condensation is visible till the temperature of contact surfaces equals the hot steam temperature. Later on during exhaust stroke when cooler steam exits through valve then cold steam comes in contact with hot surfaces (surfaces are hot due to hot steam admission). Due to cold steam contact with hot surfaces reevaporation of condensed steam occurs. This reevaporation of steam slightly increases work and reduces wetness of steam. During admission of steam the condensation causes loss of steam without doing work. Condensation during intake generally causes increase in steam consumption by up to 40%. Condensation of steam can be prevented to some extent by one or more of the following ways. (a) Superheated steam may be supplied to engine thereby offering great margin upto the condensation state. It will allow for only reduction in degree of superheat due to contact with low temperature surfaces and thus condensation is prevented. (b) Condensation can also be controlled by providing steam jacket around cylinder wall so as to maintain engine\u2019s contact surfaces at high temperatures thereby, preventing condensation. (c) Valve timings can be modified so as to result in greater degree of compression prior to admission of steam. This increased compression yields increased temperature of residual steam therefore, causing increased temperature of engine surfaces. Thus during the 520 ________________________________________________________ Applied Thermodynamics subsequent admission of steam condensation of admitting steam shall be limited due to less temperature difference. (d) Condensation can be reduced by increasing the speed of engine for definite output because for particular power output higher speed shall reduce size of engine, thereby reducing the contact surface area. Hence with smaller contact surface area the condensation gets reduced. If the multiple expansion is employed in place of single stage expansion then also condensation gets reduced as the temperature range in every expansion stage gets lowered. Smaller temperature range in any expansion causes lower temperature variations and so reduced condensation. Thus it is seen that condensation and reevaporation are processes which shall always be there whether in large quantity or small quantity in an actual engine. Because of this condensation the actual volume occupied by steam will be less than hypothetical (theoretical) volume. This difference between the actual volume of steam and theoretical volume of steam at any point is known as \u2018missing quantity\u2019. If we assume all steam states during expansion to be dry and saturated and obtain the expansion curve throughout the stroke, then such curve is called saturation curve. Figure 12.16 shows the saturation curve and missing quantity of steam. Fig. 12.16 Saturation curve Here bc shows the actual expansion curve and bc is the expansion curve considering all states of steam during expansion to be dry and saturated. At any point during expansion the amount of steam condensed can be accounted by missing quantity. At some pressure, line ghi shows the volume occupied by steam in actual engine as Vgh and theoretical volume occupied being Vgi. Length \u2018gh\u2019 and \u2018gi\u2019 indicate these volumes. Using these volume values dryness fraction neglecting leakage and other losses can be given by, xh = gh gi. Similarly dryness fraction at \u2018k\u2019, xk = jk j l. During return stroke there shall occur reevaporation of expanded steam leaving cylinder and this reduces the missing quantity of steam in this stroke. This missing quantity of steam causes the loss of work as evident from p-V diagram. Loss of work can be quantified by the area between saturation curve (bc) and actual expansion curve (bc). Missing quantity can be quantified as, At point \u2018h\u2019 the volume of missing quantity = \u2018hi\u2019 SFV where SFV is scaling factor for volume axis of indicator diagram. Similarly at other pressure, the volume of missing quantity of steam at point \u2018k\u2019 = \u2018kl\u2019 S FV Steam Engine _________________________________________________________________ 521 12.7 HEAT BALANCE AND OTHER PERFORMANCE PARAMETERS Heat balance for steam engine is the accounting of the total heat supplied to engine. Heat balance sheet is documentation of heat supplied to engine and the different items in which this heat supplied gets consumed. For preparing heat balance sheet the engine is run for known time at constant load and constant steam supply rate and observations taken for different parameters. Table 12.1 Heat balance sheet for steam engine Heat supplied 1. Heat added in steam,", " Q1, 100% Heat utilization (a) Heat equivalent to bhp, Qa (b) Heat lost in condensate, Qb (c) Heat lost in cooling water, Qc (d) Unaccounted heat lost in radiation, friction etc. Qd = Q1 \u2013 (Qa + Qb + Qc) Fig. 12.17 Schematic for trial of steam engine 1. Heat supplied with steam; Q1 = ms hs1, where ms is mass flow rate of steam in kg/min and hs1 is enthalpy of steam at inlet to engine. (a) Heat equivalent to B.H.P.: Brake power of engine is the power available at crank shaft of engine. Brake power of engine is always less than indicated power by the amount of power lost in overcoming friction in moving parts. Brake power of engine is generally measured employing dynamometers which are of many types. Power output of steam engines was determined originally by use of brake, hence the power output of steam engines was termed as brake power. This term \u2018brake power\u2019 has persisted since then and is now used for all kind of engines. A mechanical friction dynamometer of \u2018Rope brake\u2019 type is shown in Fig. 12.18. Schematic of rope brake dynamometer has a rope wound around the flywheel of steam engine as shown. Rope is hung from a spring balance suspended from rigid support. On the other end of rope a known weight \u2018W\u2019 is suspended. Rope wound round engine flywheel absorbs energy from flywheel. Friction between rope and flywheel can be increased by increasing the weight suspended. For the different parameters having values as denoted in Fig. 12.18 the torque acting on flywheel can be given by T = Net force or frictional force Radius. where, Net force = (W \u2013 S) g newton 522 ________________________________________________________ Applied Thermodynamics Fig. 12.18 Schematic of rope brake dynamometer Radius = Hence, torque T = (W \u2013 S). Power absorbed or brake power = T 2N Brake horse power = Brake power in kW, BP = \uf8ee \uf8ef \uf8ef \uf8f0 \uf8f1 \uf8f2 \uf8f3 \uf8ee \uf8ef \uf8ef \uf8f0 \uf8f1 \uf8f2 \uf8f3 ( \u2212 W S ) 4500 \uf8fc \uf8eb \uf8fd\uf8ec \uf8ed \uf8fe +.P \u00d7 1000 60 \uf8fc \uf8fd \uf8fe 2 ) ( \u22c5 \u03c0 2 N \uf8f9 ) \uf8fa \uf8fa \uf8fb, kW Heat equivalent of Brake power may be obtained by converting it from kW to kcal/min, i.e. BP 60 4 18., kcal/min Qa = (b) Heat lost in condensate, Qb = ms hs3 (c) Heat lost in cooling water, Qc = mw cp,w (Tw, out \u2013 Tw, in), where mw is mass flow rate of cooling water in kg/min. Steam Engine _________________________________________________________________ 523 (d) Unaccounted heat lost, Qd = Q1 \u2013 Qa \u2013 Qb \u2013 Qc Other Performance Parameters: Other performance parameters of steam engine comprise of different efficiencies, steam consumption etc. For getting different efficiencies of engine the estimation of indicated power becomes necessary. (i) Calculation of indicated power: Indicated power of steam engine refers to the power developed inside the engine. It is called indicated power because it is obtained using a mechanism called indicator diagram arrangement. Mechanism is explained along with internal combustion engines. Indicator diagram has the actual pressure and volume variation plotted on p-V diagram. For a single acting steam engine, the indicated power can be given by, p mep1 L A N 1 4500, H.P. IP1 = where pmep1 = Mean effective pressure, i.e. A Cd L d, here Ad and Ld are area and length of diagram and C is indicator spring constant in N/cm2 cm travel. Dp 4 for bore diameter being Dp A1 = Piston surface area, i.e. 2 L = Stroke length N = Speed of rotation in r.p.m. For a double acting engine, the indicated power shall be sum of the indicated power developed on cylinder head side and crank end side both. Power developed on cylinder head side, IP1 = p mep1 L A N 1 4500, H.P. Similarly, Power developed on crank end side, IP2 = Total indicated power, IP = IP1 + IP2 p mep2 L A N 2 4500, H.P. where pmep2 = Mean effective pressure on crank end side of piston. A2 = Piston surface area on crank end side i.e. (A1 \u2013 Area", " of piston rod) 2 dR 4, where dR is diameter of piston rod (D2 p \u2013 d2 R ) = A1 \u2013 A2 = 4 If piston rod area is neglected then A1 = A2 and also pmep1 = pmep2 Thus in such case IP1 = IP2 and total indicated power, IP = 2 IP1 Here this indicated power is expressed in horse power so may be called \u2018indicated horse power\u2019. 524 ________________________________________________________ Applied Thermodynamics (ii) Mechanical efficiency: Mechanical efficiency of engine is given by the ratio of brake power to indicated power. Thus it quantifies that how effectively power developed inside cylinder can be transferred to crank shaft. Brake power i.e. power available at crank shaft is less than that developed inside cylinder due to frictional losses etc. Mechanical efficiency, mech = Brake power Indicated power or, mech = Brake horse power Indicated horse power Also, it can be said that brake power is less than indicated power by an amount of friction power, thus Indicated power = Brake power + Friction power Brake power = Indicated power \u2013 Friction power Indicated power \u2013 Friction power Indicated power mech = So, (iii) Thermal efficiency: Thermal efficiency is the ratio of work done to the energy supplied to the steam engine cylinder. Thermal efficiency may also be defined as the ratio of work done on brake power basis to the energy supplied to steam engine cylinder. When work done per minute is taken on IHP basis then it is called \u2018indicated thermal efficiency\u2019 and when work done per minute is taken on BHP basis then it is called brake thermal efficiency.\u2019 Thus for steam flow rate of ms, kg/min, enthalpy of steam entering being hs1 and enthalpy of condensate leaving to hotwell being hs3 the thermal efficiency may be given as, Energy supplied to engine = ms (hs1 \u2013 hs3) kcal/min \u00d7 IHP 4500 J Work done per minute on IHP basis =, kcal/min Work done per minute on BHP basis = \u00d7 BHP 4500 J, kcal/min Hence, Indicated thermal efficiency, ith = IHP J m h s s 4500 h s 1 3 Brake thermal efficiency, bth = BHP J m h s s 4500 h s 1 3 or, Mechanical efficiency = Brake thermal efficiency Indicated thermal efficiency (iv) Specific steam consumption: Specific steam consumption refers to steam consumption per hour per unit power. When this is defined in respect to indicated horse power then it is called specific steam consumption on IHP basis and when it is defined in respect to brake horse power then it is called specific steam consumption on BHP basis. Specific steam consumption on IHP basis = Specific steam consumption on BHP basis = ms 60 IHP sm \u00d7 BHP 60, kg per hour per IHP, kg per hour per BHP Steam Engine _________________________________________________________________ 525 (v) Overall efficiency: This is given by the ratio of brake work (shaft work) available to the energy supplied to the boiler. Say, for mf being mass of fuel burnt in kg/min and calorific value of fuel, being CV, kcal/kg. The total energy input to boiler, Qinput = mf CV, kcal. 4500 Overall efficiency, overall = BHP J Q input = BHP J m f 4500 CV (vi) Rankine efficiency: This efficiency is given by the ratio of Rankine work available from engine to the energy supplied to steam engine. Rankine efficiency, Rankine = Nomenclature is similar to that used in heat balance. Ratio of thermal efficiency to Rankine efficiency is given as the relative efficiency. Thermal efficiency Rankine efficiency Relative efficiency, Relative =. 12.8 GOVERNING OF SIMPLE STEAM ENGINES In a steam engine operating at varying loads it is desired to have some regulating mechanism so that constant speed is available at engine shaft even when the loads are varying. When load on engine is increased then the speed may get decreased and in order to have constant speed running of engine the steam supply rate should be increased. Similarly, when load on engine is decreased then the speed may get increased and in order to have constant speed of engine the steam supply rate should be decreased. This arrangement by which steam engine output is altered so as to meet fluctuating load while maintaining constant speed at engine shaft is called governing of steam engine. Generally, there are two basic principles of governing as described ahead. Governing of steam engine can be either \u2018throttle governing\u2019 or \u2018cut-off governing\u2019. Fig. 12.19 Effect of throttle governing shown on p-V diagram (i) Throttle governing: Throttle governing employs throttling of steam at inlet to steam engine thereby varying the output from engine. In throttle governing the cut-off point in engine is kept constant and only pressure of steam entering steam engine is", " altered. Throttling of steam at inlet is shown on p-V diagram in Fig. 12.19. In throttle governing the throttle action causes wastage of steam energy as 526 ________________________________________________________ Applied Thermodynamics the available heat drop gets reduced due to it. Thermal efficiency of steam engine gets reduced due to throttling of steam at inlet to steam engine. p-V diagram shows that at full load cycle is abcde. At reduced loads the throttling causes reduction in steam engine inlet pressure resulting in cycles given by ab c de, ab cde, abcde. Due to throttle action the work output gets reduced as evident from modified cycles shown on p-V diagram. Pressure \u2018pa\u2019 at full load is throttled to pa, pa, pa etc. depending upon load on engine. In throttle governing of steam engines the steam consumption is directly proportional to indicated power. Therefore, graphical pattern between steam consumption and indicated power is a straight line as shown in Fig. 12.20. Straight line variation between steam consumption and indicated power is called \u2018Willan\u2019s law\u2019 and straight line is called Willan\u2019s line. Willan\u2019s law is given by, m = (S ip) + C. Here \u2018S\u2019 is slope of Willan\u2019s line, \u2018C\u2019 is steam consumption at zero indicated power and \u2018ip\u2019 is indicated power at any point. Fig. 12.20 Steam consumption vs. indicated power, Willan\u2019s law (ii) Cut-off governing: In cut-off governing the regulation is made by varying the point of cut-off. Cut-off point is modified and thus the steam engine output is altered due to alteration in period of admission of steam. Principle behind cut-off governing is to change the quantity of steam admitted by changing the cut-off point depending on the load on engine. For full load the cut-off point is located such that the maximum amount of steam is admitted. Here the inlet pressure and back pressure do not change due to governing. Figure 12.21 shows the cut-off governing on p-V diagram. Fig. 12.21 Effect of cut off governing shown on p-V diagram Steam Engine _________________________________________________________________ 527 P-V diagram shows that at full load the cycle abcde gives the maximum output which varies upon change in cut-off point to b, b, b etc. When cut off points are changed then it results in modified outputs from engine as given by area enclosed in abcde, abcde, abcde etc. Cutoff governing employs special kind of valve which has provision for changing the cut-off point. This may be realized by employing Mayer\u2019s expansion valve or suitable linkages for getting the changed cut off points. Cut-off governing is successfully used in steam engine based locomotive where torque available at crank shaft should match with load in the train. Cut-off governing is efficient and economical as compared to throttle governing due to optimal steam utilization. 12.9 COMPOUND STEAM ENGINE Performance of steam engine can be increased by increasing the pressure and temperature of steam entering the steam engine and lowering the back pressure. Since the lowering of back pressure is limited by the atmospheric pressure in case of non condensing engine and by the condenser pressure in case of condensing engines. Therefore, increasing pressure and temperature of steam at inlet to steam engine can be used for improving efficiency of steam engine. But the high pressure at inlet and low back pressure result into number of practical difficulties such as; the size of cylinder has to be of large volume and should have heavy structure in order to bear with large pressure difference, large volume after expansion and higher mechanical stresses, tendency of condensation loss and leakage loss gets increased, large temperature difference causes excessive thermal stresses, heavy structure poses the difficulty in balancing etc. In order to overcome these difficulties instead of realizing expansion in single steam engine more than one steam engine cylinders are employed for achieving total expansion from high inlet pressure to low back pressure. Thus for any given expansion ratio (i.e. ratio of high pressure to low pressure) total expansion occurs in more than one engine cylinders successively. The arrangement in which expansion occurs in two or more engine cylinders successively is called compound steam engine. Compounding is thus the phenomenon of realizing expansion in more than one steam engine cylinders in succession. In case of two cylinder engines first cylinder is called high pressure cylinder (HP cylinder) and second cylinder is called low pressure cylinder (LP cylinder). In case of three cylinders the engine cylinders shall be called as high pressure (HP), intermediate pressure (IP) and low pressure (LP) cylinders. 12.10 METHODS OF COMPOUNDING Compounding of steam engines can be done by combining more than one cylinders of steam engines such that expansion of steam occurs successively in these cylinders. There are basically two approaches of compounding, namely \u2018Tandem type\u2019 and \ufffd", "\ufffdCross type\u2019. Cross type compounding can be of \u2018Woolf type\u2019 and \u2018Receiver type\u2019. Thus compound steam engines can be classified as below based on method of compounding. (i) Tandem compound engines (ii) Woolf compound engines (iii) Receiver compound engines Cross type compounding. Arrangement in each type of compounding is described ahead. (i) Tandem compound engines: Tandem type compounding of steam engines has the in line cylinders having pistons mounted on the same piston rod which is further having crosshead and connecting rod providing power output at crankshaft. Figure 12.22 shows schematic of tandem compound 528 ________________________________________________________ Applied Thermodynamics steam engine having two in line cylinders. Steam at high pressure first enters the HP cylinder and after getting partially expanded enters the LP cylinders. Two pistons of HP and LP cylinders are mounted such that the two cylinders produce power output without any phase difference. Maximum output and minimum output from both cylinders is available at same crank angle position and output is of additive type i.e. maximum torque from compound engine. Fig. 12.22 Tandem compound engine Thus there exists very large variation of torque during a cycle and for compensating this large variation heavy flywheel and large balancing weights are required. (ii) Woolf compound engine: It is a cross type compounding having two cylinders having pistons at 180\u00b0 phase difference i.e. at some position one cylinder may have piston at inner dead centre and other cylinder has piston at outer dead centre. The cylinders are arranged parallel to one another and steam leaving HP cylinder enters into LP cylinder directly thereby expansion remains continuous throughout the stroke. Since the power output from two cylinders is available with phase difference of 180\u00b0 so the torque variation is very large similar to that in case of tandem compounding. Therefore, in woolf compound engine also large flywheel is required. Fig. 12.23 Woolf compound engine. Steam Engine _________________________________________________________________ 529 (iii) Receiver compound engine: Receiver compound engine is also a cross compound engine having two cylinders with out of phase pistons and receiver in between. In this the cranks for two cylinders are at 90\u00b0 as shown in Fig. 12.24. Here steam leaving HP cylinder passes into receiver and subsequently at correct time steam enters LP cylinder. Receiver is required because the two cylinders have out of phase pistons and exhaust from HP does not go directly into LP cylinder, so some kind of storage is a necessity. The maximum torque and minimum torques are not at the same time so the torque variation is not too much in this compounding. Hence the size of flywheel required is smaller in this arrangement. Also balancing requirement is not too much in case of receiver compounding. Since steam is stored in reservoir so there are chances of pressure drop and heat loss taking place in receiver. These losses reduce efficiency of receiver compound engine and so require some jacketing/ steam heating of receiver to prevent heat loss. Advantages of compounding: Compounding of steam engine offers number of advantages over the simple steam engine as given below: Fig. 12.24 Receiver compound engine (a) Thermal stresses and phenomenon of condensation and reevaporation is reduced due to lesser temperature variation in different steam engine cylinders. (b) Loss of work due to condensation of steam is reduced as the condensation and reevaporation is utilised in cylinders subsequent to first cylinder. Loss due to condensation is only there with last cylinder. (c) Due to reduced pressure difference the chances of leakage are reduced. (d) Due to expansion occurring in parts in different cylinders the reheating can be employed, if desired to control condensation of steam. 530 ________________________________________________________ Applied Thermodynamics (e) In compound steam engine the HP cylinder is fed with high pressure steam so the sturdy structure requirement is there only for HP cylinder. LP cylinders need not be of very sturdy type due to low pressure steam entering it although volume of LP cylinder has to be large. Due to lighter reciprocating parts the vibrations are also minimized. (f) Compound steam engine is capable of getting started at any crank position due to phase difference between different cylinders. (g) Compound steam engines are better balanced and more uniform torque is available, thus requiring light flywheel. (h) Due to multi cylinder engine the reliability of compound steam engine is better as in the event of failure of even one cylinder the reduced power will be available from other cylinder. (i) Compound steam engines are cheaper than simple steam engine for same power output and efficiency. (j) Steam economy is increased by 10 to 25% at rated load for non-condensing engine and by 15 to 40% for condensing engine. Disadvantages of compound steam engine: Disadvantages with compounding are as under: (a) Compound steam engine requires more maintenance and attention due to complexity. (b) Lubrication requirements are excessive because of increased wear and tear in engine parts. (c)", " Heat losses are more due to large surface area of compound engines. (d) Compound steam engines are bulky and require large space. 12.11 INDICATOR DIAGRAM FOR COMPOUND STEAM ENGINE Combined indicator diagram for the compound steam engine having HP and LP cylinder can be obtained similar to that in simple steam engine. Figure 12.25 shows the actual indicator diagram for compound steam engine. For getting the actual indicator diagram the indicator diagram arrangement is used separately and subsequently the diagrams for two cylinders are combined together. Actually the indicator diagrams obtained from respective cylinders shall be on separate p-V scales with different spring constants, different masses of cushion steam etc. Combined indicator diagrams are obtained after getting the average indicator diagrams for HP and LP cylinders (both sides in case of double acting) being reduced to common pressure and volume scales. These individual indicator diagrams are plotted with definite clearance volumes in respective cylinders. Saturation curves shall be different for HP and LP cylinders because of different masses of cushion steam in clearance volume, pressure drop during transfer of steam from HP to LP and condensation etc. Hypothetical diagram shown has been drawn with common saturation curve as it has been drawn assuming zero clearance volume, no pressure loss and no condensation etc. In hypothetical indicator diagrams for HP and LP when combined show loss of work due to unresisted expansion at interface of HP and LP cylinders, as shown by hatched area on p-V diagram Fig. 12.25 (a). Hypothetical indicator diagram drawn on common scale is generally used for all calculations. Combined indicator diagram shown has HP cylinder having steam entering at p1, getting expanded up to p2 and exhausted at p3 for entering into LP cylinder, getting expanded upto p4 and finally exhausted at pressure p5 from LP cylinder. Hypothetical cycle for HP cylinder is shown by \u2018abhog\u2019 and for LP cylinder is given by \u2018gcdef \u2019. Steam Engine _________________________________________________________________ 531 Fig. 12.25 Indicator diagram for compound steam engine 12.12 CALCULATIONS FOR COMPOUND STEAM ENGINES Compound steam engine cylinders may be designed based upon the assumption of \u2018equal initial piston loads on all pistons\u2019 or \u2018equal power developed in each cylinder for getting uniform torque\u2019 or \u2018equal temperature drop in every cylinder for optimal use of steam. The hypothetical work output (indicator work) for HP and LP cylinders can be given as below; Hypothetical indicator work from HP cylinder = [p1 Vb (1 + ln rHP) \u2013 p3 Vh] Actual indicator work from HP cylinder = [p1 Vb (1 + ln rHP) \u2013 p3 Vh] dfHP where rHP and dfHP are expansion ratio and diagram factor for HP cylinder. So, rHP = V h V b Hypothetical indicator work from LP cylinder = [p3 Vc (1 + ln rLP) \u2013 p5 Vd] 532 ________________________________________________________ Applied Thermodynamics Actual indicator work from LP cylinder = [p3 Vc (1 + ln rLP) \u2013 p5 Vd] dfLP where rLP and dfLP are expansion ratio and diagram factor for LP cylinder. Here rLP = V d V c. Total actual indicator work from compound steam engine = [p1 Vb ( 1 + ln rHP) \u2013 p3 Vh] dfHP + [p3 Vc (1 + ln rLP) \u2013 p5 Vd] dfLP Expansion ratio and diagram factors for two cylinders may be same or different. Loss of work due to unresisted expansion can be quantified by the difference between total hypothetical output and actual output from compound engine Loss of work = Area of total hypothetical diagram \u2013 (Area of hypothetical HP diagram + Area of hypothetical LP diagram) = Aabdef \u2013 (Aabhog + Agcdef) Mean effective pressures of both cylinders may be given separately or with reference to common cylinder. Mean effective pressure of HP cylinder = Area of indicator diagram for HP cylinder Stroke volume for HP cylinder i.e. length of diagram for HP mepHP = W HP V HP Similarly, Mean effective pressure of LP cylinder = Area of indicator diagram for LP cylinder Stroke volume for LP cylinder i.e. length of diagram for LP mepLP = W LP V LP Mean effective pressures can be mathematically given as below, mepHP = 1 1 ln HP p r HP r \u2013 p3 mepLP = 3 1 ln LP p r LP r \u2013 p5 In case of multi cylinder engines as in case of compound steam engines mean effective pressures when defined separately offer difficulty in comparing them. Therefore, mean effective pressure is defined in reference to LP cylinder, which means it is assumed that the stroke volume of LP cylinder is used for all cylinders and the mean effective pressure values can be directly used for comparing work output from respective cylinders. In case of compound engine having HP and LP cylinders the mean effective pressure of HP", " cylinder referred to LP cylinder can be given as below. Steam Engine _________________________________________________________________ 533 Mean effective pressure of HP referred to LP cylinder, mepHP ref LP = mepHP Stroke volume of HP cylinder Stroke volume of LP cylinder or, mepHP ref LP = mep HP V HP V LP Mathematical conditions for different basis of compounding steam engine cylinders can be given as below. (a) For equal initial piston loads on all pistons the loads on both HP and LP cylinder pistons can be equated. Mathematically, Load on HP piston = Load on LP piston (p1 \u2013 p3) AHP = (p3 \u2013 p5) ALP where AHP and ALP are cross section areas of HP and LP piston respectively (b) For equal power developed in each cylinder the power developed in HP and LP cylinders can be given as, Power developed in HP cylinder = Power developed in LP cylinder Work output in HP cylinder = Work output in LP cylinder or mepHP VHP = mepLP VLP mepLP = mep HP V HP V LP mepLP = mepHP ref. LP In case of actual indicator diagram of compound steam engine, (Mean effective pressure of combined diagram) df combined = (mepHP ref. LP dfHP) + (mepLP dfLP) or Mean effective pressure of combined diagram = mep HP ref LP mep LP df LP HP df df combined where \u2018df \u2019 refers to diagram factor. 12.13 GOVERNING OF COMPOUND STEAM ENGINE Governing of compound steam engines is done based on the principles similar to that of simple steam engine i.e. throttle governing and cut-off governing. Throttle governing is based on qualitative regulation while cut-off governing is based on quantitative regulation of steam. (i) Throttle governing: Throttle governing of compound steam engine is done by throttling steam before being admitted to HP cylinder, thereby yielding reduced output from both HP and LP cylinders. In fact throttling reduces the heat potential of steam due to its throttling at constant heat, thereby lowering capacity of producing work by throttled steam. Figure 12.26 shows throttle governing being used in compound steam engine. Reduction in steam inlet pressure at inlet of HP cylinder without varying cut- 534 ________________________________________________________ Applied Thermodynamics off point alters the discharge pressure from HP and subsequently steam with reduced enthalpy enters LP cylinder. Thus total output from engine is reduced because of less output available from both HP and LP cylinders. Throttle governing involves wastage of steam. Fig. 12.26 Throttle governing of compound steam engine (ii) Cut-off governing: In cut-off governing the point of cut-off of steam admission is modified in HP cylinder thereby modifying quantity of steam admitted. Here in cut-off governing the work producing capacity is not modified instead its quantity available for doing work in engine is modified. Figure 12.27 shows the cut-off governing. Mayer\u2019s expansion valve is generally used for altering the cut-off point in HP cylinder. This expansion valve is operated independently depending upon load on engine. Work output from both HP and LP cylinders is modified due to change in cut-off point. However, change in LP cylinder work is much more than change in HP cylinder work due to varying cut off point. This offers unbalance in work done by HP and LP cylinders. Fig. 12.27 Cut-off governing of compound steam engine Steam Engine _________________________________________________________________ 535 Therefore, in order to have balance in the work outputs from HP and LP cylinders i.e. maintaining the right proportion of works from two cylinders for getting balance, the cut off point is changed in both HP and LP cylinders. Variation of cut off point in LP and HP both is done not to change total work output but to change the proportion of work from two. Thus in some cases where unbalance in work is not problem the cut-off governing is employed for only HP cylinder while when unbalance is serious problem then cut-off governing is employed for both HP and LP cylinders. 12.14 UNIFLOW ENGINE Uniflow engines are also called as straight flow engines. Uniflow engines have flow of steam through inlet and exit ports and passages in one direction only. Generally in a steam engine same port/passage is for inlet of steam and exit of steam from cylinder thereby causing severe problem of condensation and reevaporation. In uniflow engine steam enters at the ends of cylinder and leaves out as exhaust from centre. Thus the hot steam does not come in contact with low temperature parts and cold steam does not come in contact with hot parts of engine, similar to compound steam engine. Uniflow engines offer higher diagram factor in the absence of condensation and receiver loss. Uniflow engines of horizontal type are available in sizes up to 1200 hp while vertical type uniflow engines are available in sizes up to 7000 hp. Figure 12.28 shows the schematic of unif", "low engine in its two sectional views. It has mechanically operated valves at two ends of cylinder for admitting steam and exhaust port is located at centre of cylinder. Let us consider piston to be on inner dead centre and inlet valve gets opened. Thus live steam (high pressure and temperature steam) enters cylinder causing piston displacement towards crank end. Piston movement is caused by pushing of piston by steam till cut-off point and subsequent expansion of steam till exhaust port gets uncovered. Piston subsequently reaches to the outer dead centre when inlet valve at this end gets opened thereby causing inlet of live steam. Thus again piston moves towards inner dead centre partly due to push of live steam and partly due to its expansion. Again in this stroke exhaust shall occur as the exhaust ports get uncovered. Hence it is seen that there is straight flow of steam, eliminating problem of condensation. Here exhaust stroke is very small compared to stroke length in cylinder. Again piston reaches inner dead centre with cushion steam in clearance volume. This cushion steam gets compressed in clearance volume before inlet of live steam. Thus temperature of cushion steam gets increased before it comes in contact with live steam and condensation is regulated. Fig. 12.28 Sectional view of uniflow engine. 536 ________________________________________________________ Applied Thermodynamics Live steam again enters cylinder upon opening of inlet valve causing piston to move toward outer dead centre and thus engine operates. Uniflow engine offers advantages of \u2018reduced condensation of steam\u2019, \u2018less specific steam consumption and so higher thermal efficiency\u2019, \u2018simple and ease of design and fabrication\u2019 and \u2018reduced friction loss due to elimination of D-slide or other complex valves\u2019 etc. Disadvantages of the engine are, \u2018large size flywheel requirement to smoothen torque fluctuations compared to compound engine\u2019, \u2018poor mechanical balancing of engine\u2019, \u2018robustness of cylinder\u2019 and \u2018reduced output because of early compression of steam\u2019 etc. EXAMPLES 1. A steam engine operates with steam being supplied at 0.2 MPa, 250C and expanding upto 0.3 bar. Steam is finally released out at 0.05 bar. Determine the modified Rankine cycle efficiency and compare it with the efficiency of Carnot cycle operating between given limits of pressure. Neglect pump work. Solution: Here states 1, 2, and 3 refer to cut-off point, state at the end of expansion and after release from engine. Fig. 12.29 p-V and T-s representation for modified Rankine cycle At inlet to engine, from steam tables, h1 = 2971 kJ/kg s1 = 7.7086 kJ/kgK Let dryness fraction at 2 be x2. s1 = s2; s2 = 7.7086 = sf at 0.3 bar + x2 sfg at 0.3 bar x2 = 0.99 v2 = vf at 0.3 bar + x2 vfg at 0.3 bar = 5.1767 m3/kg h2 = 2601.97 kJ/kg Work output from engine cycle per kg of steam = (h1 \u2013 h2) + v2(p2 \u2013 p3) = (2971 \u2013 2601.97) + 5.1767 (0.3 \u2013 0.05) 102 Steam Engine _________________________________________________________________ 537 = 511.95 kJ/kg Heat input per kg of steam = h1 \u2013 hf at 0.05 bar = 2971 \u2013 137.82 = 2833.18 kJ/kg Efficiency of modified Rankine cycle =.95 511. 2833 18 = 18.07 % = 0.1807 In order to find out Carnot cycle efficiency let us find out saturation temperature corresponding to 0.2 MPa and 0.05 bar. Tmax = Tsat at 0.2 MPa = 120.23C or 393.23 K Tmin = Tsat at 0.05 bar = 32.88C or 305.88 K T T max = 0.2221 or 22.21% min Carnot cycle efficiency = 1 \u2013 Modified Rankine cycle efficiency = 18.07%, Carnot efficiency = 22.21% Ans. 2. A steam engine has steam being supplied at 10 bar, 100C into a cylinder of 30 cm diameter and stroke to bore ratio of 2. Steam is expanded upto 0.75 bar and then released into condenser at 0.25 bar. Determine (i) the modified Rankine efficiency and (ii) the modified stroke length considering that the same steam is expanded from inlet state to the condenser pressure. Neglect pump work. Solution: States 1, 2 and 3 refer to point of cut off, end of expansion and release. At 10 bar, 100C h1 = 2676.2 kJ/kg From steam table,", " s1 = 7.3614 kJ/kg K Fig. 12.30 538 ________________________________________________________ Applied Thermodynamics At state 2, s1 = s2 s2 = 7.3614 = sf at 0.75 bar + x2 sfg at 0.75 bar x2 = 0.9848 v2 = vf at 0.75 bar + x2 vfg at 0.75 bar v2 = 2.1833 m3/kg h2 = hf at 0.75 bar + x2 hfg at 0.75 bar h2 = 2628.35 kJ/kg h4 = hf at 0.25 bar = 271.93 kJ/kg Work output from steam engine per kg of steam, wnet = (h1 \u2013 h2) + v2(0.75 \u2013 0.25) 102 = 157.015 kJ/kg Heat added, qadd = (h1 \u2013 h4) = 2676.2 \u2013 271.93 = 2404.27 kJ/kg Efficiency of modified Rankine cycle = net w q add = 0.0653 or 6.53% Let us now consider the expansion to occur up to condenser pressure i.e. state 6 s1 = s2 = s6 = 7.3614 kJ/kg K s6 = 7.3614 = sf at 0.25 bar + x6 sfg at 0.25 bar x6 = 0.9323 h6 = hf at 0.25 bar + x6 hfg at 0.25 bar = 2459.38 kJ/kg v6 = vf at 0.25 bar + x6 vfg at 0.25 bar = 5.784 m3/kg Given volume of cylinder, V = 4 d2 L = 4 (0.3)2 0.6 = 0.0424 In a stroke the mass of steam at 2, m = V v2 = 0.0194 kg Let the modified stroke length be L m. Volume requirement at 6, V = m v6 = 0.0194 5.784 V = 0.1122 m 3 or V = 4 d2 L = 0.1122 L = 1.5873 m or 158.73 cm Modified Rankine cycle efficiency = 6.53% New stroke length = 158.73 cm Ans. 3. A double acting steam engine has bore of 30 cm and stroke to bore ratio of 2 with cut-off occurring at 40% of stroke. Steam enters the engine cylinder at 7.5 bar and exhausts at 0.1 bar. Engine runs at 180 rpm. Neglecting clearance volume and considering diagram factor of 0.6 determine the indicated horse power. Solution: Cut off occurs at 40% of stroke, so 0.4 = V 1 V 2 Steam Engine _________________________________________________________________ 539 Fig. 12.31 Indicator diagram (hypothetical) Expansion ratio. = 2.5 Hypothetical mean effective pressure mep = = p 1 (1 + ln r) \u2013 p3 r 7.5.5 2 (1 + ln 2.5) \u2013 0.1 mep = 5.65 bar Actual mean effective pressure = mep diagram factor = 5.65 0.6 = 3.39 bar Indicated power =. 3 39. 0 6 2 180 10. 0 09 60 4 2, (1 hp = 0.7457 kW) = 86.26 kW or 115.67 hp Indicated power = 115.67 hp Ans. 4. In a single acting steam engine, steam is admitted at 15 bar, 200C and exhausts at 0.75 bar with cut-off occurring at 25% of stroke. Engine produces 150 hp at 240 rpm. The mechanical efficiency of engine is 85%, diagram factor is 0.7, brake thermal efficiency is 20% and stroke to bore ratio is 1.5. Determine the specific steam consumption and cylinder dimensions. Neglect the cross-section area of piston rod and clearance volume. Solution: Neglecting clearance volume, Expansion ratio = 1 0.25 = 4 540 ________________________________________________________ Applied Thermodynamics P1 = 15 bar, P3 = 0.75 bar, mephypothetical = P 1 (1 + ln r) \u2013 P3 r 15 1 ln 4 \u2013 0.75 = 4 mephypothetical = 8.19 bar Fig. 12.32 Actual mean effective pressure = 8.19 0.7 = 5.733 bar Indicated horse power = Brake horse power/ Let diameter of cylinder be \u2018d\u2019, Stroke = (1.5 d), {1 hp = 0.7457 kW} mech = 176.47 hp Indicated horse power = 176.47 = mep L A N. 0 7457 60. 5 733 10 4 2 d.5 1 0 7457 60. 2 d 240 Ans. d = 0", ".3652 m L = 1.5 0.3652 = 0.5478 m Bore = 36.52 cm Stroke = 54.78 cm Heat added per kg of steam = h15 bar, 200C \u2013 hf at 0.75 bar qadd = 2803.3 \u2013 384.39 = 2418.91 kJ/kg Let specific steam consumption be m kg/hp. hr So Brake thermal efficiency = 1hp brake output m q. 1 0 7457 3600 2418 add.91 m m = 5.55 kg/hp. hr 0.2 = Specific steam consumption = 5.55 kg/hp. hr Ans. Steam Engine _________________________________________________________________ 541 5. A single cylinder steam engine has steam consumption of 18 kg/min and yields indicated power of 100 kW while running at 240 rpm. Engine cylinder has bore of 30 cm and stroke of 40 cm. Steam is admitted at 10 bar, 200C and exhausted at 0.75 bar with cut-off occurring at 25% of stroke. Determine the diagram factor and indicated thermal efficiency. Neglect clearance volume. Solution: p1 = 10 bar, p3 = 0.75 bar Enthalpy at inlet to cylinder, At 10 bar, 200C, h1 = 2875.3 kJ/kg At 0.75 bar, hf = 384.39 kJ/kg Heat added per kg of steam, qadd = h1 \u2013 hf at 0.75 bar qadd = 2490.91 kJ/kg 1 0.25 p 1 (1 + ln r) \u2013 p3 = 5.215 bar r = 4 Hypothetical mep = Expansion ratio, r = Fig. 12.33 Theoretically, Indicated power output = mep L A N 60 =.215 5. 0 40. 0 3 2 240 10 2 60 = 235.92 kW or 316.37 hp. Diagram factor = Actual indicated power Theoretical indicated power = 100 235.92 = 0.4238 Indicated thermal efficiency = Indicated work Heat added Here, steam supplied is 18 kg/min or 0.3 kg/s. 542 ________________________________________________________ Applied Thermodynamics Indicated thermal efficiency = 235.92 m q = 235.92 0 3 2490..91 add = 0.3157 or 31.57% Diagram factor = 0.4238 Ans. Indicated thermal efficiency = 31.57% 6. A steam engine has steam being supplied at 10 bar, 0.9 dry and exhausted at 1 bar. Cut-off occurs at 60% of stroke. Determine (i) (ii) (iii) the fraction of work obtained by expansion working of engine. the efficiency of cycle the % increase in work and efficiency when engine works as non-expansive engine. Neglect clearance volume and assume expansion to be hyperbolic. Solution: Let us analyze for 1 kg of steam. At inlet to engine cylinder, h1 = hf at 10 bar + 0.9 hfg at 10 bar h1 = 2576.58 kJ/kg, v1 = vf at 10 bar + 0.9 vfg at 10 bar v1 = 0.1751 m3/kg At 1 bar, hf = 417.46 kJ/kg Heat added per kg of steam, qadd = h1 \u2013 hf at 1 bar qadd = 2159.12 kJ/kg Fig. 12.34 For given cut-off volume, v1 = 0.6 v2 or v2 = 0.2918 m3/kg, Expansion ratio, r = 1 0 6. = 5 3 Steam Engine _________________________________________________________________ 543 Non expansive work in cycle 1-2-3-4-5 can be given by area 1-3-4-5. Non expansive work per kg of steam wne = v1 (10 \u2013 1) 102 = 157.59 kJ/kg Expansive work in cycle 1-2-3-4-5 can be given by area 1-2-3-3. Expansive work per kg of steam. we = [p1 v1 ln r \u2013 p3(v2 \u2013 v1)] we = [10 102 0.1751 ln we = 77.77 kJ/kg \u2013 1 102 (0.2918 \u2013 0.1751) 5 3 Total work per kg of steam = wne + we = 235.36 kJ/kg Fraction of work obtained by expansive working = w e + w n e w e Thermal efficiency of cycle = =. 77 77. 235 36 = 0.3304 or 33.04% 235 36. qadd =. 235 36 2159 12. = 0.1090 or 10.90% Fraction of expansive work = 33.04% of total output Ans. Thermal efficiency = 10.90% For cycle having complete non expansive work:", " Such modified cycle is shown by 1-3-4-5. This refers to the situation when cut-off shall become unity. Therefore, the mass of steam admitted shall be increased as compared to previous case in which 1 kg steam is admitted up to 60% cut-off point. Steam admitted per cycle when cut off is at 60% = 1 kg Steam admitted per cycle when cut off becomes unity = 1 0 6. = 1.66 kg Total work per cycle = Area 1-3-4-5 = (p1 \u2013 p3) v1 1.66 = 261.59 kJ 100 = 11.14% % increase in work = 235 36. 261 Modified thermal efficiency = = 0.0729 or 7.29%.59. 235 36.59 2159 12 261.. 1 66 There is reduction in thermal efficiency. % decrease in efficiency = 10.9 10 7.29.9 100 = 33.12% Ans. % increase in work = 11.14% % decrease in efficiency = 33.12% 7. A non condensing double acting steam engine produces 60 bhp when steam at 12 bar pressure is admitted and exhausted at 1 bar. Engine runs at 240 rpm and piston speed of 2 m/s. The piston rod has diameter of 4 cm. Cut off occurs at 60% of stroke. Clearance volume is 5% of stroke volume. Considering diagram factor as 0.8 and mechanical efficiency of 90% determine the bore of the cylinder. 544 ________________________________________________________ Applied Thermodynamics Solution: p1 = 12 bar, p3 = 1 bar V1 = 0.6 (V2 \u2013 V5) V5 = 0.05 (V2 \u2013 V5) V5 = 0 05.. 1 05 V2 Expansion ratio, r = V 2 V 1 Substituting V1 and V2 from above as, V2 =. 1 05 0 05. V 0 6. 5. 0 05 r = 1.75 V1 = V5 = 21 V5 = 12 V5 Hypothetical mean effective pressure ln 1.75 ( 21 ) P 1 P V 5 3 ) \u2212 ) ( \u2212 \u00d7 12 1 V 5 V 5 ( \u2212 \u00d7 1 21 ) \u2212 V V 5 5 P V 1 1 1 ln r mep = \u00d7 12 12 V 5 = mep = 9.63 bar Fig. 12.35 Actual mean effective pressure mepactual = Hypothetical mean effective pressure diagram factor = 9.63 0.8 = 7.704 bar 7.70 bar Indicated power = mepactual Effective area Piston speed 60 = 7.70 102 A 2. 0 7457.9 0 A = 0.03228 m2 Steam Engine _________________________________________________________________ 545 Let diameter of piston be D meter. Since it is double acting engine so this area A calculated shall include area on both sides of piston. A = 2 A = 2 Apiston + Apiston rod D 4 10 + D 2 2 4 4 4 0.03228 = 2 2 2 4 10 2 2 4 D = 0.1405 m or 14.05 m Bore = 14.05 cm Ans. 8. A double acting steam engine has cylinder diameter of 20 cm and stroke of 30 cm. Clearance volume is 2 103 cm3. Engine uses 0.05 kg steam per stroke effectively i.e. after correction for leakage. Compression starts at 80% of stroke with pressure of 1 bar and steam is dry at the point of starting of compression. Cut-off point is at 10% of stroke and release occurs at 90% of stroke. The pressures at cut off point and point of release are 15 bar and 3 bar. Determine total mass of steam present during expansion and dryness at cut-off and release. Consider the expansion to follow PV n = constant and find heat leaking through cylinder walls per kg of steam during expansion. Solution: Clearance volume, Fig. 12.36 p-V diagram V6 = V5 = 2 103 cm3 = 2 10\u20133 m3 Stroke volume, V3 \u2013 V6 = (0.2)2 0.3= 9.42 10\u20133 m3 4 Total volume = V3 = 2 10\u20133 + 9.42 10\u20133 = 11.42 10\u20133 m3 546 ________________________________________________________ Applied Thermodynamics At the point of beginning of compression, i.e. \u20184\u2019 V4 = V3 \u2013 0.8 (V3 \u2013 V6) = 3.884 10\u20133 m3 At the start of compression, v4 = vg at 1 bar, dry saturated = 1.6940 m3/kg Therefore, mass of steam at state 4, i.e. cushion steam = = 2.293 10\u20133 kg V 4 v 4 Total mass of steam during expansion = Cushion steam + Steam used per stroke = 2.293 103 + 0.05 = 0.052293 kg Volume", " at cut-off point, V1 = V6 + 0.1 (V3 \u2013 V6) V1 = 2.942 10\u20133 m3 Dryness fraction at cut-off point, x1 = Volume m v\u22c5 g at15bar = 3. 0 052293 2 49 10.. 0 13177 Volume at point of release, x1 = 0.4269 V2 = V6 + 0.9 (V3 \u2013 V6) = 0.010478 m3 Dryness fraction at point of release, x2 =,V 2 Volume m v g at 3bar =. 0 010478. 0 052293. 0 6058 x2 = 0.3307 Let us find out index of expansion n for process 1\u20132. 1 = P2 V n 2 P1 V n 15 (2.942 10\u20133)n = 3 (0.010478)n Work done in a stroke = n = 1.267 P V 2 2 1 15 2.942 3 10.267 1 0 010478. 3 1 P V 1 1 n = 102 Work done per kg of steam =. 4 755 Steam available per stroke = 4.755 kJ Change in internal energy during expansion = u1 \u2013 u2 u1 = uf at 15 bar + x1 ufg at 15 bar = 1590.79 kJ/kg = 4 755. 0.052293 = 90.93 kJ/kg Steam Engine _________________________________________________________________ 547 u2 = uf at 3 bar + x2 ufg at 3 bar u2 = 1216.73 kJ/kg Applying first law of thermodynamics, q = u + w = (u2 \u2013 u1) \u2013 w = (1216.73 \u2013 1590.79) \u2013 90.93 = \u2013 464.99 kJ/kg Total mass of steam during expansion = 0.052293 kg Dryness fraction at cut off and release = 0.4269, 0.3307 Heat leakage = 464.99 kJ/kg steam Ans. 9. Determine dryness fraction and missing quantity of steam in kg/hr at point of cut-off and point of release for the following data available from steam engine; Point of cut off: 30% of stroke At any point on compression curve: pressure = 4 bar, indicated volume = 0.15m3 At any point on expansion curve immediately after cut off: pressure = 12 bar Pressure at release: 5 bar Indicated volume at release: 0.5 m3 Bore and stroke: 60 cm and 120 cm Clearance volume: 10% of stroke volume Mass of steam admitted = 1.5 kg/stroke. Number of working strokes = 180 per minute. Also determine the percentage re-evaporation during expansion. Solution: Fig. 12.37 p-V diagram 548 ________________________________________________________ Applied Thermodynamics Let us approximate a point immediately after cut-off to be cut-off point. Vs = Stroke volume = 4 (0.6)2 1.2 = 0.339 m3 Clearance volume, V5 = 0.1 Vs = 0.0339 m3 Total volume of cylinder, V3 = V5 + Vs = 0.3729 m3 V1 = Volume at cut-off point V1 = V5 + 0.3 Vs V1 = 0.1356 m3 Pressure at cut-off point = 12 bar Volume at release,V2 = 0.5 m3 At the point on compression curve, point 4 The steam present in cylinder at state 4 shall act as cushion steam. Mass of cushion steam = 0 15. vg at 4 bar = 0 15.. 0 4625 = 0.3243 kg Total mass of steam during expansion = Cushion steam + Mass of steam admitted = 0.3243 + 1.5 = 1.8243 kg At cut-off point Dryness fraction at cut-off point, x1 = V 1 m v g at12 bar = 0 1356... 1 8243 0 16333 = 0.455 Missing quantity per hour = (1.8243 \u2013 1.8243 0.455) 180 60 = 10737.83 kg At point of release Dryness fraction at point of release, x2 = V 2 m v gat 5bar = 0.5.. 1 8243 0 3749 x2 = 0.731 Missing quantity per hour = (1.8243 \u2013 1.8243 0.731) 180 60 = 5299.96 kg Percentage re-evaporation during expansion \u2212 Missing quantity at cut off Missing quantity at release Missing quantity at cut-off = Percentage re-evaporation = 10737 83 5299.96. 10737 83. = 0.5064 or 50.64% Steam Engine _________________________________________________________________ 549 Dryness fraction at cut-off = 0.", "455 Dryness fraction at release = 0.731 Missing quantity at cut-off = 10737.83 kg/hr Missing quantity at release = 5299.96 kg/hr Percentage re-evaporation = 50.64% Ans. 10. A double acting compound steam engine has two cylinders. Steam is supplied at 1.5 MPa and 0.9 dry. Each cylinder has equal initial piston loads and the exhaust from engine occurs at 40 kPa into condenser. Diagram factor referred to low pressure cylinder is 0.8 and length of stroke of both cylinders is 38 cm. Bore of cylinders of HP and LP cylinders are 20 cm and 30 cm. Expansion occur completely in HP cylinder and engine runs at 240 rpm. Neglecting clearance volume determine. (i) intermediate pressure, (ii) indicated power output and (iii) steam consumption in kg/hr. Solution: On P-V diagram, HP engine: 1267, LP engine: 23456 For equal initial piston load, (P1 \u2013 P2) AHP = (P2 \u2013 P4) ALP = (P2 \u2013 40).2 0 2 4 4 (1.5 103 \u2013 P2) Solving, we get 0 3 2. Ans. Intermediate pressure = 192 kPa P2 = 192 kPa For expansion process 1-2-3, for hyperbolic expansion, Fig. 12.38 P-V diagram P1 V1 = P2 V2 V1 = P V 2 2 P 1 Due to no clearance, 550 ________________________________________________________ Applied Thermodynamics V2 = 4 (0.2)2 0.38 m3 192.2 0 2. 0 38 V1 = 4.5 1 Cut-off volume in HP cylinder = 1.53 10\u20133 m3 10 3 = 1.53 10\u20133 m3 Volume of LP cylinder = 4 (0.3)2 0.38 = 0.0268 m3 Expansion ratio throughout the engine = Volume of LP cylinder Cut-off volumein HP cylinder = 0.0268 1.53 10 3 = 17.52 Mean effective pressure for compound engine referred to LP cylinder mep = P 1 (1 + ln r) \u2013 P4 r 3 =.5 1 17 mep = 290.76 kPa 10.52 (1 + ln 17.52) \u2013 40 Actual mep = Hypothetical mep diagram factor referred to LP = 290.76 0.8 = 232.608 kPa Indicated power = mepactual L A N = 232.608 0.38 = 49.98 kW 4 (0.3)2 240 60 2 Volume of steam admitted per hour = V1 240 2 60 = 1.53 10\u20133 240 2 60 = 44.064 m3 Specific volume of steam being admitted, v1 = vf at1.5 MPa + 0.9 vfg at 1.5 MPa v1 = 0.1187 m3/kg Steam consumption, kg/hr =. 44 064. 0 1187 = 371.22 kg/hr Indicated power = 49.98 kW, Steam consumption = 371.22 kg/hr Ans. 11. A double acting compound steam engine has two cylinders. Steam is admitted at 1.4 MPa and is exhausted at 25 kPa. Engine runs at 240 rpm. Low pressure cylinder has bore of 60 cm and stroke of 60 cm. Diagram factor referred to LP cylinder is 0.8 while expansion is hyperbolic throughout. Expansion is complete in high pressure cylinder and the clearance volume may be Steam Engine _________________________________________________________________ 551 neglected. Total expansion ratio is 8 throughout the engine. Considering equal work to be produced by each cylinder determine, the indicated power output the diameter of high pressure cylinder considering same stroke length for both cylinders. the intermediate pressure. (i) (ii) (iii) Solution: Overall expansion ratio, r = 8 Hypothetical mep referred to LP cylinder = P 1 (1 + ln r) \u2013 P4 r. 1 4 10 8 = 513.9 kPa = 3 (1 + ln 8) \u2013 25 Fig. 12.39 P-V diagram Actual mep = Hypothetical mep d.f. referred to LP mepactual = 513.9 0.8 = 411.12 kPa Indicated power output = mepactual L A N = 411.12 0.6 4 (0.6)2 240 60 2 = 557.96 kW Total area of diagram = mepactual Volume of LP cylinder Since each HP and LP perform equal work so work done in HP cylinder = Total area of diagram 2 = 411 12 2 0 6. 4 0 6. 2 = 34.87 kJ 552 ________________________________________________________ Applied Thermodynamics HP cylinder work = P1 V1 ln V 2 V 1 = 34.87 kJ Also from expansion ratio, V 3 V 1 = 8 V1 = 4", " V3 V1 = 0.0212 m3 Substituting P1 = 1.4 103 kPa and V1 = 0.0212 m3, we get V2 0 0212 (1.4 103 0.0212) ln = 34.87. Thus, the volume of HP cylinder shall be 0.0686 m3. Let diameter of HP cylinder be \u2018d\u2019 so. V2 = 0.0686 m3 0.0686 = 4 (d2) 0.6 d = 0.3815 m or 38.15 cm From hyperbolic process consideration on process 1\u20132. P1 V1 = P2 V2 (1.4 103 0.0212) = P2 0.0686 P2 = 432.65 kPa Indicated power = 557.96 kW Diameter of HP cylinder = 38.15 cm Intermediate pressure = 432.65 kPa Ans. 12. A double acting compound steam engine has HP and LP cylinders to produce output of 250 kW. Steam is admitted at 1.5 MPa and exhausted at 25 kPa. Overall expansion ratio referred to LP cylinder is 12. Diameter of LP cylinder is 40 cm and stroke length in both cylinders is 60 cm each. Diagram factor is 0.75. Expansion ratio in H.P. cylinder is 2.5. Neglecting clearance volume and considering expansion to be hyperbolic, determine the speed of engine in rpm and diameter of HP cylinder. Solution: Hypothetical mep referred to LP cylinder,.5 1 12 mephypothetical = (1 + ln 12) \u2013 25 10 3 mephypothetical = 410.61 kPa Actual mep = 410.61 0.75 mepactual = 307.96 kPa Power output = 250 = mepactual L A N 2 N 60 250 = 307.96 0.6 (0.4)2 4 N = 323 rpm 2 Steam Engine _________________________________________________________________ 553 Speed of engine = 323 rpm Ans. Fig. 12.40 P-V diagram Volume of low-pressure cylinder = 4 (0.4)2 0.6 = 0.0754 m3 = V3 = V4 Cut-off volume in HP cylinder = = 6.28 10\u20133 m3 Total volume in HP cylinder = 6.28 10\u20133 2.5 = 0.0157 m3 Let diameter of HP cylinder be \u2018d\u2019 meter, then. 0 0754 12 4 d2 0.60 = 0.0157 d = 0.1825 m or 18.25 cm Diameter of HP cylinder = 18.25 cm Ans. 13. A triple expansion steam engine has HP, IP and LP cylinder diameters of 25 cm, 40 cm and 85 cm with actual mean effective pressures of 0.5 MPa, 0.3 MPa and 0.1 MPa. Steam is supplied at 1.5 MPa and the exhaust finally occurs at 25 kPa. The cut-off in HP cylinder occurs at 60% of the stroke. Considering all the cylinders to have some stroke length determine, (i) (ii) (iii) the actual and hypothetical mep referred to LP cylinder, the overall diagram factor the contribution of each cylinder in overall output. Neglect clearance volume. Solution: In HP cylinder, mep of HP referred to LP = 0.5 103 A A HP LP = 0.5 103 4 4 0.25 2 0 85. 2 = 43.25 kPa 554 ________________________________________________________ Applied Thermodynamics In IP cylinder, mep of IP referred to LP = 0.3 103 A A IP LP = 0.3 103 4 4 0 40. 2 0 85. 2 = 66.44 kPa For complete engine the overall mep referred to LP cylinder = 43.25 + 66.44 + 0.1 103 = 209.69 kPa Volume of LP cylinder = Volume of HP cylinder = 4 4 (0.85)2 L = 0.567 L (0.25)2 L = 0.049 L Cut-off volume of HP cylinder = 0.6 Volume of HP cylinder = 0.0294 L For zero clearance volume, Overall expansion ratio = Volume of LP cylinder Cut off volume of HP cylinder = 0.567 L 0.0294 L r = 19.28 Hypothetical m.e.p overall = 3 1.5 10 19.28 (1 + ln 19.28) \u2013 25 = 283.02 kPa Overall diagram factor = Actual m.e. p. Hypothetical m.e. p. = 209 69.. 283 02 = 0.741 Percentage of total indicated power developed in HP, IP and LP cylinder. Since stroke length for each cylinder is same so the ratio of indicated power output of each cylinder to total output shall be the ratio of m.e.p of specific cylinder referred to LP cylinder and", " overall mep referred to LP cylinder % of HP cylinder output = 100 = 20.63% % of IP cylinder output = % of LP cylinder output = 100 = 31.68% 100 = 47.69%.25 43 209 69. 66.44 209 69. 0 1 10.. 209 69 3 Actual mep referred to LP = 209.69 kPa Hypothetical mep referred to LP = 283.02 kPa Overall diagram factor = 0.741 % of HP, IP and LP cylinder outputs = 20.63%, 31.68%, 47.69% Ans. Steam Engine _________________________________________________________________ 555 14. A double acting compound steam engine has steam supplied at 7 bar and exhausted at 0.25 bar. HP cylinder: diameter = 25 cm cut-off point = 30% stroke clearance volume = 10% of swept volume LP cylinder: diameter = 50 cm cut-off point = 45% stroke clearance volume = 5% of swept volume Fig. 12.41 Diagram factors of HP and LP cylinder are 0.8 and 0.7 respectively. Considering hyperbolic expansion and neglecting cushioning effect estimate mep in each cylinder and total hp developed for engine running at 100 rpm. Take account of clearance volumes. Solution: Let the common stroke of cylinders be \u2018L\u2019 m. Assuming the expansion curve to be continu- ous the P-V diagram is shown here. Volume of HP cylinder = 4 (0.25)2 L = 0.04908 L m3 Clearance volume = 0.1 0.04908 L = 0.004908 L m3 = V9 Total volume of HP = 0.04908 L + 0.004908 L = 0.053988 L m3 = V2 Volume at cut-off in HP cylinder = (0.004908 L) + (0.3 0.04908 L) = 0.019632 L m3 = V1 For LP cylinder Volume of LP cylinder = 4 (0.5)2 L = 0.1963 L m3 Clearance volume of LP cylinder = 0.05 0.1963 L = 9.815 10\u20133 L m3 = V7 556 ________________________________________________________ Applied Thermodynamics Total volume of LP cylinder = (0.1963 L + 9.815 10\u20133 L) = 0.206115 L m3 = V5 Volume at cut-off of LP cylinder = 9.815 10\u20133 L + (0.45 0.1963 L) = 0.09815 L m3 = V3 Expansion ratio for HP cylinder, rHP = Expansion ratio for LP cylinder, rLP = 0 053988.. 0 019632 L L = 2.75.206115 0. 0 09815 L L = 2.1 P1 V1 = P2 V2 = P3 V3 = P4 V4 P1 V1 = P3 V3 (7 102 0.019632 L) = (P3 0.09815 L) or, Actual m.e.p. for HP cylinder p3 = 140.01 kPa ln HP r P V P 1 2 3 Stroke volume of HP 0 019632 1 ln. 7 10 2 75 2. 0 04908. L 1 P V 1 1 = 0.8 7 10 2 = 0.8 = 282.59 kPa P 3 V 9 140 01 0 053988. 140 01 0 004908.. L. Actual mep of LP cylinder = 0.7 = 0.7 P V 3 3 P 3 V 7 ln LP r 1 P 5 Stroke volume of LP ln. 2 1 140 01 25. 0.1963 L 140 01 0 09815 1.. 25 L.206115 0 9 815 10 3. L = 62.96 kPa Actual mep of HP referred to LP cylinder = 282.59.. 0 1963 0 04908 L L Total mep = 62.96 + 70.65 = 133.61 kPa Total output = Total mep L A N = 70.65 kPa. 133 61 L = 4 60 0.5 2 100 = 43.72 L kW Steam Engine _________________________________________________________________ 557 mep of HP referred to LP = 70.65 kPa mep of LP = 62.96 kPa Total output = 43.72 L kW where L is stroke length Ans. 15. During the trial of a single-cylinder double acting steam engine following is observed. Duration of trial = 15 min, Bore = 25 cm, Stroke = 30 cm, Brake diameter = 1.5 m, Net brake load = 300 N, Speed = 240 rpm, Steam pressure = 10 bar, Dryness fraction = 0.9, Cover end mean effective pressure = 0.9 bar, Crank end mean effective pressure = 0.9 bar, Steam utilized = 15 kg. Condens", "ate temperature = 45C. Amount of water circulating = 450 kg/hr. Rise in cooling water temperature = from 20C to 30C. Determine, (i) steam used per ihp. hr, (ii) mechanical efficiency, (iii) brake thermal efficiency, (iv) heat balance sheet on one hour basis. Solution: From given data; Average value of mean effective pressure = 1 2 (mepcover end + mep crank end) mep = 0.9 bar Steam consumption per hour = Indicated horse power = 60 = 60 kg/hr 15 15 mep L A N. 0 7457 60 2,{as 1 HP = 0.7457kW} 0.9 2 10. 0 30 60 =.25 0 2 240 2 4. 0 7457 = 14.22 hp or 10.604 kJ/s or 38174.4 kJ/hr Therefore, steam used per ihp. hr = 60 14.22 = 4.22 kg/ihp. hr Steam used per ihp. hr = 4.22 kg/ihp. hr Ans. Brake power can be obtained using the dynamometer data given i.e. brake load, brake drum diameter etc. Brake horse power = \u03c0 2 \u22c5 N T \u00d7 \u00d7 60 0.7457 10 = 3 2 60 240 300 0 7457 103. 0 75. Brake horse power = 7.58 hp or 5.65 kJ/s or 20340 kJ/hr Mechanical efficiency = Brake horse power Indicated horse power = 0.5331 or 53.31% Heat available with steam/hr = 60 [hat 10 bar, 0.9 dry \u2013 hat 45C] = 60 [(762.81 + 0.9 2015.3) \u2013 (4.18 45)] = 143308.8 kJ/hr 558 ________________________________________________________ Applied Thermodynamics Brake thermal efficiency = Heat equivalent of bhp Heat available with steam / sec = 7.58. 0 7457. 143308 8 3600 Heat carried away by circulating water in condenser = 0.14199 or 14.19% = 450 4.18 [30 \u2013 20] = 18810 kJ/hr Heat balance sheet on one hour basis Input k J % Output Heat available with steam/hr 143308.8 100% Total 143308.8 100% bhp fhp(= ihp - bhp) Heat carried away by circulating water Unaccounted losses (by difference) Total k J 20340 17834.4 18810 86324.4 % 14.19 12.44 13.13 60.24 143308.8 100% 16. During trial of steam engine following observation are made; Bore:38 cm, Stroke 50 cm, Piston rod diameter: 5 cm, Speed:150 rpm, Steam consumption: 36 kg/min, Brake load = 7kN at 2 m brake diameter. Area of indicator diagram at cover end: 28 cm2 Area of indicator diagram at crank end: 26cm2 Length of indicated diagram: 7 cm Spring scale: 15 kPa/mm Determine brake power, indicated power, mechanical efficiency, brake specific steam consumption and indicated specific steam consumption. Solution: M.E.P.crank = M.E.P.cover = \u00d7 2600 15 70 \u00d7 2800 15 70 = 557.14 kPa = 600 kPa Indicated power at crank end, IPcrank = MEPcrank \u00d7 L \u00d7 Acrank \u00d7 N At crank end piston area will be reduced due to piston rod, IPcrank = 557.14 \u00d7 50 \u00d7 10\u20132 \u00d7 \u03c0 4 {(0.38) 2 \u2212 (0.05) } 2 \u00d7 150 60 IPcrank = 77.62 kW Indicated power at cover end, IPcover = MEPcover \u00d7 L \u00d7 Acover \u00d7 N IPcover = 600 \u00d7 50 \u00d7 10\u20132 \u00d7 \u03c0 4 \u00d7 (0.38) 2 150 \u00d7 60 Steam Engine _________________________________________________________________ 559 IPcover = 85.06 kW Indicated power = IPcrank + IPcover = 162.68 kW Brake power, BP = 2\u03c0NT = 2 \u00d7 \u03c0 \u00d7 150 60 \u00d7 \u00d7 = 109.96 kW 7 1 Mechanical efficiency, \u03b7 mech = BP IP = 109.96 162.68 = 0.6759 = 67.59% Ans. Ans. Indicated specific steam consumption, ISFC = = Steam consumption IP \u00d7 36 60 162.68 Indicated specific steam consumption = 13.28 kg/kWh Ans. Brake specific steam consumption, BSFC = Steam consumption BP Brake specific steam consumption = \u00d7 36 60 109.96 = 19.64 kg/kwh 17. During trial of single cylinder, double acting steam engine of condensing type following observations were made; Bore: 24 cm, Stroke: 34 cm; Engine speed: 150 rpm, Piston rod diameter: 5cm, Bra", "ke load: 120kg. Spring balance reading: 100N. Brake wheel drum diameter: 100cm; Steam inlet state: 15 bar, 0.98 dry, Mean effective pressure at cover end: 1.8 bar Mean effective pressure at crank end: 1.6 bar Cooling water flow through condenser: 42 kg/min Rise in temperature of cooling water: 20\u00b0C Condensate discharged from condensor: 4 kg/min Temperature of condensate: 50\u00b0C Determine brake power, indicated power, mechanical efficiency, brake thermal efficiency, indicated steam consumption and energy balance sheet on minute basis. Solution: Brake power = 2\u03c0N(Brake load \u2013 Spring balance reading) \u00d7 Drum radius Brake power = \u03c0 \u00d7 2 \u00d7 150(120 9.81 100) \u00d7 1000 60 \u2212 \u00d7 (100 / 2) 10 \u00d7 \u2212 2 = 8.46 kW Indicated power, cover end = MEPcover \u00d7 L \u00d7 Acover \u00d7 N = \u00d7 1.8 10 2 \u00d7 0.34 \u03c0 \u00d7 \u00d7 4 (0.24) 4 150 \u00d7 60 560 ________________________________________________________ Applied Thermodynamics Indicated power, crank end = MEPcrank \u00d7 L \u00d7 Acrank \u00d7 N = 6.92 kW = \u00d7 1.6 10 2 \u00d7 0.34 \u03c0 \u00d7 \u00d7 4 {(0.24) 2 \u2212 (0.05) } 2 \u00d7 150 60 = 5.89 kW Total indicated power = 6.92 + 5.89 = 12.81 kW Mechanical efficiency = Brake power Indicated power = 8.46 12.81 = 0.6604 or 66.04% Enthalpy of steam at inlet, Enthalpy of condensate, hs,in = hf at 15 bar + 0.98. hfg at 15 bar = 844.89 + (0.98 \u00d7 1947.3) hs,in = 2753.24 kJ/kg hcond = hf at 50\u00b0C = 209.33 kJ/kg Energy supplied by steam = hs,in \u2013 hcond = 2753.24 \u2013 209.33 = 2543.91 kJ/kg Steam consumption rate = 4 kg/min = 240 kg/hr Brake thermal efficiency = 3600 m steam BP \u00d7 ( h s in, \u2212 h cond ) = \u00d7 3600 8.46 240 2543.91 \u00d7 = 0.0498 or 4.98% Ans. Indicated steam consumption = steamm IP = 240 12.81 = 18.74 kg/kWh Ans. Energy balance sheet on per minute basis. Energy supplied per minute, Qin = 4(hs,in \u2013 hcond) = 4 \u00d7 2543.91 = 10175.64 kJ/min Energy consumed by brake or available at shaft Energy consumed by cooling water in condenser, Q1 = 8.46 \u00d7 60 = 507.6 kJ/min Q2 = mwater \u00d7 Cp,water \u00d7 \u2206T = 42 \u00d7 4.18 \u00d7 20 = 3511.2 kJ/min Energy going alongwith condensate, Q3 = msteam \u00d7 Cp \u00d7 \u2206T = 4 \u00d7 4.18 \u00d7 50 = 836 kJ/min Steam Engine _________________________________________________________________ 561 Engery loss to surroundings, Q4 = Qin \u2013 (Q1 + Q2 + Q3) = 10175.64 \u2013 (507.6 + 3511.2 + 836) = 5320.84 kJ/min Energy Balance sheet on per minute basis Energy in, kJ/min Energy out, kJ/min Energy supplied = 10175.64 Energy available at shaft = 507.6 Energy consumed by cooling water = 3511.2 Energy lost with condensate = 836 Energy loss to surroundings = 5320.84 Total: 10175.64, kJ/min Total: 10175.64 kJ/min 12.1 Classify steam engines. 12.2 Give a neat sketch of simple steam engine and explain its working. 12.3 Describe with sketches any five different parts of steam engines. 12.4 Why is it necessary to provide guides for a cross-head? 12.5 Describe the function of crank. 12.6 What is the eccentric? Explain its utility? 12.7 Describe the working of D-slide valve. 12.8 Describe hypothetical and actual indicator diagrams for a simple steam engine. 12.9 What are factors responsible for actual indicator diagram differing from hypothetical diagram? 12.10 Describe significance of diagram factor. 12.11 What do you understand by missing quantity of steam? 12.12 Why the cylinder of steam engine is generally steam jacketed? 12.13 What do you understand by compound steam engines? 12.14 Describe various types of compounding in steam engines and their relative merits and demerits. 12.15 Show the hypothetical and actual P-V diagram for compound steam engine having", " HP and LP cylinders. 12.16 Discuss the relevance of mep referred to LP cylinder in compound steam engines. 12.17 Discuss different methods of governing of steam engines. 12.18 Explain Willan\u2019s line. 12.19 What do you understand by unresisted expansion in reference to compound steam engines? 12.20 Discuss different types of governing of compound steam engines with p-V diagrams. 12.21 A single cylinder double acting steam engine is supplied with steam at 5 bar and steam is exhausted at 0.34 bar. Cut-off occurs at one-third of stroke. The swept volume of cylinder is 0.05 m3 and diagram factor is 0.85. The engine delivers brake power of 37.5 kW at the speed of 120 rpm. Determine the mechanical efficiency of engine. [85%] 12.22 In a steam engine the steam is supplied at 20 bar, 340C and is expanded adiabatically up to 1.5 bar after which there occurs instantaneous pressure drop at constant volume up to 0.07 bar. Determine the work done per kg of steam supplied and the modified Rankine cycle efficiency. [694.72 kJ, 23.5%] 12.23 Determine the indicated power output from a double acting single cylinder steam engine having stroke of 30 cm and stroke to bore ratio of 1.3. The steam is admitted at 67.45 bar and the back 562 ________________________________________________________ Applied Thermodynamics pressure is 1.3 bar with expansion ratio of 2.5. The engine runs at 210 rpm and the diagram factor [27 kW] is 0.8. 12.24 A single cylinder double acting steam engine has steam being admitted at 10 bar, 0.96 dry and exhausted at 0.5 bar. The expansion occurs up to 1.64 bar. Power output from engine is 60 hp at 210 rpm. The cylinder stroke volume is 0.021 m3. Determine the diagram factor. [0.75] 12.25 A single cylinder double acting steam engine has bore of 25 cm and stroke to bore ratio of 1.2. What shall be the indicated power output from engine if steam is admitted at 10.35 bar and discharged at 0.34 bar. The diagram factor is 0.81 and cut off occurs at 50% of stroke. [94.27 hp] 12.26 In a double acting steam engine the steam is supplied at 6.5 bar and 0.85 dry and exhausted at 0.28 bar. The engine runs with speed of 150 rpm. The diameter of piston is 500 mm and piston rod has diameter of 64 mm. The expansion ratio at the front end and back end of piston are 10 and 8 respectively. Stroke length is 600 mm. Steam is being supplied at 6.8 kg per ihp. hr. Steam leaving engine is condensed and the feed water at 35C is sent back. Determine thermal efficiency [19%, 159.6 hp] and total indicated horse power of engine. 12.27 A steam engine has throttle governing such that it develops 37.5 kW with steam consumption of 1000 kg/hr. The steam consumption at no load is 125 kg/hr. Determine the steam consumption [750 kg/hr] for the indicated power developed being 25.8 kW. 12.28 Determine the missing quantity of steam at cut-off, release and percentage reevaporation for the double acting steam engine having following details. Steam pressure at cut-off = 6 bar, the cut-off point is at distance of 38 cm from zero. Steam pressure at release = 3 bar, the point of release is at distance of 8.3 cm from zero. Steam pressure at beginning of compression = 1.05 bar, the point of start of compression is at distance of 1.3 cm from zero. Clearance length = 0.86 cm Speed of engine = 150 rpm Steam consumption = 270 kg/hr Steam is supplied dry saturated. Mean stroke volume = 7300 cm3 [0.005 kg, 0.003 kg, 32.4%] 12.29 In a two cylinder double acting compound steam engine the stroke in each cylinder is 50 cm and the indicated power output is 220 kW at speed of 270 rpm. Steam is supplied at 12 bar and exhausted at 0.28 bar. The total expansion ratio through the engine is 10 and the two cylinders develop equal power. Neglecting clearance volume and assuming expansion to be complete in HP cylinder and hyperbolic throughout determine the diameters of HP and LP cylinder. [37.3 cm, 47.5 cm] 12.30 A double acting compound steam engine has HP and LP cylinder with steam being admitted at 17.25 bar and exhausted at 0.415 bar. Two cylinders have common stroke length with LP cylinder having bore of 45 cm and stroke of 40 cm and HP cylinder bore is 28.5 cm. Engine runs at 270 rpm and the total expansion ratio is 9. The diagram factor of engine referred to LP cylinder is 0.78. Considering hyperbolic expansion", ", zero clearance and each cylinder having equal initial piston loads determine the intermediate pressure and indicated power of engine. [5.23 bar, 364.75 hp] 12.31 A double acting compound steam engine has stroke length of 40 cm for both HP and LP cylinders. The bore of HP and LP cylinders are 30 cm and 60 cm respectively. The steam is supplied at 1 3 rd of stroke. The engine develops brake power of 125 kW while running at 160 rpm. Considering mechanical efficiency of 0.78 determine the actual and hypothetical mean effective pressures, [2.66 bar, 3.73 bar, 0.7] overall diagram factor. pressure of 13.8 bar and the back pressure is 0.28 bar. The cut-off in HP cylinder occurs at Steam Engine _________________________________________________________________ 563 12.32 Determine the overall diagram factor and the actual mean effective pressure of a triple expansion engine having following details: Steam admission pressure = 13.8 bar, Back pressure = 0.28 bar, Cut-off in HP = 60% of stroke Bore (cm) Indicator Diagram area (mm2) Indicator diagram length (mm) Indicator constant (kg/mm2/m) HP 30 1805 104 40 IP 45 1875 105 16 LP 75 2000 103 5.35 12.33 A compound steam engine has following details; [0.76, 3.18 bar] No. of cylinders = 2, HP and LP. HP cylinder piston area = 480 cm2 front and 450 cm2 back end. LP cylinder piston area = 1920 cm2 front and 1890 cm2 back end. Mean effective pressure of HP = 2.8 bar front and 2.67 back end. Mean effective pressure of LP = 0.77 bar front and 0.84 back end. Stroke length = 92 cm for both HP and LP. Steam admission = 10.5 bar, dry saturated. Temperature of hot well = 48C Speed of engine = 198 rpm Steam consumption = 1225 kg per hour Condenser circulating water requirement = 23400 kg/hr Circulating water temperature rise = From 11C to 36C Draw up a heat balance for engine on per minute basis and also determine thermal efficiency. [Thermal efficiency = 17.7%] 564 ________________________________________________________ Applied Thermodynamics 13 13.1 INTRODUCTION A nozzle is a flow passage of varying cross sectional area in which the velocity of fluid increases and pressure drops in the direction of flow. Thus in nozzle the fluid enters the variable cross section area duct with small velocity and high pressure and leaves it with high velocity and small pressure. During flow through nozzle the enthalpy drops and heat drop in expansion is spent in increasing the velocity of fluid. Similar to nozzle a duct with variable cross-section area will be called diffuser if the fluid gets decelerated, causing a rise in pressure along the direction of flow. Nozzles are generally used in turbines, jet engines, rockets, injectors, ejectors etc. Fig. 13.1 General arrangement in nozzle and diffuser Here in this chapter the one-dimensional analysis of nozzle has been considered. Different types of nozzles, thermodynamic analysis and different phenomenon in nozzles are discussed ahead. Momentum transfer across the control volume may be accounted for as, [Time rate of momentum transfer across the control volume] = mC Newton\u2019s law states that resultant force F acting on the control volume equals the difference between the rates of momentum leaving and entering the control volume accompanying mass flow. Momentum equation says; F = m Since at steady state, m 2 C2 \u2013 m 2 = m F = m (C2 \u2013 C1) 1 C1 1 i.e. continuity equation being satisfied The resultant force F includes forces due to pressure acting at inlet and exit, forces acting on the portion of the boundary through which there is no mass flow, and force due to gravity. Nozzles _______________________________________________________________________ 565 13.2 ONE DIMENSIONAL STEADY FLOW IN NOZZLES Here one dimensional steady flow analysis in nozzle is carried out assuming the change in crosssectional area and axis to be gradual and thermodynamic properties being uniform across planes normal to axis of duct. In general real flow through nozzle is not truly one-dimensional but this assumption offers fairly correct analysis. Flow through nozzle occurs at comparatively high velocities in many engineering applications and so exhibits changes in fluid density. Therefore, it is necessary to first look at the compressible flow preliminaries. Compressible flow preliminaries: Let as consider compressible flow and obtain momentum equa- tion for one dimensional steady flow. Fig. 13.2 The one dimensional steady flow through a duct is shown above. For control volume shown the principle of conservation of mass, energy and Newton\u2019s law of motion may be applied. By Newton\u2019s law of motion, F = m a where F is the resultant force acting on system of mass \u2018m\u2019 and acceleration \u2018a\ufffd", "\ufffd. Pressure waves and Sound waves: Let us consider a cylindrical duct having piston on one end for generating the pressure wave. Figure 13.3 shows the arrangement for producing a pressure wave moving to right with velocity \u2018a\u2019. Sound wave is also a small pressure disturbance that propagates through a gas, liquid or solid with velocity \u2018a\u2019 that depends on the properties of medium. Fig. 13.3 Propagation of pressure wave (sound wave) Figure 13.3 shows how the generation of pressure wave causes displacement of fluid thereby causing rise in pressure, density and temperature by p, and T in respect to the region on the right of wave (undisturbed region). In the undisturbed region say pressure, density, temperature and fluid velocity be p,, T and C = 0 respectively. Due to piston movement fluid velocity increases by C and other properties also change by elemental values as shown. For analysing there are two approaches available as shown in Figs. 13.3 (a) and (b). One approach considers observer to be stationary and gas moving and second approach considers observer to be moving along with wave i.e. relative velocity of observer with respect to wave is zero. 566 ________________________________________________________ Applied Thermodynamics Respective values of fluid velocity, wave propagation velocity, pressure, density and temperature are labelled on figure. For an observer at rest relative to wave (observer moving with wave) it seems as if the fluid is moving towards the stationary wave from right with velocity a, pressure p, density and temperature T and moves away on left with velocity \u2018a \u2013 C\u2019, pressure \u2018p + p\u2019, density \u2018 + \u2019 and temperature \u2018T + T\u2019. From conservation of mass, applying continuity equation upon control volume we get m 1 = m 2 = m Aa = ( + )A(a \u2013 C) where A is constant cross section area of duct. Aa = (Aa) \u2013 (AC) + (Aa) \u2013 (AC) Upon neglecting higher order terms and rearranging we get, (a) \u2013 (C) = 0 C = a or, Applying momentum equation to the control volume; (pA) \u2013 ((p + p)A) = (m (a \u2013 C)) \u2013 (ma) \u2013 p A = m(\u2013 C) for mass flow rate m we can write, m = Aa so, pA = AaC or, C = p a Equating two values obtained for \u2018C\u2019 we get a = p a a = p Thus, velocity of wave propagation comes out as the square root of the ratio of change in pressure and change in density. In case of sound waves the magnitude of changes in pressure, density and temperature are infinitesimal and so these may also be called as infinitesimal pressure wave. It is also seen that thermodynamic process occurring across an infinitesimal pressure wave may be considered nearly isentropic. Therefore the velocity of sound can be given as square root of derivative of pressure with respect to density across the wave under isentropic conditions. a = \uf8eb \u2202 p \uf8ec \u2202 \u03c1\uf8ed \uf8f6 \uf8f7 \uf8f8 = s constt. in terms of specific volume values; d = \u2212 dv 2 v so, a = \u2212 2 v \uf8eb \u2202 \uf8ec \u2202\uf8ed p v \uf8f6 \uf8f7 \uf8f8 = s constt Nozzles _______________________________________________________________________ 567 Let us consider fluid to be a perfect gas following isentropic process given by pvk = constt. Taking log of both sides and then partially differentiating we get, p v s = k p v Substituting in expression for sound velocity a = k p v For ideal gas, a = k R T. In case of air, a = RT Using the velocity of sound and fluid velocity a non dimensional parameter called Mach number can be defined. Mach number is given by the ratio of velocity of fluid (object) to the velocity of sound. It is generally denoted by M. M = C a Based upon Mach no. value flow can be classified as given below. For M < 1 M = 1 M > 1 flow is called subsonic flow. flow is called sonic flow. flow is called supersonic flow. Nozzle flow analysis: Let us consider one dimensional steady flow in nozzles. Let us take a varying cross-section area duct such that velocity increases and pressure decreases from inlet to exit. From conservation of mass, upon applying continuity equation, it can be given that, AC = constant Fig. 13.4 Taking log of both the sides, ln + ln A + ln C = ln constant Differentiating partially we get Let us now apply steady flow energy equation assuming no change in potential energy, no heat interaction and no work interaction", " during the flow across control volume. S.F.E.E. yields, dq = dh + dw + d(KE) + d(PE) Applying assumptions, 568 ________________________________________________________ Applied Thermodynamics dh + d(KE) = 0 It can be rewritten for section 1 and 2 as, KE C2 2 or h1 + = h2 + 2 C1 2 ho1 = ho2 2 C2 2 Stagnation enthalpy at section 1 = Stagnation enthalpy at section 2. C2 2 From differential form, dh + d = 0 dh + 2 C dC 2 = 0 or, dh = \u2013 CdC From first and second law combined we know, dh = Tds + vdp Using the adiabatic flow considerations, ds = 0, so dh = vdp = d p Above shows that with increase or decrease in pressure along the direction of flow the specific enthalpy also change in same way. From thermodynamic property relations pressure can be given as function of density and entropy i.e. p = p(, s). or, dp = For isentropic flow considerations dp = We know from sound velocity a = p p d + p s ds d s s p so, s dp = a2d Combining two expressions for dh we get \u2013 C dC = d p. This shows that as pressure increases in direction of flow then velocity must decrease. Substituting from dp as obtained above, it yields, \u2013 C dC = a d2 Nozzles _______________________________________________________________________ 569 or, dC C = d d 2 = C 2 a dC C 2 2 a C Substituting above in the equation available from continuity equation, d d A A dC C = 0 dA A \u2212 = = or, or d A A = As Mach no. M = dC C C a so, \u03c1 dC dC C 2 C 2 a 1 d A A = dC C {M 2 \u2013 1} Using above relation the effect of area variation upon the flow can be seen in subsonic, sonic and super sonic flow regimes. Case 1 For subsonic flow i.e. M < 1 Nozzle: For positive velocity gradient i.e. velocity increases along the direction of flow as in case \uf8eb \uf8ec \uf8ed of nozzle d C C along the direction of flow. + is ve \uf8f6 \uf8f7 \uf8f8, it yields d A A as \u2013 ve. Negative area gradient means cross section area decreases Diffuser: For negative velocity gradient tion of flow as in case of diffuser, it yields d A A cross section area along the direction of flow. Fig. 13.5 \uf8eb \uf8ec \uf8ed i.e., d C C \u2212 is ve \uf8f6 \uf8f7 \uf8f8 the velocity decreases along the direc- as + ve. Positive area gradient means duct has diverging 570 ________________________________________________________ Applied Thermodynamics Case 2 For supersonic flow i.e. M > 1 Nozzle: For positive velocity gradient i.e. dC C supersonic flow the nozzle duct shall have diverging cross-sectional area along the direction of flow. being + ve, it yields as + ve. It means that in d A A Fig. 13.6 Diffuser: For negative velocity gradient i.e. dC C being \u2013 ve it yields d A A as \u2013 ve. It means in supersonic flow the diffuser duct shall have converging cross-sectional area along the direction of flow. From above discussion it can be concluded that (i) Nozzle must be of convergent duct type in subsonic flow region and such nozzles are called subsonic nozzles or convergent nozzles. (ii) Nozzle must be of divergent duct type in supersonic flow region and such nozzles are called supersonic nozzles or divergent nozzles. (iii) For acceleration of fluid flow from subsonic to supersonic velocity the nozzle must be first of converging type till flow becomes sonic and subsequently nozzle should be of diverging type in supersonic flow. The portion of duct at which flow becomes sonic (M = 1) and dA is zero i.e. duct is constant cross-section area duct, is called throat. Thus in this type of flow from subsonic to supersonic the duct is of converging type followed by throat and a diverging duct. Such nozzles are also called convergent-divergent nozzles. Throat gives the minimum cross-section area in convergent-divergent nozzles. Let us consider the expansion through a nozzle between sections 1 and 2. In nozzle the velocity of fluid is so high that there is no time available for heat exchange with the surroundings and the expansion may be considered", " adiabatic. Also the change in potential energy may be negligible if the elevation does not change from inlet to exit. Work done during flow is absent. Application of steady flow energy equation yields, h1 + 2 C1 2 = h2 + 2 C2 2 Velocity at exit from nozzle: C2 = 2 1 h h 2 For negligible velocity of fluid at inlet to nozzle, C1 0, m/s 2 C 1 C2 = 2 1 h, m/s, where h1 and h2 are enthalpy in J/kg at sections 1 and 2 respectively. h 2 Nozzles _______________________________________________________________________ 571 Expansion of fluid on p-v diagram is shown below. Expansion of gases on T-s diagram is as shown in Fig. 13.8. Fig. 13.7 P-V diagram for flow through nozzle Fig. 13.8 T-s diagram for flow through nozzle Expansion of steam on T-s and h-s diagram for superheated steam and wet steam is shown by 1\u20132 and 3\u20134 respectively under ideal conditions. Fig. 13.9 T-s and h-s representation for steam flow through nozzle In above representations the isentropic heat drop shown by 1\u20132 and 3\u20134 is also called \u2018Rankine heat drop\u2019. Mass flow through a nozzle can be obtained from continuity equation between sections 1 and 2. m = A C 1 1 v 1 Mass flow per unit area; m A 2 = From different from of S.F.E.E or, dh + d(K.E.) = 0 dq = dh + dw + d(K.E.) + d(P.E.) 572 ________________________________________________________ Applied Thermodynamics du + pdv + vdp + d(K.E.) = 0 also as dq = du + pdv = 0, so d(K.E.) = \u2013 vdp or C 2 2 C2 For the expansion through a nozzle being governed by process pvn = constt., C2 2 \u2013 C1 2 = 2 n n 1 p1 v1 1 p v 2 2 p v 1 1 or, Velocity at exit from nozzle C2 = For negligible inlet velocity, say C1 0 Velocity at exit from nozzle C2 = If the working fluid is perfect gas then n = and for air = 1.4. However, if working fluid is steam a good approximation for n can be obtained from some polytropic considerations. For steam being dry saturated initially and process of expansion occurring in wet region the index n can be approximated as 1.135. For steam being initially superheated and expanded in superheated region the index n can be approximated as 1.3. Looking at mathematical expression for exit velocity it could be concluded that maximum exit velocity is possible only when fluid is expanded upto zero pressure. The maximum velocity is, Cmax = 2 n 1 n p v 1 1 Mass flow rate, m = A C 2 2 v 2 Mass flow rate per unit area, 2 \uf8eb n \uf8ec \u2212\uf8ed \uf8eb \u2212\uf8ec = From expansion\u2019s governing equation, p1 v1 v2 = or = p2 v2 Nozzles _______________________________________________________________________ 573 or, or This expression for mass flow rate through nozzle depends upon inlet and exit pressures, initial specific volume and index of expansion. It has been seen earlier that the mass flow per unit area is maximum at throat and nozzle should be designed for maximum discharge per unit area. Thus there will be some value of throat pressure (p2) which offers maximum discharge per unit area. Mathematically this pressure value can be obtained by differentiating expression for mass flow per unit area and equating it to zero. This pressure at throat for maximum discharge per unit area is also called \u2018critical pressure\u2019 and pressure ratio with inlet pressure is called \u2018critical pressure ratio\u2019. Let pressure ratio p 2 p 1 = 2 = r, then mass flow per unit area can be re-written as dr dr Here p1, v1 are inlet conditions and remain constant. Also n being index of expansion remains constant so differentiating and putting equal to zero. 2 n r n 2 n or/ or, Critical pressure ratio, n \u2212 )1 ( n r = \uf8eb 2 \uf8ec +\uf8ed n \uf8f6 \uf8f7 \uf8f8 1 Let critical pressure at throat be given by pc or pt then, n \u2212 )1 ( n cp p = 1 \uf8eb 2 \uf8ec +\uf8ed n \uf8f6 \uf8f7 \uf8f8 1 n \u2212 )1 ( n tp p = 1 \uf8eb 2 \uf8ec +\uf8ed n \uf8f6 \uf8f7 \uf8f8 1 574 ________________________________________________________ Applied Thermodynamics Here subscript \u2018c\u2019 and \u2018t\u2019 refer to critical and throat", " respectively. While designing a nozzle the critical pressure ratio at throat is equal to the one obtained above. Critical pressure ratio value depends only upon expansion index and so shall have constant value. Value of adiabatic expansion index and critical pressure ratio are tabulated ahead; Table 13.1: Adiabatic expansion index and critical pressure ratio for selected fluids Fluid Adiabatic expansion index, n Wet steam 1.135 (n = 1.035 + 0.1x, where x is dryness fraction of wet steam) Superheated steam Air 1.3 1.4 Critical pressure ratio.577 0.545 0.528 The maximum discharge per unit area can be obtained by substituting critical pressure ratio in expression for mass flow per unit area at throat section. m At At At Maximum discharge per unit area = For this maximum discharge per unit area at throat the velocity at throat can be obtained for critical pressure ratio. This velocity may also be termed as \u2018critical velocity\u2019. At throat C2 = 2 \uf8eb n \uf8ec \u2212\uf8ed n 1 Ct = 2 \uf8eb n \uf8ec \u2212\uf8ed \uf8f6 \uf8f7 \uf8f8 Nozzles _______________________________________________________________________ 575 2 n 1 n p v t t Ct = Substituting critical pressure ratio n 1 n p v t t Ct = Ct = n p vt t Hence, critical velocity = n p vt t For perfect gas; Ct = n RTt For n =, Ct = R Tt = a = Velocity of sound. Thus it can be concluded that for maximum discharge per unit area at throat the fluid velocity (critical velocity) equals to the sonic velocity. At the throat section mach no. M = 1 for critical pressure ratio. For perfect gas: All the above equations obtained for the flow through nozzle can also be obtained for perfect gas by substituting n = and pv = RT Velocity at exit from nozzles or, or, or C2 = 2 C2 = 2 c T p 1 T 2 C2 = 2 h 1 h 2 C2 = 2 as cp = R 1 Critical velocity at throat, Ct = R Tt Mass flow rate per unit area / Maximum discharge per unit area at throat for critical conditions, = 1 1 2 1 m At p 1 v 1 576 ________________________________________________________ Applied Thermodynamics Critical pressure ratio, 2 1 1 p c p 1 = 13.3 CHOKED FLOW Let us consider a converging nozzle as shown in Fig. 13.10 with arrangement for varying back pressure. A valve is provided at exit of nozzle for regulating the back pressure at section 2-2. Let us denote back pressure by pb. Expansion occurs in nozzle from pressure p1 to pb. Initially when back pressure pb is equal to p1 there shall be no flow through the nozzle but as back pressure pb is reduced the mass flow through nozzle increases. With the reduction in back pressure a situation comes when pressure ratio equals to critical pressure ratio (back pressure attains critical pressure value) then mass flow through nozzle is found maximum. Further reduction in back pressure beyond critical pressure value does not affect the mass flow i.e. mass flow rate does not increase beyond its\u2019 limiting value at critical pressure ratio. Thus under these situations flow is said to be choked flow or critical flow. Fig. 13.10 Flow through a convergent nozzle A nozzle operating with maximum mass flow rate condition is called choked flow nozzle. At the critical pressure ratio the velocity at exit is equal to the velocity of sound. If the back pressure is reduced below critical pressure then too the mass flow remains at maximum value and exit pressure remains as critical pressure and the fluid leaving nozzle at critical pressure expands violently down to the reduced back pressure value. Graphical representation of mass flow rate with pressure ratio and variation of pressure along length of nozzle explain the above phenomenon. State a refers to the state having back pressure more than critical pressure, state c refers to the state having back pressure equal to critical pressure and state d refers to state having back pressure less than critical pressure. Nozzles _______________________________________________________________________ 577 In case of convergent-divergent nozzle also the maximum mass flow through such nozzle shall be obtained when pressure ratio at throat section equals critical pressure ratio and velocity at throat equals sonic velocity. The cross-sectional area of throat decides the mass flow through nozzle for given inlet conditions. 13.4 OFF DESIGN OPERATION OF NOZZLE Design operation of nozzle refers to the nozzle operating with pressure ratio equal to critical pressure ratio and maximum discharge rate per unit area then nozzle is said to be operating under design conditions. If the nozzle does not operate under design conditions then it is called off design operation of nozzle. Depending upon the back pressure value in reference to design value of pressure at exit of nozzle, the nozzle can be classified as under-expanding, over-expanding nozzles. Fig. 13.11 Off design operation of converging nozzle Nozzle is called under-exp", "anding if the back pressure of nozzle is below the designed value of pressure at exit of nozzle. Nozzle is called over expanding if the back pressure of a nozzle is above designed value of pressure at exit of nozzle. Detail discussion about the off design operation of nozzle is given ahead for convergent and convergent-divergent nozzle. Convergent nozzle: Let us look at convergent nozzle having arrangement for varying back pressure. Fluid enters the nozzle at state 1, say pressure p1. Variation of back pressure using valve at exit of nozzle shows the pressure and velocity variation as shown in Fig. 13.11. Following significant operating states are shown here. 578 ________________________________________________________ Applied Thermodynamics (i) When back pressure pb = p1, there is no flow. (ii) When back pressure is such that back pressure is more than critical pressure i.e. pc < pb < p1, there is flow through nozzle. Here pc is critical pressure at exit. This operating state of nozzle having back pressure higher than critical pressure is called over expanding nozzle. In this over expanding nozzle the mass flow rate through nozzle is less than designed value. (iii) When back pressure is such that back pressure is equal to critical pressure i.e. pb = pc. In this situation the mass flow through nozzle is maximum and nozzle is said to be choked. (iv) When back pressure is further lowered such that back pressure is less than critical pressure i.e. pb < pc, the nozzle is said to be under expanding nozzle. In underexpanding nozzle there is no change in specific volume, velocity and mass flow rate through exit as that at choked flow state of nozzle. Since back pressure at exit is less than critical pressure while fluid leaves nozzle at critical pressure so fluid expands violently and irreversibly upto back pressure outside the nozzle. Fig. 13.12 Off design operation of convergent-divergent nozzle. Convergent-divergent nozzle: Converging-diverging nozzles are generally used for accelerating flow up to supersonic velocity. Arrangement for varying back pressure is as shown in Fig. 13.12. Nozzles _______________________________________________________________________ 579 Different operating regime of nozzle are obtained by varying back pressure using valve at exit. Let us start with back pressure equal to inlet pressure. Following operating states are shown here; (i) When back pressure pb is equal to inlet pressure p1 then there is no flow as shown by state e on pressure-nozzle length plot. (ii) When back pressure is reduced and is slightly below p1, then there is some flow through nozzle shown by state f. The maximum velocity and minimum pressure occurs at throat section. With further reduction in back pressure from e to h, the flow rate increases. Flow remains subsonic for back pressure between e to h. In subsonic region the diverging portion of nozzle acts as diffuser, thereby increasing pressure and decreasing velocity in the direction of flow. In this regime the convergentdivergent nozzle is also used as venturimeter for flow rate measurement. Nozzle is said to be over expanding nozzle. (iii) With further reduction in back pressure the throat pressure decreases and the throat velocity increases. Back pressure at which throat velocity becomes sonic and throat pressure equals to critical pressure pc is shown by state h. Here maximum velocity occurs at throat so the diverging portion of nozzle still acts as diffuser. Mass flow rate through nozzle has become maximum as the sonic flow conditions are obtained at throat. Thus it can be seen that flow through converging duct with subsonic velocity at inlet can never result in the velocity higher than sonic velocity and pressure less than critical pressure. This is choked flow state of nozzle. (iv) Further lowering of back pressure less than critical pressure causes no effect on the flow in converging portion of nozzle and the pressure at throat remains equal to critical pressure and velocity at throat remains sonic. Also the flow rate through nozzle does not change. However, the nature of flow in diverging section of the duct changes as the breakage of flow occurs. As pressure is reduced to i and j the fluid passing through the throat continues to expand and accelerate in diverging portion of nozzle. Flow velocity beyond throat is supersonic in diverging portion of nozzle. At the section downsteam of throat there occurs discontinuity in the flow due to abrupt irreversible increase in pressure accompanied by deceleration from supersonic to subsonic velocity. This discontinuity in flow is called shock and generally plane of discontinuity is normal to direction of flow so it may also be called normal shock. Flow through shock is of irreversible and steady adiabatic type. Beyond shock the fluid undergoes further isentropic deceleration as diverging section acts as a subsonic diffuser. With further reduction in back pressure pb the shock moves downstream till it approaches nozzle exit plane and pb approaches the pressure given by state k. For the back pressure", " equal to pressure given by point k i.e. pk the normal shock reaches at exit end of nozzle. Here flow within nozzle is isentropic, subsonic in converging portion, sonic at throat and supersonic in diverging portion. Due to shock the flow leaving nozzle becomes subsonic. These are all over expanding states of nozzle. When back pressure is further lowered and back pressure becomes equal to exit plane pressure as at state l i.e. pb = pl, the isentropic expansion occurs throughout nozzle and no shock is found during flow. Fluid leaving nozzle is supersonic. (v) Further lowering of back pressure below pl flow remains same as for pressure upto point j, but the back pressure being less than design pressure causes breaking of flow at downstream of nozzle exit. An abrupt expansion of irreversible type occurs at nozzle exit. Irrespective of reduced back pressure the pressure at nozzle exit does not go below design pressure pl and mass flow rate and exit velocity also do not change. This operating state of nozzle is also called under expanding nozzle. 580 ________________________________________________________ Applied Thermodynamics 13.5 EFFECT OF FRICTION ON NOZZLE Inspite of the inside surface of nozzle being smooth the frictional losses always prevail due to friction between fluid and nozzle surface and friction within fluid itself. Due to friction prevailing during fluid flow through nozzle the expansion process through nozzle becomes irreversible. Expansion process since occurs at quite fast rate and time available is very less for heat transfer to take place so it can be approximated as adiabatic. Friction prevailing during flow through nozzle causes heat drop by about 10\u201315% and reduces the exit velocity. For the flowing fluid to be gas the T-S diagram representation is as follows: Fig. 13.13 T-s representation for expansion through nozzle Non ideal operation of nozzle causes reduction in enthalpy drop. This inefficiency in nozzle can be accounted for by nozzle efficiency. Nozzle efficiency is defined as ratio of actual heat drop to ideal heat drop. Nozzle efficiency, Nozzle = Actual heat drop Ideal heat drop Nozzle = Nozzle = In case of working fluid being steam the friction causes heating of steam flowing through nozzle thereby increasing dryness fraction. The volume of steam at exit also increases due to increase in dryness fraction. T\u2013s and h-s representation of steam expanding through nozzle is given in Fig. 13.14. Fig. 13.14 T-s and h-s representation for steam expanding through nozzle Nozzles _______________________________________________________________________ 581 Due to friction the velocity at exit from nozzle gets modified by nozzle efficiency as given below. Velocity at exit, C2 = 2 1 h In case of nozzle with friction the enthalpy drop, (h1 \u2013 h2) gives velocity at exit as,, for no friction C2 = 2 2 2 C2 or, (h1 \u2013 h2) = and (h1 \u2013 h2) = Substituting in nozzle efficiency, Nozzle = For negligible inlet velocity i.e. C1 0 2 C 2 2 C 2 Nozzle efficiency, Nozzle = Thus it could be seen that friction loss will be high with higher velocity of fluid. Generally frictional losses are found to be more in the downstream after throat in convergent-divergent nozzle because of simple fact that velocity in converging section upto throat is smaller as compared to after throat. Expansion upto throat may be considered isentropic due to small frictional losses. Apart from velocity considerations the significantly high frictional loss in diverging portion of nozzle compared to converging portion can be attributed to the contact surface area. Length of converging section upto throat is quite small compared to length of diverging portion after throat as it has subsonic acceleration which can be completed in short length. Diverging section of nozzle is designed comparatively longer than converging section so as to avoid flow separation due to adverse duct geometry (diverging type). Turbulence losses are also significant in diverging portion compared to converging portion. Due to the different factors discussed above the frictional losses are found to be more in diverging portion compared to converging portion. Normally angle of divergence in divergent portion is kept between 10 and 25 so as to avoid flow separation. But small divergence angle causes increase in length of diverging portion therefore increasing frictional losses. Thus a compromise should be striked in selecting angle of divergence as very small angle is desirable from flow separation point of view but undesirable due to long length and larger frictional losses point of view. Length of diverging portion of nozzle can be empirically obtained as below L = 15 tA\u22c5 where At is cross-sectional area at throat. While designing the nozzle parameters due care should be taken for smoothness of nozzle profile, surface finish for minimum friction and ease of manufacturing etc. Thus finally, it can be concluded that nozzle efficiency depends upon nozzle material, size and shape of nozzle, angle of divergence, nature of fluid flowing and", "xpands upto the back pressure such that exit state of steam lies in wet region, then during expansion steam vapours expand isentropically and slowly get condensed up to exit state. During such expansion steam also passes across saturated steam line or saturation line having unity dryness fraction. Thus it is obvious that expansion of steam is accompanied by simultaneous state change from superheated state to wet state. At every point along expansion line there exists a mixture of vapour and liquid in equilibrium. An expansion process starting at 1 goes up to state 2 in thermal equilibrium as shown on T-S and h-s diagram. Fig. 13.15 Expansion of steam on T-s and h-s diagram under equilibrium Superheated steam undergoes continuous change in state and becomes dry saturated steam at state 3 and subsequently wet steam leaving steam turbine at state 2. Some times expansion of steam occurs in metastable equilibrium or in equilibrium in which change of steam state could not maintain its pace with expanding steam. This phenomenon in which change of steam state could not occur simultaneously with expanding steam in nozzle is called phenomenon of supersaturation and flow is called supersaturated flow or metastable flow. Fig. 13.16 Metastable flow through steam nozzle In supersaturated flow the condensation of steam lags behind the expansion and so steam does not condense at the saturation temperature corresponding to the pressure. It may be understood as the shift in state 2 from 2 to 2 i.e. condensation gets extended up to 2. The dry saturated steam state which should be attained at state 3 cannot be realized at 3, but below 3 on vertical expansion line 1\u20132. This Nozzles _______________________________________________________________________ 583 delayed phase transformation of steam causing supersaturation phenomenon may be attributed to the following. (i) Steam flow through nozzle may be so fast that sufficient time is not available for heat transfer to take place and so the phase change lags behind the expansion. Generally time available may be of the order of 10\u20132 second for steam to flow through nozzle along with its condensation. (ii) Also the condensation of steam may have inherent requirement of nuclei of condensation which act as initiators for condensation. These nuclei of condensation may be provided by foreign particles, solid boundary etc. In the absence of nuclei of condensation the phase change of steam may get delayed and lags behind. These could be the factors responsible for supersaturation. Phenomenon of supersaturation or metastable equilibrium continues up to generally 94\u201395% dryness fraction. Beyond this the condensation of steam occurs suddenly at very fast rate irreversibly and the expansion process attains stable equilibrium. The locci of points up to which metastable equilibrium is observed is called Wilson line. Law of expansion for supersaturated flow is considered as pv1.3 = constant. Phenomenon of supersaturation causes increase in discharge by 2\u20135% because of increase in density at throat and also the heat drop gets slightly reduced thereby causing reduced velocity at exit. Supersaturation causes slight increase in dryness fraction and entropy. Figure 13.17 shows the supersaturated flow and Wilson line. Fig. 13.17 Supersaturated flow on T-s and h-s diagram. Region between saturation line and Wilson line is called supersaturated field. Here 1\u20132 line shows isentropic process. In the absence of supersaturated flow expansion occurs as 1\u20132 while with metastable flow it gets extended up to 1\u20132 as shown on T\u2013s diagram. Meta stable equilibrium gets settled and stable equilibrium is attained as shown by 2\u20132 and then normal expansion in stable equilibrium continues from 2 to 3. Thus it is obvious that in supersaturated flow the expansion occurs as if there is no saturated steam line and state 2 lies on the extended constant pressure line. The temperature at 2 is less than saturation pressure corresponding to p2 due to excess kinetic energy of steam at the cost of sensible heat. Metastable flow is characterized by a parameter called \u201cdegree of supersaturation\u201d and \u201cdegree of undercooling\u201d. \u2018Degree of supersaturation\u2019 refers to the ratio of saturation pressures corresponding to temperatures of states in stable equilibrium and metastable equilibrium (i.e. saturation pressures corresponding to 2 and 2 states). Degree of super saturation has value more than unity. \u2018Degree of 584 ________________________________________________________ Applied Thermodynamics undercooling\u2019 refers to the difference of two temperatures i.e. saturation temperature at state in stable equilibrium and temperature of the state in unstable equilibrium. 13.7 STEAM INJECTOR Steam injector refers to the device for injecting water into boiler using steam which may be available from boiler or exhaust steam from engine. Fig. 13.18 Steam injector Schematic for steam injector is shown in Fig. 13.18. Here high pressure steam enters a converging steam nozzle and leaves with high velocity and low pressure. Water gets entrained from feed tank and is carried by steam through combining", " nozzle. In due course steam gets condensed and the resulting mixture enters the divergent feed water nozzle where kinetic energy of water gets transformed into pressure head. Pressure head available in feed water nozzle is sufficiently above boiler pressure so that water can be fed to the boiler. Surplus water, if any gets discharged from over flow. The steam injector works on its own. It may be noted that the potential energy removed from live steam is many times more than the potential energy returned. Potential energy removed from live steam = Boiler steam pressure Volume of steam. Potential energy returned = Boiler pressure Volume of condensate and boiler feed Difference in the two potential energy exists due to large decrease in volume as steam condenses and this difference is only used for pumping water. EXAMPLES 1. Dry steam at 10 bar and 100 m/s enters a nozzle and leaves it with velocity of 300 m/s at 5 bar. For 16 kg/s of steam mass flow rate determine heat drop in nozzle and final state of steam leaving nozzle assuming heat loss to surroundings as 10 kJ/kg. Solution: Given,C1 = 100 m/s, C2 = 300 m/s p1 = 10 bar = 1 MPa, p2 = 0.5 MPa By steady flow energy equation between sections 1 and 2. h1 + 2 C1 2 + q = h2 + 2 C2 2 + w Nozzles _______________________________________________________________________ 585 Fig. 13.19 Nozzle Here there is no work so w = 0 and heat loss, q = \u2013 10 kJ/kg q = (h2 \u2013 h1) + C 2 2 C2 1 2 From steam table, (h2 \u2013 h1) = q + (h2 \u2013 h1) = \u2013 104 + C h1 = hg at 1 MPa = 2778.1 kJ/kg 2 2 C1 2 2 2 2 C1 2 2 = \u2013 104 + 100 300 C 2 2 2 h2 \u2013 h1 = \u2013 30,000 Heat drop in nozzle = 30 kJ/kg Total heat drop = 480 kJ/s Ans. or, At 5 bar, pressure, h2 = 2748.1 kJ/kg hf = 640.23 kJ/kg, hfg = 2108.5 kJ/kg Let dryness fraction at exit be x2, 2748.1 = 640.23 + x2 2108.5 x2 = 0.99 Dryness fraction at exit = 0.99 Ans. 2. Determine the mass flow rate of steam through a nozzle having isentropic flow through it. Steam enters nozzle at 10 bar, 500C and leaves at 6 bar. Cross-section area at exit of nozzle is 20 cm2. Velocity of steam entering nozzle may be considered negligible. Show the process on h-s diagram also. Solution: At inlet section (1), h1 = 3478.5 kJ/kg, s1 = 7.7622 kJ/kg. K Due to isentropic expansion, s1 = s2 At exit section (2), s2 = 7.7622 kJ/kg. K Enthalpy at rate (2) can be directly looked from mollier diagram by drawing vertical line from state (1) till it intersects 6 bar line. Else from steam table it can be obtained as under, looking at steam table for 6 bar it indicates that the state (2) shall be superheated state because s2 > sg at 6 bar. 586 ________________________________________________________ Applied Thermodynamics Fig. 13.20 Flow through nozzle on h-s diagram Degree of superheat can be determined by interpolation. Entropy 7.7622 kJ/kg K, s2 lies between temperature of 400C (s400C, 6 bar = 7.7079 kJ/kg K) and 500C (s500C, 6 bar = 8.0021 kJ/kg K). 7.7622 = 7.7079 + T2 = 418.45C. 8 0021 7 7079. 400 500 (T2 \u2013 400) Hence enthalpy at (2), h2 = h418.45C, 6 bar h2 = h400C, 6 bar + h 500 (418.45 \u2013 400) C,6 bar 500 h 400 400 3270 3., C bar 6 18.45 Neglecting inlet velocity, C1 0, assuming no heat loss, = 3270.3 +. 3482 8 100 h2 = 3309.51 kJ/kgK h C2 = 2 1 h 2 C2 = 581.36 m/s A 2 C 2 Mass flow rate, = v 2 Specific volume at exit (2) v2 = v400C, 6 bar + v 500, C bar 6 500 v 400 400, C bar 6 (418.45 \u2013 400) = 0.", "5137 + 0.5920 0 100.5137 18.45 v2 = 0.5281 kJ/kgK Nozzles _______________________________________________________________________ 587 Mass flow rate = 4. 581 36 20 10 0.5281 = 2.2017 Mass flow rate = 2.202 kg/s Ans. 3. In a nozzle steam expands from 12 bar and 300C to 6 bar with flow rate of 5 kg/s. Determine throat and exit area if exit velocity is 500 m/s and velocity at inlet to nozzle is negligible. Also find coefficient of velocity at exit. Coefficient of velocity is the ratio of actual velocity of fluid at nozzle exit to the velocity at exit considering isentropic flow through nozzle. Solution: Fig. 13.21 Nozzle and T-s representation of process Here velocity at exit is 500 m/s which is more than sonic velocity thus nozzle shall be converg- ing-diverging nozzle as shown. At inlet to nozzle, h1 = 3045.8 kJ/kg, s2 = 7.0317 kJ/kg K Due to isentropic flow through nozzle, s1 = s2 = s3 = 7.0317 kJ/kg K Pressure at throat section i.e. (2) can be estimated using critical pressure ratio, which is given as For superheated steam, Throat pressure, From steam table; At 6.54 bar, Velocity at throat.3 = p2 12 2 1 1 3. p2 = 6.54 bar / 1 3 0 3.. = 0.545 h2 = 2900.05 kJ/kg T2 = 224.48C, v2 = 0.3466 m3/kg C2 = h 2 h 1 2 = 3 2 10 3045 8. 2900 05. C2 = 539.9 m/s From continuity equation, m1 = m2 = 5 kg/s m2 = A C 2 2 v 2 588 ________________________________________________________ Applied Thermodynamics.9 A2 5 = 539 0 3466. A2 = 3.209 10\u20133 m2 Cross-sectional area at throat = 3.209 10\u20133 m2 Ans. Velocity at exit of nozzle, C3 = 2 1 h From steam tables, Temperature at exit of nozzle, T3 = 215.15C Enthalpy at exit of nozzle, h3 = 2882.55 kJ/kg Specific volume at exit of nozzle, v3 = 0.3647 m3/kg 3 2 10 3045 8. Ideal velocity at exit, h 3 2882.55 C3 = C3 = 571.40 m/s Actual velocity at exit, C3, actual = 500 m/s Area at exit, m1 = m2 = m3 = 5 kg/s = A 3 C 3 v 3, actual A3 = 5. 0 3647 500 = 3.647 10\u20133 m2 Cross-sectional area at exit = 3.647 10\u20133 m2 Ans. Coefficient of velocity = C 3, actual C 3 = 500 571 40. = 0.875. Coefficient of velocity = 0.875 Ans. 4. In a steam nozzle steam expands from 16 bar to 5 bar with initial temperature of 300C and mass flow of 1 kg/s. Determine the throat and exit areas considering (i) expansion to be frictionless and, (ii) friction loss of 10% throughout the nozzle. Solution: Let us assume velocity at inlet to nozzle to be negligible. From steam table, At inlet to the nozzle, enthalpy, h1 = 3034.8 kJ/kg at 16 bar, 300C s1 = 6.8844 kJ/kgK v1 = 0.15862 m3/kg (i) Considering expansion to be frictionless Pressure at throat of nozzle, p2 = p 1 2 1 n / n n 1, for n = 1.3, p2 = 8.73 bar Enthalpy at throat and exit of nozzle can be looked from steam table as, h2 = 2891.39 kJ/kg h3 = 2777.0 kJ/kg v2 = 0.2559 m3/kg v3 = 0.3882 m3/kg Nozzles _______________________________________________________________________ 589 Fig. 13.22 Heat drop up to throat section = h1 \u2013 h2 Velocity at throat, Heat drop from throat to exit = h2 \u2013 h3 q12 = 143.41 kJ/kg C2 = 2 1 h h 2 = 535.56 m/s Velocity at exit, q23 = 114.39 kJ/kg 2 2 h h C3 = 3 C3 = 718.06 m/s m v 2 = 2 C 2 Throat area, A2 = Exit area, A3 = 2559", " 1 0.56 535. 1 0 3882. 718 06 = 4.78 10\u20134 m2 = 5.41 10\u20134 m2 For frictionless expansion, Throat area = 4.78 cm2 Exit area = Ans. 5.41 cm2 (ii) Considering expansion to have 10% friction loss: Due to frictional heat loss actual heat drop shall get reduced. Actual heat drop upto throat, q q 12 = 0.9 q12 12 = 129.07 kJ/kg h2, actual = 2905.73 kJ/kg Thus, enthalpy at throat, from steam table, v2, actual = 0.2598 m3/kg corresponding to the modified state as indicated by enthalpy, h2, actual. This new state at throat shall have temperature of 231.35C, which can be calculated from known enthalpy h2, actual at respective pressure. C2, actual = 2 129 07 103. C2, actual = 508.1 m/s.2598. 508 1 Actual velocity at throat, Actual throat area, A2, actual = 1 0 Actual throat area = 5.11 10\u20134 m2 or, 5.11 cm2 Ans. 590 ________________________________________________________ Applied Thermodynamics Actual heat drop from throat to exit of nozzle, q q Thus, enthalpy at exit, 23 = 0.9 114.39 23 = 102.95 kJ/kg h3, actual = 2905.73 \u2013 102.95 = 2802.78 kJ/kg From steam table, corresponding to modified state as given by h3, actual the specific volume can be given as, v3, actual = 0.40023 m3/kg Actual velocity at exit, Actual area at exit 3 10. 508 1 2 2 102 C3, actual = C3, actual = 681.22 m/s 1 0 40023.95. 681.22 A3, actual = A3, actual = 5.875 10\u20134 m2 Actual area at exit, = 5.875 10\u20134 m2 or 5.875 cm2 Ans. 5. An impulse turbine of 1 MW has steam entering at 20 bar and 300C and steam consumption of 8 kg per kW hour. Steam leaves at 0.2 bar and 10% of total heat drop is lost in overcoming friction in diverging portion of nozzle. If throat diameter of each nozzle is 1 cm then determine (i) the number of nozzles required (ii) exit diameter of each nozzle. Solve using mollier diagram. Solution: In impulse turbine the impulse action of steam is used for getting shaft work. A stage of simple impulse turbine has steam nozzle and a ring of symmetrical blade. Total pressure drop in the stage thus occurs in nozzle itself. Throat pressure = 0.545 20 = 10.9 bar From mollier diagram, Heat drop up to throat, q12 = h1 \u2013 h2 q12 = 142 kJ/kg v2 = 0.20 m3/kg Considering negligible inlet velocity, the velocity at throat, Fig. 13.23 3 2 10 C2 = C2 = 532.9 m/s 142 Nozzles _______________________________________________________________________ 591 Mass flow rate through nozzle = A C 2 2 v 2 2 2 = 10 4 0 = 0.209 kg/s.20 532.9 The number of nozzles can be determined by using the capacity of turbine, steam consumption and mass flow rate through a nozzle. Number of nozzles = 3 1 10 3600 8.209 0 = 10.63 11 nozzles Number of nozzles required = 11 Ans. (ii) Total heat drop in nozzle, from mollier chart, (h1 \u2013 h3) = 807 kJ/kg Useful heat drop, \u2206q13, useful = 0.90 \u00d7 807 = 726.3 kJ/kg Specific volume at exit, Velocity at exit = 726.3 v3 = 7.2 m3/kg 3 C3 = 1205.2 m/s 2 10 Therefore, from continuity equation, A3 = 3 m v 3 C 3 0 =.209 1205 A3 = 1.25 \u00d7 10\u20133 m2 7.2.2 or Area at exit = 12.5 cm2 Ans. 6. A nozzle is supplied with steam at 0.7 mPa and 275C. Determine temperature and velocity at throat considering no losses. If diverging portion of nozzle is 6 cm long and throat diameter of 0.5 cm, determine the angle of cone in nozzle so that steam leaves nozzle at 0.1 MPa. Assume heat utilization in diverging portion to be 85%. Solve using mollier diagram. Solution: Pressure at throat section of nozzle p2 = 0.545 0.7 = 0.38 MPa From mollier", " diagram, Heat drop in nozzle up to throat, Specific volume, h1 \u2013 h2 = q12 = 138 kJ/kg v2 = 0.58 m3/kg 592 ________________________________________________________ Applied Thermodynamics Temperature = 203C Considering velocity at inlet to be negligible, velocity at throat, Fig. 13.24 2 138 103 C2 = C2 = 525.35 m/s 4 0.5 2 2 10 0.58 525 35. Mass flow rate = From mollier diagram, heat drop in diverging portion i.e. from throat to exit of nozzle shall be; m1 = 0.0178 kg/s = m2 = m3 q23 = h2 \u2013 h3 = 247 kJ/kg q23, actual = 209.95 kJ/kg Specific volume, v3, actual = 1.7 m3/kg Total heat drop = 138 + 209.95 = 347.95 kJ/kg Velocity at exit of nozzle =.95 103 347 2 C3 = 834.2 m/s m v 3 3 C 3 Area at exit, A3 =.. = 0 0178 1 7 834 = 3.63 10\u20135 m2.2 Diameter at exit = 5 3 63 10. = 6.8 10\u20133 m or 6.8 mm 4 For Cone angle to be 2, tan = 6.8 2 5 60, = 0.86 Cone angle, 2 = 1.72 Nozzles _______________________________________________________________________ 593 With no losses, Temperature at throat = 203 Velocity at throat = 525.35 m/s With losses, Cone angle = 1.72 Ans. 7. An impulse steam turbine generating 5000 hp requires 6 kg of steam per hp-hr at 16 bar, dry saturated. In the first stage the steam is expanded through nozzle with an efficiency of 0.90 to a pressure 1 3 of 8 bar. These nozzles are placed so as to extend over approximately rd of the circumference with the pitch circle diameter of 600 mm and pitch of 50 mm. Considering nozzle angle as 12 to the plane of wheel and plates dividing being 3 mm thick, determine total length of nozzle arc and radial height of nozzles. Solution: At inlet to nozzle, from steam table, h1 = hg at 16 bar = 2794 kJ/kg s1 = sg at 16 bar = 6.4218 kJ/kgK Fig. 13.25 At exit of nozzle, s1 = s2 = 6.4218 kJ/kgK Since s2 < sg at 8 bar so the state at the end of expansion lies in wet region. Let dryness fraction at state 2 be x2, then, s2 = sf at 8 bar + x2 sfg at 8 bar 6.4218 = 2.0462 + x2 4.6166 or, x2 = 0.9478 h2 = hf at 8 bar + x2 hfg at 8 bar = 721.11 + (0.9478 2048) h2 = 2662.2 kJ/kg Theoretically, change in enthalpy, h12 = 2794 \u2013 2662.2 h12 = 131.8 kJ/kg 12 h actual = 0.90 or h12 actual = 0.9 131.8 = 118.62 kJ/kg 12 h h2 actual = 2794 \u2013 118.62 = 2675.38 kJ/kg h2, actual = 2675.38 = hf at 8 bar + x2 actual hfg at 8 bar 594 ________________________________________________________ Applied Thermodynamics 2675.38 = 721.11 + (x2 actual 2048) x2 actual = 0.9542 v2 actual = vf at 8 bar + x2 actual vfg at 8 bar v2 actual = 0.001115 + (0.9542 (0.2404 \u2013 0.001115)) v2 actual = 0.2294 m3/kg Specific volume, Neglecting the velocity at inlet to nozzle, h actual 2 C2 = 12 C2 = 487.1 m/s By continuity equation, m = A C 2 2 v 2 actual A2 0. 487 1 5000 6 3600 =.2294 A2 = 3.925 10\u20133 m2 Area at exit of nozzle 39.25 cm2 Approximate length of nozzle arc, = Number of nozzles = 60 3 = 62.83 cm Length of nozzle arc Pitch = 62.83 5 13 nozzles. Correct length of nozzle arc Flow area at exit of each nozzle = (5 sin 12 \u2013 0.3) h = 13 5 = 65 cm 39.25 = (5 sin 12 \u2013 0.3) h 13 or, h = 4.08 cm Length of nozzle = 65 cm Ans. Radial height of nozzle = 4.", "08 cm 8. Air is expanded reversibly and adiabatically in a nozzle from 13 bar and 150C to a pressure of 6 bar. The inlet velocity of the nozzle is very small and the process occurs under steady flow conditions. [U.P.S.C., 1992] Calculate the exit velocity of the nozzle. Solution: Let inlet and exit states in nozzle be given as 1 and 2 respectively. Assuming no heat interaction and work interaction, no change in potential energy during flow across the nozzle. Steady flow energy equation gets modified as under; h1 + 2 C1 2 = h2 + 2 C2 2 Neglecting velocity at inlet to nozzle C1. h 2 1 h 2 C2 = Nozzles _______________________________________________________________________ 595 T 2 Assuming adiabatic expansion in nozzle, between states 1 and 2. C2 = = T2 = T1 P 2 P 1 1 or, 1 4 1.. 1 4 6 13 = (273 + 150) T2 = 339.16 K Substituting in expression for exit velocity, C2, 423. C2 = 2 1 005 C2 = 12.98 m/s Exit Velocity = 12.98 m/s. 339 16 Ans. 9. During a test on steam nozzle steam impinges a stationary flat plate which is perpendicular to the direction of flow and the force on the plate is measured. The force is found to be 350 N when dry saturated steam at 8 bar is expanded to 1 bar. Throat cross-section area is 5 cm2 and exit area is such that the complete expansion is achieved under these conditions. Determine (i) (ii) the discharge at throat. the efficiency of nozzle assuming that all the losses occur after throat and that n = 1.13 for isentropic expansion. Solution: For isentropic expansion up to throat P2 = 4.62 bar P2 = 8 2 113 1. 1 13.. 0 13 From steam table, Enthalpy at 1, h1 = hg at 8 bar = 2769.1 kJ/kg s1 = sg at 8 bar = 6.6628 kJ/kg.K Fig. 13.26 596 ________________________________________________________ Applied Thermodynamics At throat section let the dryness fraction be x2, then s1 = s2 = 6.6628 = sf at 4.62 bar + sfg at 4.62 bar x2 6.6628 = 1.7872 + x2 5.0178 x2 = 0.9717 or 97.17% h2 = hf at 4.62 bar + x2 hfg at 4.62 bar = 627.33 + (0.9717 2117.77) h2 = 2685.17 kJ/kg Specific volume at throat, v2 = vf at 4.62 bar + x2 vfg at 4.62 bar = 0.001089 + (0.9717 0.4035) v2 = 0.3932 m3/kg Enthalpy change up to throat h12 = h1 \u2013 h2 h12 = 83.93 kJ/kg 2 Velocity at throat, C2 = Discharge at throat, m2 = h 12 A C 2 2 v 2 = 2 83.93 103 = 409.71 m/s = 5 10 4. 409 71. 0 3932 m2 = 0.521 kg/s Discharge at throat = 0.521 kg/s Ans. Force acting on plate F = 350 N = m (C3 C1 0 = 671.78 m/s Exit actual velocity C3 For diverging section between states 1 and 3, s1 = s2 = s3 \u2013 C1) s3 = 6.6628 = sf at 1 bar + x3 sfg at 1 bar 6.6628 = 1.3026 + (x3 6.0568) x3 = 0.8238 or 82.38% Enthalpy at exit h3 = hf at 1 bar + x3 hfg at 1 bar = 417.46 + (0.8238 2258) h3 = 2277.6 kJ/kg Theoretical enthalpy drop, h23 = h2 \u2013 h3 = 407.57 kJ/kg Actual enthalpy drop, h23 = Actual velocity h23 N C 3 = 671.78 = 2 h 23 2 N 407.5 103 Nozzles _______________________________________________________________________ 597 N = 0.5537 or 55.37% Nozzle efficiency = 55.37% Ans. 10. A convergent-divergent nozzle operates with 5 kg of steam per minute being discharged at 1 bar. For the steam supplied to nozzle being at 10 bar and 200C and supersaturation occurring up to throat and normal afterwards, determine, (i) (ii) (iii", ") the diameter of nozzle at exit. the maximum degree of supersaturation. the amount of undercooling at throat. For supersaturation take Pv1.3 = constant and Solution: Pressure at throat = constant P2 = P1 n \u2212 1) ( n. \uf8eb 2 \uf8ec +\uf8ed n \uf8f6 \uf8f7 \uf8f8 1 n = 1.3 P2 = 5.45 bar From steam tables, Enthalpy at inlet, h1 = 2827.9 kJ/kg s1 = 6.6940 kJ/kgK v1 = 0.2060 m3/kg Enthalpy drop up to throat, Fig. 13.27 h1 \u2013 ) 1 \uf8f6 \uf8f7 \uf8f8 P 2 P 1 P1v1 \uf8eb \u2212 \uf8ec \uf8ed \uf8ee \uf8ef 10 102 0.2060) 1 = 1 3. 0 3. = 116.67 kJ/kg = 2711.23 kJ/kg h1 \u2013 h2 h2. 5 45 10 /. 0 3 1 3. Let dryness fraction be x2 at throat, so using steam tables it can be obtained as under, = hf at 5.45 bar + (x2 hfg at 5.45 bar) h2 598 ________________________________________________________ Applied Thermodynamics 2711.23 = 663.36 + (x2 2098.15) x2 = 0.9760 = sf at 5.45 bar + (x2 sfg at 5.45 bar) s2 = (1.8936 + (0.9760 4.8989) = 6.6749 kJ/kgK Let dryness fraction at exit be x3, so = 6.6749 kJ/kgK s3 = s2 s3 = sf at 1 bar + x3 sfg at 1 bar 6.6749 = 1.3026 + x3 6.0568 Enthalpy at exit, Specific volume at exit, x3 = 0.8869 h3 = hf at 1 bar + x3 hfg at 1 bar = 417.46 + (0.8869 2258) h3 = 2420.08 kJ/kg v3= vf at 1 bar + x3 vfg at 1 ba = (0.001043 + 0.8869 1.6929) Velocity at exit, Mass flow rate, = 903.13 m/s v3 = 1.5025 m3/kg h C3 = 2 13 A C 3 3 v 3 m3 = = A3. 903 13.5025 1 5 60 A3 = 1.386 10\u20134 m2 or 1.386 cm2 Diameter of nozzle at exit d 3 = 0.0133 m or 1.33 cm Diameter of nozzle at exit = 1.33 cm Ans. Theoretically up to throat545 Saturation pressure corresponding to 411.18 K, Psat, 138.18C = 3.44 bar T2 = 473 0 Degree of supersaturation = P 2 P sat. 138 18, C 1 3 1.. 1 3 = 411.18 K = 138.18C =. 5 45. 3 44 = 1.58 Amount of undercooling = (Tsat at 5.45 bar \u2013 T2) = (155.12 \u2013 138.18) = 16.94C Nozzles _______________________________________________________________________ 599 Degree of supersaturation = 1.58 Amount of undercooling = 16.94C Ans. 11. Steam undergoes expansion from 4 bar, 180C to a pressure of 1.5 bar. If the expansion is supersaturated and occurs with friction loss of 5 percent, determine the actual heat drop and degree of undercooling. Following equations may be used for supersaturated steam, Pv 102 = (h \u2013 2614), (Here P is in bar, v is specific volume in m3/kg, h is enthalpy in kJ/kg) P 1 3.. 0 3 T = constant and Pv1.3 = constant Consider specific heat as 2.174 kJ/kgK. Solution: At inlet for expansion, state 1, specific volume can be seen from steam table, v1 = 0.5088 m3/kg Enthalpy at state 1 from empirical relation, 4 0.5088 102 = (h1 \u2013 2614) h1 = 2817.52 kJ/kg Using governing equation given, Fig. 13.28 P1 v1 1.3 = P2 v2 1.3 4 102 (0.5088)1.3 = 1.5 102 v2 1.3 v2 = 1.082 m3/kg Enthalpy at 2, or, P2 v2 102 = (h2 \u2013 2614) (1.5", " 1.082 102) = (h2 \u2013 2614) h2 = 2776.3 kJ/kg N (h1 \u2013 h2) Actual heat drop = = 0.95 (2817.52 \u2013 2776.3) = 39.16 kJ/kg Actual heat drop = 39.16 kJ/kg Ans. 600 ________________________________________________________ Applied Thermodynamics Temperature at state 2 10 1 3 0 3 /. 453 2. = constant =.5 1 10 T2 = 361.24 K Supersaturation causes rise in temperature due to reheat =. 0 05 2776.3.52 2817. 2 174 = 0.95 K or 0.95C = 361.24 + 0.95 = 362.19 K T2 Amount of undercooling = T3 \u2013 T2 = Tsat at 1.5 bar \u2013 T2 = ((273 + 111.37) \u2013 362.19) = 22.18C Degree of undercooling = 22.18C Ans. 12. Steam is supplied at 14 bar, 400C to set of 16 nozzles of impulse turbine. Pressure of steam at exit of nozzle is 10 bar and the discharge is 5 kg/s. Nozzle efficiency is 90%. Determine the crosssectional area at exit of each nozzle. Also determine the percentage increase in discharge if steam is supplied to nozzle with velocity of 100 m/s. Solution: Let inlet and exit states be 1 and 2. Therefore, enthalpy at state 1, h1 = hat 14 bar, 400C From steam tables, h1 = 3257.5 kJ/kg s1 = 7.3026 kJ/kgK Since steam is superheated at inlet, so n = 1.3 For maximum discharge, throat pressure = 14 2 1 3 1. = 7.64 bar P2 P2 1 1 3.. 0 3 The actual pressure given at exit of nozzle is 10 bar which means that the nozzle is of convergent type. Actual pressure at exit of nozzle, P2. For isentropic expansion from 14 bar to 10 bar, s1 = s2 = 7.3026. State of steam at exit is superheated because s2 is more than sg at 10 bar (= 6.5865 kJ/kgK). This entropy s2 lies between superheat steam temperature of 350C and 400C at 10 bar. Nozzles _______________________________________________________________________ 601 s at10bar,, 350 C 400 C 400 s at10bar 350 (T2 \u2013 350). 7 4651 7 3011. 50 s2 = sat 10 bar, 350C + 7.3026 = 7.3011 + T2= 350.46C (T2 \u2013 350) At 350.46C and 10 bar from steam tables by interpolation, h at h2 = hat 10 bar, 350C + bar 10, C 400 400 h at10 bar,350 C 350 (T2 \u2013 350) (350.46 \u2013 350) = 3157.7 + 3263.9 3157 7. 50 h2 = 3158.7 kJ/kg Specific volume at exit, Actual enthalpy change, v2 = 0.2827 m3/kg h12 = (h1 \u2013 h2) N = (3257.5 \u2013 3158.7) 0.9 = 88.92 kJ/kg Velocity at exit of nozzle, C2 = 2 h 12 3 2 10 = = 421.71 m/s 88.92 Cross-sectional area at exit of nozzle A2 = = \u00d7 m v 2 \u00d7 N C 2.2827 5 0. 421 71 16 104 = 2.13 cm2 Cross-sectional area at exit of nozzle, = 2.13 cm2 Ans. Inlet velocity being C1 = 100 m/s Modified velocity at nozzle exit C2 2 C 1 12 h = 2 = = 433.41 m/s 2 10 3 C2 100 2 88.92 Discharge with modified velocity = C2 A 2 v 2 16 16 = 4. 433 41. 2 13 10 0.2827 = 5.22 kg/s 602 ________________________________________________________ Applied Thermodynamics Percentage increase in discharge =.22 5 5 = 4.4% 5 100 % Increase in discharge = 4.4% Ans. 13. In a nozzle the steam enters at 20 bar, dry saturated and expands up to 5 bar. Considering the expansion to be frictionless throughout and steam remaining in dry state during expansion upto throat, determine the degree of supersaturation and degree of undercooling. Also determine the change of entropy, the loss due to undercooling and percentage loss if steam has to revert instantaneously to saturated state at constant enthalpy and if the further expansion takes place in thermal equilibrium. Solution: At inlet to nozzle, P1 = 20 bar, T1 = Tsat at 20 bar", " T1 = 212.42C or, Pressure at throat, P2 = P1 0.58 P2 = 11.6 bar / n T P 2 2 T P 1 1 T2 = 485.42 0.882 T2 = 428.14K or, = 1 n T2 = T1 ( \u2212 )(1.3 1) 1.3 0.58 Fig. 13.29 But saturation temperature corresponding to throat pressure, Tsat 11.6 bar = 186.43C Saturation pressure corresponding to 428.14K (155.14C) Psat at 155.14C = 0.5452 MPa Psat at 155.14C = 5.452 bar Degree of supersaturation =. 11 6. 5 452 = 2.13 Degree of undercooling = Tsat at 11.6 bar \u2013 T2 = 186.43 \u2013 155.14 = 31.29C Nozzles _______________________________________________________________________ 603 Degree of supersaturation = 2.13 Degree of undercooling = 31.29C Ans. Enthalpy at state 1, Isentropic enthalpy drop, h1 = hf at 20 bar = 2799.5 kJ/kg v1 = vg at 20 bar = 0.009963 m3/kg s1 = sg at 20 bar = 6.3409 kJ/kgK ( n \u2212 ) \uf8f6 \uf8f7 \uf8f8 \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb h1 \u2013 h2 = \uf8eb n \uf8ec \u2212\uf8ed n \uf8f6 \uf8f7 \uf8f8 1 \u22c5 Pv 1 1 h1 \u2013 h2 = \uf8eb \uf8ec \uf8ed 1.3 \u2212 1.3 1 \uf8f6 \uf8f7 \uf8f8 20 102 0.009963 1. 11 6 20 1 3 1.. 1 3 or, h12 = h1 \u2013 h2 = 10.19 kJ/kg h2 = 2789.31 kJ/kg In the absence of supersaturation the state at throat will be 2. Thus for getting enthalpy at 2the dryness fraction may be assumed x2. The entropy at inlet and throat shall be equal, = s1 = 6.3409 s2 s2 sfg at 11.6 bar 4.3338) 6.3409 = sf at 11.6 bar + x2 6.3409 = 2.2016 + (x2 = 0.955 = hf at 11.6 bar + x2 x2 h2 hfg at 11.6 bar = 791.73 + (0.955 1991.88) Enthalpy, = 2693.98 kJ/kg h2 Heat drop with no supersaturation upto throat, h12 h12 = 105.52 = h1 \u2013 h2 \u2013 h12 = 95.33 kJ/kg =. 95 33. 459 43 Increase in entropy, s12 Loss of available heat drop = h12 h 12 T sat at s12 = 0.2075 kJ/kgK = s1 + s12 s2 = 6.5484 kJ/kgK s2 Hence, entropy at 2 = h 12 11 6. bar With super saturation For isentropic expansion after throat, s2 = s3 604 ________________________________________________________ Applied Thermodynamics Let dryness fraction at exit be x3 sfg at 5 bar) 4.9606). So, = sf at 5 bar + (x3 s3 6.5484 = 1.8607 + (x3 = 0.945 hfg at 5 bar) = hf at 5 bar + (x3 = 640.23 + (0.945 2108.5) = 2632.76 kJ/kg h3 x3 Enthalpy at 3, h3 Without supersaturation Let dryness fraction be x3 at state 3, which will be obtained in the absence of supersaturation. s1 = s3 s3 = 6.3409 = sf at 5 bar + (x3 sfg at 5 bar) 6.3409 = 1.8607 + (x3 4.9606) x3 = 0.903 Enthalpy at 3, h3 = hf at 5 bar + (x3 hfg at 5 bar) = 640.23 + (0.903 2108.5) h3 = 2544.21 kJ/kg Thus, loss due to undercooling, h3 \u2013 h3 = 2632.76 \u2013 2544.21 3 = 88.55 kJ/kg 100 h3 h 3 3 h h 1 3 Percentage loss = = 88.55.5 2544.21 2799 100 = 34.68% Entropy change = 0.2075 kJ/kgK Ans", ". Loss due to undercooling = 88.55 kJ/kg Percentage loss = 34.68% 14. A steam injector delivers 150 kg of water per minute from a water tank where the constant water level of 5 m below the axis of injector is maintained. Injector injects into the boiler having steam at 20 bar. Water level in boiler is 0.8 m above the injector. Steam for injector is taken from boiler and has dryness fraction of 0.95. Water is supplied at 25C and velocity in delivery pipe is 20 m/s. Determine, (i) (ii) (iii) (iv) the mass of water pumped per kg of steam the diameter of throat of mixing nozzle the diameter of the steam nozzle considering pressure at throat to be 0.7 times of supply pressure. the temperature of water coming out of injector. Nozzles _______________________________________________________________________ 605 Solution: Fig. 13.30 Pressure at inlet of nozzle, P1= 20 bar, Temperature Pressure at throat, Enthalpy at state T1 = Tsat at 20 bar = 212.42C P2 = 0.7 20 = 14 bar 1, h1 = hf at 20 bar + x1 hfg at 20 bar = (908.79) + (0.95 1890.7) h1 = 2704.95 kJ/kg s1 = sf at 20 bar + x1 sfg at 20 bar = 2.4474 + (0.95 3.8935) At throat Let dryness fraction at throat be x2 s1 = 6.1462 kJ/kgK s2= s1 = 6.1462 kJ/kgK s2 = sf at 14 bar + x2 sfg at 14 bar 6.1462 = 2.2842 + (x2 4.1850) x2 = 0.923 h2 = hf at 14 bar + x2 hfg at 14 bar = 830.30 + (0.923 1959.7) Specific volume Velocity of steam at throat, C2 = h2 = 2639.10 kJ/kg v2 = vf at 14 bar + x2 vfg at 14 bar = 0.13 m3/kg 2 1 h h 2 = 2 2704.95 2639 10. 103 606 ________________________________________________________ Applied Thermodynamics Applying Bernoulli\u2019s equation between states 3 and 4, C2 = 362.9 m/s gZ3 + P 3 2 C 3 2 = gZ4 + P 4 2 C 4 2 2 C3 2 + 5. 1 013 10 10 3 = 5 20 10 10 3 + (0.8 9.81) + 2 20 2 C3 = 64.91 m/s Let mass of water lifted by per kg of steam be m kg, i.e. m kg of water per kg of steam. C2 \u2013 m C2 \u2013 m 2 2 gh w 2 9 81 5. = (m + 1) C3 = (m + 1) C3 362.9 \u2013 9.90 m = (m + 1) 64.91 m = 3.98 kg of water per kg steam Mass of water pumped per kg of steam = 3.98 kg Ans. Mass of mixture passing through state 3 for given water flow, 1.98 3 = 150 60 150 60 = 2.5 + 0.628 m3 = 3.128 kg/s Area of throat of mixing nozzle = m 3 C 3 = 3 128. 3 64 10.91 = 0.482 cm2 Diameter of throat of mixing nozzle, d3 = 0.783 cm Diameter of throat of mixing nozzle = 0.783 cm Ans. Mass of steam required for given water flow rate = 150 60 1.98 3 msteam = 0.628 kg/s C A msteam = steam v steam \u00d7 steam msteam = 2 \u00d7 A C 2 v 2 A2 362 9 0.628 =. 0 13 A2 = 2.25 cm2 d2 = 1.69 cm Diameter of throat of steam nozzle = 1.69 cm Ans. Nozzles _______________________________________________________________________ 607 In order to determine temperature of water coming out of injector the heat balance may be written assuming no heat loss. For unit mass of steam entering injector x1 hfg1 + cps (T1 \u2013 T3) = m cpw(T3 \u2013 Tw) (0.95 1890.7) + 4.18 (212.42 \u2013 T3) = 3.98 4.18 (T3 \u2013 25) T3 = 148.92C Temperature of water coming out of injector = 148.92C Ans. 15. A steam injector is used for maintaining supply of the feed water to boiler producing steam at 20 bar. The pressure of exhaust steam for operating injector is 1.5", " bar and dryness fraction is 0.9. The mass of water taken from feed water tank is 5000 kg/hr and temperature is 17C. Determine the mass of water that can be pumped per kg of steam, area of steam and water discharge orifices. Delivery pressure may be assumed to be 20% more than boiler pressure so as to overcome frictional losses and positive delivery. Neglect change in elevation. Solution: At inlet to the steam nozzle pressure, P1 = 1.5 bar Pressure at throat, P2 = P1 2 1 n n n 1, here n = 1.13, P2 = 0.87 bar Fig. 13.31 Temperature at state 1, T1 = Tsat at 1.5 bar = 111.37C = 384.37 K Enthalpy at state 1, h1 = hf at 1.5 bar + x1 hfg at 1.5 bar = 467.11 + (0.9 2226.5) h1 = 2470.96 kJ/kg s1 = 1.4336 + (0.9 5.7897) s1 = 6.6443 kJ/kgK s2 = s1 = 6.6443 = sf at 0.87 bar + x2 sfg at 0.87 bar 6.6443 = 1.2560 + x2 6.1538 At state 2, x2 = 0.88 h2 = hf at 0.87 bar + x2 hfg at 0.87 bar = 400.26 + (0.88 2268.71) h2 = 2396.72 kJ/kg v2 = vf at 0.87 bar + x2 vfg at 0.87 bar Specific volume, 608 ________________________________________________________ Applied Thermodynamics Steam velocity, = 0.001039 + (0.88 1.9649), v2 = 1.7302 m3/kg h C2 = 2 2 C2 = 385.33 m/s h 1 By Bernoulli\u2019s equation, 2 C3 2 = P4.2 1 C3 = 5 2 10 20 1000 C3 = 69.28 m/s Let mass of water entrained per kg of steam be m kg water per kg of steam. Neglecting momentum of entering water the momentum balance can be given as under. 1 C2 = (m + 1) C3 385.33 = (m + 1) 69.28 m = 4.56 kg water per kg of steam Mass of steam supplied per second = 5000 3600 4.56 = 0.3046 kg/s Area of steam nozzle, A2 = m v 2 C 2. = 0 3046 1 7302.. 385 33 A2 = 13.68 cm2 5000 3600. Total discharge from injector = 0 3046 = 1.693 kg/s Area of discharge orifice, = 1 693..28 103 = 0.2444 cm2 69 = 0.2444 10\u20134 m2 Mass of water pumped per kg of steam = 4.56 kg water/kg of steam Ans. Area of steam nozzle = 13.68 cm2 Area of discharge orifice = 0.244 cm2 13.1 What do you understand by nozzle? Discuss different types of nozzles. 13.2 Derive the expression of velocity of fluid leaving nozzle, considering flow to be frictionless and adiabatic. 13.3 Obtain expression for discharge through nozzle. Nozzles _______________________________________________________________________ 609 13.4 For a convergent-divergent nozzle sketch variation of specific volume, velocity, area and pressure along the nozzle axis. 13.5 Show the following statement, \u201cConverging duct behaves as nozzle for subsonic flows and the diverging duct shall behave as nozzle when flow is supersonic.\u201d 13.6 Prove that the maximum discharge of fluid per unit area through a nozzle shall occur when the ratio of fluid pressure at throat to the inlet pressure is n \u2212 ( n )1 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \uf8eb 2 \uf8ec +\uf8ed n 1 \uf8eb \uf8ec \uf8f6 \uf8ec \uf8ed \uf8f7 \uf8f8, where n is the index of adiabatic expansion. Also obtain the expression for maximum mass flow through a convergent-divergent nozzle having isentropic expansion starting from rest. 13.7 For isentropic flow through a nozzle prove that area on p-v diagram gives the heat drop during expansion. Also show processes on T-s and h-s diagram. 13.8 Explain the significance of choked flow. 13.9 Describe the \u2018over expansion\u2019 and \u2018under expansion\u2019 in nozzles. 13.10 Give the pressure and velocity variations along the length of \u2018convergent nozzle\ufffd", "\ufffd and \u2018convergentdivergent nozzle\u2019 when back pressure is reduced gradually from inlet pressure to the pressure below designed back pressure value. 13.11 Discuss the effects of friction on the performance of nozzles. Also show these effects on T-s and h-s diagrams. 13.12 What do you understand by supersaturation of steam flowing through nozzles? Explain the phenomenon and factors responsible for it. 13.13 Describe the physical significance of supersaturation in steam nozzle. 13.14 Describe the working of steam injector and its applications. 13.15 Write short notes on the following: (a) Nozzle efficiency (c) Discharge coefficient (b) Velocity coefficient (d) Critical velocity m/s. Determine the velocity at exit considering isentropic flow through nozzle. 13.16 Steam at 10 bar and 250C is admitted in convergent-divergent nozzle with initial velocity of 75 [886.35 m/s] 13.17 Air enters into a nozzle at 4 bar and leaves at 1 bar, 350C and 700 m/s. Determine temperature of air entering nozzle, nozzle efficiency and nozzle exit area for air flow rate of 4 kg/s. Take cp = 1.003 kJ/kgK. [594.27C, 86.04%, 10.4 cm2] 13.18 In a steam nozzle the expansion pressure ratio of 7.5 is achieved with steam entering at 300C and leaving at 2 bar. For the steam mass flow rate of 1.2 kg/s determine the throat area considering [4.45 cm2] adiabatic expansion in the absence of friction loss. 13.19 Steam at dry saturated state enters the nozzle at 10 bar and 90 m/s velocity. Steam leaves the nozzle at 6 bar, 435 m/s. Considering heat loss of 1.51 kcal/kg of steam flow determine dryness [0.92, 3.22] fraction at exit of nozzle and ratio of inlet to exit cross-section area. 13.20A convergent-divergent nozzle has steam entering at 10 bar and 0.97 dry and leaving at 0.6 bar. For the steam mass flow rate of 0.2 kg/s determine the ratio of throat diameter to exit diameter considering that 10% of overall isentropic heat drop is the frictional heat generated in divergent [0.516] portion of nozzle. 13.21 A steam nozzle has steam entering at 6 bar, 300C and expansion occurs upto steam pressure of 1.5 bar. For the mass flow rate being 5 kg/s and nozzle exit area being 6.75 cm2 determine nozzle [90%] efficiency. 610 ________________________________________________________ Applied Thermodynamics 13.22 In a convergent-divergent nozzle steam enters at 7 bar, 0.96 dry and back pressure is 1 bar. Throat area of nozzle is 6.5 cm2. In the diverging portion the friction causes reduction in total enthalpy drop by 15% compared to ideal enthalpy drop. Determine mass flow rate, velocity and condition of steam leaving nozzle assuming negligible inlet velocity. [0.684 kg/s, 736 m/s, 0.925] 13.23 Convergent-divergent nozzle operates with steam entering at 30 bar, 300C and leaving at 5 bar, 800 m/s. Assuming negligible inlet velocity and the friction to occur in only divergent portion of nozzle with mass flow rate of 10 kg/s, determine, (a) the efficiency for expansion in divergent portion of nozzle, (b) the velocity of steam at throat. [90%, 530 m/s] 13.24 Steam enters a convergent-divergent nozzle at 7 bar, 773 K and isentropic expansion occurs up to pressure of 3 bar. Determine the rate of steam mass flow considering (i) equilibrium flow (ii) supersaturated flow with pv1.3 = constant Also determine degree of supersaturation and degree of undercooling. Assume inlet velocity to [96 kg/s, 1.71, 17.5C] be negligible. 13.25 Steam expands from 20 bar, 250C to pressure of 3 bar. Nozzle efficiency is 90% and exit area is 3.15 cm2. Assuming supersaturated flow to be governed by pv1.3 = constant and mean value of specific heat for supersaturated steam being 1.93 kJ/kgK determine the mass flow rate of steam [36 kg/min] on per minute basis. 13.26 Steam at 8 bar, dry saturated enters a convergent-divergent nozzle to expand upto 1.5 bar. Determine the ratio of cross-section area at exit and throat considering flow to be isentropic and [1.94] expansion index being 1.135. 13.27 Convergent nozzle has steam entering at 4", " bar, 150C and leaving at atmospheric pressure. The supersaturated expansion occurs following the equation, Pv 103 = 2.308 (h \u2013 1943) where v is in m3/kg, P is in bar, h is in kJ/kg. Determine the mass flow rate of steam for nozzle efficiency of 90% and throat area of 16.75 cm2. [1 kg/s] 13.28 A convergent-divergent nozzle has steam entering at 10 bar, 270C and leaving at 1.2 bar. Loss in nozzle occurs only in diverging portion because of friction and amounts to 15% of total enthalpy drop. For the cone angle being 5 in diverging portion, throat diameter being 6 mm [32 mm] determine the length of diverging portion of nozzle. 13.29 Steam at 1.5 bar, dry saturated is used in steam injector for feeding water into a boiler. Water enters into boiler at pressure being 20% more than boiler pressure of 20 bar. Water is taken from feed water tank at 20C and rate of water supply is 1.67 kg/s. Determine (a) (b) (c) Also state the assumptions made for analysis, if any. [5.2 kg/kg of steam, 7.3, 121C] 13.30 A steam injector lifts water from 4 m below the injector axis and feeds 150 kg per minute water into the boiler which is 1 m above the injector axis. Boiler produces steam at 15 bar, dry saturated and for steam injector the steam is taken from boiler itself. Water in the tank is at 30\u00b0C and the velocity of water being delivered into boiler is 20 m/s. Determine (a) the mass of steam required per hour for operating injector, (b) diameter of throat of mixing nozzle, (c) temperature of water leaving injector, (d) diameter of throat of steam nozzle for the maximum steam flow and nozzle efficiency of 92%. [1560 kg/hr, 8.15 mm, 124C, 16.35 mm] the mass of water injected per kg of steam if coefficient of steam nozzle is 0.95. ratio of diameters of steam nozzle to water discharge nozzle. temperature of water entering the boiler. Steam Turbines ________________________________________________________________ 611 14 14.1 INTRODUCTION Steam turbines date back to 120 B.C. when the first steam turbine was developed by Hero of Alexandria. Subsequently number of steam turbines came up but the practically successful steam turbine appeared at the end of nineteenth century when Gustaf De Laval designed a high speed turbine built on the principle of reaction turbine in 1883. Before this in 1629 G. Branca developed the first impulse turbine. Branca\u2019s impulse turbine and Hero\u2019s reaction turbine are shown in Fig. 14.1. Fig. 14.1 Hero\u2019s reaction turbine and Branca\u2019s impulse turbine In nineteenth century some more steam turbines were developed by Sir Charles A. Parsons and C.G. Curtis which gave a filip to the development to the modern steam turbine. Over the period of time the modern steam turbines evolved with capacity from few kilowatts to 350,000 kW and in speed from 1000 rpm to 40,000 rpm. Steam turbines offer the advantages over other prime movers in terms of simplicity, reliability and low maintenance costs. Reciprocating steam engines use pressure energy of steam while steam turbines use dynamic action of the steam. Steam turbines require less space as compared to diesel engine or steam engine and also the absence of reciprocating parts & reciprocating motion in steam turbine results in lesser vibrations and lighter foundation. In steam turbine the expanding steam does not come into contact with lubricant and so exhaust steam leaves uncontaminated. 612 ________________________________________________________ Applied Thermodynamics 14.2 WORKING OF STEAM TURBINE A steam turbine has basically nozzle and ring of moving blades mounted on a shaft called rotor. The shaft motion of turbine depends solely upon the dynamic action of steam. Steam is injected through nozzles over to the ring of moving blades. Thermal energy of steam is partly converted into kinetic energy due to static pressure drop in nozzle. High velocity steam leaving nozzle enters the moving blade and the direction of steam flow gets changed from inlet to exit. This change in direction of steam flow causes change of momentum, which results in dynamic force acting as driving thrust for rotation of shaft. Steam turbine can be impulse turbine or reaction turbine based on mechanism of driving thrust creation. If the static pressure drop occurs principally in stationary nozzle with little or no static pressure drop occurring in rotor blade passage, then turbine is called an \u2018impulse turbine\u2019. If the substantial static pressure drop occurs in stationary nozzle and rotor blade passage both then turbine is called \u2018reaction turbine\u2019. Thus in case of impulse turbine driving thrust is available due to change in momentum because of change in velocity direction while moving across the blade from inlet to exit. In", " case of reaction turbine the static pressure drop in rotor blade passage causes further conversion of thermal energy into kinetic energy and hence generation of resultant reactive force. Driving thrust in reaction turbine comprises of reactive force and force associated with change in momentum due to change in direction of velocity. Impulse Turbine has basically following components which are also described in Fig. 14.2. Stationary nozzle: Impulse turbine has stationary passage comprising of one or more stationary nozzles in which steam at high static pressure and low velocity is expanded so as to increase velocity of steam at exit of nozzle. High velocity stream leaves nozzle so as to deliver steam with correct direction for smooth entry into moving blade. Moving blades: Moving blades are fixed around the circumference of a rotor shaft with top of blades connected together for rigidity by means of a blade shroud ring. Blades actually cause change in direction of steam and so the momentum change occurs and thus impulse is generated. Steam turbine blades are made of alloy steel and manufactured by machining from bar stock. Typical blade material called nickel based super alloy may have Ni, Co, W, Cr, Al, Tn, Ti, Hf, Fe, Mo, C, Si, Mn, Cu, Za, B, S, Pb in percentage by weight of 59, 10, 10, 9, 5.5, 2.5, 1.5, 1.5, 0.25, 0.25, 0.15, 0.1, 0.1, 0.05, 0.05, 0.015, 0.008 and 0.0005, respectively. Guide blades: Turbine rotor has guide blades prior to moving blades so as to guide steam in proper direction for smooth entry into moving blade. Guide blades are stationary guides mounted between the rotor blade rings. Guide blades perform function of reversing the direction of steam leaving the preceding moving blade row so that direction of steam entering moving blade rows is similar. Casing: Casing refers to the outer enclosure housing nozzles and fixed blades. Casing is also referred to as shell or cylinder. Casing confines steam to flow passages and also provides the structural frame. Casings of large steam turbines are normally horizontally split for convenience of repair and inspection. For high pressures and temperatures casings are made of cast carbon steel while low pressure turbine casings are made from rolled flat steel plate. For very high temperatures stainless steel casings are good. Shaft: Shaft or rotor or spindle refers to the rotating member upon which moving blade ring is mounted. Ring or Wheel: Wheel refers to the ring upon which moving blades are mounted. Wheel is keyed on to the shaft. This ring is also called disc. Steam Turbines ________________________________________________________________ 613 Fig. 14.2 Impulse turbine Diaphragm: Diaphragm is attached to the casing containing the nozzles and performs function of confining steam flow to nozzle passage. Packing: Packing is provided for preventing the leakage across the annular space between the diaphragm and shaft, casing and shaft. Packing is provided in the form of carbon rings or labyrinth glands. Carbon rings present an effective seal against shaft leakage in small turbines and are extensively used with labyrinth and water seals for preventing shaft leakage in larger turbines. Carbon ring consists of a ring of carbon rectangular in cross section and usually divided into four segments. Rings snugly fit into a recess in casing and are kept tight against shaft by means of garter spring. Labyrinth seals consist of series of thin strips fixed to the casing or other stationary member and arranged so as to maintain the smallest possible clearance with the shaft. The labyrinth seals have small restrictions that increase the velocity of leaking fluid only to have it dissipated in the pockets, thereby throttling the fluid. Figure 14.3 shows the straight and stepped labyrinth seals. The tips of the strips are extremely thin so that if rubbing occurs the tip gets worn away without damaging the shaft. These labyrinth seals do not prevent complete leakage of fluid. These are effective in only reducing leakage. Complete leakage prevention is done in association with other seals as carbon seals described earlier. Steam chest: This is the steam supply chamber which houses steam before being supplied to Fig. 14.3 Labyrinth seals nozzles. 614 ________________________________________________________ Applied Thermodynamics Exhaust hood: The portion of casing which collects and delivers the exhaust steam to exhaust pipe or condenser is called exhaust hood. Throttle valve: Throttle valve, also called stop valve is located in steam supply line of the steam turbine. Throttle valve may be manually operated or hydraulically operated for regulating steam flow during start and stop of turbine. Governor: Governor is also provided in steam turbine for controlling the steam mass flow so as to maintain constant speed with load fluctuations. An overspeed governor with trip mechanism is also provided to shut off the supply of steam. Bearings: Turbine has main bearings to support the shaft. Along with these thrust bearings are also provided to support the axial thrust. Turning gear: Turning gear is generally used", " with large turbines and consists of a gear integral with turbine shaft driven by electric motor through necessary speed reduction. Turning gear is used so as to keep the turbine shaft rotating at about 1\u201320 rpm in order to avoid springing of shaft occurring due to unequal expansions and contractions during warming and cooling of turbine. Trip mechanism: Steam turbine is provided with a trip mechanism operating through an overspeed governor to shut off supply of steam to turbine. Trip mechanism is actually safety device which gets activated upon number of other adverse operating conditions of turbine such as loss of lubricating oil pressure or condenser vacuum or excessive axial thrust etc. 14.3 CLASSIFICATION OF STEAM TURBINES Steam turbines may be classified into different categories based on various attributes as given below. (a) Based on the blade flow passage: In steam turbine thermal energy available with steam is converted into kinetic energy which in turn produces driving thrust on the shaft. Based upon the rotor blades the blade flow passage may be of (i) constant cross section area type from blade inlet to exit, or, (ii) varying cross section area type from blade inlet to exit. Turbines having former type blading are called impulse turbines while later type are in reaction turbines. Figure 14.4 shows the impulse and reaction turbine blades. Fig. 14.4 Impulse turbine and Reaction turbine blades The mechanism of impulse and reaction forces getting generated is discussed here. From New- ton\u2019s second law we know, F = m\u00b7a d V dt F = (m) For steady flow, F = (mass flow rate) (change in velocity) Similarly,Tangential force = (mass flow rate) (change in tangential component of velocity) The impulse force can be defined as the force because of change in tangential component of velocity of fluid which may be due to change in direction or magnitude. Figure 14.5 shows the impulse force generating because of change in velocity of fluid. Steam Turbines ________________________________________________________________ 615 Fig. 14.5 Impulse force In case of Fig. 14.5 (a) the impulse force is available due to change in magnitude of velocity and shall be given by the product of mass flow rate and change in velocity. In case (b) the impulse force is generated due to change in direction of velocity and if the blade is stationary and frictionless then there shall be no decrease in magnitude of velocity. Reaction force is available when the tangential velocity of fluid is increased and is opposite in reference to the direction of velocity. In the case shown in Fig 14.5 (b), the total force exerted on blade is actually a combination of impulse and reaction. Impulse force is available in the entrance half of the blade where jet impinges causing a force to right. While in the exit half, the leaving jet exerts a reactive force on the blade which also acts to the right. Combined effect of the two forces on the impulse blade is arbitrarily called impulse force. Reaction force available due to increase in tangential velocity of fluid can be seen in case of nozzle due to acceleration of fluid. Fig. 14.6 Reaction force Detailed discussion on the reaction turbines running on shaft work resulting from reaction forces is given along with \u201creaction turbines\u201d. (b) Based on the cylinder flow arrangement: Steam turbines may be classified based upon the flow arrangement into following types. (i) Single flow single casing turbine (ii) Double flow single casing turbine (iii) Cross flow compound turbine with single flow (iv) Cross flow compound turbine with double flow (v) Triple cross flow compound turbine with double flow Various arrangements in the above listed turbines are shown by line diagram in Fig 14.7 Single cylinder turbines are the one which have all the stages enclosed in one cylinder while in multi cylinder 616 ________________________________________________________ Applied Thermodynamics turbines the stages are accommodated in more than one cylinders, say two or three. Flow in these can be single flow, double flow, cross flow or reversed flow. Fig. 14.7 Classification based on flow arrangement (c) Based on direction of flow: Steam turbines can be classified based on the direction of flow by which steam flows through turbine blading. Steam turbines can be: (i) Radial flow turbine (ii) Tangential flow turbine (iii) Axial flow turbine Radial flow turbines were developed by B.F. Ljungstorm of Sweden as shown in Fig. 14.8. Radial flow turbine incorporates two shafts end to end and can be of suitably small sizes. Radial flow turbines can be started quickly and so well suited for peak load and used as stand by turbine or peak load turbines. These are also termed as Ljungstrom turbines. Steam Turbines ________________________________________________________________ 617 Fig. 14.8 Radial flow turbine (Ljungstrom turbine) In radial flow turbines the steam is injected in middle near shaft and steam flows radially outwards through the successive moving blades placed concentrically. In radial flow turbines there are no stationary blades", " so pressure drop occurs in moving blade passage. Concentric moving blades rings are designed to move in opposite directions. In tangential flow turbines the nozzle directs steam tangentially into buckets at the periphery of single wheel and steam reverses back and re-enters other bucket at its\u2019 periphery. This is repeated several times as steam follows the helical path. Tangential flow turbines are very robust but less efficient. In axial flow turbines steam flows along the axis of turbine over blades. These axial flow turbines are well suited for large turbo generators and very commonly used presently. (d) Based on number of stages: Steam turbines can also be classified based upon the number of stages in steam turbines i.e. depending upon the amount of heat drop. It can be: (i) Single stage turbine (ii) Multi stage turbine. Single stage turbines have the expansion occurring in single stage while in multi stage turbines the expansion occurs in more than one stages of turbine. When expansion occurs in two stages it is called double stage turbine and with expansion occurring in three stages it is called triple stage turbine. 618 ________________________________________________________ Applied Thermodynamics (e) Based on the application of turbine: Depending upon application the steam turbine can be classified as below: (i) Condensing turbine (ii) Non-condensing turbine (iii) Back pressure turbine (iv) Pass out turbine Condensing steam turbines are those in which steam leaving turbine enters into condenser. Such type of steam turbines permit for recirculation of condensate leaving condenser. Also the pressure at the end of expansion can be lowered much below atmospheric pressure as the expanded steam is rejected into condenser where vacuum can be maintained. Condensing turbines are frequently used in thermal power plants. Non-condensing steam turbines are those in which steam leaving turbine is rejected to atmos- phere and not to condenser as in case of condensing turbine. Back pressure turbines reject steam at a pressure much above the atmospheric pressure and steam leaving turbine with substantially high pressure can be used for some other purposes such as heating or running small condensing turbines. Fig. 14.9 Condensing and non-condensing steam turbines Pass out turbines are those in which certain quantity of steam is continuously extracted for the purpose of heating and allowing remaining steam to pass through pressure control valve into the low pressure section of turbine. Pressure control valve and control gear is required so as to keep the speeds of turbine and pressure of steam constant irrespective of variations of power and heating loads. (f) Based on speed of turbine: Steam turbines can be classified based upon the steam turbine as low speed, normal speed and high speed turbines as given below. (i) Low speed steam turbine. (ii) Normal speed steam turbine. (iii) High speed steam turbine. Low speed turbines are those steam turbines which run at speed below 3000 rpm. Normal speed steam turbines are those turbines which run at speed of about 3000 rpm while high speed steam turbines are the one which run at more than 3000 rpm. (g) Based on pressure in steam turbines: Steam turbines can also be classified based upon the inlet pressure of steam turbine as follows: Steam Turbines ________________________________________________________________ 619 (i) Low pressure steam turbine (ii) Medium pressure steam turbine (iii) High pressure steam turbine (iv) Super pressure steam turbine Low pressure steam turbines have pressure of inlet steam less than 20 kg/cm2 while medium pressure steam turbines have steam inlet pressure between 20 kg/cm2 to 40 kg/cm2. High pressure steam turbines have steam inlet pressure lying between 40 kg/cm2 to 170 kg/cm2 while turbines having inlet steam pressure more than 170 kg/cm2 are called super pressure steam turbines. 14.4 IMPULSE TURBINE Schematic of impulse steam turbine is shown in Fig. 14.10. It has single-stage having a nozzle fitted in the casing followed by ring of moving blades mounted on the shaft. Variation of velocity and pressure along the axis of turbine is also shown here. Here pressure drop occurs only in the nozzle and ideally no pressure drop occurs in blades. Fig. 14.10 Schematic of simple impulse steam turbine stage High pressure steam from boiler enters the nozzle through pipings and leaves nozzle at predefined angle so as to smoothly flow over the moving blades. Steam velocity gets increased during its flow through nozzle due to its expansion occurring in it. During the passage of steam over the moving blades steam undergoes change in its\u2019 direction while losing the velocity and thus causing rotation of moving blade ring mounted on shaft. Simple impulse turbine is used where small output at very high speed is required or only a small pressure drop is available. These are not suited for applications requiring conversion of large thermal energy into work. Speed of operation of turbine can be regulated by \u2018compounding\u2019 of impulse turbine discussed ahead. Compounding of steam turbine is required as in case of simple impulse turbine, the single stage may offer speed of the order of 30,000 rpm which can not be", " directly used for any engineering application and needs to be reduced. Also such a high speed shall induce large stresses in the blades. Compounding is a thermodynamic means for reducing the speed of turbine where speed reduction is realized without employing a gear box. 620 ________________________________________________________ Applied Thermodynamics Compounding can be of following three types: (i) Pressure compounded impulse turbine (ii) Velocity compounded impulse turbine (iii) Pressure-velocity compounded impulse turbine Detailed discussion upon the above three types of compoundings is given below: (i) Pressure compounded impulse turbine: Pressure compounded impulse turbine is also called as \u2018Rateau\u2019 turbine. Here pressure staging is done to utilize high velocity steam at acceptable shaft speed. In this the entire pressure drop is realized in parts instead of taking it in single stroke. This segmentation of pressure drop results in moderate steam velocities and thus yielding acceptable rotational speed. In case of pressure compounding there is a ring of fixed nozzles followed by ring of moving blades and subsequently there is again a ring of nozzles followed by a ring of moving blades. Thus pressure compounded impulse turbine consists of a series of simple impulse stages or De Laval turbine stages. Discharge from each moving blade row is supplied to stationary nozzle ring of the subsequent stage. In pressure compounding high pressure steam enters the first ring of stationary nozzles where part expansion occurs causing drop in pressure and increase in steam velocity. Steam subsequently enters moving blade ring where no pressure drop occurs due to symmetrical blading but velocity drops. Steam leaving moving blade ring enters the stationary nozzle ring where remaining part of expansion occurs and expanded steam subsequently enters the moving blade ring. Pressure and velocity variation in a pressure compounded impulse turbine stage are shown in Fig. 14.11 along with the schematic of such compounding. Fig. 14.11 Pressure compounded impulse turbine stage (Rateau turbine) Steam Turbines ________________________________________________________________ 621 In pressure compounding as the pressure drop occurs in parts so the steam velocities are not very large and hence the turbine velocity gets reduced to low value. Turbine velocity may be further lowered if number of stages is increased. Therefore, pressure compounded impulse turbine has large number of stages which make it most expensive. This type of compounding is of most efficient type as in this ratio of blade velocity to steam velocity remains constant. Pressure compounding is more prone to leakage of steam from one section to other section at the shaft and outer casing as all pressure drop occurs in the nozzles. (ii) Velocity compounded impulse turbine: Velocity compounded impulse turbine is called \u2018Curtis\u2019 turbine. Here velocity staging is employed in order to utilize the high velocity steam jet with acceptable rotational speed. In velocity compounded impulse turbine instead of absorbing all kinetic energy in a single moving blade ring it is divided into two or more moving blade rings with guide blades in between the rows. Schematic of velocity compounded impulse turbine stage with pressure and velocity distribution is shown in Fig. 14.12. Fig. 14.12 Velocity compounded impulse turbine or \u2018Curtis turbine\u2019 In velocity compounded impulse turbine the high velocity steam from boiler enters the first ring of stationary nozzles and undergoes the complete pressure drop as desired in a stage along with increase in velocity. Low pressure and high velocity steam leaving nozzle enters the moving blade ring where a part of velocity drop takes place while pressure drop does not occur due to symmetrical blade profile. 622 ________________________________________________________ Applied Thermodynamics Steam leaves moving blade ring and enters the fixed blades which act as guide blades. Steam is smoothly guided by fixed blade ring and passes on to ring of moving blades in which remaining part of velocity drop takes place. Thus in a stage of velocity compounded impulse turbine there is stationary nozzle ring followed by moving blade ring and subsequently a fixed blade ring and moving blade ring. Here pressure drop occurs only in nozzle and the velocity drop occurs in two parts in two moving blade rings respectively. For the smooth and symmetrical impulse turbine blades used as fixed guide blades there is no drop in velocity of steam passing through fixed blade ring. Velocity compounded impulse turbine offers advantages such as less number of stages compared to pressure compounding and so less cost. It also requires less space and is relatively more reliable and easy to start. In multi stage velocity compounded impulse turbine the first stage has large pressure drop and remaining turbine stages are subjected to lower pressure range, thus lesser number of stages. In velocity compounded impulse turbine since pressure drop occurs in nozzle itself so the rest of turbine and its\u2019 casing need not be manufactured very strong. But the efficiency is low due to large frictional losses due to large initial velocity and \u2018non optimum value of ratio of blade velocity to steam velocity for all blade rings\u2019. Efficiency of velocity compounded impulse turbine goes on decreasing with increase in number of stages. (iii) Pressure-velocity compounded impulse turbine: Pressure-velocity compounded impulse turbine is a combination of the two types of compoundings described earlier. In this, steam coming from boiler enters the stationary nozzle ring", " followed by moving blade ring and subsequently fixed blade ring followed by moving blade ring. Steam leaving moving blade ring enters the stationary nozzle ring followed by moving blade, fixed blade and moving blade ring respectively. Schematic of pressurevelocity compounded impulse turbine stage is shown in Fig. 14.13 along with pressure and velocity variation across the different sections. Fig. 14.13 Pressure-velocity compounded impulse turbine Steam Turbines ________________________________________________________________ 623 Here both pressure drop and velocity drop are divided into different sections as shown in Fig. 14.13. Thus here one or more \u2018Curtis stage\u2019 (velocity compound) followed by \u2018Rateau stage\u2019 (pressure compound) are provided. Curtis stages reduce pressure to a moderate level with high proportion of work per stage and then the highly efficient \u2018Rateau stages\u2019 absorb the remaining energy available. Here it is possible to reduce over all-length of turbine and thus there is saving in initial cost which more than offsets the lower efficiency. 14.5 VELOCITY DIAGRAM AND CALCULATIONS FOR IMPULSE TURBINES Velocity diagram gives an account of velocity of fluid entering and leaving the turbine. Velocity diagrams for single stage of simple impulse turbine and compound steam turbine are described here. Simple impulse turbine: Single stage of simple impulse turbine is shown in Fig. 14.14. It comprises of a row of nozzle followed by moving blade row. Pressure and velocity variations along the stage in nozzle ring and moving blade ring are also shown. Subscript 0, 1 and 2 refer to nozzle inlet, nozzle exit or moving blade inlet and moving blade exit respectively. Fig. 14.14 Single stage of simple impulse turbine Figure 14.15 gives the inlet and outlet velocity diagrams at inlet edge and outlet edge of moving blade along with the combined inlet and outlet velocity diagram for a stage of simple impulse turbine. The notations used for denoting velocity angles and other parameters during calculations are explained as under, (SI system of units is used here). U = Linear velocity of blade = d N 60, m/s where d = mean diameter of wheel in \u2018m\u2019 N = Speed in rpm. C1 = Absolute velocity of steam at inlet to moving blade or velocity of steam leaving nozzle. (Absolute velocity is the velocity of an object relative to the earth) C2 = Absolute velocity of steam at exit of moving bade. 624 ________________________________________________________ Applied Thermodynamics C 1 w Fig. 14.15 C1w = Whirl velocity at inlet to moving blade or tangential component of absolute velocity at inlet to moving blade. C2w = Whirl velocity at exit of moving blade or tangential component of absolute velocity at exit of moving blade. C1a = Flow velocity at inlet to moving blade or axial component of absolute velocity at inlet to moving blade. C2a = Flow velocity at exit of moving blade or axial component of absolute velocity at exit of moving blade. V1 = Relative velocity of steam at inlet of moving blade (Blade velocity at inlet) (Relative velocity is the absolute velocity of one moving object compared with absolute velocity of other object.) V2 = Relative velocity of steam at exit of moving blade (Blade velocity at exit). m = Mass of steam flowing over blade = Ratio of linear velocity of blade and absolute velocity at inlet of moving blade = U C1 Steam Turbines ________________________________________________________________ 625 k = Blade velocity coefficient (Ratio of relative velocity at exit and inlet) = V 2 V 1 = Angle of absolute velocity with respect to the direction of blade motion. 1 = Angle of absolute velocity at inlet to moving blade or nozzle angle. 2 = Angle of absolute velocity at exit of moving blade or inlet angle of fixed blade in next stage. = Angle of relative velocity with respect to the direction of blade motion. 1 = Angle of relative velocity at inlet or inlet angle of moving blade. 2 = Angle of relative velocity at exit or exit angle of moving blade. Here steam enters the nozzle and leaves so as to smoothly glide into the ring of moving blades. Steam leaves nozzle with absolute velocity C1 and at an angle of 1. This steam stream will be delivered to moving blade with velocity C1 and angle 1 but due to linear velocity of moving blade the steam stream actually glides over the moving blade with velocity V1 and blade angle at inlet 1. This velocity V1 is actually the result of two velocity vectors C1 and U. V1 is called the relative velocity of steam at inlet of moving blade. Steam stream leaves the moving blade with velocity V2 which is relative velocity of steam at exit of moving blade. Thus relative velocity is the actual velocity with which steam flows over the moving blade. For a perfectly smooth and frictionless blade this relative velocity should not change from inlet to exit as there is no expansion of steam in moving blade (blades are symmetrical and passage between two consecutive", " moving blades is of constant area type from inlet to exit). Actually there always exist some friction over the blade so the relative velocity at outlet will be smaller than the relative velocity at inlet, i.e. V2 < V1. This reduction in relative velocity is quantified by parameter called blade velocity coefficient (K). Blade velocity coefficient (K) is defined as, K = V 2 V 1. 1 in this case. If we look at inlet section 1, then it is obvious that for the maximum change in momentum steam should be delivered to the moving blade horizontally i.e. 2 = 0 with the semi-circular shaped moving blade. This semi-circular moving blade is not possible practically as the moving blade wheel (ring) has series of blades and each blade has to receive steam from series of nozzles one after the other. This is the reason due to which nozzles are placed at some angle to the blade, say angle 1 = 0 and also leave horizontally i.e. Due to injection of steam at angle 1 with velocity C1 over the blade, steam shall have two components of velocity i.e. one tangential component and other axial component. Tangential component of velocity is parallel to the direction of rotation of blades and is also called as whirl velocity. Axial component of velocity is perpendicular to the direction of rotation of blade and is also called flow velocity. Axial component or flow velocity is responsible for maintaining flow of steam across the moving blade row. Volume flow rate of steam across the moving blade ring can be given by the product of flow velocity and effective passage area available for flow. The magnitude of flow velocity influences the size of wheel for given steam volume flow rate. The whirl component of velocity is responsible for generation of thrust force due to change in momentum. Both whirl velocity and flow velocity being the two perpendicular components of absolute velocity depend largely upon the angle of absolute velocity i.e,. At inlet the angle 1 should be selected depending upon the thrust requirement and maintenance of flow across the blade row. With increase in angle 1, increases. Thus the 1, decreases while the flow velocity, C1 sin 1 the whirl velocity, C1 cos 626 ________________________________________________________ Applied Thermodynamics maximization of one leads to minimization of the other and so the compromise should be had for selecting the angle of absolute velocity. Similarly at exit of blade again there shall be whirl velocity and flow velocity components. For absolute velocity at exit being C2 and angle of absolute velocity at exit being 2 (this shall be the inlet angle for subsequent nozzle if more than one similar stages are there), the whirl velocity shall be C2 cos 2. In the simple impulse turbine stage this whirl velocity component at exit is a kind of loss at exit. Therefore in order to minimize loss at exit this component should be minimum. For minimizing the whirl velocity at exit i.e. C2 cos 2, the angle should be made minimum. Minimum loss could be reduced to zero if angle 2 is equal to 90. This becomes a typical case in which the turbine discharges axially (at 90) and such turbines are also called axial discharge turbine. 2 and flow velocity as C2 sin 2 Thus it is obvious that in case of impulse turbine stage the moving blade merely deflects the steam and the change in direction of steam from inlet to exit causes change of momentum and thus thrust is generated. Moving blades being of symmetrical type offer a constant cross-section area between two consecutive blades from inlet to exit and so no expansion occurs in the moving blade. The steam expansion only occurs in the nozzle. Also in the absence of expansion across moving blade the pressure of steam remains constant from inlet to exit under ideal conditions. For symmetrical blades the 1 = inlet and exit angles of blade are same i.e. 2. Velocity diagrams are separately drawn at inlet of moving blade and at exit of moving blade as shown in Fig. 14.15. Combined velocity diagram for the stage is also given here. Using the velocity diagram various parameters can be estimated as discussed ahead. The driving thrust on the rotor is produced as reaction of the force estimated by the rate of change of momentum. Tangential force will be the force component producing driving thrust while axial force will be the force component producing axial thrust. Driving thrust causes rotation of rotor while axial thrust is to be absorbed by the thrust bearings provided for this purpose. Let the mass flow rate be m, kg/s. Tangential force FT = m (change of tangential component of velocity or whirl velocity) 2 \u2013 C1 cos 1) = m (\u2013 C2 cos FT = \u2013 m (C2 cos 2 + C1 cos Driving thrust on rotor will be reaction of this force and will be equal and opposite. Driving thrust, FD = \u2013 FT 1) = \u2013 m Cw FD = m (C2 cos FD = mCw 2 + C1 cos a1) From velocity triangles", ", (C1 cos 1 + C2 cos or, Cw = Vw 2) = (V1 cos 1 + V2 cos 2) Hence FD = m (C2 cos = m (V1 cos 2 + C1 cos 1) 1 + V2 cos 2) This driving thrust can be used for getting the rate of work done on the rotor. Rate of work done = W = FD U W = mCw.U Work done per unit of steam mass flow, w = UCw Steam Turbines ________________________________________________________________ 627 Rate of work done will be the power produced by the turbine stage. W = mU (C2 cos W = mU (V1 cos = mU V1 cos 2 + C1 cos 1) 1 + V2 cos 2) cos V 2 1 1 cos V 1 2 1 = mU (C1 cos W = m U (C1 cos 1 \u2013 U) 1 cos cos 1 \u2013 U ) {1 + K C} V 2 V 1 2 1 where K = Blade velocity coefficient = V 2 V 1 C = Ratio of cosines of blade angles = cos cos 2 1 For perfectly smooth and symmetrical blade both K and C shall have unity value. i.e. K = 1, C = 1 Therefore for simple impulse turbine stage having perfectly smooth and symmetrical blade, rate of work done, W = 2 mU (C1 cos 1 \u2013 U) From velocity diagram also the rate of work done per unit of steam mass flow \u2018w\u2019, can be estimated as below, w = U Cw Combined velocity diagram if drawn to the scale gives, U = AB, Cw = EF. Thus length EF and AB shall give Cw and U respectively. Hence, w = AB EF, rate of work done per unit of steam mass flow Rate of work done W = m AB EF. The suitable scale factors are to be used while using lengths. The velocity diagram at inlet, outlet or combined velocity diagram can be suitably drawn to the scale and various parameters be obtained by measuring respective geometrical entity i.e. length and angle. The work available at rotor can also be obtained using steady flow energy equation between section 1 and 2. Assuming no change in potential energy from inlet to exit across the moving blade and no heat interaction across the stage, the S.F.E.E. can be given as In case of impulse stage the change in enthalpy from section 1 to 2 can be given by the change in kinetic energy associated with relative velocity from 1 to 2. 628 ________________________________________________________ Applied Thermodynamics (h1 \u2013 h2 or, (h1 \u2013 h2) = 1 2 Substituting in W, Rate of work done in a stage For perfectly smooth moving blade, V1 = V2 and (h1 \u2013 h2) = 0. Hence for stage with smooth blade From velocity triangle at inlet C1 2 = V1 2 + U 2 + 2UV1 cos and from velocity triangle at outlet, C2 Combining above two we get, 2 \u2013 C2 C1 1 + V2 cos Vw = Cw 2) = (V1 2 \u2013 V2 2) = 2UCw It is already seen, (V1 cos or, or, or, 2 \u2013 C2 2 \u2013 V1 Substituting in the expression for rate of work done, (C1 2) + (V2 2 = (V1 2 \u2013 C2 2 \u2013 V2 (C1 2) + 2UCw 1 2 = V2 2 + U2 \u2013 2UV2 cos 2 2) + 2U(V1 cos 1 + V2 cos 2) 2) = (C1 cos 1 + C2 cos 2) W = m 2 {2U Cw} or, W = mUCw Axial thrust: Axial component of velocity or flow velocity change causes creation of axial thrust. W = mU (C1 cos 1 + C2 cos 2) Axial thrust due to change in momentum because of change in flow velocity. Axial thrust = m (C1a \u2013 C2a) = m (C1 sin Axial thrust, Fa = m Ca From velocity diagrams the change in axial velocity is given by length DG. Length, DG = EC \u2013 FD 1 \u2013 C2 sin 2) Axial thrust = m DG = m (EC \u2013 FD) Diagram efficiency or blading efficiency: Diagram efficiency or blading efficiency refers to the ratio of work done to the energy supplied to rotor in a stage. In case of impulse turbine this energy supplied to rotor in a stage can be given by the kinetic energy supplied at inlet of moving blade. Assum- ing no expansion in moving blades the energy supplied to rotor can be given by m 2 1 C 2. Rate of work Steam Turbines ________________________________________________________________ 629 done per unit mass of steam is also called diagram work. Hence, diagram", " efficiency = D = Rate of work done Energy supplied to rotor Diagram or blading efficiency = 2 U C 2 C 1 w Stage efficiency or Gross efficiency: Stage efficiency refers to the ratio of rate of work done and energy supplied to the stage. Energy supplied to the stage can be accounted by the change in enthalpy between section 0 and 1 i.e. inlet of nozzle to exit of nozzle. Stage efficiency is thus the output of stage divided by the available energy for the stage. Energy supplied to stage = m (h0 \u2013 h1) Stage efficiency, s = = Rate of work done Energysupplied tostage m U C w m h h 1 0 Stage efficiency = U C w h h 1 0 Nozzle efficiency: Nozzle efficiency refers to the ratio of kinetic energy available and the enthalpy change occurring across the nozzle i.e. between inlet and outlet (sections 0 and 1). Nozzle efficiency, N = Nozzle efficiency = Combining the stage efficiency, diagram efficiency and nozzle efficiency, it can be given that; Stage efficiency = Diagram efficiency Nozzle efficiency Overall efficiency: The overall efficiency of stage can be given by the ratio of work delivered at turbine shaft to the energy supplied to the stage. Overall efficiency, o = Work delivered at shaft Energysupplied to thestage Overall efficiency = Nozzle efficiency Diagram efficiency Mechanical efficiency 630 ________________________________________________________ Applied Thermodynamics Optimization of turbine stage: Turbine being work producing machine is designed with the aim of providing maximum work output. The diagram efficiency of turbine should be maximized as it indicates the rate of work done per unit of energy supplied to the rotor. Work output per unit time in a simple impulse turbine stage is given by, W = mU (C1 cos 1 \u2013 U) (1 + KC). In this expression for work the output is seen to depend upon m, C1, U, 1. Work output shall be maximum when angle 1, K and C. For all other conditions fixed work output is seen to vary inversely with the angle 1 i.e. the work output increases with decrease in nozzle exit angle, 1 is reduced to zero but nozzle exit angle 1 can not be reduced to zero as it shall not allow steam injection to successive 1 also decides the axial or flow velocity at inlet, Ca, i.e. smaller the value of blades. Angle 1 smaller will be axial velocity or flow velocity. As described earlier that the flow velocity maintains the flow across the stage so smaller flow velocity requires larger steam flow passage area for the same mass flow of steam. Small value of angle 1 will thus yield larger blade surface area resulting into larger surface friction loss. Therefore the angle 1 is chosen as compromise between above considerations. Generally 1 lies between 15 and 30\u00b0. Thus the work can not be maximized by only selecting the minimum value of angle 1 and so requires optimization of turbine stage performance with respect to some other parameter. The diagram efficiency should be maximized with respect to suitable parameter. Diagram efficiency cos 1 2 C 1 cos 1 U 1 KC U C 1 (1 + KC) Here U C1, is non-dimensional form of velocity. Let us denote by i.e. U C1 Diagram efficiency can be optimized with respect to non-dimensional velocity denoted by. This non-dimensional velocity is also called as \u2018blade-steam velocity ratio\u2019 or \u2018blade speed-steam velocity ratio\u2019 or \u2018blade-steam speed ratio\u2019. 1 \u2013 ) (1 + KC) D = 2 (cos, so, = Differentiating D with respect to, we get, d D d = 2 (cos 1 \u2013 2) (1 + KC) For a perfectly smooth and symmetrical blade, K = 1cos 1 \u2013 2) Equating first differential to zero; cos 1 \u2013 2 = 0 = cos 2 1 Steam Turbines ________________________________________________________________ 631 Second order differential of D with respect to indicates that the diagram efficiency is maximum corresponding to the blade speed-steam velocity ratio given as cos 2 1. Hence, maximum diagram efficiency, = D,max cos2 1 1 2 KC For perfectly smooth and symmetrical blade, maximum diagram efficiency, D,max = cos2 1 Corresponding to = cos 2 1 done. Mathematically,, the rate of work done can be given as the maximum rate of work Maximum rate of work done can be obtained by substituting Rate of work, = U C1 = cos 2 1 or C1 = 2 U cos 1 W = mU \u00d7 (C1 cos a1 \u2013 U) (1 + KC) Wmax = mU U (1 + KC) Wmax = mU 2 (1 + KC) For perfectly smooth and symmetrical blade, K = 1, C = 1, Wmax = 2 mU 2 The variation of diagram efficiency can be plotted with varying blade-steam velocity ratio as given below. Fig. 14.16 Diagram efficiency with blade speed-steam", " velocity ratio. 632 ________________________________________________________ Applied Thermodynamics 14.6 IMPULSE TURBINE BLADE HEIGHT Impulse turbine blade height is a very important parameter. Blade height is a function of total annular area required for the flow of fluid. It depends upon the mass flow rate through the section, specific volume of steam at that section, Fig. 14.17 Nozzle and blade passage area of section through which steam is passing and flow velocity at the section. In estimating the net annular area available for flow, edge thickness and blade angles are taken into account. The nozzle passage and two row Curtis stage are shown in Fig 14.17. Here estimation of both nozzle height and blade height has been described ahead. For the nozzle passage shown the area available for flow at exit of one nozzle passage can be given as A; A = hn, where is the width of flow passage at exit at mean nozzle height and hn is nozzle height. The width of flow passage at exit at mean nozzle height can be given as, = (p sin \u2013 t), where \u2018p\u2019 is nozzle pitch at mean nozzle height and \u2018t\u2019 is edge thickness of nozzle and \u2018\u2019 is nozzle angle. Therefore, A = (p sin \u2013 t)hn In the absence of information about edge thickness considerations above equation can be used to get nozzle height. For full peripheral admission when nozzle diaphragm is occupied by nozzles, the number of nozzles can be obtained as, d p n = where \u2018d\u2019 is the diameter of mean nozzle ring and \u2018n\u2019 is number of nozzles. Edge thickness in nozzle calculations can be accounted by edge thickness factor (k) denoted as, k = p sin p sin t Thus, A = k hn p sin, can be used to get nozzle height. Considering the total number of nozzles, the total nozzle area can be given as, An = nA An = d hn k sin Steam Turbines ________________________________________________________________ 633 Figure 14.17(b) shows the two stage Curtis stage having blade heights as hb1, hb2, hb3. Theoretically blade height at entrance is equal to nozzle height at exit but in order to avoid spilling of the fluid issuing from nozzle passage the blade height is slightly increased at entrance. In case of impulse turbine blade the pressure on both sides of blade remains constant and by continuity equation the volume flow rate shall also remain same from inlet to exit. From continuity equation for nozzle to moving blade, the volume flow rate shall be, k sin = V1 d hb where hb is blade height, kb is blade edge thickness factor and is blade angle. C1 dhn sin kb Generally it is assumed, k = kb, hence sin sin = From velocity diagram at blade inlet it can be seen that, C1 sin = C1a = V1 sin = V1a. Hence hb = hn, ideally but as described earlier the height at blade entrance is slightly increased, which is also called \u2018blade step-up\u2019 or \u2018overlap\u2019. Similarly for two row Curtis stage considered here the blade height can be related as, hb1 = hb2 = hb3 = Step up or overlap is arbitrarily taken and varies from 0.2 cm in high pressure stages to 2 cm in low pressure stages of large turbines. Here in this calculation no consideration is made for increase in specific volume of fluid in blade passage due to reheat although it shall be there but may be neglected on account of being small. Maximum height of blade is restricted by the stresses getting generated due to bending and centrifugal forces. Alternatively, the blade height can be estimated by the volume flow rate considerations. Moving blade height at exit can be given as below; l pb (pb sin mv = 2 \u2013 tb) lhb1 V2 where, l is the length of arc covered by nozzle and pb is the distance between two consecutive blades, l pb gives the number of exit passages in tb is edge thickness of blade as shown in Fig 14.18. Here nozzles through which steam arrives over the moving blade. 634 ________________________________________________________ Applied Thermodynamics Steam turbines are called full admission turbines when the nozzle coverage of blades is complete i.e. in such case l = d, where \u2018d\u2019 is mean blade diameter of the rotor. Fig. 14.18 Blade height description 14.7 CALCULATIONS FOR COMPOUNDED IMPULSE TURBINE Velocity diagram for simple impulse turbine is described earlier. Similar to the velocity diagram described, the velocity diagram for the compounded impulse turbine stage can be drawn. Velocity diagram for velocity compounded impulse turbine stages is shown in Fig. 14.19. Fig. 14.19 Stage of velocity compounded impulse turbine and velocity diagram for the stage. Here high pressure steam first enters the nozzle and steam leaving nozzle", " enters the row of moving blades. Steam leaving moving blades enters the fixed blade row. Fixed blades just act as guide Steam Turbines ________________________________________________________________ 635 blades and are fixed to casing. Steam is guided by fixed blade to smoothly enter into next row of moving blades. Thus nozzle, moving blade row, fixed blade row and moving blade row together constitute stage of velocity compounded impulse turbine. Let us consider section at exit of nozzle as 1 \u2013 1, at exit of first moving blade row as 2 \u2013 2, at exit of fixed blade row as 3 \u2013 3 and at exit of second row of moving blades as 4 \u2013 4. Absolute velocity, relative velocity and different angles are shown in velocity diagrams at sections 1 \u2013 1, 2 \u2013 2, 3 \u2013 3, and 4 \u2013 4. Steam enters first row of moving blades at section 1 \u2013 1 with absolute velocity C1 and leaves 2. Steam passes through fixed blades and leaves it with absolute 3. Steam finally leaves second row of moving blades with absolute velocity C4 at 4. Let us assume that the moving blades and fixed blades are symmetrical and the blades are moving blade with velocity C2 at angle velocity C3 at angle angle perfectly smooth i.e. no friction loss occurs on moving blades and fixed blades. Mathematically the velocity and angles can be related as follows for the above assumptions. Constant relative velocity across moving blades due to smooth blades surface; and V1 = V2 V3 = V4 Across the fixed blade also there shall be no friction loss resulting in; C2 = C3 Due to symmetry of moving blades; 1 = 2 & 3 = 4 Due to symmetry of fixed blades; Let us also assume that the stream finally leaving the stage goes out with minimum energy for yielding maximum efficiency. Stream should leave the stage axially for above assumption, i.e. 4 = 90\u00b0. In this case when discharge from last blade row is axial then the velocity of whirl at exit is zero. Such turbines are also called axial discharge turbine. 2 = 3 The total work available from turbine stage shall be algebraic summation of work available from first row of moving blades and second row of moving blades. Diagram efficiency shall be the ratio of total work and energy supplied to the stage. Thus, Total work from stage = Work from first row of moving blades Work from first row of moving blades can be written as already described for simple impulse + Work from second row of moving blades stage. WI = 2 m U (C1 cos Similarly, work from second row of moving blades, 1 \u2013 U) or WII = 2 m U (C3 WII = 2 m U (C2 cos 3 \u2013 U) cos 2 \u2013 U) From velocity diagram, C2 Substituting, cos 2 = V2 cos 2 \u2013 U 636 ________________________________________________________ Applied Thermodynamics WII = 2 m U (V2 cos WII = 2 m U (V1 cos 2 \u2013 2U) 1 \u2013 2U) from velocity diagram, V1 cos Substituting, 1 = C1 cos 1 \u2013 U WII = 2m U (C1 cos 1 \u2013 3U) Total work from stage, WT = WI + WII WT = 2m U (C1 cos WT = 2m U (2C1 cos 1 \u2013 4U) 1 \u2013 U) + 2m U (C1 cos 1 \u2013 3U) Diagram efficiency, d = = Total work Energy supplied m U 2 2 cos 2C1 cos 1 \u2013 4U) Non dimensional form of velocity, = i.e. blade speed to steam velocity ratio or, d = 4 (2cos U C1 1 \u2013 4) Differentiating diagram efficiency with respect to and applying conditions to get the optimum value of for maximum diagram efficiency; d d d = d d (8cos 1 \u2013 162) Substituting or, d d d = 0 Maximum efficiency, For the optimum value of or 4U = C1 cos 1 1 opt = cos 4 d,max = cos2 opt the work may be estimated as; 1 Work from Ist row, WI = 6m U 2 Work from IInd row, WII = 2m U 2 Total work from velocity compounded turbine stage, WT = WI + WII Steam Turbines ________________________________________________________________ 637 WT = 8mU 2 Here it is obvious that for optimum value of i.e. opt the work available from second row is 1 4 th of the total work from stage. It can be extrapolated from here that if there are more than two rows of moving blades in a velocity compounded turbine stage i.e. three, four or more number of rows, then the optimum blade speed to steam velocity ratio can be generalised for \u2018n\u2019 number of moving blade rows. opt = cos 2n 1 i.e. for three rows, opt = cos 6 1 Similarly, work from the last row of moving blades shall be, Wlast row = 1 2 n of total work. Thus from a", " three row stage, work from last row shall be row stage, work from last row shall be 1 16 th of the total work and from a four 1 8 th of the total work. Here it is obvious that as the number of rows increases the work from last row goes on giving diminishing work. Therefore, with increase in number of rows the work obtained from last row decreases, so the number of rows is generally limited to two. If we look at diagram efficiency then it shows that the efficiency also diminishes with increase in number of stages. Graphical pattern between the diagram efficiency versus blade speed to steam velocity ratio is shown in Fig. 14.20. Fig. 14.20 Diagram efficiency vs. blade speed to steam velocity ratio for velocity compounded impulse turbine 14.8 REACTION TURBINES In a reaction turbine the pressure drop occurs in both stationary and moving rows contrary to the impulse turbine where the total pressure drop occurs in stationary nozzles alone. The difference in blading of reaction and impulse has already been described earlier. In reaction turbine the passage 638 ________________________________________________________ Applied Thermodynamics between two consecutive blades is of converging type as compared to impulse turbine blading which has constant cross-sectional area passage between two consecutive blades. Fig. 14.21 Reaction turbine stage Figure 14.21 shows the schematic of a reaction turbine stage having fixed blades followed by moving blades row. Due to the varying cross section area for steam flow the pressure drop occurs in both stationary (fixed) blades row and moving blades row. The velocity increases in stationary blades which act as nozzles. Thus the passage formed in the stationary blades in reaction turbine are of nozzle type although they do not have conventional nozzle shape. Steam stream leaving stationary blades impinges upon the moving blades. This impinging stream exerts a force to the right as evident from the velocity diagrams of reaction blading. Velocity diagram of reaction turbine is similar in principle to the velocity diagram in impulse turbine. Steam Turbines ________________________________________________________________ 639 Fig. 14.22 Velocity and force diagram of reaction blading Steam entering moving blades is subjected to pass through converging area passage along with change in direction. Thus there is increase in velocity (V2 > V1) from inlet to exit in moving blade which results in a reaction force. Change in direction of velocity is accompanied by change in momentum thus an impulse force. It shows that the rotation of shaft is caused by the combination of impulse and reaction forces. The magnitude of impulse force depends upon the pressure drop in fixed blades. It may be noted that due to shaft rotation being caused by combination of impulse and reaction forces these reaction turbines are also termed as impulse-reaction turbine. These are also called full admission turbines as the steam enters through fixed blade row over complete annulus. The enthalpy drop over the reaction turbine stage shows that heat drop occurs in both fixed blades and moving blades rows. If the total enthalpy drop in stage is equally divided between the stationary and moving blades then the stage is called 50% reaction stage. A mathematical parameter called \u2018degree of reaction\u2019 is used to quantify the proportion of enthalpy drops occurring in stationary and moving blades. The \u2018degree of reaction\u2019 is defined as the ratio of enthalpy drop in moving blades row (rotor blades) to the total enthalpy drop in the stage. Mathematically it can be given as, Degree of reaction = Enthalpy drop in moving blades rotor blades Total enthalpy drop in thestage Degree of reaction In case of 50% degree of reaction, = 0.5 and h1 \u2013 h2 = 1 2 (h0 \u2013 h2) or, (h0 \u2013 h1) = (h1 \u2013 h2) = 1 2 (h0 \u2013 h2) Such turbines having 50% degree of reaction are called \u2018Parson\u2019s turbine\u2019. Parson\u2019s turbine has symmetrical blades for moving and stationary blades i.e. inlet angles of stationary and moving blades are equal and also the exit angles of stationary and moving blades are equal. Term symmetrical blading in reaction turbine refers to the 50 per cent reaction stage. 640 ________________________________________________________ Applied Thermodynamics A typical degree of reaction value of zero refers to the condition when there is no enthalpy drop in moving blades. This refers to the impulse turbine stage where no heat drop occurs in moving blades. In reaction turbines also the compounding is done similar to that in impulse turbine for reducing the speed of rotation. The compounding in reaction turbine is shown in Fig. 14.23. where pressure drop gradually occurs in the stationary and moving blade rows. In reaction turbine blading there is possibility of steam leakage across the tip of blade and casing as there is large clearance here. This leakage of pressure from the tip clearance is more vulnerable in high pressure region as there occurs large pressure drop in high pressure region for given enthalpy drop. Due to this fact the initial stages in high pressure region are taken to be of impulse", " type and subsequent stages are reaction stages. Above fact is generally interpreted as \u201cthe inefficient reaction blading in high pressure stage\u201d. Generally, the modern turbines have first stage as velocity compounded (Curtis stage) followed by stages of pressure compounded type (Rateau stages) or reaction stages. Fig. 14.23 Reaction turbine having compounding Velocity diagram for 50% reaction turbine: 50% degree of reaction turbine stage has symmetrical stationary and moving blades. The combined velocity diagram for a moving blade having section 1 \u2013 1 as inlet section and section 2 \u2013 2 as exit section is shown in Fig. 14.24. Steam Turbines ________________________________________________________________ 641 Fig. 14.24 Combined velocity diagram for a moving blade of 50% reaction turbine. From earlier discussions it is clear that 50% degree of reaction turbine has equal enthalpy drops occuring in stationary blade row and moving blade rows. Steady flow energy equation (S.F.E.E.) when applied over turbine stage yields the following assuming no heat interaction, no change in potential energy. Let us assume that turbine has more than one symmetrical stage such that absolute velocity of steam leaving moving blade row equals to the velocity of steam entering fixed blade. Thus, from given velocity diagram steam enters fixed blade with velocity C2 while leaves it with velocity C1. In case of moving blades the relative velocities are to be considered i.e. velocity increasing from V1 to V2 from inlet to exit. For stationary blade row the S.F.E.E. gives, h0 \u2013 h1 = C 2 2 C2 1 2 For moving blade row the S.F.E.E. gives, h1 \u2013 h2 = 2 Total enthalpy drop in stage, h0 \u2013 h2 = (h0 \u2013 h1) + (h1 \u2013 h2) and for 50% degree of reaction as C1 = V2 and V1 = C2 so, 2 V 2 2 V 1 h0 \u2013 h1 = h1 \u2013 h2 h0 = 2h1 \u2013 h2 or, i.e. (h0 \u2013 h2) = 2 (h1 \u2013 h2 = Hence it proves that symmetrical blading results in 50% degree of reaction turbine. Rate of work done from reaction stage can be estimated similar to that of impulse stage. Work done = m U Cw where For symmetrical blading, or, 1 + C2 cos 2) 1, so, Cw = C1 cos 1 + V1 cos 1 Cw= Cw2 + Cw1 = (C1 cos 2 = C2 = V1, Cw = C1 cos Cw = 2C1, cos 1 + (C1 cos 1 \u2013 U 1 \u2013 U) Therefore, work done, W = m U (2C1 cos 1 \u2013 U) 642 ________________________________________________________ Applied Thermodynamics Axial thrust in turbine can be quantified by the sum of thrust due to change in axial velocity and due to pressure difference across the moving blade row. Mathematically the axial thrust can be given as, Fa = m (C1a \u2013 C2a) + (p1A1 \u2013 p2A2), where A1 and A2 are cross-sectional areas available as flow passage at inlet and exit of moving blade row and p1 & p2 are pressures of steam at inlet and exit sections of moving blade rows. In case of 50% reaction turbine C1a = C2a, therefore axial thrust is only due to pressure difference across moving blade row. Diagram efficiency of reaction stage can be estimated by knowing the energy input to the moving blades and taking ratio of work done to energy input. Energy input to moving blades, Ein = m 2 C1 Ein = = m 2 m 2 (C1 2 + V2 2) 2 \u2013 V1 (C1 2 + C1 2 \u2013 V1 2), as C1 = V2 From velocity diagram, V1 Substituting value of V1 in energy input expression, U Ein = m (C1 2 = C1 ) 2 \u2013 2 V1 2 2 + U 2 \u2013 2C1U cos 2 1 2 C 1 2 1 Ein = m C Ein = m C 2 1 C U 1 cos 1 2 2 cos 2 UC 1 2 1 2 U Thus, diagram efficiency, d = W Ein U U 2 1 1 cos cos UC 1 1 2 2 1 2 cos cos cos cos 2 1 d = Substituting = U C1, d = The maximum value of diagram efficiency and the optimum value of blade speed to steam velocity ratio, can be estimated by differentiating with respect to and equating to zero cos or, = cos or, Substituting '' for getting maximum value of diagram efficiency, 1 d,max Steam Turbines ________________________________________________________________ 643 2 d,max = cos 2 cos 1 2 1", " 2 cos 1 cos cos 2 1 1 1 or, d,max = 2 2 cos cos 1 1 2 1 Work output for the condition of maximum diagram efficiency can be obtained by substituting = cos 1 or, U C1 = cos 1 Substituting in expression for work, or, U = C1 1 (2C1 cos For d,max work; W = m C1 cos cos 1 1 \u2013 C1 cos 1) W = m C1 2 cos2 1 W = m U 2 for maximum diagram efficiency. Optimum blade speed to steam velocity ratio yields the condition U = C1cos 1 which shall modify the velocity diagram for such optimum reaction stage as shown in Fig. 14.25. Fig. 14.25 Velocity diagram for optimum reaction turbine stage having \u03c1 = cos \u03b1 1 Height of blade in reaction turbine can be estimated from the volume flow rate as given below. Reaction turbine stage is full admission stage. Taking number of blades as \u2018N\u2019, thickness of blade as \u2018t\u2019, the mass flow rate as \u2018m\u2019, average specific volume as \u2018v\u2019 it can be written that, m v = (d \u2013 Nt) l V2 sin 2 Height of blade \u22c5 \u2212 \u22c5 sin \u03b2 2 It is observed that the minimum height for reaction turbine blade should not be less than 5 per cent of mean blade ring diameter and the maximum height should not exceed beyond 20 per cent of mean blade ring diameter. If the blade is of tapered type then the maximum height may go up to 30 per cent of mean blade ring diameter. The lower and upper limits of blade height are to be maintained while deciding blade height as shorter blades have increased energy losses while longer blades may have excessive stresses in them. Generally in reaction stages of turbine the blade height is allowed to increase gradually so as to give increasing annular area for expanding fluid. 644 ________________________________________________________ Applied Thermodynamics 14.9 LOSSES IN STEAM TURBINES Steam turbine being work producing device running at quite high speed has number of losses occurring in it. These losses when put together result into substantial loss of energy. Therefore, while selecting a turbine due attention should be paid to the losses in turbine. Some of the losses occur within turbine stages while some are external to stage. These losses are described ahead. 1. Losses in nozzles: Steam turbine nozzle is designed for isentropic expansion so as to result in increase in velocity from inlet to exit. Practically in a nozzle the steam leaving nozzle may not have velocity equal to the designed velocity value. This deviation in operating state of nozzle may occur because of non-isentropic expansion. The reasons for non-isentropic expansion may be friction losses between the steam and nozzle wall, viscous friction resistance to flow in the steam particles, boundary layer formation and separation, heat loss during flow etc. Mathematically, this shift from isentropic expansion to non-isentropic expansion is quantified using the parameter called \u2018nozzle efficiency\u2019. Nozzle efficiency as described earlier is defined by the ratio of \u2018actual enthalpy drop\u2019 to the \u2018isentropic enthalpy drop\u2019 between inlet and exit of nozzle. 2. Losses in moving blades: In steam turbine stage steam is supposed to glide smoothly over the moving blades after leaving nozzles or fixed blades. In actual turbine stage during flow of steam over moving blade, there may be number of factors causing loss of energy as given under: (i) Blade friction may incur frictional loss which is taken into account by the blade friction factor. This friction factor largely depends upon the Reynolds number, although it is earlier defined as the ratio of relative velocity leaving blades to the relative velocity of steam entering blades. This loss may be termed as \u201cpassage loss\u201d. (ii) \u201cBoundary layer separation\u201d may occur due to sharp deflection of fluid within the blade passage. Deflection results in centrifugal force which causes compression near concave surface and the rarefaction near the convex surface of blade, thus resulting in separation of boundary layer. (iii) Loss of energy may be due to turbulence at outlet of preceding row of nozzles due to finite thickness of nozzle exit edge. There is mixing of steam jet leaving nozzles and entering moving blade. Due to this transition of flow from nozzle passage to blade passage there is formation of eddies and turbulence gets set in. This turbulence is generally in the form of trailing vortices which keep on disappearing at high velocities. These cause the reduction of kinetic energy delivered to blades and are called \u201cwake losses\u201d. Wake losses are visible at the trailing edge of fixed blades too due to thickness of trailing edge. (iv) Loss of energy is also there due to breakage of flow which occurs upon the impingement of steam upon the leading edge of moving blade. This is", " also termed as \u2018impingement loss\u2019. These losses are less if the flow is laminar as compared to the turbulent flow. (v) Loss of energy also occurs during passage of steam from one stage to other i.e. rows. This loss is also termed as \u2018carry over loss\u2019. This carry over loss is minimum if spacing between consecutive rows is kept small. The different losses as described above are accounted by taking the profile loss coefficient (kp), incidence loss coefficient (ki) and carry over loss coefficient (kc). Profile loss coefficient takes into account the losses due to turbulence, friction, fluid deflection in blade passage, curvature of blade and actual exit angle being different from blade exit angle. Incidence loss coefficient takes care of losses due to turbulence introduced by angle of incidence. The carry over loss coefficient takes care of losses due to kinetic energy loss during transition of flow between the rows. Therefore, the actual relative velocity leaving blade shall be, V2 = kp ki kc V1 Steam Turbines ________________________________________________________________ 645 3. Disc friction loss: This is a kind of loss of energy visible whenever any object say disc is rotated in air or other medium. The disc would cut the atmosphere and impart motion to surrounding air. There shall always exist relative motion between solid wall of object and the air or surrounding fluid. Due to this relative motion surrounding medium always exerts a resistance to motion of moving object. This may result in loss of energy due to friction which may be felt by the increase in enthalpy of surrounding fluid. In case of steam turbines too the rotor is completely surrounded by the steam which offers resistance to the rotor motion. The loss of energy of rotor may go into the steam enveloping it. This loss of energy is termed as \u2018disc friction loss\u2019. The disc friction loss may cause heating of steam surrounding the rotor i.e. a portion of kinetic energy is transferred from the rotor disc to steam by heating of steam. Disc friction loss is substantial in case of impulse stages as compared to reaction stages where it is very small and can be neglected. 4. Windage loss: Windage loss occurs when the rotor blades come in contact with near stationary fluid (steam). In case of partial admission turbines i.e. generally impulse turbines there is churning of steam in the region having no active steam in steam turbine. When moving rotor blades come in contact with inactive steam then there is transfer of energy from blade to steam. This loss of energy from rotor to fluid is termed as \u2018windage loss\u2019. In case of full admission turbine the region inside turbine having inactive steam is negligible and so the windage loss is nearly negligible. To minimize windage loss the turbine should be filled with moving steam (active steam). Windage losses are very small in case of low pressure stages. Reaction turbines have negligible windage losses as they are full admission turbines. 5. Loss due to leakage: Steam leakage may occur across the turbine shaft and between stages. Leakage of steam may result in availability of less work from stage as steam is not fully utilized for producing work. Leakage occurs during flow from one stage to other or from one row to other through the clearance space between diaphragm and shaft. Leakage also occurs across the blade tip. Leakage across diaphragm occurs in both impulse turbine and reaction turbine stages. Leakage across tip is not prominent in case of impulse turbine as the pressure difference is very small. Tip leakage is prominent in reaction turbine stages. Due to this diaphragm and tip leakage effective mass flow rate for doing work gets reduced and is consequently a loss of energy. Leakage is accompanied by the increase in entropy and so the decrease in availability of work due to throttling of steam which is an irreversible process. Leakage loss can be minimized by reducing the clearances as much as possible after providing for expansion of turbine parts so that the metal-to-metal rubbing is avoided. Different seals such as labyrinths, carbon rings, water, steam or air seals are used to prevent this leakage through clearance. Also in order to reduce leakage loss the drum type construction is preferred to diaphragm and wheel type construction in reaction turbines. Another type of leakage may be of balance-piston leakage which refers to leakage between balance piston and casing. Here fluid leaks out in high pressure region of turbine and atmospheric air bleeds into casing in low pressure sub atmospheric region (condensor side). Generally this is not a total loss as the leakage out and leakage in are not varying too much. This kind of loss can also be prevented by employing labyrinth packing. At low pressure/subsonic region (condenser side) of turbine the labyrinth packing is fed with low pressure steam so that steam leaks in instead of air in case of unavoidable leakage. 6. Losses in bearings: Turbine bearings are critical parts to support high speed rotation of shaft. Generally, a loss to the tune of 1% of", " turbine output occurs in bearings. Although this loss depends upon bearing load, oil viscosity, speed of shaft, bearing surface area and film thickness etc. 7. Losses at inlet and exit: Loss at inlet of steam turbine occurs at regulating valves at entry. At these valves which may be stop valve or governor valves the throttling loss generally occurs causing lowering of entering steam pressure. 646 ________________________________________________________ Applied Thermodynamics At the exit of steam turbine the steam becomes wet and the fluid now is mixture of water droplet and steam. Due to wetness of steam at exit end the water particles being heavier cause loss of kinetic energy. These water particles, if in excessive amount may also endanger the turbine blade. The kinetic energy of fluid at exit of steam turbine is a total loss and theoretically the fluid leaving turbine must have the lowest possible energy in it. This loss of energy may be of the order of 10 \u2013 13 per cent. 8. Losses due to radiations: Radiation losses also occur in steam turbines, although they are very small compared to other losses and may be neglected. In case of steam turbines the high temperature steam is limited to small part of casing so losses are small. But the radiation losses are quite significant in gas turbines. In order to prevent radiation losses the pipings, turbine casing etc. carrying hot fluid should be well insulated. 14.10 REHEAT FACTOR In case of steam turbine the expansion process must occur isentropically under ideal conditions. But in actual practice due to large number of losses such as friction, leakage, entry and exit losses etc. the expansion inside does not remain isentropic. These factors are actually resulting into the situation having incomplete conversion of available energy into the shaftwork. Thus, the actual steam leaving turbine will be leaving with higher enthalpy than the enthalpy of steam at exit under isentropic expansion. Figure. 14.26 shows the ideal and actual expansion process in case of single stage and multi stage steam turbine. Single stage expansion shows that ideally steam enters at state 1 and leaves at 2. Fig. 14.26 Ideal and actual expansion in steam turbine showing reheat effect on h\u2013s diagram While actually steam enters at state 1 and leaves at state 2 i.e. enthalpy at state 2 is higher than at 2. Comparison of expansion process 1\u20132 and 1\u20132 shows that due to various losses the total available energy (h1 \u2013 h2) is not completely converted into shaft work. Portion of available energy given by {(h1 \u2013 h2) \u2013 (h1 \u2013 h2)} remains with expanding fluid i.e. steam itself. On h\u2013s diagram the shift in state after expansion from 2 to 2 due to non-ideal operation is understood as reheating of steam at constant pressure. For the sake of simplicity this reheating from 2 to 2 is considered to be of constant pressure Steam Turbines ________________________________________________________________ 647 type, whereas actually it may not be so. If the state of steam after expansion lies in wet region then it can be seen that non-isentropic expansion causes increase in dryness of steam at exit i.e. dryness fraction at 2, (x2) will be less than that at 2(x2 ), x2 < x2. Reheating of steam is accompanied by increase in entropy i.e. s2 > s2. Figure 14.27 shows the ideal and actual expansion in steam turbine on T\u2013s diagram. For single stage expansion the T\u2013 s diagram shows the amount of heat lost/dissipated due to the friction etc. and subsequently recovered by steam expanding in steam turbine in the form of reheating by the area 22 BA. Fig. 14.27 Ideal and actual expansion in steam turbine on T-S diagram In case of multi stage expansion the h \u2013 s diagram shown in Fig. 14.26 indicates the four stage expansion starting from state 1. In the first stage theoretically the ideal expansion is given by the line 1\u20132 while actually it is 1\u20132. Second stage expansion shall begin from state 2 and isentropic and nonisentropic expansion processes are shown by 2\u20133 and 2\u20133 respectively. Actual state after expansion in second stage is 3 and the subsequent stage expansion begins from this point. Expansion in third stage results in theoretical and actual states as 4 and 4 respectively. In the last stage starting from 4 the ideal and actual states after expansion are at 5 and 5 respectively. Thus for multi stage expansion the actual states at the end of each stage are 2, 3, 4 and 5. For any stage, say first stage the isentropic and non-isentropic enthalpy drop may be related to stage efficiency as below, Stage efficiency, s for Ist stage = h 1 h 1 h 2 h 2 Here the locus of actual state points 1\u20132 \u2013 3 \u2013 4 \u2013 5 is also called the \u2018condition\u2019 line as", " it approximates actual state of steam during its exit through turbine stages. If we look at the enthalpy drops occurring in ideal and actual conditions then it is seen that the sum of isentropic enthalpy drops in stages i.e. (h12 + h2 3 + h34 + h4\u20135) shall be different from the actual enthalpy drop (h12 + h23 + h34 + h45). This sum of isentropic enthalpy drop (heat drop) is generally termed as cumulative heat drop (hc). Total enthalpy drop for isentropic expansion can be given by h1-5 and this is also termed as \u2018Rankine heat drop\u2019 (hRankine). Let us look actual and ideal expansion in a stage, say second stage where expansion begins at 2 and should go up to 3 while it actually ends at 3. The shift in state from 3 to 3 indicates reheating of steam during expansion. This phenomenon is called reheating of steam during expansion in steam turbine and is quantified by \u2018Reheat Factor\u2019. \u2018Reheat factor\u2019 is defined by the ratio of \u2018cumulative heat 648 ________________________________________________________ Applied Thermodynamics drop\u2019 to \u2018Rankine heat drop\u2019 i.e. cumulative heat drop (hc) to direct isentropic heat drop (hRankine). Reheat factor = Cumulative heat drop Rankine heat drop = \u2206 h 12 + \u2206 RF = h 12 Rankine Reheat factor value is always greater than unity because cumulative heat drop shall always be more than Rankine heat drop. This may be seen from h\u2013s diagram which has diverging constant pressure lines because of which the cumulative heat drop becomes more than Rankine heat drop. Reheat factor value increases with increase in number of stages for a given pressure range. If the stage efficiency is reduced then also the value of reheat factor increases for given pressure range and number of stages. From designer\u2019s point of view the reheat factor must have value of unity under ideal conditions. But, since unity reheat factor value can not be attained so efforts must be made to reduce the reheat factor value to make it close to unity. Unity reheat factor indicates that actual expansion is similar to ideal expansion and \u2018cumulative heat drop\u2019 equals to \u2018Rankine heat drop.\u2019 Reheat factor may be related to internal efficiency of turbine as detailed below. Internal efficiency is given by the ratio of \u2018sum of actual outputs of stages\u2019 and \u2018the available energy in turbine\u2019. Internal efficiency may also be given by the ratio of internal work done of turbine to the Rankine heat drop. Mathematically, internal efficiency, or, int = int = int = h actual h Rankine h 12 h 15 h 15 15 If the stage efficiency for turbine stages is given by S and stage efficiency remains same for all stages, then stage efficiency can be given as, S = h 12 h 12 h 2 3 h 2 3 h15 = h12 + h23 + h34 + h45 Also, Let us assume the available energy in each stage to be equal as, hstage h12 = h23 = h34 = h45 = hstage or, h12 = hstage = h23 = h34 = h45 S Substituting in internal turbine efficiency, for \u2018n\u2019 number of stages h15 = S int = hstage h stage h 15 S n n S = h stage h Rankine n From definition of reheat factor, RF = h c h Rankine Steam Turbines ________________________________________________________________ 649 hRankine = ch\u2206 RF Substituting hRankine in h stage h c int = int, S n RF S int = n stage RF h h c For single stage turbine, int = RF h 12 h 12 For the four stage turbine as considered here, h 15 h 15 RF int = Thus it is obvious that the internal efficiency of a turbine having multiple stages is always more than stage efficiency as Reheat factor is more than unity. 14.11 STEAM TURBINE CONTROL In actual operation of steam turbine the speed and output of turbine are to be constantly monitored and controlled as rarely any turbine operates at constant load. This operation of turbine at off-design conditions also deteriorates performance of turbine. Therefore, for safe and optimum operation of turbine it has certain control and monitoring instruments. Number and size of these control and monitoring instruments depends upon size of turbine and its\u2019 application. Some of typical controls and instruments used are described below for monitoring pressure, temperature, speed, noise and vibration etc. (i) Pressure measurement is done for oil pressure to bearing, water/steam pressure to gland seals, steam pressure at stop valves, steam chest, inlet and exit of each stage in turbine. Pressure gauges are mounted at suitable locations for each one of above. (ii) Temperature measurement are done at almost", " all the locations of pressure measurement. (iii) Speed measurement is required to know speed of turbine in rpm during starting and shutdown of turbine. Turbine speed is obtained from the generator frequency recorder. (iv) Eccentricity of shaft is also to be monitored for which an eccentricity recorder is provided to indicate and record eccentricity of shaft at high pressure end of turbine. Eccentricity recorder has a detector coil at low pressure end of shaft, a transfer switch on turning gear, and a power unit and recorder mounted on control panel. Eccentricity of shaft causes variation in air space between shaft and coil detector and therefore change in current flowing through coil, which is being sensed appropriately. (v) Vibrations at main bearings of turbine and generator are being monitored using vibration amplitude recorder. This type of recorder has a detector at each bearing and detector voltage is proportional to the bearing vibrations. Voltage variation is sent for suitable signal transformation and vibration amplitude is recorded. (vi) Due to excessive temperature the turbine casing may undergo axial expansion for which an expansion indicator is put. Expansion is monitored with respect to some selected mark on 650 ________________________________________________________ Applied Thermodynamics foundation. Steady expansion of casing is permitted while unsteady expansion may be harmful. (vii) Noise of turbine operation is being measured using noise meter as any unusual sound level may indicate possibility of failure. (viii) Flow rate of steam at all crucial points in steam turbine is being measured employing flow meters and signals monitored at control panel. (ix) Control panel also has wattmeters, voltmeters, ammeters for respective measurements. Control panel also has the digital control of various valves, governor control etc. Now-a-days the control panel of turbines has all digital controls and monitoring employing suitable microprocessors. 14.12 GOVERNING OF STEAM TURBINES Governing of steam turbine is required for controlling and regulating the output of a turbine according to variable demands in service. This regulation of turbine is done by maintaining accurate and positive control of speed, pressure and flow by employing governors. Governing of turbine maintains the speed of turbine at constant level irrespective of load. Governing of steam turbines can be of following types depending upon the mechanism of regulation. (i) Throttle governing (ii) Nozzle control governing (iii) By-pass governing (iv) Combined governing i.e., Combination of above as \u2018throttle and nozzle control governing\u2019 or \u2018throttle and by-pass governing\u2019. (i) Throttle Governing: Throttle governing of steam turbine bases upon the throttling of steam up to suitable pressure and regulate the mass flow of steam through the turbine so as to control the output of turbine. Here in throttle governing, aim is to alter the mass flow rate as for example by reducing it, the available energy gets reduced and hence lowering of output. Fig. 14.28 Throttle governing on h-s diagram In throttle governing the steam entering is regulated by opening and closing of valve. As the valve is closed, the throttling or constant enthalpy process occurs across the valve with an increase in entropy and corresponding decrease in availability of energy per unit mass flow of steam. Also due to throttling Steam Turbines ________________________________________________________________ 651 the state of steam entering turbine stage gets modified and the modified expansion line for each load is obtained. It may be noted that even when the governor valve is full open the pressure drop does occur and thus it can be said that throttling is evident at all loads on turbine. Representation of throttle governing on h-s diagram shows that the steam is available at state \u20180\u2019 at p0 pressure in the main steam line. At the inlet from main steam line when the governor valve is full open the throttling results in modified state 1 from where expansion occurs following path 1\u20133 under insentropic expansion and non-isentropic expansion occurs following path 1\u20133. When mass flow rate is reduced for reduced load on turbine then this partial closing of valve causes throttling as shown by 1\u20132. Now as a result of this throttling from 1 to 2 the modified expansion paths are 2\u20134 and 2\u20134 for isentropic and non-isentropic expansion in turbine. Thus it shows that as a result of throttling the available energy gets reduced from h1-3 to h2-4 and h2-4 < h1-3. Hence this lowering of available energy causes reduced output from turbine. Fig. 14.29 Schematic of simple throttle governing Schematic of simple throttle governing in steam turbines is shown in Fig. 14.29. Here a centrifgal governor is used to sense the change in speed of shaft. The relay system has a pilot valve and servomotor. The displacement of servomotor piston either upward or downward decides the opening of throttle valve \u2018C\u2019. Servomotor piston is actuated by the high pressure oil entering from pilot valve to upper or", " lower half of servomotor piston D. Under normal operation the servomotor piston occupies middle position and pilot valves keep the inlet and exit ports in closed position. When oil enters the upper half of servomotor then servomotor piston lowers down and the throttle valve starts closing causing reduction of steam flow rate and so the output till the speed is maintained to normal running speed. Simultaneously, the oil from lower half of servomotor gets drained out through pilot valve port. When oil under pressure enters lower half of servomotor then servomotor piston gets lifted up causing lift of throttle valve. For throttle governing of steam turbine the steam consumption rate may be plotted with load resulting into characteristic line called Willans line as shown in Fig. 14.30. 652 ________________________________________________________ Applied Thermodynamics Fig. 14.30 Characteristics of throttle governing During throttle governing the Willan\u2019s line is straight line making an intercept on y-axis. Math- ematically, it can be given as, M = K\u00b7L + M0 where M is steam consumption in kg/h at any load L. M0 is steam consumption in kg/h at no load i.e. L = 0 L is any load on turbine in kW. K is the constant and gives slope of Willan\u2019s line. Here it shows that even at no load the steam consumption shall be M0 (kg/h) which is graphically given by intercept on y-axis. Above equation can also be written as, M L = K + M 0 L where M L is the specific steam consumption at any load, kg/kWh and M 0 is the specific steam consumption at no load, kg/kWh. L Throttle governing offers following disadvantages due to throttle action at inlet: (a) Throttling increases initial superheat at inlet and the greatest variations in steam velocity occur in the later stages. (b) The wetness of steam gets reduced in later stages due to throttling. Due to this reduced wetness there occurs reduction in stage efficiency at part load operation of turbine. (ii) Nozzle Control Governing: Nozzle control governing is the one in which steam flowing through nozzles is regulated by valves. Nozzle control governing is generally employed at first stage of turbine due to practical limitations. The nozzle areas in remaining stages remain constant. If some how the nozzle governing is provided for all nozzles in each and every stage then an ideal condition of turbine flow passage areas conforming to mass flow rate at all loads shall exist. Under such ideal conditions the pressure, velocities and nozzle and blade efficiencies would be constant with load. For such ideal condition the Willan\u2019s line would be straight line as indicated for throttle governing of turbine. Steam Turbines ________________________________________________________________ 653 In nozzle governing the nozzles of turbines are grouped in two, three or more groups upto six or eight groups. When nozzle governing is employed then the pressure and temperature of steam entering first stage nozzles are independent of load. Figure 14.31 shows the schematic of nozzle control governing. Fig. 14.31 Schematic of nozzle control governing As the valves are being regulated for actuating nozzle control governing so there occurs some throttling of steam at each valve. However, the amount of throttling is considerably lesser and the decrease in availability of energy to turbine is not too much. In order to avoid this occurrence of throttling very large number (infinite) of nozzle and governing valves may be put. (iii) By Pass Governing: In case of by-pass governing arrangement is made for by-passing surplus quantity of steam without allowing total steam quantity to contribute in turbine output when load reduces. Arrangement of by pass governing is shown in Fig. 14.32. Fig. 14.32 Arrangement in by pass governing 654 ________________________________________________________ Applied Thermodynamics Diagram shows that steam from main line enters the main valve which is controlled by speed governor. Steam from main valve enters the nozzle box or steam chest. By pass valve is also provided on the nozzle box. By pass valve is connected to a passage which delivers steam being by passed to later end of turbine. By pass valve is actuated when load varies, thus allowing only part of steam entering main valve to contribute in power output. By pass valve is controlled by speed governor for all loads within its\u2019 range. In this kind of governing depending upon turbine and its\u2019 application there may be more than one by pass valves. (iv) Combined governing: Some times when the governing requirements are not met by any one kind of arrangements of governing i.e. throttle, nozzle control and by pass governing, then the combination of two governing mechanisms may be employed. These popular combinations are \u2018throttle and nozzle control combined governing\u2019 and \u2018throttle and by pass combined governing\u2019. 14.13 DIFFERENCE BETWEEN THROTTLE GOVERNING AND NOZZLE", " CONTROL GOVERNING Differences between throttle governing and nozzle control governing are as given; Throttle governing Nozzle control governing (a) This governing is realized by throttling of steam at inlet to turbine so there occur severe throttling losses. (b) It has smaller losses due to partial admission of steam. (c) This governing is generally employed in both impulse turbines and reaction turbines. (a) This governing is realized by nozzle control so the degree of throttling losses in valves are negligible. (b) It has large partial admission losses. (c) This governing is employed in impulse turbine and reaction turbines having initial impulse stages. 14.14 DIFFERENCE BETWEEN IMPULSE AND REACTION TURBINES Differences between the impulse turbine and reaction turbines are as under. Impulse turbines Reaction turbines (a) (b) (c) (d) Impulse turbine has profile type blades and has constant area between two consecutive blades. Impulse turbine stage has pressure drop occurring only in nozzles. No pressure drop occurs in moving blade row. Impulse turbines have incomplete admission of steam or steam being admitted at selected positions around the motor. Impulse turbine diaphragm has nozzles mounted on it. Velocity of (a) Reaction turbine has airfoil type blades and has converging area between two consecutive blades. (b) Reaction turbine stage has pressure drop occurring in both fixed as well as moving blades. (c) Reaction turbines have complete admission of steam or steam being admitted all around the rotor through fixed blade ring. (d) Reaction turbine has fixed blade ring attached to casing to serve Cont. Steam Turbines ________________________________________________________________ 655 steam is quite large. as nozzle and guide blades for entering steam. Velocity of steam is comparatively smaller. (e) Impulse turbine has lesser efficiency and much power can not be developed from it. (f) These occupy less space for same power output. (g) These are not very costly as the manufacturing of impulse turbine blades is much simpler. Impulse turbines are suitable for small power requirement. (h) (e) Reaction turbine has higher efficiency and is capable of producing large power output. (f) These occupy large space for same power output. (g) Reaction turbines are costly as their blades are very difficult to be manufactured. (h) Reaction turbines are suited for medium and higher power requirements. EXAMPLES 1. Single stage impulse turbine has equal blade angles and nozzle angle of 15\u00b0. Determine the maximum possible blade efficiency if the blade velocity coefficient is 0.85. Determine the blade speed to steam velocity ratio if the actual blade efficiency is 90% of maximum blade efficiency. Solution: Given, 1 = 15\u00b0, 1 = 2, K = 0.85 Maximum blade efficiency, b,max = = 1 K cos cos 1 2 (1 + 0.85) 2 1 cos 2 cos2 15 2 Maximum blade efficiency = 86.3% Ans. = 0.863 or 86.3% Actual blade efficiency = 0.90 0.863 = 0.7767 Blade efficiency for single stage impulse turbine can be given as, b = 2(1 + K) ( cos 0.7767 = 2(1 + 0.85) ( cos 15 \u2013 2) 0.2099 = ( 0.966 \u2013 2) 1 \u2013 2) or, 2 \u2013 0.966 + 0.21 = 0 = 0.653 or 0.347 Blade speed to steam velocity ratio = 0.653 or 0.347 Ans. 2. A single stage of simple impulse turbine produces 120 kW at blade speed of 150 m/s when steam mass flow rate is 3 kg/s. Steam enters moving blade at 350 m/s and leaves the stage axially. Considering velocity coefficient of 0.9 and smooth steam entry without shock into blades, determine the nozzle angle and blade angles. Solve using velocity diagram. 656 ________________________________________________________ Applied Thermodynamics Solution: Given: U = 150 m/s, C1 = 350 m/s, For impulse turbine single stage, V 2 V 1 = 0.9, 2 = 90\u00b0 Output = 120 103 = 150 Cw 3 Cw = 266.67 m/s. Let scale be 1 cm = 50 m/s for drawing velocity diagram using given information. V1 = 255 m/s BD = 5.1 cm C1 = 350 m AD = 7 cm V2 = 229.5 m/s BE = 4.59 4.6 cm Fig. 14.33 Steps: (i) Draw AB = 3 cm = 150 50 corresponding to U. (ii) Draw perpendicular from A as 2 = 90\u00b0 for axial discharge from turbine. Mark point F along AB corresponding to Cw (= 266.67 m/s), AF = 5.33 cm (iii) Draw perpendicular from F along FD. Open compass for C1 (= 350 m/s) i.e. 7 cm and mark point D keeping needle at", " A. Join AD and DB corresponding to C1 and V1 respectively. Measure V1 (= 5.1 cm or 255 m/s) by length BD. Estimate V2 = V1 0.9 = 229.5 m/s. (iv) Open compass for V2 (= 229.5 m/s) i.e. 4.59 cm and mark point E keeping needle on B. Join points E and A, E, B to show C2 and V2 respectively. This completes velocity diagram. Measure, BAD = FBD = ABE = 1 = 39\u00b0 1 = 61.5\u00b0 2 = 47\u00b0 Nozzle angle = 39\u00b0 Blade angles at inlet and exit = 61.5\u00b0 and 47\u00b0 Ans. 3. In a single stage simple impulse turbine the steam flows at rate of 5 kg/s. It has rotor of 1.2 m diameter running at 3000 rpm. Nozzle angle is 18\u00b0, blade speed ratio is 0.4, velocity coefficient is 0.9, outlet angle of blade is 3\u00b0 less than inlet angle. Determine blade angles and power developed. Steam Turbines ________________________________________________________________ 657 Solution: Given: = U C1 = 0.4, V 2 V 1 = 0.9, 1 = 18\u00b0, 2 = 1 \u2013 3 U = 2 3000 1 2 60.2 = 188.5 m/s C1 = 471.25 m/s Using these parameters the velocity diagram is drawn. Scale 1 cm = 50 m/s. Fig. 14.34 From velocity diagram, 1 = 29\u00b0, Power developed = m U Cw 2 = 26\u00b0, Cw = 495 m/s (= length FG) = 5 188.5 495 = 466537.5 J/s = 466.54 kW Blade angles at inlet and exit = 29\u00b0 and 26\u00b0 Power developed = 466.54 kW Ans. 4. In a single stage impulse turbine the isentropic enthalpy drop of 200 kJ/kg occurs in the nozzle having efficiency of 96% and nozzle angle of 15\u00b0. The blade velocity coefficient is 0.96 and ratio of blade speed to steam velocity is 0.5. The steam mass flow rate is 20 kg/s and velocity of steam entering is 50 m/s. Determine (a) (b) (c) (d) the blade angles at inlet and outlet if the steam enters blades smoothly and leaves axially. the blade efficiency the power developed in kW the axial thrust. Solve using velocity diagram. Solution: Let the state at inlet to nozzle, leaving nozzle and leaving blade be denoted using subscripts 0, 1 and 2. Using nozzle efficiency the actual enthalpy drop may be calculated as. Actual enthalpy drop in nozzle, hN = 200 0.96 = 192 kJ/kg Velocity of steam leaving nozzle, C1 = 2 h N C 0 2 Steam Turbines ________________________________________________________________ 659 Axial thrust, Fa =m (Va1 \u2013 Va2) = 20 (170 \u2013 75) = 1900 N Axial thrust = 1900 N Ans. 5. A De Laval turbine runs with steam supplied at 16 bar, 250\u00b0C. The nozzle efficiency is 90%, blade velocity coefficient is 0.98, mechanical efficiency is 95%, nozzle angle is 15\u00b0, symmetrical blades with 30\u00b0 angle, mean diameter of wheel is 80 cm, back pressure is 0.15 bar. Determine (a) (b) (c) (d) the speed of rotation, the steam consumption per-bhp-hr, the stage efficiency, the percentage energy loss at exit. Solution: From steam table at inlet to turbine nozzle, h0 = 2919.2 kJ/kg s0 = 6.6732 kJ/kg\u00b7K At exit i.e. back pressure 0.15 bar, s1 = s0 = 6.6732 kJ/kg\u00b7K s1 = sf at 0.15 bar + x1 6.6732 = 0.7549 + x1 7.2536 sfg at 0.15 bar x1 = 0.8159 Enthalpy h1 = hf at 0.15 bar + x1 hfg at 0.15 bar = (225.94 + (0.8159 2373.1)) h1 = 2162.15 kJ/kg Isentropic heat drop = h0 \u2013 h1 = (2919.2 \u2013 2162.15) = 757.05 kJ/kg Velocity of steam C1 = 2 757 05 0..90 103 C1 = 1167.34 m/s Nozzle angle 1 = 15\u00b0, Blade angle 1 = Blade velocity coefficient, = 0.98 2 = 30\u00b0 V 2 V 1 The velocity diagram is drawn using above information. Taking scale 1 cm = 100 m/s. Fig. 14.", "36 660 ________________________________________________________ Applied Thermodynamics From velocity diagram, V1 = 630 m/s, given by length BD, Length AB gives U = 570 m/s Using blade velocity coefficient, V2 = 0.98 V1 = 617.4 m/s Cw = Given by length GF = 1080 m/s Length AE gives C2 velocity, C2 = 320 m/s U = 2 N r 570 = 2 N 22 7. 0 80 2 N = 226.7 rps N = 13602 rpm Speed of rotation = 13602 rpm Ans. Work done per kg of steam, W = U Cw = 570 1080 = 615600 W/kg = 615.6 kW/kg 1hp = 0.7457 kW hp per kg of steam per second = 826.86 hp Using mechanical efficiency, bhp can be obtained as = 0.95 826.86 = 785.52 hp Steam consumption per bhp-hr = 3600 785.52 = 4.58 kg Steam consumption per bhp-hr = 4.58 kg Ans. Stage efficiency = Work done Energysupplied tostage = 615.6 757.05 = 0.8132 Stage efficiency = 81.32% Ans. Loss at turbine exit per kg of steam = 2 C2 2 Percentage energy loss at exit = = 51200 J/kg 51200 757 05 103. = 6.76% 100 Percentage energy loss at exit = 6.76% Ans. 6. Following data refer to a De Laval steam turbine having equiangular blades; Steam entering nozzle = 100 m/s 662 ________________________________________________________ Applied Thermodynamics Horse power developed = m U Cw 745 7. 3 200 485 = 745 7. = 390.23 hp HP developed = 390.23 hp Ans. 7. In a simple impulse steam turbine stage steam enters the nozzle at 15 bar, dry saturated with velocity of 150 m/s. Nozzle angle is 20\u00b0 and steam leaves nozzle at 8 bar and enters into smooth blades. Considering nozzle velocity coefficient of 0.90 and blades to be equiangular determine the following for maximum diagram efficiency. the blade angles, the blading efficiency, the stage efficiency. (a) (b) (c) Solution: From steam tables, enthalpy at inlet to nozzle, h0 = hg at 15 bar, = 2792.2 kJ/kg entropy, s0 = sg at 15 bar = 6.4448 kJ/kg.K Let dryness fraction at exit of nozzle be x1 at 8 bar. For isentropic expansion, s1 = s0 = 6.4448 kJ/kg\u00b7K s1 = 6.4448 = sf at 8 bar + x1 sfg at 8 bar 6.4448 = 2.0462 + (x1 4.6166) x1 = 0.9528 h1 = hf at 8 bar + x1 hfg at 8 bar = 721.11 + (0.9528 2048) h1 = 2672.44 kJ/kg 2 C 0 h 1 2 0 h 2 119 76 10. C1 = 487.92 m/s = 3 0.9 150 2 Isentropic heat drop in nozzle = (h0 \u2013 h1) = 119.76 kJ/kg Velocity at exit of nozzle C1 = For maximum diagram efficiency the blade velocity can be obtained by, = cos 2 1 U C1 = cos 2 1 U = 487.92 cos15 2 U = 229.25 m/s Velocity diagram is drawn considering C1 = 487.92 m/s, U = 229.25 m/s Scale 1 cm = 50 m/s. 1 = 20\u00b0, 1 = 2. Steam Turbines ________________________________________________________________ 663 From velocity diagram, length BD = 5.7 cm i.e. V1 = 285 m/s Since blades are smooth so V1 = V2 Fig. 14.38 C2 = 150 m/s (= length AE) Cw = 455 m/s (= length AF) V2 = 285 m/s 1 = 36\u00b0 = 2 Blade angle at inlet = 36\u00b0 Blade angle at exit = 36\u00b0 Ans. Work done per kg of steam = U Cw = (229.25 455) = 104.31 103 J/kg 2 Kinetic energy supplied per kg of steam = = 2 C1 2.92 487 2 Blading efficiency = = 0.8763 = 119.03 103 J/kg 104 31 10 119 03 10.. 3 3 Stage efficiency = Work done per kgof steam Energysupplied per kgof steam tostage = Blading efficiency = 87.63% Stage efficiency = 79.62% Ans. 104.31 119 76 10. 3 3 10 150 2 = 0.7962 2 8. Single", " row impulse turbine operates between 10 bar and 5 bar with expansion efficiency of 95%. 10 kg of dry saturated steam per second enters into nozzle and leaves nozzle at angle of 20\u00b0 to the axis of rotation of blades. The blade velocity coefficient is 0.90, blade speed is 200 m/s and internal losses due to disc friction and windage losses is 0.5 kJ/kg of steam. Consider that there is no axial thrust on the blades. Determine, (a) Blade angles, (b) Stage efficiency, (c) Stage output in hp and prepare heat balance sheet. Also reason out the error if any by solving using velocity diagram. 664 ________________________________________________________ Applied Thermodynamics Solution: From steam tables: Enthalpy of steam entering, Entropy. For isentropic expansion Let dryness fraction at exit of nozzle be x1 at pressure of 5 bar. s1 = sf at 5 bar + x1 sfg at 5 bar 6.5865 = 1.8607 (x1 4.9606) h0 = hg at 10 bar = 2778.1 kJ/kg s0 = sg at 10 bar = 6.5865 kJ/kg\u00b7K s1 = s0 = 6.5865 kJ/kg\u00b7K Enthalpy at exit of nozzle, x1 = 0.9527 h1 = hf at 5 bar + x1, hfg at 5 bar = 640.23 + (x1 2108.5) h1 = 2648.99 kJ/kg Velocity of steam leaving nozzle considering negligible inlet velocity, Isentropic heat drop = (h0 \u2013 h1) = 129.11 kJ/kg Actual heat drop = 0.95 (h0 \u2013 h1) = 122.65 kJ/kg 0.95 2 h 0 h 1 C1 = = 2 122 65 103. C1 = 495.28 m/s Given: U = 200 m/s, 1 = 20\u00b0, V 2 V 1 = 0.90 C1a = C2a (As there is no axial thrust) From velocity diagram Fig. 14.39 V1 = 320 m/s, V2 = 0.90 V1 = 288 m/s, C2 = 180 m/s, Cw = 485 m/s angle FBD = angle ABE = 1 = 32\u00b0 2 = 39\u00b0 Work done per kg of steam = U Cw = (200 485) = 97000 J/kg, Actual work = (97 \u2013 0.5) = 96.5 kJ/kg Steam Turbines ________________________________________________________________ 665 Stage efficiency = = Work done per kgof steam Energy input per kgof steam tostage 97000.5 0 = 0.7474 or 74.74% 10 3 129 11..5 0 7457 10. Output in hp = 97 0 = 1294.09 hp Energy lost in nozzle = (h0 \u2013 h1) 0.05 = 6.46 kJ/kg 2 V 288 1 = 320 2 V 2 Energy lost in blade friction = 2 2 2 2 = 9.73 kJ/kg Exit energy loss = 2 C2 2 = 2 180 2 = 16.2 kJ/kg Heat balance sheet in kJ/kg of steam Input, kJ/kg % Expenditure, kJ/kg Heat supplied = 129.11 100% (i) Useful work = 96.5 (ii) Nozzle friction loss = 6.4 (iii) Blade friction loss = 9.7 (iv) Windage & disc friction loss = 0.5 (v) Exit energy loss = 16.2 129.11 129.3 kJ/kg, Error = 0.19 kJ/kg % 74.63 4.95 7.50 0.38 12.55 Error of 0.19 kJ/kg is due to cumulative error creeping in from construction and rounding off measurements from velocity diagram. 9. Following data refer to the velocity compounded impulse turbine with two rows of moving blades: Steam supply: 60 bar, 500\u00b0C Speed of rotation: 3600 rpm Mean radius of blades: 60 cm Steam supply: 6 kg/s Blade speed to steam velocity ratio: 0.2 Nozzle efficiency: 90% Nozzle angle: 20\u00b0 Exit angles: 25\u00b0, 25\u00b0 and 40\u00b0 for first moving, fixed and second moving blades respectively Blade velocity coefficient: 0.85 for all blades Disc friction and Windage loss: 20 hp. Determine, (a) (b) (c) (d) the stage pressure the blading efficiency the power output the state of steam leaving stage. 666 ________________________________________________________ Applied Thermodynamics Solution: Blade speed, U = r = 60 10 2 2 60 3600 = 226.2 m/s = 0.2 = U C1 so C1 = 1131 m/s Actual heat drop in stage", ", Theoretical heat drop Enthalpy at state 1, Enthalpy at inlet to nozzle, Entropy = 639.58 kJ/kg = 710.64 kJ/kg h h01 = 2 = 01 = 2 C1 1131 2 2 h 01 0.9 h1 = h0 \u2013 h01 h0 = hat 60 bar, 500\u00b0C h0 = 3422.2 kJ/kg s0 = 6.8803 kJ/kg\u00b7K Fig. 14.40 hence, h1 = 2711.56 kJ/kg, s1 = s0 = 6.8803 kJ/kg\u00b7K From mollies diagram, P1 = 3.8 bar corresponding to h1 vertically below 60 bar and 500\u00b0C state. Stage pressure = 3.8 bar Ans. Velocity diagram is drawn using scale of 1 cm = 100 m/s and following, U = 226.2 m/s, C1 = 1131 m/s, 4 = 40\u00b0 V 2 V 1 2 = 25\u00b0, 1 = 20\u00b0, 3 = 25\u00b0, = 0.85 V 4 V 3 From diagram, V1 = 935 m/s, V2 = 794.75 m/s, C2 = 610 m/s Fig. 14.41 Steam Turbines ________________________________________________________________ 667 = 0.85, C3 = 0.85 C2 = 518.5 m/s C 3 C 2 V3 = 335 m/s V4 = 284.75 m/s Velocity diagram is drawn in following steps: (i) Draw AB corresponding to U (= 226.2 m/s) (ii) Draw AD at angle 20\u00b0 corresponding to C1 (= 1131 m/s) (iii) (iv) Draw BE corresponding to V2 = (794.75 m/s) at angle (v) 2 (= 25\u00b0) Join D with B and measure BD (= 9.35 cm) or V1 = 935 m/s, Estimate V2 = 0.85 V1 Join E with A and measure AE (= 6.1 cm) or C2 = 610 m/s. For fixed blade C3 = 0.85 C2, Estimate C3. (vi) Draw AG at angle (vii) 3 (= 25\u00b0) corresponding to C3 (= 518.5 m/s) Join G with B and measure BG (= 3.35 cm) or V3 = 335 m/s. Estimate V4 = 0.85 V3 for IInd moving blade. (viii) Draw BF at angle 40\u00b0 corresponding to V4 (= 284.75 m/s) (ix) Join F with A and measure AF (= 2.8 cm) or C4 = 280 m/s. Length KH (= 15.6 cm) or Cw Ist moving = 1560 m/s Length JI (= 5.45 cm) or Cw II moving = 545 m/s Total Work done = (U Cw Ist moving) + (U Cw II moving) per kg of steam = 226.2 (1560 + 545) = 476.15 kJ/kg Blading efficiency = Net power output = = 0.7444 or 74.44% 476.15 639.58 476.15 6 0 7457. \u2013 20 Energy loss at exit per kg = = 3811.16 hp = 2 C4 2 2 280 2 = 39.2 kJ/kg Heat equivalent of net work =. 6 Hence h = 473.66 + 39.2 = 512.86 kJ/kg Corresponding to enthalpy drop of 512.86 kJ/kg from h0 the final condition of steam can be seen = 473.66 kJ/kg 381116 0 7457. from mollier diagram. The state is at 225\u00b0 (superheated steam at 3.8 bar (approximately) Blading efficiency = 74.44% Net output = 3811.16 hp State of steam leaving stage = 225\u00b0C Ans. 10. Determine blade speed, blade and nozzle angles and work done per kg of steam, ratio of height required for moving blades rows and fixed blade row with respect to nozzle height. Increase in 668 ________________________________________________________ Applied Thermodynamics volume of steam during expansion in velocity compounded impulse turbine having two moving blade rows may be neglected. Further data required for analysis is given as under, Velocity of steam leaving nozzle = 500 m/s Angle at exit of second moving blade = 30\u00b0 from plane of rotation of blades Velocity coefficient = 0.90 for each blade. There is no axial thrust on any of moving blade row and the discharge from stage is axial. The axial velocity in first moving blade is twice that of second moving blade. Solve using velocity diagram. Solution: Let us draw the velocity diagram for the", " stages in backward manner starting from exit of second moving blade and coming gradually to inlet of Ist moving Blade. Given, C1 = 500 m/s, 4 = 30\u00b0,.90 4 = 90\u00b0, C2a = C1a and C2a = 2 C3a C3a = C4a Steps: (i) Start with any length AB designating U, say AB = 2 cm (ii) Draw line BG at angle 4 (= 30\u00b0) such that G lies vertically below A, there is axial discharge from stage. Measure BG = 2.7 cm which gives V4. Estimate V3, as V3 = V4 0.9 = 3 cm (iii) Draw a line GF parallel to AB as C3a = C4a. Using compass for 3 cm width mark state F with compass needle at B. Join B and F. Measure AF (= 4.8 cm) which gives C3. Estimate C2 = C3 0.9 = 5.33 cm (iv) Draw a line ED parallel to AB as C2a = 2 C3a i.e. 3.5 cm below AB and using compass for 5.33 cm width mark state E with compass needle at A. Join E and A to show C2. Join E and B which shows V2. Measure BE (= 7 cm) which gives V2. Estimate V1 = V2 0.9 = 7.78 cm. (v) Using compass open for 7.78 cm mark point D with compass needle at B. Join line BD and AD which show V1 and C1 respectively. Measure length AD = 9.6 cm. Thus C1 of 500 m/s velocity is shown on length of 9.6 cm. Scale comes out, 9.6 cm = 500 m/s, 1cm = 52.08 m/s Fig. 14.42 Steam Turbines ________________________________________________________________ 669 Hence using scale, 1 cm = 52.08 m/s. So AB = 2 cm U = 2 52.08 m/s, U = 104.16 m/s CwI moving = HI = 13.05 cm = 679.64 m/s CwII moving = 4.55 cm or 236.96 m/s AG = C4 = C3a = 1.75 cm = 91.14 m/s HE = C2a = C1a = 3.5 cm = 182.28 m/s Angles BAD = IBD = ABE = HAE = BAH = ABF = 1 = 21\u00b0 1 = 26\u00b0 2 = 28.5\u00b0 2 = 39\u00b0 3 = 20\u00b0 3 = 33\u00b0 Blade speed = 104.16 m/s Ist moving blade inlet and exit angle = 26\u00b0, 28.5\u00b0 Fixed blade inlet and exit angle = 39\u00b0, 20\u00b0 IInd moving blade inlet angle = 33\u00b0 Nozzle angle = 21\u00b0 Ans. Total work done = U CwI moving + U CwII moving per kg of steam Using continuity equation, = 104.16 (679.64 + 236.96) = 95473.056 J/kg 95.47 kJ/kg Area Flow velocity or axial velocity Specific volume =constant, Let height of nozzle first moving, fixed blade and second moving blades be ln, lm1, lf, lm2 respec- tively = ln Ca1 = lm1 Ca2 182 182.28.28 = 1 Ratio of height of Ist moving blade to nozzle = 1 Ratio of height of fixed blade to nozzle = Ratio of height of IInd moving blade to nozzle = 182.28. 9114 = 2 Ratio of height of Ist moving blade to nozzle = 1 Ratio of height of fixed blade to nozzle = 1 Ratio of height of IInd moving blade to nozzle = 2 Ans. 670 ________________________________________________________ Applied Thermodynamics 11. In a Parson\u2019s reaction turbine the rotor of 1m diameter runs at 3000 rpm. Determine the isentropic enthalpy drop in the stage considering stage efficiency of 0.80, = 0.7, blade outlet angle = 20\u00b0. Solution: Given, = U C1 = 0.7, stage = 0.80, 2 = 20\u00b0 U = r = 2 60 or, C1 = 224.4 m/s 3000 1 2 = 157.08 m/s Drawing velocity diagram considering given information. 1 = As it is Parson\u2019s reaction turbine so Scale 1 cm = 50m/s 2 and 1 = 2 Steps: Fig. 14.43 (i) Draw AB corresponding to U = 157.08 m/s. (ii) Draw AD at angle (iii) (iv) Draw C2 along AE at angle 1 = Join point D with B which gives V1 (= 95 m/s). Measure angle FBD = 2 =", " It intersects at E. Lines AE and BE give C2 and V2 respectively. 1 = 56\u00b0. Also draw EB at angle 20\u00b0. 2 = 20\u00b0 corresponding to C1 (= 224.4 m/s). (v) Length FG gives Cw (= 260 m/s) of 5.2 cm. 1 = 56\u00b0 Work output = U Cw (per kg of steam) = 157.8 260 = 40840.8 J/kg State efficiency stage = 0.80 = 40840 8. hisentropic hisentropic = 51051 J/kg or 51.05 kJ/kg Enthalpy drop in stage = 51.05 kJ/kg Ans. 12. A Parson\u2019s reaction turbine has mean diameter of blades as 1.6 m and rotor moving at 1500 rpm. The inlet and outlet angles are 80\u00b0 and 20\u00b0 respectively. Turbine receives steam at 12 bar, 200\u00b0C and has isentropic heat drop of 26 kJ/kg. 5% of steam supplied is lost through leakage. Determine the following considering horse power developed in stage to be 600 hp. (a) (b) the stage efficiency the blade height. Steam Turbines ________________________________________________________________ 671 Solution: Given: 50% reaction turbine with N = 1500 rpm, d = 1.6 m. 1 6 2 2 60 U = r = 1500 1 = 80\u00b0 & 2 = 20\u00b0 2 = 80\u00b0 and 1 = 20\u00b0 = 125.66 m/s Velocity diagram for turbine stage is drawn as below using above data. Fig. 14.44 From velocity diagram, Length AD = C1 = 168.75 m/s Length FG = Cw = 170 m/s Length FE = C2a = 52.5 m/s = C1a Work done = U Cw = 125.66 170 = 21362.2 J/kg or 21.36 kJ/kg per kg of steam. Since 5% of steam gets leaked so only 95% of steam supplied will be used for production of work Net work per kg of steam = 0.95 21.36 = 20.29 kJ/kg Stage efficiency = Work done Heat supplied = 20.29 26 = 0.7804 or 78.04% Let mass of steam supplied be m kg per second to produce 600 hp. 0.7457 600 = m 20.29 m = 22.05 kg/s Assuming heat drop per stage to be small so that the specific volume remains nearly constant throughout in the stage. Specific volume at inlet from steam table, v1 = vat 12 bar, 200\u00b0c v1 = 0.16930 m3/kg Let height of blades be \u2018h\u2019. Volume of steam flow rate = d h C1a 22.05 0.16930 = 1.2 h 52.5 h = 0.01886 m or 1.89 cm Stage efficiency = 78.04% Height of blade = 1.89 cm Ans. 13. In a reaction turbine 6 kg/s steam is admitted at 15 bar dry saturated in the first stage. Turbine has eight pairs on mean diameter of 50 cm and run at 3000 rpm with mean blade speed to steam velocity 672 ________________________________________________________ Applied Thermodynamics ratio of 0.8. There occurs tip leakage of steam at all rows amounting to 10% of total and efficiency of working steam is 85%. Considering blade outlet angles for both fixed and moving blades to be 20\u00b0, determine the following analytically. (i) (ii) (iii) the output from turbine in hp. the pressure of steam leaving turbine, the mean blade height. Solution: Let the degree of reaction be 50% for the reaction turbine. Mean blade speed U = 50 10 2 2 2 3000 60 = 78.54 m/s Steam velocity, = 98.18 m/s = C1= U 1 = 20\u00b0, Cw = 2C1 cos 78 54. 0 8. 2 = 20\u00b0 1 \u2013 U Angles, = {(2 98.18 cos 20) \u2013 78.54)} Cw = 105.98 m/s Actual mass of steam flowing through rows, mactual = 6 0.90 = 5.4 kg/s Power output = U Cw mactual no. of pairs 3 78.54 105 5 4 8 10..98. 0 7454 = = 482.4 hp Output from turbine = 482.4 hp Ans. Theoretical heat drop in turbine can be obtained using efficiency of working steam, h = 482 4 0 7454. 6 0 85.. = 70.51 kJ/kg At inlet to steam turbine the enthalpy may be seen from steam table as, h1 = hat 15 bar, dry saturated = hg at 15 bar = 2792.2 kJ/kg s1 = sg at 15 bar = 6.448 kJ", "/kgK, v1 = 0.13177 m3/kg Hence, enthalpy at exit of turbine hexit = h1 \u2013 h = 2792.2 \u2013 70.51 hexit = 2721.69 kJ/kg Pressure at exit of turbine can be approximated from the mollier diagram corresponding to hexit and sexit (= s1) vertically below state 1. Pressure at exit = 12 bar (approx.) Pressure at exit = 12 bar Ans. Dryness fraction at exit of turbine, is obtained using entropy values, s1 = sexit Steam Turbines ________________________________________________________________ 673 = sf at 12 bar + xexit sfg at 12 bar 6.448 = 2.2166 + xexit 4.3067 xexit = 0.98 vexit = vf at 12 bar + xexit\u2019 vfg at 12 bar = 0.001139 + 0.98 (0.16333 \u2013 0.001139) v Mean specific volume = 1 vexit = 0.16009 m3/kg v+ 2 = 0.14593 m3/kg = exit.. 0 13177 0 16009 2 Steam mass flow rate = 6 0.92 0.14593 = 50 h C a 1 100 Axial component of velocity C1a = C1sin \u03b1 1 = 98.18 sin 20 C1a = 33.58 m/s This axial component of velocity, C1a is approximated as average flow velocity Substituting C1a, h is estimated as under 6 0 h =.92. 50 33 0 14593 100.58 = 0.0153 m Mean blade height = 1.53 cm Ans. 14. A reaction turbine has mean blade speed of 180 m/s, blade speed to steam velocity ratio of 0.8, outlet angles of fixed and moving blades as 20\u00b0 and 30\u00b0, specific volume at outlet of fixed blade as 0.5 m3 and at moving blade outlet as 0.6 m3. Areas at exit of fixed blade and moving blades are same, Consider the efficiency of blades as 90% when considered as nozzles and K2 = 0.88 where K is blade velocity coefficient. Determine the degree of reaction, (i) (ii) adiabatic heat drop, (iii) the overall stage efficiency. Solution: Given: 1 = 20\u00b0, 2 = 30\u00b0, = 0.8, U = 180 m/s. Steam velocity C1 = Mean blade speed Blade speed to steam velocity ratio = 180 0.8 C1 = 225 m/s Axial velocity component, at inlet C1a = C1 sin \u03b1 1 = 225 sin 20 C1a = 76.95 m/s From continuity equation, the steam mass flow rate at exit of fixed blades and moving blades shall be same, so 674 ________________________________________________________ Applied Thermodynamics aC 1 v at exit of fixed blades = aC 2 v at exit of moving blades 76.95 0.5 = C a2 0 6. C2a = 92.34 m/s From geometry of velocity diagrams, C2a = V2 sin 2. 92 34 sin 30 = V2 = C a2 sin V2 = 184.68 m/s 2 = C1 V1 2 2 + U 2 \u2013 2C1U cos 1 = (225)2 + (180)2 \u2013 (2 225 180 cos 20) = 6309.9 Heat drop in fixed blades = Heat drop in moving blades = Here nozzle efficiency is given as 0.9 for both fixed and moving blades. C2 2 + U 2 \u2013 2UV 2 cos 2 = V2 2 = (184.68)2 + (180)2 \u2013 (2 180 184.68 cos 30) 2 = 8929.18 C2 C2 = 94.49 m/s 2 225. 0 88 8929 18 2 = 23.76 kJ/kg.9 0. 184 68. 2 2 0 88 6309. 0.9.9 Heat drop in fixed blades = Heat drop in moving blades = = 15.86 kJ/kg Therefore, degree of reaction = Heat drop in moving blades Total heat drop = 15 86. 23 76 15 86 = 0.4003 or 40.03%.. Degree of reaction = 40.03% Ans. Adiabatic heat drop in one pair of blades = (23.76 + 15.86) = 39.62 kJ/kg. Steam Turbines ________________________________________________________________ 675 Heat drop in a pair of blades = 39.62 kJ/kg Ans. Work done = U Cw = U (V2 cos 2 + C1 cos = 180 (184.68 cos 30 + 225 cos 20 \u2013 180) = 34446.31 J/kg = 34.45 kJ/kg 1 \u2013 U) Stage efficiency = 34 45.. 39 62 = 0.8695 or", " 86.95% Stage efficiency = 86.95% Ans. 15. A multi stage steam turbine has steam entering at 20 bar, 300\u00b0C and leaving at 0.05 bar and 0.95 dry. Determine reheat factor, condition of steam at exit from each of five stages considering efficiency ratio of 0.555 and all stages doing equal work. Solution: At inlet to turbine, the enthalpy Enthalpy at exit, h1 = hat 20 bar, 300\u00b0C h1 = 3023.5 kJ/kg s1 = 6.7664 kJ/kgK h6 = hf at 0.05 bar + x hfg at 0.05 bar = 137.82 + (0.95 2423.7) h6 = 2440.34 kJ/kg Given: Stage efficiency or efficiency ratio = 0.55 Since all the stages do equal work, so the useful enthalpy drop in each of stage shall be equal. h1 \u2013 h6 = 583.16 kJ/kg = Actual heat drop Fig. 14.45 676 ________________________________________________________ Applied Thermodynamics Enthalpy drop in each stage = h 1 h 6 5 = 116.63 kJ/kg For first stage, h1 \u2013 h2 = 116.63 kJ/kg Stage efficiency = h1 \u2013 h2 = h 1 h 1 h 2 h 2 116.63 0.555 = 210.05 kJ/kg For second stage, h2 \u2013 h3 = 116.63 kJ/kg Stage efficiency = 0.555 = h2 \u2013 h3 = h 2 h 2 h 3 h 3 116.63 0.555 = 210.05 kJ/kg Similarly, for third stage h3 \u2013 h4 = 210.05 kJ/kg for fourth stage, h4 \u2013 h5 = 210.05 kJ/kg for fifth stage, h5 \u2013 h6 = 210.05 kJ/kg After isentropic heat drop up to state 6, s1 = s6 = 6.7664 kJ/kgK s6 = sf at 0.05 bar + x6 sfg at 0.05 bar. 6.7664.9187 7 x6 = 0.7943 0 4764 x6 = h6 = 2062.96 kJ/kg Enthalpy, h6 = hf at 0.05 bar + x6 hfg at 0.05 bar = 137.82 + (0.7943 2423.7) h 4 Adialatic heat drop Cumulative heat drop h 1 Reheat factor, RF = 210 05. 2062.5 3023.96 = RF = 1.09 Reheat factor = 1.09 Ans. Plotting the actual heat drop shown by 1-6 on Mollier diagram state of steam leaving each stage can be given as under. No. of stage Pressure at exit of stage, bar Condition of steam 1 6.9 235\u00b0C 2 2.6 160\u00b0C 3 0.8 dry sat. 4 0.2 0.975 dryness factor 5 0.05 0.95 dryness factor Ans. Steam Turbines ________________________________________________________________ 677 16. In a five stage turbine the steam is supplied at 35 bar, 450\u00b0C and leaves at 0.07 bar, 0.89 dry. Determine the following for equal work between the stages: the state of steam at inlet to each stage, the stage efficiency or efficiency ratio for all stages, the reheat factor, the overall turbine efficiency. (i) (ii) (iii) (iv) Solve using Mollier diagram. Fig. 14.46 Solution: From mollier diagram, h1 \u2013 h6 = 1028 kJ/kg Considering equal work the enthalpy drop in each stage = h 1 h 6 5 = 205.6 kJ/kg Plotting the condition line as straight line between 1 and 6, the following states are obtained at inlet of each stage No. of stage Pressure, at inlet bar State of steam 1 35 5 3 0.35 5.5 450\u00b0C 342\u00b0C 235\u00b0C 125\u00b0C 0.95 2 14.5 4 1.7 The enthalpy drops as evident from mollier diagrams, h1 \u2013 h2 = 264 kJ/kg, h2 \u2013 h3 = 246 kJ/kg, h3 \u2013 h4 = 240 kJ/kg, 678 ________________________________________________________ Applied Thermodynamics h4 \u2013 h5 = 240 kJ/kg, h5 \u2013 h6 = 235 kJ/kg and h1 \u2013 h2 = h2 \u2013 h3 = h3 \u2013 h4 = h4 \u2013 h5 = h5 \u2013 h6 = 205.6 kJ/kg. 205 6 Stage efficiency for different stages, 264 stage,.7788 stage,II = stage,III = stage,IV = stage", ", =. 205 6 246 = = =. 205 6 240. 205 6 240. 205 6 235 = 0.8358 = 0.8567 = 0.8567 = 0.8749 Stage efficiencies = 77.88%, 83.58%, 85.67%, 85.67%, 87.49% Ans. From mollier chart. h1 \u2013 h6 = 1162 kJ/kg ( + \u2212 + \u2212 + + \u2212 \u2212 ( ) ( ) ) ( Reheat factor, RF = ( ( \u2212 235 264 246 = 240 240 1162 RF = 1.05 Ans. Reheat factor = 1.05 Overall efficiency 1028 1162 = 0.8847 Overall efficiency = 88.47% Ans. 17. One stage of an impulse turbine consists of a converging nozzle ring and one ring of moving blades. The nozzles are inclined at 22\u00b0 to the blades whose tip angles are both 35\u00b0. If the velocity of steam at exit from the nozzle is 660 m/s, find the blade speed so that the steam passes without shock. Find diagram efficiency neglecting losses if the blades are run at this speed. [U.P.S.C., 1992] Solution: Diagram efficiency of impulse turbine, when losses are neglected, shall be equal to the maximum diagram efficiency obtained for optimum blade speed-steam velocity ratio (). diag, max = 1 cos2 2 (1 + K\u00b7C) Considering turbine to have smooth and symmetrical blades i.e. K = 1,C = 1 and Blade angles, 1 = 2 Nozzle angle = 1 = 22\u00b0 Steam Turbines ________________________________________________________________ 679 1 = 0.8597 or 85.97% diag, max = cos2 cos 2 Velocity of steam leaving nozzle is C1 and given as 660 m/s. optimum = Also U C1 = 1 cos22 2 = U 660 U = 305.97 m/s Blade speed = 305.97 m/s Diagram efficiency = 85.97% Ans. 18. A simple impulse turbine has a mean blade speed of 200 m/s. The nozzles are inclined at 20\u00b0 to the plane of rotation of the blades. The steam velocity from nozzles is 600 m/s. The turbine uses 3500 kg/hr of steam. The absolute velocity at exit is along the axis of the turbine. Determine (i) the inlet and exit angles of the blades, (ii) the power output of the turbine, (iii) the diagram efficiency, and (iv) the end thrust (per kg steam per second) and its\u2019 direction. [U.P.S.C., 1998] Solution: Given, U = 200 m/s, scale: 1 cm = 50 m/s 1 = 20\u00b0, C1 = 600 m/s, 2 = 90\u00b0, m = 3500 kg/hr From geometry, C1 sin 1 = 600 sin 20 = 205.2 m/s Fig. 14.47 tan 1 = C 1 1 = tan\u20131 U sin C 1 1 cos 1.2 205 20 600.cos 200 For symmetrical blade, 1 = 29.42\u00b0 1 = 2 = 29.42\u00b0 Blade inlet angle = Blade exit angle = 29.42\u00b0 Ans. Power output = m U Cw 200 600 3500 cos 20 = 60 60 1000 = 109.63 kW Power output = 109.63 kW Ans. 680 ________________________________________________________ Applied Thermodynamics Diagram efficiency = = = = cos U C 1 2 C 1 cos C 1 200 U 1 2 2 2 cos 20 600 = 0.6265 Diagram efficiency = 62.65% Ans. From velocity diagram, V2 cos = 229.61m/s 2 = cos 2 = U, V2= 200. 29 42 U cos End thrust = (C2 \u2013 C1 sin 1) = (V2 sin 2 \u2013 C1 sin 1) = (229.61 (sin 29.42) \u2013 600 sin 20) = \u2013 92.43 N/kg\u00b7s End thrust = 92.43 N/kgs in reverse direction Ans. EXERCISE 14.1 Describe the principle of steam turbine operation. 14.2 Classify the steam turbines and differentiate between steam turbine and steam engines. 14.3 Differentiate between impulse and reaction turbines. 14.4 What do you understand by compounding of steam turbines? Describe different types of compounding of steam turbines. 14.5 Describe the velocity diagram for single stage impulse turbine. Also obtain the expressions for force, work done, diagram efficiency, gross stage efficiency and axial thrust. 14.6 Obtain the condition for maximum blade efficiency in single stage impulse turbine. Also show how this efficiency varies with blade speed to steam velocity ratio. 14.7 Compare different types of compounding of steam turbines with one another. 14.8 Sketch the velocity diagram of a two stage velocity compounded turbine. 14.9 Obtain the", " optimum blade speed to steam velocity ratio for two stage velocity compounded impulse turbine. Also plot the variation of diagram efficiency with blade speed to steam velocity ratio. 14.10 What do you understand by axial discharge turbine? 14.11 Discuss the significance of \u201cwhirl component of velocity\u201d and \u201cflow component or axial component of velocity\u201d in steam turbine. 14.12 Describe the working of reaction turbine. Explain, why pure reaction turbine is not used in practice. 14.13 How does reaction turbine blades differ from impulse turbine. 14.14 Discuss different losses in steam turbines. Steam Turbines ________________________________________________________________ 681 14.15 Sketch the velocity diagram for a single stage reaction turbine having 50% degree of reaction. 14.16 Obtain the condition for maximum diagram efficiency of a reaction turbine in terms of absolute angle at inlet to moving blade. 14.17 What is meant by reheat factor? Also discuss its\u2019 relevance with the help of T\u2013s and h\u2013s diagrams 14.18 Describe the governing in steam turbines. Give different methods for governing in steam turbines. 14.19 Differentiate amongst different methods of governing of steam turbine. 14.20 Give the reason for following statement; \u201cVelocity compounded impulse stages are generally placed at the high pressure inlet end of steam turbine\u201d. 14.21 Show that the reheat factor in steam turbine having steam entering dry and saturated at absolute temperature T1, and expanding up to the absolute temperature T2 with stage efficiency s, shall be given as, RF = The number of stages may be considered infinite and the condition line is straight line. 14.22 A steam turbine stage is supplied with steam at 50 bar, 350\u00b0C and leaves turbine at pressure of 5 bar. Determine steam supply rate if the isentropic efficiency is 82% and power output is 15224 [37.8 kg/s] kW 14.23 A De Laval steam turbine has steam leaving nozzle with velocity of 1200 m/s at angle of 20\u00b0 and mean blade speed of 400 m/s. Considering blades to be symmetrical, blade velocity coefficient of 0.8 and steam flowing at rate of 1000 kg/hr, determine the following using velocity diagram, (a) (b) (c) (d) the blade angle the power developed the relative velocity of steam entering blades the blade efficiency. [30\u00b0, 30\u00b0, 145.5 kW, 830 m/s, 72.8%] 14.24 Determine the turbine output from a single stage impulse turbine having smooth blades, steam flow rate of 20 kg/s, steam velocity of 600 m/s, blade speed of 250 m/s, nozzle angle of 20\u00b0 and [3275 kW, 800N] blade outlet angle of 25\u00b0. Also calculate axial thrust. 14.25 Determine nozzle angle, blade angles at inlet and exit for a single stage impulse steam turbine developing 132 kW, speed of 3340 rpm, mean rotor diameter of 1 m, steam flow rate of 2 kg/s, blade velocity coefficient of 0.9, steam leaving nozzle at 400 m/s and steam discharges axially from [21\u00b0, 36\u00b0, 32\u00b0] turbine. Solve using velocity diagram. 14.26 De-Laval turbine has nozzle angle of 20\u00b0, symmetrical blades and blade velocity coefficient of 0.85. Determine maximum blade efficiency. Also find out the blade speed to steam velocity ratio if the [81.6%, 0.6 or 0.34] actual blade efficiency is 92% of maximum blade efficiency. 14.27 What shall be the nozzle angle for a De-Laval turbine offering maximum possible efficiency of [18.44\u00b0] 14.28 A stage of simple impulse turbine has 10 kg/s steam entering the rotor having mean diameter of 105 cm, running at 50 r.p.s.. Nozzle angle is 18\u00b0, blade speed to steam velocity ratio is 0.42, blade velocity coefficient is 0.84 and the exit angle of blade 3\u00b0 less than inlet angle. Draw the velocity diagram and obtain (a) the axial thrust on blades, (b) the tangential thrust on blades, (c) the [250 N, 3900 N, 3908 N, 641.5 kW] resultant thrust on blades (d) the power developed. 14.29 A singe stage impulse steam turbine has mean diameter of rotor as 1 m and runs at 3000 rpm. Steam leaves nozzle at 20\u00b0 with velocity of 300 m/s. Determine the power developed if the axial thrust on 90%? 682 ________________________________________________________ Applied Thermodynamics blades is 98 N, blades are symmetrical and there occurs 19% frictional loss of kinetic energy of [181.2 kW] relative velocity at inlet to blades. 14.30 Single row impulse turbine has blade speed of 175 m/s and steam leaves nozzle at 400 m/s. Steam flows at the rate of", " 163.2 kg/min and leaves turbine axially. Considering blade velocity coefficient of 0.9 determine nozzle angle, blade angles at inlet and exit, axial thrust, energy loss at exit, energy loss in blades and diagram efficiency if turbine produces 180 kW output. [19\u00b0, 32.6\u00b0, 36.3\u00b0, 4.35 N, 8.27 kJ/kg, 5.53 kJ/kg. 41.36%] 14.31 Following data refer to a stage of velocity compounded impulse steam turbine; Nozzle angle: 16\u00b0 Angle at exit of first moving blade: 20\u00b0 Angle at exit of fixed blade: 25\u00b0 Angle at exit of second moving blade: 30\u00b0. Velocity of steam leaving nozzle: 800 m/s Blade speed: 150 m/s Blade velocity coefficient for first moving, fixed and second moving blades: 0.8, 0.85 and 0.85. Steam flow rate: 5 kg/s Determine, (a) diagram efficiency, (b) the energy carried at exit, (c) the axial thrust on each moving [66.2%, 3.79 kJ/kg, 20.9 N, 27 N, 1059 kW] blade ring, (d) the power developed. 14.32 A two stage velocity compound impulse steam turbine has mean blade speed of 150 m/s, steam leaves nozzle with 675 m/s velocity at angle of 20\u00b0. The exit angles of first row moving blade, fixed blade and second row moving blade are 25\u00b0, 25\u00b0 and 30\u00b0 respectively. For the blade velocity coefficient of 0.9 for all blades and flow rate of4.5 kg/s determine, (a) the power output, (b) the diagram efficiency. [80.7 kW, 78.5%] 14.33 A simple impulse turbine has nozzles at angle of 20\u00b0 to the direction of motion of moving blades and steam leaves nozzle at 375 m/s, blade velocity coefficient is 0.85, steam flow rate is 10 kg/ s and blade speed is 165 m/s. Determine, (a) (b) The power output. Inlet and outlet angles for blades such that there is no axial thrust. [34.4\u00b0, 41.6\u00b0, 532 kW] 14.34 A Parson\u2019s reaction turbine has blade speed of 78.5 m/s, stage efficiency of 80%, blade speed to steam velocity ratio of 0.7 and blade outlet angle of 20\u00b0. Estimate the isentropic enthalpy drop [12.32 kJ/kg] in the stage. 14.35 In a 50% reaction turbine the inlet and exit angles are 80\u00b0 and 20\u00b0 respectively. Blade speed is 113 m/s and steam is admitted in turbine at 12 bar, 200\u00b0C and leaves after adiabatic enthalpy drop of 17.5 kJ/kg with only 95% of heat drop being utilized for producing work of 620 hp. Calculate [78.5%, 2.29 cm] the stage efficiency and blade speed. 14.36 A reaction turbine has mean blade speed of 220 m/s and blade speed to steam velocity ratio of 0.7. Determine the angle at inlet of blade and work done per kg of steam flow if angle at exit is 20\u00b0. Also determine the percentage increase in diagram efficiency if turbine is run for optimum blade speed to steam velocity ratio. [55\u00b0, 81.5 kJ/kg, 93.8%] 14.37 A reaction turbine has eight pairs with mean blade speed of 68 m/s and average blade speed to steam velocity ratio of 0.8. Steam at 15 bar, dry saturated is supplied at the rate of 300 kg/min. Steam Turbines ________________________________________________________________ 683 Angle at exit of fixed and moving blade is 20\u00b0. Determine (a) the horse power, (b) the average blade [312.6 hp, 1.35 cm] height. 14.38 In a four stage velocity compounded impulse turbine, steam is supplied at 35 bar, 420\u00b0C and leaves at 0.07 bar. Obtain the initial pressure, quality at inlet of each stage and internal efficiency of turbine if the average efficiency is 70%. Also find reheat factor. [11.3 bar, 304\u00b0C, 2.9 bar, 189\u00b0C, 0.55 bar, 0.99, 74.6% 1.07] 14.39 A reaction turbine stage has steam leaving fixed blade at 3 bar, 0.94 dry and velocity of 143 m/s. Ratio of axial velocity of flow to blade velocity is 0.7 at entry and 0.75 at exit from moving blades. Angles at exit of fixed and moving blades are same. The height of blade is 1.8 cm and steam flow is 150 kg/min. For blade velocity of 70 m/s determine the degree of reaction. [58", ".2%] 14.40 A three stage turbine has steam entering at 30 bar, 350\u00b0C and leaves first, second and third stages at 7 bar, 1 bar and 0.1 bar respectively. For equal stage efficiency of 70% determine (a) the final condition of steam, (b) the reheat factor, (c) the overall thermal efficiency. [0.94, 1.05, 24.5%] 14.41 A four stage turbine has steam supplied at 20 bar, 350\u00b0C and leaves at 0.05 bar. Considering overall efficiency to be 80% and equal work from the stages and straight condition line determine, (a) the steam pressure at exit of each stage, (b) the each stage efficiency, (c) the reheat factor. [9 bar, 2.4 bar, 0.5 bar, 0.05 bar, 72.5%, 76%, 78.4%, 82.5%, 1.04] 684 ________________________________________________________ Applied Thermodynamics 15 15.1 INTRODUCTION Condenser is one of the essential components of steam power plants as it facilitates condensation of steam at given conditions with minimum expenditure of energy and minimum loss of heat and finally gives condensate which can be recirculated by feed pump to boiler for steam generation. Condenser generally operates at pressure less than atmospheric pressure. In the steam power plant the use of condenser permits expansion in steam turbine even upto less than atmospheric pressure and subsequently condensing steam to yield condensate for recirculation thus improving plant efficiency and output. Expansion in steam turbine/engine can not be extended to pressures less than atmospheric in the absence of condenser. \u201cCondenser can be defined as device used for condensation of steam at constant pressure; generally pressure is less than atmospheric pressure\u201d. Condenser is thus a closed vessel which is generally maintained at vacuum and cold fluid is circulated for picking heat from steam to cause its condensation. Use of Condenser offers advantages such as hotter feed water for being sent to boiler\u2019, \u2018removal of air and non condensable dissolved gases from feed water\u2019, \u2018recovery of condensate reduces treated water requirement\u2019, \u2018expansion upto subatmospheric conditions and capital cost is reduced by recycling of feed water\u2019 etc. Increase in expansion work due to use of condenser is shown in Fig. 15.1 on p-V diagram. Fig. 15.1 p-V diagram showing how condenser increases work output in steam engine Steam power plant employing condenser and the condensing plant are shown in Fig. 15.2. Steam Condenser _______________________________________________________________ 685 Fig. 15.2 Schematic for steam power plant having condensing plant Discharge from steam turbine passes into condenser where it is condensed using cooling water being circulated employing coolant pump. Condensate being at pressure less than atmospheric pressure is to be sucked out using condensate extraction pump. Condensate is extracted and sent to hot well from where it is pumped to boiler using feed pump. Dissolved gases and air etc. if any are extracted out from condenser using air extraction pump. This air or vapour may be present because of air leaking into vacuum system and air coming with steam. Cooling water for supply to condenser is taken either from some river or from cooling tower. Cooling water requirement may be up to 100 kg water per kg of steam or even more depending upon the type of condenser and its capacity. Cooling tower cools the hot cooling water leaving condenser to get cooled by evaporation of water and heat exchange with air. Water evaporated or lost in cooling tower is compensated by the make up treated water available from feed water treatment plant. 15.2 CLASSIFICATION OF CONDENSER Condenser can be broadly classified on the basis of type of heat exchange i.e. direct or indirect contact condensers. (i) Direct contact type or Mixing type or Jet condenser (ii) (iii) Evaporative condenser Indirect contact type or Non-mixing type or Surface condenser Jet condensers have direct contact between steam and cooling fluid thereby causing contamination of condensate. Surface condensers have indirect heat exchange through metal interface and the two 686 ________________________________________________________ Applied Thermodynamics fluids do not come in direct contact to each other. Evaporative condensers use evaporation of water for heat extraction and is well suited for dry weather so that evaporation is not difficult. Due to direct contact of two fluids the circulating water requirement is much less in jet condenser as compared to other types of condensers. Space requirement and size of condenser etc. are also less with jet condensers. Surface condenser is advantageous over direct contact type condensers because any type of cooling fluid can be used in it and also there is no scope of contamination etc. Different types of condensers are discussed ahead. (i)", " Jet condenser: In jet condenser the steam to be condensed and cooling water are intimately mixed by breaking up of water in the form of spray and allowing small sized water particles to fall down through the body of steam. The water may also be discharged out through suitably shaped nozzles into body of steam. Thus it is desired to atomize water into small sized particles so that increased surface area is available for heat exchange between hot and cold fluid. Number of arrangements for flow of steam and water are available such as; counter flow type having steam entering from bottom and flowing upwards while water enters from top and falls downwards with air pump connected on top where air is colder etc. Jet condenser may be further classified based on relative movement of two fluids, and based on arrangement used for removal of condensate. Based on relative moment of two fluids jet condenser can be, (a) Counter flow jet condenser (b) Parallel flow jet condenser Based on arrangement for removal of condensate jet condenser can be, (a) Low level jet condenser (b) High level jet condenser (c) Ejector condenser Fig. 15.3 Schematic of low level jet condenser Steam Condenser _______________________________________________________________ 687 (a) Low level jet condenser: Low level jet condenser is the one which is placed at low level such that vacuum inside condenser draws cooling water into condenser from river/pond/cooling tower. Difference between atmospheric pressure (at which cooling water is available) and condenser pressure causes flow of cooling water from cooling water reservoir to condenser i.e. (patm \u2013 pcond). Flow of steam and cooling water could be parallel flow or counter flow type. Counter flow type and parallel flow type low level jet condensers are shown in Fig. 15.3. There is provision for extraction of air and dissolved gases from top of condenser by using air extraction pump. Condensate extraction pump is used for taking out condensate from condenser and sending it to hot well. Cooling water supplied to jet condenser has generally a large percentage of dissolved air which gets liberated due to atomization of water, vacuum and heating of water and is extracted out. Low level jet condenser suffers from inherent drawback that in the event of failure of condensate extraction pump condenser shall be flooded with cooling water. (b) High level jet condenser: High level jet condenser is the one which is placed at a height more than that of water and water is to be injected into condenser using a pump and the condensate will flow out of condenser because of gravity. Here no condensate extraction pump is required, instead pump is required for pumping water upto condenser inlet. High level jet condenser is also called as \u2018barometric condenser\u2019. High level jet condenser is placed at suitable height depending upon efficient drainage and capacity of sump (hot well) into which tail pipe of condenser discharges out. Mathematically, it could be said that jet condenser placed above hotwell by 10.36 m shall be high level jet condenser or barometric condenser. High level jet condenser may also be of counterflow type or parallel flow type depending upon the direction of flow of steam and cooling water. Figure 15.4 shows counterflow high level jet condenser. Fig. 15.4 Schematic of high level condenser 688 ________________________________________________________ Applied Thermodynamics * Fig. 15.5 Ejector condenser High level jet condenser do not pose problem of flooding of condenser in the event of failure of pump as it is in case of low level jet condensers. But high level jet condensers are costlier than low level jet condenser. Also there is loss of vacuum between turbine and condenser in case of high level jet condenser. Ejector condenser: Ejector condenser has water jet discharging through the series of guide cones which guide steam on to the surface of water jet. Discharge of water through these convergent nozzles causes partial vacuum due to conversion of potential energy into kinetic energy. Subsequently water jet enters the diffuser nozzle where kinetic energy is converted into the pressure head and water is discharged against the vacuum pull. Ejector condensers are well suited for moderate vacuum only. Steam is injected in condenser with non return valve in between and is condensed by the mixing with cooling water. Condensation of steam further increases vacuum. Ejector condenser do not require air pump because of air entraining effect of water jet itself. Here condensing jet has number of nozzles arranged concentrically and have their axis inclined at such an angle that water jet assumes the form of inverted cone. Around the water jet the guide cones are arranged with increasing area from bottom to top. Water will be colder in upper part of condensing cone as compared to lower down. The temperature difference between steam and water at top will be greater than at", " lower end and so the condensation is greatest at top and gradually diminishes to zero at bottom. In case of failure of cooling water supply water may be sucked from hot well to go into steam pipe, but this is prevented by non-return valve in steam supply line. (ii) Non mixing type or surface condensers: Surface condensers are the most common types of condenser and offer great advantage in terms of no contamination of feed water. In these condensers the steam to be condensed and cooling fluid (water) do not come in contact with one another, instead Steam Condenser _______________________________________________________________ 689 the heat transfer occurs between two fluids through surface in between. Generally, cooling water flows through the pipes/tubes and steam surrounds them. These condensers are preferred in the locations where large quantity of poor quality cooling fluid (impure water) is available and condensate is to be recirculated. Surface condensers can be classified based on number of passes of condenser i.e. single pass or multipass. Number of times the cooling water crosses any transverse section is called a pass. Surface condensers may be of \u2018down flow type\u2019 or \u2018central flow type\u2019 depending on the type of flow of condensate and tube arrangement. Typical surface condenser having two passes, down flow type and central flow type arrangement are shown in Fig. 15.6. Fig. 15.6 Surface condenser Two pass surface condenser has cooling water entering from one end and coming out after twice traversing through the tubes (generally, brass) containing water and surrounded by steam to be condensed. Condensate gets collected at bottom and is subsequently sucked by condensate extraction pump. Steam is admitted from the top. Cooling water may be picked directly from river/pond/cooling 690 ________________________________________________________ Applied Thermodynamics tower. For extraction of air the provision is made for air pump. Thus, this type of condenser has three pumps i.e. one for circulating cooling water, second for condensate extraction and third for air extraction. In surface condenser the space occupied by tubes in shell is about 10% of shell volume. Steam is not passed through the tubes because at this steam pressure the specific volume of steam is large requiring large number of tubes. Down flow condenser has steam and air entering from top and flowing downwards across the bundle of tubes having cooling water flowing through them. Air is extracted from bottom and before being handled by air pump it is flown through air cooler so as to reduce the temperature of air. Low temperature of air enhances the air handling capacity of pump. With the flow of steam down and simultaneous heat exchange the condensate is taken out by condensate extraction pump. Central flow condenser has air cooling section in the centre of condenser. Steam enters from top and passes over the tube banks of similar type as in case of down flow condenser. As air is being sucked from centre so the flow of steam is radially inwards towards the centre. During this flow steam passes over tubes. Condensate is collected from bottom. In this type of condenser there is better contact between steam and tubes because of radial flow of steam in whole of condenser, thus arrangement is better as compared to down flow condenser. In different designs of condenser it is always attempted to have maximum heat transfer between two fluids. Also air extraction should be done effectively. Thus designer of condenser should keep following things in consideration for making a better design surface condenser. (i) Steam should be uniformly distributed over cooling water tubes. i.e. cooling surface. (ii) Distribution of steam should be such that there is minimum pressure loss. (iii) Number of tubes should be minimum. Water must be flown inside tubes and steam should surround them. (iv) Tubes should be cleaned from inside and outside both. Although on external surface the steam surrounding tubes prevents deposition. For internal cleaning of tubes mechanical or chemical means of cleaning be used at frequent intervals. (v) Leakage of air into condenser (due to vacuum) should be prevented as it reduces the work output. Also this reduces the heat transfer rate. Even if there is leakage of air, arrangement should be made for quick and effective removal of air with minimum work input. (vi) Air should be cooled to maximum extent inside condenser before being thrown out as this shall cause condensation if possible within condenser and thus reduce loss of condensate. Also the cool air shall enhance air handling capacity of pump. (vii) Rate of circulation of cooling water should be such that the range of temperature variation in cooling water lies near the optimum temperature range. Generally, the cooling water temperature rise is limited to 10\u00b0C for having maximum heat exchange between two fluids. (viii) Material of tubes is generally taken as brass. Tube material should be such as to offer maximum heat transfer rate i.e. high thermal conductivity. Generally, surface condensers", " are bulky and require large space. (ix) Cost of surface condenser should be kept low. Capital cost, running and maintenance cost should be maintained as low as possible. Generally, these costs are high in case of surface condenser as compared to other types of condensers. (iii) Evaporative condenser: Evaporative condensers are generally used where the availability of water is very poor. Figure 15.7 shows the schematic of such type of condenser where water falls from top through the nozzles over the condenser coil. Water picks up heat from the steam flowing through condenser coil and gets warmed up. This water is recirculated by circulation pump. Air flow inside Steam Condenser _______________________________________________________________ 691 condenser is maintained by using exhaust fan. This flow of air across condenser coil may be natural or forced to enhance the cooling rate. Fig. 15.7 Evaporative condenser Water gets evaporated and evaporated vapours are taken by air leaving condenser. Heat required for evaporation is extracted finally from the steam flowing inside tubes and thus causing its phase transformation. For preventing the exit of water vapours with air going out the separator/eliminator is put on the top before the final exit by which water vapour are recovered upto certain extent. Evaporative condensers are named so because the technique of evaporation is used for realizing the cooling. Amount of water to be sprinkled on condenser tubes should be just sufficient to maintain tube surface in thoroughly wet state. In case of air being humid the vapourizing capacity of wet air gets reduced compared to dry air and so the performance of evaporative condenser deteriorates when humidity in atmosphere is high. Evaporative condenser is advantageous over the surface condenser as the vacuum maintained in evaporative condenser is not very high and the water requirement is small. These condensers are generally used in small capacity power plants where shortage of water supply is there. 15.3 AIR LEAKAGE Generally, inside the condenser pressure less than atmospheric pressure is maintained, thereby increasing the chances of air leakage into condenser. Leakage of air occurs due to leaking joints, packings, glands etc. along with air in dissolved form coming with feed water. This leakage of air accounts up to 0.005% and 0.5% of steam condensed in case of jet type condenser and surface condensers respectively. Thus, leakage of air is practically always there in the condensers. Air leakage causes the reduction in work done per kg of steam as it increases the back pressure. Also the quantity of water required for condensation of steam is increased due to lowering of partial pressure of steam due to pressure of air. At low pressure the latent heat of steam to be released is more than at higher pressure. Air (having lower conductivity) when present between water and steam hampers the heat exchange and also takes away a portion of 692 ________________________________________________________ Applied Thermodynamics heat. Because of this reason also the more quantity of cooling water is required. Hence, leakage of air reduces the condenser efficiency and auxiliary devices such as reciprocating pump, rotary pump, steam ejector or air pumps etc. are required. Also the presence of air increases corrosive action as the corrosion depends largely upon the oxygen content. Since air leakage in condenser is quite damaging to the performance of condenser so air leakage should be detected and subsequent extraction of air being done. Air leakage is detected by isolating condenser from the rest of plant after the steady states are attained in it i.e. pressure and temperature become steady. After isolation of condenser from plant by stopping the steam and cooling fluid pump, if the vacuum gauge and thermometer readings change then it shows that their is air leakage. For identifying location of leak points, the soap bubble test is carried out in carried out in which the bubble formation occurs at leak point if soapy water is put on that after filling condenser with air. Thus, for soap bubble test condenser needs to be emptied and filled with high pressure air. This test has drawback that condenser is to be made non functional and running of plant suffers till the test is performed. For locating the sources of air leakage during the operational state of condenser peppermint oil test may be used. In this peppermint oil is applied at suspected sources of leak point and in case of leakage at the joint peppermint oil fumes enter the condenser and will come out with the air. Odour of peppermint could be felt in the air leaving condenser. Leak joints can also be detected by passing candle flame over the probable joint. In case of leakage the flame gets distorted. 15.4 CONDENSER PERFORMANCE MEASUREMENT The vacuum inside the condenser can be quantified by looking at barometer reading which gives atmospheric pressure and vacuum gauge reading and taking their difference to get absolute pressure inside condenser. Thus, Absolute pressure (in cm) in condenser", " = (Barometric head in cm of Hg) \u2013 (Vacuum pressure in cm of Hg) Generally, this barometric head depends upon the atmospheric conditions and so the absolute pressure also keeps on changing depending upon it. In order to take care for these variations a pressure head called as corrected vacuum in condenser is being defined. This corrected vacuum pressure is defined in reference to 76 cm of mercury which is the standard barometric head as below, Corrected vacuum pressure (in cm of Hg) = 76 \u2013 Absolute pressure in condenser (in cm of Hg). Therefore, the corrected vacuum pressure is used in cases where barometric head differs from 76 cm of mercury. By the Dalton\u2019s law of partial pressures, the absolute pressure inside condenser is the sum of partial pressures of steam and air inside it. The partial pressure of steam shall be equal to the saturation pressure corresponding to entering steam temperature. This partial pressure of steam could be seen from steam table. Mathematically, absolute pressure in condenser (pc), as per Dalton\u2019s law; pc = pa + ps where pa is partial pressure of air and ps is partial pressure of steam. Theoretically the vacuum in condenser can be given as, pv, th = pb \u2013 ps where pv, th is theoretical vacuum is condenser and pb is barometric pressure. It could be understood that the leakage of air into condenser shall disturb the vacuum inside the condenser and actually due to this air leaked into condenser the condenser pressure is always greater than the theoretical condenser pressure. In the absence of air leakage and with air leakage their is loss in performance of condenser and so we need to quantify this effect. \u2018Vacuum efficiency\u2019 of condenser is such parameter which is defined by the ratio of actual vacuum to theoretical vacuum inside condenser. Actually vacuum in condenser in the Steam Condenser _______________________________________________________________ 693 presence of air leakage can be given by; pv, act = pb \u2013 (pa + ps) Thus, vacuum efficiency of condenser = Vacuum efficiency = Actual vacuum in condenser ( p Theoretical vacuum in condenser, ( ) v, act ) p v, th Here, partial pressure of air (pa) shall be zero in the absence of air leakage and vacuum efficiency shall be 100%. If we look at mathematical expression of efficiency, it is obvious that vacuum efficiency increases with reduction in partial pressure of air. Also the vacuum efficiency shall increase with decrease in barometric pressure for constant exit steam pressure and condenser actual pressure. In case of less cooling water the condenser pressure increases and reduces the vacuum efficiency of condenser for other pressures remaining same. Designer always wishes to have condenser with highest vacuum efficiency i.e. close to 100%. \u2018Condenser efficiency\u2019 is another condenser performance parameter. It is given by the ratio of actual rise in cooling water temperature to the maximum possible temperature rise. Condenser facilitates heat exchange between two fluids and under ideal conditions the steam should only reject latent heat to cooling water so as to yield condensate at saturated liquid condition. Thus, there should be no undercooling of condensate in ideal condenser. Therefore, ideal condenser may be defined as condenser in which steam rejects only latent heat to cooling fluid and condensate is available without any undercooling. Ideal condenser requires minimum quantity of cooling water and shows maximum gain in cooling water temperature so as to condense the steam. Mathematically, Condenser efficiency = Actual rise in cooling water temperature Maximum possible temperature rise Here, the maximum possible temperature rise = {(Saturated temperature corresponding to condenser pressure) \u2013 (Cooling water inlet temperature)} 15.5 COOLING TOWER Cooling tower is similar to evaporative condenser where water used for cooling is being cooled effectively. Water used for cooling becomes hotter after extracting heat from condenser steam and needs to be cooled down if it is to be recycled. Cooling towers are preferably used where the water supply is limited and cooling water has to be recirculated without being thrown out. Cooling tower is such an arrangement made of wood or metal structure having baffles inside to facilitate better heat exchange between hot water falling down and atmospheric air blowing across it. Generally, hot water is admitted from top and is broken into small size (atomized) while falling down. Air enters tower at bottom and flows upward either due to natural draught or forced draught as the case may be. Air picks up heat by intimate contact with hot water particles and leaves cooling tower from exit passage at top. Cooled water falls down and is collected in a tank at bottom of cooling tower. The heat transfer from hot water to air occurs due to evaporative cooling of water and convective heating of air both. The effectiveness of cooling tower diminishes in humid weather conditions due to reduced capacity of air. Dry air shall offer better cooling effectiveness as compared to moist air. During cooling there occurs some loss of water as it", " is carried away by air. This water loss may be from 1 to 4% due to evaporation and drift losses. 694 ________________________________________________________ Applied Thermodynamics Typical values for a 500 MW steam power plant indicate that this plant has exhaust of steam at the rate of 3.6 kg/kWh at full load, requiring 18 107 kg per hour of cooling water and to cool this cooling water air requirement is about 30 106 kg per hour in cooling tower. Fig. 15.8 Cooling tower Figure 15.8 shows schematic of different cooling towers. The performance of cooling tower depends largely upon the duration of contact between water particle and air, surface area of contact between water particle and air, humidity of air and relative velocity of air and water flow etc. Steam Condenser _______________________________________________________________ 695 1. In a condenser the vacuum of 71 cm of Hg is maintained with barometer reading of 76 cm in Hg. Temperature in condenser is 35\u00b0C while hot well is at temperature of 30\u00b0C. The cooling water is circulated at the rate of 800 kg/min and condensate is available at 25 kg/min. The temperature of cooling water at inlet and outlet are 15\u00b0C and 25\u00b0C. Determine the mass of air in kg/m3 of condenser volume, dryness fraction of steam entering condenser and vacuum efficiency. Take mercury density as 0.0135951 kg/cm3, g = 9.81 m/s2. Solution: Absolute pressure in the condenser pt = (76 \u2013 71) 10\u20132 0.0135951 106 9.81 = 6668.396 N/m2 = 6.67 kPa Partial pressure of steam in condenser = Saturation pressure of steam corresponding to 35\u00b0C (from steam table) ps = 5.62 kPa Partial pressure of air, pa = pt \u2013 ps = 6.67 \u2013 5.63 = 1.04 kPa Mass of air per m3 of condenser volume can be obtained from gas equation, ma = p V a RT. 1 04 35 1. 0 287 273 = 0.012 kg/m3 Let the enthalpy of steam entering condenser be hs so by heat balance, Tc) mw Cpw \u2013 Twi mw= 800 kg/min, ms = 25 kg/min, Two ) = ms = 25\u00b0C, Twi (hs \u2013 Cpw (Two 800 4.18 (25 \u2013 15) = 25 (hs \u2013 4.18 30) = 15\u00b0C, Tc = 30\u00b0C Let dryness fraction of steam entering be x. hs= 1463 kJ/kg hs = 1463 = hf at 35\u00b0C + x hfg at 35\u00b0C 1463 = 146.68 + x 2418.6 x = 0.5442 Vacuum efficiency = 76 5 76 0 0135951 10. 0 0135951 10 4 9 81.. 4 9 81. 5 63 10. 3 = 0.9891 or 98.91% Mass of air in kg/m3 of condenser volume = 0.012 kg/m3, dryness fraction of steam entering = 0.5442, Vacuum efficiency = 98.91% Ans. 2. A condenser has vacuum of 70 cm of Hg when barometer reading is 76 cm. Condenser has temperature of 30\u00b0C. Air leaks into condenser at the rate of 1 kg air per 2500 kg steam. Calculate (i) the capacity of air pump per kg of steam for removal of air from steam entering condenser, and (ii) the mass of water vapour accompanying this air. 696 ________________________________________________________ Applied Thermodynamics Solution: Absolute pressure in condenser = (76 \u2013 70) cm of Hg = 6 0.0135951 104 9.81 = 8002.67 Pa pt = 8.003 kPa Partial pressure of steam, ps = Saturation pressure corresponding to 30\u00b0C from steam table ps = 4.246 kPa Partial pressure of air, pa = Total pressure in condenser \u2013 Partial pressure of steam pa = pt \u2013 ps = 3.757 kPa Mass of air accompanying per kg steam due to leakage = Using gas equation, pa V = mRT 1 2500 = 0.0004 kg Volume of air per kg of steam = mRT pa 0 0004.. 0 287. 3 757 273 30 = 9.26 10\u20133 m3/kg Capacity of air pump = 9.26 10\u20133 m3 per kg steam Ans. Volume of water vapour accompanying air shall be equal to the volume of air. So volume of water vapour accompanying air = 9.26 10\u20133 m3/kg Specific volume of dry steam at condenser temperature of 30\u00b0C = vg at 30\u00b0C = 32.89 m3/kg. 9 26 3 Mass of water vapour accompanying air = 10. 32 89", " Mass of water vapour accompanying air = 2.82 10\u20134 kg/kg of steam Ans. = 2.82 10\u20134 kg/kg of steam 3. During the trial on a condenser it is seen to have vacuum of 67 cm of Hg while barometer reading is 75 cm of Hg. The mean condenser temperature is 40\u00b0C and temperature of hot well is 35\u00b0C. Circulating water flows at 1000 kg/min for giving condensate at the rate of 50 kg/min. Temperature of cooling water at inlet and exit are 10\u00b0C and 25\u00b0C. Determine, (i) the vacuum corrected to standard barometer reading of 76 cm. (ii) the vacuum efficiency of condenser, (iii) the undercooling of condensate, (iv) the condenser efficiency, (v) the state of steam entering condenser, (vi) the mass of air per m3 of condenser volume, and (vii) the mass of air per kg of uncondensed steam. Solution: Vacuum corrected to standard barometer reading of 76 cm = 76 \u2013 (75 \u2013 67) = 68 cm Hg Corrected vacuum = 68 cm Hg Ans. Absolute pressure in condenser = 75 \u2013 67 = 8 cm of Hg or (8 0.0135951 104 9.81 = 10.67 kPa) or, pt = 10.67 kPa Partial pressure of steam, ps = Saturation pressure at 40\u00b0C from steam table Steam Condenser _______________________________________________________________ 697 ps = 7.384 kPa Partial pressure of air, pa = pt \u2013 ps = 3.286 kPa Actual vacuum Theoretical vacuum Vacuum efficiency = 75 8. 0 0135951 10. 9 81. 4 4 9 81. 7 384 10. 3 75 0 0145951 10 = 0.9645 or 96.45% Ans.Vacuum efficiency = 96.45% Undercooling of condensate = 40 \u2013 35 = 5\u00b0C Ans. Undercooling = 5\u00b0C Condenser efficiency = T w out T w T p sat corresponding to t in T w in Saturation temperature corresponding to absolute pressure in condenser, 40.01 kPa. T sat corresponding to pt = 46.9\u00b0C, from steam table. Tw out = 25\u00b0C, Tw in = 10\u00b0C 10 10 25 46 9. Condenser efficiency = = 0.4065 or 40.65% Let us consider enthalpy of steam entering be h kJ/kg and dryness fraction be x. By applying heat balance, ms (h \u2013 cpw Tc) = mw cpw (Two ms = 50 kg/min, mw = 1000 kg/min \u2013 Twi ) 50(h \u2013 4.18 40) = 1000 4.18 (25 \u2013 10) h = 1421.2 kJ/kg = hf at 40\u00b0C + x hfg at 40\u00b0C 1421.2 = 167.57 + x2406.7 x = 0.5209 Dryness fraction of steam entering = 0.5209 Ans. Mass of air per m3 of condenser volume m = p V a RT c. 3 286 273 1 40 = 0.0366 kg/m3. 0 287 Volume of per kg of uncondensed steam = vg at 40\u00b0C = 19.52 m3/kg Mass of air in one kg of uncondensed steam = Mass of air in 19.52 m3 volume =. 3 286. 0 287 40. 19 52 273 = 0.714 kg Mass of air/m3 of condenser volume = 0.0366 kg/m3 Mass of air in per kg of uncondensed steam = 0.714 kg Ans. 4. In a surface condenser operating with steam turbine the vacuum near inlet of air pump is 69 cm of Hg when barometer reading is 76 cm of Hg. Temperature at inlet of vacuum pump is 30\u00b0C. Air leakage occurs at the rate of 60 kg/hr. Determine, 698 ________________________________________________________ Applied Thermodynamics (i) (ii) the capacity of air pump in m3/hr and mass of vapour extracted with air in kg/hr. the dimension of reciprocating air pump cylinder if it runs at 240 rpm and L/D ratio is 1.5. Solution: Absolute pressure at inlet to air pump = (76 \u2013 69) = 7 cm Hg, or = 7 0.0135951 104 9.81 pt = 9.336 kPa Partial pressure of vapour at 30\u00b0C = Saturation pressure at 30\u00b0C Partial pressure of air, Volume of 60 kg air at pressure of 89.112 kPa ps = 4.246 kPa pa = pt \u2013 ps = 5.09 kPa V = 60. 0 287. 5 09 = 1025.1 m3/hr 273 30 Capacity of air pump", " in m3/hr = 1025.1 m3/hr Ans. Volume handled, m3/hr = 1025.1 = 4 4 D2 L N 60 = 1025.1 1.5 D3 240 60 D = 0.3924 m or 39.24 cm L = 58.86 cm Mass of water vapour going with air = V V g at 30 C \u00b0 = 1025.1 32.89 = 31.17 kg/hr Bore = 39.24 cm, Stroke = 58.86 cm Mass of water vapour extracted with air = 31.17 kg/hr Ans. 5. A steam condenser is supplied with 1000 kg/min steam in 0.9 dry state. The pressure at suction of air extraction pump on condenser is 70 cm of Hg and barometer reads 77 cm of Hg. Temperature in suction pipe is 30\u00b0C and air leaks at the rate of 5 10\u20134 kg per kg of steam. Cooling water temperature gets increased by 15\u00b0C. Determine the mass handled by dry air extractor and cooling water circulation rate in kg/min. Solution: Absolute pressure in condenser = (77 \u2013 70) = 7 cm of Hg or (7 0.0135951 104 9.81) Partial pressure of steam, Partial pressure of air, pt = 9.34 kPa ps = Saturation pressure corresponding to 30\u00b0C = 4.246 kPa pa = pt \u2013 ps = 5.094 kPa Rate of air extraction per minute = 5 10\u20134 1000 = 0.5 kg/min 273 30. 0 5 Volume of air extracted per minute = 0 287.. 5 094 mRT pa = 8.54 m3/min Steam Condenser _______________________________________________________________ 699 Specific volume corresponding to 30\u00b0C, vg = 32.89 m3/kg Volume of air extracted = Volume of mixture sucked per minute = 8.54 m3/min Mass of steam extracted in mixture handled per minute =.54 8. 32 89 = 0.2596 kg/min Therefore, mass handled by air extraction pump = 0.5 + 0.2596 = 0.7596 kg/min Mass handled by air pump = 0.7596 kg/min Enthalpy of steam entering condenser, h = hf at 30\u00b0C + 0.9 (hfg at 30\u00b0C) = 125.79 + (0.9 2430.5) = 2313.24 kJ/kg Ans. Mass flow rate of circulating water can be obtained by energy balance on condenser, mw Cpw (Tw) = ms (h \u2013 Cpw Tc) mw 4.18 15 = 1000 (2313.24 \u2013 4.18 303) mw = 16693.78 kg/min Water circulation rate = 16693.78 kg/min Ans. 6. In a surface condenser vacuum of 70 cm Hg is maintained when the barometric pressure is 76 cm Hg. Steam enters 0.85 dry into condenser at the rate of 300 kg/min. Temperature of condensate is 30\u00b0C and the rise in circulating water temperature is 20\u00b0C. For sending water through condenser and piping a pressure head of 5 m is required. For surface condenser determine, (i) (ii) (iii) the flow surface area required when water flows at 50 m/min, the cooling surface area required when heat transfer rate is 15 105 kJ/m2 hr \u00b0C. the total head required to be developed by pump Solution: Absolute pressure in condenser = 76 \u2013 70 = 6 cm of Hg, or, (6 0.0135951 104 9.81) Pa pt = 8.002 kPa Partial pressure of steam, ps = Saturation pressure corresponding to 30\u00b0C = 4.246 kPa Applying heat balance on condenser, mw 4.18 20 = (hf at 30\u00b0C + 0.85 hfg at 30\u00b0C \u2013 4.18 30) 300 Volume flow of water = mw = 7415 kg/min m3/min = 7.415 m3/min 7415 1000 Flow surface area requirement = Velocity flow rateof water Velocity of water flow = 7.415 50 = 0.1483 m2 Flow surface area required = 0.1483 m2 Ans. Cooling surface area required = = Total heat given by steam Heat transfer rate h f at 30 C h. 0 85 300 fg at. 4 18 5 C 30 15 10 30 60 700 ________________________________________________________ Applied Thermodynamics = 0.413 60 = 24.79 m2 Cooling surface area required = 24.79 m2 Ans. Velocity head present = 1 2 50 60 2 1. 9 81 = 0.0354 m Total head required = Pressure head + Velocity head = 5 + 0.0354 = 5.0354 m Head required = 5.", "0354 m Ans. 7. A jet condenser has steam entering at 350 kg/min when vacuum of 680 mm is maintained in it and the barometer reads 760 mm. Air mass going into condenser is 0.05% of steam mass entering. Water at 20\u00b0C enters the condenser to condense the steam such that temperature of condensate is 30\u00b0C. Volume of water required is 0.02 m3 per kg steam. The volume of air dissolved in the water injected may be considered as 5% of volume of water at atmospheric pressure. Determine the volume handling capacity of air pump for removing air and condensate when pump has volumetric efficiency of 90%. Solution: Absolute pressure in condenser = (76 \u2013 68) = 8 cm of Hg or, = (8 0.0135951 9.81 104 10\u20133) Partial pressure of steam, pt = 10.67 kPa ps = Saturation pressure corresponding to 30\u00b0C ps = 4.246 kPa pa = pt \u2013 ps = 10.67 \u2013 4.246 = 6.424 kPa Partial pressure of air, Volume of cooling water required per minute = 350 0.02 = 7m3/min 0 05 Mass of air going into condenser with steam per minute = 350. 100 = 0.175 kg/min Volume of air entering per minute with cooling water = 7 5 100 = 0.35 m3/min Mass of air with cooling water, using pV = mRT, m = pV RT here p = atmospheric pressure = 101.3 kPa, V = 0.35 m3, R = 0.287 kJ/kgK T = 273 + 20 = 293 K Mass of air with cooling water = 101 3 0 35. 293 0..287 = 0.422 kg/min Thus, total mass of air inside condenser per minute = (Mass of air with steam + Mass of air with cooling water) = 0.175 + 0.422 = 0.597 kg/min Steam Condenser _______________________________________________________________ 701 Volume of air corresponding to 0.597 kg/min 0.597 0 =.287 6.424 273 30 = 8.08 m3/min Volume of steam condensed (condensate) = 350 vf at 30\u00b0C = 350 0.001004 = 0.3514 m3/min Total capacity of air pump (wet air pump) = Volume of condensate/min + Volume of air/min + Volume of cooling water/min. = 0.3514 + 8.08 + 7 = 15.4314 m3/min Actual capacity of air pump = 15 4314..9 0 = 17.15 m3/min Capacity of air pump = 17.15 m3/min Ans. 8. A barometric jet condenser has steam entering at the rate of 20 kg/min and 12 kg cooling water per kg of steam is supplied into it for condensation. After condensation the cooling water and condensate leave at 40\u00b0C. Cooling water enters the condenser at 20\u00b0C. Vacuum of 650 mm Hg is maintained inside condenser while barometer read 760 mm Hg. An air pump is put for extracting air from condensate at the rate of 2m3 of wet mixture per minute. At the suction side of air pump a vacuum of 660 mm Hg and temperature of 35\u00b0C is observed. Determine, (i) (ii) (iii) the dryness fraction of steam entering condenser, the mass of air entering condenser per minute, the effective pressure head at tail of barometric condenser. Solution: Absolute pressure in condenser, or, pt = (760 \u2013 650) 10\u20131 = 11 cm Hg pt = 11 104 0.0135951 9.81 10\u20133, kPa pt = 14.67 kPa Partial pressure of steam, ps = Saturation pressure corresponding to 40\u00b0C = 7.384 kPa Partial pressure of air, pa = pt \u2013 ps = 7.286 kPa Cooling water required = 12 20 = 240 kg/min Let the dryness fraction of steam entering condenser be x and enthalpy h. Heat balance on condenser yields; ms h + mw cpw Twi = (ms + mw) cpw Two 20h + (240 4.18 20) = (20 + 240) 4.18 40 h = 1170.4 kJ/kg = hf at 40\u00b0C + xhfg at 40\u00b0C 1170.4 = 167.57 + (x2406.7) x = 0.4167 Dryness fraction of steam entering = 0.4167 Ans. 702 ________________________________________________________ Applied Thermodynamics Absolute pressure at suction of air pump = (760 \u2013 660) 10\u20131 = 10 cm Hg or (10 0.0135951 104 9.81 10\u20133 =", " 13.34 kPa) Partial pressure of steam at suction of air pump = Saturation pressure at 35\u00b0C. p s = 5.628 kPa Partial pressure of air, p a = 13.34 \u2013 5.628 = 7.712 kPa Now, at suction of pump volume of air will be equal to the volume of mixture. Volume of mixture = Volume of air = 2 m3 Mass of air entering = p V a RT a = 0 7 712 2. 273 35.287 Effective pressure head tail of barometric condenser = 0.1745 kg/min = Head corresponding to (Barometric pressure \u2013 Absolute pressure in condenser) = 10 9 81 0 0135951 10 101 3 14 67.... 3 = 0.649 m 6 Mass of air entering = 0.1745 kg/min Ans. Head = 0.649 m 9. A steam condenser has steam entering at 35\u00b0C and condensate being removed at 34\u00b0C. Condenser has two pumps one for extracting air and other for extraction of condensate. Air is removed at temperature of 33\u00b0C. The air leaks into condenser at the rate of 3 kg/hr. Consider the pressure inside condenser to remain uniform and neglect change in pressure due to air at steam inlet. Determine the volume of air handled by air pump in kg/hr and also determine the volume to be handled if a combined air and condensate pump is being used. Solution: Partial pressure of steam at 35\u00b0C = Saturation pressure corresponding to 35\u00b0C. ps = 5.628 kPa If the pressure of air at inlet is neglected then the total pressure in condenser, At the suction of air pump, partial pressure of steam pt = ps = 5.628 kPa. Partial pressure of air, = Saturation pressure corresponding to 33\u00b0C ps= 5.075 kPa pa = pt \u2013 ps = 5.628 \u2013 5.075 = 0.553 kPa Volume of air handled by air pump = mRT a p 3 0 =.287 0 273 33.553 a = 467.43 m3/hr In case the air and condensate mixture is to be handled by same pump then, Partial pressure of steam = Saturation pressure corresponding to 34\u00b0C Partial pressure of air, ps = 5.352 kPa pa = pt \u2013 ps = 5.628 \u2013 5.352 = 0.276 kPa Steam Condenser _______________________________________________________________ 703 Volume of air = Volume of mixture handled by pump = 3 0.287 0 = 957.71 m3/hr Volume of air handled = 467.43 m3/hr Ans. Volume of mixture handled = 957.71 m3/hr 273 34.276 10. In a surface condenser the vacuum at inlet is seen to be 72 cm Hg and at outlet it is 73 cm Hg. The barometer reading is 76 cm and the dryness fraction of steam at inlet is 0.92. Cooling water entering the condenser is at 20\u00b0C. Considering no air in the condenser and the temperature rise in cooling water to be maximum determine. (i) (ii) the minimum amount of undercooling. the amount of cooling water required per kg of steam. Solution: Inlet pressure in condenser = (76 \u2013 72) cm Hg = 4 cm Hg. = 4 104 0.0135951 9.81 10\u20133 = 5.335 kPa Outlet pressure in condenser = (76 \u2013 73) cm Hg = 3 cm Hg = 3 104 0.0135951 9.81 10\u20133 = 4.001 kPa Since there is no air in condenser so the 5.335 kPa and 4.001 kPa will be the pressure of steam. Saturation temperature corresponding to above pressures give temperature at inlet and outlet respectively. Saturation temperature at inlet =33.87\u00b0C, (from steam table at 5.335 kPa) Saturation temperature at outlet =28.96\u00b0C, (from steam table at 4.001 kPa) Thus, steam will leave at maximum temperature of 28.96\u00b0C The minimum amount of undercooling = 33.87 \u2013 28.96 = 4.91\u00b0C For maximum temperature rise of cooling water the temperature of cooling water outlet will be equal to the temperature of steam at inlet of 33.87\u00b0C. Therefore, the maximum rise in cooling water temperature = 33.87 \u2013 20 = 13.87\u00b0C Enthalpy of steam entering, h = hf at 33.87\u00b0C + 0.92 hfg at 33.87\u00b0C h = 141.97 + (0.92 2421.33) = 2369.59 kJ/kg Let mass of cooling water required be m kg per kg steam. Heat balance on condenser yields, m 4.18 13.87 =", " 1 (2369.59 \u2013 4.18 28.96) m = 38.78 kg water per kg of steam Undercooling = 4.91\u00b0C Cooling water requirement = 38.78 kg/kg steam Ans. 704 ________________________________________________________ Applied Thermodynamics 15.1 What do you understand by condenser? Discuss its significance. 15.2 How does condenser improve performance of steam power plant? 15.3 Discuss different types of condenser briefly. 15.4 Differentiate between surface condenser and jet condenser. 15.5 Give a sketch of barometric jet condenser and explain its working. 15.6 Discuss the effect of air leakage upon the performance of condenser. 15.7 How the air leaking into condenser is extracted out? Explain. 15.8 Describe the factors affecting the efficiency of condensing plant. 15.9 Discuss the relevance of Dalton\u2019s law of partial pressures in condenser calculations. temperature of 32\u00b0C. The barometer reading is 765 mm of Hg. 15.10 What do you understand by cooling towers? Explain their utility. 15.11 Determine the vacuum efficiency of a surface condenser having vacuum of 715 mm of Hg and [98%] 15.12 A surface condenser having vacuum of 715 mm Hg and temperature of 32\u00b0C has cooling water circulated at 800 kg/min. The cooling water entering condenser becomes warmer by 14\u00b0C. The condensate is available from condenser at 25 kg/min. The hot well temperature is 30\u00b0C. Barometer reading is 765 mm of Hg. Determine the mass of air in kg/m3 of condenser volume and dryness [0.022 kg/m3, 0.84] fraction of steam entering. 15.13 A surface condenser has vacuum of 71 cm Hg and mean temperature of 35\u00b0C. The barometer reading is 76.5 cm Hg. The hot well temperature is 28\u00b0C. Steam enters condenser at 2000 kg/hr and requires cooling water at 8\u00b0C at the rate of 1000 kg/min. Cooling water leaves condenser at 24\u00b0C. Determine (i) (ii) (iii) the vacuum efficiency of condenser, the undercooling in condenser corrected vacuum in reference to standard barometer reading, (iv) the condenser efficiency. [0.982, 7\u00b0C, 70.5 cm Hg, 0.505] 15.14 In a surface condenser steam enters at 40\u00b0C and dryness fraction of 0.85. Air leaks into it at 0.25 kg/min. An air pump is provided upon the condenser for extracting out air. Temperature at suction of air pump is 32\u00b0C while condensate temperature is 35\u00b0C. Determine. (i) (ii) (iii) the reading of vacuum gauge the volume handling capacity of air pump in m3/hr the loss of condensate in kg/hr. [705 mm Hg, 500 m3/hr, 16.9 kg/hr] 15.15 A steam turbine discharges steam into a surface condenser having vacuum of 700 mm Hg. The barometer reading is 760 mm Hg. Leakage into condenser is seen to be 1.4 kg/min. The air pump is employed for extracting out air leaking in. Temperature at the inlet of air pump is 20\u00b0C. The air pump is of reciprocating type running at 300 rpm and has L : D ratio of 2 : 1. Determine, (i) (ii) (iii) the capacity of air pump is m3/hr the dimensions of air pump the mass of vapour going out with air in air pump, kg/hr. [1250 m3/hr, bore: 35.36 cm, stroke: 70.72 cm, 21.5 kg/hr] 15.16 A surface condenser handles condensate at 70.15 cm Hg when barometer reads 76 cm Hg. Steam entering at 2360 kg/hr requires cooling water at 6.81 102 kg/hr, 10\u00b0C. Cooling water leaves condenser at 27.8\u00b0C while condenser has mean temperature of 37\u00b0C. Air leaks into condenser at 0.3 kg/min. Determine, (i) (ii) the mass of vapour going out with air per hour the state of steam entering. [119 kg/hr, 0.89] Steam Condenser _______________________________________________________________ 705 15.17 A steam condenses working at 71.5 cm Hg vacuum and temperature of 32.28\u00b0C. Condenser has steam entering at 13750 kg/hr. The leakage of air into condenser is 0.4 10\u20133 kg/kg steam entering. Considering volumetric capacity of air pump as 80% determine the capacity of air pump, handling [10.", "3 m3/min] mixture of air and water vapour. 15.18 A jet condenser has pressure at inlet as 0.07 bar. It is supplied steam at 4000 kg/hr. Water is supplied for condensation at 16 104 kg/hr. The volume of air dissolved in water at 1 bar, 15\u00b0C is 0.05 of that water. Air leaks in with steam at 0.05 kg/min. Temperature at suction of air pump is 30\u00b0C and volumetric efficiency is 80%. Determine the capacity of air pump in m3/min for extracting out air. [519 kg/hr] 15.19 A surface condenser has steam entering at 0.09 bar, 0.88 dry at the rate of 16000 kg/hr. The air leakage into condenser is 8 kg/hr. Temperature at condensate and air extraction pipe is 36\u00b0C. For the average heat transfer rate of 133760 kJ/m2hr, determine the surface required. [254 m2] 15.20 A surface condenser has vacuum of 581 mm of Hg and steam enters into it at 57.4\u00b0C, 3100 kg/ hr. The barometer reads 726 mm of Hg. The temperature at suction of air pump is 50\u00b0C. The cooling water is supplied at 5.6 104 kg/hr, 15\u00b0C and leaves at 43.6\u00b0C. Determine, (i) (ii) the mass of air entering condenser per kg of steam the vacuum efficiency. [0.136 kg/kg steam, 0.981] 706 ________________________________________________________ Applied Thermodynamics 16 16.1 INTRODUCTION Compressors are work absorbing devices which are used for increasing pressure of fluid at the expense of work done on fluid. The compressors used for compressing air are called air compressors. Compressors are invariably used for all applications requiring high pressure air. Some of popular applications of compressor are, for driving pneumatic tools and air operated equipments, spray painting, compressed air engine, supercharging in internal combustion engines, material handling (for transfer of material), surface cleaning, refrigeration and air conditioning, chemical industry etc. Compressors are supplied with low pressure air (or any fluid) at inlet which comes out as high pressure air (or any fluid) at outlet, Fig. 16.1. Work required for increasing pressure of air is available from the prime mover driving the compressor. Generally, electric motor, internal combustion engine or steam engine, turbine etc. are used as prime movers. Compressors are similar to fans and blowers but differ in terms of pressure ratios. Fan is said to have pressure ratio up to 1.1 and blowers have pressure ratio between 1.1 and 4 while compressors have pressure ratios more than 4. Compressors can be classified in the following different ways. (a) Based on principle of operation: Based on the principle of operation compressors can be Fig. 16.1 Compressor classified as, (i) Positive displacement compressors (ii) Non-positive displacement compressors In positive displacement compressors the compression is realized by displacement of solid boundary and preventing fluid by solid boundary from flowing back in the direction of pressure gradient. Due to solid wall displacement these are capable of providing quite large pressure ratios. Positive displacement compressors can be further classified based on the type of mechanism used for compression. These can be Reciprocating and Rotary Compressor ______________________________________________ 707 (i) Reciprocating type positive displacement compressors (ii) Rotary type positive displacement compressors Reciprocating compressors generally, employ piston-cylinder arrangement where displacement of piston in cylinder causes rise in pressure. Reciprocating compressors are capable of giving large pressure ratios but the mass handling capacity is limited or small. Reciprocating compressors may also be single acting compressor or double acting compressor. Single acting compressor has one delivery stroke per revolution while in double acting there are two delivery strokes per revolution of crank shaft. Rotary compressors employing positive displacement have a rotary part whose boundary causes positive displacement of fluid and thereby compression. Rotary compressors of this type are available in the names as given below; (i) Roots blower (ii) Vaned type compressors Rotary compressors of above type are capable of running at higher speed and can handle large mass flow rate than reciprocating compressors of positive displacement type. Non-positive displacement compressors, also called as steady flow compressors use dynamic action of solid boundary for realizing pressure rise. Here fluid is not contained in definite volume and subsequent volume reduction does not occur as in case of positive displacement compressors. Nonpositive displacement compressor may be of \u2018axial flow type\u2019 or \u2018centrifugal type\u2019 depending upon type of flow in compressor. (b) Based on number of stages: Compressors may also be", " classified on the basis of number of stages. Generally, the number of stages depend upon the maximum delivery pressure. Compressors can be single stage or multistage. Normally maximum compression ratio of 5 is realized in single stage compressors. For compression ratio more than 5 the multi-stage compressors are used. Typical values of maximum delivery pressures generally available from different types of com- pressor are, (i) Single stage compressor, for delivery pressure up to 5 bar (ii) Two stage compressor, for delivery pressure between 5 and 35 bar (iii) Three stage compressor, for delivery pressure between 35 and 85 bar (iv) Four stage compressor, for delivery pressure more than 85 bar (c) Based on capacity of compressors: Compressors can also be classified depending upon the capacity of compressor or air delivered per unit time. Typical values of capacity for different compressors are given as; (i) Low capacity compressors, having air delivery capacity of 0.15 m3/s or less (ii) Medium capacity compressors, having air delivery capacity between 0.15 and 5 m3/s. (iii) High capacity compressors, having air delivery capacity more than 5 m3/s. (d) Based on highest pressure developed: Depending upon the maximum pressure available from compressor they can be classified as low pressure, medium pressure, high pressure and super high pressure compressors. Typical values of maximum pressure developed for different compressors are as under; (i) Low pressure compressor, having maximum pressure up to 1 bar (ii) Medium pressure compressor, having maximum pressure from 1 to 8 bar (iii) High pressure compressor, having maximum pressure from 8 to 10 bar (iv) Super high pressure compressor, having maximum pressure more than 10 bar. 708 ________________________________________________________ Applied Thermodynamics 16.2 RECIPROCATING COMPRESSORS Reciprocating compressor has piston cylinder arrangement as shown in Fig. 16.2. Fig. 16.2 Line diagram of reciprocating compressor Reciprocating compressor has piston, cylinder, inlet valve, exit valve, connecting rod, crank, piston pin, crank pin and crank shaft. Inlet valve and exit valves may be of spring loaded type which get opened and closed due to pressure differential across them. Let us consider piston to be at top dead centre (TDC) and move towards bottom dead centre (BDC). Due to this piston movement from TDC to BDC suction pressure is created causing opening of inlet valve. With this opening of inlet valve and suction pressure the atmospheric air enters the cylinder. Air gets into cylinder during this stroke and is subsequently compressed in next stroke with both inlet valve and exit valve closed. Both inlet valve and exit valves are of plate type and spring loaded so as to operate automatically as and when sufficient pressure difference is available to cause deflection in spring of valve plates to open them. After piston reaching BDC it reverses its motion and compresses the air inducted in previous stroke. Compression is continued till the pressure of air inside becomes sufficient to cause deflection in exit valve. At the moment when exit valve plate gets lifted the exhaust of compressed air takes place. This piston again reaches TDC from where downward piston movement is again accompanied by suction. This is how reciprocating compressor keeps on working as flow device. In order to counter for the heating of piston-cylinder arrangement during compression the provision of cooling the cylinder is there in the form of cooling jackets in the body. Reciprocating compressor described above has suction, compression and discharge as three prominent processes getting completed in two strokes of piston or one revolution of crank shaft. Reciprocating and Rotary Compressor ______________________________________________ 709 16.3 THERMODYNAMIC ANALYSIS Compression of air in compressor may be carried out following number of thermodynamic processes such as isothermal compression, polytropic compression or adiabatic compression. Figure 16.3 shows the thermodynamic cycle involved in compression. Theoretical cycle is shown neglecting clearance volume but in actual cycle clearance volume can not be negligible. Clearance volume is necessary in order to prevent collision of piston with cylinder head, accommodating valve mechanism etc. Compression process is shown by process 1\u20132, 1\u20132, 1\u20132 following adiabatic, polytropic and isothermal processes. On p-V diagram process 4\u20131 shows the suction process followed by compression during 1\u20132 and discharge through compressor is shown by process 2\u20133. Air enters compressor at pressure p1 and is compressed up to p2. Compression work requirement can be estimated from the area below the each compression process. Area on p\u2013V diagram shows that work requirement shall be minimum with isothermal process 1\u20132. Work requirement is maximum with process 1\u20132 i.e. adiabatic process. As a designer one shall be interested in a compressor having minimum compression work requirement. Therefore, ideally compression should occur isothermally", " for minimum work input. In practice it is not possible to have isothermal compression because constancy of temperature Fig. 16.3 Compression cycle on p-V diagram during compression can not be realized. Generally, compressors run at substantially high speed while isothermal compression requires compressor to run at very slow speed so that heat evolved during compression is dissipated out and temperature remains constant. Actually due to high speed running of compressor the compression process may be assumed to be near adiabatic or polytropic process following law of compression as pV n = C with value of \u2018n\u2019 varying between 1.25 and 1.35 for air. Compression process following three processes is also shown on T-s diagram in Fig. 16.4. It is thus obvious that actual compression process should be compared with isothermal compression process. A mathematical parameter called isothermal efficiency is defined for quantifying the degree of deviation of actual compression process from ideal compression process. Isothermal efficiency is defined by the ratio of isothermal work and actual indicated work in reciprocating compressor. 710 ________________________________________________________ Applied Thermodynamics Fig. 16.4 Compression process on T-s diagram Isothermal efficiency = Isothermal work Actual indicated work Practically, compression process is attempted to be closed to isothermal process by air/water cooling, spraying cold water during compression process. In case of multi-stage compression process the compression in different stages is accompanied by intercooling in between the stages. Mathematically, for the compression work following polytropic process, pV n = C. Assuming negligible clearance volume the cycle work done, Wc = Area on p-V diagram \u2013 p1V1 = = Wc = Wc = (p1V1p2V2 \u2013 p1V1] (mRT1) (p1V1 or, Wc = n n 1 mR (T2 \u2013 T1) In case of compressor having isothermal compression process, n = 1, i.e. p1V1 = p1V2 Wc, iso = p2V2 + p1V1 ln r \u2013 p1V1 Reciprocating and Rotary Compressor ______________________________________________ 711 Wc, iso = p1V1 ln r, where r = V 1 V 2 In case of compressor having adiabatic compression process, n = Wc, adiabatic = 1 mR (T2 \u2013 T1) or, Wc, adiabatic = mCp (T2 \u2013 T1) Wc, adiabatic = m (h2 \u2013 h1) Hence, isothermal efficiency, iso = ln r 1 n n 1 p 2 p 1 The isothermal efficiency of a compressor should be close to 100% which means that actual compression should occur following a process close to isothermal process. For this the mechanism be derived to maintain constant temperature during compression process. Different arrangements which can be used are: (i) Faster heat dissipation from inside of compressor to outside by use of fins over cylinder. Fins facilitate quick heat transfer from air being compressed to atmosphere so that temperature rise during compression can be minimized. (ii) Water jacket may be provided around compressor cylinder so that heat can be picked by cooling water circulating through water jacket. Cooling water circulation around compressor regulates rise in temperature to great extent. (iv) (iii) The water may also be injected at the end of compression process in order to cool the air being compressed. This water injection near the end of compression process requires special arrangement in compressor and also the air gets mixed with water and needs to be separated out before being used. Water injection also contaminates the lubricant film on inner surface of cylinder and may initiate corrosion etc. The water injection is not popularly used. In case of multistage compression in different compressors operating serially, the air leaving one compressor may be cooled up to ambient state or somewhat high temperature before being injected into subsequent compressor. This cooling of fluid being compressed between two consecutive compressors is called intercooling and is frequently used in case of multistage compressors. Considering clearance volume: With clearance volume the cycle is represented on Fig. 16.3 (b). The work done for compression of air polytropically can be given by the area enclosed in cycle 1\u20132\u20133\u2013 4. Clearance volume in compressors varies from 1.5% to 35% depending upon type of compressor. Wc, with CV = Area1234 = n n 1 (p1V1p4V4) Here p1 = p4, p2 = p3 (p1V1) 712 ________________________________________________________ Applied Thermodynamics Wc, with CV = p1V4 Wc, with CV = In the cylinder of reciprocating compressor (V1 \u2013 V4) shall be the actual volume of air delivered per cycle. Vd = V1 \u2013 V4. This (", "V1 \u2013 V4) is actually the volume of air inhaled in the cycle and delivered subsequently. (V1 \u2013 V4) p1 Wc, with CV = 1 If air is considered to behave as perfect gas then pressure, temperature, volume and mass can be inter related using perfect gas equation. The mass at state 1 may be given as m1, mass at state 2 shall be m1, but at state 3 after delivery mass reduces to m2 and at state 4 it shall be m2. p1Vd p 2 p 1 n 1 1 So, at state 1, p1V1 = m1RT1 at state 2, p2V2 = m1RT2 at state 3, p3V3 = m2RT3 or, p2V3 = m2RT3 at state 4, p4V4 = m2RT4, or p1V4 = m2RT4 Ideally there shall be no change in temperature during suction and delivery i.e., T4 = T1 and T2 = T3. From earlier equation, n n 1 p1 (V1 \u2013 V4) n 1 n p 2 p 1 1 Wc, with CV = = Temperature and pressure can be related as, n n 1 and Substituting Wc, with CV = (m1RT1 \u2013 m2RT4) 1 Substituting for constancy of temperature during suction and delivery Wc, with CV = (m1RT1 \u2013 m2RT1) or, Wc, with CV = n n 1 (m1 \u2013 m2) R(T2 \u2013 T1) Reciprocating and Rotary Compressor ______________________________________________ 713 Thus, (m1 \u2013 m2) denotes the mass of air sucked or delivered. For unit mass of air delivered the work done per kg of air can be given as, Wc, with CV = n n 1 R(T2 \u2013 T1), per kg of air Thus from above expressions it is obvious that the clearance volume reduces the effective swept volume i.e. the mass of air handled but the work done per kg of air delivered remains unaffected. From the cycle work estimated as above the theoretical power required for running compressor shall be as given ahead. For single acting compressor running with N rpm, power input required, assuming clearance volume. Power required = n 1 n p V 1 1 V 4 for double acting compressor, power = 2N Volumetric efficiency: Volumetric efficiency of compressor is the measure of the deviation from volume handling capacity of compressor. Mathematically, the volumetric efficiency is given by the ratio of actual volume of air sucked and swept volume of cylinder. Ideally the volume of air sucked should be equal to the swept volume of cylinder, but it is not so in actual case. Practically the volumetric efficiency lies between 60 and 90%. Volumetric efficiency can be overall volumetric efficiency and absolute volumetric efficiency as given below: Overall volumetric efficiency = Volume of free air sucked into cylinder Swept volume of LP cylinder or (Volumetric efficiency referred to free air conditions) = Mass of air delivered per unit time Mass of air corresponding toswept volume of LP cylinder per unit time for free air conditions Here free air condition refers to the standard conditions. Free air condition may be taken as 1 atm or 1.01325 bar and 15\u00b0C or 288 K. Consideration for free air is necessary as otherwise the different compressors can not be compared using volumetric efficiency because specific volume or density of air varies with altitude. It may be seen that a compressor at datum level (sea level) shall deliver large mass than the same compressor at high altitude. This concept is used for giving the capacity of compressor in terms of \u2018free air delivery\u2019 (FAD). \u201cFree air delivery is the volume of air delivered being reduced to free air conditions.\u201d In case of air the free air delivery can be obtained using perfect gas equation as, 714 ________________________________________________________ Applied Thermodynamics where subscript a or pa, Va, Ta denote properties at free air conditions or, Va = = FAD per cycle This volume Va gives \u2018free air delivered\u2019 per cycle by the compressor. Absolute volumetric efficiency can be defined, using NTP conditions in place of free air condi- tions. Absolute volumetric efficiency = = Volume of air sucked into cylinder at NTP Swept volume of LP cylinder Mass of air delivered per unit time Massof air corresponding to swept volume of LP cylinder per unit time estimated at NTP Thus, volumetric efficiency referred to free air conditions. vol. = Volume of air sucked referred to freeair conditions FAD Swept volume Vs is swept volume, Vs = V1 \u2013 V3 Vc is clearance volume, Vc = V3 vol. = vol Here, and Here, V 4 = Vs = V V 4 c Vc Let the ratio", " of clearance volume to swept volume be given by C.i.e. or, vol vol. Reciprocating and Rotary Compressor ______________________________________________ 715 Volumetric efficiency depends on ambient pressure and temperature, suction pressure and temperature, ratio of clearance to swept volume, and pressure limits. Volumetric efficiency increases with decrease in pressure ratio in compressor. 16.4 ACTUAL INDICATOR DIAGRAM Theoretical indicator diagram of reciprocating compressor as shown in earlier discussion refers to the ideal state of operation of compressor. The practical limitations, when considered in the indicator diagram yield actual indicator diagram as shown in Fig. 16.5. Fig. 16.5 Actual indicator diagram Actual p-V diagram varies from theoretical p-V diagram due to following: Compressor has mechanical types of valves so the instantaneous opening and closing of valves can never be achieved. Also during suction and discharge there occurs throttling due to reduction in area of flow across inlet and exit valve. 123 4 shows theoretical indicator diagram and actual indicator diagram is shown by 1234 on p-V diagram. Compression process 1\u20132 ends at state 2. At state 2 exit valve should open instantaneously which does not occur and also due to restricted opening there shall be throttling causing drop in pressure. Due to time lag in opening of exit valve compression process is continued up to 2. Thus, additional work is done during delivery from compressor as shown by hatched area 2 23. After delivery stroke the inlet valve should theoretically open at 4 but does not open at this point instead is opened fully at 4. Shift from state 4 to 4 is there due to inertia in opening of valve throttling, gradual opening, and friction losses etc. Thus it is seen that during suction there occurs intake depression as shown in actual indicator diagram. Work required as shown in actual indicator diagram is more than theoretical diagram. In order to have compressor close to ideal compressor with minimum losses it shall be desired to have actual indicator diagram close to theoretical diagram, which requires less inertia and efficient operation of valves. Friction losses in pipings and across valves should be minimized. 716 ________________________________________________________ Applied Thermodynamics 16.5 MULTISTAGE COMPRESSION Multistage compression refers to the compression process completed in more than one stage i.e. a part of compression occurs in one cylinder and subsequently compressed air is sent to subsequent cylinders for further compression. In case it is desired to increase the compression ratio of compressor then multi-stage compression becomes inevitable. If we look at the expression for volumetric efficiency then it shows that the volumetric efficiency decreases with increase in pressure ratio. This aspect can also be explained using p-V representation shown in Fig. 16.6. Fig. 16.6 Multistage compression Figure 16.6 a, shows that by increasing pressure ratio i.e. increasing delivery pressure the volume of air being sucked goes on reducing as evident from cycles 1234 and 12 34. Let us increase pressure from p2 to p2 and this shall cause the suction process to get modified from 4\u20131 to 4\u20131. Thus volume, sucked reduces from (V1 \u2013 V4) to (V1 \u2013 V4) with increased pressure ratio from to p 2 p 1 p 2 p 1 thereby reducing the free air delivery while swept volume remains same. Therefore, the volumetric efficiency reduces with increasing pressure ratio in compressor with single stage compression. Also for getting the same amount of free air delivery the size of cylinder is to be increased with increasing pressure ratio. The increase in pressure ratio also requires sturdy structure from mechanical strength point of view for withstanding large pressure difference. The solution to number of difficulties discussed above lies in using the multistage compression where compression occurs in parts in different cylinders one after the other. Figure 16.6 b, shows the multistage compression occurring in two stages. Here first stage of compression occurs in cycle 12671 and after first stage compression partly compressed air enters second stage of compression and occurs in cycle 2345. In case of multistage compression the compression in first stage occurs at low temperature and subsequent compression in following stages occurs at higher temperature. The compression work requirement depends largely upon the average temperature during compression. Higher average temperature during compression has larger work requirement compared to low temperature so it is always desired to keep the low average temperature during compression. Apart from the cooling during compression the temperature of air at inlet to compressor can be reduced so as to reduce compression work. In multistage compression the partly compressed air leaving first stage is cooled up to ambient air temperature in intercooler and then sent to subsequent cylinder (stage) for compression. Thus, intercoolers when put between the stages reduce the compression work and compression is called intercooled compression. Intercooling is called perfect when temperature at Reciprocating and Rotary Compressor ______________________________________________ 717 inlet to subsequent stages of compression is reduced to ambient temperature. Figure 16.6 c, shows multistage (two stage", ") intercooled compression. Intercooling between two stages causes temperature drop from 2 to 2 i.e. discharge from first stage (at 2) is cooled up to the ambient temperature state (at 2) which lies on isothermal compression process 1\u20132\u20133. In the absence of intercooling the discharge from first stage shall enter at 2. Final discharge from second stage occurs at 3 in case of intercooled compression compared to discharge at 3 in case of non-intercooled compression. Thus, intercooling offers reduced work requirement by the amount shown by area 223 3 on p-V diagram. If the intercooling is not perfect then the inlet state to second/subsequent stage shall not lie on the isothermal compression process line and this state shall lie between actual discharge state from first stage and isothermal compression line. Fig. 16.7 Schematic for two-stage compression (Multistage compressor) Figure 16.7 shows the schematic of multi stage compressor (double stage) with intercooler between stages. T\u2013s representation is shown in Fig. 16.8. The total work requirement for running this shall be algebraic summation of work required for low pressure (LP) and high pressure (HP) stages. The size of HP cylinder is smaller than LP cylinder as HP cylinder handles high pressure air having smaller specific volume. Fig. 16.8 T-s representation of multistage compression Mathematical analysis of multistage compressor is done with following assumptions: (i) Compression in all the stages is done following same index of compression and there is no pressure drop in suction and delivery pressures in each stage. Suction and delivery pressure remains constant in the stages. 718 ________________________________________________________ Applied Thermodynamics (ii) There is perfect intercooling between compression stages. (iii) Mass handled in different stages is same i.e. mass of air in LP and HP stages are same. (iv) Air behaves as perfect gas during compression. From combined p-V diagram the compressor work requirement can be given as, Work requirement in LP cylinder, WLP = Work requirement in HP cylinder, WHP = p1V1 p2V2 For perfect intercooling, p1V1 = p2V2 and WHP = n n 1 p2V2 n 1 n p 2 p 2 1 Therefore, total work requirement, Wc = WLP + WHP, for perfect intercooling Wc = = Wc = p1V1 Power required can be given in HP as below, considering speed to be N rpm. Power = W Nc 75 60, HP If we look at compressor work then it shows that with the initial and final pressures p1 and p2 remaining same the intermediate pressure p2 may have value floating between p1 and p2 and change the work requirement Wc. Thus, the compressor work can be optimized with respect to intermediate pressure p2. Mathematically, it can be differentiated with respect to p2. d d p2 Reciprocating and Rotary Compressor ______________________________________________ 719 yields Equating or or or p 2 p 1 = p or or, 2 = p1 p2, p2 = p p2 p1 2 Pressure ratio in Ist stage = Pressure ratio in IInd stage Thus, it is established that the compressor work requirement shall be minimum when the pressure ratio in each stage is equal. In case of multiple stages, say i number of stages, for the delivery and suction pressures of pi +1 and p1 the optimum stage pressure ratio shall be, 1 i 1/ Optimum stage pressure ratio = p i p 1 pi \u2013 1, pi, pi + 1 for pressures at stages being p1, p2, p3, p4, p5, Minimum work required in two stage compressor can be given by n n 1 p1V1 2 Wc, min = For i number of stages, minimum work, p 2 p 1 n 1 n 1 Wc, min = n n 1 p1V1 or, Wc, min = i Wc, min = p1V1 It also shows that for optimum pressure ratio the work required in different stages remains same for the assumptions made for present analysis. Due to pressure ratio being equal in all stages the temperature ratios and maximum temperature in each stage remains same for perfect intercooling. 720 ________________________________________________________ Applied Thermodynamics Cylinder dimensions: In case of multistage compressor the dimension of cylinders can be estimated basing upon the fact that the mass flow rate of air across the stages remains same. For perfect intercooling the temperature of air at suction of each stage shall be same. If the actual volume sucked during suction stroke is V1, V2, V3 for different stages then by perfect gas law, p1V1 = RT1, p2V2 = RT2, p3V3 = RT3 For perfect intercooling (T1 = T2 =", " T3 = ) so or If the volumetric efficiency of respective stages in p1V1 = RT1, p2V2 = RT1, p3V3 = RT1, p1V1 = p2V2 = p3V3 =,,, V1 V2 V3 Then theoretical volume of cylinder 1, V1, th = V 1 V 1 ; V1 = V1 V1, th cylinder 2, cylinder 3, Substituting, p1 V2, th = V3, th= V 2 V 2 V 3 V 3 ; V2 = V2 V2, th ; V3 = V3 V3, th V1, th = p2 V2 V2, th = p3 V3, th = V3 V1 Theoretical volumes of cylinder can be given using geometrical dimensions of cylinder as diam- eters D1, D2, D3, and stroke lengths L1, L2, L3. or or, V1, th = V2, th = V3, th = 4 4 4 D1 2L1 D2 2L2 D3 2L3 p1 D2 D1 2L2 = p3 D1 V1 If the volumetric efficiency is same for all cylinders, i.e. 2 L1 = p2 2 L1 = p2 4 p1 2 L2 = p3 4 D2 V2 V3 Vi V2 D3 V1 = V3 D3 4 2L3 = V2 = 2L3 = V3 = and stroke for all cylinder is same i.e. L1 = L2 = L3 = Then, D1 2 p1 = D2 2 p2 = D3 2 p3 = These generic relations may be used for getting the ratio of diameters of cylinders of multistage compression. Energy balance: Energy balance may be applied on the different components constituting multistage compression. Reciprocating and Rotary Compressor ______________________________________________ 721 For LP stage the steady flow energy equation can be written as below: mh1 + WLP = mh2 + QLP QLP = WLP \u2013 m (h2 \u2013 h1) QLP = WLP \u2013 mCp (T2 \u2013 T1) For intercooling (Fig. 16.10) between LP and HP stage steady flow energy equation shall be; Fig. 16.9 or, mh2 = mh2 + QInt QInt = m (h2 \u2013 h2) QInt = mCp (T2 \u2013 T2) For HP stage (Fig. 16.11) the steady flow energy equation yields. mh2 + WHP = mh3 + QHP Fig. 16.10 QHP = WHP + m (h2 \u2013 h3) QHP = WHP + mCp (T2 \u2013 T3) = WHP \u2013 mCp (T3 \u2013 T2) Fig. 16.11 In case of perfect intercooling and optimum pressure ratio, T2 = T1 and T2 = T3 Hence for these conditions, QLP = WLP \u2013 mCp (T2 \u2013 T1) 722 ________________________________________________________ Applied Thermodynamics QInt = mCp (T2 \u2013 T1) QHP = WHP \u2013 mCp (T2 \u2013 T1) Total heat rejected during compression shall be the sum of heat rejected during compression and heat extracted in intercooler for perfect intercooling. Heat rejected during compression for polytropic process = n 1 Work 16.6 CONTROL OF RECIPROCATING COMPRESSORS Output from the compressors can be controlled by different measures which regulate the compressor output. In practical applications the compressors are fitted with air receiver to store the high pressure air and supply as and when required. Therefore, the compressors are run only for the duration required to maintain the limiting pressure inside receiver. When the pressure inside receiver starts dropping down then the compressor again starts supplying compressed air till the level is restored. Different ways for this control are based on throttle, clearance, blow off control and speed control. (i) Throttle control has the regulation of opening/closing of inlet valve so that the quantity of air entering can be varied. With partial opening of inlet valve throttling occurs at valve and quantity of air entering is reduced while the pressure ratio gets increased. (ii) Clearance control is the arrangement in which the clearance volume is increased when pressure ratio exceeds the limit. For this cylinder has clearance openings which are closed by spring loaded valves. Whenever pressure exceeds, then the clearance openings get opened and the increased clearance volume reduces maximum pressure. (iii) Blow-off control has spring loaded safety valve or by pass valve for blowing out excess air in receiver. After release of excess air automatically the valve gets closed on its own. (iv) Speed of", " compressor can also be controlled by regulating the prime mover, thereby regulating compressor output. Thus, compressor is run on variable speed for its\u2019 control. 16.7 RECIPROCATING AIR MOTOR Air motors are actually prime movers run using the compressed air. It is used extensively in the applications where electric motor/IC engine/Gas turbine etc. can not be used due to fire hazard, specially in coal/oil mining applications. Apart from use of air motors in coal/oil mines these are also used for running pneumatic tools in workshops and manufacturing/assembly lines. Air motors have a reciprocating piston-cylinder arrangement where compressed air is admitted in cylinder with inlet valve open for limited period in suction stroke and this causes piston movement to yield shaft work. Thus air motors are reverse of air compressors. Schematic for air motor and p\u2013V diagram for cycle used is shown in Fig. 16.12. Compressed air enters the cylinder up to state 1 and expands up to state 2. Expansion process is generally polytropic and yields expansion work. Expansion is terminated even before the atmospheric pressure is reached because in later part of expansion the positive work is less than negative work. The exhaust valve opens causing drop in pressure from 2 to 3. Exhaust pressure is slightly more or nearly equal to atmospheric pressure. The exhaust of air occurs at constant pressure up to state 4 where the closing of exhaust valve occurs. The inlet valve is opened at state 5 causing sudden rise in pressure upto state 6 after which piston displacement begins and is continued with compressed air inlet up to state 1 after which expansion occurs with inlet valve closed. Here point of cut-off is at state 1 which can be Reciprocating and Rotary Compressor ______________________________________________ 723 suitably varied depending upon output required from air motor. Here, total work available is the result of non-expansive work during 6\u20131 and expansion work during 1\u20132. Work available from air motor can be quantified by the area enclosed on p-V diagram. Wmotor = p1 (V1 \u2013 V6V3 \u2013 V4) p3 \u2013 Fig. 16.12 Reciprocating air motor Air motor suffers from the drawback that in case of expansion the air temperature may even go below the dew point temperature and moisture may condense or even get frozen also inside the cylinder. This phenomenon may be prevented by increasing the temperature of air entering cylinder so that temperature after expansion does not go near the dew point temperature. Some times air is heated before being supplied to air motors in order to avoid condensation. 16.8 ROTARY COMPRESSORS Rotary compressors are those compressors in which rotating action is used for compression of fluid. Rotary air compressors have capability of running at high speeds up to 40,000 rpm and can be directly coupled to any prime mover such as electric motor, turbine etc. due to compact design, no balancing problem and less no. of sliding parts. Comparative study of rotary compressor with reciprocating compressor shows that rotary compressors can be used for delivering large quantity of air but the maximum pressure at delivery is less compared to reciprocating compressors. Generally, rotary compressors can yield delivery pressure up to 10 bar and free air delivery of 3000 m3/min. Rotary compressors are less bulky, and offer uniform discharge compared to reciprocating compressor even in the absence of big size receiver. Lubrication requirement and wear and tear is less due to rotary motion of parts in rotary compressors compared to reciprocating compressors. Rotary compressors may work on the principle of positive displacement and dynamic action both. Rotary compressors having positive displacement may be of following types: (i) Roots blower (ii) Screw type or Helical type compressor (iii) Vane type compressor 724 ________________________________________________________ Applied Thermodynamics Rotary compressors employing dynamic action may be of centrifugal type or axial type depending upon the direction of flow. These centrifugal type or axial compressors may also be termed as nonpositive displacement type steady flow compressors. (i) Roots blower: Roots blower is a positive displacement type rotary compressor. It has two rotors having two or three lobes having epicycloid and hypocycloid or involute profiles such that they remain in proper contact. Figure 16.13 shows two lobe rotors in a roots blower. To prevent wear and tear two rotors have clearance in between. Out of two rotors one is driven by prime mover while other one is driven by first rotor. When two rotors rotate then their typical geometry divides the region inside casing into two regions i.e. high pressure region and low pressure region. Although there occurs slight leakage across the mating parts which can only be minimised not eliminated completely. Fig. 16.13 Roots blower Figure 16.13 b shows the general arrangement in roots blower. It has inlet at section", " 1\u20131 and exit at 2\u20132. Air at atmospheric pressure enters the casing and is trapped between rotor A and the casing. When the rotor rotate then air trapped in volume space V is displaced towards high pressure region due to rotation of rotor. Exit end is connected to receiver in which air is gradually transferred and the pressure inside receiver increases due to cumulative effect of air being transferred from atmospheric pressure region to receiver region. In one revolution this positive displacement of air trapped between rotor and casing from inlet end to receiver end shall occur four times in case of two lobe rotor as shown. While in case of three lobes rotor this transfer shall occur six times. Every time when V volume of air is displaced without being compressed to the receiver side high pressure region, then the high pressure air rushes back from receiver and mixes irreversibly with this air until the pressure gets equalized. Thus, gradually air pressure builds up and say this pressure becomes p2. For inlet air pressure being p1, the work done per cycle, Reciprocating and Rotary Compressor ______________________________________________ 725 W = (p2 \u2013 p1) V In one revolution the total work required shall depend on number of lobes in rotor and in case of two lobes rotor, Wrevolution = (p2 \u2013 p1) (4 V) Let volume thus compressed in one revolution, be Vc = 4V Wrevolution = (p2 \u2013 p1)Vc = p1 p 2 p 1 Vc 1 The efficiency of compression can be defined by taking the ratio of work required in compressing the air isentropically and actual work done. This efficiency is also called roots efficiency. Roots efficiency = Work required in reversible adiabatic isentropic compression ( ) Actual work required in compression For compressing volume Vc from pressure p1 to p2 the work required in reversible adiabatic (isentropic) compression shall be; Wisen = 1 1 W 1 p1 Vc actual revolution roots = roots = Let pressure ratio be given by, r = so, 1 roots = r 1 r 1 The expression for roots blower efficiency shows that the efficiency depends upon pressure ratio and the increase in pressure ratio yields decrease in roots efficiency. At low pressure ratios the roots efficiency is quite high. Reduced efficiency with high pressure ratios is attributed to the mechanism involved in pressure rise. With high pressure ratio there occurs the reversal of flow during positive displacement of air from casing to receiver. This reversal causes the loss of energy thereby lowering efficiency due to increased actual work requirement. 726 ________________________________________________________ Applied Thermodynamics Roots blower are available in the capacity ranging from 0.15 m3/min to 1500 m3/min and the pressure ratios are of the order of 1 to 3.5. These are generally used for scavenging, supercharging of IC engines etc. Generally the roots efficiency is around 50\u201360% but high efficiency values up to 80% can also be achieved with low pressure ratios. (ii) Screw type or Helical type compressor: Screw type compressor is very much similar to roots blower. Fig. 16.14 Line sketch of screw compressor These may have two spiral lobed rotors, out of which one may be called male rotor having 3\u20134 lobes and other female rotor having 4\u20136 lobes which intermesh with small clearance. Meshing is such that lobes jutting out of male rotor get placed in matching hollow portion in female rotors. Initially, before this intermeshing the hollows remain filled with gaseous fluid at inlet port. As rotation begins the surface in contact move parallel to the axis of rotors toward the outlet end gradually compressing the fluid till the trapped volume reaches up to outlet port for getting discharged out at designed pressure. Since the number of lobes are different so the rotors operate at different speed. The material of casing may be cast iron or cast steel while rotors may be of steel and generally internally cooled by circulation of lubrication oil. Surface of lobes are smooth and the shaft is sealed by carbon rings at oil pressure. Two rotors are brought into synchronization by the screw gears. Thrust upon rotors is taken care of by oil lubricated thrust bearings. These compressors are capable of handling gas flows ranging from 200 to 20000 m3/h under discharge pressures of 3 bar gauge in single stage and up to 13 bar gauge in two stages. Even with increase in number of stages pressures up to 100 bar absolute have been obtained with stage pressure ratio of 2. Reciprocating and Rotary Compressor ______________________________________________ 727 Mechanical efficiency of these compressors is quite high and their isothermal efficiencies are even more than vane blowers and may be compared with centrifugal and axial compressors. But these are very noisy, sensitive to dust and fragile due to small clearances. (iii) Vane type compressor: Schematic of vane type compressor is shown in Fig. 16.15", ". It has cylindrical casing having an eccentrically mounted rotor inside it. The rotor has number of slots in it with rectangular vanes of spring loaded type mounted in slots. These vanes are generally non metallic and made of fibre or carbon composites or any other wear resistant material. These vanes remain in continuous contact with casing such that leakage across the vane-casing interface is minimum or absent. It has one end as inlet end and other as the delivery end connected to receiver. Upon rotation the eccentric rotor has the vanes having differential projection out of rotor depending upon their position. Air is trapped between each set of two consecutive blades in front of inlet passage and is positively displaced to the delivery end after compressing the volume V1 initially to V2, V3 and V4. When compressed volume comes in front of delivery passage and further rotation results in the situation when partly compressed air is forced to enter the receiver as their is no other way out. This cumulative transfer of partly compressed air in receiver causes irreversible compression resulting in gradual pressure rise. The p-V representation shown in Fig. 16.15 (b) indicates that the total pressure rise is due to the combined effect of reversible pressure rise inside casing and irreversible pressure rise inside receiver. Generally, the contribution of reversible pressure rise and irreversible pressure rise is in proportion of 50 : 50. Fig. 16.15 Vane type compressor Vane compressors are available for capacity up to 150 m3/min and pressure ratios up to 8 and efficiency up to 75%. For higher pressure ratios the efficiency of vane compressors is more than that of roots blower but the vane compressors have maximum speed up to 2500 rpm as compared to 7500 rpm in case of roots blower. Vane compressors have large power requirement as compared to roots blower 728 ________________________________________________________ Applied Thermodynamics due to large number of vanes on rotor. However, with small number of vanes the power required by vaned compressor may be comparable to roots blower. Vane compressors require large maintenance due to wear of vanes and other moving parts. Work required per revolution for vane compressor with n number of vanes shall be partly in reversible and partly in irreversible process. Let the inlet air be at pressure p1 and volume trapped is V1. This volume is partly compressed up to volume V4 and pressure rises from to p1 to p2, p3 and p4. Thus, the irreversible compression occurs for volume V4 and pressure rising from p4 to pf. Total work required, W = n + n (pf \u2013 p4) V4 p1V1 1 1 1 p 4 p 1 16.9 CENTRIFUGAL COMPRESSORS Centrifugal compressor is a radial flow machine compressing the fluid due to the dynamic action of impeller. Centrifugal compressors have impeller mounted on driving shaft, diffuser and volute casing as shown in Fig. 16.16. Centrifugal compressors have air inlet at the centre of impeller. The portion of impeller in front of inlet passage is called impeller eye. Fig. 16.16 Centrifugal compressor Impeller is a type of disc having radial blades mounted upon it. Compressor casing has a diffuser ring surrounding impeller and the air enters the impeller eye and leaves from impeller tip to enter diffuser ring. Volute casing surrounds the diffuser ring. Volute casing has cross section area increasing gradually up to the exit of compressor. These impellers of centrifugal compressors may also be of double sided type such that air can enter from two sides (both) of impeller. Thus double sided impeller shall have double impeller eye compared to single impeller eye as shown in Fig. 16.17. Reciprocating and Rotary Compressor ______________________________________________ 729 Fig. 16.17 Single sided impeller and double sided impeller Air enters the impeller eye axially and flows radially outwards after having entered compressor. Radial flow of air inside compressor is due to impeller (blades) rotating about its axis. These impeller blades impart momentum to the air entering, thereby rising its pressure and temperature. Subsequently the high pressure fluid leaving impeller enters the diffuser ring where the velocity of air is lowered with further increase in pressure of air. Thus in diffuser ring the kinetic energy of air is transformed into pressure head. High pressure air leaving diffuser is carried by volute casing to the exit of compressor. Due to increased cross section area of volute casing some velocity is further reduced causing rise in its pressure, although this is very small. Total pressure rise in compressor may be due to \u2018impeller action\u2019 and \u2018diffuser action\u2019 both. Generally, about half of total pressure rise is available in impeller and remaining half in diffuser. Pressure and velocity variation in centrifugal compressor is shown in Fig. 16.", "18. Fig. 16.18 Pressure and velocity variation in compressors Centrifugal compressors are used in aircrafts, blowers, superchargers, etc. where large quantity of air is to be supplied at smaller pressure ratios. Generally, pressure ratio up to 4 is achieved in single 730 ________________________________________________________ Applied Thermodynamics stage centrifugal compressors while in multistage compressors the pressure ratio up to 12 can be achieved. These compressors run at speed of 20,000\u201330,000 rpm. Thermodynamic analysis: In case of centrifugal compressors the fluid is being compressed following reversible adiabatic process ideally. In actual case the frictionless adiabatic compression is not possible due the frictional losses between fluid and casing. Irreversibilities also creep into compression process due to the entry and exit losses such as throttling, shock at inlet and exit and eddy formation etc. Therefore, the actual compression occurs following polytropic compression process. Figure 16.19 shows the T\u2013s representation for ideal and actual compression process. Fig. 16.19 T-s representation for compression in centrifugal compressor Compression process may also be of cooled type and this cooling during compression modifies the process to 1\u20132. Actual compression process may have the index of compression value (generally, n = 1.5 to 1.7) more than that of adiabatic process (for air, = 1.4). But the cooling reduces the index value even less than the adiabatic index. In centrifugal compressor the fluid moves with significant velocity during compression i.e. it has considerable kinetic energy which can not be neglected. For correct thermodynamic analysis the kinetic energy of fluid should be considered. The stagnation properties are evolved for considering the effect of fluid velocity. These considerations are not required in analysis of reciprocating compressors because the velocities are quite small and so negligible. Stagnation property refers to thermodynamic property of fluid upon being decelerated to zero velocity isentropically as described in earlier chapter of gas power cycles. Velocity diagram for centrifugal compressor: Figure 16.20 shows the velocity diagram for impeller blade of centrifugal compressor. Air enters the blade at 1 and leaves at 2. The air velocity at inlet and exit corresponding to the blade angles are shown on velocity diagrams for axial entry i.e. 1 = 90\u00b0. Tangential force acting on blade can be estimated by change in whirl velocity value; = m (Cw2 \u2013 Cw1); for axial entry Cw1 = 0 FT = m Cw2. This tangential force shall be responsible for rotation of shaft. The work done on impeller can be estimated using tangential force and distance travelled along the direction of force. This work is also called Euler\u2019s work. W = mCw2 U2 Reciprocating and Rotary Compressor ______________________________________________ 731 Fig. 16.20 Velocity diagrams for centrifugal compressor For radial blades on impeller the whirl velocity Cw2 shall coincide with U2 i.e. velocity diagram at exit of impeller blade; and, 2 W = mU2 This work required can also be estimated in terms of thermodynamic properties. By steady flow energy equation between sections 1 and 2 for no heat interaction in case of adiabatic compression process for no change in potential energy. h1 + 2 C1 2 2 C2 2 + W = h2 + W = m h 2 W = mCp \u2013 mCp T 2 C 2 2 Cp 1 2 C 1 2 Cp From definition of stagnation temperature, T 2 2 C 2 2 Cp = To2 and T 1 = T01 2 C 1 2 Cp 732 ________________________________________________________ Applied Thermodynamics or W = mCp (T02 \u2013 T01) perature from inlet to exit. Thus, impeller work can be given by the product of specific heat and change in stagnation tem- 1 W = mCp To1 = mCp To1 p 02 Cp To1 1 1 is ratio of stagnation pressure across impeller. For radial flow blades, substituting 1 2 = mCpT01 mU2 1 1 p 02 p 01 Here, for W, or, = p 02 p 01 2 U 2 C Tp 01 p 02 p 01 1 1 Thus, it shows that stagnation pressure ratio depends on the square of blade tip speed. Here it has been assumed that in case of radial blades blade tip speed is equal to the whirl velocity component, which is not valid condition in reference to actual operation. Actually, due to turbulence and secondary flow effects the Cw2 U2 and the Cw2 < U2. Difference between U2 and Cw2 i.e. (U2 \u2013 Cw2) is called the slip. This slip is quantified by the parameter called slip factor. Slip factor is the ratio", " of actual whirl velocity component to the ideal whirl velocity component. Slip factor = C w2 U 2 = Actual whirl velocity Ideal whirl velocity C U w2 2 Losses in centrifugal compressor: Centrifugal compressor has losses at inlet. Every compressor offers its optimum performance at design operation. Any deviation from design condition increases the different losses occuring in it. Due to friction, turbulence, eddy formation etc. there are flow losses inside the compressor. These losses are proportional to the square of flow velocity. Compressibility of fluid at high Mach nos. also offers losses. Losses in diffuser, volute casing are also there due to friction. Friction and secondary flow losses also cause slip and actual whirl component is less than theoretical value. 16.10 AXIAL FLOW COMPRESSORS Axial flow compressors have the fixed blades and moving blades mounted along the axis of compressor. Air enters axially and leaves axially. It has primarily two components i.e. rotor and casing. The rotor has Reciprocating and Rotary Compressor ______________________________________________ 733 blades mounted on it constituting moving blade ring. Blades are also mounted on the inner side of casing thereby constituting stages as fixed blade ring followed by moving blade ring followed by fixed blade ring, moving blade ring and so on. Due to the reduction in volume the volume space for compressed air may be gradually reduced. Gradual reduction in volume can be done by flaring the rotor while keeping stator diameter uniform or by flaring the stator while keeping rotor diameter constant as shown in fig. 16.21. The pressure of fluid entering the axial flow compressor increases upon passing through the fixed and moving blades. This flow of fluid over moving blades is accompanied by enthalpy rise while the fixed blades merely deflect the fluid so as to facilitate smooth entry into moving blades. Absolute velocity of air increases along axis of rotor due to work input from the prime mover. Relative velocity of air decreases during its flow through rotor. Blades have aerofoil section so as to have minimum losses due to turbulence, boundary layer formation and separation, eddy formation etc. Fig. 16.21 Axial flow compressor 16.11 SURGING AND CHOKING Let us look at the variation of theoretical pressure ratio with mass flow rate in case of a centrifugal compressor running at constant speed, as shown in Fig. 16.22. 734 ________________________________________________________ Applied Thermodynamics Let the compressor be run with completely closed delivery valve thus the mass flow rate is zero and pressure ratio available is shown by point a. This operating state of compressor shall not have any amount of air entering into it instead the air trapped inside the impeller and casing shall be subjected to impeller action (compression) causing its pressure rise upto the state a. Now upon opening of delivery valve the flow of air into compressor and its discharge begins. With this gradual opening of delivery valve the pressure ratio increases, say from state a to b and c along with increase in mass flow rate. The pressure ratio reaches to some maximum value at point c. Further opening of delivery valve shows that the mass flow rate increases but the pressure ratio decreases after point c i.e. beyond the point having maximum pressure ratio. The decrease in pressure ratio is due to the dominance of pressure loss due to friction etc. over the pressure rise in diffuser. After point c pressure losses are more than the pressure rise thereby, showing net decrease in pressure ratio as evident from points d and e on the delivery pressure vs mass flow rate characteristic curve. Fig. 16.22 Surging and choking Theoretically, the characteristic curve may be extrapolated and the maximum discharge (mass flow rate) is evident from point f at which the pressure ratio becomes unity and efficiency is zero. Maximum efficiency is available at the point (c) corresponding to the maximum pressure ratio. At the state f all work input to compressor is used in overcoming frictional resistance. In actual case the state \u2018e\u2019 shall indicate the maximum mass flow rate practically possible. This maximum mass flow state is known as choked mass flow. State of operation of compressor at point e is called \u2018choking\u2019 of compressor. Choking gives the maximum mass flow from compressor and the pressure ratio corresponding to this is less than maximum pressure ratio. Reciprocating and Rotary Compressor ______________________________________________ 735 Let us open delivery valve for increase in mass flow rate from mb to mb1. With this opening of valve the delivery pressure increases from pb to pb1. At point b1 the pressure on upstream and down stream do not rise simultaneously instead the pressure at upstream becomes pb1 while in downstream it remains pb for some time, due to inertia till pressure is not equalized. Since pb < pb1 so there is no problem in terms of flow reversal. Similarly as the delivery valve is gradually opened up to state c there is no problem of flow reversal as upstream pressure is always more than downstream pressure", " up to c. Upon partly closing the delivery valve, say from b to b2 the mass flow rate decreases from mb to mb2. This is accompanied by decrease in pressure at both upstream and downstream. Pressure at upstream gets changed (lowered) from pb to pb2 (pb2 < pb) instantaneously while the downstream pressure remains equal to pb for some time due to inertia. Pressure gets equalized to pb2 after some time lag. During this time lag the reversal of flow (stoppage of discharge from compressor) occurs from downstream to upstream as in transient period downstream pressure pb is more than upstream pressure pb2. Reversed flow stabilizes to normal flow after the pressure gets equalized. This reversal of flow occurs during part closing of valve in the operating range from a to c. Reversal of flow causes pulsations to the compressor and the noise, vibration and, jerks are felt during this transition period. This phenomenon is called \u2018surging\u2019. Similar, study of part closing and opening of valve when carried out in between c and e shows that no such reversed flow is observed. This changed behaviour is seen as even in case of closing of valve the mass flow rate shall decrease but the pressure ratio increases and in case of opening of valve the mass flow rate increases although pressure ratio decreases. Region lying between c and e gives the safe operating zone where surging phenomenon is not observed and also it is within the limits of choking of compressor. Compressor should be therefore run between c and e for its\u2019 smooth running. The excessive surging may even lead to failure (fracture) of rotating parts in compressor and is therefore completely avoided. For every compressor the safe zone for its operation is obtained and compressors are run in that region. Surging generally occurs in diffuser passages where the flow is slightly obstructed and gets retarded due to friction near vanes. Surging tends to increase with increasing number of diffuser vanes as with large no. of vanes it becomes difficult to divide air equally in all the passages. To avoid this the number of diffuser vanes is kept less than the impeller blades so that each of diffuser vane passage is provided with flow from more than one impeller blade channel. Thus, the condition of flow remain nearly same for all diffuser passages and surging due to flow reversal does not occur. 16.12 STALLING Stalling refers to the instability in the flow occurring due to flow separation from blade surfaces. Stalling may occur due to the non-uniform flow passing through the channels/passage between consecutive blades/vanes. In case of centrifugal compressor the flow output from impeller passes into the diffuser vanes and subsequently to exit end. In case of breakdown of flow in any one channel in diffuser, which may be due to reduced angle of incidence/increased angle of incidence, the flow pattern gets disturbed. For example the Fig. 16.23 shows the reduced angle of incidence in channel B of diffuser section in centrifugal compressor. Since the diffuser vanes are supposed to handle the similar amount of total flow therefore, inspite of reduced flow through the channel B (due to reduced angle of incidence) the mass flow rate does not change. Due to reduced flow through B the remaining channels have to carry additional flow in order to compensate for flow reduction in one of channels. These different angles of incidence into different channels result into non-uniform mass of air being handled by these channels. Such as in present case channel A has higher angle of incidence and channel C has reduced angle of incidence due to reduced flow through channel B. It results into flow separation due to increase and decrease in angle of incidence. Flow separation disturbs the flow pattern and results into excessive 736 ________________________________________________________ Applied Thermodynamics vibrations. This stall phenomenon passes from one channel to the other gradually. Stall may rotate in direction opposite to that of impeller rotation and is termed as rotating stall. This phenomenon of flow separation from blade surface, called as stalling also occurs in axial flow compressors, in the manner similar to that of centrifugal compressor. Stalling results in reduction of delivery pressure at compressor exit which may eventually lead to reversal of flow or surging. Stalling also leads to reduction in stage efficiency along with vibrations in compressor which may even rise upto the level causing failure of compressor parts when frequency matches with natural frequency. Fig.16.23 Stall propagation 16.13 CENTRIFUGAL COMPRESSOR CHARACTERISTICS Compressor characteristics are the set of curves obtained between delivery pressure and temperature plotted against mass flow for different values of rotational speed. Characteristics curves are machine specific i.e. every compressor shall have its own unique characteristic curves. These characteristic curves are plotted using non-dimensional parameters. Non-dimensional parameters used as the performance indicator of compressor. depend upon entry pressure, temperature, physical properties of fluid being compressed, mass flow rate and rotational speed etc. In order to", " obtain characteristic curve excessively large number of experimental observations are needed which can not be presented concisely. Use of non-dimensional parameters helps in combining number of parameters in order to get dimensionless groups which can be used for getting compressor characteristic curves. The non- dimensional groups thus used are, p 02 p 01, T 02 T 01, 01 m RT 2 D p 01, ND RT 01 where p, T, m, R, D, N are pressure, temperature, mass flow rate, gas constant, characteristics linear dimension and rotational speed respectively. Here for a particular compressor running with definite fluid the R and D shall be fixed and non- dimensional groups can be modified as, p 02 p 01, T 02 T 01, 01 m T p 01, N T 01. Different non-dimensional groups given here are termed as described below inspite of the fact that some of these are no more dimensionless due to omission of R and D terms. p 02 p 01 = Dimensionless pressure Reciprocating and Rotary Compressor ______________________________________________ 737 T 02 T 01 = Dimensionless temperature 01 m T p 01 = Dimensionless mass flow N T01 = Dimensionless rotational speed Characteristic curves p 02 p (i) \u201cdimensionless pressure\u201d 01 for different \u201cdimensionless rotational speeds with \u201cdimensionless mass flow for centrifugal compressor are obtained by plotting; relative to design value\u201d relative to design value\u201d and, (ii) \u201cisentropic relative to design value\u201d for different m T p 01 01 01 N T01 m T p 01 efficiency\u201d with \u201cdimensionless mass flow \u201cdimensionless rotational speeds N T01 relative to design value.\u201d Fig. 16.24 Centrifugal compressor characteristic curve Figures 16.24 and 16.25 shows the compressor characteristic curves. First curve is plot between pressure ratio and mass flow rate. Similar type of plot has been referred while discussing surging and choking. Here only the portion between surge line and choke line has been shown for different dimensionless rotational speeds relative to design value of 0.6, 0.7, 0.8, 0.9, 1.0. The locus of points having maximum efficiency is also shown on characteristic curve. This curve identifies the two extreme operating states between which the compressor runs. 738 ________________________________________________________ Applied Thermodynamics Fig. 16.25 Isentropic efficiency of a centrifugal compressor (characteristic curve) Similar to the plot between dimensionless pressure and dimensionless mass flow rate the plot between dimensionless temperature and dimensionless mass flow rate can be obtained. Using these patterns the plot between isentropic efficiency and dimensionless mass flow rate for different dimensionless rotational speeds can be obtained as shown in Fig. 16.25. These characteristic curves give the operating range of compressor and are also used for matching of compressors with other devices such as turbine etc. operating together. Characteristics of compressor and turbine operating together in any plant should be matching else inefficiencies are observed. Looking at the centrifugal compressor characteristic curves it can be concluded that, \u201cfor a given pressure ratio the increase in speed shows increase in flow rate and simultaneous reduction in efficiency,\u201d \u201cfor any particular speed of compressor the decrease in mass flow rate is accompanied by increase in pressure ratio\u201d and \u201cat all speeds the compressor operation is limited by surging on one end and choking on other end\u201d etc. Figure 16.26 shows a typical characteristic curve for some centrifugal compressor. Fig. 16.26 Typical centrifugal compressor characteristic curve 740 ________________________________________________________ Applied Thermodynamics Here it may be noted that since axial flow compressor comprises of number of stages so each stage shall also have its\u2019 own characteristic curve, although stage characteristic curves are similar to over all characteristic curve but with small pressure ratio. Characteristic curves show that the flow rate does not change much with increase in pressure ratio near design conditions. The efficiency is seen to decrease with decrease in flow rate and pressure ratios. Characteristic curves for axial flow compressor can also be shown as in Fig. 16.28. Fig. 16.28 Performance characteristic of axial flow compressor It is evident from characteristic curves that when compressor is run at less than design speed then rise in pressure and temperature is less than design value and also the choking decides the limiting operating point for compressor. 16.15 COMPARATIVE STUDY OF COMPRESSORS Comparative study of different types of compressors is presented below: (i) Comparison of reciprocating compressors with rotary compressors Reciprocating compressors Rotary compressors 1. These are suitable for low discharge rate at higher delivery pressures. Speed of rotation of reciprocating compressors is quite small due to higher inertia of reciprocating parts. 2. Reciprocating compressor has intermittent delivery of high pressure air thereby requiring receiver for maintaining continuous delivery. 1. These are suitable for high discharge rate at smaller delivery pressures. Speed of rotation of rotary", " compressors is quite high up to 40,000 rpm. 2. Rotary compressor has continuous delivery of high pressure air therefore no receiver is required. 3. Reciprocating compressors have 3. Rotary compressors have better balancing problems. balancing as compared to reciprocating compressors. (Contd.) Reciprocating and Rotary Compressor ______________________________________________ 741 Reciprocating compressors Rotary compressors 4. Quality of air delivered from 4. Air delivered is comparatively cleaner. reciprocating compressors is poor due to its contamination with lubricant. 5. Lubrication requirements are severe 5. Lubricating requirements are less in these compressors. 6. Reciprocating compressors can not be directly coupled to high speed prime movers such as turbines etc. compared to reciprocating compressors. 6. Rotary compressors can be directly coupled to high speed prime movers. 7. Reciprocating compressors are desi- 7. Rotary compressors have isentropic gned to achieve compression close to isothermal compression by cooling etc. compression process with index more process ideally but actually the compression occurs following polytropic compression with index more than adiabatic index. (ii) Comparison between the centrifugal compressor and axial flow compressor is given as under. Centrifugal Compressor Axial flow compressor (i) Centrifugal compressors generally have stage pressure ratio of around 5 and have radial flow inside with axial entry of fluid. Maximum pressure up to 400 bar can be achieved by multistaging, although multistaging is slightly difficult. (ii) Centrifugal compressors have good part load efficiency and wide operating range between surging and choking limits. (iii) Centrifugal compressors have small starting torque requirement. (iv) Due to large frontal area these are well suited for land applications. (v) These have simple construction and are comparatively cheap. (vi) These have poor isentropic efficiency ( 70%). Power required per kg of air flow rate is more. (vii) Efficiency vs. speed characteristics for centrifugal compressor is flat. It has higher efficiency in large speed range. (i) Axial flow compressors generally have stage pressure around 1.5 and have axial inlet and axial exit with flow inside compressor being parallel to the compressor axis. Maximum pressure up to 10 bar is possible to be achieved through multistaging which is very convenient in axial compressors. (ii) Axial flow compressors have poor part load efficiency and narrow operating range between surging and choking limits. (iii) Axial flow compressors have large starting torque requirement. (iv) Due to small frontal area axial flow compressors are well suited for aviation applications. (v) These have complex construction and are costly. (vi) These have better isentropic efficiency ( 85 to 90%). Power required per kg of air flow rate is less in axial flow compressors. (vii) Efficiency vs. speed characteristics is slightly bell shaped. It means higher efficiency in narrow range of speed. (viii) Centrifugal compressor\u2019s (viii) Axial compressor\u2019s performance performance does not alter with any kind of deposition on blade surface. deteriorates with any kind of deposition on blade surface. 742 ________________________________________________________ Applied Thermodynamics EXAMPLES 1. A reciprocating air compressor has cylinder with 24 cm bore and 36 cm stroke. Compressor admits air at 1 bar, 17\u00b0C and compresses it up to 6 bar. Compressor runs at 120 rpm. Considering compressor to be single acting and single stage determine mean effective pressure and the horse power required to run compressor when it compresses following the isothermal process and polytropic process with index of 1.3. Also find isothermal efficiency when compression is of polytropic and adiabatic type. Solution: Compression ratio = P 2 P 1 = 6 = r From cylinder dimensions the stroke volume = 4 (0.24)2 (0.36) = 0.01628 m3 Volume of air compressed per minute = 0.01628 120 = 1.954 m3/min Let us neglect clearance volume. Work done in isothermal process Wiso = P1 V1 ln r Mean effective pressure in isothermal process mepiso = r ln p V 1 1 V 1 = P1 ln r = 1 102 ln 6 = 179.18 kPa Work done in polytropic process with index n = 1.3, i.e. PV 1.3 = C 1 n n 1 Wpoly = P1 V1 P 2 P 1 n n 1 Mean effective pressure in polytropic process 102 meppoly = meppoly = = meppoly = 221.89 kPa Reciprocating and Rotary Compressor ______________________________________________ 743 P 2 P 1 1 1 W adiabatic V 1 Fig. 16.29", " p-V representation Work done in adiabatic process, Wadibatic = 1 P1 V1 Mean effective pressure in adiabatic process, mepadiabatic = 1 mepadiabatic = 1 P 2 P 1 P1 1 102 mepadiabatic = 233.98 kPa Horse power required for isothermal process, HPiso = (As 1 hp = 0.7457 kW) mepiso Volume per minute. 0 7457 60 = 179 18 1.954.. 0 7457 60 HPiso = 7.825 hp Horse power required for polytropic process, HPpoly = = meppoly Volume per minute. 0 7457 60.954. 221 89 1 0 7457 60. HPpoly = 9.69 hp 744 ________________________________________________________ Applied Thermodynamics Horse power required for adiabatic process, HPadiabatic = mepadiabatic Volume per minute. 0 7457 60 =.98 233. = 10.22 hp 0 7457 60 1.954 Isothermal efficiency = Isothermal process power required Actual power required Isothermal efficiencypoly = iso HP HP poly = 7 825.. 9 69 = 0.8075 or 80.75% Isothermal efficiencyadiabatic = HP adiabatic HP poly =. 7 825.22 10 = 0.7657 or 76.57% mep: 179.18 kPa for isothermal, 221.89 kPa for polytropic process HP required: 7.825 HP for isothermal, 9.69 HP for polytropic Isothermal efficiency: 80.75% for polytropic process, 76.57% for adiabatic process Ans. 2. A single stage single acting reciprocating air compressor has air entering at 1 bar, 20\u00b0C and compression occurs following polytropic process with index 1.2 upto the delivery pressure of 12 bar. The compressor runs at the speed of 240 rpm and has L/D ratio of 1.8. The compressor has mechanical efficiency of 0.88. Determine the isothermal efficiency and cylinder dimensions. Also find out the rating of drive required to run the compressor which admits 1 m3 of air perminute. Solution: Using perfect gas equation the mass of air delivered per minute can be obtained as, m = = P V 1 1 RT 1 2 1 10.287 0 1 293 = 1.189 kg/min Compression process follows PV1.2 = constt. Temperature at the end of compression; T2 = T1 T2 = 293 12 1 T2 = 443.33 K Reciprocating and Rotary Compressor ______________________________________________ 745 Work required during compression process, W = mR (T2 \u2013 T1) n n 1.189 0.287 (443.33 \u2013 293) 1 1.2.2 1 = 1 W = 307.79 kJ/min =. 307 79. 0 7457 60 hp W = 6.88 hp Capacity of drive required to run compressor = Isothermal work required for same compression, 6.88 0 88. = 7.82 hp P 2 P 1 Wiso = m RT1 ln = (1.189 0.287 293) ln Wiso = 248.45 kJ/min 12 1 Isothermal efficiency = Isothermal work Actual work = 248.45 307.79 = 0.8072 Volume of air entering per cycle = 1 240 = 4.167 10\u20133 m3/cycle Volume of cylinder = 4.167 10\u20133 = = 4.167 10\u20133 = 4 4 D2 L D2 1.8D Bore, D = 0.1434 m or 14.34 cm Stroke length L = 1.8 D = 1.8 14.34 = 25.812 cm Isothermal efficiency = 80.72 % Cylinder dimension, D = 14.34 cm L = 25.812 cm Rating of drive = 7.82 hp Ans. 3. A reciprocating compressor of single stage, double acting type delivers 20 m3/min when measured at free air condition of 1 bar, 27\u00b0C. The compressor has compression ratio of 7 and the conditions at the end of suction are 0.97 bar, 35\u00b0C. Compressor runs at 240 rpm with clearance volume of 5% of swept volume. The L/D ratio is 1.2. Determine the volumetric efficiency and dimensions of cylinder and isothermal efficiency taking the index of compression and expansion as 1.25. Also show the cycle on P-V diagram. Solution: P1 = 0.97 bar, T1 = 273 + 35 = 308 K, P 2 P 1 = 7, N = 240 rpm 746 ________________________________________________________ Applied Thermodynamics V3 = 0.05 Vs, Vs = V1 \u2013 V3, V1 = 1.05 Vs For free air condition of 1 bar, 27\u00b0C the mass of air delivered", ", = m = PV RT 1 10 0.287 m = 23.23 kg/min 2 20 300 For compression process 1\u20132, PV 1.25 = Constant Fig. 16.30 P-V representation. 1 25 1. 1 25 T2 = T1 = 308 (7)0.2 P 2 P 1 T2 = 454.54 K For expansion process, 3\u20134, PV 1.25 = Constant 1 1 25 P 3 P 4. V4 = V3 V4 = 0.05 Vs (7)1/1.25 V4 = 0.273 Vs V1 \u2013 V4 = 1.05 Vs \u2013 0.237 Vs = 0.813 Vs Volume of air corresponding to the suction conditions can be obtained using the volume of free air delivered 300 (V1 \u2013 V4) = 0 300.97 1 308 0.813 Vs 0 =.97 1 308 = 0.768 Vs V. 0 768 s V s Volumetric efficiency = = 0.768 or 76.8% Volumetric efficiency = 76.8% Ans. Reciprocating and Rotary Compressor ______________________________________________ 747 In order to find out the dimension of cylinder the volume of air sucked in one cycle. Volume of air sucked in a cycle = 20 2 240 = 0.0208 m3/cycle. Volume per cycle = Swept volume 0.0208 = D2 L = 4 D 2 (1.2 D) 4 D = 0.2805 m or 28.05 cm L = 33.66 cm Bore = 28.05 cm Stroke = 33.66 cm Ans. Work required in reciprocating compressor W = mR (T2 \u2013 T1) n n 1 W =.25 1.25 1 = 4884.92 kJ/min 1 = 4884..92 0 7457 60 W = 109.18 hp 23.23 0.287 (454.54 \u2013 308), {as 1 hp = 0.7457 kW] P 2 P 1 Work required when compression is isothermal, Wiso = m RT1 ln = 23.23 0.287 308 ln (7) = 3995.81 kJ/min =. 3995 81. 0 7457 60 Wiso = 89.31 hp W iso = W. 89 31. 109 81 = 0.8133 or 81.33% Isothermal efficiency = Isothermal efficiency = 81.33% Ans. 4. A reciprocating compressor of single stage and double acting type is running at 200 rpm with mechanical efficiency of 85%. Air flows into compressor at the rate of 5 m3/min measured at atmospheric condition of 1.02 bar, 27\u00b0C. Compressor has compressed air leaving at 8 bar with compression following polytropic process with index of 1.3. Compressor has clearance volume of 5% of stroke volume. During suction of air from atmosphere into compressor its temperature rises by 10\u00b0C. There occurs pressure loss of 0.03 bar during suction and pressure loss of 0.05 bar during discharge passage through valves. Determine the dimensions of cylinder, volumetric efficiency and power input required to drive the compressor if stroke to bore ratio is 1.5. 748 ________________________________________________________ Applied Thermodynamics Solution: Considering the losses at suction and discharge, the actual pressure at suction and delivery shall be as under. Atmospheric pressure, Pressure at suction, Pa = 1.02 bar, Ta = 273 + 27 = 300 K, Va = 5 m3/min P1 = 1.02 \u2013 0.03 = 0.99 bar T1 = 300 + 10 = 310 K P2 = 8 + 0.05 = 8.05 bar Pressure at delivery, Volume corresponding to suction condition of P1, T1,. 1 02 310 Therefore, work required for compression, V1 = 300.99 = 0 = 5.32 m3/min. P1 V1 32.99 2 10 60 0 3.. 1 3. 8 05.99 0 1 = 23.66 kW or 31.73 hp Power input required =. 31 73. 0 85 = 37.33 hp Power input = 37.33 hp Ans. Volumetric efficiency, Here, vol =.05, so vol =.99 300 0 1 02 310. = 0.7508 or 75.08% 1 0 05 0 05.. 1. 1 3. 8 05.99 0 Stroke volume per cycle = 5 2 200 = 0.0125 m3/cycle Actual stroke volume taking care of volumetric efficiency = 0 0125.. 0 7508 = 0.0167 m3/cycle Stroke volume = 4 D2L = 0.0167 4 D2 1.5D = 0.0167 D = 0.2420 m or 24.20 cm Stroke L = 1.5 D = 36.3 cm Reciprocating", " and Rotary Compressor ______________________________________________ 749 Cylinder dimensions = bore of 24.20 cm, stroke of 36.3 cm. Ans. Volumetric efficiency = 75.08% 5. In a reciprocating air compressor the air is compressed at the rate of 4 m3/min at 1 bar, 27\u00b0C up to the pressure of 8 bar following index of compression as 1.2. The compression occurs in two stages with intercooling at optimum intercooler pressure and perfect intercooling. Compare the work input required if the same compression occurs in single stage. Also compare the work input if same compression occurs in two stages with imperfect intercooling up to 30\u00b0C at the optimum intercooling pressure. Consider Cp = 1.0032 kJ/kg K and R = 0.287 kJ/kg K. Solution: Mass of air compressed per minute; m = 1 10 0.287 2 4 300 = 4.65 kg/min Fig. 16.31 P-V diagram Different types of compression are shown on P\u2013V diagram For single stage compression from 1 bar to 8 bar, process 1\u20132\u20133, PV 1.2 = Constt. Work input, WI = 2 1.2 1 = 1. 994 113. 60 0 7457 8 1 P1V1 1 102 4 1 WI = 994.113 kJ/min or. 1 2 1 1 2. 1 = 22.22 hp Optimum intercooling pressure = 8 1 = 2.83 bar For two stage compression with perfect intercooling; Work input, WII, I/C = 2 n n 1 P1V1 n 2 1 n P 3 P 1 1 750 ________________________________________________________ Applied Thermodynamics = 2 1.2.2 1 1 1 102 WII, I/C = 908.19 kJ/min or 20.29 hp When there is two stage compression with imperfect intercooling: Intercooler pressure, P2 = 2.83 bar, T2 = 273 + 30 = 303 K Volume of air at inlet of HP cylinder, V2 = V2 = 1 4 303. 2 83 300 = 1.43 m3/min Hence, work required for two stage compression, W W = II,I/C = W II,I/C = n HP + W P 2 P 1 LP P V 1 1 1 n 1.2 1.2 1 1 10 83.2 1. 83 10. 2. 1 43 8. 2 83 1 2 1. 1 2. 1 % saving in work when compression occur with perfect intercooling as compared to single stage = [454.3] + [459.2] = 913.5 kJ/min or 20.42 hp compression = 100 W W I II I C /, W I % excess work to be done when two stage compression occurs with imperfect intercooling as = 8.69% Ans. compared to two stage compression with perfect intercooling: = 100 W II I C /, W W II I C /, II I C /, = 0.636% Ans. 6. A reciprocating air compressor has four stage compression with 2 m3/min of air being delivered at 150 bar when initial pressure and temperature are 1 bar, 27\u00b0C. Compression occur polytropically following polytropic index of 1.25 in four stages with perfect intercooling between stages. For the optimum intercooling conditions determine the intermediate pressures and the work required for driving compressor. Solution: Here there is four stage compression with perfect intercooling at optimum intercooling conditions. / So optimum stage pressure ratio = 150 1 4 = 3.499 3.5 Reciprocating and Rotary Compressor ______________________________________________ 751 Intermediate pressure shall be as follows: Between Ist and IInd stage = 3.5 bar Between IInd and IIIrd stage = 12.25 bar Between IIIrd and IVth stage = 42.87 bar Intermediate pressure: 3.5 bar, 12.25 bar, 42.87 bar Ans. Since it is perfect intercooling so temperature at inlet of each stage will be 300 k. So temperature at the end of fourth stage, T = T1 P 2 P 1 n 1 n = 300 3 T = 385.42 K.5 1 25 1.. 1 25 Mass of air, kg/min, m = PV RT = 2 150 10 0.287 2 385 42. = 271.21 kg/min Work required for driving compressor, W = = n n 1 1.25.25 RT1 271.21 0.287 300 3 4 1 = 132978.04 kJ/min or 2972.11 hp 1 25 1.. 1 25.5. 132978 04. 60 0 7457 4 1. Work input = 2972.11 hp Ans. 7. A two stage reciprocating compressor has air", " entering at 1 bar, 17\u00b0C into LP compressor and leaving HP stage at 16 bar. An intercooler working at 4 bar pressure is provided between the HP and LP stages. The compression process follows the process given by PV1.3 = constant. The bore diameters of HP and LP cylinder are 6 cm and 12 cm respectively while stroke lengths are equal. For perfect intercooling determine the work done in compressing per unit mass of air. Also state whether the intercooler pressure will rise, fall or no change if the volumetric efficiency is taken as 0.90 for LP cylinder. Neglect clearance volumes. Solution: For perfect intercooling the amount of work required shall be; W = n m n 1 RT.287 (273 + 17 16 4 752 ________________________________________________________ Applied Thermodynamics = 271.94 kJ/kg. Work = 271.94 kJ/kg Ans. The ratio of volumes handled by HP and LP stages. Using the perfect intercooling. Fig. 16.32 Two stage compression on P-V diagram State at inlet of LP cylinder = P1 V1 State at inlet of HP cylinder = P2 V2 Since state 2 lies on isothermal compression process line so, P1 V1 = P2 V2 or = = Volume at inlet to LP Volume at inlet to HP = 4 From given cylinder dimensions, for given bore diameters and common stroke length, Ratio of effective cylinder volumes = Effective volume of LP cylinder Effective volume of HP cylinder 4 2 D HP 0.90 4 0.9 0 12. 2. 0 06 2 D L LP L 2 = 3.6 = = Theoretically, the volume ratio is 4 while considering volumetric efficiency the ratio of effective cylinder volumes comes out to be 3.6 which is less than the theoretical volume ratio. Therefore, it can be concluded that less amount of air is given into HP cylinder than its\u2019 capacity. Thus, the HP cylinder would inhale volume equal to its\u2019 capacity and since it has larger capacity than volume available so the pressure of intercooler shall drop. 8. In a two stage reciprocating air compressor running at 200 rpm the air is admitted at 1 bar, 17\u00b0C and discharged at 25 bar. At low pressure stage suction conditions the rate of air flow is 4 kg/minute. The low pressure cylinder and high pressure cylinders have clearance volumes of 4% and 5% of respective cylinder stroke volumes. The index for compression and expansion processes in two stages are same Reciprocating and Rotary Compressor ______________________________________________ 753 as 1.25. Considering an optimum and perfect intercooling in between two stages determine the power required, isothermal efficiency, free air delivered, heat transferred in each cylinder and the cylinder volumes. Solution: For the optimum intercooling the pressure ratio in each stage = 25 1 = 5 Fig. 16.33 P-V representation Perfect intercooling indicates, T1 = T5 = 273 + 17 = 290 K T2 = T1 T6 = T5 = 400.12 K = 400.12 K Actual compression work requirement, n 1 = 2 W = WHP + WLP m RT1 4 0.287 290 25 1.25 1 1 1 n W = 1264.19 kJ/min or 28.25 hp. 1 25 1. 1 25 1 Work requirement if the process is isothermal compression, P 6 P 1 = 4 0.287 290 ln (25) Wiso = m RT1 ln Wiso = 1071.63 kJ/min 754 ________________________________________________________ Applied Thermodynamics Isothermal efficiency = W iso = 0.8477 or 84.77% W Free air delivered = mRT 1 P 1 4 =.287 0 1 102 290 = 3.33 m3/min Heat transferred in HP cylinder = Heat transferred in LP cylinder = Q (Due to optimum and perfect intercooling) \u2013 4 1.0032 (400.12 \u2013 290) \u2013 m Cp (T2 \u2013 T1). = Q = W 2 1264 19 2 Q = 190.21 kJ/min Volumetric efficiency, vol = Here the ambient conditions and suction conditions are same so expression gets modified as, vol = Volumetric efficiency of HP, vol, HP = 1 + CHP \u2013 CHP P 6 P 5 1 n CHP = 0.04 = 1 + 0.04 \u2013 0.04 (5)1/1.25 vol, HP = 0.895 or 89.5% Volumetric efficiency of LP, vol, LP = 1 + CLP \u2013 CLP CLP = 0.05.05 \u2013 0.05 (5)1/1.25 = 0.8688 or 86.88% Stroke volume of HP cylinder = Free air delivery speed Pressure ratio vol, HP Vs, HP = 3 33. 5 200 Clearance volume, Vc, HP =", " 0.05 3.721 10\u20133 = 1.861 10\u20134 m3 Total HP cylinder volume, VHP = Vs, HP + Vc, HP = 3.907 10\u20133 m3 ; Vc, HP = Clearance volume of HP = 3.721 10\u20133 m3. 0 895 Stroke volume of LP cylinder = Free air delivery Speed vol, LP =. 3 33. 200 0 8688 Reciprocating and Rotary Compressor ______________________________________________ 755 Vs, LP = 0.01916 m3 Clearance volume, Vc, LP = 0.04 Vs, LP = 7.664 10\u20134 m3 Total LP cylinder volume, VLP = Vs, LP + Vc, LP = 0.019926 m3 Power required = 28.25 hp, Isothermal efficiency = 84.77% Free air delivered = 3.33 m3/min, Heat transfer in HP cylinder = 190.21 kJ/min Heat transferred in LP cylinder = 190.21 kJ/min, HP cylinder volume = 3.907 10\u20133 m3 LP cylinder volume = 0.019926 m3 Ans. 9. A two stage double acting reciprocating air compressor running at 200 rpm has air entering at 1 bar, 25\u00b0C. The low pressure stage discharges air at optimum intercooling pressure into intercooler after which it enters at 2.9 bar, 25\u00b0C into high pressure stage. Compressed air leaves HP stage at 9 bar. The LP cylinder and HP cylinder have same stroke lengths and equal clearance volumes of 5% of respective cylinder swept volumes. Bore of LP cylinder is 30 cm and stroke is 40 cm. Index of compression for both stages may be taken as 1.2. Determine, (i) (ii) (iii) the heat rejected in intercooler, the bore of HP cylinder, the hp required to drive the HP cylinder. Solution: Optimum intercooling pressure = 9 = 3 bar LP stage pressure ratio = HP stage pressure ratio = 3 From the given dimensions of LP cylinder, the volume of LP cylinder, in m3/min (0.30)2 (0.40) 200 2 VLP = VLP = 11.31 m3/min 4 Volumetric efficiency of LP compressor, here ambient and suction conditions are same, vol, LP = vol, LP = 0.9251 or 92.51% = 1 + 0.05 \u2013 0 05. 1. 1 2 3 1 Fig. 16.34 P-V diagram 756 ________________________________________________________ Applied Thermodynamics Volume of air inhaled in LP stage = VLP vol, LP = 11.31 0.9251 = 10.46 m3/min Mass of air per minute, m = P V 1 1 RT 1 = 2 1 10 0.287 10 46. 298 = 12.23 kg/min Temperature after compression in LP stage, T2 = T1 P 2 P 1 = 298 3 1 T2 = 357.88 Volume of air going into HP cylinder V5 = mRT 5 P 5 After intercooling, T5 = 298 K, P5 = 2.9 bar,.23 2 V5 = 12 0.287 10 2.9 V5 = 3.61 m3/min 298 Since the clearance volume fraction and pressure ratio for both HP and LP stages are same so the volumetric efficiency of HP stage referred to LP stage suction condition shall be same Hence, the volume of HP cylinder/min = = 3.902 m3/min vol, HP = vol, LP = 0.9251 V5 = vol,HP 3 61..9251 0 Let bore of HP cylinder be DHP, 3.902 = 4 (DHP)2 0.40 2 200 DHP = 0.1762 m or 17.62 cm Heat rejected in intercooler, Q = mCp (T2 \u2013 T5) = 12.23 1.0032 (357.88 \u2013 298) = 734.68 kJ/min In HP stage, n 1 T6 = 298 T6 = 359.91 K Reciprocating and Rotary Compressor ______________________________________________ 757 Work input required for HP stage, WHP = 12.23 0.287 (359.9 \u2013 298) n n 1 mR (T6 \u2013 T5) = 1.2.2 1 1 or WHP = 1303.62 kJ/min WHP = 29.14 hp Heat rejected in intercooler = 734.68 kJ/min Ans. Bore of HP cylinder = 17.62 cm Horse power required to drive HP stage = 29.14 hp 10. During an experiment on reciprocating air compressor the following observations are being taken; Barometer reading = 75.6 cm Hg, Manometer reading across orifice = 13 cm Hg. Atmospheric", " temperature = 25\u00b0C. Diameter of orifice = 15 mm. Coefficient of discharge across the orifice = 0.65 Take density of Hg = 0.0135951 kg/cm3 Determine the volume of free air handled by compressor in m3/min. Solution: Cross-sectional area of orifice, A = 4 (15 10\u20133)2 = 1.77 10\u20134 m2 Atmospheric pressure = 75.6 0.0135951 9.81 104 10\u20133 Specific volume of air per kg at atmospheric conditions, = 100.83 kPa RT P Density of air = 1 v Pressure difference across orifice = 13 0.0135951 9.81 104 10\u20133 = 1.18 kg/m3.287. 100 83 298 v = = = 0.848 m3/kg 1 0 Height of air column for pressure difference across orifice. = 17.34 kPa Put, a ha g = 17.34 103 a = 1.18 kg/m3 ha = 1497.95 m Free air delivery = Cd A 2gha = 0.65 1.77 10\u20134 2. 9 81 1497 = 0.01972 m3/s or 1.183 m3/min.95. Free air delivery = 1.183 m3/min Ans. 11. During a trial on single acting single stage compression the following observations are made; Dimensions of cylinder: 10 cm bore and 8 cm stroke. Speed of rotation: 500 rpm. 758 ________________________________________________________ Applied Thermodynamics Barometer reading: 76 cm Hg Atmospheric temperature: 27\u00b0C Delivery air temperature = 130\u00b0C Free air delivery = 15 m3/hr Spring balance of dynamometer type (electric motor) reading: 10 kg Radius of arm of spring balance: 30 cm Take mechanical efficiency = 0.90. Determine the volumetric efficiency, shaft output per m3 of free air per minute. Solution: Free air delivery = 15 m3/hr = 0.25 m3/min Volume of cylinder = (0.10)2 (0.08) = 6.28 10\u20134 m3 4 Volumetric efficiency = = 0.7962 or 79.62% Ans. 500 15 60 4 6.28 10 N T 60 2 2 Shaft output = Shaft output = 500 10 9 81 0 30 10.. 3 60 = 15.41 kJ/s or 20.66 hp Shaft output per m3 of free air per minute = 20 66..25 0 = 82.64 hp per m3 of free air per minute. Ans. 12. Determine the minimum number of stages required in an air compressor which admits air at 1 bar, 27\u00b0C and delivers at 180 bar. The maximum discharge temperature at any stage is limited to 150\u00b0C. Consider the index for polytropic compression as 1.25 and perfect and optimum intercooling in between the stages. Neglect the effect of clearance. Solution: Let there be \u2018i\u2019 number of stages. So the overall pressure ratio considering inlet state as Pa and Ta and delivery state pressure as Pi When perfect and optimum intercooling is considered then pressure ratio in each stage will be samer)i, for any stage, say second stage, T1 = 273 + 27 = 300 K Reciprocating and Rotary Compressor ______________________________________________ 759 and and T2 = 273 + 150 = 423 r)i = in n 1 T 2 T 1 180 1 = ln 180 = 423 300 i.25 1.25 0 i 1 25. 1 25 1., Taking log for solving, ln 423 300 Upon solving, i = 3.022 say 3 stages 3 stages Ans. 13. In a triple stage reciprocating compressor of single acting type the air enters at 1 bar, 27\u00b0C. The compressor has low pressure cylinder with bore of 30 cm and stroke of 20 cm. Clearance volume of LP cylinder is 4% of the swept volume. The final discharge from compressor takes place at 20 bar. The expansion and compression index may be taken uniformly as 1.25 for all the stages. The intercooling between the stages may be considered to be at optimum intercooling pressure and perfect intercooling. Determine, the interstage pressures, effective swept volume of low pressure cylinder, temperature and volume of air delivered in each stroke and the work done per kg of air. Solution: Here P1 = 1 bar, T1 = 300 K, C = 0.04, P10 = 20 bar, n = 1.25, See Fig. 16.35 For optimum and perfect intercooling 10 P 6 = 20 1 1 3 = 2.714 P2 = 2.714 bar, T5 = T1 = 300 K P6 = 7.366 bar T9 = T1 = 300 K Volumetric efficiency of LP stage, 1 1 25. vol, LP =", ".04 \u2013 0.04 (2.714)1/1.25 = 0.9511 or 95.11% 4 D2 L = (0.30)2 0.20 = 0.01414 m3 LP swept volume, V1 \u2013 V3 = Effective swept volume of LP cylinder, V1 \u2013 V4 = V1 \u2013 V4 = 0.9511 0.01414 = 0.01345 m3 Temperature of air delivered, T10 = T9 vol, LP (V1 \u2013 V3) = 300 2 714 1 25 1. 1 25. 1 25 1.. 1 25 4. P 10 P 6 = 366.31 K 760 ________________________________________________________ Applied Thermodynamics Fig. 16.35 P-V representation For the compression process of air as perfect gas 10 V 1 V 11 V 10 T 10 T 10 P 10 V 4 T 1 P 1. 0 01345 366.31 1 10 300 20 10 2 2 V10 \u2013 V11 = = Total Work done per kg air, Volume of air delivered = V10 \u2013 V11 = 8.2115 10\u20134 m3 n W = 3 1 25 1.. 1 25 = 3.287 0 P 2 P 1 RT 1 n 1.25 1.25 1 = 285.44 kJ/kg of air 1 300 1. 2 714 1 25 1.. 1 25 1 Effective swept volume of LP cylinder = 0.01345 m3 Intermediate pressure = 2.714 bar, 7.366 bar Ans. Temperature of air delivered = 366.31 K Volume of air delivered = 8.2115 10\u20134 m3 Work done = 285.44 kJ/kg of air 14. A two stage reciprocating air compressor has air being admitted at 1 bar, 27\u00b0C and delivered at 30 bar, 150\u00b0C with interstage pressure of 6 bar and intercooling up to 35\u00b0C. Compressor delivers at the rate of 2 kg/s. Clearance volumes of LP and HP cylinders are 5% and 7% of stroke volume respectively. The index of compression and expansion are same throughout. Determine the swept volume of both cylinders in m3/min, amount of cooling required in intercooler and total power required. Also estimate the amount of cooling required in each cylinder. Reciprocating and Rotary Compressor ______________________________________________ 761 Solution: Given: P1 = 1 bar, T1 = 300 K, P2 = 6 bar, P6 = 30 bar, T6 = 273 + 150 = 423 K, T5 = 273 + 35 = 308 K, CLP = 0.05, CHP = 0.07, m = 2 kg/s For process 5\u20136, P2 = P5 P 6 P 5 = = 30 6 Taking log of both sides, Fig. 16.36 P-V representation T 6 T 5 423 308 n 1 n n 1 n ln (5) = n n 1 ln (1.3734) Upon solving we get, Volumetric efficiency of LP cylinder, n = 1.245 vol, LP = 1 + CLP \u2013 CLP = 1 + 0.05 \u2013 0.05 6 1 = 0.8391 or 83.91% 1 1 245.. 1 1 245 P 2 P 1 1 1 245.. 1 1 245 P 6 P 5 30 6 Volumetric efficiency of HP cylinder, vol, HP = 1 + CHP \u2013 CHP = 1 + 0.07 \u2013 0.07 vol, HP = 0.815 or 81.50% 762 ________________________________________________________ Applied Thermodynamics For suction of LP cylinder 300 = 1.722 m3/s or 103.32 m3/min P1 (V1 \u2013 V4) = mRT1 2 0 (V1 \u2013 V4) =.287 1 10 2 V 4 V 3 103.32 0.8391 V 1 V 1 vol, LP = = = 0.8391 (V1 \u2013 V3) = 123.13 m3/min Swept volume of LP cylinder = 123.13 m3/min Ans. For HP cylinder, P2 (V5 \u2013 V8) = m RT5 (V5 \u2013 V8) = = mRT 5 P 2 2 0 308 = 0.2946 m3/s or 17.676 m3/min vol, HP = V 5 V 6 (V6 \u2013 V7) = 17 676.. 0 815 = 21.69 m3/min.287 6 102 V 8 V 7 Swept volume of HP cylinder = 21.69 m3/min For compression in LP stage, Ans. T2 = T1 \uf8f6 \uf8f7 \uf8f8 = 300 6 1 T2 = 426.83 K. 1 245 1. 1 245 Cooling required in intercooler, QI/C = m Cp (T2 \u2013 T", "5) = 2 1.0032 (426.83 \u2013 308) QI/C = 238.42 kJ/s Heat picked in intercooler = 238.42 kW Ans. Work input required = WLP + WHP = = n n 1 n n 1 m RT1 mR RT5 Reciprocating and Rotary Compressor ______________________________________________ 763 2 0.287 300 1.245.245 1 =. 1 245 1. 1 245. 1 245 1. 1 245 30 6 308 6 1 1 1 1 = 704.71 kJ/s Total work required = 704.71 kW Ans. Heat transferred in LP cylinder = Amount of cooling required in LP cylinder n 1 Cv (T2 \u2013 T1) QLP = m n = 2 1 4 1..245 1 = 115.55 kJ/s.245 1 0.72 (426.83 \u2013 300). Amount of cooling required in LP cylinder = 115.55 kW Ans. Heat transferred in HP cylinder = Amount of cooling required in HP cylinder QHP = m Cv (T6 \u2013 T5) n n 1 1 4 1..245 1.245 1 = 2 0.72 (423 \u2013 308) = 104.77 kJ/s Amount of cooling required in HP cylinder = 104.77 kW Ans. 15. A roots blower handles free air of 0.5 m3/s at 1 bar and 27\u00b0C and delivers air at pressure of 2 bar. Determine indicated power required to drive compressor and isentropic efficiency. Solution: Indicated power required = (P2 \u2013 P1) V1 Indicated power when isentropic compression occurs, Wroots = (2 \u2013 1) 102 0.5 = 50 kJ/s Wroots = 50 kW or 67.05 hp P1 V1 1 102 0. Wisentropic =. 1 4 1 1 4. 1 = 38.33 kW or 51.4 hp Fig. 16.37 764 ________________________________________________________ Applied Thermodynamics Isentropic efficiency of roots blower = W isentropic W roots =. 38 33 50 = 0.7666 or 76.66% Indicated power of roots blower = 67.05 hp, Ans. Isentropic efficiency = 76.66% 16. A vaned compressor handles free air of 0.6 m3/s at 1 bar and compresses up to 2.3 bar. There occurs 30% reduction in volume before the back flow occurs. Determine the indicated power required and isentropic efficiency. Solution: Here on P\u2013V diagram the state 2 indicates the point at which delivery occurs. While 2 is the point up to which air is compressed inside. V2 = 0.7 V1 V 1 V 2 V 1 0 7 P2 = 1 = 1.65 bar Fig. 16.38 P-V diagram Indicated power required for vaned compressor Wvane = = P1 V1 1 102 0.6 1 65. = 59.60 kJ/s or 79.925 hp + (P2 \u2013 P2) V2 1 4 1.. 1 4 + (2.3 \u2013 1.65) 102 (0.7 0.6) 1 Power requirement when compression occurs isentropically, Wisentropic = 1 P1 V1 1 P 2 P 1 1 Reciprocating and Rotary Compressor ______________________________________________ 765 1 102 0. = 56.42 kJ/s or 75.66 hp Isentropic efficiency of Vane compressor = = 0.9466 or 94.66% W isentropic W vane = 75 66..925 79 Indicated power required = 79.925 hp, isentropic efficiency = 94.66% Ans. 17. A centrifugal compressor delivers free air of 18 kg/min. Air is sucked at static states of 1 bar, 27\u00b0C with inlet velocity of 50 m/s. The total head pressure ratio is 4 and isentropic efficiency of compressor is 0.75. The mechanical efficiency of motor attached to it is 0.90. Determine total head temperature of air at exit of compressor and brake power required to drive compressor. Solution: Stagnation temperature at inlet, T01= T1 + 2 V 1 2 Cp T01 = 300 + Isentropic efficiency of compressor, 2 50 2 1 0032 10. = 301.25 K 3 For process 1\u20132, isen = T 02 T 02 T 01 T 01 1 P 02 P 01 T = 02 T 01 T02 = T01 4 T02 = 301.25. /. = 447.66 K Substituting temperature values in expression of isentropic efficiency, Fig. 16.39 766 ________________________________________________________ Applied Thermodynamics T 02 + T01 T02 = T 01 isen.", " = 447 66 301. 0 75 T02 = 496.46 K.25 + 301.25 Total head temperature at exit = 496.46 K Ans. Brake power of drive required = = m C p T 02 T 01 mech. 18 1 0032 60 301.25 496.46.9 0 Brake power required = 87.54 hp Ans. = 65.28 kW or 87.54 hp 18. A double-acting single cylinder reciprocating air compressor has a piston displacement of 0.015 m3 per revolution, operates at 500 r.p.m. and has a 5% clearance. The air is received at 1 bar and delivered at 6 bar. The compression and expansion are polytropic with n = 1.3. Determine, (i) (ii) (iii) the volumetric efficiency the power required the heat transferred and its direction, during compression if inlet temperature of air is 20\u00b0C. [U.P.S.C., 1998] Solution: V = 0.015 m3 per revolution, N = 500 r.p.m., C = 5% or 0.05, n = 1.3, P1 = 1 bar, P2 = 6 bar, T1 = 20\u00b0C Volumetric efficiency, vol =.05 \u2013 0.05 6 1 = 0.8516 or 85.16% 1 1 3. Ans. Volumetric efficiency = 85.16% Power required = 2N Here swept volume in cylinder = Vs = V = 0.015 2 500 Vs = 15 m3/min Actual air inhaled = V1 = 15 0.85 V1 = 12.75 m3/min Reciprocating and Rotary Compressor ______________________________________________ 767 Mass of air entering, m = P V 1 1 RT 1 = 2 1 10 0.287 12 75. 293 = 15.16 kg/min. Power required = 1 3. 1 3 1. 1 102 12.75 = 2829.21 kJ/min Ans. Power required = 2829.21 kJ/min Heat transferred during compression, (this is heat rejected), for a polytropic process n 1 Q = m Cv (T2 \u2013 T1) n During compression process T2 = 298 6 1 T2 = 450.59 K. 1 3 1. 1 3 Substituting in heat transferred Q, Q = 15.16 0.718 Q = 571.78 kJ/min 450.59 \u2013 293) Heat rejected during compression = 571.78 kJ/min Ans. 16.1 Classify the compressors. 16.2 Discuss the applications of compressed air to highlight the significance of compressors. 16.3 Describe the working of single stage reciprocating compressor. 16.4 Discuss the indicator diagram for reciprocating compressor. Also describe the factors responsible for deviation of hypothetical indicator diagram to actual indicator diagram. 16.5 Obtain the volumetric efficiency of single stage reciprocating compressor with clearance volume and without clearance volume. 16.6 Discuss the effects of clearance upon the performance of reciprocating compressor. 16.7 Define isothermal efficiency. Also discuss its significance. 16.8 What do you understand by multistage compression? What are its\u2019 merits over single stage compression? 16.9 Show that the volumetric efficiency with respect to free air delivery is given by, vol \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb where all the terms of expression have their usual meanings. 768 ________________________________________________________ Applied Thermodynamics 16.10 Discuss the significance of intercooling upon the performance of multi-stage compression. 16.11 What is the optimum pressure ratio for perfect intercooling in between two stages of compression? The inlet and outlet pressures may be taken as P1 and P3. 16.12 Discuss the control of reciprocating air compressor. 16.13 Discuss the working of positive displacement rotary compressors. 16.14 Describe the working of centrifugal compressors. 16.15 What do you understand by surging and choking phenomenon? 16.16 Explain the stalling and its effect on the compressor performance. 16.17 Describe the characteristics of centrifugal compressor. 16.18 Compare the axial flow compressor with centrifugal compressors. 16.19 Show that the heat rejected in each stage of a reciprocating compressor with perfect intercooling is given by \u2212 \uf8ec \u2212\uf8ed T2 \u2013 T1) 16.20 Write short notes on the following: (i) Free air delivery (ii) Volumetric efficiency (iii) Axial flow compressors (iv) Air flow rate measurement in reciprocating compressors. 16.21 A single stage single cylinder reciprocating compressor has 60 m3/hr air entering at 1.013 bar, 15\u00b0C and air leaves at 7 bar. Compression follows polytropic process with index of 1.35. Considering negligible", " clearance determine mass of air delivered per minute, delivery temperature, indicated power and isothermal efficiency. [1.225 kg/min, 202.37\u00b0C, 4.23 kW, 77.1%] 16.22 A reciprocating compressor of single stage and double acting type has free air delivered at 14 m3/min measured at 1.013 bar, 288 K. Pressure and temperature at suction are 0.95 bar and 305 K. The cylinder has clearance volume of 5% of swept volume. The air is delivered at pressure of 7 bar and expansion and compression follow the common index of 1.3. Determine the indicated power required and volumetric efficiency with respect to free air delivery. [63.55 kW, 72.4%] 16.23 A single stage double acting reciprocating compressor delivers 14 m3/min measured at suction states of 1 bar and 20\u00b0C. Compressor runs at 300 rpm and air is delivered after compression with compression ratio of 7. Compressor has clearance volume of 5% of swept volume and compression follows polytropic process with index 1.3. Determine the swept volume of cylinder and indicated [0.028 m3, 76.86 hp] power in hp. 16.24 A single stage single acting reciprocating air compressor handles 0.5 m3/min of free air measured at 1 bar. Compressor delivers air at 6.5 bar while running at 450 rpm. The volumetric efficiency is 0.75, isothermal efficiency is 0.76 and mechanical efficiency is 0.80. Determine indicated mean [0.185 MPa, 3.44 hp] effective pressure and power required to drive the compressor. 16.25 A single stage single acting reciprocating air compressor compresses air by a ratio of 7. The polytropic index of both compression and expansion is 1.35. The clearance volume is 6.2% of cylinder volume. For volumetric efficiency of 0.8 and stroke to bore ratio of 1.3 determine the dimensions of cylinder. [14.67 cm and 19.08 cm] 16.26 A single stage single acting reciprocating air compressor runs with air entering at 1 bar and leaving at 7 bar following PV1.3 = constant. Free air delivery is 5.6 m3/minute and mean piston speed is 150 m/min. Take stroke to bore ratio of 1.3 and clearance volume to be 1 15 th of swept Reciprocating and Rotary Compressor ______________________________________________ 769 volume per stroke. The suction pressure and temperature are equal to atmospheric air pressure and temperature. Determine volumetric efficiency, speed of rotation, stroke and bore. Take mean piston speed = 2 stroke rpm. [76.88%, 164 rpm, 45.7 cm, 35.1 cm] 16.27 A reciprocating compressor of single acting type has air entering at 1.013 bar, 15\u00b0C and leaving at 8 bar. Compressor is driven by electric motor of 30.84 hp and the mechanical efficiency is 0.87. The clearance volume is 7% of swept volume and the bore is equal to stroke. The compression and expansion follow PV1.3 = constant. Determine (i) free air delivered in m3/min, (ii) volumetric efficiency, and (iii) cylinder dimensions. [4.47 m3/min, 72.68%, L = D = 29.7 cm] 16.28 A reciprocating compressor has two stages with inlet air going into LP stage at 1 bar, 16\u00b0C and at the rate of 12 m3/min. Air is finally delivered at 7 bar and there is perfect intercooling at optimum pressure between the stages. The index for compression is 1.25 and compressor runs at 600 rpm. Neglecting clearance volume determine intermediate pressure, total volume of each cylinder and total work required. [2.645 bar, 0.02 m3, 0.0075 m3, 57.6 hp] 16.29 A two stage reciprocating air compressor delivers 4.2 kg of free air per min at 1.01325 bar and 15\u00b0C. The suction conditions are 0.95 bar, 22\u00b0C. Compressor delivers air at 13 bar. Compression throughout occurs following PV1.25 = C. There is optimum and perfect intercooling between the two stages. Mechanical efficiency is 0.75. Neglecting clearance volume determine (i) (ii) (iii) the heat transfer in intercooler per second. the capacity of electric motor. the % saving in work if two stage intercooling is compared with single stage compressor between same limits. [7.6 kJ/s, 44.65 hp, 13%] 16.30 A single acting reciprocating air compressor has two stages with the optimum and perfect intercooling in between. Compressor has air sucked at 1 bar and at the rate of 2.4 m3/min when measured at 1.013", " bar, 288 K. Compressor delivers air at 70 bar. Temperature at the end of suction stroke is 32\u00b0C. The compression and expansion follows polytropic process PV1.25 = C uniformly. The clearance volume is 3% of swept volume in each HP and LP cylinder. Compressor runs at 750 rpm. If the mechanical efficiency is 0.85 then determine the power of drive required, swept volumes of each cylinder, % saving in power as compared to single stage compression within limits. [35.8 hp, 3963 cm3, 473 cm3, 20.89%] 770 ________________________________________________________ Applied Thermodynamics 17 17.1 INTRODUCTION History of internal combustion engines dates back to year 1680 when Christian Huygens developed an engine using gun powder for explosion inside a cylindrical tube. Subsequently, in year 1860 a non compression engine utilising coal gas was developed by \u2018Lenoir\u2019 in France. Engine was called Lenoir engine and it was operationally similar to steam engine. In 1866 \u2018Otto-Langen free piston engine\u2019 was developed in Germany and it had thermal efficiency more than \u2018Lenoir engine.\u2019 In 1876 four stroke engine based on Otto cycle was developed by Nikolous Otto in Germany which revolutionised the developments of internal combustion engines and are even used till date. The compression ignition engine technology and engines based on it evolved in year 1892. Such engines are credited to Rudolf Diesel another German engineer and are so named as Diesel engine. Two stroke engines came up in 1881 in Scotland and were first developed by Dugald Clerke. \u2018Brayton engine\u2019 based on Brayton cycle and \u2018Atkinson engine\u2019 based on Atkinson cycle were developed in year 1873 and 1885 respectively. Apart from conventional two stroke and four stroke engines described earlier the \u2018Wankel engine\u2019 which was rotary engine and the \u2018Stirling engine\u2019 which was external combustion engine came up in years 1957 and 1938 respectively. In 1923 an engine called as free piston engine was developed by Pateras Pescara in France. Free piston engine is actually a combination of reciprocating piston-cylinder engine and turbine. \u201cEngine refers to a device which transforms one form of energy into the other form\u201d. \u201cHeat engine is a modified form of engine used for transforming chemical energy of fuel into thermal energy and subsequently for producing work\u201d. Based on the mechanism used for adding thermal energy they can be classified into the following: (a) External combustion engine (b) Internal combustion engine. External combustion engine have combustion occuring outside engine and adding heat to the working fluid used in the engine. Thus, in external combustion engines heat released during combustion is indirectly utilized by the working fluid in external combustion engine. Internal combustion engines have combustion occuring in engine itself and heat released during combustion is directly utilized for getting shaft work. Internal combustion engines have numerous advantages over external combustion engines such as lower weight to power output ratio, simplicity, smaller initial cost, and higher efficiency etc. Due to indirect heat transfer the external combustion engines permit for the use of any cheaper fuel such as coal, wood, oil etc. for combustion. Internal combustion engines are exhaustively used in automobiles, gas turbine etc. and external combustion engines are used in steam turbine, steam engine, nuclear power plant etc. Internal combustion Introduction to Internal Combustion Engines ________________________________________ 771 engines are efficient than external combustion engines and also are smaller than external combustion engine of similar capacity. 17.2 CLASSIFICATION OF IC ENGINES Internal combustion engines can be classified on the following basis. (a) Based on number of strokes : Number of strokes involved in a cycle of IC engine can be two strokes or four strokes. Such engine can be; (i) Two stroke engines (ii) Four stroke engines (b) Based on thermodynamic cycle : Depending upon thermodynamic cycle used in the internal combustion engines these can be classified as: (i) Engines based on Otto cycle (\u2018Spark-Ignition engine\u2019) (ii) Engines based on Diesel or Dual cycle (\u2018Compression-Ignition engine\u2019) (c) Based on mechanism of ignition: Internal combustion engines have combustion as the basic process. Combustion process may be initiated using externally assisted ignition (spark ignition) or it may get initiated on its\u2019 own due to excessive compression (compression ignition). Such engines are called: (i) Spark ignition engines (ii) Compression ignition engines. The spark ignition engines may have \u201cmagneto ignition system\u201d or \u201cbattery ignition system\u201d for creating necessary electric potential for producing spark. (d) Based on type of fuel used: IC engines may be classified depending upon the type of fuel being used. These can be: (i) Petrol engines (petrol being used as fuel) (ii) Gas engines (gaseous fuel", " being used) (iii) Diesel engines (diesel being used as fuel) (iv) Multi-fuel engines (more than one fuel being used) (e) Based on fuel admission: IC engines can be of different types depending upon arrangement used for fuel admission: (i) Carburettor type engines (use carburettor fuel metering) (ii) Injection type engines (use fuel injector and injection system) (f) Based on type of cooling: IC engines have inherent requirement of continuous cooling of engine. Based on type of cooling these can be classified as: (i) Air cooled engines (Generally used in small sized engines) (ii) Water cooled engines (Generally used in large sized engines) (g) Based on type of motion: IC engines may have reciprocating motion of piston or it may also have rotary motion. Such engines can be: (i) Reciprocating engines (ii) Rotary engines Reciprocating engines may have different cylinder arrangements such as: (i) Opposed cylinder engines (ii) Inclined cylinder engines (iii) V-shaped cylinder arrangement. Rotary engines may be further classified as single rotor engines or multirotor engines i.e. (i) Single rotor engine (ii) Multi rotor engine 772 ________________________________________________________ Applied Thermodynamics 17.3 IC ENGINE TERMINOLOGY Some of the basic components and generally used terms in internal combustion engines are given as under. 1. Cylinder: It is a cylindrical block having cylindrical space inside for piston to make reciprocating motion. Upper portion of cylinder which covers it from the top is called cylinder head. This is manufactured by casting process and materials used are cast iron or alloy steel. 2. Piston and Piston rings: Piston is a cylindrical part which reciprocates inside the cylinder and is used for doing work and getting work. Piston has piston rings tightly fitted in groove around piston and provide a tight seal so as to prevent leakage across piston and cylinder wall during piston\u2019s reciprocating motion. Pistons are manufactured by casting or forging process. Pistons are made of cast iron, aluminum alloy. Piston rings are made of silicon, cast iron, steel alloy by casting process. 3. Combustion space: It is the space available between the cylinder head and top of piston when piston is at farthest position from crankshaft (TDC). 4. Intake manifold: It is the passage/duct connecting intake system to the inlet valve upon cylinder. Through intake manifold the air/air-fuel mixture goes into cylinder. 5. Exhaust manifold: It is the passage/duct connecting exhaust system to the exhaust valve upon cylinder. Through exhaust manifold burnt gases go out of cylinder. 6. Valves: Engine has both intake and exhaust type of valves which are operated by valve operating mechanism comprising of cam, camshaft, follower, valve rod, rocker arm, valve spring etc. Valves are generally of spring loaded type and made out of special alloy steels by forging process. 7. Spark plug: It is the external ignitor used for initiating combustion process. Spark plug is activated by electrical energy fed by electrical system with engine. It delivers spark with suitable energy to initiate combustion at appropriate time for suitable duration. 8. Bearing: Bearings are required to support crank shaft. Bearings are made of white metal leaded bronze. 9. Connecting rod: It is the member connecting piston and crankshaft. It has generally I section and is made of steel by forging process. 10. Crank: It is the rigid member connecting the crankshaft and connecting rod. Crank is mounted on crankshaft. Crank transfers motion from connecting rod to crankshaft as it is linked to connecting rod through crank pin. 11. Crankshaft: It is the shaft at which useful positive work is available from the piston-cylinder arrangement. Reciprocating motion of piston gets converted into rotary motion of crankshaft. Crankshaft are manufactured by forging process from alloy steel. 12. Crankcase: Crankcase actually acts like a sump housing crank, crankshaft, connecting rod and is attached to cylinder. These are made of aluminium alloy, steel, cast iron etc. by casting process. 13. Gudgeon pin: It is the pin joining small end of the connecting rod and piston. This is made of steel by forging process. 14. Cams and Camshafts: Cams are mounted upon camshaft for opening and closing the valves at righ timings and for correct duration. Camshaft gets motion from crankshaft through timing gears. 15. Carburettor: Carburettor is device to prepare the air fuel mixture in right proportion and supply at right time. 16. Bore: It is nominal inner diameter of the cylinder. 17. Piston area: It is the area of a circle of diameter", " equal to bore. 18. Stroke: It is the nominal distance travelled by the piston between two extreme positions in the cylinder. Introduction to Internal Combustion Engines ________________________________________ 773 Fig. 17.1 Internal combustion engine 19. Dead centre: It refers to the extreme end positions inside the cylinder at which piston reverses it\u2019s motion. Thus, there are two dead centres in cylinder, called as \u2018top dead centre\u2019 or \u2018inner dead centre\u2019 and \u2018bottom dead centre\u2019 or \u2018outer dead centre\u2019. Top dead centre (TDC) is the farthest position of piston from crankshaft. It is also called inner dead centre (IDC). Bottom dead centre (BDC) refers to the closed position of piston from crankshaft. It is also called outer dead center (ODC). 20. Swept volume : It is the volume swept by piston while travelling from one dead centre to the other. It may also be called stroke volume or displacement volume. Mathematically, Swept volume = Piston area \u00d7 Stroke 21. Clearance volume: It is the volume space above the piston inside cylinder, when piston is at top dead centre. It is provided for cushioning considerations and depends, largely upon compression ratio. 22. Compression ratio: It is the ratio of the total cylinder volume when piston is at BDC to the clearance volume. Compression ratio = Swept volume + Clearance volume Clearance volume 17.4 4-STROKE SI ENGINE Spark ignition (SI) engines employ external ignition system for initiating the combustion process. Spark plug is the most commonly used ignition method used in spark ignition engines. Let us first understand 774 ________________________________________________________ Applied Thermodynamics working of 4-stroke SI engine. Figure 17.3 shows the different strokes and processes associated with them. Valve position Action Work Heat State 1 1\u20132 2 2\u20133 3 3\u20134 4 4\u20135 5 Inlet Just opened Open Just closed Closed Closed Closed Closed Closed Closed 5\u20136 Closed 6 Closed 6\u20131 Closed Exhaust Just closed Closed Closed Closed Closed Closed Closed Closed Just opened Open Open Open Air-fuel intake Compression Ignition Combustion Max. T, P Expansion Exhaust blowdown Pressure drop to atmosphere Exhaust Air to piston Piston to air Gas to piston None None Piston to gas None None Input to gas None Gas to atmosphere Gas to atmosphere Gas to atmosphere Fig. 17.2 P-V and T-s diagram for SI engine cycle (Otto cycle) Let us start with piston at TDC, state 1. As piston moves for TDC to BDC, the inlet vlave gets opened and fresh air-fuel mixture prepared in carburettor enters the cylinder. This supply of air-fuel mixture into cylinder is called suction process or suction stroke during which inlet valve is open while exhaust valve remains closed. After the piston reaches BDC, it reverses its motion and moves towards TDC. During this piston travel both inlet and exhaust valves remain closed. Thus, the air-fuel mixture inside cylinder gets compressed till piston reaches TDC. This is the second stroke and called compression stroke or compression process (2\u20133). Introduction to Internal Combustion Engines ________________________________________ 775 IVC : Inlet valve closed, IVO: Inlet valve open, EVC: Exhaust valve closed, EVO: Exhaust valve open. Fig. 17.3 Working of 4-stroke S.I. engine Now highly compressed air-fuel mixture is available inside the cylinder and ready for combustion. With piston at TDC, the spark plug is activated and it releases spark for igniting air-fuel mixture. Spark plug used in the SI engines gets activated by suitable mechanism in the engine and provides suitable amount of energy in the form of spark for initiating combustion process. This burning of mixture is accompanied by sudden increase in pressure and temperature while piston passes the TDC position. Here constant volume heat addition takes place and state 4 is attained. Due to this release of fuel energy, the combustion products try to expand and piston moves from 4\u20135 i.e. TDC to BDC. During this travel the inlet and exhaust valves remain closed. This stroke is called expansion stroke or power stroke or expansion process. This is the stroke accompanied by positive work available at shaft. Now while piston is at BDC the exhaust valve gets opened and combustion products are exhausted out during 5\u20136. Cylinder is further emptied and made ready for being recharged while piston travels from BDC to TDC, 6\u20131 pushing out burnt gases. This is called exhaust stroke. Here we have seen that out of suction, compression, expansion and exhaust strokes only expansion stroke is accompanied by the production of positive work, rest three strokes are work absorbing strokes. Work requirement for the three strokes is met from the work available during expansion stroke. For 776 ________________________________________________________ Applied Thermodynamics storing", " the excess energy and releasing it when required, a flywheel is mounted over the crankshsaft. Cycle gets completed in two revolutions of crankshaft. Examples of four stroke engines are petrol engines used in cars etc. 17.5 2-STROKE SI ENGINE 2-stroke SI engine is a modified form of 4-stroke engine where all the four processes required for completion of one cycle of SI engine get completed in two strokes. Thus, obviously in each stroke two processes get completed. Figure 17.4 shows the line diagram of 2-stroke SI engine. General arrangement shows that here there are no valves as in case of 4-stroke engines, instead it has exhaust and suction ports. Piston has a projection on its top, which acts like deflector. Mixture of air-fuel goes into crank case first and then gets transferred to top of piston at appropriate time. Let us start piston movement from TDC to BDC. When piston reverses its motion from BDC to TDC then the suction port gets uncovered and fresh mixture enters and goes into crank case. With piston moving from TDC to BDC and during covered position of suction port the mixture gets transferred to the top of piston through transfer port. Upon reversal during piston travelling from BDC to TDC, the air fuel mixture on top of piston gets compressed and subsequently gets ignited by spark from spark plug. The combustion of fuel-air mixture results in release of excessive energy which forces piston to move from TDC to BDC. Simultaneously as piston uncovers exhaust port the burnt gases go out through exhaust port. Again when piston reaches BDC it reverses its motion and during travel from BDC to TDC the suction takes place as explained above at the bottom of piston while compression of fuel air mixture takes place on top of piston. Thus, suction and compression, both processes get completed during travel of piston from BDC to TDC. Expansion and exhaust processes occur during travel of piston from TDC to BDC along with transfer of fresh fuel air mixture from crankcase to top of piston. Fig. 17.4 2-stroke SI engine Here all four processes occur during two strokes and one revolution of crank shaft. Thermodynamic cycle followed by 2-stroke SI engine is Otto cycle. Scooter engines are generally two stroke engines. 2-stroke SI engines are used for smaller applications. 17.6 4-STROKE CI ENGINE Compression ignition (CI) engines operate generally on \u201cDiesel\u201d/\u201cDual\u201d cycle. In these engines the combustion is realized due to excessive compression and are so called compression ignition engines. Here air alone is sucked inside the cylinder during suction stoke and compressed. Degree of compression is much more than that of spark ignition (SI) engines. After compression of air the fuel is injected into the high pressure and high temperature compressed air. Due to high temperature of air the combustion of fuel gets set on its\u2019 own. Self ignition of fuel takes place due to temperature of air-fuel mixture being higher than self ignition temperature of fuel. Thus in CI engines, larger amount of compression causes high temperature, therefore unassisted combustion. Introduction to Internal Combustion Engines ________________________________________ 777 Schematic of 4-stroke CI engine is quite similar to that of 4-stroke SI engine with the only major difference that spark plug is replaced by fuel injector for injecting fuel at high pressure into compressed air. 4-Stroke CI engine works with following four processes getting completed in separate strokes. General arrangement in CI engine is similar to that of SI engine with spark plug replaced by fuel injector. Stroke 1: Piston travels from TDC to BDC and air is sucked. Stroke 2: Piston travels from BDC to TDC, while air is compressed with inlet and exit passages closed. Stroke 3: Piston reaches TDC and air gets compressed. Fuel injector injects fuel into compressed air for certain duration. Ignition of fuel also takes place simultaneously as air temperature is much higher than self ignition temperature of fuel. Burning of fuel results in release of fuel chemical energy, which forces piston to travel from TDC to BDC. Contrary to SI engine where heat addition gets completed near instantaneously, in CI engines fuel injection and thus heat addition is spread in certain stroke travel of piston i.e. heat addition takes place at constant pressure during which piston travels certain stroke length as decided by cut-off ratio. This is expansion process and piston comes down to BDC with both inlet and exit valves closed. Stroke 4: After expansion piston reverses its motion upon reaching BDC and travels up to TDC with exit passage open. During this piston travel burnt gases are expelled out of cylinder i.e. exhaust stroke. Completion of above four stroke requires two revolutions of crankshaft. 17.7 2-STROKE CI ENGINE Fig. 17.5 shows the general arrangement in 2-stroke CI engine.", " Here the structure of CI engine is very much similar to that of SI engine with the major difference that spark plug is replaced by fuel injector, structure is made more sturdy to withstand high compression ratio. Working of 2-stroke compression ignition engine is shown in (a), (b), (c), (d) explaining the suction, compression, expansion and exhaust processes. During piston travel from BDC to TDC air enters crankcase. When piston reaches TDC and reverses its motion to BDC air in crankcase gets partly compressed and is transferred from crankcase to top of piston through transfer port. Upon reversal of piston motion from BDC to TDC the compression of air occurs by the top side of piston while on the bottom side of piston air again enters into crankcase. Upon piston reaching TDC fuel is injected into compressed air which is at high temperature and pressure. As fuel is injected into compressed air the fuel ignition gets set on its own due to temperature being more than self ignition temperature of fuel, i.e. compression ignition. Fuel injection is continued for some duration along with its ignition which causes release of excessive fuel energy. This energy release causes piston to go back from TDC to BDC, i.e. the expansion process as shown in (c). As piston reaches BDC it simultaneously forces air in crank case to get transferred to cylinder space and forces burnt gases out of cylinder i.e. exhaust process. Here also cycle gets completed in single revolution of crankshaft i.e two processes occurring simultaneously in each stroke. 778 ________________________________________________________ Applied Thermodynamics Fig. 17.5 2-stroke CI engine 17.8 THERMODYNAMIC CYCLES IN IC ENGINES Thermodynamic cycles used in IC engines are Otto cycle, Diesel cycle and Dual cycle and are described in Chapter 9 under title of \u2018Air standard cycles\u2019 However, the comparison of these cycles with one another for different reference conditions is presented here. (a) Comparison of Otto, Diesel and Dual cycles for same compression ratio and heat input: Figure 17.6 shows the three cycles on P-V and T-s diagram for same compression ratio and heat input. For same heat input, T-s diagram shows; Area bcef = Area bc\" e\" f = Area bb' c'e'f (Dual) (Diesel) (Otto) Introduction to Internal Combustion Engines ________________________________________ 779 Fig. 17.6 Comparison of Otto Diesel and Dual cycles on P-V and T-s diagram for same compression ratio and heat input. Comparison can be made by looking at these cycle representations. Heat rejected can be given as; For Otto = Area a d e f For Diesel = Area a d\"e\" f For Dual = Area a d' e' f A look at areas on T-s diagram shows Area a d e f < Area a d'e'f < Area a d\" e\" f. Since heat input is same therefore based on the heat rejected one can say that as heat rejected for Otto cycle is smallest so this has highest efficiency. It is followed by Dual cycle and smallest efficiency is for Diesel cycle. (b) Comparison of Otto, Diesel and Dual cycles for same maximum pressure and same heat input: For carrying out comparison the P-V and T-s representations for three cycles are shown in Fig. 17.7. Fig. 17.7 Comparison of Otto, diesel and Dual cycles on P-V and T-s diagram for same maximum pressure and same heat input. 780 ________________________________________________________ Applied Thermodynamics Here heat input is same so, Area bcfg = Area b' c' f' g = Area b\" c\" d\" f\" g. (Otto) (Diesel) (Dual) Heat rejected shall be given by; Area adfg for Otto cycle Area ad' f' g for Diesel cycle Area ae\" f\" g for Dual cycle Area adfg > Area ae\" f\" g > Area ad' f' g \u21d2 WDiesel > WDual > WOtto Here W refers to net work in respective cycle. Hence, it can be seen that heat rejected is highest in Otto cycle followed by Dual cycle and Diesel cycle. Thus, efficiency shall be maximum for Diesel cycle due to largest work available followed by Dual cycle and minimum for Otto cycle at same maximum pressure and same heat input. 17.9 INDICATOR DIAGRAM AND POWER MEASUREMENT Internal combustion engines have combustion taking place inside and power is available at crankshaft. The shaft work available is less than the total energy released inside the cylinder due to frictional and other losses. For performance evaluation of internal combustion engine one is interested in following different powers. (a) Indicated power: It refers to the power available inside the cylinder i.e. the power provided to piston. Mathematically: Indicated power = (Energy in fuel) \u2013 (Energy loss in exhaust, coolant,", " radiation etc.) It is measured from the indicator diagram which is obtained using indicator mechanism. (b) Brake power: It refers to the power available at crankshaft i.e. it is the useful shaft work. Mathematically: Brake power = (Indicated power) \u2013 (Energy loss in friction, pumping and unaccounted losses etc.) Brake power is usually measured by absorption or transmission type dynamometers. It can be given as: Brake power = NT\u03c0 2 60 Watt where {N is speed of rotation of shaft in rpm, T is torque, N. m.} (c) Friction power: It refers to the power lost due to friction and other reasons. It is quantified by the difference between indicated power and brake power. Friction power = Indicated power \u2013 Brake power Indicator diagram: Indicator diagram is the graphical description of pressure and volume variations occurring inside cylinder. An indicator diagram for a four-stroke internal combustion engine is shown in Fig. 17.8 along with the four different processes. Introduction to Internal Combustion Engines ________________________________________ 781 (a) Four processes in 4-stroke IC engine (Otto cycle) (b) Indicator diagram of 4-stroke IC engine (Otto cycle). Fig 17.8 Indicator diagram and processes in IC engine For getting the indicator diagram, \u201cindicator diagram mechanism\u201d is put over the IC engine and real variation of pressure and volume states obtained. Indicator mechanism shown in Fig. 17.9, has number of links \u2018g\u2019 and chord for transferring linear displacement of piston rod into the rotation of drum \u2018f \u2019. A stylus \u2018d \u2019 is operated over the drum through a parallel motion link \u2018e\u2019. Displacement of \u2018e\u2019 is caused along y-axis by an indicator piston rod \u2018h\u2019, which is connected to indicator piston \u2018b\u2019 put in indicator cylinder \u2018a\u2019 as shown. Indicator piston is resisted by an indicator spring \u2018c\u2019. In between indicator cylinder and engine cylinder a cock is provided to isolate it after indicator diagram is obtained. Indicator mechanism has thus drum \u2018f \u2019 rotating about its axis and stylus \u2018d\u2019 operated by pressure variation inside cylinder along y-axis. Thus, displacement along volume axis is made available by rotation of drum and along pressure axis it is made available by vertical reciprocatory movement of stylus depending upon pressure inside engine cylinder. 782 ________________________________________________________ Applied Thermodynamics Fig. 17.9 Indicator diagram mechanism For getting indicated power we need mean effective pressure (average pressure) and volume displaced. From the indicator diagram obtained the mean effective pressure can be obtained, knowing area of diagram, length of diagram and indicator spring constant. Area of diagram can be estimated using planimeter and length of diagram measured. Mean effective pressure, mep = A d L d where Ad is area enclosed in diagram \u00d7 C Ld is length of diagram C is indicator spring constant given as N/cm2 \u00d7 cm travel. Using mean effective pressure the indicated power can be obtained by product of pressure and volume displaced. Indicated power for 2-stroke engine, = \u00d7 \u00d7 \u00d7 mep A L N n 60 \u00d7 Watt where mep = mean effective pressure in Pa A = cross-sectional area of cylinder. A = \u03c0 4 D2, where D is bore. L = length of stroke. N = rpm n = no. of cylinders. Introduction to Internal Combustion Engines ________________________________________ 783 In case of 4-stroke engines, 2 revolutions are required for completing one cycle, therefore effective rpm shall be N/2. Indicated power for 4-stroke engine = \u00d7 \u00d7 \u00d7 mep A L N ( / 2) \u00d7 n 60 Watt Based on indicated power and brake power, the mechanical efficiency of engine can be obtained as below. Mechanical efficiency of engine \u03b7 mech = Brake power Indicated power Mechanical efficiency is thus indicator of how efficiently indicated power is converted into brake power. 17.10 COMBUSTION IN SI ENGINE Combustion is the process of oxidation of fuel resulting into the release of energy equivalent to calorific value of fuel. Energy released in combustion is in the form of heat. Combustion process in spark ignition engine has requirement of the \u2018mixture of fuel and air in right proportion,\u2019 \u2018mechanism for initiation of combustion process\u2019 and \u2018stabilization and propagation of flame for complete burning.\u2019 For complete combustion of every fuel there is chemically correct fuel-air ratio also called stoichiometric fuel-air ratio. This fuel air ratio may be rich or lean depending upon the proportion of fuel and air present in mixture. In SI engine this fuel air ratio generally varies between 1 : 7 to 1 : 30 with lean mixture at 1 : 30 and rich mixture at 1 : 7. Stoichiometric fuel-air ratio is around 1", " : 14 to 1 : 15 for hydrocarbon fuel. The extreme values of fuel-air ratio permissible in SI engine on both rich and lean ends put limits as \u2018lower ignition limit\u2019 and \u2018upper ignition limit\u2019. Varying fuel-air ratio is required in SI engine due to varying loads on engine between no load to full load on engine. The ratio of actual fuel-air ratio to stoichiometic fuel-air ratio is given by \u2018equivalence ratio\u2019 or \u2018relative fuel-air ratio\u2019. Appropriate fuel-air ratio is maintained in SI engines through \u2018carburettor\u2019 (the fuel metering system). Combustion in SI engine may be described to be occurring in following significant phase: (i) After compression of fuel-air mixture in cylinder the high temperature spark is delivered by spark plug in the compressed fuel-air mixture. Temperature at the tip of spark plug electrode may go even more than 10,000\u00baC at the time of release of spark. Sparkles released have sufficiently high temperature to initiate the combustion of fuel. For complete combustion of fuel mere initiation of combustion does not serve the purpose instead a sustainable combustion process is required. After setting up of combustion a sustainable flame front or flame nuclei is needed so that it proceeds across the combustion space to ensure complete combustion. Thus, this phase in which spark is first released followed by setting up of sustainable flame front is called \u201cpreparation phase\u201d and may consume around 10\u00ba of crank angle rotation. Crank angle rotation consumed in \u201cpreparation phase\u201d depends upon the speed of engine, constructional feature of cylinder, piston, location of spark plug, strength of spark, characteristics of fuel, fuel-air ratio etc. Designer always wishes to complete combustion in smallest possible time as, theoretically Otto cycle requires combustion to occur instantaneously. Therefore, this first phase termed as preparation phase should be as small as possible. Figure 17.10 shows different phases of combustion. Preparation phase is shown to occur from \u2018a\u2019 to \u2018b\u2019 with small or negligible pressure rise as initially rate of burning is very small. (ii) After sustainable combustion flame is set up then the flame nuclei get scattered due to excessive turbulence in combustion space causing pressure to rise from \u2018b\u2019 to \u2018c\u2019. This phase of combustion 784 ________________________________________________________ Applied Thermodynamics I = Preparation phase II = Flame propagation phase III = After burning phase Non-firing : Piston is made to reciprocate due to cranking in the absence of combustion. Engine runs with combustion occurring. Firing : Fig. 17.10 Phases of combustion in SI engine depends upon the turbulence inside cylinder, strength of compbustion nuclei, fuel-air ratio, strength of spark, cylinder geometry, fuel properties etc. This phase of combustion is called as \u201cflame propagation phase\u201d and is accompanied by the excessive pressure rise. Flame propagation phase should also be as small as possible. (iii) After the maximum amount of fuel-air mixture is burnt, the residual gets burnt after the piston has moved across the TDC. This last phase is termed as \u201cafter burning phase\u201d and occurs during the expansion stroke. Hence, it can be summarised that the complete combustion in SI engine occurs in three distinct zones i.e. preparation phase, flame propagation phase and after burning phase. In order to complete combustion in smallest possible time the flame propagation phase and preparation phase should be shortened. It is seen that out of total distance travelled in combustion space in first phase i.e. preparation phase about 10% of combustion space length is covered in about 20\u201330% of total time for combustion. Flame propagation phase is spread in about 80% of combustion space length and is covered in 60\u201370% of total time of combustion. \u2018After burning\u2019 occurs in less than 10% of combustion space in less than 10% of total combustion time as evident from Fig. 17.11. Combustion may also sometimes occur abnormally. \u201cAbnormal combustion\u201d is said to occur when combustion begins inside the cylinder on its\u2019 own before the stipulated time for it. This abnormal combustion may be due to preignition (i.e. ignition of fuel even before spark plug ignites it) and results in uncontrolled pressure rise. Abnormal combustion is also termed as detonation or knocking and can be felt by jerky operation of engine, excessive noise, reduced power output etc. Fig. 17.11 Duration of combustion phases Introduction to Internal Combustion Engines ________________________________________ 785 17.11 COMBUSTION IN CI ENGINES Combustion in CI engines differ from SI engine due to the basic fact that CI engine combustion is unassisted combustion occurring on its\u2019 own. In CI engine the fuel is injected into combustion space after the compression of air is completed. Due to excessively high temperature and pressure of air", " the fuel when injected in atomised form gets burnt on its\u2019 own and burning of fuel is continued till the fuel is injected. Theoretically this injection of fuel and its\u2019 burning should occur simultaneously upto the cutoff point, but this does not occur in actual CI engine. Different significant phases of combustion are explained as under. (i) Injection of fuel in atomized form is initiated into the combustion space containing compressed air. Fuel upon injection does not get burnt immediately instead some time is required for preparation before start of combustion. Fuel droplet injected into high temperature air first gets transformed into vapour (gaseous form) and then gets enveloped around by suitable amount of oxygen present so as to form combustible mixture. Subsequently, if temperature inside is greater than self ignition temperature at respective pressure then ignition gets set. Thus, the delay in start of ignition may be said to occur due to \u2018physical delay\u2019 i.e. time consumed in transformation from liquid droplet into gaseous form, and \u2018chemical delay\u2019 i.e. time consumed in preparation for setting up of chemical reaction (combustion). The duration of ignition delay depends upon fuel characteristic, compression ratio (i.e. pressure and temperature after compression), fuel injection, ambient air temperature, speed of engine, and geometry of combustion chamber etc. Ignition delay is inevitable stage and in order to accommodate it, the fuel injection is advanced by about 20\u00ba before TDC. Ignition delay is shown by a \u2013 b in Fig. 17.12, showing pressure rise during combustion. Fuel injection begins at \u2018a\u2019 and ignition begins at \u2018b\u2019. Theoretically, this ignition delay should be as small as possible. I = Ignition delay II = Uncontrolled combustion III = Controlled combustion IV = After burning Fig. 17.12 Pressure rise during combustion in CI engines (ii) During the ignition delay period also the injection of fuel is continued as it has begun at point \u2018a\u2019 and shall continue upto the point of cut-off. For the duration in which preparation for ignition is made, the continuous fuel injection results in accumulation of fuel in combustion space. The moment when ignition just begins, if the sustainable flame front is established then this accumulated fuel also gets burnt rapidly. This burning of accumulated fuel occurs in such a manner that combustion process becomes uncontrolled resulting into steep pressure rise as shown from \u2018b\u2019 to \u2018c\u2019. The uncontrolled burning 786 ________________________________________________________ Applied Thermodynamics continues till the collected fuel gets burnt. During this \u2018uncontrolled combustion\u2019 phase if the pressure rise is very abrupt then combustion is termed as \u2018abnormal combustion\u2019 and may even lead to damage of engine parts in extreme conditions. Thus, it is obvious that \u2018uncontrolled combustion\u2019 depends upon the \u2018ignition delay\u2019 period as during ignition delay itself the accumulation of unburnt fuel occurs and its\u2019 burning results in steep pressure rise. Hence in order to have minimum uncontrolled combustion the ignition delay should be as small as possible. During this uncontrolled combustion phase about one-third of total fuel heat is released. (iii) After the \u2018uncontrolled combustion\u2019 is over then the rate of burning matches with rate of fuel injection and the combustion is termed as \u2018controlled combustion\u2019. Controlled combustion is shown between \u2018c\u2019 to \u2018d\u2019 and during this phase maximum of heat gets evolved in controlled manner. In controlled combustion phase rate of combustion can be directly regulated by the rate of fuel injection i.e. through fuel injector. Controlled combustion phase has smooth pressure variation and maximum temperature is attained during this period. It is seen that about two-third of total fuel heat is released during this phase. (iv) After controlled combustion, the residual if any gets burnt and the combustion is termed as \u2018after burning\u2019. This after burning may be there due to fuel particles residing in remote position in combustion space where flame front could not reach. \u2018After burning\u2019 is spread over 60 \u2013 70\u00ba of crank angle rotation and occurs even during expansion stroke. Thus, it is seen that the complete combustion in CI engines may be comprising of four distinct phases i.e. \u2018ignition delay\u2019 followed by \u2018uncontrolled combustion,\u2019 \u2018controlled combustion\u2019 and \u2018after burning\u2019. Combustion generally becomes abnormal combustion in CI engines when the ignition delay is too large resulting into large uncontrolled combustion and zig-zag pressure rise. Abnormal combustion in CI engines may also be termed as \u2018knocking\u2019 in engines and can be felt by excessive vibrations, excessive noise, excessive heat release, pitting of cylinder head and piston head etc. In order to control the knocking some additives are put in CI engine fuel so as to reduce its\u2019 self ignition temperature and accelerate ignition process. Also, the combustion chambers are properly designed so as to have", " reduced physical and chemical delay. 17.12 IC ENGINE FUELS Fuels used in SI engines and CI engines are different as the nature of combustion process is different in the two engines. Normally hydrocarbon fuels are used for both the applications and should possess desirable properties such as high calorific value, suitable combustion characteristics, ease of handling, environment friendly etc. In SI engines the fuel used is generally \u2018gasoline\u2019 also called as petrol and is mixture of different hydrocarbons. Gasoline is available from refining process and its\u2019 exact composition depends upon the degree of refining and source etc. Gasoline should have capability of mixing rapidly with air, suitable volatility, resistance to abnormal combustion etc. SI engine fuels are characterised in respect to its\u2019 resistance to abnormal combustion by its\u2019 rating in terms of \u201cOctane number\u201d. Octane number of fuel is determined by comparing the combustion performance of actual fuel with that of reference fuel. Octane number is defined as the percentage by volume of iso-octane in a mixture of iso-octane (C8H18) and n-heptane (C7H16). Iso octane is low boiling point fuel having very high resistance to abnormal combustion and is arbitrarily assigned octane number of 100. \u2018n-heptane\u2019 is seen to have very poor resistance to abnormal combustion and is so assigned octane number of 0. These \u2018iso-octane\u2019 and \u2018n-heptane\u2019 are also called as \u2018primary reference fuels\u2019. In order to determine octane number of any fuel its combustion characteristics are matched with those of a variable mixture of \u2018iso-octane\u2019 and \u2018n-heptane\u2019. The composition of iso-octane fraction and n-heptane fraction which yields the similar combustion characteristics is used to know the octane number. Say, for example in a particular test the mixture of 80% by volume of iso-octane and 20% by volume of n- Introduction to Internal Combustion Engines ________________________________________ 787 heptane offers the combustion similar to that of test fuel. Then the test fuel shall have octane number of 80. Octane number generally lies between 0 and 100 but can be extended beyond 100 by using certain additives. \u2018Tetra ethyl lead\u2019 is the most popular additive used in SI engines which further increases resistance to abnormal combustion i.e. even more than that of iso-octane. In CI engines the Diesel fuel is generally used which is also available during pertroleum refining but is impure as compared to SI engine fuel. CI engine fuels are also characterized by a rating given in terms of \u2018Cetane number\u2019 which also indicates the resistance of fuel to knocking. Cetane number for any fuel is given by percentage by volume of cetane (C16 H34) in a mixture of cetane and \u03b1-methyl napthalene (C10H7CH3) which offers the combustion characteristics similar to that of test fuel. Cetane is arbitrarily assigned cetane number of 100 as it offers maximum knock resistance and \u03b1methyl napthalene is assigned cetane number of 0 as it has mimimum knock resistance. Cetane rating is also obtained by the test similar to that for octane rating. 17.13 MORSE TEST In IC engines there are three powers namely \u2018indicated power\u2019 which is developed inside cylinder, \u2018brake power\u2019 which is available at crank shaft and \u2018friction power\u2019 which is lost in overcoming friction and other losses. Brake power can be measured by using dynamometer at the crank shaft of engine. Friction power can be experimentally determined by number of methods namely \u2018Willan\u2019s line method\u2019, \u2018Motoring test\u2019 and by the numerical difference between indicated power and brake power. Morse test is experimental method for estimating the friction power of the multi cylinder engine. Here engine is first run upto certain speed of rotation. Subsequently, one by one each cylinder of the engine is isolated from contributing to shaft work either by switching off electric supply to spark plug or by stopping fuel supply. When one cylinder is not producing power i.e. due to its isolation then also piston reciprocates in this cylinder similar to other cylinders thus the frictional losses remain constant for the shaft running at same speed. When second cylinder is cut then also frictional losses remain same but the brake power and indicated power get reduced accordingly. This process of cutting off each cylinder one by one is carried out till last cylinder and the brake power is measured by dynamometer attached to crank shaft which is maintained to run at same speed of rotation when cylinders are cut off one by one. Mathematically, if there are n number of cylinders in a multicylinder engine. Then initially when all cylinders are working. Indicated power from n cylinders, (ip)", "n = (bp)n + (fp)n; here ip, bp and fp refer to indicated power, brake power and friction power respectively. When one cylinder is cut then, When second cylinder is cut then, (ip)n \u20131 = (bp)n \u2013 1 + (fp)n (ip)n \u20132 = (bp)n \u2013 2 + (fp)n From above the power produced from first cylinder being cut off can be obtained as: (ip)nth cylinder = (bp)n \u2013 (bp)n \u2013 1 Total indicated power from engine: (ip)n = \u2211 (ip)nth cylinder Thus, after knowing indicated power of engine the friction power can be determined as under, because brake power is also known through dynamometer. (fp)n = (ip)n \u2013 (bp)n Friction power estimated using Morse test gives fairly correct estimate. Although, there may be slight deviation because when each cylinder is cut then there is substantial increase in engine vibrations and the friction losses may change due to change in back pressure etc. 788 ________________________________________________________ Applied Thermodynamics 17.14 COMPARATIVE STUDY OF IC ENGINES Comparative study of IC engines with other prime movers is described here. (a) Difference between SI and CI engine SI engine CI engine It works on Otto cycle 1. 2. Compression ratio is from 5 to 15. 3. Petrol (Gasoline) is used as fuel. 4. Fuel should have high self ignition temperature. 5. A mixture of air and fuel is sucked inside the cylinder during suction process. 6. Carburettor is required for preparing air-fuel mixture. 2. It works on Diesel/Dual cycle 2. Compression ratio is from 12 to 30. 3. Diesel is used as fuel. 4. Fuel should have low self ignition temperature. 5. Air alone is sucked during suction process. 6. Carburettor is not required. 7. Fuel goes into cylinder alongwith air 7. Fuel is injected after compression. For during suction. injection of fuel a fuel pump and injector is required 8. Spark plug is required to initiate 8. Ignition of fuel takes place on its\u2019 own due to combustion. 9. Combustion takes place isochorically. 10. Due to low compression ratio the engine structure is light. 11. Engine speed is higher. 12. Efficiency is low. 13. Generally used in 2-wheeler automobiles high temperature of air inside cylinder. 9. Combustion takes place isobarically. 10. To withstand high Compression ratio the engine structure is sturdy. 11. Engine speed is lower. 12. Efficiency is high. 13. Generally used in 4 wheelers and bigger and smaller and lighter engines. engines. (b) Comparison of 2-stroke engine with 4-stroke engine 2-stroke engine 4-stroke engine 1. One cycle is completed in one revolution 1. One cycle is completed in two revolution of of crankshaft. crankshaft. 2. Power stroke is available in each 2. Power stroke is available in alternate revolution of crankshaft. 3. Thermal efficiency is low. 4. Volumetric efficiency is less due to less time available for suction. revolution of crankshaft. 3. Thermal efficiency is high. 4. Volumetric efficiency is high due to large time available for suction. 5. Lighter flywheel is required due to more 5. Heavier flywheel is required due to less uniform torque available. uniform torque available. 6. Engine is compact and light in weight. 7. Cooling requirements are less. 8. Valves are not required. 6. Engine is heavy. 7. Cooling requirements are high. 8. Complex valve operating mechanism is 9. Cost is low. 10. Used in light vehicles and small engines. required. 9. Cost is high. 10. Used in heavy duty vehicles and bigger engines. Introduction to Internal Combustion Engines ________________________________________ 789 (c) Comparison of IC engines with gas turbine IC engines Gas turbine 1. 2. IC engine have good part load efficiency. These have internal combustion. IC engines are well suited for smaller applications such as automobiles. IC engines can be used for small capacity power plants. 1. Gas turbines have poor part load efficiency. These may have external combustion or internal combustion both as per requirement. 2. Gas turbines are not well suited for smaller applications, specially automobiles. These are well suited for aviation applications and land applications in power plants etc. as prime movers. It has small specific weight and small size capability which make these suitable for aeroplanes. 3. Balancing is poor due to reciprocating 3. Balancing is good due to absence of parts. Mechanical efficiency is also poor. 4. Maintenance requirement is high. 5. Type of fuels which can be used is limited in number. reciprocating parts. Mechanical efficiency is good due to rotary parts and", " no sliding members 4. Poor maintenance requirement. 5. Large number of different types of fuels can be used. 6. Operation is noisy and large vibrations 6. Operation is silent and smooth. are present. 7. No modification is possible for improving efficiency and output in existing IC engines. 7. Modifications such as reheating, regeneration and intercooling etc. are possible in order to improve efficiency and output. (d) Comparison of gas turbine with steam turbines Gas turbine Steam turbines 1. These are generally internal combustion type, but may be of external combustion type too. 1. These are of external combustion type. 2. High quality fuel is being used. 2. Any type of poor or good quality fuel can be 3. Thermal efficiency is better. 4. It does not require boiler, feed water supply, condensing plant etc. Number of components involved is small. utilized. 3. Thermal efficiency is poor than gas turbines. 4. It requires boiler, feed water supply, feed pump, condensing plant etc. instead of compressor, combustion chamber as in gas turbines. Number of components involved is large. 5. These can be started quickly. 5. These take time to start because of time 6. Maintenance cost is small. 7. These have smaller specific weight and size i.e. small weight per kW output and so these can be used in aviation applications. Initial cost and running cost are small. 8. required in generating steam. 6. Maintenance cost is high. 7. These have large specific weight and size and are exclusively suited for land applications. 8. Initial cost and running cost are large in steam turbines. 790 ________________________________________________________ Applied Thermodynamics (e) Comparison between Diesel engine, Gas turbine and Steam turbine based plants: (Parametric form) Parameter Diesel engine Steam turbine plant Gas turbine 1. Power generation 10 kW to 10 MW 1 MW to 1000 MW 100 kW to 100 MW capacity 2. Efficiency 3. Fuel 0.35 to 0.40 High grade fuel, liquid Medium to low grade High to medium grade, fuel of diesel grade 0.30 to 0.40 0.20 to 0.40 fuel, may be gaseous, may be liquid or gas liquid or solid fuel. Large $US 140\u20131200 per kW Negligible fuel. Small $US 80\u2013300 per kW Oil topping up. 4. Size and weight 5. Capital cost 6. Lubrication requirement Medium $US 110\u2013180 US$ per kW Frequent oil and filter changes 7. Maintenance period Weekly and monthly 8. Specific installation Very heavy foundations, Coal handling facility, Daily Monthly and annual Intake and exhaust needs vibration isolation, enclosures for making its operation quiet etc. water, exhaust Ash handling facility, mufflers etc. enough supplies of 9. Plant life 10\u201312 years emission control etc. 25 to 35 years 15 to 30 years EXAMPLES 1. A four cylinder diesel engine of 4-stroke type has stroke to bore ratio as 1.2 and the cylinder diameter is 12 cm. Estimate indicated power of the engine using the indicator diagram arrangement. Indicator card shows the diagram having area of 30 cm2 and length as half of stroke. Indicator spring constant is 20 \u00d7 103 kN/m2 and engine is running at 2000 rpm. Also find out mechanical efficiency of engine if 10% of power is lost in friction and other losses. Solution: From stroke to bore ratio i.e. L D = 1.2 and cylinder diameter = bore, i.e. D = 12 cm, so L = 14.4 cm Area of indicator diagram = 30 cm2 = 30 \u00d7 10\u20134 m2 Length of indicator diagram = 1 2 \u00d7 stroke = 7.2 cm or 7.2 \u00d7 10\u20132 m Mean effective pressure = \u00d7 30 10 \u2212 4 \u00d7 \u00d7 20 10 \u2212 \u00d7 2 7.2 10 3 \u00d7 3 10 m.e.p. = 8333.333 \u00d7 102 N/m2 Cross-sectional area of piston = \u03c0 4 2 D = 0.01131 m2 Introduction to Internal Combustion Engines ________________________________________ 791 For one cylinder indicated power 8333.333 10 \u00d7 2 \u00d7 \u00d7 0.01131 14.4 10 \u00d7 \u2212 2 \u00d7 2000 \u00d7 2 60 = = 22619.9 W For four cylinders total indicated power = 90479.6 W Frictional power loss = 0.10 \u00d7 90479.6 = 9047.96 W Brake power available = Indicated power \u2013 Frictional power = (90479.6 \u2013 9047.96) = 81431.64 W Therefore, mechanical efficiency of engine = brake power indicated power = 81431.64 90479.6 = 0.90 Indicated power = 90479.6 W Mechanical efficiency = 90% Ans. 2. Determine the power required to drive a double acting reciprocating pump having indicator diagram with area 40 cm2 and length 8 cm. Bore and stroke of the pump are 15 cm and 20 cm. The pump motor runs at", " 100 rpm. Indicator spring constant is 1.5 \u00d7 108 Pa per m. Solution: Reciprocating Pump is work absorbing machine having its mechanism similar to the piston-cylinder arrangement in IC engine. Mean effective pressure = \u00d7 40 10 \u2212 4 \u00d7 \u00d7 8 1.5 10 \u2212 2 \u00d7 8 10 m.e.p. = 7.5 \u00d7 106 Pa (It is double acting so let us assume total power to be double of that in single acting) Indicated power = (Piston area \u00d7 Stroke \u00d7 mep \u00d7 Speed \u00d7 No. of cylinder) \u03c0\uf8eb \uf8ec \uf8ed 4 = \u00d7 (0.15) 2 \u00d7 = 88357.29 W 0.20 7.5 10 \u00d7 \u00d7 6 100 \u00d7 60 \u00d7 \u00d7 1 2 \uf8f6 \uf8f7 \uf8f8 Power required to drive = 88.36 kW Ans. 3. An engine with 90% mechanical efficiency has rating of 38 kW brake power. Estimate its indicated power and frictional power loss. Also determine the mechanical efficiency at quarter load assuming frictional power to remain same. Solution: Indicated power = Brake power Mechanical efficiency = 38 0.9 792 ________________________________________________________ Applied Thermodynamics = 42.22 kW Frictional power loss = Indicated power \u2013 Brake power = 42.22 \u2013 38 = 4.22 kW Brake power at quarter load = 0.25 \u00d7 38 = 9.5 kW Mechanical efficiency = Brake power Indicated power = 9.5 9.5 + 4.22 = 0.6924 or 69.24% Indicated power = 42.22 kW Frictional power loss = 4.22 kW Mechanical efficiency = 69.24% Ans. 4. An eight cylinder diesel engine of two stroke type has specific fuel consumption of 0.25 kg/kWh. The brake mean effective pressure of each cylinder is 1.5 MPa and engine run at 100 rpm. The bore and stroke of cylinder are 85 cm and 220 cm respectively. Considering the calorific value of diesel as 43 MJ/ kg determine the brake power of engine, fuel consumption in kg/hr and brake thermal efficiency of engine. Solution: Brake power of engine = Pb mep \u00d7 L \u00d7 A \u00d7 N 1.5 \u00d710 3 \u00d7 (2.2) \u00d7 = \u00d7 (0.85) 2 \uf8f6 \uf8f7 \uf8f8 \u00d7 100 \u03c0\uf8eb \uf8ec \uf8ed 4 60 = 3120.97 kW or 31.21 MW Brake power = 31.21 MW Ans. The fuel consumption in kg/hr = 0.25 \u00d7 3120.97 = 780.24 kg/hr In order to find out brake thermal efficiency the heat input from fuel per second is required. Heat from fuel, kJ/s = \u00d7 780.24 43 10 3600 Energy to brake power = 3120.97 kW \u00d7 3 = 9319.53 kJ/s Brake thermal efficiency = 3120.97 9319.53 = 0.33488 or 33.49% Fuel consumption = 780.24 kg/hr, Brake thermal efficiency = 33.49% Ans. 5. A four cylinder, four stroke diesel engine has brake mean effective pressure of 6 bar at full load speed of 600 rpm and specific fuel consumption of 0.25 kg/kWh. The cylinder has bore of 20 cm and stroke length of 30 cm. The air fuel ratio is measured as 26 from the exhaust gas analysis. The ambient conditions are 1 bar, 27\u00baC. Assuming the calorific value of fuel as 43 MJ/kg determine the brake thermal efficiency and the volumetric efficiency. Also find out brake power. Introduction to Internal Combustion Engines ________________________________________ 793 Solution: Brake thermal efficiency = 3600 (brake specific fuel consumption \u00d7 Calorific value) = 3600 \u00d7 \u00d7 3 (0.25 43 10 ) = 0.3349 or 33.49% Brake thermal efficiency = 33.49% Ans. Brake power = \u00d7 \u00d7 4 6 10 2 \u00d7 0.30 \u00d7 (0.20) 2 \uf8f6 \uf8f7 \uf8f8 \u00d7 600 \u00d7 \u03c0\uf8eb \uf8ec \uf8ed 4 60 = 226.19 kW Brake power = 226.19 kW Ans. Brake specific fuel consumption = fm brake power \u21d2 mf = 56.55 kg/hr Air consumption from given air fuel ratio = 26 \u00d7 56.55 = 1470.3 kg/hr or 0.25 = fm 226.19 Using perfect gas equation, ma = 24.505 kg/min P \u00b7 Va = ma RT 1 \u00d7 102 \u00d7 Va = 24.505 \u00d7 0.287 \u00d7 300 Va = 21.098 m3/min \u03c0 \u03c0 Swept volume, Vs = 4 4 D2 L = \u00d7 (0.30)2 \u00d7 0.4 = 0.02827 m3 Volumetric efficiency", ", \u03b7 vol number or cylinders \u00d7 = 21.09 2 \u00d7 0.02827 600 4 = 0.6217 or 62.17% \u00d7 Volumetric efficiency = 62.17% Ans. 6. A two stroke two cylinder engine runs with speed of 3000 rpm and fuel consumption of 5 litres/hr. The fuel has specific gravity of 0.7 and air-fuel ratio is 19. The piston speed is 500 m/min and indicated mean effective pressure is 6 bar. The ambient conditions are 1.013 bar, 15\u00baC. The volumetric efficiency is 0.7 and mechanical efficiency is 0.8. Determine brake power output considering R for gas = 0.287 kJ/kg \u00b7 K (Take piston speed, m/min = 2 LN where L is stroke (m) and N is rpm) 794 ________________________________________________________ Applied Thermodynamics Solution: Let the bore of cylinder be \u2018D\u2019 meter. Using piston speed, 500 = 2 \u00d7 L \u00d7 3000 \u21d2 L = 0.0833 m Volumetric efficiency, \u03b7 vol. = Air sucked (m /min) Swept volume \u00d7 \u00d7 number of cylinders N 3 0.7 = \u03c0\uf8eb \uf8ec \uf8ed 4 Air sucked \uf8f6 \uf8f7 \uf8f8 2 \u00b7 D L \u00d7 3000 2 \u00d7 \u21d2 Air sucked = 274.78 D2 m3/min Air requirement, kg/min = A/F ratio \u00d7 Fuel consumption per minute = 19 \u00d7 \u00d7 \u00d7 3 5 0.7 10 60 \u00d7 \u2212 3 10 Using perfect gas equation, PVa = ma RT \u00d7 ma = 1.11 kg/min \u00d7 1.11 0.287 288 \u00d7 Va = 1.013 10 Va = 0.906 m3/min 2 Ideally, Air sucked = Va 274.78 D2 = 0.906 \u21d2 D = 0.0574 m or 57.4 mm Indicated power = Pimep \u00d7 L \u00d7 A \u00d7 N \u00d7 no. of cylinders \u00d7 \u03c0 2 (0.0574) = 6 \u00d7 102 \u00d7 0.0833 \u00d7 \u00d7 4 60 3000 2 \u00d7 = 12.93 kW Brake power = Indicated power \u00d7 Mechanical efficiency = 12.93 \u00d7 0.8 = 10.34 kW Brake power = 10.34 kW Ans. 7. During trial of four stroke single cylinder engine the load on dynamometer is found 20 kg at radius of 50 cm. The speed of rotation is 3000 rpm. The bore and stroke are 20 cm and 30 respectively. Fuel is supplied at the rate of 0.15 kg/min. The calorific value of fuel may be taken as 43 MJ/kg. After some time the fuel supply is cut and the engine is rotated with motor which required 5 kW to maintain the same speed of rotation of engine. Determine the brake power, indicated power, mechanical efficiency, brake thermal efficiency, indicated thermal efficiency, brake mean effective pressure, indicated mean effective pressure. Solution: After switching off fuel supply the capacity of motor required to run engine will be the friction power required at this speed of engine. Friction power = 5 kW Introduction to Internal Combustion Engines ________________________________________ 795 Brake power = 2\u03c0 NT = \u00d7 \u00d7 \u03c0 2 = 30.82 kW 3000 20 9.81 0.5 10 \u00d7 \u00d7 \u00d7 \u2212 3 \u00d7 60 Indicated power = Brake power + Friction power = 35.82 kW Mechanical efficiency = Brake specific fuel consumption = Brake power Indicated power = 30.82 35.82 Specific fuel consumption Brake power = 0.8604 or 86.04% = 0.15 \u00d7 60 30.82 = 0.292 kg/kW hr Brake thermal efficiency = 3600 Brake specific fuel consumption \u00d7 calorific value = 3600 0.292 \u00d7 43\u00d710 3 = 0.2867 or 28.67% Also, Mechanical efficiency = Indicated thermal efficiency = Brake thermal efficiency Indicated thermal efficiency 0.2867 0.8604 = 0.3332 or 33.32% Indicated power = Pimep \u00d7 L \u00d7 A \u00d7 N \u03c0 35.82 = Pimep \u00d7 0.30 \u00d7 \u00d7 (0.20)2 \u00d7 3000 60 4 Also, Mechanical efficiency = Pimep = 76.01 kPa P b P i mep mep \u21d2 \u21d2 Brake mean effective pressure = Pimep \u00d7 Mechanical efficiency Pimep = 76.01 \u00d7 0.8604 = 65.39 kPa Brake power = 30.82 kW, Indicated power = 35.82 kW Mechanical efficiency = 86.04%. Brake thermal efficiency = 28.67% Indicated thermal efficiency = 33.32%, Brake mean effective pressure = 65.39 kPa, Indicated mean effective pressure = 76.01 kPa Ans. 8. A four stroke four cylinder diesel engine running at 300 rpm produces 250 kW of brake power. The cylinder", " dimensions are 30 cm bore and 25 cm stroke. Fuel consumption rate is 1 kg/min while air fuel ratio is 10. The average indicated mean effective pressure is 0.8 MPa. Determine indicated power, mechanical efficiency, brake thermal efficiency and volumetric efficiency of engine. The calorific value of fuel is 43 MJ/kg. The ambient conditions are 1.013 bar, 27\u00baC. 796 ________________________________________________________ Applied Thermodynamics Solution: Given, D = 0.3 m, L = 0.25 m, N = 300 rpm, mf = 1 kg/min, F/A = 20, Pimep = 0.8 MPa, Brake power = 250 kW Indicated power kW = Pimep \u00d7 L \u00d7 A \u00d7 = 282.74 kW 4 N \u00d7 60 = 0.8 \u00d7 0.25 \u00d7 \u03c0 4 \u00d7 (0.3)2 \u00d7 300 60 \u00d7 4 \u00d7 103 Mechanical efficiency = Brake power Indicated power = 250 282.74 = 0.8842 or 88.42% Brake specific fuel consumption = Brake thermal efficiency = 1\u00d7 60 250 = 0.24 kg/kW hr 3600 bsfc\u00d7 CV = 3600 0.24\u00d7 43000 = 0.3488 or 34.88% Swept volume, Vs = \u03c0 4 D2 L = \u03c0 4 \u00d7 (0.30)2 \u00d7 0.25 = 0.01767 m3 Mass of air corresponding to above swept volume, using Perfect gas equation PVs = ma RT \u00d7 \u00d7 2 1.013 10 0.017 ma = \u00d7 0.287 300 = 0.02 kg Volumetric efficiency, \u03b7 vol = Mass of air taken/minute Mass corresponding to swept volume per minute Mass of air taken per minute = 1 \u00d7 10 = kg/min Mass corresponding to swept volume per minute = 0.02 \u00d7 4 \u00d7 300 2 = 12 kg/min Volumetric efficiency = 10 12 = 0.8333 or 83.33% Indicated power = 282.74 kW, Mechanical efficiency = 88.42% Brake thermal efficiency = 34.88%, Volumetric efficiency = 83.33% Ans. 9. During an experiment on four stroke single cylinder engine the indicator diagram obtained has average height of 1 cm while indicator constant is 25 kN/m2 per mm. The engine run at 300 rpm and the swept volume is 1.5 \u00d7 104 cm3. The effective brake load upon dynamometer is 60 kg while the effective brake drum radius is 50 cm. The fuel consumption is 0.12 kg/min and the calorific value of fuel oil is 42 MJ/kg. The engine is cooled by circulating water around it at the rate of 6 kg/min. The cooling water enters at 35\u00ba C and leaves at 70\u00baC. Exhaust gases leaving have energy of 30 kJ/s with them. Take specific heat of water as 4.18 kJ/kg K. Determine indicated power output, brake power output and mechanical efficiency. Also draw the overall energy balance in kJ/s. Introduction to Internal Combustion Engines ________________________________________ 797 Solution: Indicated mean effective pressure = 10 \u00d7 25 = 250 kPa Indicated power = Pimep \u00d7 L \u00d7 A \u00d7 \uf8f1 \uf8f2 here \uf8f3 N 2 = N N 2 as it is four stroke \uf8fc \uf8fd \uf8fe = 250 \u00d7 1.5 \u00d7 104 \u00d7 300 2 60\u00d7 \u00d7 10\u20136 = 9.375 kW Brake power = 2\u03c0NT = 2 \u00d7 \u03c0 \u00d7 300 2 60\u00d7 \u00d7 (60 \u00d7 9.81 \u00d7 0.50) \u00d7 10\u20133 = 4.62 kW Mechanical efficiency = Brake power Indicated power = = 0.4928 or 49.28% 4.62 9.375 Indicated power = 9.375 kW, Brake power = 4.62 kW, Mechanical efficiency = 49.28% Ans. Energy liberated from fuel = 3 42 \u00d710 \u00d7 0.12 60 = 84 kJ/s Energy available as brake power = 4.62 kW Energy to coolant = 6 60 \u00d7 4.18 \u00d7 (70 \u2013 35) = 14.63 kW Energy carried by exhaust gases = 30 kJ/s Unaccounted energy loss = 84 \u2013 4.62 \u2013 14.63 \u2013 30 = 34.75 kW Energy Balance on per second basis Energy input, kJ/s Energy consumed, kJ/s Heat from fuel = 84 kJ/s, 100% Energy consumed as brake power = 4.62 kW Energy carried by coolant = 14.63 kW Energy carried by exhaust gases = 30 kW Unaccounted losses = 34.75 kW in % 5.5% 17.42% 35.71% 41.37% 10. During 15 minutes trial of an internal combustion engine of 2-stroke single cylinder type the total 4 kg fuel is consumed while the engine is run at", " 1500 rpm. Engine is cooled employing water being circulated at 15 kg/min with its inlet and exit temperatures as 27\u00baC and 50\u00baC. The total air consumed is 150 kg and the exhaust temperature is 400\u00baC. The atmospheric temperature is 27\u00baC. The mean specific heat of exhaust gases may be taken as 1.25 kJ/kg K. The mechanical efficiency is 0.9. Determine, the brake power, brake specific fuel consumption and indicated thermal efficiency. Also draw energy balance on per minute basis. Brake torque is 300 Nm and the fuel calorific value is 42 MJ/kg. Solution: Brake power = 2\u03c0NT = \u03c0\u00d7 \u00d7 2 \u00d7 1500 300 60 = 47.124 kW 798 ________________________________________________________ Applied Thermodynamics Brake power = 47.124 kW Ans. Brake specific fuel consumption = \u00d7 4 60 \u00d7 15 47.124 = 0.339 kg/kW hr. Brake specific fuel consumption = 0.339 kg/kW hr Ans. Indicated power = Brake power Mechanical efficiency = 47.124 0.9 = 52.36 kW Indicated thermal efficiency = 52.36 \uf8eb \uf8ec \uf8ec \uf8ed 3 4\u00d7 42\u00d710 \u00d7 15 60 \uf8f6 \uf8f7 \uf8f7 \uf8f8 = 0.2805 or 28.05% Indicated thermal efficiency = 28.05% Ans. Heat available from fuel, kJ/min = 4 15 \u00d7 42 \u00d7 103 = 11200 kJ/min Energy consumed as brake power = 47.124 \u00d7 60 = 2827.44 kJ/min Energy carried by cooling water = 15 \u00d7 4.18 \u00d7 (50 \u2013 27) = 1442.1 kJ/min Energy carried by exhaust gases = \uf8eb \uf8ec \uf8ed + 150 4 15 \uf8f6 \uf8f7 \uf8f8 \u00d71.25\u00d7 (400 \u2013 27) = 4786.83 kJ/min Unaccounted energy loss = 11200 \u2013 (2827.44 + 1442.1 + 4786.83) = 2143.63 kJ/min. Heat balance on per minute basis Heat input, kJ/min % Heat consumed, kJ/min Heat from fuel = 11200 100% Total = 11200 kJ/min 100% 1. Brake power = 2827.44 2. Cooling water = 1442.1 3. Exhaust gases = 4786.83 4. Unaccounted loss = 2143.63 (by difference) Total = 11200 kJ/min % 25.24% 12.88% 42.74% 19.14% 100% 11. During Morse Test experiment on a six cylinder petrol engine the brake power output was found 50 kW when all cylinders run at full load. When one by one each cylinder is cut and load is reduced to bring engine back to original speed, the measured brake power outputs are as under. Determine the indicated power of engine and mechanical efficiency of engine. No. of cylinders cut Brake power, kW 1 2 3 4 5 6 40.1 39.5 39.1 39.6 39.8 40 Introduction to Internal Combustion Engines ________________________________________ 799 Solution: Indicated power of Ist cylinder = 50 \u2013 40.1 = 9.9 kW Indicated power of IInd cylinder = 50 \u2013 39.5 = 10.5 kW Indicated power of IIIrd cylinder = 50 \u2013 39.1 = 10.9 kW Indicated power of IVth cylinder = 50 \u2013 39.6 = 10.4 kW Indicated power of Vth cylinder = 50 \u2013 39.8 = 10.2 kW Indicated power of VIth cylinder = 50 \u2013 40 = 10 kW Total indicated power = 61.9 kW Mechanical efficiency = Brake power Indicated power = 50 61.9 = 0.8077 or 80.77% Indicated power = 61.9 kW Mechanical efficiency = 80.77% Ans. 12. During trial of a four cylinder four stroke petrol engine running at full load it has speed of 1500 rpm and brake load of 250 N when all cylinders are working. After some time each cylinder is cut one by one and then again brought back to same speed of engine. The brake readings are measured as 175 N, 180 N, 182 N and 170 N. The brake drum radius is 50 cm. The fuel consumption rate is 0.189 kg/min with the fuel whose calorific value is 43 MJ/kg and A/F ratio of 12. Exhaust gas temperature is found to be 600\u00baC. The cooling water flows at 18 kg/min and enters at 27\u00baC and leaves at 50\u00baC. The atmospheric air temperature is 27\u00baC. Take specific heat of exhaust gas as 1.02 kJ/kg K. Determine the brake power output of engine, its indicated power and mechanical efficiency. Also draw a heat balance on", " per minute basis. Solution: Brake power output of engine = NT\u03c0 2 60 \u03c0\u00d7 2 = \u00d7 1500 250 0.50 10 \u00d7 \u00d7 60 \u20133 = 19.63 kW Brake power when cylinder 1 is cut = \u03c0\u00d7 \u00d7 2 1500 175 0.5 10 \u00d7 \u00d7 \u00d7 60 Indicated power of first cylinder = 19.63 \u2013 13.74 = 5.89 kW Brake power when cylinder 2 is cut = \u03c0\u00d7 \u00d7 2 1500 180 0.5 10 \u00d7 \u00d7 \u00d7 60 Indicated power of second cylinder = 19.63 \u2013 14.14 = 5.49 kW Brake power when cylinder 3 is cut = \u03c0\u00d7 \u00d7 2 1500 182 0.5 10 \u00d7 \u00d7 \u00d7 60 Indicated power of third cylinder = 19.63 \u2013 14.29 = 5.34 kW Brake power when cylinder 4 is cut = \u03c0\u00d7 \u00d7 2 1500 170 0.5 10 \u00d7 \u00d7 \u00d7 60 \u20133 \u20133 \u20133 \u20133 = 13.74 kW = 14.14 kW = 14.29 kW = 13.35 kW Indicated power of fourth cylinder = 19.63 \u2013 13.35 = 6.28 kW Total indicated power output = 5.89 + 5.49 + 5.34 + 6.28 = 23 kW 800 ________________________________________________________ Applied Thermodynamics Engine\u2019s mechanical efficiency = Brake power Indicated power = 19.63 23 = 0.8535 or 85.35% B.P. = 19.63 kW, I.P. = 23 kW, \u03b7 mech = 83.35% Ans. Heat liberated by fuel, kJ/min = 0.189 \u00d7 43 \u00d7 103 = 8127 kJ/min Heat carried by exhaust gases = (12 + 1) \u00d7 0.189 \u00d7 1.02 \u00d7 (600 \u2013 27) = 1436.02 kJ/min Heat carried by cooling water = 18 \u00d7 4.18 \u00d7 (50 \u2013 27) = 1730.52 kJ/min Energy to brake power = 19.63 \u00d7 60 = 1177.8 kJ/min Unaccounted losses = 8127 \u2013 (1436.02 + 1730.52 + 1177.8) = 3782.66 kJ/min Heat balance sheet on per minute basis Energy in, kJ/min % Energy out, kJ/min Heat released by fuel = 8127 100% (a) Brake power = 1177.8 (b) Exhaust gases = 1436.02 (c) Cooling water = 1730.52 (d) Unaccounted = 3782.66 (by difference) Total = 8127, kJ/min 100% Total = 8127 kJ/min % 14.50 17.67 21.29 40.54 100% 13. During the trial of a single acting oil engine, cylinder diameter is 20 cm, stroke 28 cm, working on two stroke cycle and firing every cycle, the following observations were made: Duration of trial :1 hour Total fuel used :4.22 kg Calorific value :44670 kJ/kg Proportion of hydrogen in fuel : 15% Total number of revolutions : 21000 Mean effective pressure : 2.74 bar Net brake load applied to a drum of 100 cm diameter : 600 N Total mass of cooling water circulated : 495 kg Total mass of cooling water : inlet 13\u00baC, outlet 38\u00baC Air used : 135 kg Temperature of air in test room : 20\u00baC Temperature of exhaust gases : 370\u00baC Assume Cp, gases = 1.005 kJ/kg K, Cp, steam at atmospheric pressure = 2.093 kJ/kg K Calculate thermal efficiency and draw up the heat balance. Solution: Brake power = 2\u03c0 NT N = 21, 000 60 rpm = 350 rpm T = 600 \u00d7 0.5 = 300 N. m [U.P.S.C. 1997] Introduction to Internal Combustion Engines ________________________________________ 801 Brake power = \u03c0 2 \u00d7 350\u00d7 300 60 \u00d71000 = 10.996 kW Indicated power = = mep \u22c5 \u00d7 \u22c5 L A N \u00d7 60 1000 kW \u00d7 2.74 10 5 \u00d7 (0.2) 2 \u00d7 300 0.28 \u00d7 \u00d7 \u00d7 \u03c0 \u00d7 60 1000 4 = 14.06 kW (A) Heat added = 4.22 3600 \u00d7 44670 = 52.36 kJ/s or 3141.79 kJ/min Thermal efficiency, \u03b7 = Indicated power Heat added = 14.06 4.22 3600 \u00d7 44670 = 0.2685 or 26.85% (B) Heat equivalent of brake power = 10.996 \u00d7 60 = 659.76 kJ/min Heat lost to cooling water = mw \u00d7 Cpw \u00d7 \u2206T (C) = 495 60 \u00d7 4.18 \u00d7 (38 \u2013 13) Heat carried by exhaust gases", " = 862.13 kJ/min = Heat carried by steam in exhaust gases + Heat carried by fuel gases (dry gases) in exhaust gases Mass of exhaust gases = mass of air + mass of fuel + (135 4.22) 60 = = 2.32 kg/min kg/min Mass of steam in exhaust gases = 9 \u00d7 \uf8eb \uf8ec \uf8ed 0.15 \u00d7 4.22 60 \uf8f6 \uf8f7 \uf8f8 = 0.095 kg/min Mass of dry exhaust gases = 2.32 \u2013 0.095 = 2.225 kg/min (D) Heat carried by steam in exhaust = 0.095 \u00d7 {4.18 (100 \u201320) + 2256.9 + 2.093 \u00d7 (370 \u2013 100)} (E) Heat carried by fuel gases (dry gases) in exhaust gases = 299.86 kJ/min (F) Unaccounted loss =.225 \u00d7 1.005 (370 \u2013 20) = 782.64 kJ/min = (3141.79) \u2013 (659.76) \u2013 (862.13) \u2013 (299.86) \u2013 (782.64) = 537.4 kJ/min 802 ________________________________________________________ Applied Thermodynamics Heat balance sheet (on per minute basis) Input Expenditure Heat, kJ/min %, Percentage Heat, kJ/min (A) Heat added by fuel = 3141.79 100 (B) Heat equivalent to brake power = 659.76 %, Percentage 21% (C) Heat lost to cooling water = 862.13 (D) Heat carried by steam in exhaust 27.44% 9.54% gases = 299.86 (E) Heat carried by dry fuel gases in 24.92% exhaust gases = 782.64 (F) Unaccounted losses = 537.4 17.10% Ans. 17.1 Give classification of internal combustion engines. 17.2 Give a sketch of simple one cylinder internal combustion engine and label important parts. 17.3 What is meant by spark ignition and compression ignition engines? 17.4 Explain the working of 4-stroke SI engine. 17.5 Describe the working of 4-stroke CI engine. 17.6 Explain working of 2-stroke CI engine. 17.7 Explain working of 2-stroke SI engine. 17.8 What is meant by indicator diagram? Describe the arrangement for getting it? 17.9 Explain the power measurement using indicator diagram arrangement. 17.10 Explain the combustion in SI engines. 17.11 Explain the combustion in CI engines. 17.12 Describe some IC engine fuels. 17.13 What is More test? How is it carried out? 17.14 Compare SI engines with CI engines. 17.15 Compare 2 stroke SI engine with 4-stroke SI engines. 17.16 Write short notes on the following: Brake power, Indicated power, Brake specific fuel consumption, Indicated specific fuel consumption, Brake mean effective pressure, Indicated mean effective pressure, Mechanical efficiency. Brake thermal efficiency, Indicated thermal efficiency. 17.17 A single cylinder four stroke CI engine runs at 200 rpm and the cylinder has bore and stroke of 10 cm and 12.5 cm respectively. The indicator diagram taken has length of 10 cm and area of 20 cm2. The indicator constant is 30 kPa per mm. Considering mechanical efficiency of 0.8 determine indicated [600 kPa, 5.89 kW] mean effective pressure and indicated power. 17.18 Determine the dimensions of cylinder, brake thermal efficiency and indicated mean effective pressure for the single cylinder four stroke diesel engine producing 100 kW of brake power. Engine runs at speed of 400 rpm, brake mean effective pressure of 850 kPa, brake specific fuel consumption of 0.335 kg/kW hr. The mechanical efficiency is 0.8 and stroke to bore ratio is 1.25. The calorific value of fuel may be taken as 43.5 MJ/kg. [D = 33 cm, L = 41.25 cm, 24.7%, 10.62 bar] Introduction to Internal Combustion Engines ________________________________________ 803 17.19 Determine thermal efficiency and mean effective pressure of thermodynamic cycle used by a 4-stroke petrol engine. Details of cycle are as follows. Compression ratio = 7 Initial state = 100 kPa and 90\u00baC Swept volume = 0.1 m3 Heat added to cycle at constant volume = 100 kJ/cycle. Consider air as working fluid. [54%, 530 kPa] 17.20 A 4-stroke four cylinder petrol engine has bore and stroke as 0.15 m and 0.17 m respectively. Indicator diagram arrangement when put on a cylinder yields indicator diagram having area of 25 cm2 and length of diagram as 6 cm. The speed of engine is 2500 rpm. Determine the indicated power and frictional power if mechanical efficiency is 88%. Take indicator spring constant", " as 20 \u00d7 103 kN/m3. [208.6 kW, 25 kW] 17.21 Determine the cylinder diameter and stroke length for a 8 cylinder, single acting, 2-stroke diesel engine for which details of engine and indicator diagram are as under; Indicated power = 16266.7 kW Speed = 150 rpm L : D ratio = 1.5 Area of indicator diagram = 550 mm2 Length of diagram = 60 mm Indicator spring constant = 14.7 \u00d7 107 Pa per m [80 cm, 120 cm] 17.22 Determine the capacity of motor required for running a single-cylinder, double acting reciprocating water pump at 50 rpm. Indicator diagram arrangement gives an indicator diagram with area of 37.5 cm2 and length 7.5 cm. Indicator spring constant is 14.7 \u00d7 107 Pa per m. Pump cylinder diameter and stroke [33.66 kW] are 15 cm and 20 cm Mechanical efficiency of arrangement is 90%. 17.23 During a test on single cylinder four stroke diesel engine having bore of 15 cm, stroke of 18 cm gives indicator diagram of 10 cm length and area of 9 cm2. Engine has speed of 480 rpm, fuel consumption of 0.05 kg/min and cooling water flow rate of 250 kg/hr. Cooling water enters at 27\u00baC and leaves at 67\u00baC. The calorific value of fuel is 44000 kJ/kg. Brake torque is 200 Nm. Determine the indicator constant, brake specific fuel consumption and brake thermal efficiency. Take mechanical efficiency as 87.8%. [0.298 kg.kW hr, 27.4%] 17.24 A two cylinder four stroke engine runs at 240 rpm developing a torque of 5.16 kN. m. The bore and stroke of cylinder are 30 cm and 58.5 cm respectively. Engine runs with gaseous fuel having calorific value of 16.8 MJ/m3. The gas and air mixture is supplied in proporation of 1 : 7 by volume. The volumetric efficiency is 0.85. Determine (i) the brake power (ii) the mean piston speed in m/s (iii) the brake mean effective pressure (iv) the brake thermal efficiency. [129.68 kg, 4.68 m/s, 4.89 bar, 27.37%] 17.25 A four cylinder petrol engine has output of 52 kW at 200 rpm. During the Morse test being performed each cylinder is cut one by one. The brake load readings are 35.47 kg, 36.08 kg, 34.25 kg and 34.66 kg when one by one all cylinders are cut and the engine is brought back to original running speed. The brake drum radius may be taken as 50 cm. The specific fuel consumption at this speed is 364 gm/kW hr. The calorific value of fuel may be taken as 42 MJ/kg. Determine the mechanical efficiency, brake [81.61%, 23.54%, 28.87%] thermal efficiency, and indicated thermal efficiency of engine. 17.26 In the testing of two stroke engine the following observations are made. Bore and stroke = 20 cm, 25 cm Brake drum radius = 60 cm 804 ________________________________________________________ Applied Thermodynamics Speed = 360 rpm Brake load = 500 N Fuel consumption = 0.06 kg/min Calorific value of fuel = 42000 kJ/kg Cooling water flow rate = 350 kg/hr Cooling water temperature rise = 30\u00b0C Area of indicator diagram = 2.8 cm2 Length of indicator diagram = 10 cm Spring constant = 100 kPa/min Determine indicated power, brake power and thermal efficiency. [13.19 kW, 11.31 kW, 26.93%] 17.27 A four stroke engine has single cylinder with bore of 18 cm and stroke of 34 cm. Engine has indicated mean effective pressure of 555 kPa at 390 rpm. At this the brake load is 510 N at brake radius of 56.5 cm. Cooling water is flown at the rate of 270 kg/hr with its inlet and exit temperatures being 18\u00baC and 58\u00baC. The fuel has calorific value of 43000 kJ/kg and the fuel is supplied at 3.6 kg/ hr. The exhaust gases carry 38% of fuel energy. Determine mechanical efficiency and brake thermal efficiency. Also prepare a heat [75.54%, 27.36%] balance sheet on per second basis. 17.28 Trial on a four stroke single cylinder engine results in following observations: Bore = 16.5 cm, Stroke = 30.5 cm Speed = 300 rpm Brake load = 360 N Brake diameter = 0.9 m Gas consumption = 4.95 m3/hr Calorific value of gaseous fuel = 18850 kJ/m3 Cooling water supply = 204 kg/hr Cooling water temperature rise = 30\u00baC Indicated mean effective pressure =", " 455 kPa Determine (i) the mechanical efficiency. (ii) the indicated thermal efficiency (iii) the brake thermal efficiency Draw the energy balance on per second basis. [68.86%, 28.51%, 19.63%] Introduction to Refrigeration and Airconditioning ____________________________________ 805 18 18.1 INTRODUCTION Refrigeration refers to the use of mechanical devices or heat activated devices for producing and maintaining the temperature in a region less than that of surroundings. American Society of Refrigerating Engineers defines refrigeration as \u201cthe science of providing and maintaining temperature below that of the surrounding atmosphere\u201d. Refrigerating a region in reference to the human comfort is called air conditioning. Thus, air conditioning refers to maintaining a space for the human comfort i.e. temperature, humidity and ventilation are the parameters to be controlled. When refrigeration provides temperature below \u2013 150\u00baC then this is known as cryogenics. Equipment used for removing the heat continuously for maintaining low temperature in a space is called \u2018refrigerator\u2019 and cycles on which it operates are called \u2018refrigeration cycles\u2019. The working fluids used for carrying away heat are called \u2018refrigerants\u2019 which are used in both refrigeration and air conditioning equipments. Equipment used for air conditioning are called airconditioners. The air conditioning of space may also be the requirement of various goods, equipments and processes etc. When air conditioning is done for human comfort then it is called comfort air conditioning while when it is done for industrial purposes then it is called \u2018industrial air conditioning\u2019. Comfort air conditioning can be further categorized as summer air conditioning and winter air conditioning. Summer air conditioning is used in summer season and generally requires cooling and dehumidification. Winter air conditioning used in winter generally requires heating and humidification. Comfort air conditioning is extensively used in residential buildings, hospitals, offices, working spaces, vehicles, trains, aeroplanes etc. Industrial air conditioning is used for spot cooling/heating, environmental laboratories, printing industry, textile industry, precision parts manufacturing, photographic product handling, computer rooms, control rooms of power plants etc. Refrigeration is extensively used for increasing the storage life of perishable items specially food products, vegetables, fruits, milk, beverages, chilling of water, ice formation etc. along with industrial applications in chemical manufacturing, petroleum refinery, petrochemical plants, paper and plup industry etc. History of refrigeration system dates back to 1834 when Jacob Perkins had got first patent on mechanical type refrigeration system using \u2018ether\u2019 as refrigerant. First successful commercial refrigeration system was produced in 1857 by James Harrison and D.E. Siebe using ethyl ether as refrigerant. Refrigeration can be achieved by using refrigeration cycle which may be open cycle type or closed cycle type. In open cycle the refrigerant passes through the system once for refrigerating the space and is thrown away as it gets contaminated. In closed cycle the refrigerant flows inside tubes while 806 ________________________________________________________ Applied Thermodynamics refrigerating the space and does not get contaminated. Thus, in closed cycle a definite mass of refrigerant can be used for quite long time while it is not so in case of open cycle where for every run fresh refrigerant is to be picked up. Different methods for refrigeration are described below: (i) Refrigeration by evaporation: Refrigeration effect can be realized by evaporation of liquid. Evaporation of liquid requires latent heat of evaporation which is provided by the surroundings of region where liquid is being evaporated. This causes the cooling of the region supplying latent heat for evaporation. Examples of this type of refrigeration are: \u2013 Cooling of water kept in earthen pot. Earthen pot has porosity thereby allowing evaporation of water and cooling down the inside remaining water. \u2013 Cooling of water kept in desert bag (made of thick canvas i.e. tight woven fabric). Here some quantity of water always keeps on leaking out across the cloth surface. This water which appears on the surface of bag gets evaporated after absorbing heat from atmospheric air and water inside the bag, thus cooling the water kept in bag. \u2013 Sweating from the human skin is also example of such refrigeration which keeps the human being cool. \u2013 Evaporative condensers also use this concept for cooling. (ii) Refrigeration by ice: Ice is being popularly used for refrigeration. In this kind of refrigeration ice is kept in between the walls of an insulated cabinet. Apart from above there can be some other arrangements too such as ice may be kept in a tray upon the insulated box having flow of air across ice blocks to the refrigerated space. Temperature inside refrigerated space is controlled by flow rate of air and the temperature upto 5\u00baC can be attained like this. (iii) Refrigeration by expansion of air: Generally, the expansion of air is accompanied by the cooling", " air due to reduction in its temperature. Subsequently cool air is passed into the region where refrigeration is to be done. Generally the adiabatic expansion of air is used in air refrigeration system described ahead. (iv) Refrigeration by throttling process: Throttling of certain gases show the reduction in temperature of gas after throttling. Effect of throttling is evident by Joule-Thomson coefficient which shows whether the throttling is accompanied by cooling or heating of gases. Throttling is constant enthalpy process for which Joule-Thomson coefficient (\u00b5) is given as: \u00b5 = \u2202\uf8eb p \uf8ec \u2202\uf8ed T \uf8f6 \uf8f7 \uf8f8 h ; for \u00b5 > 0 the cooling is seen. (v) Refrigeration by dry ice: Dry ice is oftenly used in packaging of frozen foods for maintaining them at low temperature. Carbon dioxide in solid phase is termed as dry ice. It has typical property of getting transformed directly from solid phase to vapour phase at its sublimation temperature of \u201378\u00baC at atmospheric pressure. Dry ice blocks in suitable sizes are stacked around the frozen food for keeping them in frozen state. (vi) Refrigeration using liquid gases: Liquid gases such as liquid nitrogen, liquid carbon dioxide are used for refrigeration in refrigerated cargo vehicles. Liquid gases are expanded thereby causing its transformation from liquid to gas accompanied by lowering of temperature of refrigerated space. Liquid gases are sprayed inside the cooling area and it gets transformed into vapour which is exhausted out from vehicle. Nitrogen and carbon dioxide being cheaply available are generally used as liquid refrigerant. It facilitates attainable temperature up to \u201330\u00baC. Introduction to Refrigeration and Airconditioning ____________________________________ 807 (vii) Vapour Refrigeration system: Vapour refrigeration system based on vapour compression refrigeration system and vapour absorption refrigeration system are discussed ahead. 18.2 PERFORMANCE PARAMETER Refrigerator\u2019s performance is given by coefficient of performance (COP), COP for a refrigerator shown in block diagram, Fig. 18.1 can be given as under. COP is the ratio of refrigeration effect and net work done upon refrigerator. COP is dimensionless quantity so the two quantities should have same units. COP of refrigeration = Desired effect Net work done Here the desired effect is the continuous removal of heat at the rate of Q2 for maintaining body at lower temperature T2. Net work done for getting this refrigeration effect is W. 2Q W Q 2 \u2212 Q Q 1 2 COP of refrigerator may have any magnitude i.e. less than unity or greater than unity. COP = = ) ( High temp, T1, surroundings Q1 T T2 1< R W Q2 Body,T2 Fig. 18.1 Refrigerator Heat pump is the device which runs exactly similar to the refrigerator but its purpose is to maintain the body at temperature higher than that of surroundings. Therefore, heat pump supplies heat at constant rate to the body to be maintained at higher temperature. Performance of heat pump is also given by the parameter defined exactly like COP but instead of calling it COP, it is called as energy performance ratio. Different name of EPR is assigned to differentiate it from COP of refrigerator. Thus, COP of heat pump = Energy performance ratio (EPR) 1Q W = or, EPR = Q 1 \u2212 Q Q 2 EPR = COP 1+ 1 EPR shall always have its magnitude greater than unity. Body,T1 Q1 T T2 1< HP W Q2 Surroundings at low temp, T2 Fig. 18.2 Heat pump 808 ________________________________________________________ Applied Thermodynamics 18.3 UNIT OF REFRIGERATION Refrigeration effect is the amount of heat extracted by refrigerator from the refrigerated space. The higher capacity of refrigerator means higher shall be the refrigeration effect. This refrigeration effect is defined by the unit of refrigeration called \u2018Ton\u2019 of refrigeration. \u2018Ton\u2019 as unit of refrigeration has been defined based on formation of ice as described ahead. \u201cOne \u2018Ton\u2019 of refrigeration can be defined by the amount of heat being removed from one ton of water at 0\u00baC to form one ton of ice at 0\u00baC within 24 hours.\u201d Thus, a Ton of refrigeration shall quantify the latent heat required to be removed for solidification of water at 0\u00baC. Latent heat of solidification from water to ice at 0\u00baC = 334.5 kJ/kg 1 Ton of refrigeration = mass of water \u00d7 latent heat at 0\u00baC from water to ice (in SI units) = \u00d7 (1000 334.5) 24 kJ/hr 334.5 10 \u00d7 24 3600 Generally, 1 Ton is taken as 3.5 kJ/sec. This deviation from 3.87 to 3.", "5 kJ/sec is there because Ton was originally defined in fps units and the approximations during conversion yielded numerical value of 3.5 kJ/sec which is now universally accepted. = 3.87 kJ/sec = \u00d7 3 1 Ton of refrigeration (in MKS units) = \u00d7 1000 80 \u00d7 24 60 = 50 kcal/min In MKS units this is commonly taken as 50 kcal/min. Therefore, = 55.4 kcal/min and 1 Ton = 3.5 kJ/sec 1 Ton = 50 kcal/min 18.4 CARNOT REFRIGERATION CYCLE Carnot cycle being the most efficient and ideal cycle can also be used for getting the refrigeration effect upon its reversal. Reversed carnot cycle is shown in Fig. 18.3. Surroundings Th Qh I> T Th R W QI Body,TI a ( ) Refrigerator 3 P 4 Rev. lsothermal process 2 Qh,heat rejected T 3 Th 2 Rev. adiabatic process 1 TI 1 4 Ql, heat absorbed s4 s1 b p-V ( ) - representation c T-s ( ) representation V s Fig. 18.3 Reversed Carnot cycle for refrigeration Introduction to Refrigeration and Airconditioning ____________________________________ 809 Here, the refrigerated space/body is to be maintained at low temperature Tl for which heat Ql should be removed at constant rate and rejected to surroundings at high temperature, Th. Amount of heat rejected to surroundings is Qh while the net work done upon refrigerator is W. Different processes in reversed Carnot cycle are, 1\u20132 Reversible adiabatic compression 2\u20133 Reversible isothermal heat rejection, Qh at temperature, Th 3\u20134 Reversible adiabatic expansion 4\u20131 Reversible isothermal heat absorption, Ql at temperature, Tl Refrigeration effect is seen during process 4\u20131. For cyclic process; Net work W = Qh \u2013 Ql W = Th (S1 \u2013 S4) \u2013 Tl (S1 \u2013 S4) W = (S1 \u2013 S4) (Th \u2013 Tl) For reversed Carnot cycle COP = lQ \u2212 )( The coefficient of performance of refrigerator depends upon the two temperature values i.e. low temperature Tl and high temperature Th. For COP value to be high the low temperature Tl should be high while higher temperature Th should be small. With increasing low temperature value the COP goes on increasing, while with decreasing high temperature Th value the COP goes on increasing. Practically, the lower temperature requirement is decided by the user while higher temperature is generally fixed by the atmospheric temperature value. Thus, it could be said that for certain low temperature to be maintained, COP of refrigerator shall be more during cold days as compared to hot days. COPcold days > COPhot days because Th, cold days < Th, hot days Similar to the carnot power cycles, reversed carnot cycle for refrigeration is not feasible in practice. Therefore, number of modified cycles are practically used for refrigeration as discussed ahead. But the reversed carnot cycle provides the basis for comparison and provides the benchmark for achieving maximum COP. 18.5 AIR REFRIGERATION CYCLES Refrigeration cycles using air as refrigerant are termed as air refrigeration cycles. Air refrigeration cycles offer distinct advantage of refrigerant being available in abundance and free of cost. Bell-Coleman cycle is one such cycle running on air as refrigerant. Bell-Coleman cycle is modified form of reversed carnot cycle in order to make it feasible cycle. Here isothermal heat addition and release are replaced by isobaric processes, Bell-Coleman cycle based refrigerators are the ancient refrigeration systems used in cargo ships. Figure 18.4 shows the schematic of refrigeration system based on this cycle. It has a compressor, heat exchanger, expander and refrigeration unit. 810 ________________________________________________________ Applied Thermodynamics Inlet cold water Heat exchanger Cool high pressure air 3 Low pressure air with low temperature Expander 4 Refrigeration unit Exit, hot water 2 High pressure air 1 Low pressure air Compressor Piston Piston (a) Schematic Bell Coleman cycle based refrigeration system p2 p2 3 3 2 2 p1 p1 4 4 V V T T Adiabatic Adiabatic process process 1 1 3 3 4 4 p2 p2 2 2 p1 p1 1 1 Isobaric process Isobaric process S S (b) p-V diagram (c) T-s diagram Fig. 18.4 Bell Coleman cycle Bell-Coleman cycle is actually the reversed Brayton cycle. Different processes in Bell-Coleman cycle are as under. 1\u20132: Adiabatic compression of air causing rise in its pressure and temperature inside compressor. Ideally, this compression process is considered isentropic whereas actually it may not be so", ". 2\u20133: Isobaric heat rejection causing cooling of high pressure and high temperature air coming from compressor making it low temperature and high pressure air inside heat exchanger. Water may be used as cooling fluid inside heat exchanger. 3\u20134: Adiabatic expansion inside expander causing further cooling of high pressure low temperature air entering expander. This very cool air enters the refrigeration unit subsequently. Ideally expansion can be considered to be of isentropic type whereas actually it may not be isentropic. 4\u20131: Low temperature cool air enters refrigeration space and picks up heat from space thereby showing refrigeration effect. Subsequently low temperature air gets heated up and leaves the space after showing desired refrigeration to enter in to compressor. Thus, process 4\u20131 is the actual process which is giving the desired refrigeration effect. Thermodynamically, the refrigeration effect, net work supplied and COP estimated are as under. Heat picked up from refrigerated space: Introduction to Refrigeration and Airconditioning ____________________________________ 811 Net work done upon system, W = (Heat rejected \u2013 heat absorbed), as for a cycle. Qabsorbed = mCp (T1 \u2013 T4) heat rejected, Hence, \u2211Wcycle = \u2211Qcycle Qrejected = mCp (T2 \u2013 T3) W = mCp (T2 \u2013 T3) \u2013 mCp (T1 \u2013 T4) COP = Desired effect Net work = ( mC T p 2 ( mC mC T p 1 ( \u2212 T 4 ) Theoretical COP for ideal cycle, COP = = COP ideal Above expression for COP is correct till the different processes are of isobaric or isentropic type. In actual practice the expansion and compression processes are of polytropic type and given by pV n = constant i.e. both expansion and compression are assumed to follow same index. Actual cycle shall have heat absorption and heat rejection occurring with some pressure loss thereby deviating from ideal isobaric process. Work required in compressor, Wc = Wc = Work available from expander, We = 1 n n \u2212 [ p2 V2 \u2013 p1V1] \u00d7 m \uf8eb n \uf8ec \u2212\uf8ed n \uf8f6 \uf8f7 \uf8f8 R[T2 \u2013 T1] \u00d7 m 1 1 n n \u2212 [ p2 V3 \u2013 p1 V4] \u00d7 m \uf8eb n \uf8ec \u2212\uf8ed n \uf8f6 \uf8f7 \uf8f8 R[T3 \u2013 T4] \u00d7 m Net work, Wc = 1 Wactual = Wc \u2013 We \uf8f6 \uf8eb n \uf8f7 \uf8ec \uf8f8 \u00b7 m \u00b7 R [(T2 \u2013 T1) \u2013 (T3 \u2013 T4)] \u2212\uf8ed n Wactual = 1 as R\u03b3 \u03b3 \u2212 1) Cp = (, so Wactual = \uf8eb n \uf8ec \u2212\uf8ed n 1 \uf8f6 \uf8f7 \uf8f8 \u00b7 m \u00b7 Cp \u00b7 \uf8eb \uf8ec \uf8ed \uf8f6\u2212 1\u03b3 \uf8f7 \u03b3 \uf8f8 [(T2 \u2013 T1) \u2013 (T3 \u2013 T4)] Therefore, COPactual = mC p \uf8eb n \uf8ec \u2212\uf8ed n 1 \uf8eb \uf8f6 \uf8f7\uf8ec \uf8f8\uf8ed \u2212 T 4 ) mC T ( p 1 \uf8f6\u2212 \u03b3 1 \uf8f7 \u03b3 \uf8f8 [( )] 812 ________________________________________________________ Applied Thermodynamics COPactual = \uf8eb \uf8ec \uf8ed n \u2212 1 \uf8eb \uf8f6 \uf8f7\uf8ec \uf8f8\uf8ed \uf8fb For isentropic process, \u03b3 = n and COPideal = COPactual 1 Using isentropic considerations, COPideal = COPideal = \uf8ee \uf8ef \uf8f0 \uf8eb \uf8ec \uf8ed \u2212 \uf8eb \uf8ec \uf8ec \uf8ec \u2212\uf8ec \uf8fa \uf8fa \uf8fb For isentropic process between states 1 and 2, 3 and 4 and \uf8f6 \uf8f7 \uf8f8 = Upon equating we get, or, Therefore, upon substitution or COPideal = \uf8f6 \uf8f7 \uf8f8 COP ideal \uf8eb = \uf8ec \uf8ed T 1 \u2212 T 2 T 1 \uf8f6 \uf8f7 \uf8f8 COP actual = \u03b3 \uf8eb n \uf8ec \u2212\uf8ed n 1 \uf8eb \uf8f6 \uf8f7\uf8ec \uf8f8\uf8ed T 1 \uf8f6\u2212 Comparing the COP of ideal Bell Coleman cycle with reversed Carnot cycle COP we find that COP of Bell Coleman cycle will be less than that", " of COP of reversed Carnot cycle. COPBell Coleman < COPCarnot Introduction to Refrigeration and Airconditioning ____________________________________ 813 Carnot cycle operating between the temperature limits of Bell Coleman cycle shall have COP given by, COPcarnot = T 1 T\u2212 1 T 3 Since T2 > T3 so, COPBell Coleman < COPCarnot. Constant pressure heat transfer results in operating range which is greater than minimum range given by temperature of those parts of the surroundings with which heat is exchanged. Air refrigeration systems offer advantages as these run on air as refrigerant which is cheap, noninflammable and non toxic. Also, air refrigeration systems are light weight per ton of refrigeration compared to other refrigeration systems and is used in aircraft applications. Leakage of air refrigerant may not be dangerous problem. Air refrigeration systems are disadvantageous due to lower COP compared to other refrigeration systems. There is significant possibility of freezing of moisture accompanied by atmospheric air due to low temperature in refrigeration space. Air may also pick up moisture while passing through refrigeration space. Quantity of air refrigerant required is more compared to other refrigerants for similar refrigeration capacity. This is because of fact that air only carries heat in the form of sensible heat and so its large quantity is required compared to other refrigerants. 18.6 VAPOUR COMPRESSION CYCLES Vapour compression cycle based refrigeration systems are extensively used in refrigeration systems. These cycles are used for most of small domestic and large industrial applications. The vapour compression cycle has the refrigerant being circulated in closed circuit through compressor, condenser, throttle valve or expansion valve and evaporator as shown in Fig. 18.5. Refrigerant (in gas/vapour) is compressed isentropically in compressor from state 1 to 2. High pressure and high temperature refrigerant enters the condenser at state 2 where its condensation occurs and refrigerant is available in liquid form at state 3. Refrigerant in the form of saturated liquid at high pressure is passed through expansion valve where isenthalpic expansion occurs. Refrigerant leaving expansion valve at state 4 is in the form of low pressure wet mixture of liquid and vapour. Low pressure liquid-vapour mixture is passed through evaporator section in which refrigerant picks up heat from surroundings thereby showing refrigeration effect. As a result of this heat absorption the liquid-vapour mixture refrigerant gets transformed into dry gaseous refrigerant in case of dry compression as shown in Fig. 18.5 (b). Compression of dry and saturated or superheated refrigerant is called dry compression. Compression of dry refrigerant yields superheated state of refrigerant as shown by state 2. (Fig. 18.5 (b)) 3 3' Qrejected Expansion valve Compressor 4 4' Evaporator 1 1' Qabsorbed (a) Schematic for vapour compression cycle Wcompressor Compressor 1\u20132 or 1'\u20132' : - Isentropic compression 2\u20133 or 2'\u20133' : - Isobaric heat rejection 3\u20134 or 3'\u20134' : - Isenthalpic expansion 2 2' process or throttling process 4\u20131 or 4'\u20131' : - Isobaric heat absorption 814 ________________________________________________________ Applied Thermodynamics b) T-s and p-h diagram for dry compression 2' 1' 3' 4' s p 3' 4' h 2' 1' (c) T-s and p-h diagram for wet compression Fig. 18.5 Vapour compression cycle Wet compression: It is also possible that the refrigerant is in wet state i.e. liquid-vapour mixture at inlet of compressor, state 1'. Compression of wet mixture gets transformed into dry refrigerant (gaseous form) as shown by state 2'. Dry refrigerant at high pressure and high temperature is passed through condenser where refrigerant gets condensed and condensate is available in saturated liquid form at high pressure. Subsequently refrigerant is throttled from high pressure to low pressure inside expansion valve from state 3' to 4'. Low pressure refrigerant in wet state is passed through evaporator from state 4' to 1' where it picks up heat from surroundings and some of its liquid fraction gets transformed into vapour but it does not become dry (gas) refrigerant at inlet to compressor. This wet refrigerant is compressed inside compressor. In case of dry compression the compressor efficiency is found to be more than that of wet compression due to higher volumetric efficiency with dry refrigerant and also chances of damage by liquid refrigerant are absent. During dry compression the temperature of compressed refrigerant becomes more than condensation temperature. Due to this high temperature after compression the compressor becomes very hot therefore requiring cooling. Coolant used in condenser may also be used for cooling compressor as its temperature is lower than compressor temperature. This cooling of compressor", " reduces compression work. T-s and p-h diagram for vapour compression cycle shows that the refrigeration capacity can be increased by subcooling the condensate before it enters the expansion valve and also by increasing the degree of expansion in expansion valve. Here in second option the condensation temperature is fixed and evaporator temperature is lowered which increases expansion ratio and compression ratio. This increase in compression ratio shows poor Introduction to Refrigeration and Airconditioning ____________________________________ 815 volumetric efficiency in single stage dry compression. The temperature after compression also gets increased with increased compression ratio which may be harmful to refrigerant properties. To regulate the excessive rise in refrigerant temperature the intercooling during compression with multistage compression is used. This intercooled compression may be devised using refrigerant itself as intercooling medium. Thermodynamic analysis: For simple vapour compression cycle shown in Fig. 18.6 the COP, refrigeration effect and work input can be estimated based on following assumptions: (i) All process of refrigeration cycle are internally reversible except the expansion through valve which is throttling process and is irreversible. (ii) Compressor and expansion valve have no heat interaction with surroundings during their operation i.e. they operate adiabatically. (iii) Refrigerant leaving condenser is saturated liquid. (iv) Refrigerant entering compressor is saturated vapour in case of \u2018dry compression\u2019 and liquid- vapour mixture in case of \u2018wet compression\u2019. (v) Changes in kinetic energy and potential energy are negligible. For mass flow rate of refrigerant being given by \u2018m\u2019 kg/s the refrigeration capacity or refrigeration effect. Qabsorbed = m(h1 \u2013 h4), in dry compression Qabsorbed = m(h1\u2032 \u2013 h4') in wet compression Net work = Work input, Wcompressor = m(h2 \u2013 h1), in dry compression Therefore, Wcompressor = m(h2\u2032 \u2013 h1\u2032), in wet compression. COP = ( ) ) \uf8fc \uf8fd \uf8fe, in dry compression COP = ( m h 1' ) ) \uf8fc \uf8fd \uf8fe, in wet compression. Some calculations in vapour compression cycle: Let us take three different cases of vapour compression cycle similar to the dry and wet compression. Case 1: Compression yields dry and saturated vapour at the end of compression. Vapour compression cycle in which the compression process produces vapour dry and saturated, is shown on T-s diagram in Fig. 18.6. Different processes are, 1\u20132 : Isentropic compression 2\u20133 : Heat rejection in condenser 3\u20134 : Isenthalpic expansion in expansion valve (throttling) 4\u20131 : Heat absorption in evaporator or refrigerated space. Work done in the cycle can be given by the area enclosed 1\u20132\u20133\u20135\u20131 on T-s diagram. The work required for compression process. The refrigeration effect in evaporator can be given by area enclosed 1\u20134\u20139\u201310\u20131 on T\u2013s diagram. Cycle : 1 2 3 4 2 1 T T2 3 6 4 5 T1 O 7 8 9 10 s Fig. 18.6 816 ________________________________________________________ Applied Thermodynamics Compressor work, W = Area Area 2 \u2013 3 \u2013 5 \u2013 7 \u2013 10 \u2013 2) \u2013 (Area 1 \u2013 5 \u2013 7 \u2013 10 \u2013 1) = (Enthalpy at 2) \u2013 (Enthalpy at 1) Here, W = h2 \u2013 h1 h h2 = g T at 2 Refrigeration effect = Area 1 \u2013 4 \u2013 9 \u2013 10 \u2013 1 = (Area 1 \u2013 5 \u2013 7 \u2013 10 \u2013 1) \u2013 (Area 4 \u2013 5 \u2013 7 \u2013 9 \u2013 4) = (Enthalpy at 1) \u2013 (Enthalpy at 4) Refrigeration effect = h1 \u2013 h4 Enthalpy at 3 = Enthalpy at 4 For throttling process 3 \u2013 4, h3 = h4 (Area 3 \u2013 7 \u2013 8 \u2013 3) = (Area 4 \u2013 5 \u2013 7 \u2013 9 \u2013 4) or, or, (Area 3 \u2013 5 \u2013 6 \u2013 3) = (Area 6 \u2013 5 \u2013 7 \u2013 8 \u2013 6) = (Area 6 \u2013 5 \u2013 7 \u2013 8 \u2013 6) \u2013 (Area 4 \u2013 6 \u2013 8 \u2013 9 \u2013 4) (Area 3 \u2013 5 \u2013 6 \u2013 3) = (Area 4 \u2013 6 \u2013 8 \u2013 9 \u2013 4) (Area 3 \u2013 5 \u2013 7 \u2013 8 \u2013 3) \u2013 (Area 5 \u2013 7 \u2013 0 \u2013 5) \u2013 (Area 6 \u2013 5 \u2013 0 \u2013 8 \u2013 6) = (Area 4 \u2013 6 \u2013 8 \u2013 9 \u2013 4) (Enthalpy at 3) \u2013 (Enthalpy at 5) \u2013 (T1 \u00d7 \u2206s56) = (T1", " \u00d7 \u2206s46) s\u2212 \u21d2 (Area 3 \u2013 5 \u2013 6 \u2013 3) = {( h\u2212 )} h s f at T 2 f at T 1 ) \u2013 T1 \u00d7 ( f at T 2 f at T 1 Here enthalpy at 3 and enthalpy at 5 are enthalpies for saturated liquid states at temperature T2 and T1 respectively. Enthalpy at 3 = h f T at 2 = h3 Enthalpy at 5 = h f State 6 lies vertically below 3 so, s3 = s6 and, s Entropy at 5 = = h5 T at 1 f T at 1 = s5 Entropy at 3 = s f T at 2 = s3 = s6 \u2206s56 = s6 \u2013 s5 = (s3 \u2013 s5) = ( s f T at 2 \u2013 s f ) T at 1 So, for known h3, h5, \u2206s56 the entropy change \u2206s46 can be obtained as under; \u2206s46 = 1 T 1 ( h f at T 2 h\u2212 f at T 1 ) \u2013 \u2206s56 Therefore, for getting work, W = Area 1 \u2013 2 \u2013 3 \u2013 5 \u20131 = (Area 3 \u2013 5 \u2013 6 \u2013 3) + Area (2 \u2013 3 \u2013 6 \u2013 1 \u2013 2) = {( h f at T 2 h\u2212 f at T 1 ) \u2013 T1 \u00d7 ( s f at T 2 s\u2212 f at T 1 )} + {(T2 \u2013 T1) \u00d7 \u2206s23} h\u2212 W = {( h f at T 2 T 2 s\u2212 h\u2212 The refrigeration effect = (Area 1 \u2013 4 \u2013 9 \u2013 10 \u2013 1) ) \u2013 T1 \u00d7 ( W = {( at at at at at f f f f f ) \u2013 T1 \u00d7 ( s s\u2212 f at T 1 f at T 1 = T1 \u00d7 \u2206s14 = T1 \u00d7 (\u2206s16 \u2013 \u2206s46) )} + {(T2 \u2013 T1) \u00d7 (s2 \u2013 s3)} s\u2212 )} + {(T2 \u2013 T1) \u00d7 ( T 2 s at g f )} at T 2 Introduction to Refrigeration and Airconditioning ____________________________________ 817 here, \u2206s16 = s1 \u2013 s6 = Substituting \u2206s16 and \u2206s46 values in expression for refrigeration effect. T 2 T 2 s at at g f s\u2212 Refrigeration effect = T1 {( s g at T 2 s\u2212 f at at T 2 h\u2212 f at T 1 ) \u2013 \u2206s56)} = T1 {( s g at T 2 s\u2212 f at T 2 ) \u2013 = T1 {( s g at T 2 s\u2212 f at at T 2 s\u2212 f at T 1 ) + ( s f at T 2 s\u2212 f at T 1 )} ( h f at T 2 h\u2212 f at T 1 )} Here, COP of cycle = COP = Refrigeration effect Work done { ( s T 1 g { ( h f at T 2 \u2212 h f at T 1 \u2212 s at T 2 \u2212 ( T s 1 ) f at T 1 f at T 2 \u2212 ) ( h f at \u2212 s f at T 1 \u2212 h ( T 2 T 2 + ) at f \u2212 T 1 )\u00b7( T 1 s g at T 2 \u2212 s f at T 2 } ) } ) T 1 )(1/ Case 2 : Compression yields wet vapour at the end of compression Figure 18.7 gives T-s representation of the vapour compression cycle having compression process yielding wet vapour after compression i.e. at state 2. Let wetness after compression be x2. Work done = Area Area 3 \u2013 5 \u2013 6 \u2013 3) + (Area 2 \u2013 3 \u2013 6 \u2013 1 \u20132) From earlier analysis, Area 3 \u2013 5 \u2013 6 \u2013 3 = Area (4 \u2013 6 \u2013 8 \u2013 9 \u2013 4) Area 3 \u2013 5 \u2013 6 \u2013 3 = {( h h\u2212 f at T 2 f at T 1 ) \u2013 T1 ( s f at T 2 s\u2212 f at T 1 )} T T2 3 2 5 T1 6 4 1 7 0 8 9 10 s Fig. 18.7 Work = {Area 3 \u2013 5 \u2013 6 \u2013 3} + {Area 2 \u2013 3 \u2013 6 \u2013 1 \u2013 2} Work = {( h f at T 2 h\u2212 f at T 1 ) \u2013 T1 ( s f at T 2 s\u2212 f at T 1 )} + \uf8f1 \uf8f2 \uf8f3 ( T 2 \u2212 T 1 ) \u00d7 T at 2 \u00b7 x h 2 fg T 2 \uf8fc \uf8fd \uf8fe Refrigeration effect = Area 1 \u2013 4 \u2013 9 \u2013 10 \u2013 1 818 ________________________________________________________ Applied Ther", "modynamics = {Area 1 \u2013 6 \u2013 8 \u2013 10 \u2013 1}\u2013 {Area 4 \u2013 6 \u2013 8 \u2013 9 \u2013 4} = {Area 1 \u2013 6 \u2013 8 \u2013 10 \u2013 1} \u2013 {Area 3 \u2013 5 \u2013 6 \u2013 3} s fg = {T1 \u00b7 (x2 \u00b7 \u22c5 \u22c5 (( x 2 { T 1 s at 2 T + s )} \u2013 {( h f at T 2 h\u2212 f at T 1 s f at T 2 s\u2212 f at T 1 )} ) \u2013 T1 ( \u2212 h Refrigeration effect = fg f at COP may be obtained by taking the ratio of refrigeration effect and work. Case 3: Compression yields superheated vapour after compression Vapour compression cycle having the compression process yielding superheated vapour at the end at at at at f f f } \u2212 ) s \u2212 )} { h of compression is shown on T-s diagram in Fig. 18.8. Cycle is shown as 1\u20132\u20133\u20134\u20131. T5 T2 T 3 2' 5 T1 6 4 11 2 1 7 0 8 9 12 10 s Fig. 18.8 In this cycle additional work shown by area 2 \u2013 2' \u2013 11 \u2013 1 \u2013 2 is done and also the additional refrigeration effect shown by area 1 \u2013 11 \u2013 12 \u2013 10 \u2013 1 is available. Additional work W' = Area 2 \u2013 2' \u2013 11 \u2013 1 \u2013 2 = (Area 2 \u2013 2' \u2013 12 \u2013 10 \u2013 2) \u2013 (Area 1 \u2013 11 \u2013 12 \u2013 10 \u2013 1) W' = Cp (T5 \u2013 T2) \u2013 \uf8eb \uf8ec \uf8ec \uf8ed T C 1 \u00b7 lnp \uf8eb \uf8ec \uf8ed T 5 T 2 \uf8f6 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f8 Additional refrigeration effect = (Area 1 \u2013 11 \u2013 12 \u2013 10 \u2013 1) = T1 Cp ln \uf8eb \uf8ec \uf8ed T 5 T 1 \uf8f6 \uf8f7 \uf8f8 Therefore, COP of system could be obtained by ratio of total refrigeration effect and total work done. COP = Refrigeration effect (= Area 1 \u2013 4 \u2013 9 \u2013 10 \u2013 1) Work done (= Area )'This ratio of mass fraction of gas (vapour) to liquid i.e. \uf8eb \uf8ec \uf8ed m 9 m 7 \uf8f6 \uf8f7 \uf8f8 can be estimated by energy balance on flash chamber. State 3 can be determined by the energy balance on mixing box. Introduction to Refrigeration and Airconditioning ____________________________________ 819 Thermodynamic analysis Thermodynamic analysis of multistage vapour compression cycle can also be carried out for the dry compression considered here based on the assumptions as under (i) All processes of refrigeration cycle are internally reversible except the processes in expansion valve, flash chamber and mixing box. (ii) The processes in expansion valve, flash chamber, mixing box and compressors operate adiabatically. (iii) Refrigerant leaving condenser is saturated liquid and the refrigerant entering low pressure compressor is saturated vapour. (iv) There are no changes in kinetic and potential energies in different processes. For mass flow rate of m kg/s and dryness fraction of refrigerant at inlet of flash chamber being x, the refrigeration effect, net work and COP can be given as under. Refrigeration capacity or refrigeration effect, Net work or Work input, Wcompressor = {(m \u2013 x \u00b7 m) (h2 \u2013 h1) + m (h4 \u2013 h3)}kJ/s Qabsorbed = (m \u2013 x \u00b7 m) \u00b7 (h1 \u2013 h8) kJ/s COP = {( \u2212 ( m xm h 2 \u2212 m xm h 1 + \u2212 )( ) )( )} = {(1 \u2212 \u2212 (1 )( x h 2 )( )} 18.7 MULTISTAGE VAPOUR COMPRESSION CYCLE Intercooled compression using flash chamber is shown in Fig. 18.9. Here compression for total compression ratio occurs in two stages i.e. LP compressor and HP compressor. Qrejected m, kg/s 5 Condenser Expansion valve 1 m, kg/s 6 Flash chamber (m-xm), kg/s 7 9 x.m, kg/s Expansion valve 2 4 m, kg/s 3 High pressure compressor Mixing box 2 (m-xm), kg/s Evaporator 1 8 Low pressure compressor Qabsorbed (a) Two stage intercooled vapour compression cycle with flash chamber 820 ________________________________________________________ Applied Thermodynamics 1\u20132, 3\u20134 Reversible adiabatic 5\u20136, 7\u20138 6\u20137, 6\u20139 4\u20135 9", "\u20133\u20132 compression Irreversible throttling process In flash chamber Isobaric condensation Mixing of refrigerant from LP compressor with cool vapour from flash chamberb) T-s representation of two stage intercooled vapour compression cycle Fig. 18.9 Multistage vapour compression cycle Refrigerant is partly compressed in LP stage from 1 to 2. Compressed refrigerant leaving compressor is cooled by mixing it with the cool vapour leaving flash chamber at state 9 and finally yielding low pressure refrigerant at low temperature, state 3 (Temperature at 3 < Temperature at 2). This reduced temperature of refrigerant from 2 to 3 results in smaller compressor work requirement. Subsequently, refrigerant is compressed in high pressure compressor upto state 4. High pressure refrigerant is passed through condenser where it gets condensed into saturated liquid form of refrigerant and passed through expansion valve (expansion valve 1) from state 5 to state 6. Expansion is carried out till the refrigerant\u2019s pressure gets reduced to interstage pressure and sent to flash chamber. Liquid vapour mixture entering flash chamber is separated such that liquid fraction of refrigerant passes into next expansion valve (expansion valve 2) at state 7 and the gaseous (vapour) fraction of refrigerant leaves flash chamber at state 9 for intercooling refrigerant in mixing box between state 2 and 3. Fraction of liquid refrigerant entering expansion valve is expanded from state 7 to 8. As a result of expansion the refrigerant\u2019s pressure gets reduced and low pressure liquid refrigerant is passed through evaporator from state 8 to 1 and its phase transformation occurs from liquid to vapour. Inside evaporator this phase transformation from liquid to vapour (gas) occurs by absorbing latent heat of vaporization from the refrigerated space thereby showing cooling affect. 18.8 ABSORPTION REFRIGERATION CYCLE In vapour compression cycle it is seen that the power input is made for compressing refrigerant from low pressure to high pressure. For constant refrigerating capacity the COP of refrigeration system can be improved by reducing the work input to cycle. Here in order to reduce work input to cycle the compressor is being replaced by pump if the refrigerant is in liquid form instead of being in gas (vapour) form. Since the average specific volume of liquid solution is much less than that of refrigerant in vapour form so considerably less amount of work is required for increasing its pressure in pump. The absorption refrigeration cycle is based on this premise of employing pump for raising pressure of refrigerant in order to take advantage of less work requirement. For realizing it entirely different arrangement is employed as shown in Fig. 18.10. Such arrangement is called absorption refrigeration system. Here aqua-ammonia absorption refrigeration system has been discussed. Ammonia works as refrigerant and water acts as carrier for refrigerant in this cycle. Ammonia vapour leaving evaporator enter the absorber where ammonia vapours are released through perforated pipe into water for getting Introduction to Refrigeration and Airconditioning ____________________________________ 821 absorbed in water. This dissolving of NH3 in water result in strong aqua-ammonia solution. Mixing of NH3 in water is an exothermic process where heat is released similar to condensation process of pure substance. Due to heat release in absorber some arrangement should be made so that heat could be extracted out for maintaining its constant temperature. This mixing of ammonia into water is more when temperature of water mixing is low. Therefore, it is always tried to maintain absorber at low temperature and for this cooling water may be circulated in absorber. This strong aqua-ammonia solution prepared in absorber is pumped employing a pump for being sent to generator at high pressure i.e. condenser pressure. Generator is used for separating out NH3 vapour. For driving out ammonia from strong-aquaammonia solution heat is added to generator. The separation of NH3 from strong solution is an endothermic process similar to evaporation process. Generator may have arrangement for heat addition employing burners. Ammonia vapours leaving generator at high pressure are sent to condenser at state 2 where the condensation of ammonia takes place yielding high pressure ammonia in liquid form at state 3. High pressure ammonia liquid is passed through expansion valve from state 3 to 4. Weak solution of aquaammonia left inside generator is sent to absorber at state d. Thus, it is seen that the combination of absorber, pump and generator helps in getting high pressure refrigerant (ammonia) without using compressor. A Qrejected Qadd High pr. NH liquid 3 3 Condenser 2 NH3 vapour Heat added for removal of NH from strong solution 3 Generator Expansion valve Low pr. NH3 4 Evaporator b, strong liquid solution of (H O + NH ) 3 2 c, Weak liquid solution of (H O + NH ) 3 2 Valve Absor", "ber d 1 NH 3 vapour Qabsorbed A Wpump a Pump Cooled water, in Out Fig 18.10 Ammonia-water absorption system Low pressure ammonia leaving expansion valve may be pure liquid or liquid-vapour mixture for being sent to evaporator at state 4. Ammonia leaving evaporator after picking up heat from refrigerating space gets transformed into vapour at state 1. Ammonia vapours enter into absorber for getting mixed with weak aqua-ammonia solution to result in strong aqua-ammonia solution at state a. It is obvious from the arrangement that work input to absorption cycle gets considerably reduced but the cost incurred in replacing compressor by absorber, pump and generator and heat source etc. is quite large and nullifies the advantage of small work input. Here in absorption system the arrangement of components 822 ________________________________________________________ Applied Thermodynamics between stages 2, 3, 4 and 1 is similar to that of vapour compression system i.e. condenser followed by expansion valve and evaporator as shown on left of section A\u2013A on schematic of ammonia water absorption system (Fig. 18.10). Performance of aqua-ammonia absorption depends largely upon the characteristics of binary mixtures i.e. water and ammonia in present case. The generic phase equilibrium diagram for ammonia-water mixture is shown in Fig. 18.11. Vapour at p2 2 C Vapour at p1 Temp. b1 a b2 Liquid at p2 a1 Liquid at p1 Saturated liquid line Saturated vapour line Here x = mNH3 m + m NH 3 H O 2 0 x 1.0 Fig. 18.11 Ammonia-water mixture equilibrium diagram Phase equilibrium diagram has temperature on y axis and mass fraction of ammonia, x on x-axis. The saturated liquid states and saturated vapour states are shown on saturated liquid line and saturated vapour lines given by firm lines and dotted lines respectively. Saturation curves are shown for two pressures p1 and p2 between which vapour absorption cycle operates. Here when x = 0 then mixture shall be complete water and the temperature shall be saturation temperature of water at respective pressure. When x = 1 then mixture shall be complete ammonia and temperature is saturation temperature of ammonia at respective pressure. Phase equilibrium diagram shows that the saturated liquid and saturated vapour can exist together in equilibrium as evident from states a and a1 respectively at any temperature. It also shows that at any temperature the concentration of ammonia is more in vapour than ). For the absorption cycle shown in Fig. 18.11 the saturated liquid at state a liquid i.e. ( leaving absorber at pressure p1 enters the pump for being pumped upto pressure p2. For causing evaporation of NH3 the mixture is heated in generator upto state b. Further heat addition in generator causes evaporation of liquid to vapour yielding state b2. Transformation from b1 to b2 results in formation of ammonia rich vapour thereby reducing concentration of ammonia in liquid. If the pressure and temperature are not changed then evaporation gets stopped as lower concentration liquid is not at its boiling point. With further heat addition the evaporation of mixture continues and the mixture\u2019s temperature is raised showing new liquid and vapour states as C and 2. Liquid at C is weak aqua-ammonia mixture and is passed through valve back to absorber. NH3 vapour (gas) leaves at state 2 from generator to the condenser for being condensed. x> a x a 1 ( ) 18.9 MODIFIED ABSORPTION REFRIGERATION CYCLE In the absorption cycle discussed earlier it is seen that the strong ammonia solution preparation requires low temperature in absorber while vaporization of NH3 from strong aqua-ammonia solution needs heating up. Thus, weak solution leaving generator shall be at higher temperature while strong solution leaving absorber shall be at lower temperature. It may be noted that it will be desirable to send strong solution Introduction to Refrigeration and Airconditioning ____________________________________ 823 into generator at high temperature so that less external heat addition is required. Also, it is desirable to send weak solution into absorber at lower possible temperature so that less external cooling is needed for effective absorption. Using above two facts the absorption cycle can be modified by employing a heat exchanger between generator and absorber as shown in Fig. 18.12. Fig. 18.12 Modified ammonia-water absorption system Ammonia vapour leaving generator are outcome of evaporation from aqua-ammonia mixture due to its heating and there is possibility of water vapour accompanying the ammonia vapour leaving generator. This moisture in refrigerant may undergo freezing (probably at expansion valve or evaporator) and choke the refrigerant flow circuit which is not desirable. Therefore, for preventing moisture accompanying refrigerant from entering into circuit the device called rectifier is put after generator where traces of water are separated out and retained. Rect", "ifier is shown in schematic of modified absorption cycle, Fig. 18.12. Thus, modified absorption cycle employing heat exchanger and rectifier permits to take advantage of heat of weak solution and make refrigerant free of moisture content. Such type of absorption cycle can also be run using \u2018lithium-bromide\u2019 as absorbent and \u2018water\u2019 as refrigerant on the principle similar to \u2018water-ammonia system\u2019 discussed here. This lithium bromide and water system may not offer very low temperature refrigeration and so this cycle may be cascaded with another cycle using refrigerant having good low temperature characteristic, such as ammonia. This combination is termed as cascade refrigeration system. 18.10 HEAT PUMP SYSTEMS Heat pump refers to the device which is used for maintaining a region or body at temperature more than that of surroundings. Heat pumps have their working similar to refrigerator with only difference of objective i.e. heat pump maintains temperature more than surroundings contrary to refrigerator which maintains temperature less than surroundings temperature. Heat pump systems can be based upon vapour compression cycle, absorption cycle etc. and are extensively used for space heating applications in domestic and industrial premises. Reversed Carnot cycle is the ideal thermodynamic cycle for heat pumps similar to the refrigerators. Heat pump system based on vapour compression cycle is described here in Fig. 18.13. 824 ________________________________________________________ Applied Thermodynamics Qrejected Heat is continously rejected to space being heated 3 Condenser 2 Expansion valve 4 Evaporator 1 Qabsorbed Heat is absorbed from surroundings Wcompressor Compressor Fig. 18.13 Heat pump system based on vapour compression cycle Here compressor compresses the refrigerant from low pressure state 1 to high pressure state 2. High pressure gaseous refrigerant is sent to condenser where the condensation of refrigerant occurs rejecting heat equal to the latent heat of condensation to the space being heated up. Subsequently refrigerant in saturated liquid form passes into expansion valve from state 3 to 4. Low pressure liquid refrigerant leaving expansion valve subsequently enters the evaporator where its evaporation occurs up to state 1. Evaporation of refrigerant occurs due to heat being provided by surroundings, water in rivers/ ponds, wells or solar collectors etc. Hence the performance of heat pump can be given by: COP of heat pump = Desired effect (heating of space) Net work ( ) ) = COP of heat pump = \uf8f1 \uf8f2 \uf8f3 ( ) ) \uf8fc \uf8fd \uf8fe Here, heat pump works between the low temperature surroundings or other heat sources giving heat of Qabsorbed and the region/body acting as high temperature body where heat Qrejected is rejected to region/body. These heat pumps are not very attractive options when used for only direct heating but they become very attractive when used with some other application. 18.11 REFRIGERANTS Refrigerant is the working fluid used in refrigeration/air conditioning equipments having capability of carrying heat/rejecting heat in the form of sensible heat or latent heat. Refrigerants carrying heat/rejecting heat in the form of latent heat are more effective and for this refrigerant should possess suitable properties to get transformed from liquid to gas and vice-versa. During selection of refrigerant its chemical, physical and other general properties are being looked into along with refrigeration cycle requirements and application. Refrigerants can be classified into two main categories as \u2018Primary refrigerants\u2019 and \u2018Secondary refrigerants\u2019. The primary refrigerants are those refrigerants which are directly involved in Introduction to Refrigeration and Airconditioning ____________________________________ 825 refrigeration system while secondary refrigerants are first cooled by primary refrigerants and then used for imparting refrigeration. The primary refrigerants are used in vapour compression systems while secondary refrigerants are liquids used for transporting low-temperature heat energy from one place to another as done by \u2018brine\u2019, antifreeze agents etc. Commonly used refrigerants are air, ammonia, carbondioxide, sulphur-dioxide, fluorinated hydrocarbons and Freons etc. All refrigerants are assigned with internationally acceptable number such as R\u201312, R\u2013717 etc. (i) Primary refrigerants: Primary refrigerants can be further classified into following categories depending upon their characteristics. (a) Halocarbon compounds: Halocarbon group includes refrigerants containing one or more of three halogens such as chlorine, fluorine and bromine. These refrigerants are traded in market under the brand names of Freon, Genetron, Isotron and Arctron. Halocarbons, specially chlorine containing halocarbons (chloro fluoro carbons, CFCs) were commonly used from about 1940 to 1990s in most of vapour compression systems. But due to disastrous effect of chlorine in refrigerant upon the earth\u2019s", " protective ozone layer the efforts are being made to replace the use of CFCs by the class of refrigerants having hydrogen in place of chlorine. Such new ecofriendly refrigerants are called hydrofluoro carbons (HFCs) for example R\u2013134a (CF3 CH2F) can replace R-12 (CCl2F2). Name of refrigerant R\u201311 R\u201312 R\u201313 R\u201321 R\u201322 R\u201330 R\u201340 R\u2013100 R\u2013113 R\u2013114 R\u2013152 Table 18.1 Halocarbon refrigerants Chemical name Chemical formula Trichloro mono fluoro-methane Dichloro difluoro-methane Monochloro trifluoro-methane Dichloro monofluoro-methane Monochloro difluoro-methane Methylene-chloride Methly chloride Ethyl chloride Trichloro tetrafluoro-ethane Dichloro tetrafluoro-ethane Difluoro-ethane CCl3F CCl2F2 CClF3 CHCl2F CHClF2 CH2Cl2 CH3Cl CH3CH2Cl C2F3Cl3 CClF2CClF2 CH3CH3F2 The numbering of halocarbon group refrigerants is based on; \u2013 first digit on right gives number of fluorine atoms. \u2013 second digit from right is one more than the number of hydrogen atoms. \u2013 third digit from right is one less than the number of carbon atoms. Third digit is not given when it is zero. Some halocarbon refrigerants are as given in Table 18.1. (b) Inorganic compounds: Refrigerants used in olden days were inorganic compounds. Some of such refrgerants are still used in different applications such as ice plants, steam refrigeration, aeroplanes and ship refrigeration etc. due to their inherent thermodynamic and physical properties. These inorganic compounds are listed in Table 18.2. 826 ________________________________________________________ Applied Thermodynamics Table 18.2 Inorganic compound refrigerants Name of refrigerant Chemical name Chemical formula R\u2013717 R\u2013718 R\u2013729 R\u2013744 R\u2013764 Ammonia Water Air Carbondioxide Sulfurdioxide NH3 H2O CO2 SO2 (c) Hydrocarbons: Some of hydrocarbons as given in Table 18.3, are also used as refrigerants particularly in petroleum and petrochemical industry. Table 18.3 Hydrocarbon refrigerants Name of refrigerant Chemical name Chemical formula R\u201350 R\u2013170 R\u2013290 Methane Ethane Propane CH4 C2H6 C3H8 (d) Azeotropes: Azeotropes are those mixture of different refrigerants which do not separate into their components by distillation or with the change in pressure and temperature. Azeotrope refrigerants behave differently from their constituents such as azeotrope evaporates and condenses as single substance having properties that are different from those of either constituents. Examples of azeotrope are; R\u2013502 is azeotrope which is mixture of 48.8%, R\u201322 and 51.2% of R\u2013115. R\u2013500 is azeotrope which is mixture of 73.8% R\u201312 and 26.2% of R\u2013152. (e) Unsaturated organic compounds: These are refrigerants from hydrocarbon group and have ethylene or propylene as the base. Table 18.4 gives-such refrigerants. Table 18.4 Unsaturated organic compounds Name of refrigerant Chemical name Chemical formula R\u20131120 R\u20131130 R\u20131150 R\u20131270 Trichloro ethylene Dichloro ethylene Ethylene Propylene C2H4Cl3 C2H4Cl2 C2H4 C3H6 (ii) Secondary refrigerants: Secondary refrigerants are indirect refrigerants that carry heat from the substance being cooled to the evaporator. Secondary refrigerants undergo change in their temperature by absorbing heat and rejecting it at evaporator without any phase change. Examples of secondary refrigerants are water, brine and antifreezes etc. Antifreezing agents may be solution of water and ethylene glycol, propylene glycol or calcium chloride, with freezing temperature well below 0\u00baC. Secondary refrigerants offer advantages of ease of handling and control, ease of maintaining temperature of large buildings by controlling flow of secondary refrigerants such as brine etc. Secondary refrigerants also eliminate requirement of long primary refrigerant lines. Introduction to Refrigeration and Airconditioning ____________________________________ 827 18.12 DESIRED PROPERTIES OF REFRIGERANTS A refrigerant being used in refrigeration should have following thermodynamic properties, physical properties and working properties. (i) Boiling", " temperature of refrigerant should be quite low at atmospheric conditions for effective refrigeration. For refrigerants having higher boiling temperatures at atmospheric conditions the compressor is run at higher vacuum. (ii) For an ideal refrigerant the freezing temperature of refrigerant should be quite low so as to prevent its freezing at evaporator temperature. Freezing point temperature should be less than evaporator temperature. For example, refrigerant R\u201322 has freezing point of \u2013 160\u00baC and normally most of refrigerants have freezing point below \u2013 30\u00baC. (iii) Critical temperature of the ideal refrigerant should be higher than the condenser temperature for the ease of condensation. (iv) Refrigerant should have large latent heat at evaporator temperature as this shall increase the refrigerating capacity per kg of refrigerant. (v) Refrigerant should have small specific volume at inlet to compressor as this reduces compressor size for same refrigeration capacity. (vi) Specific heat of refrigerant in liquid form should be small and it should be large for refrigerant in vapour form, since these increase the refrigerating capacity per kg of refrigerant. (vii) Thermal conductivity of refrigerant should be high. (viii) Viscosity of refrigerants should be small for the ease of better heat transfer and small pumping work requirement. (ix) Refrigerant should be chemically inert and non toxic. (x) Refrigerant should be non-flammable, non explosive and do not have any harmful effect upon coming in contact with material stored in refrigeration space. (xi) Refrigerant may have pleasant distinct odour so as to know about its leakage. (xii) Refrigerant should be readily available at lesser price. Choice of refrigerant: Selection of suitable refrigerant depends upon the number of parameters as there can not be single refrigerant well suited for all kinds of refrigeration systems. Thermodynamic properties, physical properties and other properties are considered in reference to the refrigeration system during choice of suitable refrigerant. Apart from these properties, due consideration should be given to the working pressure and temperature range and pressure ratio, space restrictions, corrosiveness and inflammability, oil miscibility etc. before selecting refrigerant with suitable compressor and other equipments of refrigeration system. 18.13 PSYCHROMETRY Psychrometry or Psychrometrics refers to the study of system involving dry air and water. Properties of mixture of air and water are called psychrometric properties. Psychrometry becomes very important from air conditioning point of view as in air conditioning the human comfort conditions defined in terms of temperature, humidity and air circulation/ventilation are being considered. Different psychrometric properties, charts and processes generally used are defined ahead. (i) Dry air: Atmospheric air having 79% nitrogen and 21% oxygen by volume is considered dry air. Its, molecular weight is taken as 29. (ii) Moist air: Moist air is the mixture of dry air and water vapour in which dry air is treated as if it were pure component. Quantity of water vapour present in the mixture depends upon the temperature of air and it may vary from zero in dry air to the maximum quantity when mixture is 828 ________________________________________________________ Applied Thermodynamics saturated of water vapour (called saturation capacity of air). Moist air is assumed to behave as ideal gas for the purpose of analysis. Mixture pressure is the sum of partial pressures of dry air and water vapour. When the partial pressure of water vapour corresponds to the saturation pressure of water at mixture temperature then mixture is said to be saturated. Saturated air is the mixture of dry air and saturated water vapour. When the temperature of mixture of air and vapour is above the saturation temperature of water vapour then the vapour is called superheated vapour. (iii) Humidity ratio or specific humidity: This is given by the ratio of the mass of water vapour to the mass of dry air. It can also be defined as mass of water vapour present in per kg of dry air. It is given in terms of grams per kg of dry air. Mathematically, humidity ratio or specific humidity, \u03c9 = m v m a where mv and ma are mass of water vapour and dry air respectively. Humidity ratio can also be given in terms of partial pressures and molecular weights. mv = ma = \u00b7 M p V \u00b7 v v RT \u00b7M p V \u00b7 a a RT where pv and pa are partial pressures of water vapour and dry air respectively, Mv and Ma are molecular weights of water vapour and dry air respectively. For total pressure of mixture being p, the Dalton\u2019s law of partial pressure says, or, or, Hence, upon substitution, p = pa + pv pa = p \u2013 pv \u03c9 = \u00b7 M p v = v \u00b7\u2212 \u00b7( M p a ) v \u03c9 = 0.622 \uf8eb p \uf8ec \u2212\uf8ed p v p v", " \uf8f6 \uf8f7 \uf8f8 Humidity ratio is measured using an instrument called \u2018hygrometer\u2019 in which moist air sample is processed till the moisture present in it gets absorbed and moisture absorbed can be quantified. (iv) Absolute humidity: Absolute humidity refers to the weight of water vapour present in unit volume of air. (v) Relative humidity: Relative humidity gives an account of moisture content in an actual mixture as compared to the mixture in saturated state at same temperature and pressure. It can be given by the ratio of actual mass of water vapour in given volume to the mass of water vapour if the air is saturated at the same temperature and pressure. Mathematically, relative humidity (\u03c6) can be given as: \u03c6 = ( (Actual mass of water vapour in given volume of mixture) Maximum mass of water vapour in same volume of mixture being staurated at same temperature and pressure ) Introduction to Refrigeration and Airconditioning ____________________________________ 829 \u03c6 = m v m v, sat, where mv and mv, sat are the actual and maximum mass of water vapour in same volume of mixture as defined earlier. \u00b7 M p V \u00b7 v v RT \uf8eb \uf8ec \uf8ed \u00b7 M p v v, sat \uf8f6 \uf8f7 \uf8f8 \u00b7 V RT \uf8f6 \uf8f7 \uf8f8 \u03c6= \uf8eb \uf8ec \uf8ed or, \u03c6 = p v p v, sat, where pv and pv, sat are the partial pressures of \u2018actual mass of water vapour\u2019 and \u2018maximum mass of water vapour in same volume of mixture as defined earlier\u2019. (vi) Enthalpy of moist air or Mixture enthalpy: The enthalpy of moist air can be estimated by addition of the enthalpy contribution of each constituent component i.e water vapour and dry air at the state of mixture. Thus, enthalpy of mixture (moist air) of dry air and water can be given as, H = Ha + Hv = ma ha + mv hv, or, H m a = ha + \uf8eb \uf8ec \uf8ed m v m a \uf8f6 \uf8f7 \uf8f8 hv (Enthalpy of moist air per unit mass of dry air) = (Enthalpy of unit mass of dry air) + (Enthalpy of water vapour associated with unit mass of dry air) Here H and h refer to total enthalpy and specific enthalpy values respectively. Subscripts a and v refer to dry air and water vapour respectively. (vii) Dry bulb temperature: Dry bulb temperature refers to the temperature of air measured with ordinary thermometer having its bulb open. (viii) Wet bulb temperature: Wet bulb temperature refers to the temperature measurement with the thermometer having its bulb covered with wet cloth and exposed to air stream whose temperature is being measured. For getting wet bulb temperature thermometer bulb is covered with water wet cloth/wick moistened with water and then temperature is measured. Difference between dry bulb temperature and wet bulb temperature is called wet bulb depression. Wet bulb depression is zero in case of saturated air as the dry bulb temperature and wet bulb temperatures are equal. (ix) Dew point temperature: It is the temperature at which condensation of water vapour present in moist air just begins. It could be understood from different examples that if the moist air temperature is reduced then water vapour gets condensed. Examples for such condensation is formation of dew on grass, condensation of water vapour on exterior surface of steel tumbler having cold water inside, condensation of water vapour on glass window-panes during winter season etc. Dew point temperature shall be equal to the saturation temperature corresponding to partial pressure of water vapour in moist air. Difference between dew point temperature and dry bulb temperature is called dew point depression. (x) Humid specific volume: It refers to the volume of mixture per kg of dry air in mixture and is given by m3/kg of dry air. 830 ________________________________________________________ Applied Thermodynamics (xi) Psychrometric charts: Psychrometric chart gives the graphical representation of different important properties of moist air. These charts are readily available for different mixture pressure. Fig. 18.14 shows psychrometric chart. Abscissa (x-axis) of chart gives the dry bulb temperature (\u00baC) and the ordinate (y-axis) has humidity ratio (\u03c9) in kg or gram of water vapour per kg of dry air. Fig. 18.14 Psychrometric chart Since humidity ratio can be directly related to partial pressure of water vapour so vapour pressure (pv) can also be shown on ordinate. Constant relative humidity (\u03c6) curves are also shown on psychrometric chart for different \u03c6 values such as \u03c6 = 10", "%,..., 60%,..., 100%. Mixture enthalpy per unit mass of dry air (kJ/kg of dry air) is also available on psychrometric chart. Dew point temperature for the moist air can be known by following the line of constant \u03c9 (or constant pv) up to saturation line, \u03c6 = 100% as the dew point refers to the state where mixture becomes saturated when cooled at constant vapour pressure. Wet bulb temperature is also available on psychrometric chart where constant wet bulb temperature is also available on psychrometric chart where constant wet bulb temperature lines run from upper left to the lower right of chart. These constant wet bulb temperature lines approximate to the lines of constant mixture enthalpy per unit mass of dry air. Pscychrometric chart also has lines representing volume per unit mass of dry air (m3/kg). These specific volume lines can be approximated as state giving volume of dry air or water vapour per unit mass of dry air since each component of mixture i.e. air and water vapour occupy the same volume. (xii) Psychrometric processes: Different psychrometric processes commonly encountered in air conditioning are as follows. (a) Sensible heating or sensible cooling: Sensible heating or cooling refers to the heating or cooling without phase change i.e. heating or cooling of air without increase or decrease of moisture content. For sensible heating the air is passed over heating coils (electrical resistance type or steam type) while for sensible cooling the air is passed over cooling coils (such as evaporator coil of refrigeration cycle), as shown in Fig. 18.15. Introduction to Refrigeration and Airconditioning ____________________________________ 831 1 2 1 2 Air in Air out Air in Air out (a) Sensible heating (without by-pass) (b) Sensible cooling (without by-pass) Heating coil Cooling coil 1 3 2 2 3 1 TDBT TDBT (c) Sensible heating on psychrometric chart (d) Sensible cooling on psychrometric chart By pass By pass B (1\u2013B) 1kg Air in 1 B kg 3 1 kg Air out Heating coil Air in 1 Air out 3 Cooling coil (e) Sensible heating with by-pass (f ) Sensible cooling with by-pass Fig. 18.15 Sensible heating and sensible cooling This heating and cooling may be for complete amount of air being passed over the coil as shown in Fig. 18.15 a and b, or the portion of air flowing can be by passed through by-pass passage thus, only a fraction is passed over coil. Later arrangement is called sensible heating or sensible cooling with by pass as shown in Fig. 18.15 e and f. Sensible heating and sensible cooling without by pass is shown on psychrometric charts Fig. 18.15 c and d, respectively between states 1 and 2. Thus, for inlet air temperature being T1, the heating/cooling causes change in temperature up to T2. When some amount of air is by passed then the temperature of air coming out shall be different from T2. In case of sensible heating with by pass, temperature of air coming out is less than temperature of air wihout by pass i.e. T3 < T2. Similarly in case of sensible cooling with by pass temperature of air leaving is more than temperature of air without by pass i.e. T3 > T2. Heat added can be quantified by the amount of heat gained by dry air and water vapour between dry bulb temperature T1 and T2. Qsensible heating = Cpa (T2 \u2013 T1) + \u03c9 \u00b7 Cpv (T2 > T1) = (Cpa + \u03c9 Cpv) (T2 \u2013 T1) Qsensible heating = Cpm \u00b7 (T2 \u2013 T1), Hence Cpm is mean specific heat. 832 ________________________________________________________ Applied Thermodynamics In case of sensible heating with by pass, if the amount of air by passed is B kg per kg of air then heat balance can be given as below. Here (1 \u2013 B) kg of air passes over the coil and later on mixes with by passed air at exit. B \u00b7 (Cpa + \u03c9 Cpv) T1 + (1 \u2013 B) (Cpa + \u03c9 Cpv) T2 = 1 \u00d7 (Cpa + \u03c9 Cpv) \u00b7 T3 or, By pass factor = on psychrometric chart length length on psychrometric chart 2-3 2-1 This amount of air by passed per kg of air flowing is also called as By-pass factor of the coil. Factor (1 \u2013 B) is called contact factor. (b) Cooling with dehumidification: This process refers to the reducing air temperature along with removal of water vapour from air. Removal of moisture content can be done by", " reducing the temperature of air below dew point temperature so as to allow condensation of water vapour. Arrangement for cooling with humidification is shown in Fig. 18.16 where moist air entering is first passed over cooling coil having cool refrigerant inside. Water vapour present in moist air gets condensed and saturated moist air leaves dehumidification section at state 2. (a) Schematic (b) Representation on psychrometric chart Fig. 18.16 Cooling with dehumidification Introduction to Refrigeration and Airconditioning ____________________________________ 833 Condensate also gets cooled up to temperature T2 before leaving dehumidification section. Moist air at state (2) may be excessively cooled so that it can be subsequently heated by passing it over heating coil. Thus, at state (3) the available air shall be at least temperature and in dehumidified form. States (1), (2) and (3) are also shown on psychrometric chart. Mass balance shows, Dry air at section (1) and (2) ma1= ma2 = ma Vapour content at section (1) and (2); mv1= mw + mv2, or mw= mv1 \u2013 mv2, It can also be given: mv1 = \u03c9 1 \u00b7 ma ; mv2 = \u03c9 Mass of water condensed per unit mass of dry air; \u03c9 Between section (2) and (3), \u03c9 1 \u2013 \u03c9 m w m a 2 \u00b7 ma 2 = 2 = \u03c9 3. (c) Humidification: This is the process of adding moisture to the air. Humidification may be required during air conditioning when air being circulated in occupied space may have little moisture in it. Available means for increasing humidity are to inject steam or spray liquid water into air. Normally the steam being injected shall be at temperature more than that of air and humidification shall be accompanied by increase in humidity ratio and dry-bulb temperature. When liquid water is sprayed then the moist air shall leave humidification section with lower temperature than at inlet and increased humidity ratio. Figure 18.17 shows schematic for humidification process along with representation on psychrometric chart for humidification using steam and liquid water spray both. (a) Schematic for humidifier (b) Psychrometric chart representation for humidification Fig. 18.17 Humidification (d) Evaporator cooling: This type of cooling is needed in hot and dry climates. Evaporative cooling has arrangement for spray of spraying liquid water into air or passing air through a pad soaked with water. Due to less humidity of air it shall evaporate some amount of water in its contact and thus reduce its temperature because of heat extracted for evaporation of water. This air leaving evaporative cooler shall have temperature less than inlet air temperature and also due to moisture being picked up the humidity ratio gets increased. Arrangement for such type of cooling and its representation on psychrometric chart is shown in Fig. 18.18. 834 ________________________________________________________ Applied Thermodynamics (a) Evaporative cooler (b) Representation of evaporative cooling on psychrometric chart Fig. 18.18 Evaporative cooling (e) Adiabatic mixing of moist air stream: Adiabatic mixing of moist air streams is the one in which there is no heat interaction during mixing. Obviously there is no work interaction during mixing. This process of mixing of moist air streams is used very frequently in air conditioning systems. Figure 18.19 shows adiabatic mixing of two streams entering at (1) and (2) and leaving at (3). If dry air entering be ma1 & ma2 and water vapour flow rate be mv1 & mv2 with streams (1) and (2) respectively at steady state then mass balances and energy balances can be given as under: Fig. 18.19 Adiabatic mixing For dry air, ma1 + ma2 = ma3 For water vapour, mv1 + mv2 = mv3 Since mv1= \u03c9 2 \u00b7 ma2, mv3 = \u03c9 1 \u00b7 ma1, mv2 = \u03c9 1 \u00b7 ma1 + \u03c9 \u03c9 1 \u00b7 ma1 + \u03c9 \u03c9 2 \u00b7 ma2 = \u03c9 2 \u00b7 ma2 = \u03c9 or, Energy balance yields; following as Q = 0, W = 0. 3 \u00b7 ma3 3 (ma1 + ma2) 3 \u00b7 ma3, then substitution yields; ma1 \u00b7 (ha1 + \u03c9 1 hg1) + ma2 \u00b7 (ha2 + \u03c9 2 hg2) = ma3 \u00b7 (ha3 + \u03c9 3 hg3) Introduction to Refrigeration and Airconditioning ____________________________________ 835 18.14 AIR CONDITIONING SYSTEMS Air conditioning systems require basic arrangement for", " getting refrigeration effect through cooling coil followed by subsequent humidification/dehumidification and heating etc. in order to provide air conditioned space with air at desired temperature and humidity. Air conditioning systems require different arrangements depending upon the atmospheric air condition and comfort condition requirement. Such as summer air conditioning systems and winter air conditioning systems are different. These systems have different arrangement if outdoor conditions are hot and humid, hot and dry etc. Summer air conditioning system for hot and dry outdoor condition is given in Fig. 18.20. Here the comfort conditions may require delivery of air to air-conditioned space at about 25\u00baC DBT and 60% relative humidity where the outdoor conditions may be up to 40\u201344\u00ba C DBT and 20% relative humidity in Indian conditions. Generic arrangement has air blower which blows air across the air filter between (1) and (2). Air coming out from filter passes over cooling coils and is subsequently sent for humidification between states (3) and (4). Large size water particles carried by air are retained by water eliminator. Air finally coming out at state (5) is sent to air conditioned space. Here psychrometric representation is made considering negligible change in humidity in water eliminator. (a) Schematic summer airconditioning (b) Representation on psychrometric chart Fig. 18.20 Summer air conditioning 836 ________________________________________________________ Applied Thermodynamics Winter air conditioning has slightly different arrangement as compared to summer air conditioning. Figure 18.21 shows the schematic for winter air-conditioning along with representation on psychrometric chart. (a) Schematic for winter airconditioning Change in enthalpy b) Representation on psychrometric chart Fig. 18.21 Winter air conditioning Let us assume air from atmosphere to be available at 10\u00baC and 80% relative humidity while comfort conditions are 25\u00baC DBT and 60% relative humidity. Here atmospheric air is blown into air filter and it comes out at (2). Air is passed across heating coil for being heated. Hot air at (3) passes through humidifier section where its humidity increases as shown on psychrometric chart by state (4). It is further passed through water eliminator to retain large sized water particles. Air may again be passed through heating coil section so as to compensate for temperature reduction in humidifier section and attain desired temperature. Introduction to Refrigeration and Airconditioning ____________________________________ 837 18.15 COMPARISON OF VARIOUS REFRIGERATION METHODS Merits and demerits of Vapour compression system over Air-refrigeration system. Vapour compression system when compared with air refrigeration system possess following merits and demerits. Merits Thermodynamic cycle involved in vapour compression system is very close to Carnot cycle hence its COP is quite high of the order of 3\u20134 while for air refrigeration COP is generally less than unity. Refrigerant required is smaller in vapour compression system because the heat is carried away by latent heat of vapour and so the amount of liquid refrigerant circulated per ton is less. Also size of evaporator is small. Operating cost of vapour compression system is very less as compared to air refrigeration system on ground. Refrigerated space temperature can be easily regulated by controlling the throttle valve. Demerits Initial investment is too large as compared to air refrigeration system. There are chances of leakage of refrigerant as large pressure differential exists throughout the cycle. Air leakage into refrigerant offers reduced performance of vapour compression system and so it should be prevented. Merits and demerits of Vapour absorption system over Vapour compression system. The merits and demerits of vapour absorption system over vapour compression system are listed as under: Merits Vapour absorption system has no moving component throughout cycle, so it works quietly and has reduced maintenance requirement. Lesser work input is required for pump as compared to compression in vapour compression system. Vapour absorption system may be run with some installation rejecting waste heat, which may be used to run generator, thus energy conservation. Vapour absorption system is well suited for large refrigeration capacity applications such as even above 1000 tons, which is difficult in vapour compression systems. Vapour absorption systems offer smaller variation with load whereas COP of vapour compression system varies largely with load. At smaller loads the COP of absorption system is more than that of vapour absorption system. Demerits Size of absorption system is large and is bulky to handle. If heating source (such as electric heating) is costly then the vapour absorption system becomes costly than vapour compression system. At very high loads the COP of vapour compression system is more than that of vapour absorption system. 838 ________________________________________________________ Applied Thermodynamics EXAMPLES 1. A refrigerator operating on reversed Carnot cycle extracts 500 kJ/min heat from a refrigerated space being maintained at \u201316\u00baC and rejects heat to the atmosphere at 27\u00baC. Deter", "mine the work input required to run the refrigerator. Solution: For refrigerator working on reversed Carnot cycle. Q 1 + \u2212 ( 16 273) = Q 2 + (27 273) Also it is given that Work input required, Q 1 Q 2 = 257 300 Q1 = 500 kJ/min Q2 = 583.66 kJ/min W = Q2 \u2013 Q1 = 83.66 kJ/min Work input = 83.66 kJ/min Ans. \u201316\u00b0C Refrigerated space W R Q1 Q2 27\u00b0C Atmosphere Fig. 18.22 2. A refrigerator operating between \u20135\u00baC and 27\u00b0C has the capacity of 800 tons. Determine the mass rate of ice formation considering water being supplied at atmospheric temperature of 27\u00baC. Also find the minimum hp required. Consider the refrigerator to work upon reversed Carnot cycle. Latent heat for ice formation from water may be taken as 335 kJ/kg. Solution: Refrigeration capacity = 800 \u00d7 3.5 = 2800 kJ/s = Heat extraction rate Let the ice formation rate be \u2018m\u2019 kg/s. Heat to be removed from per kg of water for its transformation into ice. = 4.18 \u00d7 (27 \u2013 0) + 335 kJ/kg = 447.86 kJ/kg Ice formation rate = Refrigeration capacity Heat removed for getting per kg of ice = 2800 447.86 = 6.25 kg/s Mass rate of ice formation = 6.25 kg/s Ans. COP of refrigerator, = T 1 T\u2212 1 T 2 = Refrigeration capacity Work done Here T1 = 273 \u2013 5 = 268 K, T2 = 27 + 273 = 300 K COP = 268 300 \u2013 268 = 2800 W W = 334.33 kJ/s Introduction to Refrigeration and Airconditioning ____________________________________ 839 = 334.33 0.7457 = 448.34 hp hp {as 1 hp = 0.7457 kW} HP required = 448.34 hp Ans. 3. 3 hp per ton of refrigeration is required to maintain temperature of \u201327\u00baC in refrigerator working on Carnot cycle. Determine COP of cycle and temperature of surroundings to which heat is rejected by refrigerator. Solution: Also, Equating COP, COP = T 1 T\u2212 1 T 2 = 246 T \u2212 2 246 COP = Refrigeration effect Work input = 1\u00d7 3.5 3\u00d7 0.7457 COP = 1.56 COP = 1.56 Ans. 1.56 = 246 \u00d7 246 T 2 T2 = 403.69 K Fig. 18.23 Temperature of surroundings = 403.69 K Ans. 4. An air refrigeration cycle has pressure ratio of 8, temperature of air at beginning of compression is \u201330\u00baC and air is cooled up to 27\u00baC after compression. Compression and expansion processes have isentropic efficiency of 85% and 90% respectively. Determine the refrigeration capacity and coefficient of performance. Consider Cp = 1.005 kJ/kg \u00b7K, \u03b3 = 1.4 and air flow rate of 1 kg/s. Solution: Given: T1 = \u201330\u00baC, T3 = 27\u00baC or 300 K During process 1\u20132\u2032, \u03b3 \u03b3 T1 = \u201330 + 273 = 240 \uf8f6 \uf8f7 \uf8f8 = 840 ________________________________________________________ Applied Thermodynamics T2' = 240 \u00d7 ( )(1.4 1) Theoretical temperature after compression T2' = 434.75 K For compression process, 1.4 8 \u2212 \u03b7 isen,compr. = 0.85 = T2 \u2013 240) = \u2212 (434.75 240) 0.85 T2 = 469.12 \uf8f6 \uf8f7 \uf8f8 For expansion process, 3\u20134', \u2212 1.4 1 1.4 \uf8eb \uf8ec \uf8ed \uf8f6 \uf8f7 \uf8f8 T4' = 300 \u00d7 1 8 T4' = 165.61 K T 3 T 3 T4 = 179.05 K Wc = 1 \u00d7 Cp \u00d7 (T2 \u2013 T1 isen, expansion = 0.90 = T\u2212 4 300 165.61 300 \u2212 or, Work during compression, Work during expansion = 1 \u00d7 1.005 (469.12 \u2013 240) = 230.27 kJ/s WT = 1 \u00d7 Cp \u00d7 (T3 \u2013 T4) WT = 1 \u00d7 1.005 \u00d7 (300 \u2013 179.05) = 121.55 kJ/s Fig. 18.24 Refrigeration effect is realized during process, 4\u20131. So refrigeration effect shall be, Qref = 1 \u00d7 Cp \u00d7 (T1 \u2013 T4) = 1 \u00d7 1.005 (240 \u2013 179.05) Qref = 61.25 kJ/s or", " 17.5 ton Net work required = WC \u2013 WT = 230.27 \u2013 121.55 = 108.72 kJ/s COP = Q ref W W\u2212 C T = 61.25 108.72 = 0.56 Refrigeration capacity = 17.5 ton or 61.25 kJ/s COP = 0.56 Ans. 5. A refrigeration system has air leaving refrigerated space at 7\u00baC and 1 bar and subsequently air is compressed isentropically to 5 bar. After being compressed air is cooled up to 27\u00baC at this pressure and then expanded isentropically up to 1 bar and discharged into refrigerated space. Determine COP of system. Introduction to Refrigeration and Airconditioning ____________________________________ 841 Solution: Given T1 = 7\u00baC or 280 K T3 = 27\u00baC or 300 K P1 = 1 bar, P2 = 5 bar For isentropic compression process \u2212 T2 = 280 \u00d7 ( )(1.4 1) T2 = 443.47 K 1.4 5 For isentropic expansion process \uf8eb \uf8ec \uf8ed \u2212 = ( )(1.4 1) 5 \uf8f6 T 3 \uf8f7 T \uf8f8 4 T4 = 189.42 K 1.4 Heat rejected during process 2\u20133, Q23 = m \u00d7 Cp \u00d7 (T2 \u2013 T3) Fig. 18.25 = m \u00d7 1.005 \u00d7 (443.47 \u2013 300) Q23 = 144.19 m Refrigeration process, heat picked during process 4\u20131, Q41 = m \u00d7 Cp \u00d7 (T1 \u2013 T4) = m \u00d7 1.005 \u00d7 (280 \u2013 189.42) Q41 = 91.03 m Net work = Q23 \u2013 Q41 = (144.19 \u2013 91.03) m W = 53.16 m kJ COP = Refrigeration effect Net work = 91.03m 53.16m = 1.71 COP = 1.71 Ans. 6. In a refrigerator working on Bell Coleman cycle the air enters compressor at 1 bar and \u201310\u00baC and gets compressed up to 5.5 bar. Compressed air is cooled to 27\u00baC at same pressure before being sent to expander for expansion up to 1 bar and then passes through refrigerated space. Determine refrigeration capacity, hp required to run compressor and COP of system if air flow rate is 0.8 kg/s. Solution: Given: T1 = \u201310\u00baC or 263 K, P1 = 1 bar, T3 = 27\u00baC or 300 K, P2 = 5.5 bar, m = 0.8 kg/s. 842 ________________________________________________________ Applied Thermodynamics For process 1\u20132, For process 3\u20134 T2 = 263 \u00d7 ( T2 = 428.04 K 5. \u2212 )(1.4 1) 1.4 Fig. 18.26 \u2212 \u03b3 1 \u03b3 \uf8f6 \uf8f7 \uf8f8 = T3 \u00d7 \u2212 \uf8f6 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ed T4 = T3 \u00d7 P 4 P 3 T4 = 184.33 K Refrigeration capacity = mCp (T1 \u2013 T4) = 0.8 \u00d7 1.005 (263 \u2013 184.33) = 63.25 kJ/s or 18.07 ton Refrigeration capacity = 18.07 ton Ans. Work required to run the compressor = \u00d7 m n \u2212 (p2v2 \u2013 p1v1) 1 n \u00d7 m n \u2212 R(T2 \u2013 T1) 1 n \u00d7 0.8 1.4 (0.4) \u00d7 0.287 (428.04 \u2013 263) = = = 132.63 kJ/s or HP required to run compressor = 132.63 0.7457 = 177.86 hp HP required to run compressor = 177.86 hp Ans. Net work input = mCp {(T2 \u2013T3) \u2013 (T1 \u2013 T4)} = 0.8 \u00d7 1.005 {(428.04 \u2013 300) \u2013 (263 \u2013 184.33)} W = 39.69 kJ/s COP = Refrigeration capacity Work = 63.25 39.69 = 1.59 COP = 1.59 Ans. Introduction to Refrigeration and Airconditioning ____________________________________ 843 7. Air refrigeration system for an aircraft has two cooling turbines, one intercooler and a fan. First cooling turbine is supplied with ram air at 1.2 bar, 15\u00baC and delivers it after expansion to intercooler at 0.9 bar for cooling the air coming from compressor at 4 bar. The cooling air from intercooler is sucked by fan and discharged to atmosphere and the cooled air from intercooler is expanded upto 1 bar in second turbine and discharged into passenger", " cabin. The air is exhausted from cabin at 25\u00baC and 10 ton of refrigeration is required in cabin. Index of compression and expansion are 1.45 and 1.30 respectively. Determine the air mass flow rate in cabin, ram air mass flow rate, and COP of system if compressed air is cooled to 50\u00baC in intercooler and temperature rise in intercooler for ram air is limited to 30\u00baC. Solution: C : Compressor T : Turbine CT1 = Cooling turbine 1 CT2 = Cooling turbine 2 I/C = Intercooler Fig. 18.27 Schematic air refrigeration cycle Given: P1 = P6 = 1.2 bar, T1 = T6 = 15\u00baC = 288 K, P7 = 0.9 bar P3 = P2 = 4 bar, P4 = 1 bar, T5 = 25\u00baC, T3 = 50\u00baC or 323 K, T8 = 30\u00baC For process 1\u20132, n = 1.45 For process 3\u20134, n' = 1.3, \u2212 1.45 1 1.45 \uf8f6 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8ed T2 = 288 4 1.2 T2 = 418.47 K \u2212\uf8eb'844 ________________________________________________________ Applied Thermodynamics T4 = 323 \u2212 1.3 1 1.3 \uf8eb \uf8ec \uf8ed 1 4 \uf8f6 \uf8f7 \uf8f8 Refrigeration effect in passenger cabin with m kg/s mass flow rate of air. T4 = 234.57 K 10 \u00d7 3.5 = m \u00d7 CP (T5 \u2013 T4) 10 \u00d7 3.5 = m \u00d7 1.005 (298 \u2013 234.57) m = 0.55 kg/s Air mass flow rate in cabin = 0.55 kg/s Ans. Let the mass flow rate of air through intercooler be m1 kg/s then the energy balance upon intercooler yields, m1 \u00d7 CP \u00d7 (T8 \u2013 T7) = m \u00d7 CP (T2 \u2013 T3) During process 6 \u2013 7, \u2212 1.3 1 1. T7 = 288 \u2212 1.3 1 1.3 \uf8eb \uf8ec \uf8ed 0.9 1.2 \uf8f6 \uf8f7 \uf8f8 = 269.5 K Substituting T2, T3, T7, T8 and m in energy balance on intercooler, m1 = 1.005 \u00d7 (303 \u2013 269.5) = 0.55 \u00d7 1.005 \u00d7 (418.47 \u2013 323) m1 = 1.57 kg/s Total ram air mass flow rate = m + m1 = 0.55 + 1.57 = 2.12 kg/s Ram air mass flow rate = 2.12 kg/s Ans. Work input to compressor = m \u00d7 CP (T2 \u2013 T1) = 0.55 \u00d7 1.005 (418.47 \u2013 288) = 72.12 kJ/s COP = Refrigeration effect Work input = 10 \u00d7 3.5 72.12 COP = 0.485 COP = 0.485 Ans. 8. An air cooling system for aeroplane has refrigeration capacity of 15 tons. Atmospheric air is available at 0.9 bar, 3\u00baC and due to ramming action pressure rises up to 1 bar. The ram air leaves main compressor at 4 bar and subsequently 66% of heat of compressed air is removed in air cooled heat exchanger and then passed through an evaporator for further cooling. Temperature of air gets lowered by 10\u00baC in evaporator and this cool air is then expanded in cooling turbine before being supplied to passenger cabin maintained at 1.03 bar. Air leaves cabin at 25\u00baC. Considering isentropic efficiency of compressor and turbine as 0.9 and 0.8 determine, (a) HP required to run the system, (b) COP of system. Introduction to Refrigeration and Airconditioning ____________________________________ 845 Solution: Given: P0 = 0.9 bar, T0 = 3\u00baC or 276 K, P2 = P3 = P4 = 4 bar, P5 = 1.03 bar. T3 \u2013 T4 = 10, T6 = 25\u00baC or 298 K Schematic or air refrigeration system Fig. 18.28 Considering index of compression and expansion as 1.4. During ramming action, process 0\u20131 T1 = 276 \u00d7 \u2212 1.4 1 1.4 \uf8eb \uf8ec \uf8ed 1 0.9 \uf8f6 \uf8f7 \uf8f8 During compression process 1\u20132' T1 = 284.43 \uf8f6 \uf8f7 \uf8f8 T2' = 284.43 \u2212 1.4 1 1.4 \uf8eb \uf8ec \uf8ed 4 1 \uf8f6", " \uf8f7 \uf8f8 T2' = 422.66 K 846 ________________________________________________________ Applied Thermodynamics T 1 T 1 In heat exchanger 66% of heat loss shall result in temperature at exit from heat exchanger to be, or, T2 = 438.02 K \u03b7 isen, c = 0.9 = T 2'T 2 \u2212 \u2212 T3 = 0.34 \u00d7 438.02 = 148.93 K. Subsequently for 10\u00baC temperature drop in evaporator, T4 = 148.93 \u2013 10 T4 = 138.93 K Expansion in cooling turbine during process 4\u20135', \uf8f6 \uf8f7 \uf8f8 T5' = 138.93 \u2212 1.4 1 1.4 \uf8eb \uf8ec \uf8ed 1.03 4 \uf8f6 \uf8f7 \uf8f8 T5' = 94.29 K T T 5 4 T T 5' 4 \u2212 \u2212 \u03b7 isen, T = T\u2212 5 0.8 = 138.93 \u2212 138.93 94.29 T5 = 103.22 K Let the mass flow rate of air through cabin be m kg/s. Using refrigeration capacity heat balance yields. 15 \u00d7 3.5 = m \u00d7 CP (T6 \u2013 T5) m = \u00d7 15 3.5 \u2212 1.005 (298 103.22) \u00d7 m = 0.268 kg/s Work input to compressor = 0.268 \u00d7 1.005 (T2 \u2013 T1) = 41.37 kJ/s or 55.48 hp COP = Refrigeration effect Work input = 15 \u00d7 3.5 41.37 COP = 1.27 COP = 1.27 HP required = 55.48 hp Ans. 9. A refrigerator working on simple vapour compression cycle operates between the temperature of 25\u00baC and \u2013 15\u00baC with NH3, refrigerant. Ammonia is found to be dry after compression and no under cooling of liquid refrigerant occurs in cycle. Using following data obtain COP of system. Introduction to Refrigeration and Airconditioning ____________________________________ 847 Properties of NH3 Temperature \u2013 15\u00baC 25\u00baC Solution: Enthalpy, kJ/kg hf \u2013 54.51 99.94 hg 1303.74 1317.95 Entropy, kJ/kg \u00b7 K sg 5.0536 4.4809 sf \u2013 0.2132 0.3386 T1 = \u201315\u00baC or 258 K, T2 = 25\u00baC or 298 K h2 = hg at 25\u00baC = 1317.95 kJ/kg s2 = sg at 25\u00baC = 4.4809 kJ/kg\u00b7K h3 = hf at 25\u00baC = 99.94 kJ/kg s3 = sf at 25\u00baC = 0.3386 kJ/kg\u00b7K Fig. 18.29 T-s representation Here work done, W = Area 1\u2013 2 \u2013 3 \u2013 9 \u2013 1 Refrigeration effect = Area 1\u2013 5 \u2013 6 \u2013 4 \u2013 1 Area 3 \u2013 8 \u2013 9 = (Area 3 \u2013 11 \u2013 7) \u2013 (Area 9 \u2013 11 \u2013 10) \u2013 (Area 9 \u2013 8 \u2013 7 \u2013 10) = (h3) \u2013 (h9) \u2013 T1 \u00d7 (s3 \u2013 s9) here h9 = hf at \u2013 15\u00baC = \u2013 54.51 kJ/kg, s9 = hf at \u2013 15\u00baC = \u2013 0.2132 kJ/kg \u00b7 K Substituting Area 3 \u2013 8 \u2013 9 = (99.94 \u2013 (\u2013 54.51) \u2013 258 (0.3386 + 0.2132) = 12.09 kJ/kg During throttling process between 3 and 4. h3 = h4 (Area 3 \u2013 11 \u2013 7 \u2013 3) = (Area 4 \u2013 9 \u2013 11 \u2013 6 \u2013 4) (Area 3 \u2013 8 \u2013 9) + (Area 8 \u2013 9 \u2013 11 \u2013 7 \u2013 8) or (Area 3 \u2013 8 \u2013 9) = (Area 4 \u2013 6 \u2013 7 \u2013 8 \u2013 4) = (Area 4 \u2013 6 \u2013 7 \u2013 8 \u2013 4) + (Area 8 \u2013 9 \u2013 11 \u2013 7 \u2013 8) 848 ________________________________________________________ Applied Thermodynamics (Area 4 \u2013 6 \u2013 7 \u2013 8 \u2013 4) = 12.09 kJ/kg Also, (Area 4 \u2013 6 \u2013 7 \u2013 8 \u2013 4) = T1 \u00d7 (s4 \u2013 s8) 12.09 = 258 \u00d7 (s4 \u2013 s8) s4 \u2013 s8 = 0.0469 kJ/kg \u00b7 K s3 = s8 = 0.3386 kJ/kg \u00b7 K s4 = 0.3855 kJ/kg \u00b7 K s1 = s2 = 4.4809 kJ/kg Also, therefore, Also Refriger", "ation effect = Area (1 \u2013 5 \u2013 6 \u2013 4 \u2013 1) = T1 \u00d7 (s1 \u2013 s4) = 258 \u00d7 (4.4809 \u2013 0.3855) = 1056.61 kJ/kg Work done = Area Area 3 \u2013 8 \u2013 9) + ((T2 \u2013 T1) \u00d7 (s1 \u2013 s8)) = 12.09 + ((298 \u2013 258) \u00d7 (4.4809 \u2013 0.3386)) = 177.782 kJ/kg COP = Refrigeration effect Work done = 1056.61 177.782 = 5.94 COP = 5.94 Ans. 10. A vapour compression cycle works on Freon-12 refrigerant with condensation temperature of 40\u00baC and evaporator temperature of \u2013 20\u00baC. Refrigeration effect of 2.86 ton is desired from the cycle. The compressor runs with 1200 rpm and has clearance volume of 2%. Considering compression index of 1.13 and following data determine. (a) the COP (b) the piston displacement in the reciprocating compressor used for compression. Properties of Freon - 12 Temperature \u00baC \u2013 20 40 Saturation pressure, bar 1.51 9.61 Specific volume, vgm3/kg 0.1088 \u2013 Enthalpy, kJ/kg Entropy, kJ/kg\u00b7K hf hg sf sg 178.61 17.8 74.53 203.05 0.0730 0.2716 0.7082 0.682 specific heat kJ/kg\u00b7K Cpf \u2013 0.976 Cpg 0.605 0.747 Solution: Given T1 = \u2013 20\u00baC = 253 K. During expansion (throttling) between 3 and 4, h3 = h4, h3 = hf at 40\u00baC = 74.53 kJ/kg = h4 Process 1\u20132 is adiabatic compression so and s1 = s2 s1 = sg at \u2013 20\u00baC = 0.7082 kJ/kg \u00b7 K T\uf8eb 2 \uf8ec \uf8ed 313 0.7082 = sg at 40\u00baC + Cpg at 40\u00baC \u00b7 ln \uf8f6 \uf8f7 \uf8f8 Introduction to Refrigeration and Airconditioning ____________________________________ 849 0.7082 = 0.682 + 0.747 ln T\uf8eb 2 \uf8ec \uf8ed 313 \uf8f6 \uf8f7 \uf8f8 Temperature after compression T2 = 324.17 K Enthalpy after compression, h2 = hg at 40\u00baC + Cpg at 40\u00baC \u00b7 (324.17 \u2013 313) = 203.05 + (0.747 \u00d7 (324.17 \u2013 313)) h2 = 211.39 kJ/kg Compression work required, per kg = h2 \u2013 h1 Fig. 18.30 = 211.39 \u2013 hg at \u2013 20\u00baC = 211.39 \u2013 178.61 Wc = 32.78 kJ/kg Refrigeration effect during cycle, per kg = h1 \u2013 h4 = 178.61 \u2013 74.53 = 104.08 kJ/kg Mass flow rate of refrigerant, m = \u00d7 2.86 3.5 104.08 m = 0.096 kg/s COP = 104.08 32.78 = 3.175 Volumetric efficiency of reciprocating compressor, given C = 0.02 \u03b7 vol = 1 + C \u2013 C 1 \uf8f6 nP.02 \u2013 0.02 1 1.13 \uf8eb \uf8ec \uf8ed 9.61 1.51 \uf8f6 \uf8f7 \uf8f8 \u03b7 vol = 0.9171 Let piston displacement by V, m3. 850 ________________________________________________________ Applied Thermodynamics Mass flow rate, 0.096 = 0.096 = V \u00d7 60 \u00d7 0.9171 1200 \u00d7 v g at\u201320\u00baC V \u00d7 \u00d7 0.9171 1200 60 0.1088 \u00d7 V = 569.45 cm3 COP = 3.175 Piston displacement = 569.45 cm3 Ans. 11. A vapour compression refrigeration cycle operates between the condensation temperature of 20\u00baC and evaporator temperature of \u201310\u00baC with carbon dioxide as refrigerant. Temperature after isentropic compression is 40\u00baC and condensate leaves at 10\u00baC before being passed through expansion valve. Determine COP of system and mass flow rate of CO2 required to get refrigeration effect of 2 kW. Properties of CO2 are given as under: Properties of CO2 Temperature \u00baC Specific Saturation Pressure, Volume,vg Specific heat kJ/kg\u00b7K Enthalpy, kJ/kg Entropy, kJ/kg\u00b7K bar 57.27 26.49 m3/kg hf hg sf sg \u2013 0.", "014 144.11 60.78 0.523 299.62 322.28 0.2381 1.0527 1.2324 Cpf 2.889 \u2013 Cpg 2.135 \u2013 20 \u2013 10 Solution: Processes of vapour compression cycle are shown on T-s diagram 1 \u2013 2 : Isentropic compression process 2 \u2013 3 \u2013 4: Condensation process 4 \u2013 5: Isenthalpic expansion process 5 \u2013 1: Refrigeration process in evaporator h1 = hg at \u2013 10\u00baC = 322.28 kJ/kg h2 = hg at 20\u00baC + Cpg at 20\u00baC \u00d7 (40 \u2013 20) = 299.62 + (2.135 \u00d7 20) h2 = 342.32 kJ/kg Entropy at state 2, s2 = sg at + 20\u00baC + Cpg at 20\u00baC \u00d7 ln \uf8eb \uf8ec \uf8ed + 273 40 + 273 20 \uf8f6 \uf8f7 \uf8f8 = 1.0527 + 2.135 \u00d7 ln \uf8eb \uf8ec \uf8ed 313 293 \uf8f6 \uf8f7 \uf8f8 Entropy during isentropic process, s1 = s2 s2 = 1.1937 kJ/kg\u00b7K 1.1937 = sf at \u2013 10\u00baC + x1 \u00d7 sfg at \u2013 10\u00baC 1.1937 = 0.2381 + x1 \u00d7 (1.2324 \u2013 0.2381) x1 = 0.961 Fig. 18.31 T-s representation Introduction to Refrigeration and Airconditioning ____________________________________ 851 Enthalpy at state 1, h1 = hf at \u2013 10\u00baC + x1 \u00d7 hfg at \u2013 10\u00baC = 60.78 + (0.961 \u00d7 (322.28 \u2013 60.78)) h1 = 312.08 kJ/kg h3 = hf at 20\u00baC = 144.11 kJ/kg Since undercooling occurs up to 10\u00baC, so, h4 = h3 \u2013 Cpf \u00b7 \u2206T = 144.11 \u2013 (2.889 \u00d7 (20 \u2013 10)) h4 = 115.22 kJ/kg Also, h4 = h5 = 115.22 kJ/kg Refrigeration effect per kg of refrigerant = (h1 \u2013 h5) = (312.08 \u2013 115.22) = 196.86 kJ/kg Let refrigerant flow rate be m kg/s. Refrigeration effect \u21d2 2 = m \u00d7 196.86 m = 0.1016 kg/s Compressor work, wc= h2 \u2013 h1 = 342.32 \u2013 312.08 wc = 30.24 kJ/kg COP = Refrigeration effect per kg Compressor work per kg = 196.86 30.24 = 6.51 COP = 6.51, Mass flow rate = 0.1016 kg/s Ans. 12. Determine partial pressure of vapour and relative humidity in the atmospheric air having specific humidity of 16 gm/kg of air and 25\u00baC DBT. Solution: Here pressure of atmospheric air may be taken as 1.013 bar Specific humidity, \u03c9 = 0.622 \uf8eb \uf8ec \uf8ed p \u2212 v p v \uf8f6 \uf8f7 \uf8f8 p or, \uf8eb \uf8ec \uf8ed p 1.013 v \u2212 \uf8f6 \uf8f7 \uf8f8 0.016 = 0.622 p v pv = Partial pressure of vapour = 0.0254 bar Relative humidity, sat \uf8f6 \uf8f7 \uf8f7 \uf8f8 From psychrometric properties of air table pv, sat at 25\u00baC = 0.03098 bar \u03c6 = 0.0254 0.03098 = 0.8198 or 81.98% Partial pressure of vapour = 0.0254 Relative humidity = 81.98% Ans. 852 ________________________________________________________ Applied Thermodynamics 13. For the atmospheric air at room temperature of 30\u00baC and relative humidity of 60% determine partial pressure of air, humidity ratio, dew point temperature, density and enthalpy of air. Solution: At 30\u00baC from steam table, saturation pressure, pv, sat = 0.0425 bar Partial pressure of vapour = Relative humidity \u00d7 pv, sat = 0.6 \u00d7 0.0425 = 0.0255 bar Partial pressure of air = Total pressure of mixture \u2013 Partial pressure of vapour = 1.013 \u2013 0.0255 = 0.9875 bar Partial pressure of air = 0.9875 bar Ans. Humidity ratio, \u03c9 = 0.622 = 0.622 p \uf8eb \uf8ec \uf8ed 0.0255 1.013 - 0.0255 \ufffd", "\ufffd \uf8f7 \uf8f8 p v p\u2212 v \u03c9 = 0.01606 kg/kg of dry air. Humidity ratio = 0.01606 kg/kg of dry air Ans. Dew point temperature may be seen from the steam table. The saturation temperature corresponding to the partial pressure of vapour is 0.0255 bar. Dew point temperature can be approximated as 21.4\u00baC by interpolation. Dew point temperatrure = 21.4\u00baC Ans. Density of mixture = Density of air (\u03c1 a) + Density of vapour (\u03c1 v1 + \u03c9) = \u03c1 P a RT a = (1 + 0.01606) 1.01606 2 \u00d7 = 1.013 10 \u00d7 \u00d7 0.287 303 = 1.1835 kg/m3 Density = 1.1835 kg/m3 Ans. Enthalpy of mixture, h = Cp \u00b7 T + \u03c9 (hg at 30\u00baC + 1.860 (30 \u2013 21.4)) = (1.005 \u00d7 30) + (0.01606 \u00d7 (2540.1 + 1.86 \u00d7 8.6)) = 71.2 kJ/kg of dry air Enthalpy of mixture = 71.2 kJ/kg of dry air Ans. 14. Determine the mass of water added and heat transferred for conditioning atmospheric air at 15\u00baC and 80% relative humidity to temperature of 25\u00baC and relative humidity of 50%. Final volume of conditioned air is 0.8 m3/s. Introduction to Refrigeration and Airconditioning ____________________________________ 853 Solution: Initial state at 15\u00baC and 80% relative humidity is shown by point 1 and final state at 25\u00baC and 50% relative humidity is shown by point 2 on psychrometric chart. \u03c9 1 = 0.0086 kg/kg of air and h1 = 37 kJ/kg \u03c9 2 = 0.01 kg/kg of air and h2 = 50 kJ/kg v2 = 0.854 m3/kg Mass of water added between states 1 and 2 = \u03c9 2 \u2013 \u03c9 1 = 0.01 \u2013 0.0086 = 0.0014 kg/kg of air Fig. 18.32 Psychrometric chart Mass flow rate of air = 0.8 0.854 Total mass of water added = 0.9368 \u00d7 (\u03c9 0.8 v = 2 = 0.9368 kg/s 2 \u2013 \u03c9 1) = 0.9368 \u00d7 0.0014 = 0.001312 kg/s Heat transferred = ma (h2 \u2013 h1) = 0.9368 \u00d7 (50 \u2013 37) = 12.18 kJ/s Mass of water added = 0.001312 kg/s, Heat transferred = 12.18 kW Ans. 15. Two streams of moist air, one having flow rate of 3 kg/s at 30\u00baC and 30% relative humidity, other having flow rate of 2 kg/s at 35\u00baC and 85% relative humidity get mixed adiabatically. Determine specific humidity and partial pressure of water vapour after mixing. Take Cp, stream = 1.86 kJ/kg \u00b7 K Solution: Let temperature after mixing be T\u00baC. For getting final enthalpy after adiabatic mixing the enthalpy of two streams are required. For moist air stream at 30\u00baC and 30% relative humidity. 854 ________________________________________________________ Applied Thermodynamics \u03c6 1 = vp 1 pv sat at 30\u00baC p\u03bd sat at 30\u00baC = 0.04246 bar, so, pv1 = 0.3 \u00d7 0.04246 = 0.01274 bar Corresponding to vapour pressure of 0.01274 bar the dew point temperature shall be 10.5\u00baC. Specific humidity, \u03c9 \u00d7 1 = 0.622 \u2212 (.007922 kg/kg of air = 0.622 0.01274 (1.013 0.01274) \u00d7 \u2212 \u03c9 At dew point temperature of 10.5\u00baC, enthalpy, hg at 10.5\u00baC = 2520.7 kJ/kg. h1 = Cp, air \u00d7 T1 + \u03c9 1{hg at 10.5\u00baC \u2013 Cp, stream \u00d7 (T1 \u2013 Tdp1)} = (1.005 \u00d7 30) + 0.007922 {2520.7 \u2013 1.86 \u00d7 (30 \u2013 10.5)} For second moist air stream at 35\u00baC and 85% relative humidity. h1 = 49.83 kJ/kg of dry air \u03c6 2 = 0.85 = p v 2 p v sat at 35\u00baC = pv 2 0.005628 or, Specific humidity \u03c9 pv2 = 0", ".004784 bar 0.622 \u2212.002951 kg/kg of air \u03c9 0.622 0.004784 (1.013 0.004784) \u00d7 \u2212 = Corresponding to vapour pressure of 0.004784 bar the dew point temperature is 32\u00baC. So, enthalpy of second stream, h2 = Cp, air \u00b7 T2 + \u03c9 2 {hg at 32\u00baC + Cp, stream \u00d7 (T2 \u2013 Tdp2)} = 1.005 \u00d7 35 + 0.002951 {2559.9 + 1.86 \u00d7 (35 \u2013 32)} Enthalpy of mixture after adiabatic mixing h2 = 42.75 kJ/kg of dry air = 1 m m1 2 \uf8fc \uf8fd \uf8fe = 1 5 \u00d7 \uf8f1 \uf8eb \uf8f2 \uf8ec \uf8ed \uf8f3 \u00d7 49.83 3 1.007922 \uf8f6 \uf8f7 \uf8f8 + \uf8eb \uf8ec \uf8ed \u00d7 42.75 2 1.002951 \uf8f6 \uf8f7 \uf8f8 \uf8fc \uf8fd \uf8fe Mass of vapour per kg of moist air = = 46.71 kJ/kg of moist air \u00d7 \u03c9 m 1 1 + \u03c9 ) (.007922 3 1.007922 \uf8f6 \uf8f7 \uf8f8 + \uf8eb \uf8ec \uf8ed \u00d7 0.002951 2 1.002951 \uf8f6 \uf8f7 \uf8f8 \uf8fc \uf8fd \uf8fe = 0.005893 kg per kg of moist air Specific humidity of mixture = 0.005893 \u2212 1 0.005893 Introduction to Refrigeration and Airconditioning ____________________________________ 855 = 0.00593 kg/kg of dry air \u03c9 = \u00d7 0.622 p v \u2212 p p v 0.00593 = 0.622 1.013 \u00d7 \u2212 p v p v Partial pressure of water vapour = 0.00957 bar Specific humidity of mixture = 0.00593 kg/kg dry air Partial pressure of water vapour in mixture = 0.00957 bar Ans. 16. Calculate the amount of heat added to moist air entering heating coil at 15\u00baC dbt and 80% relative humidity at a rate of 3m3/s. Air leaves the heating coil at 30\u00baC and atmospheric pressure is 1.013 bar. Solution: The type of heating involved is sensible heating. Locating state 1 on psychrometric chart corresponding to 15\u00baC dbt and 80% relative humidity the other property values shall be, h1 = 36.4 kJ/kg, \u03c9 Final state 2 has, h2 = 52 kJ/kg 1 = 0.0086 kg/kg of air, v1 = 0.825 m3/kg Mass of air = 3 0.825 = 3.64 kg/s Amount of heat added = 3.64 \u00d7 (52 \u2013 36.4) = 56.78 kJ/s. Heat added = 56.78 kJ/s Ans. 18.1 Define refrigeration and compare it with air conditioning. 18.2 Explain reversed Carnot cycle and its limitations. 18.3 Define unit of refrigeration. 18.4 Explain working of a Carnot refrigerator and state why it cannot be used in practice. 18.5 Describe air-refrigeration cycles. 18.6 Explain simple air refrigeration cycle being used in aeroplanes. 18.7 What do you understand by vapour compression cycle? Explain the processes on T-s and p-h diagrams. 18.8 Compare Carnot refrigeration cycle with vapour compression cycle. 18.9 What is meant by wet compression and dry compression in vapour compression cycles? Explain. 18.10 What is there in multistage vapour compression cycle? List its advantages over simple vapour compression cycle. 18.11 Describe the vapour absorption cycle for ammonia-water system and its working. 18.12 Explain the modified absorption-refrigeration cycle and its working. 18.13 Describe heat pump system based on vapour compression cycle. 18.14 Classify refrigerants and list desired properties of refrigerants. 18.15 Explain sensible heating and sensible cooling on psychrometric chart. 18.16 Write short notes on relative humidity, specific humidity, dry bulb temperature, dew point temperature, wet bulb temperature. 18.17 Discuss the working of summer air-conditioning system. 18.18 Describe working of winter air-conditioning system. 856 ________________________________________________________ Applied Thermodynamics 18.19 Show that the moisture content \u2018\u03c9\u2019 of the atmospheric air is given as, \u03c9 = \u00d7 p v 0.622 p a where pv and pa are partial pressure of", " vapour and air respectively. 18.20 Determine power required to run the compressor and heat rejected by compressor per second from a refrigerator working on reversed Carnot cycle between temperatures of 266 K and 300 K. Refrigerator produces 0.139 kg/s ice at \u20135\u00baC from water at 287 K. Also determine COP of refrigerator. required for machine. [9.76 hp, 63.19 kJ/s, 7.82] 18.21 A refrigerator working between \u20135\u00baC and 20\u00baC has capacity of 600 tons. Determine the minimum hp [266 hp] 18.22 An air refrigeration cycle has pressure ratio of 4 and the air enters compressor at 15\u00baC and expander at 30\u00baC. Consider compression and expansion index to be 1.25 and 1.35 respectively. Determine COP of [1.7] refrigeration cycle. 18.23 Show that the ratio of maximum COP of Bell-Coleman cycle and COP of Carnot cycle is given by: COP Bell\u2013Colemen COP carnot \uf8eb \uf8ec \uf8ed = T 1 \u2212 T 2 T 2 2 \uf8f6 \uf8f7 \uf8f8 where T1 and T2 are the temperature of surroundings and refrigerated space. Assume that the rise in temperature of air passing through heat exchanger (cooler) is same in both cases and is constant. 18.24 A Bell-coleman refrigeration cycle of 75.8 kJ/s refrigeration capacity runs between pressures of 1 bar and 5 bar. Refrigerated space is maintained at 0\u00baC and ambient temperature is 300 K. For isentropic efficiency of compression and expansion being 0.85 and 0.90 respectively. Determine COP of system. [0.815] 18.25 An air refrigeration system working on reversed Brayton cycle has refrigeration capacity of 10 ton and pressure ratio of 3. Pressure in refrigerated space is 1.4 bar, temperature of air at inlet to compressor is 253 K and air is cooled to a temperature of 323 K in cooler. Considering Cp = 1.07 kJ/kg\u00b7K determine COP of system, air circulation per minute, theoretical piston displacement and power required. [2.43, 114 kg/min, 59.13 m3/min, 13 kW] 18.26 An air refrigeration cycle works between 1 bar and 12 bar to produce refrigeration capacity of 25 tons when air leaves cooler at 25\u00baC and air leaves refrigerated space at 0\u00baC. Considering compression and expansion processes to have index of 1.35 determine. (i) the COP of system (ii) the HP required (iii) the mass of air circulated per minute (iv) the piston displacements of compressor and expanders. [1.94, 116 hp, 133 kg/min, 35.5 m3/min, 20.35 m3/min] 18.27 Air cooled aeroplane system has pressure of air getting increased from atmospheric pressure of 0.9 bar to 1 bar due to ram action. Compressed air leaves compressor at 3.5 bar and temperature of compressed air leaving heat exchanger is 355 K and subsequently leaving the evaporator is 348 K. High pressure cool air leaving evaporator enters the cooling turbine and subsequently to passenger cabin at 1.03 bar. Considering isentropic efficiencies of compression and expansion to be 0.8 and 0.75 determine hp required for desired cooling and COP of system when stale air leaves passenger cabin at [596 hp, 0.16] 298 K. Ambient air temperature is 296 K. 18.28 The passenger cabin of an aircraft is maintained at 298 K when ambient air is at 0.35 bar and \u201315\u00baC. Compressor in aircraft has pressure ratio of 3 and aircraft is moving at 1000 kmph. Compressed air leaving compressor is passed through a heat exchanger and cooled there upto ambient air temperature. Subsequently this cool air passes through cooling turbine so as to leave it at 1.06 bar. Considering cooling load to be 200 ton determine. (i) the stagnation temperature and pressure Introduction to Refrigeration and Airconditioning ____________________________________ 857 (ii) the COP of system (iii) the mass flow rate of air Take Cpair = 1 kJ/kg\u00b7K, \u03b3 = 1.4 [296.5 K, 0.57 bar, 0.31, 2 kg/s] 18.29 Calculate COP of vapour compression system running on ammonia (NH3) as refrigerant between 25\u00baC and \u2013 10\u00baC such that the refrigerant is dry saturated at the end of compression and there is no undercooling. Properties of NH3 are; Temperature, Enthalpy, kJ/kg \u00baC 25 \u201310 h f 100.32 \u2013 33.52 h fg 1228.92 1345.96 Entropy, kJ/kg\u00b7K sf 0.34", "69 \u2013 0.1379 [7.16] 18.30 A vapour compression cycle running on NH3 had refrigeration capacity of 1990.5 ton while operating between temperature limits of 298 K and 258 K. The vapour is dry saturated at the end of compression and actual COP of cycle is only 60% of theoretical COP, then determine power required to drive compressor. Properties of NH3 Temperature, Enthalpy, kJ/kg \u00baC 25\u00baC \u201315\u00baC h f 99.94 \u201354.51 h g 1318 1303.7 Entropy, kJ/kg\u00b7K sg sf 0.0347 \u2013 0.2132 4.4809 5.0536 [43.8 hp] 18.31 Determine the COP of vapour compression system running with CO2 and operating between temperature of 25\u00baC and \u2013 5\u00baC such that carbon dioxide entering compressor is 0.6. Properties of CO2 Temperature, Enthalpy, kJ/kg \u00baC 25 \u2013 5 h f 81.09 \u2013 7.52 h g 202.31 236.59 Entropy, kJ/kg\u00b7K sg sf 0.2508 \u2013 0.0418 0.6287 0.8402 [3.3] 18.32 A water cooler working on Freon\u2013 12 as refrigerant operates between 299 K and 275 K such that the vapour leaves evaporator dry saturated. Considering volumetric efficiency of compressor as 80%, mechanical efficiency of compressor as 85%, 20% of useful cooling getting lost and cooler cooling water from 299 K to 279 K at the rate of 100 kg/hr determine. (i) the hp required in compressor (ii) the volumetric capacity of compressor Properties of Freon-12 Temperature, \u00baC Sp. volume vg, m3/kg 26 2 0.027 0.053 Enthalpy, kJ/kg Entropy, kg/kg\u00b7 K h f 443.12 419.88 h g 583.95 573.54 sf 4.2665 4.1867 sg 4.737 4.745 Specific heat kg/kg\u00b7 K Cpf 60.9948 60.9405 C pg 0.6729 0.618 [0.442 hp, 0.085 m3/min.] 858 _________________________________________________________ Applied Thermodynamics 19 19.1 INTRODUCTION World War-II witnessed the use of a new gas turbine engine type aircraft being developed independently in Germany and in England. These engines were called as jet engines. In these aircraft thrust generated by propulsion system caused movement of aircraft in air. Underlying principle for the creation of thrust was Newton\u2019s third law of motion. In jet propulsion system working fluid is accelerated by the system and reaction to this acceleration produces force on the system. Amount of thrust created is directly proportional to the mass flow rate through engine and exit velocity of gas. Jet engines have atmospheric air being inducted into compressor for being compressed and subsequently used for burning fuel in combustor. High temperature, high pressure burnt gases are then expanded in gas turbine and hot exhaust gases are finally passed through a nozzle for creation of thrust. Rocket engines also work on similar principle but unlike jet engine these rocket engines carry oxygen with them for combustion. History of jet propulsion dates back to 100 B.C. when Hero of Alexandria first developed engine called \u2018aeolipile\u2019 which was basically reaction turbine working on steam. In 1500 A.D. Leonardo Vinci used hot air coming out from chimney to drive rotor called chimney jack. Steam jet propelled wagon was developed in 1687 A.D. in Dutch. Major breakthrough was achieved when the gas turbine was patented in the name of John Barbar in 1791 A.D. This evolution of gas turbine paved way for development of gas based engines for use in gas turbine power plants and aircraft etc. By the year 1903 A.D. number of gas turbine based plants came up and gas turbine was found to be successful. 19.2 PRINCIPLE OF JET PROPULSION Jet propulsion refers to the imparting of forward motion to the object as a reaction to exit of high velocity gas/liquid stream from the rear end of object. Jet propulsion devices are popularly used in highspeed, high altitude aircraft/missile/spacecrafts etc. Simple example of jet propulsion is forward motion of an inflated balloon when air is suddenly released from it. Jet propulsion is based on the principle of Newton\u2019s second law and third law of motion. In a jet propulsion engine the objective is to get the propelling thrust for the engine and for getting it the momentum change occurs in fluid stream such that the reaction to the impulse created by momentum change gives propelling thrust. Thus, the change of momentum of the fluid stream flowing across engine and the reaction to the impulse due to momentum change are responsible for jet propulsion. For realizing momentum change the high temperature and", " pressure gas stream is expanded through a nozzle so that Jet Propulsion and Rocket Engines ________________________________________________ 859 the gas stream comes out in atmosphere with significantly high velocity, thus giving change of momentum. This momentum change yields impulse force whose reaction produces propelling thrust. Figure. 19.1 gives block diagram for jet propulsion engine. Fig. 19.1 Principle of jet propulsion engine Let us consider jet propulsion device as shown in Fig. 19.1. Let the inlet and exit of device be indicated by subscript \u2018i\u2019 and \u2018e\u2019. Cross-sectional area, velocity of gas, pressure of gas at inlet and exit are given by Ai, Ci, pi and Ae, Ce, pe respectively. Let us analyze for air entering device and flowing through device such that velocity changes from inlet \u2018Ci\u2019 to exit \u2018Ce\u2019. Atmospheric air enters device at atmospheric pressure pa, velocity Ci. Consider mass of air entering into engine be at the rate of ma kg/s. Mass of gas leaving engine will be at the rate of (ma + mf ) kg/s where mf is mass flow rate of fuel required for running. Expansion occurring in engine shall be complete when the pressure at exit pe equals to atmospheric pressure, pa. In actual case this expansion may not be complete. For simplicity the mass flowing through device may be assumed constant i.e., fuel added is negligible, then the thrust T may be estimated by the rate of change of momentum. This is called as momentum thrust. considering mass of fuel added. Tm = ma (Ce \u2013 Ci) Fm = {(ma + mf )\u00b7 Ce \u2013 ma\u00b7Ci} In case of incomplete expansion the exhaust gas may leave device with pressure pe which is greater than pa. This incomplete expansion would cause additional pressure thrust exerted over the device exit area Ae equal to Ae\u00b7 (pe \u2013 pa). This total thrust will have two components i.e., momentum thrust and pressure thrust. Total thrust, T = Momentum thrust, Tm + Pressure thrust, Tp T = ma (Ce \u2013 Ci) + Ae (pe \u2013 pi) Above mathematical equation shows that the total thrust may thus be maximized by increasing momentum thrust and pressure thrust both or any one of these. Key parameters responsible for increase in total thrust could be high velocity of jet leaving (Ce), high mass flow rate (ma) high pressure difference at exit and higher cross sectional area at exit. But in view of the interdependency of these parameters the required thrust may be said to be available either by high velocity jet of small mass flow or a low velocity jet of high mass flow. While considering the fuel mass addition the total thrust may be given as ) } Basic arrangement required in a jet propulsion engine comprises of a compressor for compressing air, followed by combustor (combustion chamber) for generating high pressure and high temperature 860 _________________________________________________________ Applied Thermodynamics gases followed by a turbine for generating shaft work to drive compressor and a nozzle at end to finally exhaust the gas stream. Figure 19.2 shows the schematic of turbojet engine. The jet leaving engine has quite high velocity and the reaction of this jet propels the engine (device) in direction opposite to that of jet. Fig. 19.2 Turbojet, Turbofan and Turboprop engines Total thrust } + ); (neglecting fuel addition) ( A p e e \u2212 p a ); (considering fuel addition) 19.3 CLASSIFICATION OF JET PROPULSION ENGINES Jet propulsion engines can be broadly classified based on type of suction. In case of jet engine sucking atmospheric air it is called air-breathing jet engines or atmospheric jet engines. Jet engines may be fed with oxygen carried separately in engine and such engines do not induct atmospheric air. These engines which carry their own oxidizer for combustion of fuel are called non-air breathing jet engines or rocket engines. Thus, jet propulsion engines can be; (i) Air breathing jet engines or atmospheric jet engines (ii) Non-air breathing jet engines or rocket engines (i) Air breathing jet engines or atmospheric jet engines can be further classified depending on the type of air flow as, (a) Continuous air flow engines or steady combustion type engines. (b) Intermittent air flow engines or intermittent combustion type engines. Continuous air flow type engines can be further classified as, (a.1) Turbojet engines (a.2) Turbofan engines (a.3) Turbojet with afterburner (a.4) Turboprop or propjet (a.5) Ramjet or Athodyds and Lorin tube (ii) Non-air breathing jet engine or rocket engine can be classified based on the type of propelling fluid used. It could be liquid propellant type rocket engine and solid propellant type rocket engine. Jet Propulsion and Rocket Engines ________________________________________________ 861", " 19.4 PERFORMANCE OF JET PROPULSION ENGINES Performance of jet propulsion engine can be judged by the performance parameters given below. Let us consider turbojet engine as shown in Fig. 19.1, moving with velocity through air as Ca. Thus the velocity of air entering engine would be Ca (approach velocity) if the atmospheric air is at rest. Thrust obtained per kg of air (N/kg of air/s) flowing can be given as : i) Thrust power (TP): Thrust power indicates the actual power available for propulsion. It refers to the work done per unit time by the engine. This thrust power can be expressed by the product of thrust and velocity with which engine moves (flight velocity). TP = T \u00d7 Ca \uf8ee \uf8ef \uf8ef \uf8f0 \uf8f1 \uf8f4 \uf8f2 \uf8f4 \uf8f3 \uf8eb \uf8ec \uf8ed = + Neglecting pressure thrust it could be given as: TP = \uf8ee \uf8ef \uf8ef \uf8f0 \uf8eb \uf8ec \uf8ed + \uf8fa \uf8fb Neglecting mass of fuel added; TP = (Ce \u2013 Ca)\u00b7Ca, W/kg of air (ii) Propulsive power (PP): Propulsive power indicates the total energy available for propulsion. It can be estimated by the difference between the rate of kinetic energy entering with air and leaving with jet of exit gases. Mathematically; \u21d2 PP = K.E.leaving \u2013 K.E.entering = \u2206KE 2 C e \u2206KE = /kg of air 862 _________________________________________________________ Applied Thermodynamics \u21d2 PP = /kg of air Neglecting fuel mass addition, propulsive power can be given by, PP = /kg of air (iii) Propulsive efficiency (\u03b7\u03b7\u03b7\u03b7\u03b7 Prop): Propulsive efficiency is measure of effectiveness by which propulsive power is transformed into thrust power i.e., how efficiently propelling duct can propel the engine. Mathematically, it can be given by ratio of thrust power (TP) to propulsive power (PP). Propulsive efficiency is also called Froude efficiency. \u03b7 prop = = TP PP \uf8ee \uf8ef \uf8ef \uf8f0 \uf8eb \uf8ec \uf8ed \uf8f9 \uf8fa \uf8fa \uf8fb Neglecting fuel mass addition, propulsive efficiency can be given as, \u03b7 prop = The above expression indicates that propulsive efficiency would be unity when exit jet velocity equals to approach velocity, but this is not desired as it decreases thrust available. th): Thermal efficiency can be given by the ratio of kinetic energy (iv) Thermal efficiency (\u03b7\u03b7\u03b7\u03b7\u03b7 available and total heat supplied. \u03b7 th = = f 1 2 m Kinetic energy available Heat supplied ( Calorific value of fuel f m a 2 C a 2 C e \u2212 \u00d7 Calorific value of fuel, m f m a or, \u03b7 th = 2 where FA is fuel-air ratio. = 2 \u00d7 ( 1 \u00d7 FA + 2 ) FA C e \u2212 2 C a Calorific value of fuel Jet Propulsion and Rocket Engines ________________________________________________ 863 (v) Overall efficiency (\u03b7\u03b7\u03b7\u03b7\u03b7 to the energy supplied. overall): Overall efficiency can be given by the ratio of useful workdone \u00d7 Calorific value of fuel \u03b7 overall = m 1 + f m a th \u00d7 \u03b7 = \u03b7 prop (vi) Jet efficiency (\u03b7\u03b7\u03b7\u03b7\u03b7 jet): Jet efficiency is given by the ratio of final kinetic energy in jet to the total of isentropic heat drop in jet pipe and carry over from turbine. Comparative performance of turbojet and other propulsion engines in respect to the propulsive efficiency and jet velocity is shown in Fig. 19.3. Figure 19.4 shows comparison of specific fuel consumption with jet plane speed for these different devices. Fig. 19.3 Comparison of propulsive efficiency Fig. 19.4 Comparison of specific fuel consumption 19.5 TURBOJET ENGINE Fig. 19.5 shows the schematic of turbojet engine. It has a diffuser section at inlet for realizing some compression of air passing through this section. Due to this air reaching compressor section has pressure more than ambient pressure. This action of partly compressing air by passing it through diffuser section is called \u201cramming action\u201d or \u201cram effect\u201d. Subsequently compressor section compresses air which is fed to combustion chamber and fuel is added to it for causing combustion. Combustion products available at high pressure and temperature are then passed through turbine and expanded there. Thus, turbine yields positive work which is used for driving compressor. Expanding gases leaving turbine are passed through exit nozzle where it is further expanded and results in high velocity jet at exit", ". This high velocity jet leaving nozzle is responsible for getting desired thrust for propulsion. Pressure variation along the length of turbojet is shown in Fig. 19.5. Velocity and temperature variations are also shown there in. Different salient states shown are plotted on temperature-entropy (T\u2013s) diagram as given in Fig. 19.5. Ambient air at state \u2018a\u2019 enters the diffuser section and ram effect is seen from \u2018a\u2019 to \u20181\u2019 resulting into small pressure rise. Desired compression occurs between \u20181\u2019 to \u20182\u2019. Subsequent combustion occurs in combustor section resulting into combustion products at state \u20183\u2019. Expansion occurs in turbine from \u20183\u2019 to \u20184\u2019 and subsequently expanded from \u20184\u2019 to \u20185\u2019. 864 _________________________________________________________ Applied Thermodynamics Between states \u2018a\u2019 and \u20181\u2019 diffusion of air entering occurs isentropically with velocity varying from Ca to zero for diffuser section having 100% efficiency. Actually diffuser efficiency may be 0.9\u20130.95. T\u2013s diagram depicts both theoretical states and actual states. Actual states have been shown as 1\u2032, 2\u2032, 4\u2032 and 5\u2032, while theoretical states are 1, 2, 3, 4 and 5. Fig. 19.5 Turbojet engine Jet Propulsion and Rocket Engines ________________________________________________ 865 Considering the states shown on T\u2013s diagram to be static states the mathematical equations for different sections can be given as under. In diffuser section (a \u2013 1) the energy equation yields + wa \u2212 1 For ideal diffuser, C1 = 0, qa\u20131 = 0 and wa\u20131 = 0, which means after ram compression air gets decelerated to C1 = 0, diffuser section is perfectly insulated and no work interaction occurs in this section. or, h1 = h a T1 = T a 2 C a+ 2 2 C a 2 C + p Deviation from ideal diffusion process to actual can be quantified through diffuser efficiency. T a T a \u03b7 diffuser = \u2212 \u2212 \u2212 \u2212 = \u21d2 h1 diffuser \u21d2 = T \u2032 1 T a Pressure rise in diffuser can be given by pressure ratio 2 C a diffusor + \u03b7 2. p 1 pa p 1 pa = \uf8ee \uf8ef \uf8ef \uf8f0 diffuser + \u03b3 \u2212 \u03b3 ( 1) ; \uf8f9 \uf8fa 1 \uf8fa \uf8fb here substituting, Cp = \u03b3 R ( \u2212 1) and \u03b3 velocity of sound =,RT\u03b3 \u00b7 or, p 1 pa = 1 + \u03b7 diffuser \u2212 \u03b3 2 1 2 Ma \u03b3 \u2212 1 ) \u03b3 ( ; Here Ma is Mach no. at \u2018a\u2019. Similar to diffuser section the energy equation when applied to compressor section between states 1 and 2 yields, (neglecting change in potential energy wcomp = + ) comp + \u2212( \u2212 w h 1 + 2 C 1 2 or, substituting stagnation property, h02 = h 2 + 2 C 2 2, h 01 = h 1 + 2 C 1 2 866 _________________________________________________________ Applied Thermodynamics Neglecting change in kinetic energy and potential energy the work required by compressor shall wcomp = h02 \u2013 h01 be, Actual state at the end of compressor lies at state 2\u2032 instead of 2. This shift from 2 to 2\u2032 can be quantified by compressor efficiency (\u03b7 wcomp = h2 \u2013 h1 = Cp (T2 \u2013 T1) wcomp = h2 \u2013 h1 = Cp (T2 \u2013 T1) comp) as given below \u2212 \u2212 = \u2032 Ideal compressor work Actual compressor work \u03b7 comp = Heat addition in combustion chamber occurs between states 2 and 3. Neglecting mass of fuel added, the heat supplied can be estimated as; Actual heat supplied can be given as: q2\u20133 = h3 \u2013 h2 = Cp (T3 \u2013 T2) Q2\u2013 } ) After combustion the hot gases enter turbine for expansion between 3 and 4. Energy equation can be written for turbine, as + Considering expansion in turbine to be occurring adiabatically; q3\u20134 = 0 \u2212 wturbine = turbine or In terms of stagnation properties; Neglecting change in kinetic energy across turbine, wturbine = h03 \u2013 h04 \u2212 h 4 wturbine = h 3 wturbine = CpT3 \u2013 CpT4 turbine = (h3 \u2013 h4\u2032) w = w \u2032 Actual turbine work, w\u2032 Turbine efficiency, ( h 3 ( h 3 At the exit of turbine gases pass through exit nozzle from 4 to 5. Energy equation can be given as, turbine = ( turbine turbine ) ) )", " ) = \u2032 \u2032 \u2212 \u2212 \u2212 \u2212 + for actual expansion in exit nozzle nozzle = Nozzle efficiency Considering above T-s diagram thermal efficiency can be mathematically given as \u2212 th = \u03b7 th = \u2212 = \u2212 \u2212 ) ) ( ) ( \u2032 \u2032 \u2032 \u2032 \u2032 Jet Propulsion and Rocket Engines ________________________________________________ 867 19.6 TURBOFAN ENGINE Turbofan engine is the modified turbojet engine in which additional thrust is realised by putting fan at the entry of the engine casing. Fan blades propel by pass air around engine core between inner and outer engine casing. This air does not participate in combustion but provides additional thrust while leaving through exit nozzle. Figure. 19.6 shows the schematic of turbofan engine. The propeller fan put at inlet to engine sucks air and it passes through by pass passage as shown up to the exit nozzle end. Thus there are two streams of air flowing, one air stream gets rammed, compressed, burnt, expanded in turbine and finally passes through exit nozzle and other air stream passes through passage between outer and inner casings from inlet to nozzle exit. Total thrust created will be due to two jet streams one due to cold air or fan air and other due to burnt gases leaving turbine. Fig. 19.6 Turbofan engine Arrangement in turbofan engine indicates that the proportion of air flowing through bypass and engine play significant role, this is mathematically defined as bypass ratio. Bypass ratio is given by the ratio of flow through bypass duct and flow through compressor. Generally bypass ratio has value in the range 0.3 to 0.8. Thus, for bypass mass flow rate being m\u2032 b and mass flow rate through compressor being m\u2032 c the bypass factor (B) shall be total air entering turbofan, m\u2032 B = m m a = m and \u2032 b \u2032 c \u2032 + \u2032 mb c or m Let the velocity of bypass duct air leaving jet be given by Cb and velocity of gas leaving jet be Ce, then the thrust can be given as; Thrust = ( \u22c5 \u2032 868 _________________________________________________________ Applied Thermodynamics Turbo fan engine\u2019s performance primarily depends on the fan pressure ratio, bypass ratio, overall pressure ratio, turbine inlet temperature, cruising speed and altitude. Amongst these thermodynamic parameters. For optimum performance of turbofan engine the plot of specific fuel consumption for the varying bypass ratio is shown in Fig. 19.7. Fig. 19.7 Specific fuel consumption vs. specific thrust for turbofan engine 19.7 TURBOPROP ENGINE Turboprop (Turbo-propeller) engine, also called turboshaft engine or propjet engine differs slightly from turbofan engine. It uses thrust to turn a propeller. It consists of a compressor for compressing the inlet air, combustion chamber and turbine followed by exit nozzle. A part of turbine output is used to drive the compressor and remaining for driving propeller. Thus some stages of turbine give shaft work for driving compressor and some stages produce shaft work for driving propeller as shown in Fig. 19.8. It can also be understood as if the gases expand through main turbine which drives compressor and also expands through power turbine which drive propeller through suitable reduction gear box. Turboprop engines are used in small passenger planes, cargo planes etc. Fig. 19.8 Turboprop engine 19.8 TURBOJET ENGINE WITH AFTERBURNER Turbojet engine with afterburner is similar to turbojet engine with afterburner in tail pipe. Figure 19.9 shows schematic for turbojet engine with afterburner where exhaust leaving turbine is passed through afterburner in which combustion causes rise in exhaust gas temperature. Thus exhaust gas have high potential to produce thrust for propulsion. Jet Propulsion and Rocket Engines ________________________________________________ 869 Fig. 19.9 Turbojet engine with afterburner 19.9 RAMJET ENGINE Ramjet engine is the simplest of jet engines having no moving parts. Ramjet is a typically shaped duct open at both ends with air being compressed merely due to forward motion of engine. Fuel is subsequently added for combustion and thus high pressure, high temperature gases exit from exhaust nozzle. High pressure air is continuously available as engines keeps on moving forward. These ramjets are extensively used for propulsion in number of high speed aircrafts. Fig. 19.10 Ramjet engine Ramjet is also called Athodyd Lorin tube based on the name of its inventor Rene Lorin, a French engineer. First ramjet engine was developed in 1913 and it had steady flow through inlet diffuser, combustion chamber and outlet nozzle. These can not operate under stationary condition as sole compression of air occurs due to ram compression. Ram pressure ratio is relatively small initially as forward speed is slow but increases as the speed increases. Ramjet is boosted up to substantially high speed using turbojet", " or rocket engine for getting desired thrust by ramjet. Ramjet are boosted up to speed of 300 km/hr. The efficiency of ramjet largely depends upon the design of diffuser section. Fig. 19.11 Schematic and T\u2013s diagram of ramjet 870 _________________________________________________________ Applied Thermodynamics Advantages of ramjet engine are: (i) (ii) (iii) (iv) It has no moving parts and hence ramjet are better balanced. It yields greater thrust per unit mass as compared to any other propulsion engine at supersonic speed. It is much simpler in construction and light in weight. It yields much greater thrust per unit frontal area at supersonic speeds. Best performance can be had at 1700 km/hr to 2200 km/hr speed range. (v) Variety of fuels can be used in ramjet. (vi) These are ideal propulsion device for aircraft missiles. Disadvantages of ramjet engine are: (i) Forward motion is very much necessary to realize ram compression. (ii) Ram pressure ratio increases gradually. (iii) Ramjets are unable to work at low flight speeds. Ram compression results in decrease in velocity and increase of pressure of air passing through this diffuser section. At the end of diffuser section a pressure barrier is created and after this point fuel is injected through nozzles and ignited using spark plug. Combustion results in expansion of gases which is restrained due to pressure barrier on one side and so move out through exit nozzle with high velocity. 19.10 PULSE JET ENGINE Pulse jet engine was developed by German scientist Paul Schmidt in 1930. It is quite similar to ramjet engine except the difference that pulse jet employs a non-return type mechanical valve of V-type for preventing flow of hot gases through diffuser. Pulse jet engine has diffuser section in which ram compression occurs and after diffuser section a grid of non-return valves is put for maintaining intermittent flow of compressed air. In combustion section fuel is atomized during injection and burnt using spark plug or igniters. Combustion of air and fuel results in combustion products having high pressure and temperature. Due to high pressure of combustion product non-return valve remains closed and causes flow through tail pipe so as to produce thrust. When combustion gases move out from combustion chamber then the pressure lowers down in this side as compared to pressure available in diffuser section. Due to this pressure differential the flow of compressed air again occurs into the combustion chamber which again burns and flows out from exit nozzle. Thus there occur the processes of suction, combustion and exhaust one after the other. Pulse jet engine tube shown in Fig. 19.12 has temperature varying continuously from inlet to exit. This pulse jet engine has advantages of being cheap compared to turbojet engine and produces static thrust. Pulse jet engine produces thrust more than drag at lower speed compared to ramjet. Pulse jet engines have disadvantages of noise, maintenance in view of mechanical operation of valves, high fuel consumption rate and vibration etc. Fig. 19.12 Pulse jet engine Jet Propulsion and Rocket Engines ________________________________________________ 871 19.11 PRINCIPLE OF ROCKET PROPULSION Rocket engines are non-air breathing engines and carry their own oxidiser for burning of fuel. Rocket propulsion is realized by the thrust produced by combustion products leaving exit nozzle. It has injection system for fuel and oxidizer followed by combustion chamber and exit nozzle as shown in Fig. 19.13. Fig. 19.13 Layout for rocket engine In rocket engines the combustion products get discharged from the exit nozzle with supersonic velocity and thus have very high kinetic energy. Rocket gets desired thrust by the reaction available from the nozzle stream. Thrust is available due to change of momentum and pressure with which jet comes out. Net thrust available; where m\u2032 exit jet pe and atmospheric pressure is pa. T = m\u2032 p Ce + Ae ( pe \u2013 pa) p is mass flow rate of propellant, jet exit velocity Ce, area of exit nozzle Ae, pressure of Above expression shows that for maximizing thrust exit velocity should be maximized, pressure difference at exit (pe \u2013 pa) should be maximized. Thus rocket would get maximum thrust when atmospheric pressure is not there i.e., pa = 0, which means maximum thrust would be available in vacuum. Thrust could also be given in terms of rocket performance parameter called effective jet velocity. p Ce + Ae ( pe \u2013 pa) = m\u2032 T = m\u2032 p\u00b7Cej Here Cej is effective jet velocity which could be given as, Specific impulse can be given as Cej = ) Isp = = Ce j T \u2032 m p i.e., Specific impulse is the thrust produced per unit mass flow rate of propellant. Thrust power in case of rocket engine can be given as; TP = T\u00b7Ca = m\u2032 p \u00b7Cej \u00b7Ca here Ca is the velocity with which rocket moves forward. Propulsive efficiency: For rocket engine \u03b7 prop", " = TP TP + loss of kinetic energy 872 _________________________________________________________ Applied Thermodynamics = ( \u2032 m C C ej a p \u22c5 \u2032 \u22c5 \u22c5 m C C ej a \u2032 ( m C p p + ( ) ej \u2212 2 ) /2) C a \u03b7 prop = 2 + 1 ( / C C ej a C C / ej a ) 2 ) ( 19.12 ROCKET ENGINE Rocket engines as described earlier can be classified based on type of propellant carried by them. These could be solid propellant and liquid propellant resulting into, (i) Solid propellant rocket engine (ii) Liquid propellant rocket engine. Since Rocket engines carry propellant fuel and oxidizer along with, therefore, the characteristics of propellant have to be such as to give best performance with minimum oxidizer requirement. Desired properties of propellant for rocket engine are, (i) propellant should have large heating value. (ii) propellant should have high density so that storage space required is small. (iii) propellant should be capable of having smooth ignition. (iv) propellant should have stability and ease of handling and storage. (v) propellant should be non-toxic and non-corrosive. (vi) propellant should be environment friendly. Applications of rocket engines have proved boon for our civilization as satellites in space which are serving air communication and other requirements are rocket engines. Rocket engines are also extensively used in spacecrafts, missiles, jet assisted air planes, pilotless aircraft, etc. 19.13 SOLID PROPELLANT ROCKET ENGINES These rocket engines use solid propellants which burns using oxidizer present within it. There are two types of solid propellant rockets depending upon type of burning i.e., restricted burning and unrestricted burning. Figure 19.14 shows the different arrangements in solid propellant rocket engines. Solid propellant has composition such that all essential requirements for combustion are met. Specific requirements of solid propellant are, (i) Propellant should have sufficient compressive and impact strength at low temperature. (ii) Propellant should give uniform burning. (iii) Propellant should give high specific impulse. Fig. 19.14 Solid propellant rocket engines Jet Propulsion and Rocket Engines ________________________________________________ 873 Arrangement shows that it has seamless sturdy steel tube closed at one end. At the open end exit nozzle is provided and this may be single nozzle type or multi nozzle type exit. Propellant is filled in tube while in liquid state and gets solidified gradually upon cooling and fits tube completely. Solid propellant filled inside tube is burnt gradually so as to give combustion products for producing desired thrust. Restricted burning rocket engines have the propellant burnt only on its exposed surface towards exit nozzle. Gradually complete propellant is burnt and combustion products move out through exit nozzle. Thus restricted burning nozzles are very similar to cigarette whose burning begins at one end and goes till end. Restricted burning rockets are suited for small thrust applications for longer duration. Unrestricted burning in rocket engine occurs such that all exposed surfaces of solid propellant are burnt simultaneously inside tube. Since burning of whole solid fuel is initiated so it gets burnt quickly and produces large thrust. 19.14 LIQUID PROPELLANT ROCKET ENGINES Figure 19.15 shows the schematic of liquid propellant rocket engines. These contain liquid propellants stored in containers outside combustion chamber. Liquid propellant fuel is being fed using either pressure feed or pump feed to transfer fuel from storage tank to combustion chamber. Pressure feed system has pump driven by small turbine. This feed system is relatively cheap and simple compared to other feed systems. Fig. 19.15 Liquid propellant rocket engines There could be mono-propellant or bi-propellant type rocket engines. Mono-propellant systems are heavier in view of additional quantity of fuel required. Schematic shows two liquid bi-propellant rocket system with rocket motor and propellant system etc. Rocket motor has propellant injector for injecting liquid fuel and oxidizer, ignition systems, combustion chamber and exit nozzle as shown. Liquid propellant is heated up by circulating it around the combustion chamber and nozzle walls for cooling it before being injected for combustion. Cooling is done so as to maintain temperature of walls low, so that excessive thermal stresses are not developed. Liquid propellants are selected based on specific impulse, availability, cost, ease of handling and density etc. There are number of oxidizer and liquid propellants available, namely, liquid oxygen + liquid hydrogen, liquid oxygen + ethyl alcohol, liquid oxygen + high quality gasoline, nitric acid + aniline etc. EXAMPLES 1. A jet propulsion engine moves with speed of 200 m/s at an altitude where total pressure and total temperature at inlet to compressor are 0.5 bar and 272 K. Isentropic efficiency of compressor and turbine are 0.84 and 0.82 respectively. Total", " pressure and temperature of gases at inlet to turbine are 3 bar and 1000 K and static back pressure of propulsion nozzle is 0.4 bar and nozzle efficiency based on total 874 _________________________________________________________ Applied Thermodynamics pressure drop is 92%. Determine (i) power required to run compressor per kg of air, (ii) the air-fuel ratio, (iii) the total pressure of gas leaving turbine, (iv) thrust per unit air flow rate, (v) Plot the processes on h\u2013s diagram. Take pC gas = 1.13 kJ/kg.K, \u03b3 gas = 1.33, pC air = 1.005 kJ/kg.K, \u03b3 air = 1.4 and Calorific value = 41.84 MJ/kg of fuel. Solution: We know, = T 02 T 01 T02 = 453.83 K p 02 p 01 \u2212\u03b3 \u03b3 1 \u21d2 T02 = 272. Considering compressor efficiency, =. \u03b7 compr = 0 84 Wcompr = Cpair ( \u2212 ( T 02 \u2212\u2032 ( T 02 \u2032 \u2212 T 01 T 01 T 01 ) ) ) T 02 \u21d2 T02\u2032 = 488.46 K Power required for compressor = 217.54 kW/kg Ans. = 1.005 (488.46 \u2013 272) = 217.54 kW/kg Fig. 19.16 In combustion chamber 03 \u2212 \u2032 m C a \u22c5 T 02 p air p gas = \u2032 \u00d7 m\u2032 f C.V or \uf8f1 \uf8f4 \uf8f2 \uf8f4 \uf8f3 CV \u2212 C C p gas \u22c5 T 03 \u22c5 T 03 p gas \u2212 C T \u2032 02 p air \uf8fc \uf8f4 \uf8fd \uf8f4 \uf8fe ( 41840 1 13 1000 (. \u00d7 \u2212 \u2212. 1 13 1000 \u00d7 1 005 488 46 \u00d7. ). ) = = = 63.69 Air fuel ratio : In this arrangement the power developed in turbine is used for driving compressor so, 63.69 Ans. Jet Propulsion and Rocket Engines ________________________________________________ 875 C pair ( T 02 \u2032 \u2212 T 01 ) = C p gas ( T 03 \u2212 T 04 \u2032 ) \u21d2 \u21d2 1.005 (488.46 \u2013 272) = 1 13 1000 (. \u2212 \u2032T 04 ) T04\u2032 = 810.46. 63 69 1 From turbine isentropic efficiency, \u03b7 turb = or 0.82 = ( T 03 ( T 03 \u2032 T 04 T 04 ) ) \u2212 \u2212 \u2212 T 03 T 03. 810 46 \u2212 T 04 \u21d2 For isentropic expansion, T04 = 768.85 K \u03b3 ( gas \u03b3 / gas \u2212 1) p 04 p 03 = \uf8eb \uf8ec \uf8ed T 04 T 03 \uf8f6 \uf8f7 \uf8f8 Pressure of gas leaving turbine, \u21d2 p04 = 3. 768 85 1000 1 33.. 0 33 = 1.04 bar Ans. Considering T04 and p04 as static values, Ts = =. 810 46 0 4.. 1 04 0 33. 1 33. =. 639 4 K For exit nozzle, \u03b7 nozzle = \u2032 ) ) \u21d2 Exit jet velocity, Thrust per kg of air per second \u2212 \u2032T ) 5 639 4. ) 0.92 = 810 46 (. \u2212 810 46. ( T5\u2032 = 653.08 K C5\u2032 = 2 h 4 ( \u2032 \u2032\u2212 h 5 ) \u2032\u2212 T 5 ) = 2 C ( T 4 \u2032 pgas = \u00d7. 2 1 13 C5\u2032 = 596.4 m/s = Ce \u00d7 ( 63.69 \uf8f6 \uf8f7 \uf8f8 \u00d7 596.4 \u2212 200 Thrust = 405.73 N/kg/s Ans.. 810 46 \u2212. 653 08 ) \u00d7 1000 876 _________________________________________________________ Applied Thermodynamics 2. A jet propulsion engine has compressor with pressure ratio 4 and compressed air enters into combustion chamber where combustion occurs so as to yield temperature of 500\u00b0C at turbine inlet. Actual temperature at inlet to combustion chamber is 10% more than that of isentropic compressor temperature rise. Exhaust from turbine is expanded up to atmospheric pressure of 1 bar. The ambient temperature is 285 K. Determine, (i) power required to drive compressor, (ii) air fuel ratio if calorific value of fuel is 43100 kJ/kg, (iii) static thrust developed per kg of air per second. Solution: T1 = 285 K, p1 = 1 bar, T3 = 773 K p 2 p 1 = 4 \u21d2 p2 = 4 bar, p 2 p 1 = 285 \u00d7 (4)0.286 \u2212\u03b3 \u03b3 1 T2 = T 1 T2 = 423.7", " K Actual temperature rise, ( \u21d2 \u2032 \u2212 T 2 = Fig. 19.17 \u2212 T ( 2 1 1+. 1 1. T ) 1 T2\u2032 = T 1 T 1 T 2 = 285 + 1.1 (423.7 \u2013 285) ) \u2212 T 1 ) ( Work required in compressor, Wcompr = C T2\u2032 = 437.57 K ( ) pair T 1 T 2 \u2032 \u2212 = 1.005 (437.57 \u2013 285) = 153.3 kJ/kg of air. Power required to drive compressor = 153.3 kW/kg of air Ans. Heat added in combustion chamber, qadd = Cpair (T3 \u2013 T2\u2032) = 1.005 (773 \u2013 437.57) Jet Propulsion and Rocket Engines ________________________________________________ 877 qadd = 337.11 kJ/kg of air for air-fuel mass ratio being \u2032 \u00d7 add = C.V. = \u2032 m \u2032 m a f 43100. 337 11 =. 127 85 kg of air / kg of fuel Air-fuel ratio = 127.85 Ans For isentropic expansion, \u21d2 Enthalpy drop in nozzle = ( T5 = T 3 p 1 p 2 T5 = 519.\u03b3 \u03b3 1 = 773 1 4 0 286. \u2212 ( T 5 ) h 2 \u2212 ) \u2212\u2032 h 1 C T ( p 2 \u2212\u2032 T 1 ) = 1 005. ) 519 9 (. = 101.03 kJ/kg of air 773 \u2212 \u2212 (. 437 57 \u2212 ) 285 = 2 101 03 1000. Ce = 449.5 m/s Ans. \u00d7 \u00d7 Velocity of exit gas from nozzle, Ce Thrust, T = m a for negligible velocity at inlet, i.e., Ca = 127 85. Thrust = 453.02 N/kg of air/s Ans. 449 5. \u2212 1 0 3. Following data refer to double spool turbofan engine in which low pressure turbine drives fan and high pressure turbine drives compressor. Ambient state : : Fan pressure ratio Overall pressure ratio By pass ratio, m \u2032 \u2032 p 1 bar, 15\u00b0C 1.5 : 28 : 5.5 c m Mass flow rate of air : 200 kg/s. Stagnation temperature at turbine inlet : 1300\u00b0C Mechanical efficiency \u03b7 mech. = 96% 878 _________________________________________________________ Applied Thermodynamics Polytropic efficiency of fan and compressor = \u03b7 Polytropic efficiency of turbine, \u03b7 poly, turb = 95% Pressure loss in combustion chamber = 1.5 bar Nozzle efficiency, \u03b7 Lower calorific value of fuel : 43100 kJ/kg nozzle = 95% poly, fan = \u03b7 poly, compr. = 90% Cpair = 1.005 kJ/kg\u00b7K Cpgas = 1.14 kJ/kg\u00b7K \u03b3 gas = 1.33 expansion can be obtained. Determine air-fuel ratio, specific fuel consumption and specific thrust. Solution: Using the polytropic efficiency for compression and expansion the index of compression and = For compression: \u2212 1. 0 328 poly, compr = 87. 1 4. 1 4 For expansion: \u2212 n T n T 1 = \u03b7 poly, turb \uf8eb \uf8ec \uf8ec \uf8ed \u03b3 \uf8f6\u2212 1 \uf8f7 \uf8f7 \uf8f8 gas gas \u03b3 = 0.90 \uf8eb \uf8ec \uf8ed 0.33 1.33 \uf8f6 \uf8f7 \uf8f8 In fan, \u2032 T 02 T 01 \u2212 n c n c 1 ; Take, p01 = pa and T01 = Ta = = 0.223 p 02 p 01. 0 328 T02\u2032 = 288 1 5. 1 T02\u2032 = 328.96 K n T n T T \u2032 03 T 02 p 03 p 02 = \u2032 \u2212 1 In compressor T03\u2032 = 328.96. 0 223 28. 1 5 =. 631 82 K For flow of bypass atmospheric air through nozzle the nozzle may be checked for whether it is in choked state or not. Critical pressure ratio for by pass air nozzle, \u2212 1) \uf8eb 2 = \uf8ec +\uf8ed \u03b3 1 \u03b3 \u03b3 /( \uf8f6 \uf8f7 \uf8f8 cp p 02 p c p 02 = 02 =. 0 528 Jet Propulsion and Rocket Engines ________________________________________________ 879 \u21d2 p 02 pc = 1.89 Actual pressure ratio for this nozzle, p 02 pa =. 1 5 1.= 1 5 Since critical pressure ratio (1.89) is more than actual expansion pressure ratio (1.5) so it means that by pass air nozzle is not choked and p8 may be taken equal to pa. Fig. 19.18 Twin spool turbofan engine. Using nozzle", " efficiency T 02 \u2032 \u2032\u2212 T 8 = \u03b7 nozzle T 02 1 02 = 0 95 328 96 Velocity at exit of nozzle, T02\u2032 \u2013 T8\u2032 = 34.22 K C8 = 3 2 1 005 10 \u00d7. \u00d7 \u00d7. 34 22 = 26.2 m/s Mass flow rate of bypass air ) ( = \u00d7. 200 5 5 6 5 (. ) = 169.23 kg/s Mass flow rate of hot gases, mc\u2032 = 200 \u2013 169.23 = 30.77 kg/s Thrust available due to by pass air, Tb = m\u2032 b \u00d7 C8 = 169.23 \u00d7 262.26 Applying energy balance on combustion chamber we get, Tb = 44.382 kN \u2032 pgas 04 = C p air \u22c5 T \u2032 03 + \u2032 14 1573 = (. 1 005 631 82 \u00d7. + ) LCV m m \u2032 \u2032 f c \u00d7 43100 880 _________________________________________________________ Applied Thermodynamics m f \u2032 \u2032 = 0.028 m c work required in HP compressor. For HP turbine driving the HP compressor the work output from HP turbine may be equated to = \u03b7 p\u00d7 C T 04 T 05 T 02 T 03 mech m pair C \u2212 + \u2212 1 ( ) ( ) gas \u2032 \u2032 f \u2032 \u2032 \u2032 m c 1.005 (631.82 \u2013 328.96) = 0.96 (1 + 0.028) \u00d7 1.14 \u00d7 (1573 \u2013 T05\u2032) T05\u2032 = 1302.45 K LP turbine is used for driving fan so, output of LP turbine goes for driving fan. \u22c5 \u03b7 \u2212 gas pC 1 = ( \u21d2 1.14 \u00d7 0.96 (1302.45 \u2013 T06\u2032) = (1 + 55) \u00d7 1.005 (328.96 \u2013 288) B C ) T \u2032 06 T 02 T 01 mech ( T \u2032 05 pair ) ) ( \u2212\u2032 + T06\u00a2 = 1057.96 K Pressure at inlet to HP turbine, p04 = p03 \u2013 \u2206ploss = 28 \u2013 1.5 = 26.5 bar Pressure at exit of LP turbine, (p06) may be estimated as under, 1302 45. 1573 p05 = p 04 p06 = p 05 1057 96. 1302 45. T \u2032 05 T 04 11 37. 26 5. and 223 /. T 06 T 05 \u2032 1 0 223 /. = 11 37. bar = 4 476. bar Critical expansion pressure ratio of nozzle handling hot gases, \u03b3 gas \u2212 = \u03b3 \u03b3 ( = + 1 gas 1) 1 33.. 0 33 \uf8eb \uf8ec \uf8ed gas 2. 2 33 2 \uf8f6 \uf8f7 \uf8f8 Critical pressure ratio = 1.85 Since critical pressure ratio is less than expansion ratio in nozzle for hot gas so nozzle will be in choked state. so, For exit nozzle \u21d2 > Critical pressure ratio p 06 pa p 06 p 7 p7 = 2.42 bar = 1.85 ( T 06 \u2032 \u2032\u2212 T 7 ) = \u03b7 nozzle T \u2032 06 \uf8ee \uf8ef \u2212\uf8ef 06 \u03b3 gas \u03b3 \uf8f6\u2212 1 \uf8f7 \uf8f7 \uf8f8 gas \uf8f9 \uf8fa \uf8fa \uf8fa \uf8fb \uf8ee \uf8ef 0.95 1057.96.85 0.33 1.33 \uf8f6 \uf8f7 \uf8f8 \uf8eb \uf8ec \uf8f6 \uf8ed \uf8f7 \uf8f8 \uf8f9 \uf8fa \uf8fa \uf8fb = T \u2032 06 Velocity of hot gases at exit of nozzle, = 142.28 K ) ( \u2032\u2212 T 7 C7 = 2 C pgas \u00d7 ( T 06 \u2032 \u2032\u2212 T 7 ) Jet Propulsion and Rocket Engines ________________________________________________ 881 2 1 14 1000 142 28 \u00d7. \u00d7. C7 = \u00d7 C7 = 569.55 m/s Thrust due to hot gases = m Cc\u2032 \u00d7 7 = 30.77 \u00d7 569.55 = 17.525 kN Total thrust = thrust due to bypass air + thrust due to hot gases = 44.382 + 17.525 = 61.91 kN Specific thrust = Specific fuel consumption = = 61 91. 200 m m \u2032 \u2032 f = 0.309 kN/kg/s Ans. \u00d7 3600 \u2032\u00d7 m c c Total thrust \u00d7. \u00d7. 0 028 30 77 61 90 103 \u00d7. 3600 = 0.05 kg/h.N Ans. 4. A turbojet plane flies with speed of 1000 kmph and inducts air at the rate of 50 kg/s. Air fuel ratio is 52 and the expansion of gases occurs", " up to ambient pressure. Lower calorific value of fuel is 43100 kJ/kg. For maximum thrust the flight to jet velocity ratio is 0.5. Determine (i) jet velocity, (ii) maximum thrust, (iii) specific thrust, (iv) thrust power, (v) propulsive, thermal and overall efficiencies, (vi) Specific fuel consumption. Solution: Velocity of turbojet plane = 1000 kmph = 277.78 m/s For maximum thrust power, Velocity of jet at exit, Ce = 277 78.. 0 5 Ce = 555.56 m/s Jet velocity = 555.56 m/s Ans. Air will be admitted in plane with the velocity with which plane moves forward. Thrust 50 50 52 \u00d7 + \u2212 \u2032 \u22c5 m C a a \u2212 555 56. ( 50 \u00d7 277 78. ) = T = 14423.19 ~\u2212 14.42 kN Ans. Specific thrust =. 14423 19 50 = 288.46 N/kg/s Ans. Propulsive efficiency \u03b7 Thrust power = 14.42 \u00d7 277.78 = 4005.58 kW Ans. 66.66% Ans. prop = 0 6666. 2 \u2248 = + 1 1 0 5. Thermal efficiency, \u03b7 th = ( \u00d7 + 1 FA 2 2 ) FA C e \u00d7 \u2212 2 C a Calorific value 882 _________________________________________________________ Applied Thermodynamics + 1 1 52 ( 555 56 ). 2 \u2212 ( 277 78. ) 2 = 1 \u00d7 52 = 0.1432 \u2248 14.32% Ans. 3 43100 10 \u00d7 \u00d7 2 Overall efficiency \u03b7 overall = \u03b7 prop \u00d7 \u03b7 = 0.1432 \u00d7 0.6666 th = 0.0954 = 9.54% Ans. Specific fuel consumption = = m f \u2032 \u00d7 3600 Thrust \u00d7 50 52 14423 19. 3600 = 0.2399 kg/h N Ans. 5. A turboject engine has air entering at 2.2 bar, \u2013 53\u00b0C and 260 m/s. Turbojet has compressor pressure ratio of 12, turbine inlet temperature of 1000\u00b0C, pressure at exit of nozzle as 2.2 bar, isentropic processes in diffuser & nozzle and isentropic efficiecies of compressor and turbine as 85% and 88% respectively. Neglecting pressure drop in combustion chamber determine the pressure and temperature at all salient points and the velocity at exit of nozzle. Also consider air as working fluid throughout. Solution: p1 = 2.2 bar T1 = 220 K or \u2013 53\u00b0C C1 = 260 m/ : Diffuser, CC : Combustion chamber, C : Compressor T : Turbine, N : Nozzle, 3' and 5' are actual states. Fig. 19.19 Stagnation properties at inlet state can be estimated for knowing the states at point 2 which is inlet of compressor. In diffuser section kinetic energy will be getting transformed to modify pressure and temperature at inlet to diffuser leading to negligible kinetic energy at inlet to compressor. T2 = T 1 + 2 C 1 2 C p = 220 + 2 (260) \u00d7 3 (2 1.005 10 ) \u00d7 = 253.63K Jet Propulsion and Rocket Engines ________________________________________________ 883 \u03b3 \u03b3\u2212 1) /( p2 =.2 \uf8eb \uf8ec \uf8ed 253.63 220 3.5 \uf8f6 \uf8f7 \uf8f8 = 3.62 bar p3 = p2 \u00d7 12 = 3.62 \u00d7 12 = 43.44 bar = p4 T3 = T 2 ( 1) \u03b3 = 253.63(12) \u2212 (1.4 1) 1 = 516.23K Compressor isentropic efficiency, 0.85 = \u2212 T T 2 3 \u2212 T\u2032 T 3 2 0.85 = \u2212 516.23 253.63 \u2212 T \u2032 3 253.63 \u21d2 T3\u2032 = 562.57K In case of turbojet engine, under ideal condition the work requirement of compressor should be equal, Wcomp = Wtrubine Cp(T3\u2032 \u2013 T2) = Cp(T4 \u2013 T5\u2032) Given turbine inlet temperature, T4 = 1273K (562.57 \u2013 253.63) = (1273 \u2013 T5\u2032) \u21d2 T5\u2032 = 964.06 K Turbine iseutropic efficiency, 0.88 = ( T 4 ( T 4 \u2032\u2212 T 5 \u2212 T 5 ) ) 0.88 = \u2212 (1273 964.06) (1273 \u2212 )T 5 \u21d2 T5 = 921.93 K Corresponding to this temperature T5, pressure p5 can be determined \u03b3 /( \u03b3\u2212 1) p5 = p 4 \uf8eb \uf8ec \uf8ed T", " 5 T 4 \uf8f6 \uf8f7 \uf8f8 = 43.44 3.5 \uf8eb \uf8ec \uf8ed 921.93 1273 \uf8f6 \uf8f7 \uf8f8 p5 = 14.042 bar T2 = 260 + 2 (200) \u00d7 2 1005 T2 = 279.9 K Considering diffuser efficiency, 0.\u2032 2 1 \u21d2 T2\u2032 = 282.11 K Fig. 19.20 884 _________________________________________________________ Applied Thermodynamics p2 = p 1 ( 1) \u03b3\u2212 \u03b3 \uf8eb \uf8ec \uf8ed T 2 T 1 \uf8f6 \uf8f7 \uf8f8 = 282.11(8) 0.286 Considering compressor efficiency, 0..8 = \u2212 511.33 282.11 \u2212 ( T \u2032 3 282.11) Turbine inlet temperature, T3\u2032 = 568.635 K T4 = 1000K Considering energy balance on combustion chamber; mf \u00d7 CV + m3 \u00d7 Cp \u00d7 T3\u2032 = (m3 + mf) \u00d7 Cp \u00d7 T4 CV \uf8eb + \uf8ec \uf8ec \uf8ed ) = CpT4 \u2013 CV Air\u2013Fuel ratio ( \u2212 CV) \u2212 C T p 4 ) Temperature at exit of nozzle for given exit pressure of 2.2 bars p6 = 2.2 bar; T6 = T 5 ( 1) \u03b3 T6 = 921.93 \u2212 (1.4 1) 1.4 \uf8eb \uf8ec \uf8ed 2.2 14.042 \uf8f6 \uf8f7 \uf8f8 T6 = 542.83 K Velocity at exit of nozzle = T5 C6 = 2( T 5 \u2212 ) T C 6 p C6 = 872.92 m/s Velocity of exit of nozzle = 872.92 m/s. Ans. Jet Propulsion and Rocket Engines ________________________________________________ 885 6. A jet engine plane is moving at 200 m/s and admits air at 0.7 bar, 260 K in diffuser section having inlet cross section area of 1000 cm2 from where it enters compressor with negligible velocity. Compressor pressure ratio is 8 and turbine inlet temperature is 1000\u00b0C. Combustion gases expand in turbine and subsequently leave nozzle at 0.7 bar. Efficiencies of nozzle section, diffuser section, compressor and turbine are 0.9, 0.9, 0.8 and 0.8 respectively. Fuel used in combustion chamber has calorific value of 45000 kJ/kg of fuel and pressure drop in combustion chamber is 0.15 bar. Determine the fuel air ratio, specific thrust and total thrust available. Neglect change in specific heat of air when transformed into combustion products and treat it as air. Take Cp = 1.005 kJ/kg\u00b7K. Solution: Temperature at inlet to compressor T2 = 45000 (1000 1.005) \u00d7 \u2212 \u00d7 ((1.005 1000) \u2212 (1.005 568.635)) \u00d7 Air-Fuel ratio = 101.48 : 1 Turbine inlet pressure, p4 = p3 \u2013 0.15 = 7.248 \u2013 0.15 = 7.098 bar We know that in jet engines normally the turbine output should be just sufficient to run compres- sor; so Wcompressor = Wturbine Cp(T3\u2032 \u2013 T2\u2032) = Cp(T4 \u2013 T5\u2032) (568.635 \u2013 282.11) = (1000 \u2013 T5\u2032) \u21d2 T5\u2032 = 713.475K Considering turbine efficiency 0.8 = T 4 T 4 \u2032\u2212 T 5 \u2212 T 5 = 1000 713.475 \u2212 1000 \u2212 T 5 \u21d2 T5 = 641.84K Considering adiabatic expansion p5 = p 4 \u03b3 \u03b3\u2212 1.098 3.5 \uf8eb \uf8ec \uf8ed 641.84 1000 \uf8f6 \uf8f7 \uf8f8 = 1.5036 bar In nozzle exit pressure is given, so, p6 = 0.7 bar T6 = T \u2032 5 ( 1) \u03b3 = 713.475 0.286 \uf8eb \uf8ec \uf8ed 0.7 1.5036 \uf8f6 \uf8f7 \uf8f8 = 573.35K Considering nozzle efficiency; 0. = T6\u2032 = 587.36 K \u2212 (713.475 \u2212 (713.475 573.35) T 6 ) \u2032 886 _________________________________________________________ Applied Thermodynamics Velocity at exit of nozzle, C6 = \u2032\u2212 2 ( T\u2032 pC T 5 6 ) = 2 1005(713.475 587.36) \u00d7 \u2212 C6 = 503.48 m/s Volume flow rate of air = 200 \u00d7 1000 \u00d7 10\u20134", " = 20 m3/s Mass flow rate = \u22c5 p v RT = 2 \u00d7 \u00d7 0.7 10 20 \u00d7 0.287 260 = 18.76 kg/s Specific thrust = \uf8eb \uf8ec \uf8ed + 1 fm m 3 \uf8f6 \uf8f7 \uf8f8 \u00d7 C 6 Specific thrust = 508.44 N/kg of air/s Ans. Total thurst = 18.76 \u00d7 508.44 = 9538.33N Ans. 7. An aeroplane flying at speed of 250 m/s, running on jet engine inducts air at 0.3 bar and \u201320\u00b0C from atmosphere, compressor pressure ratio is 6 and the turbine inlet temperature is 700 \u00b0C. Velocity at exit of turbine is 180 m/s, turbine and compressor efficiencies are 0.85 each and jet engine efficiency is 90%. Thurst power produced is 800kW, and calorific value of fuel is 43000 kJ/kg. Considering following properties for air and gas determine overall efficiency, rate of air consumption, power developed by turbine, and exit area of jet. For air, Cp,air = 1.005 kJ/kg\u00b7K, \u03b3 For gas, Cp,gas = 1.087 kJ/kg\u00b7K, \u03b3 Solution: Let mass flow rate of air and fuel be mair & mfuel kg/s Writing down energy balance on combustion chamber; air = 1.4, R = 0.287 kJ/kg\u00b7K gas = 1.33 (mfuel \u00d7 CV) + (mair\u00b7Cpair\u00b7T2\u2032) = (mfuel + mair)\u00b7Cpgas.T3 In diffuser T1 = T 0 + 2 C 0 2 C = 253 + 2 (250) \u00d7 2 1005 = 284.09K pair \u03b3 /( \u03b3\u2212 1) p1 =.31 bar Let, \uf8eb \uf8ec \uf8ed m m fuel air = FA \uf8f6 \uf8f7 \uf8f8 Thus energy balance on combustion chamber; FA \u00d7 CV = (1 + FA)Cp,gas\u00b7T3 \u2013 Cp,air T2\u2032 In compressor, T2 = T1(6)0.286 = 474.25K Fig. 19.21 Jet Propulsion and Rocket Engines ________________________________________________ 887 T2 compr. = + 284.09 (474.25 284.09) 0.85 \u2212 T2\u2032 = 507.81K, T3 = 973 K p2 = p1 \u00d7 pressure ratio = 0.31 \u00d7 6 = 1.86 bar = p3 Substituting in energy balance on combustion chamber; FA \u00d7 43000 = {(1 + FA) \u00d7 1.005 \u00d7 973} \u2013 (1.087 \u00d7 507.81) FA = 0.01013 or Air \u2013 fuel ratio = 98.67 : 1 Ans. Ideally, Wcompr = Wturb Cp,air = (T2\u2032 \u2013 T1) = cpgas (1 + FA) \u00d7 (T3 \u2013 T4\u2032) 1.005(507.81 \u2013 284.09) = 1.087(1 + 0.01013) (973 \u2013 T4\u2032) Isentropic expansion in turbine yields, T4\u2032 = 768.23 K \u03b7 turb = T 3 T 3 \u2032\u2212 T 4 \u2212 T 4 \u21d2 0.85 = \u2212 (973 768.23) \u2212 (973 )T 4 \u21d2 T4 = 732.09 K \u03b3 gas \u03b3 \u2212 ( gas 1) p4 = \uf8f6 \uf8f7 \uf8f8 p4 = 1.86 \uf8eb \uf8ec \uf8ed 732.09 973 \uf8f6 \uf8f7 \uf8f8 1.33 \u2212 (1.33 1) p4 = 0.5909 bar In exit nozzle jet, T5 = \uf8f6\u2212 1 \uf8f7 \uf8f7 \uf8f8 gas \u03b3 gas T5 = 768.23 \uf8eb \uf8ec \uf8ed 0.3 0.5909 \uf8f6 \uf8f7 \uf8f8 \u2212 (1.33 1) 1.33 T5 = 649.3K Jet engine efficiency = 0.9 = ( C p gas ( ) 888 _________________________________________________________ Applied Thermodynamics Given C4\u2032 = 180 m/s \u21d2 0.9 = {1087(786.23 649.3) + 2 (180 / 2)} / 2) 2 ( C 5 \u2212 Nozzle exit velocity. C5 = 511.72 m/s. + {(1 Overall efficiency = = ) \u2212 FA C C C 5 0 \u00d7 { FA CV 0 } } \u22c5 {(1 0.01013) 511.72 250} 250 \u2212 \u00d7", " + \u00d7 3 (0.01013 43000) 10 \u00d7 \u00d7 It is given that engine develops thrust power of 800 kW. = 0.1532 or 15.32% Hence, Thrust power = 800 = {(1 + \u2212 ) }, C m FA C 0 5 1000 air \u00d7 C 0 \u21d2 mair = 11.99 kg/s Rate of air consumption = 11.99 kg/s Ans. Power produced by turbine = mair (1 + FA)Cp gas(T3 \u2013 T4\u2032) Power produced by turbine = 2695.83 kW Ans. = 11.99(1 + 0.01013) \u00d7 1.087(973 \u2013 768.23) The actual state at exit of nozzle jet may not be result of isentropic expansion in nozzle, let actual state at exit of nozzle be 5\u2032, Cp,gas(T4\u2032 \u2013 T5\u2032) =.087 \u00d7 103(768.23 \u2013 T5\u2032) = (511.72) 2 \u2212 (180) 2 2 T5\u2032 = 662.68K Density of exhaust gases, 5\u2032\u03c1 = p \u2032 5 RT \u2032 5 = \u00d7 2 0.3 10 \u00d7 0.287 662.68 = 0.1577 m3/kg Let cross-sectional area of jet at exit be Aj m2. Aj \u00d7 C5 \u00d7 5\u2032\u03c1 = mair(1 + FA) Aj = + 11.99(1 0.01013) \u00d7 511.72 0.1577 = 0.15m2 Jet exit area = 0.15m2 Ans. Jet Propulsion and Rocket Engines ________________________________________________ 889 8. A jet aeroplane flies at speed of 900 km/h. The density of air at this altitude is 0.15 kg/m3 and drag on plane is 6800kW. Considering jet plane to have 2 jets and engine working on turbo-prop system with propulsive efficiency of 56% determine the absolute velocity of jet and diameter of jet. Solution: Velocity of aeroplane = Propulsive efficiency = \u00d7 900 1000 3600 2 C a C+ a C e ( ) = 250 m/s Here Ca = 250 m/s, Ce = Relative velocity of jet leaving 0.56 = \u00d7 2 250 + ( eC Ce = 642.86 m/s 250) Absolute velocity of jet = Ce \u2013 Ca = 642.86 \u2013 250 = 392.86 m/s Drag on plane or thrust = m\u2032(Ce \u2013 Ca).6800 = m\u2032(642.86 \u2013 250) m\u2032 = 17.31 kg/s Volume flow rate = 17.31 0.15 = 115.4 m3/s Let jet diameter of jet be dj; Volume flow rate = 115. \u21d2 dj = 0.338 m Jet diameter = 33.8 cm Ans. 9. An aeroplane is flying at speed of 900 km/hr. It is run by turbojet engine. The plane has wing area of 2 \u00d7 105 cm2, drag coefficient is 0.018 and jet velocity is 550 m/s. Considering Cp = 1.005 kJ/kg. K for air and density ratio between STP and flight altitude being 0.4 determine specific thrust. Solution: Density at STP = 2 \u00d7 1.01325 10 \u00d7 0.287 288 = 1.226 kg/m3 Density of air at altitude = 0.4 \u00d7 1.226 = 0.488 kg/m3 Thrust on Aeroplane = A.018 (2 10 \u00d7 5 \u00d7 \u2212 4 \u00d7 10 ) 0.488 \u00d7 \u00d7 3 (900 10 / 3600) 2 2 = 5490 N or 5.49 kN 890 _________________________________________________________ Applied Thermodynamics Jet velocity, Ce = 550 m/s Thurst can also be given as, = m\u2032(Ce \u2013 Ca) 5490 = \u2032 m \uf8eb \uf8ec \uf8ec \uf8ed 550 \u2212 \uf8eb \uf8ec \uf8ec \uf8ed 3 \u00d7 900 10 3600 \uf8f6 \uf8f7 \uf8f7 \uf8f8 \uf8f6 \uf8f7 \uf8f7 \uf8f8 m\u2032 = 18.3 kg/s Specific thrust = 5490 18.3 = 300 N/kg of air Ans. 10. An air craft running with speed of 900 km/hr uses turbo\u2013 jet engine taking air at rate of 55 kg/s. The air-fuel ratio is 85 and combustion efficiency is 0.96. Fuel has lower calorific value of 43000 kJ/kg. The turbo jet engine nozzle has isentropic enthalpy change of 220 kJ/kg and velocity coefficient is 0.95. Calculate propulsive power, thrust power, propulsive efficiency, overall efficiency, fuel flow rate and thrust neglecting fuel mass addition. Solution: Given Ca = 900 km/", "hr = \u00d7 900 1000 3600 = 250 m/s Jet velocity = Velocity coeff \u00d7 2 Enthalpy change in nozzle \u00d7 \u00d7 0.95 Ce = 615.67 m/s Specific thrust per kg of air = 1 \u00d7 (615.67 \u2013 250) = 400.67 N/kg air 3 2 220 10 \u00d7 \u00d7 = For given Air-Fuel ratio, fuel flowrate = \uf8eb \uf8ec \uf8ed 1 85 \uf8f6 \uf8f7 \uf8f8 \u00d7 3600 55 \u00d7 Fuel flow rate = 2329.4 kg/hr Specific fuel consumption = 2329.4 400.67 55\u00d7 = 0.1057 kg/N\u00b7hr Thrust power = m\u2032 (Ce \u2013 Ca) \u00d7 Ca = 55 \u00d7 250 \u00d7 (615.67 \u2013 250) = 5027962.5 W or 5027.96 kW Propulsive power = \u2212 \u2032 615.67) 2 \u00d7 55 \u2212 (250) 2 } 2 = 8705112.59 W or 8705.11 kW Ans. Propulsive efficiency = Thrust power Propulsive power = 5027.96 8705.11 = 0.5776 or 57.76% Ans. Overall efficiency = Thrust power Heat supplied = \u00d7 55 \uf8eb \uf8ec \uf8ed 5027.96 \uf8f6 \uf8f7 \uf8f8 \u00d7 1 85 \u00d7 43000 0.96 = 0.1882 or 18.82% Ans. Jet Propulsion and Rocket Engines ________________________________________________ 891 19.1 Describe the principle of jet propulsion. 19.2 Classify the jet propulsion engines. 19.3 Define and give mathematical expressions for thrust power, propulsive power and propul- sive efficiency. 19.4 Compare turbojet engine with other jet propulsion engines. 19.5 Compare between solid propellent rocket engine and liquid propellant rocket engines. 19.6 Discuss the working of turboprop, ramjet and pulse jet engines. 19.7 Determine specific fuel consumption and propulsive efficiency for the following turbojet engine. Forward speed of engine : 862.5 km/hr Ambient pressure and temperature : 0.265 bar, \u201349.8\u00b0C Compressor pressure ratio : 10 Stagnation temperature at turbine inlet : 1400 K Lower calorific value of fuel : 4.31 \u00d7 105 kJ/kg Isentropic efficiency of intake : 0.93 Isentropic efficiency of compressor : 0.87 Isentropic efficiency of turbine : 0.90 Isentropic efficiency of exit nozzle : 0.95 Mechanical efficiency : 0.98 Combustion efficiency : 0.98 Combustion pressure loss : 4% of compressor exit pressure Cpa = 1.005 kJ/kg\u00b7k, Cpg = 1.148 kJ/kg\u00b7K, \u03b3 = 1.33 [0.13 kg/h\u00b7N, 0.57] 19.8 Considering a jet propulsion unit to have isentropic compression and expansion and heat supply at constant pressure show that thrust developed per kg of air per second for negligible velocity of approach can be given by; [2Cp\u00b7Ta (t \u2013 1) 1) \u03b3 \u2212 \u03b3 ( ( pr \u2212 ] 1) Here t is the ratio of absolute temperature after combustion and before combustion, rp is compression ratio and Ta is absolute atmospheric temperature. 892 ________________________________________________________ Applied Thermodynamics 20 1. Thermodynamic system refers to a portion of (a) Universe (c) Control surface (b) Surroundings (d) None of these 2. Units of temperature, mass, time, length and energy in SI system of units are given by (a) Celsius, kilogram, second, metre and watt (b) Kelvin, kilogram, second, metre and joule (c) Kelvin, gram, hour, metre and joule (d) Kelvin, kilogram, second, centimetre, joule 3. Which of the following set has all intensive properties? (a) Pressure, volume, energy, heat, specific volume (b) Pressure, volume, energy, heat density (c) Pressure, volume, temperature, density, specific volume (d) Pressure, volume, heat, density, specific volume 4. Which of the following sets has all open systems? (a) Window air conditioner, scooter engine, thermometer (b) Boiler, gas turbine, compressor (c) Pump, thermosflask, refrigerator (d) None of these 5. Locii of the series of state change during heat addition to a gas shall be, 6. (a) Process (c) State change In engineering thermodynamics the macroscopic approach is followed and it may also be called as: (a) Statistical thermodynamics (c) Microscopic study (b) Classical thermodynamics (d) None of these (b) Path (d) All of the above 7", ". Time, length, mass and temperature are: (a) Primary dimensions (c) Both of above (b) Secondary dimensions (d) None of these 8. Difference between the pressure of a fluid and the pressure of atmosphere is called as: (a) Absolute pressure (c) Gauge pressure (b) Barometric pressure (d) None of these 9. Which of the following sets has all properties as point functions? (a) Pressure, temperature, heat (b) Entropy, enthalpy, work (c) Temperature, enthalpy, internal energy (d) Heat, work, enthalpy Multiple Answer Type Questions __________________________________________________ 893 10. Which of the following is not property of thermodynamic system? (a) Temperature (c) Pressure (b) Work (d) Internal energy 11. At a given temperature and pressure all molecules of gas occupy the same volume. This is called (a) Charle\u2019s law (c) Avagadro\u2019s law (b) Boyle\u2019s law (d) Regnault\u2019s law 12. At constant pressure the volume of gas is directly proportional to its temperature. This refers to (a) Charle\u2019s law (c) Dalton\u2019s law If temperature in degree Celsius is 37\u00b0 then the equivalent temperature in Fahrenheit will be: (b) 99 (a) 98.6 (d) 98 (c) 98.3 (b) Boyle\u2019s law (d) None of these 13. 14. The characteristic gas constant R and universal gas constant R can be related through molecular weight M as under, (a) R = R M (c) R = R + M \u2018Work\u2019 done upon the system and \u2018heat\u2019 added to the system shall have sign convention as shown respectively. (a) (\u2013) and (\u2013) (c) (+) and (+) In a constant volume gas thermometer the thermometric property used is (c) mass (a) length (b) R = R M (d) R = R + M (b) (\u2013) and (+) (d) (+) and (\u2013) (d) resistance (b) volume 15. 16. 17. A car\u2019s tyre become hotter due to their exposure to sunlight. Considering air in tyre as system determine sign of heat and work interactions. (a) (+) and (+) (c) (\u2013) and (\u2013) (b) (\u2013) and (+) (d) (+) and (\u2013) 18. Specific heats at constant pressure (Cp) and at constant volume (Cv) can be related by 19. (a) Cp = Cv (c) Cp > Cv In reference to thermodynamic equilibrium it is required to have, (a) Mechanical equilibrium (b) Chemical equilibrium (c) Thermal equilibrium (d) Mechanical, chemical and thermal equilibrium (b) Cp < Cv (d) Cp + Cv = 0 20. The real gas follows the modified form of gas equation involving compressibility factor Z and may be given by: (a) P = Z R T (c) P = R T (b) P = R T Z (d) None of these 21. If ideal gas equation p = R T is used to model a real gas then modeling will becoming more accurate when 894 ________________________________________________________ Applied Thermodynamics (a) Pressure decreases, temperature decreases and molar mass decreases (b) Pressure decreases, temperature increases and molar mass decreases (c) Pressure increases, temperature increases and molar mass increases (d) Pressure increases, temperature decreases and molar mass increases. 22. Compressibility factor (correction factor for real gas) is function of following properties of 23. real gas (a) V and (c) P and T In the van der Waals equation for real gas the constant b refers to (a) Co-volume (b) Mutual attraction amongst molecules (c) Both (d) None of these (b) U and s (d) T and U 24. Which of the following represent the real gas law? (a) nRT = (P + a/v2) (v \u2013 b) (c) nRT = (P \u2013 a/v2) (v + nb) (b) nRT = (P \u2013 a/v2) (v + b) (d) nRT = (P + n2a/v2) (v \u2013 nb) 25. Gas constants for hydrogen and carbon dioxide shall be: (a) 4.124 and 0.189 (c) 0.260 and 0.287 (b) 0.287 and 2.077 (d) 0.461 and 0.260 26. Steam point and ice point temperatures in Fahrenheit scale can be given as: (a) 491.67 and 671.67 (", "c) 373.15 and 273.15 (b) 212 and 32 (d) 100 and 0 27. Triple point of water temperature in Rankine scale can be given by: (a) 491.69 (b) 0.01 (c) 273.16 (d) 32.02 28. Solid, liquids and gases have how many values for specific heat of respective phases. (a) 1, 1 and 1 (c) 1, 1 and 2 (b) 2, 2 and 2 (d) 0, 1 and 2 29. A polytropic process pVn = C having the value of index \u2018n\u2019 as infinity refers to (a) Isobaric process (c) Isothermal process (b) Isochoric process (d) None of these 30. For an ideal gas the hyperbolic process will be similar to: (a) Isobaric process (c) Isothermal process (b) Isochoric process (d) Isentropic process 31. As per the Zeroth law of thermodynamics if the two bodies A and B are in thermal equilibrium with a third body C separately then the two bodies A and B shall also be in thermal equilibrium with each other. Using this law as basis of temperature measurement of the body A the thermometer shall be refering to (a) Body A (c) Body C (b) Body B (d) None of these 32. Which of the following temperature and volume relation is correct for an adiabatic process between state 1 and 2a) (cb) (d = Multiple Answer Type Questions __________________________________________________ 895 33. For a hyperbolic process between the states 1 and 2, the heat interaction can be given as, (a) q = Cv (T2 \u2013 T1) (b) q = 0 n n 1 (c) q = Cp (T2 \u2013 T1) (d) q = Cv (T2 \u2013 T1) + RT1 ln 34. The specific heat for polytropic process can be given as, (a) Cn = Cv (c) Cn = Cv n n n n 1 1 (b) Cn = Cv (d) Cn = Cv 35. In case of free expansion between the states 1 and 2 following is correct, considering no heat interaction. (a) U1 = U2 (c) Q1\u20132 = 0 (b) W1\u20132 = 0 (d) All of these 2 P\u00d7dV refers to, 36. The work interaction between two states 1 and 2, if given by W1\u20132 = (a) Flow work (c) Steady flow work (b) Non-flow work (d) None of these 1 37. Enthalpy remains constant during which of the following process? (a) Throttling process (c) Isothermal process (b) Free expansion (d) None of these 38. Amongst the following which cannot be reversible process (a) Isothermal (c) Adiabatic process (b) Throttling (d) None of these 39. For a perfect gas the internal energy is the function of, 40. Specific heat at constant pressure can be related to R and as under, (a) Pressure only (c) Volume only (a) R 1 (c) R (b) Temperature only (d) None of these (b) R 1 (d) None of these 41. The work done by 2m3 of air at 5N/m2 while getting expanded freely to 10 m3 shall be, (a) 40 J (c) 10 J (b) Zero (d) 50 J 42. For the expansion of air occurring between volume state V1 and V2 the adiabatic process and isothermal process lines shall be as under, (a) A : adiabatic, B : isothermal (b) A : isothermal, B : adiabatic (c) Any one may refer to any of process (d) None of these 896 ________________________________________________________ Applied Thermodynamics 43. Change in internal energy in a reversible process occurring in closed system will be equal to heat transferred if process occurs at constant (c) Temperature (b) Volume (a) Pressure 44. Flow work between the state 1 and 2 is given by (d) Enthalpy 2 pdv (b) 2 pdv 1 (d) None of these (a) (c) 2 vdp 2 vdp + 1 1 1 (a) q = h2 \u2013 h1 (c) q = 0 45. Steady flow energy equation when applied to boiler shall yield, (b) q = w (d) None of these 46. Law of energy conservation is also stated by, (a) Zeroth law of thermodynamics (c) Second law of thermodynamics (b) First law", " of thermodynamics (d) Third law of thermodynamics 47. Which of the following is/are correct? (a) h = u + pv (c) h = 2 Cp dT 1 (b) dq = du + d (d) All of these 48. Heat interaction in polytropic process can be given by, 1 n n n (a) (c) w w (b) (d) 1 n n n w w 49. If a Carnot engine rejects heat at temperature of 400 K and accepts at 750 K. What shall be heat absorbed, if heat rejected is 1000 kJ, (c) 1875 kJ (b) 800 kJ (a) 946 kJ (d) 750 kJ 50. A perpetual motion machine of second kind refers to a heat engine having 100% efficiency. This heat engine will not be possible as it violates (a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Third law of thermodynamics Multiple Answer Type Questions __________________________________________________ 897 51. Efficiency of Carnot engine depends upon the highest and lowest (a) Temperatures (c) Volumes (b) Pressures (d) None of these 52. Efficiency of Carnot engine having the limiting temperatures Tmax and Tmin, shall be minimum when, (a) Tmax > Tmin (c) Tmin = 0 (b) Tmax = Tmin (d) None of these 53. Which of the following is correct in reference to first law of thermodynamics? (a) No difference between heat and work (b) No indication of direction of process (c) Law of energy conservation (d) All of these 54. Which amongst the following is not necessary requirement for a device to be heat engine? (a) Cyclic processes (b) Operation between high temperature source and low temperature sink (c) Reversible processes (d) Converting heat into work 55. A heat engine has efficiency of 60% and operates between temperatures of T1 and 300 K. What shall be the temperature T1? (a) 120 K (c) 120 K or 750 K Joule\u2019s law states that the internal energy of an ideal gas is function of only (c) Pressure (a) Temperature (b) Volume (b) 750 K (d) None of these (d) None of these 56. 57. Coefficient of performance for heat pump may have value: (a) Slightly more than 0 (c) Greater than 1 (b) Equal to 1 (d) All of these 58. Difference between the COP of heat pump and COP of refrigerator is (a) 0 (b) 1 (c) > 1 (d) < 1 59. For reversible heat engine operating between source and sink having temperatures T1 and T2 and heat interactions being Q1 and Q2 respectively thermodynamic temperature scale states (a) (cbd) None of these 60. In case of adiabatic process which of the following may not be true. (a) s = 0 (b) Q = 0 (cd) None of these 61. For a reversible adiabatic process the following is/are true. (a) s = 0 (b) Q = 0 (cd) All of these 898 ________________________________________________________ Applied Thermodynamics 62. Carnot cycle has following processes: (a) Two adiabatics and two isothermal processes (b) Two reversible adiabatics and two reversible isothermal processes (c) Two isentropic processes and two isothermal processes (d) None of these 63. C.O.P. of a reversible heat pump is 1.2. If it is reversed to run as reversible heat engine then its efficiency shall be: (a) 0.833 (b) 0.2 (c) 1.2 (d) None of these 64. For a reversible engine cycle the Clausius inequality says, dQ T T dQ > 0 = 0 (a) (c) T dQ (b) < 0 (d) None of these 65. For an irreversible process in isolated system the entropy change can be given as, (a) dSisolated > 0 (c) dSisolated = 0 (b) dSisolated < 0 (d) None of these 66. For a reversible process in isolated system the entropy change can be given as, (a) dSisolated > 0 (c) dSisolated = 0 (b) dSisolated < 0 (d) None of these 67. The entropy generated and the total entropy change for a process to be impossible shall be, (b) Sgen or Stotal > 0 (d) None of these 68. The entropy change in polytropic process can be given by, (b) Cv ln 1 T 2 T 1 n n 1 (a) S", "gen or Stotal = 0 (c) Sgen or Stotal < 0 T 2 T 1 (c) Cv ln (a) Cv ln T 2 T 1 n n (d) None of these 69. Entropy change for an adiabatic process can be given by, 70. Which of the following relation(s) is/are correct? P 2 P 1 (a) ( ) pC T dT T 2 \u222b 1 \u2013 R ln (c) Cp ln T 2 T 1 (a) Tds = du + pdv (c) Tds = Cp dT \u2013 v dp line (mv) can be related as, (a) mp = mv (c) mp > mv T 2 T 1 \u2013 R ln 2 1 (b) Cv ln (d) Cv ln T 2 T 1 (b) dh = Tds + vdp (d) All of these (b) mv > mp (d) mp mv = 1 72. Exergy can not be less than zero as the maximum work interaction can not be (a) Less than zero (c) Zero (b) Greater than zero (d) None of these 71. On T-s diagram the slope of constant pressure line (mp) and the slope of constant volume Multiple Answer Type Questions __________________________________________________ 899 73. Law of degradation of energy says that the unavailable energy is gradually increasing due to (a) Increase in reversible processes (b) Increase in irreversible processes (c) Increase in unavailable energy (d) None of these 74. Which of the following is/are correct? I: Irreversibility, A: Availability Wc, max: Maximum work, T0: Dead state temperature, Sc: Combined system entropy change. (a) I = T0 Sc (c) A = Wc, max (b) Wc = A \u2013 I (d) All of these 75. The efficiency of reversible and irreversible heat engine cycles can be related as, (a) (c) rev = rev > irrev irrev (b) rev < (d) None of these irrev 76. Which of the following is/are correct? (a) Entropy is used as a convenient coordinate in thermodynamics. (b) Entropy is used to decide whether a process is reversible, irreversible or impossible. (c) Entropy is used to determine the heat transfer in reversible process (d) All of these. 77. Equation G = H \u2013 TS tells us that an exothermic reaction will be associated with which of the following: (i) \u2013 ve H (iii) More disordered, +ve s (v) More ordered \u2013 ve s Possible answers (a) (i) and (v) (c) (ii) and (iii) (ii) + ve H (iv) Spontaneous reaction (b) (ii) and (v) (d) (i), (iii) and (iv) 78. Availability of a closed system at state 1 and dead states referred with subscript 0 is given by (a) (E1 \u2013 E0) + p0(V1 \u2013 V0) \u2013 T0(S1 \u2013 S0) (b) (E1 \u2013 E2) + p0(V1 \u2013 V2) \u2013 T0(S1 \u2013 S2) (c) (h1 \u2013 h0) + g(Z1 \u2013 Z0) \u2013 T0(S1 \u2013 S0) + (d) None of these 2 V1 2 79. The availability of steadily flowing fluid, also called stream availability is given by; (the current state is 1 and dead state is given by 0). (a) (E1 \u2013 E0) + p0(V1 \u2013 V0) \u2013 T0(S1 \u2013 S0) (b) (h1 \u2013 h2) + g(Z1 \u2013 Z2) \u2013 T0(S1 \u2013 S2c) (h1 \u2013 h0) + g(Z1 \u2013 Z0) \u2013 T0(S1 \u2013 S0) + (d) None of these 2 V1 2 80. A characteristic function is one from which all properties of a substance can be obtained by differentiation alone, so that no arbitrary functions requiring supplementary data for their evaluation are required. Characteristic function is/are out of the following: (a) f (u, v, s) = 0 (c) f (g, p, t) = 0 (b) f (h, p, s) = 0 (d) All of these 900 ________________________________________________________ Applied Thermodynamics 81. Which of the following is/are incorrect Maxwell equations? = (b) (a) T p s v s p T sv p vT c) (d) None of these 82. The equation relating the following", " three measurable properties is called: (i) The slope of saturation pressure-temperature line (ii) The latent heat (iii) The change in volume during phase transformation (a) Joule\u2019s equation (b) Clapeyron equation (c) Maxwell relation (d) None of these Joule Thomoson coefficient has its value changing on two sides of inversion line as under, (a) + ve to the right and \u2013 ve on the left of inversion line (b) \u2013 ve to the right and + ve on the left of inversion line (c) Zero on the right and zero on the left of inversion line (d) None of these 83. 84. Critical point pressure and temperature for water are, (a) 22.12 MPa and 374.15\u00b0C (c) 18.2 MPa and 899\u00b0C (b) 0.23 MPa and \u2013268\u00b0C (d) None of these 85. On T-s diagram the state change during transformation of 0\u00b0C water to 0\u00b0C steam at atmospheric pressure due to heat addition can be shown by (a) Slant line (c) Horizontal line (b) Vertical line (d) None of these 86. On T-s diagram the state change during transformation of 0\u00b0C water to 100\u00b0C water at atmospheric pressure due to heat addition can be shown by (a) Line with positive slope (c) Horizontal line (b) Vertical line (d) None of these 87. Slope of constant pressure line on enthalpy-entropy diagram of steam with increasing pressure. (a) Increases (c) Remains unaltered (b) Decreases (d) None of these 88. The value of dryness fraction at critical point for water-steam phase transformation may be, (c) Either 0 or 1 89. During the dryness fraction measurement using throttling calorimeter the wet state of steam (d) All of these (b) 1 (a) 0 is throttled so as to lie in (a) Superheated state (c) Subcooled state (b) Wet state (d) None of these 90. The expression for air standard efficiency of Diesel cycle is (a) = 1 \u2013 1 1 r (c) = 1 \u2013 1 Here r,, are compression ratio, cut-off ratio and pressure ratio. (d) None of these 1 1 1 r 1 (b) = 1 \u2013 1 1 1 1 r Multiple Answer Type Questions __________________________________________________ 901 91. The mean effective pressure for Diesel cycle is, 1 (a) mep = b) mep = c) Both of these Here r,, p1 and are compression ratio, cut-off ratio pressure at beginning of compression and pressure ratio. (d) None of these 92. Thermodynamic processes constituting a Rankine cycle are isobaric heat addition, adiabatic expansion, isobaric heat release and..., (a) Adiabatic pumping (c) Polytropic pumping (b) Isothermal pumping (d) None of these 93. Rankine cycle efficiency can be improved by (a) Reduced heat addition in boiler (c) Reduced feed pump work (b) Increasing expansion work (d) All of these 94. For the same compression ratio and same heat input the Otto, Diesel and Dual cycle efficiencies can be given as, (a) (c) Dual > Dual > Otto > Diesel > Diesel Otto Otto > (b) (d) None of these Diesel > Dual 95. Carnot vapour power cycle comprises of following processes. (a) Two reversible isothermals and two reversible adiabatic processes (b) Two isothermals and two adiabatic processes (c) Two isothermals and two isentropic processes (d) Two isobarics and two adiabatic processes 96. The steam is superheated in boiler at (a) Isothermal process (c) Isochoric process (b) Isobaric process (d) None of these 97. Throttling process can be shown on Mollier diagram by, (a) Vertical line (c) Inclined line (b) Horizontal line (d) None of these 98. The heat required for vaporization of water to steam decreases with (a) Increasing steam generation pressure (b) Decreasing steam generation pressure (c) Decreasing steam generation temperature (d) None of these 99. Adding heat to wet steam causes increase in its dryness fraction while the temperature and pressure remain constant in (a) Subcooled region (c) Wet region (b) Superheated region (d) None of these 100. Latent heat of vaporization of water at critical point is (a) Zero (c) 1 kcal/kg (b) 539 kcal/kg (d) None of these 101. The combustion analysis in", " which the fuel is separated into arbitrary constituents such as moisture, volatile matter, fixed carbon and ash etc. is called as (b) Proximate analysis (a) Ultimate analysis (d) None of these (c) Stoichiometric analysis 902 ________________________________________________________ Applied Thermodynamics 102. A combustion has all the oxygen supplied with air to the reactants being used and no free oxygen appears in products. Such air supplied is called as (a) Stoichiometric air (c) Dry air (b) Excess air (d) None of these 103. Composition of dry air on mass basis is (a) 0.232 kg O2 and 0.768 kg N2 (c) 1 kg O2 and 3.31 kg N2 (b) 0.21 kg O2 and 0.79 kg N2 (d) None of these 104. Enthalpy of combustion of any fuel containing hydrogen in it and the H2O present in liquid form in products is termed as, (a) Higher heating valve (c) Enthalpy of hydration (b) Lower heating valve (d) None of these 105. When the temperature of a pure substance approaches zero on Kelvin scale then the entropy of substance approaches zero. This is stated by (a) First law of thermodynamics (c) Third law of thermodynamics (b) Second law of thermodynamics (d) None of these 106. The efficiency of Carnot cycle may be equal to which of the following cycle running between same temperature limits, (a) Rankine cycle (c) Brayton cycle (b) Stirling cycle (d) Otto cycle 107. Which of the following is not vapour power cycle? (a) Rankine cycle (c) Otto cycle (b) Carnot vapour power cycle (d) None of these 108. The amount of energy added by heat transfer to the cycle to produce unit of net work output is called (a) Heat rate (c) Back work ratio (b) Work ratio (d) Thermal efficiency 109. Difference between temperature of feed water outlet and saturation temperature of steam entering the heater is called (a) Pinch point (c) LMTD (b) Terminal temperature difference (d) None of these 110. Deaerator is feed water heater of which type (a) Closed type feed water heater (c) Open type feed water heater (b) Surface type feed water heater (d) None of these 111. Effectiveness by second law of thermodynamics can be given by (a) Availability output Availability in (c) Availability used Availability supplied 112. Which of the following is incorrect? (b) thermal thermal, rev (d) All of these (a) Availability follows the law of conservation (b) Availability is function of states of matter under consideration and environment (c) Availability always depends upon pressure (d) Availability increases with temperature drop at low temperatures 113. For a reversible isothermal and isochoric process the Helmholtz free energy, (a) Increases (c) Remains constant (b) Decreases (d) Nothing can be said Multiple Answer Type Questions __________________________________________________ 903 114. Clausius-Clapeyron equation given by is valid for = hfg sat d T RT 2 d p p sat (a) Low pressure phase transformation of liquid-vapour type (b) High pressure phase transformation of liquid-vapour type (c) High pressure phase transformation of solid-vapour type (d) None of these 115. Expression given by refers to v T P 1 v (a) Isothermal compressibility (c) Isothermal bulk modulus (b) Coefficient of volume expansion (d) None of these 116. Mayer\u2019s relation is given by; (Here and are volume expansivity and isothermal compressibility). \u22c5 v \u22c5 T \u03b2 2 (a) Cp \u2013 Cv = \u03b1 (c) Cp \u2013 Cv = v\u00b7T\u00b7 2 (b) Cp \u2013 Cv = \u22c5v T \u03b1 (d) None of these 117. Fuel-air mixture will be called lean when the equivalence ratio has its value (c) 1 (d) None of these (b) > 1 (a) < 1 118. Which of the following is/are type of coal? (a) Peat (c) Bituminous coal (b) Lignite (d) All of these 119. Natural gas has highest fraction of which of the following in it, (a) CH4 (b) C2H4 (c) C2H6 (d) CO 120. Minimum air required per kg of fuel for complete combustion can be given by, (considering mass fraction being C, H, O and S of respective constituent), (a) (c) 1 23 1 23 b) 1 21 3 C 8 H O 8 S S (d) None of these 121. Equipment used for carrying out volum", "etric analysis of dry products of combustion is called, (a) Bomb calorimeter (c) Gas calorimeter (b) Orsat analyzer (d) None of these 122. Lancashire boiler has how many number of fire tubes in it? (a) Three (b) Four (c) Two (d) One 123. Cornish boiler has how many flue tubes in it? (a) One (b) Two 124. Which of the following is/are super critical boiler? (c) Three (d) None (a) Cochran boiler (c) Benson boiler (b) Locomotive boiler (d) Lancashire boiler 125. Which of the following is/are fire tube boiler(s)? (a) Velox boiler (c) Locomotive boilers (b) Nestler boiler (d) All of these 904 ________________________________________________________ Applied Thermodynamics 126. Water level indicator has how many vertical tubes? (b) Two (c) Three (a) One (d) Four 127. Fusible plug has plug made of copper and the body is made of (b) Aluminium (c) Iron (a) Gun metal (d) Bronze 128. Feed pump in boilers may be of (a) reciprocating pump type (c) injector type (b) centrifugal pump type (d) All of these 129. The maximum discharge through a natural draught chimney can be given as function of m (mass of air supplied per kg of fuel), Ta (atmospheric temperature) and Tg (hot gas temperature). T a 1 (a) Tg = (c) Tg = a) Pinduced = Pforced (c) Pinduced > Pforced (b) Tg = 2 T m a 1 m (d) None of these (b) Pinduced < Pforced (d) None of these 130. The power requirement in forced draught and induced draught shall be related as, 131. State which of the following is/are incorrect? (a) Combustion is better in forced draught (b) Fan size of induced draught is larger than forced draught fan (c) Maintenance in F.D. fan is easy as compared to ID fan (d) None of these 132. Steam jet draught is not of following type (a) Natural draught (c) Forced draught (b) Induced draught (d) Artificial draught 133. Boiler efficiency is given by the the ratio of heat used in steam generation and (a) Heat recovered using artificial draught (b) Heat lost due to use of natural draught (c) Heat available due to fuel burning (d) None of these 134. Mean temperature of heat addition gets increased and shows increase in cycle thermal efficiency. This cycle is called, (a) Regenerative cycle (c) Carnot cycle (b) Reheat cycle (d) None of these 135. At a given compression ratio the Otto cycle efficiency for the working fluids having = 1.2, 136. 1.4 and 1.67 can be related as, (a) = 1.2 > = 1.4 > = 1.67 (c) = 1.2 = = 1.4 = = 1.67 Intercooling in between the compression process of gas turbine cycle causes (a) Reduction in expansion work (c) Increase in expansion work (b) Increase in compression work (d) Reduction in compression work (b) = 1.2 < = 1.4 < = 1.67 (d) None of these 137. Stagnation temperature can be related to the static temperature and dynamic temperature as under (a) Stagnation temperature = Static temperature \u2013 Dynamic temperature (b) Stagnation temperature = Static temperature + Dynamic temperature (c) Stagnation temperature = Dynamic temperature \u2013 Static temperature (d) None of these Multiple Answer Type Questions __________________________________________________ 905 138. In case of multistage compressor inspite of constant stage efficiency the isentropic compression efficiency goes on reducing with increasing pressure ratio due to (a) Preheating effect (b) Increase in temperature due to friction (c) Friction in a stage results in more work requirement in next stage (d) All of above 139. Modified index of compression (n) accounting for non-isentropic compression can be given n (a) as function of compressor\u2019s polytropic efficiency, 1 poly,b) (c) poly,c n n (d) None of these poly,c as = 1 poly,c 140. Modified index of expansion (n), accounting for non-isentropic expansion can be given as function of turbine\u2019s polytropic efficiency, (a) (c poly,T 1 poy,T poly,T. 1 n n (b) = poly,T 1 (d) None of these 141. Which of the following cycles may", " have highest efficiency under ideal conditions? 142. (a) Ericsson cycle (c) Carnot cycle In a steam engine eccentric the ratio of \u2018throw of eccentric\u2019 and \u2018radius of eccentric\u2019 is equal to (a) 1 (b) Stirling cycle (d) All of these (d) None of these (c) 0.5 (b) 2 143. Distance by which valve uncovers the steam port when piston is at dead centre is called (a) Lead (c) Clearance (b) Stroke (d) None of these 144. Expansion in steam engine cylinder is following type of thermodynamic process (a) Adiabatic process (c) Isochoric process (b) Isobaric process (d) Hyperbolic process 145. Mean effective pressure for steam engine without clearance volume can be given as function of inlet pressure p1, back pressure pb, expansion ratio \u2018r\u2019 as under: (a) (c) ln r ln r 1 1 p r 1 1 p r \u2013 pb \u2013 pb (b) ln r 1 1 p r + pb (d) None of these 146. Actual indicator diagram differs from the hypothetical indicator diagram in steam engine due to (a) Pressure drop because of friction (b) Non instantaneous operation of inlet and exit valves (c) Expansion not occurring hyperbolically in actual engine (d) All of these 906 ________________________________________________________ Applied Thermodynamics 147. Diagram factor in steam engines has its value (a) Less than unity (c) Equal to unity (b) Greater than unity (d) None of these 148. Condensation can be prevented in steam engines by (a) Supplying superheated steam (b) Providing steam jacket around cylinder wall (c) Increasing speed of engine for certain output (d) All of these If all the steam states during expansion are found to be dry and saturated then the expansion curve is called (a) Saturation curve (c) Missing curve (b) Willan\u2019s line (d) None of these 149. 150. A special kind of valve employed for cut-off governing of steam engine is called (a) D-slide valve (c) Meyer\u2019s expansion valve (b) Piston valve (d) None of these 151. Compounding of steam engine offers the advantages of (a) Possibility of increased pressure and temperature of steam at inlet (b) Reduced thermal stresses (c) Reduced condensation losses (d) All of these 152. Two pistons of HP and LP cylinders are mounted in a compound steam engine with zero phase difference. This type of engine is called (a) Tandem compound engine (c) Receiver compound engine (b) Woolf compound engine (d) None of these 153. The mean effective pressure of HP referred to LP cylinder can be given by (a) (c) mep HP V HP mep LP V LP V LP V HP mep (b) V HP HP V LP (d) None of these 154. For equal power developed in each cylinder of compound steam engine having LP and HP cylinders it can be said that (a) mepLP = mepHP referred to LP (c) mepLP VHP = mepHP VLP (b) mepLP mepHP = 1 (d) None of these 155. The variation of cross-sectional area (A), velocity of gas flowing (C), and Mach number (M) during flow through a variable cross section area duct can be related as, (a) dA A = (c) d C C = d C C dA A M1 2 (b) dA A = d C C M 2 1 (M2 \u2013 1) (d) None of these 156. For a subsonic flow the increase in velocity from inlet to exit may be obtained from a duct of, (a) Diverging cross-sectional area type (b) Diverging-converging cross-sectional area type (c) Converging cross sectional area type (d) None of these Multiple Answer Type Questions __________________________________________________ 907 157. For supersonic flow the increase in velocity from inlet to exit may be obtained from a duct of, (a) Diverging cross-sectional area type (b) Converging-diverging cross-sectional area type (c) Converging cross-sectional area type (d) None of these 158. The critical pressure ratio for maximum discharge through nozzle can be given asa) (cb) (d) 159. The critical pressure ratio for air flowing through nozzle shall be, (d) none of these (b) 0.545 (c) 0.577 (a) 0.528 160. The maximum discharge per unit area at throat for critical conditions for perfect gas flowing through nozzle can be given by, (a) p1 v 1 (", "c) p1 b) 1 1 2 1 p1 v 1 (d) None of these 161. When the back pressure of nozzle is below the designed value of pressure at exit of nozzle then the nozzle is called, (a) Inder-expanding nozzle (c) Designed operation nozzle (b) Over-expanding nozzle (d) None of these 162. When the back pressure of a nozzle is equal to critical pressure then such operating state of nozzle is called, (a) Under-expanding nozzle (c) Choked operation of nozzle If the velocity at inlet to nozzle is C1 and the velocity at exit of nozzle during isentropic expansion and non isentropic expansion are C2 and C 2 respectively, then the nozzle efficiency shall be; (b) Over-expanding nozzle (d) None of these 163. (ab) 2 C'2 2 C 2 (cd 164. Phenomenon of supersaturation, when occurring in nozzle causes, (a) Increase in discharge (c) Increase in dryness fraction and entropy (d) All of these (b) Reduced velocity at exit 165. The ratio of saturation pressure corresponding to the temperatures of steam states in stable equilibrium and metastable equilibrium during steam flow through nozzle is called (a) Degree of supersaturation (c) Critical pressure ratio (b) Degree of undercooling (d) None of these 908 ________________________________________________________ Applied Thermodynamics 166. The difference between the saturation temperature corresponding to the steam states in 167. 168. (b) Degree of undercooling (d) None of these stable equilibrium and metastable equilibrium during steam flow through nozzle is called (a) Degree of supersaturation (c) Critical temperature difference In a steam turbine the substantial static pressure drop occurs in stationary nozzle and rotor blade passage, the type of turbine may be (a) Impulse turbine (c) Branca\u2019s turbine In a steam turbine if the leakage along the mating surface of shaft and casing is reduced by using the principle of steam throttling then the type of sealing arrangement may be (a) Carbon rings (c) Labyrinth glands (b) Reaction turbine (d) None of these (b) Packings (d) None of these 169. Radial flow turbines are also called as (a) Rateau turbine (c) Curtis turbine (b) Ljungstrom turbine (d) None of these 170. A steam turbine rejects steam at a pressure much above the atmospheric pressure and this steam is being used for some purposes such as heating etc. This could be: (a) Back pressure turbine (c) Mixed pressure turbine (b) Passout turbine (d) None of these 171. Which of the following is not a part of impulse steam turbine? (a) Nozzle (c) Aerofoil type blades (b) Symmetrical moving blades ring (d) None of these 172. A steam turbine stage has nozzle followed by moving blade ring followed by fixed blade ring and lastly moving blade ring. This could be (a) Rateau turbine (b) Curtis turbine (c) Pressure compounded impulse turbine (d) None of these 173. The ratio of blade velocity to steam velocity remains constant in which type of compounding of impulse steam turbines (a) Pressure compounding (c) Pressure-Velocity compounding (b) Velocity compounding (d) None of these 174. The efficiency of a compound impulse turbine goes on decreasing with increase in the number of stages. This could be, (a) Pressure compounding (c) Pressure-Velocity compounding In case of an impulse turbine the blade velocity coefficient shall have its value for actual case as, (a) < 1 (b) Velocity compounding (d) None of these (d) None of these (b) > 1 (c) = 1 175. 176. The component of velocity which is responsible for maintaining the flow of steam across the moving blade row is called, (a) Relative velocity (c) Tangential component of velocity In an impulse turbine the absolute angle at inlet should be selected depending upon (b) Maintaining flow across the blade row (a) Thrust requirement (d) None of these (c) Both (a) and (b) (b) Absolute velocity (d) Axial component of velocity 177. Multiple Answer Type Questions __________________________________________________ 909 178. The ratio of work done to the energy supplied to rotor in a turbine stage is called, (a) Blading efficiency (c) Nozzle efficiency (b) Stage efficiency (d) None of these 179. The ratio of rate of work done and energy supplied to the stage in a turbine is called (a) Blading efficiency (c) Nozzle efficiency (b) Stage efficiency (d) None of these 180. Stage efficiency, nozzle efficiency and diagram efficiency can be related as, (a) (c) stage = diagram stage nozzle = nozzle diagram (b) stage = diagram nozzle (d) None of", " these 181. The diagram efficiency is highest for a simple impulse turbine stage having smooth and symmetrical blade when blade\u2014steam speed ratio can be given as, (a) cos 1 (b) cos 4 1 (c) cos 2 1 (d) none of these 182. Steam turbines having the complete nozzle coverage of blades are called, (a) Full admission turbine (c) Reaction turbines (b) Partial admission turbine (d) None of these 183. The optimum value of blade to steam speed ratio can be generalised for \u2018n\u2019 number of moving blade rows in impulse turbine as, (a) cos 2n 1 (b) cos 2 n 1 (c) cos 1 (d) None of these 184. If the degree of reaction is 0.75 then the enthalpy changes occurring in rotor and stator blades in a stage can be related as, (a) hstator = (c) hrotor = hrotor hstator 1 3 1 3 (b) 3 hstator = 4hrotor (d) None of these 185. Which of the following may be reaction turbine? (a) Curtis turbine (c) Parson\u2019s turbine (b) Rateau turbine (d) None of these 186. Reheat factor for steam turbine stages has its value, (a) > 1 (b) < 1 (c) = 1 (d) None of these 187. The ratio of sum of actual output of stages and the available energy in turbine is called (a) Reheat factor (c) Internal efficiency (b) Stage efficiency (d) None of these 188. Use of condenser in a steam power plant offers advantages such as (a) It permits the recirculation of condensate (b) It allows for lower back pressure (c) It increases the turbine output (d) All of these 189. Condenser generally operates at (a) Sub-atmospheric pressure (c) Above atmospheric pressure (b) Atmospheric pressure (d) None of these 910 ________________________________________________________ Applied Thermodynamics 190. Which of the following can not be achieved by using condenser? (a) Removal of air and other non-condensable dissolved gases (b) Reduced load on water treatment plant (c) Availability of hotter feed water (d) None of these 191. The condenser in which there is direct contact between the steam and cooling fluid could be (a) Surface condenser (c) Evaporative condenser (b) Jet condenser (d) None of these 192. A condenser is placed at low level such that the vacuum inside condenser draws cooling water into condenser from river. Such condenser may be (a) Low level jet condenser (c) Surface condenser (b) High level jet condenser (d) None of these 193. A condenser is placed at a height more than that of water and water is to be injected into condenser using a pump and condensate flows out of condenser due to gravity. This condenser may be, (a) Low level jet condenser (c) Surface condenser (b) High level condenser (d) None of these 194. Ejector condenser does not require air pump because of (a) No air present (b) Air entraining effect of water jet itself (c) Air being separated and released separately (d) None of these In ejector condenser in the event of failure of cooling water supply water may be sucked from hot well to go into steam pipe. But this is prevented due to (a) Non return valve in steam line (c) Low pressure (b) High pressure (d) None of these 195. 196. While designing the condenser it is desired to have, (a) Maximum heat transfer between two fluids (b) Effective air extraction (c) No air leakage (d) All of these 197. A surface condenser should preferably have, (a) Uniform distribution of steam over cooling water tubes (b) Minimum pressure loss of steam (c) Minimum number of tubes (d) All of these 198. Leakage of air into condenser causes, (a) Reduction in work done per kg of steam due to increase in back pressure (b) Increased heat exchange between water and steam (c) Increased efficiency of auxiliary devices (d) None of these 199. Air leakage into condenser could be identified by (a) Drop in thermometer reading, after isolation of condenser from plant (b) Soap bubble test (c) Peppermint oil test (d) All of above Multiple Answer Type Questions __________________________________________________ 911 200. Vacuum efficiency of condenser can be given by, (a) (c) Theoretical vacuum Actual vacuum Condenser pressure Ideal condenser pressure 201. Condenser efficiency is given by, (b) Actual vacuum Theoretical vacuum (d) None of these (a) (b) (c) Actual cooling water temperature rise Maximum", " possible temperature rise Actual vacuum Theoretical vacuum Actual cooling water required Minimum cooling water requirement (d) none of these 202. Effectiveness of cooling tower diminishes due to, (a) Humid weather conditions (c) Poor atomization of hot water (b) Drift losses (d) All of these 203. The performance of cooling tower depends upon, (a) Duration of contact between water and air (b) Area of contact between water and air (c) Relative velocity of air and water flow (d) All of these 204. For delivery pressure more than 85 bar the number of stages desired in compressor may be (a) 1 (b) 2 (c) 3 (d) 4 205. The compression work requirement is minimum in case of the compression following process of, (a) Adiabatic type (c) Isothermal type (b) Isochoric type (d) Hyperbolic type 206. The ratio of volume of free air sucked into cylinder and the swept volume of LP cylinder is called, (a) Volumetric efficiency (c) Compression efficiency (b) Diagram factor (d) None of these 207. Volumetric efficiency referred to free air conditions in case of reciprocating air compressor can be given bya) (cbd) None of these 208. What will happen to the volumetric efficiency with increasing pressure ratio in case of single stage compression in compressors? (a) Decrease (c) Remains unaffected (b) Increase (d) None of these 912 ________________________________________________________ Applied Thermodynamics 209. In order to get the same free air delivery from a compressor how should the size of cylinder be changed with increasing pressure ratio? (a) Increased (d) None of these (c) No change (b) Reduced 210. For the multistage compression with perfect intercooling the work requirement shall be minimum when, (a) Pressure ratio in each stage are equal (b) Pressure ratio in first stage is greater than second stage (c) Pressure ratio in second stage is double of pressure ratio in first stage (d) None of these 211. Which of the following is not a positive displacement type compressor? 212. (a) Roots blower (c) Vane type compressor In vane type compressor the contribution of reversible pressure rise and irrversible pressure rise is generally in the proportion of (a) 3 : 1 (b) Screw type compressor (d) None of these (d) None of these (b) 1 : 3 (c) 1 : 1 213. Slip factor for a centrifugal compressor may be given by the ratio of (a) Actual whirl velocity and ideal whirl velocity (b) Actual tangential velocity and ideal absolute velocity (c) Relative velocity and absolute velocity (d) None of these 214. For the centrifugal compressor having radial flow blades the stagnation pressure may be given by (a) 1 (c) 1 01 Tp 01 1 1 1 1 C T p 01 2 U 2 (b) (d) None of these 215. The index of compression value in case of air compression process of cooled type in centrifugal compressor may (a) Be equal to adiabatic index (c) Be more than from adiabatic index (d) None of these (b) Be even less than adiabatic index 216. What shall happen to surging if the number of diffuser vanes is increased in a centrifugal compressor? (a) Increases (c) Remains unchanged In order to prevent surging due to flow reversal the number of diffuser vanes (nd) and the number of impeller blades (ni) shall be as under in case of centrifugal compressor, (a) nd = ni (b) Decreases (d) None of these (d) None of these (b) ni < nd (c) ni > nd 217. 218. Which of the following occurs due to stalling? (a) Reduced stage efficiency (c) Reduced delivery pressure (b) Increased vibrations (d) All of these 219. Which of the following refers to centrifugal compressor characteristics? (a) For a given pressure ratio the increase in speed shows increase in flow rate and simultaneous reduction in efficiency. (b) For a particular speed of compressor the decrease in mass flow rate is accompanied by increase in pressure ratio. Multiple Answer Type Questions __________________________________________________ 913 (c) At all speeds the compressor operation is limited by surging and choking at the two ends. (d) All of these 220. Which of the following refers to axial flow compressor characteristics? 221. (a) Flow rate changes too much with increase in pressure ratio at design conditions (b) Efficiency decreases with decreasing flow rate (c) Efficiency increases with decreasing pressure ratio (d) All of these Isentropic efficiencies for the centrifugal compressor and the axial flow compressor can be related as, (a) (c) isen, cent > (b) (d", ") None of these isen, cent = isen, cent < isen, axial isen, axial isen, axial 222. Starting torque requirements of centrifugal compressor and axial flow compressor can be related as, (a) Tcentrifugal > Taxial (c) Tcentrifugal = Taxial (b) Tcentrifugal < Taxial (d) None of these 223. Morse test is used for measurement of, (a) Torque (c) Brake power (b) Specific fuel consumption (d) Friction power 224. Which of the following are phases of combustion in CI engine? (a) Ignition delay (c) Controlled combustion (b) Uncontrolled combustion (d) All of these 225. One ton of refrigeration is equivalent to (a) 3.87 kJ/s (c) 56 kcal/min (b) 3 kJ/s (d) None of these 226. For a definite low temperature to be maintained, the COP of refrigerator can be related for cold and hot days as, (a) COPhot days > COPcold days (c) COPhot days = COPcold days (b) COPhot days < COPcold days (d) None of these 227. Bell-Coleman cycle based refrigeration system has (a) Isobaric heat addition (c) Isochoric heat addition (b) Isothermal heat release (d) Isothermal heat addition 228. COP of Bell-Coleman cycle and reversed Carnot cycle can be related as (b) COPBell-Coleman < COPCarnot (d) None of these (a) COPBell-Coleman > COPCarnot (c) COPBell-Coleman = COPCarnot 229. For the compression of refrigerant beginning with wet state of refrigerant in compressor, the compression process is called, (a) Dry compression (c) Adiabatic compression (b) Wet compression (d) None of these 230. Throttling of certain gases may be used for getting the refrigeration. The value of Joule- Thomson coefficient for such throttling processes shall be, (a) = 1 (d) = 0 (b) < 0 (c) > 0 231. Energy performance ratio (EPR) and Coefficient of performance (COP) can be related by, (a) EPR = COP + 1 (c) EPR = COP (b) EPR + COP = 1 (d) None of these 914 ________________________________________________________ Applied Thermodynamics 232. The absorption refrigeration cycle employs following devices for raising pressure of refrigerant, (a) Compressor (b) Pump (c) Generator (d) Absorber 233. The coolant used in ammonia-LiBr, water absorption refrigeration system is 234. (a) Ammonia only (c) Ammonia-LiBr In ammonia-water absorption system the separation of ammonia from strong aqua-ammonia solution requires, (a) Heat addition to it (c) No energy interaction (b) Heat extraction from it (d) None of these (b) Water only (d) Ammonia-water 235. Absorber used in ammonia-water absorption system should be kept at low temperature because it, (a) Helps in generating high pressure ammonia (b) Facilitates better mixing of ammonia into water (c) Drives out ammonia quickly (d) None of these 236. The low pressure ammonia leaving expansion valve in ammonia-water absorption system is (a) Pure vapour (c) Liquid-vapour mixture (b) Pure liquid (d) Pure liquid or liquid-vapour mixture 237. Device used in absorption system for preventing the moisture accompanying refrigerant from entering into the circuit of absorption cycle is called, (a) Rectifier (b) Evaporator (c) Generator (d) Absorber 238. Which of the following are secondary refrigerants? (b) Water (d) All of these (a) Brine (c) Anti-freeze agents 239. For increasing the refrigerating capacity the refrigerant should have small value of specific heat in liquid form and its specific heat value in vapour form should be, (a) Small (c) Nothing can be said (b) Large (d) None of these 240. Which of the following is/are desired properties of refrigerants? (a) High viscosity (c) Large specific volume (b) Low thermal conductivity (d) None of these 241. Humidity ratio can be given in terms of partial pressures of dry air (pa) and water vapour (pv) as, (a) 0.622 (c) 0.622 p v p a p v p v pa (b) 0.622 p v p a (d) None of these 242. Wet bulb depression in case of saturated air shall be", ", w bt d bt (c) wbt (a) 0 (b) (d) dbt 243. The saturation temperature corresponding to the partial pressure of water vapour present in moist air can be given as, (a) Dry bulb temperature (c) Dew point temperature (b) Wet bulb temperature (d) None of these Multiple Answer Type Questions __________________________________________________ 915 244. 245. If air is passed over the cooling coils then this process is termed as (a) Sensible cooling (c) Cooling with dehumidification If in a process the humidity ratio and dry bulb temperature increase then the process may be, (a) Sensible heating (c) Humidification (b) Cooling with humidification (d) None of these (b) Dehumidification (d) None of these 246. What processes could be there in summer air conditioning? (b) Humidification (d) All of these (a) Cooling (c) Dehumidification 247. The generally used comfort conditions required in air conditioned space in India are (a) 30\u00b0C DBT and 60% relative humidity (b) 15\u00b0C DBT and 50% relative humidity (c) 25\u00b0C DBT and 60% relative humidity (d) 25\u00b0C DBT and 50% relative humidity 248. Which of the following is demerit with vapour compression system over air refrigeration system (a) Large initial investment (c) Smaller operating cost (b) Small refrigerant requirement (d) All of these 249. Which of the following is advantage with vapour absorption system over the vapour compression system? (a) No moving component (c) Utilization of waste energy (b) Less work input requirement (d) All of these 250. Which of the following gives demerit with vapour absorption system over vapour compression system? (a) Large size (c) Poor COP at high loads (b) Requirement of heating source (d) All of these 1. (a) 7. (a) 13. (a) 19. (d) 25. (a) 31. (c) 37. (a) 43. (b) 49. (c) 55. (c) 61. (d) 67. (c) 73. (b) 79. (c) 85. (c) 91. (b) 97. (b) 2. (b) 8. (c) 14. (b) 20. (a) 26. (b) 32. (b) 38. (b) 44. (a) 50. (c) 56. (a) 62. (b) 68. (a) 74. (d) 80. (d) 86. (a) 92. (a) 98. (a) ANSWERS 3. (c) 9. (c) 15. (b) 21. (b) 27. (a) 33. (d) 39. (b) 45. (a) 51. (a) 57. (d) 63. (a) 69. (a) 75. (c) 81. (d) 87. (a) 93. (d) 99. (c) 4. (b) 10. (b) 16. (a) 22. (c) 28. (c) 34. (a) 40. (a) 46. (b) 52. (b) 58. (b) 64. (c) 70. (d) 76. (d) 82. (b) 88. (d) 94. (a) 100. (a) 5. (b) 11. (c) 17. (a) 23. (a) 29. (b) 35. (d) 41. (b) 47. (d) 53. (d) 59. (a) 65. (a) 71. (b) 77. (d) 83. (a) 89. (a) 95. (a) 101. (b) 6. (b) 12. (a) 18. (c) 24. (d) 30. (c) 36. (b) 42. (b) 48. (a) 54. (c) 60. (a) 66. (c) 72. (a) 78. (a) 84. (a) 90. (b) 96. (b) 102. (a) 916 ________________________________________________________ Applied Thermodynamics 103. (a) 109. (b) 115. (b) 121. (b) 127. (a) 133. (c) 139. (a) 145. (c) 151. (d) 157. (a) 163. (d) 169. (b) 175. (a) 181. (c) 187. (c) 193. (b) 199. (d", ") 205. (c) 211. (d) 217. (c) 223. (d) 229. (b) 235. (b) 241. (b) 247. (c) 104. (b) 110. (c) 116. (a) 122. (c) 128. (d) 134. (a) 140. (b) 146. (d) 152. (a) 158. (d) 164. (d) 170. (a) 176. (d) 182. (a) 188. (d) 194. (b) 200. (b) 206. (a) 212. (c) 218. (d) 224. (d) 230. (c) 236. (d) 242. (a) 248. (a) 105. (c) 111. (d) 117. (a) 123. (a) 129. (a) 135. (b) 141. (d) 147. (a) 153. (b) 159. (a) 165. (a) 171. (c) 177. (c) 183. (b) 189. (a) 195. (a) 201. (a) 207. (a) 213. (a) 219. (d) 225. (a) 231. (a) 237. (a) 243. (c) 249. (d) 106. (b) 112. (a) 118. (d) 124. (c) 130. (c) 136. (d) 142. (b) 148. (d) 154. (a) 160. (b) 166. (b) 172. (b) 178. (a) 184. (a) 190. (d) 196. (d) 202. (d) 208. (a) 214. (c) 220. (b) 226. (b) 232. (b) 238. (d) 244. (a) 250. (d) 107. (c) 113. (c) 119. (a) 125. (d) 131. (d) 137. (b) 143. (a) 149. (a) 155. (b) 161. (a) 167. (b) 173. (a) 179. (b) 185. (c) 191. (b) 197. (d) 203. (d) 209. (a) 215. (b) 221. (c) 227. (a) 233. (b) 239. (b) 245. (c) 108. (a) 114. (a) 120. (a) 126. (b) 132. (a) 138. (d) 144. (d) 150. (c) 156. (c) 162. (c) 168. (c) 174. (b) 180. (b) 186. (a) 192. (a) 198. (a) 204. (d) 210. (a) 216. (a) 222. (b) 228. (b) 234. (a) 240. (d) 246. (d) APPENDIX Table 1 Ideal-gas specific heats of various common gases at 300 K Gas Air Argon Butane Carbon dioxide Carbon monoxide Ethane Ethylene Helium Hydrogen Methane Neon Nitrogen Octane Oxygen Propane Steam Formula \u2013 Ar C4H10 CO2 CO C2H6 C2H4 He H2 CH4 Ne N2 C8H18 O2 C3H8 H2O Gas constant R kJ/(kg.K) 0.2870 0.2081 0.1433 0.1889 0.2968 0.2765 0.2964 2.0769 4.1240 0.5182 0.4119 0.2968 0.0729 0.2598 0.1885 0.4615 c p kJ/(kg.K) 1.005 0.5203 1.7164 0.846 1.040 1.7662 1.5482 5.1926 14.307 2.2537 1.0299 1.039 1.7113 0.918 1.6794 1.8723 cv kJ/(kg.K) 0.718 0.3122 1.5734 0.657 0.744 1.4897 1.2518 3.1156 10.183 1.7354 0.6179 0.743 1.6385 0.658 1.4909 1.4108 \u03b3 1.400 1.667 1.091 1.289 1.400 1.186 1.237 1.667 1.405 1.299 1.667 1.400 1.044 1.395 1.126 1.327 Table 2 Saturated Steam (temperature) table Specific volume m3/kg Internal", " energy kJ/kg Enthalpy kJ/kg Sat. liquid vapour liquid Evap. vapour liquid Evap. vapour Sat. Sat. Sat. Sat. Sat. vf vg u f u fg u g h f h fg h g Entropy kJ/(kg.K) Sat. Sat. liquid Evap. vapour sg sfg sf Sat. Temp. press. kPa Psat \u00b0C T 0.01 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 0.0 2375.3 2375.3 20.97 2361.3 2382.3 42.00 2347.2 2389.2 62.99 2333.1 2396.1 83.95 2319.0 2402.9 0.01 2501.4 2501.4 0.000 9.1562 9.1562 0.6113 0.001000 206.14 20.98 2489.6 2510.6 0.0761 8.9496 9.0257 0.8721 0.001000 147.12 42.01 2477.7 2519.8 0.1510 8.7498 8.9008 1.2276 0.001000 106.38 62.99 2465.9 2528.9 0.2245 8.5569 8.7814 77.93 1.7051 0.001001 57.79 2.339 0.001002 83.96 2454.1 2538.1 0.2966 8.3706 8.6672 43.36 104.88 2409.8 2309.8 104.89 2442.3 2547.2 0.3674 8.1905 8.5580 3.169 0.001003 32.89 125.78 2290.8 2416.6 125.79 2430.5 2556.3 0.4369 8.0164 8.4533 4.246 0.001004 25.22 146.67 2276.7 2423.4 146.68 2418.6 2565.3 0.5053 7.8478 8.3531 5.628 0.001006 19.52 167.56 2262.6 2430.1 167.57 2406.7 2574.3 0.5725 7.6845 8.2570 7.384 0.001008 9.593 0.001010 15.26 188.44 2248.4 2436.8 188.45 2394.8 2583.2 0.6387 7.5261 8.1648 12.03 209.32 2234.2 2443.5 209.33 2382.7 2592.1 0.7038 7.3725 8.0763 12.349 0.001012 9.568 230.21 2219.9 2450.1 230.23 2370.7 2600.9 0.7679 7.2234 7.9913 15.758 0.001015 7.671 251.11 2205.5 2456.6 251.13 2358.5 2609.6 0.8312 7.0784 7.9096 19.940 0.001017 6.197 272.02 2191.1 2463.1 272.06 2346.2 2618.3 0.8935 6.9375 7.8310 25.03 0.001020 5.042 292.95 2176.6 2469.6 292.98 2333.8 2626.8 0.9549 6.8004 7.7553 31.19 0.001023 4.131 313.90 2162.0 2475.9 313.93 2321.4 2635.3 1.0155 6.6669 7.6824 38.58 0.001026 3.407 334.86 2147.4 2482.2 334.91 2308.8 2643.7 1.0753 6.5369 7.6122 47.39 0.001029 2.828 355.84 2132.6 2488.4 355.90 2296.0 2651.9 1.1343 6.4102 7.5445 57.83 0.001033 2.361 376.85 2117.7 2494.5 376.92 2283.2 2660.1 1.1925 6.2866 7.4791 70.14 0.001036 1.982 397.88 2102.7 2500.6 397.96 2270.2 2668.1 1.2500 6.1659 7.4159 84.55 0.001040 918 ________________________________________________________ Applied Thermodynamics Table 2 Saturated Steam (tem", "perature) table Specific volume m3/kg Internal energy kJ/kg Enthalpy kJ/kg Sat. liquid vapour liquid Evap. vapour liquid Evap. vapour Sat. Sat. Sat. Sat. Sat. vf vg u f u fg u g h f h fg h g Entropy kJ/(kg.K) Sat. Sat. liquid Evap. vapour sg sfg sf Te mp. \u00b0C T Sat. press. kPa Psat 418.94 2087.6 2506.5 419.04 2257.0 2676.1 1.3069 6.0480 7.3549 1.6729 100 0.10135 0.001044 440.02 2072.3 2512.4 440.15 2243.7 2683.8 1.3630 5.9328 7.2958 1.4194 105 0.12082 0.001048 461.14 2057.0 2518.1 461.30 2230.2 2691.5 1.4185 5.8202 7.2387 1.2102 110 0.14327 0.001052 482.30 2041.4 2523.7 482.48 2216.5 2699.0 1.4734 5.7100 7.1833 1.0366 115 0.16906 0.001056 503.50 2025.8 2529.3 503.71 2202.6 2706.3 1.5276 5.6020 7.1296 0.8919 120 0.19853 0.001060 524.74 2009.9 2534.6 524.99 2188.5 2713.5 1.5813 5.4962 7.0775 0.7706 125 0.2321 0.001065 546.02 1993.9 2539.9 546.31 2174.2 2720.5 1.6344 5.3925 7.0269 0.6685 130 0.2701 0.001070 567.35 1977.7 2545.0 567.69 2159.6 2727.3 1.6870 5.2907 6.9777 0.5822 135 0.3130 0.001075 588.74 1961.3 2550.0 589.13 2144.7 2733.9 1.7391 5.1908 6.9299 0.5089 140 0.3613 0.001080 610.18 1944.7 2554.9 610.63 2129.6 2740.3 1.7907 5.0926 6.8833 0.4463 145 0.4154 0.001085 631.68 1927.9 2559.5 632.20 2114.3 2746.5 1.8418 4.9960 6.8379 0.3928 150 0.4758 0.001091 653.24 1910.8 2564.1 653.84 2098.6 2752.4 1.8925 4.9010 6.7935 0.3468 155 0.5431 0.001096 674.87 1893.5 2568.4 675.55 2082.6 2758.1 1.9427 4.8075 6.7502 0.3071 160 0.6178 0.001102 696.56 1876.0 2572.5 697.34 2066.2 2763.5 1.9925 4.7153 6.7078 0.2727 165 0.7005 0.001108 718.33 1858.1 2576.5 719.21 2049.5 2768.7 2.0419 4.6244 6.6663 0.2428 170 0.7917 0.001114 740.17 1840.0 2580.2 741.17 2032.4 2773.6 2.0909 4.5347 6.6256 175 0.8920 0.001121 0.2168 762.09 1821.6 2583.7 763.22 2015.0 2778.2 2.1396 4.4461 6.5857 180 1.0021 0.001127 0.19405 784.10 1802.9 2587.0 785.37 1997.1 2782.4 2.1879 4.3586 6.5465 185 1.1227 0.001134 0.17409 806.19 1783.8 2590.0 807.62 1978.8 2786.4 2.2359 4.2720 6.5079 190 1.2544 0.001141 0.15654 828.", "37 1764.4 2592.8 829.98 1960.0 2790.0 2.2835 4.1863 6.4698 195 1.3978 0.001149 0.14105 850.65 1744.7 2595.3 852.45 1940.7 2793.2 2.3309 4.1014 6.4323 200 1.5538 0.001157 0.12736 873.04 1724.5 2597.5 875.04 1921.0 2796.0 2.3780 4.0172 6.3952 205 1.7230 0.001164 0.11521 895.53 1703.9 2599.5 897.76 1900.7 2798.5 2.4248 3.9337 6.3585 210 1.9062 0.001173 0.10441 918.14 1682.9 2601.1 920.62 1879.9 2800.5 2.4714 3.8507 6.3221 2.104 0.001181 0.09479 215 940.87 1661.5 2602.4 943.62 1858.5 2802.1 2.5178 3.7683 6.2861 2.318 0.001190 0.08619 220 963.73 1639.6 2603.3 966.78 1836.5 2803.3 2.5639 3.6863 6.2503 2.548 0.001199 0.07849 225 2.795 0.001209 0.07158 230 986.74 1617.2 2603.9 990.12 1813.8 2804.0 2.6099 3.6047 6.2146 3.060 0.001219 0.06537 1009.89 1594.2 2604.1 1013.62 1790.5 2804.2 2.6558 3.5233 6.1791 235 3.344 0.001229 0.05976 1033.21 1570.8 2604.0 1037.32 1766.5 2803.8 2.7015 3.4422 6.1437 240 3.648 0.001240 0.05471 1056.71 1546.7 2603.4 1061.23 1741.7 2803.0 2.7472 3.3612 6.1083 245 3.973 0.001251 0.05013 1080.39 1522.0 2602.4 1085.36 1716.2 2801.5 2.7927 3.2802 6.0730 250 4.319 0.001263 0.04598 1104.28 1596.7 2600.9 1109.73 1689.8 2799.5 2.8383 3.1992 6.0375 255 4.688 0.001276 0.04221 1128.39 1470.6 2599.0 1134.37 1662.5 2796.9 2.8838 3.1181 6.0019 260 5.081 0.001289 0.03877 1152.74 1443.9 2596.6 1159.28 1634.4 2793.6 2.9294 3.0368 5.9662 265 5.499 0.001302 0.03564 1177.36 1416.3 2593.7 1184.51 1605.2 2789.7 2.9751 2.9551 5.9301 270 5.942 0.001317 0.03279 1202.25 1387.9 2590.2 1210.07 1574.9 2785.0 3.0208 2.8730 5.8938 275 6.412 0.001332 0.03017 1227.46 1358.7 2586.1 1235.99 1543.6 2779.6 3.0668 2.7903 5.8571 280 6.909 0.001348 0.02777 1253.00 1328.4 2581.4 1262.31 1511.0 2773.3 3.1130 2.7070 5.8199 285 7.436 0.001366 0.02557 1278.92 1297.1 2576.0 1289.07 1477.1 2766.2 3.1594 2.6227 5.7821 290 1305.2 1264.7 2569.9 1316.3 1441.8 2758.1 3.2062 2.5375 5.7437 7.993 0.001384 0.", "02354 295 1332.0 1231.0 2563.0 1344.0 1404.9 2749.0 3.2534 2.4511 5.7045 8.581 0.001404 0.02167 300 1359.3 1195.9 2555.2 1372.4 1366.4 2738.7 3.3010 2.3633 5.6643 9.202 0.001425 0.019948 305 1387.1 1159.4 2546.4 1401.3 1326.0 2727.3 3.3493 2.2737 5.6230 9.856 0.001447 0.018350 310 Appendix _____________________________________________________________________ 919 Table 2 Saturated Steam (temperature) table Specific volume m3/kg Internal energy kJ/kg Enthalpy kJ/kg Entropy kJ/(kg.K) Sat. liquid vapour liquid Evap. vapour liquid Evap. vapour Sat. Sat. Sat. Sat. Sat. vf vg u f u fg u g h f h fg h g Sat. Sat. liquid Evap. vapour sfg sg sf Sat. Temp. press. \u00b0C MPa P sat T 315 10.547 0.001472 0.016867 320 11.274 0.001499 0.015488 330 12.845 0.001561 0.012996 340 14.586 0.001638 0.010797 350 16.513 0.001740 0.008813 360 18.651 0.001893 0.006945 370 21.03 0.002213 0.004925 374.14 22.09 0.003155 0.003155 1415.5 1121.1 2536.6 1431.0 1283.5 2714.5 3.3982 2.1821 5.5804 1444.6 1080.9 2525.5 1461.5 1238.6 2700.1 3.4480 2.0882 5.5362 993.7 2498.9 1525.3 1140.6 2665.9 3.5507 1.8909 5.4417 1505.3 894.3 2464.6 1594.2 1027.9 2622.0 3.6594 1.6763 5.3357 1570.3 893.4 2563.9 3.7777 1.4335 5.2112 776.6 2418.4 1670.6 1641.9 720.3 2481.0 3.9147 1.1379 5.0526 626.3 2351.5 1760.5 1725.2 441.6 2332.1 4.1106 0.6865 4.7971 384.5 2228.5 1890.5 1844.0 0 4.4298 0 2029.6 2099.3 2029.6 0 2099.3 4.4298 Sat. Press. Temp. kPa P \u00b0C Tsat Specific volume m3/kg Sat. liquid v f Sat. vapour v g Table 3 Saturated steam (pressure) table Internal energy kJ/kg Enthalpy kJ/kg Entropy kJ/(kg.K) Sat. Sat. liquid Evap. vapour u fg u g u f Sat. Sat. liquid Evap. vapour h fg h g h f Sat. Sat. liquid Evap. vapour sfg sg sf 0.6113 1.0 1.5 2.0 2.5 3.0 4.0 5.0 7.5 10 15 20 25 30 40 50 75 0.01 0.001000 6.98 0.001000 13.03 0.001001 17.50 0.001001 21.08 0.001002 24.08 0.001003 28.96 0.001004 32.88 0.001005 40.29 0.001008 45.81 0.001010 53.97 0.001014 60.06 0.001017 64.97 0.001020 69.10 0.001022 75.87 0.001027 81.33 0.001030 91.78 0.001037 206.14 129.21 87.98 67.00 54.25 45.67 34.80 28.19 19.24 14.67 10.02 7.649 6.204 5.229 3.993 3.240 2.217 0.01 2501.3 2501.4 0.0000 9.1562 9.1562 0.00 2375.3 2375.3 29.30 2484.9 2514", ".2 0.1059 8.8697 8.9756 29.30 2355.7 2385.0 54.71 2470.6 2525.3 0.1957 8.6322 8.8279 54.71 2338.6 2393.3 73.48 2460.0 2533.5 0.2607 8.4629 8.7237 73.48 2326.0 2399.5 88.48 2315.9 2404.4 88.49 2451.6 2540.0 0.3120 8.3311 8.6432 101.04 2307.5 2408.5 101.05 2444.5 2545.5 0.3545 8.2231 8.5776 121.45 2293.7 2415.2 121.46 2432.9 2554.4 0.4226 8.0520 8.4746 137.81 2282.7 2420.5 137.82 2423.7 2561.5 0.4764 7.9187 8.3951 168.78 2261.7 2430.5 168.79 2406.0 2576.8 0.5764 7.6750 8.2515 191.82 2246.1 2437.9 191.83 2392.8 2584.7 0.6493 7.5009 8.1502 225.92 2222.8 2448.7 225.94 2373.1 2599.1 0.7549 7.2536 8.0085 251.38 2205.4 2456.7 251.40 2358.3 2609.7 0.8320 7.0766 7.9085 271.90 2191.2 2463.1 271.93 2346.3 2618.2 0.8931 6.9383 7.8314 289.20 2179.2 2468.4 289.23 2336.1 2625.3 0.9439 6.8247 7.7686 317.53 2159.5 2477.0 317.58 2319.2 2636.8 1.0259 6.6441 7.6700 340.44 2143.4 2483.9 340.49 2305.4 2645.9 1.0910 6.5029 7.5939 384.31 2112.4 2496.7 384.39 2278.6 2663.0 1.2130 6.2434 7.4564 920 ________________________________________________________ Applied Thermodynamics Sat. Press. Temp. MPa P \u00b0C Tsat Specific volume m3/kg Sat. liquid v f Sat. vapour vg Table 3 Saturated steam (pressure) table Internal energy kJ/kg Enthalpy kJ/kg Entropy kJ/(kg.K) Sat. Sat. liquid Evap. vapour u fg u g u f Sat. Sat. liquid Evap. vapour h fg h g h f Sat. Sat. liquid Evap. vapour sfg sg sf 417.36 2088.7 2506.1 417.46 2258.0 2675.5 1.3026 6.0568 7.3594 1.6940 0.100 99.63 0.001043 444.19 2069.3 2513.5 444.32 2241.0 2685.4 1.3740 5.9104 7.2844 1.3749 0.125 105.99 0.001048 466.94 2052.7 2519.7 467.11 2226.5 2693.6 1.4336 5.7897 7.2233 1.1593 0.150 111.37 0.001053 486.80 2038.1 2524.9 486.99 2213.6 2700.6 1.4849 5.6868 7.1717 1.0036 0.175 116.06 0.001057 504.49 2025.0 2529.5 504.70 2201.9 2706.7 1.5301 5.5970 7.1271 0.8857 0.200 120.23 0.001061 520.47 2013.1 2533.6 520.72 2191.3 2712.1 1.5706 5.5173 7.0878 0.7933 0.225 124.00 0.001064 535.10 2002.1 2537.2 535.37 2181.5 2716.9 1.6072 5.4455 7.0527 0.7187 0.250 127.44 0.001067 548.59 1991.9 2540.5 548.89 2172.3 2721.3 1.6408 5.", "3801 7.0209 0.6573 0.275 130.60 0.001070 561.15 1982.4 2543.6 561.47 2163.8 2725.3 1.6718 5.3201 6.9919 0.6058 0.300 133.55 0.001073 572.90 1973.5 2546.4 573.25 2155.8 2729.0 1.7006 5.2646 6.9652 0.5620 0.325 136.30 0.001076 583.95 1965.0 2548.9 584.33 2148.1 2732.4 1.7275 5.2130 6.9405 0.5243 0.350 138.88 0.001079 594.40 1956.9 2551.3 594.81 2140.8 2735.6 1.7528 5.1647 6.9175 0.4914 0.375 141.32 0.001081 604.31 1949.3 2553.6 604.74 2133.8 2738.6 1.7766 5.1193 6.8959 0.4625 0.40 143.63 0.001084 622.77 1934.9 2557.6 623.25 2120.7 2743.9 1.8207 5.0359 6.8565 0.4140 0.45 147.93 0.001088 639.68 1921.6 2561.2 640.23 2108.5 2748.7 1.8607 4.9606 6.8213 0.3749 0.50 151.86 0.001093 655.32 1909.2 2564.5 665.93 2097.0 2753.0 1.8973 4.8920 6.7893 0.3427 0.55 155.48 0.001097 669.90 1897.5 2567.4 670.56 2086.3 2756.8 1.9312 4.8288 6.7600 0.3157 0.60 158.85 0.001101 683.56 1886.5 2570.1 684.28 2076.0 2760.3 1.9627 4.7703 6.7331 0.2927 0.65 162.01 0.001104 696.44 1876.1 2572.5 697.22 2066.3 2763.5 1.9922 4.7158 6.7080 0.2729 0.70 164.97 0.001008 708.64 1866.1 2574.7 709.47 2057.0 2766.4 2.0200 4.6647 6.6847 0.2556 0.75 167.78 0.001112 720.22 1856.6 2576.8 721.11 2048.0 2769.1 2.0462 4.6166 6.6628 0.2404 0.80 170.43 0.001115 731.27 1847.4 2578.7 732.22 2039.4 2771.6 2.0710 4.5711 6.6421 0.2270 0.85 172.96 0.001118 741.83 1838.6 2580.5 742.83 2031.0 2773.9 2.0946 4.5280 6.6226 0.2150 0.90 175.38 0.001121 751.95 1830.2 2582.1 753.02 2023.1 2776.1 2.1172 4.4869 6.6041 0.95 177.69 0.001124 0.2042 761.68 1822.0 2583.6 762.81 2015.3 2778.1 2.1387 4.4478 6.5865 1.00 179.91 0.001127 0.19444 780.09 1806.3 2586.4 781.34 2000.4 2781.7 2.1792 4.3744 6.5536 017753 1.10 184.09 0.001133 797.29 1791.5 2588.8 798.65 1986.2 2784.8 2.2166 4.3067 6.5233 1.20 187.99 0.001139 0.16333 813.44 1777.5 2591.0 814.93 1972.7 2787.6 2.2515 4.2438 6.4953 1.30 191.64 0.001144 0.15125 828.70 1764.1 2592.8 830.30 1959.", "7 2790.0 2.2842 4.1850 6.4693 1.40 195.07 0.001149 0.14084 843.16 1751.3 2594.5 844.89 1947.3 2792.2 2.3150 4.1298 6.4448 1.50 198.32 0.001154 0.13177 876.46 1721.4 2597.8 878.50 1917.9 2796.4 2.3851 4.0044 6.3896 1.75 205.76 0.001166 0.11349 906.44 1693.8 2600.3 908.79 1890.7 2799.5 2.4474 3.8935 6.3409 2.00 212.42 0.001177 0.09963 933.83 1668.2 2602.0 936.49 1865.2 2801.7 2.5035 3.7937 6.2972 2.25 218.45 0.001187 0.08875 2.5 223.99 0.001197 0.07998 959.11 1644.0 2603.1 962.11 1841.0 2803.1 2.5547 3.7028 6.2575 3.0 233.90 0.001217 0.06668 1004.78 1599.3 2604.1 1008.42 1795.7 2804.2 2.6457 3.5412 6.1869 3.5 242.60 0.001235 0.05707 1045.43 1558.3 2603.7 1049.75 1753.7 2803.4 2.7253 3.4000 6.1253 4 250.40 0.001252 0.04978 1082.31 1520.0 2602.3 1087.31 1741.1 2801.4 2.7964 3.2737 6.0701 5 263.99 0.001286 0.03944 1147.81 1449.3 2597.1 1154.23 1640.1 2794.3 2.9202 3.0532 5.9734 6 275.64 0.001319 0.03244 1205.44 1384.3 2589.7 1213.35 1571.0 2784.3 3.0267 2.8625 5.8892 7 285.88 0.001351 0.02737 1257.55 1323.0 2580.5 1267.00 1505.1 2772.1 3.1211 2.6922 5.8133 8 295.06 0.001384 0.02352 1305.57 1264.2 2569.8 1316.64 1441.3 2758.0 3.2068 2.5364 5.7432 9 303.40 0.001418 0.02048 1350.51 1207.3 2557.8 1363.26 1378.9 2742.1 3.2858 2.3915 5.6722 10 311.06 0.001452 0.018026 1393.04 1151.4 2544.4 1407.56 1317.1 2724.7 3.3596 2.2544 5.6141 Contd. Appendix _____________________________________________________________________ 921 Sat. Press. Temp. MPa P \u00b0C Tsat Specific volume m3/kg Sat. liquid vapour Sat. v f v g 11 318.15 0.001489 0.015987 12 324.75 0.001527 0.014263 13 330.93 0.001567 0.012780 14 336.75 0.001611 0.011485 15 342.24 0.001658 0.010337 16 347.44 0.001711 0.009306 17 352.37 0.001770 0.008364 18 357.06 0.001840 0.007489 19 361.54 0.001924 0.006657 20 365.81 0.002036 0.005834 21 369.89 0.002207 0.004952 22 373.80 0.002742 0.003568 22.09 374.12 0.003155 0.003155 Table 3 Saturated steam (pressure) table Internal energy kJ/kg Enthalpy kJ/kg Entropy kJ/(kg.K) Sat. Sat. liquid Evap. vapour u fg u g u f Sat. Sat. liquid Evap. vapour h fg h g h f Sat. Sat. liquid Evap. vapour", " sfg sg sf 1433.7 1096.0 2529.8 1450.1 1255.5 2705.6 3.4295 2.1233 5.5527 1473.0 1040.7 2513.7 1491.3 1193.3 2684.9 3.4962 1.9962 5.4924 985.0 2496.1 1531.5 1130.7 2662.2 3.5606 1.8718 5.4323 1511.1 928.2 2476.8 1571.1 1066.5 2637.6 3.6232 1.7485 5.3717 1548.6 869.8 2455.5 1610.5 1000.0 2610.5 3.6848 1.6249 5.3098 1585.6 930.6 2580.6 3.7461 1.4994 5.2455 809.0 2431.7 1650.1 1622.7 856.9 2547.2 3.8079 1.3698 5.1777 744.8 2405.0 1690.3 1660.2 777.1 2509.1 3.8715 1.2329 5.1044 675.4 2374.3 1732.0 1698.9 688.0 2464.5 3.9388 1.0839 5.0228 598.1 2338.1 1776.5 1739.9 583.4 2409.7 4.0139 0.9130 4.9269 507.5 2293.0 1826.3 1785.6 446.2 2334.6 4.1075 0.6938 4.8013 388.5 2230.6 1888.4 1842.1 143.4 2165.6 4.3110 0.2216 4.5327 125.2 2087.1 2022.2 1961.9 0 4.4298 0 2029.6 2099.3 2029.6 0 2099.3 4.4298 Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h kJ/kg s v kJ/(kg\u00b7K) m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u h s kJ/kg kJ/kg kJ/(kg\u00b7K) P = 0.01 MPa (45.81 \u00baC) Sat. 14.674 50 14.869 100 17.196 150 19.512 200 21.825 250 24.136 300 26.445 400 31.063 500 35.679 600 40.295 700 44.911 800 49.526 900 54.141 1000 58.757 1100 63.372 1200 67.987 1300 72.602 2437.9 2483.9 2515.5 2587.9 2661.3 2736.0 2812.1 2968.9 3132.3 3302.5 3479.6 3663.8 3855.0 4053.0 4257.5 4467.9 4683.7 P = 0.10 MPa (99.63 \u00baC) P = 0.05 MPa (81.33 \u00baC) 3.240 2483.9 2645.9 7.5939 1.6940 2506.1 2675.5 7.3594 8.1502 2584.7 8.1749 2592.6 3.418 2511.6 2682.5 7.6947 1.6958 2506.7 2676.2 7.3614 8.4479 2687.5 3.889 2585.6 2780.1 7.9401 1.9364 2582.8 2776.4 7.6134 8.6882 2783.0 2.172 2658.1 2875.3 7.8343 4.356 2659.9 2877.7 8.1580 8.9038 2879.5 2.406 2733.7 2974.3 8.0333 4.820 2735.0 2976.0 8.3556 9.1002 2977.3 2.639 2810.4 3074.3 8.2158 5.284 2811.3 3075.5 8.5373 9.2813 3075.5 3.103 2967.9 3278.2 8.5435 6.209 2968.5 3278.9 8.8642 9.6077 3279.6 3.565 3131.6 3488.1 8.8342 7.134 3132.0 3488", ".7 9.1546 3489.1 9.8978 4.028 3301.9 3704.4 9.0976 8.057 3302.2 3705.1 9.4178 3705.4 10.1608 4.490 3479.2 3928.2 9.3398 8.981 3479.4 3928.5 9.6599 3928.7 10.4028 4.952 3663.5 4158.6 9.5652 4159.0 10.6281 9.904 3663.6 4158.0 9.8852 5.414 3854.8 4396.1 9.7767 4396.4 10.8396 10.828 3854.9 4396.3 10.0967 5.875 4052.8 4640.3 9.9764 4640.6 11.0393 11.751 4052.9 4640.5 10.2964 6.337 4257.3 4891.0 10.1659 4891.2 11.2287 12.674 4257.4 4891.1 10.4859 6.799 4467.7 5147.6 10.3463 5147.8 11.4091 13.597 4467.8 5147.7 10.6662 7.260 4683.5 5409.5 10.5183 5409.7 11.5811 14.521 4683.6 5409.6 10.8382 922 ________________________________________________________ Applied Thermodynamics Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h kJ/kg s v kJ/(kg\u00b7K) m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u kJ/kg h s kJ/kg kJ/(kg\u00b7K) P = 0.20 MPa (120.23 \u00baC) Sat. 0.8857 150 0.9596 200 1.0803 250 1.1988 300 1.3162 400 1.5493 500 1.7814 2.013 600 2.244 700 2.475 800 2.705 900 2.937 1000 3.168 1100 3.399 1200 3.630 1300 2529.5 2576.9 2654.4 2731.2 2808.6 2966.7 3130.8 3301.4 3478.8 3663.1 3854.5 4052.5 4257.0 4467.5 4683.2 P = 0.30 MPa (133.55 \u00baC) P = 0.40 MPa (143.63 \u00baC) 7.1272 0.6058 2543.6 2725.3 6.9919 0.4625 2553.6 2738.6 6.8959 2706.7 7.2795 0.6339 2570.8 2761.0 7.0778 0.4708 2564.5 2752.8 6.9299 2768.8 7.5066 0.7163 2650.7 2865.6 7.3115 0.5342 2646.8 2860.5 7.1706 2870.5 7.7086 0.7964 2728.7 2967.6 7.5166 0.5951 2726.1 2964.2 7.3789 2971.0 7.8926 0.8753 2806.7 3069.3 7.7022 0.6548 2804.8 3066.8 7.5662 3071.8 8.2218 1.0315 2965.6 3275.0 8.0330 0.7726 2964.4 3273.4 7.8985 3276.6 8.5133 1.1867 3130.0 3486.0 8.3251 0.8893 3129.2 3484.9 8.1913 3487.1 8.7770 1.3414 3300.8 3703.2 8.5892 1.0055 3300.2 3702.4 8.4558 3704.0 9.0194 1.4957 3478.4 3927.1 8.8319 1.1215 3477.9 3926.5 8.6987 3927.6 9.2449 1.6499 3662.9 4157.8 9.0576 1.2312 3662.4 4157.3 8.9244 4158.2 9.4566 1.8041 3854.2 4395.4 9.2692 1.3529 3853.9 4395.1 9.1362", " 4395.8 9.5663 1.9581 4052.3 4639.7 9.4690 1.4685 4052.0 4639.4 9.3360 4640.0 4890.7 9.8458 2.1121 4256.8 4890.4 9.6585 1.5840 4256.5 4890.2 9.5256 5147.5 10.0262 2.2661 4467.2 5147.1 9.8389 1.6996 4467.0 5146.8 9.7060 5409.3 10.1982 2.4201 4683.0 5409.0 10.0110 1.8151 4682.8 5408.8 9.8780 Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h kJ/kg s v kJ/(kg\u00b7K) m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u kJ/kg h s kJ/kg kJ/(kg\u00b7K) P = 0.50 MPa (151.86 \u00baC) Sat. 0.3749 200 0.4249 250 0.4744 300 0.5226 350 0.5701 400 0.6173 500 0.7109 600 0.8041 700 0.8969 800 0.9896 900 1.0822 1000 1.1747 1100 1.2672 1200 1.3596 1300 1.4521 2561.2 2642.9 2723.5 2802.9 2882.6 2963.2 3128.4 3299.6 3477.5 3662.1 3853.6 4051.8 4256.3 4466.8 4682.5 2748.7 2855.4 2960.7 3064.2 3167.7 327l.9 3483.9 3701.7 3925.9 4156.9 4394.7 4639.1 4889.9 5146.6 5408.6 P = 0.60 MPa (158.85 \u00baC) P = 0.80 MPa (170.43 \u00baC) 6.8213 0.3157 2567.4 2756.8 6.7600 0.2404 2576.8 2769.1 6.6628 7.0592 0.3520 2638.9 2850.1 6.9665 0.2608 2630.6 2839.3 6.8158 7.2709 0.3938 2720.9 2957.2 7.1816 0.2931 2715.5 2950.0 7.0384 7.4599 0.4344 2801.0 3061.6 7.3724 0.3241 2797.2 3056.5 7.2328 7.6329 0.4742 2881.2 3165.7 7.5464 0.3544 2878.2 3161.7 7.4089 7.7938 0.5137 2962.1 3270.3 7.7079 0.3843 2959.7 3267.1 7.5716 8.0873 0.5920 3127.6 3482.8 8.0021 0.4433 3126.0 3480.6 7.8673 7.3522 0.6697 3299.1 3700.9 8.2674 0.5018 3297.9 3699.4 8.1333 8.5952 0.7472 3477.0 3925.3 8.5107 0.5601 3476.2 3924.2 8.3770 8.8211 0.8215 3661.8 4156.5 8.7367 0.6181 3661.1 4155.6 8.6033 9.0329 0.9017 3853.4 4394.4 8.9486 0.6761 3852.8 4393.7 8.8153 9.2328 0.9788 4051.5 4638.8 9.1485 0.7340 4051.0 4638.2 9.0153 9.4224 1.0559 4256.1 4889.6 9.3381 0.7919 4255.6 4889.1 9.2050 9.6029 1.1330 4466.5 5146.3 9.5185 0.8497 4466.1 5145.9 9.3855 9.7749 1.2101 4682.3 5408.3 9.6906 0.9076 4681.8 5407.9 9.", "5575 Appendix _____________________________________________________________________ 923 Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h s v u kJ/kg kJ/(kg\u00b7K) m3/kg kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u kJ/kg h s kJ/kg kJ/(kg\u00b7K) P = 1.00 MPa (179.91 \u00baC) P = 1.20 MPa (187.99 \u00baC) Sat. 0.19444 2583.6 2778.1 2621.9 2827.9 200 0.2060 2709.9 2942.6 250 0.2327 2793.2 3051.2 300 0.2579 2875.2 3157.7 350 0.2825 2957.3 3263.9 400 0.3066 3124.4 3478.5 500 0.3541 3296.8 3697.9 600 0.4011 3475.3 3923.1 700 0.4478 3660.4 4154.7 800 0.4943 3852.2 4392.9 900 0.5407 4050.5 4637.6 1000 0.5871 4255.1 4888.6 1100 0.6335 4465.6 5145.4 1200 0.6798 4681.3 5407.4 1300 0.7261 6.5865 0.16333 2588.8 2784.8 6.5233 0.14084 2592.8 2790.0 6.6940 0.16930 2612.8 2815.9 6.5898 0.14302 2603.1 2803.3 6.9247 0.19234 2704.2 2935.0 6.8294 0.16350 2698.3 2927.2 2789.2 3045.8 7.0317 0.18228 2785.2 3040.4 7.1229 0.2138 2869.2 3149.5 2872.2 3153.6 7.2121 0.2003 7.3011 0.2345 2952.5 3257.5 2954.9 3260.7 7.3774 0.2178 7.4651 0.2548 3121.1 3474.1 3122.8 3476.3 7.6759 0.2521 7.7622 0.2946 3294.4 3694.8 3295.6 3696.3 7.9435 0.2860 8.0290 0.3339 3473.6 3920.8 3474.4 3922.0 8.1881 0.3195 8.2731 0.3729 3659.0 4153.0 3659.7 4153.8 8.4148 0.3528 8.4996 0.4118 3851.1 4391.5 3851.6 4392.2 8.6272 0.3861 8.7118 0.4505 4049.5 4636.4 4050.0 4637.0 8.8274 0.4192 8.9119 0.4892 4254.1 4887.5 4254.6 4888.0 9.0172 0.4524 9.1017 0.5278 4464.7 5144.4 4465.1 5144.9 9.1977 0.4855 9.2822 0.5665 4680.4 5406.5 4680.9 5407.0 9.3698 0.5186 9.4543 0.6051 P = 1.40 MPa (195.70 \u00baC) 6.4693 6.4975 6.7467 6.9534 7.1360 7.3026 7.6027 7.8710 8.1160 8.3431 8.5556 8.7559 8.9457 9.1262 9.2984 Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u kJ/kg h s kJ/kg kJ/(kg\u00b7K) P = 1.60 MPa (201.41 \u00baC) Sat 0.12380 2596.0 2794.0 6.4218 225 0.13287 2644.7 2857.3 6.5518 250 0.14184 2692.3 2919.2 6.6732 300 0.15862 2781.", "1 3034.8 6.8844 350 0.17456 2866.1 3145.4 7.0694 400 0.19005 2950.1 3254.2 7.2374 3119.5 3472.0 7.5390 500 0.2203 3293.3 3693.2 7.8080 600 0.2500 3472.7 3919.7 8.0535 700 0.2794 3658.3 4152.1 8.2808 800 0.3086 3850.5 4390.8 8.4935 900 0.3377 4049.0 4635.8 8.6938 1000 0.3668 4253.7 4887.0 8.8837 1100 0.3958 4464.2 5143.9 9.0643 1200 0.4248 4679.9 5406.0 9.2364 1300 0.4538 P = 1.80 MPa (207.15 \u00baC) p = 2.00 MPa (212.42 \u00baC) 0.11042 2598.4 2797.1 6.3794 0.09963 2600.3 2799.5 6.3409 0.11673 2636.6 2846.7 6.4808 0.10377 2628.3 2835.8 6.4147 0.12497 2686.0 2911.0 6.6066 0.11144 2679.6 2902.5 6.5453 0.14021 2776.9 3029.2 6.8226 0.12547 2772.6 3023.5 6.7664 0.15457 2863.0 3141.2 7.0100 0.13857 2859.8 3137.0 6.9563 0.16847 2947.7 3250.9 7.1794 0.15120 2945.2 3247.6 7.1271 0.19550 3117.9 3469.8 7.4825 0.17568 3116.2 3467.6 7.4317 3292.1 3691.7 7.7523 0.19960 3290.9 3690.1 7.7024 0.2220 3470.9 3917.4 7.9487 3471.8 3971.8 7.9983 0.2232 0.2482 3657.0 4150.3 8.1765 3657.6 4151.2 8.2258 0.2467 0.2742 3849.3 4389.4 8.3895 3849.9 4390.1 8.4386 0.2700 0.3001 4048.0 4634.6 8.5901 4048.5 4635.2 8.6391 0.2933 0.3260 4252.7 4885.9 8.7800 4253.2 4886.4 8.8290 0.3166 0.3518 4463.3 5142.9 8.9607 4463.7 5143.4 9.0096 0.3398 0.3776 4679.0 5405.1 9.1329 4679.5 5405.6 9.1818 0.3631 0.4034 924 ________________________________________________________ Applied Thermodynamics Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u kJ/kg h s kJ/kg kJ/(kg\u00b7K) v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) P = 2.50 MPa (223.99 \u00baC) Sat. 0.07998 2603.1 2803.1 225 0.08027 2605.6 2806.3 250 0.08700 2662.6 2880.1 300 0.09890 2761.6 3008.8 350 0.10976 2851.9 3126.3 400 0.12010 2939.1 3239.3 450 0.13014 3025.5 3350.8 500 0.13993 3112.1 3462.1 600 0.15930 3288.0 3686.3 700 0.17832 3468.7 3914.5 800 0.19716 3655.3 4148.2 900 0.21590 3847.9 4387.6 4046.7 4633.1 1000 0.2346 4251.5 4884.6 1100 0.2532 4462.1 5141.7 1200 0.2718 4677.8 5404.0 1300 0.2905 P = 3.00", " MPa (233.90 \u00baC) P = 3.50 MPa (242.60 \u00baC) 6.2575 0.06668 2604.1 2804.2 6.1869 0.05707 2603.7 2803.4 6.1253 6.2639 6.4085 0.07058 2644.0 2855.8 6.2872 0.05872 2623.7 2829.2 6.1749 6.6438 0.08114 2750.1 2993.5 6.5390 0.06842 2738.0 2977.5 6.4461 6.8403 0.09053 2843.7 3115.3 6.7428 0.07678 2835.3 3104.0 6.6579 7.0148 0.09936 2932.8 3230.9 6.9212 0.08453 2926.4 3222.3 6.8405 7.1746 0.10787 3020.4 3344.0 7.0834 0.09196 3015.3 3337.2 7.0052 7.3234 0.11619 3108.0 3456.5 7.2338 0.09918 3103.0 3450.9 7.1572 7.5960 0.13243 3285.0 3682.3 7.5085 0.11324 3282.1 3678.4 7.4339 7.8435 0.14838 3466.5 3911.7 7.7571 0.12699 3464.3 3908.8 7.6837 8.0720 0.16414 3653.5 4145.9 7.9862 0.14056 3651.8 4143.7 7.9134 8.2853 0.17980 3846.5 4385.9 8.1999 0.15402 3845.0 4384.1 8.1276 8.4861 0.19541 4045.4 4631.6 8.4009 0.16743 4044.1 4630.1 8.3288 8.6762 0.21098 4250.3 4883.3 8.5912 0.18080 4249.2 4881.9 8.5192 8.8569 0.22652 4460.9 5140.5 8.7720 0.19415 4459.8 5139.3 8.7000 9.0291 0.24206 4676.6 5402.8 8.9442 0.20749 4675.5 5401.7 8.8723 Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h kJ/kg s v kJ/(kg\u00b7K) m3/kg u kJ/kg h s v u h s kJ/kg kJ/(kg\u00b7K) m3/kg kJ/kg kJ/kg kJ/(kg\u00b7K) P = 4.0 MPa (250.40 \u00baC) P = 4.5 MPa (257.49 \u00baC) P = 5.0 MPa (263.99 \u00baC) Sat. 0.04978 275 0.05457 300 0.05884 350 0.06645 400 0.07341 450 0.08002 500 0.08643 600 0.09885 700 0.11095 800 0.12287 900 0.13469 1000 0.14645 1100 0.15817 1200 0.16987 1300 0.18156 2602.3 2667.9 2725.3 2826.7 2919.9 3010.2 3099.5 3279.1 3462.1 3650.0 3843.6 4042.9 4248.0 4458.6 4674.3 2801.4 2886.2 2960.7 3092.5 3213.6 3330.3 3445.3 3674.4 3905.9 4141.5 4382.3 4628.7 4880.6 5138.1 5400.5 6.0701 0.04406 2600.1 2798.3 6.0198 0.03944 2597.1 2794.3 5.9734 6.2285 0.04730 2650.3 2863.2 6.1401 0.04141 2631.3 2838.3 6.0544 6.3615 0.05135 2712.0 2943.1 6.2828 0.04532 2698.0 2924.5 6.2084 6.5821 0.05840 2817.8 3080.6", " 6.5131 0.05194 2808.7 3068.4 6.4493 6.7690 0.06475 2913.3 3204.7 6.7047 0.05781 2906.6 3195.7 6.6459 6.9363 0.07074 3005.0 3323.3 6.8746 0.06330 2999.7 3316.2 6.8186 7.0901 0.07651 3095.3 3439.6 7.0301 0.06857 3091.0 3433.8 6.9759 7.3688 0.08765 3276.0 3670.5 7.3110 0.07869 3273.0 3666.5 7.2589 7.6198 0.09847 3459.9 3903.0 7.5631 0.08849 3457.6 3900.1 7.5122 3646.6 4137.1 7.7440 4139.3 7.7942 0.09811 7.8502 0.10911 3648.3 3840.7 4378.8 7.9593 4380.6 8.0091 0.10762 8.0647 0.11965 3842.2 4040.4 4625.7 8.1612 4627.2 8.2108 0.11707 8.2662 0.13013 4041.6 4245.6 4878.0 8.3520 4879.3 8.4015 0.12648 8.4567 0.14056 4246.8 4456.3 5135.7 8.5331 5136.9 8.5825 0.13587 8.6376 0.15098 4457.5 4672.0 5398.2 8.7055 5399.4 8.7549 0.14526 8.8100 0.16139 4673.1 Appendix _____________________________________________________________________ 925 Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u kJ/kg h s kJ/kg kJ/(kg\u00b7K) v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) P = 6.0 MPa (275.64 \u00baC) P = 7.0 MPa (285.88 \u00baC) P = 8.0 MPa (295.06 \u00baC) Sat. 0.03244 2589.7 300 0.03616 2667.2 350 0.04223 2789.6 400 0.04739 2892.9 450 0.05214 2988.9 500 0.05665 3082.2 550 0.06101 3174.6 600 0.06525 3266.9 700 0.07352 3453.1 800 0.08160 3643.1 900 0.08958 3837.8 1000 0.09749 4037.8 1100 0.10536 4243.3 1200 0.11321 4454.0 1300 0.12106 4669.6 2784.3 2884.2 3043.0 3177.2 3301.8 3422.2 3540.6 3658.4 3894.2 4132.7 4375.3 4622.7 4875.4 5133.3 5396.0 5.8892 0.02737 2580.5 2772.1 5.8133 0.02352 2569.0 2758.0 5.7432 6.0674 0.02947 2632.2 2838.4 5.9305 0.02426 2590.9 2785.0 5.7906 6.3335 0.03524 2769.4 3016.0 6.2283 0.02995 2747.7 2987.3 6.1301 6.5408 0.03993 2878.6 3158.1 6.4478 0.03432 2863.8 3138.3 6.3634 6.7193 0.04416 2978.0 3287.1 6.6327 0.03817 2966.7 3272.0 6.5551 6.8803 0.04814 3073.4 3410.3 6.7975 0.04175 3064.3 3398.3 6.7240 7.0288 0.05195 3167.2 3530.9 6.9486 0.04516 3159.8 3521.0 6.8778 7.1677 0.05565 3260.7 3650.3 7.08", "94 0.04845 3254.4 3642.0 7.0206 7.4234 0.06283 3448.5 3888.3 7.3476 0.05481 3443.9 3882.4 7.2812 7.6566 0.06981 3639.5 4128.2 7.5822 0.06097 3636.0 4123.8 7.5173 7.8727 0.07669 3835.0 4371.8 7.7991 0.06702 3832.1 4368.3 7.7351 8.0751 0.08350 4035.3 4619.8 8.0020 0.07301 4032.8 4616.9 7.9384 8.2661 0.09027 4240.9 4872.8 8.1933 0.07896 4238.6 4870.3 8.1300 8.4474 0.09703 4451.7 5130.9 8.3747 0.08489 4449.5 5128.5 8.3115 8.6199 0.10377 4667.3 5393.7 8.5475 0.09080 4665.0 5391.5 8.4842 Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) P = 9.0 MPa (303.40 \u00baC) P = 10.0 MPa (311.06 \u00baC) P = 12.5 MPa (327.89 \u00baC) Sat. 0.02048 2557.8 2742.1 5.6772 0.018026 2544.4 2724.7 5.6141 0.013495 2505.1 2673.8 5.4624 325 0.02327 2646.6 2856.0 5.8712 0.019861 2610.4 2809.1 5.7568 350 0.02580 2724.4 2956.6 6.0361 0.02242 400 0.02993 2848.4 3117.8 6.2854 0.02641 450 0.03350 2955.2 3256.6 6.4844 0.02975 500 0.03677 3055.2 3386.1 6.6576 0.03279 550 0.03987 3152.2 3511.0 6.8142 0.03564 600 0.04285 3248.1 3633.7 6.9589 0.03837 650 0.04574 3343.6 3755.3 7.0943 0.04101 700 0.04857 3439.3 3876.5 7.2221 0.04358 800 0.05409 3632.5 4119.3 7.4596 0.04859 900 0.05950 3829.2 4364.8 7.6783 0.05349 1000 0.06485 4030.3 4614.0 7.8821 0.05832 1100 0.07016 4236.3 4867.7 8.0740 0.06312 1200 0.07544 4447.2 5126.2 8.2556 0.06789 1300 0.08072 4662.7 5389.2 8.4284 0.07265 2699.2 2923.4 5.9443 0.016126 2624.6 2826.2 5.7118 2832.4 3096.5 6.2120 0.02000 2789.3 3039.3 6.0417 2943.4 3240.9 6.4190 0.02299 2912.5 3199.8 6.2719 3045.8 3373.7 6.5966 0.02560 3021.7 3341.8 6.4618 3144.1 3500.9 6.7561 0.02801 3125.0 3475.2 6.6290 3241.7 3625.3 6.9029 0.03029 3225.4 3604.0 6.7810 3338.2 3748.2 7.0398 0.03248 3324.4 3730.4 6.9218 3434.7 3870.5 7.1687 0.03460 3422.9 3855.3 7.0536 3628.9 4114.8 7.4077 0.03869 3620", ".0 4103.6 7.2965 3826.3 4361.2 7.6272 0.04267 3819.1 4352.5 7.5182 4027.8 4611.0 7.8315 0.04658 4021.6 4603.8 7.7237 4234.0 4865.1 8.0237 0.05045 4228.2 4858.8 7.9165 4444.9 5123.8 8.2055 0.05430 4439.3 5118.0 8.0937 4460.5 5387.0 8.3783 0.05813 4654.8 5381.4 8.2717 926 ________________________________________________________ Applied Thermodynamics Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u h s kJ/kg kJ/kg kJ/(kg\u00b7K) P = 17.5 MPa (354.75 \u00baC) P = 15.0 MPa (342.24 \u00baC) P = 20.0 MPa (365.81 \u00baC) Sat. 0.010337 2455.5 2610.5 5.3098 0.007920 2390.2 2528.8 5.1419 0.005834 2293.0 2409.7 4.9269 350 0.011470 2520.4 2692.4 5.4421 400 0.015649 2740.7 2975.5 5.8811 0.012447 2685.0 2902.9 5.7213 0.009942 2619.3 2818.1 5.5540 450 0.018445 2879.5 3156.2 6.1404 0.015174 2844.2 3109.7 6.0184 0.012695 2806.2 3060.1 5.9017 2996.6 3308.6 6.3443 0.017358 2970.3 3274.1 6.2383 0.014768 2942.9 3238.2 6.1401 500 0.02080 3104.7 3448.6 6.5199 0.019288 3083.9 3421.4 6.4230 0.016555 3062.4 3393.5 6.3348 550 0.02293 3191.5 3560.1 6.5866 0.018178 3174.0 3537.6 6.5048 3208.6 3582.3 6.6776 0.02106 600 0.02491 3296.0 3693.9 6.7357 0.019693 3281.4 3675.3 6.6582 3310.3 3712.3 6.8224 0.02274 650 0.02680 3386.4 3809.0 6.7993 3324.6 3410.9 3840.1 6.9572 0.02434 700 0.02861 398.7 6.8736 0.02113 3592.7 4069.7 7.0544 3601.8 4081.1 7.1244 0.02385 3610.9 4092.4 7.2040 0.02738 800 0.03210 3797.5 4326.4 7.2830 3804.7 4335.1 7.3507 0.02645 3811.9 4343.8 7.4279 0.03031 900 0.03546 4003.1 4582.5 7.4925 4009.3 4589.5 7.5589 0.02897 4015.4 4596.6 7.6348 0.03316 1000 0.03875 4211.3 4840.2 7.6874 4216.9 4846.4 7.7531 0.03145 4222.6 4852.6 7.8283 0.03597 1100 0.04200 4422.8 5101.0 7.8707 4428.3 5106.6 7.9360 0.03391 4433.8 5112.3 8.0108 0.03876 1200 0.04523 4638.0 5365.1 8.0442 4643.5 5370.5 8.1093 0.03636 4649.1 5376.0 8.1840 0.04154 1300 0.04845 Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h s kJ", "/kg kJ/(kg\u00b7K) v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u h s kJ/kg kJ/kg kJ/(kg\u00b7K) P = 25.0 MPa P = 30.0 MPa P = 35.0 MPa 375 0.0019731 1798.7 1848.0 4.0320 400 0.006004 2430.1 2580.2 5.1418 425 0.007881 2609.2 2806.3 5.4723 450 0.009162 2720.7 2949.7 5.6744 500 0.011123 2884.3 3162.4 5.9592 550 0.012724 3017.5 3335.6 6.1765 600 0.014137 3137.9 3491.4 6.3602 650 0.015433 3251.6 3637.4 6.5229 700 0.016646 3361.3 3777.5 6.6707 800 0.018912 3574.3 4047.1 6.9345 900 0.021045 3783.0 4309.1 7.1680 3990.9 4568.5 7.3802 4200.2 4828.2 7.5765 4412.0 5089.9 7.7605 4626.9 5354.4 7.9342 1000 0.02310 1100 0.02512 1200 0.02711 1300 0.02910 0.0017892 1737.8 1791.5 3.9305 0.0017003 1702.9 1762.4 3.8722 1914.1 1987.6 4.2126 0.002790 2253.4 2373.4 4.7747 0.005303 2498.7 2672.4 5.1962 0.006735 2751.9 2994.4 5.6282 0.008678 2921.0 3213.0 5.9026 0.010168 3062.0 3395.5 6.1179 0.011446 3189.8 3559.9 6.3010 0.012596 3309.8 3713.5 6.4631 0.013661 3536.7 4001.5 6.7450 0.015623 3754.0 4274.9 6.9386 0.017448 3966.7 4541.1 7.2064 0.019196 4178.3 4804.6 7.4037 0.020903 4390.7 5068.3 7.5910 0.022589 4605.1 5333.6 7.7653 0.024266 2067.4 2151.1 4.4728 0.002100 2455.1 2614.2 5.1504 0.003428 2619.3 2821.4 5.4424 0.004961 2820.7 3081.1 5.7905 0.006927 2970.3 3275.4 6.0342 0.008345 3100.5 3443.9 6.2331 0.009527 3221.0 3598.9 6.4058 0.010575 3335.8 3745.6 6.5606 0.011533 3555.5 4024.2 6.8332 0.013278 3768.5 4291.9 7.0718 0.014883 3978.8 4554.7 7.2867 0.016410 4189.2 4816.3 7.4845 0.017895 4401.3 5079.0 7.6692 0.019360 4616.0 5344.0 7.8432 0.020815 Appendix _____________________________________________________________________ 927 Table 4 Superheated steam table T \u00baC v m3/kg u kJ/kg h kJ/kg s kJ/(kg\u00b7K) v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u h s kJ/kg kJ/kg kJ/(kg\u00b7K) P = 60.0 MPa P = 40.0 MPa P = 50.0 MPa 375 0.0016407 1677.1 1742.8 3.8290 0.0015594 1638.6 1716.6 3.7639 0.0015028 1609.4 1699.5 3.7141 400 0.0019077 1854.6 1930.9 4.1135 0.0017309", " 1788.1 1874.6 4.0031 0.0016335 1745.4 1843.4 3.9318 425 0.002532 2096.9 2198.1 4.5029 0.002007 1959.7 2060.0 4.2734 0.0018165 1892.7 2001.7 4.1626 450 0.003693 2365.1 2512.8 4.9459 0.002486 2159.6 2284.0 4.5884 0.002085 2053.9 2179.0 4.4121 500 0.005622 2678.4 2903.3 5.4700 0.003892 2525.5 2720.1 5.1726 0.002956 2390.6 2567.9 4.9321 550 0.006984 2869.7 3149.1 5.7785 0.005118 2763.6 3019.5 5.5485 0.003956 2658.8 2896.2 5.3441 600 0.008094 3022.6 3346.4 6.0144 0.006112 2942.0 3247.6 5.8178 0.004834 2861.1 3151.2 5.6452 650 0.009063 3158.0 3520.6 6.2054 0.006966 3093.5 3441.8 6.0342 0.005595 3028.8 3364.5 5.8829 700 0.009941 3283.6 3681.2 6.3750 0.007727 3230.5 3616.8 6.2189 0.006272 3177.2 3553.5 6.0824 800 0.011523 3517.8 3978.7 6.6662 0.009076 3479.8 3933.6 6.5290 0.007459 3441.5 3889.1 6.4109 900 0.012962 3739.4 4257.9 6.9150 0.010283 3710.3 4224.4 6.7882 0.008508 3681.0 4191.5 6.6805 1000 0.014324 3954.6 4527.6 7.1356 0.011411 3930.5 4501.1 7.0146 0.009480 3906.4 4475.2 6.9121 1100 0.015642 4167.4 4793.1 7.3364 0.012496 4145.7 4770.5 7.2184 0.010409 4124.1 4748.6 7.1195 1200 0.016940 4380.1 5057.7 7.5224 0.013561 4359.1 5037.2 7.4058 0.011317 4338.2 5017.2 7.3083 1300 0.018229 4594.3 5323.5 7.6969 0.014616 4572.8 5303.6 7.5808 0.012215 4551.4 5284.3 7.4837 Table 5 Compressed liquid water table T \u00baC v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u kJ/kg h s kJ/kg kJ/(kg\u00b7K) v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) P = 5 MPa (263.99 \u00baC) P = 10 MPa (311.06 \u00baC) P = 15 MPa (342.24 \u00baC) Sat. 0.0012859 1147.8 1154.2 2.9202 0.0014524 1393.0 1407.6 10.04 0.0002 0.0009928 93.33 0.2945 0.0009950 0.04 5.04 0.0001 0.0009952 83.65 83.65 0.2956 0.0009972 3.3596 0.0016581 l585.6 1610.5 3.6848 0.15 15.05 0.0004 0.09 0 0.0009977 20 0.0009995 83.06 97.99 0.2934 83.36 40 0.0010056 166.95 171.97 0.5705 0.0010034 166.35 176.38 0.5686 0.0010013 165.76 180.78 0.5666 60 0.0010149 250.23 255.30 0.8285 0.0010127", " 249.36 259.49 0.8258 0.0010105 248.51 263.67 0.8232 80 0.0010268 333.72 338.85 1.0720 0.0010245 332.59 342.83 1.0688 0.0010222 331.48 346.81 1.0656 100 0.0010410 417.52 422.72 1.3030 0.0010385 416.12 426.50 1.2992 0.0010361 414.74 430.28 1.2955 120 0.0010576 501.80 507.09 1.5233 0.0010549 500.08 510.64 1.5189 0.0010522 498.40 514.19 1.5145 140 0.0010768 586.76 592.15 1.7343 0.0010737 584.68 595.42 1.7292 0.0010707 582.66 598.72 1.7242 160 0.0010988 672.62 678.12 1.9375 0.0010953 670.13 681.08 1.9317 0.0010918 667.71 684.09 1.9260 180 0.0011240 759.63 765.25 2.1341 0.0011199 756.65 767.84 2.1275 0.0011159 753.76 770.50 2.1210 2.3178 0.0011433 841.0 858.2 2.3104 856.0 200 0.0011530 848.1 853.9 2.3255 0.0011480 844.5 2.5039 0.0011748 929.9 947.5 2.4953 220 0.0011866 938.4 944.4 2.5128 0.0011805 934.1 945.9 2.6872 0.0012114 1020.8 1039.0 2.6771 240 0.0012264 1031.4 1037.5 2.6979 0.0012187 1026.0 1038.1 2.8699 0.0012550 1114.6 1133.4 2.8576 260 0.0012749 1127.9 1134.3 2.8830 0.0012645 1121.1 1133.7 3.0548 0.0013084 1212.5 1232.1 3.0393 0.0013216 1220.9 1234.1 280 3.2469 0.0013770 1316.6 1337.3 3.2260 300 0.0013972 1328.4 1342.3 0.0014724 1431.1 1453.2 3.4247 320 0.0016311 1567.5 1591.9 3.6546 340 928 ________________________________________________________ Applied Thermodynamics Table 5 Compressed liquid water table T \u00baC v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) m3/kg v u kJ/kg h s kJ/kg kJ/(kg\u00b7K) v m3/kg u kJ/kg h s kJ/kg kJ/(kg\u00b7K) P = 20.0 MPa (365.81 \u00baC) P = 30 MPa P = 50 MPa Sat. 0.002036 1785.6 1826.3 4.0139 0.25 29.82 0.0001 0.0009766 82.17 111.84 0.2899 0.0009804 0.19 20.01 0.0004 0.0009856 82.77 102.62 0.2923 0.0009886 0.20 49.03 0.0014 0 0.0009904 20 0.0009928 81.00 130.02 0.2848 40 0.0009992 165.17 185.16 0.5646 0.0009951 164.04 193.89 0.5607 0.0009872 161.86 211.21 0.5527 60 0.0010084 247.68 267.85 0.8206 0.0010042 246.06 276.19 0.8154 0.0009962 242.98 292.79 0.8052 80 0.0010199 330.40 350.80 1.0624 0.0010156 328.30 358.77 1.0561 0.0010073 324.34 374.70 1.0440 100 0.0010337 413.39 434.06 1.2917 0.0010290 410.78 441.66 1.2844 0.0010201", " 405.88 456.89 1.2703 120 0.0010496 496.76 517.76 1.5102 0.0010445 493.59 524.93 1.5018 0.0010348 487.65 539.39 1.4857 140 0.0010678 580.69 602.04 1.7193 0.0010621 576.88 608.75 1.7098 0.0010515 569.77 622.35 1.6915 160 0.0010885 665.35 687.12 1.9204 0.0010821 660.82 693.28 1.9096 0.0010703 652.41 705.92 1.8891 180 0.0011120 750.95 773.20 2.1147 0.0011047 745.59 778.73 2.1024 0.0010912 735.69 790.25 2.0794 865.3 2.2893 0.0011146 819.7 875.5 2.2634 860.5 200 0.0011388 837.7 2.3031 0.0011302 831.4 2.4870 0.0011590 918.3 220 0.0011695 925.9 953.1 2.4711 0.0011408 904.7 961.7 2.4419 949.3 2.6674 0.0011920 1006.9 1042.6 2.6490 0.0011702 990.7 1049.2 2.6158 240 0.0012046 1016.0 1040.0 2.8459 0.0012303 1097.4 1134.3 2.8243 0.0012034 1078.1 1138.2 2.7860 260 0.0012462 1108.6 1133.5 3.0248 0.0012755 1190.7 1229.0 2.9986 0.0012415 1167.2 1229.3 2.9537 280 0.0012965 1204.7 1230.6 3.2071 0.0013304 1287.9 1327.8 3.1741 0.0012860 1258.7 1323.0 3.1200 300 0.0013596 1306.1 1333.3 3.3979 0.0013997 1390.7 1432.7 3.3539 0.0013388 1353.3 1420.2 3.2868 320 0.0014437 1415.7 1444.6 3.6075 0.0014920 1501.7 1546.5 3.5426 0.0014032 1452.0 1522.1 3.4557 340 0.0015684 1539.7 1571.0 3.8772 0.0016265 1626.6 1675.4 3.7494 0.0014838 1556.0 1630.2 3.6291 360 0.0018226 1702.8 1739.3 0.0018691 1781.4 1837.5 4.0012 0.0015884 1667.2 1746.6 3.8101 380 Table 6 Saturated ice-steam (temperature) table Sat. Temp.Press. kPa \u00b0C Specific volume m3/kg Sat. ice Sat. vapour T Tsat \u00d7 103 vi vg Internal energy kJ/kg Enthalpy kJ/kg Entropy kJ/(kg.K) Sat. ice u i Sat. vapour u g Sat. ice h i Sat. Subl. vapour h ig h g Subl. u ig Sat. ice si Sat. Subl. v a p o u r sig sg 0.01 0.6113 0 0.6108 \u20132 0.5176 \u20134 0.4375 \u20136 0.3689 \u20138 0.3102 \u201310 0.2602 \u2013l2 0.2176 \u201314 0.1815 \u201316 0.1510 0.l252 \u201318 \u201320 0.1035 \u201322 0.0853 \u201324 0.0701 \u201326 0.0574 1.0908 1.0908 1.0904 1.0901 1.0898 1.0894 1.0891 1.0888 1.0884 1.0881 1.0878 1.0874 1.0871 1.0868 1.0864 9.156 206.1 \u2013333.40 2708.7 2375.3 \u2013333.40 2834.8 2501.4 \u20131.221 10.378 9.157 206.3 \u2013333.43", " 2708.8 2375.3 \u2013333.43 2834.8 2501.3 \u20131.221 10.378 9.219 241.7 \u2013337.62 2710.2 2372.6 \u2013337.62 2835.3 2497.7 \u20131.237 10.456 9.283 283.8 \u2013341.78 2711.6 2369.8 \u2013341.78 2835.7 2494.0 \u20131.253 10.536 9.348 334.2 \u2013345.91 2712.9 2367.0 \u2013345.91 2836.2 2490.3 \u20131.268 10.616 9.414 394.4 \u2013350.02 2714.2 2364.2 \u2013350.02 2836.6 2486.6 \u20131.284 10.698 9.481 466.7 \u2013354.09 2715.5 2361.4 \u2013354.09 2837.0 2482.9 \u20131.299 10.781 9.550 553.7 \u2013358.14 2716.8 2358.7 \u2013358.14 2837.3 2479.2 \u20131.315 10.865 9.619 658.8 \u2013362.15 2718.0 2355.9 \u2013362.15 2837.6 2475.5 \u20131.331 10.950 9.690 786.0 \u2013366.14 2719.2 2353.1 \u2013366.14 2837.9 2471.8 \u20131.346 11.036 9.762 940.5 \u2013370.10 2720.4 2350.3 \u2013370.10 2838.2 2468.1 \u20131.362 11.123 9.835 1128.6 \u2013374.03 2721.6 2347.5 \u2013374.03 2838.4 2464.3 \u20131.377 11.212 9.909 1358.4 \u2013377.93 2722.7 2344.7 \u2013377.93 2838.6 2460.6 \u20131.393 11.302 1640.1 \u2013381.80 2723.7 2342.0 \u2013381.80 2838.7 2456.9 \u20131.408 11.394 9.985 1986.4 \u2013385.64 2724.8 2339.2 \u2013385.64 2838.9 2453.2 \u20131.424 11.486 10.062 Contd. Appendix _____________________________________________________________________ 929 Sat. Temp.Press. kPa \u00b0C Specific volume m3/kg Sat. ice Sat. vapour T Tsat \u00d7 103 vi vg Internal energy kJ/kg Enthalpy kJ/kg Entropy kJ/(kg.K) Sat. ice u i Sat. vapour u g Sat. ice h i Sat. Subl. vapour h ig h g Subl. u ig Sat. ice si Sat. Subl. vapour sig sg \u201328 0.0469 \u201330 0.0381 \u201332 0.0309 \u201334 0.0250 \u201336 0.0201 \u201338 0.0161 0.0l29 \u201340 1.0861 1.0858 1.0854 1.0851 1.0848 1.0844 1.0841 2413.7 \u2013389.45 2725.8 2336.4 \u2013389.45 2839.0 2449.5 \u20131.439 11.580 10.141 2943 \u2013393.23 2726.8 2333.6 \u2013393.23 2839.0 2445.8 \u20131.455 11.676 10.221 3600 \u2013396.98 2727.8 2330.8 \u2013396.98 2839.1 2442.1 \u20131.471 11.773 10.303 4419 \u2013400.71 2728.7 2328.0 \u2013400.71 2839.1 2438.4 \u20131.486 11.872 10.386 5444 \u2013404.40 2729.6 2325.2 \u2013404.40 2839.1 2434.7 \u20131.501 11.972 10.470 6731 \u2013408.06 2730.5 2322.4 \u2013408.06 2839.0 2430.9 \u20131.517 12.073 10.556 8354 \u2013411.70 2731.3 2319.6 \u2013411.70 2839.9 2427.2 \u20131.532 12.176 10.644 Table 7 Critical Point Data For Some Substances Substance Formula Molecular Weight Ammonia Argon Bromine Carbon Dioxide Carbon Monoxide Chlorine Deuterium Helium Helium3 Hydrogen Krypton Neon Nitrogen Nitrous Oxide Oxygen Sulfur Dioxide Water Xenon Benzene n-Butane Carbon T", "etrachloride Chloroform Dichlorodifluoromethane Dichlorofluoromethane Ethane Ethyl Alcohol Ethylene n-Hexane Methane Methyl Alcohol Methyl Chloride Propane Propene Propyne Trichlorofluoromethane NH3 Ar Br2 CO2 CO Cl2 D2 He He H2 Kr Ne N2 N2O O2 SO2 H2O Xe X2H6 C4H10 CCl4 CHCl2 CCl2F2 CHCl2F C2H6 C2H5OH C2H14 C6H4 CH4 CH3OH CH3Cl C3H8 C3H6 C3H4 CCl3F 17.03 39.948 159.808 44.01 28.001 70.906 4.00 4.003 3.00 2.016 83.80 20.183 28.013 44.013 31.999 64.063 18.015 131.30 78.115 58.124 153.82 119.38 120.91 102.92 30.070 46.07 28.054 86.178 16.043 32.042 50.488 44.097 42.081 40.065 137.37 Temperature Tc\u2019 K 405.5 151 584 304.2 133 417 38.4 5.3 3.3 33.3 209.4 44.5 126.2 309.7 154.8 430.7 647.3 289.8 562 425.5 556.4 536.6 384.7 451.7 305.5 516 282.4 507.9 191.1 513.2 416.3 370 365 401 471.2 Pressure pc\u2019 MPa 11.28 4.86 10.34 7.39 3.50 7.71 1.66 0.23 0.12 1.30 5.50 2.73 3.39 7.27 5.08 7.88 22.09 5.88 4.92 3.80 4.56 5.47 4.01 5.17 4.88 6.38 5.12 3.03 4.64 7.95 6.68 4.26 4.62 5.35 4.38 Volume m3/kmol.0724.0749.1355.0943 0.930.1242 \u2014 0.578 \u2014.0649.0924.0417.0899.0961.0780.1217.0568.1186.2603.2547.2759.2403.2179.1973.1480.1673.1242.3677.993.1180.1430.1998.1810 \u2014.2478 930 ________________________________________________________ Applied Thermodynamics Table 8 Saturated Ammonia Table Specific volume m3/kg Enthalpy kJ/kg Entropy kJ/(kg.K) Temp. \u00b0C Abs. Press. k p a P 40.88 \u201350 45.96 \u201348 51.55 \u201346 57.69 \u201344 64.42 \u201342 71.77 \u201340 79.80 \u201338 88.54 \u201336 98.05 \u201334 108.37 \u201332 119.55 \u201330 131.64 \u201328 144.70 \u201326 158.78 \u2013 24 173.93 \u201322 190.22 \u201320 207.71 \u201318 226.45 \u201316 246.51 \u201314 267.95 \u201312 290.85 \u201310 315.25 \u20138 341.25 \u20136 368.90 \u20134 398.27 \u20132 429.44 0 462.49 2 497.49 4 534.51 6 573.64 8 614.95 10 658.52 12 407.44 14 752.79 16 803.66 18 857.12 20 913.27 22 24 972.19 26 1033.97 28 1098.71 30 1166.49 32 1237.41 34 1311.55 36 1389.03 38 1469.92 40 1554.33 42 1442.35 44 1734.09 46 1829.65 48 1929.13 50 2032.62 Sat. Liquid v f 0.001 424 0.001 429 0.001 434 0.001 439 0.001 444 0.001 449 0.001 454 0.001 460 0.001 465 0.001 470 0.001 476 0.001 481 0.001 487 0.001 492 0.001 498 0.001 504 0.001 510 0.001 515 0.001 521 0.001 528 0.001 534 0.001 540 0.001 546 0.001 553 0.001 559 0.001 566 0.001 573 0", ".001 580 0.001 587 0.001 594 0.001 601 0.001 608 0.001 616 0.001 623 0.001 631 0.001 639 0.001 647 0.001 655 0.001 663 0.001 671 0.001 680 0.001 689 0.001 698 0.001 707 0.001 716 0.001 726 0.001 735 0.001 745 0.001 756 0.001 766 0.001 777 Evap. v fg Sat. Vapor v g Sat. Liquid h f Evap. h fg Sat. Vapor h g Sat. Liquid sf Sat. Evap. Vapor sfg sg 2.6239 2.3518 2.1126 1.9018 1.7155 1.5506 1.4043 1.2742 1.1582 1.0547 0.9621 0.8790 0.8044 0.7373 0.6768 0.6222 0.5728 0.5280 0.4874 0.4505 0.4169 0.3863 0.3583 0.3328 0.3094 0.2879 0.2683 0.2502 0.2335 0.2182 0.2040 0.1910 0.1789 0.1677 0.1574 0.1477 0.1388 0.1305 0.1228 0.1156 1.1089 0.1027 0.0969 0.0914 0.0863 0.0815 0.0771 0.0728 0.0689 0.0652 0.0617 2.6254 2.3533 2.1124 1.9032 1.7170 1.5521 1.4058 1.2757 1.1597 1.0562 0.9635 0.8805 0.8059 0.7388 0.6783 0.6237 0.5743 0.5296 0.4889 0.4520 0.4185 0.3878 0.3599 0.3343 0.3109 0.2895 0.2698 0.2517 0.2351 0.2198 0.2056 0.1926 0.1805 0.1693 0.1590 0.1494 0.1405 0.1322 0.1245 0.1173 0.1106 0.1044 0.0986 0.0931 0.0880 0.0833 0.0788 0.0746 0.0707 0.0669 0.0635 \u201344.3 \u201335.5 \u201326.6 \u201317.8 \u20138.9 0.0 8.9 17.8 26.8 35.7 44.7 53.6 62.6 71.6 80.7 89.7 98.8 107.8 116.9 126.0 135.2 144.3 153.5 162.7 171.9 181.1 190.4 199.6 208.9 218.3 227.6 237.0 246.4 255.9 265.4 274.9 284.4 294.0 303.6 313.6 322.9 332.6 342.3 352.1 361.9 371.7 381.6 391.5 401.5 411.5 421.7 1416.7 1411.3 1405.8 1400.3 1394.7 1389.0 1383.3 1377.6 1371.8 1365.9 1360.0 1354.0 1347.9 1341.8 1335.6 1329.3 1322.9 1316.5 1310.0 1303.5 1296.8 1290.1 1283.3 1276.4 1269.4 1262.4 1255.2 1248.0 1240.6 1233.2 1225.7 1218.1 1210.4 1202.6 1194.7 1186.7 1178.5 1170.3 1162.0 1153.6 1145.0 1136.4 1127.6 1118.7 1109.7 1100.5 1091.2 1081.7 1072.0 1062.2 1052.0 1372.4 1375.8 1379.2 1382.5 1385.8 1389.0 1392.2 1395.4 1398.5 1401.6 1404.6 1407.6 1410.5 1413.4 1416.2 1419.0 1421.7 1424.4 1427.0 1429.5 1432.0 1434.4 1436.8 1439.1 1441.3", " 1443.5 1445.6 1447.6 1449.6 1451.5 1453.3 1455.1 1456.8 1458.5 1460.0 1461.5 1462.9 1464.3 1465.6 1466.8 1467.9 1469.0 1469.9 1470.8 1471.5 1472.2 1472.8 1473.2 1473.5 1473.7 1473.7 \u20130.1942 \u20130.1547 \u20130.1156 \u20130.0768 \u20130.0382 0.0000 0.0380 0.0757 0.1132 0.1504 0.1873 0.2240 0.2605 0.2967 0.3327 0.3684 0.4040 0.4393 0.4744 0.5093 0.5440 0.5785 0.6128 0.6469 0.6808 0.7145 0.7481 0.7815 0.8148 0.8479 0.8808 0.9136 0.9463 0.9788 1.0112 1.0434 1.0755 1.1075 1.1394 1.1711 1.2028 1.2343 1.2656 1.2969 1.3281 1.3591 1.3901 1.4209 1.4518 1.4826 1.5135 6.3502 6.2696 6.1902 6.1120 6.0349 5.9589 5.8840 5.8101 5.7372 5.6652 5.5942 5.5241 5.4548 5.3864 5.3188 5.2520 5.1860 5.1207 5.0561 4.9922 4.9290 4.8664 4.8045 4.7432 4.6825 4.6223 4.5627 4.5037 4.4451 4.3871 4.3295 4.2725 4.2159 4.1597 4.1039 4.0486 3.9937 3.9392 3.8850 3.8312 3.7777 3.7246 3.6718 3.6192 3.5669 3.5148 3.4630 3.4112 3.3595 3.3079 3.2561 6.1561 6.1149 6.0746 6.0352 5.9967 5.9589 5.9220 5.8858 5.8504 5.8156 5.7815 5.7481 5.7153 5.6831 5.6515 5.6205 5.5900 5.5600 5.5305 5.5015 5.4730 5.4449 5.4173 5.3901 5.3633 5.3369 5.3108 5.2852 5.2599 5.2350 5.2104 5.1861 5.1621 5.1385 5.1151 5.0920 5.0692 5.0467 5.0244 5.0023 4.9805 4.9589 4.9374 4.9161 4.8950 4.8740 4.8530 4.8322 4.8113 4.7905 4.7696 Appendix _____________________________________________________________________ 931 Table 9 Superheated Ammonia Table Abs. Press. kPa Temperature, \u00b0C 0 \u201320 v 50 h s v 75 h s v 100 h \u201310 2.4474 \u20132.5481 2.6482 1435.8 1457.0 1478.1 6.3256 6.4077 6.4865 1.6233 1.6915 1.7591 1433.0 1454.7 1476.1 6.1190 6.2028 6.2828 1.2110 1.2631 1.3145 1430.1 1452.2 1474.1 s 5.9695 6.0552 6.1366 0.9635 1.0059 1.0476 v 1427.2 1449.8 1472.0 125 h 5.8512 5.9389 6.0217 s 0.7984 0.8344 0.8697 v 1424.1 1447.3 1469.8 150 h 5.7526 5.8424 5.9266 s 0.6199 0.6471 v 1442.0 1465.5 200 h 5.6863 5.7737 s 0.4910 0.5135 v 1436.6 1461.0 250 h 5.5609 5.6517 s 0.4243 v 1456.3 h 5.5493 300 s", " 0.3605 v 1451.5 350 h 5.4600 s 0.3125 v 1446.5 400 h 5.3803 s 0.2752 v 1441.3 450 h 5.3078 s 10 30 20 2.7479 2.8473 2.9464 1499.2 1520.4 1541.7 6.5625 6.6360 6.7073 1.8263 1.8932 1.9597 1497.5 1518.9 1540.3 6.3597 6.4339 6.5058 1.3654 1.4160 1.4664 1495.7 1517.3 1538.9 6.2144 6.2894 6.3618 1.0889 1.1297 1.1703 1493.9 1515.7 1537.5 6.1006 6.1763 6.2494 0.9045 0.9388 0.9729 1492.1 1514.1 1536.1 6.0066 6.0831 6.1568 0.6738 0.7001 0.7261 1488.4 1510.9 1533.2 5.8559 5.9342 6.0091 0.5354 0.5568 0.5780 1484.5 1507.6 1530.3 5.7365 5.8165 5.8928 0.4430 0.4613 0.4792 1480.6 1504.2 1527.4 5.6366 5.7186 5.7963 0.3770 0.3929 0.4086 1476.5 1500.7 1524.4 5.5502 5.6342 5.7135 0.3274 0.3417 0.3556 1472.4 1497.2 1521.3 5.4735 5.5597 5.6405 0.2887 0.3017 0.3143 1468.1 1493.6 1518.2 5.4042 5.4926 5.5752 40 3.0453 1563.0 6.7766 2.0261 1561.8 6.5756 1.5165 1560.5 6.4321 1.2107 1559.3 6.3201 1.0068 1558.0 6.2280 0.7519 1555.5 6.0813 0.5989 1552.9 5.9661 0.4968 1550.3 5.8707 0.4239 1547.6 5.7890 0.3692 1544.9 5.7173 0.3266 1542.2 5.6532 50 60 70 80 100 3.1441 3.2427 3.3413 3.4397 1584.5 1606.1 1627.8 1649.7 6.8441 6.9099 6.9743 7.0372 2.0923 2.1584 2.2244 2.2903 1583.4 1605.1 1626.9 1648.9 6.6434 6.7096 6.7742 6.8373 1.5664 1.6163 1.6659 1.7155 1.8145 1582.2 1604.1 1626.0 1648.0 1692.6 6.5003 6.5668 6.6316 6.6950 6.8l77 1.2509 1.2909 1.3309 1.3707 1.4501 1581.1 1603.0 1625.0 1647.2 1691.8 6.3887 6.4555 6.5206 6.5842 6.7072 1.0405 1.0740 1.1074 1.1408 1.2072 1580.0 1602.0 1624.1 1646.3 1691.1 6.2970 6.3641 6.4295 6.4933 6.6167 0.7774 0.8029 0.8282 0.8533 0.9035 1577.7 1599.9 1622.2 1644.6 1689.6 6.1512 6.2189 6.2849 6.3491 6.4732 0.6196 0.6401 0.6605 0.6809 0.7212 1575.4 1597.8 1620.3 1642.8 1688.2 6.0368 6.1052 6.1717 6.2365 6.3613 0.5143 0.5316 0.5488 0.5658 0.5997 1573.0 1595.7 1618.4 1641.1 1686.7 5.", "9423 6.0114 6.0785 6.1437 6.2693 0.4391 0.4541 0.4689 0.4837 0.5129 1570.7 1593.6 1616.5 1639.3 1685.2 5.8615 5.9314 5.9990 6.0647 6.1910 0.3826 0.3959 0.4090 0.4220 0.4478 1568.3 1591.5 1614.5 1637.6 1683.7 5.7907 5.8613 5.9296 5.9957 6.1228 0.3387 0.3506 0.3624 0.3740 0.3971 1565.9 1589.3 1612.6 1635.8 1682.2 5.7275 5.7989 5.8678 5.9345 6.0623 20 30 40 50 60 70 80 100 120 140 160 180 v 500 h s v 600 h 0.2698 0.2813 0.2926 1489.9 1515.0 1539.5 5.4314 5.5157 5.5950 0.2217 0.2317 0.2414 1482.4 1508.6 1533.8 0.3036 0.3144 0.3251 1563.4 1587.1 1610.6 5.6704 5.7425 5.8120 0.2508 0.2600 0.2691 1558.5 1582.7 1606.6 0.3357 1634.0 5.8793 0.2781 1630.4 0.3565 0.3771 0.3975 1680.7 1727.5 1774.7 6.0079 6.1301 6.2472 0.2957 0.3130 0.3302 1677.7 1724.9 1772.4 932 ________________________________________________________ Applied Thermodynamics s 5.3222 5.4102 5.4923 v 0.1874 0.1963 0.2048 1474.5 1501.9 1528.1 5.2259 5.3179 5.4029 0.1615 0.1696 0.1773 1466.3 1495.0 1522.2 5.1387 5.2351 5.3232 0.1488 0.1559 1488.0 1516.2 5.1593 5.2508 0.1321 0.1388 1480.6 1510.0 5.0889 5.1840 0.1129 1497.1 5.0629 0.0944 1483.4 4.9534 700 h s v 800 h s v 900 h s v 1000 h s v 1200 h s v 1400 h s v 1600 h s v 1800 h s v 2000 h s 5.7826 5.5697 5.6436 5.7144 0.2369 0.2131 0.2212 0.2291 1626.8 1553.4 1578.2 1602.6 5.6997 5.4826 5.5582 5.6303 0.2060 0.1848 0.1920 0.1991 1623.1 1548.3 1573.7 1598.6 5.6268 5.4053 5.4827 5.5562 0.1820 0.1627 0.1693 0.1757 1619.4 1543.0 1569.1 1594.4 5.5614 5.3354 5.4147 5.4897 0.1627 0.1450 0.1511 0.1570 1615.6 1537.7 1564.4 1590.3 5.5021 5.2713 5.3525 5.4292 0.1338 0.1185 0.1238 0.1289 1608.0 1526.6 1554.7 1581.7 5.3970 5.1560 5.2416 5.3215 0.1132 0.0995 0.1042 0.1088 1600.2 1515.1 1544.7 1573.0 5.3053 5.0530 5.1434 5.2270 0.0977 0.0851 0.0895 0.0937 1592.3 1502.9 1534.4 1564.0 5.2232 4.9584 5.0543 5.1419 0.0856 0.0739 0.0781 0.0820 1490.0 1523.5 1554.6 1584.1 4.8693 4.9715 5.0635 5.14825 5.9129 6.", "0363 6.1541 0.2522 0.2672 0.2821 1674.6 1722.4 1770.2 5.8316 5.9562 6.0749 0.2196 0.2329 0.2459 0.2589 1671.6 1719.8 1768.0 1816.4 5.7603 5.8861 6.0057 6.1202 0.1942 0.2061 0.2178 0.2294 1668.5 1717.1 1765.7 1814.4 5.6968 5.8237 5.9442 6.0594 0.1739 0.1847 0.1954 0.2058 0.2162 1665.4 1714.5 1763.4 1812.4 1861.7 5.6392 5.7674 5.8888 6.0047 6.1159 0.1434 0.1526 0.1616 0.1705 0.1792 1659.2 1709.2 1758.9 1808.5 1858.2 5.5379 5.6687 5.7919 5.9091 6.0214 0.1216 0.1297 0.1376 0.1452 0.1528 1652.8 1703.9 1754.3 1804.5 1854.7 5.4501 5.5836 5.7087 5.8273 5.9406 0.1053 0.1125 0.1195 0.1263 0.1330 1646.4 1698.5 1749.7 1800.5 1851.2 5.3722 5.5084 5.6355 5.7555 5.8699 0.0926 0.0992 0.1055 0.1116 0.1177 1639.8 1693.1 1745.1 1796.5 1847.7 5.4409 5.5699 5.6914 5.8069 3018 0.0648 0.0688 0.0725 0.0760 0.0824 0.0885 0.0943 0.0999 0.1054 1476.1 1512.0 1544.9 1575.6 1633.2 1687.6 1740.4 1792.4 1844.1 4.7834 4.8930 4.9902 5.0786 5.2371 5.3793 5.5104 5.6333 5.7499 Appendix _____________________________________________________________________ 933 Table 10 Saturated Freon - 12 Table Specific volume m3/kg Enthalpy kJ/kg Entropy kJ/(kg.K) Sat. Liquid v f Evap. v fg Sat. Vapor v g Sat. Liquid h f Evap. h fg Sat. Vapor h g Sat. Liquid sf Evap. sfg Sat. Vapor sg Temp. Abs. \u00b0C Press. MPa P 0.0028 0.000 608 4.414 937 4.415 545 \u201343.243 \u201390 0.0042 0.000 612 3.036 704 3.037 316 \u201338.968 \u201385 0.0062 0.000 617 2.137 728 2.138 345 \u201334.688 \u201380 0.0088 0.000 622 1.537 030 1.537 651 \u201330.401 \u201375 0.0123 0.000 627 1.126 654 1.127 280 \u201326.103 \u201370 0.0168 0.000 632 0.840 534 0.841 166 \u201321.793 \u201365 0.0226 0.000 637 0.637 274 0.637 910 \u201317.469 \u201360 0.0300 0.000 642 0.490 358 0.491 000 \u201313.129 \u201355 \u20138.772 \u201350 0.0391 0.000 648 0.382 457 0.383 105 \u20134.396 \u201345 0.0504 0.000 654 0.302 029 0.302 682 \u20130.000 0.0642 0.000 659 0.241 251 0.241 910 \u201340 4.416 0.0807 0.000 666 0.194 732 0.195 398 \u201335 0.1004 0.000 672 0.158 703 0.159 375 \u201330 8.854 13.315 0.1237 0.000 679 0.130 487 0.131 166 \u201325 17.800 0.1509 0.000 685 0.108 162 0.108 847 \u201320 22.312 0.1826 0.000 693 0.090 326 0.091 018 \u201315 26.", "851 0.2191 0.000 700 0.075 946 0.076 646 \u201310 31.420 0.2610 0.000 708 0.064 255 0.064 963 \u20135 36.022 0.3086 0.000 716 0.054 673 0.055 389 0 40.659 0.3626 0.000 724 0.046 761 0.047 485 5 45.337 0.4233 0.000 733 0.040 180 0.040 914 10 50.058 0.4914 0.000 743 0.034 671 0.035 413 15 54.828 0.5673 0.000 752 0.030 028 0.030 780 20 59.653 0.6516 0.000 763 0.026 091 0.026 854 25 64.539 0.7449 0.000 774 0.022 734 0.023 508 30 69.494 0.8477 0.000 786 0.019 855 0.020 641 35 74.527 0.9607 0.000 798 0.017 373 0.018 171 40 79.647 1.0843 0.000 811 0.015 220 0.016 032 45 84.868 1.2193 0.000 826 0.013 344 0.014 170 50 90.201 1.3663 0.000 841 0.011 701 0.012 542 55 1.5259 0.000 858 0.010 253 0.011 111 95.665 60 1.6988 0.000 877 0.008 971 0.009 847 101.279 65 1.8858 0.000 897 0.007 828 0.008 725 107.067 70 2.0874 0.000 920 0.006 802 0.007 723 113.058 75 2.3046 0.000 946 0.005 875 0.006 821 119.291 80 2.5380 0.000 976 0.005 029 0.006 005 125.818 85 2.7885 0.001 012 0.004 246 0.005 258 132.708 90 3.0569 0.001 056 0.003 508 0.004 563 140.068 95 3.3440 0.001 113 0.002 790 0.003 903 148.076 l00 3.6509 0.001 197 0.002 045 0.003 242 157.085 105 3.9784 0.001 364 0.001 098 0.002 462 168.059 110 4.1155 0.001 792 0.000 005 0.001 797 174.920 112 189.618 187.608 185.612 183.625 181.640 179.651 177.653 175.641 173.611 171.558 169.479 167.368 165.222 163.037 160.810 158.534 156.207 153.823 151.376 148.859 146.265 143.586 140.812 137.933 134.936 131.805 128.525 125.074 121.430 117.565 113.443 109.024 104.255 99.068 93.373 87.047 79.907 71.658 61.768 49.014 28.425 0.151 146.375 \u20130.2084 148.640 \u20130.1854 150.924 \u20130.1630 153.224 \u20130.1411 155.536 \u20130.1197 157.857 \u20130.0987 160.184 \u20130.0782 162.512 \u20130.0581 164.840 \u20130.0384 167.163 \u20130.0190 169.479 \u20130.0000 0.0187 171.784 0.0371 174.076 0.0552 176.352 0.0730 178.610 0.0906 180.846 0.1079 183.058 0.1250 185.243 0.1418 187.397 0.1585 189.518 0.1750 191.602 0.1914 193.644 0.2076 195.641 0.2237 197.586 0.2397 199.475 0.2557 201.299 0.2716 203.051 0.2875 204.722 0.3034 206.298 0.3194 207.766 0.3355 209.109 0.3518 210.303 0.3683 211.321 0.3851 212.126 0.4023 212.665 0.4201 212.865 0.4385 212.614 0.4579 211.726 0.4788 209.8", "43 0.5023 206.099 0.5322 196.484 0.5651 175.071 1.0352 0.9970 0.9609 0.9266 0.8940 0.8630 0.8334 0.8051 0.7779 0.7519 0.7269 0.7027 0.6795 0.6570 0.6352 0.6141 0.5936 0.5736 0.5542 0.5351 0.5165 0.4983 0.4803 0.4626 0.4451 0.4277 0.4104 0.3931 0.3758 0.3582 0.3405 0.3224 0.3038 0.2845 0.2644 0.2430 0.2200 0.1946 0.1655 0.1296 0.0742 0.0004 0.8268 0.8116 0.7979 0.7855 0.7744 0.7643 0.7552 0.7470 0.7396 0.7329 0.7269 0.7214 0.7165 0.7121 0.7082 0.7046 0.7014 0.6986 0.6960 0.6937 0.6916 0.6897 0.6879 0.6863 0.6848 0.6834 0.6820 0.6806 0.6792 0.6777 0.6760 0.6742 0.6721 0.6697 0.6667 0.6631 0.6585 0.6526 0.6444 0.6319 0.6064 0.5655 934 ________________________________________________________ Applied Thermodynamics Table 11 Superheated Freon - 12 Table Temp. \u00baC v m3/kg h kJ/kg s kJ/kg K v m3/kg h kJ/kg s kJ/kg K v m3/kg h kJ/kg s kJ/kg K 0.05 MPa 0.341 857 181.042 0.356 227 186.757 0.370 508 102.567 0.384 716 198.471 0.398 863 204.469 0.412 959 210.557 0.427 012 216.733 0.441 030 222.997 0.455 017 229.344 0.468 978 235.774 0.482 917 242.282 0.496 838 248.868 0.7912 0.8133 0.8350 0.8562 0.8770 0.8974 0.9175 0.9372 0.9565 0.9755 0.9942 1.0126 0.20 MPa 0.088 608 189.669 0.092 550 195.878 0.096 418 202.135 0.100 228 208.446 0.103 989 214.814 0.107 710 221.243 0.111 397 227.735 0.115 055 234.291 0.118 690 240.910 0.122 304 247.593 0.125 901 254.339 0.129 483 261.147 0.40 MPa 0.045 836 198.762 0.047 971 205.428 0.050 046 212.095 0.052 072 218.779 0.054 059 225.488 0.056 014 232.230 0.057 941 239.012 0.059 846 245.837 0.061 731 252.707 0.063 600 259.624 0.065 455 266.590 0.067 298 273.605 0.7320 0.7543 0.7760 0.7972 0.8178 0.8381 0.8578 0.8772 0.8962 0.9149 0.9332 0.9512 0.7199 0.7423 0.7639 0.7849 0.8054 0.8253 0.8448 0.8638 0.8825 0.9008 0.9187 0.9364 \u201320.0 \u201310.0 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0 110.0 120.0 130.0 140.0 150.0 160.0 170.0 180.0 190.0 200.0 210.0 220.0 230.0 0.167 701 0.175 222 0.182 647 0.189 994 0.197 277 0.204 506 0", ".211 691 0.218 839 0.225 955 0.233 044 0.240 111 0.247 159 0.069 752 0.073 024 0.076 218 0.079 350 0.082 431 0.085 470 0.088 474 0.091 449 0.094 398 0.097 327 0.100 238 0.103 114 0.035 646 0.037 464 0.039 214 0.040 911 0.042 565 0.044 184 0.045 774 0.047 340 0.048 886 0.050 415 0.051 929 0.053 430 0. 10 MPa 179.861 185.707 191.628 197.628 203.707 209.866 216.104 222.421 228.815 235.285 241.829 248.446 0.25 MPa 188.644 194.969 201.322 207.715 214.153 220.642 227.185 233.785 240.443 247.160 253.936 260.770 0.7401 0.7628 0.7849 0.8064 0.8275 0.8482 0.8684 0.8883 0.9078 0.9269 0.9457 0.9642 0.7139 0.7366 0.7587 0.7801 0.8010 0.8214 0.8413 0.8608 0.8800 0.8987 0.9171 0.9352 0.15 MPa 184.619 190.660 196.762 202.927 209.160 215.463 221.835 228.277 234.789 241.371 248.020 0.30 MPa 187.583 194.034 200.490 206.969 213.480 220.030 226.627 233.273 239.971 246.723 253.530 260.391 0.114 716 0.119 866 0.124 932 0.129 930 0.134 873 0.139 768 0.144 625 0.149 450 0.154 247 0.159 020 0.163 774 0.057 150 0.059 984 0.062 734 0.065 418 0.068 049 0.070 635 0.073 185 0.075 705 0.078 200 0.080 673 0.083 127 0.085 566 0.50 MPa 0.60 MPa 196.935 203.814 210.656 217.484 224.315 231.161 238.031 244.932 251.869 258.845 265.862 772.923 0.6999 0.7230 0.7452 0.7667 0.7875 0.8077 0.8275 0.8467 0.8656 0.8840 0.9021 0.9198 0.030 422 0.031 966 0.033 450 0.034 887 0.036 285 0.037 653 0.038 995 0.040 316 0.041 619 0.042 907 0.044 181 202.116 209.154 216.141 223.104 230.062 237.027 244.009 251.016 258.053 265.124 272.231 0.7318 0.7543 0.7763 0.7977 0.8186 0.8390 0.8591 0.8787 0.8980 0.9169 0.9354 0.6984 0.7216 0.7440 0.7658 0.7869 0.8075 0.8276 0.8473 0.8665 0.8853 0.9038 0.9220 0.7063 0.7291 0.7511 0.7723 0.7929 0.8129 0.8324 0.8514 0.8700 0.8882 0.9061 Appendix _____________________________________________________________________ 935 0.70 MPa 0.80 MPa 0.90 MPa 40.0 50.0 60.0 70.0 80.0 90.0 100.0 110.0 120.0 130.0 140.0 150.0 50.0 60.0 70.0 80.0 90.0 100.0 110.0 120.0 130.0 140.0 150.0 160.0 70.0 80.0 90.0 100.0 110.0 l20.0 130.0 140.0 150.0 160.0 170.0 180.0 0.026 761 207.580 0.028 100 214.745 0.029 387 221.854 0.030 632 228.931 0.031 843 235.", "997 0.033 027 243.066 0.034 189 250.146 0.035 332 257.247 0.036 458 264.374 0.037 572 271.531 0.038 673 278.720 0.039 764 285.946 1.00 MPa 0.018 366 0.019 410 0.020 397 0.021 341 0.022 251 0.023 133 0.023 993 0.024 835 0.025 661 0.026 474 0.027 275 0.028 068 210.162 217.810 225.319 232.739 240.101 247.430 254.743 262.053 269.369 276.699 284.047 291.419 1.60 MPa 0.011 208 216.650 0.011 984 225.177 0.012 698 233.390 0.013 366 241.397 0.014 000 249.264 0.014 608 257.035 0.015 195 264.742 0.015 765 272.406 0.016 320 280.044 0.016 864 287.669 0.017 398 295.290 0.017 923 302.914 0.7148 0.7373 0.7590 0.7799 0.8002 0.8199 0.8392 0.8579 0.8763 0.8943 0.9119 0.9292 0.7021 0.7254 0.7476 0.7689 0.7895 0.8094 0.8287 0.8475 0.8659 0.8839 0.9015 0.9187 0.6959 0.7204 0.7433 0.7651 0.7859 0.8059 0.8253 0.8440 0.8623 0.8801 0.8975 0.9145 0.022 830 0.024 068 0.025 247 0.026 380 0.027 477 0.028 545 0.029 588 0.030 612 0.031 619 0.032 612 0.033 592 0.034 563 0.014 483 0.015 463 0.016 368 0.017 221 0.018 032 0.018 812 0.019 567 0.020 301 0.021 018 0.021 721 0.022 412 0.023 093 0.009 406 0.010 187 0.010 884 0.011 526 0.012 126 0.012 697 0.013 244 0.013 772 0.014 284 0.014 784 0.015 272 0.015 752 205.924 213.290 220.558 277.766 234.941 242.101 249.260 256.428 263.613 270.820 278.055 285.320 1.20 MPa 206.661 214.805 222.687 230.398 237.995 245.518 252.993 260.441 267.875 275.307 282.745 290.195 1.80 MPa 213.049 222.198 230.835 239.155 247.264 255.228 263.094 270.891 278.642 286.364 294.069 301.767 0.7016 0.7248 0.7469 0.7682 0.7888 0.8088 0.8283 0.8472 0.8657 0.8838 0.9016 0.9189 0.6812 0.7060 0.7293 0.7514 0.7727 0.7931 0.8129 0.8320 0.8507 0.8689 0.8867 0.9041 0.6794 0.7057 0.7298 0.7524 0.7739 0.7944 0.8141 0.8332 0.8518 0.8698 0.8874 0.9046 0.019 744 0.020 912 0.022 012 0.023 062 0.024 072 0.025 051 0.026 005 0.026 937 0.027 851 0.028 751 0.029 639 0.030 515 0.012 579 0.013 448 0.014 247 0.014 997 0.015 710 0.016 393 0.017 053 0.017 695 0.018 321 0.018 934 0.019 535 0.008 704 0.009 406 0.010 035 0.010 615 0.011 159 0.011 676 0.012 172 0.012 651 0.013 116 0.013 570 0.014 013 204.170 211.765 219.212 226.564 233.856 241.113 248.355 255.", "593 262.839 270.100 277.381 284.687 1.40 MPa 211.457 219.822 227.891 235.766 243.512 251.170 258.770 266.334 273.877 281.411 288.946 2.00 MPa 218.859 228.056 236.760 245.154 253.341 261.384 269.327 277.201 285.027 292.822 300.598 0.6982 0.7131 0.7358 0.7575 0.7785 0.7987 0.8184 0.8376 0.862 0.8745 0.8923 0.9098 0.6876 0.7123 0.7355 0.7575 0.7785 0.7988 0.8183 0.8373 0.8558 0.8738 0.8914 0.6909 0.7166 0.7402 0.7624 0.7835 0.8037 0.8232 0.8420 0.8603 0.8781 0.8955 936 ________________________________________________________ Applied Thermodynamics Table 11 Superheated Freon - 12 Table Temp. \u00baC v m3/kg h kJ/kg s kJ/kg K v m3/kg h kJ/kg s kJ/kg K v m3/kg h kJ/kg s kJ/kg K 2.50 MPa 3.00 MPa 3.50 MPa 0.005 231 0.005 886 0.006 419 0.006 887 0.007 313 0.007 709 0.008 083 0.008 439 0.008 782 0.009 114 0.009 436 220.529 232.068 242.208 251.632 260.620 269.319 277.817 286.171 294.422 302.597 310.718 0.6770 0.7075 0.7336 0.7573 0.7793 0.8001 0.8200 0.8391 0.8575 0.8753 0.8927 0.004 324 0.004 959 0.005 456 0.005 884 0.006 270 0.006 626 0.006 961 0.007 279 0.007 584 0.007 878 222.121 234.875 245.661 255.524 264.846 273.817 282.545 291.100 299.528 307.864 0.6750 0.7078 0.7349 0.7591 0.7814 0.8023 0.8222 0.8413 0.8597 0.8775 90.0 100.0 110.0 120.0 130.0 140.0 150.0 160.0 170.0 180.0 190.0 200.0 120.0 130.0 140.0 150.0 160.0 170.0 180.0 190.0 200.0 210.0 220.0 230.0 0.006 595 219.562 0.007 264 229.852 0.007 837 239.271 0.008 351 248.192 0.008 827 256.794 0.009 273 265.180 0.009 697 273.414 0.010 104 281.540 0.010 497 289.589 0.010 879 297.583 0.011 250 305.540 0.011 614 313.472 4.00 MPa 0.003 736 224.863 0.004 325 238.443 0.004 781 249.703 0.005 172 259.904 0.005 522 269.492 0.005 845 278.684 0.006 147 287.602 0.006 434 296.326 0.006 708 304.906 0.006 972 313.380 0.007 228 321.774 0.007 477 330.108 0.6823 0.7103 0.7352 0.7582 0.7798 0.8003 0.8200 0.8390 0.8574 0.8752 0.8926 0.9095 0.6771 0.7111 0.7386 0.7630 0.7854 0.8063 0.8262 0.8453 0.8636 0.8813 0.8985 0.9152 Appendix _____________________________________________________________________ 937 Table 12 Enthalpies of Formation. Gibbs Function of Formation and Absolute Entropy at 25\u00b0C and 1 atm Pressure Substance Acetylene C2H2(g) Ammonia NH3(g) Benzene C6H6(g) Butane C4H10(g) Carbon C graphite Carbon Dioxide CO2(g) Carbon Monoxide CO(g) Ethane C2H6", "(g) Ethene C2H4(g) Hydrazine N2H4(g) Methane CH4(g) Octane C8H18(g) Octane C8H18(l) Propane C3H8(g) Water H2O(g) Water H2O(l) Dodecane C12H26(g) Dodecane C12H26(l) M 26.038 17.032 78.108 58.124 12.011 44.01 28.001 30.07 28.054 32.048 16.043 114.23 114.23 44.097 18.015 18.015 170.328 170.328 Enthalpy of formation (kJ/kgmol) Gibbs function of formation (kJ/kgmol) Absolute Entropy (kJ/kgmol.K) 209 290 \u201316 390 129 732 \u201317 164 0 \u2013344 631 \u2013137 242 \u201332 905 681 59 159 260 \u201350 844 16 536 6 614 \u201323 502 \u2013228 729 \u2013237 327 200.98 192.72 269.38 310.32 5.69 213.83 197.68 229.64 219.58 238.77 186.27 467.04 361.03 270.09 188.85 69.98 226 866 \u201345 926 82 976 \u2013126 223 0 \u2013393 757 \u2013110 596 \u201384 718 52 315 95 410 \u201374 917 \u2013208 581 \u2013250 102 \u2013103 909 \u2013241 971 \u2013286 010 \u2013290 971 \u2013394 199 Appendix _____________________________________________________________________ 939 Chart 2 : Mollier diagram This page intentionally left blank A \u03b1-methyl napthalene 787 Absolute humidity 828 Absolute volumetric efficiency 714 Absorption Refrigeration cycle 820 Accessories 459 Actual indicator diagram 715 Adiabatic expansion index 574 Adiabatic flame temperature 400 Adiabatic mixing 834 Air breathing jet engines or atmospheric jet engine 860 Air compressors 706 Air conditioning systems 835 Air cooled engines 771 Air fuel ratio 398 Air leakage 691 Air preheater 465 Air refrigeration cycles 809 Air requirement 409 Air standard cycle 330, 778 Air standard efficiency 331 Ammonia-water absorption system 821 Angle of advance 514 Artificial draught 474 Axial flow compressor characteristics 739 Axial flow compressors 732 Axial flow turbine 616 Axial thrust 628 Azeotropes 826 B Babcock and Wilcox boiler 448 Back pressure turbine 273, 618 Back work ratio 251 Balanced draught 476 Bearings 614, 772 Bell-Coleman 810 Benson boiler 451 Bent water tube boilers 440 Binary vapour cycle 268 Blade height 632 Blade velocity coefficient 625, 627 Blow off cock 463 Blowers 706 Boiler 436 Boiler efficiency 478 Boiler mountings 459 Boiler trial 481 Bomb calorimeter 402 Bore 772 Bottom dead centre (BDC) 708, 773 Brake power 780, 787 Brake thermal efficiency 524 Branca\u2019s impulse turbine 611 Brayton cycle 340 By pass factor 833 By-pass governing 650, 653 C Calorific value 402 Cams and Camshafts 772 Carburettor 772 Carnot cycle 121 Carnot engine 121 942 ________________________________________________________ Applied Thermodynamics Carburettor type engines 771 Carnot gas power cycle 250 Carnot refrigeration cycle 808 Carnot vapour power cycle 250, 251 Cascade method 265 Casing 612 Central flow type 689 Centrifugal compressor characteristics 736 Centrifugal compressors 728, 730 Cetane 788 Cetane number 788 Chimney 468 Choice of refrigerant 827 Choked flow 576 Choking 733 Classification of jet propulsion engines 860 Classification of steam engines 506 Classification of steam turbines 614 Clearance volume 519, 773 Cloud point 400 Cochran boiler 443 Coefficient of performance 807 Coefficient of velocity 581 Cogeneration 272 Combined cycle 270 Combined governing 650 Combined heat and power 272 Combustion analysis 407 Combustion in CI engines 785 Combustion in SI engine 783 Comfort air conditioning 805 Composite boilers 442 Composition of air 400 Compound steam engine 508, 527, 530 Compounded impulse turbine 634 Compressible flow 565 Compression ignition engines 771, 772 Compression ratio 331, 773 Compressor and turbine efficiency 358 Compressors 706 Condenser 684 Condenser efficiency 693 Condensing engine 508 Condensing plant 685 Condensing turbine 618 Continuum 3 Connecting rod 510, 772 Control of Reciprocating Compressors 722 Convergent nozzle 577 Convergent-divergent nozzle 578 Cooling tower 693 Cooling with dehumidification 832", " COP 807 Cornish boiler 446 Corrected vacuum 692 Counter flow jet condenser 686 Crank 509, 512, 772 Crank pin 510 Crank shaft 509, 512, 772 Crankcase 772 Critical pressure ratio 573, 575 Cross flow compound turbine with double flow 615 Cross flow compound turbine with single flow 615 Cross head 509, 512 Cumulative heat drop 647 Curtis turbine 621 Cut-off governing 508, 525, 526, 534 Cylinder 772 D D-slide valve 514 Damper 449 Dead centre 773 Dead state 233 Dead steam 509 Dead weight safety valve 460 Deaerator 262 Degree of reaction 639 Degree of supersaturation 583 Degree of undercooling 583 Dew point temperature 829 Diagram efficiency 629 Diagram efficiency or blading efficiency 628 Index _________________________________________________________________________ 943 Diagram factor 519 Diaphragm 613 Diesel cycle 250, 334, 779 Diesel engines 771, 790 Diffuser 569, 570 Dimensionless mass flow 737 Dimensionless pressure 737 Dimensionless rotational speeds 737 Direct contact heaters 262 Direct contact type 685 Disc friction loss 645 Dissociation 400 Double acting engine 507 Double flow single casing turbine 615 Down flow type 689 Draught 467 Driving thrust 626 Dry bulb temperature 829 Dry compression 813 Dry ice 806 Dual cycle 250, 337, 779 E Eccentric 513 Economizer 464 Efficiency 192 Efficiency of chimney 473 Ejector condenser 686, 688 Energy performance ratio 807 Enthalpy of combustion 400 Enthalpy of formation 400 Enthalpy of moist air or mixture enthalpy 829 Equivalence ratio 399 Equivalent Evaporation 477, 478 Ericsson cycle 362, 250 Evaporative condenser 685, 690 Evaporator cooling 834 Excess air 399 Exhaust hood 614 Exhaust manifold 772 Expansive type 507 External combustion engine 770 Externally fired boilers 437 F Fans 706 Feed check valve 462 Feed pump 465 Feed water 684 Feed water heaters 261 Fire bridge 444 Fire point 399 Fire tube 438 Fire tube boilers 437, 438 Fired boilers 438 Flash point 399 Flow velocity 624 Flue gas analysis 411 Fluidized bed boiler 454 Forced circulation boilers 438 Forced draught 474 Four stage compressor 707 Four stroke engines 771 Free air condition 713 Free air delivery 713 Friction power 780 Friction on nozzle 580 Friction power 787 Fuel 401 Fuel cells 413 Fuel feed hopper 449 Fusible plug 461 G Gas calorimeter 404, 405 Gas engines 771 Gas fired boilers 437 Gas power cycle 250, 330 Gas turbine 790 Gas turbine irreversibilities 355 944 ________________________________________________________ Applied Thermodynamics Gaseous fuels 402 Governor 614 Gross efficiency 629 Gudgeon pin 772 Guide blades 612 Guide ways 509 Guideways 512 Gustaf de laval 611 H Halocarbon refrigerants 825 Heat Pump Systems 823 Heat rate 251 Heat recovery steam generators 456 Height of blade 643 Helical type compressor 723 Hero\u2019s reaction turbine 611 HHV = LHV 400 High capacity compressors 707 High level jet condenser 686 High pressure boiler 450 High pressure compressor 707 High speed engine 507 High speed steam turbine 618 High steam and low water safety valve 460 Higher heating value (HHV) 400 Horizontal boiler 437 Humid specific volume 829 Humidification 833 Humidifier 833 Humidity ratio or specific humidity 828 Hydrocarbons 826 Hyperbolic expansion 515 Hypothetical indicator diagram 518 I IC engine fuels 786 IC engines 789 Impulse turbines 614, 619, 654 Inclined boiler 437 Inclined cylinder engines 771 Indicated power 780, 787 Indicated thermal efficiency 524 Indicator diagram 518, 780 Indirect contact type 685 Induced draught 475 Industrial air conditioning 805 Injection type engines 771 Inorganic compounds 825 Intake manifold 772 Intercooling 351 Internal combustion engine 770 Internal efficiency 648 Internally fired 437 Isentropic compressor efficiency 358 Isentropic efficiency 737 Isentropic turbine efficiency 358 Iso-octane 786 Isothermal efficiency 710 J Jet condenser 685, 686 Jet efficiency 863 Jet propulsion 858 L La Mont boiler 453 Labyrinth seals 613 Lancashire Boiler 444 Lead 514 Liquid fuel fired boilers 437 Liquid fuels 401 Liquid propellant rocket engines 873 Live steam 509, 514 Locomotive boilers 446 Locomotive engine 508 Loeffler boiler 452 Loss due to leakage 645 Losses at inlet and exit 646 Losses due to radiations 646 Index _________________________________________________________________________ 945 Losses in bearings 645 Losses", " in moving blades 644 Losses in nozzles 644 Low capacity compressors 707 Low level jet condenser 686 Low pressure compressor 707 Low speed steam turbine 618 Lower heating value 400 M Mach number 567 Manhole and mud box 463 Marine engine 508 Maximum discharge 473, 574 Mean effective pressure 333 Mechanical draught 474 Mechanical efficiency 524 Mechanical stoker 449 Medium capacity compressors 707 Medium pressure compressor 707 Medium speed engine 507 Metastable flow 583 Methods of Compounding 527 Missing quantity 519, 520 Mixed pressure turbine 274 Modified absorption refrigeration cycle 822 Modified rankine cycle 515 Morse test 787 Moving blades 612 Multi pressure HRSG 457 Multi rotor engine 771 Multi stage turbine 616 Multi-fuel engines 771 Multistage compression 716 Multistage intercooled compression 352 Multistage vapour compression cycle 819 N N-heptane 786 Natural circulation boilers 437 Natural draught 467 Nestler boilers 448 Non condensing engine 508 Non-air breathing jet engines or rocket engines 860 Non-condensing turbine 618 Non-expansive steam engine 507 Non-positive displacement compressors 706 Normal speed steam turbine 618 Nozzle 564 Nozzle control governing 650, 654 Nozzle efficiency 580, 629 Nozzle flow analysis 567 O Octane number 786 Off design operation 577 Open type heaters 262, 264 Opposed cylinder engines 772 Optimization 630 Optimum blade speed to steam velocity ratio 643 Orsat analyzer 412 Otto cycle 250, 331, 779 Over-expanding nozzles 577 Overall efficiency 525, 629, 863 Overall volumetric efficiency 713 P Package boilers 437 Packing 613 Parallel flow jet condenser 686 Pass out or extraction turbine 274 Pass out turbine 618 Petrol engines 771 Piston 772 Piston and piston rod 509 Piston area 772 Piston rings 510, 772 Piston valve 515 Polytropic compression 730 Polytropic efficiency 358, 361 946 ________________________________________________________ Applied Thermodynamics Portable boiler 437 Positive displacement compressors 706 Pour point 400 Power 476 Power cycles 250 Pressure compounded impulse turbine 620 Pressure gauge 462 Pressure waves 565 Pressure-velocity compounded impulse turbine 620, 622 Primary reference fuels 786 Primary refrigerants 825 Process 6 Property 18 Propulsive efficiency 862 Propulsive power 861 Proximate analysis 401 Psychrometric charts 830 Psychrometric processes 830 Psychrometry 827 Pulse jet engine 870 R Radial flow turbine 616 Radius of eccentric 513 Ramjet engine 869 Ramjet or athodyds and lorin tube 860 Rankine cycle 250, 253 Rankine efficiency 525 Rankine heat drop 648 Rateau turbine 620 Reaction force 615 Reaction turbines 614, 637, 654 Receiver compound engines 527 Reciprocating air motor 722 Reciprocating compressors 708 Reciprocating engines 771 Reciprocating type positive displacement compresso 707 Refrigerants 824 Refrigeration 355, 805 Refrigeration by evaporation 806 Refrigeration by ice 806 Refrigeration cycle 805 Regenerative cycle 250, 260 Regenerative cycle efficiency 266 Regenerative gas turbine cycle 345, 247 Reheat and Intercooling 354, 355 Reheat cycle 250, 258 Reheat factor 646, 648 Relative efficiency 525 Relative humidity 828 Relative velocity 624 Ring or wheel 612 Rocket engine 872 Rocket propulsion 871 Roots blower 707, 723 Roots efficiency 725 Rotary compressors 723 Rotary engines 771 Rotary type positive displacement compressors 707 S Safety valve 460 Saturation curve 519 Saturation curve 520 Screw type 723 Secondary refrigerants 825, 826 Sensible cooling 830 Sensible heating 830 Shaft 612 Simple impulse turbine 623 Simple steam engine 508 Single acting engine 507 Single flow single casing turbine 615 Single pressure HRSG 457 Single rotor engine 771 Single stage 508 Single stage compressor 707 Single stage turbine 617 Slide valve 514 Index _________________________________________________________________________ 947 Slip factor 732 Slow speed engine 507 Smoke box 449 Solid fuel 401 Solid fuel fired boilers 437 Solid propellant rocket engines 872 Sonic flow 567 Sound waves 565 Spark ignition engines 771 Spark plug 772 Spark-Ignition engine 771 Specific steam consumption 251, 524 Stack 468 Stage efficiency 629, 647 Stage efficiency 359 Stagnation enthalpy 568 Stagnation properties 356 Stalling 735 Standard reference state 400 Stationary engine 508 Stationary nozzle 612 Stationery boiler 437 Steam chest 509, 613 Steam engine 506 Steam generator 436 Steam injector 584 Steam jet draught 477 Steam trap 466 Steam turbine control 649 Steam turbine plant 790 Steam turbines 611 Stirling boiler 4", "49 Stirling cycle 250, 364 Stop valve 462 Straight water tube boilers 440 Stroke 773 Stuffing box 512 Subatmospheric 684 Subsonic flow 567, 569 Summer air conditioning systems 835 Super high pressure compressor 707 Superheater 463 Supersaturation 582 Supersonic flow 567 Supplementary fired boilers 438 Surface condenser 685 Surface type heaters method 263 Surging 733 Swept volume 773 Symmetrical blades 626 T Tandem compound engines 527 Tangential flow turbine 616 Tangential force 626 Terminal temperature difference 262 Theoretical air 399 Thermal efficiency 250, 524, 862 Three stage 508 Three stage compressor 707 Throat 573 Throttle governing 508, 525, 533, 650, 654 Throttle valve 614 Throttling process 806 Throw of eccentric 513 Thrust power 861 Ton 808 Top dead centre 773 Top dead centre (TDC) 708 Trip mechanism 614 Triple cross flow compound turbine with double flow 615 Turbofan engines 860, 867 Turbojet engine with afterburner 868 Turbojet engines 860, 863 Turbojet with afterburner 860 Turboprop engine 868 Turboprop or propjet 860 Turning gear 614 Two stage 508 Two stage compressor 707 Two stroke engines 771 948 ________________________________________________________ Applied Thermodynamics U Ultimate analysis 401 Under-expanding 577 Unfired boilers 438 Uniflow Engine 535 Unsaturated organic compounds 826 V V-shaped cylinder arrangement 771 Vacuum efficiency 693 Valves 772 Vaned type compressors 707, 723 Vapour compression cycles 813 Vapour power cycle 250 Vapour refrigeration system 807 Velocity compounded impulse turbine 620, 621 Velocity diagram 623, 730 Velox boiler 452 Vertical boiler 437 Volume handled 476 Volumetric and gravimetric analysis 400 Volumetric efficiency 714 W Waste geat boiler 456 Water cooled engines 771 Water level indicator 459 Water tube boilers 437, 438 Wet and dry analysis of combustion 400 Wet bulb temperature 829 Wet compression 814 Whirl velocity 624 Windage loss 645 Winter air conditioning systems 835, 836 Woolf compound engines 527 Work ratio 251 Z Zeroth law 40 opportunity arose to establish experimentally the existence of the all-pervasive ether by measuring the speed of light relative to Earth as Earth moved relative to the ether at speed v, as would be predicted by Equation 1-3. The value of c was given by the Maxwell equations, and the speed of Earth relative to the ether, while not known, was assumed to be at least equal to its orbital speed around the Sun, about 30 km s. Since the maximum observable effect is of the order and given this assumption v2>c2 108, an experimental accuracy of about 1 part in 108 is necessary in order to detect Earth\u2019s motion relative to the ether. With a single exception, equipment and v2>c2 c > y q S y1 y\u00b4 S\u00b4 v x x\u00b4 z z\u00b4 y\u0153 Figure 1-4 The observers in S and S see identical electric fields 2k 1 from an infinitely long wire carrying uniform charge per unit length. Observers in both S and S measure a force 2kq y1 on q due to the line of charge; however, the S observer measures > v2q>(2y1) an additional force arising from the motion of 0 the wire in the x direction. Thus, the electromagnetic force does not have the same form in different inertial systems, implying that Maxwell\u2019s equations are not invariant under a Galilean transformation. due to the magnetic field at y1 at a distance y\u0153 1 > y1 Classical Concept Review The concepts of classical relativity, frames of reference, and coordinate transformations \u2014 all important background to our discussions of special relativity \u2014 may not have been emphasized in many introductory courses. As an aid to a better understanding of the concepts of modern physics, we have included the Classical Concept Review on the book\u2019s Web site. As you proceed through Modern Physics, the icon in the margin will alert you to potentially helpful classical background pertinent to the adjacent topics. 1-1 The Experimental Basis of Relativity 7 techniques available at the time had an experimental accuracy of only about 1 part in 104, woefully insufficient to detect the predicted small effect. That single exception was the experiment of Michelson and Morley.5 Questions 1. What would the relative velocity of the inertial systems in Figure 1-4 need to be in order for the S observer to measure no net electromagnetic force on the charge q? 2. Discuss why the very large value for the speed of the electromagnetic waves would imply that the ether be rigid, i.e., have a large bulk modulus. The Michelson-Morley Experiment All waves that were known to", " nineteenth-century scientists required a medium in order to propagate. Surface waves moving across the ocean obviously require the water. Similarly, waves move along a plucked guitar string, across the surface of a struck drumhead, through Earth after an earthquake, and, indeed, in all materials acted upon by suitable forces. The speed of the waves depends on the properties of the medium and is derived relative to the medium. For example, the speed of sound waves in air, i.e., their absolute motion relative to still air, can be measured. The Doppler effect for sound in air depends not only on the relative motion of the source and listener, but also on the motion of each relative to still air. Thus, it was natural for scientists of that time to expect the existence of some material like the ether to support the propagation of light and other electromagnetic waves and to expect that the absolute motion of Earth through the ether should be detectable, despite the fact that the ether had not been observed previously. Michelson realized that although the effect of Earth\u2019s motion on the results of any \u201cout-and\u2013back\u201d speed of light measurement, such as shown generically in Figure 1-5, would be too small to measure directly, it should be possible to measure v2 c2 by a difference measurement, using the interference property of the light waves as a sensitive \u201cclock.\u201d The apparatus that he designed to make the measurement is called the Michelson interferometer. The purpose of the Michelson-Morley experiment was to measure the speed of light relative to the interferometer (i.e., relative to Earth), thereby detecting Earth\u2019s motion through the ether and thus verifying the latter\u2019s existence. To illustrate how the interferometer works and the reasoning behind the experiment, let us first describe an analogous situation set in more familiar surroundings. > Light source Mirror Observer Figure 1-5 Light source, mirror, and observer are moving with speed v relative to the ether. According to classical theory, the speed of light c, relative to the ether, would be c v relative to the observer for light moving from the source toward the mirror and c v for light reflecting from the mirror back toward the source. Albert A. Michelson, here playing pool in his later years, made the first accurate measurement of the speed of light while an instructor at the U.S. Naval Academy, where he had earlier been a cadet. [AIP Emilio Segr\u00e8 Visual Archives.] 8 Chapter 1 Relativity I (a) (b) L B A v 1 2 Ground River v L C Ground c c2 \u2013 v2 c c2 \u2013 v2 A\u2192B v B\u2192A Figure 1-6 (a) The rowers both row at speed c in still water. (See Example 1-1.) The current in the river moves at speed v. Rower 1 goes from A to B and back to A, while rower 2 goes from A to C and back to A. (b) Rower 1 must point the bow upstream so that the sum of the velocity vectors c v results in the boat moving from A directly to B. His speed relative to the banks (i.e., points A and B) is then The same is true on the return trip. (c2 v2)1>2. EXAMPLE 1-1 A Boat Race Two equally matched rowers race each other over courses as shown in Figure 1-6a. Each oarsman rows at speed c in still water; the current in the river moves at speed v. Boat 1 goes from A to B, a distance L, and back. Boat 2 goes from A to C, also a distance L, and back. A, B, and C are marks on the riverbank. Which boat wins the race, or is it a tie? (Assume c v.) SOLUTION The winner is, of course, the boat that makes the round trip in the shortest time, so to discover which boat wins, we compute the time for each. Using the classical velocity transformation (Equations 1-3), the speed of 1 relative to the ground is (c2 v2)1>2, as shown in Figure 1-6b; thus the round-trip time t1 for boat 1 is t1 tASB tBSA 2L c A1 v2 c2 2L c L 2c2 v2 a1 v2 c2 1/2 b L 2c2 v2 2L c 2L 2c2 v2 a1 1 2 v2 c2 \u00c1 b 1-4 where we have used the binomial expansion. Boat 2 moves downstream at speed c v, also relative to the ground. The round-trip time t2 is thus relative to the ground and returns at c v t2 L c v L c v 2L c 1 1 v2 c2 2L c 2Lc c2 v2 a1 v", "2 c2 \u00c1 b 1-5 1-1 The Experimental Basis of Relativity 9 which, you may note, is the same result obtained in our discussion of the speed of light experiment in the Classical Concept Review. The difference \u00a2t between the round-trip times of the boats is then a1 v2 c2 b 2L c a1 1 2 2L c \u00a2t t2 Lv2 c3 t1 v2 c2 b 1-6 The quantity Lv2>c3 is always positive; therefore, t2 t1 and rower 1 has the faster average speed and wins the race. The Results Michelson and Morley carried out the experiment in 1887, repeating with a much-improved interferometer an inconclusive experiment that Michelson alone had performed in 1881 in Potsdam. The path length L on the new interferometer (Figure 1-7) was about 11 meters, obtained by a series of multiple reflections. Michelson\u2019s interferometer is shown schematically in Figure 1-8a. The field of view seen by the observer consists of parallel alternately bright and dark interference bands, called fringes, as illustrated in Figure 1-8b. The two light beams in the interferometer are exactly analogous to the two boats in Example 1-1, and Earth\u2019s motion through the ether was expected to introduce a time (phase) difference as given by Light source Mirrors Adjustable mirror Silvered glass plate Unsilvered glass plate Mirrors Mirrors Telescope 1 2 3 5 4 Figure 1-7 Drawing of Michelson-Morley apparatus used in their 1887 experiment. The optical parts were mounted on a 5 ft square sandstone slab, which was floated in mercury, thereby reducing the strains and vibrations during rotation that had affected the earlier experiments. Observations could be made in all directions by rotating the apparatus in the horizontal plane. [From R. S. Shankland, \u201cThe Michelson-Morley Experiment,\u201d Copyright \u00a9 November 1964 by Scientific American, Inc. All rights reserved.] 10 Chapter 1 Relativity I (a) (b) 1 Fringe width B M1 1 M\u00b42 L v Rotation Beam splitter Compensator Sodium light source (diffuse) A O 2 L M2 C Figure 1-8 Michelson interferometer. (a) Yellow light from the sodium source is divided into two beams by the second surface of the partially reflective beam splitter at A, at which point the two beams are exactly in phase. The beams travel along the mutually perpendicular paths 1 and 2, reflect from mirrors M1 and M2, and return to A, where they recombine and are viewed by the observer. The compensator\u2019s purpose is to make the two paths of equal optical length, so that the lengths L contain the same number of light waves, by making both beams pass through two thicknesses of glass before recombining. M2 is then tilted slightly so that it is not quite perpendicular to M1. Thus, the observer O sees M1 and splitter, forming a thin wedge-shaped film of air between them. The interference of the two recombining beams depends on the number of waves in each path, which in turn depends on (1) the length of each path and (2) the speed of light (relative to the instrument) in each M\u0153 path. Regardless of the value of that speed, the wedge-shaped air film between M1 and 2 results in an increasing path length for beam 2 relative to beam 1, looking from left to right across the observer\u2019s field of view; hence, the observer sees a series of parallel interference fringes as in (b), alternately yellow and black from constructive and destructive interference, respectively. the image of M2 formed by the partially reflecting second surface of the beam M\u0153 2, Equation 1-6. Rotating the interferometer through 90\u00b0 doubles the time difference and changes the phase, causing the fringe pattern to shift by an amount \u00a2N. An improved system for rotating the apparatus was used in which the massive stone slab on which the interferometer was mounted floated on a pool of mercury. This dampened vibrations and enabled the experimenters to rotate the interferometer without introducing mechanical strains, both of which would cause changes in L and hence a shift in the fringes. Using a sodium light source with 590 nm and assuming v 30 km s (i.e., Earth\u2019s orbital speed), \u00a2N was expected to be about 0.4 of the width of a fringe, about 40 times the minimum shift (0.01 fringe) that the interferometer was capable of detecting. > 1-2 Einstein\u2019s Postulates 11 To Michelson\u2019s immense disappointment and that of most scientists of the time, the expected shift in the fringes did not occur. Instead, the shift observed was only about 0.01 fringe, i", ".e., approximately the experimental uncertainty of the apparatus. With characteristic reserve, Michelson described the results thus:6 The actual displacement [of the fringes] was certainly less than the twentieth part [of 0.4 fringe], and probably less than the fortieth part. But since the displacement is proportional to the square of the velocity, the relative velocity of the earth and the ether is probably less than one-sixth the earth\u2019s orbital velocity and certainly less than one-fourth. Michelson interferometers with arms as long as 4 km are currently being used in the search for gravity waves. See Section 2-5. > Michelson and Morley had placed an upper limit on Earth\u2019s motion relative to the ether of about 5 km s. From this distance in time it is difficult for us to appreciate the devastating impact of this result. The then-accepted theory of light propagation could not be correct, and the ether as a favored frame of reference for Maxwell\u2019s equations was not tenable. The experiment was repeated by a number of people more than a dozen times under various conditions and with improved precision, and no shift has ever been found. In the most precise attempt, the upper limit on the relative velocity was lowered to 1.5 km s by Georg Joos in 1930 using an interferometer with light paths much longer than Michelson\u2019s. Recent, high-precision variations of the experiment using laser beams have lowered the upper limit to 15 m s. > > More generally, on the basis of this and other experiments, we must conclude that Maxwell\u2019s equations are correct and that the speed of electromagnetic radiation is the same in all inertial reference systems independent of the motion of the source relative to the observer. This invariance of the speed of light between inertial reference frames means that there must be some relativity principle that applies to electromagnetism as well as to mechanics. That principle cannot be Newtonian relativity, which implies the dependence of the speed of light on the relative motion of the source and observer. It follows that the Galilean transformation of coordinates between inertial frames cannot be correct but must be replaced with a new coordinate transformation whose application preserves the invariance of the laws of electromagnetism. We then expect that the fundamental laws of mechanics, which were consistent with the old Galilean transformation, will require modification in order to be invariant under the new transformation. The theoretical derivation of that new transformation was a cornerstone of Einstein\u2019s development of special relativity. More A more complete description of the Michelson-Morley experiment, its interpretation, and the results of very recent versions can be found on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Figures 1-9 through 1-11 here, as well as Equations 1-7 through 1-10. 1-2 Einstein\u2019s Postulates In 1905, at the age of 26, Albert Einstein published several papers, among which was one on the electrodynamics of moving bodies.11 In this paper, he postulated a more general principle of relativity that applied to the laws of both electrodynamics and mechanics. A consequence of this postulate is that absolute motion cannot be detected 12 Chapter 1 Relativity I by any experiment. We can then consider the Michelson apparatus and Earth to be at rest. No fringe shift is expected when the interferometer is rotated 90\u00b0, since all directions are equivalent. The null result of the Michelson-Morley experiment is therefore to be expected. It should be pointed out that Einstein did not set out to explain the Michelson-Morley experiment. His theory arose from his considerations of the theory of electricity and magnetism and the unusual property of electromagnetic waves that they propagate in a vacuum. In his first paper, which contains the complete theory of special relativity, he made only a passing reference to the experimental attempts to detect Earth\u2019s motion through the ether, and in later years he could not recall whether he was aware of the details of the Michelson-Morley experiment before he published his theory. The theory of special relativity was derived from two postulates proposed by Einstein in his 1905 paper: Postulate 1. The laws of physics are the same in all inertial reference frames. Postulate 2. The speed of light in a vacuum is equal to the value c, independent of the motion of the source. Postulate 1 is an extension of the Newtonian principle of relativity to include all types of physical measurements (not just measurements in mechanics). It implies that no inertial system is preferred over any other; hence, absolute motion cannot be detected. Postulate 2 describes a common property of all waves. For example, the speed of sound waves does not depend on the motion of the sound source. When an approaching car sounds its horn, the frequency heard increases according to the Doppler effect, but the speed of the waves traveling through the air does not depend on the speed of the car", ". The speed of the waves depends only on the properties of the air, such as its temperature. The force of this postulate was to include light waves, for which experiments had found no propagation medium, together with all other waves, whose speed was known to be independent of the speed of the source. Recent analysis of the light curves of gamma-ray bursts that occur near the edge of the observable universe have shown the speed of light to be independent of the speed of the source to a precision of one part in 1020. > Although each postulate seems quite reasonable, many of the implications of the two together are surprising and seem to contradict common sense. One important implication of these postulates is that every observer measures the same value for the speed of light independent of the relative motion of the source and observer. Consider a light source S and two observers R1, at rest relative to S, and R2, moving toward S with speed v, as shown in Figure 1-12a. The speed of light measured by R1 is c 3 108 m s. What is the speed measured by R2? The answer is not c v, as one would expect based on Newtonian relativity. By postulate 1, Figure 1-12a is equivalent to Figure 1-12b, in which R2 is at rest and the source S and R1 are moving with speed v. That is, since absolute motion cannot be detected, it is not possible to say which is really moving and which is at rest. By postulate 2, the speed of light from a moving source is independent of the motion of the source. Thus, looking at Figure 1-12b, we see that R2 measures the speed of light to be c, just as R1 does. This result, that all observers measure the same value c for the speed of light, is often considered an alternative to Einstein\u2019s second postulate. This result contradicts our intuition. Our intuitive ideas about relative velocities are approximations that hold only when the speeds are very small compared with the speed of light. Even in an airplane moving at the speed of sound, it is not possible to measure the speed of light accurately enough to distinguish the difference between the results c and c v, where v is the speed of the plane. In order to make such a (a) (b) S S v R1 R2 v R1 v R2 Figure 1-12 (a) Stationary light source S and a stationary observer R1, with a second observer R2 moving toward the source with speed v. (b) In the reference frame in which the observer R2 is at rest, the light source S and observer R1 move to the right with speed v. If absolute motion cannot be detected, the two views are equivalent. Since the speed of light does not depend on the motion of the source, observer R2 measures the same value for that speed as observer R1. 1-2 Einstein\u2019s Postulates 13 distinction, we must either move with a very great velocity (much greater than that of sound) or make extremely accurate measurements, as in the Michelson-Morley experiment, and when we do, we will find, as Einstein pointed out in his original relativity paper, that the contradictions are \u201conly apparently irreconcilable.\u201d Events and Observers In considering the consequences of Einstein\u2019s postulates in greater depth, i.e., in developing the theory of special relativity, we need to be certain that meanings of some important terms are crystal clear. First, there is the concept of an event. A physical event is something that happens, like the closing of a door, a lightning strike, the collision of two particles, your birth, or the explosion of a star. Every event occurs at some point in space and at some instant in time, but it is very important to recognize that events are independent of the particular inertial reference frame that we might use to describe them. Events do not \u201cbelong\u201d to any reference frame. Events are described by observers who do belong to particular inertial frames of reference. Observers could be people (as in Section 1-1), electronic instruments, or other suitable recorders, but for our discussions in special relativity we are going to be very specific. Strictly speaking, the observer will be an array of recording clocks located throughout the inertial reference system. It may be helpful for you to think of the observer as a person who goes around reading out the memories of the recording clocks or receives records that have been transmitted from distant clocks, but always keep in mind that in reporting events, such a person is strictly limited to summarizing the data collected from the clock memories. The travel time of light precludes him from including in his report distant events that he may have seen by eye! It is in this sense that we will be using the word observer in our discussions. Each inertial reference frame may be thought of as being formed by a cubic th", "reedimensional lattice made of identical measuring rods (e.g., meter sticks) with a recording clock at each intersection as illustrated in Figure 1-13. The clocks are all identical, and we, of course, want them all to read the \u201csame time\u201d as one another at any instant; i.e., they must be synchronized. There are many ways to accomplish synchronization of the clocks, but a very straightforward way, made possible by the second postulate, is to use one of the clocks in the lattice as a standard, or reference clock. For convenience we will also use the location of the reference clock in the lattice as the coordinate origin for the reference frame. The reference clock is started with its indicator (hands, pointer, digital display) set at zero. At the instant it starts, it also sends out a flash of light that spreads out as a spherical wave in all directions. When the flash from the reference clock reaches the lattice clocks 1 meter away (notice that in Figure 1-13 there are six of them, two of which are off the edges of the figure), we want their indicators to read the time required for light to travel 1 m ( 1 299,792,458 s). This can be done simply by having an observer at each clock set that time on the indicator and then having the flash from the reference clock start them as it passes. The clocks 1 m from the origin now display the same time as the reference clock; i.e., they are all synchronized. In a similar fashion, all of the clocks throughout the inertial frame can be synchronized since the distance of any clock from the reference clock can be calculated from the space coordinates of its position in the lattice and the initial setting of its indicator will be the corresponding travel time for the reference light flash. This procedure can be used to synchronize the clocks in any inertial frame, but it does not synchronize the clocks in reference frames that move with respect to one another. Indeed, as we shall see shortly, clocks in relatively moving frames cannot in general be synchronized with one another. > (Top) Albert Einstein in 1905 at the Bern, Switzerland, patent office. [Hebrew University of Jerusalem Albert Einstein Archives, courtesy AIP Emilio Segr\u00e8 Visual Archives.] (Bottom) Clock tower and electric trolley in Bern on Kramstrasse, the street on which Einstein lived. If you are on the trolley moving away from the clock and look back at it, the light you see must catch up with you. If you move at nearly the speed of light, the clock you see will be slow. In this, Einstein saw a clue to the variability of time itself. [Underwood & Underwood/CORBIS.] 14 Chapter 1 Relativity I Figure 1-13 Inertial reference frame formed from a lattice of measuring rods with a clock at each intersection. The clocks are all synchronized using a reference clock. In this diagram the measuring rods are shown to be 1 m long, but they could all be 1 cm, 1 or 1 km as required by the scale and precision of the measurements being considered. The three space dimensions are the clock positions. The fourth spacetime dimension, time, is shown by indicator readings on the clocks. m, y Reference clock x z When an event occurs, its location and time are recorded instantly by the nearest clock. Suppose that an atom located at x 2 m, y 3 m, z 4 m in Figure 1-13 emits a tiny flash of light at t 21 s on the clock at that location. That event is recorded in space and in time or, as we will henceforth refer to it, in the spacetime coordinate system with the numbers (2,3,4,21). The observer may read out and analyze these data at his leisure, within the limits set by the information transmission time (i.e., the light travel time) from distant clocks. For example, the path of a particle moving through the lattice is revealed by analysis of the records showing the particle\u2019s time of passage at each clock\u2019s location. Distances between successive locations and the corresponding time differences make possible the determination of the particle\u2019s velocity. Similar records of the spacetime coordinates of the particle\u2019s path can, of course, also be made in any inertial frame moving relative to ours, but to compare the distances and time intervals measured in the two frames requires that we consider carefully the relativity of simultaneity. Relativity of Simultaneity Einstein\u2019s postulates lead to a number of predictions about measurements made by observers in inertial frames moving relative to one another that initially seem very strange, including some that appear paradoxical. Even so, these predictions have been experimentally verified; and nearly without exception, every paradox is resolved by an understanding of the relativity of simultaneity, which states that Two spatially separated events simultaneous in one reference frame are not, in general, simultaneous in another inertial frame moving relative to the first. 1", "-2 Einstein\u2019s Postulates 15 A corollary to this is that Clocks synchronized in one reference frame are not, in general, synchronized in another inertial frame moving relative to the first. What do we mean by simultaneous events? Suppose two observers, both in the inertial frame S at different locations A and B, agree to explode bombs at time to (remember, we have synchronized all of the clocks in S). The clock at C, equidistant from A and B, will record the arrival of light from the explosions at the same instant, i.e., simultaneously. Other clocks in S will record the arrival of light from A or B first, depending on their locations, but after correcting for the time the light takes to reach each clock, the data recorded by each would lead an observer to conclude that the explosions were simultaneous. We will thus define two events to be simultaneous in an inertial reference frame if the light signals from the events reach an observer halfway between them at the same time as recorded by a clock at that location, called a local clock. Einstein\u2019s Example To show that two events that are simultaneous in frame S are not simultaneous in another frame S moving relative to S, we use an example introduced by Einstein. A train is moving with speed v past a station platform. We have observers located at A, B, and C at the front, back, and middle of the train. (We consider the train to be at rest in S and the platform in S.) We now suppose that the train and platform are struck by lightning at the front and back of the train and that the lightning bolts are simultaneous in the frame of the platform (S; Figure 1-14a). That is, an observer located at C halfway between positions A and B, where lightning strikes, observes the two flashes at the same time. It is convenient to suppose that the (a) (b) (c) (d\u00b4 B\u00b4 v C\u00b4 C A\u00b4 A S\u00b4 B\u00b4 C\u00b4 A\u00b4 v C A S\u00b4 B\u00b4 C\u00b4 A\u00b4 v C S\u00b4 B\u00b4 C A C\u00b4 A A\u00b4 v Figure 1-14 Lightning bolts strike the front and rear of the train, scorching both the train and the platform, as the train (frame S) moves past the platform (system S) at speed v. (a) The strikes are simultaneous in S, reaching the C observer located midway between the events at the same instant as recorded by the clock at C as shown in (c). In S the flash from the front of the train is recorded by the C clock, located midway between the scorch marks on the train, before that from the rear of the train (b and d, respectively). Thus, the C observer concludes that the strikes were not simultaneous. 16 Chapter 1 Relativity I lightning scorches both the train and the platform so that the events can be easily located in each reference frame. Since C is in the middle of the train, halfway between the places on the train that are scorched, the events are simultaneous in S only if the clock at C records the flashes at the same time. However, the clock at C records the flash from the front of the train before the flash from the back. In frame S, when the light from the front flash reaches the observer at C, the train has moved some distance toward A, so that the flash from the back has not yet reached C, as indicated in Figure 1-14b. The observer at C must therefore conclude that the events are not simultaneous, but that the front of the train was struck before the back. Figures 1-14c and 1-14d illustrate, respectively, the subsequent simultaneous arrival of the flashes at C and the still-later arrival of the flash from the rear of the train at C. As we have discussed, all observers in S on the train will agree with the observer C when they have corrected for the time it takes light to reach them. The corollary can also be demonstrated with a similar example. Again consider the train to be at rest in S that moves past the platform, at rest in S, with speed v. Figure 1-15 shows three of the clocks in the S lattice and three of those in the S lattice. The clocks in each system\u2019s lattice have been synchronized in the manner that was described earlier, but those in S are not synchronized with those in S. The observer at C midway between A and B on the platform announces that light sources at A and B will flash when the clocks at those locations read to (Figure 1-15a). The observer at C, positioned midway between A and B, notes the arrival of the light flash from the front of the train (Figure 1-15b) before the arrival of the one from the rear (Figure 1-15d). Observer C thus concludes that if the flashes were each emitted at to on the local clocks, as announced, then the clocks at A and B", " are not synchronized. All observers in S would agree with that conclusion after correcting for the time of light travel. The clock located at C records the arrival of the two flashes simultaneously, of course, since the clocks in S are synchronized (Figure 1-15c). (a) S\u00b4 S S S S (b) (c) (d) B\u00b4 B S\u00b4 B B B C\u00b4 C v A\u00b4 A B\u00b4 C\u00b4 A\u00b4 v C A S\u00b4 B\u00b4 C\u00b4 A\u00b4 v S\u00b4 C B\u00b4 C A C\u00b4 A A\u00b4 v Figure 1-15 (a) Light flashes originate simultaneously at clocks A and B, synchronized in S. (b) The clock at C, midway between A and B on the moving train, records the arrival of the flash from A before the flash from B shown in (d). Since the observer in S announced that the flashes were triggered at to on the local clocks, the observer at C concludes that the local clocks at A and B did not read to simultaneously; i.e., they were not synchronized. The simultaneous arrival of the flashes at C is shown in (c). (a) Earth view of Earth clocks (b) Spaceship view of spaceship clocks 1-3 The Lorentz Transformation 17 B A B A (c) Earth view of spaceship clocks (d) Spaceship view of Earth clocks B A B A v v Figure 1-16 A light flash occurs on Earth midway between two Earth clocks. At the instant of the flash the midpoint of a passing spaceship coincides with the light source. (a) The Earth clocks record the lights\u2019 arrival simultaneously and are thus synchronized. (b) Clocks at both ends of the spaceship also record the lights\u2019 arrival simultaneously (Einstein\u2019s second postulate) and they, too, are synchronized. (c) However, the Earth observer sees the light reach the clock at B before the light reaches the clock at A. Since the spaceship clocks read the same time when the light arrives, the Earth observer concludes that the clocks at A and B are not synchronized. (d) The spaceship observer similarly concludes that the Earth clocks are not synchronized. Notice, too, in Figure 1-15 that C also concludes that the clock at A is ahead of the clock at B. This is important, and we will return to it in more detail in the next section. Figure 1-16 illustrates the relativity of simultaneity from a different perspective. Questions 3. In addition to that described above, what would be another possible method of synchronizing all of the clocks in an inertial reference system? 4. Using Figure 1-16d, explain how the spaceship observer concludes that Earth clocks are not synchronized. 1-3 The Lorentz Transformation We now consider a very important consequence of Einstein\u2019s postulates, the general relation between the spacetime coordinates x, y, z, and t of an event as seen in reference frame S and the coordinates x, y, z, and t of the same event as seen in reference frame S, which is moving with uniform velocity relative to S. For simplicity we will 18 Chapter 1 Relativity I Figure 1-17 Two inertial frames S and S with the latter moving at speed v in x direction of system S. Each set of axes shown is simply the coordinate axes of a lattice like that in Figure 1-13. Remember, there is a clock at each intersection. A short time before, the times represented by this diagram O and O were coincident and the lattices of S and S were intermeshed. y S O z (xa, ta ) x O\u00b4 z\u00b4 y\u00b4 S\u00b4 v (xb, tb ) x\u00b4 consider only the special case in which the origins of the two coordinate systems are coincident at time t t 0 and S is moving, relative to S, with speed v along the x (or x) axis and with the y and z axes parallel, respectively, to the y and z axes, as shown in Figure 1-17. As we discussed earlier (Equation 1-2), the classical Galilean coordinate transformation is x x vt y y z z t t 1-2 which expresses coordinate measurements made by an observer in S in terms of those measured by an observer in S. The inverse transformation is x x vt y y z z t t and simply reflects the fact that the sign of the relative velocity of the reference frames is different for the two observers. The corresponding classical velocity transformation was given in Equation 1-3 and the acceleration, as we saw earlier, is invariant under a Galilean transformation. (For the rest of the discussion we will ignore the equations for y and z, which do not change in this special case of motion along the x and x axes.) These equations are consistent with experiment as long as v is much less than c. It should be clear that the classical velocity transformation is not consistent with the Einstein postulates of special relativity. If light moves", " along the x axis with speed \u0153 c c in S, Equation 1-3 implies that the speed in S is ux. The Galilean transformation equations must therefore be modified to be consistent with Einstein\u2019s postulates, but the result must reduce to the classical equations when v is much less than c. We will give a brief outline of one method of obtaining the relativistic transformation that is called the Lorentz transformation, so named because of its original discovery by H. A. Lorentz.12 We assume the equation for x to be of the form \u0153 c v ux rather than x (x vt) 1-11 where is a constant that can depend upon v and c but not on the coordinates. If this equation is to reduce to the classical one, must approach 1 as v c approaches 0. The inverse transformation must look the same except for the sign of the velocity: > x (x vt) 1-12 With the arrangement of the axes in Figure 1-17, there is no relative motion of the frames in the y and z directions; hence y y and z z. However, insertion of the as yet unknown multiplier modifies the classical transformation of time, t t. To see 1-3 The Lorentz Transformation 19 this, we substitute x from Equation 1-11 into Equation 1-12 and solve for t. The result is t ct (1 2) 2 d x v 1-13 Now let a flash of light start from the origin of S at t 0. Since we have assumed that the origins coincide at t t 0, the flash also starts at the origin of S at t 0. The flash expands from both origins as a spherical wave. The equation for the wave front according to an observer in S is x2 y2 z2 c2t2 and according to an observer in S, it is x2 y2 z2 c2t2 1-14 1-15 where both equations are consistent with the second postulate. Consistency with the first postulate means that the relativistic transformation that we seek must transform Equation 1-14 into Equation 1-15 and vice versa. For example, substituting Equations 1-11 and 1-13 into 1-15 results in Equation 1-14 if 1 A1 v2 c2 1 21 2 1-16 where v c. Notice that 1 for v 0 and is illustrated in Example 1-2 below. > S for v c. How this is done EXAMPLE 1-2 Relativistic Transformation Multiplier Show that must be given by Equation 1-16 if Equation 1-15 is to be transformed into Equation 1-14 consistent with Einstein\u2019s first postulate. SOLUTION Substituting Equations 1-11 and 1-13 into Equation 1-15 and noting that z z in this case yields y y 2(x vt)2 y2 z2 c22 ct 1 2 2 2 d x v and 1-17 To be consistent with the first postulate, Equation 1-15 must be identical to term in Equation 1-17 be equal to 1-12. This requires that the coefficient of the, and that of the xt term be equal to 0. Any of 1, that of the those conditions can be used to determine, and all yield the same result. Using, for example, the coefficient of, we have from Equation 1-17 that term be equal to x2 x2 c2 t2 2 c22 (1 2)2 4v2 1 20 Chapter 1 Relativity I which can be rearranged to c2 (1 2)2 2v2 (1 2) Canceling 1 2 on both sides and solving for yields 1 A1 v2 c2 With the value for found in Example 1-2, Equation 1-13 can be written in a somewhat simpler form, and with it the complete Lorentz transformation becomes and the inverse with x (x vt) t at vx c2 b y y z z x (x vt) t at vx c2 b y y z z 1 21 2 1-18 1-19 EXAMPLE 1-3 Transformation of Time Intervals The arrivals of two cosmic-ray leptons (muons) are recorded by detectors in the laboratory, one at time at locain the laboratory reference frame, S tion in Figure 1-17. What is the time interval between those two events in system S, which moves relative to S at speed v? and the second at time at location xa xb ta tb SOLUTION Applying the time coordinate transformation from Equation 1-18, t\u0153 b t\u0153 a atb vxb c2 vxa c2 b t\u0153 b t\u0153 a (tb ta) xa) b ata v c2 (xb 1-20 We see that the time interval", " measured in S depends not just on the corresponding time interval in S, but also on the spatial separation of the clocks in S that measured the interval. This result should not come as a total surprise, since we have already discovered that although the clocks in S are synchronized with each other, they are not, in general, synchronized for observers in other inertial frames. 1-3 The Lorentz Transformation 21 ta) Special Case 1 If the two events happen to occur at the same location in S, i.e., then (tb, the time interval measured on a clock located at the events, is called the proper time interval. Notice that since 1 for all frames moving relative to S, the proper time interval is the minimum time interval that can be measured between those events. xb, xa Special Case 2 Does an inertial frame exist for which the events described above would be measured as being simultaneous? Since the question has been asked, you probably suspect that the answer is yes, and you are right. The two events will be simultaneous in a system S for which, i.e., when 0 t\ufb02 b t\ufb02 a or when (tb ta) v c2 (xb xa) v c a tb xb ta xa bc 1-21 xa)>c (xb Notice that time for a light beam to travel from thus we can characterize S as being that system whose speed relative to S is that fraction of c given by the time interval between the events divided by the travel time of light between them. (Note, too, that, a nonphysical situation that we will discuss in Section 1-4.) ta) (xb implies that xa) 1 c(tb to xb; xa While it is possible for us to get along in special relativity without the Lorentz transformation, it has an application that is quite valuable: it enables the spacetime coordinates of events measured by the measuring rods and clocks in the reference frame of one observer to be translated into the corresponding coordinates determined by the measuring rods and clocks of an observer in another inertial frame. As we will see in Section 1-4, such transformations lead to some startling results. Relativistic Velocity Transformations The transformation for velocities in special relativity can be obtained by differentiation of the Lorentz transformation, keeping in mind the definition of the velocity. Suppose a dx>dt, dy>dt, uy particle moves in S with velocity u whose components are dx\u0153>dt\u0153, u\u0153 An observer in S would measure the components and x u\u0153 u\u0153 z y Using the transformation equations, we obtain dz>dt. and dy\u0153>dt\u0153, dz\u0153>dt\u0153. ux uz dx (dx vdt) dt adt vdx c2 b dy dy dz dz 22 Chapter 1 Relativity I from which we see that u\u0153 x is given by u\u0153 x dx dt (dx vdt) adt vdx c2 b (dx>dt v) dx 1 v c2 dt or u\u0153 x ux 1 v vux c2 1-22 and, if a particle has velocity components in the y and z directions, it is not difficult to find the components in S in a similar manner. u\u0153 y uy a1 vux c2 b u\u0153 z uz a1 vuz c2 b Remember that this form of the velocity transformation is specific to the arrangec, i.e., when, the relativistic velocity transforms reduce to the classical velocity ment of the coordinate axes in Figure 1-17. Note, too, that when v v> c 0 addition of Equation 1-3. Likewise, the inverse velocity transformation is V u\u0153 x ux a1 v vu\u0153 x c2 u\u0153 y uy b a1 vu\u0153 x c2 b uz u\u0153 z a1 vu\u0153 x c2 b 1-23 EXAMPLE 1-4 Relative Speeds of Cosmic Rays Suppose that two cosmic ray protons approach Earth from opposite directions, as shown in Figure 1-18a. The speeds. What is Earth\u2019s relative to Earth are measured to be velocity relative to each proton, and what is the velocity of each proton relative to the other? 0.8c 0.6c and v2 v1 (a) v1 1 v2 Earth (b) S\u00b4 S 2 S\u00b4\u00b4 x\u00b4 x x\u00b4\u00b4 Figure 1-18 (a) Two cosmic ray protons approach Earth from opposite directions at speeds v1 and v2 with respect to Earth. (b) Attaching an inertial frame to each particle and Earth enables one to visualize the several relative speeds involved and apply the velocity transformation correctly. SOLUTION Consider each particle and Earth to be inertial reference frames S, S", ", and S with their 0.6c v1 respective x axes parallel as in Figure 1-18b. With this arrangement v\u0153 0.8c. Thus, the speed of Earth measured in S is and and Ex the speed of Earth measured in S is 0.8c. u1x 0.6c u2x v2 v\ufb02 Ex 1-3 The Lorentz Transformation 23 To find the speed of proton 2 with respect to proton 1, we apply Equation 1-22 i.e., the speed of particle 2 in S. Its speed in S has been measured to, where the S system has relative speed with respect to S. to compute u2x be Thus, substituting into Equation 1-22, we obtain u\u0153 2x, 0.8c 0.6c v1 u\u0153 2x 0.8c (0.6c) 1 (0.6c)(0.8c)>c2 1.4c 1.48 0.95c and the first proton measures the second to be approaching (moving in the direction) at 0.95c. The observer in S must of course make a consistent measurement, i.e., find the speed of proton 1 to be 0.95c in the x direction. This can be readily shown by a second application of Equation 1-22 to compute x u\ufb02 1x : u\ufb02 1x 0.6c (0.8c) 1 (0.6c)(0.8c)>c2 1.4c 1.48 0.95c Questions 5. The Lorentz transformation for y and z is the same as the classical result: y y and z z. Yet the relativistic velocity transformation does not give the classical result Explain. and uz u\u0153 z. uy u\u0153 y 6. Since the velocity components of a moving particle are different in relatively moving frames, the directions of the velocity vectors are also different in general. Explain why the fact that observers in S and S measure different directions for a particle\u2019s motion is not an inconsistency in their observations. Spacetime Diagrams The relativistic discovery that time intervals between events are not the same for all observers in different inertial reference frames underscores the four-dimensional character of spacetime. With the diagrams that we have used thus far, it is difficult to depict and visualize on the two-dimensional page events that occur at different times, since each diagram is equivalent to a snapshot of spacetime at a particular instant. Showing events as a function of time typically requires a series of diagrams, such as Figures 1-14, 1-15, and 1-16, but even then our attention tends to be drawn to the space coordinate systems rather than the events, whereas it is the events that are fundamental. This difficulty is removed in special relativity with a simple yet powerful graphing method called the spacetime diagram. (This is just a new name given to the t vs. x graphs that you first began to use when you discussed motion in introductory physics.) On the spacetime diagram we can graph both the space and time coordinates of many events in one or more inertial frames, albeit with one limitation. Since the page offers only two dimensions for graphing, we suppress, or ignore for now, two of the space dimensions, in particular y and z. With our choice of the relative motion of inertial frames along the x axis, y y and z z anyhow. (This is one of the reasons we made that convenient choice a few pages back, the other reason being mathematical simplicity.) This means that for the time being, we are limiting our attention to one space dimension and to time, i.e., to events that occur, regardless of 24 Chapter 1 Relativity I ct (m) 3 2 1 0 B \u20133 \u20132 \u20131 C \u20131 A D 1 2 3 x (m) Figure 1-19 Spacetime diagram for an inertial reference frame S. Two of the space dimensions (y and z) are suppressed. The units on both the space and time axes are the same, meters. A meter of time means the time required for light to travel one meter, i.e., 3.3 109 s. when, along one line in space. Should we need the other two dimensions, e.g., in a consideration of velocity vector transformations, we can always use the Lorentz transformation equations. In a spacetime diagram the space location of each event is plotted along the x axis horizontally and the time is plotted vertically. From the three-dimensional array of measuring rods and clocks in Figure 1-13, we will use only those located on the x axis, as in Figure 1-19. (See, things are simpler already!) Since events that exhibit relativistic effects generally occur at high speeds, it will be convenient to multiply the time scale by the speed of", " light (a constant), which enables us to use the same units and scale on both the space and time axes, e.g., meters of distance and meters of light travel time.13 The time axis is, therefore, c times the time t in seconds, i.e., ct. As we will see shortly, this choice prevents events from clustering about the axes and makes possible the straightforward addition of other inertial frames into the diagram. As time advances, notice that in Figure 1-19 each clock in the array moves vertically upward along the dotted lines. Thus, as events A, B, C, and D occur in spacetime, one of the clocks of the array is at (or very near) each event when it happens. Remember that the clocks in the reference frame are synchronized, and so the difference in the readings of clocks located at each event records the proper time interval between the events. (See Example 1-3.) In the figure, events A and D occur at the same place (x 2 m) but at different times. The time interval between them measured on clock 2 is the proper time interval since clock 2 is located at both events. Events A and B occur at different locations but at the same time (i.e., simultaneously in this frame). Event C occurred before the present since ct 1 m. For this discussion we will consider the time that the coordinate origins coincide,, to be the present. ct ct 0 Worldlines in Spacetime Particles moving in space trace out a line in the spacetime diagram called the worldline of the particle. The worldline is the \u201ctrajectory\u201d of the particle on a ct versus x graph. To illustrate, consider four particles moving in space (not spacetime) as shown in Figure 1-20a, which shows the array of synchronized clocks on the x axis and the space trajectories of four particles, each starting at x 0 and moving at some constant speed during 3 m of time. Figure 1-20b shows the worldline for each of the particles in spacetime. Notice that constant speed means that the worldline has constant slope; i.e., it is a straight line (slope \u00a2t \u00a2x 1 (\u00a2x \u00a2t) 1 speed). > > > > 1-3 The Lorentz Transformation 25 Figure 1-20 (a) The space trajectories of four particles with various constant speeds. Note that particle 1 has a speed of zero and particle 2 moves in the x direction. The worldlines of the particles are straight lines. (b) The worldline of particle 1 is also the ct axis since that particle remains at x 0. The constant slopes are a consequence of the constant speeds. (c) For accelerating particles 5 and 6 [not shown in (a)], the worldlines are curved, the slope at any point yielding the instantaneous speed. (a) (b) #2 \u20131 \u20131 #3 #4 #1 0 1 2 3 x (m) ct (m) #2 #1 #3 #m) (c) ct (m) #6 #5 3 2 1 0 \u20131 1 2 3 x (m) That was also the case when you first encountered elapsed time versus displacement graphs in introductory physics. Even then, you were plotting spacetime graphs and drawing worldlines! If the particle is accelerating \u2014 either speeding up like particle 5 in Figure 1-20c or slowing down like particle 6 \u2014 the worldlines are curved. Thus, the worldline is the record of the particle\u2019s travel through spacetime, giving its speed ( 1 slope) and acceleration ( 1 rate at which the slope changes) at every instant. > > EXAMPLE 1-5 Computing Speeds in Spacetime Find the speed u of particle 3 in Figure 1-20. SOLUTION The speed u \u00a2x \u00a2t 1 slope, where we have \u00a2ct c # \u00a2t 3.0 0 3.0 m (3.0>3.0 108) u 1.5 m>108 s 0.5c. 108 s and > > (from Figure 1-20). Thus, \u00a2x 1.5 0 1.5 m \u00a2t (3.0>c) and The speed of particle 4, computed as shown in Example 1-5, turns out to be c, the speed of light. (Particle 4 is a light pulse.) The slope of its worldline \u00a2(ct) \u00a2x x 3 m 3 m 1. Similarly, the slope of the worldline of a light pulse moving in the. Since relativity limits the speed of particles with mass to less than c, direction is 1 > > 26 Chapter 1 Relativity I Figure 1-21 The speed-of-light limit to the speeds of particles limits the slopes of worldlines for particles that move through x 0 at ct 0 to the shaded area of spacetime,", " i.e., to slopes 1 and 1. The dashed lines are worldlines of light flashes moving in the x and x directions. The curved worldline of the particle shown has the same limits at every instant. Notice that the particle\u2019s speed 1/slope. ct (m) 2 1 0 A 1 2 x (m) \u20132 \u20131 ct 2 m as we will see in Chapter 2, the slopes of worldlines for particles that move through x 0 at ct 0 are limited to the larger shaded triangle in Figure 1-21. The same limits to the slope apply at every point along a particle\u2019s worldline, such as point A on the curved spacetime trajectory in Figure 1-21. This means that the particle\u2019s possible must lie within the heavily shaded triangle. worldlines for times greater than Analyzing events and motion in inertial systems that are in relative motion can now be accomplished more easily than with diagrams such as Figures 1-14 through 1-18. Suppose we have two inertial frames S and S with S moving in the x direction of S at speed v as in those figures. The clocks in both systems are started at t t 0 (the present) as the two origins x 0 and x 0 coincide, and, as before, observers in each system have synchronized the clocks in their respective systems. The spacetime diagram for S is, of course, like that in Figure 1-19, but how does S appear in that diagram, i.e., with respect to an observer in S? Consider that as the origin of S (i.e., the point where x 0) moves in S, its worldline is the ct axis since the ct axis is the locus of all points with x 0 (just as the ct axis is the locus of points with x 0.) Thus, the slope of the ct axis as seen by an observer in S can be found from Equation 1-18, the Lorentz transformation, as follows: x (x vt) 0 for x 0 or and x vt (v c)(ct) ct > ct (1 )x > which says that the slope (in S) of the worldline of the point x 0, the ct axis, is 1. (See Figure 1-22a.) In the same manner, the x axis can be located using the fact that it is the locus of points for which ct 0. The Lorentz transformation once again provides the slope: > t at vx c2 b 0 or t vx c2 and ct v c x x Thus, the slope of the x axis as measured by an observer in S is, as shown in Figure 1-22a. Don\u2019t be confused by the fact that the x axes don\u2019t look parallel anymore. They are still parallel in space, but this is a spacetime diagram. It shows motion in both (a) ct\u00b4 (m) ct (m) 4 3 2 1 1-3 The Lorentz Transformation 27 Figure 1-22 Spacetime diagram of S showing S moving at speed v 0.5c in the x direction. The diagram is drawn with t t 0 when the origins of S and S coincided. The dashed line shows the worldline of a light flash that passed through the point x 0 at t 0 heading in the x direction. Its slope equals 1 in both S and S. The ct and x 1> 2 axes of S have slopes of 0.5, respectively. (a) Calibrating the axes of S as described in the text allows the grid of coordinates to be drawn on S. Interpretation is facilitated by remembering that (b) shows the system S as it is observed in the spacetime diagram of S. and x\u00b4 (m) \u20131 1 2 3 4 x (m) (b) \u20131 ct (m) ct\u00b4 (m) 4 4 3 2 1 0.866 \u20131 \u20131 \u20131 \u20131 3 2 1 2 1 1 2 0.866 x\u00b4 (m) 4 3 3 4 x (m) space and time. For example, the clock at x 1 m in Figure 1-22b passed the point x 0 at about ct 1.5 m as the x axis of S moved both upward and to the right in S. Remember, as time advances, the array of synchronized clocks and measuring rods that are the x axis also move upward, so that, for example, when ct 1, the origin of S (x 0, ct 0) has moved vt (v c)ct ct to the right along the x axis. > Question 7. Explain how the spacetime diagram in Figure 1-22b would appear drawn by an observer in S. 28 Chapter 1 Relativity I Figure 1-23 Spacetime equivalent of Figure 1-15, showing the spacetime diagram for", " the system S in which the platform is at rest. Measurements made by observers in S are read from the primed axes. ct ct\u00b4 2 Train (S\u00b4 frame) ct\u00b4 (B\u00b4) 1 x\u00b4 x A\u00b4 A Platform (S frame) 2 1 C\u00b4 C \u20131 \u20132 B B\u00b4 \u20131 ct\u00b4 (A\u00b4) \u20132 EXAMPLE 1-6 Simultaneity in Spacetime Use the train-platform example of Figure 1-15 and a suitable spacetime diagram to show that events simultaneous in one frame are not simultaneous in a frame moving relative to the first. (This is the corollary to the relativity of simultaneity that we first demonstrated in the previous section using Figure 1-15.) SOLUTION Suppose a train is passing a station platform at speed v and an observer C at the midpoint of the platform, system S, announces that light flashes will be emitted at clocks A and B located at opposite ends of the platform at t 0. Let the train, system S, be a rocket train with v 0.5c. As in the earlier discussion, clocks at C and C both read 0 as C passes C. Figure 1-23 shows this situation. It is the spacetime equivalent of Figure 1-15. Two events occur, the light flashes. The flashes are simultaneous in S since both occur at ct 0. In S, however, the event at A occurred at ct(A) (see Figure 1-23), about 1.2 ct units before ct 0, and the event at B occurred at ct(B), about 1.2 ct units after ct 0. Thus, the flashes are not simultaneous in S and A occurs before B, as we also saw in Figure 1-15. EXPLORING Calibrating the Spacetime Axes By calibrating the coordinate axes of S consistent with the Lorentz transformation, we will be able to read the coordinates of events and calculate space and time intervals between events as measured in both S and S directly from the diagram, in addition to calculating them from Equations 1-18 and 1-19. The calibration of the S axes is 1-4 Time Dilation and Length Contraction 29 straightforward and is accomplished as follows. The locus of points, e.g., with x 1 m, is a line parallel to the ct axis through the point x 1 m, ct 0, just as we saw earlier that the ct axis was the locus of those points with x 0 through the point x 0, ct 0. Substituting these values into the Lorentz transformation for x, we see that the line through x 1 m intercepts the x axis, i.e., the line where ct 0 at x (x vt) (x ct) 1 x or x 1> 21 2 1-24 or, in general, x x 21 2 In Figure 1-22b, where 0.5, the line x 1 m intercepts the x axis at x 0.866 m. Similarly, if x 2 m, x 1.73 m; if x 3 m, x 2.60 m; and so on. The ct axis is calibrated in a precisely equivalent manner. The locus of points with ct 1 m is a line parallel to the x axis through the point ct 1 m, x 0. Using the Lorentz transformation, the intercept of that line with the ct axis (where x 0) is found as follows: t (t vx>c2) which can also be written as ct (ct x) 1-25 # 1>2 1>2 or ct ct for x 0. Thus, for ct 1 m, we have 1 ct or ct (1 2) again in general, ct ct(1 2) and,. The x ct coordinate grid is shown in Figure 1-22b. Notice in Figure 1-22b that the clocks located in S are not found to be synchronized by observers in S, even though they are synchronized in S. This is exactly the conclusion that we arrived at in the discussion of the lightning striking the train and platform. In addition, those with positive x coordinates are behind the S reference clock and those with negative x coordinates are ahead, the differences being greatest for those clocks farthest away. This is a direct consequence of the Lorentz transformation of the time coordinate \u2014 i.e., when ct 0 in Equation 1-25, ct x. Note, too, that the slope of the worldline of the light beam equals 1 in S as well as in S, as required by the second postulate. 1-4 Time Dilation and Length Contraction The results of correct measurements of the time and space intervals between events do not depend upon the kind of apparatus used for the measurements or on the events themselves. We are free therefore to choose any events", " and measuring apparatus that will help us understand the application of the Einstein postulates to the results of measurements. As you have already seen from previous examples, convenient events in relativity are those that produce light flashes. A convenient, simple such clock is a light clock, pictured schematically in Figure 1-24. A photocell detects the light pulse and sends a voltage pulse to an oscilloscope, which produces a vertical deflection of the oscilloscope\u2019s trace. The phosphorescent material on the face of the oscilloscope tube gives a persistent light that can be observed visually, photographed, or recorded electronically. The time between two light flashes is determined by measuring the 30 Chapter 1 Relativity I Figure 1-24 Light clock for measuring time intervals. The time is measured by reading the distance between pulses on the oscilloscope after calibrating the sweep speed. L1 Detector Signal L2 L2 \u0394t Vertical input Time base Mirror L1 = ct1 2L2 = ct2 \u0394t = t2 \u2013 t1 = 2L2 \u2013 L1 \u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013c distance between pulses on the scope and knowing the sweep speed. Such clocks can easily be calibrated and compared with other types of clocks. Although not drawn as in Figure 1-24, the clocks used in explanations in this section may be thought of as light clocks. Time Dilation (or Time Stretching) We first consider an observer A at rest in frame S a distance D from a mirror, also in S, as shown in Figure 1-25a. She triggers a flash gun and measures the time interval \u00a2t between the original flash and the return flash from the mirror. Since light travels with speed c, this time is \u00a2t (2D) c. > Now consider these same two events, the original flash of light and the returning flash, as observed in reference frame S, with respect to which S is moving to the right with speed v. The events happen at two different places, in frame S because between the original flash and the return flash observer A has moved a horizontal distance v\u00a2t, where \u00a2t is the time interval between the events measured in S. and x2, x1 (a) y\u00b4 (b) y Mirror S\u00b4 A\u00b4 x1\u00b4 D A\u00b4 A\u00b4 A\u00b4 x\u00b4 S x1 (c) v x c \u0394t \u2013\u2013\u2013\u2013 2 v \u0394t \u2013\u2013\u2013\u2013 2 D Mirror x2 Figure 1-25 (a) Observer A and the mirror are in a spaceship at rest in frame S. The time it takes for the light pulse to reach the mirror and return is measured by A to be frame S, the spaceship is moving to the right with speed v. If the speed of light is the same in 2D>c both frames, the time it takes for the light to reach the mirror and return is longer than S because the distance traveled is greater than 2D. (c) A right triangle for computing the time t in frame S. 2D>c. (b) In in 1-4 Time Dilation and Length Contraction 31 In Figure 1-25b, a space diagram, we see that the path traveled by the light is longer in S than in S. However, by Einstein\u2019s postulates, light travels with the same speed c in frame S as it does in frame S. Since it travels farther in S at the same speed, it takes longer in S to reach the mirror and return. The time interval between flashes in S is thus longer than it is in S. We can easily calculate \u00a2t in terms of \u00a2t. From the triangle in Figure 1-25c, we see that or 2 b a c\u00a2t 2 D2 a v\u00a2t 2 2 b \u00a2t 2D 2c2 v2 2D c 1 21 v2>c2 Using \u00a2t 2D c, we have > \u00a2t \u00a2t 21 v2>c2 \u00a2t 1-26 where \u00a2t is the proper time interval that we first encountered in Example 1-3. Equation 1-26 describes time dilation; i.e., it tells us that the observer in frame S always measures the time interval between two events to be longer (since 1) than the corresponding interval measured on the clock located at both events in the frame where they occur at the same location. Thus, observers in S conclude that the clock at A in S runs slow since that clock measures a smaller time interval between the two events. Notice that the faster S moves with respect to S, the larger is, and the slower the S clocks will tick. It appears to the S observer that time is being stretched out in S. Be careful! The same clock must be located at each event for \u00a2t to be the proper time interval. We can see why this is true by noting that Equation 1-26 can be obtained directly from", " the inverse Lorentz transformation for t. Referring again to Figure 1-25 and calling the emission of the flash event 1 and its return event 2, we have that \u00a2t t2 t1 at\u0153 2 vx\u0153 2 c2 b at\u0153 1 vx\u0153 1 c2 b \u00a2t (t\u0153 2 t\u0153 1) v c2 (x\u0153 2 x\u0153 1) or \u00a2t \u00a2t v c2 \u00a2x 1-27 If the clock that records is located at the events, then \u00a2x 0. If that is not the case, however,, though certainly a valid measurement, is not a proper time interval. Only a clock located at an event when it occurs can record a proper time interval. t\u0153 and 1 \u00a2t and t\u0153 2 \u00a2x 0 32 Chapter 1 Relativity I ct (m) ct\u00b4 (m) 3 2 1 ct = 2\u03b3 2 1 1 x\u00b4 (m) 2 1 2 x (m) Figure 1-26 Spacetime diagram illustrating time dilation. The dashed line is the worldline of a light flash emitted at x 0 and reflected back to that point by a mirror at x 1 m. 0.5. EXAMPLE 1-7 Spatial Separation of Events Two events occur at the same point at times separation of these events measured in S? x\u0153 0 in S, which moves with speed v relative to S. What is the spatial and t\u0153 2 t\u0153 1 SOLUTION 1. The location of the events in S is given by the Lorentz inverse transformation Equation 1-19: x (x vt) 2. The spatial separation of the two events x1 is then \u00a2x x2 terms cancel: 3. The x\u0153 0 4. Since \u00a2t is the proper time interval, Equation 1-26 yields \u00a2x (x\u0153 0 vt\u0153 2) (x\u0153 0 vt\u0153 1) \u00a2x v(t\u0153 2 t\u0153 1) v\u00a2t\u0153 \u00a2x v v\u00a2t 5. Using the situation in Figure 1-26 as a numerical example, where 0.5 and 1.15, we have \u00a2x v c \u00a2(ct) (1.15)(0.5)(2) 1.15 m EXAMPLE 1-8 The Pregnant Elephant14 Elephants have a gestation period of 21 months. Suppose that a freshly impregnated elephant is placed on a spaceship and sent toward a distant space jungle at v 0.75c. If we monitor radio transmissions from the spaceship, how long after launch might we expect to hear the first squealing trumpet from the newborn calf? SOLUTION 1. In S, the rest frame of the elephant, the time interval from launch to birth, is 21 months. In the Earth frame S the time interval is given by Equation 1-26: \u00a2t1, \u00a2t1 1 21 2 1 21 (0.75)2 (21 months) 31.7 months 2. At that time the radio signal announcing the happy event starts toward Earth at speed c, but from where? Using the result of Example 1-7, since launch the spaceship has moved \u00a2x in S, given by 3. Notice that there is no need to con- vert \u00a2x into meters since our interest is in how long it will take the radio signal to travel this distance in S. \u00a2t2, That time is given by 4. Thus, the good news will arrive at Earth at time \u00a2t after launch where \u00a2x vt ct (1.51)(0.75)(21 c # months) 23.8 c # months # where c month is the distance light travels in one month. \u00a2t2 \u00a2x>c 23.8 c # months>c 23.8 months \u00a2t \u00a2t1 \u00a2t2 31.7 23.8 55.5 months 1-4 Time Dilation and Length Contraction 33 Remarks: This result, too, is readily obtained from a spacetime diagram. Figure 1-27 illustrates the general appearance of the spacetime diagram for this example, showing the elephant\u2019s worldline and the worldline of the radio signal. Question 8. You are standing on a corner and a friend is driving past in an automobile. Both of you note the times when the car passes two different intersections and determine from your watch readings the time that elapses between the two events. Which of you has determined the proper time interval? The time dilation of Equation 1-26 is easy to see in a spacetime diagram such as Figure 1-26, using the same round trip for a light pulse used above. Let the light flash leave x 0 at ct 0 when the S and S origins", " coincided. The flash travels to x 1 m, reflects from a mirror located there, and returns to x 0. Let 0.5. The dotted line shows the worldline of the light beam, reflecting at (x 1, ct 1) and returning to x 0 at ct 2 m. Note that the S observer records the latter event at ct 2 m ; i.e., the observer in S sees the S clock running slow. ct (c \u2022 mo) ct\u00b4 Radio signal x\u00b4 Elephant 55.5 31.7 0 0 23.8 (c \u2022 mo) x Figure 1-27 Sketch of the spacetime diagram for 0.75. Example 1-8. The colored line is the worldline of the pregnant elephant. The worldline of the radio signal is the dashed line at 45\u00b0 toward the upper left. Experimental tests of the time dilation prediction have been performed using macroscopic clocks, in particular, accurate atomic clocks. In 1975, C. O. Alley conducted a test of both general and special relativity in which a set of atomic clocks were carried by a U.S. Navy antisubmarine patrol aircraft while it flew back and forth over the same path for 15 hours at altitudes between 8000 m and 10,000 m over Chesapeake Bay. The clocks in the plane were compared by laser pulses with an identical group of clocks on the ground. (See Figure 1-13 for one way such a comparison might be done.) Since the experiment was primarily intended to test the gravitational effect on clocks predicted by general relativity (see Section 2-5), the aircraft was deliberately flown at the rather sedate average speed of 270 knots (140 m s) 4.7 107c to minimize the time dilation due to the relative speeds of the clocks. Even so, after Alley deducted the effect of gravitation as predicted by general relativity, the airborne clocks lost an average of 5.6 109 s due to the relative speed during the 15-hour flight. This result agrees with the prediction of special relativity, 5.7 109 s, to within 2 percent, even at this low relative speed. The experimental results leave little basis for further debate as to whether traveling clocks of all kinds lose time on a round trip. They do. ct (m) > 2 1 Length Contraction (x1) 1 L ct\u2032 (m) x\u2032 (m) Lp 2 (x\u2032)2 1 (x\u2032)1 (x2) 2 x (m) A phenomenon closely related to time dilation is length contraction. The length of an object measured in the reference frame in which the object is at rest is called its proper length Lp. In a reference frame in which the object is moving, the measured length parallel to the direction of motion is shorter than its proper length. Consider a rod at rest in the frame S with one end at and the as illustrated in Figure 1-28. The length of the rod other end at Some care must in this frame is its proper length x\u0153 1, x\u0153 2 x\u0153 2 x\u0153 1. Lp and 2 m 1 m. x\u0153 2 0.79 Figure 1-28 A measuring rod, a meter stick in this case, lies at rest in S between x\u0153 System S moves with 1 relative to S. Since the rod is in motion, S must measure the locations of the ends of the rod x2 and x1 simultaneously in order to have made a valid length measurement. L is obviously shorter than Lp. By direct measurement from the diagram (use a millimeter scale) L>Lp 0.61 1>. 34 Chapter 1 Relativity I L x2 be taken to find the length of the rod in frame S. In this frame, the rod is moving to the right with speed v, the speed of frame S. The length of the rod in frame S is x1, where x2 is the position of one end at some time t2 and x1 defined as is the position of the other end at the same time as measured in frame S. Since t\u0153 the rod is at rest in S, Equation 1-18 is convenient to use to cal2 x1 at some time t because it relates x, x, and t, whereas Equation 1-19 is culate not convenient because it relates x, x, and t: need not equal t2 t\u0153 1. x2 t1 x\u0153 2 (x2 vt2) and x\u0153 1 (x1 vt1) Since t2 t1, we obtain x\u0153 2 x\u0153 1 (x2 x1) x2 x1 or 1 (x\u0153 2 x\u0153 1) A1 v2 c2 (x\u0153 2 x\u0153 1) L 1 Lp A1 v2 c2 Lp 1-28 Thus, the length of a rod is smaller when it is measured in a frame with respect", " to which it is moving. Before Einstein\u2019s paper was published, Lorentz and G. FitzGerald had independently shown that the null result of the Michelson-Morley experiment could be explained by assuming that the lengths in the direction of the interferometer\u2019s motion contracted by the amount given in Equation 1-28. For that reason, the length contraction is often called the Lorentz-FitzGerald contraction. EXAMPLE 1-9 Speed of S A stick that has a proper length of 1 m moves in a direction parallel to its length with speed v relative to you. The length of the stick as measured by you is 0.914 m. What is the speed v? SOLUTION 1. The length of the stick measured in a frame relative to which it is moving with speed v is related to its proper length by Equation 1-28: 2. Rearranging to solve for : 3. Substituting the values of Lp and L: 4. Solving for v: L Lp Lp L 1 m 0.914 m 1 21 v2>c2 21 v2>c2 0.914 1 v2>c2 (0.914)2 0.835 v2>c2 1 0.835 0.165 v2 0.165c2 v 0.406c 1-4 Time Dilation and Length Contraction 35 Figure 1-29 The appearance of rapidly moving objects depends on both length contraction in the direction of motion and the time when the observed light left the object. (a) The array of clocks and measuring rods that represents S as viewed by an observer in S with (b) When S approaches the S observer with 0.9, apparent. This is what an observer on a cosmic-ray proton might see as it passes into the lattice of a face-centered-cubic crystal such as NaCl. [P.-K. Hsiung, R. Dunn, and C. Cox. Courtesy of C. Cox, Adobe Systems, Inc., San Jose, CA.] the distortion of the lattice becomes 0. It is important to remember that the relativistic contraction of moving lengths occurs only parallel to the relative motion of the reference frames. In particular, observers in relatively moving systems measure the same values for lengths in the y and y and in the z and z directions perpendicular to their relative motion. The result is that observers measure different shapes and angles for two- and three-dimensional objects. (See Example 1-10 and Figures 1-29 and 1-30.) (a) y\u00b4 S\u00b4 v = 0.5c (b) y S B\u00b4 A\u00b4 B A A\u00b4 \u03c6\u00b4 = 60\u00b0 B\u00b4 \u03b8\u00b4 = 30\u00b0 A \u03c6 = 63\u00b0 B \u03b8 = 34\u00b0 1 x\u00b4 1 x Figure 1-30 Length contraction distorts the shape and orientation of two- and threedimensional objects. The observer in S measures the square shown in S as a rotated parallelogram. 36 Chapter 1 Relativity I xy EXAMPLE 1-10 The Shape of a Moving Square Consider the square in the plane of S with one side making a 30\u00b0 angle with the x axis, as in Figure 1-30a. If S moves with 0.5 relative to S, what is the shape and orientation of the figure in S? SOLUTION The S observer measures the x components of each side to be shorter by a factor than those measured in S. Thus, S measures 1> A [cos2 30 sin2 30>2]1>2A 0.968A B [sin2 30 cos2 30>2]1>2B 0.901B Since the figure is a square in S, A B. In addition, the angles between B and the x axis and between A and the x axis are given by, respectively, tan 1 c B sin 30 B cos 30> d tan1 c sin 30 cos 30 d 33.7\u00b0 tan 1 c A cos 30 A sin 30> d tan1 c cos 30 sin 30 d 63.4\u00b0 Thus, S concludes from geometry that the interior angle at vertex 1 is not 90\u00b0, but 180\u00b0 (63.4\u00b0 33.7\u00b0) 82.9\u00b0 \u2014 i.e., the figure is not a square, but a parallelogram whose shorter sides make 33.7\u00b0 angles with the x axis! Its shape and orientation in S are shown in Figure 1-30b. Muon Decay An interesting example of both time dilation and length contraction is afforded by the appearance of muons as secondary radiation from cosmic rays. Muons decay according to the statistical law of radioactivity: N(t) N0e(t>) 1-29 where N0 is the original number of muons at time t 0, N(t) is the number remaining at time t, and is the mean lifetime (a proper time interval", "), which is about 2 s for muons. Since muons are created (from the decay of pions) high in the atmosphere, usually several thousand meters above sea level, few muons should reach sea level. A typical muon moving with speed 0.998c would travel only about 600 m in 2 s. However, the lifetime of the muon measured in Earth\u2019s reference frame is increased according to time dilation (Equation 1-26) by the factor, which is 15 for this particular speed. The mean lifetime measured in Earth\u2019s reference frame is therefore 30 s, and a muon with speed 0.998c travels about 9000 m in this time. From the muon\u2019s point of view, it lives only 2 s, but the atmosphere is rushing past it with a speed of 0.998c. The distance of 9000 m in Earth\u2019s frame is thus contracted to only 600 m in the muon\u2019s frame, as indicated in Figure 1-31. 1>(1 v2>c2)1>2 It is easy to distinguish experimentally between the classical and relativistic predictions of the observations of muons at sea level. Suppose that we observe 108 muons at an altitude of 9000 m in some time interval with a muon detector. How many would (a) Muon 9000 m (b) Muon 600 m (c) p L / L 1.0 0.8 0.6 0.4 0.2 0 0.001 0.01 0.1 1.0 \u03b2 s Figure 1-31 Although muons are created high above Earth and their mean lifetime is only about 2 when at rest, many appear at Earth\u2019s surface. (a) In Earth\u2019s reference frame, a typical muon moving at 0.998c has a mean lifetime of 30 and travels 9000 m in this time. (b) In the reference frame of the muon, the distance traveled by Earth is only 600 m in the muon\u2019s lifetime of. (c) L varies only slightly 2 from Lp until v is of the order of 0.1c. as v S c. L S 0 s s 1-4 Time Dilation and Length Contraction 37 we expect to observe at sea level in the same time interval? According to the nonrel> ativistic prediction, the time it takes for these muons to travel 9000 m is (9000 m) 108 and t 15 into 0.998c Equation 1-29, we obtain 30 s, which is 15 lifetimes. Substituting N0 N 108e15 30.6 We would thus expect all but about 31 of the original 100 million muons to decay before reaching sea level. According to the relativistic prediction, Earth must travel only the contracted distance of 600 m in the rest frame of the muon. This takes only 2 s 1. Therefore, the number of muons expected at sea level is N 108e1 3.68 107 Thus relativity predicts that we would observe 36.8 million muons in the same time interval. Experiments of this type have confirmed the relativistic predictions. The Spacetime Interval We have seen earlier in this section that time intervals and lengths ( space intervals), quantities that were absolutes, or invariants, for relatively moving observers using the classical Galilean coordinate transformation, are not invariants in special relativity. The Lorentz transformation and the relativity of simultaneity lead observers in inertial frames to conclude that lengths moving relative to them are contracted and time intervals are stretched, both by the factor. The question naturally arises: Is there any quantity involving the space and time coordinates that is invariant under a Lorentz transformation? The answer to that question is yes, and as it happens, we have already dealt with a special case of that invariant quantity when we first obtained the correct form of the Lorentz transformation. It is called the spacetime interval, or usually just the interval, \u00a2s, and is given by (\u00a2s)2 (c\u00a2t)2 [\u00a2x2 \u00a2y2 \u00a2z2] 1-30 or, specializing it to the one-space-dimensional systems that we have been discussing, (\u00a2s)2 (c\u00a2t)2 (\u00a2x)2 1-31 It may help to think of Equations 1-30 and 1-31 like this: [interval]2 [separation in time]2 [separation in space]2 The interval \u00a2s is the only measurable quantity describing pairs of events in spacetime for which observers in all inertial frames will obtain the same numerical value. The negative sign in Equations 1-30 and 1-31 implies that (\u00a2s)2 may be positive, negative, or zero depending on the relative sizes of the time and space separations. With the sign of (\u00a2s)2, nature is telling us about the causal", " relation between the two events. Notice that whichever of the three possibilities characterizes a pair for one observer, it does so for all observers since \u00a2s is invariant. The interval is called timelike if the time separation is the larger and spacelike if the space separation predominates. If the two terms are equal, so that \u00a2s 0, then it is called lightlike. Experiments with muons moving near the speed of light are performed at many accelerator laboratories throughout the world despite their short mean life. Time dilation results in much longer mean lives relative to the laboratory, providing plenty of time to do experiments. 38 Chapter 1 Relativity I Timelike Interval Consider a material particle15 or object, e.g., the elephant in Figure 1-27, that moves relative to S. Since no material particle has ever been measured traveling faster than light, particles always travel less than 1 m of distance in 1 m of light travel time. We saw that to be the case in Example 1-8, where the time interval between launch and birth of the baby was 31.7 months on the S clock, during which time the elephant had moved a distance of 23.8c months. Equation 1-31 then yields (c\u00a2t)2 (\u00a2x)2 (31.7c)2 (23.8c)2 (21.0c)2 (\u00a2s)2, and the interval in S is \u00a2s 21.0 c months. The time interval term being the larger, \u00a2s is a timelike interval and we say that material particles have timelike worldlines. Such worldlines lie within the shaded area of the spacetime diagram in Figure 1-21. Note that in the elephant\u2019s frame S the separation in space between the launch and birth is zero and \u00a2t is 21.0 months. Thus \u00a2s 21.0 c months in S, too. That is what we mean by the interval being invariant: observers in both S and S measure the same number for the separation of the two events in spacetime. # # # The proper time interval between two events can be determined from Equations 1-31 using space and time measurements made in any inertial frame since we can write that equation as \u00a2s c 2(\u00a2t)2 (\u00a2x>c)2 Since \u00a2t when \u00a2x 0 \u2014 i.e., for the time interval recorded on a clock in a system moving such that the clock is located at each event as it occurs \u2014 in that case 2(\u00a2t)2 (\u00a2x>c)2 22 0 \u00a2s c 1-32 Notice that this yields the correct proper time 21.0 months in the elephant example. Spacelike Interval When two events are separated in space by an interval whose square is greater than the value of (c\u00a2t)2, then \u00a2s is called spacelike. In that case it is convenient for us to write Equation 1-31 in the form (\u00a2s)2 (\u00a2x)2 (c\u00a2t)2 1-33 so that, as with timelike intervals, (\u00a2s)2 is not negative.16 Events that are spacelike occur sufficiently far apart in space and close together in time that no inertial frame could move fast enough to carry a clock from one event to the other. For example, suppose two observers in Earth frame S, one in San Francisco and one in London, agree to each generate a light flash at the same instant, so that c\u00a2t 0 m in S and \u00a2x 1.08 107 m. For any other inertial frame (c\u00a2t)2 0, and we see from Equation 1-33 that (\u00a2x)2 must be greater than (1.08 107)2 in order that \u00a2s be invariant. In other words, 1.08 107 m is as close in space as the two events can be in any system; consequently, it will not be possible to find a system moving fast enough to move a clock from one event to the other. A speed greater than c, in this case infinitely greater, would be needed. Notice that the value of \u00a2s Lp, the proper length. Just as with the proper time interval, measurements of space and time intervals in any inertial system can be used to determine Lp. 1-4 Time Dilation and Length Contraction 39 Lightlike (or Null) Interval The relation between two events is lightlike if s in Equation 1-31 equals zero. In that case c\u00a2t \u00a2x 1-34 and a light pulse that leaves the first event as it occurs will just reach the second as it occurs. The existence of the lightlike interval in relativity has no counterpart in the world of our everyday experience, where the geometry of space is Euclidean. In order for the distance between two points in space", " to be zero, the separation of the points in each of the three space dimensions must be zero. However, in spacetime the interval between two events may be zero, even though the intervals in space and time may individually be quite large. Notice, too, that pairs of events separated by lightlike intervals have both the proper time interval and proper length equal to zero since s 0. Things that move at the speed of light17 have lightlike worldlines. As we saw earlier (see Figure 1-22), the worldline of light bisects the angles between the ct and x axes in a spacetime diagram. Timelike intervals lie in the shaded areas of Figure 1-32 and share the common characteristic that their relative order in time is the same for observers in all inertial systems. Events A and B in Figure 1-32 are such a pair. Observers in both S and S agree that A occurs before B, although they of course measure different values for the space and time separations. Causal events, i.e., events that depend upon or affect one another in some fashion, such as your birth and that of your mother, have timelike intervals. On the other hand, the temporal order of events with spacelike intervals, such as A and C in Figure 1-32, depends upon the relative motion of the systems. As you can see in the diagram, A occurs before C in S, but C occurs first in S. Thus, the relative order of pairs of events is absolute in the shaded areas but elsewhere may be in either order. ct Absolute future ctB B ct\u00b4 ct\u00b4B ctC A ct\u00b4C x\u00b4 C x Absolute past Worldline of light moving in +x direction Worldline of light moving in \u2013x direction Figure 1-32 The relative temporal order of events for pairs characterized by timelike intervals, such as A and B, is the same for all inertial observers. Events in the upper shaded area will all occur in the future of A; those in the lower shaded area occurred in As past. Events whose intervals are spacelike, such as A and C, can be measured as occurring in either order, depending on the relative motion of the frames. Thus, C occurs after A in S but before A in S. 40 Chapter 1 Relativity I Question 9. In 1987 light arrived at Earth from the explosion of a star (a supernova) in the Large Magellanic Cloud, a small companion galaxy to the Milky Way, located about 170,000 c y away. Describe events that together with the explosion of the star would be separated from it by (a) a spacelike interval, (b) a lightlike interval, and (c) a timelike interval. # EXAMPLE 1-11 Characterizing Spacetime Intervals Figure 1-33 is the spacetime diagram of a laboratory showing three events, the emission of light from an atom in each of three samples. 1. Determine whether the interval between each of the three possible pairs of events is timelike, spacelike, or lightlike. 2. Would it have been possible in any of the pairs for one of the events to have been caused by the other? If so, which? Event 3 Event Event 1 0 0 2 4 6 x (m) 8 10 Figure 1-33 A spacetime diagram of three events whose intervals s are found in Example 1-11. SOLUTION 1. The spacetime coordinates of the events are event 1 2 3 ct 2 5 8 x 1 9 6 and for the three possible pairs 1 and 2, 2 and 3, and 1 and 3 we have pair 1 & 2 2 & 3 1 & 3 ct 5\u20132 8\u20135 8\u20132 x 9\u20131 6\u20139 6\u20131 (ct )2 (x )2 9 9 36 64 9 25 spacelike lightlike timelike 2. Yes, event 3 may possibly have been caused by either event 1 since 3 is in the absolute future of 1, or event 2, since 2 and 3 can just be connected by a flash of light. 1-5 The Doppler Effect 41 1-5 The Doppler Effect In the Doppler effect for sound the change in frequency for a given velocity v depends on whether it is the source or receiver that is moving with that speed. Such a distinction is possible for sound because there is a medium (the air) relative to which the motion takes place, and so it is not surprising that the motion of the source or the receiver relative to the still air can be distinguished. Such a distinction between motion of the source or receiver cannot be made for light or other electromagnetic waves in a vacuum as a consequence of Einstein\u2019s second postulate; therefore, the classical expressions for the Doppler effect cannot be correct for light. We will now derive the relativistic Doppler effect equations that are correct for light", ". Consider a light source moving toward an observer or receiver at A in Figure 1-34a at velocity v. The source is emitting a train of light waves toward receivers A and B while approaching A and receding from B. Figure 1-34b shows the spacetime (a) (b) c v c B A ct ct\u00b4 Worldline of light wave toward B c \u0394t c \u0394t\u00b4 Worldline of light wave toward A B c \u0394t v \u0394t c \u0394t 0 A x (c) y Observer x (in S) y\u00b4 S S\u00b4 \u03b8 (measured in S) x Source x\u00b4 (d) K\u00fcndig\u2019s experiment Source \u03c9 v Gamma rays Receiver Figure 1-34 The Doppler effect in light, as in sound, arises from the relative motion of the source and receiver; however, the independence of the speed of light on that motion leads to different expressions for the frequency shift. (a) A source approaches observer A and recedes from observer B. The spacetime diagram of the system S in which A and B are at rest and the source moves at velocity v illustrates the two situations. (b) The source located at x 0 (the x axis is omitted) moves along its worldline, the ct axis. The N waves emitted toward A in time t occupy space x ct vt, whereas those headed for B occupy x ct vt. In three dimensions the observer in S may see light emitted at some angle with respect to the x axis as in (c). In that case a transverse Doppler effect occurs. (d) K\u00fcndig\u2019s apparatus for measuring the transverse Doppler effect. 42 Chapter 1 Relativity I diagram of S, the system in which A and B are at rest. The source is located at x 0 (x axis is not shown), and, of course, its worldline is the ct axis. Let the source emit a train of N electromagnetic waves in each direction beginning when the S and S origins were coincident. First, let\u2019s consider the train of waves headed toward A. During the time t over which the source emits the N waves, the first wave emitted will have traveled a distance ct and the source itself a distance vt in S. Thus, the N waves are seen by the observer at A to occupy a distance ct vt and, correspondingly, their wavelength is given by c\u00a2t v\u00a2t N and the frequency f c> is f c cN (c v)\u00a2t 1 1 N \u00a2t c> N>\u00a2t, The frequency of the source in S, called the proper frequency, is given by where t is measured in S, the rest system of the source. The f0 time interval t is the proper time interval since the light waves, in particular the first and the Nth, are all emitted at x 0; hence x 0 between the first and the Nth in S. Thus, t and t are related by Equation 1-26 for time dilation, so t t, and when the source and receiver are moving toward each other, the observer A in S measures the frequency f 1 1 \u00a2t f0 \u00a2t f0 1 1 1-35 or f 21 2 1 f0 A 1 1 f0 (approaching) 1-36 This differs from the classical equation only in the addition of the time dilation factor. Note that f fo for the source and observer approaching each other. Since for visible light this corresponds to a shift toward the blue part of the spectrum, it is called a blueshift. Suppose the source and receiver are moving away from each other, as for observer B in Figure 1-34b. Observer B, in S, sees the N waves occupying a distance c\u00a2t v\u00a2t, and the same analysis shows that observer B in S measures the frequency f 21 2 1 f0 A 1 1 f0 (receding) 1-37 The use of Doppler radar to track weather systems is a direct application of special relativity. Notice that f f0 for the observer and source receding from each other. Since for visible light this corresponds to a shift toward the red part of the spectrum, it is called a redshift. It is left as a problem for you to show that the same results are obtained when the analysis is done in the frame in which the source is at rest. (i.e., ), as is often the case for light sources moving on Earth, useful (and easily remembered) approximations of Equations 1-36 and 1-37 can be obtained. Using Equation 1-36 as an example and rewriting it in the form In the event that V 1 v V c f f0(1 )1>2(1 )1>2 1-5 The Doppler Effect 43 the two quantities in parentheses can be expanded by the binomial theorem to yield", " f f0 a1 Multiplying out and discarding terms of higher order than yields f>f0 1 (approaching) and, similarly, f>f0 1 (receding) and \u0192\u00a2f>f0 \u0192 in both situations, where f f0 f. EXAMPLE 1-12 Rotation of the Sun The Sun rotates at the equator once in about 25.4 days. The Sun\u2019s radius is 7.0 108 m. Compute the Doppler effect that you would expect to observe at the left and right limbs (edges) of the Sun near the equator for light of wavelength 550 nm 550 109 m (yellow light). Is this a redshift or a blueshift? SOLUTION The speed of limbs v (circumference)/(time for one revolution) or v 2R T 2(7.0 108) m 25.4 d # 3600 s>h # 24 h>d 2000 m>s \u00a2f fo > so we may use the approximation equations. Using v V c, that c fo tional change in frequency of receding limb, a blueshift for the approaching one. we have Hz. Since (3 108 m s) (550 109) 5.45 1014 Hz, f represents a fracor about one part in 105. It is a redshift for the \u00a2f 2000>550 109 3.64 109 c> o, > or > \u00a2f>f0,, o Doppler Effect of Starlight In 1929 E. P. Hubble became the first astronomer to suggest that the universe is expanding.18 He made that suggestion and offered a simple equation to describe the expansion on the basis of measurements of the Doppler shift of the frequencies of light emitted toward us by distant galaxies. Light from distant galaxies is always shifted toward frequencies lower than those emitted by similar sources nearby. Since, the general expression connecting the frequency f and wavelength of light is c f the shift corresponds to longer wavelengths. As noted above, the color red is on the longer-wavelength side of the visible spectrum (see Chapter 4), so the redshift is used to describe the Doppler effect for a receding source. Similarly, blueshift describes light emitted by stars, typically stars in our galaxy, that are approaching us. Astronomers define the redshift of light from astronomical sources by the frequency measured in the frame of the star or expression z (fo galaxy and f frequency measured at the receiver on Earth. This allows us to write f) f, where fo v c in terms of z as > > (z 1)2 1 (z 1)2 1 1-38 44 Chapter 1 Relativity I Equation 1-37 is the appropriate one to use for such calculations, rather than the approximations, since galactic recession velocities can be quite large. For example, the quasar 2000-330 has a measured z 3.78, which implies from Equation 1-38 that it is receding from Earth due to the expansion of space at 0.91c. (See Chapter 13.) EXAMPLE 1-13 Redshift of Starlight The longest wavelength of light emitted by hydrogen in the Balmer series (see Chapter 4) has a wavelength of 656 nm. In light from a distant galaxy, this wavelength is measured as 1458 nm. Find the speed at which the galaxy is receding from Earth. o SOLUTION 1. The recession speed is the v in o, Since Equation 1-37 applies: this is a redshift and v> c. 2. Rewriting Equation 1-37 in terms of the wavelengths: 3. Squaring both sides and substituting val- ues for o and : 4. Solving for : 5. The galaxy is thus receding at speed v, where f 1 1 f0 A f 1 1 A o f f0 1 1 a 2 b 0 a 656 nm 1458 nm 2 b 0.202 1 (0.202)(1 ) 1.202 1 0.202 0.798 0.798 1.202 v c 0.664c 0.664 EXPLORING Transverse Doppler Effect Our discussion of the Doppler effect in Section 1-5 involved only one space dimension, wherein the source, observer, and the direction of the relative motion all lie on the x axis. In three space dimensions, where they may not be colinear, a more complete analysis, though beyond the scope of our discussion, makes only a small change in Equation 1-35. If the source moves along the positive x axis but the observer views the light emitted at some angle with the x axis, as shown in Figure 1-34c, Equation 1-35 becomes f f0 1 1 cos 1-35a, When 0, this becomes the equation for the source and receiver approaching, and the equation becomes that for the source and receiver", " receding. Equation when 1-35a also makes the quite surprising prediction that even when viewed perpendicular 1-6 The Twin Paradox and Other Surprises 45 >2, the observer will still see a frequency shift, to the direction of motion, where the so-called transverse Doppler effect, f f0 >. Note that f f0 since 1. It is sometimes referred to as the second-order Doppler effect and is the result of time dilation of the moving source. [The general derivation of Equation 1-35a can be found in the French (1968), Resnick (1992), and Ohanian (2001) references at the end of the chapter.] Following a suggestion first made by Einstein in 1907, K\u00fcndig in 1962 made an excellent quantitative verification of the transverse Doppler effect.19 He used 14.4-keV gamma rays emitted by a particular isotope of Fe as the light source (see Chapter 11). The source was at rest in the laboratory, on the axis of an ultracentrifuge, and the receiver (an Fe absorber foil) was mounted on the ultracentrifuge rim, as shown in Figure 1-34d. Using the extremely sensitive frequency measuring technique called the M\u00f6ssbauer effect (see Chapter 11), K\u00fcndig found a transverse Doppler effect in agreement with the relativistic prediction within 1 percent over a range of relative speeds up to about 400 m s. > 1-6 The Twin Paradox and Other Surprises The consequences of Einstein\u2019s postulates \u2014 the Lorentz transformation, relativistic velocity addition, time dilation, length contraction, and the relativity of simultaneity \u2014 lead to a large number of predictions that are unexpected and even startling when compared with our experiences in a macroscopic world where and geometry obeys the Euclidean rules. Still other predictions seem downright paradoxical, with relatively moving observers obtaining equally valid but apparently totally inconsistent results. This chapter concludes with the discussion of a few such examples that will help you hone your understanding of special relativity. 0 Twin Paradox Perhaps the most famous of the paradoxes in special relativity is that of the twins or, as it is sometimes called, the clock paradox. It arises out of time dilation (Equation 1-26) and goes like this. Homer and Ulysses are identical twins. Ulysses travels at a constant high speed to a star beyond our solar system and returns to Earth while his twin, Homer, remains at home. When the traveler Ulysses returns home, he finds his twin brother much aged compared to himself \u2014 in agreement, we will see, with the prediction of relativity. The paradox arises out of the contention that the motion is relative and either twin could regard the other as the traveler, in which case each twin should find the other to be younger than he and we have a logical contradiction \u2014 a paradox. Let\u2019s illustrate the paradox with a specific example. Let Earth and the destination star be in the same inertial frame S. Two other frames S and S move relative to S at v 0.8c and v 0.8c, respectively. Thus, in both cases. The spaceship carrying Ulysses accelerates quickly from S to S, then coasts with S to the star, again accelerates quickly from S to S, coasts with S back to Earth, and brakes to a stop alongside Homer. 5> 3 It is easy to analyze the problem from Homer\u2019s point of view on Earth. Suppose, according to Homer\u2019s clock, Ulysses coasts in S for a time interval t 5 y and in S for an equal time. Thus, Homer is 10 y older when Ulysses returns. The time interval in S between the events of Ulysses\u2019 leaving Earth and arriving at the star is shorter because it is a proper time interval. The time it takes to reach the star by Ulysses\u2019 clock is \u00a2t \u00a2t 5 y 5>3 3 y 46 Chapter 1 Relativity I Since the same time is required for the return trip, Ulysses will have recorded 6 y for the round trip and will be 4 y younger than Homer upon his return. The difficulty in this situation seems to be for Ulysses to understand why his twin aged 10 y during his absence. If we consider Ulysses as being at rest and Homer as 1.8 y, and it moving away, Homer\u2019s clock should run slow and measure only appears that Ulysses should expect Homer to have aged only 3.6 years during the round trip. This is, of course, the paradox. Both predictions can\u2019t be right. However, this approach makes the incorrect assumption that the twins\u2019 situations are symmetrical and interchangeable. They are not. Homer remains in a single inertial frame, whereas Ulysses changes inertial frames, as illustrated in Figure 1-35a, the spacetime diagram for", " Ulysses\u2019 trip. While the turnaround may take only a minute fraction of the total time, it is absolutely essential if the twins clocks are to come together again so that we can compare their ages (readings). 3> (a) ct ct (b) Ulysses\u2019 returning worldline \u03b2 = \u20130.8 (S\u02dd frame) Turnaround point (at star) Ulysses\u2019 outgoing worldline \u03b2 = 0.8 (S\u00b4 frame) x B Light worldline (through A) Ulysses\u2019 returning worldline Turnaround point A Light worldline (through O) Ulysses\u2019 outgoing worldline x Figure 1-35 (a) The spacetime diagram of Ulysses\u2019 journey to a distant star in the inertial frame in which Homer and the star are at rest. (b) Divisions on the ct axis correspond to years on Homer\u2019s clock. The broken lines show the paths (worldlines) of light flashes transmitted by each twin with a frequency of 1/year on his clock. Note the markedly different frequencies at the receivers. 1-6 The Twin Paradox and Other Surprises 47 A correct analysis can be made using the invariant interval s from Equation 1-31 rewritten as 2 b a \u00a2s c (\u00a2t)2 a 2 b \u00a2x c where the left side is constant and equal to the proper time interval squared, and the right side refers to measurements made in any inertial frame. Thus, Ulysses along each of his worldlines in Figure 1-35a has x 0 and, of course, measures t 3 y, or 6 y for the round trip. Homer, on the other hand, measures ()2, (\u00a2t)2 ()2 a 2 b \u00a2x c > and since (x c)2 is always positive, he always measures t. In this situation x 0.8ct, so or (\u00a2t)2 (3 y)2 (0.8c\u00a2t>c)2 (\u00a2t)2(0.36) (3)2 \u00a2t 3 0.6 5 y or 10 y for the round trip, as we saw earlier. The reason that the twins\u2019 situations cannot be treated symmetrically is because the special theory of relativity can predict the behavior of accelerated systems, such as Ulysses at the turnaround, provided that in the formulation of the physical laws we take the view of an inertial, i.e., unaccelerated, observer such as Homer. That\u2019s what we have done. Thus, we cannot do the same analysis in the rest frame of Ulysses\u2019 spaceship because it does not remain in an inertial frame during the round trip; hence, it falls outside of the special theory, and no paradox arises. The laws of physics can be reformulated so as to be invariant for accelerated observers, which is the role of general relativity (see Chapter 2), but the result is the same: Ulysses returns younger than Homer by just the amount calculated above. EXAMPLE 1-14 Twin Paradox and the Doppler Effect This example, first suggested by C. G. Darwin,20 may help you understand what each twin sees during Ulysses\u2019 journey. Homer and Ulysses agree that once each year, on the anniversary of the launch date of Ulysses\u2019 spaceship (when their clocks were together), each twin will send a light signal to the other. Figure 1-35b shows the light signals each sends. Based on our discussion above, Homer sends 10 light flashes (the ct axis, Homer\u2019s worldline, is divided into 10 equal interval corresponding to the 10 years of the journey on Homer\u2019s clock) and Ulysses sends 6 light flashes (each of Ulysses\u2019 worldlines is divided into 3 equal intervals corresponding to 3 years on Ulysses\u2019 clock). Note that each transmits his final light flash as they are reunited at B. Although each transmits light signals with a frequency of 1 per year, they obviously do not receive them at that frequency. For example, Ulysses sees no signals from Homer during the first three years! How can we explain the observed frequencies? 48 Chapter 1 Relativity I SOLUTION The Doppler effect provides the explanation. As the twins (and clocks) recede from each other the frequency of their signals is reduced from the proper frequency f0 according to Equation 1-37 and we have A 1 1 A 1 0.8 1 0.8 1 3 f f0 which is exactly what both twins see (refer to Figure 1-35b): Homer receives 3 flashes in the first 9 years and Ulysses 1 flash in his first 3 years; i.e., f (1 3)f0 for both. > After the turnaround they are approaching each other and Equation 1-38 yields A 1", " 1 A 1 0.8 1 0.8 3 f f0 and again this agrees with what the twins see: Homer receives 3 flashes during the final (10th) year and Ulysses receives 9 flashes during his final 3 years; i.e., f 3fo for both. Question 10. The different ages of the twins upon being reunited are an example of the relativity of simultaneity that was discussed earlier. Explain how that accounts for the fact that their biological clocks are no longer synchronized. More It is the relativity of simultaneity, not their different accelerations, that is responsible for the age difference between the twins. This is readily illustrated in The Case of the Identically Accelerated Twins, which can be found on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Figure 1-36 here. The Pole and Barn Paradox An interesting problem involving length contraction, reported initially by W. Rindler, involves putting a long pole into a short barn. One version, from E. F. Taylor and J. A. Wheeler,22 goes as follows. A runner carries a pole 10 m long toward the open front door of a small barn 5 m long. A farmer stands near the barn so that he can see both the front and the back doors of the barn, the latter being a closed swinging door, as shown in Figure 1-37a. The runner carrying the pole at speed v enters the barn, and at some instant the farmer sees the pole completely contained in the barn and closes the 1-6 The Twin Paradox and Other Surprises 49 (a) (b) ct Front door Rear door Back of pole Front of pole (c) ct\u00b4 10 Pole entirely within barn (ct = 5.8 m) Back end of pole S Rear door of barn 10 Front end of pole Back of pole enters barn Front door of barn S\u00b4 5 Front of pole leaves barn \u20135 0 5 10 x (m) \u201310 \u20135 0 Pole Front end of pole enters barn door at ct = 0 Pole 5 x\u00b4 (m) Front of pole enters barn door at ct\u00b4 = 0 Figure 1-37 (a) A runner carrying a 10-m pole moves quickly enough so that the farmer will see the pole entirely contained in the barn. The spacetime diagrams from the point of view of the farmer\u2019s inertial frame (b) and that of the runner (c). The resolution of the paradox is in the fact that the events of interest, shown by the large dots in each diagram, are simultaneous in S but not in S. front door, thus putting a 10-m pole into a 5-m barn. The minimum speed of the runner v that is necessary for the farmer to accomplish this feat can be computed from Equation 1-28, giving the relativistic length contraction L Lp proper length of the pole (10 m) and L length of the pole measured by the farmer, to be equal to the length of the barn (5 m). Therefore, we have, where Lp > 1 21 v2>c2 Lp L 10 5 1 v2>c2 (5>10)2 v2>c2 1 (5>10)2 0.75 v 0.866c or 0.866 50 Chapter 1 Relativity I A paradox seems to arise when this situation is viewed in the rest system of the runner. For him the pole, being at rest in the same inertial system, has its proper length of 10 m. However, the runner measures the length of the barn to be > 5 21 2 L Lp L 2.5 m How can he possibly fit the 10-m pole into the length-contracted 2.5-m barn? The answer is that he can\u2019t, and the paradox vanishes, but how can that be? To understand the answer, we need to examine two events \u2014 the coincidences of both the front and back ends of the pole, respectively, with the rear and front doors of the barn \u2014 in the inertial frame of the farmer and in that of the runner. These are illustrated by the spacetime diagrams of the inertial frame S of the farmer and barn (Figure 1-37b) and that of the runner S (Figure 1-37c). Both diagrams are drawn with the front end of the pole coinciding with the front door of the barn at the instant the clocks are started. In Figure 1-37b the worldlines of the barn doors are, of course, vertical, while those of the two ends of the pole make an angle 49.1\u00b0 with the x axis. Note that in S the front of the pole reaches the rear door of the barn at ct 5 m 0.866 5.8 m simultaneous with the arrival of the back end of the pole at the front door; i.e., at that instant in S the pole is entirely contained in the barn. tan1 (1>)", " > In the runner\u2019s rest system S it is the worldlines of the ends of the pole that are vertical, while those of the front and rear doors of the barn make angles of 49.1\u00b0 with the x axis (since the barn moves in the x direction at v). Now we see that the rear door passes the front of the pole at ct 2.5 m 0.866 2.9 m, but the front door of the barn doesn\u2019t reach the rear of the pole until ct 10 m 0.866 11.5 m. Thus, the first of those two events occurs before the second, and the runner never sees the pole entirely contained in the barn. Once again, the relativity of simultaneity is the key \u2014 events simultaneous in one inertial frame are not necessarily simultaneous when viewed from another inertial frame. > > 0.866. Now let\u2019s consider a different version of this paradox, the one initially due to W. Rindler. Suppose the barn\u2019s back wall were made of thick, armor-plate steel and had no door. What do the farmer and the runner see then? Once again, in the farmer\u2019s (and the barn\u2019s) rest frame, the instant the front of the pole reaches the armor plate, the farmer shuts the door and the 10-m pole is instantaneously contained in the 5-m barn. However, in the next instant (assuming that the pole doesn\u2019t break) it must either bend (i.e., rotate in spacetime) or break through the armor plate. Since this is relativity, the runner must come to the same conclusion in his rest frame as the 2.5-m barn races toBut now when the armor plate back wall contacts the front of ward him at the pole, the barn continues to move at taking the front of the pole with it and leaving at that instant 7.5 m of the pole still outside the barn. Yet like the farmer, the runner must also see the 10-m pole entirely contained within the 2.5-m barn. How can that be? Like this: the instant the tip of the pole hits the steel plate, that information (an elastic shock wave) begins to propagate down the pole. Even if the wave were to 10 m>3.0 108 m>s 3.33 108 s propagate at the speed of light c, it will take to reach the back of the pole. In the meantime, the barn door must move only 7.5 m to 7.5 m>(0.866 3.0 108 m>s) reach the back of the pole and does so in only 2.89 108 s. Thus, the runner, in agreement with the farmer, sees the 10-m pole entirely contained within the 2.5-m barn \u2014 at least briefly! 0.866, 1-6 The Twin Paradox and Other Surprises 51 Question 11. In the discussion where the barn\u2019s back wall was made from armor-plate steel and had no door, do the farmer and the runner both see the pole entirely contained in the barn, no matter what their relative speed is? Explain. Headlight Effect We have made frequent use of Einstein\u2019s second postulate asserting that the speed of light is independent of the source motion for all inertial observers; however, the same is not true for the direction of light. Consider a light source in S that emits light uniformly in all directions. A beam of that light emitted at an angle with respect to the x axis is shown in Figure 1-38a. During a time t the x displacement of the beam is x, and these are related to by 1-39 1-40 \u00a2x c\u00a2t \u00a2x \u00a2(ct) cos The direction of the beam relative to the x axis in S is similarly given by \u00a2x \u00a2(ct) cos Applying the inverse Lorentz transformation to Equation 1-40 yields cos \u00a2x c\u00a2t (\u00a2x v\u00a2t) c(\u00a2t v\u00a2x> c2 Dividing the numerator and denominator by t and then by c, we obtain cos (\u00a2x>\u00a2t v) ca1 v c2 \u00a2x>\u00a2tb \u00a2x>\u00a2(ct) v>c 1 v c # \u00a2x \u00a2(ct) and substituting from Equation 1-39 yields cos cos 1 cos 1-41 (a) y\u00b4 (b) y S\u00b4 v = 0.7c \u03b8\u00b4 = 36\u00b0 x\u00b4 S \u03b8 = 15\u00b0 x Figure 1-38 (a) The source at rest in S moves with 0.7 with respect to S. (b) Light emitted uniformly in S appears to S concentrated into a cone in the forward direction. Rays shown in (a) are 18\u00b0 apart. Rays shown in (b) make angles calculated from Equation 1-41. The", " two colored rays shown are corresponding ones. 52 Chapter 1 Relativity I In determining the brightness of stars and galaxies, a critical parameter in understanding them, astronomers must correct for the headlight effect, particularly at high velocities relative to Earth. between 0.5, 60\u00b0, is very small, e.g., Considering the half of the light emitted by the source in S into the forward >2, hemisphere, i.e., rays with note that Equation 1-41 restricts the angles measured in S for those rays (50 percent of all the light) to lie between cos1. the observer in S would see half of the total For example, for light emitted by the source in S to lie between i.e., in a cone of half angle 60\u00b0 whose axis is along the direction of the velocity of the source. For values of near yields 8.1\u00b0. This means that the observer in S unity sees half of all the light emitted by the source to be concentrated into a forward cone with that half angle. (See Figure 1-38b.) Note, too, that the remaining 50 percent of the emitted light is distributed throughout the remaining 344\u00b0 of the two-dimensional diagram.23 As a result of the headlight effect, light from a directly approaching source appears far more intense than that from the same source at rest. For the same reason, light from a directly receding source will appear much dimmer than that from the same source at rest. This result has substantial applications in experimental particle physics and astrophysics..99 Question 12. Notice from Equation 1-41 that some light emitted by the moving source into the rear hemisphere is seen by the observer in S as having been emitted into the forward hemisphere. Explain how that can be, using physical arguments. EXPLORING Superluminal Speeds We conclude this chapter with a few comments about things that move faster than light. The Lorentz transformations (Equations 1-18 and 1-19) have no meaning in the event that the relative speeds of two inertial frames exceed the speed of light. This is generally taken to be a prohibition on the moving of mass, energy, and information faster than c. However, it is possible for certain processes to proceed at speeds greater than c and for the speeds of moving objects to appear to be greater than c without contradicting relativity theory. A common example of the first of these is the motion of the point where the blades of a giant pair of scissors intersect as the scissors are quickly closed, sometimes called the scissors paradox. Figure 1-39 shows the situation. A long straight rod (one blade) makes an angle with the x axis (the second blade) and moves in the y direction at constant speed vy > the intersection of the blades, point P, moves to the right a distance x. Note from the figure that y t. During time t, y P \u0394x vy \u0394y vy \u03b8 x Figure 1-39 As the long straight rod moves vertically downward, the intersection of the \u201cblades,\u201d point P, moves toward the right at speed vp, and x t. In terms of vy > >tan. vp vy 1-6 The Twin Paradox and Other Surprises 53 y x > tan. The speed with which P moves to the right is \u00a2x>\u00a2t vp \u00a2x \u00a2y>vy vy \u00a2x \u00a2x tan 1-42 or vp vy tan Since tan S 0 as enough to zero so that vp the angle gets progressively smaller, so in principle all that one needs for vp are long blades so that S 0, close it will always be possible to find a value of c for any (nonzero) value of vy. As real scissors are closed, c S 0. Question 13. Use a diagram like Figure 1-32 to explain why the motion of point P cannot be used to convey information to observers along the blades. v Meteorite first glow The point P in the scissors paradox is, of course, a geometrical point, not a material object, so it is perhaps not surprising that it could appear to move at speeds greater than c. As an example of an object with mass appearing to do so, consider a tiny meteorite moving through space directly toward you at high speed v. As it enters Earth\u2019s atmosphere, about 9 km above the surface, frictional heating causes it to glow and the first light from the glow starts toward your eye. After some time t the frictional heating has evaporated all of the meteorite\u2019s matter, the glow is extinguished, and its final light starts toward your eye, as illustrated in Figure 1-40. During the time between the first and the final glow, the meteorite traveled a distance vt. During that same time interval light from the first glow has traveled toward your eye a distance ct. Thus, the space interval between the first and final glows is", " given by \u00a2y c\u00a2t v\u00a2t \u00a2t(c v) and the visual time interval at your eye teye between the arrival of the first and final light is v \u0394t c \u0394t Last glow wave front First glow wave front (c \u2013 v) \u0394t Eye \u00a2teye \u00a2y>c \u00a2t(c v) c \u00a2t(1 ) and, finally, the apparent visual speed va that you record is va v\u00a2t \u00a2teye v\u00a2t \u00a2t(1 ) c 1 1-43 Clearly, 0.5 yields va c and any larger yields va orite approaching you at v 0.8c is perceived to be moving at va c. For example, a mete 4c. Certain galactic Figure 1-40 A meteorite moves directly toward the observer\u2019s eye at speed v. The spatial distance between the wave fronts is (c v)t as they move at c, so the time interval between their arrival at the observer is not t, but teye, which is (1 )\u00a2t, speed of approach is v\u00a2t>\u00a2teye va and the apparent (c v)\u00a2t>c c>(1 ). 54 Chapter 1 Relativity I Figure 1-41 Superluminal motion has been detected in a number of cosmic objects. This sequence of images taken by the Hubble Space Telescope shows apparent motion at six times the speed of light in galaxy M87. The jet streaming from the galaxy\u2019s nucleus (the bright round region at the far left in the bar image at the top) is about 5000 is enlarged. The slanting lines track the moving features and indicate the apparent speeds in each region. [John Biretta, Space Telescope Science Institute.] long. The boxed region c # y Superluminal Motion in M87 Jet 1994 1995 1996 1997 1998 6.0c 5.5c 6.1c 6.0c structures may also be observed to move at superluminal speeds, as the sequence of images of the jet from galaxy M87 in Figure 1-41 illustrates. As a final example of things that move faster than c, it has been proposed that particles with mass might exist whose speeds would always be faster than light speed. One basis for this suggestion is an appealing symmetry: ordinary particles always have v c, and photons and other massless particles have v c, so the existence of particles with v c would give a sort of satisfying completeness to the classification of particles. Called tachyons, their existence would present relativity with serious but not necessarily insurmountable problems of infinite creation energies and causality paradoxes, e.g., alteration of history. (See the next example.) No compelling theoretical arguments preclude their existence and eventual discovery; however, to date, all experimental searches for tachyons24 have failed to detect them, and the limits set by those experiments indicate that it is highly unlikely they exist. EXAMPLE 1-15 Tachyons and Reversing History Use tachyons and an appropriate spacetime diagram to show how the existence of such particles might be used to change history and, hence, alter the future, leading to a paradox. SOLUTION In a spacetime diagram of the laboratory frame S the worldline of a particle with v c created at the origin traveling in the x direction makes an angle less than 45\u00b0 with the x axis; i.e., it is below the light worldline, as shown in Figure 1-42. After some time the tachyon reaches a tachyon detector mounted on a spaceship moving rapidly away at v c in the x direction. The spaceship frame S is shown in the figure at P. The detector immediately creates a new tachyon, sending it off in the x direction and, of course, into the future of S, i.e., with ct 0. The second tachyon returns to the laboratory at x 0 but at a time ct before the first tachyon was emitted, having traveled into the past of S to point M, where ct 0. Having sent an object into our own past, we would then have the ability to alter events that occur after M and produce causal contradictions. For example, the laboratory tachyon detector could be coupled to equipment that created the first tachyon via a computer programmed to Summary 55 ct S O M Light worldline ct\u00b4 S\u00b4 P x\u00b4 x Figure 1-42 A tachyon emitted at O in S, the laboratory frame, catches up with a spaceship moving at high speed at P. Its detection triggers the emission of a second tachyon at P back toward the laboratory at x 0. The second tachyon arrives at the laboratory at ct 0, i.e., before the emission of the first tachyon. cancel emission of the first tachyon if the second tachyon is detected. (Shades of The Terminator!) It is logical that contradictions such as this, together", " with the experimental results referred to above, lead to the conclusion that faster-than-light particles do not exist. As mentioned above, one attraction (or specter) associated with objects moving faster than light is the prospect of altering history via time travel. We close this chapter on relativity by illustrating one such paradox in Figure 1-43. January 1, 1906: Aristotle leaves for the past. Aristotle studies the new paper. March 1, 1905: Aristotle finds the famous paper entitled \u201cOn the Electrodynamics of Moving Bodies,\u201d by Albert Einstein, published in the journal Annals of Physics earlier in 1905. February 1, 1905: Aristotle arrives in the future. Figure 1-43 The knowledge creation paradox illustrates a causality problem associated with time travel, one possible consequence of material objects moving faster than light speed. [The authors thank Costas Efthimiou for this example.] January 1, 1905: Einstein publishes the paper. March 1, 1904: Aristotle explains the paper to Einstein. February 1, 1904: Aristotle meets Einstein, and they start discussing physics. January 1, 1904: Aristotle returns before the pubblication of the paper. January 1, 350 B.C.: Time traveler Aristotle leaves for the future. Where did the knowledge come from? Aristotle travels to the future Aristotle travels to the past Summary TOPIC RELEVANT EQUATIONS AND REMARKS 1. Classical relativity Galilean transformation x x vt y y z z t t 1-2 Newtonian relativity Newton\u2019s laws are invariant in all systems connected by a Galilean transformation. 56 Chapter 1 Relativity I TOPIC RELEVANT EQUATIONS AND REMARKS 2. Einstein\u2019s postulates The laws of physics are the same in all inertial reference frames. The speed of light is c, independent of the motion of the source. 3. Relativity of simultaneity Events simultaneous in one reference frame are not in general simultaneous in any other inertial frame. 4. Lorentz transformation x (x vt)y y z z 1-18 t (t vx>c2) with (1 v2>c2)1>2 5. Time dilation 6. Length contraction Proper time is the time interval between two events that occur at the same space point. If that interval is t then the time interval in S is, t t where (1 v2>c2)1>2 1-26 The proper length of a rod is the length Lp measured in the rest system of the rod. In S, moving at speed v with respect to the rod, the length measured is L Lp > 1-28 7. Spacetime interval All observers in inertial frames measure the same interval s between pairs of events in spacetime, where (\u00a2s)2 (c\u00a2t)2 (\u00a2x)2 1-31 8. Doppler effect Source/ observer approaching Source/ observer receding f 1 1 f0 B f 1 1 f0 B General References The following general references are written at a level appropriate for readers of this book. Bohm, D., The Special Theory of Relativity, W. A. Benjamin, New York, 1965. (trans. W. Perrett and J. B. Jeffery), Dover, New York, 1923. A delightful little book containing Einstein\u2019s original paper [\u201cOn the Electrodynamics of Moving Bodies,\u201d Annalen der Physik, 17 (1905)] and several other original papers on special relativity. French, A. P., Special Relativity, Norton, 1968. Includes an Ohanian, H. C., Special Relativity: A Modern Introduction, excellent discussion of the historical basis of relativity. Physics Curriculum & Instruction, 2001. Gamow, G., Mr. Tompkins in Paperback, Cambridge University Press, Cambridge, 1965. Contains the delightful Mr. Tompkins stories. In one of these Mr. Tompkins visits a dream world where the speed of light is only about 10 mi/h and relativistic effects are quite noticeable. Lorentz, H. A., A. Einstein, H. Minkowski, and W. Weyl, The Principle of Relativity: A Collection of Original Memoirs on the Special and General Theory of Relativity Pais, A., Subtle Is the Lord..., Oxford University Press, Oxford, 1982. Resnick, R., and D. Halliday, Basic Concepts in Relativity and Early Quantum Theory, 2d ed., Macmillan, 1992. Rindler, W., Essential Relativity, Van Nostrand Reinhold, New York, 1969. Taylor, E. F., and J. A. Wheeler, Spacetime Physics, 2d ed., W. H. Freeman and Co. 1992. This is a good book with many examples, problems, and diagrams. Notes 1. Polish astronomer, 1473\u20131543. His book describing heliocentric", " (i.e., Sun-centered) orbits for the planets was published only a few weeks before his death. He had hesitated to release it for many years, fearing that it might be considered heretical. It is not known whether or not he saw the published book. 2. Events are described by measurements made in a coordinate system that defines a frame of reference. The question was, Where is the reference frame in which the law of inertia is valid? Newton knew that no rotating system, e.g., Earth or the Sun, would work and suggested the distant \u201cfixed stars\u201d as the fundamental inertial reference frame. 3. The speed of light is exactly 299,792,458 m s. The value is set by the definition of the standard meter as being the distance light travels in 1 299,792,458 s. > > 4. Over time, an entire continuous spectrum of electromagnetic waves has been discovered, ranging from extremely lowfrequency (radio) waves to extremely high-frequency waves (gamma rays), all moving at speed c. 5. Albert A. Michelson (1852\u20131931), an American experimental physicist whose development of precision optical instruments and their use in precise measurements of the speed of light and the length of the standard meter earned him the Nobel Prize in 1907. Edward W. Morley (1838\u20131923), an American chemist and physicist and professor at Western Reserve College when Michelson was a professor at the nearby Case School of Applied Science. 6. Albert A. Michelson and Edward W. Morley, The American Journal of Science, XXXIV, no. 203, November 1887. 7. Note that the width depends on the small angle between M \u0153 and M1. A very small angle results in relatively few wide 2 fringes, a larger angle in many narrow fringes. 8. Since the source producing the waves, the sodium lamp, was at rest relative to the interferometer, the frequency would be constant. 9. T. S. Jaseja, A. Javan, J. Murray, and C. H. Townes, Physical Review, 133, A1221 (1964). 10. A. Brillet and J. Hall, Physical Review Letters, 42, 549 (1979). 11. Annalen der Physik, 17, 841(1905). For a translation from the original German, see the collection of original papers by Lorentz, Einstein, Minkowski, and Weyl (Dover, New York, 1923). 12. Hendrik Antoon Lorentz (1853\u20131928), Dutch theoretical physicist, discovered the Lorentz transformation empirically while investigating the fact that Maxwell\u2019s equations are not invariant under a Galilean transformation, although he did not recognize its importance at the time. An expert on electromagnetic Notes 57 theory, he was one of the first to suggest that atoms of matter might consist of charged particles whose oscillations could account for the emission of light. Lorentz used this hypothesis to explain the splitting of spectral lines in a magnetic field discovered by his student Pieter Zeeman, with whom he shared the 1902 Nobel Prize. 13. One meter of light travel time is the time for light to travel 1 m, i.e., ct 1 m, or t 1 m/3.00 108 m s 3.3 109 s. Similarly, 1 cm of light travel time is ct 1 cm, or t 3.3 1011 s, and so on. > 14. This example is adapted from a problem in H. Ohanian, Modern Physics (Englewood Cliffs, NJ: Prentice Hall, 1987). 15. Any particle that has mass. 16. Equation 1-31 would lead to imaginary values of s for spacelike intervals, an apparent problem. However, the geometry of spacetime is not Euclidean, but Lorentzian. While a consideration of Lorentz geometry is beyond the scope of this chapter, suffice it to say that it enables us to write (s)2 for spacelike intervals as in Equation 1-33. 17. There are only two such things: photons (including those of visible light), which will be introduced in Chapter 3, and gravitons, which are the particles that transmit the gravitational force. 18. Edwin P. Hubble, Proceedings of the National Academy of Sciences, 15, 168 (1929). 19. Walter K\u00fcndig, Physical Review, 129, 2371 (1963). 20. C. G. Darwin, Nature, 180, 976 (1957). 21. S. P. Boughn, American Journal of Physics, 57, 791 (1989). 22. E. F. Taylor and J. A. Wheeler, Spacetime Physics, 2d ed. (New York: W. H. Freeman and Co., 1992). 23. Seen in three space dimensions by the observer in S, 50 percent of the light is concentrated in 0", ".06 steradian of 4 steradian solid angle around the moving source. 24. T. Alv\u00e4ger and M. N. Kreisler, \u201cQuest for Faster-ThanLight Particles,\u201d Physical Review, 171, 1357 (1968). 25. Paul Ehrenfest (1880\u20131933), Austrian physicist and professor at the University of Leiden (the Netherlands), longtime friend and correspondent of Einstein, about whom, upon his death, Einstein wrote, \u201c[He was] the best teacher in our profession I have ever known.\u201d 26. This experiment is described in J. C. Hafele and R. E. Keating, Science, 177, 166 (1972). Although not as accurate as the experiment described in Section 1-4, its results supported the relativistic prediction. 27. R. Shaw, American Journal of Physics, 30, 72 (1962). 58 Chapter 1 Relativity I Problems Level I Section 1-1 The Experimental Basis of Relativity > > 1-1. In episode 5 of Star Wars, the Empire\u2019s spaceships launch probe droids throughout the > galaxy to seek the base of the Rebel Alliance. Suppose a spaceship moving at 2.3 108 m s toward Hoth (site of the rebel base) launches a probe droid toward Hoth at 2.1 108 m s relative to the spaceship. According to Galilean relativity, (a) What is the speed of the droid relative to Hoth? (b) If rebel astronomers are watching the approaching spaceship through a telescope, will they see the probe before it lands on Hoth? 1-2. In one series of measurements of the speed of light, Michelson used a path length L of 27.4 km (17 mi). (a) What is the time needed for light to make the round trip of distance 2L? (b) What is the classical correction term in seconds in Equation 1-5, assuming Earth\u2019s speed is v 104c? (c) From about 1600 measurements, Michelson arrived at a result for the speed of light of 299,796 4 km s. Is this experimental value accurate enough to be sensitive to the correction term in Equation 1-5? 1-3. A shift of one fringe in the Michelson-Morley experiment would result from a difference of one wavelength or a change of one period of vibration in the round-trip travel of the light when the interferometer is rotated by 90\u00b0. What speed would Michelson have computed for Earth\u2019s motion through the ether had the experiment seen a shift of one fringe? 1-4. In the \u201cold days\u201d (circa 1935) pilots used to race small, relatively high-powered airplanes around courses marked by a pylon on the ground at each end of the course. Suppose two such evenly matched racers fly at airspeeds of 130 mph. (Remember, this was a long time ago!) Each flies one complete round trip of 25 miles, but their courses are perpendicular to each other and there is a 20-mph wind blowing steadily parallel to one course. (a) Which pilot wins the race and by how much? (b) Relative to the axes of their respective courses, what headings must the two pilots use? 1-5. Paul Ehrenfest25 suggested the following thought experiment to illustrate the dramatically different observations that might be expected, dependent on whether light moved relative to a stationary ether or according to Einstein\u2019s second postulate: Suppose that you are seated at the center of a huge dark sphere with a radius of 3 108 m and with its inner surface highly reflective. A source at the center emits a very brief flash of light that moves outward through the darkness with uniform intensity as an expanding spherical wave. > What would you see during the first 3 seconds after the emission of the flash if (a) the sphere moved through the ether at a constant 30 km s and (b) if Einstein\u2019s second postulate is correct? 1-6. Einstein reported that as a boy he wondered about the following puzzle. If you hold a mirror at arm\u2019s length and look at your reflection, what will happen as you begin to run? In particular, suppose you run with speed v 0.99c. Will you still be able to see yourself? If so, what would your image look like, and why? 1-7. Verify by calculation that the result of the Michelson-Morley experiment places an upper limit on Earth\u2019s speed relative to the ether of about 5 km s. 1-8. Consider two inertial reference frames. When an observer in each frame measures the following quantities, which measurements made by the two observers must yield the same results? Explain your reason for each answer. > (a) The distance between two events (b) The value of the mass of a proton (c) The speed of light (d) The time interval between two events (", "e) Newton\u2019s first law ( f ) The order of the elements in the periodic table (g) The value of the electron charge Problems 59 Section 1-2 Einstein\u2019s Postulates > 1-9. Assume that the train shown in Figure 1-14 is 1.0 km long as measured by the observer at C and is moving at 150 km h. What time interval between the arrival of the wave fronts at C is measured by the observer at C in S? 1-10. Suppose that A, B, and C are at rest in frame S, which moves with respect to S at speed v in the x direction. Let B be located exactly midway between A and C. At t 0 a light flash occurs at B and expands outward as a spherical wave. (a) According to an observer in S, do the wave fronts arrive at A and C simultaneously? (b) According to an observer in S, do the wave fronts arrive at A and C simultaneously? (c) If you answered no to either (a) or (b), what is the difference in their arrival times and at which point did the front arrive first? Section 1-3 The Lorentz Transformation >. t\u0153 2 t\u0153 1 x\u0153 0 t\u0153 2. and v> c. 1> (1 v2>c2)1>2 as a function of (1 v2>c2)1>2. in frame S at times ranging from 0 up to 0.995. (a) Use Equation t\u0153 by a 1 (b) Why is Equation 1-18 less convenient than Equation 1-19 for this problem? 1-11. Make a graph of the relativistic factor Use at least 10 values of 1-12. Two events happen at the same point 1-19 to show that in frame S, the time interval between the events is greater than factor 1-13. Suppose that an event occurs in inertial frame S with coordinates x 75 m, y 18 m, z 4.0 m at t 2.0 105 s. The inertial frame S moves in the x direction with v 0.85c. The origins of S and S coincided at t t 0. (a) What are the coordinates of the event in S? (b) Use the inverse transformation on the results of (a) to obtain the original coordinates. 1-14. Show that the null effect of the Michelson-Morley experiment can be accounted for if the interferometer arm parallel to the motion is shortened by a factor of 1-15. Two spaceships are approaching each other. (a) If the speed of each is 0.9c relative to Earth, what is the speed of one relative to the other? (b) If the speed of each relative to Earth is 30,000 m s (about 100 times the speed of sound), what is the speed of one relative to the other? 1-16. Starting with the Lorentz transformation for the components of the velocity (Equation 1-23), derive the transformation for the components of the acceleration. 1-17. Consider a clock at rest at the origin of the laboratory frame. (a) Draw a spacetime diagram that illustrates that this clock ticks slow when observed from the reference frame of a rocket moving with respect to the laboratory at v 0.8c. (b) When 10 s have elapsed on the rocket clock, how many have ticked by on the lab clock? 1-18. A light beam moves along the y axis with speed c in frame S, which is moving to the right with speed v relative to frame S. (a) Find ux and uy, the x and y components of the velocity of the light beam in frame S. (b) Show that the magnitude of the velocity of the light beam in S is c. 1-19. A particle moves with speed 0.9c along the x axis of frame S, which moves with speed 0.9c in the positive x direction relative to frame S. Frame S moves with speed 0.9c in the positive x direction relative to frame S. (a) Find the speed of the particle relative to frame S. (b) Find the speed of the particle relative to frame S. Section 1-4 Time Dilation and Length Contraction 1-20. Use the binomial expansion to derive the following results for values of when applicable in the problems that follow. v V c and use (a) (b) (c v2 c2 v2 c2 1 2 v2 c2 1-21. How great must the relative speed of two observers be for their time-interval measurements to differ by 1 percent (see Problem 1-20)? 60 Chapter 1 Relativity I > 1-22. A nova is the sudden, brief brightening of a star (see Chapter 13). Suppose Earth astronomers see two novas occur simultaneously, one in", " the constellation Orion (the Hunter) and the other in the constellation Lyra (the Lyre). Both nova are the same distance from Earth, 2.5 103 c # y, and are in exactly opposite directions from Earth. Observers on board an aircraft flying at 1000 km h on a line from Orion toward Lyra see the same novas, but note that they are not simultaneous. (a) For the observers on the aircraft, how much time separates the nova? (b) Which one occurs first? (Assume Earth is an inertial reference frame.) 1-23. A meter stick moves parallel to its length with speed v 0.6c relative to you. (a) Compute the length of the stick measured by you. (b) How long does it take for the stick to pass you? (c) Draw a spacetime diagram from the viewpoint of your frame with the front of the meter stick at x 0 when t 0. Show how the answers to (a) and (b) are obtained from the diagram. 1-24. The proper mean lifetime of mesons (pions) is 2.6 10\u20138 s. If a beam of such particles has speed 0.9c, (a) What would their mean life be as measured in the laboratory? (b) How far would they travel (on the average) before they decay? (c) What would your answer be to part (b) if you neglected time dilation? (d) What is the interval in spacetime between creation of a typical pion and its decay? 1-25. You have been posted to a remote region of space to monitor traffic. Near the end of a quiet shift, a spacecraft streaks past. Your laser-based measuring device reports the spacecraft\u2019s length to be 85 m. The identification transponder reports it to be the NCXXB-12, a cargo craft of proper length 100 m. In transmitting your report to headquarters, what speed should you give for this spacecraft? 1-26. The light clock in the spaceship in Figure 1-25 uses a light pulse moving up the y-axis to reflect back from a mirror as the ship moves along the x-axis. Suppose instead the light pulse moves along the x-axis between x 0 and a mirror at x L. (a) What is the time required for the pulse to make a round trip in the rest system of the spaceship? (b) What is the round-trip time in the laboratory frame? (c) Does the result in (b) agree with that expected from time dilation? Justify your answer. 1-27. Two spaceships pass each other traveling in opposite directions. A passenger on ship A, which she knows to be 100 m long, notes that ship B is moving with a speed of 0.92c relative to A and that the length of B is 36 m. What are the lengths of the two spaceships measured by a passenger in B? 1-28. A meter stick at rest in S is tilted at an angle of 30\u00b0 to the x axis. If S moves at how long is the meter stick as measured in S and what angle does it make with the x axis? 1-29. A rectangular box at rest in S has sides a 2 m, b 2 m, and c 4 m and is oriented as shown in Figure 1-44. S moves with with respect to the laboratory frame S. (a) Compute the volume of the box in S and in S. (b) Draw an accurate diagram of the box as seen by an observer in S. 0.65 0.8, Section 1-5 The Doppler Effect 650 0)> 525 ( nm)? blue ( nm)? green ( nm) for it to appear yellow 1-30. How fast must you be moving toward a red light ( 590 460 ( nm)? 1-31. A distant galaxy is moving away from us at speed 1.85 107 m s. Calculate the fractional red shift 1-32. The light from a nearby star is observed to be shifted toward the blue by 2 percent, i.e., fobs Is the star approaching or receding from Earth? How fast is it moving? (Assume motion is directly toward or away from Earth to avoid superluminal speeds.) 1-33. Stars typically emit the red light of atomic hydrogen with wavelength 656.3 nm (called the spectral line). Compute the wavelength of that light observed at Earth from stars receding directly from us with relative speed v 103c, v 102c, and v 101c. of the light from this galaxy. 1.02f0. H > 0 Section 1-6 The Twin Paradox and Other Surprises 1-34. Heide boards a spaceship and travels away from Earth at a constant velocity 0.45c toward Betelgeuse (a red giant star in the constellation Orion). One year later on Earth clocks,", " Heide\u2019s twin, Hans, boards a second spaceship and follows her at a constant velocity of 0.95c in the y\u00b4 S\u00b4 a\u00b4 b\u00b4 25\u00b0 x\u00b4 z\u00b4 c\u00b4 Figure 1-44 Problems 61 same direction. (a) When Hans catches up to Heide, what will be the difference in their ages? (b) Which twin will be older? 1-35. You point a laser flashlight at the Moon, producing a spot of light on the Moon\u2019s surface. At what minimum angular speed must you sweep the laser beam in order for the light spot to streak across the Moon\u2019s surface with speed v c? Why can\u2019t you transmit information between research bases on the Moon with the flying spot? 1-36. A clock is placed in a satellite that orbits Earth with a period of 108 min. (a) By what time interval will this clock differ from an identical clock on Earth after 1 y? (b) How much time will have passed on Earth when the two clocks differ by 1.0 s? (Assume special relativity applies and neglect general relativity.) 1-37. Einstein used trains for a number of relativity thought experiments since they were the fastest objects commonly recognized in those days. Let\u2019s consider a train moving at 0.65c along a straight track at night. Its headlight produces a beam with an angular spread of 60\u00b0 according to the engineer. If you are standing alongside the track (rails are 1.5 m apart), how far from you is the train when you see its approaching headlight suddenly disappear? > Level II 1-38. In 1971 four portable atomic clocks were flown around the world in jet aircraft, two eastbound and two westbound, to test the time dilation predictions of relativity.26 (a) If the westbound plane flew at an average speed of 1500 km h relative to the surface, how long would it have had to fly for the clock on board to lose 1 second relative to the reference clock on the ground at the U.S. Naval Observatory? (b) In the actual experiment the planes circumflew Earth once and the observed discrepancy of the clocks was 273 ns. What was the average speed of each plane? 1-39. \u201cEther drag\u201d was among the suggestions made to explain the null result of the Michelson-Morley experiment (see More section). The phenomenon of stellar aberration refutes this proposal. Suppose Earth moves relative to the ether at velocity v and a light beam (e.g., from a star) approaches Earth at an angle with respect to v. (a) Show that the angle of approach in Earth\u2019s reference frame is given by tan sin cos v> c s 90\u00b0, differ from 90\u00b0? by how much does is the stellar aberration angle. If (b) 1-40. A friend of yours who is the same age as you travels to the star Alpha Centauri, which is c # y away, and returns immediately. He claims that the entire trip took just 6 years. (a) How 4 fast did he travel? (b) How old are you when he returns? (c) Draw a spacetime diagram that verifies your answers to (a) and (b). 1-41. A clock is placed in a satellite that orbits Earth with a period of 90 min. By what time interval will this clock differ from an identical clock on Earth after 1 year? (Assume that special relativity applies.) after event A and at x 1.5 km from event A. (a) How 1-42. In frame S, event B occurs 2 fast must an observer be moving along the x axis so that events A and B occur simultaneously? (b) Is it possible for event B to precede event A for some observer? (c) Draw a spacetime diagram that illustrates your answers to (a) and (b). (d) Compute the spacetime interval and proper distance between the events. 1-43. A burst of mesons travels down an evacuated beam tube at Fermilab moving at 0.92 for this group of pions. (b) The proper mean lifetime of pions is 2.6 108 s. What mean lifetime is measured in the lab? (c) If the burst contained 50,000 pions, how many remain after the group has traveled 50 m down the beam tube? (d) What would be the answer to (c) ignoring time dilation? 1-44. H. A. Lorentz suggested 15 years before Einstein\u2019s 1905 paper that the null effect of the Michelson-Morley experiment could be accounted for by a contraction of that arm of the interL Lp(1 v2>c2)1>2. ferometer lying parallel to Earth\u2019s motion through the ether to a length He thought of this, incorrectly, as an actual shrinking of matter. By about how many", " atomic diameters would the material in the parallel arm of the interferometer have had to shrink in order with respect to the laboratory. (a) Compute 62 Chapter 1 Relativity I s s m.) 1010, \u00a2f>f0 later at x2 8 107. to account for the absence of the expected shift of 0.4 of a fringe width? (Assume the diameter of atoms to be about 480 m. A second explo1-45. Observers in reference frame S see an explosion located at x1 1200 m. In reference frame S, which is moving along the x axis sion occurs 5 at speed v, the explosions occur at the same point in space. (a) Draw a spacetime diagram describing this situation. (b) Determine v from the diagram. (c) Calibrate the ct axis and deterbetween the two explosions as measured in S. (d) Verify your mine the separation in time in results by calculation. 1-46. Two spaceships, each 100 m long when measured at rest, travel toward each other with speeds of 0.85c relative to Earth. (a) How long is each ship as measured by someone on Earth? (b) How fast is each ship traveling as measured by an observer on the other? (c) How long is one ship when measured by an observer on the other? (d) At time t 0 on Earth, the fronts of the ships are together as they just begin to pass each other. At what time on Earth are their ends together? (e) Sketch accurately scaled diagrams in the frame of one of the ships showing the passing of the other ship. 1-47. If v is much less than c, the Doppler frequency shift is approximately given by \u00a2f>f0 both classically and relativistically. A radar transmitter-receiver bounces a signal What off an aircraft and observes a fractional increase in the frequency of is the speed of the aircraft? (Assume the aircraft to be moving directly toward the transmitter.) 1-48. The null result of the Michelson-Morley experiment could be explained if the speed of light depended on the motion of the source relative to the observer. Consider a binary eclipsing star system, that is, a pair of stars orbiting their common center of mass with Earth lying in the orbital plane of the system, as is very nearly the case for the binary system Algol (see More section about the Michelson-Morley experiment). Assume that the stars in the system have circular orbits with a period of 115 days and that one of the stars\u2019 orbital speed is 32 km s (about the same as Earth\u2019s orbital speed around the Sun). If the suggestion above were true, astronomers would simultaneously see two images of the star in opposition, i.e., on opposite sides of its orbit. What is the minimum distance L from Earth to the binary for this phenomenon to occur? 1-49. Frames S and S are moving relative to each other along the x and x axes. They set their # clocks to t t 0 when their origins coincide. In frame S, event 1 occurs at x 1 1 c y and # 0.5 y. These events occur simultaneously 2.0 c y and t2 t1 in frame S. (a) Find the magnitude and direction of the velocity of S relative to S. (b) At what time do both of these events occur as measured in S? (c) Compute the spacetime interval s between the events. (d) Is the interval spacelike, timelike, or lightlike? (e) What is the proper distance Lp between the events? 1-50. Do Problem 1-49 parts (a) and (b) using a spacetime diagram. 1-51. An observer in frame S standing at the origin observes two flashes of colored light separated spatially by x 2400 m. A blue flash occurs first, followed by a red flash 5 later. An observer in S moving along the x axis at speed v relative to S also observes the flashes 5 s apart and with a separation of 2400 m, but the red flash is observed first. Find the magnitude and direction of v. 1-52. A cosmic-ray proton streaks through the lab with velocity 0.85c at an angle of 50\u00b0 with the x direction (in the xy plane of the lab). Compute the magnitude and direction of the proton\u2019s velocity when viewed from frame S moving with 1 y and event 2 occurs at x2 0.72. s > measured in S. Level III 1-53. A meter stick is parallel to the x axis in S and is moving in the y direction at constant speed vy. From the viewpoint of S show that the meter stick will appear tilted at an angle with respect to the x axis of S moving in the x direction at Compute the angle 1-54. The equation for the spherical", " wave front of a light pulse that begins at the origin at time t 0 is Using the Lorentz transformation, show that such a light pulse also has a spherical wave front in S by showing that in S. 1-55. An interesting paradox has been suggested by R. Shaw27 that goes like this. A very thin steel plate with a circular hole 1 m in diameter centered on the y axis lies parallel to the xz plane x\u01532 y\u01532 z\u01532 (ct)2 0 x2 y2 z2 (ct)2 0. 0.65. Problems 63 y \u03b2 = v/c z vy vy x Figure 1-45 t2 D>c t1 x1 t1. T t2 and a time v> c. D x2 (T vD>c2). in frame S and moves in the y direction at constant speed vy as illustrated in Figure 1-45. A meter stick lying on the x axis moves in the x direction with The steel plate arrives at the y 0 plane at the same instant that the center of the meter stick reaches the origin of S. Since the meter stick is observed by observers in S to be contracted, it passes through the 1-m hole in the plate with no problem. A paradox appears to arise when one considers that an observer in S, the rest system of the meter stick, measures the diameter of the hole in the plate to be contracted in the x dimension and, hence, becomes too small to pass the meter stick, resulting in a collision. Resolve the paradox. Will there be a collision? 1-56. Two events in S are separated by a distance (a) Use the Lorentz transformation to show that in frame S, which is moving with speed v relative to S, the time separation is (b) Show that the events can be simultaneous in frame S only if D is greater than cT. (c) If one of the events is the cause of the other, the separation D must be less than cT since is the smallest time that a signal can take to travel from x1 to x2 in frame S. Show that if D is less than cT, in all reference frames. (d) Suppose that a signal could be sent with speed c c so that in frame S the cause precedes Show that there is then a reference frame moving with speed the effect by the time v less than c in which the effect precedes the cause. 1-57. Two observers agree to test time dilation. They use identical clocks and one observer in frame S moves with speed v 0.6c relative to the other observer in frame S. When their origins coincide, they start their clocks. They agree to send a signal when their clocks read 60 min and to send a confirmation signal when each receives the other\u2019s signal. (a) When does the observer in S receive the first signal from the observer in S. (b) When does he receive the confirmation signal? (c) Make a table showing the times in S when the observer sent the first signal, received the first signal, and received the confirmation signal. How does this table compare with one constructed by the observer in S? 1-58. The compact disk in a CD-ROM drive rotates with angular speed There is a clock at the center of the disk and one at a distance r from the center. In an inertial reference frame, the u r. clock at distance r is moving with speed Show that from time dilation in special relativity, time intervals to for the clock at rest and tr for the moving clock are related by is greater than T D>c.. t\u0153 2 t\u0153 1 \u00a2tr \u00a2to \u00a2to r22 2c2 if r V c 1-59. Two rockets, A and B, leave a space station with velocity vectors vA and vB relative to the station frame S, perpendicular to each other. (a) Determine the velocity of A relative to B, vBA. (b) Determine the velocity of B relative to A, vAB. (c) Explain why vAB and vBA do not point in opposite directions. 1-60. Suppose a system S consisting of a cubic lattice of meter sticks and synchronized clocks, e.g., the eight clocks closest to you in Figure 1-13, moves from left to right (the x direction) at high speed. The meter sticks parallel to the x direction are, of course, contracted and the cube 64 Chapter 1 Relativity I would be measured by an observer in a system S to be foreshortened in that direction. However, recalling that your eye constructs images from light waves that reach it simultaneously, not those leaving the source simultaneously, sketch what your eye would see in this case. Scale contractions and show any angles accurately. (Assume the moving cube to be farther than 10 m from your eye.) 1-61. Figure 1-", "11b (in the More section about the Michelson-Morley experiment) shows an eclipsing binary. Suppose the period of the motion is T and the binary is a distance L from Earth, where L is sufficiently large so that points A and B in Figure 1-11b are a half orbit apart. Consider the motion of one of the stars and (a) show that the star would appear to move from A to B in time assuming classical velocity addition applies to light, i.e., that emission theories of light were correct. (b) What rotational period would cause the star to appear to be at both A and B simultaneously? 1-62. Show that if a particle moves at an angle with respect to the x axis with speed u in system S, it moves at an angle with the x axis in S given by T>2 2Lv>(c2 v2), T>2 2Lv>(c2 v2) and from B to A in time tan sin (cos v> u) 1-63. Like jets emitted from some galaxies (see Figure 1-41), some distant astronomical objects can appear to travel at speeds greater than c across our line of sight. Suppose distant galaxy AB15 moving with velocity v at an angle with respect to the direction toward Earth emits two bright flashes of light separated by time t on the galaxy AB15 local clock. Show that (a) the and (b) the apparent speed of AB15 measured by obtime interval sin 1 cos. \u00a2t(1 cos ) \u00a2xEarth \u00a2tEarth compute the value of 0.75, \u00a2tEarth (c) For vapp servers on Earth is which vapp c. for CHAPTER 2 Relativity II In the opening section of Chapter 1, we discussed the classical observation that, if Newton\u2019s second law, F ma, holds in a particular reference frame, it also holds in any other reference frame that moves with constant velocity relative to it, i.e., in any inertial frame. As shown in Section 1-1, the Galilean transformation (Equation 1-2) leads to the same accelerations in both frames, and forces such as those due to stretched springs are also the same in both frames. However, according to the Lorentz transformation, accelerations are not the same in two such reference frames. If a particle has acceleration ax and velocity ux in frame S, its acceleration in S, obtained by computing from Equation 1-22, is ax a\u0153 x du\u0153 >dt x a\u0153 x ax 3(1 vux >c2)3 2-1 F>m Thus, does not hold. must transform in a similar way, or else Newton\u2019s second law, F ma, It is reasonable to expect that F ma does not hold at high speeds, for this equation implies that a constant force will accelerate a particle to unlimited velocity if it acts for a long time. However, if a particle\u2019s velocity were greater than c in some reference frame S, we could not transform from S to the rest frame of the particle because becomes imaginary when v c. We can show from the velocity transformation that, if a particle\u2019s velocity is less than c in some frame S, it is less than c in all frames moving relative to S with v c. This result leads us to expect that particles never have speeds greater than c. Thus, we expect that Newton\u2019s second law F ma is not relativistically invariant. We will, therefore, need a new law of motion, but one that resince F ma is consistent duces to Newton\u2019s classical version when with experimental observations when (v>c) S 0, V 1. In this chapter we will explore the changes in classical dynamics that are dictated by relativity theory, directing particular attention to the same concepts around which classical mechanics was developed, namely, mass, momentum, and energy. We will 2-1 Relativistic Momentum 2-2 Relativistic Energy 2-3 Mass/Energy Conversion and Binding Energy 2-4 Invariant Mass 2-5 General Relativity 66 70 81 84 97 65 66 Chapter 2 Relativity II find these changes to be every bit as dramatic as those we encountered in Chapter 1, including a Lorentz transformation for momentum and energy and a new invariant quantity to stand beside the invariant spacetime interval s. Then, in the latter part of the chapter, we will briefly turn our attention to noninertial, or accelerated, reference frames, the realm of the theory of general relativity. 2-1 Relativistic Momentum Among the most powerful fundamental concepts that you have studied in physics until now are the ideas of conservation of momentum and conservation of total energy. As we will discuss a bit further in Chapter 12, each of these fundamental laws arises because of a particular symmetry that exists in the laws of physics. For example, the conservation of total energy in classical physics is", " a consequence of the symmetry, or invariance, of the laws of physics to translations in time. As a consequence, Newton\u2019s laws work exactly the same way today as they did when he first wrote them down. The conservation of momentum arises from the invariance of physical laws to translations in space. Indeed, Einstein\u2019s first postulate and the resulting Lorentz transformation (Equations 1-18 and 1-19) guarantee this latter invariance in all inertial frames. The simplicity and universality of these conservation laws leads us to seek equations for relativistic mechanics, replacing Equation 1-1 and others, that are consistent with momentum and energy conservation and are also invariant under a Lorentz transformation. However, it is straightforward to show that the momentum, as formulated in classical mechanics, does not result in relativistic invariance of the law of conservation of momentum. To see that this is so, we will look at an isolated collision between two masses, where we avoid the question of how to transform forces because the net external force is zero. In classical mechanics, the total momentum p miui is conserved. We will see that relativistically, conservation of the quantity miui is an approximation that holds only at low speeds. S Consider one observer in frame S with a ball A and another in with ball B. The balls each have mass m and are identical when measured at rest. Each observer throws his ball along his y axis with speed u0 (measured in his own frame) so that the balls collide.1 Assuming the balls to be perfectly elastic, each observer will see his ball rebound with its original speed u0. If the total momentum is to be conserved, the y component must be zero because the momentum of each ball is merely reversed by the collision. However, if we consider the relativistic velocity transformation, we can see that the quantity muy does not have the same magnitude for each ball as seen by either observer. Let us consider the collision as seen in frame S (Figure 2-1a). In this frame ball u0. Ball B has x component of velocity A moves along the y axis with velocity uyA uxB v and y component uyB u\u0153 yB > u0 21 v2>c2 2-2 and u0 u\u0153 Here we have used the velocity transformation (Equation 1-22) and the facts that yB We see that the y component of the velocity of ball B is is just smaller in magnitude than that of ball A. The quantity comes from the time dilation factor. The time taken for ball B to travel a given distance along the y axis in S is greater than the time measured in S for the ball to travel this same distance. (1 v2>c2)1>2 0. u\u0153 xB 2-1 Relativistic Momentum 67 Thus, in S the total y component of classical momentum is not zero. Since the y components of the velocities are reversed in an elastic collision, momentum as defined by p mu is not conserved in S. Analysis of this problem in S leads to the same conclusion (Figure 2-1b), since the roles of A and B are simply interchanged.2 In the 1 classical limit u0. uyB and The reason for defining momentum as mu in classical mechanics is that this quantity is conserved when there is no external force, as in our collision example. We now see that this quantity is conserved only in the approximation We will define the relativistic momentum p of a particle to have the following properties: momentum is conserved, of course, because in that limit v V c, v V c. 1. p is conserved in collisions. 2. p approaches mu as u>c approaches zero. Let\u2019s apply the first of these conditions to the collision of the two balls that we just discussed, noting two important points. First, for each observer in Figure 2-1, the (for the obspeed of each ball is unchanged by the elastic collision. It is either server\u2019s own ball) or (for the other ball). Second, the failure of the conservation of momentum in the collision we described can\u2019t be due to the velocities because we used the Lorentz transformation to find the y components. It must have something to do with the mass! Let us write down the conservation of the y component of the momentum as observed in S, keeping the masses of the two balls straight by writing for the S observer\u2019s own ball and v2)1>2 u observer\u2019s ball. for the m(u) (u2 y u0 S m(u0) m(u0)u0 m(u)uyB m(u0)u0 m(u)uyB (before collision) (after collision) Equation 2-3 can be readily rewritten as m(u) m", "(u0) u0 uyB 2-3 2-4 (a) y\u00b4 v B S\u00b4 x\u00b4 u0 A u0 B x\u00b4 y S (b) y\u00b4 S\u00b4 y v S A x x Figure 2-1 (a) Elastic collision of two identical balls as seen in frame S. The vertical component of the > velocity of ball B is in S if it is u0 in S. (b) The same collision as seen in S. In this frame, ball A has vertical component of velocity u0 >. u0 is small compared to the relative speed v of the reference frames, then it follows u0 If from Equation 2-2 that uyB V v and, therefore, If we can now imagine the limiting case where i.e., where each ball is at rest in its \u201chome\u201d frame so that the collision becomes a \u201cgrazing\u201d one as B moves past A at speed v u, then we conclude from Equations 2-2 and 2-4 that in order for Equation 2-3 to hold, i.e., for the momentum to be conserved, u v. u0 S 0, m(u v) 0) m(u0 u0 21 v2>c2 u0 or m(u) m 21 v2>c2 2-5 Equation 2-5 says that the observer in S measures the mass of ball B, moving relative to him at speed u, as equal to times the rest mass of the ball, or its mass measured in the frame in which it is at rest. Notice that observers always measure the mass of an object that is in motion with respect to them to be larger than the value measured when the object is at rest. 1>(1 v2>c2)1>2 The design and construction of large particle accelerators throughout the world, such as CERN\u2019s LHC, are based directly on the relativistic expressions for momentum and energy. 68 Chapter 2 Relativity II Thus, we see that the law of conservation of momentum will be valid in relativity, provided that we write the momentum p of an object with rest mass m moving with velocity u relative to an inertial system S to be p mu 21 u2>c2 2-6 where u is the speed of the particle. We therefore take this equation as the definition of relativistic momentum. It is clear that this definition meets our second criterion because the denominator approaches 1 when u is much less than c. From this definition, the momenta of the two balls A and B in Figure 2-1 as seen in S are pyA mu0 21 u2 0 >c2 pyB muyB 21 (u2 xB u2 yB)>c2 u0(1 v2>c2)1>2 uyB pyA. where pyB Lorentz transformation, Equation 2-6 is often written uxB Because of the similarity of the factor v. and It is similarly straightforward to show that 1> 11 u2>c2 in the and p mu with 1 21 u2>c2 2-7 This use of the symbol for two different quantities causes some confusion; the notation is standard, however, and simplifies many of the equations. We will use this notation except when we are also considering transformations between reference frames. Then, to avoid confusion, we will write out the factor and where v is the relative speed of the frames. Figure 2-2 reserve The quantity m(u) in shows a graph of the magnitude of p as a function of Equation 2-5 is sometimes called the relativistic mass; however, we will avoid using the term or a symbol for relativistic mass: in this book m always refers to the mass 1>(1 v2>c2)1>2, 1>(1 u2>c2)1>2 u>c. for 4mc p 3mc 2mc mc 0 0 Relativistic momentum 0.2 0.4 0.6 u/c 0.8 1.0 u>c, Figure 2-2 Relativistic momentum as given where u speed by Equation 2-6 versus of the object relative to an observer. The magnitude of the momentum p is plotted in units of mc. The fainter dashed line shows the classical momentum mu for comparison. 2-1 Relativistic Momentum 69 measured in the rest frame. In this we are following Einstein\u2019s view. In a letter to a colleague in 1948 he wrote:3 of a It is not good to introduce the concept of mass body for which no clear definition can be given. It is better to introduce no other mass than \u201cthe rest mass\u201d m. Instead of introducing M, it is better to mention the expression for the momentum and energy of a body in motion. M m>(1 v 2>c 2)", "1>2 EXAMPLE 2-1 Measured Values of Moving Mass For what value of measured mass of an object u>c exceed the rest mass by a given fraction f? m will the SOLUTION From Equation 2-5 we see that m m m f Solving for u>c, 1 1 21 u2>c2 1 1 u2>c2 1 (f 1)2 \u00a1 u2>c2 1 1 (f 1)2 or u>c 2f(f 2) f 1 for any from which we can compute the table of values below or the value of other f. Note that the value of that results in a given fractional increase f in the measured value of the mass is independent of m. A diesel locomotive moving at a will be observed to have the same f as a proton moving with that u>c. particular u>c u>c u>c f 1012 5 109 0.0001 0.01 (1%) 1.0 (100%) 10 100 u>c Example 1.4 106 jet fighter aircraft 0.0001 0.014 0.14 0.87 0.996 Earth\u2019s orbital speed 50-eV electron quasar 3C273 quasar 0Q172 muons from cosmic rays 0.99995 some cosmic ray protons EXAMPLE 2-2 Momentum of a Rocket A high-speed interplanetary probe with a mass m 50,000 kg has been sent toward Pluto at a speed u 0.8c. What is its momentum as measured by Mission Control on Earth? If, preparatory to landing on Pluto, the probe\u2019s speed is reduced to 0.4c, by how much does its momentum change? 70 Chapter 2 Relativity II SOLUTION 1. Assuming that the probe travels in a straight line toward Pluto, its momentum along that direction is given by Equation 2-6: mu 21 u2>c2 p 6.7 104 c # kg 2.0 1013 kg # m>s (50,000 kg)(0.8c) 21 (0.8c)2>c2 2. When the probe\u2019s speed is reduced, the momentum declines along the relativistic momentum curve in Figure 2-2. The new value is computed from the ratio: p0.4c p0.8c m(0.4c)> 21 (0.4)2 m(0.8c)> 21 (0.8)2 3. The reduced momentum p0.4c 21 (0.8)2 21 (0.4)2 1 2 0.33 is then given by p0.4c 0.33p0.8c (0.33)(6.7 104 c # kg) 2.2 104 c # kg 6.6 1012 kg # m>s Remarks: Notice from Figure 2-2 that the incorrect classical value of p0.8c # have been 4.0 104 c kg. Also, while the probe\u2019s speed was decreased to initial value, the momentum was decreased to of the initial value. 1>3 would 1>2 its Question 1. In our discussion of the inelastic collision of balls A and B, the collision was a \u201cgrazing\u201d one in the limiting case. Suppose instead that the collision is a \u201chead-on\u201d one along the x axis. If the speed of v 0.1c, what would a spacetime diagram of the collision look like? (i.e., ball B) is low, say, S 2-2 Relativistic Energy As noted in the preceding section, the fundamental character of the principle of conservation of total energy leads us to seek a definition of total energy in relativity that preserves the invariance of that conservation law in transformations between inertial systems. As with the definition of the relativistic momentum, Equation 2-6, we will require that the relativistic total energy E satisfy two conditions: 1. The total energy E of any isolated system is conserved. 2. E will approach the classical value when u>c approaches zero. Let us first find a form for E that satisfies the second condition and then see if it also satisfies the first. We have seen that the quantity mu is not conserved in collisions We have also noted that Newton\u2019s second but that 1> (1 u2>c2)1>2. is, with mu 2-2 Relativistic Energy 71 Aerial view of the Jefferson Laboratory\u2019s Continuous Electron Beam Accelerator Facility (CEBAF) in Virginia. The dashed line indicates the location of the underground accelerator, where electrons are accelerated to 6 GeV and reach speeds of more than 99.99 percent of the speed of light. The circles outline the experiment halls, also underground. [Thomas Jefferson National Accelerator Facility/U.S. Department of Energy.] law in", " the form F ma cannot be correct relativistically, one reason being that it leads to the conservation of mu. We can get a hint of the relativistically correct form of the second law by writing it F dp dt. This equation is relativistically correct if the relativistic momentum p is used. We thus define the force in relativity to be > F dp dt d(mu) dt 2-8 as the work done by Now, as in classical mechanics, we will define kinetic energy a net force in accelerating a particle from rest to some velocity u. Considering motion in one dimension only, we have Ek Ek u u0 F dx u 0 d(mu) dt dx u 0 u d(mu) using u dx dt. The computation of the integral in this equation is not difficult but requires a bit of algebra. It is left as an exercise (Problem 2-2) to show that > d(mu) ma1 3>2 du b u2 c2 Substituting this into the integrand of the equation for Ek above, we obtain u d(mu) u u ma1 u du 3>2 b Ek u2 c2 0 0 mc2a 1 21 u2>c2 1b or mc2 mc2 Ek 2-9 72 Chapter 2 Relativity II Figure 2-3 Experimental confirmation of the relativistic relation for kinetic energy. Electrons were accelerated to energies up to several MeV in large electric fields, and their velocities were determined by measuring their time of flight over 8.4 m. Note that when the velocity the relativistic and nonrelativistic (i.e., classical) relations are indistinguishable. [W. Bertozzi, American Journal of Physics, 32, 551 (1964).] u V c, Nonrelativistic 1 Ek = mu2 \u2013\u2013 2 Relativistic Ek = mc2 1 \u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013 1 \u2013 (u/c)2 \u2013 1 1 2 3 4 5 Kinetic energy (MeV) 2 u\u2013\u2013c 1.0 0.5 0.0 0 is not mu2>2 or even Equation 2-9 defines the relativistic kinetic energy. Notice that, as we warned earlier, Ek This is strikingly evident in Figure 2-3. However, consistent with our second condition on the relativistic total energy E, Equation 2-9 does approach We can check this assertion by noting that for u>c V 1, u V c. expanding by the binomial theorem yields mu2>2 when mu2>2. and thus a1 1>2 b u2 c2 1 1 2 u2 c2 \u00c1 Ek mc2a1 1 2 u2 c2 \u00c1 1b 1 2 mu2 The expression for kinetic energy in Equation 2-9 consists of two terms. One mc2, depends on the speed of the particle (through the factor ), and the other term, mc2, is called the rest energy of term, is independent of the speed. The quantity the particle, i.e., the energy associated with the rest mass m. The relativistic total energy E is then defined as the sum of the kinetic energy and the rest energy. mc2 E Ek mc2 mc2 mc2 21 u2>c2 2-10 Thus, the work done by a net force increases the energy of the system from the rest (or increases the measured mass from m to energy mc2 mc2 m to ). For a particle at rest relative to an observer, and Equation 2-10 becomes perhaps the most widely recognized equation in all of physics, Einstein\u2019s famous E mc2. When Equation 2-10 can be written as u V c, Ek 0, E 1 2 mu2 mc2 2-2 Relativistic Energy 73 Before the development of relativity theory, it was thought that mass was a conserved quantity;4 consequently, m would always be the same before and after an interaction or event and would therefore be constant. Since the zero of energy is arbitrary, we are always free to include an additive constant; therefore, our definition of the relativistic total energy reduces to the classical kinetic energy for and our second condition on E is satisfied.5 u V c mc2 Ek mc2. and not Be very careful to understand Equation 2-10 correctly. It defines the total energy E, and E is what we are seeking to conserve for isolated systems in all inertial frames, not Remember, too, the distinction between conserved quantities and invariant quantities. The former have the same value before and after an interaction in a particular reference frame. The latter have the same value when measured by observers in different reference frames. Thus, we are not requiring observers in relatively moving inertial frames to measure the same values for E, but rather that E be unchanged in interactions as measured in each frame. To assist us in showing that E as defined by", " Equation 2-10 is conserved in relativity, we will first see how E and p transform between inertial reference frames. Lorentz Transformation of E and p Consider a particle of rest mass m that has an arbitrary velocity u with respect to frame S, as shown in Figure 2-4. System is a second inertial frame moving in the x direction. The particle\u2019s momentum S and energy are given in the S and systems, respectively, by, S y S y\u00b4 S\u00b4 v In S: where In S: where E mc2 mux px muy py muz pz 1> 21 u2>c2 E mc2 p\u0153 mu\u0153 x x mu\u0153 p\u0153 y y mu\u0153 p\u0153 z z u x, x\u00b4 z z\u00b4 Figure 2-4 Particle of mass m moves with velocity u measured in S. System S moves in the x direction at speed v. The Lorentz velocity transformation makes possible determination of the relations connecting measurements of the total energy and the components of the momentum in the two frames of reference. 2-11 2-12 1> 21 u2>c2 Developing the Lorentz transformation for E and p requires that we first express in terms of quantities measured in S. (We could just as well express in terms of primed quantities. Since this is relativity, it makes no difference which we choose.) The result is 1 21 u2>c2 >c2) (1 vux 21 u2>c2 where now 1 21 v2>c2 2-13 74 Chapter 2 Relativity II Substituting Equation 2-13 into the expression for E in Equation 2-12 yields E mc2 21 u2>c2 c mc2 21 u2>c2 >c2 mc2vux 21 u2>c2 d The first term in the brackets you will recognize as E, and the second term, canceling the from Equation 2-11. Thus, we have factors, as c2 vpx E (E vpx) Similarly, substituting Equation 2-13 and the velocity transformation for expression for in Equations 2-12 yields p\u0153 x p\u0153 x mu\u0153 x 21 u2>c2 c mux 21 u2>c2 mv 21 u2>c2 d 2-14 u\u0153 x into the px The first term in the brackets is vE>c2. E>c2, Thus, we have the second term is from Equation 2-11, and, because m(1 u2>c2)1>2 p\u0153 x (px vE>c2) 2-15 Using the same approach, we can show (Problem 2-46) that p\u0153 y and p\u0153 z py pz Together these relations are the Lorentz transformation for momentum and energy: p\u0153 x (px vE>c2) p\u0153 y p\u0153 z E (E vpx) The inverse transformation is (p\u0153 px x E (E vp \u0153 vE>c2) py x) pz py pz p\u0153 y p\u0153 z 2-16 2-17 with 1 21 v2>c2 1 21 2 Note the striking similarity between Equations 2-16 and 2-17 and the Lorentz transformation of the space and time coordinates, Equations 1-18 and 1-19. The moand the total enmentum ergy E transforms like the time t. We will return to this remarkable result and related matters shortly, but first let\u2019s do some examples and then, as promised, show that the energy as defined by Equation 2-10 is conserved in relativity. transforms in relativity exactly like p(px, py, pz) r(x, y, z), EXAMPLE 2-3 Transforming Energy and Momentum Suppose a micrometeorite of mass 109 kg moves past Earth at a speed of 0.01c. What values will be measured for the energy and momentum of the particle by an observer in a system moving relative to Earth at 0.5c in the same direction as the micrometeorite? S 2-2 Relativistic Energy 75 SOLUTION Taking the direction of the micrometeorite\u2019s travel to be the x axis, the energy and momentum as measured by the Earth observer are, using the approximation of Equation 2-10: u V c E 1 2 mu2 mc2 109 kg[(0.01c)2>2 c2] E 1.00005 109 c2 J and px For this situation momentum will be mux 1.1547, (109 kg)(0.01c) 1011c kg # m>s so in S the measured values of the energy and E (E vpx) (1.15", "47)[1.00005 109c2 (0.5c)(1011c)] E (1.1547)(1.00005 109 0.5 1011)c2 E 1.14898 109 c2 J and p\u0153 x p\u0153 x p\u0153 x vE>c2) (1.1547)[1011c (0.5c)(1.00005 109c2)>c2] (px (1.1547)(1011 5.00025 1010)c 5.66 1010 c kg # m>s 56.6 1011 c kg # m>s Thus, the observer in measures a total energy nearly 15 percent larger and a momentum more than 50 times greater and in the x direction. S EXAMPLE 2-4 A More Difficult LorentzTransformation of Energy Suppose that a particle with mass m and energy E is moving toward the origin of a system S such that its velocity u makes an angle with the y axis, as shown in Figure 2-5. Using the Lorentz transformation for energy and momentum, determine the energy of the S, which moves relative to S so that the partiparticle measured by an observer in cle moves along the axis. E y SOLUTION S System moves in the x direction at speed Thus, Lorentz velocity transformation for u sin, v u sin. 0. E mc2> 21 u2>c2 p mu> 21 u2>c2 u\u0153 x Also, as determined from the and from the latter, p Amu> 21 u2>c2Bsin In S the energy will be E (E vpx) 1 21 v2>c2 [E (u sin )Am > 21 v2>c2Bsin ] 1 21 u2 sin2 >c2 [E Am> 21 u2>c2Bu2 sin2 ] y m \u03b1 u S x Figure 2-5 The system discussed in Example 2-4. 76 Chapter 2 Relativity II Multiplying the second term in the brackets by terms yield c2>c2 and factoring an E from both E E 21 (u2>c2)sin and u c Since in which case S and E S E cos. sin2 1, S we see that E E, except for are the same system. Note, too, that for 0, if 0 when E E, u S c, As we will see later, this is the case for light. Question 2. Recalling the results of the measurements of time and space intervals by observers in motion relative to clocks and measuring rods, discuss the results of corresponding measurements of energy and momentum changes. Conservation of Energy As with our discussion of momentum conservation in relativity, let us consider a collision of two identical particles, each with rest mass m. This time, for a little variety, we will let the collision be completely inelastic\u2014i.e., when the particles collide, they called the zero momentum frame, the particles apstick together. In the system axis with equal speeds u\u2014hence equal and opposite proach each other along the momenta\u2014as illustrated in Figure 2-6a. In this frame the collision results in the formation of a composite particle of mass M at rest in If moves with respect to a second frame S at speed v u in the x direction, then the particle on the right before the collision will be at rest in S and the composite particle will move to the right at speed u in that frame. This situation is illustrated in Figure 2-6b. S, x S. S (a) (b) Figure 2-6 Inelastic collision of two particles of equal rest mass m. (a) In the zero momentum frame S the particles have equal and opposite velocities and hence momenta. After the collision, the composite particle of mass M is at rest in S. The diagram on the far right is the spacetime diagram of the collision from the viewpoint of S. (b) In system S the frame S is moving to the right at speed u so that the particle on the right is at rest in S, while the left one moves at After collision, the composite particle moves to the right at speed u. Again, the spacetime diagram of the interaction is shown on the far right. All diagrams are drawn with the collision occurring at the origin. 2u>(1 u2>c2). y\u00b4 S\u00b4 (before) y\u00b4 S\u00b4 (after) u \u2013u m m x\u00b4 M x\u00b4 y S (before) y S (after) 2u/(1+u2/c2) m m x u M x ct\u00b4 S\u00b4 Worldline of M M x\u00b4 m m ct S M Worldline of M m x m 2-2 Relativistic Energy 77 Using the total energy as defined by Equation 2-10, we have in S : Before collision: mc2 21 u", "2>c2 E\u0153 before mc2 21 u2>c2 2mc2 21 u2>c2 After collision: E\u0153 after Mc2 Energy will be conserved in S if E\u0153 before E\u0153 after, i.e., if 2mc2 21 u2>c2 Mc2 2-18 2-19 2-20 This is ensured by the validity of conservation of momentum, in particular by Equation 2-5, and so energy is conserved in (The validity of Equation 2-20 is important and not trivial. We will consider it in more detail in Example 2-7.) To see if energy as we have defined it is also conserved in S, we transform to S using the inverse transform, Equation 2-17. We then have in S: S. Before collision: Ebefore (E\u0153 before vp\u0153 x) Ebefore a up\u0153 x b 2mc2 21 u2>c2 2mc2 21 u2>c2 Ebefore a b since p\u0153 x 0 2-21 After collision: Eafter (Mc2 up\u0153 x) Mc2 since again p\u0153 x 0 2-22 The energy will be conserved in S and, therefore, the law of conservation of energy Eafter, will hold in all inertial frames if Ebefore i.e., if a 2mc2 21 u2>c2 b Mc2 2-23 which, like Equation 2-20, is ensured by Equation 2-5. Thus, we conclude that the energy as defined by Equation 2-10 is consistent with a relativistically invariant law of conservation of energy, satisfying the first of the conditions set forth at the beginning of this section. While this demonstration was not a general one, since that is beyond the scope of our discussions, you may be assured that our conclusion is quite generally valid. Question 3. Explain why the result of Example 2-4 does not mean that energy conservation is violated. 78 Chapter 2 Relativity II EXAMPLE 2-5 Mass of Cosmic Ray Muons In Chapter 1, muons produced as secondary particles by cosmic rays were used to illustrate both the relativistic length contraction and time dilation resulting from their high speed relative to observers on Earth. That speed is about 0.998c. If the rest energy of a muon is 105.7 MeV, what will observers on Earth measure for the total energy of a cosmic ray\u2013produced muon? What will they measure for its mass? SOLUTION The electron volt (eV), the amount of energy acquired by a particle with electric charge equal in magnitude to that on an electron (e) accelerated through a potential difference of 1 volt, is a convenient unit in physics, as you may have learned. It is defined as 1.0 eV 1.602 1019 C 1.0 V 1.602 1019 J 2-24 Commonly used multiples of the eV are the keV (103 eV), the MeV (106 eV), the GeV (109 eV), and the TeV (1012 eV). Many experiments in physics involve the measurement and analysis of the energy and/or momentum of particles and systems of particles, and Equation 2-10 allows us to express the masses of particles in energy units rather than the SI unit of mass, the kilogram. That and the convenient size of the eV facilitate6 numerous calculations. For example, the mass of an electron is 9.11 1031 kg. Its rest energy is given by E mc2 9.11 1031 kg # c2 8.19 1014 J or or E 8.19 1014 J 1 1.602 1019 J>eV 5.11 105 eV E 0.511 MeV rest energy of the electron The mass of the particle is often expressed with the same number thus: m E c2 0.511 MeV>c2 mass of the electron Now, applying the above to the muons produced by cosmic rays, each has a total energy E given by E mc2 1 21 (0.998c)2>c2 105.7 MeV c2 c2 E 1670 MeV and a measured mass (see Equation 2-5) of m E>c2 1670 MeV>c2 The dependence of the measured mass on the speed of the particle has been verified by numerous experiments. Figure 2-7 illustrates a few of those results. 6.0 5.0 4.0 3.0 2.0 m m/ \u03b3 2-2 Relativistic Energy 79 The data points are Figure 2-7 A few of the many experimental measurements of the mass of electrons as a u>c. function of their speed plotted onto Equation 2-5, the solid line. The data points represent the work of W. Kaufmann (, 1901), A. H. Bucherer (", ", 1908), and W. Bertozzi (\u25cf, 1964). Note that Kaufmann\u2019s work preceded the appearance of Einstein\u2019s 1905 paper on special relativity. Kaufmann used an incorrect mass for the electron and interpreted his results as support for classical theory. [Adapted from Figure 3-4 in R. Resnick and D. Halliday, Basic Concepts in Relativity and Early Quantum Theory, 2d ed. (New York: Macmillan, 1992).] 1.0 0 0.2 0.4 0.6 0.8 1.0 u/c EXAMPLE 2-6 Change in the Solar Mass Compute the rate at which the Sun is losing mass, given that the mean radius R of Earth\u2019s orbit is 1.50 108 km and the intensity of solar radiation at Earth (called the solar constant) is 1.36 103 W m2. > SOLUTION 1. The conversion of mass into energy, a consequence of conservation of energy in relativity, is implied by Equation 2-10. With u 0 that equation becomes E mc2 2. Assuming that the Sun radiates uniformly over a sphere of radius R, the total power radiated by the Sun is given by P (area of the sphere)(solar constant) (4R2)(1.36 103 W>m2) 4(1.50 1011 m)2(1.36 103 W>m2) 3.85 1026 J>s 3. Thus, every second the Sun emits 3.85 1026 J, which, from Equation 2-10, is the result of converting an amount of mass given by m E>c2 3.85 1026 J (3.00 108 m>s)2 4.3 109 kg Remarks: Thus, the Sun is losing 4.3 109 kg of mass (about 4 million metric tons) every second! If this rate of mass loss remains constant (which it will for the next few billion years) and with a fusion mass-to-energy conversion efficiency of about 1 percent, the Sun\u2019s present mass of about 2.0 1030 kg will \u201conly\u201d last for about 1011 more years! 80 Chapter 2 Relativity II EXPLORING From Mechanics, Another Surprise One consequence of the fact that Newton\u2019s second law, F ma, is not relativistically invariant is yet another surprise\u2014the lever paradox. Consider a lever of mass m at rest in S (Figure 2-8). Since the lever is at rest, the net torque due to the forces Fx and Fy is zero, i.e. (using magnitudes): net net Fxb Fya 0 y x and therefore Fxb Fya An observer in system moving with lever moving in the S x direction and measures the torque to be 0.866 ( 2) with respect to S sees the \u0153 net \u0153 x \u0153 y F \u0153 xb F \u0153 y a Fxb (Fy >2)(a>2) Fxb Fxb>4 (3>4)Fxb 0 where F \u0153 x Fx and F \u0153 y Fy >2 (see Problem 2-53) and the lever is rotating! The resolution of the paradox was first given by the German physicist Max von Laue (1879\u20131960). Recall that the net torque is the rate of change of the angular momentum L. The observer measures the work done per unit time by the two forces as S For For xv Fxv zero, since F \u0153 y is perpendicular to the motion and the change in mass m per unit time of the moving lever is \u00a2m \u00a2t \u00a2E>c2 \u00a2t 1 c2 \u00a2E \u00a2t 1 c2 Fxv (a) y (b) y\u00b4 Pin a Fy Pin a\u00b4 F\u00b4y S b Fx 0 x 0\u00b4 b\u00b4 F\u00b4x S\u00b4 v \u03b3 = 2 \u03b2 = 0.866 x\u00b4 Figure 2-8 (a) A lever in the xy plane of system S is free to rotate about the pin P but is S held at rest by the two forces Fx and Fy. (b) The same lever as seen by an observer in that is moving with instantaneous speed v in the x direction. For the is moving in the direction. x observer the lever S 2-3 Mass/Energy Conversion and Binding Energy 81 S The time given by observer measures a change in the magnitude of angular momentum per unit net \u00a2L \u00a2t b\u00a2p \u00a2t bv\u00a2m \u00a2t Substituting for \u00a2m>\u00a2t from above yields Fxv c2 bFx bv \u0153 net \u00a2L \u00a2t v2 c2 bFx 2 3 4 Fxb F \u0153 x an observer in that system sees the As a result of the motion of the", " lever relative to force doing net work on the lever, thus changing the angular momentum over time, and the paradox vanishes. (The authors thank Costas Efthimiou for bringing this paradox to our attention.) S, m u m u\u00b4 (a) m u S (b) m S\u00b4 u u M M > Figure 2-9 Two objects colliding with a massless spring that locks shut. The total rest mass of the system M is greater than that of the parts 2m by the amount Ek c2, where Ek is the internal energy, which in this case is the original kinetic energy. (a) The event as seen in a reference frame S in which the final mass M is at rest. (b) The same event as seen in a frame S moving to the right at speed u relative to S, so that one of the initial masses is at rest. 2-3 Mass/Energy Conversion and Binding Energy > The identification of the term mc2 as rest energy is not merely a convenience. Whenever additional energy E in any form is stored in an object, the mass of the object is increased by E c2. This is of particular importance whenever we want to compare the mass of an object that can be broken into constituent parts with the mass of the parts (for example, an atom containing a nucleus and electrons, or a nucleus containing protons and neutrons). In the case of the atom, the mass changes are usually negligibly small (see Example 2-8). However, the difference between the mass of a nucleus and that of its constituent parts (protons and neutrons) is often of great importance. Ek 2A 1 the original kinetic energy As an example, consider Figure 2-9a in which two particles, each with mass m, are moving toward each other with speeds u. They collide with a spring that compresses and locks shut. (The spring is merely a device for visualizing energy storage.) In the Newtonian 2 mu2B mechanics description, is converted into potential energy of the spring U. When the spring is unlocked, the potential energy reappears as kinetic energy of the par U, ticles. In relativity theory, the internal energy of the system, Ek appears as an increase in the rest mass of the system. That is, the mass of the system M is now greater than 2m by Ek c2. (We will derive this result in the next example.) This change in mass is too small to be observed for ordinary-size masses and springs, but it is easily observed in transformations that involve nuclei. For example, in the fission of a nucleus, the energy released as kinetic energy of the fission fragments is an appreciable fraction of the rest energy of the original nucleus. (See Example 11-19.) This energy can be calculated by measuring the difference between the mass of the original system and the total mass of the fragments. Einstein was the first to point out this possibility in 1905, even before the discovery of the atomic nucleus, at the end of a very short paper that followed his famous article on relativity. 7 After deriving the theoretical equivalence of energy and mass, he wrote: 235U > It is not impossible that with bodies whose energy content is variable to a high degree (e.g., with radium salts) the theory may be successfully put to the test. The relativistic conversion of mass into energy is the fundamental energy source in the nuclear reactor\u2013based systems that produce electricity in 30 nations and in large naval vessels and nuclear submarines. 82 Chapter 2 Relativity II EXAMPLE 2-7 Change in the Rest Mass of the Two-Particle and Spring System of Figure 2-9 Derive the increase in the rest mass of a system of two particles in a totally inelastic collision. Let m be the mass of each particle so that the total mass of the system is 2m when the particles are at rest and far apart, and let M be the rest mass of the system when it has internal energy Ek. The original kinetic energy in the reference frame S (Figure 2-9a) is 2mc2( 1) Ek 2-25 SOLUTION In a perfectly inelastic collision, momentum conservation implies that both particles are at rest after collision in this frame, which is the center-of-mass frame. The total kinetic energy is therefore lost. We wish to show that if momentum is to be conserved in any reference frame moving with a constant velocity relative to S, the total mass of the system must increase by m, given by \u00a2m Ek c2 2m( 1) 2-26 We therefore wish to show that the total mass of the system with internal energy is M, given by M 2m \u00a2m 2m 2-27 To simplify the mathematics, we chose a second reference frame moving to the right with speed v u relative to frame S so that one of the particles is initially at rest, as shown in Figure 2-9b. The initial speed of the other particle in", " this frame is S u u v 1 uv>c2 2u 1 u2>c2 2-28 After collision, the particles move together with speed u toward the left (since they are at rest in S). The initial momentum in S is The final momentum is p\u0153 i mu 21 u2>c2 to the left p\u0153 f Mu 21 u2>c2 to the left Using Equation 2-28 for gives u, squaring, dividing by c2, and adding 1 to both sides Then 1 u2 c2 1 4u2>c2 (1 u2>c2)2 (1 u2>c2)2 (1 u2>c2)2 p\u0153 i m[2u>(1 u2>c2)] (1 u2>c2)>(1 u2>c2) 2mu (1 u2>c2) 2-3 Mass/Energy Conversion and Binding Energy 83 Conservation of momentum in frame S requires that p\u0153 f p\u0153 i, or Solving for M, we obtain Mu 21 u2>c2 2mu 1 u2>c2 M 2m 21 u2>c2 2m which is Equation 2-27. Thus, the measured value of M would be 2m. If the latch in Figure 2-9b were to suddenly come unhooked, the two particles would fly apart with equal momenta, converting the rest mass m back into kinetic energy. The derivation is similar to that in Example 2-7. Mass and Binding Energy When a system of particles is held together by attractive forces, energy is required to break up the system and separate the particles. The magnitude of this energy Eb is called the binding energy of the system. An important result of the special theory of relativity that we will illustrate by example in this section is The mass of a bound system is less than that of the separated particles by Eb c 2, where Eb is the binding energy. > In atomic and nuclear physics, masses and energies are typically given in atomic mass units (u) and electron volts (eV) rather than in standard SI units of kilograms and joules. The u is related to the corresponding SI units by 1 u 1.66054 1027 kg 931.494 MeV>c2 2-29 (The eV was defined in terms of the joule in Equation 2-24.) The rest energies of some elementary particles and a few light nuclei are given in Table 2-1, from which you can see by comparing the sums of the masses of the constituent particles with the nuclei listed that the mass of a nucleus is not the same as the sum of the masses of its parts. 2H, The simplest example of nuclear binding energy is that of the deuteron, which consists of a neutron and a proton bound together. Its rest energy is 1875.613 MeV. The sum of the rest energies of the proton and neutron is 938.272 939.565 1877.837 MeV. Since this is greater than the rest energy of the deuteron, the deuteron cannot spontaneously break up into a neutron and a proton without violating conservation of energy. The binding energy of the deuteron is 1877.837 1875.613 2.224 MeV. In order to break up a deuteron into a proton and a neutron, at least 2.224 MeV must be added. This can be done by bombarding deuterons with energetic particles or electromagnetic radiation. If a deuteron is formed by combination of a neutron and a proton (fusion; see Chapter 11), the same amount of energy is released. 84 Chapter 2 Relativity II Table 2-1 Rest energies of some elementary particles and light nuclei Particle Symbol Rest energy (MeV) Photon Neutrino (antineutrino) Electron (positron) Muon Pi mesons (pions) Proton Neutron Deuteron Helion Alpha () e or e (e) (0) p n 2H or d 3He or h 4He or 0 2.8 106 0.5110 105.7 139.6 (135) 139.6 938.272 939.565 1875.613 2808.391 3727.379 EXAMPLE 2-8 Binding Energy of the Hydrogen Atom The binding energies of atomic electrons to the nuclei of atoms are typically of the order of 106 times those characteristic of particles in the nuclei; consequently, the mass differences are correspondingly smaller. The binding energy of the hydrogen atom (the energy needed to remove the electron from the atom) is 13.6 eV. How much mass is lost when an electron and a proton form a hydrogen atom? SOLUTION The mass of the proton plus that of the electron must be greater than that of the hydrogen atom by 13", ".6 eV 931.5 MeV>u 1.46 108 u This mass difference is so small that it is usually neglected. 2-4 Invariant Mass In Chapter 1 we discovered that as a consequence of Einstein\u2019s relativity postulates, the coordinates for space and time are linearly dependent on one another in the Lorentz transformation that connects measurements made in different inertial reference frames. Thus, the time t became a coordinate, in addition to the space coordinates x, y, and z, in the four-dimensional relativistic \u201cworld\u201d that we call spacetime. We note in passing that the geometry of spacetime was not the familiar Euclidean geometry of our three-dimensional world, but the four-dimensional Lorentz geometry. The difference became apparent when one compared the computation of the distance r between two points in space with that of the interval between two events in spacetime. The forr2 x2 y2 z2. mer is, of course, the vector r, whose magnitude is given by The vector r is unchanged (invariant) under a Galilean transformation in space, and quantities that transform like r are also vectors. The latter we called the spacetime interval s, and its magnitude, as we have seen, is given by (\u00a2s)2 (c\u00a2t)2 [(\u00a2x)2 (\u00a2y)2 (\u00a2z)2] 2-30 2-4 Invariant Mass 85 The interval s is the four-dimensional analog of r and therefore is called a fourvector. Just as x, y, and z are the components of the three-vector r, the components of the four-vector s are x, y, z, and ct. We have seen that s is also invariant under a Lorentz transformation in spacetime. Correspondingly, any quantity that transforms like s\u2014i.e., is invariant under a Lorentz transformation\u2014will also be a four-vector. The physical significance of the invariant interval s is quite profound: for timelike (the proper time interval), for spacelike intervals s Lp (the intervals proper length), and the proper intervals can be found from measurements made in any inertial frame.8 \u00a2s>c In the relativistic energy and momentum we have components of another fourvector. In the preceding sections we saw that the momentum and energy, defined by Equations 2-6 and 2-10, respectively, were not only both conserved in relativity, but also together satisfied the Lorentz transformation, Equations 2-16 and 2-17, with p(px, py, pz) the components of the momentum transforming like the space components of r(x, y, z) and the energy transforming like the time t. The questions then are, What invariant four-vectors are they components of? and, What is its physical significance? The answers to both turn out to be easy to find and yield for us yet another relativistic surprise. By squaring Equations 2-6 and 2-10, you can readily verify that E2 (pc)2 (mc2)2 This very useful relation we will rearrange slightly to (mc2)2 E2 (pc)2 2-31 2-32 Comparing the form of Equation 2-32 with that of Equation 2-30 and knowing that E and p transform according to the Lorentz transformation, we see that the magnitude of the invariant energy/momentum four-vector is the rest energy of the mass m! Thus, observers in all inertial frames will measure the same value for the rest energy of isolated systems and, since c is constant, the same value for the mass. Note that only in the rest frame of the mass m, i.e., the frame where p 0, are the rest energy and the total energy equal. Even though we have written Equation 2-31 for a single particle, we could as well have written the equations for momentum and energy in terms of the total momentum and total energy of an entire ensemble of noninteracting particles with arbitrary velocities. We would only need to write down Equations 2-6 and 2-10 for each particle and add them together. Thus, the Lorentz transformation for momentum and energy, Equations 2-16 and 2-17, holds for any system of particles and so therefore does the invariance of the rest energy expressed by Equation 2-32. We may state all of this more formally by saying that the kinematic state of the system is described by the four-vector s where (\u00a2s)2 (c\u00a2t)2 [(\u00a2x)2 (\u00a2y)2 (\u00a2z)2] and its dynamic state is described by the energy/momentum four-vector mc2, given by (mc2)2 E2 (pc)2 The next example illustrates how this works", ". 86 Chapter 2 Relativity II EXAMPLE 2-9 Rest Mass of Moving Object A particular object is observed to move through the laboratory at high speed. Its total energy and the components of its momentum are measured by lab workers to be (in SI units) E 4.5 1017 J, 3.8 108 kg # m>s, 3.0 108 kg # m>s, 3.0 108 kg # m>s. px What is the object\u2019s rest mass? and py pz SOLUTION A From Equation 2-32 we can write (mc2)2 (4.5 1017)2 [(3.8 108c)2 (3.0 108c)2 (3.0 108c)2] (4.5 1017)2 [1.4 1017 9.0 1016 9.0 1016]c2 2.0 1035 2.9 1034 1.74 1035 J2 m (1.74 1035 J2)1>2>c2 4.6 kg SOLUTION B A slightly different but sometimes more convenient calculation that doesn\u2019t involve carrying along large exponentials makes use of Equation 2-32 divided by c4: m2 a 2 b E c2 a 2 b p c 2-33 Notice that this is simply a unit conversion, expressing each term in (mass)2 units\u2014 e.g., kg2 when E and p are in SI units: m2 a 4.5 1017 c a 3.8 108 a 3.0 108 a 3.0 108 c2 (5.0)2 [(1.25)2 (1.0)2 (1.0)2] 25 3.56 m (21.4)1>2 4.6 kg In the example, we determined the rest energy and mass of a rapidly moving object using measurements made in the laboratory without the need to be in the system in which the object was at rest. This ability is of enormous benefit to nuclear, particle, and astrophysicists, whose work regularly involves particles moving at speeds close to that of light. For particles or objects whose rest mass is known, we can use the invariant magnitude of the energy/momentum four-vector to determine the values of other dynamic variables, as illustrated in the next example. EXAMPLE 2-10 Speed of a Fast Electron The total energy of an electron produced in a particular nuclear reaction is measured to be 2.40 MeV. Find the electron\u2019s momentum and speed in the laboratory frame. (The rest mass of an electron is 9.11 1031 kg and its rest energy is 0.511 MeV.) SOLUTION The magnitude of the momentum follows immediately from Equation 2-31: pc 2E2 (mc2)2 2(2.40 MeV)2 (0.511 MeV)2 2.34 MeV p 2.34 MeV>c 2-4 Invariant Mass 87 where we have again made use of the convenience of the eV as an energy unit. The resulting momentum unit MeV c can be readily converted to SI units by converting the MeV to joules and dividing by c, i.e., 1 MeV>c 1.602 1013 J 2.998 108 m>s 5.34 1022 kg # m>s > Therefore, the conversion to SI units is easily done, if desired, and yields 5.34 1022 kg # m>s 1 MeV>c p 2.34 MeV>c p 1.25 1021 kg # m>s The speed of the particle is obtained by noting from Equation 2-32 or from Equations 2-6 and 2-10 that or u c pc E 2.34 MeV 2.40 MeV 0.975 u 0.975c 2-34 It is extremely important to recognize that the invariant rest energy in Equation 2-32 is that of the system and that its value is not the sum of the rest energies of the particles of which the system is formed, if the particles move relative to one another. Earlier we used numerical examples of the binding energy of atoms and nuclei that illustrated this fact by showing that the masses of the atoms and nuclei were less than the sum of the masses of their constituents by an amount mc2 that equaled the observed binding energy, but those were systems of interacting particles\u2014i.e., there were forces acting between the constituents. A difference exists, even when the particles do not interact. To see this, let us focus our attention on specifically what mass is invariant. 3c # kg, Consider two identical noninteracting particles, each of rest mass m 4 kg movas illustrated ing toward each other along the x axis of S with momentum in Figure 2-10a. The energy of each particle, using Equation 2-33, is px E>c2", " 2m2 (p>c)2 2(4)2 (3)2 5 kg kg, since the energy is a Thus, the total energy of the system is scalar. Similarly, the total momentum of the system is since the momentum is a vector and the momenta are equal and opposite. The rest mass of the system is then 3c # kg 3c # kg 0, 5c2 5c2 10c2 m 2(E>c2)2 (p>c)2 2(10)2 02 10 kg Hence, the system mass of 10 kg is greater than the sum of the masses of the two particles, 8 kg. (This is in contrast to bound systems, such as atoms, where the system mass is smaller than the total of the constituents.) This difference is not binding energy, since the particles are noninteracting. Neither does the 2 kg \u201cmass difference\u201d reside equally with the two particles. In fact, it doesn\u2019t reside in any particular place but is a property of the entire system. The correct interpretation is that the mass of the system is 10 kg. Although the invariance of the energy/momentum four-vector guarantees that observers in other inertial frames will also measure 10 kg as the mass of the system, let e.g., the one shown us allow for a skeptic or two and transform to another system in Figure 2-10c, just to be sure. S, 88 Chapter 2 Relativity II (a) y S u = 0.6c u = \u20130.6c m = 4 kg m = 4 kg x (b) y S System m = 10 kg x Figure 2-10 (a) Two identical particles with rest mass 4 kg approach each other with equal but oppositely directed momenta. The rest mass of the system made up of the two particles is not 4 kg 4 kg because the system\u2019s rest mass includes the mass equivalent of its internal motions. That value, 10 kg (b), would be the result of a measurement of the system\u2019s mass made by an observer in S, for whom the system is at rest, or by observers in any other inertial frames. (c) Transforming to S moving at v 0.6c with respect to S, as described in Example 2-11, also yields m 10 kg. (c) y\u00b4 S\u00b4 System v = 0.6c m = 10 kg x\u00b4 EXAMPLE 2-11 Lorentz Transformation of System Mass For the system illustrated S, 0.6, in Figure 2-10, show that an observer in also measures the mass of the system to be 10 kg. which moves relative to S at SOLUTION 1. The mass m measured in S is given by Equation 2-33, which in this case is m [(E>c2)2 (p\u0153 >c)2]1>2 x 2. E is given by Equation 2-16: E (E vpx) 1 21 (0.6)2 (1.25)(10c2) 12.5 c2 # kg is also given by Equation 2-16: 3. p\u0153 x (10c2 0.6c 0) p\u0153 x vE>c2) (px (1.25)[0 (0.6c)(10c2)>c2] 7.5 c # kg into Equation 2-33 yields m [(12.5c2>c2)2 (7.5c>c)2]1>2 p\u0153 x and 4. Substituting E [(12.5)2 (7.5)2]1>2 10 kg Remarks: This result agrees with the value measured in S. The speed of chosen for this calculation, v 0.6c, is convenient in that one of the particles consti; however, that has no effect on the generality of tuting the system is at rest in the solution. S S 2-4 Invariant Mass 89 Thus, we see that it is the rest energy of any isolated system that is invariant, whether that system is a single atom or the entire universe. And based on our discussions so far, we note that the system\u2019s rest energy may be greater than, equal to, or less than the sum of the rest energies of the constituents depending on their relative velocities and the detailed character of any interactions between them. Questions 4. Suppose two loaded boxcars, each of mass m 50 metric tons, roll toward each other on level track at identical speeds u, collide, and couple together. Discuss the mass of this system before and after the collision. What is the effect of the magnitude of u on your discussion? 5. In 1787 Count Rumford (1753\u20131814) tried unsuccessfully to measure an increase in the weight of a barrel of water when he increased its temperature from 29", "\u00b0F to 61\u00b0F. Explain why, relativistically, you would expect such an increase to occur, and outline an experiment that might, in principle, detect the change. Since Count Rumford preceded Einstein by about 100 years, why might he have been led to such a measurement? Massless Particles (mc2)2, Equation 2-32 formally allows positive, negative, and zero values for just as was the case for the spacetime interval (s)2. We have been tacitly discussing positive cases so far in this section; a discussion of possible negative cases we will defer until Chapter 12. Here we need to say something about the mc2 0 possibility. Note first of all that the idea of zero rest mass has no analog in classical physics since classically If m 0, then the momentum and kinetic energy are alEk ways zero too, and the \u201cparticle\u201d seems to be nothing at all, experiencing no secondlaw forces, doing no work, and so forth. However, for mc2 0, Equation 2-32 states that in relativity mu2>2 p mu. and E pc (for m 0) 2-35 and, together with Equation 2-34, that u c, i.e., a particle whose mass is zero moves at the speed of light. Similarly, a particle whose speed is measured to be c will have m 0 and satisfy E pc. We must be careful, however, because Equation 2-32 was obtained from the rel- ativistic definitions of E and p: E mc2 mc2 21 u2>c2 p mu mu 21 u2>c2 u S c, 1> 21 u2>c2 S ; m, which is tending toward however, since m is also approaching zero, the quanAs 0>0, tity can (and does) remain defined. Indeed, there is ample experimental evidence for the existence of particles with mc2 0. Current theories suggest the existence of three such particles. Perhaps the most important of these and the one thoroughly verified by experiment is the photon, a particle of electromagnetic radiation (i.e., light). Classically, electromagnetic radiation was interpreted via Maxwell\u2019s equations as a wave phenomenon, its energy and momentum being distributed continuously throughout the space occupied by the wave. 90 Chapter 2 Relativity II It was discovered around 1900 that the classical view of light required modification in certain situations, the change being a confinement of the energy and momentum of the radiation into many tiny packets or bundles, which were referred to as photons. Photons move at light speed, of course, and, as we have noted, are required by relativity to have mc2 0. Recall that the spacetime interval s for light is also zero. Strictly speaking, of course, the second of Einstein\u2019s relativity postulates prevents a Lorentz transformation to the rest system of light since light moves at c relative to all inertial frames. Consequently, the term rest mass has no operational meaning for light. EXAMPLE 2-12 Rest Energy of a System of Photons Remember that the rest energy of a system of particles is not the sum of the rest energies of the individual particles if they move relative to one another. This applies to photons too! Suppose two photons, one with energy 5 MeV and the second with energy 2 MeV, approach each other along the x axis. What is the rest energy of this system? SOLUTION px and The momentum of the 5-MeV photon is (from Equation 2-35) 2 MeV>c. px Thus, the energy of the system is that of the 2-MeV photon is E 5 MeV 2 MeV 7 MeV and its momentum is p 5 MeV c 2 MeV c > 3 MeV c. From Equation 2-32 the system\u2019s rest energy is 5 MeV>c > > mc2 2(7 MeV)2 (3 MeV)2 6.3 MeV!! A second particle whose rest energy is zero is the gluon. This massless particle transmits or carries the strong interaction between quarks, which are the \u201cbuilding blocks\u201d of all fundamental particles, including protons and neutrons. The existence of gluons is well established experimentally. We will discuss quarks and gluons further in Chapter 12. Finally, there are strong theoretical reasons to expect that gravity is transmitted by a massless particle called the graviton, which is related to gravity in much the same way that the photon is related to the electromagnetic field. Gravitons, too, move at speed c. While direct detection of the graviton is beyond our current and foreseeable experimental capabilities, major international cooperative experiments are currently under way to detect gravity waves. (See Section 2-5.) Until about the beginning of this century a fourth particle, the neutrino, was also thought to have zero rest mass. However, substantial experimental evidence collected by", " the Super-Kamiokande (Japan) and SNO (Canada) imaging neutrino detectors, among others, made it clear that neutrinos are not massless. We discuss neutrino mass and its implications further in Chapters 11 and 12. Creation and Annihilation of Particles The relativistic equivalence of mass and energy implies still another remarkable prediction that has no classical counterpart. As long as momentum and energy are conserved in the process, 9 elementary particles with mass can combine with their antiparticles, the masses of both being completely converted to energy in a process called annihilation. An example is that of an ordinary electron. An electron can orbit briefly with its antiparticle, called a positron,10 but then the two unite, mutually annihilating and producing two or three photons. The two-photon version of this process is shown schematically in Figure 2-11. Positrons are produced naturally by cosmic rays in the upper atmosphere and as the result of the decay of certain radioactive nuclei. P. A. M. Dirac predicted their existence in 1928 while investigating the invariance of the energy/momentum four-vector. 2-4 Invariant Mass 91 (a) (b) + \u2013 Figure 2-11 (a) A positron orbits with an electron about their common center of mass, shown by the dot between them. (b) After a short time, typically of the order of 1010 s for the case shown here, the two annihilate, producing two photons. The orbiting electron-positron pair, suggestive of a miniature hydrogen atom, is called positronium. u V c If the speeds of both the electron and the positron (not a requirement for the process, but it makes the following calculation clearer), then the total energy of each particle is E mc2 0.511 MeV. Therefore, the total energy of the system in Figure 2-11a before annihilation is 2mc2 1.022 MeV. Also from the diagram, the momenta of the particles are always opposite and equal, and so the total momentum of the system is zero. Conservation of momentum then requires that the total momentum of the two photons produced also be zero, i.e., that they move in opposite directions relative to the original center of mass and have equal momenta. Since E pc for photons, then they must also have equal energy. Conservation of energy then requires that the energy of each photon be 0.511 MeV. (Photons are usually called gamma rays when their energies are a few hundred keV or higher.) Notice from Example 2-12 that the magnitude of the energy/ momentum four-vector (the rest energy) is not zero, even though both of the final particles are photons. In this case it equals the rest energy of the initial system. Analysis of the three-photon annihilation, although the calculation is a bit more involved, is similar. By now you will not be surprised to learn that the reverse process, the creation of mass from energy, can also occur under the proper circumstances. The conversion of mass and energy works both ways. The energy needed to create the new mass can be provided by the kinetic energy of another massive particle or by the \u201cpure\u201d energy of a photon. In either case, in determining what particles might be produced with a given amount of energy, it is important to be sure, as was the case with annihilation, that the appropriate conservation laws are satisfied. As we will discuss in detail in Chapter 12, this restricts the creation process for certain kinds of particles (including electrons, protons, and neutrons) to producing only particle-antiparticle pairs. This means, for example, that the energy in a photon cannot be used to create a single electron but must produce an electron-positron pair. Decay of a Z into an electron-positron pair in the UA1 detectors at CERN. This is the computer image of the first Z event recorded (30 April 1983). The newly created pair leave the central detector in opposite directions at nearly the speed of light. [CERN.] 92 Chapter 2 Relativity II (a) S \u2013 (b) + \u2013 \u2013 S u = 0.8c Figure 2-12 (a) A photon of energy E and momentum p E c encounters an electron at rest. The photon produces an electron-positron pair (b), and the group move off together at speed u 0.8c. > To see how relativistic creation of mass goes, let us consider a particular situation, the creation of an electron-positron pair from the energy of a photon. The photon moving through space encounters, or \u201chits,\u201d an electron at rest in frame S, as illustrated in Figure 2-12a.11 Usually the photon simply scatters, but occasionally a pair is created. Encountering the existing electron is important since it is not possible for the photon to produce spontaneously the two rest masses of the pair", " and also conserve momentum. (See Problem 2-45.) Some other particle must be nearby, not to provide energy to the creation process, but to acquire some of the photon\u2019s initial momentum. In this case we have selected an electron for this purpose because it provides a neat example, but almost any particle would do. (See Example 2-13.) While near the electron, the photon suddenly disappears, and an electron-positron pair appears. The process must occur very fast since the photon, moving at speed c, will travel across a region as large as an atom in about 1019 s. Let\u2019s suppose that the details of the interaction that produced the pair are such that the three particles all move off together toward the right in Figure 2-12b with the same speed u\u2014i.e., they which moves to the right with speed u relative to S.12 What must are all at rest in the energy of the photon be for this particular electron-positron pair to be created? To answer this question, we first write the conservation of energy and momentum: S, E Before pair creation mc2 Ef Ei pi E E c Ei After pair creation mc2 E E c pi pf where mc2 rest energy of an electron. In the final system after pair creation the total rest energy is 3mc2 in this case. We know this because the invariant rest energy equals the sum of the rest energies of the constituent particles (the original electron and the S pair) in the system where they do not move relative to one another, i.e., in we have for the system after pair creation So in S. (3mc2)2 E2 (pc)2 9(mc2)2 (E mc2)2 a Ec c 2 b 9(mc2)2 E2 2Emc2 (mc2)2 E2 Noting that the E2 terms cancel and dividing the remaining terms by mc2, we see that E 4mc2 2-4 Invariant Mass 93 Thus, the initial photon needs energy equal to 4 electron rest energies in order to create 2 new electron rest masses in this case. Why is the \u201cextra\u201d energy needed? they must also Because the three electrons in the final system share momentum have kinetic energy Ek given by >c, E Ek E 3mc2 (E 4mc2 mc2 3mc2 2mc2 mc2) 3mc2 or the initial photon must provide the 2mc2 necessary to create the electron and positron masses and the additional 2mc2 of kinetic energy that they and the existing electron share as a result of momentum conservation. The speed u at which the group of particles moves in S can be found from pc E (Equation 2-34): u>c > a E c (E cb mc2) u>c 4mc2 5mc2 0.8 The portion of the incident photon\u2019s energy that is needed to provide kinetic energy in the final system is reduced if the mass of the existing particle is larger than that of an electron and, indeed, can be made negligibly small, as illustrated in the following example. EXAMPLE 2-13 Threshold for Pair Production What is the minimum or threshold energy that a photon must have in order to produce an electron-positron pair? SOLUTION The energy E of the initial photon must be E mc2 Ek mc2 Ek EkM where mc2 electron rest energy, Ek and are the kinetic energies of the elec- kinetic energy of the existing particle of tron and positron, respectively, and mass M. Since we are looking for the threshold energy, consider the limiting case where the pair is created at rest in S, i.e., and correspondingly p Therefore, momentum conservation requires that Ek EkM Ek p 0. 0 Ek pinitial E >c pfinal Mu 21 u2>c2 > 105 103 to MeV c2 and the value of E where u speed of recoil of the mass M. Since the masses of single atoms are in the range of at the threshold is clearly less than 2 MeV (i.e., it must be less than the value 4mc2 2.044 MeV, the speed with which M recoils from the creation event is quite small compared with c, even for the smallest M available, a single proton! (See Table 2-1.) Thus, the kinetic energy becomes negligible, and we conclude that the minimum energy of the initial photon that can produce an electron-positron pair is 2mc2, i.e., that E needed just to create the two rest masses. 2 mu2 EkM 1 E 94 Chapter 2 Relativity II E = (pc)2 + (mc2)2 Some Useful Equations and Approximations pc mc2 Figure 2-13 Triangle showing the relation between energy, momentum, and rest mass in special relativity. Caution: Remember that E and pc are not relat", "ivistically invariant. The invariant is mc2. E2 (pc)2 (mc2)2 2-31 Extremely Relativistic Case The triangle shown in Figure 2-13 is sometimes useful in remembering this result. If the energy of a particle is much greater than its rest energy mc2, the second term on the right of Equation 2-31 can be neglected, giving the useful approximation E pc for E W mc2 2-36 This approximation is accurate to about 1 percent or better if E is greater than about 8mc2. Equation 2-36 is the exact relation between energy and momentum for particles with zero rest mass. From Equation 2-36 we see that the momentum of a high-energy particle is simply its total energy divided by c. A convenient unit of momentum is MeV c. The momentum of a charged particle is usually determined by measuring the radius of curvature of the path of the particle moving in a magnetic field. If the particle has charge q and a velocity u, it experiences a force in a magnetic field B given by > F qu B where F is perpendicular to the plane formed by u and B and hence is always perpendicular to u. Since the magnetic force is always perpendicular to the velocity, it does no work on the particle (the work-energy theorem also holds in relativity), so the energy of the particle is constant. From Equation 2-10 we see that if the energy is constant, must be a constant, and therefore the speed u is also constant. So F qu B dp dt d(mu) dt m du dt u \u22a5 B, > the particle moves in a circle with centripetal acceleration For the case u2 R. (If u is not perpendicular to B, the path is a helix. Since the component of u parallel to B is unaffected, we will only consider motion in a plane.) We then have quB m ` du dt ` ma b u2 R or BqR mu p 2-37 This is the same as the nonrelativistic expression except for the factor of Figure 2-14 u>c. shows a plot of BqR mu versus It is useful to rewrite Equation 2-37 in terms of practical but mixed units; the result is >. p 300 BR a b q e 2-38 where p is in MeV c, B is in tesla, and R is in meters. > 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 1 \u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013 1 \u2013 u2/c2 0 0.1 0.2 0.3 0.5 0.6 0.7 0.8 0.4 u/c 2-4 Invariant Mass 95 > > Figure 2-14 BqR mu versus u c for particle of charge q and mass m moving in a circular orbit of radius R in a magnetic field B. The agreement of the data with the curve predicted by relativity theory supports the assumption that the force equals the time rate of change of relativistic momentum. [Adapted from I. Kaplan, Nuclear Physics, 2d ed., Reading, MA: Addison-Wesley Publishing Company, Inc., 1962; by permission.] R q B \u2013 \u2013 \u2013 \u2013 u m EXAMPLE 2-14 Electron in a Magnetic Field What is the approximate radius of the path of a 30-MeV electron moving in a magnetic field of 0.05 tesla ( 500 gauss)? SOLUTION 1. The radius of the path is given by rearranging Equation 2-38 and substituting q e: R p 300 B 2. In this situation the total energy E is much greater than the rest energy mc2: E 30 MeV W mc2 0.511 MeV 3. Equation 2-36 may then be used to determine p: p E>c 30 MeV>c 4. Substituting this approximation for p into Equation 2-38 yields R 30 MeV>c (300)(0.05) 2 m Remarks: In this case the error made by using the approximation, Equation 2-36, rather than the exact solution, Equation 2-31, is only about 0.01 percent. Nonrelativistic Case Nonrelativistic expressions for energy, momentum, and other quantities are often easier to use than the relativistic ones, so it is important to know all the relativistic expressions when these expressions are accurate enough. As approach the classical ones. In most situations, the kinetic energy or the total energy is is, from Equation 2-10, given, so that the most convenient expression for calculating S 1, E mc2 1 Ek mc2 2-39 96 Chapter 2 Relativity II When the kinetic energy is much less than the rest energy, is approximately 1 and nonrelat", "ivistic equations can be used. For example, the classical approximation (1>2)mu2 p2>2m Ek if can be used instead of the relativistic is much less than mc2. We can get an idea of the accuracy of these expressions by Ek expanding using the binomial expansion as was done in Section 2-2, and examining the first term that is neglected in the classical approximation. We have ( 1)mc 2 Ek, and Then a1 1>2 b u2 c2 1 1 2 u2 c2 3 8 u4 c4 \u00c1 Ek ( 1)mc 2 1 2 mu2 3 2 2 mu2B 2 A 1 mc2 \u00c1 Ek 1 2 mu2 Ek 3 2 Ek mc2 Ek >mc2 1 percent, the error in using the approximation if is about 1.5 percent. For example, (1>2)mu2 Ek At very low energies, the velocity of a particle can be obtained from its kinetic energy just as in classical mechanics. At very high energies, the velocity of a particle is very near c and the following approximation is sometimes useful (see Problem 2-28): (1>2)mu2 Ek u c 1 1 22 for W 1 2-40 An exact expression for the velocity of a particle in terms of its energy and momentum was obtained in Example 2-10. u c pc E 2-41 This expression, of course, is not useful if the approximation been made. E pc has already EXAMPLE 2-15 Different Particles, Same Energy An electron and a proton are each accelerated through 10 106 V. Find the momentum, and the speed for each., SOLUTION Since each particle has a charge of e, each acquires a kinetic energy of 10 MeV. This is much greater than the 0.511 MeV rest energy of the electron and much less than the 938.3 MeV rest energy of the proton. We will calculate the momentum and speed of each particle exactly and then by means of the nonrelativistic (proton) or the extreme relativistic (electron) approximations. 1. We first consider the electron. From Equation 2-39 we have 1 Ek mc2 1 10 MeV 0.511 MeV 20.57 2-5 General Relativity 97 Since the total energy is magnitude of the energy/momentum four-vector (Equation 2-31), mc2 10.511 MeV, we have, from the Ek pc 2E2 (mc2)2 2(10.511)2 (0.511)2 10.50 MeV The exact calculation then gives p 10.50 MeV c. The high-energy or extreme relativistic approximation agreement with the exact result. If we use Equation 2-34, we obtain for the speed On the other hand, the approximation of Equation 2-40 gives pc>E 10.50 MeV>10.51 MeV 0.999. p E>c 10.50 MeV is in good u> 20.57 2 b 0.999 2. For the proton, the total energy is Ek mc2 10 MeV 938.3 MeV 948.3 MeV. 1 10>938.3 1.01. From Equation 2-39 we obtain >mc2 Equation 2-31 gives for the momentum 1 Ek pc 2E2 (mc2)2 2(948.3)2 (938.3)2 p 137.4 MeV>c The nonrelativistic approximation gives Ek 1 2 mu2 (mu)2 2m p2 2m p2c2 2mc2 or pc 22mc2Ek p 137.0 MeV>c 2(2)(938.3)(10) The speed can be determined from Equation 2-34 exactly or from p mu approximately. From Equation 2-34 we obtain u c pc E 137.4 948.3 0.1449 From p mu, the nonrelativistic expression for p, we obtain u c pc mc2 137.0 938.3 0.1460 2-5 General Relativity The generalization of relativity to noninertial reference frames by Einstein in 1916 is known as the general theory of relativity. This theory is much more difficult mathematically than the special theory of relativity, and there are fewer situations in which it can be tested. Nevertheless, its importance in the areas of astrophysics and cosmology and the need to take account of its predictions in the design of such things as global navigation systems13 calls for its inclusion here. A full description of the general theory uses tensor analysis at a quite sophisticated level, well beyond the scope of this book, 98 Chapter 2 Relativity II The sensitivity of modern electronic devices is so exceptional that general relativistic effects are included in the design of such systems as the Global Positioning System. so we", " will be limited to qualitative or, in some instances, semiquantitative discussions. An additional purpose to the discussion that follows is to give you something that few people will ever have, namely, an acquaintance with one of the most remarkable of all scientific accomplishments and a bit of a feel for the man who did it. Einstein\u2019s development of the general theory of relativity was not motivated by any experimental enigma. Instead, it grew out of his desire to include the descriptions of all natural phenomena within the framework of the special theory. By 1907 he realized that he could accomplish that goal with the single exception of the law of gravitation. About that exception he said,14 I felt a deep desire to understand the reason behind this [exception]. The \u201creason\u201d came to him, as he said later, while he was sitting in a chair in the patent office in Bern. He described it like this:15 Then there occurred to me the happiest thought of my life, in the following form. The gravitational field has only a relative existence in a way similar to the electric field generated by electromagnetic induction. Because for an observer falling freely from the roof of a house there exists\u2014at least in his immediate surroundings\u2014no gravitational field [Einstein\u2019s italics].... The observer then has the right to interpret his state as \u201cat rest.\u201d Out of this \u201chappy thought\u201d grew the principle of equivalence that became Einstein\u2019s fundamental postulate for general relativity. Principle of Equivalence The basis of the general theory of relativity is what we may call Einstein\u2019s third postulate, the principle of equivalence, which states: A homogeneous gravitational field is completely equivalent to a uniformly accelerated reference frame. This principle arises in a somewhat different form in Newtonian mechanics because of the apparent identity of gravitational and inertial mass. In a uniform gravitational field, all objects fall with the same acceleration g independent of their mass because the gravitational force is proportional to the (gravitational) mass while the acceleration varies inversely with the (inertial) mass. That is, the mass m in and that in F ma (inertial m) F GMm r2 rN (gravitational m) appear to be identical in classical mechanics, although classical theory provides no ex GMm>r2 FG planation for this equality. For example, near Earth\u2019s surface, mgrav mgrav g minertial a F. to better than one part in 1012. Recent experiments have shown that minertial To understand what the equivalence principle means, consider a compartment in space far away from any matter and undergoing uniform acceleration a as shown in Figure 2-15a. If people in the compartment drop objects, they fall to the \u201cfloor\u201d with acceleration g a. If they stand on a spring scale, it will read their \u201cweight\u201d of magnitude ma. No mechanics experiment can be performed within the compartment (a) a (b) 2-5 General Relativity 99 Figure 2-15 Results from experiments in a uniformly accelerated reference frame (a) cannot be distinguished from those in a uniform gravitational field (b) if the acceleration a and gravitational field g have the same magnitude. g Planet that will distinguish whether the compartment is actually accelerating in space or is at rest (or moving with uniform velocity) in the presence of a uniform gravitational field g a. Einstein broadened the principle of equivalence to apply to all physical experiments, not just to mechanics. In effect, he assumed that there is no experiment of any kind that can distinguish uniformly accelerated motion from the presence of a gravitational field. A direct consequence of the principle is that is a requirement, not a coincidence. The principle of equivalence extends Einstein\u2019s first postulate, the principle of relativity, to all reference frames, noninertial (i.e., accelerated) as well as inertial. It follows that there is no absolute acceleration of a reference frame. Acceleration, like velocity, is only relative. mgrav minertial Transparent plastic sphere Small brass ball String Weak spring Question 6. For his 76th (and last) birthday Einstein received a present designed to demonstrate the principle of equivalence. It is shown in Figure 2-16. The object is, starting with the ball hanging down as shown to put the ball into the cup with a method that works every time (as opposed to random shaking). How would you do it? (Note: When it was given to Einstein, he was delighted and did the experiment correctly immediately.) Broomstick \u2248 4 ft Figure 2-16 Principle of equivalence demonstrator given to Einstein by E. M. Rogers. The object is to put the hanging brass ball into the cup by a technique that always works. The spring is weak, too weak to pull the ball in as it stands, and is stretched even when the ball is in the cup. The transparent sphere, about 10 cm in diameter, does not open.", " [From A. P. French, Albert Einstein: A Centenary Volume, Harvard University Press, Cambridge, Mass. (1979).] 100 Chapter 2 Relativity II Some Predictions of General Relativity In his first paper on general relativity, in 1916, Einstein was able to explain quantitatively a discrepancy of long standing between the measured and (classically) computed values of the advance of the perihelion of Mercury\u2019s orbit, about 43 arc seconds/century. It was the first success of the new theory. A second prediction, the bending of light in a gravitational field, would seem to be more difficult to measure owing to the very small effect. However, it was accurately confirmed less than five years later when Arthur Eddington measured the deflection of starlight passing near the limb of the Sun during a total solar eclipse. The theory also predicts the slowing of light itself and the slowing of clocks\u2014i.e., frequencies\u2014in gravitational fields, both effects of considerable importance to the determination of astronomical distances and stellar recession rates. The predicted slowing of clocks, called gravitational redshift, was demonstrated by Pound and co-workers in 1960 in Earth\u2019s gravitational field using the ultrasensitive frequency measuring technique of the M\u00f6ssbauer effect (see Chapter 11). The slowing of light was conclusively measured in 1971 by Shapiro and co-workers using radar signals reflected from several planets. Two of these experimental tests of relativity\u2019s predictions, bending of light and gravitational redshift, are discussed in the Exploring sections that follow. The perihelion of Mercury\u2019s orbit and the delay of light are discussed in More sections on the book\u2019s Web page. Many other predictions of general relativity are subjects of active current research. Two of these, black holes and gravity waves, are discussed briefly in the concluding paragraphs of this chapter. EXPLORING Deflection of Light in a Gravitational Field This relativistic effect results in gravitational lenses in the cosmos that focus light from extremely distant galaxies, greatly improving their visibility in telescopes, both on Earth and in orbit. FG GMm>r2 With the advent of special relativity, several features of the Newtonian law of gravitation became conceptually troublesome. One of these was the implication from the relativistic concept of mass-energy equivalence that even particles with zero rest mass should exhibit properties such as weight and inertia, thought of classically as masslike; classical theory does not include such particles. According to the equivalence principle, however, light, too, would experience the gravitational force. Indeed, the deflection of a light beam passing through the gravitational field near a large mass was one of the first consequences of the equivalence principle to be tested experimentally. To see why a deflection of light would be expected, consider Figure 2-17, which shows a beam of light entering an accelerating compartment. Successive positions of the compartment are shown at equal time intervals. Because the compartment is accelerating, the distance it moves in each time interval increases with time. The path of the beam of light, as observed from inside the compartment, is therefore a parabola. But according to the equivalence principle, there is no way to distinguish between an accelerating compartment and one with uniform velocity in a uniform gravitational field. We conclude, therefore, that a beam of light will accelerate in a gravitational field as do objects with rest mass. For example, near the surface of Earth light will fall with acceleration 9.8 m s2. This is difficult to observe because of the enormous speed of light. For example, in a distance of 3000 km, which takes about 0.01 second to cover, a beam of light should fall about 0.5 mm. Einstein pointed out that the deflection of a light beam in a gravitational field might be observed when light from a distant star passes close to the Sun.16 The deflection, or bending, is computed as follows. > (a) Light beam a 2-5 General Relativity 101 (b) t1t2t3t4 t1 t2 t3 t4 Figure 2-17 (a) Light beam moving in a straight line through a compartment that is undergoing uniform acceleration. The position of the light beam is shown at equally spaced times t1, t2, t3, t4. (b) In the reference frame of the compartment, the light travels in a parabolic path, as would a ball were it projected horizontally. Note that in both (a) and (b) the vertical displacements are greatly exaggerated for emphasis. Rewriting the spacetime interval s (Equation 2-32) in differential form and converting the space Cartesian coordinates to polar coordinates (in two dimensions, since the deflection occurs in a plane) yields ds 2 c2 dt 2 (dr 2 r2 d2) 2-42 Einstein showed that this expression is slightly modified in the presence of a (spherical, nonrotating) mass M to become ds 2 (r)2", "c2 dt 2 dr 2>(r)2 r2 d2 2-43 with G universal gravitational constant and (r) (1 2GM>c2r)1>2, where r distance from the mass M. The factor (r) is roughly analogous to the of special relativity. In the following Exploring section on gravitational redshift, we will describe can be thought of as correcting for gravitational time how dilation (the first term on the right of Equation 2-43) and gravitational length contraction (the second term). arises. For now, (r) (r) in the vicinity of the mass M since This situation is illustrated in Figure 2-18, which shows the light from a distant star just grazing the edge of the Sun. The gravitational deflection of light (with mass m E>c2 ) can be treated as a refraction of the light. The speed of light is reduced to (r)c (see Equation 2-43), thus bending the wave fronts, and hence the beam, toward M. This is analogous to the deflection of starlight toward Earth\u2019s surface as a result of the changing density\u2014hence index of refraction\u2014of the atmosphere. By integrating Equation 2-43 over the entire trajectory of the light beam (recall that ds 0 for light) as it passes by M, the total deflection is found to be17 (r) 1 4GM>c2R 2-44 where R distance of closest approach of the beam to the center of M. For a beam just solar radius 6.96 108 m. Substituting the values for G grazing the Sun, and the solar mass (M 1.99 1030 kg) yields 1.75 arc second.18 R R} Ordinarily, of course, the brightness of the Sun prevents astronomers (or anyone else) from seeing stars close to the limbs (edges) of the Sun, except during a total eclipse. Einstein completed the calculation of in 1915, and in 1919 expeditions were organized by Eddington19 at two points along the line of totality of a solar eclipse, both of which were 1>R } successful in making measurements of dependence of for several stars and testing the predicted for grazing beams at the two sites were:. The measured values of At Sobral (South America): At Principe Island (Africa): 1.98 0.12 1.61 0.30 arc seconds arc seconds Star Light path Apparent position of star Apparent light path \u03b1 R Sun, M Earth Figure 2-18 Deflection (greatly exaggerated) of a beam of starlight due to the gravitational attraction of the Sun. 102 Chapter 2 Relativity II Expected Einstein effect Actual measurements ) 1.8 s d n o c e s 1..2 1 2 3 4 5 6 R Distance from center of the Sun (solar radii) R} according to Figure 2-19 The deflection angle depends on the distance of closest approach Equation 2-44. Shown here is a sample of the data for 7 of the 13 stars measured by the Eddington expeditions. The agreement with the relativistic prediction is apparent. their average agreeing with the general relativistic prediction to within about 2 percent. Figure 2-19 illustrates the agreement of the dependence with Equation 2-44. (Einstein learned of the successful measurements via a telegram from H. A. Lorentz.) Since 1919, many measurements of have been made during eclipses. Since the development of radio telescopes, which are not blinded by sunlight and hence don\u2019t require a total eclipse, many more measurements have been made. The latest data agree with the deflection predicted by general relativity to within about 0.1 percent. 1>R } The gravitational deflection of light is being put to use by modern astronomers via the phenomenon of gravitational lensing to help in the study of galaxies and other large masses in space. Light from very distant galaxies passing near or through other galaxies or clusters of galaxies between the source and Earth can be bent so as to reach Earth in much the same way that light from an object on a bench in the laboratory can be refracted by a glass lens and thus reach the eye of an observer. An intervening galaxy or cluster of galaxies can thus produce images of the distant source, even ones magnified and distorted, just as the glass lens can. Figure 2-20a will serve as a reminder of a refracting lens in the laboratory, while Figure 2-20b illustrates the corresponding action (a) Observer Galaxy Observed image Object Image Lens Galaxy cluster (b) Figure 2-20 (a) Ordinary refracting lens bends light, causing many rays that would not otherwise have reached the observer\u2019s eye to do so. Their apparent origin is the image formed by the lens. Notice that the image is not the same size as the object (magnification) and, although not shown here, the shape of the lens", " can cause the image shape to be different from that of the object. (b) Gravitational lens has the same effects on the light from distant galaxies seen at Earth. Viewer 2-5 General Relativity 103 of a gravitational lens. The accompanying photograph shows the images of several distant galaxies drawn out into arcs by the lens effect of the cluster of galaxies in the center. The first confirmed discovery of images formed by a gravitational lens, the double image of the quasar QSO 0957, was made in 1979 by D. Walsh and his co-workers. Since then, astronomers have found many such images. Their discovery and interpretation is currently an active area of research. Gravitational lensing was recently used to help image the first apparent findings of dark matter in this cosmos. (See Chapter 13.) EXPLORING Gravitational Redshift S A B ) as illustrated in Figure 2-21. The spaceship S A second prediction of general relativity concerns the rates of clocks and the frequencies of light in a gravitational field. As a specific case that illustrates the gravitational redshift as a direct consequence of the equivalence principle, suppose we consider two identical light sources (A and ) and detectors (B and ) located in identical spaceS ships (S and in Figure 2-21b is located far from any mass. At time t 0, begins to accelerate, and simultaneously an atom f0. During the time in the source acquires a speed v at gh c, and t( h>c) A, the detector receding from the original location of measures the frequency of the v V c. incoming light to be f redshifted by a fractional amount (See Section 1-5.) Thus, emits a light pulse of its characteristic frequency to for the light to travel from B, f)>f0 B, (f0 for A A B > (f0 f)>f0 \u00a2f>f v>c gh>c2 2-45 (a) g S B (b) S\u00b4 B\u00b4 h a (= \u2013g) A A\u00b4 Planet Figure 2-21 (a) System S is at rest in the gravitational field of the planet. (b) Spaceship S, far from any mass, accelerates with a g. 104 Chapter 2 Relativity II \u00a2 gh Notice that the right side of Equation 2-45 is equal to the gravitational potential (i.e., the between A and B, divided by c2. gravitational potential energy per unit mass) According to the equivalence principle, the detector at B in S must also measure the frequency of the arriving light to be f, even though S is at rest on the planet and, therefore, the shift cannot be due to the Doppler effect! Since the vibrating atom that produced the light pulse at A can be considered to be a clock and since no \u201ccycles\u201d of the vibration are lost on the pulse\u2019s trip from A to B, the observer at B must conclude that the clock at A runs slow, compared with an identical clock (or an identical atom) located at B. Since A is at the lowest potential, the observer concludes that clocks run more slowly the lower the gravitational potential. This shift of clock rates to lower frequencies, hence longer wavelengths, in lower gravitational potentials is the gravitational redshift. In the more general case of a spherical, nonrotating mass M, the change in gravitational potential between the surface at some distance R from the center and a point at infinity is given by \u00a2 R GM r2 dr GM(1>r) ` R GM R 2-46 and the factor by which gravity shifts the light frequency is found from or \u00a2f>f0 (f0 f)>f0 GM>c2R f>f0 1 GM>c2R (gravitational redshift) 2-47 Notice that if the light is moving the other way, i.e., from high to low gravitational potential, the limits of integration in Equation 2-46 are reversed and Equation 2-47 becomes f>f0 1 GM>c2R (gravitational blueshift) 2-48 Images of distant galaxies are drawn out into arcs by the massive cluster of galaxies Abell 2218, whose enormous gravitational field acts as a lens to c # y magnify, brighten, and distort the images. Abell 2218 is about 2 billion Earth. The arcs in this January 2000 Hubble Space Telescope photograph are c # y away. [NASA, A. Fruchter; ERO Team.] images of galaxies 10 to 20 billion from 2-5 General Relativity 105 Analyzing the frequency of starlight for gravitational effects is exceptionally difficult because several shifts are present. For example, the light is gravitationally redshifted as it leaves the star and blueshifted as it arrives at Earth. The blueshift near Earth is negligibly small with current measuring technology; however, the redshift due to the receding of nearby stars and", " distant galaxies from us as a part of the general expansion of the universe is typically much larger than gravitational effects and, together with thermal frequency broadening in the stellar atmospheres, results in large uncertainties in measurements. Thus, it is quite remarkable that the relativistic prediction of Equation 2-48 has been tested in the relatively small gravitational field of Earth. R. V. Pound and his co-workers,20 first in 1960 and then again in 1964 with improved precision, measured the shift in the frequency of 14.4-keV gamma rays emitted by falling through a height h of only 22.5 m. Using the M\u00f6ssbauer effect, an extremely sensitive frequency shift measuring technique developed in 1968, Pound\u2019s measurements agreed > with the predicted fractional blueshift gh c2 2.45 1015 to within 1 percent. Equations 2-47 and 2-48 have been tested a number of times since then\u2014using atomic clocks carried on aircraft, as described in Section 1-4, and, in 1980, by R. F. C. Vessot and his co-workers using a precision microwave transmitter carried to 10,000 km from Earth by a space probe. The results of these tests, too, agree with the relativistically predicted frequency shift, the latter to 1 part in 14,000. 57Fe Question 7. The frequency f in Equation 2-47 can be shifted to zero by an appropriate value of M R. What would be the corresponding value of R for a star with the mass of the Sun? Speculate on the significance of this result. > More The inability of Newtonian gravitational theory to correctly account for the observed rate at which the major axis of Mercury\u2019s orbit precessed about the Sun was a troubling problem, pointing as it did to some subtle failure of the theory. Einstein\u2019s first paper on general relativity, the Perihelion of Mercury\u2019s Orbit, quantitatively explained the advance of Mercury\u2019s orbit, setting the stage for general relativity to supplant the old Newtonian theory. A clear description of the relativistic explanation is on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Equations 2-49 through 2-51 here, as well as Figure 2-22 and Table 2-2. More General relativity includes a gravitational interaction for particles with zero rest mass, such as photons, which are excluded in Newtonian theory. One consequence is the prediction of a Delay of Light in a Gravitational Field. This phenomenon and its subsequent observation are described qualitatively on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Equation 2-52 here, as well as Figures 2-23 and 2-24. Black Holes Black holes were first predicted by J. R. Oppenheimer and H. Snyder in 1939. According to the general theory of relativity, if the density of an object such as a star is great enough, the gravitational attraction will be so large that nothing can escape 106 Chapter 2 Relativity II from its surface, not even light or other electromagnetic radiation. It is as if space itself were being drawn inward faster than light could move outward through it. A remarkable property of such an object is that nothing that happens inside it can be communicated to the outside world. This occurs when the gravitational potential at the surface of the mass M becomes so large that the frequency of radiation emitted at the surface is redshifted to zero. From Equation 2-47 we see that the frequency will be zero when This result is a consequence the radius of the mass has the critical value approximation. A precise of the principle of equivalence, but Equation 2-47 is a called the Schwarzschild radius, yields derivation of the critical value of the radius GM>c2. v V c RG RG, 2MG c2 RG 2-53 For an object with mass equal to that of our Sun to be a black hole, its radius would be about 3 km. A large number of black holes have been identified by astronomers in recent years, one of them in the center of the Milky Way. (See Chapter 13.) An interesting historical note is that Equation 2-53 was first derived by nineteenth-century French physicist Pierre Laplace using Newtonian mechanics to compute the escape velocity ve from a planet of mass M before anyone had ever heard of Einstein or black holes. The result, derived in first-year physics courses by setting the kinetic energy of the escaping object equal to the gravitational potential energy at the surface of the planet (or star), is ve 2GM r A c gives Equation 2-53. Laplace obtained the correct result by making two Setting ve fundamental errors that just happened to cancel each other! Gravitational Waves Einstein\u2019s formulation of general relativity in 1916 explicitly predicted the existence of gravitational radiation. He showed that, just as accelerated electric charges generate time-dependent electromagnetic fields in space\u2014i", ".e., electromagnetic waves\u2014accelerated masses would create time-dependent gravitational fields in space\u2014i.e., gravitational waves\u2014that propagate from their source at the speed of light. The gravitational waves are propagating ripples, or distortions of spacetime. Figure 2-25 illustrates gravitational radiation emitted by two merging black holes distorting the otherwise flat \u201cfabric\u201d of spacetime. The best experimental evidence that exists so far in support of the gravitational wave prediction is indirect. In 1974 R. A. Hulse and J. H. Taylor24 discovered the first binary pulsar, i.e., a pair of neutron stars orbiting each other, one of which was emitting periodic flashes of electromagnetic radiation (pulses). In an exquisitely precise experiment they showed that the gradual decrease in the orbital period of the pair was in good agreement with the general relativistic prediction for the rate of loss of gravitational energy via the emission of gravitational waves. Experiments are currently under way in several countries to directly detect gravitational waves arriving at Earth. One of the most promising is LIGO (Laser Interferometer Gravitational-Wave Observatory), a pair of large Michelson interferometers with FabryPerot cavities at the Livingston Observatory in Louisiana and the Figure 2-25 Gravitational waves, intense ripples in the fabric of spacetime, are expected to be generated by a merging binary system of neutron stars or black holes. The amplitude decreases with distance due to the 1 R falloff and because waves farther from the source were emitted at an earlier time, when the emission was weaker. [Courtesy of Patrick Brady.] > 2-5 General Relativity 107 Figure 2-26 The LIGO detectors are equal-arm Michelson interferometers. The mirrors, each 25 cm in diameter by 10 cm thick and isolated from Earth\u2019s motions, are also the test masses of the gravitational wave detector. Arrival of a gravitational wave would change the length of each arm by about the diameter of an atomic nucleus and result in light signal at the photodetector. M 4 km Light storage arm M M are mirrors/test masses M M Light storage arm Laser Beam splitter 4 km Photodetector > Hanford Observatory 3002 km away in Washington, operating in coincidence. Figure 2-26 illustrates one of the LIGO interferometers. Each arm is 4 km long. The laser beams are reflected back and forth in the cavities, making about 75 round trips along each arm and recombining at the photodetector, making the effective lengths of the arms about 400 km. (A half-size but equally sensitive instrument using Fabry-Perot cavities is also housed at the Hanford Observatory.) The arrival of a gravitational wave would stretch one arm of the interferometer by about 1 1000 of the diameter of a proton and squeeze the other arm by the same minuscule amount! Nonetheless, that tiny change in the lengths is sufficient to change very slightly the relative phase of the recombining laser beams and produce a signal at the detector. The two LIGO interferometers must record the event within 10 ms of each other for the signal to be interpreted as a gravitational wave, that being the travel time between the two observatories for a gravitational wave moving at speed c. LIGO completed its two-year, low-sensitivity initial operational phase and went online in mid-2002. By 2005 LIGO had completed two science runs that included observations on 28 pulsars. (See Chapter 13.) At this writing a third science run with improved sensitivity is under way, operating jointly with GEO 600, a similar gravitational wave interferometer in Germany. A third gravitational wave observatory, Virgo, in Italy, is currently being commissioned and will soon join the search. These instruments are by far the most sensitive scientific instruments ever built. So far, none of the half-dozen or so experiments under way around the world has detected a gravitational wave.25 An enormous amount remains to be learned about the predictions and implications of general relativity\u2014not just about such things as black holes and gravity waves, but also, for example, about gravity and spacetime in the very early universe, This application of Michelson\u2019s interferometer may well lead to the first direct detection of \u201cripples\u201d or waves in spacetime. 108 Chapter 2 Relativity II Aerial view of the LIGO gravitational wave interferometer near Hanford, Washington. Each of the two arms is 4 km long. [CalTech/LIGO.] when forces were unified and the constituents were closely packed. These and other fascinating matters are investigated more specifically in the areas of astrophysics and cosmology (Chapter 13) and particle physics (Chapter 12), fields of research linked by general relativity, perhaps the grandest of Einstein\u2019s great scientific achievements. Question 8. Speculate on what the two errors made by Laplace in deriving Equation 2-53 might have been. Summary TOP", "IC RELEVANT EQUATIONS AND REMARKS 1. Relativistic momentum p mu The relativistic momentum is conserved and approaches mu for (1 u2>c2)1>2 in Equation 2-7, where u particle speed in S. v V c. 2. Relativistic energy E mc2 Total energy Kinetic energy The relativistic total energy is conserved. mc2 mc2 Ek mc2 is the rest energy. (1 u2>c2)1>2 in Equations 2-9 and 2-10. 3. Lorentz transformation for E and p. p\u0153 x (px vE>c2) p\u0153 y p\u0153 z py pz E (E vpx) where v relative speed of the systems and (1 v2>c2)1>2 2-7 2-10 2-9 2-16 4. Mass/energy conversion Whenever additional energy E in any form is stored in an object, the rest mass of the object is increased by m E c2. > 5. Invariant mass (mc2) E2 (pc)2 2-32 The energy and momentum of any system combine to form an invariant four-vector whose magnitude is the rest energy of the ma\u02dcss m. TOPIC RELEVANT EQUATIONS AND REMARKS 6. Force in relativity The force F ma is not invariant in relativity. Relativistic force is defined as F dp dt d(mu) dt 2-8 Notes 109 7. General relativity Principle of equivalence General References A homogeneous gravitational field is completely equivalent to a uniformly accelerated reference frame. The following general references are written at a level appropriate for readers of this book. Depend upon Its Energy Content?\u201d [Annalen der Physik, 17 (1905)]. Bohm, D., The Special Theory of Relativity, W. A. Benjamin, New York, 1965. French, A. P., Albert Einstein: A Centenary Volume. Harvard University Press, Cambridge, Mass., 1979. An excellent collection of contributions from many people about Einstein\u2019s life and work. Lorentz, H. A., A. Einstein, H. Minkowski, and W. Weyl, The Principle of Relativity: A Collection of Original Memoirs on the Special and General Theory of Relativity (trans. W. Perrett and J. B. Jeffery), Dover, New York, 1923. Two of Einstein\u2019s papers reprinted here are of interest in connection with this chapter: \u201cOn the Electrodynamics of Moving Bodies\u201d [Annalen der Physik, 17 (1905)], and \u201cDoes the Inertia of a Body Ohanian, H. C., Special Relativity: A Modern Introduction, Physics Curriculum & Instruction, 2001. Pais, A., Subtle Is the Lord..., Oxford University Press, Oxford, 1982. Resnick, R., Introduction to Relativity, Wiley, New York, 1968. Resnick, R., and D. Halliday, Basic Concepts in Relativity and Early Quantum Theory, 2d ed., Macmillan, New York, 1992. Rosser, W. G. V., The Theory of Relativity, Butterworth, London, 1964. Taylor, E. F., and J. A. Wheeler, Spacetime Physics, 2d ed., W. H. Freeman and Co., 1992. A good book with many examples, problems, and diagrams. Notes 1. This Gedankenexperiment (thought experiment) is based on one first suggested by G. N. Lewis and R. C. Tolman, Philosophical Magazine, 18, 510, (1909). 2. You can see that this is so by rotating Figure 2-1a through 180\u00b0 in its own plane; it then matches Figure 2-1b exactly. 3. C. G. Adler, American Journal of Physics, 55, 739 (1987). 4. This idea grew out of the results of the measurements of masses in chemical reactions in the nineteenth century, which, within the limits of experimental uncertainties of the time, were always observed to conserve mass. The conservation of energy had a similar origin in the experiments of James Joule (1818\u20131889) as interpreted by Hermann von Helmholtz (1821\u20131894). This is not an unusual way for conservation laws to originate; they still do it this way. 5. The approximation of Equation 2-10 used in this discussion was, of course, not developed from Newton\u2019s equations. The rest energy mc2 has no classical counterpart. 6. \u201cFacilitates\u201d means that we don\u2019t have to make frequent unit conversions or carry along large powers of 10 with nearly every factor in many calculations. However, a word of caution is in order. Always remember that the", " eV is not a basic SI unit. When making calculations whose results are to be in SI units, don\u2019t forget to convert the eV! i (1)1>2. 7. A. Einstein, Annalen der Physik, 17, 1905. 8. Strictly speaking, the time component should be written ict, where The i is the origin of the minus sign in the spacetime interval, as well as in Equation 2-32 for the energy/momentum four-vector and other four-vectors in both special and general relativity. Its inclusion was a contribution of Hermann Minkowski (1864\u20131909), a Russian-German mathematician who developed the geometric interpretation of relativity and who was one of Einstein\u2019s professors at Zurich. Consideration of the four-dimensional geometry is beyond the scope of our discussions, so we will not be concerned with the i. 110 Chapter 2 Relativity II 9. Other conservation laws must also be satisfied, e.g., elec- tric charge, angular momentum. 10. The positron is a particle with the same mass as an ordinary electron but with a positive electric charge of the same magnitude as that carried by the electron. It and other antiparticles will be discussed in Chapters 11 and 12. 11. Since electrons are thought to be point particles, i.e., they have no space dimensions, it isn\u2019t clear what it means to \u201chit\u201d an electron. Think of it as the photon close to the electron\u2019s location, hence within its strong electric field. 12. Such a system is called a polyelectron. It is analogous to an ionized hydrogen molecule much as positronium is analogous to a hydrogen atom. (See Figure 2-12 caption.) 13. Satellite navigation systems, e.g., the Global Positioning System, are now so precise that the minute corrections arising primarily from the general relativistic time dilation must be taken into account by the system\u2019s programs. 14. From Einstein\u2019s lecture in Kyoto in late 1922. See A. Pais, Subtle Is the Lord... (Oxford: Oxford University Pres, 1982). 15. From an unpublished paper now in the collection of the Pierpont Morgan Library in New York. See Pais (1982). 16. Einstein inquired of the astronomer George Hale (after whom the 5-m telescope on Palomar Mountain is named) in 1913 whether such minute deflections could be measured near the Sun. The answer was no, but a corrected calculation two years later doubled the predicted deflection and brought detection to within the realm of possibility. 17. This is not a simple integration. See, e.g., Adler et al., Introduction to General Relativity (New York: McGraw-Hill, 1963). h>c2 by the solar mass R}, 18. Both Newtonian mechanics and special relativity predict half this value. The particle-scattering formula used in Chapter 4 to obtain Equation 4-3, applied to the gravitational M} deflection of a photon of mass at impact parameter b equal to the solar radius shows how this value arises. 19. A copy of Einstein\u2019s work (he was then in Berlin) was smuggled out of Germany to Eddington in England so that he could plan the project. Germany and England were then at war. Arthur S. Eddington was at the time director of the prestigious Cambridge Observatory. British authorities approved the eclipse expeditions to avoid the embarrassment of putting such a distinguished scientist as Eddington, a conscientious objector, into a wartime internment camp. 20. See, e.g., R. V. Pound and G. A. Rebka, Jr., Physical Review Letters, 4, 337 (1960). 21. These values are relative to the fixed stars. 22. A. Einstein, \u201cThe Foundation of the General Theory of Relativity,\u201d Annalen der Physik, 49, 769 (1916). 23. I. I. Shapiro et al., Physical Review Letters, 26, 1132 (1971). 24. R. A. Hulse and J. H. Taylor, Astrophysical Journal, 195, L51 (1975). 25. Gravity wave detectors outside the U.S. are the TAMA 300 (Japan), GEO600 (Germany), and Virgo (Italy). NASA and the European Space Agency are designing a space-based gravity-wave detector, LISA, that will have arms 5 million kilometers long. The three satellites that LISA will comprise are scheduled for launch in 2011. Problems Level I Section 2-1 Relativistic Momentum and Section 2-2 Relativistic Energy pyB, 2-1. Show that balls in Figure 2-1, given by pyA where pyA and pyB are the relativistic momenta and speeds of the pyA mu0 21 u2 0 >c2 py", "B muyB 21 (u2 xB u2 yB)>c2 uyB u0 21 v2>c2 uxB v, mc2 d(mu) m(1 u2>c2)3>2 du. 0.511 MeV moves with respect to the laboratory at 2-2. Show that 2-3. An electron of rest energy speed u 0.6c. Find (a) (b) p in units of MeV c, (c) E, and (d) 2-4. How much energy would be required to accelerate a particle of mass m from rest to a speed of (a) 0.5 c, (b) 0.9 c, and (c) 0.99 c? Express your answers as multiples of the rest energy. 2-5. Two 1-kg masses are separated by a spring of negligible mass. They are pushed together, compressing the spring. If the work done in compressing the spring is 10 J, find the change in mass of the system in kilograms. Does the mass increase or decrease? Ek. > Problems 111 u>c does the measured mass of a particle exceed its rest mass by 2-6. At what value of (a) 10%, (b) a factor of 5, and (c) a factor of 20? 2-7. A cosmic ray proton is moving at such a speed that it can travel from the Moon to Earth in 1.5 s. (a) At what fraction of the speed of light is the proton moving? (b) What is its kinetic energy? (c) What value would be measured for its mass by an observer in Earth\u2019s reference frame? (d) What percent error is made in the kinetic energy by using the classical relation? (The Earth-Moon distance is 3.8 105 km. Ignore Earth\u2019s rotation.) 2-8. How much work must be done on a proton to increase its speed from (a) 0.15c to 0.16c? (b) 0.85c to 0.86c? (c) 0.95c to 0.96c? Notice that the change in the speed is the same in each case. 2-9. The Relativistic Heavy Ion Collider (RHIC) at Brookhaven is colliding fully ionized gold (Au) nuclei accelerated to an energy of 200 GeV per nucleon. Each Au nucleus contains 197 nucleons. (a) What is the speed of each Au nucleus just before collision? (b) What is the momentum of each at that instant? (c) What energy and momentum would be measured for one of the Au nuclei by an observer in the rest system of the other Au nucleus? 2-10. (a) Compute the rest energy of 1 g of dirt. (b) If you could convert this energy entirely into electrical energy and sell it for 10 cents per kilowatt-hour, how much money would you get? (c) If you could power a 100-W lightbulb with the energy, for how long could you keep the bulb lit? 2-11. An electron with rest energy of 0.511 MeV moves with speed u 0.2c. Find its total energy, kinetic energy, and momentum. 2-12. A proton with rest energy of 938 MeV has a total energy of 1400 MeV. (a) What is its speed? (b) What is its momentum? 2-13. The orbital speed of the Sun relative to the center of the Milky Way is about 250 km s. By what fraction do the relativistic and Newtonian values differ for (a) the Sun\u2019s momentum and (b) the Sun\u2019s kinetic energy? 2-14. An electron in a hydrogen atom has a speed about the proton of what percent do the relativistic and Newtonian values of momentum values differ? 2-15. Suppose that you seal an ordinary 60-W lightbulb and a suitable battery inside a transparent enclosure and suspend the system from a very sensitive balance. (a) Compute the change in the mass of the system if the lamp is on continuously for one year at full power. (b) What difference, if any, would it make if the inner surface of the container were a perfect reflector? 2.2 106 m s. (a) By differ? (b) By what percent do the Ek > > Section 2-3 Mass/Energy Conversion and Binding Energy 2H 3H 3H 4He atom to yield a atom, leaving a atom plus a neutron. atom plus a proton and an electron. 2-16. Use Appendix A and Table 2-1 to find how much energy is needed to remove one proton from a 2-17. Use Appendix A and Table 2-1 to find how much energy is required to remove one", " of the neutrons from a 2-18. The energy released when sodium and chlorine combine to form NaCl is 4.2 eV. (a) What is the increase in mass (in unified mass units) when a molecule of NaCl is dissociated into an atom of Na and an atom of Cl? (b) What percentage of error is made in neglecting this mass difference? (The mass of Na is about 23 u and that of Cl is about 35.5 u.) (a) Calculate 2-19. In a nuclear fusion reaction two the decrease in rest mass in unified mass units. (b) How much energy is released in this reaction? (c) How many such reactions must take place per second to produce 1 W of power? 2-20. An elementary particle of mass M completely absorbs a photon, after which its mass is 1.01M. (a) What was the energy of the incoming photon? (b) Why is that energy greater than 0.01Mc2? 2-21. When a beam of high-energy protons collides with protons at rest in the laboratory (e.g., in a container of water or liquid hydrogen), neutral pions are produced by the reaction p p S p p 0. Compute the threshold energy of the protons in the beam for this reaction to occur. (See Table 2-1 and Example 2-11.) atoms are combined to produce one (0) 4He. 2H 112 Chapter 2 Relativity II 235U 2-22. The energy released in the fission of a mass (in kg) is converted to energy in this fission? 2-23. The temperature of the Sun\u2019s core is about K. Assuming the core to consist of atomic hydrogen gas and recalling that temperature measures the average kinetic energy of the atoms, compute (a) the thermal energy of 1 kg of the gas and (b) the mass associated with this where k is the Boltzmann constant; see Chapter 3). energy ( nucleus is about 200 MeV. How much rest 1.5 107 3kT>2, Ek Section 2-4 Invariant Mass E mc2(1 p2>m2c2)1>2 2-24. Compute the force exerted on the palm of your hand by the beam from a 1.0-W flashlight (a) if your hand absorbs the light, and (b) if the light reflects from your hand. What would be the mass of a particle that exerts that same force in each case if you hold it at Earth\u2019s surface? 2-25. An electron-positron pair combined as positronium is at rest in the laboratory. The pair annihilate, producing a pair of photons (gamma rays) moving in opposite directions in the lab. Show that the invariant rest energy of the gamma rays is equal to that of the electron pair. 2-26. Show that Equation 2-31 can be written expansion to show that, when pc is much less than mc2, 2-27. An electron of rest energy 0.511 MeV has a total energy of 5 MeV. (a) Find its momentum in units of MeV c. (b) Find 2-28. Make a sketch of the total energy of an electron E as a function of its momentum p. (See Equations 2-36 and 2-39 for the behavior of E at large and small values of p.) 2-29. What is the speed of a particle that is observed to have momentum 500 MeV c and energy 1746 MeV? What is the particle\u2019s mass (in MeV c2)? 2-30. An electron of total energy 4.0 MeV moves perpendicular to a uniform magnetic field along a circular path whose radius is 4.2 cm. (a) What is the strength of the magnetic field B? (b) By what factor does 2-31. A proton is bent into a circular path of radius 2 m by a magnetic field of 0.5 T. (a) What is the momentum of the proton? (b) What is its kinetic energy? E mc2 p2>2m. and use the binomial exceed m? u>c. m > > > Section 2-5 General Relativity 5.98 1024 for light from a distant star that would, according to gen- 2-32. Compute the deflection angle eral relativity, be measured by an observer on the Moon as the light grazes the edge of Earth. 2-33. A set of twins work in the Sears Tower, a very tall office building in Chicago. One works on the top floor and the other works in the basement. Considering general relativity, which twin will age more slowly? (a) They will age at the same rate. (b) The twin who works on the top floor will age more slowly. (c) The twin who works", " in the basement will age more slowly. (d) It depends on the building\u2019s speed. (e) None of the previous choices is correct. 2-34. Jupiter makes 8.43 orbits/century and exhibits an orbital eccentricity 0.048. Jupiter is 5.2 AU from the Sun (see footnote for Table 2-2 in the More section) and has a mass 318 times the Earth\u2019s kg. What does general relativity predict for the rate of precession of Jupiter\u2019s perihelion? (It has not yet been measured.) (The astronomical unit AU the mean Earth-Sun distance 2-35. A synchronous satellite \u201cparked\u201d in orbit over the equator is used to relay microwave transmissions between stations on the ground. To what frequency must the satellite\u2019s receiver be tuned if the frequency of the transmission from Earth is exactly 9.375 GHz? (Ignore all Doppler effects.) 2-36. A particular distant star is found to be 92 from Earth. On a direct line between us c # y from the distant star is a dense white dwarf star with a mass equal to and the star and 35 M} 3 times the Sun\u2019s mass and a radius of 104 km. Deflection of the light beam from the distant star by the white dwarf causes us to see it as a pair of circular arcs like those shown in Figure 2-20(b). Find the angle formed by the lines of sight to the two arcs. 1.50 1011 c # y m.) 2 Level II 2-37. A clock is placed on a satellite that orbits Earth with a period of 90 min at an altitude of 300 km. By what time interval will this clock differ from an identical clock on Earth after 1 year? (Include both special and general relativistic effects.) Problems 113 106 103 c>2 m kg of fuel in a very short time at a speed of 2-38. Referring to Example 2-11, find the total energy E as measured in S where p 0. 2-39. In the Stanford linear collider, small bundles of electrons and positrons are fired at each other. In the laboratory\u2019s frame of reference, each bundle is about 1 cm long and 10 in diameter. In the collision region, each particle has energy of 50 GeV, and the electrons and positrons are moving in opposite directions. (a) How long and how wide is each bundle in its own reference frame? (b) What must be the minimum proper length of the accelerator for a bundle to have both its ends simultaneously in the accelerator in its own reference frame? (The actual length of the accelerator is less than 1000 m.) (c) What is the length of a positron bundle in the reference frame of the electron bundle? (d) What are the momentum and energy of the electrons in the rest frame of the positrons? 2-40. The rest energy of a proton is about 938 MeV. If its kinetic energy is also 938 MeV, find (a) its momentum and (b) its speed. 2-41. A spaceship of mass kg is coasting through space when suddenly it becomes necessary relative to the to accelerate. The ship ejects ship. (a) Neglecting any change in the rest mass of the system, calculate the speed of the ship in the frame in which it was initially at rest. (b) Calculate the speed of the ship using classical Newtonian mechanics. (c) Use your results from (a) to estimate the change in the rest mass of the system. 2-42. A clock (or a light-emitting atom) located at Earth\u2019s equator moves at about 463 m s relative to one located at the pole. The equator clock is also about 21 km farther from the center of Earth than the pole clock due to Earth\u2019s equatorial bulge. For an inertial reference frame centered on Earth, compute the time dilation effect for each clock as seen by an observer at the other clock. Show that the effects nearly cancel and that, as a result, the clocks read very close to the same time. (Assume that g is constant over the 21 km of the equatorial bulge.) 2-43. Professor Spenditt, oblivious to economics and politics, proposes the construction of a circular proton accelerator around Earth\u2019s circumference using bending magnets that provide a magnetic field of 1.5 T. (a) What would be the kinetic energy of protons orbiting in this field in a circle of radius RE? (b) What would be the period of rotation of these protons? 2-44. In ancient Egypt the annual flood of the Nile was predicted by the rise of Sirius (the Dog Star). Sirius is one of a binary pair whose companion is a white dwarf. Orbital analysis of the pair indicates that the dwarf\u2019s mass is kg (i.e", "., about one solar mass). Comparison of spectral lines emitted by the white dwarf with those emitted by the same element on Earth shows Assuming this to be due to a gravitational redshift, coma fractional frequency shift of pute the density of the white dwarf. (For comparison, the Sun\u2019s density is 1409 kg m3.) 2-45. Show that the creation of an electron-positron pair (or any particle-antiparticle pair, for that matter) by a single photon is not possible in isolation, i.e., that additional mass (or radiation) must be present. (Hint: Use the conservation laws.) 2-46. With inertial systems S and S arranged with their corresponding axes parallel and S moving in the x direction, it was apparent that the Lorentz transformation for y and z would be y y and z z. The transformations for the y and z components of the momentum are not pz. so apparent, however. Show that, as stated in Equations 2-16 and 2-17, 7 104. 2 1030 py and p\u0153 y p\u0153 z > > Level III 2-47. Two identical particles of rest mass m are each moving toward the other with speed u in frame S. The particles collide inelastically with a spring that locks shut (see Figure 2-9) and come to rest in S, and their initial kinetic energy is transformed into potential energy. In this problem you are going to show that the conservation of momentum in reference frame S, in which one of the particles is initially at rest, requires that the total rest mass of the system after the collision be (a) Show that the speed of the particle not at rest in frame S is 2m>(1 u2>c2)1>2. and use this result to show that u 2u 1 u2>c2 B1 u2 c2 1 u2>c2 1 u2>c2 114 Chapter 2 Relativity II Ek 6mc2. 2mc2. u p 2mu> (1 u2>c2). M 2m>(1 u2>c2)1>2. (b) Show that the initial momentum in frame S is (c) After the collision, the composite particle moves with speed u in S (since it is at rest in S). Write the total momentum after the collision in terms of the final rest mass M, and show that the conservation (d) Show that the total energy is conserved of momentum implies that in each reference frame. p 2-48. An antiproton has the same rest energy as a proton. It is created in the reaction p p S p p p p. In an experiment, protons at rest in the laboratory are bombarded with protons of kinetic energy Ek, which must be great enough so that kinetic energy equal to 2mc2 can be converted into the rest energy of the two particles. In the frame of the laboratory, the total kinetic energy cannot be converted into rest energy because of conservation of momentum. However, in the zero-momentum reference frame in which the two initial protons are moving toward each other with equal speed u, the total kinetic energy can be converted into rest energy. (a) Find the speed of each proton u such that the total kinetic energy in the zero(b) Transform to the laboratory\u2019s frame in which one proton is at momentum frame is of the other proton. (c) Show that the kinetic energy of the moving rest, and find the speed proton in the laboratory\u2019s frame is 2-49. In a simple thought experiment, Einstein showed that there is mass associated with electromagnetic radiation. Consider a box of length L and mass M resting on a frictionless surface. At the left wall of the box is a light source that emits radiation of energy E, which is absorbed at the right wall of the box. According to classical electromagnetic theory, this radiation carries momentum of magnitude p E c. (a) Find the recoil velocity of the box such that momentum is conserved when the light is emitted. (Since p is small and M is large, you may use classical mechanics.) (b) When the light is absorbed at the right wall of the box, the box stops, so the total momentum remains zero. If we neglect the very small velocity of the box, the time it takes for the radiation to travel across the box is t L c. Find the distance moved by the box in this time. (c) Show that if the center of mass of the system is to remain at the same place, the radiation must carry mass 2-50. A pion spontaneously decays into a muon and a muon antineutrino according to (among Recent experimental evidence indicates that the mass m of the other processes) is no larger than about 190 keV c2 and may be as small as zero. Assuming that the pion decays at rest in the laboratory", ", compute the energies and momenta of the muon and muon antineutrino (a) if the mass of the antineutrino were zero and (b) if its mass were 190 keV c2. The mass of the pion is 139.56755 MeV c2 and the mass of the muon is 105.65839 MeV c2. (See Chapters 11 and 12 for more on the neutrino mass.) 2-51. Use Equation 2-47 to obtain the gravitational redshift in terms of the wavelength Use that result to determine the shift in wavelength of light emitted by a white dwarf star at 720.00 nm. Assume the white dwarf has the same mass as the Sun ( kg) but a radius equal to 6.96 108 only 1 percent of the solar radius 2-52. For a particle moving in the xy plane of S, show that the y component of the acceleration is given by 1.99 1030 m). S m E>c2. R} R}.. > > > > > > ( a\u0153 y ay 2(1 uxv>c2)2 axuyv>c2 2(1 uxv>c2)3 S F\u0153 y Fy 2-53. Consider an object of mass m at rest in S acted upon by a force F with components Fx and Fy. System moves with instantaneous velocity v in the x direction. Defining the force with Equation 2-8 and using the Lorentz velocity transformation, show that (a) and (b) 2-54. An unstable particle of mass M decays into two identical particles, each of mass m. Obtain an expression for the velocities of the two decay particles in the lab frame (a) if M is at 4mc2 rest in the lab and (b) if M has total energy when it decays and the decay particles move along the direction of M. (Hint: See Problem 2-52.) Fx >. F\u0153 x CHAPTER 3 Quantization of Charge, Light, and Energy 3-1 Quantization of Electric Charge 3-2 Blackbody Radiation 3-3 The Photoelectric Effect 3-4 X Rays and the Compton Effect 115 119 127 133 The idea that all matter is composed of tiny particles, or atoms, dates to the specu- lations of the Greek philosopher Democritus1 and his teacher Leucippus in about 450 B.C. However, little attempt was made to correlate such speculations with observations of the physical world until the seventeenth century. Pierre Gassendi, in the mid-seventeenth century, and Robert Hooke, somewhat later, attempted to explain states of matter and the transitions between them with a model of tiny, indestructible solid objects flying in all directions. But it was Avogadro\u2019s hypothesis, advanced in 1811, that all gases at a given temperature contain the same number of molecules per unit volume, that led to great success in the interpretation of chemical reactions and to development of kinetic theory in about 1900. Avogadro\u2019s hypothesis made possible quantitative understanding of many bulk properties of matter and led to general (though not unanimous) acceptance of the molecular theory of matter. Thus, matter is not continuous, as it appears, but is quantized (i.e., discrete) on the microscopic scale. Scientists of the day understood that the small size of the atom prevented the discreteness of matter from being readily observable. In this chapter, we will study how three additional great quantization discoveries were made: (1) electric charge, (2) light energy, and (3) energy of oscillating mechanical systems. The quantization of electric charge was not particularly surprising to scientists in 1900; it was quite analogous to the quantization of mass. However, the quantization of light energy and mechanical energy, which are of central importance in modern physics, were revolutionary ideas. 3-1 Quantization of Electric Charge e Early Measurements of and e>m The first estimates of the order of magnitude of the electric charges found in atoms were obtained from Faraday\u2019s law. The work of Michael Faraday (1791\u20131867) in the early to mid-1800s stands out even today for its vision, experimental ingenuity, and thoroughness. The story of this self-educated blacksmith\u2019s son who rose from errand boy and 115 116 Chapter 3 Quantization of Charge, Light, and Energy bookbinder\u2019s apprentice to become the director of the distinguished Royal Institution of London and the foremost experimental investigator of his time is a fascinating one. One aspect of his work concerned the study of the conduction of electricity in weakly connow called the ducting solutions. His discovery that the same quantity of electricity, faraday and equal to about 96,500 C, always decomposes one gram-ionic weight, that NA,", " is, Avogadro\u2019s number of monovalent ions leads to the reasonable conclusion that each monovalent ion carries the same electric charge, and therefore F, e, F NAe 3-1 e NA Equation 3-1 is called Faraday\u2019s law of electrolysis. While was readily measurable, neither nor could be experimentally determined at the time. Faraday was aware of this but could not determine either quantity. Even so, it seemed logical to expect that electric charge, like matter, was not continuous, but consisted of particles of some discrete minimum charge. In 1874, G. J. Stoney2 used an estimate of from e from Equation 3-1 to be about kinetic theory to compute the value of howe ever, direct measurement of the value of had to await an ingenious experiment conducted by R. A. Millikan a third of a century later. NA 1020 C; F Meanwhile, Pieter Zeeman, in 1896, obtained the first evidence for the existence of atomic particles with a specific charge-to-mass ratio by looking at the changes in the discrete spectral lines emitted by atoms when they were placed in a strong magnetic field. He discovered that the individual spectral lines split into three very closely spaced lines of slightly different frequencies when the atoms were placed in the magnetic field. (This phenomenon is called the Zeeman effect and will be discussed further in Chapter 7.) Classical electromagnetic theory relates the slight differences in the frequencies of adjacent lines to the charge-to-mass ratio of the oscillating charges producing the light. From his measurements of the splitting, Zeeman calculated 1.6 1011 C>kg, 1.759 1011 C>kg Zeeman concluded that the oscillating particles were negatively charged. to be about which compares favorably with the presently accepted value, (see Appendix D). From the polarization of the spectral lines, q>m Discovery of the Electron: J. J. Thomson\u2019s Experiment q>m The year following Zeeman\u2019s work, J. J. Thomson3 measured the value for the socalled cathode rays that were produced in electrical discharges in gases and pointed out that, if their charge was Faraday\u2019s charge as determined by Stoney, then their mass was only a small fraction of the mass of a hydrogen atom. Two years earlier J. Perrin had collected cathode rays on an electrometer and found them to carry a negative electric q>m charge.4 Thus, with his measurement of for the cathode rays, Thomson had, in fact, discovered the electron. That direct measurement of of electrons by J. J. Thomson in 1897, a little over a century ago, can be justly considered to mark the beginning of our understanding of atomic structure. e>m e Measurement of e/m When a uniform magnetic field of strength is established perpendicular to the direction of motion of charged particles, the particles move in a R of the path can be obtained from Newton\u2019s second law by circular path. The radius quB setting the magnetic force times the centripetal acceleration u2>R, u is the speed of the particles: equal to the mass where m B quB mu2 R or R mu qB and q m u RB 3-2 3-1 Quantization of Electric Charge 117 + \u2013 C A B D E B A Figure 3-1 J. J. Thomson\u2019s tube for measuring through the slits at and an electric field between the plates direction is perpendicular to the electric field between measured on a scale on the tube at the screen, \u201cCathode Rays,\u201d Philosophical Magazine (5), 44, 293 (1897).] e>m D D e>m. and strike a phosphorescent screen. The beam can be deflected by and or by a magnetic field (not shown) whose E. From the deflections and E can be determined. [From J. J. Thomson, Electrons from the cathode pass C e>m experiments of somewhat different designs. The second, Thomson performed two more reproducible of the two has become known as the J. J. Thomson experiment (Figure 3-1). In this experiment he adjusted perpendicular fields so that the particles were undeflected. This allowed him to determine the speed of the electrons by equating the magnitudes of the magnetic and electric forces and then to compute e>m( q>m) from Equation 3-2: and B e quB qe or u e B 3-3 Thomson\u2019s experiment was remarkable in that he measured for a subatomic particle using only a voltmeter, an ammeter, and a measuring rod, obtaining the result 0.7 1011 C>kg. Present-day particle physicists routinely use the modern equivalent of Thomson\u2019s experiment to measure the momenta of elementary particles. e>m Thomson\u2019s technique for controlling the direction of the electron beam with \u201ccrossed\u201d electric and magnetic fields", " was subsequently applied in the development of cathode ray tubes used in oscilloscopes and the picture tubes of television receivers. J. J. Thomson in his laboratory. He is facing the screen end of an ray tube is visible in front of his left shoulder. [Courtesy of Cavendish Laboratory.] tube; an older cathode e>m 118 Chapter 3 Quantization of Charge, Light, and Energy Thomson repeated the experiment with different gases in the tube and different metals for cathodes and always obtained the same value for within his experimental uncertainty, thus showing that these particles are common to all metals. The agreement of these results with Zeeman\u2019s led to the unmistakable conclusion that these particles\u2014called corpuscles by Thomson and later called electrons by Lorentz\u2014 which have one unit of negative charge and a mass about 2000 times less than the mass of the lightest known atom were constituents of all atoms. e>m e Questions 1. One advantage of Thomson\u2019s evidence over others\u2019 (such as Faraday\u2019s or Zeeman\u2019s) was its directness. Another was that it was not just a statistical inference. How does the Thomson experiment show that large number of the particles? e>m is the same for a 2. Thomson noted that his values for were about 2000 times larger than those for the lightest known ion, that of hydrogen. Could he tell from his data whether this was the result of the electron having either a greater charge or a smaller mass than the hydrogen ion? e>m Measuring the Electric Charge: Millikan\u2019s Experiment e>m The fact that Thomson\u2019s measurements always yielded the same results regardless of the materials used for the cathodes or the kind of gas in the tube was a persuaof negative electric charge. sive argument that the electrons all carried one unit e. Thomson initiated a series of experiments to determine the value of The first of these experiments, which turned out to be very difficult to do with high precision, was carried out by his student J. S. E. Townsend. The idea was simple: a small (but visible) cloud of identical water droplets, each carrying a single charge was observed to drift downward in response to the gravitational force. The total charge on the was measured, as were the mass of the cloud and the radius of a single cloud N, drop. Finding the radius allowed calculation of the total number of drops in the cloud, and, hence, the value of Q Ne e. e, e The accuracy of Thomson\u2019s method was limited by the uncertain rate of evaporation of the cloud. In addition, the assumption that each droplet contained a single charge could not be verified. R. A. Millikan tried to eliminate the evaporation problem by using a field strong enough to hold the top surface of the cloud stationary so that he could observe the rate of evaporation and correct for it. That, too, turned out to be very difficult, but then he made a discovery of enormous importance, one that allowed him to measure directly the charge of a single electron! Millikan described his discovery in the following words: It was not found possible to balance the cloud as had been originally planned, but it was found possible to do something much better: namely, to hold individual charged drops suspended by the field for periods varying from 30 to 60 seconds. I have never actually timed drops which lasted more than 45 seconds, although I have several times observed drops which in my judgment lasted considerably longer than this. The drops which it was found possible to balance by an electric field always carried multiple charges, and the difficulty experienced in balancing such drops was less than had been anticipated.5 3-2 Blackbody Radiation 119 Atomizer Telescope Figure 3-2 Schematic diagram of Millikan\u2019s oil drop experiment. The drops are sprayed from an atomizer and pick up a static charge, a few falling through the hole in the top plate. Their fall due to gravity and their rise due to the electric field between the capacitor plates can be observed with the telescope. From measurements of the rise and fall times, the electric charge on a drop can be calculated. The charge on a drop could be changed by exposure to x rays from a source (not shown) mounted opposite the light source. (+) (\u2013) (\u2013) (+) Light source The discovery that he could see individual droplets and that droplets suspended in a vertical electric field sometimes suddenly moved upward or downward, evidently because they had picked up a positive or negative ion, led to the possibility of observing the charge of a single ion. In 1909, Millikan began a series of experiments that not only showed that charges occurred in integer mule, e tiples of an elementary unit but measured the value of to about 1 part in 1000. To eliminate evaporation, he used oil drops sprayed into dry air between the plates of a capacitor (Figure 3-2). These drops were already charged by the spraying process, i.e., by friction in the spray", " nozzle, and during the course of the observation they picked up or lost additional charges. By switching the direction of the electric field between the plates, a drop could be moved up or down and observed for several hours. When the charge on a drop changed, the velocity of the drop with the field \u201con\u201d changed also. Assuming only that the terminal velocity of the drop was proportional to the force acting on it (this assumption was carefully checked experimentally), Millikan\u2019s oil drop experiment6 gave conclusive evidence that electric charges always occur in integer multiples of a fundamental unit whose value he determined to be The currently accepted value7 is An expanded discussion of Millikan\u2019s experiment is included in the Classical Concept Review. 1.60217653 1019 C. 1.601 1019 C. e, 3-2 Blackbody Radiation The first clue to the quantum nature of radiation came from the study of thermal radiation emitted by opaque bodies. When radiation falls on a opaque body, part of it is reflected and the rest is absorbed. Light-colored bodies reflect most of the visible radiation incident on them, whereas dark bodies absorb most of it. The absorption part of the process can be described briefly as follows. The radiation absorbed by the body increases the kinetic energy of the constituent atoms, which oscillate about their equilibrium positions. Since the average translational kinetic energy of the atoms determines the temperature of the body, the absorbed energy causes the temperature to rise. However, the atoms contain charges (the electrons), and they are accelerated by the oscillations. Consequently, as required by electromagnetic theory, the atoms emit electromagnetic radiation, which reduces the kinetic energy of the oscillations and tends to reduce the temperature. When the rate of absorption equals the rate of emission, the temperature is constant, and we say that the body is in thermal equilibrium with its surroundings. A good absorber of radiation is therefore also a good emitter. The electromagnetic radiation emitted under these circumstances is called thermal radiation. At ordinary temperatures (below about ) the thermal radiation emitted by a body is not visible; most of the energy is concentrated in wavelengths much longer than those of visible light. As a body is heated, the quantity of thermal radiation emitted increases, and the energy radiated extends to shorter and shorter wavelengths. At about there is enough energy in the visible spectrum so that the body glows and becomes a dull red, and at higher temperatures it becomes bright red or even \u201cwhite hot.\u201d 600\u00b0\u2013700\u00b0C 600\u00b0C 120 Chapter 3 Quantization of Charge, Light, and Energy T that passes through the slit is Figure 3-3 Radiation emitted by the object at temperature dispersed according to its wavelengths. The prism shown would be an appropriate device for that part of the emitted radiation in the visible region. In other spectral regions other types of devices or wavelength-sensitive detectors would be used. Radiation T Prism Slit Dispersed radiation Object Detector A body that absorbs all radiation incident on it is called an ideal blackbody. In 1879 Josef Stefan found an empirical relation between the power radiated by an ideal blackbody and the temperature: R T4 3-4 T R is the power radiated per unit area, where is the absolute temperature, and 5.6703 108 W>m2K4 is a constant called Stefan\u2019s constant. This result was also derived on the basis of classical thermodynamics by Ludwig Boltzmann about five years later, and Equation 3-4 is now called the Stefan-Boltzmann law. Note that the power per unit area radiated by a blackbody depends only on the temperature and not on any other characteristic of the object, such as its color or the material of which tells us the rate at which energy is emitted by the obit is composed. Note, too, that ject. For example, doubling the absolute temperature of an object, e.g., a star, increases the energy flow out of the object by a factor of An object at room will double the rate at which it radiates energy as a result of a temperature Thus, the Stefan-Boltzmann law has an enormous temperature increase of only effect on the establishment of thermal equilibrium in physical systems. 24 16. (300\u00b0C) 57\u00b0C. R Objects that are not ideal blackbodies radiate energy per unit area at a rate less than that of a blackbody at the same temperature. For those objects the rate does depend on properties in addition to the temperature, such as color and the composition of the surface. The effects of those dependencies are combined into a factor called the emissivity which multiplies the right side of Equation 3-4. The values of which is itself temperature dependent, are always less than unity., R, R() d Like the total radiated power the spectral distribution of the radiation emitted T. by a blackbody is found empirically to depend only on the absolute temperature The spectral distribution is determined experimentally as illustrated schematically in Figure 3-3. With d", ", verand sus The curves in Figure 3-4 are quite remarkable in several respects. One is that the wavelength at which the distribution has its maximum value varies inversely with the temperature: Figure 3-4 shows the measured spectral distribution function the power emitted per unit area with wavelength between for several values of ranging from 6000 K. 1000 K R() R() to T m 1 T or mT constant 2.898 103 m # K 3-5 3-2 Blackbody Radiation 121 R() R () axis is in measured at different Figure 3-4 Spectral distribution function temperatures. The arbitrary units for comparison only. in the visible Notice the range of spectrum. The Sun emits radiation very close to that of a blackbody at is indicated for the 5000-K curves. m and 6000-K 5800 K. 0.3 \u03bb m ) \u03bb ( R 0.001 0.2 ) \u03bb ( R 0.1 6000 K Visible \u03bb m 5000 K 0 0 4000 K 3000 K 2000 K 1500 K 1000 K 2000 4000 \u03bb (nm) 0 0 500 1500 1000 Wavelength \u03bb (nm) 2000 This result is known as Wien\u2019s displacement law. It was obtained by Wien in 1893. Examples 3-1 and 3-2 illustrate its application. EXAMPLE 3-1 How Big Is a Star? Measurement of the wavelength at which the from a certain star is maximum indicates that the star\u2019s spectral distribution If the star is also found to radiate 100 times the surface temperature is power ) The Sun\u2019s surface temperature is 5800 K. radiated by the Sun, how big is the star? (The symbol } Sun. 3000 K. R() P} SOLUTION If the Sun and the star both radiate as blackbodies (astronomers nearly always make that assumption, based on, among other things, the fact that the solar spectrum is very nearly that of an ideal blackbody), their surface temperatures from Equation respectively. Measurement also indicates that 3-5 are Pstar 100 P}. 3000 K, 5800 K and Thus, from Equation 3-4 we have Pstar (area)star 100 P} 4r2 Rstar star T4 star and So we have R} P} (area)} P} 4r2 } T4 } a T} Tstar T} Tstar b 4 b 2 10a 5800 3000 2 b r} r2 star 100 r2 } rstar rstar 10 r} a 37.4 r} r} 6.96 108 m, Since the radius of the orbit of Mercury. This star is a red giant (see Chapter 13). this star has a radius of about 2.6 1010 m, or about half 122 Chapter 3 Quantization of Charge, Light, and Energy Rayleigh-Jeans Equation Figure 3-5 A small hole in the wall of a cavity approximating an ideal blackbody. Radiation entering the hole has little chance of leaving before it is completely absorbed within the cavity. R() The calculation of the distribution function involves the calculation of the energy density of electromagnetic waves in a cavity. Materials such as black velvet or lampblack come close to being ideal blackbodies, but the best practical realization of an ideal blackbody is a small hole leading into a cavity (such as a keyhole in a closet door; Figure 3-5). Radiation incident on the hole has little chance of being reflected back out of the hole before it is absorbed by the walls of the cavity. The power radiated out of the hole is proportional to the total energy density (the energy per unit volume of the radiation in the cavity). The proportionality constant can be shown to be is the speed of light.8 where c>4, U c R 1 4 cU 3-6 Similarly, the spectral distribution of the power emitted from the hole is proportional is the fracto the spectral distribution of the energy density in the cavity. If d, R() tion of the energy per unit volume in the cavity in the range are related by u() d then u() and R() 1 4 cu() 3-7 The energy density distribution function can be calculated from classical physics in a straightforward way. The method involves finding the number of modes of oscillation of the electromagnetic field in the cavity with wavelengths in the interand multiplying by the average energy per mode. The result is that the number val of modes of oscillation per unit volume, is independent of the shape of the cavity and is given by n(), d u() n() 84 3-8 According to classical kinetic theory, the average energy per mode of oscillation is the same as for a one-dimensional harmonic oscillator, where constant. Classical theory thus predicts for the energy density distribution function kT, is the Boltzmann k u() kT n() 8kT 4 3-9 This prediction, initially derived by Lord Rayleigh,9 is called the Rayleigh-Jeans equation. It", " is illustrated in Figure 3-6. u() At very long wavelengths the Rayleigh-Jeans equation agrees with the experimentally determined spectral distribution, but at short wavelengths this equation prewhereas experiment dicts that becomes large, approaching infinity as S 0. This shows (see Figure 3-4) that the distribution actually approaches zero as u() enormous disagreement between the experimental measurement of and the prediction of the fundamental laws of classical physics at short wavelengths was called the ultraviolet catastrophe. The word catastrophe was not used lightly; Equation 3-9 implies that S 0, 0 u() d \u00a1 3-10 That is, every object would have an infinite energy density, which observation assures us is not true. Planck\u2019s Law 3-2 Blackbody Radiation 123 u() In 1900 the German physicist Max Planck10 announced that by making somewhat that agreed with the experimenstrange assumptions, he could derive a function tal data. He first found an empirical function that fit the data and then searched for a way to modify the usual calculation so as to predict his empirical formula. We can see the type of modification needed if we note that, for any cavity, the shorter the wave S 0 length, the more standing waves (modes) will be possible. Therefore, as the number of modes of oscillation approaches infinity, as evidenced in Equation 3-8. to approach zero, we expect In order for the energy density distribution function the average energy per mode to depend on the wavelength and approach zero as predicted by classical theory. Parenthetically, we should note that those working on the ultraviolet catastrophe at the time\u2014and there were many besides Planck\u2014had no a priori way of knowing whether the number of modes (or both) was the source of the problem. Both were correct classically. Many attempts were made to rederive each so as to solve the problem. As it turned out, it was the average energy per mode (that is, kinetic theory) that was at fault. approaches zero, rather than be equal to the value or the average energy per mode n() u() kT kT Classically, the electromagnetic waves in the cavity are produced by accelerated electric charges in the walls of the cavity vibrating as simple harmonic oscillators. Recall that the radiation emitted by such an oscillator has the same frequency as the oscillation itself. The average energy for a one-dimensional simple harmonic oscillator is calculated classically from the energy distribution function, which in turn is found from the Maxwell-Boltzmann distribution function. That energy distribution function has the form (see Chapter 8) A is a constant and where The average energy f(E) AeE>kT 3-11 f(E) is the fraction of the oscillators with energy equal to E. E is then found, as is any weighted average, from E E f (E) dE 0 0 EAeE>kT dE 3-12 with the result E kT, as was used by Rayleigh and others. u(\u03bb) Rayleigh-Jeans law Planck\u2019s law 0 2000 4000 6000 \u03bb, nm T 1600 K Figure 3-6 Comparison of Planck\u2019s law and the Rayleigh-Jeans equation with experimental data at obtained by W. W. Coblenz in about 1915. The from F. K. Richmyer, E. H. Kennard, and J. N. Cooper, Introduction to Modern Physics, 6th ed., McGraw-Hill, New York (1969), by permission.] axis is linear. [Adapted u() 124 Chapter 3 Quantization of Charge, Light, and Energy E Planck found that he could derive his empirical formula by calculating the average energy assuming that the energy of the oscillating charges, and hence the radiation that they emitted, was a discrete variable; i.e., that it could take on only the is an integer; and further, that was proportional to values the frequency of the oscillators and, hence, to that of the radiation. Planck therefore wrote the energy as 0,, 2, \u00c1, n where E n n nhf n 0, 1, 2, \u00c1 En 3-13 where the proportionality constant Boltzmann distribution (Equation 3-11) then becomes h is now called Planck\u2019s constant. The Maxwell- AeEn >kT Aen>kT fn 3-14 A where must, of course, equal 1, i.e., is determined by the normalization condition that the sum of all fractions fn a n0 fn A a n0 en>kT 1 3-15 The average energy of an oscillator is then given by the discrete-sum equivalent of Equation 3-12: E a n0 En fn a n0 EnAeEn >kT 3-16 Calculating the sums in Equations 3-15 and 3-16 (see Problem 3-58) yields the result E e>kT 1", " hf ehf>kT 1 hc> ehc>lkT 1 3-17 d Multiplying this result by the number of oscillators per unit volume in the interval given by Equation 3-8, we obtain for the energy density distribution function of the radiation in the cavity u() 8hc5 hc>kT 1 3-18 This function, called Planck\u2019s law, is sketched in Figure 3-6. It is clear from the figure that the result fits the data quite well. For very large ex 1 x \u00c1, for the exponential in Equation 3-18 can be expanded using x V 1, Then x hc>kT. where and ehc>kT 1 hc kT u() S 84kT which is the Rayleigh-Jeans formula. For short wavelengths, we can neglect the 1 in the denominator of Equation 3-18, and we have u() S 8hc5ehc>kT S 0 3-2 Blackbody Radiation 125 S 0. as Planck\u2019s law, as you will show in Problem 3-23. The value of the constant in Wein\u2019s displacement law also follows from The value of Planck\u2019s constant, can be determined by fitting the function given by Equation 3-18 to the experimental data, although direct measurement (see Section 3-3) is better but more difficult. The presently accepted value is h, h 6.626 1034 J # s 4.136 1015 eV # s 3-19 Planck tried at length to reconcile his treatment with classical physics but was unable to do so. The fundamental importance of the quantization assumption implied by Equation 3-13 was suspected by Planck and others but was not generally appreciated until 1905. In that year Einstein applied the same ideas to explain the photoelectric effect and suggested that, rather than being merely a mysterious property of the oscillators in the cavity walls and blackbody radiation, quantization was a fundamental characteristic of light energy. 5800 K, EXAMPLE 3-2 Peak of the Solar Spectrum The surface temperature of the Sun is about and measurements of the Sun\u2019s spectral distribution show that it radiates very nearly like a blackbody, deviating mainly at very short wavelengths. Assuming that the Sun radiates like an ideal blackbody, at what wavelength does the peak of the solar spectrum occur? SOLUTION 1. The wavelength at the peak, or maximum intensity, of an ideal blackbody is given by Equation 3-5: mT constant 2.898 103 m # K 2. Rearranging and substituting the Sun\u2019s surface temperature yield m (2.898 103 m # K)>T 2.898 103 m # K 2.898 106 nm # K 5800 K 499.7 nm 5800 K where 1 nm 109 m. Remarks: This value is near the middle of the visible spectrum. EXAMPLE 3-3 Average Energy of an Oscillator What is the average energy of an according to Planck\u2019s calculation? oscillator that has a frequency given by hf kT E SOLUTION E e>kT 1 kT e1 1 0.582 kT Remarks: Recall that according to classical theory, frequency. E kT regardless of the The electromagnetic spectrum emitted by incandescent bulbs is a common example of blackbody radiation, the amount of visible light being dependent on the temperature of the filament. Another application is the pyrometer, a device that measures the temperature of a glowing object, such as molten metal in a steel mill. 126 Chapter 3 Quantization of Charge, Light, and Energy EXAMPLE 3-4 Stefan-Boltzmann from Planck Show that the total energy density in a blackbody cavity is proportional to in accordance with the StefanBoltzmann law. T4 SOLUTION The total energy density is obtained from the distribution function (Equation 3-18) by integrating over all wavelengths: U u() d 8hc5 ehc>kT 1 d 0 0 x hc>kT. Then dx (hc>2kT) d or Define the dimensionless variable d 2(kT>hc) dx. U Then 8hc3 ex 1 a kT hc 0 b dx 8hc a kT hc 4 b 0 dx x3 ex 1 U The value of the integral can be obtained from tables; it is Since the integral is now dimensionless, this shows that T4. U (85k4>15h3c3)T4. express Stefan\u2019s constant in terms of is proportional to 4>15. Then This result can be combined with Equations 3-4 and 3-6 to (see Problem 3-13)., k, h, and c A dramatic example of an application of Planck\u2019s law on the current frontier of", " physics is in tests of the Big Bang theory of the formation and present expansion of the universe. Current cosmological theory holds that the universe originated in an extremely high-temperature explosion of space, one consequence of which was to fill the infant universe with radiation whose spectral distribution must surely have been that of an ideal blackbody. Since that time, the universe has expanded to its present size Tnow. However, it should still be filled with radiaand cooled to its present temperature tion whose spectral distribution should be that characteristic of a blackbody at Tnow. In 1965, Arno Penzias and Robert Wilson discovered radiation of wavelength 7.35 cm reaching Earth with the same intensity from all directions in space. It was soon recognized that this radiation could be a remnant of the Big Bang fireball, and measurements were subsequently made at other wavelengths in order to construct an experimental energy density graph. The most recent data from versus the Cosmic Background Explorer (COBE) satellite, shown in Figure 3-7, and by the u() 1500 1000 500 ) 100 200 300 400 500 600 700 Frequency (\u00d7 109 Hz) Figure 3-7 The energy density spectral distribution of the cosmic microwave background radiation. The solid line is Planck\u2019s law T 2.725 K. with These measurements (the black dots) were made by the COBE satellite. 3-3 The Photoelectric Effect 127 Wilkinson Microwave Anisotropy Probe (WMAP) have established the temperature of the background radiation field at The excellent agreement of the data with Planck\u2019s equation, indeed, the best fit that has ever been measured, is considered to be very strong support for the Big Bang theory (see Chapter 13). 2.725 0.001 K. 3-3 The Photoelectric Effect It is one of the ironies in the history of science that in the famous experiment of Heinrich Hertz11 in 1887, in which he produced and detected electromagnetic waves, thus confirming Maxwell\u2019s wave theory of light, he also discovered the photoelectric effect, which led directly to the particle description of light. Hertz was using a spark gap in a tuned circuit to generate the waves and another similar circuit to detect them. He noticed accidentally that when the light from the generating gap was shielded from the receiving gap, the receiving gap had to be made shorter in order for the spark to jump the gap. Light from any spark that fell on the terminals of the gap facilitated the passage of the sparks. He described the discovery with these words: In a series of experiments on the effects of resonance between very rapid electric oscillations that I had carried out and recently published, two electric sparks were produced by the same discharge of an induction coil, and therefore simultaneously. One of these sparks, spark was the discharge spark of the induction coil, and served to excite the primary oscillation. I occasionally enclosed spark in a dark case so as to make observations more easily, and in so doing I observed that the maximum spark length became decidedly smaller inside the case than it was before.12 B, B The unexpected discovery of the photoelectric effect annoyed Hertz because it interfered with his primary research, but he recognized its importance immediately and interrupted his other work for six months in order to study it in detail. His results, published later that year, were then extended by others. It was found that negative particles were emitted from a clean surface when exposed to light. P. Lenard in 1900 deflected them in a magnetic field and found that they had a charge-to-mass ratio of the same magnitude as that measured by Thomson for cathode rays: the particles being emitted were electrons. Albert A. Michelson, Albert Einstein, and Robert A. Millikan at a meeting in Pasadena, California, in 1931. [AP/Wide World Photos.] 128 Chapter 3 Quantization of Charge, Light, and Energy V C Pump W A B L \u03b2 \u03b1 W L Figure 3-8 Schematic diagram of the apparatus used by P. Lenard to demonstrate the photoelectric effect and to show that the particles emitted in the process were electrons. Light from the source are recorded by the electrometer connected to A magnetic field, indicated by the circular pole piece, could deflect the particles to a second electrometer connected to making possible the establishment of the sign of the charges and their 359 (1900).], ratio. [P. Lenard, Annalen der Physik, 2, Photoelectrons going through the hole in anode strikes the cathode e>m. C. A L V A C is incident on a clean metal surface (cathode Figure 3-8 shows a schematic diagram of the basic apparatus used by Lenard. ), electrons are emitted. When light pass through the small hole, a curIf some of these electrons that reach the anode. The number of rent results in the external electrometer circuit connected to the emitted electrons reaching the anode can be increased or decreased by making the anode positive or negative with respect to the cathode. Letting be", " the potential difference between the cathode and anode, Figure 3-9a shows the current versus for is positive, the two values of the intensity of light incident on the cathode. When electrons are attracted to the anode. At sufficiently large all the emitted electrons reach the anode and the current reaches its maximum value. Lenard observed that the maximum current was proportional to the light intensity, an expected result since doubling the energy per unit time incident on the cathode should double the number of electrons emitted. Intensities too low to provide the electrons with the energy necessary to escape from the metal should result in no emission of electrons. However, in contrast with the classical expectation, there was no minimum intensity below which is negative, the electrons are repelled from the anode. the current was absent. When can reach the Then only electrons with initial kinetic energy anode. From Figure 3-9a we see that if no electrons reach the V0 is called the stopping potential. It is related to the maximum anode. The potential kinetic energy of the emitted electrons by greater than V0, mv2>2 is less than e\u0192V\u0192 V V V V V a 1 2 mv2b eV0 3-20 The experimental result, illustrated by Figure 3-9a, that is independent of the incident light intensity was surprising. Apparently, increasing the rate of energy falling on the cathode does not increase the maximum kinetic energy of the emitted electrons, contrary to classical expectations. In 1905 Einstein offered an explanation of this result in a remarkable paper in the same volume of Annalen der Physik that contained his papers on special relativity and Brownian motion. V0 3-3 The Photoelectric Effect 129 1 \u2013\u2013 2 mv2 \u03c6 (a) i (\u03bcA) (c) Bright light Dim light I2 > I1 I1 Energy hf \u20135 0 5 10 15 V hf 0 \u2013V0 (b) i (\u03bcA) f2 > f1 f1 ft \u20135 \u2013V01 \u2013V02 0 5 10 15 V Filled electron states Distance Inside metal Outside metal Surface V I1 i I1. I2, and where Figure 3-9 (a) Photocurrent versus anode voltage I2 intensities The stopping voltage I, constant Einstein\u2019s explanation of the photoelectric effect indicates that the magnitude of the f1, as observed, and that there should be a than stopping voltage should be greater for threshold frequency below which no photoelectrons were seen, also in agreement with experiment. (c) Electron potential energy curve across the metal surface. An electron with the hf. highest energy in the metal absorbs a photon of energy that its kinetic energy after leaving the surface be hf. for light of frequency with two V0 is the same for both. (b) For Conservation of energy requires f2 ft f Einstein assumed that the energy quantization used by Planck in solving the blackbody radiation problem was, in fact, a universal characteristic of light. Rather than being distributed evenly in the space through which it propagated, light energy consisted of discrete quanta, each of energy When one of these quanta, called a photon, penetrates the surface of the cathode, all of its energy may be absorbed completely by a single electron. If is the energy necessary to remove an electron from is called the work function and is a characteristic of the metal), the maxthe surface ( imum kinetic energy of an electron leaving the surface will be as a consequence of energy conservation; see Figure 3-9c. (Some electrons will have less than this amount because of energy lost in traversing the metal.) Thus, the stopping potential should be given by hf hf. eV0 a 1 2 mv2b hf max 3-21 Equation 3-21 is referred to as the photoelectric effect equation. As Einstein noted, If the derived formula is correct, then when represented in Cartesian coordinates as a function of the frequency of the incident light, must be a straight line whose slope is independent of the nature of the emitting substance.13 V0, As can be seen from Equation 3-21, the slope of At the time of this prediction there was no evidence that Planck\u2019s constant had anything to do with the photoelectric effect. There was also no evidence for the dependence of versus should equal V0 f h>e. Among the many applications of the photoelectric effect is the photomultiplier, a device that makes possible the accurate measurement of the energy of the light absorbed by a photosensitive surface. The SNO and Kamiokande neutrino observatories (see Chapter 12) use thousands of photomultipliers. Hundreds more are being deployed inside the Antarctic ice cap in the Ice Cube particle physics experiment. 130 Chapter 3 Quantization of Charge, Light, and Energy Figure 3-10 Millikan\u2019s data for stopping potential versus frequency for the photo", "electric effect. The data fall h>e, on a straight line with slope as predicted by Einstein a decade before the experiment. The intercept on the stopping potential axis is [R. A. Millikan, Physical Review, 7, 362 (1915).] >e1 40 50 70 60 ft = 43.9 \u00d7 1013 80 90 100 110 120 \u00d7 1013 Frequency (Hz) V0 on the frequency. Careful experiments by Millikan, reported the stopping potential in 1914 and in more detail in 1916, showed that Equation 3-21 was correct and that measurements of from it agreed with the value obtained by Planck. A plot taken from this work is shown in Figure 3-10. h The minimum, or threshold, frequency for the photoelectric effect, labeled this plot and in Figure 3-9b, and the corresponding threshold wavelength V0 to the work function by setting in Equation 3-21: 0 ft in are related t hf t hc t 3-22 t (and therefore having wavelengths greater Photons of frequencies lower than ) do not have enough energy to eject an electron from the metal. Work functhan tions for metals are typically on the order of a few electron volts. The work functions for several elements are given in Table 3-1. ft Table 3-1 Photoelectric work functions Element Work function (eV) Na C Cd Al Ag Pt Mg Ni Se Pb 2.28 4.81 4.07 4.08 4.73 6.35 3.68 5.01 5.11 4.14 EXAMPLE 3-5 Photoelectric Effect in Potassium The threshold frequency of potassium is 558 nm. What is the work function for potassium? What is the stopping potential when light of 400 nm is incident on potassium? 3-3 The Photoelectric Effect 131 SOLUTION 1. Both questions can be answered with the aid of Equation 3-21: a 1 2 hf e hf mv2b eV0 V0 max e 2. At the threshold wavelength the photoelectrons have just enough energy to overa 1 2 come the work function barrier, so mv2b hence 0, 0, and V0 max e hc e t hfte e 1240 eV # nm 558 nm 2.22 eV 3. When 400-nm light is used, is given by Equation 3-21: V0 hc e e V0 hf e e 1240 eV # nm 400 nm 2.22 eV 3.10 eV 2.22 eV 0.88 V Another interesting feature of the photoelectric effect that is contrary to classical physics but is easily explained by the photon hypothesis is the lack of any time lag between the turning on of the light source and the appearance of photoelectrons. Classically, the incident energy is distributed uniformly over the illuminated surface; the time required for an area the size of an atom to acquire enough energy to allow the emission of an electron can be calculated from the intensity (power per unit area) of the incident radiation. Experimentally, the incident intensity can be adjusted so that the calculated time lag is several minutes or even hours. But no time lag is ever observed. The photon explanation of this result is that although the rate at which photons are incident on the metal is very small when the intensity is low, each photon has enough energy to eject an electron, and there is some chance that a photon will be absorbed immediately. The classical calculation gives the correct average number of photons absorbed per unit time. EXAMPLE 3-6 Classical Time Lag Light of wavelength 400 nm and intensity 102 W>m2 is incident on potassium. Estimate the time lag for the emission of photoelectrons expected classically. SOLUTION According to Example 3-5, the work function for potassium is r 1010 in time is If we assume m to be the typical radius of an atom, the total energy falling on the atom 2.22 eV. t E (102 W>m2)(r2)t (102 W>m2)(1020 m2)t (3.14 1022 J>s)t 132 Chapter 3 Quantization of Charge, Light, and Energy Setting this energy equal to 2.22 eV gives (3.14 1022 J>s)t (2.22 eV)(1.60 1019 J>eV) (2.22 eV)(1.60 1019 J>eV) (3.14 1022 J>s) t 1.13 103 s 18.8 min According to the classical prediction, no atom would be expected to emit an electron until 18.8 min after the light source was turned on. According to the photon model of light, each photon has enough energy to eject an electron immediately. Because of the low intensity, there are few photons incident per second, so the chance of any particular atom absorbing a photon and emitting an electron in any given time interval", " is small. However, there are so many atoms in the cathode that some emit electrons immediately. EXAMPLE 3-7 Incident Photon Intensity In Example 3-6, how many photons are incident per second per square meter? SOLUTION The energy of each photon is E hf hc> (1240 eV # nm)>(400 nm) (3.10 eV)(1.60 1019 J>eV) 4.96 1019 J Since the incident intensity is per second per square meter is 102 W>m2 102 J>s # m2, the number of photons N 102 J>s # m2 4.96 1019 J>photon 2.02 1016 photons>s # m2 This is, of course, a lot of photons, not a few; however, the number per atom at n 2.02 1016 photons>s # m2 (1010)2 m2>atom n the surface is quite small. 6.3 104 photons>s # atom, or about 1 photon for every 1000 atoms. Questions 3. How is the result that the maximum photoelectric current is proportional to the intensity explained in the photon model of light? 4. What experimental features of the photoelectric effect can be explained by classical physics? What features cannot? The photoemission of electrons has developed into a significant technique for investigating the detailed structure of molecules and solids, making possible discoveries far beyond anything that Hertz may have imagined. The use of x-ray sources (see Section 3-4) and precision detectors has made possible precise determination of valence electron configurations in chemical compounds, leading to detailed understanding of chemical bonding and the differences between the bulk and surface atoms of solids. Photoelectric-effect microscopes will show the chemical situation of each element in a specimen, a prospect of intriguing and crucial importance in molecular biology and microelectronics. And they are all based on a discovery that annoyed Hertz\u2014at first. 3-4 X Rays and the Compton Effect 133 3-4 X Rays and the Compton Effect Further evidence of the correctness of the photon concept was furnished by Arthur H. Compton, who measured the scattering of x rays by free electrons and, by his analysis of the data, resolved the last lingering doubts regarding special relativity (see Chapter 1). Before we examine Compton scattering in detail, we will briefly describe some of the early work with x rays since it provides a good conceptual understanding of x-ray spectra and scattering. X Rays The German physicist Wilhelm K. Roentgen discovered x rays in 1895 when he was working with a cathode-ray tube. Coming five years before Planck\u2019s explanation of the blackbody emission spectrum, Roentgen\u2019s discovery turned out to be the first significant development in quantum physics. He found that \u201crays\u201d originating from the point where the cathode rays (electrons) hit the glass tube, or a target within the tube, could pass through materials opaque to light and activate a fluorescent screen or photographic film. He investigated this phenomenon extensively and found that all materials are transparent to these rays to some degree and that the transparency decreases with increasing density. This fact led to the medical use of x rays within months after the publication of Roentgen\u2019s first paper.14 (a) (b) (c) Pyrex glass envelope Electron beam Filament + Anode Cathode \u2013 Tungsten target X rays (a) Early x-ray tube. [Courtesy of Cavendish Laboratory.] (b) X-ray tubes became more compact over time. This tube was a design typical of the mid-twentieth century. [Courtesy of Schenectady Museum, Hall of Electrical History, Schenectady, NY.] (c) Diagram of the components of a modern x-ray tube. Design technology has advanced enormously, enabling very high operating voltages, beam currents, and x-ray intensities, but essential elements of the tubes remain unchanged. 134 (a) X rays (b) Chapter 3 Quantization of Charge, Light, and Energy Crystal Photographic plate with Laue spots An x ray of Mrs. Roentgen\u2019s hand taken by Roentgen shortly after his discovery. Figure 3-11 (a) Schematic sketch of a Laue experiment. The crystal acts as a threedimensional grating, which diffracts the x-ray beam and produces a regular array of spots, called a Laue pattern, on photographic film or an x-ray-sensitive chargecoupled device (CCD) detector. (b) Laue x-ray diffraction pattern using a niobium boride crystal and molybdenum x rays. 20-keV [General Electric Company.] Roentgen was unable to deflect these rays in a magnetic field, nor was he able to observe refraction or the interference phenomena associated with waves. He thus gave the rays the somewhat mysterious name of x rays. Since classical", " electromagnetic theory predicts that accelerated charges will radiate electromagnetic waves, it is natural to expect that x rays are electromagnetic waves produced by the acceleration of the electrons when they are deflected and stopped by the atoms of a target. Such radiation is called bremsstrahlung, German for \u201cbraking radiation.\u201d The slight diffraction broadening of an x-ray beam after passing through slits a few thousandths of a millimeter wide indicated the wavelength of x rays to be of the order of In 1912 Max von Laue suggested that since the wavelengths of x rays were of the same order of magnitude as the spacing of atoms in a crystal, the regular array of atoms in a crystal might act as a threedimensional grating for the diffraction of x rays. Experiments (Figure 3-11) soon confirmed that x rays are a form of electromagnetic radiation with wavelengths in the range of about and that atoms in crystals are arranged in regular arrays. 1010 m 0.1 nm. 0.01 to 0.10 nm W. L. Bragg, in 1912, proposed a simple and convenient way of analyzing the diffraction of x rays by crystals.15 He examined the interference of x rays due to scattering from various sets of parallel planes of atoms, now called Bragg planes. Two sets of Bragg planes are illustrated in Figure 3-12 for NaCl, which has a cubic structure called face-centered cubic. Consider Figure 3-13. Waves scattered from the two successive atoms within a plane will be in phase and thus interfere constructively, independent of the wavelength, if the scattering angle equals the incident angle. (This condition is the same as for reflection.) Waves scattered at equal angles from atoms in two Figure 3-12 A crystal of NaCl showing two sets of Bragg planes. 3-4 X Rays and the Compton Effect 135 \u03b8 \u03b8 d sin \u03b8 d Figure 3-13 Bragg scattering from two successive planes. The waves from the two atoms shown have a path length difference of 2d sin m is met. They will be in phase if the Bragg condition 2d sin. different planes will be in phase (constructive interference) only if the difference in path length is an integral number of wavelengths. From Figure 3-13 we see that this condition is satisfied if 2d sin m where m an integer 3-23 Equation 3-23 is called the Bragg condition. Measurements of the spectral distribution of the intensity of x rays as a function of the wavelength using an experimental arrangement such as that shown in Figure 3-14 produces the x-ray spectrum and, for classical physics, some surprises. Figure 3-15a shows two typical x-ray spectra produced by accelerating electrons through two voltand bombarding a tungsten target mounted on the anode of the tube. In this figages. ure Figure 3-15b shows the short wavelength lines produced with a molybdenum target and 35-keV electrons. Three features of the spectra are of immediate interest, only one of which could be explained by classical physics. (1) The spectrum consists of a series of sharp lines, called the characteristic spectrum, superimposed on (2) the continuous is the intensity emitted within the wavelength interval for each value of V I() d X rays Lead collimator Crystal Anode Electron beam + \u2013 X-ray tube Ionization chamber Figure 3-14 Schematic diagram of a Bragg crystal spectrometer. A collimated x-ray beam is incident on a crystal and scattered into an ionization chamber. The crystal and ionization chamber can be rotated to keep the angles of incidence and scattering equal as both are varied. By measuring the ionization in the chamber as a function of angle, the spectrum of the x rays is the separation of the can be determined using the Bragg condition d Bragg planes in the crystal. If the wavelength d is known, the spacing can be determined. 2d sin m, where 136 (a Chapter 3 Quantization of Charge, Light, and Energy L series K series V = 80 kV V = 40 kV 0 \u03bb 0.2 m \u03bb 0.4 0.6 m 0.8 \u03bb, \u00c5 1.0 1.2 1.4 1.6 (b) K\u03b1 ( K\u03b2 V = 35 kV 0 0 0.2 \u03bb 0.4 0.6 0.8 1.0 1.2 m \u03bb, \u00c5 Figure 3-15 (a) X-ray spectra from tungsten at two accelerating voltages and (b) from molybdenum at one. The names of the line series ( 0.5 nm. m is independent of the target element and is related to the voltage on the x-ray tube by The wavelengths of the lines are characteristic of the element. and ) are historical and explained in Chapter 4. The The cutoff wavelength lines for molybdenum (not shown)", " are at about hc>eV. L-series m K V L bremsstrahlung spectrum. The line spectrum is characteristic of the target material and varies from element to element. (3) The continuous spectrum has a sharp cutoff wavewhich is independent of the target material but depends on the energy of length, the bombarding electrons. If the voltage on the x-ray tube is volts, the cutoff wavelength is found empirically to be given by m, V m 1.24 103 V nm 3-24 m 20,000 eV (a few eV) o r hc>eV 1.2407 106 V1 m Equation 3-24 is called the Duane-Hunt rule, after its discoverers. It was pointed out rather quickly by Einstein that x-ray production by electron bombardment was an insimverse photoelectric effect and that Equation 3-21 should apply. The Duane-Hunt ply corresponds to a photon with the maximum energy of the electrons, that is, the photon emitted when the electron losses all of its kinetic energy in a single collision. Since the kinetic energy of the electrons in an x-ray tube is or higher, the work is negligible by comparison. That is, Equation 3-21 becomes function eV hf hc> 1.24 103 V1 nm. m m Thus, the Duane-Hunt rule is explained by Planck\u2019s quantum hypothesis. (Notice that h>e. ) the value of The continuous spectrum was understood as the result of the acceleration (i.e., \u201cbraking\u201d) of the bombarding electrons in the strong electric fields of the target atoms. Maxwell\u2019s equation predicted the continuous radiation. The real problem for classical physics was the sharp lines. The wavelengths of the sharp lines were a function of the target element, the set for each element being always the same. But the sharp lines never appeared if was such that was larger than the particular line, as can be seen m from Figure 3-15a, where the shortest-wavelength group disappears when is reduced from becomes larger. The origin of the sharp lines was a mystery that had to await the discovery of the nuclear atom. We will explain them in Chapter 4. can be used to determine 40 keV 80 keV so that to V V m m Well-known applications of x rays are medical and dental x rays (both diagnostic and treatment) and industrial x-ray inspection of welds and castings. Perhaps not so well known is the use of x rays in determining the structure of crystals, identifying black holes in the cosmos, and \u201cseeing\u201d the folded shapes of proteins in biological materials. Compton Effect 3-4 X Rays and the Compton Effect 137 >c hf1 hf1 hf2 It had been observed that scattered x rays were \u201csofter\u201d than those in the incident beam, that is, were absorbed more readily. Compton16 pointed out that if the scattering process were considered a \u201ccollision\u201d between a photon of energy (and mo) and an electron, the recoiling electron would absorb part of the incimentum of the scattered photon would therefore be less dent photon\u2019s energy. The energy hf2 than the incident one and thus of lower frequency (The fact and momentum E was known from that electromagnetic radiation of energy classical theory and from the experiments of Nichols and Hull in 1903. This relation is also consistent with the relativistic expression for a particle with zero rest energy.) Compton applied the laws of conservation of momentum and energy in their relativistic form (see Chapter 2) to the collision of a photon with an isolated electron to obtain the change in the wavelength of the photon as a function of the scattering angle The result, called Compton\u2019s equation and derived in a More section on the home page, is E2 p2c2 (mc2)2 carried momentum 1 E>c >c. 2 f2. Molybdenum K \u03b1 line primary (a) 2 1 h mc (1 cos ) 3-25 Scattered by graphite at 45\u00b0 The change in wavelength is thus predicted to be independent of the original wavehas the dimensions of length and is called the Compton length. The quantity wavelength of the electron. Its value is h>mc (b) c h mc hc mc2 1.24 103 eV # nm 5.11 105 eV 0.00243 nm (c) (d) Because 1 2 1)> fractional change generally only observed for x rays and gamma radiation. is small, it is difficult to observe unless ( is very small so that the 1 is appreciable. For this reason the Compton effect is 2 1 Compton verified his result experimentally using the characteristic x-ray line of wavelength 0.0711 nm from molybdenum for the incident monochromatic photons and scattering these photons from electrons in graphite. The wavelength of the scattered photons was measured", " using a Bragg crystal spectrometer. His experimental arrangement is shown in Figure 3-16; Figure 3-17 shows his results. The first peak at Defining slit S2 Calcite crystal Bragg spectrometer S1 R \u03c6 X-ray tube (Mo target) Shutter Ionization chamber R Figure 3-16 Schematic sketch of Compton\u2019s apparatus. X rays from the tube strike the and are scattered into a Bragg-type crystal spectrometer. In this diagram, carbon block the scattering angle is and Although the entire spectrum is being scattered by molybdenum. S1 the spectrometer scanned the region around the The beam was defined by slits 30\u00b0. S2. K R, line of Scattered at 90\u00b0 135\u00b0 6\u00b030\u00b4 7\u00b0 Angle from calcite 7\u00b030\u00b4 Figure 3-17 Intensity versus wavelength for Compton scattering at several angles. The left peak in each case results from photons of the original wavelength that are scattered by tightly bound electrons, which have an effective mass equal to that of the atom. The separation in wavelength of the peaks is given by Equation 3-25. The horizontal scale used by the Compton \u201cangle from calcite\u201d refers to the calcite analyzing crystal in Figure 3-16. 138 Chapter 3 Quantization of Charge, Light, and Energy Arthur Compton. After discovering the Compton effect, he became a world traveler seeking an explanation for cosmic rays. He ultimately showed that their intensity varied with latitude, indicating an interaction with Earth\u2019s magnetic field, and thus proved that they are charged particles. [Courtesy of American Institute of Physics, Niels Bohr Library.] each scattering angle corresponds to scattering with no shift in the wavelength due to scattering by the inner electrons of carbon. Since these are tightly bound to the atom, it is the entire atom that recoils rather than the individual electrons. The expected shift in this case is given by Equation 3-25, with being the mass of the atom, which is about times that of the electron; thus, this shift is negligible. The varia\u00a2 tion of with was found to be that predicted by Equation 3-25. 104 m 1 2 We have seen in this section and the preceding two sections that the interaction of electromagnetic radiation with matter is a discrete interaction that occurs at the atomic level. It is perhaps curious that after so many years of debate about the nature of light, we now find that we must have both a particle (i.e., quantum) theory to describe in detail the energy exchange between electromagnetic radiation and matter and a wave theory to describe the interference and diffraction of electromagnetic radiation. We will discuss this so-called wave-particle duality in more detail in Chapter 5. More Derivation of Compton\u2019s Equation, applying conservation of energy and momentum to the relativistic collision of a photon and an electron, is included on the home page: www.whfreeman.com/tiplermodern physics5e. See also Equations 3-26 and 3-27 and Figure 3-18 here. Questions 5. Why is it extremely difficult to observe the Compton effect using visible light? 6. Why is the Compton effect unimportant in the transmission of television and radio waves? How many Compton scatterings would a typical FM signal have before its wavelengths were shifted by 0.01 percent? 3-4 X Rays and the Compton Effect 139 EXAMPLE 3-8 X Rays from TV The acceleration voltage of the electrons in a typical television picture tube is What is the minimum wavelength x ray produced when these electrons strike the inner front surface of the tube? 25 keV. SOLUTION From Equation 3-24, we have m 1.24 103 V nm 1.24 103 25,000 0.050 nm These x rays penetrate matter very effectively. Manufacturers provide essential shields to protect against the hazard. EXAMPLE 3-9 Compton Effect In a particular Compton scattering experiment it is when the scattering found that the incident wavelength angle of the shifted photon when the scattering angle is (b) What will be the wavelength is shifted by 1? (a) What is the value of 120\u00b0. 1.5 percent 75\u00b0? 2 1 SOLUTION 1. For question (a), the value of 1 is found from Equation 3-25: 2 1 \u00a2 h mc (1 cos ) 0.00243(1 cos 120\u00b0) 2. That the scattered wavelength 3. Combining these yields: 2 \u00a2 1 0.015 is shifted by 1.5 percent from means that 1 1 \u00a2 0.00243(1 cos 120) 0.015 0.015 0.243 nm 4. Question (b) is also solved with the aid of Equation 3-25, rearranged as 5. Substituting 75\u00b0 2 2 1 0.00243(1 cos ) 1 and from above yields 0.243 0.00243(1 cos 75) 0.243 0.002", " 0.245 nm A Final Comment In this chapter, together with Section 2-4 of the previous chapter, we have introduced and discussed at some length the three primary ways by which photons interact with matter: (1) the photoelectric effect, (2) the Compton effect, and (3) pair production. As we proceed with our explorations of modern physics throughout the remainder of the book, we will have many occasions to apply what we have learned here to aid in our understanding of myriad phenomena, ranging from atomic structure to the fusion \u201cfurnaces\u201d of stars. 140 Chapter 3 Quantization of Charge, Light, and Energy Summary TOPIC RELEVANT EQUATIONS AND REMARKS 1. J. J. Thomson\u2019s experiment Thomson\u2019s measurements with cathode rays showed that the same particle (the electron), with about 2000 times that of ionized hydrogen, exists in all elements. e>m 2. Quantization of electric charge e 1.60217653 1019 C 3. Blackbody radiation Stefan-Boltzmann law Wein\u2019s displacement law Planck\u2019s radiation law Planck\u2019s constant R T4 mT 2.898 103 m # K u() 8hc5 ehc>kT 1 h 6.626 1034 J # s 4. Photoelectric effect eV0 hf 5. Compton effect 2 1 h mc (1 cos ) 3-4 3-5 3-18 3-19 3-21 3-25 6. Photon-matter interaction The (1) photoelectric effect, (2) Compton effect, and (3) pair production are the three ways of interaction. General References The following general references are written at a level appropriate for readers of this book. Millikan, R. A., Electrons ( and ) Protons, Photons, Neutrons, Mesotrons, and Cosmic Rays, 2d ed., University of Chicago Press, Chicago, 1947. This book on modern physics by one of the great experimentalists of his time contains fascinating, detailed descriptions of Millikan\u2019s oil drop experiment and his verification of the Einstein photoelectric-effect equation. Mohr, P. J., and B. N. Taylor, \u201cThe Fundamental Physical Constants,\u201d Physics Today (August 2004). Also available at http://www.physicstoday.org/guide/fundcont.html. Richtmyer, F. K., E. H. Kennard, and J. N. Cooper, Introduction to Modern Physics, 6th ed., McGraw-Hill, New York, 1969. This excellent text was originally published in 1928, intended as a survey course for graduate students. Shamos, M. H. (ed.), Great Experiments in Physics, Holt, Rinehart, and Winston, New York, 1962. This book contains 25 original papers and extensive editorial comment. Of particular interest for this chapter are papers by Faraday, Hertz, Roentgen, J. J. Thomson, Einstein (photo electric effect), Millikan, Planck, and Compton. Thomson, G. P., J. J. Thomson, Discoverer of the Electron, Doubleday/Anchor, Garden City, NY, 1964. An interesting study of J. J. Thomson by his son, also a physicist. Virtual Laboratory (PEARL), Physics Academic Software, North Carolina State University, Raleigh, 1996. Computer simulation software allows the user to analyze blackbody radiation emitted over a wide range of temperatures and investigate the Compton effect in detail. Weart, S. R., Selected Papers of Great American Physicists, American Institute of Physics, New York, 1976. The bicentennial commemorative volume of the American Physical Society. Notes 1. Democritus (about 470 B.C. to about 380 B.C.). Among his other modern-sounding ideas were the suggestions that the Milky Way is a vast conglomeration of stars and that the Moon, like Earth, has mountains and valleys. 2. G. J. Stoney (1826\u20131911). An Irish physicist who first called the fundamental unit of charge the electron. After Thomson discovered the particle that carried the charge, the name was transferred from the quantity of charge to the particle itself by Lorentz. 3. Joseph J. Thomson (1856\u20131940). English physicist and director for more than 30 years of the Cavendish Laboratory, the first laboratory in the world established expressly for research in physics. He was awarded the Nobel Prize in 1906 for his work on the electron. Seven of his research assistants also won Nobel Prizes. 4. Much early confusion existed about the nature of cathode rays due to the failure of Heinrich Hertz in 1883 to observe any deflection of the rays in an electric field. The failure was later found to be the result of ionization of the gas in the tube; the ions quickly neutralized the charges on the deflecting", " plates so that there was actually no electric field between the plates. With better vacuum technology in 1897, Thomson was able to work at lower pressure and observe electrostatic deflection. 5. R. A. Millikan, Philosophical Magazine (6), 19, 209 (1910). Millikan, who held the first physics Ph.D. awarded by Columbia University, was one of the most accomplished experimentalists of his time. He received the Nobel Prize in 1923 for the measurement of the electron\u2019s charge. Also among his many contributions, he coined the phrase cosmic rays to describe radiation produced in outer space. 6. R. A. Millikan, Physical Review, 32, 349 (1911). Problems 141 7. Mohr, P. J., and B. N. Taylor, \u201cThe Fundamental Physical Constants,\u201d Physics Today (August 2004). 8. See pp. 135-137 of F. K. Richtmyer, E. H. Kennard, and J. N. Cooper (1969). 9. John W. S. Rayleigh (1842\u20131919). English physicist, almost invariably referred to by the title he inherited from his father. He was Maxwell\u2019s successor and Thomson\u2019s predecessor as director of the Cavendish Laboratory. 10. Max K. E. L. Planck (1858\u20131947). Most of his career was spent at the University of Berlin. In his later years his renown in the world of science was probably second only to that of Einstein. 11. Heinrich R. Hertz (1857\u20131894), German physicist, student of Helmholtz. He was the discoverer of electromagnetic \u201cradio\u201d waves, later developed for practical communication by Marconi. 12. H. Hertz, Annalen der Physik, 31, 983 (1887). 13. A. Einstein, Annalen der Physik, 17, 144 (1905). 14. A translation of this paper can be found in E. C. Watson, American Journal of Physics, 13, 284 (1945), and in Shamos (1962). Roentgen (1845\u20131923) was honored in 1901 with the first Nobel Prize in Physics for his discovery of x rays. 15. William Lawrence Bragg (1890\u20131971), Australian-English physicist. An infant prodigy, his work on x-ray diffraction performed with his father, William Henry Bragg (1862\u20131942), earned for them both the Nobel Prize in Physics in 1915, the only father-son team to be so honored thus far. In 1938 W. L. Bragg became director of the Cavendish Laboratory, succeeding Rutherford. 16. Arthur H. Compton (1892\u20131962), American physicist. It was Compton who suggested the name photon for the light quantum. His discovery and explanation of the Compton effect earned him a share of the Nobel Prize in Physics in 1927. Problems Level I Section 3-1 Quantization of Electric Charge H2 The beam then enters a region of uniform magnetic field 3-1. A beam of charged particles consisting of protons, electrons, deuterons, and singly ionmolecules all pass through a velocity selector, all emerging with ized helium atoms and 2.5 106 m>s. B 0.40 T speeds of directed perpendicular to their velocity. Compute the radius of curvature of the path of each type of particle. 3-2. Consider Thomson\u2019s experiment with the electric field turned \u201coff.\u201d If the electrons enter l, a region of uniform magnetic field show that the electrons are deflected through eB>mu (Assume that the electrons are moving at nonan angle relativistic speeds.) for small values of and length. B 142 Chapter 3 Quantization of Charge, Light, and Energy 5.0 104 eV. What magnetic field will be needed? 3-3. Equation 3-3 suggests how a velocity selector for particles or mixtures of different particles all having the same charge can be made. Suppose you wish to make a velocity selector that allows undeflected passage for electrons whose kinetic energy is The elec2.0 105 V>m. tric field available to you is 3-4. A cosmic ray proton approaches Earth vertically at the equator, where the horizontal 3.5 105 T. 3.0 106 m>s, component of Earth\u2019s magnetic field is If the proton is moving at what is the ratio of the magnetic force to the gravitational force on the proton? 3-5. An electron of kinetic energy netic field of 0.325 T. the motion. 3-6. traverse a distance of nent of their velocity after passing between the plates if the electric field is 3-7. per was about 2000-eV 3-8. On drop #16, Millikan measured the following total charges, among others, at different times: In", " J. J. Thomson\u2019s first method (see Problem 3-44), the heat capacity of the beam stopHow many moves in a circular orbit perpendicular to a mag(a) Compute the radius of the orbit. (b) Find the period and frequency of find (a) their speed, (b) the time needed to in Figure 3-1, and (c) the vertical compo- If electrons have kinetic energy of 5 cm and the temperature increase was about electrons struck the beam stopper? 2000 eV, E and 3.33 103 V>m. 5 103 cal>\u00b0C between plates 45 keV 2\u00b0C. D 25.41 1019 C 20.64 1019 C 17.47 1019 C 19.06 1019 C 12.70 1019 C 14.29 1019 C e e 2mg>q. What value of the fundamental quantized charge do these numbers imply? 3-9. Show that the electric field needed to make the rise time of the oil drop equal to its fieldfree fall time is 3-10. One variation of the Millikan oil drop apparatus arranges the electric field horizontally, rather than vertically, giving charged droplets acceleration in the horizontal direction. The result is that the droplet falls in a straight line that makes an angle with the vertical. Show that sin qe>bv\u0153 t v\u0153 t is the terminal speed along the angled path. where 3-11. A charged oil droplet falls tric field. The specific gravity of air is 1.80 105 N # s>m2. is two units of electric charge and is in an electric field of electric force to the gravitational force on the droplet? at terminal speed in the absence of an elecThe viscosity of air and that of oil is 0.75. (a) What are the mass and radius of the drop? (b) If the droplet carries what is the ratio of the 20.0 s 1.35 103 2.5 105 V>m, 5.0 mm in Section 3-2 Blackbody Radiation m for blackbody radiation at (a) 3-12. Find 3-13. Use the result of Example 3-4 and Equations 3-4 and 3-6 to express Stefan\u2019s constant in terms of 3-14. Show that Planck\u2019s law, Equation 3-18, expressed in terms of the frequency and Using the known values of these constants, calculate Stefan\u2019s constant. and (c) h, c, (b) f, k. is T 3000 K. T 300 K, T 3 K, u(f) 8f2 c3 hf ehf>kT 1 2.7 K. 3-15. As noted in the chapter, the cosmic microwave background radiation fits the Planck (a) What is the wavelength at the maximum intensity of the equation for a blackbody at spectrum of the background radiation? (b) What is the frequency of the radiation at the maximum? (c) What is the total power incident on Earth from the background radiation? 3-16. Find the temperature of a blackbody if its spectrum has its peak at (a) (visible), (b) (microwave region), and (c) 3-17. If the absolute temperature of a blackbody is doubled, by what factor is the total emitted power increased? 3-18. Calculate the average energy 10 hc>kT, (b) a short wavelength kT sical prediction rem discussed in Chapter 8.) per mode of oscillation for (a) a long wavelength and compare your results with the clas(see Equation 3-9). (The classical value comes from the equipartition theo- m (FM radio waves). E 0.1 hc>kT, 700 nm 3 cm 3 m m m Problems 143 20,000 K. 27.0 m 3-19. A particular radiating cavity has the maximum of its spectra distribution of radiated power at a wavelength of (in the infrared region of the spectrum). The temperature is then changed so that the total power radiated by the cavity doubles. (a) Compute the new temperature. (b) At what wavelength does the new spectral distribution have its maximum value? (a) Assuming 3-20. A certain very bright star has an effective surface temperature of is maximum? (b) In what that it radiates as a blackbody, what is the wavelength at which part of the electromagnetic spectrum does the maximum lie? 1.36 103 W>m2, 3-21. The energy reaching Earth from the Sun at the top of the atmosphere is called the solar constant. Assuming that Earth radiates like a blackbody at uniform temperature, what do you conclude is the equilibrium temperature of Earth? 3-22. A 3300 K. fm Assuming that the bulb radiates like a blackbody,", " (a) what are the frequency and the waveis a good approximation of the length average frequency of the photons emitted by the bulb, about how many photons is the bulb radiating per second? (c) If you are looking at the bulb from 5 m away, how many photons enter your eye per second? (The diameter of your pupil is about 3-23. Use Planck\u2019s law, Equation 3-18, to derive the constant in Wein\u2019s law, Equation 3-5. incandescent bulb radiates from a tungsten filament operating at at the maximum of the spectral distribution? (b) If 5.0 mm. 40-W u() m fm ) Section 3-3 The Photoelectric Effect 1.9 eV, 380 nm 750 nm. to about 4.87 eV and (c) 100-W (in eV) What is the energy of an FM photon of the frequency? (a) What is in visible light? (b) A typical FM radio station\u2019s broadcast 3-24. The wavelengths of visible light range from about the range of photon energies frequency is about 100 MHz. 3-25. The orbiting space shuttle moves around Earth well above 99 percent of the atmosphere, yet it still accumulates an electric charge on its skin due, in part, to the loss of electrons caused by the photoelectric effect with sunlight. Suppose the skin of the shuttle is coated with Ni, which has a relatively large work function at the temperatures encountered in orbit. (a) What is the maximum wavelength in the solar spectrum that can result in the emission of photoelectrons from the shuttle\u2019s skin? (b) What is the maximum fraction of the total power falling on the shuttle that could potentially produce photoelectrons? 3-26. The work function for cesium is the lowest of any metal. (a) Find the threshold frequency and wavelength for the photoelectric effect. Find the stopping potential if the wavelength of the incident light is (b) 300 nm 400 nm. 3-27. (a) If 5 percent of the power of a bulb is radiated in the visible spectrum, how many visible photons are radiated per second? (b) If the bulb is a point source radiating equally in all directions, what is the flux of photons (number per unit time per unit area) at a distance of 3-28. The work function of molybdenum is the photoelectric effect in molybdenum? (b) Will yellow light of wavelength tion of photoelectrons from molybdenum? Prove your answer. 3-29. The NaCl molecule has a bond energy of that is, this energy must be supplied in order to dissociate the molecule into neutral Na and Cl atoms (see Chapter 9). (a) What are the minimum frequency and maximum wavelength of the photon necessary to dissociate the molecule? (b) In what part of the electromagnetic spectrum is this photon? 3-30. A photoelectric experiment with cesium yields stopping potentials for 546.1 nm frequency and work function for cesium and the value of 3-31. Under optimum conditions, the eye will perceive a flash if about 60 photons arrive at the cornea. How much energy is this in joules if the wavelength of the light is 3-32. The longest wavelength of light that will cause emission of electrons from cesium is (ultraviolet) were to 653 nm. shine on cesium, what would be the energy of the ejected electrons? and respectively. Using these data only, find the threshold (a) What is the threshold frequency for cause ejec- (a) Compute the work function for cesium. (b) If light of 435.8 nm 550 nm? 4.26 eV; 4.22 eV. 560 nm 300 nm 0.38 V, 0.95 V to be 2 m? and h. Section 3-4 X Rays and the Compton Effect 3-33. Use Compton\u2019s equation (Equation 3-25) to compute the value of To what percent shift in the wavelength does this correspond? \u00a2 in Figure 3-17d. 144 Chapter 3 Quantization of Charge, Light, and Energy 110\u00b0 What 3 cm, eV>c and in 80 kV. and (d) 400 nm, 0.511-MeV 0.511-MeV kg # m>s 1 \u00c5 0.1 nm, if the wavelength is (a) photon from a positron-electron annihilation scatters at 3-34. X-ray tubes currently used by dentists often have accelerating voltages of is the minimum wavelength of the x rays they produce? 3-35. Find the momentum of a photon in (b) (c) 2 nm. 3-36. Gamma rays emitted by radioactive nuclei also exhibit measurable Compton scattering. Suppose a from a free electron", ". What are the energies of the scattered photon and the recoiling electron? Relative to the initial direction of the photon, what is the direction of the recoiling electron\u2019s velocity vector? 3-37. The wavelength of Compton-scattered photons is measured at 1 percent, what should the wavelength of the incident photon be? 3-38. Compton used photons of wavelength tons? (b) What is the wavelength of the photons scattered at 180\u00b0? of the photons scattered at 3-39. Compute for photons scattered at (c) 3-40. Compton\u2019s equation (Equation 3-25) indicates that a graph of 2 h. should be a straight line whose slope length of 1 graph the results versus 3-41. (a) Compute the Compton wavelength of an electron and a proton. (b) What is the energy of a photon whose wavelength is equal to the Compton wavelength of (1) the electron and (2) the proton? (a) What is the energy of these pho(c) What is the energy 180\u00b0? from (a) free protons, (b) free electrons, and (1 cos ) allows a determination of Given that the wavefor each scattering angle in the figure and 2 What is the slope of the line? (d) What is the recoil energy of the electrons if in Figure 3-17 is molecules in air. (1 cos ). 180\u00b0? 0.0711 nm, 0.0711 nm. 90\u00b0. compute is to be versus h>mc \u00a2> 120\u00b0 N2 \u00a2 If 1.90 V. 450 nm are emitted. If the wavelength of the incident light is changed to Level II 3-42. When light of wavelength is incident on potassium, photoelectrons with stopping 0.52 V 300 nm, potential of Using only these numbers together with the values of the speed the stopping potential is of light and the electron charge, (a) find the work function of potassium and (b) compute a value for Planck\u2019s constant. 3-43. Assuming that the difference between Thomson\u2019s calculated in his second experiment (Figure 3-19) and the currently accepted value was due entirely to his neglecting the horizontal component of Earth\u2019s magnetic field outside the deflection plates, what value for that component e 1.5 104 V>m, does the difference imply? (Thomson\u2019s data: x1 B 5.5 104 T, 5 cm, y2 8>110.) e>m >x2 Deflection plates uy \u03b8 ux y2 ux y1 x1 x2 Figure 3-19 Deflection of the electron beam in Thomson\u2019s apparatus. The deflection plates are in Figure 3-1. Deflection is shown with magnetic field off and the top plate positive. The magnetic field is applied perpendicular to the plane of the diagram and directed into the page. and D E e>m \u00a2V 3-44. In his first experiment, Thomson determined the speed of electrons accelerated by collecting them in an insulated beam stopper and measuring both through a potential of the beam stopper. (a) Show that the total collected charge with those measurements he could obtain an expression for in terms of the speed of the electrons and the directly measured quantities. (b) Show that the expression obtained in (a) toin terms of directly gether with the result of Problem 3-2 enabled Thomson to compute measured quantities. and the temperature rise e>m e>m \u00a2T Q 3-45. Data for stopping potential versus wavelength for the photoelectric effect using sodium are, nm V0, V 200 4.20 300 2.06 400 1.05 500 0.41 600 0.03 Problems 145 h>e. Plot these data in such a way as to be able to obtain (a) the work function, (b) the threshold frequency, and (c) the ratio 3-46. Prove that the photoelectric effect cannot occur with a completely free electron, i.e., one not bound to an atom. (Hint: Consider the reference frame in which the total momentum if the electron and the incident photon is zero.) 3-47. When a beam of monochromatic x rays is incident on a particular NaCl crystal, Bragg d 0.28 nm. reflection in the first order (i.e., with What is the minimum voltage at which the x-ray tube can be operating? 3-48. A The blackbody is initially at rest in a frictionless space. (a) Compute the total energy and momentum absorbed by the blackbody from the light beam, (b) calculate the blackbody\u2019s velocity at the end of the period of illumination, and (c) compute the final kinetic energy of the blackbody. Why is the latter less than the total energy of the absorbed photons? 3-49. Show that the maximum kinetic", " energy electron can carry away from a Compton scattering event is given by beam of light is shone onto a blackbody of mass called the Compton edge, that a recoiling 2 103 kg The value of ) occurs at 20\u00b0. m 1 100-W 104 s. Ek, for Ek hf 1 mc2>2hf 2E2 mc2 2E 3-50. The x-ray spectrometer on board a satellite measures the wavelength at the maximum intensity emitted by a particular star to be Assuming that the star radiates like a blackbody, (a) compute the star\u2019s surface temperature. (b) What is the ratio of the intensity radiated at 3-51. Determine the fraction of the energy radiated by the Sun in the visible region of the spectrum 3-52. Millikan\u2019s data for the photoelectric effect in lithium are shown in the table: Assume that the Sun\u2019s surface temperature is (350 nm to 700 nm). to that radiated at 100 nm 82.8 nm. 70 nm 5800 K. and at m. m Incident (nm) 253.5 312.5 365.0 404.7 433.9 Stopping voltage V0 (V) 2.57 1.67 1.09 0.73 0.55 (a) Graph the data and determine the work function for lithium. (b) Find the value of Planck\u2019s constant directly from the graph in (a). (c) The work function for lead is Which, if any, of the wavelengths in the table would not cause emission of photoelectrons from lead? 4.14 eV. where a for which du>d 0 a hc>kT. for various values of (b) Show that the value of u C5(ea> 1)1, Level III 3-53. This problem is to derive the Wein displacement law, Equation 3-5. (a) Show that the C energy density distribution function can be written is a consatisfies the equation stant and 5(1 ea>) a. (c) This equation can be solved with a calculator by the trial-and-error >a is determined to four significant method. Try figures. (d) Show that your solution in (c) implies and calculate the value of the constant. 3-54. This problem is one of estimating the time lag (expected classically, but not observed) of light energy. for the photoelectric effect. Assume that a point light source emits (a) Assuming uniform radiation in all directions, find the light intensity in at a disfrom the light source. (b) Assuming some reasonable size for an atom, find the entance of ergy per unit time incident on the atom for this intensity. (c) If the work function is how long does it take for this much energy to be absorbed, assuming that all of the energy hitting the atom is absorbed? until mT constant eV>s # m2 1 W 1 J>s 2 eV, 1 m 146 Chapter 3 Quantization of Charge, Light, and Energy 15 MeV>c. 3-55. A photon can be absorbed by a system that can have internal energy. Assume that a 15-MeV photon is absorbed by a carbon nucleus initially at rest. The recoil momentum of the (a) Calculate the kinetic energy of the carbon nucleus. What carbon nucleus must be is the internal energy of the nucleus? (b) The carbon nucleus comes to rest and then loses its internal energy by emitting a photon. What is the energy of the photon? 3-56. The maximum kinetic energy given to the electron in a Compton scattering event plays a role in the measurement of gamma-ray spectra using scintillation detectors. The maximum is referred to as the Compton edge. Suppose that the Compton edge in a particular experiment is found to be 520 keV. 3-57. An electron accelerated to in an x-ray tube has two successive collisions in being brought to rest in the target, emitting two bremsstrahlung photons in the process. The second longer than the first. (a) What are the wavelengths photon emitted has a wavelength of the two photons? (b) What was the energy of the electron after emission of the first photon? 3-58. Derive Equation 3-17 from Equations 3-15 and 3-16. What were the wavelength and energy of the incident gamma rays? 0.095 nm 50 keV CHAPTER 4 The Nuclear Atom Among his many experiments, Newton found that sunlight passing through a small opening in a window shutter could be refracted by a glass prism so that it would fall on a screen. The white sunlight thus refracted was spread into a rainbow-colored band\u2014a spectrum. He had discovered dispersion, and his experimental arrangement was the prototype of the modern spectroscope (Figure 4-1a). When, 150 years later, Fraunh", "ofer1 dispersed sunlight using an experimental setup similar to that shown in Figure 4-1b to test prisms made of glasses that he had developed, he found that the solar spectrum was crossed by more than 600 narrow, or sharp, dark lines.2 Soon after, a number of scientists observed sharp bright lines in the spectra of light emitted by flames, arcs, and sparks. Spectroscopy quickly became an important area of research. It soon became clear that chemical elements and compounds emit three general types of spectra. Continuous spectra, emitted mainly by incandescent solids, show no lines at all, bright or dark, in spectroscopes of the highest possible resolving power. Band spectra consist of very closely packed groups of lines that appear to be continuous in instruments of low resolving power. These are emitted when small pieces of solid materials are placed in the source flame or electrodes. The line spectra mentioned above arise when the source contains unbound chemical elements. The lines and bands turned out to be characteristic of individual elements and chemical compounds when excited under specific conditions. Indeed, the spectra could be (and are today) used as a highly sensitive test for the presence of elements and compounds. 4-1 Atomic Spectra 148 4-2 Rutherford\u2019s Nuclear Model 150 4-3 The Bohr Model of the Hydrogen Atom 4-4 X-Ray Spectra 159 169 4-5 The Franck-Hertz Experiment 174 Voltaire\u2019s depiction of Newton\u2019s discovery of dispersion. [El\u00e9mens de la Philosophie de Newton, Amsterdam, 1738.] 147 148 Chapter 4 The Nuclear Atom (a) Prism Screen Slit Spectrum Source (b) Lens Source of wavelengths 1 and \u03bb \u03bb 2 > \u03bb (\u03bb 1) 2 Prism Screen \u03bb 2 \u03bb 1 Slit Figure 4-1 (a) Light from the source passes through a small hole or a narrow slit before falling on the prism. The purpose of the slit is to ensure that all the incident light strikes the prism face at the same angle so that the dispersion by the prism causes the various frequencies that may be present to strike the screen at different places with minimum overlap. (b) The source emits only two wavelengths, 1. lens so that parallel light passes through the narrow slit, projecting a narrow line onto the face of the prism. Ordinary dispersion in the prism bends the shorter wavelength through the larger total angle, separating the two wavelengths at the screen. In this arrangement each wavelength appears on the screen (or on film replacing the screen) as a narrow line, which is an image of the slit. Such a spectrum was dubbed a \u201cline spectrum\u201d for that reason. Prisms have been almost entirely replaced in modern spectroscopes by diffraction gratings, which have much higher resolving power. The source is located at the focal point of the 2 Line spectra raised an enormous theoretical problem: although classical physics could account for the existence of a continuous spectrum (if not its detailed shape, as we saw with blackbodies), it could in no way explain why sharp lines and bands should exist. Explaining the origin of the sharp lines and accounting for the primary features of the spectrum of hydrogen, the simplest element, was a major success of the so-called old quantum theory begun by Planck and Einstein and will be the main topic in this chapter. Full explanation of the lines and bands requires the later, more sophisticated quantum theory, which we will begin studying in Chapter 5. 4-1 Atomic Spectra The characteristic radiation emitted by atoms of individual elements in a flame or in a gas excited by an electrical discharge was the subject of vigorous study during the late nineteenth and early twentieth centuries. When viewed or photographed through a spectroscope, this radiation appears as a set of discrete lines, each of a particular 4-1 Atomic Spectra 149 The uniqueness of the line spectra of the elements has enabled astronomers to determine the composition of stars, chemists to identify unknown compounds, and theme parks and entertainers to have laser shows. color or wavelength; the positions and intensities of the lines are characteristic of the element. The wavelengths of these lines could be determined with great precision, and much effort went into finding and interpreting regularities in the spectra. A major breakthrough was made in 1885 by a Swiss schoolteacher, Johann Balmer, who found that the lines in the visible and near ultraviolet spectrum of hydrogen could be represented by the empirical formula n 364.6 n2 n2 4 nm 4-1 where n is a variable integer that takes on the values n 3, 4, 5,.... Figure 4-2a shows the set of spectral lines of hydrogen (now known as the Balmer series) whose H wavelengths are given by Balmer\u2019s formula. For example, the wavelength of the line could be found by letting n = 3 in Equation 4-1 (try it!), and other integers each predicted a line that was found in the spectrum.", " Balmer suggested that his formula might be a special case of a more-general expression applicable to the spectra of other elements when ionized to a single electron, i.e., hydrogen-like elements Hydrogen (a) Ha Hb Hg Hd He b) (c) (d Sodium Mercury Sodium f D2 and H, Figure 4-2 (a) Emission line spectrum of hydrogen in the visible and near ultraviolet. The lines appear dark because the spectrum was photographed; hence, the bright lines are exposed (dark) areas on the film. The names of the first five lines are shown, as is the point beyond which no lines appear, called the limit of the series. (b) Part of the emission spectrum of sodium. The two very close bright lines at D1 589 nm are the lines. They are the principal radiation from sodium street lighting. (c) Part of the emission spectrum of mercury. (d) Part of the dark line (absorption) spectrum of sodium. White light shining through sodium vapor is absorbed at certain wavelengths, resulting in no exposure of the film at those points. Note that the line at 259.4 nm is visible here in both the bright and dark line spectra. Note, too, that frequency increases toward the right, wavelength toward the left in the four spectra shown. 150 Chapter 4 The Nuclear Atom Such an expression, found independently by J. R. Rydberg and W. Ritz and thus called the Rydberg-Ritz formula, gives the reciprocal wavelength 3 as 1 mn Ra 1 m2 1 n2 b for n m 4-2 where m and n are integers and R, the Rydberg constant, is the same for all series of spectral lines of the same element and varies only slightly, and in a regular way, from 1.096776 107 m1. For element to element. For hydrogen, the value of R is very heavy elements, R approaches the value of 1.097373 107 m1. Such empirical expressions were successful in predicting other series of spectral lines, such as other hydrogen lines outside the visible region. RH R EXAMPLE 4-1 Hydrogen Spectral Series The hydrogen Balmer series reciprocal wavelengths are those given by Equation 4-2, with m 2 and n 3, 4, 5,.... For example, the first line of the series, would be for m 2, n 3: 1 23 Ra 1 22 1 32 or H, b 5 36 R 1.523 106 m1 656.5 nm Other series of hydrogen spectral lines were found for m 1 (by Theodore Lyman) and m 3 (by Friedrich Paschen). Compute the wavelengths of the first lines of the Lyman and Paschen series. 23 SOLUTION For the Lyman series (m 1), the first line is for m 1, n 2: 1 12 12 Ra 1 12 1 22 b 3 4 R 8.22 106 m1 121.6 nm (in the ultraviolet) For the Paschen series (m 3), the first line is for m 3, n 4: 1 34 34 Ra 1 32 1 42 b 7 144 R 5.332 105 m1 1876 nm (in the infrared) All of the lines predicted by the Rydberg-Ritz formula for the Lyman and Paschen series are found experimentally. Note that no lines are predicted to lie beyond > 1 R 91.2 nm for the Lyman series and 9 R 820.6 nm for the Paschen se- ries and none are found by experiments. > 4-2 Rutherford\u2019s Nuclear Model Many attempts were made to construct a model of the atom that yielded the Balmer and Rydberg-Ritz formulas. It was known that an atom was about 1010 m in diameter (see Problem 4-6), that it contained electrons much lighter than the atom (see Section 3-1), and that it was electrically neutral. The most popular model was J. J. Thomson\u2019s model, already quite successful in explaining chemical reactions. Thomson 4-2 Rutherford\u2019s Nuclear Model 151 (a) (b) \u03b1 \u03b8 Figure 4-3 Thomson\u2019s model of the atom: (a) A sphere of positive charge with electrons embedded in it so that the net charge would normally be zero. The atom shown would have been phosphorus. (b) An particle scattered by such an atom would have a scattering angle much smaller than 1\u00b0. attempted various models consisting of electrons embedded in a fluid that contained most of the mass of the atom and had enough positive charge to make the atom electrically neutral. (See Figure 4-3a.) He then searched for configurations that were stable and had normal modes of vibration corresponding to the known frequencies of the spectral lines. One difficulty with all such models was that electrostatic forces alone cannot produce stable equilibrium. Thus, the charges were required to move and, if they stayed within the", " atom, to accelerate; however, the acceleration would result in continuous emission of radiation, which is not observed. Despite elaborate mathematical calculations, Thomson was unable to obtain from his model a set of frequencies of vibration that corresponded with the frequencies of observed spectra. The Thomson model of the atom was replaced by one based on the results of a set of experiments conducted by Ernest Rutherford4 and his students H. W. Geiger and E. Marsden. Rutherford was investigating radioactivity and had shown that the radia. and He tions from uranium consisted of at least two types, which he labeled showed, by an experiment similar to that of Thomson, that for the was half that of the proton. Suspecting that the particles were doubly ionized helium, Rutherford and his co-workers in a classic experiment let a radioactive substance decay in a previously evacuated chamber; then, by spectroscopy, they detected the spectral lines of ordinary helium gas in the chamber. Realizing that this energetic, massive particle would make an excellent probe for \u201cfeeling about\u201d within the interiors of other atoms, Rutherford began a series of experiments with this purpose. q>m Hans Geiger and Ernest Rutherford in their Manchester Laboratory. [Courtesy of University of Manchester.] 152 Chapter 4 The Nuclear Atom In these latter experiments, a narrow beam of particles fell on a zinc sulfide screen, which emitted visible light scintillations when struck (Figure 4-4). The distribution of scintillations on the screen was observed when various thin metal foils were particles were either undeflected or placed between it and the source. Most of the (a) Radioactive source R Au foil F D \u03b8 Pb shield \u03b1 beam Scintillation screen S Microscope M Observer Rotation (b) F S D R M Figure 4-4 Schematic diagram of the apparatus used by Geiger and Marsden to test Rutherford\u2019s atomic model. (a) The beam of particles is defined by the small hole D in the shield surrounding the radioactive source R of 214Bi (called RaC in Rutherford\u2019s day). beam strikes an ultrathin gold foil F (about 2000 atoms thick), and the particles are The individually scattered through various angles. Those scattering at the angle shown strike a small screen S coated with a scintillator, i.e., a material that emits tiny flashes of light (scintillations) when struck by an particle. The scintillations were viewed by the observer through a small microscope M. The scintillation screen\u2013microscope combination could be rotated about the center of the foil. The region traversed by the beam is evacuated. The experiment consisted of counting the number of scintillations as a function of of the actual apparatus as it appeared in Geiger and Marsden\u2019s paper describing the results. The letter key is the same as in (a). [Part (b) from H. Geiger and E. Marsden, Philosophical Review, 25, 507 (1913).]. (b) A diagram 4-2 Rutherford\u2019s Nuclear Model 153 deflected through very small angles of the order of 1\u00b0. Quite unexpectedly, however, a few particles were deflected through angles as large as 90\u00b0 or more. If the atom consisted of a positively charged sphere of radius 1010 m, containing electrons as in the Thomson model, only a very small deflection could result from a single encounter between an particle penetrated into the atom. Indeed, calculations showed that the Thomson atomic model could not possibly account for the number of large-angle scatterings that Rutherford saw. The unexpected scatterings at large angles were described by Rutherford with these words: particle and an atom, even if the It was quite the most incredible event that ever happened to me in my life. It was as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. Rutherford\u2019s Scattering Theory and the Nuclear Atom The question is, then, Why would one obtain the large-angle scattering that Rutherford saw? The trouble with the Thomson atom is that it is too \u201csoft\u201d\u2014the maximum force experienced by the is too weak to give a large deflection. If the positive charge of the atom is concentrated in a more compact region, however, a much larger force will occur at near impacts. Rutherford concluded that the large-angle scattering obtained experimentally could result only from a single encounter of the particle with a massive charge confined to a volume much smaller than that of the whole atom. Assuming this \u201cnucleus\u201d to be a point charge, he calculated the expected angular distribution for the scattered particles. His predictions of the dependence of scattering probability on angle, nuclear charge, and kinetic energy were completely verified in a series of experiments carried out in his laboratory by Geiger and Marsden. We will not go through Rutherford\u2019s derivation in detail, but merely outline the assumptions and conclusions. Figure 4-5", " shows the geometry of an particle being scattered by a nucleus, which we take to be a point charge Q at the origin. Initially, particle approaches with speed v along a line a distance b from a parallel line the COA through the origin. The force on the given by particle is F kqQ>r2, z\u00b4 \u03c6 m\u03b1 \u03c6 0 F r O b \u03b8 v m\u03b1 B A \u03c6 0 b m\u03b1 v C Figure 4-5 Rutherford scattering geometry. The nucleus is assumed to be a point charge Q at the origin O. At any distance r the particle experiences a repulsive force The particle travels along a hyperbolic path that is initially parallel to line COA a distance b from it and finally parallel to line OB, which makes an angle with OA. The scattering angle can be related to the impact parameter b by classical mechanics. kqQ>r2. 154 Chapter 4 The Nuclear Atom F F \u221d Q/r2 r R r r R Coulomb\u2019s law (Figure 4-6). After scattering, when the particle is again far from the nucleus, it is moving with the same speed v parallel to the line OB, which makes an angle with line COA. (Since the potential energy is again zero, the final speed must be equal to the initial speed by conservation of energy, assuming, as Rutherford did, that the massive nucleus remains fixed during the scattering.) The distance b is called the impact parameter and the scattering angle. The path of the particle can be the angle shown to be a hyperbola, and the scattering angle can be related to the impact parameter b from the laws of classical mechanics. The result is, b kqQ mv2 cot 2 4-3 Q>r2, Figure 4-6 Force on a point charge versus distance r from the center of a uniformly charged sphere of radius R. Outside the sphere the force is where Q proportional to is the total charge. Inside the sphere, the force is proportional to QrR>R3, where is the charge within a sphere of radius r. The maximum force occurs at r R. q>r2 q Q(r> R)3 Of course, it is not possible to choose or know the impact parameter for any particular particle, but when one recalls the values of the cotangent between 0\u00b0 and 90\u00b0, all such particles with impact parameters less than or equal to a particular b will be scattered through an angle greater than or equal to that given by Equation 4-3; i.e., the smaller the impact parameter, the larger the scattering angle (Figure 4-7). particles per second per unit Let the intensity of the incident area. The number per second scattered by one nucleus through angles greater than equals the number per second that have impact parameters less than b ( ). This number is particle beam be I0 which has the dimensions of an area, is called the cross section is thus defined as the number scattered per nucleus per unit time divided by the incident intensity. for scattering through angles greater than The cross section. b2I0. The quantity b2, The particle-scattering technique devised by Rutherford to \u201clook\u201d at atoms now has wide application throughout physics. Scattering of high-energy electrons from protons and neutrons provided our first experimental hint of the existence of quarks. Rutherford back-scattering spectroscopy is widely used as a highly sensitive surface analysis technique. \u03b1 2 \u03b1 1 b2 b1 +Ze Area \u03c0b1 2 Area \u03c0b2 2 \u03b8 1 \u03b8 2 Figure 4-7 Two particles with equal kinetic energies approach the positive charge Q Ze with impact parameters b1 and b2, where b1 the angle is scattered will be larger than 1 with impact parameters smaller than a particular value of b will have scattering angles larger is called the cross section than the corresponding value of for scattering with angles greater than. b2. According to Equation 4-3, In general, all particles from Equation 4-3. The area through which b2 2. 1 The total number of particles scattered per second is obtained by multiplying by the number of nuclei in the scattering foil (this assumes the foil to be thin enough to make the chance of overlap negligible). Let n be the number of nuclei per unit volume: b2I0 n (g>cm3)NA(atoms>mol) M(g>mol) NA M atoms cm3 4-4 For a foil of thickness t, the total number of nuclei \u201cseen\u201d by the beam is nAt, where A is the area of the beam (Figure 4-8). The total number scattered per second through angles greater than is b2I0ntA. If we divide this by the number of particles incithus I0A, we get the fraction f scattered through angles dent per second greater than : f b2nt 4-5 4-2 Rutherford\u2019s Nuclear", " Model 155 t Area A of beam Number of foil nuclei in beam is nAt Figure 4-8 The total number of nuclei of foil atoms in the area covered by the beam is nAt, where n is the number of foil atoms per unit volume, A is the area of the beam, and t is the thickness of the foil. EXAMPLE 4-2 Scattered Fraction f Calculate the fraction of an incident beam of particles of kinetic energy 5 MeV that Geiger and Marsden expected to see for!90\u00b0 from a gold foil (Z 79) 106 m thick. SOLUTION 1. The fraction f is f b2nt related to the impact parameter b, the number density of nuclei n, and the thickness t by Equation 4-5: 2. The particle density n is given by Equation 4-4: n NA M (19.3 g>cm3)(6.02 1023 atoms>mol) 197 gm>mol 5.90 1022 atoms>cm3 5.90 1028 atoms>m3 3. The impact parameter b is related to by Equation 4-3: b kqQ (2)(79)ke2 mv2 cot 2K 2 (2)(79)(1.44 eV # nm) (2)(5 106 eV) cot 90\u00b0 2 2.28 105 nm 4. Substituting these into Equation 4-5 yields f: 2.28 1014 m f (2.28 1014 m)2a5.9 1028 atoms m3 b(106 m) 9.6 105 104 Remarks: This outcome is in good agreement with Geiger and Marsden\u2019s measurement of about 1 in 8000 in their first trial. Thus, the nuclear model is in good agreement with their results. On the strength of the good agreement between the nuclear atomic model and the measured fraction of the incident particles scattered at angles Rutherford derived an expression, based on the nuclear model, for the number of particles N that would be scattered at any angle That number, which also depends on the atomic!90\u00b0,. 156 Chapter 4 The Nuclear Atom (a Silver Gold (b) 160 Gold i n m N \u0394 / 120 80 40 0 0 Silver Copper Aluminum 0.4 0.8 1.2 1.6 2.0 Foil thickness t, cm of air equiv. 2 3 4 5 6 log \u0394N Figure 4-9 (a) Geiger and Marsden\u2019s data for the data over several orders of magnitude. Note that scattering angle increases downward along the vertical axis. (b) Geiger and Marsden also measured the dependence of N on t predicted by Equation 4-6 for foils made from a wide range of elements, this being an equally critical test. Results for four of the elements used are shown. scattering from thin gold and silver foils. The graph is a log-log plot to show number Z and thickness t of the scattering foil, on the intensity particles and their kinetic energy detector area and r is the foil-detector distance), is given by Ek, and on the geometry of the detector ( I0 Asc of the incident is the \u00a2N a I0Ascnt r2 b a kZe2 2Ek b 2 1 sin4 2 4-6 Within the uncertainties of their experiments, which involved visually observing several hundred thousand particles, Geiger and Marsden verified every one of the predictions of Rutherford\u2019s formula over four orders of magnitude of N. The excellent agreement of their data with Equation 4-6 firmly established the nuclear atomic model as the correct basis for further studies of atomic and nuclear phenomena. (See Figure 4-9.) More Rutherford\u2019s derivation of Equation 4-6 was based on his atomic model and the well-known Coulomb scattering process of charged particles. Rutherford\u2019s Prediction and Geiger and Marsden\u2019s Results are described on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Equations 4-7 through 4-10 here, as well as Figures 4-10 through 4-12. The Size of the Nucleus The fact that the force law is shown to be correct, confirming Rutherford\u2019s model, does not imply that the nucleus is a mathematical point charge, however. The force law would be the same even if the nucleus were a ball of charge of some radius particle did not penetrate the ball. (See Figures 4-6 and 4-13.) For a given scattering angle, the distance of closest approach of the particle to the nucleus can be calculated from the geometry of the collision. For the largest angle, near 180\u00b0, the collision is nearly \u201chead-on.\u201d The corresponding distance of closest approach rd is thus an experimental as long as the R0 upper limit on the size of the target nucleus. We can calculate the distance of closest approach for a", " head-on collision rd by noting that conservation of energy requires the potential energy at this distance to equal the original kinetic energy: (a) 4-2 Rutherford\u2019s Nuclear Model 157 (V Ek)large r a0 1 2 mv2b large r 0b a (V Ek)rd kqQ rd kqQ rd 1 2 mv2 \u03b1 rd (b) or rd kqQ 1 2 mv2 4-11 \u03b1 For the case of 7.7-MeV collision is (2)(79)(1.44 eV # nm) 7.7 106 eV rd 3 105 nm 3 1014 m particles, the distance of closest approach for a head-on F kqQ>r2, For other collisions, the distance of closest approach is somewhat greater, but for particles scattered at large angles it is of the same order of magnitude. The excellent agreement of Geiger and Marsden\u2019s data at large angles with the prediction of Equation 4-6 thus indicates that the radius of the gold nucleus is no larger than about 3 1014 m. If higherenergy particles could be used, the distance of closest approach would be smaller, and as the energy of the particles increased, we might expect that eventually the particles would penetrate the nucleus. Since, in that event, the force law is no longer the data would not agree with the point-nucleus calculation. Rutherford did not have higherenergy particles available, but he could reduce the distance of closest approach by using targets of lower atomic numbers.9 For the case of aluminum, with Z 13, when the most energetic particles that he had available (7.7 MeV from 214Bi) scattered at large angles, they did not follow the predictions of Equation 4-6. However, when the kinetic energy of the particles was reduced by passing the beam through thin mica sheets of various thicknesses, the data again followed the prediction of Equation 4-6. Rutherford\u2019s data are shown in Figure 4-14. The value of rd (calculated from Equation 4-11) at which the data begin to deviate from the prediction can be thought of as the surface of the nucleus. From these data, Rutherford estimated the radius of the aluminum nucleus to be about 1.0 1014 m. (The radius of the Al nucleus is actually about 3.6 1015 m. See Chapter 11.) \u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013.0 0.5 0 0 i 0.6 Figure 4-13 (a) If the particle does not penetrate the nuclear charge, the nucleus can be considered a point charge located at the center. (b) If the particle has enough energy to penetrate the nucleus, the Rutherford scattering law does not hold but would require modification to account for that portion of the nuclear charge \u201cbehind\u201d the penetrating particle. Aluminum 0.8 1.4 1.6 1.0 1.2 rd, 10 \u201314 m 1.8 A unit of length convenient for describing nuclear sizes is the fermi, or femtometer (fm), defined as 1 fm 1015 m. As we will see in Chapter 11, the nuclear radius varies from about 1 to 10 fm from the lightest to the heaviest atoms. Figure 4-14 Data from Rutherford\u2019s group showing observed angle versus values of rd computed from Equation 4-11 for various kinetic energies. scattering at a large fixed EXAMPLE 4-3 Rutherford Scattering at Angle particles from 226Ra are scattered at 45\u00b0 from a silver foil and 450 particles are counted each minute at the scintillation detector. If everything is kept the same except that the detector is moved to observe particles scattered at 90\u00b0, how many will be counted per minute? In a particular experiment, 158 Chapter 4 The Nuclear Atom SOLUTION Using Equation 4-6, we have that N 450 when but we don\u2019t have any of the other parameters available. Letting all of the quantities in the parenthesis equal a constant C, we have that 45\u00b0, or \u00a2N 450 C sin4 45\u00b0 2 C 450 sin4a 45\u00b0 2 b When the detector is moved to 90\u00b0, \u00a2N C sin4a 90\u00b0 2 38.6 39 particles>min b 450 sin4a 45\u00b0 2 the value of C is unchanged, so b sin4a 90\u00b0 2 b 6.0 MeV impinges EXAMPLE 4-4 Alpha Scattering A beam of particles with Ek on a silver foil thick. The beam current is 1.0 nA. How many particles will be counted by a small scintillation detector of area equal to 5 mm2 located 2.0 cm from the foil at an angle of 75\u00b0? (For silver Z 47, 10.5 gm> cm3, and M 108.)", " 1.0 m SOLUTION 1. The number counted N is given by Equation 4-6: \u00a2N a I0Ascnt r2 b a kZe2 2Ek b 2 1 sin4 2 2. Since each particle 2e, has is: q I0 3. The kinetic energy of each is 4. For silver, n is given by I0 (1.0 109A)(2 1.60 1019 C>)1 3.12 109 >s Ek (6.0 MeV)(1.60 1013 J>MeV) 9.60 1013 J n NA >M (10.5 g>cm3)(6.02 1023 atoms>mol) 108 g>mol 5.85 1022 atoms>cm3 5.85 1028 atoms>m3 5. Substituting the given values and computed results into Equation 4-6 gives N: \u00a2N (3.12 109 >s)(5 106 m2)(5.85 1028 atoms>m3)(106 m) (2 102)2 sin4 (75\u00b0>2) c (9 109)(47)(1.60 1019)2 (2)(9.60 1013) 2 d 528 >s EXAMPLE 4-5 Radius of the Au Nucleus The radius of the gold (Au) nucleus has been measured by high-energy electron scattering as 6.6 fm. What kinetic energy particle particles would Rutherford have needed so that for 180\u00b0 scattering, the would just reach the nuclear surface before reversing direction? SOLUTION From Equation 4-11, we have kqQ rd mv2 1 2 4-3 The Bohr Model of the Hydrogen Atom 159 (9 109)(2)(79)(1.60 1019)2 6.6 1015 5.52 1012 J 34.5 MeV Alpha particles of such energy are not emitted by naturally radioactive materials and so were not accessible to Rutherford. Thus, he could not have performed an experiment for Au equivalent to that for Al illustrated by Figure 4-14. Questions 1. Why can\u2019t the impact parameter for a particular particle be chosen? 2. Why is it necessary to use a very thin target foil? 3. Why could Rutherford place a lower limit on the radius of the Al nucleus but not on the Au nucleus? 4. How could you use the data in Figure 4-9a to determine the charge on a silver nucleus relative to that on a gold nucleus? 5. How would you expect the data (not the curve) to change in Figure 4-9 if the foil were so thick that an appreciable number of gold nuclei were hidden from the beam by being in the \u201cshadow\u201d of the other gold nuclei? 4-3 The Bohr Model of the Hydrogen Atom In 1913, the Danish physicist Niels H. D. Bohr10 proposed a model of the hydrogen atom that combined the work of Planck, Einstein, and Rutherford and was remarkably successful in predicting the observed spectrum of hydrogen. The Rutherford model assigned charge and mass to the nucleus but was silent regarding the distribution of the charge and mass of the electrons. Bohr, who had been working in Rutherford\u2019s laboratory during the experiments of Geiger and Marsden, made the assumption that the electron in the hydrogen atom moved in an orbit about the positive nucleus, bound by the electrostatic attraction of the nucleus. Classical mechanics allows circular or elliptical orbits in this system, just as in the case of the planets orbiting the Sun. For simplicity, Bohr chose to consider circular orbits. Such a model is mechanically stable because the provides the cen- V kZe2>r Coulomb potential tripetal force Niels Bohr explains a point in front of the blackboard (1956). [American Institute of Physics, Niels Bohr Library, Margrethe Bohr Collection.] F kZe2 r2 mv2 r 4-12 necessary for the electron to move in a circle of radius r at speed v, but it is electrically unstable because the electron is always accelerating toward the center of the circle. 160 Chapter 4 The Nuclear Atom The laws of electrodynamics predict that such an accelerating charge will radiate light of frequency f equal to that of the periodic motion, which in this case is the frequency of revolution. Thus, classically, f v 2r a kZe2 rm b 1>2 1 2r a kZe2 42m b 1>2 1 r3>2 1 r3>2 4-13 The total energy of the electron is the sum of the kinetic and the potential energies: E 1 2 mv2 a kZe2 r b 1 From Equation 4-12, we see that tion in any inverse-square force field), so the total energy can be written as 2 mv2 kZe2>2r (a result that holds for circular mo- (a) \ufffd", "\ufffd (b) \u03b3 \u03b3 \u03b3 \u03b3 Figure 4-15 (a) In the classical orbital model, the electron orbits about the nucleus and spirals into the center because of the energy radiated. (b) In the Bohr model, the electron orbits without radiating until it jumps to another allowed radius of lower energy, at which time radiation is emitted. E kZe2 2r kZe2 r kZe2 2r 1 r 4-14 Thus, classical physics predicts that, as energy is lost to radiation, the electron\u2019s orbit will become smaller and smaller while the frequency of the emitted radiation will become higher and higher, further increasing the rate at which energy is lost and ending when the electron reaches the nucleus. (See Figure 4-15a.) The time required for the electron to spiral into the nucleus can be calculated from classical mechanics and electrodynamics; it turns out to be less than a microsecond. Thus, at first sight, this model predicts that the atom will radiate a continuous spectrum (since the frequency of revolution changes continuously as the electron spirals in) and will collapse after a very short time, a result that fortunately does not occur. Unless excited by some external means, atoms do not radiate at all, and when excited atoms do radiate, a line spectrum is emitted, not a continuous one. Bohr \u201csolved\u201d these formidable difficulties with two decidedly nonclassical postulates. His first postulate was that electrons could move in certain orbits without radiating. He called these orbits stationary states. His second postulate was to assume that the atom radiates when the electron makes a transition from one stationary state to another (Figure 4-15b) and that the frequency f of the emitted radiation is not the frequency of motion in either stable orbit but is related to the energies of the orbits by Planck\u2019s theory hf Ei Ef 4-15 where h is Planck\u2019s constant and Ei and Ef are the energies of the initial and final states. The second assumption, which is equivalent to that of energy conservation with the emission of a photon, is crucial because it deviated from classical theory, which requires the frequency of radiation to be that of the motion of the charged particle. Equation 415 is referred to as the Bohr frequency condition. In order to determine the energies of the allowed, nonradiating orbits, Bohr made a third assumption, now known as the correspondence principle, which had profound implications: In the limit of large orbits and large energies, quantum calculations must agree with classical calculations. 4-3 The Bohr Model of the Hydrogen Atom 161 Thus, the correspondence principle says that whatever modifications of classical physics are made to describe matter at the submicroscopic level, when the results are extended to the macroscopic world, they must agree with those from the classical laws of physics that have been so abundantly verified in the everyday world. While Bohr\u2019s detailed model of the hydrogen atom has been supplanted by modern quantum theory, which we will discuss in later chapters, his frequency condition (Equation 4-15) and the correspondence principle remain as essential features of the new theory. 2, nh>2, In his first paper,11 in 1913, Bohr pointed out that his results implied that the angular momentum of the electron in the hydrogen atom can take on only values that are integral multiples of Planck\u2019s constant divided by in agreement with a discovery made a year earlier by J. W. Nicholson. That is, angular momentum is quantized; it can assume only the values where n is an integer. Rather than follow the intricacies of Bohr\u2019s derivation, we will use the fundamental conclusion of angular momentum quantization to find his expression for the observed spectra. The development that follows applies not only to hydrogen, but to any atom of nuclear charge Ze with a single orbital electron\u2014e.g., singly ionized helium He or doubly ionized lithium Li2. If the nuclear charge is Ze and the electron charge e, we have noted (Equation 4-12) that the centripetal force necessary to move the electron in a circular orbit is provided by the Coulomb force Solving Equation 4-12 for the speed of the orbiting electron yields kZe2>r2. v a 1>2 kZe2 mr b Bohr\u2019s quantization of the angular momentum L is L mvr nh 2 nU n 1, 2, 3, \u00c1 4-16 4-17 where the integer n is called a quantum number and read \u201ch-bar,\u201d is often more convenient to use than h itself, just as the angular frequency 2f is often more convenient than the frequency f.) Combining Equations 4-16 and 4-17 allows us to write for the circular orbits (The constant U h>2. U, Squaring this relation gives r nU", " mv 1>2 nU m a rm kZe2 b r2 n2U2 m2 a rm kZe2 b and canceling common quantities yields where rn n2U2 mkZe2 n2a0 Z a0 U2 mke2 0.529 \u00c5 0.0529 nm 4-18 4-19 is called the Bohr radius. The, a unit commonly used in the early days of spectroscopy, equals 1010 m or 101 nm. Thus, we find that the stationary orbits of Bohr\u2019s first postulate have quantized radii, denoted in Equation 4-18 by the subscript on rn. \u00c5 162 Chapter 4 The Nuclear Atom for hydrogen (Z 1) corresponds to the orbit radius Notice that the Bohr radius with n 1, the smallest Bohr orbit possible for the electron in a hydrogen atom. Since the Bohr orbits for single-electron atoms with Z 1 are closer to the nucleus rn than the corresponding ones for hydrogen. Z1, a0 The total energy of the electron (Equation 4-14) then becomes, upon substitution of rn from Equation 4-18, kZe2 2 a mkZe2 n2U2 b En En kZe2 2rn mk2Z2e4 2U2n2 E0 Z2 n2 n 1, 2, 3, \u00c1 4-20 mk2e4>2U2. E0 where Thus, the energy of the electron is also quantized; i.e., the stationary states correspond to specific values of the total energy. This means that energies Ei and Ef that appear in the frequency condition of Bohr\u2019s second postulate must be from the allowed set En, and Equation 4-15 becomes hf Eni Enf E0 Z2 n2 i aE0 b Z2 n2 f or f E0Z2 h a 1 n2 f 1 n2 i b 4-21 which can be written in the form of the Rydberg-Ritz equation (Equation 4-2) by substituting f c and dividing by c to obtain > or where 1 E0Z2 hc a 1 n2 f 1 n2 i b 1 Z2Ra 1 n2 f 1 n2 i b R E0 hc mk2e4 4cU3 4-22 4-23 is Bohr\u2019s prediction for the value of the Rydberg constant. U Using the values of m, e, c, and known in 1913, Bohr calculated R and found his result to agree (within the limits of uncertainties of the constants) with the value obtained from spectroscopy, 1.097 107 m1. Bohr noted in his original paper that this equation might be valuable in determining the best values for the constants e, m, because of the extreme precision possible in measuring R. This has indeed and turned out to be the case. U The possible values of the energy of the hydrogen atom predicted by Bohr\u2019s model are given by Equation 4-20 with Z 1: En mk2e4 2U2n2 E0 n2 4-24 where E0 mk2e4 2U2 2.18 1018 J 13.6 eV 4-3 The Bohr Model of the Hydrogen Atom 163 E1( E0) is the magnitude of En with n 1. is called the ground state. It is convenient to plot these allowed energies of the stationary states as in Figure 4-16. Such a plot is called an energy-level diagram. Various series of transitions between the stationary states are indicated in this diagram by vertical arrows drawn between the levels. The frequency of light emitted in one of these transitions is the energy difference divided by h according to Bohr\u2019s frequency condition, Equation 4-15. The energy required to remove the electron from the atom, 13.6 eV, is called the ionization energy, or binding energy, of the electron. (a) 0 \u20130.54 \u20130.85 \u20131.51 \u20133.40 13.60 Paschen Balmer Lyman series n \u221e 5 4 3 2 1 (b) Lyman Balmer Paschen 100 200 500 1000 2000 \u03bb, nm 13.6>n2 eV, Figure 4-16 Energy-level diagram for hydrogen showing the seven lowest stationary states and the four lowest energy transitions each for the Lyman, Balmer, and Paschen series. There are an infinite number of levels. Their energies are given by En where n is an integer. The dashed line shown for each series is the series limit, corresponding to the energy ) in a transition to the state with n nf for that series. that would be radiated by an electron at rest far from the nucleus ( The horizontal spacing between the transitions shown for each series is proportional to the wavelength spacing between", " the lines of the spectrum. (b) The spectral lines corresponding to the transitions shown for the three series. Notice the regularities within each series, particularly the short-wavelength limit and the successively smaller separation between adjacent lines as the limit is approached. The wavelength scale in the diagram is not linear. n S 164 Chapter 4 The Nuclear Atom A bit different sort of application, the Bohr/Rutherford model of the nuclear atom and electron orbits is the picture that, for millions of people, provides their link to the world of the atom and subatomic phenomena. 2, ni At the time Bohr\u2019s paper was published, there were two spectral series known for 3, 4, 5,..., and a series hydrogen: the Balmer series, corresponding to nf 4, 5, 6,.... named after its discoverer, Paschen (1908), corresponding to nf Equation 4-22 indicates that other series should exist for different values of nf. In 1916 1, and in 1922 and 1924 F. S. Brackett Lyman found the series corresponding to nf 5. As can and A. H. Pfund, respectively, found series corresponding to nf be easily determined by computing the wavelengths for these series, only the Balmer series lies primarily in the visible portion of the electromagnetic spectrum. The Lyman series is in the ultraviolet, the others in the infrared. 4 and nf 3, ni EXAMPLE 4-6 Wavelength of the H spectral line, i.e., the second line of the Balmer series predicted by Bohr\u2019s model. The H Line Compute the wavelength of the line is emitted in the transition from ni 4 to nf 2. H SOLUTION 1. Method 1: The wavelength is given by Equation 4-22 with Z 1: 1 Ra 1 n2 f 1 n2 i b 2. Substituting R 1.097 107 m1 and the values of ni and nf: or 1 (1.097 107 m1)a 1 22 1 42 b 4.86 107 486 nm 3. Method 2: The wavelength may also be computed from Equation 4-15: hf hc> E i Ef or 1 1 hc (Ei Ef) 4. The values of Ei and Ef are given by Equation 4-24: Ei Ef a 13.6 eV n2 i a 13.6 eV n2 f b a 13.6 eV 42 b a 13.6 eV 22 b 0.85 eV b 3.4 eV 5. Substituting these into Equation 4-15 yields 1 [0.85 eV (3.4 eV)](1.60 1019 J>eV) (6.63 1034 J # s)(3.00 108 m>s) 2.051 106 m1 or 4.87 107 m 487 nm Remarks: The difference in the two results is due to rounding of the Rydberg constant to three decimal places. Reduced Mass Correction 4-3 The Bohr Model of the Hydrogen Atom 165 R, The assumption by Bohr that the nucleus is fixed is equivalent to the assumption that it has infinite mass. In fact, the Rydberg constant in Equation 4-23 is normally writas we will do henceforth. If the nucleus has mass M, its kinetic energy will ten a 1 be is the momentum. If we assume that the total momentum of the atom is zero, conservation of momentum requires that the momenta of the nucleus and electron be equal in magnitude. The total kinetic energy is then 2 Mv 2 p2>2M, p Mv where Ek p2 2M p2 2m M m 2mM p2 p2 2 where mM m M m 1 m>M 4-25 called the This is slightly different from the kinetic energy of the electron because reduced mass, is slightly different from the electron mass. The results derived above for a nucleus of infinite mass can be applied directly for the case of a nucleus of mass M if we replace the electron mass in the equations by reduced mass defined by Equation 4-25. (The validity of this procedure is proved in most intermediate and advanced mechanics books.) The Rydberg constant (Equation 4-23) is then written,, R k2e4 4cU3 mk2e4 4cU3 a 1 1 m>M b R a 1 1 m>M b 4-26 This correction amounts to only 1 part in 2000 for the case of hydrogen and to even less for other nuclei; however, the predicted variation in the Rydberg constant from atom to atom is precisely that which is observed. For example, the spectrum of a singly ionized helium atom, which has one remaining electron, is just that predicted by Equation 4-22 and 4-26 with Z 2", " and the proper helium mass. The current value for the Rydberg constant from precision spectroscopic measurements12 is R R 1.0973732 107 m1 1.0973732 102 nm1 4-27 Urey13 used the reduced mass correction to the spectral lines of the Balmer series to discover (in 1931) a second form of hydrogen whose atoms had twice the mass of ordinary hydrogen. The heavy form was called deuterium. The two forms, atoms with the same Z but different masses, are called isotopes. EXAMPLE 4-7 Rydberg Constants for H and for H and mp He 1.6726 1027 kg, Compute the Rydberg constants applying the reduced mass correction. (m 9.1094 1031 kg, 5.0078 1027 kg) m He SOLUTION For hydrogen: RH a R 1 1 m>MH 1.09677 107 m1 b R a 1 1 9.1094 1031>1.6726 1027 b 166 Chapter 4 The Nuclear Atom For helium: Since M in the reduced mass correction is the mass of the nucleus, for this calculation we use M equal to the particle mass. RHe a R 1 1 m>MH 1.09752 107 m1 b R a 1 1 9.1094 1031>5.0078 1027 b Thus, the two Rydberg constants differ by about 0.04 percent. Correspondence Principle According to the correspondence principle, which applies also to modern quantum mechanics, when the energy levels are closely spaced, quantization should have little effect; classical and quantum calculations should give the same results. From the energy-level diagram of Figure 4-16, we see that the energy levels are close together when the quantum number n is large. This leads us to a slightly different statement of Bohr\u2019s correspondence principle: In the region of very large quantum numbers (n in this case) quantum calculation and classical calculation must yield the same results. To see that the Bohr model of the hydrogen atom does indeed obey the correspon n and dence principle, let us compare the frequency of a transition between level ni n 1 for large n with the classical frequency, which is the frequency of level nf revolution of the electron. From Equation 4-22 we have f c Z2mk2e4 4U3 c 1 (n 1)2 1 n2 d Z2mk2e4 4U3 2n 1 n2(n 1)2 For large n we can neglect the ones subtracted from n and 2n to obtain f Z2mk2e4 4U3 2 n3 Z2mk2e4 2U3n3 4-28 The classical frequency of revolution of the electron is (see Equation 4-13) frev v 2r v nU>mr Using we obtain from Equation 4-17 and r n2U2>mkZe2 from Equation 4-18, frev (nU>mr) 2r nU 2mr2 nU 2m(n2U2>mkZe2)2 frev m2k2Z2e4nU 2mn4U4 mk2Z2e4 2U3n3 4-29 which is the same as Equation 4-28. Fine Structure Constant nf The demonstration of the correspondence principle for large n in the preceding para 1; however, we have seen (see Figure 4-16) that transigraph was for n ni \u00a2n! 1 when n is small, and such transitions tions occur in the hydrogen atom for should occur for large n too. If we allow n 2, 3,... for large values of n, then the frequencies of the emitted radiation would be, according to Bohr\u2019s model, integer multiples of the frequency given in Equation 4-28. In that event, Equations 4-28 and 4-29 would not agree. This disagreement can be avoided by allowing elliptical orbits.14 A result of Newtonian mechanics, familiar from planetary motion, is that in an inverse-square force 4-3 The Bohr Model of the Hydrogen Atom 167 4-30 4-31 field, the energy of an orbiting particle depends only on the major axis of the ellipse and not on its eccentricity. There is consequently no change in the energy at all unless the force differs from inverse square or unless Newtonian mechanics is modified. A. Sommerfeld considered the effect of special relativity on the mass of the electron in the Bohr model in an effort to explain the observed fine structure of the hydrogen spectral lines.15 Since the relativistic corrections should be of the order of v2 c2 (see Chapter 2), it is likely that a highly eccentric orbit would have a larger correction because v becomes greater as the electron moves nearer the nucleus. The Sommerfeld calculations are quite complicated, but we can estimate the order of magnitude of", " the effect of special relativfor the first Bohr orbit in hydrogen. For n 1, we have from ity by calculating Equation 4-17 that U2>mke2, Then, using we have U. > v>c mvr1 and v v c ke2 Uc r1 a0 U m(U2>mke2) U mr1 1.44 eV # nm 197.3 eV # nm ke2 U 1 137 where we have used another convenient combination Uc 1.24 103 eV # nm 197.3 eV # nm 2 > is called the fine-structure constant because The dimensionless quantity of its first appearance in Sommerfeld\u2019s theory, but as we will see, it has much more fundamental importance. ke2>Uc Though v2 c2 is very small, an effect of this magnitude is observable. In Sommerfeld\u2019s theory, the fine structure of the hydrogen spectrum is explained in the following way. For each allowed circular orbit of radius rn and energy En, a set of n elliptical orbits is possible of equal major axes but different eccentricities. Since the velocity of a particle in an elliptical orbit depends on the eccentricity, so then will the mass and momentum, and therefore the different ellipses for a given n will have slightly different energies. Thus, the energy radiated when the electron changes orbit depends slightly on the eccentricities of the initial and final orbits as well as on their major axes. The splitting of the energy levels for a given n is called fine-structure splitting, and its > value turns out to be of the order of v2 c2 just as Sommerfeld predicted. However, the agreement of Sommerfeld\u2019s prediction with the observed fine-structure splitting was quite accidental and led to considerable confusion in the early days of quantum theory. Although he had used the relativistic mass and momentum, he computed the energy using classical mechanics, leading to a correction much larger than that actually due only to relativistic effects. As we will see in Chapter 7, fine structure is associated with a completely nonclassical property of the electron called spin. 2, A lasting contribution of Sommerfeld\u2019s effort was the introduction of the finea0 structure constant and the quantized energies of the Bohr model in a particularly elegant form. Equations 4-24 and 4-19 for hydrogen become With it we can write the Bohr radius ke 2>Uc 1>137. En mk2e4 2U2n2 c2 c2 mc2 2 2 1 n2 a0 U2 mke2 c c U mc 1 4-32 4-33 168 Chapter 4 The Nuclear Atom is a dimensionless number formed of universal constants, all observers will Since measure the same value for it and find that energies and dimensions of atomic systems 1>, are proportional to respectively. We will return to the implications of this intriguing fact later in the book. and 2 EXPLORING Giant Atoms rn n2a0 Giant atoms called Rydberg atoms, long understood to be a theoretical possibility and first detected in interstellar space in 1965, are now being produced and studied in the laboratory. Rydberg atoms are huge! They are atoms that have one of the valence electrons in a state with a very large quantum number n. (See Figure 4-17.) Notice in and n can be any Equation 4-18 that the radius of the electron orbit positive integer, so the diameter of a hydrogen atom (or any other atom, for that matter) could be very large, a millimeter or even a meter! What keeps such giant atoms from being common is that the energy difference between adjacent allowed energy states is extremely small when n is large and the allowed states are very near the 0 level where ionization occurs, because For example, if n 1000, E En 0.1 mm, but both E1000 and the difthe diameter of a hydrogen atom would be r1000 \u00a2E E1001 are about 105 eV! This energy is far below the ference in energy average energy of thermal motion at ordinary temperatures (about 0.025 eV), so random collisions would quickly ionize an atom whose electron happened to get excited to a level with n equal to 20 or so with r still only about 108 m. E1000 >Z n2 1>n2. The advent of precisely tunable dye lasers in the 1970s made it possible to nudge electrons carefully into orbits with larger and larger n values. The largest Rydberg atoms made so far, typically using sodium or potassium, are 10,000 times the diameter of ordinary atoms, about 20 m across or the size of a fine grain of sand, and exist for several milliseconds inside vacuum chambers. For hydrogen, this corresponds to quantum number n 600. An electron moving so far from the nucleus is bound by a miT 1>f n3 nuscule force. It also", " moves rather slowly since the classical period of and follows an elliptical orbit. These characteristics of very large n orbits provide several intriguing possibilities. For example, very small electric fields might be studied, making possible the tracking of chemical reactions that proceed too quickly to be followed otherwise. More dramatic is the possibility of directly testing Bohr\u2019s correspondence principle by directly observing the slow (since ) movement of the electron around the large n orbits\u2014the transition from quantum mechanics to classical mechanics. Computer simulations of the classical motion of a Rydberg electron \u201cwave\u201d (see Chapter 5) in orbit around a nucleus are aiding the design of experiments to observe the correspondence principle. v 1>n Electron Nucleus Figure 4-17 A lithium (Z 3) Rydberg atom. The outer electron occupies a small volume and follows a nearly classical orbit with a large value of n. The two inner electrons are not shown. 4-4 X-Ray Spectra 169 Questions 6. If the electron moves in an orbit of greater radius, does its total energy increase or decrease? Does its kinetic energy increase or decrease? 7. What is the energy of the shortest-wavelength photon that can be emitted by the hydrogen atom? 8. How would you characterize the motion and location of an electron with E 0 and n S in Figure 4-16? 4-4 X-Ray Spectra The extension of the Bohr theory to atoms more complicated than hydrogen proved difficult. Quantitative calculations of the energy levels of atoms of more than one electron could not be made from the model, even for helium, the next element in the periodic table. However, experiments by H. Moseley in 1913 and J. Franck and G. Hertz in 1914 strongly supported the general Bohr-Rutherford picture of the atom as a positively charged core surrounded by electrons that moved in quantized energy states relatively far from the core. Moseley\u2019s analysis of x-ray spectra will be discussed in this section, and the Franck-Hertz measurement of the transmission of electrons through gases will be discussed in the chapter\u2019s concluding section. Using the methods of crystal spectrometry that had just been developed by W. H. Bragg and W. L. Bragg, Moseley16 measured the wavelengths of the characteristic x-ray line spectra for about 40 different target elements. (Typical x-ray spectra are shown in Figure 3-15.) He noted that the x-ray line spectra varied in a regular way from element to element, unlike the irregular variations of optical spectra. He surmised that this regular variation occurred because characteristic x-ray spectra were due to transitions involving the innermost electrons of the atoms. (See Figure 4-18.) Because the inner electrons are L\u03b1 x ray \u2013 Ejected electron Henry G.-J. Moseley. [Courtesy of University of Manchester.] K\u03b1 x ray n = 1 n = 2 Nucleus +Ze n = 3 n = 4! rn Figure 4-18 A stylized picture of the Bohr circular orbits for n 1, 2, 3, and 4. The radii n2. In a high-Z element (elements with Z 12 emit x rays), electrons are distributed over all the orbits shown. If an electron in the n 1 orbit is knocked from the atom, e.g., by being hit by a fast electron accelerated by the voltage across an x-ray tube, the vacancy thus produced is filled by an electron of higher energy (i.e., n 2 or higher). The difference in energy between the two orbits is emitted as a photon, according to the Bohr frequency condition, whose wavelength will be in the x-ray region of the spectrum if Z is large enough. 170 Chapter 4 The Nuclear Atom shielded from the outermost electrons by those in intermediate orbits, their energies do not depend on the complex interactions of the outer electrons, which are responsible for the complicated optical spectra. Furthermore, the inner electrons are well shielded from the interatomic forces that are responsible for the binding of atoms in solids. According to the Bohr theory (published earlier the same year, 1913), the energy of an electron in the first Bohr orbit is proportional to the square of the nuclear charge (see Equation 4-20). Moseley reasoned that the energy, and therefore the frequency, of a characteristic x-ray photon should vary as the square of the atomic number of the target element. He therefore plotted the square root of the frequency of a particular characteristic line in the x-ray spectrum of various target elements versus the atomic number Z of the element. Such a plot, now called a Moseley plot, is shown in Figure 4-19. Z 68 5 4 3 Wavelength, \u00c5 2 1.5 1 0.9 0.8 0.7 0.6 78Pt 76Os 74W 72Lu 70TmII 68Er 66Ho 64Gd 62Sm 60", "Nd 58Ce 56Ba 54Xe 52Te 50Sn 48Cd 46Pd 44Ru 42Mo 40Zr 38Sr 36Kr 34Se 32Ge 30Zn 28Ni 26Fe 24Cr 22Ti 20Ca 18A 16S 14Si 79Au 77Ir 75 73Ta 71Yi 69TmI 67Ds 65Tb 63Eu 61 59Pr 57La 55Cs 53I 51Sb 49In 47Ag 45Rh 43 41Nb 39Y 37Rb 35Br 33As 31Ga 29Cu 27Co 25Mn 23V 21Sc 19K 17Cl 15P 13Al 6 8 \u03b1 \u03b2 \u03b3 L series \u03b2 K series 12 20 10 Square root of frequency, 108 Hz1/2 18 14 16 \u03b1 22 24 Figure 4-19 Moseley\u2019s plots of the square root of frequency versus Z for characteristic x rays. When an atom is bombarded by high-energy electrons, an inner atomic electron is sometimes knocked out, leaving a vacancy in the inner shell. The K-series x rays are produced by atomic transitions to vacancies in the n 1 (K) shell, whereas the L series is produced by transitions to the vacancies in the n 2 (L) shell. [From H. Moseley, Philosophical Magazine (6), 27, 713 (1914).] 4-4 X-Ray Spectra 171 These curves can be fitted by the empirical equation f1>2 An(Z b) 4-34 where An and b are constants for each characteristic x-ray line. One family of lines, called the K series, has b 1 and slightly different values of An for each line in the graph. The other family shown in Figure 4-19, called the L series,17 could be fitted by Equation 4-34 with b 7.4. If the bombarding electron in the x-ray tube causes ejection of an electron from the innermost orbit (n 1) in a target atom completely out of the atom, photons will be emitted corresponding to transitions of electrons in other orbits (n 2, 3,...) to fill the vacancy in the n 1 orbit. (See Figure 4-18.) (Since these lines are called the K series, the n 1 orbit came to be called the K shell.) The lowest-frequency line corresponds n 1). This line is called the to the lowest-energy transition (n 2 line. The S n 1 is called the transition n 3 line. It is of higher energy, and hence higher K K frequency, than the x ray may then be filled by an electron of higher energy, e.g., one in the n 3 orbit, resulting in the emission of a line in the L series, and so on. The multiple L lines in the Moseley plot (Figure 4-19) are due in part to the fact that there turn out to be small differences in the energies of electrons with a given n that are not predicted by the Bohr model. Moseley\u2019s work gave the first indication of these differences, but the explanation will have to await our discussion of more advanced quantum theory in Chapter 7. 1 and line. A vacancy created in the n 2 orbit by emission of a Using the Bohr relation for a one-electron atom (Equation 4-21) with nf S K using (Z 1) in place of Z, we obtain for the frequencies of the K series K f mk2e4 4U3 (Z 1)2a 1 12 1 n2 b cR(Z 1)2a1 1 n2 b 4-35 R where is given by is the Rydberg constant. Comparing this with Equation 4-34, we see that An A2 n cR a1 1 n2 b The wavelengths of the lines in the K series are then given by c f c n(Z 1)2 A2 1 R(Z 1)2a1 1 n2 b 4-36 4-37 K EXAMPLE 4-8 line of molybdenum (Z 42), and compare the result with the value 0.0721 nm measured by Moseley and with the spectrum in Figure 3-15b. for Molybdenum Calculate the wavelength of the K SOLUTION Using n 2, 1.097 107 m1, and Z 42, we obtain R c(1.097 107 m1)(41)2a1 1 4 1 b d 7.23 1011 m 0.0723 nm This value is within 0.3 percent of Moseley\u2019s measurement and agrees well with that in Figure 3-15b. 172 Chapter 4 The Nuclear Atom The fact that f is proportional to (Z 1)2 rather than to Z is explained by the partial shielding of the nuclear charge by the other electron remaining in the K shell as \u201cseen\u201d by electrons in the n 2 (L", ") shell.18 Using this reasoning, Moseley concluded that, since b 7.4 for the L series, these lines involved electrons farther from the nucleus, which \u201csaw\u201d the nuclear charge shielded by more inner electrons. Assuming that the L series was due to transitions to the n 2 shell, we see that the frequencies for this series are given by f cR a 1 22 1 n2 b(Z 7.4)2 4-38 where n 3, 4, 5,.... A>2. Before Moseley\u2019s work, the atomic number was merely the place number of the element in Mendeleev\u2019s periodic table of the elements arranged by weight. The experiments of Geiger and Marsden showed that the nuclear charge was approximately A>2, while x-ray scattering experiments by C. G. Barkla showed that the number of These two experiments are consiselectrons in an atom was also approximately tent since the atom as a whole must be electrically neutral. However, several discrepancies were found in the periodic table as arranged by weight. For example, the 18th element in order of weight is potassium (39.102), and the 19th is argon (39.948). Arrangement by weight, however, puts potassium in the column with the inert gases and argon with the active metals, the reverse of their known chemical properties. versus Z, argon had to Moseley showed that for these elements to fall on the line have Z 18 and potassium Z 19. Arranging the elements by the Z number obtained from the Moseley plot rather than by weight, gave a periodic chart in complete agreement with the chemical properties. Moseley also pointed out that there were gaps in the periodic table at Z 43, 61, and 75, indicating the presence of undiscovered elements. All have subsequently been found. Figure 4-20 illustrates the discovery of promethium (Z 61). f1>2 (a) (b) (c) K and Figure 4-20 Characteristic x-ray spectra. (a) Part of the spectra of neodymium (Z 60) and samarium (Z 62). The two pairs of bright lines K lines. (b) Part of the are the spectrum of the artificially element promethium (Z 61). This element was first positively identified in 1945 at the Clinton Laboratory (now Oak Ridge). Its and those of neodymium and samarium, just as Moseley predicted. (c) Part of the spectra of all three of the elements neodymium, promethium, and samarium. [Courtesy of J. A. Swartout, Oak Ridge National Laboratory.] K K lines fall between Auger Electrons 4-4 X-Ray Spectra 173 The process of producing x rays necessarily results in the ionization of the atom since an inner electron is ejected. The vacancy created is filled by an outer electron, producing the x rays studied by Moseley. In 1923 Pierre Auger discovered that, as an alternative to x-ray emission, the atom may eject a third electron from a higher-energy outer shell via a radiationless process called the Auger effect. In the Auger (pronounced oh-zhay) process, the energy difference E E2 E1 that could have K x ray is removed from the atom by the third electron, resulted in the emission of a E, the n 3 electron would e.g., one in the n 3 shell. Since the magnitude of E3 leave the atom with a characteristic kinetic energy which is determined by the stationary-state energies of the particular atom.19 Thus, each element has a characteristic Auger electron spectrum. (See Figure 4-21a.) Measurement of the Auger electrons provides a simple and highly sensitive tool for identifying impurities on clean surfaces in electron microscope systems and investigating electron energy shifts associated with molecular bonding. (See Figure 4-21b.) \u00a2E \u0192E3 \u0192, Question 9. Why did Moseley plot f1>2 versus Z rather than f versus Z? (a) Copper Atomic number Cu 29 (b) E d / ) E N( d E Elemental Al Al oxide 679 734 778 1086 969 942 842 Ar 110 E d / ) E N( d E 66 0 200 400 600 800 1000 Kinetic energy (eV) 922 1280 1300 1320 1340 1360 1380 1400 1420 Kinetic energy (eV) Figure 4-21 (a) The Auger spectrum of Cu bombarded with 10-keV electrons. The energy of the Auger electrons is more precisely determined by plotting the weighted derivative E dN(e) dE of the electron intensity rather than the intensity N(e). (b) A portion of the Auger spectrum of Al from elemental Al and Al oxide. Note the energy shift in the largest peaks resulting from adjustments in", " the Al electron shell energies in the Al2O3 molecule. > 174 Chapter 4 The Nuclear Atom 4-5 The Franck-Hertz Experiment We conclude this chapter with discussion of an important experiment that provided strong support for the quantization of atomic energies, thus helping to pave the way for modern quantum mechanics. While investigating the inelastic scattering of electrons, J. Franck and G. Hertz20 made a discovery that confirmed by direct measurement Bohr\u2019s hypothesis of energy quantization in atoms. First done in 1914, it is now a standard undergraduate laboratory experiment. Figure 4-22a is a schematic diagram of the apparatus. A small heater heats the cathode. Electrons are ejected from the heated cathode and accelerated toward a grid, which is at a positive potential V0 relative to the cathode. Some electrons pass through the grid and reach the plate P, V. The tube is filled with a which is at a slightly lower potential Vp low-pressure gas of the element being investigated (mercury vapor, in Franck and Hertz\u2019s experiment). The experiment involves measuring the plate current as a function of V0. As V0 is increased from 0, the current increases until a critical value (about 4.9 V for Hg) is reached, at which point the current suddenly decreases. As V0 is increased further, the current rises again. V0 The explanation of this result is a bit easier to visualize if we think for the moment of a tube filled with hydrogen atoms instead of mercury. (See Figure 4-22b.) Electrons accelerated by V0 that collide with hydrogen electrons cannot transfer en 10.2 eV ergy to the latter unless they have acquired kinetic energy eV0 since the hydrogen electron according to Bohr\u2019s model cannot occupy states with energies intermediate between E1 and E2. Such a collision will thus be elastic; i.e., the incident electron\u2019s kinetic energy will be unchanged by the collision, and thus it can overcome the potential V and contribute to the current I. However, if eV0 10.2 eV, E2 E1! (a) V = 0 C H V = V0 G Vp = V0 \u2013 \u0394V P V0 I + \u2013 \u2013 + (b) Incoming electron \u2013 Proton Electron after scattering \u2013 n = 1 n = 2 Figure 4-22 (a) Schematic diagram of the Franck-Hertz experiment. Electrons ejected from the heated cathode C at zero potential are drawn to the positive grid G. Those passing through the holes in the grid can reach the plate P and thereby contribute to the current I if they have sufficient kinetic energy to overcome the small back potential V. The tube contains a low-pressure gas of the element being studied. (b) Results for hydrogen. If the incoming electron does not have sufficient energy to transfer E E2 If the incoming electron does have at least E kinetic energy, then an inelastic collision can occur in which E is transferred to the n 1 electron, moving it to the n 2 orbit. The excited electron will typically return to the ground state very quickly, emitting a photon of energy E. E1 to the hydrogen electron in the n 1 orbit (ground state), then the scattering will be elastic. 4-5 The Franck-Hertz Experiment 175 I 350 300 250 200 150 100 50 0 0 5 10 15 V0, V Figure 4-23 Current versus accelerating voltage in the Franck-Hertz experiment. The current decreases because many electrons lose energy due to inelastic collisions with mercury atoms in the tube and therefore cannot overcome the small back potential indicated in Figure 4-21a. The regular spacing of the peaks in this curve indicates that only a certain quantity of energy, 4.9 eV, can be lost to the mercury atoms. This interpretation is confirmed by the observation of radiation of photon energy 4.9 eV emitted by the mercury atoms, when V0 is greater than this energy. [From J. Franck and G. Hertz, Verband Deutscher Physiklischer Gesellschaften, 16, 457 (1914).] then the incoming electron can transfer 10.2 eV to the hydrogen electron in the ground state (n 1 orbit), putting it into the n 2 orbit (called the first excited state). The incoming electron\u2019s energy is thus reduced by 10.2 eV; it has been inelastically scattered. With insufficient energy to overcome the small retarding potential V, the incoming electrons can no longer contribute to the plate current I, and I drops sharply. The situation with Hg in the tube is more complicated since Hg has 80 electrons. Although Bohr\u2019s theory cannot predict their individual energies, we still expect the energy to be quantized with a ground state, first excited state, and so on, for the atom. Thus, the explanation of the observed 4", ".9-V critical potential for Hg is that the first excited state is about 4.9 eV above the lowest level (ground state). Electrons with energy less than this cannot lose energy to the Hg atoms, but electrons with energy greater than 4.9 eV can have inelastic collisions and lose 4.9 eV. If this happens near the grid, these electrons cannot gain enough energy to overcome the small back voltage V and reach the plate; the current therefore decreases. If this explanation is correct, the Hg atoms that are excited to an energy level of 4.9 eV above the ground state should return to the ground state by emitting light of wavelength c f hc hf hc eV0 253 nm There is indeed a line of this wavelength in the mercury spectrum. When the tube is viewed with a spectroscope, this line is seen when V0 is greater than 4.9 eV, while no lines are seen when V0 is less than this amount. For further increases in V0, additional sharp decreases in the current are observed, corresponding either to excitation of other levels in Hg (e.g., the second excited state of Hg is at 6.7 eV above the ground state) or to multiple excitations of the first excited state, i.e., due to an electron losing 4.9 eV more than once. In the usual setup, multiple excitations of the first level are observed and dips are seen every 4.9 V.21 The probability of observing such multiple first-level excitations, or excitations of other levels, depends on the detailed variation of the potential of the tube. For example, a second decrease in the current at 2 4.9 9.8 V results when electrons have inelastic collisions with Hg atoms V0 about halfway between the cathode and grid (see Figure 4-22a). They are reaccelerated, reaching 4.9 eV again in the vicinity of the grid. A plot of the data of Franck and Hertz is shown in Figure 4-23. The Franck-Hertz experiment was an important confirmation of the idea that discrete optical spectra were due to the existence in atoms of discrete energy levels that could be excited by nonoptical methods. It is particularly gratifying to be able to detect the existence of discrete energy levels directly by measurements using only voltmeters and ammeters. Electron Energy Loss Spectroscopy The Franck-Hertz experiment was the precursor of a highly sensitive technique for measuring the quantized energy states of atoms in both gases and solids. The technique, called electron energy loss spectroscopy (EELS), is particularly useful in solids, where it makes possible measurement of the energy of certain types of lattice vibrations and other processes. It works like this. Suppose that the electrons in an incident beam all have energy Einc. They collide with the atoms of a material, causing them to undergo some process (e.g., vibration, lattice rearrangement, electron excitation) that requires energy El. Then, if a beam electron initiates a single such process, El \u2014i.e., it has been inelastically scattered. it will exit the material with energy Einc 176 Chapter 4 The Nuclear Atom Figure 4-24 Energy-loss spectrum measurement. (a) A well-defined electron beam impinges upon the sample. Electrons inelastically scattered at a convenient angle enter the slit of the magnetic spectrometer, whose B field is directed out of the paper, and turn through radii R determined by their energy Einc Equation 3-2 written in the form E1)]1>2>eB. R [2m(Einc energy-loss spectrum for a thin Al film. [From C. J. Powell and J. B. Swan, Physical Review, 115, 869 (1954).] E1 via (b) An (a) Incident electron beam (b) 10 Sample Detector plane Slits Entrance slit Spectrometer 20 40 60 80 100 120 Electron energy loss, eV The exit energy can be measured very accurately with, e.g., a magnetic spectrometer designed for electrons.22 Figure 4-24a illustrates a typical experimental arrangement for measuring an energy-loss spectrum. As an example of its application, if an incident beam of electrons with Einc 2 keV is reflected from a thin Al film, the scattered electron energies measured in the magnetic spectrometer result in the energy-loss spectrum shown in Figure 4-24b, which directly represents the quantized energy levels of the target material. The loss peaks in this particular spectrum are due to the excitation of harmonic vibrations in the thin film sample, as well as some surface vibrations. The technique is also used to measure the vibrational energies of impurity atoms that may be absorbed on the surface and, with higher incident electron energies, to measure energy losses at the atomic inner", " levels, thus yielding information about bonding and other characteristics of absorbed atoms. Inelastic scattering techniques, including those using particles in addition to electrons, provide very powerful means for probing the energy structure of atomic, molecular, and nuclear systems. We will have occasion to refer to them many times throughout the rest of the book. More Here and in Chapter 3 we have discussed many phenomena that were \u201cexplained\u201d by various ad hoc quantum assumptions. A Critique of Bohr Theory and the \u201cOld Quantum Mechanics\u201d contrasts some of its successes with some of its failures on the Web page: whfreeman.com/modphysics5e. Summary 177 RELEVANT EQUATIONS AND REMARKS 1 mn R a 1 m2 1 n2 b n m 4-2 This empirical equation computes the correct wavelengths of observed spectral lines. The Rydberg constant R varies in a regular way from element to element. b kqQ mv2 cot 2 f b2nt for a scattering foil with n nuclei/unit volume and thickness t \u00a2N a I0Ascnt r2 b a kZe2 2Ek b 2 1 sin4 2 rd kqQ 1 2 mv2 4-3 4-5 4-6 4-11 Summary TOPIC 1. Atomic spectra 2. Rutherford scattering Impact parameter Scattered fraction f Number of scattered alphas observed Size of nucleus 3. Bohr model Bohr\u2019s postulates 1. Electrons occupy only certain nonradiating, stable, circular orbits selected by quantization of the angular momentum L. L mvr nh 2 nU for integer n 4-17 2. Radiation of frequency f occurs when the electron jumps from an allowed orbit of energy Ei to one of lower energy Ef. f is given by the frequency condition hf Ei Ef 4-15 Correspondence principle In the region of very large quantum numbers classical and quantum calculations must yield the same results. Bohr radius Allowed energies a0 U2 mke2 U mc 0.0529 nm En Z2E0 n2 for n 1, 2, 3, \u00c1 where E0 mk2e2>2U2 13.6 eV Reduced mass Fine-structure constant mM m M ke2 Uc 1>137 4-19 4-20 4-25 4-30 178 Chapter 4 The Nuclear Atom TOPIC RELEVANT EQUATIONS AND REMARKS 4. X-ray spectra Moseley equation 5. Franck-Hertz experiment f1>2 An(Z b) 4-34 Supported Bohr\u2019s theory by verifying the quantization of atomic energies in absorption. General References The following general references are written at a level appropriate for the readers of this book. Boorse, H., and L. Motz (eds.), The World of the Atom, Basic Books, New York, 1966. This two-volume, 1873-page work is a collection of original papers, translated and edited. Much of the work referred to in this chapter and throughout this book can be found in these volumes. Cline, B., The Questioners: Physicists and the Quantum Theory, Thomas Y. Crowell, New York, 1965. Gamow, G., Thirty Years That Shook Physics: The Story of the Quantum Theory, Doubleday, Garden City, NY, 1965. Herzberg, G., Atomic Spectra and Atomic Structure, Dover Publications, New York, 1944. This is without doubt one of the all-time classics of atomic physics. Melissinos, A., and J. Napolitano, Experiments in Modern Physics, 2d ed., Academic Press, New York, 2003. Many of the classic experiments that are now undergraduate laboratory experiments are described in detail in this text. Notes 1. Joseph von Fraunhofer (1787\u20131826), German physicist. Although he was not the first to see the dark lines in the solar spectrum that bear his name (Wollaston had seen seven, 12 years earlier), he systematically measured their wavelengths, named the prominent ones, and showed that they always occurred at the same wavelength even if the sunlight were reflected from the moon or a planet. 2. To date, more than 10,000 Fraunhofer lines have been found in the solar spectrum. 3. Although experimentalists preferred to express their measurements in terms of wavelengths, it had been shown that the many empirical formulas being constructed to explain the observed regularities in the line spectra could be expressed in simpler form if the reciprocal wavelength, called the wave number and equal to the number of waves per unit length, was used instead. Since c f this was equivalent to expressing the formulas in terms of the frequency., 4. Ernest Rutherford (1871\u20131937), English physicist, an exceptional experimentalist and a student of J. J. Thomson. He was an early researcher in the field of radioactivity and received the Nobel Prize in", " 1908 for his work in the transmutation of Mohr, P. J., and B. N. Taylor, \u201cThe Fundamental Physical Constants,\u201d Physics Today (August 2004). Also available at http://physicstoday.org/guide/fundcont.html. Shamos, M. H. (ed.), Great Experiments in Physics, Holt, Rinehart & Winston, New York, 1962. Virtual Laboratory (PEARL), Physics Academic Software, North Carolina State University, Raleigh, 1996. Includes an interactive model of the Bohr atom. Virtual Spectroscope, Physics Academic Software, North Carolina State University, Raleigh, 2003. Several sources can be viewed with a spectroscope to display the corresponding spectral lines. Visual Quantum Mechanics, Kansas State University, Manhattan, 1996. The atomic spectra component of this software provides an interactive construction of the energy levels for several elements, including hydrogen and helium. elements. He bemoaned the fact that his prize was awarded in chemistry, not in physics, as work with the elements was considered chemistry in those days. He was Thomson\u2019s successor as director of the Cavendish Laboratory. 5. Alpha particles, like all charged particles, lose energy by exciting and ionizing the molecules of the materials through which they are moving. The energy lost per unit path length (dE dx) is a function of the ionization potential of the molecules, the atomic number of the atoms, and the energy of the particles. It can be computed (with some effort) and is relatively simple to measure experimentally. > 2 sin d d\u00c6, 6. Notice that the differential solid angle subtended at the scattering nucleus by the surface in Figure 4-11. Since the cross section b2, then d 2 b db and Equation 4-9 can be rewritten as d d\" a kZe2 mv2 b 1 sin4 (>2) d>d\" 7. H. Geiger and E. Marsden, Philosophical Magazine (6), is called the differential cross section. 25, 605 (1913). Problems 179 spaced sets of lines, i.e., fine structure. We will discuss this topic in detail in Chapter 7. 16. Henry G.-J. Moseley (1887\u20131915), English physicist, considered by some the most brilliant of Rutherford\u2019s students. He would surely have been awarded the Nobel Prize had he not been killed in action in World War I. His father was a naturalist on the expedition of the HMS Challenger, the first vessel ever devoted to the exploration of the oceans. 17. The identifiers L and K were assigned by the English physicist C. G. Barkla, the discoverer of the characteristic xray lines, for which he received the Nobel Prize in 1917. He discovered two sets of x-ray lines for each of several elements, the longer wavelength of which he called the L series, the other the K series. The identifiers stuck and were subsequently used to label the atomic electron shells. 18. That the remaining K electron should result in b 1, i.e., shielding of exactly 1e, is perhaps a surprise. Actually it was a happy accident. It is the combined effect of the remaining K electron and the penetration of the electron waves of the outer L electrons that resulted in making b 1, as we will see in Chapter 7. 19. Since in multielectron atoms the energies of the stationary states depend in part on the number of electrons in the atom (see Chapter 7), the energies En for a given atom change slightly when it is singly ionized, as in the production of characteristic x-ray lines, or doubly ionized, as in the Auger effect. 20. James Franck (1882\u20131964), German-American physicist; Gustav L. Hertz (1887\u20131975), German physicist. Franck won an Iron Cross as a soldier in World War I and later worked on the Manhattan Project. Hertz was a nephew of Heinrich Hertz, discoverer of the photoelectric effect. For their work on the inelastic scattering of electrons, Franck and Hertz shared the 1925 Nobel Prize in Physics. 21. We should note at this point that there is an energy state in the Hg atom at about 4.6 eV, slightly lower than the one found by Franck and Hertz. However, transitions from the ground state to the 4.6-eV level are not observed, and their absence is in accord with the prediction of more advanced quantum mechanics, as we will see in Chapter 7. 22. Since for electrons is much larger than for ionized atoms, the radius for an electron magnetic spectrometer need not be as large as for a mass spectrometer, even for electron energies of several keV. (See Equation 3-2.) q>m 8. The value of Z could not be measured directly in this experiment; however, relative values for different foil materials could be", " found and all materials heavier than aluminum had Z approximately equal to half the atomic weight. 9. This also introduces a deviation from the predicted N associated with Rutherford\u2019s assumption that the nuclear mass was much larger than the particle mass. For lighter-atomicweight elements that assumption is not valid. Correction for the nuclear mass effect can be made, however, and the data in Figure 4-9b reflect the correction. 10. Niels H. D. Bohr (1885\u20131962), Danish physicist and firstrate soccer player. He went to the Cavendish Laboratory to work with J. J. Thomson after receiving his Ph.D.; however, Thomson is reported to have been impatient with Bohr\u2019s soft, accented English. Happily, the occasion of Thomson\u2019s annual birthday banquet brought Bohr in contact with Rutherford, whom he promptly followed to the latter\u2019s laboratory at Manchester, where he learned of the nuclear atom. A giant of twentieth-century physics, Bohr was awarded the Nobel Prize in 1922 for his explanation of the hydrogen spectrum. On a visit to the United States in 1939, he brought the news that the fission of uranium atoms had been observed. The story of his life makes absolutely fascinating reading. 11. N. Bohr, Philosophical Magazine (6), 26, 1 (1913). 12. P. J. Mohr and B. N. Taylor, \u201cThe Fundamental Physical Constants,\u201d Physics Today (August 2004). Only 8 of the 14 current significant figures are given in Equation 4-27. The relative uncertainty in the value is about 1 part in 1012! 13. Harold C. Urey (1893\u20131981), American chemist. His work opened the way for the use of isotopic tracers in biological systems. He was recognized with the Nobel Prize in 1934. 14. The basic reason that elliptical orbits solve this problem is that the frequency of the radiation emitted classically depends on the acceleration of the charge. The acceleration is constant for a circular orbit but varies for elliptical orbits, being dependent on the instantaneous distance from the focus. The energy of a particle in a circular orbit of radius r is the same as that of a particle in an elliptical orbit with a semimajor axis of r, so one would expect the only allowed elliptical orbits to be those whose semimajor axis was equal to an allowed Bohr circular orbit radius. 15. Viewed with spectrographs of high resolution, the spectral lines of hydrogen in Figure 4-2a\u2014and, indeed, most spectral lines of all elements\u2014are found to consist of very closely Problems Level I Section 4-1 Atomic Spectra 4-1. Compute the wavelength and frequency of the series limit for the Lyman, Balmer, and Paschen spectral series of hydrogen. 4-2. The wavelength of a particular line in the Balmer series is measured to be 379.1 nm. What transition does it correspond to? 4-3. An astronomer finds a new absorption line with 164.1 nm in the ultraviolet region of the Sun\u2019s continuous spectrum. He attributes the line to hydrogen\u2019s Lyman series. Is he right? Justify your answer. 180 Chapter 4 The Nuclear Atom 4-4. The series of hydrogen spectral lines with m 4 is called Brackett\u2019s series. Compute the wavelengths of the first four lines of Brackett\u2019s series. 4-5. In a sample that contains hydrogen, among other things, four spectral lines are found in the infrared with wavelengths 7460 nm, 4654 nm, 4103 nm, and 3741 nm. Which one does not belong to a hydrogen spectral series? Section 4-2 Rutherford\u2019s Nuclear Model > >, 4-6. A gold foil of thickness 2.0 m is used in a Rutherford experiment to scatter particles with energy 7.0 MeV. (a) What fraction of the particles will be scattered at angles greater than 90\u00b0? (b) What fraction will be scattered at angles between 45\u00b0 and 75\u00b0? (c) Use NA, and M for gold to compute the approximate radius of a gold atom. (For gold, 19.3 g cm3 and M 197 g mol.) 4-7. (a) What is the ratio of the number of particles per unit area on the screen scattered at 10\u00b0 to those at 1\u00b0? (b) What is the ratio of those scattered at 30\u00b0 to those at 1\u00b0? 4-8. For particles of 7.7 MeV (those used by Geiger and Marsden), what impact parameter will result in a deflection of 2\u00b0 for a thin gold foil? 4-9. What will be the distance of closest approach rd to a gold nucleus for an 5.0 MeV? 7.7 MeV? 12 MeV? 4-10. What energy particle would be needed to just reach the surface of an Al", " nucleus if its radius is 4 fm? 4-11. If a particle is deflected by 0.01\u00b0 in each collision, about how many collisions would be necessary to produce an rms deflection of 10\u00b0? (Use the result from the one-dimensional random walk problem in statistics stating that the rms deflection equals the magnitude of the individual deflections times the square root of the number of deflections.) Compare this result with the number of atomic layers in a gold foil of thickness 106 m, assuming that the thickness of each atom is 0.1 nm 1010 m. 4-12. Consider the foil and particle energy in Problem 4-6. Suppose that 1000 of those particles suffer a deflection of more than 25\u00b0. (a) How many of these are deflected by more than 45\u00b0? (b) How many are deflected between 25\u00b0 and 45\u00b0? (c) How many are deflected between 75\u00b0 and 90\u00b0? particle of > a0 2mc2 1 2 c 2 Section 4-3 The Bohr Model of the Hydrogen Atom 4-13. The radius of the n 1 orbit in the hydrogen atom is a0 0.053 nm. (a) Compute the radius of the n 6 orbit. (b) Compute the radius of the n 6 orbit in singly ionized helium (He), which is hydrogen-like, i.e., it has only a single electron outside the nucleus. 4-14. Show that Equation 4-19 for the radius of the first Bohr orbit and Equation 4-20 for the magnitude of the lowest energy for the hydrogen atom can be written as hc mc2 E1 h mc is the Compton wavelength of the electron and ke2 hc is the fine-structure where c constant. Use these expressions to check the numerical values of the constants a0 and E1. 1) in nm and indicate 4-15. Calculate the three longest wavelengths in the Lyman series (nf their position on a horizontal linear scale. Indicate the series limit (shortest wavelength) on this scale. Are any of these lines in the visible spectrum? 4-16. If the angular momentum of Earth in its motion around the Sun were quantized like a hydrogen electron according to Equation 4-17, what would Earth\u2019s quantum number be? How much energy would be released in a transition to the next lowest level? Would that energy release (presumably as a gravity wave) be detectable? What would be the radius of that orbit? (The radius of Earth\u2019s orbit is 1.50 1011 m.) 4-17. On average, a hydrogen atom will exist in an excited state for about 108 sec before making a transition to a lower energy state. About how many revolutions does an electron in the n 2 state make in 108 sec? 4-18. An atom in an excited state will on average undergo a transition to a state of lower energy in about 108 seconds. If the electron in a doubly ionized lithium atom (Li2, which is hydrogenlike) is placed in the n 4 state, about how many revolutions around the nucleus does it make before undergoing a transition to a lower energy state? > Problems 181 4-19. It is possible for a muon to be captured by a proton to form a muonic atom. A muon is identical to an electron except for its mass, which is 105.7 MeV c2. (a) Calculate the radius of the first Bohr orbit of a muonic atom. (b) Calculate the magnitude of the lowest energy state. (c) What is the shortest wavelength in the Lyman series for this atom? 4-20. In the lithium atom (Z 3) two electrons are in the n 1 orbit and the third is in the n 2 orbit. (Only two are allowed in the n 1 orbit because of the exclusion principle, which will be discussed in Chapter 7.) The interaction of the inner electrons with the outer one can be approximated by writing the energy of the outer electron as > E Z2(E1 >n2). Z H Z. 13.6 eV, n 2, and where E1 is the effective nuclear charge, which is less than 3 because of the screening effect of the two inner electrons. Using the measured ionization energy of 5.39 eV, calculate 4-21. Draw to careful scale an energy-level diagram for hydrogen for levels with n 1, 2, 3, Show the following on the diagram: (a) the limit of the Lyman series, (b) the 4, line, (c) the transition between the state whose binding energy ( energy needed to remove the electron from the atom) is 1.51 eV and the state whose excitation energy is 10.2 eV, and (d) the longest wavelength line of the Paschen series. radiation. (a) Compute the", " 4-22. A hydrogen atom at rest in the laboratory emits the Lyman recoil kinetic energy of the atom. (b) What fraction of the excitation energy of the n 2 state is carried by the recoiling atom? (Hint: Use conservation of momentum.) 4-23. (a) Draw accurately to scale and label completely a partial energy-level diagram for C5, including at minimum the energy levels for n 1, 2, 3, 4, 5, and (b) Compute the wavelength of the spectral line resulting from the n 3 to the n 2 transition, the C5 line. (c) In what part of the EM spectrum does this line lie? 4-24. The electron-positron pair that was discussed in Chapter 2 can form a hydrogenlike system called positronium. Calculate (a) the energies of the three lowest states and (b) the wavelength of the Lyman 4-25. With the aid of tunable lasers, Rydberg atoms of sodium have been produced with n 100. 600. (a) What would be The resulting atomic diameter would correspond in hydrogen to n the diameter of a hydrogen atom whose electron is in the n 600 orbit? (b) What would be the speed of the electron in that orbit? (c) How does the result in (b) compare with the speed in the n lines. (Detection of those lines is a \u201csignature\u201d of positronium formation.) 1 orbit? and H. Section 4-4 X-Ray Spectra K K K line) of x-ray line for an element is measured to be 0.0794 nm. What 4-26. (a) Calculate the next two longest wavelengths in the K series (after the molybdenum. (b) What is the wavelength of the shortest wavelength in this series? 4-27. The wavelength of the is the element? 4-28. Moseley pointed out that elements with atomic numbers 43, 61, and 75 should exist and (at that time) had not been found. (a) Using Figure 4-19, what frequencies would Moseley\u2019s x ray for each of these elements? (b) Compute the graphical data have predicted for the wavelengths for these lines predicted by Equation 4-37. 4-29. What is the approximate radius of the n 1 orbit of gold (Z 79)? Compare this with the radius of the gold nucleus, about 7.1 fm. 4-30. An electron in the K shell of Fe is ejected by a high-energy electron in the target of an xray tube. The resulting hole in the n 1 shell could be filled by an electron from the n 2 shell, the L shell; however, instead of emitting the characteristic Fe x ray, the atom ejects an Auger electron from the n 2 shell. Using Bohr theory, compute the energy of the Auger electron. 4-31. In a particular x-ray tube, an electron approaches the target moving at 2.25 108 m s. It > slows down on being deflected by a nucleus of the target, emitting a photon of energy 32.5 keV. Ignoring the nuclear recoil, but not relativity, compute the final speed of the electron. 4-32. (a) Compute the energy of an electron in the n 1 (K shell) of tungsten, using Z 1 for the effective nuclear charge. (b) The experimental result for this energy is 69.5 keV. Assume that the effective nuclear charge is Z where is called the screening constant, and calculate from the experimental result. K, 182 Chapter 4 The Nuclear Atom 4-33. Construct a Moseley plot similar to Figure 4-19 for the below (the x-ray energies are given in keV): K x rays of the elements listed Al 1.56 Ar 3.19 Sc 4.46 Fe 7.06 Ge 10.98 Kr 14.10 Zr 17.66 Ba 36.35 Determine the slope of your plot, and compare it with the K line in Figure 4-19. Section 4-5 The Franck-Hertz Experiment 4-34. Suppose that, in a Franck-Hertz experiment, electrons of energy up to 13.0 eV can be produced in the tube. If the tube contained atomic hydrogen, (a) what is the shortest-wavelength spectral line that could be emitted from the tube? (b) List all of the hydrogen lines that can be emitted by this tube. 4-35. Using the data in Figure 4-24b and a good ruler, draw a carefully scaled energy-level diagram covering the range from 0 eV to 60 eV for the vibrational states of this solid. What approximate energy is typical of the transitions between adjacent levels corresponding to the larger of each pair of peaks? 4-36. The transition from the first excited state to", " the ground state in potassium results in the emission of a photon with 770 nm If potassium vapor is used in a Franck-Hertz experiment, at what voltage would you expect to see the first decrease in current? 4-37. If we could somehow fill a Franck-Hertz tube with positronium, what cathode-grid voltage would be needed to reach the second current decrease in the positronium equivalent of Figure 4-23? (See Problem 4-24.) 4-38. Electrons in the Franck-Hertz tube can also have elastic collisions with the Hg atoms. If such a collision is a head-on, what fraction of its initial kinetic energy will an electron lose, assuming the Hg atom to be at rest? If the collision is not head-on, will the fractional loss be greater or less than this? Level II 4-39. Derive Equation 4-8 along the lines indicated in the paragraph that immediately precedes it. 4-40. Geiger and Marsden used particles with 7.7-MeV kinetic energy and found that when they were scattered from thin gold foil, the number observed to be scattered at all angles agreed with Rutherford\u2019s formula. Use this fact to compute an upper limit on the radius of the gold nucleus. 4-41. (a) The current i due to a charge q moving in a circle with frequency frev is qfrev. Find the current due to the electron in the first Bohr orbit. (b) The magnetic moment of a current loop is iA, where A is the area of the loop. Find the magnetic moment of the electron in the first Bohr orbit in units A-m2. This magnetic moment is called a Bohr magneton. 4-42. Use a spreadsheet to calculate the wavelengths (in nm) of the first five spectral lines of the Lyman, Balmer, Paschen, and Brackett series of hydrogen. Show the positions of these lines on a linear scale and indicate which ones lie in the visible. 4-43. Show that a small change in the reduced mass of the electron produces a small change in a spectral line given by in the Balmer red line 656.3 nm between hydrogen and deuterium, which has a nucleus with twice the mass of hydrogen. 4-44. Consider the Franck-Hertz experiment with Hg vapor in the tube and the voltage between the cathode and the grid equal to 4.0 V, i.e., not enough for the electrons to excite the Hg atom\u2019s first excited state. Therefore, the electron-Hg atom collisions are elastic. (a) If the kinetic energy of the electrons is Ek, show that the maximum kinetic energy that a recoiling Hg atom can have where M is the Hg atom mass. (b) What is the approximate maxiis approximately mum kinetic energy that can be lost by an electron with Ek 4-45. The Li2 ion is essentially identical to the H atom in Bohr\u2019s theory, aside from the effect of the different nuclear charges and masses. (a) What transitions in Li2 will yield emission lines whose wavelengths are very nearly equal to the first two lines of the Lyman series in hydrogen? Use this to calculate the difference 2.5 eV? 4mEk >. >M, > \u00a2 Problems 183 S M K, L, line of hydrogen and the x rays are emitted in the n 2 (b) Calculate the difference between the wavelength of the Lyman emission line from Li2 that has very nearly the same wavelength. scattering experiment, the area of the particle detector is 0.50 cm2. The detector 4-46. In an is located 10 cm from a 1.0- m-thick silver foil. The incident beam carries a current of 1.0 nA, and the energy of each particle is 6.0 MeV. How many particles will be counted per second by the detector at (a) 60\u00b0? (b) 120\u00b0? n 2, and 4-47. The and n 4 n 3 transitions, respectively. For calcium (Z 20) the energies of these transitions are 3.69 keV, 0.341 keV, and 0.024 keV, respectively. Suppose that energetic photons impinging on a calcium surface cause ejection of an electron from the K shell of the surface atoms. Compute the energies of the Auger electrons that may be emitted from the L, M, and N shells (n 2, 3, and 4) of the sample atoms, in addition to the characteristic x rays. K 4-48. Figure 3-15b shows the characteristic x rays emitted by a molybdenum (Mo) target in an x-ray tube whose accelerating potential is 35 kV. The wavelengths are 0.071 nm and 0.063 nm.", " (a) Compute the corresponding energies of these photons. (b) Suppose we K x rays by passing the molybdenum x rays wish to prepare a beam consisting primarily of x rays more strongly than through a material that absorbs x rays by photoelectric effect on K-shell electrons of the material. Which of the materials listed in the accompanying table with their K-shell binding energies would you choose? Explain your answer. n 1, n 3 and K K K K K S S Element Z Zr 40 Nb 41 Mo 42 Tc 43 Ru 44 EK (keV) 18.00 18.99 20.00 21.04 22.12 Level III 4-49. A small shot of negligible radius hits a stationary smooth, hard sphere of radius R, making an angle with the normal to the sphere, as shown in Figure 4-25. It is reflected at an equal angle to the normal. The scattering angle is 180\u00b0 2, as shown. (a) Show by the geometry of (b) If the incoming inthe figure that the impact parameter b is related to by? tensity of the shot is I0 particles/s area, how many are scattered through angles greater than (c) Show that the cross section for scattering through angles greater than 0\u00b0 is R2. (d) Discuss the implication of the fact that the Rutherford cross section for scattering through angles greater than 0\u00b0 is infinite. b R cos Figure 4-25 Small particle scattered by a hard sphere of radius R. 4-50. Singly ionized helium He is hydrogenlike. (a) Construct a carefully scaled energy-level diagram for He similar to that in Figure 4-16, showing the levels for n 1, 2, 3, 4, 5, and. (b) What is the ionization energy of He? (c) Compute the difference in wavelength between each of the first two lines of the Lyman series of hydrogen and the first two lines of the He Balmer series. Be sure to include the reduced mass correction for both atoms. (d) Show that for every spectral line of hydrogen, He has a spectral line of very nearly the same wavelength. (Mass of He 6.65 1027 kg.) 184 Chapter 4 The Nuclear Atom 4-51. Listed in the table are the x-ray wavelengths for several elements. Construct a Moseley plot from these data. Compare the slope with the appropriate one in Figure 4-19. Determine and interpret the intercept on your graph, using a suitably modified version of Equation 4-35. L Element Z P 15 Ca 20 Co 27 Kr 36 Mo 42 I 53 Wavelength (nm) 10.41 4.05 1.79 0.73 0.51 0.33 4-52. In this problem you are to obtain the Bohr results for the energy levels in hydrogen without using the quantization condition of Equation 4-17. In order to relate Equation 4-14 to the n2r0, where n is Balmer-Ritz formula, assume that the radii of allowed orbits are given by rn an integer and r0 is a constant to be determined. (a) Show that the frequency of radiation for a n 1 f kZe2>hr0n3 for large n. (b) Show that the frequency transition to of revolution is given by is given by nf f2 rev kZe2 42mr3 0n6 (c) Use the correspondence principle to determine r0 and compare with Equation 4-19. 4-53. Calculate the energies and speeds of electrons in circular Bohr orbits in a hydrogenlike atom using the relativistic expressions for kinetic energy and momentum. 4-54. (a) Write a computer program for your personal computer or programmable calculator that will provide you with the spectral series of H-like atoms. Inputs to be included are ni, nf, Z, and the nuclear mass M. Outputs are to be the wavelengths and frequencies of the first six lines and the series limit for the specified nf, Z, and M. Include the reduced mass correction. (b) Use the program to compute the wavelengths and frequencies of the Balmer series. (c) Pick 100, name the series the [your name] series, and use your program to compute the an nf wavelengths and frequencies of the first three lines and the limit. 4-55. Figure 4-26 shows an energy loss spectrum for He measured in an apparatus such as that shown in Figure 4-24a. Use the spectrum to construct and draw carefully to scale an energylevel diagram for He 21.21 23.07 20.61 19.82 0 10 20 Energy loss, eV Figure 4-26 Energy-loss spectrum of helium. Incident electron energy was 34 eV. The elastically scattered electrons cause the peak at 0 eV. H 4-56. If electric charge did not exist and electrons were bound to protons by the", " gravitational force to form hydrogen, derive the corresponding expressions for a0 and En and compute the enline and the limit of the Balmer series. Compare these with the ergy and frequency of the corresponding quantities for \u201creal\u201d hydrogen. 4-57. A sample of hydrogen atoms are all in the n 5 state. If all the atoms return to the ground state, how many different photon energies will be emitted, assuming all possible transitions occur? If there are 500 atoms in the sample and assuming that from any state all possible downward transitions are equally probable, what is the total number of photons that will be emitted when all of the atoms have returned to the ground state? 4-58. Consider muonic atoms (see Problem 4-19). (a) Draw a correctly scaled and labeled partial energy level diagram including levels with n 1, 2, 3, 4, 5, and for muonic hydrogen. (b) Compute the radius of the n 1 muon orbit in muonic H, He1, Al12, and Au78. (c) Compare the results in (b) with the radii of these nuclei. (d) Compute the wavelength of the photon emitted in the n 2 to n 1 transition for each of these muonic atoms. CHAPTER 5 The Wavelike Properties of Particles In 1924, a French graduate student, Louis de Broglie,1 proposed in his doctoral disser- tation that the dual\u2014i.e., wave-particle\u2014behavior that was by then known to exist for radiation was also a characteristic of matter, in particular, electrons. This suggestion was highly speculative, since there was yet no experimental evidence whatsoever for any wave aspects of electrons or any other particles. What had led him to this seemingly strange idea? It was a \u201cbolt out of the blue,\u201d like Einstein\u2019s \u201chappy thought\u201d that led to the principle of equivalence (see Chapter 2). De Broglie described it with these words: After the end of World War I, I gave a great deal of thought to the theory of quanta and to the wave-particle dualism.... It was then that I had a sudden inspiration. Einstein\u2019s wave-particle dualism was an absolutely general phenomenon extending to all physical nature.2 Since the visible universe consists entirely of matter and radiation, de Broglie\u2019s hypothesis is a fundamental statement about the grand symmetry of nature. (There is currently strong observational evidence that ordinary matter makes up only about 4 percent of the visible universe. About 22 percent is some unknown form of invisible \u201cdark matter,\u201d and approximately 74 percent consists of some sort of equally mysterious \u201cdark energy.\u201d See Chapter 13.) 5-1 The de Broglie Hypothesis De Broglie stated his proposal mathematically with the following equations for the frequency and wavelength of the electron waves, which are referred to as the de Broglie relations: f E h h p 5-1 5-2 where E is the total energy, p is the momentum, and length of the particle. For photons, is called the de Broglie wavethese same equations result directly from 5-1 The de Broglie Hypothesis 185 5-2 Measurements of Particle Wavelengths 5-3 Wave Packets 5-4 The Probabilistic Interpretation of the Wave Function 5-5 The Uncertainty Principle 5-6 Some 187 196 202 205 Consequences of the Uncertainty Principle 208 5-7 Wave-Particle Duality 212 185 186 Chapter 5 The Wavelike Properties of Particles Einstein\u2019s quantization of radiation E h\u0192 and Equation 2-31 for a particle of zero rest energy E pc as follows: \u03bb E pc hf hc 3 Figure 5-1 Standing waves around the circumference of a circle. In this case the circle is in circumference. If the vibrator were, for example, a steel ring that had been suitably tapped with a hammer, the shape of the ring would oscillate between the extreme positions represented by the solid and broken lines. By a more indirect approach using relativistic mechanics, de Broglie was able to demonstrate that Equations 5-1 and 5-2 also apply to particles with mass. He then pointed out that these equations lead to a physical interpretation of Bohr\u2019s quantization of the angular momentum of the electron in hydrogenlike atoms, namely, that the quantization is equivalent to a standing-wave condition. (See Figure 5-1.) We have mvr nU nh 2 for n integer 2r nh mv nh p n circumference of orbit 5-3 The idea of explaining discrete energy states in matter by standing waves thus seemed quite promising. De Broglie\u2019s ideas were expanded and developed into a complete theory by Erwin Schr\u00f6dinger late in 1925. In 1927, C. J. Davisson and", " L. H. Germer verified the de Broglie hypothesis directly by observing interference patterns, a characteristic of waves, with electron beams. We will discuss both Schr\u00f6dinger\u2019s theory and the Davisson-Germer experiment in later sections, but first we have to ask ourselves why wavelike behavior of matter had not been observed before de Broglie\u2019s work. We can understand why if we first recall that the wave properties of light were not noticed either until apertures or slits with dimensions of the order of the wavelength of light could be obtained. This is because the wave nature of light is not evident in experiments where the primary dimensions of the apparatus are large compared with the wavelength of the light used. For example, if A represents the diameter of a lens or the width of a slit, then diffraction effects3 (a manifestation of wave properties) 0\u00b0 are limited to angles around the forward direction ( In >A S 0, geometric (ray) optics too. However, if a characteristic sin S 0, sin >A. ) where so Louis V. de Broglie, who first suggested that electrons might have wave properties. [Courtesy of Culver Pictures.] 5-2 Measurements of Particle Wavelengths 187 the wavelength dimension of the apparatus becomes of the order of (or smaller than) of light passing through the system, then is readily observable, and the wavelike properties of light become apparent. Because Planck\u2019s constant is so small, the wavelength given by Equation 5-2 is extremely small for any macroscopic object. This point is among those illustrated in the following section. In that event >A S 1. >A, 5-2 Measurements of Particle Wavelengths Although we now have diffraction systems of nuclear dimensions, the smallest-scale systems to which de Broglie\u2019s contemporaries had access were the spacings between the planes of atoms in crystalline solids, about 0.1 nm. This means that even for an ex) moving tremely small macroscopic particle, such as a grain of dust ( through air with the average kinetic energy of the atmospheric gas molecules, the smallest diffraction systems available would have resulted in diffraction angles only of the order of 1010 radian, far below the limit of experimental detectability. The small mag- nitude of Planck\u2019s constant ensures that will be smaller than any readily accessible aperture, placing diffraction beyond the limits of experimental observation. For objects whose momenta are larger than that of the dust particle, the possibility of observing particle, or matter waves, is even less, as the following example illustrates. m 0.1 mg EXAMPLE 5-1 De Broglie Wavelength of a Ping-Pong Ball What is the de Broglie wavelength of a Ping-Pong ball of mass 2.0 g after it is slammed across the table with speed 5 m s? > SOLUTION 6.63 1034 J # s (2.0 103 kg)(5 m>s) h mv 6.6 1032 m 6.6 1023 nm This is 17 orders of magnitude smaller than typical nuclear dimensions, far below the dimensions of any possible aperture. The case is different for low-energy electrons, as de Broglie himself realized. At his soutenance de th\u00e8se (defense of the thesis), de Broglie was asked by Perrin4 how his hypothesis could be verified, to which he replied that perhaps passing particles, such as electrons, through very small slits would reveal the waves. Consider an electron that has been accelerated through V0 volts. Its kinetic energy (nonrelativistic) is then E p2 2m eV0 Solving for p and substituting into Equation 5-2, Using hc 1.24 103 eV # nm 1.226 V1>2 0 h p hc pc hc (2mc2eV0)1>2 and mc2 0.511 106 eV, we obtain nm for eV0 mc2 5-4 The following example computes an electron de Broglie wavelength, giving a measure of just how small the slit must be. 188 Chapter 5 The Wavelike Properties of Particles EXAMPLE 5-2 De Broglie Wavelength of a Slow Electron Compute the de Broglie wavelength of an electron whose kinetic energy is 10 eV. SOLUTION 1. The de Broglie wavelength is given by Equation 5-2: h p 2. Method 1: Since a 10-eV electron is nonrelativistic, we can use the classical relation connecting the momentum and the kinetic energy: Ek p2 2m or p 22mEk 4(2)(9.11 1031 kg)(10 eV)(1.60 1019 J>eV) 1.71 1024 kg # m>s 3. Substituting this result into Equ", "ation 5-2: 6.63 1034 J # s 1.71 1024 kg # m>s 3.88 1010 m 0.39 nm Electron gun Ionization chamber \u03d5 Ni crystal # from Figure 5-2 The DavissonGermer experiment. Low-energy electrons scattered at angle a nickel crystal are detected in an ionization chamber. The kinetic energy of the electrons could be varied by changing the accelerating voltage on the electron gun. 4. Method 2: The electron\u2019s wavelength can also be computed from Equation 5-4 with V0 10 V: 1.226 V1>2 0.39 nm 1.226 210 Remarks: Though this wavelength is small, it is just the order of magnitude of the size of an atom and of the spacing of atoms in a crystal. The Davisson-Germer Experiment In a brief note in the August 14, 1925, issue of the journal Naturwissenschaften, Walter Elsasser, at the time a student of J. Franck\u2019s (of the Franck-Hertz experiment), proposed that the wave effects of low-velocity electrons might be detected by scattering them from single crystals. The first such measurements of the wavelengths of electrons were made in 1927 by C. J. Davisson5 and L. H. Germer, who were studying electron reflection from a nickel target at Bell Telephone Laboratories, unaware of either Elsasser\u2019s suggestion or de Broglie\u2019s work. After heating their target to remove an oxide coating that had accumulated during an accidental break in their vacuum system, they found that the scattered electron intensity as a function of the scattering angle showed maxima and minima. The surface atoms of their nickel target had, in the process of cooling, formed relatively large single crystals, and they were observing electron diffraction. Recognizing the importance of their accidental discovery, they then prepared a target consisting of a single crystal of nickel and extensively investigated the scattering of electrons from it. Figure 5-2 illustrates their experimental arrangement. Their data for 54-eV electrons, shown in Figure 5-3, indicate a Consider the scattering from a set of Bragg strong maximum of scattering at # 50\u00b0. 5-2 Measurements of Particle Wavelengths 189 (a) \u03d5 = 0\u00b0 (b) 100 \u03d5 = 50 80 60 40 20 0 0 \u03d5 = 90\u00b0 60 20 40 Detector angle \u03d5 80 # Figure 5-3 Scattered intensity vs. detector angle for 54-eV electrons. (a) Polar plot of the data. The intensity at each angle is indicated by the distance of the point from the origin. Scattering angle on a Cartesian graph. The intensity scales are arbitrary but the same on both graphs. In each plot there is maximum intensity at h> p. wavelength \u00a9 Nobel Foundation, 1964).] [From Nobel Prize Lectures: Physics (Amsterdam and New York: Elsevier, is plotted clockwise starting at the vertical axes. (b) The same data plotted as predicted for Bragg scattering of waves having # 50\u00b0, planes, as shown in Figure 5-4. The Bragg condition for constructive interference is n 2d sin 2d cos. The spacing of the Bragg planes d is related to the spacing of the atoms D by d D sin ; thus or n 2D sin cos D sin 2 n D sin # 5-5 where # 2 is the scattering angle. The spacing D for Ni is known from x-ray diffraction to be 0.215 nm. The waveby Davisson length calculated from Equation 5-5 for the peak observed at and Germer is, for n 1, # 50\u00b0 0.215 sin 50\u00b0 0.165 nm The value calculated from the de Broglie relation for 54-eV electrons is 1.226 (54)1>2 0.167 nm Incident beam \u03d5 = 2\u03b1 D \u03b1 d \u03b1 \u03b1 \u03b8 \u03b8 Intense reflected beam Figure 5-4 Scattering of electrons by a crystal. Electron waves are strongly scattered if the Bragg n 2d sin condition met. This is equivalent to the condition n D sin #. is 190 Chapter 5 The Wavelike Properties of Particles \u00c5, \u03bb 2.0 1.5 1.0 0.5 0 0 0.05 0.10 0.15 0.20 0.25 \u20131/2 V0 The agreement with the experimental observation is excellent! With this spectacular result Davisson and Germer then conducted a systematic study to test the de Broglie relation using electrons up to about 400 eV and various experimental arrangements. Figure 5-5 shows a plot of measured wavelengths versus The wavelengths measured by diffraction are slightly lower than the theoretical predictions because the refraction of the electron waves at the crystal surface has been neglected. We have seen from the photoelectric effect that it takes work of the order of several eV to remove an electron from a metal", ". Electrons entering a metal thus gain kinetic energy; therefore, their de Broglie wavelength is slightly less inside the crystal.6 V1>2 0. V1>2 0 h> p. Figure 5-5 Test of the de Broglie formula wavelength is computed from a plot of the diffraction where V0 is the accelerating data plotted against, 1.226V1>2 voltage. The straight line is as predicted nm h(2mE) 1>2. from These are the data referred to in the quotation from Davisson\u2019s Nobel lecture. ( From observations with diffraction apparatus; particularly reliable; observations with reflection apparatus.) [From Nobel Prize Lectures: Physics (Amsterdam and New York: Elsevier, \u00a9 Nobel Foundation, 1964).] same, grazing beams. same, } n 0, The From A subtle point must be made here. Notice that the wavelength in Equation 5-5 depends only on D, the interatomic spacing of the crystal, whereas our derivation of that equation included the interplane spacing as well. The fact that the structure of the crystal really is essential shows up when the energy is varied, as was done in collecting the data for Figure 5-5. Equation 5-5 suggests that resulting from a change in the energy, would a change in mean only that the diffraction maximum would occur at some other value of such that the equation remains satisfied. However, as can be seen from examination of the angle Figure 5-4, the value of of the planes determined by the crystal structure. Thus, if there are no crystal planes # sin 1(>D) making an angle will not result in constructive interference and strong reflection for that value of even though Equation 5-5 is satisfied. This is neatly illustrated by Figure 5-6, which shows a series of polar graphs (like Figure 5-3a) for electrons of energies from 36 eV through 54 V, as we 68 eV. The building to a strong reflection at have already seen. But Equation 5-5 by itself would also lead us to expect, for exam 40 V, which obviously does not occur. ple, a strong reflection at with the surface, then setting the detector at is evident for V0 is determined by #>2 # 64\u00b0 # 50\u00b0,, # # when V0 36 V 50\u00b0 40 V 44 V 48 V 54 V 60 V 64 V 68 V Figure 5-6 A series of polar graphs of Davisson and Germer\u2019s data at electron accelerating potentials from 36 V to 68 V. Note the development of the peak at # 50\u00b0 when V0 to a maximum 54 V. 5-2 Measurements of Particle Wavelengths 191 Clinton J. Davisson (left) and Lester H. Germer at Bell Laboratories, where electron diffraction was first observed. [Bell Telephone Laboratories, Inc.] In order to show the dependence of the diffraction on the inner atomic layers, # fixed and varied the accelerating voltage Davisson and Germer kept the detector angle rather than search for the correct angle for a given Writing Equation 5-5 as D sin # n. D sin (2) n 5-6 0, #>2 V 1>2 0 ( 1>) V1>2 a graph of intensity versus # and noting that should yield (1) a series of equally spaced peaks corresponding to successive values of the integer n, if is an existing angle for atomic planes, or (2) no diffraction #>2 is not such an angle. Their measurements verified the dependence upon peaks if the interplane spacing, the agreement with the prediction being about percent. Thus, Davisson and Germer showed conFigure 5-7 illustrates the results for clusively that particles with mass moving at speeds do indeed have wavelike properties, as de Broglie had proposed. for a given angle # 50\u00b0. v V c 1 0.7 0.6 0.5 0.4 0.3 0.2 0..5 4th order 3rd order \u03d5 = 50\u00b0 5th order 6th order 7th order 8th order 2.0 1.0 1.5 1/\u03bb The diffraction pattern formed by high-energy electron waves scattered from nuclei provides a means by which nuclear radii and the internal distribution of the nuclear charge (the protons) are measured. See Chapter 11. Figure 5-7 Variation of the scattered electron intensity with wavelength for constant #. The incident beam in this case was 10\u00b0 from the normal, the resulting refraction causing the measured peaks to be slightly shifted from the positions computed from Equation 5-5, as explained in note 6. [After C. J. Davisson and L. H. Germer, Proceedings of the National Academy of Sciences, 14, 619 (1928).] 192 Chapter 5 The Wavelike Properties of Particles Here is Davisson\u2019s account of the connection between de Brogl", "ie\u2019s predictions and their experimental verification: Perhaps no idea in physics has received so rapid or so intensive development as this one. De Broglie himself was in the van of this development, but the chief contributions were made by the older and more experienced Schr\u00f6dinger. It would be pleasant to tell you that no sooner had Elsasser\u2019s suggestion appeared than the experiments were begun in New York which resulted in a demonstration of electron diffraction \u2014 pleasanter still to say that the work was begun the day after copies of de Broglie\u2019s thesis reached America. The true story contains less of perspicacity and more of chance... It was discovered, purely by accident, that the intensity of elastic scattering [of electrons] varies with the orientations of the scattering crystals. Out of this grew, quite naturally, an investigation of elastic scattering by a single crystal of predetermined orientation... Thus the New York experiment was not, at its inception, a test of wave theory. Only in the summer of 1926, after I had discussed the investigation in England with Richardson, Born, Franck and others, did it take on this character.7 A demonstration of the wave nature of relativistic electrons was provided in the same year by G. P. Thomson, who observed the transmission of electrons with energies in the range of 10 to 40 keV through thin metallic foils (G. P. Thomson, the son of J. J. Thomson, shared the Nobel Prize in 1937 with Davisson). The experimental arrangement (Figure 5-8a) was similar to that used to obtain Laue patterns with x rays (see Figure 3-11). Because the metal foil consists of many tiny crystals randomly oriented, the diffraction pattern consists of concentric rings. If a crystal is oriented at an angle with the incident beam, where satisfies the Bragg condition, this crystal will thus there will be a scattered beam making an strongly scatter at an equal angle angle with the incident beam. Figure 5-8b and c show the similarities in patterns produced by x rays and electron waves. 2 ; (b) (c) Figure 5-8 (a) Schematic arrangement used for producing a diffraction pattern from a polycrystalline aluminum target. (b) Diffraction pattern produced by x rays of wavelength 0.071 nm and an aluminum foil target. (c) Diffraction pattern produced by 600-eV electrons (de Broglie wavelength of about 0.05 nm) and an aluminum foil target. The pattern has been enlarged by 1.6 times to facilitate comparison with (b). [Courtesy of Film Studio, Education Development Center.] (a) Screen or film Incident beam (x rays or electrons) \u03b8 \u03b8 Al foil target Circular diffraction ring 5-2 Measurements of Particle Wavelengths 193 Diffraction of Other Particles The wave properties of neutral atoms and molecules were first demonstrated by O. Stern and I. Estermann in 1930 with beams of helium atoms and hydrogen molecules diffracted from a lithium fluoride crystal. Since the particles are neutral, there is no possibility of accelerating them with electrostatic potentials. The energy of the molecules was that of their average thermal motion, about 0.03 eV, which implies a de Broglie wavelength of about 0.10 nm for these molecules, according to Equation 5-2. Because of their low energy, the scattering occurs just from the array of atoms on the surface of the crystal, in contrast to Davisson and Germer\u2019s experiment. Figure 5-9 illustrates the geometry of the surface scattering, the experimental arrangement, and the results. Figure 5-9c indicates clearly the diffraction of He atom waves. Since then, diffraction of other atoms, of protons, and of neutrons has been observed (see Figures 5-10, 5-11, and 5-12 on page 194). In all cases the measured wavelengths agree with de Broglie\u2019s prediction. There is thus no doubt that all matter has wavelike, as well as particlelike, properties, in symmetry with electromagnetic radiation. The diffraction patterns formed by helium atom waves are used to study impurities and defects on the surfaces of crystals. Being a noble gas, helium does not react chemically with molecules on the surface nor \u201cstick\u201d to the surface. \u03bb Incident He beam (plane wave) \u03b8 \u03d5 \u03b8 (a) d Li + F \u2013 (b) \u03d5 (c Detector He atomic beam Slits \u03b8 \u03b8 LiF crystal \u201320\u00b0 \u201310\u00b0 0\u00b0 10\u00b0 20\u00b0 Detector setting \u03d5 Figure 5-9 (a) He atoms impinge upon the surface of the LiF crystal at angle in Estermann and Stern\u2019s experiment). The reflected beam also makes the same angle with relative to an axis perpendicular to the surface but is also scattered at azimuthal angles the surface. (b) The detector views the surface", " at angle but can scan through the angle d sin # (c) At angle where the path difference ( interference, i.e., a diffraction peak, occurs. The n 1 peaks occur on either side of the n 0 maximum. ) between adjacent \u201crays\u201d is ( 18.5\u00b0 constructive n, #. # # 194 Chapter 5 The Wavelike Properties of Particles Figure 5-10 Diffraction pattern produced by 0.0568-eV neutrons (de Broglie wavelength of 0.120 nm) and a target of polycrystalline copper. Note the similarity in the patterns produced by x rays, electrons, and neutrons. [Courtesy of C. G. Shull.] Figure 5-11 Neutron Laue pattern of NaCl. Compare this with the x-ray Laue pattern in Figure 3-14. [Courtesy of E. O. Wollan and C. G. Shull.] 101 100 10\u20131 10\u20132 10\u20133 10\u20134 10\u20135 10\u20136 Figure 5-12 Nuclei provide scatterers whose dimensions are of the order of 1015 m. Here the diffraction of 1-GeV protons from oxygen nuclei result in a pattern similar to that of a single slit. 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 Scattering angle, degrees An Easy Way to Determine de Broglie Wavelengths It is frequently helpful to know the de Broglie wavelength for particles with a specific kinetic energy. For low energies where relativistic effects can be ignored, the equation leading to Equation 5-4 can be rewritten in terms of the kinetic energy as follows: h p h 22mEk 5-7 5-2 Measurements of Particle Wavelengths 195 To find the equivalent expression that covers both relativistic and nonrelativistic speeds, we begin with the relativistic equation relating the total energy to the momentum: E2 (pc)2 (mc2)2 2-31 Writing E0 for the rest energy mc2 of the particle for convenience, this becomes E2 (pc)2 E2 0 Since the total energy Equation 5-8 becomes Ek, E E0 Ek)2 (pc)2 E2 0 (E0 5-8 c \u03bb / \u03bb that, when solved for p, yields (2E0Ek p k)1>2 E2 c from which Equation 5-2 gives \u03bb c = h/mc E0 = mc2 103 102 10 1 10\u20131 10\u20132 10\u20133 hc E2 k)1>2 (2E0Ek 5-9 10\u20134 10\u20136 10\u20134 10\u20132 1 102 104 Ek/E0 This can be written in a particularly useful way applicable to any particle of any energy by dividing the numerator and denominator by the rest energy as follows: mc2 E0 hc>mc2 E2 (2E0Ek k)1>2>E0 [2(Ek h>mc >E0) (Ek >E0)2]1>2 Recognizing h mc as the Compton wavelength mass m (see Section 3-4), we have that, for any particle, of the particle of c > > c 1 >E0) (Ek [2(Ek >E0)2]1>2 Figure 5-13 The de Broglie wavelength expressed in units of the Compton wavelength c for a particle of mass m versus the kinetic energy of the particle Ek expressed in units of its rest energy E0 E0 E0 mc2. For protons and neutrons 1.32 fm. 0.00234 nm. 0.938 GeV and c 0.511 MeV and c For electrons 5-10 > c A log-log graph of nearly constant slope, one for a curved portion lying roughly between 0.1 illustrates the use of Figure 5-13. Ek versus Ek >E0 V mc2 is shown in Figure 5-13. It has two sections of Ek connected by 10. The following example and the other for >E0 Ek W mc2, EXAMPLE 5-3 The de Broglie Wavelength of a Cosmic Ray Proton Detectors on board a satellite measure the kinetic energy of a cosmic ray proton to be 150 GeV. What is the proton\u2019s de Broglie wavelength, as read from Figure 5-13? SOLUTION The rest energy of the proton is mc2 0.938 GeV and the proton\u2019s mass is 1.67 1027 kg. Thus, the ratio is Ek >E0 Ek E0 150 GeV 0.938 GeV 160 196 Chapter 5 The Wavelike Properties of Particles This value on the curve corresponds to about 6 103 on the Compton wavelength of the proton is >", " c axis. The c h mc 6.63 1034 J # s (1.67 1027 kg)(3 108 m>s) 1.32 1015 m and we have then for the particle\u2019s de Broglie wavelength (6 103)(1.32 1015 m) 7.9 1018 m 7.9 103 fm Questions 1. Since the electrons used by Davisson and Germer were low energy, they penetrated only a few atomic layers into the crystal, so it is rather surprising that the effects of the inner layers shows so clearly. What feature of the diffraction is most affected by the relatively shallow penetration? 2. How might the frequency of de Broglie waves be measured? 3. Why is it not reasonable to do crystallographic studies with protons? 5-3 Wave Packets In any discussion of waves the question arises, \u201cWhat\u2019s waving?\u201d For some waves the answer is clear: for waves on the ocean, it is the water that \u201cwaves\u201d; for sound waves and the B. So what in air, it is the molecules that make up the air; for light, it is the is waving for matter waves? For matter waves as for light waves, there is no \u201cether.\u201d As will be developed in this section and the next, for matter it is the probability of finding the particle that waves. e Classical waves are solutions of the classical wave equation $2y $x2 1 v2 $2y $t2 5-11 Important among classical waves is the harmonic wave of amplitude y0, frequency \u0192, and period T, traveling in the direction as written here: x y(x, t) y0cos (kx t) y0cos 2 a x t T b y0cos 2 (x vt) 5-12 where the angular frequency and the wave number8 k are defined by and 2f 2 T k 2 5-13a 5-13b and the velocity v of the wave, the so-called wave or phase velocity vp, is given by f vp 5-14 A familiar wave phenomenon that cannot be described by a single harmonic wave is a pulse, such as the flip of one end of a long string (Figure 5-14), a sudden noise, or the brief opening of a shutter in front of a light source. The main characteristic of a pulse is localization in time and space. A single harmonic wave is not localized in either time or space. The description of a pulse can be obtained by the superposition of a group of harmonic waves of different frequencies and wavelengths. Such a group is called a wave packet. The mathematics of representing arbitrarily shaped pulses by sums of sine or cosine functions involves Fourier series and Fourier integrals. We will illustrate the phenomenon of wave packets by considering some simple and somewhat artificial examples and discussing the general properties qualitatively. Wave groups are particularly important because a wave description of a particle must include the important property of localization. Consider a simple group consisting of only two waves of equal amplitude and nearly equal frequencies and wavelengths. Such a group occurs in the phenomenon of beats and is described in most introductory textbooks. The quantities k, and v are related to one another via Equations 5-13 and 5-14. Let the wave numbers be k1 and and the speeds v1 and v2. The sum of the two k2, the angular frequencies waves is 2, and, 1 y(x, t) y0cos (k1x 1t) y0cos (k2x 2t) which, with the use of a bit of trigonometry, becomes y(x, t) 2y0cos a \u00a2k 2 x \u00a2 2 tb cos a k1 k2 2 x 1 2 2 tb 5-3 Wave Packets 197 (a) (b) Figure 5-14 (a) Wave pulse moving along a string. A pulse has a beginning and an end; i.e., it is localized, unlike a pure harmonic wave, which goes on forever in space and time. (b) A wave packet formed by the superposition of harmonic waves. k1 \u00a2k k2, and where of k and we will write The sum is then \u00a2 2 k (k1 1. k2)>2 Since the two waves have nearly equal values 2)>2 for the mean values. and ( 1 y(x, t)) 2y0cos a 1 2 \u00a2kx 1 2 \u00a2tb cos (kx t) 5-15 Figure 5-15 shows a sketch of y (x, t0) versus x at some time t0. The dashed curve is the envelope of the group of two waves, given by the first cosine term in Equation 5-15. the phase velocity vp due to The wave within the envelope moves with the speed the second cosine term. If we write the first", " (amplitude-modulating) term as cos {1 The 2 speed of the envelope is called the group velocity vg. we see that the envelope moves with speed \u00a2k[x (\u00a2>\u00a2k) t]}, \u00a2>\u00a2k. >k, (a) y (b) y x x1 x x2 \u0394x y(x) at a given instant for each of the two Figure 5-15 Two waves of slightly different wavelength and frequency produce beats. (a) Shows waves. The waves are in phase at the origin, but because of the difference in wavelength, they become out of phase and then in phase again. (b) The sum of these waves. The spatial extent of the group x is inversely proportional to the difference in wave numbers k, where k is related to the wavelength by if y is plotted versus time t at a fixed point x. In that case the extent in time t is inversely proportional to the frequency difference \u00a2. Identical figures are obtained k 2>. 198 Chapter 5 The Wavelike Properties of Particles A more general wave packet can be constructed if, instead of adding just two sinusoidal waves as in Figure 5-15, we superpose a larger, finite number with slightly different wavelengths and different amplitudes. For example, Figure 5-16a illustrates 1>9 1>15 the superposing of seven cosines with wavelengths from ) at time t0. The waves are all in phase at (wave numbers from 30 18 9 to to 15 k9 x 0 and again at k15 x 12, x 24, \u00c1. Their sum 15 y(x, t0) a i9 yi (x, t0) oscillates with maxima at those values of x, decreasing and increasing at other values as a result of the changing phases of the waves (see Figure 5-16b). Now, if we superpose an infinite number of waves from the same range of wavelengths and wave numbers as in Figure 5-16 with infinitesimally different values of k, the central group around x 0 will be essentially the same as in that figure. However, the additional groups will no longer be present since there is now no length along the x axis into which an exactly integral number of all of the infinite number of component waves can fit. Thus, we have formed a single wave packet throughout this (one-dimensional) space. (a) y9 y10 y11 y12 y13 y14 y15 k 18\u03c0 20\u03c0 22\u03c0 24\u03c0 26\u03c0 28\u03c0 30\u03c0 \u20136 \u20135 \u20134 \u20133 \u20132 \u2013 10 11 12 x(units of 1/2) (b) y = \u03a3 yi i (c) 1 y0 3/4 1/2 1/3 1/4 16\u03c0 20\u03c0 24\u03c0 28\u03c0 32\u03c0 k 4\u03c0 Figure 5-16 (a) Superposition of seven sinusoids k (2)15 from k (2)9 with t to yk(x, t) y0k cos (kx t) 0. The maximum amplitude is 1 at the center of the range to 1 2, 1 3, and 1 4, respectively, for the waves on each side of the central wave. (b) The sum > > > with uniformly spaced wave numbers ranging decreasing (k (2)12), 15 y(x, 0) a i9 yi(x, 0) is maximum at x 0 with additional maxima equally spaced along the x axis. (c) Amplitudes of the sinusoids yi versus wave number k. This packet moves at the group velocity The mathematics needed to demonstrate the above involves use of the Fourier integral described in the Classical Concept Review. vg d>dk. The phase velocities of the individual harmonic waves are given by Equation 5-14: f a vp 2 b a 2 k b k Writing this as by Equation 5-16: kv p, the relation between the group and phase velocities is given vg d dk vp k dvp dk 5-16 > If the phase velocity is the same for all frequencies and wavelengths, then dvp dk 0, and the group velocity is the same as the phase velocity. A medium for which the phase velocity is the same for all frequencies is said to be nondispersive. Examples are waves on a perfectly flexible string, sound waves in air, and electromagnetic waves in a vacuum. An important characteristic of a nondispersive medium is that, since all the harmonic waves making up a packet move with the same speed, the packet maintains its shape as it moves; thus, it does not change its shape in time. Conversely, if the phase velocity is different for different frequencies, the shape of the pulse will change as it travels. In that case the group velocity and phase velocity are not the same. Such a medium is called a dispersive medium; examples are water waves", ", waves on a wire that is not perfectly flexible, light waves in a medium such as glass or water in which the index of refraction has a slight dependence on frequency, and electron waves. It is the speed of the packet, the group velocity vg, that is normally seen by an observer. Classical Uncertainty Relations Notice that the width of the group9 x of the superposition just a bit larger than k has width \u00a2k 4, 1>12. in Figure 5-16b is Similarly, the graph of the amplitude of these waves versus y(x, t0) which is a bit more than 12 (Figure 5-16c), so we see that \u00a2k\u00a2x 1 By a similar analysis, we would also conclude that \u00a2\u00a2t 1 5-17 5-18 The range of wavelengths or frequencies of the harmonic waves needed to form a wave packet depends on the extent in space and duration in time of the pulse. In general, if the extent in space x is to be small, the range k of wave numbers must be large. Similarly, if the duration in time t is small, the range of frequencies must be large. We have written these as order-of-magnitude equations because the exact depends on how these ranges are defined, as value of the products well as on the particular shape of the packets. Equation 5-18 is sometimes known as the response time\u2013bandwidth relation, expressing the result that a circuit component such as an amplifier must have a large bandwidth ( ) if it is to be able to respond to signals of short duration. \u00a2k\u00a2x \u00a2\u00a2t and \u00a2 \u00a2 5-3 Wave Packets 199 The classical uncertainty relations define the range of signal frequencies to which all kinds of communications equipment and computer systems must respond, from cell phones to supercomputers. 200 Chapter 5 The Wavelike Properties of Particles There is a slight variation of Equation 5-17 that is also helpful in interpreting the relation between x and k. Differentiating the wave number in Equation 5-13b yields dk 2d 2 5-19 Replacing the differentials by small intervals and concerning ourselves only with magnitudes, Equation 5-19 becomes \u00a2k 2\u00a2 2 which when substituted into Equation 5-17 gives \u00a2x\u00a2 2 2 5-20 Equation 5-20 says that the product of the spatial extent of a classical wave x will always and the uncertainty (or \u201cerror\u201d) in the determination of its wavelength be of the order of The following brief examples will illustrate the meaning of Equations 5-17 and 5-18, often referred to as the classical uncertainty relations, and Equation 5-20. 2>2. \u00a2 \u00a2 EXAMPLE 5-4 for Ocean Waves Standing in the middle of a 20-m-long pier, you notice that at any given instant there are 15 wave crests between the two ends of the pier. Estimate the minimum uncertainty in the wavelength that could be computed from this information. SOLUTION 1. The minimum uncertainty \u00a2 in the wavelength is given by Equation 5-20: \u00a2x\u00a2 2 2 2. The wavelength of the waves is 20 m 15 waves 1.3 m 3. The spatial extent of the waves used for this calculation is: \u00a2x 20 m 4. Solving Equation 5-20 for \u00a2 and substituting these values gives \u00a2 2 2\u00a2x 0.013 m (1.3)2 2 20 \u00a2 0.01 m 1 cm Remarks: This is the minimum uncertainty. Any error that may exist in the measurement of the number of wave crests and the length of the pier would add further uncertainty to the determination of. 5-3 Wave Packets 201 EXAMPLE 5-5 Frequency Control The frequency of the alternating voltage produced at electric generating stations is carefully maintained at 60.00 Hz. The frequency is monitored on a digital frequency meter in the control room. For how long must the frequency be measured and how often can the display be updated if the reading is to be accurate to within 0.01 Hz? SOLUTION Since 2\u0192, then \u00a2 2\u00a2\u0192 2(0.01) rad/s and \u00a2t 1>\u00a2 1> 2(0.01) \u00a2t 16 s Thus, the frequency must be measured for about 16 s if the reading is to be accurate to 0.01 Hz and the display cannot be updated more often than once every 16s. Questions 4. Which is more important for communication, the group velocity or the phase velocity? 5. What are x and k for a purely harmonic wave of a single frequency and wavelength? Particle Wave Packets e(x, t) The quantity analogous to the displacement y(x, t) for waves on a string, to the pressure P(x, t) for a sound wave, or to the electric field for electromagnetic waves \u00b0(x,", " t) is called the wave function for particles and is usually designated that we will relate to the probability of finding the particle and, as we alerted you earlier, it is the probability that waves. Consider, for example, an electron wave consisting of a single frequency and wavelength; we could represent such a wave by any of \u00b0(x, t) A cos (kx t), the following, exactly as we did the classical wave: \u00b0(x, t) A sin (kx t), or \u00b0(x, t) Aei(kxt). The phase velocity for this wave is given by \u00b0(x, t). It is f (E> h)(h>p) E>p vp where we have used the de Broglie relations for the wavelength and frequency. Using the nonrelativistic expression for the energy of a particle moving in free space, i.e., no potential energy) with no forces acting upon it, E 1 2 mv2 p2 2m we see that the phase velocity is E>p (p2>2m)>p p>2m v>2 vp i.e., the phase velocity of the wave is half the velocity of an electron with momentum p. The phase velocity does not equal the particle velocity. Moreover, a wave of a single frequency and wavelength is not localized but is spread throughout space, which makes it difficult to see how the particle and wave properties of the electron could be related. 202 Chapter 5 The Wavelike Properties of Particles An application of phase and particle speeds by nature: produce a wave on a still pond (or in a bathtub) and watch the wavelets that make up the wave appear to \u201cclimb over\u201d the wave crest at twice the speed of the wave. Thus, for the electron to have the particle property of being localized, the matter waves \u00b0(x, t) must be of the electron must also be limited in spatial extent\u2014i.e., realistically,. a wave packet containing many more than one wave number k and frequency It is that we expect to move at a group velocity equal to the partithe wave packet cle velocity, which we will show below is indeed the case. The particle, if observed, \u00b0(x, t), we will expect to find somewhere within the spatial extent of the wave packet precisely where within that space being the subject of the next section. \u00b0(x, t) To illustrate the equality of the group velocity vg and the particle velocity v, it is convenient to express de Broglie\u2019s relations in a slightly different form. Writing Equation 5-1 as follows, E hf h>2 or E U and Equation 5-2 as p h h 2>k hk 2 or p Uk 5-21 5-22 The group velocity is then given by d>dk (dE>U)>(dp>U) dE>dp vg Again using the nonrelativistic expression E p2 2m, we have that > dE>dp p>m v vg \u00b0(x, t) and the wave packet moves with the velocity of the electron. This was, in fact, one of de Broglie\u2019s reasons for choosing Equations 5-1 and 5-2. (De Broglie used the relativistic expression relating energy and momentum, which also leads to the equality of the group velocity and particle velocity.) 5-4 The Probabilistic Interpretation of the Wave Function Let us consider in more detail the relation between the wave function and the location of the electron. We can get a hint about this relation from the case of light. the electric field, The wave equation that governs light is Equation 5-11, with e2, as the wave function. The energy per unit volume in a light wave is proportional to but the energy in a light wave is quantized in units of hf for each photon. We expect, a contherefore, that the number of photons in a unit volume is proportional to nection first pointed out by Einstein. \u00b0(x, t) y e, e2, Consider the famous double-slit interference experiment (Figure 5-17). The pattern observed on the screen is determined by the interference of the waves from the slits. At a point on the screen where the wave from one slit is 180\u00b0 out of phase with that from the other, the resultant electric field is zero; there is no light energy at this point, and this point on the screen is dark. If we reduce the intensity to a very low value, we can still observe the interference pattern if we replace the ordinary screen by a scintillation screen or a two-dimensional array of tiny photon detectors (e.g., a CCD camera) and wait a sufficient length of time. 5-4 The Probabilistic Interpretation of the Wave Function 203 Figure 5-17 Two-source interference pattern. If the sources are coherent", " and in phase, the waves from the sources interfere constructively at points for which the path difference (d number of wavelengths. ) is an integral sin I S1 S2 S1 \u03b8 d \u03b8 S2 d sin \u03b8 The interaction of light with the detector or scintillator is a quantum phenomenon. If we illuminate the scintillators or detectors for only a very short time with a lowintensity source, we do not see merely a weaker version of the high-intensity pattern; we see, instead, \u201cdots\u201d caused by the interactions of individual photons (Figure 5-18). At points where the waves from the slits interfere destructively, there are no dots, and at points where the waves interfere constructively, there are many dots. However, (a) (b) (c) (d) Figure 5-18 Growth of two-slit interference pattern. The photo (d) is an actual two-slit electron interference pattern in which the film was exposed to millions of electrons. The pattern is identical to that usually obtained with photons. If the film were to be observed at various stages, such as after being struck by 28 electrons, then after about 1000 electrons, and again after about 10,000 electrons, the patterns of individually exposed grains would be similar to those shown in (a), (b), and (c) except that the exposed dots would be smaller than the dots drawn here. Note that there are no dots in the region of the interference minima. The probability of any point of the film being exposed is determined by wave theory, whether the film is exposed by electrons or photons. [Parts (a), (b), and (c) from E. R. Huggins, Physics 1, \u00a9 by W. A. Benjamin, Inc., Menlo Park, California. Photo (d) courtesy of C. Jonsson.] 204 Chapter 5 The Wavelike Properties of Particles when the exposure is short and the source weak, random fluctuations from the average predictions of the wave theory are clearly evident. If the exposure is long enough that many photons reach the detector, the fluctuations average out and the quantum nature of light is not noticed. The interference pattern depends only on the total number of photons interacting with the detector and not on the rate. Even when the intensity is so low that only one photon at a time reaches the detector, the wave theory predicts the correct average pattern. For low intensities, we therefore interpret as proportional to the probability of detecting a photon in a unit volume of space. At points on the detector where is zero, photons are never observed, whereas they are most likely to be observed at points where is large. e2 e2 e2 \u00b0 It is not necessary to use light waves to produce an interference pattern. Such patterns can be produced with electrons and other particles as well. In the wave theory of \u00b0. electrons the de Broglie wave of a single electron is described by a wave function at any point is related to the probability of finding the particle at The amplitude of \u0192\u00b0\u0192 2 is proporthe quantity that point. In analogy with the foregoing interpretation of \u0192\u00b0\u0192 2 \u00b0*\u00b0, tional to the probability of detecting an electron in a unit volume, where \u0192\u00b0\u0192 2dx is the In one dimension, the function probability of an electron being in the interval dx.10 (See Figure 5-19.) If we call this probability P(x)dx, where P(x) is the probability distribution function, we have being the complex conjugate of \u00b0* e2, \u00b0. P(x)dx \u0192\u00b0\u0192 2dx 5-23 y x |\u03a8(x, y, t)|2 t = 0 y x |\u03a8(x, y, t)|2 t = \u0394t Figure 5-19 A three-dimensional wave packet representing a particle moving along the x axis. The dot indicates the position of a classical particle. Note that the packet spreads out in the x and y directions. This spreading is due to dispersion, resulting from the fact that the phase velocity of the individual waves making up the packet depends on the wavelength of the waves. y x |\u03a8(x, y, t)|2 t = 2\u0394t 5-5 The Uncertainty Principle 205 In the next chapter we will more thoroughly discuss the amplitudes of matter waves associated with particles, in particular developing the mathematical system for computing the amplitudes and probabilities in various situations. The uneasiness that you may feel at this point regarding the fact that we have not given a precise physical interpretation to the amplitude of the de Broglie matter wave can be attributed in part to the complex nature of the wave amplitude; i.e., it is in general a complex function with a real part and an imaginary part, the latter proportional to i (1)1>2. We cannot directly measure or physically interpret complex numbers in our", " world of real numbers. However, as we will see, defining the probability in which is always real, presents no difficulty in its physical interpretaterms of tion. Thus, even though the amplitudes of the wave functions have no simple meaning, the waves themselves behave just as classical waves do, exhibiting the wave characteristics of reflection, refraction, interference, and diffraction and obeying the principles of superposition. \u0192\u00b0\u0192 2, \u00b0 5-5 The Uncertainty Principle The uncertainty relations for classical wave packets (Equations 5-17 and 5-18) have very important matter wave analogs. \u00b0(x, t) Consider a wave packet representing an electron. The most probable position of the electron is the value of x for which is a maximum. Since \u0192\u00b0(x, t)\u0192 2 \u0192\u00b0(x, t)\u0192 2 is proportional to the probability that the electron is located at x and is nonzero for a range of values of x, there is an uncertainty in the position of the electron (see Figure 5-19). This means that if we make a number of position measurements on identical electrons\u2014electrons with the same wave function\u2014we will not always obtain the same result. In fact, the distribution function for the results of such measurements will be given by If the wave packet is very narrow, the uncertainty in position will be small. However, a narrow wave packet must contain a wide range of wave numbers k. Since the momentum is related to the wave number by p hk, a wide range of k values means a wide range of momentum values. We have seen that for all wave packets the ranges x and k are related by \u0192\u00b0(x, t)\u0192 2. \u0192\u00b0(x, t)\u0192 2 \u00a2k\u00a2x 1 5-17 Similarly, a packet that is localized in time t must contain a range of frequencies where the ranges are related by \u00a2, \u00a2\u00a2t 1 5-18 Equations 5-17 and 5-18 are inherent properties of waves. If we multiply these equaU tions by and use E U, we obtain p Uk and and \u00a2x\u00a2p U \u00a2E\u00a2t U 5-24 5-25 Equations 5-24 and 5-25 provide a statement of the uncertainty principle first enunciated in 1927 by Werner K. Heisenberg.11 Equation 5-24 expresses the fact that the distribution functions for position and momentum cannot both be made arbitrarily narrow simultaneously (see Figure 5-16); thus, measurements of position and momentum will have similar uncertainties, which are related by Equation 5-24. Of course, because of 206 Chapter 5 The Wavelike Properties of Particles Heisenberg\u2019s uncertainty principle is the key to the existence of virtual particles that hold the nuclei together (see Chapter 11) and is the root of quantum fluctuations that may have been the origin of the Big Bang (see Chapter 13). U. inaccurate measurements, the product of x and p can be, and usually is, much larger than The lower limit is not due to any technical problem in the design of measuring equipment that might be solved at some later time; it is instead due to the wave and particle nature of both matter and light. If we define precisely what we mean by the uncertainty in the measurements of position and momentum, we can give a precise statement of the uncertainty principle. is the For example, if k has its standard deviation for measurements of the wave number, the product minimum value of 1 2 when the distribution functions are Gaussian. If we define x and p to be the standard deviations, the minimum value of their product is Thus, is the standard deviation for measurements of position and k U>2. x x > \u00a2x\u00a2p! 1 2 U \u00a2E\u00a2t! 1 2 U 5-26 5-27 Similarly, Question 6. Does the uncertainty principle say that the momentum of a particle can never be precisely known? EXPLORING The Gamma-Ray Microscope Let us see how one might attempt to make a measurement so accurate as to violate the uncertainty principle. A common way to measure the position of an object such as an electron is to look at it with light, i.e., scatter light from it and observe the diffraction pattern. The momentum can be obtained by looking at it again a short time later and computing what velocity it must have had the instant before the light scattered from it. Because of diffraction effects, we cannot hope to make measurements of length (position) that are smaller than the wavelength of the light used, so we will use the shortestwavelength light that can be obtained, gamma rays. (There is, in principle, no limit to how short the wavelength of electromagnetic radiation can be.) We also know that light carries momentum and energy, so that when it scatters off the electron, the motion of the electron will be disturbed, affecting the momentum. We", " must, therefore, use the minimum intensity possible so as to disturb the electron as little as possible. Reducing the intensity decreases the number of photons, but we must scatter at least one photon to observe the electron. The minimum possible intensity, then, is that corresponding to one photon. The scattering of a photon by a free electron is, of course, a Compton scattering, which was discussed in Section 3-4. The momentum of the photon is The that is used to measure the position, the more the photon will disturb the elecsmaller tron, but we can correct for that with a Compton-effect analysis, provided only that we know the photon\u2019s momentum and the scattering angles of the event. hf>c h>. Figure 5-20 illustrates the problem. (This illustration was first given as a gedankenexperiment, or thought experiment, by Heisenberg. Since a single photon doesn\u2019t form a diffraction pattern, think of the diffraction pattern as being built up by photons from 5-5 The Uncertainty Principle 207 many identical scattering experiments.) The position of the electron is to be determined by viewing it through a microscope. We will assume that only one photon is used. We can take for the uncertainty in position the minimum separation distance for which two objects can be resolved; this is12 \u00a2x 2 sin where is the half angle subtended by the lens aperture, as shown in Figure 5-20a and b. (a) Photons that go through lens are restricted to this region Screen Lens \u03b8 Electron Light source (b) Scattered photon x component of photon\u2019s recoil momentum (h/\u03bb) sin \u03b8\u00b4 \u03b8\u00b4 \u03b8 x component of electron\u2019s recoil momentum (h/\u03bb) sin \u03b8\u00b4 Incident photon p\u03b3 = hf/c = h/\u03bb (c) Intensity \u0394x Diffraction pattern seen on screen x Figure 5-20 \u201cSeeing an electron\u201d with a gamma-ray microscope. (b) Because of the size of the lens, the momentum of the scattered photon is uncertain by \u00a2px p sin h sin >. Thus the recoil momentum of the electron is also uncertain by at least this amount. (c) The position of the electron cannot be resolved better than the width of the central maximum of the diffraction pattern \u00a2x >sin. uncertainties px the order of Planck\u2019s constant h. x is therefore of The product of the 208 Chapter 5 The Wavelike Properties of Particles Let us assume that the x component of momentum of the incoming photon is known precisely from a previous measurement. To reach the screen and contribute to the diffraction pattern in Figure 5-20c, the scattered photon need only go through the lens aperture. Thus, the scattered photon can have any x component of momentum from 0 to px where p is the total momentum of the scattered photon. By conservation of momentum, the uncertainty in the momentum of the electron after the scattering must be greater than or equal to that of the scattered photon (it would be equal, of course, if the electron\u2019s initial momentum were known precisely); therefore, we write sin, p and \u00a2px! p sin h sin \u00a2x\u00a2px! 2 sin h sin 1 2 h Thus, even though the electron prior to our observation may have had a definite position and momentum, our observation has unavoidably introduced an uncertainty in the measured values of those quantities. This illustrates the essential point of the uncertainty principle\u2014that this product of uncertainties cannot be less than about h in principle, that is, even in an ideal situation. If electrons rather than photons were used to locate the object, the analysis would not change since the relation is the same for both. h>p 5-6 Some Consequences of the Uncertainty Principle In the next chapter we will see that the Schr\u00f6dinger wave equation provides a straightforward method of solving problems in atomic physics. However, the solution of the Schr\u00f6dinger equation is often laborious and difficult. Much semiquantitative information about the behavior of atomic systems can be obtained from the uncertainty principle alone without a detailed solution of the problem. The general approach used in applying the uncertainty principle to such systems will first be illustrated by considering a particle moving in a box with rigid walls. We then use that analysis in several numerical examples and as a basis for discussing some additional consequences. Minimum Energy of a Particle in a Box An important consequence of the uncertainty principle is that a particle confined to a finite space cannot have zero kinetic energy. Let us consider the case of a one-dimensional \u201cbox\u201d of length L. If we know that the particle is in the box, x is not larger than L. This implies that p is at least (Since we are interested in orders of magnitude, we 1>2 in the minimum uncertainty product. In general, distributions are not will ignore the Gaussian anyway, so px will be larger than", " ) Let us take the standard deviation as a measure of p, U>L. U>2. (\u00a2p)2 (p p)2 av (p2 2pp p 2)av p2 p 2 If the box is symmetric, will be zero since the particle moves to the left as often as to the right. Then p (\u00a2p)2 p2! a 2 b U L 5-6 Some Consequences of the Uncertainty Principle 209 and the average kinetic energy is E p2 2m! U2 2mL2 5-28 Thus, we see that the uncertainty principle indicates that the minimum energy of a particle (any particle) in a \u201cbox\u201d (any kind of \u201cbox\u201d) cannot be zero. This minimum energy given by Equation 5-28 for a particle in a one-dimensional box is called the zero point energy. EXAMPLE 5-6 A Macroscopic Particle in a Box Consider a small but macroscopic particle of mass m 106 g confined to a one-dimensional box with L 106 m, e.g., a tiny bead on a very short wire. Compute the bead\u2019s minimum kinetic energy and the corresponding speed. SOLUTION 1. The minimum kinetic energy is given by Equation 5-28: (1.055 1034 J # s)2 (2)(109 kg)(106 m)2 2mL2 E U2 5.57 1048 J 3.47 1029 eV 2. The speed corresponding to this kinetic energy is: v 2E m A 2(5.57 1048 J) 109 kg C 1.06 1019 m>s Remarks: We can see from this calculation that the minimum kinetic energy implied by the uncertainty principle is certainly not observable for macroscopic objects even as small as 106 g. EXAMPLE 5-7 An Electron in an Atomic Box If the particle in a one-dimensional box of length L 0.1 nm (about the diameter of an atom) is an electron, what will be its zero-point energy? SOLUTION Again using Equation 5-28, we find that (Uc)2 2mc2L2 E (197.3 eV # nm)2 2(0.511 106 eV)(0.1 nm)2 3.81 eV This is the correct order of magnitude for the kinetic energy of an electron in an atom. Size of the Hydrogen Atom The energy of an electron of momentum p a distance r from a proton is E p2 2m ke2 r 210 Chapter 5 The Wavelike Properties of Particles If we take for the order of magnitude of the position uncertainty x r, we have The energy is then (\u00a2p)2 p2! U2 r2 2mr2 There is a radius rm at which E is minimum. Setting dE dr 0 yields rm and Em: > U2 E ke2 r and rm U2 ke2m a0 0.0529 nm Em k2e4m 2U2 13.6 eV The fact that rm came out to be exactly the radius of the first Bohr orbit is due to the judicious choice of x r rather than 2r or r 2, which are just as reasonable. It should be clear, however, that any reasonable choice for x gives the correct order of magnitude of the size of an atom. > Widths of Spectral Lines Equation 5-27 implies that the energy of a system cannot be measured exactly unless an infinite amount of time is available for the measurement. If an atom is in an excited state, it does not remain in that state indefinitely but makes transitions to lower energy states until it reaches the ground state. The decay of an excited state is a statistical process. We can take the mean time for decay, the time available to determine the energy of the state. For atomic transitions, order of 108 s. The uncertainty in the energy corresponding to this time is called the lifetime, to be a measure of is of the 6.58 1016 eV # s \u00a2E! U 107 eV 108 s \u00a2 This uncertainty in energy causes a spread in the wavelength of the light emitted. For transitions to the ground state, which has a perfectly certain energy E0 because of its infinite lifetime, the percentage spread in wavelength can be calculated from E E0 dE hc hc d 2 \u0192\u00a2E\u0192 hc \u0192\u00a2\u0192 2 \u00a2 \u00a2E E E0 thus, 5-6 Some Consequences of the Uncertainty Principle 211 % \u00a2E U> % 0, is called the natural line width and is represented by 0. The energy width Other effects that cause broadening of spectral lines are the Doppler effect, the recoil of the emitting atom, and atomic collisions. For optical spectra in the eV energy range, the Doppler width D is about 106 eV at", " room temperature, i.e., roughly 10 times the natural width and the recoil width is negligible. For nuclear transitions in the MeV range, both the Doppler width and the recoil width are of the order of eV, much larger than the natural line width. We will see in Chapter 11 that in some special cases of atoms in solids at low temperatures, the Doppler and recoil widths are essentially zero and the width of the spectral line is just the natural width. This effect, called the M\u00f6ssbauer effect after its discoverer, is extremely important since it provides photons of well-defined energy that are useful in experiments demanding extreme precision. For example, the 14.4-keV photon from 57Fe has a natural width of the order of 1011 of its energy. Questions 7. What happens to the zero-point energy of a particle in a one-dimensional box as the length of the box L S? 8. Why is the uncertainty principle not apparent for macroscopic objects? EXAMPLE 5-8 Emission of a Photon Most excited atomic states decay, i.e., emit a photon, within about following excitation. What is the minimum uncertainty in the (1) energy and (2) frequency of the emitted photon? 10 8 s SOLUTION 1. The minimum energy uncertainty is the natural line width \u2260 0 U>; therefore, 6.63 1034 J # s 4.14 1015 eV # s 2 10 8 s 2 10 8 s \u2260 0 6.6 108 eV 2. From de Broglie\u2019s relation E U we have \u00a2E U\u00a2 U(2\u00a2f) h\u00a2f so that Equation 5-27 can be written as \u00a2E\u00a2t h\u00a2f\u00a2t! U and the minimum uncertainty in the frequency becomes \u00a2f! 1 2\u00a2t \u00a2f! 1.6 107 Hz 1 2 10 8 Remarks: The frequency of photons in the visible region of the spectrum is of the order of 1014 Hz. 212 Chapter 5 The Wavelike Properties of Particles 5-7 Wave-Particle Duality We have seen that electrons, which were once thought of as simply particles, exhibit the wave properties of diffraction and interference. In earlier chapters we saw that light, which we previously had thought of as a wave, also has particle properties in its interaction with matter, as in the photoelectric effect or the Compton effect. All phenomena\u2014electrons, atoms, light, sound\u2014have both particle and wave characteristics. It is sometimes said that an electron, for example, behaves both as a wave and a particle. This may seem confusing since, in classical physics, the concepts of waves and particles are mutually exclusive. A classical particle behaves like a pellet or BB shot from an air-powered rifle. It can be localized and scattered, it exchanges energy suddenly in a lump, and it obeys the laws of conservation of energy and momentum in collisions, but it does not exhibit interference and diffraction. A classical wave behaves like a water wave. It exhibits diffraction and interference patterns and has its energy spread out continuously in space and time, not quantized in lumps. Nothing, it was thought, could be both a classical particle and a classical wave. We now see that the classical concepts do not adequately describe either waves or particles. Both matter and radiation have both particle and wave aspects. When emission and absorption are being studied, it is the particle aspects that are dominant. When matter and radiation propagate through space, wave aspects dominate. Notice that emission and absorption are events characterized by exchange of energy and discrete locations. For example, light strikes the retina of your eye and a photon is absorbed, transferring its energy to a particular rod or cone: an observation has occurred. This illustrates the point that observations of matter and radiation are described in terms of the particle aspects. On the other hand, predicting the intensity distribution of the light on your retina involves consideration of the amplitudes of waves that have propagated through space and been diffracted at the pupil. Thus, predictions, i.e., a priori statements about what may be observed, are described in terms of the wave aspects. Let\u2019s elaborate on this just a bit. \u00b0 e(x, t) \u00b0(x, t). Every phenomenon is describable by a wave function that is the solution of a wave equation. The wave function for light is the electric field (in one space dimension), which is the solution of a wave equation like Equation 5-11. We have We will study the wave equation of called the wave function for an electron is the solution, called the Schr\u00f6dinger equation, in the next chapter. The which magnitude squared of the wave function gives the probability per unit volume that the electron, if looked for, will be found in a given volume or region. The wave function exhibits the classical wave properties of interference and diffraction. In order to predict where an electron, or", " other particle, is likely to be, we must find the wave function by methods similar to those of classical wave theory. When the electron (or light) interacts and exchanges energy and momentum, the wave function is changed by the interaction. The interaction can be described by classical particle theory, as is done in the Compton effect. There are times when classical particle theory and classical wave theory give the same results. If the wavelength is much smaller than any object or aperture, particle theory can be used as well as wave theory to describe wave propagation because diffraction and interference effects are too small to be observed. Common examples are geometrical optics, which is really a particle theory, and the motion of baseballs and jet aircraft. If one is interested only in time averages of energy and momentum exchange, the wave theory works as well as the particle theory. For example, the wave theory of light correctly predicts that the total electron current in the photoelectric effect is proportional to the intensity of the light. Summary 213 More That matter can exhibit wavelike characteristics as well as particlelike behavior can be a difficult concept to understand. A wonderfully clear discussion of wave-particle duality was given by R. P. Feynman, and we have used it as the basis of our explanation on the home page of the TwoSlit Interference Pattern for electrons: whfreeman.com/tiplermodern physics5e/. See also Figures 5-21 and 5-22 and Equation 5-29 here. Summary TOPIC 1. De Broglie relations 2. Detecting electron waves Davisson and Germer 3. Wave packets Wave equation Uncertainty relations Wave speed Group (packet) speed Matter waves RELEVANT EQUATIONS AND REMARKS f E>h h> p Electrons and all other particles exhibit the wave properties of interference and diffraction Showed that electron waves diffracted from a single Ni crystal according to Bragg\u2019s equation n D sin # d2y dx2 1 v2 d2y dt 2 \u00a2k\u00a2x 1 \u00a2\u00a2t 1 f >k vp vg d dk vp k dvp dk The wave packet moves with the particle speed; i.e., the particle speed is the group speed vg. 5-1 5-2 5-5 5-11 5-17 5-18 5-16 4. Probabilistic interpretation The magnitude square of the wave function is proportional to the probability of observing a particle in the region dx at x and t. P(x)dx \u0192\u00b0\u0192 2dx 5-23 5. Heisenberg uncertainty principle \u00a2x\u00a2p! 1 2 U \u00a2E\u00a2t! 1 2 U 5-26 5-27 where each of the uncertainties is defined to be the standard deviation. 214 Chapter 5 The Wavelike Properties of Particles TOPIC RELEVANT EQUATIONS AND REMARKS Particle in a box Energy of H atom E! U2 2mL2 5-28 The minimum energy of any particle in any \u201cbox\u201d cannot be zero. The Heisenberg principle predicts the Bohr model. Emin 13.6 eV in agreement with General References The following general references are written at a level appropriate for the readers of this book. De Broglie, L., Matter and Light: The New Physics, Dover, New York, 1939. In this collection of studies is de Broglie\u2019s lecture on the occasion of receiving the Nobel Prize, in which he describes his reasoning that led to the prediction of the wave nature of matter. Feynman, R., \u201cProbability and Uncertainty\u2014The QuantumMechanical View of Nature,\u201d filmed lecture, available from Educational Services, Inc., Film Library, Newton, MA. Feynman, R. P., R. B. Leighton, and M. Sands, Lectures on Physics, Addison-Wesley, Reading, MA, 1965. Notes 1. Louis V. P. R. de Broglie (1892\u20131987), French physicist. Originally trained in history, he became interested in science after serving as a radio engineer in the French army (assigned to the Eiffel Tower) and through the work of his physicist brother Maurice. The subject of his doctoral dissertation received unusual attention because his professor, Paul Langevin (who discovered the principle on which sonar is based), brought it to the attention of Einstein, who described de Broglie\u2019s hypothesis to Lorentz as \u201cthe first feeble ray of light to illuminate the worst of our physical riddles.\u201d He received the Nobel Prize in Physics in 1929, the first person so honored for work done for a doctoral thesis. \u00c1 2. L. de Broglie, New Perspectives in Physics, Basic Books, New York, 1962. 3. See, e.g., Tipler, Physics for Scientists and Engineers, 5th ed. (New", " York: W. H. Freeman and Co., 2008), Section 35-5. 4. Jean-Baptiste Perrin (1870\u20131942), French physicist. He was the first to show that cathode rays are actually charged particles, setting the stage for J. J. Thomson\u2019s measurement of their > q m ratio. He was also the first to measure the approximate size of atoms and molecules and determined Avogadro\u2019s number. He received the Nobel Prize in Physics for that work in 1926. Fowles, G. R., Introduction to Modern Optics, Holt, Rinehart & Winston, New York, 1968. Hecht, E., Optics, 2d ed., Addison-Wesley, Reading, MA, 1987. Jenkins, F. A., and H. E. White, Fundamentals of Optics, 4th ed., McGraw-Hill, New York, 1976. Mehra, J., and H. Rechenberg, The Historical Development of Quantum Theory, Vol. 1, Springer-Verlag, New York, 1982. Resnick, R., and D. Halliday, Basic Concepts in Relativity and Early Quantum Theory, 2d ed., Wiley, New York, 1992. Tipler, P. A., and G. Mosca, Physics for Scientists and Engineers, 6th ed., W. H. Freeman and Co., New York, 2008. Chapters 15 and 16 include a complete discussion of classical waves. 5. Clinton J. Davisson (1881\u20131958), American physicist. He shared the 1937 Nobel Prize in Physics with G. P. Thomson for demonstrating the diffraction of particles. Davisson\u2019s Nobel Prize was the first ever awarded for work done somewhere other than at an academic institution. Germer was one of Davisson\u2019s assistants at Bell Telephone Laboratory. 6. Matter (electron) waves, like other waves, change their direction in passing from one medium (e.g., Ni crystal) into another (e.g., vacuum) in the manner described by Snell\u2019s law and the indices of refraction of the two media. For normal incidence Equation 5-5 is not affected, but for other incident angles it is altered a bit, and that change has not been taken into account in either Figure 5-6 or 5-7. 7. Nobel Prize Lectures: Physics (Amsterdam and New York: Elsevier, 1964). 8. In spectroscopy, the quantity is called the wave number. In the theory of waves, the term wave number is used for k 2>. k 1 9. Following convention, the \u201cwidth\u201d is defined as the full width of the pulse or envelope measured at half the maximum amplitude. Problems 215 \u0192\u00b0\u0192 2. \u0192\u00b0\u0192 2 10. This interpretation of was first developed by German physicist Max Born (1882\u20131970). One of his positions early in his career was at the University of Berlin, where he was to relieve Planck of his teaching duties. Born received the Nobel Prize in Physics in 1954, in part for his interpretation of 11. Werner K. Heisenberg (1901\u20131976), German physicist. After obtaining his Ph.D. under Sommerfeld, he served as an assistant to Born and to Bohr. He was the director of research for Germany\u2019s atomic bomb project during World War II. His work on quantum theory earned him the Nobel Prize in Physics in 1932. 12. The resolving power of a microscope is discussed in some detail in Jenkins and White, Fundamentals of Optics, 4th ed. (New York: McGraw-Hill, 1976), pp. 332\u2013334. The expression for x used here is determined by Rayleigh\u2019s criterion, which states that two points are just resolved if the central maximum of the diffraction pattern from one falls at the first minimum of the diffraction pattern of the other. 13. Richard P. Feynman (1918\u20131988), American physicist. This discussion is based upon one in his classic text Lectures on Physics (Reading, MA: Addison-Wesley, 1965). He shared the 1965 Nobel Prize in Physics for his development of quantum electrodynamics (QED). It was Feynman who, while a member of the commission on the Challenger disaster, pointed out that the booster stage O-rings were at fault. A genuine legend in American physics, he was also an accomplished bongo drummer and safecracker. Problems Level I Section 5-1 The de Broglie Hypothesis 3kT>2, E pc (a) What is the de Broglie wavelength of a 1-g mass moving at a speed of 1 m per year? If the kinetic energy of a particle is much greater than its rest energy, the relativistic apholds. Use this approximation to find the de Broglie wavelength of an 5-", "1. (b) What should be the speed of such a mass if its de Broglie wavelength is to be 1 cm? 5-2. proximation electron of energy 100 MeV. 5-3. Electrons in an electron microscope are accelerated from rest through a potential difference V0 so that their de Broglie wavelength is 0.04 nm. What is V0? 5-4. Compute the de Broglie wavelengths of (a) an electron, (b) a proton, and (c) an alpha particle of 4.5-keV kinetic energy. 5-5. According to statistical mechanics, the average kinetic energy of a particle at temperature T is where k is the Boltzmann constant. What is the average de Broglie wavelength of nitrogen molecules at room temperature? 5-6. Find the de Broglie wavelength of a neutron of kinetic energy 0.02 eV (this is of the order of magnitude of kT at room temperature). 5-7. A free proton moves back and forth between rigid walls separated by a distance L 0.01 nm. (a) If the proton is represented by a one-dimensional standing de Broglie wave with a node at 2L> n each wall, show that the allowed values of the de Broglie wavelength are given by where n is a positive integer. (b) Derive a general expression for the allowed kinetic energy of the proton and compute the values for n 1 and 2. 5-8. What must be the kinetic energy of an electron if the ratio of its de Broglie wavelength to its Compton wavelength is (a) 102, (b) 0.2, and (c) 103? 5-9. Compute the wavelength of a cosmic-ray proton whose kinetic energy is (a) 2 GeV and (b) 200 GeV. Section 5-2 Measurements of Particle Wavelengths 5-10. What is the Bragg scattering angle energy is (a) 75 eV, (b) 100 eV? 5-11. Compute the kinetic energy of a proton whose de Broglie wavelength is 0.25 nm. If a beam of such protons is reflected from a calcite crystal with crystal plane spacing of 0.304 nm, at what angle will the first-order Bragg maximum occur? for electrons scattered from a nickel crystal if their # 216 Chapter 5 The Wavelike Properties of Particles 5-12. (a) The scattering angle for 50-eV electrons from MgO is 55.6\u00b0. What is the crystal spacing D? (b) What would be the scattering angle for 100-eV electrons? 5-13. A certain crystal has a set of planes spaced 0.30 nm apart. A beam of neutrons strikes the # 42\u00b0. crystal at normal incidence and the first maximum of the diffraction pattern occurs at What are the de Broglie wavelength and kinetic energy of the neutrons? 5-14. Show that in Davisson and Germer\u2019s experiment with 54-eV electrons using the D 0.215 nm planes, diffraction peaks with n 2 and higher are not possible. 5-15. A beam of electrons with kinetic energy 350 eV is incident normal to the surface of a KCl crystal that has been cut so that the spacing D between adjacent atoms in the planes parallel to the surface is 0.315 nm. Calculate the angle at which diffraction peaks will occur for all orders possible. # Section 5-3 Wave Packets 5-16. Information is transmitted along a cable in the form of short electric pulses at 100,000 pulses/s. (a) What is the longest duration of the pulses such that they do not overlap? (b) What is the range of frequencies to which the receiving equipment must respond for this duration? 5-17. Two harmonic waves travel simultaneously along a long wire. Their wave functions are 0.002 cos (7.6x 380t), where y and x are in meters and y1 t in seconds. (a) Write the wave function for the resultant wave in the form of Equation 5-15. (b) What is the phase velocity of the resultant wave? (c) What is the group velocity? (d) Calculate the range x between successive zeros of the group and relate it to k. 5-18. (a) Starting from Equation 5-16, show that the group velocity can also be expressed as 0.002 cos (8.0x 400t) and y2 vg vp (dvp >d) s dvp >d in glass? Is positive or negative? (b) The phase velocity of each wavelength of white light moving through ordinary glass is a funcvp tion of the wavelength; i.e., glass is a dispersive medium. What is the general dependence of on 5-19. A radar transmitter", " used to measure the speed of pitched baseballs emits pulses of 2.0-cm in duration. (a) What is the length of the wave packet produced? wavelength that are 0.25 (b) To what frequency should the receiver be tuned? (c) What must be the minimum bandwidth of the receiver? 5-20. A certain standard tuning fork vibrates at 880 Hz. If the tuning fork is tapped, causing it to vibrate, then stopped a quarter of a second later, what is the approximate range of frequencies contained in the sound pulse that reached your ear? 5-21. If a phone line is capable of transmitting a range of frequencies f 5000 Hz, what is the approximate duration of the shortest pulse that can be transmitted over the line? 5-22. (a) You are given the task of constructing a double slit experiment for 5-eV electrons. If you wish the first minimum of the diffraction pattern to occur at 5\u00b0, what must be the separation of the slits? (b) How far from the slits must the detector plane be located if the first minima on each side of the central maximum are to be separated by 1 cm? Section 5-4 The Probabilistic Interpretation of the Wave Function 5-23. A 100-g rigid sphere of radius 1 cm has a kinetic energy of 2 J and is confined to move in a force-free region between two rigid walls separated by 50 cm. (a) What is the probability of finding the center of the sphere exactly midway between the two walls? (b) What is the probability of finding the center of the sphere between the 24.9- and 25.1-cm marks? 5-24. A particle moving in one dimension between rigid walls separated by a distance L has the wave function Since the particle must remain between the walls, what must be the value of A? \u00b0(x) A sin (x>L). 5-25. The wave function describing a state of an electron confined to move along the x axis is given at time zero by \u00b0(x, 0) Aex2>42 Problems 217 Find the probability of finding the electron in a region dx centered at (a) x 0, (b) (c) (d) Where is the electron most likely to be found? x 2. x, and Section 5-5 The Uncertainty Principle f0 vibrates for a time t and sends out a waveform that looks 5-26. A tuning fork of frequency like that in Figure 5-23. This wave function is similar to a harmonic wave except that it is confined to a time t and space x vt, where v is the phase velocity. Let N be the approximate number of cycles of vibration. We can measure the frequency by counting the cycles and dividing by t. (a) The number of cycles is uncertain by approximately cycle. Explain why (see the figure). What uncertainty does this introduce in the determination of the frequency f? (b) Write an expression for the wave number k in terms of x and N. Show that the uncertainty in N of leads to an uncertainty in k of \u00a2k 2>\u00a2x. 1 1 y \u0394t t Figure 5-23 5-27. If an excited state of an atom is known to have a lifetime of 107 s, what is the uncertainty in the energy of photons emitted by such atoms in the spontaneous decay to the ground state? 5-28. A ladybug 5 mm in diameter with a mass of 1.0 mg being viewed through a low-power magnifier with a calibrated reticule is observed to be stationary with an uncertainty of 102 mm. How fast might the ladybug actually be walking? 5-29. 222Rn decays by the emission of an particle with a lifetime of 3.823 days. The kinetic energy of the particle is measured to be 5.490 MeV. What is the uncertainty in this energy? Describe in one sentence how the finite lifetime of the excited state of the radon nucleus trans- lates into an energy uncertainty for the emitted particle. 5-30. If the uncertainty in the position of a wave packet representing the state of a quantumsystem particle is equal to its de Broglie wavelength, how does the uncertainty in momentum compare with the value of the momentum of the particle? 5-31. In one of G. Gamow\u2019s Mr. Tompkins tales, the hero visits a \u201cquantum jungle\u201d where h is A cheetah runs past you a very large. Suppose that you are in such a place where few meters away. The cheetah is 2 m long from nose to tail tip and its mass is 30 kg. It is moving at 40 m s. What is the uncertainty in the location of the \u201cmidpoint\u201d of the cheetah? Describe in one sentence how the cheetah would look different to you than when h has its actual value. 5-32", ". In order to locate a particle, e.g., an electron, to within 5 1012 m using electromagnetic waves (\u201clight\u201d), the wavelength must be at least this small. Calculate the momentum and energy of a photon with If the particle is an electron with \u00a2x 5 1012 m, 5-33. The decay of excited states in atoms and nuclei often leave the system in another, albeit lower-energy, excited state. (a) One example is the decay between two excited states of the nucleus of 48Ti. The upper state has a lifetime of 1.4 ps, the lower state 3.0 ps. What is the fracin the energy of 1.3117-MeV gamma rays connecting the two states? tional uncertainty what is the corresponding uncertainty in its momentum? h 50 J # s. 5 1012 m. \u00a2E>E > 218 Chapter 5 The Wavelike Properties of Particles H (a) Another example is the line of the hydrogen Balmer series. In this case the lifetime of both states is about the same, 108 s. What is the uncertainty in the energy of the photon? 5-34. Laser pulses of femtosecond duration can be produced but for such brief pulses it makes no sense to speak of the pulse\u2019s color. To see this, compute the time duration of a laser pulse whose range of frequencies covers the entire visible spectrum (4.0 1014 Hz to 7.5 1014 Hz). H Section 5-6 Some Consequences of the Uncertainty Principle 250 MeV 1200 2000 1600 Energy (MeV) 2400 Figure 5-24 m where 135 MeV>c2. 5-35. A neutron has a kinetic energy of 10 MeV. What size object is necessary to observe neutron diffraction effects? Is there anything in nature of this size that could serve as a target to demonstrate the wave nature of 10-MeV neutrons? 5-36. Protons and neutrons in nuclei are bound to the nucleus by exchanging pions ( mesons) with each other (see Chapter 11). This is possible to do without violating energy conservation provided the pion is reabsorbed within a time consistent with the Heisenberg uncertainty relap S p (a) Ignoring tions. Consider the emission reaction kinetic energy, by how much is energy conservation violated in this reaction? (b) Within what time interval must the pion be reabsorbed in order to avoid violation of energy conservation? \u00a2s U 5-37. Show that the relation for a particle moving in a is the linear momentum tangential to the circle, s is the arc circle about the z axis, where length, and L is the angular momentum. How well can the angular position of the electron be specified in the Bohr atom? 5-38. An excited state of a certain nucleus has a half-life of 0.85 ns. Taking this to be the uncertainty t for emission of a photon, calculate the uncertainty in the frequency using Equation 5-25. If 5-39. The lifetimes of so-called resonance particles cannot be measured directly but is computed from the energy width (or uncertainty) of the scattering cross section versus energy graph (see Chapter 12). For example, the scattering of a pion ( meson) and a proton can produce a short-lived resonance particle with a mass of 1685 MeV c2 and an energy width of 250 MeV as shown in Figure 5-24: Compute the lifetime of the 0.01 nm, p S \u00a2. can be written \u00a2L\u00a2# U \u00a2ps ps \u00a2f>f. > find \u00a2f, \u00a2. Section 5-7 Wave-Particle Duality There are no problems for this section. Level II 5-40. Neutrons and protons in atomic nuclei are confined within a region whose diameter is about 1015 m. (a) At any given instant, how fast might an individual proton or neutron be moving? (b) What is the approximate kinetic energy of a neutron that is localized to within such a region? (c) What would be the corresponding energy of an electron localized to within such a region? 5-41. Using the relativistic expression E2 p2c2 m2c4, (a) show that the phase velocity of an electron wave is greater than c. (b) Show that the group velocity of an electron wave equals the particle velocity of the electron. 5-42. Show that if y1 and y2 are solutions of Equation 5-11, the function y3 also a solution for any values of the constants C1 and C2. 5-43. The London \u201cBobby\u201d whistle has a frequency of 2500 Hz. If such a whistle is given a 3.0-s blast, (a) what is the uncertainty in the frequency? (b) How long is", " the wave train of this blast? (c) What would be the uncertainty in measuring the wavelength of this blast? (d) What is the wavelength of this blast? 5-44. A particle of mass m moves in a one-dimensional box of length L. (Take the potential energy of the particle in the box to be zero so that its total energy is its kinetic energy C2y2 is C1y1 Problems 219 > n(>2) L, > where f \u00a2E>h n2E1 where E1 p2 2m.) Its energy is quantized by the standing-wave condition is the de Broglie wavelength of the particle and n is an integer. (a) Show that the allowed energies h2 8mL2. (b) Evaluate En for an electron in a box of size are given by En L 0.1 nm and make an energy-level diagram for the state from n 1 to n 5. Use Bohr\u2019s to calculate the wavelength of electromagnetic radiation emitted second postulate when the electron makes a transition from (c) n 2 to n 1, (d) n 3 to n 2, and (e) n 5 to n 1. 5-45. (a) Use the results of Problem 5-44 to find the energy of the ground state (n 1) and the first two excited states of a proton in a one-dimensional box of length L 1015 m 1 fm. (These are of the order of magnitude of nuclear energies.) Calculate the wavelength of electromagnetic radiation emitted when the proton makes a transition from (b) n 2 to n 1, (c) n 3 to n 2, and (d) n 3 to n 1. 5-46. (a) Suppose that a particle of mass m is constrained to move in a one-dimensional space between two infinitely high barriers located A apart. Using the uncertainty principle, find an expression for the zero-point (minimum) energy of the particle. (b) Using your result from (a), compute the minimum energy of an electron in such a space if A 1010 m and A 1 cm. (c) Calculate the minimum energy for a 100-mg bead moving on a thin wire between two stops located 2 cm apart. 5-47. A proton and a 10-g bullet each move with a speed of 500 m s, measured with an uncertainty of 0.01 percent. If measurements of their respective positions are made simultaneous with the speed measurements, what is the minimum uncertainty possible in the position measurements? > Level III 5-48. Show that Equation 5-11 is satisfied by y f(#), where # x vt for any function f. > 5-49. An electron and a positron are moving toward each other with equal speeds of 3 106 m s. The two particles annihilate each other and produce two photons of equal energy. (a) What were the de Broglie wavelengths of the electron and positron? Find the (b) energy, (c) momentum, and (d) wavelength of each photon. > p n where the n represents a neutron. The 5-50. It is possible for some fundamental particles to \u201cviolate\u201d conservation of energy by creating and quickly reabsorbing another particle. For example, a proton can emit a according has a mass of 140 MeV c2. The to reabsorption must occur within a time t consistent with the uncertainty principle. (a) Considering the example shown, by how much E is energy conservation violated? (Ignore kinetic energy.) (b) For how long t can the is moving at nearly the speed of light, how far from the nucleus could it get in the time t? (As we will discuss in Chapter 11, this is the approximate range of the strong nuclear force.) (d) Assuming that as soon as one pion is reabsorbed another is emitted, how many pions would be recorded by a s? \u201cnucleon camera\u201d with a shutter speed of 1 5-51. De Broglie developed Equation 5-2 initially for photons, assuming that they had a small 30 m but finite mass. His assumptions was that RF waves with traveled at a speed of at 500 nm. Beginning with the relativistic exleast 99 percent of that of visible light with verify de Broglie\u2019s calculation that the upper limit of the rest mass of a pression photon is 1044 g. (Hint: Find an expression for v c in terms of hf and mc2, and then let mc2 V hf exist? (c) Assuming that the 1> 21 v2>c2). hf mc2, ) ( > 5-52. Suppose that you drop BBs onto a bull\u2019s-eye marked on the floor. According to the uncertainty principle, the BBs do not necessarily fall straight down from the release point to", " the center of the bull\u2019s-eye but are affected by the initial conditions. (a) If the location of the release point is uncertain by an amount x perpendicular to the vertical direction and the horizontal component of the speed is uncertain by vx, derive an expression for the minimum spread X 220 Chapter 5 The Wavelike Properties of Particles of impacts at the bull\u2019s-eye if it is located a distance y0 below the release point. (b) Modify your result in (a) to include the effect on X of uncertainties y and vy at the release point. 5-53. Using the first-order Doppler-shift formula calculate the energy shift of a 1-eV photon emitted from an iron atom moving toward you with energy (3 2)kT at T 300 K. Compare this Doppler line broadening with the natural line width calculated in Example 5-8. Repeat the calculation for a 1-MeV photon from a nuclear transition. 5-54. Calculate the order of magnitude of the shift in energy of a (a) 1-eV photon and (b) 1-MeV photon resulting from the recoil of an iron nucleus. Do this by first calculating the momentum of the photon and then by calculating p2 2m for the nucleus using that value of momentum. Compare with the natural line width calculated in Example 5-8. f f 0(1 v>c), > > CHAPTER 6 The Schr\u00f6dinger Equation The success of the de Broglie relations in predicting the diffraction of electrons and other particles, and the realization that classical standing waves lead to a discrete set of frequencies, prompted a search for a wave theory of electrons analogous to the wave theory of light. In this electron wave theory, classical mechanics should appear as the short-wavelength limit, just as geometric optics is the short-wavelength limit of the wave theory of light. The genesis of the correct theory went something like this, according to Felix Bloch,1 who was present at the time:... in one of the next colloquia, Schr\u00f6dinger gave a beautifully clear account of how de Broglie associated a wave with a particle and how he [i.e., de Broglie] could obtain the quantization rules... by demanding that an integer number of waves should be fitted along a stationary orbit. When he had finished Debye2 casually remarked that he thought this way of talking was rather childish... [that to] deal properly with waves, one had to have a wave equation. In 1926, Erwin Schr\u00f6dinger3 published his now-famous wave equation, which governs the propagation of matter waves, including those of electrons. A few months earlier, Werner Heisenberg had published a seemingly different theory to explain atomic phenomena. In the Heisenberg theory, only measurable quantities appear. Dynamical quantities such as energy, position, and momentum are represented by matrices, the diagonal elements of which are the possible results of measurement. Though the Schr\u00f6dinger and Heisenberg theories appear to be different, it was eventually shown by Schr\u00f6dinger himself that they were equivalent, in that each could be derived from the other. The resulting theory, now called wave mechanics or quantum mechanics, has been amazingly successful. Though its principles may seem strange to us, whose experiences are limited to the macroscopic world, and though the mathematics required to solve even the simplest problem is quite involved, there seems to be no alternative to describe correctly the experimental results in atomic and nuclear physics. In this book we will confine our study to the Schr\u00f6dinger theory because it is easier to learn and is a little less abstract than the Heisenberg theory. We will begin by restricting our discussion to problems in one space dimension. 6-1 The Schr\u00f6dinger Equation in One Dimension 6-2 The Infinite Square Well 6-3 The Finite 222 229 Square Well 238 6-4 Expectation Values and Operators 6-5 The Simple Harmonic Oscillator 6-6 Reflection and Transmission of Waves 242 246 250 221 222 Chapter 6 The Schr\u00f6dinger Equation 6-1 The Schr\u00f6dinger Equation in One Dimension The wave equation governing the motion of electrons and other particles with mass, which is analogous to the classical wave equation (Equation 5-11), was found by Schr\u00f6dinger late in 1925 and is now known as the Schr\u00f6dinger equation. Like the classical wave equation, the Schr\u00f6dinger equation relates the time and space derivatives of the wave function. The reasoning followed by Schr\u00f6dinger is somewhat difficult and not important for our purposes. In any case, it must be emphasized that we can\u2019t derive the Schr\u00f6dinger equation just as we can\u2019t derive Newton\u2019s laws of motion. Its validity, like that of any fundamental equation, lies in its agreement with experiment. Just as Newton\u2019s second law", " is not relativistically correct, neither is Schr\u00f6dinger\u2019s equation, which must ultimately yield to a relativistic wave equation. But, as you know, Newton\u2019s laws of motion are perfectly satisfactory for solving a vast array of nonrelativistic problems. So, too, will be Schr\u00f6dinger\u2019s equation when applied to the equally extensive range of nonrelativistic problems in atomic, molecular, and solidstate physics. Schr\u00f6dinger tried without success to develop a relativistic wave equation, a task accomplished in 1928 by Dirac. Although it would be logical merely to postulate the Schr\u00f6dinger equation, we can get some idea of what to expect by first considering the wave equation for photons, which is Equation 5-11 with speed v c and with y(x, t) replaced by the electric field (x, t). e $2e $x2 1 c2 $2e $t2 6-1 As discussed in Chapter 5, a particularly important solution of this equation is the har0 cos (kx t). monic wave function Differentiating this function twice, we obtain e(x, t) e $2e $t2 2e 0 cos (kx t) 2e(x, t) and $2e $x2 k2e (x, t) Substitution into Equation 6-1 then gives or k2 2 c2 kc Using E> U and p Uk for electromagnetic radiation, we have E pc 6-2 6-3 which, as we saw earlier, is the relation between the energy and momentum of a photon. 6-1 The Schr\u00f6dinger Equation in One Dimension 223 Erwin Schr\u00f6dinger. [Courtesy of the Niels Bohr Library, American Institute of Physics.] Now let us use the de Broglie relations for a particle such as an electron to find the and k that is analogous to Equation 6-2 for photons. We can then relation between use this relation to work backward and see how the wave equation for electrons must differ from Equation 6-1. The total energy (nonrelativistic) of a particle of mass m is E p2 2m V 6-4 where V is the potential energy. Substituting the de Broglie relations (Equations 5-21 and 5-22) in Equation 6-4, we obtain U U2k2 2m V 6-5 This differs from Equation 6-2 for a photon because it contains the potential energy V and because the angular frequency does not vary linearly with k. Note that we get a factor of when we differentiate a harmonic wave function with respect to time and a factor of k when we differentiate with respect to position. We expect, therefore, that the wave equation that applies to electrons will relate the first time derivative to the second space derivative and will also involve the potential energy of the electron. & & 2(x, t) 1(x, t) a2 This ensures that, if Finally, we require that the wave equation for electrons will be a differential equa&(x, t), and tion that is linear in the wave function & are both solutions of the wave equation for the same potential energy, then any arbitrary linear combination of these solutions is also a solution\u2014i.e., & &(x, t) a1 is a solution, with a1 and a2 being arbitrary constants. Such a combination is called linear because both appear only to the first power. Linearity guarantees that the wave functions will add together to produce constructive and destructive interference, which we have seen to be a characteristic of matter waves, as well as all other wave phenomena. Note, in particular, that (1) the linearity requirement means that every term in the wave equation must be linear in and (2) that any derivative of is linear in &(x, t). &(x, t) &(x, t) 1(x, t) 2(x, t) 2(x, t) 1(x, t) and & & 4 224 Chapter 6 The Schr\u00f6dinger Equation The Schr\u00f6dinger Equation We are now ready to postulate the Schr\u00f6dinger equation for a particle of mass m. In one dimension, it has the form U2 2m $2&(x, t) $x2 V(x, t)&(x, t) iU $&(x, t) $t 6-6 We will now show that this equation is satisfied by a harmonic wave function in the special case of a free particle, one on which no net force acts, so that the potential cos(kx t) energy is constant, V(x, t) V0. First note that a function of the form does not satisfy this equation because differentiation with respect to time changes the cosine to a sine, but the second derivative with", " respect to x gives back a cosine. Similar reasoning rules out the form However, the exponential form of the harmonic wave function does satisfy the equation. Let sin(kx t). &(x, t) Aei(kxt) 6-7 A[cos(kx t) i sin(kx t)] where A is a constant. Then and $& $t $2& $x2 iAei(kxt) i& (ik)2Aei(kxt) k2& Substituting these derivatives into the Schr\u00f6dinger equation with V(x, t) V0 gives or U2 2m (k2&) V0 & iU(i)& U2k2 2m V0 U which is Equation 6-5. An important difference between the Schr\u00f6dinger equation and the classical i (1)1>2. wave equation is the explicit appearance5 of the imaginary number The wave functions that satisfy the Schr\u00f6dinger equation are not necessarily real, as we see from the case of the free-particle wave function of Equation 6-7. Evidently the wave function which solves the Schr\u00f6dinger equation is not a directly measurable function as the classical wave function y(x, t) is since measurements always yield real numbers. However, as we discussed in Section 5-4, the probability of &(x, t) 6-1 The Schr\u00f6dinger Equation in One Dimension 225 & finding the electron in dx is certainly measurable, just as is the probability that a flipped coin will turn up heads. The probability P(x)dx that the electron will be found in the volume dx was defined by Equation 5-23 to be equal to This probabilistic interpretation of was developed by Max Born and was recognized, over the early and formidable objections of both Schr\u00f6dinger and Einstein, as the appropriate way of relating solutions of the Schr\u00f6dinger equation to the results of physical measurements. The probability that an electron is in the region dx, a real number, can be measured by counting the fraction of time it is found there in a very large number of identical trials. In recognition of the complex nature of we must modify slightly the interpretation of the wave function discussed in Chapter 5 to accommodate Born\u2019s interpretation so that the probability of finding the electron in dx is real. We take for the probability \u0192&\u0192 2 dx. &(x, t), P(x, t) dx &*(x, t)&(x, t) dx \u0192&(x, t)\u0192 2 dx 6-8 & & &, &*, is obtained from the complex conjugate of by replacing i with i where wherever it appears.6 The complex nature of serves to emphasize the fact that, in reality, we should not ask or try to answer the question, \u201cWhat is waving in a matter wave?\u201d or inquire as to what medium supports the motion of a matter wave. The wave function is a computational device with utility in Schr\u00f6dinger\u2019s theory of wave mechanics. Physical significance is associated not with itself, but with the product &*& \u0192&\u0192 2, which is the probability distribution P(x, t) or, as it is often called, the probability density. In keeping with the analogy with classical waves and wave funcis also sometimes referred to as the probability density amplitude, or just tions, the probability amplitude. &(x, t) & The probability of finding the electron in dx at x1 or in dx at x2 is the sum of separate probabilities, Since the electron must certainly be somewhere in space, the sum of the probabilities over all possible values of x must equal 1. That is7 P(x1) dx P(x2) dx. &*& dx 1 6-9 Equation 6-9 is called the normalization condition. This condition plays an important role in quantum mechanics, for it places a restriction on the possible solutions of the Schr\u00f6dinger equation. In particular, the wave function must approach zero so that the integral in Equation 6-9 remains finite. If it sufficiently fast as does not, then the probability becomes unbounded. As we will see in Section 6-3, it is this restriction together with boundary conditions imposed at finite values of x that leads to energy quantization for bound particles. x S &(x, t) In the chapters that follow, we are going to be concerned with solutions to the Schr\u00f6dinger equation for a wide range of real physical systems, but in what follows in this chapter, our intent is to illustrate a few of the techniques of solving the equation and to discover the various, often surprising properties of the solutions. To this end we will focus our attention on one-dimensional problems, as noted earlier, and use some potential energy functions with unrealistic physical characteristics, e.g., infinitely rigid walls, which will", " enable us to illustrate various properties of the solutions without obscuring the discussion with overly complex mathematics. 226 Chapter 6 The Schr\u00f6dinger Equation Separation of the Time and Space Dependencies of &(x, t) Schr\u00f6dinger\u2019s first application of his wave equation was to problems such as the hydrogen atom (Bohr\u2019s work) and the simple harmonic oscillator (Planck\u2019s work), in which he showed that the energy quantization in those systems can be explained naturally in terms of standing waves. We referred to these in Chapter 4 as stationary states, meaning they did not change with time. Such states are also called eigenstates. For such problems that also have potential energy functions that are independent of time, the space and time dependence of the wave function can be separated, leading to a greatly simplified form of the Schr\u00f6dinger equation.8 The separation is accomplished by first assuming that can be written as a product of two functions, one of x and one of t, as &(x, t) &(x, t) \u2019(x)(t) 6-10 If Equation 6-10 turns out to be incorrect, we will find that out soon enough, but it turns out that if the potential function is not an explicit function of time, i.e., if the potential is given by V(x), our assumption turns out to be valid. That this is true can be seen as follows: Substituting &(x, t) Schr\u00f6dinger equation (Equation 6-6) yields from Equation 6-10 into the general, which is U2 2m $2\u2019(x)(t) $x2 V(x)\u2019(x)(t) iU $\u2019(x)(t) $t U2 2m (t) d2\u2019(x) dx2 V(x)\u2019(x)(t) iU\u2019(x) d(t) dt time-dependent 6-11 6-12 where the derivatives are now ordinary rather than partial ones. Dividing Equation \u2019 6-12 by in the assumed product form gives & U2 2m 1 \u2019(x) d2\u2019(x) dx2 V(x) iU 1 (t) d(t) dt 6-13 Notice that each side of Equation 6-13 is a function of only one of the independent variables x and t. This means that, for example, changes in t cannot affect the value of the left side of Equation 6-13, and changes in x cannot affect the right side. Thus, both sides of the equation must be equal to the same constant C, called the separation constant, and we see that the assumption of Equation 6-10 is valid\u2014the variables have been separated. We have thus replaced a partial differential equation containing two independent variables, Equation 6-6, with two ordinary differential equations each a function of only one of the independent variables: V(x) C U2 2m 1 \u2019(x) iU 1 (t) d2\u2019(x) dx2 d(t) dt C 6-14 6-15 6-1 The Schr\u00f6dinger Equation in One Dimension 227 Let us solve Equation 6-15 first. The reason for doing so is twofold: (1) Equation (t) 6-15 does not contain the potential V(x); consequently, the time-dependent part of all solutions to the Schr\u00f6dinger equation will have the same form when the potential is not an explicit function of time, so we only have to do this once. (2) The separation constant C has particular significance that we want to discover before we tackle Equation 6-14. Writing Equation 6-15 as &(x, t) d(t) (t) C iU dt iC U dt The general solution of Equation 6-16 is (t) eiCt>U which can also be written as (t) eiCt>U cos a Ct U b i sin a Ct U b cos a2 Ct h b i sin a2 Ct h b 6-16 6-17a 6-17b (t), which describes the time variation of is an oscillatory Thus, we see that However, according to the de Broglie relation function with frequency &(x, t) there(Equation 5-1), the frequency of the wave represented by fore, we conclude that the separation constant C E, the total energy of the particle, and we have f E>h; f C>h, is &(x, t), (t) eiEt>U 6-17c for all solutions to Equation 6-6 involving time-independent potentials. Equation 6-14 then becomes, on multiplication by \u2019(x), U2 2m", " d2\u2019(x) dx2 V(x)\u2019(x) E\u2019(x) 6-18 Equation 6-18 is referred to as the time-independent Schr\u00f6dinger equation. The time-independent Schr\u00f6dinger equation in one dimension is an ordinary differential equation in one variable x and is therefore much easier to handle than the general form of Equation 6-6. The normalization condition of Equation 6-9 can be expressed in terms of since the time dependence of the absolute square of the wave function cancels. We have \u2019(x), &*(x, t)&(x, t) \u2019*(x)eiEt>U\u2019(x)eiEt>U \u2019*(x)\u2019(x) 6-19 and Equation 6-9 then becomes \u2019*(x)\u2019(x)dx 1 6-20 Conditions for Acceptable Wave Functions The form of the wave function that satisfies Equation 6-18 depends on the form of the potential energy function V(x). In the next few sections we will study some simple but important problems in which V(x) is specified. Our example potentials will be approximations to real physical potentials, simplified to make calculations easier. \u2019(x) 228 Chapter 6 The Schr\u00f6dinger Equation In some cases, the slope of the potential energy may be discontinuous, e.g., V(x) may have one form in one region of space and another form in an adjacent region. (This is a useful mathematical approximation to real situations in which V(x) varies rapidly over a small region of space, such as at the surface boundary of a metal.) The procedure in such cases is to solve the Schr\u00f6dinger equation separately in each region of space and then require that the solutions join smoothly at the point of discontinuity. \u2019(x) Since the probability of finding a particle cannot vary discontinuously from point must be continuous.9 Since the Schr\u00f6dinger equation to point, the wave function (which is the slope) involves the second derivative versus x must be smooth. (In a must also be continuous. That is, the graph of special case in which the potential energy becomes infinite, this restriction is relaxed. Since no particle can have infinite potential energy, must be zero in regions where V(x) is infinite. Then, at the boundary of such a region, may be discontinuous.) the first derivative d2\u2019>dx2 \u2019, \u2019(x) \u2019(x) \u2019 \u2019 \u2019(x) and d\u2019>dx or d&>dx. If either &(x, t) were not finite or not single valued, the same would be true of As we will shortly see, the predictions of wave mechanics regarding the results of measurements involve both of those quantities and would thus not necessarily predict finite or definite values for real physical quantities. Such results would not be acceptable since measurable quantities, such as angular momentum and position, are never infinite or multiple valued. A final restriction on the form of the wave function must x S so that normalization is preserved. For fuapproach zero sufficiently fast as ture reference, we may summarize the conditions that the wave function must meet in order to be acceptable as follows: is that in order to obey the normalization condition, \u2019(x) \u2019(x) \u2019(x) must exist and satisfy the Schr\u00f6dinger equation. 1. 2. 3. 4. 5. and and \u2019(x) \u2019(x) \u2019(x) \u2019(x) and \u2019(x) S 0 6-20, remains bounded. d\u2019>dx d\u2019>dx d\u2019>dx fast enough as must be continuous. must be finite. must be single valued. x S so that the normalization integral, Equation Note that, given Equation 6-10, the acceptability conditions above ultimately apply to &(x, t). Questions 1. Like the classical wave equation, the Schr\u00f6dinger equation is linear. Why is this important? 2. There is no factor i (1)1>2 in Equation 6-18. Does this mean that \u2019(x) must be real? 3. Why must the electric field e(x, t) be real? Is it possible to find a nonreal wave function that satisfies the classical wave equation? 4. Describe how the de Broglie hypothesis enters into the Schr\u00f6dinger wave equation. 5. What would be the effect on the Schr\u00f6dinger equation of adding a constant rest energy for a particle with mass to the total energy E in the de Broglie relation f E h? > 6. Describe in words what is meant by normalization of the wave function. EXAMPLE 6-1 A Solution to the Schr", "\u00f6dinger Equation Show that for a free particle of mass m moving in one dimension, the function is a solution to the time-independent Schr\u00f6dinger equation for any values of the constants A and B. \u2019(x) A sin kx B cos kx 6-2 The Infinite Square Well 229 SOLUTION A free particle has no net force acting upon it, e.g., V(x) 0, in which case the Differentiating kinetic energy equals the total energy. Thus, \u2019(x) p Uk (2mE)1>2. gives d\u2019 dx kA cos kx kB sin kx and differentiating again, d2\u2019 dx2 k2A sin kx k2B cos kx k2(A sin kx B cos kx) k2\u2019(x) Substituting into Equation 6-18, U2 2m [(k2)(A sin kx B cos kx)] E(A sin kx B cos kx) U2k2 2m \u2019(x) E\u2019(x) and, since U2k2 2mE, we have and the given \u2019(x) is a solution of Equation 6-18. E\u2019(x) E\u2019(x) 6-2 The Infinite Square Well A problem that provides several illustrations of the properties of wave functions and is also one of the easiest problems to solve using the time-independent, one-dimensional Schr\u00f6dinger equation is that of the infinite square well, sometimes called the particle in a box. A macroscopic example is a bead moving on a frictionless wire between two massive stops clamped to the wire. We could also build such a \u201cbox\u201d for an electron using electrodes and grids in an evacuated tube, as illustrated in Figure 6-1a. The walls of the box are provided by the increasing potential between the grids G and the electrode C as shown in Figures 6-1b and c. The walls can be made arbitrarily high and steep by increasing the potential V and reducing the separation between each grid-electrode pair. In the limit such a potential energy function looks like that in Figure 6-2, which is a graph of the potential energy of an infinite square well. For this problem the potential energy is of the form V(x) 0 0 x L V(x) x 0 and x L 6-21 Although such a potential is clearly artificial, the problem is worth careful study for several reasons: (1) exact solutions to the Schr\u00f6dinger equation can be obtained 230 Chapter 6 The Schr\u00f6dinger Equation Figure 6-1 (a) The electron placed between the two sets of electrodes C and grids G experiences no force in the region between the grids, which are at ground potential. However, in the regions between each C and G is a repelling electric field whose strength depends upon the magnitude of V. (b) If V is small, then the electron\u2019s potential energy versus x has low, sloping \u201cwalls.\u201d (c) If V is large, the \u201cwalls\u201d become very high and steep, becoming infinitely high for V S. V(x) 0 L x Figure 6-2 Infinite square well potential energy. For 0 x L, the potential energy V(x) is zero. Outside this region, V(x) is infinite. The particle is confined to the region in the well 0 x L. (a) (b) (c) Electron C G \u2013 G C \u2013 V V Potential energy Potential energy without the difficult mathematics that usually accompanies its solution for more realistic potential functions, (2) the problem is closely related to the vibrating-string problem familiar in classical physics, (3) it illustrates many of the important features of all quantum-mechanical problems, and finally, (4) this potential is a relatively good approximation to some real situations; e.g., the motion of a free electron inside a metal. Since the potential energy is infinite outside the well, the wave function is required to be zero there; that is, the particle must be inside the well. (As we proceed through this and other problems, keep in mind Born\u2019s interpretation: the probability ) We then need only to solve density of the particle\u2019s position is proportional to subject to the condition that Equation 6-18 for the region inside the well must be zero at x 0 and x L. since the wave function must be continuous, Such a condition on the wave function at a boundary (here, the discontinuity of the potential energy function) is called a boundary condition. We will see that, mathe\u2019(x) S 0 matically, it is the boundary conditions together with the requirement that that lead to the quantization of energy. A classic example is the case of a as vibrating string fixed at both ends. In that case", " the wave function y(x, t) is the displacement of the string. If the string is fixed at x 0 and x L, we have the same boundary condition on the vibrating-string wave function: namely, that y(x, t) be zero at x 0 and x L. These boundary conditions lead to discrete allowed frequencies of vibration of the string. It was this quantization of frequencies (which always occurs for standing waves in classical physics), along with de Broglie\u2019s hypothesis, that motivated Schr\u00f6dinger to look for a wave equation for electrons. \u0192\u2019\u0192 2. 0 x L, \u2019(x) x S The standing-wave condition for waves on a string of length L fixed at both ends is that an integral number of half wavelengths fit into the length L. n 2 L n 1, 2, 3, \u00c1 6-22 6-2 The Infinite Square Well 231 We will show below that the same condition follows from the solution of the Schr\u00f6dinger equation for a particle in an infinite square well. Since the wavelength is and the related to the momentum of the particle by the de Broglie relation (see Figure total energy of the particle in the well is just the kinetic energy 6-2), this quantum condition on the wavelength implies that the energy is quantized and the allowed values are given by p h> p2>2m E p2 2m h2 2m2 h2 2m(2L>n)2 n2 h2 8mL2 6-23 Since the energy depends on the integer n, it is customary to label it En. In terms of U h>2 the energy is given by n2 En 2U2 2mL2 n2E1 n 1, 2, 3, \u00c1 where E1 is the lowest allowed energy10 and is given by E1 2U2 2mL2 6-24 6-25 We now derive this result from the time-independent Schr\u00f6dinger equation (Equation 6-18), which for V(x) 0 is or U2 2m d2 \u2019(x) dx2 E\u2019(x) \u2019(x) 2mE U2 \u2019(x) k2\u2019(x) where we have substituted the square of the wave number k since k2 a 2 b p U 2mE U2 6-26 6-27 and we have written lutions of the form \u2019(x) for the second derivative d2\u2019(x)>dx2. Equation 6-26 has so- and \u2019(x) A sin kx \u2019(x) B cos kx 6-28a 6-28b where A and B are constants. The boundary condition cos 0 1, cosine solution (Equation 6-28b) because at x L gives boundary condition \u2019(x) 0 \u2019(x) 0 at x 0 rules out the so B must equal zero. The \u2019(L) A sin kL 0 6-29 232 Chapter 6 The Schr\u00f6dinger Equation This condition is satisfied if kL is any integer times values kn given by, i.e., if k is restricted to the n kn L n 1, 2, 3, \u00c1 6-30 If we write the wave number k in terms of the wavelength we see that Equation 6-30 is the same as Equation 6-22 for standing waves on a string. The quantized energy values, or energy eigenvalues, are found from Equation 6-27, replacing k by kn as given by Equation 6-30. We thus have 2> k, En U2k2 n 2m n2 U2 2 2mL2 n2E1 which is the same as Equation 6-24. Figure 6-3 shows the energy level diagram and the potential energy function for the infinite square well potential. The constant A in the wave function of Equation 6-28a is determined by the nor- malization condition: \u2019*n n dx L \u2019 0 n sin2a A2 nx L b dx 1 6-31 Energy 25E1 16E1 9E1 4E1 E1 0 V \u2192 \u221e En = n2E1 \u03c0 2 2 \u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013 2mL2 E1 = Figure 6-3 Graph of energy vs. x for a particle in an infinitely deep well. The potential energy V(x) is shown with the colored lines. The set of allowed values for the particle\u2019s total energy En as given by Equation 6-24 form the energy-level diagram for the infinite square well potential. Classically, a particle can have any value of energy. Quantum mechanically, only the values given by En yield well-behaved solutions of the Schr\u00f6dinger equation. As we become more familiar with energy-level diagrams", ", the x axis will be omitted. (U22>2mL2) n2 6-2 The Infinite Square Well 233 for Figure 6-4 Wave functions \u2019 n(x) and probability Pn(x) \u20192 n(x) densities n 1, 2, and 3 for the infinite square well potential. Though not shown, \u2019 x 0 n(x) 0 x L. and for \u03c8 3 2 /L \u03c8 2 3 2 /L 0 L/3 2L/3 L x 0 L/3 2L/3 L x \u03c8 2 2 /L \u03c8 2 2 2 /L 0 L/2 L x 0 L/2 L x \u03c8 1 2 /L 0 \u03c8 2 1 2 /L 0 L x L x Since the wave function is zero in regions of space where the potential energy is infinite, the contributions to the integral from to 0 and from L to will both be zero. Thus, only the integral from 0 to L needs to be evaluated. Integrating, we obtain An independent of n. The normalized wave function solutions for this problem, also called eigenfunctions, are then (2>L)1>2 n(x) \u2019 2 L A sin nx L n 1, 2, 3, \u00c1 6-32 These wave functions are exactly the same as the standing-wave functions yn(x) for the vibrating-string problem. The wave functions and the probability distribution functions Pn(x) are sketched in Figure 6-4 for the lowest energy state n 1, called the ground state, and for the first two excited states, n 2 and n 3. (Since these wave ) Notice in Figure 6-4 that the maximum functions are real, Pn(x), 2>L. amplitudes of each of the \u2019 n(x) They just happen to be zero for Note, too, that both x 0 and x L in this case. \u20192 n. are the same, extend to \u2026\u2019 Pn(x) \u2019 n n \u2019 n(x) and as are those of (2>L)1>2, Pn(x). The number n in the equations above is called a quantum number. It specifies both the energy and the wave function. Given any value of n, we can immediately write down the wave function and the energy of the system. The quantum number n at x 0 and x L. We will see occurs because of the boundary conditions in Section 7-1 that for problems in three dimensions, three quantum numbers arise, one associated with boundary conditions on each coordinate. \u2019(x) 0 Comparison with Classical Results Let us compare our quantum-mechanical solution of this problem with the classical solution. In classical mechanics, if we know the potential energy function V(x), we can d2x>dt2 find the force from from Newton\u2019s second law. We can then find the position x as a function of time t if we know the initial position and velocity. In this problem there is no force when the and thereby obtain the acceleration dV>dx ax Fx 234 Chapter 6 The Schr\u00f6dinger Equation particle is between the walls of the well because V 0 there. The particle therefore moves with constant speed in the well. Near the edge of the well the potential energy rises discontinuously to infinity\u2014we may describe this as a very large force that acts over a very short distance and turns the particle around at the wall so that it moves away with its initial speed. Any speed, and therefore any energy, is permitted classically. The classical description breaks down because, according to the uncertainty principle, we can never precisely specify both the position and momentum (and therefore velocity) at the same time. We can therefore never specify the initial conditions precisely and cannot assign a definite position and momentum to the particle. Of course, for a macroscopic particle moving in a macroscopic box, the energy is much larger than E1 of Equation 6-25, and the minimum uncertainty of momentum, which is of the order of U>L, is much less than the momentum and less than experimental uncertainties. Then the difference in energy between adjacent states will be a small fraction of the total energy, quantization will be unnoticed, and the classical description will be adequate.11 Let us also compare the classical prediction for the distribution of measurements of position with those from our quantum-mechanical solution. Classically, the probability of finding the particle in some region dx is proportional to the time spent in dx, which is dx v, where v is the speed. Since the speed is constant, the classical distribution function is just a constant inside the well. The normalized classical distribution function is PC(x) 1 L > In Figure 6-4 we see that for the lowest energy states, the quantum distribution function is very different from this. According to Bohr\u2019s correspondence principle, the quantum distributions", " should approach the classical distribution when n is large, that is, at large energies. For any state n, the quantum distribution has n peaks. The distribution for n 10 is shown in Figure 6-5. For very large n, the peaks are close together, and if there are many peaks in a small distance x, only the average value will sin2 knx Thus be observed. But the average value of over one or more cycles is 1>2. c\u20192 n(x) d c 2 L av sin2 knx d 2 L 1 2 1 L av which is the same as the classical distribution. \u03c8 2 0 Quantum-mechanical distribution Classical distribution P = 1 \u2013\u2013 L 0 < x < L L x Figure 6-5 Probability distribution for n 10 for the infinite square well potential. The dashed line is the classical probability density P 1 L, which is equal to the quantummechanical distribution averaged over a region x containing several oscillations. A physical measurement with resolution x will yield the classical result if n is so large that many oscillations in x. \u20192(x) has > 6-2 The Infinite Square Well 235 The Complete Wave Function The complete wave function, including its time dependence, is found by multiplying the space part by eit ei(En >U)t according to Equation 6-17c. As mentioned previously, a wave function corresponding but the probability to a single energy oscillates with angular frequency is independent of time. This is the wave-mechanical justifidistribution cation for calling such a state a stationary state or eigenstate, as we have done earlier. It is instructive to look at the complete wave function for a particular state n. n(x, t)\u0192 2 En >U, \u0192& n If we use the identity & n(x, t) 2 L A sin knx ei nt sin knx (eiknx eiknx) 2i we can write this wave function as n(x, t) 1 & 2i A 2 L [ei(knx nt) ei(knx nt)] Just as in the case of the standing-wave function for the vibrating string, we can consider this stationary-state wave function to be the superposition of a wave traveling to the right and a wave of the same frequency and amplitude traveling to the left. Since measurable quantities are related to is complex is not a problem. the fact that P \u0192&\u0192 2, & EXAMPLE 6-2 An Electron in a Wire An electron moving in a thin metal wire is a reasonable approximation of a particle in a one-dimensional infinite well. The potential inside the wire is constant on average but rises sharply at each end. Suppose the electron is in a wire 1.0 cm long. (a) Compute the ground-state energy for the electron. (b) If the electron\u2019s energy is equal to the average kinetic energy of the molecules in a gas at T 300 K, about 0.03 eV, what is the electron\u2019s quantum number n? SOLUTION 1. For question (a), the ground-state energy is given by Equation 6-25: E1 2U2 2mL2 2(1.055 1034 J # s)2 (2)(9.11 1031 kg)(102 m)2 6.03 1034 J 3.80 1015 eV 2. For question (b), the electron\u2019s quantum number is given by Equation 6-24: En n2E1 236 Chapter 6 The Schr\u00f6dinger Equation 3. Solving Equation 6-24 for n and substituting En yields 0.03 eV and E1 from above or n2 En E1 n C En E1 2.81 106 A 0.03 eV 3.80 1015 eV Remarks: The value of E1 computed above is not only far below the limit of measurability, but also smaller than the uncertainty in the energy of an electron confined into 1 cm. For a value of n this large, the correspondence principle applies. EXAMPLE 6-3 Calculating Probabilities Suppose that the electron in Example 6-2 could be measured while in its ground state. (a) What would be the probability of finding it somewhere in the region 0 x L 4? (b) What would be the probability of finding it in a very narrow region x 0.01L wide centered at x 5L 8? > > SOLUTION (a) The wave function for the n 1 level, the ground state, is given by Equation 6-32 as 1(x) \u2019 2 L A sin x L The probability that the electron would be found in the region specified is L>4 0 P1(x) dx L>4 0 2 L sin2a x L b dx u x>L,", " Letting limits on the integral, we have that hence dx L du>, and noting the appropriate change in the >4 0 2 sin2 u du 2 a u 2 sin 2u 4 b ` >4 0 2 a 8 1 4 b 0.091 Thus, if one looked for the particle in a large number of identical searches, the electron would be found in the region 0 x 0.25 cm about 9 percent of the time. This probability is illustrated by the shaded area on the left side in Figure 6-6. (b) Since the region x 0.01L is very small compared with L, we do not need to integrate but can calculate the approximate probability as follows: P P(x) x 2 L x L Substituting x 0.01L and x 5L 8, we obtain sin2 x > sin2 (5L>8) L (0.01L) (0.854)(0.01L) 0.017 P 2 L 2 L 6-2 The Infinite Square Well 237 x \u03c0 \u2013\u2013\u2013\u2013\u2013L i 2 n s 2\u2013\u2013L = 2 \u03c8 0 L/4 L/2 3L/4 L x \u20192(x) Figure 6-6 The probability density ground state of an infinite square well potential. The probability of finding > the particle in the region 0 x L 4 is represented by the larger shaded area. The narrow shaded band illustrates the probability of finding the particle within x 0.01L around the point where x 5L 8.> versus x for a particle in the > This means that the probability of finding the electron within 0.01L around x 5L 8 is about 1.7 percent. This is illustrated in Figure 6-6, where the area of the shaded narrow band at x 5L 8 is 1.7 percent of the total area under the curve. > EXAMPLE 6-4 An Electron in an Atomic-Size Box (a) Find the energy in the ground state of an electron confined to a one-dimensional box of length L 0.1 nm. (This box is roughly the size of an atom.) (b) Make an energy-level diagram and find the wavelengths of the photons emitted for all transitions beginning at state n 3 or less and ending at a lower energy state. SOLUTION (a) The energy in the ground state is given by Equation 6-25. Multiplying the nuwe obtain an expression in terms of hc and merator and denominator by mc2, the energy equivalent of the electron mass (see Chapter 2): c2>42, E1 (hc)2 8mc2L2 Substituting hc 1240 eV # nm and mc2 0.511 MeV, we obtain (1240 eV # nm)2 8(5.11 105 eV)(0.1 nm)2 E1 37.6 eV This is of the same order of magnitude as the kinetic energy of the electron in the ground state of the hydrogen atom, which is 13.6 eV. In that case, the wavelength of the electron equals the circumference of a circle of radius 0.0529 nm, or about 0.33 nm, whereas for the electron in a one-dimensional box of length 0.1 nm, the wavelength in the ground state is 2L 0.2 nm. (b) The energies of this system are given by En n2E1 n2(37.6 eV) 238 Chapter 6 The Schr\u00f6dinger Equation Figure 6-7 Energy-level diagram for Example 6-4. Transitions from the state n 3 to the states n 2 and n 1, and from the state n 2 to n 1, are indicated by the vertical arrows. n 5 4 3 2 1 E E5 = 25E1 = 940 eV E4 = 16E1 = 601.6 eV E3 = 9E1 = 338.4 eV E2 = 4E1 = 150.4 eV E1 = 37.6 eV E2 4 # (37.6 eV) 150.4 eV, 9 # (37.6 eV) 338.4 eV. Figure 6-7 shows these energies in an energy-level diagram. The energy of the first excited state is and that of the second excited state The possible transitions from level 3 to level 2, is from level 3 to level 1, and from level 2 to level 1 are indicated by the vertical arrows on the diagram. The energies of these transitions are E3 E3S2 E3S1 E2S1 338.4 eV 150.4 eV 188.0 eV 338.4 eV 37.6 eV 300.8 eV 150.4 eV 37.6 eV 112.8 eV The photon wavelengths for these transitions are 3S2 3S1 2S1 hc E3", "S2 hc E3S1 hc E2S1 188.0 eV 1240 eV # nm 1240 eV # nm 1240 eV # nm 300.8 eV 112.8 eV 6.60 nm 4.12 nm 11.0 nm 6-3 The Finite Square Well The quantization of energy that we found for a particle in an infinite square well is a general result that follows from the solution of the Schr\u00f6dinger equation for any particle confined in some region of space. We will illustrate this by considering the qualitative behavior of the wave function for a slightly more general potential energy function, the finite square well shown in Figure 6-8. The solutions of the Schr\u00f6dinger equation for this type of potential energy are quite different, depending on whether the total energy E is greater or less than V0. We will defer discussion of the case E V0 to Section 6-5 except to remark that in that case, the particle is not confined and any value of the energy is allowed; i.e., there is no energy quantization. Here we will assume that E V0. 6-3 The Finite Square Well 239 (a) V(x) (b) V(x) V0 V0 0 L x \u2013a 0 +a x Figure 6-8 (a) The finite square well potential. (b) Region I is that with x a, II with a x a, and III with x a. Inside the well, V(x) 0 and the time-independent Schr\u00f6dinger equation (Equation 6-18) becomes Equation 6-26, the same as for the infinite well: \u2019(x) k2\u2019(x) k2 2mE U2 The solutions are sines and cosines (Equation 6-28) except that now we do not require \u2019(x) be continuous at these points. Outside the well, i.e., for 0 x L, Equation 6-18 becomes to be zero at the well boundaries, but rather we require that \u2019(x) \u2019(x) and where \u2019(x) 2m U2 (V0 E)\u2019(x) 2\u2019(x) 2 2m U2 (V0 E) 0 6-33 6-34 \u2019(x) The straightforward method of finding the wave functions and allowed energies outside the well and then require for this problem is to solve Equation 6-33 for be continuous at the boundaries. The solution of Equation 6-33 is that for positive x], but applying the boundary not difficult [it is of the form conditions involves a method that may be new to you; we describe it in the More section on the Graphical Solution of the Finite Square Well. \u2019(x) Cex \u2019(x) \u2019(x) and \u2019 \u2019 and at the boundaries and the need for First, we will explain in words unencumbered by the mathematics how the condix S tions of continuity of leads to the selection of only certain wave functions and quantized energies for values of E within the well; i.e., 0 E V0. The important feature of Equation 6-33 is that the second derivative which is the curvature of the wave function, has the same sign \u2019 is also positive and the wave function as the wave function curves away from the axis, as shown in Figure 6-9a. Similarly, if is neg\u2019 curves away from the axis. This behavior is different from that inside ative and, again, the well, where 0 x L. There, always curves toward the axis like a sine or cosine function. Because of this behavior outside the well, have opposite signs so that is positive, is negative, \u2019, \u2019. \u2019 S 0 and as \u2019 \u2019 If \u2019 \u2019 \u2019 \u2019 f(x) f(x) x x (a) (b) Figure 6-9 (a) Positive function with positive curvature; (b) negative function with negative curvature. 240 Chapter 6 The Schr\u00f6dinger Equation \u03c8(x) \u03bb = 4L \u20130.5L 0 L 1.5L x Figure 6-10 The function that satisfies the Schr\u00f6dinger equation with 4L inside the well is not an acceptable wave function because it becomes infinite at large x. Although at x L, the function is heading toward zero (slope is negative), the rate of increase of the slope great that the slope becomes positive before the function becomes zero, and the function then increases. Since slope always increases and the function increases without bound. [This computer-generated plot courtesy of Paul Doherty, The Exploratorium.] has the same sign as is so the \u2019, \u2019 \u2019 x S ; for most values of the energy the wave function becomes infinite as is not well behaved", ". Such functions, though satisfying the Schr\u00f6dinger equation, are not proper wave functions because they cannot be normalized. \u2019(x) i.e., for 4L. Figure 6-10 shows the wave function for the energy E p2>2m h2>2m2 Figure 6-11 shows a wellbehaved wave function corresponding to wavelength 1, which is the ground state wave function for the finite well, and the behavior of the wave functions for two nearby energies and wavelengths. The exact determination of the allowed energy levels in a finite square well can be obtained from a detailed solution of the problem. Figure 6-12 shows the wave functions and the probability distributions for the ground state and for the first two excited states. From this figure we see that the wavelengths inside the well are slightly longer than the corresponding wavelengths for the infinite well of the same width, so the corresponding energies are slightly less than those of the infinite well. Another feature of the finite well problem is that there are only a finite number of allowed energies, depending of the size of V0. For very small V0 there is only one allowed energy level; i.e., only one bound state can exist. This will be quite apparent in the detailed solution in the More section. Note that, in contrast to the classical case, there is some probability of finding the particle outside the well, in the regions x L or x 0. In these regions, the total energy is less than the potential energy, so it would seem that the kinetic energy must be negative. Since negative kinetic energy has no meaning in classical physics, it is interesting to speculate about the meaning of this penetration of wave function beyond the well boundary. Does quantum mechanics predict that we could measure a negative kinetic energy? If so, this would be a serious defect in the theory. Fortunately, we are saved by the uncertainty principle. We can understand \u03c8(x is slightly greater than 1, Figure 6-11 Functions satisfying the Schr\u00f6dinger equation with wavelengths near the critical If wavelength 1. Figure 6-10. At the wavelength an acceptable wave function corresponding to the energy than more negative because its rate of change infinity at large x. [This computer-generated plot courtesy of Paul Doherty, The Exploratorium.] the function and its slope approach zero together. This is is slightly less the function crosses the x axis while the slope is still negative. The slope becomes the function approaches infinity like that in is now negative. This function approaches negative h2>2m2 1. 1, E1 \u2019 If 1, 6-3 The Finite Square Well 241 with ex, to be negligible beyond becomes very small in a distance of the order of x L 1, this qualitatively as follows (we will consider the region x L only). Since the wave given by Equation 6-34, the probability density function decreases as \u20192 e2x x 1. If we consider \u2019(x) we can say that finding the particle in the region x L is roughly equivalent to localizing it in a region Such a meaand surement introduces an uncertainty in momentum of the order of a minimum kinetic energy of the order of This kinetic energy is just enough to prevent us from measuring a negative kinetic energy! The penetration of the wave function into a classically forbidden region does have important consequences in tunneling or barrier penetration, which we will discuss in Section 6-6. Much of our discussion of the finite well problem applies to any problem in which E V(x) in some region and E V(x) outside that region. Consider, for example, the potential energy V(x) shown in Figure 6-13. Inside the well, the Schr\u00f6dinger equation is of the form x 1. p h>x h (p)2>2m h22>2m V0 E. \u2019(x) k2\u2019(x) 6-35 k2 2m[E V(x)]>U2 now depends on x. The solutions of this equation are where no longer simple sine or cosine functions because the wave number varies with x, but since have opposite signs, will always curve toward the axis and the solutions will oscillate. Outside the well, will curve away from the axis so there will be only certain values of E for which solutions exist that approach zero as x approaches infinity. k 2> \u2019 \u2019 and \u2019 \u2019 Figure 6-12 Wave functions \u2019 n(x) and probability \u20192 for n 1, n(x) distributions 2, and 3 for the finite square well. Compare these with Figure 6-4 for the infinite square well, where the wave functions are zero at x 0 and x L. The wavelengths are slightly longer than the corresponding ones for the infinite well, so the allowed energies are somewhat smaller. V(x) x E and Figure 6-13 Arbitrary welltype potential with possible energy E. Inside the well \u2019(x) \u2019(x)", " [E V(x)], have opposite signs, and the wave function will oscillate. Outside the well, \u2019(x) have the same sign and, except for certain values of E, the wave function will not be well behaved. \u2019(x) and More In most cases the solution of finite well problems involves transcendental equations and is very difficult. For some finite potentials, however, graphical solutions are relatively simple and provide both insights and numerical results. As an example, we have included the Graphical Solution of the Finite Square Well on the home page: www.whfreeman.com/tiplermodern physics5e. See also Equations 6-36 through 6-43 and Figure 6-14 here. 242 Chapter 6 The Schr\u00f6dinger Equation (a) Energy (b) Energy Potential well L1 0 L2 L1 0 L2 Figure 6-15 (a) Two infinite square wells of different widths L1 and L2, each containing the same number of electrons, are put together. An electron from well 1 moves to the lowest empty level of well 2. (b) The energies of the two highest electrons are equalized, but the unequal charge in the two wells distorts the energy-level structure. The distortion of the lowest empty levels in each well results in a potential well at the junction between the wells. The orientation of the newly formed well is perpendicular to the plane of the figure. Quantum Wells Development of techniques for fabricating devices whose dimensions are of the order of nanometers, called nanostructures, has made possible the construction of quantum wells. These are finite potential wells of one, two, and three dimensions that can channel electron movement in selected directions. A one-dimensional quantum well is a thin layer of material that confines particles to within the dimension perpendicular to the layer\u2019s surface but does not restrict motion in the other two dimensions. In the case of three-dimensional wells, called quantum dots, electrons are restricted entirely to quantized energy states within the well. A ubiquitous current application of quantum wells is the diode lasers that read CDs, DVDs, and bar codes. Quantum dots have potential applications in data storage and quantum computers, devices that may greatly enhance computing power and speed. One-dimensional quantum wells, called quantum wires, offer the possibility of dramatically increasing the speed that electrons move through a device in selected directions. This in turn would increase the speed with which signals move between circuit elements in computer systems. Figure 6-15 is an outline of how such a well might be formed. 6-4 Expectation Values and Operators Expectation Values The objective of theory is to explain experimental observations. In classical mechanics the solution of a problem is typically specified by giving the position of a particle or particles as a function of time. As we have discussed, the wave nature of matter prevents us from doing this for microscopic systems. Instead, we find the wave function 6-4 Expectation Values and Operators 243 and the probability distribution function &(x, t) The most that we can know about a particle\u2019s position is the probability that a measurement will yield various values of x. The expectation value of x is defined as \u0192&(x, t)\u0192 2. 8x9 &* (x, t)x&(x, t) dx 6-44 The expectation value of x is the average value of x that we would expect to obtain from a measurement of the positions of a large number of particles with the same wave function As we have seen, for a particle in a state of definite energy, the probability distribution is independent of time. The expectation value of x is then given by &(x, t). 8x9 \u2019*(x)x\u2019(x) dx 6-45 For example, for the infinite square well, we can see by symmetry (or by direct calculation) that is L 2, the midpoint of the well. 8x9 > In general, the expectation value of any function f(x) is given by 8f(x)9 \u2019*f(x)\u2019 dx 6-46 For example, It is left as an exercise (see Problem 6-56) to show that can be calculated as above, for the infinite square well of width L. 8x29 8x29 L2 3 L2 2n22 6-47 You may recognize the expectation values defined by Equations 6-45 and 6-46 as being weighted average calculations, borrowed by physics from probability and statistics. We should note that we don\u2019t necessarily expect to make a measurement whose result equals the expectation value. For example, for even n, the probability of measuring x L 2 in some range dx around the midpoint of the well is zero because the because the probability wave function sin density function is symmetrical about that point. Remember that the expectation value is the average value that would result from many measurements. is zero there. We get 8x9 L>", "2 (nx>L) \u2019*\u2019 > Operators 8p9 If we knew the momentum p of a particle as a function of x, we could calculate the expectation value from Equation 6-46. However, it is impossible in principle to find p as a function of x since, according to the uncertainty principle, both p and x canwe need to know the distribution funcnot be determined at the same time. To find 8p9 tion for momentum. If we know it can be found by Fourier analysis. The \u2019(x), 8p9 also can be found from Equation 6-48, where is the mathematical operator i that produces the x component of the momentum (see also Equation 6-6). acting on & U a $ $x b 8p9 &*a U i $ $x b& dx 6-48 244 Chapter 6 The Schr\u00f6dinger Equation Similarly, 8p29 can be found from 8p29 &*a U i $ $x b a U i $ $x b& dx &. not on &(x, t), Notice that in computing the expectation value the operator representing the physical &*(x, t); &* quantity operates on This is not important to the outcome when the operator is simply some f(x), and but it is critical when the operator includes a differentiation, as in the case of 8p29 is simply 2mE since, for the infinite square the momentum operator. Note that U $ $x which operates on the wave function in i.e., its correct position is between E p2>2m. The quantity well, b, a i Equation 6-48, is called the momentum operator pop: pop U i $ $x 6-49 EXAMPLE 6-5 Expectation Values for p and p2 Find for the groundstate wave function of the infinite square well. (Before we calculate them, what do you think the results will be?) and 8p29 8p9 SOLUTION We can ignore the time dependence of &, in which case we have 8p9 L a 2 L A sin b a nx L U i $ $x b a 2 L A sin nx L b dx 0 U i 2 L L L 0 sin x L cos x L dx 0 The particle is equally as likely to be moving in the x as in the x direction, so its average momentum is zero. Similarly, since U i $ $x a U i $ $x U2a 2 L2 A 2 L sin x L b b\u2019 U2 $2\u2019 $x2 U22 L2 \u2019 we have 8p29 U22 L2 L 0 \u2019*\u2019 dx U22 L2 The time-independent Schr\u00f6dinger equation (Equation 6-18) can now be written conveniently in terms of pop: a 1 2m bp2 op \u2019(x) V(x)\u2019(x) E\u2019(x) 6-50 where p2 op \u2019(x) U i $ $x c U i $ $x \u2019(x) d U2 $2\u2019 $x2 6-4 Expectation Values and Operators 245 In classical mechanics, the total energy written in terms of the position and momentum variables is called the Hamiltonian function H p2 2m V. If we replace the momentum by the momentum operator pop and note that V V(x), we obtain the Hamiltonian operator Hop: > Hop p2 op 2m V(x) The time-independent Schr\u00f6dinger equation can then be written Hop \u2019 E\u2019 6-51 6-52 The advantage of writing the Schr\u00f6dinger equation in this formal way is that it allows for easy generalization to more-complicated problems such as those with several particles moving in three dimensions. We simply write the total energy of the system in terms of position and momentum and replace the momentum variables by the appropriate operators to obtain the Hamiltonian operator for the system. Table 6-1 summarizes the several operators representing physical quantities that we have discussed thus far and includes a few more that we will encounter later on. Table 6-1 Some quantum-mechanical operators Symbol Physical quantity Operator f(x) Any function of x\u2014e.g., the position x, the potential energy V(x), etc. x component of momentum y component of momentum z component of momentum Hamiltonian (time independent) Hamiltonian (time dependent) kinetic energy z component of angular momentum px py pz E E Ek Lz Questions f(x) U i U i U i $ $x $ $y $ $z p2 op 2m V(x) i U $ $t U2 2m iU $2 $x2 $ $ 7. Explain (in words) why 8. Can in Example 6-5 are not both zero. ever have a value that has zero probability of being measured?", " and 8x9 8p9 8p29 246 Chapter 6 The Schr\u00f6dinger Equation More In order for interesting things to happen in systems with quantized energies, the probability density must change in time. Only in this way can energy be emitted or absorbed by the system. Transitions Between Energy States on the home page (www.whfreeman.com/tiplermodern physics5e) describes the process and applies it to the emission of light from an atom. See also Equations 6-52a\u2013e and Figure 6-16 here. 6-5 The Simple Harmonic Oscillator One of the problems solved by Schr\u00f6dinger in the second of his six famous papers was that of the simple harmonic oscillator potential, given by V(x) 1 2 Kx2 1 2 m2x2 where K is the force constant and the angular frequency of vibration defined by (K> m)1>2 2f. The solution of the Schr\u00f6dinger equation for this potential is particularly important, as it can be applied to such problems as the vibration of molecules in gases and solids. This potential energy function is shown in Figure 6-17, with a possible total energy E indicated. In classical mechanics, a particle in such a potential is in equilibrium at the origin x 0, where V(x) is minimum and the force Fx dV dx is zero. If disturbed, the particle will oscillate back and forth between x A and x A, the points at which the kinetic energy is zero and the total energy is just equal to the potential energy. These points are called the classical turning points. The distance A is related to the total energy E by > E 1 2 m2A2 6-53 Classically, the probability of finding the particle in dx is proportional to the time spent in dx, which is dx v. The speed of the particle can be obtained from the conservation of energy: > 1 2 mv2 1 2 m2x2 E The classical probability is thus Pc(x) dx dx v dx B(2>m) aE 1 2 m2x2b 6-54 Any value of the energy E is possible. The lowest energy is E 0, in which case the particle is at rest at the origin. The Schr\u00f6dinger equation for this problem is U2 2m $2\u2019(x) $x2 1 2 m2x2\u2019(x) E\u2019(x) 6-55 V(x) m\u03c92x2 1 \u2013\u2013 2 E \u2013A 0 +A x Figure 6-17 Potential energy function for a simple harmonic oscillator. Classically, the particle is confined between the \u201cturning points\u201d A and A. 6-5 The Simple Harmonic Oscillator 247 The mathematical techniques involved in solving this type of differential equation are standard in mathematical physics but unfamiliar to most students at this level. We will, therefore, discuss the problem qualitatively. We first note that since the potential is symmetric about the origin x 0, we expect the probability distribution \u0192\u2019(x)\u0192 2 function also to be symmetric about the origin, i.e., to have the same value at x as at x. \u0192\u2019(x)\u0192 2 \u0192\u2019(x)\u0192 2 \u2019(x) \u2019(x) \u2019(x) & \u2019(x) \u2019(x). must then be either symmetric or antiThe wave function symmetric We can therefore simplify our discussion by considering positive x only and find the solutions for negative x by symmetry. (The symmeis discussed further in the Exploring section, Parity; see page 250.) try of Consider some value of total energy E. For x less than the classical turning point A defined by Equation 6-53, the potential energy V(x) is less than the total energy E, whereas for x A, V(x) is greater than E. Our discussion in Section 6-3 applies directly to this problem. For x A, the Schr\u00f6dinger equation can be written \u2019(x) k2\u2019(x) k2 2m U2 [E V(x)] curves toward the axis and oscillates. For x A, the Schr\u00f6dinger equation where \u2019(x) and becomes with \u2019(x) 2\u2019(x) 2 2m U2 [V(x) E] \u2019(x) and curves away from the axis. Only certain values of E will lead to solutions that are well behaved, i.e., that approach zero as x approaches infinity. The allowed values of E for the simple harmonic oscillator must be determined by solving the Schr\u00f6dinger equation; in this case they are given by En an 1 2 b U n 0, 1, 2, \u00c1 6-56 U 1 2 Thus, the ground", "-state energy is and the energy levels are equally spaced, each excited state being separated from the levels immediately adjacent by U. The wave functions of the simple harmonic oscillator in the ground state and in the first two excited states (n 0, n 1, and n 2) are sketched in Figure 6-18. The ground-state wave function has the shape of a Gaussian curve, and the lowest energy E 1 is the minimum energy consistent with the 2 uncertainty principle Figure 6-18 Wave functions for the ground state and the first two excited states of the simple harmonic oscillator potential, the states with n 0, 1, and 2. 248 Chapter 6 The Schr\u00f6dinger Equation The allowed solutions to the Schr\u00f6dinger equation, the wave functions for the simple harmonic oscillator, can be written n(x) Cnemx2>2UHn(x) \u2019 6-57 where the constants Cn are determined by normalization and the functions Hn(x) are polynomials of order n called the Hermite polynomials.13 The solutions for n 0, 1, and 2 (see Figure 6-18) are 0(x) A0emx2>2U \u2019 1(x) A1A \u2019 m U xemx2>2U 2(x) A2 \u2019 a1 2mx2 U bemx2>2U 6-58 Notice that for even values of n, the wave functions are symmetric about the origin; for odd values of n, they are antisymmetric. In Figure 6-19 the probability distribu\u20192 are sketched for n 0, 1, 2, 3, and 10 for comparison with the classical n(x) tions distribution. \u03c8 2 n \u20133 \u20132 \u20131 0 1 2 3 \u20133 \u20132 \u20131 0 1 2 3 \u20133 \u20132 \u20131 0 1 2 3 \u20133 \u20132 \u2013 = 10 \u20135 \u20134 \u20133 \u20132 \u20131 1 2 3 4 5 0 u Molecules vibrate as harmonic oscillators. Measuring vibration frequencies (see Chapter 9) enables determination of force constants, bond strengths, and properties of solids. \u20192 n Figure 6-19 Probability for the simple density harmonic oscillator plotted against the dimensionless (m>U)1>2x variable u, for n 0, 1, 2, 3, and 10. The dashed curves are the classical probability densities for the same energy, and the vertical lines indicate the classical turning points x A. 6-5 The Simple Harmonic Oscillator 249 V(x) 1 Kx2 = m\u03c92x2 \u2013\u2013 2 V(x) =V(x) = 1 \u2013\u2013 2 E5 = (5 + ) \u03c9 1 \u2013\u2013 2 E4 = (4 + ) \u03c9 1 \u2013\u2013 2 E3 = (3 + ) \u03c9 1 \u2013\u2013 2 E2 = (2 + ) \u03c9 1 \u2013\u2013 2 E1 = (1 + ) \u03c9 1 \u2013\u2013 2 E0 = \u03c9 1 \u2013\u2013 2 0 x Figure 6-20 Energy levels in the simple harmonic oscillator potential. Transitions obeying the selection rule n 1 are indicated by the arrows (those pointing up indicate absorption). is emitted or absorbed in all allowed Since the levels have equal spacing, the same energy transitions. For this special potential, the frequency of the emitted or absorbed photon equals the frequency of oscillation, as predicted by classical theory. U A property of these wave functions that we will state without proof is that nx\u2019 \u2019\u2026 m dx 0 unless n m 1 6-59 This property places a condition on transitions that may occur between allowed states. This condition, called a selection rule, limits the amount by which n can change for (electric dipole) radiation emitted or absorbed by a simple harmonic oscillator: The quantum number of the final state must be 1 less than or 1 greater than that of the initial state. This selection rule is usually written n 1 6-60 Since the difference in energy between two successive states is this is the energy of the photon emitted or absorbed in an electric dipole transition. The frequency of the photon is therefore equal to the classical frequency of the oscillator, as was assumed by Planck in his derivation of the blackbody radiation formula. Figure 6-20 shows an energy level diagram for the simple harmonic oscillator, with the allowed energy transitions indicated by vertical arrows. U, More Solution of the Schr\u00f6dinger equation for the simple harmonic oscillator (Equation 6-55) involves some rather advanced differential equation techniques. However, a simpler exact solution is also possible using an approach invented by Schr\u00f6dinger himself that we will call Schr\u00f6dinger\u2019s Trick. With the authors\u2019 thanks to Wolfgang Lorenzon for bringing it to our attention, we include it on the home page www.whfreeman.", "com/ tiplermodernphysics5e so that you, too, will know the trick. 250 Chapter 6 The Schr\u00f6dinger Equation EXPLORING Parity We made a special point of arranging the simple harmonic oscillator potential symmetrically about x 0 (see Figure 6-17), just as we had done with the finite square well in Figure 6-8b and will do with various other potentials in later discussions. The usual purpose in each case is to emphasize the symmetry of the physical situation and to simplify the mathematics. Notice that arranging the potential V(x) symmetrically about the origin means that V(x) V(x). This means that the Hamiltonian operator Hop, defined in Equation 6-51, is unchanged by a transformation that changes Such a transformation is called a parity operation and is usually denoted by the operator P. Thus, if is a solution of the Schr\u00f6dinger equation x S x. \u2019(x) Hop \u2019(x) E\u2019(x) 6-52 then a parity operation P leads to Hop \u2019(x) E\u2019(x) \u2019(x) and is also a solution to the Schr\u00f6dinger equation and corresponds to the same energy. When two (or more) wave functions are solutions corresponding to the same value of the energy E, that level is referred to as degenerate. In this case, where two wave func\u2019(x) are both solutions with energy E, we call the energy level doubly tions, degenerate. \u2019(x), and It should be apparent from examining the two equations above that \u2019(x) and \u2019(x) can differ at most by a multiplicative constant C; i.e., \u2019(x) C\u2019(x) \u2019(x) C\u2019(x) or \u2019(x) C\u2019(x) C 2\u2019(x) from which it follows that C 1. If C 1, i.e., If C 1, then \u2019(x) \u2019(x). Parity is used in quantum mechanics to describe the symmetry properties of wave functions under a reflection of the space coordinates in the origin, i.e., under a parity operation. The terms even and odd parity describe the symmetry of the wave functions, not whether the quantum numbers are even or odd. We will have more on parity in Chapter 12. is an even function, \u2019(x) \u2019(x). is an odd function, i.e., \u2019(x) \u2019(x) 6-6 Reflection and Transmission of Waves Up to this point, we have been concerned with bound-state problems in which the potential energy is larger than the total energy for large values of x. In this section, we will consider some simple examples of unbound states for which E is greater than V(x) \u2019(x) as x gets larger in one or both directions. For these problems and have opposite signs for those regions of x where E V(x), so in those regions \u0192x\u0192. Any value curves toward the axis and does not become infinite at large values of \u2019(x) of E is allowed. Such wave functions are not normalizable since does not approach zero as x goes to infinity in at least one direction and, as a consequence, d2\u2019(x)>dx2 \u2019(x) \u0192\u2019(x)\u0192 2 dx \u00a1 6-6 Reflection and Transmission of Waves 251 A complete solution involves combining infinite plane waves into a wave packet of finite width. The resulting finite packet is normalizable. However, for our purposes it is sufficient to note that the integral above is bounded between the limits a and b, Such wave functions are most frequently encounprovided only that tered, as we are about to do, in the scattering of beams of particles from potentials, so it is usual to normalize such wave functions in terms of the density of particles in the beam. Thus, \u0192b a\u0192. b a \u0192\u2019(x)\u0192 2 dx b dx b dN N a a where dN is the number of particles in the interval dx and N is the number of particles in the interval (b a).14 The wave nature of the Schr\u00f6dinger equation leads, even so, to some very interesting consequences. Step Potential Consider a region in which the potential energy is the step function V(x) 0 for x 0 V(x) V0 for x 0 as shown in Figure 6-21. We are interested in what happens when a beam of particles, each with the same total energy E, moving from left to right encounters the step. The classical answer is simple. For x 0, each particle moves with speed v (2E>m)1>2. At x 0, an impulsive force acts on", " it. If the total energy E is less than V0, the particle will be turned around and will move to the left at its original speed; that is, it will be reflected by the step. If E is greater than V0, the particle will continue We might moving to the right but with reduced speed, given by picture this classical problem as a ball rolling along a level surface and coming to a V0. If its original kinetic energy is less than steep hill of height y0, given by mgy0 V0, the ball will roll partway up the hill and then back down and to the left along the level surface at its original speed. If E is greater than V0, the ball will roll up the hill and proceed to the right at a smaller speed. v [2(E V0)>m]1>2. The quantum-mechanical result is similar to the classical one for E V0 but quite different when E V0, as in Figure 6-22a. The Schr\u00f6dinger equation in each of the two space regions shown in the diagram is given by Region I Region II (x 0) d2\u2019(x) dx2 k2 1 \u2019(x) (x 0) d2\u2019(x) dx2 k2 2 \u2019(x) k1 22mE U and k2 22m(E V0) U 6-61 6-62 V(x) V0 0 x Figure 6-21 Step potential. A classical particle incident from the left, with total energy E greater than V0, is always transmitted. The potential change at x 0 merely provides an impulsive force that reduces the speed of the particle. However, a wave incident from the left is partially transmitted and partially reflected because the wavelength changes abruptly at x 0. 252 Chapter 6 The Schr\u00f6dinger Equation (a) Energy V(x) = 0 (b) 0 \u03c8(x) 0 I I E V(x) = V0 x x II II Figure 6-22 (a) A potential step. Particles are incident on the step from the left toward the right, each with total energy E V0. (b) The wavelength of the incident wave (Region I) is shorter than that of the transmitted wave (Region II). Since k2 however, the transmission coefficient T 1. \u0192C\u0192 2 \u0192A\u0192 2; k1, The general solutions are Region I Region II (x 0) \u2019 I(x) Aeik1x Beik1x (x 0) \u2019 II(x) Ceik2x Deik2x 6-63 6-64 &(x, t), eit, by the time part of Specializing these solutions to our situation where we are assuming the incident beam of particles to be moving from left to right, we see that the first term in Equation 6-63 Aeik1x represents that beam since multiplying yields a plane wave (i.e., a beam of free particles) moving to the right. The second term, represents particles moving to the left in Region I. In Equation 6-64, D 0 Beik2x, since that term represents particles incident on the potential step from the right and there are none. Thus, we have that the constant A is known or at least obtainable (determined by normalization of in terms of the density of particles in the beam as explained above) and the constants B and C are yet to be found. We find them by ap\u2019(x) at x 0, i.e., by requiring that plying the continuity condition on and II(0)>dx I(0) \u2019 \u2019 I(0)>dx d\u2019 at x 0 yields and I(0) A B \u2019 \u2019 d\u2019(x)>dx. Continuity of II(0) C Aeik1x II(0) d\u2019 \u2019 or Continuity of d\u2019>dx at x 0 gives A B C k1A k1B k2C 6-65a 6-65b 6-6 Reflection and Transmission of Waves 253 Solving Equations 6-65a and b for B and C in terms of A (see Problem 6-47), we have B C k1 k1 k1 k2 k2 2k1 k2 A A E1>2 (E V0)1>2 E1>2 (E V0)1>2 A 2E1>2 E1>2 (E V0)1>2 A 6-66 6-67 where Equations 6-66 and 6-67 give the relative amplitude of the reflected and transmitted waves, respectively. It is usual to define the coefficients of reflection R and transmission T, the relative rates at which particles are reflected and transmitted, in terms of the squares of the amplitudes A, B", ", and C as15 R \u0192B\u0192 2 \u0192A\u0192 2 a k1 k1 k2 k2 2 b T k2 k1 \u0192C\u0192 2 \u0192A\u0192 2 4k1k2 k2)2 (k1 from which it can be readily verified that T R 1 6-68 6-69 6-70 Among the interesting consequences of the wave nature of the solutions to Schr\u00f6dinger\u2019s equation, notice the following: 1. Even though E V0, R is not 0; i.e., in contrast to classical expectations, some of the particles are reflected from the step. (This is analogous to the internal reflection of electromagnetic waves at the interface of two media.) 2. The value of R depends on the difference between k1 and k2 but not on which is larger; i.e., a step down in the potential produces the same reflection as a step up of the same size. k p>U 2>, Since the wavelength changes as the beam passes the step. We \u2019 might also expect that the amplitude of will be less than that of the incident wave; II \u0192\u2019\u0192 2 is proportional to the particle density. Since particles move however, recall that the \u0192 2 k1), more slowly in Region II (k2 Figure 6-22b may be larger than illustrates these points. Figure 6-23 shows the time development of a wave packet incident on a potential step for E V0. \u23aa\u03a8(x, t)\u23aa2 \u0192\u2019 II \u0192\u2019 I \u0192 2. x Figure 6-23 Time development of a one-dimensional wave packet representing a particle incident on a step potential for E V0. The position of a classical particle is indicated by the dot. Note that part of the packet is transmitted and part is reflected. The sharp spikes that appear are artifacts of the discontinuity in the slope of V(x) at x 0. t 254 Chapter 6 The Schr\u00f6dinger Equation Figure 6-24 (a) A potential step. Particles are incident on the step from the left moving toward the right, each with total energy E V0. (b) The wave transmitted into region II is a decreasing exponential. However, the value of R in this case is 1 and no net energy is transmitted. (a) Energy V(x) = 0 V(x) = V0 E (b) 0 \u03c8(x) 0 x x Now let us consider the case shown in Figure 6-24a, where E V0. Classically, we expect all particles to be reflected at x 0; however, we note that k2 in Equation 6-64 is now an imaginary number since E V0. Thus, II(x) Ceik2x Cex \u2019 22m(V0 is a real exponential function where (We choose the positive S 0 root so that ) This means that the numerator and denominator of the right side of Equation 6-66 are complex conjugates of each other; hence \u0192B\u0192 2 \u0192A\u0192 2 and R 1 and T 0. Figure 6-25 is a graph of both R and T versus energy for a potential step. In agreement with the classical prediction, all of the particles (waves) are reflected back into Region I. However, another interesting result of our solution of Schr\u00f6dinger\u2019s equation is that the particle waves do not all reflect at x 0. x S. E)>U. 6-71 \u2019 II as T R, 1.0 0.8 0.6 0.4 0./V0 Top of step Figure 6-25 Reflection coefficient R and transmission coefficient T for a potential step V0 high versus energy E (in units of V0). 6-6 Reflection and Transmission of Waves 255 \u2019 II Since Region II is proportional to is an exponential decreasing toward the right, the particle density in \u0192\u2019 II \u0192 2 \u0192C\u0192 2e2x 6-72 Figure 6-24b shows the wave function for the case E V0. The wave function does not go to zero at x 0 but decays exponentially, as does the wave function for the bound state in a finite square well problem. The wave penetrates slightly into the classically forbidden region x 0 but eventually is completely reflected. (As discussed in Section 6-3, there is no prediction that a negative kinetic energy will be measured in such a region because to locate the particle in such a region introduces an uncertainty in the momentum corresponding to a minimum kinetic E.) This situation is similar to that of total internal reflecenergy greater than V0 tion in optics. EXAMPLE 6-6 Reflection from a Step with E V0 A beam of electrons, each with energy E 0.1 V", "0, is incident on a potential step with V0 2 eV. This is of the order of magnitude of the work function for electrons at the surface of metals. of particles penetrating the step up to a distance Graph the relative probability x 109 m, or roughly five atomic diameters. \u0192\u2019\u0192 2 SOLUTION For x 0, the wave function is given by Equation 6-71. The value of Equation 6-67, \u0192C\u0192 2 is, from \u0192C\u0192 2 ` 2(0.1 V0)1>2 (0.1 V0)1>2 (0.9 V0)1>2 2 ` 0.4 where we have taken to 109 m gives, with 6-2. Taking in Figure 6-26. e2x \u0192A\u0192 2 1. 2 2[2m(0.9 V0)]1>2>U, Computing e2x for several values of x from 0 the first two columns of the Table which is graphed yields \u0192\u2019\u0192 2, \u0192C\u0192 2 0.4 and then multiplying by Table 6-2 \u0192\u2019\u0192 2 x (m) 0 0.1 1010 1.0 1010 2.0 1010 5.0 1010 10.0 1010 2 x 0 0.137 1.374 2.748 6.869 13.74 \u0192\u2019\u0192 2 0.40 0.349 0.101 0.026 0.001 0 2 \u23aa \u03c8 \u23aa 0.4 0.3 0.2 0.1 0 0 2 4 6 8 x 10 (10\u201310 m) Figure 6-26 256 Chapter 6 The Schr\u00f6dinger Equation Barrier Potential Now let us consider one of the more interesting quantum-mechanical potentials, the barrier, illustrated by the example in Figure 6-27. The potential is V0 0 for 0 x and x a for 0 x a V(x) e 6-73 Classical particles incident on the barrier from the left in Region I with E V0 will all be transmitted, slowing down while passing through Region II but moving at their original speed again in Region III. For classical particles with E V0 incident from the left, all are reflected back into Region I. The quantum-mechanical behavior of particles incident on the barrier in both energy ranges is much different! First, let us see what happens when a beam of particles, all with the same energy E V0, as illustrated in Figure 6-27a, are incident from the left. The general solutions to the wave equation are, following the example of the potential step, I(x) Aeik1x Beik1x x 0 \u2019 II(x) Cex Dex 0 x \u2019 III(x) Feik1x Geik1x x a \u2019 6-74 III k1 and \u2019 22mE>U \u2019 and I 22m(V0 involves where, as before, contain complex exponentials. Since the partireal exponentials, whereas cle beam is incident on the barrier from the left, we can set G 0. Once again, the value of A is determined by the particle density in the beam and the four constants B, C, D, and F are found in terms of A by applying the continuity condition on and d\u2019>dx at x 0 and at x a. The details of the calculation are not of concern to us here, but several of the more interesting results are. E)>U. Note that \u2019 II \u2019 As we discovered for the potential step with E V0, the wave function incident from the left does not decrease immediately to zero at the barrier but instead will decay exponentially in the region of the barrier. Upon reaching the far wall of the barrier, the wave function must join smoothly to a sinusoidal wave function to the right of the barrier, as shown in Figure 6-27b. This implies that there will be some probability of the particles represented by the wave function being found on the far right side of the (a) Energy V0 E I 0 a II III (b) \u03c8(x) 0 a x x Figure 6-27 (a) Square barrier potential. (b) Penetration of the barrier by a wave with energy less than the barrier energy. Part of the wave is transmitted by the barrier even though, classically, the particle cannot enter the region 0 x a in which the potential energy is greater than the total energy. 6-6 Reflection and Transmission of Waves 257 barrier, although classically they should never be able to get through; i.e., there is a probability that the particles approaching the barrier can penetrate it. This phenomenon is called barrier penetration or tunneling (see Figure 6-28). The relative probability of its occurrence in any given", " situation is given by the transmission coefficient. The coefficient of transmission T from Region I into Region III is found to be (see Problem 6-64) T \u0192F\u0192 2 \u0192A\u0192 2 1 1 sinh2 a E E a1 V0 V0 4 J b K If a W 1, Equation 6-75 takes on the somewhat simpler form to evaluate T 16 a1 E V0 E V0 be2a 6-75 6-76 Scanning Tunneling Microscope In the scanning tunneling microscope (STM), developed in the 1980s by G. Binnig and H. Rohrer, a narrow gap between a conducting specimen and the tip of a tiny probe acts as a potential barrier to electrons bound in the specimen, as illustrated in Figure 6-29. A small bias voltage applied between the probe and the specimen causes the electrons to tunnel through the barrier separating the two surfaces if the surfaces are close enough together. The tunneling current is extremely sensitive to the size of the gap, i.e., the width of the barrier, between the probe and specimen. A change of only 0.5 nm (about the diameter of one atom) in the width of the barrier can cause the tunneling current to change by as much as a factor of 104. As the probe scans the specimen, a constant tunneling current is maintained by a piezoelectric feedback system that keeps the gap constant. Thus, the surface of the specimen can be mapped out by the vertical motions of the probe. In this way, the surface features of a specimen can be measured by STMs with a resolution of the order of the size of a single atom (see Figure 6-29). Figure 6-28 Optical barrier penetration, sometimes called frustrated total internal reflection. Because of the presence of the second prism, part of the wave penetrates the air barrier even though the angle of incidence in the first prism is greater than the critical angle. This effect can be demonstrated with two 45\u00b0 prisms and a laser or a microwave beam and 45\u00b0 prisms made of paraffin. An important application of tunneling is the tunnel diode, a common component of electronic circuits. Another is field emission, tunneling of electrons facilitated by an electric field, now being used in wide-angle, flat-screen displays on some laptop computers. Microtip \u2022 \u2022 \u2013 + \u0394V + \u2013 e\u2013 Figure 6-29 Schematic illustration of the path of the probe of an STM (dashed line) scanned across the surface of a sample while maintaining constant tunneling current. The probe has an extremely sharp microtip of atomic dimensions. Tunneling occurs over a small area across the narrow gap, allowing very small features (even individual atoms) to be imaged, as indicated by the dashed line. 258 Chapter 6 The Schr\u00f6dinger Equation Room temperature UHVSTM images of gold (Au) nanoparticles supported on TiC after annealing at 500\u00b0C. Images are (a) 375 nm 375 nm, (b) 200 200 nm, and (c) 100 100 nm. (d) A 3-D image of a 70 nm 70 nm section of (c). [The authors thank Beatriz Rold\u00e1n Cuenya for permission to use these STM images.] (a) (b) (c) (d) EXPLORING Alpha Decay Barrier penetration was used by G. Gamow, E. U. Condon, and R. W. Gurney in 1928 to explain the enormous variation in the mean life for decay of radioactive nuclei and the seemingly paradoxical very existence of decay.16 While radioactive decay will be discussed more thoroughly in Chapter 11, in general, the smaller the energy of the emitted particle, the larger the mean life. The energies of particles from natural radioactive sources range from about 4 to 7 MeV, whereas the mean lifetimes range from about 1010 years to 106 s. Gamow represented the radioactive nucleus by a potential well containing an particle, as shown in Figure 6-30a. For r less than the nuclear radius R, the particle is attracted by the nuclear force. Without knowing much about this force, Gamow and his co-workers represented it by a square well. Outside the nuparticle is repelled by the Coulomb force. This is represented by the cleus, the Coulomb potential energy kZze2 r, where z 2 for the particle and Ze is the re- maining nuclear charge. The energy E is the measured kinetic energy of the emitted particle, since when it is far from the nucleus its potential energy is zero. We see from the figure that a small increase in E reduces the relative height of the barrier V E and also reduces the thickness. Because the probability of transmission varies exponentially with the relative height and barrier thickness, as indicated by Equation 6-76, a small increase in E leads to a large increase in the probability of transmission and in turn to a shorter lifetime.", " Gamow and his co-workers were able to derive an expression decay rate and the mean lifetime as a function of energy E that was in good for the agreement with experimental results as follows: > 6-6 Reflection and Transmission of Waves 259 (a) V(r) (b) Po212 E r R r1 10 5 100 10 \u20135 10 \u201310 10 \u201315 ) 10 \u201320 0.3 0.4 E\u20131/2(MeV \u20131/2) U238 0.5 x Figure 6-30 (a) Model of potential-energy function for an particle and a nucleus. The strong attractive nuclear force for r less than the nuclear radius R can be approximately described by the potential well shown. Outside the nucleus the nuclear force is negligible, and the potential is given by Coulomb\u2019s law, V(r) kZze2 r, where Ze is the nuclear charge and ze is the charge of the particle. An particle inside the nucleus oscillates back and forth, being reflected at the barrier at R. Because of its wave properties, when the particle hits the barrier, there is a small chance that it will penetrate and appear outside the well at r r1. The wave function is similar to that shown in Figure 6-27b. (b) The decay rate for the emission of particles from radioactive nuclei. The solid curve is the prediction of Equation 6-79; the points are experimental results. > The probability that an proach is given by T from Equation 6-76. In fact, in this case ponential dominates the expression and particle will tunnel through the barrier in any one apis so large that the ex- a T e222m(V0 E)a>U 6-77 which is a very small number; i.e., the times per second N that the particle approaches the barrier is given roughly by particle is usually reflected. The number of N v 2R 6-78 where v equals the particle\u2019s speed inside the nucleus. Thus, the decay rate, or the probability per second that the nucleus will emit an particle, which is also the reciprocal of the mean life is given by, decay rate 1 v 2R e222m(V0 E)a>U 6-79 Figure 6-30b illustrates the good agreement between the barrier penetration calculation and experimental measurements. as a E>V0 1, In the event that there is no reflected wave a, 2, \u00c1 for result of destructive interference. For electrons incident on noble gas atoms the resulting 100 percent transmission is called Ramsauer-Townsend effect and is a way of measuring atomic diameters for those elements. 260 Chapter 6 The Schr\u00f6dinger Equation EXPLORING NH3 Atomic Clock Barrier penetration also takes place in the case of the periodic inversion of the ammonia molecule. The NH3 molecule has two equilibrium configurations, as illustrated in Figure 6-31a. The three hydrogen atoms are arranged in a plane. The nitrogen atom oscillates between two equilibrium positions equidistant from each of the H atoms above and below the plane. The potential energy function V(x) acting on the N atom has two minima located symmetrically about the center of the plane, as shown in Figure 6-31b. The N atom is bound to the molecule, so the energy is quantized and the lower states lie well below the central maximum of the potential. The central maximum presents a barrier to the N atoms in the lower states through which they slowly tunnel back and forth.17 The oscillation frequency f 2.3786 1010 Hz when the atom is in the state characterized by the energy E1 in Figure 6-31b. This frequency is quite low compared with the frequencies of most molecular vibrations, a fact that allowed the N atom tunneling frequency in NH3 to be used as the standard in the first atomic clocks, devices that now provide the world\u2019s standard for precision timekeeping. (a) x (b) V(x) E2 E1 x 0 Figure 6-31 (a) The NH3 molecule oscillates between the two equilibrium positions shown. The H atoms form a plane; the N atom is colored. (b) The potential energy of the N atom, where x is the distance above and below the plane of the H atoms. Several of the allowed energies, including the two lowest shown, lie below the top of the central barrier through which the N atom tunnels. More Quantum-mechanical tunneling involving two barriers is the basis for a number of devices such as the tunnel diode and the Josephson junction, both of which have a wide variety of useful applications. As an example of such systems, the Tunnel Diode is described on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Equation 6-80 and Figure 6-32 here. Summary TOPIC RELEVANT EQUATIONS AND REMARKS 1. Schr\u00f6dinger", " equation Time dependent, one space dimension U2 2m $2\u2019(x, t) $x2 V(x, t)\u2019(x, t) iU $\u2019(x, t) $t 6-6 TOPIC RELEVANT EQUATIONS AND REMARKS Time independent, one space dimension U2 2m d2\u2019(x) dx2 V(x)\u2019(x) E \u2019(x) Normalization condition Acceptability conditions and &*(x, t)&(x, t) dx 1 \u2019*(x)\u2019(x) dx 1 Summary 261 6-18 6-9 6-20 must exist and satisfy the Schr\u00f6dinger equation. must be continuous. 1. 2. 3. 4. 5. and and \u2019(x) \u2019(x) \u2019(x) \u2019(x) and \u2019(x) S 0 x S fast enough as Equation 6-20, remains bounded. d\u2019>dx d\u2019>dx d\u2019>dx must be finite. must be single valued. so that the normalization integral, 2. Infinite square well Allowed energies Wave functions 3. Finite square well 4. Expectation values and operators 5. Simple harmonic oscillator Allowed energies 6. Reflection and transmission n2 En 2U2 2mL2 n2E1 n 1, 2, 3, \u00c1 \u2019 n(x) 2 L A sin nx L n 1, 2, 3, \u00c1 6-24 6-32 For a finite well of width L the allowed energies En in the well are lower than the corresponding levels for an infinite well. There is always at least one allowed energy (bound state) in a finite well. The expectation or average value of a physical quantity represented by an operator, such as the momentum operator pop, is given by 8p9 \u2019 dx b\u2019 dx \u2019*a U \u2019* pop 6-48 $ $x i En an 1 2 b U n 0, 1, 2, \u00c1 6-56 When the potential changes abruptly in a distance small compared to the de Broglie wavelength, a particle may be reflected even though E V(x). A particle may also penetrate into a region where E V(x). 262 Chapter 6 The Schr\u00f6dinger Equation General References The following general references are written at a level appropriate for the readers of this book. French, A. P., and E. F. Taylor, An Introduction to Quantum Physics, Norton, New York, 1978. Brandt, S., and H. D. Dahmen, The Picture Book of Quantum Mechanics, Wiley, New York, 1985. Eisberg, R., and R. Resnick, Quantum Physics, 2d ed., Wiley, New York, 1985. Feynman, R. P., R. B. Leighton, and M. Sands, Lectures on Physics, Addison-Wesley, Reading, MA, 1965. Ford, K. W., The Quantum World, Harvard University Press, Cambridge, MA, 2004. Mehra, J., and H. Rechenberg, The Historical Development of Quantum Theory, Vol. 1, Springer-Verlag, New York, 1982. Park, D., Introduction to the Quantum Theory, 3d ed., McGraw-Hill, New York, 1992. Visual Quantum Mechanics, Kansas State University, Manhattan, 1996. Computer simulation software allows the user to analyze a variety of one-dimensional potentials, including the square wells and harmonic oscillator discussed in this chapter. Notes 1. Felix Bloch (1905\u20131983), Swiss American physicist. He was a student at the University of Zurich and attended the colloquium referred to. The quote is from an address before the American Physical Society in 1976. Bloch shared the 1952 Nobel Prize in Physics for measuring the magnetic moment of the neutron, using a method that he invented that led to the development of the analytical technique of nuclear magnetic resonance (NMR) spectroscopy. 2. Peter J. W. Debye (1884\u20131966), Dutch American physical chemist. He succeeded Einstein in the chair of theoretical physics at the University of Zurich and received the Nobel Prize in Chemistry in 1936. 3. Erwin R. J. A. Schr\u00f6dinger (1887\u20131961), Austrian physicist. He succeeded Planck in the chair of theoretical physics at the University of Berlin in 1928 following Planck\u2019s retirement and two years after publishing in rapid succession six papers that set forth the theory of wave mechanics. For that work he shared the Nobel Prize in Physics with P. A. M. Dirac in 1933. He left Nazi-controlled Europe in 1940, moving his household to Ireland. of multiplying 4. To see that this is indeed the case, consider the effect on $2 &(x, t)>$x2 by a factor C.", " Then $2C &(x, t)>$x2 C$2 &(x, t)>$x2, and the derivative is increased by the same factor. Thus, the derivative is proportional to the first power of the function; i.e., it is linear in &(x, t). &(x, t) 5. The imaginary i appears because the Schr\u00f6dinger equation relates a first time derivative to a second space derivative as a consequence of the fact that the total energy is related to the square of the momentum. This is unlike the classical wave equation (Equation 5-11), which relates two second deriva&(x, t) tives. The implication of this is that, in general, the will be complex functions, whereas the y(x, t) are real. & 6. The fact that is in general complex does not mean that its imaginary part doesn\u2019t contribute to the values of measurements, which are real. Every complex number can be i (1)1>2. written in the form z a bi, where a and b are real numThe magnitude or absolute value of z bers and The complex conjugate of z is is defined as z*z (a bi)(a bi) a2 b2 \u0192z\u0192 2; z* a bi, so \u0192&\u0192 2 thus, the value of will contain a contribution from its imaginary part. (a2 b2)1>2. 7. Here we are using the convention of probability and sta- tistics that certainty is represented by a probability of 1. 8. This method for solving partial differential equations is called separation of variables, for obvious reasons. Since most potentials in quantum mechanics, as in classical mechanics, are time independent, the method may be applied to the Schr\u00f6dinger equation in numerous situations. 9. We should note that there is an exception to this in the U2>2mL2. quantum theory of measurement. 10. E 0 corresponding to n0 is not a possible energy for a particle in a box. As discussed in Section 5-6, the uncertainty principle limits the minimum energy for such a particle to values 11. Recalling that linear combinations of solutions to Schr\u00f6dinger\u2019s equation will also be solutions, we should note here that simulation of the classical behavior of a macroscopic particle in a macroscopic box requires wave functions that are the superpositions of many stationary states. Thus, the classical particle never has definite energy in the quantum mechanical sense. 12. To simplify the notation in this section, we will sometimes omit the functional dependence and merely write for \u2019 n(x) 13. The Hermite polynomials are known functions that are tabulated in most books on quantum mechanics. 14. It is straightforward to show that the only difference benormalized in terms of the particle density and tween a one for which is the probability density is a multiplicative constant. n(x, t). \u0192\u2019(x)\u0192 2 \u2019(x) & n and for & \u2019 n Problems 263 15. T and R are derived in terms of the particle currents, i.e., particlesNunit time, in most introductory quantum mechanics books. 16. Rutherford had shown that the scattering of 8.8-MeV particles from the decay of 212Po obeyed the Coulomb force law down to distances of the order of 3 1014 m, i.e., down to about nuclear dimensions. Thus, the Coulomb barrier at that distance was at least 8.8 MeV high; however, the energy of particles emitted by 238U is only 4.2 MeV, less than half the barrier height. How that could be possible presented classical physics with a paradox. 17. Since the molecule\u2019s center of mass is fixed in an inertial reference frame, the plane of H atoms also oscillates back and forth in the opposite direction to the N atom; however, their mass being smaller than that of the N atom, the amplitude of the plane\u2019s motion is actually larger than that of the N atom. It is the relative motion that is important. 18. See, for example, F. Capasso and S. Datta, \u201cQuantum Electron Devices,\u201d Physics Today, 43, 74 (1990). Leo Esaki was awarded the Nobel Prize in Physics in 1973 for inventing the resonant tunnel diode. Problems Level I Section 6-1 The Schr\u00f6dinger Equation in One Dimension &(x, t) Aekxt is given by \u2019(x) Aex2>2L2 &(x, t) Aei(kxt) does not satisfy the time-dependent satisfies both the time-dependent Schr\u00f6dinger equation In a region of space, a particle has a wave function given by and energy where L is some length. (a) Find the potential", " energy as a function of x, and sketch V 6-1. Show that the wave function Schr\u00f6dinger equation. 6-2. Show that and the classical wave equation (Equation 6-1). 6-3. U2>2mL2, versus x. (b) What is the classical potential that has this dependence? (a) For Problem 6-3, find the kinetic energy as a function of x. (b) Show that x L is the 6-4. classical turning point. (c) The potential energy of a simple harmonic oscillator in terms of its 2 m 2x2. Compare this with your answer to part (a) of angular frequency Problem 6-3, and show that the total energy for this wave function can be written 6-5. (a) Show that the wave function dependent Schr\u00f6dinger equation. (b) Show that does satisfy this equation. 6-6. The wave function for a free electron, i.e., one on which no net force acts, is given by \u2019(x) A sin(2.5 1010x), where x is in meters. Compute the electron\u2019s (a) momentum, (b) total energy, and (c) de Broglie wavelength. 6-7. A particle with mass m and total energy zero is in a particular region of space where its (a) Find the potential energy V(x) versus x and (b) make a wave function is sketch of V(x) versus x. 6-8. Normalize the wave function in Problem 6-2 between a and a. Why can\u2019t that wave function be normalized between does not satisfy the time&(x, t) A cos(kx t) iA sin(kx t) &(x, t) A sin(kx t) \u2019(x) Cex2>L2. V(x) 1 E 1 2 h.? and Section 6-2 The Infinite Square Well 6-9. A particle is in an infinite square well of width L. Calculate the ground-state energy if (a) the particle is a proton and L 0.1 nm, a typical size for a molecule; (b) the particle is a proton and L 1 fm, a typical size for a nucleus. 6-10. A particle is in the ground state of an infinite square well potential given by Equation 6-21. Find the probability of finding the particle in the interval x 0.002 L at (a) x L 2, (b) x 2L 3, and (c) x L. (Since x is very small, you need not do any integration.) 6-11. Do Problem 6-10 for a particle in the second excited state (n 3) of an infinite square well potential. 6-12. A mass of 106 g is moving with a speed of about 101 cm s in a box of length 1 cm. Treating this as a one-dimensional infinite square well, calculate the approximate value of the quantum number n. > > > 264 Chapter 6 The Schr\u00f6dinger Equation > > (xp)>U? 6-13. (a) For the classical particle of Problem 6-12, find x and p, assuming that x L 0.01 percent and p p 0.01 percent. (b) What is > 6-14. A particle of mass m is confined to a tube of length L. (a) Use the uncertainty relationship to estimate the smallest possible energy. (b) Assume that the inside of the tube is a forcefree region and that the particle makes elastic reflections at the tube ends. Use Schr\u00f6dinger\u2019s equation to find the ground-state energy for the particle in the tube. Compare the answer to that of part (a). 6-15. (a) What is the wavelength associated with the particle of Problem 6-14 if the particle is in its ground state? (b) What is the wavelength if the particle is in its second excited state (quantum number n 3)? (c) Use de Broglie\u2019s relationship to find the magnitude for the momentum of the particle in its ground state. (d) Show that p2 2m gives the correct energy for the ground state of this particle in the box. 6-16. The wavelength of light emitted by a ruby laser is 694.3 nm. Assuming that the emission of a photon of this wavelength accompanies the transition of an electron from the n 2 level to the n 1 level of an infinite square well, compute L for the well. 6-17. The allowed energies for a particle of mass m in a one-dimensional infinite square well are given by Equation 6-24. Show that a level with n 0 violates the Heisenberg uncertainty principle. 6-18. Suppose a macroscopic bead with a mass of 2", ".0 g is constrained to move on a straight frictionless wire between two heavy stops clamped firmly to the wire 10 cm apart. If the bead is moving at a speed of 20 nm y (i.e., to all appearances it is at rest), what is the value of its quantum number n? 6-19. An electron moving in a one-dimensional infinite square well is trapped in the n 5 state. (a) Show that the probability of finding the electron between x 0.2 L and x 0.4 L is 1 5. (b) Compute the probability of finding the electron within the \u201cvolume\u201d x 0.01 L at > x L 2. 6-20. In the early days of nuclear physics before the neutron was discovered, it was thought that the nucleus contained only electrons and protons. If we consider the nucleus to be a onedimensional infinite well with L 10 fm and ignore relativity, compute the ground-state energy for (a) an electron and (b) a proton in the nucleus. (c) Compute the energy difference between the ground state and the first excited state for each particle. (Differences between energy levels in nuclei are found to be typically of the order of 1 MeV.) 6-21. An electron is in the ground state with energy En of a one-dimensional infinite well with L 1010 m. Compute the force that the electron exerts on the wall during an impact on either >dL. wall. (Hint: Why?) How does this result compare with the weight of an electron at the surface of Earth? 6-22. The wave functions of a particle in a one-dimensional infinite square well are given by \u2019 Equation 6-32. Show that for these functions are orthogonal. m(x) dx 0, F dEn i.e., that \u2019 m(x) n(x)\u2019 \u2019 n(x) and > > Section 6-3 The Finite Square Well 6-23. Sketch (a) the wave function and (b) the probability distribution for the n 4 state for the finite square well potential. 6-24. A finite square well 1.0 fm wide contains one neutron. How deep must the well be if there are only two allowed energy levels for the neutron? 6-25. An electron is confined to a finite square well whose \u201cwalls\u201d are 8.0 eV high. If the ground-state energy is 0.5 eV, estimate the width of the well. 6-26. Using arguments concerning curvature, wavelength, and amplitude, sketch very carefully the wave function corresponding to a particle with energy E in the finite potential well shown in Figure 6-33. Figure 6-33 Problem 6-26. Problems 265 Energy V2 V2 V1 V = 0 0 x 6-27. For a finite square well potential that has six quantized levels, if a 10 nm (a) sketch the finite well, (b) sketch the wave function from x 2a to x 2a for n 3, and (c) sketch the probability density for the same range of x. Section 6-4 Expectation Values and Operators > 8x9 8x9 and 8x29 8x29 and (b) 92U2>2mL2. for a classical particle in such a well. for the second excited state (n 3) in an infinite square well 6-28. Compute the expectation value of the x component of the momentum of a particle of mass m in the n 3 level of a one-dimensional infinite square well of width L. Reconcile your answer with the fact that the kinetic energy of the particle in this level is 6-29. Find (a) potential. 6-30. (a) Show that the classical probability distribution function for a particle in a onedimensional infinite square well potential of length L is given by P(x) 1 L. (b) Use your result in (a) to find 6-31. Show directly from the time-independent Schr\u00f6dinger equation that 82m[E V(x)]9 8p29 82mE9 8p29 to compute x in general and that for the ground state of the infinite square well. x by symmetry and 6-32. Find function of an infinite square well. (Use the fact that from Problem 6-31.) 8x9 6-33. Compute A0 6-34. Use conservation of energy to obtain an expression connecting x2 and p2 for a harmonic oscillator, then use it along with the result from Problem 6-33 to compute for the harmonic oscillator ground state. 6-35. (a) Using A0 from Problem 6-33, write down the total wave function ground state of a harmonic oscillator. (b) Use the operator for 8p29 for the infinite square well. Use this result for the ground state", " of a harmonic oscillator (Equation 6-58). Use for the ground-state wave 8p29 82mE9 from Table 6-1 to compute 28p29 8p92, 28x29 8x92, and 8p9 0 (m>U)1>4. for the 8p29. 0(x, t) 8p29 8x29 and p p & px Section 6-5 The Simple Harmonic Oscillator 6-36. For the harmonic oscillator ground state n 0 the Hermite polynomial Hn(x) in Equation 8V(x)9 1. Find (a) the normalization constant C0, (b) 6-57 is given by H0 for this state. (Hint: Use the Probability Integral in Appendix B1 to compute the needed integrals.) 6-37. For the first excited state, H1(x) x. Find (a) the normalization constant C1, (b) (c) 6-38. A quantum harmonic oscillator of mass m is in the ground state with classical turning points at A. (a) With the mass confined to the region compute p for this state. (b) Compare the kinetic energy implied by p with (1) the ground-state total energy and (2) the expectation value of the kinetic energy. for this state (see Problem 6-36). x 2A, 8V(x)9 and (c) 8x29, 8x29, 8x9, (d) 266 Chapter 6 The Schr\u00f6dinger Equation > and 8x29 n S. for (a) the ground state and (b) the first excited state of the 6-39. Compute the spacing between adjacent energy levels per unit energy, i.e., En En, for the quantum harmonic oscillator and show that the result agrees with Bohr\u2019s correspondence principle (see Section 4-3) by letting 8x9 6-40. Compute harmonic oscillator. 6-41. The period of a macroscopic pendulum made with a mass of 10 g suspended from a massless cord 50 cm long is 1.42 s. (a) Compute the ground-state (zero-point) energy. (b) If the pendulum is set into motion so that the mass raises 0.1 mm above its equilibrium position, what will be the quantum number of the state? (c) What is the frequency of the motion in (b)? 6-42. Show that the wave functions for the ground state and the first excited state of the simple 1(x) dx 0. harmonic oscillator, given in Equation 6-58, are orthogonal; that is, show that 0(x)\u2019 \u2019 and Section 6-6 Reflection and Transmission of Waves 6-43. A free particle of mass m with wave number k1 is traveling to the right. At x 0, the potential jumps from zero to V0 and remains at this value for positive x. (a) If the total energy is >2m 2V0, E U2k2 what is the wave number k2 in the region x 0? Express your answer in 1 V0. k1 (b) Calculate the reflection coefficient R at the potential step. (c) What is the terms of transmission coefficient T? (d) If one million particles with wave number k1 are incident upon the potential step, how many particles are expected to continue along in the positive x direction? How does this compare with the classical prediction? 6-44. In Problem 6-43, suppose that the potential jumps from zero to V0 at x 0 so that the free particle speeds up instead of slowing down. The wave number for the incident particle is again k1, and the total energy is 2V0. (a) What is the wave number for the particle in the region of positive x? (b) Calculate the reflection coefficient R at the potential step. (c) What is the transmission coefficient T? (d) If one million particles with wave number k1 are incident upon the potential step, how many particles are expected to continue along in the positive x direction? How does this compare with the classical prediction? 6-45. In a particular semiconductor device an oxide layer forms a barrier 0.6 nm wide and 9 V high between two conducting wires. Electrons accelerated through 4 V approach the barrier. (a) What fraction of the incident electrons will tunnel through the barrier? (b) Through what potential difference should the electrons be accelerated in order to increase the tunneling fraction by a factor of 2? 6-46. For particles incident on a step potential with E V0, show that T 0 using Equation 6-70. 6-47. Derive Equations 6-66 and 6-67 from those that immediately precede them. 6-48. A beam of", " electrons, each with kinetic energy E 2.0 eV, is incident on a potential bar 6.5 eV and width 5.0 1010 m. (See Figure 6-26.) What fraction of the elecrier with V0 trons in the beam will be transmitted through the barrier? 6-49. A beam of protons, each with kinetic energy 40 MeV, approaches a step potential of 30 MeV. (a) What fraction of the beam is reflected and transmitted? (b) Does your answer change if the particles are electrons? Level II 6-50. A proton is in an infinite square well potential given by Equation 6-21 with L 1 fm. (a) Find the ground-state energy in MeV. (b) Make an energy-level diagram for this system. Calculate the wavelength of the photon emitted for the transitions (c) n 2 to n 1, (d) n 3 to n 2, and (e) n 3 to n 1. 6-51. A particle is in the ground state of an infinite square well potential given by Equation 2 L, 6-21. Calculate the probability that the particle will be found in the region (a) 4 L. (b) 6-52. (a) Show that for large n, the fractional difference in energy between state n and state n 1 for a particle in an infinite square well is given approximately by 0 x 1 0 x 3 0 x 1 and (c) 3 L, En1 En En 2 n (b) What is the approximate percentage energy difference between the states n1 n2 1001? (c) Comment on how this result is related to Bohr\u2019s correspondence principle. 1000 and Problems 267 6-53. Compute the expectation value of the kinetic energy of a particle of mass m moving in the n 2 level of a one-dimensional infinite square well of width L. 6-54. A particle of mass m is in an infinite square well potential given by V x L>2 V 0 L>2 x L>2 V L>2 x Since this potential is symmetric about the origin, the probability density \u2019(x) \u2019(x). symmetric. (a) Show that this implies that either that the proper solutions of the time-independent Schr\u00f6dinger equation can be written \u2019(x) \u2019(x) or must also be (b) Show \u0192\u2019(x)\u0192 2 and \u2019(x) 2 L A cos nx L n 1, 3, 5, 7, \u00c1 \u2019(x) 2 L A sin nx L n 2, 4, 6, 8, \u00c1 (c) Show that the allowed energies are the same as those for the infinite square well given by Equation 6-24. 6-55. The wave function represents the ground-state energy of a harmonic oscillator. (a) Show that is also a solution of Schr\u00f6dinger\u2019s equation. (b) What is the energy of this new state? (c) From a look at the nodes of this wave function, how would you classify this excited state? 6-56. For the wave functions 0(x) Aex2>2L2 \u2019 1(x) L d\u2019 \u2019 0(x)>dx \u2019(x) 2 L A sin nx L n 1, 2, 3, \u00c1 corresponding to an infinite square well of width L, show that 8x29 L2 3 L2 2n22 6-57. A 10-eV electron is incident on a potential barrier of height 25 eV and width 1 nm. (a) Use Equation 6-76 to calculate the order of magnitude of the probability that the electron will tunnel through the barrier. (b) Repeat your calculation for a width of 0.1 nm. 6-58. A particle of mass m moves in a region in which the potential energy is constant V V0. (a) Show that neither satisfies the timenor then C1 and dependent Schr\u00f6dinger equation. (Hint: If C2 must be zero.) (b) Show that does satisfy the time-independent Schr\u00f6dinger equation providing that k, V0, and are related by Equation 6-5. &(x, t) A[cos(kx t) i sin(kx t)] Aei(kxt) C1 sin C 2 cos 0 &(x, t) A cos(kx t) &(x, t) A sin(kx t) for all values of, Level III 6-59. A particle of mass m on a table at z 0 can be described by the potential energy V mgz for z 0 V for z 0 For some positive value of total energy E, indicate the classically allowed region on a sketch of V(z", ") versus z. Sketch also the kinetic energy versus z. The Schr\u00f6dinger equation for this problem is quite difficult to solve. Using arguments similar to those in Section 6-3 about the curvature of a wave function as given by the Schr\u00f6dinger equation, sketch your \u201ceducated guesses\u201d for the shape of the wave function for the ground state and the first two excited states. 268 Chapter 6 The Schr\u00f6dinger Equation 6-60. Use the Schr\u00f6dinger equation to show that the expectation value of the kinetic energy of a particle is given by 8Ek 9 \u2019(x)a U2 2m d2\u2019(x) dx2 b dx 1>2 6-61. An electron in an infinite square well with L 1012 m is moving at relativistic speed; hence, the momentum is not given by p (2mE). (a) Use the uncertainty principle to verify that the speed is relativistic. (b) Derive an expression for the electron\u2019s allowed energy levels and (c) compute E1. (d) By what fraction does E1 computed in (c) differ from the nonrelativistic E1? 6-62. (a) Derive Equation 6-75. (b) Show that, if Equation 6-76 follows from Equation 6-75 as an approximation. 6-63. A beam of protons, each with energy E 20 MeV, is incident on a potential step 40 MeV high. Graph the relative probability of finding protons at values of x 0 from x 0 to x 5 fm. (Hint: Take and refer to Example 6-6.) x W 1, \u0192A\u0192 2 1 CHAPTER 7 Atomic Physics In this chapter we will apply quantum theory to atomic systems. For all neutral atoms except hydrogen, the Schr\u00f6dinger equation cannot be solved exactly. Despite this, it is in the realm of atomic physics that the Schr\u00f6dinger equation has had its greatest success because the electromagnetic interaction of the electrons with one another and with the atomic nucleus is well understood. With powerful approximation methods and high-speed computers, many features of complex atoms, such as their energy levels and the wavelengths and intensities of their spectra, can be calculated, often to whatever accuracy is desired. The Schr\u00f6dinger equation for the hydrogen atom was first solved in Schr\u00f6dinger\u2019s first paper on quantum mechanics, published in 1926. This problem is of considerable importance not only because the Schr\u00f6dinger equation can be solved exactly in this case, but also because the solutions obtained form the basis for the approximate solutions for other atoms. We will therefore discuss this problem in some detail. Although the mathematics that arises in solving the Schr\u00f6dinger equation is a bit difficult in a few places, we will be as quantitative as possible, presenting results without proof and discussing important features of these results qualitatively only when necessary. Whenever possible, we will give simple physical arguments to make important results plausible. 7-1 The Schr\u00f6dinger Equation in Three Dimensions In Chapter 6 we considered motion in just one dimension, but of course the real world is three-dimensional. Although in many cases the one-dimensional form brings out the essential physical features, there are some considerations introduced in threedimensional problems that we want to examine. In rectangular coordinates, the timeindependent Schr\u00f6dinger equation is U2 2m a $2\u2019 $x2 $2\u2019 $y2 $2\u2019 $z2 b V\u2019 E\u2019 7-1 The wave function and the potential energy are generally functions of all three coordinates x, y, and z. 7-1 The Schr\u00f6dinger Equation in Three Dimensions 269 7-2 Quantization of Angular Momentum and Energy in the Hydrogen Atom 272 7-3 The Hydrogen Atom Wave Functions 7-4 Electron Spin 7-5 Total Angular 281 285 Momentum and the Spin-Orbit Effect 291 7-6 The Schr\u00f6dinger Equation for Two (or More) Particles 7-7 Ground States of Atoms: The Periodic Table 7-8 Excited States and Spectra of Atoms 295 297 301 269 270 Chapter 7 Atomic Physics Infinite Square Well in Three Dimensions Let us consider the three-dimensional version of a particle in a cubical box. The potential energy function V(x, y, z) 0 for 0 x L, 0 y L, and 0 z L. V is infinite outside this region. For this problem, the wave function must be zero at the walls of the box and will be a sine function inside the box. In fact, if we consider just one coordinate such as x, the solution will be the same as in the one-dimensional box discussed in Section 6-2. That is, the x dependence of the wave function will be of the, form sin k1x with the restriction where n1", " is an integer. The complete wave \u2019(x, y, z) can be written as a product of a function of x only, a function of y function only, and a function of z only. k1L n1 \u2019(x, y, z) \u2019 1(x)\u2019 2(y)\u2019 3(z) 7-2 where each of the functions For example, if we try the solution \u2019 n is a sine function as in the one-dimensional problem. \u2019(x, y, z) A sin k1x sin k2y sin k3z 7-3 we find by inserting this function into Equation 7-1 that the energy is given by which is equivalent to E U2 2m (k2 1 k2 2 k2 3) E p2 z) (p2 x p2 y 2m px Uk1 and so forth. Using the restrictions on the wave numbers >L with from the boundary condition that the wave function be zero at the walls, we obtain for the total energy ni ki En1n2n3 U22 2mL2 (n2 1 n2 2 n2 3) 7-4 where n1, n2, and n3 are integers greater than zero, as in Equation 6-24. Notice that the energy and wave function are characterized by three quantum numbers, each arising from a boundary condition on one of the coordinates. In this case the quantum numbers are independent of one another, but in more-general problems the value of one quantum number may affect the possible values of the others. For example, as we will see in a moment, in problems such as the hydrogen atom that have a spherical symmetry, the Schr\u00f6dinger equation is most readily solved in spherical coordinates r, and The quantum numbers associated with the boundary conditions on these coordinates are interdependent. #., The lowest energy state, the ground state for the cubical box, is given by Equation 1. The first excited energy level can be obtained in three differ 1 since 2, n1 E112. Each has a different wave function. n2 7-4 with n1 ent ways: either n1 we see from Equation 7-4 that E211 1 or n2 E121 n3 2, n2 1 or n3 2, n1 n3 n2 n3 7-1 The Schr\u00f6dinger Equation in Three Dimensions 271 E122 = E212 = E221 = 9E1 E211 = E121 = E112 = 6E1 L1 = L2 = L3 L1 < L2 < L3 E221 E212 E122 E211 E121 E112 E111 = 3E1 (a) (b) Figure 7-1 Energy-level diagram for (a) cubic infinite square well potential and (b) noncubic infinite square well. In the cubic well, the energy levels above the ground state are threefold degenerate; i.e., there are three wave functions having the same energy. The degeneracy is removed when the symmetry of the potential is removed, as in (b). The diagram is only schematic, and none of the levels in (b) necessarily has the same value of the energy as any level in (a). For example, the wave function for n1 2 and n2 n3 y 2x L L sin sin 1 is of the form z L \u2019 211 A sin \u2019(x, y, z) An energy level that has more than one wave function associated with it is said to be degenerate. In this case there is threefold degeneracy because there are three wave functions corresponding to the same energy. The degeneracy is related to the symmetry of the problem, and anything that destroys or breaks the symmetry will also destroy or remove the degeneracy.1 If, for example, we considered a noncubical box V 0 for 0 x L1, 0 y L2, and 0 z L3, the boundary condition at the walls would lead to the quantum conditions and k3L3 and the total energy would be n1 n2 n3 k1L1 k2L2,,, En1n2n3 U22 2m a n2 1 L2 1 n2 2 L2 2 b n2 3 L2 3 7-5 Figure 7-1 shows the energy levels for the ground state and first two excited states L3, for which the excited states are degenerate, and when L1, L2, and when L1 L3 are slightly different, in which case the excited levels are slightly split apart and the degeneracy is removed. L2 The Schr\u00f6dinger Equation in Spherical Coordinates In the next section we are going to consider another, different potential, that of a real atom. Assuming the proton to be at rest, we can treat the hydrogen atom as a single particle, an electron moving with kinetic energy p2", " 2me and a potential energy V(r) due to the electrostatic attraction of the proton: > V(r) Zke2 r 7-6 As in the Bohr theory, we include the atomic number Z, which is 1 for hydrogen, so we can apply our results to other similar systems, such as ionized helium He, where Z 2. We also note that we can account for the motion of the nucleus by replacing 272 Chapter 7 Atomic Physics Figure 7-2 Geometric relations between spherical (polar) and rectangular coordinates. z P \u03b8 r Range of variables Cartesian x, y, z: \u2013\u221e \u2192 +\u221e Spherical r: 0 \u2192 +\u221e \u03b8: 0 \u2192 \u03c0 \u03c6: 0 \u2192 2\u03c0 z = r cos \u03b8 y \u03c6 r sin\u03b8 x = r sin \u03b8 cos \u03c6 y = r sin \u03b8 sin \u03c6 x Hydrogenlike atoms, those with a single electron, have been produced from elements up to and including U91. Highly ionized atomic beams are used to further our understanding of relativistic effects and atomic structure. Collision of two completely ionized Au atoms, each moving at nearly the speed of light, produced the \u201cstar\u201d of thousands of particles reproduced on page 562 in Chapter 12. the electron mass me by the reduced mass where MN is the mass of the nucleus. The time-independent Schr\u00f6dinger equation for a particle of mass moving in three dimensions is Equation 7-1, with m replaced by : me >(1 me >MN), U2 2 a $2\u2019 $x2 $2\u2019 $y2 $2\u2019 $z2 b V\u2019 E\u2019 7-7 Since the potential energy V(r) depends only on the radial distance the problem is most conveniently treated in spherical coordinates related to x, y, and z by r (x2 y2 z2)1>2, r,,. These are and x r sin cos y r sin sin z r cos 7-8 These relations are shown in Figure 7-2. The transformation of the three-dimensional Schr\u00f6dinger equation into spherical coordinates is straightforward but involves much tedious calculation, which we will omit. The result is U2 2 1 r2 $ $r ar2 $\u2019 $r b U2 2r2 c 1 sin $ $ asin $\u2019 $ b 1 sin2 $2\u2019 $2 d V(r)\u2019 E\u2019 7-9 Despite the formidable appearance of this equation, it was not difficult for Schr\u00f6dinger to solve because it is similar to other partial differential equations that arise in classical physics, and such equations had been thoroughly studied. We will present the solution of this equation in detail, taking care to point out the origin of the quantum number associated with each dimension. As was the case with the threedimensional square well, the new quantum numbers will arise as a result of boundary conditions on the solution of the wave equation, Equation 7-9 in this case. 7-2 Quantization of Angular Momentum and Energy in the Hydrogen Atom In this section we will solve the time-independent Schr\u00f6dinger equation for hydrogen and hydrogenlike atoms. We will see how the quantization of both the energy and the angular momentum arise as natural consequences of the acceptability conditions on the wave function (see Section 6-1) and discover the origin and physical meaning of the quantum numbers n, and m., 7-2 Quantization of Angular Momentum and Energy in the Hydrogen Atom 273 The first step in the solution of a partial differential equation such as Equation 7-9 is to search for separable solutions by writing the wave function as a product of functions of each single variable. We write \u2019(r,, ) \u2019(r,, ) R(r)f()g() 7-10,. \u2019(r,, ) and g depends where R depends only on the radial coordinate r, f depends only on only on When this form of is substituted into Equation 7-9, the partial differential equation can be transformed into three ordinary differential equations, one for R(r), one for \u0192( ), and one for g( ). Most of the solutions of Equation 7-9 are, of course, not of this separable product form; however, if enough product solutions of the form of Equation 7-10 can be found,2 all solutions can be expressed as superpositions of them. Even so, the separable solutions given by Equation 7-10 turn out to be the most important ones physically because they correspond to definite values (eigenvalues) of energy and angular momentum. When Equation 7-10 is substituted into Equation 7-9 and the indicated differentiations are performed, we obtain U2 2 fg 1 r2 d dr ar2 dR dr b U2 2r2 R", "g 1 sin d d asin U2 2r2 b df d Rf sin2 d2g d2 VRfg ERfg 7-11 since derivatives with respect to r do not affect \u0192( ) and g( ), derivatives with respect to do not affect R(r) and \u0192( ). do not affect R(r) and g( ), and those with respect to Separation of the r-dependent functions from the - and -dependent ones is accomplished by multiplying Equation 7-11 by and rearranging slightly to obtain 2r2>(U2Rfg) 1 R(r) d dr ar2 dR(r) dr b 2r2 U2 cE V(r) d c 1 f()sin d d asin df() d b 1 g()sin2 d2g() d2 d 7-12 Note two points about Equation 7-12: (1) The left side contains only terms that are functions of r, while the right side has only terms depending on and Since the variables are independent, changes in r cannot change the value of the right side of the equation, have any effect on the left side. Thus, the two sides of the nor can changes in and ( 1). equation must be equal to the same constant, which we will call, with foresight, (2) The potential is a function only of r so the solution of the right side, the angular part, of Equation 7-12 will be the same for all potentials that are only functions3 of r.. In view of the second point above, we will first solve the angular equation so that its results will be available to us as we consider solutions to the r-dependent equation, referred to usually as the radial equation, for various V(r). Setting the right side of Equation sin2, 7-12 equal to ( 1)sin2 sin f() and rearranging slightly, we obtain d2g() d2 multiplying by df () d ( 1), 1 g() csin d d 7-13 d Once again we see that the two sides of the relation, Equation 7-13, are each a function of only one of the independent variables hence both sides must be equal to the same constant, which we will, again with foresight, call m2. Setting the left side of Equation 7-13 equal to m2 and solving for g( ) yields gm() eim 7-14 274 Chapter 7 Atomic Physics \u2019 The single valued condition on which in turn requires that m be a positive or negative integer or zero. (see Section 6-1) implies that g( 2) g(), Now letting the right side of Equation 7-13 equal m2 and solving for \u0192( ), we obtain (not intended to be obvious; for the detailed solution see Weber and Arfken, Chapter 11): fm() (sin )\u0192m\u0192 2! c d d(cos ) d \u0192m\u0192 (cos2 1) 7-15 \u2019 be finite requires that \u0192( ) be finite at The condition that which restricts the values of to zero and positive integers and limits tion reflects the link between values of m ranging from 0 up to called the associated Legendre functions. The subset of those with as the Legendre polynomials. fm(), The notaand m, namely, that each value of has associated. given by Equation 7-15, are is referred to which describes the angular dependence of for all spherically symmetric potentials, forms an often-encountered family The product of m. The functions fm(), gm(), m 0 0 and and \u2019(r,, ) of functions Ym(, ), Ym(, ) fm()gm() 7-16 called the spherical harmonics. The first few of these functions, which give the combined angular dependence of the motion of the electron in the hydrogen atom, are given in Table 7-1. The associated Legendre functions and the Legendre polynomials (m 0) can, if needed, be easily taken from the same table. (Extended tables of both functions can be found in Weber and Arfken.) In the following section we will discover the physical significance of and m. Table 7-1 Spherical harmonics 0 1 2 m 1 Y11 Y00 A Y11 A Y10 A Y22 A Y21 A Y20 A Y21 Y22 A A 1 4 3 8 sin ei 3 4 cos 3 8 sin ei 15 32 sin2 e2i 15 8 sin cos ei 5 16(3 cos2 1) 15 8 sin cos ei 15 32 sin2 e2i 7-2 Quantization of Angular Momentum and Energy in the Hydrogen Atom 275 Quantization of the Angular Momentum The definition of the angular momentum L of a mass m moving with velocity v, hence momentum p, at", " some location r relative to the origin, given in most introductory physics textbooks, is L r p where the momentum p m(dr dt). In cases where V V(r), such as the electron in the hydrogen atom, L is conserved (see Problem 7-15) and the classical motion of the mass m lies in a fixed plane perpendicular to L, which contains the coordinate origin. The momentum p has components (in that plane) pr along r and pt perpendicular to r, as illustrated in Figure 7-3, whose magnitudes are given by > pr a dr dt b and pt r a dA dt b and the magnitude of the conserved (i.e., constant) vector L is L rp sin A rpt The kinetic energy can be written in terms of these components as p2 2 p2 r p2 t 2 p2 r 2 L2 2r2 from which the classical total energy E is given by p2 r 2 L2 2r2 V(r) E 7-17 Rewriting Equation 7-17 in terms of the \u201ceffective\u201d potential as is often done, we obtain Veff(r) L2>2r2 V(r), p2 r 2 Veff (r) E 7-18 which is identical in form to Equation 6-4, which we used as a basis for our introduction to the Schr\u00f6dinger equation. pt p A pr r Orbit Figure 7-3 The orbit of a classical particle with V V(r) lies in a plane perpendicular to L. The components of the momentum p parallel and perpendicular to r are pr and pt, respectively. The momentum p makes an angle A with the displacement r. 276 Chapter 7 Atomic Physics Equation 7-17 can be used to write the Schr\u00f6dinger equation, just as we did in Chapter 6 by inserting de Broglie\u2019s relation and the appropriate differential operators Doing so is a lengthy though not particularly in spherical coordinates for the operator turns out to be difficult exercise whose details we will omit here. For and L2. p2 r p2 r (p2 r)op U2 1 r2 $ $r ar2 $ $r b 7-19 2 which, divided by you recognize as the first term (kinetic energy) of the Schr\u00f6dinger equation in spherical coordinates (Equation 7-9). Similarly, the operator for L2 turns out to be and operating on \u2019, (L2)op U2c 1 sin $ $ asin $ $ b 1 sin2 $2 $2 d 7-20 2r2 and operating on is the second term of the Schr\u00f6dinger which, divided by equation in spherical coordinates (Equation 7-9). The right side of Equation 7-12, U2f()g(), which equals remembering that can now be written as follows when multiplied by fm()gm() Ym(, ): ( 1), \u2019, U2c 1 sin $ $ asin $ $ b 1 sin2 $2 $2 dYm(, ) ( 1)U2Ym(, ) 7-21a or (L2)opYm(, ) ( 1)U2Ym(, ) or, since \u2019(r,, ) R(r)Y(, ), (L2)op \u2019(r,, ) ( 1)U2\u2019(r,, ) 7-21b 7-21c Thus, we have the very important result that, for all potentials where V V(r), the angular momentum is quantized and its allowed magnitudes (eigenvalues) are given by \u0192L\u0192 L 2( 1)U for 0, 1, 2, 3, \u00c1 7-22 where tum number. is referred to as the angular momentum quantum number or the orbital quan- In addition, if we use the same substitution method on L, we find that the z component of the angular momentum is also quantized and its allowed values are given by the z component of Lz, mU for m 0, 1, 2, \u00c1, Lz 7-23 The physical significance of Equation 7-23 is that the angular momentum L, whose magnitude is quantized with values can only point in those directions in space such that the projection of L on the z axis is one or another of Thus, L is also space quantized. The quantum number m is the values given by referred to as the magnetic quantum number. (Why \u201cmagnetic\u201d? See Section 7-4.) 1( 1)U, mU. 7-2 Quantization of Angular Momentum and Energy in the Hydrogen Atom 277 L = l(l + 1) = 2(2 + 1) = 6 Figure 7-4 Vector model illustrating the possible orientations of L in space and the possible values of Lz for the case where 2", ". z 2 1 0 \u20131 \u20132 mU is always less than the magnitude Figure 7-4 shows a diagram, called the vector model of the atom, illustrating the possible orientations of the angular momentum vector. Note the perhaps unexpected result that the angular momentum vector never points in the z direction, since the maximum z component This is a consequence of the uncertainty principle for angular momentum (which we will not derive) that implies that no two components of angular momentum can be precisely known simultaneously,4 except in the case of zero angular momentum. It is worth noting that for a given value of in integral steps. Operators for Lx and Ly can also be obtained by the substitution method; however, operating with them on does not produce eigenvalues. This is mainly because specifying rotation about the x and y axes requires measurement of both and. possible values of m, ranging from 1( 1)U. there are 2 1 to \u2019 EXAMPLE 7-1 Quantized Values of L If a system has angular momentum characterwhat are the possible values of Lz, what is the ized by the quantum number magnitude L, and what is the smallest possible angle between L and the z axis? 2, SOLUTION 1. The possible values of Lz are given by Equation 7-23: 2. The values of m for are 3. Thus, allowed values of Lz are 2 4. The magnitude of L is given by Equation 7-22. For 2 5. From Figure 7-4 the angle between L and the z axis is given by: mU Lz 2U, 1U, 0, U, 2U m 0, 1, 2 Lz \u0192L\u0192 2( 1)U 26U 2.45U cos Lz L mU 2( 1)U m 2( 1) 6. The smallest possible angle between L and the z axis is that for m which for 2 gives, cos 2 26 0.816 or 35.5\u00b0 278 Chapter 7 Atomic Physics Quantization of the Energy The results discussed so far apply to any system that is spherically symmetric, that is, one for which the potential energy depends on r only. The solution of the radial equation for R(r), on the other hand, depends on the detailed form of V(r). The new quantum number associated with the coordinate r is called the principal quantum number n. This quantum number, as we will see, is related to the energy in the hydrogen atom. Figure 7-5 shows a sketch of the potential energy function of Equation 7-6. If the total energy is positive, the electron is not bound to the atom. We are interested here only in bound-state solutions, for which the values of E are negative. For this case, the potential energy function becomes greater than E for large r, as shown in the figure. As we have discussed previously, for bound systems only certain values of the energy E lead to well-behaved solutions. These values are found by solving the radial equation, ( 1). which is formed by equating the left side of Equation 7-12 to the constant For V(r) of the hydrogen atom, given by Equation 7-6, the radial equation is b c kZe2 r U2( 1) 2r2 dR(r) ER(r) $R(r) $r U2 2r2 $ $r 7-24 ar2 The radial equation can be solved using standard methods of differential equations whose details we will omit here, except to note that (1) we expect a link to appear between the principal quantum number n and the angular momentum quantum number (since the latter already appears in Equation 7-24) and (2) in order that the solutions of Equation 7-24 be well behaved, only certain values of the energy are allowed, just as we discovered for the square well and the harmonic oscillator. The allowed values of E are given by a kZe2 U b En 2 2n2 Z2E1 n2 7-25 Energy 0 E\u00b4 r E V(r) = \u2013 kZe2 \u2013\u2013\u2013\u2013 r Figure 7-5 Potential energy of an electron in a hydrogen atom. If the total energy is greater than zero, as E, the electron is not bound and the energy is not quantized. If the total energy is less than zero, as E, the electron is bound. Then, as in one-dimensional problems, only certain discrete values of the total energy lead to well-behaved wave functions. 7-2 Quantization of Angular Momentum and Energy in the Hydrogen Atom 279 Table 7-2 Radial functions for hydrogen er>a0 R10 R20 R21 R30 2 2a3 0 1 22a3 0 1 226a3 0 2 1 2 R31 R32 a1 r 2a0 b er>2a0 er>2a0 r a0 a1 2r 3a", "0 a1 323a3 0 8 2726a3 0 4 81230a3 0 r a0 r2 a2 0 er>3a0 2r2 27a2 0 b er>3a0 r 6a0 b er>3a0 E1 (1>2)(ke2>U)2 13.6 eV and the principal quantum number n can take where. on the values n 1, 2, 3,..., with the further restriction that n must be greater than These energy values are identical to those found from the Bohr model. The radial functions resulting from the solution of Equation 7-24 for hydrogen are given by Equation 7-26, where the are standard functions called Laguerre polynomials. l n(r>a0) Rn(r) An er>a0nrl n(r>a0) 7-26 for n 1, 2, and 3 are and the Bohr radius given in Table 7-2. (For a detailed solution of Equation 7-24 and an extended table of Laguerre polynomials, see Weber and Arfken, Chapter 13.) The radial functions U2>(ke2). Rn(r) a0 Summary of the Quantum Numbers The allowed values of and restrictions on the quantum numbers n, with the variables are summarized as follows: r,, and, and m associated n 1, 2, 3, \u00c1 0, 1, 2, \u00c1, (n 1) m, (1), \u00c1, 0, 1, 2, \u00c1, 7-27 The fact that the energy of the hydrogen atom depends only on the principal quantum number n and not on is a peculiarity of the inverse-square force. It is related to the result in classical mechanics that the energy of a mass moving in an elliptical orbit in an inverse-square force field depends only on the major axis of the orbit and not on the eccentricity. The largest value of angular momentum ( n 1) corresponds most nearly to a circular orbit, whereas a small value of corresponds to a highly eccentric orbit. (Zero angular momentum corresponds to oscillation along a line through the force center, i.e., through the nucleus, in the case of the hydrogen atom.) For central forces that do not obey an inverse-square law, the energy does depend on the angular momentum (both classically and quantum mechanically) and thus depends on both n and. 280 Chapter 7 Atomic Physics The quantum number m is related to the z component of angular momentum. Since there is no preferred direction for the z axis for any central force, the energy cannot depend on m. We will see later that if we place an atom in an external magnetic field, there is a preferred direction in space (the direction of the field) and the energy then does depend on the value of m. (This effect, called the Zeeman effect, is discussed in a More section on the Web site. See page 303.) Figure 7-6 shows an energy-level diagram for hydrogen. This diagram is similar to Figure 4-16a except that states with the same n but different are shown separately. These states are referred to by giving the value of n along with a code letter: S stands for 0, P for 1, D for 2, and F for 3. These code letters are remnants of the spectroscopist\u2019s descriptions of various series of spectral lines as Sharp, Principal, Diffuse, and Fundamental. (For values of greater than 3, the letters follow alphabetically; thus G for 4, etc.) The allowed electric dipole transitions between energy levels obey the selection rules \u00a2m 0 or 1 \u00a2 1 7-28 That the quantum number of the atom must change by 1 when the atom emits or absorbs a photon results from conservation of angular momentum and the fact that the photon itself has an intrinsic angular momentum of 1 For the principal quantum number, n is unrestricted. U. Energy, eV 0.00 \u20130.85 \u20131.51 \u20133.40 \u201313.6 eV Figure 7-6 Energy-level diagram for the hydrogen atom, showing transitions obeying the selection rule 1. States with the same n value but different value have the same energy, E1 n2, where E1 The wavelengths of the Lyman (n 2 S n 1) and Balmer (n 3 S n 2) lines are shown in nm. Note that the latter has two possible transitions due to the degeneracy. 13.6 eV, as in the Bohr theory. > 7-3 The Hydrogen Atom Wave Functions 281 Questions 1. Why wasn\u2019t quantization of angular momentum noticed in classical physics? 2. What are the similarities and differences between the quantization of angular momentum in the Schr\u00f6dinger theory and in the Bohr model? 3. Why doesn\u2019t the energy of the hydrogen atom depend on? Why doesn", "\u2019t it depend on m? 7-3 The Hydrogen Atom Wave Functions satisfying the Schr\u00f6dinger equation for the hydroThe wave functions. gen atom are rather complicated functions of In this section we will write some of these functions and display some of their more important features graphically. nm(r,, ) \u2019 r,, and eim. As we have seen, the dependence of the wave function, given by Equation 7-14, is simply The dependence is described by the associated Legendre funcfm() given by Equation 7-15. The complete angular dependence is then given tions Ym(, ), by the spherical harmonic functions as indicated by Equation 7-16 and, for the first few, tabulated in Table 7-1. The solutions to the radial equation are of the form indicated by Equation 7-26 and are listed in Table 7-2 for the three lowest values of the principal quantum number n. Referring to Equation 7-10, our assumed product solutions of the time-independent Schr\u00f6dinger equation, we have that the complete wave function of the hydrogen atom is the product of fm() gm() Rn(r) and nm(r,, ) CnmRn(r)fm()gm() \u2019 7-29 is a constant determined by the normalization condition. where Cnm, We see from the form of this expression that the complete wave function depends and m that arose because of the boundary conditions on the quantum numbers n, on R(r), \u0192( ), and g( ). The energy, however, depends only on the value of n. From Equation 7-27 we see that for any value of n, there are n possible values of ( 0, 1, 2, \u00c1, n 1), there are 2 1 possible values ). Except for the lowest energy level (for which n 1 of and therefore and m can only be zero) there are generally many different wave functions corresponding to the same energy. As discussed in the previous section, the origins of this degeneracy are the 1 r dependence of the potential energy and the fact that there is no preferred direction in space. m (m, 1, \u00c1, and for each value of, > The Ground State Let us examine the wave functions for several particular states beginning with the lowest-energy level, the ground state, which has n 1. Then and m must both be zero. The Laguerre polynomial l 10 in Equation 7-26 is equal to 1, and the wave function is \u2019 100 C100eZr>a0 7-30 282 Chapter 7 Atomic Physics z x r sin \u03b8 r sin \u03b8 d\u03c6 d\u03c6 dr r d\u03b8 r \u03c6 \u03b8 d\u03b8 y d\u03c4 = (r sin \u03b8 d\u03c6)(r d\u03b8) dr = r2 sin \u03b8 dr d\u03b8 d\u03c6 Figure 7-7 Volume element d in spherical coordinates. The constant C100 is determined by normalization: \u2019*\u2019d 2 \u2019*\u2019r2 sin d d dr 1 0 0 0 using for the volume element in spherical coordinates (see Figure 7-7) d (r sin d)(r d)(dr) \u2019*\u2019 for this state is spherically symmetric, the integration over angles gives Because 4. Carrying out the integration over r gives5 C100 3>2 1 2 a Z a0 b 1 2 a 1 a0 b 3>2 for Z 1 7-31 The probability of finding the electron in the volume d is \u2019*\u2019d. \u2019*\u2019 The probability density is illustrated in Figure 7-8. The probability density for the ground state is maximum at the origin. It is often of more interest to determine the probability of finding the electron in a spherical shell between r and r dr. This probability, P(r)dr, is just the probability density times the volume of the spherical shell of thickness dr: \u2019*\u2019 The angular dependence of the electron probability distributions is critical to our understanding of the bonding of atoms into molecules and solids (see Chapters 9 and 10). P(r) dr \u2019*\u2019 4r2 dr 4r2C2 100e2Zr>a0 dr 7-32 Figure 7-9 shows a sketch of P(r) versus It is left as a problem (see Problem 7-21) to show that P(r) has its maximum value at In contrast to the Bohr model for hydrogen, in which the electron stays in a well-defined orbit at r a0, we see that it is possible for the electron to be found at any distance from the nucleus. r a0 >Z. r>a0. 7-3 The Hydrogen Atom Wave Functions 283 (a) z 5a0 (b) \u23aa\u03c8 \u23aa2 100", " \u20135a0 5a0 x 0 5 r/a0 \u20135a0 \u2019*\u2019 for the ground state in hydrogen. The quantity e\u2019*\u2019 Figure 7-8 Probability density can be thought of as the electron charge density in the atom. (a) The density is spherically symmetric, is greatest at the origin, and decreases exponentially with r. This computergenerated plot was made by making hundreds of \u201csearches\u201d for the hydrogen electron in the x-z plane (i.e., for \u0192 2 graph of the probability density 100 [This computer-generated plot courtesy of Paul Doherty, The Exploratorium.] ), recording each finding with a dot. (b) The more conventional Compare the two graphs carefully. r>a0. 0 vs. \u0192\u2019 However, the most probable distance is a0, and the chance of finding the electron at a much different distance is small. It is useful to think of the electron as a charged cloud of charge density (We must remember, though, that the electron is always observed as one charge.) Note that the angular momentum in the ground state is zero, contrary to the Bohr model assumption of 1 r e\u2019*\u2019. U. The Excited States In the first excited state, n 2 and can be either 0 or 1. For 0, m 0, and again we have a spherically symmetric wave function, given by \u2019 200 C200 a2 Zr a0 beZr>2a0 7-33 For 1, m can be 1, 0, or 1. The corresponding wave functions are (see Tables 7-1 and 7-2) \u2019 210 C210 eZr>2a0 cos \u2019 211 C211 eZr>2a0 sin ei Zr a0 Zr a0 7-34 7-35 P(r) P(r) r r2\u23aa\u03c8\u23aa2 0 1 2 3 4 5 6 r/a0 Figure 7-9 Radial probability density P(r) versus > r a0 for the ground state of the hydrogen atom. P(r) is proportional to The most probable distance r is the Bohr radius a0. r2\u0192\u2019 \u0192 2. 100 284 Chapter 7 Atomic Physics Figure 7-10 (a) Radial probability density P(r) vs. > r a0 for the n 2 states in hydrogen. P(r) for 1 has a maximum at the Bohr value 22a0. For 0, there is a maximum near this value and a smaller submaximum near the origin. The markers on > the r a0 axis denote the 8r>a0 values of > r a0 for the n 3 states in hydrogen. (b) P(r) vs. 9. (a) P(r) P(r) \u221d r2\u23aa\u03c8\u23aab) P(r) 0 P(r) 0 P(r) 0 5 5 5 5 10 15 r/a0 n = 3 l = 0 10 n = 3 l = 1 15 r/a0 10 15 r/a0 n = 3 l = 2 10 15 r/a0 Figure 7-10a shows P(r) for these wave functions. The distribution for n 2, 1 is maximum at the radius of the second Bohr orbit, rmax 22a0 while for n 2 and 0, P(r) has two maxima, the larger of which is near this radius. Radial probability distributions can be obtained in the same way for the other excited states of hydrogen. For example, those for the second excited state n 3 are shown in Figure 7-10b. The main radial dependence of P(r) is contained in the factor eZr>na0, except near the origin. A detailed examination of the Laguerre polynomials is greatest near the origin when shows that is small. as r S 0. Thus, for a given n, \u2019 S r nm \u2019 An important feature of these wave functions is that for 0, the probability densities are spherically symmetric, whereas for 0, they depend on the angle. The probability density plots of Figure 7-11 illustrate this result for the first excited state n 2. These angular distributions of the electron charge density depend only on the value of and not on the radial part of the wave function. Similar charge distributions for the valence electrons in more complicated atoms play an important role in the chemistry of molecular bonding. Question 4. At what value of r is \u2019*\u2019 maximum for the ground state of hydrogen? Why is P(r) maximum at a different value of r? z z z 7-4 Electron Spin 285 1 \u2019*\u2019 for the n 2 states in hydrogen. The probability is sin2 Figure 7", "-11 Probability densities spherically symmetric for 0. It is proportional to for 1, m 1. The probability densities have rotational symmetry about the z axis. Thus, the three-dimensional charge density for the 1, m 0 state is shaped roughly like a dumbbell, while that for the 1, m 1 states resembles a doughnut, or toroid. The shapes of these distributions are typical for all atoms in S states ( 0) and P states ( 1) and play an important role in molecular bonding. [This computer-generated plot courtesy of Paul Doherty, The Exploratorium.] for 1, m 0, and to cos2 7-4 Electron Spin As was mentioned in Chapter 4, when a spectral line of hydrogen or other atoms is viewed with high resolution, it shows a fine structure; that is, it is seen to consist of two or more closely spaced lines. As we noted then, Sommerfeld\u2019s relativistic calculation based on the Bohr model agrees with the experimental measurements of this fine structure for hydrogen, but the agreement turned out to be accidental since his calculation predicts fewer lines than are seen for other atoms. In order to explain fine structure and to clear up a major difficulty with the quantum-mechanical explanation of the periodic table (Section 7-6), W. Pauli6 in 1925 suggested that in addition to the, quantum numbers n, and m, the electron has a fourth quantum number, which could take on just two values. As we have seen, quantum numbers arise from boundary conditions on some coordinate (see Equations 7-14 and 7-15). Pauli originally expected that the fourth quantum number would be associated with the time coordinate in a relativistic theory, but this idea was not pursued. In the same year, S. Goudsmit and G. Uhlenbeck,7 graduate students at Leiden, suggested that this fourth quantum number was the z component, ms, of an intrinsic angular momentum of the electron, euphemistically called spin. They represented the spin vector S with the same form that Schr\u00f6dinger\u2019s wave mechanics gave for L: \u0192S\u0192 S 2s(s 1)U 7-36 286 Chapter 7 Atomic Physics S, L. 1 2 is described by a quantum number s Since this intrinsic spin angular momentum like the orbital angular momentum quantum number we expect 2s 1 possible values of the z component just as there are 2 1 possible z components of the If ms is to have only two values, as Pauli had sugorbital angular momentum gested, then s could only be and ms only In addition to explaining fine structure and the periodic table, this proposal of electron spin explained the unexpected results of an interesting experiment that had been preformed by O. Stern and W. Gerlach in 1922, which is described briefly in an Exploring section later on (see pages 288\u2013289). To understand why the electron spin results in the splitting of the energy levels needed to account for the fine structure, we must consider the connection between the angular momentum and the magnetic moment of any charged particle system. 1 2. Magnetic Moment If a system of charged particles is rotating, it has a magnetic moment proportional to its angular momentum. This result is sometimes known as the Larmor theorem. Consider a particle of mass M and charge q moving in a circle of radius r with speed v and frequency this constitutes a current loop. The angular momentum of the particle is L Mvr. The magnetic moment of the current loop is the product of the current and the area of the loop. For a circulating charge, the current is the charge times the frequency, f v>2r; i qf qv 2r and the magnetic moment is8 iA qa v 2r b(r2) 1 2 qa b L M 7-37 7-38 From Figure 7-12 we see that, if q is positive, the magnetic moment is in the same direction as the angular momentum. If q is negative, and L point in opposite directions; i.e., they are antiparallel. This enables us to write Equation 7-38 as a vector equation: q 2M L 7-39 Equation 7-39, which we have derived for a single particle moving in a circle, also holds for a system of particles in any type of motion if the charge-to-mass ratio q M is the same for each particle in the system. > L v M r \u03bc Applying this result to the orbital motion of the electron in the hydrogen atom and substituting the magnitude of L from Equation 7-22, we have for the magnitude of i Figure 7-12 A particle moving in a circle has angular momentum L. If the particle has a positive charge, the magnetic moment due to the current is parallel to L. e 2me L eU 2me 2( 1) 2( 1) B and, from Equation 7-23, a z component", " of z eU 2me m m B 7-40 7-41 7-4 Electron Spin 287 The orbital motion and spin of the electrons are the origin of magnetism in metals, such as iron, cobalt, and nickel (see Chapter 10). Devices ranging from giant electricity transformers to decorative refrigerator magnets rely on these quantum properties of electrons. (a where me is the mass of the electron, tum, and B has the value is the z component of the angular momenis a natural unit of magnetic moment called the Bohr magneton, which mU B eU 2me 9.27 1024 joule>tesla 5.79 109 eV>gauss 5.79 105 eV>tesla 7-42 The proportionality between and L is a general property of rotating charge distributions; however, the particular relation expressed by Equation 7-39 is for a single charge q rotating in a circle. To allow the same mathematical form to be used for other, more complicated situations, it is customary to express the magnetic moment in terms and a dimensionless quantity g called the gyromagnetic ratio, or simply the g of factor, where the value of g is determined by the details of the charge distribution. 1 and Equation In the case of the orbital angular momentum L of the electron, gL 7-39 would be written B B L gL U and Equations 7-40 and 7-41 as 2( 1) gL z mgL B B 7-43 7-44 7-45 There are minus signs in Equations 7-43 and 7-45 because the electron has a negative charge. The magnetic moment and the angular momentum vectors associated with the orbital motion are therefore oppositely directed, and we see that quantization of angular momentum implies quantization of magnetic moments. Other magnetic moments and g factors that we will encounter will have the same form. Finally, the behavior of a system with a magnetic moment in a magnetic field can be visualized by considering a small bar magnet (Figure 7-13). When placed in an external magnetic field B there is a torque that tends to align the magnet with the field B. If the magnet is spinning about its axis, the effect of the torque is to make the spin axis precess about the direction of the external field, just as a spinning top or gyroscope precesses about the direction of the gravitational field. B To change the orientation of the magnet relative to the applied field direction d, (whether or not it is spinning), work must be done on it. If it moves through angle the work required is S (b) B dW d B sin d d(B cos ) d( # B) S The potential energy of the magnetic moment be written in the magnetic field B can thus U # B If B is in the z direction, the potential energy is U zB 7-46 7-47 Figure 7-13 Bar-magnet model of magnetic moment. (a) In an external magnetic field, the moment experiences a torque that tends to align it with the field. If the magnet is spinning (b), the torque causes the system to precess around the external field. 288 Chapter 7 Atomic Physics Applying these arguments to the intrinsic spin of the electron results in the pre- dictions (with s 1 2 ) 2s(s 1) B 23 4 and z B ms B 1 2 B 7-48 Since the atomic electron is in a magnetic field arising from the apparent motion of the nuclear charge around the electron, the two values of ms correspond to two different energies, according to Equation 7-47. It is this splitting of the energy levels that results in the fine structure of the spectral lines. ms 1 2 The restriction of the spin, and hence the intrinsic magnetic moment, to two orientations in space with is another example of space quantization. The magnitude of the magnetic moment due to the spin angular momentum can be determined from quantitative measurement of the deflection of the beam in a Stern-Gerlach experiment. The result is not Bohr magneton, as predicted by Equation 7-41 1 2, with but twice this value. (This type of experiment is not an accurate way to measure magnetic moments, although the measurement of angular momentum this way is accurate because that involves simply counting the number of lines.) The g factor for the electron, gs in Equation 7-49, has been precisely measured to be gs 2.002319. m ms 1 2 z msgs B 7-49, This result, and the fact that s is a half integer rather than an integer like the orbital quantum number makes it clear that the classical model of the electron as a spinning ball is not to be taken literally. Like the Bohr model of the atom, the classical picture is useful in describing results of quantum-mechanical calculations, and it often gives useful guidelines as to what to expect from an experiment. The phenomenon of spin, while not a part of Schr\u00f6dinger\ufffd", "\ufffds wave mechanics, is included in the relativistic wave mechanics formulated by Dirac. In its nonrelativistic limit, Dirac\u2019s 2, which is approximately correct. The exact value of gs wave equation predicts gs is correctly predicted by quantum electrodynamics (QED), the relativistic quantum theory that describes the interaction of electrons with electromagnetic fields. Although beyond the scope of our discussions, QED is arguably the most precisely tested theory in physics. EXPLORING Stern-Gerlach Experiment If a magnetic moment the will feel an external force that depends on cause the force F is the negative gradient of the potential, so is placed in an inhomogeneous external magnetic field B, and the gradient of B. This is be- z F \u2207U \u2207( # B) 7-50 from Equation 7-46. If we arrange the inhomogeneous B field so that it is homogeneous 0 and F has only a z comin the x and y directions, then the gradient has only ponent, i.e., $B>$z Fz z(dB>dz) mgL B(dB>dz) 7-51 This effect was used by Stern and Gerlach9 in 1922 (before spin) to measure the possible orientations in space, i.e., the space quantization, of the magnetic moments of silver atoms. The experiment was repeated in 1927 (after spin) by Phipps and Taylor using hydrogen atoms. Magnet > z Oven Collimator The experimental setup is shown in Figure 7-14. Atoms from an oven are collimated and sent through a magnet whose poles are shaped so that the magnetic field Bz increases slightly with z, while Bx and By are constant in the x and y directions, respectively. The atoms then strike a collector plate. Figure 7-15 illustrates the effect of the dB dz on several magnetic moments of different orientations. In addition to the torque, which merely causes the magnetic moment to precess about the field direction, there is the force Fz in the positive or negative z direction, depending on whether is positive or negative, since dB dz is always positive. This force deflects the magnetic moment up or down by an amount that depends on the magnitudes of both dB dz and the z component of the magnetic moment Classically, one would expect a continuum of possible orientations of the magnetic moments. However, since the magnetic moment is proportional to L, which is quantized, quantum mechanics predicts that z also can have only the 2 1 values corresponding to the 2 1 possible values of m. We therefore expect 2 1 deflections (counting 0 as a deflection). For example, for 0 there should be one line on the collector plate corresponding to no deflection, and for 1 there should be three lines corresponding to the three values m 1, m 0, and m 1. The 1 case is illustrated in Figure 7-15. z. x y > > z 7-4 Electron Spin 289 S N Collector plate Figure 7-14 In the SternGerlach experiment, atoms from an oven are collimated, passed through an inhomogeneous magnetic field, and detected on a collector plate. (a) z Atomic beam \u03bc m = \u20131 \u03bc m = 0 m = +1 \u03bc (b) (c) S N m = \u20131 m = +1 \u03bc \u03bc \u03bc m = 0 Collector plate Figure 7-15 (a) In an inhomogeneous magnetic field the magnetic moment depends on the direction of the z component from an oven (not shown) is collimated into a horizontal line. (b) The pattern for the l case illustrated in (a). The three images join at the edges and have different detailed shapes due to differences in the field inhomogeneity. (c) The pattern observed for silver and hydrogen. Fz and dB dz. The beam and whose magnitude depends on those of experiences a force whose direction z z of > 290 Chapter 7 Atomic Physics Using neutral silver atoms, Stern and Gerlach expected to see only a single line, the middle line in Figure 7-15b, because the ground state of silver was known to be an 0 state; therefore, m 0 and 0. The force Fz would then be zero, and no deflection of the atomic beam should occur. However, when the experiment was done with either silver or hydrogen atoms, there were two lines, as shown in Figure 7-15c. Since the ground state of hydrogen also has 0, we should again expect only one line, were it not for the electron spin. If the electron has spin angular momentum of magnitude \u0192 S \u0192 1s(s 1)U, Since the orbital angular momentum is zero, the total internal angular momentum of the atom is simply the spin10 and two lines would be expected. Stern and Gerlach had made the first direct observation of electron spin and space quantization. the z", " component can be either s 1 2, U>2. U>2 where or The Complete Hydrogen Atom Wave Functions Our description of the hydrogen atom wave functions in Section 7-3 is not complete because we did not include the spin of the electron. The hydrogen atom wave functions are also characterized by the spin quantum number ms, which can be (We need not include the quantum number s because it always has the value s ) A general where we have included the subscript on m/ to wave function is then written distinguish it from ms. There are now two wave functions for the ground state of the hydrogen atom, corresponding to an atom with its electron spin \u201cparallel\u201d or \u201cantiparallel\u201d to the z axis (as defined, for example, by a external magnetic field). In general, the ground state of a hydrogen atom is a linear combination of these wave functions: \u2019 1001>2, or 1 2. 1 2. 1001>2 nmms 1 2 and \u2019 \u2019, \u2019 \u2019 C 1 ms 1 2 10012 C2 \u2019 10012 (for example, by observing to which spot The probability of measuring the atom goes in the Stern-Gerlach experiment) is Unless atoms have been preselected in some way (such as by passing them through a previous inhomogeneous will each magnetic field or by their having recently emitted a photon), and be and measuring the spin \u201cdown\u201d (ms so that measuring the spin \u201cup\u201d 1 2, 1 2) are equally likely. 1 2) (ms \u0192C1 \u0192C1 \u0192C2 \u0192 2. \u0192 2 \u0192 2 Questions 5. Does a system have to have a net charge to have a magnetic moment? 6. Consider the two beams of hydrogen atoms emerging from the magnetic field in the Stern-Gerlach experiment. How does the wave function for an atom in one beam differ from that of an atom in the other beam? How does it differ from the wave function for an atom in the incoming beam before passing through the magnetic field? (a) (b) Photographs made by Stern and Gerlach with an atomic beam of silver atoms. (a) When the magnetic field is zero, all atoms strike in a single, undeviated line. (b) When the magnetic field is nonzero, the atoms strike in upper and lower lines, curved due to differing inhomogeneities. [From O. Stern and W. Gerlach, Zeitschr. f. Physik 9, 349 (1922).] 7-5 Total Angular Momentum and the Spin-Orbit Effect 291 7-5 Total Angular Momentum and the Spin-Orbit Effect In general, an electron in an atom has both orbital angular momentum characterized by the quantum number and spin angular momentum characterized by the quantum number s. Analogous classical systems that have two kinds of angular momentum are Earth, which is spinning about its axis of rotation in addition to revolving about the Sun, or a precessing gyroscope, which has angular momentum of precession in addition to its spin. Classically the total angular momentum J L S 7-52 is an important quantity because the resultant torque on a system equals the rate of change of the total angular momentum, and in the case of central forces, the total angular momentum is conserved. For a classical system, the magnitude of the total angular momentum J can have any value between L S and We have already seen that in quantum mechanics, angular momentum is more complicated: both L and S are quantized and their relative directions are restricted. The quantum-mechanical rules for combining orbital and spin angular momenta or any two angular momenta (such as for two particles) are somewhat difficult to derive, but they are not difficult to understand. For the case of orbital and spin angular momenta, the magnitude of the total angular momentum J is given by \u0192L S\u0192. \u0192J\u0192 2j(j 1)U where the total angular momentum quantum number j can be either j s or j \u0192 s\u0192 and the z component of J is given by Jz mj U where mj j, j 1, \u00c1, j 1, j 7-53 7-54 7-55 3 2 j 1 1 2 (If 0, the total angular momentum is simply the spin, and j s.) Figure 7-16a is a simplified vector model illustrating the two possible combinations for the case of an electron with 1. The lengths j 1 1 1 and 2 2 [( 1)]1>2, [s(s 1)]1>2, of the vectors are proportional to The j s spin and orbital angular momentum vectors are said to be \u201cparallel\u201d when A quantum mechanically more accurate vector and \u201cantiparallel\u201d when addition is shown in Figure 7-16b. The quantum number m", "j can take on 2j 1 possible values in integer steps between j and j, as indicated by Equation 7-55. Equation 7-55 also implies that mj [j(j 1)]1>2. and ms since Jz j \u0192 s\u0192. Sz. m/ Lz Equation 7-54 is a special case of a more-general rule for combining two angular momenta that is useful when dealing with more than one particle. For example, there are two electrons in the helium atom, each with spin, orbital, and total angular momentum. The general rule is If J1 is one angular momentum (orbital, spin, or a combination) and J2 is another, the resulting total angular momentum J J1 J2 has the value [j(j 1)]1>2 U for its magnitude, where j can be any of the values j1 j2, j1 j2 1, \u00c1, \u0192j1 j2 \u0192 292 Chapter 7 Atomic Physics (a) S (b) L J L S J 1\u2013\u2013 2 1 1 1 \u2013 1\u2013\u2013 2 \u2013 1\u2013\u2013 2 1\u2013\u2013 2 1\u2013\u2013 2 3\u2013\u2013 2 1\u2013\u2013 2 1\u2013\u2013 2 j = 3 \u2013\u2013 2 j = 3 \u2013\u2013 2 j = 1 \u2013\u2013 2 Figure 7-16 (a) Simplified vector model illustrating the addition of orbital and spin angular momenta. The case shown is for 1 and s 1 2. quantum number for the total angular momentum: j s and (b) Vector addition of the orbital and spin angular momenta, also for the case 1 and s 1 According to the uncertainty principle, the vectors can lie anywhere on the cones, 2. corresponding to the definite values of their z components. Note in the middle sketch that there 2, mj are two ways of forming the states with There are two possible values of the j s 1 2. and j 1 j 3 2, mj 1 2. 1 2 3 2 EXAMPLE 7-2 Addition of Angular Momenta I Two electrons each have zero orbital angular momentum. What are the possible quantum numbers for the total angular momentum of the two-electron system? (For example, these could be the He atom electrons in any of the S states.) SOLUTION The general rule then gives two possible results, j 1 and In this case j 0. These combinations are commonly called parallel and antiparallel, respectively. j2 1 2. j1 EXAMPLE 7-3 Addition of Angular Momenta II An electron in an atom has orbital 2, and a second electron has angular momentum L1 with quantum number 1 3. What are the possible orbital angular momentum L2 with quantum number 2 L2? quantum numbers for the total orbital angular momentum L L1 SOLUTION Since 1 2 5 and \u0192 1 2 \u0192 1, the possible values of are 5, 4, 3, 2, and 1. Spectroscopic Notation Spectroscopic notation, a kind of shorthand developed in the early days of spectroscopy to condense information and simplify the description of transitions between states, has since been adopted for general use in atomic, molecular, nuclear, and 7-5 Total Angular Momentum and the Spin-Orbit Effect 293 particle physics. The notation code appears to be arbitrary,11 but it is easy to learn and, as you will discover, convenient to use. For single electrons we have: 1. For single-electron states the letter code s p d f g h... is used in one-to-one correspondence with the values of the orbital angular momentum quantum number d electron or in a d state. : 0 1 2 3 4 5.... For example, an electron with 2 is said to be a 2. The single-electron (Bohr) energy levels are called shells, labeled K L M N O... in one-to-one correspondence with the values of the principal quantum number n: 1 2 3 4 5.... For example, an electron with n 3 in an atom is said to be in the M shell. (This notation is less commonly used.) For atomic states that may contain one or more electrons, the notation includes the principal quantum number and the angular momenta quantum numbers. The total orbital angular momentum quantum number is denoted by a capital letter in the same sequence as in rule 1 above, i.e., S P D F... correspond to values 0 1 2 3.... The value of n is written as a prefix and the value of the total angular momentum quantum number j by a subscript. The magnitude of the total spin quantum number s appears as a left superscript in the form 2s 1.12 Thus, a state with 1, a P state, would be written as n2s1Pj 12 S1>2, For example, the ground state of the hydrogen atom (n 1, 0, s 1 2", ") is read \u201cone doublet S one-half.\u201d The n 2 state can have 0 or written 22P1>2. (The principal quantum number and spin superscript are sometimes not included if they are not needed in specific situations.) 1, so the spectroscopic notation for these states is 22 S1>2, 22 P3>2, and > Spin-Orbit Coupling Atomic states with the same n and values but different j values have slightly different energies because of the interaction of the spin of the electron with its orbital motion. This is called the spin-orbit effect. The resulting splitting of the spectral lines such as the one that results from the splitting of the 2P level in the transition 2P S 1S in hydrogen is called fine-structure splitting. We can understand the spin-orbit effect qualitatively from a simple Bohr model picture, as shown in Figure 7-17. In this picture, the electron moves in a circular orbit with speed v around a fixed proton. In the figure, the orbital angular momentum L is up. In the frame of reference of the electron, the proton moves in a circle around it, thus constituting a circular loop current that produces a magnetic field B at the position of the electron. The direction of B is also up, parallel to L. Recall that the potential energy of a magnetic moment in a magnetic field depends on its orientation relative to the field direction and is given by U # B zB 7-56 The potential energy is lowest when the magnetic moment is parallel to B and highest when it is antiparallel. Since the intrinsic magnetic moment of the electron is directed opposite to its spin (because the electron has a negative charge), the spin-orbit energy is highest when the spin is parallel to B and thus to L. The energy of the state in 2P3>2 (a) L p v e (b) p v e B Figure 7-17 (a) An electron moving about a proton with angular momentum L up. (b) The magnetic field B seen by the electron due to the apparent (relative) motion of the proton is also up. When the electron spin is parallel to L, the magnetic moment is antiparallel to L and B, so the spin-orbit energy has its largest value. 294 Chapter 7 Atomic Physics Figure 7-18 Fine-structure energy-level diagram. On the left, the levels in the absence of a magnetic field are shown. The effect of the magnetic field due to the relative motion of the nucleus is shown on the right. Because of the spin-orbit interaction, the magnetic field splits the 2P level into two energy levels, with the j 3 level having slightly greater energy than the 2 j 1 level. The spectral line due to the transition 2 2P S 1S is therefore split into two lines of slightly different wavelengths. (Diagram is not to scale.) 2P 1S 2P3/2 2P1/2 \u0394U + \u03bcB \u2013 \u03bcB B L B L S \u03bc \u03bc S \u0394U = 2\u03bcB hydrogen, in which L and S are parallel, is therefore slightly higher than the state, in which L and S are antiparallel (Figure 7-18).13 The measured splitting is about 4.5 105 eV for the levels in hydrogen. For other atoms, the finestructure splitting is larger than this. For example, for sodium it is about 2 103 eV, as will be discussed in Section 7-7. Recalling that transitions resulting in spectral lines in the visible region are of the order of 1.5 to 3.0 eV, you can see that the fine-structure splitting is quite small. 2P1>2 2P3>2 2P1>2 and and levels in hydrogen is 4.5 105 eV. From this, estimate the magnetic field EXAMPLE 7-4 Fine-Structure Splitting The fine-structure splitting of the 2P1>2 that the 2p electron in hydrogen experiences. Assume B is parallel to the z axis. 2P3>2 SOLUTION 1. The energy of the 2p electrons is shifted in the presence of a magnetic field by an amount given by Equation 7-56: 2. U is positive or negative depending on the relative orientation of so the total energy difference E between the two levels is: and B, 3. Since the magnetic moment of the electron is B, z B and U # B zB \u00a2E 2U 2 zB \u00a2E 2 BB 4. Solving this for B substituting for the energy splitting \u00a2E gives B and B \u00a2E 2 B 4.5 105 eV (2)(5.79 105 eV>T) 0.39 T Remarks: This is a substantial magnetic field, nearly 10,000 times Earth\u2019s average magnetic field. 7-6 The Schr\u00f6dinger Equation for Two (or More", ") Particles 295 When an atom is placed in an external magnetic field B, the total angular momentum J is quantized in space relative to the direction of B and the energy of the atomic state characterized by the angular momentum quantum number j is split into 2j 1 energy levels corresponding to the 2j 1 possible values of the z component of J and therefore to the 2j 1 possible values of the z component of the total magnetic moment. This splitting of the energy levels in the atom gives rise to a splitting of the spectral lines emitted by the atom. The splitting of the spectral lines of an atom placed in an external magnetic field was discovered by P. Zeeman and is known as the Zeeman effect. (See the second More section on page 303 and Section 3-1.) Zeeman and Lorentz shared the Nobel Prize in Physics for the discovery and explanation of the Zeeman effect. 7-6 The Schr\u00f6dinger Equation for Two (or More) Particles Our discussion of quantum mechanics so far has been limited to situations in which a single particle moves in some force field characterized by a potential energy function V. The most important physical problem of this type is the hydrogen atom, in which a single electron moves in the Coulomb potential of the proton nucleus. This problem is actually a two-body problem, as the proton also moves in the Coulomb potential of the electron. However, as in classical mechanics, we can treat this as a one-body problem by considering the proton to be at rest and replacing the electron mass with the reduced mass. When we consider more complicated atoms we must face the problem of applying quantum mechanics to two or more electrons moving in an external field. Such problems are complicated by the interaction of the electrons with each other and also by the fact that the electrons are identical. The interaction of the electrons with each other is electromagnetic and essentially the same as that expected classically for two charged particles. The Schr\u00f6dinger equation for an atom with two or more electrons cannot be solved exactly, and approximation methods must be used. This is not very different from the situation in classical problems with three or more particles. The complication arising from the identity of electrons is purely quantum mechanical and has no classical counterpart. The indistinguishability of identical particles has important consequences related to the Pauli exclusion principle. We will illustrate the origin of this important principle in this section by considering the simple case of two noninteracting identical particles in a one-dimensional infinite square well. The time-independent Schr\u00f6dinger equation for two particles of mass m is U2 2m $2\u2019(x1, x2) $x2 1 U2 2m $2\u2019(x1, x2) $x2 2 V\u2019(x1, x2) E\u2019(x1, x2) 7-57 where x1 and x2 are the coordinates of the two particles. If the particles are interacting, the potential energy V contains terms with both x1 and x2, which cannot usually be separated. For example, if the particles are charged, their mutual electrostatic potential en\u0192. ergy (in one dimension) is If they do not interact, however, we can write V as V1(x1) V2(x2). For the case of an infinite square well potential, we need solve the Schr\u00f6dinger equation only inside the well where V 0 and require the wave function to be zero at the walls of the well. Solutions of Equation 7-57 can be written as products of single-particle solutions and linear combinations of such solutions. ke2> \u0192x2 x1 296 Chapter 7 Atomic Physics The single-particle product solutions are nm(x1, x2) \u2019 \u2019 n(x1)\u2019 m(x2) 7-58 \u2019 n(x1) \u2019 m(x2) where well given by Equation 6-32. Thus, for n 1 and m 2, and are the single-particle wave functions for an infinite square \u2019 12 C sin x1 L sin 2x2 L 7-59 \u0192\u2019(x1, x2)\u0192 2 dx1 dx2, The probability of finding particle 1 in dx1 and particle 2 in dx2 is \u0192\u2019(x2)\u0192 2 dx2. which is just the product of the separate probabilities However, even though we have labeled the particles 1 and 2, if they are identical, we cannot distinguish which is in dx1 and which is in dx2. For identical particles, therefore, we must construct the wave function so that the probability density is the same if we interchange the labels: \u0192\u2019(x1)\u0192 2 dx1 and Equation 7-60 holds if change of particles\u2014that is, \u0192\u2019(x1, x2)\u0192 2 \u0192\u2019(x2", ", x1)\u0192 2 \u2019(x1, x2) is either symmetric or antisymmetric on ex- 7-60 \u2019(x2, x1) \u2019(x1, x2) symmetric \u2019(x2, x1) \u2019(x1, x2) antisymmetric We note that the general wave function of the form of Equation 7-58 and the example (Equation 7-59) are neither symmetric nor antisymmetric. If we interchange x1 and x2, we get a different wave function, implying that the particles can be distinguished. These forms are thus not consistent with the indistinguishability of identical particles. However, from among all of the possible linear combination solutions of the single \u2019 product functions, we see that, if are added or subtracted, we form symmetric or antisymmetric wave functions necessary to preserve the indistinguishability of the two particles: and \u2019 mn nm \u2019 S \u2019 A CC\u2019 n(x1)\u2019 m(x2) \u2019 n(x2)\u2019 CC\u2019 n(x1)\u2019 m(x2) \u2019 n(x2)\u2019 m(x1)D symmetric m(x1)D antisymmetric Pauli Exclusion Principle There is an important difference between the antisymmetric and symmetric combinations. If n m, the antisymmetric wave function is identically zero for all x1 and x2, whereas the symmetric function is not. More generally, it is found that electrons (and many other particles, including protons and neutrons) can only have antisymmetric total wave functions, that is & nmms RnYm Xms 7-61 Rn is the radial wave function, is the spin where for two wave function. Thus, single-particle wave functions such as such particles cannot have exactly the same set of values for the quantum numbers. Xms is the spherical harmonic, and \u2019 m(x1) and \u2019 n(x1) Ym 7-7 Ground States of Atoms: The Periodic Table 297 This is an example of the Pauli exclusion principle. For the case of electrons in atoms and molecules, four quantum numbers describe the state of each electron, one for each space coordinate and one associated with spin. The Pauli exclusion principle for electrons states that No more than one electron may occupy a given quantum state specified by a particular set of single-particle quantum numbers n,, m, ms. The effect of the exclusion principle is to exclude certain states in the manyelectron system. It is an additional quantum condition imposed on solutions of the Schr\u00f6dinger equation. It will be applied to the development of the periodic table in the following section. Particles such as particles, deuterons, photons, and mesons have symmetric wave functions and do not obey the exclusion principle. 7-7 Ground States of Atoms: The Periodic Table We now consider qualitatively the wave functions and energy levels for atoms more complicated than hydrogen. As we have mentioned, the Schr\u00f6dinger equations for atoms other than hydrogen cannot be solved exactly because of the interaction of the electrons with one another, so approximate methods must be used. We will discuss the energies and wave functions for the ground states of atoms in this section and consider the excited states and spectra for some of the less complicated cases in the following section. We can describe the wave function for a complex atom in terms of singleparticle wave functions. By neglecting the interaction energy of the electrons, that description can be simplified to products of the single-particle wave functions. These wave functions are similar to those of the hydrogen atom and are characterized by the ms. The energy of an electron is determined mainly by quantum numbers n, the quantum numbers n (which is related to the radial part of the wave function) and (which characterizes the orbital angular momentum). Generally, the lower the value of n and the lower the energy of the state. (See Figure 7-19.) The specification of n for each electron in an atom is called the electron configuration. Customarily, and the value of and the various electron shells are specified with the same code defined in the subsection \u201cSpectroscopic Notation\u201d (page 292) in Section 7-5. The electron configurations of the atomic ground states are given in Appendix C. m,,, The Ground States of the Atoms Helium (Z 2) The energy of the two electrons in the helium atom consists of the kinetic energy of each electron, a potential energy of the form for each electron corresponding to its attraction to the nucleus, and a potential energy of interaction Vint corresponding to the mutual repulsion of the two electrons. If r1 and r2 are the position vectors for the two electrons, Vint is given by kZe2>ri Vint", " ke2 r1 \u0192 \u0192r2 7-62 Energy 6p 5d 4f 6s 5p 4d 5s 4p 3d 4s 3p 3s 2p 2s 1s Because this interaction term contains the position variables of the two electrons, its presence in the Schr\u00f6dinger equation prevents the separation of the equation into separate equations for each electron. If we neglect the interaction term, however, the Figure 7-19 Relative energies of the atomic shells and subshells. 298 Chapter 7 Atomic Physics Schr\u00f6dinger equation can be separated and solved exactly. We then obtain separate equations for each electron, with each equation identical to that for the hydrogen atom except that Z 2. The allowed energies are then given by E Z2E0 n2 1 Z2E0 n2 2 where E0 13.6 eV 7-63 The lowest energy, E1 case, 1 2 the form 1. For this 108.8 eV, occurs for n1 0. The total wave function, neglecting the spin of the electrons, is of 2(2)2E0 n2 \u2019 \u2019 100(r1, 1, 1)\u2019 100(r2, 2, 2) 7-64, m The quantum numbers n, the fourth quantum number ms is different, i.e., if one electron has and the other has therefore be zero. 1 2. and ms The resulting total spin of the two electrons must can be the same for the two electrons only if 1 2 ms We can obtain a first-order correction to the ground-state energy by using the approximate wave function of Equation 7-64 to calculate the average value of the interaction energy Vint, which is simply the expectation value The result of this calculation is 9. 8Vint 8Vint 9 34 eV With this correction, the ground-state energy is E 108.8 34 74.8 eV 7-65 7-66 This approximation method, in which we neglect the interaction of the electrons to find an approximate wave function and then use this wave function to calculate the interaction energy, is called first-order perturbation theory. The approximation can be continued to higher orders: for example, the next step is to use the new ground-state energy to find a correction to the ground-state wave function. This approximation method is similar to that used in classical mechanics to calculate the orbits of the planets about the Sun. In the first approximation the interaction of the planets is neglected and the elliptical orbits are found for each planet. Then, using this result for the position of each planet, the perturbing effects of the nearby planets can be calculated. The experimental value of the energy needed to remove both electrons from the helium atom is about 79 eV. The discrepancy between this result and the value as 74.8 eV is due to the inaccuracy of the approximation used to calculate indicated by the rather large value of the correction (about 30 percent). (It should be pointed out that there are better methods of calculating the interaction energy for helium that give much closer agreement with experiment.) The helium ion He, formed by removing one electron, is identical to the hydrogen atom except that Z 2; so the ground-state energy is 9, 8Vint Z2(13.6) 54.4 eV The energy needed to remove the first electron from the helium atom is 24.6 eV. The corresponding potential, 24.6 V, is called the first ionization potential of the atom. The ionization energies are given in Appendix C. The configuration of the ground state of the helium atom is written 1s2. The 1 signifies n 1, the s signifies 0, and the 2 signifies that there are two electrons in this state. Since can only be zero for n 1, the two electrons fill the K shell (n 1). 7-7 Ground States of Atoms: The Periodic Table 299 George Gamow and Wolfgang Pauli in Switzerland in 1930. [Courtesy of George Gamow.] Lithium (Z 3) Lithium has three electrons. Two are in the K shell (n 1), but the third cannot have n 1 because of the exclusion principle. The next-lowest energy state for this electron has n 2. The possible values are 1 or 0. In the hydrogen atom, these values have the same energy because of the degeneracy associated with the inverse-square nature of the force. This is not true in lithium and other atoms because the charge \u201cseen\u201d by the outer electron is not a point charge.14 The positive charge of the nucleus Ze can be considered to be approximately a point charge, but the negative charge of the K shell electrons 2e is spread out in space over >Z. a0 a volume whose radius is of the order of We can in fact take for the charge denr e\u0192\u2019\u0192 2, \u2019 is a hydrogenlike 1s wave function where sity of each inner electron (", "neglecting the interaction of the two electrons in the K shell). The probability distribution for the outer electron in the 2s or 2p states is similar to that shown in Figure 7-10. We see that the probability distribution in both cases has a large maximum well outside the inner K-shell electrons but that the 2s distribution also has a small bump near the origin. We could describe this by saying that the electron in the 2p state is nearly always outside the shielding of the two 1s electrons in the K shell so that it sees an effective 1, whereas in the 2s state the electron penetrates this \u201cshieldcentral charge of Zeff ing\u201d more often and therefore sees a slightly larger effective positive central charge. The energy of the outer electron is therefore lower in the 2s state than in the 2p state, and the lowest energy configuration of the lithium atom is 1s22s. The total angular momentum of the electrons in this atom is due to the spin of the outer electron since each of the electrons has zero orbital angular momentum and the inner K-shell electrons are paired to give zero spin. The first ionization potential for lithium is only 5.39 V. We can use this result to calculate the effective positive charge seen by the 2s electron. For Z Zeff and n 2, we have Z2E0 n2 eff(13.6 eV) 22 5.39 eV E Z2 U 1 2 1.3. It is generally true that the smaller the value of which gives Zeff the greater the penetration of the wave function into the inner shielding cloud of electrons: The result is that in a multielectron atom, for given n, the energy of the electron increases with increasing (See Figure 7-19.),. Beryllium (Z 4) The fourth electron has the least energy in the 2s state. The exclusion principle requires that its spin be antiparallel to the other electron in this state so that the total angular momentum of the four electrons in this atom is zero. The electron configuration of beryllium is 1s22s2. The first ionization potential is 9.32 V. This is greater than that for lithium because of the greater value of Z. m Boron to Neon (Z 5 to Z 10) Since the 2s subshell is filled, the fifth electron must go into the 2p subshell; that is, n 2 and 1. Since there are three possible (1, 0, and 1) and two values of ms for each, there can be up to six values of electrons in this subshell. The electron configuration for boron is 1s22s22p. Although it might be expected that boron would have a greater ionization potential than beryllium because of the greater Z, the 2p wave function penetrates the shielding of the core electrons to a lesser extent and the ionization potential of boron is actually about 8.3 V, slightly less than that of beryllium. The electron configuration of the elements carbon (Z 6) to neon (Z 10) differs from boron only by the number of electrons in the 2p subshell. The ionization potential increases slightly with Z for these elements, reaching the value of 21.6 V for the last element in the group, neon. Neon has the 300 Chapter 7 Atomic Physics maximum number of electrons allowed in the n 2 shell. The electron configuration of neon is 1s22s22p6. Because of its very high ionization potential, neon, like helium, is chemically inert. The element just before this, fluorine, has a \u201chole\u201d in this shell; that is, it has room for one more electron. It readily combines with elements such as lithium, which has one outer electron that is donated to the fluorine atom to make an F ion and a Li ion, which bond together. This is an example of ionic bonding, to be discussed in Chapter 9. Sodium to Argon (Z 11 to Z 18) The eleventh electron must go into the n 3 shell. Since this electron is weakly bound in the Na atom, Na combines readily with atoms such as F. The ionization potential for sodium is only 5.14 V. Because of the lowering of the energy due to penetration of the electronic shield formed by the other 10 electrons\u2014similar to that discussed for Li\u2014the 3s state is lower than the 3p or 3d states. (With n 3, can have the values 0, 1, or 2.) This energy difference between subshells of the same n value becomes greater as the number of electrons increases. The configuration of Na is thus 1s22s22p63s. As we move to higher-Z elements, the 3s subshell and then the 3p subshell begin to fill up. These two subshells can accommodate 2 6 8 electrons. The configuration of argon (Z 18) is 1s22", "s22p63s23p6. There is another large energy difference between the eighteenth and nineteenth electrons, and argon, with its full 3p subshell, is stable and inert. Atoms with Z 18 One might expect that the nineteenth electron would go into the 3d subshell, but the shielding or penetration effect is now so strong that the energy is lower in the 4s shell than in the 3d shell. The nineteenth electron in potassium (Z 19) and the twentieth electron in calcium (Z 20) go into the 4s rather than the 3d subshell. The electron configurations of the next 10 elements, scandium (Z 21) through zinc (Z 30), differ only in the number of electrons in the 3d subshell except for chromium (Z 24) and copper (Z 29), each of which has only one 4s electron. These elements are called transition elements. Since their chemical properties are mainly due to their 4s electrons, they are quite similar chemically. Kr Xe He 30 25 20 15 10 Ne Ar 5 2s Li 3s Na 0 0 2 10 4s K 20 18 5s Rb 6s Cs Rn 7s Fr 30 36 40 50 54 60 70 80 86 90 100 Z Figure 7-20 First ionization energy vs. Z up to Z 90. The energy is the binding energy of the last electron in the atom. This energy increases with Z until a shell is closed at Z values of 2, 10, 18, 36, 54, and 86. The next electron must go into the next higher shell and hence is farther from the center of core charge and thus less tightly bound. The ionization potential (in volts) is numerically equal to the ionization energy (in eV). Figure 7-20 shows a plot of the first ionization potential of an atom versus Z up to Z 90. The sudden decrease in ionization potential after the Z numbers 2, 10, 18, 36, and 54 marks the closing of a shell or subshell. A corresponding sudden increase occurs in the atomic radii, as illustrated in Figure 7-21. The ground-state electron configurations of the elements are tabulated in Appendix C. 7-8 Excited States and Spectra of Atoms 301 Cs Rb K 0..2 Na Li 0.1 0 0 3 10 11 20 19 30 40 50 60 70 80 90 100 37 55 Z Figure 7-21 The atomic radii versus Z shows a sharp rise following the completion of a shell as the next electron must have the next larger n. The radii then decline with increasing Z, reflecting the penetration of wave functions of the electrons in the developing shell. The recurring patterns here and in Figure 7-20 are examples of the behavior of many atomic properties that give the periodic table its name. Questions 7. Why is the energy of the 3s state considerably lower than that of the 3p state for sodium, whereas in hydrogen these states have essentially the same energy? 8. Discuss the evidence from the periodic table of the need for a fourth quantum number. How would the properties of He differ if there were only three quantum numbers, n, and m?, The concept of shell structure for the electrons in the atomic systems was a significant aid to the later understanding of molecular bonding (see Chapter 9) and the complex structure of the atomic nuclei (see Chapter 11). 7-8 Excited States and Spectra of Atoms Alkali Atoms In order to understand atomic spectra, we need to understand the excited states of atoms. The situation for an atom with many electrons is, in general, much more complicated than that of hydrogen. An excited state of the atom usually involves a change in the state of one of the electrons or more rarely two or even more electrons. Even in the case of the excitation of only one electron, the change in state of this electron changes the energies of the others. Fortunately, there are many cases in which this effect is negligible, and the energy levels can be calculated accurately from a relatively simple model of one electron plus a stable core. This model works particularly well for the alkali metals: Li, Na, K, Rb, and Cs. These elements are in the first column of the periodic table. The optical spectra of these elements are similar in many ways to the spectrum of hydrogen. Another simplification is possible because of the wide difference between excitation energy of a core electron and the excitation energy of an outer electron. Consider the case of sodium, which has a neon core (except Z 11 rather than Z 10) and an outer 3s electron. If this electron did not penetrate the core, it would see an effective 1 resulting from the 11e nuclear charge and the 10e of the nuclear charge of Zeff completed electron shells. The ionization energy would be the same as the energy of the n 3 electron in hydrogen, about 1.5 eV. Penetration into the core increases Zeff and so lowers the energy of", " the outer electron, i.e., binds it more tightly, thereby increasing the ionization energy. The measured ionization energy of sodium is about 5 eV. The energy needed to remove one of the outermost core electrons, a 2p electron, is about 31 eV, whereas that needed to remove one of the 1s electrons is about 1041 eV. 302 Chapter 7 Atomic Physics Among the many applications of atomic spectra is their use in answering questions about the composition of stars and the evolution of the universe (see Chapter 13). An electron in the inner core cannot be excited to any of the filled n 2 states because of the exclusion principle. Thus, the minimum excitation of an n 1 electron is to the n 3 shell, which requires an energy only slightly less than that needed to remove this electron completely from the atom. Since the energies of photons in the visible range (about 400 to 800 nm) vary only from about 1.5 to 3 eV, the optical (i.e., visible) spectrum of sodium must be due to transitions involving only the outer electron. Transitions involving the core electrons produce line spectra in the ultraviolet and xray regions of the electromagnetic spectrum. Figure 7-22 shows an energy-level diagram for the optical transitions in sodium. Since the spin angular momentum of the neon core adds up to zero, the spin of each 1 2. state in sodium is have a slightly lower energy than those with Each state is therefore a doublet (except for the S states). The doublet splitting is very small and is not evident on the energy scale of Figure 7-22 but is shown in Figure 7-18. The states are labeled by the usual spectroscopic notation, with the superscript 2 before the letter indicating read as \u201cdoublet P three-halves,\u201d denotes a state that the state is a doublet. Thus, Because of the spin-orbit effect, the states with j 1 2. j 1 2 2P3>2, 2S1/2 2P3/2 2P1/2 2D5/2, 3/2 2F7/2, 5/2 0 7d 6d 5d 4d 1267.8 1845.9 3d 7p 6p 5p 4p 0 5.3 9 8 0.2 2 3 3 9 8.2 9 4 6 7.8 9 4 568.82 3 8.3 1 7 8 8.2 814.91 6 5 3p 9 9. 8 8 5 ) (D2 9 9.5 (D1 7s 6s \u20131 5s 4s \u20132 V e E, \u20133 \u20134 \u20135 \u20135.14 3s Figure 7-22 Energy-level diagram for sodium (Na) with some transitions indicated. Wavelengths shown are in nanometers. The spectral lines labeled D1 and D2 are very intense and are responsible for the yellow color of lamps containing sodium. The energy splittings of the D and F levels, also doublets, are not shown. 7-8 Excited States and Spectra of Atoms 303 j 5/2 3/2 3/2 1/2 f Figure 7-23 The transitions between a pair of doublet energy states in singly ionized calcium. The transition represented by the dotted line is forbidden by the j 1, 0 selection rule. The darkness of the lines indicates relative intensity. Under low resolution the faint line on the left of the spectrum at the bottom merges with its neighbor and the compound doublet (or triplet) looks like a doublet. 2D 2P S 1 2 j 3 2. in which 1 and (The S states are customarily labeled as if they were doublets even though they are not. This is done because they belong to the set of levbut, unlike the others, have 0 and are thus not split. The number els with indicating the n value of the electron is often omitted.) In the first excited state, the outer electron is excited from the 3s level to the 3p level, which is about 2.1 eV above the ground state. The spin-orbit energy difference between the states due to the spin-orbit effect is about 0.002 eV. Transitions from these states to the ground state give the familiar sodium yellow doublet P3>2 P1>2 and 3p(2P1>2) S 3s(2S1>2) 589.6 nm 3p(2P3>2) S 3s(2S1>2) 589.0 nm The energy levels and spectra of other alkali atoms are similar to those for sodium. It is important to distinguish between doublet energy states and doublet spectral lines. All transitions beginning or ending on an S state give double lines because they involve one doublet state and one singlet state (the selection rule 1 rules out transitions between two S states). There are four", " possible energy differences between two doublet states. One of these is ruled out by a selection rule on j, which is15 \u00a2j 1 or 0 (but no j 0 S j 0) 7-67 Transitions between pairs of doublet energy states therefore result in three spectral lines, i.e., a triplet. Under relatively low resolution the three lines look like two, as illustrated in Figure 7-23, because two of them are very close together. For this reason, they are often referred to as a compound doublet to preserve the verbal hint that they involve doublet energy states. More Atoms with more than one electron in the outer shell have more complicated energy-level structures. Additional total spin possibilities exist for the atom, resulting in multiple sets of nearly independent energy states and multiple sets of spectral lines. Multielectron Atoms and their spectra are described on the home page: www.whfreeman.com/ tiplermodernphysics5e. See also Equations 7-68 and 7-69 and Figures 7-24 through 7-27 here. More Tradition tells us that Mrs. Bohr encountered an obviously sad young Wolfgang Pauli sitting in the garden of Bohr\u2019s Institute for Theoretical Physics in Copenhagen and asked considerately if he was unhappy. His reply was, \u201cOf course I\u2019m unhappy! I don\u2019t understand the anomalous Zeeman effect!\u201d On the home page we explain The Zeeman Effect so you, too, won\u2019t be unhappy: www.whfreeman.com/ tiplermodernphysics5e. See also Equations 7-70 through 7-74 and Figures 7-28 through 7-31 here. 304 Chapter 7 Atomic Physics EXPLORING Frozen Light Using the quantum properties of atomic energy states, tunable lasers, and a BoseEinstein (BE) condensate of sodium atoms (see Chapter 8), physicists have been able to slow a light pulse to a dead stop, then regenerate it some time later and send it on its way. Here is how it\u2019s done. Consider the 3s and 3p energy levels of sodium in Figure 7-22. L-S coupling does not cause splitting of the 3s state because the orbital angular momentum of that state is zero; however, we will discover in Chapter 11 (see also Problem 7-72) that protons and neutrons also have intrinsic spins and magnetic moments, resulting in a nuclear spin and magnetic moment. Although the latter is smaller than the electron\u2019s magnetic moment by a factor of about 1000, it causes a very small splitting of the 3s level exactly analogous to that due to L-S coupling in states with nonzero orbital angular momenta. Called hyperfine structure (because it\u2019s smaller than the fine-structure splitting discussed earlier), the 3s level is split into two levels spaced about 3.5 106 eV above and below the original 3s state. Producing the BE condensate results in a cigar-shaped \u201ccloud\u201d about one centimeter long suspended by a magnetic field in a vacuum chamber. The cloud contains several million sodium atoms all with their spins aligned and all in the lower of the two 3s hyperfine levels, the new ground state. (See Figure 7-32a.) The light pulse that we wish to slow (the probe beam) is provided by a laser precisely tuned to the energy difference between the lower of the 3s hyperfine levels (the new ground state) and the 3p state. A second laser (the coupling beam) is precisely tuned to the energy difference between the higher of the 3s hyperfine levels and the 3p state and illuminates the BE condensate perpendicular to the probe beam. If the probe beam alone were to enter the sample, all of the atoms would be excited to the 3p level, absorbing the beam completely. As the atoms relaxed back to the ground state, sodium yellow light would be emitted randomly in all directions. If the coupling beam alone entered the sample, no excitation of the 3p level would result because the coupling beam photons do not have enough energy to excite electrons from the ground state to the 3p state. However, if the coupling beam is illuminating the sample with all atoms in the ground state and the probe beam is turned on, as the leading edge of the probe pulse enters the sample (Figure 7-32b), the two beams together shift the sodium atoms into a quantum superposition of both states, meaning that in that region of the sample each atom is in both hyperfine states (Figure 7-32c). Instead of both beams now being able to excite those atoms to the 3p level, the two processes cancel, a phenomenon called quantum interference, and the BE condensate becomes transparent to the probe beam, as in Figure 7-32c. A similar cancellation causes the index of refraction of the sample to change very steeply over", " the narrow frequency range of the probe pulse, slowing the leading edge from As the rest 3 108 m>s ) enters the sample and slows, it piles of the probe pulse (still moving at up behind the leading edge, dramatically compressing the pulse to about 0.05 mm in length, which fits easily within the sample. Over the region occupied by the compressed pulse, the quantum superposition shifts the atomic spins in synchrony with the superposition, as illustrated in Figure 7-32d. 3 108 m>s 15 m>s. to about At this point the coupling beam is turned off. The BE condensate immediately becomes opaque to the probe beam; the pulse comes to a stop and turns off. The light has \u201cfrozen\u201d! The information imprinted on the pulse is now imprinted like a hologram on the spins of the atoms in the superposition states. (See Figure 7-32e.) When the coupling pulse is again turned on, the sample again becomes transparent to the probe pulse. 7-8 Excited States and Spectra of Atoms 305 3 108 m>s. The \u201cfrozen\u201d probe pulse is regenerated carrying the original information, moves (See Figure 7-32f.) slowly to the edge of the sample, then zooms away at The ability to slow and stop light raises new opportunities in many areas. For example, it may make possible the development of quantum communications that cannot be eavesdropped upon. Building large-scale quantum computers may depend upon the ultra-high-speed switching potential of quantum superpositions in slow light systems. Astrophysicists may be able to use BE condensates in vortex states, already achieved experimentally, with slow light to simulate in the laboratory the dragging of light into black holes. Stay tuned! Coupling beam (a) Probe beam (b) (c) (d) (e) (f) Figure 7-32 (a) The coupling beam illuminates the sodium Bose-Einstein condensate, whose atoms are in the ground state with spins aligned. (b) The leading edge of the probe beam pulse enters the sample. (c) Quantum superposition shifts the spins, and the rapidly changing refractive index dramatically slows and shortens the probe beam inside the condensate. (d) Now completely contained inside the sample, the speed of the probe pulse is about 15 m s. (e) The coupling beam is turned off and the probe pulse stops, its information stored in the shifted spins of the atoms. (f) The coupling beam is turned back on and the probe pulse regenerates, moves slowly to the edge of the sample, then leaves at 3 108 m s. [Courtesy of Samuel Velasco.] > > 306 Chapter 7 Atomic Physics Summary TOPIC RELEVANT EQUATIONS AND REMARKS 1. Schr\u00f6dinger equation in three dimensions 2. Quantization Angular momentum z component of L Energy 3. Hydrogen wave functions The equation is solved for the hydrogen atom by separating it into three ordinary, differential equations, one for each coordinate and m arise from the boundary conditions to the solutions of these equations. The quantum numbers n, r,,. \u0192L\u0192 2( 1)U for 0, 1, 2, 3, \u00c1 Lz En mU for m 0, 1, 2, \u00c1, a kZe2 U b 2 2n2 13.6 Z2 n2 eV 7-22 7-23 7-24 & nm CnmRn(r)Ynm(, ) Cnm where spherical harmonics. are normalization constants, Rn are the radial functions, and Ym are the 4. Electron spin The electron spin is not included in Schr\u00f6dinger\u2019s wave equation. Magnitude of S z component of S \u0192S\u0192 2s(s 1)U s 1 2 Sz ms U ms 1 2 Stern-Gerlach experiment This was the first direct observation of the electron spin. 5. Spin-orbit coupling L and S add to give the total angular momentum J L S, whose magnitude is given by \u0192J\u0192 2j(j 1)U where j s or the energy levels. \u0192 s\u0192. This interaction leads to the fine-structure splitting of 7-36 7-53 No more than one electron can occupy a given quantum state specified by a particular set of the single-particle quantum numbers n,, m, and ms. 6. Exclusion principle General References The following general references are written at a level appropriate for the readers of this book. Brehm, J. J., and W. J. Mullin, Introduction to the Structure of Matter, Wiley, New York, 1989. Kuhn, H. G., Atomic Spectra, Academic Press, New York, 1962. Mehra, J., and H. Rechenberg, The Historical Development of Quantum Theory, Vol.", " 1, Springer-Verlag, New York, 1982. Pauling, L., and S. Goudsmit, The Structure of Line Spectra, Eisberg, R., and R. Resnick, Quantum Physics, 2d ed., Wiley, McGraw-Hill, New York, 1930. New York, 1985. Herzberg, G., Atomic Spectra and Atomic Structure, Dover, New York, 1944. Weber, H. J., and G. B. Arfken, Essential Mathematical Methods for Physicists, Elsevier Academic Press, New York, 2004. Notes 1. Degeneracy may arise because of a particular symmetry of the physical system, such as the symmetry of the potential energy described here. Degeneracy may also arise for completely different reasons and can certainly occur for nonproduct wave functions. The latter are sometimes called accidental degeneracies, and both types can exist in the same system. 4. Lz \u0192L\u0192 2. \u201cEnough\u201d means a complete set in the mathematical sense. 3. Such potentials are called central field or, sometimes, conservative potentials. The Coulomb potential and the gravitational potential are the most frequently encountered examples. 0. All three components of L would then be known exactly, a violation of the uncertainty principle. Ym 5. The functions Cnm and listed in Tables 7-1 and 7-2 are are simply the products of those corre- would mean that Lx normalized. The sponding normalization constants. Ly Rn 6. Wolfgang Pauli (1900\u20131958), Austrian physicist. A bona fide child prodigy, while a graduate student at Munich he wrote a paper on general relativity that earned Einstein\u2019s interest and admiration. Pauli was 18 at the time. A brilliant theoretician, he became the conscience of the quantum physicists, assaulting \u201cbad physics\u201d with an often devastatingly sharp tongue, one of his oft-quoted dismissals of a certain poor paper being, \u201cIt isn\u2019t even wrong.\u201d He belatedly won the Nobel Prize in Physics in 1945 for his discovery of the exclusion principle. 7. Samuel A. Goudsmit (1902\u201378) and George E. Uhlenbeck (1900\u201388), Dutch-American physicists. While graduate students at Leiden, they proposed the idea of electron spin to their thesis adviser Paul Ehrenfest, who suggested that they ask H. A. Lorentz his opinion. After some delay, Lorentz pointed out that an electron spin of the magnitude necessary to explain the fine structure was inconsistent with special relativity. Returning to Ehrenfest with this disturbing news, they found that he had already sent their paper to a journal for publication. 8. Since the same symbol is used for both the reduced mass and the magnetic moment, some care is needed to keep these unrelated concepts clear. The symbol m is sometimes used to designate the magnetic moment, but there is confusion enough between the symbol m of the quantum number for the z component of angular momentum and me as the electron mass. 9. Otto Stern (1888\u20131969), German-American physicist, and Walther Gerlach (1899\u20131979), German physicist. After working as Einstein\u2019s assistant for two years, Stern developed the atomic/molecular beam techniques that enabled him and Gerlach, an excellent experimentalist, to show the existence of space quantization in silver. Stern received the 1943 Nobel Prize in Physics for his pioneering molecular beam work. 10. The nucleus of an atom also has angular momentum and therefore a magnetic moment, but the mass of the nucleus is about 2000 times that of the electron for hydrogen and greater still for other atoms. From Equation 7-39 we expect the magof a netic moment of the nucleus to be on the order of 1>2000 Notes 307 12S1>2 22P1>2 22S1>2 s 1 2, so 2s 1 2. ground state via a transition to the Bohr magneton since M is now mp rather than me. This small effect does not show up in the Stern-Gerlach experiment. 11. The letters first used, s, p, d, f, weren\u2019t really arbitrary. They described the visual appearance of certain groups of spectral lines. After improved instrumentation vastly increased the number of measurable lines, the letters went on alphabetically. As we noted in Chapter 4, the K, L, etc., notation was assigned by Barkla. 12. This particular form for writing the total spin was chosen because it also corresponded to the number of lines in the fine structure of the spectrum; e.g., hydrogen lines were doublets and 13. A more precise interpretation is that the electron, possessing an intrinsic magnetic moment due to its spin, carries with it a dipole magnetic field. This field varies in time due to the orbital motion of the electron, thus generating a time-varying electric field", " at the (stationary) proton, which produces the energy shift. 14. Actually, it\u2019s not quite true for hydrogen either. W. Lamb showed that the 2S and 2P levels of hydrogen differ slightly in energy. That difference together with the spin-orbit splitting of level 4.4 106 eV below the the 2P state puts the 22S1>2 level, an energy difference called the Lamb shift. It enables the state, which would otherwise have been metastable due to the 1 selection rule, to deactivate to level. The the Lamb shift is accounted for by relativistic quantum theory. 15. We can think of this rule in terms of the conservation of angular momentum. The intrinsic spin angular momentum of a photon has the quantum number s 1. For electric dipole radiation, the photon spin is its total angular momentum relative to the center of mass of the atom. If the initial angular momentum quantum number of the atom is j1 and the final is j2, the rules for combining angular momenta imply that j2 16. This is true for nearly all two-electron atoms, such as He, Be, Mg, and Ca, except for the triplet P states in the very heavy atom mercury, where fine-structure splitting is of about the same order of magnitude as the singlet-triplet splitting. 17. Pieter Zeeman (1865\u20131943), Dutch physicist. His discovery of the Zeeman effect, which so enlightened our understanding of atomic structure, was largely ignored until its importance was pointed out by Lord Kelvin. Zeeman shared the 1902 Nobel Prize in Physics with his professor H. A. Lorentz for its discovery. 18. The terminology is historical, arising from the fact that the effect in transitions between singlet states could be explained by Lorentz\u2019s classical electron theory and hence was \u201cnormal,\u201d while the effects in other transitions could not and were thus mysterious or \u201canomalous.\u201d 19. This calculation can be found in Herzberg (1944). 20. After Alfred Land\u00e9 (1888\u20131975), German physicist. His collaborations with Born and Heisenberg led to the correct interpretation of the anomalous Zeeman effect. 0, j2 must be 1. 1, j1, or j1 0. If j1 1, if j1 22P1>2 j1 308 Chapter 7 Atomic Physics 2L1, and 3L1. Give the sets of quantum numbers n1, n2, and n3 that correspond to the lowest 10 en- Problems Level I Section 7-1 The Schr\u00f6dinger Equation in Three Dimensions 7-1. Find the energies E311, E222, and E321 and construct an energy-level diagram for the threedimensional cubic well that includes the third, fourth, and fifth excited states. Which of the states on your diagram are degenerate? 7-2. A particle is confined to a three-dimensional box that has sides L1, L2 L3 ergy levels of this box. 7-3. A particle moves in a potential well given by V(x, y, z) 0 for L 2 x L 2, 0 y L, and 0 z L and V outside these ranges. (a) Write an expression for the ground-state wave function for this particle. (b) How do the allowed energies compare with those for a box having V 0 for 0 x L rather than for L 2 x L 2? 7-4. Write down the wave functions for the 5 lowest energy levels of the particle in Problem 7-2. 4L1. (b) What sets of quantum 7-5. numbers correspond to degenerate energy levels? 7-6. Write down the wave functions for the lowest 10 quantized energy states for the particle in Problem 7-5. 7-7. Suppose the particle in Problem 7-1 is an electron and L 0.10 nm. Compute the energy of the transitions from each of the third, fourth, and fifth excited states to the ground state. 7-8. Consider a particle moving in a two-dimensional space defined by V 0 for 0 x L and 0 y L and V elsewhere. (a) Write down the wave functions for the particle in this well. (b) Find the expression for the corresponding energies. (c) What are the sets of quantum numbers for the lowest-energy degenerate state? (a) Repeat Problem 7-2 for the case L2 2L1 and L3 > > > > Section 7-2 Quantization of Angular Momentum and Energy in the Hydrogen Atom #. 4 2.? (b) For each value of >2 735 rev>min If n 3, (a) what are the possible values of 7-9. in (a), list the possible values of m. (c) Using the fact that there are two quantum states for each combination and m because of electron", " spin, find the total number of electron states with of values of n 3. 7-10. Determine the minimum angle that L can make with the z axis when the angular momentum quantum number is (a) and (b) 7-11. The moment of inertia of a compact disc is about 105 kg m2. (a) Find the angular momentum L I when the disc rotates at and (b) find the approximate value of the quantum number 7-12. Draw an accurately scaled vector model diagram illustrating the possible orientations of the angular momentum vector L for (a) 1, (b) 2, (c) 4. (d) Compute the magnitude of L in each case. 7-13. For 2, (a) what is the minimum value of 2 Ly 2 Ly 2? Lx Lx of this? (d) What is the minimum value of n that this state can have? 7-14. For 1, find (a) the magnitude of the angular momentum L and (b) the possible values of m. (c) Draw to scale a vector diagram showing the possible orientations of L with the z axis. (d) Repeat the above for 3. 7-15. Show that, if V is a function only of r, then dL dt 0, i.e., that L is conserved. 7-16. What are the possible values of n and m if (a) 3, and (b) 4, and (c) 0? (d) Compute the minimum possible energy for each case. 7-17. A hydrogen atom electron is in the 6f state. (a) What are the values of n and? (b) Compute the energy of the electron. (c) Compute the magnitude of L. (d) Compute the possible values of Lz in this situation. 7-18. At what values of r a0 is the radial function R30 equal to zero? (See Table 7-2.) (b) What is the maximum value for 2 and m 1? Can either Lx or Ly be determined from 2 Ly 2? Lx (c) What is > > 2 Section 7-3 The Hydrogen Atom Wave Functions Problems 309 \u20192, (b) and (c) the \u2019, 7-19. For the ground state of the hydrogen atom, find the values of (a) radial probability density P(r) at r a0. Give your answers in terms of a0. 7-20. For the ground state of the hydrogen atom, find the probability of finding the electron in the range r 0.03a0 at (a) r a0 and at (b) r 2a0. 7-21. The radial probability distribution function for the hydrogens ground state can be written where C is a constant. Show that P(r) has its maximum value at r a0 Z. P(r) Cr2e2Zr>a0, 7-22. Compute the normalization constant C210 in Equation 7-34. 7-23. Find the probability of finding the electron in the range r 0.02a0 at (a) r a0 and (b) r 2a0 for the state n 2, 0, m 0 in hydrogen. (See Problem 7-25 for the value of C200.) 7-24. Show that the radial probability density for the n 2, 1, m 0 state of a oneelectron atom can be written as > P(r) A cos2 r2eZr>a0 where A is a constant. 7-25. The value of the constant C200 in Equation 7-33 is 3>2 C200 1 22 b a Z a0 \u2019, \u20192, (b) and (c) the radial probability density P(r) at r a0 for the state Find the values of (a) n 2, 0, m 0 in hydrogen. Give your answers in terms of a0. 7-26. Show that an electron in the n 2, 1 state of hydrogen is most likely to be found at r 4a0. 7-27. Write down the wave function for the hydrogen atom when the electron\u2019s quantum numbers are n 3, 2, and 1. 7-28. Verify that the wave function equation, Equation 7-9. Check to be sure that the wave function is normalized. \u2019 is a solution of the time-independent Schr\u00f6dinger m 100 Section 7-4 Electron Spin 7-29. If a classical system does not have a constant charge-to-mass ratio throughout the system, the magnetic moment can be written g Q 2M L > (I 1 2 MR2) where Q is the total charge, M is the total mass, and g 1. (a) Show that g 2 for a solid cylinthat spins about its axis and", " has a uniform charge on its cylindrical surface. der (b) Show that g 2.5 for a solid sphere (I 2MR2 5) that has a ring of charge on the surface at the equator, as shown in Figure 7-33. 7-30. Assuming the electron to be a classical particle, a sphere of radius 1015 m and a uniform \u0192S\u0192 s(s ]1)]1>2U (3>4)1>2U mass density, use the magnitude of the spin angular momentum to compute the speed of rotation at the electron\u2019s equator. How does your result compare with the speed of light? 7-31. How many lines would be expected on the detector plate of a Stern-Gerlach experiment (see Figure 7-15) if we use a beam of (a) potassium atoms, (b) calcium atoms, (c) oxygen atoms, and (d) tin atoms? 7-32. The force on a magnetic moment with z component moving in an inhomogeneous magnetic field is given by Equation 7-51. If the silver atoms in the Stern-Gerlach experiment traveled horizontally 1 m through the magnet and 1 m in a field-free region at a speed of 250 m s, what must have been the gradient of Bz, dBz dz in order that the beams each be deflected a maximum of 0.5 mm from the central, or no-field, position? 7-33. (a) The angular momentum of the yttrium atom in the ground state is characterized by the quantum number j 3 2. How many lines would you expect to see if you could do a SternGerlach experiment with yttrium atoms? (b) How many lines would you expect to see if the beam consisted of atoms with zero spin but 1? z > > > Figure 7-33 Solid sphere with charge Q uniformly distributed on ring. 310 Chapter 7 Atomic Physics Section 7-5 Total Angular Momentum and the Spin-Orbit Effect d3>2 and (b) the states of atomic 7-34. The spin-orbit effect removes a symmetry in the hydrogen atom potential, splitting the energy levels. (a) Considering the state with n 4, write down in spectroscopic notation the identification of each state and list them in order of increasing energy. (b) If a weak external magnetic field is applied to the atoms, into how many levels will each state in (a) be split? 7-35. Suppose the outer electron in a potassium atom is in a state with 2. Compute the magnitude of L. What are the possible values of j and the possible magnitudes of J? 7-36. A hydrogen atom is in the 3D state (n 3, 2). (a) What are the possible values of j? (b) What are the possible values of the magnitude of the total angular momentum? (c) What are the possible z components of the total angular momentum? d5>2 7-37. Compute the angle between L and S in (a) the hydrogen. 7-38. Write down all possible sets of quantum numbers for an electron in a (a) 4f, (b) 3d, and (c) 2p subshell. 7-39. Consider a system of two electrons, each with 1 and (a) What are the possible values of the quantum number for the total orbital angular momentum L L1 L2? (b) What are the possible values of the quantum number S for the total spin S S1 S2? (c) Using the results of parts (a) and (b), find the possible quantum numbers j for the combination J L S. (d) What are the possible quantum numbers j1 and j2 for the total angular momentum of each particle? (e) Use the results of part (d) to calculate the possible values of j from the combinations of j1 and j2. Are these the same as in part (c)? 7-40. The prominent yellow doublet lines in the spectrum of sodium result from transitions from the states to the ground state. The wavelengths of these two lines are 589.0 nm and 589.6 nm. (a) Calculate the energies in eV of the photons corresponding to these wavelengths. (b) The difference in energy of these photons equals the difference in energy E states. This energy difference is due to the spin-orbit effect. Calculate E. of the (c) If the 3p electron in sodium sees an internal magnetic field B, the spin-orbit energy splitting BB, is the Bohr magneton. Estimate B from the will be of the order of B energy difference E found in part (b). \u00a2E 2 s 1 2. where 3P1>2 3P3>2 3P3>2 3P1>2 and and Section 7-6 The Schr\u00f6dinger", " Equation for Two (or More) Particles 7-41. Show that the wave function of Equation 7-59 satisfies the Schr\u00f6dinger equation (Equation 7-57) with V 0 and find the energy of this state. 7-42. Two neutrons are in an infinite square well with L 2.0 fm. What is the minimum total energy that the system can have? (Neutrons, like electrons, have antisymmetric wave functions. Ignore spin.) 7-43. Five identical noninteracting particles are placed in an infinite square well with L 1.0 nm. Compare the slowest total energy for the system if the particles are (a) electrons and (b) pions. Pions have symmetric wave functions and their mass is 264 me. Section 7-7 Ground States of Atoms: The Periodic Table 7-44. Write the electron configuration of (a) carbon, (b) oxygen, and (c) argon. 7-45. Using Figure 7-34, determine the ground-state electron configurations of tin (Sn, Z 50), neodymium (Nd, Z 60), and ytterbium (Yb, Z 70). Check your answers with Appendix C. Are there any disagreements? If so, which one(s)? 7-46. In Figure 7-20 there are small dips in the ionization potential curve at Z 31 (gallium), Z 49 (indium), and Z 81 (thallium) that are not labeled in the figure. Explain these dips, using the electron configuration of these atoms given in Appendix C. 7-47. Which of the following atoms would you expect to have its ground state split by the spinorbit interaction: Li, B, Na, Al, K, Ag, Cu, Ga? (Hint: Use Appendix C to see which elements have 0 in their ground state and which do not.) Problems 311 Figure 7-34 Energy of electron ground-state configurations versus Z. \u0394f 6s 5d 6p 4d 5s 5p 3d 4p 4s 3p 3s y g r e n E 2p 2s 1s 10 20 30 40 z 50 60 70 80 90 > 7-48. If the 3s electron in sodium did not penetrate the inner core its energy would be 13.6 eV 32 1.51 eV. Because it does penetrate, it sees a higher effective Z and its energy is lower. Use the measured ionization potential of 5.14 V to calculate Zeff for the 3s electron in sodium. 7-49. What elements have these ground-state electron configurations? (a) 1s22s22p63s23p2 and (b) 1s22s22p63s23p64s2? 7-50. Give the possible values of the z component of the orbital angular momentum of (a) a d electron, (b) an f electron, and (c) an s electron. Section 7-8 Excited States and Spectra of Atoms 7-51. Which of the following elements should have an energy-level diagram similar to that of sodium and which should be similar to mercury: Li, He, Ca, Ti, Rb, Ag, Cd, Mg, Cs, Ba, Fr, Ra? 7-52. The optical spectra of atoms with two electrons in the same outer shell are similar, but they are quite different from the spectra of atoms with just one outer electron because of the interaction of the two electrons. Separate the following elements into two groups such that those in each group have similar spectra: lithium, beryllium, sodium, magnesium, potassium, calcium, chromium, nickel, cesium, and barium. 7-53. Which of the following elements should have optical spectra similar to that of hydrogen and which should have optical spectra similar to that of helium: Li, Ca, Ti, Rb, Ag, Cd, Ba, Hg, Fr, Ra? 7-54. The quantum numbers n, respectively, in the ground state; 4, 1, and and second excited state. Make a table giving the n, in potassium (see Figure 7-24). 7-55. Which of the following transitions in sodium do not occur as electric dipole transitions? (Give the selection rule that is violated.) and j for the outer electron in potassium have the values 4, 0, 3 in the in the first excited state; and 4, 1, and 2, and j values for the 12 lowest-energy states 1 2 1 2,, 4S1>2 4D3>2 S 3S1>2 S 3P1>2 4S1>2 4D3>2 S 3P3>2 S 3S1>2 4P3>2 5D3>2 S 3S1>2 S 4S1>", "2 4D5>2 5P1>2 S 3P1>2 S 3S1>2 7-56. Transitions between the inner electron levels of heavier atoms result in the emission of characteristic x rays, as was discussed in Section 4-4. (a) Calculate the energy of the electron in the K shell for tungsten using Z 1 for the effective nuclear charge. (b) The experimental result for this energy is 69.5 keV. Assume that the effective nuclear charge is (Z ), where is called the screening constant, and calculate from the experimental result for the energy. 7-57. Since the P states and the D states of sodium are all doublets, there are four possible energies for transitions between these states. Indicate which three transitions are allowed and which one is not allowed by the selection rule of Equation 7-67. 312 Chapter 7 Atomic Physics > E C Z2 eff(13.6) eVD>n2 7-58. The relative penetration of the inner-core electrons by the outer electron in sodium can be described by the calculation of Zeff from and comparing with E 13.6 eV n2 for no penetration (see Problem 7-45). (a) Find the energies of the outer electron in the 3s, 3p, and 3d states from Figure 7-22. (Hint: An accurate method is to use 5.14 eV for the ground state as given and find the energy of the 3p and 3d states from the photon energies of the indicated transitions.) (b) Find Zeff for the 3p and 3d states. (c) Is the approximation 13.6 eV n2 good for any of these states? 0.55 T. 7-59. A hydrogen atom in the ground state is placed in a magnetic field of strength Bz (a) Compute the energy splitting of the spin states. (b) Which state has the higher energy? (c) If you wish to excite the atom from the lower to the higher energy state with a photon, what frequency must the photon have? In what part of the electromagnetic spectrum does this lie? 7-60. Show that the change in wavelength of a transition due to a small change in energy is \u00a2 > \u00a2 2 hc \u00a2E E hc>. ) (Hint: Differentiate 7-61. (a) Find the normal Zeeman energy shift for a magnetic field of strength B 0.05 T. (b) Use the result of Problem 7-57 to calculate the wavelength changes for the sin 579.07 nm. (c) If the smallest wavelength change glet transition in mercury of wavelength that can be measured in a spectrometer is 0.01 nm, what is the strength of the magnetic field needed to observe the Zeeman effect in this transition? \u00a2E eUB>2me Zeff e Level II 7-62. If the outer electron in lithium moves in the n 2 Bohr orbit, the effective nuclear charge 1e and the energy of the electron would be 13.6 eV 22 3.4 eV. would be However, the ionization energy of lithium is 5.39 eV, not 3.4 eV. Use this fact to calculate the effective nuclear charge Zeff seen by the outer electron in lithium. Assume that r 4a0 for the outer electron. 7-63. Show that the expectation value of r for the electron in the ground state of a one-electron atom is 7-64. If a rigid body has moment of inertia I and angular velocity 8r9 (3>2)a0 its kinetic energy is >Z., > E 1 2 I2 (I)2 2I L2 2I where L is the angular momentum. The solution of the Schr\u00f6dinger equation for this problem leads to quantized energy values given by E ( 1)U2 2I E1 I 1 r 0.074 nm U2>I. (c) The moment of inertia of the H2 molecule is (a) Make an energy-level diagram of these energies, and indicate the transitions that obey the selection rule 1. (b) Show that the allowed transition energies are E1, 2E1, 3E1, 4E1, etc., where mp is the where is the distance between the protons. Find the energy of mass of the proton and the first excited state 1 for H2, assuming it is a rigid rotor. (d) What is the wavelength of the radiation emitted in the transition 1 to 0 for the H2 molecule? 7-65. In a Stern-Gerlach experiment hydrogen atoms in their ground state move with speed 14.5 km s. The magnetic field is in the z direction, and its maximum gradient is given by dBz > vx dz 600 T m. (a) Find the maximum acceleration of the hydrogen atoms. (b", ") If the region of the magnetic field extends over a distance x 75 cm and there is an additional 1.25 m from the edge of the field to the detector, find the maximum distance between the two lines on the detector. 7-66. Find the minimum value of the angle between the angular momentum L and the z axis for a general value of and show that for large values of, 1>1>2. 2 mpr2,, > > min Problems 313 and 4P3>2 4P1>2 7-67. The wavelengths of the photons emitted by potassium corresponding to transitions from states to the ground state are 766.41 nm and 769.90 nm. (a) Calculate the enthe ergies of these photons in electron volts. (b) The difference in energies of these photons equals the difference in energy E between the states in potassium. Calculate E. (c) Estimate the magnetic field that the 4p electron in potassium experiences. 7-68. The radius of the proton is about R0 side the volume occupied by the proton is given by 1015 m. The probability that the electron is in- 4P1>2 4P3>2 and P R0 0 P(r) dr V for r e2r>a0 1 where P(r) is the radial probability density. Compute P for the hydrogen ground state. (Hint: a0 is valid for this calculation.) Show that 7-69. (a) Calculate the Land\u00e9 g factor (Equation 7-74) for the levels in a oneelectron atom and show that there are four different energies for the transition between these levels in a magnetic field. (b) Calculate the Land\u00e9 g factor for the level and show that there are six different energies for the transition 7-70. (a) Show that the function 2P3>2 in a magnetic field. S 2S1>2 2S1>2 2P1>2 2P3>2 and \u2019 A r a0 er>2a0 cos is a solution of Equation 7-9, where A is a constant and a0 is the Bohr radius. (b) Find the constant A. K K Level III 7-71. Consider a hypothetical hydrogen atom in which the electron is replaced by a partiis a meson with spin 0, hence, no intrinsic magnetic moment. The only magnetic cle. The moment for this atom is that given by Equation 7-43. If this atom is placed in a magnetic field 1.0 T, (a) what is the effect on the 1s and 2p states? (b) Into how many lines does the with Bz 2p S 1s spectral line split? (c) What is the fractional separation between adjacent lines? (See Problem 7-57.) The mass of the 7-72. If relativistic effects are ignored, the n 3 level for one-electron atoms consists of the 32S1>2, 32P1>2, 32P3>2, 32D3>2, states. Compute the spin-orbit-effect splittings of 3P and 3D states for hydrogen. 7-73. In the anomalous Zeeman effect, the external magnetic field is much weaker than the internal field seen by the electron as a result of its orbital motion. In the vector model (Figure 7-30) the vectors L and S precess rapidly around J because of the internal field and J precesses slowly around the external field. The energy splitting is found by first calculating the compoin nent of the magnetic moment in the direction of J and then finding the component of is 493.7 MeV c2. 32D5>2 \u00a2> and K > J z the direction of B. (a) Show that J can be written # J J (b) From J2 (L S) in part (b) into that of part (a) to obtain (L S) show that # J B UJ (L2 2S2 3S # L) S # L 1 2 (J2 L2 S2 J B 2UJ (3J2 S2 L2) (d) Multiply your result by >J Jz to obtain z B a1 J2 S2 L2 2J2 b Jz U ). (c) Substitute your result 314 Chapter 7 Atomic Physics f f \u0192I J\u0192. ranges from I J to 7-74. If the angular momentum of the nucleus is I and that of the atomic electrons is J, the total angular momentum of the atom is F I J, and the total angular momentum quantum numShow that the number of possible values is 2I 1 if I J ber or 2J 1 if J I. (If you can\u2019t find a general proof, show it for enough special cases to convince yourself", " of its validity.) (Because of the very small interaction of the nuclear magnetic moment with that of the electrons, a hyperfine splitting of the spectral lines is observed. When I J, the value of I can be determined by counting the number of lines.) 7-75. Because of the spin and magnetic moment of the proton, there is a very small splitting of the ground state of the hydrogen atom called hyperfine splitting. The splitting can be thought of as caused by the interaction of the electron magnetic moment with the magnetic field due to the magnetic moment of the proton, or vice versa. The magnetic moment of the proton is paris called the nuclear magneton. (a) The allel to its spin and is about magnetic field at a distance r from a magnetic moment varies with angle, but it is of the order 107 in SI units. Find B at r a0 if (b) Calculate the of BB, order of magnitude of the hyperfine splitting energy is the Bohr magneton and B is your result from part (a). (c) Calculate the order of magnitude of the wavelength of radiation emitted if a hydrogen atom makes a \u201cspin flip\u201d transition between the hyperfine levels of the ground state. [Your result is greater than the actual wavelength of this transition, making the energy E found in part 21.22 cm, because (b) greater. The detection of this radiation from hydrogen atoms in interstellar space is an important part of radio astronomy.] 2.8 N. where B is appreciably smaller than eU>2mp B 2km \u00a2E 2 where km where >r3, 8r39 3, a0 2.8 N N, CHAPTER 8 Statistical Physics The physical world that we experience with our senses consists entirely of macro- scopic objects, i.e., systems that are large compared with atomic dimensions and thus are assembled from very large numbers of atoms. As we proceed to the description of such systems from our starting point of studying single-electron atoms, then multielectron atoms and molecules, we expect to encounter increasing complexity and difficulty in correctly explaining their observed properties. Classically, the behavior of any macroscopic system could, in principle, be predicted in detail from the solution of the equation of motion for each constituent particle, given its state of motion at some particular time; however, the obvious problems with such an approach soon become intractable. For example, consider the difficulties that would accompany the task of accounting for the measured properties of a standard liter of any gas by simultaneously solving the equations of motion for all of the 1022 molecules of which the system is composed. Fortunately, we can predict the values of the measurable properties of macroscopic systems without the need to track the motions of each individual particle. This remarkable shortcut is made possible by the fact that we can apply general principles of physics, such as conservation of energy and momentum, to large ensembles of particles, ignoring their individual motions, and determine the probable behavior of the system from statistical considerations. We then use the fact that there is a relation between the calculated probable behavior and the observed properties of the system. This successful, so-called microscopic approach to explaining the behavior of large systems is called statistical mechanics. It depends critically on the system containing a sufficiently large number of particles so that ordinary statistical theory is valid.1 In this chapter we will investigate how this statistical approach can be applied to predict the way in which a given amount of energy will most likely be distributed among the particles of a system. You may have already encountered kinetic theory, the first successful such microscopic approach, in introductory physics. Since the assumptions, definitions, and basic results of kinetic theory form the foundation of classical statistical physics, we have included a brief review of kinetic theory in the Classical Concept Review. We will see how, in an isolated system of particles in thermal equilibrium, the particles must be able to exchange energy, one result of which is that the energy of any individual particle may sometimes be larger and sometimes smaller than the average value for a particle in the system. Classical statistical mechanics requires 8-1 Classical Statistics: A Review 316 8-2 Quantum Statistics 328 8-3 The Bose-Einstein Condensation 335 8-4 The Photon Gas: An Application of Bose-Einstein Statistics 344 8-5 Properties of a Fermion Gas 351 315 316 Chapter 8 Statistical Physics that the values of the energy taken on by an individual particle over time, or the values of the energy assumed by all of the particles in the system at any particular time, be determined by a specific probability distribution, the Boltzmann distribution. In the first section of the chapter we will briefly review the principal concepts of classical statistical physics, noting some of the successful applications and some of the serious failures. We will then see how quantum considerations require modification of the procedures used for classical particles, obtaining in the process the quantum-mechanical FermiDirac distribution for particles with antisymmetric wave functions, such as electrons, and", " the Bose-Einstein distribution for particles with symmetric wave functions, such as helium atoms. Finally, we will apply the distributions to several physical systems, comparing our predictions with experimental observations and gaining an understanding of such important phenomena as superfluidity and the specific heat of solids. 8-1 Classical Statistics: A Review Statistical physics, whether classical or quantum, is concerned with the distribution of a fixed amount of energy among a large number of particles, from which the observable properties of the system may then be deduced. Classically, the system consists of a large ensemble of identical but distinguishable particles. That is, the particles are all exactly alike, but in principle they can be individually tracked during interactions. Boltzmann2 derived a distribution relation that made possible prediction of the probable numbers of particles that will occupy each of the available energy states in such a system in thermal equilibrium. Boltzmann Distribution The Boltzmann\u2019s distribution tion function of classical statistical physics: fB(E) given by Equation 8-1 is the fundamental distribu- fB(E) AeE>kT 8-1 where A is a normalization constant whose value depends on the particular system being considered, is called the Boltzmann factor, and k is the Boltzmann constant: eE>kT k 1.381 1023 J>K 8.617 105 eV>K Boltzmann\u2019s derivation was done to establish the fundamental properties of a distribution function for the velocities of molecules in a gas in thermal equilibrium that had been obtained by Maxwell a few years before and to show that the velocity distribution for a gas that was not in thermal equilibrium would evolve toward Maxwell\u2019s distribution over time. Boltzmann\u2019s derivation is more complex than is appropriate for our discussions, but in the Classical Concept Review we present a straightforward numerical derivation that results in an approximation of the correct distribution and then show by a simple mathematical argument that the form obtained is exact and is the only one possible. Here we will illustrate application of the Boltzmann distribution with some examples by way of providing a basis for comparing classical and quantum statistical physics later in the chapter. The number of particles with energy E is given by n(E) g(E)fB(E) A g(E)eE>kT 8-2 where g(E) is the statistical weight (degeneracy) of the state with energy E. 8-1 Classical Statistics: A Review 317 Figure 8-1 n(E) versus E for data from Table 1 in the CCR Boltzmann distribution derivation. The solid curve is the exponential n(E) BeE>Ec constants B and Ec have been adjusted to give the best fit to the data points., where the \u0394E 2\u0394E 3\u0394E 4\u0394E 5\u0394E 6\u0394E 7\u0394E 8\u0394E ( Classically, the energy E is a continuous function and so is n(E) (Figure 8-1). Consequently, g(E) and fB(E) are also continuous functions, in which case g(E) in Equation 8-2 is referred to as the density of states, meaning that g(E)dE is the number of states with energy between E and EdE. The next two examples illustrate how to apply the Boltzmann distribution and how the results explain observations of physical systems. EXAMPLE 8-1 The Law of Atmospheres Consider an ideal gas in a uniform gravitational field. (a) Find how the density of the gas depends upon the height above ground. (b) Assuming that air is an ideal gas with molecular weight 28.6, compute the density of air 1 km above the ground when T 300 K. (The density at the ground is 1.292 kg m3 at 300 K.) > SOLUTION (a) Let the force of gravity be in the negative z direction and consider a column of gas of cross-sectional area A. The energy of a gas molecule is then E p2 x 2m p2 y 2m p2 z 2m mgz p2 2m mgz p2 z p2 p2 x p2 where y at height z above the ground. The density proportional to N, the number of molecules in a unit volume at height z, and N is proportional to fB. and mgz is the potential energy of a molecule is is proportional to fB since From Equation 8-1 we have Aep2>2mkTemgz>kT fB Since we are interested only in the dependence on z, we can integrate over the other variables px, py, and pz. The integration merely gives a new normalization constant A; i.e., the result is equivalent to ignoring these variables. The fraction of the molecules between z and z dz is then fB(z) dz Aemgz>kT d", "z 8-3 The constant A is obtained from the normalization condition The result is Amg kT. The density, therefore, also decreases exponentially with the distance above the ground. Equation 8-3 is known as the law of atmospheres. 0 fB(z) dz 1. > 318 Chapter 8 Statistical Physics (b) The ratio of the density at z 1000 m to that at z 0 m is the same as where is given by Equation 8-3. Thus, fB(1000)>fB(0), (1000) (0) fB(z) fB(1000) fB(0) emg(1000)>k(300) emg(0)>k(300) emg(1000)>k(300) Substituting m 28.6 1.67 1027 kg and g 9.8 m s2 yields > (1000) (0)e0.113 1.292 0.893 1.154 kg>m3 EXAMPLE 8-2 H Atoms in the First Excited State The first excited state E2 of the hydrogen atom is 10.2 eV above the ground state E1. What is the ratio of the number of atoms in the first excited state to the number in the ground state at (a) T 300 K and (b) T 5800 K? The latter is the temperature at the surface of the Sun. SOLUTION 1. The number of atoms in a state with energy E is given by Equation 8-2: n(E) g(E)fB(E) A g(E)eE>kT 2. The ratio of the number in the first excited state to the number in the ground state is then n2 n1 Ag2eE2 Ag1 eE1 >kT >kT g2 g1 e(E2 E1)>kT 3. The statistical weight ( degeneracy) of the ground state g1, including spin, is 2; states, and three 0 1 the degeneracy of the first excited state g2 is 8 (one each with two spin states). Therefore: 8 2 4 g2 g1 and n2 n1 4e(E2 E1)>kT 4. For question (a), at T 300 K, kT 0.026 eV. Substituting this and E2 E1 10.2 eV from above gives n2 n1 4e(10.2)>(0.026) 4e392 10171 0 5. For question (b), at the surface of the Sun where T 5800 K, kT 0.500. Substituting this and E2 n2 n1 10.2 eV gives E1 4e(10.2)>(0.500) 4e20.4 e19 108 Remarks: The result in step 4 illustrates that, because of the large energy difference between the two states compared with kT, very few atoms are in the first excited state. Even fewer would be in the higher excited states, which explains why a container of hydrogen sitting undisturbed at room temperature does not spontaneously emit the visible Balmer series. At the surface of the Sun (step 5 above) about 1015 atoms of every mole of atomic hydrogen are in the first excited state at any given time. 8-1 Classical Statistics: A Review 319 More In learning about systems containing large numbers of particles, the meaning of the temperature needs to be more carefully defined. It is closely related to another descriptor of such systems, the entropy. To help you understand both concepts better, we have included > Temperature and Entropy on the home page: www.whfreeman.com tiplermodernphysics5e. See also Equations 8-4 a, b, c, and d here. Maxwell Distribution of Molecular Speeds The Boltzmann distribution is a very fundamental relation from which many properties of classical systems, both gases and condensed matter, can be derived. Two of the most important are Maxwell\u2019s distribution of the speeds of molecules in a gas and the equipartition theorem. Considering the first of these, Maxwell derived the velocity and speed distributions of gases in 1859, some five years before Boltzmann derived Equation 8-1. As with the Boltzmann distribution, we will present the results here, illustrating their application with examples and including fuller descriptions and derivations in the Classical Concept Review. Maxwell obtained the velocity distribution, F(vx, vy, vz), which can also be used to obtain the speed distribution, by assuming that the components vx, vy, and vz of the velocity were independent and that therefore the probability of a molecule having a certain vx, vy, vz could be factored into the product of the separate probabilities of its having vx, vy, and vz. He also assumed that the distribution could depend only on the speed;", " i.e., the velocity components could occur only in the combination vx 2. He thus wrote for the distribution function for 2 vy 2 vz vx f(vx) Cemvx 2>2kT 8-5 where \u0192(vx) is the distribution function for vx only; i.e., \u0192(vx)dvx is the fraction of the total number of molecules that have their x component of velocity between vx and dvx.3 Similar expressions can be written for \u0192(vy) and \u0192(vz). The constant C is vx determined by the normalization condition. The complete normalized velocity distribution is F(vx, vy, vz) f(vx)f(vy)f(vz) a 3>2 m 2kT b em(vx 2 vy 2vz 2)>2kT 8-6 The utility of distribution functions is that they make possible the calculation of average or expectation values of physical quantities; i.e., they allow us to make predictions regarding the physical properties of systems. For example, the observation from Figure 8-2 that the average value of vx is zero can be verified by computing as indicated by Equation 8-7. 9 8vx f(vx) vx Figure 8-2 The distribution function f(vx) for the x component of velocity. This is a Gaussian curve symmetric about the origin. 9 8vx vxf(vx) dvx vx a m 2kT b 1>2 emvx 2>2kT dvx 8-7 Writing m> 2kT, we have 9 (>) 8vx 2 vxevx dvx 320 n(v) Chapter 8 Statistical Physics vm = 2kT/m \u2329v\u232a = 8kT/\u03c0m vrms = 3kT/m From Table B1-1 we see that the value of the integral is zero, so as expected. 9 0, 8vx The probability distribution function for the speeds of the molecules in a classical ideal gas can be derived from the Boltzmann distribution. The result is the famous Maxwell distribution of molecular speeds: n(v) dv 4Na 3>2 m 2kT b v2emv2>2kT dv 8-8 vm vrms \u2329v\u232a v Figure 8-3 Maxwell speed distribution function n(v). The most probable speed vm, the average speed and the rms speed vrms are indicated. 8v9, 8v9, The distribution of speeds is shown graphically in Figure 8-3. The most probable speed vm, the average and the rms speed vrms are indicated in the speed figure. Although the velocity distribution function F (see Equation 8-6 and Figures 8-4 and 8-5) is a max 0), the imum at the origin (where vx speed distribution function n(v) approaches zero as 4v2 dv v S 0 because the latter is proportional to the volume of the spherical shell (see Equation 8-8), which approaches zero. At very high speeds, the speed distribution function again approaches zero because of the exponential factor emv2>2kT. vy vz The most probable speed vm is that where n(v) has its maximum value. It is left as an exercise (see Problem 8-9) to show that its value is 1>2 vm a 2kT m b 8-9 The average speed example. 8v9 is obtained in general and for a specific situation in the next vy 2v0 v0 dv v v0 = kT/m v0 2v0 vx Figure 8-5 Two-dimensional representation of velocity distribution in velocity space. Each molecular velocity with components vx, vy, and vz is represented by a point in velocity space. The velocity distribution function is the density of points in this space. The density is maximum at the origin. The speed distribution is found by multiplying the density times the volume of the spherical shell computer-generated plot courtesy of Paul Doherty, The Exploratorium.] 4v2 dv. [This One of the ways used to separate 235U from the far more abundant 238U isotope is to react the uranium metal with fluorine, forming UF6, a gas. 235UF diffuses through a membrane just a bit faster than 238UF since both molecules have the same average kinetic energy. After several stages of diffusion, the concentration of 235U is high enough for making nuclear reactor fuel (see Chapter 11). vz dvx dvy dvz vy vx Figure 8-4 Velocity vectors in velocity space. The velocity distribution function gives the fraction of molecular velocities whose vectors end in a cell of volume d", "vxdvydvz. 8-1 Classical Statistics: A Review 321 EXAMPLE 8-3 Average Speed of N2 Molecules Obtain the average speed of the Maxwell distribution and use it to compute the average speed of nitrogen molecules at 300 K. The mass of the N2 molecule is 4.68 1026 kg. SOLUTION 1. The average speed is found by multiplying the distribution of speeds 8v9 8v9 (Equation 8-8) by v, integrating over all possible speeds, and dividing by the total number of molecules N: vn(v) dv 8v9 1 N 0 A 4(m>2kT)3>2. and 0 Av3ev2 dv where m> 2kT 2. Writing this as where 8v9 AI3 v3ev2 0 I3 dv 3. Using Table B1-1 for evaluating I3, we have a 8v9 A2>2 4 2 a 8kT m b m 2kT 1>2 3>2 b 2 b a 2kT m 8-10 4. The 8v9 found in step 3 can now be used to find the average speed of nitrogen molecules at T 300 K. Substituting the mass of a nitrogen molecule into Equation 8-10 yields 8v9 c 8 1.38 1023 300 4.68 1026 d 1>2 475 m>s 1700 km>h The average speed is about 8 percent less than as indicated in Figure 8-3. The rms speed can be computed from the speed distribution following the same procedure as in Example 8-3 or, as we will see below, from the equipartition theorem. Figure 8-6, a plot of Equation 8-8 for H2 and O2 molecules at 300 K, illustrates the effect of mass on the speed distribution. vrms (3kT>m)1>2, ) ( v n \u2013N \u2013 \u2013 0.002 0.001 0 0 O2 H2 1000 2000 Molecular speed v, m/s Figure 8-6 Graph of n(v)/N versus v from Equation 8-8 for O2 and H2 molecules, both at T 300 K. 322 Chapter 8 Statistical Physics Figure 8-7 Schematic sketch of apparatus of Miller and Kusch for measuring the speed distribution of molecules. Only one of the 720 helical slits in the cylinder is shown. For a given angular velocity only molecules of a certain speed from the oven pass through the helical slits to the detector. The straight slit is used to align the apparatus. [From R. C. Miller and P. Kusch, Physical Review, 99, 1314 (1955).] Detector Oven source \u03c6 \u03c9 Maxwell\u2019s speed distribution has been precisely verified by many experiments, so there is little incentive to perform additional measurements. One of the more recent experiments, that of R. C. Miller and P. Kusch, illustrated in Figures 8-7 and 8-8, is applicable to the measurement of any sort of molecular speed distribution, and variations of it are used to measure the speeds in jet or nozzle molecular beams. Evaporation is a cooling process, even at very low temperatures! The sample from which a BE condensate will form, confined at about 1 mK, is cooled further by allowing the atoms in the high-speed \u201ctail\u201d of the Maxwell distribution to \u201cleak\u201d from the sample, taking kinetic energy with them and thus reducing the temperature (see Section 8-3). Figure 8-8 Data of Miller and Kusch showing the distribution of speed of thallium atoms from an oven at 870 K. The data have been corrected to give the distribution inside the oven since the faster molecules approach the exit slit more frequently and skew the external distribution slightly. The measured value for vm at 870 K is curve is that predicted by the Maxwell speed distribution. [From R. C. Miller and P. Kusch, Physical Review, 99, 1314 (1955).] 376 m>s. The solid 20 15 10.2 Run 99 Run 97 0.6 1.0 Reduced velocity, v/vm 1.4 1.8 Questions 1. How does vrms for H2 molecules compare with vrms for O2 molecules under standard conditions? Maxwell Distribution of Kinetic Energy 8-1 Classical Statistics: A Review 323 Maxwell\u2019s distribution of molecular speeds also provides, as a bonus, the distribution of the molecular translational kinetic energy and the average kinetic energy of a molv2 2E m, ecule. These can also be determined from Equation 8-2. Since and dv (2mE)1>2 dE, g(E) dE is E 1 2 mv2, > g(E) dE 4C(2E>m)(2mE)1>2 dE Substit", "uting the above into Equation 8-2, we have n(E) dE 4A(2>m3)1>2E1>2eE>kT dE 8-11 8-12 Evaluating A using the fact that 0 n(E) dE N, the total number of particles, allows us to write Maxwell\u2019s distribution of kinetic energy as n(E) dE 2N (kT)3>2 E1>2eE>kT dE 8-13 The kinetic energy distribution is sketched in Figure 8-9. The average kinetic energy is computed in the same manner as the average speed; i.e., the distribution is multiplied by E (the quantity being averaged), and the result is integrated4 over all values of E (from ) and divided by the number of molecules N: 0 S E n(E) dE 2 (kT)3>2 0 E3>2eE>kT dE 3 2 kT 8-14 8E9 1 N 0 ) E n ( \u2329E\u232a = kT 3 \u2013\u2013 2 0 kT 2kT 3kT E Figure 8-9 Maxwell distribution of kinetic energies for the molecules of an ideal gas. The average energy 8E9 3kT>2 is shown. EXAMPLE 8-4 Escape of H2 from Earth\u2019s Atmosphere A rule of thumb used by astrophysicists is that a gas will escape from a planet\u2019s atmosphere in 108 years if the average speed of its molecules is one-sixth of the escape velocity. Compute the average speed from the average kinetic energy and show that the absence of hydrogen in Earth\u2019s atmosphere suggests that Earth must be older than 108 years (mass of H2 molecules 3.34 1027 kg). 324 Chapter 8 Statistical Physics SOLUTION The escape speed at the bottom of the atmosphere, i.e., Earth\u2019s surface, is 11.2 km s, and one-sixth of that value is 1.86 km s. If we assume T 300 K, the average energy of a hydrogen molecule (or any other molecule since is independent of mass) is kT 3 1.38 1023 300 6.21 1021 J 8E9 > > 8E9 3 2 2 Thus, or, for hydrogen molecules, 1 2 mv2 6.21 1021 J v2 2 6.21 1021 3.34 1027 3.72 106 m2>s2 Therefore, v 1.93 km>s Remarks: Since v (1 6)vesc sphere suggests that the age of Earth is greater than 108 years. 1.86 km s, the absence of hydrogen in the atmo- > > Questions 2. How does 9 8Ek standard conditions? for He molecules compare with 8Ek 9 for Kr molecules under 3. H2 molecules can escape so freely from Earth\u2019s gravitational field that H2 is not found in Earth\u2019s atmosphere. (See Example 8-4.) Yet the average speed of H2 molecules at ordinary atmospheric temperatures is much less than the escape speed. How, then, can all of the H2 molecules escape? 4. Why wouldn\u2019t you expect all molecules in a gas to have the same speed? Heat Capacities of Gases and Solids The second important property of classical systems derivable from the Boltzmann distribution is one that applies to both gases and solids. Called the equipartition theorem, it states that In equilibrium, each degree of freedom contributes energy per molecule. 1 2 kT to the average A degree of freedom is a coordinate or a velocity component that appears squared in the expression for the total energy of a molecule. For example, the one-dimensional harmonic oscillator has two degrees of freedom, x and vx; a monatomic gas molecule has three degrees of freedom, vx, vy, and vz. More That each degree of freedom in a classical material should have the same average energy per molecule is not at all obvious. On the home page we have included A Derivation of the Equipartition Theorem for a special case, the harmonic oscillator, to illustrate how the more general result arises: www.whfreeman.com/tiplermodernphysics5e. See also Equations 8-15 through 8-23 here. CV for Gases The power of the equipartition theorem is its ability to accurately predict the heat capacities of gases and solids, but therein is also found its most dramatic failures. As an example, consider a rigid-dumbbell model of a diatomic molecule (Figure 8-10a) that can translate in the x, y, and z directions and can rotate about axes x and y through the center of mass and perpendicular to the z axis along the line joining the two atoms.5 The total energy for this rigid-dumbbell model molecule", " is then E 1 2 mv2 x 1 2 mv2 y 1 2 mv2 z 1 2 Ix 2 x 1 2 Iy 2 y 8-1 Classical Statistics: A Review 325 (a) y y\u00b4 z\u00b4 (b) y CM x\u00b4 x x z z Figure 8-10 (a) Rigid-dumbbell model of a diatomic gas molecule that can translate along the x, y, or z axis and rotate about the x or y axis fixed to the center of mass. If the spheres are smooth or are points, rotation about the z axis can be neglected. (b) Nonrigid-dumbbell model of a diatomic gas molecule that can translate, rotate, and vibrate. Iy Ix and are the moments of inertia about the x and where y axes. Since this molecule has 5 degrees of freedom, 3 translational and 2 rotational, the equipartition theorem predicts the average energy to be (5 2)kT per molecule. The > > energy per mole U is then (5 2)NAkT (5 2)RT and the molar heat capacity at constant is (5 2)R. The observation that CV for both nitrogen and volume oxygen is about (5 2)R enabled Rudolf Clausius to speculate (in about 1880) that these gases must be diatomic gases, which can rotate about two axes as well as translate. (See Table 8-1.) ($U>$T)V CV > > > Table 8-1 CV for some gases at 15\u00b0C and 1 atm >R > CV (cal mol-deg) CV Gas Ar He CO H2 HCl N2 NO O2 Cl2 CO2 CS2 H2S N2O SO2 2.98 2.98 4.94 4.87 5.11 4.93 5.00 5.04 5.93 6.75 9.77 6.08 6.81 7.49 R 1.987 cal mol-deg > 1.50 1.50 2.49 2.45 2.57 2.49 2.51 2.54 2.98 3.40 4.92 3.06 3.42 3.76 From J. R. Partington and W. G. Shilling, The Specific Heats of Gases (London: Ernest Benn, Ltd., 1924). 326 Chapter 8 Statistical Physics If a diatomic molecule is not rigid, the atoms can also vibrate along the line joining them (Figure 8-10b). Then, in addition to the translational energy of the center of mass and rotational energy, there can be vibrational energy. The vibration, a simple harmonic motion, adds two more squared terms to the energy, one for the potential energy and one for kinetic energy. For a diatomic molecule that is translating, rotating, and vibrating, the > equipartition theorem thus predicts a molar heat capacity of (3 2 2), or (7 2)R. However, measured values of CV for diatomic molecules (see Table 8-1) show no contribution from the vibrational degrees of freedom. The equipartition theorem provides no explanation for their absence. 1 2 R Experimental values of CV for several diatomic gases are included in Table 8-1. For all of these except Cl2, the data are consistent with the equipartition theorem prediction, assuming a rigid, nonvibrating molecule. The value for Cl2 is about halfway between that predicted for a rigid molecule and that predicted for a vibrating molecule. The situation for molecules with three or more atoms, several of which are also listed in Table 8-1, is more complicated and will not be examined in detail here. > The equipartition theorem in conjunction with the point-atom, rigid-dumbbell model was so successful in predicting the molar heat capacity for most diatomic molecules that it was difficult to understand why it did not do so for all of them. Why should some diatomic molecules vibrate and not others? Since the atoms are not points, the moment of inertia about the line joining the atoms, while small, is not zero, and there are three terms for rotational energy rather than two. Assuming no vibration, CV should then be (6 2)R. This agrees with the measured value for Cl2 but not for the other diatomic gases. Furthermore, monatomic molecules would have three terms for rotational energy if the atoms were not points, and CV should also be (6 2)R for these atoms rather than the (3 2)R that is observed. Since the average energy is calculated by counting terms, it should not matter how small the atoms are as long as they are not merely points. In addition to these difficulties, it is found experimentally that the molar heat capacity depends on temperature, contrary to the predictions from the equipartition theorem. The most spectacular case is that of H2, shown in Figure 8-11", ". It seems as if at very low temperatures, below about 60 K, H2 behaves like a monatomic molecule and does not rotate. It seems to undergo a transition, and be (5 2)R, thus behaving like a rotating rigid tween about 250 K and 700 K it has CV > > > v C 7 \u2013\u2013 2 5 \u2013\u2013 2 3 \u2013\u2013 2 1 \u2013\u2013 2 R R R R Vibration Rotation Translation 25 50 100 250 500 1000 2500 5000 T, K > Figure 8-11 Temperature dependence of molar heat capacity of H2. Between about 250 and 750 K, CV is (5 2)R, as predicted by the rigid-dumbbell model. At low temperatures, CV is only (3 2)R, as predicted for a nonrotating molecule. At high temperatures CV seems to be approaching (7 2)R, as predicted for a dumbbell model that rotates and vibrates, but the molecule dissociates before this plateau is reached. > > 8-1 Classical Statistics: A Review 327 > dumbbell. At very high temperatures H2 begins to vibrate, but the molecule dissociates before CV reaches (7 2)R. Other diatomic gases show similar behavior except that at low temperatures they liquefy before CV reaches (3 2)R. The failure of the equipartition theorem to account for these observations occurs because classical mechanics itself fails when applied to atoms and molecules. As we will see, it must be replaced by quantum mechanics. > CV for Solids The equipartition theorem is also useful in understanding the heat capacity of solids. In 1819 P. Dulong and A. Petit pointed out that the molar heat capacity of most solids was very nearly equal to 6 cal K-mol 3R. This result was used by them to obtain unknown molecular weights from the experimentally determined heat capacities. The empirical Dulong-Petit law is easily derived from the equipartition theorem by assuming that the internal energy of a solid consists entirely of the vibrational energy of the molecules (see Figure 8-12). If the force constants in the x, y, and z directions are k1, k2, respectively, the vibrational energy of each molecule is and k3, > E 1 2 mv2 x 1 2 mv2 y 1 2 mv2 z 1 2 k1x2 1 2 k2y2 1 2 k3z2 1 2 Since there are six squared terms, the average energy per molecule is 6( kT), and the total energy of 1 mole is 3NAkT 3RT, giving CV 3R. At high temperatures, all solids obey the Dulong-Petit law. For temperatures below some critical value, CV drops appreciably below the value of 3R and approaches zero as T approaches zero. The critical temperature is a characteristic of the solid. It is lower for soft solids such as lead than for hard solids such as diamond. The temperature dependence of CV for several solids is shown in Figure 8-13. The fact that CV for metals is not appreciably different from that for insulators is puzzling. The classical model of a metal is moderately successful in describing the conduction of electricity and heat. It assumes that approximately one electron per atom is free to move about the metal, colliding with the atoms much as the molecules do in a gas. According to the equipartition theorem, this \u201celectron gas\u201d should have an average kinetic energy of (3 2)kT per electron; thus, the molar heat capacity should be about (3 2)R greater for a conductor than for an insulator. Although the molar heat capacity for metals is slightly greater than 3R at very high temperatures, the difference is much less than the (3 2)R predicted for the contribution of the electron gas. > > > Figure 8-12 Simple model of a solid consisting of atoms connected to each other by springs. The internal energy of the solid then consists of kinetic and potential vibrational energy Lead Aluminum Silicon Carbon (diamond) 200 400 600 800 1000 1200 Absolute temperature, K Figure 8-13 Temperature dependence of molar heat capacity of several solids. At high temperatures CV is 3R, as predicted by the equipartition theorem. However, at low temperatures CV approaches zero. The critical temperature at which CV becomes nearly 3R is different for different solids. 328 Chapter 8 Statistical Physics The Boltzmann distribution and statistical mechanics were enormously successful in predicting the observed thermal properties of physical systems; however, the failure of the theory to account correctly for the heat capacities of gases and solids was a serious problem for classical physics, constituting as it did a failure of classical mechanics itself. The search for an understanding of specific heats was instrumental in the discovery of energy quantization in the early years of the twentieth century. The following sections show how quantum mechanics provides a basis for the complete understanding of the experimental observations. EXAMPLE", " 8-5 Broadening of Spectral Lines In Chapter 5 we saw that spectral lines emitted by atoms had a certain natural width due to the uncertainty principle. However, in luminous gases, such as sodium and mercury vapor lamps and the visible surface of the Sun, the atoms are moving with the Maxwell velocity distribution. The velocity distribution results in a Doppler effect that Rayleigh showed was proportional to the Boltzmann factor and led to a broadening of spectral lines equal to \u00a2 0.72 1062T>M is the wavelength of the line, T is the absolute temperature, and M is the where molecular weight. From this, compute the velocity (Doppler) broadening of the line emitted by H atoms at the surface of the Sun, where T 5800 K. hydrogen H SOLUTION The wavelength of the H line is 656.3 nm and the atomic weight of H is 1, so \u00a2 0.72 106 656.325800>1 0.036 nm H For comparison, the natural width of the line is about 0.0005 nm. Note that the effect of the pressure of the gas in causing spectral line broadening via collisions is also an important factor and, in fact, at high pressures, is the dominant cause. Collisions reduce the level lifetime, hence broaden the energy width (uncertainty principle). This is the reason that the Sun\u2019s visible spectrum is a continuous one. 8-2 Quantum Statistics Bose-Einstein and Fermi-Dirac Distributions The classical systems that were the subject of Section 8-1 consisted of identical but distinguishable particles. They were treated like billiard balls: exactly the same as one another but with numbers painted on their sides. Indeed, that was the point of the first assumption on the first page of the kinetic theory review in the Classical Concept Review on the Web site. However, the wave nature of particles in quantum mechanics prevents identical particles from being distinguished from one another. The finite extent and the overlap of wave functions makes identical particles indistinguishable. Thus, if two identical particles 1 and 2 pass within a de Broglie wavelength of each other in some event, we cannot tell which of the emerging particles is 1 and which is 2\u2014i.e., we cannot distinguish between the several possible depictions of the event in Figure 8-14. The treatment of classical particles that led to the Boltzmann distribution can be extended to systems containing large numbers of identical indistinguishable particles. 8-2 Quantum Statistics 329 Particle 1 Particle 2 Particle 2 Particle 1 Particle 1 Particle 2 Particle 2 Particle 1 Figure 8-14 The wave nature of quantum-mechanical particles prevents us from determining which of the four possibilities shown actually occurred when the two identical, indistinguishable particles passed within a de Broglie wavelength of each other. The first such theoretical treatment for particles with zero or integer spins\u2014i.e., those that do not obey the exclusion principle, such as helium atoms (spin 0) and photons (spin 1), was done by Bose6 in 1924, when he realized that the Boltzmann distribution did not adequately account for the behavior of photons. Bose\u2019s new statistical distribution for photons was generalized to massive particles by Einstein shortly thereafter. The resulting distribution function, called the Bose-Einstein distribution fBE(E), is given by fBE(E) 1 eeE>kT 1 8-24 e where tributions are given by Equation 8-24 are called bosons. is a system-dependent normalization constant. Particles whose statistical dis- Following the discovery of electron spin and Dirac\u2019s development of relativistic wave mechanics for spin- particles, Fermi7 and Dirac8 completed the statistical mechanics for quantum mechanical particles by deriving the probability distribution for large ensembles of identical indistinguishable particles that obey the exclusion principle. The result is called the Fermi-Dirac distribution fFD(E) and is given by 1 2 fFD(E) 1 eeE>kT 1 8-25 where, again, havior is described by Equation 8-25 are called fermions or Fermi-Dirac particles. is a system-dependent normalization constant. Particles whose be- e Comparison of the Distribution Functions We can write the Boltzmann distribution (Equation 8-1) in the form fB(E) 1 eeE>kT 8-26 e. where the normalization constant A in Equation 8-1 is replaced by After doing so, one is immediately struck by the very close resemblance between the three distributions (Equations 8-24, 8-25, and 8-26), the Fermi-Dirac and Bose-Einstein probability functions differing from that of Boltzmann only by the 1 in the denominator. The question immediately arises as to the significance of this seemingly small difference. In particular, since integrals of the form", " require the use of numerical methods for their solutions, it would be helpful to know if and under what conditions the Boltzmann distribution can be used for indistinguishable quantum-mechanical particles. 0 F(E)fBE(E) dE 0 F(E)fFD(E) dE and Enrico Fermi on a picnic in Michigan in July 1935. The bandage covers a cut on his forehead received when he accidentally hit himself with his racket while playing tennis. 330 Chapter 8 Statistical Physics Let us first examine the physical meaning of the difference between the distributions. Consider a system of two identical particles, 1 and 2, one of which is in state n and the other in state m. As we discussed in Section 7-6, there are two possible singleparticle-product solutions to the Schr\u00f6dinger equation. They are \u2019 \u2019 nm(1, 2) \u2019 nm(2, 1) \u2019 n(1)\u2019 n(2)\u2019 m(2) m(1) 8-27a 8-27b where the numbers 1 and 2 represent the space coordinates of the two particles. If the two identical particles are distinguishable from each other, i.e., if they are classical particles, then we can tell the difference between the two states represented by Equations 8-27a and 8-27b. However, for indistinguishable particles we have seen that the solutions must be the symmetric or antisymmetric combinations given in Section 7-6: \u2019 S \u2019 A 1 22 1 22 [\u2019 n(1)\u2019 m(2) \u2019 n(2)\u2019 m(1)] [\u2019 n(1)\u2019 m(2) \u2019 n(2)\u2019 m(1)] 8-28a 8-28b 1> 22 \u2019 A Writing is the normalization constant. As we have discussed earlier (see describes particles that obey the exclusion describes indistinguishable parti- The factor Section 7-6), the antisymmetric function principle, i.e., fermions. The symmetric function cles that do not obey the exclusion principle, i.e., bosons. \u2019 S to keep us reminded of the probability distributions followed by the fermions and bosons, respectively, let us now consider the probability that, if we look for the two particles, we will find them both in the same state, say state n. For two distinguishable particles Equations 8-27a and 8-27b both become n(1)\u2019 nn(2, 1) \u2019 n(2) \u2019 n(1) \u2019 n(2)\u2019 8-29 \u2019 \u2019 and \u2019 A \u2019 S FD BE B where we have written to remind us that distinguishable particles follow the Boltzmann distribution. Thus, the probability density of finding both distinguishable particles in state n is B nn(1, 2) \u2019 nn(1, 2) \u2019 \u2019 \u2019 n(2)\u2019 Turning to indistinguishable particles, the wave function for two bosons both oc- n(1)\u2019\u2026 n(1)\u2019 \u2019\u2026 n(2) 8-30 \u2019\u2026 B \u2019 B cupying state n is, from Equation 8-28a, \u2019 BE 1 22 [\u2019 n(1)\u2019 n(2) \u2019 n(2)\u2019 n(1)] 2 22 n(1)\u2019 \u2019 n(2) 8-31 and the probability density of finding both bosons in state n is then BE \u2019 \u2019\u2026 2\u2019\u2026 n(2)\u2019 Thus, the probability that both bosons would be found by an experiment to be occupying the same state is twice as large as for a pair of classical particles. This surprising discovery can be generalized to large ensembles of bosons as follows: n(2) 2\u2019\u2026 n(1)\u2019\u2026 n(1)\u2019 \u2019 B B 8-32 BE The presence of a boson in a particular quantum state enhances the probability that other identical bosons will be found in the same state. It is as if the presence of the boson attracts other identical bosons. Thus, the 1 that appears in the denominator of Equation 8-24 results physically in an increased 8-2 Quantum Statistics 331 probability that multiple bosons will occupy a given state, compared with the probability for classical particles in the same circumstances. The laser is the most common example of this phenomenon (see Chapter 9). We will consider another result of this intriguing behavior in Section 8-3. If the two indistinguishable particles are fermions, the wave function for both occupying the same state is, as we have previously discussed in Section 7-6, \u2019 FD 1 22 [\u2019 n(1)\u2019 n", "(2) \u2019 n(2)\u2019 n(1)] 0 8-33 0, And, of course, the probability density eralized to large ensembles of fermions as follows: \u2019\u2026 \u2019 FD FD also. This result, too, can be gen- The presence of a fermion in a particular quantum state prevents any other identical fermions from occupying the same state. It is as if identical fermions actually repel one another. The 1 in the denominator of Equation 8-25 is thus due to the exclusion principle. We will consider consequences of this peculiar property of fermions further in Chapter 10. Figure 8-15 compares the distributions of bosons and fermions. With the physical discussion above in mind, now let\u2019s compare the three functions. Figure 8-16 shows a comparison of the three distributions for 0 over the energy range from zero up to 5kT. Notice that for any given energy the fBE curve for bosons lies above that for fB for classical particles, reflecting the enhanced probability pointed out by Equation 8-32. Similarly, the fFD curve for fermions lies below those for both fBE and fB, a consequence of the exclusion of identical fermions from states that are already occupied. Notice that Equations 8-24 and 8-25 both approach the Boltzmann distribution fFD(E) fB(E) V 1. when Thus, fBE(E) and fFD(E) both approach the classical Boltzmann distribution when the probability that a particle occupies the state with energy E is much less than 1. The same is also clearly the case when, for a given E fBE(E) fB(E) V 1 kT, as Figure 8-16 illustrates. For this situation e W eE>kT. W, and ) E n ( 3 2 1 nBE (E) nFD (E) 1.0 \u221e ) E ( f 0.5 Bose-Einstein fBE Boltzmann fB 0 2\u0394E 4\u0394E E 6\u0394E 8\u0394E 0 kT 2kT 3kT 4kT 5kT E Fermi-Dirac fFD Figure 8-15 n(E) versus E for a system of six identical, indistinguishable particles. or integer spin (bosons). spin (fermions). Compare with Figure 8-1. is for particles with is for particles with zero nFD(E) nBE(E) 1 2 - integer Figure 8-16 Graph of the distributions fB, fBE, fFD fBE fB, always lies above versus energy for the value which in turn is always above All three distributions are approximately equal for energies larger than about 5 kT. 0. fFD. and 332 Chapter 8 Statistical Physics At the beginning of this section we noted that identical quantum particles were rendered indistinguishable from one another by the overlap of their de Broglie waves. This provides another means of determining for a given system when the Boltzmann distribution may be used that can be shown to be equivalent to the fB(E) 1 condi- tion above but that is sometimes easier to apply. If the de Broglie wavelength is 8d9 much smaller than the average separation of the particles, then we can neglect the overlap of the de Broglie waves, in which case the particles can be treated as if they were distinguishable: V where V 8d9 h p h 22mEk h 22m(3kT>2) h 23mkT 8-34 8-35 where N V is the number The average separation of the particles is of particles per unit volume in the system. Thus, the condition stated by Equation 8-34 becomes 8d9 (V>N)1>3, > h 23mkT V a 1>3 b V N which when cubed and rearranged becomes a b N V h3 (3mkT)3>2 V 1 8-36 Equation 8-36 gives the condition under which the Boltzmann distribution can be used. Note that in general the condition requires low particle densities and high temperatures for particles of a given mass. The next example illustrates the application of the condition. EXAMPLE 8-6 Statistical Distribution of He in the Atmosphere He atoms have spin 0 and hence are bosons. He makes up 5.24 106 of the molecules in the atmosphere. (a) Can the Boltzmann distribution be used to predict the thermal properties of atmospheric helium at T 273 K? (b) Can it be used for liquid helium at T 4.2 K? SOLUTION (a) NA atoms of air occupy 2.24 102 m3 at standard conditions. The number of He atoms per unit volume is then NA V 6.02 1023 5.24 106 2.24 102 m3 1.41 10", "20 molecules He>m3 The left side of Equation 8-36 is then (1.41 1020)(6.63 1034)3 (3 1.66 1027 4 1.38 1023 273)3>2 6.3 1011 V 1 The behavior of the helium in the atmosphere can therefore be described by the Boltzmann distribution. 8-2 Quantum Statistics 333 (b) The density of liquid helium at its boiling point T 4.2 K is 0.124 g cm3. > The particle density N V is then > N V NA molecules 4 g (0.124 g>cm3) (102 cm>m)3 1.87 1028 He atoms>m3 The left side of Equation 8-36 is then (1.87 1028)(6.63 1034)3 (3 1.66 1027 4 1.38 1023 4.2)3>2 4.39 V which is not 1. Therefore, the Boltzmann distribution does not adequately describe the behavior of liquid helium, so the Bose-Einstein distribution must be used. Using the Distribution: Finding n(E ) In order to find the actual number of particles n(E) with energy E, each of the three distribution functions given by Equations 8-24, 8-25, and 8-26 must be multiplied by the density of states, as indicated by Equation 8-2. nB(E) gB(E)fB(E) nBE(E) gBE(E)fBE(E) nFD(E) gFD(E)fFD(E) 8-37b 8-37a 8-37c Finding g(E) enables the constant to be determined for particular systems from the normalization condition that we have used several times, namely, the total number of particles N 0 n(E) dE. e Density of States As an example of determining g(E), consider an equilibrium system of N classical particles confined in a cubical volume of side L. Treating the cube as a three-dimensional infinite square well, in Chapter 7 we found the energy of a particle in such a well to be En1n2n3 U22 2mL2(n2 1 n2 2 n2 3) which we will for the convenience of our present discussion write as En E0(n2 x n2 y n2 z) R nz 7-4 8-38 U22>2mL2. E0 where x, y, and z replace 1, 2, and 3 and The three quantum numbers nx, ny, and nz specify the particular quantum state of the system. Recalling that g(E) is the number of states with energy between E and (E dE), our task is to find an expression for the total number of states from zero energy up to E, then differentiate that result to find the number within the shell dE. This is made quite straightforward by (1) observing that Equation 8-38 is the equation of a sphere of radius R (E>E0)1>2 in nx ny nz \u201cspace\u201d and (2) recalling that the quantum numbers must be integers, each combination of which represents a particular energy and corresponds to a point in the \u201cspace.\u201d (See Figure 8-17.) Since the quantum numbers must all be positive, the \u201cspace\u201d is confined to that octant of the sphere, as Figure 8-17 shows. ny nx Figure 8-17 A representation of the allowed quantum states for a system of particles confined in a threedimensional infinite square well. The radius R E1>2. 334 Chapter 8 Statistical Physics The number of states N within radius R (equal to the number of different combinations of the quantum numbers) in the volume is given by N a 1 8 b a 4R3 3 b 3>2 6 a b E E0 The density of states in nx ny nz \u201cspace\u201d is g(E) dN dE 4 0 E1>2 E3>2 (2m)3>2L3 42U3 E1>2 or g(E) (2m)3>2V 42U3 E1>2 2(2m)3>2V h3 E1>2 8-39 8-40 8-41 where the volume V L3. If the particles were electrons, then each state could accommodate two (one with spin up and one with spin down) and the density of states ge(E) would be twice that given by Equation 8-41, or ge(E) 4(2me)3>2V h3 E1>2 8-42 We can compute the constant from the normalization condition e in the Boltzmann distribution for these two cases N nB(E) dE", " gB(E)fB(E) dE 0 0 0 gB(E)eeE>kT dE 8-43 If the distinguishable particles are electrons, gB(E) ge(E) and we have that 4(2me)3>2V h3 E1>2eE>kT dE N e 0 which, when evaluated, yields N 2(2mekT)3>2V h3 e or e Nh3 2(2mekT)3>2V or e 2(2mekT)3>2V Nh3 8-44 For particles that do not obey the exclusion principle, the 2 multiplying the parentheses in Equation 8-44 is not present. Note that depends upon the number density of particles N V. Note too that is essentially the quantity on the left side of Equation 8-36, which was obtained from de Broglie\u2019s relation for classical particles. Thus, the test for when the Boltzmann distribution may be used given by Equation 8-36 is equivalent to the condition that e V 1. e e > 8-3 The Bose-Einstein Condensation 335 Questions 5. How can identical particles also be distinguishable classically? 6. What are the physical conditions under which the Boltzmann distribution holds for a system of particles? 7. Do the opposite spins of two electrons in the same state make them distinguishable from each other? 8. How would you characterize a boson? A fermion? 8-3 The Bose-Einstein Condensation We saw in Section 8-2 that, for ordinary gases, the Bose-Einstein distribution differs very little from the classical Boltzmann distribution, basically because there are many quantum states per particle due to the low density of gases and the large mass of the particles. However, for liquid helium, there is approximately one particle per quantum state at very low temperatures, and the classical distribution is invalid, as was illustrated in Example 8-6. The somewhat daring idea that liquid helium can be treated as an ideal gas obeying the Bose-Einstein distribution was suggested in 1938 by F. London in an attempt to understand the amazing properties of helium at low temperatures. When liquid helium is cooled, several remarkable changes take place in its properties at a temperature of 2.17 K. In 1924, H. Kamerlingh Onnes and J. Boks measured the density of liquid helium as a function of temperature and discovered a cusp in the curve at that temperature, as illustrated in Figure 8-18. In 1928, W. H. Keesom and M. Wolfke suggested that this discontinuity in the slope of the curve was an indication of a phase transition. They used the terms \u201chelium I\u201d for the liquid above 2.17 K, called the lambda point (see Figure 8-19), and \u201chelium II\u201d for the liquid below that temperature. In London\u2019s theory, called the two-fluid model, helium II is imagined to consist of two parts, a normal fluid with properties similar to helium I and a superfluid (i.e., a fluid with viscosity 0) with quite different properties. The density of liquid helium II is the sum of the densities of the normal fluid and the superfluid: 0.144 0.136 0.148 0.140 3 m c / g p, n s 8-45 As the temperature is lowered from the lambda point, the fraction consisting of the superfluid increases and that of the normal fluid decreases until, at absolute zero, only the superfluid remains. The superfluid corresponds to the helium atoms being in the lowest possible quantum state, the ground state. These atoms are not excited to higher states, so the superfluid cannot contribute to viscosity. When the viscosity of helium II is measured by the rotating disk method (a standard technique for measuring the viscosity of liquids), only the normal-fluid component exerts a viscous force on the disk. As the 0.132 0.128 0.124 0 1 \u03bb 2 3 4 5 T, K Figure 8-18 Plot of density of liquid helium versus temperature, by Kamerlingh Onnes and Boks. Note the discontinuity at 2.17 K. [From F. London, Superfluids (New York: Dover Publications, Inc., 1964). Reprinted by permission of the publisher.] 336 Chapter 8 Statistical Physics temperature is lowered, the fraction of helium in the normal component decreases from 100 percent at the lambda point to 0 percent at T 0 K; thus, the viscosity decreases rapidly with temperature in agreement with experiment. It is not at all obvious that liquid helium should behave like an ideal gas, because the atoms do exert forces on each other. However, these are weak van der Waals forces (to be discussed in Chapter 9), and the fairly low density of", " liquid helium (0.145 g cm3 near the lambda point) indicates that the atoms are relatively far apart. The ideal gas model is therefore a reasonable first approximation. It is used mainly because it is relatively simple and because it yields qualitative insight into the behavior of this interesting fluid. > EXPLORING Liquid Helium In a classic experiment conducted in 1908, H. Kamerlingh Onnes9 succeeded in liquefying helium, condensing the last element that had steadfastly remained in gaseous form and culminating a determined effort that had consumed nearly a quarter of a century of his life. Even then, he nearly missed seeing it. After several hours of cooling, the temperature of the helium sample, being measured by a constant-volume helium gas thermometer, refused to fall any further. The liquid hydrogen being used to precool the system was gone, and it appeared that the experiment had failed when one of the several interested visitors gathered in Kamerlingh Onnes\u2019s lab suggested that perhaps the temperature was steady because the thermometer was immersed in boiling liquid that was so completely transparent as to be very hard to see. At the visitor\u2019s suggestion, a light was shined from below onto the glass sample vessel and the gas-liquid interface became clearly visible! Condensation to the very low-density, transparent liquid had occurred at 4.2 K. H. Kamerlingh Onnes and J. D. Van der Waals by the helium liquefier in the Kamerlingh Onnes Laboratory in Leiden in 1911. [Courtesy of the Kamerlingh Onnes Laboratory.] Liquid helium, because of its extremely low boiling temperature, is the standard coolant for superconducting magnets throughout the world. Medical diagnostic MRI systems use such magnets. The large particle accelerators at, e.g., CERN and Fermilab use hundreds of them (see Chapter 11). The liquid helium must have been boiling vigorously. Soon afterward Kamerlingh Onnes was able to reduce the temperature further, passing below 2.17 K, at which point the vigorous boiling abruptly ceased. He must have observed the sudden cessation of the violent boiling, yet he made no mention of it then or in the reports of any of his many later experiments. Indeed, it was another quarter century before any mention of this behavior would appear in the literature,10 even though many investigators must have surely seen it. The abrupt halt in boiling at 2.17 K signaled a phase transition in which helium changed from a normal fluid to a superfluid, that is, bulk matter that flows essentially without resistance (viscosity 0). Of all the elements, only the two naturally occurring isotopes of helium exhibit this property. The transition to the superfluid phase in 4He occurs at 2.17 K. In 3He, which accounts for only 1.3 104 percent of natural helium, the transition occurs at about 2 mK. This transition should not be interpreted as due in some way to a peculiarity in the structure of helium. Liquid phases of other bosons do not become superfluids because all other such systems solidify at temperatures well above the critical temperature for Bose-Einstein condensation. Only helium remains liquid under its vapor pressure at temperatures approaching absolute zero.11 The fundamental reason that it does not solidify is that the interaction potential energy (see Section 9-3) between helium molecules is quite weak. Since helium atoms have small mass, their zero-point motion (i.e., their motion in the lowestallowed energy level\u2014see Section 5-6) is large, in fact, so large that its kinetic energy exceeds the interaction potential energy, thus melting the solid at low pressure. It is the superfluid phase of 4He that we will be referring to throughout the remainder of this section. It turns out that 3He becomes a superfluid for a different reason. (Hint: 4He has spin 0, hence is a boson; 3He has spin and is thus a fermion.) 1 2 8-3 The Bose-Einstein Condensation 337, Figure 8-19 Specific heat of liquid helium versus temperature. Because of the resemblance of this curve to the Greek letter transition point is called the lambda point. [From F. London, Superfluids (New York: Dover Publications, Inc., 1964). Reprinted by permission of the publisher.] the 3.0 2.5 2.0 1.5 1.0 0.5 K \u00b7 g / l a c C, 0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 T, K Experimental Characteristics of Superfluid 4He, In 1932 W. Keesom and K. Clusius measured the specific heat as a function of temperature and made a dramatic discovery of an enormous discontinuity, obtaining the curve shown in the transition temFigure 8-19. Because of the similarity of this curve to", " the Greek letter perature 2.17 K is called the lambda point. Figure 8-20 shows this same curve measured with much greater resolution. Just above the lambda point, He boils vigorously as it evaporates. The bubbling immediately ceases at the lambda point, although evaporation continues. This effect is due to the sudden large increase in the thermal conductivity at the lambda point. In normal liquid helium, like other liquids, the development of local hot spots causes local vaporization, resulting in the formation of bubbles. Below the lambda point the thermal conductivity becomes so large, dissipating heat so rapidly, that local hot spots cannot form. Measurements of thermal conductivity show that helium II conducts heat better than helium I by a factor of more than a million; in fact, helium II is a better heat conductor than any metal, exceeding that of copper at room temperature by a factor of 2000. This conduction process is different from ordinary heat conduction, for the rate of conduction is not proportional to the temperature difference. Bubble formation ceases (even though evaporation continues) because all parts of the fluid are at exactly the same temperature. K \u00b7 g / J C, 24 22 20 18 16 14 12 10 8 6 4 2 0 \u20131.0 0 1.0 \u20134 T\u2013T\u03bb, K Figure 8-20 The lambda point with high resolution. The specific heat curve maintains its shape as the scale is expanded. [From M. J. Buckingham and W. M. Fairbank, \u201cThe Nature of the -Transition,\u201d Progress in Low Temperature Physics, edited by C. J. Gorter, Vol. III, (Amsterdam: North-Holland Publishing Company, 1961).] 30 6 \u201320 \u201310 10 20 0 T\u2013T\u03bb, \u03bcK \u20132 2 0 4 T\u2013T\u03bb, mK 338 Chapter 8 Statistical Physics This lambda-point transition is clearly visible on the surface of the liquid shown in Figures 8-21a and b, which also illustrates the phenomenon largely responsible for applying the name superfluid to helium II. The small container of liquid helium suspended above the surface has a bottom made of tightly packed, ultrafine powder (fine emery powder or jeweler\u2019s rouge). The microscopic channels through the powder are too small for the ordinary liquid to pass through, but when the temperature drops below the lambda point, the superfluid flows through essentially unimpeded, the viscosity suddenly dropping at that point by a factor of about one million.12 Figures 8-22a and b illustrate the creeping film effect. A container containing liquid helium has a thin film (several atomic layers thick) of helium vapor coating the walls, just as is the case with any other enclosed liquid. However, if the level of liquid helium in the container is raised above the general level in the reservoir, such as the cup in the photo of Figure 8-22a, the superfluid film on the walls creeps up the inner walls, over the top, and down the outside and returns to the reservoir until both surfaces are level or the cup is empty! In the thermomechanical effect, which involves two containers of liquid helium II connected by a superleak, if heat is added to one side, e.g., by a small heater as illustrated in Figure 8-23a, the superfluid on the other side migrates toward the heated side, where the level of liquid (still superfluid) rises. If the system is suitably arranged, as in Figure 8-23b, the rising liquid can jet out a fine capillary in the so-called fountain effect.13 1 2 Superfluid 3He Physicists thought for a long time that 3He could not form a superfluid since its nucleus consists of two protons and a neutron. It thus has -integer spin and obeys Fermi-Dirac statistics, which prohibits such particles from sharing the same energy state. However, early in the 1970s D. M. Lee, D. D. Osheroff, and R. C. Richardson showed that when cooled to 2.7 mK, the spins of pairs of 3He atoms can align parallel, creating, in effect, a boson of spin 1 and allowing the liquid to condense to a superfluid state. Two additional superfluid states were subsequently discovered, a spin-0 state (antiparallel spins) at 1.8 mK and a second spin-1 state that is created when an external magnetic field aligns the spins of the 3He pairs. The three scientists received the 1996 Nobel Prize for their discovery. (a) (b) Figure 8-22 (a) The creeping film. The liquid helium in the dish is at a temperature of about 1.6 K. A thin film creeps up the sides of the dish, over the edge, and down the outside to form the drop shown, which then falls into the reservoir below. [Courtesy of A. Leitner", ", Rensselaer Polytechnic Institute.] (b) Diagram of creeping film. If the dish is lowered until partially submerged in the reservoir, the superfluid creeps out until the levels in the dish and reservoir are the same. If the level in the cup is initially lower than that of the reservoir, superfluid creeps into the dish. (a) (b) Figure 8-21 (a) Liquid helium being cooled by evaporation just above the lambda point boils vigorously. (b) Below the lambda point the boiling ceases and the superfluid runs out through the fine pores in the bottom of the vessel suspended above the helium bath. [Courtesy of Clarendon Laboratory. From K. Mendelssohn, The Quest for Absolute Zero: The Meaning of Low Temperature Physics, World University Library (New York: McGraw-Hill Book Company, 1966).] 8-3 The Bose-Einstein Condensation 339 (b) (c) Fountain (a) Heater Superleak Superfluid reservoir Liquid helium II Superleak Heater Figure 8-23 (a) Diagram of the thermomechanical effect. The level of the fluid rises in the container where the heat is being added. (b) A bulb containing liquid helium is in a cold bath of liquid helium II at 1.6 K. When light containing infrared radiation is focused on the bulb, liquid helium rises above the ambient level. The height of the level depends on the narrowness of the tube. If the tube is packed with powder and the top drawn out into a fine capillary, the superfluid spurts out in a jet as shown, hence the name \u201cfountain effect.\u201d (c) Diagram showing the components in the photograph in (b). [Photo courtesy of Helix Technology Corporation.] In the Bose-Einstein distribution the number of particles in the energy range dE is given by n(E)dE, where we have from Equation 8-37b n(E) g(E) eeE>kT 1 8-46, where g(E) is given by Equation 8-41. The constant which is determined by normalization, cannot be negative, for if it were, n(E) would be negative for low values of E. This situation would make no sense physically since, if were negative for small energies (i.e., is the number of particles in the state with energy E, and a negative value would be meaningless. The normalization condition is n(E) dE 2V would be negative. But \u0192\u0192 E>kT 2V N fBE(E) fBE(E) ), then 8-47 h3 (2mkT)3>2 h3 (2m)3>2 E1>2 dE eeE>kT 1 x1>2 dx ex 1 0 0 0 where x E kT and the integral in this equation is a function of >. The usual justification for using a continuous energy distribution to describe a quantum system with discrete energies is that the energy levels are numerous and closely spaced. In this case, as we have already seen, for a gas of N particles in a macroscopic box of volume V (the container), this condition holds, as you can demonstrate for yourself by computing the spacing using Equation 7-4 for a three-dimensional box. 340 Chapter 8 Statistical Physics g(E) E1>2 However, in replacing the discrete distribution of energy states by a continuous distribution, we ignore the ground state. This is apparent from Equation 8-41, where we see therefore, if E 0, g(E) 0 also. This has little effect for a gas conthat sisting of Fermi-Dirac particles since there can be only two particles in any single state, and ignoring two particles out of 1022 causes no difficulty. In a Bose-Einstein gas, however, there can be any number of particles in a single state. If we ignore the ground state, as we have up to now, the normalization condition expressed by Equation 8-47 cannot corresponding to the minimum be satisfied below some minimum critical temperature, 0. possible value of This implies that at very low temperatures there are a significant number of particles in the ground state. Tc The critical temperature Tc can be found by evaluating Equation 8-47 numerically. has its minimum value of 0. The integral has a maximum value of 2.315 when This results in a maximum value for N V given by > N V 2 h3 (2mkT)3>2(2.315) > Since N V is determined by the density of liquid helium, this implies a value for the critical temperature, given by T! h2 2mk c N 2(2.315)V d 2>3 Tc 8-48 Inserting the known constants and the density of helium, we find for the critical", " temperature 3.1 K Tc 8-49 For temperatures below 3.1 K the normalization Equation 8-47 cannot be satisfied for Evidently at these temperatures there are a significant number of any value of particles in the ground state, which we have not included.. We can specifically include the ground state by replacing Equation 8-47 with N N0 2V h3 (2mkT)3>2 0 x1>2 dx ex 1 8-50 where N0 is the number in the ground state. If we choose E0 ground state, this number is 0 for the energy of the N0 g0 >kT 1 eeE0 1 e 1 8-51 where g0, the density of states or statistical weight, is 1 for a single state. We see that, N0 becomes large as becomes small. With the inclusion of N0, which depends on can be computed the normalization condition (Equation 8-50) can be met and numerically for any temperature and density. For temperatures below the critical tem- perature we see from Equation 8-51 that yields and that the fraction of molecules in the ground state is given approximately by e for small Expanding 1 N0 is of the order of and we thus conclude that e 1 \u00c1, e 1 1>N0. Tc N0 N 1 a 3>2 b T Tc 8-52 8-3 The Bose-Einstein Condensation 341 /2 Tc/2 Tc T Figure 8-24 Graph of the fraction of superfluid in a sample of liquid helium as a function of temperature. T Tc In the London two-fluid model the N0 atoms that we added in Equation 8-50 have condensed to the ground state. These particles in the ground state constitute the superfluid. The remaining (N N0) atoms are the normal fluid. That fraction of the fluid is shown in Figure 8-24. that is superfluid for The value is not very different from the observed lambda-point temperature T 2.17 K, especially considering that our calculation is based on the assumption that the liquid helium is an ideal gas. The process of atoms dropping into the is called Bose-Einstein condenground state as the temperature is lowered below sation. Such an occurrence was predicted by Einstein in 1924, before there was any evidence that such a process could occur in nature. 3.1 K Tc Tc The Bose-Einstein Condensate Like all atoms, the constituents of 4He (protons, neutrons, and electrons) are fermions; however, they are assembled in such a way that the total spin of the ground state is integer (zero), so that the 4He atom is a boson. Indeed, a review of the periodic table shows that, although atoms can be either fermions or bosons, the ground-state spins are mostly integer, so in their lowest energy state most atoms are bosons. This fact is of no great consequence in determining the properties of a gas in a macroscopic container because the spacing between the quantized energies is extremely small, so the probability that any particular level is occupied by an atom is also small. For example, the spacing between adjacent levels in a cubical box with a volume of 1 cm3 containing sodium gas is about 1020 eV (see Equation 8-38), so even at relatively low temperatures the atoms in a sample of a few billion would be widely spread among the allowed levels, as in Figure 8-25a. In addition, the average distance between atoms in the box or tens would be about of thousands of atomic diameters, so the interactions between the atoms are minuscule. (106 m3>109 atoms)1>3 105 m, (b) Cold atoms (a) Hot atoms If our goal is to form a Bose-Einstein condensate (BEC) from the widely separated atomic bosons of the gas sample in the box, the obvious approach is that used to condense any gas; that is, the sample is cooled and the density is increased until the gas liquefies. However, this approach presents us with a formidable problem: as the gas liquefies, the atoms get very close together, the density approximating that of the solid. The atoms now interact strongly, mainly via their outer electrons, and thus all begin to act like fermions! (This is essentially what happens in liquid helium II, where even at very low temperatures the fraction of the atoms in the ground state [superfluid phase] is only about 10 percent or so.) This problem was solved by C. E. Wieman and E. Cornell in 1995, more than 70 years after Einstein\u2019s prediction. They did it by forming the BEC directly from a supersaturated vapor, cooling the sample but never allowing it to reach ordinary thermal equilibrium.14 This was done with standard cooling methods and a very neat \u201ctr", "ick.\u201d First, a sample of rubidium vapor at room temperature was illuminated by the beams from six small diode \u03bb DB= h/mv \u03bb DB> d Bosons BEC Figure 8-25 (a) The atoms in a sample of dilute gas in any macroscopic container are distributed over a very large number of levels, making the probability of any one level being occupied quite small. (b) Cooled to the point where the de Broglie wavelength becomes larger than the interatomic spacing, atoms fall into the ground state, all occupying the same region of space. 342 Chapter 8 Statistical Physics m 2 had been polarized in the lasers of appropriate frequency. Collisions of the laser photons with atoms in the lowspeed tail of the Maxwell distribution (see Figure 8-3) slowed those atoms, and within a second or two a sample of about 107 atoms collected in the volume defined by the intersecting laser beams, about 1.5 cm in diameter. The temperature of this laser-cooled sample was about 1 mK. Then a special magnetic trap (i.e., a magnetic field shaped so as to confine the atoms) was used to \u201csqueeze\u201d the cooled sample, whose atomic spins ( 2U) direction. (Polarizing the spins was the \u201ctrick\u201d referred to above. Equilibrium is reached in the spin-polarized vapor very rapidly, long before the true thermal equilibrium state\u2014the solid\u2014can form, thus maintaining the sample as a supersaturated vapor.) The warmer atoms on the high-speed tail of the Maxwell distribution of the trapped atoms are allowed to escape through a \u201cleak\u201d in the magnetic trap, taking with them a substantial amount of the kinetic energy and evaporatively cooling the remaining few thousand atoms to less than 100 nK, just as water molecules evaporating from the surface of a cup of hot coffee cool that which remains in the cup. These remaining cold atoms fall into the ground state of the confining potential and have, within the experimental uncertainties, reached absolute zero. They are the condensate. The BEC is illustrated in Figure 8-25b. The condensate, if left undisturbed in the dark, lives for 15 to 20 seconds, its destruction eventually resulting from collisions with impurity atoms in the vacuum that are also colliding with the hot (room temperature) walls of the experimental cell. The peak in Figure 8-26 is a macroscopic quantum wave function of the condensate. Figure 8-26 Two-dimensional velocity distributions of the trapped cloud for three experimental runs with different amounts of cooling. The axes are the x and z velocities, and the third axis is the number density of atoms per unit velocity-space volume. This density is extracted from the measured optical thickness of the shadow. The distribution on the left shows a gentle hill and corresponds to a temperature of about 200 nK. The middle picture is about 100 nK and shows the central condensate spire on the top of the noncondensed background hill. In the picture on the right, only condensed atoms are visible, indicating that the sample is at absolute zero, to within experimental uncertainty. The gray bands around the peaks are an artifact left over from the conversion of false-color contour lines into the present black and white. [From C. E. Wieman, American Journal of Physics 64 (7), 853 (1996).] 8-3 The Bose-Einstein Condensation 343 Since the discovery by Wieman and Cornell, several other physicists have produced Bose-Einstein condensates. One of the largest produced (by W. Ketterle and co-workers) contained 9 107 sodium atoms, was about a millimeter long, and lived for half a minute. Its direct photograph is shown in Figure 8-27. As of this writing, the largest condensates are made of hydrogen and contain about 109 atoms. Does this discovery have any potential use? The answer is, probably many that we can\u2019t even imagine yet, but here is one possibility. The BEC can form the basis of an atomic laser. This was demonstrated in late 1996, also by Ketterle and his colleagues, and is illustrated in Figure 8-28. The condensate is coherent matter, just as the laser beam is coherent light. It could place atoms on substrates with extraordinary precision, conceivably replacing microlithography in the production of microcircuitry. Here is another potential use for the BEC: It could form the basis for atomic interferometers, making possible measurements far more precise than those made with visible lasers since the de Broglie wavelengths are much shorter than those of light. Ketterle, Cornell, and Wieman shared the 2001 Nobel Prize in Physics for their work. (a) (b) Optical laser Photons Electron Electron Atom laser Atoms Figure 8-27 Successive images show shadow of a millimeter-", "long cloud of atoms containing BoseEinstein condensate as it expands from its initial cigar shape (top). [From D. S. Durfee, Science 272, 1587 (1996).] 0 0.5 Absorption 1 Figure 8-28 (a) When the two identical condensates of sodium atoms, each containing about 5 106 atoms, were allowed to expand freely and overlapped, phase contrast imagery revealed interference fringes, the \u201csignature\u201d of coherent waves\u2014the first atomic laser. (b) Optical lasers amplify light by stimulating atoms to emit photons. Atom lasers amplify by stimulating more atoms to join the \u201cbeam.\u201d [(a) From D. S. Durfee, Science 275, 639 (1997). (b) From Science 279, 986 (1998). Courtesy of L. Carroll.] Questions 9. Explain how the escaping \u201chot\u201d rubidium atoms cool those remaining in the sample. 10. What is Bose-Einstein condensation? 11. Would you expect a gas or liquid of 3He atoms to be much different from one of 4He atoms? Why or why not? 344 Chapter 8 Statistical Physics 8-4 The Photon Gas: An Application of Bose-Einstein Statistics Photon Gas Planck\u2019s empirical expression for the energy spectrum of the blackbody radiation in a cavity (Equation 3-18) can now be derived by treating the photons in the cavity as a gas consisting of bosons. The distribution is then given by fBE(E) 1 eeE>kT 1 8-24 As we saw in Section 8-2 and in particular in the discussion of Equation 8-44, the value of is determined by the total number of particles that the system contains. However, in the case of photons contained in a cavity, which we are discussing, that seems to present a problem since the total number of photons is not constant. Photons are continually being created (emitted by the oscillators in the cavity walls) and de: stroyed (absorbed by the oscillators). Even so, this does indeed specify the value of it tells us that Equation 8-24 for photons cannot be a function of i.e., e; fph(E) 1 eE>kT 1 8-53 The fact that the total number of photons is not constant makes it necessary that We will see in a moment that this must be true. so that e 1. 0 The number of photons with energy E is found by substituting Equation 8-53 into Equation 8-37b, which yields or nph(E) gph(E)fph(E) nph(E) gph(E) eE>kT 1 8-54 gph(E) is derived in the same manner as it was for massive parThe density of states ticles in Section 8-2. The result, which we first encountered as in our discussion of Planck\u2019s derivation of the blackbody spectrum, is given in terms of the photon frequency n() 84 as f gph(E) dE 8Vf2 df c3 8VE2 dE c3h3 8-55 where V is the volume of the cavity. The energy density u(E)dE in the energy interval between E and E dE is then given by u(E) dE Egph(E)fph(E) dE V 8E3 dE c3h3(eE>kT 1) 8-56 or, in terms of the photon frequency f, using E hf for the conversion, we have u(f) df 8f2 c3 hf df ehf>kT 1 8-57 8-4 The Photon Gas: An Application of Bose-Einstein Statistics 345 f c f. to frequency as the variable using Equation 8-57 is identical to Equation 3-18 when the latter is converted from wavelength We saw in Chapter 3 that Equation 3-18 is in precise agreement with experimental observations. This agreement serves as justification for the Bose-Einstein distribution function for photons given by Equation 8-53, which resulted from our argument that for photons. Notice that Planck\u2019s derivation presented in Chapter 3, in which the radiation in the blackbody cavity was treated as a set of distinguishable standing electromagnetic waves to which he (correctly) applied the Boltzmann distribution, agrees exactly with the derivation presented here, in which the radiation is treated as indistinguishable particles to which the Bose-Einstein distribution must be applied. This is an example of the wave-particle duality of photons. 0 EXAMPLE 8-7 Photon Density of the Universe The high temperature of the early universe implied a thermal (i.e., blackbody) electromagnetic radiation field which has, over aeons, cooled to the present 2", ".7 K. This cosmic background radiation was discovered in 1965. (See Chapter 13.) Compute the number of these photons per unit volume in the universe. SOLUTION 1. The number of photons with energy E is given by Equation 8-54: nph(E) gph(E) eE>kT 1 2. The total number per unit > volume N V is then given by N V 1 V nph(E) dE 1 V 0 0 gph(E) dE eE>kT 1 3. Substituting the density of gph(E) states 8-55 yields from Equation N V 8E2 dE (ch)3(eE>kT 1) 0 8(kT)3 (ch)3 0 (E>kT)2(dE>kT) eE>kT 1 4. Letting x E kT, this can > be written N V 8a kT ch 3 b 0 x2 dx ex 1 5. Evaluating the integral from standard tables: 0 x2 dx ex 1 2.40 6. Substituting values into the expression for N V in step 4 yields > N V 8a 1.38 1023 J>K 2.7 K 3.00 108 m>s 6.63 1034 J # s 3 b (2.40) 3.97 108 photons>m3 346 Chapter 8 Statistical Physics Quantization of the Energy States of Matter CV We pointed out earlier that the molar heat capacity for solids falls appreciably below the classical Dulong-Petit value of 3R when the temperature falls below some critical value. In 1908 Einstein showed that the failure of the equipartition theorem in predicting the specific heats of solids at low temperatures could be understood if it were assumed that the atoms of the solid could have only certain discrete energy values. Einstein\u2019s calculation is closely related to Planck\u2019s calculation of the average energy of a harmonic oscillator, assuming the oscillator can take on only a discrete set of energies. The calculation itself presents no real problem, as we have seen in Chapter 3. Einstein\u2019s most important contribution in this area was the extension of the idea of quantization to any oscillating system, including matter. We will see in this subsection how the idea of quantized energy states for matter also explains the puzzling behavior of the heat capacities of diatomic gases that was pointed out in Section 8-1. In particular, we will be able to understand why the H2 molecule seems to have only 3 degrees of freedom (corresponding to translation) at low temperatures, 5 degrees of freedom at intermediate temperatures (corresponding to translation and rotation), and 7 degrees of freedom at high temperature (corresponding to translation, rotation, and vibration). Consider 1 mole of a solid consisting of NA molecules, each free to vibrate in three dimensions about a fixed center. For simplicity, Einstein assumed that all the in each direction. The problem is then molecules oscillate at the same frequency f. equivalent to 3NA distinguishable one-dimensional oscillators, each with frequency The classical distribution function for the energy of a set of one-dimensional oscillators is the Boltzmann distribution, given by Equation 8-1. Following Planck, Einstein assumed that the energy of each oscillator could take on only the values given by f nhf En 8-58 n 0,1,2, \u00c1, rather than have an average value of kT as predicted by the where equipartition theorem. He then used the Boltzmann distribution15 to compute the averfor the distinguishable oscillators, just as we have done previously, from age energy 8E9 obtaining 8E9 0 EnB(E) dE 8E9 hf ehf>kT 1 8-59 8-60 which is, of course, the same as Equation 3-17. We can expand the exponential, using ex 1 x (x2>2!) \u00c1 (see Appendix B2). At high temperatures the quantity and then, keeping only the first two terms of the expansion, x V 1 for hf>kT V 1 x hf>kT, where ehf>kT 1 a1 hf kT \u00c1 b 1 hf kT 8E9 and statistics (see Equation 8-14). approaches kT, in agreement with the equipartition theorem from classical 8-4 The Photon Gas: An Application of Bose-Einstein Statistics 347 The total energy for 3NA oscillators is now U 3NA 8E9 3NAhf ehf>kT 1 and the heat capacity is CV dU dT 3NAka 2 b hf kT ehf>kT (ehf>kT 1)2 8-61 8-62 S 3NAk 3R as T S. It is left as an exercise (see Problem 8-", "29) to show directly from Equation 8-62 that Cv S 0 as T S 0 and Cv By comparing the Einstein calculation of the average energy per molecule, Equation 8-60, with the classical one, we can gain some insight into the problem of when the classical theory will work and when it will fail. Let us define the critical temperature, called the Einstein temperature. The energy distribution in terms of this temperature is TE hf k 8-63 fB(En) AeEn >kT Aenhf>kT AenTE >T fB(En) fB(En1). For temperatures T much higher than TE, small changes in n have little effect on Then E can be treated as the exponential in the distribution; that is a continuous variable. However, for temperatures much lower than TE, even the smallest possible change in n, n 1, results in a significant change in and we would expect that the discontinuity of possible energy values becomes significant. Since hard solids have stronger binding forces than soft ones, their frequencies of molecular oscillation and therefore their Einstein temperatures are higher. For lead and gold, TE is of the order of 50 to 100 K; ordinary temperatures of around 300 K are \u201chigh\u201d for these metals, and they obey the classical Dulong-Petit law at these temperatures. For diamond, TE is well over 1000 K; in this case 300 K is a \u201clow\u201d temperature, and Cv is much less than the Dulong-Petit value of 3R at this temperature. enTE The agreement between Equation 8-62 and experimental measurements justifies Einstein\u2019s approach to understanding the molar heat capacity of solids. Figure 8-29 shows a comparison of this equation with experiments. The curve fits the experimental points well except at very low temperatures, where the data fall slightly above the curve. >T.1 0.2 0.3 0.4 0.5 0.6 T/TE 0.7 0.8 0.9 1.0 Figure 8-29 Molar heat capacity of diamond versus T>TE. reduced temperature The solid curve is that predicted by Einstein. [From Einstein\u2019s original paper, Annalen der Physik 22 (4), 180 (1907).] 348 Chapter 8 Statistical Physics R 3 / v C 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 Aluminum TD= 396 K Copper TD= 309 K Silver TD= 215 K Lead TD= 95 K 0.2 0.4 0.6 0.8 T/TD 1.0 1.2 1.4 1.6 Figure 8-30 Molar heat capacity of several solids versus reduced temperature T TD, where TD >k. The solid curve is that predicted by Debye. is the Debye temperature, defined as > The data are taken from Debye\u2019s original paper. Cv 3R 1 is the Dulong-Petit value. [From Annalen der Physik 39 (4), 789 (1912), as adapted by David MacDonald, Introductory Statistical Mechanics for Physicists (New York: John Wiley & Sons, Inc., 1963); by permission.] hfD TD > The lack of detailed agreement of the curve with the data at low T is due to the oversimplification of the model. A refinement of this model was made by P. Debye, who gave up the assumption that all molecules vibrate at the same frequency. He allowed for the possibility that the motion of one molecule could be affected by that of the others and treated the solid as a system of coupled oscillators. The effect was to allow a range of vibrational frequencies from 0 up to a maximum called the Debye frequency, used to define the Debye temperature This contrasts with the infinite range of oscillation modes in the blackbody cavity. Debye\u2019s argument was that the number of vibrational modes or frequencies cannot exceed the number of degrees of freedom of the atoms that make up the solid. Calculations with the Debye model are somewhat more involved and will not be considered here. The improvement of the Debye model over the Einstein model is shown by Figure 8-30. Note that all solids fall on the same curve. hfD fD, >k. TD f Understanding Specific Heats of Gases Let us now see if we can understand the specific heats of diatomic gases on the basis of discrete, or quantized, energies. In Section 8-1 we wrote the energy of a diatomic f molecule as the sum of translational, rotational, and vibrational energies. If is the nhf, as we asfrequency of vibration and the vibrational energy is quantized by sumed for solids, we know from the previous calculation (see Equation 8-62) that for low temperatures, the average energy of vibration approaches zero and vibration will", " not contribute to We can define a critical temperature for vibration of a diatomic gas molecule by Evib CV. Tv hf k 8-64 f is the frequency of vibration. Apparently 15\u00b0 C for all the diatomic is of the where gases listed in Table 8-1 except for Cl2. From Figure 8-11 we can see that order of 1000 to 5000 K for H2. Tv Tv 8-4 The Photon Gas: An Application of Bose-Einstein Statistics 349 The rotational energy of a diatomic molecule is 1 2 I2 ER where I is the moment of inertia and is the angular velocity of rotation. It is not obvious how the rotational energy is quantized or even if it is; however, let us make use of a result from Section 7-2, where we learned that the angular momentum is quantized. If L is the angular momentum of a diatomic molecule, and we can write the energy as L I, ER L2 2I Equation 7-22 tells us that energy becomes L2 ( 1)U2, where 0,1,2, \u00c1. Thus, the rotational ER ( 1) h2 82I 8-65 The energy distribution function will contain the factor >kT e(1)h2>82IkT eER and we can define a critical temperature for rotation similar to that for vibration as TR ER k h2 82Ik 8-66 W kT, If this procedure is correct, we expect that for temperatures the equipartition theorem will hold for rotation and the average energy of rotation will approach the average energy of rotation will approach 0. Let us examine for each axis of rotation, while for low temperatures, for some cases of interest: T V TR, T W TR, (1 2)kT i.e., ER TR 1. H2. For rotation about the x or y axis as in Figure 8-10a, taking the z axis as the line joining the atoms, the moments of inertia mass are Ix and Iy through the center of Ix Iy 1 2 MR2 The separation of the atoms is about M 940 106 eV>c2. about R 0.08 nm. kTR: We first calculate The mass of the H atom is kTR h2 82I (hc)2 42Mc2R2 (1.24 103 eV # nm)2 42(940 106 eV)(0.08 nm)2 6.4 103 eV Using k 2.6 102 eV>300 K, we obtain 6.4 103 2.6 102 300 K 74 K TR As can be seen from Figure 8-11, this is indeed the temperature region below which the rotational energy does not contribute to the heat capacity. 2. O2. Since the mass of the oxygen atom is 16 times that of the hydrogen atom and the separation is roughly the same, the critical temperature for rotation will be TR For all temperatures at which O2 exists as a gas, T (74>16) 4.6 K. TR.W 350 Chapter 8 Statistical Physics 3. A monatomic gas, or rotation of diatomic gas about the z axis. We will take the H atom for calculation. The moment of inertia of the atom is mainly due to the electron since the radius of the nucleus is extremely small (about 1015 m). The distance from the nucleus to the electron is about the same as the separation of atoms in the H2 molecule. Since the mass of the electron is about 2000 times smaller than that of the atom, we have IH 1 2000 IH2 and 2000 74 K 1.5 105 K TR This is much higher than the dissociation temperature for any diatomic gas. Thus, the line joining the atoms for all attainable temperatures. for monatomic gases and for rotation of diatomic gases about 9 0 8ER We see that energy quantization explains, at least qualitatively, the temperature dependence of the specific heats of gases and solids. EXAMPLE 8-8 Average Vibrational Energy What is the average energy of vibration of the molecules in a solid if the temperature is (a) T h\u0192 2k, (b) T 4h\u0192 k? > > SOLUTION (a) This is lower than the critical temperature for vibration h\u0192 k given by Equation 8-64, so we expect a result considerably lower than the high temperature limit of kT given by the equipartition theorem. From Equation 8-60 we have > 8E9 hf ehf>kT 1 2kT e2 1 0.31 kT (b) This temperature is four times the critical temperature, so we expect a result near the high temperature limit of kT. Using h\u0192 kT 1 4 in Equation 8-60, we have > > 8E9 0.25kT e0.25 1 0.880 kT EXAMPLE 8", "-9 Number of Oscillators At the \u201clow\u201d and \u201chigh\u201d temperatures of h\u0192 to the Example 8-8, find the ratio of the number of oscillators with energy E1 number with E0 0. SOLUTION At any temperature T, the Boltzmann distribution for the fraction of oscillators with energy En this gives \u01920 (a) For n 1 and fB(En) AeEn nh\u0192 is >f0 Ae0 A. The ratio >f0 fn fn is then kT 1 >f0 f1 2 hf, we have oscillators are in the lowest energy state E0 (b) For the higher temperature kT 4h\u0192, we get ehf>kT e0.25 0.779. At the higher temperature the states are more nearly equally populated and the average energy is larger. >kT Aenhf>kT. enhf>kT. ehf>kT e2 0.135. 0. >f0 f1 For n 0, Most of the 8-5 Properties of a Fermion Gas 351 EXAMPLE 8-10 Debye Frequency Note from Figure 8-30 that the Debye temperature of silver is 215 K. Compute the Debye frequency for silver and predict the Debye temperature for gold. Silver and gold have identical crystal structures and similar physical properties. SOLUTION 1. From the definition of the Debye temperature silver can be computed: TD, the Debye frequency fD for TD hfD k or fD kTD h 1.38 1023 J>K 215 K 6.63 1034 J # s 4.48 1012 Hz 2. We would expect the interatomic forces of silver and gold to be roughly the same, hence their vibrational frequencies to be in inverse ratio to the square root of their atomic masses: fD(Ag) fD(Au) M(Au) M(Ag) A kTD(Ag)>h kTD(Au)>h TD(Ag) TD(Au) 3. Solving this for TD(Au) yields TD(Au) TD(Ag)A M(Au) M(Ag) 215A 108 197 159 K Remarks: This estimate is in reasonable agreement with the measured value of 164 K. 8-5 Properties of a Fermion Gas The fact that metals conduct electricity so well led to the conclusion that they must contain electrons free to move about through a lattice of more or less fixed positive metal ions. Indeed, this conclusion had led to the development of a free-electron theory to explain the properties of metals within three years after the electron\u2019s discovery by Thomson and long before wave mechanics was even a glimmer in Schr\u00f6dinger\u2019s eye. The free-electron theory of metals was quite successful in explaining a number of metallic properties, as we will discuss further in Chapter 10; however, it also suffered a few dramatic failures. For example, in a conductor at temperature T the lattice ions have average energy 3kT consisting, as we have seen, of 3kT 2 of kinetic energy and 3kT 2 of potential energy, leading to a molar heat capacity 3R (rule of Dulong-Petit). Interactions (i.e., collisions) between the free electrons and lattice ions would be expected to provide the electrons with an average translational kinetic energy of 3kT 2 at thermal equilibrium, resulting in a total internal energy U for metals of 3kT 3kT 2 9kT 2. Thus, metals should have 4.5R. In fact, they do not. The heat capacity of conductors is essentially the same as that of other solids, except for a slight temperature-dependent increase that is much smaller than 3R 2. The problems with the classical free-electron theory are due mainly to the fact that electrons are indistinguishable particles that obey the exclusion principle, and as a > CV CV > > > > > 352 Chapter 8 Statistical Physics consequence they have the Fermi-Dirac distribution of energies rather than the Boltzmann distribution. In this section we will investigate the general characteristics of systems made up of fermions. In Chapter 10 we will see how the absence of a significant electron contribution to the heat capacity of conductors is explained. In the Fermi-Dirac distribution given by it is convenient to write as fFD(E) 1 eeE>kT 1 EF kT 8-25 8-67 where EF is called the Fermi energy. Doing so allows Equation 8-25 to be written as fFD(E) 1 e(EEF)>kT 1 8-68 The Fermi energy is an important quantity in systems of fermions, such as the electron gas in metals (discussed in Chapter 10) and the neutron gas in a neutron star.", " Notice in particular that for E EF the quantity for all values of the temperature greater than zero and hence If we consider a system of fermions at T 0 K, we find that For E<EF: fFD(EF) 1 2. e(EEF)>kT 1 and For E>EF: fFD(E) 1 e(EEF)>kT 1 1 fFD(E) 1 e 1 0 In other words, at absolute zero all energy states from the ground state up to the Fermi energy are occupied and all energy states above the Fermi energy are empty. This is in sharp contrast with a system of bosons, such as the rubidium BE condensate, where all particles condense to the ground state at T 0 K. This situation is illustrated in Figure 8-31a. If the system contains N fermions, we can find its Fermi energy by filling the energy states in increasing order starting with the ground state. (a) (b) (c) ) ( E D F f 1.0 0 T = 0 K E EF ) ( E D F f 1.0 0.5 0 T > 0 K EF E 2kT ) ( E D F f 1.0 0.5 0 T >> 0 K E EF 2kT Figure 8-31 Fermi-Dirac distribution function T 0 K, all levels above the Fermi energy can move to levels within about kT above kT EF, (c) For high temperatures where even particles in the lower energy states may move to higher levels so that fFD(0) 1. are unoccupied. (b) For T 0 K with for three different temperatures. (a) At some particles near kT EF, fFD(E) EF. EF 0 K 300 K 1200 K gFD(E) \u221d E1/2 8-5 Properties of a Fermion Gas 353 Figure 8-32 The distribution of fermion energies at three different temperatures for a material whose Fermi energy is 4.8 eV. Curves are plots of Equation 8-69 for the indicated values of temperature. (See text for explanation of shaded area.) 0 1 2 3 4 6 E (ev) 5 EF ) ( E D F n The energy state occupied by the Nth particle will be the Fermi energy. We can find the total energy of the system simply by adding up the energies of all N particles and their average energy by dividing that total by N. Each of these calculations will be done for electrons in Section 10-3. If the temperature of the system is increased to some temperature T 0 K but with kT remaining smaller than EF, fermions within about kT of the Fermi energy could now move to previously unoccupied levels lying within about kT above the Fermi energy in response to collisions with the lattice ions. However, fermions occupying levels much lower than kT below EF would not be able to move since the additional kT of energy that they might acquire in a collision would not be enough to move them past levels occupied by other fermions in order to reach the unoccupied levels near or above EF. Figure 8-31b illustrates this situation. At temperatures so high that kT EF, fermions in even the very low-lying energy states will be able to move drop below 1, as shown in Figure 8-31c. This to higher states. Only then can latter situation also corresponds to the lowest curve in Figure 8-16. fFD(0) The number of fermions with energy E is given by Equation 8-37c. The density of states was computed for fermions in Section 8-3 and is given by Equation 8-42, so we have for fermions that nFD(E) nFD(E) 3>2 2 a 8m h2 b VE1>2 e(EEF)>kT 1 8-69 Figure 8-32 is a graph of Equation 8-69 for three different temperatures. The T 0 K in Figure 8-31a by the curve is the result of multiplying function, The curves for T 300 K and T 1200 K result from E1>2. which increases as gFD(E) by appropriate versions of Figure 8-31b. The shaded areas for multiplying T 0 K represent those electrons near the Fermi energy, a very small number, that are able to move into the empty states above EF at each temperature. gFD(E) fFD(E) Quantum Degenerate Fermion Gas Since fermions have half-integer spins, the Pauli exclusion principle prohibits two identical fermions from occupying the same quantum state. Thus, a system of halfinteger-spin atoms cannot all occupy the ground state to form a fermion version of the Bose-Einstein condensate as is", " possible for integer-spin bosons. The fermion analog of 354 Chapter 8 Statistical Physics the BEC occurs when the atoms fill all of the energy states from the ground state up to the Fermi energy. The transition to this quantum degenerate state for a gas of fermions is a gradual one, quite unlike the sudden phase transition to the BEC. This makes it harder to detect, in addition to which the exclusion principle makes evaporative cooling that is so important in producing the BEC much less effective as the temperature of the fermion gas decreases. In 1999 these problems were solved by Deborah Jin and Brian DeMarco, four years after the first BEC was produced. They loaded a magnetic trap with 40K (total atomic spin 9 2), dividing the atoms between two magnetic substates to solve the evaporative cooling problem. One of several ways used to detect the quantum degenerate state of the 40K atoms was to determine the total energy (from the momentum distribution) of the approximately 8 105 atoms in the sample (Figure 8-33). Classically, the total energy (3 2)NAkT S 0 as T S 0. Quantum mechanically, however, the total energy should be higher than expected classically as T decreases and remain finite as T S 0. This is exactly what Jin and DeMarco observed. High on the list of new things their discovery may make possible is the study of Cooper pairs (see Section 10-8) as they condense into a superconductor. The cloud on T/TF = 0.5 > > EFERMI EFERMI T/TF = 3 Figure 8-33 Quantum degenerate state of a Fermi gas. The images show that more of the atoms of the ultracold gas lie below the Fermi energy (black circles) than above it in the right sample than in the left one. The colder cloud on the right T 0.29 K. contains 0.78 million 40K atoms at the left contains 2.5 million atoms at T 2.4 K. More recently, scientists have been successful in forming Bose-Einstein condensates from paired fermions using 6Li and 40K. The very loosely bonded 6Li-6Li and 40K-40K are bosons and dropped into their respective ground states when the temperature reached about 50 109 K. Questions 12. Why does the exclusion principle make evaporative cooling less effective as T decreases for fermions in a single-spin state? 13. Why does the total energy of the fermion gas not approach zero as T S 0? Summary TOPIC RELEVANT EQUATIONS AND REMARKS 1. Boltzmann distribution fB(E) AeE>kT where the distribution will be occupied. fB(E) is the probability that the state with energy E 8-1 Boltzmann\u2019s constant k 1.381 1023 J>K 8.617 105 eV>K General References 355 TOPIC RELEVANT EQUATIONS AND REMARKS Maxwell distribution of molecular speeds n(v) dv 4Na 3>2 m 2kT b v2emv2>2kT dv Equipartition theorem In equilibrium, each degree of freedom contributes energy per molecule. 1 2 kT to the average Average kinetic energy kT 8E9 3 2 8E9 where is the average translational kinetic energy per molecule. Dulong-Petit law 3R CV 2. Quantum statistics Bose-Einstein distribution Fermi-Dirac distribution fBE(E) fFD(E) 1 eeE>kT 1 1 eeE>kT 1 8-8 8-14 8-24 8-25 In all three distributions \u0192B, \u0192BE, and \u0192FD, is a normalization constant that depends on 1 the particle density. The FD distribution applies to particles with -integral spin, the BE 2 distribution to particles with zero or integral spin. At high energies both \u0192BE and \u0192FD approach \u0192B. e The Boltzmann distribution will be a good approximation of either \u0192BE or \u0192FD if e V 1. 3. Applications Liquid helium 4He becomes a superfluid at 2.17 K, called the lambda point. 3He, the only other naturally occurring isotope that has this property, becomes superfluid at about 2 mK. Bose-Einstein condensate Bosons undergo a phase transition, condensing to the lowest quantum state. Degenerate Fermi gas Fermions condensed to states from the ground state to the Fermi energy. General References The following general references are written at a level appropriate for the readers of this book. Blatt, F. J., Modern Physics, McGraw-Hill, New York, 1992. Brehm, J. J., and W. J. Mullin, Introduction to the Structure of Matter, Wiley, New", " York, 1989. Eisberg, R., and R. Resnick, Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles, 2d ed., Wiley, New York, 1985. An excellent but somewhat more advanced discussion of quantum statistics can be found in Chapter 11 of this book. Ford, K. W., The Quantum World, Harvard University Press, Cambridge, MA, 2005. Kittel, C., and H. Kroemer, Thermal Physics, W. H. Freeman and Co., New York, 1995. Leitner, A., Liquid Helium II: The Superfluid, Michigan State University, East Lansing, 1963. This 39-minute film is an excellent introduction to the subject of liquid helium II. London, F., Superfluids, Vol. II:Macroscopic Theory of Superfluid Helium, 2d rev. ed., Dover, New York, 1954. Mandel, F., Statistical Physics, Wiley, New York, 1988. Mendelssohn, K., The Quest for Absolute Zero: The Meaning of Low Temperature Physics, World University Library, McGraw-Hill, New York, 1966. 356 Chapter 8 Statistical Physics Notes 1. The statistical approach may also be used as an approximation in systems where the number of particles is not particularly large. For example, in Chapter 11 we will discuss briefly a statistical model of the atomic nucleus, a system containing only of the order of 100 particles. 2. Ludwig E. Boltzmann (1844\u20131906), Austrian physicist. His pioneering statistical interpretation of the second law of thermodynamics earned for him recognition as the founder of statistical mechanics. He explained theoretically the experimental observations of Stefan, whom he served as an assistant while in college, that the quantity of radiation increased with the fourth power of the temperature. He eventually succeeded Josef Stefan in the chair of physics at Vienna. A strong proponent of the atomic theory of matter, his suicide was apparently motivated in part by opposition to his views by others. 3. To avoid having to repeat this rather long phrase frequently, which will occur for E as well as v, we will hereafter use the expression \u201cthe number in dvx at vx\u201d or simply \u201cthe number in dvx.\u201d 4. Or refer to a table of integrals. 5. Historically, rotation about the z' axis of the dumbbell was ruled out by assuming either that the atoms are points and the moment of inertia about this axis is therefore zero (not true) or that the atoms are hard smooth spheres, in which case rotation about this axis cannot be changed by collisions and therefore does not participate in the exchange of energy (also not true). Either of these assumptions also rules out the possibility of rotation of a monatomic molecule. 6. Satyendra Nath Bose (1894\u20131974), Indian physicist. Following publication of his paper on the statistics of indistinguishable particles, which was translated into German for publication by Einstein himself, Bose spent two years in Europe, then returned to India to devote himself to teaching. Lacking a Ph.D., he was denied a professorship until a onesentence postcard from Einstein was received at Dacca University in his support. 7. Enrico Fermi (1901\u20131954), Italian-American physicist. An exceedingly prolific scientist and intrepid amateur tennis player whose work encompassed solid-state, nuclear, and particle physics, he is perhaps best known as the \u201cfather\u201d of the nuclear reactor. He was awarded the Nobel Prize in Physics in 1938 for his work in nuclear physics. 1 2 8. Paul A. M. Dirac (1902\u20131984), English physicist. His development of relativistic wave mechanics for spin- particles led to his prediction in 1930 of the existence of the positron. Its discovery by Anderson two years later resulted in Dirac\u2019s being awarded (along with Schr\u00f6dinger) the 1933 Nobel Prize in Physics. From 1932 until his retirement he occupied the Lucasian Chair of Mathematics at Cambridge University, which had been held 250 years earlier by Newton and is currently held by Stephen Hawking. 9. Heike Kamerlingh Onnes (1853\u20131926), Dutch physicist. His success in liquefying helium enabled him to investigate the properties of other materials at liquid helium temperatures. This in turn led to his discovery of superconductivity in 1911. His work on the behavior of materials at low temperatures earned him the Nobel Prize in Physics in 1913. 10. J. C. McLennan, H. D. Smith, and J. O. Wilhelm, Philosophical Magazine, 14, 161 (1932). 11. At very low temperatures liquid 4He does solidify at a pressure of about 25 atm, liquid 3He at about 30 atm. 12. Narrow channels that permit only the superfluid to pass are, of course, called superleaks. 13. These and many other properties are", " elegantly displayed in the film Liquid Helium II: The Superfluid, available from the Instructional Media Center, Michigan State University, East Lansing, Michigan 48824. 14. In the thermodynamic equilibrium state their sample, rubidium, is a solid metal at room temperature. 15. Einstein used the Boltzmann distribution in its discrete form fB(E) a n0 AeEn >kT. Problems Level I Section 8-1. Classical Statistics: A Review > (a) Calculate vrms for H2 at T 300 K. (b) Calculate the temperature T for which vrms for 8-1. H2 equals the escape speed of 11.2 km s. 8-2. (a) The ionization energy for hydrogen atoms is 13.6 eV. At what temperature is the average kinetic energy of translation equal to 13.6 eV? (b) What is the average kinetic energy of translation of hydrogen atoms at T 107 K, a typical temperature in the interior of the Sun? 8-3. The molar mass of oxygen gas (O2) is about 32 g mol and that of hydrogen gas (H2) about 2 g mol. Compute (a) the rms speed of O2 and (b) the rms speed of H2 when the temperature is 0\u00b0C. 8-4. Show that the SI units of (3RT>M)1>2 are m s.> > > Problems 357 (a) Find the total kinetic energy of translation of 1 mole of N2 molecules at T 273 K. 8-5. (b) Would your answer be the same, greater, or less for 1 mole of He atoms at the same temperature? Justify your answer. 8-6. Use the Maxwell distribution of molecular speeds to calculate a gas. 8-7. Neutrons in a nuclear reactor have a Maxwell speed distribution when they are in therand vm for neutrons in thermal equilibrium at 300 K. Show that n(v) mal equilibrium. Find v vm (Equation 8-8) has its maximum value at 8-8. A container holds 128 identical molecules whose speeds are distributed as follows: for the molecules of (2kT>m)1>2. 8v29 8v9 No. of molecules > Speed range (m s) 4 12 20 24 20 16 12 8 6 4 0.0\u20131.0 1.0\u20132.0 2.0\u20133.0 3.0\u20134.0 4.0\u20135.0 5.0\u20136.0 6.0\u20137.0 7.0\u20138.0 8.0\u20139.0 9.0\u201310.0 Ek 1 1 >M and and vrms. CV What are nR above the lowest energy state (g1 3) is about 1). What is the ratio of the numbers of mole- vm, 8v9, Graph these data and indicate on the graph 8-9. Show that the most probable speed vm of the Maxwell distribution of speeds is given by Equation 8-9. 8-10. Compute the total translational kinetic energy of one liter of oxygen held at a pressure of one atmosphere and a temperature of 20\u00b0C. 8-11. From the absorption spectrum it is determined that about one out of 106 hydrogen atoms in a certain star is in the first excited state, 10.2 eV above the ground state (other excited states can be neglected). What is the temperature of the star? (Take the ratio of statistical weights to be 4, as in Example 8-2.) 8-12. The first excited rotational energy state of the H2 molecule (g2 4 103 eV cules in these two states at room temperature (300 K)? 8-13. A monatomic gas is confined to move in two dimensions so that the energy of an atom 2 mv2 2 mv2 CV, CP, and the heat capacity at constant is y. x >CV. C P CV pressure, is equal to CV 8-14. Use the Dulong-Petit law that for solids to calculate the specific heat in cal g for (a) aluminum, M 27.0 g mol, (b) copper, M 63.5 g mol, and > > cV (c) lead, M 207 g mol, and compare your results with the values given in a handbook. (Include the reference in your answer.) 8-15. Calculate the most probable kinetic energy Em from the Maxwell distribution of kinetic energies (Equation 8-13). 8-16. (a) Show that the speed distribution m e(v>vm)2 41>2(v>vm)2 v1 vm where, and approximate dv by v 0.01 (b) v 0, (c) v vm, vm, 8-17. Consider a sample containing hydrogen atoms at 300", " K. (a) Compute the number of atoms in the first (n 2) and second (n 3) excited states compared to those in the ground state (n 1). Include the effects of degeneracy in your calculations. (b) At what temperature would one percent of the atoms be in the n 2 state? (c) At the temperature found in (b), what fraction of the atoms will be in the n 3 state? 8-18. Consider a sample of noninteracting lithium atoms (Li, Z 3) with the third (outer) electron in the 3p state in a uniform 4.0 T magnetic field. (a) Determine the fraction of the atom in and 1 states at 300 K. (b) In the the transition, what will be the relative intensities of the three lines of the Zeeman effect? n(v) is the most probable speed. Consider 1 mole of molecules Find the number of molecules with speeds in dv at vm. and (e) v 8 for this gas? ( ) 3R function can be written (d) v 2 1, 0, 3p S 2s CP, vm. m > > Section 8-2 Quantum Statistics > e1b2 4ac e 1 e 106. from Equation 8-44 for O2 gas at standard conditions. (b) At 8-19. Find the number density N V for electrons such that (a) 8-20. (a) Compute what temperature is 8-21. Given three containers all at the same temperature, one filled with a gas of classical molecules, one with a fermion gas, and one with a boson gas, which will have the highest pressure? Which will have the lowest pressure? Support your answer. e 1 for O2? and (b) 358 Chapter 8 Statistical Physics 0, (b) Still with 0)? for the energy in part (a)? fBE(E) to what value must the temperature change if 8-22. (a) For T 5800 K, at what energy will the Bose-Einstein distribution function equal 1 (for fBE(E) 0.5 8-23. A container at 300 K contains H2 gas at a pressure of one atmosphere. At this temperature H2 obeys the Boltzmann distribution. To what temperature must the H2 gas be cooled before quantum effects become important and the use of the Boltzmann distribution is no longer appropriate? (Hint: Equate the de Broglie wavelength at the average energy to the average spacing between molecules, using the ideal gas law to compute the density.) > > Section 8-3 The Bose-Einstein Condensation 8-24. Compute N0 N from Equation 8-52 for (a) T 3Tc 4, (b) (d) T Tc 8. 8-25. Show that 8-52. 8-26. Like 4He, the most common form of neon, 20Ne, is a rare gas and the 20Ne atoms have zero spin and hence are bosons. But unlike helium, neon does not become superfluid at low temperatures. Show that this is to be expected by computing neon\u2019s critical temperature and comparing it with the element\u2019s freezing point of 24.5 K. as asserted in the paragraph above Equation (c) T Tc 4, and for small values of 1> T 1 2 Tc, N0 > > Section 8-4 The Photon Gas: An Application of Bose-Einstein Statistics T 0.1hf>k 8-27. If the Sun were to become cooler (without changing its radius), the energy density at the surface would decrease according to Equation 8-56. Suppose the Sun\u2019s temperature were to decrease by 5 percent. Compute the fractional change in the rate at which solar energy arrives at Earth. (Assume that the Sun\u2019s surface is in equilibrium and radiates as a blackbody.) 8-28. Find the average energy of an oscillator at (a) (c) 8-29. (a) Show that the rule of Dulong-Petit follows directly from Einstein\u2019s specific heat formula (Equation 8-62) as 8-30. Using Figure 8-13, compute the (approximate) frequency of atomic oscillations in silicon and in aluminum at 200 K. 8-31. Use Equation 8-62 to calculate the value of TE and compare your results with those from the equipartition theorem. for a solid at the Einstein temperature T 10hf>k, (b) Show that T hf>k, as T S 0. hf>k. T S. S 0 (b) CV CV and Section 8-5 Properties of a Fermion Gas 8-32. Use Equation 8-69 to plot an accurate graph of nFD(E) V for electrons whose Ferm", "i energy is 4.8 eV from E 4.5 eV to E 5.1 eV at T 300 K. Determine from the graph the number of electrons per unit volume just below the Fermi energy that can move to states just above the Fermi energy. 8-33. Consider a gas of electrons (fermions) and a gas of photons (bosons). Which has more states available at T 1 K? Explain why. > Level II 8-34. The molar heat capacity data given in Table 8-2 are taken from AIP Handbook, 2d ed. (McGraw-Hill, New York, 1963). Plot the data for these solids all on one graph and sketch in the curves CV versus T. Estimate the Einstein temperature for each of the solids using the result of Problem 8-31. 8-35. Recalling that the Fermi-Dirac distribution function applies to all fermions, including 1 consider a nucleus of 22Ne consisting of 10 protons and neutrons, each of which have spin 2, protons and 12 neutrons. Protons are distinguishable from neutrons, so two of each particle (spin up, spin down) can be put into each energy state. Assuming that the radius of the 22Ne estimate the Fermi energy and the average energy of the nucleus in nucleus is 22Ne. Express your results in MeV. Do the results seem reasonable? 3.1 1015 m, Problems 359 Table 8-2 Heat capacities in cal mole K for Au, diamond, Al, and Be > # T, K 20 50 70 100 150 200 250 300 400 500 600 800 1000 Au 0.77 3.41 4.39 5.12 5.62 5.84 5.96 6.07 6.18 6.28 6.40 6.65 6.90 Diamond 0.00 0.005 0.016 0.059 0.24 0.56 0.99 1.46 2.45 3.24 3.85 4.66 5.16 Al 0.05 0.91 1.85 3.12 4.43 5.16 5.56 5.82 6.13 6.42 6.72 7.31 7.00 Be 0.003 0.04 0.12 0.43 1.36 2.41 3.30 3.93 4.77 5.26 5.59 6.07 6.51 >4. T Tc 8-36. What is the ground-state energy of 10 noninteracting bosons in a one-dimensional box of length L? 8-37. Make a plot of \u0192FD(E) versus E for (a) T 0.1TF, and (b) T 0.5TF, where TF 8-38. Compute the fraction of helium atoms in the superfluid state at (a) (b) 8-39. The depth of the potential well for free electrons in a metal can be accurately determined by observing that the photoelectric work function is the energy necessary to remove an electron at the top of the occupied states from the metal; an electron in such a state has the Fermi energy. Assuming each atom provides one free electron to the gas, compute the depth of the well for the free electrons in gold. The work function for gold is 4.8 eV. 8-40. An early method testing Maxwell\u2019s theoretical prediction for the distribution of molecular speeds is shown in Figure 8-34. In 1925 Otto Stern used a beam of Bi2 molecules emitted from an oven at 850 K. The beam defined by slit S1 was admitted into the interior of a rotating drum via slit S2 in the drum wall. The identical bunches of molecules thus formed struck and EF k. > >2 and T Tc S1 S2 Drum Oven A B w Glass plate Figure 8-34 360 Chapter 8 Statistical Physics adhered to a curved glass plate fixed to the interior drum wall, the fastest molecules striking near A, which was opposite S2, the slowest near B, and the others in between depending on their speeds. The density of the molecular deposits along the glass plate were measured with a densitometer. The density (proportional to the number of molecules) plotted against distance along the glass plate (dependent on v) made possible determination of the speed distribution. If the drum is 10 cm in diameter and is rotating at 6250 rpm, (a) find the distance from A where molwill strike. (b) The plot in (a) must be corrected slightly in ecules traveling at order to be compared with Maxwell\u2019s distribution equation. Why? (c) Would N2 molecules work as well as Bi2 molecules in this experiment? Why or why not? 8-41. The speed distribution of molecules in a container is the Maxwell distribution f(v) v2emv2>2kT. The number with speed v", " that hit the wall in a given time is proportional to the speed v and to \u0192(v). Thus, if there is a very small hole in the wall (too small to have much effect on the distribution inside), the speed distribution of those that escape is F(v) vf(v) v3emv2>2kT. Show that the mean energy of those that escape is 2kT. vm,8v9, and vrms to. 8E9 8Au29 Level III 8-42. This problem is related to the equipartition theorem. Consider a system in which the energy of a particle is given by E Au2, where A is a constant and u is any coordinate or (a) Write the probability of the particle having u in momentum that can vary from the range du and calculate the normalization constant C in terms of A. (b) Calculate the average energy and show that 8-43. Calculate the average value of the magnitude of 8-44. Show that 8-45. Carry out the integration indicated in Equation 8-43 to show that 8-44. 8-46. Consider a system of N particles that has only two possible energy states, E1 CeEi E2 8E9 average energy the heat capacity is 0 and (a) What is C for this case? (b) Compute the (c) Show that as fFD(E) S fB(E) for E W EF. from the Maxwell distribution. The distribution function is is given by Equation fi 8E9 S 0 as T S 0 and and show that 8E9 S >2 8E9 1 T S.. 2 kT. >kT. vx CV Nk a 2 b kT e>kT (1 e>kT)2 CV versus T. (d) Sketch 8-47. If the assumptions leading to the Bose-Einstein distribution are modified so that the number of particles is not assumed constant, the resulting distribution has This distribution can be applied to a \u201cgas\u201d of photons. Consider the photons to be in a cubic box of side L. The momentum components of a photon are quantized by the standing-wave conditions n3 k2 kx is the magnitude of where the momentum. (a) Show that the energy of a photon can be written where N2 n2 (b) Assuming two photons per space state because of the two possible 1 polarizations, show that the number of states between N and N dN is N2 dN. (c) Find the density of states and show that the number of photons in the energy interval dE is z)1>2 E N(Uc>L), p U(k2 x >L, ky e 1. n2 3. n2 n1 n2 2 k2 y >L, >L, and kz n(E) dE 8(L>hc)3E2 dE eE>kT 1 (d) The energy density in dE is given by u(E) dE En(E) dE L3. Use this to obtain the Planck d, blackbody radiation formula for the energy density in u() 8hc5 ehc>kT 1 is the wavelength: > where PART2 Applications of Quantum Mechanics and Relativity Part 1 introduced the principles of the special and general relativity theories and illustrated how they led to profound alterations of our classical views of space and time. We then saw how the ideas and methods of quantum mechanics developed and how their application to atomic physics provides us with an understanding of atomic structure and spectra that is in excellent accord with our observations. In Part 2 we extend the applications of quantum theory and relativity to a wider variety of physical systems and phenomena that are, like atomic physics, of great interest to engineers, chemists, and physicists. The topics we will discuss form the foundation of a broad range of theoretical and experimental research by physicists, chemists, and mathematicians and provide the basic understanding of the principles underlying many practical devices developed by engineers. They include molecular bonding and spectra (Chapter 9); the structure of solids and their thermal and electrical properties (Chapter 10); superconductors (Chapter 10); nuclear structure, radioactivity, and nuclear reactions (Chapter 11); and elementary particles, the quarks and leptons, that are the constituents of all visible matter (Chapter 12). Practical applications include the study of lasers (Chapter 9); semiconductors, semiconductor junctions, and transistors (Chapter 10), and radioactive dating and elemental analysis, nuclear fission and fusion, and reactors (Chapter 11). Many of these applications have revolutionized contemporary society. Part 2 concludes with a look outward into the cosmos from our solar system to the Big Bang, the realm of astrophysics and cosmology (Chapter 13), topics that stimulate the imagination of everyone. These chapters are independent of one", " another and can be studied in any order. 361 This page intentionally left blank CHAPTER 9 Molecular Structure and Spectra In this chapter we will study the bonding of molecules\u2014systems of two or more atoms. Properly, a molecule is the smallest constituent of a substance that retains its chemical properties. The study of the properties of molecules forms the basis for theoretical chemistry. The application of quantum mechanics to molecular physics has been spectacularly successful in explaining the structure of molecules and the complexity of their spectra and in answering such puzzling questions as why two H atoms join together to form a molecule but three H atoms do not. As in atomic physics, the detailed quantum-mechanical calculations are often difficult. When the difficulty would tend to obscure understanding of the physics, we will, as before, make our discussions semiquantitative or qualitative. In the final sections we will discuss the interaction of electromagnetic radiation with molecules, concluding with a discussion of the common general types of lasers. There are essentially two extreme views we can take of a molecule. Consider, for example, H2. We can think of it either as two H atoms somehow joined together or as a quantum-mechanical system of two protons and two electrons. The latter picture is more fruitful in this case because neither of the electrons in the H2 molecule can be considered as belonging to either proton. Instead, the wave function for each electron is spread out in space about the whole molecule. For more complicated molecules, however, an intermediate picture is useful. Consider the N2 molecule as an example. We need not consider the complicated problem of 2 nuclei and 14 electrons. The electron configuration of an N atom in the ground state is 1s22s22p3. Of the three electrons 1 state with their spins paired (that is, with spins in the 2p state, two are in an m/ antiparallel so that the resultant spin for those two is zero). The third one is in an 0 level and its spin is, of course, unpaired. Only the electron with the unpaired m/ spin is free to take part in the bonding of the N2 molecule. We therefore can consider this molecule as two N ions and two electrons that belong to the molecule as a whole. The molecular wave functions for these bonding electrons are called molecular orbitals. In many cases these molecular wave functions can be constructed from linear combinations of the atomic wave functions with which we are familiar. Another type of bonding involves the transfer of one or more electrons between atoms, the bond resulting from Coulomb attraction between the ions, an example being NaCl. Again in this case, as in all four types of molecular bonding, it is the wave properties of the spin- electrons that are the key to understanding. 1 2 9-1 The Ionic Bond 364 9-2 The Covalent Bond 369 9-3 Other Bonding Mechanisms 375 9-4 Energy Levels and Spectra of Diatomic Molecules 9-5 Scattering, Absorption, and Stimulated Emission 9-6 Lasers and Masers 379 390 396 363 364 Chapter 9 Molecular Structure and Spectra 9-1 The Ionic Bond The two principal types of bonds that join two or more atoms together to form a molecule are called ionic and covalent bonds. Other types of bonds that are important in the bonding of liquids and solids are dipole-dipole bonds and metallic bonds. In many cases the bonding is a mixture of these mechanisms. We will discuss all of these in this chapter and the next, but it is important to recognize that all types of molecular bonding arise for the same fundamental reasons: the total energy of the stable bound molecule is lower than the total energy of the constituent atoms when they are widely separated, and there is a net attractive force between constituent atoms when their separation becomes larger than some equilibrium value. The bonding mechanisms are primarily due to electrostatic forces between the atoms or ions of the system together with the wave properties of electrons and the fact that they obey the exclusion principle. The complete description of molecular bonding is in most cases quite complex, involving as it does the mutual interactions of many electrons and nuclei; consequently, we will discuss each type using simplified models consisting of two or a few atoms, then illustrate qualitatively the extension of the results to more complex molecules. The easiest type of bond to understand is the ionic bond, typically the strongest of the bonds and the one found in most salts. Consider KCl as an example. For the molecule to be stable, we must be able to show that E(KCl) E(K) E(Cl) when the K and Cl atoms are far apart and at rest. Let us define the energy of the system to be zero when the neutral atoms are widely separated. (See Figure 9-1.) The potassium atom has one 4s electron outside an argon core, 1s22s22p63s23p6. The ionization energy for K is low, as it is for all the alkali", " metals; for K only 4.34 eV is required to remove the outer electron from the atom. (See Table 9-1.) The removal of one electron from K leaves a positive ion with a spherically symmetric, closed-shell core. Chlorine, on the other hand, is only one electron short of having a closed argon core. The energy released by the acquisition of one electron is called the electron affinity, which in the case of Cl is 3.62 eV. Energy is released because the wave function of the \u201cextra\u201d electron penetrates the outer shell to a degree (see Figure 7-10b) and thus sees a net positive charge. The acquisition of one electron by chlorine results in a negative Figure 9-1 Net energy required to ionize a K and a Cl atom. An addition of 4.34 eV is required to remove the 4s electron from the neutral K atom, forming K and a free electron. That electron (or some electron) can then occupy the vacancy in the 3p shell of the Cl atom forming a Cl ion. The electron is positively bound, with the release of 3.62 eV. Formation of the widely separated K and Cl ions thus requires a net addition of 0.72 eV. V e E, 5.0 4.0 3.0 2.0 1.0 0 4.34 eV Ionization energy K 3.62 eV electron affinity Cl K, Cl atoms (far apart) K+, 1 electron and Cl atom 0.72 eV K+, Cl\u2013 (far apart) K Cl +4.34 eV K+ Cl \u20133.62 eV e\u2013 K+ Cl\u2013 9-1 The Ionic Bond 365 Table 9-1 Ionization energies of alkali metal atoms and electron affinities of halogen atoms Alkali metal Ionization energy (eV) Halogen Electron affinity (eV) Li Na K Rb Cs Fr 5.39 5.14 4.34 4.18 3.89 4.07 F Cl Br I At 3.40 3.61 3.36 3.06 2.8 Source: Data from Handbook of Chemistry and Physics, 85th ed. (New York: Chemical Rubber Co., 2004). ion with a spherically symmetric, closed-shell electron core. Thus, the formation of a K ion and a Cl ion by the donation of one electron of K to Cl requires just 4.343.62 0.72 eV. If this were the whole story, the KCl molecule would not form; however, the electrostatic potential energy of the two ions separated by a distance r is ke2 r. When the separation of the ions is less than about 2.8 nm, the negative potential energy of attraction is of greater magnitude than the energy needed to create the ions, and the ions move toward each other. > Since the electrostatic attraction increases as the ions get closer, it would seem that equilibrium could not exist. For very small separation of the ions, however, the wave function of the 3p electrons in the K ion and the 3p electrons in the Cl ion begin to overlap. Since the 3p shells in each ion contain electrons with sets of quantum numbers identical to those in the other, a strong repulsion develops due to the exclusion principle. This \u201cexclusion-principle repulsion\u201d is primarily responsible for the repulsion of the atoms in all molecules (except H2), no matter which type of bonding occurs. When the ions are very far apart, the wave function for a core electron of one ion does not overlap that of the other ion. We can distinguish the electrons by the ion to which they belong, and the electrons of one ion can have the same quantum numbers as in the other ion. However, when the ions are close, the wave functions of their core electrons begin to overlap, and some of the electrons must go into higher-energy quantum states because of the exclusion principle, thus increasing the total energy of the system. This is not a sudden process: the energy states of the electrons are gradually changed as the ions move closer together. The total potential energy U of the KCl system can be expressed in terms of the separation r of the ion centers as the sum of the electrostatic potential, the exclusion-principle repulsion, and net ionization energy. U(r) ke2 r Eex Eion 9-1 0.72 eV for K and Cl, as was found above. The exclusion-principle where Eion repulsion Eex can be written as Eex A rn 9-2 366 Chapter 9 Molecular Structure and Spectra Figure 9-2 (a) Potential energy for K and Cl ions as a function of separation distance r. The energy at infinite separation was chosen to be 0.72 eV, corresponding to the energy needed to form the ions from neutral atoms. The minimum energy for this curve is at the equilibrium separation", " 0.27 nm for the ions in r0 the molecule. (b) Potential energy for Na and Cl ions as a function of r. Differences between the two similar molecules are due to the higher ionization potential and smaller core of Na. (a) (b1 \u20132 \u20133 \u20134 \u20135 Ed 1 \u20132 \u20133 \u20134 \u20135 Exclusion-principle repulsion Eex Net ionization energy 0.72 eV r0 1.0 2.0 K+ + Cl\u2013 r, nm 3.0 K + Cl Total potential energy Electrostatic (Coulomb) potential energy \u2013 ke \u2013\u2013 r eV 0.2 0.4 0.6 1.2 0.8 1.0 Ed = 4.26 eV 1.4 1.6 r, nm Na+ + Cl\u2013 1.53 eV Na + Cl r0 = 0.236 nm where A and n are constants for each ionic molecule. Figure 9-2a is a sketch of the potential energy of the K and Cl ions versus their separation. The energy is lowest at an equilibrium separation r0 of about 0.27 nm. At smaller separations, the energy rises steeply as a result of the exclusion principle. The energy Ed required to separate the ions and form K and Cl atoms, called the dissociation energy, is about 4.40 eV. Figure 9-2b shows the total potential energy of another ionically bonded molecule, NaCl. Note the differences between the two total potential energy curves, which are due to the higher ionization potential and smaller closed-shell core of Na compared to K. Example 9-1 illustrates calculations used to construct curves like those in the diagram. Example 9-2 describes how the constants A and n in Equation 9-2 are found. EXAMPLE 9-1 Ionic Bonding in NaF The ionization potential of sodium is 5.14 eV, the electron affinity of fluorine is 3.40 eV, and the equilibrium separation of sodium fluoride (NaF) is 0.193 nm. (a) How much energy is needed to form Na and F ions from neutral sodium and fluorine atoms? (b) What is the electrostatic potential energy of the Na and F ions at their equilibrium separation? (c) The dissociation energy of NaF is 4.99 eV. What is the energy due to repulsion of the ions at the equilibrium separation? SOLUTION (a) Since the energy needed to ionize sodium is 5.14 eV and the electron affinity of F is 3.40 eV, the energy needed to form Na and F ions from neutral sodium and fluorine atoms is 5.14 eV 3.40 eV 1.74 eV Eion. (b) The electrostatic potential energy of the Na and F ions at their equilibrium separation (with ke2 r 0 at infinite separation) is > ke2 r0 (8.99 109 N # m2>C2)(1.60 1019C)2 1.93 1010 m 1.19 1018 J 7.45 eV 9-1 The Ionic Bond 367 (c) Choosing the total potential energy at infinity to be 1.74 eV (the net ionization energy needed to form Na and F from the neutral atoms), the net electrostatic (Coulomb) potential is UC UC ke2 r 1.74 eV 7.45 eV 1.74 eV At the equilibrium separation r0, this energy is UC 5.71 eV. Since the measured dissociation energy is 4.99 eV, the potential energy due to exclusion-principle repulsion Eex of the Na and F at equilibrium separation, from Equation 9-1, must be 5.71 eV 4.99 eV 0.72 eV. EXAMPLE 9-2 Contribution from Exclusion-Principle Repulsion Find the values of A and n in Equation 9-2 for NaF. SOLUTION From Example 9-1 we have that the potential energy due to exclusion-principle repulsion at equilibrium separation of the ions is Eex(r0) A rn 0 A (0.193 nm)n 0.72 eV At r r0 the net force on each ion must be zero because the potential energy has its minimum value at that point. This means that at r r0, the net Coulomb force FC is equal in magnitude and opposite in sign to the exclusion-principle repulsive force, i.e., a FC dUC dr b rr0 a nA rn1 b rr0 FC UC(r0) r0 ke2 r2 0 38.7 eV>nm At r r0 Thus, we have that nA rn1 0 n r0 A rn 0 n r0 (0.72 eV) 38.7 eV>nm or n", " 38.7 eV>nm 0.72 eV and, therefore, A 5.4 108 Eex (0.193 nm) 10.4 10 Finally, for NaF, Eex is given by eV # nm10. (5.4 108 eV # nm10) r10 368 Chapter 9 Molecular Structure and Spectra It should be emphasized that our discussion of ionic bonding and, in particular, the graphs of potential energy in Figure 9-2 apply to the ground states of the molecules. The outer (valence) electrons of molecules may occupy excited states, just as they do in atoms. Since the electron wave functions of the excited states tend to extend farther from the ions than do those of the ground state, the potential energy curve is broader and shallower than for the ground state, resulting in a slightly weaker bond and a larger equilibrium separation of the ions. In our discussion we have ignored two additional contributions to the total energy of the molecule: (1) the zero-point energy (see Section 5-6), which decreases the magnitude of Ed, and (2) the van der Waals attraction, which increases the magnitude of Ed. Both are small and tend to partially offset each other. The latter, which arises from induced dipole moments, is the only form of bonding available for certain molecules and will be discussed later in this chapter. The KCl equilibrium separation of 0.27 nm noted earlier is for gaseous diatomic KCl (which can be obtained by evaporation of solid KCl). Normally, KCl exists in a cubic crystal structure, with K and Cl at alternate corners of a cube. The separation of the ions in a crystal is somewhat larger\u2014about 0.32 nm. Because of the presence of neighboring ions of opposite charge, the Coulomb energy per ion pair is lower when where r0 is the the ions are in a crystal. This energy is usually expressed as equilibrium separation distance or bond length and called the Madelung constant, depends on the crystal structure, as will be discussed further in Chapter 10. For KCl, is about 1.75. The values of Ed and r0 listed in Table 9-2 are for several ionically bonded (gaseous) molecules. ke2>r0,, Table 9-2 Dissociation energies Ed and equilibrium separations r0 for several ionic molecules* in the gaseous state Molecule Dissociation energy (eV) Equilibrium separation (nm) NaCl NaF NaH NaBr LiCl LiH LiI KCl KBr RbF RbCl CsI 4.27 5.34 1.92 3.81 4.86 2.47 3.58 4.49 3.94 5.12 4.43 3.50 0.236 0.193 0.189 0.250 0.202 0.159 0.239 0.267 0.282 0.227 0.279 0.332 *The two entries of molecules formed by an alkali atom and a hydrogen atom may seem odd, but hydrogen atoms, like those of a number of other elements, may form molecules as either positive or negative ions. The ionization energy of H is, of course, 13.6 eV; its electron affinity is 0.75 eV. Source: Data from Handbook of Chemistry and Physics, 85th ed. (New York: Chemical Rubber Co., 2004). 9-2 The Covalent Bond 369 One final comment concerning ionic bonding: Few of the molecules in Table 9-2 are bonded exclusively by the ionic mechanism. As we will see in the next section, they may also be covalently bonded. 9-2 The Covalent Bond A completely different mechanism is responsible for the bonding of such molecules as H2, N2, H2O, and CO and also leads to bonding of many of the molecules in Table 9-2. If we calculate the energy needed to form the ions H and H by the transfer of an electron from one atom to the other, we find the net ionization energy to be more than 12 eV. Adding this energy to the electrostatic energy (including the repulsion of the protons), we find that there is no separation distance for which the total energy is negative. The bond of H2 thus cannot be ionic. The attraction of two hydrogen atoms is instead an entirely quantum-mechanical effect. The decrease in energy when two hydrogen atoms approach each other is due to the sharing of the two electrons by both atoms and is intimately connected with the symmetry properties of the electron wave functions. We can gain some insight into this phenomenon by first studying a simple one-dimensional quantum-mechanics problem\u2014that of two finite square wells each of width L. & \u0192&\u0192 2 Consider first a single electron that is equally likely to be in either well. Since the wells are identical, symmetry requires that be symmetric about the midpoint of the wells. Then must be either", " symmetric or antisymmetric about that point. These two possibilities for the ground state are shown in Figure 9-3. Previously, we did not distinguish between these two possibilities when superimposing (i.e., adding) 2U2>2mL2 wave functions because the energies for both of these wave functions are the same when the wells are far apart. Figure 9-4 shows the symmetric and antisymmetric wave functions when the wells are very close together. Now the parts of the wave function describing the electron in one well or the other overlap, and the symmetric and antisymmetric resultant wave functions are quite different. Notice that for the symmetric wave functions the probability of the electron being found in the region between the wells is much larger than for the antisymmetric wave function. In the limiting case of no separation, the symmetric approaches the ground-state wave function for a particle in a well wave function and the probability densities &2 & S (a) (b) \u03a8 S (c) \u03a8 A \u03a8 S \u03a8 A Figure 9-3 (a) Two square wells far apart. The electron wave function can be either (b) symmetric or (c) antisymmetric. The probability distributions and energies are the same for the two wave functions when the wells are far apart. Figure 9-4 Symmetric and antisymmetric space wave functions for two square wells close together. The probability distributions and energies are not the same for the two wave functions in this case. The symmetric space wave function (and therefore the probability density) is larger between the wells than the antisymmetric space wave function. 370 Chapter 9 Molecular Structure and Spectra of size 2L and the antisymmetric wave function cited state in such a well; thus has two important results: & S & A is a lower energy state than approaches that for the first exA. This discussion & 1. The originally equal energies for as the wells become close. & A and & S are split into two different energies 2. The wave function for the symmetric state is large in the region between the wells, whereas that for the antisymmetric state is small. Now consider adding a second electron to the two wells. The total wave function for the two electrons must be antisymmetric on exchange of the electrons since they obey the Pauli exclusion principle. Note that exchanging the electrons in the wells is the same as exchanging the wells; i.e., for a two-particle system, exchange symmetry is the same as space symmetry. The two electrons can therefore be in the spacesymmetric state if the spins are antiparallel (S 0) or in the space-antisymmetric state if their spins are parallel (S 1). Molecule H2 Now let us consider a real physical system with one electron, the hydrogen molecule. For a one-dimensional model, the double potential well formed by the two ion H2 protons is illustrated in Figure 9-5. The Hamiltonian (total energy) operator for this system is (see Equation 6-51 and Figure 9-5b) Hop p2 op 2m ke2a 1 r1 1 r2 1 r0 b For our In the ground state, the hydrogen atom wave function is proportional to one-dimensional model, we will write this as The symmetric and antisymmetric combinations for two values of the distance between the protons are shown in Figure 9-6. In general, e\u0192x\u0192>a0. er>a0. and & S a 1 22 & A a 1 22 (a) Proton 1 + Proton 2 + Figure 9-5 Coulomb potential for an electron resulting from two protons separated by a distance r0. The solid line is the total potential for a onedimensional model. The circled plus signs mark the locations of the protons. r0 b(& 100(r1) & 100(r2)) b(& 100(r1) & 100(r2)) (b) Electron 1 r1 r0 r2 2 \u03a8 1 S= \u03a8 1 \u2013 \u03a8 2 9-2 The Covalent Bond 371 r S= \u03a8 1 \u2013 \u03a8 2 (a) (b) (c) 2 \u03a8 S 2 \u03a8 A Figure 9-6 One-dimensional symmetric and antisymmetric electron space wave functions for (a) two protons far apart and (b) two protons close together. (c) Probability distributions for wave functions in (b). Computer-drawn electron density around the protons is shown above the probability densities. & S & A is large in the region between the The results are similar to the square-well case: protons, while is small in that region. Only in the case where the electron wave function and, hence, the probability density is large near the center of the molecule do we expect a stable molecular bond", " to form. This concentration of negative charge beholds the protons together. Similarly, we would not expect tween the protons for & to result in a stable molecule. The justification of this conclusion would be the A solution of the Schr\u00f6dinger equation and calculation of \u0192&\u0192 2 & S for. H2 The solution and calculation are quite difficult, so we will simply state the results for the energy of the molecule as a function of the separation r of the protons, describing in the process how, in general, the potential energy function arises. Referring first to Figure 9-6a, when the protons are far apart, the electron\u2019s energy is 13.6 eV. The potential energy Up (repulsion) of the protons is negligibly small for large r and, since there is only a single electron in the system, there is no exclusion-principle repulsion. As the two protons are brought closer together as in Figure 9-6b, Up increases and the energy of the electron decreases since the electron experiences a greater Coulomb force and becomes more tightly bound. Consider what is happening to the energy of the electron as the separation r of the protons is reduced. As r S 0, the electron\u2019s wave function is approaching that of an atom with Z 2. The symmetric wave function has a maximum at r 0 and thus corresponds to the 1s (ground) state of the Z 2 atom. As we have al 13.6 Z2/n2 54.4 eV. For our ready seen (Equation 7-25), its energy is E1 discussion here, let us call the electron\u2019s energy ES for the wave function & S. & S 372 Chapter 9 Molecular Structure and Spectra > & A 13.6 eV for r S and ES 54.4 eV for r S 0. The antisymmetThus, ES is zero at r 0 and thus corresponds to the 2p (first excited) ric wave function state of the Z 2 atom, this state being the lowest energy state with a wave function that vanishes at r 0. (See Equation 7-26 and Table 7-2.) The energy of this 13.6 Z 2 n2 13.6 eV. As above, if we call EA the energy of the state is E2 \u0192 2 \u0192& & A, electron for the wave function S \u0192 2 \u0192& 13.6 eV for r S 0. Recall that the smaller avare the same) and EA and A & as r S 0 implies a smaller energy for the symerage slope of compared to A metric state. The variation of both ES and EA are shown in Figure 9-7. 13.6 eV for r S (where then EA & S H 2 The potential energy of the protons as a function of their separation is, of course, ke2 r, and the total energy of the molecule is then ES or EA, depending on which of the electronic wave functions happens to exist. Up As can be seen in Figure 9-7, only one of the total energy functions has a minimolecule. The potential energy mum and therefore can result in bonding of the H function Etotal 2 0.106 nm and binding energy molecule is stable, with equilibrium separation r0 Etotal(r S ) Etotal(r0) 13.6 (16.3) 2.7 eV. In contrast, the potential EA has no minimum; therefore, the antisymmetric energy function Etotal wave function does not result in a stable molecule, as we expected at the outset of this discussion. Note that the -type bond will tend to be unstable unless the nuclei have the same Z. ES has a minimum at r 0.106 nm. This tells us that the H 2 H 2 Up Up Up Up Up > V e, y g r e n E 40 30 20 10 0 \u201310 \u201313.6 \u201320 \u201330 \u201340 \u201350 \u201360 Up Up+ EA 0.1 Up+ ES Separation 0.2 0.3 r, nm 0.106 nm EA 2.7 eV ES Figure 9-7 Dependence of the molecular potential energy on the separation of the protons and on the symmetry of the electron wave function for the system. H 2 9-2 The Covalent Bond 373. H2 H2 Molecule Formation of the H2 molecule is very similar to that of We can think of it as two H atoms in their ground states, initially far apart. Each has a 1s electronic orbital,1 i.e., with an energy of 13.6 eV. an electron the space part of whose wave function is Thus, the total energy of the H2 system for large r (i.e., r S ) is 27.2 eV. As the two atoms approach each other, the wave functions begin to overlap, again", " as illustrated by Figure 9-6a and b, so that the two atoms (protons) share both electrons. Just as was discussed above, the two wave functions may add to produce a symmetric total wave function that results in a stable bound H2 molecule or an antisymmetric one & which does not lead to a stable molecule. Since the total wave function must always be antisymmetric to an exchange of the electrons, the space wave function S( RnYm) & (see Section 7-6). Thus, must be associated with an antisymmetric spin function ( A is a triplet state (S 1). is a singlet state (S 0) and 100, & S A, & & There is a difference between the H2 molecule and the H 2 molecule that needs explanation. Just as H 2, the H2 molecule has two molecular states whose total energy at large r is, as we have seen, 27.2 eV. As r gets smaller, the molecule still has two states, but their energies separate, as sketched in Figure 9-8a. The lower energy ES is, & S, the electronic wave function of the stable molecule, as before, associated with known also as the bonding orbital. The wave function associated with the energy EA that does not result in bonding is also called the antibonding orbital. The difference is that there are now two electrons whose probability density is large in the region between the protons, both in the molecular orbital. Since electrons obey the exclusion principle, their spins must be antiparallel (S 0). Thus, a molecular orbital, just like an atomic orbital, can be occupied by no more than two electrons. For H2 both electrons can occupy the bonding orbital. Both electrons being in s states, H2 is referred to as being s-bonded. & A & S & A & S Figure 9-8b illustrates the potential energy functions for H2. The energy the bonding orbital, has a minimum of E 31.7 eV at 0.074 nm\u2014and the binding corresponding to r 0.074 nm\u2014i.e., the equilibrium separation r0 & S, (a) (b27.2 Up + EA Up + ES r 10 0 \u201310 \u201320 \u201327.2 \u201330 \u201340 0.1 0.2 Separation r, nm 0.074 nm Antibonding (Up + EA) 4.5 eV Bonding (Up + ES) Figure 9-8 (a) The two levels of the H2 system, which have the same energy for r S, have different energies as the atoms E S is for the symmetric (bonding) approach each other. (b) Potential energy versus separation for two hydrogen atoms. Up EA is for the antisymmetric (antibonding) space wave function. As the separation approaches space wave function, and Up zero, both curves approach. 374 Chapter 9 Molecular Structure and Spectra H2 energy is E(r S ) E(r0) 27.2 (31.7) 4.5 eV. The effect of adding the second electron to to form H2 is evident from a comparison of Figure 9-7 and Figure 9-8b. The increased charge concentration between the protons binds them more tightly, the binding energy increasing from 2.7 eV to 4.5 eV and the equilibrium separation decreasing by 30 percent. The sharing of the outer, or valence, electrons in a molecule, as in our H2 example, is the mechanism of the covalent molecular bond. The basic requirement for covalent bonding is that the wave functions of the valence electrons of the participating atoms overlap as much as possible. Unlike the H 2 case, the covalent bond is just as strong for nonidentical nuclei as it is for identical nuclei.2 We can now see why three H atoms do not bond to form H3. If a third H atom is brought near an H2 molecule, the third electron cannot be in a 1s state and have its spin antiparallel to both the other electrons. It must therefore occupy the higherenergy, antibonding orbital. If it is in an antisymmetric state with respect to exchange with one of the electrons, the repulsion of this atom is greater than the attraction of the other. Thus, as the three atoms are pushed together, the third electron is in effect forced into a higher quantum state by the exclusion principle. The bond between two H atoms is called a saturated bond because there is no room for another electron. The two electrons being shared essentially fill the 1s states of both atoms. This is basically the reason why covalent bonds involving three (or more) electrons are typically unstable. However, be aware that the H 3 ion is stable. Discovered by J. J. Thomson in 1911, this simplest of all polyatomic molecules provides important cosmic spectral lines for astrophysicists and", " a calculation benchmark for quantum chemists. It should also be clear now why He atoms do not bond together to form He2. There are no valence electrons that can be shared. As two He atoms approach each other, the bonding and antibonding molecular orbitals form, just as they do for H2; however, each orbital can accommodate only two electrons (with spins antiparallel), so two of the four electrons in the He2 system cannot remain in the 1s atomic states but must be in the antibonding orbital. The net effect is that He2 does not form as a stable bond. At low temperatures or high pressures, He atoms do bond together, but the bonds are very weak and are due to van der Waals forces, which we will discuss in Section 9-3. The bonding is so weak that at atmospheric pressure, He boils at 4.2 K, and it does not form a solid at any temperature unless the pressure is greater than about 20 atm. Covalent or Ionic? When two identical atoms bond, as in homonuclear diatomic molecules such as O2 or N2, the bonding is purely covalent. Since the wave functions of the two atoms are exactly alike, neither atom dominates and the electrons are completely shared between them. However, the bonding of two dissimilar atoms is often a mixture of covalent and ionic bonding. Even in NaCl, the electron donated by sodium to chlorine has some probability of being at the sodium atom because its wave function does not suddenly fall to zero. Thus, this electron is partially shared in a covalent bond, although this bonding is only a small part of the total bond, which is mainly ionic. A measure of the degree to which a bond is ionic or covalent can be obtained from the electric dipole moment of the molecule. For example, if the bonding in NaCl were purely ionic, the center of positive charge would be at the Na ion and the center of negative charge would be at the Cl ion. The electric dipole moment would have the magnitude pionic er0 9-3 H 3 ion is! Although the H3 molecule is not bound, the Discovered by J. J. Thomson in 1911, and lacking a stable excited state, is used as a probe in Jupiter\u2019s atmosphere and serves as the benchmark for quantum chemistry calculations for polyatomic molecules. H 3 Covalently bonded fullerene molecules have been assembled into nanotubes, i.e., tubes with diameters in the nanometer range. Adding a few impurity atoms per molecule turns it into a superconductor. (See Section 10-8.) 9-3 Other Bonding Mechanisms 375 where r0 is the equilibrium separation of the ions. Thus, the dipole moment of NaCl would be pionic er0 (1.60 1019 C)(2.36 1010 m) 3.78 1029 C # m The actual measured electric dipole moment of NaCl is 3.00 1029 C # m pmeasured A purely covalent molecule would be expected to have an electric dipole moment of zero. We can define the ratio of pmeasured to pionic as the fractional amount of ionic bonding. For NaCl, this ratio is 3.00/3.78 0.79. Thus, the bonding in NaCl is about 79 percent ionic and 21 percent covalent. EXAMPLE 9-3 Bonding in LiH The measured electric dipole moment of LiH is 1.96 1029 C m. This molecule is among those listed in Table 9-3 (on the home page) as being covalently s-bonded. What portion of the LiH bond is covalent? # SOLUTION The equilibrium separation of LiH from the table is 0.159 nm. If it were a purely ionically bonded molecule, its dipole moment pionic would be (1.60 1019 C)(0.159 109 m) 2.54 1029 C # m The fractional amount of the bond that is ionic is 1.96/2.54 0.77. Thus, LiH is only about 23 percent covalently s-bonded. pionic More In addition to the s-bonded H2 molecule, there are many other covalently bonded molecules involving shared pairs of s electrons, s and p electrons, and p electrons. Important among these are the s-p bonds involving carbon that are the basis for the vast array of hydrocarbon molecules and compounds. Several examples, including the remarkable fullerenes, are discussed in Other Covalent Bonds on the home page: www.whfreeman.com/tiplermodernphysics5e See also Equations 9-4 and 9-5 here, as well as Tables 9-3 through 9-6 and Figures 9-9 through 9-17. EXPLORING 9-3 Other Bonding Mechanisms", " The two bonding mechanisms that we have discussed thus far, ionic and covalent, account for a large fraction of the cases in which atoms combine to form molecules. As is described in Chapter 10, when atoms combine on a larger scale to form solids, these exact same mechanisms are responsible for the bonding in many solids. In addition to these types of bonding, two other types occur in solids. One of these, molecular bonding, or dipole-dipole bonding, also occurs in the formation of many large molecules from smaller molecules and will be discussed in this section. The second type, metallic bonding, is responsible for the structure of metals in the solid state and has no single-molecule version or counterpart. For that reason, our discussion of metallic bonding will be deferred to Chapter 10. 376 Chapter 9 Molecular Structure and Spectra Dipole-Dipole Bonding It was first suggested by J. D. van der Waals5 in 1873 that any two separated molecules will be attracted toward each other by electrostatic forces. Similarly, atoms that do not otherwise form ionic or covalent bonds will be attracted to each other by the same sort of weak electrostatic bonds. The practical result of this is that at temperatures low enough so that the disruptive effects of thermal agitation are negligible, all substances will condense into a liquid and then a solid form. (Recall that helium is the only element that does not solidify at any temperature under its own vapor pressure.) The relatively weak electrostatic forces responsible for this sort of intermolecular attraction arise because of the electrostatic attraction of electric dipoles. The electric field due to an electric dipole is illustrated in Figure 9-18a. The electric field Ed at point A due to the dipole is given by 3(p # r) r5 k c p r3 9-6 Ed r d whose magnitude for r W a is Ed kqa r3 kp1 r3 9-7 \u0192p1 \u0192 qa is the dipole moment.6 Thus, the electric field of the dipole, and hence where > the electric force on a charge, falls off as 1 r3. This result, which is correct even if the point A is not on the perpendicular, is to be compared with the 1 r2 dependence of the Coulomb force that occurs in the covalent and ionic bonds: the force on a test charge due to the dipole qa is weaker at a distance r than that due to a charge q. A second dipole p2 that happens into the vicinity of p1 will then orient itself along the Ed field lines, as illustrated in Figure 9-18b, as a result of the electric force on the charges. > The potential energy of the second dipole p2 in the field of p1 is given by # Ed U p2, the electric force 9-8 and, since Ed falls off like nent dipoles falls off as of short range. 1>r3 F ( $U>$r) between two perma. Thus, it is attractive (F is negative), relatively weak, and 1>r4 Polar Molecules It is then not hard to see physically why molecules with permanent electric dipole moments\u2014so-called polar molecules such as H2O and NaCl\u2014will attract other polar molecules. Consider the H2O molecule as an example. Although the molecule is electrically neutral, its bonding is partially ionic, so the electrons tend to be concentrated nearer the oxygen atom, making it look like the negative end of a dipole. The two protons then look like the positive end of the dipole. There will then be a mutual (b) (a) +q a \u2013q r E\u2013 A E+ Ed + \u2013 p1 \u2013 + p2 Figure 9-18 (a) The electric field Ed at a point A on a line perpendicular to the axis of an qa. (b) The field of Ed acts on a second dipole p2 to orient it along the electric dipole p1 field lines. The force on a charge due to p1 is. 1>r3 9-3 Other Bonding Mechanisms 377 (a) H + p (b) + \u2013 (c) O \u2013\u2013 + \u2013 + + \u2013 + \u2013 Figure 9-19 (a) Schematic of four H2O molecules. The water molecules\u2019 permanent dipole moments are shown by the vectors p. (b) The four polar water molecules represented as electric dipoles. Notice that the attractive dipole-dipole force tends to align the dipoles so that the nearest neighbors of each charge are charges of the opposite sign. (c) A snowflake\u2014one result of dipole-dipole bonding. attraction between the molecule and other nearby molecules with potential energy given by Equation 9-8. (See Figure 9-19.) Pairs of polar molecules will thus move closer to each other", ", decreasing their potential energy, until the combined effects of the increasing nuclear repulsion and the exclusion principle produce a minimum in the total potential energy similar to the minima in Figures 9-2 and 9-8b. For H2O the resulting bonding energy is about 0.5 eV per molecule. Although this is only about 10 percent of the strength of the bond in the water molecule, it is this dipole-dipole force that bonds H2O molecules to each other to form ice and is responsible in part for the beautiful hexagonal patterns that we see in snowflakes (Figure 9-19c). H \u00ac OH When dipole-dipole bonds between molecules with permanent dipole moments involve hydrogen, as is the case for water, the bond is referred to as a hydrogen bond. The hydrogen bond is of enormous importance since it is the bonding mechanism responsible for the cross-linking that allows giant biological molecules and polymers to hold their fixed shape. For example, it is the hydrogen bond that forms the linkage between the two strands of the double helix DNA molecule. It is the weakness of the hydrogen bonds relative to the covalent/ionic bonds along each strand that allows the two strands to unwind from each other in the DNA molecular replication process. Notice that the hydrogen bond can be viewed as the sharing of a proton by two negatively charged atoms, oxygen atoms in the case of water. (See Figure 9-19.) In this way it is similar to the sharing of electrons that is responsible for the covalent bond. Hydrogen bonding is facilitated by the small mass of the proton and the absence of inner-core electrons. Nonpolar Molecules A nonpolar molecule will be polarized by the field of a polar molecule and thus have an induced dipole moment and be attracted to the polar molecule. If p2 in Figure 9-18b is an induced dipole, then p2 Ed 9-9 is a constant characteristic of the nonpolar molecule called the polarizability. since we where In this case we expect the potential energy of the interaction to fall off as have from Equations 9-8 and 9-9 that 1>r6 U p2 E 2 d k2p2 1 >r6 9-10 # Ed Once again, the energy is negative, signifying that the force between the dipoles is attracis thus proportional to 1 r7, i.e., the force is very short range, tive. The force dropping rapidly with increasing r. Indeed, increasing the separation of the molecules by a factor of 2 reduces the attractive force between them to only 0.008 of its original value. F $U>$r > 378 Chapter 9 Molecular Structure and Spectra p p2 Perhaps surprisingly, two molecules, neither of which has a permanent dipole moment, can also attract each other via the mechanism just described. It is somewhat harder to see why an attractive force exists between two nonpolar molecules. Though the average dipole moment of a nonpolar molecule is zero, the average square dipole moment is not because the electrons are in constant motion and at any given instant there will be an excess or deficiency of them in one part or another of the molecule. A measurement that we might do in the laboratory reveals the average value (zero), not the instantaneous value. The instantaneous dipole moment of a nonpolar molecule is, in general, not zero. When two nonpolar molecules are nearby, the fluctuations in the instantaneous dipole moments tend to be correlated so as to produce attraction, as illustrated in Figure 9-20. The potential energy is again given approximately by Equation 9-10, so that the potential energy is proportional to and the attractive force is pro. This attractive force between nonpolar molecules is called the van der portional to Waals force7 or, occasionally, the London dispersion force, after Fritz London, the German physicist who in 1930 first explained the physical origin of the interaction. 1>r7 1>r6 Scientists have recently succeeded in trapping a sample of molecules in a single quantum level at a temperature in the millikelvin range. This ability raises the possibilities for, among other things, high-precision molecular spectroscopy and producing a molecular Bose-Einstein condensate. As van der Waals first suggested, dipole-dipole forces act between all molecules and, in addition, between all atoms. They are the only forces that occur between rare gas atoms, without which the atoms of these elements would not condense into liquids or form solids. (The single exception to the latter is He, whose quantum-mechanical zeropoint energy exceeds the minimum of the potential energy resulting from Equation 9-10 and core repulsion.) The dipole-dipole forces between molecules, although relatively weak, are also responsible for the physical phenomena of surface tension and friction. \u2013 + + \u2013 +\u2013 \u2013 + + \u2013 +\u2013 \u2013 + \u2013", " + +\u2013 \u2013 \u2013 \u2013 ++ + + \u2013 + \u2013 \u2013+ r \u2013 + \u2013 + +\u2013 + ++ \u2013 \u2013 \u2013 + \u2013 + \u2013 \u2013+ Figure 9-20 Nonpolar molecules have, on average, symmetric charge distributions, as illustrated by the pair of molecules at the top of the figure. However, instantaneous fluctuations in the electron distribution are asymmetric and tend to be correlated with those of nearby molecules, as shown in the other three examples. The correlated distributions lead to an attractive force > proportional to 1 r7 that draws the molecules closer to each other as shown. The 2-nm height of DNA molecules is readily imaged by an atomic force microscope (AFM). [Taken from www.di.com, Digital Instruments, Veeco metrology Group, Santa Barbara, CA.] 9-4 Energy Levels and Spectra of Diatomic Molecules 379 Questions 1. Why would you expect the separation distance between the two protons to be larger in the H 2 ion than in the H2 molecule? 2. Would you expect the NaCl molecule to be polar or nonpolar? 3. Would you expect the N2 molecule to be polar or nonpolar? 4. Does neon occur naturally as Ne or Ne2? Why? 9-4 Energy Levels and Spectra of Diatomic Molecules As is the case with an atom, a molecule often emits electromagnetic radiation when it makes a transition from an excited energy state to a state of lower energy. Conversely, a molecule can absorb radiation and make a transition from a lower energy state to a higher energy state. The study of molecular emission and absorption spectra thus provides us with information about the energy states of molecules. For simplicity, we will consider only diatomic molecules here. As might be expected, the energy levels of molecular systems are even more complex than those of atoms. The energy of a molecule can be conveniently separated into three parts: electronic, due to the excitation of its electrons; vibrational, due to the oscillations of the atoms of the molecule; and rotational, due to the rotation of the molecule about an axis through its center of mass. Fortunately, the magnitudes of these energies are sufficiently different that they can be treated separately. Electrons in molecules can be excited to higher states, just as those in atoms. For example, a 1s electron in the H2 molecule can be excited to a 2p level, emitting a photon as it returns to the ground state. The energies due to the electronic excitations of a molecule are of the order of magnitude of 1 eV, the same as for the excitation of atoms. We have already discussed such transitions and will not consider them further in this section. The energies of vibration and rotation are about times smaller and will be the focus of our attention. 1>1000 1>100 to Rotational Energy Levels Classically, the kinetic energy of rotation is E 1 2 I2 (I)2 2I L2 2I 9-11 where I is the moment of inertia, the angular momentum. The solution of the Schr\u00f6dinger equation for the rotation of a rigid body leads to the quantization of the angular momentum, with values given by the angular velocity of rotation, and L I L2 ( 1)U2 0, 1, 2, \u00c1 9-12 where / is the rotational quantum number. This is the same quantum condition on angular momentum that holds for the orbital angular momentum of an electron in an atom. 380 Chapter 9 Molecular Structure and Spectra Energy 0 = 3 = 2 = 1 = 0 Figure 9-21 Energy levels and allowed transitions for a rotating rigid body as given by Equation 9-13. r1 r2 m1 CM r0 m2 Figure 9-22 Diatomic molecule rotating about an axis through its center of mass. Note, however, that L in Equation 9-11 refers to the angular momentum of the entire molecule rotating about an axis through its center of mass. The energy levels of a rotating molecule are therefore given by E ( 1)U2 2I ( 1)E0r 0, 1, 2, \u00c1 9-13 where E0r is the characteristic rotational energy of a particular molecule, which is inversely proportional to its moment of inertia: E0r U2 2I 9-14 The rotational-energy level scheme is shown in Figure 9-21. Transitions between these levels produce the pure rotational spectrum of a molecule. While all diatomic molecules have rotational energy levels, those without permanent dipole moments (symmetric molecules such as H2, Cl2, or CO2) cannot emit or absorb electric dipole radiation by only changing the rotational quantum state and thus do not have a pure rotational spectrum. For molecules that do have dipole moments and emit pure rotational spectra, the quantum number / is subject to the selection rule / 1, just as it was for the atomic electrons. Thus, the energy separation between adjacent rotation states is", " given by \u00a2E, 1 [( 1)( 2) ( 1)]U2 2I ( 1)U2 I 9-15 A measurement of the rotational energy of a molecule from its rotational spectrum can be used to determine the moment of inertia of the molecule, which can then be used to find the equilibrium separation of the atoms in the molecule, i.e., the bond length. The moment of inertia about an axis through the center of mass of a diatomic molecule (see Figure 9-22) is I m1r2 1 m2r2 2 Using m1r1 of mass, and r0 inertia as m2r2, which relates the distances r1 and r2 from the atoms to the center r2 for the separation of the atoms, we can write the moment of r1 where, the reduced mass, is I r2 0 m1m2 m2 m2), as in H2 and O2, the reduced mass m1 If the masses are equal (m1 I 1 2 mr2 0 9-16 9-17 m> 2 and 9-18 A unit of mass convenient for discussing atomic and molecular masses is the unified mass unit u, which is defined as one-twelfth of the mass of a neutral carbon-12 (12C) atom. 9-4 Energy Levels and Spectra of Diatomic Molecules 381 The mass of one 12C atom is thus 12 u. The mass of an atom in unified mass units is therefore numerically equal to the molar mass of the atom in grams. The unified mass unit is related to the gram and kilogram by 1 u 1g NA 103 kg 6.0221 1023 1.6605 1027 kg 931.4940 106 eV>c2 9-19 where NA is Avogadro\u2019s number. EXAMPLE 9-4 The Reduced Mass of HCl Compute the reduced mass of the HCl molecule. SOLUTION 1. The reduced mass is given by Equation 9-17: m1m2 m2 m1 2. From the periodic table on the inside back cover of this book, the mass of the hydrogen atom is 1.01 u and that of the chlorine atom is 35.5 u. Substituting these gives (1.01 u)(35.5 u) 1.01 u 35.5 u 0.98 u Remarks: Note that the reduced mass of the HCl molecule is less than that of a single hydrogen atom. EXAMPLE 9-5 Equilibrium Separation in CO The energy difference E between the / 0 and / 1 rotational levels in the CO molecule is found experimentally from measurement of the wavelength of the corresponding transition. For CO, E is equal to 4.77 104 eV. Find the equilibrium separation, or bond length r0, of the CO molecule. 2.6 mm SOLUTION 1. The bond length r0 is given in terms of the moment of inertia I of the molecule by Equation 9-16: 2. I in terms of E is given by Equation 9-15: 3. Substituting / 0 and E 4.77 104 eV into step 2 gives 4. The reduced mass of the CO molecule is computed from Equation 9-17 using atomic mass values from the periodic table: I r2 0 or r0 I A \u00a2E, 1 ( 1)U2 I or I ( 1)U2 \u00a2E, 1 I U2 4.77 104 eV m1m2 m2 m1 6.86 u (12 u)(16 u) 12 u 16 u 382 Chapter 9 Molecular Structure and Spectra 5. Substituting these results into step 1 gives a r0 U2 4.77 104 eV 6.86 u b 1>2 1.055 1034 [(4.77 104 eV)(1.60 1019 J>eV)(6.86 u)(1.66 1027 kg>u)]1>2 0.133 nm The rotational energy levels are several orders of magnitude smaller than those due to electron excitation, which have energies of the order of 1 eV or higher. For example, the characteristic rotational energy of the O2 molecule, whose equilibrium separation is about 0.1 nm, is 2.59 104 eV, calculated from Equation 9-14. Transitions within a given set of rotational energy levels yield photons in the far infrared region of the electromagnetic spectrum. Notice that the rotational energies are also small compared with the typical thermal energy kT at normal temperatures. For T 300 K, for example, kT is about 2.6 102 eV. Thus, at ordinary temperatures, a molecule can easily be excited to the lower rotational energy levels by collisions with other molecules. But such collisions cannot excite the molecule to electronic energy levels above the ground state. Vibrational Energy Levels The molecular vibrational energies are a bit harder", " to estimate than were the rotational energies. Our discussion is aided by the fact that the molecular potential energy functions of Figures 9-2, 9-7, and 9-8b can be closely approximated by parabolas in the vicinity of the equilibrium point. (See Figure 9-23b.) Thus, we can use the (a) Energy (b) 0 0.1 0.2 0.3 0.4 0.5 Separation distance, nm hf hf hf hf hf 1 \u20132 \u20133 \u20134 \u20135 Figure 9-23 (a) The energy levels of the molecular vibrations are equally spaced in the vicinity of the equilibrium spacing of the atoms. (b) A harmonic oscillator potential fitted to the actual potential energy function of the NaCl molecule shown in Figure 9-2b. 9-4 Energy Levels and Spectra of Diatomic Molecules 383 results of our study of the simple harmonic oscillator in Chapter 6. The energy levels are given by ( 1> 2)hf Ev 0, 1, 2, 3, \u00c1 9-20 is the vibrational quantum number.8 where f is the frequency of the vibration and An interesting feature of this result is that the energy levels are equally spaced with as shown in Figure 9-23a. The frequency of vibration of a intervals diatomic molecule can be related to the force exerted by one atom on the other. Consider two objects of mass m1 and m2 connected by a spring of force constant K. The frequency of oscillation of this system is (see Section 6-5) \u00a2E, 1 hf, f 1 2A K 9-21 where is the reduced mass given by Equation 9-17. The effective force constant of a diatomic molecule can thus be determined from a measurement of the frequency of oscillation of the molecule. We could get a good estimate of f by fitting the one-dimensional parabolic harmonic oscillator potential energy function for the molecule as illustrated in Figure 9-23b, but for simplicity we can get a rough idea of the order of magnitude of the vibrational energies by observing that the energy of an atom of mass m in a square well of width r0 is (Figure 9-24) En n2 h2 8mr2 0 n2 42U2 8mr2 0 n2 2 U2 2 mr2 0 (and the n2), this expression is the same as the charExcept for the factor acteristic rotational energy E0r; thus we expect the vibrational energies to be somewhat larger than the rotational energies. 2>2 5 Square well approximation Simple harmonic oscillator approximation r Molecular potential Figure 9-24 Molecular potential. The simple harmonic oscillator approximation, used to calculate the energy levels, and a square well approximation, used to estimate the order of magnitude of the energy levels, are each indicated by dashed curves. 384 Chapter 9 Molecular Structure and Spectra The selection rule for transitions between vibrational states (of the same elecchange only by 1, so the energy of a photon emitted tronic state) requires that by such a transition is hf and the frequency is the same as the frequency of vibration. A typical measured frequency of a transition between vibrational states is 5 1013 Hz, which gives for the order of magnitude of vibrational energies E hf (4.14 1015 eV # s)(5 1013 s1) 0.2 eV Thus, a typical vibrational energy is actually about 1000 times greater than the typical rotational energy E0r of the O2 molecule we noted above and about 8 times greater than the typical thermal energy kT 0.026 eV at T 300 K. In contrast with the rotational levels, the molecular vibrational states are not readily excited by collisions between molecules at ordinary temperatures. EXAMPLE 9-6 Force Constant of CO The observed vibrational frequency of the CO molecule is 6.42 1013 Hz. What is the effective force constant for this molecule? SOLUTION 1. The force constant K is given in terms of the vibrational frequency \u0192 by Equation 9-21: f 1 2A K 2. The reduced mass of the CO molecule was 6.86 u computed in step 4 of Example 9-5: 3. Solving Equation 9-21 for K and substituting the values of \u0192 and gives K (2f)2 (2 6.42 1013 Hz)2(6.86 u)(1.66 1027 kg>u) 1.86 103 N>m Emission Spectra Figure 9-25 shows schematically some electronic, vibrational, and rotational energy levels of a diatomic molecule. The vibrational levels are labeled with the quantum number v and the rotational levels with the quantum number /. The lower vibrational levels are evenly spaced, with E h\u0192. For higher vibrational levels, the approximation that the vibration is simple harmonic is not valid and the levels", " are not quite evenly spaced. The actual potential spreads somewhat more rapidly, as can be seen in Figure 9-24, and the spacing of the vibrational levels becomes closer for large values of the quantum number v. Notice in Figure 9-25 that the potential energy curves representing the force between the two atoms in the molecule do not have exactly the same shape for the electronic ground and excited states. This implies that the fundamental frequency of vibration f is different for different electronic states. For transitions between vibrational states of different electronic states, the selection rule v 1 does not hold. Such transitions result in the emission of photons of wavelength in or near the visible spectrum. Energy Excited state Ground state 0 r \u03bd = 1 \u03bd = 0 9-4 Energy Levels and Spectra of Diatomic Molecules 385 Figure 9-25 Electronic, vibrational, and rotational energy levels of a diatomic molecule. The rotational levels are shown in an enlargement of the v 0 and v 1 vibrational levels of the electronic ground state Rotational levels The spacing of the rotational levels increases with increasing values of /. Since the energies of rotation are so much smaller than those of vibrational or electronic excitations of a molecule, molecular rotation shows up in molecular spectra as a fine-structure splitting of the spectral lines. When the fine structure is not resolved, the spectrum appears as bands, as shown in Figure 9-26a (page 386). Close inspection of these bands reveals that they have a fine structure due to the rotational energy levels, as shown in the enlargement in Figure 9-26b. Absorption Spectra Much molecular spectroscopy is done using infrared absorption techniques in which only the vibrational and rotational energy levels of the ground-state electronic level are excited. Consequently, we will now direct our attention to what is called the vibration-rotation spectrum. For ordinary temperatures, the vibrational energies are sufficiently large in comparison with the thermal energy kT that most of the molecules are in the lowest vibrational state v 0, for which the energy is The transition from v 0 to v 1 is the predominant transition in absorption. The rotational energies, however, are sufficiently smaller than kT that the molecules are distributed among several rotational energy states, the relative number in each state being determined by the Boltzmann factor. If the molecule is originally in a rotational state characterized by the quantum number /, its initial energy, in addition to that of the electronic state, is 2 hf. 1 E0 E 1 2 hf ( 1)E0r 9-22 where E0r is given by Equation 9-14. From this state, two transitions are permitted by the selection rules. For a transition to the next highest vibrational state v 1 and a rotational state characterized by / 1, the final energy is E1 3 2 hf ( 1)( 2)E0r 9-23 386 Chapter 9 Molecular Structure and Spectra a) 1 2 3 Upper electronic state Lower electronic state (b) 380.5 nm Figure 9-26 Part of the emission spectrum of N2. (a) These components of the band are due to transitions between the vibrational levels of two electronic states, as indicated in the diagram. (b) An enlargement of part of (a) shows that the apparent lines in (a) are in fact band heads with structure caused by rotational levels. [Courtesy of J. A. Marquisee.] For a transition to the next highest vibrational state and to a rotational state characterized by / 1, the final energy is E1 3 2 hf ( 1)E0r The energy differences are \u00a2E S 1 E1 E hf 2( 1)E0r where / 0, 1, 2,..., and 9-24 9-25 9-1 Relativity 387 \u00a2E S 1 E1 E hf 2E0r 9-26 where / 1, 2, 3,.... (In Equation 9-26, / begins at / 1 because from / 0 only the transition / S / 1 is possible.) Figure 9-27 illustrates these transitions. The frequencies of these transitions are given by f S 1 \u00a2E S 1 h f 2( 1)E0r h 0, 1, 2, \u00c1 9-27 and f S 1 \u00a2E S 1 h f 2E0r h 1, 2, 3, \u00c1 9-28 > > The frequencies for the transitions / / 1 are thus, f 2(E0r h), f 4(E0r h), f 6(E0r h), and so forth; those corresponding to the transition / S / 1 are f 2(E0r h), f 4(E0r h), f 6(E0r h), and so forth. We thus expect the absorption > spectrum to contain frequencies equally spaced by 2E0r h except for a gap of 4", "E0r h at the vibrational frequency f, as shown in Figure 9-28. A measurement of the position of the gap gives f and a measurement of the spacing of the absorption peaks gives E0r, which is inversely proportional to the moment of inertia of the molecule Energy hf 20E0r 12E0r 6E0r 2E0r 0 \u2192 + 1 \u2192 \u2013 1 Figure 9-27 Absorptive transitions between the lowest vibrational states v 0 and v 1 in a diatomic molecule. These transitions obey the selection rule / 1 and fall into two bands. The energies of the / S / 1 band are hf 2E0r, hf 4E0r, hf 6E0r, and so forth, whereas the energies of the / S / 1 band are hf 2E0r, hf 4E0r, hf 6E0r, and so forth. 388 Chapter 9 Molecular Structure and Spectra \u2192 \u2013 1 \u2192 + 1 2E0r = 3 to = 2 = 2 to = 1 = 1 to = 0 hf E = 0 to = 1 = 1 to = 2 = 2 to = 3 Figure 9-28 Expected absorption spectrum of a diatomic molecule. The right branch corresponds to the transitions / S / 1 and the left branch to the transitions / S / 1. The lines are equally spaced by 2E0r. The energy midway between the branches is hf, where f is the frequency of vibration of the molecule. Figure 9-29 shows the absorption spectrum of HCl. The double-peak structure results from the fact that chlorine occurs naturally in two isotopes, 35Cl and 37Cl, which results in slightly different moments of inertia. If all of the rotational levels were equally populated initially, we would expect the intensities of each absorption line to be equal. However, the population n(E/) of a rotational level / is proportional to the density of states g(E/), which equals the degeneracy of the level in this case, that is, the number of states with the same value of /, which is 2/ 1, and to the Boltzmann factor where E/ is the energy of the state: eE >kT, n(E) g(E)eE >kT n(E) (2 1)e[1 2 hf(1)E0r]kT 9-29 9-30 \u2192 \u2013 or = 7 = 10 = 8 8.00 8.20 8.40 8.60 8.80 9.00 \u00d71013 9.20 Frequency, Hz Figure 9-29 Absorption spectrum of the diatomic molecule HCl. The double-peak structure results from the two isotopes of chlorine, 35Cl (abundance 75.5 percent) and 37Cl (abundance 24.5 percent). The intensities of the peaks vary because the population of the initial state depends on /. 9-4 Energy Levels and Spectra of Diatomic Molecules 389 U (2 1) -fold degeneracy of the rotational state with angular momentum / makes The [( 1)U2>2 IkT]. the thermal equilibrium population proportional to (2/ 1) exp Therefore, the / 0 state is usually not the most densely populated state at room temperature. For low values of /, the population increases slightly because of the degeneracy factor, whereas for higher values of /, the population decreases because of the Boltzmann factor. The intensities of the absorption lines therefore increase with / for low values of / and then decrease with / for high values of /, as can be seen from the figure. We can find out where the maximum population of the rotational states is located and hence which lines will be the most intense by differentiating Equation 9-30 with respect to / and setting dn d/ equal to zero. The result is > max 1 2 c A 4kT h2>mr2 1 d 9-31 For a measurement made at room temperature, kT 0.026 eV and thus / too, can be seen in Figure 9-29. max L 3. This, Notice also in Figure 9-29 that the spacing between adjacent peaks, which we expected to be constant and equal to 2E0r on the basis of our calculation above, is in fact not constant. The reason for this is our assumption that the moment of inertia of the molecule is constant. The rotation of the molecule tends to increase the separation of the atoms and hence increase the moment of inertia and decrease the rotational energy. As might be expected and the figure also shows, this effect becomes larger as / increases. As mentioned above, the gap in the spectrum in Figure 9-29 is due to the absence of a transition beginning on the / 0 level in the / S / 1 group of peaks. The center of the gap is at the characteristic oscillation frequency f of the molecule given by Equ", "ation 9-21. From the figure we see that f for HCl is about 8.56 1013 Hz, or about 0.36 eV, corresponding to a force constant K of about 476 N/m. Table 9-7 lists the rotational and vibrational constants for several diatomic molecules. All diatomic Table 9-7 Rotational and vibrational constants for selected diatomic molecules Molecule Equilibrium separation r0 (nm) Frequency \u0192 (Hz) E0r (eV) H2 Li2 O2 LiH HCl35 NaCl35 KCl35 KBr79 0.074 0.267 0.121 0.160 0.127 0.251 0.279 0.294 1.32 1014 1.05 1013 4.74 1013 4.22 1013 8.97 1013 1.14 1013 8.40 1012 6.93 1012 7.56 103 8.39 105 1.78 104 9.27 104 1.32 103 2.36 105 1.43 105 9.1 106 Symmetric molecules such as H2 or O2 have no electric dipole moment. The vibration or rotation of these molecules does not involve a changing dipole moment, and there is no vibrationalrotational electric dipole absorption or radiation for these molecules. 390 Chapter 9 Molecular Structure and Spectra molecules 9 have a gap at f in their vibration-rotation spectra; however, many polyatomic molecules have more complex vibrations and rotations, one result of which is that / 0 may be allowed, i.e., vibrational energy may change without an accompanying rotational transition. In that event, a line will occur in the vibrationrotation spectrum at the frequency f. Such lines are given the rather enigmatic name of Q branch. 10 9-5 Scattering, Absorption, and Stimulated Emission Scattering In the interactions between radiation incident on atomic or molecular systems, photons may also be scattered both elastically and inelastically. The process by which photons scatter elastically, i.e., without a change in their frequency, is called elastic or Rayleigh scattering since it was first described adequately by a classical scattering theory derived by Rayleigh in about 1900. Rayleigh scattering is illustrated in Figure 9-30d. In the classical theory, the oscillating electric field of the incident radiation produces an oscillating acceleration of the atomic electrons, causing them to radiate electromagnetic waves of the same frequency as and in phase with the incident wave. Thus, the electrons of the target atoms and molecules absorb energy from the incident wave and re-emit, or scatter it, in all directions without changing its frequency. The intensity of the scattered radiation is proportional to f 4. Rayleigh scattering is the origin of the unmodified line in our discussion of the Compton scattering of x rays in Section 3-4. (See Figure 3-17.) We saw there that if the incident wavelength was 2\u2014 1, large compared with the Compton shift i.e., visible wavelengths or larger, then the scattered wave always included a wavelength equal to the incident wavelength to within experimental accuracy regardless of whether the electron mass or atomic mass is used in Equation 3-25. So as the quantum explanation of Chapter 3 and Rayleigh\u2019s classical explanation of elastic scattering agree. However, for incident wavelengths in the x-ray and gamma-ray regions of the spectrum, Compton scattering, shown in Figure 9-30e, becomes increasingly important for lowZ atoms whose electron binding energies are not large. In the gamma-ray region as S 0, the photon energy becomes so large that even the most tightly bound electrons are freed in the process and Compton effect becomes the dominant process. S, 1 The incident and scattered photons are also correlated in the inelastic scattering process illustrated in Figure 9-30f. Such scattering of light from molecules was first observed by physicist C. V. Raman11 and is known as Raman scattering, or sometimes as the Raman effect. The scattered photon may have less energy than the incident photon or it may have greater energy if the molecule is initially in an excited vibrational or rotational energy state. Both possibilities are illustrated in Figure 9-30f. Thus, the scattered frequency is not the same as the incident frequency, nor is it related to a characteristic frequency of the molecule. It is found that for incident monochromatic radiation of frequency f the scattered radiation contains not only the frequency f (Rayleigh scattering; see Figure 9-31), but also much weaker lines on either side of the Rayleigh line, with frequencies given by f f \u00a2f 9-32 9-5 Scattering, Absorption, and Stimulated Emission 391 These are the Raman lines illustrated in Figure 9-31. If the incident frequency is varied, the Raman lines are observed to move along the frequency axis at the same rate so that the difference f between f and f remains constant. It is this difference f that", " corresponds to characteristic transitions of the scattering molecule. (b) hf\u00b4 hf\u00b4\u00b4 hf\u00b4\u00b4\u00b4 hf hf Fluorescence hf Resonance absorption (d) (f) hf hf Rayleigh scattering 3 2 1 0 l Raman scattering hf \u2013 \u0394E01 = hf\u00b4 hf + \u0394E01 = hf\u00b4 Stimulated emission Compton effect (a) (c) (e) (g) Photoelectric effect Figure 9-30 Description of photon interactions with an atom. (a) The photon is absorbed and the atom, in an excited state, later emits one or more photons as it decays to a state of lower energy. This is a two-step process called fluorescence, and the emitted photons are uncorrelated with the incident photon. (b) If the energy of the incident photon matches one of the excitation energies of the atom, resonance radiation results. (c) The atom, in an excited state, is stimulated to make a transition to a lower state by an incident photon of just the right energy. The emitted and incident photons have the same energy and are coherent. The Rayleigh scattering process (d) and Raman scattering (f) differ from (a) and (b) in that they are single-step processes and there is a correlation between the incident and emitted photons. Parts (e) and (g) illustrate Compton scattering and the photoelectric effect discussed in Chapter 3. 392 Chapter 9 Molecular Structure and Spectra Figure 9-31 The rotational Raman spectrum of N2. The alternating intensities, determined by the nuclear spins, are in the ratio of 1/2, corresponding with the I 1 spin of the nitrogen nucleus. The dark central area is the result of the much more intense Rayleigh-scattered portion of the incident wave. [R. Eisberg and R. Resnick, Quantum Physics, 2d ed. (New York: Wiley, 1985), p. 436.] N2 Although the measurements of f for each line in the Raman spectrum makes possible construction of the rotational levels for a given molecule,12 its quantummechanical explanation is different from that of the rotational spectrum. In particular, the selection rule for the rotational quantum number in the Raman effect is / 0, 2. The / 0 value yields Rayleigh scattering, while / 2 yields the Raman lines. One can see how this comes about physically by studying the transitions shown in Figure 9-30f. An electron initially in the / 0 state absorbs energy E01 E12 from the incident photon of frequency f and emits energy E12. Thus, the energy of the scattered photon is hf hf (\u00a2E01 \u00a2E12) \u00a2E12 hf \u00a2E01 9-33 or f f \u00a2f where \u0192 E01 h. If the electron is initially in the / 1 state, then it absorbs E12 from the incident photon and emits E12 E01. Thus, the scattered photon has energy > hf hf \u00a2E12 (\u00a2E12 \u00a2E01) hf \u00a2E01 9-34 or f f \u00a2f Many Raman spectra have been studied. They provide a valuable source of information regarding molecular quantum states, including, as was pointed out earlier, the structure of the rotational levels for homonuclear diatomic molecules. For example, the detailed understanding of the complex vibrations and rotations of the ammonia molecule referred to in Section 6-6 that enabled the development of the first atomic clocks was made possible by studies of the Raman rotational-vibrational spectrum of the NH3 molecule, the so-called ammonia inversion spectrum. Finally, Figure 9-30g illustrates the photoelectric effect, the final example of the interaction of radiation with matter, in which the absorption of the photon ionizes the atom or molecule. Like Compton scattering, this effect was discussed in Chapter 3 and will not be considered further here. 9-5 Scattering, Absorption, and Stimulated Emission 393 Absorption Information about the energy levels of an atom or molecule is usually obtained from the radiation emitted when the atom or molecule makes a transition from an excited state to a state of lower energy. As mentioned in Section 9-4, we can also obtain information about such energy levels from the absorption spectrum. When atoms and molecules are irradiated with a continuous spectrum of radiation, the transmitted radiation shows dark lines corresponding to absorption of light at discrete wavelengths. Absorption spectra of atoms were the first line spectra observed. Fraunhofer in 1817 labeled the most prominent absorption lines in the spectrum of sunlight; it is for this reason that the two intense yellow lines in the spectrum of sodium are called the Fraunhofer D lines. Since at normal temperatures atoms and molecules are in their ground states or in low-lying excited states, the absorption spectra are usually simpler than the emission spectra. For example,", " only those lines corresponding to the Lyman emission series are seen in the absorption spectrum of atomic hydrogen because nearly all the atoms are originally in their ground states. In the Chapter 6 MORE section \u201cTransitions Between Energy States,\u201d we described how transitions between quantum states in an atomic system occur as a result of interaction with oscillating electromagnetic fields. In particular, if the frequency greater than \u019212 is present in radiation incident on an atom whose ground-state and an excited-state energies are respectively E1 and E2, then there is a probability that the atom will undergo a transition from the lower energy state E1, absorbing the energy E1 from the radiation. This absorption of energy resulting from the inh\u019212 teraction between the electric field of the radiation oscillating at \u019212 and the charge on the atomic electrons was first described quantum mechanically by Einstein, who expressed the probability of absorption per atom per unit time as B12u(\u0192), where u(\u0192) is the energy density of the radiation per unit frequency and B12 is Einstein\u2019s coefficient of absorption. E2 In addition to absorption, several other interesting phenomena occur when electromagnetic radiation\u2014i.e., photons\u2014is incident on atoms or molecules. These are illustrated in Figure 9-30. In Figure 9-30a a photon of energy hf is absorbed and the system makes a transition to the excited state. Later, the system makes a transition to a lower state and/or back to the ground state with the spontaneous emission of one or more photons via the mechanism described on the home page in the Chapter 6 MORE section \u201cTransitions Between Energy States.\u201d The radiation thus emitted is called fluorescence. If state 2 happens to be the first excited state, then this two-step process is called resonance absorption and the photon emitted is called resonance radiation, as shown in Figure 9-30b. As a result of motions that occur while the system is in state 2, there is no correlation in direction or phase between the incident and emitted photons. While in state 2 the system has definite probabilities of making spontaneous transitions to each of the lower states as determined by the probability density given by Equation 6-52d in the MORE section. For example, the probability per atom per unit time of returning to state 1 with the spontaneous emission of a photon can be expressed by the quantity A21 (transitions per unit time). Notice that the reciprocal 1 A21 has the units of time per transition; i.e., it is a measure of how long the system stays in state 2 before returning to state 1. 1 A21. For most atomic This is ts, the mean lifetime of the state, defined as ts (electric dipole) transitions this characteristic lifetime is of the order of 108 s. A21 is called Einstein\u2019s coefficient of spontaneous emission. > > 394 Chapter 9 Molecular Structure and Spectra Stimulated Emission In addition to the spontaneous emission of fluorescent and resonant radiation with probability A21, which is independent of the energy density u(f ) of the incident radiation, emission can also be induced to occur by the oscillating electromagnetic field of the incident radiation. Called stimulated emission, its probability does depend upon u(f ). This phenomenon, like absorption and spontaneous emission, was first analyzed by Einstein, in 1917. The probability of stimulated emission per atom per unit time (transition rate) can be written as B21u(f ), where B21 is called Einstein\u2019s coefficient of stimulated emission. In this process the electric field of an incident photon with en E1 in Figure 9-30c stimulates the atom or ergy h\u0192 equal to the energy difference E2 h\u0192, which is propagated molecule in state 2 to emit a photon with energy E2 in the same direction and with the same phase as the incident photon. Such photons (or radiation) are said to be coherent. E1 The relation between the three Einstein coefficients can be found as follows. Consider a system of atoms and radiation in thermal equilibrium at temperature T. Let N1 and N2 be the number of atoms occupying the states with energies E1 and E2. The ratio N2 N1 is determined by the Boltzmann factor, given by Equation 8-2, assuming the two states have the same degeneracy: > N2 N1 e(E2 E1)>kT ehf>kT 9-35 This ratio represents a dynamic equilibrium in which the number of absorption tranS E2) per unit time equals the sum of the number of spontaneous and stimsitions (E1 S E1) per unit time. Since the number of atoms making a tranulated emissions (E2 sition (of any type) is proportional to the population of the state on which the transition begins and to the probability, we can express the dynamic equilibrium as N1B12u(f) N2(A21 B21u(f)) Solving Equation 9-", "36 for the energy density u(f) of the radiation yields u(f) A21 B21 B12 B21 1 N1 N2 Inserting N1 N2 from Equation 9-35, we have that > u(f) A21 B21 B12 B21 ehf>kT 1 9-36 9-37 9-38 This expression for the energy density of radiation of frequency f in thermal equilibrium at temperature T with atoms of energies E1 and E2 must be consistent with Planck\u2019s law for a blackbody spectrum at temperature T given by Equation 8-57: u(f) 8hf3 c3 a 1 ehf>kT 1 b 9-39 9-5 Scattering, Absorption, and Stimulated Emission 395 Comparing Equation 9-38 and 9-39, we conclude that and that B12 B21 1 A21 B21 8hf 3 c3 9-40 9-41 Although this analysis gives us only the ratios of the coefficients, A21 can be computed from quantum mechanics, as was discussed in the home page MORE section \u201cTransitions Between Energy States\u201d in Chapter 6, and the other coefficients may then be computed from the result. There are several points of interest in these equations. For instance, Equation 9-40 tells us that the coefficients for absorption and stimulated emission are the same for the same pair of states. Notice, too, that Equation 9-41 says that the ratio of the spontaneous emission coefficient to that for stimulated emission is proportional to \u01923. E1, the more likely spontaneous emission will This means that the larger E E2 be comparable to stimulated emission. Rewriting Equation 9-39 as A21 B21u(f) ehf>kT 1 9-42 hf W kT. yields the result that, in equilibrium situations, spontaneous emission is far more probable than stimulated emission for Since this is usually the case for electronic transitions in both atoms and molecules, de-excitation of excited electronic states by stimulated emission is normally ignored in these transitions. Stimulated emission does become important when and may dominate de-excitation of excited states when This latter condition exists for ordinary temperatures in the microwave region of the spectrum. We will return to these matters in Section 9-6 in connection with the discussion of lasers and masers. hf V kT. hf kT EXAMPLE 9-7 Spontaneous versus Stimulated Emission Compare the relative probabilities of spontaneous and stimulated emission in an equilibrium system at room temperature (T 300 K) for transitions that occur in (a) the visible and (b) the microwave regions of the spectrum. SOLUTION Equation 9-42 gives the ratio of the probability for spontaneous emission A21 to that for stimulated emission B21u(f). At T 300 K, kT 0.026 eV. > (a) In the visible region of the spectrum, Therefore, > so hf kT 2 0.026 77. hf 2 eV, A21 B21u(f) e77 1 Clearly, under these conditions spontaneous emission is favored over stimulated emission by an enormous factor. (b) In the microwave region of the spectrum, 0.0038 1>260 and stimulated emission is rather heavily favored. hf 104 eV, so hf>kT 104>0.026 396 Chapter 9 Molecular Structure and Spectra Questions 5. How does Rayleigh scattering differ from resonance absorption? 6. How does the photoelectric effect differ from all the other processes illustrated in Figure 9-30? 7. Why is stimulated emission usually not observed? 9-6 Lasers and Masers The laser (light amplification by stimulated emission of radiation) is a device that produces a strong beam of coherent photons by stimulated emission. The maser, where microwave replaces light in the definition from which the acronym is formed, was the laser\u2019s predecessor. Both devices depend on stimulated emission for their operation. We will discuss it more fully here because of its application to these important devices. Stimulated emission occurs if the atom is initially in an excited state E1, where E2 is the and if the energy of the photon incident on the atom is just E2 excited energy of the atom and E1 is the energy of a lower state or the ground state. In this case, the oscillating electromagnetic field of the incident photon accelerates the electron(s) at a rate that matches the photon\u2019s frequency and thus, we say, stimulates the excited atom, which may then emit a photon in the same direction as the incident photon and with the same phase. We have seen that the relative probabilities of stimulated emission and absorption B21 and B12 are equal (Equation 9-40). Ordinarily, at normal temperatures, nearly all atoms will initially be in the ground state, so absorption will be the main effect. That is, N1 W N2, so N1u(f", ")B12 W N2u(f)B21 where N1 and N2 are the populations of the two states. To produce more stimulated emission transitions than absorption transitions, we must arrange to have more N1). This condition is called atoms in the excited state than in the ground state (N2 population inversion. It can be achieved if the excited state E2 is a metastable state. S E1 Once population inversion is achieved, any light emitted by a spontaneous E2 transition is amplified by stimulated emission from the excited atoms that it encounters. Population inversion is often obtained by a method called optical pumping, in which atoms are \u201cpumped\u201d up to energy levels greater than E2 by the absorption of an intense auxiliary radiation. The atoms then decay down to the metastable state E2 by either spontaneous emission or by nonradiative transitions such as those due to collisions. The maser was the first of these devices to be constructed, an accomplishment of Charles Townes and his co-workers in 1953. Currently, the most important type of maser is the hydrogen maser, which is used as an atomic-frequency standard, one type of atomic clock. The hydrogen transition used in these masers is that between the hyperfine levels of the ground state, the same transition used to map hydrogen clouds in interstellar space (see Chapter 13). This transition between the 100 1 2 states is shown in Figure 9-32a. Townes shared the 1964 Nobel Prize in Physics for his contributions to the development of masers and lasers. and 100 1 2 & & (a) Energy (eV) Spin 5.88 \u00d7 10\u2013 6 Proton Spin Spin Electron Photon \u03bb = 21 cm Electron 0 Proton Spin (b) (c) Flashtube Ruby Trigger electrode Beam 9-6 Lasers and Masers 397 Pump levels E3 Fast (nonradiative) transition Pump radiation E2 Metastable state Laser output 694.3 nm E1 Ground state Figure 9-32 (a) Hyperfine levels used in the hydrogen maser. (b) Schematic diagram of the first ruby laser. (c) Energy levels of chromium in ruby, Al2O3. The Ruby Laser Figure 9-32b shows a schematic diagram of the first laser, a ruby laser built by Theodore Maiman in 1960.13 It consists of a small rod of ruby (a few centimeters long) surrounded by a helical gaseous flashtube. The ends of the ruby rod are flat and perpendicular to the axis of the rod. Ruby is a transparent crystal of Al2O3 containing a small amount (about 0.05 percent) of chromium. It appears red because the chromium ions (Cr3) have strong absorption bands in the blue and green regions of the visible spectrum. The energy levels of chromium that are important for the operation of a ruby laser are shown in Figure 9-32c. When the mercury- or xenon-filled flashtube is fired, there is an intense burst of light lasting a few milliseconds. Absorption excites many of the chromium ions to the bands of energy levels called pump levels in Figure 9-32c. The excited chromium ions give up their energy to the crystal in nonradiative transitions and drop down to a pair of metastable states labeled E2 in the figure. These metastable states are about 1.79 eV above the ground state. If the flash is intense enough, more atoms will make the transition to the states E2 than remain in the ground state. As a result, the populations of the ground state and the metastable states become inverted. When some of the atoms in the states E2 decay to the ground state by spontaneous emission, they emit photons of energy 1.79 eV and wavelength 694.3 nm. Some of these photons then stimulate other excited atoms to emit photons of the same energy and wavelength, moving in the same direction with the same phase. (b) (c) (d) (e) 398 Chapter 9 Molecular Structure and Spectra (a) Silvered end (a) L Partially silvered end (b) Mirror 1 Beam Beam m = 1 Mirror 2 m = 2 Propagation w Reflective surface >2, Figure 9-33 Laser as a resonating optical cavity. If mirror 1 is the partially reflecting end, then (a) illustrates the longitudinal standing-wave modes, for which L m is the laser where wavelength and m is integral. If the sides of the cavity are also reflective, as in (b), then standingwave modes transverse to the long axis are also possible. Notice that the exit beam for these modes is not parallel to the long axis of the laser. The ruby laser, like other conventional lasers, acts as a resonating optical cavity. In the ruby laser, both ends of the crystal are silvered such that one end is almost totally reflecting (about 99.9 percent) and the other end is", " only partially reflecting (about 99 percent) so that some of the beam is transmitted through that slightly transparent end. If the ends are parallel, standing waves are set up, as shown in Figure 9-33, and an intense beam of coherent light emerges through the partially silvered end. Figure 9-34 illustrates the buildup of the beam inside the laser. When photons traveling parallel to the axis of the crystal strike the silvered ends, all are reflected from the back face and most are reflected from the front face, with a few escaping through the partially silvered front face. During each pass through the crystal, the photons stimulate more and more atoms so that an intense photon beam is developed. Modern ruby lasers generate intense light beams with energies ranging from 50 J to 100 J in pulses lasting a few milliseconds. This pulse length is approximately equal to that of the flash tube, whose output excites atoms into the pump levels shown in Figure 9-32c. The output of the laser during that time is actually a series of very short pulses, each of the order of a microsecond long, as illustrated in Figure 9-35. This is because the pump levels depopulate quickly compared to the pump rate. Therefore, the flash requires some time to reestablish the population inversion that generates the next short pulse. Flashlamp Short laser pulses y t i s n e t n I Threshold 0 1 2 3 Time, ms Figure 9-35 A single output pulse from a ruby laser. The pulse actually consists of a series of very short pulses each about 1 s long. Flashlamp intensities below the threshold do not produce a sufficient population inversion to initiate lasing. Not shown is a weak background of incoherent spontaneous emission that accompanies the coherent laser light. Figure 9-34 Buildup of photon beam in a laser. (a) Some of the atoms spontaneously emit photons, some of which travel to the right and stimulate other atoms to emit photons parallel to the axis of the crystal. The others are absorbed, transmitted through the walls, or otherwise lost to the lasing process. (b) Four photons strike the partially silvered right face of the laser. (c) One photon has been transmitted and the others have been reflected. As these photons traverse the laser crystal, they stimulate other excited atoms to emit photons and the beam intensity increases. By the time the beam reaches the right face again (d), it comprises many photons. (e) Some of these photons are transmitted to become part of the external laser beam and the rest are reflected to sustain the process. 9-6 Lasers and Masers 399 Notice in Figure 9-35 that the first of the very brief laser pulses, or \u201cspikes,\u201d be N1 occurs and ends when N2 falls gins very soon after the population inversion N2 back to N1 due to stimulated emissions. Extremely intense spikes can be generated via a technique called Q-switching, whereby the resonating property of the cavity is temporarily destroyed in order to sharply reduce the stimulated emissions so as to allow \u226b N1. The resonant status is then suddenly rethe pumping radiation to make N2 stored and an extremely intense laser pulse results. This is how the very high-energy pulses mentioned above are typically produced. The Q in Q-switching refers to the cavity\u2019s quality factor, or its ability to maintain the intensity of the reverberating wave. If the end mirrors are low loss and the medium very transparent to the laser frequency, then the wave will die out slowly and the cavity is of high quality, or high Q. If Q is low, then substantial light is lost in each pass and the wave will die out quickly. If Q is too low, lasing will not occur at all. Q can be made very low, for example, by replacing the totally reflecting end mirror with an external one of equal reflectivity that rotates. When the rotating mirror is not parallel to the one on the other end of the cavity, Q is very low and little stimulated emission occurs as the pumping flash builds \u226b N1. When the rotating mirror becomes paralthe population of state E2 so that N2 lel to the other, Q suddenly becomes very high (hence the name Q-switch) and the extremely intense laser pulse is generated as E2 depopulates. Sustaining laser action requires that the increase in the number of coherent photons produced by stimulated emission per round trip through the resonating cavity to be greater than or equal to the decrease resulting from all losses, such as transmission through the partially reflecting end mirror and scattering. Although its a bit difficult, we have information to calculate the population inversion density necessary for lasing with the aid of the Einstein coefficients from Section 9-5, so let\u2019s try it. To begin, let\u2019s combine all of the various ways by which photons may be lost into a single characteristic time tp. That is, the intensity of radiation I of a particular frequency f in the resonant cavity", " will decay due to the losses according to I I0et>tp 9-43 where I0 is the intensity at t 0. Thus, the rate at which intensity is lost is given by a dI dt b loss I0 tp et>tp I tp 9-44 S E2 is equal to the difference in the populations (N2 The net rate at which the intensity of the frequency f gains due to the difference S E1 and the offsetting loss from abbetween the gain from stimulated emissions E2 N1) times the insorptions E1 tensity per photon times the transition probability u(f )B21. The transition probability u(f )B21 must be corrected for the width \u0192 of the spectral line emitted in the E2 S E1 transition arising from the finite width of the level E2, as described in Chapter 5.14 The correction is a multiplicative factor approximately equal to Taking these together, we obtain 2>\u00a2f. a dI dt b gain (N2 N1) hfc V 2 \u00a2f u(f)B21 9-45 where V is the volume of the resonant cavity and hfc V is the intensity per photon. > 400 Chapter 9 Molecular Structure and Spectra Using Equation 9-38 and the fact that A21 is the reciprocal of the lifetime for spontaneous emission ts, Equation 9-45 can be written as or a dI dt b gain (N2 N1) V (hfc) c u(f)c3 42hf 3\u00a2f ts d a dI dt b gain (n2 n1) c3I 42f2\u00a2f ts 9-46 9-47 > N2 V and n1 N1 V are the population densities of the states and where n2 cu(\u0192) I, the intensity.15 If the density of states (degeneracies) g(E) of E2 and E1 are not equal, then Equation 9-47 must be modified to > a dI dt b gain cn2 n1 a g(E2) g(E1) b d c3I 42f2\u00a2f ts Thus, the condition for laser action becomes or a dI dt b gain! a dI dt b loss cn2 n1 a g(E2) g(E1) b d c3I 42f2\u00a2f ts! I tp 9-48 9-49 9-50 The equal sign provides the threshold condition for the initiation of lasing. The greater-than sign represents sustained laser action. Solving the threshold condition yields the critical population inversion density nc: 42f2\u00a2f ts c3tp \u00a2nc 9-51 where n2 n1 a g(E2) g(E1) b \u00a2nc Equation 9-51 describes the population inversion that must be established if laser action is to be achieved for a given frequency and spontaneous emission lifetime. It also points out that the only property of the cavity that affects nc is its characteristic decay lifetime tp. The ruby laser is an example of a three-level laser, referring to the energy levels in Figure 9-32c. Such lasers have a practical disadvantage for many applications in that S E3 in order to obtain the necmore than half of the atoms must be pumped from E1 essary population inversion between levels E2 and E1. In addition, the source of the excitation energy, the flashlamp, produces light over a broad range of frequencies, most of which do not contribute to exciting the level E3 and are thus wasted. The large pumping requirement and relatively low excitation efficiency mean that substantial energy must be dissipated as heat, so three-level solid-state lasers such as the ruby laser must be pulsed in order to allow the system time to cool periodically. A more advantageous 9-6 Lasers and Masers 401 system is one that does not require such a large fraction of the atoms to be excited at any one time and avoids the excess heat produced by optical pumping. Such lasers provide continuous output and are called continuous wave, or cw, lasers. Helium-Neon Lasers In 1961, the first successful operation of a cw laser, a continuous helium-neon gas laser, was announced by Ali Javan, W. R. Bennet, Jr., and D. R. Herriott.16 Figure 9-36 shows a schematic diagram of the type of helium-neon laser commonly used for physics demonstrations and laser pointers and by land surveyors and carpenters. It consists of a gas tube containing 15 percent helium gas and 85 percent neon gas. A totally reflecting flat mirror is mounted on one end of the gas tube and a partially reflecting concave mirror is placed at the other end. The concave mirror focuses parallel light at the flat mirror and also acts as a lens that transm", "its part of the light so that it emerges as a parallel beam. Population inversion is achieved somewhat differently in the helium-neon laser than in the ruby laser. Figure 9-37 shows the energy levels of helium and neon that are important for operation of the laser. (The complete energy-level diagrams for helium and neon are considerably more complicated.) Helium has excited states, the 23S and 21S levels that lie 19.72 eV and 20.61 eV, respectively, above the 11S ground state. Both are metastable because of the / 1 selection rule, the 23S level being more strongly forbidden due to the S 0 selection rule, discussed in the Chapter 7 home page MORE section \u201cMultielectron Atoms,\u201d that prohibits intercombination lines. Helium atoms are excited to these states by an electrical discharge. Neon has closely spaced groups of excited states at 19.83 eV and 20.66 eV above its ground state\u2014the energies of these neon states almost exactly match the excited states of helium. The neon atoms are excited to these levels by collisions with excited helium atoms. The kinetic energy of the helium atoms provides the extra energy, about 0.05 eV, needed to excite the neon atoms. There is another excited state of neon that is 18.70 eV above its ground state and 1.96 eV below the 20.66 eV state. Since this state is normally unoccupied, population inversion between these states is obtained immediately. The stimulated emission that occurs between these states results in photons of energy 1.96 eV and wavelength 632.8 nm, which produces a bright red light. After stimulated emission, the atoms in the 18.70 eV state decay to the ground state by spontaneous emission of a photon with a wavelength of about 600 nm followed by a nonradiative de-excitation, typically collision with the cavity wall. The collisions are an important part of the laser process since if the diameter of the tube (see Figure 9-36) is too large, the probability of collision with the wall decreases and the 600-nm radiation may re-excite the 18.70 eV level. This reduces the population inversion and decreases the laser gain. Stimulated emission also occurs from the state at 19.83 eV to the 18.70 eV level, producing laser light with a wavelength of 1100 nm (infrared). Rear Laser tube Front Parallel laser beam Flat mirror: 99.9% reflective Concave mirror: reflects 99%, transmits 1% Figure 9-36 Schematic drawing of a helium-neon laser. The use of a concave mirror rather than a second plane mirror makes the alignment of the mirrors less critical than it is for the ruby laser. The concave mirror on the right also serves as a lens that focuses the emitted light into a parallel beam. 402 Chapter 9 Molecular Structure and Spectra 0.05 eV 21S 23S 0.15 eV 1s12s1 1s12s1 2p55s1 2p54s1 Stimulated emission (laser output 632.8 nm) 2p53p1 Spontaneous emission (~600 nm) 2p53s1 Nonradiative transition 1S1 1s2 2p6 He levels Ne levels 20 15 10 Figure 9-37 Energy levels of helium and neon that are important for the helium-neon laser. The helium atoms are excited by electrical discharge to energy states 19.72 eV and 20.61 eV above the ground state. They collide with neon atoms, exciting some neon atoms to energy states 19.83 eV and 20.66 eV above the ground state. Population inversion is thus achieved between these levels and one at 18.70 eV above the ground state. The spontaneous emission of photons of energy 1.96 eV from the upper state stimulates other atoms in the upper state to emit photons of energy 1.96 eV, producing the characteristic He-Ne red laser light. Emission from the 19.83 eV neon state to the 18.70 eV level also produces laser output at about 1100 nm. Helium-neon lasers have recently been developed that lase at a number of other visible and infrared wavelengths. The several possible laser wavelengths are not present simultaneously since each device is designed to operate at a particular wavelength. Note that there are four energy levels involved in producing the 632.8-nm heliumneon laser line, whereas the ruby laser involved only three levels. In a three-level laser, population inversion is difficult to achieve because more than half the atoms in the > N1 2 in Equation 9-46. In a four-level laser, ground state must be excited, i.e., N2 population inversion is easily achieved because the state after stimulated emission is L 0. not the ground state but an excited state that is normally unpopulated, so that N1 New Lasers and Applications A laser", " beam is coherent, very narrow, and intense. Its coherence makes the laser beam useful in the production of holograms, such as those used on credit cards and \u201cheadsup\u201d displays; i.e., transparent data displays that don\u2019t obstruct the user\u2019s view. The precise direction and small angular spread of the beam make it useful as a surgical tool for destroying cancer cells or reattaching a detached retina. Lasers are also used by land surveyors and carpenters to ensure precise alignment over large distances. Distances can be accurately measured by reflecting a laser pulse from a mirror and measuring the time the pulse takes to travel to the mirror and back. The distance to the Moon is currently measured to within a few millimeters using corner reflectors placed on the Moon by Apollo astronauts for that purpose. Laser beams are also used in fusion research. 9-6 Lasers and Masers 403 Three counterpropagating pairs of orthogonal laser beams illuminate about 100 million sodium atoms at their intersection. The pressure of the laser light cools the atoms, slowing them to rms velocities comparable to those resulting from recoil due to emission or absorption of a single photon. Systems incorporating laser cooling of sodium atoms have achieved a record low temperature of 450 pK. [National Institute of Science and Technology.] An intense laser pulse is focused on tiny pellets of deuterium-tritium in a combustion chamber. The beam heats the pellets to temperatures of the order of 108 K in a very short time, causing the deuterium and tritium to fuse and release energy. (See Section 11-8.) At the other end of the temperature scale, using advanced cooling techniques that included the focusing of three orthogonal pairs of lasers (optical tweezers) on a sample containing 2500 cesium atoms, in 2003 W. Ketterle and his group achieved a record low temperature of 450 pK.17 (See the photo above.) Orthogonal pairs of laser beams, called optical traps, capable of cooling samples containing millions of atoms down to the submicrokelvin range, are used in creating Bose-Einstein condensates and the degenerate Fermi gas, discussed in Chapter 8, and antihydrogen atoms, described in Chapter 12. While cw lasers are the lasers of choice for many applications, many others require pulsed lasers, particularly those where very high power levels are important or even essential. For these applications the current state of the art is chirped pulse amplification. With this technique, invented in the 1980s by G\u00e9rhard Mourou, an input ultrashort pulse is stretched out (in time) by dispersing the wavelengths (frequencies) of the pulse with a suitable arrangement of gratings or prisms together with mirrors. After passing through the optical stretcher, the pulse duration is up to 105 times longer than the original pulse and the intensity has been correspondingly lowered as a result to avoid nonlinear effects that would damage or destroy the amplification medium. The laser pulse, amplified by a factor of 106 or more, is recompressed by an optical system that is essentially the reverse of the stretcher. Off-the-shelf chirped pulse lasers are available with peak power in the 10- to 100-gigawatt range. Several major research facilities operate chirped lasers with peak power in the petawatt range. Laser technology is advancing so fast that it is possible to mention only a few recent developments. In addition to the ruby laser, there are many other solid-state lasers with output wavelengths ranging from about 18 nm (soft x rays) to about 3900 nm (infrared). Lasers that generate more than 1 kW of continuous power have been constructed, and pulsed lasers can now deliver nanosecond pulses of power exceeding 109 W. Various gas lasers produce wavelengths ranging from the far infrared to the ultraviolet. Semiconductor lasers (also known as diode lasers or junction lasers; these will be discussed further in Chapter 10) the size of a pinhead can develop 200 mW of power. Long-range fiber optic communication lines will be enhanced by the recent development of the erbiumdoped fiber optic amplifier. Light from a diode laser \u201cpumps up\u201d a segment of erbium-doped optical fiber in the line. A signal moving down the line stimulates emission from the erbium atoms, resulting in amplification of the signal. 404 Chapter 9 Molecular Structure and Spectra In addition to the ubiquitous diode lasers used in supermarket checkout counters, compact disc players, copiers, and computer printers, through very recent developments in materials physicists have constructed reliable diode lasers that emit in the blue to ultraviolet region of the spectrum. These lasers should make possible significantly increased high-density optical storage on digital versatile discs (DVDs). Liquid lasers using chemical dyes can be tuned over a range of wavelengths (about 70 nm for continuous lasers and more than 170 nm for pulsed lasers). The", " free-electron laser extracts light energy from a beam of free electrons moving through a spatially varying magnetic field. It has the potential for very high power and high efficiency and can be tuned over a large range of wavelengths. The variety and uses of modern lasers seem limitless. Questions 8. What are the advantages of a four-level laser over a three-level laser? 9. Why is helium needed in a helium-neon laser? Why not just use neon? EXAMPLE 9-8 Critical Population Inversion Comparison Compare the critical population inversion necessary for laser action in the ruby and He-Ne lasers. Compute the corresponding power requirements. SOLUTION The critical population density nc is given by Equation 9-51. The typical parameters of these systems are as follows: Parameter Ruby laser He-Ne laser \u0192 n (refractive index) ts tp \u0192 N (Cr3 concentration) 694.3 nm 4.32 1014 s1 1.76 3 10 3 s 2.9 108 s 3.3 1011 s1 2 1019 cm3 > 632.8 nm 4.74 1014 s1 1.00 107 s 3.3 107 s 9 108 s1 ) For ruby laser: \u00a2nc 42f2\u00a2f ts c3tp 42(4.32 1014)2(3.3 1011)(3 103) (3.00 108>1.76)3(2.9 108) \u00a2nc 5.08 1022 atoms>m3 5.08 1016 atoms>cm3 For He-Ne laser: \u00a2nc 42f2\u00a2f ts c3tp 42(4.74 1014)2(9 108)(107) (3.00 108)3(3.3 107) \u00a2nc 8.96 1013 atoms>m3 8.96 107 atoms>cm3 Thus, the critical population density is far smaller for the He-Ne laser. Summary 405 The minimum power input P needed to maintain the laser action in the helium- neon system is approximately equal to nc(h\u0192 ts) since N1 > L 0, or P(He-Ne) \u00a2nc(hf>ts) (8.96 107)(6.63 1034)(4.74 1014) 107 2.8 104 W>cm3 For the ruby laser, about one-half of the Cr3 ions must be in the pumped level E3 in Figure 9-32c, and the power per unit volume necessary to maintain that population is approximately P(ruby) N 2 a hf ts b (2 1019)(6.63 1034)(4.32 1014) 2 3 103 955 W>cm3 Summary TOPIC RELEVANT EQUATIONS AND REMARKS 1. The ionic bond The bonding mechanism typical of most salts, it involves the transfer of one or more electrons to form ions that are attracted by the Coulomb force. The exclusion principle limits the close approach of the ions, resulting in a minimum in the potential energy U(r). For a diatomic molecule, U(r) ke2 r Eex Eion 9-1 where Eion is the net ionization energy and Eex is the exclusion-principle energy. The latter is given by Eex A rn where A and n are constants. 9-2 2. The covalent bond This bond is a quantum-mechanical effect arising from the sharing of one or more electrons by identical or similar atoms. The symmetry of the molecular wave functions resulting from their superposition of electron orbitals determines whether bonding will occur. The wave function for the symmetric state wells, resulting in minimum potential energy and bonding. The antisymmetric wave function of ionic and covalent bonding. is small in that region. Bonding of two nonidentical atoms is often a mixture is large between the atomic potential & A & S Other covalent bonds Covalent bonds differ in detail, depending upon which electrons are shared. For example, H2 with only s electrons, is s-bonded. O2 is p-p bonded. There are also s-p bonds, of which H2O is one example. 3. Dipole-dipole bonds Bonding between atoms and molecules may arise due to interactions between dipole moments. The interaction may involve molecules with either permanent electric dipole moments (polar molecules) or induced dipole moments (nonpolar molecules). The potential energy U of a dipole p2 in the electric field Ed of dipole p1 is given by U p2 # Ed 9-8 The force between permanent dipoles decreases as 1 r4. If one or both of the dipoles is an induced dipole, the force between them decreases as 1 r7. > > 406 Chapter 9 Molecular Structure and Spectra TOPIC RELEVANT EQUATIONS AND REMARKS 4. Molecular spectra The", " energy states of diatomic molecules consist of rotational bands superimposed on more widely spaced vibrational levels, which are in turn superimposed on the much more widely spaced atomic electron levels. Rotational energies The rotational energies of a diatomic molecule are E ( 1)U2 2I ( 1)E0r 0, 1, 2, \u00c1 U2>2I where I is the moment of inertia, and / is the rotational quantum number, which obeys the selection rule / 1. is the characteristic rotational energy, E0r Vibrational energies The vibrational energies of a diatomic molecule are E ( 1> 2)hf 0, 1, 2, \u00c1 where \u0192 is the vibrational frequency and obeys the selection rule \u00a2 1. is the vibrational quantum number, which 9-13 9-20 5. Scattering, absorption, and stimulated emission 6. Lasers and masers A photon incident on an atom can be absorbed, producing fluorescence or resonance radiation, or scattered elastically (Rayleigh scattering) or inelastically (Raman scattering). If the photon energy is greater than the ionization energy of the atom, Compton scattering or the photoelectric effect can occur. If the atom is initially in an excited state, an incident photon of the proper energy can stimulate emission of another photon of the same energy. The incident and emitted photons are in phase and travel parallel to each other. In an equilibrium system the probabilities (Einstein coefficients) for absorption and for stimulated emission between two states are equal. Lasers and masers are important applications of stimulated emission, differing only in the wavelengths of their outputs. Amplification by stimulated emission depends on the possibility of obtaining population inversion, in which there are more atoms in an excited state than in the ground state or another excited state of lower energy. Population inversion is usually obtained by optical pumping and is produced more readily in four-level systems than in three-level systems. General References The following general references are written at a level appropriate for the readers of this book. Brehm, J. J., and W. J. Mullin, Introduction to the Structure of Matter, Wiley, New York, 1989. Eisberg, R., and R. Resnick, Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles, 2d ed., Wiley, New York, 1985. Pauling, L., The Chemical Bond, Cornell University Press, Ithaca, NY, 1967. Schawlow, A. L., \u201cLaser Light,\u201d Scientific American, September 1968. This article and several other excellent articles on lasers and masers are reprinted in Lasers and Light, W. H. Freeman and Co., New York, 1969. Serway, R. A., C. J. Moses, and Curt A. Moyer, Modern Physics, 3d ed., Thomson, Belmont, CA, 2005. Herzberg, G., Atomic Spectra & Atomic Structure, Dover, Zeilinger, A., \u201cQuantum Teleportation,\u201d Scientific American, New York, 1944. April 2000. Notes 1. The term orbital is frequently used in molecular physics and in chemistry to refer to the space part of the electron wave functions, that is, the quantum numbers n, /, and m/. In molecular physics the electrons of interest are usually the outermost (valence) ones of the constituent atoms, which become associated with the entire molecule rather than their original atoms, so we speak of \u201cmolecular orbitals\u201d as well as \u201catomic orbitals.\u201d 2. Molecules whose atoms are identical, such as H2, are sometimes called homopolar or homonuclear. Those whose atoms are not identical are called heteropolar or heteronuclear. 3. C60 and the other fullerenes are named after the philosopher and engineer R. Buckminster Fuller, who invented the architectural geodesic dome structure. Such domes, as Fuller pointed out, can be considered as networks of pentagons and hexagons. 4. Leonhard Euler (1707\u20131783), Swiss mathematician. Arguably the most prolific mathematician of all time, he published 866 papers during his lifetime and, despite having lost his sight in 1766 (in part due to his earlier observations of the Sun), he left so many manuscripts at his death that it took another 35 years to get them all published. He introduced the symbol e as the base of the natural logarithms and i as the square root of 1. 5. Johannes D. van der Waals (1837\u20131923), Dutch physicist. Largely self-taught, he became interested in the fact that the ideal gas law derived from kinetic theory does not hold exactly for real gases. This led him to question the assumption that no forces act between individual gas molecules except during collisions, which resulted in his development of an equation, the van der", " Waals equation, which more accurately describes real gases. He was awarded the 1910 Nobel Prize in Physics for his work. 6. This result is derived in most introductory physics books. See, e.g., P. A. Tipler and G. Mosca, Physics for Scientists and Problems 407 Engineers, 6th ed., W. H. Freeman and Co., New York, 2008, p. 671. 7. Terminology concerning the dipole-dipole forces is a bit confused. Some textbooks use van der Waals to describe all three types of dipole-dipole forces. We will follow the more common (and traditional) use, reserving van der Waals for the attractive force between induced dipoles only. 8. We use v (the Greek letter nu) here rather than n so as not to confuse the vibrational quantum number with the principal quantum number n for electronic energy levels. 9. The nitric oxide (NO) molecule is an exception due to its S 1 S 1 group of lines are called the P group the R branch. odd electron. 10. Also, the branch and the 11. Chandrasekhara V. Raman (1888\u20131970), Indian physicist. Graduating from college at the age of 16, like Einstein he became a civil servant and worked at science in his spare time. He had predicted that visible light should be inelastically scattered even before Heisenberg had predicted and Compton had found the effect for x rays. He was awarded the 1930 Nobel Prize in Physics for his work, becoming the first Asian to be so recognized in the sciences. 12. There is also a Raman effect for the vibrational and electronic levels of molecules. 13. T. H. Maiman, \u201cStimulated Optical Radiation in Ruby,\u201d Nature, 187, 493 (1960). 14. The correction essentially accounts for the fact that, due to the finite line width, the energy density u(f) in the transition probability must include a narrow range of frequencies f rather than just the single frequency f. 15. Recall that the energy per unit volume u(f) times c is the intensity, e.g., W/m2 in SI units. 16. A. Javan, W. B. Bennet, Jr., and D. R. Herriott, Physical Review Letters, 6, 106 (1961). 17. W. Ketterle et al., Science, 301, 1513 (2003). Problems Level I Section 9-1 The Ionic Bond 9-1. The dissociation energy is sometimes expressed in kilocalories per mole. (a) Find the relation between electron volts per molecule and kilocalories per mole. (b) Find the dissociation energy of molecular NaCl in kilocalories per mole. (c) The dissociation energy of the Li2 molecule is 106 kJ/mole. Find the value in eV per molecule. 408 Chapter 9 Molecular Structure and Spectra 9-2. The dissociation energy of Cl2 is 2.48 eV. Consider the formation of an NaCl molecule by the reaction Na 1 2 Cl2 \u00a1 NaCl Is this reaction endothermic (requiring energy) or exothermic (giving off energy)? How much energy per molecule is required or given off? 9-3. Using the data in Table 9-1, compute the net energy required to transfer an electron between the following pairs of atoms: Cs to F, Li to I, and Rb to Br. 9-4. Using the data in Tables 9-1 and 9-2, estimate the dissociation energy of the three ionically bonded molecules CsI, NaF, and LiI. Your results are probably all higher than those in Table 9-2. Explain why. 9-5. The equilibrium separation of the Rb and Cl ions in RbCl is about 0.267 nm. (a) Calculate the potential energy of attraction of the ions, assuming them to be point charges. (b) The ionization energy of rubidium is 4.18 eV, and the electron affinity of Cl is 3.62 eV. Find the dissociation energy, neglecting the energy of repulsion. (c) The measured dissociation energy is 4.37 eV. What is the energy due to repulsion of the ions? 9-6. Compute the Coulomb energy of the KBr molecule at the equilibrium separation. Use that result to compute the exclusion-principle repulsion at r0. 9-7. coefficient A and the exponent n. 9-8. Compute the dissociation energy of molecular NaBr in kilocalories per mole. 9-9. Note in Table 9-2 that the equilibrium separations of the KBr and RbCl molecules are very nearly equal. Compute the exclusion-principle repulsion for these molecules. If the exclusion-principle repulsion in Problem 9", "-6 is given by Equation 9-2, compute the Section 9-2 The Covalent Bond 9-10. Hydrogen can bond covalently with many atoms besides those listed in Tables 9-3 and 9-5, including sulfur, tellurium, phosphorus, and antimony. What would you expect to be the chemical formula of the resulting molecules? (Hint: Use the table of electron configurations in Appendix C.) 9-11. What kind of bonding mechanism would you expect for (a) the KCl molecule, (b) the O2 molecule, and (c) the CH4 molecule? 9-12. The equilibrium separation of the atoms in the HF molecule is 0.0917 nm, and its mea# sured electric dipole moment is 6.40 1030 C m. What percentage of the bonding is ionic? 9-13. The equilibrium separation of CsF is 0.2345 nm. If its bonding is 70 percent ionic, what should its measured electric dipole moment be? 9-14. Ionic bonding in the BaO molecule involves the transfer of two electrons from the Ba atom. If the equilibrium separation is 0.193 nm and the measured electric dipole moment is 26.7 1030 C m, to what extent is the bond actually ionic? # Section 9-3 Other Bonding Mechanisms 9-15. Find three other elements with the same subshell electron configuration in the two outermost orbits as carbon. Would you expect the same kind of hybrid bonding for these elements as for carbon? Support your answer. 9-16. The dipole moment p of the water molecule, illustrated in Figure 9-19, is actually the vector sum of two equal dipoles p1 and p2 directed from the oxygen atom to each of the hydrogen atoms. The measured value of the angle between the two hydrogen atoms is 104.5\u00b0, the O\u2013H bond length is 0.0956 nm, and the magnitude of p is 6.46 1030 C m. Compute the fraction of the electron charge that is transferred from each hydrogen to the oxygen. # 9-17. The polarizability of Ne is 1.1 1037 m C2 N. (a) At what separation would the dipoledipole energy between a molecule of H2O and an atom of Ne in the atmosphere be sufficient to withstand collision with an N2 molecule moving with the average kinetic energy for T 300 K? (b) At what separation does this energy occur for a typically bonded molecule? (c) On the basis of these results, do you expect H2O\u2013Ne bonds to be very likely? Explain your answer. > # Problems 409 9-18. The hydrogen bonds linking the two helical strands of DNA have bond strengths of about 0.3 eV, or approximately 15 percent of the strengths of the ionic/covalent bonds along the strands. (a) What is the wavelength of a photon with sufficient energy to break this bond? (b) In what part of the spectrum does this wavelength lie? (c) Since a significant intensity exists at this wavelength in the environment, why haven\u2019t all the DNA hydrogen bonds long since broken? 9-19. Would you expect the following molecules to be polar or nonpolar? Explain your answer in each case. (a) NaCl; (b) O2. E0r in eV. 1 1 vibrational state? vibrational state to the Section 9-4 Molecular Spectra 9-20. The characteristic rotational energy E0r for N2 molecules is 2.48 104 eV. From this, find the separation distance of the nitrogen atoms in N2. 9-21. For the O2 molecule, the separation of the atoms is 0.121 nm. Calculate the characteris U2>2I tic rotational energy 0 9-22. The CO molecule undergoes a transition from the state. (a) What is the wavelength of the emitted photon? (b) At what temperature would 1 percent of the CO molecules be in the 9-23. Using data from Table 9-7, (a) compute the vibrational energy of the LiH molecule in its lowest vibrational state. (b) Compute the reduced mass of LiH. (c) Determine the force constant for LiH. (d) From those results, compute an estimate of the LiH bond length and compare your result with the value in the table. 9-24. Calculate the reduced mass in unified mass units for (a) H2, (b) N2, (c) CO, and (d) HCl. for KCl is 1.43 105 eV. (a) Find the 9-25. The characteristic rotational energy reduced mass for the KCl molecule. (b) Find the separation distance of the K and Cl ions. 9-26. Use the data from Table 9-", "7 to find the force constant for (a) the H35Cl and (b) the K79Br molecules. 9-27. The equilibrium separation of HBr is 0.141 nm. Treating the Br atom as fixed, compute the four lowest rotational energies of the HBr molecule and show them in a carefully sketched energy-level diagram. 9-28. The vibrational spectrum of Li2 consists of a series of equally spaced lines in the microwave region 1.05 1013 Hz apart. Compute the equilibrium separation for Li2. 9-29. Compute the difference in the rotational energy E0r for K35Cl and K37Cl. 9-30. What type of bonding mechanism would you expect for (a) NaF, (b) KBr, (c) N2, and (d) Ne? 9-31. For NaCl compute (a) the energy in eV necessary to excite the first rotational state and (b) the wavelength and frequency of the photon emitted in the transition back to the ground state. (Assume that the molecule is in the electronic and vibrational ground states.) U2>2I E0r 0, E2 4.30 eV, E4 7.2 eV, and E5 Section 9-5 Scattering, Absorption, and Stimulated Emission 3.80 eV, 9-32. The five lowest levels of a certain monatomic gas have the values E1 7.5 eV. (a) If the temperature is high enough that all levE3 els are occupied and the gas is illuminated with light of wavelength 2400 nm, what transitions can occur? (b) Which of those found in part (a) will still occur if the temperature is so low that only the state E1 is occupied? (c) Repeat (a) and (b) for light of 250 nm wavelength. (d) What wavelength of the incident light would stimulate emission from state E4? 9-33. A hydrogen discharge tube is operated at about 300 K in the laboratory in order to produce the Balmer series. Compute the ratio of the probability for spontaneous emission of the H 9-34. Determine the ratio of the number of molecules in the v 1 state to the number in the v 0 state for a sample of O2 molecules at 273 K. Repeat the calculation for 77 K. (Ignore rotational motion.) line to that for stimulated emission. 410 Chapter 9 Molecular Structure and Spectra 9-35. The nuclei in the F2 molecule are separated by 0.14 nm. (a) Compute the energy separations and sketch an energy-level diagram for the lowest four rotational levels with v 0. (b) What are the wavelengths of possible transitions between these levels? Section 9-6 Lasers and Masers 9-36. A pulse from a ruby laser has an average power of 10 MW and lasts 1.5 ns. (a) What is the total energy of the pulse? (b) How many photons are emitted in this pulse? 9-37. A helium-neon laser emits light of wavelength 632.8 nm and has a power output of 4 mW. How many photons are emitted per second by this laser? 9-38. A laser beam is aimed at the Moon, a distance 3.84 108 m away. The angular spread of the beam is given by the diffraction formula (Rayleigh\u2019s criterion), where D is the diameter of the laser tube or rod. (a) Calculate the size of the beam on the Moon for D 10 cm and 600 nm. (b) Repeat the calculation if the laser beam is projected toward the Moon through a 1.0-m-diameter telescope. 9-39. A particular atom has two energy levels with a transition wavelength of 420 nm. At 297 K there are 2.5 1021 atoms in the lower state. (a) How many atoms are in the upper state? (b) Suppose that 1.8 1021 of the atoms in the lower state are pumped to the upper state. How much energy could this system release in a single laser pulse? sin 1.22 >D, > Level II 9-40. (a) Calculate the electrostatic potential energy of Na and Cl ions at their equilibrium separation distance of 0.24 nm, assuming the ions to be point charges. (b) What is the energy of repulsion at this separation? (c) Assume that the energy of repulsion is given by Equation 9-2. From Figure 9-2b, this energy equals ke2 r at about r 0.14 nm. Use this and your answer to part (b) to calculate n and A. (Although this calculation is not very accurate, the energy of repulsion does vary much more rapidly with r than does the energy of attraction.) 9-41. The angular width of a ruby laser beam is", " determined by Rayleigh\u2019s criterion (see 694.3 nm. Problem 9-38). For this laser the diameter of the ruby rod is 1.0 cm and (a) What is the diameter of the spot projected by the ruby laser at a distance of 1.0 km? (b) If the laser is emitting 1018 photons/s, what is the power deposited per square centimeter on the target at 1.0 km? 9-42. The equilibrium separation of the K and Cl ions in KCl is about 0.267 nm. (a) Calculate the potential energy of attraction of the ions assuming them to be point charges at this separation. (b) The ionization energy of potassium is 4.34 eV and the electron affinity of chlorine is 3.62 eV. Find the dissociation energy for KCl, neglecting any energy of repulsion (see Figure 9-2a). (c) The measured dissociation energy is 4.40 eV. What is the energy due to repulsion of the ions at the equilibrium separation? 9-43. Use the equilibrium separation for the K and Cl ions given in Problem 9-42 and the reduced mass of KCl to calculate the characteristic rotational energy E0r of KCl. 9-44. In this problem, you are to find how the van der Waals force between a polar and a nonpolar molecule depends on the distance between the molecules. Let the dipole moment of the polar molecule be in the x direction and the nonpolar molecule be a distance x away. (a) How does the electric field due to an electric dipole depend on the distance x? (b) Use the facts that the potential energy of an electric dipole of moment p in an electric field E is U p E and that the induced dipole moment of the nonpolar molecule is proportional to E to find how the potential energy of interaction of the two molecules depends on separation distance. (c) Using Fx 9-45. The microwave spectrum of CO has lines at 0.86 mm, 1.29 mm, and 2.59 mm. (a) Compute the photon energies and carefully sketch the energy level diagram that corresponds. What molecular motion produces these lines? (b) Compute the equilibrium separation (bond length) of CO. dU dx, find the x dependence of the force between the two molecules. > # Problems 411 9-46. Carefully draw a potential energy curve for a diatomic molecule (like Figure 9-2b) and indicate the mean values of r for two vibrational levels. Show that because of the asymmetry of the curve, rav increases with increasing vibrational energy and therefore solids expand when heated. 9-47. A sample of HCl is illuminated with light of wavelength 435.8 nm. (a) Compute the wavelengths of the four lines in the rotational Raman spectrum that are closest to that of the incident light. (b) Compare the difference in their frequencies with the corresponding lines in Figure 9-29. 9-48. Use data from Table 9-7 to compute the first excited vibrational and the first excited rotational states of (a) the Li2 and (b) the K79Br molecules. 9-49. Calculate the effective force constant for HCl from its reduced mass and the fundamental vibrational frequency obtained from Figure 9-29. 9-50. Notice in Figure 9-32c that the level E2 in Cr3 is a doublet, the pair of states being separated by only 0.0036 eV. (a) Assume that all of the Cr3 ions in a certain laser are in the three states E1 and E2 (doublet) and compute the relative populations of these levels. (b) If only the lower state of the E2 doublet can produce laser light but both levels must be pumped together, determine the pumping power necessary for laser action to occur. The density of states (degeneracy) of level E1 is 4 and for each of the E2 levels is 2. 9-51. The central frequency for the absorption band of HCl shown in Figure 9-29 is at f 8.66 1013 Hz, and the absorption peaks are separated by about f 6 106 Hz. Using this information, find (a) the lowest (zero-point) vibrational energy for HCl, (b) the moment of inertia of HCl, and (c) the equilibrium separation of the atoms. Level III 9-52. The potential energy between two atoms in a molecule can often be described rather well by the Lenard-Jones potential, which can be written U(r) U0 c a 12 b a r 2a 6 d b a r where U0 and a are constants. (a) Find the interatomic separation r0 in terms of a for which the potential energy is minimum. (b) Find the corresponding value of Umin.", " (c) Use Figure 9-8b to obtain numerical values for r0 and U0 for the H2 molecule. Express your answer in nanometers and electron volts. (d) Make a plot of the potential energy U(r) versus the internuclear separation r for the H2 molecule. Plot each term separately, together with the total U(r). 9-53. (a) Find the exclusion-principle repulsion for NaCl. (b) Use Equation 9-2 to find A and n. 9-54. Show that the H\u2013H system cannot be ionically bonded. (Hint: Show that U(r) has no negative minimum.) \u00a2> > for the reduced masses of the H35Cl and 9-55. (a) Calculate the fractional difference H37Cl molecules. (b) Show that the mixture of isotopes in HCl leads to a fractional difference \u00a2f>f \u00a2>. in the frequency of a transition from one rotational state to another given by (c) Compute \u0192 \u0192 and compare your result with Figure 9-29. 9-56. For a molecule such as CO, which has a permanent electric dipole moment, radiative between two rotational energy levels of the transitions obeying the selection rule does not same vibrational energy state are allowed. (That is, the selection rule 0.113 nm, and calculate the charhold.) (a) Find the moment of inertia of CO for which r0 acteristic rotational energy E0r in electron volts. (b) Make an energy-level diagram for the rotational levels for / 0 to / 5 for some vibrational level. Label the energies in electron \u00a2 1 \u00a2 1 412 Chapter 9 Molecular Structure and Spectra volts, starting with E 0 for / 0. (c) Indicate on your diagram transitions that obey / 1 and calculate the energy of the photons emitted. (d) Find the wavelength of the photon emitted for each transition in (c). In what region of the electromagnetic spectrum are these photons? 9-57. An H2 molecule in its ground electronic, vibrational, and rotational state absorbs a photon of frequency 1.356 1014 Hz, undergoing a transition to the / 1 state while re/ 2 state, maining in the electronic ground state. It then undergoes a transition to the emitting a photon of frequency 1.246 1014 Hz. (a) Compute the moment of inertia of the H2 molecule about an axis through the center of mass. (b) Determine the vibrational frequency and r0 for H2 and compare these with the values in Table 9-7. 0, 1, CHAPTER 10 Solid State Physics The many and varied properties of solids have intrigued us for centuries. Technological developments involving metals and alloys have shaped the courses of civilizations, and the symmetry and beauty of naturally occurring, large single crystals have consistently captured our imaginations. However, the origins of the physical properties of solids were not understood even in rudimentary form until the development of quantum mechanics. The application of quantum mechanics to solids has provided the basis for much of the technological progress of modern times. We will study briefly some aspects of the structure of solids in Section 10-1 and then concentrate on the electrical and magnetic properties of solids. 10-1 The Structure of Solids In our everyday world we see matter in three phases: gases, liquids, and solids. In a gas the average distance between two atoms or molecules is large compared with the size of an atom or molecule. The molecules have little influence on one another except during their frequent but brief collisions. In a liquid or solid the atoms or molecules are close together and exert forces on one another comparable to the forces that bind atoms into molecules. (There is a fourth phase of matter, plasma, which occurs when the matter consists largely\u2014or entirely\u2014of ions and free electrons. Usually this condition exists only at very high temperatures, such as inside stars, in intense electrical discharges\u2014e.g., lightning\u2014and in the laboratory. The properties of a plasma are very different from those of an ordinary gas because of the long-range electrical and magnetic effects arising from the charges of the particles. The recently discovered low-temperature gas phase of matter, the Bose-Einstein condensate, was discussed in Chapter 8.) In a liquid, the molecules form temporary short-range bonds that are continually broken and re-formed as the result of the thermal kinetic energy of the molecules. The strength of the bonds depends on the type of molecule. For example, as we discussed in Section 9-3, the bonds between helium atoms are very weak van der Waals bonds, and He does not liquefy at atmospheric pressure until the very low temperature of 4.2 K is reached. 10-1 The Structure of Solids 413 10-2 Classical Theory of Conduction 422 10-3 Free-Elect", "ron Gas in Metals 426 10-4 Quantum Theory of Conduction 430 10-5 Magnetism in Solids 10-6 Band Theory of Solids 10-7 Impurity 434 438 Semiconductors 445 10-8 Semiconductor Junctions and Devices 452 10-9 Superconductivity 458 413 414 Chapter 10 Solid State Physics If a liquid is slowly cooled, the kinetic energy of its molecules is reduced and the molecules will arrange themselves in a regular crystalline array, producing the maximum number of bonds and leading to a minimum potential energy. However, if the liquid is cooled rapidly so that its internal energy is removed before the molecules have a chance to arrange themselves, a solid is often formed that is not crystalline but resembles a \u201csnapshot\u201d of a liquid. Such a solid is called amorphous; it displays short-range order but not the long-range order (over many atomic diameters) characteristic of a crystal. Glass is a typical amorphous solid. A characteristic of the long-range ordering of a crystal is that it has a well-defined melting point, whereas an amorphous solid merely softens as its temperature is increased. Many materials may solidify in either an amorphous or a crystalline state, depending on how they are prepared. Others exist only in one form or the other. Most common solids are polycrystalline\u2014i.e., they are collections of single crystals. The size of such single crystals is typically a fraction of a millimeter; however, large single crystals occur naturally and can be produced artificially (Figure 10-1). We will discuss only simple crystalline solids in this chapter. The most important property of a single crystal is its symmetry and regularity of structure: it can be thought of as a single unit structure repeated throughout the solid. (b) (a) Figure 10-1 (a) A single crystal of quartz, one of several naturally occurring crystalline forms of SiO2. [Courtesy of Sawyer Research Products, Inc.] (b) A synthetic silicon crystal is created beginning with a raw material containing silicon (for instance, common beach sand), purifying out the silicon, and melting it. From a seed crystal, the molten silicon grows into a cylindrical crystal, such as the one shown here. The crystals (typically about 1.3 m long) are formed under highly controlled conditions to ensure that they are flawless and sliced into thousands of thin wafers, onto which the layers of an integrated circuit are etched. [Courtesy of the Museum of Modern Art, New York City.] 10-1 The Structure of Solids 415 The smallest unit of a crystal is called the unit cell. The structure of the unit cell depends on the type of bonding between the atoms, ions, or molecules in the crystal. If more than one kind of atom is present, the structure will also depend on their relative size. The structure may also change in response to changes in pressure and/or temperature. The bonding mechanisms are those discussed in Chapter 9: ionic, covalent, and dipole-dipole, the latter including the hydrogen and van der Waals bonds. In addition, a quantum-mechanical mechanism responsible for bonding metals in the solid state, metallic bonding, will be described later in this section. Ionic and Covalent Solids Figure 10-2 shows the structure of the ionic crystal NaCl. The Na and ions are ions approximately twice as large spherically symmetric (see Section 9-1) with the as the Na ions. The minimum potential energy of this crystal occurs when an ion of either kind has six nearest neighbors of the other kind. This structure is called facecentered-cubic ( fcc) because the unit cell is a cube and an ion, in this case occupies the center of each face. Note that the Na and ions are not paired into NaCl molecules in solid NaCl. Cl, Cl Cl Cl The net attractive part of the potential energy of an ion in a crystal can be written Uatt ke2 r 10-1 Cl ions in crystalline NaCl) and where r is the separation distance between neighboring ions (which is 0.282 nm for the Na and called the Madelung constant, depends on the geometry of the crystal. If only the six nearest neighbors of each ion were important, would be 6. However, in addition to the six neighbors of opposite charge at a distance r there are 12 ions of the same charge at a distance 8 ions of opposite charge at distance 21>2r, 31>2r,, and so on. The Madelung constant is thus an infinite sum: 20 25 6 12 22 8 23 6 2 \u00c1 10-2 Unfortunately, the sum in Equation 10-2 does not converge! We are saved by the fact that NaCl crystals are not spherical, as the analysis above implies. A better physical approach is to use cubic shells rather than spherical ones; then the cubic-shell equivalent", " Na+ Cl\u2013 Cl\u2013 Na+ Figure 10-2 Structure of the face-centered-cubic (fcc) NaCl crystal. 416 Chapter 10 Solid State Physics Table 10-1 Properties of selected crystalline solids Solid Bonding Equilibrium separation (nm) Crystal symmetry Madelung constant Cohesive Melting point energy (eV>atom) (K) NaCl LiBr KCl RbF CsCl ZnO Li Fe Au Zn C Si ionic ionic ionic ionic ionic ionic metallic metallic metallic metallic covalent covalent covalent Ge H2O dipole-dipole dipole-dipole C60 Ne dipole-dipole 0.282 0.275 0.315 0.282 0.348 0.222 0.302 0.248 0.288 0.266 0.154 0.235 0.245 0.367 1.00 0.313 fcc fcc fcc fcc sc hcp bcc bcc fcc hcp fcc fcc fcc hcp fcc fcc 1.7476 1.7476 1.7476 1.7476 1.7627 1.4985 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 3.19 3.10 3.24 3.55 3.27 7.22 1.63 4.28 3.81 1.35 7.37 4.63 3.85 0.52* 1.5* 0.020 1074 823 1043 1068 918 2248 454 1811 1338 693 \u2020 1687 1211 273? 24 * eV>molecule. \u2020 Diamond transforms to graphite at high temperature. The latter then sublimes at about 3800 K. of Equation 10-2 does converge, albeit slowly. The result for face-centered-cubic strucThe geometric details of other ionic arrangements tures such as NaCl is results in slightly different values for (See Table 10-1.) 1.7476.. Cl When Na and ions are very close together, they repel each other because of the overlap of the wave functions of their electrons and the exclusion principle repulsion discussed in Section 9-1. A simple empirical expression for the potential energy associated with this repulsion that works fairly well is where A and n are constants.1 The total potential energy of an ion is then Urep A rn ke2 r > The equilibrium separation r r0 is that at which the force F dU dr is zero. Differentiating Equation 10-3 and setting dU dr 0 at r r0, we obtain U A rn 10-3 > A ke2rn1 n 0 10-4 The total potential energy of an ion in the crystal can thus be written U ke2 r0 1 n r0 r r0 r b a d c n At r r0, we have U(r0) ke2 r0 c1 1 n d 10-1 The Structure of Solids 417 10-5 10-6 If we know the equilibrium separation r0, which can be found from x-ray diffraction experiments or computed from the crystal density, the value of n can be found approximately from the dissociation energy or lattice energy of the ionic crystal, which is the energy needed to break up the crystal into its constituent ions. In the case of NaCl the measured dissociation energy is 770 kJ mol. Using 1 eV 1.602 10 19 J and the fact that 1 mol of NaCl contains NA pairs of ions, we can express the dissociation energy in electron volts per ion pair, which makes possible an easier comparison with, e.g., the binding energy per molecule. The conversion between electron volts per ion pair and kilojoules per mole is > 1 eV ion pair 6.022 1023 ion pairs mol 1.602 1019 J 1 eV 96.47 kJ mol 10-7 > Thus, 770 kJ mol 7.98 eV per ion pair. Substituting 7.98 eV for U(r0), 0.282 nm for r0 (see Example 10-1), and 1.75 for in Equation 10-6, we can solve for n. The result is n 9.35 9. The dissociation energy is also used to compute the cohesive energy of a crystal, which is the potential energy per atom or per atomic pair rather than per ion pair and is the term used for all crystalline bonding mechanisms. For the NaCl illustration above, 7.98 eV is the energy needed to remove an Na and pair from the crystal. requires the input of 3.62 eV, and forming Na from Na releases Forming Cl from 5.14 eV. Therefore, the energy necessary to remove the neutral Na and Cl pair from the crystal is 7.98 eV 3.62 eV 5.14 e", "V 6.46 eV, and the cohesive energy of NaCl is 6.46 eV per Na and Cl pair. This result is in good agreement with the observed value of 3.19 eV atom in Table 10-1. A large cohesive energy implies a high melting point and vice versa. Cl Cl > EXAMPLE 10-1 Equilibrium Spacing r0 in an NaCl Crystal Calculate the equilibrium 2.16 g>cm3. spacing r0 for NaCl from the measured density of NaCl, which is SOLUTION We consider each ion to occupy a cubic volume of side r0. The mass of 1 mol of NaCl is 58.4 g, which is the sum of the atomic masses of sodium and chlorine. The 6.02 1023 is Avogadro\u2019s number. 2NAr3 0, ions occupy a volume of The density is thus related to r0 by where NA Then m V m 2NAr3 0 r3 0 m 2NA 58.4 g 2(6.02 1023)(2.16 g>cm3) 2.82 108 cm 0.282 nm r0 2.24 1023 cm3 418 Chapter 10 Solid State Physics ( EXAMPLE 10-2 Measuring r0 from X-Ray Diffraction Molybdenum strike the diagonal Bragg planes of the NaCl crystal shown on the right 0.071 nm) in Figure 3-15 such that a diffraction maximum (a bright Laue spot) is observed for 10.25\u00b0. Determine the value of r0. x rays K SOLUTION 1. Since NaCl is a cubic crystal, the distance d between the diagonal Bragg planes is related to the equilibrium separation r0 by d r0 22 2. The x-ray diffraction maxima satisfy the Bragg condition, Equation 3-23: 3. For m 1 and substituting d from above: 2d sin m 2a r0 22 bsin 4. Solving this for r0 and substituting values from above gives r0 22 2 sin (22)(0.071 nm) (2)(sin 10.25\u00b0) 0.282 nm Remarks: This result agrees with the value calculated from the density of NaCl in Example 10-1. Most ionic crystals, such as LiF, KF, KCl, KI, AgCl, and others formed by molecules in Table 9-2, have a face-centered-cubic structure. Some elemental solids that also have this structure are Ag, Al, Au, Ca, Cu, Ni, and Pb. Figure 10-3 illustrates the structure of another ionic crystal, CsCl, which is called simple cubic (sc) because it can be considered as two interpenetrating cubic structures, one of Cs ions and the other of ions. In this structure, each ion has eight nearest-neighbor ions of the opposite charge. The Madelung constant for ionic crystals with simple cubic structure is 1.7627. Other crystals with this structure include CsI, TlI, TlBr, LiHg, and NH4Cl. Some elemental solids, such as Ba, Cs, Fe, K, Li, Mo, and Na, also crystallize with the structure shown in Figure 10-3; when the atoms are the same at the vertices and in the center of the cube, the structure is called body-centered-cubic (bcc). Cl Cl\u2013 Cl\u2013 Cs+ Cs+ Figure 10-3 Structure of the simple cubic (sc) crystal CsCl. 10-1 The Structure of Solids 419 Figure 10-4 illustrates another important crystal structure, called hexagonal closepacked (hcp). This is the structure obtained by stacking identical spheres such as Ping-Pong balls. In one layer, each ball touches six others; hence the name hexagonal. In the next layer, each ball fits into the triangular depressions of the first layer. In the third layer, each ball fits into a triangular depression of the second layer such that it lies directly over a ball in the first layer. Elements with a hexagonal close-packed crystal structure include Be, Cd, Ce, Mg, Os, Zn, and Zr. There are a total of 14 different types of three-dimensional crystal lattice structures, of which we have discussed only a few of the most common ones. (See Appendix B3.) EXAMPLE 10-3 Madelung Constant for a Two-Dimensional Crystal Calculate out to four terms in the series the Madelung constant for the hypothetical univalent, two-dimensional ionic crystal shown in Figure 10-5. Figure 10-4 Hexagonal close-packed (hcp) crystal structure. SOLUTION The net attractive potential is given by Equation 10-1. When the negative ion at the origin of Figure 10-", "5 is considered, there are four positive ions located a distance r away, as indicated by circle A in the diagram. There are four negative ions lying on circle B, whose radius is Four negative ions are located on circle C, whose distance from the ion at the origin is 2r, and finally, eight positive ions lie on circle D at from the origin. Therefore, to four terms the net attractive potential is 21>2r. 51>2r Uatt ke2a 4 r 4 22 r 4 2r 8 25 r b or Uatt ke2 r a4 4 22 b 2 8 25 The quantity in parentheses is the Madelung constant correct to four terms in the infinite expansion; thus we have that 2.749 \u2013 + \u2013 + \u2013 + \u2013 Figure 10-5 A hypothetical univalent two-dimensional ionic crystal (Example 10-3). 420 Chapter 10 Solid State Physics (a) Figure 10-6 (a) Diamond crystal structure showing how this structure can be considered to be a combination of two interpenetrating face-centeredcubic structures. (b) Synthetic diamonds magnified about 50,000 times. In diamond, each carbon atom is at the center of a tetrahedron formed by four other carbon atoms. [Courtesy of Chris Kovach Discover Publications.] > (b) 1 m In covalently bonded crystals the nature of the individual bonds is just like that in covalently bonded molecules, as was described in Section 9-2. The electron-sharing character of the bond enhances its effectiveness in crystals, for example, enabling tetravalent carbon atoms to form bonds with as many as four other carbon atoms. The crystal structure is determined by the directional nature of the bonds. Figure 10-6 illustrates the diamond structure of carbon (which is also the structure of Ge and Si), in which each atom is bonded to four others located at the vertices of a regular tetrahedron as a result of the sp3 hybridization discussed in the Chapter 9 MORE section Other Covalent Bonds on the home page. The diamond structure can be considered to be two interpenetrating face-centered-cubic structures. This arrangement with equal bond angles is particularly tightly bound and results in the carbon diamond structure having one of the largest atomic cohesive energies of all solids, about 7.37 eV per carbon atom. Carbon has two other well-defined crystalline structures, graphite and solid fullerenes,2 both the result of carbon orbitals hybridized in the sp2 configuration. In graphite, illustrated in Figure 10-7a, three of the valence electrons link each atom to (a) (b) 3 nm (c) Figure 10-7 (a) An atomic-force micrograph of graphite. The brighter spots at each vertex are single carbon atoms. In graphite, carbon atoms are arranged in sheets, each sheet made up of atoms in hexagonal rings. The sheets slide easily across one another, a property that allows graphite to function as a lubricant. [Courtesy of Srinivas Manne, University of California at Santa Barbara.] (b) This high-resolution transmission electron micrograph shows clearly the close-packed fcc arrangement of the C60 molecules in the solid fullerene. [Courtesy of P. R. Buseck, Science, 257, 215 (1992).] (c) Carbon nanotubes grown on a titanium substrate. The nanotubes are perpendicular to the substrate and range between 40 nm and 100 nm in diameter. [Courtesy of Z. F. Ren et al., Boston College.] 10-1 The Structure of Solids 421 three near neighbors via directed bonds, forming a planar hexagonal structure. The planes thus formed are connected by much weaker dipole-dipole forces. This results in a structure consisting of strong sheets that can be readily separated from one another. The structure of the fullerenes, using solid C60 as an example, is quite different from that of both diamond and graphite. As described in Section 9-2, the C60 molecule achieves its spheroid shape by incorporating 12 pentagons into the hexagonal structure, thus distorting the graphite planes into the soccer-ball configuration. The C60 molecules are then bonded to each other by dipole-dipole forces, just as are the sheets of graphite. As a result, the cohesive energy per atom is quite high, about 7.4 eV, or nearly equal to that of diamond, but the cohesive energy per molecule is low, only 1.5 eV. The C60 crystal, shown in Figure 10-7b, is face-centered-cubic. The equilibrium separation between the molecules is 1.00 nm. The nanotubes shown in Figure 10-7c are a remarkable example of carbon\u2019s possible bonding configurations. Metallic Bonding in Solids All solid metals, formed from the metal elements that make up more than half of the", " periodic table, are bonded by the metallic bond, which, as was noted earlier, has no single-molecule counterpart. It is somewhat analogous to the covalent bond, in which the atoms of the molecule share one or more electrons. In the metallic bond one or two of the valence electrons of each atom are free to move throughout the solid, and all of the atoms share all of those electrons. Thus, the metallic crystal can be pictured as a lattice of fixed, positive ions immersed in an electron gas. It is the attraction between the positively charged lattice and the negatively charged electron gas that results in bonding of the solid.3 To see how metallic bonding occurs, let us consider a specific simple example, the bonding of solid lithium. The electron structure of the lithium atom is and the radial wave function of the 2s electron, which \u201csees\u201d a hydrogenlike core consisting of the nucleus and the completed 1s shell, is 1s22s, \u2019 20 C20 a2 ber>2a0 r a0 10-8. where C20 is a normalization constant and a0 is the Bohr radius. The probability density corresponding to this wave function for a single lithium atom located at r 0 is shown in one dimension in Figure 10-8a. The probability density decays exponentially to zero as r approaches Figure 10-8b illustrates the probability density of the electrons in the metal, which must be the same around each Li ion core. The peaks of the probability density are now closer to the positive Li ion core than was the case for the isolated atom. Thus, the potential energy of the electrons has been reduced. However, the effect of assembling the atoms into a lattice has also been to effectively confine the electrons to within about 0.3 nm of the ion core rather than the larger volume of the isolated atom. The uncertainty principle then implies an increase in the momentum, hence kinetic energy, of the electrons. The metallic bond is stable because the rise in kinetic energy is more than offset by the decrease in the potential energy, thus lowering the total energy of the system of atoms. The net effect is greatest when the difference in size between the atom and the core is large (so that the magnitude of the potential energy reduction is large) and when the number of valence electrons is small (so that the increase in kinetic energy is as small as possible). These conditions are increasingly satisfied as one moves toward the left across the periodic table. Racing bicycles and Formula 1 race cars are constructed from woven carbon fibers. They absorb shock extremely well and are lighter and stronger than older bikes and race cars made of steel or aluminum. Even hydrogen becomes a metal under ultrahigh pressure. The pressure reduces the conductionvalence band gap (see Section 10-6) from about 15 eV to 0.3 eV. Understanding metallic hydrogen will be of significant benefit to fusion energy research (see Section 11-8). 422 Chapter 10 Solid State Physics Figure 10-8 (a) Probability density for the 2s electron in an isolated Li atom. (b) Probability density for the 2s electrons in a (one-dimensional) Li crystal. The large dots on the r axis represent Li nuclei. Note that \u0192\u2019\u0192 2 is compressed relative to that of the single atom and that an electron is, on the average, confined to within about 0.3 nm of a Li nucleus rather than between. (a) \u23aa\u03c8 \u23aa2 20 0.3 0.2 0.1 0 0.1 0.2 0.3 r, nm (b) \u23aa\u03c8\u23aa2 0.6 0.3 0 0.3 0.6 r, nm Questions 1. Why is r0 different for solid NaCl than for the diatomic molecule? 2. Why would you not expect NaCl to have an hcp structure? 3. How can you account for the difference in the Madelung constants of NaCl and CsCl? 4. Although it is in the same column of the periodic table as Li, why is solid hydrogen not metallically bonded? 10-2 Classical Theory of Conduction Because metals conduct electricity so readily, there must be charges in metals that are relatively free to move. The idea that metals contain electrons free to move about through a lattice of relatively fixed positive ions was proposed by the German physicist Paul Drude around 1900, just three years after Thomson\u2019s discovery of the electron, and was developed by H. A. Lorentz in about 1909. This microscopic model, now called the classical model of electrical conduction, successfully predicts Ohm\u2019s law and relates electrical conduction and heat conduction to the motion of free electrons in conductors. However, the model gives the wrong temperature dependence for electrical conductivity, and it predicts that the heat capacity of metals should be greater than )R per mole, which is not observed. Despite these failures, the that of", " insulators by ( classical free-electron theory is a good starting point for a more sophisticated treatment 3>2 10-2 Classical Theory of Conduction 423 of metals based on quantum mechanics. For that reason, a discussion of the classical theory is included in the Classical Concept Review on the Web page. In this section we will briefly outline those predictions from classical theory that are pertinent to our subsequent discussion of the quantum mechanical theory of conduction. As we will see, the main defects in the classical theory are the use of the classical Maxwell-Boltzmann distribution function for electrons in a conductor and the treatment of the scattering of electrons by the lattice as a classical particle scattering. Electrical Conduction The classical model of a metal is a regular three-dimensional array of atoms or ions with a large number of electrons free to move throughout the entire metal. In the absence of an applied electric field the average speed of these electrons is quite high. For example, at T 300 K, their average speed is v d \u0394 t e A vd 8v9 8kT me A B 8(1.38 1023 J>K)(300 K) (9.11 1031 kg) 1.08 105 m>s 10-9 Figure 10-9 In time t, all the charges in the shaded volume pass through A. If there are n charge carriers per unit volume, each with charge e, the total charge in this volume is Q nevd A t, where vd is the drift velocity of the charge carriers. The total current is then I Q t nevd A. > Applying an electric field on the free electrons that is opposite to the field direction. For n electrons per unit volume, the resulting current I in the conductor, the charge Q through a cross-sectional area A per unit time (see Figure 10-9), is superimposes a drift velocity vd e I \u00a2Q \u00a2t neAvd 10-10 In contrast to the average speed of the electrons due to their thermal motion, the drift velocity is quite low, as Example 10-4 illustrates for copper. EXAMPLE 10-4 Drift Velocity of Electrons in Copper What is the magnitude of the drift velocity of electrons in a typical copper wire of radius 0.815 mm carrying a current of 1 A? SOLUTION If we assume one free electron per copper atom, the density of free electrons is the same as the density of atoms na, which is related to the mass density Avogadro\u2019s number NA, and the molar mass M by, na NA M For copper and M 63.5 g mol. Then 8.92 g>cm3 (8.93 g>cm3)(6.02 1023 atoms>mol) 63.5 g>mol > na 8.47 1022 atoms>cm3 The density of electrons is then 8.47 1022 electrons>cm3 8.47 1028 electrons>m3 The magnitude of the drift velocity is therefore vd I Ane 1 C>s (0.000815 m)2(8.47 1028 m3)(1.60 1019 C) 3.54 105 m>s 424 Chapter 10 Solid State Physics We see that typical drift velocities are of the order of 0.01 mm s, which is quite small. Notice in particular that the magnitude of the drift velocity is very small compared with the average speed of the electrons due to their thermal energy as given by Equation 10-9. Indeed, the difference is approximately 10 orders of magnitude. > > According to Ohm\u2019s law, the current in a conducting wire segment is proportional 1>R, to the voltage drop across the segment I so that I V R. The resistance R of the wire is independent of both I and V, being proR L>A. portional to the length of the wire L divided by its cross-sectional area A: is called the resistivity of the conductor material. The constant of proportionality Combining these two expressions and recalling that the electric field in the wire is e V>L, V. The constant of proportionality is Equation 10-10 enables us to write j ne vd 10-11 where j I A is the current density. For materials that obey Ohm\u2019s law, the resistivity and, of course, its reciprocal the conductivity must be independent of e. > Mean Free Path in The objective of the classical theory of conduction is to find an expression for terms of the properties of the conductor, a task that is aided by a consideration of the average distance an electron travels in the conductor between collisions, called the. mean free path and the average time be, tween collisions It is the product of the average speed called the relaxation time: 8v9 8v9 1>na where na is the number of ions per unit volume and r is the ion radius. As an exam- ple, the resistivity and conductivity are given by for copper", " is about 0.38 nm. In terms of 10-12 r2 and 8v9 me ne2 ne2 8v9 me 10-13 In Example 10-5 we compute the classical values for the resistivity and conductivity for copper, which illustrates a basic defect in the classical theory of conduction. EXAMPLE 10-5 Conductivity and Resistivity of Copper Calculate the values of the resistivity and the conductivity of copper at 300 K. SOLUTION Using Equations 10-12 and 10-13, we have >8v9 (0.38 10 9 m)>(1.08 105 m>s) 3.52 1015 s and then 8v9 me ne2 me ne2 9.11 1031 kg (8.47 1028 electrons>m3)(1.60 1019 C)2(3.52 1015 s) 1.19 107 \" # m and 1> 8.33 106 (\" # m)1 Remarks: This value for the resistivity is about 7 times greater than the measured value of 1.7 108 V # m. 10-2 Classical Theory of Conduction 425 Defects in the Classical Theory e vd, i.e., and ee. ee>me At first glance it is surprising that any material obeys Ohm\u2019s law since in the presIf this ence of an electric field, a free electron experiences a force of magnitude were the only force acting on the electron, it would have acceleration and its velocity would steadily increase. However, the agreement of Ohm\u2019s law with experiment implies that there is a steady-state situation in which the drift velocity of e because the current I is proportional to the electron is proportional to the field j>ne e>ne. vd both In the classical model, it is assumed that a free electron is accelerated for a short time and then collides with a lattice ion. After the collision, the velocity of the electron is assumed to be completely unrelated to that before the collision. The justification for this assumption is that, as we have seen, the drift velocity is very small compared to the average thermal velocity. With the average speed given by Equation 10-9 and the mean free path by Equation 10-12, the resistivity has been expressed in terms of the properties of metals, which was the objective of the classical theory of conduction. According to Ohm\u2019s law, the resistivity is independent of the electric field The only quantities in Equation and the mean 10-13 that might depend on the electric field are the average speed free path As we have seen, the drift velocity, a result of the applied electric field, is very much smaller than the average thermal speed of the electrons in equilibrium with the lattice ions. Thus, the electric field has essentially no effect on the average speed of the electrons. The mean free path of the electrons depends on the size of the lattice ions and on the density of the ions, neither of which depends on the elece. tric field Thus the classical model predicts Ohm\u2019s law with the resistivity as given by Equation 10-13. 8v9 e.. Although successful in predicting Ohm\u2019s law, the classical theory of conduction has several defects. We saw from Example 10-5 that the magnitude of the resistivity of copper calculated from Equation 10-13 is about 7 times the measured value at T 300 K. The temperature dependence of is also not correct. Experimentally, the resistivity varies linearly with temperature over a wide range of temperatures. The temperature dependence of resistivity in Equation 10-13 is given completely by the speed Thus, this calculation does not give a linear dependence on temperature. Finally, the classical model says nothing about why some materials are conductors, others insulators, and still others semiconductors. which according to Equation 10-9 is proportional to T1>2. 8v9, In the quantum-mechanical theory of electrical conduction, which is discussed in Section 10-4, the resistivity is again given by Equation 10-13, but the average speed and the mean free path are interpreted in terms of quantum theory. We discovered in Section 8-5 that the average energy of the electrons, hence their average speed, is not but is approximately independent of T because electrons do not proportional to obey the Boltzmann distribution law, but instead obey the Fermi-Dirac distribution. Also, in the quantum-mechanical calculation of the mean free path the wave nature of the electron is important and must be taken into account. T1>2, 426 Chapter 10 Solid State Physics Heat Capacity If the electron gas in metals were a classical ensemble of identical distinguishable particles, it would obey Boltzmann statistics (see Chapter 8) and have the Maxwell dis3>2 )kT, and we tribution of speeds. It should then have an average kinetic energy ( )", "R greater than that of an would expect the molar heat capacity of a metal to be ( insulator\u2014that is, 3R from the lattice vibrations (rule of Dulong and Petit\u2014see Section 8-1) and ( )R from the electron gas: 3>2 3>2 Cv (3R)lattice vibrations (3>2)Relectron gas (9>2)R As was noted in Section 8-5, this is not observed. The molar heat capacity of metals is very nearly 3R. At higher temperatures it is slightly greater, but the increase is )R predicted by the classical theory. The increase is, nowhere near the value of ( in fact, proportional to temperature, and at T 300 K, it is only about 0.02R. 3>2 10-3 Free-Electron Gas in Metals Classically, at T 0, all the electrons in a metal would have zero kinetic energy. As a )kT, conductor is heated, the lattice ions acquire an average kinetic energy of ( which is imparted to the electron gas by interactions of the lattice with the electrons. )kT The electrons classically would be expected to have a mean kinetic energy of ( in equilibrium. Quantum mechanically, however, since the electrons are confined to the space occupied by the metal, it is clear from the uncertainty principle that even at T 0, an electron cannot have zero kinetic energy. Furthermore, the exclusion principle prevents more than two electrons (with opposite spins) from being in the lowest energy level. At T 0, we expect the electrons to have the lowest energies consistent with the exclusion principle. This is illustrated clearly by first considering a onedimensional model that provides us with the foundation needed for the quantum theory of conduction in Section 10-4. 3>2 3>2 One-Dimensional Model To simplify visualization, let us first consider N electrons in a one-dimensional infinite square well of width L. The physical analog of such a model could be a long, thin metal wire. As we have seen previously, the allowed energies are given by En n2h2 8mL2 n2E1 10-14 > h2 8mL2 is the energy of the ground state. where m is the electron mass and E1 Since two electrons can be put in the n 1 level, two in the n 2 level, etc., at T 0, > the N electrons in the system will fill N 2 levels, i.e., from the n 1 to the n N 2 state. (See Figure 10-10a.) The energy of the last filled level (or half-filled level, if N happens to be odd) is the Fermi energy, which for our one-dimensional system is > EF EN>2 (N>2)2h2 8mL2 h2 32m a N L 2 b 10-15 10-3 Free-Electron Gas in Metals 427 (a) E (b) E EF 0 n = (N/2) + 1 n = N/2 n = (N/2 EF dE 0 x L Figure 10-10 (a) A one-dimensional infinite square well for N electrons at T 0 K. Two electrons, one with spin up and one with spin down, occupy each level. The Fermi energy is the energy of the level with n N 2, the highest occupied level. (b) The levels are so closely spaced, they can be assumed to be continuous. The density of states g(E) is the number of states between E and E dE divided by dE. > We see that the Fermi energy is a function of the number of electrons per unit length, which is the number density or number per unit volume in one dimension. The number density of electrons in copper, computed in Example 10-4, is 8.47 1028 m3. In one dimension this corresponds to > N L (8.47 1028>m3)1>3 4.40 109>m 0.440>\u00c5 4.40>nm Using this value, we see that the Fermi energy for a one-dimensional copper system, such as a wire, is EF (hc)2 32mc2 a N L 2 b (1240 eV # nm)2(4.40>nm)2 (32)(5.11 105 eV) 1.82 eV This value is much larger than the room temperature value of kT, which is about 0.026 eV. The average energy of the electrons is the total energy divided by N: 8E9 1 N N>2 a n1 2n2E1 where the factor of 2 accounts for the two electrons in each energy state. Since N>2 W 1, the summation can be replaced by an integral, so we have that 8E9 E1 N N>2 0 2n2 dn E1 N", " 2 3 a N 2 3 b h2 8mL2 2 3N a N 2 3 b 1 3 EF 10-16 Our one-dimensional calculation thus gives an average energy for copper\u2019s free electrons of about 0.6 eV at T 0. This is 15 times the room-temperature average kinetic energy of molecules in the atmosphere. The temperature at which the average energy would be 0.6 eV for a one-dimensional Boltzmann distribution is about 14,000 K, obtained from 1 2 kT 0.6 eV. 428 Chapter 10 Solid State Physics The expression for the number n(E) of electrons with energy E in the one- dimensional system follows from Equation 8-37c: n(E) dE g(E) fFD(E) dE fFD(E) 1 for T 0 K and E EF and 0 for T 0 K and E EF. where The density of states g(E) is the number dn of states between E and E dE divided by dE and multiplied by 2 to account for the two spin states per space state (see Figure 10-10b): fFD(E) g(E) 2 dn dE Since E n2E1, then dE 2E1n dn 2E1>2 g(E) E1>2 1 E1>2 dn 1 E1>2 and we have that The number of electrons with energy E at T 0 K in the one-dimensional conductor is then n(E) b E1>2 1 E1>2 0 for E EF for E EF 10-17 Three-Dimensional Electron Gas Now let us extend the discussion to three-dimensional systems. The Fermi energy can be computed from the general expression for the number of fermions nFD(E) given by Equation 8-37c. The number density N V of electrons in three dimensions, where V is the volume of the metal, is > N V 1 V 0 nFD(E) dE 2 a 8m h2 3>2 b 0 E1>2 dE e(EEF)>kT 1 10-18 For arbitrary values of T, Equation 10-18 must be evaluated numerically, but for 0 or 1 as E T 0 K, the solution is straightforward since, as noted above, is greater than or less than EF. In that event, we have that fFD(E) N V 2 a 8m h2 b 3>2 EF E1>2 dE 0 2 a 8m h2 b 3>2 2 3 3>2 EF Solving for EF we have for T 0 K that EF 2>3 h2 2m a 3N 8V b 10-19 10-20 Table 10-3 lists the number density of free electrons for several elements. Notice that EF increases slowly with N V, as would be expected at T 0 K since all states up > to EF are being filled and an increasing N V requires more states to be filled, i.e., a larger value of EF. The number n(E) of electrons with energy E is then given by > n(E) 3>2 2 a 8m h2 b VE1>2 3N 2 F E1>2 E3>2 10-21 10-3 Free-Electron Gas in Metals 429 Table 10-3 Free-electron number densities, Fermi energies, and Fermi temperatures for selected elements Element N V ( 1028 m 3) > Fermi energy (eV) Fermi temperature ( 104 K) Al Ag Au Cu Fe K Li Mg Mn Na Sn Zn 18.1 5.86 5.90 8.47 17.0 1.40 4.70 8.61 16.5 2.65 14.8 13.2 11.7 5.53 5.55 7.06 11.2 2.13 4.77 7.14 11.0 3.26 10.3 9.50 13.6 6.41 6.43 8.19 13.0 2.47 5.53 8.28 12.8 3.78 11.9 11.0 and the average energy of an electron at T 0 K by EF EF E n(E) dE 3 2 8E9 1 N E3>2 F 0 0 E3>2 dE 3 5 EF 10-22 At temperatures greater than zero, some electrons will gain energy and occupy higher energy states. However, electrons cannot move to a higher or lower energy state unless the state is unoccupied. Since the kinetic energy of the lattice ions is of the order of kT, electrons cannot gain much more energy than kT in collisions with the lattice; therefore, only those electrons with energies within about kT of the Fermi energy can gain energy as the temperature is increased. At T 300", " K, kT is only 0.026 eV, so the exclusion principle prevents all but a very few electrons near the top of the energy distribution from gaining energy by random collisions. Figure 10-11 shows the small fraction of the free electrons that move at T 300 K (shaded rectangle at the Fermi energy of the T 0 K curve). Even for temperatures as high as several thousand degrees, the energy distribution of an electron gas does not differ very much from that at T 0 K. For values of T 0, we must remember that the Fermi energy is defined by Equation 8-68, since for T 0, there is no state below which all states are full and above which all states are empty. Equation 8-68 defines the Fermi energy as that energy for which fFD(E) 1 For all but extremely high temperatures, the difference 2. and that at T 0 between the Fermi energy at temperature T, K, EF(0) is essentially negligible. As is clear from Equation 8-68 and at arbitrary T differs from that at Figure 8-31b, the value of T 0 K only for those energies within about kT of the Fermi energy. fFD(E) EF(T) ) E n ( kT T > 0 EF E Figure 10-11 The fraction of the N electrons in the metal that contribute to Cv is the ratio of the shaded rectangle to the total area under the n(E) versus E curve. 430 Chapter 10 Solid State Physics Fermi Temperature It is convenient to define the Fermi temperature TF by EF 10-23 kTF For temperatures much lower than the Fermi temperature, the average energy of the lattice ions will be much less than the Fermi energy; thus the electron energy distribution will not differ greatly from that at T 0. The Fermi temperature corresponding to 7.0 eV for copper is about 81,900 K. Table 10-3 lists the Fermi temperatures for EF several elements. At temperatures much larger than the Fermi temperature (i.e., much larger than 81,900 K for copper) and the Fermi-Dirac distribution approaches the Boltzmann distribution. This result is not very important for the understanding of the behavior of conductors since there are no conductors that remain as solids or even liquids at such extreme temperatures. approaches fFD(E) eE>kT EXAMPLE 10-6 Fermi Energy and Temperature of Silver Compute (a) the Fermi energy and (b) the Fermi temperature for silver at 0 K. SOLUTION The density of silver is 10.50 gm cm3, and its molecular weight is 107.9 gm mol. If each silver atom is assumed to contribute one electron to the Fermi gas, the number density N V is computed as follows: > > > N V (10.50 g>cm3)(1>107.9 g>mol)(6.02 1023 electrons>mol) 5.86 1022 electrons>cm3 5.86 1028 electrons>m3 which agrees with the entry in Table 10-3. (a) The Fermi energy is then, from Equation 10-20, (6.63 1034 J # s)2 2(9.11 1031 kg) EF a 3 5.86 1028 8 b 2>3 8.84 1019 J 5.53 eV in agreement with the entry for Ag in Table 10-3. (b) The Fermi temperature is then TF EF k 8.84 1019 J 1.38 1023 J>K 6.41 104 K again, in agreement with Table 10-3. 10-4 Quantum Theory of Conduction With two relatively simple but important quantum-mechanical modifications of the classical free-electron theory, we can understand the electrical conductivity, heat capacity, and thermal conductivity of metals. First, we must replace the classical Boltzmann distribution with the Fermi distribution of energies in the electron gas, as was discussed in Section 8-5. Second, we must consider the effect of the wave properties of the electrons on their scattering by the lattice ions. We will discuss the latter modification qualitatively. 10-4 Quantum Theory of Conduction 431 Electrical Conduction We might expect that most of the electrons would not participate in the conduction of electricity because of the exclusion principle, but this is not the case because the electric field accelerates all the electrons together. Figure 10-12 shows the Fermi-Dirac distribution function versus velocity for some temperature T that is small compared with TF (such as T 300 K). The function is approximately 1 for uF u F, where the Fermi speed uF is the speed corresponding to the Fermi energy: vx a uF 1>2 b 2EF me 10-24 EXAMPLE 10", "-7 Fermi Speed in Al Compute the Fermi speed of electrons in aluminum. SOLUTION From Table 10-3, the Fermi energy EF of Al is 11.7 eV. Thus, uF(Al) a 2 11.7 eV 1.60 1019 J>eV 9.11 1031 kg b 1>2 2.03 106 m>s The dashed curve in Figure 10-12 shows the Fermi distribution after the electric field has been acting for some time t. Although all of the electrons have been shifted to higher velocities, the net effect is equivalent to shifting only the electrons near the Fermi level; therefore, we can use the classical equations for the resistivity and conductivity (Equations 10-13) if we use the Fermi speed uF in place of 8v9: 1 meuF ne2 10-25 We now have two problems. First, since uF is independent of temperature (to a very is independent of temperature good approximation), the above expression for unless the mean free path depends on it. The second problem concerns the magnitudes. We saw in Example 10-5 that the classical expression for yielded a result that was calculated from the Maxwell-Boltzmann distritoo small by a factor of 7, using bution. Since uF is about 19 times the value of the magnitude of predicted from Equation 10-25 will be even smaller by another factor of 19 and the magnitude of will, correspondingly, be larger than the observed value by the same factor. 8v9, and 8v9 No electric field With electric field fFD(E) 1 0 \u2013uf +uf vx fFD(E) Figure 10-12 Occupation versus probability velocity in one dimension, with no electric field and with an electric field in the x direction. The difference is greatly exaggerated. 432 Chapter 10 Solid State Physics The resolution of both of these problems lies in the way that the value of the mean free path is calculated. If we use uF from Equation 10-24 and the experimental value 1.7 108 \" # m for copper in Equation 10-25, we obtain for the mean free path 39 nm, about 100 times the value of 0.38 nm, which was noted in Section 10-2 for Cu ions. We shouldn\u2019t be too surprised that the mean free path of electrons in the copper lattice is not given correctly by classical kinetic theory. The reason for this large discrepancy between the classical calculation of the mean free path and the \u201cexperimental\u201d result calculated from Equation 10-25 is that the wave nature of the electron must be taken into account. The collision of an electron with a lattice ion is not similar to the collision of a baseball and a tree. Instead, it involves the scattering of the electron wave by the regularly spaced ions of the lattice. If the wavelength is long compared with the crystal spacing, as is approximately the case here (see Problem 10-55), Bragg scattering cannot occur. Detailed calculations of the scattering of electron waves by a perfectly ordered crystal of infinite extent show that there is no scattering, and the mean free path is infinite. Thus, the scattering of electron waves must arise from imperfections in the crystal lattice. The most common imperfections are due to impurities and to thermal vibrations of the lattice ions. r2. In Equation 10-12, for the classical mean free path, the quantity can be thought of as the cross-sectional area of the lattice ions as seen by the electron, where r is the ion radius. Figure 10-13a depicts the classical picture in which the lattice ions have area According to quantum mechanics applied to the scattering of electron waves, however, the \u201carea\u201d of the ion\u2019s cross section seen by the electron wave has nothing to do with the size of the ion. Instead, it depends upon the deviations of the lattice ions from a perfectly ordered array. We can estimate the magnitude of the deviations and thus compute a more accurate value for the mean free path in the following way. r2 r2 x2 y2 Let us assume that the lattice ions are points that are vibrating because of their r2, thermal energy. (See Figure 10-13b.) We will take for the scattering cross section where is the mean-square displacement of the point atom in a plane perpendicular to the direction of the electron\u2019s motion and represents a measure of the deviation of the ion from its equilibrium location. We can calculate from the equipartition theorem. We have r2 1 2 Kr2 1 2 M2r2 kT 10-26 where K is the force constant, M is the mass of the ion, and gular frequency of vibration. The mean free path is then (K>M)1>2 is the an- 1 nr2 M2 2nk 1 T 10-27 We thus see", " that this argument gives the correct temperature dependence for and that is, from the classical calculation. (Equation 10-25), and so ; as was obtained rather than T1>2, 1> T We can then calculate the magnitude of using the Einstein model of a solid, which is fairly accurate except at very low temperatures. In the Einstein model (see Section 8-4) all the atoms vibrate with the same frequency. The Einstein temperature was defined by Equation 8-63 as and therefore r2,, kTE hf U (a) r Area = \u03c0r2 (b) r0 Area = \u03c0r2\u23af\u23af r2 to the Figure 10-13 (a) Classical picture of the lattice ions as spherical balls of radius r that present an area electrons. (b) Quantummechanical picture of the lattice ions as points that are vibrating in three dimensions. The area presented to the r2 where r0 is electrons is 0, the amplitude of oscillation of the ions. R/R290 K 4.0 \u00d7 10 \u20133 3.0 \u00d7 10 \u20133 2.0 \u00d7 10 \u20133 1.0 \u00d7 10 \u20133 10-4 Quantum Theory of Conduction 433 Figure 10-14 Relative resistance versus temperature for three samples of sodium. The three curves have the same temperature dependence but have different magnitudes because of differing amounts of impurities in the samples. [From D. MacDonald and K. Mendelssohn, Proceedings of the Royal Society, A202, 103 (1950).] 2 4 6 8 10 14 16 18 20 12 T, K Using this for we have, r2 2kT M2 2TU2 MkT2 E 2(Uc)2 Mc2kTE T TE 10-28 The Einstein temperature for copper is about 200 K, corresponding to an energy of 0.0172 eV. Using this and Mc2 63.5 931 MeV for the mass of a copper kTE ion, the value of at T 300 K is r2 r2 2(197.3 eV # nm)2 (63.5 931 106 eV)(0.0172 eV) 300 K 200 K 1.14 104 nm2 Since this is about 100 times smaller than the area presented by a copper ion of radius 0.1 nm, the mean free path is about 100 times larger than that calculated from the classical model, in agreement with that calculated from the measured value of the conductivity. We see, therefore, that the free-electron model of metals gives a good account of electrical conduction if the classical average speed is replaced by the Fermi speed uF and if collisions are interpreted in terms of the scattering of electron waves for which only deviations from a perfectly ordered lattice are important. The presence of impurities in a metal also causes deviations from perfect regularity in the crystal. The effects of impurities on resistivity are approximately independent of temperature. The resistivity of a metal containing impurities can be written is due to impurities. Figure 10-14 shows a typical resistance versus temperature curve for a metal with impurities. As the temperature approaches zero, approaches zero and the resistivity approaches the constant is due to the thermal motion of the lattice and t I, where I t t I. Heat Capacity Next, let\u2019s estimate the contribution of the electron gas to the molar heat capacity. At T 0, the average energy of the electron, given by Equation 10-22, is ( )EF, so the total energy for N electrons is U At a temperature T, only those electrons near the Fermi level can be excited by random collisions with the lattice ions, which have (3>5)NEF. 3>5 434 Chapter 10 Solid State Physics an average energy of the order of kT. The fraction of the electrons that are excited is of the order kT EF, and their energy is increased from that at T 0 by an amount of the order of kT. We can thus write for the energy of the N electrons at temperature T, > U 3 5 NEF N kT EF kT 10-29 is some constant, which we expect to be of the order of 1 if our reasoning is where requires the use of the complete Fermi electron distribucorrect. The calculation of tion at an arbitrary temperature T and is quite difficult. Such a calculation, first carried out by A. A. Sommerfeld, shows that this equation is correct with Using this result, the contribution of the electrons to the molar heat capacity is 2>4. Understanding quantum theory of heat conduction and heat capacity has made possible such devices as \u201chigh-tech\u201d frozen yogurt/ice cream makers for consumers. Using specially designed working fluids, they are replacing the old salt/ice-ina-bucket freezers because of their convenience. Cv(electrons) dU dT 2Nk", " kT EF 2 2 R T TF 10-30 where Nk R for 1 mole and TF EF k is the Fermi temperature. We see that because of the large value of TF, the contribution of the electron gas is a small fraction of R at 81,900 K for copper, the molar heat capacity of the ordinary temperatures. Using TF electron gas at T 300 K is > Cv 2 2 a 300 81,900 bR 0.018R which is in reasonable agreement with the value estimated from the small fraction of electrons with energies greater than EF in Figure 10-11 and in good agreement with experiment. More Quantum theory readily accounts for heat conduction, predicting results in good agreement with observations. Thermal Conduction\u2014 The Quantum Model is outlined briefly on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Equations 10-31 and 10-32 here. Questions 5. When the temperature is lowered from 300 K to 4 K, the resistivity of pure copper drops by a much greater factor than that of brass. Why? 6. Explain why, physically, you would expect the mean free path of electrons in a metal to decrease as the temperature increases. 10-5 Magnetism in Solids Electron spins with their associated magnetic moments are the origin of magnetism in solids. If the atoms of the solid have unpaired spins, the solid itself may have a net magnetic moment. Since the atoms are effectively fixed in one or another of the several crystalline structures, the interactions between them may have a substantial effect on the magnetism exhibited by the solid. Several types of magnetism are observed in solids, ferromagnetism being perhaps the most familiar, though hardly the most common, among elements and compounds. In this section we will describe each of the several types. Paramagnetism 10-5 Magnetism in Solids 435 1 2 Consider a solid consisting of atoms that each have an unpaired electron spin; that is, 13>4U, of course) and the atoms do not intereach atom has a net spin of (actually act magnetically. Examples of such solids are the rare earth elements and many of the transition elements. In that event, the only magnetic energy the system may have results from interaction with an applied external field B. Such a solid, one with no net magnetic moment in the absence of an applied external field, is called paramagnetic. The magnetic moment of each atom is thus that of the unpaired electron gs ss>U. Its z component is given by Equation 7-49: z B msgs B 7-49 where gs is the g factor for the electron and is the Bohr magneton. In an applied field B, whose direction provides the atom with an external z axis, the possible energies of the magnetic moment are or U zB U msgs BB 7-47 10-33 1>2 1>2, ms ms orientation of s (called \u201cspin down\u201d because is Since the antiparallel to B) is of lower energy than the orientation (called \u201cspin up,\u201d of course). Thus, in a thermal distribution the spin-down states will contain more atoms than the spin-up states and the solid will have a net magnetic moment per unit volume M whose magnitude is given by 1>2 ms s M ( ) where tively. Since is proportional to B: and are the densities of electrons with spin-up and spin-down, respecis negative, M is positive. For sufficiently small fields M and 0M (B 10-34 ( where B V kT, is called the magnetic susceptibility. For high temperatures such that it can be shown that (see Problem 10-60) ( 0M B 2 kT 10-35 ( ) is the total electron density. Equation 10-35 is known as Curie\u2019s where law, after Pierre Curie. Thus, as T increases, the ability of the magnetic field to align the spins decreases. Many solids exhibit Curie\u2019s law behavior. For low temperatures where corresponding to the alignment of all the magnetic moments with the field. B W kT, M S T S 0, as since TF T V TF, Equation 10-35 does not apply to the magnetism arising from electrons in metals. 104 105 K for metals. Thus, the electrons are The reason is that highly degenerate, with each allowed level containing two with paired spins. When an external B field is applied, spins cannot just \u201cflip\u201d to align with the field since doing so would violate the exclusion principle. A spin flip must be accompanied by raising that electron to a higher, unoccupied energy state. Thus, even at T 0 K, metals have a finite susceptibility. This type of magnetic behavior is called Pauli paramagnetism. 436 Chapter 10 Solid State Physics Figure 10-15 Electrons 1 and 2 orbit the atomic core (not shown)", " in opposite directions. The magnetic field B is perpendicular to the plane of the orbits. The magnetic forces F1 and F2 increase the orbital magnetic moment of electron 2 and decrease that of electron 1, resulting in a net moment opposite to B. v1 F1 Electron 1 \u03bc L1 S1 v2 S2 F2 Electron 2 \u03bc L2 B Diamagnetism F e(v B). Recall that a free electron moving perpendicular to a magnetic field experiences a magnetic force The resulting circular motion produces a current loop with a magnetic moment opposite to the direction of the applied field. (To see this, use the right-hand rule.) Now consider two electrons with paired spins orbiting in opposite directions in an atom. (See Figure 10-15.) If an external B field perpendicular to Fmagnetic) on electron 1 is the plane of the orbits is turned on, the net force (FCoulomb reduced, reducing its orbital magnetic moment, which is parallel to B. The net force on electron 2 is increased, increasing its magnetic moment. The result is a net magnetic moment opposite to the direction of the applied field. This magnetic behavior is called diamagnetism. The diamagnetic effect is seen only in solids consisting of atoms whose electron spins are all paired. As we will see in Section 10-9, the \u201ctest\u201d of superconductivity is that the material exhibit perfect diamagnetism. Ferromagnetism The first magnetic effect discovered, a result of its existence in iron that led to its early use as a compass, ferromagnetism is the consequence of a phase transition in certain materials. At high temperatures a piece of iron is unmagnetized, the spin directions of the atoms having rotational symmetry\u2014all spin directions are equally probable. (In an applied B field iron is paramagnetic at high temperatures.) As the temperature decreases, at a certain temperature TC, called the Curie temperature, the magnetic interaction between the atoms exceeds the randomizing effect of thermal agitation, spontaneously breaking the rotational symmetry and causing a phase transition in the solid that tends to align the spins parallel to each other, converting the sample into a permanent magnet. Only four elements besides iron exhibit ferromagnetism: nickel, cobalt, gadolinium, and dysprosium. There are also several ferromagnetic compounds, including some that contain none of the ferromagnetic elements. In certain compounds the magnetic interaction between the atoms tends to align the spins on adjacent atoms antiparallel below a certain temperature, analogous to the Curie temperature, called the Neel temperature TN. Such materials are called antiferromagnetic. Examples are FeO, NiCl2, MnO, and MnS. In a few other materials the spins on adjacent sites are antiparallel below TN, but because they contain two different types of positive ions, the spins do not exactly cancel and the material is left with a small net magnetization. Such materials are called ferrimagnetic. The most common example is the iron ore magnetite, FeO Fe2O3. # 10-5 Magnetism in Solids 437 Spin valves, the first spintronic devices, form the read/write heads on the hard drives of essentially all modern computers. In addition, recent commercial development of magnetoresistive random access memory (MRAM) chips have read/write speeds much faster than flash memory and, like the latter, do not degrade over time. EXPLORING Spintronics A relatively new field of research with almost immediate applications, spintronics, orspin electronics, is the manipulation of electron spin currents rather than charge currents. N. F. Mott was the first to suggest the possibility of spin-polarized currents in ferromagnetic materials more than 30 years ago. Utilization of spin currents was first realized with the discovery of the giant magnetoresistance (GMR) of magnetic multilayers in 1988. A magnetic multilayer film consists of a stack of alternate ferromagnetic and nonmagnetic layers. The resistance to current flow is low when the electron spins, hence magnetic moments, of the ferromagnetic layers are aligned parallel. The resistance is high when the spins of the ferromagnetic layers are aligned antiparallel, a result of spin-dependent scattering. The resulting relative resistance change can be as large as 200 percent (although 10\u201320 percent is more typical); hence the name giant magnetoresistance. Depending on the design of the layers, the direction of the spins (magnetic moments) can be changed very quickly by an applied magnetic field of only about 106 tesla. These so-called spin valves can detect very small magnetic fields, such as those of the magnetic bits on CDs and DVDs. Another spintronic device with enormous potential applications is the magnetic tunnel junction. In these devices the ferromagnetic layers are separated by very thin insulating layers, typically aluminum oxide (see Figure 10-16). Electrons can tunnel through the insulating layer (", "see Sections 6-6 and 10-8) and, since the tunneling probability from a ferromagnetic layer depends on the spin direction, the resistance of the junction is different by as much as 75 percent between the parallel and antiparallel configurations. Extremely small junctions can be mass-produced making possible random access memory for portable permanent computer memory with write speeds three orders of magnitude faster than current flash memory devices. The magnetic tunnel junction was also the key element in the recent direct electrical detection of the potential due to the spin Hall effect (see page 452), opening opportunities for controlling spin currents with electric fields. Yet another intriguing future possibility is the application of spintronic devices to the development of quantum computers. The use of electron charge states for information storage is currently a barrier to their development since such states are readily destroyed by scattering. Spin states, on the other hand, have very long relaxation times. Current Nonmagnetic layer Ferromagnetic layers Nonmagnetic layer Ferromagnetic layers Spins parallel Low resistance Spins antiparallel High resistance Figure 10-16 A magnetic tunnel junction consists of two ferromagnetic layers separated by a very thin nonmagnetic insulating layer. (a) The probability of electrons tunneling through the barrier layer is dependent on the spin direction, being highest when the spins, hence the magnetic moments, of the electrons in the ferromagnetic layers are parallel. This is the configuration of low resistance. (b) The tunneling probability is low when the spins are antiparallel, resulting in high resistance. Thus, each junction can store one bit of data, (a) representing, e.g., \u201c0\u201d and (b) representing \u201c1.\u201d 438 Chapter 10 Solid State Physics 10-6 Band Theory of Solids We have seen that, if the electron gas is treated as a Fermi gas and the electron-lattice collisions treated as the scattering of electron waves, the free-electron model gives a good account of the electrical and thermal properties of conductors. This simple model, however, gives no indication why one material is a good conductor and another is an insulator. The conductivity (and its reciprocal, the resistivity) vary enormously from the best insulators to the best conductors. For example, the resistivity of a typical insulator (such as quartz) is of the order of whereas that of a typical conductor (most metals) 108 \" # m is of the order of 1016 \" # m, and that of a superconductor is less than 1019 \" # m. To understand why some materials conduct and others do not, we must refine the free-electron model and consider the effect of the lattice on the electron energy levels. There are two standard approaches to this problem of determining the energy levels of electrons in a crystal. One is to consider the problem of an electron moving in a periodic potential and to determine the possible energies by solving the Schr\u00f6dinger equation. The other is to determine the energy levels of the electrons in a solid by following the behavior of the energy levels of individual atoms as they are brought together to form the solid, in much the same way that we did in Section 9-2 in the explanation of the covalent bonding in the H2 molecule. Both approaches lead to the result that the energy levels are grouped into allowed and forbidden bands. The details of the band structure of a particular material determine whether that material is a conductor, an insulator, or a semiconductor. Qualitative discussion of the first of these methods is given in this section. The second is described in the MORE section Energy Bands in Solids\u2014An Alternate Approach on the home page. Kronig-Penney Model Consider first the problem of an electron moving in a periodic potential. Figure 10-17a shows a one-dimensional sketch of the potential energy function for a lattice of positive ions. The most important feature of this potential is not the shape, but the fact that it is periodic. A simpler periodic potential consisting of finite square wells is shown in Figure 10-17b. The model based on this potential is called the Kronig-Penney model. It has the important feature of periodicity and is easier to treat mathematically; however, even for this model the mathematical solution of the Schr\u00f6dinger equation is (a) (b) Figure 10-17 (a) One-dimensional potential energy of an electron in a crystal. U(x) approaches of an electron in a crystal. at the atom sites. (b) Simplified (Kronig-Penney) model of potential energy 10-6 Band Theory of Solids 439 quite involved, and we will only outline it here. For both potential functions shown in Figure 10-17, for certain ranges of energy traveling-wave-type solutions of the Schr\u00f6dinger equation exist. This result is based on an important discovery made by Felix Bloch4 that solutions to the Schr\u00f6dinger equation for periodic potentials", " must be of the form (in one dimension) \u2019(x) uk(x)eikx 10-36 where uk(x) uk(x L) uk(x nL), L is the periodic spacing of the potential wells, and n is an integer. The function eikx is a plane wave, i.e., a free electron (see Section 6-6) with wave number As Bloch himself described it: k 2>. I felt that the main problem was to explain how the electrons could sneak by all the ions in a metal.... I found to my delight the wave differed from the plane wave of free electrons only by a periodic modulation. Thus, we require that the solutions of the Schr\u00f6dinger equation U2 2m d2\u2019(x) dx2 U(x)\u2019(x) E\u2019(x) 10-37 where U(x) is the Kronig-Penney potential of periodic square wells and has the form of the Bloch function given by Equation 10-36. The solution for the region 0 x a in Figure 10-18 is \u2019(x) \u2019(x) A1eikx A2eikx 10-38 k 2> (2mE)1>2>U2. where the form In the region b x 0, the solutions are of \u2019(x) B1ex B2ex 10-39 [2m(U0 where 10-36 means that E)]1>2>U2. The requirement that \u2019(x) have the form of Equation \u2019(x a b) uk(x a b)eik(xab) \u2019(x a b) uk(x)eikxeik(ab) \u2019(x a b) \u2019(x)eik(ab) where a b is the periodic spacing of the wells. In general, \u2019(x n(a b)) \u2019(x)eikn(ab) 10-40a 10-40b U(x) U0 \u2013(a + b) \u2013b 0 a a + b x Figure 10-18 A portion of the Kronig-Penney potential of Figure 10-17b showing the width of the square wells a and their periodic spacing (a b). 440 Chapter 10 Solid State Physics and \u2019(x) d\u2019>dx As was done in Section 6-3 for solving the finite one-dimensional square well, the constants A1, A2, B1, and B2 are chosen so as to make continuous at x 0 and x a. Obtaining the constants is beyond the scope of our discussion here, but as in Chapter 6, doing so yields a conditional equation connecting k, k, and with a and b, the parameters of the lattice. The result is that, in order to satisfy the requirement of Equation 10-40, only certain ranges of electron energies are allowed. These energy ranges, called bands, are separated by forbidden energy regions called energy gaps, in which no traveling wave can exist. Figure 10-19a shows the energy versus the wave number k for a completely free electron. This is, of course, merely a Figure 10-19b shows E versus k for an electron in the perisketch of odic potential of Figure 10-18. The energy gaps occur at E U2k2>2m. ka n 10-41 where n is an integer and a is the lattice spacing.5 We can understand this result in terms of the Bragg reflection of the electron waves. Consider E to be small (near zero is large. As E increases, k eventually bein Figure 10-19b) so that k is small; hence comes large enough so that becomes small enough to suffer a Bragg reflection (constructive interference) from the lattice. (See Section 3-4.) Bragg reflection is governed by the Bragg condition (Equation 3-23): n 2a sin E E (a) (b) Allowed bands k Energy gap Eg 0 \u03c0 /a 2\u03c0 /a 3\u03c0 /a 4\u03c0 /a k Figure 10-19 (a) Energy versus k for a free electron. (b) Energy versus k for a nearly free electron in the one-dimensional periodic potential of Figure 10-18 with b 0 and S. Energy gaps occur at the k values which satisfy the Bragg scattering condition. In each case only the portions with k 0 are shown. The complete curves are symmetric about k 0. U0 10-6 Band Theory of Solids 441 Probability density \u23aa\u03c8 \u23aa2 2 \u23aa\u03c8 \u23aa2 1 a x Figure 10-20 Probability density (proportional to the charge distribution) for standing waves of wave number is a maximum at the lattice ion sites and has a", " lower potential energy than the dashed curve \u0192\u2019 1 in a one-dimensional crystal. The solid curve k >a \u0192\u2019 2 \u0192 2 \u0192 2. Since k 2>, In a one-dimensional system such as we are considering here, reflection means 90\u00b0. Equation 10-41 becomes the condition for Bragg reflection. The reason that traveling waves cannot exist for these wave numbers is that the amplitude of the reflection from one atom in the chain becomes equal to and in phase with the forward electron wave from the preceding atom, so that standing waves are set up. Figure 10-20 shows a sketch of the electron probability density for the two types of standing waves for the lowest energy gap, where the value \u0192\u2019\u0192 2 k >a: \u2019 1 sin kx sin x a \u2019 2 cos kx cos x a \u2019 1. \u2019 Since gives a higher concentration of electron charge density near the ion sites 2 \u2019 \u2019 1, than The difference in the potential than for the potential energy is less for 2 energies corresponds to the magnitude of the energy gap. Within the allowed energy bands, the energy has a continuous range if the number of atoms in the chain is infinite; for N atoms, there are N allowed energy levels in each band. Since the number of atoms is very large in a macroscopic solid, the energy bands can be considered continuous. Calculations in three dimensions are more difficult, of course, but the results are similar. The allowed ranges of the wave vector k are called Brillouin zones. Referring to Figure 10-19a, the first Brillouin zone has the second has 2>a k >a >a k >a, >a k 2>a, and so on. and Conductors, Insulators, and Semiconductors Conductors We can now understand why some solids are conductors and others are insulators. Consider sodium. There is room for two electrons in the 3s state of each atom, but each sodium atom has only one 3s electron. Therefore, when N sodium atoms are bound in a solid, the 3s energy band is only half filled. In addition, the empty 3p band overlaps the 3s band. The allowed energy bands of sodium are shown schematically in Figure 10-21. We can see that many allowed energy states are available immediately above the filled lower half of the 3s band, so the valence electrons can easily be raised to a higher energy state by an electric field. Accordingly, sodium is a good conductor. Magnesium, on the other hand, has two 3s electrons, so the 3s band is filled. However, like sodium, the empty 3p band overlaps the 3s band, so magnesium is also a conductor. The band occupied by the outer, or valence electrons, is called the valence band. The next (higher) allowed band is called the conduction band. 3p 3s 2p 2s 1s Figure 10-21 Energy-band structure of sodium. The empty 3p band overlaps the half-filled 3s band. Just above the filled states are many empty states into which electrons can be excited by an electric field, so sodium is a conductor. 442 Chapter 10 Solid State Physics (a) (b) (c) (d) Forbidden Allowed, empty Allowed, occupied Key Conductor Conductor Insulator Semiconductor Figure 10-22 Four possible band structures for a solid. (a) The allowed band is only partially full, so electrons can be excited to nearby energy states. At 0 K the Fermi level is at the top of the filled states. (a) is a conductor. (b) is a conductor because the allowed bands overlap. In (c) there is a forbidden band with a large energy gap between the filled band and the next allowed band; this is an insulator. (d) The energy gap between the filled band and the next allowed band is very small, so some electrons are excited to the conduction band at normal temperatures, leaving holes in the valence band. The Fermi level is approximately in the middle of the gap. (d) is a semiconductor. Thus, a conductor is a solid whose valence band is only partly filled or whose conduction band overlaps its valence band. There are a few elements, notably antimony, arsenic, and bismuth, whose conduction band overlaps the valence band only very slightly, limiting the number of available empty states. These materials are called semimetals. (See Figures 10-22a and b.) Insulators A solid that has a completely filled valence band is an insulator if the energy gap between the valence band and the empty conduction band is larger than about 2eV, as illustrated in Figure 10-22c. For example, ionic crystals are insulators. The band structure of an ionic crystal, such", " as NaCl, is quite different from that of a metal. The energy bands arise from the energy levels of the Na and ions. Both of these ions have a closed-shell configuration, so the highest occupied band in NaCl is completely full. The next allowed band, which is empty, arises from the excited states of Na There is a large energy gap between the filled band and this empty band. Typical electric fields applied to NaCl will be too weak to excite an electron from the upper energy levels of the filled valence band across the large gap into the lower energy levels of the empty conduction band, so NaCl is an insulator. When an applied electric field is sufficiently strong to cause an electron to be excited to the empty band, the phenomenon called dielectric breakdown occurs. Cl. Cl and Intrinsic Semiconductors If the gap between a filled valence band and an empty conduction band is small, the solid is a semiconductor. Consider carbon, which has two 2s electrons and two 2p electrons. We might expect carbon to be a conductor because of the four unfilled 2p states. However, the 2s and 2p levels mix when carbon forms covalent bonds.6 Figure 10-23 shows the splitting of the eight 2s 2p levels when carbon bonds in the diamond structure. This splitting is due to the nature of the covalent bond and is similar to the splitting of the 1s levels in hydrogen discussed in Section 9-2. The energy of the levels corresponding to the four space-symmetric wave functions (one for the 2s levels and three for the 2p levels) is lowered while the energy of the other four levels (one 2s and three 2p) is raised. The valence band therefore contains four levels per atom that are filled, and the conduction band is empty. E 7 eV 1 eV kT 0.026 eV At the diamond lattice spacing of about 0.154 nm, the energy gap between the filled valence band and the empty conduction band is about 7 eV. Since this gap is large compared to the energy that an electron might receive by thermal excitation due to scattering from the lattice ions, which on the averat T 300 K, age is of the order of very few electrons can reach the conduction band. Thus, diamond is an insulator. The band structure is similar for silicon, which has two 3s and two 3p electrons, and for germanium, which has two 4s and two 4p electrons. At the silicon lattice spacing of 0.235 nm the energy gap is about 1 eV; at the germanium lattice spacing of 0.243 nm the energy gap is only about 0.7 eV. For these gaps, at ordinary temperatures there are an appreciable number of electrons in the conduction band due to thermal excitation, although the number is still small compared with the number in a typical conductor. Solids such as these are called intrinsic semiconductors. Figure 10-22d illustrates the band structure of intrinsic semiconductors. 10-6 Band Theory of Solids 443 Empty Filled 3 states per atom 2p, 3p, or 4p 3 states per atom 1 state per atom 2s, 3s, or 4s 1 state per atom 0.4 0.6 0.8 R, nm 0.2 0 \u2248 0.15 nm RC RGe \u2248 RSi \u2248 0.24 nm Figure 10-23 Splitting of the 2s and 2p states of carbon, the 3s and 3p states of silicon, or the 4s and 4p states of germanium versus separation of the atoms. The energy gap between the four filled states in the valence band and the empty states in the conduction 0.154 nm. For band is 7 eV for the diamond-lattice spacing, RC 0.235 nm, the energy gap is 1.09 eV. The the silicon spacing RSi splitting is similar for the 4s and 4p levels in germanium, which has an atom spacing of 0.243 nm, giving an energy gap of only 0.7 eV. In the presence of an electric field, the electrons in the conduction band of an intrinsic semiconductor can be accelerated because there are empty states nearby. Also, for each electron that has been excited to the conduction band there is a vacancy, or hole, in the nearby filled valence band. In the presence of an electric field, other electrons in this band can be excited to the vacant energy level, thus filling that hole but creating another hole. This contributes to the electric current and is most easily described as the motion of a hole in the direction of the field and opposite to the motion of the electrons. The hole thus acts like a positive charge. An analogy of a two-lane, one-way road with one lane full of parked cars and the other empty may", " help in visualizing the conduction of holes. If a car moves out of the filled lane into the empty lane, it can move ahead freely. As the other cars move up to occupy the space left, the empty space propagates backward in the direction opposite to the motion of the cars. Both the forward motion of the car in the nearby empty lane and the backward propagation of the empty space contribute to a net forward propagation of the cars. An interesting characteristic of semiconductors is that the conductivity increases (and the resistivity decreases) as the temperature increases, which is contrary to the case for normal conductors. The reason is that as the temperature is increased, the number of free electrons is increased because there are more electrons in the conduction band. The number of holes in the valence band is also increased, of course. In semiconductors, the effect of the increase in the number of charge carriers, both electrons and holes, exceeds the effect of the increase in resistivity due to the increased scattering of the electrons by the lattice ions due to thermal vibrations. Semiconductors therefore have negative temperature coefficients of resistivity. Whether a solid with a filled valence band will be a semiconductor or an insulator depends critically on the width of the energy gap Eg, as Figure 10-23 suggests. A comparison of the relative numbers of electrons with various energies that could be 444 Chapter 10 Solid State Physics Table 10-4 Values of fFD (E ) for T 293 K E EF (eV) Multiple of kT fFD(E) 0.05 0.10 0.25 1.0 2.5 7.5 2 1.2 101 4 1.8 102 10 5.1 105 40 6.5 1018 100 1.1 1043 300 1.3 10129 above the Fermi level (located at the center of the band gap) at ordinary temperatures illustrates why this is true. Those numbers are given by the Fermi-Dirac distribution fFD(E) given by Equation 8-68: fFD(E) 1 E(EEF)>kT 1 8-68 At T 293 K, kT 0.025 eV. Recall that for E EF, fFD(E) 1 2 (see Section 8-5). For (E EF) 0.10, or 4(kT), we have > 0.018 fFD(E) 1 e0.10>0.025 1 Repeating this calculation for several additional values of (E EF) yields the relative numbers of electrons in Table 10-4. From the numbers in the table we see that, if a certain material has an energy gap Eg between the valence and conduction bands of 0.25 eV, for example, then approximately 105 of the electrons within kT of the Fermi level would be excited to the conduction band and thus able to participate in the conduction of electricity. This is a sizable number, given the numbers of electrons near the Fermi level, so we expect this material to have a higher electrical conductivity than materials with larger values of Eg. fFD(E) For a gap of 1.0 eV, just four times that of the previous example, the relative number of electrons excited to the conduction band decreases by more than 12 orders of magnitude, illustrating the sharp decline of as the energy gap increases. The calculation of above also illustrates the increased conductivity of semiconductors as the temperature increases described earlier. If the temperature of a material with an energy gap of increases to 1.5 1013, 1.0 eV is increased to 393 K from 293 K, as in Table 10-4, thus increasing the relative number of electrons in the conduction band by nearly four orders of magnitude. Table 10-5 lists the energy gaps for several semiconducting elements and compounds. Notice that the energy gap is slightly temperature dependent. fFD(E) fFD(E) the energy of a free electron of wavelength A concept that is helpful in understanding a number of characteristics of semiconductors is that of effective mass. As pointed out above, Figure 10-19a is a graph of E U2k2>2me, The curvature d2E>dk2 U2>me, of the E versus k graph is given by and we may say that the curva> ture is determined by 1 me, the reciprocal mass. In Figure 10-19a 1 me is of course constant; however, in regions near the energy gaps in Figure 10-19b the curvature is much higher than that for the free electron. Since the behavior of electrons near the band/gap boundary is of considerable interest, particularly in the discussion of impurity semiconductors and devices in Section 10-7, it is helpful to continue to describe the curvature of the E versus k curve near the boundary in terms of a reciprocal mass. Accordingly, we define the effective mass m* as 2>", "k. > 1 m* 1 U2 d2E dk2 10-42 10-7 Impurity Semiconductors 445 Table 10-5 Energy gap Eg and dielectric constant * for selected semiconductors Eg (eV) Material 0 K 293 K Si Ge Te GaAs InSb ZnS 1.15 0.74 \u2014 1.53 0.23 \u2014 1.11 0.67 0.33 1.35 0.16 3.54 * 11.8 15.9 \u2014 13.1 17.8 5.2 Eg (eV) Material 0 K 293 K CdTe PbS InP CdSe GaP PbTe 1.56 0.28 1.41 1.85 2.40 0.19 1.44 0.37 1.27 1.74 2.24 0.25 * 10.2 17.0 12.4 10.1 11.1 30.1 > Then, as in the case of the free electron, the curvature of E versus k for electrons bound in the crystal energy bands is also described in terms of a reciprocal mass, 1 m*. For a free electron m* me, as is also the case for electrons that are not close to the boundaries in Figure 10-19b. Close to the band/gap boundaries, however, is a different matter. Starting from k 0 in the figure, the curvature is initially constant and k >a, equal to that of a free electron, thus m* me, but near the boundary where the curvature becomes large and, very close to the boundary, negative; hence m* becomes smaller than me and also eventually negative! Just above the gap, the curvature is large and positive, so m* me and positive. For the situation where Eg is small compared to the width of the band, the values of the effective mass are typically of the order of 0.010.1 of the mass of a free electron. We will make further use of the effective mass in Section 10-7. Questions 7. How does the change in resistivity of copper compare with that of silicon when the temperature increases? 8. Suppose an electron is excited from the valence band of a semiconductor to a state several levels above the lower edge of the conduction band. Devise an explanation for why it will quickly \u201cdecay\u201d to a level at the bottom of the conduction band. More An alternative to the Kronig-Penny model of a solid is based on the molecular bonding model discussed in Section 9-2 for hydrogen. Energy Bands in Solids\u2014An Alternate Approach is described briefly on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Figures 10-24 and 10-25 here. 10-7 Impurity Semiconductors Most semiconductor devices, such as the semiconductor diode and the transistor, make use of impurity semiconductors, which are created through the controlled addition of certain impurities to intrinsic semiconductors. This process is called doping. 446 Chapter 10 Solid State Physics Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si (a) Figure 10-26 (a) A two-dimensional schematic illustration of solid silicon. Each atom forms a covalent bond with four neighbors, sharing one of its four valence electrons with each neighbor. (b) X-ray scattering measurement of electron density in the vicinity of two atoms in an Si crystal. The arrow points to the high electron density of the covalent bond. [Adapted from Y. W. Yang, P. Coppens, Solid State Comm., 15, 1555 (1974).] Si Si Si Si (b) Figure 10-26a illustrates the lattice structure of pure silicon, and Figure 10-27a is a schematic illustration of silicon doped with a small amount of arsenic such that arsenic atoms replace a few of the silicon atoms in the crystal lattice. Arsenic has five valence electrons in the n 4 shell, whereas silicon has four valence electrons in the n 3 shell. Four of the five arsenic electrons take part in covalent bonds with the four neighboring silicon atoms, and the fifth electron is very loosely bound to the atom. This extra electron occupies an energy level that is just slightly below the conduction band in the solid and is easily excited into the conduction band, where it can contribute to electrical conduction. The fifth arsenic valence electron and the arsenic ion core form a hydrogenlike system. Thus, Bohr theory (see Section 4-3) can be used to calculate the approximate values of the energies available to it, provided only that we make allowance for the fact that the electronarsenic ion system is embedded in the semiconductor crystal rather than being isolated from other atoms. First, the crystal is a medium with a high dielectric constant; thus the potential energy function in the Schr\u00f6dinger", " equation for a hydrogenlike atom (Equation 7-6) becomes7 is the dielectric constant of the materwhere ial and Second, the electron mass in the Schr\u00f6dinger equation must be replaced by the effective mass m*, which accounts for the fact that the electron \u201csees\u201d a three-dimensional version of the periodic potential of Figure 10-17. With these two modifications the solution of the Schr\u00f6dinger equation is carried out just as in Chapter 7. The results for the allowed energies and average values of the radii of the Bohr orbits for the fifth arsenic electron are given by V(r) (Zke2>R)(1>*) k 1>4 0. * En 1 2 a ke2 U b 2 me n2 m* me 1 2 * 1 *2 me m* a ke2 U b 2 m* *2 1 n2 10-43 10-44 8rn 9 a0n2 where a0 is the Bohr radius, equal to 0.0529 nm (see Equation 4-19), and n is the principal quantum number. To understand where these energy levels lie relative to the bands and gap of the silicon, consider that when the arsenic atom is ionized by removing the fifth electron, that electron is then free to move about and to participate in electrical conduction; i.e., it is then in the conduction band. Thus, we conclude that is at the bottom edge of the conduction band and the other En hydrogenlike levels lie below it in the gap. The energy of the ground state E1 can be calculated from the experimentally determined value of the electron\u2019s effective mass in silicon, about 0.2 me, and the dielectric constant of silicon given in Table 10-5. Substituting these into Equation 10-43 0.020 eV below the conduction band, which is substantially smaller yields E1 than the 13.6-eV ground state for hydrogen. Similarly, substitution into Equation or about 60 times the ground-state radius of hydrogen.8 10-44 yields 9 3.1 nm, 0 E 8r1 10-7 Impurity Semiconductors 447 (a) Extra electron (b) Empty conduction band Extra electron Si Si Si Si Si As Si Si Si Si Si Si Si Si As Si Si Si Si Si Impurity donor levels Filled valence band Figure 10-27 (a) A two-dimensional schematic illustration of silicon doped with arsenic. Because arsenic has five valence electrons, there is an extra, weakly bound electron that is easily excited to the conduction band, where it can contribute to electrical conduction. (b) Band structure of an n-type semiconductor such as silicon doped with arsenic. The impurity atoms provide filled energy levels that are just below the conduction band. These levels donate electrons to the conduction band. These energies are quite close to the conduction band, as illustrated in Figure 10-27b; thus these electrons can be easily excited to the conduction band since their ionization energy is comparable to kT at room temperature. These hydrogenlike levels just below the conduction band are called donor levels because they donate electrons to the conduction band without leaving holes in the valence band. Such a semiconductor is called an n-type semiconductor because the major charge carriers are negative electrons. The conductivity of a doped semiconductor can be controlled by controlling the amount of impurity added. The addition of just one part per million can increase the conductivity by several orders of magnitude. Another type of impurity semiconductor can be made by replacing a silicon atom in the crystal lattice with a gallium atom, which has three electrons in its valence level rather than four (see Figure 10-28a). The gallium atom accepts electrons from the valence band of the silicon in order to complete its four covalent bonds, thus creating a hole in the valence band. The effect on the band structure of silicon achieved by doping it with (a) Si Si Si Si Si Si Ga Si Hole Si Si Si Si Si Ga Si Si Hole Si Si Si Si (b) Empty conduction band Impurity acceptor levels Filled valence band Figure 10-28 (a) Two-dimensional schematic illustration of silicon doped with gallium. Because gallium has only three valence electrons, there is a hole in one of its bonds. As electrons move into the hole, the hole moves about, contributing to the conduction of electric current. (b) Band structure of a p-type semiconductor such as silicon doped with gallium. The impurity atoms provide empty energy levels just above the filled valence band that accept electrons from the valence band. 448 Chapter 10 Solid State Physics gallium is shown in Figure 10-28b. The empty levels shown just above the valence band are due to the holes resulting from the ionization of the gallium atoms", ". These levels are called acceptor levels because they accept electrons from the filled valence band when these electrons are thermally excited to a higher energy state. They arise because the holes, which act like positive charges, may be bound to the negative gallium core much like the fifth electron was bound to the positive arsenic core. Thus, the hole-gallium ion system also forms a hydrogenlike system and the energy levels of the hole can also be calculated approximately using the Bohr model with results similar to Equation 10-43. Since the energy-band diagrams like Figure 10-27b and 10-28b are drawn with electron energy increasing upward, hole energy in those diagrams increases downward. Ionizing the hole-gallium system means returning the hole to the valence band; hence these levels are just above the top of the valence band, as shown in Figure 10-28b, and their magnitudes are of the same order as those of the donor levels discussed previously. Increasing the energy of holes is equivalent to promoting electrons from the valence band into the acceptor levels. This creates holes in the valence band that are free to propagate in the direction of an electric field. Such a semiconductor is called a p-type semiconductor because the charge carriers are positive holes. The fact that conduction is due to the motion of holes can be verified by the Hall effect, described in the Exploring section on page 449. EXAMPLE 10-8 Donor Ionization Energy in Ge If phosphorus is used to dope germanium to form an n-type semiconductor, what is the ionization energy of the levels? What is the radius of the electron\u2019s orbit? Phosphorus has five valence electrons. (The effective mass for electrons in germanium is about 0.1 me.) SOLUTION 1. The magnitude of the ionization energy is computed from Equation 10-43 with n 1: E1 1 2 a ke2 U b 2 m* *2 2. The dielectric constant is given in Table 10-5: * for germanium * 15.9 3. Substituting values into Equation 10-43 gives E1 a 1 2 9 109 N # m2>C2 (1.60 1019 C)2 1.055 1034 J # s 8.6 1022 J 5.4 103 eV b 2 (0.1 9.11 1031 kg) (15.9)2 4. The orbit radius 9 8r1 rus electron is computed from Equation 10-44 with n 1: of the fifth phospho- 5. Substituting values, where the Bohr radius a0 0.0529 nm, gives 8r1 9 a0 me m* * 8r1 9 0.0529 8.4 nm me 0.1me 15.9 Remarks: The value computed above for E1 is very close to the experimental value of 12.0 103 eV, even though our calculation is a Bohr model approximation. 10-7 Impurity Semiconductors 449 EXPLORING Hall Effect The number of donated electrons in a doped n-type semiconductor, or holes in a doped p-type semiconductor, is typically much greater than the intrinsic number of electron-hole pairs created by thermal excitation of electrons from the valence band to the conduction band. In an electric field, the current will therefore consist of both majority carriers (electrons in an n-type or holes in a p-type semiconductor) and minority carriers. The reality of conduction by motion of positive holes is clearly evident in the Hall effect, illustrated in Figure 10-29a. In this figure a thin strip of a doped semiconductor is connected to a battery (not shown), so that there is a current to the right. A uniform magnetic field B is applied perpendicular to the current. For the direction of the current and magnetic field shown, the magnetic force on a moving charged particle is the drift velocity) independent of whether the current is due to a positive charge moving to the right or a negative charge moving to the left. Let us assume for the moment that the charge carriers are electrons, as in Figure 10-29b. The magnetic force will then cause the electrons to drift up to the top of the strip, leaving the bottom of the strip with an excess positive charge. This will continue until the electrostatic field caused by the charge separation produces an electric force on the charge carriers just balancing If w is the width of the the magnetic force. The condition for balance is strip, there will be a potential difference called the Hall voltage is upward (where qe qvdB. B q vd vd e VH ew vdBw 10-45 between the top and bottom of the strip. This potential difference can be measured with a high-resistance voltmeter. A measurement of the sign of the potential difference (i.e.,", " whether the top of the strip is at a higher potential due to positive charge or lower potential due to negative charge) determines the sign of the majority carriers. Such measurements reveal that, indeed, the charge carriers are negative in n-type and positive in p-type semiconductors. The value of the Hall voltage provides a measurement of the drift velocity vd. Since the current density j nqvd can be easily measured from the total current and cross-sectional area of the strip, measurement of the drift velocity determines n, the number of charge carriers per unit volume. Hall-effect probes are frequently used to measure magnetic field strengths. A current is established in a calibrated metal strip. Measuring the Hall voltage then yields the value of B (see Equation 10-46). (a) B in (b) B in ++++++++++++++++++++ \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 F +q vd w F vd \u2013 ++++++++++++++++++++ I Figure 10-29 The Hall effect. The force on the charge carriers is up whether the carriers are positive charges moving to the right (a) or negative charges moving to the left (b). The sign of the charge carriers can be determined by the sign of the potential difference between the top and bottom of the strip, and the drift velocity can be determined by the magnitude of this potential difference. The thickness t of the strip is not shown. 450 Chapter 10 Solid State Physics t 250 m EXAMPLE 10-9 Hall Effect in Aluminum A strip of aluminum of width w 1.5 cm and thickness is placed in a uniform magnetic field of 0.55 T oriented perpendicular to the plane of the strip. When a current of 25 A is established in the is measured across the width of the strip. What is the denstrip, a voltage of 1.64 sity of charge carriers in aluminum and how many charge carriers are provided, on average, by each atom? V SOLUTION Substituting for the drift velocity vd in terms of the current density (Equation 10-11) in Equation 10-45 yields VH vd Bw jBw nq iB qnt 10-46 since j i wt. The density of the charge carriers in aluminum is then given by > n iB qtVH (25 A)(0.55 T) (1.60 1019 C)(250 106 m)(1.64 106 V) 2.10 1029 carriers>m3 The density of atoms N in aluminum is given by the following, where the density (Al) 2.72 103 kg>m3 and the molar mass M 26.98 kg mol: > N NA M (6.02 1026 atoms>mol)(2.702 103 kg>m3) 26.98 kg>mol 6.02 1028 atoms>m3 Thus, each aluminum atom contributes on average n N 3.5 charge carriers. > The Quantum Hall Effect According to Equation 10-46, the Hall voltage should increase linearly with the magnetic field B for a given current. In 1980, while studying the Hall effect in thin semiconductors at very low temperatures and very large magnetic fields, Klaus von Klitzing9 discovered that a plot of VH versus B was not linear, but included a series of plateaus, as shown in Figure 10-30a. That is, the Hall voltage is quantized. More VH i, it is the Hall resistance that is specifically, if we define the Hall resistance RH quantized, taking on only the values > RH VH i RK n 10-47 where RK, called the von Klitzing constant, is related to the fundamental electron charge e and Planck\u2019s constant h by 6.626 1034 J # s (1.602 1019 C)2 h e2 RK 25,813 \" 10-48 (a) V m, H V 300 200 100 10-7 Impurity Semiconductors 451 (b) 1.5 1.0 0./2 2/3 1/2 1 2/3 5 10 15 B, T 0 0 2 4 6 8 10 12 14 0 16 B, T 4/3 5/3 4/3 5/3 2 2 3 1 Figure 10-30 (a) A plot of the Hall voltage versus applied magnetic field shows plateaus, indicating that the Hall voltage is quantized. These data were taken at a temperature of 1.39 K with the current i held fixed at 25.52 quantum Hall effect. The Hall resistance RH (the curve with the plateaus) is read on the left vertical axis, the normal resistance RN (the curve with the peaks) on the right vertical axis. [Data collected by Y. W. Suen and co-workers at Princeton University.] (b) The fractional A. The values of n found by von Klitzing were small positive integers (n 1, 2, 3,.", "..), as indicated in Figure 10-30a. Then, in 1982, Daniel Tsui and his co-workers,10 while investigating the quantum Hall effect in ultrapure semiconductors, discovered quantized values of the Hall resistance for values of that were series of rational fractions formed from small integers. Values of RH have been found thus far for more than 30 values of v where a, b, where a and b are integers with no common factors. Several of these are seen in Figure 10-30b. v n e. Von Klitzing\u2019s discovery is referred to as the integral quantized Hall effect (IQHE) and that of Tsui and his colleagues as the fractional quantized Hall effect (FQHE). The theoretical models that have been developed to explain these phenomena are as yet incomplete and, in any case, beyond the scope of our discussion here; however, we can give a brief qualitative description of the IQHE. In the \u201cnormal\u201d Hall effect the material carries a current i due to an applied electric field The electric field is perpendicular to the applied magnetic field B, and as a result, the charge carriers move in a circular path, or orbit, of radius r m*v qB. The fact that electrons obey the Pauli > exclusion principle prevents the orbits from overlapping and determines how closely the electrons can group together on the negative side of the sample. Recalling that the orbital motion of electrons is quantized with only certain radii being allowed\u2014namely those for which the orbit circumference equals an integral number of de Broglie wavelengths\u2014we see that increasing the magnetic field decreases the orbit radius, but such decreases must occur suddenly and result in another, smaller allowed radius. Thus, more electron orbits can fit without overlapping in a given area and the density of charge carriers increases on the edges of the semiconductor sample. This increases the frequency of collisions and hence the Hall resistance. Since the orbit radii change only in quantized steps, so must the Hall resistance. Surprisingly, when the Hall resistance is on one of the plateaus, the ordinary resistance R V i falls to zero, as illustrated by the multiple peaked curve in Figure 10-30b. The additional plateaus that occur in the FQHE are due to electron-electron spin interactions. > Because the von Klitzing constant can be measured with a precision of better than 1 part in 1010, the quantum Hall effect is now used to define the standard of resistance. The ohm is now defined so that RK has the value 25,812.807 exactly. \" 452 Chapter 10 Solid State Physics \u2013 electrons + holes \u2013 + \u2013 + \u2013 + \u2013 + \u2013 + \u2013 + p side n side Figure 10-31 A pn junction. Because of the difference in their concentrations, holes diffuse from the p side to the n side and electrons diffuse from the n side to the p side. As a result, there is a double layer of charge at the junction, with the p side negative and the n side positive. Energy Conduction band + + + + + + Eg Valence band EF 0 Conduction band \u2013 \u2013 \u2013 \u2013 \u2013 \u2013 \u0394E Valence band p side n side Junction or depletion region Figure 10-32 Electron energy levels for an unbiased pn junction. Spin Hall Effect A new vista for spintronics was opened in 2004 with the observation of the spin Hall effect in GeAs at 20 K by David Awschalom and his group. In the spin Hall effect the electrons of a charge current flow in a nonmagnetic material in the absence of an external magnetic field but with an applied external electric field separate perpendicular to the conventional current direction into spin-up and spin-down populations that accumulate on opposite sides of the conducting material. The same phenomenon was detected at 295 K (room temperature) in ZnSe in 2006 by the same researchers. Although the effect is small, injecting spinpolarized electrons into semiconductor devices, would be enormous if current experiments to improve the size of the effect are successful. the potential useful applications, e.g., 10-8 Semiconductor Junctions and Devices Semiconductor devices such as diodes and transistors make use of n-type and p-type semiconductors joined together as shown in Figure 10-31. In practice, the two types of semiconductors are often a single silicon crystal doped with donor impurities on one side and acceptor impurities on the other. The region in which the semiconductor changes from a p-type to an n-type is called a junction. When an n-type and p-type semiconductor are placed in contact, the initially unequal concentrations of electrons and holes result in the diffusion of electrons across the junction from the n side to the p side until equilibrium is established. The result of this diffusion is a net transport of positive charge from the p side to the n side. Unlike the case when two different metals are in contact, there are fewer electrons available", " to participate in this diffusion because the semiconductor is not a particularly good conductor. The diffusion of electrons and holes creates a double layer of charge at the junction similar to that on a parallel-plate capacitor. There is thus a potential difference V across the junction, which tends to inhibit further diffusion. In equilibrium, the n side with its net positive charge will be at a higher potential than the p side with its net negative charge. In the junction region, there will be very few charge carriers of either type, so the junction region has a high resistance. Figure 10-32 shows the energy-level diagram for a pn junction. The junction region is also called the depletion region because it has been depleted of charge carriers. Figure 10-33 A pn-junction diode. (a) Forward-biased pn junction. The applied potential difference enhances the diffusion of holes from the p side to the n side and electrons from the n side to the p side, resulting in a current I. (b) Reverse-biased pn junction. The applied potential difference inhibits the further diffusion of holes and electrons, so there is no current. (a) + \u03b5 \u2013 (b side n side Forward bias \u2013 \u2013 \u2013 No current \u2013 + + \u2013 \u2013 + \u2013 + \u2013 + \u2013 + + + + p side n side Reverse bias Diodes A semiconductor with a pn junction can be used as a simple diode rectifier. In Figure 10-33, an external potential difference has been applied across the junction by connecting a battery and resistor to the semiconductor. When the positive terminal of the battery is connected to the p side of the junction, as shown in Figure 10-33a, the diode is said to be forward biased. Forward biasing lowers the potential across the junction. The diffusion of electrons and holes is thereby increased as they attempt to reestablish equilibrium, resulting in a current in the circuit. If the positive terminal of the battery is connected to the n side of the junction, as shown in Figure 10-33b, the diode is said to be 10-8 Semiconductor Junctions and Devices 453 Figure 10-34 Current versus applied voltage across a pn junction. Note the different scales for the forward- and reverse-bias conditions. Breakdown voltage 30 20 V Reverse bias I, mA 30 20 10 Forward bias 0.2 0.6 1.0 V, V 10 20 30 \u03bcA reverse biased. Reverse biasing tends to increase the potential difference across the junction, thereby further inhibiting diffusion. Figure 10-34 shows a plot of current versus voltage for a typical semiconductor junction. Essentially, the junction conducts only in one direction, the same as a vacuum-tube diode. Junction diodes have replaced vacuum diodes in nearly all applications except when a very high current is required. We can get an idea of how the current depends on applied voltage quantitatively if we note that the electrons and holes, being at the high energy end of the distribution, are approximately described by the Maxwell-Boltzmann distribution. Let Ne be the number of conduction electrons in the n region. With no external voltage, only a small fraction given by will have enough energy to diffuse across the contact potential difference. When a forward bias Vb is applied, the number that can cross the barrier becomes NeeeV>kT Neee(VVb)>kT (NeeeV>kT )eeVb >kT I A The current due to the majority electron carriers in the n region will be I I0eeVb >kT where I0 is the current with no bias. The current due to the minority carriers, the holes from the n side, will be merely I0, the same as with no bias. (The minority carriers are swept across the junction by the contact potential V with or without a bias voltage.) The net current due to carriers from the n side will therefore be Inet I0(eeVb >kT 1) 10-49 If we now consider the current due to the majority and minority carriers from the p side, we obtain the same results. We can use Equation 10-49 for the total current if we interpret I0 as the total current due to both kinds of minority carriers, holes in the n region and electrons in the p region. For positive the exponential quickly dominates. For 0, the current is 0, and for less than zero, the current saturates at I0, due to the flow of minority carriers. Note that the current in Figure 10-35 suddenly increases in magnitude at extreme values of reverse bias. In such large electric fields, two things can happen: either electrons are stripped from their atomic bonds or the few free electrons that exist in a reversed-biased junction are accelerated across the junction and gain enough energy to cause others to break loose. The first effect is Vb Vb Vb B V Figure 10-35 Current versus applied voltage for a tunnel", " diode. Up to point A, an increase in the bias voltage enhances tunneling. Between points A and B, an increase in the bias voltage inhibits tunneling, i.e., the diode acts as if it has negative resistance. After point B, the tunneling is negligible, and the diode behaves like an ordinary pn-junction diode, shown by the dashed line. 454 Chapter 10 Solid State Physics called Zener breakdown; the second, avalanche breakdown. Although such a breakdown can be disastrous in a circuit where it is not intended, the fact that it occurs at a sharp voltage value makes it of use in a special voltage reference standard known as a Zener diode. An interesting effect that we can discuss only qualitatively occurs if both the n side and the p side of a pn-junction diode are so heavily doped that the bottom of the conduction band lies below the top of the valence band. Figure 10-36a shows the energylevel diagram for this situation. Since there are states on the p side with the same energy as states on the n side and the depletion region is now so narrow, electrons can tunnel across the potential barrier (see Section 6-6). This flow of electrons is called tunneling current, and such a heavily doped diode is called a tunnel diode. At equilibrium with no bias, there is an equal tunneling current in each direction. When a small bias voltage is applied across the junction, the energy-level diagram is as shown in Figure 10-36b and the tunneling of electrons from the n to the p side is increased, whereas that in the opposite direction is decreased. This tunneling current in addition to the usual current due to diffusion results in a considerable net current. When the bias voltage is increased slightly, the tunneling current decreases because there are fewer states on the p side with the same energy as states on the n side. Although the diffusion current is increased, the net current is decreased. At large bias voltages the energy-level diagram is as shown in Figure 10-36c, the tunneling current is completely negligible, and the total current increases with increasing bias voltage due to diffusion as in an ordinary pn-junction diode. Figure 10-35 shows the current versus voltage curve for a tunnel diode. Such diodes are used in electric circuits because of their very fast response time. When operated near the peak in the current versus voltage curve, a small change in bias voltage results in a large change in the current. Among the many applications of semiconductors with pn junctions are particle detectors called surface-barrier detectors. These consist of a pn-junction semiconductor with a large reverse bias so that there is ordinarily no current. When a high-energy particle, such as an electron, passes through the semiconductor, it excites electrons into the conduction band, creating many electron-hole pairs as it loses energy. The intrinsic electric field sweeps the electrons toward the positive (n) side of the junction and the holes toward the negative (p) side. The resulting current pulse signals the passage of the particles and records the energy lost by the particle in the detector. The pulses are of short duration (108 107 seconds), making possible high-energyresolution measurements. (a) (b) (c) Conduction band EF Empty Valence band Filled EF EF Tunneling EF EF EF p side n side p side n side p side n side Figure 10-36 Electron energy levels for a heavily doped pn-junction tunnel diode. (a) With no bias voltage, some electrons tunnel in each direction. (b) With a small bias voltage, the tunneling current is enhanced in one direction, making a sizable contribution to the net current. (c) With further increases in the bias voltage, the tunneling current decreases dramatically. 10-8 Semiconductor Junctions and Devices 455 \u2212 + \u2212 + p side n side I \u2212 \u2212 \u2212 \u2212 + +++ + + + + +++ + + + + p side R (a) (b) + \u2212 + + \u2212 \u2212 \u2212 \u2212 \u2212\u2212\u2212\u2212\u2212\u2212 \u2212 \u2212\u2212 \u2212 \u2212 n side Figure 10-37 (a) A pn junction as a solar cell. Radiation striking the junction produces electrons and holes. The electrons are swept from the p side and holes from the n side by the intrinsic electric field. The accumulated charge results in a potential difference that produces a current through an external load. (b) A pn junction as an LED. Large forward bias produces current of electrons moving to the left and holes moving to the right. When they recombine, radiation is emitted. Light-emitting and light-absorbing pn-junction semiconductors function similar to gaseous atoms emitting and absorbing light, with the conduction and valence bands analogous to the atomic energy levels. The light-absorbing pn-junction semiconductor diode, or solar cell", ", is illustrated schematically in Figure 10-37a. When photons with energy greater than the gap energy (1.1 eV in silicon) strike the pn-junction, they can excite electrons from the valence band into the conduction band, leaving holes in the valence band. This region is already rich in holes. Some of the electrons created by the photons will recombine with holes, but some will migrate to the junction. From there they are accelerated into the n-type region by the intrinsic electric field between the double layers of charge. This creates an excess negative charge in the n-type region and excess positive charge in the p-type region. The result is a potential difference, a photovoltage, between the two regions, which in practice is about 0.6 V. If a load resistance is connected across the two regions, a charge flows through the resistance. Some of the incident light energy is thus converted into electrical energy. The current in the resistor is proportional to the number of incident photons, which is in turn proportional to the intensity of the incident light. Light-emitting diodes (LEDs) are pn-junction semiconductors with a large forward bias that produces a large excess concentration of electrons on the p side and holes on the n side of the junction. (See Figure 10-37b.) Under these conditions, the diode emits light as the electrons and holes recombine. This is essentially the reverse of the process that occurs in a solar cell. Following the first practical demonstration of an LED (in 1962), the performance of LEDs has steadily improved. (See Figure 10-38.) They can be fabricated in all of the primary colors and, more recently, white light as well, portending them as a common source of white light in the future. LEDs already provide a viable alternative to filtered incandescent lighting in applications requiring monochromatic light. They are used, for example, as indicator lamps in appliances, Fluorescent lamp Incandescent lamp Edison's first bulb ) 100 10 1 0.1 1960 1970 1980 1990 2000 Year Figure 10-38 Summary of the performance improvements in LEDs over the span of their existence. The marks the current performance of small-molecule OLEDs; the marks that of the polymer OLEDs. A few performance benchmarks are indicated on the vertical axis. \u25c6 \u25a0 456 Chapter 10 Solid State Physics The development of InGaAlP HB LEDs in the early 1990s led to their rapid application to automotive rear lighting. The recent development of high-powered, large area, white GaN HB LEDs makes automobile headlights using these LEDs a possibility for the near future. I n-type p-type I Coherent light Figure 10-39 The resonant cavity is formed by cleaving the ends of the diode crystal parallel to one another and with the proper separation. Gallium arsenide and similar compounds, which have much higher photonproduction efficiency than silicon, are typically used as diode laser semiconductors. Their light-energy-out to electrical-energy-in ratios are greater than 50 percent. Figure 10-40 (a) A pnp transistor. The heavily doped emitter emits holes that pass through the thin base to the collector. (b) Symbol for a pnp transistor in a circuit. The arrow points in the direction of the conventional current, which is the same as that of the emitted holes. electronic equipment, calculators, and digital watches. In automobiles they are used in instrument panel lighting, and high-brightness (HB) LEDs are now often used for rear, stop, and turn lights. In traffic signals the red, amber, and green LED arrays now in common use require only 10 percent of the power consumed by the standard 140-W incandescent lamps, are brighter, and have a much longer lifetime. Rapid development of organic semiconductor light-emitting diodes (OLEDs) in the 1990s was catalyzed by worldwide efforts to construct large, full-color, flat-screen displays. Fabricated from small organic molecules and various polymers, OLEDs have an advantage over LEDs in that they can be produced on a large scale at very low cost. In LEDs high forward currents result in a very large population inversion, i.e., electrons on the p side and holes on the n side, so that stimulated emission dominates the light emission process and lasing results. By appropriate construction of the diode, a resonant cavity can be formed, leading to the production of a coherent beam of laser light in a selected direction (see Figure 10-39). Transistors The transistor, invented in 1948 by William Shockley, John Bardeen, and Walter Brattain,11 has revolutionized the electronics industry and our everyday world. A simple junction transistor consists of three distinct semiconductor regions called the emitter, the base, and the collector. The base is a very thin region of one type of semiconductor sandwiched between two regions of the opposite", " type. The emitter semiconductor is much more heavily doped than either the base or the collector. In an npn transistor, the emitter and collector are n-type semiconductors and the base is a p-type semiconductor; in a pnp transistor, the base is an n-type semiconductor and the emitter and collector are p-type semiconductors. In a pnp transistor holes are emitted by the emitter; in an npn transistor electrons are emitted. Figures 10-40 and 10-41 show, respectively, a pnp transistor and an npn transistor with the symbols used to represent each transistor in circuit diagrams. Notice that a transistor consists of two pn junctions. The operation of a pnp transistor is described in the MORE section How Transistors Work. The operation of an npn transistor is similar. (a) (a) (b) Collector p-type Collector n-type Base n-type Base p-type Emitter p-type Emitter n-type Base Collector Emitter (b) Collector Base Emitter Figure 10-41 (a) An npn transistor. The heavily doped emitter emits electrons that pass through the thin base to the collector. (b) Symbol for an npn transistor. The arrow points in the direction of the conventional current, which is opposite the direction of the emitted electrons. (a) 10-8 Semiconductor Junctions and Devices 457 (a) LED traffic lights use 10 percent of the power of the old 140-W signal lamps, last far longer, and are collectively brighter. (b) Automobile LED rear lighting and brake lights have become ubiquitous, as have LED traffic signals. Pictured here are examples of the scores of different colors, shapes, and sizes of vehicle clearance and safety LED lights one sees every day on the streets and highways. [Photos by Francisco Roman.] (b) More How Transistors Work, on the home page at www.whfreeman.com/ tiplermodernphysics5e, describes the way transistors function in electrical circuits. Also located here are Equations 10-50 through 10-54, Example 10-10, and Figures 10-42 and 10-43. Questions 9. Why is a semiconductor diode less effective at high temperatures? 10. Explain why adding impurities to metals decreases their conductivity but adding impurities to semiconductors increases their conductivity. 11. What would you expect to be the effect on the conductivity when impurities are added to an insulator? 458 Chapter 10 Solid State Physics 0.15 0.125 0.10 \u03a9, R 0.075 0.05 0.025 10 \u20135 \u03a9 0.00 4.00 4.10 4.20 T, K 4.30 Figure 10-44 Plot by Kamerlingh Onnes of the resistance of mercury versus temperature, showing sudden decrease at the critical temperature T 4.2 K signifying the onset of superconductivity. 4.40 10-9 Superconductivity In 1911, just a few years after he had succeeded in liquefying helium and while he was investigating the properties of materials at liquid helium temperatures, the Dutch physicist H. Kamerlingh Onnes discovered that for some materials a certain temperature exists, called the critical temperature 1> S. Tc, below which the resistivity is zero and the conductivity He called this phenomenon superconductivity. Figure 10-44 shows the plot Kamerlingh Onnes obtained of the resistance of mercury versus temperature. The critical temperature for mercury is 4.2 K. The critical temperature varies from material to material, but below this temperature the electrical resistance of the material is zero. Critical temperatures for other superconducting elements range from less than 0.1 K for hafnium and iridium to 9.2 K for niobium. The critical temperatures of several superconducting materials are given in Table 10-6. In the presence of a magnetic field B, the critical temperature is lower than it is when there is no field. As the magnetic field increases, the critical temperature decreases, as illustrated in Figure 10-45. If the magnetic field is greater than some critical field Bc, superconductivity does not exist at any temperature. The values of Tc in the table are for B 0. Many metallic compounds are also superconductors. For example, the superconducting alloy Nb3Ge, discovered in 1973, has a critical temperature of 23.2 K, which was the highest known until 1986, when the first of the complex high-Tc cuprate ceramic superconductors was discovered. More recently, in 2001 Jun Akimitsu discovered that the metal compound MgB2, available \u201coff the shelf\u201d for about $2>g, became superconducting at 39 K, as of this writing the highest Tc yet for a conventional superconductor. (See Table 10-6.) Despite the cost and inconvenience of refrigeration with expensive", " liquid helium, which boils at 4.2 K, many superconducting magnets have been built using such materials. Electromagnets wound with superconducting wire are used in applications ranging from medical diagnostic MRI systems to beam-focusing and beam-bending magnets at large particle accelerators worldwide. Table 10-6 Tc and Bc values for some type I and type II superconductors Type I element Al Cd Hg In Nb Os Pb Sn Tl Zn Tc (K) 1.175 0.517 4.154 3.408 9.25 0.66 7.196 3.722 2.38 0.85 Bc (at 0 K; T) Type II compound 0.0105 0.0028 0.0411 0.0282 0.2060 0.0070 0.0803 0.0305 0.0178 0.0054 Nb3Sn Nb3Ge NbN V3Ga V3Si PbMoS CNb MgB2 Rb3C60 Cs2RbC60 Tc (K) 18.1 23.2 16.0 16.5 17.1 14.4 8.0 39.0 29.0 33.0 Bc2 (at 0 K; T) 24.5 34.0 15.3 35.0 15.6 6.0 1.7 16?? 10-9 Superconductivity 459 The conductivity of a superconductor cannot be defined since its resistance is zero. There can be a current in a superconductor even when the electric field in the superconductor is zero. Such currents are called supercurrents. Indeed, steady currents have been observed to persist for years without apparent loss in superconducting rings in which there was no electric field. Meissner Effect Bc(0) T B, 0.10 0.08 0.06 0.04 0.02 Normal Superconducting 0 2 Consider a superconducting material that is originally at a temperature greater than the critical temperature and is in the presence of a small external magnetic field B Bc. We now cool the material below the critical temperature so that it becomes superconducting. Since the resistance is now zero, there can be no emf in the superconductor. Thus, from Faraday\u2019s law, the magnetic field in the superconductor cannot change. We therefore expect from classical physics that the magnetic field in the superconductor will remain constant. However, it is observed experimentally that when a superconductor is cooled below the critical temperature in an external magnetic field, the magnetic field lines are expelled from the superconductor and thus the magnetic field inside the superconductor is zero. (See Figure 10-46.) This effect was discovered by H. W. Meissner 4 6 8 T, K Tc Figure 10-45 Variation of the critical temperature with magnetic field for lead. Note that Bc approaches zero as T approaches Tc. (a) T > Tc T < Tc Figure 10-46 (a) The Meissner effect in a superconducting sphere cooled in a constant applied magnetic field. As the temperature drops below the critical temperature Tc, the magnetic field lines are expelled from the sphere. (b) Demonstration of the Meissner effect. A superconducting tin cylinder is situated with its axis perpendicular to a horizontal magnetic field. The directions of the field lines near the cylinder are indicated by weakly magnetized compass needles mounted in a Lucite sandwich so that they are free to turn. [Courtesy of A. Leitner, Rensselaer Polytechnic Institute.] 460 Chapter 10 Solid State Physics (a) \u03bc 0M (b) \u03bc 0M Type I superconductor Bc Bapp \u03bc 0M = \u2013Bapp Type II superconductor Bc1 Bc2 Bapp 0 Figure 10-47 Plots of times the magnetization M versus applied magnetic field for type I and type II superconductors. (a) In a type I superconductor, the resultant magnetic field is zero below a critical applied field Bc because the field due to induced currents on the surface of the superconductor exactly cancels the applied field. Above the critical field, the material is a normal conductor and the magnetization is too small to be seen on this scale. (b) In a type II superconductor, the magnetic field starts to penetrate the superconductor at a field Bc1, but the material remains superconducting up to the field Bc2, after which it becomes a normal conductor. 0, 0 where B2>2 and R. Ochsenfeld in 1933 and is now known as the Meissner effect. It, not zero resistance, is the criterion that determines if a material is a superconductor. The mechanism by which the magnetic field lines are expelled or, more specifically, canceled within", " the bulk of the superconductor is that a supercurrent (called a screening current) is induced on the surface in such a direction as to exactly cancel the external field within the material.12 Thus, the superconductor exhibits perfect diamagnetism. Establishing the supercurrent \u201ccosts\u201d the superconductor an amount of energy per unit volume is the permeability of the vacuum. When the field B beequal to comes larger than Bc, there is insufficient energy available and the material reverts to its \u201cnormal\u201d resistive state. The magnetic levitation shown in the photograph below results from the repulsion between the permanent magnet producing the external field and the magnetic field produced by the currents induced in the superconductor. Only certain superconductors called type I, or \u201csoft,\u201d superconductors exhibit the complete Meissner effect. Type I superconductors are primarily very pure metal elements. Figure 10-47a shows a plot of the magnetization M times versus the applied magnetic field Bapp for a type I superconductor. For a magnetic field less than the critical field Bc, the magnetic field induced in the superconductor is equal and opposite to the external magnetic field; that is, the superconductor is a perfect diamagnet. The values of Bc for type I superconductors are all too small for such materials to be useful in the coils of a superconducting magnet. (See Table 10-6.) 0M 0 A small, cubicle permanent magnet levitates above a disk of the superconductor yttrium-barium-copper oxide, cooled by liquid nitrogen to 77 K. At temperatures below 92 K, the disk becomes superconducting. The magnetic field of the cube sets up circulating electric supercurrents in the superconducting disk, such that the resultant magnetic field in the superconductor is zero. These currents produce a magnetic field opposite to that of the cube, and thus the cube is repelled. [Courtesy of IBM Research.] Other materials, known as type II, or \u201chard,\u201d superconductors, have a magnetization curve similar to that in Figure 10-47b. Such materials are usually alloys or metals that have large resistivities in the normal state. Type II superconductors exhibit two critical magnetic fields, Bc1 and Bc2, as shown in Figure 10-48 for tantalum. Applied fields less than Bc1 result in the Meissner effect of total magnetic flux cancellation and the entire sample is superconducting, as in type I superconductors. Applied fields greater than Bc2 result in complete penetration of the magnetic field throughout the sample, and the resistivity of 10-9 Superconductivity 461 0.20 0.15 T B, 0.10 0.05 Bc1 Bc2 0 1 2 3 4 T, K Tc = 4.47 K Figure 10-48 Critical magnetic fields Bc1 and Bc2 for Ta (99.95 percent) as a function of temperature. Below the Bc1 curve Ta exhibits the Meissner effect. Between the two curves is a mixed, or vortex, state with filaments of normal Ta penetrating the superconducting state. Above the Bc2 curve there is complete magnetic field penetration and the entire sample has normal resistivity. the material returns to normal. However, in the region between Bc1 and Bc2 there is partial penetration of the magnetic field, the field lines being confined to flux tubes, also called vortices, in which the material has normal resistivity. The surrounding material remains field free and superconducting, as illustrated schematically in Figure 10-49. Each flux tube contains one quantized unit of magnetic flux, as will be described later in this section. For many type II superconductors the critical field Bc2 may be several hundred times larger than the typical values of critical fields for type I superconductors 34 T. Such (see Table 10-6). For example, the alloy Nb3Ge has a critical field Bc2 materials can be used to construct high-field superconducting magnets. (a) B B= 0 (b) Flux tube B (c) Screening supercurrent Figure 10-49 (a) Below Bc1 the type II material shows the Meissner effect. For temperatures below superconducting and B 0 throughout the volume. (b) For Bc1 B Bc2, magnetic field lines penetrate the material but are confined to flux tubes of normally resistive material that form the so-called vortex lattice. For a given T, as the applied field B approaches Bc2, the size of the superconducting region shrinks as more flux tubes occupy the volume. When B Bc2, the entire material has normal resistivity. (c) The lattice of magnetic vortices in UPt3, a strongly type II supercond", "uctor, is shown clearly by neutron diffraction. the material is Tc Tc 462 Chapter 10 Solid State Physics Vacuum Pb 2 m Vacuum Pb (a) (b) Fluxoids penetrating a superconducting film. The image has been formed by the technique of electron holography, in which coherent electron beams are used in place of coherent light beams to create a hologram. Electrons passing by a magnetic field are phase-shifted; i.e., the phase term in their wave function changes. (The shift arises from a phenomenon known as the Aharonov-Bohm effect.) By superposing such a phase-shifted beam with an unshifted reference beam, an interference pattern is created that can be interpreted as an image of the magnetic field. For the upper images, a magnetic field was applied perpendicular to a thin superconducting lead film. When the field was weak, it was expelled by the Meissner effect. A stronger field, however, penetrated the film. The fluxoids shown arose from vortices of current set up in the superconductor\u2014not from the applied field directly. In the upper right is an isolated fluxoid; in the upper left is an antiparallel pair of fluxoids. The lower micrograph, in which the lead film is thicker, shows penetration by bundles of fluxoids. [Courtesy of Akira Tonomura, Hitachi Ltd., Saitama, Japan.] (b) A lattice of fluxoid vortices penetrating the surface of a superconductor. They were made visible for the photograph by a dusting of fine ferromagnetic particles. [Courtesy of U. Essmann.] EXPLORING Flux Quantization m BnA Consider a superconducting loop of area A carrying a current. There can be a magnetic through the loop due to the current in the loop. According to Faraday\u2019s flux law of induction, if the flux changes, an emf will be induced in the loop that is proportional to the rate of change of the flux. But for a superconductor there can be no emf in the loop because there is no resistance. Therefore, the flux through the ring is frozen and cannot change. Indeed, the quantum-mechanical treatment of superconductivity reveals that the total flux through the loop is quantized and is given by m n h 2e n 1, 2, 3, \u00c1 The quantum of flux, called a fluxoid, is 0 h 2e 2.0678 1015 T # m2 Each flux tube in a type II superconductor with Bc1 of flux. B Bc2 contains one quantum 10-9 Superconductivity 463 BCS Theory Our discussion of the classical free-electron theory in Section 10-2 considered the ions of the crystal lattice to be fixed. Resistivity was due to the interactions of the electrons with the ions of the lattice, and both electron-electron interactions and the effects of lattice vibrations, i.e., electron-phonon interactions, were ignored. In the quantum theory of conduction, lattice vibrations were explicitly taken into account (see Equations 10-41 to 10-43). Lattice vibrations are also responsible for the isotope effect13 in superconductivity, discovered in 1950. This experimental observation revealed that the critical temperature depended on the isotopic mass of the crystal according to MTc constant 10-55 where M is the average isotopic mass and varies from material to material. For ex 4.185 K for samples of average isotopic mass ample, for mercury M 199.5 u, whereas Tc 4.146 K for samples with M 203.4 u. Table 10-7 lists for a few superconductors. experimental values for 0.50 and Tc The importance of the discovery represented by Equation 10-55 is to tell us that the lattice vibrations, hence the electron-phonon interactions, cannot be ignored. The assumption of fixed lattice ions is equivalent to assuming that for electronlattice ion interactions. But if then Tc would be zero for all materials. M S, M S It had been recognized for some time that superconductivity is due to a collective behavior of the conducting electrons, and discovery of the isotope effect pointed to the crucial interaction as being with the phonons. In 1957, John Bardeen, Leon Cooper, and Bob Schrieffer published a successful theory of superconductivity now known as the BCS theory.14 According to this theory, the electrons in a superconductor are coupled in pairs at low temperatures. The coupling comes about because of the interaction between electrons and the crystal lattice. An electron moving through the lattice of positive ions interacts with and perturbs it, as illustrated in Figure 10-50. The electron attracts the positive ions nearby, displacing them slightly, resulting in a region of increased positive charge density. Because the ions are bound to the lattice", " by elastic forces, this region of increased charge density propagates through the material as a vibrational wave in the lattice, i.e., a phonon. The momentum of the phonon has been provided by the electron, and we can think of the electron as having emitted a phonon. Table 10-7 Experimental values of for a few superconductors Material Cd Hg Pb 0.32 0.50 0.49 Material Nb3Sn Os Zn 0.08 0.15 0.45 Data from C. Kittel, Introduction to Solid State Physics, 8th ed. (New York: Wiley, 2005). + + + + + \u2013 + + + v + + Figure 10-50 An electron traveling through the lattice of positive ions generates a wave of increased charge density, shown in two dimensions by the dashed lines. The momentum of the wave comes at the expense of the electron\u2019s momentum. A second electron may encounter the wave and absorb its momentum. The net effect is an attraction between the two electrons and the production, for T, of a Cooper pair. Tc 464 Chapter 10 Solid State Physics A second electron that encounters the wave of increased positive charge concentration is attracted toward it by the Coulomb interaction and can absorb the momentum carried by the wave; i.e., it may absorb the phonon. Thus, the two electrons can interact via the phonon and (very important) the interaction is an attractive one since both electrons experience an attractive force toward the region of increased positive charge density. At low temperatures (T Tc) the attraction between the two electrons can exceed the Coulomb repulsion between them. Then the electrons can form a bound state called a Cooper pair, provided the temperature is low enough so that the number and energy of randomly generated thermal phonons will not disrupt its formation. The electrons in a Cooper pair have opposite spins and equal and opposite linear momenta. Thus, they form a system with zero spin and zero momentum. Each Cooper pair may be considered as a single particle with zero spin. Such a particle does not obey the Pauli exclusion principle, so any number of Cooper pairs may be in the same quantum state with the same energy.15 In the ground state of a superconductor (at T 0), all the electrons are in Cooper pairs and all the Cooper pairs are in the same energy state. In the superconducting state, the Cooper pairs are correlated, so that they all act together. In order for the electrons in a superconducting state to absorb or emit energy, the binding of the Cooper pairs must be broken. The energy needed to break up a Cooper pair is analogous to that needed to break up a molecule into its constituent atoms. This energy is called the superconducting energy gap Eg. In the BCS theory, this energy at absolute zero is predicted to be 3.5kTc In agreement with experimental observations, BCS theory also predicts the flux quantization described in the Exploring section on page 462 and the temperature dependence of Bc. 10-56 Eg Bc(T)>Bc(0) 1 (T>Tc)2 EXAMPLE 10-11 Energy Gap of Cadmium (a) Calculate the superconducting energy gap at T 0 K predicted by the BCS theory for cadmium and compare the result with the measured result of 1.50 104 eV. (b) Compute the wavelength of a photon whose energy is just sufficient to break up a Cooper pair in cadmium. SOLUTION (a) From Table 10-6, we have that Tc of the energy gap is then 0.517 K for cadmium. The BCS prediction Eg 3.5kTc 3.5(1.38 1023 J>K)(0.517 K) (1.60 1019 J>eV) This differs from the measured values of 1.50 104 eV by about 4 percent. 1.56 104 eV (b) Eg hf hc>, hc>Eg or we have that (6.63 1034 J # s)(3.00 108 m>s) (1.56 104 eV)(1.60 1019 J>eV) 7.97 103 m Remarks: This wavelength is in the short-wavelength microwave region of the electromagnetic spectrum. Note that the energy gap for a typical superconductor is much smaller than the energy gap for a typical semiconductor, which is of the order of 1 eV. As the temperature is increased from T 0, some of the Cooper pairs are broken. The resulting individual (unpaired) electrons interact with the remaining Cooper pairs, reducing the energy the energy gap is zero (see Figure 10-51). Notice, too, that the gap gap until at T Tc energy is typically larger than that available from the thermal energy of the T 0.5 Tc, Eg(T) (0.95) Eg(0)", " (3.3) kTc, system. For example, for whereas the thermal energy kT (0.5)kTc. The Cooper pairs that we have discussed so far have zero momentum, so there are as many electrons traveling in one direction as the other and there is no current. Cooper pairs can also be formed with a net momentum p rather than zero momentum, but all the pairs have the same momentum. In this state, current is carried by the Cooper pairs. In ordinary conductors, resistance is present because the current carriers can be scattered with a change in momentum. As we have discussed, this scattering may be due to impurity atoms or thermal vibrations of the lattice ions. In a superconductor, the Cooper pairs are constantly scattering each other, but since the total momentum remains constant in this process, there is no change in the current. A Cooper pair cannot be scattered by a lattice ion because all the pairs act together. The only way that the current can be decreased by scattering is if a pair is broken up, which requires energy greater than or equal to the energy gap Eg. At reasonably low currents, scattering events in which the total momentum of a Cooper pair is changed are completely prohibited, so there is no resistance. 10-9 Superconductivity 465 ).0 0.8 0.6 0.4 0.2 BCS curve Tin Tantalum Niobium 0 0 0.2 0.4 0.6 0.8 1.0 T/Tc Figure 10-51 Ratio of the energy gap at temperature T to that at T 0 as a function of the relative temperature T Tc. The solid curve is that predicted by the BCS theory. > EXAMPLE 10-12 How Big Is a Cooper Pair? Calculate an estimate of the separation x of the electrons forming a Cooper pair, assuming that the binding energy of the pair equals the gap energy Eg and that, like semiconductors, the gap is centered on the Fermi energy EF. SOLUTION The energy of either electron is, with the aid of the de Broglie relation, given by E U2k2 p2 2m* 2m* \u00a2E 2kU2 \u00a2k 2m* and If we associate E with the Fermi energy and E with the gap, then \u00a2E E Eg EF 2kU2 \u00a2k 2m* 2m* U2k2 2 \u00a2k k Since the Fermi energy is typically of the order of 1 eV and the gap of the order 104 eV, as computed in Example 10-12, then 104 and Eg \u00a2k 0.5 104k >EF k >a EF. where k refers to the value at the Fermi level. As was discussed in Section 10-6 and at the top of the first allowed band, where the illustrated in Figure 10-19, a 0.1 nm, energy is approximately so we have that k >0.1 nm1 From the uncertainty relation (Equation 517), we then have that the uncertainty in the location of either electron, i.e., the extent of their wave functions in space, is \u00a2x 1 \u00a2k \u00a2k 103 nm1. The lattice spacing 103 nm and or roughly equal to 10,000 atomic diameters and approximately equal to the wavelength of visible light. 466 Chapter 10 Solid State Physics High-Temperature Superconductivity For many years, the highest known critical temperature for a superconductor was 23.2 K for the alloy Nb3Ge. Then in 1986 J. G. Bednorz and K. A. M\u00fcller found that an oxide of lanthanum, barium, and copper became superconducting at 30 K. Soon afterward, in 1987, superconductivity with a critical temperature of 92 K was found in a ceramic of copper oxide containing yttrium and barium (YBa2Cu3O7). Since then, several copper oxides have been found with critical temperatures as high as 164 K. Table 10-8 lists some of the high-temperature superconductors along with their critical temperatures. These discoveries have revolutionized the study of superconductivity because relatively inexpensive liquid nitrogen, which boils at 77 K, can be used for a coolant. However, there are many problems, such as the brittleness of ceramics, that thus far make these new superconductors difficult to use. High-temperature superconductors are all type II superconductors with very high upper critical fields. For some, Bc2 is estimated to be as high as 100 T. Although the BCS theory appears to be the correct starting place for understanding these new superconductors, they have many features that are not clearly understood, and the theoretical explanation of the phenomenon is not yet in hand. Thus, there is much work, both experimental and theoretical, to be done. Table 10-8 Critical temperatures of some high Tc", " superconductors Material LaBaCuO La2CuO4 YBa2Cu3O7 DyBa2Cu3O7 C60(CHBr3) BiSrCaCuO Tl2Ba2Ca2Cu3O10 Hg.8Tl.2Ba2Ca2Cu3O8.33 Tc, K 30 40 92 92.5 117 120 120 138 EXPLORING Josephson Junction In Section 6-6, we discussed barrier penetration\u2014the tunneling of a single particle through a potential barrier. The tunneling of electrons from one metal to another can be observed by separating the two metals with a thin layer only a few nanometers thick of an insulating material such as aluminum oxide. When both metals are normal metals (not superconductors), the current resulting from the tunneling of (a) I (b) I 10-9 Superconductivity 467 V Vc V Figure 10-52 Tunneling current versus voltage for a junction of two metals separated by a thin oxide layer. (a) When both metals are normal metals, the current is proportional to the voltage as predicted by Ohm\u2019s law. (b) When one metal is a normal metal and one is a superconductor, the current is approximately zero until the applied voltage exceeds the critical voltage Vc > Eg 2e. > electrons through the insulating layer obeys Ohm\u2019s law for low applied voltages (see Figure 10-52a). When one of the metals is a normal metal and the other is a superconductor, there is no current (at absolute zero) unless the applied voltage V is Eg 2e, where Eg is the superconductor energy greater than a critical voltage Vc gap. Figure 10-52b shows the plot of current versus voltage for this situation. The current jumps abruptly when V is great enough to break up a Cooper pair. (At temperatures above absolute zero, there is a small current because some of the electrons in the superconductor are thermally excited above the energy gap and therefore are not paired.) The superconducting energy gap can thus be accurately measured by measuring the critical voltage Vc. In 1962, Brian Josephson 16 proposed that when two superconductors form a junction, now called a Josephson junction, Cooper pairs could tunnel from one superconductor to the other with no resistance. The current is observed with no voltage applied across the junction and is given by I Imaxsin ( 2 1) 10-57 where Imax is the maximum current, which depends on the thickness of the barrier, 1 is the phase of the wave function for the Cooper pairs in one of the superconductors, is the phase of the corresponding wave function in the other superconductor. and (The phase of a wave function is the exponent of the time part of the total wave function. See Section 6-1.) This result has been observed experimentally and is known as the dc Josephson effect. Et>U t 2 Josephson also predicted that if a dc voltage were applied across a Josephson junction, there would be a current that alternates with frequency f given by f 2eV h 10-58 This result, known as the ac Josephson effect, has also been observed experimentally, and careful measurement of the frequency allows a precise determination of the ratio > e h. Because frequency can be measured so accurately, the ac Josephson effect is also used to establish precise voltage standards. The inverse effect, in which the application of an alternating voltage across a Josephson junction results in a dc current, has also been observed. 468 Chapter 10 Solid State Physics h EXAMPLE 10-13 AC Josephson Effect Using e 1.602 1019 C and 6.626 1034 J s, calculate the frequency of the Josephson current if the applied voltage is 1 V. # SOLUTION From Equation 10-58, we obtain f 2eV h a 2(1.602 1019 C)(106 V) 6.616 1034 J # s b 4.836 108 Hz 483.6 MHz There is a third effect observed with Josephson junctions. When a dc magnetic field is applied through a superconducting ring containing two Josephson junctions, the total supercurrent shows interference effects that depend on the intensity of the magnetic field (Figure 10-53). This effect can be used to measure very weak magnetic fields. It is the basis for a device called a SQUID (for Superconducting Quantum Interference Device) that can detect magnetic fields as low as 1014 T. Such a device can detect the magnetic fields produced by the tiny currents flowing in the heart and brain. I1 I2 Itotal Junctions Itotal Superconducting ring Figure 10-53 A superconducting ring with two Josephson junctions. When there is no applied magnetic field through the ring, the currents I1 and I2 are in phase. A very small applied magnetic field produces a phase difference in the two", " currents that produces interference in the total current exiting the ring. Summary TOPIC RELEVANT EQUATIONS AND REMARKS 1. Structure of solids Solids are often found in crystalline form, in which a small structure called the unit cell is repeated over and over. The structure of the unit cell depends on the type of bonding between the atoms, ions, or molecules forming the crystal. Ionic and covalent solids The attractive part of the potential energy of an ion in an ionic crystal is Uatt ke2 r 10-1 where r is the separation between neighboring ions and which depends on the crystal geometry. The constant cubic crystals. is the Madelung constant, is 1.7476 for face-centered- Summary 469 TOPIC RELEVANT EQUATIONS AND REMARKS In covalently bonded crystals the individual bonds are just like those in covalently bonded molecules. The metallic bond has no single-molecule counterpart. One or more valence electrons are free to move throughout the solid, and all of the atoms share all of the free electrons, making this bond roughly analogous to the covalent bond. 2. Classical free-electron theory 3. Electron gas in metals 4. Quantum theory of conduction Electrical resistivity and conductivity are given by ne2 8v9 me and is the mean speed of the electrons and 8v9 me ne2 8v9 where collisions. The latter is given by 1 na r2 10-13 is the mean free path between 10-12 where na is the ion density. These yield Ohm\u2019s law correctly but result in the wrong temperature dependence of the resistivity. The average energy of the electrons at ordinary temperatures is much larger than kT: 8E9 3 5 10-22 EF where typical values of the Fermi energy EF are 1 to 2 eV. This theory results from making two important corrections to the classical freeelectron theory. First, the Fermi-Dirac distribution of electron energies is used rather than the Maxwell-Boltzmann distribution. Second, the effect of the wave characteristics of the electrons is considered in their scattering from the lattice ions. The resulting theory is in good agreement with observations. 5. Magnetism in solids The origin of magnetism in solids is the electron spins and their associated magnetic moments. 6. Band theory of solids Kronig-Penney model When many atoms are brought together to form a solid, the individual energy levels are split into bands of allowed energies. The splitting depends on the type of bonding and the lattice separation. In a conductor, the uppermost band containing electrons is only partially full, so there are many available states for excited electrons. In an insulator, the uppermost band containing electrons, the valence band, is completely full and there is a large energy gap between it and the next allowed band, the conduction band. In a semiconductor, the energy gap between the filled valence band and the empty conduction band is small, so at ordinary temperatures an appreciable number of electrons are thermally excited into the conduction band. The solid is modeled as a periodic potential. The wave functions are then \u2019(x) uk(x)eikx where the function uk(x) is periodic with a period equal to that of the spacing of the potential wells and ka n is a free electron, i.e., a plane wave. The energy gaps occur at 10-36 10-41 eikx for integer n and a equal to the lattice spacing. 470 Chapter 10 Solid State Physics TOPIC RELEVANT EQUATIONS AND REMARKS 7. Impurity semiconductors 8. Superconductivity The conductivity of a semiconductor can be greatly increasing by doping. In an n-type semiconductor, the doping adds electrons just below the conduction band. In a p-type semiconductor, holes are added just above the valence band. A junction between an n-type and p-type semiconductors has applications in many devices, such as diodes, solar cells, and light-emitting diodes. A transistor consists of a very thin semiconductor of one type sandwiched between two semiconductors of the opposite type. Transistors are used in amplifiers because a small variation in the base current results in a large variation in the collector current. In a superconductor the resistance drops suddenly to zero below a critical temperature Tc. Magnetic field lines are expelled and B 0 inside a type I semiconductor, a phenomenon called the Meissner effect. Superconductivity at low temperatures is described by BCS theory, in which free electrons form Cooper pairs. Recently discovered high-temperature semiconductors are only partially understood in terms of BCS theory. General References The following general references are written at a level appropriate for the readers of this book. Anderson, B., Fundamentals of Semiconductor Devices, McGraw-Hill, New York, 2005.", " Blatt, F., Modern Physics, McGraw-Hill, New York, 1992. Burns, G., Solid State Physics, Academic Press, Orlando, FL, 1985. Eisberg, R., and R. Resnick, Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles, 2d ed., Wiley, New York, 1985. Fermi, E., Molecules, Crystals, and Quantum Statistics (trans. M. Ferro-Luzzi), W. A. Benjamin, New York, 1966. Notes 1. The constant n is often called the Born exponent. 2. Carbon also has a fourth solid form, charcoal, which has no well-defined crystalline structure. 3. Notice that this view of the metal fits the definition of a plasma set forth in the opening paragraph of the chapter. Though not usually thought of in that way, metals are indeed low-temperature plasmas. 4. Felix Bloch (1905\u20131983), Swiss American physicist. He devised a method for measuring atomic magnetic fields in liquids and solids that led to the development of nuclear magnetic resonance (NMR) spectroscopy and earned for him a share (with E. M. Purcell) of the 1952 Nobel Prize in Physics. He was the first director-general of CERN, the European Center for Nuclear Research. Holden, A., The Nature of Solids, Columbia University Press, 1968. An excellent nonmathematical treatment of the properties of solids. Kittel, C., Introduction to Solid State Physics, 8th ed., Wiley, New York, 2005. Leitner, A., Introduction to Superconductivity, Michigan State University, East Lansing, 1965. This excellent film, running 48 minutes, is probably one of the best introductions to superconductivity available. Shockley, W., Electrons and Holes in Semiconductors, Van Nostrand, Princeton, NJ, 1950. 5. The graph of the energy bands and gaps of Figure 10-19b results from a simplified version of the conditional equation connecting k, k, and In that limit the lattice spacing is a, rather than a b, as in Figure 10-18. in which S. b S 0 and U0 6. This mixing, called hybridization, was discussed in Section 9-2. 7. See, e.g., Section 25-5 in P. Tipler and G. Mosca, Physics for Scientists and Engineers, 6th ed. (W. H. Freeman and Co., New York, 2008). 8. The fact that the radius of the bound electron is several times the equilibrium spacing of the atoms helps justify our tacit assumption that the fifth electron sees a uniform dielectric constant in the crystal. Problems 471 9. Klaus von Klitzing (b. 1943), German physicist. He re- ceived the 1985 Nobel Prize in Physics for this discovery. 10. Daniel C. Tsui (b. 1939), Chinese American physicist. He received the 1998 Nobel Prize in Physics with H. L. Stormer and R. B. Laughlin for their discovery. 11. William B. Shockley (1910\u20131989), John Bardeen (1908\u2013 1991), and Walter H. Brattain (1902\u20131987), American physicists. Shockley discovered that doped germanium crystals were excellent rectifiers, and subsequently the three Bell Laboratories colleagues discovered that two such \u201csolid-state rectifiers\u201d combined would amplify current. The discovery of this device, the transistor, earned them the 1956 Nobel Prize in Physics. 12. Actually, the field decreases exponentially across the surface, reaching zero at a depth of about 10 nm. 13. Isotopes are atoms with the same atomic number Z but different atomic mass numbers A. Isotopes will be discussed in Chapter 11. 14. John Bardeen (1908\u20131991), Leon N. Cooper (b. 1930), and J. Robert Schrieffer (b. 1931), American physicists. Developed at the University of Illinois, the BCS theory earned the collaborators the 1972 Nobel Prize in Physics, and Bardeen became the only person thus far to win two physics Nobel Prizes (see note 11). 15. This may make it seem like the Cooper pair is a boson and superconductivity another example of Bose-Einstein condensation (see Section 8-3); however, the large size of the Cooper pair (see Example 10-12) means many pairs overlap and that the symmetry of the pair with respect to an exchange of electrons must also take into account exchanges involving electrons in different pairs. The result is that the Cooper pair is neither a pure boson nor a pure fermion. 16. Brian D. Josephson (b. 1940), Welsh physicist. For this discovery, made while he was still a graduate student, he shared the 1973 Nobel Prize in Physics with L. Esaki and I. Giaever.", " Bardeen had strongly opposed Josephson\u2019s tunneling prediction until experiments, led by those of Giaever (also done while he was a graduate student), confirmed tunneling by Cooper pairs. Problems Level I Section 10-1 The Structure of Solids > > > > Cl Cl 0.257 nm. ions in LiCl is 0.257 nm. Use this and the molec- 10-1. Find the value of n in Equation 10-6 that gives the measured dissociation energy of 741 kJ mol for LiCl, which has the same structure as NaCl and for which r0 10-2. Calculate the distance r0 between the K and ions in KCl, assuming that each ion occupies a cubic volume of side r0. The molar mass of KCl is 74.55 g mol and its density is 1.984 g cm3. 10-3. The distance between the Li and ular mass of LiCl (42.4 g mol) to compute the density of LiCl. 10-4. The crystal structure of KCl is the same as that of NaCl. (a) Calculate the electrostatic potential energy of attraction of KCl, assuming that r0 is 0.314 nm. (b) Assuming that n 9 in Equation 10-6, calculate the dissociation energy in eV per ion pair and in kcal mol. (c) The measured dissociation energy is 165.5 kcal mol. Use this to determine n in Equation 10-6. 10-5. The observed dissociation energy of solid LiBr is 788 kJ mol. Compute the cohesive energy of LiBr and compare the result with the value in Table 10-1. (Ionization energies for Li and Br are in Table 9-1.) 10-6. The density of NaCl (an fcc crystal) is 2.16 g cm3. Find the distance between ions that are nearest neighbors. 10-7. The separation of nearest neighbor ions in the KCl crystal (an fcc structure) is 0.315 nm. Use this information to determine the density of KCl. 10-8. Using the data for ionic and metallic crystals from Table 10-1, (a) graph cohesive energy versus melting point and put the best straight line through the points. (b) Determine the cohesive energies of cobalt, silver, and sodium, whose melting temperatures are 1495\u00b0C, 962\u00b0C, and 98\u00b0C, respectively. (The measured values are cobalt 4.43 eV, silver 2.97 eV, and sodium 1.13 eV.) 10-9. Figure 10-54 shows a one-dimensional ionic lattice consisting of doubly charged positive ions and twice as many singly charged negative ions. Compute the Madelung constant for this \u201ccrystal\u201d to within 1 percent Figure 10-54 472 Chapter 10 Solid State Physics Section 10-2 Classical Theory of Conduction 0.4 nm (a) Given a mean free path 8v9 8v9 1.17 105 m>s and a mean speed 10-10. for the current flow in copper at a temperature of 300 K, calculate the classical value for the resistivity of copper. (b) The classical model suggests that the mean free path is temperature independent and that depends on temperature. From this model, what would be at 100 K? 10-11. Find (a) the current density and (b) the drift velocity if there is a current of 1 mA in a No. 14 copper wire. (The diameter of No. 14 wire, which is often used in household wiring, is 0.064 in 0.163 cm.) 10-12. A measure of the density of the free-electron gas in a metal is the distance rs, which is defined as the radius of the sphere whose volume equals the volume per conduction electron. (a) Show that where na is the free-electron number density. (b) Calculate rs for copper in nanometers. 10-13. Calculate the number density of free electrons in (a) Ag ( 19.3 g> cm3), ues listed in Table10-3. 10-14. Calculate the number density of free electrons for (a) Mg 7.1 g> cm3 ( ues listed in Table 10-3. 10-15. copper from Equations 10-13. Using the same value of (c) T 100 K. and (b) Zn ), assuming two free electrons per atom, and compare your results with the val- and (b) Au assuming one free electron per atom, and compare your results with the val- at T 300 K, calculate and find 8v9 1.08 105 m>s, for at (b) T 200 K and ( 10.5 g> cm3) ( 1.74 g> cm", "3) (3>4na), 0.37 (a) Using nm and and rs 4.77 eV). Section 10-3 Free-Electron Gas in Metals 10-16. Find the average energy of the electrons at T 0 K in (a) copper (EF (b) Li (EF 10-17. Calculate the Fermi energy for magnesium in a long, very thin wire. 10-18. Compute (a) the Fermi energy and (b) the Fermi temperature for silver and for iron and compare your results with the corresponding values in Table 10-3. 10-19. Show that for T 300 K about 0.1 percent of the free electrons in metallic silver have an energy greater than EF. 7.06 eV) and Section 10-4 Quantum Theory of Conduction 4.2 \" # cm, 10-20. What is the Fermi speed, i.e., the speed of a conduction electron whose energy is equal to the Fermi energy EF, for (a) Na, (b) Au, and (c) Sn? (See Table 10-3.) 2.04 \" # cm, 10-21. The resistivities of Na, Au, and Sn at T 273 K are and 10.6 \" # cm, respectively. Use these values and the Fermi speeds calculated in Problem 10-20 to find the mean free paths 10-22. At what temperature is the heat capacity due to the electron gas in copper equal to 10 percent of that due to lattice vibrations? 10-23. Use Equation 10-29 with to calculate the average energy of an electron in copper at T 300 K. Compare your result with the average energy at T 0 and the classical result of ( 10-24. Compute the maximum fractional contribution to the heat capacity of solid iron that can be made by the electrons. for the conduction electrons in these elements. 2>4 3>2 )kT. Section 10-5 Magnetism in Solids 10-25. The magnetic polarization of any material is defined as Compute the high-temperature polarization of a paramagnetic solid at T 200 K in a magnetic field of 2.0 T. 10-26. Show that the magnetic susceptibility is a dimensionless quantity. ( P P ( )>. Problems 473 Section 10-6 Band Theory of Solids 10-27. (a) The energy gap between the valence band and the conduction band in silicon is 1.14 eV at room temperature. What is the wavelength of a photon that will excite an electron from the top of the valence band to the bottom of the conduction band? Do the same calculation for (b) germanium, for which the energy gap is 0.72 eV, and (c) for diamond, for which the energy gap is 7.0 eV. 10-28. (a) The energy band gap in germanium is 0.72 eV. What wavelength range of visible light will be transmitted by a germanium crystal? (Think about it carefully!) (b) Now consider a crystal of an insulator whose energy band gap is 3.6 eV. What wavelength range of visible light will this crystal transmit? (c) Justify each of your answers to (a) and (b). m 10-29. A photon of wavelength 3.35 has just enough energy to raise an electron from the valence band to the conduction band in a lead sulfide crystal. (a) Find the energy gap between these bands in lead sulfide. (b) Find the temperature T for which kT equals this energy gap. 10-30. Consider a small silicon crystal measuring 100 nm on each side. (a) Compute the total number N of silicon atoms in the crystal. (The density of silicon is 2.33 g cm3.) (b) If the conduction band in silicon is 13 eV wide and recalling that there are 4N states in this band, compute an approximate value for the energy spacing between adjacent conduction band states for the crystal. > Section 10-7 Impurity Semiconductors 10-31. Arsenic has five valence electrons. If arsenic is used as a dopant in silicon, compute (a) the ionization energy and (b) the orbit radius of the fifth arsenic electron. The effective mass for electrons in silicon is 0.2 me. (c) What is the ratio of the ionization energy of the fifth electron to the energy gap in silicon? 10-32. Gallium has three valence electrons. If gallium is used to dope germanium, compute (a) the ionization energy of the hole and (b) the orbit radius of the hole. The effective mass of holes in germanium is 0.34 me. 10-33. What type of semiconductor is obtained if silicon is doped with (a", ") aluminum and (b) phosphorus? (See Appendix C for the electron configurations of these elements.) 10-34. The donor energy levels in an n-type semiconductor are 0.01 eV below the conduction band. Find the temperature for which kT 0.01 eV. 10-35. A strip of tin is 10 mm wide and 0.2 mm thick. When a current of 20 A is established in the strip and a uniform magnetic field of 0.25 T is oriented perpendicular to the plane of the is measured across the width of the strip. Compute (a) the denstrip, a Hall voltage of 2.20 sity of charge carriers in tin and (b) the average number of charge carriers contributed by each tin atom. The density of tin is 5.75 103 kg m3 and its molecular mass is 118.7. V > Section 10-8 Semiconductor Junctions and Devices 10-36. For a temperature of 300 K, use Equation 10-49 to find the bias voltage Vb for which the exponential term has the value (a) 10 and (b) 0.1. 10-37. For what value of bias voltage Vb does the exponential in Equation 10-49 have the value (a) 5, and (b) 0.5 for T 200 K? 10-38. Compute the fractional change in the current through a pn junction diode when the forward bias is changed from 0.1 V to 0.2 V. 10-39. For T 300 K, use Equation 10-49 to find the bias voltage Vb for which the exponential term had the value (a) 10 and (b) 0.1. 10-40. When light of wavelength no larger than 484 nm illuminates a CdS solar cell, the cell produces electric current. Determine the energy gap in CdS. Section 10-9 Superconductivity 10-41. Three naturally occurring isotopes of lead are 206Pb, 207Pb, and 208Pb. Using the value of from Table 10-7 and the isotopic masses from Appendix A, compute the critical temperatures of these isotopes. 474 Chapter 10 Solid State Physics 10-42. Compute (a) the superconducting energy gap for indium and (b) the wavelength of a photon that could just break up a Cooper pair in indium at T 0 K. 10-43. (a) Use Equation 10-56 to calculate the superconducting energy gap for tin and compare your result with the measured value of 6 104 eV. (b) Use the measured value to calculate the wavelength of a photon having sufficient energy to break up a Cooper pair in tin at T 0 K. 10-44. Use the BCS curve in Figure 10-51 to estimate the energy gaps in (a) tin, (b) niobium, (c) aluminum, and (d) zinc, all at T 0.5Tc. 10-45. Expressing the temperature T as a fraction of the critical temperature Tc, according to BCS theory at what temperature is (a) Bc 0.1Bc(0), (b) Bc 0.5Bc(0), (c) Bc 0.9Bc(0)? Level II 10-46. Estimate the fraction of free electrons in copper that are in excited states above the Fermi energy at (a) room temperature of 300 K and (b) 1000 K. 10-47. A one-dimensional model of an ionic crystal consists of a line of alternating positive and negative ions with distance r0 between adjacent ions. (a) Show that the potential energy of attraction of one ion in the line is (b) Using the result that V 2ke2 r0 a1 ln(1 x) x x2 2 x3 3 x4 4 \u00c1 A 2 ln 2 1.386. (2>3) (EF)3>2 g(E) AE1>2, show that the Madelung constant for this one-dimensional model is 10-48. Estimate the Fermi energy of zinc from its electronic molar heat capacity of (3.74 104 J>mol # K)T. 10-49. The density of the electron states in a metal can be written where A is a constant and E is measured from the bottom of the conduction band. (a) Show that the total. (b) About what fraction of the conduction electrons are within number of states is kT of the Fermi energy? (c) Evaluate this fraction for copper at T 300 K. 10-50. High-purity germanium (HPGe) crystals are used as detectors for x rays and gamma rays. On interacting with the crystal, incoming photons produce electron-hole pairs, exciting many electrons across the 0.72-", "eV energy gap into the conduction band. The decay of the radioisotope 60Co results in the emission of two gamma rays with energies 1.17 MeV and 1.33 MeV (see Chapter 11). (a) Compute the numbers of electrons N1 and N2 excited across the energy gap by each of the two gamma rays. (b) The numbers N1 and N2 are subject to statistical Compute the fractional uncertainties in N1 and N2. fluctuations of (c) Compute the corresponding fractional uncertainties in the energies of the two gamma rays. This is a measure of the energy resolution of the HPGe crystal. 10-51. A doped n-type silicon sample with 1016 electrons per cubic centimeter in the conduction band has a resistivity of at 300 K. Find the mean free path of the electrons. Use 0.2 me for the effective mass of the electron. Compare your result with the mean free path of electrons in copper at 300 K. 10-52. A \u201cgood\u201d silicon diode has a current-voltage characteristic given by 5 103 \" # m 2N2. 2N1 and I I0(eeVb >kT 1) Let kT 0.025 eV (room temperature) and the saturation current I0 1 nA. (a) Show that for (Hint: Do a Taylor expansion of the exposmall reverse-bias voltages, the resistance is 25 nential function or use your calculator and enter small values for Vb.) (b) Find the dc resistance for a reverse bias of 0.5 V. (c) Find the dc resistance for a 0.5-V forward bias. What is the current in this case? (d) Calculate the ac resistance dV dI for a 0.5-V forward bias. M\". > Problems 475 10-53. The relative binding of the extra electron in the arsenic atom that replaces an atom in silicon or germanium can be understood from a calculation of the first Bohr orbit of this electron in these materials. Four of arsenic\u2019s outer electrons form covalent bonds, so the fifth electron sees a singularly charged center of attraction. This model is a modified hydrogen atom. In the Bohr model of the hydrogen atom, the electron moves in free space at a radius a0 given by 0h2 mee2 a0 * * 0 and me with an effective mass for the electron. For silicon * When an electron moves in a crystal, we can approximate the effect of the other atoms by re- is 12 and the placing with 0 effective mass is about 0.2me, and for germanium is 16 and the effective mass is about 0.1me. Estimate the Bohr radii for the outer electron as it orbits the impurity arsenic atom in silicon and germanium. InSb is a semiconductor. The energy gap Eg between its valence and conduction bands 10-54. is 0.23 eV, and its dielectric constant In the InSb crystal the electron\u2019s effective mass m* 0.015me. (a) Compute the ionization energy for an electron donor in InSb. (b) What is the radius of the ground-state orbit? (c) At approximately what donor concentration will the orbits of adjacent donor atoms begin to overlap? 10-55. The mean free path of an electron in a metal depends on both the lattice oscillations of 1> the metal ions and those of any impurity ions according to The resistivity i. 1.2 108 \" # m by the addition of 1 percent (by number of pure copper is increased by about of atoms) of a certain impurity dispersed evenly throughout the metal. (a) Estimate from this information. (b) The impurity atoms are \u201cseen\u201d by the electrons to have an effective diameter d. Estimate the scattering cross section d2 from Equation 10-12, where d 2r. 1> 1> m * 18. i Level III 10-56. When arsenic is used to dope silicon, the fifth arsenic electron and the As ion act like a hydrogen atom system except that the potential function V(r) and the electron mass must be modified as described in Section 10-7 to account for the crystal lattice. With these modifications, (a) solve the Schr\u00f6dinger equation, using the solution in Chapter 7 as a guide. (b) Obtain Equation 10-43, and (c) sketch a properly scaled energy-level diagram for the fifth electron for n 1 through 5. 10-57. The quantity K is the force constant for a \u201cspring\u201d consisting of a line of alternating positive and negative ions. If these ions are displaced slightly from their equilibrium separation r0, they will vibrate with a frequency f 1 2 A K m, n, and r0", " for NaCl and the reduced mass for the NaCl molecule to cal(a) Use the values of culate this frequency. (b) Calculate the wavelength of electromagnetic radiation corresponding to this frequency, and compare your result with the characteristic strong infrared absorption bands in the region of about 10-58. Consider a model for a metal in which the lattice of positive ions forms a container for a classical electron gas with n electrons per unit volume. In equilibrium, the average electron velocity is zero, but the application of an electric field produces an acceleration of the electrons. If to account for the electron-lattice collisions, then we have the equation we use a relaxation time that are observed for NaCl. 61 m m dv dt m v eE (a) Solve the equation for the drift velocity in the direction of the applied electric field. (b) Verify that Ohm\u2019s law is valid, and find the resistivity as a function of n, e, m, and the relaxation time. 476 Chapter 10 Solid State Physics e(1)r # (1)s # (1)t e(1)rst, 10-59. Imagine a cubic crystal like NaCl, with a negative charge at the center of a Cartesian coordinate system with scale units equal to the interatomic distance (a) Show that an ion at a position r units along the x axis, s units along the y axis, and t units along the z axis has a charge where e is the electron charge. (b) Using Equation 10-2 of as a guide, calculate the Madelung constant for a cube 2 units on a side. Do the same for cubes of sides 4, 6, 8, 10, 12, 16, and 20 units (You will probably want to use a computer spreadsheet to write a program to do the calculations for the larger cubes). Are your answers approaching the value 10-60. the magnetization per unit volume M is given by (a) Show that for a paramagnetic solid with electron energies given by Equation 10-33 1.7476? (b) For B V kT show that the susceptibility is given by Equation 10-35. M tanh (B>kT) CHAPTER 11 Nuclear Physics The first information about the atomic nucleus came with the discovery of radioactivity by A. H. Becquerel1 in 1896. Intrigued by Roentgen\u2019s discovery of x rays the previous year, Becquerel was investigating the possibility that minerals that exhibit fluorescence after exposure to sunlight might also be emitting x rays. He used the simple technique of placing a sample of such a mineral, potassium uranyl sulfate, on top of a photographic plate wrapped in black paper lying in the sunlight on a window ledge. Sure enough, an image of the sample appeared on the developed plate, and Becquerel concluded that x rays had indeed been emitted. But when a similar sample lying on a wrapped photographic plate in a drawer without exposure to sunlight during a period of cloudy weather produced an image just as dark, he investigated further and found that the sample was spontaneously emitting a previously unknown penetrating radiation. He had discovered radioactivity.2 according to their ability to penetrate matter and to ionize air: The rays emitted by radioactive nuclei were first classified by Rutherford as,,, radiation penand etrates the least and produces the most ionization, radiation penetrates the most with the least ionization, and radiation is intermediate between them. In a classic exper- iment, Rutherford soon found that rays are nuclei. It was also quickly discov- rays are very short-wavelength electromagnetic ered that rays are electrons and -particle-scattering experiments in 1911 (see radiation. Geiger and Marsden\u2019s Section 4-2) and the successes of the Bohr model of the atom led to the modern view of an atom as consisting of a tiny, massive nucleus with a radius of 1 to 10 femtometer (fm; 1 fm 1015 m) surrounded by a cloud of electrons at a relatively great distance, of the order of 0.1 nm ( 100,000 fm) from the nucleus. 4He in rapid succession in 1932, In 1928, the correct explanation of radioactivity as a quantum-mechanical, barrier-penetration phenomenon was given by G. Gamow, R. W. Gurney, and E. U. the neutron was discovered by Condon. Then, J. Chadwick and the positron by C. D. Anderson, and the first nuclear reaction using artificially accelerated particles (protons) was observed by J. D. Cockcroft and E. T. S. Walton.3 Thus, it is quite reasonable to mark that year as the beginning of modern nuclear physics. Much of the information about nuclei is obtained by bombarding them with various particles and observing the results. The advent of particle accelerators, the Van de Graaff", " electrostatic generator in 1931 and the cyclotron in 1932, 11-1 The Composition of the Nucleus 478 11-2 Ground-State Properties of Nuclei 11-3 Radioactivity 480 492 11-4 Alpha, Beta, and Gamma Decay 495 11-5 The Nuclear Force 11-6 The Shell Model 11-7 Nuclear Reactions 11-8 Fission and Fusion 11-9 Applications 506 513 516 526 537 477 478 Chapter 11 Nuclear Physics made many experimental studies possible without the severe limitations on particle type and energy imposed by naturally radioactive sources. Since then, an enormous technology has been developed for accelerating and detecting particles, and many nuclear reactions and fundamental particle interactions have been studied. Among the myriad nuclear reactions that have been investigated are two types of special interest: fission and fusion. Both are processes by which nuclear mass is converted into other forms of energy, such as thermal energy, just as some atomic mass is converted in chemical reactions such as oxidation. Fission reactions currently provide a significant, albeit controversial, means of producing electrical energy in 30 countries, accounting for 6.0 percent of the world\u2019s total consumption of energy in 2005.4 The similar potential of fusion reactions has not yet been realized at a practical level; however, of far more intrinsic importance is the role of fusion in the production of energy in stars. The grim reality that both fission and fusion are also the basis for weapons of enormous destructive power means that this application of nuclear reactions influences political debate to a greater degree than has perhaps any other scientific discovery in history. In this chapter we will discuss some of the general properties of atomic nuclei and the important features of radioactivity. While our discussions will of necessity be only semiquantitative, we will consider the nature of the nuclear force as it is currently understood and describe one of the most useful models in terms of which many nuclear properties may be explained. The applications of radioactivity and nuclear reactions are by no means limited to fission and fusion. The radiations emitted by radioisotopes have long been used in medical diagnosis and treatment. These contributions were measurably enhanced with the development of computer assisted tomography5 (CAT) in the 1970s, which made possible not only x-ray CAT scans, but also the more recent development of positron emission tomography, called PET. Neutron-induced nuclear reactions provide an extremely sensitive technique, called neutron activation analysis (NAA), for measuring trace amounts of certain isotopes for most elements in the periodic table. These and many other applications will be discussed in this chapter. 2 A Z 1 11-1 The Composition of the Nucleus The experiments of Moseley (see Section 4-4) showed that the nuclear charge is Z times the proton charge, where Z is the atomic number, which is about half the atomic mass number A (except for hydrogen, for which Z A). Thus, the nucleus has a mass about equal to that of A protons but a charge of only protons. Before the discovery of the neutron, it was difficult to understand this unless there were A\u2013Z electrons in the nucleus to balance the charge without changing the mass very much. The idea that the nucleus contained electrons was supported by the observation of decay, in which electrons are ejected by certain radioactive nuclei. However, there were serious difficulties with this model. A relatively simple calculation from the uncertainty principle (see Problem 11-2) shows that an electron has a minimum kinetic energy of about 100 MeV if it is confined in the region of r 1014 m; however, the energies of the electrons emitted in decay are only of the order of 1 or 2 MeV. In addition, there is no evidence for such a strong attractive force between nuclei and electrons as would be implied by a negative potential energy of 50 to 100 MeV inside the nucleus. Furthermore, since the electrostatic potential energy of the electron and nucleus is negative, there is no barrier to be overcome, as there is in decay (Figure 11-1). If the electron\u2019s total energy were positive, as required for decay, the electron should esemitters should cape from the nucleus immediately and most naturally occurring 11-1 The Composition of the Nucleus 479 Figure 11-1 (a) Potential barrier for an particle compared with (b), potential for a negative electron. Because there is no barrier for the electron, it will not be bound at all unless the total energy is negative, in which case it can never escape. The very narrow steep rise to the potential in (a) as represents the \u201chard core\u201d of the nucleus. r S 0 (a) Energy (b) Positive Coulomb barrier E\u03b1 r Ee > 0 r < 0 Ee Nuclear potential well in which \u03b1 particle is confined Negative Coulomb potential between electron and nucleus have long since disappeared. A further difficulty is the observation that the magnetic eU>2mp, about 2000 moments of nuclei are", " of the order of nuclear magnetons, which would be expected if there times smaller than a Bohr magneton were electrons inside the nucleus. eU>2me, N B A further convincing argument against electrons existing in the nucleus concerns and as such angular momentum. Protons and neutrons are fermions with spins of both obey the exclusion principle. The angular momentum of the nitrogen nucleus has a quantum number of 1, which can be inferred from a very small splitting of atomic spectral lines called hyperfine structure (see Section 11-2). It is also known (from molecular spectra\u2014see Section 9-4) that the nitrogen nucleus obeys Bose-Einstein contained 14 protons and 7 electrons, each rather than Fermi-Dirac statistics. If with spin etc., and the nucleus would obey Fermi-Dirac statistics. the resultant angular momentum would have to be 1>2, 3>2, 5>2, 1>2, 1>2 14N In 1920 Rutherford suggested that there might be a neutral particle, possibly a proton and an electron tightly bound together, which he called a neutron. When such a particle was found by Chadwick in 1932, the idea that electrons were permanent constituents of nuclei was abandoned. Instead, the nucleus was assumed to contain N neutrons and Z protons, a total of A N Z particles. N is referred to as the neutron number. The notion of the neutron being a proton and electron bound together has also been abandoned since the spin of the neutron is Thus, the nucleus is composed of protons and neutrons, the nucleons, which collectively occupy a volume whose radius is of the order of 1 to 10 fm. All of the large variety of nuclei with their broad diversity of properties are assembled from various numbers of these two particles. The fundamental properties of the individual nucleons are given in Table 11-1. We should note at this point that the nucleons are not fundamental particles. Each of the two types of nucleons is composed of a set of three quarks, fundamental particles that interact with each other via the strong force, which accounts for the fact that the nucleons also feel that force. Quarks and their interactions will be discussed in Chapter 12. 1>2. Table 11-1 Fundamental properties of atomic constituents Particle Charge Mass (u) Mass (kg) Spin Magnetic moment Proton Neutron Deuteron Electron e 0 e e 1.007276 1.008665 2.013553 5.4858 104 1.6726 1027 1.6749 1027 3.3436 1027 9.1094 1031 1>2 1>2 1 1>2 2.79285 N 1.91304 N 0.85744 N 1.00116 B 480 Chapter 11 Nuclear Physics 11-2 Ground-State Properties of Nuclei Understanding nuclei, like atoms, requires the application of quantum theory. It was the study of nuclear spectra, the energy and particles emitted spontaneously by radioactive nuclei, that provided the first indication of the existence of quantized energy levels, angular momenta, and magnetic moments in nuclei, just as the regularities in atomic spectra had earlier pointed the way to Bohr\u2019s theory and, ultimately, to wave mechanics. Interpreting the nuclear studies presents more complex problems due to the existence of two nucleons, the possible emission of several different particles in addition to photons from excited energy states, and our incomplete knowledge of the nuclear potential function. In this section we will discuss some of the properties of nuclei in the ground state and mention a few methods of determining these properties. In Section 11-3 we will study radioactivity, which provides information about the excited states of nuclei. Several of the general references at the end of this chapter contain good discussions of the experimental methods used in measuring nuclear properties. We will use the following standard terminology: the letter N stands for the number of neutrons in a nucleus and Z for the number of protons (the atomic number); A N Z is the total number of nucleons, the mass number. The mass number is an integer approximately equal to the atomic weight. A particular nuclear species is called a nuclide. Nuclides are denoted by the chemical symbol with a presuperscript 16O 15O. Sometimes Z is giving the value of A, such as or 15 given as a pre-subscript, such as although this is not 8O, necessary because each element (Z number) has a unique chemical symbol. Occasionally, N is also given as a subalthough this, too, is unnecessary script, such as since N A Z. Nuclides with the same Z, such as 15O 16O, are called isotopes. Nuclides with the same N, and 13 are called isotones, while nuclides such as 6C7 are called isobars. with the same A, such as Rutherford formula 15 8O7,", " 14 7N7, 14N, and and 14C 25 20 35 \u03b1-particle energy, MeV 30 40 45 Figure 11-2 Rutherford\u2019s -scattering formula (Equation 4-6) is shown by the dashed line. particles of increasing energy incident on the nuclei of a Pb target scattered as would be expected by the Rutherford formula until their energy reached about 27 MeV. At greater energies the particles approach the Pb nuclei closely enough so that the nucleons of the and the Pb interact via the attractive nuclear force and the scattered intensity falls below that predicted by the Rutherford equation. [Data from R. M. Eisberg and C. E. Porter, Rev. Mod. Phys., 33, 190 (1961).] Size and Shape of Nuclei Nuclear Radii All of the methods for measuring nuclear radii agree that the radii are proportional to the cube root of the mass number. The nuclear radius can be determined by scattering experiments similar to the first ones of Rutherford or in some cases from measurements of radioactivity. Indeed, as we discussed in Section 4-2 and as illustrated in Figure 11-2, Rutherford\u2019s original -particle-scattering experiment furnished the first measurement of the nuclear radius. An interesting, nearly classical method of determining the nuclear radius involves the measurement of the energy of decay between mirror nuclides, which are nuclides whose Z and N numbers are interchanged (Figure 11-3). For example, 15O, with with eight protons and seven neutrons, and 15N, 103 102 10 10 15 11-2 Ground-State Properties of Nuclei 481 eight neutrons and seven protons, are mirror nuclides. If one assumes that the nuclear force between nucleons is independent of the kind of nucleons, the only difference in energy between is electrostatic. The electrostatic energy of a ball of and uniform charge is given by 15N 15O U 3 5 1 4 0 q2 R 11-1 where q is the charge and R is the radius. we will discuss further in a later section, decays to positron and a neutrino. The energy difference between the beta-decay energy, is then is radioactive and, as by emitting a 15O 15N, and 15N 15O \u00a2U 3 5 1 4 0 e2 R cZ2 (Z 1)2 d 11-2 with Z 8. A measurement of the energy of decay, equal to U, thus gives a measurement of R. If one assumes a uniform charge distribution, measurements of the positron-decay energies (see Section 11-4) for 18 pairs of mirror nuclides give for the nuclear radius R R0A1>3 with R0 1.2 0.2 fm 11-3 where A is the atomic mass number. The value of R0 in Equation 11-3 includes the effect of a quantum-mechanical correction using a charge distribution calculated from the nuclear shell model discussed in Section 11-6. The consistency of these results with other methods of determining R is a strong indication that the nuclear part of the potential energy is the same for each pair of mirror nuclei. n p n p 15N 15O Figure 11-3 Mirror nuclides. If all the neutrons are changed to protons and all the protons are 15N changed to neutrons, becomes its mirror, 15O. The ground-state energy of mirror pairs differs only in the electrostatic energy. The most extensive measurements of nuclear radii were carried out by Robert Hofstadter and his co-workers in a series of experiments begun in 1953.6 In these experiments at the Stanford Linear Accelerator (SLAC), nuclei were bombarded with electrons having energies of about 200 to 500 MeV. The wavelength of a 500-MeV electron is about 2.5 fm, which is smaller than the radius of heavy nuclei. It is thus possible to learn something about the detailed structure of the charge distribution of nuclei by analyzing the diffraction pattern that results from the scattering of these electrons. The analysis is fairly complicated because the electrons are relativistic. Figure 11-4 shows the diffrac16O tion pattern of high-energy electrons scattered by nuclei. If we consider the incoming electron and beam to be a plane wave of wavelength the scattering process is similar to the diffraction of light from a circular hole of radius R, discussed in most introductory physics textbooks, where R in this case is the nuclear radius. The first minimum of the diffraction pattern is then given approximately by 12C, sin 0.16>R 11- 16O, 360 MeV 10 \u20131 10 \u20132 16O, 420 MeV 10 \u20133 10 \u20134 12C, 420 MeV 10 \u20135 30 40 50 Scattering angle, degrees 60 70 Figure 11-4 Diffraction pattern of high-energy electrons scattered by The angle at which the minimum occurs in each pattern", " is given by Equation 11-4. and 12C. 16O 482 Chapter 11 Nuclear Physics Figure 11-5 (a) Charge density versus distance for several nuclei as determined by high-energy electron scattering experiments. (b) Definitions of parameters R0 and t used to describe nuclear charge density. The skin thickness t is measured from 10 percent to 90 percent of the central core density. [From R. Hofstadter, Annual Review of Nuclear Science, 7, 231 (1957).] (a.50 2.25 2.00 1.75 1.50 1.25 1.00 0.75 0.50 0.25 0 0 (b) He C Mg Co He C Mg Ca Sr In Au Bi He \u00d7 1 1 \u2013\u2013\u2013\u2013\u2013 10 2 3 1.00 0.90 0.50 0.10 4 5 Radial distance r, fm 6 7 8 9 10 Surface thickness t R0 ).0 2.0 3.0 4.0 5.0 r, fm 6.0 7.0 8.0 9.0 10.0 Example 11-2 shows how the nuclear radius can be calculated from Equation 11-4 with the aid of Figure 11-4. Figure 11-5a shows some charge distributions obtained from detailed analysis of these experiments. The mean electromagnetic radius R and the surface thickness t, defined in Figure 11-5b, are given by R (1.07 0.02)A1>3 fm t 2.4 0.3 fm These results are consistent with those obtained from the nuclides. 11-5 -decay studies of mirror 11-2 Ground-State Properties of Nuclei 483 4He and 238U Use Equation 11-3 to compute the EXAMPLE 11-1 Nuclear Radii of radii of 238U. and 4He SOLUTION For 4He: For 238U: RHe 1.2(4)1>3 1.90 fm RU 1.2(238)1>3 7.42 fm Thus, the nuclear radius varies only by a factor of about 4 from the lightest nuclides to the heaviest. EXAMPLE 11-2 Nuclear Radius of tered from 16O in Figure 11-4, compute a value for the radius of the 16O nucleus. 16O Using the data for 420-MeV electrons scat- SOLUTION 1. The radius R of the 16O nucleus is computed from Equation 11-4: 2. The angle in Equation 11-4 is the first minimum of the diffraction pattern. From Figure 11-4 we see that the first minimum occurs at about: sin 0.61 R or R 0.61 sin 44\u00b0 3. The de Broglie wavelength of the electrons is h p 4. The momentum p of the 420-MeV electrons is computed from the relativistic expression, Equation 2-32: p2c2 E2 (mc2)2 (420)2 (0.511)2 (420 MeV)2 or p 420 MeV>c 2.24 1019 kg # m>s 6.63 1034 J # s 2.24 1019 kg # m>s 2.96 1015 m 2.96 fm R (0.61)(2.96 fm) sin 44\u00b0 2.60 fm 5. Substituting this value in from step 3 gives 6. The radius R is computed by substituting the values for into Equation 11-4: and Remarks: This result agrees well with the values of R0 for the low-Z nuclei in Figure 11-5a. 484 Chapter 11 Nuclear Physics A different kind of measurement of the nuclear radius can be made using the attenuation of a beam of fast neutrons as it moves through a sample. The total cross section for attenuation can be shown to be 2 aR 2 b 2 11-6 where R is the nuclear radius and is the de Broglie wavelength of the neutron. The >2 R in order to gain information about R neutrons must be fast enough so that from measurement of These experiments do not measure the charge distribution but instead measure the \u201cradius\u201d of the nuclear force between a neutron and the nucleus. The results of these measurements are. R R0A1>3 with R0 1.4 fm 11-7 A1>3 These different types of experiments thus give comparable but not identical results, depending on whether the particular experiment measures the nuclear force radius (neutrons) or the nuclear charge radius (electrons). The fact that the radius is proporimplies that the volume of the nucleus is proportional to A. Since the tional to mass of the nucleus is also approximately proportional to A, the densities of all nuclei are approximately the same. A drop of liquid also has a constant density independent of its size, and this fact has led to a", " model in which the nucleus is viewed as analogous to a liquid drop. This model has been helpful in computing nuclear masses and in understanding certain types of nuclear behavior, particularly the fission of heavy elements. The numerical value of the nuclear density is about 1017 kg m3. This fantas> tically high density, compared with 103 kg m3 for atoms, is a consequence of the fact that nearly all the mass of the atom is concentrated in a region whose radius is only about 105 that of the atom. A cubic millimeter of nuclear matter has a mass of about 200,000 metric tons, or about the same mass as a supertanker filled with petroleum! > EXAMPLE 11-3 Radius of a Neutron Star In certain supernova events, the outer layer of the star is blown away, leaving a core consisting entirely of neutrons. This stellar remnant is called a neutron star, and its density is approximately the same as that of atomic nuclei. Compute the radius of a neutron star whose mass is equal to that of the Sun, 1.99 1030 kg. SOLUTION The mass of the neutron star is is approximately 1017 kg m3. Assuming the neutron star to be a sphere, we have that M 1.99 1030 kg V (1017 kg>m3)(4R3>3) where V is the volume and the density M V, > where R is the radius of the star in meters. Solving for R3 yields R3 (3)(1.99 1030 kg) (4)(1017 kg>m3) 4.75 1012 m3 and taking the cube root R 1.68 104 m 16.8 km Remarks: By way of comparison, the mean diameter of the Sun is 1.39 106 km. 11-2 Ground-State Properties of Nuclei 485 (a) z z x (b) x y y Figure 11-6 Nonspherical nuclear shapes. Nuclei with positive quadrupole moments have (z2)av greater than (x2)av or (y2)av and are of watermelon shape, as in (a). Nuclei with negative quadrupole moments have (z2)av less than (x2)av or (y2)av and are shaped like flattened pumpkins, as in (b). Nuclear Shape With a few exceptions, nuclei are nearly spherical. Most of the exceptions occur in the rare earth elements (the transition region in the periodic table, Z 57 to Z 71), in which the shape is ellipsoidal, with the major axis differing from the minor axis by about 20 percent or less. In these heavy nuclides, the inner atomic electron wave functions penetrate the nucleus, and deviations from spherical shape, which correspond to deviations in the nuclear charge distributions, show up as small changes in the atomic energy levels. In direct analogy with the fact that the potential at points outside a static distribution of charges is determined by the dimensions of the distribution7 and, conversely, that measuring the potential yields information about the distribution, measuring these small changes in the atomic energy levels yields information about the nuclear charge distribution, even though it can\u2019t be measured directly. If the nucleus is shaped like a watermelon (Figure 11-6a), with the extent of the distribution larger along the z axis than along the x and y axes, the average value of z2 is larger than the average values of x2 and y2. In this case the electric quadrupole moment Q, which is proportional to is positive. This is the most common case for nonspherical nuclei. Nuclei with negative quadrupole moments are shaped more like flattened pumpkins, with the two equal axes longer than the third axis, as in Figure 11-6b. The average value of the electric quadrupole moment is given by (x2 y2 z2)av, 3(z2)av (x2 y2 z2)av Dc dV 8Q9 Zc*C3(z2)av 0 0 0 for z2 x2, y2 for z2 x2 y2 for z2 x2, y2 (Figure 11-6a) (spherical) (Figure 11-6b) 11-8 Figure 11-7 shows the measured values of the electric quadrupole moment for the odd A nuclei, i.e., those for which either Z or N is odd. Equation 11-8 is evaluated for wave functions corresponding to the nuclear charge distributions of various theoretical models of the nucleus and compared with the values in Figure 11-7. As you might imagine, the calculations are formidable! ) 2 R Z Q ( / 167Er 176Lu 50 82 126 0.30 0.20 0.10 0 \u20130.10 8 28 20 123Sn 0 40 80 Number of odd nucleons 120 Figure 11-7 The electric quadrupole moment Q divided by Z and R2,", " where R is the average nuclear radius, is plotted versus the number of nucleons of the odd type (Z or N). The arrows indicate the points where Q ZR2 0, corresponding to spherical shape. > 486 Chapter 11 Nuclear Physics Nuclear Stability Among the more than 3000 known nuclides, there are only 266 whose ground states are stable. All of the rest have unstable ground states, which eventually undergo radioactive decay, that is, transition to some lower-energy state of a different element. Figure 11-8 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 Line of stability N = Z 0 0 10 20 30 40 50 60 70 80 90 100 110 Proton number Z Figure 11-8 Plot of neutron number N versus proton number Z for the known nuclides. The 266 stable nuclides are indicated by the black dots. The area between the irregular colored lines represents the known unstable, or radioactive, nuclides whose lifetimes are longer than about a millisecond. The curved line through the stable nuclides is called the line of stability. 11-2 Ground-State Properties of Nuclei 487 (a) n 4 En 16E1 (b) n 4 En 16E1 3 2 1 n levels 9E1 4E1 E1 p levels 3 2 1 n levels 9E1 4E1 E1 p levels Figure 11-9 (a) Seven neutrons in an infinite square well. In accordance with the exclusion principle, only two neutrons can be in a given space state. The total energy is (2 9E1) (2 4E1) (2 1E1) 44E1. (b) Four neutrons and three protons in 16E1 the same infinite square well. Because protons and neutrons are not identical, four particles (two neutrons and two protons) can be in the state n 1. The total energy is (3 4E1) (4 1E1) 16E1. This is much less than in (a). The integers on the left of each well are infinite square well principal quantum numbers. shows a plot of the neutron number N versus the proton number Z for the stable nuclides and the known unstable ones whose lifetimes are longer than about a millisecond. The straight line is N Z. The general shape of the line of stability, shown by the light curve tracing through the stable nuclides in Figure 11-8, can be understood in terms of the exclusion principle and the electrostatic energy of the protons. Consider the kinetic energy of A particles in a one-dimensional square well, which is an adequate model for demonstrating this point. The energy is smallest if A 2 are neutrons and A 2 are protons and greatest if all the particles are of one type (Figure 11-9). Hence there is a tendency, due to the exclusion principle, for N and Z to be equal. If we include the electrostatic energy of repulsion of the protons, the result is changed somewhat. This potential energy is proportional to Z 2. At large A, the energy is increased less by adding two neutrons than by adding one neutron and one proton, and so the difference N Z increases with increasing Z. > > There is also a tendency for nucleons to pair with other identical nucleons. Of the 266 nuclides whose ground states are stable, 159 have even Z and even N, 50 have odd Z and even N, 53 have even Z and odd N, and only 4 have both odd N and Z. (See Table 11-2.) Table 11-2 N versus Z for stable isotopes Z Even Odd 159 53 50 4 N Even Odd 488 Chapter 11 Nuclear Physics Since there are about 100 different elements and about 270 stable nuclides, there is an average of about 2.7 stable isotopes per element. There is a larger than the average number of stable isotopes for nuclei with Z equal to 20, 28, 50, and 82. For example, tin, with Z 50, has 10 stable isotopes. Similarly, nuclides with these same numbers of neutrons have a larger than average number of isotones. These numbers, called magic numbers, are a manifestation of shell structure in very much the same way that the atomic \u201cmagic numbers\u201d 2, 10, 18, and 36 correspond to closed-electronshell structure. As we will discuss further in Section 11-6, the nuclear magic numbers, which also include 2, 8, and 126, represent configurations of particular stability. An island of stability is hypothesized to exist around Z 126. In the search for it thus far, a few atoms with atomic numbers up to 118 have been created in that region. (Atoms with atomic number 117 have not yet been synthesized.) Nuclides that fall between the irregular colored lines in Figure 11-8, except those marked by the black dots, are radioactive. We will discuss radioactivity in", " Section 11-3. Masses and Binding Energies > The mass of an atom can be accurately measured in a mass spectrometer, which measures q M for ions by bending them in a magnetic field.8 The mass of an atom is slightly smaller than the mass of the nucleus plus the mass of the electrons because of the binding energy of the electrons. The binding energy of the electrons is defined by Batomic MNc2 Zmec2 MAc2 \u00a2mc2 11-9 where MN is the mass of the nucleus, MA is the mass of the atom, me is the mass of an electron, and m is the mass equivalent of Batomic. (See Section 2-3.) Because the binding energies of atoms are only of the order of keV, compared with nuclear binding energies of many MeV, atomic binding energies are usually neglected in nuclear physics. The binding energy of a nucleus with Z protons and N neutrons is defined as Bnuclear Zmpc2 Nmnc2 MAc2 11-10 where mp is the mass of a proton, mn the mass of a neutron, and MA the mass of the nucleus of mass number A. Since the mass of an atom is very nearly equal to the mass of the nucleus plus the mass of the electrons (neglecting the atomic binding energy), the nuclear binding energy can be accurately computed from Bnuclear ZMHc2 Nmnc2 MAc2 11-11 where MA is the atomic mass and MH is the mass of a hydrogen atom. Note that the masses of the Z electrons cancel out. This expression is more convenient to use because it is the mass of the atom that is usually measured in mass spectrometers. The atomic masses of all stable nuclides and of many unstable ones are listed in Appendix A. Once the mass of a nucleus or atom is determined, the binding energy can be computed from Equation 11-10 or 11-11. The binding energy per nucleon B A is plotted against A for the most stable isotope of each element in Figure 11-10. The mean value is about 8.3 MeV nucleon. The fact that this curve is approximately constant (for A 16) indicates that the nuclear force is a saturated force. This is partially explained by the short range of the nuclear force (see Section 11-5). If each nucleon interacted with every other nucleon, there would be A 1 interactions for each nucleon, and the binding energy per nucleon would be proportional to A 1 rather than constant. > > 20Ne 16O 12C 8 Ca Fe Zn Kr 24Mg Mo Te 11-2 Ground-State Properties of Nuclei 489 Sm Lu Hg Ra Figure 11-10 The binding energy per nucleon versus atomic mass number A. The solid curve represents the Weizs\u00e4cher semiempirical binding-energy formula, Equation 11-12. 11B 7Li 6Li 3He 2H 7 4He 6 V e M A 50 100 150 200 A Figure 11-10 indicates that instead there is a fixed number of interactions per nucleon, as would be the case if each nucleon were attracted only to its nearest neighbors. Such a situation also leads to a constant nuclear density, consistent with the radius measurements. If the binding energy per nucleon were instead proportional to the number of nucleons, then the nuclear radius would be approximately constant, as is the case for the atomic radii. More Of the several models of the nucleus physicists have developed, the liquid-drop model has been one of the most useful. It has been successful in describing the fission process and nuclear reactions and, in particular, predicting the binding energies (i.e., masses) of isotopes and individual nucleons within the nucleus. These topics are discussed in Liquid-Drop Model and the Semiempirical Mass Formula on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Equations 11-12 through 11-14 and Table 11-3 here, as well as Examples 11-4 through 11-6. Nuclear Angular Momenta and Magnetic Moments 1>2, The spin quantum number of both the neutron and the proton is which means that the nucleons are fermions. The angular momentum of the nucleus is a combination of the spin angular momenta of the nucleons plus any orbital angular momentum due to the motion of the nucleons. This resultant angular momentum is usually called nuclear spin and designated by the symbol I. The nucleons individually have magnetic moments, which also combine to produce the nuclear magnetic moment. Evidence for nuclear spin and magnetic moment was first found in atomic spectra. The nuclear spin adds to the angular momentum J L S of the electrons to form a total angular momentum F: F I J 11-15 490 Chapter 11 Nuclear Physics F 0 S F 0. (I J), (I J 1), \u00c1, \u0192I J\u0192, The possible quantum numbers for F", " are according to the usual rule for combining angular momenta. F obeys the selection rule F 1, 0, The number of possible values of F is (2J 1) or (2I 1), but no whichever is smaller. Because of the energy of the interaction between the electronic magnetic moment and the nuclear magnetic moment associated with I, each atomic spectral line is split into (2J 1) or (2I 1) components. This splitting is one of several effects that are the result of interactions of the nuclear spins and moments with the environment of the nucleus, including its own atomic electrons, collectively called hyperfine structure.10 The hyperfine splitting of the spectral lines associated with the nuclear magnetic moment occurs for a reason that is exactly analogous to the spin-orbit coupling discussed in Section 7-5 that is the origin of the fine structure of the atomic spectral lines. The coupling between I and J expressed by Equation 11-15 results in a splitting of the atomic energy levels by an amount E, in addition to the spin-orbit splitting of Equation 7-68, given by the analogous relation \u00a2E gNmI NBe 11-16 N eU>2mp where gN is the nuclear Land\u00e9 factor, mI is the magnetic quantum number of the z comis the nuclear magneton, and Be is the magnetic field at the ponent of I, nucleus produced by the electrons (Table 11-4). The product is the nuclear N, the quantities on the right side of Equation 11-16 are magnetic moment. Except for all of the same order of magnitude as the corresponding ones in Equation 7-68; however, the ratio Thus, the hyperfine splitting for a given atom is very small, about 103 times the fine-structure splitting. Using as an example the sodium shown in Figure 7-30, which produce the yellow D doublet levels lines, Figure 11-11 illustrates the hyperfine splitting of these levels resulting from I J coupling. It can be observed only with extremely high resolution. The use of tunable dye lasers and atomic beam fluorescence spectroscopy has made high precision measurements of these extremely small energy splittings possible in recent years. 2P1>2, 2P3>2, and 2S1>2 103. gNmI > B N N I J, For the case there are (2I 1) different F states; thus the nuclear spin can be determined by counting the number of lines in the hyperfine splitting. The spin of all even-even nuclides (those with even Z and even N) is zero in the ground state. Table 11-4 Magnetic field Be at the nucleus due to electron for selected alkali elements Element H Li Na K Rb Cs n 1 2 3 4 5 6 Be, 2S1>2(T) Be, 2P1>2(T) Be, 2P3>2(T) 17 13 44 63 130 210 \u2014 \u2014 4.2 7.9 16 28 \u2014 \u2014 2.5 4.6 8.6 13 Source: Data are from E. Segr\u00e8, Nuclei and Particles, 2d ed. (Menlo Park, CA: Benjamin/Cummings 1977), p. 259. 11-2 Ground-State Properties of Nuclei 491 2P3>2 2P1>2 Figure 11-11 (a) Transitions between the sodium doublet levels produce the yellow D S 2S1>2 being D1 lines, S 2S1>2 being D2. and Coupling between the atomic angular momentum J I 3>2 and the nuclear spin results in the hyperfine splitting, each level having total angular momentum F I J. Note that the hyperfine splitting of each of the doublet levels is about 103 times that of the fine-structure splitting of the rule the D2 line being split into six components. The D1 line is correspondingly split into four components (not shown). 2P \u00a2F 1, 0 level. (b) The selection leads to (a) 0.5 \u00d7 10\u20136 eV (b) 2 \u00d7 10 \u20133 eV 2.1 eV 2P3/2 2P1/2 2S1/.8 \u00d7 10\u20136 eV 7.3 \u00d7 10\u20136 eV Evidently the nucleons couple together in such a way that their angular momenta add to zero in pairs, as is often the case for electrons in atoms. There is no such simple rule for other nuclides with either odd N or odd Z or both. Some of the successes of the shell model to be discussed in Section 11-6 are the correct prediction of nuclear spins for many nuclei. N gNmI eU>2mp, The magnetic moment of the nucleus is of the order of the nuclear since the magnitude of gN is typically between 1 and 5 magneton, N \u0192m1 and the maximum value", " of The exact value is difficult to predict because it depends on the detailed motion of the nucleons. If the proton and neutron obeyed the Dirac relativistic wave equation, as does the electron, the magnetic moment due to spin would be 1 nuclear magneton for the proton because its charge is e and 0 for the neutron because it has no charge. 11 The experimentally determined moments of the nucleons are \u0192 I. ( ( p)z n)z 2.79285 N 1.91304 N As we will see in Chapter 12, the proton and neutron are more complex particles than the electron. It is interesting that the deviations of these moments from those predicted by the Dirac equation are about the same magnitude, 1.91 for the neutron and 1.79 for the proton. The reason that the magnetic moments of the nucleons have these partic- ular values is not yet completely understood; the current theoretical predictions of p agree with high-precision, experimentally measured values only to within and about 1 percent. n The degeneracy of the hyperfine levels in nuclei with nonzero spins, e.g., the 1H, proton in is removed by an external B field, a nuclear analog of the Zeeman effect. Transitions between these levels, separated (in by 2 precession rate. Detection of the resulting absorption or emission of radiation allows \u201cmapping\u201d of the hydrogencontaining soft tissue, the basis for medical magnetic resonance imaging (MRI). oscillate at the spin pB, 1H) 492 Chapter 11 Nuclear Physics EXAMPLE 11-7 Nuclear Spin of Thallium-205 High-resolution spectroscopic study S of the spectrum of 2S1>2 (377.7 nm), 2P3>2 consists of three hyperfine components. This requires that there be two hyperfine levels for each J. Determine the spin of the 205Tl S 2S1>2 (535.2 nm) reveals that each component of the doublet nucleus. 2P1>2 205Tl SOLUTION I J, then there are (2I 1) different F levels, and if I J, there are (2J 1) If different F levels. Since the hyperfine spectrum indicates that there are two levels for each J, then for the level either 2P3>2 2I 1 2 or 2J 1 2 But we already know that J (2I 1) 2 and the spin of the (3>2), so (2J 1) cannot equal 2; therefore 205Tl nucleus (in its ground state) must be 1>2. Note that for the and since in these two cases I J. 2P1>2 2S1>2 levels both of the equations above are satisfied Questions 1. Why is N approximately equal to Z for stable nuclei? Why is N greater than Z for heavy nuclei? 2. Why are there no stable isotopes with Z 83? 3. The mass of 12C, which contains 6 protons and 6 neutrons, is exactly 12.000 u by the definition of the unified mass unit. Why isn\u2019t the mass of contains 8 protons and 8 neutrons, exactly 16.000 u? 16O, which 11-3 Radioactivity Of the more than 3000 nuclides known, only 266 are stable. All of the rest are radioactive; that is, they decay into other nuclides by emitting radiation. The term radiation here refers to particles as well as electromagnetic radiation. In 1900 Rutherford discovered that the rate of emission of radiation from a substance was not constant but decreased exponentially with time. This exponential time dependence is characteristic of all radioactivity and indicates that it is a statistical process. Because each nucleus is well shielded from others by the atomic electrons, pressure and temperature changes have no effect on nuclear properties.12 For a statistical decay (in which the decay of any individual nucleus is a random event), the number of nuclei decaying in a time interval dt is proportional to dt and to the number of nuclei present. If N(t) is the number of radioactive nuclei at time t and dN is the number that decay in dt (the minus sign is necessary because N decreases), we have dN N dt 11-17 where the constant of proportionality, is the probability per unit time of the decay of any given nucleus. The solution of this equation is is called the decay constant., N(t) N0et 11-18 where N0 is the number of nuclei present at time t 0. The decay rate is N0 11-3 Radioactivity 493 R dN dt N0et R0et 11-19 Note that both the number of nuclei and the rate of decay decrease exponentially with time. It is the decrease in the rate of decay that is determined experimentally. Figure 11-12 shows N versus t. If we multiply the numbers on the N", " axis by this becomes a graph of R versus t., We can calculate the mean lifetime from Equation 11-18. The number of nuclei N dt; with lifetimes between t and t dt is the number that decay in dt, which is thus the fraction with lifetimes in dt is f(t) dt N dt N0 et dt When we use this distribution function, the mean lifetime is tf(t) dt 0 0 tet dt 1 11-20 11-21 ) t ( N N0 1 \u2013\u2013 2 e\u20131 N0 t1/2 \u03c4 t Figure 11-12 Exponential radioactive decay law. The number of nuclei remaining at time t decreases exponentially with time t. The half-life mean life indicated. The decay rate R(t) N(t) has the same time dependence. t1>2 1> and the are which is the reciprocal of the decay constant The half-life is defined as the time after which the number of radioactive nuclei has decreased to half its original value. From Equation 11-18, t1>2. 1 2 N0 N0et1>2 or et1>2 2 t1>2 ln 2 (ln 2) 0.693 0.693 11-22 After each time interval of one half-life, the number of nuclei left in a given sample and the decay rate have both decreased to half of their previous values. For example, (1>2) if the decay rate is R0 initially, it will be R0 after two half-lives, and so forth. During one mean lifetime, the number of nuclei remaining in the sample and the decay rate have decreased to 1 e of their previous values. Thus, if the initial decay rate is R0, it will be has elapsed, (1>e)(1>e) 2, and so on. The SI unit of radioactivity is the becquerel (Bq), which is defined as one decay per second: R0 after one half-life, R0 after time R0 after time (1>2)(1>2) (1>e) > 1 Bq 1 decay>s 11-23 A historical unit of activity, the curie (Ci), is also frequently used. The curie is defined as 1 Ci 3.7 1010 decay>s 3.7 1010 Bq 11-24 The curie is the disintegration rate of 1 g of radium. Since this is a very large unit, the millicurie (mCi), microcurie ( Ci), and picocurie (pCi) are also often used. 494 Chapter 11 Nuclear Physics EXAMPLE 11-8 Counting Rate of a Radioactive Sample A radioactive source has a half-life of 1 minute. At time t 0 it is placed near a detector and the counting rate (the number of decay particles detected per unit time) is observed to be 2000 counts s. (a) Find the mean life and the decay constant. (b) Find the counting rate at times t 1 min, 2 min, 3 min, and 10 min. > SOLUTION 1. For question (a), the mean is related to the halfby Equation 11-22: life life t1>2 (ln 2) t1>2 or t1>2 ln 2 1 min 0.693 1.44 min 86.6 s 2. From Equation 11-21, the decay constant is given by 1 1 86.6 s 1.16 102 s1 3. Method 1. For question R 2000 et 2000 e(1.1610 2)t 2000 e(ln 2)t>t1>2 2000 e(0.693)t where t is now expressed in minutes (b), the counting rate is proportional to the decay rate R in Equation 11-19. The counting rate at t 0 has the same proportionality to R0, so we can write the counting rate as R, substitut- ing values for and for t1>2: 4. The counting rate R can now be computed for each of the times t 1 min, 2 min, 3 min, and 10 min as follows: R(1 min) 2000 e(0.693)(1) 2000 0.50 1000 counts>s R(2 min) 2000 e(0.693)(2) 500 counts>s and similarly R(3 min) 250 counts>s R(10 min) 1.95 counts>s 5. Method 2. Since the halflife is 1 min, the counting rate at t 1 min will be half that at t 0; at t 2 min it will be half of that at t 1 min, and so on. In general, at t n min the count rate will be R (1>2)nR0 and again R(1 min) (1>2)", "12000 1000 counts>s and R(10 min) (1>2)102000 (0.0010)(2000) 1.95 counts>s 11-4 Alpha, Beta, and Gamma Decay 495 More Very often the decay of a radioactive nucleus results in a new nucleus that is also radioactive and that, in general, has a different decay constant. In some cases such sequential decays may result in a dozen or more different radioactive isotopes. Production and Sequential Decays, on the home page (www.whfreeman.com/tiplermodernphysics5e), describes the way to calculate the total activity and the net rate at which new isotopes are produced. See Equations 11-25 through 11-29 and Figures 11-13 and 11-14 here, as well as Examples 11-9 and 11-10 and Questions 4 and 5. (a) 11-4 Alpha, Beta, and Gamma Decay From the time when Becquerel\u2019s discovery of radioactivity gave the first hint of the existence of the nucleus, much of what physicists have learned about nuclear structure has resulted from studies of radioactive nuclides, that is, by studying the transitions of nuclei from one quantum state to another of lower energy. As noted earlier, of the approximately 3000 known isotopes, all but 266 are radioactive. In addition, nearly all of a theoretically estimated 2000 more possible isotopes that have yet to be created are radioactive. The radioisotopes decay by one or another of at least nine different modes; however, most decays occur via one or sometimes two of the most common modes: alpha, beta, and gamma. Others occur by more unusual routes, such as emission of a proton or neutron or spontaneous fission. A few may decay by modes that are exceedingly rare, such as a double beta decay, which is the focus of considerable current theoretical and experimental interest. The fundamental purpose of the studies of radioactive nuclei is to obtain information about nuclear structure, the nature of the strong nuclear force, and the properties of elementary particles. Proton dripline 150 100 (b) 82 50 50 28 20 8 2 In the subsections that follow we will discuss the three most common types of decay in some detail, touching on certain of the others when pertinent. In these discussions it will be helpful to keep two points in mind. The first of these is that the line of stability in Figure 11-8 is the floor of an energy valley formed by plotting the binding energy for each isotope on an energy scale perpendicular to the N and Z axes, as illustrated in Figure 11-15a. In Figure 11-15a the energy is artificially truncated; however, there are theoretical limitations placed on the numbers of protons and neutrons that can be assembled into a nucleus, even a highly unstable one. These limits, given the whimsical name driplines, are shown in Figure 11-15b and define the N-Z boundaries within which lie the 5000 or so isotopes that may, in principle, exist. The limits are set by the energies at which the nuclei will spontaneously emit a proton or neutron. 2 8 20 28 50 82 Neutron number 126 Figure 11-15 (a) The graph of Z versus N with the nuclear binding energy B (in MeV) plotted upward. The surface thus formed is truncated at 100 MeV to make the energy valley more clearly visible. (b) More than 5000 theoretically predicted nuclei lie between the proton and neutron driplines. Only about 3000 (those between the inner irregular colored lines) are found in nature or have been created in the laboratory, and only 266 of those are stable (black dots). The edges of the truncation in (a) are analogous to artificial driplines. B 100 100 50 Z Neutron dripline 496 Chapter 11 Nuclear Physics The second point to bear in mind is that radioactive-decay processes conform to the same conservation laws that are obeyed by all physical processes. In particular, (1) relativistic mass-energy, (2) electric charge, (3) linear momentum, (4) angular momentum, (5) nucleon number, and (6) lepton number13 are all conserved quantities. The first four of these are already familiar to you from your previous study of physics. The last two relate specifically to the interactions and decays of fundamental particles and will be discussed in Chapter 12. As we discuss the three most common modes of radioactive decay, consequences of each of the conservation laws will be illustrated. Alpha Decay are 241Am The particles emitted by a tiny amount of used to ionize the air inside smoke detectors. When smoke is present, the ionized air molecules stick to the smoke particles, reducing a trickle current maintained in the ionized air, thereby triggering an alarm. Ionization-type smoke detectors are considerably more sensitive than those using photoelectric sensors. 4.5 109 In order for a radioactive substance to be found in nature,", " either it must have a halflife that is not much shorter than the age of Earth (about years) or it must be continually produced by the decay of another radioactive substance or by a nuclear reaction. For a nucleus to be radioactive at all, its mass must be greater than the sum of the masses of the decay products. Many heavy nuclei are unstable to decay. Because the Coulomb barrier inhibits the decay process (the particle must \u201ctunnel\u201d through a region in which its energy is less than the potential energy, as shown in Figure 11-1a), the half-life for decay can be very long if the decay energy is small, that is, if the width of the barrier to be tunneled through is large. Indeed, the relation between the half-life of an emitter and the energy of the particle is so striking that it was first noticed by two research assistants in Rutherford\u2019s laboratory, Geiger and Nuttall, in 1911, the same year that Rutherford discovered the nucleus. The general relation, called the Geiger-Nuttall rule, is illustrated in Figure 11-16 and given by Equation 11-30. log t1>2 AE1>2 B 11-30 is the kinetic energy of the emitted particle and A and B are experimen- E where tally determined constants 1020 1015 1010 105 100 10 \u20135 10 \u201310 4 5 109 y 106 y 103 ms 1 \u03bcs 1 ns 9 10 6 7 \u03b1 kinetic energy, MeV 8 Figure 11-16 The Geiger-Nuttall relation is illustrated by the semilogarithmic graph of the decay half-life versus the kinetic energy of the emitted particle for the naturally occurring emitters. The broken line represents the empirical Geiger-Nuttall rule given by Equation 11-30. 11-4 Alpha, Beta, and Gamma Decay 497 V(r) E\u03b11 E\u03b12 R r E1 Figure 11-17 Schematic representations of the wave functions of two particles with energies and E2 within the nuclear potential well. 1 The probability of penetrating the 2 barrier is larger than that for since the barrier is narrower at Thus, the amplitude of outside the nucleus than that of (1) (2) Hence, t1>2(1) t1>2(2). E1. is larger \u2019(2). and therefore \u2019(1) E Subsequently, an expression for the half-life of an emitter was derived from the Schr\u00f6dinger equation treating decay as a barrier penetration phenomenon. Its good agreement with experimental results was one of the earliest successful applications of wave mechanics. Briefly, the derivation considered an particle confined within the nucleus with energy as was shown in Figure 11-1. The wave functions for two such particles are illustrated in Figure 11-17. The potential for r R is taken to be the > Coulomb function V(r) zZe2 4 particle, with a smooth transition to the nuclear potential. The probability that the particle will penetrate the barrier on any one approach is the transmission coefficient T that was derived in Section 6-6, Equations 6-75 and 6-76. The decay constant is then given by the product of the transmission coefficient T and the frequency with which the nuclear particle approaches the barrier. The latter, given by Equation 6-78, determined by its kinetic energy for r R in particle\u2019s speed depends on the Figure 11-17 and the value of the nuclear radius. Thus, 0r where z 2 for the 1> 0.693>t1>2 v, Tv 2R 11-31 The result of the wave mechanical derivation, done by B. Taagepera and M. Nurmia, is log t1>2 1.61(ZE1>2 Z2>3) 28.9 11-32 t1>2 where the dependence of radii that is independent of the methods mentioned in Section 11-2. E is in MeV, and Z refers to the daughter nucleus. Notice that on the nuclear radius provides a method of measuring nuclear is in years, t1>2 Alpha-Decay Chains All very heavy nuclei (Z 83) are theoretically unstable to decay because the mass of the parent radioactive nucleus is greater than the sum of the masses of the decay products\u2014the daughter nucleus and an particle. When a nucleus emits an particle, both N and Z decrease by 2, and A decreases by 4. There are -decay chains or sequences, depending on whether A equals 4n, four possible (4n 1), (4n 2), or (4n 3), where n is an integer. For the longest-lived nucleus in each sequence, n 58 for the first and fourth and n 59 for the second and third. All but one of these series are found in nature. The (4n 1) series is not, because its has a half-life longest-lived member (other than the", " stable end product 237Np of only present when Earth was formed has long since decayed away. years, which is much shorter than the age of Earth; hence 209Bi), 237Np, 2 106 498 Chapter 11 Nuclear Physics Thorium series (4n) 232Th \u03b1 228Ra 228Ac \u03b2\u2013 228Th \u03b2\u2013 \u03b1 224Ra \u03b1 220Rn 140 135 N \u03b1 216Po \u03b1 \u03b2\u2013 \u03b1 212Bi \u03b2\u2013 212Po \u03b1 130 212Pb 208Tl \u03b2\u2013 208Pb 125 80 85 Z 90 Figure 11-18 The thorium (4n) -decay series. The broken line is the line of stability (floor of the energy valley) shown in Figures 11-8 and 11-15. 232Th Figure 11-18 illustrates the thorium series, which has A 4n and begins with an decay from to 228Ra. Decreasing n successively by 1 generates A for possible daughter nuclides until a stable one is reached. The daughter nuclide of an decay is on the left, or neutron-rich, side of the stability curve (dashed line), so it often decays by decay, in which one neutron changes to a proton by emitting an electron. In 228Ac, decays by Figure 11-18 228Th. which in turn decays to 212Pb, There is a which decays to 212Bi, which decays either by decay branch point at The branches meet at to The (4n 2) series the stable lead isotope The (4n 3) se238U begins with 235U Figure 11-19 ries starts with and ends with shows a typical -decay spectrum. 212Po. 208Pb. and ends with decay to There are then four decays to decay to 206Pb. 209Pb. 212Bi. or by 228Ra 208Tl 232Th More The energy spectrum of the alpha particles emitted by a heavy nucleus such as shows a number of sharp peaks with energies less than the decay energy Q. The highest energy of these corresponds to the transition from the parent\u2019s ground state to that of the daughter. The others are the result of alpha transitions to excited states of the daughter. In Energetics of Alpha Decay, on the home page (www.whfreeman.com/tiplermodernphysics5e), we describe how they can be used to construct the excited levels of the daughter nucleus. See also Figures 11-19 and 11-20, Equations 11-33 through 11-36, and Example 11-11 here. 223Ra The highest-energy Figure 11-19 Alphaparticle spectrum from 227Th. particles correspond to decay to the ground state of with a transition energy of Q 6.04 MeV. The next highest energy particles, 30, transitions to the first 223Ra, excited state of 30 keV above the ground state. The energy levels of the daughter nucleus, 223Ra, by measurement of the can be determined -particle energies. result from 1600 1200 800 400 334 \u03b1 330 \u03b1 376 \u03b1 \u03b1 342 350 \u03b1 286 \u03b1 61 \u03b1 0 \u03b1 247 \u03b1 280 \u03b1 235 \u03b1 174 \u03b1 124 \u03b1 80 \u03b1 30 5.70 5.80 5.90 6.00 Energy, MeV 11-4 Alpha, Beta, and Gamma Decay 499 223Ra 227Th, determined by Figure 11-20 Energy levels of measurement of -particle energies from as shown in Figure 11-19. Only the lowest-lying levels and some of the -ray transitions are shown. 6.04 MeV 227Th \u03b1 286 \u03b1 280 \u03b1 247 \u03b1 235 \u03b1 174 \u03b1 124 \u03b1 80 \u03b1 61 \u03b1 30 \u03b1 0 286 keV 280 keV 247 keV 235 keV 174 keV 124 keV 80 keV 61 keV 30 keV 0 \u03b3 \u03b3 \u03b3 \u03b3 \u03b3 223Ra Beta Decay There are three radioactive-decay processes in which the mass number A remains unchanged while Z and N change by 1. These are decay, in which a neutron inside a nucleus changes into a proton with the emission of an electron; decay, in which a proton inside a nucleus changes into a neutron with the emission of a positron; and electron capture (EC), in which a proton in a nucleus changes to a neutron by capturing an atomic electron, usually a 1s electron from the K shell since these have the highest probability density in the vicinity of the nucleus. Those nuclei on the neutronemission, while rich side of the energy valley in Figure 11-15 will tend to decay by those on the proton-rich side will most probably decay by emission or electron capture. We will discuss each of these processes briefly. Decay The simplest example of decay is that of the free neutron, which decays into a proton plus an electron with a half-life of about 10.8 minutes. The energy of decay is 0.78 MeV, which is the difference between the rest energy of the neutron (939.57 MeV", ") and that of the proton plus electron (938.28 0.511 MeV). More gendecay, a nucleus of mass number A, atomic number Z, and neutron erally, in number N changes into one with mass number A, atomic number Z Z 1, and neutron number N N 1, conserving charge with the emission of an electron. 500 N Chapter 11 Nuclear Physics c2 The energy of decay Q is times the difference between the mass of the parent nucleus and that of the decay products. If we add the mass of Z electrons to both the parent nucleus and the decay products, we can write Q in terms of the atomic masses of the parent and daughter atoms: Q c2 MP MD 11-37 Ek Emax Figure 11-21 Energy spectrum of electrons emitted in decay. The number of electrons per unit energy interval N is plotted versus kinetic energy. The fact that all the electrons do not have the same kinetic energy Emax suggests that there is another particle emitted that shares the energy available for decay. decay an electron of Another way of understanding this result is to note that in mass me leaves the atom, which is now a daughter ion of nuclear charge (Z 1) and Z atomic electrons. To obtain the mass of the neutral daughter atom, we must add the mass of an electron me so the total mass change is just the difference in mass between the parent and daughter atoms. If the decay energy Q were shared only by the daughter atom and the emitted electron, the energy of the electron would be uniquely determined by conservation of energy and momentum, just as in decay. Experimentally, however, the energies of the electrons emitted in decay are observed to vary from Emax. zero to the maximum energy available A typical energy spectrum is shown in Figure 11-21; compare this with the discrete spectrum of -particle energies of Figure 11-19. Thus, in a particular decay event in which the electron carried away less than the energy it would appear that energy was not conserved since in that decay Q>c2 MP A moment of reflection will persuade you that linear momentum would not be conserved either and, recalling that the neutron, proton, and electron are all spin particles, neither would the angular momentum. Emax, MD. 1>2 A solution to this apparent multiple failure of conservation laws was first suggested by Wolfgang Pauli in 1930. He proposed that a third particle was emitted in decay that carried the energy, linear momentum, and angular momentum needed to conserve these quantities in each individual decay. It would carry no electric charge since charge was already conserved in decay. Its mass would be much less than that of the electron since the maximum energy of electrons emitted in decay is observed to be very nearly equal to the value of Q, the total energy available for the decay. In 1933 Enrico Fermi developed a highly successful quantum theory of decay that incorporated Pauli\u2019s proposed particle, which Fermi called the neutrino (\u201clittle neutral one\u201d in Italian) to distinguish it from the massive neutron that had been discovered by Chadwick earlier that same year. It was not until 1956, in an experiment performed by Clyde Cowan and Frederick Reines, that neutrinos were first observed in the labo( ratory. It is now known that there are six kinds of neutrinos, one associated with e) ( associated with the tau particle, and associated with muons, one electrons, one ) e, the antiparticles of each of those, written and The electrons, muons, and taus, together with the neutrinos constitute a family of particles called leptons, which will be discussed further in Chapter 12. The decay of the free neutron is then expressed by ( ). and that of 198Au, a more or less typical emitter, by n \u00a1 p e 198Au \u00a1 198Hg e 11-38 11-39 where the lepton conservation law (see Section 12-3) dictates the emission of an electron antineutrino to accompany a decay. Presently the subject of intense experimental and theoretical research, current results place the upper limit of the electron neutrino\u2019s mass at about 2.2 eV c2, or no more than about times the mass of the electron. 4 106 > 11-4 Alpha, Beta, and Gamma Decay 501 Decay In decay, a proton changes into a neutron with the emission of a positron and a neutrino. A free proton cannot decay by positron emission because of conservation of energy (the rest energy of the neutron is greater than that of the proton), but because of binding energy effects, a proton inside a nucleus may emit a positron. A typical decay is 13 7N \u00a1 13 6C e 11-40 The only naturally occurring positron emitter known to exist is decay by emission or electron capture! The 40K, -decay equation is which also may 40 19K \u00a1 40 18", "Ar e 11-41 As in all nuclear transformations, the decay energy Q is related to the difference in mass between the parent nucleus and the decay products. Note that if we add the mass of Z electrons to the nuclear masses (Z 7 in the case of Equation 11-40 and Z 19 in Equation 11-41), we obtain on the right side of each equation the mass of the daughter atom plus two extra electron masses (the positron and electron have identical mass). The decay energy for decay is thus related to the atomic mass of the parent and daughter atoms by Q c2 MP (MD 2me) 11-42 decay, a positron of mass me leaves Again, we can understand this by noting that in the system, which is now a negative daughter ion of nuclear charge (Z 1) and Z atomic electrons. To obtain the mass of the neutral daughter atom, we must subtract the mass of another electron, giving a net change of 2me in addition to the difference decay cannot occur unless that in mass of the parent and daughter atoms. Thus, energy difference is at least 2mec2 1.022 MeV. As we have mentioned, neither electrons nor positrons exist inside the nucleus prior to the decay. They are created in the process of decay by the conversion of energy to mass, just as photons are created when an atom makes a transition from a higher to a lower energy state. In this regard decay differs from decay. There is, however, a fundamental difference between the emission of electrons (and neutrinos) that de-excite the bound states of nucleons that compose a nucleus and the emission of photons accompanying the de-excitation of the electrons bound to a nucleus. The latter bonding is due to the electromagnetic interaction, whereas the nucleons are bound by the strong nuclear force. However, electrons and neutrinos are not affected by the strong nuclear force and, since the neutron is uncharged, the electromagnetic interaction is not involved in its decay. Thus, in order to explain -decay lifetimes are typically quite long compared to the characteristic nuclear time scale the time for a particle moving at near the speed of light to cross the nucleus), the new interaction must act for a long (nuclear) time in order to generate the decay. In other words, it is weaker than the strong attractive force between the nucleons and is therefore called the weak interaction or the weak force. So we now have two nuclear forces, a strong one and a weak one. Like the strong nuclear force, the weak nuclear force also has a short range. decay, we must invoke a new interaction. Since ( 1023 s, PET scanners used in medical diagnosis detect photons emitted in electron-positron annihilation following decay. 502 Chapter 11 Nuclear Physics Electron Capture In electron capture, a proton inside a nucleus captures an atomic electron and changes into a neutron with the emission of a neutrino; thus the effect on the atomic number is the same as in decay. The energy available for this process is given by Q c2 MP MD 11-43 Whenever the mass of an atom of atomic number Z is greater than that of the adjacent atom with atomic number (Z 1), electron capture is possible. If the mass difference is greater than 2me, decay is also possible, and these two processes compete. The probability of electron capture is negligible unless the atomic electron is in the immediate vicinity of the nucleus. This probability is proportional to the square of the electron wave function integrated over the volume of the nucleus. It is significant only for the 1s electrons of the K shell or, with much lower probability, the 2s electrons of the L shell. A typical example of electron capture is 51 24Cr \u00a1 51 23V e 11-44 which has Q 0.751 MeV. Note that the emission of the neutrino conserves leptons since the captured electron has disappeared. Further understanding of the -decay processes can be gained by considering their relation to the energy valley of the N versus Z graphs shown in Figure 11-15, with the energy scale computed from the Weizs\u00e4cker formula (Equation 11-14). Cuts through Figure 11-15a at constant mass number A yield parabolas since Equation 11-14 is qua 12 MeV c2 dratic in Z, one parabola for a5 (even A). Figure 11-22 illustrates an example of each case. The decays always proceed down the sides of the energy valley toward the lowest-energy, stable isotope on to 60Ni. the valley floor. Notice in Figure 11-22b the possible double decay from 0 (odd A) and two parabolas for a5 60Fe > (a) 101.916 u, s s a M 101.914 101.912 101.910 101.908 101.906 \u03b2 \u2013 \u03b2 \u2013 A = 101 \u03b2+ \u03b2+ \u03b2 \u2013 E.C. 101.904 40 42 44 Z 46 48 (b) 59.9", "44 59.942 59.940 \u03b2 \u2013 u, s s a M 59.938 59.936 59.934 59.932 \u03b2 \u2013 \u03b2 \u2013 Possible double \u03b2\u2013 decay A = 60 \u03b2+ \u03b2+ 59.930 24 26 28 Z 30 32 Figure 11-22 Profiles of constant atomic mass show the cross section of the energy valley of the N versus Z graph. The energy axes are expressed in mass units as computed from the Weizs\u00e4cher mass formula. (a) Odd-A nuclei, such as A 101 shown, 12 MeV c2 or odd-odd with have Z and N odd-even or even-odd and a5 a5 12 MeV c2. The even-even parabola lies below the odd-odd one. 0. (b) Even-A nuclei are either even-even with a5 > > 11-4 Alpha, Beta, and Gamma Decay 503 Since decay proceeds via the weak interaction, the probability of the weak force producing two particles simultaneously is indeed small, as you might imagine. Prior to 1985 its existence had been inferred only indirectly by abundance measurements on decay products in geologic materials. In 1985 Steven Elliott and his co-workers made the first direct observation of double beta decay using as the source. The decay equation is 82Se 36Kr e e 11-45 82 34Se \u00a1 82 The half-life for the double decay measured by Elliott is 1.1 1020 years! Double beta decay has subsequently been observed in several other nuclides. Since recent experiments show that the neutrino has a very small mass, current theory (the Standard Model\u2014see Chapter 12) would allow the decay in Equation 11-45 to proceed without the emission of neutrinos, albeit with an even lower probability. The implications of a neutrinoless double beta decay are profound for both particle physics and cosmology. Although active searches are currently under way, no such decays have yet been observed. EXAMPLE 11-12 Maximum decay modes of imum energy of the positrons? 40K Energy from We noted earlier that one of the is positron emission, shown in Equation 11-41. What is the max- 40K SOLUTION 1. The maximum energy Q of the positrons is is the given by Equation 11-42, where is the daughter: parent and 40Ar 40K 2. The atomic masses are given in Appendix A: Q c2 MP (MD 2me) M(40K) 39.964000 u M(40Ar) 39.962384 u me 5.4858 104 u 3. Substituting these into Equation 11-42 yields Q c2 39.964000 u [39.962384 2 3.4858 104] u 0.000519 u 931.5 MeV>c2 # u 0.483 MeV>c2 Remarks: Neglecting the recoil of the Ar nucleus, the decay energy Q 0.483 MeV is the maximum energy of the emitted positrons. EXAMPLE 11-13 The Decay of among decay and the three types of decay are allowed for Determine which decay mode or modes 233 93Np 233 93Np. SOLUTION The four decays whose possibility of occurrence we are to find are decay: 233 decay: 233 decay: 233 electron capture: 233 93Np \u00a1 229 93Np \u00a1 233 93Np \u00a1 233 93Np \u00a1 233 91Pa 94Pu 92U 92U e e e 504 Chapter 11 Nuclear Physics The decay energy Q for each of these is computed as follows: decay (Equation 11-34): Q c2 233.040805 (229.032085 4.002603) 0.006117 u 5.70 MeV>c2 which is greater than zero; therefore, decay is allowed. decay (Equation 11-37): Q c2 233.040805 233.042963 0.002158 u 2.01 MeV>c2 decay is forbidden. decay (Equation 11-42): Q c2 233.040805 (233.039630 2 5.4858 104) 0.000078 u 0.073 MeV>c2 decay is allowed. Electron capture (Equation 11-43): Q c2 233.040805 233.039630 0.001175 u 1.09 MeV>c2 Electron capture is allowed. Thus, the available decay energy would allow decay, decay is very small. decay, and electron decay is forbidden. decays more than 99 percent of the time by electron capture capture, although the energy for Experimentally, and about 0.3 percent of the time by decay. decay has not been observed. 233Np Gamma Decay In decay, a nucleus in an excited state decays to a lower energy state of the same isotope by the emission of a photon.", " This decay is the nuclear analog of the emission of light by atoms. Since the spacing of the nuclear energy levels is of the order of MeV (as compared with eV in atoms), the wavelength of the emitted photons are of the order of hc E 1240 eV # nm 1 MeV 1.24 103 nm Gamma-ray emission usually follows beta decay or alpha decay. For example, if a radioactive parent nucleus decays by beta decay to an excited state of the daughter nucleus, the daughter nucleus often decays to its ground state by emission of one or more rays. The mean life for decay is usually very short. Direct measurements of mean lives as short as 1011 s are possible. Measurements of lifetimes smaller than 1011 s are difficult but can sometimes be accomplished by determining the natural emitters have very line width and using the uncertainty relation U> %. A few % 11-4 Alpha, Beta, and Gamma Decay 505 long lifetimes, of the order of hours and even, in a few cases, years. Nuclear energy states with such long lifetimes are called isomers or metastable states. The differences in -ray lifetimes are a consequence of the quantum-mechanical selection rules that govern transitions between the energy levels of nuclei, just as they do between atomic energy levels. For example, large angular momentum (spin) changes are forbidden for transitions; that is, they have very low probability. This is the major reason that, for an isomer, decays to the ground state with a instance, the first excited state of 1>2, while that of the ground half-life of 13.6 years. The spin of the isomeric state is 4U state is of angular momentum, a ray to carry away very unlikely occurrence that accounts for the long half-life. The decay requires the 93Nb, 9>2. The energy hf of a gamma-ray photon is the difference in energy of the states be- tween which the transition occurs. That is, hf Ehigh Elow Elow 11-46 hf p = hf/c Mc2 p Ehigh is the energy of the upper level and is that of the lower level. Several where 223Ra gamma decays are shown in Figure 11-20 between some of the excited states of ray is emitted from the 174that resulted from the decay of 223Ra, reducing the excitation energy of that nucleus to 61 keV above the keV level of ray is equal to 174 keV 61 ground state. From Equation 11-46, the energy of that keV 113 keV. To be more precise, conservation of momentum requires that the 223Ra nucleus carry a small part of this energy as it recoils from the emission of the photon. (See Figure 11-23.) The energy of the nuclear recoil Er is given by For example, a 227Th. Figure 11-23 A nucleus of rest energy Mc2 emits a photon of energy hf and momentum p hf c. Conservation of momentum requires that the nucleus also recoil with momentum p. > Er p2 2M (hf)2 2Mc2 11-47 where M is the nuclear mass. All gamma-ray energies are small compared with atomic hf V Mc2 V hf. and nuclear rest energies; that is, Thus, Equation 11-46 is an excellent approximation of the gamma ray\u2019s energy. hf>Mc2 V 1; therefore, Er or Internal Conversion An important alternative to gamma-ray emission for the deexcitation of an excited nuclear state, particularly low-lying states rather than being emitted as a photon, is the process of internal conversion. In this process the excitation energy of the state is transferred to an orbital electron, which is ejected from the atom. Those electrons with the highest probabilities of being close to the nucleus, the K and L electrons, are the ones most likely to be emitted. The ejected electron has kinetic energy equal to the nuclear transition energy minus the electron\u2019s binding energy. Since the latter are accurately known for nearly all elements, measuring the kinetic energies of the conversion electrons makes possible determination of many nuclear excited states. While internal conversion is quantum mechanically a one-step process, it was initially pictured as the emission of a photon followed by a photoelectric-effect interaction with an orbital electron of the same atom, hence the name internal conversion. More In 1958 Rudolf M\u00f6ssbauer14 observed a remarkable feature of gamma A discovery decay, the recoilless emission of gamma rays from made while he was still a graduate student, it made possible high-precision-frequency measurements, leading to a host of applications. It is described in the section The M\u00f6ssbauer Effect on the home page: www.whfreeman.com/tiplermodernphysics5e. See Figures 11-24 through 11-26 here. 191Ir. The exceptional precision of frequency measurements made possible by the M\u00f6ssbauer effect has applications in", " a broad range of areas, such as measurements of gravitational red shift, impurities and imperfections in crystalline solids, and the transverse Doppler effect (see Section 1-5), to name just three. 506 Chapter 11 Nuclear Physics Questions 6. De-excitation of the first excited state of 93Nb requires the gamma ray to carry of angular momentum. Since the gamma ray\u2019s intrinsic angular 4U away momentum is 1U, how could it carry away 4U? 7. Why is the decay series A (4n 1) not found in nature? 8. A decay by emission is often followed by a decay. When this occurs, it is decay. Why? usually a 7. How can the application of very high pressure affect the lifetime of a sample that decays by electron capture? Why are other types of decay not affected? V(r) ke2>r 11-5 The Nuclear Force The study of nuclear physics is quite different from that of atomic physics. The simplest atom, the hydrogen atom, can be completely understood by solving the Schr\u00f6dinger equation using the known potential energy of interaction between the electron and proton, (though, as we have seen, the mathematics needed is fairly complicated). The simplest nucleus (other than a single proton) is the deuteron, consisting of a proton and a neutron. We cannot solve the Schr\u00f6dinger equation for this problem and then compare with the experiment because, although many of its characteristics have been determined, the exact mathematical form of the potential energy of interaction V is not known. There is no macroscopic way to measure the force between a neutron and a proton. It is clear from the fact that many nuclei are stable that there are other forces much stronger than electromagnetic or gravitational forces between nucleons. Considering as an example, we see that the electrostatic potential energy of two protons separated by 1 fm is 4He V ke2 r 1.44 MeV # fm 1 fm 1.44 MeV and note that it is positive\u2014i.e., the electrostatic force between the protons is, of course, repulsive. However, the energy needed to remove a proton or neutron from 4He is about 20 MeV. The force responsible for such a large binding energy must be attractive and significantly stronger than the electrostatic force. This must certainly be the case since the neutrons are electrically neutral and hence do not feel the Coulomb force and the protons are all positively charged and thus feel a repulsive electrostatic force. Nor can we appeal to the gravitational attractive force between the protons to offset their Coulomb repulsion since, as Example 11-14 illustrates, the gravitational force between pairs of protons in the nucleus is insignificantly small compared to their Coulomb repulsion. Thus, the attractive force that holds the nucleons together must be strong, stronger even than the electromagnetic interaction. It is called the nuclear or hadronic force or often simply the strong force. Determination of the characteristics of the nuclear force is one of the central problems of nuclear physics. Much information about this force can be and has been obtained from scattering experiments involving protons, neutrons, and other particles. Although the results of a scattering experiment can be predicted unambiguously from knowledge of the force law, the force law cannot be completely determined from the results of such experiments. The results of scattering experiments do indicate that 11-5 The Nuclear Force 507 (1) the nuclear force has the same strength between any two nucleons\u2014that is, n-n, p-p, or n-p; (2) the force is strong when the particles are close together and drops rapidly to zero when the particles are separated by a few fm; and (3) it is a saturated force. The potential energy of the nucleon-nucleus interaction can be roughly represented by a square well of about 40 MeV depth and a few fm width. EXAMPLE 11-14 Ratio of Fgrav FCoul Between Protons Compare the gravitational attractive force between two protons in an atomic nucleus (or anywhere else, for that matter) with the electrostatic repulsion between them. > SOLUTION The electrostatic repulsion for two protons separated by a distance r is FCoul 1 4 0 e2 r2 (1.60 1019 C)2 4 0r2 and the gravitational attraction between them is FGrav G m2 p r2 G(1.67 1027 kg)2 r2 The ratio is independent of r and equal to Gm2 p (1>4 0)e2 (6.67 1011 N # m2>kg2)(1.67 1027 kg)2 (8.99 109 N # m2>C2)(1.60 1019)2 Fgrav FCoul Fgrav FCoul 8.1 1037 1036 2H Solution of the nuclear-wave equation presents all of the mathematical complexities", " of our earlier studies of atomic and molecular systems plus some truly monu1H mental new ones. Like the atomic and molecular systems, the nucleus (except for ) is a many-body system with all of the accompanying computational difficuland ties. In addition, the nuclear interaction is far more complex than the electromagnetic interaction and, even worse, it is not yet known how the nuclear interaction can be expressed in closed, analytic form; that is, we do not know the nuclear force law equivalent of Coulomb\u2019s law for the electrostatic force. This means that we cannot yet write down the exact form of the nuclear potential function that must be included in the wave equation in order to solve for the nuclear-wave functions and allowed energies. Substantial progress has been made in recent years toward obtaining the analytic expression for the interaction. For instance, an estimate of the depth of the nuclear potential can be made by assuming its shape to be approximated by a square well and computing the ground-state energy of a nucleon, based on a reasonable assumption of the well width. Using 2 fm as a typical width for light nuclei (see Figure 11-5), the potential V for a nucleon is approximately (6.63 1034 J # s)2 (8)(1.67 1027 kg)(2 1015 m)2(1.60 1013 J>MeV) V E1 h2 8ma2 V 50 MeV 508 Chapter 11 Nuclear Physics Two protons separated by that same distance experience an electrostatic Coulomb repulsive potential given by VCoulomb VCoulomb e2 a 1 4 0 0.72 MeV (9 109 N # m2>C2)(1.60 1019 C)2 (2 1015 m)(1.60 1013 J>MeV) Thus, our square well approximation suggests that at 2 fm, the attractive nuclear potential exceeds the Coulomb repulsion experienced by a proton by nearly two orders of magnitude.15 More detailed understanding of the nature of the nuclear force and the shape and depth of the potential is provided by two types of experiments. First, just as atomic spectroscopy yielded information that made possible the determination of such things as the energies, spins, and magnetic moments of the electronic structure of the atoms, nuclear spectroscopy\u2014i.e., the study of the emission and absorption of particles and radiation by the nuclei\u2014yields valuable information concerning the ground and excited states of nuclei, including energies, magnetic moments, electric quadrupole moments, and spins. The second source of our detailed information comes from the analysis of scattering experiments. These are experiments in which particles that feel the nuclear force, such as protons or alpha particles, are used as projectiles \u201cfired\u201d at target nuclei. The de Broglie wavelength of projectile protons with kinetic energies of 20 MeV (or more) are of the order of nuclear dimensions: 6.63 1034 J # s 22 1.67 1027 kg 20 MeV 1.60 1013 J>MeV h p h 22mE 6 1015 m 6 fm Thus, such protons will experience considerable diffraction in collisions with the target nuclei. Analysis of the resulting diffraction pattern yields detailed information concerning the interaction between the particles. Many such experiments, particularly protons scattered from protons, called p-p scattering, and neutrons scattered from protons, or n-p scattering, reveal that the nuclear potential for proton-proton pairs and neutron-proton pairs are of the form sketched in Figure 11-27. Although the (a) (b) Neutron-proton Neutron-neutron Proton-proton 2 4 6 r, fm V e M, ) r ( V 20 0 \u201320 \u201340 V e M, ) r ( V 20 0 \u201320 \u201340 2 4 6 r, fm Figure 11-27 (a) The approximate shape of the potential between n-p and n-n pairs. The hard core suggested by the nearly constant central density of the nucleus has a radius of about 0.5 fm. (b) The p-p potential differs from those in (a) by the added Coulomb repulsion, which dominates beyond about 3 fm. Notice the n-p and n-n potential well is slightly deeper than the p-p potential due to the absence of the Coulomb repulsion. 11-5 The Nuclear Force 509 shape of the potential for neutron-neutron pairs can only be determined indirectly since free neutrons are radioactive and we do not know how to make targets consisting only of neutrons (such as the matter of neutron stars), it appears to be identical to that of n-p pairs. In fact, when the Coulomb repulsion component of the p-p pair potential in Figure 11-27b is subtracted from the total potential V(r), the remaining nuclear p-p", " potential is also the same as those for n-p and n-n pairs. This leads to the very important conclusion that the nuclear force is independent of the charge of the nucleons. This suggests that the proton and neutron can be considered as different charge states of the same particle, the nucleon. We will pursue this suggestion further in Chapter 12. 1>2 As described in Section 11-2, the charge radius of the proton is about 1 fm. The neutron is approximately the same size. As Figure 11-27 illustrates, two nucleons experience the attractive nuclear force as long as they are within about 2.5 fm of each other, but the force diminishes rapidly over the next fm of separation and is essentially zero beyond 3 fm. Thus, we also conclude that the nuclear force is a shortrange force. Nucleon pairs also experience an extremely strong repulsive component of the nuclear force when they approach within about 0.5 fm. This hard core is consistent with the observation that the central density is nearly the same for all nuclei (see Figure 11-27). That is, as more and more nucleons are added, the size of the nucleus increases in such a way that the density remains approximately constant, so something must prevent the nucleons from crowding too closely together. The short range of the nuclear force together with the repulsion of the hard core means that, as the size of the nucleus increases beyond the 2.5- to 3-fm range of the force, an individual nucleon will be able to interact with only a limited number of the other particles in the nucleus, namely its nearby neighbors that are within range of its force. This is analogous to the limited number of bonds associated with each atom in the covalent bonding of solids. For example, each carbon atom in diamond bonds with only four of its nearest neighbors, and we could describe the carbon covalent bond as being a saturated bond. Similarly, the nuclear force is a saturated force. The Nuclear Exchange Force Without knowing the analytic form of the nuclear potential function, we have been able to conclude that the nuclear force is a short-range, saturated, charge-independent, spin-dependent force with a hard core and a small noncentral component and is about two orders of magnitude stronger than the electrostatic force. What could be the origin or mechanism for such a force was first suggested by H. Yukawa16 in 1935. Yukawa proposed that the nuclear force resulted from an exchange of particles between the nucleons. He based his theory on an analogy with the quantum-mechanical explanation of the electrostatic interaction, one of two exchange mechanisms that you have previously studied, though perhaps not by that name. Classically, any distribue and the force felt by another charge q lotion of charges produces an electric field cated in the field is the product q Any change in the charge distribution changes e; however, the information that a change has occurred does not appear instantaneously throughout the field but is propagated outward at the speed of light. Timee, dependent changes in the charge distribution create time-dependent changes in that is, electromagnetic radiation, or waves. 17 We have seen that the particle representation of the electromagnetic radiation is the photon. Quantum mechanically, every charge is continually emitting and absorbing photons, even when it is not moving. They are called virtual photons, meaning that they are not directly observable. e. 510 Chapter 11 Nuclear Physics A charge can emit a virtual photon of energy hf without changing its energy or recoiling, i.e., without violating conservation of energy and momentum, provided that \u00a2t U>\u00a2E, where E hf, as required by the the photon exists for no longer than uncertainty principle. The distance that the virtual photon can travel during the time t, called the range R, is given by and substituting for E, R c \u00a2t cU>\u00a2E R cU>hf c>2f >2 11-48 11-49 A second charge located up to a distance R from the first can absorb the photon and a similar photon emitted by the second charge may be absorbed by the first, all without violating energy and momentum conservation. It is this exchange of virtual photons that results in the electrostatic Coulomb force between the two stationary charges in quantum mechanics. Note that there is no limit to the wavelength of the photon in Equation 11-49 since the energy of the photon may be arbitrarily small, the photon having no rest mass. Thus, the distance separating the two charges, the range R of the Coulomb force, may also be infinite, as you have already learned. An exchange mechanism was also used in BCS theory to account for the attractive force between the electrons of the Cooper pairs. (See Section 10-8.) In that case, the exchange particles were the phonons and the range of the \u00a2E force was not infinite but determined by the fact that the energy gap. t Yukawa proposed that the nuclear force could", " also be explained in terms of the exchange of virtual particles by the nucleons. These particles, which he called mesons, were pictured as the analogs of the virtual photons in the electromagnetic interaction and established the meson field in analogy with the electromagnetic field. The mechanism for the nuclear force was proposed to be an exchange of a meson between a pair of nucleons, as illustrated by Figure 11-28. Yukawa accounted for the observed short range of the nuclear force by assigning mass to the meson. Thus, the energy uncertainty E in Equation 11-48 would be \u00a2E! mc2 x where m is the mass of the meson and mc2 is its rest energy. The range R of the meson and therefore the nuclear force that it mediates cannot be larger than R c \u00a2t cU>\u00a2E U>mc 11-50 > since the speed of the meson must be less than the speed of light. Recall that h mc is the Compton wavelength of the particle c >2. R whose mass is m, so The range of the nuclear force c was known to be about 1 fm, which allowed an approximation of the meson\u2019s expected mass from Equation 11-50: m 3.5 1028 kg 380 me 200 MeV>c2 \u0394t Nucleon Meson Nucleon Figure 11-28 Schematic representation of the exchange of a meson by a pair of nucleons. The meson is emitted by the nucleon on the left, which recoils as a result and is absorbed after a time t by the nucleon on the right, which also recoils. The effect on the nucleons is as if they had interacted with each other. This kind of spacetime diagram of the interaction of fundamental particles is called a Feynman diagram (see Section 12-1). The x and t axes are normally omitted. 11-5 The Nuclear Force 511 t \u03c0 Meson Nucleus Nucleon x Figure 11-29 A Feynmanlike diagram of a nucleon emitting a virtual pion in the vicinity of a nucleus. If the nucleus can provide at least the pion\u2019s rest energy and participate in the conservation of momentum, the pion may become real, i.e., visible in the laboratory. The observed charge independence of the nuclear force was incorporated by Yukawa into the theory by allowing the mesons to carry e, 0, or e charge. Thus, the exchange of a neutral meson would leave both of the nucleons with their original charge, while the exchange of a charged meson would interchange their charges (see Figure 11-28). Note that m 0 for photons in Equation 11-50 implies the infinite range of the electromagnetic force. If the nucleon that emits the meson happens to interact with another particle (or nucleus) that has sufficient kinetic energy in the emitting nucleon\u2019s rest system to supply the meson\u2019s rest energy and also provide the recoil momentum to the emitting nucleon, thus conserving both energy and momentum, the virtual meson can become real and be observable in the laboratory. Such a situation is shown schematically in Figure 11-29. Note the analogy to the emission of photons (bremsstrahlung) by accelerated electrons in an x-ray tube (see Section 3-4). It was interactions such as shown in Figure 11-29 in which Yukawa\u2019s mesons, now called mesons or pions, were first seen in cosmic rays in 1947, more than a decade after they were proposed.18 The mass measured for the pions is in quite good agreement and all three with Yukawa\u2019s predicted approximate value of about charge versions were subsequently discovered, providing beautiful confirmation of Yukawa\u2019s theory. Since then additional mesons have been discovered and our understanding of the nuclear force has been modified to include the effect of their being exchanged by nucleons as well, but the pions remain as the dominant carrier of the force between nucleons and the cornerstone of our understanding of it. As we will discuss further in Chapter 12, the Standard Model of particle physics describes the nucleons and the mesons as composites of other, more fundamental particles, called quarks. The interaction between quarks to form these particles is mediated by a field particle, the gluon, carrying the strong force between quark pairs in analogy with our discussion above. 140 MeV>c2, 200 MeV>c2, EXAMPLE 11-15 Range of the Nuclear Force Using the experimentally measured mass of the pion, 140 MeV c2, estimate the range of the nuclear force. > SOLUTION The range R cannot be larger than have that Uc>mc2 according to Equation 11-50. We then R Uc mc2 (1.06 1034 J # s)(3.00 108 m>s) (140 MeV>c2)(c2)(", "1.60 1013 J>MeV) R 1.4 1015 m 1.4 fm Questions 10. What property of the nuclear force is indicated by the fact that all nuclei have about the same density? 11. How does the nuclear force differ from the electromagnetic force? 12. Mesons that have been discovered in recent years are all more massive than the pion. What does that mean regarding the range of the force that they mediate? 512 Chapter 11 Nuclear Physics EXPLORING Probability Density of the Exchange Mesons A nucleon continually emits and absorbs virtual mesons. The time t during which a virtual meson exists can be estimated from Equation 11-50: (1.055 1034 J # s) (140 MeV>c2)(c2)(1.60 1013 J>MeV) \u00a2t mc2 U \u00a2t 5 1024 s This is not a very long time! Thus, a 1020-second time-exposure \u201csnapshot\u201d of a nucleon would show a cloud consisting of more than 10,000 mesons surrounding the nucleon! The probability density of the mesons can be determined using the results that we obtained from relativity and wave mechanics in Chapters 2 and 6, respectively. The relativistic expression connecting the total energy E and momentum p, the magnitude of the energy/momentum four-vector, is (mc2)2 E 2 (pc)2 With the appropriate operator substitutions from Table 6-1, E \u00a1 iU $ $t p2 \u00a1 U2\u22072 Equation 2-32 can be written as \u22072+(r, t) 1 c2 $2+(r, t) $t2 a 2 b mc U +(r,t) 2-32 11-51 11-52 +(r,t) where is the wave function of the meson. Equation 11-52 is a relativistic wave equation. It was first obtained by Oskar Klein and Walter Gordon in 1926, the same year that Schr\u00f6dinger developed his nonrelativistic wave equation. That the extent of the meson field is related to the range of the nuclear force given by Equation 11-50 can be illustrated by computing the probability density of the meson \u0192+\u0192 2 for a static, or time-independent, distribution. This is roughly analogous to the virtual photon distribution, or the electric field intensity, for a stationary charge. In this case, the time derivative of vanishes and Equation 11-52 can be written as + whose solution is \u22072+(r) a 2 b mc U +(r) 1 R +(r) +(r) Aer>R r 11-53 11-54 where A is a factor determined by the normalization condition. The probability density is then \u0192 +(r) \u0192 2 \u0192 A \u0192 2 e2r>R r2 11-55 and we see that the probability density of the mesons falls off exponentially at a rate determined by R. In other words, R determines the range of the exchange mesons as we had interpreted it in Equation 11-50. Figure 11-30 illustrates the probability distribution for the virtual mesons. For values of r greater than about 0.5R, function P(r) \u0192+\u0192 2r2 11-6 The Shell Model 513 \u0192+\u0192 2r2, for the virtual Figure 11-30 Probability P(r), equal to mesons emitted by a nucleon, and the range the Compton wavelength of the mesons divided by mesons beyond about 3R. There are essentially no R U>mc, 2.0 0.8 0.6 0.4 0.2 0 R 2R 3R r the curve agrees well with experimental results; however, for small values of r the measured meson density is much lower than Figure 11-30 would suggest. Indeed, if the predicted values at very small r values actually existed, they would lead to some quite unusual nuclear properties that are, in fact, not observed. Nuclear theorists conclude that the number of pions at very small r is somehow suppressed, likely as a result of the quark-gluon interaction mentioned above. This is an area of active current research. 11-6 The Shell Model Although the general features of the binding energy of nuclei are well accounted for by the semiempirical mass formula that was based on modeling the nucleus as a liquid drop, the binding energy and other properties do not vary with perfect smoothness from nucleus to nucleus. It is not surprising that the smooth curve predicted by Equation 11-12 does not fit the data for very small A, for which the addition of a single proton or a neutron makes a drastic difference. However, even for medium and large A there are some substantial fluctuations of nuclear properties in neighboring", " nuclei. Consider the binding energy of the last neutron in a nucleus. (Note that this is not the same as the average binding energy per nucleon.) We can calculate this from the semiempirical mass formula by computing M(A, Z). the difference in mass M[(A 1), Z] mn Figure 11-31 shows a plot of the difference between the experimentally measured binding energy and that calculated from Equation 11-12 as a function of the neutron number N. There are large fluctuations near N 20, 28, 50, 82, and 126. These are also the neutron numbers of the nuclei that have an unusually large number of isotones. Nuclei with these proton numbers (except that no element with Z 126 has been observed) have an unusually large number of isotopes. 3 2 1 0 \u20131 \u20132 n B \u0394 V e M 50 60 40 28, These numbers are the \u201cmagic numbers\u201d that were referred to in Section 11-2. In the regions between these magic numbers, the binding energy of the last neutron is predicted quite accurately by the semiempirical mass Existence of 126 as a magic number has prompted searches for unusually stable (but still radioactive ) isotopes with Z in the vicinity of 126. Finding them will strengthen our understanding of nuclear structure. Thus far, the highest Z discovered is 118. 80 82 100 120 140 N 126 Figure 11-31 Difference in the measured binding energy of the last neutron and that calculated from mass formula versus neutron number. Note the similarity of this curve and the ionization energy of atoms versus Z (Figure 7-20). The neutron numbers 28, 50, 82, and 126 correspond to closed shells. These data show that the neutron with N magic number 1 is much less tightly bound than that with N magic number. 514 Chapter 11 Nuclear Physics Figure 11-32 The capture cross section measures the probability that a neutron approaching a nucleus will be captured, or bound to the nucleus. The solid line traces the average value. Notice the sharp drop in capture probability of nearly two orders of magnitude at N 50, 82, and 126. 100.0 10.0 b m, \u03c3 1.0 Lu Re Re Ta Au W Hg Pt Bi Pb In Ag Rh Ag I Sb Lu Nd Nd Ru Rb Br Nb Ru Mo Sn Mo Y Br As Zn Ga Cu Co Kr Cu Ni Mn Sr Kr Rb Ca V K Si Ar Cl Mg Al Na Sn Sn La Pr Ce Ce Ba Xe 10 30 50 70 N 90 110 130 formula. Figure 11-31 should be compared with Figure 7-20, which shows the binding energy of the last electron in an atom as a function of the atomic number Z. The similarity of these two figures suggests a shell structure of the nucleus analogous to the shell structure of atoms. There is considerable additional evidence for these magic numbers, such as the electric quadrupole moments (see Figure 11-7), the neutron capture cross sections illustrated in Figure 11-32, and the binding energies of the last neutron for isotopes of a given Z as shown in Figure 11-33. Additional evidence of nuclear shell structure is discussed in Mayer and Jensen (1955). Although the unusual stability of the nuclei with N or Z equal to one of the magic numbers was noticed in the 1930s, there was no successful explanation in terms of shell structure until 1949. In the discussion of atoms in Chapter 7, we started with a fixed positive charge Ze and computed the energies of individual electrons, assuming first that each electron was independent of the others as long as the exclusion principle was not violated. The interaction of the outer electrons with the inner core could be taken care of by assuming an effective nuclear charge that is less than Z because of the screening of the nuclear charge by the inner electrons. This works quite well since the electrons are fairly far from each other in an atom. We could therefore use the individual electron quantum states of the hydrogen atom described by n, l, ml, and ms as a first approximation for the electrons in complex atoms. The atomic magic numbers come about naturally due to the large energy difference between one shell or subshell and the next. The actual calculations of atomicwave functions and atomic energies require powerful approximation or numerical techniques, but they can be done reliably because the forces involved are well known. 90 N The situation is not the same for the nuclear-shell model. In the first place, there is no central potential analogous to the fixed positive charge of the atom. The interaction of the nucleons with one another is the only interaction present. In addition to being noncentral, the situation is further complicated by the fact that we know little about the strong force between nucleons beyond what we have V e M B, 15 10 5 0 70 80 Figure 11-33 Binding energy B for the last neutron of the isotopes of Ce (Z 58). These data are typical of nuclei with Z 20. B decreases sharply (about 2 MeV) for N 82", " 1. This graph also shows the pairing energy associated with a5 in the Weizs\u00e4cher formula (see Section 11-2), where the last neutron is more tightly bound if N is even than if N is odd. 11-6 The Shell Model 515 (ab) x1 x x1 x2 x3 x4 x Figure 11-34 (a) A single nucleon moving in one dimension sees the potential due to a second nucleon located at x1. (b) The potential seen by the single nucleon due to four other nucleons located along the x axis fluctuates rapidly; however, the average of the four potentials can be reasonably well approximated by the dashed curve, a finite well with sloping sides. discussed: that it is saturated, has a short range, is charge independent, and is spin dependent. At first sight, it is difficult to imagine a neutron or proton moving almost freely in a well-defined orbit when there are A 1 particles nearby exerting very strong forces on it. Despite these difficulties, the observed properties, such as are illustrated in Figures 11-7, 11-31, 11-32, and 11-33, give strong motivation to try a model in which each nucleon moves about more or less freely in an average potential field produced by the other nucleons. Figure 11-34 shows how such an average potential could be produced. The assumption that the nucleon can move in an orbit without making many collisions can be rationalized by using the exclusion principle. Consider N neutrons in some potential well. In the ground state, the N lowest energy levels will be filled. A collision between two neutrons that does not result in their merely exchanging states is forbidden by the exclusion principle if there are no accessible unfilled states. A collision involving the exchange of identical particles has no effect. Thus, only those nucleons in the highest filled levels, where there are empty states available nearby, can collide with one another. This is analogous to the result that most of the free electrons in a metal cannot absorb energy in random collisions with the lattice because all the nearby energy levels are full. Like the electrons, the nucleons also have a Fermi level. The first shell-model calculations attempted to use a square well about 40 MeV deep to fit the nuclear energy levels, but they failed to produce the correct magic numbers. In 1949, M. Mayer and J. H. D. Jensen 19 independently showed that, with a modification in these calculations, the magic numbers do follow directly from a relatively simple shell model. Mayer and Jensen resolved the problem by proposing that the spin dependence of the nuclear force results in a very strong spin-orbit interaction, coupling the spin of each nucleon to its own orbital angular momentum. Thus, the nuclear spinorbit effect depends on j-j coupling20 rather than L-S coupling that characterizes the electron spin-orbit interaction (see Section 7-5). This strong spin-orbit interaction results in a decrease in the energy if the spin and the orbital angular momentum of the nucleon are parallel and an increase if they are antiparallel. Figure 11-35 illustrates the nuclear-shell model of Mayer and Jensen that yields the correct magic numbers. Depending on the details of the spin-orbit interaction in the superheavy elements, the island of stability may begin to 2g 1i 3p 2f 1h 3s 2d 1g 2p 1f 2s 1d 1p 1s 1i11/2 2g9/2 3p1/2 1i13/2 2f7/2 3s1/2 1h11/2 1g7/2 2p1/2 2p3/2 1d3/2 1d5/2 1p1/2 1s1/2 12 [148] 10 [136] 2 [126] 126 4 [124] 14 [120] 6 [106] 8 [100] 10 [92] 2 [82] 82 4 [80] 12 [76] 6 [64] 8 [58] 10 [50] 50 2 [40] 6 [38] 4 [32] 3p3/2 2f5/2 1h9/2 2d3/2 2d5/2 1g9/2 1f5/2 1f7/2 8 [28] 28 2s1/2 4 [20] 20 2 [16] 6 [14] 1p3/2 2 [8] 8 4 [6] 2 [2] 2 Figure 11-35 Energy levels for a single particle in a nuclear well, including spin-orbit splitting. The maximum number of particles in each level is given at the right, followed by the total number through that level in brackets. The total numbers just before the large energy gaps are the magic numbers. The spacing shown here is for protons; the spacing for neutrons is slightly different (lower). 516 Chapter 11 Nuclear Physics (a) (b", ") (a) The Cockcroft-Walton accelerator. Walton is sitting in the shielded enclosure in the foreground. J. D. Cockcroft and E. T. S. Walton produced the first transmutation of nuclei with artificially accelerated particles in 1932, for which they received the Nobel Prize (1951). (b) M. S. Livingston and E. O. Lawrence standing in front of their 27-in. cyclotron in 1934. Lawrence won the Nobel Prize (1939) for the invention of the cyclotron. [(a) Courtesy of Cavendish Laboratory. (b) Courtesy of Lawrence Radiation Laboratory, University of California, Berkeley.] be evident at Z 114 or 120 with metastable states whose lifetimes may be as long as hours or days. In the MORE section on the Web site we consider some of the more detailed qualitative aspects of the nuclear-shell model. Detailed calculations of energies and wave functions require many approximations, the understanding of which is a major area of continuing study in nuclear physics. More Finding the \u201cCorrect\u201d Shell Model describes some of the qualitative aspects of the several approaches to developing the nuclear-shell model and its successes (and some failures) in predicting nuclear spins and magnetic moments. It is on the home page: www.whfreeman.com/ tiplermodernphysics5e. See also Equations 11-56 and 11-57 and Figures 11-36 through 11-38. 11-7 Nuclear Reactions When a particle is incident on a nucleus, any of several different things can happen. The particle may be scattered elastically or inelastically (in which case the nucleus is left in an excited state and decays by emitting photons or other particles) or the original particle may be absorbed and another particle or particles emitted. Figure 11-39 illustrates schematically the several possible stages of a nuclear reaction. Elastic scattering refers to the reflection of the incident particle\u2019s wave at the edge of the nuclear potential well. This is the kind of scattering for particles that 11-7 Nuclear Reactions 517 Direct interaction Multiple collisions and collective effects Y Decay of compound nucleus Compound nucleus y X Absorption x x Elastic scattering x Independent particle system Compound system Final system Figure 11-39 Schematic representation of the several possible stages of the nuclear reaction X (x, y) Y, according to the theory developed by V. Weisskopf and H. Feshbach. was described by Rutherford\u2019s theory in Section 4-2. If the incident particle interacts with a single nucleon in the nucleus so that the nucleon leaves the nucleus, the reaction is called a direct interaction. Direct interactions are more probable at high energies since the incident particle can penetrate deeper into the nucleus. If the nucleon does not leave the nucleus but interacts with several other nucleons, complicated excited states can be formed in the nucleus. In such a case, when the energy carried by the incident particle is shared by many nucleons, the excited nucleus is called a compound nucleus. The compound nucleus can decay by emitting a particle identical to the incident particle and with the same kinetic energy (also elastic scattering) or by emission of one or more other particles (including photons). The decay of the compound nucleus can be treated as a statistical process independent of the detailed manner of formation, just as in the case of a radioactive nucleus. In this section we will examine some of the systematics of nuclear reactions and some typical reactions produced by incident neutrons, protons, or deuterons. We will limit the discussion to energies of less than 140 MeV. At higher energies, mesons and other particles can be created. The study of higher-energy reactions is generally undertaken to reveal the properties of fundamental particles and of the nuclear force rather than the structure of the nucleus and will be discussed further in Chapter 12. Energy Conservation Consider a general reaction of particle x incident on nucleus X resulting in nucleus Y and particle y. The reaction may be written x X \u00a1 Y y Q or, as we will usually write it, X(x, y)Y. The quantity Q, defined by Q (mx mX my mY)c2 11-58 is the energy released in the reaction and is called the Q value of the reaction. When energy is released by a nuclear reaction, the reaction is said to be exothermic. In an exothermic reaction, the total mass of the initial particles is greater than that of the final particles and the Q value is positive. If the total mass of the initial particles is 518 (a) Chapter 11 Nuclear Physics Negative-ion beam (b) Steel pressure tank Metal rings +Very high voltage metal terminal Charging belt + + + 50 kV Positive-ion beam Gas inlet Charge exchange canal Accelerating tube (a) Schematic diagram of a two-stage, or tandem, Van de Graaff accelerator. Negative ions at ground potential (atoms of a large fraction of the elements in the periodic table form stable negative ions)", " enter the beam tube at the top and are accelerated to the positive high-voltage terminal in the center, acquiring eV of kinetic energy. In the charge exchange canal, electrons are stripped from the negative ions in collisions with gas molecules, producing positive ions with charges up to Ze. The positive ions are accelerated back to ground potential, acquiring an additional kinetic energy as large as ZeV. Large Van de Graaff accelerators have terminal voltages V over 16 million volts. Thus, for example, oxygen atoms stripped of all their electrons may be accelerated to energies of 100 MeV or more. (b) A portion of the tandem Van de Graaff laboratory at Purdue University. The high-voltage terminal is in the tank at the right rear, insulated from the surroundings by inert gas under high pressure. The beam travels in the tube and is deflected to experimental areas by the bending magnets. The Purdue accelerator is used extensively in accelerator mass spectrometry. [Courtesy of David Elmore, Purdue University.] less than that of the final particles, the Q value is negative and energy is required for the reaction to take place. The reaction is then endothermic. Examples are n 1H \u00a1 2H 2.22 MeV (exothermic) 2H \u00a1 1H n 2.22 MeV (endothermic) Thus, an endothermic reaction cannot take place unless a certain threshold energy is supplied to the system. In the reference frame in which the total momentum is zero However, many reactions (the center-of-mass frame), the threshold energy is just occur with nucleus X at rest relative to the laboratory. In this frame, called the laboratory frame, the incident particle x must have energy greater than because, by conservation of momentum, the kinetic energy of y and Y cannot be zero. Consider the nonrelativistic case of x, of mass m, incident on X, of mass M (Figure 11-40). \u0192Q\u0192. \u0192Q\u0192 11-7 Nuclear Reactions 519 v = p/m M m V = p/M CM system v + V M m Lab system p = mv = MV ECM = p2/2m + p2/2M = (m + M)p2/2mM plab = m(v + V) = mv(1 + m/M) = M + m \u2013\u2013\u2013\u2013\u2013 M p Elab = 2 pL \u2013\u2013\u2013 2m = p2 \u2013\u2013\u2013 2m M + m \u2013\u2013\u2013\u2013\u2013 M 2 = M + m \u2013\u2013\u2013\u2013\u2013 M ECM Figure 11-40 Energetics of nuclear reaction in center-of-mass system and laboratory system. The energies are related by Elab [(M m) M]ECM. > In the center-of-mass frame, both particles have momenta of equal magnitude and the total kinetic energy is ECM p2 2m p2 2M 1 2 p2a m M mM b 11-59 where p mv MV. We transform to the lab frame by adding V to each velocity so that M is at rest and m has velocity The momentum of m in the lab frame is then v V. plab m(v V) mva1 m M b pa m M M b and its energy is Elab p2 lab 2m p2 2m a m M M 2 b m M M ECM The threshold for an endothermic reaction in the lab frame is thus Eth m M M \u0192Q\u0192 11-60 11-61 (If the incident particle is a photon, the Lorentz transformation must be used. For low energies, the momentum of a photon is small and approximate methods can be used. For a photon, pc E, whereas for a proton or neutron, for E V 140 MeV.) pc (2mc2E)1>2 W E EXAMPLE 11-16 Q Value of a Nuclear Reaction Find the Q value of the reaction p 7Li \u00a1 4He 4He and state whether the reaction is exothermic or endothermic. The atomic mass of is 7.016003 u. 7Li 520 Chapter 11 Nuclear Physics SOLUTION Using 1.007825 u for the mass of Appendix A, we have for the total mass of the initial particles 1H and 4.002602 u for the mass of 4He from 1.007825 u 7.016003 u 8.023828 u mi and for the total mass of the final particles mf 2(4.002602 u) 8.005204 u Since the initial mass is greater than the final mass by \u00a2m mi mf 8.023828 u 8.005204 u 0.018624 u mass is converted into energy and the reaction is exothermic. The Q value is positive and given by Q (\u00a2m)c2 (0.018624 u)c2(931.", "5 MeV>u # c2) 17.35 MeV Note that we used the mass of atomic hydrogen rather than that of the proton and the atomic masses of the atoms rather than the masses of the individual nuclei so that the masses of the four electrons on each side of the reaction cancel. and 4He 7Li EXAMPLE 11-17 Threshold Energy in Lab Frame Compute the minimum kinetic energy of protons incident on nuclei at rest in the laboratory that will produce the endothermic reaction 13C 13C(p, n)13N. SOLUTION 1. The minimum, or threshold, energy of the incident protons in the lab frame is given by Equation 11-61: 2. The magnitude of the Q value of the reaction is Eth m M M \u0192Q\u0192 \u0192Q\u0192 c2 minitial mfinal [M(13N) mn] [M(13C) M(1H)] 3. The masses of the particles involved are tabulated in Appendix A: M(13C) 13.003355 u M(1H) 1.007825 u M(13N) 13.005738 u mn 1.008665 u 4. Substituting these into the expression for \u0192Q\u0192 gives 5. Substituting this value, m M(1H), M M(13C) 11-61 gives and into Equation \u0192Q\u0192 (14.014403 14.011180) u # c2 0.003223 u # c2 931.5 MeV>u # c2 3.00 MeV Eth 1.007825 13.003355 13.003355 3.00 3.23 MeV 11-7 Nuclear Reactions 521 Cross Section The probability that a particle incident on a nucleus will scatter or induce a reaction depends on the particle\u2019s energy and what particular particle and nucleus are involved. It is as if different kinds of particles approaching a given nucleus \u201csee\u201d targets of different sizes. Similarly, identical particles with different energies \u201csee\u201d the same target nucleus larger or smaller than actual size. This effect is a consequence of the detailed arrangement of the allowed energy states of the target nucleus. A useful measure of the effective size of a nucleus for a particular scattering or nuclear reaction is the cross section If I is the number of particles incident per unit time per unit area (the incident intensity) and R is the number of reactions per unit time per nucleus, the cross section is defined as. R I 11-62 by protons. A number of reactions Consider, for example, the bombardment of the first p indicates an incident might occur. Elastic scattering is written proton, the second indicates that the particle that leaves is also a proton. If the scattering is inelastic, the outgoing proton is indicated by p and the nucleus in the resulting excited state by Some other possible reactions are 13C 13C(p, p)13C; 13C(p, p)13C*. and one writes 13C* (p, n) 13C(p, n)13N capture 13C(p, )14N (p, ) 13C(p, )10B Each possible scattering or reaction has its own cross section, called the partial cross section. The partial cross section is also defined by Equation 11-62 with R equal to the number of events of the specific kind per unit time per nucleus. The total cross section is the sum of the partial cross sections: p, p p, p p, n p, p, \u00c1 Cross sections have the dimensions of area. Since nuclear cross sections are of the order of the square of the nuclear radius, that is, a convenient unit for them is the barn, defined by (1014 m)2, 1 barn 1024 cm2 1028 m2 11-63 The cross section for a particular reaction is a function of energy. For an endothermic reaction, it is zero for energies below the threshold. Compound Nucleus In 1936, Niels Bohr pointed out that many low-energy reactions could be described as two-stage processes\u2014the formation of a compound nucleus and its subsequent decay. In this description, the incident particle is absorbed by the target nucleus and the energy is shared by all the nucleons of the compound nucleus. After a time that is long compared with the time necessary for the incident particle to cross the nucleus, enough of the excitation energy of the compound nucleus becomes concentrated in one particle for it to escape. The emission of a particle is a statistical process that depends only on the state of the compound nucleus and not on how it was produced. 522 Chapter 11 Nuclear Physics 209Bi 16 MeV \u03b3 209Bi* \u201cHot\u201d or excited compound nucleus n \u03b3 n Neutrons and photons with energies of a few MeV boil off Figure 11-", "41 Nuclear reaction via formation of compound nucleus. The 16-MeV photon is 209Bi absorbed by the nucleus, producing an excited nucleus that lives so long that excitation energy is shared by many nucleons. The excited nucleus then decays by emitting neutrons and photons, each with energy of the order of a few MeV. An incident 1-MeV proton has a speed of about 107 m s, so that it takes time R>v 1014>107 1021 s to cross a nucleus. The lifetime of a compound nucleus can be inferred to be about 1016 s. This is too short to be measured directly, but it is so long compared with 1021 s that it is reasonable to assume that the decay is independent of how it was formed. > 13C The compound nucleus for the reactions on cleus can be formed by many other reactions, such as shown above is 14N*. This nu- 10B \u00a1 14N* \u00a1 12C d 13N n \u00a1 14N* \u00a1 10B 14N \u00a1 14N* \u00a1 14N 12C d \u00a1 14N* \u00a1 13N n 11-64 The reactions on the left are called the entrance channels and the decays on the right are called the exit channels. 14N* Since the decay of is independent of the formation, we can write the cross as the product of the cross secand the relative probability of 13C(p, n)13N c, section for a particular reaction such as tion for the formation of the compound nucleus, decay by neutron emission, Pn: p, n cPn 11-65 An illustration of the statistical decay of the compound nucleus is afforded by the energy distribution of neutrons from reactions such as (Figure 11-41) 209Bi \u00a1 208Bi n where to the ground state. shows a broad peak at 14 to 20 MeV and neutrons \u201cevaporate\u201d as 209Bi decays Excited States of Nuclei from Nuclear Reactions (E) The excited states of a nucleus can be determined in two ways from nuclear reacas a function of energy indicates an excited tions. A peak in the cross section state of the compound nucleus, corresponding to the relatively large probability of the incident particle giving up all its energy in the single event of exciting an allowed energy level. (Think of the Franck-Hertz experiment as an analogy.) Information about the lifetimes of the excited states of the compound nucleus is obtained by of these peaks, or resonances, and using the uncermeasuring the energy width 14N tainty principle -particle energy. The peaks by the reaction nucleus. The Q value for this reaction in this curve indicate energy levels in the M(10B)c2 M()c2 M(14N)c2 The Q value is the binding enis ergy of the incident particle in the compound nucleus, which is always of the order of 6 to 10 MeV; thus levels with energy less than 6 MeV cannot be reached in the compound nucleus. Figure 11-42 shows the cross section for formation of as a function of the 14N % U. 10B S 14N* 11.61 MeV. % The kinetic energy in the center-of-mass frame is related to the lab energy of the particle by ECM M M m Elab 10 14 Elab \u03c3 11-7 Nuclear Reactions 523 Figure 11-42 Cross section for the reaction 10B S 14N* versus resonances indicate energy levels in the compound nucleus 14N*. energy. The 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 E\u03b1, MeV The peak in Figure 11-42 at Elab (10>14) of energy E 11.61 12C 2H S 14N*. the reaction deuteron energy in the lab must be 1.63 MeV corresponds to an excited state in 14N (1.63) 12.77 MeV. The same level can be excited by For this case, the Q value is 10.26 MeV. Thus, the Ed a 14 12 b(12.77 10.26) 2.93 MeV p 14N S 14N* p A second way to determine the energy levels in a nucleus is to observe the energies of particles scattered inelastically. In this case, the energy levels of the target nucleus are determined. Figure 11-43 shows the energy spectrum of protons from the reaction using 6.92-MeV protons. (The horizontal scale in this figure is proportional to the momentum of the protons since this is what is measured experimentally.) The two peaks in the curve correspond to energy losses of 2.31 of 2.31 and 3.75 MeV. The excited and 3.75 MeV, which indicated energy levels in product nucleus decays from these states by emission. The method of inelastic scattering can", " determine energy levels of the target nucleus lying relatively close to the ground state, whereas the levels excited in the compound nucleus must be much higher because of the Q values for formation of the compound nucleus. 14N 14N(p,p\u2032)14N 14N*(3.75 MeV) 14N*(2.31 MeV) Momentum Figure 11-43 Spectrum of 14N, protons scattered from indicating energy levels in 14N. 524 Chapter 11 Nuclear Physics Neutrons are used to dope silicon with phosphorus more uniformly than the conventional diffusion method. Irradiating Si with neutrons produces the P dopant via the reaction and subsequent decay 31Si S 31P. doped with P this way can operate at higher power levels in rectifier applications than diffusion-doped silicon. 30Si S Silicon n Reactions with Neutrons Nuclear reactions involving neutrons are important for understanding the elemental analytical technique of neutron activation analysis and the operation of nuclear reactors. The most likely reaction with a nucleus for a neutron of more than about 1 MeV is scattering. However, even if the scattering is elastic, the neutron loses some energy to the nucleus because conservation of momentum requires that the nucleus recoil. If a neutron is scattered many times in a material, its energy decreases until it is of the order of the energy of thermal motion kT, where k is the Boltzmann constant and T is the absolute temperature. (At ordinary room temperatures, kT is about 0.025 eV.) The neutron is then equally likely to gain or lose energy from a nucleus when it is elastically scattered. A neutron with energy of the order of kT is called a thermal neutron. At low energies, a neutron is more likely to be captured, with the emission of a ray from the excited nucleus: n A ZM \u00a1 A1 ZM For example, n 107 47Ag \u00a1 108 47Ag ) 1>v, Since the binding energy of a neutron is of the order of 6 to 10 MeV and the kinetic energy of the neutron is negligible by comparison, the excitation energy of the compound nucleus is from 6 to 10 MeV, and rays of this energy are emitted. Figure 11-44 shows the neutron capture cross section for silver as a function of the energy of the neutron. Except for the resonances, the cross section (n, varies smoothly with energy, deis the speed of the neucreasing with increasing energy approximately as tron. This energy dependence can be understood as follows. Consider a neutron moving with speed near a nucleus of diameter 2R. The time it takes the neutron to pass the nucleus is Thus, the neutron capture cross section is proportional to the time spent by the neutron in the vicinity of the nucleus. The dashed line in Figure 11-44 independence.21 At the maximum of the large resonance, the value of dicates this the cross section is very large 5000 barns) compared with a value of only about 10 barns just past the resonance. Many elements show similar resonances in the neuis about tron capture cross sections. For example, the maximum cross section for 57,000 barns. Thus, is a strong absorber, which makes it very useful as a shield against low-energy neutrons.22 v 2R>v. where 113Cd 113Cd 1>v ( v 10,000 s n r a b, \u03c3 1000 100 10 1 0.01 0.1 10 100 1 En, eV Figure 11-44 Neutron capture cross section for Ag versus energy. The dashed-line extension would be expected if there were no resonances and the cross section were merely proportional to the time spent near the nucleus, i.e., proportional to The resonance widths of a few eV indicate states with lifetimes of the order of h>% 1016 s. 1>v. Questions 13. What is meant by the cross section for a nuclear reaction? Why is that term used to describe it? 11-7 Nuclear Reactions 525 14. Why is the neutron capture cross section (excluding resonances) proportional to 1>v? 15. What is meant by the Q value of a reaction? Why is the reaction threshold not equal to Q? 16. Why can\u2019t low-lying energy levels (1 to 2 MeV above the ground state) be studied using neutron capture? 91Zr(n,)92Zr EXAMPLE 11-18 Determination of Reaction Rates The cross section for the reaction is 900 millibarns for thermal neutrons. This reaction is produced in the so-called thermal column of a reactor where the flux of thermal neutrons is 6.5 1012 neutrons>cm2 # s. The sample of natural Zr is a circular foil 1.0 cm in diameter and 20.0 makes up 11.27 percent of natural Zr. Compute the rate of this reaction. thick. The density of Zr is 6.506 g cm3, and m", " 91Zr > SOLUTION First we need to compute the number of given by 91Zr atoms in the sample. This number is N(91Zr) Zr NAV MZr 0.1127 where the volume of the sample weight of Zr, MZr > 91.22 g mol. Thus, V 2.00 103 (>4)cm3 and the molecular N(91Zr) (6.02 1023 atoms>mol)a2.00 103 4 cm3b 6.506 g>cm3 91.22 gm>mol N(91Zr) 1.04 1019 atoms ) reactions per unit time per From the definition of the cross section given by Equation 11-62, the number of (n, nucleus is R I (900 103 barns 1024 cm2>barn) 6.5 1012 neutrons>cm2 # s R 5.85 1012 s1 per 91Zr nucleus 91Zr The rate r at which the reaction 91Zr(n, )92Zr proceeds is then r N(91Zr)R (1.04 1019 91Zr nuclei)(5.85 1012 s1 per 91Zr nucleus) r 6.08 107 s1 Remarks: This is a low reaction rate, given the high neutron flux. It is the result of the low neutron capture cross section of 91Zr and the other naturally occurring Zr isotopes. This is the principal reason why zirconium is used to enclose nuclear reactor fuel elements. 526 Chapter 11 Nuclear Physics 11-8 Fission and Fusion Two nuclear reactions, fission and fusion, are of particular importance. In the fission for example, the uranium nucleus is excited by the capture of a neutron and of splits into two nuclei, each with very roughly half of the original total mass. A typical fission reaction is 235U, 235U n \u00a1 92Kr 142Ba 2n 179.4 MeV 11-66 The Coulomb force of repulsion drives the fission fragments apart, giving them very large kinetic energies. As a result of collisions with other atoms, this energy eventually shows up as thermal energy. In fusion, two light nuclei such as those of deuterium plus and tritium a neutron). A typical reaction is fuse together to form a heavier nucleus (in this case and 4He 3H) (2H 2H 3H \u00a1 4He n 17.6 MeV 11-67 > (A 200) (M Zmp and very light nuclides Figure 11-45 shows a plot of the mass difference per nucleon Nmn)>A versus A in units of MeV c2. This curve is just the negative of the binding energy curve of Figure 11-10. From Figure 11-45 we see that the rest energy per particle for both very heavy nuclides is more than that for nuclides of intermediate mass. Thus, in both fission and fusion the total mass decreases and energy is released. Since for A 200 the rest energy is about 1 MeV per nucleon greater than for A 100, about 200 MeV is released in the fission of a heavy nucleus. The energy release in fusion depends on the particular reaction. For the reaction in Equation 11-67, 17.6 MeV is released. Although this is less than the energy released in a single fission, it is a greater amount of energy per unit mass, as Example 11-19 illustrates. In this section, we will look at some of the features of fission and fusion that are important for their application in reactors to generate electricity. (A 20) 2H 3H / 1 2H \u20132 \u20133 \u20134 \u20135 \u20136 \u20137 \u20138 7Li 4He 23Na 27Al 56Fe 98Mo 130Xe 208Pb 238U \u20139 0 20 40 60 80 100 140 60 180 200 220 240 120 A Figure 11-45 Plot of mass difference per nucleon (M Zmp Nmn) A in units of MeV c2 versus A. The rest energy per nucleon is smaller for intermediate-mass nuclei than for either very light or very heavy nuclei. > > EXAMPLE 11-19 Energy Release in Fission and Fusion Compare the energy release per unit mass in the fusion of deuterium and tritium (Equation 11-67) with that of a 235U typical fission reaction, such as that of given by Equation 11-66. 11-8 Fission and Fusion 527 235U SOLUTION (a) A quick approximate comparison can be made by noting that the energy difference per nucleon between and its fission products is about 1.0 MeV. In the 2H 3H, it is 17.6 MeV 5 nucleons 3.5, or about 3.5 times larger. fusion of Thus, the energy released per kilogram will also be about 3", ".5 times larger in the fusion reaction. (b) The mass differences per nucleon for and the two fission products in Equation 11-66 can be estimated from Figure 11-45. A more accurate calculation of the total binding energy can be made with the aid of Equation 11-12 and used to compute the total mass differences as follows: 235U > 235U: 7.6 MeV>c2 per nucleon \u00a1 1797.1 MeV>c2 per nucleus 92Kr: 8.7 MeV>c2 per nucleon \u00a1 800.9 MeV>c2 per nucleus 142Ba: 8.4 MeV>c2 per nucleon \u00a1 1189.5 MeV>c2 per nucleus 235U The difference between the mass of and the sum of masses for the fission products is 193.3 MeV c2. Thus, the energy release per fission event (with these particular products) is 193.3 MeV. The mass of (see Appendix A) is 235.043924 u 3.9030 1025 kg. Therefore, the energy release per kilogram in the fission of 235U 235U is > b a 193.3 MeV 235U 3.903 1025 kg 235U a b 4.95 1026 MeV>kg The energy release in the deuterium/tritium fusion reaction is 17.6 MeV Mt Md 17.6 MeV 8.353 1027 kg 2.11 1027 MeV>kg Thus, the fusion reaction releases about 4.3 times the energy kg released by the fission reaction. > Fission The fission of uranium was discovered in 1938 by O. Hahn and F. Strassmann,23 who found, by careful chemical analysis, that medium-mass elements (in particular, barium) were produced in the bombardment of uranium with neutrons. The discovery that several neutrons are emitted in the fission process led to speculation concerning the possibility of using these neutrons to cause further fissions, thereby producing a chain reaction. When nucleus undergoes fission about 85 percent of the time and emits gamma rays as it de-excites to the ground state about 15 percent of the time. The fission process is somewhat analogous to the oscillation of a liquid drop, as shown in Figure 11-46. If the oscillations are violent enough, the drop splits in two. Using the liquid-drop model, A. Bohr and J. Wheeler calculated the critical energy Ec needed by the nucleus to undergo 235U. ) fission. is the compound nucleus formed by the capture of a neutron by captures a thermal neutron, the resulting (236U 236U 235U 236U 235U 236U 236U 236U (a) n (b) (c) (d) (e) n n n Fission fragments 235U in an excited state. Figure 11-46 Schematic illustration of nuclear fission. (a) The absorption of a leads to neutron by 236U (b) (c) Oscillation deforms the excited (d) The oscillation of has become unstable. (e) The nucleus splits apart into two nuclei of medium mass and emits several neutrons that can produce fission in other nuclei. nucleus. 236U 236U 528 Chapter 11 Nuclear Physics Figure 11-47 The nucleus may exist instantaneously as two fragments as shown on the left; however, the Coulomb potential barrier prevents their fission. To overcome the barrier, energy equal to the critical energy must be provided. Separation Critical energy Total energy of fragments y g r e n E Nuclear potential Separation of the fragments 239U 238U 235U 238U 235U to form The critical energy is the magnitude of the Coulomb barrier seen by the fragments, as illustrated in Figure 11-47. For this nucleus, the critical energy is about 6.2 MeV, which is less than the 6.5 MeV of excitation energy produced when captures a therefore produces an excited state of the neutron. The capture of a neutron by 236U nucleus that has more than enough energy to break apart. On the other hand, the critical energy for the fission of the nucleus is 5.9 MeV. The capture of a neutron nucleus produces an excitation energy of only 5.2 MeV. Therefore, when a by a 239U, the excitation energy is not great thermal neutron is captured by 239U enough for fission to occur. In this case the excited nucleus de-excites by or emission. Nuclides that may fission upon capturing a slow neutron are called fissile. We noted earlier that all nuclei with Z 83 are radioactive. Among the possible decay modes of the very heavy nuclei (Z 90) is that of spontaneous fission. These nuclei may break apart into two nuclei even if left to themselves without absorbing a neutron. We can also understand spontaneous fission using the analogy of", " a liquid drop of positive charges. If the drop is not too large, surface tension can overcome the repulsive forces of the charges and hold the drop together. There is, however, a certain maximum size beyond which the drop will be unstable and will spontaneously break apart since the repulsive force is proportional to the number of protons, which is proportional to the volume, hence to R3, whereas the surface tension is proportional to the surface area and so increases only as R2. (See Section 11-2.) Spontaneous fission puts an upper limit on the size of a nucleus and therefore on the number of elements that are possible. It should be noted that the probability for spontaneous fission in naturally occurring nuclides is quite low compared with the other possible decay modes. for decay is 4.5 109 years, while that for sponFor example, the half-life of taneous fission is about 1016 years. The reason is that fission, like decay, is inhibited by the Coulomb barrier. Even though the process is energetically possible, the large positively charged fission fragments have a very low probability of tunneling through the Coulomb barrier part of the nuclear potential. 238U A fissioning nucleus can break into two medium-mass fragments in many different ways, as shown in Figure 11-48. Depending on the particular reaction, one, two, or three neutrons may be emitted. The average number of neutrons emitted in the thermal-neutron-induced fission of is about 2.4. Equation 11-66 is a typical fission reaction. The reason that several neutrons are emitted is that the fission fragments are typically neutron rich and far off the line of stability, as shown in Figure 11-49. 235U 10.0 1..10 0.01 70 80 90 100 110 120 Mass number 130 11-8 Fission and Fusion 529 235U. Symmetric Figure 11-48 Distribution of fission fragments from the thermal-neutroninduced fission of fission, in which the uranium nucleus splits into two nuclei of nearly equal mass, is much less probable than asymmetric fission, in which the fragments have unequal masses. Note the symmetry of the light and heavy lobes of the distribution, including the small variations in the tops of the peaks and the convex outer edges. [Data from G. J. Dilorio, Direct Physical Measurement of Mass Yields in Thermal Fission of Uranium235, Garland, New York, 1979.] 140 150 160 The fission fragments and their decay products that build up in reactors are the source of many radioisotopes used in medical diagnosis, treatment, and research. Important among these is 99Mo, the source of the most widely used radioisotope in nuclear medicine. 99Tc, As a result, neutrons are spontaneously emitted during fission and the fragments decay toward stability. The Coulomb force of repulsion drives the fission fragments apart with very large kinetic energies. This energy is transferred to other nearby atoms via collisions, eventually showing up as thermal energy of the surroundings. We have seen that about 200 MeV per nucleus is released in such a fission, a large amount of energy. By contrast, in the chemical combustion reaction, only about 4 eV is released per molecule of oxygen consumed. 236U Line of stability N = Z Fission fragments 160 120 N 80 40 0 0 40 80 Z 120 236U Figure 11-49 Fission of the left of the line of stability. As a result, the fission is accompanied by the prompt emission of one or more of the excess neutrons followed by reduce their neutron numbers. n) produces fragments that are neutron-rich and well to decay of the fission fragments to further (235U 530 Chapter 11 Nuclear Physics EXAMPLE 11-20 Kilowatt-hours from hours released in the fission of 1 g of fission. 235U 235U, Calculate the total energy in kilowattassuming that 200 MeV is released per SOLUTION Since 1 mol of 235U number of has a mass of 235 g and contains 235U nuclei in 1 g is NA 6.02 1023 nuclei, the N 6.02 1023 nuclei>mol 235 g>mol 2.56 1021 nuclei>g The energy released per gram is then 200 MeV nucleus 2.56 1021 nuclei 1 g 1.6 1019 J 1 eV 1 h 3600 s 1 kW 1000 J>s 2.28 104 kW # h>g Remark: This is approximately equal to the amount of electrical energy used by a typical U.S. household in 15 months. The discovery that several neutrons were emitted in the fission process led to speculation concerning the possibility of using these neutrons to initiate other fissions, thereby producing a chain reaction. On December 2, 1942, less than four years after Hahn and Strassmann\u2019s discovery of fission, a group", " led by Enrico Fermi produced the first self-sustaining chain reaction in a nuclear reactor that they had constructed at the University of Chicago.24 The application of both fission and fusion to the development of nuclear weapons has had a profound effect on our lives for more than 60 years. The peaceful application of these reactions to the development of energy resources may well have an even greater effect in the future, provided that satisfactory solutions are found to problems concerning safety, environmental protection, and the spread of nuclear weapons technology. Indeed, as world demand for energy increases, the diminishing finite reserves of fossil fuels will undoubtedly result in increasing use of nuclear reactors to provide the primary energy for the generation of electricity. The MORE section Nuclear Power is a comprehensive primer on fission reactors and closely related issues. More Nuclear fission reactors provided 6.0 percent of the energy consumed worldwide in 2005. Nuclear Power, on the home page at www. whfreeman.com/tiplermodernphysics5e, is a thorough review of existing and possible future types of fission reactors, the nuclear fuel cycle, reactor control, and safety issues. See also Equations 11-68 through 11-70 here, as well as Tables 11-5 and 11-6, Figures 11-50 through 11-54, and Examples 11-21 through 11-23. 11-8 Fission and Fusion 531 (a) (c) (a) A sketch of the world\u2019s first nuclear reactor, the CP-1 (for Chicago Pile number 1). Projecting from the near face next to the top of the ladder is one of the cadmium-plated rods used to control the chain reaction by absorbing neutrons. The cubical balloon surrounding the reactor, open on the near side, was to contain neutron-activated radioactive air. News of the reactor\u2019s successful test was transmitted by A. H. Compton, one of those present, to President Franklin Delano Roosevelt\u2019s adviser (and Harvard University president) J. B. Conant in a phone call thus: \u201cThe Italian navigator [i.e., Fermi] has landed in the New World,\u201d said Compton. \u201cHow were the natives?\u201d asked Conant. \u201cVery friendly,\u201d was Compton\u2019s reply. (b) The only photograph of CP-1 known to exist, taken during addition of the 19th layer of graphite. Alternate layers of graphite, containing uranium metal and/or uranium oxide, were separated by layers of solid-graphite blocks. Layer 18, almost covered, contained uranium oxide. (c) Enrico Fermi, leader of the group of scientists who succeeded in initiating the first man-made nuclear chain reaction, on December 2, 1942. [(a) and (b) American Institute of Physics, Emilio Segr\u00e8 Visual Archives; courtesy of Argonne National Laboratory, University of Chicago. (c) Courtesy of Argonne National Laboratory.] (b) Fusion The production of power from the fusion of light nuclei has the potential for future use because of the relative abundance of the fuel and the absence of some of the hazards presented by fission reactors. In fusion, two light nuclei such as deuterium and tritium (2H) fuse together to form a heavier nucleus. A typical fusion reaction is (3H) 2H 3H \u00a1 4He n 17.6 MeV 11-71 > As was shown in Example 11-19, the energy released in this fusion reaction is (17.6 MeV) (5 nucleons) 3.52 MeV per nucleon, or about 3.5 times as great as the 1 MeV per nucleon released in fission. The technology necessary to make fusion a practical source of energy has not yet been developed. We will consider the fusion reaction of Equation 11-71; other reactions present similar problems. 532 Chapter 11 Nuclear Physics 3H 2H and Because of the Coulomb repulsion between the nuclei, very large kinetic energies, of the order of 1 MeV, are needed to get the nuclei close enough together for the attractive nuclear forces to become effective and cause fusion. Such energies can be obtained in an accelerator, but since the scattering of one nucleus by the other is much more probable than fusion, the bombardment of one nucleus by the other in an accelerator requires the input of more energy than is recovered. Therefore, to obtain energy from fusion, the particles must be heated to a temperature great enough for the fusion reaction to occur as the result of random thermal collisions. Because a significant number of particles have kinetic energies greater than the mean kT and because some particles can tunnel through the Coulomb kinetic energy barrier, a temperature T corresponding to is adequate to ensure that a reasonable number of fusion reactions will occur if the density of particles is sufficiently high. The temperature corresponding to kT 10 keV is of the order of 108 K. Such temperatures occur in", " the interiors of stars, where such reactions are common. At these temperatures, a gas consists of positive ions and negative electrons called a plasma (see Chapter 10). One of the problems arising in attempts to produce controlled fusion reactions is that of confining the plasma long enough for the reactions to take place. In the interior of the Sun the plasma is confined by the enormous gravitational field of the Sun. In a laboratory on Earth, confinement is a difficult problem. kT 10 keV (3>2) The energy required to heat a plasma is proportional to the density of its ions n, whereas the fusion rate is proportional to n2, the square of the density (since the rate is the product of the Maxwell energy distribution and the fusion cross section, both of which are proportional to n). If is the confinement time, the output energy is thus proportional to n2 If the output energy is to exceed the input energy, we must have. C1n2 C2n where C1 and C2 are constants. In 1957, the British physicist J. D. Lawson evaluated these constants from estimates of the efficiencies of various hypothetical fusion reactors and derived the following relation between density and confinement time, known as Lawson\u2019s criterion: n 1020 s # particles>m3 11-72 If Lawson\u2019s criterion is met and the thermal energy of the ions is great enough (kT 10 keV), the energy released by a fusion reactor will just equal the energy input; that is, the reactor will just break even. For the reactor to be practical, much more energy must be released. Two schemes for achieving Lawson\u2019s criterion are currently under investigation. In one scheme, magnetic confinement, a magnetic field is used to confine the plasma.25 In the most common arrangement, first developed in Russia and called the tokamak, the plasma is confined in a large toroid. The magnetic field is a combination of the doughnut-shaped magnetic field due to the current in the windings of the toroid and the self-field due to the current of the circulating plasma. An international consortium of nations is currently constructing the International Thermonuclear Experimental Reactor (ITER) tokamak in France. Production of the first plasma is scheduled for 2016. The Chinese Experimental Advanced Superconducting Tokamak (EAST) which uses superconducting windings recognized as essential for continuous energy production, began operation late in 2006. The break-even point using magnetic confinement was achieved a few years ago, but we are still a long way from building a practical fusion reactor. Central solenoid Outer intercoil structure Toroidal field coil Poloidal field coil Machine gravity supports 11-8 Fission and Fusion 533 Blanket module Vacuum vessel Cryostat Port plug (IC heating) Divertor Torus cryopump Schematic of the ITER tokamak experimental fusion reactor. The toroidal field coils, encircling the 6.2-m maximum diameter doughnut-shaped tritium-deuterium plasma contained in the vacuum vessel, are designed to conduct current for 300 s up to, eventually, steady state. The design plasma current is 15 106 A, producing a magnetic field of 5.3 T. This field is the principal means of confining the deuterium-tritium plasma that circulates within the vacuum vessel. Sets of poloidal field coils, perpendicular to the toroidal coils, carry an oscillating current that generates a current through the confined plasma itself, heating it ohmically. Additional poloidal fields help stabilize the confined plasma. Design total fusion power is 15 MW. ITER\u2019s first plasma is expected to be produced in 2016. Follow the development of ITER at www.iter.org. [ITER Organization.] In a second scheme, called inertial confinement, a pellet of frozen-solid deuterium and tritium is bombarded from all sides by intense pulsed laser beams of energies of the order of 106 J lasting about 108 s. (Intense ion and electron beams are also used.) Computer simulation studies indicate the momentum absorbed by the hydrogen nuclei from the beams should compress the pellet to about 104 times its normal density and heat it to a temperature greater than 108 K. This should produce about 106 J of fusion energy in 1010 s, which is so brief that confinement is achieved by inertia alone. (See Figure 11-55a and b.) In theory, after this burst of fusion energy is radiated away from the site to be absorbed by a heat-transfer fluid, such as liquid lithium, another pellet is injected at the confluence of the beams and the process repeats. Because the break-even point has barely been achieved in magnetic confinement fusion and because the building of a fusion reactor involves many practical problems that have not yet been solved, for example, activation of the reactor walls, the availability of fusion to meet world energy needs is not expected for several decades. 534 (a) Chapter 11 Nuclear Physics", " Injector 3H 2H Still Water cooled by evaporation Reaction zone Neutrons P ure lithiu m Cool lithium Steam Steam turbine Electrical generator + \u2013 Condenser Cooling tower Pump 20\u00b0C Pump Laser, ion, or electron beam Hot lithium Lake or river (source of cool water) e+ ve ve e+ 2H 2H p p p p p p (c) \u03b3 3He 3He \u03b3 p p 4He (b) 6Li (slow neutron) or Figure 11-55 (a) Schematic diagram of a possible fusion reactor using inertial confinement 2H 3H S 4He n reaction. This reaction produces 17.6 MeV per fusion, and the and the 7Li neutron produced reacts with either (fast neutron) to produce the needed for the reaction. The latter reaction produces an additional slow neutron; thus, every two neutrons produced by fusion have the potential for generating three nuclei; that is, this system may also be a tritium breeder. (b) The Nova inertial confinement fusion reactor uses 10 powerful laser beams focused on a hydrogen-containing pellet 0.5 mm in diameter. The resulting fusion reaction, visible here as a tiny bright star, lasts 1010 s and releases 1013 neutrons. (c) The proton-proton reaction is the primary source of the Sun\u2019s energy. The neutrino produced in the initial reaction escapes from the core. The net energy produced per cycle is about 26.7 MeV. [(b) Courtesy of Lawrence Livermore National Laboratory, U.S. Department of Energy.] 3H 3H 11-8 Fission and Fusion 535 1H 1H S 2H e EXAMPLE 11-24 Fusion Temperature for The fusion of two protons requires that two particles be separated by no more than about 1014 m in order for the attractive force of the nuclear potential to overcome the repulsive force of the Coulomb potential. Compute (a) the minimum temperature of a hydrogen plasma that will enable a proton with the average energy of those in the plasma to overcome the Coulomb barrier and (b) the energy released in the fusion. SOLUTION (a) The height of the potential energy barrier seen by the protons is given by U 1 4 0 e2 r (9 109 N # m2>C2)(1.60 1019 C)2 3.0 1015 m U 7.68 1014 J 0.48 MeV In order to overcome this barrier, the average energy of the protons in the kT, must equal at least half this amount; that is, each of the two plasma, fusing protons must have 3.84 1014 (3>2) J. (3>2)kT 3.84 1014 J where k is Boltzmann\u2019s constant. Thus, T 2 3.84 1014 J 3 1.38 1023 J>K 1.9 109 K (b) The energy released, equal to the Q value of the fusion reaction, is Q [2m(1H) m(2H) 2me]c2 [2 1.007825 u 2.014102 u 0.001097 u]c2 0.000451 u # c2 931.5 MeV>u # c2 0.42 MeV 1H 1H where the atomic mass values are given in Appendix A. Thus, the energy release per fusion illustrated in Figure 11-55a is 17.6 MeV, which explains why the latter reaction is used in controlled fusion experiments. fusion is 0.42 MeV. That of the 2H 3H The Source of the Sun\u2019s Energy The present energy content of the Sun as calculated from thermodynamics would be radiated away in about 3 107 years. Since life has existed on Earth for approximately 100 times that long, we can conclude that the Sun has been radiating at close to its present rate for at least 3 109 years. Therefore, the Sun must have a supply of energy far larger than that represented by the hot plasma and the observed radiation field. The source of the Sun\u2019s energy is nuclear fusion. Current theory proposes that, as the young Sun contracted, its temperature rose. Eventually the temperature of the core reached about 1.5 107 K, which is high enough for the hydrogen nuclei (protons) in the plasma to have sufficient energy on the average (about 1 keV) to fuse into helium nuclei. This reaction, actually a chain of reactions, was first proposed by H. A. Bethe and is referred to as the proton-proton cycle. The first reaction in the chain is (see Example 11-24) 1H 1H \u00a1 2H e 0.42 MeV e 536 Chapter 11 Nuclear Physics The probability for this reaction is very low except for those protons in the highenergy tail of the Maxwell-Boltzmann", " distribution. This sets a limit on the rate at which the Sun can produce energy and thus ensures a long lifetime for the Sun and similar stars. This limit is sometimes called the \u201cbottleneck\u201d of the solar fusion cycle. Once (deuterium) is formed, the following reaction becomes very probable: 2H 2H 1H \u00a1 3He 5.49 MeV It is followed by: 3He 3He \u00a1 4He 2 1H 12.86 MeV This process by which hydrogen nuclei are \u201cburned\u201d to helium nuclei is shown 3He schematically in Figure 11-55c. There are other possible reactions for converting all of which have the same net Q value. Their rates, however, differ depending to on the composition and temperature of the interior. 4He, 4He The neutrinos produced in the proton-proton cycle escape from the core, providing our only means for direct observation of the Sun\u2019s interior. The measured value of the total power radiated by the Sun and the known total Q value of the proton-proton cycle allow a calculation of the total reaction rate. In addition, the alternative reactions have different neutrino energy spectra, thus providing a way of determining for the relative contributions of each reaction and gaining information about the composition and temperature of the core. However, the measured rate at which solar neutrinos arrive at Earth is less than half that predicted by theoretical calculations based on the standard solar model. This discrepancy is referred to as the solar-neutrino problem. Solving this problem was the focus of a recent major international research effort. Results from the Sudbury (Canada) and Super-Kamiokande (Japan) neutrino observatories show that neutrinos have a small mass and may transform from one type to another, leading to the observed discrepancy. (See Section 12-5.) Questions 17. Explain why water is more effective than lead in slowing down fast neutrons. 18. What happens to the neutrons produced in fission that do not produce another fission? 19. Why does fusion occur spontaneously in the Sun but not on Earth? More The Interaction of Particles and Matter is of central importance in understanding the biological effects of ionizing radiation, in the development and use of nuclear radiation detectors, and in protecting the environment from potential radiation hazards. This topic is discussed for charged particles, neutrons, and photons on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Equations 11-73 through 11-83 here, as well as Figures 11-56 through 11-61 and Example 11-25. 11-9 Applications 537 11-9 Applications Certainly among the most important of the applications of nuclear reactions and interactions have been those developed in the field of nuclear medicine, particularly in the area of diagnosis, but also including the treatment of cancer and certain other diseases. State-of-the-art detectors and computer-based data analysis have made critical contributions to these developments. Also important to a broad spectrum of disciplines ranging from art through chemistry and geology to zoology are the precision isotope-specific analytical techniques of accelerator mass spectrometry and neutron activation analysis. Anthropologists, archeologists, and geologists routinely rely on the decay properties of a number of radioisotopes to determine the age of artifacts and samples. Examples of these applications will be discussed briefly in this concluding section of the chapter. Neutron Activation Analysis This isotope-specific analytical method for elements is capable of very high sensitivity and accuracy. While some elements are more readily analyzed by activation analysis than others, it is particularly useful for the many elements that cannot be conveniently assayed by the more standard chemical methods of trace analysis. It has a wide range of applications from identifying trace pollutants in the environment through semiconductor processing and materials science to the analysis and authentication of works of art. 9Be and 241Am Potential oil- and gasbearing regions in exploratory oil wells are identified by lowering an intense source of neutrons 239Pu (usually a mixture of 9Be) and or into the well along with a gamma-ray detector. The neutrons produce gamma X (n, )Y rays via the surrounding rock. Analysis of the gamma-ray spectra identifies elements in the rock that are typical indicators of the presence of oil and natural gas. reactions in (a) (b) (c) An application of neutron activation analysis. Hidden layers in paintings are analyzed by bombarding the painting with neutrons and observing the radiative emissions from nuclei that have captured a neutron. Different elements used in the painting have different half-lives. (a) Van Dyck\u2019s painting Saint Rosalie Interceding for the Plague-Stricken of Palermo. The black-and-white images in (b) and (c) were formed using a special film sensitive to electrons emitted by the radioactively decaying elements. Image (b", "), taken a few hours after the neutron irradiation, reveals the presence of manganese, found in umber, a dark earth pigment used for the painting\u2019s base layer. (Blank areas show where modern repairs, free of manganese, have been made.) The image in (c) was taken four days later, after the umber emissions had died away and when phosphorus, found in charcoal and boneblack, was the main radiating element. Upside down is revealed a sketch of Van Dyck himself. The self-portrait, executed in charcoal, had been overpainted by the artist. (See Problem 11-83.) [(a) Courtesy of Metropolitan Museum of Art, New York City. (b) and (c) Courtesy of Paintings Conservation Department, Metropolitan Museum of Art, New York City.] 538 Chapter 11 Nuclear Physics The method consists of exposing the sample to be analyzed to a high flux of slow ZM(n, )A1 A A ZM ZM, neutrons. Isotope of the element of interest undergoes the reaction as described in Section 11-7, where can be identified by ZM its half-life and the energy of its beta- and gamma-ray emissions. The activity R(t) after the beginning of the neutron irradiation is given by is radioactive. ZM A1 A1 R(t) N(t) R0(1 et) 11-84 is the decay constant of ZM where isotope. R(t) is measured and R0 is computed from Equation 11-62 since R0 where N0 is the number of and R0 is the constant production rate of that N0R, nuclei in the sample. Thus, A ZM A1 R0 N0R N0 I 11-85 where tron flux in neutrons>s # cm2. is the cross section for the reaction Equation 11-84 can then be written A ZM(n, ) A1 ZM in cm2 and I is the neu- R(t) N0 I(1 et) 11-86 When the half-life is short enough, irradiation is usually continued to saturation, that R(t) R() N0 is, until for a few isotopes. The number of atoms of in the sample is, at the saturation activity, Table 11-7 gives saturation activities per I. g A ZM N0 R() I and the mass of A ZM in the sample is m(A ZM) N0W NA R()W I NA 11-87 where W is the atomic weight of the element and NA is Avogadro\u2019s number. Table 11-7 Selected saturation activities (I 1012 neutrons>min # cm2) R() Saturation activity decays>m # g A Z M 55Mn 63Cu 127I A1 Z M 56Mn 64Cu 128I 197Au 198Au 8.8 106 1.7 106 1.6 106 1.7 107 11-9 Applications 539 1010 neutrons>s # cm2 EXAMPLE 11-26 The \u201cGold\u201d Chain After buying a chain advertised as 10 percent pure gold, the suspicious purchaser irradiates one 25-mg link in a constant neutron for a time long enough for any gold activity to saturate. flux of with a detector She then measures the activity of the link to be whose efficiency is 12 percent. What is the percent by weight of gold in the link? ( 7.5 104 decays>s is 98.8 barns.) 197Au for SOLUTION Since the detector efficiency is 12 percent, the actual value of R() is R() 7.5 104 decays>s 0.12 6.3 105 decays>s From Equation 11-87 we can then compute m(197Au) (6.3 105 decays>s)(197 g>mol) (6.02 1023 atoms>mol)(98.8 1024 cm2)(1010 neutrons>s>cm2) m(197Au) 2.1 104 g The weight percent of gold in the link is then %Au a 2.1 104 g 25 103 g b 100 0.8% or less than 1>10 of the advertised amount. Nuclear Magnetic Resonance In Section 7-7, we saw that the energy levels of the atom were split in the presence of an external magnetic field (the Zeeman effect) because of the interaction of the atomic magnetic moment and the field. Since nuclei also have magnetic moments, the energy levels of a nucleus is also split in the presence of a magnetic field. We can readily understand this by considering the simplest case, the hydrogen atom, for which the nucleus is a single proton. given by The potential energy of a magnetic moment U # B in an external magnetic field B is 11-88 The potential energy is lowest when the magnetic", " moment is aligned with the field and highest when it is in the opposite direction. Since the spin quantum number of the proton is the proton\u2019s magnetic moment has two possible orientations in an external magnetic field: parallel to the field (spin up) or antiparallel to the field (spin down). The difference in energy of these two orientations (Figure 11-62) is 1>2, \u00a2E 2( z)pB 11-89 When hydrogen atoms are irradiated with photons of energy E, some of the nuclei are induced to make transitions from the lower state to the upper state by resonance absorption. These nuclei then decay back to the lower state, emitting photons of energy E. The frequency of the photons absorbed and emitted is found from hf \u00a2E 2( z)pB \u0394E = 2\u03bcB B \u03bc B \u03bc Figure 11-62 A proton has two energy states in the presence of a magnetic field, corresponding to whether the magnetic moment of the proton is aligned parallel or antiparallel to the field. 540 Chapter 11 Nuclear Physics In a magnetic field of 1 T, this energy is \u00a2E 2( z)pB 2(2.79 N)a 3.15 108 eV>T 1 N b(1 T) 1.76 107 eV and the frequency of the photons is f \u00a2E U 1.76 107 eV 4.14 1015 eV # s 4.25 107 Hz 42.5 MHz This frequency is in the radio band of the electromagnetic spectrum; hence the radiation is called RF (radio-frequency) radiation. The measurement of this resonance frequency for free protons can be used to determine the magnetic moment of the proton. When a hydrogen atom is in a molecule, the magnetic field at the proton is the sum of the external magnetic field and the local magnetic field due to the electrons and nuclei of the surrounding material. Since the resonance frequency is proportional to the total magnetic field seen by the proton, a measurement of this frequency can give information about the internal magnetic field seen in the molecule. This is called nuclear magnetic resonance. It is a sensitive tool for probing the internal magnetic structure of materials. Nuclear magnetic resonance is also used as an alternative to x rays or ultrasound for medical imaging, in which case it is called magnetic resonance imaging (MRI). A patient can be placed in a magnetic field (provided by superconducting magnets) that is constant in time but not in space. When the patient is irradiated by a broadband RF source, the resonance frequency of the absorbed and emitted RF photons is then dependent on the value of the magnetic field, which can be related to specific positions in the body of the patient. Since the energy of the photons is much less than the energy of molecular bonds and the intensity used is low enough so that it produces negligible heating, the RF photons produce little, if any, biological damage. Diagnosis with MRI requires no surgical procedure and is more sensitive than other methods in detecting tumors in soft tissue. Computer-Assisted Tomography Wilhelm Roentgen received the first physics Nobel Prize in 1901 for his discovery of x rays in 1895, an event that also marked the beginning of radiography, the use of radiation and particle beams to produce images that are otherwise inaccessible. For half a century x rays were the probing beam of medical imaging. Then in the late 1940s the introduction of radioisotopes into a patient\u2019s body made it possible for physicians to target particular organs and produce images that recorded their behavior, a technique now a part of the specialty of nuclear medicine. The isotopes used are typically and particles have ranges in biologirelatively short-lived gamma emitters since cal tissue that are too short to be useful. The detector normally employed is a collimated (to provide directional information) scintillation crystal viewed by a photomultiplier. (See Figure 11-63.) The image is then constructed by a computer from the output of the photomultiplier. Focus NaI(Tl) crystal 11-9 Applications 541 Figure 11-63 Schematic drawing of a scintillation crystal with a Pb collimator to define a focus, a gamma camera. As the detector is moved around the patient, the intensity of the gamma radiation yields information about the location and concentration of the source radioisotope in the body, which can be used by a computer to produce an image of the distribution. Actual gamma cameras incorporate collimators with hundreds or even thousands of tiny channels for the gamma rays to reach the crystal. Pb collimator/shield Just as with ordinary x-ray radiographs, the images formed by the gamma camera are two-dimensional projections of a three-dimensional distribution. Thus, radiographs provide no depth information, a very serious disadvantage. G. Hounsfield and A. Cormack solved this problem in 1972 with the invention of the computer-assisted tomography (CT or CAT) scanner.26 A", " fan-shaped x-ray beam collimated to a thickness of a few millimeters is rotated about the patient and the transmitted fan beam is recorded by an arc of detectors opposite the source, as illustrated in Figure 11-64. The measurements are then reconstructed into an image of a two-dimensional image (not a projection) of a transverse slice of the body\u2014a tomograph. By simultaneously making a series of two-dimensional projections with a gamma camera and combining the results with the CT scan, the distribution of the trace radioisotopes in two-dimensional transverse sections can be constructed. The combination system is called single-photon emission computer tomography, or SPECT. It had been recognized early on that the collimators that were essential to the operation of CT scanners and gamma cameras placed a serious restriction on their sensitivity. It was also recognized that the collimators could be eliminated and the sensitivity X-ray tube Collimator Detectors Figure 11-64 Sections of the patient\u2019s body transverse to the long axis are imaged by the CT scanner. The fan-shaped x-ray beam, a few millimeters thick, and the bank of detectors, typically proportional or wire counters, rotate about the long axis to produce each complete image. The patient is moved slowly along the axis while the scanner produces successive images, their sum constituting a full three-dimensional composite. 542 Chapter 11 Nuclear Physics (b) (a) 0.511-MeV photon 1 4 B A 3 2 Figure 11-65 (a) Nuclei emit positrons at A and B. The oppositely directed 0.511-MeV photons from each annihilation are detected by a pair of BGO crystal detectors in the annular ring around the subject (not shown). Electronic coincidence circuits establish the line along which each pair of photons traveled. (b) The pattern of coincidence measurements is used by a computer to construct an image of the distribution of the radioisotope in the plane of the detector ring. This sequence of PET scans shows the utilization of glucose in the brain, traced by 7 mCi of a positron emitter. The sequence begins in the upper left. [Courtesy of D. W. Townsend, Division of Nuclear Medicine, University Hospital of Geneva, Geneva, Switzerland.] significantly enhanced if the trace radioisotope employed was a positron emitter. The reason is that the positron is stopped within a few millimeters in the tissue and its subsequent annihilation results in two 0.511-MeV photons emitted in opposite directions. Detection of the photons by counters 180\u00b0 apart whose outputs are analyzed by a timeof-flight coincidence spectrometer yields a precise location for the decay. (See Figure 11-65.) However, this idea did not find its way into a useful diagnostic scanner until the mid-1980s because of the absence of detectors with good efficiency for the 0.511-MeV photons and small enough to localize the incident photons to within a millimeter or so. This problem was solved with the invention by C. Thompson and his co-workers of the bismuth germanate (BGO) crystal. Currently, nearly all commercial positron emission tomography (PET) scanners rely on detector rings made of BGO crystals, as illustrated in Figure 11-65a. A PET scan of brain activity made with BGO detectors is shown in Figure 11-65b. The availability of PET scans is limited to locations in the proximity of cyclotron facilities because most biologically useful positron emitters, those that readily participate in reactions in the body, are They have short half-lives of 20 min, 10 min, 2 min, and 110 min, respectively, and supplies must be regularly replenished by nuclear reactions. 11C, 13N, 15O, and 18F. Radioactive Dating Technetium (Z 43) does not occur in nature. Predicted by Moseley in 1914 and first produced by Segr\u00e8 in 1937, 99Tc is by far the most widely used radioisotope in nuclear medicine research and diagnosis. Its decay produces a 140-keV gamma ray easily detected by scintillation and germanium detectors. Radioactivity occurs in nature as a result of (1) decays within the three decay chains originating with long-lived emitters discussed in Section 11-4, (2) the existence of isolated long-lived primordial radioisotopes such as and (3) the production of isolated radioisotopes due to reactions between cosmic ray protons and neutrons and nuclei in the atmosphere. Each of these provides a means by which the age of materials, such as rocks and archeological artifacts, can be measured. As one might guess, the very long-lived isotopes, such as are used in determining the ages of \u201cold\u201d rocks, while shorter-lived isotopes are employed in determining the ages of \u201cyounger\u201d rocks, other inorganic materials, and archeological", " samples containing carbon, such as charcoal. 1.24 1010 y), 1.25 109 y), 232Th (t1>2 (t1>2 and 40K 40K 11-9 Applications 543 Table 11-8 Selected naturally occurring isolated radioactive nuclides Nuclide t1>2 (y) Abundance (%) Daughter 14C 40K 87Rb 147Sm 176Lu 187Re 5730 1.25 109 4.88 1010 1.06 1011 3.59 1010 4.30 1010 1.35 1010 0.0117 27.83 15.0 2.59 62.60 14N 40A 87Sr 143Nd 176Hf 187Os The general technique used in determining the age of a sample by radioactive dating is to measure the present abundance ratio of two isotopes, at least one of which is either radioactive or the stable end product of a radioactive decay, relative to the abundance ratio that is known (or assumed) to have existed at the time when the material was formed. Table 11-8 lists the present isotopic abundances of a few of the naturally occurring isolated radioisotopes used in dating. Dating An important example, used in dating archeological materials contain14C>12C. is continuously produced in the atmosphere by the reaction emitter that 14C ing carbon such as bone and charcoal, measures the abundance ratio Radioactive 14N(n, p)14C. decays back to 14C The neutrons are produced by cosmic rays. via the reaction is a 14N 14C with t1>2 5730 years.27 14C \u00a1 14N e 11-90 14C \u201cGroup of Stags\u201d from the Lascaux Caves in France. Prehistoric paintings such as -dated, the oldest this are found so far having been painted 33,000 to 38,000 B.C., 14C>12C depending on the ratio used for that period. [Art Resource.] 544 Chapter 11 Nuclear Physics 14C 14C The chemical behavior of atoms is the same as that of ordinary atoms. For nuclei combine with oxygen to form CO2 molecules. Since livexample, atoms with ing organisms continually exchange CO2 with the atmosphere, the ratio of in a living organism is the same as the equilibrium ratio in the atmosphere, which is presently about 1.35 1012. When an organism dies, from the it no longer absorbs 14C>12C atmosphere. The ratio in a dead sample continually decreases due to the 14C. A measurement of the decay rate per gram of carbon thus allows radioactive decay of the calculation of the time of death of the organism, as illustrated by Example 11-28. 14C 12C 14C to 12C per gram of carbon in a living organism, assuming the ratio Decay Rate in Living Organisms Calculate the decay rate of 14C>12C 14C EXAMPLE 11-27 14C 1.35 1012. The half-life of 14C is 5730 years. SOLUTION 1. Combining Equation 11-19 with Equation 11-22, the decay rate R can be written in terms of the half-life and the number of radioactive atoms N as 2. N is computed from the 14C>12C ratio by first computing the 12C number of e.g., in 1 g: in a unit mass, R dN dt N 0.693 t1>2 N N12C NA M 6.02 1023 atoms>mol 12 g>mol 5.02 1022 nuclei>g 3. The number N of 14C gram is then given by nuclei per N14C 1.35 1012 N12C (1.35 1012)(5.02 1022) 6.78 1010 nuclei>g 4. The decay rate is then (0.693)(6.78 1010 g1)(60 s>min) (5730 y)(3.16 107 s>y) R 15.6 decays>min # g Remarks: Thus, the decay rate for a living organism is 15.6 decays per minute per gram of carbon. EXAMPLE 11-28 Age of a Bone Fragment A bone fragment found in central Mexico was thought to be associated with the army of Hern\u00e1n Cort\u00e9s, who conquered the Aztecs in the early 1500s. The fragment contains 200 g of carbon and > has a -decay rate of 400 decays min. Could the sample have come from a person who died during the 16th century? SOLUTION First we obtain a rough estimate. If the bone were from a living organism, we would expect the decay rate to be 200 g 15.6 decays>min # g 3120 decays>min. 11-9 Applications 545 400>3120 is roughly Since the sample must have decayed for about 3 half-lives, or be about 3 5730 years old. To find", " the age more accurately, we note that after n half-lives, the decay rate has decreased by a factor of We therefore find n from (1>2)n. 3 (actually 1>8 1>2 1>7.8), or n b a 1 2 400 3120 2n 3120 400 n ln 2 ln 7.8 n ln 7.8 ln 2 7.8 2.96 2.96(5730 years) 16,980 years. Thus, the bone The age is therefore fragment is much older than 500 years and cannot be related to Cort\u00e9s\u2019s conquests. Instead, it places early humans in Mesoamerica at least 17,000 years ago. t nt1>2 14N 14C Note that the calculation in Example 11-28 assumes that the concentration in the atmosphere and the cosmic ray intensity 17,000 years ago were essentially the same as they are today. Actually, neither has remained unchanged over that period. Accurate measurements must include corrections for (1) the variations of Earth\u2019s magnetic field, which affects the cosmic ray intensity, and (2) the changing composition of the atmosphere, which depends on global geological and chemical activity and on the average temperature of the atmosphere. For example, current evidence suggests that just prior to 9000 years ago, the ratio was about 1.5 times as large as the current value. The ratio has also been significantly altered over the past century by the -free carbon to the atmosphere, and by atmoburning of fossil fuels, which adds spheric testing of hydrogen weapons, which added during the 1950s. Accelerator mass spectrometry, which was originally developed for just this purpose, makes possible determination of the ratio with sufficient accuracy to extend the dating back 50,000 years before the present with samples as small applicability of as a few milligrams. Calibration of the ratio for earlier periods requires cross-dating with other methods, such as U-Th dating. 14C>12C 14C>12C 14C 14C 14C Dating Ancient Rocks Starting with Equation 11-18, a useful relation can be derived for the age of a sample that initially contains N0 radioactive parent nuclei that decay to a stable daughter with a half-life Assuming there are no daughter nuclei present initially, after a time t has elapsed, there will be NP parent nuclei and ND daughter nuclei in the sample. From Equation 11-18, t1>2. t 1 ln a N0 NP b t1>2 ln 2 ln a b N0 NP Since NP ND N0 at any time, Equation 11-91 can be written as t t1>2 ln 2 ln a1 b ND NP where ND NP is the isotopic ratio at age t. > 11-91 11-92 546 Chapter 11 Nuclear Physics and the dual ratio 238U>206Pb, 87Rb>87Sr, 40K>40Ar, Several isotopic abundance ratios are used as \u201crock clocks\u201d for samples of geologic 238U>234U>230Th. age. These include These have been used to determine the age of Earth rocks, Moon rocks, meteorites, and, by inference, the solar system itself. The oldest rocks on Earth have been dated at about 4.5 109 years. At that time the molten surface froze, fixing the isotopic ratios, which thereafter changed only as a result of decay. Surprisingly, perhaps, all meteorites turn out to be about the same age, 4.5 109 years, regardless of their composition or when they collided with Earth. This suggests that they originated in or are the debris of other bodies within the solar system that formed at the same time as Earth. This value for the age of Earth is supported by a number of independent ratio measurements, initially the ratio and corroborated more rerelative abundances of and ratios. cently by measurements of the 238U>206Pb 87Rb>87Sr 235U 40K>40Ar and the and 238U EXAMPLE 11-29 to be 40.0. How old is the rock? 87Rb>87Sr Dating The 87Rb>87Sr ratio for a particular rock is found SOLUTION Note first that in Equation 11-92, the radioactive parent appears in the denominator 1 0.025. of the ratio; therefore, Substituting this value and the half-life of from Table 11-8 into Equation 11-92, we have > in this case ND NP 87Rb >(87Rb>87Sr) 1>40.0 t 4.88 1010 y ln 2 ln(1 0.025) 4.88 1010 y 0.693 0.0247 1.74 108 y This is a young rock, considerably younger than the 4.5 109 y age of Earth. Rocks found on Earth\u2019", "s surface have a range of ages from zero up to 3.7 109 years. None are older. In contrast, rocks brought back from the surface of the Moon by Apollo astronauts have ages ranging from 3.1 to 4.5 109 years; none are younger. The implications of these results from radioactive dating are (1) Earth surface rocks older than 3.7 109 years have weathered, eroded, and been recycled into other rocks or into the mantle and (2) the Moon\u2019s internal heat source (gravity and radioactivity) cooled sufficiently to solidify all its material and fix the initial isotopic ratios about 1.5 109 years after it was formed. The Earth\u2019s internal heat source has not yet reached that point. Accelerator Mass Spectrometry 14C Originally developed to extend the usable time span and improve the accuracy of dating of archeological materials, accelerator mass spectrometry (AMS) is an ultrasensitive analytical technique in which the atoms of interest in a sample are counted directly rather than irradiating the sample with slow neutrons, then counting the gamma rays emitted by the radioactive daughter produced or measuring the radiations emitted by long-lived, naturally occurring radionuclides. To understand how AMS works, we will use its application to dating as an illustration. At the present time ratio in living organic material is about 1012. Thus, a 1-g sample of carthe bon contains about 5 1010 14C atoms. Since the half-life of is 5730 y, a 1.0-g per minute. To record sample of 20,000-year-old charcoal would emit about one 10,000 decays (the number needed for a statistical accuracy of 1 percent) would require counting for one week and involve only 2 106 of the atoms present in the sample, a very inefficient method. 14C>12C 14C 14C 14C 11-9 Applications 547 Table 11-9 Radioisotopes measurable with AMS Nuclide Half-life (y) Stable isobar Sensitivity 3H 10Be 14C 26Al 36Cl 41Ca 129I 12.3 1.5 106 5.730 103 7.40 105 3.01 105 1.0 105 1.6 107 3He 10B 14N 26Mg 36S 41K 129Xe 1014 1015 2 1015 1015 2 1015 1015 1014 14C 14N and 13CH, 12CH2 14C>12C these include Mass spectrometry, which records every atom in the sample, provides a possible alternative (see Section 3-1). However, conventional mass spectrometers do not have the capability of measuring isotope ratios at the level of (or the other radioisotopes listed in Table 11-9) due to the presence of isobars and molecules with from residual air within the nearly the same mass. In the case of spectrometer and both from the sample itself or contamination. AMS works in part like a conventional mass spectrometer but reduces background due to mass ambiguities by taking advantage of the operational characteristics of mediumenergy accelerators, particularly cyclotrons and tandem van de Graaffs. Using the latter as the basis of our discussion and referring to the photograph and diagram on page 518 and to Figure 11-66, the positive high-voltage terminal is in the middle of the accelerator with the two ends of the beam tube essentially at ground potential. The atoms of the sample are converted to negative ions in the ion source. The atoms of most elements can form stable negative ions, a notable exception being nitrogen. Thus, AMS 14N. immediately removes background due to A bending magnet deflects the ions according to their radii of curvature. The negative ions are accelerated to the positive terminal, where a stripper removes several electrons, forming positive ions. If more than three electrons are removed, most molecules break apart. The ions are then accelerated further (to 50\u2013100 MeV), emerging from the machine into another bending magnet, which effectively removes the molecular fragments that in general do not have the same radii of curvature as do the atomic ions. After passing through another 90\u00b0 bending magnet that cleans the beam of any residual molecular fragments, the high-energy beam enters the detector, a so-called E-E counting telescope (see High-energy analyzer magnet Second analyzer magnet Beam \u0394E E Tandem Van de Graaff accelerator Low-energy bending magnet Stripper Particle detector Negative-ion source Figure 11-66 Schematic drawing of a tandem Van de Graaff accelerator configured as an accelerator mass spectrometer. 548 Chapter 11 Nuclear Physics Figure 11-66). The very thin E detector measures the energy loss by the atoms, which for particles with the same energy is approximately proportional to Z 2, thus rejecting atoms with a different atomic number than that of interest. The high-energy ions are then stopped in the E detector, which measures the energy of each one. The product", " E E for each atom is approximately proportional to mZ 2. Thus, requiring sample masses of only a few milligrams, AMS measures the mass and atomic number of each atom, and it does so with very high precision and extremely low background. The perfectly preserved mummy of \u00d6tzi the Iceman was found in the Tyrolean Alps in 1991. Accelerator mass spectrometry places his date of death between 3300 and 3200 years B.C. Recent (2003) measurements of oxygen isotopic ratios in his teeth and bones have pinpointed the area where he lived. [\u00a9 South Tyrol Museum of Archeology, Italy. www.iceman.it.] Table 11-9 lists several long-lived radioisotopes that can be effectively assayed with AMS. For example, the technique has been used to time the migration of surface water into deep aquifers by measuring the concentration of 36Cl, produced by cosmic ray bombardment of argon in the atmosphere. Using only a few -dated the famous Shroud of strands, AMS Turin as having been made in the Middle Ages, around 1300. \u00d6tzi (the Iceman) discovered in 1991 in the Tyrolean Alps, was found to have lived during the late Neolithic age, about 5200 years ago. Some meteorites have been found to contain relatively short-lived in excess of the concentration attributable to cosmic ray production, raising the intriguing question of its origin in the cosmos. 26Al 14C Particle-Induced X-ray Emission An elemental analysis technique similar to neutron activation analysis (NAA), particle-induced x-ray emission (PIXE) involves bombarding the material of interest with low-energy (a few MeV) ions, such as protons or alpha particles. Coulomb interaction between the ions and the target atoms ionize the latter by ejecting K- or L-shell electrons. Since the interactions occur over atomic dimensions, the cross sections are quite high, as much as 1000 b for low-Z atoms, decreasing smoothly to about 1 b at Z 82 (Pb). The vacancies produced are quickly filled by electrons from higher-energy shells, emitting K and L x rays or Auger electrons in the process that are characteristic of the elements in the target (see Section 4-4). Since the bombarding particles are relatively low energy, they do not penetrate far into matter, so the interactions occur near the surface. That fact, together with the low energy of the emitted x rays, 10 to 100 keV, dictates the use of thin samples. Figure 11-67a is a schematic of a typical PIXE experimental arrangement. The sensitivity of PIXE is comparable to NAA and has the advantage of being applicable to all elements above Z 20, whereas NAA is restricted to those nuclides with sufficiently large thermal neutron absorption cross sections. The main disadvantage of PIXE is x-ray energy ambiguities. For example, the energy of the x ray from Pb is 10.55 keV, while that of the line of As is 10.54 keV. The resolution of the cooled Si(Li) detectors used for x rays is about 100 eV, insufficient to resolve the two lines. Figure 11-67b shows a typical PIXE spectrum. K L 11-9 Applications 549 Figure 11-67 (a) Schematic drawing of a typical particle-induced x-ray emission system. (b) PIXE spectrum from an aerosol bombarded with 2-MeV protons. [S. A. E. Johansson and T. B. Johansson, Nuclear Instruments and Methods, 137, 473, 1976.] (a) Detector X rays Thin window (Be) Beam (b) 104 Sample Beam cup Si K 103 102 101 Cl Ca Fe Zn V Mn Cr Fe Pb Zn Ni Cu Pb Br Br Pb 0 100 200 300 400 500 Channel number Questions 20. If the 14C>12C ratio was 1.5 times larger than that used in Example 11-28, is the calculated age too large or too small? Explain. 21. Some meteorites are found to contain measurable amounts of t1>2 whose only 7.4 105 years. Devise a scenario that would account for its presence. 40Ar is a gas at ordinary temperatures. Explain why solid rocks can be accurately dated using the ratio in spite of that fact. 40K>40Ar 22. 26Al, is 23. Explain why accelerator mass spectrometry can achieve reliable results using samples of only 1 mg. More The biological effects of ionizing radiation were largely unknown in the early days of atomic and nuclear physics. It took such things as the plight of the radium-watch dial painters, x-ray crystallographers with missing fingertips, and young cyclotron physicists with cataracts to focus scientific attention on the risks that attend exposure to ionizing radiation in the home, the workplace, and the environment. Questions of", " Radiation Dosage, its definition, origin, and effects, are discussed on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Equations 11-93 through 11-95 here, as well as Tables 11-10 through 11-13. 550 Chapter 11 Nuclear Physics Summary TOPIC 1. The composition of the nuclei 2. Ground-state properties of nuclei Size and shape Binding energy and mass Magnetic moments RELEVANT EQUATIONS AND REMARKS Nuclei have Z protons, N neutrons, and mass number A Z N. Nuclei with the same Z but different N (and A) are called isotopes. The nucleons are particles and both have intrinsic magnetic moments. Fermi-Dirac (spin 1>2) The mean radius of the nuclear charge distribution is R (1.07 0.02)A1>3 fm 11-5 The radii thus vary from about 1 fm for the proton to about 10 fm for the heaviest nuclei. With few exceptions, nuclei are nearly spherical. The binding energy of the nucleus is given by ZMHc2 Nmnc2 MAc2 Bnuclear 11-11 The moments of the proton and neutron are ( ( p)z n)z 2.79285 N 1.91304 N eU>2mp N is the nuclear magneton. where 3. Radioactivity The decay rate R of radioactive nuclei is R N0et R0et dN dt is the decay constant. N0 and R0 are the number of nuclei present 11-19 where and the decay rate at t 0. Half-life Units t1>2 ln 2 0.693 where 1> is the mean life. 1 decay>s 1 becquerel 1 Bq 11-22 11-23 4. Alpha, beta, and gamma decay These are the three most common forms of radioactive decay. Alpha particles are nuclei, beta particles are electrons and positrons, and gamma rays are very short wavelength electromagnetic radiation. 4He 5. The nuclear force The nuclear force is 0 (a) About 102 stronger than the Coulomb force (b) Short-range ( (c) Charge independent (d) Saturated (e) Dependent on spin orientation beyond 3 fm) The nuclear force is considered to be an exchange force in which the attraction between a pair of nucleons is due to an exchange of virtual pions. The range R of the force, determined by the uncertainty principle, is R c \u00a2t cU>\u00a2E U>mc 11-50 where m is the mass of the virtual pion. TOPIC RELEVANT EQUATIONS AND REMARKS General References 551 6. The shell model 7. Nuclear reactions Cross section 8. Fission and fusion 9. Applications An independent particle model, similar to that used for assigning energy states to the atomic electrons, but one that makes use of a strong spin-orbit coupling for each nucleon accounts for the shell-like structure of the protons and neutrons. It explains the magic numbers 2, 8, 20, 28, 50, 82, and 126 in terms of the completion of the shells. Shell-model calculations are relatively successful in predicting nuclear spins and magnetic moments, particularly in the vicinity of closed shells. The Q value of a reaction X(x, y)Y determines if energy is released or must be supplied. Q is given by Q (mx my The cross section measures the effective size of a nucleus for a particular nuclear reaction. mY)c2 mX 11-58 R I 11-62 where R is the number of reactions per unit time per nucleus and I is the incident particle intensity. 239Pu Fission is the process by which heavy elements such as capture a neutron and split into two medium-mass nuclei. Each event releases about 1 MeV nucleon. 3H, Fusion is the reaction in which two light nuclei, such as fuse together to produce a heavier nucleus. Each event releases 1 to > 4 MeV nucleon. 235U and and 2H > The applications of nuclear reactions in medicine include the use of nuclear radiation in the treatment of diseases and the use of nuclear-based imaging techniques in diagnosis and research. Nuclear magnetic resonance imaging (MRI) is an alternative to x-ray imaging with the advantage that the RF photons involved produce little damage to biological tissue. Computerassisted tomography using short-lived positron emitters (PET) provides rapid, three-dimensional images. Radioactive dating employs a number of naturally occurring radioisotopes to determine the age of rocks and artifacts. Accelerator mass spectrometry and neutron activation analysis are highly sensitive means of measuring the concentration of particular isotopes of nearly every element in the periodic table. General References The following general references are written at a level appropriate for the readers of this book. Beyer, R. (ed.), Foundations of Nuclear Physics, Dover", ", New York, 1949. This paperback contains 13 original papers, eight in English\u2014by Anderson, Chadwick, Cockroft and Walton, Fermi, Lawrence and Livingston, Rutherford (two papers), and Yukawa\u2014the others are in German or French. Biological Effects of Ionizing Radiation (BEIR III), National Academy of Sciences/National Research Council, Washington, D.C., 1988. Das, A., and T. Ferbil, Introduction to Nuclear and Particle Physics, Wiley, New York, 1994. Fermi, E., Nuclear Physics, rev. ed., University of Chicago Press, Chicago, 1974. Frauenfelder, H., The M\u00f6ssbauer Effect, W. A. Benjamin, New York, 1962. 552 Chapter 11 Nuclear Physics Frauenfelder, H., and E. M. Henley, Subatomic Physics, 2d Priest, J., Energy: Principles, Problems, and Alternatives, 5th ed., Prentice-Hall, Englewood Cliffs, N.J., 1991. ed., Kendall/Hunt, Dubuque, Iowa, 2000. Krane, K. S., Introductory Nuclear Physics, Wiley, New York, Segr\u00e8, E., Nuclei and Particles, 2d ed., Benjamin Cummings, 1987. Lilley, J., Nuclear Physics: Principles and Applications, Wiley, Chichester, UK, 2001. Mayer, M., and J. H. D. Jensen, Elementary Theory of Nuclear Shell Structure, Wiley, New York, 1955. Nero, A. V., Jr., A Guidebook to Nuclear Reactors, University of California Press, Berkeley, 1979. Notes 1. Antoine Henri Becquerel (1852\u20131908), French physicist. He held the scientific post at the Museum of Natural History in Paris that had been held by his father and grandfather before him, and his research on the fluorescence of potassium uranyl sulfate was a continuation of work that his father had begun. His discovery of radioactivity, which revolutionized existing theories of atomic structure, earned him a share of the 1903 Nobel Prize in Physics, together with Marie and Pierre Curie. 2. The phenomenon was named radioactivity by Marie Curie in 1898. 3. These accomplishments were of such importance to the development of nuclear physics that all four men were subsequently awarded the Nobel Prize, James Chadwick in 1935, Carl Anderson in 1936 (shared with Victor Hess, the discoverer of cosmic rays), and John Cockcroft and Ernest Walton in 1951. 4. The United States produced about 30 percent of the world\u2019s nuclear-generated electric power in 2005. 5. The term tomography is from the Greek tomos, meaning \u201cslice,\u201d and graph\u00e9, meaning \u201cpicture.\u201d Thus, a tomograph is the pictorial representation of a slice through the object or body being studied. 6. Robert Hofstadter (1915\u20131990), American physicist. His electron-scattering measurements also revealed that the proton and neutron possessed internal structure, opening the way to a more fundamental understanding of the structure of matter. For his work he shared the 1961 Nobel Prize in Physics with Rudolf M\u00f6ssbauer. 7. See, for example, Section 23-2 in P. A. Tipler and G. Mosca, Physics for Scientists and Engineers, 6th ed., W. H. Freeman and Co., New York, 2008. 8. See, for example, Section 26-2 in P. A. Tipler and G. Mosca, Physics for Scientists and Engineers, 6th ed., W. H. Freeman and Co., New York, 2008. 9. See P. A. Seeger, Nuclear Physics, 25, 1 (1961). 10. The electric quadrupole moment of the nucleus, discussed earlier in this section, also causes hyperfine splitting, as do externally applied magnetic and electric fields. The effect of the reduced mass (isotope effect) mentioned in Chapter 4 is also considered a hyperfine effect. 11. Actually, the electron\u2019s magnetic moment deviates slightly from that predicted by the Dirac wave equation, one Bohr magneton. Quantum electrodynamics is able to account Menlo Park, Calif., 1977. Walecka, J. D., Theoretical Nuclear and Subnuclear Physics, 2d ed., Imperial College Press/World Scientific, London, 2004. Part 1 is appropriate reading for this chapter. for the small deviation observed experimentally with an error of less than 1 part in 108, one of the most remarkable agreements between quantum theory and experiment in physics. 12. This statement requires a small qualification. An alternadecay, discussed in Section 11-4, is electron captive to ture, in which an orbital electron may be captured by the nucleus. The probability of its occurrence depends on the probability density of the electrons, which can be affected slightly by very high external pressures", ". 13. Leptons include the electrons and neutrinos that are emitted in decay. (See Chapter 12.) 14. Rudolf Ludwig M\u00f6ssbauer (b. 1929), German physicist. His discovery of the recoilless emission and absorption of gamma rays, made while he was a graduate student in Munich, made possible the verification (by R. V. Pound and G. A. Rebka in 1960) of the gravitational red shift predicted by general relativity. M\u00f6ssbauer shared the 1961 Nobel Prize in Physics with Robert Hofstadter. 15. Note that this electrostatic potential corresponds to a force of nearly 60 N, or the weight of a 6-kg mass! It is acting not on 6 kg, however, but on only 16. Hideki Yukawa (1907\u20131981), Japanese physicist. His paper presenting the exchange meson theory of the nuclear force was his first publication. He was awarded the 1949 Nobel Prize in Physics for the discovery. 17. See, for example, Section 30-3 in P. A. Tipler and G. Mosca, Physics for Scientists and Engineers, 6th ed., W. H. Freeman and Co., New York, 2008. 18. Previously unknown particles had been observed in cosmic rays at about the same time that Yukawa proposed the meson exchange theory. He sent an article to the journal Nature in 1937 suggesting that they might be the mesons, but the journal rejected the article. Those particles were later found to be muons, a product of the decay of Yukawa\u2019s pi mesons. 19. Maria Goeppert-Mayer (1906\u20131972), German-American physicist, and Johannes Hans Daniel Jensen (1907\u20131973), German physicist. Goeppert-Mayer\u2019s antecedents for many generations had been university professors, while Jensen was the son of a gardener. They co-authored a famous (among physicists, at least) book explaining their nuclear-shell model and for that work shared the 1963 Nobel Prize in Physics with Eugene Wigner. 1.67 1027 kg. Problems 553 ) 20. In j-j coupling the spin and orbital angular momentum of each particle add to give a total angular momentum j for that particle, and then J equals the sum of the individual j vectors. In L-S coupling the spins of all the particles and the orbital angular momenta of all the particles add to yield total S and total L, which then add to yield J. 21. This dependence, which occurs only for (n, reactions with relatively low-energy neutrons, was first measured by Emilio Segr\u00e8 in 1935. 22. The first such resonance was observed unexpectedly in the results of a neutron irradiation of silver conducted by Edoardo Amaldi and others on the morning of October 22, 1934. By 3:00 p.m. that day, Enrico Fermi had developed the correct explanation of the strange phenomenon. The paper describing the discovery was written that evening and delivered to the scientific journal Ricerca Scientifica the next morning, less than 24 hours after the discovery! 23. Otto Hahn (1879\u20131968), German physical chemist, and Fritz Stassmann (1902\u20131980), German chemist. Hahn recognized that uranium nuclei bombarded with neutrons were breaking apart but carefully avoided characterizing the event as fission since no such thing had been recorded before. He received the 1944 Nobel Prize in Chemistry for the discovery. 24. Actually, Fermi\u2019s reactor was the first constructed fission reactor. About 2 billion years ago several deposits of natural uranium located in what is now Gabon, West Africa, began chain reactions that continued for 150 million years at an average power of 100 kW before naturally shutting themselves off. The evidence that verified the discovery of the first of these, a fascinating example of scientific detective work, can be found in G. A. Cowan, \u201cA Natural Fission Reactor,\u201d Scientific American, July 1976. The sites are being mined, and efforts to preserve one of the natural reactors as an international historic site are currently under way. 25. An elementary discussion of a magnetic bottle can be found in Section 26-2 in P. A. Tipler and G. Mosca, Physics for Scientists and Engineers, 6th ed., W. H. Freeman and Co., New York, 2008. 26. Godfrey Hounsfield (1919\u20132004), English engineer, and Allan Cormack (1924\u20131998), American physicist. They shared the 1979 Nobel Prize in Medicine for the invention of the CT scanner. 27. The radiocarbon-dating technique was developed by Willard F. Libby (1908\u20131980), an American chemist. He received the 1960 Nobel Prize in Chemistry for his work. Problems Level I Section 11-1 The Composition of the Nucleus and 222Rn. 18F, 25Na, 51V,", " 84Kr, 120Te, 148Dy, 175W, 11-1. What are the number of protons and the number of neutrons in each of the following isotopes? 11-2. Electrons emitted in decay have energies of the order of 1 MeV or smaller. Use this fact and the uncertainty principle to show that electrons cannot exist inside the nucleus. 11-3. The spin of the ground state of which constitutes 7.5 percent of natural lithium, is zero. Show that this value is not compatible with a model of the nucleus that consists of protons and electrons. 11-4. The magnetic moment of a model of the nucleus that consists of protons and electrons. 11-5. Suppose that the deuteron really did consist of two protons and one electron. (It doesn\u2019t!) Compute the spin and magnetic moment of such a deuteron\u2019s ground state and compare the results with the values in Table 11-1. Show that this value is not compatible with is 0.4035 6Li, 14N N. Section 11-2 Ground-State Properties of Nuclei (b) (b) 18F, 14O, 63Ni, 120Sn. 208Pb, and (c) and (c) 11-6. Give the symbols for at least two isotopes and two isotones of each of the following nuclides: (a) 11-7. Give the symbols for at least two isobars and one isotope of each of the following nuclides: (a) 11-8. Approximating the mass of a nucleus with mass number A as A u and using Equation 11-3, compute the nuclear density in SI units. 11-9. Use the masses in the table in Appendix A to compute the total binding energy and the binding energy per nucleon of the following nuclides: (a) and (c) 11-10. Use Equation 11-3 to compute the radii of the following nuclei: (a) (c) and (d) 238U. 57Fe. 16O, 197Au, 236Np. 56Fe, 9Be, 13C, (b) (b) 554 Chapter 11 Nuclear Physics 4He, 11-11. Find the energy needed to remove a neutron from (a) 23Na. 11-12. Use the Weizs\u00e4cker formula to compute the mass of ference between the result and the value in the table in Appendix A. 11-13. Compute the \u201ccharge distribution radius\u201d from Equation 11-5 and the \u201cnuclear force radius\u201d from Equation 11-7 for the following nuclides: (a) 11-14. to compute the radius of (b) Use Equations 11-1 and 11-2 (b) Compute the percent dif- are a mirror pair, decaying into 16O, 39K. and (c) and (c) 208Pb. 63Cu, 40Ca. 39Ca 39Ca 14N. and 7Li, 39K Section 11-3 Radioactivity 11-15. The counting rate from a radioactive source is 4000 counts per second at time t 0. After 10 s, the counting rate is 1000 counts per second. (a) What is the half-life? (b) What is the counting rate after 20 s? 11-16. A certain source gives 2000 counts per second at time t 0. Its half-life is 2 min. (a) What is the counting rate after 4 min? (b) After 6 min? (c) After 8 min? 11-17. A sample of a radioactive isotope is found to have an activity of 115.0 Bq immediately after it is pulled from the reactor that formed it. Its activity 2 h 15 min later is measured to be 85.2 Bq. (a) Calculate the decay constant and the half-life of the sample. (b) How many radioactive nuclei were there in the sample initially? 11-18. The half-life of radium is 1620 years. (a) Calculate the number of disintegrations per second of 1 g of radium and show that the disintegration rate is approximately 1 Ci. (b) Calculate the approximate energy of the particle in the decay assuming the energy of recoil of the Rn nucleus is negligible. (Use the mass table of Appendix A.) 11-19. The counting rate from a radioactive source is 8000 counts per second at time t 0. Ten minutes later the rate is 1000 counts per second. (a) What is the half-life? (b) What is the decay constant? (c) What was the counting rate after 1 minute? 11-20. The counting rate from a radioactive source is measured every minute. The resulting numbers of counts per second are 1000, 820, 673, 552, 453,", " 371, 305, 250,... (a) Plot the counting rate versus time and (b) use your graph to estimate the half-life. (c) What would be the approximate result of the next measurement after the 250 counts per second? 11-21. energy x-ray beam] and decays by take to produce 90 percent of the equilibrium value of 235U 11-22. The decay constant of many decays occur each second in a 1.0-g main in the 22Na 11-23. The decay constant of tivity of a sample containing 1.0 g of have passed? (d) How many is 9.8 1010 y1. (a) Compute the half-life. (b) How 1.0-g atoms will resample after 106 years? is 0.266 y1. (a) Compute the half-life. (b) What is the ac22Na? (c) What is the activity of the sample after 3.5 years n) reaction on placed in a highdecay with a half-life of about 10 min. How long does it is produced at a constant rate [e.g., by the ( atoms remain in the sample at the time? 226Ra S 222Rn, (c) How many sample of 62Cu? 235U? 62Cu 63Cu 22Na 235U, Section 11-4 Alpha, Beta, and Gamma Decay 23Na. 22Na and (b) What kind of radioactivity would you expect of 11-24. The stable isotope of sodium is 24Na? (a) 11-25. Using Figure 11-16, find the parameters A and B in Equation 11-30. 11-26. Make a diagram like Figure 11-18 for the (4n 1) decay chain that begins with a nuclide that is no longer present in nature. (Use Appendix A.) 232Th 11-27. Show that the particle emitted in the decay of percent, of the total decay energy. 11-28. characteristics of the decay be altered and, if so, how if (a) a sample of 7Be very high pressure or (b) all four electrons were stripped from each 11-29. Compute the energy carried by the neutrino in the electron capture decay of ground state of 11-30. Compute the maximum energy of the 11-31. In Example 11-13 we saw that decay by emission of a nucleon of either type is forbidden for this nuclide. decays exclusively by electron capture to particle emitted in the decay of could decay by emitting an particle. Show that with a half-life of 53.3 d. Would the were placed under 7Be atom in the sample? carries away 4.01 MeV, or 98 237Np, 233Np to the 72Zn. 67Zn. 67Ga 7Be 7Li 223Ra 11-32. With the aid of Figures 11-19 and 11-20, list the energies of all of the possible that may be emitted by 8Be 11-33. is unstable toward decay. why 80Br 11-34. case. (b) Compute the decay energy for each case. is very unusual among low-Z nuclides; it decays by emitting two particles. Show can undergo all three types of decay. (a) Write down the decay equation in each following the decay of 227Th. 8Be Problems 555 rays Section 11-5 The Nuclear Force 12C 11-35. Assuming that the average separation between two protons in is equal to the nuclear diameter, compute the Coulomb force of repulsion and the gravitational force of attraction between the protons. If the nuclear potential seen by the protons is 50 MeV for separations up to 3 fm, compare the nuclear force to the other two forces. 11-36. Suppose the range of the nuclear force was 5 fm. Compute the mass (in MeV c2) of an exchange particle that might mediate such a force. 11-37. The repulsive force that results in the \u201chard core\u201d of the nucleus might be due to the exchange of a particle, just as the strong attractive force is. Compute the mass of such an exchange particle if the range of the repulsive force equals about 0.25 fm, the radius of the core. > Section 11-6 The Shell Model 1d5>2 has the 11-38. The nuclei listed below have filled j shells plus or minus one nucleon. (For example, 29 2s5>2 shell.) 14Si Use the shell model to predict the orbital and total angular momentum of these nuclei: 31Ga 59 shell filled for both neutrons and protons, plus one neutron in the 27Co 73 32Ge 33 17Cl 71 14Si 37 16S 87 38Sr 29 and 14N 145Eu? exhibit a hyperfine structure, indicating that the ground", " 11-39. Use the shell model to predict the nuclear magnetic moments of the isotopes listed in Problem 11-38. 11-40. The atomic spectral lines of state is split into three closely spaced levels. What must be the spin of the 11-41. Which of the following nuclei have closed neutron shells: 82Ge, 88Sr, 93Ru, 94Ru, 131In, 11-42. Sketch diagrams like Figure 11-9 for the ground states of and 11-43. Which of the following nuclei have closed proton shells: 60Ni, 60Cu, 90Zr, 124Sn, 166Yb, and 11-44. (b) What value would you predict for the value of j? (c) What value would you predict for j for the first excited state? (d) Draw a diagram like Figure 11-9 for the first excited state. (Is there only one possible?) 11-45. Use Figure 11-35 to predict the values of j for the ground states of and (a) Use Figure 11-35 to draw a diagram like Figure 11-9 for 3He, 19F, 12C, 40Ca, 50Ti, 56Fe, 3H, 3He, 14N, 14C, 15N, 15O, 36S, 50V, 50Ca, 53Mn, 61Ni, 30Si, 37Cl, 55Co, 90Zr, ground state? 204Pb? 107In. 16O. 13N. 14N Section 11-7 Nuclear Reactions 1H 3H (b) 3He 4He, (d,p) and (c) nuclei are bombarded with 3H 1H S 3He n Q. 2H 2H S 3H 1H Q, 6Li n S 3H 4He Q. (a) Find the Q value for the reaction 11-46. Using data from Appendix A, find the Q values for the following reactions: (a) 11-47. for this reaction if stationary (c) Find the threshold for this reaction if stationary from an accelerator. 11-48. What is the compound nucleus for the reaction of deuterons on sible product nuclei and particles for this reaction? 11-49. Using data from Appendix A, compute the Q value for the reaction (a) and (b) 11-50. The cross section for the reaction ple of natural As in the form of a crystal 1 cm 2 cm that is 30 mal neutron flux of ceeds. (Natural arsenic is 100% is 4.5 b for thermal neutrons. A samthick is exposed to a therCompute the rate at which this reaction pro- (b) Find the threshold nuclei from an accelerator. nuclei 0.95 1013 neutrons>cm2 # s. 75As. Its density is 5.73 g>cm3.) nuclei are bombarded with What are the pos- 75As(n, )76As 16O(p,d)17O. 12C(, 14N? m 15N, 1H 3H p) 556 Chapter 11 Nuclear Physics and (c) p 14C, 11-51. Write three different reactions that could produce the products (a) n (b) 11-52. Write down the correct symbol for the particle or nuclide represented by the x in the 9Be(x, n)12C; following reactions: (a) (e) 208Pb (n, x)208Pb; (h) 14N(n, p) x; (g) (b) x(d, n)4He; 162Dy(, 6n)x; 90Zr(d, x)91Zr. 16O(d, )x; (, p)61Cu; d 31P. 23Na, (c) x (d) (f) Section 11-8 Fission and Fusion n p S d. 11-53. A few minutes after the Big Bang the first fusion reaction occurred in the early uniCompute the Q for this reaction. verse. It was 11-54. Assuming an average energy release of 200 MeV per fission, calculate the number of fissions per second needed for a 500-MW reactor. If the reproduction factor of a reactor is k 1.1, find the number of generations needed 11-55. for the power level to (a) double, (b) increase by a factor of 10, and (c) increase by a factor of 100. Find the time needed in each case if (d) there are no delayed neutrons, so the time between generations is 1 ms, and (e) there are delayed neutrons that make the average time between generations 100 ms. 11-56. Write down the several reactions possible when 1n, 2n, 3n, or 4n is produced. 11-57. Assuming an average energy release", " of 17.6 MeV fusion, calculate the rate at which 2H 11-58. From Figure 11-52, the cross section for the capture of 1.0-MeV neutrons by 238U is 0.02 b. A 5-gm sample of per m2. Compute the number of 11-59. Compute the total energy released in the following set of fusion reactions. This is the proton-proton cycle, the primary source of the sun\u2019s energy. 238U is exposed to a total flux of 1.0-MeV neutrons of 5.0 1011 239U must be supplied to a 500-MW fusion reactor. captures a thermal neutron and atoms produced. 235U > 1H 1H \u00a1 2H e e 2H 1H \u00a1 3He 3He 3He \u00a1 4He 21H 11-60. A particular nuclear power reactor operates at 1000 MWe (megawatts electric) with an overall efficiency in converting fission energy to electrical energy of 30 percent. What mass of 235U must fission in order for the power plant to operate for (a) one day, (b) one year? (c) If the 235U, what would be the answers to (a) and energy were provided by burning coal instead of 3.15 107 J>kg.) (b)? (Burning coal produces approximately (a) Assuming that the natural abundance of deuterium given in Appendix A is reflected 11-61. in the formation of water molecules, compute the energy that would be released if all the deuterons in 1.0 m3 of water were fused via the reaction (b) Given that the world\u2019s 5.9 109 people used 3.58 1020 J in 1999, how long (in hours) would the result in part (a) have lasted a \u201ctypical\u201d person? 11-62. Consider the possible fission reaction 92U \u00a1 120 2H 1H S 3He. 44Ru 3n 48Cd 112 n 235 (a) Compute the energy released in the reaction. (b) Is this reaction likely to occur? Explain. Section 11-9 Applications 14C be for this bone? 11-63. A bone claimed to be 10,000 years old contains 15 g of carbon. What should the decay rate of 11-64. A sample of animal bone unearthed at an archeological site is found to contain 175 g of in the sample is measured to be 8.1 Bq. How old is the bone? carbon, and the decay rate of 11-65. The ratio for a particular rock is measured to be 36.5. How old is the rock? In a PIXE experiment, an element with A 80 forms 0.001 percent by weight of a thin 11-66. foil whose mass is 0.35 mg cm2. The foil is bombarded with a 250 nA proton beam for 15 minutes. The cross section for exciting the L shell is 650 b. If the probability that the excited atom will emit an L x ray is 0.60 and the overall efficiency of the x-ray detector is 0.0035, how many counts will the detector record during the 15-minute bombardment? 87Rb>87Sr 14C > Problems 557 232Th 232Th 208Pb. Compute the age of the rock. and eventually ends on and 0.88 g (See Figure 11-18.) A particular rock is measured to contain 4.11 g of 11-67. The naturally occurring A 4n decay series begins with 208Pb. of 11-68. Compute the resonance frequency of free protons in a magnetic field of (a) 0.5 104 T (approximate strength of Earth\u2019s field), (b) 0.25 T, and (c) 0.5 T. 11-69. A small piece of papyrus is to be -dated using AMS. During a 10-minute run with the 3 ions, the beam 14C, system set to record ratio, assuming both isotopes are transmitted with the current is 12 same efficiency. (b) If the entire sample is consumed in 75 minutes, what was the mass of it contained? (Assume a constant consumption rate and an efficiency of 0.015. (c) How old is the sample? 14C 11-70. A wooden spear found in the mountains of southeastern Spain was found to have activity of 2.05 disintegrations per minute per gram. How old is it? (The activity of live wood is 15.6 disintegrations per minute per gram.) 1500 ions are counted. With the system set to transmit (a) Compute the 14C>12C A. 14C 12C 12C 14C and 65Cu, 114Nd 139La? g>cm2, 27Al, particle? (b) Express this range in using 1.29 103 ( 2.70", " g>cm3), is the same as that of aluminum has a mass of 0.05394 kg and emits an average of 2.36 particles Level II dM>dZ 0, 11-71. Using Equation 11-14 and the constants in Table 11-3, find the Z for which i.e., the minimum of curves like Figure 11-22a for (a) A 27, (b) A 65, and (c) A 139. Do these calculations give the correct stable isobars 11-72. An empirical expression for distance that particles can travel in air, called the range, is for E in MeV and 4 E 7 MeV. (a) What is the range in air of a 5-MeV R(cm) (0.31)E3>2 g>cm3 for air. (c) Assuming g>cm2 the range in find the range in aluminum in cm for a 5-MeV particle. 11-73. Show that the average electrostatic energy of a proton-proton pair is about 6ke2 5R, k 1>4 where R is the separation of the pair and 11-74. A sample of per second. Determine the decay constant and the half-life. > 11-75. A sample of radioactive material is found initially to have an activity of 115.0 decays minute. After 4 d 5 h, its activity is measured to be 73.5 decays minute. (a) Calculate the halflife of this material. (b) How long (after t 0) will it take for the sample to reach an activity of 10.0 decays minute? (c) How long after the time in (b) will it take for the activity to reach 2.5 > decays minute? 11-76. The half-life of an emitter whose at time t 0. 223Ra half-life is 11.43 days. A particular sample contains 106 atoms of (a) How many atoms of each type will be in the sample at t 15 days? (b) At what time will the number of atoms of each type be equal? 11-77. The M\u00f6ssbauer effect was discovered using the decay of the 0.12939-MeV second excited The lifetime of this isomer is 0.13 ns. (a) Compute the width state of of this level. atom that emits the 0.12939-MeV photon. (c) Resonant (b) Compute the recoil energy of a free (recoilless) absorption occurs when is bound into a lattice. If a Doppler shift equal to destroys the resonance absorption, show that the Doppler velocity necessary is given by is 18.72 days. It decays by emission to 223Ra, and no 227Th 227Th 191Ir. 191Ir 191Ir 0. % % v > > > v c% e 3H 3He and and 47Ca 48Ca. Let the protons be separated by the radius of the helium nucleus. 11-78. are a pair of mirror nuclei. Compute the difference in total binding energy between the two nuclides and compare the result to the electrostatic repulsion of the protons in 3He. 11-79. Use the masses in Appendix A to compute the energy necessary to separate a neutron From those results determine a value for a5 in the Weizs\u00e4cker formula from (Equation 11-14) and compare it with the value in Table 11-3. 11-80. The centripetal force of a nucleus with Figure 11-1a and a (classical) argument to show why this is the case. 141 from Equation 11-4. (b) Assume that after the 11-81. 56Ba the two nuclei are momentarily separated by a distance r equal fission of to the sum of the radii found in (a), and calculate the electrostatic potential energy for these two nuclei at this separation. Compare your result with the measured fission energy of 175 MeV. (a) Calculate the radii of 235U 141Ba 92Kr, makes it more stable toward decay. Use I 0 92 36Kr into and and 558 Chapter 11 Nuclear Physics 11-82. Consider a neutron of mass m moving with speed and colliding head-on with a nucleus of mass M. (a) Show that the speed of the center of mass in the lab frame is V mvL (m M). (b) What is the speed of the nucleus in the center-of-mass frame before the collision? After the collision? (c) What is the speed of the nucleus in the original lab frame after the collision? (d) Show that the energy of the nucleus after the collision is vL > 1 2 M(2V)2 c 4mM (m M)2 d 1 2 mv2 L )", " 31P 235U is 0.180 b and for 1012 neutrons>cm2 # s and use this to obtain Equation 11-82. 11-83. Suppose that the Van Dyck painting shown in the photographs on page 537 was irradiated with a thermal neutron flux of for 2 h. In terms of the numbers of manganese and phosphorus atoms initially present, determine the activity (a) 2 hours and (b) 2 days after the irradiation stopped. The (n, is 13.3 b. cross section for (Both isotopes are 100 percent of the naturally occurring elements.) 11-84. The total energy consumed in the United States in 1 y is about 7.0 1019 J. How many kilograms of would be needed to provide this amount of energy if we assume that 200 MeV of energy is released by each fissioning uranium nucleus, that 3 percent of the uranium atoms undergo fission, and that all of the energy-conversion mechanisms used are 25 percent efficient? 11-85. The rubidium isotope into early animals contain a ratio of when the rocks were formed, calculate the age of these fossils. 11-86. In 1989, researchers claimed to have achieved fusion in an electrochemical cell at room temperature. They claimed a power output of 4 W from deuterium fusion reactions in the palladium electrode of their apparatus. (a) If the two most likely reactions are emitter with a half-life of 4.9 1010 y that decays It is used to determine the age of rocks and fossils. Rocks containing the fossils of present of 0.010. Assuming that there were no 55Mn 87Rb 87Rb 87Sr. is a 87Sr 87Sr to and 2H 2H \u00a1 3He n 3.27 MeV 2H 2H \u00a1 3H 1H 4.03 MeV 1>10 with 50 percent of the reactions going by each branch, how many neutrons per second would we expect to be emitted in the generation of 4 W of power? (b) If th of these neutrons were absorbed by the body of an 80.0-kg worker near the device, and if each absorbed neutron carries an average energy of 0.5 MeV with an RBE of 4, to what radiation dose rate in rems per hour would this correspond? (c) How long would it take for a person to receive a total dose of 500 rems? (This is the dose that is usually lethal to half of those receiving it.) 11-87. Neutron activation analysis is used to study a small sample of automotive enamel found at the scene of a hit-and-run collision. The sample was exposed to a thermal neutron flux of 3.5 1012 neutrons>cm2 # s for 2.0 minutes. Placed immediately in a gamma-ray detector, it was Compute the total amount found to have an activity of 35 Bq due to 50Ti is 0.15 b.) of each metal in the original sample. (The cross section for 11-88. A fusion reactor using only deuterium for fuel would have the following two reactions taking place in it: and 115 Bq due to 59Co 51Ti. is 19 b; that for 60Co and 2H 2H \u00a1 3He n 3.27 MeV 2H 2H \u00a1 3H 1H 4.03 MeV The 3H produced in the second reaction reacts immediately with another 2H to produce 3H 2H \u00a1 4He n 17.7 MeV to 2H 1H atoms in naturally occurring hydrogen is The ratio of would be produced from 4 liters of water if all of the 11-89. (a) Using the Compton-scattering result that the maximum change in wavelength is \u00a2 2hc> Mc2 show that for a photon to lose an amount of energy Ep to a proton, the energy of the photon must be at least E [(1>2)Mc2Ep]1>2. 1.5 104. nuclei undergo fusion? and the approximation \u00a2E hc \u00a2>2, How much energy 2H Problems 559 14N (b) Calculate the photon energy needed to produce a 5.7-MeV proton by Compton scattering. (c) Calculate the energy given a nucleus in a head-on collision with a 5.7-MeV neutron. (d) Calculate the photon energy needed to give a nucleus this energy by Compton scattering. 11-90. A photon of energy E is incident on a deuteron at rest. In the center-of-mass reference frame, both the photon and the deuteron have momentum p. Prove that the approximation p E>c is good by showing that the deuteron with this momentum has energy much less than E. If the binding energy of the deuteron is 2.22 MeV, what is the", " threshold energy in the lab for photodisintegration? 14N (7Li, 7Be), (11B, 11C), Level III (a) Compute the binding-energy differences between the two nuclides of the mirror 11-91. (b) From each value computed in (a), determine a pairs value of the constant a3 in Equation 11-14. Compare each value and their average with the value given in Table 11-3. 11-92. (a) Differentiate the Weizs\u00e4cker empirical mass formula with respect to Z, as in Problem 11-71, and show that the minima of the constant A curves that result, i.e., Z values for the most stable isotopes, are given by (15N, 15O). and Z E A 2 (mn 1 1 mp )c2 4a4 a3A2>3 4a4 U is 11B 17O, 5>2. 1>2. (a) Use Figure 11-35 to make a diagram like Figure 11-9 for the ground state of (a) Determine the atomic number for the most stable nuclides for A 29, 59, 78, 119, and 140. (c) Compare the results in (b) with the data in Appendix A and discuss any differences. 11B. 11-93. What do you predict for the value of j for this state? (b) The first excited state of involves excitation of a proton. Draw the diagram for this state and predict its j value. (c) The j value Draw a diagram of the nucleons like Figure 11-9 that could for the second excited state is account for that value. (d) Repeat parts (a) and (b) for where the excitation of the first ex17O cited state involves a neutron. (e) The j value for the second excited state of Draw a diagram like Figure 11-9 that would explain that value. 11-94. Approximately 2000 nuclides remain to be discovered between the proton and neutron driplines in Figure 11-15b. Consider those that lie on the energy parabola (see Figure 11-22a) for A 151, whose only stable isotope is 151Eu. (a) From the data in Appendix A, draw an accurate diagram of the A 151 parabola showing known nuclides and those yet to be discovered between Z 50 and Z 71. (b) Determine where the edges of the driplines lie for A 151, i.e., the lowest mass isotopes for which spontaneous proton or neutron emission becomes possible. 11-95. There are theoretical reasons to expect that a cluster of relatively long-lived nuclides will exist in the neighborhood of the doubly magic nucleus with Z 126 and N 184, the latter being the next magic number beyond 126 predicted by the shell model. (a) Compute the mass of this exotic nucleus using Equation 11-14. (b) Computing the necessary masses of the nearby nuclei, predict the decay modes that would be available to the doubly magic nucleus. 11-96. Assume that a neutron decays into a proton plus an electron without the emission of a neutrino. The energy shared by the proton and electron is then 0.782 MeV. In the rest frame of the neutron, the total momentum is zero, so the momentum of the proton must be equal and opposite that of the electron. This determines the relative energies of the two particles, but because the electron is relativistic, the exact calculation of these relative energies is somewhat difficult. (a) Assume that the kinetic energy of the electron is 0.782 MeV and calculate the momentum p of the electron in units of MeV c. (Hint: Use Equation 2-32.) (b) From your result for (a), calof the proton. (c) Since the total energy of the electron plus culate the kinetic energy proton is 0.782 MeV, the calculation in (b) gives a correction to the assumption that the energy of the electron is 0.782 MeV. What percentage of 0.782 MeV is this correction? > p2>2mp 560 Chapter 11 Nuclear Physics 11-97. Radioactive nuclei with a decay constant of are produced in an accelerator at a con N. stant rate Rp. The number of radioactive nuclei N then obeys the equation dN dt Rp > (a) If N is zero at t 0, sketch N versus t for this situation. (b) The isotope 62Cu is produced at in a beam of high-energy photons. a rate of 100 per second by placing ordinary copper The reaction is (63Cu) 63Cu \u00a1 62Cu n 62Cu decays by how many nuclei are there? decay with a half-life of 10 minutes. After a time long enough so", " that 62Cu dN>dt 0, 11-98. The (4n 3) decay chain begins with (a) How many decays are there in the chain? (b) How many decays are there? (c) Compute the total energy released atom decays through the complete chain. (d) Assuming no energy escapes, when one determine the approximate temperature rise of 1 kg of 11-99. Energy is generated in the Sun and other stars by fusion. One of the fusion cycles, the proton-proton cycle, consists of the following reactions: metal over the period of 1 year. and ends on 207Pb. 235U 235U 235U followed by either or 1H 1H \u00a1 2H e 1H 2H \u00a1 3He 3He 3He \u00a1 4He 1H 1H 1H 3He \u00a1 4He e (a) Show that the net effect of these reactions is 41H \u00a1 4He 2 2 e (b) Show that the rest mass energy of 24.7 MeV is released in this cycle, not counting the 2 0.511 MeV released when each positron meets an electron and is annihilated according to (c) The Sun radiates energy at the rate of about 4 1026 W. Assuming that this e e S 2. is due to the conversion of four protons into helium plus rays and neutrinos, which releases 26.7 MeV, what is the rate of proton consumption in the Sun? How long will the Sun last if it continues to radiate at its present level? (Assume that protons constitute about half the total mass of the Sun, which is about 2 1030 kg.) 11-100. The fusion reaction between and 2H 3H is 3H 2H \u00a1 4He n 17.7 MeV 235U (a) A particular light-water Using the conservation of momentum and the given Q value, find the final energies of both 4He the nucleus and the neutron, assuming that the initial momentum of the system is zero. 11-101. -fueled reactor had a reproduction factor of 1.005 and an average neutron lifetime of 0.08 s. By what percentage will the rate of energy production by the reactor increase in 5 s? (b) By what fraction must the neutron flux in the reactor be reduced in order to reduce the reproduction factor to 1.000? 11-102. Compute the reproduction factor for uranium enriched to (a) 5 percent and (b) 95 percent in Compute the corresponding fission rate doubling time in each case. Assuming no loss of neutrons and the release of 200 MeV fission, at what rate will energy be produced in each case 1.0 s after the first fission occurs? 235U. > CHAPTER 12 Particle Physics Notwithstanding the speculations of the ancient Greek natural philosopher Democritus (about 450 BC) and Dalton\u2019s atomic theory of matter (1808),1 the story of particle physics really began with the discovery of the electron by Thomson in 1897 (see Section 3-1). That event was followed in 1913 by Rutherford\u2019s discovery of the atomic nucleus whose lightest example, that of hydrogen, he named the proton (see Section 4-2). As one moved upward through the periodic table of the elements, a dilemma arose, caused by the more-rapid increase of the atomic mass compared to that of the nuclear charge, even though both were presumably due to the protons bound together in the nucleus. That problem was solved in 1932 by Chadwick\u2019s discovery of the neutron (see Section 11-1). In the meantime Einstein had proposed (in 1905) that Planck\u2019s quantization of blackbody radiation was in fact a quite general property of the electromagnetic field (see Sections 3-2 and 3-3). Einstein\u2019s suggestion was not widely accepted until, over the next 20 years, Millikan\u2019s thorough experimental investigation of the photoelectric effect and Compton\u2019s discovery and explanation of the Compton effect provided incontrovertible evidence for the quantization of electromagnetic radiation, the field quantum being a particle we now call the photon. For a brief time, it was thought these four were the \u201celementary\u201d particles from which all matter was formed. But then Anderson discovered the positron, or antielectron, later in 1932. Shortly thereafter, the muon, pion, and many other particles were discovered in searches that have intensified and continued down to the present. During the past 50 years several nations and international consortia have constructed increasingly larger and more sophisticated particle accelerators capable of producing greater and greater energies with the goals of testing the predictions of current theories and searching for additional particles predicted by them. Initially, an important consideration in such complex experiments, which often involve hundreds of scientists from many nations, was the question of how to tell if a particle is truly elementary or composed of a combination of other", " particles. For example, both the proton and the neutron were once thought to be elementary, but probing with high-energy (short-wavelength) electron beams revealed that the nucleons have internal structure, just as do atoms and nuclei. Each of the nucleons was found to be a composite particle consisting of three, still more fundamental particles called quarks. Several hundred particles have at one time or another been considered to be elementary, but a series of 12-1 Basic Concepts 562 12-2 Fundamental Interactions and the Force Carriers 12-3 Conservation Laws and Symmetries 12-4 The Standard Model 12-5 Beyond the Standard Model 570 580 591 605 561 562 Chapter 12 Particle Physics The construction of large particle accelerators in various countries has, over the years, been an impetus for developing bigger and better superconducting electromagnets. The LHC at CERN has 1232 large, superconducting dipole magnets. High-field, efficient superconducting magnets are used in applications ranging from medical diagnostic magnetic resonance imaging (MRI) systems to magnetically levitated (maglev) trains. brilliant theoretical achievements over that same 50-year period vastly expanded our understanding of the \u201cparticle zoo.\u201d The culmination of these achievements is the Standard Model, which has been spectacularly successful in explaining and predicting the properties and interactions of particles by describing them in terms of a relatively small number of truly (for now, at least) elementary particles. Research at universities and at the giant accelerator laboratories around the world continues to strengthen our understanding of the structure of matter. In addition to the usual particle properties of mass, charge, and spin, research has unveiled new properties that have no classical analogs, some given whimsical names such as strangeness, charm, and color. Coincident with the construction of the large accelerators has been the development and deployment of larger and more sensitive particle detectors at the big machines and, for neutrinos, detectors deep underground, in the oceans, and buried in the polar ice cap. In this chapter, we will first look at a few basic concepts that will enable us to classify and describe particles. We will then consider the fundamental interactions between particles and the conservation laws that apply to them. Central to our discussions will be the current theory of elementary particles, the Standard Model, in which all matter in nature\u2014from the most exotic particles produced in the giant accelerator laboratories to ordinary grains of sand\u2014is constructed from just three groups of elementary particles: leptons, quarks, and the particles that mediate interactions between them. 12-1 Basic Concepts Antiparticles The Positron In the same year that the neutron was discovered, the positron was discovered (and named) by Carl Anderson.2 This particle has the same mass and intrinsic angular momentum as the electron but has positive charge; therefore, its intrinsic magnetic moment is parallel, rather than antiparallel, to its spin. It is the antiparticle of the electron and is represented by the symbol e, or sometimes in radioactive decay equations by The existence of the positron had been predicted by Dirac from his relativistic wave equation,3 though there was some difficulty about the interpretation of this prediction. (See Section 2-4.). (a) (b) (c) (d) (e) The Relativistic Heavy Ion Collider (RHIC) at Brookhaven National Laboratory began colliding gold nuclei (fully ionized gold atoms) late in 2000, with each of the ions moving at 99.99 percent of the speed of light. (a) through (d) are simulations of the accelerating Au nuclei at several stages. (a) Two Lorentz-contracted ions approach each other. (b) The collision \u201cmelts\u201d the protons and neutrons and (c) for an instant releases the quarks and gluons from which the nucleons were formed. (d) From the enormous energy of the collision thousands more are created, creating in turn thousands of particles. (e) Computer construction of the tracks of the thousands of particles created in a single collision of two gold ions. [Courtesy Brookhaven National Laboratory, STAR experiment.] 12-1 Basic Concepts 563 Air view of the European Laboratory for Particle Physics (CERN) just outside of Geneva, Switzerland. The large circle shows the Large Hadron Collider (LHC) tunnel, which is 27 kilometers in circumference. The irregular dashed line is the border between France and Switzerland (in the foreground). The LHC occupies the tunnel formerly used by the Large ElectronPositron (LEP) collider, which was retired in 2000. The LHC began operations in 2008. [CERN.] The energy of a relativistic particle is given by Equation 2-31: from which we can write E2 (pc)2 (mc2)2 E [(pc)2 (mc2)2]1>2 2-31 12-1", " Though we can usually choose the plus sign and ignore the negative-energy solution with a \u201cphysical argument,\u201d the mathematics of the Dirac equation requires the existence of wave functions corresponding to these negative-energy states. Dirac postulated that all the negative-energy states were filled with electrons. Electrons in the negative-energy states would exert no net force on anything and thus would not be observable. Dirac invoked the exclusion principle to suggest that only holes in this \u201cinfinite sea\u201d of negative-energy states would be observable. The holes would act as positive charges with positive energy. Anderson\u2019s discovery of a particle with mass identical to that of the electron but with positive charge seemed to indicate that this interpretation was reasonable, since the positron is produced simultaneously with an electron in pair production (see Figure 12-1). Antiparticles The notion that we are immersed in an infinite sea of negative-energy electrons is an unsettling one, however. It was rendered unnecessary with the development of quantum electrodynamics (QED) by Feynman4 and others in the late 1940s. The negative energy solutions of the Dirac equation were re-expressed as positive energy solutions of a new particle\u2014the positron. And the need for the invisible \u201csea\u201d of electrons with its mysterious \u201choles\u201d vanished. However, Dirac\u2019s prediction of an antielectron turned out to be farsighted. QED, whose predictions have been verified to the highest precision of any physical theory, requires that every particle must have a corresponding antiparticle with the same mass but opposite electric charge. For example, particles the theory predicts that protons and neutrons, which are both spinwhose wave functions are solutions of the Dirac equation, should have antiparticles. 1>2 564 Chapter 12 Particle Physics \u03b3 \u03b3 \u03b3 e+ e\u2013 e+ e\u2013 e+ e\u2013 Figure 12-1 Tracks of electron-positron pairs produced by 300-MeV synchrotron x rays at the Lawrence Livermore Laboratory. The magnetic field in the chamber points out of the page. [Photo courtesy of Lawrence Radiation Laboratory, University of California, Berkeley.] The creation of a proton-antiproton pair requires at least 2mpc2 1877 MeV, which was not available except in cosmic rays until the development of high-energy accelerators in the 1950s. The antiproton (designated ) was discovered by Segr\u00e85 and Chamberlain at Berkeley in 1955 using a beam of protons with kinetic energy 6.2-GeV from the Bevatron particle accelerator. (See Figure 12-2.) The antineutron ( ) (a particle with the same mass as the neutron but with a positive magnetic moment) was discovered two years later. (The standard notation for an antiparticle is the overbar; however, in many cases it is customary to specify the charge instead, as we did for the positron.) p n Particles with integral spin, whose wave functions are not solutions of the Dirac equation, also have antiparticles. For example, those with zero spin, which are described by the Klein-Gordon relativistic wave equation (see Equation 11-52), include the pions, thought in the early days (circa 1940) to be the mediating exchange particle, or force carrier, of the nuclear force. In general, an antiparticle has exactly the same mass as the particle but with electric charge, baryon number, and strangeness (see Section 12-4) opposite in sign to that of the particle. p \u2013 p p p e+ e\u2013 Figure 12-2 Bubble chamber tracks showing creation of proton-antiproton pair in the collision of an incident 25-GeV proton from the Brookhaven Alternating Gradient Synchrotron with a liquid hydrogen nucleus (stationary proton). The reaction is The energy necessary to create the pair is 2mpc2 1.877 GeV mass system. A relativistic calculation in the laboratory frame shows that the beam protons must have at least 6mpc2 5.6 GeV to reach the reaction threshold. [Photo courtesy of R. Ehrlich.] p p S p p p p. in the center of 12-1 Basic Concepts 565 The tunnel of the protonantiproton collider at CERN. The same bending magnets and focusing can be used for protons or antiprotons moving in opposite directions. The rectangular box in the foreground is a focusing magnet; the next four boxes are bending magnets. [CERN.] Although the positron is stable, it has only a short-term existence in our universe because of the large supply of electrons in matter. The fate of the positron is annihilation according to the reaction e e \u00a1 or e e \u00a1 12-2 1S 3S Whether bound (as positronium\u2014see Section 2-4", ") or unbound, annihilation occurs from S states (zero orbital angular momentum), the antiparallel spins state producing two state producing three phoquanta as on the left in Equation 12-2, the parallel spins tons. The fact that we call electrons particles and positrons antiparticles does not imply that positrons are less fundamental than electrons, but was initially merely an arbitrary choice reflecting the nature of our part of the universe. If our matter were made up of negative protons, positive electrons, and neutrons with positive magnetic moments, then particles such as positive protons, negative electrons, and neutrons with negative magnetic moments would suffer quick annihilation and would probably be called the antiparticles. Antihydrogen atoms (an antiproton and a positron) were first produced \u201chot\u201d in the antiproton beam at the European Center for Nuclear Research (CERN) in 1995. Subsequently, the CERN ATHENA project has produced substantial amounts of \u201ccold\u201d (slow) antihydrogen and is conducting definitive comparisons of its physical properties with those of hydrogen. The cover of this book illustrates the annihilation of an antihydrogen atom recorded by the ATHENA project. The matter-antimatter asymmetry of the universe, that is, why our universe consists of matter with essentially no antimatter despite the prediction of QED and the symmetry of the relativistic wave equation, is a question we will return to later in this chapter and in Chapter 13. EXAMPLE 12-1 Proton-Antiproton Annihilation A proton and an antiproton at rest annihilate according to the reaction (standard particle physics notation typically omits the signs in reaction equations): Find the energies and wavelengths of the photons. pp \u00a1 SOLUTION Since the proton and the antiproton are at rest, conservation of momentum requires that the two photons created in their annihilation have equal and opposite momenta 566 Chapter 12 Particle Physics and therefore equal energies. Since the total energy on the left side of the reaction is 2mpc2, the energy of each photon is The wavelength is hc hf c f E mpc2 938 MeV 1240 eV # nm 9.38 108 eV hc E 1.32 1015m 1.32 fm Feynman Diagrams As a part of quantum electrodynamics Feynman developed a wonderfully clear yet powerful technique for describing all electromagnetic phenomena. Like QED itself, the technique of Feynman diagrams is so good that it is used as a model by other quantum field theories, notably quantum chromodynamics (QCD), which we will discuss in Section 12-4. The detailed rules for drawing Feynman diagrams are directly related to the equations of QED and are beyond the scope of our discussions here; however, a brief description of a simplified version of the diagrams and a few basic rules will be ample for our use in illustrating the phenomena of interest in this chapter. (For a more complete discussion of Feynman diagrams refer to D. J. Griffith, Chapter 2, cited in the General References section.) Feynman diagrams are spacetime diagrams, that is, ct versus x graphs, similar to those developed and used in Chapters 1 and 2. In particle physics Feynman diagrams are used to describe interactions at the level of quarks, leptons, and the mediators of the interactions and to compute lifetimes and cross sections for events. As noted in Figure 11-28, where a Feynman-like diagram was used to illustrate the early view of the meson as the mediator of the nuclear force, the ct and x axes are normally not drawn. In this chapter, as in the earlier relativity chapters, time (ct) is positive upward. (Particle physicists often draw the diagrams with time flowing horizontally toward the right; there is no convention.) Particles are represented by straight lines with an arrow. A particle line whose arrow points backward in time is interpreted as the corresponding antiparticle moving forward in time. The arrows allow us to omit the overbars in the diagrams. The lines are symbolic and do not represent the particle trajectories. The rules for analyzing the diagrams, the details of which are beyond the scope of our discussions, force conservation of energy and momentum at each vertex. It is the interactions that we are interested in describing. Particles that are their own antiparticles, like the photon, have no arrows and are represented by wiggly or broken lines of various sorts. All electromagnetic phenomena can be represented by combinations of the process illustrated in Figure 12-3, called the primitive vertex. Interactions occur at the vertices. This diagram is read as follows: a moving charged particle enters, emits (or absorbs) a photon, and leaves. The primitive vertex is not itself a complete Feynman diagram, but rather the basic unit from which complete diagrams are constructed. Let\u2019s examine the Feynman diagrams for a few familiar events. In", " Figure 12-4a, two electrons enter, exchange a photon, and then leave. That is Coulomb repulsion of like charges.6 Figure 12-4b represents Coulomb attraction of opposite charges. These serve to illustrate one more rule: Particle lines that both begin and end within the diagram are virtual particles, that is, like Yukawa\u2019s exchange pion in Section 11-5, they are not, indeed, cannot be observed in the laboratory. Note that a virtual particle need not have the same mass as the corresponding real particle; it is energy and momentum that are conserved at vertices, not mass. Only lines that enter or leave the diagram represent real, observable particles, and these do, of course, have the proper mass. The diagram makes clear why we say that the electromagnetic force is mediated by photons. Richard Feynman, who called himself a \u201ccurious character,\u201d shared the 1965 Nobel Prize in Physics for his contributions to the development of quantum electrodynamics. [American Institute of Physics, Emilio Segr\u00e8 Visual Archives, Physics Today Collection.] \u03b3 e e Figure 12-3 The primitive vertex of the Feynman diagram. The particle, shown as an electron, could be a proton or any other particle that feels the electromagnetic force. Note that the photon line has no arrow. The primitive vertex should be thought of as a \u201cbuilding block,\u201d combinations of which form complete Feynman diagrams. (a) (ba) (c 12-1 Basic Concepts 567 (b) e e \u03b3 e e Figure 12-4 Feynman diagrams describing (a) Coulomb repulsion of charges of the same sign and (b) Coulomb attraction between charges of opposite signs. Figure 12-5 (a) The Compton effect. A photon enters and is absorbed by an electron, which then emits a photon and leaves. (Note that time, ct, is positive to the right in this diagram.) (b) At the lower vertex an electron and a positron enter and annihilate, producing a photon. At the upper vertex the photon creates a particle-antiparticle pair. (c) Another possible pair annihilation process. Figure 12-5a illustrates Compton scattering. Figure 12-5b is another diagram that describes electron-positron scattering (Coulomb attraction) and includes both pair production (upper part) and pair annihilation (lower part). This points up the fact that there may be many diagrams representing any given reaction.7 For example, Figure 12-5c is also a possible pair annihilation. With this introduction we will now use simple Feynman diagrams throughout the remainder of this chapter to visualize interactions that might otherwise be very difficult to understand. EXAMPLE 12-2 Feynman Diagram of Particle-Antiparticle Creation In Section 2-4 we described the production of an electron-positron pair. Construct a Feynman diagram that illustrates this process. SOLUTION Consider the primitive vertex as an electron-photon interaction as below, left. Using the rules outlined above and noting that the now virtual electron exists for too short a time to be measured, we draw its line horizontal, i.e., with (ct) 0. The positron interacts with a photon at the second primitive vertex. Together the two diagrams depict the creation of an electron-positron pair Other, so-called higher-order diagrams representing pair production are also pos- sible. Can you draw one? 568 Chapter 12 Particle Physics Questions 1. What problem might arise in using Dirac\u2019s filled infinite sea of negative energy states to explain the existence of particle-antiparticle pairs of pions, whose spins are zero? 2. Why do electron-positron pairs annihilate mainly from S states? Leptons and Quarks Since Thomson discovered the electron, theoretical and experimental research in particle physics has revealed the existence of 62 fundamental particles and antiparticles, fundamental in the sense that they have no internal structure, as far as we can tell with current technology. This is not to say that this is all that exist. In fact, an important task of CERN\u2019s Large Hadron Collider is to search for the predicted Higgs boson, which may be the key to explaining the origin of mass, and to test current theoretical predictions of supersymmetry (SUSY) that suggest the existence of a \u201csuperpartner\u201d for each of the known fundamental particles (see Section 12-5). Many particles with electric charge were first \u201cseen\u201d in the particle detectors of experimental searches. The existence of many electrically neutral particles was deduced indirectly by applying conservation laws, particularly energy and momentum, to interactions that included charged particles recorded by particle detectors. Still others, both charged and neutral, remain unseen directly or indirectly. These are the quarks and the force carriers that bind them together, the gluons. Nevertheless, we are confident of their existence because", " their properties and interactions are so successfully explained by the Standard Model of particle physics, which is second only to QED in the precision of its predictions. We will be discussing the Standard Model and its relation to the fundamental interactions and conservation laws throughout the rest of this chapter. In this section we will introduce the classifications of the quarks and leptons in generations (or families) and flavors and list a few of their physical properties. Once you are familiar with general characteristics, we will discuss their properties and interactions more thoroughly. Leptons There are three generations of leptons,8 each consisting of a charged lepton and its related neutrino, as shown in Table 12-1. The electron is the most familiar of the charged leptons and the only one that is stable. Each charged lepton has a distinct antiparticle. The Standard Model assigns each lepton a weak isospin Tz, the z component of a quantum-mechanical property represented by the vector T that is loosely analogous to spin (see Section 12-3). For each neutrino there is also an antineutrino, although at this point in time it is possible that the two are not distinct; that is, each neutrino may be its own antiparticle (a so-called Majorana neutrino), much as the photon is its own antiparticle. Investigating that possibility is an active area of current research. Unlike the quarks, as we will see, there are no lepton-lepton bound states. We also refer to leptons as having three flavors: electron, muon, and tau. We will use this terminology in Section 12-5 in a discussion of neutrino mass. Quarks As with leptons, there are six quarks grouped into three generations. All have fractional electric charge and distinct antiparticles. As we will learn in the following sections, it is the quarks and antiquarks that bind together in a multitude of ways to form more than 200 particles, accounting for the vast majority of the visible mass of the universe. The bound states of the quarks and antiquarks are called hadrons (from 12-1 Basic Concepts 569 Table 12-1 The leptons Lepton l Symbol Charge (e) Weak isospin Tz Mass MeV>c 2 ( ) Lifetime (s) Spin ( U ) 1st generation electron electron neutrino 2nd generation muon muon neutrino 3rd generation tau tau neutrino e ve v v 1 0 1 0 1 0 1>2 1>2 1>2 1>2 1>2 1>2 0.5110 2.2 eV>c2 105.659 3.5 eV>c2 1,784 8.4 eV>c2 stable stable 2.197 106 stable 3.3 1013 stable 1>2 1>2 1>2 1>2 1>2 1>2 the Greek hadros, meaning \u201crobust\u201d). There are two subgroups of hadrons. Threequark combinations are called baryons (from the Greek barys, meaning \u201cheavy\u201d), of which the proton and neutron are the two most common examples. Quark-antiquark pairs form the mesons. The term meson, derived from the Greek mesos, meaning \u201cmiddle,\u201d was chosen because the first mesons discovered (the pions) had masses intermediate between those of the electron and the proton; however, many mesons heavier than the proton were subsequently discovered, so the name is no longer an indicator of the masses of these hadrons. For reasons we will discuss in Section 12-4, single, or \u201cfree\u201d quarks have not been nor seem likely to be observed. The recently reported five-quark combination has not been independently confirmed. Table 12-2 records basic descriptions of the quarks. Each quark in the table also has an additional property, analogous to electric charge, called color, or color charge. Color has three possible values: red, blue, and green. So, for example, there are three different u quarks: ur, ub, and ug. The antiquarks have antiub, color, just as they have opposite electric charge, so the three anti-u quarks are the and Of course, these terms have nothing to do with the usual meanings of the words color, red, blue, and green. They are simply labels that are used to describe a particular ug. ur, Table 12-2 The quarks Quark (q) Symbol Charge (e) Weak isospin Tz Mass MeV>c 2 ( ) Spin ( U ) Baryon number 1st generation 2nd generation 3rd generation up down charm strange top bottom u d c s t b 2>3 1>3 2>3 1>3 2>3 1>3 1>2 1", ">2 1>2 1>2 1>2 1>2 336 338 1,500 540 170,900 5,000 1>2 1>2 1>2 1>2 1>2 1>2 1>3 1>3 1>3 1>3 1>3 1>3 570 Chapter 12 Particle Physics > > quantum-mechanical property of the particles, a choice that will turn out, perhaps unexpectedly, to be very convenient (see Section 12-4). Like electric charge, color charge is conserved. Quarks with 2e 3 electric charge (see Table 12-2) are up-type quarks (up, charm, and top), and those with e 3 are referred to as down-type quarks (down, strange, and bottom). As with the leptons, the Standard Model also assigns each quark Tz a weak isospin Tz. The up-type quarks have the down-type quarks have Tz Notice in Table 12-2 that each of the quark generations is an isospin doublet. The Standard Model provides for an equal number of lepton and quark generations, as you see are contained in Tables 12-1 and 12-2. We refer to the quarks as having six flavors; e.g., the down quark and antiquark are of the \u201cdown flavor.\u201d Altogether, Table 12-2 represents 36 quarks and antiquarks. Like the leptons, the quarks are all fermions. 1>2. 1>2; 12-2 Fundamental Interactions and the Force Carriers All the different forces observed in nature, from ordinary friction to the tremendous forces involved in supernova explosions, can be understood in terms of the four basic interactions that occur among elementary particles. In order of decreasing strength, these are 1. The strong interaction 2. The electromagnetic interaction 3. The weak interaction 4. The gravitational interaction Molecular forces and most of the everyday forces that we observe between macroscopic objects (for example, friction, contact forces, and forces exerted by springs and strings) are complex manifestations of the electromagnetic interaction, which occurs between all particles that carry electric charge. Although gravity, the interaction between all particles with mass, plays an important role in our lives, it is so weak compared with other forces that its role in the interactions between elementary particles is essentially negligible. The weak interaction acts between particles that carry weak charge and describes, among others, the interaction between electrons or positrons and nucleons that results in beta decay, which we discussed in Chapter 11. The strong interaction acts between particles that carry color charge and describes, for example, the force between nucleons that holds nuclei together. Some particles participate in all four interactions, whereas others participate in only some of them. In 1979, S. L. Glashow, A. Salam, and S. Weinberg shared the Nobel Prize in Physics for development of the electroweak theory, successfully unifying theories of the electromagnetic and the weak interactions. This event, which came exactly 100 years after Maxwell had accomplished unification of the theories of electricity and magnetism, was a major advance toward achieving unification of the theoretical descriptions of the four basic interactions. Developing such a unified field theory has been a goal of physics for a long time, one that was vigorously sought without success by Einstein, among many others. As we will discuss in Section 12-4, the electroweak unification occurs only at high particle energies. Current efforts to unify the electroweak, strong, and gravitational interaction will be discussed in Section 12-5. The term \u201cstrength\u201d of the interactions refers specifically to the relative magnitudes of the dimensionless coupling constants9 that multiply the fundamental space-dependent part of the potential energy function whose gradient determines the particular force. 12-2 Fundamental Interactions and the Force Carriers 571 The relative strengths stated in the paragraphs below and in Table 12-4 are only approximate as there is no unambiguous method of comparison, particularly for the weak interaction. As an example, the electric (Coulomb) potential energy of two 1>r charges is is made dimensionless by dividing both sides of the equation by the quantity The multiplier of the space-dependent function Uc: U(r) (1>4 0)e2>r. V(r) U(r)>Uc e2 4 0 Uc 1 r 12-3 Uc) (e2>4 0 1>137, where V(r) is in m1. The quantity you will recognize as the fine-structure constant first encountered in our discussion of Bohr\u2019s model of the hydrogen atom (see Section 4-3). The fine-structure constant is thus the coupling constant of the electromagnetic interaction. As we discovered in Chapter 4, energies resulting from this interaction are proportional to and characteristic dimensions (e.g., the Bohr radius a0) are proportional to (See Equations 4-32 and", " 4-33.) Moreover, the probability densities for atomic phenomena discussed in Chapter 7 are all directly dependent on the value of (See Equation 7-32.) 1>.. 2 Just as Yukawa postulated the pion as the mediator, or carrier, of the force between nucleons (see Section 11-5), the Standard Model postulates one or more particles as the force carrier, or mediator, of each fundamental interaction. Each of these mediators, all of which the theory requires to be bosons, will be introduced in the following paragraphs, concerned with each of the interactions. Strong Interaction 1>2 -integral spins ( 1>2, 3>2, 5>2, All hadrons interact via the strong interaction. Of the two subgroups of hadrons, baryons (the three-quark combinations) have etc.). Mesons (the two-quark combinations) have zero or integral spins. The range of the strong force is about 1015 m, or one fm. (See Chapter 11.) The coupling constant s of the strong interaction is approximately 1, or about 102 larger than the fine-structure constant of the electromagnetic force. Within the framework of the Standard Model, the strong force is due to color charge, analogous to the electromagnetic force being due to electric charge. The mediator of the strong force is the gluon. Like the quarks, the gluons carry color charge, but with a difference. Each quark carries one unit of one of the three color charges, but each gluon carries one unit of one of the three color charges and one unit of one of the three anticolor charges. Since there are nine possible combinations of r, b, and g with we expect nine different gluons; however, a technicality reduces that number to eight. One consequence of color-charged gluons is that the emission of a gluon by a quark can change the color (but not the flavor) of the quark. Another is that gluons can couple to other gluons (see Section 12-3). Since leptons don\u2019t carry color charge, they don\u2019t participate in the strong interaction. Note, too, that the photon, the electromagnetic interaction\u2019s counterpart to the gluon, does not carry electric charge. r, b, and g, The characteristic interaction time of the strong interaction is extremely short, only about 1023 s, meaning that an event caused by this interaction \u201chappens\u201d in this length of time. Thus, if the probability is to be high that two particles will interact via the strong force by exchanging a virtual particle, the two must remain within the range of the force for at least 1023 s. Similarly, particles that change into another particle or particles, that is, decay due to the action of the strong force, do so within about 1023 s. This is about the time it takes light to travel a distance equal to the diameter of a nucleus. 572 Chapter 12 Particle Physics e+ e+ e\u2013 e\u2013 Lead sheet \u03b3 \u03b3 \u03c0 \u2013 \u03c00 K\u2013 K ) enters a bubble chamber from the bottom and decays into a A negative kaon ( which immediately decays into two photons, whose paths are moves off to the right, and a indicated by the dashed lines in the drawing. Each photon interacts in the lead sheet, producing an electron-positron pair. The spiral at the right is an electron that has been knocked out of an atom in the chamber. (Other, extraneous tracks have been removed from the photograph.) which 0,, Table 12-3 lists some of the properties of the hadrons that are stable against decay via the strong interaction, that is, those with lifetimes significantly longer than 1023 s. Those that decay via the electromagnetic and weak interactions have much longer lifetimes, typically of the order of 1018 s and 1010 s, respectively. Note that all baryons ultimately decay to a proton. Note, too, that the baryons cluster into \u201ccharge multiplets\u201d of about the same mass: the nucleons (n and p) of mass about 939 MeV, the of mass about 1116 MeV, the particles of mass about 1190 MeV, the particles of mass about 1315 MeV, and the of mass 1672 MeV. The differences in masses within multiplets (such as between the neutron and proton) are due primarily to differences in the masses of the constituent quarks (see Section 12-4). The energy of the electromagnetic field also makes a contribution to the mass differences. There are six mesons in Table 12-3: three pions, two kaons, and the eta particle. The mesons also cluster into charge multiplets. As with the baryons, the mass differences within each multiplet are due primarily to the mass differences of the", " constituent quarks. Note that the mass of the as it must be since these particles are antiparticles of each other. is exactly equal to that of the \",, Being complex particles composed of other, more fundamental particles (quarks), the hadrons each have a ground state and a set of quantized excited states directly analogous to the allowed energy levels of atoms and nuclei, which are of course also complex particles composed of other, more fundamental particles. These excited hadron states usually decay via the strong interaction and thus have large energy widths, as 12-2 Fundamental Interactions and the Force Carriers 573 Table 12-3 Hadrons that are stable against decay via the strong interaction Symbol Mass MeV>c 2 ( ) Spin U ) ( Charge (e) Antiparticle Mean lifetime(s) Name Baryons Nucleon p (proton) or N 938.3 n (neutron) or N0 939.6 Lambda Sigma Xi* Omega Charmed lambda Mesons Pion Kaon Eta -0 0,0, \" c 0 K K0.0 1116 1189 1193 1197 1315 1321 1672 2285 139.6 135 139.6 493.7 497.7 549 1>2 1>2 1>2 1>2 1>2 1>2 1>2 1>2 3>2 1>0 0,0, \" c self K K0 self Typical decay products\u2020 p e ve p n -0 n -0 0 -0,0 1032 y 930 2.5 1010 0.8 1010 1020 1.7 1010 3.0 1010 1.7 1010 1.3 1010 1.8 1013 p K - 2.6 108 0.8 1016 2.6 108 1.24 108 0.88 1010 and 5.2 108\u2021 2 1019 v v 0 e ve \u2020 Other decay modes also occur for most particles. \u2021 The * The has two distinct lifetimes, sometimes referred to as particle is sometimes called the cascade. K 0, K 0 short and K 0 long. All other particles have a unique lifetime. (\u00a2E U>\u00a2t) required by the uncertainty principle and in contrast to the much slower atomic transitions and nuclear decays. Excited hadron states are usually observed as resonances in the cross section for scattering of one hadron on another and are therefore also called resonance particles. We describe resonance particles more thoroughly on the home page (see page 591). 574 Chapter 12 Particle Physics Electromagnetic Interaction \u03c0 \u2013 n p n. that the positive proton and exist, they can interact Figure 12-6 A neutron emits During the time a virtual \u00a2t the with other charged particles. \u00a2t After time reabsorbed by the proton. the is The need to transfer rapidly enormous volumes of data collected by detectors at the major particle physics laboratories throughout the world to the thousands of collaborating scientists in many countries led to the development of the World Wide Web at CERN. This is the dominant interaction at scales larger than subatomic, the realm of the strong interaction, and smaller than astronomical, where the gravitational interaction rules. All particles that carry electric charge or have a magnetic moment participate in the electromagnetic interaction. In addition, neutral particles without magnetic moments may also participate in the interaction if the emission of virtual particles results in charged particles. A neutron emitting and reabsorbing a virtual as shown in Figure 12-6 is an example of a neutral particle involved in an electromagnetic interaction. The range of the electromagnetic force is infinite, and its strength is about 1>137 times that of the strong interaction, as we discussed earlier. Its characteristic interaction time is about 1018 s. According to QED, the mediator of the electromagnetic force is the photon. In contrast to the gluon, the photon does not carry electric charge. Decays via the electromagnetic interaction generally result in the emission of one or more photons, although there are a few exceptions, e.g., Notice in.0 Table 12-3 that the 0 S ee. decay via the electromagnetic interaction. and 0, 0, Weak Interaction All quarks and leptons participate in the weak interaction. The range of the weak force is about 1018 m or about 103 fm, considerably smaller than that of the strong force. Example 12-3 shows how the range of the weak force is determined. Its characteristic interaction time varies from about 1016 s to about 1010 s. No particular name is given to the source of the weak force, although it is occasionally called the weak charge or flavor charge, in analogy with electric charge. The strength of the weak interaction relative to the strong interaction is about 105. The weak force is carried by three particles, the charged weak force by the W and W (W for \u201cweak\u201d) and the neutral weak force by the Z0 (Z for \u201c", "zero\u201d). All three have spin 1 and thus are bosons. A very important aspect of the weak force is that interactions mediated by the W turn one quark flavor into another. The weak interaction does not, however, change the lepton flavor. The mediation of three typical weak interactions, the scattering of a muon neutrino by an electron, the scattering of an electron neutrino and a muon, and the inverse beta decay of a proton are illustrated in Figure 12-7. The mediators of the weak interaction were all discovered in 1983 by C. Rubbia collider and a large international team of co-workers after a long search using the pp (a) \u03bd\u03bc \u03bd\u03bc Z0 e e (b) \u03bc W+ \u03bd e \u03bd e \u03bc (c) e \u03bd e W\u2013 p n Figure 12-7 (a) The scattering of a muon neutrino from an electron involves the exchange of a is called a neutral current interaction. The interaction does not convert the electron into a muon neutrino. (b) The scattering of an electron neutrino from a muon may also occur via a neutral current interaction as in (a), but a charged current interaction in which a charged W is exchanged is also possible, and both would contribute to the cross section. Measuring the cross sections thus provides a means of testing the standard model. (c) Inverse beta decay also proceeds via a charged current interaction. Such an exchange Z0. (a) (b) 12-2 Fundamental Interactions and the Force Carriers 575 360 300 240 180 \u03c6 120 60 0 90 140 40 \u03b8 SppS Figure 12-8 (a) The production and subsequent decay of one of the first W bosons ever detected was recorded W by the UA1 detector at the CERN is proton-antiproton collider. A The tau decays into charged particles clearly seen as the thicker produced, which decays by pencil-jet in the central detector directed nearly vertically downward. Conservation of energy and momentum for all particle tracks produced yields results consistent with a missing UA1\u2019s energy detectors surrounding the beam pipe recorded the energetic to the decay event. Energy is plotted vertically upward. from the decay. [CERN.] (b) The collision occurs at the center. A and its angular position relative W S pp. at CERN10 that was specifically designed for the task. (See Figures 12-8 and 12-9.) GeV>c2 The Z 0 is the second-heaviest elementary particle known, with a mass of 91 GeV>c2, or nearly 100 times that of the proton. The W, with masses of 80 are the next heaviest. (a) (b) e2 10 GeV e1 360\u00b0 270\u00b0 180\u00b0 \u03c6 90\u00b0 140\u00b0 0\u00b0 90\u00b040\u00b0 \u03b8 Figure 12-9 (a) This computer reconstruction of the CERN UA1 detector shows the first decay ever recorded, obtained by Rubbia\u2019s group in 1983. Millions more such events have since been seen. [CERN Courier, 33, 4 (1993).] (b) The energy plot of the electron-positron pair from the \u201cwrap around\u201d the central cylinder of the UA1. The angular locations of the recorded electron and positron are measured relative to the position of the decay. Energy, plotted vertically, is measured by individual detectors that Graphs like this are called \u201cLego plots.\u201d Z0 S e e Z0. Z0 576 Chapter 12 Particle Physics EXAMPLE 12-3 Range of the Weak Interaction The mass of the Z 0 has been accurately measured to be What range does that value imply for the neutral current weak interaction mediated by the Z 0? 91.16 GeV>c2. SOLUTION 1. The range, the distance R t U>\u00a2E traveled in time by a particle moving at about c, is given by Equation 11-50: 2. Substituting the mass of into this expression Z0 the for R gives 3. An alternate calculation R of R: R U mc Uc mc2 (1.055 1034 J # s)(3.00 108 m>s) (91.16 GeV>c2)(1.60 1010 J>GeV) R 2.17 1018 m 2.17 103 fm (197.3 eV # nm) (91.16 GeV)(107 eV>GeV)(109 nm>m) 2.17 103 fm Gravitational Interaction All particles participate in the gravitational interaction, but this interaction is so weak as to be unimportant in the discussion of elementary particles. As we have seen previously, its strength relative to the strong interactions is about 1038. The interaction has infinite range, with the force decreasing as as does the electrostatic force. The mediating particle", " for this force is the graviton, which is expected to be uncharged, massless, and have spin 2. This particle has not yet been observed, nor does the experimental capability to do so yet exist. Experiments with the objective of detecting gravity waves are currently under way. (See Section 2-5.) The gravitational interaction is produced by mass, which is the \u201cgravitational charge\u201d corresponding to the color charge, electric charge, and weak charge of the strong, electromagnetic, and weak interactions, respectively. Table 12-4 summarizes the characteristics of the four fundamental interactions. 1>r2, Table 12-4 Characteristics of the fundamental interactions Interaction Force carrier Mass GeV>c 2 ( ) Spin U ) ( Source Particles carrying charge Strong gluon Electromagnetic photon 0 0 1 1 color charge electric charge q, e, q, g,, W Weak W, Z0 80, 91 1, 1 weak charge Gravity graviton 0 2 mass,, q, e, W, Z0 q, e,,, v, W, Z0 Range (m) Interaction time (s) Coupling constant 1015 1018 1023 1018 1016 1010 to 1 s 1/137 w 105? g 1038 12-2 Fundamental Interactions and the Force Carriers 577 EXPLORING A Further Comment About Interaction Strengths At the beginning of this section we defined the strengths of the interactions in terms of the coupling constants, relating their approximate values to the most familiar one, e 2>4 which is the fine-structure constant In QED the electric charge Uc. 0 e 24 0 Uc 2 is the amplitude of the coupling of the photon (the mediating boson) to the electron (the particle). Thus, the probability of events involving that coupling, such as the Compton effect, illustrated in Figure 12-5(a), is proportional to e2. The time-independent solution to the Klein-Gordon equation (Equation 11-52) can also be interpreted as the static potential U(r) of the field of a point charge represented by the exchange particles. We then have U(r) Aer>R r 12-4 where A is a constant of integration and the Compton wavelength action the range R is infinite and U(r) becomes >2 c R U>mc is both the range of the force and of the mediating boson. For the electromagnetic inter- U(r) A r 12-5 0r, U(r) q>4 Recalling from classical electromagnetism that the electrostatic potential of a point charge q is we see that the constant A in Equation 12-5 plays the same role as the charge. In this manner a coupling constant proportional to A2, just as e 2, can be obtained for each of the interactions, albeit not without some difficulty. Doing so for the strong and weak interactions involves mathematics beyond the scope of our discussions, but as we will see in Section 12-4, this use of QED as a model is a powerful aid in understanding both the weak and the strong interaction. The coupling constants and other characteristics of the four interactions are given in Table 12-4. One last comment before we leave this topic: The coupling constants are not actually constants. Again, this can be most clearly illustrated using the electromagnetic interaction. Consider a positive point charge q embedded in a dielectric as shown in Figure 12-10a. The charge q polarizes the nearby molecules of the dielectric. As a result, the charge q is partially screened by the negative ends of the polarized molecules and the electric field of q at a distance r away is correspondingly reduced. Thus, the value measured for q is the effective charge qeff, which depends on how far from q the measurement is made, where qeff is given by qeff q 12-6 is the dielectric constant of the material, which you remember is a measure of and how difficult it is to polarize the material. Only by measuring very close to q, roughly speaking within the molecular equilibrium separation r0 (closer than the closest molecule so that there is no screening), will you actually measure the value q. This is shown q>, in Figure 12-10b. Notice also that (1) measurements made at large values of r yield not q, and (2) the value of qeff increases for very small values of r. 578 Chapter 12 Particle Physics (a +\u2013 \u2013 + (b) qeff q q/\u03b5 (c) \u03b3 e\u2013 e+ \u03b3 r0 r q Figure 12-10 (a) A positive charge q placed in a dielectric material polarizes the dielectric by orienting the nearby molecules with their negative ends closest to q. An observer at some distance r from q sees a reduced electric field because of the screen of negative charges. (b) The value qeff is measured for the charge. At small", " distances, those less than the equilibrium separation r0 of the molecules of the dielectric, the value of qeff approaches the value of q. (c) The vacuum also polarizes like a dielectric due to production of virtual electron-positron pairs by virtual photons. The effect is to increase the value of the finestructure constant at very short interaction distances. The production and absorption of virtual particles in QED results in the vacuum behaving like a dielectric. The positive charge q (or any charge) is continually emitting and absorbing virtual photons. Some of the photons occasionally create electronpositron pairs, which then annihilate, as the Feynman diagram in Figure 12-10c illustrates. The virtual electron and positron are attracted and repelled, respectively, by q, resulting in vacuum polarization, which partially screens q, just as it was screened when embedded in the dielectric. And just as in the dielectric, the full value of the charge q is not seen, or measured, until you get inside the screen. In vacuum polarization the role of the equilibrium separation r0 is played by the Compton wavelength of h>mc 2.43 1012 m. Thus, even in a vacuum the \u201cactual\u201d value the electron of q can only be measured at distances closer than about What we measure experimentally and refer to as \u201cthe charge of the electron\u201d is actually the completely screened effective charge. Thus, the fine-structure constant which is proportional to the square of the electric charge, will increase at very small distances from q. A corresponding discussion can be given for the weak and strong interactions, but there are significant differences. The photon, which mediates the electromagnetic interaction, does not carry electric charge. However, the W and Z 0, which mediate the weak interaction, have mass and carry weak charge. (The W also carry electric charge.) The gluons, which mediate the strong force, carry color charge. This latter difference results in an important characteristic of the strong force called quark confinement, which we will discuss further in Section 12-4. 2.43 1012 m., c 12-2 Fundamental Interactions and the Force Carriers 579 Questions 3. How are baryons and mesons similar? How are they different? 4. What properties do all leptons have in common? 5. The mass of the muon is nearly equal to that of the pion. How do these particles differ? 6. The bonding of the electrons to nuclei to form atoms is an example of the electromagnetic interaction. Use the interaction\u2019s properties to explain why the dimensions of atoms are of the order of 1010 m. 7. Describe a way the world would be different if electrons felt the strong interaction. 8. What might be the \u201ctechnicality\u201d that results in there being eight gluons instead of the expected nine? EXAMPLE 12-4 Neutron Decay The free neutron decays via the weak interaction with a half-life of 10.4 min according to the reaction n \u00a1 p e e Use a Feynman diagram to illustrate the details of this decay. SOLUTION Since this decay involves a change in the charge of the hadron, the mediating boson is a W. The W then decays to the e and The Feynman diagram describing these events is therefore as shown below, where particles shown moving backward in time are to be interpreted as the corresponding antiparticle moving forward in time. Notice that, as mentioned earlier, the charged weak interaction changes the quark flavor. e. proton: u d u v e W neutron: u d d In words, this diagram is read like this: A d quark in the neutron emits a W, changing (decaying) to a u quark, thus changing the neutron into a proton. The W then decays to an e and an e. Remarks: The mediating boson could also be a W. What would the diagram look like in that case? 580 Chapter 12 Particle Physics EXAMPLE 12-5 Estimate of Cross Section for Strong Interaction Obtain a rough estimate for the cross section of a typical strong interaction scattering of two hadrons, such as pions by protons or protons by protons. SOLUTION 1. The cross section for an interaction or reaction is given approximately by the area of a circle whose radius is the range of the interaction. (See Section 11-7.) For the strong interaction we can write, therefore, that 2. From Example 11-15 we found the range of the strong interaction RS to be about 1015 m. Therefore, is equal to S S R2 S S (1015 m)2 3.1 1030 m2 31 mb Remarks: The cross section, as noted in Section 11-7, is actually dependent on the collision energy, but typical values are of the order of tens of millibarns, in agreement", " with our approximation. 12-3 Conservation Laws and Symmetries One of the maxims of nature, sometimes referred to as the totalitarian principle, is \u201canything that can happen does happen.\u201d If a conceivable decay or reaction does not occur, then there must be a reason. The reason is usually expressed in terms of a conservation law. You are already familiar with several such laws. The conservation of energy rules out the decay of any particle for which the total mass of the decay products would be greater than the initial mass of the particle before decay. The conservation of linear momentum requires that when an electron and positron annihilate, two photons (at least) must be emitted. Angular momentum must also be conserved in a reaction or decay. A fourth conservation law that restricts the possible particle decays and reactions is that of electric charge. The net electric charge before a decay or reaction must equal the net charge after the decay or reaction. Every conservation law is a consequence of a particular symmetry in the laws of physics that govern the universe. This is a paraphrased statement of a theorem proven in 1918 by Emmy Noether11 for conjugate variables in classical mechanics. For instance, the laws of physics are symmetric (i.e., invariant) with respect to translations in time. That means they work the same today as they have in the past. Noether\u2019s theorem relates this particular invariance of the physical laws to the conservation of energy. The fact that the physical laws are symmetric under translations in space leads to the conservation of linear momentum. If a system is symmetric to rotations about a point, then the angular momentum is conserved. The conservation of electric charge is a consequence of the invariance of the laws of electrodynamics under a gauge (i.e., scale) transformation. There is a quantum theory analog of Noether\u2019s theorem; however, as was the case in classical physics, the conservation law is often discovered empirically before the symmetry that is its origin is identified. For example, Herman von Helmholtz set forth the law of conservation of energy primarily on the basis of James Joules\u2019s experiments long before Emmy Noether had proven her theorem. This is also the situation today in particle physics. Most of the conservation laws discussed in this section are empirical discoveries since no symmetry has yet been identified that provides their foundation. We will point out a few of these as we proceed. 12-3 Conservation Laws and Symmetries 581 Baryon Number In Section 11-4 we mentioned two conservation laws in our discussion of radioactive decay: conservation of nucleon number and of lepton number. We now need to state these more explicitly. The first is a special case of the following more general law: The baryon number is conserved. All baryons have baryon quantum number B 1, all antibaryons have B 1, and all other particles are assigned B 0. Conservation of baryon number requires that the total B for all particles before a decay or reaction occurs must be equal to that for all particles afterward. As an example of baryon conservation, consider the production of the antiproton in Figure 12-2 again. The reaction is pp \u00a1 ppp p 12-7 The total baryon number before the reaction is B 1 1 2. That after the reaction is B 1 1 1 1 2. Thus, conservation of B requires that three protons appear on the right side of Equation 12-7; that is, the production of an antiproton is always accompanied by the production of a proton. Conservation of baryon number together with the conservation of energy implies that the least massive baryon, the proton, must be stable. Whether that is in fact true is currently an active experimental question among particle physicists. There is no known symmetry requiring conservation of baryon number. There are several conceivable proton decay modes, all involving the proton decaying to a lepton and a meson, both of which have B 0. To date, nonconservation of baryon number has never been observed. Current experiments place the lower limit of the proton lifetime at about 1032 years. We will return to this matter later in this chapter. Lepton Number In the original version of the Standard Model neutrinos have no mass and are polarized. Experiments had shown that neutrinos were left-handed; that is, their spin direction was antiparallel to their momentum, and antineutrinos were right-handed, their spin being parallel to their momentum (see Figure 12-11a). In the Standard Model mass arises from interaction with the Higgs boson. That interaction also changes right-handedness to lefthandedness and vice versa, so the fact that neutrinos were always detected as left-handed meant that, like the photon, they did not interact with the Higgs and therefore had no mass. This in turn means that lepton number", " is conserved in the weak interactions and conservation of leptons applies independently to each of the three flavors. The lepton number for each flavor of leptons is independently conserved. The lepton quantum number for the electron and the electron neutrino is Le that for the positron and electron antineutrino is Le the other leptons, have Le assigned for the muon generation and tion of lepton number works, consider the following decays: 1 and 1. All other particles, including are for the tau generation. To see how conserva- 0. In a similar fashion the lepton quantum numbers L L p \u00a1 0e \u00a1 e e \u00a1 e n \u00a1 pe e 12-8 12-9a 12-9b 12-10 582 Chapter 12 Particle Physics The decay shown in Equation 12-8 (one of the conceivable proton decay modes) would conserve energy, charge, angular momentum, and linear momentum, but it has not been observed. It conserves neither baryon number B nor lepton number Le. The decay, given by Equation 12-9a, results in both an electron neutrino and a muon of the 1 antineutrino. The decay given in Equation 12-9b had ) and Le (the been searched for by many groups without success for many years. Its absence was the first indicator that Le and were independently conserved. Equation 12-10, the decay of the neutron, conserves both B and Le. Conservation of lepton number implies that the neutrino emitted in the beta decay of a free neutron is an electron antineutrino. 1 1 1 0 (the e and ve). The 0. The decay products also have and Le has L L L v However, during the past several years experiments at the Sudbury Neutrino Observatory and Super-Kamiokande have shown that neutrinos do in fact have mass and oscillate, albeit slowly, from one flavor to another as they travel. This discovery was the first experimental evidence that the Standard Model is an approximation of a more comprehensive, unknown theory. We will speculate briefly on what that theory might be in Section 12-5, but for our discussion here the implications are considerable. Since neutrinos have mass, their speeds are less than c. This implies that a left-handed neutrino can become a right-handed neutrino with respect to an observer and vice versa. Since we know of no fundamental symmetry that requires conservation of leptons, if lepton number is not conserved, then we have no way to distinguish between neutrinos and antineutrinos; the neutrino may be a Majorana particle, as was alluded to in Section 12-1. A number of theoretical extensions of the Standard Model have been suggested to deal with this problem, but as yet there is no clear solution. With this caveat in mind, we will for the remainder of this section use the lepton and baryon conservation laws stated above and defer discussion of the possible consequences of their violation until Section 12-5. EXAMPLE 12-6 Conservation Laws What conservation laws (if any) are violated by the following reactions? (a) n \u00a1 p (b) -0 \u00a1 p (c) \u00a1 SOLUTION (a) There are no leptons in this decay, so there is no problem with the conservation of lepton number. The net charge is zero before and after the decay, so charge is conserved. Also, the baryon number is 1 before and after the decay. However, the rest energy of the proton (938.3 MeV) plus that of the pion (139.6 MeV) is greater than the rest energy of the neutron (939.6 MeV). Thus, this decay violates the conservation of energy. -0 -0, 1 (b) Again, there are no leptons involved, and the net charge is zero before and after the decay. Also, the rest energy of the (1116 MeV) is greater than the rest energy of the antiproton (938.3 MeV) plus that of the pion (139.6 MeV), thus energy is conserved with the loss in rest energy equaling the gain in kinetic energy of the decay products. However, this decay does not conserve baryon number, which is 1 for the (c) There are no baryons involved, so conservation of baryon number is not a problem. The net charge is 1 before and after the decay, so charge is conserved. Also, the rest v, energy of the so energy is conserved, the difference appearing as kinetic energy of the muon and on the right side, thereon the left side and neutrino. Finally, fore lepton number is also conserved. This is the reaction by which the (139.6 Me", "V) is greater than that of the for the antiproton, and 0 for the pion. (105.7 MeV) and the 1 1 0 decays. 0 L L 12-3 Conservation Laws and Symmetries 583 More Each conservation law results from a particular symmetry in the laws that govern the physical universe. Since it is not necessarily obvious under what mathematical operations the laws of physics will be symmetric, on a pragmatic level it is fair to ask, quantum mechanically, When Is a Physical Quantity Conserved? We provide an answer to this question on the home page: www.whfreeman.com/tiplermodernphysics5e. See also Equations 12-11 through 12-22 and Note 12 here, as well as Example 12-7. More Conservation Laws The quantum numbers and corresponding conservation laws of the hadrons described in this section arise logically from combinations of so-called internal quantum numbers of the quarks. These are listed in Table 12-5. They are the electric charge Q, baryon number B, strangeness S, charm C, bottom B, and top T. As we have noted above, electric charge and baryon number are conserved in all interactions. Strangeness, charm, bottom, and top are conserved in the strong and electromagnetic interactions but are not conserved in the weak interaction. Strangeness There are some conservation laws that are not universal but apply only to certain kinds of interactions. In particular, there are quantities that are conserved in decays and reactions that occur via the strong interaction but not in decays or reactions that occur via the weak interaction. This is somewhat analogous to the selection rules discussed in atomic transitions. For example, the selection rule \u00a2 1 holds for electric dipole transitions from one atomic state to another. via An atom in a state with electric dipole radiation because of this selection rule, but it can decay via an electric quadrupole transition, which is generally much slower than electric dipole transitions. One of the particularly important quantities conserved in strong interactions is strangeness. This quantity was introduced by M. Gell-Mann 13 and K. Nishijima in 1952 to explain the seemingly strange behavior of the heavy baryons and mesons. cannot decay to a lower energy state with 0 2 Table 12-5 Internal quantum numbers of the quarks Quark Q B u d c s t b 2>3 1>3 2>3 1>3 2>3 1>3 1>3 1>3 1>3 1>3 1>3 1> 584 Chapter 12 Particle Physics (a) p s p s \u2013 \u03bd \u03bd (b) \u03c0 \u2013 \u03c0 \u2013 \u2192 + p \u039b0 + K0 \u03c0 \u2013 K0 \u03c0 + \u039b0 \u03c0 \u2013 p Figure 12-11 (a) The spin of antineutrinos is parallel to their momentum. The spin of neutrinos is antiparallel to their momentum. Described in terms of helicity ms s with the z axis in the direction of p, antineutrinos have > helicity 1 and neutrinos have helicity 1. (b) An early photograph of bubble chamber tracks at the Lawrence Berkeley Laboratory, showing the production, represented by Equation 12-23, and decay of two strange particles, the These neutral particles are identified by the tracks of their decay particles. The lambda particle was so named because of the similarity of the tracks of its decay particles and the Greek letter meson had energy of 1 GeV. [(b) Lawrence Berkeley Laboratory Photo Researchers.] The incident and the -0. -. K0 > Consider the reaction in which a high-energy interacts with a proton, p \u00a1 -0K0 12-23 -0 shown in Figure 12-11b. The cross section for this reaction is large, as would be expected since it takes place via the strong interaction (see Example 12-5). However, the and K0 are of the order of 1010 s, which is characteristic of decay times for both the weak interaction. When first discovered, their unexpectedly long lifetimes were very strange, so these and other particles showing similar behavior were called strange particles. An early success of the quark model was the explanation of these is a uds quark combination, which unexpectedly long lifetimes. As an example, the corresponds to a particular set of the internal quark quantum numbers. If no lighter (i.e., lower-energy) hadron with that exact set of quantum numbers exists, then decay via the strong or electromagnetic interaction is not possible. Decay can only occur via the much slower weak interaction. -0 These particles are always produced in pairs and never singly, even when all other conservation laws are met. This behavior is described by assigning to them a new quantum number called strangeness. The strangeness of the ordinary hadrons\u2014the nucleons and pions\u2014was arbitrarily chosen to be zero", ". The strangeness of the K 0 was particle must be 1 arbitrarily chosen to be 1. Therefore, the strangeness of the so that strangeness is conserved in the reaction of Equation 12-23. The strangeness of other particles could then be assigned by looking at their various reactions and decays. In reactions and decays that occur via the strong and electromagnetic interactions, strangeness is conserved. In those that occur via the weak interaction, strangeness is not conserved but can only change by 1. -0 12-3 Conservation Laws and Symmetries 585 Isospin As pointed out earlier, a striking feature of the hadrons is that they cluster into charge multiplets, groups of particles with nearly the same mass, such as the multiplet consisting of the proton ( p uud) and neutron (n udd). Within each multiplet all of the particles have the same spin, parity (see below), baryon number, strangeness, charm, and bottom but differ in their electric charges. In addition, we learned in Section 11-5 that the strong (nuclear) force is independent of electric charge. Were it not for the masses of the quarks and the electromagnetic interaction, the masses of the particles in a given charge multiplet would be the same. We are thus led to the view that the members of the multiplet are simply different charge states of the same particle. The \u201csplitting\u201d of particle mass states is analogous to the splitting of atomic energy states due to the spin-orbit interaction. (See Section 7-5.) Because of the analogy with isotopes (atoms with the same Z but slightly different masses) and with the splitting of different spin states, the term isospin is used to describe this multiplicity. The isospin I is treated as a vector in a three-dimensional \u201ccharge space,\u201d just as the orbital angular momentum L is a vector in real space. The component of I in the \u201cz direction\u201d is called I3 and is quantized, just as the z components of the orbital and intrinsic angular momenta of atomic electrons are quantized. Similarly, there are (2I 1) different I3 states. The charge q on a particle is related to its value of I3 by q eQ eaI3 B S 2 b 12-24 1>2 for the proton and 1>2, where Q is the charge quantum number. The value of I of the nucleon is with the I3 I3 for the neutron. The two possible values 1>2 isospin I is also for the xi doublet and 0 for the lambda and omega singlets. It is 1 ). In the case of the 1 for the triplet (with mesons, the pion isospin triplet has I 1, the kaon doublet and the eta singlet I 0. The rules for combining isospin are the same as those for combining real spin or angular momentum. If only the strong interaction is present, then Iop and Hop commute and I is conserved. Decays and reactions in which the total isospin of the system is not conserved do not proceed via the strong interaction. I 1>2, and 1 for 1>2, 0 0, for for I3 Hypercharge Four of the quantum numbers that we have discussed thus far turn out to be related to one another. These are strangeness, charge, isospin, and baryon number. The relation is S 2(Q I3) B 12-25 Strangeness is now used less frequently than a simpler quantity called hypercharge Y, which is defined as Y B S C B T 12-26 With the aid of Equation 12-25 the hypercharge quantum number Y is then given by Y S B 2(Q I3) 12-27 Stated simply, the hypercharge is twice the average charge of a given multiplet. For example, the average charge of the nucleon multiplet is Thus, for the nucleon, Y 1, as given by Equation 12-27. Since baryon number is strictly conserved and strangeness is conserved only in strong interactions, hypercharge, too, is conserved only in strong interactions. Since S 1 or 0 in weak interactions, (1e 0)>2 (1>2)e. 586 Chapter 12 Particle Physics Table 12-6 Some quantum numbers of the hadrons that are stable against decay via the strong interaction Particle Spin ( U ) p n -0 0,0, \" 0 K K0.0 1>2 1>2 1>2 1>2 1>2 1>2 1>2 1>2 3>2 0 0 0 0 0 0 I 1>2 1>2 0 1 1 1 1>2 1>2 0 1 1 1 1>2 1>2 0 I3 1>2 1>2 0 1 0", " 1 1>2 1>2 0 1 0 1 1>2 1> changes in hypercharge are similarly restricted to Y 1 or 0. Table 12-6 lists the values of these additional quantum numbers for those hadrons that are stable against decay via the strong interaction. Note that, if it were not for the conservation of strangeness or hypercharge in the strong interaction, all the baryons except the nucleons would decay via the strong interaction and live only for about 1023 s. The singlet, doublet, and triplet charge multiplets discussed above are clearly represented in graphs of Y versus I3. Studies of the regularities apparent in such graphs (see Figure 12-12) were instrumental in the development of the quark model of fundamental particles to be discussed in Section 12-4. The regularities are analogous to those observed in the multiplet structure of atomic energy states that ultimately led to the understanding of atomic structure. The conservation laws and the properties of charge Q, lepton number L, baryon number B, and strangeness S give us some insight into the relation between particles and their antiparticles. A particle and its antiparticle must have opposite signs for the values of each of these properties. Any particle that has a nonzero value for any of these properties will therefore have a distinct antiparticle. The photon, graviton, and have Q 0, L 0, B 0, and S 0 and are therefore in some sense their own the antiparticles. The and mesons are somewhat special because they have charge but have zero values for L, B, and S. They are therefore antiparticles of each other, but since there is no conservation law for mesons, it is impossible to say which is the particle and which is the antiparticle. 0 12-3 Conservation Laws and Symmetries 587 \", 3>2. these are Figure 12-12 Graphs of I3 hypercharge Y vs. the component of the isospin. 1>2. (a) Baryons with spin (b) Mesons with spin 0. (c) Baryons with spin Except for the resonance particles, as discussed in Section 12-3 and on the home page. Masses in parentheses are in Notice in each case that particles of like charge lie along downwardsloping diagonals and particles of like hypercharge (and strangeness) lie along horizontal lines. MeV>c2. (a) Y +1 n (b) p K0(498) Y +1 K+(494) \u03a3 0(1193) \u039b0(1116) \u03c0 0(135) \u03b7(549) \u03a3 \u2013(1197) \u20131 \u03a3 +(1189) \u03c0 \u2013(140) \u03c0 +(140) +1 I3 \u20131 \u03b7\u2032(958) +1 I3 \u039e \u2013(1321) \u20131 \u039e0(1315) K\u2013(494) \u20131 \u2013 K0(498) (c) Y +2 \u0394\u2013(1232) \u03940(1232) \u0394+(1232) \u0394++(1232) +1 \u03a3 0(1385) \u03a3 \u2013(1385) \u03a3+(1385) \u20132 \u20131 +1 +2 I3 \u039e \u2013(1530) \u20131 \u039e0(1530) \u20132 \u03a9 \u2013(1672) EXAMPLE 12-8 Applying the Conservation Laws State whether the following decays can occur via the strong interaction, via the electromagnetic interaction, via the weak interaction, or not at all: (a) \u00a1 p 0 (b) 0 \u00a1 -0 (c),0 \u00a1 n0 SOLUTION We first note that the mass of each decaying particle is greater than that of the decay products, so there is no problem with energy conservation in any of the decays. In addition, there are no leptons involved in any of the decays, and charge and baryon number are both conserved in all the decays. (a) From Figure 12-12, we can see that the hypercharge of the is 0, whereas the hypercharge of the proton is 1 and that of the pion is 0. This decay is possible via the weak interaction but not the strong interaction. It is, in fact, one of the decay particle, with a lifetime of the order of 1010 s. modes of the 588 Chapter 12 Particle Physics 0 -0 and (b) Since the hypercharge of both the electromagnetic interaction. It is the dominant mode of decay of the with a lifetime of about 1020 s. is 1, whereas that of the neutron is 1 and that of (c) The hypercharge of the the pion is 0. Since hypercharge cannot change by 2 in a decay or reaction, this decay cannot occur. is 0, this decay can proceed via the particle,,0 0 Questions 9.", " How can you tell if a decay proceeds via the strong, electromagnetic, or weak interactions? 10. Can the strangeness or hypercharge of a new particle be determined even if the number of particles in the multiplet is unknown? How or why not? Parity The parity of a nucleus or particle is defined in the same way as for an atom. (See Section 6-5.) The parity operation reflects the space variables in the coordinate origin. If the parity operator acting on the wave function changes the sign of the wave function, the parity is said to be odd, or 1. If the wave function does not change sign, the parity is even, or 1. The parity quantum number P is different from the other quantum numbers we have been considering. It can have only the values 1 or 1. If the value of the parity of a system changes, the new value is 1 times the old value. Parity is therefore a multiplicative property rather than an additive property like baryon number, strangeness, or hypercharge. The parity of an atomic wave function is related P (1). to the orbital angular momentum by The parity is odd or even depending on whether is odd or even. In our discussion of radiation from atoms, we saw that the parity of an atom can change just as the angular momentum of the atom changes when so the parity and anguthe atom emits light. For electric dipole transitions, lar momentum quantum numbers always change. However, if the complete system, including the photon, is considered, the total angular momentum and the total parity do not change in atomic transitions; that is, parity is conserved in electromagnetic interactions. Parity is also conserved in the strong interaction. \u00a2 1, Until 1956 it was assumed that parity is conserved in all nuclear reactions and radioactive decays. In that year, T. D. Lee and C. N. Yang suggested that parity might not be conserved in weak interactions. This suggestion grew out of attempts to understand the peculiar behavior of what were then known as the and mesons. These particles were identical in every way except that the meson decayed into two pions with positive parity, whereas the decayed into three pions with negative parity. (Each elementary particle can be assigned an intrinsic parity. That of the pion is negative.) The puzzle was this: Are there two different particles with all properties identical except parity, or is it possible that parity is not conserved in some reactions? After careful study Lee and Yang found that all the experimental evidence for parity conservation pertained to strong or electromagnetic interactions and not for weak interactions. They suggested that the nonconservation of parity could be observed experimentally by measuring the angular distribution of electrons emitted in decay of nuclei that have their spins aligned. Such an experiment was performed in December 1956 by a group led by C. S. Wu and E. Ambler. The results confirmed Lee and Yang\u2019s predictions, for which they received the Nobel Prize in Physics in 1957. The and mesons are a single particle, now known as the meson, which has two distinct modes of decay. The signifdecay will be discussed further in the TCP Invariance section below. icance of the K0 K0 12-3 Conservation Laws and Symmetries 589 Figure 12-13 (a) The mirror image of a right-handed coordinate system (x y in the z direction) is a left-handed coordinate system (x y in the \u2013z direction). No combination of translation and rotation can change a right-handed coordinate system into a lefthanded system. (b) Spinning nucleus emitting an electron in the direction of its spin. In the mirror, the image nucleus is emitting the electron in the direction opposite to its spin because the mirror reverses the direction of the spin vector. z z y y x Mirror (a) x (b) Spin Spin x\u0153 y\u0153 z z. is in the negative x, y, z z\u0153 The conservation of parity essentially means that a process described by the coorx x, y y, dinates x, y, and z appears the same if described by the coordinates The system x, y, z is called a right-handed coordinate system because and x y is in the z direction. Similarly, the system is called a left-handed coordinate system because direction. No rotation can change a right-handed coordinate system into a left-handed one, but reflection in a mirror does, as shown in Figure 12-13a. We can thus state the law of conservation of parity in more physical terms: If parity is conserved, the mirror image of a process cannot be distinguished from the process itself. Figure 12-13b shows a spinning nucleus emitting an electron in the direction of its spin. In the mirror, the nucleus appears to be emitting the electron in the direction opposite to that of its spin. If parity is conserved in decay, the chance of emission in the direction of the nuclear spin must equal the", " chance of emission in the opposite direction; i.e., there can be no preferred direction. Whether or not one direction is actually preferred in decay is usually not observable because the nuclear spins are randomly oriented. Wu and Ambler aligned the nuclei in 60Co by placing their sample in a magnetic field at a very low temperature (about 0.01 K). They found that more particles were emitted opposite to the spin of the nucleus than in the direction of the spin, indicating that parity is not conserved in weak interactions. Table 12-7 summarizes the conservation laws discussed in this section. TCP Invariance It is a property of any relativistic quantum theory in which signal speeds cannot exceed the vacuum speed of light, that the combined operations of Parity (r S r), (t S t), (particle S antiparticle) Time reversal leave any wave function unchanged: and Charge conjugation TCP &(r, t) 1 &(r, t) or TCP 1 12-28 590 Chapter 12 Particle Physics Table 12-7 Conserved quantities in fundamental particle interactions Conserved quantity Strong Electromagnetic Weak Interaction Energy Momentum Charge (Q) Baryon number (B) Lepton number (L) Isospin (I) Hypercharge (Y) Strangeness (S) Parity (P) \u2202 Yes Yes Yes Yes Yes Yes No Yes Yes Yes Yes No No No (\u00a2I 1, 0) (\u00a2Y 1, 0) (\u00a2S 1, 0) No It makes no difference in what order the operations are performed. Invariance under these combined operations requires that particles and their antiparticles have the same masses and lifetimes. It was long thought that the invariance under the combined operations was the result of invariance of physical laws under each of the operations independently, i.e., T 1, P 1, and C 1. However, as we described in the previous section, it was discovered that parity was not conserved in weak interactions. For the weak interaction the parity operation yields P 1. That immediately implies that one of the other operations must not be conserved in the weak interaction. Lee and Yang\u2019s solution to the puzzle mesons (kaons) with nearly identical masses but revealed that there are two (S for \u201cshort\u201d) decays to two pions very different decay modes and lifetimes. The (L for \u201clong\u201d) decays to three The with a lifetime of about pions with a lifetime of about 0.9 1010 s. (See Table 12-3.) 0.5 107 s. K0 S K0 L K0 Then in 1964 J. H. Christenson and his collaborators showed that in about 1 of also decayed into just two pions. This result means that for decay, the combined operation CP 1 because the two-pion final system every 1000 decays, the the has CP 1 and the three-pion final system has CP 1. K0 L K0 L The implications of this result are enormous and continue to be a focus of intense theoretical and experimental research. For example, within the framework of the Standard Model, CP violation requires that there be three generations of quarks and, correspondingly, the very large number of hadrons that can be assembled from them. The observed matter-antimatter asymmetry in the universe also requires CP violation. If TCP 1 and CP 1, T 1 also, which establishes an absolute direction for the flow of time. Questions 11. Suppose a new uncharged meson is discovered. What condition is necessary for it to have a distinct antiparticle? 12. How might Table 12-7 be different if strangeness were not conserved in interactions between hadrons? 12-4 The Standard Model 591 More Particles and excited states of particles that decay via the strong interaction have mean lives of only 1023 s or so, not nearly long enough to be tracked by a particle detector. Such particles are instead detected by measuring resonances in the scattering cross sections in a way analogous to J. Franck and G. Hertz\u2019s detection of the first excited state of the Hg atom by measuring the resonances in the electron scattering from Hg atoms. Many fundamental particles have been found in this way, as is described on the home page (www.whfreeman.com/tiplermodernphysics5e), Resonances and Excited States, which also includes a partial list of meson and baryon resonances. See also Figures 12-14 through 12-16, Table 12-8, Equations 12-29 through 12-31, and Examples 12-9 and 12-10 here. 12-4 The Standard Model The Standard Model is currently (since 1978) the accepted theory of elementary particle physics. It includes the quark model of particle structure that had been developed earlier, the unified theory of electromagnetic and weak interactions called the electroweak theory, and the strong-interaction analog of quantum", " electrodynamics called quantum chromodynamics (QCD). It has been remarkably though not totally successful in explaining the characteristics of fundamental particles and the interactions between them and is second only to QED in the accuracy of its predictions. In our discussions thus far in this chapter we have had occasion to allude to a number of specific features of the Standard Model. In this section we will consider the three of its constituents noted above in further detail. Since the complexity of the Standard Model\u2019s mathematical detail is beyond the level of this book, our discussion will be largely qualitative and conceptual. Searches for experimental support for the Standard Model led to the development of many new types of particle detectors. Several have found applications beyond particle physics, one example being BGO crystal detectors, used in medical diagnostic PET scanners (see Figure 11-65). Quark Model of the Hadrons The Eightfold Way The construction of large high-energy particle accelerators beginning in the 1950s made possible the production of a flood of previously unseen hadrons.14 Among the many attempts at understanding and classifying the jumble of hadrons, the most successful scheme is known as the eightfold way.15 It was suggested independently by Gell-Mann and Y. Ne\u2019eman in 1961 and subsequently justified by the Standard Model. In this scheme, hadrons that make up the charge multiplets were arranged in groups, called supermultiplets, in which each member had the same intrinsic spin and parity, J P, where J is the intrinsic spin and P is the parity. Three of Gell-Mann\u2019s supermultiplets are shown in Figure 12-12: (a) the eight lightest baryons, called the baryon octet; (b) the eight lightest mesons, the meson octet (actually a nonet); and (c) the next 10 heavier baryons, the baryon decuplet. Figure 12-17 shows the energies of the baryon octet in a diagram analogous to the fine-structure splitting of atomic states. The energy splittings between the isospin multiplets (from 78 to 176 MeV) are about 20 times the splitting within the multiplets. There are no completed baryon supermultiplets beyond the octet and decuplet, although there are several partially completed ones. The known mesons complete six nonets. Note that there is also an antibaryon octet, decuplet, etc., but for the mesons their antiparticles are members of the same nonet. Gell-Mann\u2019s accomplishment is the elementary 592 Chapter 12 Particle Physics JP = + (1/2) \u039e 125 MeV \u03a3 78 MeV \u039b 176 MeV N MeV 1318.4 1311 \u039e \u2013 \u039e 0 \u03a3 \u2013 \u03a3 0 \u03a3 + 1196 1191.5 1189.4 \u039b0 1115.4 n p 939.5 938.2 JP (1>2) Figure 12-17 The energy-level diagram of the baryon octet, the supermultiplet of the hadronically stable baryons. In the absence of any interactions, all these particles should have the same mass. The different numbers of s quarks splits the mass state into four states, corresponding to the nucleon (N), lambda ( ), sigma ( ), and xi ( ) particles. The different quark masses and the weaker electromagnetic interaction further splits the particles into the N doublet, triplet, and doublet.,, - particle analog of Mendeleev\u2019s development of the periodic table of the chemical elements, which was first published in 1869, nearly 100 years earlier, far in advance of the theoretical foundation for the periodic table provided by atomic theory and quantum mechanics. The eightfold way is based on part of a mathematical theory known as the theory of continuous groups that was developed by the Norwegian mathematician S. Lie, among others. The simplest Lie group is known as SU(2), for special unitary group of 2 2 matrices. A special condition on the 2 2 arrays reduces the number of components from 4 to 3. The three independent components of these arrays correspond to the three components of angular momentum (or isospin). As we have seen previously, the various possible values of angular momentum J have corresponding states that occur in multiplets having 1, 2, 3, 4,..., (2J 1) elements, which we describe as having angular momentum of units. The next higher Lie group is known as SU(3), for special unitary group of 3 3 arrays. Again, a special condition reduces the number of components from 9 to 8 (hence the name eightfold way). The eight quantities in the application of SU(3) group theory to hadrons consist of the three components of isospin, the hypercharge, and four that are yet to be named.", " Without going into the details of group theory, we will merely state that the SU(3) group leads to multiplets of 1, 3, 8, 10,... elements. Rather than assigning a single number to these multiplets analogous to the angular momentum quantum number of SU(2), it is more useful to make two-dimensional diagrams called weight diagrams, which are the geometric patterns of points, triangles, and hexagons shown in Figure 12-18. In the application of SU(3) to particle theory, the axes are Y and I3, as in Figure 12-12. 0, 1>2, 1, 3>2, \u00c1 U \" JP 3>2 In the plot of Y versus I3 for the baryons (the decuplet, shown in Figure 12-12c), neither the nor the had been discovered prior to 1961. Note that the difference in rest energy between each line of the decuplet is about 140 MeV. A constant energy difference between successive multiplets in the decuplet is predicted by SU(3) theory. The prediction of the particle by Gell-Mann in 1961 and its discovery in 1964 with just the mass and spin Gell-Mann had predicted was one of the spectacular successes of the pre\u2013Standard \" is the only particle in the decuplet that is not Model eightfold way. Note that the a resonance particle. The mass of the is just small enough that energy conservation prevents it from decaying via a strangeness-conserving strong interaction such as \" S,0 K. \" \" Y Y Y Y I3 I3 I3 I3 Figure 12-18 Weight diagrams occurring in SU(3) group theory. The circle and dot at the origin in the hexagon indicate two particles at the origin, making this pattern an octet. 12-4 The Standard Model 593 2/3 S = \u20131 s S = 0 u\u2013 \u2013 d Q = \u20131/3 Q = \u20132/3 Q = +1/3 Figure 12-19 The SU(3) weight diagrams (Y vs. I3) for the three light quarks and their antiquarks. As in the supermultiplet diagrams of the eightfold way, the downward-sloping lines are constant charge and the horizontal lines are constant strangeness. Other supermultiplets can be formed from the unstable baryons and mesons, but there are no observed groups of three particles corresponding to the triplet allowed by SU(3) theory, illustrated second from the left in Figure 12-18. This fact and the absence of a reason for the supermultiplets of the eightfold way led Gell-Mann and G. Zweig in 1964 to independently propose that all hadrons are composed of even more fundamental constituents called quarks.16 Their proposal is the basis of the quark model, arguably the most important advance in our understanding of elementary particles. 1>2. Each quark has and that of the d and s quarks is In the original quark model, quarks came in three types labeled u, d, and s (for up, down, and strange). Later discoveries, as we have seen, added three more quarks, labeled c, b, and t for charm, bottom, and top. Recall that the charge of the u 2e>3 e>3. B 1>3. quark is The strangeness of the u and d quark is 0 and that of the s Each quark has spin quark is 1. Each quark has an antiquark with the opposite electric charge, baryon number, and strangeness. The three types up, down, and strange form the triangular SU(3) weight diagram of Figure 12-18, as shown in detail in Figure 12-19. The properties of the quarks are listed in Table 12-2. The basic assertion of the quark model is that all baryons consist of three quarks (or three antiquarks for antiparticles), whereas mesons consist of a quark and an antiquark. The mesons thus have baryon number B 0, as required. The proton consists of the combination uud and the neutron udd. Baryons with a strangeness S 1 contain one s quark. All the particles listed in Table 12-3 can be constructed from the three original quarks and the corresponding three antiquarks.17 Murray Gell-Mann, who proposed the existence of strangeness, developed the classification system for hadrons [SU(3)] and postulated the existence of fractionally charged particles, which he called quarks. He won the Nobel Prize in Physics in 1969. [American Institute of Physics, Neils Bohr Library.] A great success of the quark model was that all of the allowed combinations of the three quarks and quark-antiquark pairs resulted in known hadrons. Strong evidence for the existence of quarks inside the nucleons is provided", " by high-energy scattering experiments called deep inelastic scattering. In these experiments, a nucleon is bombarded with electrons or muons of energies from 15 to 200 GeV. Analyses of particles scattered at large angles indicate the presence within the nuparticles of sizes much smaller than that of the nucleon. These excleon of spinperiments are analogous to Rutherford\u2019s scattering of particles by atoms in which the presence of a tiny nucleus in the atom was inferred from the large-angle scattering of the particles. 1>2 594 Chapter 12 Particle Physics Table 12-9 Properties of three-quark combinations Combination uuu uud udd uus uss uds ddd dds dss sss Spin U ) ( 3>2 1>2, 3>2 1>2, 3>2 1>2, 3>2 1>2, 3>2 1>2, 3>2 3>2 1>2, 3>2 1>2, 3>2 3>2 Charge (e) Baryon number Strangeness Hypercharge I3 3>2 1>2 1>2 1 1>2 0 3>2 1 1>2 0 Since the conservation laws represented by the several quantum numbers in Table 12-5 are additive, it is simply a matter of arithmetic to determine the properties of the hadrons. For example, a particle formed by the combination uds can have a spin of either and baryon number or B 1>3 1>3 1>3 1. Table 12-9 lists the possible three-quark combinations (baryons), and Table 12-10 lists the possible quark-antiquark combinations (mesons). 2>3 1>3 1>3 0, charge equal to 3>2, 1>2 Table 12-10 Properties of quark-antiquark combinations for three quarks Combination Spin Charge ( (e) U ) Baryon number Strangeness Hypercharge uu ud us du dd ds su sd ss 0, 1 0, 1 0, 1 0, 1 0, 1 0, 1 0, 1 0, 1 0 I3 0 1 1>2 1 0 1>2 1>2 1>2 0 12-4 The Standard Model 595 1>2 three-quark Figure 12-20 (a) The graph of Y vs. I3 for the spincombinations\u2014the baryon octet. (b) The graph of Y vs. I3 for the quark-antiquark combinations that form the lightest meson nonet. (a) Y (b) u d +1 d u u d Y +1 d s\u2013 u s\u2013 d d s \u20131 \u20131/2 u d s du s +1/2 u u s +1 I3 u\u2013 d \u20131 s s\u2013 \u2013 d u\u2013 d \u20131/2 u \u2013 d u +1 I3 +1/2 s s d \u20131 s s u s\u2013 s u\u2013 \u20131 s \u2013 d 1>2 The eight spin- baryons make up the baryon octet of Figure 12-12a. The three quarks of which each member is composed are shown in Figure 12-20a. Notice that Table 12-10 lists nine quark-antiquark combinations, rather than eight, as given by the eightfold way. The ninth meson identified by the quark model as a part of this group, had already been found but had been thought to be a singlet in the eightfold the way. Figure 12-20b shows the quark-antiquark composition of the first of the several meson nonets, the one illustrated in Figure 12-12b.., EXPLORING Where Does the Proton Get Its Spin? 1>2U 1>2U In the quark model of the hadrons the proton consists of two up quarks and a down quark, uud. The electric charge and quantum numbers of the proton, as with the neutron and other composite particles, are correctly given by summing the corresponding the proton\u2019s charge is quantities for the constituent quarks. For example, (2>3)e (2>3)e (1>3)e 1e, combination of the and its spin is the three spin quarks. However, a series of deep inelastic scattering experiments of electrons and muons on protons have yielded a surprising result. Begun in 1987 at CERN and continued up to the present there and at Brookhaven National Laboratory (Relativistic Heavy Ion Collider, RHIC) and the Thomas Jefferson National Accelerator Facility, the experiments consist of scattering extremely high-energy ( very short wavelength) muons or electrons whose spins are polarized from protons and nuclei whose spins are also polarized. Measuring the exit angles and energies of the scattered particles is a", " rich source of information concerning the spin structure of both the proton and the neutron. Surprisingly, the experimental results indicate that the spins of the three constituent, or \u201cvalence,\u201d quarks account for only 20 to 30 percent of the proton\u2019s spin! The experiments show that the spins of the u quarks are aligned parallel with the proton\u2019s spin, but this is not true for the spin of the d quark, which is as one would expect it to be. These results suggest that the orbital angular momentum of the valence quarks and perhaps the spins of the gluons and the \u201csea\u201d quarks make a significant contribution to the nucleon spin. Aptly called \u201cthe proton spin crisis,\u201d the results have underscored that our understanding of nucleon structure and quantum chromodynamics (QCD) is incomplete in some important respects. 3He 596 Chapter 12 Particle Physics Nor does the spin crisis stop there. The results also show that the \u201csea\u201d of virtual quark-antiquark pairs that surround the valence quarks (just as virtual pions surround the nucleons themselves in the nucleus) is strongly polarized with its collective spin direction opposite to the proton\u2019s net spin. Even more mysterious, the \u201csea\u201d turns out to contain a significant number of strange (s) quarks. As one scientist put it, there is no simple \u201cgee whiz\u201d explanation for the spin crisis. Several theories have been advanced to account for the discovery, but thus far none have been successful. The spin crisis is currently the focus of vigorous experimental and theoretical research. EXAMPLE 12-11 Predicting the Properties of Particles What are the properties of (c) dds, and (d) uss? the particles made up of the following quarks: (a) ud, ud, (b) SOLUTION (a) Since ud is a quark-antiquark combination, it has baryon number 0 and is therefore a meson. There is no strange quark here, so the strangeness of the meson is zero. The charge of the up quark is 1>3e, quark is nation of the meson. and that of the anti-down so the charge of the meson is 1e. This is the quark combi- 2>3e (b) The particle ud 2>3e (1>3e) 1e. is also a meson with zero strangeness. Its electric charge is This is the quark combination of the meson. (c) The particle dds is a baryon with strangeness since it contains one strange 1 1>3e 1>3e 1>3e 1e. This is the quark quark. Its electric charge is combination for the particle. (d) The particle uss is a baryon with strangeness 2. Its electric charge is,0 This is the quark combination for the 2>3e 1>3e 1>3e 0. particle. Color In Section 12-1 we briefly introduced the concept of the color charge of the quarks. In this section we will extend that discussion to include some important quark and hadron properties related to color. The quark model as described thus far in Section 12-4, essentially that developed over the decade following the introduction of Gell-Mann\u2019s quark hypothesis, contained two significant problems: despite numerous experimental searches, no free quarks had been found, and the model\u2019s construction of baryons was inconsistent with the Pauli exclusion principle. For example, the (1232) has spin and thus contains three u quarks (fermions) with exactly the same set of quantum numbers. 3>2 The solution to the exclusion-principle dilemma came from O. W. Greenberg, who postulated that each quark flavor (u, d, and s) came in three colors in addition to their other properties. The color charge of a quark has three possible values: red, blue, and green. Thus, a blue quark would have blueness 1, redness 0, and greenness 0, and its antiquark would have blueness 1, and so on. The terms \u201ccolor\u201d and \u201ccolor charge\u201d are, of course, simply labels to describe a quark property analogous to electric charge and are in no way related to the usual meanings of the words. The use of the three primary colors for this purpose did, however, provide a very simple rule to ensure that the exclusion principle was obeyed: All particles that occur in nature are colorless. 12-4 The Standard Model 597 The term \u201ccolorless\u201d means that either 1. The total amount of color (i.e., the sum of the color quantum numbers) is zero, or", " 2. There are equal amounts of all three colors present (in analogy with the combin- ing of the three primary colors to produce white). Thus, for example, the three up quarks that compose the (1232) are one each of ur, ub, and ug. J>\u2019 \u2019. Puzzle The solution provided by color seemed an artificial one, as did the The explanation for seeing no free quarks described in the next subsection, but strong support for the model came late in 1974 from an unexpected quarter. Two groups independently discovered a new meson. The first, S. Ting and his co-workers at Brookhaven, called it the J, while the second group, B. Richter and his co-workers at SLAC,18 called it the Now referred to as the the new meson had three times the mass of the proton and a lifetime of 1020 s, extraordinarily long for a strongly interacting particle. The exceptionally long lifetime pointed to new physics,19 and within months after its discovery it was recognized that the was composed of a fourth quark and its antiquark. The fourth quark, which had been proposed by S. Glashow and others for compelling theoretical reasons some years earlier so as to make equal numbers of quarks and lep20 v tons (before the discovery of the and Discovery of the first charm baryon, the ), is called the charm quark and C, is shown in Figure 12-21. Figure 12-22 J>\u2019 (cc). J>\u2019, J>\u2019 e+e\u2013 hadrons (b) 1000 b n, \u03c3 100 10 (a) 3.050 3.090 3.100 3.110 3.120 3.130 Energy ECMS, GeV Figure 12-21 (a) Discovery of the first charm baryon, the S -0 baryon decays via National Laboratory.] (b) A portion of the experimental data obtained by B. Richter and his co-workers at SLAC showing the J>\u2019 p S c too soon to leave a track, but the subsequent decay of the. -0 is easily seen. [Brookhaven The reaction is resonance. c. The charm c 598 Chapter 12 Particle Physics, v /, shows some supermultiplets formed with four quarks. This discovery made two nicely, ) and four quarks (u, d, s, c) and, of course, symmetric sets of four leptons (e, ve, their antiparticles. Then in 1975 a new lepton was found! The new lepton, the prev, and the numerical symmetry of the genersumably had an associated neutrino, the ations of particles was again upset. But within two years a new heavy meson, the upsilon was discovered and quickly recognized as being composed of a fifth quark-antiquark pair. The fifth quark is called the bottom (or sometimes beauty) quark, and / (bb ). The theory then predicts a sixth quark, called, as you might guess, the top (or truth) quark. The t quark was found in 1995 by two groups at Fermilab, thus restoring Glashow\u2019s symmetry of fundamental quarks and leptons and completing the new periodic table of the constituents of fundamental particles. (See Figure 12-23.) At 176 MeV>c2, the t quark is the most massive fundamental particle that has been discovered. There are substantial theoretical and experimental reasons to believe that there are no more quarks or leptons to be found. (See Figure 12-24.) Table 12-5 lists the up, down, strangeness, charm, top, and bottom internal quantum numbers for the six flavors of quarks and their antiquarks. (a) (b) \u2013 dd \u2013 cc \u2013 ss Mesons \u2013 du (\u03c1) \u2013 cu \u2013 dc \u2013 cs Charm = +1 \u2013 cd \u2013 ds (K*0) Charm = 0 \u2013 ud (\u03c1*) (c) cds \u2013 \u2013 sd (K*0) \u2013 uc Charm = \u20131 \u2013 sc Mesons cud Charm = +1 cus Charm = 0 uds Baryons (d) (e) ccc Charm = +3 ccd ccs cdd cud ccu Charm = +2 cuu Charm = +1 cds cus css dds (\u03a3 \u2013) dss (\u039e \u2013) uss (\u039e*) Baryons spin = 1/2 uus (\u03a3\u00b0) Charm = 0 ccd ccu Charm = +2 ccs cud cus uud (\u0394+) cdd cds css udd ddd (\u0394\u2013) cuu Charm = +1 uuu (\u0394++) Charm = 0 uus (\u03a3*+) dss (\u039e*\u2013) uss (\u039e*0", ") sss (\u03a9\u2013) Baryons spin = 3/2 Figure 12-22 Supermultiplets formed from u, d, s, and c quarks. The circles indicate that there are two particles with the same quark composition and different energies. 12-4 The Standard Model 599 First generation Second generation Third generation Charge +2/3 \u20131/3 \u20131 0 ur dr ub db e \u03bd e Quarks cr sr cb sb cg sg ug dg tr br Leptons \u03bc \u03bd\u03bc tg bg tb bb \u03c4 \u03bd\u03c4 Figure 12-23 Periodic table of elementary particle constituents. 40 35 30 25 20 15 10 Two flavors Three flavors Four flavors 0 88 89 90 91 92 Energy, GeV 93 94 95 96 Figure 12-24 Both the shape and height of Z 0 resonance are theoretically related to the number of flavors of the leptons and quarks. As that number increases, the maximum cross section decreases and the energy width (at half the maximum height) becomes larger. Current measurements, shown by the black circles, are fully consistent with three flavors, or generations, excluding both two and four. Quantum Chromodynamics Quantum chromodynamics (QCD) is the modern theory that describes the strong interaction between quarks and gluons. It is directly analogous to quantum electrodynamics (QED), which so successfully accounts for the electromagnetic interaction. Indeed, QCD was modeled on QED. As stated earlier, the particle (boson) that mediates the strong quark-quark interaction is the gluon and the fundamental process (analogous 600 Chapter 12 Particle Physics g (b) g (a) q q g(g, r )\u2013 dr dg g g Figure 12-25 (a) The fundamental vertex of QCD, in which a quark emits a virtual gluon. (b) Since gluons carry a color and an anticolor, the emission of the gluon may also change the color of the quark. g q Figure 12-26 Feynman diagram of a quark emitting a gluon, which then creates two gluons that recombine, the resulting gluon being absorbed by the quark. QCD potential V = \u2013 \u03b1 s\u2013\u2013\u2013\u2013 r 4 \u2013\u2013 3 + kr e S e q S q g to shown in Figure 12-4) is illustrated by the Feynman diagram in Figure 12-25a. The gluons are the QCD analog of the photon in QED. Like photons, they are massless and have spin 1; however, there is one crucial difference between the two particles. The gluons carry color charge, whereas the photon is electrically neutral. In fact, the gluons are bicolored, carrying one unit of a color charge and one of an anticolor the quark may charge and hence are not color neutral. Thus, in the process change color (but not flavor), as shown in Figure 12-25b. Since the gluons carry net color charge, they can also interact with each other via the strong interaction and they form an octet in the SU(3) group theory representation, just as do the mesons.21 This means that, in addition to the increase in the strong interaction coupling constant at very short distances analogous to the vacuum polarization in QED discussed in s Section 12-2 and shown in Figure 12-10, there are also gluon-gluon loops, as shown in Figure 12-26. The effect of such gluon loops is to, in a sense, dilute the strong force and decrease the value of the strong interaction coupling constant As it turns out, this lats ter effect predominates at very small quark separations. As a pair of quarks move extremely close to each other, their coupling decreases, a condition called asymptotic freedom. The result is that inside the nucleon, the quarks move more or less as free particles, a phenomenon that has been confirmed by electron deep scattering experiments. Indeed, hundreds of experiments have confirmed the property of asymptotic freedom. at extremely short distances (1018 m). V = \u2013 \u03b1 s\u2013\u2013\u2013\u2013 r 4 \u2013\u2013 3 One of the possible potential functions for the strong inter- action has the approximate form VQCD(r) 4 S 3r kr 12-32 \u03b1 s = 0.3 k = 1 GeV \u00b7 fm \u20131 0.5 r, fm 1.0 Figure 12-27 The potential seen by quarks in QCD. In this diagram the strong coupling constant > s and k 1 GeV fm in Equation 12-29. 0.3 FQCD VQCD It has been reasonably well tested experimentally at short distances. Notice that VQCD increases indefinitely with r (see Figure 12-27), constant, that is, the strong force at large", " r, rather than going to zero, as do the Coulomb and gravitational forces. The result is to prevent the quarks from getting too far apart, effectively containing them inside the hadrons, a phenomenon called quark confinement. This is the QCD explanation for why free quarks have not yet been found. When a large amount of energy is added to a quark system such as a nucleon, a quarkantiquark pair is created and the original quarks remain confined within the original system. This is the origin of the virtual pions that were postulated by the Yukawa model of the nuclear force as the mediator of that interaction. (See Figure 12-28.) 1 0 V e G V, \u20131 \u20132 0 12-4 The Standard Model 601 n d u d d u Figure 12-28 Shown is one possible illustration of quark confinement. If energy is added to remove a d quark from a neutron, a ( combine to form a proton and no free quark appears. while the u from the pair and the original u and d combine to produce a ) pair is created. The and one of the d quarks, u, u u During particle decays and interactions quarks transform into one another. For example, the decay of the neutron given by Equation 11-38 proceeds according to the quark model in which a d quark turns into a u quark, as illustrated by Example 12-4. All baryons eventually decay in one or more steps to the lightest (lowest-energy) baryon, the proton. The decay of the proton is prohibited by conservation of energy and baryon number, but searches for proton decay are continuing. Example 12-12 illustrates the decay of another baryon, the -0. Questions 13. How can you tell whether a particle is a meson or a baryon by looking at its quark content? 14. Are there any quark-antiquark combinations that result in nonintegral electric charge? 15. What experimental evidence exists to support the assertion that natural particles are colorless? EXAMPLE 12-12 Decay of involved in the decay of the -0 -0, Draw a Feynman diagram that shows the quarks which goes according to -0 \u00a1 pe e p u d u u-0 d s ve e\u2013 W SOLUTION is composed of a u quark, a d quark, and an s From Table 12-10 we see that the quark. The proton consists of two u quarks and a d quark. The decay results from the weak interaction, the s quark decaying to a u quark and a W. Note that strangeness is not conserved in the weak interaction that transforms the s quark into the u quark. -0 602 Chapter 12 Particle Physics Table 12-11 Quark composition of selected particles Baryons Quarks Mesons Quarks p n -0 \u00a2 0,0, \" c c c, c uud udd uds uuu uus uds dds uss dss sss udc uuc udc usc K K0 K0 K J>\u2019 D D0 D s B B0 B0 B ud ud us ds sd su cc cd cu cs ub db db ub The Electroweak Theory ( W 100 GeV), In the electroweak theory, the electromagnetic and weak interactions are considered to be two different manifestations of a more fundamental electroweak interaction. the electroweak interaction is mediated by four At very high energies bosons. From symmetry considerations, these would be a triplet consisting of W, W 0, and W, all of equal mass, and a singlet boson B 0 of some other mass. Neither the W 0 nor the B 0 would be observed directly, but one linear combination of the W 0 and the B 0 would be the Z 0, and another would be the photon. At ordinary energies, the symmetry is spontaneously broken. By \u201cspontaneously broken symmetry\u201d we mean the following: the Hamiltonian Hop retains the complete symmetry, but the ground state computed from that Hop does not, or, as we say, the symmetry is broken. For example, magnetism in solids arises due to interaction of the spins of the atoms of the crystal lattice. For a ferromagnet, such as iron, the Hop describing that interaction is invariant under rotation, but in the ground state magnetic domains are spontaneously formed in the sample. The spin direction changes from domain to domain but is the same inside each domain. A domain is certainly not invariant to a rotation of the spins. Thus, the ground state spontaneously breaks the rotational symmetry. (To further help you visualize what \u201cspontaneously broken symmetry\u201d means, think of a small plastic strip, like a short ruler, gripped at the ends between your thumb and index finger. As you squeeze, the strip", " will snap into a curve to one side or the other, breaking the original left-right symmetry. See Figure 12-29.) (a) (b) (c) Figure 12-29 The plastic strip in (a) has left-right symmetry. Increased vertical force on the ends of the strip breaks the symmetry, causing the strip to take one or the other of the positions (b) or (c), neither of which has leftright symmetry. 12-4 The Standard Model 603 100 GeV>c2 The broken symmetry in the electroweak interaction leads to the separation of the electromagnetic interaction mediated by the photon and the weak interaction mediated by the W, W, and Z 0 particles. The fact that the photon is massless and that the W and Z particles have masses of the order of shows that the symmetry assumed in the electroweak theory does not exist at lower energies. The symmetrybreaking agent is called a Higgs field, which requires a new boson, the Higgs boson, whose rest energy is expected to be of the order of 1 TeV (1 TeV 1012 eV). According to the Standard Model, it is by interacting with the Higgs field that particles acquire their masses. The Higgs boson has not yet been observed. Calculations show that Higgs bosons (if they exist) should be produced in a head-on collision between protons of energies of the order of 20 TeV. While such energies are not available with existing accelerators, the Large Hadron Collider, which began operation at CERN in 2008, will be able to reach and exceed that energy by accelerating beams of a variety of nuclei. Searching for the Higgs boson is a primary goal for the LHC. The Standard Model\u2014A Summary 1>2 The Standard Model is the theoretical model of elementary particles and their interactions. It is based on a combination of the quark model, electroweak theory, and quantum chromodynamics. In this model, the fundamental particles are the leptons and quarks, each of which comes in three generations, as shown in Tables 12-1 and 12-2. The force carriers are the photon, the W and Z 0 particles, and eight types of gluons. The leptons and quarks are all spinfermions, which obey the Pauli exclusion principle. The force carriers are integral-spin bosons, which do not obey the Pauli exclusion principle. Every force in nature is due to one of the four basic interactions: strong, electromagnetic, weak, and gravitational. A particle experiences one of the basic interactions if it carries a charge associated with that interaction. Electric charge is the familiar charge that you have studied previously. It is carried by the quarks and charged leptons. Weak charge, also called flavor charge, is carried by leptons and quarks. The charge associated with the strong interaction is called color charge and is carried by quarks and gluons but not by leptons. The charge associated with the gravitational force is mass. It is important to note that the photon, which mediates the electromagnetic interaction, does not carry electric charge. The W and Z 0 particles, which mediate the weak interaction, do carry weak charge, and the gluons, which mediate the strong interaction, carry color charge. This latter fact is related to the confinement of quarks. All matter is made up of leptons and quarks. There are no known composite particles consisting of leptons bound together by the weak force. Leptons exist only as isolated particles. Hadrons (baryons and mesons) are composite particles consisting of quarks bound together by the color charge. A result of the QCD theory is that only color-neutral combinations of quarks are allowed. Three quarks of different colors can combine to form color-neutral baryons, such as the neutron and proton. Mesons contain a quark and an antiquark and are also color neutral. Excited states of hadrons are considered to be different particles. For example, the particle is produced by p S \u00a2. The must have the exact same set of internal quantum numbers as which from Equations 12-26 and 12-27 the means Y 1 and The three u quarks can be in the same spin state in the without violating the exclusion principle because they have different colors. p: B 1, C S T B 0, 3>2. I3 The strong interaction can manifest itself in two ways: the fundamental, or color, interaction and the residual strong interaction. The fundamental interaction is responsible for the force exerted by one quark on another and is mediated by gluons. 604 Chapter 12 Particle Physics The residual strong interaction is responsible for the force between color-neutral nucleons, such as the neutron and proton. This force is due to the residual strong interactions between the color-charged quarks that make up the nucleons and can be viewed as being mediated by the", " exchange of mesons. The residual strong interaction between color-neutral nucleons is analogous to the residual electromagnetic interaction between neutral atoms that bind them together to form molecules. L L, and For each particle there is an antiparticle. A particle and its antiparticle have identical mass and spin but opposite electric charge. For leptons, the leptons numbers Le, of the antiparticles are the negatives of the corresponding num 1 bers for the particles. For example, the lepton number for the electron is Le 1. For hadrons, the baryon number, strangeness, and that for the positron is Le charm, topness, and bottomness are the sums of those quantities for the quarks that make up the hadron. The number for each antiparticle is the negative of the -0, number for the corresponding particle. For example, the lambda particle which is made up of the uds quarks, has B 1 and S 1, whereas its antiparquarks, has B 1 and S 1. Particles ticle, and the Z 0 that have zero electric charge, B 0, L 0, such as the photon S 0 and zero charm, top, and bottom, are each their own antiparticle. Note that has a zero value for all of these quantities except strangeness, the meson which is 1. Its antiparticle, the meson has strangeness 1, which makes K0. have electric charge but zero and it distinct from the values for L, B, and S. They are antiparticles of each other, but since there is no conservation law for mesons, it is impossible to say which is the particle and which is the antiparticle. Similarly, the W and W are antiparticles of each other. Table 12-11 lists the quark compositions of several particles. which is made up of the K0 (ud) (ds), (ud) The (ds) -0, uds K0 EXAMPLE 12-13 Decay of the \" The \" decays according to the equation \" \u00a1 -0 K and the resulting -0 and K usually decay according to -0 \u00a1 p and K \u00a1 Write each of these reactions in terms of quarks. SOLUTION Using Table 12-11, the \" decay is given by sss \u00a1 uds su in which an s is changed to a d and a according to, for the -0, uu pair is created. The \"0 and K decay uds \u00a1 uud ud where again an s is changed to a d and a uu pair is created, and for the meson, K where the s and u annihilate, producing a W, which decays to the leptons. su \u00a1 12-5 Beyond the Standard Model 605 12-5 Beyond the Standard Model Grand Unification Theories At the beginning of Section 12-4 we noted that the Standard Model of particle physics, while correctly accounting for a wide range of observations, has left a number of fundamental questions unanswered. Of premier importance among these are why nature requires four interactions rather than one and why their strengths and properties should be so different. The successful unification of the electromagnetic and weak interactions into the electroweak theory discussed earlier has led to a number of efforts to include the strong interaction and, ultimately, the gravitational interaction into a single, so-called grand unification theory, or GUT.22 As in the electroweak theory, the different strengths at energies well below the rest energies of the mediating bosons would be accounted for by spontaneous symmetry breaking. GUTs also explain the equality of the electron and proton charges. A central feature of current GUTs is that the coupling constants of all four inter, actions approach the same value, approximately that of the fine-structure constant at some very high energy. It is a remarkable experimental observation that the measured values of the coupling constants do appear to be tending toward a common value. Unfortunately, extrapolation to the common point must be made over an extraordinarily large energy range, that point of common value being at about 1016 GeV, compared with about 104 GeV that can be reached with the largest existing accelerator, the LHC at CERN. (See Figure 12-30.) To assume that nature has no surprises or new physics to await us somewhere in that colossal energy range ignores the lessons of history. Supersymmetry (SUSY) A number of GUTs include a proposed new symmetry in addition to the symmetries we have discussed. Called supersymmetry (with the acronym SUSY), it assigns to each elementary particle a superpartner. The superpartner is in every way identical to fermions, have the particle except for its spin. The leptons and quarks, both spin- 1>2 (Quarks) Strong EM (Leptons) Weak Current accelerators log Energy Unification energy (Leptoquarks) Gravity Ex Figure 12-30 The coupling constants of the four interactions appear", " to be approaching a common value at some energy in the range 1016 to 1018 GeV. Since the largest existing accelerator, the Large Hadron Collider at CERN, can reach only about 14 TeV, the extrapolation to the unification energy Ex is highly uncertain. 606 Chapter 12 Particle Physics Table 12-12 Elementary particle and their superpartners Particle Symbol Spin Superpartner Symbol Spin Quark Electron Muon Tau W Z Photon Gluon Higgs q e W Z g H 1>2 1>2 1>2 1>2 1 1 1 1 0 Squark Selectron Smuon Stau Wino Zino Photino Gluino Higgsino >2 1>2 1>2 1>2 1>2 superpartners. The supersuperpartners with spin 0. The spin-1 bosons have spinpartners of the fermions are given the same names with a prefix s; e.g., the electron\u2019s superpartner is the selectron. The superpartners of the bosons have the same names with a suffix ino added (sort of), e.g., the gluon\u2019s superpartner is the gluino. The particles and their superpartners are listed in Table 12-12. 1>2 Exact supersymmetry would equate the masses of the particles and their superpartners. However, this is apparently not true in nature or the superpartners would have been detected long ago. So SUSY is modified to account for that absence by postulating that the mass of the lightest superpartner would be very large indeed, of the order of the masses of the W and the Z 0 bosons. Doing so ultimately predicts the GUT unification energy in the vicinity of the current extrapolated projections, predicts the proton lifetime in agreement with current experimental limits, and keeps the GUT unification coupling constant in line with current extrapolations. An important goal of the Large Hadron Collider is to test predictions of supersymmetry. SUSY is also a component of current theories designed to include gravity within GUTs. These are the string and superstring theories. String theories replace pointlike elementary particles with tiny, quantized strings and require 10 or more dimensions. Their purpose is to surmount current theoretical problems in quantizing gravity. Currently, particle physicists are sharply devided over string theories, some heralding them as the \u201ctheory of everything,\u201d others dismissing them as \u201cnot even wrong.\u201d As of this date, there is no experimental evidence supporting any of the string theories. Proton Decay In GUTs the quarks and leptons are states of one particle, the leptoquark, and occur symmetrically in the same multiplet. This would account for why there are equal numbers of quark and lepton flavors and also lead to the prediction that each type of particle can be changed into the other. If that is the case, then baryon number is no longer a conserved quantity and the proton should not be stable. Current versions of GUTs place the lifetime of the proton at about 1030 to 1033 years, the long lifetime being the result of the large energy at which unification of the interactions occurs. 12-5 Beyond the Standard Model 607 (a) (b) The Sudbury Neutrino Observatory (SNO) in Canada and the Super-Kamiokande (Super-K) neutrino detector in Japan collected data on neutrino interactions that confirmed neutrino oscillations. (a) The SNO neutrino detector is a spherical acrylic vessel 12 m in diameter and contains 1000 metric tons of ultrapure heavy water (D2O) located 2000 m below ground. herenkov light produced by neutrino reactions in the water is viewed by 9456 photomultipliers, each 20 cm in diameter. (b) The Super-K neutrino detector, a cylindrical structure 41 m tall and 39 m in diameter, contains 45,000 metric tons of pure water (H2O) viewed by 11,200 photomultipliers. [(a) Courtesy of Sudbury Neutrino Observatory; (b) courtesy of Kamioka Observatory, University of Tokyo.] C\u02c7 Current experiments have placed the lower limit on the proton lifetime at about 1032 years; to date, no proton decays have been detected. Searches for proton decay, such as that at Super-Kamiokande, generally involve monitoring very large volumes of pure water, watching for one of the several possible proton decay \u201csignatures\u201d; e.g., p S 0e The nonconservation of baryon number in the early universe when energies were very high provides an explanation of a major cosmological problem, namely, why the present universe has many more baryons than antibaryons. p S", " or e. Lepton numbers would no longer be conserved at the unification energy, and cur S e rently forbidden reactions such as would be allowed. Experimental searches have been made, but no lepton-number-nonconserving events have been found. S e e e and Massive Neutrinos From the time Pauli first suggested their existence in 1930, neutrinos were thought to have zero mass. Then, based on Bahcall\u2019s theoretical calculation of the solar neutrino flux and Davis\u2019s remarkable measurement23 of the flux at only about 30 percent of Bahcall\u2019s prediction, the solar neutrino problem emerged (see Section 11-8). Its recent solution through the detection of neutrino oscillations by the Sudbury Neutrino 608 Chapter 12 Particle Physics Observatory (SNO) and Super-Kamiokande (Super-K) experiments gives support to GUTs since most GUTs require that neutrinos have mass. The theories predict their mass to be given approximately by m M2 eW Mx 12-33 Ex >c2 1016 GeV>c2 and Mx is the unification mass where MeW is a characteristic mass of the electroweak interaction, roughly 102 GeV>c2, (see Figure 12-30). Nearly all GUTs project Mx values of this order of magnitude, which in turn means that all neutrinos would have mv less than about 1 eV. The theories also predict m(ve) V m(v) V m(v). The impact of massive neutrinos on both the solar neutrino problem described briefly in Chapter 11 and the universe\u2019s \u201cmissing energy\u201d (or \u201cmissing mass\u201d) problem discussed in Chapter 13 is substantial. S. Mikheyev, A. Smirnov, and L. Wolfenstein proposed a solution to the solar neutrino problem in v particular in which an ve can oscillate to a or while propagating through the Sun\u2019s mass. For this complex process, called the MSW effect, to occur, the neutrino wave functions must each consist of superpositions of the three mass states. Similar processes can be described for neutrinos moving through space and the atmosphere. (See Exploring, Neutrino Oscillations and Mass.) The relative phases of the mass states may change for two reasons. (1) In passing through the solar matter (electrons and protons), the three mass states scatter differently; hence their relative phases change. (2) While propagating through space and the atmosphere, the mass states move at different speeds, which also results in a change in the relative phase. The phase changes result in interference of the neutrino matter waves, and as a consequence a neutrino emitted in the Sun as a ve may oscillate to a before reaching Earth and therefore not be detected by experiments searching for electron neutrinos. Experimental evidence supporting the existence of oscillations was provided by the SNO and Super-K measurements. \u2019(ve), \u2019(v), \u2019(v) and or v v v Magnetic Monopoles Magnetic monopoles, first suggested by Dirac in 1929, are also proposed by GUTs. Dirac showed that their existence in relativistic quantum mechanics leads to the quantization of both the electric charge e and the magnetic charge qm. The magnetic charge of a monopole would be n qm Uc 2e for n 1, 2, 3, \u00c1 12-34 >Uc Uc>e21>. q2 m 1016 GeV>c2, In the unified theories the quantizaIt is important to note that tion of electric charge occurs naturally in units of e, and magnetic monopoles of charge qm and mass Mm are then predicted. The predicted values of Mm are very large, about far beyond the energy achievable in any accelerator. Cosmic ray searches for monopoles place an upper limit on their flux at about 1015 cm2 s1 per unit solid angle. Coincidentally, this value corresponds approximately to the maximum flux that could exist in the Milky Way without having long since destroyed the galactic magnetic field. As of this writing, only a single possible observation of a magnetic monopole has been reported in the literature, by B. Cabrera in 1982. This observation is inconsistent with current limits. 12-5 Beyond the Standard Model 609 Quantum Gravity The addition of quantum gravity to grand unified theories is a formidable task. Called superstring theories because of their basic view of fundamental particles as strings rather than points, perhaps the most promising of the current versions is based on a 10-dimensional universe (nine space and one time dimension) in which six of the space dimensions have been collapsed or curled up on themselves. The string \u201clengths\u201d are much shorter than can be measured, about 1035 m. Besides the inclusion of quantum gravity, superstring theories also produce the gauge", " theories24 with the correct mediating bosons; however, although they are the subject of considerable interest to theoretical physicists, there is as yet no experimental support for these theories, and it is not clear to what extent, if any, they represent physical reality. Many questions are still unanswered. For example, do the quarks have internal structure? What is the origin of isospin? There is some indication that hadrons are surrounded by a \u201csea\u201d of virtual quark-antiquark pairs. What is their role? Is the fractional charge of the quarks related to color? Investigating these problems experimentally will require new, higher-energy accelerators and more advanced detectors than currently exist anywhere in the world. Obviously, there is much to be done. EXPLORING Neutrino Oscillations and Mass Quantum mechanics requires that, if neutrinos oscillate from one type, or flavor, to another, then they must have mass, whether they originate in the Sun, the atmosphere, a nuclear power reactor, an accelerator, or somewhere else in the cosmos. While the detailed justification of this requirement is beyond the scope of our discussions, an outline of why this must be true is presented here. The relationship between neutrino (flavor) wave functions and the mass eigenfunctions is given by \u2019 Ui \u2019 12-35 i 3), \u2019 and, 1, ), \u2019 (i ( are the wave functions of the neutrino flavors 2, e, are the where Ui (and its complex conjugate) is a function mass eigenfunctions that describes the extent to which mixing of the flavors or masses occurs, i.e., the \u2019 phases of the oscillations. Note that, if there were no oscillations, would equal the mass eigenfunction for however, the experiments noted above The states with different mass eigenfunctions propagate at different show that speeds, the less massive moving faster than the more massive ones. Since the mass eigenfunctions are combinations of neutrino (flavor) wave functions, the difference in speeds results in interference between the neutrino waves in each mass eigenfunction. When, eventually, constructive interference occurs, one neutrino flavor has changed into another. and the 1. 1 ; Ui Ui The mass eigenfunctions are plane wave solutions of the time-dependent Schr\u00f6dinger equation. (See Equation 6-7.) \u2019 i(x, t) ei(pxEt)>U 12-36 610 Chapter 12 Particle Physics The dependence of the energy on the mass is given by Equation 2-31 and the extremely relativistic approximation of Equation 2-36 E 2(pc)2 (mc 2)2 2-31 written as E (pc) c1 1>2 (mc2)2 (pc)2 d Since for each of the neutrinos under discussion the total energy mial expansion allows us to write E W mc2, the bino- E pc (mc2)2 2pc \u00c1 12-37 v c, After time t moving at mass wave functions become the neutrino has traveled a distance x ct L and the \u2019 i(L) e iL Uc (mc2)2 2E \u2019 i(0) 12-38 and the probability that a neutrino of flavor at t 0 will be observed to have changed or oscillated to flavor is given by * iUie i iL Uc (mc2)2 2E ` 2 12-39 If we confine our attention to the two-neutrino case (the three-neutrino case is more complex to describe), Equation 12-39 becomes (after some work!) P S, sin2 2 sin2 a 1.267 \u00a2m2L E GeV eV 2km b 12-40 4 is the neutrino mixing angle and \u00a2m where is the difference in the masses of the two neutrinos. Although Equation 12-40 applies to a two-neutrino world, its is a decent oscillations in the atmosphere since electron neutrinos approximation for the do not contribute significantly in this case. It is also reasonable for solar electron neutrinos oscillating to superpositions of and From Equation 12-40 we now can see why flavor-changing neutrinos must have and measure for solar neutrinos ) 45\u00b0, then in both \u00a2m2 and for atmospheric neutrinos Thus, neutrinos have mass. The current values of mass. Since experiments show that e, ( 33.9\u00b0 (0) cases ) \u00a2m2 0. P S, 0 ( are. \u00a2m2 solar \u00a2m2 atmo 8.0 105 eV 2 2.4 103 eV 2 More GUTs aren\u2019t the only avenue being actively explored by physicists in", " search of a deeper understanding of the structure of matter than we now have. The so-called Theories of Everything that seek to account for all of physics within a single theoretical construct are highly speculative, and none yet have any experimental support. On the home page at www.whfreeman.com/tiplermodernphysics5/e. Summary TOPIC 1. Particles and antiparticles Feynman diagrams Leptons and quarks 2. Fundamental interactions Force carriers General References 611 RELEVANT EQUATIONS AND REMARKS Each fundamental particle found in nature has an antiparticle; some are distinct, e.g., the electron and positron; some are the antiparticles of themselves, e.g., the photon. These are spacetime diagrams that provide a useful way of visualizing interactions between particles\u2014for example, Coulomb repulsion of like charges. All visible matter is made of two types of elementary particles, leptons and quarks, each consisting of three generations. 1. Strong interaction\u2014gluons 2. Electromagnetic interaction\u2014photons 3. Weak interaction\u2014W, Z 0 4. Gravitational interaction\u2014graviton Interaction \u201cstrengths\u201d This term refers to the magnitudes of the dimensionless coupling constants that multiply the space-dependent part of the potential energy functions. 3. Conservation laws and symmetries 4. Standard Model Color QCD 5. Beyond the standard model General References Every symmetry of the particle Hamiltonians leads to a conservation law and vice versa (Noether\u2019s theorem). Energy, momentum, electric charge, and angular momentum are conserved in all interactions. Some quantities are conserved in some interactions but not in others. For example, isospin is conserved in the strong interaction but not in the weak interaction. The Standard Model seeks to explain all matter in terms of the interactions among three types of elementary particles: quarks, leptons, and force carriers. All quarks and gluons have color charge with one of three possible values: red, blue, and green. The exclusion principle requires that all particles that occur in nature are colorless. The potential function of the strong interaction has the approximate form VQCD(r) 4 s 3r kr 12-32 Grand unification theories (GUTs) attempt to unify all four basic interactions mathematically. While thus far unsuccessful, some predict, among other things, proton decay, magnetic monopoles, and massive neutrinos. The latter has been verified by experiments. The following general references are written at a level appropriate for the readers of this book. Das, A., and T. Ferbel, Introduction to Nuclear and Particle Physics, Wiley, New York, 1994. Eisberg, R., and R. Resnick, Quantum Physics, 2d ed., Wiley, New York, 1985. Griffiths, D. J., Introduction to Elementary Particles, Prentice Hall, New York, 1987. Martin, B. R., and G. Shaw, Particle Physics, 2d ed., Wiley, New York, 1997. Rubbia, C., and M. Jacob, \u201cThe Z0,\u201d American Scientist, 78, 502 (1990). Frauenfelder, H., and E. M. Henley, Subatomic Physics, 2d Veltman, M., Facts and Mysteries in Elementary Particle ed., Prentice Hall, Englewood Cliffs, N.J., 1991. Physics, World Scientific, London, 2003. Fritzch, H., Elementary Particles, World Scientific, Singapore, Williams, W. S. C., Nuclear and Particle Physics, Oxford, 2005. New York, 1997. 612 Chapter 12 Particle Physics Notes 1. The word atom comes from the Greek word atomos, meaning indivisible, which was coined by the philosopher Democritus, a contemporary of Socrates, about 2400 years ago. In addition to suggesting that matter consisted of a variety of tiny atoms, he also suggested that the Milky Way was made of a large number of individual stars and that the Moon had mountains and valleys just like Earth. 2. Carl David Anderson (1905\u20131991), Swedish-American physicist. His discovery of the positron in cosmic ray cloud chamber tracks was followed three years later by his discovery of the muon in cloud chamber tracks recorded on Pikes Peak in Colorado. The former earned him a share of the 1936 Nobel Prize in Physics. 3. The Dirac equation for particles with -integral spin, like the electron, is the relativistic analog of the Schr\u00f6dinger equation; however, it is not obtained by operator substitution into Equation 2-31 since the resulting wave function does not include the effects of spin. 1>2 4. Richard Phillips Feynman (1918\u20131988), American physicist who described himself as a \u201ccurious character.\u201d An almost legendary figure among physicists in the United States, he was one of many who worked", " on the Manhattan Project at Los Alamos during World War II, where he also became an accomplished safecracker. An excellent bongo drummer and a passable artist, he shared the 1965 Nobel Prize in Physics with Julian Schwinger and Sin-itiro Tomonaga, all of whom independently contributed to the development of quantum electrodynamics. His books Surely You\u2019re Joking, Mr. Feynman! and What Do You Care What People Think? provide delightful insights into his life. 5. Emilio Gino Segr\u00e8 (1905\u20131989), Italian-American physicist. A lifelong friend and colleague of Fermi, Segr\u00e8 shared the 1959 Nobel Prize in Physics with Owen Chamberlain, a member of his Berkeley research group, for the discovery of the antiproton. Of greater interest to most people might be his discovery of technetium (Z 43), the first chemical element to be artificially created. The isomeric state of the isotope of technetium, 99Tc, is by far the most widely used radioisotope in medical diagnosis, treatment, and research. 6. This process is called M\u00f8ller scattering in QED. 7. In fact, an infinite number. The contribution that each possible diagram makes to the total process decreases sharply as the number of vertices increases, so complex diagrams may typically be ignored. 8. The name lepton, which means \u201clight particle,\u201d was originally selected to reflect the small mass of these particles relative to that of the hadrons; however, the (discovered by M. Perl in 1975) has a mass nearly twice that of the proton, so the name is no longer an indicator of the mass of these particles. 9. The reason for making the coupling constants dimensionless is so all observers will measure comparable values, independent of the units they may have used. 10. Carlo Rubbia (b. 1934), Italian physicist. He shared the 1984 Nobel Prize in Physics with Simon van der Meer for their contributions to the discovery of the W and Z 0. 11. Emmy A. Noether (1882\u20131935), German mathematician. Dismissed from her position at G\u00f6ttigen by the Nazi regime, she came to the United States in 1933. Her obituary in the New York Times was written by Einstein. 12. Operators, like Hop, that result in real (i.e., observable) values are called Hermitian operators. They obey the rule a(x)Fopb(x) dx b(x)F\u2026 opa(x) dx 13. Murray Gell-Mann (b. 1929), American physicist. He received the 1969 Nobel Prize in Physics for this and other work on fundamental particles and their interactions. 14. The rate of discovery became so large that one physicist quipped that \u201cby 1990 all physicists would be famous because (30,000). there would be a particle named for each physicist\u201d Most \u201cdiscoveries\u201d turned out to be spurious. 15. From a saying attributed to the Buddha: \u201cNow this, O monks, is the noble truth of the way that leads to the cessation of pain: this is the noble Eightfold Way: namely, right views, right intention, right speech, right action, right living, right effort, right mindfulness, and right concentration.\u201d 16. The name \u201cquark\u201d was suggested to Gell-Mann by a quotation from Finnegans Wake, by James Joyce: \u201cThree quarks for Muster Mark.\u201d Joyce did not tell us and the context does not make clear exactly what a quark is. 17. The correct quark combinations of hadrons are not always obvious because of the symmetry requirements on the total wave function. For example, the meson is represented by a linear combination of 18. Samuel Chao Chung Ting (b. 1936), American physicist, and Burton Richter (b. 1931), American physicist, shared the 1976 Nobel Prize in Physics for this important discovery. 19. One physicist put the long lifetime of the in biological terms by comparing it with someone coming upon a remote mountain village where the average age of the inhabitants was 70,000 years. That would be a definite indication of new biology. 20. Particle physicists call the discovery of the the \u201cNovember revolution,\u201d referring to the enormous support of the quark model that its November 1974 publication provided. 21. They form nine combinations, just like the mesons, but for the gluons the ninth combination is really a singlet and hence is independent. 22. Since no theory of quantum gravity complementing QED and QCD exists, current efforts to develop GUTs include only the strong and electroweak interactions. 23. Raymond Davis, Jr. (1914\u20132006), American physicist, and John Bahcall (1934\u20132005) American physicist. Davis\u2019s measurements won him a share of the 2002 Nobel Prize", " in Physics. J>\u2019 J>\u2019 and dd 0 uu 24. Theories in which the interaction is determined by the invariance of the theory (i.e., its mathematical equations) under particular transformations are called gauge theories. For example, classical electrodynamics is a gauge theory (although not usually referred to as such), as are QED and QCD. Historically, interactions were \u201cfigured out\u201d by clever physicists on the basis of experimental evidence. A bit of a surprise, Schr\u00f6dinger\u2019s wave mechanics is not a gauge theory. Problems 613 Problems Level I Section 12-1 Basic Concepts (b) (b) S. and (c) S p p; (a) Why 12-1. Two pions at rest annihilate according to the reaction must the energies of the two gamma rays be equal? (b) Find the energy of each gamma ray. (c) Find the wavelength of each gamma ray. 12-2. Find the minimum energy of the photon needed for the following reactions: (a) 12-3. Draw two different Feynman diagrams for each of the following events: (a) 12-4. Draw a Feynman diagram illustrating each of the following scattering events: (a) electron-electron, (b) electron-positron, and (c) Compton effect. 12-5. Find (a) the energy of the electron, (b) the energy of the mentum of each in the decay (n S p e). (The rest mass of p p S. 12-6. The fate of an antiproton is usually annihilation via the reaction Assume that the proton and antiproton annihilate at rest. (a) Why must there be two photons rather than just one? (b) What is the energy of each photon? (c) What is the wavelength of each photon? (d) What is the frequency of each photon? 12-7. Figure 12-2 shows the production of the first antiproton. It was produced by the reaction p p S p p p p and required a minimum kinetic energy of 5.6 GeV. (The proton beam energy was actually 25 GeV.) Less energy would be required by either of the following reactions. Why is neither of them a possible alternative? Justify your answer. (a) (b) 12-8. Positronium is a bound state of an electron and a positron (see Section 2-4). Its lifetime expressed in natural units used by particle physicists ( where m electron mass and the fine-structure constant. Use dimensional analysis (a) to include U nucleus, and (c) the moassuming no neutrino in the final state and c in the expression for and (b) to compute the value of 32P S 32S e, 32P is 31.973762 u.) 2> m5, U c 1 ) is 32S. Section 12-2 Fundamental Interactions and the Force Carriers 12-9. Name the interaction responsible for each of the following decays: (a) (b) (c) (d) 12-10. Which of the following decays\u2014 pect to have the longer lifetime? Why? 12-11. Of the reactions below, which are allowed to proceed via the weak interaction and which are forbidden? Justify your answer. (a) (b) (c) (d0 \u00a1 e e \u00a1 n 0 S or S \u2014would you ex- 614 Chapter 12 Particle Physics 12-12. Which of the four fundamental interactions is most likely responsible for the following reactions? (a) (b) (c) (d) (e) ( f ) 12-13. Using the information concerning the neutrinos from SN1987A, including Figure 12-31, compute an upper limit to the mass of the electron neutrino. 16O (excited state) \u00a1 16O (ground state) e e \u00a1 ve ve 3H \u00a1 3He 40 30 20 10 0 Figure 12-31 Problem 12-13. Electron antineutrino energy versus arrival time in the Kamiokande detector in Japan for antineutrinos emitted by the supernova 1987A. The spread in arrival times (about 13 s) permits a calculation of an upper limit to the mass of the e. 0 5 t, s 10 are slightly different, but those of the and are and 12-14. The rest energies of the exactly the same. Explain this difference in behavior. 12-15. Draw Feynman diagrams of the following decays: (a) (b) (c) \u00a1 e ve \u00a1 \u00a1 v Section 12-3 Conservation Laws and Symmetries (c) (b) (2030), \u00a2(1232). -(1670), ve p \u00a1 e e 12-16. What is", " the uncertainty in the rest energies of the following particles? (a) 12-17. State which of the decays or reactions that follow violate one or more of the conservation laws, and give the law or laws violated in each case. (a) (b) (c) (d) (e) ( f ) 12-18. Determine the change in strangeness in each reaction that follows, and state whether the reaction can proceed via the strong interaction, the electromagnetic interaction, the weak interaction, or not at all: (a) (b) (c) 12-19. Determine the change in strangeness for each decay, and state whether the decay can proceed via the strong interaction, the electromagnetic interaction, the weak interaction, or not at all: (a) \" \u00a1,0,0 \u00a1 p 0 -0 \u00a1 p and (c) 0 \u00a1 -0 \u00a1 p 0, e, \" \u00a1 -0 e (b) Problems 615 1 1>2 T 1>2 n n n p p n n 12-20. The rules for determining the isospin of two or more particles are the same as those for combining angular momentum. For example, since for nucleons, the combination of two nucleons can have either T 1 or T 0 or may be a mixture of these isospin states. Since for the proton, the combination p p has and therefore must have T 1. T3 T3 Find T3 and the possible values of T for the following: (a) (b) (c) (d) (e) 12-21. Which of the following decays are allowed and which are forbidden? If the decay is allowed, state which interaction is responsible. If it is forbidden, state which conservation law its occurrence would violate. \u00a1 e (a) 0 \u00a1 e e ve (b) \u00a1 e e v (c) -0 \u00a1 (d) n \u00a1 p e (e) e 12-22. For each of the following particles, write down two possible decays that satisfy all con\", servation laws: (a) (d) 12-23. Consider the following reactions: and (e) e K., -0, 0, (b) (c) K p \u00a1 K0 K \" followed by \" \u00a1,0,0. and the Given that B 1 for the proton and B 0 for mesons and that baryon number is conserved, \" determine the baryon number of the 12-24. Which of the following decays and reactions conserve strangeness? (a) (b) (c) (d) (e) p p \u00a1, \u00a1 -0 \u00a1 -0 p \u00a1 \" \u00a1, 0 Section 12-4 The Standard Model ss, (c) (d) (b) us, ud, ud, and (e) 12-25. Find the baryon number, charge, isospin, and strangeness for the following quark combinations and identify the corresponding hadron: (a) uud, (b) udd, (c) uuu, (d) uss, (e) dss, (f) suu, and (g) sdd. 12-26. Find the baryon number, charge, isospin, and strangeness for the following quark combinations and identify the corresponding hadron (the charge and strangeness of the antiquarks are the negatives of those of the corresponding quarks, as with any other particle-antiparticle ds pair): (a) 12-27. Draw two Feynman diagrams that represent the decay of the antibottom quark. 12-28. Some quark combinations can exist in two or more isospin states, with each state corresponding to a different hadron. One such combination is uds. (a) What is the value of I3 for this combination? (b) What are the possible values of total isospin I for this combination? (c) Find the baryon number, charge, and strangeness of this combination, and identify the hadron corresponding to each isospin state. 12-29. The particle is a baryon that decays via the strong interaction. Its strangeness, charm, topness, and bottomness are all zero. What combination of quarks gives a particle with these properties? 12-30. Compute the approximate range of a weak interaction mediated by a W. 616 Chapter 12 Particle Physics 12-31. One mode of weak decay of the K0 is K0 \u00a1 -0 -0 -0. 00(1193) -0 S p. Showing the quark content undergoes a weak decay as follows: Showing the quark content of the particles, draw the Feynman diagram of this so-called semileptonic decay. -", "0 12-32. The of the particles, draw the Feynman diagram of this so-called nonleptonic decay. 12-33. Show that the neutron cannot undergo the weak decay shown for the in Problem 12-32. 12-34. The decay of the shown in Problem 12-32 can also proceed via the strong interaction. Showing the quark content of the particles, draw the Feynman diagram that illustrates the strong decay of the 12-35. The baryon, strangeness, charm, and bottom quantum numbers of the swer to (a), what is the quark content of the 12-36. Find a possible combination of quarks that gives the correct values for electric charge, baryon number, and strangeness for (a) and (b) 12-37. The D meson has strangeness 0, but it has charm of 1. (a) What is a possible quark combination that will give the correct properties for this particle? (b) Repeat (a) for the D meson, which is the antiparticle of the D. 12-38. The lifetime of the of the tween members of the same isospin multiplet be explained? is 0.8 1010 s and that is 1.48 1010 s, nearly twice as long. How can these differences in lifetimes be- K p S 0 00. 00(1193). (a) Determine the (b) From your an- can be produced by the reaction The lifetime of the 6 1020 s. 00(1193)? K K 0. 0 is Section 12-5 Beyond the Standard Model 12-39. Grand unification theories predict that the proton is unstable. If that turns out to be true, why does it mean that baryon number is not conserved? If leptons and quarks are interchangeable at the unification energy, does this mean that there is a new, conserved \u201cleptoquark number\u201d? 12-40. GUTs predict a lifetime of about 1032 y for the proton. If that is the case, how many protons will decay each year in the world\u2019s oceans? (Assume the average depth of the oceans to be 1 km and that they cover 75 percent of Earth\u2019s surface.) 12-41. Protons might decay via a number of different modes. What conservation laws are violated by the following possibilities? (a) (b) (c) p \u00a1 e -0 ve p \u00a1 p \u00a1 K0 \", and (e) Level II 12-42. Find a possible quark combination for the following particles: (a) (d) 12-43. State the properties of the particles made up of the following quarks: (a) ddd, (b) (c) and (d) 12-44. Show that the 12-45. Consider the following decay chain: cannot decay into two identical zero-spin particles.,.,0, sss. ub, (b) (c) Z0 n,, uc,,0 \u00a1 -0 0 -0 \u00a1 p 0 \u00a1 \u00a1 v \u00a1 e ve v (a) Are all the final products shown stable? If not, finish the decay chain. (b) Write the overall to the final products. (c) Check the overall decay reaction for the conserdecay reaction for vation of electric charge, baryon number, lepton number, and strangeness. (d) In the first step of the chain, could the have been a 0?,0 -0 Problems 617 12-46. There are six hadrons with quantum numbers (Q,U,S,C,B) (2,1,0,1,0); (0,1,2,1,0); (0,0,1,0,1); (0,1,1,0,0); (0,1,1,1,0); (1,1,3,0,0). Determine the quark content of each hadron. 12-47. Show that the following decays conserve all lepton numbers: (a) (b) (c) (d) v v \u00a1 e ve \u00a1 v n \u00a1 p e ve \u00a1 v 0 12-48. A with energy 850 MeV decays in flight via the reaction angles made by the momenta of the gammas with the original direction of the 12-49. Test the following decays for violation of the conservation of energy, electric charge, baryon number, and lepton number: (a) (b) (c) -0 \u00a1 p \u00a1 n p \u00a1 e ve 0 S. 0. Compute the v Assume that linear and angular momentum are conserved. State which conservation laws (if any) are violated in each decay. 12-50", ". Consider the following decay chain: \" \u00a1,0,0 \u00a1 e ve \u00a1 v \u00a1 n \u00a1 v \u00a1 e ve \u00a1 e ve v v (a) Are all the final products shown stable? If not, finish the decay chain. (b) Write the overall to the final products. (c) Check the overall decay reaction for the condecay reaction for servation of electric charge, baryon number, lepton number, and strangeness. \" 0 -0 is MeV>c2. particle at rest decays into a Level III 12-51. The mass of the hydrogen atom is smaller than the sum of the masses of the proton and 139.6 MeV>c2; the electron, the difference being the binding energy. The mass of the however, the masses of the quarks of which it is composed are only a few How can that be explained? -0 S p p. 12-52. (a) Calculate the total kinetic energy of the decay products for the decay is initially at rest. (b) Find the ratio of the kinetic energy of the pion to the kiAssume the netic energy of the proton. (c) Find the kinetic energies of the proton and the pion for this decay. plus a photon. (a) What is the total energy of the 12-53. A decay products? (b) Assuming that the kinetic energy of the is negligible compared with the energy of the photon, calculate the approximate momentum of the photon. (c) Use your result for (b) to calculate the kinetic energy of the (d) Use your result for (c) to obtain a better estimate of the momentum and the energy of the photon. 12-54. In this problem, you will calculate the difference in the time of arrival of two neutrinos of different energy from a supernova that is 170,000 light-years away. Let the energies of the neutrinos be MeV, and assume that the mass of a neutrino is eV>c2. Because their total energy is so much greater than their rest energy, the neutrinos 2.2 (a) If have speeds that are very nearly equal to c and energies that are approximately t1 to travel a distance x, show that are the times it takes for neutrinos of speeds E pc. MeV and 20 5 and and -0. E2 E1 -0 -0 t2 \u00a2t t2 t1 x u1 u1 u2 u1u2 u2 x \u00a2u c2 618 Chapter 12 Particle Physics (b) The speed of a neutrino of mass Show that when E W m0c2, m0 the speed u is given by and total energy E can be found from Equation 2-10. u c 1 1 2 a m0c2 E 2 b u1 x 170,000 c # y. u2 (d) Repeat the calculation in (c) using for the energies and mass given, and calculate m0c2 40 (c) Use the result for (b) to calculate from the result (a) for for the rest energy of a neutrino. 12-55. There are three possible decay modes for the each mode. (b) Which mode is the most probable? Explain why. 12-56. In a large accelerator, such as the Large Hadron Collider at CERN, the momentum of a proton in a circular orbit of radius R is given by p 0.3 RB where B is the magnetic field. Derive this expression. (a) Draw the Feynman diagrams for GeV>c, \u00a2t eV. CHAPTER 13 Astrophysics and Cosmology Physics is an experimental science. The formulation and acceptance of our current understanding of the physical world, from Newton\u2019s laws and Maxwell\u2019s equations to relativity theory and quantum mechanics, are based on countless experimental observations. In this chapter, we look outward from Earth into the cosmos and apply the principles and techniques of physics first to the composition and evolution of stars, a branch of physics called astrophysics, and then to the large scale structure and evolution of the universe, a field called cosmology. In doing so the scale of our discussions expands from the nanometer and femtometer dimensions of the molecules, atoms, and nuclei to the light-year and parsec dimensions of galaxies and space, a span of more than 40 orders of magnitude. When observing stars and galaxies, astrophysicists and cosmologists are limited to examining the electromagnetic radiation and occasional particles emitted at times past that happen to have traveled to the vicinity of Earth so as to arrive at the moment of observation. The information thus gained, together with the fundamental assumption that the laws of physics discovered here on Earth are also valid throughout the universe, forms the basis for their work. During most of history, the instrument used for studying the", " cosmos was the human eye. Though well adapted to life on Earth, the eye is a relatively poor instrument for the scientific examination of the sky because it stores information for only a small fraction of a second before transmitting it to the brain for analysis, is sensitive to a very limited portion of the electromagnetic spectrum, and has limited resolution and light-gathering capacity. Today, most of our information about the distant universe is received through telescopes. 13-1 The Sun As we look outward from Earth and beyond the Moon, the most obvious object in the sky is, of course, the Sun. It is important to us for several reasons. First, the light that reaches us from the Sun is responsible for life on Earth. It sustains a comfortable average temperature on Earth\u2019s surface and is the ultimate source of virtually all of our energy. Since the Sun contains nearly all of the mass of the solar system, it also provides the gravitational force that binds our planet to the system. But most important 13-1 The Sun 13-2 The Stars 13-3 The Evolution of Stars 13-4 Cataclysmic Events 13-5 Final States of Stars 13-6 Galaxies 619 630 639 644 647 653 13-7 Cosmology and Gravitation 662 13-8 Cosmology and the Evolution of the Universe 664 619 620 Chapter 13 Astrophysics and Cosmology for our purposes in this section, the Sun is the only star of the 100 billion or so in the Milky Way that is close enough for us to examine its surface in detail. The others are so far away that they appear only as point sources when viewed by even the largest telescopes. (The very recent exceptions are Alpha Orionis\u2014Betelgeuse\u2014which has been imaged by the Hubble Space Telescope, and Altar, which has been imaged by long-baseline interferometry.) What we learn from studies of our star not only provides a more complete understanding of the processes taking place in it, but surely applies to other stars as well. The Surface and Atmosphere of the Sun We can see only the thin outer layer of the Sun, the photosphere, which emits the light that makes the Sun visible. The photosphere is generally considered to be the surface of the Sun. The energy per second per square meter that arrives from the Sun at the top of Earth\u2019s atmosphere is called the solar irradiance or the solar constant f. It has been measured to be f 1.365 103 W>m2 13-1 The corresponding quantity for stars other than the Sun is called the radiant flux, as we shall see in Section 13-2. Using the solar constant, the Earth-Sun distance of 1 astronomical unit (AU) 1.496 108 km, and conservation of energy, we can calculate the luminosity L, which is the total power radiated by the Sun or by any star. The area A of a sphere of 1 AU radius is A 4r2 4(1.496 1011)2 m2 At that radius, each square meter receives energy from the Sun at the rate given by the solar constant. Therefore, the Sun\u2019s luminosity is given by L} L} L} Af 4(1.496 1011 m)2(1.365 103 W>m2) 3.84 1026 W 13-2 This is truly enormous power. If we could put a 1000-MW electricity-generating plant on each square meter of Earth\u2019s surface, all of them combined would only produce 0.1 percent of the power produced by the Sun. If we assume that the Sun radiates as a blackbody, we can use the luminosity of the Sun along with its radius (6.96 108 m) to calculate the effective temperature at the surface of the Sun from the Stefan-Boltzmann law. It states that the power per unit area R ( intensity) radiated by a blackbody in thermal equilibrium is proportional to the fourth power of its surface temperature: R T4 3-4 where Stefan\u2019s constant ture. If the radius of the Sun is R}, and T is the absolute temperathe intensity radiated at the surface of the Sun is 5.67 108 W>m2 # K4 R L} 4R2 } 13-3 Take care not to confuse the intensity R with the solar radius The effective temperature Te for the surface of the Sun is defined as the temperature for which the intensity radiated satisfies the Stefan-Boltzmann law for a blackbody: R}. R L} 4R2 } T4 e Solving for Te, we obtain a Te 1>4 R b a L} 4R2 } 1>4 b 13-4 EXAMPLE 13-1 The Temperature of the Sun\u2019s Photosphere Use the StefanBoltzmann law to calculate the effective temperature of the photosphere. SOLUTION L} Using 3.84 1026 W from Equation 13-2 in Equ", "ation 13-4, we have a Te L} 4R2 } 1>4 b c 3.84 1026 W 4(5.67 108 W>m2 # K4)(6.96 108 m)2 d 1>4 5780 K 104 102 1 10\u20132 10\u20134 10\u2013 Sun 5800-K blackbody 10\u201312 10\u201310 10\u20138 The intensity of solar radiation has been measured at wavelengths ranging from about 1013 m in the gamma-ray region to nearly 10 m in the radio region, a range accounting for over 99 percent of the Sun\u2019s emitted power. Over much of this span, the solar spectrum is quite well predicted by Planck\u2019s law of blackbody radiation (see Chapter 3) with T 5800 K, as shown in Figure 13-1. The distribution peaks in about the middle (yellow) region of the visible spectrum. This agreement between the measured and theoretical spectra is very constant and is one of the characteristics of the quiet Sun. If we examine the edge of the solar disk, called the limb, we see that it is sharply demarcated and darker than the rest of the Sun. From the sharpness of the limb, we conclude from the following reasoning that the photosphere is very thin. Atmospheric turbulence during daylight limits the angular resolution of optical telescopes to 1>3600 of a degree). At the about 1 arc second ( distance of the Sun, this corresponds to about 700 km. As we look at the Sun, the angle over Figure 13-1 The spectral distribution of energy emitted by the Sun closely matches that of a blackbody at 5800 K. The discrepancy between the measured and theoretical curves in the region illustrated is due mainly to the fact that the photosphere is not in thermal equilibrium. The hump at short wavelengths is due to x rays emitted by the corona, which is at a much higher temperature. (In this figure 10 6 m. ) 13-1 The Sun 621 m u r t c e p s Sun and 5800-K blackbody l i e b s V i 10\u20136 10\u20138 10\u201310 10\u201312 10\u201314 10\u201316 10\u201318 10\u20134 10\u20132 \u03bb, m 1 10 622 Chapter 13 Astrophysics and Cosmology (b) (a) Protective filter High T Low T A B C D Less than 1 arc second Figure 13-2 (a) Two sight lines of equal optical path length, A and B. Along B the observer sees deeper ( hotter brighter) into the Sun than along A; therefore, path B looks brighter than path A, so the limb looks darker than the disk of the Sun. For paths C and D, the angle over which the gas of the photosphere changes from transparent to opaque is smaller than we can resolve, so the limb looks sharp. (b) A full disk image taken in the visible spectrum or white light at the National Solar Observatory/Sacramento Peak, New Mexico, on October 28, 2003. This image shows sunspot groups and evidence of limb darkening. [SOHO (ESA & NASA).] which the gas of the photosphere changes from rarified and transparent to optically dense and opaque is smaller than we can resolve. Therefore, the photosphere must be less than 700 km thick, which is only about 0.1 percent of the solar radius. The relatively dark appearance of the limb tells us about the temperature gradient in the Sun\u2019s atmosphere. Figure 13-2 shows two paths, A and B, for viewing the Sun. Because the photosphere is more transparent when viewed at normal incidence than when viewed at a grazing angle, the light traveling along path B originates deeper in the Sun than light traveling along path A. Since the interior is hotter than the outer layers, the light traveling path B originates in a hotter (brighter) part of the Sun than light traveling path A. Thus, the light from the limb appears less intense, hence darker (cooler). By measuring the change in brightness from path A to path B, we can determine the temperature gradient in the photosphere. It is shown in the left portion of Figure 13-3. Notice in the right portion of Figure 13-3 that the temperature begins to rise sharply, accompanying the transition from the Sun\u2019s surface, the photosphere, into the solar atmosphere. 6000 5000 4000 Photosphere Chromosphere 0 0 200 400 600 800 Height above base of photosphere, km Figure 13-3 The temperature of the Sun decreases from the base of the photosphere outward to a minimum at about 500 km, then increases rather rapidly to about 15,000 K in the chromosphere. Outside the photosphere are two layers of the Sun\u2019s atmosphere that are not generally seen because of the brightness of the photosphere. The inner most of the two layers of the solar atmosphere, the chromosphere, is visible for a few seconds just before totality during a solar eclipse. Under high resolution", ", the chromosphere resembles a field of burning grass, although each burning \u201cblade\u201d is about 700 km thick and 7000 to 10000 km high and lasts for only 5 to 15 minutes. Spectral examination indicates that the temperature of the chromosphere increases with distance above the photosphere, averaging about 15,000 K (Figure 13-4). 13-1 The Sun 623 Figure 13-4 This ultraviolet image shows a loop in the magnetic field, seen circling back toward the Sun, trapping hot gas in the chromosphere. [SOHO (ESA & NASA).] When the totality of the eclipse blocks out the chromosphere, the outer layer of the Sun\u2019s atmosphere, the corona, becomes visible. It is decidedly nonuniform in thickness, consisting of faint white streamers extending 2 to 3 solar diameters into space. The temperature of the corona is approximately 2,000,000 K. Radiation from the corona would overpower that from the 5800 K photosphere, except for the fact that the gas of the corona is so rarified that the total energy it emits is minuscule compared to that of the photosphere. It does, however, account for the relatively high intensity of x rays emitted by the Sun, which shows up in Figure 13-1 as a deviation from the spectral distribution of the blackbody at short wavelengths. It is thought that the extreme temperatures in the corona are produced by acoustic waves generated in the Sun\u2019s interior that build into shock waves in the corona. These shock waves heat the gases of the Sun\u2019s outer atmosphere and give the particles so much energy that even the Sun\u2019s enormous gravity cannot confine them. These high-energy particles, mostly protons and electrons, stream outward from the corona continuously. They form the solar wind that pervades the entire solar system. The Sun\u2019s Interior We cannot see through the photosphere into the interior of the Sun. Consequently, our understanding of the processes there is purely theoretical. With the single exception of solar neutrinos, no radiation or particles originating in the interior reach us directly. To understand the principle features of the current theory, we need first to determine the Sun\u2019s mass, as we can do easily with the aid of Newton\u2019s law of universal gravitation 1.99 1030 kg. and the second law of motion. The result is that the Sun\u2019s mass For simplicity, theoretical models usually consider the Sun to be a nonrotating star in hydrostatic equilibrium. This means that the outward pressure at any point, which is presumed to be due to energy conversion processes occurring in the Sun\u2019s interior, is exactly balanced by the inward pressure of gravity. Although the mean denthe sity of the Sun enormous pressures that exist in the solar interior substantially exceed those that correspond to the electrodynamic forces that bind the electrons to the nuclei. Thus, the matter in the interior of the Sun\u2014and certainly within the core, the central region in which temperatures are high enough to allow hydrogen fusion\u2014must surely be in the plasma (ionized) state. is not much different from that of Earth (5.5 g>cm3), (1.4 g>cm3) M} 624 Chapter 13 Astrophysics and Cosmology EXAMPLE 13-2 Hydrogen in the Sun\u2019s Core Show that neutral hydrogen is unlikely to exist in the Sun\u2019s interior. SOLUTION The pressure at the center of the Sun Pc is of the order and mass> unit surface area a g 1 2 Ga M} R2 } b g, where b Pc M} R2 } 1015 N>m2. is the average acceleration due to gravity in the Sun. The pressure turns out to be about This is the pressure pushing on the surface of a hydrogen atom near the Sun\u2019s center. The resistance to this gravitational pressure would come from the Coulomb force tending to hold the atom together. That pressure is given by the Coulomb attraction between the proton and electron per unit surface area of the atom. Using the Bohr radius a0 for hydrogen, we have F A ke2>a2 0 4a2 0 ke2 4a4 0 (9 109 N # m2>C2)(1.6 1019 C)2 4(0.5 1010 m)4 2.9 1012 N>m2 Thus, the gravitational pressure in the Sun\u2019s interior, at least near the center, exceeds that tending to hold the hydrogen atoms together by a factor of about 1000\u2014making it unlikely that neutral hydrogen atoms could exist there. Remarks: Given the Sun\u2019s density, even the particles in the depths of the core are still relatively far apart, so that the plasma behaves much like an ideal gas. This allows calculation of the core temperature from the ideal-gas law. It is found to be about 1.5 107 K. The Source of the Sun", "\u2019s Energy Using the value for the luminosity of the Sun that we computed earlier, Lord Kelvin was the first to point out that the present energy content of the Sun as calculated from thermodynamics would be radiated away in about 3 107 years. Since life has existed on Earth for approximately 100 times that long, we can conclude that the Sun has been radiating at close to its present luminosity for at least 3 109 years. Therefore, the Sun must have a supply of energy far larger than that represented by gravitational potential energy, the hot plasma, and the observed radiation field. The source of the Sun\u2019s energy is nuclear fusion. Current theory proposes that, as the young Sun contracted, its temperature rose. To understand why the Sun contracted and why that caused its temperature to rise, we start with Newton\u2019s law of universal gravitation. Fgrav G Mm R2 where Fgrav is the attractive gravitational force between the masses M and m which are separated by a distance R, and G is the constant of gravitation.. Notice that, as R becomes smaller, Fgrav becomes larger, which means that the masses M and m move 13-1 The Sun 625 Surface atom Core atom Fg Fg Fg Fg Surface atom Figure 13-5 Atoms in the outer areas of stars feel a net gravitational force directed toward the core. The net gravitational force on the atoms in the core is essentially zero. In the absence of an outward-directed pressure, the star collapses. toward one another with increasing acceleration. Conservation of energy requires that the resulting increase in kinetic energy must come from somewhere. That \u201csomewhere\u201d is the gravitational potential energy Ugrav of the masses in their original positions; Ugrav must decrease correspondingly. How might this account for the energy emitted by the Sun (and other stars)? A star is a huge ball of gas. The gas atoms near the surface feel the gravitational force attracting them toward the inner atoms of the star\u2019s core. The core atoms feel that force, too, but they are also attracted in the opposite direction by the gas atoms near the surface on the other side; hence, the core atoms don\u2019t move. (See Figure 13-5.) However, the entire surface of the star accelerates toward the core\u2014the star undergoes gravitational contraction. The increasing kinetic energy of the accelerated atoms (heat) increases the star\u2019s temperature, radiating energy into space. Eventually the temperature of the core reached about 1.5 107 K, which is high enough for the hydrogen nuclei (protons) in the plasma to have sufficient energy on the average (about 1 keV) to fuse into helium nuclei. This reaction, actually a chain of reactions, was first proposed by H. A. Bethe1 and is referred to as the proton-proton cycle. The first reaction in the chain is 1H 1H \u00a1 2H e e 0.42 MeV 13-5 Due to the height of the Coulomb barrier the probability for this reaction is very low except for those protons in the high-energy tail of the Maxwell-Boltzmann distribution. Fusion is possible only because of quantum mechanical tunneling. This sets a limit on the rate at which the Sun can produce energy and thus ensures a long lifetime for the Sun and similar stars. This limit is sometimes called the \u201cbottleneck\u201d of the solar fusion cycle. Once (deuterium) is formed via Equation 13-5, the following reaction becomes very probable: 2H 2H 1H \u00a1 3He 5.49 MeV It is followed by 3He 3He \u00a1 4He 2 1H 12.86 MeV 13-6 13-7 This process by which hydrogen nuclei are \u201cburned\u201d to helium nuclei is shown sche4He, matically in Figure 13-6a. There are other possible reactions for converting all of which have the same net Q value. Their rates, however, differ depending on the composition and temperature of the Sun\u2019s interior. The most important of these is the CNO (carbon-nitrogen-oxygen) cycle, which accounts for about 1.5 percent of the total solar luminosity. The CNO cycle is very temperature dependent and is the dominant H-fusion cycle for stars slightly more massive than the Sun. 1H to L} The neutrinos produced in the proton-proton cycle escape from the core, providing our only means for direct observation of the Sun\u2019s interior. The measured luminosand the known total Q value of the proton-proton cycle enable a calculation of ity the total reaction rate. In addition, the alternative reactions for have different neutrino energy spectra, thus providing a way of determining the relative contributions of each reaction and gaining further information about the composition and temperature of the core. The neutrino flux arriving at Earth from all reactions in the proton-", "proton fusion cycle determined by John Bahcall\u2019s2 definitive theoretical analysis of the solar neutrino spectrum based on the standard solar model is shown in Figure 13-6b. 3He 626 Chapter 13 Astrophysics and Cosmology e+ \u03bd e \u03bd e e+ 2H 2H p p p p p p (a) \u03b3 \u03b3 3He 3He p p 4He 1012 1011 1010 109 108 107 106 105 104 103 102 101 pp \u2192 \u00b11% 7Be\u2192 \u00b110.5% pep\u2192 \u00b12% 13N\u2192 15O\u2192 17F\u2192 7Be\u2192 \u00b110.5% aB \u2192 \u00b116% hep \u2192 \u00b116% (b) 0.1 1 Neutrino energy in MeV 10 Figure 13-6 (a) The proton-proton cycle is the primary source of the Sun\u2019s energy. The neutrino created in the initial step escapes from the core. The net energy produced, including that released from orbital electron binding and 26.7 MeV per are shown by solid lines; those produced in the CNO cycle are shown by dotted lines. [J. Bahcall and A. Serenilli, ApJ, 621, L85 (2005).] produced. (b) Neutrino flux at Earth predicted by the standard solar model. Neutrinos produced in the p-p cycle annihilation, is about ee 4He 8B p S 8Be* e ne For those neutrinos resulting from the reaction (Table 13-1) the predicted intensity is 8.1 1.3 Solar Neutrino Units (1 SNU 1 event per 1036 target atoms per second). However, experiments conducted by Ray Davis and his co-workers over more than 30 years using a chlorine radiochemical detector located deep inside a gold mine in South Dakota which was sensitive primarily to the 8B-produced neutrinos3 found the measured rate at which solar neutrinos from this reaction arrive at Earth to be 2.56 0.16 SNU, only about 32 percent of the expected rate. Subsequently, experiments sensitive to other reactions in the p-p cycle performed at six other laboratories around the world confirmed this discrepancy. This discrepancy is referred to as the solar-neutrino problem. Davis shared the 2002 Nobel Prize in Physics for this discovery. The discrepancy between the theoretical prediction of the standard solar model and the experimental results presented a very serious problem for both astrophysics and particle physics. In the words of John Bahcall, whose calculations provided the solar model prediction: Is the solar neutrino problem caused by unknown properties of neutrinos or by a lack of understanding of the interior of the Sun? In other words, is this a case of new physics or faulty astrophysics? It turned out to be a case of \u201cnew physics.\u201d As was described in Section 12-5, the recent discovery of neutrino oscillations that enables neutrinos of one flavor to change into neutrinos of another flavor, means that electron neutrinos emitted in the Sun\u2019s p-p fusion cycle may oscillate to muon or tau neutrinos during their trip from the Sun to Earth. Davis\u2019s neutrino telescope was sensitive only to electron neutrinos. This accounts for the discrepancy and also, according to the Standard Model of particle physics, requires that neutrinos have a nonzero mass. 13-1 The Sun 627 Table 13-1 Proton-proton nuclear fusion cycle Reaction % of events energy (MeV) 1. 2. 3. 4. 5. 6. 7. p p \u00a1 2H e e or p e p \u00a1 2H e 2H p \u00a1 3He 3He 3He \u00a1 4He 2p or 3He 4He \u00a1 7Be 7Be e \u00a1 7Li e 7Li p \u00a1 2 4He or 7Be p \u00a1 8B 8B \u00a1 8Be* e e 8Be* \u00a1 2 4He or 3He p \u00a1 4He e e 99.96 0.04 100 85 15 15 0.02 0.423 1.445 0.863 (90%) 0.385 (10%) 15 0.00003 18.8 Source: Data from J. Bahcall, Phys. Rev. C, 56, 3391 (1997). The Active Sun In addition to the relatively stable phenomena that we have discussed, the Sun exhibits a number of transient phenomena, most of which are associated with its magnetism. We noted earlier that the solar interior must be primarily a plasma composed of protons and electrons. The Sun rotates with different angular velocities at different latitudes. At any given latitude, it probably has different angular velocities at different distances from the spin axis as well. The complex motions resulting from this differential rotation and from the rise and fall of charged particles in the", " convection zone between the core and the photosphere are probably the source of the Sun\u2019s chaotic magnetic field structure (Figure 13-7). This transient structure may have localized magnetic field strengths exceeding 1 T on occasion. The transient structure is superimposed onto a general average magnetic field of about 104 T, roughly twice Earth\u2019s average magnetic field. The origin of this general field is not known, except that it is not a remnant of the Sun\u2019s formation, since a primeval field would likely have decayed away by now. Its presence poses formidable problems for any theoretical solar model. Not only must the model explain the origin of the general field, but it must also account for the fact that its polarity reverses every 11 years, in step with the sunspot cycle. Figure 13-7 The field lines that describe the Sun\u2019s magnetic field structure at any given time are derived from ground-based measurements, e.g., Zeeman effect and the transport of charged matter. The highintensity, chaotic structure is superimposed onto a constant general magnetic field of about 104 T. 628 Chapter 13 Astrophysics and Cosmology N N N S S S (a) (b) (c) Figure 13-8 (a) Magnetic flux lines are distorted by the Sun\u2019s differential rotation and pushed up out of the surface by motion in the convection zone. (b) A sunspot occurs where a bundle of field lines leaves and reenters the surface. The areas where they leave and return to the surface become sunspots. This very large spot on this part of the Sun\u2019s surface is about 20 arc seconds in diameter. The granular appearance of the Sun\u2019s surface is very apparent. Sunspot activity on 29 March 2001. [(b) Institute for Solar Physics of the Royal Swedish Academy of Sciences. (c) SOHO (ESA & NASA).] Sunspots, dark blemishes on the solar disk, were first reported in pretelescope times and were observed by Galileo in 1610. They originate in the following way, according to one of the current models: As shown in Figure 13-8, the Sun\u2019s magnetic field lines are distorted into bundles or tubes by the Sun\u2019s differential rotation. Occasionally vertical movements in the convection zone may push a bundle through the surface. The area where it leaves the surface and the area where it returns to the surface become the sunspots. They appear darker than the adjacent photosphere, which means that they are cooler, typically around 3800 K. One of the pair of spots will have a magnetic north pole, the other a south pole. If the bundle of field lines does not protrude completely through the photosphere, only a single sunspot is formed. The number of spots per year varies regularly from about 20 to about 150 in a cycle of 11 years, as can be seen in Figure 13-9. Early in each new cycle, the sunspots form at a latitude of about 30\u00b0. As the Sun progresses through its 11-year cycle, the spots form progressively closer to the equator. There is an additional cyclical variation in the annual number of sunspots with a period of about 100 years that is also apparent in Figure 13-9. Currently, the theoretical explanation for these regularities, while in agreement with some features of the observations, is not complete. Solar flares are violent, stormlike phenomena that appear to be associated with the large magnetic fields in the vicinity of sunspots. There is, however, no generally accepted model to explain them. Solar flares erupt explosively, ejecting particles and emitting radiation ranging from the x-ray through the radio regions of the spectrum. They last anywhere from a few minutes to a few hours and can have temperatures as high as 5 106 K. The particles ejected by solar flares reach Earth within a day or so and often produce auroras as they accelerate in Earth\u2019s magnetic field. Solar flares can disrupt some types of radio transmissions, and on rare occasions can generate surges in high-voltage transmission lines. A flare that happened to be directed toward Earth in 1996 caused, among other things, the failure of a communications satellite. 13-1 The Sun 629 200 180 160 140 120 100 80 60 40 20 1600 1640 1680 1720 1760 1840 1880 1920 1960 2000 1800 Year Figure 13-9 The number of sunspots that occur each year has varied regularly on an 11-year cycle for more than 270 years. The unexplained absence of spots between about 1650 and 1700, referred to as the Maunder minimum, approximately coincides with a period of unusually low temperatures in Europe referred to as the \u201cLittle Ice Age.\u201d Whether there exists a causal connection between the two phenomena is a matter of scientific debate. Sunspot cycle 23 began in 1996 and ended in 2007. (Counting the cycles began in 1775.) Sunspot activity is next expected to peak in 2011\u20132012. Two other", " transient solar phenomena related to sunspots are plages and filaments. Plages are bright (hotter) areas adjacent to the dark sunspots. The evolution of plages suggests that they are areas of increased mass density, resulting perhaps from the movement of the magnetic field bundles generated from the sunspots. Filaments are dark, thin lines that seem to thread their way across the disk, sometimes for thousands of kilometers. They do not lie on the surface but extend out into space, sometimes more than 100,000 kilometers, in graceful loops and swirls. Filaments that are seen projecting into space at the Sun\u2019s edge are called prominences. They may erupt and disappear quickly or persist for several weeks. While prominences appear closely related to the shape of the magnetic field, as with other transient features, there is as yet no model that fully accounts for them. The huge handle-shaped prominence shown on the upper right was photographed in 304 \u00c5 [30.4 nm] light on 14 September 1999. It consists of charged particles confined by the magnetic field of the sun. [SOHO (ESA & NASA).] 630 Chapter 13 Astrophysics and Cosmology EXPLORING Is There Life Elsewhere? We are certainly not the first to ask that question. The Greek philosophers beginning with Thales and continuing through Plato and his student Aristotle thought and wondered about the structure of the heavens and the mysteries they might contain. Many scientists in the nineteenth century assumed that the other planets of the solar system, particularly Mars and Venus, were inhabited. In the twentieth century the term Martian became synonymous with \u201cbeings from outer space.\u201d An entire movie and television genre has flourished based on time travel, spaceships, aliens from outer space, and a plethora of weird science\u2014pseudoscience. The real issue, however, is much more serious: Is life \u201cout there\u201d possible? The answer is surely, Yes. Consider: With the development and evolution of telescopes we have learned that the motions of stars and galaxies obey the same laws of physics that have been discovered on Earth. Our location is in no way special. The physical processes that occur on Earth occur throughout the universe. All of the chemical elements discovered in our studies of the near and distant universe also occur on Earth. The relativistic and quantum physics we have developed works in the cosmos, too. So, then, must our biology\u2014an application of physics and chemistry\u2014work throughout the universe. On Earth we have learned that life-forms can survive and prosper in seemingly hostile environments. Sea animals thrive in the scalding hot water and enormous pressures of the deep ocean volcanic vents. Other organisms live in the rocks of deep mines and in the permanent ice of the Antarctic. Other creatures have been discovered that use sulfur in their metabolism, rather than oxygen. And the complex organic molecules that are the building blocks of life as we know it have been found in meteorites and identified in interstellar gas clouds. Thus, it would appear that the development of life elsewhere in the Galaxy and the universe may not be all that unusual. If intelligent life has developed elsewhere, where is it and how do we find it? Since 1995 new technology and techniques have led to the indirect discovery (e.g., via stellar motion perturbations) of 229 extrasolar planets (called exoplanets) orbiting 194 relatively nearby stars (as of April 2007). While most of the discovered planets are hot and large, Neptune to Jupiter in size, at least two are cool and about 5 times Earth\u2019s mass orbiting a faint star in the Milky Way\u2019s central bulge. As of now, only one exoplanet has been imaged directly. If intelligent life on these planets (or the millions of others that must exist throughout the universe) are sending electromagnetic signals into space, as we are, detecting those signals could provide the first clue that we are not the only intelligent life that has existed in the universe. Listening for those signals is the objective of the Search for Extraterrestrial Intelligence (SETI) project. Look for SETI online to learn more. 13-2 The Stars On clear, dark nights, we can see about 6000 stars without the aid of telescopes. The sight is incredibly beautiful and must surely have been just as awesome to our forebears. A cursory glance at the night sky reveals the following features: the distribution of stars is not uniform, the stars do not all have the same brightness, and there is a dim irregular band of light bisecting the sky. In this section, we will investigate these features. The hazy band of light that stretches across the entire sky is the Milky Way. With the aid of a small telescope or even binoculars, the band is resolved into a mass of individual stars. It is part of a huge Galaxy4 containing an estimated 1011 stars that are 13-2 The Stars 631 bound together gravitationally in our region of the universe. Most of the stars visible to", " the unaided eye seen in any direction are simply those members of the Galaxy close enough to Earth to be individually resolved by the eye. Constellations Chance groupings in the celestial pattern, usually among the brighter individual stars, are called constellations. Ancient peoples associated them with persons, gods, and objects from their histories, religions, and myths, probably as mnemonic devices. The constellations, as well as several prominent stars, have always had practical uses. For centuries, seafarers have used the Pole Star (in the northern hemisphere) and the Southern Cross (in the southern hemisphere) as aids in navigation. In ancient Egypt the pharaoh\u2019s advisors learned to predict the life-sustaining annual flooding of the Nile by watching for the first appearance of the bright star Sirius5 above the horizon in the early spring. Today, 88 constellations (see Figure 13-10 for some of them) are used by astronomers to identify sections of the sky. For example, the center of the Milky Way is said to be \u201cin Sagittarius,\u201d meaning it is in the direction of the constellation Sagittarius. (The center of the Galaxy is actually more than 10 times farther from the Sun than are the stars that form that constellation.) Figure 13-10 Star chart of the sky as it appears on a spring evening at latitude 40\u00b0 north, showing many of the constellations visible. During the night, the entire pattern revolves about 120\u00b0 about Polaris, the Pole Star. To use the chart, hold it (or a copy) in front of you with the S (south) at the bottom while you face south. Match the lower half to the stars that you see. Then rotate the chart, putting the W at the bottom, face west and again match the lower half to the stars you see, and so on. [R. A. Freedman and W. J. Kaufmann III, Universe, 8e (New York: W. H. Freeman and Co., 2008), p. S-6.] 632 Chapter 13 Astrophysics and Cosmology Globular cluster G1 in galaxy M31 contains more than 300,000 stars. G1 orbits the Andromeda galaxy, the nearest large spiral to Earth. The two bright stars with \u201cspikes\u201d are in the Milky Way. [Michael Rich, Kenneth Mighell, and James D. Neill (Columbia University), and Wendy Freedman (Carnegie Observatories), and NASA.] Stellar Populations One characteristic of our Galaxy is that certain regions of it have many more stars than other nearby regions. Such concentrations are called star clusters. There are two types of star clusters. Galactic clusters, also called open clusters, may contain from about 20 to several hundred stars. All stars in galactic clusters appear to have very similar compositions, as inferred from studies of their optical spectra. About 70 percent of their mass is hydrogen, another 28 percent or so is helium, and 2 to 3 percent consists of elements heavier than helium. Stars with this characteristic composition, like our Sun, are referred to as population I stars. Globular clusters may consist of 103 to 106 stars in a compact, roughly spherical group. Their concentrations of elements heavier than helium are all very similar and much lower than that of population I stars, typically 0.1 to 0.01 percent. These are called population II stars. One such cluster, photographed by the Hubble Space Telescope, is shown in the photo at the left. Population I stars are thought to be current generation stars that formed after the gas and dust that exists between these stars had been enriched by the products of ancient fusion reactions in the early universe. The lower concentration of heavier elements in the population II stars suggests that they are of a previous generation, hence older than those of population I. The fact that they are found in regions of space where there is little dust or gas tends to support that interpretation. Classification ofStars Stars are grouped into classes based primarily on the spectral lines each emits and absorbs. That different stars have different spectra was discovered nearly 200 years ago by Joseph Fraunhofer who also measured numerous absorption lines in the solar spectrum. Over the years, advances in spectroscopy, instrumentation, and atomic theory enabled astrophysicists Edward Pickering and Annie Jump Cannon6 to rearrange systematically the earlier classification scheme into a temperature sequence. Stars are grouped according to temperature categories (or spectral types) ranging from hot blue, so-called O stars, to cool red, so-called M stars. The seven categories are: O B A F G K M. Generations of students have memorized the classifications by using the phrase \u201cOh Be A Fine Girl/Guy, Kiss Me.\u201d Table 13-2 lists some of their important characteristics. Cannon also added ten subdivisions (0 to 9) within each category to provide for finer distinctions between the stars in each group. For example, B0 stars are hotter (called early type) than B9 stars (called late type", "). The physical basis for the distinction between the groups and early/late types lies in the quantum mechanical details of the spectra and the atomic electron excitations and ionizations of the elements comprising each star. Improved observational techniques and analytical methods have led to a number of additional classifications including several for hot blue emission stars and classes L, T, and Y for cool red and brown dwarfs. Stellar Magnitudes The Greek astronomer Hipparchus7 devised the first classification of stars based on how bright each appeared. Called apparent magnitude and represented with the letter m, the values he assigned ranged from m 1 for the brightest stars to m 6 for the dimmest visible to his eye. (The telescope had not yet been invented.) As time passed and technology was developed and improved, astronomers extended and refined the apparent magnitude scale. The modern definition of the 13-2 The Stars 633 Table 13-2 Characteristics of star categories Spectral type Important characteristics O B A F G K M Hottest blue-white stars; helium absorption lines Hot blue-white stars; helium and hydrogen absorption lines White stars; hydrogen and calcium absorption lines Yellow-white stars; calcium and some metal absorption lines Yellow stars; solar-type spectra with calcium and iron absorption lines (The Sun is a G2 star.) Cool orange stars; strong metal absorption lines Coolest red stars; strong metal absorption lines apparent magnitude scale is that a difference of 5 in the value of m corresponds to a factor of 100 in brightness; that is, a difference of 1 in the value of m between two stars means the ratio of their respective brightness is Thus, star A with m 2 is 2.51 times brighter than star B with m 3, 2.51 2.51 6.31 times brighter than star C with m 4, and so on. (Note that smaller m-values mean brighter, larger m-values mean dimmer.) Modern technology enables scientists to measure apparent magnitude with an accuracy of 0.01 and has vastly extended the range of m-values. For example, the brightest star in the sky, the Sun, has m 26.81 and the faintest objects that can be observed have about m 29. 1001>5 2.51. Of course, apparent magnitudes are not the whole story. Two stars with the same luminosity, but located at different distances from us will have different m-values, the furthest away being the dimmer and, therefore, having the larger m-value. So we define a new, more basic quantity, the absolute magnitude M, in terms of the radiant flux F which includes both the star\u2019s luminosity L and its distance R from Earth (see Section 13-1). The radiant flux F is defined as F L 4R2 13-8 4R2 is the surface area of a sphere of radius R and has SI units of meter Recall that J>s # m2. squared (m2). The units of radiant flux are then Using Equation 13-8, the absolute magnitude M of a star is defined as being equal to the apparent magnitude the star would have, if it were located at a distance of 10 parsecs (pc; see Section 13-3) from Earth. Using the expression for the radiant flux and the definitions for the apparent and absolute magnitudes, one can eventually obtain the expression below connecting F, m, M, and R: 100(mM)>5 F10 pc F a 2 b R 10 pc 13-9 where R is the actual distance between the star and Earth (measured in pc). 634 Chapter 13 Astrophysics and Cosmology Figure 13-11 Infrared view of the Milky Way taken by the COBE satellite showing the disk and the central bulge. [The COBE Project, DIRBE/NASA.] The Structure of the Milky Way Figure 13-11 is a map of the Milky Way viewed from the location of the Sun. The size and shape of the Galaxy are not at all obvious in the picture\u2014hardly surprising from the perspective of an observer inside the Galaxy itself.8 However, painstaking counts of the number of stars per unit volume in various directions have revealed that the Milky Way is basically a huge disk. Up until the early 1900s, astronomers thought that the Sun was located at the disk\u2019s center. The true size and shape of the Galaxy (Figure 13-12) were deduced by H. Shapley9 through a brilliant analysis of the distribution of globular clusters. He discovered that 200 or so globular clusters are distributed approximately spherically in space and proposed that the center of that distribution coincided with the center of our Galaxy. That center lies about 28,000 c y (8 kpc; see Equation 13-12) from the Sun, which is approximately one-third of the way out from the center. The Milky in diameter. It has been said that Shapley Way is roughly dethroned the Sun from the center of the Galaxy in much the same way that", " Copernicus had dethroned Earth from the center of the universe. # 1.63 105 c # y Disk Globular cluster 2000 c\u00b7 y Sun Central bulge 28,000 c\u00b7 y Figure 13-12 A diagram of the presently accepted structure of the Milky Way based on the work of Harlow Shapley. The Milky Way is brighter in the summer night sky in the northern hemisphere than in the winter because the summer night sky looks toward the center of the Galaxy, while the winter night sky is toward its outer edge. Following Shapley\u2019s work, astronomers studying other nearby galaxies with the aid of new, high-resolution telescopes found that the distribution of stars within those systems, many of which had open spiral structures such as shown in Figure 13-24b (page 655), depends in part upon the ages and compositions of the stars, with open clusters being found mainly in the arms of the spirals. Making the reasonable assumption that such distribution patterns would also hold for the Milky Way and with meticulous measurements of the distances to about 200 galactic clusters, astronomers have identified three spiral arms and a bar associated with the central bulge for the Milky Way. Thus, if we could look down on the Milky Way from the Galactic north pole, it would look much like Figure 13-13a. 13-2 The Stars 635 (a) (b) # Figure 13-13 (a) The combination of observations in the visible and radio regions of the spectrum reveal a spiral structure with a faint bar for the Milky Way. To an observer looking down on the Galaxy from about 1 million parsecs, the Milky Way might look something like this. The Sun is 28,000 c y from the center in one of the spiral arms. (b) Viewed from Earth, the center of the Galaxy is obscured by clouds of dust and gas that prevent most visible light from reaching us; however, it contains several areas of strong radio emission, the strongest of which is Sagittarius A*, a compact radio source that appears to dominate the large-scale motion of the galactic center. This radio image (taken at 6 cm # wavelength) is of the inner 8 c y of the Milky Way. The dark spot at the very center is Sagittarius A, which is very likely a huge black hole (see Section 13-5). This image was made using the Very Large Array, a radio frequency interferometer made of 27 synchronized antennae with an effective diameter of about 40 km, located in New Mexico. Its resolution is better than that of the best ground-based optical telescopes by about a factor of five. [(a) Gemini Observatory-GMOS Team. (b) HST Astronomy Imaging Workbench/Farhad Yusef-Sadeh/ Northwestern University.] 636 Chapter 13 Astrophysics and Cosmology EXPLORING The Celestial Sphere Copernicus\u2019s refined heliocentric model of the solar system and its subsequent verification by observations laid to rest for all time the geocentric model of Aristotle and Ptolemy. However, we still use one feature of the latter. Putting aside the motions of the planets and a few interplanetary space probes, our observations of the stars and specification of their locations in the sky are normally referred to a coordinate system centered on Earth, not the Sun. The distances to the stars are so great that the stars appear to us to be fixed relative to one another and collectively form the surface of a huge sphere\u2014the celestial sphere\u2014with Earth at its center. The celestial sphere rotates regularly each night from east to west, its axis of rotation coinciding with Earth\u2019s rotational axis and its north and south poles oriented in the same way as Earth\u2019s poles. The locations of the stars on the celestial sphere, like towns on a road map and points on the surface of Earth, are specified with two coordinates. For locations on Earth the coordinates are called latitude and longitude. The former specifies how many degrees north or south of the equator (which is defined as zero degrees, 0\u00b0 latitude) the point is located; the latter tells how many degrees west of zero degrees longitude the point lays. The longitude line (also called a meridian) that passes through Greenwich, United Kingdom, is defined as 0\u00b0. (See Figure 13-14a.) Since there are 360\u00b0 around Earth and our planet rotates on its axis once every 24 hours (1440 min), longitude is often expressed in time units (hours) rather than angle units (degrees or radians). 1440 min 360 deg 4.0 min deg For example, Orlando, Florida, is located at 28.4\u00b0 N latitude, 81.3\u00b0 W longitude. In time units 81.3\u00b0 W longitude is 81.3 deg 4.0 min>deg 325.2 min 5.42 h Thus, Orlando is 5.42 hours west of (i.e., earlier than) Greenwich. On the celestial sphere the locations of", " stars are described in an exactly analogous way. The analog of longitude on the celestial sphere is right ascension.10 It is represented by the lower case Greek letter Right ascension is measured in hours, rather than degrees, from zero up to 24 as on Earth. The analog of latitude is declination, represented by the lower case Greek letter 1. Declination is measured in degrees north () or south () of the celestial equator. (See Figure 13-14b.). Figure 13-14 (a) Diagrammatic definition of longitude and latitude on Earth. (b) Corresponding definition of right ascension and declination on the celestial sphere. marks the celestial sphere analog of Earth\u2019s 0\u00b0 longitude, the meridian through Greenwich, United Kingdom. 1 / Orlando, U.S.A. Latitude (a) Earth NP Greenwich, U.K. 0\u02da Equator (b) Celestial sphere NCP Star Earth \u03b4 \u03b1 / SP Longitude SCP Celestial equator 13-2 The Stars 637 Choosing the analog of the Greenwich meridian, i.e., the 0\u00b0 longitude, for the celestial sphere requires a bit of explanation. Since Earth rotates on its axis from west toward east, the stars fixed on the celestial sphere continually move across the sky from east toward west. Also, Earth orbits the Sun once every 365.26 days. This means that while Earth rotates on its axis once each 24 hours, it also advances along its orbit around the Sun slightly less than 1\u00b0 during that 24-hour period. To bring the Sun directly over the same meridian as on the day before, Earth must rotate very nearly 361\u00b0; however, doing the same thing with a star on the celestial sphere requires only a 360\u00b0 rotation, because the distances to the stars are vastly greater than Earth\u2019s daily motion in its orbit. As we noted above, 1\u00b0 corresponds to 4.0 minutes, so a given star rises in the east each night 4.0 minutes earlier than it did the night before, as a result of Earth\u2019s orbital motion around the Sun. In addition to the nightly advance of star rise, there is a gradual change in the orientation of the celestial sphere that varies with the seasons. This change is due to the fact that Earth\u2019s rotational axis is tilted at about 23.5\u00b0 with respect to the plane of our orbit around the Sun. This means that over the course of one year as viewed from Earth, the Sun follows a path, called the ecliptic, on the celestial sphere that ranges from 23.5\u00b0 north to 23.5\u00b0 south of the celestial equator. Thus, the ecliptic intersects 0\u00b0 declination twice each year (Figure 13-15), once on about March 20, the vernal (or spring) equinox, and again on about September 23, the autumnal equinox. The Sun reaches its maximum north declination of 23.5\u00b0 N on about June 21, the summer solstice and its maximum south declination of 23.5\u00b0 S on about December 21, the winter solstice11 (see Figure 13-15). By international agreement, the point at which the Sun\u2019s path projected against the celestial sphere (the ecliptic) crosses the celestial equator (0\u00b0 declination) in the spring, the vernal equinox, is defined as zero hours (and 0\u00b0 ) right ascension. On the celestial (see Figure 13-14b). sphere the vernal equinox is designated with the Greek capital letter There are a number of other small motions of Earth that affect the appearance of the celestial sphere over very long periods of time, e.g., the slow wander of Earth\u2019s poles; however, those are beyond the scope of our discussions in this chapter 30.0 23.5 20.0 10.0 0.0 \u221210.0 \u221220.0 \u221223.5 \u221230.0 Dec 21 Sept 23 Jun 21 Mar 20 Dec 21 Summer solstice Sun\u2019s path Vernal equinox Autumnal equinox Celestial equator Winter solstice 18 Winter solstice 12 6 Right ascension, h 0 18 Figure 13-15 The Sun\u2019s path as projected onto the celestial sphere. 638 Chapter 13 Astrophysics and Cosmology The Mass (and Missing Mass) of the Milky Way m>s Using the Doppler effect, J. Oort and B. Lindblad first demonstrated in 1926 that the Galaxy is rotating. The Sun is apparently moving in a circular orbit at a speed of about 2.5 105 toward the constellation Cygnus. Assuming that the Sun\u2019s speed is constant, we can compute the time for the Sun to complete one revolution around the center of # the Milky Way (a \u201cSun year\u201d) and the mass of the Galaxy. Since the Sun is 28,000", " c y from the galactic center, a Sun year is 2.1 108 Earth years. (See Problem 13-4.) EXAMPLE 13-3 The Mass of the Galaxy Calculate an approximate value for the mass of the Galaxy. Include in the calculation the mass that lies inside the Sun\u2019s orbit in the Milky Way. SOLUTION 1. Using Newton\u2019s law of gravitation, where the gravitational M} force acting on the solar mass by the mass of the Galaxy MG is given by: 2. This gravitational force provides the centripetal force that holds the Sun in its galactic orbit of radius R. Thus, F G M}MG R2 GM}MG R2 M}v2 R 3. Solving for MG gives: MG Rv2 G 4. Substituting values for the Sun\u2019s orbital radius R and speed v and for the uni- versal gravitational constant G gives: (28,000 c # y)(9.46 1015 m>c # y)(2.5 105 m>s)2 6.67 1011 Nm2>kg2 MG 2.48 1041 kg Remarks: Thus, if the Sun\u2019s mass is a representative average for the stars of the Milky Way, the Galaxy contains about 1.3 1011 stars. A problem arises in that, if we add together the masses of all of the visible stars in the Galaxy, including those beyond the Sun\u2019s orbit, plus all of the dust and gas clouds, we can account for only about 4 percent of the gravitational mass necessary to hold the Galaxy together. This discrepancy is referred to as the missing mass or dark matter problem. It exists for all galaxies and, indeed, for the universe itself. The first hint of the problem came in 1933. Based on his studies of the motions of the galaxies in the Coma Cluster of galaxies (see Section 13-6), F. Zwicky found that the mass of the cluster, estimated from the brightness and number of galaxies it contained, was too small by a factor of about 400 to account for the observed motions. He inferred that there must be some kind of unseen gravitational mass in the cluster\u2014dark matter. Various solutions to the problem, such as black holes and dark matter, are under intense investigation and debate. Possible dark matter candidates include massive neutrinos and weakly interacting massive particles (called WIMPs). Among the WIMP 13-3 The Evolution of Stars 639 candidates (out of many suggested possibilities) are axions and neutralinos, hypothetical elementary particles that many astrophysicists and cosmologists think may be the best choices. Although it is now certain that neutrinos have mass, they probably do not contribute significantly to solution of the dark matter problem because, being relativistic, they don\u2019t clump together into clouds like cold interstellar hydrogen does. Axions were postulated more than 30 years ago as part of an elegant theoretical solution to a problem with quantum chromodynamics, namely that QCD predicted a large electric dipole moment for the neutron (that experiments show it does not have). The axion resulted from the breaking of a symmetry (see Section 12-4) in the theory that explained the absence of an electric dipole moment for the neutron. As proposed the axion would have no electric charge and interact only minimally with ordinary matter. As a contributor to the solution of the dark matter problem, its value may be limited, since recent experiments place an upper limit to its mass, if it exists, of 106 eV>c2. The neutralino is considered by many astrophysicists and cosmologists as perhaps the best candidate to solve the dark matter problem. The neutralinos (there may be four of them) are the mass eigenstates that result from the quantum mixing of the supersymmetry partners of the W, Z, and Higgs bosons, the Wino, Zino, and Higgsino (see Table 12-11). One of the neutralinos may be the lightest possible supersymmetric particle and would, therefore, be stable. In some cosmological models it was produced copiously in the early universe and, with no decay channel available, it may have a relic abundance that could account for the dark matter. The lightest neutralinos mass is estimated at 10 to ) It would couple to other particles only via the weak interaction, so its behavior would be similar to that of the neutrino in that it would not be directly observable in existing detectors at the big accelerators. A significant portion of the experimental runs of the Large Hadron Collider will be searches for supersymmetry particles, including the energy/momentum discrepancy signature of the neutralinos. (The proton mass is 0.938 GeV>c2. 104 GeV>c2. 13-3 The Evolution of Stars While no universally accepted theory of stellar formation exists, it is generally agreed that stars are formed from massive clouds of dust", " and gas that exist throughout space. At some point in the swirling cloud, gravitational attraction begins to cause aggregations of matter to collect. These contract further due to gravity, attracting still more matter to them and eventually\u2014if the cloud has sufficient mass\u2014increasing the temperature to that necessary to initiate fusion as was described earlier, and a star is born. In this section, we discuss how stars evolve once they have been formed. Two characteristics of stars are important for this discussion, the luminosity L and the effective temperature Te. The effective temperature of a star is difficult to measure. It is usually inferred from a comparison of the spectral distribution of its radiation with that of a blackbody or from measurements of the absorption lines of hydrogen and helium in the atmosphere of the star. The luminosity is the total power radiated by the star. It is determined from the radiant flux F of the star at Earth (remember, called the solar constant f for the Sun) and the distance r from Earth to the star (see Equation 13-2): L 4r2F 13-10 Determining the distance to a star is generally a very difficult task. For stars that are relatively close, the distance can be determined from the apparent motion of the star in the sky due to the motion of Earth around the Sun. During one complete revolution 640 Chapter 13 Astrophysics and Cosmology Star \u03b8 r Parallax angle of Earth, a star appears to move in an ellipse of angular radius along the major axis called the parallax angle as shown in Figure 13-16. The parallax angle is given by 1 AU r 13-11 Earth Sun 1 AU Orbit Figure 13-16 The parallax method of finding distances to nearby stars. A parsec is the distance r for which the parallax angle subtended by 1 AU is 1 arc second. Astronomical distances are measured in parsecs or light-years. One parsec (1), (pc) is that distance at which 1 AU subtends an angle of 1 arc second in Equation 13-11, we obtain which equals of a degree. Setting 1>3600 1 1 parsec 1 AU 1 3600 1\u00b0 180\u00b0 rad 2.60 105 AU 13-12 # Using 1 AU 1.496 1011 m and 1 c y 9.461 1015 m, we can express the parsec in terms of meters or light-years: 1 pc 3.086 1016 m 3.26 c # y 13-13 EXAMPLE 13-4 Distance to Proxima Centauri Proxima Centauri is the star closest to the Sun. By measuring the maximum apparent change in the direction to Proxima Centauri between two observations made six months apart, the parallax angle is found to be. How far is it to Proxima Centauri? (Proxima Centauri\u2019s location: R. A. 14h29m43s, Dec. 16\u00b04158.) 0.77233 SOLUTION Since 1 AU 1 1 parsec, we have for > 0.77233, 1 AU 0.77233 r 1 AU 4.22 c # y 1 AU 1 1 0.77233 1 pc 1 0.77233 1.2948 pc 0.001 Parallax angles as small as can be measured, which means that the parallax method of Example 13-4 can be used to measure stellar distances from the Sun out to about 1 kpc. Since it is about 8 kpc to the center of the Galaxy, the method can be used for only about 10,000 stars that are relatively close to the Sun and, thus, in the Milky Way. For the rest, the parallax angle is immeasurably small. In other situations, more indirect measurements of distance are necessary. One involves complex analyses of intensity variations over time for particular types of pulsating stars (Cepheid variables) found primarily in star clusters. Distances to clusters as far away as about 29 Mpc have been measured by this method. Supernovae (see Section 13-4) provide several methods for determining distances to the galaxies in which they are located. The most important of these makes use of the similarity of the light curves, that is, the emitted light intensity versus time, of so-called Type Ia supernovae. Such measurements enable calculations of the distances to supernovae that are accurate to within about 5 percent and recently provided the crucial evidence that the expansion of the universe is accelerating. Hertzsprung-Russell Diagram The various states of stars can be conveniently displayed by plotting the luminosity L versus the effective temperature Te. The result is called the Hertzsprung-Russell (H-R) diagram. Figure 13-17a shows an H-R diagram for some stars of representative masses. The large majority of stars on an H-R diagram fall in the broad central band called the main sequence. Main sequence stars are normal in that they are homogeneous", " mixtures, except in the core, they have essentially the same chemical composition, and they are fusing hydrogen into helium via one or another of the nuclear reactions discussed earlier. Stars expand as they leave the main sequence. For that reason, stars in the main sequence are often called main sequence dwarfs. Between 80 and 90 percent of all stars are on the main sequence. M} M} M} to about 60 The location of a star along the main sequence in the H-R diagram depends on its luminosity, which is primarily dependent on the mass of the star. The masses of stars M}, range from about 0.08 is the mass of the Sun. Gaseous where objects with less than about 0.08 do not have enough gravity for their central cores to be compressed sufficiently to generate the temperature necessary to sustain the nuclear fusion reactions needed for energy emission. Objects with masses greater would generate such enormous than 60 internal temperatures that the outward radiation pressure would exceed the gravitygenerated inward pressure. Such a system would be very unstable, if indeed it could form at all. M} Evaluation of the masses of binary stars has shown that the luminosity of a star is approximately proportional to the fourth power of its mass: L M4 13-14 } L L 106 104 102 1 10\u22122 10\u22124 13-3 The Evolution of Stars 641 O B A F G K M Red supergiants Red giants Main sequence Sun White dwarfs 40000 20000 102 10 1 } L L 6300 5000 2500 10000 Temperature, K (a) 6 4 5 5000 4000 Temperature, K (b) Figure 13-17 (a) The Hertzsprung-Russell (H-R) diagram for stars in the solar neighborhood. Most stars (80 to 90 percent) fall on the main sequence. Stars in the lower right end of the main sequence are cool and dim, those in the upper left are hot and bright. (b) The Sun\u2019s evolutionary track from the time it entered the main sequence at point 1. The Sun is currently between points 1 and 2. It will leave the main sequence at point 4. The time between successive points is approximately 109 years. The lifetime of a star tL is proportional to the total available energy, which is proportional to the star\u2019s mass (E = Mc2), and inversely proportional to the rate of energy emission, which is the luminosity: tL E L Mc2 M4 M3 13-15 642 Chapter 13 Astrophysics and Cosmology Thus, more massive stars burn their hydrogen more quickly than do less massive stars. 1>8 Thus, a star with twice the Sun\u2019s mass would be expected to have a lifetime only as long as that of the Sun. (Equation 13-15 doesn\u2019t work for very small or very large stars because the luminosity-mass relationship of Equation 13-14 is only an average result. The exponent in Equation 13-15 is larger in magnitude for very small stars and smaller for very large stars.) Considerations of energy balance for stars on the main sequence lead to the approximate proportionality of the radius and the mass, as can be demonstrated using the data in Table 13-3 (see Problem 13-6): R M 13-16 Combining this with Equation 13-4, which relates the effective temperature to the luminosity per unit area, we can relate the effective temperature to the mass of the star: a Te L 4R2 b 1>4 a 1>4 M4 M2 b M1>2 13-17 Thus, stars with larger masses have higher effective temperatures and, hence, higher luminosities than those with lower masses. It is on the basis of Equations 13-15 and 13-17 that the stellar masses were plotted on the H-R diagram in Figure 13-17a. Table 13-3 lists properties of stars by spectral type. The following values for the Sun\u2019s characteristics will enable calculation of numerical values for the correspond 6.96 108 m; L} ing characteristics of individual stars: M} 1.99 1030 kg. As the star ages, it consumes its primary fuel, hydrogen. What happens to it as the hydrogen supply in the core becomes exhausted depends on its initial mass. Low-mass and high-mass stars follow somewhat different evolutionary paths. In either case, however, the fundamental processes involved are successive nuclear reactions fueled by the product of the previous cycle. Thus, after the hydrogen in the core has fused to helium, the star must begin fusing helium in a cycle that eventually forms carbon. 3.83 1026 J>s; R} Table 13-3 Selected properties of stars Spectral type Surface temperature (K) L>L} R>R} M>M} O5 B0 A0 F0 G0 Sun (G2) K0 M0 M8 44,500 30,000 9,520 7,200 6,030 5,800 5,300 3", ",900 2,600 790,000 15 52,000 54 6 1.5 1.0 0.4 0.08 0.001 8 3 2 1.1 1.0 0.8 0.6 60 18 3 2 1.1 1.0 0.8 0.5 0.17 0.06 13-3 The Evolution of Stars 643 Before this can occur, the core must heat up still further to the 108 K necessary to initiate helium fusion. The chain of events involved in this process is complex and beyond the scope of this book. However, its result for low-mass stars is that the radius (and therefore the surface area) increases while luminosity remains nearly constant. Thus, the intensity (luminosity per unit area) and, consequently, the effective temperature decrease and the radiation emitted shifts to longer wavelengths as the star expands to become a red subgiant. The photosphere rapidly becomes more transparent as Te and the density decrease, increasing the luminosity and effectively limiting the decrease in temperature. The star is then a red giant. The track of a typical evolving low mass star such as the Sun is shown on the H-R diagram in Figure 13-17b. > Helium ignition results in the star again increasing its effective temperature and moving to the horizontal branch. When the helium in the core is exhausted, the star begins fusing carbon and ascends the red giant branch again, becoming a red supergiant. Betelgeuse, the bright star in the shoulder of the constellation Orion, is a red supergiant. Its density is about 1.5 105 kg m3, a hundred thousand times less than the air we breathe! What happens after this is not completely clear. Through a combination of events that includes the loss of considerable mass, perhaps including the ejection of an expanding shell of gas (called a planetary nebula), such as that shown in Figure 13-18, the star may become a white dwarf, slowly cooling toward thermal equilibrium with the universe. We will discuss white dwarfs further in Section 13-5. High mass stars\u2014those with masses greater than about \u2014evolve much more quickly than low-mass ones, as predicted by Equation 13-15. In addition, they have sufficient initial mass to generate gravitationally the high pressures and temperatures necessary to ignite the fusion reactions with oxygen, neon, and then silicon to produce, ultimately, iron. These reactions occur with phenomenal speed and lead to catastrophic events that will be discussed in the next section. An extremely massive star, such as Betelgeuse, may become a supernova via core collapse (see Section 13-4). 6M} (a) (b) Figure 13-18 The nebula 30 Doradus (a), also known as the Tarantula nebula, is believed to be older than nebula NGC 7293 (b), also known as the Helix nebula. The Tarantula nebula\u2019s rapidly expanding gas cloud consequently shows a greater degree of diffusion. Located in the Large Magellanic Cloud, the Tarantula contains one of the most massive stars known, as well as supernova SN1987A, the very bright star slightly below the center (of the left-hand photo). Ultraviolet radiation from stars heats the gas of a nebulae, causing it to radiate. [(a) The Hubble Heritage Team (AURA/STScI/NASA). (b) NASA, NOAO, ESA, the Hubble Helix Nebula Team, M. Meixner (STScI), and T.A. Rector (NRAO).] 644 Chapter 13 Astrophysics and Cosmology 13-4 Cataclysmic Events Huge explosions and other sorts of cataclysmic events are a natural part of the life cycle of stars. Stars formed in swirling clouds of gas move along the H-R diagram, incorporating such occurrences into their evolution and forming in the process the elements needed to form new stars. Why these cataclysmic events occur is the subject of this section. M2 CM M1 L Novae Figure 13-19 Cross sections of two gravitational equipotential surfaces for stars M1 and M2. The point labeled L is one of five Lagrangian points where the net gravitational potential is an extremum (a saddle point in this case) and the net force is zero. M1 M2 L Accretion disk Matter falling from M1 toward M2 Figure 13-20 Material from M1 pouring through the Lagrangian point into the Roche lobe of M2 forms an accretion disk in M2\u2019s equatorial plane. Material arriving later hits the disk, generating a hightemperature impact area. This causes novae to flicker irregularly. More than half of all stars are members of binary pairs or even larger associations. These stars orbit their common center of mass as the group moves with the rotation of the Galaxy. The periods of binaries vary from a few hours for those with the companions very close to", " each other to millions of years for those with the companions separated by thousands of astronomical units. Here, we are interested in close binaries. A complete analysis of the interactions between the two stars forming a close binary is beyond the scope of this book, but a qualitative explanation will suffice. Consider a binary whose stars of masses M1 and M2 rotate about their common center of mass in circular orbits. An observer at rest in the rotating system experiences a net force that is the sum of the gravitational forces due to the two stars and the pseudoforces due to the rotation. Figure 13-19 shows an equipotential surface about a binary pair. It is easy to visualize that there is a point along the line joining the centers of the two stars where the net potential is a minimum. At this point, the net force due to the combined effects of the rotation and the gravitational attraction by the masses M1 and M2 is zero. This point is a Lagrangian point. The three-dimensional equipotential surface that includes the Lagrangian point L forms an envelope around each star called the Roche lobe.12 Now consider what happens when, through natural evolution, one of the stars, say M1, begins expanding and fills its Roche lobe. The photosphere of the star feels a vacuum outside the surface, the outward pressure at any point being balanced by gravity. But at the Lagrangian point there is no gravity. Thus, material from M1 pours through the Langrangian point into the Roche lobe of M2. Once inside it is gravitationally attracted toward M2. Since the system is rotating, the material from M1 doesn\u2019t simply move directly toward M2 but, because of the Coriolis effect, forms a spiraling accretion disk (Figure 13-20). If M2 is a normal star, nothing of great consequence occurs, but if it is a white dwarf, then cataclysmic events called novae can occur. We will mention two possibilities. Material flowing through the Lagrangian point into the accretion disk is stored there until some instability occurs in the disk that results in the dumping of material onto the surface of the white dwarf. The impact heats the surface, causing a sudden luminosity increase by a factor of 10 to 100. Such events recur at intervals of a few weeks for dwarf novae to hundreds or thousands of years for recurrent novae. Between these sudden bursts in intensity, the novae flicker as described in the caption of Figure 13-20. For classical novae, which eject substantial material into space and can brighten by a factor of a million within a few days, astrophysicists suggest that the sudden dumping of material from the accretion disk onto the hot surface may result in the buildup of sufficient hydrogen to initiate a thermonuclear explosion. After the blast the system returns to a more quiescent state, pending the accumulation of more hydrogen in the disk. The theoretical problems involved in explaining such an event are formidable, however, and no general agreement on the mechanism exists. 13-4 Cataclysmic Events 645 Supernovae A supernova\u2014the catastrophic explosion of an entire star\u2014is, perhaps surprisingly, somewhat more clearly understood than the nova. Supernovae are classified as Type I or Type II mainly on the basis of their spectra (Figure 13-21). The spectra of Type I supernovae do not contain hydrogen lines, indicating that they are devoid of hydrogen, or nearly so. In contrast, Type II supernovae exhibit strong hydrogen lines. Type I supernovae are further divided into Type Ia, which show a strong line of singly ionized Si at 615 nm, Type Ib whose spectra include strong He lines, and Type Ic whose spectra do not include He lines. Recalling that H and He are in that order the most abundant elements in the universe, these spectral differences indicated that there are significant differences in the progenitors of the Type I and Type II supernovae. Type II supernovae are subdivided into two groups based on the shape of their light intensity versus time curves following peak brightness. The light curves of Type II-L supernovae decline linearly with time; the light curves of Type II-P exhibit an intermediate plateau lasting 30 days or more before the intensity decline resumes. Supernovae are not just big novae. Their origin is completely different. In Section 13-3, we saw what occurs in a star as it uses up the hydrogen in the core and begins moving off the main sequence of the H-R diagram. The star begins to fuse helium, then carbon. If it were a low-mass star, it would have insufficient gravitational energy to ignite the fusion of heavier nuclei in quantity. For massive stars, however, the situation is different. Type Ia supernovae originate in binary systems where one star is a massive white dwarf rich in carbon and oxygen (see Section 13-5). An expanding companion that fills its Roche lobe may dump a huge amount", " of gas directly onto the dwarf\u2019s surface, increasing the gravitational pressure in the core. If the pressure increases the core temperature enough to trigger carbon fusion, a runaway fusion reaction results, producing a massive thermonuclear explosion\u2014a huge \u201ccarbon bomb.\u201d Depending on the characteristics of the particular binary, it may be possible that Type Ib and Type Ic supernovae result from gravitational core collapse. Type I supernovae occur among population II stars. (Don\u2019t be confused by the apparent inconsistency in nomenclature.) As noted earlier, Type Ia supernovae all have very similar maximum intensities and light curves which makes it possible for them to be identified even at very large distances, enabling them to be used as \u201cstandard candles\u201d by astronomers. It is these features that have made it possible for astronomers to measure much greater distances to host galaxies than was previously possible. No plateau Type II-L H lines Type II Plateau Type II-P Supernova Si lines Type Ia No H lines Type I He lines Type Ib No Si lines No He lines Type Ic Figure 13-21 Schematic of the classifications of supernovae. Type I subgroups are based on their spectra at maximum brightness. Type II subgroups are based on the existence or absence of a plateau in the light curve. 646 Chapter 13 Astrophysics and Cosmology 8M}, If a star\u2019s mass is greater than about evolution toward a Type II supernova proceeds approximately as follows. Gravity is strong enough to continue to draw mass from the middle layers into the core as the core uses up fuel. The increasing temperatures, exceeding 108 K, are sufficient to ignite fusion in neon and silicon ultimately producing iron. As we saw in Chapter 11, the specific binding energy of iron is the highest in the periodic table. Fusing elements above iron doesn\u2019t emit energy, it absorbs energy. Thus, when the core has been fused to iron, there is nowhere else to go via thermonuclear reactions. With no counteracting outward pressure from nuclear reactions, gravitational contraction continues even more rapidly and the core consequently continues to heat up until it exceeds 109 K. At that point, the radiation within the star is intense and iron nuclei undergo photodisintegration into helium and neutrons, absorbing energy from the core and accelerating the gravitational collapse: 56 26Fe \u00a1 13 4 2He 4n 13-18 The helium nuclei then begin to photodisintegrate, absorbing enormous amounts of energy to overcome the nuclear binding energy of helium: 2He \u00a1 2p 2n The core is now in gravitational free fall, compressing the electrons and protons into neutrons via inverse beta decay: 13-19 4 p e \u00a1 n e 13-20 What happens next to the core is a matter of intense theoretical conjecture that we will explore further in Section 13-5. # What happens to the envelope of the star\u2014the material outside the core\u2014although unclear theoretically, is certainly apparent visually. The entire envelope is blown away in an incredibly massive explosion. This is a Type II supernova. Supernovae are extremely rare, but scientists were fortunate enough to observe one in 1987 only 170,000 c y away in the Large Magellanic Cloud, a small irregular galaxy that is a companion to the Milky Way. Called SN1987A (see Figure 13-18 and the photo at the left), it was the first to occur close enough to be visible to the unaided eye since 1604, when both Kepler and Galileo saw one. Two others were recorded earlier, in 1006 and 1054, the latter documented by Chinese astronomers and still visible as the Crab Nebula. Several others have been observed with telescopes. As a result of the enormous number of nuclear reactions and decays initiated by the supernova, the radiation emitted in the explosion is accompanied by a flood of neutrinos. Neutrinos emitted by SN1987A were detected by the Kamiokande neutrino observatory, bringing with them information about the core-collapse model of supernovae and a hint of the nonzero neutrino mass (see Problem 12-13). At its peak light output, a supernova typically shines more brightly than the entire galaxy in which it is located. The spectra of supernovae reveal the presence of elements throughout the entire periodic table. This indicates that some of the energy removed from the core following the production of iron is used to produce elements of even higher atomic numbers. The supernova ejects some of this material into space, where it eventually contributes to the formation of a new generation of stars and their planets via condensation. Such events undoubtedly preceded the birth of the Sun and the formation of Earth. We are, as has been said before, \u201cmade of the stuff of stars.\u201d Supernova 1987A developed a set of rings some weeks after it was first seen. The rings are likely caused by a beam of high-energy radiation or particles sweeping across", " the gas. The source of the beam may be a previously unseen companion of the star that exploded. This Hubble Space Telescope photo was made with hydrogen Balmer alpha light. [NASA, ESA, P. Challis and R. Kirshner (Harvard-Smithsonian Center for Astrophysics).] 13-5 Final States of Stars 647 13-5 Final States of Stars The cataclysmic events that occur near the end of the life of a star lead to one of only three possible final states: a white dwarf, a neutron star, or a black hole. The mass of the star, particularly that of the core, appears to be the primary factor in determining the final state. White Dwarfs 6M} Stars whose masses are less than about follow an evolutionary track on the H-R diagram that takes them through one or more periods of substantial mass loss from the outer layers of gas. How this occurs is not clear, but the ejected mass, which is heated to a glowing planetary nebula by the hot core, leaves behind a white dwarf, a term used because many, though by no means all, are literally white hot. Its mass is typically about and its radius of the order of 107 m, which is about the same as the radius of Earth. M} 1 Thus, the density of a typical white dwarf is about 5 105 g cm3 compared to Earth\u2019s average density of about 5.5 g cm3. A coin the size of a penny made from white dwarf material would have a mass of over 200 kg, one the size of a Euro over 400 kg. > > Thermonuclear reactions have ceased in the white dwarf, leaving it with a core consisting primarily of carbon and oxygen, so there is no outward pressure due to them from within the star. The star therefore collapses because of the inward gravitational pressure until the exclusion principle prevents the atomic electrons from coming any closer together. This effect is similar to the exclusion-principle repulsion between atoms in a molecule that we discussed in Chapter 9. It results in an outward pressure that is larger even than the thermal pressure of the hot core. It is this electron degeneracy pressure that supports the white dwarf. When the outward electron degeneracy pressure equals the inward pressure due to gravity, the star stops contracting. Explicit derivation of the expression for the electron degeneracy pressure leads to a nonrelativistic relation between the dwarf\u2019s radius R and mass M: R (3.1 1017 m # kg1>3)a 5>3 b Z A M1>3 13-21 M}! 1M} where Z is the atomic number and A is the atomic mass number of the material of the star. Note the interesting result that the larger the mass, the smaller the radius, a consequence of the gravitational contraction that was discussed earlier. For example, a white dwarf with a mass of will have a radius smaller than one with a mass of 0.5 Equation 13-21 raises the interesting question of whether, when the electrons become relativistic, the mass might become large enough for the radius of the dwarf to shrink to zero. Although Equation 13-21 does not formally allow that possibility until M approaches infinity, S. Chandrasekhar13 derived the corresponding relativistic relaM}. tion and found that the radius would go to zero when the mass reached about 1.4 This quantity is called the Chandrasekhar limit. Its validity is strongly supported by the fact that the masses of all white dwarfs that have been measured are less than that value. A lone white dwarf continually radiates heat to space and, without a nuclear furnace, slowly cools and dims. When it is no longer visible, it has become a black dwarf. It continues to cool toward thermal equilibrium with the universe. It is not likely that any white dwarfs have yet reached this final stage. However, if the white dwarf is a part of a binary, then in addition to the possibility of a nova described earlier, mass may flow from the companion directly onto the surface of the white dwarf. When the degenerate electron pressure can no longer support the white dwarf (at the Chandrasekhar limit), the star implodes, suddenly raising the core temperature and 648 Chapter 13 Astrophysics and Cosmology White dwarfs identified by the Hubble Space Telescope in M4, the globular cluster # closest to Earth (7000 c y). M4 contains more than 100,000 stars. [(left) Kitt Peak National Observatory 0.9-meter telescope, National Optical Astronomy Observatories; courtesy M. Bolte (University of California, Santa Cruz). (right) Harvey Richer (University of British Columbia, Vancouver, Canada) and NASA.] Ground HST detonating fusion in the carbon/oxygen core. The sudden energy release causes the M}, white dwarf to explode as a Type Ia supernova. The common mass limit, 1.4 at which the white dwarfs explode, is a major factor", " in the resulting similarity of the Type Ia supernovae light curves that enables their use as a luminosity standard for measuring astronomical distances. Following the supernova explosion, about half of the residual core of the star is iron. Neutron Stars In the discussion of supernovae, we saw that the enormous pressures developed in the core forced inverse beta decay to occur, converting the core into neutrons. If the mass of the core following the explosion is greater than the Chandrasekhar limit, what happens? We can get an idea by considering the neutrons to be an ideal gas of fermions and derive a nonrelativistic expression for the mass-radius relation analogous to Equation 13-21. The result is R (1.6 1014 m # kg1>3)M1>3 13-22 where M is the mass in kilograms and R is the radius of the core in meters. Such a star is called a neutron star, since the envelope was blown away in the supernova and all that is left is the core consisting of neutrons. For Equation 13-22 yields the radius R 1.27 104 m 12.7 km. M 1.0M}, The density of the neutron star is about 1.2 1014 g cm3. This is only slightly less than the density of the neutron itself, which is about 4 1014 g cm3. Thus, we can conclude that the gravitational pressure of the neutron star is balanced by the repulsive component (due to the exclusion principle) of the strong nuclear force between the neutrons. As you might guess from our earlier discussion, gravity can overcome even this resisting pressure. The mass corresponding to the gravity at which that occurs would be the maximum mass possible for a neutron star, a mass analogous to the Chandrasekhar limit for white dwarfs. Current theory puts the maximum mass of a neutron star between The few neutron stars that have been tentatively identified and measured all have masses below this limit. 1.7 M} 3M}. and > > 13-5 Final States of Stars 649 A lone neutron star, the first seen in visible light, is very hot (about 650,000 K at the surface) and may be no larger than 28 km in diameter. [Fred Walter (State University of New York at Stony Brook) and NASA.] Regularly pulsing radio sources, called pulsars, discovered in 1967 in nebulae such as the Crab Nebula that are remnants of supernovae, are thought to be neutron stars. Current theory suggests that the radiation is emitted as the result of charged particles emitted by the neutron star that are accelerated along the star\u2019s magnetic field lines as a consequence of the star\u2019s rapid rotation as illustrated in Figure 13-22. (a) Rotational axis + \u2013 \u2013 + \u03c9 \u2013 + + \u2013 Magnetic axis Cone of radiation (a) (b) (c) Figure 13-22 (a) The neutron star acquires much of the original star\u2019s angular momentum and magnetic field, causing it to rotate rapidly while dragging along a distorted magnetosphere. Accelerated charged particles radiate in a cone about the rotating magnetic axis like a cosmic lighthouse. (b) The pulsar in the Crab Nebula. As the cone of radiation swings to face Earth, light emitted from accelerated electrons becomes visible (the bright spot in the image). (c) A fraction of a second later, the pulsar has turned and this light is no longer directed toward Earth. Currently rotating about 30 times a second, the pulsar has a period that is increasing by about 105 s per year. [(b) and (c) Harvard/Smithsonian Center for Astrophysics.] 650 Chapter 13 Astrophysics and Cosmology The Crab pulsar also corresponds to an optical variable as illustrated in Figures 13-22b and c. It emits energy at an incredible 3 1031 W. Its period is equally incredible, 0.033 seconds, one of the shortest known. As it emits energy into space, the neutron star also slowly cools, approaching thermal equilibrium with the universe. Black Holes to 1.7M} What happens when the mass of the remaining 3M} core of a supernova exceeds the upper limit for the formation of a neutron star? The velocity necessary for an object with mass to escape from an object of mass M is found by equating the gravitational potential energy at the surface of M to the kinetic energy necessary to escape. This results in the escape velocity This composite image of the Crab Nebula was made from an x-ray image recorded by the Chandra X-Ray Observatory and an optical image from the Hubble Space Telescope. The inner ring is about 1 c y across. [Credits for X-ray Image: NASA/CXC/ASU/J. Hester et al. Credits for Optical Image: NASA/HST/ASU/J. Hester et al.] # 1>2 ve a 2GM R b 13-23 M 1", ".0M}, For a neutron star with > 1.3 108 m s, more than 40 percent of the speed of light. If there were no relativistic and quantum mechanical effects, the escape velocity would equal c when ve RS 2GM c2 13-24 RS is called the Schwarzschild radius. Thus, if an incipient neutron star is so where no object with mass can escape from its surmassive that its radius is less than emitted at some distance R from mass M face. In addition, radiation of wavelength 0 is shifted to a longer wavelength according to the gravitational redshift described in Section 2-5; the ratio is given by RS, 0 a1 1>2 v2 e c2 b a1 2GM c2R b 1>2 a1 1>2 RS R b 13-25 If R shrinks to the Schwarzschild radius, then approaches infinity and the energy E hf hc> ( no energy can escape the surface as radiation, either. Such an object is called a black hole, because it neither emits nor reflects radiation or mass and, hence, appears absolutely black.14 ) approaches zero. Thus, if R is less than RS, M}, The radius of a black hole with a mass of 1 if there is such an object, would be only about 3 km. Many astrophysicists currently believe that a massive black hole is located at the center of the Milky Way and may account for a very small part of the \u201cmissing mass\u201d of the Galaxy. During the past decade astronomers at the Max Planck Institute (Germany) have tracked about a dozen individual stars (of more than 300) orbiting an unseen object at Sagittarius A*, a radio source near the center of the Milky Way (see Figure 13-13). One of these, for example, is called S2. Determining S2\u2019s orbital period to be 15.2 y enabled calculation of the mass of the unseen object to 13-5 Final States of Stars 651 The series of Chandra X-Ray Observatory images on the left show jets of high-energy particles being produced near a black hole in a binary system, first on the left (top image), then on the right (middle image). The jets are moving away from each other at about 0.5 c. In the lower image the left jet has disappeared. The schematic on the right illustrates how the jets originate. The black hole draws mass from the normal companion, then intense electromagnetic forces in the accretion disk expel the jets of high-energy particles. [Left: X-ray (NASA/CXC); Right: Illustration (CXC/M.Weiss).] M}. be about 3 106 S2\u2019s orbital speed of approximately 4 106 m s \u201cconfines\u201d the unseen mass of Sagittarius A* to such a small volume that there can be little doubt that it is an enormous black hole. Unlike white dwarfs and neutron stars, black holes are not cooling toward thermal equilibrium with the universe. > # Black hole in the center of galaxy NGC 7052. The disk is 3700 c y in diameter. The black hole, whose mass is about 300 million solar masses, will swallow the disk in a few billion years. [Roeland P. van der Marel (STScI), Frank C. van den Bosch (Univ. of Washington), and NASA.] 652 Chapter 13 Astrophysics and Cosmology Gamma-Ray Bursts Flashes of gamma rays (and x rays) that occur about once a day at apparently random locations in the sky were discovered by Vela military satellites in 1967. They are short, lasting from less than a second to as long as a few minutes. During the burst they are by far the brightest gamma-ray sources in the sky, their fluxes exceeding those of the brightest steady sources, such as the Sun and the Crab Nebula, by a factor of a 1000 or more. Their brief lifetime makes it difficult to attempt to identify the bursts with individual stars or galaxies because of the inherent delay in processing the burst information and re-aiming the large telescopes. With a bit of good fortune, that problem was first solved in 1997 when the Dutch/Italian BeppoSAX satellite detector discovered an x-ray afterglow following burst GRB970228.15 Then in 1999 burst GRB990510 was seen simultaneously by the Compton Gamma Ray Observatory and BeppoSAX satellites within the field of view of the Very Large Telescope (VLT) in Chile, the largest telescope in the southern hemisphere. The subsequent x-ray, optical, and radio wavelength afterglows told astronomers what to look for and, since then, many bursts have been studied more thoroughly via their afterglows, some lasting for several months. Since that time the all-sky surveys of the Burst and Transient Source Experiment (BATSE) and Swift satellites have recorded more than 8000 gamma-ray bursts distributed isotropically over the", " sky (see the photo below). The uniform distribution of the GRBs across the sky is strong evidence that they occur in the distant universe, since only at great distances does the cosmos appear uniform. The VLT measured the redshift z of GBR990510 to be 1.61 implying a recession speed that places the source about halfway to the edge of the visible universe. The origin of the bursts and the mechanism for the enormous energy release implied by the gamma-ray flux are not yet clear. As of this writing approximately 100 afterglows have been located and for most of these a host galaxy has been identified. The GRBs in many cases appear to be the result of the supernova collapse of very large stars becoming neutron stars or black holes. This is an area of active current research. This map shows the locations of more than 2700 gamma-ray bursts recorded by BATSE aboard the Compton Gamma-Ray Observatory during its 9 years of operation. The projection is in galactic coordinates, the plane of the Milky Way being the horizontal line through the middle of the figure. The burst locations are color-coded based on the integrated energy over the duration of the burst. [Image courtesy of the BATSE team, http://gammaray.nsstc.nasa.gov.] 13-6 Galaxies 653 13-6 Galaxies In Section 13-2 we saw that the Milky Way is shaped like a spiral disk with a central bulge located about 28,000 c y from the Sun. The disk is surrounded by a roughly spherical \u201chalo\u201d of globular clusters made up mostly of population II stars, which are also part of our Galaxy. We will now look at some of the characteristics of galaxies. # Material Between the Stars \u201cHoles in the sky\u201d\u2014regions where no stars are seen\u2014have been observed since the early days of astronomy and were assumed to be empty space. However, studies of open clusters about 70 years ago led to the discovery of a more or less continuous distribution of tiny dust particles, called interstellar dust, between the stars. Consisting of solid specks of silicates and carbides averaging only a few hundred nanometers in diameter (approximately matching the wavelength of visible light), the interstellar dust both absorbs and scatters some of the starlight striking it. Thus, dust in the interstellar medium (ISM) dims starlight coming toward Earth and, since blue light scatters more efficiently than red light, starlight is reddened on its trip to us, just as sunlight is reddened at sunset. Although the dust seems to pervade the entire Galaxy, the concentrations are very low and its total mass makes only a very small contribution to the total mass of the ISM. The vacuum in interstellar space is far better than the best obtainable in the laboratory. The ISM consists primarily of hydrogen and helium. Hydrogen as atomic hydrogen, ionized hydrogen (protons), and hydrogen molecules (H2) makes up about 70 percent of the mass of the ISM. Atomic helium is most of the rest, while the carbide and silicate dust contribute only a few percent of the ISM\u2019s total mass. Spectroscopic studies of binaries reveal some absorption lines that are not Doppler-shifted. In 1904, J. F. Hartmann reasoned correctly, although not to universal acceptance, that the unshifted lines result from absorption of light from the binary by an intervening gas cloud, rather than by gas in the atmosphere of the star. Though still difficult to demonstrate conclusively in all cases, the existence of interstellar gas clouds is now generally accepted. As a result of temperature variations in the early universe and the subsequent continuous action of gravity, huge clouds of primarily hydrogen have formed throughout the ISM. They range in mass from about 1 to 1000 times the mass of the Sun and have temperatures of about 30 to 150 K. At these temperatures the hydrogen atoms in the clouds are in their ground states. Even though the atoms are in their ground states, it is possible to see the clouds as a result of the hyperfine splitting of the hydrogen ground state due to the spins (i.e., magnetic moments) of the electrons and protons. If the spins of the electron and the proton in a particular atom are antiparallel, the atom\u2019s ground state energy is very slightly lower that it would be if the spins were parallel. (See Section 7-4 and Figure 13-23.) The energy difference between the two states is so small, 5.9 106 eV, that a collision with another atom or a dust grain can result in the atom absorbing enough energy to \u201cflip\u201d the electron\u2019s spin into the parallel configuration. Once the electron\u2019s spin is flipped, the atom has only two ways to return to the ground state: it can have another collision 5.88 \u00d7 10-6 Energy (eV) Proton Proton 0 Electron Photon \u03bb = 21 cm Electron Figure 13-", "23 The hyperfine spitting of the hydrogen atom ground state is the origin of the 21-cm radiation used to map gas clouds in the ISM. 654 Chapter 13 Astrophysics and Cosmology enabling the atom to release the excess energy as heat or it can spontaneously flip the spin back antiparallel to that of the proton, emitting a photon with wavelength 21 cm in the process. The likelihood of either of these occurring is very small. For a given atom, collisions only occur every few hundred years and spontaneous return to the lower state may take millions of years. Even so, the number of atoms in the huge clouds is so large that there is a faint continuous emission of the 21-cm photons that enables mapping of the clouds by radiotelescopes. Together, the interstellar dust and the clouds of gas account for an estimated 2 to 3 percent of the total mass (ordinary and dark matter) of our Galaxy. It is nearly certain that there is not enough unseen gas and dust to account for the Galaxy`s dark matter. Gaseous Nebulae Though most gas clouds, or nebulae, in interstellar space are irregular in shape, a few are circular, leading to speculation that they are self-gravitating and represent the very early stages of new star formation. Some large hydrogen clouds have spherical inner regions of ionized hydrogen, with a quite sharp demarcation between the H and H regions. Astrophysicists believe that the ionized region is maintained by ultraviolet photons with frequencies above the Lyman limit emitted by a hot, newly formed star at the center of the region. The view that new stars form in the nebulae in an ongoing process is strongly supported by the observation that, although the Galaxy is of the order of 1010 years old, our Galaxy contains main sequence stars that are no more than 2 to 3 106 years old. Furthermore, high-resolution radioastronomy has in recent years located numerous newly forming stars embedded in clouds of dust and gas that are completely opaque to optical wavelengths. Classification of Galaxies Although fuzzy, extended objects, at one time called \u201cnebulae\u201d (not to be confused with planetary nebulae defined in Section 13-3), that were obviously not stars had been observed in the night sky since the 1700s, what and where they were was a matter of active scientific debate until well into the twentieth century. The answer had to await the development of telescopes with sufficient resolution and light gathering power and a theoretical means of computing distances from observations made with them. These came together in the mid-1920s when Edwin Hubble16 used the 2.5 m telescope on Mount Wilson, the largest in existence at the time, to measure the intensities of rare stars, called Cepheid variables17 that he discovered in three \u201cnebulae.\u201d One of those nebulae, the great spiral Andromeda, he measured to be 2 106 lightyears away. In one stroke, he was able to demonstrate that the \u201cnebulae\u201d were in fact galaxies much like our own, as had first been suggested by the philosopher Emmanuel Kant 150 years earlier, and that they were far outside the Milky Way. Exploring Hubble\u2019s discovery will take us into the realm of cosmology, the study of the largescale structure of the universe. Following his discovery that these \u201cnebulae\u201d were in reality distant galaxies, Hubble conducted a systematic study of the enormous number that were visible. He found that all but a very few fit into four general categories. Most had regular geometrical shapes and occur in two varieties: ellipticals which are roundish, rather like a football, and disks. The disks in turn had two subgroups, ordinary spirals and barred spirals (i.e., spirals with a \u201cbar\u201d of stars across the center). The small percentage that did not have regular shapes he called irregular galaxies. Figure 13-24 shows an example of each type of galaxy. The Milky Way is a large barred spiral. 13-6 Galaxies 655 (a) (c) (b) (d) Figure 13-24 In Hubble\u2019s galaxy classification scheme, (a) is an example of an elliptical galaxy, (b) illustrates an ordinary spiral, (c) is a barred spiral, and (d) is an irregular galaxy. The Milky Way is thought to be a spiral with a faint bar. [(a) NASA, ESA, and The Hubble Heritage Team (STScI/AURA). (b) NASA and ESA. (c) NASA, ESA, and The Hubble Heritage Team (STScI/AURA). (d) NASA, ESA, and The Hubble Heritage Team (STScI/AURA).] In addition to their geometrical differences, the four types of galaxies have other dissimilarities. A large fraction of the motion of the stars in spirals is rotational about the galactic center, whereas the motion of stars in ellipticals is", " generally random with only a relatively small rotational component. Ellipticals also seem to have very little interstellar gas and dust, while spirals and many irregulars have a substantial amount. The fact that most ellipticals have no young stars is probably a consequence of that absence. With a few exceptions, ellipticals are much smaller than spirals, typically having only about 20 percent of the diameter of an average spiral and only a thousandth of the mass. 656 Chapter 13 Astrophysics and Cosmology Quiet and Active Galaxies Most of the approximately 1010 galaxies in the observable universe appear to be quiet galaxies\u2014i.e., there is very little activity other than what might be expected for such dynamic systems. The vast majority of these galaxies are so distant that our instruments cannot resolve internal details. Therefore, only the composite spectra and radiant flux F for the entire galaxy can be observed. The range of velocities v that exists in the stars of the regular galaxies, measured by the Doppler broadening of the spectral lines, turns out to be related to the total luminosity L by L (\u00a2v)4 13-26 Since L is related to F and r, the distance to the galaxy, by Equation 13-8, the distance r can be found from measurements of the redshift and the apparent brightness of the galaxy, assuming that L is known. In a very small percentage of galaxies something extremely violent, even by comparison with stellar supernovae, is occurring. Such systems are called active galaxies. There are several distinct types, some of which may not even be galaxies at all. The first discovered were Seyfert galaxies, named after Carl Seyfert, who first identified many of them. They are spirals with extremely bright, central starlike cores, or nuclei. In many of them, light coming from the core exceeds that from all of the stars in the galaxy and may vary in intensity by a factor of two or more in less than a year. Such a rapid variation in the total intensity means that the source must be less than one light-year in extent while producing as much energy as 1011 stars. Even more incredible, if possible, is the fact that the light emitted by a Seyfert galaxy consists of broad emission lines originating in both allowed and forbidden transitions in highly ionized atomic systems superimposed on a continuum, but without the absorption lines typical of stars. That suggests that its enormous energy is not coming from thermonuclear reactions. The source is not yet understood. A similar sort of extreme activity occurs in a few ellipticals called N galaxies and BL Lac objects. N galaxies are elliptical counterparts of Seyfert galaxies, that is, they have very bright centers. BL Lac objects seem to be like N galaxies, but exhibit substantial short-term intensity variations. In these, an intensity variation of a factor of two can occur within a week and a complete reversal of the polarization of the emitted light within one day, suggesting that the energy source is only one light-day in diameter. BL Lac objects are now thought to be giant ellipticals about 109 c y from Earth. # Some of the giant ellipticals are also strong emitters in the radio region of the spectrum. Study of these radio galaxies has been intense, and the results have been astonishing. For example, the radio source Centaurus A is double-lobed with a small radio-emitting nucleus midway between the lobes. It is one of the brightest radioemitting objects in the universe. Analyses of its spectra indicate that the initial energy release represented by the radiation that we now see amounted to 1056 J, which is about the equivalent of all the stars in the Milky Way undergoing supernova explosions simultaneously! The nature of such a colossal event is not currently understood. In a universe of strange phenomena, quasars, short for quasi-stellar radio sources, are among the strangest. Discovered as radio sources, their optical images look like stars; that is, they have no resolved structure. Their spectra resemble that of a Seyfert galaxy. Resolved radio images of some quasars show that a few of them are doublelobed, like the radio galaxies, which makes their identification ambiguous. The Sloan Digital Sky Survey (SDSS) has catalogued 46,420 quasars; of those, 520 have redshifts! 4, the most distant at a redshift of z 5.41. There is also a group of objects about 10 times more numerous than quasars, radio sources that were earlier called Radio galaxy 3C368. The contours show the centers of strong radio emission. The bright knots may be regions of star formation in this elliptical galaxy. [NASA, NRAO, VLA, HST, WFPC 2, M. Longair (U. Cambridge).] 13-6 Galaxies 657 # Debris from the catastrophic collision of two galaxies may be fueling quasar IRAS045052958. The quasar", " is about 3 109 c y from Earth. Astronomers believe the collision ripped out the core of a spiral galaxy (bottom of the picture). The ring lies in front of the quasar (the bright object in the middle) at a distance of 15,000 c y (one-seventh of the diameter of the Milky Way). The bright object just above the quasar is a foreground star. [John Bahsall (Institute for Advanced Study, Princeton), Mike Disney (University of Wales), and NASA.] # quasi-stellar objects. These are like quasars in every major way, except that they are not radio-emitters. Current terminology refers to both types as quasars, the radio sources as radio-loud quasars (QSRs) and the others as radio-quiet quasars (QSOs). Perhaps the strangest thing about the quasars is the magnitude of the redshift of their spectra, which is very large. It implies that some quasars are receding directly away from us at greater than 0.95 c. This would make them the most distant massive objects, of the order of 1010 c y from Earth. Their radiant flux F together with the great distance imply power outputs of 1040 W, greater than that of 1012 Suns. Not only that, but the intensities of some quasars vary over only a few hours, suggesting dimensions of only a few light-hours. # s / m k 10 4 Hubble\u2019s law E. P. Hubble was the first astronomer to recognize that there is a relation between the redshifts of the spectra of galaxies and their distances from us. This relation is illustrated in Figure 13-25 for a group of spiral galaxies used by astronomers for calibrating distances. The recession velocity v of a galaxy is related to its distance r from us by Hubble\u2019s law: v H0r 13-27 where H0 is the Hubble constant 103 102 6 60 600 Apparent distance, 106 c\u00b7 y Figure 13-25 A plot of the recession velocities of individual galaxies versus apparent distance illustrates Hubble\u2019s law. The slope of the line is H0. 658 Chapter 13 Astrophysics and Cosmology In principle, the value of H0 is easy to obtain, since it relies on the direct calculation of v from redshift measurements. However, recall that astronomical distances are very difficult to obtain and that they have been measured for only a miniscule fraction of the 1010 or so galaxies in the observable universe. Thus, the value of H0 changes as the interpretation of distance calibration data is refined. The currently accepted value of the Hubble constant is 71 4 km>s per Mpc 22 2 km>s per 106 c # y H0 13-28 Notice that the basic unit of H0 is reciprocal time. The quantity is called the Hubble time and equals about 1.3 1010 years. This would correspond to the age of the universe if gravitational pull on the receding galaxies were ignored. 1>H0 The redshift z, defined in Section 1-5, is given by z f0 f f 0 0 13-29 0 where f0 and are measured in the rest system of the star or galaxy (the emitter) and f and are measured at Earth (the observer). Figure 13-26 shows the redshifted spectra of five galaxies whose distances from us range from 2.6 to 287.5 Mpc. NGC 221 v 210 km/s d 2.6 Mpc NGC 4473 v 2 300 km/s d 28.8 Mpc NGC 379 v 5 500 km/s d 68.8 Mpc Galaxy in the Ursa Major Cluster v 15 000 km/s d 187.5 Mpc Galaxy in the Gemini Cluster v 23 000 km/s d 287.5 Mpc Figure 13-26 The redshifts of the Ca, H, and K absorption spectral lines are shown for five galaxies at different distances from us. The line spectra above and below the absorption spectrum are standards used for determining the amount of shift accurately. [California Institute of Technology.] 13-6 Galaxies 659 EXAMPLE 13-5 Distance to a Galaxy in Virgo Redshift measurements on a galaxy in the constellation Virgo yields a recession velocity of 1200 km s. How far is it to Virgo? > SOLUTION Using Hubble\u2019s law, we obtain r v H0 1200 km>s 21 km>s per 106 c # y 17.5 Mpc (1200 km>s)(106 c # y) 21 km>s 57 106 c # y Remarks: Compare this result with distance measurements to Virgo made by some of the standard astronomical distance-measuring methods given in Table 13-4. Hubble\u2019s law tells us that the galaxies are all receding from us, with those the farthest away moving the fastest. However, there is no reason why our location in the universe", " should be special. An observer in any galaxy would make the same observations and compute the same Hubble constant. (See Problem 13-25.) Thus, Hubble\u2019s law states > that all of the galaxies are receding from each other at an average speed of 71 km s per Mpc of separation. In other words, the universe\u2014space itself\u2014is expanding. This is a profound discovery with enormous theoretical implications. All galaxies participate in the general expansion of the universe. As a result, the wavelengths of light emitted toward Earth by galaxies (and stars and anything else out there) is lengthened or stretched along with the space through which it is moving, producing the cosmological redshift. It is the cosmological redshift that is described by Hubble\u2019s law. This redshift is not related to the galaxy\u2019s recessional velocity by the relativistic Doppler effect equation that we developed in Chapter 1 (Equation 1-38), even though astronomers often use that equation to express a measured redshift z as the radial velocity of a galaxy as if it were moving through space, rather than the actual velocity with which it receding from us due to the expansion of space. That this practice provides a reasonable estimate of relatively nearby distances can be seen as follows: Substituting Equation 13-27 into Equation 13-29 and solving for r yields r (z 1)2 1 (z 1)2 1 c H0 13-30 z 2, Equation 13-30 yields values of r within about 5 percent of the values meaFor z V 1, sured by the methods listed in Table 13-4. If then the nonrelativistic equation r cz>H0 z 0.1 the error is about the same as above. However, always remember that the cosmological redshift has nothing to do with the Doppler effect. can be used and, e.g., for Table 13-4 Distance measurements to Virgo Method Cepheids Novae Brightness fluctuations Type Ia supernovae Distance to Virgo (Mpc) Maximum useful distance (Mpc) 15 \u2013 25 29 21.1 3.9 20 15.9 0.9 50 19.4 5.0 1000 660 Chapter 13 Astrophysics and Cosmology The fractional change in the wavelength for the cosmological redshift is equal to the fractional change in the \u201csize\u201d or scale R of the universe since the time when the light was emitted. This allows us to also write the redshift as z 0 0 R R0 R0 Robserved Remitted Remitted \u21d2 Robserved Remitted 1 z 13-31 Equation 13-31 says, for example, that a galaxy with z 2 is now three times farther from Earth than it was when the observed light was emitted. Looking at regions in space that are cosmologically nearby or close to us provides a \u201csnapshot\u201d of what the universe looks like everywhere now. At Earth at the present time (t0), z 0. That is, galaxies close by the Milky Way have no (measurable) cosmological redshift. Looking at objects with higher z values corresponds to looking back in time. Thus, at redshift z we are seeing the universe as it appeared when it was of its size now. 1>(1 z) The discovery and analysis of the redshifts and luminosities of the quasars has significantly furthered our understanding of the expansion and evolution of the universe. Observations show that bright quasars are more numerous at large z than at small z, that is, the space or volume density of bright quasars was larger at earlier times than it is now. This could be because there where more of them in earlier times or their luminosities could have been higher or both. Or it could be that the observations are simply the result of the general expansion of space and the volume density and luminosities of the quasars have not changed over time. To remove the complicating effect of the expansion, astrophysicists and cosmologists define comoving coordinates and, correspondingly, for our purposes here comoving space density. The former we will return to in Section 13-8. The latter removes the effect of the expanding universe by dividing the number density of objects per cubic Mpc at redshift z by (1 z)3. This converts the number density of objects to the value it would have at z 0 (today). Thus, if the number density and/or brightness have been constant over time, their comoving space density and brightness will be constant. Changes signal a change in the number density or an evolution of the quasars or both. Observations of the comoving space density of bright quasars show that they are more than 1000 times more numerous at z 2 than they are today (at z 0), but the total number of quasars has not changed back to about z 2. Therefore", ", observations indicate that the luminosity of quasars evolves over time, but not their comoving space density. That is, at least back to z 2, there appears to be a constant number of quasars growing dimmer as the universe expands. Further back than z 2, measurements have reached about z 6. In this range the picture is more complicated, diminthe comoving space density declining after about ishing by approximately a factor of 10 by z 4. The meaning of the decrease is a focus of continuing research. z 3, An obvious question is whether there are other observational results that support Hubble\u2019s conclusion. For example, is the observed expansion general, or could it be a statistical accident\u2014a consequence of our having measured distances to only a fraction of the 250,000 (out of the 1010 galaxies in the observable universe) whose redshifts have been measured to date? Thus, redshift surveys of the universe are an important first step in studying Hubble\u2019s expansion. 13 109 c # y The oval encircles the most distant galaxy yet discovered (as of 2007). The faint streaks are the galaxy\u2019s image as gravitationally lensed by galaxy cluster Abell 2218. Its redshift of 6.6 puts from Earth moving away it about from us at about 0.97 c. The light that formed this image left the galaxy when the universe was only 700 million years old. [NASA, Andrew Fruchter and the ERO Team [Sylvia Baggett (STScI), Richard Hook (ST-ECF), Zoltan Levay (STScI)] (STScI).] 13-6 Galaxies 661 Redshift surveys establish the existence of large scale structure in the universe and, together with independent distance measurements for individual galaxies, determine the Hubble constant. Such surveys have been underway for several years and about 105 of the volume of the visible universe has now been mapped. These surveys have yielded several unexpected discoveries, but have not yet answered the question above conclusively. There are huge voids in space\u2014regions where the density of galaxies is only 20 percent or so of the average for the universe. In addition, the galaxies themselves tend to be grouped into clusters and the clusters into superclusters. The Milky Way is a part of the Local Cluster that contains about a dozen galaxies. The galaxies also tend to lie on thin, sheetlike structures. How such structures might have evolved in the general expansion described by Hubble\u2019s law presents a challenge to cosmological models. The most successful cosmological model thus far has been the Cold Dark Matter ( CDM) is Einstein\u2019s cosmological constant. One of the largest of the galaxy model, where mapping projects has been the 2dF Galaxy Redshift Survey (2dF GRS; Figure 13-27). The project, completed in 2002, obtained high-quality spectra and redshifts for 245,591objects, mainly galaxies. Still underway is the largest survey, the Sloan Digital Sky Survey (SDSS). The SDSS first phase was completed in 2005 after measuring the spectra of 675,000 galaxies, 90,000 quasars, and 185,000 individual stars. The SDSS second phase of measurements is currently underway at the project\u2019s dedicated 2.5-m telescope in New Mexico. - - - Figure 13-27 The 2dF Galaxy Redshift Survey (2dF GRS) is a major spectroscopic survey utilizing unique facilities built by the Anglo-Australian Observatory. By the survey\u2019s completion in 2002 it had recorded precise spectra for 245,591 objects enabling a wide range of new analyses including, for example, the first direct comparison with the microwave background anisotropy on the same spatial scale and studies of galaxy clustering to test inflationary cosmological models of the early universe. The survey is integrated with the 2dF Quasi Stellar Objects survey. [Matthew Colless and the 2dF Galaxy Redshift Survey Team.] 662 Chapter 13 Astrophysics and Cosmology 1>H0, 13-7 Cosmology and Gravitation We have seen that applying Hubble\u2019s law to the observations of galaxies leads inescapably to the conclusion that the universe is expanding and provides us with a measure, of how long ago that expansion began. In this section, we will examine the basic theoretical framework that suggests possible tests of that conclusion. The basis for this discussion is the philosophical view that at large scale the universe is homogeneous and isotropic at any instant in time. That is, at any given instant the universe has the same physical properties everywhere and looks the same in all directions from every location. This is called the cosmological principle. Note that Hubble\u2019s law is consistent with the cosmological principle. We have already seen that the cosmological principle clearly does not hold on a local scale. Galaxies are clustered into local groups.", " Even on a scale of 108 c y, the dimension typical of galactic superclusters, the universe is neither homogeneous nor isotropic. However, when maps of very distant space are examined (Figure 13-28), the distribution does appear to be statistically homogeneous and isotropic. Redshift survey maps like Figure 13-27, which extend to about 4 109 c y, do indeed show homogeneity and isotropy in a statistical sense. # # The Critical Energy Density of the Universe 1>H0 1.3 1010 Noting that the Hubble age years ignores the effect of gravity and ignoring for the moment the recently discovered acceleration of the expansion, the expectation is that gravity tends to slow the expansion over time. Is the gravity in the universe strong enough to eventually reverse the expansion and cause the universe to collapse? Or will the expansion continue forever? The answer depends upon the mass density of the universe. We can understand this by considering the motion of a single galaxy of mass m at a very large distance R from Earth. Let M be the total mass of all the galaxies within the spherical volume of radius R. The gravitational potential energy of our single galaxy is The total energy of the galaxy is GMm>R. 0 E K U 1 2 mv2 GMm R 13-32 Figure 13-28 A map showing approximately 2 million galaxies ranging # up to 2 109 c y away. The distribution of the galaxies looks essentially homogeneous and isotropic. This is a composite of 185 contiguous photos taken by the Schmidt telescope at the European Southern Observatory. The south Galactic pole is at the bottom center. [S. Maddox (Nottingham U.) et al., APM Survey, Astrophys. Dept. Oxford U.] 13-7 Cosmology and Gravitation 663 If we project an object with some speed v from Earth, the object will escape if its total energy is greater than or equal to zero, but if the total energy is negative, the particle will eventually stop and fall back to Earth. Similarly, if the total energy of the galaxy is greater than or equal to zero, it will continue to move away from Earth forever, but if the total energy is negative, the galaxy will eventually stop moving away from Earth and start moving back toward Earth. We can see from Equation 13-32 that the total energy of the galaxy depends on the mass density We can find the critical mass density of the universe by setting the total energy in Equation 13-32 equal to zero: M> (4>3R3). c 1 2 mv2 GMm R Substituting v H0R from Hubble\u2019s law (Equation 13-27), we obtain 1 2 m(H0R)2 GMm R Then 1 2 H2 0 GM R3 c M a 4 3 bR3 3H2 0 8G 13-33 Substituting the values for H0 and G, we obtain for the critical mass density of the universe: c 1026 kg>m3 This corresponds to about five hydrogen atoms per cubic meter of space. c 0 0 0 0 c, c, is greater than Determining the present mass density based on the visible (baryonic matter) universe is only about 4 percent of of the universe is thus an important goal. the expansion will reverse and the universe will collapse. If it is If it is larger than smaller, then the expansion will continue forever. If it should happen that the expansion will coast to a stop, but will not begin to contract. It should also be clear that if now, it will always be so because it is actually the conservation of energy that determines whether contraction or continued expansion will occur. Since must decline over time as expansion progresses, the Hubble constant In other words, must also decline over time to ensure that the Hubble constant must be a function of time H(t), that is, The value c, of suggesting that the universe will expand forever. However, the dark matter of the Together, the visible matter and universe discussed earlier affects the value of c. dark matter account for about 26 percent of the mass necessary to make Examination of the recession rates of the Type Ia \u201cstandard candle\u201d supernovae suggests that dark energy provides the additional 74 percent needed. Recalling that they have very similar brightness and light curves, the discovery that for a given brightness the redshifts of distant Type Ia supernovae are less than expected implies that the universe was expanding at less than the expected rate in the past. Therefore, the universe is expanding at an accelerated rate today. The implication is that dark energy corresponds to a repulsive force that is speeding up the expansion. remains larger than H0 c. H(t0). 0. 0 0 0 664 Chapter 13 Astrophysics and Cosmology Cosmic background spectrum at the north Galactic pole 2 4 6 10 8 Wave number, mm\u20131 12 14 16 18 20 Figure 13-29 (a) The spectrum of the cosmic background radiation measured by NASA\u2019", "s Cosmic Background Explorer (COBE) spacecraft. The dots are the data points. The solid curve is the Planck radiation curve for a blackbody at 2.725 K. [CERN Courier, June 1991, p. 2, courtesy of NASA.] (b) This detailed allsky picture of the infant universe includes three years of WMAP data. It shows 13.7 109-years-old temperature fluctuations that were the seeds that eventually became the galaxies and provides new clues regarding events that occurred in the first trillionth of a second following the Big Bang. The range of the temperature 200 K. fluctuations is [NASA/WMAP Science Team.].4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 (a) (b) This view is independently supported by very recent research that compares the temperature fluctuations in the cosmic microwave background shown in Figure 13-29b with their origin in the \u201clumpiness\u201d of matter in the 2dF GRS (see Figure 13-27). 13-8 Cosmology and the Evolution of the Universe Following his completion of general relativity in 1915, Einstein turned to cosmology. He based his early work on the assumption that the universe was not only homogeneous and isotropic, but also constant in time. This is sometimes called the perfect cosmological principle. He quickly discovered that, like a universe described by Newton\u2019s gravitational theory, only an empty (no mass) universe can be static. He found that a static universe could be metastable if it contained mass and a cosmological constant, thereby committing what he later described as the biggest blunder of his life. On learning of Hubble\u2019s discovery of the expansion of the universe, he abandoned the cosmological constant. However, dark energy is essentially a revision of the cosmological constant for an expanding universe. Since the constant, in effect, generated the mass of the universe, adjusting its value could shift some of the predicted mass into energy, thereby accounting for the unseen mass. 13-8 Cosmology and the Evolution of the Universe 665 One difficulty with the steady-state model is a problem known as Olber\u2019s paradox, first posed by Edmund Halley in 1720, but named after the nineteenth-century physician-astronomer Wilhelm Olbers who publicized it widely. If there is a uniform distribution of stars throughout an infinite space, then no matter in which direction you look, you will eventually see a star. Since stars are bright, the night sky should look as bright as the surface of the average star. (This is analogous to standing in an infinitely large forest in which all the trees are painted white. Along any line of sight, you will eventually see a white tree, so you should see white in all directions.) Why then is the night sky dark? The solution offered by Olber himself was that interstellar dust absorbs the light from distant stars. But this is no help since the dust would eventually be heated to glowing, so the night sky should still be bright. The solution to this problem came in part with Hubble\u2019s discovery of the expansion of the universe. The point is not that light is redshifted out of the visible region,18 but that the energy of every photon is diminished, since However, redshift can account for only a very small part of the solution. The key is that, since the velocity of light is finite, looking into space means looking back in time. Looking deeper into space we eventually would be looking at a time before the stars began to form, that is, at a time greater than the Hubble age. (In terms of our forest analogy, the distant trees have not yet been painted white; therefore, if the separation of the trees is great enough, many lines of sight will end on dark trees.) E hc>. A Simple Cosmology Model19 To a considerable extent, descriptions of the origin and evolution of the universe depend on the cosmology model that is used to interpret observations. The appearance of galaxies at cosmological distances is directly affected by the curvature of spacetime through which the light travels on its trip to Earth. One would reasonably expect that the distortion of spacetime would be more complex at higher redshifts when the visible universe was smaller and the mass density larger than now, which is understandably a region of high interest to cosmologists. A proper interpretation of observations at high redshifts necessarily requires the use of the general theory of relativity. Such an application is beyond the scope of our discussions; however, we can develop a useful, albeit approximate view of the expansion of the universe with the aid of cosmological a model based on Newtonian mechanics and the cosmological principle, then follow up with a very brief look at the current state of the theory based on general relativity and measurements from the Wilkinson Microwave Anisotropy Project (WMAP). Consider a Universe Mass m (a) r Expansion r(t2) > r(t1", ") Universe r(t1) r(t2) (b) Mass m thin spherical shell of radius r in our homogeneous, isotropic universe (Figure 13-30a). The shell contains and uniformly distributed total mass m. Figure 13-30 (a) Cross section of a thin spherical shell containing mass m in an isotropic, homogeneous universe. (b) The same shell as it has expanded from its size at time t1 to time t2. The thickness of the shell also expands; however, the mass in the shell is constant. The shell\u2019s comoving coordinate r(t0) remains unchanged. 666 Chapter 13 Astrophysics and Cosmology Our shell, like all such shells, expands along with the general expansion of the universe, becoming both large and thicker; however, m remains constant. Assuming gravity to be the only interaction present, the total energy of the mass m within the shell is the kinetic energy plus the gravitational potential energy: E K U 1 2 mv2(t) GMrm r(t) 13-34 (4>3)r3 where v(t) is the recessional velocity of the shell and Mr is the mass within the sphere that is enclosed by the shell. Like m, Mr also remains constant because Mr and, although r and are both functions of time, The mass of our universe outside the shell exerts no net gravitational force on m. (Why not?) As the shell expands the gravitational force due to Mr causes the kinetic energy of m to decrease and the gravitational potential energy to increase, i.e., to become less negative. Conservation of energy requires that the total energy E be unchanged, so we will, with remarkable foresight, write the total energy as r 3. E 1 2 kmc2r2(t0) 13-35 now and k is a constant with units where r(t0) is the radius of the shell at t0 (length)2. As we will see, the constant k determines the geometry of the universe. Combining Equations 13-34 and 13-35, substituting for Mr, and cancelling m yield v2(t) 8 3 G(t) r2(t) kc2r2(t0) 13-36 Referring to Equation 13-36, note that: \u2022 If k 0, then the total energy of the mass m is negative. In that event there is a radius r(t) beyond which the shell cannot expand because v is (instantaneously) zero and we say that the universe is closed or bounded. The shell will then begin to contract due to the mass Mr interior to the shell and undergo a time-reversed copy of the expansion back to what is sometimes called the \u201cBig Crunch.\u201d \u2022 If k 0, then the total energy of the mass m is zero. In that case as t S, r(t) S coast forever toward a halt. We refer to such a universe as flat. the recession velocity v (t) S 0 and the shell (and universe) \u2022 If k 0, then the total energy of the mass m is positive. In that event, as r(t) increases, the gravitational potential energy becomes steadily more negative. But v2(t) must continually increase in order to keep the total energy positive. We then say that the universe is open and will continue to expand forever. Since the cosmological principle requires that all shells, including ours in Figure 13-30a, must expand in the same way, that is, the time required for the radii of all shells to, say, triple, must be the same, we can express the radius r(t) of our shell (or any shell) as r(t) R(t)r(t0) 13-37 13-8 Cosmology and the Evolution of the Universe 667 where r(t) is the distance from the coordinate origin to the shell (see Figure 13-30a) and R(t) is the scale factor first introduced in Equation 13-31 which describes the expansion or contraction of the universe. Since there is nothing special about our shell, R(t) is the same for all shells. The constant r(t0) that in effect labels the shell is called the comoving coordinate (see Figure 13-30b). Assuming that the present is t0, R(t0) 1 and, as we have noted, the present radius of our shell is r(t0). In Equation 13-31 R refers to the rest frame of the observer and R0 to that of the emitting star or galaxy. Since R(t0) 1, the scale factor R(t) and the redshift z are related by 1 z 1 R(t) 13-38 For example, looking back in time to redshift z 2, the scale factor R 1 3, that is, the visible universe was one-third of", " its present size. Hubble\u2019s law can now be written as > v (t) H(t)r(t) H(t)R(t)r(t0) 13-39 Differentiating Equation 13-37 with respect to time, substituting into Equation 13-36, and cancelling r2(t0) yield aH2(t) 8 3 G (t)bR2(t) kc2 13-40 Cosmologists define the density parameter present value is \u00c6 (t)> c(t) whose \u00c6 0 0 c0 8G 0 3H3 0 13-41 and since, as you will show in Problem 13-26, (z) (1 z)3, \u00c6 \u00c6 0 (1 z)3 H2 0 H2 13-42 \u00c6 \u00c6 \u00c6 \u00b6. rel, In relativistic cosmology models, universes are described in m, terms of three components: matter (including dark matter) and dark energy (the relativistic particles (e.g., neutrinos) cosmological constant) Current data suggests that relativistic particles do not contribute significantly to the energy density at the present time. Cosmologists, like all scientists, use graphical representations whenever possible, in this situation employing a twodimensional graph of to assess the current state of the universe. It is reproduced as Figure 13-31. The coordinates (\u00c6 for the best values of the density parameters based on (0.73 0.04, 0.27 0.04) current WMAP observations are. Delving further into this and related issues is on the leading edge of cosmological research, but regretfully beyond the scope of this book. \u00b60, \u00c6 versus m0) \u00c6 \u00c6 \u00b60 m0 No Big Bang 3.0 2.5 2.0 1.5 1.0 0 \u039b \u03a9 0.5 0.0 \u22120.5 \u22121.0 \u22121.5 \u22122.0 g a ti n a ti iv u n iv e r s e e r s e 0.0 0.5 1.0 2.0 2.5 3.0 1.5 \u03a9 m0 \u00c6 \u00c6 \u00b60 Figure 13-31 Every point in the m0 plane represents a possible universe. The lines drawn are based on various parameters from general relativity analyses. The line with negative slope starting in the lower right corner corresponds to the geometry parameter k 0 (see Equation 13-36). 668 Chapter 13 Astrophysics and Cosmology The Big Bang Over the past century cosmologists have developed a well-defined standard model of the universe that fits a comprehensive set of very precise, constraining measurements and observations. The foundation of the standard model is the Big Bang theory. It is the observational foundation of the theory that (13.7 0.2) 109 years ago20 the universe was in a hot, dense state and at that particular time a single event, the Big Bang,21 initiated an expansion and cooling that has continued to the present time. Two major astrophysical discoveries made in the 1960s were the first of several that have convinced most scientists that the model is correct. The first of the two discoveries that supported the evolving universe model was Martin Ryle\u2019s22 observations revealing that there is a higher comoving space density of distant radio galaxies than nearby ones. Since distant observations correspond to earlier times, this meant that the universe had looked different at earlier times than it does now; that is, it has evolved over time. The second discovery was monumental, as important as Hubble\u2019s discovery of the expansion of the universe itself. In investigating ways of accounting for the cosmic abundance of elements heavier than hydrogen, cosmologists recognized that nucleosynthesis in stars could explain the abundance of those heavier than helium but not that of helium itself. (See Figure 13-32.) Helium must therefore have been formed during the Big Bang. Synthesizing the amount of helium that would account for its present abundance requires that the Big Bang occurred at an extremely high initial temperature to provide the necessary reaction rate before the fusion was shut down by the decreasing density due to the very rapid initial expansion. The high temperature implies a corresponding thermal (blackbody) radiation field that would cool as the expansion progressed. Theoretical analysis predicted that from the Big Bang to the present, the remnants of the radiation field should have cooled to a temperature of Hydrogen Helium Carbon Oxygen Neon Magnesium Silicon Sulfur Argon Iron Calcium Nickel Nitrogen Boron Lithium Beryllium 1 10\u22122 10\u22124 10\u22126 10\u22128 10\u221210 10\u221212 10 20 30 40 50 Atomic number, Z Figure 13-32 The abundances, relative to hydrogen, of elements in the Milky Way up to Z 50 (tin). Note the peak at Z 26 (iron), the sharp decline in abundances after iron, and the extremely", " low relative abundances of lithium, beryllium, and boron. 13-8 Cosmology and the Evolution of the Universe 669 max about 3 K, corresponding to a blackbody spectrum with peak wavelength in the microwave region. In 1965, the predicted Cosmic Microwave Background (CMB) radiation was discovered by Arno Penzias and Robert Wilson23 at Bell Labs. Since this landmark discovery, analysis of data from the Cosmic Background Explorer (COBE) satellite by John Mather and George Smoot24 and by the Wilkinson Microwave Anisotropy Project (WMAP) collaborators have established the temperature of the background field at 2.725 0.001 K with deviations from that value of no more than a few thousandths of a percent. These results show that the CMB has the isotropic distribution in space that is absolutely essential for a universe that satisfies the cosmological principle. Indeed, the Cosmic Microwave Background is the most precise blackbody known in nature (see Figure 13-29a). In addition, the WMAP detection of (see Figure 13-29b) provided the first temperature fluctuations in the range of 30 evidence for density inhomogeneities that cosmologists believe seeded all of the galactic structure of the universe. K The Very Early History of the Universe What was the Big Bang like? The singular event that initiated the expansion of the universe must have been a huge explosion of space that occurred throughout the entire hot, dense state. Most cosmologists favor the standard model as the theoretical description of the evolution of the universe following the Big Bang. It relies heavily on recent experimental discoveries and theoretical advances in particle physics and reflects the increasing overlap of frontier research in those areas of physics over the past several years. The standard model\u2019s account of how the universe evolved from t 0 to now, when t 1010 years, is outlined in the following discussion and illustrated in Figure 13-33. In the beginning the universe was dominated by energy at negative pressure which led to an early exponentially accelerated expansion referred to as inflation. The theoretical basis for inflation comes from general relativity and the cosmological principle which together give the acceleration equation (not intended to be obvious) below. 4G 13-43 1 R(t) d2R dt2 3c2 (c2 3P) c2 where R(t) is the dimensionless scale factor discussed earlier, is the energy density of the universe, and P is the pressure. Notice that in situations where negative pressure dominates, the expansion has positive acceleration. This very early period of inflation for which, bear in mind, we have no direct evidence, is nonetheless successful in resolving several cosmological questions, including (1) Why is the CMB temperature so uniform in every direction? (2) Why is the geometry of the universe so close to being flat? (3) Why do we not see magnetic monopoles? (4) What is the origin of the anisotropies measured by WMAP? Following that brief, but extremely rapid inflation, the universe was dominated by radiation, then subsequently by matter. Recently, cosmologically speaking, it has again become dominated by a negative energy pressure which is driving a new, but slower acceleration of the general expansion. Today, as we have alluded to earlier, matter accounts for only about 26 percent of the energy density of the universe, only 4 percent being ordinary matter. (See Figure13-34). The other 22 percent is the cold dark matter (CDM) discussed earlier that neither emits nor reflects light or is affected by radiation pressure, but does participate in the gravitational interaction. Figure 13-34c shows what may be the first indirect observation of dark matter. Dark energy, the remaining 74 percent of the energy density of the universe, is apparently driving the new acceleration. 6 7 0 Quantum gravity \u2022 Supergravity? \u2022 Extra dimensions? \u2022 Supersymmetry? \u2022 Superstrings? End of grand unification \u2022 Origin of matter-antimatter asymmetry \u2022 Monopoles \u2022 Inflation End of electroweak unification \u2022 End of supersymmetry? Quark/hadron transition Big Bang nucleosynthesis Matter domination \u2022 Formation of structure begins \u2022 Formation of atoms \u2022 Decoupling of matter and radiation B ig Ban g 1030 K 10 25 K 10 20 K 1015 K 1010 K 105 K 1 K Temperature 1018 GeV 1015 GeV 1012 GeV 109 GeV 106 GeV 1 TeV 1 GeV 1 MeV 1 keV 1 eV 1 meV Energy Rest energy of flea KE of sprinter Highest-energy cosmic rays CM energy TeV I TeV II Nuclear binding energy Atomic binding energy 10\u201330 10\u201325 10\u201320 10\u201315 10\u201310 10\u20135 1 Size 1097 g/cm3 1081 g/cm3 1065 g/cm3 1049 g/cm3 1033 g/cm3 1017 g/cm3 10 g/cm3 10\u201315 g/cm3", " 10\u201330 g/cm3 Density 10\u201342 s 10\u201336 s 10\u201330 s 10\u201324 s 10\u201318 s 10\u201312 s 10\u20136 s 1 s 106 s 1012 s 1018 Seconds Nuclear matter Air Water 1 atom/cm3 1 103 106 109 Years Time Constituents Leptons and quarks \u03bd e e\u2013 u d \u03bd\u03bc \u03bc\u2013 c s \u03bd\u03c4 \u03c4 \u2013 t b?????? Gauge bosons Photons GLUONS W\u00b1, Z X, Y, \u2026?? \u03b3 Galaxy forms NOW Solar system forms \u03bd \u03bd \u23af e\u00b1 n, p H+, D+, 3He2+, 4He2+, 7Li3+, e\u2013 H, D, 3He, 4He, 7Li 2-K neutrino background Ratio of matter/radiation = 5 \u00d7 10\u201310 3-K microwave background Figure 13-33 The evolution of the universe from time t 0 to the present, according to the standard model. 13-8 Cosmology and the Evolution of the Universe 671 Atoms 4% Dark matter 22% Dark energy 74% (a) (c) (b) Figure 13-34 (a) The mass-energy content of the universe. Recent observations indicate that the dark energy is driving a renewed acceleration of the general expansion of the universe. (b) Stellar velocities in M31. Newton\u2019s law of M r. Thus, in M31 much of the mass lies well beyond the visible gravitation requires that constant velocity implies extent of the galaxy, a very early indicator of dark matter. (c) Galaxy cluster 1E0657-558 resulted from a small cluster passing through a larger one some time in the past. Using this cluster as a gravitational lens for more distant galaxies made possible the mapping of the gravitational potential of 1E0657-588 (the large fuzzy \u201ccloud\u201d). X-ray emission recorded by the Chandra X-Ray Observatory of the two central, darker portions of the \u201ccloud\u201d reveal the hot gases (ordinary matter) of the two original colliding clusters. The lighter portions to the outside of the \u201ccloud\u201d are inferred to be dark matter. [The authors thank Vera Rubin for permission to use image (b). (b) Vera Rubin and Janice Dunlap. (c) X-ray courtesy NASA/CXC/CfA/M Markevitch et al.; optical courtesy NASA/STScl; Magellan/U Arizona/D Clowe et al.; lensing map courtesy NASA/STScl; ESO WFI; Magellan/U Arizona/D Clowe et al.)] Initially, the four forces of nature (strong, electromagnetic, weak, and gravity) were unified into a single force. Physicists have been successful in developing theoretical descriptions that unify the first three, but a theory of quantum gravity, needed for the extreme densities of the single-force period, does not yet exist. Consequently, until the cooling universe \u201cfroze\u201d or \u201ccondensed out\u201d the gravitational force about 1043 seconds after the Big Bang when the temperature was still 1032 K, we have no means of describing what was occurring. At this point the average energy of the particles would have been about 1019 GeV. As the universe continued to cool below 1032 K the three forces other than gravity remained unified and are described by grand unification theories (GUTs). Quarks and leptons were indistinguishable and particle quantum numbers were not conserved. It was during this period that a slight excess of quarks over antiquarks occurred, roughly 1 in 109, that ultimately resulted in the matter that we now observe in the universe. 672 Chapter 13 Astrophysics and Cosmology At 1035 seconds, the universe had expanded sufficiently to cool to about 1027 K at which point another phase transition occurred as the strong force condensed out of the grand unified group, leaving only the electromagnetic and weak forces still unified as the electroweak force. During this period the previously free quarks in the dense mixture of roughly equal numbers of quarks, leptons, their antiparticles, and photons began to combine into hadrons and their antiparticles, including the nucleons. By the time the universe had cooled to about 1013 K, at about t 106 s, the hadrons had mostly disappeared through annihilation. This is because 1013 K corresponds to kT 1 GeV, which is the minimum energy needed to create nucleons and antinucleons from the photons present via the reactions and \u00a1 p p \u00a1 n n 13-44a 13-44b The particle-antiparticle pairs annihilated and there was no new production to replace them. Only the slight earlier excess of quarks led to a slight excess of protons and neutrons over their antiparticles. The annihilations resulted in photons and leptons and after about t 104 seconds,", " those particles in roughly equal numbers dominated the universe. This was the lepton era. At about t 10 seconds the temperature had fallen to 1010 K ( ). Further expansion and cooling dropped the average photon energy below that needed to form an electron-positron pair. Annihilation then removed all of the positrons as it had the antiprotons and antineutrons earlier, leaving only the small excess of electrons arising from charge conservation, and the radiation era began. The particles present were primarily photons and neutrinos. kT 1 MeV Within a few more minutes, the temperature dropped sufficiently to enable fusing protons and neutrons to form nuclei that were not immediately photodisintegrated. Deuterium, helium, and a bit of lithium were produced in this nucleosynthesis period, but the rapid expansion soon dropped the temperature too low for the fusion to continue and the formation of heavier elements had to await the birth of stars. 1>1000 A long time later, when the temperature dropped to about 3000 K as the universe of its present size, kT dropped below typical atomic ionization grew to about energies and atoms were formed. By then the expansion had cooled the radiation field so that the total radiation energy was now about equal to the energy represented by the remaining mass. This occurred when the scale factor R(t) reached about 2.8 104. As expansion and cooling continued, the energy of the steadily redshifting radiation declined until matter came to dominate the universe, its energy density exceeding that of today\u2019s 2.725 K radiation remaining from the Big Bang by about a factor of 1000. Now, once again, energy at negative pressure appears to dominate. Unanswered Questions and the Limits of Knowledge The standard model of the evolution of the universe and the current theories of stellar and galactic genesis and evolution have been amazingly successful. Still, some fundamental questions that have arisen during our discussions remain unanswered. Will the universe expand forever or rebound to its initial state and repeat the Big Bang? The answer depends on whether the present average matter density is greater or less than the critical density of about 1026 kg m3. The uncertainty in the current measurements would allow either possibility, but the value is tantalizingly close to the critical value. If it does equal the critical value, an intriguing additional question is, Why? We have noted the serious problem of the dark matter and how it might be explained. Answering some of these questions requires that we probe at the current limits of physical knowledge. > 13-8 Cosmology and the Evolution of the Universe 673 For example, near a mass m, general relativity prevents our seeing events occurring at dimensions less than the Schwarzschild radius, the event horizon, RS 2Gm c2 13-45 On the other hand the uncertainty principle in quantum theory places this limit at the Compton wavelength c: c h mc 13-46 m 2hc>2G, m 108 an expression for m that depends only on universal Equating these, kg. That relation for m together with Equation 13-46 alconstants, where lows the corresponding definition of a length unit dependent only on universal constants. That length and the time for light to travel across that length can be similarly expressed as 1035 m L c t a 1>2 Gh c5 b 1.35 1043 s 13-47 In terms of these units the mass density of the universe is such that a mass m is contained within a volume of dimensions The definition of the units of mass, length, and time in terms of fundamental constants was originally pointed out by Planck25 and are the basis for Planck units, the topic of the Exploring section that follows. L3 (1035)3 m3. Some cosmologists have suggested that, if the universe had evolved even slightly differently than it has, perhaps due to a slightly different value for h or e or some other fundamental constant, life on Earth and maybe Earth itself would be impossible. This can be attributed to the anthropic principle, that the universe looks as it does because we are here to see it. EXPLORING \u201cNatural\u201d Planck Units Not long after Max Planck had introduced the constant h in fitting physical theory to the emission spectrum of a blackbody, he pointed out that a system of \u201cnatural\u201d units for the fundamental quantities of mass, length, time, and temperature could be constructed from c speed the three fundamental constants of light, and G Newton\u2019s gravitation constant. c, and G, where Planck\u2019s constant 2 >, U, U Planck mass: Planck length: Planck time: Planck energy: MP (Uc>G)1>2 2.2 108 kg (UG>c3)1>2 1.6 1035 m (UG>c5)1>2 5.4 1044 s lP tP EP MPc2 2.0 109 J 1.2 1022 MeV Planck temperature: TP EP", " kB U1>2c5>2 G 1>2kB 1.4 1032 K 674 Chapter 13 Astrophysics and Cosmology If length, mass, temperature, and time are measured in Planck units, the result is the \u201cnatural\u201d units often used by particle physicists, astrophysicists, and cosmologists: c kB U G 1. When first proposed, Planck\u2019s suggested units had little basis in fundamental physics, but over time that has changed. As Frank Wilczek26 has pointed out, Planck\u2019s proposal has now become compelling: the constant is now the fundamental unit of action and c the fundamental unit of velocity. These are the primary units of measurement in the two great theories of modern physics, quantum mechanics and special relativity. The corresponding unit in general relativity is G (actually 1 Gc4). U > Summary TOPIC 1. The Sun As Wilczek speculates, with the natural units of measure it may soon be possible to understand why, compared to the other forces in nature, the gravitational force is so weak and how we can account for the value of the proton\u2019s mass. In addition, he suggests that we have the beginnings of a quantum theory of gravity that agrees accurately with all existing experimental data. Thus, Einstein\u2019s goal of the unification of the four natural forces may be just over the horizon. RELEVANT EQUATIONS AND REMARKS The solar energy received at the top of Earth\u2019s atmosphere, called the solar constant, is f 1.365 103 W>m2 The rate at which the Sun emits energy is the luminosity L}: L} 3.85 1026 W 13-1 13-2 Surface temperature Assuming that the Sun radiates as a blackbody, its effective surface temperature computed from the Stefan-Boltzmann law: Te can be 5780 K Te Source of Sun\u2019s energy The source of the Sun\u2019s energy in nuclear fusion, mainly via the proton-proton cycle which starts with the reaction 1H 1H \u00a1 2H e e 0.42 MeV 13-5 2. The stars The Milky Way 3. Evolution of the stars Stars are classed as either population I or population II, based on their composition. The former have 2 to 3 percent of their mass composed of elements heavier than helium; the latter are nearly devoid of those elements. Our galaxy, the Milky Way, consists of about 1010 stars. The Sun is about 28,000 c y from the center of the Galaxy, which is in the direction of the constellation Sagittarius from us. Only about 4 percent of the mass of the Galaxy is accounted for by the visible stars and the gas and dust of the ISM. # The Hertzsprung-Russell diagram displays the evolution of stars, relating their luminosities to their effective temperatures. Both quantities are related to the stellar mass. L M4 1>4 13-14 a Te L 4R2 b M1>2 13-17 Stars \u201cburning\u201d hydrogen to helium fall on the main sequence of the H-R diagram. TOPIC RELEVANT EQUATIONS AND REMARKS Final states of stars Following exhaustion of their hydrogen fuel, stars evolve to one of three possible final states, dependent on their mass: white dwarf, neutron star, or black hole. It is in cataclysmic events accompanying evolution to these states that elements heavier than Fe are formed. Notes 675 4. Galaxies Hubble\u2019s law Edwin Hubble grouped galaxies into four general categories: ellipticals, spirals, barred spirals, and irregulars. Hubble showed that the universe was expanding and, using spectral redshifts to determine the velocities of galaxies, that the recession velocities v were proportional to the distance r from us according to v H0r where the Hubble constant H0 is H0 71 4 km>s per Mpc 22 2 km>s per 106 c # y 13-28 13-27 > 1.3 1010 y is the Hubble age. It would correspond to the age of The quantity 1 H0 the universe under a constant velocity expansion if gravitational pull on receding galaxies were ignored. 5. Gravitation and cosmology Inflation The cosmological principle states that the universe has the same physical properties everywhere and looks the same in every direction from every location. The current theory of cosmology, called the standard model, describes the universe as having begun with the Big Bang 13.7 1010 years ago. It has substantial theoretical and observational support. The standard model holds that the very early universe underwent a period of exponentially accelerated growth which explains many features of the current universe. After a long period of slowing, the expansion of the universe is again accelerating. General References The following general references are written at a level appropriate for the readers of this book. Akerlof, C. W., and M. A. Srednicki, Relativistic", " Astrophysics and Particle Cosmology, New York Academy of Sciences, New York, 1993. Bahcall, J. N., Neutrino Astrophysics, Cambridge University Press, New York, 1989. Bennett, J., M. Donahue, N. Schneider, and M. Voit, Stars, Galaxies, and Cosmology, 2d ed., Addison Wesley, San Francisco, 2002. Carroll, B. W., and D. A. Ostlie, An Introduction to Modern Astrophysics, 2d ed., Pearson Addison Wesley, San Francisco, 2007. Notes Comins, N. F., and W. J. Kaufmann III, Discovering the Universe, 6th ed., W. H. Freeman, New York, 2003. Frauenfelder, H., and E. M. Henley, Subatomic Physics, 2d ed., Prentice Hall, Englewood Cliffs, N.J., 1991. Pasachoff, J. M., Contemporary Astronomy, Saunders, Philadelphia, 1977. Rees, M., Before the Beginning, Addison Wesley, San Francisco, 1997. Ryden, B., Introduction to Cosmology, Addison Wesley, San Francisco, 2003. Shu, F. H., The Physical Universe, University Science Books, Mill Valley, Calif., 1982. Weinberg, S., The First Three Minutes, updated ed., Basic Books, New York, 1993. 1. Hans Albrecht Bethe (1906\u20132005). He made the proposal concerning stellar energy sources in 1938. One of those who worked on the Manhattan Project during World War II, he received the Nobel Prize for his work on the Sun\u2019s energy source in 1967. 2. John Bahcall (1934\u20132005), American physicist. His definitive theoretical analysis of the solar neutrino spectrum provided the benchmark for experimentalists whose measurements ultimately confirmed neutrino oscillations. 676 Chapter 13 Astrophysics and Cosmology 37 17Cl 3. The reaction by which Davis\u2019s detector detected neutriSeventy-seven percent of decay, 18Ar e. nos is the neutrinos producing this reaction were from the step 6 of the p-p cycle shown in Table 13-1. S 37 8B e 4. The term galaxy is derived from the Greek word for milk. 5. Apart from the Sun, Sirius is the brightest star in the sky. 6. Annie Jump Cannon (1863\u20131941). An astronomer at the Harvard Observatory, her work on stellar classification systems forms the basis of the Henry Draper Catalogue which contains the spectral classifications of 225,300 stars. 7. Hipparchus (circa 190 B.C.\u2013circa 120 B.C.). The greatest of the Greek astronomers, he created the stellar magnitude system of classifying stars by brightness. He measured the size and distance to the Sun and Moon and made the first accurate star map showing the positions of about 1000 of the brightest stars. 8. Astronomers customarily capitalize the word Galaxy when it refers to the Milky Way. 9. Harlow Shapley (1885\u20131972). A long-time director of the Harvard Observatory, he was an early and vocal supporter of civil liberties and peace movements in the United States. 10. The term \u201cright ascension\u201d for the celestial longitude apparently comes from the appearance of the stars as rising vertically, i.e., at a right angle to the line of sight to the horizon each night. 11. Equinox means day and night are equal; solstice means \u201cstanding Sun.\u201d 12. Edouard A. Roche (1820\u20131883). A French astronomer, he also showed that a small body orbiting a large body would be broken up by tidal forces if it comes within 2.5 times the radius of the larger body. The distance is referred to as Roche\u2019s limit. It corresponds approximately to the outer limit of planetary ring systems in the solar system. 13. Subrahmanyan Chandrasekhar (1910\u20131995). He received his Ph.D. under P. A. M. Dirac and spent most of his career at the University of Chicago. He shared the 1983 Nobel Prize in Physics for his work on the evolution of stars. The Chandra X-Ray Observatory is named for him. 14. The possibility of black holes was first suggested by Rev. John Mitchell, an English amateur astronomer, in 1783. He observed that a star with the same density, but 500 times the radius of the Sun would have an escape velocity greater than the speed of light. He speculated that light could not leave such a star. The name black hole was coined by physicist John Wheeler. Problems Level I Section 13-1 The Sun 15. Gamma-ray bursts are named according to the first date of observation: GRB yymmdd 16. Edwin P. Hubble (1889\u20131953). Trained as a lawyer, he was influenced to take", " up astronomy partly by R. A. Millikan. In recognition of his many contributions, he was accorded the honor of being the first user of the 5-m Hale telescope on Mount Palomar. 17. Cepheid variables are rare stars for which a relation exists connecting the period of intensity variation to the brightness and, hence, to the distance from the Sun. They were one of the earliest means of measuring astronomical distances. Polaris, the current Pole Star, is a Cepheid variable. 18. Light beyond the blue of the visible region would be shifted into the visible. Indeed, the visible region might even get brighter! 19. This discussion is based on the development in the early part of Chapter 29 of Carroll and Ostlie (see General References). 20. This age of the universe results from measurements made by the Wilkinson Microwave Anisotropy Project. 21. The term Big Bang was coined by the eminent astronomer Fred Hoyle, a steadfast proponent of the steady-state universe, intending the term as derision of the expanding universe cosmology. 22. Sir Martin Ryle (1918\u20131984). His invention of longbaseline radiointerferometry resulted in his sharing the Nobel Prize in 1974. 23. Arno Allan Penzias (b. 1933), German-American physicist, and Robert Woodrow Wilson (b. 1936), American radioastronomer. Their discovery of the cosmic microwave background radiation, first predicted by George Gamow 20 years earlier, earned each of them a share of the 1978 Nobel Prize in Physics. 24. Frank C. Mather (b. 1946) and George F. Smoot (b. 1945), American physicists, shared the 2006 Nobel Prize in Physics for this work which provides very strong support for the Big Bang theory. 25. M. Planck, Sitzungsber. \u201cDtsh. Akad. Wiss,\u201d Berlin, Math-Phys. Tech. Kl., 440 (1899). 26. Frank Wilczek, American physicist, in the series \u201cScaling Mount Planck\u201d in Physics Today, June 2001, November 2001, and August 2002. He shared the Nobel Prize in Physics in 2004 for his contributions to quark theory. 13-1. Measurement of the Doppler shift of spectral lines in light from the east and west limbs > of the Sun at the solar equator reveal that the tangential velocities of the limbs differ by 4 km s. 6.96 105 km Use this result to compute the approximate period of the Sun\u2019s rotation. ( ) 13-2. The gravitational potential energy U of a self-gravitating spherical body of mass M and >R}. radius R is a function of the details of the mass distribution. For the Sun What would be the approximate lifetime of the Sun, radiating at its present rate, if the source of 1.99 1030 kg ) its emitted energy were entirely derived from gravitational contraction? ( 2GM2 } R} U} M} Problems 677 > Section 13-2 The Stars 13-3. Lithium, beryllium, and boron (Z 3, 4, and 5, respectively) have very low abundances in the cosmos compared to many heavier elements. (See Figure 13-32.) Considering the fusion of He to C, explain these low abundances. 13-4. The Sun is moving with speed 2.5 105 m s in a circular orbit about the center of the Galaxy. How long (in Earth years) does it take to complete one orbit? How many orbits has it completed since it was formed? 13-5. The reason that massive neutrinos were considered as a candidate for solving the missing mass problem is that, at the conclusion of the lepton era, the universe contained about equal numbers of photons and neutrinos. They are still here, for the most part. The former can be observed and their density is measured to be about 500 photons cm3; thus, there must be about that number density of neutrinos in the universe, too. If neutrinos have a nonzero mass and if the cosmological expansion has reduced their average speed so that their energy is now primarily mass, what would be the individual neutrino mass (in eV c2) necessary to account for the missing mass of the universe? Recall that the observed mass of the stars and galaxies (including the dust and gas) accounts for only about 4 percent of that needed to close the universe. 13-6. Using data from Table 13-3, construct a graph that demonstrates the validity of Equation 13-17. 13-7. Recalling that the light-year c y is the distance light travels in one year, compute in meters the distance equivalent to 1 light-second, 1 light-minute, 1 light-hour, and 1 light-day. > > # Section 13-3 The Evolution of Stars > 13-8. A unit of", " length often used by astronomers to measure distances in \u201cnearby\u201d space is the parsec (pc), defined as the distance at which a star subtends a parallax angle of one arc second due to Earth\u2019s orbit around the Sun. (See Equation 13-11 and Example 13-4.) The practical limit of such measurements is 0.01 arc second. (a) How many light-years is 1 pc? (b) If the density of stars in the Sun\u2019s region of the Milky Way is 0.08 stars pc3, how many stars could, in principle, have their distances from us measured by the trigonometric parallax method? 13-9. Astronomers often use the apparent magnitude m as a means of comparing the visual brightness of stars and relating the comparison to the luminosity and distance to \u201cstandard\u201d stars, such as the Sun. (See Equation 13-9.) The difference in the apparent magnitudes of two stars m1 >f2), and m2 is defined as a relation based on the logarithmic response of the human eye to the brightness of objects. Pollux, one of the \u201ctwins\u201d in the constellation Gemini, has apparent magnitude 1.16 and is 12 pc away. Betelgeuse, the star at Orion\u2019s right shoulder, has apparent magnitude 0.41. How far away is Betelgeuse, if they have the same luminosity? 13-10. Using the H-R diagram (Figure 13-17), determine the effective temperature and the luminosity of a star whose mass is (a) 0.3 (c) Compute the radius of each star. (d) Determine their expected lifetimes, relative to that of the Sun. 13-11. Two stars in a binary system are located 100 c y from Earth and separated from each other by 108 km. What is the angular separation of the stars in arc seconds? In degrees? 2.5 log (f1 M}. # and (b) 3 m1 M} m2 Section 13-4 Cataclysmic Events 13-12. Compute the energy required (in MeV) to produce each of the photodisintegration reactions in Equations 13-19 and 13-20. 13-13. The gas shell of a planetary nebula shown at the right in Figure 13-18 is expanding at 24 km s. Its diameter is 1.5 c y. (a) How old is the gas shell? (b) If the central star of the planetary nebula is 12 times as luminous as the Sun and 15 times hotter, what is the radius of the central star in units of? > # R} Section 13-5 Final States of Stars 13-14. Calculate the Schwarzschild radius of a star whose mass is equal to that of (a) the Sun, (b) Jupiter, (c) Earth. (The mass of Jupiter is approximately 318 times that of Earth.) 13-15. Consider a neutron star whose mass equals (a) Compute the star\u2019s radius. (b) If the neutron star is rotating at 0.5 rev s and assuming its density to be uniform, what is its rotational kinetic energy? (c) If its rotation slows by 1 part in 108 per day and the lost kinetic energy is all radiated, what is the star\u2019s luminosity? 2M}. > 678 Chapter 13 Astrophysics and Cosmology 13-16. If the 90 percent of the Milky Way\u2019s mass which is \u201cmissing\u201d resides entirely in a large black hole at the center of the Galaxy, what would be the black hole\u2019s (a) mass and (b) radius? Section 13-6 Galaxies > 13-17. Redshift measurements for a particular galaxy indicate that it has a recession velocity of 72,000 km s. (a) Compute the distance to the galaxy. (b) The value of Hubble\u2019s constant depends critically on calibration distance measurements which are difficult to make. If the calibration distance measurements are in error by 10 percent, by how much is the age calculated from Equation 13-28 in error? 13-18. The bright core of a certain Seyfert galaxy had a luminosity of The luminosity increased by 100 percent in a period of 18 months. Show that this means that the energy source of the core is less than 9.45 104 AU in diameter. How does this compare to the diameter of the Milky Way? 1010L}. Section 13-7 Cosmology and Gravitation 13-19. Evaluate Equation 13-33 for the critical density of the universe. Section 13-8 Cosmology and the Evolution of the Universe 13-20. Cosmological theory suggests that the average separation of galaxies, i.e., the scale of the universe, is inversely proportional to the absolute temperature.", " If that is true, relative to the present size, how large was the universe compared to the scale today (a) 2000 years ago, (b) 106 years ago, (c) t 10 s after the Big Bang, (d) when t 1 s and, (e) when t 106 s? 13-21. Determine the value of the mass density of the universe for t Planck time. How does this compare to the density of the proton? Of osmium? 13-22. At what wavelength is the blackbody radiation distribution of the cosmic microwave background at a maximum? 13-23. How long after the Big Bang did it take the universe to cool to the threshold temperature for the formation of muons? What would be the mass of a particle-antiparticle pair that could be formed by the average energy of the current 2.725 K background radiation? 13-24. Show that the present mass density of the universe R(t)(t). 0 > M} 2 1030 kg Level II 13-25. If Hubble\u2019s Law is true for an observer in the Milky Way (i.e., us), prove that it must also be true for observers in other galaxies. (Hint: Use the vector property of the velocity.) (z) (1 z)3. 13-26. Show that the mass density of the universe at redshift z is given by 13-27. When the Sun was formed about 75 percent of its mass was hydrogen, of which only about 13 percent ever becomes available for fusion. (The rest is in regions of the Sun where the temperature is too low for fusion reactions to occur.) and the Sun fuses about 6 1011 kg s. (a) Compute the total mass of hydrogen available for fusion during the Sun\u2019s lifetime. (b) How long (in years) will the Sun\u2019s initial supply of hydrogen last? (c) Since the solar system is currently about 4.6 109 years old, when should we begin to worry about the Sun running out of hydrogen for fusion? 13-28. Supernova SN1987A was first visible at Earth in 1987. (a) How many years B.P. (before present) did the explosion occur? (b) If protons with 100 GeV of kinetic energy were produced in the event, when should they arrive at Earth? 13-29. Assume that the Sun when it first formed was composed of 70 percent hydrogen. How many hydrogen nuclei were there in the Sun at that time? How much energy would ultimately be released if all of the hydrogen nuclei fused into helium? Astrophysicists have predicted that the Sun can radiate energy at its current rate until about 23 percent of the hydrogen has been \u201cburned.\u201d What total lifetime for the Sun does that prediction imply? Compare these results with the corresponding ones from Problem 13-27. 13-30. Kepler\u2019s third law states that the square of a planet\u2019s orbital speed is proportional to the cube of its average orbital radius. Use Kepler\u2019s third law to answer each of the following questions. (a) The Moon\u2019s orbital radius is 3.84 105 km and it orbits Earth once every 27.3 d. Neglecting the Moon\u2019s mass, compute the mass of Earth. (b) Io (one of Jupiter\u2019s moons) orbits Jupiter once every 42.5 h in a near-circular orbit of average radius 4.22 105 km. Neglecting Problems 679 Io\u2019s mass, compute the mass of Jupiter. (c) Compute the orbital period of the International Space Station as it orbits 300 km above Earth\u2019s surface. (d) Charon, a moon of Pluto, orbits that planet once every 6.4 d at an average distance of 1.97 104 km. Compute the total mass of Pluto and Charon. What fraction of Earth\u2019s mass is this? (e) Using the data for the star S2, compute the volume (upper limit) that confines the black hole at the center of the Milky Way. Compare the result with the volume of the Sun. 13-31. Consider an eclipsing binary whose orbital plane is parallel to our line of sight. Doppler measurements of the radial velocity of each component of the binary are shown in Figure 13-35. m2 and that the orbits of each component about the center of mass are Assume that the mass m1 circular. (a) What is the period T and the angular frequency of the binary? (b) Show that in this where r separation of the binary. (c) Compute the values of m1, case m2, and r from the data in the v versus t graph. m2) (2 r3)>G, (m1 v, km/s 200 0 \u2013200 m2 5 10 t, days 15 m1 Figure", " 13-35 13-32. Prove that the total energy of Earth\u2019s orbital motion is where r is Earth\u2019s orbit radius. equal to one-half of its gravitational potential energy 13-33. Given the currently accepted value of the Hubble constant and the fact that the average matter density of the universe is one H atom m3, what creation rate of new H atoms would be necessary in a steady-state model to maintain the present mass density, even though the universe is expanding? (Give your answer in H atoms m3 per 106 years.) Would you expect such a spontaneous creation rate to be readily observable? E (mv2>2) (GM} m>r) (GM} m>r), > > Level III 13-34. The ability of a planet to retain particular gases in an atmosphere depends on the temperature that its atmosphere has (or would have) and the escape velocity for the planet. In general, if the average speed of a particular gas molecule exceeds of the escape velocity, that gas will disappear from the atmosphere in about 108 years. (a) Graph the average speed of H2O, CO2, O2, CH4, H2, and He from 50 K to 1000 K. On the same graph show the points representing of the escape velocity versus average temperature of the atmosphere for the planets in Table 13-5 below. (b) Show that the escape speed v from a planet is given by 1>6 1>6 v vEarth A (M>MEarth) (R>REarth) Table 13-5 Atmospheric temperatures Average Tatmo(K) Planet > M MEarth > R REarth 300 390 600 150 60 290 Earth Venus Mercury Jupiter Neptune Mars 1.00 0.81 0.06 318.00 17.00 0.11 1.00 0.95 0.38 11.00 3.90 0.53 680 Chapter 13 Astrophysics and Cosmology (c) Which of the six gases plotted probably would and would not currently be found in the atmospheres of the solar system bodies in the table? Explain each answer briefly. 13-35. Using the parallax technique, compute the distance to (a) Alpha Centauri (parallax angle 0.742 arc seconds) and (b) Procyon (parallax angle 0.0286 arc second. Express each answer in both light-years and parsecs. 13-36. As the Sun evolves into a red giant star, suppose that its luminosity increases by a factor of 102. Show that Earth\u2019s oceans will evaporate, but that the water vapor will not escape from the atmosphere. 13-37. The approximate mass of dust in the Galaxy can be computed from the observed extinction of starlight. Assuming the mean radius of dust grains to be R with a uniform number density n grains cm3, (a) show that the mean free path d0 of a photon in interstellar dust is given (b) Starlight traveling toward an Earth observer a distance d from the star by has intensity 1>(nR2). d0 > I I0ed>d0 # M} >300 (c # y)3. M} 3000 c y. If R 105 cm, calculate n. In the vicinity of the Sun, a measurement of I yields d0 > (c) The average mass density of solid material in the Galaxy is 2 g cm3 and in the disk the denCompute the ratio of the mass density of dust to the mass sity of stars is about 1 density of stars, assuming 1 13-38. The supernova SN1987A certainly produced some heavy elements. Compared to the atom starting from the proton-proton cycle, energy released in fusing 56 112Cd how much energy would be required to fuse two 13-39. Current theory suggests that black holes evaporate by the emission of Hawking radiation in a time t that depends on the mass M of the black hole according to the following relation: atoms into one atoms into one in 300 (c y)3. atom? 56Fe 56Fe 1H # t (1.024 1042 m3>s2)G2M2>hc4 (a) Explain without calculating anything why the formula implies that high-mass black holes have longer lifetimes than low-mass ones and why the rate of evaporation accelerates as the 1M}. black hole loses mass. (b) Compute the lifetime of a black hole whose mass equals Compare this time with the current age of the universe. (c) According to some theories, the largest black hole that could conceivably form would have a mass 1012 of the order of the mass of an entire galaxy. What would be the lifetime of a black hole that large? M}, APPENDIXA Table of Atomic Masses Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Z 0 1 Element Symbol (Neutron) Hydrogen Deuterium Trit", "ium n H D T 1.00798 2 Helium He 4.00260 3 Lithium Li 6.941 4 Beryllium Be 9.0122 5 Boron B 10.811 6 Carbon C 12.011 1* 1 2 3* 3 4 6* 8* 6 7 8* 9* 11* 7* 9 10* 11* 12* 14* 8* 10 11 12* 13* 14* 15* 9* 10* 11* 12 13 14* 15* 16* 17* 99.985 0.015 0.00014 99.99986 7.5 92.5 100 19.9 80.1 98.90 1.10 1.008665 1.007825 2.014102 3.016049 3.016029 4.002602 6.018886 8.033922 6.015121 7.016003 8.022486 9.026789 11.043897 7.016928 9.012174 10.013534 11.021657 12.026921 14.042866 8.024605 10.012936 11.009305 12.014352 13.017780 14.025404 15.031100 9.031030 10.016854 11.011433 12.000000 13.003355 14.003242 15.010599 16.014701 17.022582 Half-life and decay mode (if unstable) 10.4 m 12.33 y 0.81 s 0.12 s 0.84 s 0.18 s 8.7 ms 53.3 d ec 1.5 106 y 13.8 s 23.6 ms 4.3 ms 0.77 s 0.0202 s 17.4 ms 13.8 ms 10.3 ms 0.13 s 19.3 s 20.4 m 5730 y 2.45 s 0.75 s 0.20 s (Continued) AP-1 AP-2 Appendix A Element Symbol Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Half-life and decay mode (if unstable) Z 7 Nitrogen N 14.0067 8 Oxygen O 15.9994 9 Fluorine F 18.99840 10 Neon Ne 20.180 11 Sodium Na 22.98977 12 Magnesium Mg 24.3051 12* 13* 14 15 16* 17* 18* 19* 13* 14* 15* 16 17 18 19* 20* 21* 17* 18* 19 20* 21* 22* 23* 18* 19* 20 21 22 23* 24* 25* 21* 22* 23 24* 25* 26* 23* 24 25 26 27* 28* 29* 12.018613 13.005738 14.003074 15.000108 16.006100 17.008450 18.014082 19.017038 13.024813 14.008595 15.003065 15.994915 16.999132 17.999160 19.003577 20.004076 21.008595 17.002094 18.000937 18.998404 19.999982 20.999950 22.003036 23.003564 18.005710 19.001880 19.992435 20.993841 21.991383 22.994465 23.993999 24.997789 20.997650 21.994434 22.989767 23.990961 24.989951 25.992588 22.994124 23.985042 24.985838 25.982594 26.984341 27.983876 28.375346 99.63 0.37 99.71 0.039 0.20 100 90.48 0.27 9.25 100 78.99 10.00 11.01 0.0110 s 9.96 m 7.13 s 4.17 s 0.62 s 0.24 s 8.6 ms 70.6 s 122 s 26.9 s 13.6 s 3.4 s 64.5 s 109.8 m 11.0 s 4.2 s 4.2 s 2.2 s 1.67 s 17.2 s 37.2 s 3.38 m 0.60 s 22.5 s 2.61 y 14.96 h 59.1 s 1.07 s 11.3 s 9.46 m 20.9 h 1.30 s Element Symbol Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Half-life and decay mode (if unstable) Appendix A AP-3 Z 13 Aluminum Al 26.98154 14 Silicon Si 28.086 15 Phosphorus P 30.97376 16 Sulfur S 32.066 17 Chlorine Cl 35.453 18 Argon Ar 39", ".948 19 Potassium K 39.0983 25* 26* 27 28* 29* 30* 27* 28 29 30 31* 32* 33* 30* 31 32* 33* 34* 31* 32 33 34 35* 36 34* 35 36* 37 38* 36 37* 38 39* 40 42* 39 40* 41 42* 43* 24.990429 25.986892 26.981538 27.981910 28.980445 29.982965 26.986704 27.976927 28.976495 28.973770 30.975362 31.974148 32.977928 29.978307 30.973762 31.973762 32.971725 33.973636 30.979554 31.972071 32.971459 33.967867 34.969033 35.967081 33.973763 34.968853 35.968307 36.965903 37.968010 35.967547 36.966776 37.962732 38.964314 39.962384 41.963049 38.963708 39.964000 40.961827 41.962404 42.960716 100 92.23 4.67 3.10 100 95.02 0.75 4.21 0.02 75.77 24.23 0.337 0.063 99.600 93.2581 0.0117 6.7302 7.18 s 7.4 105 y 2.24 m 6.56 m 3.60 s 4.16 s 2.62 h 172 y 6.13 s 2.50 m 14.26 d 25.3 d 12.43 s 2.57 s 87.5 d 32.2 m 3.0 105 y 37.3 m 35.04 d ec 269 y 33 y 1.28 109 y, ec, 12.4 h 22.3 h (Continued) AP-4 Appendix A Element Symbol Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Z 20 Calcium Ca 40.078 40 41* 42 43 44 46 48 41* 43* 45 46* 44* 46 47 48 49 50 48* 50* 51 48* 50 52 53 54 53* 54* 55 56* 54 55* 56 57 58 60* 57* 58* 59 60* 61* 96.941 0.647 0.135 2.086 0.004 0.187 100 8.0 7.3 73.8 5.5 5.4 0.25 99.75 4.345 83.79 9.50 2.365 100 5.9 91.72 2.1 0.28 100 39.962591 40.962279 41.958618 42.958767 43.955481 45.953687 47.952534 40.969250 42.961151 44.955911 45.955170 43.959691 45.952630 46.951765 47.947947 48.947871 49.944792 47.952255 49.947161 50.943962 47.954033 49.946047 51.940511 52.940652 53.938883 52.941292 53.940361 54.938048 55.938908 53.939613 54.938297 55.934940 56.935396 57.933278 59.934078 56.936294 57.935755 58.933198 59.933820 60.932478 Half-life and decay mode (if unstable) 1.0 105 y ec 0.596 s 3.89 h 83.8 d 49 y ec 15.97 d 1.5 1017 y 21.6 h ec 3.74 106 y 312.1 d 2.58 h ec ec 2.7 y ec 1.5 106 y 271.8 d 70.9 h ec ec, 5.27 y 1.65 h 21 Scandium Sc 44.9559 22 Titanium Ti 47.88 23 Vanadium V 50.9415 24 Chromium Cr 51.996 25 Manganese Mn 54.93805 26 Iron Fe 55.847 27 Cobalt Co 58.93320 Element Symbol Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Half-life and decay mode (if unstable) Appendix A AP-5 Z 28 Nickel Ni 58.693 29 Copper Cu 63.546 30 Zinc Zn 65.39 31 Gallium Ga 69.723 32 Germanium Ge 72.61 33 Arsenic As 74.9216 34 Selenium Se 78.96 58 59* 60 61 62 63* 64 63 64* 65 66* 64 66 67 68 70 69 70* 71", " 72* 69* 70 72 73 74 76 77* 73* 74* 75 76* 77* 74 76 77 78 79* 80 82* 57.935346 58.934350 59.930789 60.931058 61.928346 62.929670 63.927967 62.929599 63.929765 64.927791 65.928871 63.929144 65.926035 66.927129 67.924845 69.925323 68.925580 69.926027 70.924703 71.926367 68.927969 69.924250 71.922079 72.923462 73.921177 75.921402 76.923547 72.923827 73.923928 74.921594 75.922393 76.920645 73.922474 75.919212 76.919913 77.917307 78.918497 79.916519 81.916697 68.077 26.223 1.140 3.634 0.926 69.17 30.83 48.6 27.9 4.1 18.8 0.6 60.108 39.892 21.23 27.66 7.73 35.94 7.44 100 0.89 9.36 7.63 23.78 49.61 8.73 7.5 104 y ec, 100 y 12.7 h ec 5.1 m 21.1 m 14.1 h 39.1 h ec, 11.3 h 80.3 d 17.8 d ec ec, 1.1 d 38.8 h 6.5 104 y 1.4 1020 y 2 (Continued) AP-6 Appendix A Element Symbol Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Z 35 Bromine Br 79.904 36 Krypton Kr 83.80 37 Rubidium Rb 85.468 38 Strontium Sr 87.62 39 Yttrium Y 88.9058 40 Zirconium Zr 91.224 41 Niobium Nb 92.9064 42 Molybdenum Mo 95.94 Half-life and decay mode (if unstable) 17.7 m 35.3 h 2.11 105 y ec 10.76 y 18.6 d 4.75 1010 y 17.8 m 29.1 y 106.6 d ec, 2.67 d 1.5 106 y 6.8 102 y 3.5 107 y ec ec 2 104 y 3.5 103 y ec 79 80* 81 82* 78 80 81* 82 83 84 85* 86 85 86* 87* 88* 84 86 87 88 90* 88* 89 90* 90 91 92 93* 94 96 91* 92* 93 94* 92 93* 94 95 96 97 98 100 78.918336 79.918528 80.916287 81.916802 77.920400 79.916377 80.916589 81.913481 82.914136 83.911508 84.912531 85.910615 84.911793 85.911171 86.909186 87.911325 83.913428 85.909266 86.908883 87.905618 89.907737 87.909507 88.905847 89.914811 89.904702 90.905643 91.905038 92.906473 93.906314 95.908274 90.906988 91.907191 92.906376 93.907280 91.906807 92.906811 93.905085 94.905841 95.904678 96.906020 97.905407 99.907476 50.69 49.31 0.35 2.25 11.6 11.5 57.0 17.3 72.17 27.83 0.56 9.86 7.00 82.58 100 51.45 11.22 17.15 17.38 2.80 100 14.84 9.25 15.92 16.68 9.55 24.13 9.63 Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Z 43 Element Symbol Technetium Tc 44 Ruthenium Ru 101.07 45 Rhodium Rh 102.9055 46 Palladium Pd 106.42 47 Silver Ag 107.868 48 Cadmium Cd 112.41 49 Indium In 114.82 97* 98* 99* 96 98 99 100 101 102 104 102* 103 104* 102 104 105 106 107* 108 110 107 108* 109 110* 106 108 109* 110 111 112 113* 114 116 113 114* 115* 116* 96.906363 97.907215 98.906254 95.907597 97.", "905287 98.905939 99.904219 100.905558 101.904348 103.905428 101.906794 102.905502 103.906654 101.905616 103.904033 104.905082 105.903481 106.905126 107.903893 109.905158 106.905091 107.905953 108.904754 109.906110 105.906457 107.904183 108.904984 109.903004 110.904182 111.902760 112.904401 113.903359 115.904755 112.904060 113.904916 114.903876 115.905258 5.54 1.86 12.7 12.6 17.1 31.6 18.6 100 1.02 11.14 22.33 27.33 26.46 11.72 51.84 48.16 1.25 0.89 12.49 12.80 24.13 12.22 28.73 7.49 4.3 95.7 Appendix A AP-7 Half-life and decay mode (if unstable) 2.6 106 y 4.2 106 y 2.1 105 y ec 207 d ec 42 s 6.5 106 y 2.39 m ec,, 24.6 s 462 d ec 9.3 1015 y 1.2 m 4.4 1014 y 54.4 m (Continued) AP-8 Appendix A Element Symbol Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Half-life and decay mode (if unstable) Z 50 Tin Sn 118.71 112 114 115 116 117 118 119 120 121* 122 124 121 123 125* 120 122 123* 124 125 126 128* 130* 126* 127 128* 129* 124 126 128 129 130 131 132 134 136 133 134* 135* 137* 0.97 0.65 0.36 14.53 7.68 24.22 8.58 32.59 4.63 5.79 57.36 42.64 0.095 2.59 0.905 4.79 7.12 18.93 31.70 33.87 100 0.10 0.09 1.91 26.4 4.1 21.2 26.9 10.4 8.9 100 111.904822 113.902780 114.903345 115.901743 116.902953 117.901605 118.903308 119.902197 120.904237 121.903439 123.905274 120.903820 122.904215 124.905251 119.904040 121.903052 122.904271 123.902817 124.904429 125.903309 127.904463 129.906228 125.905619 126.904474 127.905812 128.904984 123.905894 125.904268 127.903531 128.904779 129.903509 130.905069 131.904141 133.905394 135.907215 132.905436 133.906703 134.905891 136.907078 55 y 2.7 y 1.3 1013 y ec 8 1024 y 1.2 1021 y 2 2 13 d ec,, 25 m 1.6 107 y, ec, 2.1 y 2 106 y 30 y 51 Antimony Sb 121.76 52 Tellurium Te 127.60 53 Iodine I 126.9045 54 Xenon Xe 131.29 55 Cesium Cs 132.9054 Element Symbol Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Z 56 Barium Ba 137.33 Appendix A AP-9 Half-life and decay mode (if unstable) 10.5 y ec 6 104 y 1.05 1011 y ec ec, 3.39 m ec, 25.0 m 2.3 1015 y 265 d 17.7 y 5.5 y 2.623 y ec ec ec 1.0 108 y 1.06 1011 y 7 1015 y 90 y (Continued) 130 132 133* 134 135 136 137 138 137* 138* 139 136 138 140 142 140* 141 142* 142 143 144* 145 146 148 150 143* 145* 146* 147* 144 146* 147* 148* 149 150 151* 152 154 129.906289 131.905048 132.905990 133.904492 134.905671 135.904559 136.905816 137.905236 136.906462 137.907105 138.906346 135.907139 137.905986 139.905434 141.909241 139.909071 140.907647 141.910040 141.907718 142.909809 143.", "910082 144.912568 145.913113 147.916888 149.920887 142.910928 144.912745 145.914698 146.915134 143.911996 145.913043 146.914894 147.914819 148.917180 149.917273 150.919928 151.919728 153.922206 0.106 0.101 2.42 6.593 7.85 11.23 71.70 0.0902 99.9098 0.19 0.25 88.43 11.13 100 27.13 12.18 23.80 8.30 17.19 5.76 5.64 3.1 15.0 11.3 13.8 7.4 26.7 22.7 57 Lanthanum La 138.905 58 Cerium Ce 140.12 59 Praseodymium Pr 140.9076 60 Neodymium Nd 144.24 61 Promethium Pm 62 Samarium Sm 150.36 AP-10 Appendix A Element Symbol Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Z 63 Europium Eu 151.96 64 Gadolinium Gd 157.25 65 Terbium Tb 158.9253 66 Dysprosium Dy 162.50 Half-life and decay mode (if unstable) 13.5 y ec, 8.59 y 4.7 y 75 y 1.8 106 y 1.1 1014 y 180 y ec,, 72.3 d 151 152* 153 154* 155* 148* 150* 152* 154 155 156 157 158 160 158* 159 160* 156 158 160 161 162 163 164 150.919846 151.921740 152.921226 153.922975 154.922888 147.918112 149.918657 151.919787 153.920862 154.922618 155.922119 156.923957 157.924099 159.927050 157.925411 158.925345 159.927551 155.924277 157.924403 159.925193 160.926930 161.926796 162.928729 163.929172 47.8 52.2 0.20 2.18 14.80 20.47 15.65 24.84 21.86 100 0.06 0.10 2.34 18.9 25.5 24.9 28.2 67 Holmium Ho 164.9303 165 166* 164.930316 165.932282 100 1.2 103 y 68 Erbium Er 167.26 162 164 166 167 168 170 161.928775 163.929198 165.930292 166.932047 167.932369 169.935462 0.14 1.61 33.6 22.95 27.8 14.9 69 Thulium Tm 168.9342 169 171* 168.934213 170.936428 100 1.92 y 70 Ytterbium Yb 173.04 168 170 171 172 173 174 176 167.933897 169.934761 170.936324 171.936380 172.938209 173.938861 175.942564 0.13 3.05 14.3 21.9 16.12 31.8 12.7 Element Symbol Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Z 71 Lutetium Lu 174.967 Appendix A AP-11 Half-life and decay mode (if unstable) 1.37 y ec 3.8 1010 y 2.0 1015 y 4.4 1010 y 2.0 1015 y 6.0 y 6.5 1011 y 2.70 d 3.14 d (Continued) 173* 175 176* 174* 176 177 178 179 180 180 181 180 182 183 184 186 185 187* 184 186* 187 188 189 190 192 194* 191 193 190* 192 194 195 196 198 197 198* 199* 172.938930 174.940772 175.942679 173.940042 175.941404 176.943218 177.943697 178.945813 179.946547 179.947542 180.947993 179.946702 181.948202 182.950221 183.950929 185.954358 184.952951 186.955746 183.952486 185.953834 186.955744 187.955744 188.958139 189.958439 191.961468 193.965172 190.960585 192.962916 189.959926 191.961027 193.962655 194.964765 195.964926 197.967867 196.9665", "43 197.968217 198.968740 97.41 2.59 0.162 5.206 18.606 27.297 13.629 35.100 0.012 99.988 0.12 26.3 14.28 30.7 28.6 37.40 62.60 0.02 1.58 1.6 13.3 16.1 26.4 41.0 37.3 62.7 0.01 0.79 32.9 33.8 25.3 7.2 100 72 Hafnium Hf 178.49 73 Tantalum 74 Tungsten (Wolfram) Ta W 180.9479 183.85 75 Rhenium Re 186.207 76 Osmium Os 190.2 77 Iridium 78 Platinum Ir Pt 192.2 195.08 79 Gold Au 196.9665 AP-12 Appendix A Element Symbol Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Half-life and decay mode (if unstable) Z 80 Mercury Hg 200.59 0.15 9.97 16.87 23.10 13.10 29.86 6.87 29.524 70.476 1.4 24.1 22.1 52.4 100 196 198 199 200 201 202 204 203 204* 205 206* 207* 208* 210* 202* 204 205* 206 207 208 210* 211* 212* 214* 207* 208* 209 210* 211* 212* 214* 215* 209* 210* 211* 212* 214* 215* 216* 218* 215* 218* 219* 195.965806 197.966743 198.968253 199.968299 200.970276 201.970617 203.973466 202.972320 203.973839 204.974400 205.976084 206.977403 207.981992 209.990057 201.972134 203.973020 204.974457 205.974440 206.975871 207.976627 209.984163 210.988734 211.991872 213.999798 206.978444 207.979717 208.980374 209.984096 210.987254 211.991259 213.998692 215.001836 208.982405 209.982848 210.986627 211.988842 213.995177 214.999418 216.001889 218.008965 214.998638 218.008685 219.011297 3.78 y 4.2 m 4.77 m 3.053 m 1.30 m 5 104 y 1.5 107 y 22.3 y 36.1 m 10.64 h 26.8 m ec ec 32.2 y 3.7 105 y ec, ec 5.01 d 2.14 m 60.6 m 19.9 m 7.4 m,, 102 y 138.38 d 0.52 s 0.30 s 164 s 0.0018 s 0.145 s 3.10 m 100 s 1.6 s 0.9 m 81 Thallium Tl 204.383 (Ra E) (Ac C) (Th C) (Ra C) 82 Lead Pb 207.2 (Ra D) (Ac B) (Th B) (Ra B) 83 Bismuth Bi 208.9803 84 Polonium (Ra E) (Th C) (Ra C) Po (Ra F) (Ac C) (Th C) (Ra C) (Ac A) (Th A) (Ra A) 85 Astatine At Element Symbol Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance Z 86 Radon 87 Francium 88 Radium 89 Actinium 90 Thorium 91 Protactinium Rn (An) (Tn) (Rn) Fr (Ac K) Ra (Ac X) (Th X) (Ra) (MsTh1) Ac (Ms Th2) Th (Rd Ac) (Rd Th) (Io) (UY) (Th) (UX1) Pa (UZ) 232.0381 92 Uranium U 238.0289 (UII) (Ac U) (UI) 93 Neptunium Np Appendix A AP-13 Half-life and decay mode (if unstable) 3.96 s 55.6 s 3.823 d 4.18 m 14.2 m 22 m 29 s 11.43 d 3.66 d 14.9 d 1600 y 5.75 y 10 d 21.77 y 6.15 h 1.04 h 100 0.0055 0.720 99.2745 18.72 d 1.913 y 7300 y 75,000 y 25.52 h 1.40 1010 y 24.1 d, sf 32,760 y 6.7 h 4.2 d", " 69 y 1.59 105 y 2.45 105 y 7.04 108 y 2.34 107 y 4.47 109 y 23.5 m 396 d 1.54 105 y 2.14 106 y ec (Continued) 219* 220* 222* 221* 222* 223* 221* 223* 224* 225* 226* 228* 225* 227* 228* 229* 227* 228* 229* 230* 231* 232* 234* 231* 234* 231* 232* 233* 234* 235* 236* 238* 239* 235* 236* 237* 219.009477 220.011369 222.017571 221.01425 222.017585 223.019733 221.01391 223.018499 224.020187 226.025402 228.031064 227.027749 228.031015 227.027701 228.028716 229.031757 230.033127 231.036299 232.038051 234.043593 231.035880 234.043300 231.036264 232.037131 233.039630 234.040946 235.043924 236.045562 238.050784 239.054290 235.044057 236.046559 237.048168 AP-14 Appendix A Z 94 Element Symbol Plutonium Pu 95 Americium Am 96 Curium Cm 97 Berkelium Bk 98 Californium Cm 99 Einsteinium Es 100 Fermium Fm 101 Mendelevium Md 102 Nobelium No 103 Lawrencium Lr 104 Rutherfordium Rf 105 Dubnium Db 106 Seaborgium 107 Bohrium 108 Hassium Sg Bh Hs 109 Meitnerium Mt Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance 236* 238* 239* 240* 241* 242* 244* 240* 241* 247* 248* 247* 249* 250* 251* 252* 253* 253* 254* 256* 258* 257* 259* 259* 260* 260* 261* 261* 262* 236.046033 238.049555 239.052157 240.053808 241.056846 242.058737 244.064200 240.055285 241.056824 247.070347 248.072344 247.070300 249.074979 250.076400 251.079580 252.082974 253.084817 253.085173 254.086849 256.093988 258.098594 257.096855 259.100932 259.102888 260.105346 260.160302 261.108588 261.111830 262.113763 263* 263.118310 262* 262.123081 265* 267* 266* 268* 265.129984 267.131770 266.137789 268.138820 Half-life and decay mode (if unstable) 2.87 y 87.7 y 24,120 y 6560 y 14.4 y 3.7 105 y 8.1 107 y 2.12 d 432 y 1.56 107 y 3.4 105 y, sf, sf, sf, sf, sf, sf ec, sf, sf 1380 y 327 d 13.1 y 898 y 1.29 y 2.02 d, sf, sf 3.00 d 3.24 h ec, sf 75.6 m 55 d ec, 25 s 58 m 6.14 s 3.0 m, sf, sf, sf 24 ms 65 s sf, sf 1.8 s 35 s 0.78 s, sf 0.10 s, sf 1.8 ms 60 ms 3.4 ms 70 ms, sf Z Element Symbol 110 Darmstadtium Ds Chemical atomic weight Mass number (* indicates radioactive) Atomic mass Percent abundance 269* 271* 273* 269.145140 271.146080 272.153480 111 Roentgenium Rg 272* 272.153480 112 Ununbium Unb 113 Ununtrium Unt 114 Ununquadium Unq 115 Ununpentium Unp 116 Ununhexium Unh 117 118 Ununoctium Uno 277* 284* 289* 288* 292* 294*?????? Appendix A AP-15 Half-life and decay mode (if unstable) 0.17 ms 1.1 ms 8.6 ms 1.5 ms 0.2 ms????? APPENDIXB1 Probability Integrals When calculating various average values using the Maxwell-Boltzmann distribution, integrals of the following type occur: 0 In xnex2 dx where n is an integer. These can be obtained from and by differentiation. Consider In to be a function", " of and take the derivative with respect to I0 I1 : dIn d x2xnex2 0 dx In2 B1-1 I0 is known, all the In for odd n can be obtained from Equation B1-1. for even n can be obtained, and if is known, all can easily be evaluated. I1 Thus, if In the Using the substitution then u x2, xex2 du 2x dx 1 dx 1 2 0 0 I1 and eu du 1 2 1 I1 Then I3 and I5 are 1b da 1 2 d 1 2 I3 2 and I5 dI3 d 3 The evaluation of evaluate I2 0 : I0 is more difficult, but it can be done using a trick. We 0 I2 0 ex2 dx 0 ey2 dy 0 0 e(x2y2) dx dy B1-2 where we have used y as the dummy variable of integration in the second integral. If we now consider this to be an integration over the xy plane, we can change to polar coordinates becomes r dr d# r 0 to and the integration over positive x and y becomes an integration from The element of area r2 x2 y2 tan # y> x. Then we have and from r # 0 dx dy and to # >2. >2 er2 0 0 r dr d# 2 I1 4 1 I2 0 and AP-16 I0 1 2 2 1>2 Appendix B1 AP-17 We then obtain I2, I4, \u00c1 by differentiation. For example, I2 dI0 d 1 4 2 3>2 Table B1-1 lists the values of the integral from 0 to 5. In calculated as above for values of n Table B1-1 Values of the integral for n 0 to n 5 0 In x nex 2 dx n 0 1 2 3 4 5 If n is even If n is odd In 1 2 1 4 3 8 1>21>2 1 1 2 1>23>2 2 1 2 1>25>2 3 xnex2 dx 2In xnex2 dx 0 APPENDIXB2 Binomial and Exponential Series Binomial Series (1 x)m 1 mx x2 m(m 1) 2! m(m 1)(m 2) 3! m(m 1)(m 2) \u00c1 (m n 2) (n 1)! xn1 Rn x3 \u00c1 where Rn m(m 1)(m 2) \u00c1 (m n 1) n! xn(1 ax)mn for all cases where 0 a 1. ` Rn m(m 1)(m 2) \u00c1 (m n 1) n! xn ` if x 0 ` m(m 1)(m 2) \u00c1 (m n 1) n! Rn xn (1 x)nm ` for x 0, n m \u0192xn\u0192(1 x)m if 1 m 0 Rn If m is a negative integer or a positive or negative fraction, the binomial expanExcept when m is a positive integer, a binomial such \u0192x\u0192 1. must be written in one of the following forms before expanding it: sion is valid only when as (a b) m ama1 m b b a if a b bma1 m b a b if b a Exponential Series ex 1 x x2 2! x3 3! \u00c1 xn1 (n 1)! xn n! \u00c1 ax 1 x log a (x log a)2 2! \u00c1 (x log a)n1 (n 1)! (x log a)n n! \u00c1 AP-18 APPENDIXB3 Diagrams of Crystal Unit Cells \u2192 c Crystalline solids are classified according to their symmetries into 7 crystal systems and 14 lattices. A lattice is defined as an infinite array of points each of which has surroundings identical to those of all other points. In three dimensions this definition is expressed by three translation vectors a, b, and c, such that the array of atoms in the crystal when viewed from point r looks the same when viewed from any other point r is reached by translations of integer multiples of the, where that is, r a i, r r m1a m2b m3c mi where ify the three axes of the crystal\u2019s unit cell. The volume of the unit cell is and no cell of smaller volume can serve as the unit to assemble the crystal. are integers. The translation vectors are usually (but not always) used to speca # (b c), Figure B3-1 illustrates the orientations of the translation vectors and the standard designations of the angles between them. Figure B3-2 illustrates diagrams of the 14 lattices. Cubic Tetragonal \u03b2 \u03b1 \u03b3 \u2192 b \u2192 a Figure B3-1 The directions of", " the translation vectors are often used to define the directions of the crystal axes, the angles between each axis pair defining the shape of the cell. primitive body-centered face-centered primitive body-centered Orthorhombic Trigonal primitive body-centered face-centered base-centered rhombohedral Monoclinic Triclinic Hexagonal primitive base-centered primitive primitive Figure B3-2 The 14 crystal lattices. AP-19 APPENDIXC Electron Configurations Electron configurations of the atoms in their ground states. For a few of the rare earth elements the heavy elements the configurations are not firmly established. (Z 89), (Z 57 to 71) and Z Element Ionization energy (eVhydrogen) He (helium) Li (lithium) Be (beryllium) B (boron) C (carbon) N (nitrogen) O (oxygen) F (flourine) 10 Ne (neon) 11 12 13 14 15 16 17 18 Na (sodium) Mg (magnesium) Al (aluminum) Si (silicon) P (phosphorus) S (sulfur) Cl (chlorine) Ar (argon) AP-20 13.6 24.5 5.4 9.3 8.3 11.3 14.5 13.6 17.4 21.6 5.1 7.6 6.0 8.1 10.5 10.4 13.0 15. 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 Element K (potassium) Ca (calcium) Sc (scandium) Ti (titanium) V (vanadium) Cr (chromium) Mn (manganese) Fe (iron) Co (cobalt) Ni (nickel) Cu (copper) Zn (zinc) Ga (gallium) Ge (germanium) As (arsenic) Se (selenium) Br (bromine) Kr (krypton) Rb (rubidium) Sr (strontium) Y (yttrium) Zr (zirconium) Nb (niobium) Mo (molybdenum) Tc (technetium) Ionization energy (eV) K n : 1 l : s 4.3 6.1 6.5 6.8 6.7 6.8 7.4 7.9 7.9 7.6 7.7 9.4 6.0 7.9 9.8 9.8 11.8 14.0 4.2 5.7 6.4 6.8 6.9 7.1 7. Appendix C AP-21 10 2 6 10 2 6 10 2 6 10 2 6 10 2 6 10 2 6 10 2 6 10 10 2 6.. 2 6 10 2 6.. 2 6 10 2 6 1. 2 6 10 2 6 2. 2 6 10 2 6 4. 2 6 10 2 6 5. 2 6 10 Continued) AP-22 Appendix C Z 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 Element Ru (ruthenium) Rh (rhodium) Pd (palladium) Ag (silver) Cd (cadmium) In (indium) Sn (tin) Sb (antimony) Te (tellurium) I (iodine) Xe (xenon) Cs (cesium) Ba (barium) La (lanthanum) Ce (cerium) Pr (praseodymium) Nd (neodymium) Pm (promethium) Sm (samarium) Eu (europium) Gd (gadolinium) Tb (terbium) Dy (dysprosium) Ho (holmium) Ionization energy (eV) K n : 1 l : s 7.4 7.5 8.3 7.6 9.0 5.8 7.3 8.6 9.0 10.5 12.1 3.9 5.2 5.6 5.6 5.5 5.5 5.5 5.6 5.7 6.2 6.0 6.8 6. 10 2 6 7. 2 6 10 2 6 8. 2 6 10 2 6 10. 2 6 10 2 6 10. 2 6 10 2 6 10. 2 6 10 2 6 10. 2 6 10 2 6 10. 2 6 10 2 6 10. 2 6 10 2 6 10. 2 6 10 2 6 10. 2 6 10 2 6 10. 10 2 6 10. 2 6 10 2 6 10. 10 2 6 10. 2 6 1. 2 6 10 2 6 10 1 2 6 1. 2 6 10 2 6 10 3 2 6.. 2 6 10 2 6", " 10 4 2 6.. 2 6 10 2 6 10 5 2 6.. 2 6 10 2 6 10 6 2 6.. 2 6 10 2 6 10 7 2 6.. 2 6 10 2 6 10 7 2 6 1. 2 6 10 2 6 10 9 2 6.. 2 6 10 2 6 10 10 2 6.. 2 6 10 2 6 10 11 Ionization energy (eV) K n : 1 l : s Z 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 Element Er (erbium) Tm (thulium) Yb (ytterbium) Lu (lutetium) Hf (hafnium) Ta (tantalum) W (tungsten) Re (rhenium) Os (osmium) Ir (iridium) Pt (platinum) Au (gold) 6.1 5.8 6.2 5.1 7.0 7.9 8.0 7.9 8.5 9.0 9.0 9.2 Hg (mercury) 10.4 Tl (thallium) Pb (lead) Bi (bismuth) Po (polonium) At (astatine) 6.1 7.4 7.3 8.4 9.5 Rn (radon) 10.7 Fr (francium) Ra (radium) Ac (actinium) Th (thorium) Pa (protactinium) 4.0 5.3 6.9 7.0 Appendix C AP-23 10 2 6 10 12 2 6.. 2 6 10 2 6 10 13 2 6.. 2 6 10 2 6 10 14 2 6.. 2 6 10 2 6 10 14 2 6 1. 2 6 10 2 6 10 14 2 6 2. 2 6 10 2 6 10 14 2 6 3. 2 6 10 2 6 10 14 2 6 4. 2 6 10 2 6 10 14 2 6 5. 2 6 10 2 6 10 14 2 6 6. 2 6 10 2 6 10 14 2 6 7. 2 6 10 2 6 10 14 2 6 9. 2 6 10 2 6 10 14 2 6 10. 2 6 10 2 6 10 14 2 6 10. 2 6 10 2 6 10 14 2 6 10. 2 6 10 2 6 10 14 2 6 10. 2 6 10 2 6 10 14 2 6 10. 2 6 10 2 6 10 14 2 6 10. 2 6 10 2 6 10 14 2 6 10. 2 6 10 2 6 10 14 2 6 10. 10 2 6 10 14 2 6 10. 2 6. 2 6 10 2 6 10 14 2 6 10. 2 6. 2 6 10 2 6 10 14 2 6 10. 2 6 1 2 6 10 2 6 10 14 2 6 10. 2 6 2 2 6 10 2 6 10 14 2 6 10 Continued AP-24 Appendix C Z 92 93 94 95 96 97 98 99 Element U (uranium) Np (neptunium) Pu (plutonium) Am (americium) Cm (curium) Bk (berkelium) Cf (californium) Es (einsteinium) 100 Fm (fermium) 101 Md (mendelevium) 102 No (nobelium) 103 Lw (lawrencium) 104 Rf (rutherfordium) 105 Du (dubnium) 106 Sg (seaborgium) 107 Bh (bohrium) 108 Hs (hassium) 109 Mt (meitnerium) 110 Ds (darmstadtium) 111 Rg (roentgenium) 112 Unb (ununbium) 113 Unt (ununtrium) 114 Unq (ununquadium) 115 Unp (ununpentium) Ionization energy (eV) K n : 1 l : s 6.1 5.8 6. 10 2 6 10 14 2 6 10 3 2 6 1 2 6 10 2 6 10 14 2 6 10 4 2 6 1 2 6 10 2 6 10 14 2 6 10 6 2 6 10 2 6 10 14 2 6 10 7 2 6. 2 6. 2 6 10 2 6 10 14 2 6 10 7 2 6 1 2 6 10 2 6 10 14 2 6 10 8 2 6 1 2 6 10 2 6 10 14 2 6 10 10 2 6. 2 6 10 2 6 10 14 2 6 10 11 2 6. 2 6 10 2 6 10 14 2 6 10 12 2 6. 2 6 10 2 6 10 14 2 6 10 13 2 6. 2 6 10 2 6 10 14 2 6 10 14 2 6. 2 6 10 2 6 10 14 2 6 10 14 2 6 1 2 6 10 2 6 10 14 2 6 10 14 2 6 2 2 6 10 2 6 10 14 2 6 10 14 2 6", " 3 2 6 10 2 6 10 14 2 6 10 14 2 6 4 2 6 10 2 6 10 14 2 6 10 14 2 6 5 2 6 10 2 6 10 14 2 6 10 14 2 6 6 2 6 10 2 6 10 14 2 6 10 14 2 6 7 2 6 10 2 6 10 14 2 6 10 14 2 6 9 2 6 10 2 6 10 14 2 6 10 14 2 6 10 2 6 10 2 6 10 14 2 6 10 14 2 6 10 2 6 10 2 6 10 14 2 6 10 14 2 6 10 2 6 10 2 6 10 14 2 6 10 14 2 6 10 2 6 10 2 6 10 14 2 6 10 14 2 6 10 Appendix C AP-25 Z Element 116 Unh (ununhexium) 117 118 Uno (ununoctium) Ionization energy (eV 10 2 6 10 14 2 6 10 14 2 6 10 2 4 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 10 2 6 APPENDIXD Fundamental Physical Constants This set of fundamental physical constants consists of selected values recommended by CODATA, the Committee on Data for Science and Technology of the International Council of Scientific Unions, resulting from the most recent (2002) compilation and computations. The digits in parentheses are the one-standard-deviation uncertainties in the last digits. (Reference: P. J. Mohr and B. N. Taylor, http://www.physicstoday.org/guide/fundconst.pdf.) Quantity Symbol Value Units Universal constants Speed of light in vacuum (exact) Permeability of vacuum (magnetic constant) (exact) Permittivity of vacuum (electric constant) (exact) Newtonian constant of gravitation Planck constant in electron volts, h>2 in electron volts, h>{e} U>{e} (Uc>G)1>2 Planck mass, (Uc5>G)1>2>k Planck temperature, U>mpc (UG>c3)1>2 Planck length, >c) (UG>c5)1>2 (lP Planck time, Electromagnetic constants Elementary charge h>2e Magnetic flux quantum, 2e>h Josephson constant von Klitzing constant, Bohr magneton, eV>T eU>2me in h>e2 0c>2 Nuclear magneton, in eV>T eU>2mp Atomic constants Fine-structure constant, e2>4 0 inverse fine-structure constant Uc mec2>2h Rydberg constant, in hertz, in joules, in eV, Bohr radius Rc Rhc Rhc>{e} Electron Mass in electron volts, mec2>{e} AP-26 0 c 0 G h U mp Tp lp tp e e>h \u03a6 0 KJ RK B N 1 R a0 me 0c2 8.854187817 299,792,458 4 10 7 12.566370614 107 1> 6.6742 (10) 6.6260693 (11) 4.13566743 (35) 1.05457168 (18) 6.58211915 (56) 2.17654 (16) 1.41679 (11) 1.61624 (12) 5.39121 (40) m # s1 N # A2 1012 F # m1 1011 m3 # kg1 # s2 1034 J # s 1015 eV # s 1034 J # s 1016 eV # s 108 kg 1032 K 1035 m 1044 s 1.60217653 (14) 2.41798940 (21) 2.06783372 (18) 483597.879 (41) 25812.807449 (86) 927.400949 (80) 5.788381804 (39) 5.05078343 (43) 3.152451259 (24) 7.297352568 (24) 137.03599911 (46) 10,973,731.568525 (73) 3.289841960360 (22) 2.17987209 (37) 13.6056923 (12) 0.5291772108 (18) 9.1093826 (16) 5.4857990945 (24) 0.510998918 (44) 1019 C 1014 A # J1 1015 Wb 109 Hz # V1 \u00c6 1024 J # T1 105 eV # T1 1027 J # T1 108 eV # T1 103 m1 1015 Hz 1018 J eV 1010 m 1031 kg 104 u MeV Quantity Symbol Value Units Appendix D AP-27 Electron-muon mass ratio Electron-", "tau mass ratio Electron-proton mass ratio Electron-deuteron mass ratio Electron- -particle mass ratio Specific charge Molar mass Compton wavelength, >2 a0 Classical radius, Thomson cross section, Magnetic moment h>mec 2>4R 2a0 C (8>3)re 2 in Bohr magnetons in nuclear magnetons Magnetic moment anomaly, 2(1 ae) g factor, Electron-muon magnetic moment ratio Electron-proton magnetic moment ratio \u0192 > B \u0192 e 1 Muon Mass in electron volts, mc2>{e} Muon-electron mass ratio Muon-tau mass ratio Molar mass Magnetic moment in Bohr magnetons in nuclear magnetons Magnetic moment anomaly, 2(1 a) g factor, Muon-proton magnetic moment ratio \u0192 \u0192 >(eU>2 m) 1 Tau Mass in electron volts Proton Mass in electron volts Proton-electron mass ratio Proton-muon mass ratio Specific charge Molar mass Compton wavelength, h>mpc C,p >2 >m me >m me >mp me >md me >m me e>me M(e) C \u2013C re e e e e ae ge me >m m m M() a g > B > N > p 4.83633167 (13) 2.87564 (47) 5.4461702173 (25) 2.7244371095 (13) 1.37093355575 (61) 1.75882012 (15) 5.4857990945 (24) 2.426310238 (16) 386.1592678 (26) 2.817940325 (28) 0.665245873 (13) 928.476412 (80) 1.0011596521859 (38) 1838.28197107 (85) 1.1596521859 (38) 2.0023193043718 (75) 206.7669894 (54) 658.2106862 (66) 1.88353140 (33) 0.1134289264 (30) 105.6583692 (94) 206.7682838 (54) 5.94592 (97) 1.134289264 (34) 4.49044799 (40) 4.84197085 (15) 8.89059770 (27) 1.16591981 (62) 2.0023318396 (12) 3.183345118 (89) m mp >me >m mp mp e>mp M(p) C,p \u2013C 3.16777 (52) 1.90768 (31) 1776.99 (29) 1.67262171 (29) 1.00727646688 (13) 938.272029 (80) 1836.15267261 (85) 8.88024333 (23) 9.57883376 (82) 1.00727646688 (13) 1.3214098555 (88) 2.103089104 (14) 103 104 104 104 104 1011 C # kg1 107 kg # mol1 1012 m 1015 m 1015 m 1028 m2 1026 J # T1 103 1028 kg u MeV 104 kg # mol1 1026 J T1 103 103 1027 kg u MeV 1027 kg u MeV 107 C # kg1 103 kg # mol1 1015 m 1016 m (Continued) AP-28 Appendix D Quantity Magnetic moment in Bohr magnetons in nuclear magnetons Diamagnetic shielding correction for protons \u0153 > 25\u00b0C), 1 p p (H2O spherical sample, Shielded proton moment spherical sample, ( H2O in Bohr magnetons in nuclear magnetons Gyromagnetic ratio 25\u00b0C) Symbol p p p > B > N H2O 2 Value 1.41060671 (12) 1.521032206 (15) 2.792847351 (28) 25.687 (15) 1.41057047 (12) 1.520993132 (16) 2.792775604 (30) 26,752.2205 (23) 42.5774813 (37) uncorrected ( H2O, spherical sample, 25\u00b0C) \u0153 p \u0153 >2 p 26,751.5333 (23) 42.5763875 (37) Neutron Mass mn in electron volts, mnc2>{e} Neutron-electron mass ratio Neutron-proton mass ratio Molar mass Compton wavelength, h>mnc >2 Magnetic moment C,n in Bohr magnetons in nuclear magnetons Neutron-electron magnetic moment ratio", " Neutron-proton magnetic moment ratio Deuteron Mass in electron volts, mdc2>{e} Deuteron-electron mass ratio Deuteron-proton mass ratio Molar mass Magnetic moment in Bohr magnetons in nuclear magnetons Deuteron-electron magnetic moment ratio Deuteron-proton magnetic moment ratio Alpha particle Mass in electron volts Physiochemical constants Avogadro constant Atomic mass constant, in electron volts, muc2>{e} m(C12)>12 1.67492728 (29) 1.00866491560 (55) 939.565360 (81) 1838.6836598 (13) 1.00137841870 (58) 1.00866491560 (55) 1.3195909067 (88) 2.100194157 (14) 0.96623645 (24) 1.04187563 (25) 1.91304273 (45) 1.04066882 (25) 0.68497934 (16) 3.34358335 (57) 2.01355321270 (35) 1875.61282 (16) 3670.4829652 (18) 1.99900750082 (41) 2.01355321270 (35) 0.433073482 (38) 0.4669754567 (50) 0.8574382329 (92) 0.4664345548 (50) 0.3070122084 (45) >me mn >mp mn M(n) C,n \u2013C, md >me md >mp md M(d.6446565 (11) 3727.37917 (32) NA, L mu 6.0221415 (10) 1.66053886 (28) 931.494043 (80) Units 1026 J # T1 103 106 1026 J # T1 103 104 sec1 # T1 MHz # T1 104 sec1 # T1 MHz # T1 1027 kg u MeV 103 kg # mol1 1015 m 1016 m 1026 J # T1 103 103 1027 kg u MeV 103 kg # mol1 1026 J # T1 103 103 1027 kg MeV 1023 mol1 1027 kg MeV Quantity Symbol Value Units Appendix D AP-29 Faraday constant Molar Planck constant Molar gas constant Boltzmann constant, in electron volts, in hertz, in wavenumbers, k>h k>hc Molar volume (ideal gas), R>NA k>{e} RT>p (at 273.15 K, 101 325 Pa) NA >Vm Loschmidt constant, Stefan-Boltzmann constant, 2hc2 First radiation constant, hc>k Second radiation constant, Wien displacement law constant, (2>60)k4>h3c2 max T c2 >4.96511423 \u00c1 Conversion factors and units (e>C)J {e}J Electron volt, mu Atomic mass unit (unified), Standard atmosphere Standard acceleration of gravity m(C12)>12 F NAh NAhc R k Vm n0 c1 c2 b eV u atm gn 96,485.3383 (83) 3.990312716 (27) 0.11962656572 (80) 8.314472 (15) 1.3806505 (24) 8.617343 (15) 2.0836644 (36) 69.50356 (12) 22.413996 (39) 2.6867773 (47) 5.670400 (40) 3.74177138 (64) 1.4387752 (25) 2.8977686 (51) 1.60217653 (14) 1.66053886 (28) 101,325 9.80665 C # mol1 1010 J # s # mol1 J # m # mol1 J # mol1 # K1 1023 J # K1 105 eV # K1 1010 Hz # K1 m1 K1 103 m3 # mol1 1025 m3 108 W # m2 # K4 1016 W # m2 102 m # K 103 m # K 1019 J 1027 kg Pa m # s2 APPENDIXE Conversion Factors Conversion factors are written as equations for simplicity; relations marked with an asterisk are exact. Length 1 km 0.6215 mi 1 mi 1.609 km 1 m 1.0936 yd 3.281 ft 39.37 in *1 in 254 cm *1 ft 12 in 30.48 cm *1 yd 3 ft 91.44 cm 1 light-year 1 c # y 9.467 1015 m *1 \u00c5 0.1 nm Area *", "1 m2 104 cm2 1 km2 0.3861 mi2 247.1 acres 1 hectare 104 m2 2.471 acres *1 in2 6.4516 cm2 1 ft2 9.29 102 m2 1 m2 10.76 ft2 *1 acre 43,560 ft2 1 mi2 640 acres 2.590 km2 Volume *1 m3 106 cm3 *1 L 1000 cm3 103 m3 1 gal 3.786 L 1 gal 4 qt 8 pt 128 oz 231 in3 1 in3 16.39 cm3 1 ft3 1728 in3 28.32 L 2.832 104 cm3 Time *1 h 60 min 3.6 ks *1 d 24 h 1440 min 86.4 ks 1 y 365.24 d 31.56 Ms Speed 1 km>h 0.2778 m>s 0.6215 m>h 1 mi>h 0.4470 m>s 1.609 km>h 1 mi>h 1.467 ft>s Angle and angular speed * rad 180\u00b0 1 rad 57.30\u00b0 1\u00b0 1.745 102 rad 1 rev>min 0.1047 rad>s 1 rad>s 9.549 rev>min Mass *1 kg 1000 g *1 metric ton 1000 kg 1 Mg AP-30 1 u 1.6606 1027 kg 1 kg 6.022 1026 u 1 slug 14.59 kg 1 kg 6.852 102 slug 1 u 931.50 MeV>c2 Density *1 g>cm3 1000 kg>m3 1 kg>L (1 g>cm3)g 62.4 lb>ft3 Force 1 N 0.2248 lb 105 dyn 1 lb 4.4482 N (1 kg)g 2.2046 lb Pressure *1 Pa 1 N>m2 *1 atm 101.325 kPa 1.01325 bars 1 atm 14.7 lb>in2 760 mmHg 29.9 inHg 33.8 ftH2O 1 lb>in2 6.895 kPa 1 torr 1 mmHg 133.32 Pa 1 bar 100 kPa Energy *1 kW # h 3.6 MJ *1 cal 4.1840 J 1 ft # lb 1.356 J 1.286 103 Btu *1 L # atm 101.325 J 1 L # atm 24.217 cal 1 Btu 778 ft # lb 252 cal 1054.35 J 1 eV 1.602 1019 J 1 u # c2 931.50 MeV *1 erg 107 J Power 1 horsepower 550 ft # lb>s 745.7 W 1 Btu>min 17.58 W 1 W 1.341 103 horsepower 0.7376 ft # lb>s 1 W 1 J>s Magnetic field *1 G 104 T *1 T 104 G Thermal conductivity 1 W>m # K 6.938 Btu # in>h # ft2 # F\u00b0 1 Btu # in>h # ft2 # F\u00b0 0.1441 W>m # K APPENDIXF Nobel Laureates in Physics Listed are the names and a brief quotation from the award citation for all Nobel laureates in physics. Included, too, are a few Nobel laureates in chemistry whose work was very closely related to physics, this latter group with a (C) following their names. (The Royal Swedish Academy of Sciences, which awards the prizes, has generally considered the discovery of new elements to be chemistry rather than physics.) Year Nobel laureate Citation for 1901 Wilhelm Konrad Roentgen 1845\u20131923 discovery of X rays 1902 Hendrik Antoon Lorentz Pieter Zeeman 1853\u20131928 1865\u20131943 their researches into the influence of magnetism upon radiation phenomena 1903 Antoine Henri Bequerel Pierre Curie Marie Sklowdowska-Curie 1852\u20131908 1859\u20131906 1867\u20131934 Henri Bequerel his discovery of spontaneous radioactivity their joint researches on the radiation phenomena discovered by 1904 Lord Rayleigh (John William Strutt) 1842\u20131919 Sir William Ramsay (C) 1851\u20131939 investigations of the densities of the most important gases and his discovery of argon his discovery of the inert gaseous elements in air and his determination of their place in the periodic system 1905 Philipp Eduard Anton von Lenard 1862\u20131947 his work on cathode rays 1906 Joseph John Thomson 1856\u20131940 his theoretical and experimental investigations on the conduction of electricity by gases 1907 Albert Abraham Michelson 1852\u20131931 his optical precision instruments and the spectroscopic and metrological investigations carried out with their aid 1908 Gabriel Lippman Ernest Rutherford (C) 1845\u20131921 1871\u20131937 his method of reproducing colors photographically based on the phenomena of interference his investigations into the disintegration of the elements", " and the chemistry of radioactive substances 1909 Guglielmo Marconi Carl Ferdinand Braun 1874\u20131937 1850\u20131918 their contributions to the development of wireless telegraphy 1910 Johannes Diderik van der Waals 1837\u20131923 his work on the state of equations of gases and liquids 1911 Wilhelm Wien Marie Curie (C) 1864\u20131928 1867\u20131934 his discoveries regarding the laws governing the radiation of heat her services to the advancement of chemistry by the discovery of the elements radium and polonium and by the isolation of radium and the study of its nature and compounds 1912 Nils Gustaf Dal\u00e9n 1869\u20131937 his invention of automatic regulators for use in conjunction with gas accumulators for illuminating lighthouses and buoys (Continued) AP-31 AP-32 Appendix F Year Nobel laureate Citation for 1913 Heike Kamerlingh Onnes 1853\u20131926 his investigations of the properties of matter at low temperatures, which led, inter alia, to the production of liquid helium 1914 Max von Laue 1879\u20131960 his discovery of the diffraction of X rays by crystals 1917 Charles Glover Barkla 1877\u20131944 his discovery of the characteristic X rays of the elements 1918 Max Planck 1858\u20131947 his discovery of energy quanta 1919 Johannes Stark 1874\u20131957 his discovery of the Doppler effect in canal rays and of the splitting of spectral lines in electric fields 1920 Charles-\u00c9douard Guillaume 1861\u20131938 the service he has rendered to precise measurement in physics by his discovery of anomalies in nickel steel alloys 1921 Albert Einstein Frederick Soddy (C) 1922 Neils Bohr Francis W. Aston (C) 1879\u20131955 1877\u20131956 his services to theoretical physics, and especially for his discovery of the law of the photoelectric effect his contributions to our knowledge of the chemistry of radioactive substances, and his investigations into the origin and nature of isotopes 1885\u20131962 1877\u20131945 his investigation of the structure of atoms and the radiation emanating from them his discovery, by means of his mass spectrograph, of isotopes in a large number of nonradioactive elements, and for his enunciation of the whole-number rule 1923 Robert Andrews Millikan 1868\u20131953 his work on the elementary charge of electricity and on the photoelectric effect 1924 Karl Manne Georg Siegbahn 1886\u20131978 his discoveries and researches in the field of X-ray spectroscopy 1925 James Franck Gustav Hertz 1882\u20131964 1887\u20131975 their discovery of the laws governing the impact of an electron upon an atom 1926 Jean-Baptiste Perrin 1870\u20131942 his work on the discontinuous structure of matter, and especially for his discovery of sedimentation equilibrium 1927 Arthur Holly Compton Charles Thomson Rees Wilson 1892\u20131962 1869\u20131959 his discovery of the effect named after him his method of making the paths of electrically charged particles visible by condensation of vapor 1928 Owen Willans Richardson 1879\u20131959 his work on the thermionic phenomenon, and especially for the discovery of the law named after him 1929 Prince Louis-Victor de Broglie 1892\u20131987 his discovery of the wave nature of electrons 1931 Werner Heisenberg 1901\u20131976 the creation of quantum mechanics, the application of which has, inter alia, led to the discovery of the allotropic forms of hydrogen 1933 Erwin Schr\u00f6dinger Paul Adrien Maurice Dirac 1887\u20131961 1902\u20131984 their discovery of new productive forms of atomic theory Appendix F AP-33 Year Nobel laureate Citation for 1934 Harold C. Urey (C) 1893\u20131991 his discovery of heavy hydrogen 1936 Victor Franz Hess Carl David Anderson Peter Debye (C) 1883\u20131964 1905\u20131991 1884\u20131966 his discovery of cosmic radiation his discovery of the positron his contributions to our knowledge of molecular structure through his investigations on dipole moments and on the diffraction of X rays and electrons in gases 1937 Clinton Joseph Davisson George Paget Thomson 1881\u20131958 1892\u20131975 their experimental discovery of the diffraction of electrons by crystals 1938 Enrico Fermi 1901\u20131954 1939 Ernest Orlando Lawrence 1901\u20131958 his demonstrations of the existence of new radioactive elements produced by neutron irradiation, and for his related discovery of nuclear reactions brought about by slow neutrons the invention and development of the cyclotron and for results obtained with it, especially with regard to artificial radioactive elements 1943 Otto Stern 1944 Isidor Issac Rabi Otto Hahn (C) 1945 Wolfgang Pauli 1888\u20131969 his contributions to the development of the molecular ray method and his discovery of the magnetic moment of the proton 1898\u20131988 1879\u20131968 his resonance method for recording the magnetic properties of atomic nuclei his discovery of the fission of heavy nuclei 1900\u20131958 his discovery of the exclusion principle, also called the Pauli principle 1946 Percy Williams Bridgman 1882\u20131961 the invention", " of an apparatus to produce extremely high pressures and for the discoveries he made in the field of high pressure physics 1947 Sir Edward Victor Appleton 1892\u20131965 his investigations of the physics of the upper atmosphere, especially for the discovery of the Appleton layer 1948 Patrick Maynard Stuart Blackett 1897\u20131974 his development of the Wilson cloud chamber method and his discoveries therewith in nuclear physics and cosmic radiation 1949 Hideki Yukawa 1907\u20131981 his prediction of the existence of mesons on the basis of theoretical work on nuclear forces 1950 Cecil Frank Powell 1903\u20131969 1951 Sir John Douglas Cockcroft Ernest Thomas Sinton Walton Edwin M. McMillan (C) Glenn T. Seaborg (C) 1897\u20131967 1903\u20131995 1907\u20131991 1912\u20131999 his development of the photographic method of studying nuclear processes and his discoveries regarding mesons made with this method their pioneer work on the transmutation of atomic nuclei by artificially accelerated atomic particles their discoveries in the chemistry of the transuranium elements 1952 Felix Bloch Edward Mills Purcell 1905\u20131983 1912\u20131997 measurements and discoveries in connection therewith the development of new methods for nuclear magnetic precision (Continued) AP-34 Appendix F Year Nobel laureate Citation for 1953 Frits Zernike 1954 Max Born Walter Bothe 1888\u20131966 his demonstration of the phase contrast method, especially for his invention of the phase contrast microscope 1882\u20131970 1891\u20131957 his fundamental research in quantum mechanics, especially his statistical interpretation of the wave function the coincidence method and his discoveries made therewith 1955 Willis Eugene Lamb Jr. b. 1913 Polykarp Kusch 1956 William Shockley John Bardeen Walter Houser Brattain 1911\u20131993 1910\u20131989 1908\u20131991 1902\u20131987 his discoveries concerning the fine structure of the hydrogen spectrum his precision determination of the magnetic moment of the electron their investigations on semiconductors and their discovery of the transistor effect 1957 Chen Ning Yang Tsung Dao Lee b. 1922 b. 1926 their penetrating investigation of the parity laws, which led to important discoveries regarding elementary particles 1958 Pavel Alekseyevich Cherenkov Ilya Mikhaylovich Frank Igor Yevgenyevich Tamm 1959 Emilio Gino Segr\u00e8 Owen Chamberlain 1960 Donald Arthur Glaser Willard F. Libby (C) 1904\u20131990 1908\u20131990 1895\u20131971 1905\u20131989 1920\u20132006 b. 1926 1908\u20131980 1961 Robert Hofstadter 1915\u20131990 Rudolf Ludwig M\u00f6ssbauer b. 1929 their discovery and interpretation of the Cherenkov effect their discovery of the antiproton the invention of the bubble chamber his method to use 14C for age determination in several branches of science his pioneering studies of electron scattering in atomic nuclei and for his discoveries concerning the structure of the nucleon achieved thereby his researches concerning the resonance absorption of rays his discovery in this connection of the effect that bears his name 1962 Lev Davidovich Landau 1908\u20131968 his pioneering theories of condensed matter, especially liquid helium 1963 Eugene Paul Wigner 1902\u20131995 Maria Goeppert Mayer J. Hans D. Jensen 1906\u20131972 1907\u20131973 his contributions to the theory of the atomic nucleus and the elementary particles, particularly through the discovery and application of fundamental symmetry principles their discoveries concerning nuclear shell structure 1964 Charles H. Townes Nikolai G. Basov Alexander M. Prokhorov b. 1915 1922\u20132001 1916\u20132002 fundamental work in the field of quantum electronics, which has led to the construction of oscillators and amplifiers based on the maser-laser principle 1965 Shin\u2019ichiro Tomonaga Julian Schwinger Richard P. Feynman 1906\u20131979 1918\u20131994 1918\u20131988 their fundamental work in quantum electrodynamics, with profound consequences for the physics of elementary particles Appendix F AP-35 Year Nobel laureate Citation for 1966 Alfred Kastler 1902\u20131984 the discovery and development of optical methods for studying Hertzian resonance in atoms 1967 Hans Albrecht Bethe 1906\u20132005 his contributions to the theory of nuclear reactions, especially his discoveries concerning the energy production in stars 1968 Luis W. Alvarez 1911\u20131988 his decisive contributions to elementary particle physics, in particular the discovery of a large number of resonance states made possible through his development of the techniques of using the hydrogen bubble chamber and data analysis 1969 Murray Gell-Mann b. 1929 his contributions and discoveries concerning the classification of elementary particles and their interactions 1970 Hannes Alfv\u00e9n 1908\u20131995 Louis-Eug\u00e8ne-F\u00e9lix N\u00e9el 1904\u20132000 fundamental work and discoveries in magnetohydrodynamics with fruitful applications in different parts of plasma physics fundamental work and discoveries concerning antiferromagnetism and ferrimagnetism, which have led to important applications in solid-state physics 1971 Dennis Gabor 1900\u20131979 his invention and development of the holographic method 1972 1973 John Bardeen Leon N. Cooper J. Robert Schrieffer Leo Esaki Ivar Giaever Brian D. Josephson 1908\u20131991 b. 1930 b. 1931 b", ". 1925 b. 1929 b. 1940 their theory of superconductivity, usually called the BCS theory their experimental discoveries of tunneling phenomena in semiconductors and superconductors, respectively his theoretical predictions of the properties of a supercurrent through a tunnel barrier, in particular those phenomena which are generally known as Josephson effects 1974 Antony Hewish Sir Martin Ryle b. 1924 1918\u20131984 the discovery of pulsars his observations and inventions in radio astronomy 1975 Aage Bohr Ben R. Mottleson L. James Rainwater b. 1922 b. 1926 1917\u20131986 the discovery of the connection between collective motion and particle motion in atomic nuclei and for the theory of the structure of the atomic nucleus based on this connection 1976 Burton Richter Samuel Chao Chung Ting b. 1931 b. 1936 their pioneering work in the discovery of a heavy elementary particle of a new kind 1977 Philip Warren Anderson Nevill Francis Mott John Hasbrouck Van Vleck 1978 Pyotr L. Kapitza Arno A. Penzias Robert Woodrow Wilson b. 1923 1905\u20131996 1899\u20131980 1894\u20131984 b. 1933 b. 1936 their fundamental theoretical investigations of the electronic structure of magnetic and disordered systems his basic inventions and discoveries in the area of low-temperature physics their discovery of cosmic microwave background radiation (Continued) AP-36 Appendix F Year 1979 Nobel laureate Citation for Sheldon Lee Glashow Abdus Salam Steven Weinberg b. 1932 1926\u20131996 b. 1933 their contributions to the theory of the unified weak and electromagnetic interaction between elementary particles, including, inter alia, the prediction of the weak neutral current 1980 James W. Cronin Val L. Fitch b. 1931 b. 1923 the discovery of violations of fundamental symmetry principles in the decay of neutral K-mesons 1981 Nicolaas Bloembergen Arthur L. Schawlow Kai M. Siegbahn b. 1920 1921\u20131999 b. 1918 their contributions to the development of laser spectroscopy his contribution to the development of high-resolution electron spectroscopy 1982 Kenneth G. Wilson b. 1936 his theory for critical phenomena in connection with phase transitions 1983 Subrahmanyan Chandrasekhar 1910\u20131995 William A. Fowler 1911\u20131995 his theoretical studies of the physical processes of importance to the structure and evolution of the stars his theoretical and experimental studies of the nuclear reactions of importance in the formation of the chemical elements in the universe 1984 Carlo Rubbia Simon van der Meer b. 1934 b. 1925 their decisive contributions to the large project, which led to the discovery of the field particles W and Z, communicators of the weak interaction 1985 Klaus von Klitzing b. 1943 the discovery of the quantized Hall effect 1986 Ernst Ruska 1906\u20131988 Gerd Binnig Heinrich Rohrer 1987 J. Georg Bednorz Karl Alex M\u00fcller 1988 Leon M. Lederman Melvin Schwartz Jack Steinberger 1989 Hans G. Dehmelt Wolfgang Paul Norman F. Ramsey 1990 Jerome I. Friedman Henry W. Kendall Richard E. Taylor his fundamental work in electron optics and for the design of the first electron microscope their design of the scanning tunneling microscope their important breakthrough in the discovery of superconductivity in ceramic materials b. 1947 b. 1933 b. 1950 b. 1927 b. 1922 1932\u20132006 b. 1921 the neutrino beam method and the demonstration of the doublet structure of the leptons through the discovery of the muon neutrino b. 1922 1913\u20131993 b. 1915 their development of the ion trap technique the invention of the separated oscillatory fields method and its use in the hydrogen maser and other atomic clocks b. 1930 1926\u20131999 b. 1929 their pioneering investigations concerning deep inelastic scattering of electrons on protons and bound neutrons which have been of essential importance for the development of the quark model in particle physics Appendix F AP-37 Year Nobel laureate Citation for 1991 Pierre-Gilles de Gennes 1932\u20132007 1992 Georges Charpak 1993 Joseph H. Taylor, Jr. Russell A. Hulse b. 1924 b. 1941 b. 1950 his discovering that methods developed for studying ordered phenomena in simple systems can be generalized to more complex forms of matter, in particular, to liquid crystals and polymers his invention and development of particle detectors, particularly multi-wire proportional counters their discovery of rare binary pulsars 1994 Bertram N. Brockhouse Clifford G. Shull 1918\u20132003 1915\u20132001 their pioneering contributions to the development of neutron scattering techniques for studies of condensed matter 1995 Martin Perl Frederick Reines 1996 David Lee Douglas Osheroff Robert Richardson 1997 Steven Chu Claude Cohen-Tannoudji William Phillips 1998 Robert B. Laughlin Horst L. St\u00f6rmer Daniel C. Tsui 1999 Gerardus \u2018t Hooft Martinus J. G. Veltman 2000 2001 Zhores I. Alferov Herbert Kroemer Jack S. Kilby Eric A. Cornell Wolfgang Ketterle Carl E. Weiman 2002 Raymond Davis, Jr. Masatoshi Koshiba Riccardo", " Giacconi b. 1927 1918\u20131998 for his discovery of the tau lepton for his discovery of the neutrino b. 1931 b. 1945 b. 1937 b. 1948 b. 1933 b. 1948 b. 1950 b. 1949 b. 1939 b. 1946 b. 1931 for their discovery of the superfluid phase of 3He for their development of techniques to chill atoms to millionths of a kelvin above absolute zero and to trap them with laser light for their discovery of a new form of quantum fluid with fractiionally charged excitations for elucidating the quantum structure of electroweak interactions in physics b. 1930 b. 1928 1923\u20132005 for basic work on information technology for his part in the invention of the integrated circuit b. 1961 b. 1957 b. 1951 for the achievement of Bose-Einstein condensation in dilute gases of alkali atoms and early fundamental studies of the properties of the condensates 1914\u20132006 b. 1926 b. 1931 their pioneering contributions to astrophysics, in particular the detection of cosmic neutrinos his pioneering contributions to astrophysics, which have led to the discovery of cosmic x-ray sources 2003 Alexei A. Abrikosov Vitaly L. Ginzburg Anthony J. Leggett b. 1928 b. 1916 b. 1938 for pioneering contributions to the theory of superconductors and superfluids (Continued) AP-38 Appendix F Year Nobel laureate Citation for 2004 David J. Gross H. David Politzer Frank Wilczek 2005 Roy J. Glauber John L. Hall Theodor W. H\u00e4nsch 2006 John C. Mather George F. Smoot b. 1941 b. 1949 b. 1951 b. 1925 b. 1934 b. 1941 b. 1946 b. 1945 for the discovery of asymptotic freedom in the theory of the strong interaction for his contribution to the quantum theory of optical coherence for their contributions to the development of laser-based precision spectroscopy, including the optical frequency combo technique for their discovery of the blackbody form and anisotropy of the cosmic microwave background radiation Answers These results are usually rounded to three significant figures. Differences in the third significant figure may result from rounding and are not important. When the solution is a diagram, graph, derivation, or proof, reference is made to the Students Solution Manual (SSM) where the solution appears. Chapter 1 Chapter 2 1-1. 1-5. 1-9. 1-13. 1-17. 1-21. 1-25. 1-29. 1-33. 1-37. 1-41. 1-45. 1-49. 1-53. 1-57. 1-61. 2-1. 2-5. 2-9. 2-13. 2-17. 2-21. 4.4 108 m>s (a) relative to Hoth. (b) No, since the droid is moving faster than light speed, a bright circle reflected from great circle perpendicular to the motion. (a) At t 2 s t 2 s (b) At \u00a2t 4.63 1013 s 3.76 105 s (a), the entire interior lights up. (b) 2.0 105 s (a) See Students Solution Manual (SSM) clock, only 6 s have passed on the lab clock. (b) When 10 s have passed on the rocket\u2019s 0.14c 0.527c (a) In S: V 16 m 3 ; in S: V 12.2 m3 (b) See SSM 657.0 nm, 662.9 nm, 725.6 nm 3.0 m 9.6 ms in the v 1.44 108 m>s x direction (b) (a) See SSM (b) v 0.5c (a) 0.866 c # y (e) 0.494 vy (a) 120 min (b) 240 min (c) identical >c (c) 4.39 s (c) (d) 0.866 c # y 4.39 s (d) spacelike 0.58 y (a) A S B: T>2 2vL>(c2 v2); B S A: T>2 2vL>(c2 v2) (b) 4vL>(c2 v2) See SSM 1.1 1016 kg 0.99998898c (a) 3.5 107 (a) 6.26 MeV 280 MeV, mass increases (b) 3.94 104 GeV>c (c) 3.31 109 GeV 3.31 109 GeV>c ; (b) 0.0079 See SSM 2-25. 2-29. m 1673 MeV>c2, u 0.286c AN-1 AN-2 Answers Chapter 3 Chapter 4 2-33. 2-37. 2", "-41. 2-45. 2-49. 2-53. 3-1. 3-5. 3-9. 3-13. 3-17. 3-21. 3-25. 3-29. 3-33. 3-37. 3-41. 3-45. 3-49. 3-53. 3-57. 4-1. 4-5. 4-9. 4-13. 4-17. 4-21. 4-25. 4-29. 4-33. 4-37. 4-41. 4-45. 4-49. 4-53. 4-57. (c) is correct. 8.62 ms (a) 1.73 105 m>s (b) 1.5 105 m>s (c) 155 kg See SSM (a) v E>Mc (b) \u00a2x EL>Mc2 (c) m E>c2 (a) See SSM (b) See SSM 6.5 102 m proton helium 0.26 m ; electron 3.6 105 m ; deuteron 0.13 m; H2 0.13 m; (a) 2.2 mm (b) 9.1 109 Hz, 1.1 1010 s See SSM 5.67 108 W>m2 K4 16 R1 278.3 K (5.3 \u00b0C) 1.4 104 (b) ultraviolet (a) (b) 255 nm 1.03 1015 Hz (a) 4.14 103 nm; 5.8% 0.243 nm (a) electron 0.00243 nm; proton 1.32 fm (b) electron 0.510 MeV; proton 939 MeV (a) 2.08 eV (b) 4.95 1014 Hz (c) 4.19 1015 eV>Hz See SSM See SSM (a) 0.0309 nm, 0.1259 nm (b) 9.90 keV Lyman 91.16 nm; Balmer 364.6 nm; Paschen 820.4 nm 4103 nm 45.5 fm; 29.5 fm; 19.0 fm (a) 1.91 nm (b) 0.95 nm 8.22 1014 Hz, 8.22 106 revolutions See SSM (a) 19.0 m (b) 3.65 103 m>s 680 fm 1.90 108 Hz1>2 10.2 V 9.27 1024 A # m2 (b) 1.054 103 A Lyman : n 6 S n 3 Lyman : n 9 S n 3 b R cos(>2) (c) n 1: v 0.0075cZ1>2; E 14.4Z eV I0R2 cos2 (>2) R2 (b) (a) (a) (a) For (b) \u00a2 0.056 nm (d) See SSM 10; 1,042 Chapter 5 Chapter 6 Chapter 7 5-1. 5-5. 5-9. 5-13. 5-17. 5-21. 5-25. 5-29. 5-33. 5-37. 5-41. 5-45. 5-49. 5-53. 6-1. 6-5. 6-9. 6-13. 6-17. 6-21. 6-25. 6-29. 6-33. 6-37. 6-41. 6-45. 6-49. 6-53. 6-57. 6-61. 7-1. 7-5. 7-9. 7-13. 7-17. (a) 2.1 1023 m (b) 2.1 1021 m>y 0.0276 nm Answers AN-3 6.18 103 fm 3.0 103 eV 50 m>s (c) 50 m>s (d) \u00a2x 5 m; \u00a2k 0.4 m1 0.445 fm (b) (a) l 0.523 nm; Ek (a) See SSM (b) 3.2 105 s A2 dx (a) (b) 1.99 1021 eV 5.3 1010 (a) 0.61A2 dx (c) 0.14A2 dx (d) x 0 (b) 1.32 107 eV See SSM (a) See SSM (b) See SSM (a) 1840 MeV (b) 2.02 fm (c) 1.22 fm (d) 0.76 fm (a) 0.243 nm (b) 0.511 MeV (c) 1.2 106 eV, 1.2 eV 0.511 MeV>c (d) 2.43", " 103 nm See SSM See SSM (a) 0.021 eV (b) 205 MeV \u00a2x 106 m; \u00a2p 1016 kg # m>s (a) (b) 9 1011 See SSM 1.21 107 N 0.87 nm L>2 (a) (b) 0.320L2 (a) 8x9 0; 8x29 U>(2m) 8x9 0 (a) See SSM (b) 2.33 1034 J (b) 4.3 106 R 0.111; T 0.889 9 U2>(2mL2) 4.95 1013 (a) 8Ek (a) (a) (b) 0.197 (c) 2.1 1028 8x29 3U>(2m) 0.70 Hz (c) (d) 8V(x)9 3U>4 is the transmitted fraction (b) See SSM R 0.111; T 0.889 (b) (a) 0.39c (b) See SSM (c) 8.03 105 eV (d) 3.76 105 eV The 1st, 2nd, 3rd, and 5th excited states are degenerate. 11E0, 12E0, 14E0, (a) See SSM (b) 1, 1, 4 and 1, 2, 2 0, m 0; for (a) (a) 0, 1, 2 (b) For (c) 18 2U2 6U2 (b) n 6, 3 (b) 3U, 2U, U, 0 (a) (d) (c) 1, m 1, 0; for 2, m 2, 1, 0 3.65 1034 J # s 5U2 (d) 0.38 eV n 3 (c) 7-21. See SSM AN-4 Answers Chapter 8 Chapter 9 7-25. 7-29. 7-33. 7-37. 7-41. 7-45. 7-49. 7-53. 7-57. 7-61. 7-65. 7-69. 7-73. 8-1. 8-5. 8-9. 8-13. 8-17. 8-21. 8-25. 8-29. 8-33. 8-37. 8-41. 8-45. 9-1. 9-5. 9-9. 9-13. 9-17. 9-21. 9-25. 9-29. 9-33. 9-37. 9-41. (a) 0.606(1>a0)3>2> 232 (b) 0.368(1>a0)3>(32) (c) 0.368>(8a0) See SSM (a) 4 (b) 3 See SSM See SSM Sn: Nd: Yb: 1s22s22p63s23p63d104s24p64d105s25p2 1s22s22p63s23p63d104s24p64d105s25p64f46s2 1s22s22p63s23p63d104s24p64d104f145s25p66s2 (a) silicon, Si (b) calcium, Ca Similar to H: Li, Rb, Ag, and Fr Similar to He: Ca, Ti, Cd, Ba, Hg, and Ra D5>2 (a) is j-forbidden S P1>2 2.90 106 eV 1.67 106 m>s2 7.83 104 nm 1.95 cm (b) (b) (a) (c) 0.638 T (a) See SSM (b) See SSM (a) through (e) See SSM (a) 1930 m>s (b) 1.01 104 K (a) 3400 J temperature. (b) One mole of any gas has the same translational energy at the same See SSM Cv (a) R, Cp >n1 n2 See SSM 2R >n1 n3, and 2 are both and 0 See SSM See SSM See SSM fFD(E) [e10(EEF)>EF 1] 1 (a) 8EK(escape) See SSM 9 2kT (b) 19,760 K (c) 0.7% (b) See SSM (a) (a) 23.06 kcal>mol 5.39 eV (b) (b) 98.5 kcal>mol (c) 0.46 eV 4.83 eV (c) 1.08 eV>molecule For KBr: 0.19 eV For R", "bCl: 0.23 eV 2.63 1029 C # m (a) 0.67 nm (b) 55 nm (c) See SSM 1.78 104 eV (a) 18.6 u (b) 0.280 nm 0.04 105 eV 5.5 1031 about 1.28 1016>s (a) 8.47 105 radians (b) 5.08 103 W>cm2 Answers AN-5 4.79 104 eV, vibrational 9-45. 9-49. 9-53. 9-57. E3 1.44 103 eV, E2 9.61 104 eV, E1 (a) states (note equal spacing); See SSM (b) 0.215 nm 480 N>m (a) (b) 0.31 eV 4.58 1048 kg # m2 8.9 1014 eV # nm20 (b) 1.32 1014 Hz (a) Chapter 10 10-1. 10-5. 10-9. 4.64 4.09 eV>atom 4.18 10-13. (a) 5.86 1022>cm3 (b) 5.90 1022>cm3 10-17. 1.87 eV 10-21. 10-25. 10-29. (a) 34.2 nm (b) 41.4 nm (c) 43.1 nm 6.7 103 (a) 0.37 eV (b) 4300 K 10-33. See SSM 10-37. (a) 27.8 mV 206Pb (b) 12.0 mV 207Pb: 10-41. For : 7.217 K; for 7.200 K; for 208Pb: 7.183 K 10-45. (a) 0.95 (b) 0.71 (c) 0.32 10-49. (a) See SSM (b) See SSM (c) 0.0055 10-53. silicon: 3.17 nm; germanium: 8.46 nm 10-57. (a) 1.28 1013 Hz (b) 23.4 m Chapter 11 11-1. 11-5. 11-9. See SSM See SSM (a) 6.46 MeV>nucleon (b) 7.47 MeV>nucleon (c) 8.77 MeV>nucleon 11-13. (a) 2.70 fm; 3.53 fm (b) 4.26 fm; 5.57 fm (c) 6.34 fm; 8.30 fm 11-17. (a) 5.21 h (b) 3.11 106 atoms 11-21. 33.2 min 11-25. A 191; B 72.2 11-29. 1.00 MeV 11-33. 93 keV 11-37. 11-41. 11-45. 11-49. 789 MeV>c2 36S, 53Mn, 82Ge, 88Sr, 94Ru, 131In, 145Eu 14Si, j 0; 37 17Cl, j 3>2; 55 30 (b) (a) 4.97 MeV 1.92 MeV 27Co, j 7>2; 90 40Zr, j 0; 107 49In, j 9>2 11-53. 224 MeV 11-57. 11-61. 11-65. 11-69. 1.78 1014 atoms>s 8.78 1012 J (a) 1.90 109 y 1013 (a) (b) 0.149 mg (b) 145 y (c) 2.15 104 y 11-73. See SSM 11-77. (a) 5.06 106 eV (b) 4.71 102 eV (c) 1.17 cm>s AN-6 Answers 11-81. 11-85. 11-89. 141Ba : 6.24 1015 m; 92Kr : 5.42 1015 m (a) 7.03 108 y (a) See SSM (b) 51.7 MeV (c) 1.43 MeV (c) 96.5 MeV (b) 249 MeV 11-93. (a) See SSM (b) See SSM (c) See SSM (d) See SSM (e) See SSM 11-97. (a) See SSM (b) 8.66 104 11-101. (a) 137% (b) 0.00498 Chapter 12 12-1. 12-5. 12-9. 12-13. 12-17. 12-21. (a) See SSM (b) 139.6 MeV (c) 8.", "88 fm (a) 1.711 MeV (b) 78.5 eV (c) 2.16 MeV>c (c) strong (d) weak (a) weak (b) electromagnetic m 22 eV>c2 (a) Conservation of energy and lepton number are violated. energy is violated. (d) No conservation laws are violated. violated. (f) Conservation of baryon number is violated. (c) Conservation of linear momentum is violated. (e) Conservation of lepton number is (b) Conservation of (a) Conservation of lepton number is violated. (d) Conservation of baryon number and angular momentum are violated. (e) Allowed. (b) Allowed. (c) Allowed. 12-25. See SSM 12-29. uuu 12-33. 12-37. 12-41. 12-45. 12-49. 138.3 MeV cd (a) (b) cd Energy is not conserved. (a) Energy is not conserved. (c) Angular momentum is not conserved. (b) Angular momentum is not conserved. (a) The final products are all stable. not conserved. (b) See SSM (c) See SSM (d) No. Energy is (a) No conservation laws are violated. number are violated. (c) No conservation laws are violated. (b) Conservation of energy and baryon 12-53. (a) 1193 MeV (b) 77 MeV>c (c) 2.66 MeV (d) 74.3 MeV>c Chapter 13 13-1. 13-5. 13-9. 13-13. 13-17. 13-21. 25.3 d 10.4 eV>c2 17.0 pc (a) (b) Rstar 9400 y 3.27 109 c # y (a) 5.5 1097 kg>m3; 1.96R} (b) 10% using the Te relation 1.67 1018 kg>m3; 2.45 104 kg>m3 13-25. See SSM 13-29. 13-33. 13-37. 8.33 1056 \u201cnew\u201d (a) 0.001 (a) See SSM (b) 8.89 1044 J (b) H atoms>m3 # 106 y; 1.1 1012 cm3 (c) 1.7 1010 y No. (c) 0.1% M} INDEX boldface indicates a definition italics indicates a figure t indicates a table * indicates a MORE section on the home page: www.whfreeman.com/tiplermodernphysics5e # indicates a Classical Concept Review section on the home page: www.whfreeman.com/tiplermodernphysics5e Absolute magnitude (M), 633 Absorption, 393 Absorption spectra. See Spectra ac Josephson effect, 467\u2013468 Accelerator mass spectroscopy (AMS), 537, 545, 546\u2013548, 548 Approximations first-order perturbation theory, 298 in nuclear physics, 507, 514, 519 in relativity, 94\u201397, 610 and the Schr\u00f6dinger equation, 228, 230, 269, 295, 297 Accelerators, particle, 71, 477, 516, 561\u2013562 in statistical physics, 356 Avalanche breakdown, 454 Average speed \u2039v\u203a, 320, 320 Avogadro\u2019s hypothesis, 115 Avogadro\u2019s number (NA), 116, 214 Awschalom, David, 452 Axions, 639 and liquid helium, 336 and the quark model of hadrons, 591 and relativistic momentum, 67 and superconductivity, 458 Acceptor levels, 448 Accretion disks, 644, 644 Active galaxies, 656, 656\u2013657, 657 Active Sun, 627\u2013629 Akimitsu, Jun, 458 Alkali atoms, excited states of, 301\u2013303, 302 Alley, C. O., 33 Alpha decay, 258\u2013259, 496\u2013499 and decay chains, 497\u2013498, 498 energetics of, 498* Alpha particles (), 178, 477 scattering of, and atomic models, 151, 151\u2013159, 152, 153, 154, 156, 157, 477 Amaldi, Edoardo, 553 Ambler, E., 588, 589 Ammonia inversion spectra, 392 Amorphous solids, 414 AMS. See Accelerator mass spectroscopy (AMS) Anderson, Carl David, 356, 477, 552, 561, 562\u2013563, 612 Anglo-Australian Observatory, 661 Angular frequency (), 196 Angular momentum (", "L), 379 conservation of, 275 nuclear (I), 489\u2013492 quantization of, 161, 275\u2013277 relativistic (L), 80\u201381 spin (S), 285\u2013290 and spin-orbit effect, 293\u2013295 total (J), 291\u2013293 Angular momentum operators, 245t Angular momentum quantum number ( ), 276 spin (ms), 285 total (j), 291 Annihilation of particles, 90\u201393, 91, 565\u2013566 Feynman diagram for, 567 Anthropic principle, 673 Antibaryons, 591 Antibonding orbitals, 373, 373 Antielectrons. See Positrons (e) Antiferromagnetism, 436 Antihydrogen, 565 and lasers, 403 Antineutrons, 564 Antiparticles, 90, 500, 562\u2013566, 573t, 604 Antiprotons, 114, 612 Apparent magnitude (m), 632\u2013633, 677 Aristotelian physics, 1, 3 Aristotle, 1, 6, 630, 636 Artwork authentication of, 537, 537\u2013539 dating of, 543, 543\u2013544 Associated Legendre functions, 274, 281 Astronomical units (AU), 620 Astrophysics, 619\u2013661. See also Cosmology; Stars; Sun and cataclysmic events, 644\u2013646 and galaxies, 653\u2013661 and radiometric dating, 545\u2013546 and stars, 630\u2013643, 647\u2013652 and the Sun, 619\u2013630 and surface temperatures of stars, 121 Asymptotic freedom, 600 Atomic beam fluorescence spectroscopy, 490 Atomic clocks, 260, 392, 396 testing relativity with, 33 Atomic lasers, 343, 343 Atomic mass number (A), 478 Atomic mass units (u), 83 Atomic models, 147\u2013184 and atomic spectra, 147\u2013150 Bohr, 159\u2013169, 176* and the Franck-Hertz experiment, 174\u2013176 Rutherford, 150\u2013159 Sommerfeld, 167 Thomson, 150\u2013151, 151 and x-ray spectra, 169\u2013173 Atomic number (Z), 478 Atomic physics, 269\u2013314 and atomic spectra, 301\u2013305 and electron spin, 285\u2013290 and hydrogen atom wave functions, 281\u2013285 and the periodic table, 297\u2013301 and quantization of angular momentum and energy, 272\u2013281 and the Schr\u00f6dinger equation for multiple particles, 295\u2013297 and the Schr\u00f6dinger equation in three dimensions, 269\u2013272 and the spin-orbit effect, 291\u2013295 Atomic radii, 300, 301 Atomic spectra, 148\u2013150, 149 and excited states, 301\u2013303, 302 Auger, Pierre, 173 Auger effect, 173, 173 Auger electrons, 173, 548 Auroras, 628 Autumnal equinox, 637, 637 BL Lac objects, 656 Bahcall, John, 607, 612, 625, 626, 675 Balmer, Johann, 149 Balmer series, 149, 149, 163, 164 and density of states, 318 redshift of, 44 Band spectra, 147 Band theory of solids, 438*\u2013445*, 440, 440 and conductors, 441\u2013442, 442 and impurity semiconductors, 445\u2013452 and insulators, 442, 442 and intrinsic semiconductors, 442, 442\u2013445 Kronig-Penney model of, 438, 438\u2013441, 439 Bands. See Band theory of solids Bardeen, John, 456, 463, 471 Barkla, Charles G., 172, 179, 307 Barns, 521 Barred spiral galaxies, 654\u2013655, 655 Barrier penetration. See Tunneling Barrier potentials, 256, 256\u2013257 Baryon decuplet, 587, 591 Baryon number, 564, 569t, 583, 583t, 586t conservation of, 581 nonconservation of, 607 Baryon octet, 587, 591, 592, 595 Baryons, 569, 573t, 587, 594t supermultiplets of, 598 Bases, transistor, 456, 456 BATSE. See Burst and Transient Source Experiment (BATSE) BCS theory of superconductivity, 463\u2013465, 510 BEC. See Bose-Einstein condensates (BEC) Becquerel, Antoine Henri, 477, 495, 552 becquerels (Bq), 493 Bednorz, Johannes G., 466 Bennet, W. R., Jr., 401 BeppoSAX satellite, 652 Beta decay, 499\u2013504, 501, 499\u2013500 double, 503 electron capture, 502\u2013504 inverse, 646, 648", " Beta particles (), 151, 477 Betelgeuse, 620, 643 Bethe, Hans Albrecht, 535, 625, 675 Big Bang, 668\u2013672, 670, 676 and blackbody radiation, 126, 126\u2013127, 664, 664, 668\u2013669 fusion reactions during, 556, 668, 668 and inflation, 669 and quantum fluctuations, 206 I-1 I-2 Index Big Crunch, 666 Binary pairs, 644, 644 Binding energy (Eb), 81\u201384, 83, 163 nuclear (B), 488\u2013489, 489, 495, 495, 513, 514 Binnig, Gerd, 257 Black dwarfs Black holes, 105\u2013106, 650\u2013651, 651 at center of the Milky Way, 635, 650\u2013651 simulating, 305 Blackbody radiation, 119\u2013127, 129 from Big Bang, 664, 664, 668\u2013669 energy density in, 122 ideal, 120 as photon gas, 344\u2013351 solar, 620\u2013621, 621 Bloch, Felix, 221, 262, 439, 470 Bloch functions, 439 Blueshift, 42, 43 gravitational, 104\u2013105 Body-centered cubic (bcc) crystal symmetry, 418 Bohr, Aage N., 527 Bohr, Niels H. D., 159, 179, 215 and atomic model, 159\u2013162, 164, 165, 170, 176*, 226 and nuclear reactions, 521 Bohr frequency condition, 160 Bohr magneton ( B), 182, 287, 552 and magnetism, 435 Bohr radius (a0), 161, 279 and probability density, 282\u2013283 Boks, J., 335 Boltzmann, Ludwig, 356 and kinetic theory, 316, 319 and thermodynamics, 120 Boltzmann constant (k), 122, 316 Boltzmann distributions (fB(E)), 316 and conduction, 425, 430 and quantization of energy states, 346 vs. other distributions, 329\u2013333, 331 eE>kT Boltzmann factor ( ), 316 and energy-level populations, 385, 388\u2013389, 394 Bond lengths (r0), 366, 368t and rotational spectra, 380 Bonding orbitals, 373, 373 Bonds, molecular, 416t. See also Covalent bonds; Ionic bonds dipole-dipole, 364, 375\u2013379, 377, 378 and electron configurations, 363 hydrogen, 377, 378 metallic, 364, 375, 415, 421\u2013422, 422 other types, 375*\u2013379 saturated, 374 and wave functions, 284, 285 Born, Max, 192, 215, 225, 307 Born exponents (n), 470 Bose, Satyendra Nath, 329, 356 Bose-Einstein condensates (BEC), 335\u2013343, 341, 343, 354 fermion analog of, 353\u2013354, 354 and lasers, 403 and photon gases, 344\u2013351 and slowing of light, 304\u2013305, 305 Bose-Einstein distributions (fBE(E)), 316, 328\u2013329 and composition of the nucleus, 479 vs. other distributions, 329\u2013333, 331 Bosons, 329\u2013332 as force mediators, 571, 602 Higgs, 568, 581, 603 W and Z0, 575, 575, 576t, 602\u2013603 Bottom (B), 570, 583, 583t Boundary conditions, 230 and energy quantization, 225 in three dimensions, 270 Brackett, F. S., 164 Brackett series, 180 Bragg, William H., 141, 169 Bragg, William L., 134, 141, 169 Bragg condition, 135, 135 and band theory of solids, 440\u2013441 for electrons, 188\u2013189, 189 Bragg planes, 134, 134 Brattain, Walter H., 456, 471 Bremsstrahlung, 134, 136, 136, 511 Brillouin zones, 441 Brookhaven National Laboratory, 564, 595, 597 Bubble chambers, 564, 572, 584 Burst and Transient Source Experiment (BATSE), 652 Cabrera, B., 608 Cannon, Annie Jump, 632, 676 Carbon dating, 543, 543\u2013545 and accelerator mass spectroscopy, 546\u2013547 Carnot, Nicolas L\u00e9onard Sadi, 1 CAT. See Computer assisted tomography (CAT) Cathode-ray tubes (CRTs), 117 Causality, and special relativity, 37, 39, 39, 54\u201355, 55 CDs. See Compact discs (CDs) CEBAF. See Continuous Electron Beam Accelerator Facility (CEBAF) Celestial sphere, 636, 636\u2013637 Center-of-mass reference frames, 519, 519, 522", " Cepheid variables, as standard candles, 640, 654, 676 CERN. See European Organization for Nuclear Research (CERN) Chadwick, James, 477, 479, 500, 552, 561 Chain reactions, 527, 530 Chamberlain, Owen, 564, 612 Chandra X-Ray Observatory, 650, 651 Chandrasekhar, Subrahmanyan, 647, 676 Chandrasekhar limit, 647, 648 Characteristic rotational energy (E0r), 380, 389t Characteristic spectra, 135, 136 Charge color, 569\u2013570, 576t, 596\u2013597 density of, in nucleus, 482 electric, 118\u2013119, 569t, 576t, 583, 583t flavor, 568, 569t, 574 gravitational, 576, 576t magnetic, 608 weak, 570, 574, 576t Charge multiplets, 572, 585\u2013586, 606 Charged current, 574 Charged weak force, 574 Charm (C), 570, 583, 583t, 597, 597 Chirped pulse amplification, 403 Christenson, J. H., 590 Chromosphere, 622, 622\u2013623, 623 Classical Concept Reviews. See also Exploring; More charge-to-mass ratio (e/m) of electrons, 116# conduction, 423# Fourier integrals, 199# Galilean transformation of coordinates, 6# inertial reference frames, 6# interference fringes, 9# kinetic theory, 315#, 316#, 319#, 328# Millikan oil-drop experiment, 119# relativity, 6# Classical novae, 644 Classical physics. See also Newtonian physics conduction, 422\u2013426, 423# dynamics, 65\u201366, 81 failures of, 325, 327, 346\u2013347, 422, 425 history of, 1\u20132 particles, 212 and relativistic approximations, 95\u201396 statistical, 316\u2013328 uncertainty relations, 199\u2013201, 200 wave equation, 196, 202, 212 Classical relativity, 4\u20136#, 8\u20139, 11 Clausius, Rudolf, 325 Clocks. See also Time atomic, 33, 260, 392, 396 in gravitational fields, 103\u2013105 light, 29, 30 local, 15 reference, 13, 14 Closed universe, 666 Clusius, K., 337 COBE. See Cosmic Background Explorer (COBE) satellite Coblenz, W. W., 123 Cockcroft, John D., 477, 516, 552 Coefficient of reflection (R), 253, 254 Coefficient of transmission (T), 253, 254 and alpha decay, 259, 497 and tunneling, 257 Coherent radiation, 391, 394 Cohesive energy, 417, 420 Collectors, transistor, 456, 456 Collisions. See also Scattering and atomic energy levels, 174, 174\u2013175 elastic, 66\u201367, 67 between electromagnetic radiation and matter, 137\u2013139 inelastic, 76, 76\u201377 and relativistic energy, 76, 76\u201377 and relativistic momentum, 66\u201367, 67 Color charge, 569\u2013570, 576t and quark model of hadrons, 596\u2013597 and strong interaction, 570, 571 Comoving coordinates, 660, 665, 667 Comoving space density, 660, 668 Compact discs (CDs), 404, 437 Complex conjugates, 204, 225, 262 Complex numbers, 109, 205, 224\u2013225, 262, 407 Compound doublets, 303 Compound nucleus, 517, 521\u2013522, 522 Compton, Arthur H., 133, 137, 138, 141, 407, 531, 561 Compton edge, 146 Compton effect, 137, 137\u2013139, 138, 390, 391 Feynman diagram for, 567 Compton Gamma Ray Observatory, 652 Compton wavelength ( c), 137, 195 and limits of knowledge, 673 and range of strong interaction, 510, 578 Compton\u2019s equation, 137, 138* Computer assisted tomography (CAT), 478, 540\u2013542, 541, 542 Conant, James B., 531 Condon, Edward U., 258, 477 Conduction and band theory of solids, 441\u2013442, 442 classical, 422\u2013426, 423# electrical, 423\u2013424 quantum, 430\u2013434 thermal, 337, 434* Conduction bands, 441, 441, 442 Conductivity (), 424\u2013425. See also Superconductivity Conservation laws, 583* of electric charge, 580 of energy, 76\u201381, 517\u2013520, 580, 663, 666 and invariant quantities, 66, 73, 580 of linear and angular momentum, 580 in particle physics, 500, 568, 580\u2013591,", " 590t, 594 in radioactive decay, 496, 500 Constellations, 631, 631 Continuity conditions, 228, 252 Continuous Electron Beam Accelerator Facility (CEBAF), 71 Continuous spectra, 147\u2013148 Continuous wave lasers, 400\u2013401 Conversion electrons, 505 Cooper, Leon N., 463, 471 Cooper pairs, 463, 464\u2013465 and exchange forces, 510 in fermion gases, 354 in Josephson junctions, 467 Coordinate systems celestial, 636, 636\u2013637 comoving, 660, 665, 667 polar, 101 right-handed vs. left-handed, 589, 589 spherical, 271\u2013272, 272, 282 Copernicus, Nicolaus, 3, 57, 636 Core, solar, 623 Cormack, Allan, 541, 553 Cornell, Eric A., 341, 343 Corona, 623 Corpuscles, 118 Correspondence principle, 160\u2013161 and Bohr atoms, 166 for infinite square wells, 233\u2013234, 236 and Rydberg atoms, 168 for simple harmonic oscillators, 248, 248 Cosmic Background Explorer (COBE) satellite, 126, 664, 669 Cosmic background radiation, 126, 126\u2013127, 664, 664 anisotropies in, 664, 664, 669 Cosmic rays, 141 and carbon dating, 543\u2013545 as charged particles, 138 de Broglie wavelength of, 195\u2013196 and pair-production, 90 relativistic speeds of, 22\u201323 Cosmological constant (-), 661, 664, 667 Cosmological principle, 662, 662, 666\u2013667, 669 perfect, 664 Cosmological redshift, 659\u2013661 Cosmology, 619, 662\u2013674. See also Astrophysics and the Big Bang, 126\u2013127, 206 and evolution of the universe, 664\u2013674 and expansion of the universe, 43\u201344 and general relativity, 101\u2013105, 104 and gravitation, 662\u2013664 and headlight effect, 52 Coulomb\u2019s law and alpha decay, 258\u2013259, 259 and atomic models, 159\u2013160, 161 and covalent bonds, 370, 370\u2013371 Feynman diagrams for, 566\u2013567, 567 and fission, 528, 528 and fusion, 625 and phonons, 464 and scattering of charged particles, 153\u2013154, 156* Coupling constants, 570\u2013571, 577\u2013578, 605, 612 for electromagnetic interactions (), 571, 577 for strong interactions ( Covalent bonds, 364, 369\u2013375, 374 s), 571, 600 other types, 375*, 420* vs. ionic bonds, 374 Covalent solids, 415\u2013421 Cowan, Clyde, 500 CP violations, 590 Crab nebula, 646, 649, 649\u2013650, 650 Creation of particles. See Pair production of particles Creeping films, 338, 338 Critical magnetic fields (Bc), 458, 458t, 459, 460, 0) of the universe, Critical population inversion density (nc), 400, 461 Critical mass density ( 662\u2013663, 672 404\u2013405 Critical temperature (Tc) for Bose-Einstein condensates, 340\u2013341 for superconductivity, 458, 458t, 459, 466t Cross sections (), 154, 154, 484, 521, 523 ds>d\u00c6 ), 178 differential ( for neutron capture, 514, 514 partial, 521 for strong interaction, 580 CRT. See Cathode-ray tubes (CRTs) Crystallography, 134\u2013136 Crystals, 414, 414\u2013415, 416t symmetry of, 415, 415\u2013421, 416t, 418, 419, 420 Curie, Marie, 552 Curie, Pierre, 435, 552 Curie temperature (TC), 436 curies (Ci), 493 Curie\u2019s law, 435 Current, charged vs. neutral, 574 Current density (j), 424 Current loops, 286 Curvature of wave functions, 239, 239 Cyclotrons, 477, 516 and accelerator mass spectroscopy, 547 and medical isotopes, 542 dc Josephson effect, 467 Dalton, John, 561 Dark energy, 185, 663, 664, 667, 669, 671 Dark matter, 185, 638\u2013639, 663, 667, 671 and black holes, 650 and interstellar dust, 654 Darwin, C. G., 47 Davis, Raymond, Jr., 607, 612, 626, 676 Davisson, Clinton J., 186, 188\u2013193, 191, 214 Davisson-Germer experiment, 188, 188\u2013192 de Broglie, Louis, 2,", " 185\u2013188, 186, 191\u2013193, 214 de Broglie, Maurice, 214 de Broglie relations, 185\u2013187 and Bose-Einstein condensates, 341, 343 and distinguishability of particles, 332 and Schr\u00f6dinger equation, 223 de Broglie wavelength (), 185, 195\u2013196 Debye, Peter J. W., 221, 262, 348 Debye frequency (fD), 348, 348, 351 Debye temperature (TD), 348, 348 Decay, radioactive, 495*. See also Radioactivity alpha, 496\u2013499 beta, 499\u2013504 electron capture, 502\u2013504 gamma, 504\u2013506 hadron, 572, 572 internal conversion, 505 Decay constants (), 492 Declination (1), 636, 636 Deep inelastic scattering, 593 Degeneracy, 250, 271, 316 of energy eigenvalues, 250 in fermion gases, 353\u2013354, 354 and molecular spectra, 388\u2013389 Degeneracy pressure, electron, 647 Degrees of freedom, 324, 346, 348 DeMarco, Brian, 354 Democritus, 115, 141, 561, 612 Density. See also Probability density (P(x, t)) comoving space, 660, 668 critical (nc), of population inversions, 400 critical mass ( 0), of the universe, 662\u2013663, 672 current (j), 424 energy, 122, 123 nuclear, 484, 489 number, 427, 428, 429t photon, 345 Density of states (g(E)), 317 and distribution functions, 333\u2013335 and electron gases, 428\u2013429 Density parameter (\"), 667, 667 Depleted regions, 452, 452 Deuterium, 165 fusion of, 526, 533, 536 Deuterons, 479t binding energy of, 83 Diamagnetism, 436 and superconductivity, 460 Dielectric breakdown, 442 Dielectric constant (* or ), 445t, 577 e ), 178 Differential cross section ( Diffraction. See also Interference; Superposition s>d\u00c6 of atoms, 193\u2013196, 194 of electrons, 186\u2013188\u2013192, 192, 481, 481 of x rays, 134, 134\u2013135 Index I-3 Digital versatile discs (DVDs), 404, 437 Diode lasers, 403, 456, 456 Diodes, 452\u2013456 light-emitting (LEDs), 455, 455\u2013456, 456, 457 solar cells, 455, 455 tunnel, 257, 260*, 453, 454 Zener, 453, 454 Dipole moment. See Electric dipole moment (p) Dipole-dipole bonds, 364, 375\u2013379, 377, 378 Dirac, Paul A. M., 262, 356, 676 and Fermi-Dirac distributions, 329 and magnetic monopoles, 608 and positrons, 90, 562\u2013563 and relativistic wave equation, 222, 288, 491 Direct interactions, 517 Disk galaxies, 654\u2013655, 655 Dispersion, 147, 147 of wave packets, 204, 204 Dispersive media, 199 Dissociation energy (Ed), 366, 368t, 417 Distortion of shapes, relativistic, 35, 35\u201336 Distribution functions. See Boltzmann distributions (fB(E)); Bose-Einstein distributions (fBE(E)); Energy density distributions; Fermi-Dirac distributions (fFD(E)); Maxwell distributions; Maxwell-Boltzmann distributions; Probability distribution functions (P(x)) DNA molecules, 377, 378 Donor levels, 447, 447 Doped semiconductors. See under Semiconductors Doping, 445, 447, 447\u2013448 Doppler effect, 41\u201345 applications of, 42 and gravitational redshift, 104 for light, 12 and rotation of the Milky Way, 638 for sound, 7, 41 transverse, 41, 44\u201345 and twin paradox, 47\u201348 Down (D), 570, 583, 583t Down-type quarks, 569t, 570 Drift velocity (vd), 423\u2013424 and Hall effect, 449, 449 Driplines, 495, 495 Drude, Paul, 422 Duane-Hunt rule, 136 Dulong, P., 327 Dulong-Petit law, 327, 347, 351, 426 DVDs. See Digital versatile discs (DVDs) Dwarf novae, 644 Dynamic states, 85 Weak interaction and beta decay, 501 charged vs. neutral, 574 mediation of, 574 Early type stars, 632 EAST. See Experimental Advanced Superconducting Tokamak (EAST) Ecliptic, 637, 637 Eddington, Arthur, 100, 101, 102, 110 EELS. See Electron", " energy loss spectroscopy (EELS) Effective mass (m*), 444\u2013445 Effective nuclear charge (Zeff), 299, 301 Ehrenfest, Paul, 57, 58, 307 Eigenfunctions (\u2019 Eigenstates, 226 Eigenvalues, 273 n(x)), 233, 609 angular momentum, 273 energy, 232, 273 Eightfold way, 591\u2013596, 592, 612 Einstein, Albert, 13, 127 and absorption, 393 and atomic spectra, 148, 159 and Bose-Einstein condensates, 341 I-4 Index Einstein, Albert, (continued) and Bose-Einstein distributions, 329 career of, 109, 141, 262, 307, 356, 407, 612 and cosmological constant, 661, 664 and de Broglie relations, 214 and general relativity, 97\u2013102, 105*, 108, 110, 114 and heat capacities, 346\u2013347, 356 and mass-energy equivalence, 81\u201382 and photoelectric effect, 125, 128\u2013129, 130, 136, 561 and relativistic mass, 69 and special relativity, 2, 3, 11\u201312, 15, 34 and stimulated emission, 394 and transverse Doppler effect, 45 and unified field theories, 570, 674 and wave equations, 202, 225 Einstein temperature (TE), 347, 347, 432\u2013433 Einstein\u2019s coefficients of absorption and emission, 393, 394 and lasers, 399 Einstein\u2019s postulates, 11\u201317 Elastic collisions and relativistic momentum, 66\u201367, 67 Elastic scattering, 390, 391, 516\u2013517, 517 Electric charge (e), 116, 118\u2013119, 576t of hadrons, 573t quantization of, 115\u2013119 of quarks, 569t, 583, 583t Electric dipole moment (p), 376, 376\u2013378 average, 378 average square, 378 and bond character, 374\u2013375 and rotational spectra, 380 e Electric field ( ), 423 Electric quadrupole moment (Q), 485, 485, 552 Electrical conduction, 423\u2013424, 431\u2013433. See also Conduction Electrodynamics, and special relativity, 11 Electromagnetic interaction, 509\u2013510, 511, 574, 576t. See also Fundamental interactions and cosmology, 671 Electromagnetic radiation detection of, 127 particle description of, 127 slowing, 304\u2013305, 305 transmission of, 89\u201390 wave description of, 127 Electron affinity, 364, 365t Electron capture, 502\u2013504, 552 Electron configurations, 297\u2013301 and bonding, 363 Electron degeneracy pressure, 647 Electron energy loss spectroscopy (EELS), 175\u2013176, 176 Electron gases, 352, 421, 422. See also Band theory of solids Electron holography, 462 Electron spin. See Spin angular momentum (S) Electron volts (eV), 78 Electrons (e), 116, 118, 479t, 561 Auger, 173 charge-to-mass ratio (e/m)of, 116#\u2013118 conversion, 505 diffraction of, 186, 188\u2013192 intrinsic angular momentum of, 285\u2013290 and photoelectric effect, 127\u2013132 relativistic mass of, 78, 79 speed of relativistic, 86\u201387 Electroweak interaction, 672 Electroweak theory, 570, 591, 602\u2013603 Elliott, Steven, 503 Elliptical galaxies, 654\u2013655, 655 Elsasser, Walter, 188, 192 Emission. See also Radiation field, 257 spontaneous, 393 stimulated, 391, 394\u2013396 Emission spectra. See Spectra e Emissivity ( ), 120 Emitters, transistor, 456, 456 Endothermic reactions, 518 Energy bands. See Band theory of solids Energy density distributions, 122 and Planck\u2019s law, 123 Energy (E). See also Kinetic energy (Ek); Potential energy (U) binding (Eb), 81\u201384, 83 characteristic rotational (E0r), 380, 389t cohesive, 417, 420 conservation of, 76\u201381, 517\u2013520, 663, 666 density, blackbody, 122 dissociation (Ed), 366, 368t, 417 Fermi (EF), 352, 426\u2013429, 429t first ionization, 298, 300 lattice, 417 Lorentz transformation of, 73\u201376, 74, 85 quantization of, 124\u2013125, 129, 225, 230, 238, 278\u2013279 relativistic (E), 70\u201381, 563 rest (mc2), 72\u201373, 84t, 86, 90 in Schr\u00f6dinger equation, 226\u2013227 zero point, 208\u2013209 Energy eigenvalues, 232 degenerate, 250 Energy gaps", " (Eg), 440, 440\u2013441 and conductivity, 443\u2013444, 445t superconducting (Eg), 464\u2013465, 465 Energy levels for diatomic molecules, 379\u2013384 rotational, 379\u2013382, 380 vibrational, 382, 382\u2013384 Energy-level diagrams, 163 and alpha decay, 499 for atomic spectra, 302 for atoms, 297, 311 for the baryon octet, 592 for fine-structure splitting, 294 for helium-neon lasers, 402 for hydrogen, 163, 278, 280 for infinite square wells, 232, 238 for nuclei, 515 for quantum wells, 242 for simple harmonic oscillators, 249 Entrance channels, 522 Entropy, 319* Equilibrium separations (r0), 366, 368t and rotational spectra, 380, 381\u2013382, 389t in solids, 416\u2013418, 416t Equinoxes, 637, 637, 676 Equipartition theorem, 324* Equivalence, principle of, 98\u201399, 99 and gravitational redshift, 103\u2013104 Esaki, Leo, 263, 471 Escape velocity (ve) and planetary atmospheres, 323\u2013324, 679 and Schwarzschild radius, 106, 650 Estermann, I., 193 Ether, 6, 7\u201311 Ether drag, 61 Euler, Leonhard, 407 European Organization for Nuclear Research (CERN), 91, 336, 563, 565, 575 European Southern Observatory, 662 Evaporation, 322 Event horizons, 673 Events, in special relativity, 13\u201314, 23, 32, 50 Excited states (En), 175, 223 and atomic spectra, 301\u2013303, 302 and fission, 528 of hadrons, 572\u2013573 of the hydrogen atom, 281\u2013283, 318 of molecules, 368 of nuclei, 504\u2013506, 521\u2013523, 522 Exclusion principle. See Pauli exclusion principle Exclusion-principle repulsion, 365, 367, 416 Exit channels, 522 Exoplanets, 630 Exothermic reactions, 517 Expectation values \u2039x\u203a, 242\u2013243 in kinetic theory, 319\u2013320 Experimental Advanced Superconducting Tokamak (EAST), 532 Experiments. See Burst and Transient Source Experiment (BATSE); Davisson-Germer experiment; Franck-Hertz experiment; Gedanken experiments; Ice Cube experiment; Michelson-Morley experiment; Millikan oil-drop experiment; Stern-Gerlach experiment; Thomson experiment Exploring. See also Classical Concept Reviews; MORE alpha decay, 258\u2013259 atomic clocks, 260 calibration spacetime axes, 28\u201329 celestial sphere, 636, 636\u2013637 deflection of light in a gravitational field, 100\u2013103, 101, 102 extraterrestrial life, 630 fluxoids, 462, 462 frozen light, 304\u2013305, 305 gamma-ray microscopes, 206\u2013208, 207 gravitational redshift, 103, 103\u2013105, 104 Hall effect, 449, 449\u2013452 interaction strengths, 577\u2013578 Josephson junctions, 466\u2013468, 467 liquid helium, 336\u2013339, 338, 339 neutrino oscillations and mass, 609\u2013610 other bonding mechanisms, 375\u2013379 parity, 250 Planck units, 673\u2013674 probability density of exchange mesons, 512\u2013513, 513 proton spin, 595\u2013596 Rydberg atoms, 168, 168 spintronics, 437, 437 Stern-Gerlach experiment, 288\u2013290, 289, 290 superluminal speeds, 52\u201355, 54 transverse Doppler effect, 44\u201345 Fabry-P\u00e9rot cavities, 106, 107 Face-centered cubic (fcc) crystal symmetry, 415, 415, 420 Face-centered cubic structure, 134 Faraday, Michael, 115\u2013116 faradays (F), 116 Faraday\u2019s law, 115\u2013116 and flux quantization, 462 and superconductivity, 459 Fermi, Enrico, 329, 329, 356, 500, 530, 531, 553, 612 Fermi energy (EF), 352 and band theory of solids, 442, 444 and free-electron gas in metals, 426\u2013429, 429t of nucleons, 515 Fermi speed (uF), 431 Fermi temperature (TF), 430 Fermi-Dirac distributions (fFD(E)), 316, 328\u2013329 and band theory of solids, 444, 444t and conduction, 425, 430, 431 and fermion gases, 352, 352\u2013353, 353 vs. other distributions, 329\u2013333, 331 Fermi-Dirac particles. See Fermions Fermilab, 336, 598 Fermion gases", ", 351\u2013354, 354 and lasers, 403 neutron stars as, 648 Fermions, 329\u2013332 Ferrimagnetism, 436 Ferromagnetism, 434, 436 Feynman, Richard P., 213*, 215, 566, 612 and quantum electrodynamics, 563 Feynman diagrams, 566\u2013568, 567 examples of, 510, 511, 574, 578, 579, 600, 601 Field emission, 257 Filaments, 629, 629 Fine structure, 167, 285, 288 Fine-structure constant (), 166\u2013168, 167, 571, g factor (g), 287\u2013288 and magnetism, 435 577 Fine-structure splitting, 167, 293\u2013295, 294 and emission spectra, 385, 385 Finite square wells, 238\u2013242, 239 and alpha decay, 258\u2013259, 259 and covalent bonding, 369, 369 graphical solution of, 241* and Kronig-Penney model, 438, 438, 439 and quantum wells, 242 and strong interaction, 507\u2013508, 508 and vibrational energy levels, 382, 382\u2013383, 383 First ionization energies, 298, 300 First-order perturbation theory, 298 Fissile nuclides, 528 Fission, 478, 527, 527\u2013531 and energy conversion, 81 FitzGerald, George F., 34 Flash memory, 437 Flat universe, 666 Flavor charge, 568, 569t, 574, 599, 609 Fluorescence, 391, 393 Flux lines, 628 Flux quantization, 462 Flux tubes, 461, 461, 462 ), 462, 462 Fluxoids ( Force carriers. See Mediation of forces Force constants (K) 0 and vibrational energy levels, 383, 384, 389 Forces (F). See also Electromagnetic interaction; Fundamental interactions; Gravitational interaction; Strong interaction; Weak interaction coupling constants for, 570\u2013571 inverse-square, 279 relativistic, 70\u201371 saturated, 488, 507, 509 short-range, 509 Forward biasing, 452, 452 Fountain effect, 338, 339 Fourier analysis, 197\u2013198# Four-vectors, 85 FQHE. See Fractional quantized Hall effect (FQHE) Fractional quantized Hall effect (FQHE), 451 Frames of reference. See Reference frames Franck, James, 169, 174\u2013175, 179, 188, 192, 591 Franck-Hertz experiment, 174\u2013176 Fraunhofer, Joseph von, 147, 178, 393, 632 Fraunhofer D lines, 393, 490, 491. See also Sodium (Na) Free-electron lasers, 404 Free-electron theory, 426\u2013430 failures of, 351 and Fermi temperature, 430 one-dimensional, 426\u2013428 three-dimensional, 428\u2013429 Frequency (f) control of, 201 Debye (fD), 348, 348, 351 proper (f0), 42 quantization of, 230 Friction, 378 Fringes, interference, 9#, 10. See also Interference Fuller, R. Buckminster, 407 Fullerenes, 375*, 407, 420, 420\u2013421 Fundamental interactions, 570\u2013580, 576t. See also Electromagnetic interaction; Gravitational interaction; Strong interaction; Weak interaction unification of, 671 Fundamental particles, 479 Fusion, 478, 531\u2013536. See also Proton-proton cycle during the early universe, 556, 668, 668, 672 and metallic hydrogen, 421 as stellar energy source, 624, 642\u2013643 and supernovae, 645\u2013646, 648 Galactic clusters, 632 Galaxies, 630, 653\u2013661, 676. See also Milky Way classification of, 654\u2013661, 656 and interstellar medium, 653\u2013654 quiet vs. active, 656, 656\u2013657, 657 2dF Galaxy Redshift Survey, 661, 661 Galilean transformation of coordinates, 5\u20136#, 11 Galilei, Galileo, 1, 3, 4, 646 Gamma decay, 504\u2013506 Gamma rays, 91, 477 Gamma-ray bursts (GRBs), 652, 652 Gamma-ray microscopes, 206\u2013208, 207 Gamow, George, 217, 258, 299, 477, 676 Gaseous nebulae, 654 Gases electron, 352, 421, 422 fermion, 351\u2013354, 354 free-electron, 426\u2013430 heat capacities (CV) of, 325\u2013327, 325t, 348\u2013351 kinetic theory of, 115, 316\u2013328 photon, 344\u2013351 Gassendi, Pierre, 115 Gauge theories,", " 580, 609, 613 Gedanken experiments, 109 flash of light inside a sphere, 58 gamma-ray microscopes, 206\u2013208, 207 headlight effect, 51, 51\u201352 knowledge creation paradox, 54\u201355, 55 lever paradox, 80, 80\u201381 lightning strikes train, 15, 15\u201317, 16, 28, 28\u201329 Olbers\u2019 paradox, 665 pole and barn paradox, 48\u201351, 49 pregnant elephant, 32\u201333 scissors paradox, 52\u201354, 54 twin paradox, 45\u201348*, 46 Geiger, Hans W., 151 and alpha decay, 496 and nuclear charge, 172 and Rutherford scattering, 151\u2013153, 155\u2013156, 159, 477 and size of nucleus, 157 Geiger-Nuttall rule, 496, 496 Gell-Mann, Murray, 583, 591\u2013593, 593, 596, 612 General relativity, 47, 97\u2013108 and cosmology, 665 Generations of leptons, 568, 569t, 599 Geochemistry, 545\u2013548 Geometry of space, 39, 57, 84, 666 Geosynchronous satellites, 5, 112 Gerlach, Walther, 286, 289\u2013290, 307 Germer, Lester H., 186, 188\u2013191, 191, 193, 214 Giaever, I., 471 Giant magnetoresistance (GMR), 437 Glashow, Sheldon L., 570, 597, 598 Glass, 414 Global positioning systems (GPS), 97, 98, 110 Globular clusters, 632, 632 and structure of the Milky Way, 634 Gluon-gluon loops, 600, 600 Gluons, 90, 511, 568 and force mediators, 571, 576t, 603 and quantum chromodynamics, 599\u2013600, 600 GMR. See Giant magnetoresistance (GMR) Goeppert-Mayer, Maria, 552; see also Mayer, M. Gordon, Walter, 512 Goudsmit, Samuel A., 285, 307 GPS. See Global positioning systems (GPS) Grand unification theories (GUTs), 605\u2013610 and the early universe, 671 Graphite, 420, 420 Gravitational blueshift, 104\u2013105 Gravitational charge, 576, 576t. See also Mass (m) Gravitational interaction, 576, 576t. See also Fundamental interactions and astrophysics, 623, 624\u2013625 Index I-5 and cosmology, 662\u2013664, 671 deflection of light by, 100\u2013103, 101, 102 quantum, 609, 671, 674 transmission of, 90 Gravitational length contraction, 101 Gravitational lensing, 100, 102, 102, 104 Gravitational redshift, 103, 103\u2013105, 104, 104, 650 Gravitational time dilation, 101, 105* Gravitational waves, 106, 106\u2013108, 576 Gravitons, 57, 90, 576, 576t GRBs. See Gamma-ray bursts (GRBs) Greenberg, D. W., 596 Ground states (E0), 163, 175, 233 of hadrons, 572 of the hydrogen atom, 281\u2013283 of molecules, 368 of nuclei, 480\u2013492 and the periodic table, 297\u2013301 Group theory, 592 Group velocity (vg), 197, 199 for particle waves, 202 Gurney, R. W., 258, 477 GUTs. See Grand unification theories (GUTs) Gyromagnetic ratio (g), 287\u2013288 U h-bar ( ), 161. See also Planck\u2019s constant (h) H-R diagrams. See Hertzsprung-Russell (H-R) diagrams Hadronic force. See Strong interaction Hadrons, 568\u2013569, 573t quantum numbers of, 586t quark model of, 591\u2013599 and the strong interaction, 571\u2013573 Hahn, Otto, 527, 530, 553 Hale, George, 110 Half-life ( ), 493, 543t, 547t t1>2 and charge-carriers, 449 quantum, 450\u2013451, 451 spin, 437, 452 Hall effect, 449, 449\u2013452 Hall resistance (RH), 450, 451 Halley, Edmund, 665 Hamiltonian operators (Hop), 245, 245t, 370 and symmetry breaking, 602 Hanford Observatory, 107 Hard core, 509 Hard superconductors, 460, 460 Harmonic waves, 196, 222, 224 Hartmann, J. F., 653 Hawking, Stephen, 356 Hawking radiation, 680 Heat capacities (CV), 324\u2013328 for gases, 325\u2013327, 325t, 348\u2013351 for metals, 422, 426, 433\u2013434 and phase transitions, 337, 337 for sol", "ids, 327, 327\u2013328, 347\u2013348, 348 Heisenberg, Werner K., 205, 206, 215, 221, 307, 407 Helicity, 584 Helium (He) formation of, 668, 668 in interstellar medium, 653 liquid, 336\u2013339, 338, 339 wave functions for, 297\u2013298 Helium-neon lasers, 401, 401\u2013402 Helix nebula, 643 Helmholtz, Hermann von, 109, 141, 580 Hermite polynomials, 248 Hermitian operators, 612 Herriott, D. R., 401 Hertz, Gustav L., 169, 174\u2013175, 179, 591 Hertz, Heinrich R., 127, 132, 141, 179 Hertzsprung-Russell (H-R) diagrams, 641, 641\u2013643 Hess, Victor, 552 Heteropolar (heteronuclear) molecules, 407 Hexagonal close-packed (hcc) crystal symmetry, 419, 419 I-6 Index Higgs bosons, 568, 581, 603 Higgs field, 603 High-temperature superconductivity, 460, 466, 466t Hipparchus, 632, 676 Hofstadter, Robert, 481, 552 Holes, 443, 448 Holography, electron, 462 Homopolar (homonuclear) molecules, 407 Hooke, Robert, 115 Horizontal branch, in stellar evolution, 643 Hounsfield, Godfrey, 541, 553 Hoyle, Fred, 676 Hubble, Edwin P., 43, 654, 657, 676 Hubble constant (H0), 657\u2013658, 661 Hubble Space Telescope, 620, 632, 646, 648, 650 Hubble time, 658 Hubble\u2019s law, 657, 657\u2013661 Hulse, R. A., 106 Hybridization of orbitals, 420, 470 Hydrogen bonds, 377, 378 Hydrogen (H) atoms binding energy of, 83 Bohr model of, 159\u2013169, 176* energy-level diagrams for, 163, 278, 280 excited states of, 281\u2013283, 318 fine-structure splitting in, 293\u2013295, 294 probability density in, 282\u2013285, 283, 284, 285 quantization of angular momentum in, 275\u2013277 quantization of energy in, 278\u2013279 Schr\u00f6dinger equation for, 272\u2013274 size of, 209\u2013210 spectra of, 149, 150 wave functions for, 281\u2013285, 290 Hydrogen (H2) molecules escape of, from Earth\u2019s atmosphere, 323\u2013324 heat capacity of, 326, 326 in interstellar medium, 653 wave functions for, 370\u2013374, 371 Hypercharge (Y), 585\u2013586, 586t, 587 and group theory, 592 Hyperfine splitting, 314 and interstellar dust, 653, 653\u2013654 and masers, 396, 397 Hyperfine structure, 304, 479, 490 Ice, dipole-dipole bonds in, 377, 377 Ice Cube experiment, 129 Ideal blackbodies, 120 Imaginary numbers. See Complex numbers Impact parameters (b), 154, 154 Impurity semiconductors. See under Semiconductors Indistinguishable particles, 295\u2013296, 328, 329, 330, 332 Inelastic collisions and relativistic energy, 76, 76\u201377 Inelastic scattering, 390, 391 Inertial confinement, 533 Inertial reference frames, 4, 4\u20136# and simultaneity, 14\u201317 and special relativity, 12\u201314, 14, 47 Infinite square wells, 229\u2013238, 230 complete wave function for, 235\u2013238 and correspondence principle, 233\u2013234, 236 energy levels of, 231\u2013232, 232 and free-electron gas in metals, 426, 427 and line of stability, 487 minimum energy of, 208\u2013209 in three dimensions, 270\u2013271 Inflation, 669 Insulators, and band theory of solids, 442, 442 Integral quantized Hall effect (IQHE), 451 Intensity (I), 407, 521 of electron diffraction, 190, 191 and photoelectric effect, 131\u2013132 Interaction times, 571 Interactions. See also Fundamental interactions and Feynman diagrams, 566 Interference. See also Diffraction; Superposition double-slit, 202\u2013204, 203, 213* by electrons, 186, 188\u2013189, 203 fringes produced by, 9#, 10, 203 and linear equations, 223 quantum, 304\u2013305, 305 Interferometers, Michelson, 7, 9, 9\u201311, 10 Internal conversion, 505 Internal quantum numbers, 583, 583t International Thermonuclear Experimental Reactor (ITER), 532, 533 Interstellar dust, 653\u2013654 Interstellar medium (ISM), 653\u2013", "654 Intrinsic semiconductors. See under Semiconductors Invariance, 4 and conservation laws, 66, 73, 580 and gauge theories, 613 of mass, 84\u201397 TCP, 589\u2013590 Inverse beta decay, 646, 648 Ionic bonds, 364\u2013369 vs. covalent bonds, 374\u2013375 Ionic solids, 415\u2013421 Ionization energy, 163 and ionic bonds, 364, 364\u2013365, 365t Ionizing radiation, 536*, 549* IQHE. See Integral quantized Hall effect (IQHE) Irregular galaxies, 654\u2013655, 655 Island of stability, 488, 515\u2013516 ISM. See Interstellar medium (ISM) Isobars, 480 Isomers, 505 Isospin (I), 585, 586t, 587 and group theory, 592 weak (Tz), 568, 570 Isotones, 480, 513 Isotope effect, 463 Isotopes, 165, 471, 480 ITER. See International Thermonuclear Experimental Reactor (ITER) J/\u2019 puzzle, 597\u2013599 Javan, Ali, 401 Jefferson National Accelerator Facility, 595 Jensen, Johannes Hans Daniel, 514, 515, 552 Jin, Deborah, 354 Joos, Georg, 11 Josephson, Brian D., 467, 471 Josephson effect ac, 467\u2013468 dc, 467 Josephson junctions, 466\u2013468, 467 Joule, James, 109, 580 Junction lasers, 403 Junctions Josephson, 466\u2013468, 467 semiconductor, 452, 452 K series, 170, 171 K shell, 171 Kamerlingh Onnes, Heike, 335, 336, 336, 356, 458 Kant, Emmanuel, 654 Kaons (K0), 588, 590 Keesom, Willem H., 335, 337 Kelvin, William Thompson, Lord, 2, 307, 624 Kepler, Johannes, 646 Ketterle, Wolfgang, 343, 403 Kinematic states, 85 Kinetic energy (Ek) Maxwell distributions of, 323, 323\u2013324 minimum, 208\u2013209 negative, 240, 255 in nuclear reactions, 522 and photoelectric effect, 128 relativistic, 71\u201372, 72 of rotation, 379 vs. potential energy, 81\u201383 Kinetic energy operators, 245t Kinetic theory, 315#, 316#, 319#, 328# and blackbody radiation, 119\u2013122 of gases, 115, 316\u2013328 and Planck\u2019s law, 123 Klein, Oskar, 512 Klein-Gordon relativistic wave equation, 512, 564, 577 Klitzing, Klaus von, 450\u2013451, 471 Kronig-Penney model, 438, 438\u2013441, 439 K\u00fcndig, Walter, 45 Kusch, P., 322 L series, 170, 171 Laboratory frames of reference, 518, 519, 522 Lagrangian points, 644, 644 Laguerre polynomials, 279, 279t, 281, 284 Lamb, W., 307 Lamb shift, 307 Lambda points, 335\u2013336, 337, 337, 341 - Cold Dark Matter (-CDM), 661 Land\u00e9, Alfred, 307 Land\u00e9 factor (gN), 490 Langevin, Paul, 214 Laplace, Pierre, 106 Large Hadron Collider (LHC), 563, 568, 603, 605, 639 Larmor theorem, 286 Laser Interferometer Gravitational-Wave Observatory (LIGO), 106\u2013107, 108 Lasers, 396\u2013405 applications of, 402\u2013405 atomic, 343, 343 continuous wave, 400\u2013401 diode, 403, 456, 456 helium-neon, 401, 401\u2013402 other types of, 403\u2013404 ruby, 397, 397\u2013401, 398 tunable dye, 168, 490 Late type stars, 632 Latitude, 636, 636 Lattice energy, 417 Laue, Max von, 80, 134 Laue patterns, 134, 192, 194, 418 Laughlin, R. B., 471 Law of atmospheres, 317\u2013318 Law of inertia, 4 Lawrence, Ernest O., 516 Lawrence Livermore Laboratory, 564, 584 Laws of motion, Newtonian, 4\u20136 Lawson, J. D., 532 Lawson\u2019s criterion, 532 LEDs. See Light-emitting diodes (LEDs) Lee, David M., 338 Lee, T. D., 588, 590 Left-handed coordinate systems, 589, 589 Legendre polynomials, 274 Lenard, Philipp, 127, 128 Lenard-Jones potential, 411 Length, proper (Lp), 33, 38, 85 Length contraction, 33, 33\u201336 gravitational", ", 101 Lepton era, 672 Lepton number conservation of, 581\u2013582 nonconservation of, 607 Leptons, 500, 568\u2013570, 569t, 598, 599, 612 and weak interaction, 574 Leptoquarks, 606, 671 Leucippus, 115 LHC. See Large Hadron Collider (LHC) Libby, Willard F., 553 Lie, S., 592 Life, extraterrestrial, 630 Lifetime (), 210, 493 of excited energy states, 393 and force ranges, 572 of protons, 606\u2013607 for spontaneous emission (ts), 399\u2013400 of strange particles, 584 and tunneling, 259 Light. See Electromagnetic radiation; Speed of light (c) Light clocks, 29, 30 Light curves, 640 Light-emitting diodes (LEDs), 455, 455\u2013456, 456, 457 Lightlike spacetime intervals, 37, 39 LIGO. See Laser Interferometer Gravitational- Wave Observatory (LIGO) Magnetoresistance, giant (GMR), 437 Magnetoresistive random access memory (MRAM), 437 Magnitude, stellar absolute (M), 633 apparent (m), 632\u2013633, 677 Maiman, Theodore, 397 Main sequence dwarfs, 641, 641 Main sequence stars, 641, 641 Majorana neutrinos, 568, 582 Marsden, Ernest Limbs, solar, 621, 622 Lindblad, Bertil, 638 Line of stability, 486, 487 and fission, 528, 529 Line spectra, 147\u2013148, 148 Linear combinations, 223, 262 Liquid lasers, 404 Liquid-drop model, 489*, 513 and fission, 527, 527 Liquids helium, 336\u2013339, 338, 339 structure of, 413 surface tension of, 378 Livingston, M. S., 516 Livingston Observatory, 106 Local clocks, 15 London, Fritz, 335, 378 London dispersion forces, 378. See also van der Waals attraction Longitude, 636, 636 Lorentz, Hendrik A., 34, 102, 214 and conduction, 422 and coordinate transformations, 18 and electrons, 118, 141 and Michelson-Morley experiment, 57, 61 and Zeeman effect, 295, 307 Lorentz transformation, 18 and dynamics, 65, 73\u201376 of four-vectors, 85 of mass-energy, 88 and nuclear reactions, 519 of space and time coordinates, 17\u201329 Lorentz-FitzGerald contraction, 34 Luminosity (L), 620, 639 of galaxies, 656 and Hertzsprung-Russell diagrams, 641, 641 of quasars, 660 Lyman, Theodore, 150, 164 Lyman series, 163, 164 Macroscopic quantum wave functions, 342, 342 Madelung constant (), 368, 415, 416t, 418, 419 Magic numbers, 488, 513\u2013514 Magnetic confinement, 532 Magnetic fields (B) charged particles in, 116\u2013117 critical (Bc), 458, 458t, 459, 460, 461 and fine-structure splitting, 293\u2013295 inhomogeneous, 289 at the nucleus, 490t relativistic electrons in, 94\u201395, 95 solar, 627, 627\u2013629, 628 Magnetic moment (), 286\u2013290 nuclear, 489\u2013492 quantization of, 287 Magnetic monopoles, 608, 669 Magnetic quantum number (m), 276, 280 Magnetic resonance imaging (MRI), 491, 540 and liquid helium, 336 and superconductivity, 458, 562 Magnetic susceptibility ((), 435 Magnetic traps, 342 Magnetic tunnel junctions, 437, 437 Magnetism, 287, 434\u2013437 Magnetization (M), 476 Magnetons Bohr ( nuclear ( B), 182 N), 314 and nuclear charge, 172 and Rutherford scattering, 151\u2013153, 155\u2013156, 159, 477 and size of nucleus, 157 Masers, 396, 397. See also Lasers Mass (m) effective (m*), 444\u2013445 as gravitational charge, 576, 576t gravitational vs. inertial, 98\u201399, 99 and Higgs boson, 603 invariance of, 84\u201397 of the Milky Way, 638\u2013639 of neutrinos, 500, 536, 581\u2013582, 607\u2013608, 609\u2013610, 626 of nuclei, 488\u2013489 reduced (), 165\u2013166 relativistic (m(u)), 67\u201369, 68 rest (m), 78 stellar, 623, 641\u2013642 units of, 78, 83, 380, 673 Massless particles, 89\u201390. See also Gluons; Gravitons; Photons Mather, John C., 669, 6", "76 Matter waves, 187\u2013196 Matter-antimatter asymmetry and CP violations, 590 Max Planck Institute, 650 Maxwell, James Clerk, 1, 6, 316, 319 Maxwell distributions and Bose-Einstein condensates, 342 and electromagnetism, 570 and fusion, 532, 536, 625 of kinetic energy, 323, 323\u2013324 of molecular speeds, 319\u2013322, 320, 320, 359 Maxwell-Boltzmann distributions and diode currents, 453 and Planck\u2019s law, 123\u2013124 Maxwell\u2019s equations, 6, 6, 11 Mayer, M., 514, 515 Mean free path (), 424, 432 Mediation of forces, 570\u2013580 and Feynman diagrams, 566\u2013567, 567 and the uncertainty principle, 510 Medicine nuclear, 478, 529, 540\u2013542, 612 and x rays, 133, 136 Meissner, H. Walther, 459 Meissner effect, 459, 459\u2013462, 460, 461 Mendeleev, Dmitri, 172, 592 Mercury [element], spectra of, 149, 175 Mercury [planet], precession of orbit of, 100, 105* Meson fields, 510 Meson octet (nonet), 587, 591, 595 Mesons, 510, 569, 573t, 587, 594t as force mediators, 604 probability density of, 512\u2013513, 513 supermultiplets of, 598 Metallic bonds, 364, 375, 415, 421\u2013422, 422 Metals, free-electron gases in, 351, 426\u2013430 Metastable states, 505 and lasers, 397, 397, 401 Michelson, Albert A., 1, 7, 7, 9, 11, 57, 127 Michelson interferometer, 7, 9, 9\u201311, 10 and gravity waves, 106\u2013107, 107 Michelson-Morley experiment, 7\u201311* Microscopes Index I-7 gamma-ray, 206\u2013208, 207 photoelectric-effect, 132 resolving power of, 215 scanning tunneling (STMs), 257, 257\u2013258 Mikheyev, S., 608 Milky Way, 630, 634, 635. See also Galaxies mass of, 638\u2013639 structure of, 634, 634\u2013635, 654 Miller, R. C., 322 Millikan, Robert A., 127, 141, 676 and electron charge, 116, 118\u2013119 and photoelectric effect, 130, 561 Millikan oil-drop experiment, 118\u2013119#, 119 Minkowski, Hermann, 109 Mirror nuclides, 480, 481 Missing mass. See Dark matter Mitchell, John, 676 Molecular orbitals. See Orbitals Molecules, 363\u2013412 and covalent bonds, 369\u2013375 energy levels and spectra of, 379\u2013390 and ionic bonds, 364\u2013369 and lasers and masers, 396\u2013405 and other bonds, 375\u2013379 and scattering, absorption, and stimulated emission, 390\u2013396 Moment of inertia (I), 379 and rotational spectra, 380, 387, 389 Momentum operators (pop), 244, 245t Momentum (p) of electromagnetic radiation, 137 Lorentz transformation of, 73\u201376, 74 relativistic (p), 66\u201370, 67, 68 MORE. See also Classical Concept Reviews; Exploring alpha-decay energetics, 498* Bohr atomic model, 176* conservation laws, 583* delay of light in a gravitational field, 105* derivation of Compton\u2019s equation, 138* double-slit interference, 213* energy bands, 438*, 445* entropy, 319* equipartition theorem, 324* graphical solution of the finite square well, 241* ionizing radiation, 536* liquid-drop model, 489* Michelson-Morley experiment, 11* M\u00f6ssbauer effect, 505* multielectron atoms, 303*, 401 nuclear power, 530* other covalent bonds, 375*, 420* perihelion of Mercury\u2019s orbit, 105* radiation dosages, 549* radioactive decay, 495* relativity of simultaneity, 48* resonances and excited states, 591* Rutherford scattering, 156* Schr\u00f6dinger\u2019s differential equation trick, 249* shell model, 516* spectroscopic notation, 292* temperature, 319* theories of everything, 610* thermal conduction, 434* transcendental equations, 241* transistors, 457* transitions between energy states, 246*, 393*, 395* tunnel diodes, 260* twin paradox, 48* wave-particle duality, 213* Weizs\u00e4cher formula, 489* Zeeman effect, 303* Morley, Edward W., 7, 9, 11, 57 Mose", "ley, Henry G.-J., 169, 169\u2013170, 172\u2013173, 179, 181, 478, 542 Moseley plots, 170, 170 I-8 Index M\u00f6ssbauer, Rudolf Ludwig, 505*, 552 M\u00f6ssbauer effect, 211, 505*, 557 measuring gravitational red- and blueshift with, 105, 505 measuring natural line widths with, 211 measuring transverse Doppler effect with, 45, 505 Most probable speed (vm), 320, 320 Mott, N. F., 437 Mourou, G\u00e9rhard, 403 MRAM. See Magnetoresistive random access memory (MRAM) MRI. See Magnetic resonance imaging (MRI) MSW effect, 608 M\u00fcller, Karl A., 466 Multielectron atoms, 303*, 401 Multiplets, charge, 572, 585\u2013586, 606 Muons (mu () mesons), 561 decay of, 36, 36\u201337 mass of, 78 N galaxies, 656 npn transistors, 456, 456 n-type semiconductors, 447, 447 NAA. See Neutron activation analysis (NAA) Nanostructures, 242, 420, 421 Natural line width (% 0), 210\u2013211 and gamma decay, 504\u2013505 and nuclear resonances, 522 N\u00e9el temperature (TN), 436 Ne\u2019eman, Yuval, 591 Neutral current, 574 Neutral weak force, 574 Neutralinos, 639 Neutrinos (v), 90 as dark matter, 638 and density parameter, 667 discovery of, 500, 562 Majorana, 568, 582 mass of, 500, 536, 581\u2013582, 607\u2013608, 609\u2013610, 626 oscillations of, 582, 607\u2013608, 609\u2013610, 626 in proton-proton cycle, 536 and solar-neutrino problem, 536, 607\u2013608, 625\u2013626, 626 from supernovae, 646 types of, 500, 568, 569t Neutron activation analysis (NAA), 478, 524, 537, 537\u2013539 Neutron capture, 524, 524\u2013525 cross section for, 514, 514, 524 Neutron number, 479 Neutron stars, 352, 484, 648\u2013650, 649 Neutrons (n), 479t decay of, 579 diffraction of, 193, 194 discovery of, 477, 479, 561 reactions of, 524\u2013525 thermal, 524 Newton, Isaac, 1, 3, 4, 147, 356 Newtonian physics, 4\u20136, 70\u201371. See also Classical physics Nichols, Ernest F., 137 Nicholson, J. W., 161 Nishijima, K., 583 NMR. See Nuclear magnetic resonance (NMR) Noether, Emmy, 580, 612 Nondispersive media, 199 Nonpolar molecules, 377\u2013379, 378 Normalization of Bose-Einstein distribution, 339 and hydrogen atom wave functions, 282 of Maxwell-Boltzmann distribution, 124 of probability amplitude, 225, 227, 228, 232, 240, 512 Novae, 60, 644, 644 Nuclear binding energy (B), 488\u2013489, 489 and driplines, 495, 495 and shell model, 513, 514 Nuclear exchange force, 509\u2013511 Nuclear force. See Strong interaction Nuclear magnetic moment, 489\u2013492 Nuclear magnetic resonance (NMR), 470, 539\u2013540. See also Magnetic resonance imaging (MRI) Nuclear magneton ( Nuclear physics, 477\u2013560 N), 314 applications of, 537\u2013549 and composition of the nucleus, 478\u2013479 and fission and fusion, 526\u2013536 and ground-state properties of nuclei, 480\u2013492 history of, 477\u2013478 and nuclear decay, 495\u2013506 and nuclear reactions, 516\u2013525 and radioactivity, 492\u2013495 and the shell model, 513\u2013516 and the strong interaction, 506\u2013513 Nuclear power, 478, 530* Nuclear reactions, 516\u2013525 Nuclear reactors fission, 81, 524, 525, 530, 531 fusion, 534 natural, 553 Nuclear spectra, 480, 508, 523 Nuclear spin angular momentum (I), 304, 489 Nuclear weaponry, 478, 530 Nucleons, 479, 509 mass differences between, 526, 526 Nucleosynthesis era, 672 Nucleus composition of, 478\u2013479 compound, 517, 521\u2013522, 522 density of, 484, 489 discovery of, 153\u2013156 excited states of, 504\u2013506, 521\u2013523, 522 ground state of", ", 480\u2013492 shape of, 485 size of, 156\u2013159, 480\u2013484 stability of, 486, 486\u2013488 structure of, 191 Nuclides, 480 fissile, 528 mirror, 480, 481 Number density, 427, 428, 429t Nurmia, M., 497 Nuttall, John Mitchell, 496 Observers in special relativity, 13\u201314 and wave-particle duality, 212 Ochsenfeld, Robert Ohm\u2019s law, 422, 424, 425 and Josephson junctions, 467 Oil-drop experiment. See Millikan oil-drop experiment Olbers, Wilhelm, 665 Olbers\u2019 paradox, 665 OLEDs. See Organic semiconductor light-emitting diodes (OLEDs) Oort, Jan, 638 Open clusters, 632 Open universe, 666 Operators, 243\u2013246, 245t Oppenheimer, J. Robert, 105 Optical barrier penetration, 257 Optical pumping, 396 Optical traps, 403 Optical tweezers, 403 Orbital quantum number. See Angular momentum quantum number ( ) Orbitals, 363, 373, 407 bonding vs. antibonding, 373, 373 hybridization of, 420, 470 Orbits circular atomic, 159\u2013160, 169 elliptical atomic, 166\u2013167, 279 Ordinary spiral galaxies, 654\u2013655, 655 Organic semiconductor light-emitting diodes (OLEDs), 455, 456 Oscilloscopes, 117 Osheroff, Douglas D., 338 \u00d6tzi the Iceman, 548, 548 P branches, 407 pn junctions, 452, 452 pnp transistors, 456, 456 p-type semiconductors, 447, 448 Pair production of particles, 90\u201393, 91, 92, 564, 564 Feynman diagram for, 567, 567 Paradoxes in relativity. See Gedanken experiments Parallax angles ( ), 640, 640 Paramagnetism, 435, 476 Parity operations, 250, 588 Parity (P), 588\u2013589, 589 nonconservation of, 588 Parsecs (pc), 633, 640, 640 Partial cross sections, 521 Particle physics, 561\u2013618 basic concepts for, 562\u2013570 conservation laws and symmetries in, 580\u2013591 fundamental interactions and force carriers in, 570\u2013580 and grand unified theories, 605\u2013610 history of, 561\u2013562 Standard Model of, 591\u2013604 Particle waves, 185\u2013220, 187 and the de Broglie hypothesis, 185\u2013187 measurements of, 187\u2013196 probabilistic interpretation of, 202\u2013205 and the uncertainty principle, 205\u2013211 and wave packets, 196\u2013202 and wave-particle duality, 212\u2013213 Particle-in-a-box. See Infinite square wells Particle-induced x-ray analysis (PIXE), 548\u2013549, 549 Particles. See also Alpha particles (); Antiparticles; Beta particles (); Virtual particles annihilation of, 90\u201393, 91, 565\u2013566, 567 indistinguishable, 295\u2013296, 328, 329, 330, 332 massless, 89\u201390 pair production of, 90\u201393, 91, 92, 564, 564, 567 Paschen, Friedrich, 150, 164 Paschen series, 163, 164 Pauli, Wolfgang, 299, 303, 307 and electron spin, 285, 286 and neutrinos, 500, 607 Pauli exclusion principle, 295, 296\u2013297 and baryons, 596 and bosons, 329, 330 and fermions, 329, 331, 429 and line of stability, 487 and neutron stars, 648 and white dwarfs, 647 Pauli paramagnetism, 435 Penzias, Arno Allan, 126, 669, 676 Perfect cosmological principle, 664 Periodic table, 172 of elementary particle constituents, 599 and ground states of atoms, 297\u2013301 Perl, M., 612 Perrin, Jean-Baptiste, 116, 187, 214 Perturbation theory, first-order, 298 PET. See Positron emission tomography (PET) Petit, A., 327 Pfund, A. H., 164 Phase transitions, 336, 672 Phase velocity (vp), 196, 199 for particle waves, 201 Phonons, 463\u2013465, 510 Photodisintegration, 646 Photoelectric effect, 125, 127\u2013132, 391, 392 and de Broglie wavelength, 190 Photomultipliers, 129 Photon gases, 344\u2013351 Photons, 57, 89\u201390, 129, 141, 561 as force mediators, 574, 576t and Schr\u00f6dinger equation, 222 virtual, 509\u2013510 Photosphere, 620\u2013622,", " 622 Photovoltages, 455 Photovoltaics. See Solar cells mesons. See Pions () Pickering, Edward, 632 Pions (), 511, 561, 564, 571 in Feynman diagrams, 566 virtual, 600 PIXE. See Particle-induced x-ray analysis (PIXE) Plages, solar, 629, 629 Planck, Max K. E. L., 141, 215, 262 and energy distributions, 123\u2013125, 344\u2013345 and energy quantization, 129\u2013130, 148, 159, 226, 249, 346, 561 and fundamental constants, 673 Planck units, 673\u2013674 Planck\u2019s constant (h), 124, 125 Planck\u2019s law, 123, 123\u2013127, 124 and stellar temperatures, 621 and stimulated emission, 394 Planetary nebulae, 643, 643 Plasma, 532 confinement of, 532\u2013533 metals as, 470 solar, 623, 625 structure of, 413 Plato, 630 Polar coordinates, 101 Polar molecules, 376\u2013377, 377 Polaris, 631, 676 Polarizability (), 377 Polycrystalline solids, 414 Polyelectrons, 110 Polynomials Hermite, 248 Laguerre, 279, 279t Legendre, 274 Population densities (n), 400 Population I stars, 632 Population II stars, 632 Population inversions, 396, 401 critical density (nc) of, 400, 404\u2013405 Positron emission tomography (PET), 478, 501, 542 Positronium, 91, 181 Positrons (e), 90, 91, 110, 562\u2013563 discovery of, 477, 561 Potential energy (U), 81. See also Finite square wells; Infinite square wells; Simple harmonic oscillators and alpha decay, 497, 497 and coupling constants, 570\u2013571, 577 and covalent bonding, 370\u2013374, 372, 373 of electric dipoles, 376, 377 gravitational, 625 and ionic bonding, 365\u2013366, 366 Kronig-Penney, 438, 438\u2013439 Lenard-Jones, 411 for quarks, 600, 600 and Schr\u00f6dinger equation, 223, 227\u2013228 and solids, 415\u2013417 of strong interaction, 479, 506, 508, 515 and vibrational energy levels, 382, 382\u2013383, 383 vs. kinetic energy, 81\u201383 Pound, R. V., 100, 105, 552 Primitive vertices, 566, 566, 600 Principle of equivalence, 98\u201399, 99 and gravitational redshift, 103\u2013104 Principle quantum number (n), 161, 278 Probability, and particle waves, 196, 201\u2013205, 224\u2013225 Probability amplitude (&(x, t)), 225 Probability density (P(x, t)) in band theory of solids, 441 for electron gases, 421, 422 of exchange mesons, 512\u2013513, 513 in finite square wells, 241, 241 for the hydrogen atom, 282\u2013283, 283, 284, 285 and indistinguishable particles, 295\u2013296, 328, 329, 330, 332 in infinite square wells, 233, 234, 236\u2013237, 237 in simple harmonic oscillators, 248, 248 Probability distribution functions (P(x)), 204\u2013205 for exchange mesons, 512\u2013513, 513 for infinite square wells, 234, 234 Prominences, solar, 629, 629 Proper frequency (f0), 42 Proper length (Lp), 33, 38 as four-vector, 85 Proper time interval (), 21, 24, 31, 38 as four-vector, 85 Proton-proton cycle, 535\u2013536, 625\u2013626, 626, 627t Protons (p), 479t, 561 decay of, 606\u2013607 diffraction of, 193, 194 in interstellar medium, 653 Proxima Centauri, 640 Ptolemy, 636 Pulsars, 649, 649 Pump levels, in lasers, 397, 397 Purcell, E. M., 470 Pure rotational spectra, 380 Q branches, 390 Q-switching, 399 Q values, 517\u2013520, 535 QED. See Quantum electrodynamics (QED) QSOs. See Radio-quiet quasars (QSOs) QSRs. See Radio-loud quasars (QSRs) Quality factors (Q), 399 Quantization, 115\u2013146 of angular momentum, 161, 275\u2013277 and blackbody radiation, 119\u2013127 and the Compton effect, 133\u2013139 of electric charge, 115\u2013119 of energy, 124\u2013125, 129, 225, 238, 278\u2013279 of energy states of matter, 346\u2013351 of magnetic flux, 462 of magnetic moment,"]}